Lecture 03 on 07.10.2015 Partial Differential Equations MA201 Mathematics III
M. G. P. Prasad Department of Mathematics IIT Guwahati
Learning Outcome of this Topic
We learn Surfaces Orthogonal to a Given System of Surfaces Cauchy’s Method of Characteristics (for Quasi-Linear PDE)
Reference: Sneddon Book, Chapter 2, Sections 6 to 10
Finding Surfaces Orthogonal to
A Given System of Surfaces
Surfaces orthogonal to a given system of surfaces Consider a one-parameter family of surfaces
f (x,y,z) = c.
(1)
Problem: To find a system of surfaces z = φ(x, y) which cut each of the surfaces (1) at right angles (orthogonally). Suppose the point A(x,y,z) is on one of the surfaces. Then, the normal to the surface at the point is along (P,Q,R) := (f x , f y , f z ). Suppose a surface z = φ(x, y) cuts the surface of (1) passing through A orthogonally at A . The direction of the normal z = φ(x, y) at A is
(φx , φy , −1) =
∂z
∂z , , −1 . ∂x ∂y
Since the two normals are perpendicular, z = φ(x, y) is a solution of
∂f ∂z ∂f ∂z ∂f ∂z ∂z + = ⇒ P + Q = R where P = f x , Q = f y , R = f z . ∂x ∂x ∂y ∂y ∂z ∂x ∂y (2) It is clear that any solution of (2) cuts each of the surfaces (1) orthogonally at each intersecting point.
Example: Find the surface which intersects the surfaces of the system z(x + y) = c(3z + 1) orthogonally and which passes through the circle x2 + y 2 = 1, z = 1. z(x + y) y . The Solution: Here f = , and P = 3zz+1 , Q = 3zz+1 , R = (3xz+ +1) 3z + 1 surfaces are generated by the integral curves of 2
dx dy dz = = . z(3z + 1) z(3z + 1) x + y Taking the first two, we get u(x, y) = x − y = c 1 . Also, x dx + y dy dz = , i.e., x dx + y dy = z(3z + 1)dz , which gives (x + y)z(3z + 1) x + y
v(x, y) = x 2 + y 2 − 2z 3 − z 2 = c 2 . Thus any surface which is orthogonal to the given surfaces has equations of the form f (u, v) = 0. In particular, for any f (of one variable)
x2 + y 2 − 2z 3 − z 2 = f (x − y) is a solution. For the particular surface passing through the circle x2 + y 2 = 1, z = 1, take f = −2, and get the required surface as
x2 + y 2 = 2z 3 + z 2 − 2.
Characteristic Curves (Characteristics)
Idea behind Characteristic Curves Aim To find a solution of the first order PDE F (x,y,z,p,q ) = 0. We are looking for curves on which the solution to the PDE is constant and equal to its initial value along the curves. Such curves are called characteristic curves. Along the characteristic curves, the given PDE becomes an ODE (or Eqn. with One Deg. of Freedom Less) and it can be solved with the techniques available for solving the ODE. ∂u(x,y ) = 0. Motivating Example: Consider ∂x Then the solution to this PDE is u(x, y) = g(y) which is a function of y alone. Consider a horizontal line y = a . Along this line, the initial value of u(x, y) is specified as u(0, a) = b . Then, u(0, a) = b = g(a) implies u(x, a) = b for all x on the entire line y = a . The line y = a is called a characteristic curve of the PDE u x = 0. To find the value of u on this characteristic curve, we only need the value of u at a point on this characteristic curve. Solutions of many first order PDEs can be found by slightly generalizing the above idea.
Method of Characteristics
for Quasi-linear First Order Equations in two independent variables
Method of Characteristics for Quasi-linear Equations a(x,y,z) p + b(x,y,z) q = c(x,y,z)
(3)
Let us assume that an integral surface
F (x,y,z) = z(x, y) − z = 0 of (3) can be found. Then the gradient vector ∇F = (zx , zy , −1) is perpendicular to the integral surface F (x,y,z) = 0. That is, ∇F and the vector [a,b,c] are orthogonal, if ∇F is defined and nonzero. That is, the vector [a,b,c] lies in the tangent plane of the surface F (x,y,z) = 0, if ∇F is defined and nonzero. At each point (x,y,z) on the surface, the vector [a,b,c] determines a direction that is called characteristic direction/ Monge direction. As a result, [a,b,c] determines a characteristic direction field in xyz -space and we can construct a family of curves that have the characteristic direction at each point.
continuation of previous slide
If the parametric form of these curves is
x = x(t), then
y = y(t),
z = z(t)
dx dy dz = a(x,y,z), = b(x,y,z), = c(x,y,z) (4) dt dt dt This equation (4) is called the characteristic equations for the PDE (3) and the solutions to (4) are called the characteristic curves/ Monge Curves of the PDE (3). Assuming a , b and c are sufficiently smooth and do not all vanish at the same point, the theory of ODE guarantees that a unique characteristic curve passes through each point (x0 , y0 , z0 ). The IVP for (3) requires that z(x, y) be specified on a given curve in xy -space which determines a curve C in xyz -space referred to as initial curve. To solve this IVP, we pass a characteristic curve through each point (x0 , y0 , z0 ) on the initial curve C . Then, the integral surface is formed as the union of characteristic curves. If the characteristic curves generate a surface then it is a solution of the IVP.
Method of Characteristics - Idea
Cauchy’s Initial Value Problem Initial Value Problem of Quasi-Linear first order PDE: PDE: Initial Curve:
a(x,y,z) p + b(x,y,z) q = c(x,y,z) C : x = x(s), y = y(s), z = z(s)
for s ∈ I ⊆ R
where C is continuously differentiable curve. Find the integral surface z = z(x, y) of the PDE passing through the initial curve C . Initial Value Problem of Quasi-Linear first order PDE: PDE: Initial Condition:
a(x,y,z) p + b(x,y,z) q = c(x,y,z) z|Γ = φ
where Γ : x = x(s), y = y(s) for s ∈ I is a given smooth curve in xy -plane. Here, the initial condition is stated by prescribing the values of z = z(x(s), y(s)) = φ(s) on the curve Γ . Find the integral surface z = z(x, y) of the PDE satisfying the initial condition.
Existence & Uniqueness Result for Cauchy’s IVP Theorem Consider the Cauchy’s IVP: PDE: Initial Curve:
a(x,y,z) p + b(x,y,z) q = c(x,y,z) C : x = x(s), y = y(s), z = z(s)
for s ∈ I ⊆ R
where a , b , c have continuous partial derivatives in all three variables x , y and z. Suppose that the initial curve C has a continuous tangent vector and that
∆(s) =
dy ds
a(x(s), y(s), z(s)) −
dx ds
b(x(s), y(s), z(s)) =0
for each s ∈ I .
Then there exists a unique solution z = z(x, y) defined in some neighborhood of the initial curve C to the above mentioned IVP.
= 0 ensures that the initial curve is NOT a characteristic The condition ∆(s) curve. Note that if the initial curve C is a characteristic curve then the integral surfaces of the PDE exist, but they are not unique.
Example: Solve the Quasi-linear PDE using Method of Characteristics
Consider the IVP PDE: Initial Condition:
z p + q = 0 z(x, 0) = f (x) where f is a smooth function.
Step 1: Parameterizing the curve in the initial condition
The initial curve can be parameterized as x(s) = s , y(s) = 0 and z(x(s), y(s)) = f (x(s)) = f (s) for all s ∈ R. Step 2: Finding the Characteristic Curve γ passing through the point ( s, 0, f (s)) on the initial curve C
We find the solutions x = x(s, t), y = y(s, t) and z = z(s, t) of the characteristic equations
dx dy dz = a(x,y,z) = z, = b(x,y,z) = 1, = c(x,y,z) = 0 dt dt dt with the initial conditions x(s, 0) = s , y(s, 0) = 0 and z(s, 0) = f (s). Solving we get, x(s, t) = zt + c1 (s),
y(s, t) = t + c2 (s),
z(s, t) = c 3 (s) .
Applying initial conditions at s = 0, we get x(s, t) = zt + s, y(s, t) = t , and z(s, t) = f (s) which is the parametric equation of the characteristic curve.
Step 3: Expressing ( s, t) in terms of ( x, y ) dx Since ∆(s) = dy a − b = 0 − ds ds
1 = −1 = 0, we solve (s, t) in terms of (x, y)
and get
s = x − zt
and
t = y .
Step 4: Expressing z (s, t) as a function of z (x, y )
Since z(s, t) = f (s), we get
z(x, y) = f (x − zy) which is an implicit formula of a solution to the given PDE. Note: In case of Linear and semi-linear PDE, the first two equations (dx/dt) = a , (dy/dt) = b in the system of characteristic equations decouples from the third equation (dz/dt) = c . But it does not happen in case of quasi-linear PDE. Consequently, we may get projected characteristic curves crossing themselves and which will lead to a singularity in the solution at the intersecting point.
