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Solved Problems

satisfies the differential equation on (0 . . . l). We verify that it satisfies the
initial condition and boundary conditions.

1
ˆ
ˆ
u(x, 0) = (h(x) + h(x))
2
ˆ
u(x, 0) = h(x)

x + l x + l
u(x, 0) = sign x − 2l h x − 2l
2l 2l
u(x, 0) = h(x)
1 1
ˆ
ˆ
ˆ
ˆ
u(0, t) = (h(ct) + h(−ct)) = (h(ct) − h(ct)) = 0
2 2
1 1
ˆ
ˆ
ˆ
ˆ
u(l, t) = (h(l + ct) + h(l − ct)) = (h(l + ct) − h(l + ct)) = 0
2 2

Exercise 117. The deflection u(x, T) = φ(x) and velocity u (x, T) = ψ(x) for
t
2
an infinite string (governed by u = c u ) are measured at time T, and we are
tt
xx
asked to determine what the initial displacement and velocity profiles u(x, 0) and
u (x, 0) must have been. An alert student suggests that this problem is equivalent
t
to that of determining the solution of the wave equation at time T when initial
conditions u(x, 0) = φ(x), u (x, 0) = −ψ(x) are prescribed. Is she correct? If not,
t
can you rescue her idea?

Solution: Change of Variables. Let u(x, t) be the solution of the problem with
deflection u(x, T) = φ(x) and velocity u (x, T) = ψ(x). Define
t

v(x, τ) = u(x, T − τ).

We note that u(x, 0) = v(x, T). v(τ) satisfies the wave equation.

2
v ττ = c v xx

The initial conditions for v are

v(x, 0) = u(x, T) = φ(x), v (x, 0) = −u (x, T) = −ψ(x).
t
τ
Thus we see that the student was correct.

Direct Solution. D’Alembert’s solution is valid for all x and t. We formally
substitute t−T for t in this solution to solve the problem with deflection u(x, T) =
φ(x) and velocity u (x, T) = ψ(x).
t

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

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