Series Solutions

Section 19
The Cauchy Problem -- Series Solutions
James V. Herod*

*(c) Copyright 1993,1994,1995 by James V. Herod, All rights reserved.

Web page maintained by Evans M. Harrell, II,

To this point, we have considered several modifications of the wave equation: the infinite string, a half-infinite string, and a two point boundary problem. Each of these special problems was a particular case of a second order partial differential equation with constant coefficients. In each, there was an initial distribution along the string and an initial velocity for the string. These are special cases of a general structure we describe below.

There is a sense in which this is a general notion common to both ordinary and partial differential equations. In ordinary differential equations, by an initial value problem we mean the problem of finding the solution of a differential equation with the appropriate number of initial conditions prescribed at an initial point. For example, the second order, constant coefficient differential equation might be

y'' + 3y' + 2y = sin(t)

with initial conditions y(0) = 5, y'(0) = 7.

This equation can be solved in closed form. Also, it will be recalled that equations of this type have a series solution.

> dsolve({diff(y(x),x,x)+3*diff(y(x),x)+2*y(x)=sin(x),y(0)=-3/10,

D(y)(0)= 1/10},y(x));

> dsolve({diff(y(x),x,x)+3*diff(y(x),x)+2*y(x)=sin(x),y(0)=-3/10,

D(y)(0)= 1/10},y(x),series);

A comparable problem for the case of partial differential equations might be, for example, the wave condition with u(0,x) and ut(0,x) prescribed. Here, u(0,x) is the "initial displacement" and ut(0,x) is the "initial velocity".

We want to observe again that those initial values in the partial differential equation are not necessarily given along the x-axis. Such a line for the initial values is appropriate in the wave equation because the physical model we are using to guide our description of the model is suggestive of that idea. We suppose in the remainder of this section that u is a function of x and y, instead of x and t, and investigate what is required if we prescribe initial values along a curve y(x). Such a problem is called a Cauchy Problem.

Consider the rather general second order equation:

Auxx + Buxy + Cuyy + Dux + Euy + Fu = G(x,y). (19.1)

Suppose that two functions f(x) and g(x) are prescribed along the curve L = {x,y(x)}. The Cauchy Problem is one of determining the solution u for the equation (19.1) in a region nearby this curve in the x-y plane and which satisfies the Cauchy conditions

u(x,y(x)) = f(x),

and F([[partialdiff]]u,[[partialdiff]][[eta]]) = g(x)

on the curve L where [[eta]] is the direction of the normal to L. The functions f and g are called the Cauchy data.

Recall that in R2, a tangent to the curve {x,y(x)} is {1,y'(x)} and a normal to the curve is {-y'(x),1}. In that setting, the derivative of x and of y in the direction of the normal is given by

F(dx,d[[eta]]) = -y'(x) and F(dy,d[[eta]]) = 1.

For example, here are three curves; any one could be the curve L on which Cauchy data could be given.

Figure 19.1

Geometrically, one can now imagine a curve in R3 - {x,y,u} space - whose projection onto the plane is the curve L, and whose height is u(x,y(x)) = f(x). With this picture, we look for the solution to (19.1) having the prescribed Cauchy data of (19.2). The solution is to be a surface that contains this curve in R3 while also satisfying the second of the Cauchy conditions:

F([[partialdiff]]u,[[partialdiff]][[eta]]) = g.

This second condition specifies the direction that the surface moves off the curve in R3.

If f is differentiable, we have specified the tangent plane for the surface that is the graph of the solution to the Cauchy problem at each point {x, y(x), f(x)}. That is, we have specified a point {x,y(x),f(x)} and two vectors that are tangent to the surface:

{1,y'(x), f'(x)} and {-y'(x), 1, g(x)}.

Intuitively, this seems it should be adequate to have the surface, the solution of the Cauchy Problem, determined in a region about L. A little closer look at this idea is appropriate.

If f(x) is differentiable, then along the curve L

f'(x) = F(d,dx)u(x,y(x)) = F([[partialdiff]]u,dx) 1 + F([[partialdiff]]u,[[partialdiff]]y) y'


g(x) = F(d,d[[eta]])u(x,y(x)) = F([[partialdiff]]u,dx) (-y') + F([[partialdiff]]u,[[partialdiff]]y) 1.

These two equations must be re-written in matrix form

B(ACO2(1, y',-y',1))B(A([[partialdiff]]u/[[partialdiff]]x,[[partialdiff]u/[[partialdiff]]y)) = B(A(f'(x),g(x))).

Since this 2x2 matrix has determinant not zero, we can solve for [[partialdiff]]u/[[partialdiff]]x and [[partialdiff]]u/[[partialdiff]]y.

We now consider how we can determine the second order terms using that we have established that we can determine the first order terms. Differentiate ux and uy with respect to x:

F(d,dx) F([[partialdiff]],[[partialdiff]]x) u(x,y(x)) = F([[partialdiff]]2u,[[partialdiff]]x2) 1 + F([[partialdiff]]2u,[[partialdiff]]x[[partialdiff]]y) y'


F(d,dx) F([[partialdiff]],[[partialdiff]]y) u(x,y(x)) = F([[partialdiff]]2u,[[partialdiff]]x[[partialdiff]]y) 1 + F([[partialdiff]]2u,[[partialdiff]]y2) y'

This is two equations in the second partials; one more comes from the original problem. The second partials are related by

Auxx + Buxy + Cuyy = H(x,y,u,ux, uy).

We know all the right side because the first partials are known on L from the above development. Now we have three equations and three unknowns. The matrix system is

B(ACO3( 1, y', 0, 0, 1, y', A, B, C)) B(A(uxx,uxy,uyy)) = B(A([[partialdiff]]ux/[[partialdiff]]x,[[partialdiff]]uy/[[partialdiff]]x, H)).

This equation has solution provided the determinant of this matrix is not zero. That is

A(y'(x))2 - B y'(x) + C != 0.

This says that y(x) can not be a characteristic!

If the coefficients of our partial differential equation have all derivatives, then we can compute all the partials of u by the above process. In this case, the solution could be represented as a series solution be the two-dimensional Taylor series:

u(x,y) = ISU(n, , )ISU(k, , )F(1,k!(n-k)!) F([[partialdiff]]nu0,[[partialdiff]]x0k[[partialdiff]]y0n-k) (x-x0)k(y-y0)n-k.

Example 19.1: Consider the PDE

2 uxx + 3 uxy - 2uyy + 1 ux + 1 uy + 3 u/25

with initial conditions u(0,y) = sin(y) and ux(0,y) = 0. We solved this equation in Section 12 with other methods. Here are the 0th, 1st and 2nd order series approximations, with plots.

> u0:=(x,y)->sin(y): plot3d(u0(x,y),x=0..Pi,y=-Pi..Pi);

> u2:=(x,y)->sin(y)+x^2*(-sin(y)+cos(y)+3*sin(y)/25)/2:


Exercise 19.2:

This considers a series solution for the equation

uxx - uyy + 3ux + 5u = 0, with u(x,x2) = 1, du/d[[eta]] = 0.

The zero-th order approximation:

> u0:=(x,y)->1; plot3d(u(x,y),x=0..1,y=0..1);

To get a first order approximation:

We must solve the equations ux 1 - uy 2x = 0 and -ux 2x + uy 1 = 0. This system has solution ux = uy = 0.

> u1:=(x,y)->1 + 0*x + 0*y; plot3d(u1(x,y),x=0..1,y=0..1);

To get a second order approximation:

0 = d /dx [[partialdiff]] /[[partialdiff]]x u(x,x^2) = uxx*1 + uxy*2x,

0 = d/dx [[partialdiff]] /[[partialdiff]]y u(x,x^2) = uxy*1 + uyy*2x, and

uxx - uyy = -3ux(x,x^2) + 5 u(x,x^2)

> solve({1*a+2*x*b+0*c=0,0*a+1*b+2*x*c=0,1*a+0*b-1*c=-5},{a,b,c});

> assign("); a2:=subs(x=1/2,a);

This equation cannot be evaluated! Should we have expected this problem? Surely, the characteristics are tangent to the initial curve at the point {1/2,1/4}. We try another initial curve.

Use the same PDE, with new initial curve: y = 2x.We have the same zero-th and first order approximations. The three equations for the second order change:

0 = d /dx [[partialdiff]] /[[partialdiff]]x u(x,2x) = uxx*1 + uxy*2,

0 = d/dx [[partialdiff]] /[[partialdiff]]y u(x,2x) = uxy*1 + uyy*2, and

uxx - uyy = -3ux(x,2x) + 5 u(x,2x).

The solution provides a solution:

> solve({1*a+2*b+0*c=0,0*a+1*b+2*c=0,1*a+0*b-1*c=-5},{a,b,c});

> u2:=(x,y)->1 + 0*(x-1/2)+0*(y-1)-20/3*(x-1/2)^2/2+10/3*(x-1/2)*(y-1)


> plot3d(u2(x,y),x=0..1,y=0..1,axes=BOXED);


(19.1) Describe a curve on which the initial conditions might be given and for which it would not be possible to solve this partial differential equation:

y uxx + (x+y) uxy + x uyy = 0.

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