68 2. FOUR IMPORTANT LINEAR PDE We lastly invoke the initial conditions in (3) to compute a and b. The first initial condition in (3) gives b(x) = g(x) (x ∈ R), whereas the second initial condition and (5) imply a(x) = v(x, 0) = ut(x, 0) − ux(x, 0) = h(x) − g (x) (x ∈ R). Our substituting into (7) now yields u(x, t) = 1 2 x+t x−t h(y) − g (y) dy + g(x + t). Hence (8) u(x, t) = 1 2 [g(x + t) + g(x − t)] + 1 2 x+t x−t h(y) dy (x ∈ R, t ≥ 0). This is d’Alembert’s formula. We have derived formula (8) assuming u is a (sufficiently smooth) solu- tion of (3). We need to check that this really is a solution. THEOREM 1 (Solution of wave equation, n = 1). Assume g ∈ C2(R), h ∈ C1(R), and define u by d’Alembert’s formula (8). Then (i) u ∈ C2(R × [0, ∞)), (ii) utt − uxx = 0 in R × (0, ∞), and (iii) lim (x,t)→(x0,0) t0 u(x, t) = g(x0), lim (x,t)→(x0,0) t0 ut(x, t) = h(x0) for each point x0 ∈ R. The proof is a straightforward calculation. Remarks. (i) In view of (8), our solution u has the form u(x, t) = F (x + t) + G(x − t) for appropriate functions F and G. Conversely any function of this form solves utt −uxx = 0. Hence the general solution of the one-dimensional wave equation is a sum of the general solution of ut − ux = 0 and the general solution of ut + ux = 0. This is a consequence of the factorization (4). See Problem 19.
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