2.4. WAVE EQUATION 65 Hence (46) ¨(t)e(t) e ≥ (˙(t))2 e (0 ≤ t ≤ T ). 2. Now if e(t) = 0 for all 0 ≤ t ≤ T , we are done. Otherwise there exists an interval [t1,t2] ⊂ [0,T ], with (47) e(t) 0 for t1 ≤ t t2, e(t2) = 0. 3. Write (48) f(t) := log e(t) (t1 ≤ t t2). Then ¨(t) f = ¨(t) e e(t) − ˙(t)2 e e(t)2 ≥ 0 by (46), and so f is convex on the interval (t1,t2). Consequently if 0 τ 1, t1 t t2, we have f((1 − τ)t1 + τt) ≤ (1 − τ)f(t1) + τf(t). Recalling (48), we deduce e((1 − τ)t1 + τt) ≤ e(t1)1−τ e(t)τ , and so 0 ≤ e((1 − τ)t1 + τt2) ≤ e(t1)1−τ e(t2)τ (0 τ 1). But in view of (47) this inequality implies e(t) = 0 for all times t1 ≤ t ≤ t2, a contradiction. 2.4. WAVE EQUATION In this section we investigate the wave equation (1) utt − Δu = 0 and the nonhomogeneous wave equation (2) utt − Δu = f, subject to appropriate initial and boundary conditions. Here t 0 and x ∈ U, where U ⊂ Rn is open. The unknown is u : ¯ U × [0, ∞) → R, u = u(x, t), and the Laplacian Δ is taken with respect to the spatial variables

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