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Solved Problems

(b) Use a similar idea to explain how you could use the general solution of the
wave equation to solve the finite interval problem (0 < x < l) in which

u(0, t) = u(l, t) = 0 for all t, with u(x, 0) = h(x) and u (x, 0) = 0. Take
t
h(0) = h(l) = 0.

Solution:

(a) The solution on the interval (−∞ . . . ∞) is

1
u(x, t) = (h(x + ct) + h(x − ct)).
2

Now we solve the problem on (0 . . . ∞). We define the odd extension of h(x).

(
h(x) for x > 0,
ˆ = sign(x)h(|x|)
h(x) =
−h(−x) for x < 0,

Note that

d
ˆ 0 − (−h(−x)) = h (0 ) = h (0 ).
ˆ
0
+
0
+
h (0 ) =
dx x→0 +
ˆ
2
Thus h(x) is piecewise C . Clearly
1
ˆ
ˆ
u(x, t) = (h(x + ct) + h(x − ct))
2
satisfies the differential equation on (0 . . . ∞). We verify that it satisfies the
initial condition and boundary condition.
1
ˆ
ˆ
u(x, 0) = (h(x) + h(x)) = h(x)
2
1 1
ˆ
ˆ
u(0, t) = (h(ct) + h(−ct)) = (h(ct) − h(ct)) = 0
2 2
(b) First we define the odd extension of h(x) on the interval (−l . . . l).

ˆ
h(x) = sign(x)h(|x|), x ∈ (−l . . . l)

Then we form the odd periodic extension of h(x) defined on (−∞ . . . ∞).

x + l x + l
ˆ
h(x) = sign x − 2l h x − 2l , x ∈ (−∞ . . . ∞)
2l 2l
ˆ
ˆ
2
We note that h(x) is piecewise C . Also note that h(x) is odd about the
ˆ
ˆ
points x = nl, n ∈ Z. That is, h(nl − x) = −h(nl + x). Clearly
1
ˆ
ˆ
u(x, t) = (h(x + ct) + h(x − ct))
2
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