Elminia University Faculty of Engineering
Engineering Mathematics Part 1
Dr. Ali Mohamed Eltamaly
Preface Most complex scientific and engineering models of the real world are Differential equations. Here are some that I know of: heat flow, electrostatic potential, waves (radio, light, sound, water), metal beam bending, quantum mechanics, hydrogen bombs, electrons in telegraph wires, optics, classical mechanics, general relativity, distributions of organisms, ice sheets, tsunamis, air flow, ocean currents, weather, auroras, blood flow, plate tectonics, supernovas. For this reason we introduce this notes for students in faculty of engineer. By the end of the course the student should: • be familiar with the concept of a complex number and be able perform algebraic operations on complex numbers, both with numeric and symbolic entries, solve simple equations with complex roots, and in particular describe geometrically the roots of unity; • be familiar with the concept of a matrix and be able to perform algebraic operations on matrices, both with numeric and symbolic entries, be able to define a determinant and calculate one both directly and by using row and column operations, understand the definition and use of the inverse of a non-
singular matrix, and be able to solve simple systems both using inverses and reduction to triangular form, and be able to compute inverses using Gaussian reduction and explain the method in terms of elementary matrices; • be familiar with many topics in calculus like limits, differentiations, and all methods of integrations; • be familiar with first order differential equations (linear and nonlinear) and their solution by many techniques; • be familiar with many engineering applications of first order differential equations like falling bodies, the time rate of change in temperature of an object varies as the difference in temperature between the object and surroundings, Chemical Applications, time required for liquid tanks to get empty, Half Life Of Nuclear Materials, and Electrical Circuits; • be familiar with solution of higher order linear differential equations with constant coefficients and Cuchy differential equation and their solution by many techniques; • be familiar with many engineering applications of higher order differential equations like, Free Oscillation of suspended bodies, Bending of Beams, and Electrical Circuits; • be familiar with the benefits of Laplace and inverse Laplace transforms, using Laplace transform for solving differential equations, finding Laplace transform of any periodical and non
periodical waveforms, using Laplace transform to solve many application problems like Free Oscillation of suspended bodies, Bending of Beams, and Electrical Circuits; • be familiar with finding Fourier transform of any waveform by advanced techniques like jump technique; • be familiar with curve fitting by using least square technique and using this technique to fiend Fourier transform for any waveform numerically; • be familiar with using power series for solving linear differential equations of second order, and Bessel function; • be familiar with partial differentiation and solving partial differential equations by many techniques as separation of variables, Laplace transform, and Fourier transform; • be familiar with solving differential equations which governs the conduction of heat in solids; • be familiar with eigen values and eigen vectors and using them for solving simultaneous linear differential equations; • be familiar with special functions like Gamma and Beta functions; • be familiar with many topics in numerical analysis like Numerical solution of equations by many techniques like Simple Iteration, Bisection, false position, Newton Raphson and secant method; and
• be familiar with polynomial interpolation and numerical solution of differential equations by many techniques like Euler’s and RungeKutta’s method.
Contents
Part 1 Chapter
Title
No.
Page No.
1
Mathematical Numbers
1
2
Matrices
42
3
Calculus
74
4
Ordinary Differential
101
Equations 5
Linear Differential Equations
151
Of Higher Order 6
Laplace Transforms
190
7
Fourier Series
238
8
Least Square Technique
259
References
Contents
Part 2 Chapter
Title
No. 9
Page No.
Power Series Solution Of
285
Differential Equations 10
Partial Differential Equations
346
11
Simultaneous Linear
389
Differential Equations 12
Special Functions
459
13
Numerical Analysis
478
Appendix
543
References
574
Chapter 1 Mathematical Numbers 1.1 Natural Numbers Natural numbers known as counting numbers are the numbers beginning with 1, with each successive number greater than its predecessor by 1. If the set of natural numbers is denoted by N, then N = { 1, 2, 3, ......} 1.2 Whole Numbers Whole numbers are the numbers beginning with 0, with each successive number greater than its predecessor by 1. It combines the set of natural numbers and the number 0. If the set of whole numbers is denoted by W, then W = { 0, 1, 2, 3, .......} 1.3 Integer Numbers Integers are the numbers that are in either (1) the set of whole numbers, or (2) the set of numbers that contain the negatives of the natural numbers. If the set of integers is denoted by I, then I = {......, -3, -2, -1, 0, 1, 2, 3, ......} Positive integers are the numbers in I greater than 0. Negative numbers are the numbers in I less than 0. The number zero is neither positive nor negative, i.e., it is both non-positive and non -negative.
2 Mathematical Numbers Given the above definitions, the following statements about integers can be made: (1) N is the set of positive integers. (2) W is the union of N and the number 0. (3) The set of numbers that contain the negatives of the numbers in N is the set of negative integers. (4) I is the union of W and the set of negative integers. 1.4 Real Number Line The set of real numbers can be pictorially represented by the real number line. It is a straight line, whose "origin" is designated by the number 0, and continues in both directions. All the positive integers are ordered, in ascending order from left to right, to the right side of 0; all the negative integers are ordered, in descending order from right to left, to the left side of 0. Notches are marked to denote the position of these integers in the following figure (Fig.1).
Fig.1 Every point on the line corresponds to a real number, and every real number can be paired with a point on this number line. If the real number is an integer, its point on the number line coincides with one of the notches for an integer; otherwise, its point lies between two successive notches. All real numbers represented by points to
Chapter One
3 the right of the number 0 are positive, while all real numbers represented by points to the left of the number 0 are negative. 1.5 Absolute Values The absolute value of a real number is the distance between its corresponding point on the number line and the number 0. The absolute value of the real number a is denoted by |a|. From the diagram shown in Fig.2, it is clear that the absolute value of non-negative numbers is the number itself, while the absolute value of negative integers is the negative of the number. Thus, the absolute value of a real number can be defined as follows: For all real numbers a, (1) If a > 0, then a = a . (2) If a < 0, then a = − a .
Fig.2 Example 1: |2|=2 | -4.5 | = 4.5 |0|=0
4 Mathematical Numbers 1.6 Complex Numbers 1.6.1 Introduction The solution of a second order equation ax 2 + bx + c = 0 can be
− b ± b 2 − 4ac obtained by the famous formula, x = 2a For example, if x 2 + x − 2 = 0 , then we have:
x=
−1±
(1 + 8) 2
∴ x = 1 or
=
−1± 9 −1± 3 = 2 2
−2
As we see there is no problems with solving the above equation. But if we solve the equation 5 x 2 − 6 x + 5 = 0 in the same way, we get:
x=
6±
(36 − 100) 10
=
6 ± − 64 10
And the next stage is now to determine the square root of (-64). Is it (i) 8, (ii) -8, (iii) neither? It is, of course, neither, since + 8 and − 8 are the square roots of 64 and not of (-64). In fact,
(− 64)
cannot be represented by an
ordinary number, for there is no real number whose square is a negative quantity. However, − 64 = −1 * 64 And therefore we can write:
(− 64) = (− 1 * 64) =
− 1 * 64 = 8 − 1
Chapter One
5
(− 1) ,
Of course, we are still faced with
which cannot be
evaluated as a real number, for the same reason as before, but, if we
− 1 with the letter j, then
can replace
(− 64) = (− 1) * 8 =
j8
We now have a way of finishing off the quadratic equation we started before as following:
5x 2 − 6x + 5 = 0 x=
6±
(36 − 100) 10
∴ x = 0.6 + j 0.8
=
6 ± − 64 6 ± j8 = 10 10
or x = 0.6 − j 0.8
1.6.2 Powers of j
j = −1 j 2 = −1
( ) 2 j 4 = ( j 2 ) = (− 1)2 = 1
j 3 = j 2 * j = −1 * j = − j
Note especially the last result: j 4 = 1 . Every time a factor j 4 occurs, it can be replaced by the factor 1, so that the power of j is reduced to one of the four results above. In the same way we can replace j 2 = −1 with –1. The complex number x = 1 + j 4 , consists of two separate terms, 1, and j 4 These terms cannot be combined any further, since the second is an imaginary number (due to its having the factor j).
6 Mathematical Numbers In such an expression as x = 1 + j 4
1 is called the real part of x 4 is called the imaginary part of x The two together form what is called a complex number. So, a Complex number= (Real part) + j(Imaginary part) Complex numbers is very important especially in some engineering application like electrical and mechanical engineering. So we have to fully understand how to carry out the usual arithmetical operations. 1.6.3 Addition and Subtraction of Complex Numbers. Addition and Subtraction are quite easy as shown in the following
example: Example 2 Find the results of the following arithmetical operations.
(3 +
j 7 ) + (6 − j 2 ) .
Solution :
(3 +
j 7 ) + (6 − j 2 ) = 3 + j 7 + 6 − j 2 = (3 + 6 ) + j (7 − 2 ) = 9 + j 5 So, in general, (a + jb ) + (c + jd ) = (a + c ) + j (b + d ) 1.6.4 Multiplication of. Complex Numbers
The following example illustrate the multiplication process in complex numbers. Example 3 Find the results of the following arithmetical operations.
(2 + j3)(5 + J 7 )
Chapter One
7 Solution: These are multiplied together in just the same way as
you would determine the product (2 + j 3)(5 + j 7 ) . Form the product terms of
(2 + j3)(5 + j 7 ) = 2 * 5 +
j3 * 5 + j 2 * 7 + j 2 3 * 7
= 10 + j15 + j14 + j 2 21 = 10 + j 29 − 21 = −11 + j 29 If the expression contains more than two factors, we multiply the factors together in stages:
(2 + j3)(5 +
(
)
j 7 )(1 − j 2) = 10 + j15 + j14 + j 2 21 (1 − j 2) = (10 + j 29 − 21)(1 − j 2) = (− 11 + j 29 )(1 − j 2) = −22 + j 22 + j 29 − j 2 58 = −22 + j 51 + 58 = 36 + j 51
Example 4 Find the results of the following arithmetical
operations. (5 + j8)(5 − j8) Solution:
(5 + j8)(5 − j8) = 25 +
j 40 − j 40 − j 2 64
= 25 + 64 = 89 In spite of what we said above, here we have a result containing no imaginary term. The result is therefore entirely real. This is rather an exceptional case. If we look at the two complex numbers we can
8 Mathematical Numbers find that they are identical except for the middle sign in the brackets
are different. These two complex numbers called conjugate complex numbers and the product of two conjugate complex numbers is always entirely real. In general we can say:
(a + b )(a − b ) = a 2 − b 2
difference of two squares and there is no
any imaginary part. 1.6.5 Divison of. Complex Numbers
Division of a complex number by a real number is easy enough.
5 − j4 5 4 = − j = 1.67 − j1.33 3 3 3 But how do we manage with dividing complex number with other complex one? If we could, somehow, convert the denominator into a real number, we could divide out as in the above example. So our problem is really, how can we convert (4 + j3) into a completely real denominator and this is explained in the previous item. We know that we can convert (4 + j3) into a completely real number by multiplying it by its conjugate (4 - j3). But if we multiply the denominator by (4 − j 3) , we must also multiply the numerator by the same factor.
7 − j 4 (7 − j 4)(4 − j 3) 28 − j 37 − 12 16 − j 37 = = = 4 + j 3 (4 + j 3)(4 − j 3) 16 + 9 25
16 37 −j = 0.64 − j1.48 25 25
Chapter One
9 Then, to divide one complex number by another, therefore, we
multiply numerator and denominator by the conjugate of the denominator. This will convert the denominator into a real number and the final step can then be completed. Example 5 Simplify the following expression:
(2 + j3)(1 − j 2) 3 + j4
Solution:
(2 + j3)(1 − j 2) = 2 − 3 + j4
8 − j 3 − j4 j +6 8− j = = * 3 + j4 3 + j4 3 + j4 3 − j4 24 − j 35 − 4 20 − j 35 = = = 0.8 − j1.4 9 + 16 25
Equal Complex Numbers
Now let us see what we can find out about two complex numbers which we are told are equal. Let the numbers a + jb, and c + jd are equal ∴ a + jb = c + jd Rearranging terms, we get ∴ a − c = j (d − b ) In this last statement the quantity on the left hand side is entirely real, while that on the right hand side is entirely imaginary, i.e. a real quantity equals an imaginary quantity. This seems contradictory and in general it just cannot be true. But there is one special case for which the statement can be true. That is when each side is zero.
∴ a − c = j (d − b ) can be true only if a − c = 0, ie. a = c and if d − b = 0, ie. b = d
10Mathematical Numbers So we get this important result, If two complex numbers are equal
then, (i) the two real parts are equal (ii) the two imaginary parts are equal For example, if x + jy = 5 + j 4 , then we know x = 5 and y = 4 . 1.6.6 Graphical Representation of a Complex Numbers
Although we cannot evaluate a complex number as a real number, we can represent it diagrammatically, as we shall now see. In the usual system of plotting numbers, the number 4 could be represented by a line from the origin to the point 4 on the scale. Likewise, a line to represent (-4) would be drawn from the origin to the point (-4). These two lines are equal in length but are drawn in opposite directions. Therefore, we put an arrow head on each to distinguish between them as shown in Fig.3. -4 -4
-3
-2
4 -1
0
1
2
3
4
Fig.3 A line which represents a magnitude (by its length) and direction (by the arrow head) is called a vector. We shall be using this word quite a lot. Any vector therefore must include both magnitude (or size) and direction. If we multiply (+4) by the factor (-1), we get (-4), i.e. the factor (-1) has the effect of turning the vector through 180o as shown in Fig.4.
Chapter One
11
180 o 4
-4 -4
-3
-1
-2
0
1
2
3
4
Fig.4 Multiplying by (-1) is equivalent to multiplying by j 2 , i.e. by the factor j twice. Therefore multiply in a single factor j will have half the effect and rotate the vector through only 90o . So, the factor j always turns a vector through 90o in the positive direction measuring angles, i.e. anticlockwise. If we now multiply j 4 by a further factor j, we get j 2 4 , i.e. (-4) and the following diagram (Fig.5) agrees with this result. If we multiply (-4) by a further factor j, sketch showing this new vector (− j 4 ) is shown in Fig.6. 4 3 j4
4
2
3
1 -4
180
o
-4 -4
-3
-2
0
1
-3
-2
4 0
-1
1
-1
2
3
4
-2 -j4
4 -1
o
-4
j4 2 1
180
2
-3
3
4
Fig.5
-4
Fig.6
Let us denote the two reference lines by XX, and YY, as usual. You will see that: (i) The scale on the X-axis represents real numbers, XX1 is therefore called the real axis.
12Mathematical Numbers (ii) The scale on the Y-axis represents imaginary numbers, YY1 is
therefore called the imaginary axis. If we now wish to represent 2+ 3 as the sum of two vectors, we must draw them as a chain, the second vector starting where the first one finishes as shown in Fig.7. 5 2
3
1
0
2
3
4
5
Fig.7 The two vectors, 2 and 3 are together equivalent to a single vector drawn from the origin to the end of the final vector (giving naturally that 2+3=5). If we wish to represent the complex number (3 + j2), then we add together the vectors which represent 3 and j2. Notice that the 3 is now multiplied by a factor j which turns that vector through 90o . The equivalent single vector to represent (2 + j3) is therefore the vector from the beginning of the first vector (origin) to the end of the last one. This graphical representation constitutes an Argand
diagram as shown in Fig.8. 3 j3
2 1 2 0
1
2
Fig.8
3
Chapter One
13 Example 6 Draw an Argand diagram to represent the following
vectors: z1 = 3 + j 2 , z 2 = −3 + j1 , z3 = 2 − j 4 , and z 4 = −4 − j 4 Solution: The Argand diagram of the above vectors are shown in
the following Fig.9. 3
z1
2 j1
z2
j2
1 3
-4
-3
j4
-2
0
3 -1
1
2
-1
3
4 j4
-2 -j4 -3
z4
-4
z3
Fig.9.
1.6.7 Graphical Addition of Complex Numbers
Let us find the sum of z1 = 3 + j 2 and z 2 = 2 − j 4 by Argand diagram. If we are adding vectors, they must be drawn as a chain. We therefore draw at the end of z1 , a vector representing z2 in magnitude and direction, and is parallel to it. In the same way, we therefore draw at the end of z2 , a vector representing z1 in magnitude and direction, and is parallel to it. Therefore we have a parallelogram. Thus the sum of z1 and z2 is given by the vector joining the starting point to the end of the last vector.
14Mathematical Numbers
The complex numbers z1 and z2 can thus be added together by drawing the diagonal of the parallelogram formed by z1 and
z 2 .Thus, z1 + z 2 = 3 + j 2 + 2 − j 4 = 5 − j 2 which is clear that this results is the same as obtained from Fig10. So the sum of two vectors on an Argand diagram is given by the diagonal of the parallelogram of vectors.
z1
2
j2
1 3 0
1
-1
2
3
4
z1 + z2
5
-2 -j4 -3 -4
z2 Fig.10
Regarding to the subtraction it is quite similar to addition but the only trick is simply this: z1 − z 2 = z1 + (− z 2 ) That is, we draw the vector representing z1 and the negative vector of z2 and add them as before. The negative vector of z 2 is simply a vector with the same magnitude (or length) as z 2 but pointing in the opposite direction.
Chapter One
15
Example 7 If z1 = 3 + j 2 and z 2 = 2 − j 4 Find z1 − z 2 Solution:
It is clear from Argand diagram (Fig.11) that
z1 − z 2 = 1 + j 6 . We can now check for the above results:
Z1 − Z 2 = 3 + j 2 − (2 − j 4) = 3 + j 2 + (−2 + j 4) = 1 + j 6 6
z1 − z 2
5 4
u
Fig.11
3
− z2
z1
2
j2
1 3 3
0
3 -1
-2
1
-1
2
3
4
-2 -j4 -3 -4
z2
1.6.8 Polar Form of a Complex Numbers
Complex numbers in the form a + jb is called rectangular form. Sometimes, it is convenient to express it in a different form. On an Argand diagram shown in Fig.12, let OA be a vector a + jb . Let r = length of the vector and θ the angle made with OX. Since
z = a + jb ,
this
can
be
written
z = r cos θ + jr sin θ
or
z = r (cos θ + j sin θ ) This is called the polar form of the complex number a + jb , where: r =
(a 2 + b 2 ),
b and , θ = tan −1 a
16Mathematical Numbers
A
y r
o
b
θ
a
x
Fig.12 Example 8 Express z = 4 + j 3 in polar form. Solution: First draw a sketch diagram (that always helps). We can
3 see that: r = 4 2 + 32 = 16 + 9 = 5 and θ = tan −1 = 36.87 o 4
(
z = a + jb = r (cos θ + j sin θ ) ∴ z = 5 cos 36.87 o + j sin 36.87 o
)
(i) r is called the modulus of the complex number z and is often abbreviated to mod( z ) or indicated by z
(
)
Thus if z = 3 + j 4 , ∴ z = 32 + 4 2 = 9 + 16 = 5 (ii) θ is called the argument of the complex number and can be abbreviated to arg ( z ) . So, if z = 5 + j 5 then arg( z ) = θ = 45o Warning: In finding θ , there are of course two angles between 0o
and 360 o , the tangent of which has the value θ . We must be careful to use the angle in the correct quadrant. Always draw a sketch of the vector to ensure you have the right one. The follwing table and Fig.13 show the correct angle range and quadrant.
Chapter One
17
Value of a
Value of b
Angle range
Quuadrant
+ve
+ve
0o < θ < 90o
First
-ve
+ve
90 o < θ < 180 o
Second
-ve
-ve
180o < θ < 270o
Third
+ve
-ve
270o < θ < 360o
Fourth
y
First
Second
o
0 < θ < 90o
90 o < θ < 180o o o
180 < θ < 270
o
o
270 < θ < 360
o
x
Fourth
Third
Fig.13 Example 9 Find arg( z ) when z = −3 − j 4 Solution: The vector has been drawn as shown in Fig. 14.
θ
is measured from OX to OP. We first find E the equivalent
acute angle from the trinngle shown.
tan E =
4 = 1.3333 then E = 53.13o 3
18Mathematical Numbers
But from Fig.14. This angle is 180 o < θ < 270 o then we have to add 180o to the angle E to get angle θ .
∴θ = 180o + E = 233.13o
Fig.14 Example 10 Find arg( z ) when z = −5 + j 2 Solution: The vector has been drawn as shown in Fig.15.
Fig.15
∴ tan E =
2 = 0.4 ∴ E = 21.8o 5
In this particular case, θ = 180 − E o ∴θ = 158.2 o Complex numbers in polar form are always of the same shape and differ only in the actual values of r and θ . We often use the
Chapter One
19
shorthand version r∠θ o to denote the polar form as shown in the following examples: Then, if z = −5 + j 2 r =
θ = 158.2 .
above
(
Then,
(25 + 4) = the
29 = 5.385 and from
full
)
polar
form
is
z = 5.385 cos158.2 o + j sin 158.2 o and this can be shortened to z = 5.385∠158.2o . Example 11 express 4 − j 3 in shortened form. Solution: r =
(42 + 32 ) = 5
tan E = 0.75 , ∴ E = 36.87 o ∴θ = 360 − E = 323.13o
(
)
∴ z = 5 cos 323.13o + j sin 323.13o = 5∠323.8o Of course, given a complex number in polar form, you can convert it into the basic rectangular form a + jb simply by evaluating the cosine and the sine and multiplying by the value of r. Example 12 Find the rectangular form of the following: z = 5∠35o
(
)
z = 5 cos 35o + j sin 35o = 5(0.8192 + j 0.5736) Solution: = 4.096 + j 3.868 1.6.9 Exponential Form of a complex numbers
There is still another way of expressing a complex number which we must deal with. We shall Expalin it this way: Many functions can be expressed as series. For example,
20Mathematical Numbers ∞
xm x 2 x3 x 4 =1+ x + + + + ...... e = ∑ ! 2 ! 3 ! 4 ! m m=0 x
(1)
∞
(−1) m x 2 m x2 x4 cos x = ∑ =1− + − +......... ( 2 )! 2 ! 4 ! m m=0
(2)
(−1) m x 2 m +1 x3 x5 sin x = ∑ = x− + − +......... ( 2 + 1 )! 3 ! 5 ! m m=0
(3)
∞
If we now take the series for e x and write jθ in place of x, we get the following series
e
jθ
=1+
∴e
jθ
∴e
jθ
∴e
jθ
( jθ )2 ( jθ )3 ( jθ )4 ( jθ )5 + + + jθ + 2!
3!
4!
5!
(4)
j 2 (θ )2 j 3 (θ )3 j 4 (θ )4 j 5 (θ )5 = 1 + jθ + + + + ++ 2! 3! 4! 5! = 1+
( θ )2 j (θ )3 (θ )4 j (θ )5 jθ − − + + −− 2!
3!
(θ )2 (θ )4 = 1 − + − .... + 2! 4!
4!
5!
j (θ )3 j (θ )5 j θ− + − .... 3! 5!
(5)
It is clear from (2),(3) and (5) that the first bracket is in the form of cosine and the second bracket in the form of sine.
∴ e jθ = cos θ + j sin θ
(6)
Therefore, r (cos θ + j sin θ ) can now be written as re jθ . This is called the exponential form of the complex number. It can be
Chapter One
21 obtained from the polar form quite easily since the r value is the
same and the angle θ is the same in both. The three ways of expressing a complex number are therefore (i) z = a + jb (ii) z = r (cos θ + j sin θ )
(Rectangular form) (Polar form)
(iii) z = re jθ (Exponential form) And now a ward about negative angles. We know that:
e jθ = cos θ + j sin θ if we replace θ by − θ in this result, we get the following:
e − jθ = cos(− θ ) + j sin (− θ ) = cos θ − j sin θ So, e jθ = cos θ + j sin θ And e − jθ = cos θ − j sin θ There is one operation that we have been unable to carry out with complex numbers before this. That is to find the logarithm of a complex number. The exponential form now makes this possible, since the exponential form consists only of products and powers. For, if we have z = r.e jθ we can say: ln z = ln r + jθ Example 13 Express e1− jπ / 4 in the rectangular form: Solution: Well now, we can write π π e1− jπ / 4 = e1e − jπ / 4 = e cos − j sin 4 4
1 e 1 (1 − j ) = e −j = 2 2 2
22Mathematical Numbers Since every complex number in polar form is of the same shape,
i.e. r (cos θ + j sin θ ) = r∠θ o and differs from another complex number simply by the values of r and θ , we have a shorthand method of quoting the result in polar form. Example 14 Express z = 4 − j 3 in the polar form. Solution: The vector has been drawn as shown in Fig. 16.
r=
(42 + 32 ) = 5
Fig.16
3 = −0.7, ∴ E = −36.87 o 4 ∴θ = 360 o − 36.87 o = 323.13o
From this r = 5 , tan E = −
(
)
Then, the polar form is z = 5 cos 323.13o + j sin 323.13o . And the polar form is z = 5.e form is
(
j 323.13o
And the the shortened
z = 5∠323.13o . In this last example, we have
z = 5 cos 323.13o + j sin 323.13o
)
Chapter One
23
Fig.17. But the direction of the vector, measured from OX, could be given as − 36.87 o , the minus sign showing that we are measuring the angle in the opposite direction sense from the usual positive
( (
)
(
))
direction. We could write z = 5 cos − 36.87 o + j sin − 36.87 o . But
you
already
sin (− θ ) = − sin (θ ) .
( (
known
)
as
(
∴ z = 5 cos 36.87 o − j sin 36.87 o
cos(− θ ) = cos(θ )
and
))
i.e. very much like the polar form but with a minus sign in the middle. This comes about whenever we use negative angles. In the same way we can say the following:
(
) ( (
)
(
z = 5 cos 250o + j sin 250o = 5 cos − 110o + j sin − 110o
))
It is sometimes convenient to use this form when the value of θ is greater than 180 o , i.e. in the 3rd and 4th quadrants. In the same way we can write the following:
24Mathematical Numbers
( ) ( ) z = 4(cos 290o + j sin 290o ) = 4(cos 70 o − j sin 70 o ) = 4∠ − 70 o .
z = 3 cos 230o + j sin 230o = 3 cos130o − j sin130o = 3∠ − 130o
The polar form at first sight seems to be a complicated way of representing a complex number. However it is very useful as we shall see. Suppose we multiply together two complex numbers in this form: Let z1 = r1 (cos θ1 + j sin θ1 ) and z 2 = r2 (cos θ 2 + j sin θ 2 )
∴ z1 * z 2 = r1 (cos θ1 + j sin θ1 ) * r2 (cos θ 2 + j sin θ 2 )
∴ z1 * z 2 = r1.r2 (cos θ1 cos θ 2 + j sin θ1 cos θ 2 + j cos θ1 sin θ 2 + j 2 sin θ1 sin θ 2
)
Rearranging the terms and remmembering j 2 = −1 we get:
(cos θ1 cos θ 2 − sin θ1 sin θ 2 ) z1 * z 2 = r1.r2 + j (sin θ1 cos θ 2 + cos θ1 sin θ 2 ) Now
the
brackets
(cosθ1 cosθ 2 − sin θ1 sin θ 2 ) and
(sin θ1 cos θ 2 + cos θ1 sin θ 2 ) ought to ring the bell. What are they? (cos θ1 cos θ 2 − sin θ1 sin θ 2 ) = cos(θ1 + θ 2 )
(sin θ1 cos θ 2 + cos θ1 sin θ 2 ) = sin (θ1 + θ 2 ) ∴ z1 * z 2 = r1.r2 [cos(θ1 + θ 2 ) + j sin (θ1 + θ 2 )] Note: This is important result. Then we can say that to multiply
together two complex numbers in the polar form, (i) Multiply the r's together, (ii) Add the angles, θ , together it is just easy as that.
Chapter One
25 Example 15 Find the result of the following in the polar form:
2∠30 * 3∠40 Solution: It is easy to do that as we get:
(
)
2∠30 * 3∠40 = (2 * 3)∠ 30o + 40 o = 6∠70o Now let us see if we can discover a similar set of rules for division. We already know that to simplify 5 + j 4 we first obtain a denominator that is entirely real by multiplying top and bottom by the conjugate of the denominator i.e (5 − j 4 ) Right. Then let us do the same thing with z1 and z2 as following:
z1 r (cos θ1 + j sin θ1 ) = 1 z 2 r2 (cos θ 2 + j sin θ 2 )
∴
z1 r (cos θ1 + j sin θ1 ) (cos θ 2 − j sin θ 2 ) = 1 * z 2 r2 (cos θ 2 + j sin θ 2 ) (cos θ 2 − j sin θ 2 )
∴
z1 r1 (cosθ1 cosθ 2 + j sin θ1 cosθ 2 − j cosθ1 sin θ 2 + sin θ1 sin θ 2 ) = z 2 r2 cos 2 θ 2 + sin 2 θ 2
(
)
∴
z1 r1 (cosθ1 cosθ 2 + sin θ1 sin θ 2 ) + j (sin θ1 cosθ 2 − cosθ1 sin θ 2 ) = 1 z 2 r2
∴
z1 r1 r = (cos(θ1 − θ 2 ) + j sin (θ1 − θ 2 )) = 1 ∠(θ1 − θ 2 ) z 2 r2 r2
So, for division the rule is divide the r's and subtract the angles θ 's. Example 16 Simplify the following expressions: Solution:
10∠95o 2∠44 o
(
)
= 5∠ 95o − 44 o = 5∠51o
10∠95o 2∠44 o
,
26Mathematical Numbers 1.6.10 DeMoivre’s Theorem
There is very important rule is called DeMoivre’s Theorem. It says that to raise a complex number in polare form to any power n, we raise the r to the power n and multiply the angle by n.
∴ [r (cos θ + j sin θ )]n = r n (cos nθ + j sin nθ ) Example 17 Use DeMoivre’s Theorem to find the results of the
following expression in polar form: Solution:
[3(cos110
o
+ j sin 110o
)]3
[3(cos110 + j sin 110 )] = 3 (cos(3 *110 ) + j sin (3 *110 )) ∴ [3(cos 110 + j sin 110 )] = 27(cos(330 ) + j sin (330 )) = 27∠330 o 3
o
3
o 3
o
o
o
o
o
This is where the polar form really comes into its own. For
DeMoivre's theorem also applies when we are raising the complex number to a fractional power, i.e. when we are finding the roots of a complex number as shown in the following example. Example 18 Find the square root of z = 9∠44 o Solution:
44o = 3∠22o We have z = 9∠44 = 9∠ 2 Expansion of sin nθ and cos nθ o
By DeMoiver's theorem, we know that:
cos nθ + j sin nθ = (cos θ + j sin θ )n where n is a positive integer.
o
Chapter One
27 The method is simply to expand the right hand side as a binomial
series, after which we can equate real and imaginary parts. An example will soon show you how it is done. Example 19 To find expansions for cos 3θ and sin 3θ Solution: We have cos 3θ + j sin3θ = (cosθ + j sinθ )3 = (c + js)3
Where c = cos θ and s = sin θ just for simplicity. Now expand this by the binomial series so that:
cos 3θ + j sin 3θ = (cosθ + j sin θ )3 = c 3 + 3c 2 ( js ) + 3c( js )2 + ( js )3
cos 3θ + j sin 3θ = c 3 + j 3c 2 s − 3cs 2 − js 3
(
) (
cos 3θ + j sin 3θ = c 3 − 3cs 2 + j 3c 2 s − s 3
)
Now equating real parts and imaginary parts we get:
cos 3θ = cos3 θ − 3 cos θ sin 2 θ sin 3θ = 3 cos 2 θ sin θ − sin 3 θ If
we
wish,
(
we
)
can
replace
(
sin 2 θ = 1 − cos 2 θ
)
and
cos 2 θ = 1 − sin 2 θ . So that we could write the results above as following: cos 3θ = 4 cos3 θ − 3 cos θ , ∴ sin 3θ = 3 sin θ − 4 sin 3 θ . While these results are useful, it is really the method that counts. So now do this one in just the same way as done before, obtain an expansion for cos 4θ in terms of cos θ .
cos 4θ + j sin 4θ = (cos θ + j sin θ )4 = (c + js )4
28Mathematical Numbers
= c 4 + 4c 3 ( js ) + 6c 2 ( js )2 + 4c( js )3 + ( js )4 = c 4 + j 4c 3 s − 6c 2 s 2 − j 4cs 3 + s 4
(
) (
= c 4 − 6c 2 s 2 + s 4 + j 4c 3 s − 4cs 3 Equating real parts:
(
∴ cos 4θ = c 4 − 6c 2 s 2 + s 4
(
)
) (
= c 4 − 6c 2 1 − c 2 + 1 − c 2
)
)2
= c 4 − 6c 2 + 6c 4 + 1 − 2c 2 + c 4 = 8c 4 − 8c 2 + 1 = 8 cos 4 θ − 8 cos 2 θ + 1
(
)
(
Similarly, sin 4θ = 4c 3 s − 4cs 3 = 4cs c 2 − s 2
)
( ) = 4cs (1 − s 2 ) = 4cs * c 2 = 4c 3 s = 4cs c 2 − s 2 − c 2 − s 2 + 1
∴ sin 4θ = 4 cos3 θ * sin θ Expansions for cos n θ and sin n θ in terms of sines and cosines of muiltiples of θ .
1 = z −1 = cos θ − j sin θ z 1 1 ∴ z + = 2 cos θ and , z − = j 2 sin θ z z Also by DeMoivre's theorem z n = cos nθ + j sin nθ
z = cos θ + j sin θ , ∴
Chapter One
29
1
= cos nθ − j sin nθ zn 1 1 ∴ z n + n = 2 cos nθ And, z n − = j 2 sin nθ z zn And,
Let us collect theses four results together: z = cos θ + j sin θ
1 = 2 cos θ z 1 z n + n = 2 cos nθ z
z+
1 = j 2 sin θ z 1 z n − n = j 2 sin nθ z z−
Example 20 Expand cos 3 θ Solution: From the previous results, 3
1 1 Q z + = 2 cos θ ∴ z + = (2 cos θ )3 z z 1 1 1 ∴ (2 cos θ )3 = z 3 + 3z 2 + 3z 2 + 3 z z z 1 1 = z 3 + 3z + 3 + 3 z z Now here is the trick: we rewrite this, collecting the terms up in pairs from the two extreme ends, thus:
(2 cos θ )3 = z 3 +
1 1 + + 3 z z z3
But from the previous results
∴z +
1 1 = 2 cos θ , and z 3 + = 2 cos 3θ 3 z z
30Mathematical Numbers
∴ (2 cos θ )3 = 2 cos 3θ + 3 * 2 cos θ
∴ 8 cos3 θ = 2 cos 3θ + 6 cos θ 1 ∴ cos3 θ = (cos 3θ + 3 cos θ ) 4 Example 21 Expand sin 4 θ Solution: 1 1 Q z − = j 2 sin θ , and, z n − n = j 2 sin nθ z z
∴ ( j 2 sin θ )
4
1 = z − z
4
1 1 1 1 = z 4 − 4 z 3 + 6 z 2 2 − 4 z 3 + 4 z z z z 1 1 = z 4 + 4 − 4 z 2 + 2 + 6 z z Now, Q z n +
1 zn
= 2 cos nθ
1 1 ∴ ( j 2 sin θ )4 = z 4 + 4 − 4 z 2 + 2 + 6 z z = 2 cos 4θ − 4 * 2 cos 2θ + 6 ∴ 16 sin 4 θ = 2 cos 4θ − 4 * 2 cos 2θ + 6
∴ sin 4 θ =
1 (cos 4θ − 4 cos 2θ + 3) 8
Chapter One
31
1.6.11 The Roots of Unity
The problem here is to solve the equation z n = 1 , where n is usually a positive whole number. Write both sides of the equation in polar form. Let z have polar form
z = r (cos θ + j sin θ ) ∴ z n = r n (cos nθ + j sin nθ ) We know that 1 = 1(cos 0 + j sin 0 ) . So our equation becomes:
∴ z n = r n (cos nθ + j sin nθ ) = 1 * (cos 0 + j sin 0) Now two complex numbers in standard polar form are equal if and only if their modulus and arguments are equal. In the case of the argument this statement has to be handled with care. It means are equal if reduced to the proper range. So, for example 10o and 370 o count as equal from this point of view. So we can say that r n = 1 and that nθ and 0 are equal up to the addition of some multiple of
2π radians. r n = 1
nθ = 0 + 2kπ Where k is some whole
number. Since r is real and positive, the only possibility for r is r= 1. The other equation gives us: θ = 0 + 2π
k n
This, in principle, gives us infinitely many answers one for each possible whole number k. But not all the answers are different. Remember that changing the angle by 2π does not change the number z. The distinct solutions, of which there are n, are given by r = 1and
32Mathematical Numbers
θ = 2π
k , n
k = 0, 1, 2, 3,.........n − 1
and we can write these
solutions as following: z k = cos θ k + j sin θ k Where θ = 2π
k , n
k = 0, 1, 2, 3,.........n − 1
That looks rather complicated. It becomes a lot simpler if you think in terms of the Argand diagram. All the solutions have modulus 1 and so lie on the circle of radius 1 centered at the origin. The solution with k = 1 is just z = 1. The other solutions are just n − 1 other points equally spaced round this circle, with angle 2π / n between one and the next. This is shown in Fig.18 for n = 17 .
Fig.18 The n roots of 1. Let's look at some specific examples. The cube roots of unity are the solutions to z 3 = 1 . There are three of them and they are:
zo = 1,
z1 = cos 2π / 3 + j sin 2π / 3,
z 2 = cos 4π / 3 + j sin 4π / 3
Chapter One
33
Fig.19: The three cube roots of 1. Note that z 2 = z1 , z 2 = z12 and 1 + z1 + z 2 = 0 . The roots are shown in Fig.19. Similarly the fourth roots of unity are the solutions of z 4 = 1 and these are: z = 1 , z = j , z = −1 , and z = − j A picture for n = 4 together with those for n = 5 and n= 6 is given in Fig.20.
Fig.20 The nth roots of 1 for n = 4;5;6. We can do other equations like this in much the same way.
34Mathematical Numbers Example 22 Find the solutions of the equation z 4 = j . Solution:
Put z = r (cos θ + j sin θ ) . Then z 4 = r 4 (cos 4θ + j sin 4θ ) . We know that: j = 1(cos π / 2 + j sin π / 2 ) . So our equation becomes:
z 4 = r 4 (cos 4θ + j sin 4θ ) = 1(cos π / 2 + j sin π / 2 ) Therefore; r = 1 and 4θ =
π 2
+ 2kπ
or θ =
π 8
+
kπ 2
There are 4 distinct solutions, given by k = 0;1;2;3. They form a square on the unit circle. 1.7 Polynomials
We have learned how to manipulate complex numbers, and suggested that they will prove valuable in engineering calculations. The original motivation for introducing them was to give the equation x 2 = −1 two roots, namely j and − j , rather than it having no roots. It turns out that this is all we have to do to ensure that every polynomial has the right number of roots. We now discuss this, and a number of other basic results about polynomials that are quite useful to know. A polynomial in x is a function of the form:
p ( x ) = an x n + an −1 x n −1 + .... + a1 x + ao
Chapter One
35
where the a's are (real or complex) numbers and an ≠ 0 . For example: p ( x ) = x 3 − 2 x + 4,
q (t ) = 5t 8 − t 4 + 6t 3 − 1
The highest power in the polynomial is called the degree of the polynomial. The above examples have degrees 3 and 8. A number a (real or complex) is said to be a root of the polynomial
p( x ) if p (a ) = 0 . Thus x = 1 is a root of x 2 − 2 x + 1 = 0 The first important result about polynomials is that a number a (real or complex) is a root of the polynomial p ( x ) if and only if ( x − a ) is a factor of p(x), in the sense that we can write p ( x ) as:
p ( x ) = ( x − a )q( x ) . Where q( x ) is another polynomial. This result is often called the remainder theorem . For example, x = 2 is a root of p ( x ) = x 3 + x 2 − 7 x + 2 and it turns out that
(
)
p(x ) = (x − 2) x 2 + 3x − 1
Note that necessarily the polynomial q has degree one less than the degree of p. It may be the case that you can pull more than one factor of x − a out of the polynomial. For example, 2 is a root of
p ( x ) = x 3 − x 2 − 8 x + 12 and it turns out that p ( x ) = ( x − 2 )( x − 2 )( x + 3) In such cases a is said to be a multiple root of p ( x ) . The multiplicity of the root is the number of factors ( x − a ) that you can take out. In the above example, 2 is a root of multiplicity 2, or a double root. A
36Mathematical Numbers root is called a simple root if it produces only one factor. Multiple
roots are a considerable pain in the neck in many applications. There is a simple test for multiplicity. Suppose a is a root of
p( x ) , so that p (a ) = 0 . If, in addition, p′(a ) = 0 (derivative) then a is a multiple root. To take the above example:
Q p ( x ) = x 3 − x 2 − 8 x + 12 ∴ p′( x ) = 3 x 2 − 2 x − 8
and
p (2 ) = 0
and we have p′(2 ) = 0 , so we know that 2 is a multiple root. 2.9. Theorem (Fundamental Theorem of Algebra).
Let p be any polynomial of degree n. Then p can be factored into a product of a constant and n factors of the form ( x − a ) , where a may be real or complex. Also, the factorization is unique; you cannot find two essentially different factorizations for the same polynomial. The factors need not all be different because of multiple roots. The fact that there cannot be more than n such factors is fairly obvious, since we would have the wrong degree. What is not at all obvious is that we have all the factors that we want. Note that this result does not tell you how to find these factors; just that they must be there! The result is often stated loosely as: a polynomial of degree n must have exactly n roots. You have to allow complex roots or the theorem is not true. For example p ( x ) = x 2 + 1 has no real roots at
Chapter One
37
all. Its roots are x = ± j and it factorizes as p ( x ) = ( x − j )( x + j ) . In fact, if ω ≠ 0 then p ( z ) = z n − ω
(n ≥ 1)
always has exactly n
distinct roots because we know that it must have n roots in all and it cannot have any multiple roots because p′( z ) = nz n −1 has only 0 as a root and 0 is not a root of p ( z ) . There is one other result about roots of polynomials that is worth knowing. Suppose we have a polynomial with real , as opposed to complex, coefficients. Suppose that the complex number z is a root of the polynomial. Then the complex conjugate z is also a root. So you get two roots for the price of one. You can see this in the example of the previous paragraph. x 2 + 1 has j as a root, so it automatically must have − j as a root as well. Example 23 Let p ( z ) = z 4 − 4 z 3 + 9 z 2 − 16 z + 20 . Given that 2 +
j is a root, express p( z ) as a product of real quadratic factors and list all four roots, drawing attention to any conjugate pairs. Solution:
Since p has real coefficients, and complex roots occur in pairs consisting of a root and its complex conjugate. Given that 2 + j is a root, it follows that 2 − j must also be a root, and so the quadratic:
38Mathematical Numbers
(z − (2 + j ))(z − (2 − j )) = z 2 − 4 z + 5
must be a factor. Dividing
the given polynomial by this factor gives
(
)(
p ( z ) = z 4 − 4 z 3 + 9 z 2 − 16 z + 20 = z 2 − 4 z + 5 z 2 + 4
)
The roots of z 2 + 4 are 2j and its complex conjugate, − 2 j . Thus the given polynomial, of degree four, has two pairs of complex conjugate roots. Example 24 Express z 5 − 1 as a product of real linear and quadratic
factors. Solution:
We rely on our knowledge of the nth roots of unity from the previous section. Let
2π 2π 2π = cos + j sin 5 5 5
α = exp j
Then the roots of z 5 − 1 = 0 are α , α 2 , α 3 , α 4 , and, 1.
(
)
(
)(
)(
z 5 − 1 = ( z − 1) z 4 + z 3 + z 2 + z + 1 = ( z − 1)( z − α ) z − α 2 z − α 3 z − α 4
)
For convenience, write β = α 2 , and note that β = α 3 while
(
)
α = α 4 . Our problem is to factorize z 4 + z 3 + z 2 + z + 1 as a product of real quadratic factors. We know the roots are
α ,α , β , and β . Now construct the quadratic with roots α and α . We have: ( z − α )( z − α ) = z 2 − (α + α ) + αα = z 2 − 2ℜ(α ) + 1
Chapter One
39
where ℜ(α ) is the real part of α . Since ( z − β )(z − β ) behaves in the same way, we have:
(
)(
)
z 5 − 1 = ( z − 1) z 2 − 2ℜ(α ) + 1 z 2 − 2ℜ(β ) + 1
2π 4π ∴ z 5 − 1 = ( z − 1) z 2 − 2 cos + 1 z 2 − 2 cos + 1 5 5 and this is a product of real linear and quadratic factors. Problems
١) Express the complex number z = 1 − 3 j exactly in modulus - argument form. Hence find the modulus and principal argument of z 4 . ٢) Find all solutions w to the equation ω 3 = −27 j and mark them on an Argand diagram. ٣) Let z = 1 − j 2 ω = 3 + j be complex numbers. Express each of the following complex numbers in the rectangular form zω ,
ω z+2+ j
,
1 + 3 j − zz
٤) Express the complex number 2 + 2 j exactly in
modulus - argument form. Hence find all solutions
w to the equation ω 3 = −2 + 2 j and mark them on an Argand diagram.
40Mathematical Numbers
٥) Let z = 3 + j and ω = 1 − 7 j . Express rectangular form. Find also z ,
ω ω+z
in a
ω
ω,
z
٦) Express the complex number − 2 + 2 j in polar form. Hence solve the equation
z 3 = −2 + 2 j
expressing the solutions in polar form and marking them in the Argand Diagram. ٧) Let p ( z ) = z 5 − 5 z 4 + 8 z 3 − 2 z 2 − 8 z + 8 Show that p (2 ) = 0 . Show also that z 2 − 2 z + 2 is a factor of p ( z ) . Hence write p as a product of linear factors. ٨) Show that z − (1 + j ) is a factor of the real polynomial p ( z ) = z 3 + 2 z 2 − 6 z + 8 Hence write p as a product of linear factors. ٩) Let p ( z ) = z 4 − 3 z 3 + 5 z 2 − 27 z − 36 Show that
p (3 j ) = 0 . Hence write p as a product of linear factors. ١٠)
Express in polar z = −5 − j 3
١١)
Express in rectangular 2∠156 o and 5∠ − 37 o
١٢)
If z1 = 12 cos 125o + j sin 125o and
(
)
Chapter One
(
41
)
z 2 = 5 cos 72 o + j sin 72 o Then, find (i) z1 * z 2 and results in polar form
(
z1 giving the z2
)
If z1 = 12 cos 125o + j sin 125o , find z 3 and
١٣) 1 z3
١٤)
If z = x + jy , find the equations of the two loci
defined by: (i) z − 4 = 3 and (ii) arg( z + 2 ) = ١٥)
π 6
If z = x + jy , find the value of x and y when :
3z 3z 4 + = 1− j j 3− j ١٦)
Express 2 + j 3 and 1 − j 2 in polar form and
apply DeMoiver’s theorem to evaluate
(2 + j 3)4 . 1 − j2
Express the result in rectangular and exponential form. ١٧)
Find the fifth roots of − 3 + j 3 in polar and
exponential form. ١٨)
Express 5 + j12 in polar form and hence
evaluate the principle value of 3 (5 + j12 ) giving the results in rectangular form.
42Mathematical Numbers ١٩) Obtain the expansion of sin 7θ in terms of
sin θ .
Chapter 2 Matrices 2.1 Introduction A matrix is, by definition, a rectangular array of numeric or algebraic quantities, which are subject to mathematical operations. So a real matrix is an arrangement of real numbers into rows and columns. Matrices can be defined in terms of their dimensions (number of rows and columns). Let us take a look at a matrix with 4 rows and 3 columns (we denote it as a 4x3 matrix and call it A):
7 5 A = 2 9
6 8 12 5
1 1 0 0
The dimensions of this matrix are 4 by 3. The dimensions of a matrix tell you the size of the matrix because they tell you the number of rows and columns in the matrix. By convention, we list the number of rows before the number of columns. Definition 1 The dimensions of a matrix are the number of rows and columns (listed in that order) of the matrix. Each element of the matrix is named according to its position. Typically, capital letters represent matrices and small letters with subscripts represent elements in the matrix. Since vectors can be
Chapter Two
43 considered to be matrices with only one row or one column, they could be labeled with capital letters also. However, small letters usually represents vectors. The element 6 is in the position a12 (read a one two) because it is in row 1 and column 2. Also by convention, we list the row number of the element before the column number. An element in row i and column j would be denoted by aij . This gives us a compact way to refer to specific elements of a matrix. Can you represent the same information as before in a 3 by 4 matrix? Yes, you can. It would look like the matrix B which follows.
7 B = 6 1
5
2
8 1
12 0
9 5 0
Matrix B is the transpose of A, and A is the transpose of B. Transposing a matrix results in writing the columns as rows and the rows as columns, but what really happens is that element aij is placed in the position b ji of the new matrix. Therefore, a12 moves to the position b12 when we form the transpose of A. The transpose of A is denoted by AT (read A transpose). Therefore, matrix B is
AT . Definition 2 By the transpose of the m by n matrix A, denoted by
AT , we mean the n by m matrix, which has aij as its (i, j )th element.
44 Matrices Definition 3 We say that two m by n matrices, A and B are equal if their corresponding elements are equal. In other words, A = B if A and B have the same dimensions and
a11 = b11 , a12 = b12 , etc. Is A = AT ? Usually not, but we have a special word for a matrix which satisfies A = AT . Definition 4 A matrix is said to be symmetric if A = AT . Observe that the following matrix is symmetric:
9 2 A= 5 1
2 5 1 7 0 8 0 4 6 8 6 3
Notice that aij = a ji for all i and j; as is true for all symmetric matrices. Symmetric matrices are easy to spot because if you draw a line down the main diagonal (from 9 to 3 in this matrix), then the two halves are mirror images of each other. Symmetric matrices have many special qualities that will be used when you study matrices in more detail. The matrix A, given above, has another special property; it is a square matrix because A has the same number of rows as columns. Notice that A is a 4 by 4 square matrix. We said that the main diagonal for A runs from 9 to 3. For any square matrix, the main diagonal runs from the upper left corner to the lower right corner. Definition 5 We say that an m by n matrix is square if m = n .
Chapter Two
45
2.2 Addition and Subtractions of Matrices Definition 6 Matrices of the same dimensions are added by adding corresponding elements. For instance, aij corresponds to bij because they both lie in the ith row and jth column of their respective matrices. Therefore, we would add, aij + bij to obtain the (i, j ) th element of A + B : Example 1 Find the result of the following:
7 5 A+ B = 2 9 Solution:
6 8 12 5
1 8 1 9 + 0 5 0 11
6 6 9 4
1 0 1 0
6+6 1 + 1 15 12 2 7+8 5+9 8+6 1 + 0 14 14 1 = A+ B = 2+5 12 + 9 0 + 1 7 21 1 5+4 0 + 0 20 9 0 9 + 11 Think about the similarities between addition and subtraction. How do you think matrices are subtracted? Definition 7 Matrices of the same dimensions are subtracted by subtracting corresponding elements. 2.3 Multiplication of Matrices Multiplying a matrix by a scalar value involves multiplying every element of the matrix by that value. Here we multiply our 4x3 matrix A by a scalar value k:
46 Matrices
7 5 k * A =k * 2 9 The multiplication
1 k * 7 k *6 1 k * 5 k *8 = 12 0 k * 2 k *12 5 0 k * 9 k *5 operation on matrices differs 6 8
k *1 k *1 k * 0 k * 0 significantly from
its real counterpart. One major difference is that multiplication can be performed on matrices with different dimensions. The first restriction is that the first matrix has to have the same amount of columns as the second has rows. The reason for this will become clear shortly. Another thing to note is that matrix multiplication is not commutative i.e, (CD) does not equal (DC). The procedure for matrix multiplication is rather simple. First, we determine the dimensions of the resultant matrix. All we require is that there are as many columns in the first matrix as there are rows in the second. A simple way of determining is to look at the nearest and farthest dimensions of two matrix symbols written next to each other, for instance: C[2x3] D[3x2]. The nearest dimensions are both equal to 3, and so we know that the operation is possible. The farthest dimensions will give us the dimensions of the product matrix, so our result will be a 2x2 matrix. The general rule says that in order to perform the multiplication AB, where A is a mxn matrix and B a kxl matrix, we must have n=k. The result will be a mxl matrix.
Chapter Two
47 Performing the operation product involves multiplying the cells
of a particular rows in the first matrix by the cells of a particular column in the second matrix, adding the products, and storing the result in the cell of the resultant matrix whose coordinates correspond to the row of the first matrix and the column of the second matrix. For instance, in AB = C, if we want to find the value of c12, we must multiply the cells of row 1 in the first matrix by the cells of column 2 in the second matrix and sum the results. There are several interesting things to notice about matrix multiplication. We multiplied a 1 by 3 matrix by a 3 by 4 matrix and got a 1 by 4 matrix. The following picture expresses the requirements on the dimensions:
Let's also look closely at how we multiply the matrices because we will multiply matrices with larger dimensions later. This is a hands on activity. Take your left pointer finger and place it at the beginning of the first row of the first matrix (the only row we have in this case). Take your right pointer finger and place it on the first number of the first column of the second matrix. Multiply the two numbers to which you are pointing. Each time you move, your left
48 Matrices hand will go across the row, and your right hand will go down the column. When you reach the end of the row and column, add the numbers you have obtained from the multiplications. This number goes in the first row and first column of your product matrix. This is the same as taking the inner product of the first row of first matrix and the first column of the second matrix. Now you can move to the first row, second column doing the same thing. This number will go in the first row, second column of your product matrix. In short, position ij of your product matrix consists of the inner product of the ith row of your first matrix and the jth column of the second matrix. This is a lot easier to do than it is to describe! Your left hand will move across and your right hand will move down. Do this for every row and column combination to get your product matrix.
This
picture depicts the motions necessary to find a product: Inner product of row i with column j equals position ij Definition 8 An identity matrix is a square matrix with ones along the main diagonal and zeros elsewhere. Example 2
2 1 If S = [1 4 3] And R = 2 4 2
2 0 1
1 3 4 1 1 3
0 0 Find S * R 2
Chapter Two
49
Solution:
2 1 2 1 3 S * R = [1 4 3] 2 0 4 4 1 2 1 1 3 1 Column 1 of S , R = 1 * 2 + 4 * 2 + 3 * 2 = 17 4 Column 2 of S , R
= 1 * 2 + 4 * 0 + 3 *1 = 5
Column 3 of S , R
3 1 = 1 *1 + 4 * + 3 *1 = 8 3 4
Column 4 of S , R
= 1* 0 + 4 * 0 + 3 * 2 = 6
0 0 2
∴ S * R = [17 5 8 6] Example 3 Multiply the following matrices
2 1 2 4 2
2 0 1
1 3 4 1 1 3
0 17 8 0 * 13 2 4 2 * 17 + 2 * 8 + 1 * 13 + 0 * 4 63 1 3 = 2 * 17 + 0 * 8 + * 13 + 0 * 4 = 48 4 4 67 1 1 2 * 17 + 1 * 8 + 1 * 13 + 2 * 4 3 3
50 Matrices Example 4 Multiply the following matrices
2 1 R * F = 2 4 2
2 0 1
1 3 4 1 1 3
0 17 8 0 13 2 4
510 63 1250 70 = 48 1215 90 1 120 67 1450 3
2.4 Equations Solving equations is an important part of mathematics. If we are working with more than one unknown at a time, we need to solve systems of equations. You may already know how to solve a system of linear equations, but matrices provide a more compact way to arrive at the solution. Matrices are also easier to manipulate on a computer or calculator. Both of these facts will become more important when you work with larger systems. Example 5 Solve the following system of equations:
5 x1 + 3 x2 = 93 − 4 x1 − 2 x2 = −66 Solution: Let's look at a system of linear equations:
5 x1 + 3 x2 = 93 − 4 x1 − 2 x2 = −66 Can be written in matrix form as AX = B where
Chapter Two
5 A= − 4
51
3 x1 93 ; X = , and B = −2 − 66 x2
When you learned to solve systems of linear equations, you learned that (a) You arrive at the same solution no matter which equation you write first, (b) The solution doesn't change if you multiply an equation by a scalar other than zero, and, (c) You can replace an equation with the sum of that equation and another equation without changing the solution. These may not be exactly the words you used when you were solving a system of linear equations, but you did all these things. Experiment with the system above to convince yourself that these statements are true. We can also solve this system entirely in matrix form. We use the same rules, and we call them Elementary Row Operations (EROs). The EROs tell us that we can (a) Interchange any two rows; (b) Multiply any row by a non-zero scalar; and (c) Replace any row by the sum of that row and any other row. Proper use of EROs will leave us with a system that has the same solution as our original system, but is much easier to solve. If you were presented the system
x1 = a, x2 = b
52 Matrices You would be able to solve it instantly because you only have to read the solution. If this system were written using matrix notation, it would look like this:
1 0
0 x1 a The matrix = 1 x2 b
1 0
0 is the 2 by 2 identity 1
matrix. Because you can just read of the solution when a system is in this form, our first goal is to transform our system into this form. Let's solve the system above using matrices. We can represent this entire system with a 2 by 3 matrix, which looks like this:
5 − 4
93 . This is called an augmented matrix because we −2 − 66 3
combined 2 matrices (a matrix and a vector for this system). In this case, we combined the 2 by 2 coefficient matrix which is made of the coefficients for our unknowns and the 2 by 1 matrix from the right-hand side of the equations into one 2 by 3 matrix. In other words, we put A to the left of the bar and put b to the right of the bar. The application of an ERO to the augmented matrix does not change the solution set of the linear system that the augmented matrix represents because whatever you do to the left side of an equation, you also do to the right side. Therefore, we will arrive at the same solution whether we use augmented matrices or not, and augmented matrices are more compact to write. Using matrix notation, our goal is to transform our system into one that looks like the following form:
Chapter Two
1 0
53
0 a 1 b
In other words, we want the identity matrix to the left of the bar and the solution to the right of the bar. Remark 1 The bar is not a formal part of the matrix, so it is not necessary. It is placed there so that we can refer to the different parts of the augmented matrix and easily move back and forth between the augmented matrix and the linear system that it represents. In this book, r1 represents row 1 and so on.
5 − 4
93 Original augment matrix −2 − 66
1 − 4
0.6 18.6 r1 ÷ 5 −2 − 66
1 0
0.6 18.6 4 r1 + r2 0. 4 8. 4
1 0
0.6 18.6 1 21
1 0
0 6 1 21
3
r2 ÷ 0.4 −0.6 * r2 + r1
When we convert this from augmented matrix notation back to the algebraic notation for a system of equations, it looks like this:
54 Matrices
1x1 + 0 x2 = 6 0 x1 + 1x2 = 21 This tells us that x1 = 6 and x2 = 21 . Substitute this solution into the system to assure yourself that we are correct. If we systematically use elementary row operations (ERO) to obtain the identity matrix to the left of the bar, we call this the Gauss Jordan Elimination Method. Example 6 Now, let's solve the system using Gauss Jordan elimination.
5 x1 + 3 x2 = 70 − 4 x1 − 2 x2 = −56 5 − 4
70 Original augmented matrix. −2 − 56
1 − 4
0.6 14 −2 − 56
1 0 1 0
0.6 14 0. 4 0 0.6 14 1 0
3
r1 ÷ 5
4 * r1 + r2 r2 ÷ 4
0 14 1 − 0.6r2 + r1 0 1 0 Let's look at the scalar version of this equation, ax = b ; to help us find a general method for matrices. We know that x = a −1b if
Chapter Two
55
A ≠ 0 because a −1 = 1 / a where a −1 is called the multiplicative inverse or the reciprocal. There is something analogous to this with matrices. It is also called the inverse. With scalars, a −1a = aa −1 = 1. Definition 9 The matrix A−1 (called A inverse) is the inverse of a square matrix A if A −1 A = AA−1 = I where I is the identity matrix. Once we find A1; Ax = b can be solved by matrix multiplication rather than Gauss Jordan elimination. We follow the algebraic steps below to find an expression for x:
Ax = b
∴ A−1 Ax = A−1 b
∴ I * x = A−1 b
This means that if we find A−1 ; we only need to multiply to solve systems with the same matrix A for different b vectors. Please remember that A−1b ≠ b A−1 , so you must multiply in the correct order. Remark 2 In computational mathematics, the inverse is very seldom found because other methods exist that serve the same purpose and require fewer steps. However, the inverse will serve our needs at this level and is important in the theory of matrices. Example 7 Using the Gauss Jordan elimination method, let's find
A−1
0 2 4 where A 4 2 3 1 3 6
Solution:
56 Matrices
0 2 4 1 0 0 4 2 3 0 1 0 Original augmented matrix. 1 3 6 0 0 1
Switch r1 and r3 because we cannot have a zero on the main diagonal, and we would prefer 1 rather 4.
1 0 0 1 0 0 1 0 0 1 0 0
3 6 0 0 1 − 10 − 21 0 1 − 4 − 4r1 + r2 2 4 1 0 0 3 6 0 0 1 1 2.1 0 − 0.1 0.4 r2 / (− 10 ) 2 4 1 0 0 1 1 2.1 0 − 0.1 0.4 − 2r2 + r3 0 − 0.2 1 0.2 − 0.8 3
6
0
0 1 − 0.1 0.4 r3 / (− 0.2 ) 1 − 5 −1 4
3 6 0 1 2.1 0 0
0
Chapter 3 Calculus 3.1 Limits The concept of limits is essential to calculus. A good understanding of limits will help explain many theories in calculus. So, it is recommended to start studying calculus from limits. Consider a function f defined for values of x, as x gets close to a number a, not necessarily true for x = a . If the value of f ( x ) approaches a number b as x approaches a, then the limit of f ( x ) as x approaches a is equal to b, denoted as :
lim f ( x) = b
(1)
x→a
Example 1 Find the limit of f ( x) = 5 x + 2 as x approaches 3. Solution: It is clear that as x approaches 3, 5x approaches 15, and
5 x + 2 approaches 17. Thus; lim 5 x + 2 = 17 x →3
Example 2 Find the limits of f ( x) =
1 as x approaches 5. 2 x − 10
Solution: It is clear as x approach 5, 2 x − 10 approaches zero the
1 1 approaches which is undefined. Thus; 2 x − 10 0 1 = ∞ (undefiend ) x → 5 2 x − 10 lim
Chapter Three
75
This limit lim f ( x) = f ( x) represents a horizontal line, which x→a
says that as x approaches a, and f ( x) = f ( x) or c = c where c is a constant. Then, lim f ( x) = c x→a
Then, as x approaches a, f ( x ) also approaches c. Limits can be approached from the negative ( or left ) or the positive ( or right ) side of a number denoted as:
lim f ( x) = b or lim f ( x) = b
x→a −
x→a +
If lim f ( x) ≠ lim f ( x ) Then, x→a −
x→a +
lim f ( x) = does not exist
x→a
lim f ( x) = lim f ( x) = b
x→a −
x→a +
lim f ( x) = b
x→a
If the value of f ( x ) gets larger and larger without bound as x approaches a, then: lim f ( x) = ∞ x→a
Similarly; If the value of f ( x ) gets smaller and smaller without bound as x approaches a, then: lim f ( x) = ∞ x→a
Consider a function f defined for large positive ( or negative ) values of x, as x increases indefinitely in the positive ( or negative ) direction. If the value of f ( x ) approaches a number b as x increases
76 Calculus
(or decreases ) indefinitely, then the limit of f ( x ) as x increases (or decreases ) indefinitely is equal to b, denoted as :
lim
x → +∞
f ( x) = b
or
lim
x → −∞
f ( x) = b
A function f ( x ) is continuous at x = a if f is defined at x = a and either; f is not defined anywhere near a, or f is defined arbitrarily near x = a and, lim f ( x) = f ( x) x→a
Conversely, A function f ( x ) is discontinuous at x = a if f ( x ) is defined at x = a and f ( x ) is not continuous at x = a . 3.2 Derivatives Suppose y = f (x) is shown in Fig.1, the slope of the curve is the slope of the secant line between point A and another point P on the graph is shown in the following equation:
m AP =
( f (x + h ) − f (x )) = ( f (x + h ) − f (x )) (x + h ) − x h
Notice that h can change and with it the location of point P, therefore h is the limiting factor of the slope of the curve. As h gets close to point A, the slope of the curve becomes the tangent of the graph at point A. The tangent line of f at point A is: lim
( f (x + h ) − f (x ))
h →0
So,
the Differentiation ( f (x + h ) − f (x )) lim h h →0
of
function
h f
at
x
is:
Chapter Three
77 If this limit exists, then it is called the derivative of function f
at x, which is denoted by f ′( x) or So, f ′( x) or
dy = lim dx h →0
dy . dx
( f (x + h ) − f (x )) h
Fig.1 The Approximate slope of the curve at point A.
Fig.2 The slope of the curve at point A. So, general rules of differentiation are shown in the appendix of this book before going in the following example you have to take a look to the rules of differentiation in the appendix. Example 3 Find from the first principles
(
d tan ( x ) e dx
)
78 Calculus Solution: Put u = tan ( x ) ∴ y = eu ∴ But from chain rule,
∴
(
dy du = eu and = sec 2 x du dx
dy dy dy = . , dx du dx
)
d tan ( x ) e = e tan ( x ) * sec 2 x dx
Example 4 Find
d 3 2 − x dx
Solution: 2
1
d 3 2 2 3 −1 2 − 3 =− x − x = − x 3 3 dx Example 5 Find
d − 3 dx x 2 − 1
Solution: 1 3 − − 3 d − 3 d 2 2 = − 3 * x − 1 2 = * x − 1 2 * 2x 2 2 dx x − 1 dx
(
(
)
)
3x 3x * x 2 − 1 d − 3 ∴ = = 2 dx x 2 − 1 x 2 − 1 x 2 − 1 x2 − 1
(
Example 6 Solution:
Find
)
(
)
d (5 x + 7 )4 dx
d (5 x + 7 )4 = 4(5 x + 7)3 * 5 = 20(5 x + 7 )3 dx
Chapter Three
79
Example 7 Find Solution:
d (sin (5 x + 6)) = 5 cos(5 x + 6) dx
Example 8 Find Solution:
( ( ))
d cos x 2 dx
( ( ))
( )
d (ln(3 − 4 cos x )) dx
d 1 4 sin x (ln(3 − 4 cos x )) = * (4 sin x ) = dx 3 − 4 cos x 3 − 4 cos x
Example 10 Find Solution:
( )
d cos x 2 = − sin x 2 * 2 x = −2 x * sin x 2 dx
Example 9 Find Solution:
d (sin (5 x + 6)) dx
d (log10 (2 x − 1)) dx
d 1 2 (log10 (2 x − 1)) = *2 = (2 x − 1) ln(10) (2 x − 1) ln(10) dx
Example 11 Find
(
)
d 5x e * ln (2 x − 1) dx
Solution: Assume y = e5 x * ln (2 x − 1) , u = e5 x and v = ln (2 x − 1)
∴ y = uv , ∴
dy dv du =u +v dx dx dx
1 dy = e5 x * 2 + ln (2 x − 1) * 5 e5 x (2 x − 1) dx
dy 2e 5 x ∴ = + 5 e5 x ln (2 x − 1) dx (2 x − 1)
80 Calculus Example 12 Find
(
)
d 3 x 5 ln(sin x ) dx
Solution: Assume y = 3 x 5 ln (sin x ) , u = 3 x 5 and v = ln (sin x )
∴ y = uv ,
dy dv du =u +v dx dx dx
(
)
(
)
∴
d 1 * cos x + ln (sin x ) * 3 * 5 x 4 3x 5 ln (sin x ) = 3x 5 dx sin x
∴
d 3x 5 ln (sin x ) = 3 x 5 cot x + 15 x 4 * ln (sin x ) dx
d e 2 x Example 13 Find dx ln (3x ) e2x , u = e 2 x and v = ln(3 x ) Solution: Assume y = ln (3x ) dv du −u v dy = dx 2 dx dx v 3 2x − e2x * 2 x ln (3 x ) * 2e d e 3x = ∴ 2 dx ln (3 x ) (ln(3x )) 1 e 2 x (2 ln ( x ) + 2 ln (3) − ) e 2 x (2 x ln ( x ) + 2 x ln (3) − 1) x = = 2 (ln(3x )) x(ln (3 x ))2 d x 2 sinh( 2 x) Example 14 Find dx cosh (3 x )
Chapter Three
81
x 2 sinh( 2 x) Solution: Assume y = , u = x 2 , v = sinh (2 x ) , cosh (3 x ) and w = cosh (3x ) d uv uv 1 du 1 dv 1 dw Where + − = . dx w w u dx v dx w dx d x 2 sinh( 2 x) x 2 sinh( 2 x) 1 1 ∴ = * 2 * 2x + * 2 cosh (2 x ) dx cosh (3 x ) cosh (3 x ) x sinh (2 x ) − ∴
1 * 3 sinh (3 x ) cosh (3 x )
d x 2 sinh(2 x) x 2 sinh(2 x) 2 2 cosh(2 x ) 3 sinh (3x ) = − * + dx cosh(3 x ) cosh(3 x ) x sinh (2 x ) cosh(3 x )
d x 2 sinh(2 x) x 2 sinh(2 x) 2 2 cosh (2 x ) 3 sinh (3x ) * + ∴ = − dx cosh (3 x ) cosh (3x ) x sinh (2 x ) cosh (3x ) d x 2 sinh( 2 x) 2 x sinh(2 x) x 2 2 cosh (2 x ) = + ∴ dx cosh (3x ) cosh (3x ) cosh (3x ) 3x 2 sinh(2 x) tanh (3x ) − cosh (3x ) Example 15 Find
(
d 5 x sin 2 x cos 4 x dx
)
Solution: Assume y = uvw = x 5 sin 2 x cos 4 x , where u = x 5 , v = sin 2 x , and
w = cos 4 x Take the logarithm for both sides we get:
82 Calculus
(
) ( )
∴ ln ( y ) = ln x 5 sin 2 x cos 4 x = ln x 5 + ln ( sin 2 x ) + ln ( cos 4 x ) By differentiating both sides of the above equation we get:
1 dy 1 1 1 (− 4 sin 4 x ) 2 cos(2 x ) + = 5 5x 4 + ( cos 4 x ) sin 2 x y dx x 1 dy 5 ∴ = + 2 cot 2 x − 4 tan 4 x y dx x dy 5 ∴ = x 5 sin 2 x cos 4 x * + 2 cot 2 x − 4 tan 4 x dx x ∴
Example 16 Find
d (1 + tan 2 x )3 dx
d (1 + tan 2 x )3 = 3(1 + tan 2 x )2 * d (1 + tan 2 x ) dx dx d d ∴ (1 + tan 2 x )3 = 3(1 + tan 2 x )2 * 0 + sec 2 (2 x ) (2 x ) dx dx d ∴ (1 + tan 2 x )3 = 3(1 + tan 2 x )2 * 2 * sec 2 (2 x ) dx d ∴ (1 + tan 2 x )3 = 6 sec 2 (2 x ) * (1 + tan 2 x )2 dx d 2 x 1 − cot Example 17 Find dx 3 Solution:
(
)
Solution: 1 −1 2 x 2
d 2 x 1 − 1 cot = 1 − cot dx 3 2 3
*
d x 1 − cot 2 dx 3
Chapter Three
∴
∴
∴
d 2 x − 1 cot = dx 3
83
x d x 0 − 2 cot * cot 3 dx 3 x 2 1 − cot 2 3
d dx
x 1 x − 2 cot * − csc 2 x 3 3 3 1 − cot 2 = 3 x 2 1 − cot 2 3
d dx
x x cot * csc 2 x 3 3 1 − cot 2 = 3 x 3 * 1 − cot 2 3
3.2.1 Implicit differentiation So far, all the functions being differentiated are explicit functions, meaning that one of the variables was specifically given in terms of the other variable.
f ( x) = 5 x + 2, then f ′( x) = 5 However, not all functions are given explicitly and are only implied by an equation. Example 18 xy = 1 is an equation given implicitly, explicitly it is
y = 1 / x . Now to find dy / dx for xy = 1, simply solve for y and differentiate.
84 Calculus Solution: y =
1 = x −1 x
∴ y′ = − x − 2 =
−1 x2
But, not all equations are easily solved for y, as in the equation
2 y + xy 3 = 6 xy + y 2 This is where implicit differentiation is applied. Implicit differentiation is taking the derivative of both sides of the equation with respect to one of the variables. Most commonly, used is the derivative of y with respect to x. or dy / dx . Since we have not solved for y as a function of x, the derivative of y must be left as
dy / dx . Example 19 Find the slope of 3 x + y 3 = y 2 + 4 at point (1,3) Solution:
(
∴ 3 + 3y2
dy dy = 2y dx dx
∴ y (2 − 3 y ) ∴
)
(
d d 2 3x + y 3 = y +4 dx dx ∴ 2y
) dy dy − 3y2 =3 dx dx
dy =3 dx
dy 3 = dx y (2 − 3 y )
Then the slope of the curve at point (1,3) is ∴
dy 3 3 1 = = =− dx (1,3) y (2 − 3 y ) (1,3) 3(2 − 3 * 3) 7
Chapter Three
85
3.3 Integration 3.3.1 Introduction You are now familiar with differentiation principles and have had a lot of examples about the differentiation in the previous sections. Now we are going to do the same work with integration. Integration is the reverse of differentiation. In differentiation we start with a function and proceed to find its differential coefficient. In integration, we start with the differential coefficient and have to work back to find the function from which it has been derived. The following example clearify the meaning of Integration.
(
)
d 5 x + 4 = 5x 4 dx Therefore it is true, in this case, to say that the integral of 5x 4 , with respect to x, is the function from which it came, So,
∫
5 x 4 dx = x 5 + c
Where c is constant and always called the constant of integration. This constant is very important to be included in the result of integration. If you don t put the constant of integration the results is not genuine.
dy = f ( x ) , then y is the function whose derivative is f ( x ) dx and is called the anti-derivative of f ( x ) or the indefinite integral of So, if
f ( x ) , denoted by
∫ f (x )dx .
Similarly, if y = ∫ f (u )du , then
86 Calculus
dy = f (u ) . Since the derivative of a constant is zero, all indefinite du integrals differ by an arbitrary constant. There is some important integration rules are shown in the Appendix of this book. 3.3.2 Definite Integrals Let f ( x ) be defined in an interval a ≤ x ≤ b . Divide the interval into n equal parts of length ∆x = (b − a ) / n . Then the definite integral of f ( x ) between x = a and x = b is defined as: b [ f (a )∆x + f (a + ∆x )∆x + f (a + 2∆x )∆x + ..... + f (a + (n − 1)∆x )∆x] ∫a f (x )dx = nlim →∞
n
∑ f (xi ) ∆x
= lim
n → ∞ i =1
If f ( x ) =
d g ( x ) . Then by the fundamental theorem of the integral dx
calculus the above definite integral can be evaluated by using the result. b
∫a
d b ( g ( x ))dx = g ( x ) a = g (b ) − g (a ) a dx
f ( x )dx = ∫
b
Properties of definite integration: b
a
∫a f (x )dx = − ∫b f (x )dx b b ( ) kf x dx = k ∫a ∫a f (x )dx b b b ∫a [ f (x ) ± g (x )]dx = ∫a f (x )dx ± ∫a g (x )dx b c b ( ) ( ) f x dx = f x dx + ∫a ∫a ∫c f ( x ) dx where a < c < b
Chapter Three
87
3.3.3 Methods of Integrations: 1- Substitution Sometimes it s not easy to solve some integrals without using intermediate function and we can integrate with respect to this function and then we can substitute to get the results in terms of x. Example 20 Find the results of:
∫
(
)2
2 x x 2 + 3 dx
Solution: Substitute u = x 2 + 3 and take its derivative with respect to du = 2 xdx substitute for the values of u and du in the above integral we get:
∫
u3 u du = +c 3 2
Once the solution has been found in terms of u, substitute back into terms of x, the final solution is: 3 2 ( ) 3 x + 2 +c ∫ 2 x(x + 3) dx = 2
3
Example21 Find the results of the following integral:
∴
∫
(
)
x x − 1 dx = ∫ u 2 + 1 . u . 2udu
(
x x − 1 dx
x − 1 , then u 2 = x − 1 and x = u 2 + 1
Solution: Substitute u = So, 2udu = dx
∫
)
(
)
= ∫ x x − 1. dx = ∫ u 2 + 1 2u 2 du = ∫ 2u 4 + 2u 2 du
(
)
2 2u 5 2u 3 = + + c = u 3 3u 2 + 5 + c 5 3 15 Substitute in the above equation for u =
x −1,
88 Calculus
(
) (
)
∴ ∫ x x − 1 dx =
2 15
∴ ∫ x x − 1 dx =
2 (x − 1)3 / 2 (3(x − 1) + 5) + c 15
∴
∫
x x − 1 dx =
3 2 x − 1 3 x − 1 + 5 + c
2 (x − 1)3 / 2 (3x + 2) + c 15
2- Substitution Involving Trigonometric Integrals Example 22 Find the results of the following:
∫
sin 2 x cos x dx
Solution: Assume u = sin x then du = cos x dx substitute in the above integral, we get:
∫
2
sin x cos x dx = ∫
u3 u du = +c 3 2
Substitute in the above equation for u = sin x we get:
∫
sin 3 x sin x cos x dx = +c 3 2
In the following table there is the suitable trigonometric substitution for different forms of integrals. Terms involving in the integral
Substitution
1-
a2 − u2
u = a sin θ
2-
a2 + u2
u = a sinh θ
3-
u2 − a2
u = a cosh θ
Chapter Three
89
Example 23 Find the results of the following integral:
∫
4 − x 2 dx Solution: Assume x = 2 sin θ then dx = 2 cos θ dθ . It is clear that sin θ =
that cos θ =
x then from the following figure, it is clear 2
4 − x2 , 2
Substitute in the above integral, we get:
∫
4 − x 2 dx = ∫
But cos 2 θ = 1 − sin 2 θ then,
4 − x2
∫
4 − x 2 dx = ∫ 4 cos 2 θ dθ
1 (1 + cos 2θ ) 2
sin 2θ 4 − x 2 dx = 2θ + +c 2
∴∫
But sin 2θ = 2 sin θ cos θ ∴
∫
4 − x 2 dx = 2(θ + sin θ cos θ ) + c
As we know from our assumption that: ∴ θ = sin
−1
x
θ
= ∫ 4 1 − sin 2 θ * cos θ dθ
But cos 2 θ =
2
4 − 4 sin 2 θ * 2 cosθ dθ
x x , sin θ = , and cos θ = 2 2
4 − x2 2
90 Calculus ∴
∴
∴
∫
4 − x2 −1 x x 4 − x dx = 2 sin + * 2 2 2
∫
2 −1 x x 4 − x 4 − x dx = 2 sin + 4 2
2
2
+c
+c
x 4 − x2 +c 4 − x dx = 2 sin + 2 2 2
∫
−1 x
Example 24 Find the results of the following integral:
x 2 + 4 dx
∫
Solution: Assume x = 2 sinh θ then dx = 2 cosh θ dθ substitute in the above integral, we get:
∫
x 2 + 4 dx = ∫
But cosh 2 θ =
∫
4 sinh 2 θ + 4 2 cosh θ dθ
1 (1 + cosh 2θ ) 2
x 2 + 4 dx = 4 ∫
1 (1 + cosh 2θ ) dθ = 2θ + sinh 2θ + c 2 2
But sinh 2θ = 2 sinh θ * cosh θ . Substitute in the above equation, we get:
∫
x 2 + 4 dx = 2[θ + sinh θ cosh θ ] + c
But cosh 2 θ = 1 + sinh 2 θ , Then, cosh θ = 1 + sinh 2 θ Substitute that in the above equation we get:
Chapter Three
91
x 2 + 4 dx = 2 θ + sinh θ 1 + sinh 2 θ + c
∫ ∴
∫
2 x x x − 1 x + 4 dx = 2 sinh + 1 + + c 2 2 2 2
Example 25 Find the results of the following integral:
x 2 − 4 dx
∫
Solution: Assume x = 2 cosh θ then dx = sinh θ dθ substitute in the above integral, we get:
x 2 − 4 dx = ∫
∫
4 cosh 2 θ − 4 2 sinh θ dθ
x 2 − 4 dx = 4 ∫
∴∫
cosh 2 θ − 1 2 sinh θ dθ
But cosh 2 θ − 1 = sinh 2 θ
∴∫
x 2 − 4 dx = 4 ∫ sinh 2 θ dθ
But sinh 2 θ =
1 (cosh 2θ − 1) , 2
x 2 − 4 dx = 2 ∫ (cosh 2θ − 1) dθ
∫
But sinh 2θ = 2 sinh θ cosh θ , ∴
∫
x 2 − 4 dx = 2[sinh θ cosh θ − θ ] + c
But we know that sinh 2 θ = cosh 2 θ − 1, ∴sinh θ = cosh 2 θ − 1
92 Calculus Substitute in the above equation we get: ∴∫
cosh 2 θ − 1 * cosh θ − θ + c
x 2 − 4 dx = 2
But cosh θ =
x , 2
∴∫
x − 4 dx = 2
∴∫
x x2 − 4 −1 x +c x − 4 dx = − 2 cosh 2 2
2
x2 x −1 x +c − 1 * − cosh 4 2 2
2
3- Integration By Parts If u and v are functions of x. Then we know that:
d (uv ) = u dv + v du dx dx dx Now integrate both sides with respect to x. On the left, we get back to the function from which we started
∴∫
d (uv )dx = ∫ u dv dx + ∫ v du dx dx dx dx
∴ uv = ∫ u
dv du dx + ∫ v dx dx dx
And rearranging the terms, we have
∫
u
dv du dx = uv − ∫ v dx dx dx
Chapter Three
93 On the left-hand side, we have a product of two factors to integrate. One factor is chosen as the function u; the other is thought of as being the differential coefficient of some function v. To find v, of course, we must integrate this particular factor separately. Then, knowing u and v we can substitute in the right-hand side and so complete the routine. You will notice that we finish up with another product to integrate on the end of the line, but, unless we are very unfortunate, this product will be easier to tackle than the original one. This is the key to the routine: ∴
∫
u
dv du dx = uv − ∫ v dx or dx dx
∫
u dv = u v − ∫ v du
This method is called integration by parts. Example 26 Find the results of the following integral:
∫
x cos x dx
Solution : Assume u = x
dv = cos x
du = dx and From this formula
∫
v = sin x
∫
u dv = u v − ∫ v du , we get:
x cos x dx = x sin x − ∫ sin x dx = x sin x − (− cos x ) + c
∴ ∫ x cos x dx == x sin x + cos x + c
94 Calculus Example 27 Find the results of the following integral Solution: Assume Then,
u=x
dv = ln x
du = dx and
v = x ln x − x
From this formula
∫
∫
u dv = u v − ∫ v du we get:
x ln x dx = x 2 ln x − x 2 − ∫ ( x ln x − x )dx
x2 +c ∴ 2 * ∫ x ln x dx = x 2 ln x − x 2 + 2 x2 +c ∴ 2 * ∫ x ln x dx = x ln x − 2 2
∴
∫
x 2 ln x x 2 x ln x dx = − +c 2 4
Example 28 Find the results of: I = ∫ x 2 sin x dx Solution: Assume
u = x2
dv = sin x dx
Then,
du = 2 xdx and
v = − cos x
From this formula
∫
u dv = u v − ∫ v du we get:
I = ∫ x 2 sin x dx = − x 2 cos x + ∫ 2 x cos xdx Assume u = x
dv = cos xdx
∫
x ln x dx
Chapter Three
95
∴ du = dx and, From this formula
∫
v = sin x u dv = u v − ∫ v du we get:
I = ∫ x 2 sin x dx = − x 2 cos x + 2 x sin x − 2 ∫ sin x dx I = ∫ x 2 sin x dx = − x 2 cos x + 2 x sin x + 2 cos x + c Sometimes, when integrating by pacts, an integral comes up that is similar to the original one, if that is the case, then this expression can be combined with the original
Example 29 Find the results of the following: I = ∫ e x sin x dx Solution: Assume
u = ex
dv = sin x dx
∴ du = e x dx and From this formula
∫
v = − cos x
u dv = u v − ∫ v du we get:
I = ∫ e x sin x dx = −e x sin x + ∫ e x cos x dx + c Assume u = e x
dv = cos x dx
∴ du = e x dx and From this formula
∫
v = sin x
u dv = u v − ∫ v du we get:
96 Calculus
I = ∫ e x sin x dx = −e x sin x + e x sin x − ∫ e x sin x dx + c 142 4 43 4 I
Add
∫
e x sin x dx = I to the both sides. Then,
2 I = 2 ∫ e x sin x dx = −e x sin x + e x sin x + c I = ∫ e x sin x dx =
(
1 − e x sin x + e x sin x + c 2
)
Example 30 Find the results of the following: I = ∫ e5 x cos 3 x dx
u = e5 x
Solution: Assume
dv = cos 3 x dx
∴ du = 5e5 x dx and From this formula
∫
v=
sin 3x 3
u dv = u v − ∫ v du we get:
e 5 x sin 3 x 5 I = ∫ e cos 3 x dx = − ∫ e 5 x sin 3 x dx 3 3 u = e5 x dv = sin 3 x dx 5x
∴ du = 5e5 x dx and From this formula
I=
∫
e 5 x cos 3 x dx =
∫
v=
− cos 3x 3
u dv = u v − ∫ v du we get:
e 5 x sin 3 x 5 e 5 x cos 3 x 5 − * − + ∫ e 5 x cos 3 x dx 3 3 3 3
Chapter Three
I=∫ e
5x
97 2
e5 x sin 3 x 5 5 x 5 cos 3 x dx = + e cos 3 x − ∫ e5 x cos 3 x dx 3 9 3 1442443 I
e 25 25 5x 1 + * I = 1 + * ∫ e cos 3 x dx = 9 9
5x
sin 3x 5 5 x + e cos 3x 3 9
e 5 x sin 3 x 5 5 x 34 34 5x ∴ * I = * ∫ e cos 3 x dx = + e cos 3 x 3 9 9 9
∴I =∫ e
5x
5x 9 e sin 3 x 5 5 x cos 3 x dx = + e cos 3 x + c 3 9 34
(
)
(
)
1 ∴ I = ∫ e 5 x cos 3 x dx = 3e 5 x sin 3 x + 5e 5 x cos 3 x + c 34 1 ∴ I = ∫ e 5 x cos 3 x dx = 3e 5 x sin 3 x + 5e 5 x cos 3 x + c 34 3 sin 3 x + 5 cos 3 x ∴ I = ∫ e 5 x cos 3 x dx = e 5 x +c 34 4- Integration By Partial Fractions In case we have integration like this one
∫
7 x + 12 dx we can use x( x + 2 )
the method of partial fraction to break it to be like that
∫
3 4 + dx . So that it is easy to integrate it. So, ( ) x x + 2 3 4 ∴∫ + dx = 3 ln x + 4 ln( x + 2) + c x (x + 2)
98 Calculus The method, of course, hinges on one's being able to express the given function in terms of its partial fractions. The rules of partial fractions are as follows: (i) The numerator of the given function must be of lower degree than that of the denominator. If it is not, then first of all divide out by long division. (ii) Factorize the denominator into its prime factors. This is important, since the factors obtained determine the shape of the partial fractions. (iii) A linear factor (ax + b ) gives a partial fraction of the form
A ax + b A B + ax + b (ax + b )2
(iv) Factors (ax + b )2 gives partial fractions (v)
Factors
(ax + b )3
give
partial
fractions
A B C + + ax + b (ax + b )2 (ax + b )3 (iv) a quadratic factor
(ax 2 + bx + c)
gives partial fractions
Ax + B ax 2 + bx + c Example 31 Find the results of the following:
∫
dx (x + 1)(x + 3)
Chapter Three
99
Solution : Breaking into partial fractions we get:
1
=
A B + (x + 1) (x + 3)
(x + 1)(x + 3) Multiply ( x + 1)( x + 3) to both sides of the equation, Then: 1 = A( x + 3) + B( x + 1) = Ax + 3 A + Bx + b 1 = Ax + Bx + 3 A + b
1 = ( A + B )x + 3 A + B The coefficients on both sides of the equation must be the same that is that the coefficient of x on the left side of the equation must equal the coefficient of x on the right side of the equation.
A + B = 0 and 3 A + B = 1 Solving the above equations for A and B,
1 ∴ A= , 2
and B = −
1 2
So the integral is equal to:
∴
(− 1 / 2) dx 1/ 2 ∫ (x + 1)(x + 3) = ∫ (x + 1) + (x + 3) dx
∴∫
dx 1 1 1 dx − ∫ dx = ∫ (x + 1)(x + 3) 2 (x + 1) ( x + 3)
∴∫
dx 1 = (ln( x + 1) − ln( x + 3)) + c (x + 1)(x + 3) 2
100Calculus Problems: 1- Find the slope of the tangent line to the following function at (a) y = 3 x + 4 , at x = 2 (b) y =
x + 1 at x = −1
2- Find
dy for the following functions: dx
a)
y = (1 + tan 2 x ) 3
2
(
(b) y = sin 1 − x
)
Find the following integrals π
(a)
∫
x3 + 2 x 2 − x dx 3 x
(c)
∫
x 2 sin xdx
(d)
∫
x 2 e5 x dx
(e)
∫
e3 x sin 2 x dx
(f)
∫
e3 x ln (5 x )dx
(g)
∫ sin 2 x cos 3x dx
(h)
∫
x sin 2 x dx
(i)
∫ ln(sin 2 x ) dx
(j)
∫
e 2 sin 5 x dx
2
1 + sin x dx 3 cos x 0
(b) ∫
Chapter 4 Ordinary Differential Equations 4.1 Definition: Differential equation is an equation, which contains at least one derivatives or differential of unknown function. Many important and significant problems in the physical science when formulated in mathematical terms require the determination of a function satisfying a differential equation containing derivatives of unknown function. This function can be obtained by solving the differential equation. The following are examples of different differential equations:
dy = sin x dx d2y + ky = cos x dx 2 x 3 y ′′′ − 7 x 2 y ′′ + 24 x y ′ − 36 y = 0 2
2
( y 2 e x y + 4 x 3 )dx + (2 xy e x y − 3 y 2 )dy = 0
∂2 y
(1) (2) (3) (4)
∂2 y
∂2 y (5) + = ∂x 2 ∂t 2 ∂t∂x Variables that denote values of a function are often called dependant variables. The one may take on any value in the domain of the function which the dependant variables stand for is called independent variables. Thus in (1) and (2) y is the dependant variable. In (3) and (4) either x or y can be the dependant variable, the other variable then being independent.
102 Ordinary Differential Equations 4.2 Classifications Of Differential Equations The differential equations can be classified according to its type, order and linearity as following: 4.2.1 Classification By Type The differential equation contains only ordinary derivatives of one or more dependant variables, with respect to a single independent variable; it is called ordinary differential equation. For example (1), (2), (3) and (4) are ordinary differential equation. But, the differential equation contains partial derivatives of one or more dependent variables is called partial differential equation. Equation (5) is an example of partial differential equation: 4.2.2 Classification By Order The order of the highest derivatives in a differential equation is called the order of differential equation. It is clear that (1) and (4) are first order differential equation, (2) and (5) are second order and (3) is third order differential equation. 4.2.3 Classification By Linearity A differential equations is said to be linear if it has the following form:
f n (x)
dny n
dx
+ f n−1(x)
d n−1y n−1
dx
+ .................f1(x)
d1 y 1
dx
+ f0 (x)
d0y 0
dx
= g(x) (6)
The linear differential equations are characterized by two properties:
Chapter Four
103 ١- The dependant variable y and all off its derivatives are of the first degree. ٢- Each coefficient depends on only the independent variable x.
The differential equations not in the previous form are called nonlinear differential equations. According to the above two properties (1), (2) and (3) are linear differential equation but (4) and (5) are nonlinear. 4.3 Solution Of Differential Equation A solution of differential equations is functions which satisfy the differential equation. Whereas all variables which appear in algebraic or transcendental equation are called (unknowns). The solution of differential equation can be obtained by different methods which will be detailed in the following items: 4.3.1 Graphical Solution Of Differential Equations At each point in xy plane where f ( x, y ) is dependant, the following differential equation y ′ = f ( x, y ) provides a value of y′ , which can be thought of as the slope of linear segment through that point. The totality of all such line segments form the direction field for the given differential equations. The integral curves which, at every point are tangent to the element of the direction field associated with the point.
104 Ordinary Differential Equations The general shape of the integral curves can sometimes be visualized by drawing the elements of the direction field at a sufficiently large number of points. The task of constructing the direction field for a given differential equation y ′ = f ( x, y ) is usually carried out in the following manner: Note that the direction field has the same slope C at all points on the curve f ( x, y ) = C , where C is constant. The family of curves
f ( x, y ) = C , for all possible values of C, are the level curves of f ( x, y ) , in the present context are known as the isoclines of the differential equation y ′ = f ( x, y ) . Example 1 Solve the following differential equations graphically:
y ′ = xy Solution: For the linear differential equation y ′ = xy , the isoclines are shown in Fig.1. The direction field is constructed by drawing line segments having the appropriate slope at a number of points on each isocline. Thus on the curve y = 1 / x , corresponding to C = 1 , the line segments are drawn with slope one each point. The process can be continued until the direction field is sufficiently well exhibited. Drawing curves tangent to the direction field at each point can show the general behavior of the integral curves. The process is applicable in the same manner in the general equation
y ′ = f ( x, y ) .
Chapter Four
C=−4 C=−3 C=−2 C=−1 C=1 C =2 C=3 C=4
105
C=4 C=3 C =2 C=1 C=−1 C=−2 C=−3
C=−4
Fig.1 The isoclines and direction field of y ′ = xy . 4.3.2 Separable Differential Equations The differential equation that can be reduced to be as shown in (7) is called separable differential equation. Sometimes, this equation can be called equation with separable variables.
dy L( x) or (7) = L( x)dx = g ( y ) dy dx g ( y ) If y = f (x) is a solution of the equation, then by integrating both sides of (7) we obtain: ∫ L( x)dx = ∫ g ( y ) dy By evaluating the above integral we get the general solution of (7).
106 Ordinary Differential Equations Example 2 Solve the following differential equation:
xdx + 2 y dy = 0 Solution: By integrating both sides directly we get:
∫ xdx + ∫ 2 y dy = c1 x2 ∴ + y 2 = c1 or x 2 + 2 y 2 = c 2 Where c = 2c1 and c, and c1 are constants. Example 3 Solve the following differential equation:
xydx + (2 xy 2 + 4 y 2 − x − 2)dy = 0 Solution: The above differential equation can be reduced to be as following: xydx + (2 y 2 − 1)( x + 2)dy = 0 Dividing the above equation by y ( x + 2) we get:
(2 y 2 − 1) x 2 1 dx + 2 y − dy = 0 dx + dy = 0 Or 1 − ( x + 2) y y ( x + 2) Integrating both sides we have:
2 1 1 2 − dx + y − dy = C ∫ ( x + 2) ∫ y ∴ x − Ln( x + 2) 2 + y 2 − Ln y = LnC
(
)
∴ x + y 2 = − Ln cy ( x + 2) 2 Or
2
e x + y = cy ( x + 2) 2
Example 4 Solve the following differential equation:
4 xy 2 dx + ( x 2 + 1)dy = 0 Solution: First, divide throughout by y 2 ( x 2 + 1) .
Chapter Four
∴
4x ( x 2 + 1)
dx +
107
dy y2
=0
Now the solution can be obtained by integrating each term separately, it is as the following:
2 Ln( x 2 + 1) −
1 =C y
Example 5 Solve the following differential equation:
dy = 1 + x + y 2 + xy 2 dx Solution: dy dy = (1 + x ) + y 2 (1 + x ) ∴ = (1 + x )(1 + y 2 ) dx dx By separating variables we get the following:
dy
= (1 + x ) dx (1 + y 2 ) By integrating both sides of the above equation we can get the x2 −1 +c following: tan y = x + 2 2 −1 x ∴ y = tan + x + c 2 Example 6 Solve the following differential equation:
(x ln(x ))dy − ydx = 0 Solution: By separating the variables and take logarithm of both sides we can get the following equation:
108 Ordinary Differential Equations
∫
dy dx =∫ y x ln x
∫
or
dy 1 / x dx =∫ y ln x
∴ ln y = ln (ln x ) + ln c or ln y = ln(c ln x ) By taking the exponential of both sides we can get the following form: ∴ y = c ln ( x ) Example 7 Solve the following differential equation:
x dx − y e − x dy = 0 where, y (0) = 1 Solution: x dx − y e − x dy = 0 can be changed to be in the following form: x dx = y e − x dy By integrating both sides we get the following:
∫
ydy = ∫ xe x dx + c
(
)
(
)
(
)
(
)
y2 ∴ = − xe x − e x + c ∴ y 2 = 2 e x − xe x + c1 2 But y (0 ) = 1 , then by substituting in the above equation we can get 2 ∴ c1 = − = −1 the value of c1 , 1 = 2(1 − 0 ) + c1 2 ∴ y 2 = 2 e x − xe x − 1 ∴ y = 2 e x − xe x − 1 4.3.3 Equation Reducible To Separable Form In the case of the differential equations not separable but it can be made separable by a simple change of variable, the following procedures can be used to solve this kind of differential equations: Let the differential equation can be has the following form for example:
Chapter Four
109
y (8) y′ = g x Where g is any given function of y / x .The form of the equation suggests that we set:
y =u x ∴ y = ux , by differentiation
(9)
∴ y′ = u + u ′x By substituting (10) into (8) we have:
(10)
u + u ′x = g (u )
(11)
Now we separate the variables u and x we get the following:
du dx (12) = g (u ) − u x If we integrate (12) as we make with normal separable equation then replace u by y / x , we get the general solution of (8) Example 8 Solve the following differential equation:
(y − xy 2 )dx − (x + x 2 y )dy = 0
(13)
Solution: The above equation can be reduced to have the following form: y (1 − xy ) dx − x(1 + x y ) dy = 0 Assume u = xy Then, y = u / x and y ′ =
xu ′ − u
x2 Substitute the values of u , y, y ′ in the differential equation (13) we get:
u xu ′ − u (1 − u ) − x(1 + u ) =0 x x2
110 Ordinary Differential Equations
∴ 2u − xu ′(1 + u ) = 0 ∴ 2∫
∴x
du (1 + u ) = 2u dx
(1 + u ) du + c dx =∫ 1 x u
∴ 2 Ln ( x ) = Ln (u ) + u + Ln (c)
x2 =u ∴ Ln uc ∴ x 2 = xyce xy ∴ x = yce xy Example 9 Solve the following differential equation:
dy = ( y − 4 x )2 dx Solution: - Assume u = y − 4 x Then y = u + 4 x and
dy du du = + 4, ∴ + 4 = u2 dx dx dx ∴∫
du u2 − 4
= ∫ dx + c1
1 u − 2 u − 2 4x ∴ Ln = x + Ln(c2 ) ∴ = ce 4 u + 2 u + 2 But, u = y − 4 x
∴
y − 4x − 2 = Ce 4 x y + 4x + 2
Chapter Four
111
Example 10 Solve the following differential equation:
(x3 + y3 ) dx − 3xy 2dy = 0
Solution: - Divide both sides by x 3 2 y 3 y ∴ 1 + dx − 3 dy = 0 x x
Assume u = y / x ⇒ y = ux
∴
dy du = x+u dx dx
(
)
du ∴ 1 + u 3 − 3u 2 x + u = 0 dx ∴ 1 + u 3 − 3u 3 − 3u 2
du x=0 dx
∴ 1 − 2u 3 − 3u 2
du x=0 dx
∴ 1 − 2u 3 = 3u 2
du x dx
∴∫
3u 2 1 − 2u
du = ∫ 3
dx + c1 x
∴
1 2 * 3u 2 dx = + c1 du ∫ ∫ 2 1 − 2u 3 x
∴
1 ln 1 − 2u 3 = ln x + ln c = ln (cx ) 2
(
(
)
)
∴ ln 1 − 2u 3 = 2 ln (cx ) = ln (cx )2
112 Ordinary Differential Equations
∴ 1 − 2u 3 = (cx )2
∴ u3 =
(
1 1 − kx 2 2
)
Substitute in the above equation for u = y / x 3
(
)
(
)
1 y ∴ = 1 − kx 2 2 x ∴ y3 =
1 3 x 1 − kx 2 2
∴y=x
3
(1 − kx 2 ) 2
4.3.4 Exact Differential Equations: If there is a differential equation in the form of (14) in which separation of variables may not be possible. The following method can be used to solve this kind of equations which is called exact differential equation.
M ( x, y )dx + N ( x, y )dy = 0
(14)
Suppose that a function F ( x, y ) can be found which has its total differential the expression M ( x, y )dx + N ( x, y )dy , that is:
dF = M ( x, y )dx + N ( x, y )dy = 0
(15)
Then certainly F ( x, y ) is the general solution of (14). For form (15) it follows that dF = 0 , in view of (14) M ( x, y )dx + N ( x, y )dy = 0 as desired. Then,
F ( x, y ) = c where c is constant.
(16)
Chapter Four
113
Two things then are needed: ¾ To find out under what conditions on M and N a function F exists
such
that
its
total
differential
is
exactly
M ( x, y )dx + N ( x, y )dy = 0 , ¾ If these conditions are satisfied, actually to determine the function F. If there exists a function F such that
M ( x, y )dx + N ( x, y )dy is exactly the total differential of F, are call equation (14) an exact equation. If equation (14) is exact, then by definition F exists such that:
dF = M ( x, y )dx + N ( x, y )dy = 0 But from calculus (chain rule) if F ( x, y ) is total differential, then,
∴ dF =
∂F ∂F dx + dy ∂x ∂y
∴M =
∂F ∂x
and N =
∂F ∂y
By differentiating the first equation with respect to y and the second one with respect to x, these two equations lead to:
∂M ∂ 2 F ∂N ∂ 2 F = = and ∂y ∂y∂x ∂x ∂y∂x ∂M ∂N ∴ = ∂y ∂x Thus for (14) to be exact it is necessary that (17) be satisfied.
(17)
Let us now show that, if condition (17) is satisfied, then (14) is an exact equation.
114 Ordinary Differential Equations If (14) is exact, the function F ( x, y ) can be found by the following way:
F = ∫ M ( x, y ) dx + k ( y )
(18)
In this integration, y is to be regarded as a constant, and k ( y ) plays the role of constant of integration. To determine k ( y ) we drive where,
∂F from (18), and equate it to N, ∂y
∂k ∂F = N , Then we can get , and integrate it with respect ∂y ∂y
to y to get k ( y ) . In the same way we can make the above integration (18) with respect to y to get k ( x ) as following: If (14) is exact, the function F ( x, y ) can be found by the following way: F ( x, y ) = ∫ N ( x, y ) dy + k ( x )
(19)
In the above integration, x is to be regarded as a constant and k ( x ) plays the rule of constant of integration. To determine k ( x ) we drive then we can get
∂F from (19) and equate it to M, ∂x
∂k and integrate it with respect to x to get k ( x ). ∂x
The following example has been solved by the two different methods for students to be familiar with both of them.
Chapter Four
115
Example 11 Solve the following differential equation
3 x( xy − 2)dx + ( x 3 + 2 y )dy = 0
(20)
Solution: ∴ M = 3 x( xy − 2) and N = ( x 3 + 2 y )
∂M ∂N = = 3x 2 ∂y ∂x
(21)
Then (20) is exact. Therefore, its solution is F = c , where,
∂F = M = 3x( xy − 2) ∂x And
(22)
∂F = N = ( x3 + 2 y) ∂y
(23)
Let us attempt to determine F from (22)
(
)
∴ F = ∫ 3 x 2 y − 6 x dx + k ( y ) ∴ F = x 3 y − 3x 2 + k ( y )
(24)
In order to determine k ( y ) we use the fact that F of equation (24) must also satisfy equation (14).
∴
∂F ∂k ( y ) = x3 + = x3 + 2 y 1 424 3 ∂y ∂y N
∴
∂k ( y ) = 2y ∂y
∴ k ( y ) = ∫ 2 y dy
∴ k ( y) = y 2
∴ F = x3 y − 3x 2 + y 2 = c
116 Ordinary Differential Equations Solving this example by using the other method: Let us attempt to determine F from (23)
(
)
∴ F = ∫ x 3 + 2 y dy + k ( x ) ∴ F = x3 y + y 2 + k (x )
(25)
In order to determine k ( x ) we use the fact that F of (25) must also satisfy (14). Hence
∂F ∂k ( x ) = 3x 2 y + = 3x 2 y − 6 x 1424 3 ∂x ∂x M
∂k ( x ) = −6 x ∂x ∴ k ( x ) = −3x 2 Substitute this value in (25) we get the following ∴
result: ∴ F = x 3 y − y 2 − 3 x 2 = c It is clear that we get the same solution as we get in the above solution. Example12 Solve the following differential equation: 2
2
( y 2 e x y + 4 x 3 )dx + (2 xy e x y − 3 y 2 )dy = 0 1442443 1442443 M N 2 2 ∂M ∂N Solution: = = 2 y e x y + 2 xy 3e x y ∂y ∂x Then according to the condition (17), the equation is exact. 2 ∂F = M = y 2 e x y + 4 x3 ∂x 2 ∂F = N = 2 xy e x y − 3 y 2 And ∂y
∴
(26)
(27) (28)
Chapter Four
117
2 ∴ F = ∫ y 2 e x y + 4 x 3 dx + k ( y )
2
∴ F = e x y + x 4 + k ( y) ∴
(29)
2 ∂F ∂ k ( y) = N = 2 xy e x y + ∂y ∂y
Compare the above value with the value of N in (26) we get the value of
∴
∂ k ( y) . ∂y
∂ k ( y) = −3 y 2 ∂y
∴ k ( y) = − y 3 2
∴ F = e x y + x4 − y3 = C Example 13 Solve the following differential equation:
dy 4 − 2 x cos y − 2 y 3 sec 2 2 x = dx 3 y 2 tan 2 x − x 2 sin y
(30)
dy 4 − 2 x cos y − 2 y 3 sec 2 2 x = Solution: dx 3 y 2 tan 2 x − x 2 sin y
(
)
(
)
∴ 4 − 2 x cos y − 2 y 3 sec 2 2 x dx = 3 y 2 tan 2 x − x 2 sin y dy (31) 14444244443 144424443 M
N
∴
∂M = −2 x sin y + 6 y 2 sec 2 2 x ∂y
(32)
∴
∂N = 6 y 2 sec 2 2 x − 2 x sin y ∂x
(33)
118 Ordinary Differential Equations It is clear that,
∂M ∂N = ∂y ∂x
Then, the differential equation is exact. Integrate (33) with
(
)
respect to y. ∴ F = ∫ 3 y 2 tan 2 x − x 2 sin y dy + k ( x) 144424443 N
∴ F = y 3 tan 2 x + x 2 cos y + k ( x) By differentiating the above equation with respect to x and then equate it with M as in (31) we can obtain
∂ k ( x) . Then by ∂x
integration with respect to x we can obtain k ( x ) as follows:
(
)
∂k ( x) ∂F = 2 y 3 sec 2 2 x + 2 x cos y + = 4 − 2 x cos y − 2 y 3 sec 2 2 x 14444244443 ∂x ∂x ∴
∂ k ( x) = −4, ∂x
M
∴k ( x) = −4 x
∴ F = y 3 tan 2 x + x 2 cos y − 4 x = constant Example 14 Solve the following differential equation
y 1 3 yx 2 − 2 dx + x 3 + cos y + dy = 0 x x
(34)
Solution:
y 1 ∴ M = 3 yx 2 − 2 and N = x 3 + cos y + x x ∴
1 ∂M ∂N = 3x 2 − 2 = ∂y ∂x x
(35)
Chapter Four
119
Then (34) is exact. Therefore, its solution is F = c where,
∂F y = M = 3 yx 2 − 2 ∂x x And
(36)
∂F 1 = N = x 3 + cos y + ∂y x
(37)
Let us attempt to determine F by integrating (36) with respect to x.
y + k ( y) x 1 ∂k ( y ) 3 1 ∂F ∴ = x3 + + = x + cos y + x x ∂y ∂y 1442443 ∴ F = x3 y +
∴
∂k ( y ) = cos y ∂y
N
∴ k ( y ) = ∫ cos y dy
∴ k ( y ) = sin y ∴ F = x3 y +
y + sin y = C x
Example 15 Solve the following differential equation:
(3e3x y − 2 x)dx + e3x dy = 0 Solution: M = (3e3 x y − 2 x ) and ∴
(38)
N = e3 x
∂M ∂N = = 3e3 x ∂y ∂x
Then (38) is exact. Therefore, its solution is F = c where,
(39)
120 Ordinary Differential Equations
(
∂F = M = 3e3 x y − 2 x ∂x And
)
(40)
∂F = N = e3 x ∂y
(41)
Let us attempt to determine F by integrating (40) with respect to x.
(
)
∴ F = ∫ 3e3 x y − 2 x dx +k ( y ) = y e3 x − x 2 + k ( y )
(42)
In order to determine k ( y ) we use the fact that F of equation (42) must also satisfy N of equation (39).
∂F dk = e3 x + = e3 x ∂y dy ∂k ( y ) ∴ = 0 , ∴ k ( y ) = c1 ∂y ∴
∴ F = y e3 x − x 2 + c1 = C ∴ F = y e3 x − x 2 = c Example 16 Solve the following differential equation:
sinh x cos y dx − cosh x sin y dy = 0
(43)
Solution: ∴ M = sinh x cos y
(44)
and N = cosh x sin y
∂M ∂N = = − sinh x sin y ∂y ∂x Then (43) is exact. Therefore, its solution is F = c where,
∂F = M = sinh x cos y ∂x ∂F = N = − cosh x sin y And ∂y
(45)
(46) (47)
Chapter Four
121
Let us attempt to determine F from (46)
∴ F = cosh x cos y + k ( y )
(48)
In order to determine k ( y ) we use the fact that F of equation (48) must also satisfy equation (44). Hence,
∴
∂F ∂k ( y ) = − cosh x sin y + = − cosh x sin y 14 4244 3 ∂y ∂y N
∴
∂k ( y ) =0 ∂y
∴ k ( y ) = c1
∴ F = cosh x cos y + c1 = C ∴ F = cosh x cos y = c* Where c* = c − c1 Example 17 Solve the following differential equation
(tan y − sin x ) dx + x sec 2 y
dy = 0
Solution: ∴ M = (tan y − sin x )
(49)
and N = x sec 2 y
∂M ∂N = = sec 2 y ∂y ∂x Then the differential equation is exact. Therefore, its solution is ∂F = M = (tan y − sin x ) F = c where, ∂x ∂F = N = x sec 2 y And ∂y
∴
∴ F = ∫ (tan y − sin x ) dx + k ( y ) = x tan y + cos x + k ( y ) In order to determine k ( y ) we use the fact that F of the above equation must also satisfy equation (49). Hence,
122 Ordinary Differential Equations
∴
∂F ∂k ( y ) = x sec 2 y + 0 + = x sec 2 y 1 424 3 ∂y ∂y
∂k ( y ) = 0 , ∴ k ( y ) = c1 ∂y ∴ F = x tan y + cos x + c1 = C
N
∴
∴ F = cosh x cos y = c* Where c* = c − c1 4.3.5 Linear First Order Differential Equations First order differential equations, which are linear form, an important class of differential equations, which can always be routinely solved by the use of special formula. By definition, a linear, first-order differential equation cannot contain products, powers, or other nonlinear combinations of y or y′ . The standard form of the linear first order differential equations is:
dy + f ( x) y = r ( x) (50) dx If r ( x) = 0 in (50) so it is called homogeneous, linear, first order differential equation; otherwise it called non-homogeneous, linear, first order differential equation. For the homogeneous differential equation the solution is very simple. By separating variables we have:
dy = − f ( x) dx y
∴ Ln( y ) = − ∫ f ( x)dx + C *
∴ y ( x) = Ce − ∫ f ( x ) dx * Where, C = e c and c, and c * are constants.
(51)
Chapter Four
123 In case of non-homogeneous differential equation, using the method of exact equation can solve it. Where the general solution takes the following form:
[
]
y ( x ) = e − h ∫ e h r dx + C Where h = ∫ f ( x) dx
(52)
Example 18 Solve the following differential equation:
x y′ + y − x 4 = 0 Solution: The above differential equation can be written as: y′ +
1 y = x 3 This equation in the form of (52) x
∴ f ( x) =
1 1 , r ( x) = x 3 and h = ∫ dx = Ln( x) x x
Substitute in (52) we get the following result.
[
∴ y ( x) = e − Ln ( x ) ∫ e Ln ( x ) x 3 dx + C
1 ∴ y ( x) = x
[
]
]
x4 C 1 x5 ∫ x dx + C = x 5 + C = 5 + x 4
Example 19 Solve the following differential equation:
y′ − 2 y = x 2 e5 x Solution: This equation is linear first order differential equation from (50). Then we can use the general solution (52) to solve it.
∴ f ( x) = −2 , r ( x) = x 2 e5 x
[ [∫ x
and h = ∫ − 2 dx = −2 x
∴ y ( x) = e 2 x ∫ e − 2 x x 2 e 5 x dx + C
∴ y ( x) = e 2 x
2
e3 x dx + C
]
]
124 Ordinary Differential Equations
e5 x ∴ y ( x) = 3
2 2 2 2x x − 3 x + 9 + C e
Example 20 Solve the following differential equation:
xy′ = y + ( x + 1)2 Solution: The above equation can be reduced to be as the following form: y′ −
( 1 x + 1)2 y= x x
This equation is linear first order differential equation. Then we can use the general solution, (52) to solve it.
( 1 x + 1)2 1 1 ∴ f ( x) = − , r ( x) = And h = ∫ − dx = − ln x = ln x x x x ∴ y ( x) =
1 − ln e x
∫
1 ln e x
(x + 1)2 dx + C x
1 ( x + 1)2 ∴ y ( x) = x ∫ dx + C x x ( x + 1)2 dx C ∴ y ( x) = x ∫ + x2 2 1 ∴ y ( x) = x ∫ 1 + + 2 dx + C x x 1 ∴ y ( x) = x x + 2 ln x − + C x ∴ y ( x) = x 2 + x ln x 2 − 1 + Cx
Chapter Four
125
4.4 Engineering Applications 4.4.1 Newton’s Second Low Of Motion “The product of the mass and the acceleration equal to the external force”. In symbols,
F = ma
(53)
Where F is the external force, m is the mass of the body, and a is its acceleration in the direction of F. Equation (53) can be put in the following from:
F =m
dv dt
(54)
Where v is the velocity of the body. Example 21 Consider a vertically falling body of mass m that is falling only by gravity g. The force due to gravity given by the weight W of the body, which equal to mg. The force due to air resistance is given by − kv , where k > 0 is a constant. The minus sign is required because this force opposes the velocity. Therefore, the external force is: F = mg − kv
(55)
Then from (54) we obtain the following equation:
dv dv k = mg − kv Or + v=g (56) dt dt m Which is linear first order differential equation and in the form of m
(50). So, the solution will be in the form of (52). Thus:
∴ f (t ) =
k , m
r (t ) = g and h = ∫
k k dt = t m m
126 Ordinary Differential Equations k k kt − t m t m m m ge + C1 ∫ e g dt + C1 ∴v = e ∴v k k − t mg ∴v = 1 + Ce m k k − t mg But v(0) = 0 then C = −1 ∴v = 1− e m k k − t =e m
dx in the dt above equation and integrate and use the second initial condition x(0) = 0 , gives: k − t mg m 2 g ∴x = t − 2 1− e m k k Which is the position of the body at any time t. To obtain the position x of the body we replace v by
4.4.2 Newton’s Law Of Cooling Newton’s law of cooling states that: “the time rate of change in temperature of an object varies as the difference in temperature between the object and surroundings”. Example 22 an object cools from 100oC to 70oC in 20 min. find the o 20 C. temperature in 40 minutes if the surrounding temperature is
Solutions: Let T (t ) is the temperature of object after t minutes. dT = k (T − 20) Then, from Newton law we can say that: dt Where k is the constant of proportionality.
Chapter Four
127 By solving the above equation by using separable variable
method we get: T (t ) = 20 + Ce kt Substitute the initial conditions [T(0)=100, T(20)=70] in the above equation we get: T (0 ) = 100 = 20 + C e k *(t = 0 )
∴ C = 80 . Also, T (20 ) = 70 = 20 + 80e k *20 1 70 − 20 − 0.0235t Ln = −0.0235 And, T (t ) = 20 + 80e 20 80 To obtain the temperature after 40 minutes we can substitute in the
∴k =
above equation for t=40
∴ T (40 ) = 20 + 80e − 0.0235*40 = 51.25 o C 4.4.3 Chemical Application Chemical material A dissolves in solution at a rate proportional to the instantaneous amount of undisclosed chemical and to the difference in concentration between the actual solution Ca and saturated solution Cs. Example 23 A 10 kg of certain solid A putted into a 100 liter of water and after one hour 4 kg of that solid is dissolved. If a saturated solution contains 0.2 kg of A per liter, find (a) The amount of A which is undesolved after two hours and (b) The time to dissolve 80% of A. Solution: Let y kg be undissolved after t hours.
10 − y dy = ky (C s − Ca ) = ky 0.2 − = Ky ( y + 10) 100 dt
128 Ordinary Differential Equations Where k and K are the constant of proportionality. By solving the above equation by using separable variable method we get:
y 1 Ln = Kt + C1 10 y + 10
Substitute the initial conditions [y(0)=10, y(1)=6] in the above equation we get: C1 = −0.0693 and K = −0.02877 (b) The time to dissolve 80% of A means that the undissolved amount is 20%, then y = 2kg
Q
y 1 = −0.02877t − 0.0693 Ln 10 y + 10
1 2 Ln = −0.02877t − 0.0693 10 2 + 10 ∴ t = 3.82 hours
Q
r=10 m
4.4.4 Water Tanks Example 24 A circular cylinder
dh
dV = −π *102 dh
h=25m
of radius 10 m and height 25 m whose axis is vertical as shown in Fig.2, is filled with water. How long will it take for all the water to escape through
Fig.2 dV = π (.5)2 0.6 2gh dt
an orifice with 50 cm radius at the bottom of the tank?
v = 0.6 2gh
Assuming the velocity of escape v in terms of instantaneous height h is given by v = 0.6 2 gh .
Chapter Four
129 Solution: Assume incremental volume dV will take time dt to escape from the tank. So from the geometry of the tank the 2 incremental volume is: dV = −π * r dh
(57)
Where, dh is the height of the incremental volume dV . The minus sign has to be in the above equation because the height of the water decreases with time. The incremental volume dV will take time dt to escape from the orifice of the tank at speed v . So, the incremental Volume dV can be calculated also from the following equation:
dV = A.v.dt
(58)
Where A is the area of the orifice. So if the orifice is circle with radius of ro the A = π ro2 . And assume v = 0.6 2 gh . Then the incremental volume is given by the following equation:
dV = πro2 * 0.6 2 gh dt by equating (57) and (59) we get the following equation:
(59)
− π * r 2 dh = πro2 * 0.6 2 gh dt The above differential equation can be solved by separation of ro2 dh variables. ∴ = −0.6 2 gh 2 dt h r By integrating both sides of the above equation we get the r2 dh following: ∫ = ∫ − 0.6 2 gh o2 dt + c h r Where c is the constant of integration, ∴2 h = −0.6 2 g
ro2 r
2
t+c
(60)
130 Ordinary Differential Equations So, at time t = 0 h = H o . Substitute this condition in the above equation we get the following values for c. 2 H o = 0 + c ∴ c = 2 H o Substitute this value in (60) we get the final results: 2 h = −0.6 2 g
ro2 2
t + 2 Ho
r To get the time required for tank to get empty we can substitute in the above equation for h = 0 ,
2 Ho
∴ te =
(61) 0.6 2 g ro2 / r 2 So, we can substitute the data in our example in the above equation to obtain the time for the tank to get empty, t e .
te =
2 25 2
0.6 2 * 9.81 * 0.5 / 10
2
= 1505 sec . = 25.08 min .
Example 25 A spherical tank of radius R=100 cm which contains
R
R
R-h r
water and has outlet of radius
dh
h
ro = 5 cm at the bottom of the tank as
Fig.3
shown in Fig.3. At time t = 0 the outlet is
opened and the water flows out. Determine the time when the tank will
be
empty,
assuming
the
initial
height
of
water
h(0 ) + R = 100 cm . The velocity with which liquid issues from an orifice is v = 0.6 of the gravity.
2 gh , where ( g = 9.81m/s 2 ) is the acceleration
Chapter Four
131 Solution: Let the origin be chosen at the lowest point of the tank and let h be the instantaneous depth, V the instantaneous volume, and r the instantaneous radius of the free surface of the water as shown in Fig.3. Then in an infinitesimal time dt, the water level will fall by the amount dh , and the resultant decrease in volume of the water in the tank will be:
dV = −π r 2 dh
(62)
Now by Torricelli’s law the velocity with which a liquid issues from an orifice is: v = 0.6 2 gh where g is the acceleration of gravity and h is the instantaneous height, or head, of the liquid above the orifice. In the interval dt, then, a stream of water of length
vdt = 0.6 2 gh dt and of cross-sectional area π ro2 will emerge from the outlet. The volume of this amount of water is: (63) area * length = π ro2 * 0.6 2 gh So from the above equation and (62) we can get the following equation: − π r 2 dh = π ro2 * 0.6 2 gh dt
(64)
Before this equation be solved, r must be expressed in terms of h. This is easily done through the use of equation of the circle which describes the vertical cross section of the tank:
∴ r 2 + (h − R )2 = R 2 or
r 2 = 2hR − h 2
(65)
With this, the differential equation (65) can be written as:
(
)
∴π 2hR − h 2 dh = −π ro2 * 0.6 2 gh dt
(66)
132 Ordinary Differential Equations This is a simple separable equation.
(
)
∴ 2 Rh1 / 2 − h 3 / 2 dh = −ro2 * 0.6 2 g dt 4 2 ∴ Rh3 / 2 − h5 / 2 = −ro2 * 0.6 2 g t + c (67) 3 5 14 Since h = R at t = 0 , ∴ R 5 / 2 = c 15 4 2 14 ∴ Rh3 / 2 − h 5 / 2 = − ro2 * 0.6 2 g t + R 5 / 2 3 5 15 To fiend how long it will take the tank to empty, we must determine 14 the value of t when h=0. ∴ 0 = −ro2 * 0.6 2 g t + R 5 / 2 15 5/ 2 14 R ∴t = (68) 15 ro2 * 0.6 2 g So in our example R = 1m and ro = 0.05m also g = 9.81 m / sec 2 . Substitute these values in the above equation we get:
14 15 / 2 ∴ t= = 140.5 sec . 15 0.05 2 * 0.6 2 * 9.81 Example 26 The tank shown in Fig.4 is consists of two portions, the top one is cylinder with 20 m height and 10m radius of its base and the other portion is half sphere with 10 m radius and has 50 cm2 outlet area at the bottom. If this tank filled with water and at time t=0 the outlet is opened and the water flows out. Determine the time when the tank will be empty, assuming the tank was initially filled with water. The velocity with which liquid issues from an orifice is v = 0.6 2 gh , where ( g = 9.81m/s 2 ) is the acceleration of the
gravity.
Chapter Four
133 Solution: Let the origin be chosen at the lowest point of the tank
and let h be the instantaneous depth, V the instantaneous volume, and r the instantaneous radius of the free surface of the water as shown in Fig.4. Let us start first with finding the time required for the cylindrical portion to get empty. Then in an infinitesimal time dt, the water level will fall by the amount dh, and the resultant decrease in volume of the water in the tank will be:
dV = −π 10 2 dh
Fig.4
Now by torricelli’s law the
20m
velocity with which a liquid issues from an orifice is:
v = 0.6 2 gh
where g is the
10m
acceleration of gravity and h is the instantaneous height, or head, of the liquid above the orifice. In the interval dt. The unit volume escape from the orifice at
Area = 50 cm 2 v = 0.6 2 gh
time dt is:
dV = 0.6 2 gh * 50 * 10 − 4 dt So, by equating both volumes we get the following equation:
0.6 2 gh * 50 * 10 − 4 dt = −π 10 2 dh , ∴
dh = −4.23 * 10 − 5 dt h
134 Ordinary Differential Equations It is clear that the above equation is first order differential
equation in separable form, so it is easy to solve this equation by integrating both sides.
∫
dh = ∫ − 4.23 *10 −5 dt + c ∴ 2 h = −4.23 * 10 −5 t + c h
It is clear from the initial condition that at t = 0 , h = 20 + 10 = 30m Substitute this initial condition in the above equation we get:
2 30 = −4.23 * 10 − 5 * 0 + c
∴ c = 2 30 = 10.954
So, the solution of differential equation becomes,
2 h = −4.23 * 10 − 5 t + 10.954 At the bottom of the cylinder portion h = 10m so we can obtain the time required for the cylindrical portion to get empty if we substitute in the solution of differential equation for h = 10m . Then,
2 10 = −4.23 * 10 − 5 t + 10.954 ∴ tcylinder = 30.4 hours In the same way we can obtain the time required for spherical portion to get empty. We use also unit element in spherical portion
dV = −π r 2 dh . But r = 10 2 − (10 − h )2 = 20h − h 2
(
)
∴ dV = −π 20h − h 2 dh This volume will take time dt to escape from the orifice of the tank.
∴ dV = 0.6 2 gh * 50 * 10 − 4 dt
Chapter Four
( ) ∴ ∫ (20h 0.5 − h1.5 )dh = −0.0042298dt + k
135
∴ −π 20h − h 2 dh = 0.6 2 gh * 50 * 10 − 4 dt
20 1.5 h 2.5 h − = −0.0042298 t + k ∴ 1.5 2.5 At t = 30.4 * 3600 sec, h = 10m
20 1.5 10 2.5 10 − = −0.0042298 * 30.4 * 3600 + k ∴ 1.5 2.5 Then, k = 758.055
20 1.5 10 2.5 ∴ 10 − = −0.0042298 * 30.4 * 3600 + 758.055 1.5 2.5 The total time required for the total tank to get empty can be obtained by applying h = 0 in the above equation. Then, ∴ t = 179218 sec = 49.7827hours Another Solution
For cylindrical portion we can use (60) to get the time required for this portion to get empty
∴2 h = −0.6 2 g
ro2 2
t+c
(60)
r So, at time t = 0 h = 30m . Substitute this condition in the above
equation we get the following values for c.
∴2 30 = 0 + c ∴c = 2 30 ∴2 h = −0.6 2 g
ro2 r2
t + 2 30
136 Ordinary Differential Equations To get the time required for cylindrical portion of the tank to get
empty we can substitute in the above equation for h = 10 m . Then,
∴2 10 = −0.6 2 g
ro2
10 2
Q 50 *10 − 4 = πro2
∴2 10 = −0.6 2 * 9.81 tcylinder
t + 2 30
∴ ro = 3.99cm .0399 2 2
t + 2 30
10 = 109427 sec . = 30.4 hours
In the same way we can obtain the time required for spherical portion to get empty. We use (67) as explained before.
∴
4 3/ 2 2 5/ 2 Rh − h = −ro2 * 0.6 2 g t + c 3 5 Since h = R = 10 at t = 109427 sec . ,
4 2 ∴ 10 * 103 / 2 − 105 / 2 = −0.03992 * 0.6 2 * 9.81 * 109427 + c 3 5 ∴ c = 758.135
∴
4 3/ 2 2 5/ 2 Rh − h = −ro2 * 0.6 2 g t + 758.135 3 5
Then the time required for the hole tank to get empty is at h = 0 . Then substitute h = 0 in the above equation to get tTotal
∴ 0 = −0.03992 * 0.6 2 * 9.81 t + 758.135
∴ t = 179 184.4 sec . = 49.773 hours
Chapter Four
137
Example 27 A right circular
cylinder
of
radius
8m
and
Length of 16m as shown in
16m
Fig.5, whose horizontal axis, is filled with water. How long will Fig.5
it take for all the water to escape through circular orifice has 10
16m
cm radius at the bottom of the tank? Assume v = 0.6
2 gh
Solution: In this case we will
calculate the time required for the upper half to get empty. Then we can do the same for the lower half. First, For upper half:
The horizontal incremental volume can be taken in the upper half of the tank. It is clear from geometry of the tank, the width of the incremental volume, dV is 2r and the length L of the cylinder. L
r
dh
h-R R h
Fig.6
138 Ordinary Differential Equations
It is clear from Fig.6 that r =
R 2 − (h − R )2 = 2 Rh − h 2
So the incremental volume dV is given by the following equation:
dV = −2 * 2 Rh − h 2 * L dh
(69)
Also, this volume dV can be calculated at the orifice of the tank as following: dV = πro2 * 0.6 2 gh dt
(70)
By equating (69) and (70) we get the following result:
∴ − 2 * 2 Rh − h 2 * L dh = πro2 * 0.6 2 gh dt −π 2 ∴ 2 R − h dh = ro * 2 g dt 2L By integrating both sides of the above equation we can get the following equation:
2 (2 R − h )3 / 2 dh = − π ro2 * 2 g t + c 3 2L Where c is the constant of integration.
∴
(71)
At time t = 0, h = 2 R Substitute this condition in the above equation we get the value of the constant c,
∴c =0 Substitute from (72) into (71).
(72)
2 (2 R − h )3 / 2 = − π ro2 * 2 g t (73) 3 2L To obtain the time required for the upper half to get empty we have
∴
to put h = R in the above equation, (73).
∴ tu =
4 LR 3 / 2 3π ro2 2 g
(74)
Chapter Four
139
For the lower half
The horizontal incremental volume dV can be taken in the upper half of the tank. It is clear from geometry of the tank, the width of the incremental volume, dV is 2r and the length L of the cylinder. It is clear from Fig.7 that: r = R 2 − (R − h )2 = 2 Rh − h 2 So the incremental volume dV is given by the following equation:
dV = −2 * 2 Rh − h 2 * L dh
(75)
Also, this volume dV can be calculated at the orifice of the tank as following: dV = πro2 * 0.6 2 gh dt
(76)
By equating (75) and (76) we get the following result:
− 2 * 2 Rh − h 2 * L dh = πro2 * 0.6 2 gh dt −π 2 ∴ 2 R − h dh = ro * 2 g dt 2L L
R-h R h
dh
r
Fig.7
By integrating both sides of the above equation we can get the following equation: ∴
2 (2 R − h )3 / 2 = − π ro2 * 2 g t + c 3 2L
(77)
140 Ordinary Differential Equations Where c is the constant
t = tu =
of
integration.
At
time
4 LR 3 / 2
, h = R substitute this condition in the above 3π ro2 2 g equation we get the value of the constant c, 2 4 LR 3 / 2 3/ 2 − π 2 ∴ (2 R − R ) = ro * 2 g +c (78) 3 2L 3π ro2 2 g ∴ c = 0 , substitute for c = 0 in (78) we get the following results ∴
2 (2 R − h )3 / 2 = − π ro2 * 2 g t 3 2L
The total time required for the whole tank to get empty can be obtained by Substituting in the above equation for h = 0 .
∴ ttotal =
4 L(2 R )3 / 2 3π
ro2
(79)
2g
So the time required for the lower portion only to get empty is given by the following equation: t L = ttotal − tu
(80)
The data collected from our example is ro = 0.1m , R = 8m , and
L = 16 m . Apply these data to equation (74), (79), and, (80) respectively we get the following results:
tu =
4 * 16 * 83 / 2 3 * π * 0.12 * 2 * 9.81
ttotal =
= 3468.92 sec . = 57.82 min .,
4 * 16 * (2 * 8)3 / 2 3 * π * 0.12 * 2 * 9.81
= 9811.59 sec . = 163.52 min ., and,
∴t L = ttotal − tu = 163.52 − 57.82 = 105.7 min .
Chapter Four
141 Example 28 A conical tank shown in Fig.8, filled with water and
10 m
has outlet of radius ro = 20 cm at the bottom. At time t=0 the
Fig.8
outlet is opened and the water flows
out.
Determine
the
time
required for the tank to get empty assuming
the
tank
was
initially
dh
20 m
θ
h
completely filled with water. The velocity with which liquid issues from an orifice is
v = 0. 6
Radius = 20 cm
2 gh , where ( g = 9.81m/s 2 ) is v = 0.6 2 gh
the acceleration of the gravity. Solution: Assume the radius of the base of the tank is R and its
height is L. Assume a horizontal incremental volume dV at height h from the bottom of the tank and radius r. The thickness of the incremental volume dV is dh . It is clear from the geometry of the tank that the incremental volume looks like a cylinder with radius r and height dh . So, the incremental volume can be obtained as following: dV = −π r 2 dh
(81)
r R R ∴r = h = L h L
(82)
But
Substitute the value of r in (82) into (81) we get the following result: 2
R dV = −π * h dh L
(83)
142 Ordinary Differential Equations In the same way dV can be obtained at the bottom of the tank as
following: dV = πro2 * 0.6 2 gh dt By equating (83) and (84) we get the following result:
(84)
2
R − π * h dh = πro2 * 0.6 2 gh dt L Lro2 3/ 2 dt ∴ h dh = −0.6 2 g R By integrating both sides of the above equation we get the Lr 2 2 (85) following: ∴ h 5 / 2 = 0.6 2 g o t + c R 5 where c is the constant of integration At time t = 0 , h = L . Substitute this initial condition into (85):
2 ∴ L5 / 2 = 0 + c 5
∴c =
2 5/ 2 L 5
(86)
Substitute the value of c in (86) into (85) we get the following:
Lro2 2 5 / 2 2 5/ 2 t + L = 0.6 2 g ∴ h 5 R 5 So the time required for the tank to get empty is given by applying for h = 0 in the above equation. So, 2
R * (87) Lr o Substitute the data obtained in (87) where L = 20m , R = 5m R, ro = 0.2m
2 L5 / 2 ∴t = 5 * 0. 6 2 g
2
2 * 205 / 2 5 ∴t = * = 420.68 sec . = 7.01 min . 5 * 0.6 2 * 9.81 20 * 0.2
Chapter Four
143
4.4.5 Half Life Of Nuclear Materials Example 29 Experiment shows that radium disintegrates at a rate
proportional to the amount of radium instantly present. Its half life (that is the time in which 50% of a given amount will disappear), is 1590 years. What percent will disappear in one year? Solution: Assume the instantaneous amount and starting amount of
radium exists are y and yo respectively. dy = ky ∴ (88) dt Where k is the proportional constant. Equation (88) can be easily solved by variable separation. ∴ y = Ce kt
(89)
Where C is integration constant. C and k can be determined from initial conditions. The first initial condition is y = y0 at t = 0 substitute this condition into (89).
∴ y = y0 e kt The second initial condition is
(90) y = y0 / 2 at t = 1590 years .
Substitute this condition into (90)
∴
y0 = y0 e k *1590 2
∴ k = −4.36 *10 − 4 −4
∴ y = y0 e − 4.36 *10 t (91) To obtain the percent disappear after one year we can do the following:
y e − 4.36 *10 −4 *1 y (1) 1 − * 100 = 1 − 0 y0 y0
* 100 = 0.0436 %
144 Ordinary Differential Equations 4.4.6 Electrical Circuits
(a) E (t ) = 40 V , (b) E (t ) = 20 e
+
Fig.9. Assume I (0) = 0 R=10 Ω, L=2H if: +
− 3t
E(t) _
I(t)
Solution: From Kirchoff’s laws
∴
+
_
and (c) E (t ) = 50 sin 5 t .
RI + L
_
Example 30 Find the current in the RL circuit in vR(t)
vL(t)
dI = E (t ) dt
Fig.9
dI R E (t ) + I= L dt L
(a) I& +
10 40 I= 2 2
∴ I& + 5 I = 20 By using the general solution of the first order linear differential equations, (52) we can solve it.
∴ f (t ) = 5 r (t ) = 20 ∴ I (t ) = e − 5t
[∫ e
5t
∴ h = ∫ 5dt = 5t
]
* 20dt + k ∴ I (t ) = 4 + ke − 5t
where k is a constant of integration.
(
But I (0) = 0 ∴ k = −4 ∴ I (t ) = 4 * 1 − e − 5t (b) For E (t ) = 20 e − 3t
I (t ) = e − 5t
[∫ e
5t
*10 e − 3t dt + k
]
)
Chapter Four
[
145
]
I (t ) = e − 5t 5 e 2 t + k ∴ I (t ) = 5 e − 3 t + ke − 5t But I (0) = 0 then k = −5
(
∴ I (t ) = 5 e − 3 t − e − 5t
(c) E (t ) = 50 sin 5 t
I (t ) = e − 5t
[∫ e
5t
* 25 sin 5t dt + k
)
]
5 I (t ) = e − 5t e5t (sin 5t − cos 5t ) + k But I (0) = 0 ∴ k = 5 / 2 2 ∴ I (t ) =
[
5 (sin 5t − cos 5t ) + e −5t 2
]
Example 31
Find the current in the RC circuit in Fig.10. Assume I (0) = 0 R=5 Ω,
E (t ) = 50 sin 20 t .
Assume (a) vc (0) = 0 and (b) vc (0) = 100
_
and
+
C=.02F
vR(t) +
E(t) _
+
I(t)
_
Solution:
From Kerchief’s laws
RI + ∴
1 Idt = E (t ) C∫
dI 1 1 dE (t ) + I= dt RC R dt
1 1d ∴ I& + I= 50 sin 20 t RC R dt
Fig.10
vC(t)
146 Ordinary Differential Equations
∴ I& +
1 1 I = * 50 * 20 cos 20 t ∴ I& + 10 I = 200 cos 20t 5 * .02 5
By using the general solution of the first order linear differential equations we can solve it.
f (t ) = 10 r (t ) = 200 cos 20t ∴ h = ∫ 10 dt = 10t I (t ) = e −10t
[ ∫ e * 200 cos 20 t dt + k ] [ 4 e (cos (20 t ) + 2 sin ( 20 t )) + k ]
∴ I (t ) = e −10t
10t
10t
∴ I (t ) = 4 cos (20 t ) + 8 sin ( 20 t )) + ke −10t (a) For vc (0) =
1 Idt =0 C∫ t =0
∴ 0.2 sin( 20 t ) − .4 cos (20t ) −
k −10t e =0 10 t =0
∴ k = −4
∴ I (t ) = 4 cos (20 t ) + 8 sin ( 20 t )) − 4e −10t (b) For vc (0) =
∴
1 Idt = 100 V C∫ t =0
k −10t 1 0 . 2 sin( 20 t ) − . 4 cos ( 20 t ) − e = 100 0.02 10 t =0
∴ k = −24 ∴ I (t ) = 4 cos (20 t ) + 8 sin ( 20 t )) − 24e −10t
Chapter Four
147
Problems:
[I] Draw a good direction fields for the following differential equations and plot several approximate solution curves: 1) y ′ = y 3) y ′ =
2) y ′ = cos y
xy 1+ x
4) y ′ = 2 y / x
2
[II] By using the direction field, plot an approximate solution curve of the given differential equation satisfying the given condition: 5) y′ = x + 1,
y ( 0) = 1
6) y ′ − y = 1 − e − x ,
7) y ′ + y 2 = 0, y (5) = 0.25 8) 9 yy ′ + 4 x = 0,
y ( 0) = 0 y (3) = −4
[III] Find the general Solution of the following differential equations:
dy x 2 = 9) dx y dy + y 2 sin x = 0 11) dx
dy x2 10) = dx y (1 + x 3 ) dy 12) = cos 2 x cos 2 2 y dx
1 − y2 dy 13) = x dx dy = y tanh x 15) dx
dy x − e − x 14) = dx y + e y
(
)(
16) ( xLn( x) )dy − ydx = 0
[IV] Solve the following initial value problems: 17) sin 2 xdx + cos 3 ydy = 0, y (π / 2) = π / 3
)
148 Ordinary Differential Equations
18) xdx + ye − x dy = 0, 19)
dy Ln( x) = , dx (1 + y 2 )
20) y ′ = 2e x y 3 , 21) y ′ = 3 x 2 e − y , 2
22) yy ′ = xe y , 23) xyy ′ = y + 2,
y (0) = 1
y (1) = 0
y (0) = 0.5 y (−1) = 0 y (1) = 0
y (2) = 0
24) xydx + (2 xy 2 + 4 y 2 − x − 2) dy = 0, y (0) = 1 [V] Solve the following differential equations:
(
)
25) x 3 + y 3 dx − 3 xy 2 dy = 0 26) xdy − y + x 2 − y 2 dx = 0 Equation reducible to separable form ٢٧) ٢٨)
(2 x + 3 y )dx + ( y − x)dy = 0 ( x + y )dx + (3x + 3 y − 4)dy = 0
٢٩)
(1 − xy − x 2 y 2 )dx + (x3 y − x 2 )dy = 0
٣٠)
tan 2 ( x + y )dx − dy = 0
٣١) ٣٢) ٣٣) ٣٤) ٣٥)
(2 + 2 x 2 y1 / 2 )ydx + (x 2 y1 / 2 + 2)xdy = 0 dx + ((1 − x 2 )cot y )dy = 0 (x 2 + y 2 )dx + (x 2 − xy )dy = 0 2 x 3 ydx + (x 4 + y 4 )dy = 0
( xy ′ − y )cos (2 y / x) = −3x 4
Chapter Four
149
Exact differential equations
[VI] Solve the following differential equations by prove that it is Exact, and then solve it. ٣٦)
(4 x3 y 3 − 2 xy )dx + (3x 4 y 2 − x 2 )dy = 0
٣٧)
2 2 2 x ye x − 1dx + e x dy = 0
٣٨)
(cos y + y cos x )dx + (sin x − x sin y )dy = 0
٣٩) ٤٠)
(x 2 + y 2 + x)dx + xydy = 0 (2 x + e y )dx + xe y dy = 0
٤١)
y − y sin xy dx + (Lnx − x sin xy )dy = 0 x
٤٢)
y sinh xdx + cosh xdy = 0
Linear first order differential equations
[VII] Solve the following differential equations 43) xy ′ + y = y 2 Lnx
44) y ′ + 2 y = 6e x
45) y ′ = ( y − 1) cot x
46) 1 − x 2 y ′ + xy = x
(
)
[VIII] Solve the following initial value problem 47) y ′ + y = ( x + 1) 2 , y (0) = 0 48) y ′ − 2 y = 2 cosh 2 x + 4, y (0) = −125 49) xy′ − 3 y = x 4 (e x + cos x) − 2 x 2 , 50) 2( y + 1) y ′ −
2 ( y + 1) 2 = x 4 , x
y (π ) = eπ + 2 / π
y (1) = 2 / 3 − 1
150 Ordinary Differential Equations [VIV] Applications
51) After two days, 10 grams of a radioactive chemical is present. Three days later 5 grams is present. How much of the chemical was present initially assuming the rate of disintegration is proportional to the instantaneous amount which is present. 52) It takes 15 minutes for an object to warm up from 10 C to 20 C in a room whose temperature is 30 C. Assuming Newton’s law of cooling , how long would it take to warm up from 20 C to 25 C? 53) Chemical A is transformed into chemical B at a rate proportional to the instantaneous amount of A which is untransformed. If 20% of chemical A is transformed in two hours, (a) what percentage of A is transformed in 6 hours and (b) when will 80% of A be transformed? 54) Find the current in the RLC circuit the
following
figure.
vR(t)
Assume
(a) R=200 Ω, L=short circuit, C=0.005
and E (t ) = 500 t cos 10 t
E(t) _
+
I(t)
_ +
(b) R=160 Ω, L=20H, C=short circuit,
+
_
farads and E (t ) = 200 cos 3 t
+
I (0) = 0 and v L (0) = 0 , vc (0) = 0
_
in
vL(t)
vC(t)
Chapter 5 Linear Differential Equations Of Higher Order 5.1 Definition The general linear differential equations of order n has the form shown in (1) or (2). f n ( x)
dny dx
n
+ f n −1 ( x)
d n −1 y dx
n −1
+ ......... f1 ( x)
d1y 1
dx
+ f 0 ( x)
d0y dx
0
= r ( x) (1)
f n ( x) y (n ) + f n −1 ( x) y (n −1) + ......... f1 ( x) y ′ + f 0 ( x) y = r ( x)
(2)
Any differential equations cannot be written in this form is called nonlinear differential equation. If r(x), the right side of (1) or (2) is replaced by zero, the resulting equation is called homogeneous differential equations. If r ( x) ≠ 0 the equation called nonhomogeneous differential equation.
If
f n ( x ), f n −1 ( x ), .......... .......... ......... f1 ( x ), and
f 0 ( x)
are all
constants, the differential equation is said to have constant coefficients, otherwise it is said to have variable coefficients. A set of n functions y n ( x), y n −1 ( x), .......... y 2 ( x), and y1 ( x) is said to be linearly dependant over an interval if there exist n constants
C n , C n −1 , ...........C 2 , and C1 all of them must be not equal to zero, such that, C n y n ( x) + Cn −1 y n −1 ( x) + ........ + C 2 y 2 ( x ) + C1 y1 ( x) = 0 Otherwise, the set of functions is said to be linearly independent.
152 Linear Differential Equations Of Higher Order Example 1 2e 3 x , 5e 3 x , e 3 x
are linearly dependant over any
interval since we can find constants C3 ,.C 2 , C1 not all zero such that: C1 2e 3 x + C 2 5e 3 x + C3e 3 x = 0 Identically, for instance, C3 = 0,.C2 = 2, and C1 = 5 Example
2
ex ,
xe x
are
linearly
independent
since
C1e x + C 2 xe x = 0 Identically if and only if C1 = 0 and C2 = 0 . Theorem 1, The set of the following functions:
y n ( x ), y n −1 ( x ), .......... .......... ......... y 2 ( x ), and y1 ( x ) is linearly independent on an interval if and only if the following determinant not equal zero.
W ( y1, y2
, yn )
y1( x) y1′ ( x)
y 2 ( x) y2′ ( x)
.............
yn ( x ) yn′ ( x)
............. ≠0 ............. y1n −1( x) y2n −1( x) ............. ynn −1( x)
Called the Wronskian of y1 , y 2
(3)
, y n is different from zero on that
interval. 5.2 Characteristic Equation
Substitute y = e λx ( λ is constant) in (1) to obtain the following equation:
f n ( x)λn + f n −1 ( x)λn −1 + .............................f1 ( x)λ + f 0 ( x) = 0 (4) Which is called the auxiliary or characteristic equation. This can be factored into:
Chapter Five
153
f n ( x)(λ − λn )(λ − λn −1 ).............(λ − λ2 )(λ − λ1 ) = 0 Which has roots λn , λn −1............λ2 , and λ1 . Three cases must be considered depending on the roots of this equation. These cases has been analyzed in the following items: Case 1
Roots are real and distinct.
Then e λ n x , e λ n−1 x ,........ e λ 2 x and e λ1 x are n linearly independent solutions, so that the required solution is:
y = Cn e λ n x + C n −1e λ n−1 x + ........ C1 e λ1 x Where C n , C n −1 ,..............C1 are constants.
(5)
Case 2 Some roots are complex.
When a + jb is a root of (3) so also a − jb is also root. Then the solution corresponding to the roots a + jb and a − jb is
y = e ax [ A cos(bx) + B sin(bx)]
(6)
Where A and B are constants. Case 3 Some roots are repeated.
If λ1 is a root of multiplicity k, then a solution is given by:
y = Cneλn x +Cn−1eλn−1x +......(Ck xk −1 + Ck −1xk −2 +....C1)eλ1x (7) Example 3 Solve the following differential equation:-
y ′′ − 3 y ′ + 2 y = 0 Solution:- The auxiliary equation is
λ2 − 3λ + 2 = 0 ∴ λ1 = 1 and λ2 = 2 Then, λ1 and λ2 are real distinct roots. So, the general solution is given by (5). y ( x) = C1e x + C 2 e 2 x
154 Linear Differential Equations Of Higher Order Example 4 Solve the following differential equation:y ′′′ − 5 y ′′ + 8 y ′ − 4 y = 0 Solution:- The auxiliary equation is:∴ λ3 − 5λ2 + 8λ − 4 = 0
∴ λ1 = 1 and λ2 = λ3 = 2 . Then, λ2 and λ3 are two equal roots. Then the solution will take the form shown in (7).
∴ y = C1e x + (C2 x + C3 ) e 2 x Example 5 Solve the following differential equation:-
y ′′ + y ′ + y = 0 Solution:- The auxiliary equation is:
∴ λ2 + λ + 1 = 0
∴ λ1,2 = −0.5 ± j
Then, the solution takes the form (6). 3 3 ∴ y = e −0.5 x A cos x + B sin x 2 2
3 . 2
5.3 Nonhomogeneous Linear Differential Equation
If r (x) in (1) is not zero, then (1) and (2) are called nonhomogeneous, linear differential equations. The solution of these equations takes the following form:
y ( x) = y h ( x) + y p ( x) Where y h (x) is the solution of homogeneous equation, (sometimes it called complementary solution). And y p (x) is the particular solution of (1).
Chapter Five
155 The particular solution can be assumed depending on the form of
r (x) . The trial solution to be assumed in each form of r (x) is shown in the following table. The trial solutions in this table hold in case no terms in the assumed trial solution appear in the complementary solution. If any term of the assumed trial solution does appear in the complementary solution, we multiply this trial solution by the small positive integer power of x which is large enough so that none of the terms which are then present appear in the complementary solution.
r (x )
Assumed trial solutions
ce px
Ae px
c cos px + k sin px
A cos px + B sin px
cn x n + cn −1 x n −1 + ......c0
An x n + An −1 x n −1 + ......A0
(
e px cn x n + cn −1 x n −1 + ......c0
)
(
e px An x n + An −1 x n −1 + ......A0
( ) sin px * (k n x n + k n −1 x n −1 + ......k 0 )
( ) sin px * (Bn x n + Bn −1 x n −1 + ......B0 )
( ) qx n n −1 e sin px * (k n x + k n −1 x + ......k 0 )
( ) e qx sin px * (Bn x n + Bn −1 x n −1 + ......B0 )
cos px * cn x n + cn −1 x n −1 + ......c0 +
e qx cos px * c n x n + c n −1 x n −1 + ......c0 +
)
cos px * An x n + An −1 x n −1 + ...... A0 +
e qx cos px * An x n + An −1 x n −1 + ...... A0 +
Sum of any or some of the Sums of the corresponding trial above entries.
solutions.
156 Linear Differential Equations Of Higher Order Example 6 Solve the following differential equation:
y ′′ + y ′ − 2 y = 2 e 3 x Solution:- The auxiliary equation is λ2 + λ − 2 = 0
∴ λ1 = 1 and λ2 = −2 Then, λ1 and λ2 are real distinct roots. So the general solution is given by the following equation:
∴ y h ( x) = C1e x + C 2 e − 2 x The particular solution takes the following form:
y p ( x) = ke3 x ∴ y ′p ( x) = 3ke3 x ∴ y ′p′ ( x) = 9ke 3 x ∴ e 3 x (9k + 3k − 2k ) = 2e 3 x ∴k = 0.2 ,
∴ y p ( x) = 0.2 e 3 x
∴ y ( x) = y h ( x) + y p ( x) = C1e x + C 2 e − 2 x + 0.2 e 3 x Example 7 Solve the following differential equation:
y ′′ + y ′ − 2 y = 2 e x + e − 2 x Solution:- The auxiliary equation is λ2 + λ − 2 = 0
∴ λ1 = 1 and λ2 = −2 Then, λ1 and λ2 are real distinct roots so the general solution is given by the following equation y h ( x) = C1e x + C 2 e − 2 x
Chapter Five
157 As we see, there are some terms in the homogeneous solution has
similar terms shown in r (x) . So, the similar terms must be multiplied by x in the particular solution. So, the particular solution takes the following form:
y p ( x) = k1 xe x + k 2 xe − 2 x ∴ y ′p ( x) = k1 xe x + k1e x + k 2 e − 2 x − 2k 2 xe − 2 x ∴ y ′p′ ( x) = 2k1e x + k1 xe x − 2k 2 e − 2 x + 4k 2 xe − 2 x − 2k 2 e − 2 x
2 1 and k 2 = − 3 3 2 1 ∴ y p ( x) = xe x − xe − 2 x 3 3 ∴ k1 =
∴ y ( x) = y h ( x) + y p ( x) = C1e x + C 2 e − 2 x +
2 x 1 −2x xe − xe ] 3 3
Example 8 Solve the following differential equation:-
y ′′ + 4 y = 8 sin 2 x Solution:- The auxiliary equation is λ2 + 4 = 0
∴ λ1,2 = ± j 2 Then the general solution takes the form (6). So, the general solution is given by y h = A cos(2 x) + B sin( 2 x ) For
the
particular
solution
we
normally
assume
y p = k1 cos(2 x) + k 2 sin( 2 x) . However, since the terms appear in the homogeneous solution has similar terms appears in the particular solution. Then, we have to modify the particular solution to take the
158 Linear Differential Equations Of Higher Order
following form y p = x (k1 cos(2 x ) + k 2 sin( 2 x) ) . By differentiating the above equation and substitute the results into the differential equation we get that the final result for the particular solution is as the following:
y p = − x sin( 2 x)
∴ y ( x) = y h + y p = A cos(2 x) + B sin(2 x) − x sin(2 x) Example 9 Solve the following differential equation:-
y ′′ − 4 y ′ + 4 y = e 2 x Solution:- The auxiliary equation is λ2 − 4λ + 4 = 0
∴ λ1 = λ2 = 2 Because of λ1 = λ2 , so the homogeneous solution takes the form in (7). Then, the general solution is given by y h ( x) = (C1 + C 2 x )e 2 x . The
particular
solution
corresponding
to
r ( x) = e 2 x
is
y p ( x) = ke 2 x but there is a term in the homogeneous solution dependant with y p ( x) = ke 2 x . So we have to multiply it by x to be
y p ( x) = kxe 2 x . But, still there is another term in homogeneous solution dependant with y p ( x) = kxe 2 x , then we have to multiply again till we become sure that there is no any dependant terms between homogeneous and assumed particular solutions. Then the particular solution takes the following form:
Chapter Five
159
y p ( x) = kx 2 e 2 x ∴ y ′p ( x) = 2kx 2 e 2 x + 2kxe 2 x ∴ y ′p′ ( x) = 4kx 2 e 2 x + 4kxe 2 x + 4kxe 2 x + 2ke 2 x By substituting the above values into the differential equation and by comparing both sides of the equation we get the value of the constant k, where k = 0.5 .
∴ y p ( x ) = 0. 5 x 2 e 2 x ∴ y ( x) = y h ( x) + y p ( x) = (C1 + C 2 x )e 2 x + 0.5 x 2 e 2 x Example 10 Solve the following differential equation:-
y ′′ − 5 y ′ + 6 y = x 2 e 5 x Solution:- The auxiliary equation is λ2 − 5λ + 6 = 0
Then, λ1 = 2, and λ2 = 3
∴ y h ( x) = C1e 2 x + C2 e3 x The particular solution corresponding to r ( x) = x 2 e 5 x takes the following form:
(
)
y p ( x) = k 2 x 2 + k1 x + k 0 e 5 x
(
)
∴ y ′p ( x) = 5 k 2 x 2 + k1 x + k 0 e 5 x + (2k 2 x + k1 ) e 5 x
(
)
∴ y ′p′ ( x) = 25 k 2 x 2 + k1 x + k 0 e 5 x + 5(2k 2 x + k1 ) e 5 x +5(2k 2 x + k1 ) e 5 x + (2k 2 ) e 5 x
160 Linear Differential Equations Of Higher Order By substituting the above values into the differential equation and
by comparing both sides of the resultant equation we get the values of the constants k 2 , k1 , and k 0 .
1 5 19 ∴ k 2 = , k1 = − , and k 0 = 108 6 18 5 19 5 x 1 ∴ y p ( x) = x 2 − x + e 6 18 108
5 19 5 x 1 ∴ y( x) = yh ( x) + y p ( x) = C1e 2 x + C2e3x + x 2 − x + e 6 18 108 Example 11 Solve the following differential equation:-
y ′′ − 5 y ′ + 6 y = x 2 e 2 x Solution:- The auxiliary equation is λ2 − 5λ + 6 = 0
∴ λ1 = 2,
λ2 = 3
∴ y h ( x) = C1e 2 x + C2 e3 x The particular solution corresponding to
(
r ( x) = x 2 e 2 x
)
is
y p ( x) = k 2 x 2 + k1 x + k 0 e 2 x . But there is a term in the homogeneous solution dependant with
(
)
y p ( x) = k 2 x 2 + k1 x + k 0 e 2 x . So we have to multiply it by x to be as following:
(
)
y p ( x) = k 2 x 2 + k1 x + k 0 x e 2 x
( ) ( ) ∴ y ′p′ ( x) = 4(k 2 x 3 + k1 x 2 + k 0 x ) e 2 x + 2(3k 2 x 2 + k1 x + k 0 ) e 2 x +2(3k 2 x 2 + k1 x + k 0 ) e 2 x + (6k 2 x + 2k1 ) e 2 x ∴ y ′p ( x) = 2 k 2 x 3 + k1 x 2 + k 0 x e 2 x + 3k 2 x 2 + k1 x + k 0 e 2 x
Chapter Five
161 By substituting the above values into the differential equation and
by comparing both sides of the equation we get the value of the constants k 2 , k1 , and, k 0 .
1 ∴ k 2 = − , k1 = −1, and k 0 = −2 3 1 ∴ y p ( x ) = − x 2 + x + 2 x e 2 x 3 1 ∴ y ( x) = y h ( x) + y p ( x) = C1e 2 x + C 2 e 3 x − x 2 + x + 2 x e 2 x 3 Example 12 Solve the following differential equation:-
y ′′′ − 6 y ′′ + 12 y ′ − 8 y = x 2 + e 2 x Solution:- The auxiliary equation is: λ3 − 6λ2 + 12λ − 8 = 0
(
)
∴ y h ( x) = C1 + C2 x + C3 x 2 e 2 x
∴ λ1 = λ2 = λ3 = 2
The particular solution corresponding to r ( x) = x 2 + e 2 x
(
is
)
y p ( x) = k 2 x 2 + k1 x + k 0 + k3e 2 x . But there is a term in homogeneous solution dependant with the assumed particular solution.
(
So
we
have
)
to
multiply
it
by
x
to
be
y p ( x) = k 2 x 2 + k1 x + k 0 + k3 xe 2 x . But, still there is another term in
homogeneous
(
)
solution
dependant
with
y p ( x) = k 2 x 2 + k1 x + k 0 + k3 xe 2 x . Then we have to multiply again till we become sure that there are no any dependant terms between homogeneous and particular solutions. Then the particular solution takes the following form:
162 Linear Differential Equations Of Higher Order
(
)
y p ( x) = k 2 x 2 + k1 x + k 0 + k3 x 3e 2 x ∴ y ′p ( x) = (2k 2 x + k1 ) + 2k3 x 3 e 2 x + 3k3 x 2 e 2 x ∴ y ′p′ ( x) = (2k 2 ) + 4k3 x 3 e 2 x + 12k3 x 2 e 2 x + 6k3 xe 2 x ∴ y′p′′ ( x) = 8k3 x3 e 2 x + 12k3 x 2 e 2 x + 24 k3 x 2 e2 x + 12k3 x e 2 x + 6k3e 2 x By substituting the above values into the differential equation and by comparing both sides of the equation we get the value of the constants k3 , k 2 , k1 , and, k 0 .
3 3 1 1 ∴ k3 = , k 2 = − , k1 = − , and k 0 = − . 8 8 8 6
(
)
1 2 x3 2 x ∴ y p ( x) = − x + 3 x + 3 + e 8 6
(
)
∴ y ( x) = y h ( x) + y p ( x) = C1 + C 2 x + C3 x 2 e 2 x −
(
)
1 2 x3 2x x + 3x + 3 + e 8 6
Example 13 Solve the following differential equation:-
3 y ′′ − 6 y ′ + 36 y = e x sin (3 x ) Solution:- The auxiliary equation is: 3λ2 − 6λ + 36 = 0
∴λ1 , λ2 = 1 ± 11
[
∴ y h =e x A cos 11x + B sin 11x
]
By inspecting the right hand side of the differential equation we can get the following particular solution: y p = e x (k1 cos 3 x + k 2 sin 3 x )
Chapter Five
163
∴ y ′p = e x (− 3k1 sin 3x + 3k 2 cos 3x ) + e x (k1 cos 3x + k 2 sin x )
∴ y ′p′ = e x {− 9k1 cos 3x − 9k 2 sin 3x − 3k1 sin 3x + 3k 2 cos 3x −3k1 sin 3x + 3k 2 cos 3x + k1 cos 3x + k 2 sin x
}
Substitute y p , y ′p , and, y ′p′ into the differential equation and compare both sides we get the following equation:
3e x {− 9k1 cos 3x − 9k 2 sin 3x − 3k1 sin 3x + 3k 2 cos 3 x
{
−3k1 sin 3x + 3k 2 cos 3x + k1 cos 3x + k 2 sin x
}
− 6 e x (− 3k1 sin 3 x + 3k 2 cos 3x ) + e x (k1 cos 3x + k 2 sin x )
{
}
+ 36 e x (k1 cos 3x + k 2 sin 3x ) = e x sin 3x Coefficient of e x cos 3 x = 0
∴ 3 * −9k1 + 9k 2 + 9k 2 + 9k 2 + 3k1 − 18k 2 − 6k1 + 36k1 = 0
∴ k1 = 0 Coefficient of e x sin 3 x = 1
∴ −27 k 2 − 9k1 − 9k1 + 3k 2 + 18k1 − 6k 2 + 36k 2 = 1
1 ∴ k2 = 6
e x sin 3x ∴ yp = 6
e x sin 3 x ∴ y ( x ) = yh + y p =e A cos 11x + B sin 11x + 6 x
[
]
Example 14 Solve the following differential equation:-
y ′′ + y = − 2 sin x + 4 x cos x Solution:- The auxiliary equation is: λ2 + 1 = 0 ∴ λ1 , λ2 = ± j1 ∴ y h = A cos x + B sin x
}
164 Linear Differential Equations Of Higher Order By inspecting the right hand side of the differential equation we can
get the particular solution as following:
y p = x(k1 cos x + k2 sin x) + x 2 (k3 x + k4 )cos x + x 2 (k5 x + k6 )sin x
(
)
(
)
∴ y p = k1x cosx + k2 x sin x + k3 x3 + k4 x 2 cos x + k5 x3 + k6 x 2 sin x
(
)
(
)
∴ y p = k3 x 3 + k 4 x 2 + k1 x cos x + k5 x 3 + k 6 x 2 + k 2 x sin x
(
)
(
)
∴ y ′p = − k3 x 3 + k 4 x 2 + k1 x sin x + 3k3 x 2 + 2k 4 x + k1 cos x
(
)
(
) ∴ y ′p = [− k3 x 3 + (3k5 − k 4 ) x 2 + (2k 6 − k1 )x + k 2 ]sin x +[k5 x 3 + (3k3 + k 6 )x 2 + (k 2 + 2k 4 )x + k1 ]cos x ∴ y ′p′ = [− k3 x 3 + (3k5 − k 4 ) x 2 + (2k 6 − k1 )x + k 2 ]cos x [− 3k3 x 2 + 2(3k5 − k4 ) x + (2k6 − k1 )]sin x −[k5 x 3 + (3k3 + k 6 )x 2 + (k 2 + 2k 4 )x + k1 ]sin x +[3k5 x 2 + 2(3k3 + k 6 )x + (k 2 + 2k 4 )]cos x ∴ y′p′ = [− k3 x 3 + (6k5 − k 4 ) x 2 + (4k6 + 6k3 − k1 )x + 2(k 2 + k 4 )]cos x [− k5 x3 − (k6 + 6k3 )x2 + (6k5 − 4k4 )x + (2k6 − 2k1 )]sin x + k5 x 3 + k 6 x 2 + k 2 x cos x + 3k5 x 2 + 2k6 x + k 2 sin x
Substitute y p , y ′p , and
y ′p′ into the differential equation and
compare both sides we get the following equation:
[− k x 3
(
3
]
+ (6k5 − k 4 ) x 2 + (4k 6 + 6k3 − k1 )x + 2(k 2 + k 4 ) cos x
[− k x 5
3
]
− (k 6 + 6k3 )x 2 + (6k5 − 4k 4 )x + (2k 6 − 2k1 ) sin x
)
(
)
+ k3 x 3 + k 4 x 2 + k1 x cos x + k5 x 3 + k 6 x 2 + k 2 x sin x = −2 sin x + 4 x cos x
Chapter Five
165
Coefficient of cos x = 0
∴ 0 = 2(k 2 + k 4 ), ∴ k 2 = −k 4 Coefficient of x cos x = 4
(8)
∴ 4 = 4k 6 + 6k3 − k1 + k1
∴ 2k 6 + 3k3 = 2
(9)
Coefficient of x 2 cos x = 0
∴ 6k 5 − k 4 + k 4 = 0
∴k 5 = 0 Coefficient of x 3 cos x = 0
(10)
∴ − k3 + k3 = 0 , yield nothing
Coefficient of sin x = −2
∴ −2 = (2k6 − 2k1 ) , ∴ k1 − k6 = 1
(11)
coefficient of x sin x = 0
∴ (6k5 − 4k 4 − k 2 ) + k 2 = 0 , ∴ k 4 = 0 Coefficient of x 2 sin x = 0
(12)
∴ 0 = (k 6 − 6k3 ) + k 6
∴ k6 = 3k3 coefficient of x 3 sin x = 0
(13)
∴ − k5 + k5 = 0 , yield nothing
From (8), (10) and (12) ∴ k 2 = k 4 = k5 = 0 By solving (9), (11) and (13) together we get the following equations:
0 3 2 k1 2 1 0 − 6 k = 1 3 0 3 − 1 k6 0 By solving the above equations we get the following constants: ∴ k1 = 5, k3 = 2 / 9, and k 6 = 2 / 3
166 Linear Differential Equations Of Higher Order
2 2 ∴ y p = x 3 + 5 x cos x + x 2 sin x 3 9 2 2 ∴ y ( x ) = y h + y p = A cos x + B sin x + x 3 + 5 x cos x + x 2 sin x 3 9 Example 15 Solve the following differential equation:-
y ′′′ + y ′′ + y ′ + y = sin 2 x * cos 3 x Solution:-From the following rule:
1 (sin ( A − B ) + cos( A + B )) 2 We can modify the right hand side of the above equation to be as 1 following: y ′′′ + y ′′ + y ′ + y = (sin 5 x − sin x ) 2 3 2 The auxiliary equation is: λ + λ + λ + 1 = 0 sin A. cos B =
∴ λ1 = −1, and λ2 , λ3 = ± j1
∴ y h = c1e − x + c2 cos x + c3 sin x By inspecting the right hand side of the differential equation we can get the particular solution as following:
y p = A cos 5 x + B sin 5 x + C x cos x + D x sin x Where A, B, C , and D are constants.
∴ y ′p = −5 A sin 5 x + 5 B cos 5 x − C x sin x +C cos x + D x cos x + D sin x ∴ y ′p = −5 A sin 5 x + 5 B cos 5 x + (D − C x ) sin x + (C + D x ) cos x ∴ y ′p′ = −25 A cos 5 x − 25 B sin 5 x + (D − C x ) cos x − C sin x −(C + D x ) sin x + D cos x
Chapter Five
167 ∴ y ′p′ = −25 A cos 5 x − 25 B sin 5 x + (2 D − C x )cos x − (2C + D x )sin x ∴ y ′p′′ = 125 A cos 5 x − 125 B sin 5 x − (2 D − C x )sin x
−C cos x − (2C + D x )cos x − D sin x
∴ y ′p′′ = 125 A cos 5 x − 125 B sin 5 x − (3D − C x )sin x − (3C + D x )cos x Substitute y p , y ′p , and
y ′p′ into the differential equation and
compare both sides we get the following:
125 A cos 5 x − 125B sin 5 x − (3D − C x ) sin x − (3C + D x ) cos x
− 25 A cos 5 x − 25 B sin 5 x + (2 D − C x ) cos x − (2C + D x ) sin x − 5 A sin 5 x + 5 B cos 5 x + (D − C x ) sin x + (C + D x ) cos x + A cos 5 x + B sin 5 x + C x cos x + D x sin x =
Coefficient of cos 5 x = 0
1 (sin 5 x − sin x ) 2
∴ −125B − 25 A + 5B + A = 0
∴120 B + 24 A = 0
(14)
Coefficient of sin 5 x = 0.5
∴125 A − 25 B − 5 A + B = 0.5 ∴120 A − 24 B = 0.5 From (14), and (15)
(15)
∴A=
5 1 , and B = − 1248 1248
Coefficient of cos x = 0 ∴ −3C + 2 D + C + C = 0
∴ −C + 2 D = 0
(16)
Coefficient of sin x = −0.5 ∴ −3D − 2C + D + D = −0.5
∴ D + 2C = 0.5 From (16) and (17) we get the following:
(17)
168 Linear Differential Equations Of Higher Order
1 1 C = , and D = 10 5 ∴ yp =
1 (5 cos 5 x − sin 5 x ) + 0.2 x cos x + 0.1x sin x 1248
∴ y ( x ) = y h + y p = c1e − x + c2 cos x + c3 sin x +
1 (5 cos 5 x − sin 5 x ) + 0.2 x cos x + 0.1x sin x 1248
5.4 General Solutions Of Nonhomogeneous Equations IF we have the following differential equation:
Differential
(18) y ′′ + f ( x ) y ′ + g ( x ) y = r ( x ) We shall obtain a particular solution of (18) by using the method of variation parameters as follows:The homogeneous differential equation of (18) is
y ′′ + f ( x ) y ′ + g ( x ) y = 0 The general solution of this equation is:-
(19)
y h ( x) = C1 y1 ( x) + C2 y 2 ( x) The variation parameter method consists in replacing C1 and C2 by functions u ( x) and v( x) to be determined. So that the resulting function is given by the following equation:
y p ( x) = u ( x) y1 ( x) + v( x) y2 ( x)
(20)
is a particular solution of (18)
∴ y ′p ( x) = u ′y1 + uy1′ + v′y 2 + vy 2′
(21)
We shall see that we can determine u and v as following:
u ′y1 + v′y 2 = 0
(22)
Chapter Five
169
∴ y ′p ( x) = uy1′ + vy 2′
(23)
∴ y ′p′ ( x) = u ′y1′ + uy1′′ + v′y 2′ + vy 2′′
(24)
By substituting (20), (22) and (24) into (18) and collecting terms containing u and terms containing v we obtain the following:
u ( y1′′ + f y1′ + g y1 ) + v ( y 2′′ + f y 2′ + g y 2 ) + u ′y1′ + v′y 2′ = r
(25)
Since y1 , and y2 are solutions of the homogeneous equation (19). Then, we can equate the first two terms in (25) by zero. So, (25) can be reduced to be as following:
u ′y1′ + v′y 2′ = r
(26)
Solving (22) and (26) together we get the following equation:
u′ = −
y2 r W
and v′ =
y1r W
(27)
Where W = y1 y 2′ − y1′ y 2
y2 r yr dx and v = ∫ 1 dx (28) W W Substituting (28) into (20) we get the final form as following:-
∴u = −∫
y p ( x) = − y1 ∫
y2 r yr dx + y2 ∫ 1 dx W W
Example 16 Solve the following differential equation:-
e2x x Solution:- The auxiliary equation is ∴ λ2 − 4λ + 4 = 0 y ′′ − 4 y ′ + 4 y =
∴ λ1 = λ2 = 2,
∴ y h = (c1+ c2 x ) e 2 x
(29)
170 Linear Differential Equations Of Higher Order The particular solution can be obtained from (29)
y p ( x) = − y1 ∫
y2 r yr dx + y 2 ∫ 1 dx W W
e2x ∴W = 2 x 2e
xe 2 x 4x e = (2 x + 1)e 2 x
∴ y p ( x ) = −e
2x
∫
xe 2 x e 2 x / x e4x
dx + xe
2x
∫
e2xe2x / x e4x
dx
∴ y p ( x ) = −e 2 x x + x e 2 x ln ( x ) = e 2 x (ln x − x ) y ( x ) = (c1+ c2 x ) e 2 x + e 2 x (ln x − x ) ∴ y ( x ) = (c1+ c2 x + ln x − x ) e 2 x Example 17 Solve the following differential equation:-
y ′′ + 3 y ′ + 2 y = 2 xe x Solution:-
2
The auxiliary equation is : λ2 + 3λ + 2 = 0
∴ λ1 = −1, λ2 = −2
∴ y1 ( x) = e − x ,
y 2 ( x) = e − 2 x
∴ y h ( x) = C1 e − x + C2 e − 2 x The particular solution can be obtained from (29). Where, e− x e−2x W = = −e − 3 x −x −2x −e − 2e ∴ y p ( x ) = −e
−x
∫
e−2x * 2x e x − e −3x
2
dx + e
−2x
∫
e − x * 2 xe x − e −3x
2
dx
Chapter Five
171 2
2
ex ex x x2 ∴ y p ( x) = e + −e + 1 + 2x 1+ x 2 x ∴ y p ( x) = e x 1 + 3x + 2 x 2 2
2 x ∴ y ( x) = y h + y p = C1 e − x + C2 e − 2 x + e x 1 + 3x + 2 x 2
Example 18 Solve the following differential equation:-
y ′′ + 9 y = x cos x Solution:- The auxiliary equation is λ2 + 9 = 0
∴ λ1 , λ2 = ± j 3
∴ y h = A cos 3 x + B sin 3 x
By inspecting the right hand side of the differential equation we can get the particular solution as following:
y p = (k1 x + k o ) cos x + (k 2 x + k3 ) sin x ∴ y ′p = (k1 x + ko ) (− sin x ) + k1 cos x +(k 2 x + k3 ) cos x + k 2 sin x
y ′p′ = (k1 x + ko ) (− cos x ) − k1 sin x − k1 sin x
+(k 2 x + k3 )(− sin x ) + k 2 cos x + k 2 cos x
Substitute y p , y ′p , and y ′p′ into the differential equation and compare both sides we get the following:
(k1x + ko ) (− cos x ) − k1 sin x − k1 sin x + (k 2 x + k3 )(− sin x ) + k 2 cos x + k 2 cos x + 9[(k1 x + k o ) cos x + (k 2 x + k3 ) sin x ] = x cos x
172 Linear Differential Equations Of Higher Order Coefficient of x cos x = 1
∴ − k1 + 9k1 = 1 , ∴k1 = 1 / 8
(30)
Coefficient of cos x = 0
∴ − k o + 2k 2 + 9k o = 0 ,
∴ 2 k 2 + 8k o = 0
(31)
∴ −2k1 + 8k3 = 0
(32)
Coefficient of sin x = 0
∴ −2k1 − k3 + 9k3 = 0 ,
Substitute from (30) into (32) we get:
∴ 8k 3 = 2 / 8 ,
∴ k3 =
1 32
(33)
Coefficient of x sin x = 0
∴ − k 2 + 9k 2 = 0
k2 = 0
(34)
Substitute (7) into (4) we get the following:
ko = 0
(35)
Substitute (30), (33), (34) and (35) into y p we get:
∴ yp =
x cos x sin x + 8 32
x sin x sin x + (36) 8 32 Another solution for obtaining y p by using Wronskian determinant.
∴ y ( x ) = A cos 3x + B sin 3 x +
y1 = cos 3 x , and y 2 sin 3 x cos 3x sin 3x ∴W = = 3 cos 2 3 x + 3 sin 2 3 x = 3 − 3 sin 3 x 3 cos 3 x ∴ y p = − cos 3 x ∫
x sin 3 x cos x x cos 3 x cos x dx + sin 3 x ∫ dx 3 3
Chapter Five
∴ yp =
173
− cos 3 x − x cos 4 x sin 4 x x cos 2 x sin 2 x + − + 3 8 32 4 8
sin 3 x cos 4 x x sin 4 x cos 2 x x sin 2 x + + + 3 32 8 8 4 x cos 3x cos 4 x cos 3x sin 4 x x cos 3x cos 2 x ∴ yp = − + 3*8 3 * 32 3* 4 cos 3x sin 2 x sin 3 x cos 4 x x sin 3x sin 4 x − + + 3*8 3 * 32 3*8 sin 3 x cos 2 x x sin 3x sin 2 x + + 3*8 3* 4 x[cos 3x cos 4 x + sin 3x sin 4 x ] ∴ yp = 3*8 sin 4 xcox3x − sin 3x cos 4 x − 3 * 32 x[cos 3 x cos 2 x + sin 3x sin 2 x ] + 3* 4 sin 3x cos 2 x − cos 3x sin 2 x + 3*8 x cos x sin x x cos x sin x ∴ yp = − + + 3*8 3 * 32 3* 4 3*8 x cos x sin x ∴ yp = + 8 32 x cos x sin x ∴ y ( x ) = A cos 3x + B sin 3x + + 8 32 It is clear this is the same as we get in (36) +
Example 19 Solve the following differential equation:y ′′ − 5 y ′ + 6 y = x 2 e 2 x Solution:- The auxiliary equation is λ2 − 5λ + 6 = 0
∴ λ1 = 2,
λ2 = 3
∴ y h = c1e 2 x + c2 e3 x
(37)
174 Linear Differential Equations Of Higher Order By inspecting the right hand side of the differential equation we can
get the particular solution as following:
(
)
y p = k 2 x 2 + k1 x + k o xe 2 x
(
)
∴ y p = k 2 x 3 + k1 x 2 + k o x e 2 x
( ) ( ) ∴ y ′p′ = 4(k 2 x 3 + k1 x 2 + k o x )e 2 x + 2(3k 2 x 2 + k1 x + k o )e 2 x +2(3k 2 x 2 + k1 x + k o )e 2 x + (6k 2 x + 2k1 )e 2 x ∴ y ′p = 2 k 2 x 3 + k1 x 2 + k o x e 2 x + 3k 2 x 2 + k1 x + k o e 2 x
Substitute y p , y ′p , and y ′p′ into the differential equation and compare both sides we get: 4 k 2 x 3 + k1 x 2 + k o x e 2 x + 2 3k 2 x 2 + k1 x + k o e 2 x
(
(
)
)
(
)
+2 3k 2 x 2 + k1 x + k o e 2 x + (6k 2 x + 2k1 )e 2 x
[( + 6[(k x
)
(
) ]
− 5 2 k 2 x 3 + k1 x 2 + k o x e 2 x + 3k 2 x 2 + k1 x + k o e 2 x 2
3
) ]
+ k1 x 2 + k o x e 2 x = x 2 e 2 x
Coefficient of e 2 x = 0
∴ 2k o + 2k o + 2k1 − 5k o = 0
Then, 2k1 − k o = 0
(38)
Coefficient of xe 2 x = 0
∴ 4k o + 4k1 + 4k1 + 6k 2 − 10k o − 10k1 + 6k o = 0 ∴ 6k 2 − 2k1 = 0 Coefficient of x 2 e 2 x = 1 ∴ 4k1 + 6k 2 + 6k 2 − 10k1 − 15k 2 + 6k1 = 1 1 ∴ − 3k 2 = 1 , ∴ k 2 = − 3 Substitute from (40) into (39) we get:
(39)
(40)
Chapter Five
175
6 , ∴ k1 = −1 (3 * 2) Substitute from (41) into (38) we get: ko = −2 x2 + x + 2 x e2x ∴ y p = − 3
∴ k1 = −
(41)
x2 ∴ y ( x ) = c1e + c2 e − + x + 2 x e2x (42) 3 Another solution to obtain y p by using the Wronskian theory 2x
y1 = e
2x
3x
, and y 2 = e
∴ y p = −e
2x
∫
3x
∴W =
e3 x x 2e 2 x e
5x
(
3
dx + e
e2x 2e 3x
∫
e3x
2x
3e
3x
= e5 x
e 2 x x 2e 2 x
)
e
5x
dx
x − e3 x x 2 + 2 x + 2 e − x 3 3 2 x x ∴ y p = −e + x2 + 2x + 2 3 ∴ y p = −e 2 x *
x3 ∴ y ( x ) = c1e + c2 e − + x 2 + 2 x + 2 e 2 x 3 3 2x 3 x x ∴ y ( x ) = (c1 − 2 )e + c2 e − + x2 + 2x e2x 3 3 * 2x 3 x x ∴ y ( x ) = c1 e + c2 e − + x2 + 2x e2x 3 2x
3x
(43)
Where c1* = c1 − 2 As we see the general solution (43) is the same as we get before in (42).
176 Linear Differential Equations Of Higher Order 5.5 Cauchy Equations
The following equations are so called second order and third order Cauchy differential equations.
x 2 y′′ + axy′ + by = 0
(44)
x 3 y′′′ + ax 2 y′′ + bxy′ + cy = 0
(45)
The above two equations can be solved by substituting y = x m to obtain the auxiliary equations. For second order Cauchy equation, if we apply y = x m in (44) we get the auxiliary equation (46)
m 2 + (a − 1)m + b = 0 We have two roots of (46). So, we have three different cases,
(46)
Case 1 If m1 and m2 are two different roots of (46), then the two
functions y1 ( x) = x m1 and y 2 ( x) = x m2
(47)
Constitute a fundamental system of (44). So, the corresponding general solution is y ( x) = C1 x m1 + C 2 x m2
(48)
Case 2 If there is double roots (i.e. m1 = m2 = m ), then the solution
(49) takes the following form:- y ( x) = (C1 + C 2 Ln x )x Case 3 If there are two complex conjugate roots m1 , m2 = α ± jβ m
(
)
∴ y h ( x) = xα C1Cos ( Ln( x β )) + C 2 sin ( Ln ( x β )) (50) The same way can be used with the third order Cauchy differential equation (45). So the auxiliary equation will be as follows:-
m 3 + (a − 3)m 2 + (b − a + 2)m + c = 0
(51)
Chapter Five
177 Case 1 are three different roots of (51), so the general solution of
(45) is: y ( x) = C1 x m1 + C 2 x m2 + C3 x m3
(52)
Case 2 If there is double roots (i.e. m1 = m2 = m ), then the solution
takes the following form:- y ( x) = (C1 + C2 Ln x )x
m
+ C3 x m3 (53)
Case 3 If there are two complex conjugate roots m1 , m2 = α ± jβ
(
)
∴ y h ( x) = xα C1Cos ( Ln( x β )) + C 2 sin ( Ln ( x β )) + C3 x m3 Example 20 Solve the following differential equation:-
x 2 y ′′ + 3 x y ′ + y = 0 Solution:- From (46) the auxiliary equation is
m 2 + 2m + 1 = 0 Then m1 = m2 = −1 ∴ y ( x) = (C1 + C 2 Ln x )x −1 Example 21 Solve the following differential equation:-
( x − 2 ) 2 y ′′ + 5 ( x − 2 ) y ′ + 3 y = 0 Solution:-
Put t = x − 2,
∴ dt = dx
∴ t 2 y ′′ + 5t y ′ + 3 y = 0 From (46) the auxiliary equation is
m 2 + 4m + 3 = 0 ∴ m1 = −1,
m2 = −3 ,
∴ y ( x) = C1t −1 + C2t − 3 But, t = x − 2 C C2 ∴ y ( x) = 1 + x − 2 ( x − 2 )3
(54)
178 Linear Differential Equations Of Higher Order Example 22 Solve the following differential equation:-
x 2 y ′′ + 9 x y ′ + 25 y = 125 Solutions:- From (46) the auxiliary equation is
m 2 + 8m + 25 = 0
(
∴ m1 , m2 = −4 ± j 3
∴ y h ( x) = x 4 C1Cos ( Ln( x 3 )) + C2 sin ( Ln ( x 3 )) Assume y p = C3 .Substitute
in
the
)
equation,
C3 = 5 ∴ y p = 5 . Then the total solution is
(
we
get
)
∴ y ( x) = y h ( x) + y p ( x) = x 4 C1Cos ( Ln( x 3 )) + C 2 sin ( Ln ( x 3 )) + 5 Example 23 Solve the following differential equation:x 3 y ′′′ − 9 x 2 y ′′ + 34 x y ′ − 50 y = 0 Solutions:- From (51), the auxiliary equation is
m 3 − 12m 2 + 45m − 50 = 0
∴ m1 = 2, m2 = m3 = 5
∴ y h ( x) = C1 x 2 + (C2 + C3 Ln ( x) )x 5 5.6 Applications 5.6.1 Free Oscillation
Fig.1
Example 24 A weight W suspended from the
end of a vertical spring stretches it y meters. Let A and B represents the
y
position of the end of the spring
m
before and after the weight W is put on. B is called the equilibrium position. Call y the displacement of W at any position C from the equilibrium position. Assume that y is positive in the down ward direction. If we pull the body down a certain
Chapter Five
179 distance and then release it, it undergoes a motion. We want to
determine the motion of this mechanical system. For this purpose we consider the forces acting on the body during the motion. This will lead to a differential equation, and by solving this differential equation we shall then obtain the displacement as a function of time. The first force acting on the weight is the attraction of gravity.
F1 = mg Where m is the mass of the body and (g=9.81
(55) m/sec.) is the
acceleration of gravity. The next force to be considered is the spring force extended by the spring. From Hook s law
F2 = ky (56) Where y is the stretch, the constant of proportionality k is called the spring modulus. The last force acting on the weight is the weight upward due to the weight stretch the spring by an amount y0 . Then also from Hook s law:-
F3 = − ky0 It is clear that from equilibrium
(57)
mg = ky0 (58) So that the total force acting on the weight at any position y is FT = mg − ky0 − ky (59) Substitute (58) into (59). ∴ FT = − ky (60) By Newton s law, “Mass*acceleration=net force on the weight”
180 Linear Differential Equations Of Higher Order m&y& = − ky , or m&y& + ky = 0
(61)
Is the differential equation represents the system shown in Fig.1. The roots of (61) are complex. So, the oscillation of the system is called harmonic oscillation. If we connect the mass m to the dashpot, then we have to take the corresponding various damping into account. The damping force has direction opposite to the instantaneous motion. So, the damping force is of the form
F4 = −cy& (62) Where c is the damping constant. So the resulting force acting on the body when stretched and released is
FT = − ky − cy& (63) So the motion of the system is governed by the following differential equation
m&y& + cy& + ky = 0
c k The auxiliary equation is: λ2 + λ + = 0 m m 1 c ∴ λ1, 2 = − ± c 2 − 4mk 2m 2m c 1 ∴α = − and β = c 2 − 4mk 2m 2m Where, λ1,2 = −α ± β
(64) (65) (66) (67) (68)
The form of solution of (55) will depend on the damping. So, we have three cases:
Chapter Five
181
Case 1 c 2 > 4mk , Then the roots of auxiliary equation are distinct real roots (over damping). Case 2 c 2 < 4mk , Then the roots of auxiliary equation are complex
conjugate roots (under damping). Case 3 c 2 = 4mk . Then the roots of auxiliary equation are double
roots (critical damping). 5.6.2 Electric Circuit Example 25 Find the current in the RLC circuit in Fig.2. Assume
Ω,
L=20H,
for
vR(t)
C=0.002
Farads and E (t ) = 481cos10 t . Solution:-, From KVL
E(t) _
+
I(t)
_ +
di 1 + RI + ∫ idt = 481cos10 t dt c Differentiate both sides of the L
+
_
VL + VR + VC = E (t )
_
R=160
v L ( 0) = 0
+
I (0) = 0 and
vL(t) Fig.2
above differential equation.
1 ∴ LI&& + RI& + I = −4810 sin 10t c ∴ 20 I&& + 160 I& + 500 I = −4810 sin 10t ∴ I&& + 8 I& + 25 I = −240.5 sin 10 t The homogeneous solution:- λ2 + 8λ + 25 = 0 , ∴ λ = −4 ± j 3
(69)
vC(t)
182 Linear Differential Equations Of Higher Order
∴ I h (t ) = e − 4 x [ A cos 3t + B sin 3t ]
(70)
The Particular Solution:- I P = k1 sin 10 t + k 2 cos 10 t
∴ I&P = 10k1 cos10 t − 10 k 2 sin 10 t ∴ I&&P = −100 k1 sin 10 t − 100 k 2 cos10 t Substitute in (69) we get:- k1 = −1.5
and
k 2 = −1.6
∴ I P (t ) = −1.5 sin 10 t − 1.6 cos10 t
(71)
From (70) and (71) we get the general solution I (t ) = I h (t ) + I P (t ) = e − 4 x [ A cos 3t + B sin 3t ] − 1.5 sin 10 t − 1.6 cos 10 t (72)
Q I ( 0 ) = 0 , then 0 = A − 1.6, ∴ A = 1.6
Q v L (0) = LI&(0) = 0 ∴ I&(0) = 0 ∴ I&(0) = 0 = −4e − 4 x [ A cos 3t + B sin 3t ] + e − 4 x [−3 A cos 3t + 3B sin 3t ] − 15 cos 10 t + 16 sin 10 t t = 0
Then, B = 7.13333 . Substitute A and B into (72) we get:-
∴ I (t ) = e − 4 x [1.6 cos 3t + 7.133 sin 3t ] − 1.5 sin 10 t − 1.6 cos 10 t 5.6.3 Bending of Beams
A horizontal beam situated on the x axis of the xy coordinate system and supported in various ways, bends under the influence of vertical loads. The deflection curve of the beam often called the elastic curve shown in Fig.3., is given by y = f (x) where y is measured as positive downward. This curve may be determined from the following equation:
Chapter Five
EIy ′′
(1 + y )
3/ 2 ′2
183
= M ( x)
(73)
Where M(x) is the bending moment at x and is to algebraic sum of the moment being taken as positive for forces in the positive y and negative otherwise. For small deflections, y ′ is small and the following approximate equation can be used
EIy ′′ = M (x)
(74)
Where E is Young s modulus and I is the moment of inertia of a cross section of the beam about its central axis, is called the flexural
rigidity and is generally constant. Example 26 A beam of length L is simply supported at both ends as
shown in Fig.3. (a) Find the deflection if the beam has constant weight W per unit length and (b) determine the maximum deflection. x
A
L-x
B x Deflection, y(x)
y
Fig.3 Solution:- The total weight of the beam is WL, so each end supports
weight is
1 WL . Let x be the distance from the left end A of the 2
beam. To find the bending moment M at x, consider forces to the left of x
184 Linear Differential Equations Of Higher Order
(١) Force
1 1 WL at A has moment − WLx 2 2
(٢) Force due to weight of the beam to left of x has magnitude Wx
1 and moment Wx( x / 2) = Wx 2 2 Then the total bending moment at x is
1 2 1 Wx − WLx . 2 2
1 1 ∴ EIy ′′ = Wx 2 − WLx 2 2 By solving this equation for y (0) = 0 and y ′(0) = 0 we find,
(75)
W ( x 4 − 2 Lx 3 + L3 x) (76) 24 EI (b) The maximum deflection occurs at x = L / 2 , then substitute 5WL4 in (76) for x = L / 2 we get the maximum deflection as 384 EI y ( x) =
Example 27 A beam of length 10m is simply supported at both ends
as shown in Fig.4. (a) Find the deflection if the beam has constant weight 3000 Newton per meter and 30000 Newton concentrated load at the middle of the beam (b) determine the maximum deflection.
x
A
L-x C
y
5-x
30000 N
Deflection, y(x)
Fig.4
B x
Chapter Five
185 Solution:-(a) The total weight of the beam is 3000*10=30000
Newton then the total weight is 30000+30000=60000 Newton, so each end supports weight is 60000 / 2 = 30000 Newton. Let x be the distance from the left end A of the beam. To find the bending moment M at x, consider forces to the left of x (١) Force 30000 N at A has moment − 30000 x (٢) Force due to weight of the beam to left of x has magnitude 3000x and moment 3000 x( x / 2) = 1500 x 2 Then the total bending moment at x is 1500 x 2 − 30000 x .
∴ EIy ′′ = 1500 x 2 − 30000 x
(77)
∴ y h = c1 + c2 x
(78)
(
)
∴ y P = x 2 A1 x 2 + A2 x + A3 = A1 x 4 + A2 x 3 + A3 x 2 ∴ y ′P′ = 12 A1 x 2 + 6 A2 x + 2 A3 Substitute (79) into (77)
(
(79)
)
∴ EI 12 A1 x 2 + 6 A2 x + 2 A3 = 1500 x 2 − 30000 x ∴ A1 =
1500 125 = , 12 EI EI
A2 = −
30000 5000 =− , 6 EI EI
A3 = 0
Then the general solution of (77) is:
∴ y ( x) =
(
)
1 125 x 4 − 5000 x 3 + c2 x + c1 EI
The boundary condition is : y (0) = 0, y ( L) = 0 ,
(80)
186 Linear Differential Equations Of Higher Order
(
)
1 375000 5000 L2 − 125L3 = EI EI Substitute (81) into (89) we get:
∴ c1 = 0,
c2 =
[
(81)
]
1 125 x 4 − 5000 x 3 + 375000 x (82) EI (b) The maximum deflection occurs at x = L / 2 = 5m , then
∴ y ( x) =
substitute in (82) for x = 5 we get the maximum deflection as
Example
28
1328125 EI
10
m
cantilever beam has one end
L-x A
C
x
horizontally
imbedded in concrete and a force 12000 N
B Deflection, y(x) Fig.5
12000 N
acting on the other end as shown in Fig.5. Find (a) the deflection and (b) the maximum deflection of the beam assuming its weight is 3000 Newton/meter. Solution:-
a) Let x be the distance from the left end B of the beam. To find the bending moment M at x, consider forces to the right of x (١)
Force 12000 N at B has moment 12000 x
(٢)
Force due to weight of the beam to the right of x has
magnitude 3000x and moment 3000 x( x / 2) = 1500 x 2
Then the total bending moment at x is 1500 x 2 + 12000 x .
Chapter Five
187
EIy ′′ = 1500 x 2 + 12000 x
(83)
y h = c1 + c2 x
(84)
(
)
y P = x 2 A1 x 2 + A2 x + A3 = A1 x 4 + A2 x 3 + A3 x 2 y ′P′ = 12 A1 x 2 + 6 A2 x + 2 A3
(85)
Substitute (85) into (83) we get:
(
)
EI 12 A1 x 2 + 6 A2 x + 2 A3 = 1500 x 2 + 12000 x ∴ A1 =
1500 125 = , 12 EI EI
A2 =
12000 2000 = , 6 EI EI
A3 = 0
Then the general solution of (83) is:
y ( x) =
(
)
1 125 x 4 + 2000 x 3 + c2 x + c1 EI
The boundary condition is : y (10) = 0, 7.75 * 10 6 ∴ c1 = , EI
(86) y ′(10) = 0 ,
1.1 * 10 6 c2 = − EI
(87)
Substitute (87) into (86)
∴ y ( x) =
[
1 125 x 4 + 2000 x 3 − 1.1 * 10 6 x + 7.75 * 10 6 EI
(c) The maximum deflection occurs at
]
x = 0 , then
substitute in (88) for x = 0 we get the maximum
7.75 * 10 6 deflection as EI
(88)
188 Linear Differential Equations Of Higher Order Problems Solve the following differential equations
١)
y ′′′ + y ′′ − 2 y ′ = 0
٢)
y ′′ + 6 y ′ + 9 y = 0
٣)
y′′ − 4 y′ + 13 y = 0
٤)
y′′′ + − y′′ + 9 y′ − 9 y = 0
٥)
y ′′ − 6 y ′ + 9 y = e 2 x
٦)
y ′′ − y = 4 xe x
٧)
y′′ − y = sin 2 x
٨)
y ′′ + y = csc x
٩)
y ′′ − 3 y ′ + 2 y = sin e − x
١٠)
y ′′ + 4 y = sec 2 2 x
١١)
y′′′ − 2 y′′ − 5 y′ + 6 y = e 2 x + 3
١٢)
y′′′ − 5 y′′ + 8 y′ − 4 y = e 2 x + 2e x + 3e − x
١٣)
y ′′ + 4 y = sin 2 x + cos 2 x
١٤)
y ′′ − 8 y ′ + 25 y = 5 x 3e − x − 7e − x
١٥)
y ′′ − 9 y ′ + 14 y = 3x 2 − 5 sin 2 x + 7 xe 6 x
١٦)
y (4 )+ y ′′′ = 1 − e − x
١٧)
y ′′ + 2 y ′ + y = e − x Ln( x)
( ) (
)2
Chapter Five
189
١٨)
x 2 y ′′ + xy ′ − 4 y = 0
١٩)
x 3 y ′′′ − 3x 2 y ′′ + 6 xy ′ − 6 y = 0
٢٠)
x 3 y′′′ + xy′ − y = 0
٢١)
(2 x + 1)2 y′′ − 2(2 x + 1) y′ − 12 y = 0
٢٢) Find the current in the RLC circuit in Fig.2. Assume
I (0) = 0 and v L (0) = 1V for R=160 ohms, L=20H, C=0.002 farads and E (t ) = te − t . ٢٣) Find the current I(t) in Fig.2. Assume I (0) = 1A and
v L ( 0) = 1 V
for R=20
, L=5H, C=0.04 farads and
E (t ) = 10 cosh 5 t . ٢٤) A 20 kg weight suspended from the end of a vertical spring stretches it 10 cm. assuming no external forces, find the position of the weight at any time if initially the weight is pulled down 10 cm and released. (a) Without dashpot (b) If we connect the mass m to the dashpot with damping constant c=1, state which type of damping you have and find the position of the weight at any time.
Chapter 6 Laplace Transforms 6.1 Definition For any given function f (t ), t ≥ 0 we can define another function
F (s ) by the following definite integral: ∞
L{ f (t )} = F ( s ) = ∫ e − st f (t )dt
(1)
0
The function F (s ) is called the Laplace transform of the function f (t ) . Note that F (s ) is simply the total area under the curve f (t )
for t=0 to infinity, whereas F (s ) for s greater than 0 is a "weighted" integral of f (t ) , since the multiplier e − st is a decaying exponential function equal to 1 at t = 0 . Thus as s increases, F (s ) represents the area under f (t ) weighted more and more toward the initial region near t = 0 . Knowing the value of F (s ) for all s is sufficient to fully specify f (t ) , and conversely knowing f (t ) for all t is sufficient to determine F (s ) . The benefit of considering the Laplace transform of a function is that it sometimes enables us to solve problems easily in the "s domain" that would be difficult to solve in the "t domain". See Appendix 1 for table of general properties of Laplace transform. Now let us see some helpful examples of these transforms.
191
Chapter Six
Example 1: find Laplace transform of the following function:
f (t ) = 1 ∞
∞
e − st Solution: F ( s ) = ∫ e − st f (t )dt = ∫ e − st dt = −s 0 0
∞
= 0
0 −1 1 = −s s
Example 2: find Laplace transform of the following function:
f (t ) = e at Solution: ∞
∞
∞
e − ( s − a )t − st − st at − ( s − a )t F ( s ) = ∫ e f (t )dt = ∫ e e dt = ∫ e dt = − ( s − a) 0 0 0
∞
= 0
1 s−a
6.2 Inverse Laplace Transforms
If L{ f (t )} = F ( s ) , then we can call f (t ) the inverse Laplace transform of F (s) and write L−1{F ( s )} = f (t )
Example3 If F ( s ) =
1 s2
Solution: Since F ( s ) =
, find f (t ) −1 1 , we have L 2 = t ∴ f (t ) = t s2 s
1
Theorem 1 The Laplace transform and inverse Laplace
transform is linear operator
∴ L{c1 f1 (t ) + c2 f 2 (t )} = c1L{ f1 (t )} + c2 L{ f 2 (t )} and
L{c1F1 ( s) + c2 F2 ( s)} = c1L{F1 ( s)} + c2 L{F2 ( s)}
192 Laplace Transforms Theorem 2
Suppose that
f (t ) is continuous and its
f ′(t ) is piecewise
continuous in any interval 0 ≤ t ≤ T and f (t ) is of exponent order for
t >T .
∴ L{ f ′(t )} = sL{ f (t )} − f (0) This can be extended to be as following: L{ f
( n)
(t )} = s n L{ f (t )} − s n −1 f (0) − s n − 2 f ′(0) − s n −3 f ′′(0)... − f
n −1
(0) (2)
Theorem 3
If L−1{F ( s )} = f (t ) , ∴ L−1{F ( s − a )} = e at f (t )
(3)
Proof: −1
L {F ( s − a )} =
∞
∫ f (t ) e
− ( s − a )t
0
∞
dt = ∫ f (t ) e at e − st dt =e at f (t ) 0
Theorem 4 t
If, g (t ) = ∫ f (t ) dt 0
1 ∴ L{g (t )} = L{ f (t )} s
Example 4 Let F ( s ) =
(4)
1 s(s 2 + w2 )
, find f (t )
193
Chapter Six
1 1 Solution: We know that L−1 2 = sin ωt ( s + ω 2 ) ω 1 t ∴L 2 = ∫ sin ωt dt s ( s + ω 2 ) ω 0 −1
1
1 1 [1 − cos ωt ] ∴ L−1 2 = s ( s + ω 2 ) ω 2 Example 5 Let F ( s ) =
1 s 2 (s 2 + ω 2 )
, find f (t )
Solution: We know from the previous example that:
1 1 [1 − cos ωt ] L−1 2 = s ( s + ω 2 ) ω 2 1 t ∴L 2 2 = ∫ [1 − cosωt ]dt s ( s + ω 2 ) ω 2 0 −1
1
1 1 1 = − ∴ L−1 2 2 t sin ω t s ( s + ω 2 ) ω 2 ω Example 6
Proof that L{sin at} =
a 2
(s + a 2 )
e jat − e − jat Solution: We know that: sin at = j2 It follows from the linear character of the transform that: L{sin at} =
1 1 L{e jat } − L{e − jat } j2 j2
194 Laplace Transforms
1 1 1 1 1 j 2a a − = = j 2 ( s − ja ) j 2 ( s + ja ) j 2 ( s 2 + a 2 ) ( s 2 + a 2 ) The same procedure can be used to proof that:
∴ L{sin at} =
L{cos at} =
s 2
(s + a 2 )
Example 7
Find the Laplace transform of the following functions (a) ae − bt (b) at 2 (c) 4 cos 5t (d) − 3 / t (e) sin 2 t Solution:
(a) L{ae
− bt
∞
} = ∫ ae
− bt − st
e
0
∞
dt = a ∫ e − ( s + b)t dt = 0
∞
(b) L{at } = ∫ at 2 e − st dt = a 2
Γ(3)
0
(c) L{4 cos 5t} = 4 *
s s 2 + 25
=
s3
=
a * 2! s3
=
a s+b
2a s3
4s
s 2 + 25
(d) L{−3 / t } = −3L{t1 / 2 } = −3
Γ(1 / 2) s1 / 2
=
−3 π π = −3 s s
1 (e) L{sin 2 t} = L (1 − cos 2t ) 2 1 1 1 L{sin 2 t} = L (1 − cos 2t ) = L(1) − L{cos 2t} 2 2 2 2 1 s = − 2 = 2 2 s 2 s + 4 s ( s + 4)
(
)
195
Chapter Six
Example 8 Find the Laplace transform of the following functions:
(a) e − 5t t 3 (b) e − 5t cos 4t
{
} { }s = s −5 = s34!
Solution: (a) L e 5t t 3 = L t 3
{
}
(b) L e − 5t cos 4t = L{cos 4t} s = s + 5 =
= s = s −5
6
(s − 5)4
s s 2 + 16 s = s + 5
=
s+5 ( s + 5) 2 + 16
Theorem 5
For n = 1,2,3,.....
{
}
n
L t f (t ) = (−1)
n
dn ds n
F (s)
(6)
Where F ( s ) = L{ f (t )}
Example 9
Find Laplace transform of the following functions (a) te at (b) t sin wt (c) t 2 sin wt (d) te − t cos t (e) t 3 cos ωt Solution:
{ }
(a) L te at = −
d d 1 1 L(e at ) = − = ds ds s − a (s − a )2
(b) L {t sin wt} = −
{
}
d d w 2 ws L(sin wt ) = − = 2 2 ds ds s + w s 2 + w2
(
2 ws d d (c) L t sin wt = − L(t sin wt ) = − ds ds s 2 + w 2 2
(
=
)2
6 ws 2 − 2 w3
) (s 2 + w2 )3 2
196 Laplace Transforms
{
}
L te − t cos t = − (d)
d d L(e − t cos t ) = − L(cos t ) ds ds s → s +1
( d s + 1 s + 1)2 − 1 =− = ds (s + 1)2 + 1 (s + 1)2 + 1 2
(
)
(e) t 3 cos ωt we know that L{cos ωt} =
{
3
{
3
}
s s2 + ω 2
d3 s ∴ L t cosωt = (− 1) 2 3 2 ds s + ω 3
}
d 2 s 2 + ω 2 − 2 s 2 ∴ L t cosωt = − 2 ds s 2 + ω 2
(
)
(
2 ) ( − 2 s ) + (s 2 − ω 2 )(2 * (s 2 + ω 2 )* 2 s ) { } (s 2 + ω 2 )4 3 2 ( s 2 + ω 2 ) (6 s 2 − 6ω 2 ) − 3 * 2 s (s 2 + ω 2 ) * (2 s 3 − 6ω 2 s ) =− (s 2 + ω 2 )6 2 6(s 2 + ω 2 ) (s 2 − ω 2 ) + 12 s 2 (s 2 − 3ω 2 ) 3 ∴ L{ t cos ωt }= − (s 2 + ω 2 )5
d s2 + ω 2 ∴ L t cos ωt = ds 3
Example 10 Solve the following differential equation: y ′′ + 2 y ′ + 5 y = 0 ,
Solution:
y (0) = 2,
y ′(0) = −4
197
Chapter Six
s 2Y ( s ) − sy (0) − y ′(0) + 2[ sY ( s ) − y (0)] + 5Y ( s ) = 0 ∴ Y (s) =
2s s 2 + 2s + 5
=2
s +1 ( s + 1) 2 + 2 2
−
2 ( s + 1) 2 + 2 2
∴ y (t ) = 2e − t cos 2t − e − t sin 2t
∴ y (t ) = e − t (2 cos 2t − sin 2t ) 6.3 Partial Fractions
As explained in the previous examples in many cases the function
Y (s) is so important to express it in terms of its partial fraction. In most cases Y (s ) takes the following form:
G ( s) H ( s) Where G (s ) and H (s ) are polynomials of s. Y ( s) =
(7)
Assume G (s ) and H (s ) have no common factors and have real coefficients. The degree of G (s ) is lower than H(s). Let s=a be a root of H ( s ) = 0 . There are many cases for the roots of
H (s ) will be explained in the following cases: Case 1 Unrepeated factors
A G(s) A A2 = 1 + + .... n + W ( s) (8) H ( s) s − a1 s − a2 s − an Where W (s ) is denotes the sum of the partial fractions Y (s) =
corresponding to all the (unrepeated or repeated) linear factors of
H (s ) which are not under consideration.
198 Laplace Transforms
∴ L−1{Y ( s )} = A1 e a1t + A2 e a 2 t + ....... An e a n t + L−1{W ( s )}
(9)
The constants A1 , A2 ,....... An can be determined by multiplying both sides of (8) by (s − a1 ) and by assuming s = a1 we can obtain
A1 . Same procedure can be carried out to obtain A2 by multiplying both sides of (8) by (s − a2 ) and by assuming s = a2 we can obtain
A2 . Thin we can say, An can be obtained by multiplying both sides of (8) by (s − an ) and by assuming s = an we can obtain An . To illustrate this procedure consider we need to obtain An multiply both sides of (8) by (s − an ) we get the following equation:
(s − an ) G(s) = (s − an )A1 + (s − an )A2 + ....An + (s − an )W (s) H ( s)
s − a1
s − a2
Let s = an in the above equation then the terms involving
A1 , A2 , .... An −1 all will disappear, and An will be the only constant remains in the equation following:
An = (s − an )
G( s) H (s) s = a
(10) n
Formulas for A1 , A2 ,…… An can be obtained similarly (By cyclic permutation of the subscripts. Example 11 Find Laplace transform of the following function:
G ( s) 2s 2 − 4 Y ( s) = = H ( s) s 3 − 4s 2 + s + 6
199
Chapter Six
Solution:
Y ( s) =
A3 A1 A2 G ( s) 2s 2 − 4 = + + = 3 H ( s ) s − 4 s 2 + s + 6 ( s − 2) ( s + 1) ( s − 3)
(
)
(s − 2) * G(s) (s − 2) 2s 2 − 4 ∴ A1 = = H ( s) s = 2 (s − 2)(s + 1)(s − 3) ∴ A2 =
(
)
2
(s + 1) * G(s) (s + 1) 2s − 4 = H ( s) s=−1 ( s − 2)(s + 1)(s − 3)
(
= s =2
= s=−1
)
(s − 3) * G(s) (s − 3) 2s 2 − 4 ∴ A3 = = H ( s) s =3 ( s − 2)(s + 1)(s − 3)
G ( s) − 4 / 3 − 1 / 6 7/2 ∴ Y ( s) = = + + H ( s ) ( s − 2) ( s + 1) ( s − 3)
4 4 =− 3 * −1 3
1 −2 =− 6 − 3 * −4
= s =3
14 7 = 4 2
4 1 7 ∴ y (t ) = − e 2t − e − t + e3t 3 6 2 m Case 2 Repeated factor ( s − a ) , in this case;
Y (s) =
Am Am −1 G(s) A1 = + + ...... + W ( s) (s − a ) H ( s ) (s − a )m (s − a )m −1
t m−1 t m−2 L {Y(s)}= e Am + Am−1 + ....+ A2t + A1 + L−1W(s) (m − 2)! (m −1)! −1
at
Where: Am = (s − a )m
G (s ) H (s )
d m − k (s − a )m G ( s) 1 Ak = H ( s) (m − k ) ! ds m − k Where, k = 1,2,...., m − 1
(11)
(12) (13)
, s =a
(14)
200 Laplace Transforms Example 12 Solve the following initial value problem: y ′′ − 3 y ′ + 2 y = 4 t + e 3t ,
y ( 0 ) = 1, y ′( 0 ) = − 1
Solution:
s 2Y ( s ) − sy (0) − y′(0) − 3[ sY ( s ) − y (0)] + 2Y ( s ) =
4 s2
+
1 s −3
G ( s ) s 4 − 7 s 3 + 13s 2 + 4s − 12 = Y (s) = H (s) s 2 ( s − 3)( s 2 − 3s + 2) A A B C D = 22 + 1 + + + s ( s − 3) ( s − 2) ( s − 1) s − 12 G (s ) s 4 − 7 s 3 + 13s 2 + 4 s − 12 = =2 A2 = s * = −6 H (s ) s = 0 ( s − 3)( s 2 − 3s + 2) s =0 2
d s 4 − 7 s 3 + 13s 2 + 4 s − 12 1 =3 A1 = ( 2 − 1)! ds ( s − 3)( s 2 − 3s + 2) s =0 B = (s − 3) *
C=
1 G (s ) s 4 − 7 s 3 + 13s 2 + 4 s − 12 = = 2 H (s ) s = 3 s 2 ( s 2 − 3s + 2) s =3
s 4 − 7 s 3 + 13s 2 + 4 s − 12 s 2 ( s − 3)( s − 1)
= −2 s=2
1 G (s ) s 4 − 7 s 3 + 13s 2 + 4 s − 12 = − D = (s − 1) * = 2 H (s ) s =1 s 2 ( s − 3)( s − 2) s =1 ∴ Y (s) =
2
3 1/ 2 2 1/ 2 + + − − s 2 s ( s − 3) ( s − 2) ( s − 1)
1 1 ∴ y (t ) = 2t + 3 + e3t − 2e 2t − et 2 2
201
Chapter Six
Case 3 Unrepeated complex factors (s − a ) 2 2 Where a = α + jβ , a = α − jβ , ( s − a )( s − a ) = ( s − α ) + β
Y ( s) =
G(s) As + B = + W (s) H ( s ) (s − α )2 + β 2
L−1{Y ( s)} =
(15)
1 αt e [Ta cos β t + S a sin β t ] + L−1{W ( s)}
(16)
β
[
Where Ra ( s) = S a + jTa = ( s − α ) 2 + β 2
] HG((ss))
(17) s =a
Example 13 Solve the initial value problem y ( 0 ) = 2 , y ′( 0 ) = − 4
y ′′ + 2 y ′ + 5 y = 0 ,
Solution: S 2Y ( s ) − sy (0) − y ′(0) + 2[ sY ( s ) − y (0)] + 5Y ( s ) = 0
2s 2s G ( s) = 2 = H ( s ) s + 2 s + 5 ( s − (−1 + j 2))( s − (−1 − j 2)) ∴ a = −1 + j 2, a = −1 − j 2 ∴α = −1, β = 2 G ( s) G(s) ∴ Ra ( s ) = S a + jTa = ( s − α ) 2 + β 2 = ( s + 1) 2 + 4 H ( s) H (s) G (a) ∴ Ra (a ) = S a + jTa = (−1 + j 2 + 1) 2 + 4 = 2 s = −2 + j 4 H (a) ∴ S a = −2, Ta = 4 1 ∴ L−1{Y ( s )} = eαt [Ta cos β t + S a sin β t ] + L−1{W ( s)} Y ( s) =
[ [
]
β
1 = e − t [4 cos 2 t − 2 sin 2 t ] 2 = e − t [2 cos 2 t − sin 2 t ]
[
]
]
202 Laplace Transforms Example 14 Solve the initial value problem y ′′ − 3 y ′ + 2 y = e t ,
y ( 0 ) = 1, y ′( 0 ) = 1
Solution: s 2Y ( s ) − sy (0) − y ′(0) − 3[ sY ( s ) − y (0)] + 2Y ( s ) =
G ( s) s 2 − 3s + 3 A2 A1 B Y (s) = = = + + H ( s ) (s − 1)2 ( s − 2) (s − 1)2 (s − 1) ( s − 2)
s 2 − 3s + 3 ∴ A2 = ( s − 2)
∴ A1 =
s =1
d s 2 − 3s + 3 ds ( s − 2)
=
∴ B=
= −1
s =1
( s − 3s + 3) − ( s − 2)(2 s − 3) ( s − 2) 2
s 2 − 3s + 3
∴Y ( s) =
’
2
( s − 1)
2
−1
(s − 1)2
=1 s=2
+
1 ( s − 2)
∴ y (t ) = −tet + e 2t Example 15 Find f (t ) if F (s ) equals:
F (s) = Solution:
s 3 − 7 s 2 + 14s − 9 ( s − 1) 2 ( s − 2)3
=0 s =1
1 s −1
203
Chapter Six
F (s) =
A2 =
A2
( s − 1) 2
B3 A1 B2 B1 + + + ( s − 1) ( s − 2)3 ( s − 2) 2 ( s − 2)
+
s 3 − 7 s 2 + 14 s − 9 ( s − 2) 3
=1 s =1
d s 3 − 7 s 2 + 14s − 9 A1 = ds ( s − 2) 3 B3 =
=0 s =1
s 3 − 7 s 2 + 14s − 9 ( s − 1) 2
= −1 s=2
d s 3 − 7 s 2 + 14s − 9 B2 = ds ( s − 1) 2
=0 s =2
1 d 2 s 3 − 7 s 2 + 14s − 9 B1 = 2! ds 2 ( s − 1) 2 1
∴ F (s) =
2
−
=0 s=2
1
( s − 1) ( s − 2) 3 1 ∴ f (t ) = te t − t 2 e 2t 2 Example 16 Find f (t ) if F (s ) equals:
F (s) =
Solution:
F (s ) =
A2 s2
+
B3 B2 B1 A1 + + + 2 3 s s2 + 2 s2 + 2 s2 + 2
(
) (
) (
)
s +1 2
s ( s 2 + 2) 3
204 Laplace Transforms
s +1
∴ A2 =
∴ A1 =
( s 2 + 2) 3
= s =0
d s +1 ds ( s 2 + 2) 3
(
1 8 =
s =0
)3
Q s2 + 2 = 0
1 8
∴ s = ± j 2 = α + jβ
∴ α = 0, β = 2 ∴ B3 =
s +1 s2
s= j 2
d s +1 ∴ B2 = ds s 2 s = j
1 1 =− − j 2 2 =
s 2 − 2s (s + 1) s4
2
1 d − s 2 − 2s ∴ B1 = 2! ds s4 s= j
2
(
s= j 2
)
1 s 4 (− 2 s − 2 ) + 4 s 3 s 2 + 25 = 2 s8 s = j
1 1 =− − j 2 2 2
= −0.75 − j 0.3535 2
t 1 1 −1 2 1 1 ∴ y(t ) = + + cos 2t − sin 2t cos 2t − sin 2 t − 8 8 2 4 2 2 2 − 0.35 cos 2t − 0.75sin 2t ∴ y (t ) =
(t + 1) − (cos 8
]
2t + 1.24 sin 2t
)
205
Chapter Six
Theorem 6
f (t ) ∞ ∴ L = ∫ F ( s ) ds t s
If L{ f (t )} = F ( s )
(18)
Proof: ∞ 0
By definition: L{ f (t )} = F ( s ) = ∫ f (t )e − st dt Integrate both sides of the above equation from s to ∞ , we obtain the following equation: ∞
∞ ∞ (19) F ( s ) = ∫ ∫ f (t )e − st dt ds s 0 The function f (t ) is continuous, then, the integration with respect to
∫s
s can be performed inside the t integral. Hence performing the integration we get the following: ∞ F s
∫
∞ ∞ s 0
(s )ds = ∫ ∫
f (t )e
− st
∞ 0
ds dt = ∫
∞
e − st f (t ) dt − t s
∞ f (t ) − st f (t ) e dt = L =∫ 0 t t We can obtain the useful result from theorem 6
If L{ f (t )} = F ( s )
{
∞ s
∴ f (t ) = L−1{F (s )} = tL−1 ∫ F ( s ) ds
}
Example 17 Find Laplace transform of the following function
sin ωt t Solution: From theorem (6) we can say that:
206 Laplace Transforms ∞ ω s sin ωt ∞ L ds = tan −1 = ∫s L{sin ωt}ds = ∫s 2 2 ω t s +ω s sin ωt π −1 s ∴ L = cot −1 = − tan ω ω t 2
∞ s
Example 18 Find inverse Laplace transform of the following
function:
s
(s 2 − 1)2
Solution: From above theorem we have immediately that:
−1 − 1 f (t ) ∫ 2 2 ds = tL 2 s 2 − 1 s − 1 1 sinh t ∴ f (t ) = tL−1 2 =t 2 − s 2 1
∞ = tL−1 s
( (
s
(
)
∞
)s
)
Example 19 Find inverse Laplace transform of the following
function: F (s ) =
s+2
(s 2 + 4s + 5)2
s+2 s − 2t −1 Solution: L {F (s )} = L =e L 2 2 2 2 (s + 2 ) + 1 s + 1 ∞ s −1 − 2t −1 ∞ − 2t −1 − 1 ds = t e L 2 ∴ L {F (s )} = t e L ∫ 2 s 2 2 s 1 + s + 1 s 1 t e − 2t sin t ∴ L−1{F (s )} = t e − 2t L−1 2 = 2 + s 2 1 −1
−1
(
)
(
(
)
)
(
)
(
)
207
Chapter Six
6.4 Step Functions
In order to deal effectively with functions having jump discontinuous, it is very helpful to introduce a function known as the unit step function. This function will be denoted by uc (t ) , and is defined by:
t < c, t≥c c≥0
0, uc (t ) = 1
(20)
Its graph is shown in the following Fig.1. uc(t) 1
Fig.1 t
t=c
Example 20 Find Laplace transform of uc (t ) Solution: ∞
L{uc (t )} = ∫ e
− st
0
∴ L{uc (t )} =
c
uc (t )dt = ∫ e 0
− st
∞
(0)dt + ∫ e c
− st
e − cs (1)dt =0 + s
e − cs s
Example 21 Find Laplace transform of 8 * u3 (t ) Solution:
This
function
8e − 3s ∴ L{8 * u3 (t )} = s
has
the
following
shape
Fig.2
208 Laplace Transforms 8u3(t) 8
t
t=3
Fig.2 Example 22 Find Laplace transform of the following function 2≤t≤3 2 f (t ) = Otherwise 0 Solution:
The function f (t ) is shown in Fig.3. This function can expressed in terms
of
the
f (t ) 2
f 1 (t ) f 2 (t )
3
u 2 (t ) 2 3
u 3 (t )
two
f1 ( x ) ,and,
functions
2
Fig.3
f 2 ( x ) where : f ( x ) = f1 ( x ) + f 2 ( x ) = 2u 2 (t ) − 2u3 (t )
(
2e − 2 s 2e − 3s 2 e − 2 s − e − 3s − = ∴ F (s ) = L{ f (t )} = L{2u 2 (t ) − 2u3 (t )} = s s s We can check the above solution by using the elementary Laplace transform principles as following: 3 − st e dt 2
L{ f (t )} = 2 * ∫
e − st = 2* −s
3 2
(
e − 3s − e − 2 s 2 e − 2 s − e − 3s = = 2 s s −
)
)
209
Chapter Six
Example 23 Find Laplace transform of the following function
3 f (t ) = − 2 0
4
0≤t ≤2 2≤t ≤6 Otherwise
3
terms of unit step function. If we look deeply to the function
Fig.4
1
In this example we will try to this discontinuous function in
B
2
Solution:
get general method to convert
A
O -1
E 1
-2
2
3
4
5
C
shown in Fig.4 we can see that, this function consists of four unit step functions as following: 3uo (t ),
−5u 2 (t ) , and, 2u6 (t ) ,
where f (t ) = 3uo (t ) − 5u 2 (t ) + 2u6 (t ) If we look deeply to Fig.4 we can see that it is easy to get that by starting from the origin and take each jump (arrow) as a unit step function. It is clear from Fig.4 that we have three jumps (arrows) OA, BC, and DE. For OA its direction is up so we will take +ve sign for the unit step function and its length is 3 so the amplitude of unit step function is 3 and it happened at t = 0 , then the suffix for unit step function will be zero. Then the first arrow (OA) equivalent to OA = 3uo (t ) . In the same way we can deal with the jump (arrow) BC and DE and they equivalent to − 5u 2 (t ) and 2u6 (t ) respectively. The following table clarifies the jump method.
6
D
210 Laplace Transforms
Jump
Direction Sign
Amplitude time
Equivalent
OA
UP
+ve
3
t =0
3uo (t )
BC
DOWN
-ve
5
t=2
− 5u 2 (t )
DE
UP
+ve
2
t =6
2u6 (t )
∴ f (t ) = 3uo (t )− 5u 2 (t ) + 2u6 (t ) 3 − 5e − 2 s + 2e − 6 s ∴ L{ f (t )} = {3uo (t ) − 5u 2 (t ) + 2u6 (t )} = s Example 24 Find Laplace transform of the following function
2 f (t ) = 0 1 Solution: The function
0≤t ≤3 3≤t ≤ 4 Otherwise
3 2
A
B E
1 f (t ) is
shown in Fig.5. As explained in details in
O -1 2
C 1
2
3
4D 5 6
Fig.5
the previous example we can get the following table: Jump OA
Direction Sign UP +ve
Amplitude time 2 t =0
BC
DOWN
-ve
2
t =3
− 2u3 (t )
DE
UP
+ve
1
t=4
u 4 (t )
∴ f (t ) = 2uo (t ) − 2u3 (t ) + u 4 (t )
Equivalent 2uo (t )
2 − 2e − 3s + e − 4 s ∴ L { f (t )} = L{2uo (t ) − 2u3 (t ) + u 4 (t )} = s
7
211
Chapter Six
Theorem 7 If F (s ) is L{ f (t )}, then:
L{uc (t ) f (t − c)} = e − cs F ( s )
(21)
∴ L−1{e − cs F ( s )} = uc (t ) f (t − c)
(22)
Conversely If f (t ) = L−1{F ( s )}, then:
Proof To proof theorem 7, it is sufficient to compute the transform
of uc (t ) f (t − c) . Clearly ∞
L{uc (t ) f (t − c)} = ∫ e
− st
∞
uc (t ) f (t − c)dt = ∫ e − st f (t − c)dt c
0
Introducing, a new integration variable ψ = t − c , we have ∞
L{uc (t ) f (t − c)} = ∫ e − s (ψ + c ) f (ψ )dψ 0
∞
e − cs ∫ e − sψ f (ψ )dψ = e − cs F ( s ) 0
Example 25 Find Laplace transform of the following function
t2 0
It is clear from Fig.6 that, f (t ) = t 2 [u o (t ) − u3 (t )] ∴ L{F (t )} = (− 1)2
d 2 1 e − 3s − 2 s s ds
(
)
d − 1 s − 3e − 3s − e − 3s ∴ L{F (t )} = 2 − ds s s2
d 3se − 3s + e − 3s − 1 ∴ F (s ) = ds s2
212 Laplace Transforms
∴ F (s ) = ∴ F (s ) = ∴ F (s ) =
[
]
s 2 − 9se − 3s + 3e − 3s − 3e − 3s − 2s (3se − 3s + e − 3s − 1) s4
− 9s 2 e − 3s − 6se − 3s − 2e − 3s + 2 − e − 3s s3
s3
(9s 2 + 6s + 2)+ s23 t2
Fig.6
3
uo (t ) − u3 (t )
t 2 (uo (t ) − u3 (t )) Another solution:
This equation, f (t ) = t 2 [uo (t ) − u3 (t )] can be modified to be in the form of (21). ∴ t 2 [uo (t ) − u3 (t )] = t 2uo (t ) − (t − 3)2 + 6(t − 3) + 9 u3 (t ) It is clear the above equation is in the form of (21).
(
)
213
Chapter Six
{
(
)
}
6 9 − 3s 2 − + + e s3 s3 s 2 s Which is clear, the same result that we got from previous solution.
∴ L t 2uo (t ) − (t − 3)2 + 6(t − 3) + 9 u3 (t ) =
2
Example 26 Find Laplace transform of the following function
et
0 < t <1 Solution: As explained in the previous example the above
function can be expressed as f (t ) = et [uo (t ) − u1 (t )] e − (s −1) 1 ∴ F (s ) = L{ f (t )} = − s − 1 (s − 1) Another solution: This equation, f (t ) = et [uo (t ) − u1 (t )] can be modified to be in the
et −1 form of (21). ∴ e [uo (t ) − u1 (t )] = e uo (t ) − −1 u1 (t ) e It is clear the above equation is in the form of (21). t
t
t e − ( s −1) e− s e t −1 1 1 ∴ L e uo (t ) − −1 u1 (t ) = * e = − − s 1 s 1 s 1 s 1 − − − − e Which is clear, the same result that we got from previous solution. Example 27 Find Laplace transform of the following function:
k sin ωt
0 < t < π /ω
Solution: As explained in the previous example the above
function can be expressed as following:
f (t ) = k sin ωt [uo (t ) − uπ / ω (t )] = k sin ωt uo (t ) − k sin ωt uπ / ω (t )
214 Laplace Transforms
To put the above equation in the form of (21) we have to replace sin ωt uπ / ω (t )
with
sin (ωt − π / ω ) uπ / ω (t ) . It is clear that
sin (ωt − π / ω ) = − sin (ωt ) . Substitute this in the above equation we get the following: f (t ) = k sin ωt u o (t ) + k sin (ωt − π / ω )uπ / ω (t )
∴ F (s ) =
π s ω kωe −
kω s2 + ω 2
+
s2 + ω 2 π − s kω ∴ F (s ) = 2 1+ e ω 2 s +ω Example 28 Find Laplace transform of the following function
sin t , 0 ≤ t < 2π , f (t ) = sin t + cos t t ≥ 2π Solution: Laplace transform of f (t ) can be easily computed once we recognize that f (t ) can be written in the form:
f (t ) = sin t + u 2π (t ) cos (t ) To put the above equation in the form of (21), we have to replace
cos(t ) with cos(t − 2π ) . It is clear that cos(t ) = cos(t − 2π ) . f (t ) = sin t + u 2π (t ) cos (t ) = sin t + u 2π (t ) cos (t − 2π ) ∴ L{ f (t )} = L{sin t} + L{u 2π (t ) cos (t − 2π )} ∴ L{ f (t )} = L{sin t} + e − 2πs L{cos (t )}
∴ L{ f (t )} =
1 s2 + 1
+e
− 2πs
s s2 + 1
=
1 + s e − 2πs s2 + 1
Example 29 Find Laplace transform of the following function:
f (t ) = t 2 u 2 (t )
215
Chapter Six
Solution: Laplace transform of the above function can be
obtained by two different methods as following: First method :We First find L{u2 (t )} . Then, by using (6) we can
{
}
obtain L t 2 u 2 (t ) as following:
Q L{u 2 (t )} =
{
e − 2s s
}
d 2 e − 2 s 4 4 − 2s 2 e = + + ∴ L t u 2 (t ) = (− 1) 3 ds 2 s s2 s s The second method: We can modify t 2 u2 (t ) to take the form in 2
2
(21)as following: Qt 2 = (t − 2 )2 + 4(t − 2 ) + 4
{ } {( ) } 2e − 2 s 4e − 2 s 4e − 2 s ∴ L{t 2u 2 (t )}= + 2 + 3 s s s 4 4 2 ∴ L{t 2u 2 (t )}= 3 + 2 + e − 2 s s s s ∴ L t 2u 2 (t ) = L (t − 2 )2 + 4(t − 2) + 4 u 2 (t )
Example 30 Find the inverse transform of
1 − e− s s2
Solution: From the linearity of the transform (Theorem 7) we have:
f (t ) =
1 − e − s −1 1 −1 e − s −1 L 2 = L 2− L 2
s s s ∴ f (t ) = t − (t − 1)u1 (t ) Hence f (t ) may also be given in the form: t, f (t ) = 1
0 ≤ t < 1, t ≥1
216 Laplace Transforms Example 31 Solve the following initial value problem
y ′′ + 4 y = r (t )
1, Where, r (t ) = 0
y (0) = 1, y ′(0) = 0 π ≤ t < 2π , otherwise
(23) (24)
Solution: Laplace transform of above equation can be easily
computed once we recognize that r (t ) can be written in the form:
r (t ) = uπ (t ) − u2π (t ) Then the Laplace transform of equation (23) is:
(25)
( s 2 + 4) Y ( s ) − sy (0) − y′(0) = L{uπ (t )} − L{u 2π (t )} e −πs e − 2πs ( s + 4)Y ( s ) − s = − s s −πs s e e − 2πs + − (26) Y ( s) = 2 ( s + 4) s ( s 2 + 4) s ( s 2 + 4) To compute y (t ) we must obtain the inverse transform of each term 2
on the right side of equation (26). The inverse transform of the first term of (26) is:
s y1 (t ) = L−1 = cos 2t 2 ( s + 4) The inverse transform of the second term of (26) is:
e − πs
−1 −πs A1 A2 s + A3 = + L L e 2 s s ( s 2 + 4) + ( s 4 ) −1
∴ A1 =
e −πs
( s 2 + 4) s = 0
=
1 4
(27)
Chapter Six
217
u (t ) u (t ) ∴ y2 (t ) = π + π [Ta cos 2 (t − π ) + S a sin 2 (t − π )] 4 2
(28)
[
Where Ra ( s ) = S a + jTa = ( s − α ) 2 + β 2
]HG((ss)) = S1
= s= j2
1 j2
1 1 ∴ S a + jTa = − j , then Ta = − 2 2 Substitute (29) into (28) we get:
(29)
u (t ) uπ (t ) 1 ∴ y 2 (t ) = π + − cos 2 (t − π ) 4 2 2 u (t ) ∴ y2 (t ) = π [1 − cos 2 (t − π )] (30) 4 Similarly, the inverse transform of the third term of (26) is: u 2π (t ) [1 − cos 2 (t − 2π )] 4 Combining (27), (30) and (31) we obtain finally:
∴ y3 (t ) =
y (t ) = y1 (t ) + y 2 (t ) + y3 (t ) u (t ) u (t ) = cos 2t + π [1 − cos 2 (t − π )] − 2π [1 − cos 2 (t − 2π )] 4 4
(31)
(32)
6.5 The Transformation Of Periodic Function
The application of Laplace Transformation to the important case of general periodic function is based upon the following theorem. If a function f (t ) is periodic with period k on [0, ∞) , and piecewise regular on 0 ≤ t ≤ k , then: k
∫ f (t ) e
∴ L{ f (t )} = 0
1− e
− st
− ks
dt s>0
(33)
218 Laplace Transforms Example 32 Find Laplace transform of the rectangular wave shown
in Fig.7.
f (t ) 1
t b
2b Fig.7 Rectangular wave.
The period of the given function is 2b. Hence from (33) we can get the Laplace transform as following: L{ f (t )} =
1 1− e
∴ L{ f (t )} =
2b − 2bs 0
∫
f (t )e − st dt
1
b1 * e − st dt + 2b (− 1)e − st dt ∫ ∫b 1 − e − 2bs 0
− st e ∴ L{ f (t )} = 1 − e − 2bs − s 1
b 0
e − st − −s
2b b
(
)2
1 − 2e − bs + e − 2bs 1 − e − bs ∴ L{ f (t )} = = s 1 − e − bs 1 + e − bs s 1 − e − 2bs 1
∴ L{ f (t )} =
(
)(
(ebs / 2 − e −bs / 2 ) = 1 tanh bs 2 s (1 + e − bs ) s (e bs / 2 + e − bs / 2 ) s 1 − e − bs
=
)
219
Chapter Six
Example 33 Find Laplace transform of the saw tooth wave shown in
Fig.8 f(x) k
k
x k 2k 3k Fig.8 Saw tooth waveform.
Solution: In this case, the period of the function is k, hence;
L{ f (t )} =
1 1− e
∴ { f (t )} =
k − st te dt − ks 0
∫
1 − (1 + ks )e − ks
(
s 2 1 − e − ks
)
=
k
1 1 − e − ks
e − st ( − st − 1 ) 2 s 0
( 1 + ks ) − (1 + ks )e − ks − ks =
(
s 2 1 − e − ks
( 1 + ks )(1 − e − ks ) − ks (1 + ks ) = − ∴ L{ f (t )} = 2 − ks 2
(
)
)
(
k
)
s 1− e s s 1 − e − ks Example 34 What is the Laplace transform of the staircase function
shown in Fig.9?
f(t) 4
Fig.9
3 2 1 k
2k
3k
4k
t
220 Laplace Transforms
f(t) 4
Fig.10
3 2 1 k
2k
3k
4k
t
t+k k
2 1
k
2k
4k 3k The required transform can easily be found by direct calculation. However, it is even simpler to obtain it by considering f (t ) to be the difference between the two functions shown in Fig.10. The transform of the sawtooth function ( f1 (t )) can be obtained as in the previous example as following:
L{ f1 (t )} =
1 1− e
k t − st e dt − ks 0 k
∫
=
1/ k 1 − e − ks
k
e − st ( ) − − st 1 2 s 0
221
Chapter Six
∴ L{ f (t )} = ∴ L{ f (t )} =
1 − (1 + ks )e − ks
(
ks 2 1 − e − ks
)
=
(1 + ks ) − (1 + ks )e − ks − ks
(
ks 2 1 − e − ks
(1 + ks )(1 − e − ks ) − ks = (1 + ks ) −
(
ks 2 1 − e − ks
)
ks 2
(
)
1
s 1 − e − ks
)
t + k The transform of the linear function f 2 (t ) = can be found k (t + k ) 1 1 k as following: L = + k k s2 s Then Laplace transform of the staircase function is: L{ f (t )} =
1 1 k 1 (1 + ks ) k 1 2 + − 2 − = − ks − ks k s s k s s 1− e s 1− e
(
) (
)
6.6 Pulse Functions
In some applications it is necessary to deal with functions with pulse nature, for example, voltages or forces of large magnitude, which act over very short time intervals. Such problems often lead to differential equation in the following form:
ay ′′ + by ′ + cy = r (t )
(34)
Where r (t ) is very high during the short interval 0 < t < τ , and otherwise zero. In particular let us suppose that r (t ) is given by: 1 , r (t ) = τ 0
0 ≤ t <τ, = δ (t ) otherwise
(35)
222 Laplace Transforms ∞
∫ δ (t )dt = 1
Where
(36)
−∞
∞
∴ L{δ (t )} = ∫ e
− st
δ (t )dt = lim
1
τ →0 τ
0
τ
∫e
− st
0
1 − e −τs =1 dt = lim τ →0 τs
(37)
In the similar way unit pulse acting at the instant t = t 0 is defined by δ (t − t0 ) . Where δ (t − t0 ) can be defined as following
δ (t − t0 ) = 0
t ≠ t0
(38)
∞
And
∫ δ (t − t0 ) dt = 1
t ≠ t0
(39)
−∞
The Laplace transform of pulse function δ (t = t0 ) is given by:
L{δ (t − t 0 )} = e − st0
(40)
Example 35 Solve the following initial value problem:
y′′ + 2 y′ + 2 y = δ (t − π )
y′(0) = 0
y (0) = 0,
(41)
Solution:
The Laplace transform of (41) is given by the following equation: ( s 2 + 2 s + 2) Y ( s ) = e −πs
∴ Y (s) =
e −πs ( s 2 + 2 + 2)
= e −πs
1 ( s + 1) 2 + 1
∴ y (t ) = uπ (t ) e − (t −π ) sin(t − π )
223
Chapter Six
6.7 Applications 6.7.1 Electric circuits Example 36 By using Laplace transform, Find the current I (t ) in
Fig.11. Assume I (0) = 1 A and v L (0) = 1 V for R=20 Ω, L=5H, C=0.04 farads and E (t ) = 100 cos 5 t .
vR(t)
From KVL
+
E(t)
VL + VR + VC = E (t )
_
+
I(t)
Differentiate
both
sides
of
the
above
differential equation. Then.
vL(t) Fig.11
1 LI&& + RI& + I = −500 sin 5t c ∴ I&& + 4 I& + 5 I = −100 sin 5t
I ( 0 ) = 1, V L ( 0 ) = L I& ( 0 ) = 1 ∴ I& ( 0 ) = 0 .2 ∴ s 2 I ( s ) − s − 0.2 + 4[ sI ( s ) − 1] + 5 I ( s ) =
− 100 * 5 ( s 2 + 25)
G ( s ) − 500 + ( s + 4.2)( s 2 + 25) ∴ I ( s) = = H (s) ( s 2 + 25) ( s 2 + 4 s + 5) ∴ a1 = j 5, a1 = − j 5
∴α1 = 0, β1 = 5
+
di 1 + RI + ∫ idt = 100 cos 5t dt c
_ _
L
_
+
Solution:
vC(t)
224 Laplace Transforms
[
∴ Ra ( s ) = S a + jTa = ( s − α1 ) 2 + β12 =
]HG((ss)) = [s + 5]HG((ss)) 2
− 500 = 12.5 + j12.5 (−25 + j 20 + 5)
∴ S a = 12.5, Ta = 12.5
1 αt e [Ta cos β1 t + S a sin β1 t ] + L−1{W ( s )}
∴ L−1{I1 ( s)} = =
β1
1 [12.5 cos 5 t + 12.5 sin 5 t ] + L−1{W ( s)} 5
= 2.5[cos 5 t + sin 5 t ] + L−1{W ( s )}
a2 = −2 + j1, a2 = −2 − j1 , ∴α 2 = −2, β 2 = 1
[
∴ Ra ( s ) = S a + jTa = ( s − α 2 ) 2 + β 2 2 =
]HG((ss))
− 500 + ( s + 4.2)( s 2 + 25) ( s 2 + 25)
s = −2 + j1
∴ Ra (a ) = S a + jTa = −15.55 − j 0.24 ∴ S a = −15.5, Ta = −0.24
∴ L−1{I 2 ( s)} =
1 αt e [Ta cos β 2 t + S a sin β 2 t ] + L−1{W ( s)}
β2
= e − 2t [− 0.24 cos t − 15.5 sin t ] ∴ I (t ) = I1 (t ) + I 2 (t ) ∴ I (t ) = 2.5[cos 5 t + sin 5 t ] + e − 2t [− 0.24 cos t − 15.5 sin t ]
225
Chapter Six
Example 37 Find the current in the RLC circuit in Fig.11. Assume
I (0) = 0 and v L (0) = 5 , R=150 Ω, L=50H, C=0.01 farads and E (t ) is shown in Fig.12
Fig.12
E(t)
Solution:
100
From KVL 0.1
VL + VR + VC = E (t ) L
0.2
0.3
di 1 + RI + ∫ idt = E (t ) dt c
Differentiate both sides of the above differential equation. Then.
1 ∴ LI&& + RI& + I = E (t ) c R 1 1 d (E (t )) ∴ I&& + I& + I= L Lc L dt 1 ∴ I&& + 3I& + 2 I = * 1000 50 1 ∴ I&& + 3I& + 2 I = * 1000 50
Q v L = LI&, vL (0 ) = 5 ∴ 50 I&(0 ) = 5
∴I&(0 ) = 0.1 ∴ s 2 I ( s ) − 0.1 + 3sI ( s ) + 2 I (s ) =
[
]
∴ I ( s ) s 2 + 3s + 2 =
20 s
20 20 + 0.1s + 0.1 = s s
t,sec
226 Laplace Transforms
∴ I (s) =
G ( s) 20 + 0.1s 0.1s + 20 = = H ( s ) s ( s 2 + 3s + 2) s ( s + 1)(s + 2 )
Let I ( s ) =
G(s) A B C = + + H ( s ) s ( s + 1) (s + 2 )
∴ A=
0.1s + 20 = 10 ( s + 1)(s + 2 ) s = 0
∴ B=
0.1s + 20 = −19.9 s (s + 2 ) s = −1
∴C=
0.1s + 20 = 9.9 s ( s + 1) s = −2
∴ I (s ) =
10 19.9 9.9 − + s s +1 s + 2
∴ I (t ) = 10 − 19.9e − t + 9.9e − 2t Example 38 Find the current in the RLC circuit in Fig.11. Assume
I (0) = 0 and vL (0) = 5 , R=150 Ω, L=50H, C=0.01 Farads and : 500 sin 5t E (t ) = 500
⇒
0≤t ≤
⇒
t>
2π 5
2π 5
Solution: It is clear that E (t ) can be expressed as following:
E (t ) = 500 sin 5t − u2π / 5 (500 sin 5t − 500) From KVL , L
1 di + RI + ∫ Idt = E (t ) dt c
227
Chapter Six
R 1 1 ∴ I& + I + Idt = E (t ) ∫ L LC L u 1 ∴ I&& + 3I& + 2 I = 2500 cos 5t − 2π (2500 cos 5t ) 50 5 u ∴ I&& + 3I& + 2 I = 50 cos 5t − 2π (50 cos 5t ) 5
∴ s 2 I (s ) − sI (0) − I&(0 ) + 3[sI (s ) − I (0 )] + 2 I (s ) =
I (0) = 0, VL = LI&, VL (0) = 50 * I&(0 ) = 5 ∴ I&(0 ) = 0.1 A / sec . 2π − s 50s 1 − e 5 2 ∴ I (s ) s + 3s + 2 − 0.1 = s 2 + 25
[
]
2
−
2π s 5
∴ I (s ) =
0.1s + 50s + 2.5 − 50s e
∴ I (s ) =
A B C + + 2 (s + 1) (s + 2 ) s + 25
∴A=
(s + 1)(s + 2 )(s 2 + 25)
(
2
0.1s + 50s + 2.5 − 50s e
(s + 2)(s 2 + 25)
−
)
2π s 5
= 4.934 s = −1
50 s s 2 + 25
−
50.s.e
−
2π s 5
s 2 + 25
228 Laplace Transforms
∴B =
0.1s 2 + 50 s + 2.5 − 50 s e
(s + 1)(s
2
+ 25
−
)
2π s 5
= −39.2217 s = −2
2
For s + 25 , α = 0, β = 5 ∴ S a + jTa =
2
0.1s + 50s + 2.5 − 50s e (s + 1)(s + 2 )
−
2π s 5
= 0 + j0 s = j5
∴ I (s ) =
4.934 39.2217 − (s + 1) (s + 2 )
∴ i (t ) = 4.934 e − t − 39.2217 e − 2t
Example 39 Find the current in the RLC circuit in Fig. 1. Assume
I (0) = 0 and v L (0) = 5 , R=150 Ω, L=50H, C=0.01 farads and E (t ) = 100 sin t Solution: Fig.13 shows E (t ) and its derivatives E ′(t ) E(t) 100
t -π
E ′(t )
π
2π
100
t -π
π
Fig.13. E (t ) and its derivatives E ′(t ).
2π
229
Chapter Six
From KVL VL + VR + VC = E (t ) ∴L
di 1 + RI + ∫ idt = E (t ) dt c
Differentiate both sides of the above differential equation we get: 1 ∴ LI&& + RI& + I = E (t ) c R 1 1 d (E (t )) ∴ I&& + I& + I= L Lc L dt 1 ∴ I&& + 3I& + 2 I = * E ′(t ) 50 Q v L = LI&,
v L (0 ) = 5
∴ 50 I&(0 ) = 5
∴I&(0 ) = 0.1 ∴ s 2 I ( s ) − 0.1 + 3sI ( s ) + 2 I (s ) =
1 L{E ′(t )} 50
E ′(t ) is shown in Fig.13. This function is periodical and Laplace transform for it can be obtained from (33) as following: π
∴ L{E ′(t )} =
∫ 100 * cos t e
0
− st
dt
1 − e −πs
πs 100s / tanh 100 s 1 + e 2 ∴ L{E ′(t )} = 2 = −πs 2 s +1 1− e s +1
(
( )(
−πs
)
)
(
)
πs 100s / tanh 1 2 ∴ s 2 I ( s ) − 0.1 + 3sI ( s ) + 2 I (s ) = 2 50 s +1
(
)
230 Laplace Transforms
πs 2 s / tanh 2 ∴ I ( s ) s 2 + 3s + 2 = 0.1 + s2 + 1
(
)
(
)
πs 2 s / tanh 0.1 2 ∴ I (s) = 2 + 2 s + 3s + 2 s + 1 s 2 + 3s + 2
(
) (
)(
(
)
)
πs 2 s / tanh + 0.1 s 2 + 1 2 ∴ I (s) = s 2 + 1 s 2 + 3s + 2
(
∴ I ( s) =
)(
)
A B Cs + D + + 2 (s + 1) (s + 2) s + 1
(
)
(
)
πs 2 s / tanh + 0.1 s 2 + 1 2 ∴A= 2 s + 1 (s + 2 )
(
)
= 1.19 s = −1
(
)
πs 2s / tanh + 0.1 s 2 + 1 2 ∴B = 2 s + 1 (s + 2 )
(
)
( )
= −0.903 s = −2
)
πs 2s / tanh + 0.1 s 2 + 1 2 S a + jTa = s 2 + 3s + 2
(
=0 s = j1
∴ I (s ) =
1.19 0.903 − s +1 s + 2
∴ I (t ) = 1.19e − t − 0.903e − 2t
231
Chapter Six
6.7.2 Newton s law Example 40 The mass m of Fig.2 is suspended from the end of a
vertical spring of constant k [force required to produce unite stretch].
An
external
force
F (t ) acts on the mass as well as a resistive force proportional to the
instantaneous
Assuming
that
y
velocity. is
the
displacement of the mass at time t and that the mass starts from
y m
rest at y=0, (a) set up a differentia equation for the motion and (b) find y at any time t. Solution:
(a) The resistive force is given by − B
dy . The restoring force is dt
given by − ky . Then by Newton’s law, m
d2y dy = − B − ky + F (t ) dt dt d2y dy + B + ky = F (t ) dt dt
(42)
Where, y (0) = 0, , y& (0) = 0
(43)
∴m
(b) Taking the Laplace transform of (42), we obtain m[ S 2Y ( s ) − sy (0) − y ′(0)] + B[ sY ( s ) − y (0)] + kY ( s ) = F ( s )
232 Laplace Transforms
Y (s) = =
G ( s) F ( s) = H ( s ) ms 2 + Bs + k
F ( s) 2 2 s − − B + B − k s − − B − B − k 2m m 2m m 2m 2m
Y ( s) =
G(s) = H (s)
F (s)
(44) 2 B m s + + R 2 m k B2 Where R = − m 4m 2 Then we have three cases which can be summarized as following: Case 1 R > 0
In this case let R = w 2 . We have
Bt − 1 sin wt −1 L = e 2m 2 w B m s + 2m + R Then using the convolution theorem, we find from (44) t
− 1 y (t ) = F (u )e wm 0∫
B (t − u ) 2m
sin w(t − u )du
Case 2 R=0
Bt − 1 −1 In this case L = te 2 m 2 B m s + m 2
233
Chapter Six
And by convolution form in (44) yields: − 1 t y (t ) = F ( u )( t − u ) e wm 0∫
B (t − u ) 2m
sin w(t − u )du
In this case R = −α 2 . We have
Case 3 R<0
Bt − 2 m sinh α t 1 −1 L =e 2 α B 2 m s + 2m − α And by convolution form in (44) yields: t
− 1 y (t ) = F ( u ) e αm 0∫
B (t − u ) 2m
sinh α (t − u )du
6.7.3 Bending of Beams Example 41 A beam of length 10m is simply supported at both ends
as shown in Fig.4. (a) Find the deflection if the beam has constant weight 3000 Newton per meter and 30000 Newton concentrated load at the middle of the beam (b) determine the maximum deflection. x
A
L-x C
y
5-x
30000 N
B x
Deflection, y(x)
Fig.3
234 Laplace Transforms Solution: (a) The total weight of the beam is 3000*10=30000
Newton then the total weight is 30000+30000=60000 Newton, so each end supports weight is
60000 = 30000 Newton. Let x be the 2
distance from the left end A of the beam. To find the bending moment M at x, consider forces to the left of x (١) Force 30000 N at A has moment
− 30000 x (٢) Force due to weight of the beam to left of x has magnitude 3000x and moment 3000 x( x / 2) = 1500 x 2 Then the total bending moment at x is 1500 x 2 − 30000 x . Thus, EIy ′′ = 1500 x 2 − 30000 x
(45)
The boundary condition is y(0)=y(L=10)=0 By using Laplace transform with (45) we get:
[
]
EI s 2Y ( s ) − sy (0) − y ′(0) = ∴ Y ( s) =
1500 * 2 s3
1 3000 30000 y ′(0) + 2 − 4 EI s 5 s s
[
]
−
30000 s2
1 125 x 4 − 5000 x 3 + y ′(0) x EI But y (L = 10 ) = 0 , then substitute in (46), we get:
∴ y (t ) =
y ′(0) = 375000 / (EI )
(46)
235
Chapter Six
Substitute that in (46) we get the general solution of (45) ∴ y (t ) =
[
1 125 x 4 − 5000 x 3 + 375000 x EI
]
Problems
[I] Find the Lapalce transform of the following functions f(t) where c,k, and w are constants ١)
t 2 + 3t + 4
٢)
t 2 e − 2t
٣)
e − t cos 2t
٤)
sin( wt + k )
٥)
cosh 2 (3t )
٦)
3t sin wt + wt 2 cos wt
٧)
t 3 sin wt
٨)
t 2 sinh wt
٩)
cos wt sinh wt + sin et cosh wt
١٠)
(
1 − ct e − e − kt t
)
[II] Solve the following initial value problems by using Laplace transformation ١١)
4 y ′′ − 8 y ′ + 5 y = 0, y (0) = 0, y ′(0) = 1
١٢)
y′′ + y′ = 3 cos 2t , y (0) = 0,
y′(0) = 0
236 Laplace Transforms
١٣)
y ′′ − 4 y ′ = 8t 2 − 4, y (0) = 5, y ′(0) = 10
١٤)
4 y ′′ + 2 y ′ + 2 y = 0, y (0) = 1, y ′(0) = −3
١٥)
4 y′′ − 2 y′ + 2 y = 2 cos2t − 4 sin 2t , y(0) = 0, y′(0) = 0
١٦)
4 y ′′ + 2 y ′ + y = e − 2t , y (0) = 0, y ′(0) = 0
١٧)
y ′′ + 4 y = 2 cos t + sin t , y (0) = 0, y ′(0) = −1
١٨)
4 y′′ − 2 y′ + 5 y = 8 sin t − 4 cos t , y(0) = 1, y′(0) = 3
١٩)
y′′ + 4 y′ + 4 y = f (t ) , y (0 ) = y ′(0 ) = 0 1 f (t ) = 0
0 < t <1 1< t < 2
,
f (t )
is
periodical
function. ٢٠)
y′′ + y = f (t ) , y (0 ) = y ′(0 ) = 0
1 f (t ) = 0
0
π < t < 2π
, f (t ) is periodical function.
٢١)
y′′ + 2 y ′ + 2 y = δ (t − π ), y ( 0) = 1,
٢٢)
y ′′ + 4 y = δ (t − π ) − δ (t − 2π ), y ( 0) = 0,
٢٣)
y′′ + 2 y′ + y = δ (t ) + u 2π (t ), y ( 0) = 0,
y ′( 0 ) = 0 y ′( 0 ) = 0
y ′( 0 ) = 1
237
Chapter Six
٢٤)
y′′ − y = 2δ (t − 1), y ( 0) = 1,
٢٥)
y′′ + ω 2 y = δ (t − π / ω ), y ( 0) = 1,
٢٦)
y′′ + 2 y ′ + 3 y = sin t + δ (t − π ), y ( 0) = 0,
٢٧)
y′′ + y = δ (t − π ) cos t , y ( 0) = 0,
y ′( 0 ) = 0
y ′( 0 ) = 0 y ′( 0 ) = 1
y ′( 0 ) = 1
[III] Find f(t) for the following function F(s) ٢٢)
٢٣)
5s 2 − 15s + 7 ( s + 1)( s − 2) 3 s2 + 2 ( s 2 + 10)( s 2 + 20)
[IV] In each case graph the given function, which is assumed to be zero outside the given interval, and find Laplace transform ٢٤)
t
0
٢٥)
t2
0
٢٦)
et
0 < t <1
٢٧)
k sin wt
0 < t <π /w
٢٨)
k cos wt
0 < t <π /w
٢٩)
k cos wt
0 < t < 2π / w
٣٠)
1 − e −t
0
Find and graph the inverse Laplace transform of the following functions:
238 Laplace Transforms
٣١) ٣٢) ٣٣) ٣٤)
2e − 0.5s s −πs se s2 + 4 e −πs s 2 + 2s + 2 s (1 + e −πs ) s2 +1
٣٥) Find the current in the RLC circuit in Fig.1. Assume
I (0) = 0 and v L (0) = 0 for R=160 C=0.002
farads
E (t ) = 10 cosh 5 t
for
(a)
E (t ) = te − t
, L=20H, (b)
Chapter 7 Fourier Series 7.1 Introduction An important mathematical question raised by Joseph Fourier in 1807, arising from his practical work on heat conduction, is whether an arbitrary function f (x) with period 2 L can be represented in the form of a Fourier series:
f ( x ) = a0 +
∞
∑ an
n =1
nπx nπx cos + bn sin L L
(1)
The above representation is good for all x,−∞ < x < ∞ . If f (x) is not periodic outside the interval (− L < x < L) or if f (x) is not defined beyond this interval, the representation is good only in the restricted interval. A second question is: suppose we can indeed represent f (x) by a Fourier series of the form (1), how do we calculate Fourier Coefficients a0 , an , bn ? Before we start finding the Fourier coefficients we have to state some important notes Theorem 1 A function f ( x ) is said to be even on the interval
(− L, L) if f (− x) = f ( x) . This means that this function is symmetric about the y axis.
239
Chapter Seven
Theorem 2 A function f ( x ) is said to be odd on the interval
(− L, L )
if
f (− x) = − f ( x) . This means that this function is
symmetric about the origin in xy plan. Theorem 3 On computing the Fourier coefficients we found that; • Odd* odd =even • Even* even=even • Odd* even =odd • Even* odd =odd Theorem 4 On computing the Fourier coefficients we found that
L 2 ∫ f ( x)dx L • ∫ f ( x) dx 0 −L 0
if f ( x) is even , So,
if f ( x) is odd
L n πx m πx 1- ∫ cos cos dx = L 3 4L42 44 4 0 − L 14 L
even
for m = n for m ≠ n
L
2-
m πx n πx cos sin dx {= 0 for all m and n L L − L 144424443
∫
odd
L n πx mπx 3- ∫ sin dx * sin L L 0 − L 1444 424444 3 L
even
for m = n for m ≠ n
The above relations known as orthognality relations
240
Fourier Series
7.2 Determination Of Fourier Coefficients
Now we can obtain the an ,and bn in the following way. nπx Multiplying both sides of (1) by sin and integrate with respect L to x from –L to L to get: L
∫
−L
L nπx nπx = f ( x) sin dx a dx + 0 ∫ sin L L − L 1424 3 odd
L nπx nπx a ∑ n ∫ sin L cos L dx + bn n =1 − L 144424443 odd ∞
L
By applying theorem 4 we get:
∫
−L
nπx nπx ∫ sin L sin L dx − L 14442444 3 even L
nπx f ( x) sin dx = bn * L L
1 2L nπx (2) or bn = ∫ f ( x) sin dx L 0 L
1 L nπx ∴ bn = ∫ f ( x) sin dx, L −L L
The same way can be used to obtain a0 , an by multiplying (1) by nπx cos and integrate with respect to x from –L to L yield, L L L nπx nπx
∫
−L
f ( x) cos dx = a0 ∫ cos dx + L L − L 1424 3 even
L nπx nπx a ∑ n ∫ cos L cos L dx + bn n=1 − L 144424443 even ∞
By applying theorem 4, ∴
L
∫
−L
π π n x n x sin cos dx ∫ L L − L 14 4424443 odd L
nπx f ( x) cos dx = a0 * 2 L + a n * L L
241
Chapter Seven
1 L 1 2L f ( x) dx, or a0 = f ( x) dx ∴ a0 = 2L −∫L 2L 0∫ 1 L 1 2L nπx nπx ∴ an = ∫ f ( x) cos dx, or an = ∫ f ( x) cos dx L −L L0 L L
(3)
So that Fourier coefficients can be obtained as shown in (2) and (3). Example 1 Find Fourier series for the following function
Let
0 f ( x) = k
for − L < x < 0 for 0 < x < L
And let f (x) is periodical function with period equal to 2L as shown in Fig.1.
f(x)
k
L
-L Fig.1 Solution:-
1 L k L k a0 = f ( x) dx = dx = ∫ ∫ 2L − L 2L 0 2 1 L k L nπx nπx an = ∫ f ( x) cos dx = ∫ cos dx = 0 L0 L −L L L
242
Fourier Series
1 L k L nπx nπx bn = ∫ f ( x) sin dx = ∫ sin dx L −L L0 L L 0 , n even kL (1 − cos nπ ) = Lπn 2 k , n odd πn
k 2k f ( x) = + 2 Lπ f ( x) =
1 nπx sin n L n =1, 3, 5,....... ∞
∑
k 2 k πx 1 3πx 1 5πx + sin + sin + sin + ........... 2 π L 3 L 5 L
Fig.2 shows that the partial sum of Fourier series do approach f (x) as n increases.
1
1
0.8
0.8
terms=2
0.6
0.4
0.4
0.2
0.2
0
0
-0.2
-3
-2
-1
terms=3
0.6
0
1
2
3
-0.2
-3
-2
-1
0
1
2
Fig.2 the partial sum of Fourier series do approach f (x) as n increases.
3
243
Chapter Seven
1 1
0.8
terms=4
0.6
0.8
0.4
0.4
0.2
0.2
0
-0.2
terms=10
0.6
0
-3
-2
-1
0
1
2
3
-0.2
-3
-2
-1
0
1
2
3
1
0.8
terms=10
0.6
0.4
0.2
0
-0.2 -10
-8
-6
-4
-2
0
2
4
6
8
10
Fig.2 continue Theorem 5 If f ( x ) is an odd function of x , i.e. f ( x ) = − f ( − x )
∴ an = 0
, n = 0,1,2,3,4,.......
2L nπx and bn = ∫ f ( x) sin dx L0 L
(4) (5)
Theorem 6 If f ( x ) is an even function of x , i.e. f ( x ) = f ( − x ) then,
∴ bn = 0,
n = 1, 2, 3, 4,.......
2L nπx and an = ∫ f ( x) cos dx, L0 L
1L a0 = ∫ f ( x) dx L0
(6) (7)
244
Fourier Series
Example 2 Find Fourier transform of the following function:
f ( x) = x, − L < x < L. Let f (x) is periodic function with period equal to 2L as shown in Fig.3. f(x)
L x L
-L
Fig.3 Solution:- Since f (x) is an odd periodic function, its Fourier coefficients are: an = 0, n = 0,1,2,............................
2L 2L nπx bn = ∫ f ( x) sin (−1) n +1, dx = L0 nπ L
n = 1,2,3,.............
nπ x (−1) n +1 ∴ f ( x) = sin ∑ π n =1 n L 2L
∞
Example 3 Expand f ( x) = x 2 , 0 < x < 2π . in a Fourier series if the
period is 2π as shown in Fig.4. f(x)
x
2π
−2π
Fig.4
245
Chapter Seven
This function is not odd or even. So, the Fourier
Solution:-
coefficients can be determined as follows,
1 L 1 2π 2 4π 2 ∴ a0 = f ( x) dx = x dx = 2π 0∫ 3 2 L −∫L 1 2L 1 2π 2 nπx nπx ∴ an = ∫ f ( x) cos dx = ∫ x cos dx π 0 L 0 L π 2π
− cos nx − sin nx 1 sin nx 4 ∴ an = x 2 − 2x + = 2 π n n2 n3 0 n2
1 L 1 2π 2 nπx nπx ∴ bn = ∫ f ( x) sin dx = ∫ x sin dx π π L −L L 0 2π
1 − cos nx − sin nx cos nx − 4π ∴ bn = x 2 − 2x + 2 = n n π n2 n3 0 4π 2 ∞ 4 4π + ∑ 2 cos nx − ∴ f ( x) = x = sin nx 3 n n =1 n 2
Example 4 Expand f ( x) = sin x, 0 < x < π . Find Fourier series if
the period is 2 π . f(x)
x -π
π
Fig.5
2π
246
Fourier Series
Solution:- It is clear that this function is even function. So,
bn = 0 for n = 1, 2, 3,..............
1L 1π 2 a0 = ∫ f ( x) dx = ∫ sin x dx = π0 π L0
2L 2π nπx nπx an = ∫ f ( x) cos dx dx = ∫ sin x cos π π L L0 0 =
1π
2π
∫ sin x cos (n x ) dx = π ∫ sin ( x + nx) + cos (x − n x ) dx 0 0
π
π
1 cos(n + 1) x cos(n − 1) x + = − π n +1 n −1 0 π
1 cos(n + 1) x cos(n − 1) x + = − π n +1 n −1 0 =
1 1 − cos(n + 1)π cos(n − 1)π − 1 + π n +1 n −1
=
1 1 + cos(nπ ) 1 + cos(nπ ) + − π n +1 n −1
∴ an =
− 2(1 + cos(nπ ))
(
)
π n2 − 1
, if n ≠ 1
∴ a1 = 0 ∴ f ( x) = ∴ f ( x) =
2
2
∞
∑ π π n=2 2
π
−
−
(1 + cos(nπ ))
(n 2 − 1)
cos nx
4 cos 2 x cos 4 x cos 6 x + 2 + 2 + ............ 2 π 2 −1 4 −1 6 −1
247
Chapter Seven
Example 5 Expand the function f ( x) = x, 0 < x < 2 if f (x) is even
function (Fig.6) with period 4. f(x) 2
Fig.6 x
-2
2
Solution:-Because of the function f (x) is even function then
bn = 0 . Also, 1L 12 a0 = ∫ f ( x) dx = ∫ x dx = 1 20 L0 2L 22 nπx nπx an = ∫ f ( x) cos dx dx = ∫ x cos 20 L0 2 L 2
2 n πx n πx − 4 = x sin − 2 2 cos L 0 2 n π nπ
=
4 2 2
n π
(cos nπ − 1)
f ( x) = 1 + ∑ f ( x) = 1 −
4 n 2π
(cos nπ − 1) cos 2
nπx 2
8 3πx 1 πx 1 5πx + + + cos cos cos ...... 2 32 2 2 52 π2
248
Fourier Series
7.3 Determination Of Fourier Coefficients Without Integration
In this section we will represent Fourier coefficients of any periodic function by polynomial can be obtained in terms of jumps of the function and its derivatives. By a jump J of a function f ( x ) at a point xs we mean the difference between the right hand side and left hand side of f (x) at xs . Is: J s = f ( xs + 0) − f ( xs − 0) (8) Where f ( xs + 0) and f ( xs − 0) is the value of the function f (x)
directly after and before the jump at x s . Then f may have jumps at
x0 , x1 , x2 , ...... x s ,.... xm and the same is true for derivatives f ′, f ′′,............ we choose the following notation:-
J s = Jumps for f at xs J s′ = Jumps for f ′ at xs (s = 1,2,3,........m) (9) J s′′ = Jumps for f ′′ at xs So that in case of function f (x) having period 2 L , the Fourier transform can be obtained from the following function:
an =
n πx s 1 m L − ∑ J s sin − nπ s =1 L nπ
m
∑ J s′ cos
s =1
nπ xs + L
nπ x s L nπ x s L ′ ′ ′ ′ ′ − − + + sin cos ... J J + ∑ s ∑ s L L nπ s =1 nπ s =1 m m nπ x s nπ x s 1 L ∑ J s cos bn = J s′ sin − − ∑ nπ s =1 L nπ s =1 L 2 m
3 m
nπ x s L nπ x s L ′ ′ ′ ′ ′ cos sin ..... J J − − + + + ∑ s ∑ s L L nπ s =1 nπ s =1 2 m
3 m
(10)
(11)
249
Chapter Seven
For n = 1, 2, 3, 4,...., ∞ and n = 0 can be obtained from equation (3) as usual. In case of the function f (x) is periodical function having period
2π , the Fourier transform can be obtained from the following formula:
an =
1 nπ
m 1 m − ∑ J s sin n xs − ∑ J s′ cos n xs + n s =1 s =1
1 n2
1 bn = nπ
m
1
s =1
n3
∑ J s′′ sin n xs +
m
s =1
∑ J s′′′ cos n xs − − + +.....
m 1 m ∑ J s cos n xs − ∑ J s′ sin n xs − n s =1 s =1
1 1 ′ ′ ′ ′ ′ + + − − + + J cos n x J sin n x ..... s s 2 ∑ s 3 ∑ s n s =1 n s =1 m
m
(12)
(13)
For n = 1, 2, 3, 4,...., ∞ and n=0 can be obtained from equation (3). Example 6 Find Fourier transform without integration for the
following function (Fig.7) if the period is 2π . f(x)
π
J1 -π
J2
x
π
Fig.7
250
Fourier Series
Solution:-
f
a0 =
Jumps at x1 = 0
Jumps at x2 = π
J1 = −π
J 2 = −π
1π
1π
π
f ( x) dx = ∫ x dx = 2 π 0∫ π0
As we see this function neither even nor odd. Then an can be obtained from equation (12) and the above table as follows:
1 an = nπ
an =
m − ∑ J s sin n x s = 1 (− J1 sin n x1 − J 2 sin n x2 ) nπ s =1
1 (sin n 0 + sin n π ) = 0 for all n. n
bn can be obtained from equation (13) and the above table as follows:
1 bn = nπ
∴ bn =
1 m ∑ J s cos n x s = (J cos n x1 + J 2 cos n x2 ) nπ 1 s =1
1 (cos n 0 + cos n π ) = 2 , n n
n = 2, 4, 6, 8,.....
Otherwise bn = 0 .
1 1 1 ∴ f ( x) = 2 sin 2 x + sin 4 x + sin 6 x + ...... 4 6 2
251
Chapter Seven
Example 7 Find Fourier transform without integration for the even
waveform f ( x) = x,
0 ≤ x < π as shown in Fig.8.
Solution:-
f
f′
a0 =
1π
Jumps at x1 = 0
Jumps at x2 = π
J1 = 0
J2 = 0
J1′ = 2
J 2′ = −2
1π
π
f ( x) dx = ∫ x dx = 2 π 0∫ π0
As we see it is even function then bn = 0 , for n = 1, 2, 3
an can be obtained from equation (12) as follows : f(x)
π
x -π
π f(x) 1
π
-π
-1
Fig.8
252
Fourier Series
m m − ∑ J s sin n x s − 1 ∑ J s′ cos n x s n s =1 s =1 1 1 an = − J1 sin n x1 − J 2 sin n x2 − ( J1′ cos n x1 + J 2′ cos n x2 ) nπ n Substitute in this equation from the above table we get: 1 1 −4 an = n = 1, 3,5,7,... − (2 cos n 0 − 2 cos n π ) = 2 , nπ n n π π 4 1 1 1 f ( x) = − cos x + cos 3x + cos 5 x + cos 7 x + ...... 2 π 9 25 49
1 an = nπ
Example 8 Find Fourier transform without integration for the odd
0 ≤ x < π as shown in Fig.9.
waveform f ( x) = x, Solution:-
f
f′
Jumps at x1 = 0
Jumps at x2 = π
J1 = 0
J 2 = −2π
J1′ = 0
J 2′ = 0
As we see it is odd function then an == 0 , for n = 0,1, 2, 3 . bn can be obtained by substituting
f(x)
from the above table in
π
equation (13) as following :
-π
1 m n +1 2 ∑ J s cos n xs = ( ) ( ) n − π π = − 2 cos 1 nπ n s =1 1 1 1 f ( x) = 2 sin x − sin 2 x + sin 3x − sin 4 x + ...... 2 3 4
bn =
1 nπ
Fig.9
π
x
253
Chapter Seven
Example 9 Find Fourier transform of the waveform in Fig.10: f(x)
Fig.10
Io π 6
− Io
Jumps
f
7π π 6
11π 6 2π
5π 6
at Jumps
at Jumps
at Jumps
at
x1 = π / 6
x2 = 5π / 6
x3 = 7π / 6
x4 = 11π / 6
J1 = I o
J 2 = −Io
J3 = −Io
J 4 = Io
As we see it is odd function then an=0, for n = 0,1, 2, 3 ……
bn can be obtained by substituting from the above table in equation (13) as follows : bn =
and
I m nπ 5nπ 7 nπ 11nπ ∑ J s cos n x s = o cos − cos − cos + cos nπ 6 6 6 6 s =1 2 3I o bn = − , n = 5, 7, 17, 19 nπ
1 nπ
Otherwise bn = 0 ∴ f ( x) =
2 3 Io 1 1 1 1 sin x − sin 5 x − sin 7 x + sin 11x + sin 13 x − − + +.... π 11 13 5 7
254
Fourier Series
Example 10 Find Fourier transform of the waveform in Fig.11. 1
f (x)
cos x
Fig.11 −π
x f (x )
−π / 2
π /2
π
π /6
5π / 6
7π / 6
11π / 6
J1 = I o
J 2 = −Io
J3 = −Io
J 4 = Io
3π / 2
2π
ao = a n = 0
bn = =
1 m ∑ J s cos nxs nπ s =1 Io 5nπ 7 nπ 11nπ nπ .... − cos − cos + cos cos 6 6 6 6 nπ
∴ bn =
2 3 I o , n = 1, 11, 13, 23, 25, nπ
∴ bn = −
2 3 I o , n = 5, 7, 17, 19,.... nπ
∴ bn = 0 n = else where(2, 3, 4, 6, 8, .... ∴ f (x ) =
2 3I o 1 1 1 1 sin x − sin 5 x − sin 7 x + sin 11x + sin 13x.... 5 7 11 13 π
255
Chapter Seven
7.4 Complex Fourier Series
In this section we will discuss other form of Fourier series, which is complex Fourier series. As we know, the Fourier coefficients are:
1 L a0 = f ( x) dx 2 L −∫L 1 L nπx an = ∫ f ( x) cos dx. L −L L
(14)
1 L nπx bn = ∫ f ( x) sin dx L −L L If we change n to -n in the above definition for an and bn , we will find: a− n = an , and b− n = −bn Therefore we can rewrite the following
f ( x) =
∞
∑ an
n=0 0
nπx nπx cos + bn sin L L
− nπx − nπx a− n cos − bn sin L L n = −∞ 1 nπx nπx ∴ f ( x) = ∑ an cos + bn sin L L n = −∞ ∴ f ( x) =
(15)
∑
(16)
Therefore we can write the sum of (15) and (16) as follows: ∞ nπx nπx ∴ 2 f ( x) = ∑ an cos + bn sin L L n = −∞
∴ f ( x) =
1 ∞ nπx nπx an cos + bn sin ∑ L L 2 n = −∞
(17)
256
Fourier Series
Where now, for n = 0, ± 1, ± 2, , ± 3........., ± ∞ Form (17) can be further written more compactly using the complex notations:
∴ f ( x) = ∑ cn e n = −∞ ∞
−i
nπ x L
, − L < x < L
(18)
nπ x
i 1 L L dx Where cn = f ( x ) e ∫ 2L − L The form (18) is equivalent to normal Fourier series form but this
form is Sometimes preferred it is coefficients are easier to remember.
cn =
1 (an + jbn ) 2
Example 11 Use the complex method to find Fourier transform of
the following function:
for − L < x < 0 0 Let f ( x) = for 0 < x < L k Let f (x) is periodical function with period equal to 2L as shown in Fig.12. f(x) k
-L
L
Fig.12
257
Chapter Seven
k L k 1 L Solution:- a0 = = = f x dx dx ( ) 2 L −∫L 2 L 0∫ 2 1 L cn = f ( x) e 2 L −∫L
(
k cn = j 1− e 2nπ cn = j
j
nπ x L dx =
jnπ
)
kL e L 0∫
j
nπ x L
dx =
nπ x L j L
ke L j nπ L
x =0
k 1 − (cos nπ + j sin n3 π ) =j 1 2 2nπ =0, for all n
2k , Otherwise cn = 0 nπ
Substitute in equation (18) we get: nπ x j − L ∴ f ( x ) = ∑ cn e n = −∞ πx 3π x 5π x 2k − j L 1 − j L 1 − j L = j e + e + e + π 3 5 nπ x ∞ −j L = j 2k cos π x − j sin π x ∴ f ( x ) = ∑ cn e L L π n = −∞ 1 3π x 3π x 1 5π x 5π x + cos − j sin − j sin + cos + ... L L 5 L 3 L ∞
∴ f ( x) =
k 2 k πx 1 3πx 1 5πx + sin + sin + sin + +....... 2 π L 3 L 5 L
Compare the above result with the result of Example 1 in this chapter.
258
Fourier Series
Problems
Find Fourier series of the function f(x) which is assumed to have the period of 2 by two different techniques
٣)
−π / 2 < x < 0 − 1 if f ( x) = 1 if 0< x <π /2 0 if π /2< x <π −π < x < 0 if 0 f ( x) = − 1 if 0< x <π /2 1 if π /2< x <π f ( x) = x −π < x < π
٤)
f ( x) = x 2
٥) ٦)
f ( x) = x 3 f ( x) = x
١)
٢)
٧) ٨) ٩) ١٠) ١١) ١٢) ١٣)
−π < x < π
−π < x < π −π < x < π if −π / 2 < x < 0 π + x f ( x) = if 0< x <π π −x if −π / 2 < x < π / 2 x f ( x) = π / 2 < x < 3π / 2 π − x if x 2 if −π / 2 < x < π / 2 f ( x) = 2 π / 4 if π / 2 < x < 3π / 2 0 < x <π sin x if f ( x) = if π < x < 2π 0 f ( x) = sin x −π < x < π 0< x <π cos x if f ( x) = if π < x < 2π 0 f ( x) = cos x −π < x < π
Chapter 8 Least Square Technique 8.1 Introduction Suppose you are in a science class and you receive these instructions: Find the temperature of the water at the times 1, 2, 3, 4, and 5 seconds after you have applied heat to the container. Conduct your experiment carefully. Graph each data point with time on the x-axis and temperature on the y-axis. Your data should follow a straight line. Find the equation of this line. The data from the experiment looks like this when charted and graphed:
260
Least Square Technique
Notice that our data points don't fall exactly on a straight line as they were supposed to, so how are we going to find the slope and intercept of the line? This is a common problem with experimental sciences because the data points that we measure seldom fall on a straight line. Therefore, scientists try to find an approximation. In this case, they would try to find the line that best fits the data in some sense. The first problem is to define "best fit." It is convenient to define an error as a distance from the actual value of y for x (the value that was measured in the experiment) to the predicted value of y for x. Therefore, it seems reasonable that the "best fit" line would somehow minimize the errors, but how? You could minimize the sum of the absolute values of the errors; this is called the l1 fit. It would also be reasonable to find the biggest error for each line and choose the line that minimizes this quantity; this is called the l∞ fit. However, the fit that is used most often is the l2 fit which is called the least squares fit. This method is called the least squares fit because it finds the line that minimizes the sum of the squares of the errors. Gauss developed this method to solve a problem when he was a young man (15 years old!!) to help his friend solve a chemistry problem. This is the fit that is most often used because it is the only one that can be found by solving a system of linear equations.
Chapter Eight
261
8.2 Least Squares Fit (Second Technique) You have just read a lot of new information, so let's illustrate the concepts with our example. We have the graph of the data above. Now we need to guess which line best fits our data. If we assume that the first two points are correct and choose the line that goes through them, we get the line y = 1 + x . If we substitute our points into this equation, we get the following chart. The points and line are graphed below.
Therefore, the sum of the squares of the errors is 27. Do you think that we can do better than this? If we choose the line that goes through the points when x = 3 and 4, we get the line y = 4 + x . Will we get a better fit? Let's look at it.
262
Least Square Technique
The sum of the squares of the error is 18. That is a better fit, but can we do even better? Let's try the line that is half way between these two lines. The equation would be y = 2.5 + x . It looks like this:
263
Chapter Eight
The sum of the squares of the error is 11.25 with this line, so this is the best line yet. Can we do better? It doesn't seem very scientific or efficient to keep guessing at which line would give the best fit. Surely there is a methodical way to determine the best fit line. Let's think about what we want. A line in slope-intercept form looks like c0 + c1 x = y where c0 is the y -intercept and c1 is the slope. We want to find c0 and c1 such that c0 + c1 xi = yi is true for all our data points:
c0 + 1c1 = 2 c0 + 2c1 = 3
c0 + 3c1 = 7 c0 + 4c1 = 8
c0 + 5c1 = 9 We know that there may not exist c0 and c1 that fit all these equations, so we try to find the best fit. We can write these equations in the form X c = y (these are just new letters for our familiar equation Ax = b ) where
1 1 X = 1 1 1
1 2 3 2 c0 3 , c = , and, y = 7 c1 4 8 9 5
264
Least Square Technique
In general, we cannot solve this system because the system is usually inconsistent because it is overdetermined. In other words, we have more equations than unknowns (the unknowns are the two variables, c0 and c1 , for which we are trying to solve). There is a system of equations called the normal equations that can be used to find least squares solution to systems with more equations than unknowns. Theorem 8.1 Let X be an m by n matrix such that X T X is invertible,
then
the
solution
to
the
normal
equations,
X T X c = X T y , is the least squares approximation to c in X c = y . Remark 25 It is important to remember that the solution to the normal equations is only an approximation to c for X c = y . It is not equal to c because X c = y is inconsistent, so it has no solution. In other words, there does not exist a vector, c, that makes X c = y a true statement. Therefore, we use the normal equations to approximate c. Remark 26 For now, you don't need to check to see if X T X is invertible because most of the systems that we encounter will meet this requirement. However, if you cannot find a solution to the normal equations, you should check to see if X T X is invertible. The normal equations will give us the "best fit" line (or curve) every time according to the way we defined "best fit". Let's try applying
265
Chapter Eight
the normal equations to our system. First, we multiply so that we have a system that we can solve. X T X c = X T y
1 1 1 1 1 1 1 1 2 3 4 5 1 1 1
1 2 3 4 5
2 3 c0 1 1 1 1 1 c = 1 2 3 4 5 7 1 8 9
5 15 c0 29 15 55 c = 106 1 Using Cramer s rule or Gauss elimination can easily solve the above equation. ∴ c0 = 0.1 ,
and, c1 = 1.9
When we graph and chart the line y = 0.1 + 1.9 x , we get:
The sum of the squares of the error is 2.7. This is a great improvement over our guesses and we know that we cannot do any
266
Least Square Technique
better. In general, if we have n data points, we solve X T X c = X T y with
X =
1 1
M 1 1
x1 x2 M xn −1 xn
c0 c= c1
y1 y 2 y= M yn −1 yn
What if we are told that our data is not supposed to fit a straight line, but instead falls in the shape of a parabola? Consider the following data from another experiment:
We can find the curve that best fits our data in a similar manner. The general equation for a parabola is c0 + c1 x + c2 x 2 = y . Therefore,
267
Chapter Eight
we want to find the values of the coefficients, c0 , c1 , and c3 , so that the curve we find best fits these equations:
c0 + 1c1 + 1c2 = 3 c0 + 0c1 + 0c2 = 1
c0 + 1c1 + 1c2 = −1 c0 + 2c1 + 4c2 = 1
c0 + 3c1 + 6c2 = 3 Let us use the normal equations with
1 1 X =1 1 1
−1 0 1 2 3
1 0 1 , 4 9
c0 c = c1 , c2
3 1 y = − 1 1 3
X T Xc = X T y 1 − 1 1
1 0 0
5 5 5 15 15 35
1 1 1 1 1 1 2 3 1 1 4 9 1 1 15 c0 7 35 c1 = 7 99 c2 33
−1 0 1 2 3
1 0 c0 1 c1 4 c2 9
268
Least Square Technique
Now we can augment the matrix and solve using Gaussian elimination. We get the following results:
19 c0 35 c = − 1 5 1 7 c2 6 7 These coefficients indicate that the curve we want is
y=
5 6 19 − 1 + x2 35 7 7
Let's graph this curve and fill in our chart:
269
Chapter Eight
Expected y
error
(error) 2
4 35
−4 35
16 1225
1
19 35
16 35
256 1225
1
-1
− 11 35
− 24 35
576 1225
2
1
19 35
16 35
256 1225
3
3
4 35
−4 35
16 1225
x
y
-1
3
0
3
3
We find that the sum of the squared errors is
32 . Using our 35
definition of least squares "best fit," you will not be able to find a parabola that fits the data better than this one. In general, to find the parabola that best fits the data, you use the normal equations
X T X c = X T y with
X =
1
x1
1 M
x2 M
1 1
xn −1 xn
x12 x22 M xn2−1 xn2
c0 c = c1 c2
y1 y 2 y= M y − 1 n yn
Notice that the normal equations used to find the best fit line and the best fit parabola have the same form. Do you think that we could
270
Least Square Technique
expand this to higher degree polynomials? Yes, we can. In general, we use the normal equations X T X c = X T y with
X =
1 1
x1 x2
M 1 1
M xn −1 xn1
x12 x22
K K
M
O K K
xn2−1 xn2
c0 x1m y1 c m y x2 1 2 , y = M M , c = M c xnm−1 1 − m y − 1 n cm yn xnm
where m represents the degree of the polynomial curve that you wish to fit and n represents the number of data points. The least squares
"best
fit"
curve
for
these
equations
is
c0 + c1 x + c2 x 2 + ... + cm −1 x m −1 + cm x m . Remember that the degree is the highest power of the variable in your equation. A line is a first degree polynomial and a parabola is a second degree polynomial. If we can find the best fit curve for any degree polynomial, why don't we always use a higher degree polynomial and fit the data better? After all, if we have n data points and fit them to a polynomial of degree n − 1, we will have a perfect fit every time because our systems would not be inconsistent. However, our goal is not just to find a curve that fits the data closely. Usually, we want the curve to predict what would happen between our data points. If we choose a curve that exactly fits all our data points, we are incorporating the error in our measurements into our model unless the model fits the data exactly (which occurs only rarely).
271
Chapter Eight
Unfortunately, there is no set rule for deciding what degree polynomial should be used to fit the data. However, first and second-degree polynomials provide the simplest models and should fit most of your data until you start modeling more complicated systems. 8.3 Least Squares Fit (Second Technique) There is another simple method to get the least square approximation to fit polynomial of degree n to a set of points
( xi , yi ) . The simplest example of a least square approximation is fitting line to a set of points. The mathematical expression for straight line is as following:
y = a0 + xa1 + e
(1)
Where a0 and a1 are coefficients representing the straight line and e is the error or residual between the estimated straight line and the given data points which can be represented by the following equation
e = y − a0 − xa1
(2)
One criteria is to minimize the sum of residuals s to zero for all available data as shown in the following equation:-
s=
n
n
i =1
i =1
∑ ei = ∑ yi − a0 − a1xi
(3)
272
Least Square Technique
Where n is the total number of data points. However, this criterion suffers from some problems explained in reference [1]♠. A strategy that overcomes the shortcoming of the above criteria is to minimize the sum of squares of the residuals between actual data points and straight-line model. Then,
Sr =
n
∑
i =1
=
ei2
=
n
∑ ( yi, data − yi, mod el ) 2
i =1
n
(4)
∑ ( yi − a0 − a1xi )2
i =1
This strategy has many advantages including the fact that it gives a unique line for any given data. 8.3.1 Lest Square Fit Of Straight Line Methodology As we explained before we have get the minimum sum of squares of results between data and points this can be obtained by differentiate the sum of residuals with straight line coefficient and equate it with zero. n ∂S r ∴ = −2 ∑ ( yi − a0 − a1 xi ) = 0 ∂a0 i =1
(5)
n ∂S r ∴ = −2 ∑ [( yi − a0 − a1 xi )xi ] = 0 ∂a1 i =1
(6)
♠
Steven C. Chapman and Raymond P. Canale Numerical Method of Engineering , book, McGraw-Hill international edition, third edition,1998
273
Chapter Eight
By simplifying the above equations we can obtain the following: n
n
n
i =1
i =1
i =1
∑ yi − ∑ a0 − ∑ a1xi = 0 n
n
n
i =1
i =1
i =1
∑ yi xi − ∑ a0 xi − ∑ a1xi2 = 0
(7)
(8)
The above two equations (7) and (8) can be further simplified to be as following: n n na0 + ∑ xi a1 = ∑ yi i =1 i =1
(9)
n n n ∑ xi a0 + ∑ xi2 a1 = ∑ xi yi i =1 i =1 i =1
(10)
n
n
n
i =1
i =1
i =1
n ∑ xi yi − ∑ xi * ∑ yi a1 =
n ∑ xi2 − ∑ xi i =1 i =1 n
n
2
a0 = y − a1 x
(11)
(12)
Where y and x are the mean of yi and xi respectively. The least squares procedure can be readily extended to fit the data to a higher order polynomial. For example, suppose that we fit a second order polynomial or quadratic:
y = a0 + a1x + a2 x 2 + e For this case the sum of squares of the residuals is:
(13)
274
Least Square Technique
Sr =
∑ (yi − a0 − a1xi − a2 xi2 ) n
2
(14)
i =1
8.3.2 Lest Square Fit Of Higher Order Curves Following the procedure of the previous section, we take the derivative of the above equation with respect to each of the unknown coefficients of the polynomial as in
(
)
[(
)]
(16)
[(
) ]
(17)
n ∂S r = −2 ∑ yi − a0 − a1 xi − a2 xi 2 = 0 ∂a0 i =1
(15)
n ∂S r = −2 ∑ yi − a0 − a0 xi − a2 xi 2 xi = 0 ∂a1 i =1 n ∂S r = −2 ∑ yi − a0 − a0 xi − a2 xi 2 xi2 = 0 ∂a2 i =1
These equations can be set equal to zero rearranged to develop the following set of normal equations:
na0 n
∑ (xi ) a0
i =1 n
∑ (xi 2 )a0
i =1
( ) i =1 i =1 i =1 n n n + ∑ (xi 2 )a1 + ∑ (xi 3 )a2 = ∑ xi yi i =1 i =1 i =1 n n n 3 4 + ∑ (xi )a1 + ∑ (xi )a2 = ∑ xi 2 yi n
+ ∑ ( xi ) a1
i =1
n
2
n
+ ∑ x a2 = ∑ yi
i =1
i =1
(18)
275
Chapter Eight
Note that the above three equation are linear and have three unknowns a0 , a1 , a2 . The coefficients of the unknowns can be calculated directly from the observed data. Example 1 For the following data determine a second order equation fits this data
xi
1
2
3
4
5
6
7
yi
2
3
5
6
8
7
9
Solution:-
xi
yi
xi * yi
xi2
xi3
xi4
xi2 yi
1
2
2
1
1
1
2
2
3
6
4
8
16
12
3
5
15
9
27
81
45
4
6
24
16
64
256
96
5
8
40
25
125
625
200
6
7
42
36
216
1296
252
7
9
63
49
343
2401
441
192
140
784
4676
1048
∑ Then,
28 40
276
Least Square Technique
7 a0 28a0
28a1 140a1
140a2 = 40 140a2 = 192
784a1 467a2 = 1048
140a0
From the above equation we can get:
a0 = 3.6 *10 −12 a1 = 1.90476 a2 = −0.095238 Then the required polynomial is:
F ( x) = 3.6 * 10 −12 + 1.90476 x − 0.095238 x 2 8.4 Fourier Approximation By Using Least Square Technique The above procedure can be used to determine the Fourier coefficients for Fourier series. Assume the required Fourier series will be in the following form
y = a0 + a1 cos(ω t ) + b1 sin(ω t ) + e
(19)
Where we will take only the first three terms for simplicity and later we can extend this technique for n terms. From least square technique we can say the square of errors is n
S r = ∑ ( yi − a0 − a1 cos(ω t ) − b1 sin(ω t ) )2
(20)
i =1
We have to minimize the square of errors with respect to a0 , a1 , b1 respectively. Then we have the following conditions:
277
Chapter Eight n
+ ∑ cos(ω t ) a1
na0
i =1 n
n
∑ sin(ω t ) a0
i =1
n
n
i =1
i =1 n
+ ∑ cos 2 (ω t ) a1
+ ∑ cos(ω t )b1
= ∑ cos(ω t ) yi
+ ∑ cos(ω t ) + sin(ω t ) a1
+ ∑ sin 2 (ω t )b1
= ∑ sin(ω t ) y i
∑ cos(ω t ) a0
i =1 n
n
+ ∑ sin(ω t )b1 = ∑ yi
n
i =1
i =1
i =1 n
i =1
i =1 n
i =1
By solving the above equations we can get a0 , a1, b1 . Then, n
yi i =1 n
a0 = ∑
(21)
2 n a1 = ∑ yi * cos(ω t ) n i =1
(22)
2 n b1 = ∑ yi * sin(ω t ) n i =1
(23)
The above equations can be extended to fit n terms of Fourier series as following: If the required Fourier series be in the following form:
y = a0 + a1 cos(ω t ) + b1 sin(ω t ) + a2 cos(2ω t ) + b2 sin(2ω t ).... .............................................. + am cos(mω t ) + bm sin(mω t ) + e
(24)
Where n must be > 2m + 1, or m ≤ (n − 1) / 2 . In this case: n
yi i =1 n
a0 = ∑
(25)
2 n a j = ∑ yi * cos( jω t ) n i =1
(26)
2 n b j = ∑ yi * sin( jω t ) n i =1
(27)
278
Least Square Technique
Example 2 Use the method of least square to find the Fourier transform for the function shown in the following figure (Find DC and fundamental ( a0 , a1 , b1 ) and then draw flowchart and basic program to determine and print m harmonics (divide the function to 20 intervals in your calculations and 200 intervals in the computer program per period).
Solution: From table shown in next page ; n y ∴ a0 = ∑ i = 0 i =1 n
2 n ∴ a1 = ∑ yi * cos(ω t ) = 16.17 n i =1 2 n ∑ yi * sin(ω t ) = 102.2 n i =1 If we use the exact Fourier transform will get: 0.18 100 0.08 1 2L a0 = f ( x ) dx = dt − ∫ dt = 0 .02 0.∫02 2 L 0∫ 0.12
∴ b1 =
2L 200 0.08 πt πt a1 = ∫ f (t ) cos dt = cos dt = 0 ∫ L0 .01 0.02 0.01 L
279
Chapter Eight
2L 200 0.08 π t πx b`1 = ∫ f ( x) sin dx = sin dt = 127.324 L0 .01 0.∫02 0.01 L The following computer program can be used to get m terms of Fourier series, where we divided the number of points to n=200 points then m = (200 − 1) / 2 = 99 . Then we can get the following coefficients.
xi 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 sum
0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 0.011 0.012 0.013 0.014 0.015 0.016 0.017 0.018 0.019 0.19
y
i
0 0 100 100 100 100 100 100 0 0 0 0 -100 -100 -100 -100 -100 -100 0 0 0
y
i
cos
x
i
0 0 80.90161308 58.77834693 30.90141996 -0.000367321 -30.90211865 -58.77894126 0 0 0 0 80.90118126 58.77775259 30.90072128 -0.001101962 -30.90281733 -58.7795356 0 0 161.796153
y
i
sin x i
0 0 58.7786441 80.90182898 95.10574244 100 95.10551542 80.90139717 0 0 0 0 58.77923843 80.90226079 95.10596945 99.99999999 95.1052884 80.90096535 0 0 1021.586851
280
Least Square Technique
1 DIM A(110), B(110), T(210),Y(210) 2 T(0)=0:FOR I=1 TO 19:T(I)=T(I-1)+.02/200:Y(I)=0:NEXT I 3 FOR I=20 TO 80:T(I)=T(I-1)+.02/200:Y(I)=100:NEXT I 4 FOR I=81 TO 119:T(I)=T(I-1)+.02/200:Y(I)=0:NEXT I 5 FOR I=120 TO 180:T(I)=T(I-1)+.02/200:Y(I)=-100: NEXT I 6 FOR I=181 TO 200 T(I)=T(I-1)+.02/200:Y(I)=0:NEXT I 10 INPUT "how many data points you have",N:N=N-1 20 INPUT " what is the highest harmonics you want ",M 25 IF M > (N-1)/2 THENM=(N-1)/2 35 UMY=0:WO=2*3.14*50 40 FOR I=1 TO N 60 SUMY=SUMY+Y(I) 70 FOR J=1 TO M 80 A(J)=A(J)+2/N*Y(I)*COS(J*WO*T(I)) 90 B(J)=B(J)+2/N*Y(I)*SIN(J*WO*T(I)) 100 NEXT J 110 NEXT I 115 A(0)=SUMY/N 120 FOR J=1 TO M 130 IF ABS(A(J)) > .95 THEN PRINT "A(";J;")=";A(J), 140 IF ABS(B(J)) > .95 THEN PRINT "B(";J;")=";B(J), 150 NEXT J 200 end
281
Chapter Eight
Basic computer program to determine Fourier coefficients of function in Example 2. In the above basic program note the following: From line 2 to line 6 to enter the data points of waveform, in this case we have the waveform shown below, we divide the waveform into 200 equal intervals start from t=0 and end at t=0.02 sec. You can define the waveform in any other way. In line 25, we can not get harmonic order greater than (n-1)/2 The loops from line 40 to line 110 is to calculate the Fourier coefficients Am, Bm The loop from line 120 to line 150 is to print the Fourier coefficients Am, Bm The results of the above computer program: b( 1 )= 104.1915 b( 3 )=-11.26867 b( 5 )=-25.42907 b( 7 )=-7.446722 b( 9 )= 10.24038 b( 11 )= 10.42216 b( 13 )=-1.144581 b( 15 )=-8.346155 b( 17 )=-4.108836 b( 19 )= 4.133745 b( 21 )= 5.881331 b( 25 )=-4.849892 b( 27 )=-3.221158 b( 29 )= 2.181897 b( 31 )= 4.21746 b( 33 )= .7539896 b( 35 )=-3.292385 b( 37 )=-2.796622 b( 39 )= 1.186604 b( 41 )= 3.323474 b( 43 )= 1.065102 b( 45 )=-2.3783
b( 47 )=-2.539027 b( 49 )= .559677
282
Least Square Technique Start Input data pints T(I), Y(i) Input N, M
1
A(0)=SUMY/N
SUMY=0, WO=2*.14*f
For J=1 to M
If M>(N-1)/2 M=(N-1)/2 For I=1 to N
Is Abs(A(J)) > 0.1 Print A(J)
SUMY=SUMY+Y(I) For j=1 to m Is Abs(B(J)) > 0.1
A(J)=A(J)+2/N*Y(I)*cos(J*Wo*t(I)) B(J)=B(J)+2/N*Y(I)*sin(J*Wo*t(I))
Print B(J)
Next J,I
Next J
1
End
Flowchart of Basic computer program to determine Fourier coefficients of function in Example 2. Example 3 Use the method of least square to find
1v
the Fourier transform for the function
shown
in
the
following figure (Find DC and
1ms
2ms
fundamental ( a0 , a1 , b1 ) and then draw flowchart and basic program to determine and print m harmonics (divide the function to
283
Chapter Eight
20 intervals in your calculations and 200 intervals in the computer program per period). Solution: f = 1000 Hz , ω o = 2π * 1000 = 6283
t,ms 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00
∑
f (t ) sin ω
f (t )
f (t ) cos ω o t
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00
0.000000 0.047553 0.080903 0.088171 0.061810 0.000012 -0.092689 -0.205706 -0.323589 -0.427964 -0.500000 -0.523098 -0.485449 -0.382124 -0.216398 -0.000104 0.247101 0.499509 0.728027 0.903452 1.000000
0.000000 0.015450 0.058777 0.121350 0.190209 0.250000 0.285322 0.283169 0.235138 0.139093 0.000046 -0.169906 -0.352617 -0.525815 -0.665711 -0.750000 -0.760882 -0.687743 -0.529128 -0.293725 -0.000185
10.50
0.50
-3.16
284
Least Square Technique n
∴ a0 =
yi 10.5 = = 0.525 20 n i =1
∴ a1 =
2 n 2 yi * cos(ω t ) = * 0.5 = 0.05 ∑ n i =1 20
∴ b1 =
2 n 2 yi * sin(ω t ) = * (−3.16) = −0.316 ∑ n i =1 20
∑
If we use the exact Fourier transform will get the following results
1 2L a0 = ∫ f ( x) dx 2L 0 0.001
1 0.001 1 1000t 2 = = 1000 t dt .001 0∫ .001 2 0
= 0.5
1 2L 1 0.001 πt πt a1 = ∫ f (t ) cos dt = 1000 t * cos dt = 0 L 0 0.05 0∫ 0.0005 L 1 2L πt b1 = ∫ f (t ) sin dt = L 0 L 1 0.001 πt 1000 t * sin = dt = −0.16 ∫ 0.05 0 0.0005 Problems Solve problems of Chapter 7 numerically and compare the results.
Chapter 9 Power Series Solution Of Differential Equations 9.1 Introduction In chapter five we consider solving the linear differential equation with constant coefficients. Caushy equation was the only equation in that chapter has non-constant coefficients and has been solved by special manner. In this chapter, we shall consider solving differential equations by so-called power series method, which yields solution in the form of power series. This is a very efficient standard procedure in connection with linear differential equations whose variable coefficients. We start by showing some examples of power series of famous functions:
∞ 1 = ∑ x m = 1 + x + x 2 + x 3 + x 4 + ...... 1 − x m=0
∞
xm x 2 x3 x 4 e = ∑ =1+ x + + + + ...... m ! 2 ! 3 ! 4 ! m=0 x
cos x =
∞
(−1) m x 2m x2 x4 = 1 − + − +......... ∑ (2m)! 2 ! 4 ! m=0
(−1) m x 2m +1 x3 x5 sin x = ∑ = x− + − +......... m ( 2 + 1 )! 3 ! 5 ! m=0 ∞
See Appendix 1 for more functions.
286 Power Series Solution Of Differential Equations
The power series (in power of ( x − a ) ) is an infinite series of the
following form: ∞
∑ cm ( x − a) m = c0 + c1 ( x − a) + c2 ( x − a) 2 + c3 ( x − a)3 + .......
(1)
m=0
Where c0 , c1 , c2 ,... are constants called coefficients of the series, a is a constant, called the center of the power series and x is a variable. If a = 0 , (1) takes the following form: ∞
∑ cm x m = c0 + c1x + c2 x 2 + c3 x 3 + .......
(2)
m=0
9.2 Power Series Solution Of Second Order Differential Equations In this section we will discuss how to solve second order differential equations with non constant coefficient. The form of second order differential with non constant coefficient is as shown in (3):
d2y dx
2
+ P( x)
dy + Q( x) y = 0 dx
(3)
Where P ( x) and Q ( x) are functions in x only. We will look for solution of (3) in power series of x (or in powers of x − a ), if a = 0 , then the solution has the following form:
y=
∞
∑ cm x m = c0 + c1x + c2 x 2 + c3 x 3 + .......
(4)
m =0
∴ y′ =
∞
∑ mcm x m −1 = c1 + 2c2 x + 3c3 x 2 + 4c4 x 3 .......
m=0
(5)
287
Chapter Nine
∴ y′′ =
∞
∑ m(m −1)cm xm−2 = 2c2 + 3* 2 c3 x + 4 * 3c4 x2 + 5 * 4c5 x3. + ..(6)
m=0
Insert (4), (5) and (6) into (3) and collecting like powers of x, we may write the relating equation in the following form:
k0 + k1 x + k 2 x 2 + k3 x 3 + ....... = 0
(7)
Where the constants k0 , k1 , k 2 , k3 , …. are expressions containing the unknown coefficients c0 , c1 , c2 ,...... in (4). In order that (7) hold for all x in some interval, we must have the following conditions:
k0 = 0, k1 = 0, k 2 = 0, k3 = 0 .......
(8)
From (8) we may then determine the coefficients c0 , c1 , c2 ,...... . The following examples explain in details the above manner. Example 1 Solve the following differential equation: y ′′ + y = 0 Solution: Assume, y =
∞
∑ cm x m = c0 + c1x + c2 x 2 + c3 x 3 + .......
(9)
m=0
Then y ′ and y ′′ can be obtained as in (5) and (6) respectively. Substitute these power series in the differential equation we get the following:
(2c2 + 3 * 2 c3 x + 4 * 3c4 x 2 + 5 * 4c5 x 3 + 6 * 5c6 x 4 .....) + (c0 + c1 x + c2 x 2 + c3 x 3 + ...) = 0
288 Power Series Solution Of Differential Equations
c Coefficients of x 0 = 0 , ∴ 2c2 + c0 = 0, ∴ c2 = − 0 , 2! c Coefficients of x1 = 0 , ∴ 3 * 2 c3 + c1 = 0 , ∴ c3 = − 1 , 3! c c Coefficients of x 2 = 0 , ∴ 4 * 3 c4 + c2 = 0 , ∴ c4 = − 2 = 0 , 4 * 3 4!
c c Coefficients of x 3 = 0 , ∴ 5 * 4 c5 + c3 = 0 , ∴ c5 = − 3 = 1 , 5 * 4 5! c c Coefficients of x 4 = 0 , ∴ 6 * 5 c6 + c4 = 0 , ∴ c6 = − 4 = − 0 , 6*5 6! c c Coefficients of x 5 = 0 , ∴ 7 * 6 c7 + c5 = 0 , ∴ c7 = − 5 = − 1 , 7*6 7! Coefficients of x 6 = 0 , ∴ 8 * 7 c8 + c6 = 0 , ∴ c8 = c0 / 8! , Substitute the values of c0 , c1 , c2 ,...... into (9) we get the final form for the solution of the differential equation as following:
∴ y=
∞
∑ cm x
m=0
m
x 2 x 4 x 6 x8 = c0 1 − + − + − +........ 2! 4! 6! 8! x3 x5 x 7 x9 + c1 x − + − + − +........ 3! 5! 7! 9!
∴ y = c0 cos x + c1 sin x Example 2 Solve the following differential equation by using power series method:
y ′′ − 2 xy = 0
289
Chapter Nine
Solution: Assume, y =
∞
∑ cm x m = c0 + c1x + c2 x 2 + c3 x 3 + ...
m=0
Then y ′ and y ′′ can be easily obtained as in (5) and (6) respectively. Substitute these power series in the differential equation we get the following:
(2c2 + 3 * 2 c3 x + 4 * 3c4 x 2 + 5 * 4c5 x 3 + 6 * 5c6 x 4 .....) − 2 x(c0 + c1 x + c2 x 2 + c3 x 3 + ...) = 0 Coefficients of x 0 = 0 , ∴ c2 = 0 ,
2c Coefficients of x1 = 0 , ∴ 3 * 2 c3 − 2c0 = 0 , ∴ c3 = 0 , 3* 2 2c Coefficients of x 2 = 0 , ∴ 4 * 3 c4 − 2c1 = 0 , ∴ c4 = 1 , 4*3 3
Coefficients of x = 0 , 5 * 4 c5 − 2c2 = 0 , c5 =
2c2 = 0, 5*4
Coefficients of x 4 = 0 , ∴ 6 * 5 c6 − 2c3 = 0 ,
∴ c6 =
2c3 2 * 2c0 = , 6 *5 6 *5*3* 2
Coefficients of x 5 = 0 , ∴ 7 * 6 c7 − 2c4 = 0 ,
∴ c7 =
2c4 2 * 2c1 = , 7*6 7*6*4*3
2c Coefficients of x 6 = 0 , ∴ 8 * 7 c8 − 2c5 = 0 , ∴ c8 = 5 = 0 , 8*7 Coefficients of x 7 = 0 , ∴ 9 * 8 c9 − 2c6 = 0 ,
290 Power Series Solution Of Differential Equations
2*2 23 2 2 ∴ c9 = c6 = * c0 = c0 , 9*8 9*8 6*5*3* 2 9 *8* 6 *5*3* 2 Coefficients of x 8 = 0 , ∴ 10 * 9 c10 − 2c7 = 0 ,
∴ c10
2 2 2*2 23 = c7 = * c1 = c1 , 10 * 9 10 * 9 7 * 6 * 5 * 3 10 * 9 * 7 * 6 * 5 * 3
Coefficients of x 9 = 0 ,
∴ 11 * 10 c11 − 2c8 = 0 , and ∴ c11 =
2c8 = 0, 11 * 10
Substitute the values of c0 , c1 , c2 ,...... into (4) we get the final solution as following: 2 3 22 23 6 ∴ y = c0 1 + x + x + x 9 + ... 6*5*3* 2 9 *8* 6 *5*3* 2 3 * 2 2 4 22 23 7 + c1 x + x + x + x10 + .. 4*3 7*6*4*3 10 * 9 * 7 * 6 * 5 * 3
Example 3 Solve the following differential equation:
y ′′ + (cos( x )) y = 0 Solution: ∞
(−1) m x 2m x 2 x 4 x 6 x8 Q cos x = ∑ =1− + − + − +......... ( 2 )! 2 ! 4 ! 6 ! 8 ! m m=0 Assume, y =
∞
∑ cm x m = c0 + c1x + c2 x 2 + c3 x 3 + .......
m=0
291
Chapter Nine
Then y ′ and y ′′ can be obtained as in (5) and (6) respectively. Substitute these power series in the differential equation we get: x 2 x 4 x 6 x8 m−2 m ( m − 1 ) c x + 1 − + − + − +..... ∑ m 2! 4! 6! 8! m=0 ∞
∞
∑ cm x m = 0
m=0
c Coefficients of x 0 = 0 , ∴ 2c2 + c0 = 0, ∴ c2 = − 0 , 2 c Coefficients of x1 = 0 , ∴ 6 c3 + c1 = 0 , ∴ c3 = − 1 , 6 c c Coefficients of x 2 = 0 , ∴ 12c4 + c2 − 0 = 0 , ∴ c4 = 0 , 2 4*3 c 2c1 Coefficients of x 3 = 0 , ∴ 5 * 4 c 5 + c 3 − 1 = 0 ∴ c5 = , 2 5*4*3 Substitute the values of c0 , c1 , c2 ,...... into the assumed solution.
∴ y=
∞
∑ cm x
m=0
m
x2 1 4 1 6 = c0 1 − + x − x + −....... 2 4*3 6*5 x3 1 + c1 x − + x 5 − +....... 3* 2 5*3* 2
∴ y ( x) = c0 y1 ( x) + c1 y 2 ( x) Example 4 Solve the following differential equation by using power series method: Solution:
y ′′ − xy = 0
292 Power Series Solution Of Differential Equations Assume y =
∞
∑ cm x m
(4)
m=0
Then y ′ and y ′′ can be obtained as in (5) and (6) respectively. Substitute these power series in the differential equation we get: ∞
∑ m(m − 1)cm x m − 2 − x
m=0
∴ 2c2 + (3 * 2c3 )x +
∞
∞
∑ cm x m −1 = 0
m=0
∞
∑ m(m − 1)cm x m − 2 − co x − ∑ cm x m +1 = 0 m 1 243 m 4 442444 1=4 3 1=4 assume n = m +1 ∞ (n + 1)cn + 2 x n − cn −1 x n = n=2
assume n = m − 2 ∞
(2c2 ) + (3 * 2c3 − c0 )x + ∑ (n + 2) n=2
∴ c2 = 0 , c3 =
∴
∑
co c = o 3 * 2 3!
∞
∑ [(n + 2)(n + 1)cn + 2 − cn −1 ]x n = 0
n=2
∴ cn + 2 =
cn −1 (n + 2)(n + 1) n = 2,3,4,.... c c0 c2 = 0 , ∴ c6 = 3 = , 6 * 5 6 * 5 * 3! 5*4
∴ c4 =
c1 , 4*3
∴ c7 =
c4 c1 c1 = = , 7 * 6 7 * 6 * 4 * 3 7 * 3 * 4!
∴ c9 =
∴ c5 =
∴ c8 =
c6 co co , = = 9 * 8 9 * 8 * 6 * 5 * 3! 9 * 2 * 6!
∴ c10 =
c7 c1 c1 = = , 10 * 9 10 * 9 * 7 * 3 * 4! 9 * 7 * 6!
c5 = 0, 8*7
0
293
Chapter Nine
∴ c11 =
c8 c9 co = 0 , ∴ c12 = = 11 * 10 12 * 11 12 * 11 * 9 * 2 * 6!
x3 x6 x9 x12 + + + + ...... ∴ y ( x ) = co 1 + 3! 180 12960 12 * 11 * 9 * 2 * 6! x4 x7 x10 + c1 x + + + + ........ 12 7 * 3 * 4! 9 * 7 * 6! Example 5 Solve the following differential equation by using power series method: y ′′ − x y ′ − y = 0 ∞
Solution: Assume y =
∑ cm x m
(4)
m=0
Then y ′ and y ′′ can be obtained as in (5) and (6) respectively. Substitute these power series in the differential equation we get:
∴
∞
∑ m(m − 1)cm x
m−2
−x
m=0
∴
∞
∑ m(m − 1)cm x
∞
∑ mcm x
m −1
−
m=0 m−2
m =0
∴ 2c2 + (3 * 2c3 )x +
−
∞
∑ cm x m = 0
m=0
∞
∑ (m + 1)cm x m = 0
m=0 ∞
∑ m(m − 1)cm x m − 2 − co − 2c1x
4 442444 m 1=4 3 assume n = m − 2 ∞
− ∑ (m + 1)cm x m = 0 1 42443 m 1=4 assume n = m
294 Power Series Solution Of Differential Equations
∴ 2c 2 − c 0 = 0 , ∴ c 2 = ∴
∞
c0 c , ∴ 3 * 2 * c3 − 2c1 = 0 ∴ c3 = 1 2! 3
∑ [(n + 2)(n + 1)cn + 2 − (n + 1)cn ]x n = 0
n=2
∴ cn + 2 =
(n + 1)cn (n + 2)(n + 1) n = 2,3,4,....
∴ c4 =
c c2 = o , 4 4*2
∴ c7 =
c5 c1 c1 , = = 7 7 * 5 * 3 7 * 3 * 4!
∴ c9 =
c7 c1 = , 9 9*7*5*3
∴ c5 =
∴ cn + 2 =
cn (n + 2) n = 2,3,4,....
c3 c0 c c = 1 , ∴ c6 = 4 = 5 5*3 6 6*4*2 ∴ c8 =
∴ c10 =
c6 co = 8 8*6*4*2
c8 co = 10 10 * 8 * 5 * 4 * 2
x2 x4 x6 x8 x10 + 2 + 3 + 4 + 5 + ...... ∴ y ( x ) = co 1 + 2! 2 * 2! 2 * 3! 2 * 4! 2 * 5! x3 x5 x7 x9 + c1 x + + + + ........ 3 5*3 7*5*3 9*7*5*3 Example 6 Solve the following differential equation by using power series method: (1 − x ) y ′′ + y = 0 ∞
Solution: Assume y =
∑ cm x m
(4)
m=0
Then y ′ and y ′′ can be obtained as in (5) and (6) respectively. Substitute these power series in the differential equation we get:
∴
∞
∑ m(m − 1)cm x
m=0
m−2
−x
∞
∑ m(m − 1)cm x
m =0
m−2
+
∞
∑ cm x m = 0
m=0
295
Chapter Nine ∞
∑ m(m − 1)cm x
∴
m−2
m=0
−
∞
∑ m(m − 1)cm x
m −1
+
m=0
∞
∑ cm x m = 0
m=0
∞
∴ 2c2 + (3 * 2c3 )x + ∑ m(m − 1)cm x m − 2 − 2 * 1c3 x − 4 442444 m 1=4 3 assume n = m − 2
∞
∞
∑ m(m − 1)cm x m −1 + co + c1x + ∑ cm x m = 0 = 224 3 442444 m m 1=4 3 14 3 assume n = m −1
c ∴ 2c2 + c0 = 0 , ∴ c2 = − 0 2! ∴ c3 = − ∴
n=m
∴ 3 * 2 * c3 − 2c2 + c1 = 0 ,
co c1 − 6 6
∞
∑ [(n + 2)(n + 1)cn + 2 − n(n + 1)cn +1 + cn ]x n = 0
n=2
∴ cn + 2 =
cn (n + 1) c − (n + 2)(n + 1) n +1 (n + 2)(n + 1) n = 2,3,4,....
∴ cn + 2 =
cn n cn +1 − (n + 2) (n + 2)(n + 1) n = 2,3,4,....
∴ c4 =
c c c 2c3 c c c − 2 =− o − 1 + o =− o − 1, 4 4*3 12 12 24 24 12
∴ c5 =
co − 3c0 c1 3c4 3c3 3c1 − = − + + , 5 5 * 4 5 * 24 5 * 12 6 * 5 * 4 6 * 5 * 4
∴ c5 = −
co c − 1, 5 * 4 * 3 24
296 Power Series Solution Of Differential Equations
co − 2c 0 4c5 c c1 2c1 − 4 = − + + , 6 6 * 5 3 * 5 * 4 * 3 3 * 24 6 * 5 * 24 6 * 5 * 12 − 7c0 11c1 − ∴ c6 = 720 420 x 2 x3 x 4 x5 7 x 7 − − − − + ...... ∴ y ( x ) = co 1 − 2 6 24 60 720 c6 =
x 3 x 4 x 5 11x 6 + c1 x − − − − ........ 6 12 24 420 Example 7 Solve the following differential equation by using power series method: 2 + x 2 y ′′ − x y ′ + 4 y = 0
(
)
Solution: Assume y =
∞
∑ cm x m
m=0
(4)
Then y ′ and y ′′ can be obtained as in (5) and (6) respectively. Substitute these power series in the differential equation we get:
∴2
∞
∑ m(m − 1)cm x
m−2
m =0
−x
∑ m(m − 1)cm x m
+
m=0 ∞
∞
∑ mcm x m −1 + 4 ∑ cm x m = 0
m=0
∴2
∞
m=0 ∞
∞
∑ m(m − 1)cm x m − 2 + ∑ (m 2 − 2m + 4)cm x m = 0
m=0
∞
m=0
∴ 2 * 2c 2 + 12c3 x + 2 ∑ m(m − 1)c m x m−2 + 2co + 3c1 x + 4 4424443 m 1=4 assume n = m − 2
∑ (m 2 − 2m + 4)c m x m ∞
2 44 m 1=4 424444 3 assume n = m
=0
297
Chapter Nine
Coefficient of x o = 0 ∴ 4c2 + 4co = 0,
∴ c 2 = − co
c Coefficient of x1 = 0 ∴ 12c3 + 3C1 = 0, ∴ c3 = − 1 4 − [n(n − 2 ) + 4]cn Coefficient of x n = 0 ∴ cn + 2 = 2(n + 2 )(n + 1) 2,3,4,.... ∴ c4 = − ∴ c5 =
4c c 4c 2 = o = o, 2*4*3 4! 3!
− 7c3 7c1 = , 2 * 5 * 4 10 * 16
∴ c7 = −
(5 * 3 + 4)c5 2*7*6
=−
∴ c6 = −
(4 * 2 + 4)c4 2*6*5
=−
co , 30
19c1 1920
x4 x6 2 − + ...... ∴ y ( x ) = co 1 − x + 3! 30 x 3 7 x 5 19 x 7 + c1 x − + − ........ 4 160 1920 Example 8
Solve the following differential equation by using
power series method: y ′′ − 2 x y ′ + λ y = 0 Solution: This equation known as Hermith equation. Assume, y =
∞
∑ cm x m
(4)
m=0
(Charles Hermit (1822-1901). A French mathematician, held the chair of higher algebra at the university of Paris from 1876 to 1901. This equation is important in many branches in mathematical physics; for example, in quantum mechanics.
h
298 Power Series Solution Of Differential Equations Then y ′ and y ′′ can be obtained as in (5) and (6) respectively. Substitute these power series in the differential equation we get:
∴
∞
∑ m(m − 1)cm x m − 2 − 2 x
m=0
∞
∑ mcm x m −1 +λ
m=0
∴ (2c2 + λc0 ) + (3 * 2c3 − 2c1 + λc1 )x +
∞
∑ cm x m = 0
m=0
∞
∑ m(m − 1)cm x m − 2
4 442444 m 1=4 3 assume m = n + 2
∞
∞
m −1
− 2 x ∑ mcm x +λ ∑ cm x m = 0 2 44 = 244 m =4 14 4244m4 3 assume m = n
∴ (2c2 + λc0 ) + (3 * 2c3 − 2c1 + λc1 )x + ∞
n
− 2 ∑ ncn x +λ n=2
∞
∑ (n + 2)(n + 1)cn + 2 x n
n=2
∞
∑ cn x n = 0
n=2
∴ (2c2 + λc0 ) + (3 * 2c3 − 2c1 + λc1 )x +
∞
∑ [(n + 2)(n + 1)cn + 2 − 2ncn + λcn ]x n = 0
n=2
λc Coefficients of x 0 = 0 , then, 2c2 + λc0 = 0, ∴ c2 = − 0 2 Coefficients
∴ c3 =
of
x1 = 0 ,
∴ 3 * 2 c3 − 2c1 + λc1 = 0 ,
(2 − λ ) c1 , 3* 2
Coefficients of x n = 0 , ∴ (n + 2 )(n + 1)cn + 2 − 2ncn + λcn = 0 ,
299
Chapter Nine
∴ cn + 2 =
2n − λ c (n + 2)(n + 1) n
Substitute the values of c0 , c1 , c2 ,...... into (4) we get: ∴ y=
∞
λ
∑ cm x m = c0 1 − 2! x 2 +
(4 − λ )(−λ ) 4 (8 − λ )(4 − λ ) 6 x + x + .... + 4! 6!
(2 − λ ) 3 (6 − λ )(2 − λ ) 5 (10 − λ )(6 − λ )(2 − λ ) 7 c1 x + x + x + x + .. 3! 5! 7! m=0
∴ y ( x) = c0 y1 ( x) + c1 y 2 ( x) Example 9 Solve the following differential equation by using power series method: x 2 + 9 y ′′ − ( x + 1) y ′ + x 4 y = 0
(
)
∞
Solution: Assume y =
∑ cm x m
(4)
m=0
Then y ′ and y ′′ can be obtained as in (5) and (6) respectively. Substitute these power series in the differential equation we get: ∞
∑ m(m − 1)cm x m + 9
∴ −
m =0 ∞
∑ mcm x
m −1
m=0
∴ − c1 x +
+
∞
∞
∑ m(m − 1)cm x m − 2 −
m=0
∑ mcm x m
m=0
∞
∑ cm x m + 4 = 0
m=0
∞
∑ m(m − 2)cm x m + 9 * 2c2 + 9 * 3 * 2c3 x
2 4 m 1=4 42444 3 assume n = m
+9
∞
∞
∞
∑ m(m − 1)cm x m − 2 − c1 − 2c2 x − ∑ mcm x m −1 + ∑ cm x m + 4 = 0 34244 0 4 442444 m m 1=4 3 1=4 3 m 1=42 43 assume n = m − 2
assume n = m −1
assume n = m + 4
300 Power Series Solution Of Differential Equations
∴ (9 * 2c2 − c1 ) + (9 * 3 * 2c3 − 2c2 − c1 )x ∞
∑ n ( n − 2) c n x
+ −
n=2 ∞
n
∞
+ 9 ∑ (n + 2 )(n + 1)cn + 2 x n n=2 ∞
∑ (n + 1)cn +1x n + ∑ cn − 4 x n = 0
n=2
n=4
Coefficient of x o = 0 ∴ c2 = c1 / 9 * 2 Coefficient of x1 = 0 ∴ c3 =
c1 2c 2 5c + = 51 9 *3* 2 9 *3* 2 3
∞
∑ [n(n − 2)cn + 9(n + 2)(n + 1)cn + 2 − (n + 1)cn +1 ]x
n
∞
∑ cn − 4 x n = 0
+
n=4
n=2
∴ x [0co + 9 * 4 * 3c4 − 3c3 ] + x [3 * 1c3 + 9 * 5 * 4c5 − 4c4 ] + 2
3
∞
∑ [n(n − 2)cn + 9(n + 2)(n + 1)cn + 2 − (n + 1)cn +1 ]x
n
∞
+
n=4 ∞
∑ cn − 4 x n = 0
n=4
∑ [n(n − 2)cn + 9(n + 2)(n + 1)cn + 2 − (n + 1)cn +1 + cn − 4 ]x n = 0
n=4
Coefficient of x 2 = 0 ∴ c4 =
3c3 5 = 2 7 c1 (9 * 4 * 3) 2 * 3
4c − 3c3 13 Coefficient of x 3 = 0 ∴ c5 = 4 =− c 9 1 (9 * 5 * 4) 2*3 Coefficient of x n = 0
∴ cn + 2 =
[(n + 1)cn +1 − n(n − 2)cn − cn − 4 ] 9(n + 2)(n + 1) 4,5,6....
301
Chapter Nine
∴ ∴ ∴ ∴
5c5 − 4 * 2c4 − co 72 1 = − 2 12 c1 − c6 = co 9*6*5 2 *3 2 * 5 * 32 6c − 5 * 3c5 − c1 181 * 487 1 c7 = 6 c1 − co = − 14 9*7*6 3 *7 2 * 5 * 7 * 34 7 c − 6 * 4c 6 − c 2 23 * 4363 43 c8 = 7 c = − 14 4 c1 + 4 6 o 9 *8* 7 3 *2 2 *7*3 6 7 x x x8 y ( x ) = co 1 − ...... − + + 2 2 * 5 * 7 * 34 2 4 * 7 * 36 2 * 5 * 3 x2 5x3 5x 4 13 x 5 72 x6 + c1 x + + 5 + 2 7− − 2 12 9 9 * 2 3 2 * 3 2 * 3 2 *3 −
181 * 487 x 7 314 * 7
−
23 * 4363 x 8 2 4 * 316
.......
Example 10 Solve the following differential equation by using power series method: 1 + x 2 y ′′ − 4 x y ′ + 6 y = 0
(
)
∞
Solution: Assume y =
∑ cm x m
m=0
(4)
Then y ′ and y ′′ can be obtained as in (5) and (6) respectively. Substitute these power series in the differential equation we get:
∴
∞
∑ m(m − 1)cm x
m=0
−4
∞
+
∞
∑ m(m − 1)cm x m
m=0 ∞
∑ mcm x m + 6 ∑ cm x m = 0
m=0
∴
m−2
m=0 ∞
∑ m(m − 1)cm x m − 2 + ∑ (m 2 − 5m + 6)cm x m = 0 ∞
m=0
m=0
302 Power Series Solution Of Differential Equations
∴ 2c2 + (3 * 2c3 )x +
∞
∑ m(m − 1)cm x m − 2 + 6co + 2c1x
4 442444 m 1=4 3 assume n = m − 2
∑ (m 2 − 5m + 6) cm x m = 0 ∞
2 44 m 1=4 424444 3 assume n = m
∴ 2c2 + 6co = 0, ∴ c2 = −3co , and c3 = −
∑ [(n + 2)(n + 1)cn + 2 + (n 2 − 5n + 6)cn ]x n = 0 ∞
c1 3
n=2
( n 2 − 5n + 6 ) cn ∴ cn + 2 = − (n + 2)(n + 1)
n = 2,3, 4,....
∴ c4 = 0 , ∴ c5 = 0 , and, cn + 2 = 0 x3 2 ∴ y ( x ) = co 1 − 3x + c1 x − 3
[
]
Example 11 Solve the following differential equation:
( x 2 + 1) y ′′ + x y ′ − y = 0 Solution: Assume, y =
∞
∑ cm x m
(4)
m=0
Then y ′ and y ′′ can be obtained as in (5) and (6) respectively. Substitute these power series in the differential equation we get: 2
( x + 1)
∞
∑ m(m − 1)cm x
m=0
m−2
+x
∞
∑ mcm x
m=0
m −1
−
∞
∑ cm x m = 0
m=0
303
Chapter Nine
∴
∞
∑ m(m − 1)cm x
m
+
m=0
∞
+
∑ mcm x m −
m=0
∞
∑ m(m − 1)cm x m − 2
m=0
∞
∑ cm x m = 0
m=0
∴ (2c2 − c0 ) + (3 * 2c3 + c1 − c1 )x +
∞
∑ m(m − 1)cm x m
m 2 4 1=4 42444 3 assume m = n
+
∞
∞
∞
∑ m(m − 1)cm x m − 2 + ∑ mcm x m − ∑ cm x m = 0 m 4 442444 2 44 2 44 m =4 1=4 3 m 1=4 424 3 assume m = n + 2 ∞
∴ (2c2 − c0 ) + 6c3 x +
assume m = n
∑[n(n − 1)cn + (n + 2)(n + 1)cn + 2 + ncn − cn ]x n = 0
n=2 ∞
∴ (2c2 − c0 ) + 6c3 x +
∑ [(n + 1)(n − 1)cn + (n + 2)(n + 1)cn + 2 ]x n = 0
n=2
c Coefficients of x 0 = 0 , ∴ 2c2 − c0 = 0, ∴ c2 = 0 and 2 1 Coefficients of x = 0 , ∴ 6 c3 = 0 , ∴ c3 = 0 , Coefficients of x n = 0 , (n + 1)(n − 1)cn + (n + 2)(n + 1)cn + 2 = 0 ,
1− n cn 2+n Substitute the values of c0 , c1 , c2 ,...... into (4) we get:
∴ cn + 2 =
∴y=
∞
∑ cm x m = c1x +
m=0 2
x 1 1 * 3 6 1 * 3 * 5 8 1 * 3 * 5 * 7 10 − 2 c0 1 + x4 + 3 x − 4 x + x .... 5 2 ! 2 * 2 ! 2 * 3 ! 2 * 4 ! 2 * 5 ! ∴ y ( x) = c0 y1 ( x) + c1 y 2 ( x)
304 Power Series Solution Of Differential Equations Example 12 Solve the following differential equation about the ordinary point x = 1 : y ′′ + ( x − 1) 2 y ′ − 4 ( x − 1) y = 0
(10)
Solution: To solve an equation about the point x=a , means to obtain solutions valid in a region surrounding that point, solutions expressed in powers of ( x − a ) . We first translate the axes, putting
x −1 = v . Then the differential equation (10) becomes d2y dx 2
+ v2
dy − 4 vy = 0 dx
(11)
∞
Assume, y =
∑ cm v m = c0 + c1v + c2 v 2 + c3v 3 + .......
(12)
m=0
Then y ′ and y ′′ can be easily obtained from (12). Substitute these power series in the differential equation we get the following ∞
∑ m(m − 1)cm v m − 2 +
m=0
∞
∑ mcm v m +1 − 4
m=0
∞
∑ m(m − 1)cm v m − 2 +
∴
m=0
∞
∑ cm v m +1 = 0
m =0
∞
∑ (m − 4)cm v m +1 = 0
m=0
∴ 2c2 + (3 * 2c3 + (0 − 4)c0 )v + ∞
∞
∑ m(m − 1)cm v m − 2 + ∑ (m − 4)cm v m +1 = 0 4 442444 1 4 m 1=4 3 m 1=4 42444 3 m=n+2
m = n −1
∴ 2c2 + (3 * 2c3 + (0 − 4)c0 )v +
∞
∑ [(n + 2)(n + 1)cn + 2 + (n − 5)cn −1 ]v n = 0
n=2
305
Chapter Nine
Coefficients of v 0 = 0 , ∴ 2c2 = 0, ∴ c2 = 0 and
2c Coefficients of v1 = 0 , ∴ 6 c3 = 4c0 , ∴ c3 = 0 , 3 Coefficients of v n = 0 , ∴ (n − 5)cn −1 + (n + 2)(n + 1)cn + 2 = 0 ,
∴ cn + 2 =
5−n cn −1 (n + 1)(n + 2)
C n = 0 , for n=2, 5, 8, 11, Also C n = 0 , for n=7, 10, 13, 16, ….
∴ c4 =
c c c0 c1 Also, ∴ c6 = 0 , c9 = − 0 , c12 = 4 45 324 42768
Substitute the values of c0 , c1 , c2 ,...... into (12) we get: ∞
∴y=
∑ cm v
m=0
m
v6 v9 v12 v 4 = c0 1 + − + − +... + c1 v + 4 45 324 42768
But v = x − 1 , substitute in the above equation, we get:
( x − 1) 6 ( x − 1)9 ( x − 1)12 − + − +... y = ∑ cm ( x − 1) = c0 1 + 45 324 42768 m=0 ( x − 1) 4 + c1 ( x − 1) + 4 ∞
m
∴ y ( x) = c0 y1 ( x) + c1 y 2 ( x)
306 Power Series Solution Of Differential Equations 9.3 Power Series Basics • A power series (1) is said to converge at a point x if m
∑ cn ( x − a ) n
lim
m→∞ n =0
exists. The series certainly converges
for x = a ; it may converge for all x, or it may converges for some values of x and not for others. • A power series converges absolutely for x − a < R and diverge for
x − a > R , Where R is the radius of
convergence. • A power series represents a continuous function within its interval of convergence. • A power series can be differentiated (Integrated) term wise within its interval of convergence. • Two power series with a common interval of convergence can be added term by term. • The function (solution) is analytic at x = a if it can be expand in a power series having the form:
y ( x) =
∞
∑ cm ( x − a) m = c0 + c1 ( x − a) + c2 ( x − a) 2 + c3 ( x − a)3 + .......
m=0
If both P ( x) and Q( x) are analytic at a point x = a , this point is called an ordinary point of the differential equation (3) otherwise it is called a singular point.
307
Chapter Nine
• If a is a singular point of differential equation (3) and
( x − a) P( x), ( x − a ) 2 Q ( x) are analytic function at x = a . Then a is said to be regular singular point of the differential equation (3), otherwise it is called an irregular singular point. • If x = a is an ordinary point of differential equation (3), we can always find two distinct power series solutions of the following form: y ( x) =
∞
∑ cm ( x − a ) m .
A series
m=0
solution will converge at least for x = a < R1 , where R1 is the distant to the closest single point. Example 13 Consider the following differential equation
d2y dx 2
+x
dy + ( x 2 + 4) y = 0 dx
We have P ( x) = x and Q ( x) = x 2 + 4 which are analytic functions for all x. Thus all points x are ordinary points of the differential equation. Example 14 Consider the following differential equation:
d2y dx 2
+ ex
dy + sin( x) y = 0 dx
308 Power Series Solution Of Differential Equations We have P ( x) = e x and Q( x) = sin( x) which are analytic functions for all x. Thus all points x are ordinary points of the differential equation. Example 15 Consider the following differential equation:
d2y
+ ( Ln( x)) y = 0 dx 2 This differential equation has a singular point at x = 0 . Because Q ( x) = Ln( x) possesses no power series in x. Example 16 Consider the following differential equation
( x 2 − 1) d2y
d2y dx 2
+ 2x
dy + 6y = 0 dx
dy 6 y=0 + dx 2 ( x 2 − 1) dx ( x 2 − 1) 2x 6 We have P ( x) = and Q ( x ) = ( x 2 − 1) ( x 2 − 1)
∴
+
2x
which are
analytic except at x = 1 and x = −1. Thus x = 1 and x = −1 are singular points of the differential equation, and all other finite value of x are ordinary points. We now test P( x ) and Q( x ) at each singular point
For x = −1
2x = 1 Also, x 2 − 1 x − 1 x = −1 6 6( x + 1) = = −1 Q( x ) = ( x + 1)2 * 2 − 1 x x −1 x = −1
(x + 1) P(x ) = (x + 1) * (x + 1)2
2x
=
309
Chapter Nine
It is clear that the point x = −1 is regular singular point.
For x = 1
2x = 1 Also, 2 + 1 x x −1 x =1 ( x − 1)2 Q( x ) = ( x − 1)2 * 26 = 6( x − 1) = 1 x + 1 x =1 x −1
(x − 1) P(x ) = (x − 1) *
2x
=
It is clear that the point x = 1 is regular singular point. So, the two points x = 1, and x = −1 are regular singular points for the differential equation. Example 17 Consider the following differential equation: 2
( x − 4) d2y
2
d2y dx 2
+ ( x − 2)
dy +y=0 dx
dy 1 y=0 + 2 2 dx 2 2 dx ( x − 2)( x + 2) ( x − 2) ( x + 2) 1 1 We have P ( x) = and Q ( x ) = ( x − 2)( x + 2) 2 ( x − 2) 2 ( x + 2) 2 ∴
1
+
It is clear that x = −2 and x = 2 are singular points. We now test P (x) and Q (x) at each singular point. ¾ For x = −2
( x + 2) P ( x ) =
1 1 and ( x + 2) 2 Q ( x) = ( x − 2)( x + 2) ( x − 2) 2
It is clear that ( x + 2) P ( x) is not analytic at x = −2 and hence
x = −2 is an irregular single point of the differential equation.
310 Power Series Solution Of Differential Equations ¾ For x = 2
( x − 2) P ( x ) =
1 ( x + 2) 2
and ( x − 2) 2 Q( x) =
1 ( x + 2) 2
are analytic
function at x = 2 is a regular singular point of the differential equation.
311
Chapter Nine
9.4 Solutions Around Singular Points To solve (3) around the regular singular point we employ the following theorem: Theorem 1 If x = a is a regular singular point of equation (3) then there exists at least one series solution of the form
y ( x) = ( x − a )
∞
r
∑ cm ( x − a )
m=0
m
=
∞
∑ cm ( x − a ) m + r
(13)
m =0
Where the number r is a constant which must be determined. The series will converge at least on some intervals 0 < ( x − a ) < R . Example 18 Solve the following differential equation about x = 0 :
y ′′ +
1 1 y′ − y=0 3x 3x
(14)
Solution: From (14) we have P( x ) = analytical
x 2Q( x ) = −
at
point
1 1 and Q( x ) = − which are not 3x 3x x=0
but
xP( x ) =
1 3
and
x = 0 at x = 0 . So, It is clear that x = 0 is a regular 3
singular point of (14). So from (13) we assume the solution to be in the following form:
y=
∞
∑ cm x m + r ,
m =0
(15)
312 Power Series Solution Of Differential Equations ∞
∑ (m + r )cm x m + r −1 ,
∴ y′ =
(16)
m=0 ∞
∑ (m + r )(m + r − 1)cm x m + r − 2
∴ y ′′ =
(17)
m =0
Substitute that in the differential equation we get: ∞
3 ∑ (m + r )(m + r − 1)cm x m + r −1 m =0
+
∞
∑ ( m + r )cm x
m + r −1
−
m=0
∞
∑ cm x m + r
m=0
∞
∑ (m + r )(3m + 3r − 3 + 1)cm x
m + r −1
∞
−
m=0
∴ r (3r − 2)c0 x r −1 + x r
=0
∑ cm x m + r
=0
m =0
∞
∑ (m + r )(3m + 3r − 2)cm x m −1
m 1 44442444443 1=4 n = m −1
− xr
∞
∑ cm x m = 0
m = 024 14 3 n=m
∞ ∴ x r r (3r − 2)c0 x −1 + x n ∑ [(n + r + 1)(3n + 3r + 1)cn +1 − cn ] = 0 n =0
Which implies that r (3r − 2)co = 0 , and (n + r + 1)(3n + 3r + 1)cn +1 − cn = 0
(18)
n = 0, 1, 2, 3,......
Since nothing is gained by taking c0 = 0 , we must have
r (3r − 2) = 0 .
313
Chapter Nine
∴ cn +1 =
cn (n + r + 1)(3n + 3r + 1)
n = 0, 1, 2, 3,..
The two values of r that satisfy (18), r1 =
(19)
2 and r2 = 0 , when 3
substitute in (19), give us two different recurrence relations: For r1 =
2 , substitute in (19) we get: 3
cn +1 =
cn n = 0, 1, 2, 3,...... (3n + 5)(n + 1)
∴ c1 =
c0 , 5 *1
∴ c3 =
c0 c2 = 11 * 3 3!*5 * 8 * 11
∴ c4 =
c3 c0 = 14 * 4 4!*5 * 8 * 11 * 14
∴ cn =
c0 n = 1,2,3,...... n!*5 * 8 * 11 * ......(3n + 2)
∴ c2 =
(20)
c0 c1 = 8* 2 2 *5*8
Thus we obtain the first solution: ∞ 1 ∴ y1 ( x) = co x 2 / 3 1 + ∑ xn n =1n!*5 * 8 * 11 * ......(3n + 2)
(21)
For r2 = 0 , substitute in (19) we get:
cn +1 =
cn n = 0, 1, 2, 3,...... (n + 1)(3n + 1)
Where as iteration of (22) yields:
(22)
314 Power Series Solution Of Differential Equations
c0 c0 c , c2 = 1 = 1 *1 2 * 4 2!*1 * 4 c c0 c4 = 3 = 4 * 10 4!*1 * 4 * 7 * 10 c0 ∴ cn = n!*1 * 4 * 7 * ......(3n − 2) c1 =
, c3 =
c0 c2 = 3 * 7 3!*1 * 4 * 7
n = 1, 2, 3,......
∞ 1 ∴ y 2 ( x) = co x 0 1 + ∑ xn (23) − !* 1 * 4 * 7 * ......( 3 2 ) n n n =1 By the ratio test it can be demonstrated that both (21) and (23)
converges for all finite values of x. Also, it should be clear from the form (21) and (23) that neither series is a constant multiple of the other and therefore, y1 ( x) and y 2 ( x ) are linearly independent solution on the x-axis. Hence, by the superposition principle we get:
y ( x) = C1 y1 ( x) + C2 y2 ( x) ∞ 1 y ( x) = C1 x 2 / 3 1 + ∑ xn + n =1n!*5 * 8 * 11 * ......(3n + 2) ∞ 0 1 C2 x 1 + ∑ x n , x < ∞ n =1n!*1 * 4 * 7 * ......(3n − 2)
This is another solution of differential equation. On any interval not containing the origin (such a 0 < x < ∞ ) this combination represents the general solution of the differential equation. Although the forgoing example illustrates the general procedure for using theorem 1, we hasten to point out that we may not always be able to
fined two solution so readily, or for that matter, find two solutions which are infinite series consisting entirely of power of x.
315
Chapter Nine
9.5 Indicial Equation
Equation (18) is called the indicial equation of the problem, and the values r1 =
2 and r2 = 0 are called the indicial roots or 3
exponents of the singularity. In general, if x=0 is a regular point of (3) then the function xP(x)
and x 2Q ( x) obtained from (3) are
analytic at x = 0 . That is, the following expressions are valid on intervals that have a positive radius of convergence.
xP( x) = p0 + p1 x + p2 x 2 + .......... x 2Q( x) = q0 + q1x + q2 x 2 + .......... After substituting y =
∞
∑ cm x m + r
in (3) and simplifying, the
m=0
indicial equation is a quadratic equation in r that results from equating the total coefficients of the lowest power of x equal to zero. The indicial equation has the following form:
r (r − 1) + p0 r + q0 = 0
(24)
We then solve the latter equation for the two values of the exponents and substitute these values into a recurrence relation such as (19). Example 19 Solve the following differential equation:
x y ′′ + 3 y ′ − y = 0
316 Power Series Solution Of Differential Equations Solutions: This differential equation has a regular singular point at
x = 0 . Then, assume y =
∞
∑ cm x m + r ,
(15)
m=0
Then y ′ and y ′′ can be obtained as in (16) and (17) respectively. Substitute these power series in the differential equation we get: r
∴ x r (r + 2)c0 x
−1
+x
n
∞
n=0
∑ [(n + r + 1)(n + r + 3)cn +1 − cn ] = 0
So that, the indicial equation and exponents are r (r + 2) = 0 and
r1 = 0, r2 = −2 , respectively. Since (n + r + 1)(n + r + 3)cn +1 − cn = 0 , n = 0, 1, 2, ........ It follows that, when r1 = 0 , cn +1 =
(25)
cn (n + 1)(n + 3)
∴ c1 =
2c0 2c c0 c c , ∴ c2 = 1 = , ∴ c3 = 2 = 0 1* 3 2 * 4 2!*4! 3 * 5 3!*5!
∴ c4 =
2c c3 = 0 4 * 6 4!*6!
∴ cn =
2c 0 , n = 1, 2, 3,...... n!*(n + 1)!
Thus one series solution is as following: 0
∞
∞
1
1 ∴ y1 ( x) = c0 x 1 + ∑ xn n =1n!*(n + 2)! Or y1 ( x) = c0
∑ n!*(n + 2)! x n
n=0
x <∞
(26)
317
Chapter Nine
Now when r2 = −2 ,we get the following equation:
(n − 1)(n + 1)cn +1 − cn = 0 , n = 0, 1, 2, ........
(27)
But note here that we do not divide by (n − 1)(n + 1) immediately since this term is zero for n = 1 . However, we use recurrence relation (27) for the cases n = 0 and n = 1 :
− 1 * 1c1 − c0 = 0 and 0 * 2c2 − c1 = 0 The latter equations implies that c1 = 0 and so the former equation implies that c0 = 0 . Continuing we found:
cn +1 =
cn n = 2, 3, ..... (n − 1)(n + 1)
∴ c3 =
c2 1* 3
∴ cn =
2c 0 , n = 2, 3, 4,...... (n − 2)!*n!
, ∴ c4 =
c3 c 2c 2c = 2 , ∴ c5 = 4 = 2 2 * 4 2!*4! 3 * 5 3!*5! (28)
Thus we can write
∴ y 2 ( x ) = c2 x
−2
∞
2 xn n = 2 ( n − 2)! * n!
∑
x <∞
(29)
However, close inspection of (29) reveals that y 2 is simply a constant multiple of (26). To see this, let k = n − 2 in (29). We conclude that this method gives only one series solution of this equation x y ′′ + 3 y ′ − y = 0 . And this is not acceptable. So, we have to follow special manner in case of r1 − r2 = N .
318 Power Series Solution Of Differential Equations 9.6 Cases Of Indicial Roots
Let us suppose that r1 and r2 are the real solutions of the indicial equation and that, when appropriate, r1 denotes the largest root. So, we have the following three cases: Case I
If r1 and r2 are distinct and do not differ be an integer,
Case II
If r1 − r2 = N , where N is a positive integer.
Case III
If r1 = r2
The above three cases are analyzed in details in the following: Case I If r1 and r2 are distinct and do not differ be an integer, then
there exist two linearly independent solutions of (3) of the form (30) and (31) and as explained in Example 18 and the following example.
y1 = y2 =
∞
∑ cm x m + r1 ,
c0 ≠ 0,
(30)
∑ bm x m + r2 ,
b0 ≠ 0,
(31)
m =0 ∞
m=0
Example 20 Solve the following differential equation around x = 0 :
2 x y ′′ + (1 + x ) y ′ + y = 0 Solution: If y =
∞
∑ cm x m + r
(15)
m=0
Then y ′ and y ′′ can be obtained as in (16) and (17) respectively. Substitute these power series in the differential equation we get:
319
Chapter Nine ∞
2
∑
m=0 ∞
+
(m + r )(m + r − 1)cm x
∑ ( m + r ) cm x m + r +
m =0 ∞
∑
m=0
m + r −1
+
∞
∑ (m + r )cm x m + r −1
m =0
∞
∑ cm x m + r = 0
m =0
(m + r )(2m + 2r − 1)cm x m + r −1 +
∴ r (2r − 1)c 0 x r −1 + x r
∞
∑ (m + r + 1)cm x m + r
=0
m =0
∞
∑ (m + r )(2m + 2r − 1)c m x m−1
m 1 44442444443 1=4 n = m −1
+ xr
∞
∑ (m + r + 1)c m x m
=0
m 0 442444 1=4 3 n=m
∞ ∴ x r r(2r −1)c0 x −1 + xn ∑[(n + r + 1)(2n + 2r + 1)cn+1 + (k + r +1)cn ] = 0 n=0
Which implies that r (2r − 1) = 0 , ∴ r1 =
1 , and r2 = 0 2
(n + r + 1)(2n + 2r + 1)cn +1 + (n + r + 1)cn = 0 n = 0,1, 2, 3,...... For r1 =
∴ c1 =
− cn 1 : ∴ cn +1 = n = 2, 3, ..... 2(n + 1) 2
− c0 c − c1 , ∴ c2 = = 20 2 *1 2*2 2 *2
∴ c3 =
−c − c2 = 3 0 , 2 * 3 2 * 3!
,
320 Power Series Solution Of Differential Equations
∴ cn =
(−1) n c0
n = 1,2,3,......
2 n * n!
Thus from (30) and (31) we can write the final solution as: 1/ 2
∴ y1 ( x) = c0 x
∞
(−1) n
n
∞
(−1) n
x = c0 ∑ n x n +1 / 2 1 + ∑ n n = 02 * n! n = 02 * n!
Which converge for x ≥ 0 . As given, the series is not meaningful
x < 0 because of the presence of x1 / 2 . Now for r2 = 0 ,
cn +1 =
− cn n = 2, 3, ..... 2n + 1
∴ c1 =
− c0 1
∴ c3 =
− c0 − c2 = 5 1* 3 * 5
, ∴ c2 =
c − c1 = 0 3 1* 3 , ∴ c4 =
(−1) n c0 ∴ cn = 1 * 3 * 5 * 7.....(2n − 1)
− c3 − c0 = 7 1* 3 * 5 * 7
n = 1, 2, 3,......
Thus we can write the second solution for the differential equation as: ∞ (−1) n c0 ∴ y2 ( x) = c0 1 + ∑ xn n =11 * 3 * 5 * 7.....(2n − 1)
On the interval 0 < x < ∞ the general solution is: ∴ y ( x) = C1 y1 ( x) + C2 y 2 ( x)
321
Chapter Nine
Example 21 Solve the following differential equation:
2 x y ′′ + (1 + x ) y ′ − 2 y = 0 Solution: It is clear that this equation has singular point at x = 0 .
So the assumed solution will be in the following form: ∞
y=
∑ cm x m + r
(15)
m =0
Then y ′ and y ′′ can be obtained as in (16) and (17) respectively. Substitute these power series in the differential equation we get:
∴ 2 +
∞
∑
m=0 ∞
(m + r )(m + r − 1)cm x m + r −1 +
∑ ( m + r ) cm x
m+r
m=0
m=0
− 2 ∑ cm x m + r = 0 m=0
∑ (m + r )(2m + 2r − 1)cm x
m + r −1
m =0
∴ r (2r − 1)c0 x
∑ (m + r )cm x m + r −1
∞
∞
r −1
∞
+x
r
+
∞
∑ ( m + r − 2) c m x m + r = 0
m =0
∞
∑ (m + r )(2m + 2r − 1)cm x m −1
m 1 44442444443 1=4 n = m −1
+x
r
∞
∑ ( m + r − 2) c m x m = 0
m 0 442444 1=4 3 n=m
∞ ∴ x r r (2r − 1)c0 x −1 + x n ∑[(n + r + 1)(2n + 2r + 1)cn+1 + (n + r − 2)cn ] = 0 n=0
Which implies that r (2r − 1) = 0 ,
322 Power Series Solution Of Differential Equations
∴ (n + r + 1)(2n + 2r + 1)cn +1 + (n + r + 1)cn = 0 n = 0,1, 2, 3,......
∴ r1 =
1 , r2 = 0 2
For r1 =
1 : 2
3 − n c n 2 ∴ cn +1 = 3 n + ( 2 n + 2) 2 n = 2, 3, .....
c At n = 0 ∴ c1 = 0 2
c c , At n = 1∴ c2 = 1 = 0 20 40
At n = 2 ∴ c3 =
− c2 − c0 = 42 1680
At n = 4 ∴ c5 =
co − c4 =− 22 887040
, At n = 3 ∴ c4 =
− c3 co = 24 40320
Thus we can write
x x2 x3 x4 x5 ∴ y1 ( x) = c0 x1 / 2 1 + + − + − + −....... 2 40 1680 40320 887040 Now for r2 = 0
∴ cn +1 =
(2 − n )cn − (n − 2 )cn = n = 0, 1, 2, 3, ..... (2n + 1)(n + 1) (2n + 1)(n + 1)
At n = 0 c1 = 2co ,
c c At n = 1 ∴ c2 = 1 = o , At n = 2 c3 = 0 6 3 At n = 2 c4 = 0
323
Chapter Nine
∴ cn = 0, for n = 3,4,5,6,....
x2 ∴ y2 ( x) = c0 1 + 2 x + 3 On the interval 0 < x < ∞ the general solution is:
y ( x) = C1 y1 ( x) + C2 y 2 ( x) Case II If r1 − r2 = N , where N is a positive integer number.
When the roots of the indicial equation differ by a positive integer we may or may not be able to find two solutions of (3) having form (4). If not, then one solution corresponding to smaller root contains a logarithmic term. When the exponents are equal a second solution will always contain a logarithm. This latter situation is analogous to the solutions of the Caushy-Eulaer differential
equation when the roots of the auxiliary equation are equal. The solution of this kind of equation is shown in the following examples. Example 22 Solve the following differential equation:
x y ′′ + ( x − 6 ) y ′ − 3 y = 0 Solution: If y =
∞
∑ cm x m + r
(15)
m =0
Then y ′ and y ′′ can be obtained as in (16) and (17) respectively. Substitute these power series in the differential equation we get:
324 Power Series Solution Of Differential Equations ∞
∑
(m + r )(m + r − 1)cm x
m=0
−6
m + r −1
+
∞
∑ ( m + r ) cm x m + r
m=0
∞
∞
m=0
m=0
∑ (m + r )cm x m + r −1 − 3 ∑ cm x m + r = 0 ∞
r (r − 7)c0 x r −1 + x r ∑ (m + r )(m + r − 7)cm x m −1 m =4 1 4442444443 14 n = m −1
+ xr
∞
∑ (m + r − 3)cm x m = 0
m 0 442444 1=4 3 n=m r −1
∴ r ( r − 7 ) c0 x
+
∞ ∞ x r ∑ (n + r + 1)(n + r − 6)cn +1 + ∑ (n + r + 1)cn x n = 0 n=0 n = 0 Which implies that:
r (r − 7) = 0,
∴ r1 = 7, r2 = 0, r1 − r2 = 7 , and
(n + r + 1)(n + r − 6)cn +1 + (n + r − 3)cn = 0 n = 0,1, 2, 3,...... From the smaller root r2 = 0
∴ (n + 1)(n − 6)cn +1 + (n − 3)cn = 0 n = 0, 1, 2, 3,...... (3 − n ) c , n = 0, 1, 2, 3, 4,....... ∴ cn +1 = (n + 1)(n − 6) n c c 3 ∴ c1 = co , ∴ c2 = − 1 = 0 , 5 10 1 * (− 6) c c ∴ c3 = − 2 = − 0 , ∴ c4 = c5 = c6 = 0 , 12 120 (− 3) c = 0 = Undefined number. c7 = 6 7*0 0
325
Chapter Nine
This implies that co and c7 can be chosen arbitrarily. And for n ≥ 7 , ∴ c8 =
∴ c9 = −
−4 c7 8 *1
5 4*5 c8 = c7 9*2 2! 8 * 9
∴ c10 = −
, ,
6 − 4*5*6 c9 = c7 10 * 3 3! 8 * 9 * 10
,
(−) n +1 4 * 5 * 6......(n − 4) ∴ cn = c7 n = 8, 9, 10, ..... , (n − 7)! 8 * 9 * 10...n If we choose c7 = 0 and c0 ≠ 0 we obtain the polynomial solution:
1 1 3 1 ∴ y1 ( x) = c0 1 − x + x 2 − x , 2 10 120 But when c7 ≠ 0 and c0 = 0 , it follows that a second, through infinite series, solution is ∞ ( −1) n +1 4 * 5 * 6.....( n − 4) ∴ y2 ( x) = c7 x 7 + ∑ xn n = 8 ( n − 7)! 8 * 9 * 10.....n Finally, for x > 0 the general solution of the differential equation is
y ( x) = C1 y1 ( x) + C2 y 2 ( x) 1 1 3 1 y ( x) = C1 1 − x + x 2 − x 10 120 2 7 ∞ (−1) n +1 4 * 5 * 6.....(n − 4) n x + C2 x + ∑ n n − ( 7 )! 8 * 9 * 10 ..... n =8 It is interesting to observe that in the proceeding example the larger root r1 = 7 was not used.
326 Power Series Solution Of Differential Equations Example 23 Solve the following differential equation:
x y ′′ − ( 4 + x ) y ′ + 2 y = 0 ∞
Solution: If y =
∑ cm x m + r
(15)
m =0
Then y ′ and y ′′ can be obtained as in (16) and (17) respectively. Substitute these power series in the differential equation we get: ∞
(m + r )(m + r − 1)cm x m + r −1 − 4
∑
m=0 ∞
−
∑ ( m + r ) cm x
m+r
m =0 ∞
∞
∑ (m + r )cm x m + r −1
m =0
∞
+ 2 ∑ cm x m + r = 0 m=0
∑ (m + r )(m + r − 5)cm x
m=0
m + r −1
∞
∞
+2
∑ (m + r − 2)cm x m + r
=0
m=0
r (r − 5)c0 x r −1 + x r ∑ (m + r )(m + r − 5)cm x m −1 m =4 1 4442444443 14 n = m −1
−x
∞
r
∑ ( m + r − 2) c m x m = 0
m 0 442444 1=4 3
n=m r (r − 7)c0 x −1 ∞
(
)
∞ xr n = 0 + ∑ (n + r + 1)(n + r − 6)cn +1 + ∑ (n + r + 1)cn x n=0 n =0 So, Coefficient of x −1 r (r − 5) = 0, ∴ r1 = 5, r2 = 0, r1 − r2 = 5 , Coefficient of x n = 0
∴ (n + r + 1)(n + r − 4)cn +1 − (n + r − 2)cn = 0 n = 0,1, 2, 3,......
327
Chapter Nine
∴ cn +1 =
(n + r − 2) cn n = 0, 1, 2, 3,...... (n + r + 1)(n + r − 4)
At r2 = 0 ∴ cn +1 =
( n − 2) cn n = 0, 1, 2, 3,...... (n + 1)(n − 4)
At n = 0 , ∴ c1 =
− 2co co c − c1 = , At n = 1 , ∴ c2 = = o −4 2 2 * −3 4 * 3
At n = 2 , ∴ c3 =
(2 − 2)c2 3 * (− 2)
At n = 4 , c5 =
At n = 3 , ∴ c4 =
= 0,
2c4 0 = 5 * (4 − 4 ) 0
c3 =0 5 *1
Undefined number
Assume c5 = c5*
3c5* c5* 4c6 c5* At n = 5 , ∴ c6 = = = , At n = 6 , ∴ c7 = 6 *1 2 7*2 7 At n = 7 ,
5c7 5c5* = ∴ c8 = 8*3 8*7*3
6c8 5c5* At n = 8 , ∴ c9 = = 9 * 4 6 * 7 *8*3 At n = 9 , ∴ c10
7c8 c5* = = 10 * 5 10 * 6 * 8 * 3
x x2 x6 x7 5 x8 5x9 + + + + ... ∴ y ( x) = co 1 + + + c5* x 5 + 2 7 8 * 21 42 * 24 2 12 It is interesting to observe that in the proceeding example the larger root r1 = 5 was not used.
328 Power Series Solution Of Differential Equations Example 24 Solving the following differential equation by two
different methods. x 2 y ′′ + 5 x y ′ + 3 y = 0 Solution: It is clear the above differential equation in the form of 2
Cauchy differential equation: x y′′ + axy′ + by = 0 The auxiliary equation is m 2 + (a − 1)m + b = 0 , so substitute that in the differential equation we get: m 2 + 4m + 3 = 0
∴ m1 = −3,
m2 = −1 ,
∴ y ( x) = C1 x − 3 + C2 x −1
The second solution:
The above differential equation can be solved also by means of power series concept as following. It is clears the above differential equation has regular singular point at x = 0 , so the solution of the above differential equation has the form (15). Then y ′ and y ′′ can be obtained as in (16) and (17) respectively. Substitute (15), (16) and (17) in the differential equation we get: ∞
∞
∑
m=0
(m + r )(m + r − 1)cm x m + r + 5 ∑ (m + r )cm x m + r m =0
∞
+ 3 ∑ cm x m + r = 0
xr
m=0 ∞
∑ ((m + r )(m + r − 1 + 5) + 3)cm x m = 0
m =0
329
Chapter Nine
∞ r m ∴ x r (r + 4 ) + 3 + ∑ ((m + r )(m + r + 4) + 3)cm x = 0 m 1 4444 1=4 42444444 3 n=m ∴x
r
∞
∑ ((n + r )(n + r + 4) + 3)cn x n = 0
n =1
∴ r (r + 4 ) + 3 = 0, or , r 2 + 4r + 3 = 0 ∴ r1 = −1, and, r2 = −3 , It is clear that r1 − r2 = 2 This is the second case of the indicial equation, so, At r1 = −3 , ∴ x
∴ x −3
∞
−3
∞
∑ ((n + (− 3))(n + (− 3) + 4) + 3)cn x n = 0
n =1
∑ ((n − 3)(n + 1) + 3)cn x n = 0
m =1
∴ ((n − 3)(n + 1) + 3)cn
n =1, 2,3, 4,....
=0
At n = 1, ((1 − 3)(1 + 1) + 3)c1 = (− 2 * 2 + 3)c1 = 0 ∴ c1 = 0
0 0 Then, c2 is undefined number, so we can get the solution in terms At n = 2, ∴ ((2 − 3)(2 + 1) + 3)c2 = (− 1 * 3 + 3)c2 = 0, or , c2 =
of c2 . In the same way we can substitute for n=3, 4, 5, 6,…… to get the rest of constants, so we get the following values:
c1 = c3 = c4 = c5 = c6 = c7 = c8 = ............ = 0 Substitute the values of constants in the solution form we get:
(
)
∴ y ( x ) = x − 3 co + c2 x 2 , or y ( x ) = co x − 3 + c2 x −1 It is clear we get the same solution in both methods.
330 Power Series Solution Of Differential Equations Case III If r1 = r2 there always exist two linearity independent
solutions of equation (3) of the following form:
y1 ( x ) =
∞
∑ cm x m + r1 ,
m=0
c0 ≠ 0, and y2 ( x ) = y1 ( x) Ln x
(32)
Example 25 Solve the following differential equation
x( x − 1) y"+(3 x − 1) y '+ y = 0 Solution: It is clears the above differential equation has regular
singular point at x = 0 , so the solution of the above differential ∞
equation has the following form: y =
∑ cm x m + r
(15)
m=0
Then y ′ and y ′′ can be obtained as in (16) and (17) respectively. Substitute these power series in the differential equation we get: ∞
∑
(m + r )(m + r − 1)cm x m + r −
m=0 ∞
+ 3 ∑ (m + r )cm x m=0
m+r
−
∞
∞
∑ (m + r )(m + r − 1)cm x m + r −1
m=0
∑ (m + r )cm x
m + r −1
m=0
+
∞
∑ cm x m + r
=0
m =0
∞ ∴ x r ∑ ((m + r )(m + r − 1 + 3) + 1)cm x m m = 0 −
∞
m =0
∑ (m + r )(m + r − 1 + 1)cm x m −1 = 0
331
Chapter Nine
∞ r ∴ x ∑ ((m + r )(m + r + 2) + 1)cm x m m 0 44442444443 1=4 n=m ∞ 2 −1 m −1 − r co x − ∑ (m + r )(m + r − 1 + 1)cm x =0 =14444 1m4 42444444 3 n = m −1, or , m = n +1 The indicial equation is the coefficient of x r −1 which is r 2 co = 0 , ∴ r 2 = 0 , so we have double roots r1 = r2 = 0 ∞ ∞ x r ∑ ((n + r )(n + r + 2) + 1)cn x n − ∑ (n + r + 1) 2 cn +1 x n = 0 n =0 n = 0 Substitute in the above equation for r = 0 we get:
x
r
∑ [(n(n + 2) + 1)cn − (n + 1) 2 cn +1 ]x n = 0 ∞
n=0 ∞ r
∴x
∑ [(n + 1) 2 cn − (n + 1) 2 cn +1 ]x n = 0
n=0
∴ (n + 1) 2 cn − (n + 1) 2 cn +1 = 0 ∴ cn +1 = cn , or co = c1 = c2 = c3 = c4 = c5 = c6 = c6 = ....cn
[
]
∴ y1 ( x ) = co 1 + x + x 2 + x 3 + x 4 + x 5 + ........ 1 But, = 1 + x + x 2 + x 3 + x 4 + x 5 + ....... 1− x c ln ( x ) c So, y1 ( x ) = o , and y2 ( x ) = y1 ( x ) ln ( x ) = o 1− x 1− x 1 ∴ y ( x ) = C1 y1 ( x ) + C2 y 2 ( x ) = C1* + C2* ln ( x ) 1− x
(
)
332 Power Series Solution Of Differential Equations Example 26 Solve the following differential equation by two
different methods x 2 y ′′ + 3 x y ′ + y = 0 Solution: It is clear the above differential equation in the form of
Cauchy differential equation: x 2 y ′′ + axy ′ + by = 0 . The auxiliary equation is m 2 + (a − 1)m + b = 0 , so substitute that in the differential equation we get:
m 2 + 2m + 1 = 0 , ∴ m1 = m2 = −1 ∴ y ( x) = (C1 + C2 Ln x )x −1 The second solution:
The above differential equation can be solved also by means of power series concept as following. It is clears the above differential equation has regular singular point at x = 0 , so the solution of the above differential equation has the following form: y =
∞
∑ cm x m + r
(15)
m =0
Then y ′ and y ′′ can be obtained as in (16) and (17) respectively. Substitute these power series in the differential equation we get:
∴ +
∞
∑
m =0 ∞
(m + r )(m + r − 1)cm x
∑ cm x m + r
m=0
=0
m+r
∞
+ 3 ∑ ( m + r )cm x m + r m=0
333
Chapter Nine
∴x
r
∞
∑ ((m + r )(m + r − 1 + 3) + 1)cm x m = 0
m=0
∞ r m ∴ x r (r + 2) + 1 + ∑ ((m + r )(m + r + 2) + 1)cm x = 0 m 1 44442444443 1=4 n=m ∴x
r
∞
∑ ((n + r )(n + r + 2) + 1)cn x n = 0
n =1
∴ r (r + 2 ) + 1 = 0, or r 2 + 2r + 1 = 0 ∴ r1 = r2 = −1 It is clear that this is the third case of the indicial equation, so, At r = −1 , ∴ x
r
∞
∑ ((n − 1)(n + 1) + 1)cn x n = 0
n =1
It is clear that for any value for n, the expression
((n − 1)(n + 1) + 1) ≠ 0 , so it must be cn = 0 for n = 1, 2, 3, 4,..∞ ∴ y1 ( x ) = co x −1 ∴ y 2 ( x ) = c y1 ( x ) ln ( x ) = c* x −1 ln ( x )
(
)
∴ y ( x ) = C1 y1 ( x ) + C2 y 2 ( x ) = C1* + C2* ln ( x ) x −1
334 Power Series Solution Of Differential Equations 9.7 Bessel Differential Equation
The following equation is called Bessel’s differential equation,
x 2 y ′′ + x y ′ + ( x 2 − v 2 ) y = 0
(33)
The above equation is so important differential equation used in applied mathematics, physics and engineering. In solving (33) we shall assume v ≥ 0 , where in. Since we seek series solution of each equation about x = 0 , we observe that the origin is a regular singular point of Bessel s equation. 9.7.1 Solution of Bessel s Equation
If we assume y =
∞
∑ cm x m + r
(15)
m =0
Then y ′ and y ′′ can be obtained as in (16) and (17) respectively. Then, substitute that in the Bessel s differential equation we get: ∞
∑
m=0
(m + r )(m + r − 1)cm x
m+r
+
∞
∑ ( m + r )c m x m + r
m=0 ∞
+
∑ cm x m + r + 2 − v 2
m=0
(r 2 − r + r − v 2 )c0 x r + x
r
∑ [(m + r )(m + r − 1) + (m + r ) − v ∞
m =1
(
2
]c )
mx
m
∞
∑ cm x m + r
=0
m=0
+x
r
∞
∑ cm x m + 2 = 0
m=0
∞ ∞ ∴ (r 2 − v 2 )c0 x r + x r ∑cm (m + r ) 2 − v 2 x m + ∑cm x m+ 2 = 0 m=0 m=1
From above equation, we see that the indicial equation is:
Chapter Nine
335
r 2 − v2 = 0
(34)
So that the indicial roots are r1 = v and r2 = −v When r1 = v , (33) becomes, ∞ ∞ ∴ x v ∑ c m m ( m + 2v ) x m + ∑ c m x m + 2 = 0 m=0 m =1
∞ ∞ v +1 v m m+2 =0 ∴ (1 + 2v)c1 x + x ∑ c m m ( m + 2v ) x + ∑ c m x m 2 442444 0 1=4 3 m 1=42 43 n=m n=m+2 ∞ v n ∴ x (1 + 2v)c1 x + x ∑ (n + v) 2 − v 2 cn + cn − 2 = 0 n=2
(
)
Therefore, by the usual argument we can write: (1 + 2v ) c1 = 0 ,
(
)
∴ (n + v) 2 − v 2 cn + cn − 2 = 0 n = 2, 3,4,5,...... ∴ cn =
− cn − 2 2
(n + v) − v
2
= 0 n = 2, 3,4,5,......
(35)
The choice c1 = 0 in (35) implies c3 = c5 = c7 = .... = 0 , so for
n = 2, 4,6,8 .... , we fined, after letting n=2k, k=1,2,3,….., − c2 k − 2
, k = 1,2,3,4,...... 2 2 k (k + v) − co c0 − c2 = ∴ c2 = , c4 = , 2 2 4 2 1(1 + v) 2 * 2(2 + v) 2 * 1 * 2(1 + v)(2 + v) c0 − c4 c6 = 2 =− 6 , 2 * 3(3 + v) 2 * 1 * 2 * 3(1 + v)(2 + v)(3 + v)
∴ c2 k =
336 Power Series Solution Of Differential Equations
(−1) k co
∴ c2 k = , k = 1,2,3,.... 2 2k k! (1 + v)(2 + v)(3 + v)....(k + v)
(36)
Since this latter function possesses the convenient property
Γ(1 + v) = vΓ(v) , we can reduce the indicated product in the demonstrator of (36) to one term. For example:
Γ(1 + v + 1) = (1 + v)Γ(1 + v) Γ(1 + v + 2) = (2 + v)Γ(2 + v) = (2 + v)(1 + v)Γ(1 + v) Γ(1 + v + k ) = (k + v)(k − 1 + v).....(3 + v)(2 + v)(1 + v)Γ(1 + v) (k + v)(k − 1 + v).....(3 + v)(2 + v)(1 + v) =
Γ(1 + v + k ) Γ(1 + v)
Hence we can write (37) as:
c2 k =
(−1) k Γ(1 + v)co 2
2k
k! Γ(1 + v + k )
, k = 1,2,3,....
In the above formula co is still arbitrary, and since we are looking only for particular solution, it is convenient to choose:
co =
1 2 v Γ(1 + v)
So that finally: c2 k =
(−1) k 2 2k + v k! Γ(1 + v + k )
, k = 0,1,2,3,....
(37)
So substituting the above results into assumed solution for v ≥ 0 :
337
Chapter Nine
∴ y ( x) =
∞
∑ c2 k x 2 k + v
k =0
∞
(−1) k x ∴ y ( x) = ∑ ! ( 1 ) Γ + + k v k 2 k =0
2k + v
(38)
For each v , the function y ( x ) is called a Bessel function of the first kind of order v and is donated by the symbol J v (x) : ∞
(−1) k x J v ( x) = ∑ ! ( 1 ) Γ + + k v k 2 k =0
2k + v
(39)
Graphs of J o ( x ), and J1 ( x ) are shown in the following figure. Their resemblance to the graphs of cos x and sin x is interesting. In particular, they illustrate the important fact that for each value of v the equation J v (x) =0 has infinitely many roots.
Also, for the second exponent r2 = −v we obtain, in exactly the same manner, ∞
(−1) k x J − v ( x) = ∑ k v k ! ( 1 ) Γ − + 2 k =0
2k − v
(40)
338 Power Series Solution Of Differential Equations
The functions J v (x ) and J − v (x) are called Bessel functions of the
first kind of order v and –v, respectively, depending on the value of v, (38) may contains negative powers of x and hence converges on
0 < x < ∞ , when v is not an integer, J v ( x ), and J − v ( x ) are two linearly independent solutions of Bessel’s equation and this is a complete solution of Bessel’s equation when v.
y ( x) = C1 J v ( x) + C2 J − v ( x)
(41)
For many reason it is convenient to take the linear combination
Yv ( x) =
cos vπ J v ( x) − J − v ( x) sin vπ
(42)
Instead of J − v ( x ) as a second, independent solution of Bessel’s equation. Using Yv (t ) , which is known as the Bessel function of the second kind of order v, we can thus write a complete solution of
Bessel’s equation in the alternative form:
y ( x) = C1 J v ( x) + C2Yv ( x)
(43)
Where v is not an integer Example 27 Solve the following differential equation:
1 x 2 y ′′ + x y ′ + ( x 2 − ) y = 0 4
on 0
Solution: It is clear that v 2 = 1 / 4, and v = 1 / 2 .
r1 − r2 = 2v = 1= positive integer, so this is the second case and
v ≠ integer, so y ( x) = C1 J v ( x) + C2Yv ( x)
339
Chapter Nine
And, Yv ( x) =
cos vπ J v ( x) − J − v ( x) sin vπ
(44)
And the function J v ( x ) are linearly independent solution of the differential equation. Thus another form of the general solution of the differential equation: y ( x) = C1 J v ( x) + C2Yv ( x)
Yv ( x ) is sometimes called Neumann’s function or is called Bessel function of the second kind of order v. J1 / 2 ( x ) can be obtained by substituting in (39) for v = 1 / 2 we get:
J1 / 2 ( x ) =
2 sin x πx
In the same way we can get: J −1 / 2 ( x ) =
2 cos x πx
cos vπ J1 / 2 ( x) − J −1 / 2 ( x) sin vπ 2 2 π cos sin x − cos x π π 2 x x ∴ Y1 / 2 ( x) = π sin 2 2 ∴ Y1 / 2 ( x) = − cos x πx 2 2 ∴ y ( x) = C1 sin x + C2 cos x πx πx If v is an integer, say v = m, the situation is somewhat different. ∴ Y1 / 2 ( x) =
Again the roots of the indicial equation differ by an integer, namely, 2m, and it is to be expected that a second solution of the form (15) will not exist. In fact, when we consider J − v ( x ) as the limit of
340 Power Series Solution Of Differential Equations
J v ( x ) as v approaches − m and remember that the value of the
gamma function becomes infinite when its argument approaches any non-positive integer, then it follows that as v approaches − m , the first m terms in the series (40) approach zero and the series effectively begins with the term for which k = m : 2k − m
∞
(−1) k x (45) J − m ( x) = ∑ Γ − + ! ( 1 ) 2 k m k k =m In this, let the variable of summation be changed from k to j by the substitution k = j + m . 2 ( j + m )− m (−1) ( j + m ) x ∴ J − m ( x) = ∑ j = 0( j + m )! Γ (1 − m + ( j + m )) 2
∞
∞
(−1) j * (−1) m x ∴ J − m ( x) = ∑ ( ) ! ( 1 ) + Γ + j m j 2 j =0
2 j+m
∞
(−1) j x ∴ J − m ( x) = (−1) ∑ ( ) ! ( 1 ) 2 + Γ + j m j j = 0 123 1 424 3 = Γ ( j + m +1) !j m
∞
(−1) j j! Γ( j + m + 1)
∞
(−1) k k! Γ(k + m + 1)
2 j+m
2 j+m
x ∴ J − m ( x) = (−1) ∑ 2 j =0 In the above summation, assume the index j = k . m
∴ J − m ( x) = (−1)
m
∑
k =0
∞
x 2
2k + m
(−1) k x But we know that, J v ( x) = ∑ ! ( 1 ) Γ + + k v k 2 k =0
2k + v
Chapter Nine
341
So, J − m ( x) = (−1) m J m ( x )
(46)
Thus, when v is an integer, the function J − v ( x ) is proportional to
J v ( x ) . These two solutions are therefore not independent, and the linear combination c1 J v ( x ) + c2 J − v ( x ) is no longer a complete solution of Bessel’s equation. So the solution is not yet completely satisfactory, because the second solution is defined differently, depending on whether the order v is integer or not. To provide uniformity of formalism and numerical tabulation, it is desirable to adopt a form of the second solution which is valid for all values of the order. This is the reason for introducing a standard second solution Yv ( x ) defined for all v by the formula: (a) Yv ( x) =
cos vπ J v ( x) − J − v ( x) sin vπ
(47)
(b) Yn ( x ) = lim { Yv ( x )
(48)
v→n
This function is known as the Bessel function of the second kind of order v or Neumann’s function of order v. after some modification of equation (48) we can get the following equation for Yn ( x ) as following:
(− 1)m −1 (hm + hm + n ) 2m x ∑ 2m + n ( ) 2 ! ! + m m n m=0 (49) − n n −1 (n − m − 1)! x 2m x − ∑ 2m − n π
n x x Yn ( x ) = J n ( x ) ln + γ + π 2 π
2
∞
m=0
2
m!
342 Power Series Solution Of Differential Equations Where x > 0, n = 0,1, 2,.... and γ is the so called Euler constant and
γ = 0.577 215 664 9 and ho = 0, and hs = 1 +
1 1 1 + + .... + , (s = 1,2,....) s 2 3
(50)
And when n = 0 the last sum in (49) is to be replaced by 0. For n = 0 the representation (49) takes the form (51). Furthermore, it can be shown that Y− n ( x ) = (− 1)n Yn ( x ) m −1 2 x ∞ (− 1) hm 2m Y0 ( x ) = J 0 ( x ) ln + γ + ∑ x π 2 m =1 2 2m (m!)2
(51)
Example 28 Solve the following differential equation:
x 2 y ′′ + x y ′ + ( x 2 − 4 ) y = 0 Solution: It is clear that v 2 = 4,
on 0
and v1 = 2, v2 = −2
So v = N (i.e. v is integer) so the solution can be obtained as the following: y ( x ) = C1 J 2 ( x ) + C2 Y2 ( x ) 9.7.2 Recurrence Formulas For Bessel’s Function
J n +1 ( x ) = J n′ ( x ) =
2n J n ( x ) − J n −1 ( x ) x
1 [J n −1 (x ) − J n +1 (x )] 2
xJ n′ ( x ) = nJ n ( x ) − xJ n +1 ( x ) xJ n′ ( x ) = xJ n −1 ( x ) − nJ n ( x )
343
Chapter Nine
Example 29 Show that (a) J 3 / 2 ( x ) =
2 sin x − x cos x and x πx
(b) J − 3 / 2 ( x ) = − Solution: (a) Q J n +1 ( x ) =
2 x sin x + cos x πx x
2n J n ( x ) − J n −1 ( x ) x
1 1 ∴ J 3 / 2 ( x ) = J1 / 2 ( x ) − J −1 / 2 ( x ) x 2 2 2 Q J1 / 2 ( x ) = sin x and Q J −1 / 2 ( x ) = cos x πx πx 1 2 2 ∴ J 3 / 2 (x ) = sin x − cos x πx x πx Let n =
∴ J 3 / 2 (x ) =
2 sin x 2 sin x − x cos x − cos x = πx x πx x
(b) Q J n +1 ( x ) = Let n = −
2n J n ( x ) − J n −1 ( x ) x
1 1 ∴ J1 / 2 ( x ) = − J −1 / 2 ( x ) − J − 3 / 2 ( x ) 2 x
∴ J − 3 / 2 ( x ) = − J1 / 2 ( x ) −
Q J1 / 2 ( x ) =
1 J −1 / 2 ( x ) x
2 2 sin x and Q J −1 / 2 ( x ) = cos x πx πx
2 1 2 sin x − cos x πx x πx 2 cos x 2 x sin x + cos x ∴ J −3 / 2 (x ) = − sin x + =− πx x πx x ∴ J −3 / 2 (x ) = −
344 Power Series Solution Of Differential Equations Problems Solve the following differential equations by using power
series about x=0 ١)
y ′′ − xy ′ − y = 0
٢)
x 4 y′′ − xy ′ + 2 y = 0
٣)
(1 + x 2 ) y′′ − 2 y = 0
٤)
(1 + 2 x 2 ) y ′′ + 3 xy ′ − 6 y = 0
٥)
(1 − x 2 ) y ′′ − 10 xy ′ − 18 y = 0
٦)
3 x 2 y′′ + xy′ − (1 + x) y = 0
٧)
2 xy′′ + (1 + 2 x 2 ) y′ − xy = 0
٨)
2 xy′′ + (1 + 2 x) y′ + 4 y = 0
٩)
2 x 2 y′′ + x(4 x − 1) y′ + 2(3 x − 1) y = 0
١٠)
2 x(1 − x) y′′ + (1 − 2 x) y′ + 8 y = 0
١١)
2 x 2 y′′ − 3 x(1 − x) y′ + 2 y = 0
١٢)
x 2 y′′ + 3 xy′ + (1 − 2 x) y = 0
١٣)
xy′′ + (1 − x 2 ) y′ − xy = 0
١٤)
4 x 2 y ′′ + (3 x + 1) y = 0
١٥)
x 2 y′′ + x( x − 1) y′ + (1 − x) y = 0
١٦)
4 x 2 y′′ − 2 x(2 + x) y′ + (3 + x) y = 0
١٧)
x( x − 1) y ′′ − 3 y ′ + 2 y = 0
345
Chapter Nine
١٨)
1 x 2 y′′ + xy′ + ( x 2 − ) y = 0 9
١٩)
x 2 y′′ + xy′ + (9 x 2 − 4) y = 0
٢٠)
2 x 2 y ′′ + 2 x 2 + x y ′ − y = 0
٢١)
( ) 2 x 2 y ′′ + (2 x 2 − 3 x )y ′ + ( x + 2 ) y = 0
٢٢)
4 x 2 y ′′ + (4 x + 1) y = 0
٢٣)
xy ′′ + (1 + x ) y ′ + y = 0
٢٤)
x 2 y ′′ − xy ′ + (1 + x ) y = 0
٢٥)
9 x 2 y ′′ + 9 x 2 + x y ′ + (12 x − 1) y = 0
٢٦)
3 x 2 y ′′ + 3 x 2 + 5 x y ′ + (6 x − 1) y = 0
٢٧)
x 2 y ′′ + x 2 − x y ′ + y = 0
٢٨)
xy′′ + y′ + xy = 0
٢٩)
x 2 y ′′ + 3 xy ′ + (1 + x ) y = 0
٣٠)
x 2 y ′′ − x(2 − x ) y ′ + (2 − x ) y = 0
٣١)
x 2 y ′′ + xy ′ − 2 − x 2 y = 0
٣٢)
x 2 y ′′ − x(4 − x ) y ′ + (6 − 2 x ) y = 0
٣٣)
x 2 y ′′ + xy ′ − 6 − x 2 y = 0
٣٤)
x 2 y ′′ + 3 xy ′ + (1 + x ) y = 0
٣٥)
x 2 y ′′ − 2 x 2 y ′ + x 4 + x 2 − 6 y = 0
(
(
(
)
)
(
)
)
(
)
(
)
346 Power Series Solution Of Differential Equations
٣٦)
xy ′′ − y ′ + 4 x 5 y = 0
٣٩)
( ) x y ′′ + xy ′ + (x + 9 )y = 0 x ( x + 1) y′′ − (5x + 8x + 3)xy′ + (9 x
٤٠)
Find J1 / 2 and J −1 / 2
٣٧) ٣٨)
x 2 y ′′ + xy ′ − 6 − x 2 y = 0 2
2
2
2
2
2
)
+ 11x + 4 y = 0
Chapter 10 Partial Differential Equations 10.1 Definition “A partial differential equation (PDE) is any equation involving a function of more than one independent variable and at least one partial derivative of that function. The order of a PDE is the order of the highest order derivative that appears in the PDE” 10.2 Partial Derivatives Of Function Of Two Variables Let f be a function of two independent variables. Then the first partial derivatives of f with respect to x and y are the functions f x and f y defined by the following equations:
f ( x + ∆x, y ) − f ( x, y ) ∂f = lim = ∆x ∂x ∆x → 0 ∂f f ( x, y + ∆y ) − f ( x, y ) fy = = lim = ∂y ∆y → 0 ∆y
fx =
(1)
That is to say, to find f x , we keep y constant and differentiate with respect to x. Similarly we can define f y , we keep x constant and differentiate with respect to y. Example 1 Find f x and f y for the following functions: (a) f ( x, y ) = x 4 y 2 (b) f ( x, y ) = e x sin( 2 x + y )
347
Chapter Ten
Solution: (a) f x =
∂f = 4 x 3 y 2 , and ∂x
(b) f x =
∂f = e x sin( 2 x + y ) + 2e x cos(2 x + y ) ∂x
fy =
fy =
∂f = 2x4 y ∂y
∂f = e x cos(2 x + y ) ∂y
Example 2 Find f x and f y for the following function at point (2,1)
f ( x, y ) = cosh ( x − 6 y ) + e − xy Solution: (a) f x =
∴ fx = fy =
∂f = sinh ( x − 6 y ) − ye − xy ∂x
∂f = sinh (2 − 6) − e − 2 = − sinh 4 − e − 2 ∂x ( 2,1)
∂f = −6 sinh ( x − 6 y ) − xe − xy ∂y
∴ fy =
∂f = −6 sinh (−4) − 2e − 2 = 6 sinh 4 − 2e − 2 ∂y ( 2,1)
Example 3 If u = x y Solution:
Show that
x ∂u 1 ∂u + = 2u y ∂x Ln x ∂y
348
Partial Differential Equations
∂u x ∂u = y x y −1 , ∴ = xy = u ∂x y ∂x
(2)
∂u 1 ∂u = x y Ln x , ∴ = xy = u ∂y Ln x ∂y
(3)
Add (2) and (3) we get:
x ∂u 1 ∂u + = 2u y ∂x Ln x ∂y
10.3 Partial Derivatives of Function of More Than Two Variables A similar definition of the partial derivatives is given in three variables. Let f be a function of three variables (x, y, z), then we define, for instance
∂f f ( x + ∆x, y, z ) − f ( x, y, z ) = lim = ∆x ∂x ∆x → 0 f ( x, y + ∆y, z ) − f ( x, y, z ) ∂f = lim = fy = ∆y ∂y ∆y → 0 ∂f f ( x, y, z + ∆z ) − f ( x, y, z ) fz = = lim = ∂z ∆z → 0 ∆z
fx =
Example 4 If f ( x, y, z ) = sin 2 x * cos 4 y * sin 6 z Find: fx, f y, fz Solution: ∂f fx = = 2 cos 2 x * cos 4 y * sin 6 z ∂x ∂f fy = = 4 sin 2 x * (− sin 4 y ) * sin 6 z ∂y ∂f fz = = 6 sin 2 x * cos 4 y * cos 6 z ∂z
(4)
349
Chapter Ten
10.4 Total Differentials For function f = f ( x, y, z ) , the differential of this function df is expressed by the following form:
df =
∂f ∂f ∂f dx + dy + dz = f x dx + f y dy + f z dz ∂z ∂x ∂y
(5)
10.5 Partial Derivatives Of Higher Orders If f is a function of two variables f ( x, y ) , then its partial derivatives
f x and f y are also function of two variables, so we can consider their partial derivatives, f xx , f xy , f yx and f yy which is called the second partial derivatives of f .
f xx
f xy
∂ ∂f ∂ 2 f = = 2 ∂x ∂x ∂x ∂ ∂f ∂ 2 f = = ∂y ∂x ∂y∂x
(6) (7)
f yx
∂ ∂f ∂ 2 f = = ∂x ∂y ∂x∂y
(8)
f yy
∂ ∂f ∂ 2 f = = 2 ∂y ∂y ∂y
(9)
Thus the notation f xy or
∂2 f ∂ ∂f or means that we ∂y∂x ∂y ∂x
differentiate with respect to x then with respect to y. If the function
f xy and f yx are both continuous, then f xy = f yx , a similar
350
Partial Differential Equations
statement holds for functions of more than two variables. This result, known as cumulative property of partial derivatives, may be assumed to be true for all functions encountered at this stage. Example 5 Find all second partial derivatives of:
f ( x, y, z ) = xe y cos z Solution: f x = e y cos z ,
f y = xe y cos z and
f z = − xe y sin z
f xx = 0 ,
f yy = xe y cos z and
f zz = − xe y cos z
f xy = e y cos z
f xz = −e y sin z
f yz = − xe y sin z
10.6 The Chain Rule The Chain rule is used to provide a rule for differentiating a composite function. If y is a function of x and x is a differential function of t, then y is indirectly proportional function of t and:
dy dy dx = * dt dx dt For, y = Y ( x ), and , x = X (t ) Example 6 If y = 2 x 2 , and , x = 5t 2 Find
(10)
dy dt
Solution: From (10) we can get the following results
dy dy dx = * = 4 x * 10t = 40 xt dt dx dt dy = 200t 3 But, x = 5t 2 ∴ dx
351
Chapter Ten
Case1 Suppose that z = f ( x, y ) is a differentiable function of x and y, where x = X (t ) , y = Y (t ) are both differentiable function of t, then z is a differentiable function of t and,
∴
dz ∂ z d x ∂ z d y * * + = dt ∂x dt ∂y dt
Example 7 Suppose that z = x 2 y,
(11)
x = t3,
and
y = t2
Use the Chain rule to find dz / dt . Solution:
Q
dz ∂ z d x ∂ z d y + * * = dt ∂x dt ∂y dt = 2 xy * 3t 2 + x 2 * 2t = 2 t 3 t 2 * 3t 2 + t 6 * 2t = 8 t 7
Case 2 Suppose that z = f ( x, y ) , is a differential function of x and y, where x = X ( s, t ), y = Y ( s, t ) and partial derivatives xs , xt , y s and yt are exist. Then:
∂z ∂z ∂x ∂z ∂y + = ∂s ∂x ∂s ∂y ∂s
(12)
∂z ∂z ∂x ∂z ∂y + = ∂t ∂x ∂t ∂y ∂t
(13)
352
Partial Differential Equations
Example 8 If z = e x sin y , x = st 2 , y = s 2 t find
∂z ∂z and ∂s ∂t
Solution: From case 2 of Chain rule
∂z ∂z ∂x ∂z ∂y + = ∂s ∂x ∂s ∂y ∂s = (e x sin y ) t 2 + (e x cos y ) 2 st ∂z ∂z ∂x ∂z ∂y + = ∂t ∂x ∂t ∂y ∂t = (e x sin y ) 2 st + (e x cos y ) s 2 It can be easily to extend case 2 of Chain rules to the function f which contains three variables f = f ( x, y, z ) Where x = x( s, t ), y = y ( s, t ), z = z ( s, t ) . Then we have the extension of case 2 of Chain rules
∂f ∂f ∂x ∂f = + ∂s ∂x ∂s ∂y ∂f ∂f ∂x ∂f = + ∂t ∂x ∂t ∂y
∂y ∂f ∂z + ∂s ∂z ∂s ∂y ∂f ∂z + ∂t ∂z ∂s
(14) (15)
Example 9 If f = x 4 y + y 2 z 3 , where x = r s et , y = r s 2 e − t , and
y = r s 2 e − t . And find the value of Solution:
∴
∂f ∂f ∂x ∂f ∂y ∂f ∂z = + + ∂s ∂x ∂s ∂y ∂s ∂z ∂s
∂f where r = 2, s = 1, t = 0 ∂s
353
Chapter Ten
(
)( ) (
)(
)(
)(
∂f = 4 x 3 y ret + x 4 + 2 y z 3 2rse − t + 3 y 2 z 2 r 2 sin t ∂s
∴
)
When r = 2, s = 1, t = 0 we have x = 2, y = 2, z = 0 ∴
∂f = 64 * 2 + 16 * 4 + 0 * 0 =192 ∂s
10.7 Differentiation of Implicit Functions Suppose that the following function F ( x, y, z ) = 0 defines variable, say z, as a function of the other two variables x and y. Then z is sometimes called an implicit function of x and y, where z = f ( x, y ) . Then F ( x, y, f ( x, y )) = 0 . If F is differentiable, then we get the following rule
Fy F ∂z ∂z = − x and =− ∂x Fz Fz ∂y Example 10 If
x2 a2
+
y2 b2
+
z2 c2
(16)
= 1,
Solution: Let
x2 2
+
y2 2
+
z2 2
−1 = 0
a b c Fx ∂z 2x / a2 c2 x ∴ =− =− =− 2 ∂x Fz 2z / c2 a z Fy ∂z 2 y / b2 c2 y ∴ =− =− =− 2 2 ∂y Fz 2z / c b z
Find
∂z ∂z and ∂x ∂y
354
Partial Differential Equations
10.8 Applications In this section we will drive the differential equations which governs the conduction of heat in solids. In the following sections some
y
basic problems associated with
x
this equation are solved. In the case of heat conduction in two
x0
L
x 0 + ∆x
parallel plates of the same area A and different constant temperatures T1 and T2 respectively are separated by a small distance d, an amount of heat H per unit time will pass from the warmer to the cooler. Moreover, to a high degree of approximation, H is proportional to the area A, the temperature difference
T2 − T1 , and inversely proportional to the separation distance, d. Thus we can write:
kA T2 − T1 (17) d Where the positive proportionality factor k is called the thermal H=
conductivity and depends only on the material between the two plates. The physical law expressed by equation (17) is known as Newton’s law of cooling. Now consider a straight rod of uniform cross section and homogeneous material, oriented so x-axis lies along the axis of the rod. Let x = 0 and x = L designate the ends of the bar. We will assume that the sides of the bar are perfectly isolated so that there is
355
Chapter Ten
no passage of heat through them. We will also assume that the temperature u depends only on the axial position x and time t, and not on the lateral coordination y and z; hence u equals u ( x, t ) . Consider an element of the bar lying between the cross section
x = x0 and x = x0 + ∆x , where x = x0 is arbitrary and ∆x is so small. The instantaneous rate of heat transfer from left to the right across the cross section x = x0 is obtained by taking the limit in equation (17). As d → 0 and replacing
lim (T2 − T1 ) / d by
d →0
u x ( xo , t ) we get the following equation:
H ( x0 , t ) = −kA u x ( xo , t )
(18)
The minus sign appears in this equation since there will be positive flow of heat from left to right only if the temperature is greater to the left of x = x0 than to the right; in this case u x ( xo , t ) will be negative. In a similar manner the rate at which heat passes from left to right through the cross section x = x0 + ∆x is given by:
H ( x0 + ∆x, t ) = − kA u x ( x0 + ∆x, t )
(19)
The net rate of heat at which heat flows into the segment of the bar between x = x0 and x = x0 + ∆x is thus given by:
Q = H ( x0 , t ) − H ( x0 + ∆x, t ) = kA[ u x ( x0 + ∆x, t ) − u x ( x0 , t )] (20) The average change in temperature ∆u is proportional to the amount of heat Q ∆t introduced, and inversely proportional to the mass ∆m of the element. Thus;
356
Partial Differential Equations
∆u =
Q∆t 1 Q∆t = s ∆m sρ A∆x
(21)
Where the constant of proportionality s is known as the specific heat of the material of the bar, and ρ is its density. The average temperature change ∆u in the bar element under consideration is the temperature change at some intermediate point x = x0 + θ ∆x , where 0 < θ < 1. Thus (21) can be written as:
u ( x0 + θ ∆x, t + ∆t ) − u ( x0 + θ ∆x, t ) =
Q ∆t s ρ A ∆x
(22)
Solving equation (22) for Q and equating the resulting expression with the given by equation (20) yields the following equation:
u x ( x0 + ∆x, t ) − u x ( x0 , t ) = ∆x u ( x0 + θ ∆x, t + ∆t ) − u ( x0 + θ ∆x, t ) sρ A ∆t Ak
(23)
If we let ∆x and ∆t approach zero in equation (23), we obtain the heat conduction of diffusion equation:
α 2u xx = ut
(24)
The quantity α 2 is called the thermal diffusivity, and is a parameter depending only on the material of the bar and is defined by as following:
α2 =
k
ρs
(25)
357
Chapter Ten
Several relatively simple conditions may be imposed at the end of the bar. For example, the temperature at an end constant value T. At an end where this is done the boundary condition is:
u (0, t ) = T or
u ( L, t ) = T
(26)
Another simple boundary condition occurs if the end is insulated so that no heat passes through it. Recalling the expression (18) for the amount of heat crossing any cross section of the bar, the condition for insulation is clearly that this quantity vanishes. Thus the following condition is boundary condition at an insulated end:
ux = 0
(27A)
To determine completely the flow of heat in the bar it is necessary to state the temperature distribution at one fixed distant, usually taken as the initial time t = 0 this initial condition is of the form:
u ( x,0) = f ( x), 0 < x < L
(27B)
The problem then is to determine the solution of (24) subject to one or the other of (26) and (27A) boundary conditions at the ends, and the initial condition (27B). Example 11: State exactly the boundary value problem which determines the temperature in a copper bar 1 meter in length if the entire bar is originally at 20oC, and one end is then suddenly heated to 60oC, and held at that temperature while the other end is insulated. ( α 2 = 1.14 cm 2 / sec )
358
Partial Differential Equations
Solution: α 2 = 1.14 cm 2 / sec = 0.000114m 2 / sec From (24) we can say that: α 2u xx = ut
∴ 0.000114 u xx = ut and u ( x,0) = 20 , 0 < x < 1 ,
u (0, t ) = 60, t > 0 and u (1, t ) = 0, t > 0 10.9 Methods Of Solution Of The Partial Differential Equation In this section we are going to introduce some methods of solution of second order linear partial differential equation (28).
∂ 2u
∂ 2u ∂ 2u ∂u ∂u +C 2 +D +E + Fu = G A 2 +B ∂x∂y ∂x ∂y ∂x ∂y
(28)
Where A, B, ........, G may depend on x and y but not u. A second order equation with independent variable x and y which does not have the form (28) is called nonlinear. Some terms of the above equation can be eliminated depending on the physical application used. In the following there is some easy examples showing how to solve second order partial differential equations. Example 12 Solve the following partial differential equation:
∂2z = x 2 y where, z ( x,0) = x 2 , z (1, y ) = cos y ∂x∂y
359
Chapter Ten
Solution: The above partial differential equation can be written in the following form:
∂ ∂z = x 2 y ∂x ∂y
By integrating the above equation with respect to x we fiend:
∂z 1 3 = x y + F (y) ∂y 3
where F ( y ) is arbitrary.
Integrating the above equation with respect to y we get:
1 3 2 x y + ∫ F ( y )dy + G ( x ) where G ( x ) is arbitrary. 6 The above results can be written in the following form: z=
1 3 2 x y + H ( y ) + G(x ) 6 The above equation has two arbitrary functions and is therefore a z=
general solution. Since z ( x,0 ) = x 2 we have from the general solution
x 2 = 0 + H (0 ) + G ( x ) Or, G ( x ) = x 2 − H (0 )
∴z=
1 3 2 x y + H ( y ) + x 2 − H (0) 6
Since z (1, y ) = cos y , we have from above equation,
∴ cos y =
1 2 y + H ( y ) + 1 − H (0) 6
1 2 y − 1 + H (0 ) 6 1 1 ∴ z = x 3 y 2 + cos y − y 2 + x 2 − 1 6 6
∴ H ( y ) = cos y −
360
Partial Differential Equations
∂ 2u ∂u Example 13 Solve t = x2 +2 ∂x ∂x∂t Solution: Write the equation as
∂ ∂u + 2u = x 2 t ∂x ∂t
Integrating with respect to x, ∴ t
1 ∂u + 2u = x 3 + F (t ) ∂t 3
∂u 2 1 x 3 F (t ) ∴ + u= + ∂t t 3 t t This is a linear first order ordinary differential equation
∴u =e
− ln t 2
ln t 2 1 x 3 F (t ) dt + H ( x ) *∫ e + 3 t t
2 1 x 3 F (t ) dt + H ( x ) + ∴ t u = ∫ t 3 t t 2
1 ∴ t 2u = t 2 x 3 + G (t ) + H ( x ) 6 10.9.1 Separation of Variables In this section we will describe a method for solving the problem of heat conduction in a rod of uniform material with insulated lateral surface and will use the method of separation of variables to solve this initial value problem. “The idea behind the method of separation of variables is to convert the given partial differential equation into several ordinary differential equations”.
361
Chapter Ten
Suppose that the initial temperature distribution is given by a function f ( x) , and that the ends of the rod are held at zero temperature. Then we must solve the following partial differential equation:
α 2u xx = ut , 0 < x < L,
t>0
(29)
Subject to the following initial condition:
u ( x,0) = f ( x),
0< x
(30)
And the following boundary conditions:
u (0, t ) = 0, t > 0
(31A)
u ( L, t ) = 0, t > 0
(31B)
The choice of the boundary condition (31A) and (31B) may appear unnecessarily restrictive, however this problem is sufficiently general to fully illustrate the method of solution. The mathematical solution of equations (29) to (31) can be carried out by technique known as the method of separation of variables. This method is based on the idea of finding certain solutions of the differential equation (29) which are two functions one of the in x which is X ( x) and the another one in t which is
T (t ) . Then the general solution will take the following form:
u ( x, t ) = X ( x) T (t )
(32)
Where X is a function of x only, and T is a function of t only. Substituting (32) for u in the differential equation (29) yields
362
Partial Differential Equations
α 2 X ′′( x )T (t ) = X ( x )T ′(t )
(33)
The primes refer to ordinary differentiation with respect to the independent variable, whether x or t. Equation (33) is equivalent to:
X ′′( x ) 1 T ′(t ) = X ( x ) α 2 T (t )
(34)
In which the variables are separated in each side of equal sign, that is, the left hand side depends only on x and the right hand side only on t. In order for equation (34) to be valid for 0 < x < L, t > 0 , it is necessary that both sides of (34) be equal to same constant. Otherwise by keeping one independent variable (say x) fixed and varying the other, one side (the left in this case) of (34) would remain unchanged while the other varied, thus violating the equality. This constant Called separation constant − λ2 , then equation (34) becomes:
X ′′( x ) 1 T ′(t ) = 2 = −λ2 (35) X ( x ) α T (t ) From (35) we then obtain the following two ordinary differential equations for X ( x) and T (t ) :
X ′′( x ) + λ2 X ( x ) = 0
(36)
T ′(t ) + α 2 λ2T (t ) = 0
(37)
The partial differential equation (29) has thus been replaced by two ordinary differential equations (36) and (37). Equations (36) and (37) are second and first order linear homogeneous ordinary
Chapter Ten
363
differential equations respectively. Thus is the essence of the method of separation of variables. Equations (36) and (37) can be readily solved. According to the value of λ 2 we have two cases case 1, if λ2 > 0 or case 2, if
λ2 = 0 Case 1 If λ2 > 0 2 2 ∴ X ( x) = k1 sin λ x + k 2 cos λ x , and T (t ) = e −α λ t 2 2 ∴ u ( x, t ) = X ( x) T (t ) = e −α λ t (k1 sin λ x + k 2 cos λ x)
(38)
Is a solution of the partial differential equation (29) regardless of the values of λ and the arbitrary constants k1 and k 2 . Now from u (0, t ) = X (0 )T (t ) = 0 , so, T (t ) = 0, or X (0 ) = 0 . It is clear that X (0) = 0 otherwise it gives trivial solution. So, X (0 ) = 0 is one boundary condition. Similarly from (31B)
u (L, t ) = X (L ) * T (t ) = 0 ∴ X (L ) = 0 is the second boundary condition. The constants k1 , k 2 and λ can be partially determined from the boundary condition (31). The first boundary condition requires that 2 2 u (0, t ) = k 2e −α λ t = 0 therefore k 2 equals 0. The second boundary 2 2 condition, u ( L, t ) = k1e −α λ t sin λ t = 0 will be satisfied if sin λ t
equal 0, and hence if λ equals nπ / L , where n is a positive integer.
364
Partial Differential Equations
The allowable value of the previously arbitrary parameter λ has thus been determined by the boundary conditions. Hence any function of the form (39) will satisfy the boundary condition (31) as well as the partial differential equation (29). Moreover, since both the partial differential equation (29) and the boundary condition (31) are linear and homogeneous, any finite sum of such functions will also satisfy them. The function given by (39) is sometimes known as fundamental solution. 2 2 2 2 nπx u ( x, t ) = cn e − n π α t / L sin , L
n = 1,2,3,......
(39)
Where cn is arbitrary constants corresponding to the choice of n. Case 2 λ2 = 0 , from (36) ∴ X ′′( x ) = 0 ∴ X ( x ) = c1 x + c2 So, from the boundary condition X (0 ) = 0 ⇒ c2 = 0 ∴ X ( x ) = c1 x And, X (L ) = 0 , then 0 = c1L ⇒∴ c1 = 0 Which is trivial solution. Example 14 Let f ( x) = 3 sin
∴ u n ( x,0) = cn sin
4π x L
nπ x = f ( x) L
Provided n=4 and cn =3. The solution to the complete boundary value problem is: 2 2 2 4πx u ( x, t ) = 3e −16 π α t / L sin = 0, L
n = 1,2,3,......
365
Chapter Ten
Example 15: Use the method of separation of variables to solve the following partial differential equation:
∂u ∂ 2u , = ∂t ∂x 2 where u x (0, t ) = 0, u (2, t ) = 0 , and, u ( x,0 ) = 8 cos
3πx 9πx − 6 cos 4 4
Solution: Assume u ( x, t ) = X ( x) T (t )
∂u ∂ 2u = Q ∂t ∂x 2
∴
∴ X ( x )T ′(t ) = X ′′( x )T (t )
X ′′( x ) T ′(t ) = = −λ2 X ( x ) T (t )
∴ X ′′( x ) + λ2 X ( x ) = 0
(40)
∴ T ′(t ) + λ2T (t ) = 0
(41)
By solving equation (40) we get the following:
X ( x ) = k1 cos λx + k 2 sin λx By solving equation (41) we get the following:
T (t ) = ce − λ
2
t
Q u ( x, t ) = X ( x) T (t ) ∴ u ( x, t ) = e − λ
2
t
∴ u x ( x, t ) = e − λ
(K1 cos λx + K 2 sin λx ) 2
t
(− K1λ sin λx + K 2λ cos λx )
366
Partial Differential Equations
∴ u x ( x, t ) = λe − λ
2
t
(K 2 cos λx − K1 sin λx )
2
t
(K 2 ) = 0
Q u x (0, t ) = 0
∴ u x (0, t ) = λe − λ ∴ K2 = 0
Q u (2, t ) = 0
∴ u (2, t ) = e − λ ∴ 2λ =
2
t
( K1 cos 2λ ) = 0
nπ for n=1, 3, 5,…. 2
nπ for n=1, 3, 5,…. 4 2 nπ ∴ u ( x, t ) = e − λ t K1 cos x 4 nπ ∴ u ( x,0) = K1 cos x 4 3πx 9πx Q u ( x,0 ) = 8 cos − 6 cos 4 4
∴λ =
By comparing the above equations we get the following:
nπ 9πx 3πx − 6 cos x = 8 cos 4 4 4 From the first term we get the following:∴ K1 = 8 and n = 3
∴ K1 cos
From the second term we get the following:∴ K1 = −6 and n = 9
∴ u ( x, t ) = 8e
−
9 2 π t 3π 16 cos
81
2
− π t 9π x − 6e 16 cos x 4 4
367
Chapter Ten
Example 16: Use the method of separation of variables to solve the following partial differential equation: (b)
∂2 y ∂t 2
=4
∂2 y ∂x 2
, where y (0, t ) = 0, y ( x,0 ) = 0, yt ( x,0 ) = 5 sin πx
Solution: Assume u ( x, t ) = X ( x) T (t )
Q
∴
∂2 y ∂t
2
=4
∂ 2u ∂x
2
∴ X ( x )T ′′(t ) = 4 X ′′( x )T (t )
X ′′( x ) T ′′(t ) = = − λ2 X ( x ) 4T (t )
∴ X ′′( x ) + λ2 X ( x ) = 0
(42)
∴ T ′(t ) + 4λ2T (t ) = 0
(43)
By solving equation (42) we get the following:
X ( x ) = k1 cos λx + k 2 sin λx By solving the equation (43) we get the following:
T (t ) = k3 cos 2λt + k 4 sin 2λt
Q u ( x, t ) = X ( x) T (t ) ∴ y ( x, t ) = (k1 cos λx + k 2 sin λx ) * (k3 cos 2λt + k 4 sin 2λt )
Q y (0, t ) = 0 ∴ y (0, t ) = (k1 ) * (k3 cos 2λt + k 4 sin 2λt ) = 0 It is clear that k1 = 0 otherwise (k3 cos 2λt + k 4 sin 2λt ) which is not valid. So k1 = 0 .
Q y ( x,0 ) = 0 ∴ y ( x,0 ) = (k 2 sin λx ) * (k3 ) = 0
368
Partial Differential Equations
∴ k3 = 0 ∴ y ( x, t ) = (k 2 sin λx ) * (k 4 sin 2λt )
∴ yt ( x, t ) = (k 2 sin λx ) * (k 4 * 2λ * cos 2λt ) Q yt ( x,0 ) = 5 sin πx
∴ yt ( x,0) = (k 2 sin λx ) * (k 4 * 2λ ) = 5 sin πx ∴ λ = π ∴ (k 2 ) * (k 4 * 2π ) = 5
∴ k2 k4 =
5 2π
∴ y ( x, t ) =
5 (sin πx * sin 2π t ) 2π
Example 17: Use the method of separation of variables to solve the following partial differential equation:
∂2 y
(c)
∂t
2
=4
∂2 y ∂x
2
,
where
yt ( x,0 ) = 3 sin 2πx − 2 sin 5πx Assume u ( x, t ) = X ( x) T (t )
Q
∂2 y ∂t 2
=4
∂ 2u ∂x 2
∴ X ( x )T ′′(t ) = 4 X ′′( x )T (t )
∴
X ′′( x ) T ′′(t ) = = −λ2 X ( x ) 4T (t )
y (0, t ) = 0, y ( x,0 ) = 0,
and
Chapter Ten
369
∴ X ′′( x ) + λ2 X ( x ) = 0
(44)
∴ T ′(t ) + 4λ2T (t ) = 0
(45)
By solving equation (44) we get the following:
X ( x ) = k1 cos λx + k 2 sin λx By solving equation (45) we get the following:
T (t ) = k3 cos 2λt + k 4 sin 2λt
Q u ( x, t ) = X ( x) T (t ) ∴ y ( x, t ) = (k1 cos λx + k 2 sin λx ) * (k3 cos 2λt + k 4 sin 2λt )
Q y (0, t ) = 0 ∴ y (0, t ) = (k1 ) * (k3 cos 2λt + k 4 sin 2λt ) = 0 It is clear that k1 = 0 otherwise (k3 cos 2λt + k 4 sin 2λt ) which is not valid. So k1 = 0 .
Q y ( x ,0 ) = 0 ∴ y ( x,0 ) = (k 2 sin λx ) * (k3 ) = 0
∴ k3 = 0
∴ y ( x, t ) = (k 2 sin λx ) * (k 4 sin 2λt ) ∴ yt ( x, t ) = (k 2 sin λx ) * (k 4 * 2λ * cos 2λt )
Q yt ( x,0 ) = 3 sin 2πx − 2 sin 5πx ∴ yt ( x,0 ) = (k 2 sin λx ) * (k 4 * 2λ ) = 3 sin 2πx − 2 sin 5πx Then for the first term of yt ( x,0 ) , ∴ λ = 2π
∴ (k 2 ) * (k 4 * 4π ) = 3 ∴ k 2 k 4 =
3 4π
Then for the second term of yt ( x,0 ) ,
∴ λ = 5π
370
Partial Differential Equations
−2 10π 3 (sin 2πx * sin 4π t ) − 1 (sin 5πx * sin 10π t ) ∴ y ( x, t ) = 4π 5π ∴ (k 2 ) * (k 4 * 10π ) = −2 ∴ k 2 k 4 =
10.9.2 Solution Using Fourier Series Let f ( x) = b1 sin
πx L
+ ........ + bm sin
mπ x L
Where b1 ,......, bm are given constants. We will use the principle of superposition to solve this more complicated problem. If we choose cn = bn in (39), then we obtain a function which assumes the initial value bn sin
nπ x and also L
satisfies the partial differential equation (29) and boundary conditions (31A) and (31B). By adding the solution corresponding to n = 1,2,...m we obtain the solution. m
nπx − n 2π 2α 2 t / L2 sin .... b e ∑ n L n =1 Which satisfies the desired initial condition. u ( x, t ) =
(46)
In order to satisfy the initial condition (30) we must have:
u ( x , 0) =
m
∑ bn sin
n =1
nπx = f ( x).... L
(47)
That is, a Fourier sine series. This can be done provided f ( x ) satisfies the conditions of the Fourier theorem for 0 < x < L , and is
371
Chapter Ten
defined outside the interval (0, L ) as an odd function of period 2 L . Assuming that f ( x) does have Fourier series, the coefficients bn are given by the following equation:
2L nπx bn = ∫ f ( x) sin dx L0 L
(48)
To summarize the solution of the heat conduction problem (29), (30), (31);
α 2u xx = ut , 0 < x < L, t > 0 u ( x,0) = f ( x), 0 < x < L....
u (0, t ) = u ( L, t ) = 0,
t>0
is given by the following equation:
u ( x, t ) =
∞
2
∑ bn e − n π
2
α 2 t / L2
n =1
sin
nπx .... L
(49)
Example 18 Consider the conduction of heat in the copper rod of 100 cm in length whose ends are maintained at 0 o C for all
t > 0 . Find an expression for the temperature u ( x, t ) if the initial temperature distribution in the rod is given by: (a) u ( x,0) = 50
0 < x < 100;
x (b) u ( x,0) = 100 − x Where ( α 2 = 1.14 cm 2 / sec )
0 < x < 50, 50 < x < 100
372
Partial Differential Equations
Solution: From equation (49)
∴ u ( x, t ) =
∞
2
∑ bn e − n π
2
α 2 t / L2
sin
n =1
nπx .... , L
α 2 = 1.14 cm 2 / sec , L=100cm (a) u ( x,0) = 50
0 < x < 100;
If we draw this function as odd function outside the interval (0, L ) of period 2 L it will be as shown in the following figure:
50 L=100 cm
-L=-100 cm
The Fourier coefficients of the above waveform is given by
2L 2 100 nπx nπx 50 sin bn = ∫ f ( x) sin dx = dx ∫ 100 0 L0 L 100 100
100 nπx = − cos nπ 100 0 =
=
100(1 − cos nπ ) nπ
200 nπ n =1,3,5,7,..........
u( x, t ) =
200 −π 2 1.142 t / 10000 πx 1 −9π 2 1.142 t / 10000 3πx sin sin + e e π L L 3 2 2 1 5πx + e−25π 1.14 t / 10000sin + .... L 5
373
Chapter Ten
(b) If we draw this function as odd function outside the interval
(0, L ) of period 2L it will be as shown in this figure: 50 -L=-100 cm
L=100 cm
x
2L nπx bn = ∫ f ( x) sin dx L0 L 100 2 50 nπx nπx = dx + ∫ (100 − x) sin dx ∫ x sin 100 0 100 100 50
=
400 sin( nπ / 2)
n 2π 2 2 2 2 2 400 3πx πx 1 u ( x, t ) = 2 e −π 1.14 t / 10000 sin − e − 9 π 1.14 t / 10000 sin L 9 L π 2 2 1 5πx + e − 25π 1.14 t / 10000 sin + ..... 25 L Example 19 Solve the following partial differential equation:
∂u ∂ 2u u x (0, t ) = u x (π , t ) = 0, u ( x,0 ) = π (1 − x ) , = ∂t ∂x 2 where 0 < x < π , t > 0 Solution: Assume u ( x, t ) = X ( x ) * T (t )
374
Partial Differential Equations
∂u ∂ 2u Q = 2 ∴ X ( x ) * T ′(t ) = X ′′( x ) * T (t ) ∂t ∂x X ′′( x ) T ′(t ) ∴ = = −λ2 X ( x ) T (t )
∴ X ′′( x ) + λ2 X ( x ) = 0
(50)
∴ T ′(t ) + λ2T (t ) = 0
(51)
By solving equation X ′′( x ) + λ2 X ( x ) = 0 we get the following:
X ( x ) = k1 sin λx + k 2 cos λx By solving equation T ′(t ) + λ2T (t ) = 0 we get the following:
T (t ) = ce − λ
2
t
Q u ( x, t ) = X ( x) T (t ) ∴ u ( x, t ) = e − λ
(K1 sin λx + K 2 cos λx ) 2 ∴ u x ( x, t ) = λe − λ t (K1 cos λx − K 2 sin λx ) 2 Q u x (0, t ) = 0 ∴ u x (0, t ) = λe − λ t (K1 ) = 0 ∴ K1 = 0 2 ∴ u ( x, t ) = K 2 e − λ t (cos λx ) 2
t
(52)
It is clear from (52) that we need to compare with cosine terms, so, assume u ( x,0 ) is even function. So, bn = 0 . π
−π
u ( x ,0 )
π
375
Chapter Ten
∴ ao =
∴ an = ∴ an =
1π
π
∫ π (1 − x )dx = π −
2
0
2π
π
π2
∫ π (1 − x ) cos
0
nπx
π
dx
2π
π
∫ π (1 − x ) cos nxdx
0
u = (1 − x )
dv = cos nx
du = −dx
v=
sin nx n
(1 − x ) sin nx π 1 π + ∫ sin nxdx ∴ an = 2 n n0 0 π 1 2 4 ∴ an = 2 2 (− cos nx ) = 2 (1 − cos nx ) = 2 n n 0 n 1 π2 + 4cos x + cos 3x + ∴ u ( x,0 ) = π − 2 9
+ ∴ u ( x ,0 ) =
n = 1, 3, 5,....
1 1 cos 5 x + cos 7 x + ...... 25 49
∞
∑ k 2 cos λx
λ =0
∴ u ( x, t ) = π −
π2
1 + 4e − t cos x + e − 9t cos 3x + 2 9 1 1 + e − 25t cos 5 x + e − 49t cos 7 x + .... 25 49
376
Partial Differential Equations
Example 20 Use the method of Fourier series to solve the following boundary value problems:
∂u ∂ 2u (a) = 2 2 , u (0, t ) = u (4, t ) = 0, u ( x,0 ) = 25 x ∂t ∂x where 0 < x < 4, t > 0 Solution: Assume u ( x, t ) = X ( x ) * T (t )
∂u ∂ 2u Q = 2 2 ∴ X ( x ) * T ′(t ) = 2 X ′′( x ) * T (t ) ∂t ∂x X ′′( x ) T ′(t ) ∴ = = −λ2 X ( x ) 2T (t ) ∴ X ′′( x ) + λ2 X ( x ) = 0
(53)
∴ T ′(t ) + 2λ2T (t ) = 0
(54)
By solving equation (53) we get the following:
X ( x ) = k1 sin λx + k 2 cos λx . By solving equation (54) we get the following:
T (t ) = ce − 2λ
2
t
Q u ( x, t ) = X ( x) T (t )
∴ u ( x, t ) = e − 2 λ
Q u (0, t ) = 0
∴ u (0, t ) = e − 2λ
Q u (4, t ) = 0 ∴ u (4, t ) = e − 2λ
2
t
2
t
(K1 sin λx + K 2 cos λx )
2
t
(K 2 ) = 0
(K1 sin 4λ ) = 0
∴λ =
∴ K2 = 0
nπ 4
377
Chapter Ten
∴ u ( x, t ) = e
−
n 2π 2 t 8 K sin 1
Qu ( x,0) = 25 x = K1 sin
nπ 4
x
nπ x 4
(55)
It is clear from (55) that we need to compare with sine terms so we will complete u ( x,0 ) as an odd function as shown in the figure: u ( x ,0 ) 100
x
4 4
2L nπx ∴ bn = ∫ 25 x sin dx L0 L
50 4 nπx x sin dx ∴ bn = ∫ 4 0 4
4 25 4 2 nπx nπx 4 sin x cos ∴bn = − 2 n 2π 2 4 nπ 4 0
∴bn =
− 200 cos nπ nπ
∴bn =
− 200 , n = 2, 4, 6, ...... nπ
∴bn =
200 , n = 1, 3, 5, ...... nπ
378
Partial Differential Equations
∴ u ( x ,0 ) =
200 πx 1 2πx 1 3πx 1 4πx + sin − sin + ...... sin − sin π 4 2 4 3 4 4 4
9π 2 π2 −π 2 t − t − t πx 1 2 200 8 2πx 1 3πx 8 e ∴ u ( x, t ) = + e sin − e sin sin π 4 2 4 3 4 2 1 4πx − e − 2π t sin + ...... 4 4
Example 21 Use the method of Fourier series to solve the following boundary value problems:
∂u ∂ 2u = 2 u x (0, t ) = u x (π , t ) = 0, u ( x,0) = x 2 ∂t ∂x where 0 < x < π , t > 0 Solution: Assume u ( x, t ) = X ( x ) * T (t )
∂u ∂ 2u Q = ∂t ∂x 2
∴ X ( x ) * T ′(t ) = X ′′( x ) * T (t ) ∴
X ′′( x ) T ′(t ) = = −λ2 X ( x ) T (t )
∴ X ′′( x ) + λ2 X ( x ) = 0
(56)
∴ T ′(t ) + λ2T (t ) = 0
(57)
By solving equation (56) we get the following:
X ( x ) = k1 sin λx + k 2 cos λx .
379
Chapter Ten
By solving equation (57) we get the following:
T (t ) = ce − λ
2
t
Q u ( x, t ) = X ( x) T (t ) ∴ u ( x, t ) = e − 2 λ
2
t
(K1 sin λx + K 2 cos λx )
∴ u x ( x, t ) = λe − 2λ
2
t
(K1 cos λx − K 2 sin λx )
Q u x (0, t ) = λe − 2λ
2
t
(K1 ) = 0 ∴ K1 = 0
∴ u x (π , t ) = λe − 2λ
2
t
(− K 2 sin λπ ) = 0
∴λ = n ∴ u ( x, t ) = e − 2 λ
2
t
(K 2 cos nx )
(58)
It is clear from (58) that we need to compare with cosine terms so
u ( x ,0 ) = x 2 must be completed as even function as shown in the following figure:
u ( x ,0 ) = x 2
−π
π
x
380
Partial Differential Equations
1 L ∴ ao = f ( x )dx 2 L −∫L 1 L ∴ an = ∫ f ( x ) cos nx dx L −L
1π
π3 π2 ∴ ao = ∫ x dx = = 3π 3 π0 2
∴ an =
2π
x π 0∫
2
cos nx dx
π
2 2x 2 x2 ∴ an = 2 cos nx − 3 sin nx + sin nx π n n n 0
∴ an = ∴ an = ∴ an =
2 2π cos nx π n 2 4
cos nπ
n2 4
for n = 2, 4,6,....
n2
∴ an = −
4
for n = 1,3,5,....
n2
∴ u ( x, o ) =
π2
1 1 + 4 − cos x + cos 2 x − cos 3x + ..... 3 4 9
∴ u ( x, t ) =
π2
1 1 + 4 − e − t cos x + e − 4t cos 2 x − e − 9t cos 3x + ..... 3 4 9
381
Chapter Ten
10.9.3 Solution Using Laplace Transform Example 22 Use the method of Laplace transform to solve the
∂u ∂ 2u following boundary value problem: =4 2 ∂t ∂x Where u (0, t ) = 0, u (3, t ) = 0, u ( x,0 ) = 10 sin 2πx − 6 sin 4πx Solution: Taking the Laplace transform of the given differential equation with respect to t, we have: ∞ − st ∂u e dt 0 ∂t
∫
2 ∞ − st ∂ u e 4 2 dt 0
=∫
∂x Which can be written as:
∞ − st e udt 0
s∫
− u ( x,0 ) = 4
Or sU − u ( x,0 ) = 4
∂2
∞ − st
∫ ∂x 2 0
e
u dt
∂ 2U
(59)
∂ x2 ∞ 0
where U = U ( x, s ) = L{u ( x, t )} = ∫ e − st udu
Using the given condition u ( x,0 ) = 10 sin 2πx − 6 sin 4πx becomes: 4
∂ 2U ∂x 2
− sU = 6 sin 4πx − 10 sin 2πx
(60) (61)
Taking the Laplace Transform of the conditions
u (0, t ) = 0, u (3, t ) = 0 , we have: L{u (0, t )} = 0, Or U (0, s ) = 0,
U (3, s ) = 0
L{u (3, t )} = 0 (62)
Solving the ordinary differential equation (61) subject to condition (62) by the usual elementary methods as following:
382
Partial Differential Equations
4λ2 − s = 0 ∴ λ1 =
s / 2 and λ2 = − s / 2
∴ U h ( x, s ) = C1e
s /2 x
+ C2e −
s /2 x
∴ U p ( x, s ) = A cos 4πx + B sin 4πx + D cos 2πx + F sin 2πx
∴U ′p ( x, s ) = −4π A sin 4πx + 4π B cos 4πx − 2πD sin 2πx + 2πF cos 2πx ∴U ′p′ ( x, s ) = −4 2 π 2 A cos 4πx − 4 2 π 2 B sin 4πx − 2 2 π 2 D cos 2πx − 2 2 π 2 F sin 2πx
(
∴ 4 * − 42π 2 A cos4πx − 42π 2 B sin 4πx − 22π 2 D cos2πx − 22π 2 F sin 2πx − s( A cos4πx + B sin 4πx + D cos2πx + F sin 2πx) = 6 sin 4πx − 10sin 2πx By comparing coefficients in both sides we get:
−6
A = 0 , D = 0 ∴ − 64π 2 B − sB = 6 ∴ B = And − 16 π 2 F − sF = −10 ∴ F = ∴U ( x, s ) = U h + U P = C1e
s /2x
s + 64π 2
10 s + 16π 2
+ C2 e −
s /2 x
But from (62) U (0, s ) = 0,
+
6 sin 2πx s + 16π 2
∴ C1 + C2 = 0 And, U (3, s ) = 0 ∴ C1e3
s /2
+ C2e −3
But, C 2 = −C1
s /2
+0=0
∴ C1 e3
s /2
)
− e −3
s /2
=0
−
6 sin 4πx s + 64π 2
383
Chapter Ten
Then must be C1 = 0 ∴ C2 = 0 10 sin 2πx 6 sin 4πx ∴ U ( x, s ) = − s + 16π 2 s + 64π 2 By taking the inverse Laplace transform we find:
u ( x, t ) = 10e −16π
2
t
sin 2πx − 6e − 64π
2
t
sin 4πx
Which is the required solution:
∴ U ( x, s ) =
10 sin 2πx
−
6 sin 4πx
s + 16π 2 s + 64π 2 By taking the inverse Laplace transform we find the required
solution as following:
u ( x, t ) = 10e −16π
2
t
sin 2πx − 6e − 64π
2
t
sin 4πx
Example 23 Use the method of Laplace Transforms to solve the following partial differential equation s:
∂u ∂ 2u , = ∂t ∂x 2
u ( x,0 ) = 6 sin
where
πx 2
u (0, t ) = 0, u (4, t ) = 0 ,
and
+ 3 sin πx
Solution: As explained in the previous example
∴ sU − u ( x,0) =
∂ 2U ∂ x
(63)
2 ∞ 0
Where U = U ( x, s ) = L{u ( x, t )} = ∫ e − st udu Using the given condition becomes:
u ( x ,0 ) = 6 sin
πx 2
+ 3 sin πx , (63)
384
∂ 2U ∂x 2
Partial Differential Equations
− sU = 6 sin
πx 2
+ 3 sin πxx
(64)
Taking the Laplace Transform of the conditions:
u (0, t ) = 0, u (4, t ) = 0 , we have:
L{u (0, t )} = 0,
L{u (4, t )} = 0
Or U (0, s ) = 0,
U (4, s ) = 0
(65)
Solving the ordinary differential equation (64) subject to condition (65) by the usual elementary methods as following:
λ2 − s = 0
∴ λ1 = s and λ2 = − s
∴ U h ( x, s ) = C1e s x + C 2 e − s x
∴ U p ( x, s ) = A cos
∴U ′p ( x, s ) = −
π
π 2
x + B sin
A sin
π
x+
π
π 2
x + D cos πx + F sin πx
B cos
π
2 2 2 2 − πD sin πx + πF cosπx
∴U ′p′ ( x, s ) = −
π2
π2
π
x
π
A cos x − B sin x 4 2 4 2
− π 2 D cosπx − π 2 F sinπx −
π2
A cos
π
x−
π2
B sin
π
x − π 2 D cos πx − π 2 F sin πx
4 2 4 2 π π − s A cos x + B sin x + D cos πx + F sin πx 2 2 πx = −6 sin − 3 sin π 2
385
Chapter Ten
π
By comparing the coefficient of cos
x and cos πx we get the
2
following constants: A = 0, D = 0
π
By comparing the coefficient of sin following constants: − ∴ B=
π2
x and sin πx we get the
2
B − sB = −6
4
6
and − π 2 F − sF = −3 π2 s+ 4
∴U ( x, s ) = U h + U P = C1e +
s /2x
6 s +π 2 4
+ C2 e −
sin
πx 2
+
s /2 x
3 s +π 2
QU (0, s ) = 0,
∴ C1 + C 2 = 0
And , QU (4, s ) = 0
∴ C1e 4
s
∴F =
+ C2 e − 4
sin πx
s
+0=0
But, C 2 = −C1 ∴ C1 e 4 s − e − 4 s = 0 Then must be C1 = 0
∴ C2 = 0
∴U ( x, s ) = U h + U P =
6 s +π
2
4
sin
πx 2
+
3 s +π
2
sin πx
By taking the inverse Laplace transform we find: ∴ u ( x, t ) = 6e − (π
2
) sin πx + 3e −π
4t
2
2
t
sin πx
3
s +π 2
386
Partial Differential Equations
Problems 1)
If
f ( x, y ) = x 2 sin y, and g ( x, y ) = e y sin( x + y )
Find f x , f y , g x , g y
f ( x, y, z ) = sin 2 xy * cos 2 yz * tan 2 xz
2)
If
3)
Find all second partial derivatives of
Find , f x , f y , f z
f ( x, y, z ) = e 2 x sinh y cosh z 4)
Suppose that:
y z = sin( xy ) cos , x
x = te − t ,
Use the Chain rule to find 5)
If z = x Ln y, Find
6)
y = sinh t
dz dt
x = st 2 , y = s 2 t
∂z ∂z and ∂t ∂s
If sin 2 x + cos 2 y + tan 2 z = 1 Find
∂z ∂z and ∂x ∂y
7) Let a metallic rod 20cm long be heated to a uniform temperature of 100oC. Suppose that at t=0 the ends of the bar are plunged into an ice bath at 0 o C , and thereafter maintained at this temperature, but
387
Chapter Ten
that no heat allowed to escape through the lateral surface. Find an expression for the temperature at the center of the bar at time t=30 sec if α 2 =0.86 cm 2 / sec 8) Use the method of separation of variables to solve the following partial differential equation s:
∂u ∂ 2u (a) , where u x (0, t ) = 0, u (2, t ) = 0 , and = ∂t ∂x 2
u ( x,0 ) = 8 cos (b)
∂2 y ∂t 2
=4
3πx 9πx − 6 cos 4 4
∂2 y ∂x 2
, where
y (0, t ) = y (5, t ) = 0, y ( x,0 ) = 0, yt ( x,0 ) = 5 sin πx (c)
∂2 y ∂t
2
=4
∂2 y ∂x
2
, where
y (0, t ) = y (5, t ) = 0, y ( x,0 ) = 0,
and
yt ( x,0 ) = 3 sin 2πx − 2 sin 5πx (9) Use the method of Fourier series to solve the following boundary value problems:
∂u ∂ 2u = 2 2 , u (0, t ) = u (4, t ) = 0, u ( x,0 ) = 25 x (a) ∂t ∂x where 0 < x < 4, t > 0
388
Partial Differential Equations
∂u ∂ 2u = 2 u x (0, t ) = u x (π , t ) = 0, u ( x,0) = x 2 (b) ∂t ∂x where 0 < x < π , t > 0 , and
10) Use the method of Laplace Transforms to solve the following partial differential equation s:
∂u ∂ 2u (a) , where u (0, t ) = 0, u (4, t ) = 0 , and = ∂t ∂x 2 u ( x,0 ) = 6 sin
πx 2
+ 3 sin πx
∂u ∂ 2u , where u x (0, t ) = 0, u x (2, t ) = 0 , and (a) = ∂t ∂x 2 u ( x,0 ) = 4 cos πx − 2 cos 3πx
Chapter 11 Simultaneous Linear Differential Equations 11.1 Characteristic Value Problems A variety of practical problems having to do with mechanical vibrations, alternating currents and voltages, and other oscillatory phenomena lead to linear algebraic systems of the following type:
a11 x1 + a12 x2 + ................ + a1n xn = λx1 a21 x1 + a22 x2 + ................ + a2n xn = λx2 ........................................................................... an1 x1 + an 2 x2 + ................ + ann xn = λxn
(1)
Where λ is a parameter. The metric form of this system is as following:
AX = λX
(2)
We can use the n x n identity matrix I to write (2) as AX = λIX or, equivalently, as:
(λI − A)X
=0
(3)
The homogeneous system of this sort has a nontrivial solution if, and only if,
det (λI − A) =
λ − a11
− a12
− a21
λ − a22 ....... −a2n
......
− a1n
............................................. − an1 − an 2 λ − ann
=0
(4)
390
Simultaneous Linear Differential Equations
When expanded the determinant det (λI − A) is seen to be a polynomial of degree n in the parameter A. Equation (4) is known as the characteristic equation of the matrix A, and p (λ ) is called the characteristic polynomial of A.
p (λ ) = det (λI − A)
(5)
For values of λ which satisfy (4), and for these values only, the matrix equation (3) has nontrivial solution vectors. The n roots of
p (λ ) = 0 , which need be neither distinct nor real, are called the characteristic values of the matrix A, and the corresponding nontrivial solution vectors of systems (1), (2), or (3) are called the characteristic vectors or eigen vectors of A. The problem of finding characteristic values and characteristic vectors of a square matrix is referred to as a characteristic or eigen value problem. If A is a real matrix, clearly all coefficients of its related characteristic polynomial p (λ ) must be real. Hence, complex characteristic values must occur in conjugate pairs. The following theorem affirms that the same is true of complex characteristic vectors. THEOREM 1 If x is a characteristic vector corresponding to a complex characteristic value a + jb , b ≠ 0 , of a real matrix A, then
x is a characteristic vector corresponding to the characteristic value
a − jb .
Chapter Eleven
391 Let us find the characteristic values and the characteristic vectors
of several specific matrices. Example 1 Find the characteristic values and the corresponding
4 − 5 characteristic vectors of the following matrix: A = 1 − 2 Solution: The characteristic equation of A is:
det (λI − A) =
λ−4
5
−1
λ+2
= λ2 − 2λ − 3 = (λ + 1)(λ − 3) = 0 So the characteristic values of A are λ1 = −1 and λ2 = 3 . If λ = −1 , the equation det (λI − A) X = 0 is equivalent to the following linear system:
− 5 x1 + 5 x2 = 0 − x1 + x2 = 0
or to: − x1 + x2 = 0
1 An obvious solution of the last equation is X 1 = and, of course, 1 this vector multiplied by any number is also a solution. Thus, for all
1 nonzero numbers c1 , X = c1 is a characteristic vector of A 1 corresponding to λ1 = −1 .
392
Simultaneous Linear Differential Equations
If λ = 3 , the equation det (λI − A) X = 0 is equivalent to
5 − x1 + 5 x2 = 0 , which has X 2 = as an obvious solution. Thus, 1 5 for every nonzero number c2 , X = c2 is a characteristic vector 1 of A corresponding to λ2 = 3 . As a check on the characteristic values and vectors just found, we observe that:
− 5 5 1 0 = and that: − 1 1 1 0
[λ1I − A] = 1
1
5 − 1 5 5 0 = − 1 5 1 = 0 1
[λ2 I − A]
Example 2 Find the characteristic values and the corresponding characteristic vectors of the following matrix:
3 − 2 − 5 A = 4 − 1 − 5 − 2 − 1 − 3 Solution:
λ −3
∴ det (λI − A) = − 4 2
2
λ +1 1
5 5 = λ3 + λ2 − 16λ + 20 λ +3
= (λ + 5)(λ − 2)2 = 0
Chapter Eleven
393
and so the characteristic values of A are λ1 = −5 and λ2 = 2 . The equation (− 5 I − A) X = 0 is equivalent to the following linear system:
− 8 x1 + 2 x2 + 5 x3 = 0 − 4 x1 − 4 x2 + 5 x3 = 0 2 x1 + x2 − 2 x3 = 0 which is equivalent to Setting
−3x3 = 0
4 x1
− x3 = 0
2 x2
x3 = 4c1 , we find
x2 = 2c1 , and
x1 = 3c1 . All
characteristic vectors of A corresponding to λ1 = −5 are therefore
3 given by X = c1 2 , where c1 can be any nonzero number. 4 The equivalent component form of the matrix equation
(2 I − A)X
− x1 + 2 x2 + 5 x3 = 0 = 0 is:
− 4 x1 + 3x2 + 5 x3 = 0 2 x1 + x2 + 5 x3 = 0
This system is equivalent to
+ x3
x1 x2
=0
+3 x3 = 0
Setting x3 = −c2 , we find that the characteristic vectors of A
1 corresponding to λ2 = 2 are given by X = c2 3 , c2 ≠ 0 − 1
394 Simultaneous Linear Differential Equations Example 3 Find the characteristic values and the corresponding characteristic vectors of each of the following matrix:
6 6 4 1 3 2 − 1 − 5 − 2 Solution:
−6 λ − 4 − 6 det (λI − A) = − 1 λ − 3 − 2 = 0 1 λ + 2 5
∴ (λ − 4 )(λ − 3)(λ + 2 ) + 10 + 6(− (λ + 2 ) + 2 ) − 6(− 5 − (λ − 3)) = 0 ∴ λ3 − λ2 + 4λ − 4λ2 + 4λ − 16 + 12 = 0 ∴ λ3 − 5λ2 + 8λ − 4 = 0 The roots of the above equations (eigen values) are:
λ1 = 1 and λ2,3 = 2 For λ1 = 1
− 3 x1 − 6 x2 − 6 x3 = 0 − x1 − 2 x2 − 2 x3 = 0
x1 + 5 x2 + 3 x3 = 0 ∴ 3 x2 + x3 = 0 Let x2 = c1
∴ x3 = −3 x2 = −3c1
∴ x1 + 5c1 − 9c1 = 0
Chapter Eleven
395
x1 − 4c1 = 0
∴ x1 = 4c1 x1 4 ∴ x2 = c1 1 − 3 x 3 For λ = 2
− 2 x1 − 6 x2 − 6 x3 = 0 − x1 − x2 − 2 x3 = 0
x1 + 5 x2 + 4 x3 = 0 ∴ 4 x2 + 2 x3 = 0
∴ x3 = −2x2 Let x2 = c2
∴ x3 = −2c2
∴ x1 = 3c2
The eigen vectors for λ = 2 is shown below:
x1 3 = x c 2 2 1 − 2 x 3 Example 4 Find the characteristic values and the corresponding characteristic vectors of each of the following matrix:
7 − 2 − 4 3 0 − 2 6 − 2 − 3
396 Simultaneous Linear Differential Equations Solution:
λ −7 2 det (λI − A) = − 3 λ
4
2 =0 2 λ +3
−6
∴ λ3 − 4λ2 + 5λ − 2 = 0
∴ λ1,2 = 1 and λ3 = 2 Eigen vectors for λ1, 2 = 1 :
− 6 x1 + 2 x2 + 4 x3 = 0 − 3x1 + x2 + 2 x3
=0
− 6 x1 + 2 x2 + 4 x3 = 0 From subtracting the last two equations we get the following equation:
x3 = 0 ∴ −6 x1 + 2 x2 = 0 Then, if x1 = c1
∴ x2 = 3c1
x1 1 x c = 2 1 3 0 x 3 Eigen vectors for λ3 = 2
− 5 x1 + 2 x2 + 4 x3 = 0
− 3x1 + 2 x2 + 2 x3 = 0 − 6 x1 + 2 x2 + 5 x3 = 0
Chapter Eleven
397 From subtracting the last two equations we get the following
equation 3 x1 − 3 x3 = 0
∴ x1 = x3 = c2
1 x1 1 x 2 = c2 2 x 3 1 Example 5 Find the characteristic values and the corresponding characteristic vectors of each of the following matrix:
4 −6 2 4 2 −6 − 6 − 6 − 15 Solution:
λ−2
det (λI − A) = − 4 6
−4
λ−2 6
6 6 =0 λ + 15
∴ λ3 + 11λ2 − 144λ − 324 = 0 ∴ λ1 = −2 , λ2 = 9 , and λ3 = −18 Eigen vectors for λ1 = −2
− 4 x1 − 4 x2 + 6 x3 = 0
(6)
− 4 x1 − 4 x2 + 6 x3 = 0
(7)
6 x1 + 6 x2 + 13 x3 = 0
(8)
Divide (6) by 2, we get the following equation:
398
Simultaneous Linear Differential Equations
− 2 x1 − 2 x2 + 3 x3 = 0
(9)
Multiply (9) by 3 we get the following equation:
− 6 x1 − 6 x2 + 9 x3 = 0
(10)
Add (8) to (10) we get the following:
22 x3 = 0 Let x1 = c1
∴ x3 = 0 , and x1 = − x2
∴ x2 = −c1
x1 1 ∴ x2 = c1 − 1 0 x 3 Eigen vectors for λ2 = 9
7 x1 − 4 x2 + 6 x3 = 0
(11)
− 4 x1 + 7 x2 + 6 x3 = 0
(12)
6 x1 + 6 x2 + 24 x3 = 0
(13)
From subtracting the first two equations we get the following equation:
11x1 − 11x2 = 0
∴ x1 = x2 = c2
∴ 7c2 − 4c2 + 6x3 = 0 ∴ x3 =
−1 c2 2
Chapter Eleven
399
x1 2 1 x 2 = c2 2 x 2 − 1 3 Eigen vectors for λ3 = −18
− 20 x1 − 4 x2 + 6 x3 = 0 − 4 x1 − 20 x2 + 6 x3 = 0
6 x1 + 6 x2 − 3x3 = 0 From subtracting the first two equations we get the following equation:
− 16 x1 + 16 x2 = 0
∴ x1 = x2 = c3 ∴ −20c3 − 4c3 + 6 x3 = 0
∴ x3 = 4c3 x1 1 ∴ x2 = c3 1 4 x 3 Example 6 Find the characteristic values and the corresponding characteristic vectors of each of the following matrix:
11 − 4 − 7 7 − 2 − 5 10 − 4 − 6
400 Simultaneous Linear Differential Equations Solution:
λ − 11
det (λI − A) = − 7 − 10
4
λ+2 4
7 5 =0 λ +6
∴ λ3 − 3λ2 + 2λ = 0 ∴ λ1 = 0 , λ2 = 1 and λ3 = 2 Eigen vectors for λ1 = 0
− 11x1 + 4 x2 + 7 x3 = 0
(14)
− 7 x1 + 2 x2 + 5 x3 = 0
(15)
− 10 x1 + 4 x2 + 6 x3 = 0
(16)
From subtracting (14) and (16) we get the following equation:
∴ x1 − x3 = 0
∴ x1 = x3 = c1 ∴ −7c1 + 2 x2 + 5c1 = 0
∴ x2 = c1 x1 1 ∴ x2 = c1 1 1 x 3 Eigen vectors for λ2 = 1
− 10 x1 + 4 x2 + 7 x3 = 0
(17)
− 7 x1 + 3x2 + 5 x3 = 0
(18)
Chapter Eleven
401
− 10 x1 + 4 x2 + 7 x3 = 0
(19)
Multiply (17) by 7 and (18) by 10 we get the following two equations:
− 70 x1 + 28 x2 + 49 x3 = 0
(20)
− 70 x1 + 30 x2 + 50 x3 = 0
(21)
Subtract (20) from (21) we get the following equation:
∴ 2 x2 + x3 = 0 Let x2 = c2
∴ x3 = −2c2
∴ −7 x1 + 3c2 − 10c2 = 0
∴ x1 = −c2 x1 −1 ∴ x 2 = c 2 1 − 2 x 3 Eigen vectors for λ3 = 2
− 9 x1 + 4 x2 + 7 x3 = 0
(22)
− 7 x1 + 4 x2 + 5 x3 = 0
(23)
− 10 x1 + 4 x2 + 8 x3 = 0
(24)
Subtract (23) from (22) we get the following two equations:
∴ −2 x1 + 2 x3 = 0
(25)
∴ x1 = x3 = c3
∴ −9c3 + 4 x2 + 7c3 = 0
(26)
402
Simultaneous Linear Differential Equations
∴ x2 =
c3 2
x1 2 ∴ x2 = c3 1 2 x 3 Example 7 Find the characteristic values and the corresponding characteristic vectors of each of the following matrix:
6 6 4 1 3 2 − 1 − 4 − 3 Solution:
λ−4
det (λI − A) = − 1 1
−6
λ −3 4
−6 −2 =0 λ +3
∴ λ3 − 4λ2 − λ + 4 = 0 ∴ λ1 = 4 , λ2 = 1 , and λ3 = −1 Eigen vectors for λ1 = 4
−6 x2 − 6 x3 = 0
(27)
− x1 + x2 − 2 x3 = 0
(28)
x1 + 4 x2 + 7 x3 = 0
(29)
add (28) and (29) we get the following equation:
5 x2 + 5 x3 = 0
(30)
Chapter Eleven
403
∴ x2 = c1 and x3 = −c1
∴ x1 + 4c1 − 7c1 = 0
(31)
∴ x1 = 3c1
x1 3 ∴ x2 = c1 1 − 1 x 3 Eigen vectors for λ2 = 1
− 3 x1 − 6 x2 − 6 x3 = 0
(32)
− x1 − 2 x2 − 2 x3 = 0
(33)
x1 + 4 x2 + 4 x3 = 0
(34)
add (33) and (34) we get the following equation:
2 x2 + 2 x3 = 0
(35)
∴ x2 = c2 and x3 = −c2 ∴ x1 + 4c2 − 4c2 = 0
(36)
∴ x1 = 3c1 x1 0 ∴ x2 = c2 1 − 1 x 3 Eigen vectors for λ3 = −1 − 5 x1 − 6 x2 − 6 x3 = 0
(37)
− x1 − 4 x2 − 2 x3 = 0
(38)
x1 + 4 x2 + 2 x3 = 0
(39)
404 Simultaneous Linear Differential Equations Multiply (39) by 3 and add it to (37), we get the following equation:
3 x1 + 12 x2 + 6 x3 = 0
(40)
Add (40) to (37) we get the following equation:
∴ −2 x1 + 6 x2 = 0 ∴ x1 = 3x2 Let x2 = c3
∴ x1 = 3x2
∴ 3c3 + 4c3 + 2 x3 = 0
∴ x3 =
(41)
−7 c3 2
x1 6 1 ∴ x2 = c3 2 x 2 − 7 3 Example 8 Find the characteristic values and the corresponding characteristic vectors of the following matrix:
2 3 − 1 1 A= − 1 2 1 −2
1 6 2 − 2 2 6
1 6
Solution: A factored form of the characteristic equation of the matrix A can be obtained by applying elementary properties of determinants to det (λI − A) .
Chapter Eleven
2 3 − 1 1 − 1 2 1 −2
405
1 6 2 − 2 2 6
1 6
We first add the third row of det (λI − A) to its fourth row, which permits us to factor λ − 4 out of the new fourth row. In the remaining determinant, we subtract column 4 from column 3 and then expand. This gives us the following result:
−1 −1 λ − 3 − 2 1 −1 − 6 −6 λ = (λ − 4 )2 (λ − 2)2 + 1 = 0 1 −2 λ−2 2 2 − 2 λ − 6 −1 From the last equation, we see that λ1 = 4 , λ2 = 2 + j1 , and,
[
]
λ3 = 2 − j1 are the characteristic values of A. Using Gauss-Jordan elimination to solve the linear system (4 I − A)x = 0 , we find without difficulty that the characteristic 0 0 vectors of A corresponding to λ1 = 4 are given by X = c1 1 − 1 where c1 ≠ 0 . To find the characteristic vectors which correspond to
λ2 = 2 + j1 , we reduce the following matrix to row echelon form and then write the corresponding linear system, which we find to be as following:
406
Simultaneous Linear Differential Equations
−1 − 1 + j1 − 2 − 1 1 1 + j1 − 6 −6 [(2 + j1)I − A] = 1 −2 j1 2 2 − 2 − 4 + j1 −1
x1 +
x4 = 0 x2 + x3 +
With
1 (− 1 + j1)x 4 = 0 2 + x4 = 0
x4 = −2c2 , we find from these equations that the
characteristic vectors of A corresponding to λ2 = 2 + j1 are given by the following equation:
2 − 1 + j1 , c2 ≠ 0 . X = c2 2 −2 Having found these vectors, we know from Theorem 1 that the characteristic vectors corresponding to the characteristic value
2 − 1 − j1 , c3 ≠ 0 λ3 = 2 − j1 of A are the vectors x = c3 2 −2
Chapter Eleven
407
11.2 Systems Of Linear First Order Differential Equations In many problems in applied mathematics there are not one but several dependent variables, each a function of a single independent variable, usually time. The formulation of such a problem in mathematical terms frequently leads to a system of simultaneous differential equations. Often these equations are nonlinear and exceedingly difficult, if not impossible, to solve, even with the aid of a computer. In certain important cases, however, they are linear in the dependent variables we can solve them easily as following: The simultaneous system consists of m linear differential equations in unknown functions x1 , x2 ,....xn as shown in (42). Such a system is called a linear differential system. A linear differential equations system (42) is homogeneous if, and only if, all the functions
f1 , f 2 ,.... f m are trivial on I, and the system is
nonhomogenous otherwise.
Li1 x1 + Li 2 x2 + ..............Lin xn = f i , 1 ≤ i ≤ m (42) A solution of a linear differential system (42) on an interval I is a vector function which satisfies each equation of the system for every t in I. A linear differential system is consistent or inconsistent on I according as it has, or does not have, a solution on I, and one system is equivalent to another if, and only if, every solution of either system is a solution of the other. In particular, the constant coefficient systems we consider will all be expressible in the following form:
408
Simultaneous Linear Differential Equations
p11 (D )x1 + p12 (D )x2 + ...... + p1n (D )xn = f1 (t )
p21 (D )x1 + p22 (D )x2 + ...... + p2 n (D )xn = f 2 (t )
(43) ........................................................................... pn1 (D )x1 + pn 2 (D )x2 + ...... + pnn (D )xn = f n (t ) d Where pij is a polynomial operator in D = , for 1 ≤ i, j ≤ n dt To solve a system like this, or to determine its inconsistency, we try to find elementary operations that will reduce it to an equivalent system of the following form:
q11 (D )x1 + q12 (D )x2 + ...... + q1n (D )xn
= g1 (t )
q22 (D )x2 + ...... + q2n (D )xn = g 2 (t )
......................................................................... qnn (D )xn = g n (t )
(44)
Example 9 Find a complete solution of the following linear differential system:
dx dy + 2x + + 6 y = 2e t , dt dt 2
dx dy + 3x + 3 + 8 y = −1 dt dt
Solution: As a rule, the steps in a reduction of a constantcoefficient linear differential system to form (44) are easier to detect when each system in the reduction process is written in operator form, rather than in derivative form. The operator form of the given system is:
Chapter Eleven
409
(D + 2)x + (D + 6) y = 2et ,
(45)
(2 D + 3)x + (3D + 8) y = −1
(46)
Our first step in the elimination process is to subtract twice (45) from (46) after which we interchange the equations and have:
− x + (D − 4) y = −1 − 4et
(47)
(D + 2 )x + (D + 6 )y = 2et
(48)
Working with this new system, we operate on both members of (47) with D + 2 and add the result to (48). Then we simplify the new second equation and multiply the first equation through by − 1 . This gives us the following system of equations:
x − ( D − 4 ) y = 1 + 4e t
(49)
(D 2 − D − 2)y = −2 − 10et
(50)
Of course, this system is equivalent to the one we started with and it is in the form of (44). Since (50) is an equation in y only, it is a simple matter to solve for y and we find without difficulty the following results:
y = c1e − t + c2 e 2t + 5et + 1
(51)
With y now completely determined, we substitute into (49) to obtain the following equation:
(
)
x − (D − 4 ) c1e − t + c2 e 2t + 5et + 1 = 1 + 4et
(52)
Solving this equation for x and simplifying, we find the following equation:
410
Simultaneous Linear Differential Equations
x = −5c1e − t − 2c2 e 2t − 11et − 3
(53)
Thus, a complete solution of the original differential system is:
x − 5 − t − 2 2t − 11 t − 3 y = c1 1 e + c2 1 e + 5 e + 1
(54)
It is readily verified that both of the vector functions −t − 5 − t − 5e x1 (t ) = e = , and − t 1 e
2t − 2 2 t − 2e x2 (t ) = e = 2 t 1 e
Satisfy the homogeneous system corresponding to the given system, which is:
dx dy + 2x + + 6 y = 2e t , dt dt
2
dx dy + 3x + 3 + 8 y = −1 dt dt
and that the vector function is as following: t − 11 t − 3 − 11e − 3 v(t ) = e + 1 = t 5 5e + 1
satisfies
the
non-
homogeneous system itself. We shall soon see that this observation extends to some quite general linear differential systems. Its full significance will become clear to us once we have become familiar with linear transformations and linear operator equations.
Chapter Eleven
411 Example 10 Find a complete solution of the differential system
(2D 2 + 3D − 9)x + (D 2 + 7 D − 14)y = 4
(55)
(D + 1)x + (D + 2)y = −8e 2t
(56)
Solution: During the first stage of a reduction process, it is sometimes convenient to eliminate an unknown other than the leading one from all but one equation of a reduced system. To indicate what happens when this is done, we shall eliminate y, instead of x, from one equation of the present system, although there is no particular advantage in doing so. To begin the elimination process, we operate on (56) with
(− D − 5) ,
where the choice of
(− D − 5)
as an operational
multiplier is not just a clever guess. Actually, D + 5 is the quotient when D 2 + 7 D − 14 is divided by D + 2 ; hence, subtracting
(D + 5)(D + 2) y = (D 2 + 7 D + 10)y
from
(D 2 + 7 D − 14)y
will
eliminate both D 2 y and Dy , which is one of our objectives. Add the result to (55), and simplify. Then, we multiply the second equation by 24. This gives us the following system of equations:
(D 2 − 3D − 14)x − 24 y = 4 + 56e2t
(57)
24(D + 1)x + 24(D + 2 ) y = −192e 2t
(58)
Next, we operate on (57) with D + 2 , add the result to (58), and simplify, obtaining:
412
Simultaneous Linear Differential Equations
(D 2 − 3D − 14)x − 24 y = 4 + 56e2t (D3 − D 2 + 4D − 4)x = 8 + 32e2t
(59) (60)
As it stands, this system is not of the form (44). However, by merely interchanging the x and y terms in the first equation, the system takes on that form. Of course, there is no need to actually make this interchange because (56) can be solved at once. The roots of its characteristic equation are ± j 2, 1 . Hence a complementary function is c1 cos 2t + c2 sin 2t + c3et . It is easy to see that
− 2 + 4e 2t is a particular integral, and therefore:
x = c1 cos 2t + c2 sin 2t + c3et − 2 + 4e 2t With x completely determined, a solution for y can be found by substituting the last expression for x into (59):
(
)(
)
∴ D 2 − 3D − 14 c1 cos 2t + c2 sin 2t + c3et − 2 + 4e 2t = 4 + 56e 2t Performing the indicated differentiations, collecting like terms, simplifying, and solving for y, we have:
∴y =−
1 (3c1 + c2 ) cos 2t + 1 (c1 − 3c2 ) sin 2t − 2 c3et + 1 − 5e 2t 4 4 3
Thus, in terms of the parameters
k1 = c1 / 4, k 2 = c2 / 4, and
k3 = c3 / 3 a complete solution of system (59) and (60) is given by the following:
Chapter Eleven
413
4 cos 2t 4 sin 2t x y = k1 − 3 cos 2t + sin 2t + k 2 − cos 2t − 3 sin 2t (61) − 3 2 4 + k 3 e t + + e 2t − 2 1 − 5 But, (55) and (56) are equivalent systems; hence (61) is also a
complete solution of the original system. Before considering another example, let us return to system (43) and observe that if D were a number, instead of an operator, then:
p11 (D )
p21 (D ) ....
p12 (D ) ....
p22 (D ) .... .... ....
p1n (D )
p2 n (D ) ....
(62)
pn1 (D ) pn 2 (D ) .... pnn (D ) would be a determinant. Of course, the polynomial operators pij (D ) usually are not numbers. Nevertheless, (62) is known as the determinant of the operational coefficients of (43). Moreover, whenever the following theorem is applied, (62) is expanded using the properties that hold for ordinary determinants. THEOREM 1 If the determinant of the operational coefficients of a system of n linear differential equations with constant coefficients is not identically zero, then the total number of independent arbitrary constants in any complete solution of the system is equal to the degree of the determinant of the operational coefficients, regarded as a polynomial in D. In particular cases in which the determinant of the operational coefficients is identically
414 Simultaneous Linear Differential Equations zero, the system may have no solution or it may have solutions containing any number of independent constants. Applying this theorem to system (55), (56), in the above example, we note that the determinant of the operational coefficients is as the following:
2 D 2 + 3D − 9 D 2 + 7 D − 14 D +1
D+2
The expanded form of this determinant is the third-degree polynomial
D3 − D 2 + 4D − 4
in
D . Hence, according to
Theorem 1, there must be exactly three arbitrary constants in any complete solution of the given system. This explains why the complete solution (61) of system (55), (56) contains three arbitrary constants, even though the system has only two equations and two unknowns.
Example 11 Solve the following differential system: Dx
(D − 1)x (D + 1)x
+ (D − 1) y
+ (D + 2)z = 2et
+ Dy + (D − 2)z = aet + (D − 2) y + (D + 6)z = et Solution: It is readily verified that the determinant of the operational coefficients of this system is identically zero. Hence, according to Theorem 1, the system may have no solution or it may have solutions containing any number of arbitrary constants. Let us apply elementary operations to the system and find out what the case might be.
Chapter Eleven
415 By subtracting the second equation from both the first and third
equations, and then adding − 2 times the new first equation to the new third equation, we get the following system:
x
(D − 1)x
−y
+ 4z
+ (D − 2 )z
= (2 − a )et
= aet 0 = (a − 3) et Of course, this reduced system is equivalent to the given one. + Dy
Clearly, unless a = 3 , neither system is consistent. On the other hand, if a = 3 , we can operate on the first equation of the reduced system with (− D + 1) , add the result to the second equation, simplify, and omit the third equation which is simply
0 = 0 , to obtain: x − y + 4 z = et
(63)
(2 D − 1) y − (3D − 2)z = 3et
(64)
This system is in the form of (44), and it is equivalent to the original system. From (64), y can be found in terms of z = z (t ); then, from (63) x too can be found in terms of z. Since z is subject only to the restriction that it be differentiable, it may contain any number of arbitrary constants. Thus, consistent with the possibilities allowed for by Theorem 1, we have found that, if a ≠ 3 , the original differential system has no solution; but, if a = 3 , the system may have solutions containing any number of arbitrary constants.
416 Simultaneous Linear Differential Equations 11.3 Linear Differential Systems with Constant Coefficients We shall now consider, as an alternative to the use of elementary operations, a method of solving linear differential systems with constant coefficients which resembles the way in which we solved single linear constant coefficient differential equations. A significant feature of the method, which we are about to present, is that it puts to use what we already know about characteristic value problems. Let us first investigate the method as it applies to first order systems. As in the case of a single equation, to solve the non-homogeneous first order system
x′ = Ax + f (t )
(65)
where A is a constant matrix. We begin by looking for a complete solution of the associated homogeneous system:
x′ = Ax
(66)
Guided by our experience in solving single equations, we attempt to find a solution of (66) of the form
x = ke λt
(67)
where k is a constant vector and A is a scalar. Substituting (67) into (66), dividing out the common factor e λt , and slightly altering the resulting equation, we see that (67) yields a solution of (66) if, and only if,
(λI − A)k = 0
(68)
Chapter Eleven
417 Thus, (67) will be a nontrivial solution of (66) if, and only if, λ is a characteristic value of A and k is a corresponding characteristic vector. The remaining steps required to find a complementary function of (65) may differ somewhat according as: ١. All characteristic values of A are real and distinct. ٢.
All characteristic values of A are real, but some are
repeated roots of the characteristic equation det (λI − A) = 0 ٣.
Complex characteristic values of A occur.
Perhaps, the best way of learning how to handle these three cases is by means of specific examples. Our first example (Example 12) involves a first order system whose coefficient matrix has only distinct real characteristic values.
Example 12 Find a complete solution of the following system of linear differential equations: x1′ = x2′ = 4 x1
− x2
+ x3
− x2
−4 x3
x3′ = −3x1
− x2
+4 x3
Solution: As (67) suggests, we seek solutions of (69) of the type:
x1 a x = b e λt 2 x3 c where a , b, c are constants
(69)
Simultaneous Linear Differential Equations
418
Substituting ae λt , be λt and ce λt into the given system for
x1 , x2 , and x3 , respectively, then dividing out e λt and transposing, we obtain the characteristic value problem as following:
λa
+ b − 4a +(λ + 1)b
3a
+
b
− +
c =0 4c = 0
(70)
+(λ − 4 )c = 0
The corresponding characteristic equation is as following:
λ
−1
1
0
0
−1
− 4 λ +1 4 = 4(λ − 1) λ +5 4 3 1 λ − 4 (λ − 1)(λ − 3) λ − 3 λ − 4 4 λ +5 = −(λ − 1)(λ − 3) = (λ + 1)(λ − 1)(λ − 3) = 0 1 1 Clearly λ1 = −1, λ2 = 1, and λ3 = 3 are the characteristic values of the following coefficient matrix:
0 A= 4 − 3 If λ = λ1
−1
1 − 1 − 4 − 1 4 = −1 system (70) becomes linear system as following:
−a+b−c =0 a−c =0 ∴ − 4a +4c = 0 which is equivalent to b − 2c = 0 3a + b − 5c = 0
Chapter Eleven
419
1 Setting c = 1 we fiend that 2 is a characteristic vector of A 1 1 corresponding to λ1 = −1 . By the same procedure, we fiend 0 and 1 1 − 1 to be characteristic vectors of A corresponding to the 2 respective characteristic values λ2 = 1 and λ3 = 3 respectively. It follows that: 1 1 1 2 e − t , 0 e t , and − 1 e3t 1 1 2 are particular solutions of (69). Since, for every real number t, the Wronskian W of these three solutions has the nonzero value
W (t ) = −2e3t , a complete solution of the given first order system is as following:
1 1 1 −t t X = c1 2 e + c2 0 e + c3 − 1 e3t 1 1 2 Our next example involves a simple first order system whose coefficient matrix has a repeated characteristic value. Although the system has only two unknowns, it serves to illustrate how complete solutions of more general systems can be found when repeated characteristic values occur.
420 Simultaneous Linear Differential Equations Example 13 Find a complete solution of the following system:
x1′ = x1 − 2 x2 x2′ = 2 x1 − 3 x2
(71)
1 − 2 Solution: The coefficient matrix of this system is A = . 2 − 3 a Since we are seeking solutions of (71) of the type e λt , where b a and b are constants, the characteristic value problem to be solved is:
2 a 0 λ − 1 − 2 λ + 3 b = 0 and the characteristic equation is
(72)
2 λ − 1 2 2 − 2 λ + 3 = λ + 2λ + 1 = (λ + 1) = 0 Hence, λ1 = −1 is a repeated characteristic value of A. Setting
λ = −1 in (72), we obtain the linear system − 2a + 2b = 0 − 2a + 2b = 0
which is equivalent to
a −b = 0
1 Clearly, is a characteristic vector of A corresponding to 1
λ1 = −1 , and 1 X = e −t 1 is a particular solution of (71)
(73)
Chapter Eleven
421 Strict analogy with a single differential equation, having -1 as a
double root of its characteristic equation, would suggest that a second solution of (71) might be found by multiplying (73) by t.
1 But, te − t does not satisfy (73) Furthermore, the Wronskian of 1 this vector function and (73) is a trivial function over all real values of t. This is no insurmountable problem, however for variation of parameters, which enabled us to cope with repeated roots of characteristic equations in the first al place, is still applicable. To
a apply the method, we take λ = −1, as we must if e λt is to be a b solution of (71). Then, we vary the parameters a and b and attempt to find a solution of (71) of the following type:
u − t v e
(74)
where u and v are functions of t. Of course, we want
e− t e− t
ue− t to be a fundamental matrix of (71). −t ve
Substituting ue− t and ve− t into (71) for x1 and x2 , respectively, then canceling all occurrences of e − t and regrouping, we find that (74) will be a solution of (71) if, and only if, u and v satisfy the equations:
422
Simultaneous Linear Differential Equations
u ′ = 2(u − v ) v′ = 2(u − v )
(75)
Subtracting the second of these equations from the first, we have
u ′ − v′ = (u − v )′ = 0 which implies that: u = v + k1
(76)
Replacing u by v + k1 , in the second of (75), we get v′ = 2k1 , which integrates into v = 2k1t + k 2 . With v thus determined, the last of (76) gives u = 2k1t + k 2 + k1 and (74) becomes as following:
k1 + k 2 + 2k1t − t e k 2 + 2k1t
(77)
The Wronskian of this vector function and (73) has the following value:
k1 + k 2 + 2k1t k 2 + 2k1t
1 − 2t e = k1e − 2t ≠ 0 1
Hence, (73) and (77) are fundamental solutions of (71), provided
k1 ≠ 0 . In particular, if k1 = 1 and k 2 = 0 , (73) and (77) give us: e − t e − t
(1 + 2t )e − t
−t 2te
as a fundamental matrix, and
1 (1 + 2t ) − t X = c1 e − t + c2 e t 1 2 as a complete solution of the original differential system.
Chapter Eleven
423 Looking back on (63) and (64), a shorter method of finding a
fundamental matrix of (71) comes to our attention. As soon as (73) has been found, determine constants a, b, c , and d such that: −t a − t c − t (a + ct )e b e + d te = − t (b + dt )e
and (73) will be linearly independent solutions of (71). As one further introductory example, let us consider a simple first order system whose coefficient matrix has complex characteristic values. Example 14 Find a complete solution of the following system:
x1′ = 2 x1 − 3 x2 x2′ = 3 x1 + 2 x2
(78)
This system will have a solution of the type
a λt b e
(79)
if, and only if:
λ − 2 − 3
3 a 0 = λ − 2 b 0
Solving the characteristic equation as following:
λ − 2 3 − 3
= λ2 − 4λ + 13 = 0 λ − 2
(80)
424
Simultaneous Linear Differential Equations
we obtain λ1 = 2 + j 3 and λ2 = 2 − j 3 as conjugate complex
2 characteristic values of the coefficient matrix A = 3
−3 . 2
Replacing λ by 2 + j 3 in the characteristic value problem (80), and dividing out a factor of 3 from every term, we get the linear system
ja + b = 0 − a + jb = 0 which is obviously equivalent to ja + b = 0 Taking a equal to the coefficient of b, and b equal to minus the
1 coefficient of a, in the last equation, we obtain as a − j characteristic vector of A corresponding to the characteristic value
λ1 = 2 + j 3 . Substituting these values of a, b, and λ into (79), gives us the complex vector function
1 −
(2 + j 3)t e j
(81)
which formally satisfies system (78). Of course, what we want are real fundamental solutions of (78). To find these, we use Euler's formula to write (81) as:
1 −
j 3t 2t cos 3t + j sin 3t 2t 2t j 3t e e = e e = − j (cos 3t + j sin 3t ) e j t 3 j − je
Chapter Eleven
1 ∴ −
425
2t 2t j 3t e cos 3t e e = + 2 t j e sin 3t
e 2t sin 3t j 2 t − e cos 3t
It is now easy to verify that the real vector functions:
sin 3t 2t cos 3t 2t e and − cos 3t e sin 3t Form a basis for all real solutions of system (78). Thus, we are able to write a complete solution:
cos 3t 2t X = c1 e + c2 sin 3 t
sin 3t 2t − cos 3t e of the given differential
system using only one of the two conjugate characteristic values of A. Example 15 Find a complete solution of the following first order systems:
x1′ = −2 x1 − 13x2 x2′ = x1 + 4 x2
Solution:
− 2 − 13 A= 4 1 det (λI − A) =
λ+2
13
−1
λ −4
=0
(λ + 2 )(λ − 4 ) + 13 = 0 λ2 − 2λ + 5 = 0 For λ1 = 1 + j 2
∴ λ1,2 = 1 ± j 2
426
Simultaneous Linear Differential Equations
13 a 0 3 + j 2 −1 b = 0 − 3 + j 2
(3 +
j 2 )a + 13b = 0
(3 + j 2)a = −13b Let a = C1 .ﺧﻄﺄ! اﻹﺷﺎرة اﻟﻤﺮﺟﻌﻴﺔ ﻏﻴﺮ ﻣﻌﺮّﻓﺔ
1 1+ j 2 λt 1 13 = X 1 = C1 − 1 e C 1 − 3 − j 2 e 13 ( ) + j 3 2 13 13 1+ j 2 e = c1 − 3 − j2
13 0 ∴ X 1 = c1e t cos 2t − sin 2t − 2 − 3 0 13 ∴ X 2 = c2 et cos 2t + sin 2t − 3 − 2
13 0 ∴ X = X 1 + X 2 = c1et cos 2t − sin 2t − 2 − 3 0 13 + c2 et cos 2t + sin 2t − 3 − 2 Example 16 Find a complete solution of the following first order systems.
x1′ = −3x1 − 4 x 2 x 2′ = 2 x1 + x 2
Chapter Eleven
427
Solution:
− 3 4 A= 2 1 det (λI − A) =
λ +3
4
−2
λ −1
=0
(λ + 3)(λ − 1) + 8 = 0 λ2 + 2λ + 5 = 0
∴ λ1,2 = −1 ± j 2
For λ1 = 1 + j 2
4 a 0 2 + j 2 −2 b = 0 j − 2 + 2
(2 + J 2)a + 4b = 0 and − 2a + (− 2 + j 2)b = 0 ∴ a = (− 1 + j1)b ∴ a = (− 1 + j1)c1 Let b = c1 x1 a = c1 e (−1+ j 2 )t b x2 − 1 + j1 a = c1 1 b − 1 1 θ = , and, φ = 1 0 − 1 1 ∴ X 1 = c1e − t cos 2t − sin 2t 0 1 1 − 1 ∴ X 2 = c2 e − t cos 2t + sin 2t 1 0
428
Simultaneous Linear Differential Equations
− cos 2t − sin 2t ∴ X = X 1 + X 2 = c1e − t cos 2t cos 2t − sin 2t + c2 e − t sin 2t Example 17 Find a complete solution of the following first order systems.
x1′ = −3 x1 − 2 x2 x2′ = 2 x1 + x2 Solution:
− 3 − 2 A= 1 2 det (λI − A) =
λ +3
2
−2
λ −1
(λ + 3)(λ − 1) + 4 = 0 λ2 + 2λ + 1 = 0 ∴ λ1 = λ2 = −1 Eigen values For λ1 = −1
2a + 2b = 0
∴ a = −b Let a = c1 , and b = − c1
x 1 ∴ 1 = c1 e − t − 1 x2
=0
Chapter Eleven
429
Eigen values For λ2 = −1
x u ∴ 1 = e − t x2 v
− ue − t = u ′e − t = (− 3u − 2v )e − t ∴ u ′ = −2(u + v ) − ve − t + v′e − t = (2u + v )e − t
v′ = 2(u + v ) Let u + v = c2
u ′ = −2c2 ,
∴ u = −2c2t + c3
v = c2 − u = c2 + 2c2t − c3
− 2c2t + c3 W = c2 + 2c2t − c3
1 = 2c2t − c3 − (c2 + 2c2t − c3 ) = 0 − 1
c3 = 0 ∴ u = −2c2t , and v = 2c2t + c2
1 − 2t − t e ∴ X = c1 e − t + c2 t − + 1 2 1 Example 18: Find a complete solution of the following first order systems.
x1′ = 4 x1 − 2 x2 10 , x(0 ) = x2′ = 25 x1 − 10 x2 40
430 Simultaneous Linear Differential Equations Solution:
4 −2 QA= 25 − 10 det (λI − A) =
λ−4
2
− 25
λ + 10
=0
(λ − 4)(λ + 10) + 50 = 0 λ2 + 6λ + 10 = 0 ∴ λ1, 2 = −3 ± j1 Eigen values For λ1 = −1
(− 7 + j1)a + 2b = 0 Let a = c1
∴b =
−1 c1 (− 7 + j1) 2
1 a 1 2 = c1 7 1 = c1 − b j 7 1 2 j − 2 2 2
θ = , and 7
0
φ= − 1
2 0 X 1 = c1e − 3t cos t − sin t − 1 7 0 2 X 2 = c2 e − 3t cos t + sin t 7 − 1 2 cos t − 3t 2 sin t + ∴ X = c1e − 3t c e 2 7 cos t + sin t − cos t + 7 sin t
Chapter Eleven
431
10 2c Q X (0 ) = = 1 40 7c1 − c2 ∴ 2c1 = 10 and, 7c1 − c2 = 40 ∴ c1 = 5 and, c2 − 5 x cos t − sin t ∴ X = 1 = 10e − 3t − x 4 cos t 3 sin t 2 Example19 Find a complete solution of the following first order systems.
x1′ = 3 x1 − x3 − 1 x2′ = − 2 x1 + 2 x2 + x3 x(0 ) = 2 − 8 x3′ = 8 x1 − 3 x3 Solution: 3 0 − 1 A = − 2 2 1 8 0 − 3 λ −3
der (λI − A) =
2 −8
0
1
λ−2 0
−1 = 0 λ +3
λ3 − 2λ2 − λ + 2 = 0 ∴ λ1 = 1 , λ1 = −1 , and λ3 = 2 Eigen values for λ1 = 1
− 2a 2a − 8a
+c = 0 −b
−c = 0 +4c = 0
Simultaneous Linear Differential Equations
432
Let a = c1 ∴ c = 2c1 and b = 0
x1 1 ∴ X 1 = x2 = c1 0 et 2 x 3 Eigen values for λ2 = −1
− 4a 2a − 8a
−3b
+c = 0 −c = 0 +2c = 0
Let a = C2 ∴ c = 4C2 Substitute values of a and c into the second equation we get the following:
3b = 2C2 − 4C2 ∴b =
−2 C2 3
x1 3 ∴ X 2 = x2 = c2 − 2 e − t 12 x 3 Eigen values for λ3 = 2
−a
+c = 0
2a
−c = 0
− 8a
+5c = 0
∴a = c = 0 Let b = c3
Chapter Eleven
433
x1 0 ∴ X 2 = x2 = c3 1 e 2t 0 x 3 x1 1 3 0 t −t ∴ X = x2 = c1 0 e + c2 − 2 e + c3 1 e 2t 2 12 0 x 3 − 1 Q x(0) = 2 − 8 − 1 1 3 0 ∴ 2 == c1 0 + c2 − 2 + c3 1 2 12 0 − 8 ∴ −1 = c1 + 3c2
(82)
2=
−2c2 + c3
(83)
− 8 = 2c1 + 12c2
(84)
Multiply (82) by –2 and add the results to (83) we get the following:
− 6 = 6c2 ∴ c2 = −1 ∴ c1 = 2 , and, c3 = 0 x1 2 3 t −t ∴ X = x 2 = 0 e − − 2 e 12 x 4 3 It is not always necessary, nor advantageous, to reduce a linear differential system with constant coefficients
434
Simultaneous Linear Differential Equations
pi1 (D )x1 + pi 2 (D )x2 + .... + pin (D )xn = f i (t )
(85)
where 1 ≤ i ≤ n to a first-order system before solving. For a system like (85) can usually be solved in the same manner as a first-order system. The concepts of complementary function and particular integral carry over and, can even be extended to far more general linear systems. If we define a matrix operator P(D ) , and vector functions x and f by:
p12 (D ).... p1n (D ) p11 (D ) p (D ) ( ) ( ) p D p D .... 21 22 2 n P(D ) = .............................................. pn 2 (D ).... pnn (D ) pn1 (D ) x1 f1 x f 2 2 x = .. f = ... .. ... xn f n where the pij ' s are polynomial operators in D, which, of course, have constant coefficients, system (85) can be written in the compact matrix form:
P(D )x = f (t ) The associated homogeneous equation is:
(86)
P(D )x = 0
(87)
Chapter Eleven
435 As with first order systems, a complete solution of (87) is called a
complementary function of (86) and any particular solution of (86) is called a particular integral of that system. To find a complementary function, we assume solutions of the homogeneous system exist in the form:
x = ke λt just like we did for first order systems. Since D r e λt = λr e λt , it
( )
follows that, if we substitute x = ke λt into (87), we obtain
P(λ )ke λt = 0 , or dividing out the scalar factor eλt ,
P(λ )k = 0 We will have a nontrivial solution if, and only if,
(88)
P(λ ) = 0 (89) This equation is called the characteristic equation of both the algebraic system (88) and the original differential system. It is nothing but the determinant of the operational coefficients of (85) equated to zero, with D replaced by λ . For each root λ j of the characteristic equation (89) there will be a solution vector k j of (88) determined to within an arbitrary scalar factor c j . If (89) is a polynomial equation in λ of degree N, and if its root λ1 , λ2 ,...λ N are all distinct real numbers, a complete solution of (87), i.e., a complementary function of (86), is then:
x = c1k1e λ1t + c2 k 2e λ 2 t + ....... + c N k N e λ N t
436 Simultaneous Linear Differential Equations Example 20 Find a complete solution of the differential system:
(D + 1)x + (D + 2) y + (D + 3)z = −e − t (D + 2)x + (D + 3) y + (2 D + 3)z = e−t (4 D + 6)x + (5D + 4) y + (20 D − 12)z = 7e − t
(90)
To find a complementary function of this system, we observe that
a solutions b e λt of the homogeneous system exist if, and only if, c
(λ + 1)a + (λ + 2)b + (λ + 3)c (λ + 2)a + (λ + 3)b + (2λ + 3)c (4λ + 6)a + (5λ + 4)b + (20λ − 12)c
=0 =0 =0 The characteristic equation of this system is:
(λ + 1) (λ + 2) (4λ + 6)
(λ + 2) (λ + 3) (5λ + 4)
(λ + 3) (2λ + 3) (20λ − 12)
(91)
=0
or, expanding, collecting terms, and factoring,
− (λ − 1)(λ − 2 )(λ − 3) = 0 Substituting the roots λ1 = 1, λ2 = 2, and λ3 = 3 of this equation into (52), one at a time, we obtain the three linear systems
2a + 3b + 4c = 0 3a + 4b + 5c = 0 10a + 9b + 8c = 0 3a + 4b + 5c = 0 4a + 5b + 7c = 0 14a + 14b + 28c = 0
Chapter Eleven
437
4a + 5b + 6c = 0 5a + 6b + 9c = 0 18a + 19b + 48c = 0 When the coefficient matrices of these systems are reduced to row echelon form the systems themselves are transformed, respectively, into
a−c =0 , b + 2c = 0
a + 3c = 0 , b−c =0
a + 9c = 0 b − 6c = 0
From these systems we see that (1,-2,1), (3,.-1,-1), and (9,-6,-1) are nontrivial solutions of (91) corresponding to the distinct real values 1, 2, and 3, respectively, of the parameter λ . A complementary function of (90) is therefore
1 3 9 t 2t c1 − 2 e + c2 − 1 e + c3 − 6 e3t 1 − 1 − 1
(92)
To complete the problem we now need to find a particular integral of the given differential system. Since the scalar factor e − t
− 1 in the non-homogeneous term f (t ) = 1 e − t of (90), and the 7 scalar factors et , e 2t and e3t in the terms of the complementary function are linearly independent functions, we choose a trial solution of (90) as following:
438
Simultaneous Linear Differential Equations
a v = b e − t (93) c Substituting (93) into (90), collecting terms, and canceling a factor of e − t , we find:
b + 2c = −1 a+
2b + c = 1
2a − b − 32c = 7 which is equivalent to the following: a=3 b = −1 c=0
3 Hence, a particular integral of (51) is − 1 e − t and a complete 0 solution of the original system is: 1 3 9 3 x = c1 − 2 et + c2 − 1 e 2t + c3 − 6 e3t + − 1 e − t 1 1 − 1 0 As with first order systems and single equations, if the set of roots
{λ j } of the characteristic equation (89) includes one or more pairs
of conjugate complex numbers, it is desirable to convert all corresponding complex exponential solutions into purely real solutions. To see how this can be accomplished, let p ± jq be a pair of conjugate complex roots of (89), and let k be a particular
Chapter Eleven
439 nontrivial solution vector of (88) corresponding to the root
λ = p + jq , so that P(λ )k = p( p + jq )k = 0 Then, since all the coefficients in (88) are real, it follows by taking conjugates throughout the last equation that k is a solution vector of (88) corresponding to λ = p − jq Therefore, in a complex sense, both ke( p + jq )t and k e ( p − jq )t are particular solutions of (87). By combining these as follows and applying the Euler formulas, we obtain the two independent real solutions as following:
k +k ke ( p + jq )t + k e ( p − jq )t k −k = e pt cos qt − sin qt 2 j2 2
= e pt [θ (k ) cos qt − φ (k ) sin qt ]
(94a)
k −k ke ( p + jq )t − k e ( p − jq )t k −k = e pt cos qt + sin qt j2 2 j2
= e pt [φ (k ) cos qt + θ (k ) sin qt ] (94b) where .θ (k ) and φ (k ) denote the column vectors whose components are, respectively, the real parts of the components of k and the imaginary parts of the components of k. If p(λ ) = 0 has a double root, say λ = r , and if a is a specific nontrivial solution vector of the equation P(r )k = 0 then, of course,
ae rt is a nontrivial solution of (87). To obtain a second independent solution, we proceed as indicated at the end of Example 13. That is, we determine constant vectors b1 and b2 such that:
440
Simultaneous Linear Differential Equations
b1te rt + b2 e rt
(95)
and ae rt will be linearly independent solutions of (87). It is significant that the term b2 e rt must be retained in the matrix case. This might have been anticipated had we noticed that in general the vector b2 will not be a scalar multiple of a, and hence, in constructing a complete solution of (87), the term involving b2 r rt cannot be absorbed in the term involving ae rt , as would necessarily be the case with the term b2 e rt were it included in a trial solution of the form b1te rt + b2te rt for a single scalar differential equation. It can be shown, however that to within an arbitrary scalar factor, the vector b1 is the same as the vector a. Hence, it is really only necessary to determine the components of b2 . Similar results hold for roots of (89) of higher multiplicity. Thus, for an m-fold root r, there will be particular solutions of (87) of the form:
ae rt b1te rt + b2 e rt c1t 2 e rt + c2te rt + c3e rt ......................................................... k1t m −1e rt + k 2t m − 2 e rt + .......k m −1te rt + k m e rt
(96)
Chapter Eleven
441
Example 21 Find a complete solution of the system
(D 2 + D + 8)x1 + (D 2 + 6D + 3)x2 = 0 (D + 1)x1 + (D 2 + 1)x2 = 0
(97)
Solution: In this case the characteristic equation of this system is:
λ2 + λ + 8 λ2 + 6λ + 3 = λ4 + 2λ2 − 8λ + 5 = 0 2 λ +1 λ +1
(98)
with roots 1, 1, − 1 ± j 2 . For the root -1 + j2, (98) simplifies to:
4 − j 2 − 6 + j8 k1 0 j2 k = 0 j − 2 − 4 2
(99)
which is equivalent to jk1 − (1 + j 2 )k 2 = 0
1 + j 2 Taking k1 = 1 + j 2 and k 2 = j , we have j 1
2
θ (k ) = and φ (k ) = . 0 1 Thus, from (94), two particular solutions of the given differential system are:
1 2 x1 = e − t cos 2t − sin 2t 1 0 2 1 x2 = e − t cos 2t + sin 2t 0 1 For the repeated root λ = 1 , and a trial solution ae t , (88) becomes:
442
Simultaneous Linear Differential Equations
10 a1 0 = 2 a2 0
10 P(1)a = 2
a1 1 so we can take a = = a2 − 1 From (56), another trial solution is b1tet + b2 et or since b1 = a (as we observed earlier, without proof)
1 t b12 t − 1te + b e 22
(100)
Substituting this into the original system, we obtain two equations each of which reduces to: 2b12 + 2b22 = 1 Hence, we can take b12 = 0 and b22 =
1 . 2
Two solutions associated with the double root λ = 1 are therefore
1 x3 = aet = et − 1 0 1 t t and x4 = b1te + b2 e = te + 1 e − 1 2 t
t
A complete solution of the original system is now given by
x = c1 x1 + c2 x2 + c3 x3 + c4 x4 To find a particular integral of a non-homogeneous system (86), or (65), in the usual case in which f (t ) is a vector function having only a finite number of linearly independent derivatives, we proceed
Chapter Eleven
443 very much as in the case of a single scalar differential equation. At the outset, it is convenient to identify the independent functions
ϕ1 (t ), ϕ 2 (t ),.........., ϕ m (t ) which appear in the components of f (t ) and then express f (t ) in the form:
f (t ) = f1 ϕ1 (t ) + f 2 ϕ 2 (t ).............. + f m ϕ m (t )
(101)
where the f ' s with subscripts are appropriate constant column vectors. This expression is then compared with a complementary function of the system being solved
xh (t ) = h1 u1 (t ) + h2 u 2 (t ).............. + hm u m (t ) in
which
the
h' s
are
constant
(102) vectors
and
u1 (t ), u 2 (t ),..................u n (t ), are known linearly independent scalar functions. For such terms of (101) as do not contain a nonzero scalar times any of the functions u j (t ) in (102) among the pertinent ϕ functions, or any of their derivatives, trial particular integrals can be constructed as described before, provided that the arbitrary scalar constants appearing in the entries in the table are replaced by undetermined constant vectors: The trial solutions are then substituted into the non-homogeneous system, and the arbitrary components of the coefficient vectors are determined to make the resulting equations identically true. For any term of (101) whose relevant ϕ function has a derivative, possibly of order zero, that is proportional to a function u j (t ) of (102), not only must the usual
444 Simultaneous Linear Differential Equations choice for a trial particular integral be multiplied by the lowest positive integral power of the independent variable which will eliminate the duplication, but the products of the normal choice and all lower nonnegative integral powers of the independent variable are also to be included in the actual choice. Finally, once a particular integral for each term on the right-hand side of (101) has been found, a particular integral for the non-homogeneous system having
f (t ) as its non-homogeneous term can be formed by addition. An example should clarify the details of the procedure. Example 22 Find a particular integral of the non-homogeneous system of equations obtained by adding 2et and 2et + e − 2t to the right-hand members of the respective equations of Example 21. Solution: At the outset, let us express the non-homogeneous term of the equivalent metric equation in the form:
0 2 f (t ) = e − 2t + et 1 2
(104)
then write the complementary function, found in Example 21, as:
c1 + 2c2 − t − 2c1 + c2 − t + xh = e t cos 2 −c e sin 2t c 2 1 c3 et + c4 e − t cos 2t + 1 − c − c3 + c4 4 2
(105)
Chapter Eleven
445
Since neither e − 2t , nor any of its derivatives, is proportional to any of the linearly independent functions
e − t cos 2t
e − t sin 2t
et
tet
appearing in the terms of xh we assume as a trial particular
0 integral, corresponding to the first term e − 2t of f (t ) , Simply 1 a1 v1 = ae − 2t = e − 2t a2 where a1 , a2 constants Then, substituting, we have:
P(D )v1 = P(D )ae − 2t = P(− 2 )ae − 2t 10 − 5 a1 − 2t 0 − 2t = = e a e − 1 5 2 1 or, dividing out e − 2t ,
10 − 5 a1 0 − 1 5 a = 1 2 1 2 1 1 1 1 It follows that a1 = , a2 = and a = . Thus, e − 2t is 9 9 9 2 9 2 one term in the particular integral we are seeking. To find a particular, integral corresponding to the second term
2 t t t 2 e of f (t ) , we note that, since both e and te occur in terms of
446 Simultaneous Linear Differential Equations the complementary (unction, the normal choice for a trial particular integral, namely, bet , is to be modified by multiplying it by t 2 and including the terms ctet and de t ,
b1 c1 Where b = c = b2 c2
d d = 1 are constant vectors to be d 2
determined. Then, substituting
v2 = bt 2 et + ctet + det into the
2 equation P(D )x = et and using the following equations: 2
( )
If D m te λt = λm te λt + mλm −1e λt
p(D )te λt = p(λ )te λt + p′(λ )e λt and
P(D )te λt = P(λ )te λt + P′(λ )e λt Where p (D ) is a polynomial in the operator D and P(D ) is a matrix whose elements are polynomials in D.
(
)
If D m t 2 e λt = λm t 2 e λt + 2mλm −1te λt + m(m − 1)λm − 2 e λt Hence
p(D )t 2 e λt = p(λ )t 2 e λt + 2 p′(λ )te λt + p′′(λ )e λt and
P(D )te λt = P(λ )t 2 e λt + 2 P′(λ )te λt + p′′(λ )e λt Where p (D ) is a polynomial in the operator D and P(D ) is a matrix whose elements are polynomials in D.
Chapter Eleven
(
447
)
∴ P(D ) bt 2 et + ctet + det = P(1)bt 2 et + 2 P′(1)btet + P′′(1)bet 2 + P(1)ctet + P′(1)cet + P(1)det = et 2 Hence, equating the coefficients of like functions on the two sides of this equation, we find that
P(1)b = 0 P(1)c + 2 P′(1)b = 0 2 P(1)d + P′(1)c + P′′(1)b = 2 Expanding these equations, and simplifying slightly, we obtain the linear system
b1 + b2
=0
b1 + b2
=0
3b1 + 8b2
+5c1 +5c2
=0
b1 +
+ c1
=0
2b2
2b1 + 2b2 2b2
+ c2
+3c1 +8c2
+10d1
+10d 2
=2
+ c1
+2d1
+2d 2
=2
+2c2
(103)
When reduced to row echelon form, the augmented matrix of this system becomes:
1 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0
− 1 0 0 1 − 2 − 2 − 2 2 2 1 0 0 0 0 0 0 0
0
448 Simultaneous Linear Differential Equations which gives us the following:
b1 0 − 1 0 b 0 1 0 2 c1 2 − 2 2 = k1 + k 2 + c 2 − 2 1 − 2 d1 0 0 1 d 2 1 0 0 as a complete solution of (103). In particular, taking k1 = k 2 = 0 , we have
− 1 − 2 0 b= c= d = 1 1 0 Finally, putting our results together, we get the entire particular solution:
1 1 − 2t − 1 2 t − 2 t e + t e + te 9 2 1 1 In the following there are some different examples to help v=
students in understanding this chapter. Example 23 Find a complete solution the following system:
(D + 2 )x + (D + 4 )y = 1 (D + 1)x + (D + 5) y = 2
(104) (105)
Solution: Subtract (105) from (104) we get the following equation:
x − y = −1 Multiply (3) by (D + 1)
(106)
Chapter Eleven
∴ (D + 1)x − (D + 1) y = −(D + 1)
∴ (D + 1)x − (D + 1) y = −1
449 (107)
Subtract (107) from (105) we get the following equation:
∴ (2 D + 6 ) y = 3 2
dy + 6y = 3 dt
This is first order ordinary differential equation. The solution of this equation is as following:
1 + c1e − 3t 2
y (t ) =
Substitute y (t ) into (106)
∴ x − y = −1
∴x =
−1 + c1e − 3t 2
x ∴ = y
1 1 − 1 + c1 e − 3t 2 1 1
Example 24 Find a complete solution the following system:
(D + 1)x + (4 D − 2) y = t − 1
(108)
(D + 2 )x + (5D − 2 )y = 2t − 1
(109)
Solution: Subtract (109) from (108) we get the following equation:
− x + (− Dy ) = −t
450 Simultaneous Linear Differential Equations − x − Dy = −t
(110)
Multiply (110) by (D + 1) we get the following equation:
∴ −(D + 1)x − D(D + 1) y = −(D + 1)t
(
)
∴ −(D + 1)x + − D 2 − D y = −1 − t
(111)
Add (111) to (108) we get the following equation:
(− D 2 + 3D − 2)y = −2 (D 2 − 3D + 2)y = 2
It is second order ordinary differential equation.
∴ y h (t ) = c1e t + c2 e 2t ∴ y p (t ) = 1 y (t ) = y h + y p = c1et + c2 e 2t + 1 Substitute in (110) we get the following:
− x − c1et − 2c2 e 2t = −t ∴ x = t − c1e t − 2c2 e 2t x − 1 − 2 t = c1 et + c2 e 2t + y 1 1 1 The above example can be solved by another technique as following:
λ + 1 4λ − 2 λ + 2 5λ − 2 = 0
Chapter Eleven
451
∴ λ2 − 3λ + 2 = 0 ∴ λ1 = 1 , and λ2 = 2 For λ1 = 1
∴ a = −b Let b = c1 , a = −c1
x − 1 ∴ X 1 = = c1 et y 1 For λ2 = 2
3a + 6b = 0
∴ a = −2b Let b = c2 , a = −2c2
x − 2 ∴ X 2 = = c2 e 2t y 1 t − 1 0t Q f (t ) = e 2 − 1 t ∴λ = 0
a − 2b = t − 1
(112)
2a − 2b = 2t − 1
(113)
Subtract (112) from (113) we get a = t ∴b = 1 / 2
x − 1 − 2 t = c1 et + c2 e 2t + y 1 1 1 / 2
452 Simultaneous Linear Differential Equations Example 25 Find a complete solution the following system:
(2 D + 11)x + (D + 3) y + (D − 2)z = 14et
(114)
(D − 2)x + (D − 1) y + Dz
(115)
= −2et
(D + 1)x + (D − 3) y + (2 D − 4)z = 4et
(116)
Solution:
2λ + 11 λ + 3 λ − 2 λ − 2 λ −1 λ = 0 λ + 1 λ − 3 2λ − 4 ∴ λ3 + 6λ2 + 11λ + 6 = 0
λ1 = −1 , λ2 = −2 , and λ3 = −3 For λ1 = −1
9a + 2b − 3c = 0
− 3a − 2b − c = 0 −4b − 6c = 0 Let c = c1
∴b =
−3 c1 2
4 x 1 −t ∴ X 1 = y = c1 − 9 e z 6 6 For λ2 = −2 7 a + b − 4c = 0
− 4a − 3b − 2c = 0 − a − 5b − 8c = 0
∴a =
2 c1 3
Chapter Eleven
453 After simplification of the above equation we get the following: Let c = c2 ∴ b = −
30 14 c2 and ∴ a = c2 17 17
x 14 1 ∴ X 2 = y = c2 − 30 z 17 17 For λ3 = −3
5a − 5c = 0 Let a = c = c3
− 5a − 4b − 3c = 0 ∴ b = −2c3 1 x X 3 = y = c3 − 2 e − 3t 1 z
14e t Q f (t ) = − 2et t 4e ∴λ = 1 13a + 4b − c = 14 −a
+ c = −2
2a − 2b − 2c = 4 After simplification of the above equation we get the following:
a = 1, b = 0 , and c = −1
Simultaneous Linear Differential Equations
454
x 1 ∴ X 4 = y = 0 e t z − 1
X = X1 + X 2 + X 3 + X 4 Example 26: Find a complete solution of each of the following system:
(2 D + 1)x + (D + 2 )y = sin t (3D + 1)x + (3D + 5) y = cos t
(117) (118)
Multiply (117) by 3 then subtract the result equation from (118) we get the following equation:
(3D − 2)x − y = cos t − 3 sin t
(119)
Multiply (3) by (D + 2) ).
(
)
∴ − 3D 2 − 8D − 4 x − (D + 2 ) y = −7 sin t − cos t
(120)
Add (118) and (120).
( ) ∴ (3D 2 + 6 D + 3)x = 6 sin t + cos t
∴ − 3D 2 − 6 D − 3 x = −6 sin t − cos t (121)
It is clear that (121) is a second order ordinary differential equation and has the following solution:
xh (t ) = (c1 + c2t )e − t ∴ x p (t ) = A cos t + B sin t
∴ x′p (t ) = − A sin t + B cos t
Chapter Eleven
455
∴ x′p′ (t ) = − A cos t − B sin t
∴ 3(− A cos t − B sin t ) + 6(− Asi int + B cos t ) + 3( A cos t + B sin t ) = 6 sin t + cos t
Coefficient of sin t = 6
∴ 6 = −3B − 6 A + 3B ∴ A = −1 Coefficient of cos t = 1
∴1 = −3 A + 6 B + 3 A
∴ B = 1/ 6 ∴ x p (t ) = 1 / 6 sin t − cos t
∴ x(t ) = (c1 + c2t )e − t + 1 / 6 sin t − cos t Substitute the value of x(t ) in equation (119) we get the following equation:
∴ (3D + 2 )x + y = 3 sin t − cos t
[
∴ y (t ) = 3 sin t − cos t − 3 − (c1 + c2t )e − t + c2 e − t + 1 / 6 cos t + sin t
[
−2 (c1 + c2t )e − t + 1 / 6 sin t − cos t − sin t cos t ∴ y (t ) = + + (c1 + c2t )e − t − 3c2 e − t 3 2
]
]
456 Simultaneous Linear Differential Equations Problems: 1- Find the characteristic values and the corresponding characteristic vectors of each of the following matrix:
− 4 5 5 − 5 6 5 − 5 5 6 2- Find a complete solution of each of the following first order systems. Also fiend the solution that satisfies the given condition when one is given a)
x1′ = 3x1 + 2 x2 x2′ = − 2 x1 − x2
−3 x1 − 2 x2 2 b) 5 x2′ = 2 x1 + x2 2 x1′ =
c)
x1′ = 3 x1 − 2 x2 x2′ = 5 x1 − 3 x2
d)
x1′ = 3 x1 + 2 x2 x2′ = − 2 x1 − x2
e)
x1′ = x1 − x2 x2′ = x1 + 3 x2
f)
x1′ = −2 x1 + x2 3 X (0 ) = x2′ = − 3x1 + 2 x2 + 2 sin t 4
Chapter Eleven
x1′ = 3 x1 − 2 x3 3 g) x2′ = − x1 + 2 x2 + x3 , X (0 ) = 1 3 x3′ = 4 x1 − 3 x3
x1′ = 4 x1 − 2 x2 − 10 x3 − 3 X (0 ) = − 2 x2 h) x2′ = − 1 x3′ = 2 x1 − x2 − 5 x3 x1′ = x1 − 2 x2 − x3 i) x2′ = x1 + 2 x2 + e − t x3′ = − x1 − 3x2 − 1 3- Find a complete solution of each of the following system: a)
(2 D + 1)x + (D + 2) y = 0 (D + 3)x + (D + 6) y = −3et
b)
(2 D + 1)x + (D − 1) y = −3 cos t (D + 2)x + (D + 3) y = 5 sin t
c)
(2 D + 1)x + (D + 2) y = 6et (D + 2)x + (D + 4) y = 4e −t
2t d) (D + 5)x + (D + 7 ) y = 4e (2 D + 1)x + (3D + 1) y = 0
e)
(2 D + 1)x + (D 2 + 6 D + 1)y = 0 (D + 2)x + (D 2 + 2 D + 5)y = 6 2t
457
458
Simultaneous Linear Differential Equations
f)
(2D 2 + 5)x + (D 2 + 3)y = −8 sin 3t (D 2 + 7)x + (D 2 + 5)y = 8 sin 3t
g)
(2 D + 1)x + (D + 1) y = 0 (D − 2)x + (D − 1) y = 0
h)
(D + 5)x + (2 D + 1) y = −e − t + et (D + 7 )x + (3D + 1) y = 0
i)
(D + 5)x + (D + 7 ) y = 2et (2 D + 1)x + (3D + 1) y = et
j)
(2 D + 1)x + (D + 2) y = e −t (3D − 7 )x + (3D + 1) y = 0
k)
(2 D + 1)x + (D + 2) y = 8e −t
(D 2 + D + 9)x + (D 2 − 2D + 12)y = 6
l)
(3D + 1)x + (D + 7 ) y = e − t (2 D + 1)x + (D + 5) y = e − t
m)
(D + 1)x + (D + 2) y = −et (3D + 1)x + (4 D + 7 ) y = −7et
n)
(D + 2)x + (D + 3) y = −4 (2 D − 6)x + (3D − 4) y = 2
o)
(D + 1)x + (D + 2) y = −t + 1 (5D + 1)x + (6 D + 3) y = −2t + 1
Chapter 12 Special Functions 12.1 Gamma Function The Gamma function Γ(q ) is defined by the following integral ∞
Γ(q ) = ∫ x q −1e − x dx , Where q>0
(1)
0
Some basic formulas for Gamma function are included in the appendix. One of the most important formulas of Gamma function is the recursion or recurrence formula for the Gamma function.
Γ(q + 1) = qΓ(q) Proof:
(2)
∞
Γ(q + 1) = ∫ x q e − x dx 0
By integration by parts we obtain:
Γ(q + 1) =
∞ −e − x x q 0
∞
+ q ∫ x q −1 e − x dx = qΓ(q ) 0
In case of q is positive integer, then:
Γ(q + 1) = q!
,
q = 1, 2, 3, 4,........
(3)
This shows that the Gamma function may be regarded as a generalization of the factorial function.
460
Special Functions
Γ(1)
Example 1 Find ∞
Solution: Γ(1) = ∫ x1−1 e − x dx = − e − x 0
∞ 0
=1
Γ(1 / 2)
Example 2 Find Solution: ∞
Γ(1 / 2) = ∫ x (1 / 2 )−1e − x dx
(4)
0
Assume x = u 2 ∞
∴dx = 2u du and substitute in (4) , we get:
2
Γ(1 / 2) = 2 ∫ e − u du
(5)
0
∞
2
We can say that: Γ(1 / 2) = 2 ∫ e − v dv
(6)
0
Multiply each term of (5) and (6) , we get:
(Γ(1 / 2) )
2
∞
= 2∫e
−u 2
0 ∞∞
∞
2
du * 2 ∫ e − v dv
(7)
0
(Γ(1 / 2) )2 = 4 ∫ ∫ e − (u
2
+v2 )
du dv
(8)
00
Changing to polar coordinates Where, u = r cos θ and v = r sin θ , then (8) becomes:
(Γ(1 / 2) )
2
π /2∞
=4
∫
∫e
0 0
−r 2 ∞
0
2
r dr dθ = 4 * π / 2 ∫ e − r r dr
∴ (Γ(1 / 2) ) = 4 * π / 2 ∫ e 2
∞
0
−r 2
∞ 1 −r 2 r dr = 2π * − ∫ e (−2r ) dr 20
461
Chapter Twelve
∴ (Γ(1 / 2) ) = −π e 2
−r 2
∞ 0
=π
∴ Γ(1 / 2) = π Example 3 Find
(9)
Γ (7 ) 2Γ(4)
Solution
Γ (7 ) 6! = = 60 2Γ(4) 2 * 3! Example 4 Find Γ(4.8) Solution:
Γ(4.8) = 3.8 * 2.8 * 1.8Γ(1.8) From the Gamma function table (shown below) we get:
Γ(4.8) = 3.8 * 2.8 * 1.8 * 0.9314 = 17.8382
q 1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1.45 1.50
Γ(q) 1.0000 0.9735 0.9514 0.9330 0.9182 0.9064 0.8975 0.8912 0.8873 0.8857 0.8863
q 1.55 1.60 1.65 1.70 1.75 1.80 1.85 1.90 1.95 2.00
Γ(q) 0.8889 0.8935 0.9001 0.9086 0.9191 0.9314 0.9456 0.9618 0.9799 1.0000
462
Special Functions
Example 5 Find Γ(5.5) Solution:
Γ(5.5) = 4.5 * 3.5 * 2.5 * 1.5 * 0.5 Γ(0.5) = 4.5 * 3.5 * 2.5 * 1.5 * 0.5 * π = 52.3428 In case of q<0 (negative), equation (3) can be used Example 6 Find Γ(−3.4) Solution:
Γ(− 0.4 ) Γ(− 2.4 ) Γ(− 1.4 ) = = (− 3.4)(− 2.4) (− 3.4)(− 2.4)(− 1.4) − 3. 4 Γ(0.6 ) = (− 3.4)(− 2.4)(− 1.4)(− 0.4) Γ(1.6 ) = (− 3.4)(− 2.4)(− 1.4)(− 0.4)(0.6) 0.894 = = 0.32607 (− 3.4)(− 2.4)(− 1.4)(− 0.4)(0.6)
Γ(−3.4) =
Example 7 Find Γ(−2) Solution:
Γ(−2) =
Γ(−1) Γ ( 0) = =∞ (− 2) − 2 * −1
For any integer n the following formula can be used
π (2n )! 1 Γ n + = 2 n * 2 2 n!
(10)
463
Chapter Twelve
Proof: 1 1 3 5 3 1 1 Γ n + = n − n − n − ..... * Γ 2 2 2 2 2 2 2 3 1 2n − 1 2n − 3 2n − 5 = ...... * * π 2 2 2 2 2
= = = =
π 2
n
(2n − 1)(2n − 3)(2n − 5).....3 *1
π (2n )(2n − 1)(2n − 2 )(2n − 3)(2n − 4)(2n − 5).....3 * 1 2n
(2n )(2n − 2)(2n − 4)......4 * 2
π (2n )(2n − 1)(2n − 2 )(2n − 3)(2n − 4 )(2n − 5).....3 *1 22n
(n )(n − 1)(n − 2)......2 *1
π (2n )!
2 2n (n )!
Example 8 Find Γ(3.5) Solution
1 π (2n )! Q Γ n + = 2 n (n )! 2 2 1 π (6)! 15 π ∴ Γ(3.5) = Γ 3 + = 6 = 2 2 (3)! 8 Example 9 Evaluate the following integral: ∞
∫x
0
3 −x
e
dx
464
Special Functions ∞
Solution:
∫x
3 −x
e
0
∞
dx = ∫ x 4 −1e − x dx = Γ(4) = 3! = 6 0
∞
Example 10 Evaluate the following integral Solution: Let 2 x = y , ∴2dx = dy
∫x
6 −2x
e
dx
0
Substitute in the required integral we get: ∞
∫x
0
6 −2x
e
∞
∞
∞
6
dy 1 1 6! 45 y dx = ∫ e− y = 7 ∫ ( y )6 e− y dy = 7 Γ(7) = 7 − = 2 2 2 0 8 2 2 0
∴ ∫ x 6 e − 2 x dx = 0
45 8 ∞
Example 11 Evaluate the following integral
4
−x ∫ e dx
0
Solution: Let t = x 4 , ∴ x = t1 / 4 ,
1 ∴ dx = t − 3 / 4 dt 4
Substitute in the integral, we get ∞
−x ∫e
0
∞
4
1 Γ 1 + 1 1 1 1 4 dx = ∫ t − 3 / 4e − t dt = Γ = 40 4 4 4 1 4 = Γ(1.25) = 0.96785 ∞
4
∴ ∫ e − x dx = 0.96785 0
465
Chapter Twelve
Example 12 Evaluate the following integral 1
∫
− Ln( x ) dx
0
Solution: Assume − Ln( x ) = t ,
∴ x = e − t , dx = −e − t dt
x = 0 ⇒t = ∞ x =1 ⇒t = 0
When,
Substitute in the above integral we get: 1
∫
− Ln( x ) dx =
0
∫
∞
0
(
t −e
−t
) dt = ∫ (t )1 / 2 (et ) dt ∞ 0
= Γ(3 / 2 ) =
1 1 π Γ = 2 2 2
Example 13 Evaluate the following integral ∞
∫4 x
e−
x
dx
0
Solution: Assume u =
x
∴ x = u2
∴ dx = 2udu Then at x = 0,
x=∞
u = 0 and
u=∞
466 ∞
Special Functions 4
∴∫ x e
− x
dx =
0
∫ (u
∞
)
2 1 / 4 −u
e
0
∞
(2udu ) = 2 ∫ u 3 / 2e − u du 0
It is clear the above integral is in the form of Gamma function, where,
q −1 = 3 / 2 ∞
∴ ∫ 4 x e−
∴q = 5/ 2 x
dx = 2Γ(2.5)2 * 1.5 * 0.5 * Γ(0.5)
x
dx = 2.65868
0
∞
∴ ∫ 4 x e− 0
Example 14 Evaluate the following integral ∞
∫e
− x3
dx
0
Solution: Assume x 3 = u ∴ x = u1 / 3
1 ∴ dx = u − 2 / 3du 3 Substitute in the above integral we get the following: ∞
∴ ∫e
− x3
0
∞
∴ ∫e 0
− x3
∞
1 dx = ∫ e − u u − 2 / 3du 3 0 1 ∞ − 2 / 3 −u dx = ∫ u e du 30
467
Chapter Twelve
It is clear the above integral in the form of Gamma function, where,
q − 1 = −2 / 3 , ∴ q = 1 / 3 3 1 1 Γ(4 / 3) ∴ ∫ e − x dx = Γ(1 / 3) = = 0.8933 3 3 1 / 3 0
∞
Example 15 Evaluate the following integral ∞
∫3
−x 3
x dx
0
Solution: −x It is clear that 3− x = e ln 3 = e − x ln 3 = e − x e ln 3 = 3e − x
∞
∴ ∫3
−x 3
0
∞
x dx = 3 ∫ e − x x 3dx 0
It is clear that the above integral is in the form of Gamma function, where,
q −1 = 3
∴q = 4
∞
∴ ∫ 3− x x 3dx = 3Γ(4 ) = 18 0
Example 16 Evaluate the following integral ∞
(x − e ∫e
−x
−∞
Solution:
)dx
468
Special Functions
Assume e − x = u
∴ −e − x dx = du ∴ dx =
du − e− x
=
− du u
At x = ∞
u = 0 and,
At x = −∞
u = ∞ and,
∴
∞
∫
0 ∞ ( x −e ) − ln (u )− u − du − ln (u ) − u du e dx = e e = e −x
−∞
∴
∞
∫
∫
∞
u
∫
0
u
∞ ( x −e ) e dx = u − 2 e − u du −x
−∞
∫
0
It is clear that the above integral is in the form of Gamma function, where,
∴ q − a = −2 ∴ q = −1
∴ Γ(− 1) = ∴
∞
(x − e ∫e
−∞
Γ(0 ) =∞ −1
−x
)dx = ∞
469
Chapter Twelve
12.2 Beta Function Beta function B (m, n ) is defined by: 1
B (m, n) = ∫ x m −1 (1 − x )n −1 dx
(11)
0
Where m > 0 , n > 0
Beta function has the following relation with Gamma function:
Γ ( m) Γ ( n ) Γ( m + n)
B(m, n) =
(12)
Proof ∞
Q Γ(m) = ∫ t m −1e − t dx
(13)
0
Substitute in equation (13) for t = x 2 , dt = 2 xdx ∞
2
∴ Γ(m) = ∫ x 2m − 2e − x 2 x dx
(14)
0
∞
2
= 2 ∫ x 2 m −1e − x dx 0
Similarly, ∞
2
Γ(n) = 2 ∫ y 2 n −1e − y dy 0
Multiply both sides of (14) and (15) we get:
(15)
470
Special Functions ∞∞
Γ(m) Γ(n) = 4 ∫ ∫ x 2 m −1 y 2 n −1e − ( x
2
+ y2 )
dx dy
00
Change the double integral from cartesian to polar form, then:
x = r cos θ ,
y = r sin θ , x 2 + y 2 = r 2 , dx dy = rdr dθ ∞π / 2
∴ Γ ( m) Γ ( n) = 4 ∫
2 m −1 2 n −1 − r ∫ (r cos θ ) (r sin θ ) e
2
0 0
∞ 2 m + 2 n − 2 − r 2 e 2r dr ∴ Γ(m) Γ(n) = ∫ (r ) 0 π /2 *2 ∫ (cos θ )2 m −1 (sin θ )2 n −1 dθ 0 ∞ 2( m + n −1) − r 2 ∴ Γ(m) Γ(n) = ∫ (r ) e dr 2 0
π /2 *2 ∫ (cos θ )2 m −1 (sin θ )2 n −1 dθ 0 ∞ ( m + n −1) − t Γ ( m) Γ ( n) = ∫ t e dt 0 π /2 *2 ∫ (cos θ )2 m −1 (sin θ )2 n −1 dθ 0
∴ Γ(m) Γ(n) = Γ(m + n) * B(m, n) ∴ B ( m, n ) =
Γ ( m) Γ ( n) Γ ( m + n)
r dr dθ
471
Chapter Twelve
For all m and n we have:
B (m, n) = B (n, m)
(16)
Proof:
Substitute in (1) by
1 − x = t , Then x = 1 − t ,
dx = − dt
x = 0 ⇒ t =1 x =1 ⇒ t = 0 So we have 1
B ( m, n ) = ∫ x
m −1
(1 − x )
0 1
n −1
0
dx = ∫ (1 − t )m −1 (t )n −1 (− dt ) 1
= ∫ (t )n −1 (1 − t )m −1 (dt ) = B (n, m) 0
∴ B(m, n) = B(n, m) Example 17 Evaluate the following integral 1
3 4 ∫ x (1 − x ) dx
0
Solution: 1
Γ(5) * Γ(4) 4!*3! 1 3 4 ∫ x (1 − x ) dx = B(5,4) = Γ(9) = 8! = 280
0
472
Special Functions
Example 18 Evaluate the following integral 4
1/ 2 2 ∫ t (4 − t ) dt
0
Solution 4
∫ t (4 − t )
1/ 2
2
0
4
1/ 2
2
t dt = 2∫ t 1 − 4 0
dt
Assume x = t / 4 the t = 4 x, and dt = 4dx
t =0⇒ x=0 t = 4 ⇒ x =1 4
∴ ∫ t (4 − t )
1/ 2
2
0
4
dt = 2 ∫ (4 x )2 (1 − x )1 / 2 4dx 0
4
= 2 * 4 * 4 ∫ ( x )2 (1 − x )1 / 2 dx 2
0
Γ(3) * Γ(3 / 2) 3 = 2 7 * B 3, = 2 7 Γ(9 / 2) 2 = 27
2!*[1 / 2 * Γ(1 / 2)] = 19.505 [7 / 2 * 5 / 2 * 3 / 2 * 1 / 2 * Γ(1 / 2)]
Example 19 Evaluate the following integral 2
∫x
3
(8 − x3 ) dx
0
Solution: 2
∫x
0
3
(8 − x ) 3
x 3 dx = ∫ 2 x1 − dx 2 0 2
473
Chapter Twelve 3
x Assume that = u , then x = 2 u1 / 3 2 ∴ x = 0, u = 0 And x = 2, u = 1 1 0
(
)
1/ 3 2
∴ I = 2 * ∫ 2u1 / 3 (1 − u ) I=
(
u 3
−2 / 3
du
)
1/ 3 8 1 −1 / 3 ( ) u 1 − u du 3 ∫0
2 4 Γ Γ 8 2 4 8 3 3 ∴ I = B , = 3 3 3 3 Γ(2 ) 4 ∴ Γ = Γ(1.333) 3 We know that Γ(1.3) = 0.8975 and Γ(1.35) = 0.8912 0.8975
0.0063 x 0.8 912 0.02
1.3
1.33
1.35
474
Special Functions
from the figure shown above
x 0.02 = 0.0063 0.05
∴ x = 0.0021 ∴ Γ(1.33) = 0.8912 + x = 0.8912 + 0.0021 = 0.8933 The same way
5 Γ = Γ(1.6667 ) = 0.903 3 Also,
2 3 5 3 Γ = Γ = * 0.903 = 1.3545 3 2 3 2 2
∴∫x
3
0
(8 − x3 ) dx = 83 * 0.89331*1.3545 = 3.2266
Example 20 Evaluate the following integral
(
1
)
x 3 1 − x 2 dx
∫
0
Solution:
Assume x 2 = u , ∴ x = u and, 1
∴∫
x3
∴∫
x3
0 1 0
∴ dx =
1 −1 / 2 u du 2
(1 − x ) dx = ∫ (u1/ 2 )1/ 2 (1 − u )1/ 3 12 u −1/ 2du 2
(1 − x ) 2
1
0
1 1 −1 / 4 (1 − u )1 / 3 du dx = ∫ u 20
475
Chapter Twelve
It is clear that the above integral is in the form of Beta function, where,
m − 1 = −1 / 4 and n − 1 = 1 / 3 ∴ m = 3 / 4 and n = 4 / 3
(
1
)
∴ ∫ x 3 1 − x 2 dx = 0
1 3 4 1 Γ(3 / 4 ) * Γ(4 / 3) B , = = 0.527 2 4 3 2 Γ(25 / 12 )
Example 21 Evaluate the following integral: ∞
dx
∫
(
x 4 + x2
0
)
Solution:
Assume u
4 4 + x2
∴ x2 =
,
4 −4 u
1/ 2
4 ∴ x = − 4 u
∴ dx = −2u
−2 4
− 4 u
−1 / 2
At x = 0 ,
u = 1 , and,
At x = ∞ ,
u=0
∞
∴∫
0
dx
(
x 4 + x2
)
0
du
1 / 2 −1 / 2
4 = ∫ − 4 u 1
(
)
u −2 4 − 2u − 4 4 u
−1 / 2
du
476
Special Functions
∞
∴∫
0
0
dx
(
x 4+ x
∞
∴∫
0
0
2
dx
(
x 4 + x2
∞
∴∫
)
21 4 = ∫ u −1 − 4 40 u
(
x 4 + x2
∞
∴∫
)
2 1 −1 4 = ∫ u − 4 40 u
dx
dx
(
x 4+ x
2
)
)
−3 / 4
−3 / 4
1 1 −1 4 du = ∫ u (1 − u ) 20 u
11 4 du = ∫ u −1 (1 − u ) 20 u
−3 / 4
du
−3 / 4
du
1 − 3 / 4 1 −1 3 / 4 = 4 u u (1 − u )− 3 / 4 du ∫ 2 0
=
1 2*4
3/ 4
1
∫u
−1 / 4
(1 − u )− 3 / 4 du
0
It is clear that the above integral is in the form of Beta function, where,
m − 1 = −1 / 4 and n − 1 = −3 / 4
∴ m = 3 / 4 and n = 1 / 4 ∞
∴∫
0
∞
∴∫
0
∞
∴∫
0
dx
(
x 4 + x2 dx
(
x 4 + x2 dx
(
x 4+ x
2
B (3 / 4,1 / 4 )
)
=
)
=
)
= 0.785398
2 * 43 / 4
=
Γ(3 / 4 ) * Γ(1 / 4 ) 2 * 43 / 4 Γ(1)
Γ(1.75) * Γ(1.25) 2 * 43 / 4 * 0.75 * 0.25
=
0.9191 * 0.9064 2 * 43 / 4 * 0.75 * 0.25
Example 22 Evaluate the following integral: 1
1
∫ Ln x dx
0
477
Chapter Twelve
Solution:
1 Assume ln = u x ∴x =
1 eu
∴ − ln x = u or
= e−u
1 = eu x
∴ dx = −e − u du
At x = 0 u = ∞ and At x = ∞
u=0
1
0
( )
∞
1 ∴ ∫ Ln dx = ∫ u − eu du = ∫ ue − u du x ∞ 0 0 It is clear that the above integral is in the form of Gamma function, where,
∴ q − 1 = 1∴ q = 2 1
1 ∴ ∫ Ln dx = Γ(2 ) = 1Γ(1) = 1 x 0 Problems
(I) Evaluate the following Gamma functions
Γ(6.4 ) ,
Γ(− 4.3) ,
Γ(− 2 ) Γ(4.5) Γ(2.3)
Γ(− 2 ),
Γ(0.7 ),
(II) Evaluate the following integrals 1
1)
∫
0
1
dx 1− x
q
2)
∫
0
x3
(1 − x ) dx 2
∞
3)
∫x
0
q −x
q
dx
Chapter 13 Numerical Analysis. 13.1 Numerical Solution Of Equations 13.1.1 Introduction Equations of various kinds arise in a range of physical applications and a substantial body of mathematical research is devoted to their study. Some equations are rather simple: in the early of our mathematical education we all encountered the simple linear equation ax + b = 0 , where a and b are real numbers and a ≠ 0 , whose solution is given by the formula x = −b / a . Many equations, however, are nonlinear: a simple example is ax 2 + bx + c = 0 , involving a quadratic polynomial with real coefficients a, b, c . The two solutions to this equation, labeled x1 and x2 , are found in terms of the coefficients of the polynomial form the familiar formulae:
− b + b 2 − 4ac , x1 2a
− b − b 2 − 4ac x2 2a
It is easy likely that you have seen the more intricate formula for the solution of cubic and quadratic polynomial equations. But unfortunately there is no any closed formula for finding roots for 5th order polynomial or any nonlinear equations. Our goal is to develop simple numerical methods for the approximate solution of the equation f ( x ) = 0 . Methods of the kind
Chapter Thirteen
479
discussed here are iterative in nature and produce sequences of real numbers, which in favorable circumstances, converge to the required solution. 13.1.2 Simple Iteration Method In this method, we rearrange the function f ( x ) = 0 so that x has to be in separate side of the equation, so that x = g ( x ) where
f ( x ) = g ( x ) − x . Then we chose a starting value for x = xo and compute x1 , x2 ,......xn for the following relation:
xn +1 = g ( xn )
(3)
The solution of f ( x ) = 0 is called fixed point of g ( x ) . Let us illustrate this method with the following example. Example 1 Find the solution of the following equation by using simple iteration method: f ( x ) = x 2 − 3 x + 2 = 0 Solution: We know the solution of this equation is 1, 2. So, we can watch the error in each iteration process. The equation can be
x2 + 2 rearranged to be like this: x = 3 If we chose x0 = 0 , then we can do the following iterations:
x02 + 2 x1 = = 0.666 666 666 3
480 Numerical Analysis
x12 + 2 x2 = = 0.814 814 815 3 x22 + 2 x3 = = 0.887 974 394 3 x32 + 2 x4 = = 0.929 499 508 3 and so on till
x10
x92 + 2 = = 0.994 366 675 3
The results are very clear, where are going to the first root x = 1 . To get the other root, assume x0 = 5 .
x02 + 2 ∴ x1 = = 9.0 3 x12 + 2 ∴ x2 = = 27.666 666 667 3 x22 + 2 ∴ x3 = = 255.814 815 3 It is clear that, these results diverge and we can not get the solution. If we use another value for x0 to be between 1 and 2 the results will converge to the same root x = 1 . Then we have to try to rearrange our equation in different manner as following:
x = 3x − 2
481
Chapter Thirteen
It is clear, we can not assume xo <
2 because of the root will give 3
us imaginary part. So, chose x0 = 1.5
∴ x1 = 3x0 − 2 = 1.581 138 83 ∴ x2 = 3 x1 − 2 = 1.656 326 202 ∴ x3 = 3 x2 − 2 = 1.723 072 433 ∴ x4 = 3 x3 − 2 = 1.780 229 56 and so on till
∴ x10 = 3 x9 − 2 = 1.954 534 710 It is very clear that the results going to the second solution x = 2 . Let us see if we choose x0 = 3 in the previous example
∴ x1 = 3 x0 − 2 = 2.645 751 31 ∴ x2 = 3 x1 − 2 = 2.436 648 09 ∴ x3 = 3 x2 − 2 = 2.304 331 634 ∴ x4 = 3 x3 − 2 = 2.216 527 668 and so on till
∴ x10 = 3x9 − 2 = 2.034 098 753 It is clear also these results will converge also to the second root
x = 2. Convergence theory:
482 Numerical Analysis Let x = x0 be chosen starting point for iteration. Then, if
g ′( x0 ) < 1 , then the iteration process defined by (3) converges. Let us check the previous example
x2 + 2 Q x = g (x ) = 3 g ′( x ) =
2x 3
The assumption x0 = 0 gives g ′( x0 ) < 1 then this assumption converges as shown before. But when we chose
g ′( x0 ) =
10 > 1, 3
so
divergence
occurs.
x = g (x ) = 3x − 2 g ′( x ) =
3 <1 2 3x − 2
9 ∴ < 3x − 2 4 ∴ 3x >
9 −2 4
∴x >
9 2 − 12 3
∴x >
1 12
So, x must be greater than
1 for convergence. 12
Also
xo = 5 when
483
Chapter Thirteen
13.1.3 Bisection Method As we know the root of any function is the intersection of the graph of the function with x axis . So, it is clear that f ( x ) changes its sign on after and before the root point. This means that if f ( x ) is a continuous on the interval between point a and b and f (a ) and
f (b ) have opposite signs, then there is at least one real root between those two points. Suppose f ( x ) is continuous on [a, b] and f (a ) and f (b ) have opposite signs, i.e.,
f (a ) * f (b ) < 0 then there exists a root
x* ∈ (a, b ) . To find an approximation of x* ∈ (a, b ) , we proceed as follows: First, assume, for demonstration purposes, that a0 = a and
b0 = b and divide the interval to half by at the point c0 as following:
c0 =
a0 + b0 2
So, c0 is the midpoint of the interval [a0 , b0 ] , then compute
f (c0 ) . If f (c0 ) = 0 then the point c0 is the required root and there is no any other computation needed. If f (c0 ) ≠ 0 , then the sign of f (c0 ) consides either with the sign of f (a0 ) or with the sign of f (b0 ) . Thus at the end points of one of the two intervals
484 Numerical Analysis
[a0 , c0 ] or [c0 , b0 ] the function f (x ) has the same signs and at the end points of the other opposite signs. We retain the interval at the end points of which f ( x ) has opposite signs and reject the other interval since it does not contain the required root. We denote the retained interval by [a1 , b1 ] , where
c0 sign f (a0 ) = sign f (c0 ) a1 = a0 sign f (a0 ) ≠ sign f (c0 )
(4)
c0 sign f (b0 ) = sign f (c0 ) b1 = b0 sign f (b0 ) ≠ sign f (c0 )
(5)
Obliviously,
signs
of
f (a1 ) = sign f (a0 )
and
sign
of
f (b1 ) = sign of f (b0 ) . Therefore, f (a1 ) * f (b1 ) < 0 . The desired root is now on the interval [a1 ,b1 ] which is of half length. Further we proceed in similar way. Suppose we have found some interval [ak , bk ] ⊂ [a, b] at the end points of which the function
f ( x ) has opposite signs and, consequently, which contains the desired root x* . We fiend the midpoint of the interval [ak , bk ] :
ck =
ak + bk 2
and compute f (ck ) . If f (ck ) = 0 then x* = ck . The computation have come to an end. If f (ck ) ≠ 0 , then we set the following:
Chapter Thirteen
485
ck sign f (ak ) = sign f (ck ) ak +1 = ak sign f (ak ) ≠ sign f (ck )
(6)
ck sign f (bk ) = sign f (ck ) bk +1 = bk sign f (bk ) ≠ sign f (ck )
(7)
And so forth. This process may be finite if the midpoint of the interval obtained at some step coincide with the desired root x* , otherwise this process is infinite. Fig.1 shows several initial steps. If the computations are continued to the kth step, then it is natural to take ck as an approximate value for the desired root x* . Hence, the obvious error estimate is valid:
x * − ck ≤
b−a
2
(8)
k +1
y
a = a0
a1 a2
a3
x
b3
b2
b = b0 b1
Fig.1 Several initial steps using bisection method.
486 Numerical Analysis Example 2 Find the solution of the following equation by using bisection method: f ( x ) = x 2 − 3 x + 2 = 0 Solution:
If
we
look
to
the
previous
f (x ) = x 2 − 3x + 2 = 0 Assume a0 = 0 , and, b0 = 1.5
∴ f (a0 ) = 2 , and, ∴ f (b0 ) = −0.25 f (a0 ) * f (b0 ) < 0
a0 + b0 0 + 1.5 = = 0.75 2 2 f (c0 ) = 0.3125
∴c0 =
It is clear that sign of f (c0 ) is the same as the sign of f (a0 )
∴ a1 = c0 = 0.75 , and, b1 = b0 = 1.5
∴ f (a1 ) = 0.3125 , and, ∴ f (b1 ) = −0.25 ∴ f (a1 ) * f (b1 ) < 0
c1 =
a1 + b1 = 1.125 2
f (c1 ) = −0.109 375 It is clear that sign of f (c1 ) is the same as the sign of f (a1 )
a2 = a1 = 0.75 , and, b2 = c1 = 1.125 ∴ f (a2 ) = 0.3125 , and f (b2 ) = −0.109 375
c2 =
a2 + b2 0.75 + 1.125 = = 0.937 5 2 2
example
487
Chapter Thirteen
f (c2 ) = 0.066 406 25 It is clear that sign of f (c2 ) is the same as the sign of f (a2 )
a3 = c2 = 0.937 5 , and, b3 = b2 = 1.125 f (a3 ) = 0.066 406 25 , and, f (b3 ) = −0.109 375
c3 =
a3 + b3 = 1.031 25 2
f (c3 ) = −0.030 273 438 The process continues until we get the required error. The results will be tabulated in Table(1). Table(1) The results for Example 2. i
a
b
c
f(a)
f(b)
f(c )
0 0.000000 1.500000 0.750000 2.000000 -0.250000 0.312500 1 0.750000 1.500000 1.125000 0.312500 -0.250000 -0.109375 2 0.750000 1.125000 0.937500 0.312500 -0.109375 0.066406 3 0.937500 1.125000 1.031250 0.066406 -0.109375 -0.030273 4 0.937500 1.031250 0.984375 0.066406 -0.030273 0.015869 5 0.984375 1.031250 1.007813 0.015869 -0.030273 -0.007751 6 0.984375 1.007813 0.996094 0.015869 -0.007751 0.003922 7 0.996094 1.007813 1.001953 0.003922 -0.007751 -0.001949 8 0.996094 1.001953 0.999023 0.003922 -0.001949 0.000978 9 0.999023 1.001953 1.000488 0.000978 -0.001949 -0.000488 10 0.999023 1.000488 0.999756 0.000978 -0.000488 0.000244
488 Numerical Analysis 13.1.4 False Position Method This method is similar to the Bisection Method, since the root is bracketed by an interval. The idea is to use information at two endpoints a and b to choose a better iterative value. We start with
[a, b] where f (a ). f (b ) < 0 . Join points (a, f (a )), (b, f (b )) by a line, and let x1 be the intersection of the line with the x-axis as shown in Fig.2. Then check the sign of f ( x1 ) . If f ( x1 ) f (a ) < 0 ⇒ Take
[a, x1 ] as the new interval f (x1 ) f (a ) > 0 ⇒
Take [x1 , b] as the new
interval and repeat
Fig.2 False Position method at starting position.
489
Chapter Thirteen
Again, we have a sequence of intervals [a1 , b1 ] , [a2 , b2 ] each contains a root, however bn +1 − an +1 is not necessarily equal to
bn − an . 2 Line joining (a, f (a )), (b, f (b )) :
y = f (a ) +
x−a ( f (b ) − f (a )) b−a
y = 0 ⇒ x1 = a − ∴ c=
(9)
b−a af (b ) − bf (a ) f (a ) = f (b ) − f (a ) f (b ) − f (a )
af (b ) − bf (a ) f (b ) − f (a )
(10)
The following table (Table(2)) shows a comparison between Bisection and False Position Methods. Table(2) A comparison between Bisection and False Position Methods. Bisection
c=
a+b 2
average of a and b
False Position
c=
af (b ) − bf (a ) f (b ) − f (a )
Waited average of a and b
Does not use information about uses information about
f (x )
f (x )
which may give some idea of where root is located.
490 Numerical Analysis What is the convergence criterion? f ( x ) ≤ ε1 , but this may be impossible.
(xn − xn −1 ) xn
≤ ε 2 but this may take a long time.
How fast does it converge? In general, False position may be better, but there are exceptions. For locally convex functions, could be slower than the bisection method. See Fig.3 picture of locally convex function.
Fig.3 picture of locally convex function. We can consider a speed-up for the False Position method. The slow convergence of the False Position method is because the update looks like
[a, b] ⇒ [a, x1 ] ⇒ [a, x2 ] ⇒ [a, x3 ] ⇒ ... ..[stationary ] [a, b] ⇒ [x1 , b] ⇒ [x1 , x2 ] ⇒ [x3 , x2 ] ⇒ .....[non − stationary ]
491
Chapter Thirteen
Example 3 Find the solution of the following equation by using False Position method: f ( x ) = x 2 − 3 x + 2 = 0 Solution: Applying the above methodology to the above equation we get the result shown in Table(3): Table(3) The result of Example 3. i
a
b
x
f(a)
f(b)
f(x )
0 0.000000 1.500000 1.333333 2.000000 -0.250000 -0.222222 1 0.000000 1.333333 1.200000 2.000000 -0.222222 -0.160000 2 0.000000 1.200000 1.111111 2.000000 -0.160000 -0.098765 3 0.000000 1.111111 1.058824 2.000000 -0.098765 -0.055363 4 0.000000 1.058824 1.030303 2.000000 -0.055363 -0.029385 5 0.000000 1.030303 1.015385 2.000000 -0.029385 -0.015148 6 0.000000 1.015385 1.007752 2.000000 -0.015148 -0.007692 7 0.000000 1.007752 1.003891 2.000000 -0.007692 -0.003876 8 0.000000 1.003891 1.001949 2.000000 -0.003876 -0.001946 9 0.000000 1.001949 1.000976 2.000000 -0.001946 -0.000975 10 0.000000 1.000976 1.000488 2.000000 -0.000975 -0.000488
492 Numerical Analysis 13.1.5 Illinois Method The Illinois method (or modified position method) is to use
1 2i −1
f (c ) instead of f (c ) if c is a stagnant end point has been
repeated twice or more, where i is the number of times the end point has been repeated. See Fig.4 A picture of the Illinois method. This modification markedly improves the rate of convergence of the method in general.
Fig.4 A picture of the Illinois method. Example 4 Find the solution of the following equation by using Illinois method: f ( x ) = x 2 − 3 x + 2 = 0
493
Chapter Thirteen
Solution: Applying the Illinois methodology to the above equation we get the result shown in Table(4): Table(4) The result of Example 4. i
a
b
x
f(a)
f(b)
f(x )
0 0.000000 1.500000 1.333333 2.000000 -0.250000 -0.222222 1 0.000000 1.333333 1.200000 2.000000 -0.222222 -0.160000 2 0.000000 1.200000 1.034483 1.000000 -0.160000 -0.033294 3 0.000000 1.034483 0.969900 0.500000 -0.033294 0.031006 4 0.969900 1.034483 1.001043 0.031006 -0.033294 -0.001041 5 0.969900 1.001043 1.000030 0.031006 -0.001041 -0.000030 6 0.969900 1.000030 0.999971 0.015503 -0.000030 0.000029 7 0.999971 1.000030 1.000000 0.000029 -0.000030 0.000000
13.1.6 Newton’s Method Although the Bisection method is easy to compute, it is slow. Now, we will consider more interesting methods of root-finding. The first two methods are local methods, and the last, which will be discussed in next class, is a global method. All the methods are based on using lines to get better iterative approximations for the root of a function. One of the most widely used methods is Newton’s method. Originally, Newton used a similar method to solve a cubic equation. It has since been extended to differential equations. Over a very
494 Numerical Analysis small interval, most functions can be locally approximated by a line. This idea is the basis of Newton’s method. The idea is to start with an initial guess for the root, x1 . Then draw a line tangent to the function at the point ( x1 , f ( x1 )) . The tangent line’s intersection with the x -axis is defined to be x2 . We repeat this process to get x1 , x2 , x3 , ........ . See Fig.5 for example of Newton’s Method.
Fig.5 Example of Newton’s Method. Why a tangent line? If the function f ( x ) is a linear function, i.e.,
f ( x ) = ax + b then y = f ( x ) is a line. If we start off with any guess x1 , the tangent line at ( x1 , f ( x1 )) agrees with y = f ( x ) . Therefore, x2 = x* . I.e., for linear functions, Newton’s method yields an exact solution after one iteration.
495
Chapter Thirteen
Now, if f ( x ) is any function, we may approximate if by a linear function.
At
the
point
(x1 , f (x1 )) ,
Taylor
expansion:
f ( x ) ≈ f ( x1 ) + f ′( x1 )( x − x1 ) + .......... See Fig.6 for picture of the Taylor approx at a point x1 .
Fig.6 Picture of the Taylor approx at a point x1 . Let f ( x ) ≈ F ( x ) = f ( x1 ) + f ′( x1 )( x − x1 ) which is linear. Instead of looking for the root of f ( x ) = 0 look for a root of F ( x ) = 0 i.e.
f ( x1 ) + f ′( x1 )( x − x1 ) = 0
∴ x = x1 −
f ( x1 ) this is the root of F ( x ) = 0 f ′( x1 )
f ( x1 ) Regard it as a good approximation to x* . So, let ∴ x2 = x1 − f ′( x1 ) Repeating the process, we have f ( xn ) (11) ∴ xn +1 = xn − f ′( xn )
496 Numerical Analysis
Notes regarding Newton’s Method: • Need only one initial guess, whereas bisection needs a and b. • Need to compute the derivative f ′( x ) • Requires that
f ′( x ) ≠ 0 in the neighborhood of
x* .
Otherwise, denominator blows up. • At each iteration evaluate f ( x ) and f ′( x ) (two function
evaluations). Example 5 Find the solution of the following equation by using
Newton’s method: f ( x ) = x 2 − 3 x + 2 = 0 Solution:
Applying the Newton’s methodology to the above equation we get the result shown in Table(5): Table(5) The result of Example 5.
i
xn
f(xn)
f '(xn)
0
0.000000000
2.000000000
-3.000000000
1
0.666666667
0.444444444
-1.666666667
2
0.933333333
0.071111111
-1.133333333
3
0.996078431
0.003936947
-1.007843137
4
0.999984741
0.000015259
-1.000030518
5
1.000000000
0.000000000
-1.000000000
497
Chapter Thirteen
13.1.7 Secant Method
In Newton’s method, f ′( x ) is needed. But f ′( x ) may be difficult to compute. May not ever know f ′( x ) ; e.g. if f ( x ) is provided by a subroutine. Idea: do not compute f ′( xn ) explicitly. Instead, approximate
f ′( xn ) as follows:
f ′( xn ) ≈
f ( xn ) − f ( xn −1 ) xn − xn −1
∴ x n +1 = x n −
x f ( xn ) − xn f ( xn −1 ) f ( xn ) = n −1 f ( xn ) − f ( xn −1 ) f ( xn ) − f ( xn −1 ) xn − xn −1
(12) (13)
Fig.7 Picture comparing Newton’s and Secant lines on a function.
498 Numerical Analysis
Fig.7 Picture comparing Newton and Secant lines on a function (continue). Example 6 Find the solution of the following equation by using
Secant’s method: f ( x ) = x 2 − 3 x + 2 = 0 Solution: Applying the Secant’s methodology to the above
equation we get the result shown in Table(6):
499
Chapter Thirteen
Table(6) The result of Example 6.
i
xn
f(xn)
0
0.000000000
2.000000000
1
0.500000000
0.750000000
2
0.800000000
0.240000000
3
0.941176471
0.062283737
4
0.990654206
0.009433138
5
0.999485332
0.000514933
6
0.999995237
0.000004763
7
0.999999998
0.000000002
8
1.000000000
0.000000000
Some comments:
Two initial guesses are needed, but does not require f (a ) * f (b ) < 0 unlike bisection. But there might be problems if the root lies outside the convergence range. Must have
f ( xn ) ≠ f ( xn −1 ) (similar to f ′( x ) ≠ 0 in Newton’s
method). I.e., a very flat function. Another problem case might occur is a generated guess is the same as a previous guess, resulting in the possibility of an infinite loop that never reaches the root.
500 Numerical Analysis 13.2 Polynomial Interpolation
Suppose we have, n + 1 points
(x0 , y0 ), (x1, y1 ), (x2 , y2 ),.............. (xn , yn ) polynomial Pn ( x ) which passes through
and we need to find a these points. Thus we
could estimate the values in between the given values. This is called the interpolation of these given points. Fig.8 shows general example of an interpolation.
Fig.8 General example of interpolation. Suppose we were given two points {x0 , x1} and the values at those points. We would draw a line as shown in Fig.9.
Fig.9 Linear example.
Chapter Thirteen
501
Suppose we were given three points {x0 , x1 , x2 }. We would draw a parabola as sown in Fig.10. So, for n = 1 , we have a line and for n = 2 , we have a parabola. How about n + 1 points, {x0 , x1 , x2 , ............xn }? We would then draw a n polynomial, Pn ( x ) .
Fig.10 Parabolic interpolation example. Conditions:
( x0 , y0 ), ( x1, y1 ), M
( xn , yn ),
Pn ( x0 ) = y0 a0 + a1 x0 + ....... + an x0n = y0 Pn ( x1 ) = y1 a0 + a1 x1 + ....... + an x1n = y1 M Pn ( xn ) = yn a0 + a1 xn + ....... + an xnn = yn
In the above, a0 , a1 , ........, an are unknowns, and {xi } and {yi } are known values. We can find the polynomial, if we solve the above for a0 , a1 , ........, an
502 Numerical Analysis Example 7 Find P1 ( x ) passing through ( x0 , y0 ) and ( x1 , y1 ) . Solution:
P1 ( x )
has the form:
P1 ( x ) = a0 + a1 x
(14)
P1 ( x0 ) = a0 + a1 x0 = y0 a1 ( x0 − x1 ) = y0 − y1 P1 ( x1 ) = a0 + a1 x1 = y1 a1 =
y0 − y1 x0 − x1
if x0 ≠ x1
a0 = y0 − a1 x0 = y0 − P1 ( x ) =
y0 − y1 x y −x y x0 = 0 1 1 0 x0 − x1 x0 − x1
x0 y1 − x1 y0 y0 − y1 + x x0 − x1 x0 − x1
(15)
If x0 = y0 = 0, x1 = y1 = 1, P1 ( x ) = x I.e., the polynomial P1 ( x ) = x passes through (0, 0 ) and (1,1) . Is this the only possible solution? Yes. Why? Fact. For any given n + 1 points ( x0 , y0 ), ( x1 , y1 ),.............. ( xn , y n ) , if x0 , x1, ......., xn are distinct, i.e., xi = x j if i ≠ j then there exists a unique interpolating polynomial Pn ( x ) of degree n i.e., there is a unique
Pn ( x )
which
(x0 , y0 ), (x1 , y1 ),.............. (xn , yn ) .
passes This
can
constructing a linear system of nth order equations.
be
through proved
by
503
Chapter Thirteen
Example 8 If x0 , x1, ......., xn are distinct. Suppose we have a
polynomial Pn ( x ) of degree n, so that Pn ( xi ) = 0, i = 0, 1, 2, .......n . What is Pn ( x ) ? Solution:
Pn ( x ) = 0
i.e.,
interpolates
( x0 , 0), (x1, 0), ..........., ( xn , 0),
a0 = a1 = ....... = an = 0 .
Why?
Pn ( x ) = 0
and this is a unique
interpolation of the points. Example 9 Suppose we have the following two points (2,3), (5,6).
Find a first order polynomial passing through those points. Solution:
From (15)
P1 ( x ) =
x0 y1 − x1 y0 y0 − y1 x + x0 − x1 x0 − x1
∴ P1 ( x ) =
2*6 − 5*3 3 − 6 + x 2−5 2−5
∴ P1 ( x ) = 1 + x We can check if this equation passing through the given points or not as following:
P1 (2 ) = 1 + 2 = 3 = y1 P1 (5) = 1 + 5 = 6 = y 2
504 Numerical Analysis 13.2.1 Lagrange’s Method and Lagrange Polynomials
Given distinct x0 , x1, ......., xn there is a unique polynomial of degree n passing through
(x0 ,1), (x1 , 0), (x2 , 0),........, (xn , 0) ⇒ l0n (x ) .See
Fig.11 for picture of l0n ( x ) .
Fig.11 Picture of l0n ( x ) In fact, we can construct a whole set of these polynomials, each passing through 1 for a different xi value. l0n ( x ) ( x0 , 1), ( x1 , 0 ), ( x2 , 0 ),........, ( xn , 0 )
l0n ( x0 ) = 1
l0n ( xi ) = 0, i ≠ 0
l1n ( x ) ( x0 , 0 ), ( x1 , 1), ( x2 , 0 ),........, ( xn , 0 )
l1n ( x1 ) = 1
l0n ( xi ) = 0, i ≠ 1
l2n ( x ) ( x0 , 0 ), ( x1 , 0 ), ( x2 , 1),........, ( xn , 0 )
l0n ( x2 ) = 1 l0n ( xi ) = 0, i ≠ 2
M
lnn ( x ) ( x0 , 0 ), ( x1 , 0 ), ( x2 , 0 ),........, ( xn , 1)
l0n ( xn ) = 1 l0n ( xi ) = 0, i ≠ n
A general short form for these polynomials is lin ( x ) where n is the
{ }
degree and i is the place in the set x j where it has value 1.
505
Chapter Thirteen
i≠ j
0 lin ( x ) = 1
i= j
as an example for n = 1 . We have x0 , x1 such that:
l01 ( x0 ) = 1,
l01 ( x1 ) = 0
l11 ( x0 ) = 0, l11 ( x1 ) = 1 See Fig.12 for a picture of l01 ( x ) and l11 ( x ) .
Fig.12 A picture of l01 ( x ) and l11 ( x ) . How to find l nj ( x ) ?
l0n ( x )
0 degree n n = 1
at x1 , x2 , ........, xn at x0
Consider the following polynomial of degree n
qn ( x )
= ( x − x1 )( x − x2 ).....( x − xn ) =
qn ( x )
0
at x1 , x2 , ........, xn
is almost l0n ( x ) but :
506 Numerical Analysis
qn ( x ) = ( x − x1 )( x − x2 ).....( x − xn ) ≠ 1 in general. But
(x − x1 )(x − x2 ).....( x − xn ) qn ( x ) = qn ( x0 ) ( x0 − x1 )( x0 − x2 ).....( x0 − xn ) and is a degree n polynomial.
∴ l0n ( x ) =
(x − x1 )(x − x2 ).....(x − xn ) qn ( x ) = qn ( x0 ) ( x0 − x1 )( x0 − x2 ).....( x0 − xn )
(17)
This polynomial interpolates ( x0 , 1), ( x1 , 0 ), .....( xn , 0 ) . Similarly,
∴lin ( x ) =
(x − x0 )(x − x1 ).....(x − xi −1 )(x − xi +1 ).......( x − xn ) (18) (xi − x0 )(xi − x1 ).....(xi − xi −1 )(xi − xi +1 ).....(xi − xn )
∏ j = 0 (x − x j ) n
∴lin ( x ) =
∏
j ≠i n j =0 j ≠i
(xi − x j )
(19)
Interpolates ( x0 , 0 ), ( x1 , 0 ), ....., ( xi ,1), ......, ( xn , 0 ) Why Lagrange polynomials? For a given ( x0 , y0 ), ( x1 , y1 ), ........, ( xn , y n ), consider
pn ( x ) = y0l0n ( x ) + y1l1n ( x ) + ........ + yn lnn ( x ) Where: 1. Pn ( x ) has degree n. 2. Pn ( xi ) = yi In other words, Pn ( x ) is the interpolating polynomial for
( x0 , y0 ), ( x1, y1 ), ........, ( xn , yn )
507
Chapter Thirteen
13.2.2 Lagrange Formula
The interpolating polynomial for ( x0 , y0 ), ( x1 , y1 ), ........, ( xn , y n ) is given by:
Pn ( x ) = y0l0n ( x ) + y1l1n ( x ) + ........ + y n lnn ( x ) =
n
∑ yi lin (x )
(20)
i =0
provided xi ≠ x j , i ≠ j . What does this interpolating formula look like? Consider n = 1 (x − x1 ) + y x − x0 (21) P1 ( x ) = y0 (x0 − x1 ) 1 x1 − x0 13.2.3 Interpolating Functions By Polynomials.
If we have a complicated function f ( x ) as shown in Fig.13, we may want to approximate it by a polynomial of degree n, Pn ( x ) . See Fig.13 for a picture of general example.
Fig.13 A picture of general example. How to approximate this function, f ( x ) ?
508 Numerical Analysis
We require Pn ( x ) and f ( x ) to have the same values at some given set
{xi },
of
Pn ( xi ) = f ( xi ),
x0 , x1 , .........., xn .
i.e.
i = 0, 1, 2,........, n
Then, Pn ( x ) must interpolate ( x0 , f ( x0 )), ( x1 , f ( x1 )), .., ( xn , f ( xn )) . Use the Lagrange formula,
Pn ( x ) =
n
∑ f ( xi )lin (x )
i =0
This is a polynomial of degree n which interpolates f ( x ) at x0 , x1 , .........., xn Example 10 Suppose a function f ( x ) given by the following table
i
xi f ( xi )
0 0 3
1 1 2
2 3 1
3 4 0
Find the interpolating polynomial and use it to approximate the value of f (2.5) . Find the Lagrange polynomials. Solution:
(x − 1)(x − 3)(x − 4) (− 1)(− 3)(− 4) (x − 0)(x − 3)(x − 4) l13 ( x ) = (1 − 0)(1 − 3)(1 − 4) (x − 0)(x − 1)(x − 4) l23 ( x ) = (3 − 0)(3 − 1)(3 − 4) (x − 0)(x − 1)(x − 3) l33 ( x ) = (4 − 0)(4 − 1)(4 − 3)
l03 ( x ) =
509
Chapter Thirteen
2. Find interpolating polynomial.
P3 ( x ) = 3l03 ( x ) + 2l13 ( x ) + 1l23 ( x ) + 0l33 ( x ) P3 ( x ) = 3
P3 ( x ) =
(x3 − 8x 2 + 19x − 12) + 2 (x3 − 7 x 2 + 12x) + (x3 − 5x 2 + 4 x) − 12
6
(− x3 + 6 x 2 − 17 x + 36)
−6
12
3. Use P3 (2.5) to estimate f (2.5)
( − (2.5)3 + 6(2.5)2 − 17(2.5) + 36 ) P3 (2.5) = = 1.281 25 12
∴ f (2.5) ≈ 1.281 25 13.2.4 Newton Divided Differences
There are two problems with Lagrange’s form for the unique interpolating formula: 1. It is expensive computationally. 2. If we have Pn ( x ) , we can’t use it to find Pn +1 ( x ) The Lagrange formulation Pn ( x ) =
n
∑ f (xi )lin (x ) is simple in form,
i =0
but it is difficult to compute the coefficients. So, we will look for another form for Pn ( x ) . Note that we are not looking for another polynomial, since there is only one unique interpolating polynomial. What we are looking for? We are looking for another form to express the same polynomial, that is easier to compute.
510 Numerical Analysis
Pn ( x ) = A0 + A1 ( x − x0 ) + A2 ( x − x0 )( x − x1 ) + .... + An ( x − x0 )( x − x1 )......( x − xn −1 )
(22)
And try to determine the coefficients A0 , A1 , A2 , ....... An
( xo , f (x0 )) (x1 , f (x1 ))
(x2 , f (x2 ))
Pn ( x0 ) = f ( x0 ) Pn ( x1 ) = f ( x1 )
Pn ( x2 ) = f ( x2 )
⇒ A0 = f ( x0 )
(23)
f ( x1 ) = f ( x0 ) + A1 ( x1 − x0 ) (24) ⇒ f ( x1 ) − f ( x0 ) A = 1 x1 − x0 f ( x2 ) − f ( x1 ) f ( x1 ) − f ( x0 ) − x2 − x1 x1 − x0 (25) ⇒ A2 = x2 − x0
New Notation:
We can note in the above expressions for A1 and A2 a relationship in the forms of the expressions, which leads us to the following new notation.
f [x0 ] = f ( x0 ) f [x0 , x1 ] =
f [x1 ] − f [x0 ] x1 − x0
f [x0 , x1 , x2 ] =
f [x1 , x2 ] − f [x0 , x1 ] x2 − x0
We call f [x0 , x1 ] =
⇒ A0 = f [x0 ]
(26)
⇒ A1 = f [x0 , x1 ]
(27)
⇒ A2 = f [x0 , x1 , x2 ]
(28)
f [x1 ] − f [x0 ] divided difference at [x1 , x2 ], x1 − x0
etc. Thus, the polynomial which interpolates:
511
Chapter Thirteen
( x0 , f (x0 )), (x1 , f (x1 )), ......., (xn , f (xn )) can be written as: Pn ( x ) = f [x0 ] + f [x0 , x1 ]( x − x0 ) + f [x0 , x1 , x2 ]( x − x0 )( x − x1 ) + .... ........ + f [x0 , x1 ,...., xn ]( x − x0 )( x − x1 )..........( x − xn −1 ) n −1
∴ Pn ( x ) = Pn −1 ( x ) + f [x0 , x1 ,...., xn ]∏ ( x − xi )
(29) (30)
i =0
Where, f [ x0 ]
= f ( x0 )
f [x0 , x1 ]
=
f [x0 , x1 , x2 ]
=
f [x0 , x1 , ....., xi ]
=
f [x0 , x1 , ....., xi ]
=
f [x1 ] − f [x0 ] x1 − x0
f [x1 , x2 ] − f [x0 , x1 ] x 2 − x0
f [x1 , x2 ,...., xi ] − f [x0 , x1 ,......xi −1 ] xi − x0
f [x1 , x2 ,...., xn ] − f [x0 , x1 ,......xn −1 ] xn − x0
We can build a divided difference table very easily: x0 x1 x2 x3 x4
f [x0 ] f [x1 ] f [x 2 ] f [x3 ] f [x 4 ]
f [x0 , x1 ] f [x0 , x1 , x2 ] f [x0 , x1 , x2 , x3 ] f [x1 , x2 ] f [x0 , x1 , x2 , x3 , x4 ] f [x1 , x2 , x3 ] f [x1 , x2 , x3 , x4 ] f [x2 , x3 ] f [x2 , x3 , x4 ] f [x3 , x4 ]
Example 11 Find the interpolating function for the following table: 0 1 2 3 i 0 1 3 4 xi 2 1 0 f ( xi ) 3
512 Numerical Analysis
1. Find Newton’s divided difference. 2. Find the interpolating function. Solution:
1.
xi
f [xi ]
0 1
3 2
3 4
1 0
−1 −1/ 2 −1
1/ 6 −1/ 6
− 1 / 12
2.
P3 ( x ) = 3 + (− 1)( x − 0 ) +
1 (x − 0)( x − 1) + − 1 (x − 0)(x − 1)(x − 3) 6 12
13.2.5 Finite Difference Errors
Consider if the points {xi } are evenly spaced. Let h be the fixed distance between the points. Then we can define:
∆ f ( xi ) = f ( xi + h ) − f ( xi ) = f ( xi +1 ) − f ( xi )
Or ∆ f i = f i +1 − f i This quantity is called the forward difference of f ( x ) at xi . Since the points are evenly spaced, xi = x0 + ih, For r ≥ 0 we can further define
i = 0, 1, 2, ......, n .
513
Chapter Thirteen
∆r +1 f i = ∆r f i +1 − ∆r f i With ∆0 f i = f i For example,
∆2 f i = ∆(∆f i ) = ∆( f i +1 − f i ) = ∆f i +1 − ∆f i
= ( f i + 2 − f i +1 ) − ( f i +1 − f i ) = f i + 2 − 2 f i +1 + f i
Now, let us consider the form of the Newton Divided Difference with evenly spaced points.
f [x0 , x1 ] =
f1 − f 0 1 = ∆f 0 x1 − x0 h
f [x0 , x1 , x2 ] =
f [x1 , x2 ] − f [x0 , x1 ] x2 − x0
1 1 1 ∆f1 − ∆f 0 2h h h 1 = 2 ∆2 f 0 2h =
In general, and this can be easy proved via proof by induction,
∴ f [x0 , x1 , ..., xk ] =
1 k! h k
∆k f 0
(31)
We can now modify the Newton interpolation formula to an interpolation formula based on forward differences. Since the polynomial is defined continuously, rather than with respect to the discretely spaced points, we will define for the value x at which the polynomial is defined,
µ=
x − x0 h
(32)
514 Numerical Analysis
where µ is a continuous parameter. Therefore,
x − xi = x0 + µh − x0 − ih = (µ − i )h which leads to the following form for the interpolating formula
Pn ( x ) =
n
u
∑ i ∆i
i = 0
f0
(33)
where we have used the binomial coefficients
µ µ (µ − 1)........(µ − i + 1) = , i ! i
i>0
(34)
µ and = 1 0
(35)
For example, n = 1
P1 ( x ) = f 0 + µ∆f 0 As with Newton divided differences, we can easily construct tables to evaluate the forward differences.
xi
fi
x0
f0
x1 x2 x3
f1 f2 f3
x4 M
f4 M
∆f i ∆f 0 ∆f1 ∆f 2 ∆f 3
∆2 f i ∆2 f 0 ∆2 f1
∆2 f 2
∆3 f i
L
∆3 f 0 ∆3 f1
M
515
Chapter Thirteen
Example 12. Find the interpolating function for the following table
i xi f ( xi )
0 0 3
1 1 2
2 3 0
3 4 -1
١. Find the forward differences. ٢. Find the interpolating function. Solution:
1. xi
fi
0
3
1 3
2 0
4 2.
−1
∆f i −1 −2 −1
∆2 f i
∆3 f i
−1 1
2
P3 ( x ) = 3 + (− 1)(µ ) + (− 1)
(µ )(µ − 1) + (2) (µ )(µ − 1)(µ − 2)
2 6 Note: forward differences of order greater than three are almost entirely the result of differencing the rounding errors in the table entries; therefore, interpolation in this table should be limited to polynomials of degree less than four. As you can see, there is nothing particularly special about forward differences. We can equally define backward difference interpolating functions based on
∇f i = f i − f i −1
(36)
516 Numerical Analysis 13.3 Numerical Integration
Quadrature comes from the process of “squaring”, of finding a square equal in area to a given area, e.g., finding the area of a circle. Now means numerical integration. The problem is either Given f ( x ) defined on [a, b] , and find
b
∫a f ( x )dx , or x
Given f ( xi ) defined on {xi } , and find ∫ n f ( x )dx x0 See Fig.14 for a generic example.
Fig.14 Picture of an example of an integration. The area of the shaded region is the result of the integration. Easy
example:
b a
I = ∫ xdx =
b a
I = ∫ Pn ( x )dx b a
(
1 2 b − a2 2
Hard example I = ∫ e cos x dx = ?
)
or
more
generally
517
Chapter Thirteen
In many applications, if f ( x ) is complicated then
b
∫a f ( x )dx cannot
be calculated analytically must be approximated by a numerical value The approach:
1. Locally interpolate f ( x ) by a simple function g ( x ) , e.g., a polynomial interpolation Pn ( x ) whose analytical integral is known. 2. Use the integral of the simpler function to approximate
b a
∫ f ( x )dx
locally, summing the local results as we move along. Our goal is to get as accurate an answer as possible, with as few function evaluations as possible. Quadrature can be done with fixed or variable (adaptive) spacing. 13.3.1 Trapezoid Rule
The simplest polynomial approximation to a function is a piecewise linear interpolation. See Fig.15.
Fig.15 Picture of a piecewise linear approximation to the function, and the corresponding resulting integration.
518 Numerical Analysis
Consider a linear interpolation of f ( x ) between points xi and xi +1 Therefore,
xi +1 xi
∫
f ( x )dx ≅ ( xi +1 − xi )
f ( xi ) + f ( xi +1 ) 2
So, if the step size is h, then the area of any trapezoid is
h ( f i + f i +1 ) 2 The integral is thus approximately, for n + 1 points,
I ( f ) ≅ Tn ( f ) =
n −1h
∑ 2 ( fi +
i =0 h n −1
f i +1 )
n −1 = ∑ ( f i ) + ∑ ( f i +1 ) 2 i =0 i =0
h [( f 0 + f1 + f 2 + ..... + f n − 2 + f n −1 ) + ( f1 + f 2 + ..... + f n − 2 + f n )] 2 h = [( f 0 + 2 f1 + 2 f 2 + ..... + 2 f n − 2 + 2 f n −1 + f n )] 2
=
∴ I ( f ) ≅ Tn ( f ) =
n −1 h + + f 2 f f 0 ∑ i n 2 i =1
(37)
Where f 0 = f (a ) and f n = f (b ) Discretization Error
To perform the integration using the Trapezoid Rule, we are approximating f ( x ) on [x0 , x1 ] by a first-order polynomial P1 ( x ). Thinking of this as a Taylor expansion about x0 , we know that
519
Chapter Thirteen
f ( x ) = P1 ( x ) + R2 ( x0 )
( x − x0 )2 = P1 ( x ) +
( x − x0 )3 f ′′( x0 ) +
f ′′′( x0 ) + O h 4
( x − x0 )2 ≅ P1 ( x ) +
( x − x0 )3 f ′′( x0 ) +
f ′′′( x0 )
2
2
6
6
( )
and in particular
f ( x1 ) ≅ P1 ( x ) +
h2 h3 f ′′( x0 ) + f ′′′( x0 ) 2 6
The error E (R ) in the integral I ( f ) − I (P1 ) is: x1 x0
E (R ) ≈ ∫ ≈
(x − x0 )2 2
f ′′( x0 )dx
f ′′( x0 ) x1 (x − x0 )2 dx ∫ x 0 2
f ′′( x0 ) ( x − x0 )3 ≈ 2 3
∴ E (R ) ≈
x1 x0
f ′′( x0 ) (x1 − x0 )3 6
(38)
Thus, the total error is the Trapezoid Rule minus the integral of
P1 ( x ) minus E (R ) h h2 h3 E (R ) ≈ ( f ( x0 ) + f ( x1 )) − hf ( x0 ) − f ′( x0 ) − f ′′( x0 ) − E (R ) 2 2 6 h3 h3 ∴ E (R ) ≈ f ′′( x0 ) = M2 12 12
(39)
520 Numerical Analysis
for a bound M 2 on f ′′( x ) over [x0 , x1 ] The total possible quadrature error is the sum of all the errors for each of the panels, [xi , xi +1 ]
M 2 h3 Tn ( f ) − I ( f ) ≤ ∑ 12 i =1 n
(40)
M 2 h 3n = 12
( )
M 2 (b − a )h 2 = O h2 12 Therefore,
M 2 (b − a )h 2 Tn ( f ) − I ( f ) ≤ 12
(41)
So this is a second-order method. π
Example 13 Evaluate I = ∫ e x cos( x )dx by composite trapezoidal 0
rule using 4 subintervals (panels). Solution:
[a, b] = [0, π ], f (x ) = e x cos(x ) n = 4, h =
b−a π = n 4
Such that x0 = 0, x1 =
π 4
, x2 =
π 2
, x3 =
3π , x4 = π 4
521
Chapter Thirteen
f ( x4 ) f ( x0 ) T4 ( f ) = h + f ( x1 ) + f ( x2 ) + f ( x3 ) + 2 2 =
π 1
+ 1.5509 + 0 + (− 7.4605) + 4 2
n=8 n = 64 n = 512
(− 23.141) = −13.336 2
T8 ( f ) = −12.382
T64 ( f ) = −12.075
T512 ( f ) = −12.070
True solution is − 12.0703 13.3.2 Simpson’s Rule
Now, let’s locally approximate f ( x ) by a quadratic polynomial
P2 ( x ) . Hereafter, we will always assume that Y is even (for deep reasons). See Fig.16 the knots for P2 occur at the even points. The regions between knots are called panels. With n + 1 points, the number of panels is n / 2 .
Fig.16 Function approximated by piecewise quadratic polynomial.
522 Numerical Analysis
We can develop Simpson’s Rule by using Lagrangian interpolation to find P2 ( x ) over [xi , xi + 2 ] and then integrate it to find I (P2 ) . See Fig.17.
Fig.17 Picture of a function locally approximated by a quadratic polynomial, between the points xi and xi + 2 The interpolation function is:
P2 ( x ) = f ( xi )l02 ( x ) + f ( xi +1 )l12 ( x ) + f ( xi + 2 )l22 ( x ) where
(x − xi +1 )(x − xi + 2 ) (xi − xi +1 )(xi − xi + 2 ) (x − xi +1 )(x − xi + 2 ) l12 ( x ) = (xi +1 − xi )(xi +1 − xi + 2 ) (x − xi )( x − xi +1 ) l22 ( x ) = (xi + 2 − xi )( xi + 2 − xi +1 ) l02 ( x ) =
523
Chapter Thirteen
∴ I (P2 ) =
xi + 2
∫ P2 (x )dx
x1
∴ I (P2 ) = f ( xi )
∴
xi + 2
∫
(xi + 2 − xi ) 6
f ( x )dx ≅ I (P2 ) =
x1
+ f ( xi +1 )
4( xi + 2 − xi ) x − xi + f ( xi + 2 ) i + 2 6 6
h ( f i + 4 f i +1 + f i + 2 ) 3
(47)
where h = xi +1 − xi To get the sum over the entire interval [a, b] we sum over all the panels, noting that the end points of the panels are have even numbered indicies, with h = Sn ( f ) =
=
n−2
∑
i =0 i = even
=
b−a n
h ( f 0 + 4 f1 + f 2 ) + h ( f 2 + 4 f 3 + f 4 ) + ... + h ( f n − 2 + 4 f n −1 + f n ) 3 3 3 n / 2 −1 h ( f i + 4 f i +1 + f i + 2 ) = ∑ h ( f 2i + 4 f 2i +1 + f 2i + 2 ) 3 i =0 3
h ( f 0 + 4 f1 + 2 f 2 + 4 f 3 + 2 f 4 + 4 f 5 + .... + 4 f n − 3 + 2 f n − 2 + 4 f n −1 + f n ) 3
∴ Sn ( f ) =
n / 2 −1 n / 2−2 h f 0 + 4 ∑ f 2i +1 + 2 ∑ f 2i + 2 + f n 3 i =0 i =0
(48)
By using more advanced techniques, we can show that for even n and f ( x ) four times differentiable, the local error per panel (containing three points) is:
524 Numerical Analysis
M4 (49) 90 with M 4 being the bound on f (4 ) ( x ) . For the composite Simpson’s I (P2 ) − I ( f ) ≤ h 5
Rule over the entire domain the upper bound on the error is
M 4 (b − a ) 1 M 4 (b − a )5 Sn ( f ) − I ( f ) ≤ h = 4 180 180 n 4
(50)
Therefore, Simpson’s Rule is fourth-order. Example 14 Evaluate I = ∫ e x cos( x )dx by composite Simpson’s
rule using 2 subintervals (panels). Solution:
Again, [a, b] = [0, π ], f ( x ) = e x cos( x )
b−a π = such that: n 4 π π 3π x0 = 0, x1 = , x2 = , x3 = , x4 = π 4 2 4 h C.S .R. = [ f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) + 4 f ( x3 ) + f ( x4 )] 3 n = 4, h =
=
π 12
[1 + 4(1.5509) + 2(0) + 4(− 7.4605) + (− 23.141)] = −11.985
The exact solution is − 12.070 3 . Thus with n = 4 , the Composite Simpson’s Rule has an error of 0.085 28 , as compared to the composite Trapezoid Rule for the same n, which has an error of
− 1.265 7 . With n = 8 , the result of the CSR has the same
525
Chapter Thirteen
magnitude of error as the result using n = 512 with the CTR. Since our goal is to have an accurate a result with a few a number of function evaluations as possible, the CSR is a marked improvement for this function. 13.3.3 Composite Rules
Composite rules are constructed in the following manner: 1.
Divide
[a, b]
[a, t1 ], [t1, t2 ],……., [t p −1, b]
into
subintervals
(panels)
2. Apply the basic rule (i.e., one of the Newton-Cotes formulae above) to each of [ti , ti +1 ] 3. For convenience, we assume each [ti , ti +1 ] has equal length and one basic rule is applied to each interval. Note that if there are p panels, each using a rule using m + 1 points on each panel, then n = mp . Let h =
b−a , and xi = a + ih . For n
example, look at Fig.18. Note that ti = xim
Fig.18 Picture showing p = 2 , m = 3 , and , n = mp = 6
526 Numerical Analysis b a t1 t0
∴ I = ∫ f ( x )dx t2 t1
∴ I = ∫ f ( x )dx + ∫
tp t p −1
f ( x )dx + ..... + ∫
f ( x )dx
Composite Trapezoidal Rule t i +1 ti
∫
xi +1 xi
f ( x )dx = ∫
f ( x )dx ≅
h [ f (xi ) + f (xi +1 )] 2
(51)
b a
I = ∫ f ( x )dx ∴I =
p −1
t ∑ ∫tii+1 f (x )dx
i =0 h n −1
∑ [ f (xi ) + f (xi +1 )] 2 i =0 f ( xn ) f ( x0 ) + f ( x1 ) + ... + f ( xn −1 ) + ∴ I = h 2 2 ∴I =
(52)
Composite Simpson’s Rule h t i +1 t i +1 ∫ti f (x )dx = ∫ti f (x )dx ≅ 3 [ f (x2i ) + 4 f (x2i +1 ) + f (x2i + 2 )] b a
∴ I = ∫ f ( x )dx =
∴I =
p −1
n −1 2 x2i+2 x i = 0 2i
t ∑ ∫tii+1 f (x )dx = ∑ ∫
i =0
f ( x )dx
n −1 2 h
∑ 3 [ f (x2i ) + 4 f (x2i +1 ) + f (x2i + 2 )]
i =0
n n − 1 −2 2 2 h ∴I = f 0 + 4 ∑ f 2i +1 + 2 ∑ f 2i + 2 + f n 3 i =0 i =0
(53)
527
Chapter Thirteen
π Example 15 Evaluate I = ∫ e π cos( x ) dx by composite trapezoidal 0
rule using 4 subintervals (panels). Solution:
[a, b] = [0, π ], f (x ) = e x cos(x ) n = 4, x3 =
h=
b−a π π = such that x0 = 0, x1 = , n 4 4
x2 =
π 2
,
3π , and x4 = π 4
f ( x4 ) f ( x0 ) C.T .R. = h + f ( x1 ) + f ( x2 ) + f ( x3 ) + 2 2 =
(− 23.141) = −13.336 + + + − + 1 . 5509 0 7 . 4605 4 2 2
π 1
n=8 n = 64 n = 512
CTR = −12.382 CTR = −12.075 CTR = −12.070
Then the true solution is ≅ −12.0703 So in a composite method, as n gets larger ⇒ the error gets smaller. But how do we know which Y to take for a given accuracy? 13.3.4 Gaussian Quadrature
The Newton-Cotes rules and Composite rules: b a
n
∫ f (x )dx ≅ ∑ ω i f (xi ) i =0
(54)
528 Numerical Analysis
n is fixed. xi are fixed. In trapezoidal: n = 1, x0 = a and x1 = b .
ωi can be computed when the
{xi }
are given; i.e., they are
determined by xi In trapezoid: ω0 =
h = ω1 2
Disadvantages: xi are chosen artificially how do we know they give us the best result? Note that we are considering just one panel here. Another approach
I≅
n
∑ ωi f (xi )
i =0
∴I =
n is fixed, ωi , xi are to be determined.
n
∑ ω i f (xi ) gives the “best” result.
i =0
“best:” it gives exact result for polynomials of highest degree possible. I.e., we want I ≅
n
∑ ωi f (xi )
if f ( x ) is a polynomial of some
i =0
degree, and we want the degree to be as high as possible.
529
Chapter Thirteen
Example 16
n = 1, [a, b] = [− 1,1] x0 , x1 , ω0 , ω1 are to be determined such that
∫
Pm ( x )dx = ω0 Pm ( x0 ) + ω1 Pm ( x1 )
(55)
for m as large as possible. 1. Exact for polynomial of degree 0, i.e., 55 holds for
P0 ( x ) = 1 1
∫−11dx = ω0 + ω1 ⇒ ω0 + ω1 = 2 2. Exact for polynomial of degree 1, i.e., 55 holds for
P0 ( x ) = x 1
∫−1 xdx = ω0 x0 + ω1x1 ⇒ ω0 x0 + ω1x1 = 0 3. Exact for polynomial of degree 2, i.e., 55 holds for
P0 ( x ) = x 2 1
∫−1
xdx = ω0 x02 + ω1 x12 ⇒ ω0 x02 + ω1 x12 =
2 3
4. Exact for polynomial of degree 3, i.e., 55 holds for
P0 ( x ) = x 3 1
3
3
3
3
∫−1 xdx = ω0 x0 + ω1x1 ⇒ ω0 x0 + ω1x1 = 0
530 Numerical Analysis
Can we expect the method to be exact for still higher degree polynomials? No. We have 4 unknowns, x0 , x1 , ω0 , ω1 , and if the method is exact for polynomials of degree 3, we already have 4 equations. This is enough to determine the 4 unknowns. By solving the four equations in boxes above, we find:
x0 = − Thus
3 3 , x1 = , ω0 = 1, ω1 = 1 3 3
3 ( ) ≅ − f x dx f ∫−1 3 + 1
3 f 3
This is Gaussian Quadrature on [− 1,1] with two nodes. From above, we know that
3 + f − 3
3 is exact if f 3
f = 1, x, x 2 , x 3 . Is it exact for all polynomials of degree ≤ 3 ? Yes: 1
∫−1
1
1
1
1
−1
−1
−1
−1
f ( x )dx = a0 ∫ dx + a1 ∫ xdx + a2 ∫ x 2 dx + a3 ∫ x 3dx \
2 2 3 3 3 3 3 3 3 3 = a0 [1 + 1] + a1 − + + a2 − + 3 + a3 − 3 + 3 3 3 3
2 3 3 3 3 + a 2 − + a3 − + = a0 + a1 − 3 3 3 2 3 3 3 3 a0 + a1 3 + a 2 3 + a3 3
531
Chapter Thirteen
3 + = f1 − 3
3 f 3
polynomials of degree ≤ 3
Example 17 Evaluate
∫−1 (3 + 4 x + 8 x 1
2
)
+ 2 x 3 dx Via Gaussian
Quadrature: 2 3 3 3 3 ∫ 3 + 4 x + 8 x + 2 x dx = 3 + 4 3 + 8 3 + 2 3 2 3 − 3 − 3 − 3 + 8 + 2 + 3 + 4 3 3 3 1 −1
(
2
3
)
2 3 34 = = 23 + 8 3 3
Compare that with straight integration: 1 −1
∫
(
)
1 −1
∴∫
(3 + 4 x + 8x 2 + 2 x3 )dx = 23 + 83 = 343
13.3.5 Gaussian Quadrature in General b a
∫
1
4 x 2 8x3 8x 4 + ++ 3 + 4 x + 8 x 2 + 2 x 3 dx = 3x + 2 3 3 −1
f ( x )dx ≅
n
∑ ω i f (xi )
i =0
532 Numerical Analysis
xi , ωi are chosen so that the method is exact for
1, x, x 2 ,........, x m where m is as large as possible. What is the largest possible m for a fixed n The number of unknowns are: 2n + 2 and also m + 1 functions m + 1 equations. Unknown = Eqns ⇒ m + 1 = 2n + 2 i.e., m = 2n + 1 Conclusion: Gaussian quadrature with n + 1 nodes (function evaluations) is exact for a polynomial of degree ≤ 2n + 1 . In comparison, a Newton-Cotes rule of degree n with n + 1 nodes is exact for polynomials of degree ≤ n . When we have to determine ωi , xi , we have to solve a non-linear system.
533
Chapter Thirteen
13.4 Numerical Solution of Differential Equations 13.4.1 Initial Value Problems
Consider the first-order initial value problem:
y ′ = f ( x , y ) , y ( x0 ) = y 0
(59)
To find an approximation to the solution y ( x ) of (59) on the interval
a ≤ x ≤ b , we choose N points, a = x0 inside the interval required and construct approximations y n , to y ( xn ) , n = 0, 1, 2,....N It is important to know whether or not a small perturbation of (59) shall lead to a large variation in the solution. If this is the case, it is extremely unlikely that we will be able to find a good approximation to (59). In considering numerical methods for the solution of (59) we shall use the following notation:
h>0
denotes the integration step size
xn = x0 + nh is the nth node
y ( xn ) is the exact solution at xn y n is the numerical solution at xn
f n = f ( xn , yn ) is the numerical value of f ( x ) at ( xn , y n )
534 Numerical Analysis 13.4.2 Euler's Method.
(
)
We choose N points, xn = x0 + nh where h = x f − x0 / N . From Taylor's Theorem we get:
y ′′(ζ n ) (xn +1 − xn )2 2 for ζ n between xn and xn +1 , n = 0, 1, 2,....N − 1 . Since y ′( xn ) = f ( xn , y ( xn )) and xn +1 − xn = h , it follows that: y ′′(ζ n ) 2 y ( xn +1 ) = y ( xn ) + f ( xn , y ( xn ))h + h 2 y ( xn +1 ) = y ( xn ) + y ′( xn )( xn +1 − xn ) +
(60)
(61)
We obtain Euler's method,
y n +1 = yn + hf ( xn , yn ) ,
(62)
( )
by deleting the term of order O h 2
y ′′(ζ n ) 2 h 2 called the local truncation error. The algorithm for Euler's method is as follows.
(
)
(1) Choose h such that n = x f − x0 / h is an integer. (2) Given y0 , for n = 0, 1, 2,....N , iterate the scheme
y n +1 = yn + hf ( x0 + nh, y n ) Then, y n is as an approximation to y ( xn ) . Example 18 Use Euler's method with h = 0.1 to approximate the
solution to the initial value problem:
y ′( x ) = 0.2 xy,
y (1) = 1
on the interval 1 ≤ x ≤ 1.5.
(63)
535
Chapter Thirteen
Solution: We have:
x0 = 1 , x f = 1.5 , y0 = 1 , f ( x, y ) = 0.2 xy
∴ xn = x0 + nh = 1 + 0.1n , N =
1.5 − 1 =5 0.1
and, y n +1 = y n + 0.2(1 + 0.1n ) y n With y0 = 1 For n = 0, 1, 2, 3, 4 . The numerical results are listed in Table (7). Note that the differential equation in (63) is separable. The (unique) 2 solution of (63) is y ( x ) = e (0.1x − 0.1). This formula has been used to
compute the exact values y ( xn ) in the previous table. The next example illustrates the limitations of Euler's method. In the next subsections, we shall see more accurate methods than Euler's method. Table (7) Numerical results of Example 18.
xn
yn
y ( xn )
Absolute
Relative
error
error
١٫٠٠
١٫٠٠٠٠
١٫٠٠٠٠
٠٫٠٠٠٠
٠:٠٠
١٫١٠
١٫٠٢٠٠
١٫٠٢١٢
٠٫٠٠١٢
٠٫١٢
١٫٢٠
١٫٠٤٢٤
١٫٠٤٥٠
٠٫٠٠٢٥
٠٫٢٤
١٫٣٠
١٫٠٦٧٥
١٫٠٧١٤
٠٫٠٠٤٠
٠٫٣٧
١٫٤٠
١٫٠٩٥٢
١٫١٠٠٨
٠٫٠٠٥٥
٠٫٥٠
١٫٥٠
١٫١٢٥٩
١٫١٣٣١
٠٫٠٠٧٣
٠٫٦٤
536 Numerical Analysis Example 19 Use Euler's method with h = 0.1 to approximate the
solution to the initial value problem
y ′( x ) = 2 xy,
y (1) = 1
(64)
on the interval 1 ≤ x ≤ 1.5. Solution: As in the previous example, we have
x0 = 1
, x f = 1.5 , y0 = 1 ,
∴ xn = x0 + nh = 1 + 0.1n , N = and
y n +1 = yn + 2(1 + 0.1n ) yn
1.5 − 1 =5 0.1
With
y0 = 1
For n = 0, 1, 2, 3, 4 . The numerical results are listed in Table(8). The relative errors show that our approximations are not very good. Table (8) Numerical results of Example 19.
xn
yn
y ( xn )
Absolute
Relative
error
error
١٫٠٠
١٫٠٠٠٠
١٫٠٠٠٠
٠٫٠٠٠٠
٠٫٠٠
١٫١٠
١٫٠٢٠٠
١٫٢٣٣٧
٠٫٠٣٣٧
٢٫٧٣
١٫٢٠
١٫٤٦٤٠
١٫٥٥٢٧
٠٫٠٨٨٧
٥٫٧١
١٫٣٠
١٫٨١٥٤
١٫٩٩٣٧
٠٫١٧٨٤
٨٫٩٥
١٫٤٠
٢٫٢٨٧٤
٢٫٦١١٧
٠٫٣٢٤٤
١٢٫٤٢
١٫٥٠
٢٫٩٢٧٨
٣٫٤٩٠٤
٠٫٥٦٢٥
١٦٫١٢
537
Chapter Thirteen
13.4.3 Improved Euler's method.
The improved Euler's method takes the average if the slopes at the left and right ends of each step. It is, here, formulated by means of a predictor and a corrector:
( ) h y nC+1 = ynC + [ f (xn , ynC ) + f (xn +1 , y nP+1 )], 2 y nP+1 = ynC + hf xn , y nC ,
(65) (66)
This method is of order 2. Example 20 Use the improved Euler’s method with h =0.1 to
approximate the solution to the initial value problem of Example 19
y ′( x ) = 2 xy,
y (1) = 1 on the interval 1 ≤ x ≤ 1.5.
Solution: We have
n = x0 + hn = 1 + 0.1n , n = 0, 1, ...,5. The approximation y n to y ( xn ) is given by the predictor-corrector scheme:
y0C = 1 , y nP+1 = ynC + 0.2 xn yn ,
(
y nC+1 = y nC + 0.1 xn ynC + xn +1 y nP+1
)
for n = 0, 1, . . . , 4. The numerical results are listed in Table 5.3. These results are much better than those listed in Table(9) for Euler's method.
538 Numerical Analysis
Table (9) Numerical results of Example 20.
xn
yn
y ( xn )
Absolute
Relative
error
error
١٫٠٠
١٫٠٠٠٠
١٫٠٠٠٠
٠٫٠٠٠٠
٠٫٠٠
١٫١٠
١٫٢٣٢٠
١٫٢٣٣٧
٠٫٠٠١٧
٠٫١٤
١٫٢٠
١٫٥٤٧٩
١٫٥٥٢٧
٠٫٠٠٤٨
٠٫٣١
١٫٣٠
١٫٩٨٣٢
١٫٩٩٣٧
٠٫٠١٠٦
٠٫٥٣
١٫٤٠
٢٫٥٩٠٨
٢٫٦١١٧
٠٫٠٢٠٩
٠٫٨٠
١٫٥٠
٣٫٤٥٠٩
٣٫٤٩٠٤
٠٫٠٣٤٤
١٫١٣
We need to develop methods of order greater than one, which, in general, are more precise than Euler's method. 13.4.4 Runge-Kutta Methods
Two-stage explicit Runge-Kutta methods are given by the following formula:
k1 = hf ( xn , y n )
(67)
k 2 = hf ( xn + h, yn + k1 )
(68)
1 (k1 + k 2 ) 2
(69)
y n +1 = y n +
539
Chapter Thirteen
Fourth-order Runge-Kutta method.
The fourth-order Runge-Kutta method is the very popular among the explicit one-step methods. The four-stage Runge-Kutta method of order 4 given by its formula:
k1 = hf ( xn , y n )
(70)
1 1 k 2 = hf xn + h, y n + k1 2 2
(71)
1 1 k3 = hf xn + h, y n + k 2 2 2
(72)
k 4 = hf ( xn + h, y n + k3 )
(73)
y n +1 = yn +
1 (k1 + 2k 2 + 2k3 + k 4 ) 6
(74)
The next example shows that the fourth-order Runge-Kutta method yields better results for (64) than the previous methods. Example 22 Use the fourth-order Runge-Kutta method with
h = 0.1 to approximate the solution to the initial value problem of Example 19, y ′( x ) = 2 xy,
y (1) = 1 on the interval 1 ≤ x ≤ 1.5.
Solution:
We have f ( x, y ) = 2 xy and
xn = 1.0 + 0.1n n = 0, 1, 2,....5 With the starting value y0 = 1.0 , the approximation y n to y ( xn ) is given by the scheme
540 Numerical Analysis
y n +1 = yn +
0.1 (k1 + 2k 2 + 2k3 + k 4 ) 6
where
k1 = 0.1 * 2 * (1.0 + 0.1n ) yn , k 2 = 0.1 * 2 * (1.05 + 0.1n )( y n + k1 / 2 ) , k3 = 0.1 * 2 * (1.05 + 0.1n )( yn + k 2 / 2 ) k 4 = 0.1 * 2 * (1 + 0.1(n + 1)( y n + k3 )) and n = 0,1, 2, 3, 4. The numerical results are listed in Table(10). These results are much better than all those previously obtained. Table(10). Numerical results for Example 22.
xn
yn
y ( xn )
١٫٠٠
١٫٠٠٠٠
١٫١٠
١٫٠٠٠٠
Absolute error ٠٫٠٠٠٠
Relative error ٠٫٠
١٫٢٣٣٧
١٫٢٣٣٧
٠٫٠٠٠٠
٠٫٠
١٫٢٠
١٫٥٥٢٧
١٫٥٥٢٧
٠٫٠٠٠٠
٠٫٠
١٫٣٠
١٫٩٩٣٧
١٫٩٩٣٧
٠٫٠٠٠٠
٠٫٠
١٫٤٠
٢٫٦١١٦
٢٫٦١١٧
٠٫٠٠٠١
٠٫٠
١٫٥٠
٣٫٤٩٠٢
٣٫٤٩٠٤
٠٫٠٠٠٢
٠٫٠
Example 23 Consider the initial value problem
y ′ = ( y − x − 1)2 + 2, y (0) = 1 Compute y 4 by means of Runge-Kutta's method of order 4 with step size h = 0.1.
541
Chapter Thirteen
Solution: The solution is given as shown in Table(11).
Table(11). Numerical results for Example 23. n
xn
yn
Exact value y ( xn )
Global error y ( xn ) − y n
0
0.0
1.000 000 000
1.000 000 000
0.000 000 000
1
0.1
1.200 334 589
1.200 334 672
0.000 000 083
2
0.2
1.402 709 878
1.402 710 0'36
0.000 000157
3
0.3
1.609 336 039
1.609 336 250
0.000 000181
4
0.4
1.822 792 993
1.822 793 219
0.000 000 226
Example 24 Use Runge-kutta fourth order approximation to obtain
y at x = 0.6 if y ′ =
x + y and y = 0.41 at x = 0.4 ( h = 0.2 )
Solution:
y0 = 0.41 at x0 = 0.4 , h = 0.2 and x1 = 0.6 k1 = f ( x0 , y0 ) = h x0 + y0 = 0.2 * 0.4 + 0.41 = 0.18
1 1 k 2 = hf x0 + h, y0 + k1 = 0.2 * 0.5 + 0.41 + 0.09 = 0.2 2 2 1 1 k3 = hf x0 + h, y0 + k 2 = 0.2 0.5 + 0.41 + 0.1 = 0.201 2 2
k 4 = hf ( x0 + h, y0 + k3 ) = 0.2 0.5 + 0.41 + 0.09 = 0.2201 y1 = y0 +
1 (k1 + 2k 2 + 2k3 + k 4 ) 6
∴ y1 = 0.41 + 0.2003476 = 0.6103476
542 Numerical Analysis Problems
Solve the following differential equations by the methods indicated Euler’s method: 1- y ′ = 2 x − y , x = 0,
y = 1 , x = 0, 0.2
2- y ′ = 2 x + y 2 , x = 0 , y = 1.4, x = 0.1, 0.3
Modified Euler’s method 1- y ′ = x 2 − 2 x + y ,
(
x = 0,
y = 0.5,
x = 0.1, 0.5
)1 / 2 ,
x = 0,
y = 1 x = 0.1, 0.2
3- y ′ = ( x + y ) / xy,
x = 1,
y = 1,
2- y ′ = y − x 2
x = 1.0, 1.2
Runge-Kutta’s method 1- y ′ = (2 x + y )1 / 2 ,
(
x = 1, y = 2,
)
2- y ′ = 1 − x 3 / y , x = 0, y = 1, 3- y ′ = ( x − y ) / ( x + y ),
x = 1.0, 1.4 x = 0, 0.2
x = 0, y = 1 x = 0.2, 0.4
References [1] “Advanced Engineering Mathematics” by C. Ray Wyle and
Louis C. Barrett, Fifth Edition, 1985, by McGraw-Hill, ISBN 007-Y66643-1. [2] “Advanced Engineering Mathematics” by Erwin Kreysizig, Third Edition, 1972 by John Wiley & Sons, Inc., ISBN 0-47150728-8. [3] “ Engineering Mathematics” by K. A. Stroud, Third Edition, 1992, ELBS with Macmillan, Educational Low Priced Books Scheme. ISBN 0-333-54454-4. [4] “Theory and Problems of Engineering Mathematics for
Engineers and Scientists” by Murray R. Spiegel, Schaum’s Outline Series, McGraw-Hill, 1971 ISBN 07-060216-6. [5] “Numerical Methods for Engineers with Programing and
Software Applications” by Steven C. Chapra and Raymond P. Canale, 3rd Edition, 1998, McGraw-Hill, ISBN 0-07-115895-2. [6] “Engineering Mathematics” by Ian Craw, Stuart Dagger and John Pulham, the University of Aberdeen, 2001. [7] “ Numerical Methods” by E. A. Volkov, Translated from Russian by L. Levant, MIR PUBLISHER MOSCOW,1986. [8] “Mathematical Handbook of Formulas and Tables”, by Murray
R. Spiegel and John Liu, Schaum’s Outline Series, McGraw-Hill, Second Edition, 1999, ISBN 0-07-038203-4.
