D'Alembert's Solution

Section 14
D'Alembert's Solution for the Wave Equation
James V. Herod*

*(c) Copyright 1993,1994,1995 by James V. Herod, herod@math.gatech.edu. All rights reserved.

Web page maintained by Evans M. Harrell, II, harrell@math.gatech.edu.

NOTE: This chapter covers much the same material as the corresponding chapter of Professor Harrell's own hypertext book.
The problem we consider begins a study of how initial conditions and boundary conditions might be specified for hyperbolic problems. A first start in this direction is to consider the classical wave equation

F([[partialdiff]]2u,[[partialdiff]]t2) = [[gamma]]2 F([[partialdiff]]2u,[[partialdiff]]x2) , -* < x < *, 0 < t < *. (14.1)


u(0,x) = f(x) and F([[partialdiff]]u,[[partialdiff]]t)(0,x) = g(x), -* < x < *.

First, we should note that this is a classical equation. Anyone who has studied partial differential equations has likely investigated this problem by whatever techniques were being used. Second, the problem illustrates some of the ideas of the previous two sections.

Prior to this section, if asked to find the general solution to the partial differential equation, the first reaction would have been to write the equation in the first standard form, identify a, b, etc., and find the characteristics. (Maybe the hardest job in doing this for the above equation in the context of these notes is to adapt the notation for what has come before! )

In the process of finding the characteristics, we would get the equation in a standard form. Thus, we seek function S(t,x) and T(t,x), and look for functions x1 and x2 so that

(i) S(t,x1(t)) and T(t,x2(t)) are constant, and

(ii) the equation is in the second form.

These two requirements led in the previous section to equations for the slopes of x1 and x2:

slopes of characteristics = F(b +/- R(b2 - 4ac),2a).

In this case, where b = 0, a = 1, and c = -[[gamma]]2, we have that the characteristic lines are x1 = [[gamma]]t + [[xi]] and x2 = -[[gamma]]t + [[eta]]. The functions S and T are given by

S = x - [[gamma]]t and T = x + [[gamma]]t.

Define the function v by

u(t,x) = v(S,T) = v(x-[[gamma]]t,x+[[gamma]]t)

and v satisfies the equation

F([[partialdiff]]2v,[[partialdiff]]s[[partialdiff]]t) = 0.

The equation in v has solutions

v(S,T) = [[phi]](S) + [[theta]](T)


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

We have not yet used the initial conditions:

f(x)= u(0,x) = [[phi]](x) + [[theta]](x), (14.2)


g(x) = F([[partialdiff]]u,[[partialdiff]]t)(0,x) = -[[gamma]] [[phi]]'(x) + [[gamma]] [[theta]]'(x).

Integrating this last equation, we get

- [[gamma]] [[phi]](x) + [[gamma]] [[theta]](x) = I(0,x, )g([[xi]]) d[[xi]] + constant. (14.3)

Solve (14.2) and (14.3) simultaneously to get

[[phi]](x) = F(1,2) f(x) - F(1,2[[gamma]]) I(0,x, )g([[xi]])d[[xi]] + constant (14.4)


[[theta]](x) = F(1,2) f(x) + F(1,2[[gamma]]) I(0,x, )g([[xi]])d[[xi]] + constant. (14.5)


u(t,x) = F(1,2) [f(x- [[gamma]]t) + f(x+[[gamma]]t)] + F(1,2[[gamma]]) [I(x-[[gamma]]t,x+[[gamma]]t, )g([[xi]]) d[[xi]]].


There are three problems that we use to illustrate the character of solutions for this wave equation: with [[gamma]] = 3, we take

(1) u(0,x) = exp(-x2), ut(0,x) = 0,

(2) u(0,x) = sin(x), ut(0,x) = 0,

(3) u(0,x) = 0, ut(0,x) = cos(x).

Discussion of Example (1).

In this problem, there are two ways to visualize the traveling wave. One way is to look at the graph is as a three dimensional plot [ Figure 14.1] of

u(t,x) = F(1,2) [ exp(- (x-3t)2) + exp(-(x+3t)2)]

Such a graph shows the wave splitting into two parts and traveling along the characteristic curves S(t) = x-3t and T(t) = x+3t. This illustrates very well, and again, the importance of characteristic curves in understanding hyperbolic partial differential equuations.

Figure 14.1

An alternate visualization of the problem is to remember that this is essentially a vibrating string problem and that, in order to see the phenomena, we can view the graph as a one-dimensional wave moving in time. To see the effects of time, we take successive "snapshots." Our snapshots [Figure 14.2] are at t = 1/4, t = 1/2, and t = 1. In this visualization, one is struck by the wave moving in both directions along the x axis. Also, on can observe how far the wave moves in one unit of "time." When t = 1 the peak of the wave is at x = 3 and -3. Thus, the velocity is 3 units per time unit.

Figure 14.2

Discussion of Example 2.

In this case, the solution of the equation is

u(t,x) = F(1,2) [ sin(x-3t) + sin(x+3t)].

= sin(x) cos(3t).

Looking at the 3-dimensional plot [Figure14.3], it is at first hard to see the wave moving down the characteristics. However, seeing the 3-dimensional plots of sin(x-3t)/2 and sin(s+3t)/2 separately [Figure 14.4&5] show the movement along the characteristic curves clearly. Moreover, we understand that the reason this was not observed previously in Figure 14.3 was the cancellation of waves "colliding and cancelling" across characteristic curves.

Figure 14.3

Figure 14.4

Figure 14.5

Discussion of Example 3.

This example differs from the previous in that there is a zero initial displacement, and an initial velocity of cos(x). The solution is cos(x) sin(3t)/3. The graph in 3 - dimensions is as in Figure 14.6.

Figure 14.6

For further discussion of this method, with additional examples, see the corresponding chapter of Professor Harrell's hypertext book.


14.1 Plot "snapshots" for the case that u satisfies equation (14.1) with

u(0,x) = BLC{(A( 1 if -1 < x < 1, 0 everywhere else)).

ut(0,x) = 0.

14.2 Solve and plot solutions for equation (14.1) with

u(0,x) = 0

ut(0,x) = BLC{(A( 1 if -1 < x < 1, 0 everywhere else)).

14.3 Solve and plot solutions for equation (14.1) with

u(0,x) = 0

ut(0,x) =x exp(-x2).

14.4 Modify, solve, and plot solutions for the wave equation which account for a retarding factor:

utt = uxx - d ut

with u(0,x) = f(x)

ut(0,x) = 0.

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