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integrating first over the horizontal line segments in Figure 5 and then

vertically.
A variant of the method of characteristic coordinates is to write
(6.10) as the system of two equations

u + cu = v, v − cv = f,
t
x
t
x
the first equation being the definition of v, as above. If we first solve
the second equation, then v is a line integral of f over a characteristic
line segment x + ct = constant. The first equation then gives u(x, t) by
sweeping out these line segments over the characteristic triangle ∆. To
carry out this variant is a little tricky, however, and we leave it as an

exercise.

6.6.1 Source on a half-line

The solution of the general inhomogeneous problem on a half-line
2
DE : v tt − c v xx = f(x, t) in 0 < x < ∞,
IC : v(x, 0) = φ(x) v t (x, 0) = φ(x) (6.17)
BC : v(0, t) = h(t)

is the sum of four terms, one for each data function φ, ψ, f, and h. For x > ct > 0, the
solution has precisely the same form as in (6.12), with the backward triangle ∆ as the
domain of dependence. For 0 < x < ct, however, it is given by
ZZZZ
x
v(x, t) = φ term + ψ term + h(t − ) + 1 f (6.18)
c 2c
D
where t − x/c is the reflection point and D is the shaded region in Figure 6.5
The only caveat is that the given conditions had better coincide at the origin.That is,
0
we require that φ(0) = h(0) and ψ(0) = h (0). If this were not assumed, there would be
a singularity on the characteristic line emanating from the corner.
Let’s derive the boundary term h(t − x/c) for x < ct. To accomplish this, it is
convenient to assume that φ = ψ = f = 0. We shall derive the solution from scratch
using the fact that v(x, t) must take the form v(x, t) = j(x + ct) + g(x − ct). From
the initial conditions (φ = ψ = 0), we find that j(s) = g(s) = 0 for s > 0. From the
boundary condition we have h(t) = v(0, t) = g(−ct) for t > 0. Thus g(s) = h(−s/c) for
s < 0. Therefore, if x < ct, t > 0, we have v(x, t) = 0 + h(−(x − ct/c)) = h(t − x/c).

6.6.2 Finite interval

For a finite interval (0, l) with inhomogeneous boundary conditions v(0, t) = h(t), v(l, t) =
k(t), we get the whole series of terms

x
v(x, t) = h(t − ) − h(t + x−2l ) + h(t − x+2l ) + . . .
c c c

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