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.
