2.3. HEAT EQUATION 51 Now (15) |Jε| ≤ ( ft L∞ + D2f L∞ ) ε 0 Rn Φ(y, s) dyds ≤ εC, by the lemma. Integrating by parts, we also find (16) Iε = t ε Rn ( ∂ ∂s − Δy)Φ(y, s) f(x − y, t − s) dyds + Rn Φ(y, ε)f(x − y, t − ε) dy − Rn Φ(y, t)f(x − y, 0) dy = Rn Φ(y, ε)f(x − y, t − ε) dy − K, since Φ solves the heat equation. Combining (14)–(16), we ascertain ut(x, t) − Δu(x, t) = lim ε→0 Rn Φ(y, ε)f(x − y, t − ε) dy = f(x, t) (x ∈ Rn, t 0), the limit as ε → 0 being computed as in the proof of Theorem 1. Finally note u(·,t) L∞ ≤ t f L∞ → 0. Solution of nonhomogeneous problem with general initial data. We can of course combine Theorems 1 and 2 to discover that (17) u(x, t) = Rn Φ(x − y, t)g(y) dy + t 0 Rn Φ(x − y, t − s)f(y, s) dyds is, under the hypotheses on g and f as above, a solution of (18) ut − Δu = f in Rn × (0, ∞) u = g on Rn × {t = 0}. 2.3.2. Mean-value formula. First we recall some useful notation from §A.2. Assume U ⊂ Rn is open and bounded, and fix a time T 0.

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