58 2. FOUR IMPORTANT LINEAR PDE Proof. 1. First assume (25) 4aT 1, in which case (26) 4a(T + ε) 1 for some ε 0. Fix y Rn, μ 0, and define v(x, t) := u(x, t) μ (T + ε t)n/2 e |x−y|2 4(T +ε−t) (x Rn, t 0). A direct calculation (cf. §2.3.1) shows vt Δv = 0 in Rn × (0,T ]. Fix r 0 and set U := B0(y, r), UT = B0(y, r) × (0,T ]. Then according to Theorem 4, (27) max ¯ U T v = max ΓT v. 2. Now if x Rn, (28) v(x, 0) = u(x, 0) μ (T + ε)n/2 e |x−y| 2 4(T +ε) u(x, 0) = g(x) and if |x y| = r, 0 t T , then v(x, t) = u(x, t) μ (T + ε t)n/2 e r2 4(T +ε−t) Aea|x|2 μ (T + ε t)n/2 e r2 4(T +ε−t) by (24) Aea(|y|+r)2 μ (T + ε)n/2 e r 2 4(T +ε) . Now according to (26), 1 4(T +ε) = a+γ for some γ 0. Thus we may continue the calculation above to find (29) v(x, t) Aea(|y|+r)2 μ(4(a + γ))n/2e(a+γ)r2 sup Rn g, for r selected sufficiently large. Thus (27)–(29) imply v(y, t) sup Rn g for all y Rn, 0 t T , provided (25) is valid. Let μ 0. 3. In the general case that (25) fails, we repeatedly apply the result above on the time intervals [0,T1], [T1, 2T1, ], etc., for T1 = 1 8a .
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