2.4. WAVE EQUATION 81
THEOREM 4 (Solution of nonhomogeneous wave equation). Assume that
n ≥ 2 and f ∈
C[n/2]+1(Rn
× [0, ∞)). Define u by (41). Then
(i) u ∈
C2(Rn
× [0, ∞)),
(ii) utt − Δu = f in
Rn
× (0, ∞),
and
(iii) lim
(x,t)→(x0,0)
x∈Rn,
t0
u(x, t) = 0, lim
(x,t)→(x0,0)
x∈Rn,
t0
ut(x, t) = 0 for each point
x0
∈
Rn.
Proof. 1. If n is odd,
n
2
+ 1 =
n+1
2
. According to Theorem 2, we have
u(·, ·; s) ∈
C2(Rn
× [s, ∞)) for each s ≥ 0, and so u ∈
C2(Rn
× [0, ∞)). If n
is even,
n
2
+1 =
n+2
2
. Hence u ∈ C2(Rn ×[0, ∞)), according to Theorem 3.
2. We then compute
ut(x, t) = u(x, t; t) +
t
0
ut(x, t; s) ds =
t
0
ut(x, t; s) ds,
utt(x, t) = ut(x, t; t) +
t
0
utt(x, t; s) ds = f(x, t) +
t
0
utt(x, t; s) ds.
Furthermore
Δu(x, t) =
t
0
Δu(x, t; s) ds =
t
0
utt(x, t; s) ds.
Thus
utt(x, t) − Δu(x, t) = f(x, t) (x ∈
Rn,t
0),
and clearly u(x, 0) = ut(x, 0) = 0 for x ∈
Rn.
The solution of the general nonhomogeneous problem is consequently
the sum of the solution of (11) (given by formulas (8), (31) or (38)) and the
solution of (42) (given by (41)).
Examples. (i) Let us work out explicitly how to solve (42) for n = 1. In
this case d’Alembert’s formula (8) gives
u(x, t; s) =
1
2
x+t−s
x−t+s
f(y, s) dy, u(x, t) =
1
2
t
0
x+t−s
x−t+s
f(y, s) dyds.
That is,
(43) u(x, t) =
1
2
t
0
x+s
x−s
f(y, t − s) dyds (x ∈ R, t ≥ 0).
