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Note: 1 lecture, diﬀerent from §9.6 in [EP], part of §10.7 in [BD]

We have solved the wave equation by using Fourier series. But it is often more convenient to use the so-called d’Alembert solution to the wave equation^{3}. While this solution can be derived using Fourier series as well, it is really an awkward use of those concepts. It is easier and more instructive to derive this solution by making a correct change of variables to get an equation that can be solved by simple integration.

Suppose we have the wave equation

(4.14) |

We wish to solve the equation (4.14) given the conditions

(4.15) |

We will transform the equation into a simpler form where it can be solved by simple integration. We change variables to , . The chain rule says:

We compute In the above computations, we used the fact from calculus that . We plug what we got into the wave equation,Therefore, the wave equation (4.14) transforms into . It is easy to ﬁnd the general solution to this equation by integrating twice. Keeping constant, we integrate with respect to ﬁrst^{4} and notice that the constant of integration depends on ; for each we might get a diﬀerent constant of integration. We get . Next, we integrate with respect to and notice that the constant of integration must depend on . Thus, . The solution must, therefore, be of the following form for some functions and :

The solution is a superposition of two functions (waves) traveling at speed in opposite directions. The coordinates and are called the characteristic coordinates, and a similar technique can be applied to more complicated hyperbolic PDE.

We know what any solution must look like, but we need to solve for the given side conditions. We will just give the formula and see that it works. First let denote the odd extension of , and let denote the odd extension of . Deﬁne

We claim this and give the solution. Explicitly, the solution is or in other words:

(4.16) |

Let us check that the d’Alembert formula really works.

So far so good. Assume for simplicity is diﬀerentiable. By the fundamental theorem of calculus we have

So

Yay! We’re smoking now. OK, now the boundary conditions. Note that and are odd. Also is an even function of because is odd (to see this fact, do the substitution ). So

Note that and are periodic. We compute

And voilà, it works.

Example 4.8.1: D’Alembert says that the solution is a superposition of two functions (waves) moving in the opposite direction at “speed” . To get an idea of how it works, let us work out an example. Consider the simpler setup

Here is an impulse of height 1 centered at :The graph of this impulse is the top left plot in Figure 4.21.

Let be the odd periodic extension of . Then from (4.16) we know that the solution is given as

It is not hard to compute speciﬁc values of . For example, to compute we notice and . Now and . Hence . As you can see the d’Alembert solution is much easier to actually compute and to plot than the Fourier series solution. See Figure 4.21 for plots of the solution for several diﬀerent .

It is perhaps easier and more useful to memorize the procedure rather than the formula itself. The important thing to remember is that a solution to the wave equation is a superposition of two waves traveling in opposite directions. That is,

If you think about it, the exact formulas for and are not hard to guess once you realize what kind of side conditions is supposed to satisfy. Let us give the formula again, but slightly diﬀerently. Best approach is to do this in stages. When (and hence ) we have the solution

On the other hand, when (and hence ), we let

The solution in this case is

By superposition we get a solution for the general side conditions (4.15) (when neither nor are identically zero).

(4.17) |

Do note the minus sign before the , and the in the second denominator.

Warning: Make sure you use the odd extensions and , when you have formulas for and . The thing is, those formulas in general hold only for , and are not usually equal to and for other .

Exercise 4.8.2: Using the d’Alembert solution solve , , , , , and . Hint: Note that is the odd extension of and .

Exercise 4.8.4: Take , , , , , and . a) Solve using the d’Alembert formula. Hint: You can use the sine series for . b) Find the solution as a function of for a ﬁxed , , and . Do not use the sine series here.

Exercise 4.8.5: Derive the d’Alembert solution for , , , , , and , using the Fourier series solution of the wave equation, by applying an appropriate trigonometric identity.

Exercise 4.8.6: The d’Alembert solution still works if there are no boundary conditions and the initial condition is deﬁned on the whole real line. Suppose that (for all on the real line and ), , and , where

Solve using the d’Alembert solution. That is, write down a piecewise deﬁnition for the solution. Then sketch the solution for , , , and .

Exercise 4.8.102: Take , , , , , and . Using the D’Alembert solution ﬁnd the solution at a) , b) , c) . You may have to split your answer up by cases.

Exercise 4.8.103: Take , , , , , and . Suppose that , , , . Using the D’Alembert solution ﬁnd a) , b) , c) .

^{3}Named after the French mathematician Jean le Rond d’Alembert (1717–1783).

^{4}We can just as well integrate with ﬁrst, if we wish.