2.3. HEAT EQUATION 47 A different derivation of the fundamental solution of the heat equation appears in §4.3.1. b. Initial-value problem. We now employ Φ to fashion a solution to the initial-value (or Cauchy) problem (8) ut − Δu = 0 in Rn × (0, ∞) u = g on Rn × {t = 0}. Let us note that the function (x, t) → Φ(x, t) solves the heat equation away from the singularity at (0, 0), and thus so does (x, t) → Φ(x − y, t) for each fixed y ∈ Rn. Consequently the convolution (9) u(x, t) = Rn Φ(x − y, t)g(y) dy = 1 (4πt)n/2 Rn e− |x−y| 2 4t g(y) dy (x ∈ Rn, t 0) should also be a solution. THEOREM 1 (Solution of initial-value problem). Assume g ∈ C(Rn) ∩ L∞(Rn), and define u by (9). Then (i) u ∈ C∞(Rn × (0, ∞)), (ii) ut(x, t) − Δu(x, t) = 0 (x ∈ Rn, t 0), and (iii) lim (x,t)→(x0,0) x∈Rn, t0 u(x, t) = g(x0) for each point x0 ∈ Rn. Proof. 1. Since the function 1 tn/2 e− |x| 2 4t is infinitely differentiable, with uni- formly bounded derivatives of all orders, on Rn × [δ, ∞) for each δ 0, we see that u ∈ C∞(Rn × (0, ∞)). Furthermore (10) ut(x, t) − Δu(x, t) = Rn [(Φt − ΔxΦ)(x − y, t)]g(y) dy = 0 (x ∈ Rn, t 0), since Φ itself solves the heat equation. 2. Fix x0 ∈ Rn, ε 0. Choose δ 0 such that (11) |g(y) − g(x0)| ε if |y − x0| δ, y ∈ Rn.
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