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).