48 2. FOUR IMPORTANT LINEAR PDE Then if |x x0| δ 2 , we have, according to the lemma, |u(x, t) g(x0)| = Rn Φ(x y, t)[g(y) g(x0)] dy B(x0,δ) Φ(x y, t)|g(y) g(x0)| dy + Rn−B(x0,δ) Φ(x y, t)|g(y) g(x0)| dy =: I + J. Now I ε Rn Φ(x y, t) dy = ε, owing to (11) and the lemma. Furthermore, if |x x0| δ 2 and |y x0| δ, then |y x0| |y x| + δ 2 |y x| + 1 2 |y x0|. Thus |y x| 1 2 |y x0|. Consequently J 2 g L∞ Rn−B(x0,δ) Φ(x y, t) dy C tn/2 Rn−B(x0,δ) e− |x−y|2 4t dy C tn/2 Rn−B(x0,δ) e− |y−x0|2 16t dy = C Rn−B(0,δ/ t) e− |z|2 16 dz 0 as t 0+. Hence if |x x0| δ 2 and t 0 is small enough, |u(x, t) g(x0)| 2ε. Interpretation of fundamental solution. In view of Theorem 1 we sometimes write Φt ΔΦ = 0 in Rn × (0, ∞) Φ = δ0 on Rn × {t = 0}, δ0 denoting the Dirac measure on Rn giving unit mass to the point 0. Infinite propagation speed. Notice that if g is bounded, continuous, g 0, g 0, then u(x, t) = 1 (4πt)n/2 Rn e− |x−y|2 4t g(y) dy
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