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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