Lecture 14: A gradient estimate for the heat equation on a
ball.
1
Adapting the proof of the gradient estimate to Rn
Last time we proved a gradient estimate for solutions of the heat equation on a torus. In
this lecture we will adapt the argument to prove the same theorem on Rn .
Theorem 1.1 If u : Rn × [0, ∞) → R is positive and satisfies the heat equation then
|∇u|2 ut
n
t
−
≤ .
(1)
2
u
u
2
We will sketch the proof. We work on an interval [0, T ], andnote that the result we
2
want follows immediately from this . Define f = log u and F = t |∇u|
− uut . Let φ be a
u2
cutoff function on the ball B1 (0) with 0 < φ < 1 on the interior and φ = 0 on the boundary.
We can stretch this to get a cutoff function φr on Br (0) by taking φr (x) = φ(x/r) . If
F is non positive the result is trivially true, so we can assume that φr F has an interior
maximum without loss of generality. At this maximum
d(φr F )
≥ 0, and φr ∇F = −F ∇φr
dt
We’ll use these to get a bound on F . Calculate
4(φr F ) ≤ 0,
0 ≥
d
4−
dt
(2)
(φr F )
≥ φr 4F + 2∇F · ∇φr + F 4φr − φr
(3)
dF
dt
(4)
We need to estimate some of these. The calculations are very similar to last time. We
start with
1
4F
= t4
|∇u|2 ut
−
u2
u
(5)
= t4(|∇f |2 − ft )
2 2
∂ f
= 2t
+ 2t∇(4f ) · ∇f − t4ft
∂xi ∂xj
by Bochner. Calculate 4f =
∂ ∂u/∂xi
∂xi
u
=
4u
u
−
|∇u|2
u2
(6)
(7)
= − Ft to get
2
∂2f
4F = 2t
− 2∇F · ∇f − t4ft .
(8)
∂xi ∂xj
P
P
Recall the inequality ( Ai i)2 ≤ n (A2ii ) for all matrices A from last time, and apply it
to the hessian of f to give
4F
=
≥
2t
(4f )2 − 2∇F · ∇f − t4ft
n
2F 2
− 2∇F · ∇f − 4ft .
nt
(9)
(10)
We also need an estimate on Ft . We have
Ft = |∇f |2 − ft + t(2∇f · ∇ft ) − tftt ,
(11)
Ft = |∇f |2 − ft + t(2∇f · ∇ft ) − t(4f + |∇f |2 )t
F
− t4ft .
=
t
(12)
and 4f + |∇f |2 = ft , so
Putting 4, 10 and 13 together we get
2
2F
F
− 2∇F · ∇f −
0 ≥ φr
+ 2∇F · ∇φr + F 4φr .
nt
t
(13)
(14)
Recall that φr ∇F = −F ∇φr , so
F
2F
2F 2 2F
+
∇φr · ∇f −
−
|∇φr | + F 4φr
nt
φr
t
φr
2F
2
1
|∇φr |2 4φr
≥ F φr
+ ∇φr · ∇f − − 2
+
.
nt
φr
t
φ2r
φr
0 ≥ φr
2
(15)
(16)
2
2
∂f
1 ∂φr
r ∂f
−
for all > 0. to show
≥
−
Now use an absorbing inequality ∂φ
∂xi ∂xi
∂xi
∂xi
that
1
∇φr · ∇f ≥ − |∇φr |2 − |∇f |2
(17)
for all > 0. Consequently
2F
1
2 1
|∇φr |2 4φr
2
2
.
(18)
0 ≥ F φr
−
|∇φr | + |∇f | − − 2
+
nt
φr t
φ2r
φr
Let r → ∞ so that |∇φr | and 4φr tend to zero and φr → 1, and get
2F
1
2
0≥F
.
− 2|∇f | −
nt
t
(19)
Finally we let → 0 and recover
F
0≥
t
2F
−1 .
n
From this we getF ≤ n/2 as required.
3
(20)