62 2. FOUR IMPORTANT LINEAR PDE

Proof. 1. Fix some point in UT . Upon shifting the coordinates, we may

as well assume the point is (0, 0). Suppose ﬁrst that the cylinder C(1) :=

C(0, 0; 1) lies in UT . Let C

(

1

2

)

:= C

(

0, 0;

1

2

)

. Then, as in the proof of

Theorem 8,

u(x, t) =

C(1)

K(x, t, y, s)u(y, s) dyds ((x, t) ∈ C(

1

2

))

for some smooth function K. Consequently

(39)

|DxDtu(x,

k l

t)| ≤

C(1)

|DtDxK(x,

l k

t, y, s)||u(y, s)| dyds

≤ Ckl u

L1(C(1))

for some constant Ckl.

2. Now suppose the cylinder C(r) := C(0, 0; r) lies in UT . Let C(r/2) =

C(0, 0; r/2). We rescale by deﬁning

v(x, t) := u(rx,

r2t).

Then vt − Δv = 0 in the cylinder C(1). According to (39),

|DxDtv(x,

k l

t)| ≤ Ckl v

L1(C(1))

((x, t) ∈ C(

1

2

)).

But DxDtv(x,

k l

t) =

r2l+kDxDtu(rx, k l r2t)

and v

L1(C(1))

=

1

rn+2

u L1(C(r)).

Therefore

max

C(r/2)

|DxDtu|

k l

≤

Ckl

r2l+k+n+2

u L1(C(r)).

If u solves the heat equation within UT , then for each time 0 t ≤ T ,

the mapping x → u(x, t) is analytic. (See Mikhailov [M].) However the

mapping t → u(x, t) is not in general analytic.

2.3.4. Energy methods.

a. Uniqueness. We investigate again the initial/boundary-value problem

(40)

ut − Δu = f in UT

u = g on ΓT .

We earlier invoked the maximum principle to show uniqueness and now—

by analogy with §2.2.5—provide an alternative argument based upon inte-

gration by parts. We assume as usual that U ⊂

Rn

is open and bounded

and that ∂U is

C1.

The terminal time T 0 is given.