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Solved Problems

We substitute the solution into the initial conditions.

u(x, 0) = α(ξ) + β(ξ) = 0, ξ > 0
0
0
u (x, 0) = cα (ξ) − cβ (ξ) = 0, ξ > 0
t
We integrate the second equation to obtain the system

α(ξ) + β(ξ) = 0, ξ > 0,
α(ξ) − β(ξ) = 2k, ξ > 0,

which has the solution

α(ξ) = k, β(ξ) = −k, ξ > 0.

Now we substitute the solution into the initial condition.

u(0, t) = α(ct) + β(−ct) = γ(t), t > 0

α(ξ) + β(−ξ) = γ(ξ/c), ξ > 0
β(ξ) = γ(−ξ/c) − k, ξ < 0

This determines u(x, t) for x > 0 as it depends on α(ξ) only for ξ > 0. The
constant k is arbitrary. Changing k does not change u(x, t). For simplicity, we take
k = 0.

u(x, t) = β(x − ct)
(
0 for x − ct < 0
u(x, t) =
γ(t − x/c) for x − ct > 0

u(x, t) = γ(t − x/c)H(ct − x)

Exercise 119. Let u(x, t) satisfy the equation

2
u = c u ;
tt
xx
(with c a constant) in some region of the (x, t) plane.

(a) Show that the quantity (u − cu ) is constant along each straight line defined
x
t
by x−ct = constant, and that (u +cu ) is constant along each straight line of
t
x
the form x + ct = constant. These straight lines are called characteristics; we
will refer to typical members of the two families as C and C characteristics,
−
+
respectively. Thus the line x − ct = constant is a C characteristic.
+
(b) Let u(x, 0) and u (x, 0) be prescribed for all values of x in −∞ < x < ∞,
t
and let (x , t ) be some point in the (x, t) plane, with t > 0. Draw the C +
0
0
0
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