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).

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