50 2. FOUR IMPORTANT LINEAR PDE THEOREM 2 (Solution of nonhomogeneous problem). Define u by (13). Then (i) u ∈ C1 2(Rn × (0, ∞)), (ii) ut(x, t) − Δu(x, t) = f(x, t) (x ∈ Rn, t 0), and (iii) lim (x,t)→(x0,0) x∈Rn, t0 u(x, t) = 0 for each point x0 ∈ Rn. Proof. 1. Since Φ has a singularity at (0, 0), we cannot directly justify differentiating under the integral sign. We instead proceed somewhat as in the proof of Theorem 1 in §2.2.1. First we change variables, to write u(x, t) = t 0 Rn Φ(y, s)f(x − y, t − s) dyds. As f ∈ C1 2(Rn × [0, ∞)) has compact support and Φ = Φ(y, s) is smooth near s = t 0, we compute ut(x, t) = t 0 Rn Φ(y, s)ft(x − y, t − s) dyds + Rn Φ(y, t)f(x − y, 0) dy and uxixj (x, t) = t 0 Rn Φ(y, s)fxixj (x − y, t − s) dyds (i, j = 1,...,n). Thus ut,Dxu, 2 and likewise u, Dxu, belong to C(Rn × (0, ∞)). 2. We then calculate (14) ut(x, t) − Δu(x, t) = t 0 Rn Φ(y, s)[( ∂ ∂t − Δx)f(x − y, t − s)] dyds + Rn Φ(y, t)f(x − y, 0) dy = t ε Rn Φ(y, s)[(− ∂ ∂s − Δy)f(x − y, t − s)] dyds + ε 0 Rn Φ(y, s)[(− ∂ ∂s − Δy)f(x − y, t − s)] dyds + Rn Φ(y, t)f(x − y, 0) dy. =: Iε + Jε + K.
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