Variations on the Wave Equation

Section 15
Variations on the Wave Equation
James V. Herod*

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

Web page maintained by Evans M. Harrell, II,

We consider other boundary conditions and initial conditions with the wave equation.

The semi-infinite string is set up as a vibrating string with one end fixed at zero and with initial conditions. The partial differential equation is the same:

utt = [[gamma]]2 uxx.

However, we only consider this equation for 0 < x < * and for 0 < t < *. The result is we have a boundary at the left. ( The t-axis is a boundary because x can't be negative.) There results a necessity for a left boundary condition. We choose the following

Boundary Conditions: u(t,0) = 0, 0 < t < *.

Of course, there must be initial conditions:

Initial Conditions: A(u(0,x) = f(x), ut(0,x) = g(x)) 0 < x < *.

We now examine what must be the solution for this equation. Since the general solution for the partial differential equation was derived independent of initial conditions or boundary conditions, this general solution must find an interpretation in terms of this information. The solution was

u(t,x) = [[phi]](x-[[gamma]]t) + [[theta]](x+[[gamma]]t). (15.1)

If we substitute this general solution into the initial conditions as we did in the previous section, we come to the equations

F(1,2) [f(x-[[gamma]]t) + f(x+[[gamma]]t)] + F(1,2c) [I(0,x+[[gamma]]t, )g([[xi]]) d[[xi]] - I(0,x-[[gamma]]t, )g([[xi]]) d[[xi]]]. (15.2)

The question is, how do we interpret, for example, f(x-[[gamma]]t), when x-[[gamma]]t < 0? We use the boundary condition to answer this problem. The idea is to "extend" f and the integral of g to the negative numbers knowing them only on the positive numbers, but also knowing that the boundary condition u(t,0) = 0.

We call the extensions F and G. The equations u(t,0) = [[phi]]([[gamma]]t) + [[theta]]([[gamma]]t), together with the boundary conditions, imply that

0 = F(-[[gamma]]t) + f([[gamma]]t) + F(1,2[[gamma]]) G([[gamma]]t) - F(1,2[[gamma]]) G(-[[gamma]]t).

Since F and G are independent, we must have that

f([[gamma]]t) + F(-[[gamma]]t) = 0 and F(1,[[gamma]]) G([[gamma]]t) - F(1,[[gamma]]) G(-[[gamma]]t) = 0.

Thus, we make F to be an odd extension of f and G to be an even extension of the integral of g. This enables the formula for u to be valid for all x > 0 and all t > 0:

u(t,x) = F(1,2) [F(x-[[gamma]]t) + F(x+[[gamma]]t)] + F(1,2[[gamma]]) G(x+[[gamma]]t) - F(1,2[[gamma]]) G(x-[[gamma]]t).

This situation can be examined with several examples.

Example 15.1: Take [[gamma]] = 1 and suppose that

f(x) = (x+1) exp(-(x-3)2) - exp(-9)

and g(x) = 0.

Here's a graph of F, the odd extension of f.

Figure 15.1

And, Figure 15.2 is a graph of u given by

u(t,x) = [ F(x-t) + F(x+t)]/2

Figure 15.2

Example 15.2: Suppose that

and f(x) = 0.

g(x) = (x+1) exp(-(x-3)2).

Here's a graph of the even extension of G and of the graph of u.

Figure 15.3

Figure 15.4

Example 15.4:

We next consider a semi-infinite example without a fixed left boundary. We suppose a moving boundary on a semi-infinite string and watch the wave move to the right. We suppose the initial conditions are zero at first. The problem is

The wave equation: utt = [[gamma]]2 uxx, 0 < x, 0 < t.

The initial conditions: A( u(0,x) = 0, ut(0,x) = 0)

The moving boundary: u(t,0) = sin(t).

Think about the problem for a minute ... how it should look ... and you will see a "taut string" stretched motionless until "time" starts. Then, the left boundary goes up and down in a cyclic motion. With the speed of [[gamma]], the wave moves down the string to the right. Surely you believe this picture as much as you believe the following mathematical explanation!

Since u is a solution for the wave equation and we know the general solution for the wave equation, we can say that there are functions [[phi]] and [[theta]] so that

u(t,x) = [[phi]](x-[[gamma]]t) + [[theta]](x+[[gamma]]t), t > 0.

Since all of x, c, and t are positive, and f and g are zero, we have that [[phi]](x) and [[theta]](x) are zero for positive x. In the previous problem, we found what should happen for negative x's by using the boundary condition. We do this again. The boundary condition says that [[phi]](-[[gamma]]t) = u(t,0) = sin(t). Thus,

[[phi]]([[xi]]) = BLC{(A( 0 if [[xi]] > 0, sin(-[[xi]]/[[gamma]]) if [[xi]] < 0)).


u(t,x) = [[phi]](x-[[gamma]]t) = BLC{(A( 0 if x > [[gamma]]t, sin(([[gamma]]t-x)/[[gamma]]) if [[gamma]]t > x)).

and the graph of this is below.

Figure 15.5


15.1 The importance of linearity and superposition in all these wave equations is emphasized by considering the following problem:

The wave equation: utt = [[gamma]]2 uxx, 0 < x, 0 < t.

The initial conditions: A(u(0,x) = x exp(-x), ut(0,x) = -.32 exp(-x2))

The moving boundary: u(t,0) = sin(t).

Break the problem into three problems and solve each part. Show that the sum of the solutions of these three parts is a solution of the original problem. Sketch a 3 - dimensional plot of the solutions and several "snapshots."

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