1. THE LOCAL THEORY OF THE CAUCHY PROBLEM 11 Then, (v, ∂tv) ∈ C(I ˙ 1 × L2), (v, ∂tv)|t=0 = (0, 0) and ∂t 2 v − Δv = u5 in D (R3 × I). Let h ∈ C0 ∞ (R3 × I+). Let, for t ∈ R, H(t) = − ∞ t sin ( (t − s) √ −Δ ) √ −Δ h(s)ds, so that H ∈ C∞(R4) (with compact support in x), H(t) = 0 for large t and (∂2 t − Δ)(H) = h. Let ϕ ∈ C∞(R) be such that ϕ(σ) = 1 if σ ≥ 1 and ϕ(σ) = 0 if σ ≤ 1 2 . If a ∈ (0, 1], we let Ha(x, t) = ϕ t a H(x, t). Note that Ha ∈ C0 ∞ (R4). By the distributional equation for v, we have R4 v(x, t) ( ∂t 2 − Δ ) Ha(x, t)dxdt = R4 u5(x, t)Ha(x, t)dxdt. By dominated convergence, Fubini and the self-adjointness of sin ((t−s) √ −Δ ) √ −Δ , we have lim a→0 R4 u5(x, t)Ha(x, t)dxdt − R4 u5(x, t)H(x, t)dxdt = − ∞ 0 s 0 R3 u5(x, t) sin ( (t − s) √ −Δ ) √ −Δ h(x, s)dxdtds = − R3 ∞ 0 s 0 sin ( (t − s) √ −Δ ) √ −Δ u5(x, t)dt h(x, s)dsdx. Hence, lim a→0 R4 u5(x, t)Ha(x, t)dxdt R3 ∞ 0 t 0 sin ( (t − s) √ −Δ ) √ −Δ u5(x, s)ds h(x, t)dtdx. We next consider R4 v(x, t)(∂2 t − Δ) (Ha(x, t)) dxdt = R4 v(x, t) 1 a2 ϕ t a H(x, t) + 2 a ϕ t a ∂tH(x, t) + ϕ t a h(x, t) dxdt. We will show that lim a→0 R4 v(x, t) 1 a2 ϕ t a H(x, t)dxdt = 0 lim a→0 R4 v(x, t) 1 a ϕ t a ∂tH(x, t)dxdt = 0.
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