2nd Topic Partial Differential Equations Solution of partial differential equation
Prepared by: Dr. Sunil NIT Hamirpur (HP) (Last updated on 03-09-2007)
SOLUTIONS OF A PARTIAL-DIFFERENTIAL EQUATION:
It is clear that a partial differential equation can be obtained by elimination of arbitrary constants or by the elimination of arbitrary functions. Partial differential equation of first order
The general form of a first order partial differential equation is f (x, y, z, p, q ) = 0 , where x, y are the two independent variables, z is the dependent variable and
(i)
∂z =p , ∂x
∂z = q. ∂y Complete solution:
Any function f(x, y, z, a, b) = 0,
(ii)
involving two arbitrary constants a, b and satisfying the partial differential equation (i) is known as complete solution or complete integral or primitive. General solution or general integral:
Any arbitrary function F of specific (given) functions u, v F(u, v ) = 0 ,
(iii)
satisfying partial differential equation (i) is known as general solution or general integral.
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Partial Differential Equations: Solution of PDE Prepared by: Dr. Sunil, NIT Hamirpur (HP)
Particular solution or particular integral:
A solution obtained from the complete integral by assigning particular values to the arbitrary constants is called a particular solution or particular integral. Singular solution or singular integral:
The envelope of the family of surfaces (ii), with parameters a and b, if it exist, is called a singular solution or singular integral. Remarks: The singular integral differs from the particular integral in that it is not
obtained from the complete integral by giving particular values to the constants. A solution of a partial differential equation in a region R is a function of the independent variables, whose partial derivatives satisfy the partial differential equation at every point in R. As such, a partial differential equation may have a large number of entirely different solutions.
(
)
For example, u = x 2 − y2 , u = log x 2 − y 2 , u = sin kx cos ky are solutions of the Laplace equation
∂ 2u
∂ 2u
+
∂x 2
= 0 . The unique solution of a partial differential equation
∂y 2
corresponding to a physical problem must satisfy certain other conditions at the boundary of the region R. These are known as the boundary conditions. If these conditions are given for the time t = 0, they are known as the initial conditions.
solutionss of the equatio equation n Theorem: Show that if u1 and u 2 are two solution
∂ 2u ∂x 2
+
∂ 2u ∂y 2
+
∂ 2u ∂z 2
=A
∂ 2u ∂t 2
+B
∂u , then c1u1 + c 2 u 2 is also a solution. ∂t
Proof: Since u1 and u 2 are two two solution solutionss of the given given equat equation, ion, we we have
∂ 2u1 ∂x 2 and
+
∂ 2u1 ∂y 2
∂ 2u 2
Now
∂x
2
∂2 ∂x 2
+
+
∂ 2u1 ∂z 2
∂2u 2 ∂y
2
+
=A ∂2u 2 ∂z
2
(c1u1 + c2 u 2 ) +
∂ 2u1 ∂t 2 =A ∂2 ∂y 2
+B
∂u1 , ∂t
∂ 2u 2 ∂t
2
+B
(i)
∂u 2 . ∂t
(c1u1 + c2 u 2 ) +
(ii)
∂2 ∂z 2
(c1u1 + c 2u 2 )
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Partial Differential Equations: Solution of PDE Prepared by: Dr. Sunil, NIT Hamirpur (HP)
∂ 2u1 ∂ 2u1 ∂ 2u1 ∂ 2u 2 ∂ 2u 2 ∂ 2u 2 = c1 ∂x 2 + ∂y2 + ∂z 2 + c 2 ∂x 2 + ∂y 2 + ∂z 2 ∂ 2 u1 ∂ 2u 2 ∂u1 ∂u 2 = c1 A 2 + B ∂t + c 2 A ∂t 2 + B ∂t ∂ t =A
∂2 ∂t
(c u + c 2u 2 ) + B 2 1 1
[using (i) and (ii)]
∂ (c1u1 + c2u 2 ) . ∂t
⇒ c1u1 + c 2 u 2 is also a solution of the given equation. Generalization: If u1 , u 2 , u 3 ..........u n are n independent solutions,
then c1u1 + c 2 u 2 + c 3 u 3 + ............ + c n u n is also a solution. Problem on verifications of a solution Q.No.1.: Verify that e
−n 2t
∂u ∂ 2u = sin nx is a solution of the heat equation . ∂t ∂x 2 N
Hence, show that
∑
cne
−n 2t
sin nx , where c1 , c 2 ,................, c N are arbitrary
n =1
constants, is a solution of this equation satisfying the boundary conditions u (0, t ) = 0 and u (π, t ) = 0 . st
Sol.: 1 Part: Show that e
Here u = e
−n 2 t
−n 2t
∂u ∂ 2u sin nx is a solution of the heat equation . = ∂t ∂x 2
sin nx .
2 ∂ 2u ∂u ∂u −n 2t 2 −n 2 t 2 −n t = − Then sin nx , cos nx , n e sin nx . = −n e = ne ∂t ∂x ∂x 2
∂u ∂ 2u = So that . ∂t ∂x 2 2
∴ u = e − n t sin nx is a solution of the given equation. nd
2
N
Part: Show that
∑
cne
−n 2t
sin nx is also a solution.
n =1
For n = 1, 2, …………, N, we get N different solutions. Their linear combination
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Partial Differential Equations: Solution of PDE Prepared by: Dr. Sunil, NIT Hamirpur (HP)
u = c1e
−12 t
sin 1x + c 2e
⇒ u (x, t ) =
N
∑
−22 t
sin 2x + ........... + c Ne
−N2t
sin Nx is also a solution.
2 c n e −n t sin nx is also a solution.
n =1
Clearly, u (0, t ) = 0 and u (π, t ) = 0 , since sin nπ = 0 , where n is an integer. *** *** *** *** *** ***
Home Assignments Q.No.1.: Verify that z = f x 2 + y 2 is a solution of y
∂z ∂z −x =0. ∂x ∂y
Sol.: Q.No.2.: Verify that u = cos kx sinh ky is a solution of the Laplace equation
∂ 2u ∂x 2
+
∂ 2u ∂y 2
=0.
Sol.: Q.No.3.: Verify that e
− k 2 t
2 ∂u kx 2 ∂ u . =c sin is a solution of the heat equation 2 t ∂ c ∂x n
Hence, show that
kx − 2 A k e k t sin , where A1, A2, ………. are arbitrary c k =1
∑
constants, is a solution satisfying the boundary conditions u (0, t ) = u( cπ, t ) = 0 . Sol.:
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EQUATIONS SOLVABLE BY DIRECT INTEGRATION:
Those equations, which contain only one partial derivative, can be solved by direct integration. In place of usual constants of integration, we must, use arbitrary
functions of the variable kept constant.
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Partial Differential Equations: Solution of PDE Prepared by: Dr. Sunil, NIT Hamirpur (HP)
Now let us solve some partial differential equations, which can be solved by direct integration:
Q.No.1.: Solve the following partial differential equation:
∂ 3z 2
∂x ∂y
+ 18xy 2 + sin (2x − y ) = 0 .
Sol.: Given partial differential equation is
∂ 3z ∂x 2∂y
+ 18xy 2 + sin (2x − y ) = 0 .
Integrate twice w.r.t. x (keeping y fixed), we get
∂ 2z 1 + 9x 2 y 2 − cos(2x − y) = f ( y) , ∂x∂y 2 ∂z 1 + 3x 3y 2 − sin (2x − y) = xf ( y) + g( y) . ∂y 4 Now integrate w.r.t. y (keeping x fixed), we get 3 3
z+x y −
1 4
cos(2x − y ) = x
∫
f ( y)dy +
The result may simplified by writing Thus, z =
1 4
∫
∫
g( y)dy + w ( x ) .
f ( y)dy = u ( y)
and
∫
g ( y)dy = v( y) .
cos(2 x − y ) − x 3y3 + xu ( y) + v( y) + w ( x ) ,
where u, v, w are arbitrary functions. This is the required solution. Q.No.2.: Solve the following partial differential equation:
given that when x = 0, z = e y and
∂ 2z ∂x 2
+ z =0 ,
∂z = 1. ∂x
Sol.: If z were a function of x alone, the solution would have been z = A sin x + B cos x ,
where A and B are arbitrary constants. But here z is a function of x and y, therefore, A and B can be arbitrary functions of y, the independent variable kept constant. Hence, the solution of the given equation is z = f ( y) sin x + φ( y) cos x .
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Partial Differential Equations: Solution of PDE Prepared by: Dr. Sunil, NIT Hamirpur (HP)
∂z = f ( y) cos x − φ(y )sin x . ∂x When x = 0, z = e y . When x = 0,
∂z = 1. ∂x
∴ e y = φ(y ) . ∴1 = f ( y) .
Hence, the desired solution is z = sin x + e y cos x .
∂ 2z = sin x sin y , Q.No.3.: Solve the following partial differential equation: ∂x∂y given that
∂z = −2 sin y , when x = 0; ∂y
and z = 0, when y is an odd multiple of
Sol.: Given partial differential equation is
π 2
.
∂ 2z = sin x sin y . ∂x∂y
Integrating w.r.t. x, keeping y as constant, we get
∂z = − cos x sin y + f ( y) . ∂y When x = 0,
(i)
∂z = −2 sin y . ∂y
∴ −2 sin y = − sin y + f ( y) ⇒ f ( y) = − sin y . From (i), we get
∂z = − cos x sin y − sin y . ∂y
Integrating w.r.t. y, keeping x as constant, we get z = cos x cos y + cos y + φ(x ) . When y is an odd multiple of
(ii)
π 2
, z = 0.
∴ 0 = 0 + 0 + φ(x ) , since cos(2n + 1)
π 2
= 0 ⇒ φ( x ) = 0 .
∴ From (ii), we get z = (1 + cos x )cos y , which is the required particular solution. *** *** *** *** ***
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Partial Differential Equations: Solution of PDE Prepared by: Dr. Sunil, NIT Hamirpur (HP)
*** *** *** ***
Home Assignments x ∂ 2z = +a. Q.No.1.: Solve the following partial differential equation: ∂x∂y y Ans.: z =
x2 2
log y + axy + φ(x ) + ψ (y) .
Q.No.2.: Solve the following partial differential equation: Ans.: z =
∂ 2z ∂x 2
= xy .
1 3 x y + xf (y ) + φ(y ) . 6
∂ 2u Q.No.3.: Solve the following partial differential equation: = e − t cos x . ∂x∂t Ans.: u = −e − t sin x + φ(x ) + ψ (t ) . Q.No.4.: Solve the following partial differential equation:
Ans.: z = f (y ) + xφ(y ) + ψ (y ) −
1 12
∂ 3z ∂x 2∂y
= cos(2x + 3y ) .
sin (2 x + 3y ) .
Q.No.5.: Solve the following partial differential equation:
∂ 2z ∂y 2
= z , gives that when y = 0, z = e x and
∂z = e−x . ∂y
Ans.: z = e x cosh y + e − x sinh y Q.No.6.: Solve the following partial differential equation:
∂ 2z ∂x 2
= a 2z , gives that when x = 0,
∂z ∂z = a sin y and = 0. ∂x ∂y
Ans.: z = A cosh x + sin y sinh ax . 1 ∂ 2z = . Q.No.7.: Solve the following partial differential equation: ∂x∂y xy
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Partial Differential Equations: Solution of PDE Prepared by: Dr. Sunil, NIT Hamirpur (HP)
Sol.: z = log x log y + g(y ) + φ(x ) .
∂ 2z = 2x + 2 y . Q.No.8.: Solve the following partial differential equation: ∂x∂y Ans.: z = xy(x + y ) + g( y) + φ(x ) . Q.No.9.: Solve the following partial differential equation:
Ans.: z = −
1 x
2
∂ 2z ∂y
2
= sin (xy ) .
sin xy + yf (x ) + φ(x ) .
∂2u = 4 x sin (3xy ) . Q.No.10.: Solve the following partial differential equation: ∂y∂x Ans.: u = −
4 9y
sin (3xy ) + f (x ) + φ(y) .
∂ 2z Q.No.11.: Solve the following partial differential equation: log = x +y . ∂ ∂ x y Ans.: z = e x + y + g(y ) + φ(x ) . *** *** *** *** *** *** *** *** ***
3rd Topic Partial Differential Equations Linear partial differential equation of first order (Lagrange’s linear equation)
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