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Solved Problems

(b) We can write the initial velocity in terms of the Heaviside function.


1 for 0 < x < 1

u (x, 0) = −1 for − 1 < x < 0
t

0 for |x| > 1.

u (x, 0) = −H(x + 1) + 2H(x) − H(x − 1)
t
We integrate the Heaviside function.


0 for b < c
b 
Z
H(x − c) dx = b − a for a > c
a 
b − c otherwise

If a < b, we can express this as
b
Z
H(x − c) dx = min(b − a, max(b − c, 0)).
a
Now we find an expression for the solution.

Z x+ct
1 1
u(x, t) = (u(x + ct, 0) + u(x − ct, 0)) + u (τ, 0) dτ
t
2 2c x−ct
1 Z x+ct
u(x, t) = (−H(τ + 1) + 2H(τ) − H(τ − 1)) dτ
2c x−ct

u(x, t) = − min(2ct, max(x + ct + 1, 0)) + 2 min(2ct, max(x + ct, 0))−
− min(2ct, max(x + ct − 1, 0))

Exercise 116.
(a) Consider the solution of the wave equation for u(x, t):

2
u = c u xx
tt
on the infinite interval −∞ < x < ∞ with initial displacement of the form
(
h(x) for x > 0,
u(x, 0) =
−h(−x) for x < 0,

and with initial velocity
u (x, 0) = 0.
t
Show that the solution of the wave equation satisfying these initial conditions
also solves the following semi-infinite problem: Find u(x, t) satisfying the wave
2
equation u = c u xx in 0 < x < ∞, t > 0, with initial conditions u(x, 0) =
tt
h(x), u (x, 0) = 0, and with the fixed end condition u(0, t) = 0. Here h(x) is
t
any given function with h(0) = 0.

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