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

.