1.6. The law of reflection 29 Differentiating (1.6.1) with respect to t shows that ut(t, 0) = 0, so at t = 0 (ut + ux) = −(ut ux). The incoming wave at the boundary has amplitude equal to −1 times the amplitude of the outgoing wave. We next analyze the mixed initial boundary value problem for a function u(t, x) defined in x1 0, (1.6.2) u = 0, u x1=0 = 0, u(0,x) = f , ut(0,x) = g . If the data are supported in a compact subset of R−, d then for small time the support of the solution does not meet the boundary. When waves hit the boundary, they are reflected. We analyze this reflection process. Uniqueness of solutions and finite speed of propagation for (1.6.2) are both consequences of a local energy identity. A function is a solution if and only if the real and imaginary parts are solutions. Thus it suffices to treat the real case for which ut u = ∂te j≥1 ∂j(ut∂ju) , e := u2 t + |∇xu|2 2 . Denote by Γ a backward light cone Γ := (t, x) : |x x|2 t t and by ˜ the part in {x1 0}, ˜ := Γ x1 0 . For any 0 s t, the section at time s is denoted ˜ s) := ˜ t = s . Both uniqueness and finite speed are consequences of the following estimate. Proposition 1.6.1. If u is a smooth solution of (1.6.2), then for 0 t t, φ(t) := ˜ t) e(t, x) dx is a nonincreasing function of t. Proof. Translating the time, if necessary, it suffices to show that for s 0, φ(s) φ(0). In the identity 0 = ˜∩{0≤t≤s} ut u dt dx , integrate by parts to find integrals over four distinct parts of the bound- ary. The tops and bottoms contribute φ(s) and −φ(0), respectively. The
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