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Solved Problems

This satisfies the wave equation, because the equation is shift-invariant. It also
satisfies the initial conditions.

1 1 Z x
u(x, T) = (φ(x) + φ(x)) + ψ(τ) dτ = φ(x)
2 2c x
0
0
cφ (x + c(t − T)) − cφ (x − c(t − T))
u (x, t) = +
t
2
ψ(x + c(t − T)) + ψ(x − c(t − T))
+
2
1 1
0
0
u (x, T) = (cφ (x) − cφ (x)) + (ψ(x) + ψ(x)) = ψ(x)
t
2 2

Exercise 118. In obtaining the general solution of the wave equation the interval

was chosen to be infinite in order to simplify the evaluation of the functions α(ξ)
and β(ξ) in the general solution

u(x, t) = α(x + ct) + β(x − ct).

But this general solution is in fact valid for any interval be it infinite or finite.

We need only choose appropriate functions α(ξ), β(ξ) to satisfy the appropriate
initial and boundary conditions. This is not always convenient but there are other
situations besides the solution for u(x, t) in an infinite domain in which the general
solution is of use. Consider the “whip-cracking” problem,

2
u = c u ,
xx
tt
(with c a constant) in the domain x > 0, t > 0 with initial conditions

u(x, 0) = u (x, 0) = 0 x > 0,
t
and boundary conditions

u(0, t) = γ(t)
prescribed for all t > 0. Here γ(0) = 0. Find α and β so as to determine u for

x > 0, t > 0.
Hint: (From physical considerations conclude that you can take α(ξ) = 0. Your
solution will corroborate this.) Use the initial conditions to determine α(ξ) and

β(ξ) for ξ > 0. Then use the initial condition to determine β(ξ) for ξ < 0.
Solution: Since the solution is a wave moving to the right, we conclude that we
could take α(ξ) = 0. Our solution will corroborate this.

The form of the solution is

u(x, t) = α(x + ct) + β(x − ct).

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