University of Connecticut
Mathematics
A gradient bound and a Liouville theorem for nonlinear
Poisson equations
Mingfeng Zhao
1
Email: mingfeng.zhao@uconn.edu
May 22, 2011
1
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT 06269-3009
Contents
1.
Main ideas
1
2.
A useful lemma
2
3.
A gradient estimates for Poisson’s equation
3
4.
Important auxiliary functions
6
5.
Best Gradient Estimate
10
6.
Liouville Type Theorem
20
References
21
In this note, we will give a detailed explanation of Modica’s paper[3].
1. Main ideas
(1) ∆x u(x) = F 0 (x) is translation invariant for the statement of the Theorem;
(2) Since u is bounded in Rn , then inf n |Du(x)| = 0;
x∈R
(3) Use the standard gradient estimate for the Poisson equation, we know the function
P (x) = |∇u(x)|2 − 2F (u(x))
is bounded in Rn ;
(4) Construct a special function η,ρ0 satisfying some special properties;
(5) Let v(x) = η,ρ0 (x)P (x) for |x| ≥ ρ0 , we have
v(x) ≤ max {, max P (x)},
|x|=ρ0
for all |x| ≥ ρ0 .
(6) Consider |∇u(x)|2 for the extremum point of v(x), both bolds for the previous step 5;
(7) Take & 0, we obtain
P (x) ≤ max
0, max P (x) ,
|x|=ρ0
for all |x| ≥ ρ0 .
And take ρ0 & 0, we get
P (x) ≤ max {0, P (0)} ≤ |Du(0)|2 < δ.
(8) By the standard Taylor’s expansion when considering the minimum point of u, and use Gronwall’s inequality to
show the u(x) ≡ constant in Rn .
–1–
2. A useful lemma
Theorem 2.1. For any u ∈ C 1 (Rn ) ∩ L∞ (Rn ), then
inf
x∈Rn
|Du(x)| = 0.
Proof. If the above statement is not true, then there exists some u ∈ C 1 (Rn )
inf
x∈Rn
Let v(x) =
T
L∞ (Rn ) such that
|Du(x)| = δ0 > 0.
u(x)
, then v ∈ C 1 (Rn ) ∩ L∞ (Rn ), and
δ0
for all x ∈ Rn .
|Dv(x)| ≥ 1,
Consider the ODE:
(1)

 ẋ(t) = Dv(x(t),
 x(0) = 0.
t>0
Since Dv ∈ C(Rn ), then by the ODE’s Theory, we know the ODE (1) has a solution x(t).
Now consider the function:
f (t) = v(x(t)),
t ≥ 0.
Since v ∈ C 1 (Rn ), and x(t) ∈ C 1 ((0, ∞)), then we have
f 0 (t)
= Dv(x(t)) · ẋ(t)
= Dv(x(t) · Dv(x(t))
= |Dv(x(t))|2
≥ 1,
for all t > 0.
Since f 0 is continuous, then either f 0 (t) ≥ 1 for all t > 0; or f 0 (t) ≤ −1 for all t > 0. Without loss of generality, we
assume f 0 (t) ≥ 1 for all t > 0. Then by integrating both sides, we have
f (t) − f (0) ≥ t.
So v(x(t)) ≥ f (0) − t = v(x(0)) − t = v(0) − t for all t > 0. When take t → ∞, we know
v(x(t)) → ∞,
as t → ∞.
In particular, we know v(x) is unbounded in Rn . Since v(x) =
–2–
u(x)
δ0 ,
then u(x) is unbounded in Rn ., contradiction. Remark 2.1. Let u ∈ C 1,α (Rn ) for some 0 < α ≤ 1, then infn |Du(x)| = 0. In fact, let M = sup u(x), then
x∈R
x∈Rn
n
k
k
n
there exists some sequence {xk }∞
k=1 ⊂ R such that u(x ) → M , as k → ∞. Let vk (x) = u(x + x ) for all x ∈ R ,
then vk ∈ C 1,α (Rn ), and kvk kC 1,α (Rn ) = kukC 1,α (Rn ) for all k ≥ 1. By the Ascoli-Arzela’s Theorem, there exists some
1
v ∈ Cloc
(Rn ), vk (x) ≤ v(0) for all x ∈ Rn and all k ≥ 1, and some subsequence of {vk }∞
k=1 , without loss of generality,
1
n
n
we assume {vk }∞
k=1 itself such that vk → v in Cloc (R ), as k → ∞. In particular, we get v(x) ≤ v(0) for all x ∈ R ,
1
which implies that Dv(0) = 0. Since vk → v in Cloc
(Rn ), as k → ∞, then Dvk (0) = Du(xk ) → 0 = Dv(0), as k → ∞.
Therefore, we can conclude that infn |Du(x)| = 0.
x∈R
3. A gradient estimates for Poisson’s equation
Theorem 3.1 (Theorem 3.9, in Page 41, in Gilbarg and Trudinger [1]). For any u ∈ C 2 (Rn ), f ∈ C(Rn ) and
∆x u(x) = f (x),
for all x ∈ Rn .
Then for any x0 ∈ Rn , any d > 0, we have
|Du(x0 )| ≤
n
d
sup
|u(x)| +
sup |f (x)|.
d x∈∂Qd (x0 )
2 x∈Qd (x0 )
Where Qd (x0 ) denote the cube with center x0 , and length 2d, that is
n
Qd (x0 ) = x ∈ R : max |xi − (x0 )i | < d .
1≤i≤n
Proof. Claim I: For any u ∈ C 2 (Rn ), f ∈ C(Rn ) and
∆x u(x) = f (x),
for all x ∈ Rn .
Then for d > 0, we have
|Du(0)| ≤
n
d
sup
|u(x)| +
sup |f (x)|.
d x∈∂Qd (0)
2 x∈Qd (0)
In fact, for any d > 0, 1 ≤ k ≤ n, we consider the k-th upper half cub
n
Qk+
(0)
=
x
∈
R
:
max
|x
|
<
d,
0
<
x
<
d
.
i
k
d
1≤i≤n
Then for any x ∈ Qk+
d (0), we know |xi | < d for all i = 1, · · · , k − 1, k + 1, · · · , n, and 0 < xk < d. Then −d < −xk < 0,
so (x1 , · · · , −xk , · · · , xn ) ∈ Qd (0). And
n
∂Qk+
d (0) = {x ∈ R : |xi | = k for some 1 ≤ i ≤ n, i 6= k, 0 ≤ xk ≤ d}
[
{x ∈ Rn : |xi | ≤ d, xk = 0.}
Consider the function
v(x) =
1
[u(x1 , · · · , xk , · · · , xn ) − u(x1 , · · · , −xk , · · · , xn )],
2
for all x ∈ Qk+
d (0).
Let M = supx∈∂Qd (0) |u(x)|, and N = supx∈Qd (0) |f (x)|. Then we know
• v(x1 , · · · , xk−1 , 0, xk+1 , · · · , xn ) = 0 for all (x1 , · · · , xk−1 , 0, xk+1 , · · · , xn ) ∈ ∂Qk+
d (0);
–3–
• For all x ∈ ∂Qk+
d (0), we have
|v(x)| =
≤
≤
=
1
[u(x1 , · · · , xk , · · · , xn ) − u(x1 , · · · , −xk , · · · , xn )]
2
1
[|u(x1 , · · · , xk , · · · , xn )| + |u(x1 , · · · , −xk , · · · , xn )|]
2
1
[2M ]
2
M
So we have
|v(x)| ≤ M.
sup
x∈∂Qk+
d (0)
• For any x ∈ Qk+
d (0), we have
∆x v(x)
1
[∆x u(x1 , · · · , xk , · · · , xn ) − ∆x u(x1 , · · · , −xk , · · · , xn )]
2
1
[f (x1 , · · · , xk , · · · , xn ) − f (x1 , · · · , −xk , · · · , xn )].
2
=
=
So for any x ∈ Qk+
d (0), we get
|∆x v(x)|
≤
≤
=
Consider another function

M
w(x) = 2
d
1
[2N ]
2
1
[|f (x1 , · · · , xk , · · · , xn )| + |f (x1 , · · · , −xk , · · · , xn )|]
2
N.

X
|xi |2 + xk (nd − (n − 1)xk ) +
1≤i≤n,i6=k
N
xk (d − xk )
2
for all x ∈ Qk+
d (0).
So we have
• For all (x1 , · · · , xk−1 , 0, xk+1 , · · · , xn ) ∈ ∂Qk+
d (0), we have
w(x1 , · · · , xk−1 , 0, xk+1 , · · · , xn )
=

M
d2

X
|xi |2 
1≤i≤n,i6=k
≥ 0.
• For all x such that |xj | = d for some 1 ≤ j ≤ n, j 6= k, and 0 ≤ xk ≤ d, then we know
P
1≤i≤n,i6=k
|xi |2 ≥
|xj |2 = d2 . Since 0 ≤ xk ≤ d, then we know xk (nd − (n − 1)xk ) ≥ 0, and xk (d − xk ) ≥ 0. Hence we have
w(x) ≥
M 2
·d
d2
= M.
–4–
• For any 1 ≤ i ≤ n, i 6= k, x ∈ Qk+
d (0), we have
2M
∂ 2 w(x)
= 2
2
∂xi
d
For any x ∈ Qk+
d (0), we have
∂ 2 w(x)
∂x2k
=
M
N
· (−2(n − 1)) −
·2
d2
2
= −
2(n − 1)M
− N.
d2
Therefore, for any x ∈ Qk+
d (0), we have
∆x w(x)
=
(n − 1) ·
2(n − 1)M
2M
−
−N
d2
d2
= −N.
By the properties of v(x) and w(x), we know

 ∆x (w ± v)(x) ≤ 0, in Qk+ (0)
d
 w(x) ± v(x) ≥ 0,
on ∂Qk+
d (0).
Then by the maximum principle, we know for all x ∈ Qk+
d (0), we have
w(x) ± v(x) ≥ 0.
That is, |v(x)| ≤ w(x) for all x ∈ Qk+
d (0).
Now take x1 = · · · = xk−1 = xk+1 = · · · = xn = 0, and xk & 0, we have
∂u(0)
∂xk
=
=
≤
=
=
=
lim
xk &0
u(0, · · · , xk , · · · , 0) − u(0, · · · , −xk , · · · , 0)
2xk
lim
v(0, · · · , xk , · · · , 0)
xk
lim
w(0, · · · , xk , · · · , 0)
xk
lim
M
N
[xk (nd − (n − 1)xk )] +
xk (d − xk )
d2 xk
2xk
lim
M
N
(nd − (n − 1)xk ) + (d − xk )
2
d
2
xk &0
xk &0
xk &0
xk &0
d
n
M + N.
d
2
By the definition of M and N , we get
∂u(0)
n
d
≤
sup |u(x)| +
sup |f (x)|.
∂xk
d x∈∂Qd (0
2 x∈Qd (0
Since 1 ≤ k ≤ n is arbitrary, then we get
|Du(0)| ≤
n
d
sup
|u(x)| +
sup |f (x)|.
d x∈∂Qd (0)
2 x∈Qd (0)
–5–
Claim II: For any u ∈ C 2 (Rn ), f ∈ C(Rn ) and
∆x u(x) = f (x),
for all x ∈ Rn .
Then for any x0 ∈ Rn , any d > 0, we have
|Du(x0 )| ≤
n
d
sup
|u(x)| +
sup |f (x)|.
d x∈∂Qd (x0 )
2 x∈Qd (x0 )
In fact, for any x0 , we consider
v(x) = u(x0 + x),
for all x ∈ Rn .
Then ∆x v(x) = f (x + x0 ). And by the result of the Claim I, we get for and d > 0, we have
|Dv(0)|
≤
n
d
sup
|v(x)| +
sup |f (x + x0 )|
d x∈∂Qd (0)
2 x∈Qd (0)
=
d
n
sup
|u(x)| +
sup |f (x)|
d x∈∂Qd (x0 )
2 x∈Qd (x0 )
Since Dv(0) = Du(x0 ), so we get
|Du(x0 )| ≤
n
d
sup
|u(x)| +
sup |f (x)|.
d x∈∂Qd (x0 )
2 x∈Qd (x0 )
Remark 3.1. The key point of the proof is to find these two auxiliary functions v(x) and w(x).
4. Important auxiliary functions
Theorem 4.1. Let M > 0, N > 0 be any positive numbers and fix them, and integer n ≥ 1. Then for any α > 0, β > 0,
there exists some C 2 functions:
ηα,β : [β, ∞) −→ R.
with the following good properties:
i. For any α > 0, β > 0, we have
ηα,β (β) = 1.
ii. For any α > 0, β > 0, then for any t ≥ β, we have
ηα,β (t) > 0,
and
0
ηα,β
(t) < 0.
In particular, we know for all t ≥ β, we have ηα,β (t) ≤ 1.
iii For any α > 0, β > 0, we have
lim ηα,β (t) = 0.
t%∞
iv For any α > 0, β > 0, then for all t ≥ β, we have
lim ηα,β (t) = 1.
α&0
–6–
v For any α > 0, β > 0, then for all t ≥ β, we have
 h

i2
!2
0
0
2
η
(t)
(n − 1)ηα,β (t) 
α,β
α
ηα,β (t)
M 0

00
·
−
η (t) − ηα,β
(t) −
= .
0 (t)
ηα,β
ηα,β (t)
α α,β
t
N
Proof. For any α > 0, we know
Z
0
1
α
e− N s
ds =
s2
∞
Z
αt
t2 e− N
1
∞
Z
1
dt
t2
Let t =
1
s
αt
e− N dt
=
1
∞
N − αt
e N
α
=
−
=
N −α
e N
α
1
Since α > 0, N > 0
< ∞.
So for any α > 0, and for all 0 ≤ t ≤ 1, we can define
Z 1 −α
e Ns
gα (t) =
ds
s2
t
Z 1t
αr 1
r2 e− N 2 dr
=
r
1
Z 1t
αr
=
e− N dr
Let r =
1
s
1
1
=
=
>
N αt t
− e− N
α
1
α N −α
· e N − e− N t
α
0.
So we have
α
gα0 (t) = −
e− N t
< 0,
t2
for all 0 ≤ t ≤ 1.
α −N
That means that gα is strictly decreasing, so gα has inverse function gα−1 which is also strictly decreasing on 0, N
,
αe
and
(gα−1 )0 (t) < 0,
for all 0 ≤<
For any α > 0, β > 0, then for any ∞ ≥ t ≥ β, we define
Z
hα,β (t) =
t
β
Then we know h0α,β (t) =
Mt
e− α
tn−1
Ms
e− α
ds.
sn−1
> 0 for all t ≥ β, and
hα,β (∞)
=
∞
Ms
e− α
ds
sn−1
β
–7–
Z
N −α
e N.
α
=
=
=
<
Ms
∞
e− α
ds
β n−1
β
Z
Ms
1
e− α ds
β n−1 β
α
Ms
1
·
−
· e− α
β n−1
M
Mβ
α
e− α
n−1
Mβ
Z
≤
∞
β
∞.
So 0 < hα,β (∞) < ∞. And for any t ≥ β, we have
≤
0
=
<
gα (0)
· hα,β (t)
hα,β (∞)
α
N −N
αe
hα,β (∞)
· hα,β (t)
N −α
e N.
α
So for all t ≥ β, we can define
βα,β (t) =
gα−1
gα (0)
· hα,β (t) .
hα,β (∞)
Claim: ηα,β (t) satisfy our conditions.
i. For any α > 0, β > 0, we have
ηα,β (β)
=
gα−1
gα (0)
· hα,β (β)
hα,β (∞)
= gα−1 (0)
=
1.
ii. For any α > 0, β > 0, then for any t ≥ β, since the image of gα−1 is (0, 1], then we have
gα (0)
βα,β (t) = gα−1
· hα,β (t) > 0.
hα,β (∞)
And also we have
0
ηα,β
(t)
0
gα (0)
· hα,β (β)
hα,β (∞)
0
gα (0)
gα−1
· hα,β (β) ·
hα,β (∞)
0
gα (0)
gα−1
· hα,β (β) ·
hα,β (∞)
=
=
=
gα−1
gα (0)
· h0 (β)
hα,β (∞) α,β
< 0.
So we know ηα,β (t) is strictly decreasing, so for all t ≥ β, we have
ηα,β (t) ≤ 1.
–8–
Mt
gα (0)
e− α
· n−1
hα,β (∞) t
iii. For any α > 0, β > 0, then
lim ηα,β (t)
= ηα,β (∞)
t%∞
= gα−1
gα (0)
· hα,β (∞)
hα,β (∞)
= gα−1 (gα (0))
=
0.
iv. For any α > 0, β > 0, then for any t ≥ β, since
Z
lim hα,β (t)
α&0
=
α&0
t
Z
=
β
Ms
e− α
ds
sn−1
Ms
e− α
ds By the Monotone Convergence Theorem
α&0 sn−1
lim
β
=
t
lim
0.
Because gα−1 (0) = 1 for all α > 0, and by the continuity of gα−1 , we know
lim ηα,β (t) = 1.
α&0
v. For any α > 0, β > 0, then for any t ≥ β, since
βα,β (t) =
gα−1
gα (0)
· hα,β (t) .
hα,β (∞)
Then
gα (ηα,η (t)) =
gα (0)
· hα,β (t).
hα,β (∞)
That is, for any t ≥ β, we have
Z
α
1
ηα,β (t)
gα (0)
e− N s
ds =
·
s2
hα,β (∞)
Z
t
β
Ms
e− α
ds.
sn−1
Differentiating on both sides, we can get
−
α
Mt
e N ηα,β (t)
gα (0)
e− α
0
−
·
η
(t)
=
·
.
α,β
[ηα,β (t)]2
hα,β (∞) tn−1
0
Since ηα,β
(t) < 0, and ηα,β (t) > 0, then we can take the logarithm on both sides, then we have
0
ln(−ηα,β
(t)) −
α
N ηα,β (t)
− 2 ln ηα,β (t) = ln
gα (0)
hα,β (∞)
−
Mt
− (n − 1) ln t.
α
Differentiating on both sides, we have
0
0
00
−ηα,β
(t)
αηα,β
(t)
2ηα,β
(t)
M
n−1
+
−
=−
−
.
0
−ηα,β (t) N [ηα,β (t)]2
[ηα,β (t)]2
α
t
–9–
Both sides are multiplied by
α
=
N
ηα,β (t)
0 (t)
ηα,β
!2
[ηα,β (t)]2
0
(t) ,
ηα,β
, then for all t ≥ β, we get
 h

i2
0
0
2
η
(t)
(n − 1)ηα,β (t) 
α,β
M 0

00
·
−
η (t) − ηα,β
(t) −
.
ηα,β (t)
α α,β
t
5. Best Gradient Estimate
Theorem 5.1. Let F ∈ C 2 (Rn ), and F (t) ≥ 0 for all t ∈ R. For any u ∈ C 3 (Rn )
T
the PDEs:
∆x u(x) = F 0 (u(x)),
for all x ∈ Rn .
Then
|Du(x)|2 ≤ 2F (u(x)),
for all x ∈ Rn .
Proof. a. If u(x) ≡ C in Rn , then Du(x) = 0. But F (u(x)) ≥ 0, then we have
0 = |Du(x)|2 ≤ 2F (u(x)),
b. u(x) 6≡ Constant in Rn . Since u ∈ C 3 (Rn )
T
for all x ∈ Rn .
L∞ (Rn ), by Theorem 2.1, we know
inf
x∈Rn
|Du(x)| = 0.
So for any δ > 0, there exists some x0 ∈ Rn such that
|Du(x0 )|2 < δ.
Now let v(x) = u(x + x0 ), so ∆x v(x) = F 0 (v(x)) in Rn , and we have
|Dv(0)|2 < δ.
Define the P -function of the PDE ∆v = F 0 (v) as:
P (x) = |Dv(x)|2 − 2F (v(x)),
Since u ∈ C 3 (Rn )
T
for all x ∈ Rn .
L∞ (Rn ), so there exists some A > 0 such that
kukL∞ (Rn ) ≤ A.
So for all x ∈ Rn , we have
|F 0 (u(x))|
≤
kF 0 kL∞ ([−A,A])
= B.
– 10 –
L∞ (Rn ) which is a solution of
So by Theorem 3.1, we just take d = 1, we can get for all x ∈ Rn , we have
|Du(x)|
≤ n
|u(x)| +
sup
y∈∂Q1 (x)
≤ nA +
1
sup |F 0 (y)|
2 y∈Q1 (x)
B
.
2
So for all x ∈ Rn , we have
|P (x)|
≤ nA +
B
+B
2
= C.
Since u ∈ C 3 (Rn ), and F ∈ C 2 (R), then we know P (x) ∈ C 2 (Rn ), and for all 1 ≤ i ≤ n, all x ∈ Rn , we have
∂P (x)
∂xi
=
∂ 2 P (x)
∂xi ∂xi
=
=
n
X
∂v(x)
∂ 2 v(x)
∂v(x)
− 2F 0 (v(x)) ·
∂x
∂x
∂x
∂xi
j
i
j
j=1
"
#
2
2
n
X
∂ 2 v(x)
∂v(x)
∂ 3 v(x)
∂v(x)
∂ 2 v(x)
00
2
+
·
− 2F (v(x))
− 2F 0 (v(x))
∂xi ∂xj
∂xj ∂xi ∂xi ∂xj
∂xi
∂xi ∂xi
j=1
2
2
·
2
n 2
X
∂ v(x)
j=1
∂xi ∂xj
−2F 00 (v(x)) ·
+2
n
X
∂v(x)
j=1
∂v(x)
∂xi
2
∂xj
·
∂ 3 v(x)
∂xi ∂xi ∂xj
− 2F 0 (v(x)) ·
∂ 2 v(x)
.
∂xi ∂xi
Hence, we have
∆x P (x)
=
2
2
n X
n 2
X
∂ v(x)
i=1 j=1
00
∂xi ∂xj
−2F (v(x)) ·
+2
n X
n
X
∂v(x)
2
n X
∂v(x)
i=1
∂xj
i=1 j=1
∂xi
·
∂ 3 v(x)
∂xi ∂xi ∂xj
− 2F 0 (v(x)) ·
n
X
∂ 2 v(x)
∂xi ∂xi
i=1
2
n 2
n
n
X
X
∂ v(x)
∂v(x) X ∂ 3 v(x)
= 2
+2
·
− 2F 00 (v(x))|∇v(x)|2
∂x
∂x
∂x
∂x
∂x
∂x
i
j
j
i
i
j
i,j=1
j=1
i=1
−2F 0 (v(x)) · ∆x v(x)
2
n
n 2
X
X
∂v(x) ∂∆x v(x)
∂ v(x)
+2
·
− 2F 00 (v(x))|∇v(x)|2 − 2F 0 (v(x)) · F 0 (v(x))
= 2
∂x
∂x
∂x
∂x
i
j
j
j
j=1
i,j=1
=
2
2
n 2
n
X
X
∂ v(x)
∂v(x) ∂F 0 (v(x))
+2
·
− 2F 00 (v(x))|∇v(x)|2 − 2[F 0 (v(x))]2
∂x
∂x
∂x
∂x
i
j
j
j
i,j=1
j=1
=
2
2
n 2
n
X
X
∂ v(x)
∂v(x)
∂v(x)
+2
· F 00 (v(x)) ·
− 2F 00 (v(x))|∇v(x)|2 − 2[F 0 (v(x))]2
∂x
∂x
∂x
∂x
i
j
j
j
i,j=1
j=1
2
n 2
X
∂ v(x)
= 2
+ 2F 00 (v(x))|∇v(x)|2 − 2F 00 (v(x))|∇v(x)|2 − 2[F 0 (v(x))]2
∂x
∂x
i
j
i,j=1
2
n 2
X
∂ v(x)
= 2
− 2[F 0 (v(x))]2 .
∂x
∂x
i
j
i,j=1
– 11 –
On the other hand, we know for any 1 ≤ i ≤ n, we have
∂P (x)
∂v(x)
+ 2F 0 (v(x)) ·
∂xi
∂xi
=
2
n
X
∂v(x)
j=1
∂xj
·
∂ 2 v(x)
∂xi ∂xj
So we get
∂P (x)
∂v(x)
+ 2F 0 (v(x)) ·
∂xi
∂xi
≤ 2
n
X
∂ 2 v(x)
∂v(x)
·
∂xj
∂xi ∂xj
i=1
"
≤ 2
2
n X
∂v(x)
i=1
=
∂xj
=
·
2
n 2
X
∂ v(x)
i=1
# 12
∂xi ∂xj
By Cauchy-Schwarz’s inequality
#1
" n X ∂ 2 v(x) 2 2
2|∇v(x)|
∂xi ∂xj
i=1
=
# 12 "
2|∇v(x)|
√
∆x P (x) + 2[F 0 (v(x))]2
2
21
1
2|∇v(x)|[∆x P (x) + 2[F 0 (v(x))]2 ] 2
Hence, we get
2|∇v(x)|2 [∆x P (x) + 2[F 0 (v(x))]2 ]
=
2|∇v(x)|2 ∆x P (x) + 4|∇v(x)|2 · [F 0 (v(x))]2
n
2
X
∂v(x)
∂P (x)
+ 2F 0 (v(x)) ·
∂xi
∂xi
i=1
n
n
X ∂P (x) 2
X
∂P (x) ∂v(x)
0
=
+ 2F (v(x)) ·
·
∂x
∂xi
∂xi
i
i=1
i=1
2
n X
∂v(x)
+4[F 0 (v(x))]2 ·
∂xi
i=1
≥
= |∇P (x)|2 + 2F 0 (v(x))(∇v(x) · ∇P (x)) + 4[F 0 (v(x))]2 · |∇v(x)|2
Therefore, we have
(2)
|∇v(x)|2 ∆x P (x) ≥
1
|∇P (x)|2 + 2F 0 (v(x))(∇v(x) · ∇P (x)).
2
Let M = supx∈Rn 2|F 0 (v(x))||∇v(x)|, and N = supx∈Rn 2|∇v(x)|2 . So we know 0 ≤ M, N < ∞. If N = 0, so for
all x ∈ Rn , we have
Dv(x) = 0.
This contradicts with v(x) 6≡ constant in Rn . So we have N > 0.
If M = 0, so for all x ∈ Rn , we have
|F 0 (v(x))||∇v(x)| ≡ 0.
That is, we have ∇F (v(x)) ≡ 0. So F (v(x)) ≡ Constant for all x ∈ Rn . So ∆x v(x) = 0, that is v(x) is a bounded
harmonic function, so v(x) is constant, contradiction. Hence, we have M > 0.
– 12 –
Hence, by Theorem 4.1, for M, N > 0 and fix M and N , then for any α > 0, β > 0, there exists some C 2 functions:
ηα,β : [β, ∞) −→ R.
with the following good properties:
i. For any α > 0, β > 0, we have
ηα,β (β) = 1.
ii. For any α > 0, β > 0, then for any t ≥ β, we have
0
and ηα,β
(t) < 0.
ηα,β (t) > 0,
In particular, we know for all t ≥ β, we have ηα,β (t) ≤ 1.
iii For any α > 0, β > 0, we have
lim ηα,β (t) = 0.
t%∞
iv For any α > 0, β > 0, then for all t ≥ β, we have
lim ηα,β (t) = 1.
α&0
v For any α > 0, β > 0, then for all t ≥ β, we have
 h

i2
!2
0
0
(t)
2
η
(n − 1)ηα,β (t) 
α,β
ηα,β (t)
M 0
α

00
·
−
η (t) − ηα,β
(t) −
= .
0 (t)
ηα,β
ηα,β (t)
α α,β
t
N
Now consider
wα,β (x) = ηα,β (|x|)P (x),
for all x ∈ Rn .
Claim: For all aα > 0, β, then for all |x| ≥ β, we have
wα,β (x) ≤ max α, max wα,β (x) = max α, max P (x) .
|x|=β
|x|=β
When |x| = β, we know
wα,β (x)
= ηα,β (|x|)P (x)
= ηα,β (β)P (x)
= P (x)
Since ηα,β (β) = 1.
Hence we have
max
α, max wα,β (x)
|x|=β
= max
α, max P (x) .
|x|=β
If sup|x|≥β wα,β (x) ≤ 0, since
max
α, max P (x) ≥ α > 0.
|x|=β
– 13 –
So obviously, for all |x| ≥ β, we get
wα,β (x) ≤ 0 ≤ max
α, max P (x) .
|x|=β
If Tα,β = sup|x|≥β wα,β (x) > 0, then there exists some x1 ∈ Rn such that
wα,β (x1 ) > 0
By the property iii of ηα,β (t), we know
lim ηα,β (|x|) = 0.
|x|%∞
By the definition of wα,β (x), and kP kL∞ (Rn ) ≤ C, then we get
lim
|x|%∞
wα,β (x) = 0.
Since wα,β (x1 ) > 0, so the maximum of wα,β (x) is attained in {x ∈ Rn : |x| ≥ β}. For any x ∈ {x ∈ Rn : |x| = β}
such that wα,β (x) = T , then we know
wα,β (x) ≤ Tα,β ≤ max
α, max wα,β (x) = max α, max P (x) .
|x|=β
|x|=β
On the other hand, for any x∗ ∈ {x ∈ Rn : |x| > β} such that wα,β (x∗ ) = Tα,β , then we know x∗ is an interior
maximum point of wα,β (x) in {x ∈ Rn : |x| ≥ β}. So have
∇wα,β (x∗ ) = 0,
and D2 wα,β (x∗ ) ≤ 0.
And also, by the definition of wα,β (x) = ηα,β (|x|)P (x), then we have
0 = ∇wα,β (x∗ ) = P (x∗ )∇ηα,β (|x∗ |) + ηα,β (|x∗ |)∇P (x∗ ).
Since ηα,β (|x∗ |) > 0, so we get
∇P (x∗ ) = −P (x∗ ) ·
∇ηα,β (|x∗ |)
.
ηα,β (|x∗ |)
Since wα,β (x) = ηα,β (|x|)P (x), then we have
∆x wα,β (x) = P (x)∆x ηα,β (|x|) + 2∇P (x) · ∇ηα,β (|x|) + ηα,β (|x|)∆x P (x).
So we can get
|∇v(x∗ )|2 ∆x wα,β (x∗ )
=
|∇v(x∗ )|2 [P (x∗ )∆x ηα,β (|x∗ |) + 2∇P (x∗ ) · ∇ηα,β (|x∗ |) + ηα,β (|x∗ |)∆x P (x∗ )]
=
|∇v(x∗ )|2 P (x)∆x ηα,β (|x∗ |) + 2|∇v(x∗ )|2 (∇P (x∗ ) · ∇ηα,β (|x∗ |))
+ηα,β (|x∗ |)|∇v(x∗ )|2 ∆x P (x∗ )
≥
|∇v(x∗ )|2 P (x∗ )∆x ηα,β (|x∗ |) + 2|∇v(x∗ )|2 (∇P (x∗ ) · ∇ηα,β (|x∗ |))
– 14 –
1
|∇P (x∗ )|2 + 2F 0 (v(x∗ ))(∇v(x∗ ) · ∇P (x∗ ))
By (2)
2
∇ηα,β (|x∗ |)
∗
∗
∗ 2
∗
∗ 2
∗
· ∇ηα,β (|x |)
= P (x )|∇v(x )| ∆x ηα,β (|x |) + 2|∇v(x )| −P (x ) ·
ηα,β (|x∗ |)
"
2
∇ηα,β (|x∗ |)
1
∗
+ 2F 0 (v(x∗ ))ηα,β (|x∗ |)·
−P (x∗ ) ·
+ηα,β (|x |)
2
ηα,β (|x∗ |)
∇ηα,β (|x∗ |)
∇v(x∗ ) · −P (x∗ ) ·
ηα,β (|x∗ |)
+ηα,β (|x∗ |)
=
P (x∗ )|∇v(x∗ )|2 ∆x ηα,β (|x∗ |) − 2P (x∗ ) ·
|∇v(x∗ )|2 |∇ηα,β (|x∗ |)|2
ηα,β (|x∗ |)
[P (x∗ )]2 |∇ηα,β (|x∗ |)|2
− 2P (x∗ )F 0 (v(x∗ ))(∇v(x∗ ) · ∇ηα,β (x∗ ))
2ηα,β (|x∗ |)
|∇v(x∗ )|2 |∇ηα,β (|x∗ |)|2
0
∗
∗
∗
−
2F
P (x∗ ) |∇v(x∗ )|2 ∆x ηα,β (|x∗ |) − 2 ·
(v(x
))(∇v(x
)
·
∇η
(x
))
α,β
ηα,β (|x∗ |)
+
=
+
[P (x∗ )]2 |∇ηα,β (|x∗ |)|2
2ηα,β (|x∗ |)
Since
wα,β (x∗ ) = ηα,β (|x∗ |)P (x∗ ) = T > 0.
By the Property I of ηα,β (t), we know
ηα,β (|x∗ |) > 0.
So we get P (x∗ ) > 0. By D2 wα,β (x∗ ) ≤ 0, then ∆wα,β (x∗ ) ≤ 0. Hence, we can get
2|∇v(x∗ )|2 |∇ηα,β (|x∗ |)|2
P (x∗ )|∇ηα,β (|x∗ |)|2
≤
+ 2F 0 (v(x∗ ))(∇v(x∗ ) · ∇ηα,β (x∗ )) − |∇v(x∗ )|2 ∆x ηα,β (|x∗ |).
∗
2ηα,β (|x |)
ηα,β (|x∗ |)
If |∇v(x∗ )|2 ≤ α, then we have
Tα,β
= wα,β (x∗ )
= ηα,β (|x∗ |)P (x∗ )
≤ P (x∗ )
Since 0 < ηα,β (|x∗ |) ≤ 1, and P (x∗ ) > 0
= |∇v(x∗ )|2 − F (v(x∗ ))
≤
|∇v(x∗ )|2
Since F (v(x∗ )) ≥ 0
≤ α.
If |∇v(x∗ )|2 > α, then we have
Tα,β
= wα,β (x∗ )
= ηα,β (|x∗ |)P (x∗ )
– 15 –
=
≤
≤
≤
2[ηα,β (|x∗ )]2 P (x∗ )|∇ηα,β (|x∗ |)|2
·
|∇ηα,β (|x∗ )|2
2ηα,β (|x∗ |)
2[ηα,β (|x∗ )]2 2|∇v(x∗ )|2 |∇ηα,β (|x∗ |)|2
0
∗
∗
∗
∗ 2
∗
+
2F
(v(x
))(∇v(x
)
·
∇η
(x
))
−
|∇v(x
)|
∆
η
(|x
|)
α,β
x
α,β
|∇ηα,β (|x∗ )|2
ηα,β (|x∗ |)
2[ηα,β (|x∗ )]2 2|∇v(x∗ )|2 |∇ηα,β (|x∗ |)|2
0
∗
∗
∗
∗ 2
∗
+ 2|F (v(x ))||(∇v(x ) · ∇ηα,β (x ))| − |∇v(x )| ∆x ηα,β (|x |)
|∇ηα,β (|x∗ )|2
ηα,β (|x∗ |)
2[ηα,β (|x∗ )]2 2|∇v(x∗ )|2 |∇ηα,β (|x∗ |)|2
0
∗
∗
∗
∗ 2
∗
+
2|F
(v(x
))||∇v(x
)||∇η
(x
)|
−
|∇v(x
)|
∆
η
(|x
|)
α,β
x α,β
|∇ηα,β (|x∗ )|2
ηα,β (|x∗ |)
By the Cauchy-Schwarz’s inequality
|∇v(x∗ )|2 · 2[ηα,β (|x∗ )]2 2|∇ηα,β (|x∗ |)|2
1
0
∗
∗
∗
∗
=
· 2|F (v(x ))||∇v(x )||∇ηα,β (x )| − ∆x ηα,β (|x |)
+
|∇ηα,β (|x∗ )|2
ηα,β (|x∗ |)
|∇v(x∗ )|2
[ηα,β (|x∗ )]2 2|∇ηα,β (|x∗ |)|2
M
∗
∗
≤ N·
+
|∇η
(|x
|)|
−
∆
η
(|x
|)
α,β
x α,β
|∇ηα,β (|x∗ )|2
ηα,β (|x∗ |)
α
By the definition of M and N , and |∇v(x∗ )|2 > α
0
Since ηα,β is radial function, and ηα,β
(|x∗ |) < 0, then we know
0
|∇ηα,β (|x∗ |)| = −ηα,β
(|x∗ |)
∆x ηα,β (|x∗ |)
00
= ηα,β
(|x∗ |) +
n−1 0
η (|x∗ |).
|x∗ | α,β
So we get
Tα,β
= wα,β (x∗ )
[ηα,β (|x∗ )]2 2|∇ηα,β (|x∗ |)|2
M
∗
∗
+
|∇η
(|x
|)|
−
∆
η
(|x
|)
α,β
x α,β
|∇ηα,β (|x∗ )|2
ηα,β (|x∗ |)
α
[ηα,β (|x∗ )]2
M 0
n−1 0
2|∇ηα,β (|x∗ |)|2
∗
00
∗
∗
= N·
−
η (|x |) − ηα,β (|x |) −
η (|x |)
0 (|x∗ |)]2
[−ηα,β
ηα,β (|x∗ |)
α α,β
|x∗ | α,β
α
= N·
By the Property v of ηα,β (t)
N
≤ N·
= α.
Therefore, for all |x| ≥ β, we have
wα,β (x) ≤ Tα,β = wα,β (x∗ ) ≤ α.
In a summary, we have Therefore, for all |x| ≥ β, we have
wα,β (x) = ηα,β (|x|)P (x) ≤ max
α, max P (x) .
|x|=β
Take α & 0, by the Property iv of ηα,β (t), we know
P (x) ≤ max {0, P (0)},
for all x ∈ Rn .
Since |∇v(0)|2 < δ, then we have
P (0)
= |∇v(0)|2 − 2F (v(x))
– 16 –
≤
|∇v(0)|2
<
δ.
Since F (v(x)) ≥ 0
So we get
P (x) = |∇v(x)|2 − 2F (v(x)) < δ,
for all x ∈ Rn .
Since v(x) = u(x + x0 ), then
|∇u(x + x0 )|2 − 2F (u(x + x0 )) < δ,
for all x ∈ Rn .
By the arbitrary of x in Rn , we can get
|∇u(x)|2 − 2F (u(x)) < δ,
for all x ∈ Rn .
By the arbitrary of δ > 0, we have
|∇u(x)|2 ≤ 2F (u(x)),
for all x ∈ Rn .
Approach II of Theorem 5.1, Proposition 5.1 on Page 24 in Gui[2]. Define
1
H(x) = F (u(x)) − |∇u(x)|2 ,
2
∀x ∈ Rn .
By Theorem 3.1, we know that there exists some C1 > 0 such that
H(x) > −C1 ,
∀x ∈ Rn .
So we can define
A = infn H(x).
x∈R
n
Then there exists a sequence xk ∈ R such that
lim H(xk ) = A.
k→∞
For any k ≥ 1, let uk (x) = u(x + xk ) in Rn , by the standard elliptic estimates, we know that there exists some C2 > 0
such that
kuk kC 2,α (Rn ) ≤ C2 .
By the Ascoli-Arzela Theorem, there exists some v ∈ C 2 (Rn ) such that uk → v in C 2 (Rn ), as k → ∞, which implies
that ∆uk (x) → ∆v(x), F (uk (x)) → F (v(x)) and |∇uk (x)| → |∇v(x)| for all x ∈ Rn , as k → ∞. Therefore, we know
that v satisfies ∆v(x) = F 0 (v(x)) in Rn , and
1
A = F (v(0)) − |∇v(0)|2 = infn
x∈R
2
– 17 –
1
F (v(x)) − |∇v(x)|2 .
2
Let
1
G(x) = F (v(x)) − |∇v(x)|2 ,
2
∀x ∈ Rn .
Direct computations show that
|∇v(x)|2 ∆G(x)
=
n
X
bi (x)Gxi (x)
i=1

2
n
n
n
X
X
X

+
uxj (x)uxi xj (x) − |∇u(x)|2
|uxi xj (x)|2
i=1
≤
n
X
j=1
i,j=1
bi (x)Gxi (x),
By Cauchy-Schwarz’s inequality
i=1
That is, we get
−|∇v(x)|2 ∆G(x) +
n
X
bi (x)Gxi (x) ≥ 0,
∀x ∈ Rn .
i=1
If ∇v(0) = 0, then A = F (v(0)) ≥ 0. If ∇v(0) 6= 0, since G(x) attains its minimum at x = 0, by the Strong Maximum
Principle, we know that G(x) ≡ A for all x ∈ Rn . On the other hand, by Theorem 2.1, we know that A ≥ 0. In
summary, we have G(x) ≥ A ≥ 0 in Rn , which implies that
|∇u(x)|2 ≤
1
F (u(x)),
2
∀x ∈ Rn .
Remark 5.1. In the proof of the above theorem, we find that if we assume F (t) ≥ 0 for [−M, M ], where kukL∞ (Rn ) ≤ M ,
then the above proof still works, so the theorem holds as well.
Remark 5.2. The above gradient estimate is optimal. Look at the PDE:
∆x u(x) = sin u(x), in Rn .
Letu(x) = 4 arctan ex1 , then we know
∂u(x)
∂x1
=
4ex1
1 + e2x1
∂ 2 u(x)
∂x1 ∂x1
=
4[ex1 (1 + e2x1 ) − ex1 · 2e2x1 ]
[1 + e2x1 ]2
=
4[ex1 + e3x1 − 2e3x1 ]
[1 + e2x1 ]2
=
4[ex1 − e3x1 ]
[1 + e2x1 ]2
=
4(t − t3 )
[1 + t2 ]2
Let t = ex1 > 0
So we have
∆x u(x) =
∂u(x)
4(t − t3 )
=
.
∂x1
[1 + t2 ]2
– 18 –
On the other hand, we know
sin u(x)
=
sin(4 arctan ex1 )
=
sin(4 arctan t)
=
2 tan(2 arctan t)
1 + tan2 (2 arctan t)
For tan(2 arctan t), we have
tan(2 arctan t)
=
=
2 tan(arctan t)
1 − tan2 (arctan t)
2t
.
1 − t2
Hence, we get
sin u(x)
=
2 tan(2 arctan t)
1 + tan2 (2 arctan t)
=
2t
2 · 1−t
2
2
2t
1 + 1−t
2
=
4t[1 − t2 ]
[1 − t2 ]2 + 4t2
=
4[t − t3 ]
[1 + t2 ]2
=
∆x u(x).
Then u(x) = 4 arctan ex1 is a bounded solution of this PDE, and 0 < u(x) < 2π for all x ∈ Rn . Take F (t) = 1 − cos t,
and t = ex1 , we have
2F (u(x))
=
2 − 2 cos(4 arctan ex1 )
=
2 − 2 cos(4 arctan t)
4 sin2 (2 arctan t)
2
2 tan(arctan t)
= 4
1 + tan2 (arctan t)
2
2t
= 4
1 + t2
=
=
|∇u(x)|2
16t2
[1 + t2 ]2
4ex1
1 + e2x1
4t
=
1 + t2
– 19 –
=
2
=
16t2
[1 + t2 ]2
=
2F (u(x)).
In a summary, we can find u(x) such that
|∇u(x)|2 = 2F (u(x)),
for all x ∈ Rn .
6. Liouville Type Theorem
Theorem 6.1. Let F ∈ C 2 (Rn ), and F (t) ≥ 0 for all t ∈ R. For any u ∈ C 3 (Rn )
T
L∞ (Rn ) which is a solution of
the PDEs:
∆x u(x) = F 0 (u(x)),
for all x ∈ Rn .
Furthermore, we assume that there exists some x0 ∈ Rn such that F (u(x0 )) = 0. Then
u(x) ≡ u(x0 ),
for all x ∈ Rn .
Proof. Let a = u(x0 ) ∈ R, and define
Γa (u) = {x ∈ Rn : u(x) = a}.
Then Γa (u) 6= φ. Since u ∈ C 3 (Rn ), then we know that Γa (u) is a closed subset of Rn . Now we show that Γa (u) is
also an open subset of Rn .
For any x∗ ∈ Γa (u), since F (t) ≥ 0 for all t ∈ R, and F (a) = 0, then t = a is a minimum point of F in R. So
F 0 (a) = 0,
and F 00 (a) ≥ 0.
By the Taylor’s expansion of F , we know there exist some small δ > 0 such that whenever |t − a| ≤ δ, we have
F (t)
F 00 (ξt,a )
(t − a)2
2
=
F (a) + F 0 (a)(t − a) +
=
F 00 (ξt,a )
(t − a)2
2
≤
kF 00 kL∞ ([a−δ,a+δ]) (t − a)2
=
k(t − a)2 .
Since u ∈ C 3 (Rn ), then there exists some > 0 such that whenever |x − x∗ | < , we have
|u(x) − u(x∗ )| < δ.
For any ω ∈ Rn with |ω| = 1, we define
φ(t) = u(x∗ + tω) − u(x∗ ) = u(x∗ + tω) − a,
Then we have φ0 (t) = ∇u(x∗ + tω) · ω, so
|φ0 (t)|2
= |∇u(x∗ + tω) · ω|2
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for all |t| < .
≤
|∇u(x∗ + tω)|2
≤ 2F (u(x∗ + tω))
By Cauchy-Schwarz’s inequality, and |ω| = 1
By Theorem 5.1
≤ 2k[u(x∗ + tω) − a]2
=
2k[φ(t)]2 .
Then by the Fundamental Theorem of Calculus, and φ(0) = 0, we know for all |t| < , we have
Z t
φ0 (s)ds
|φ(t)| =
0
t
Z
|φ0 (s)|ds
≤
0
≤
Z t√
2k|φ(s)|ds
0
√
=
Z
t
|φ(s)|ds.
2k
0
By the Bellman’s inequality or the Integral form of the Gronwall’s inequality (Theorem 2, in Page 2, in [4]), all |t| < ,
we have
|φ(t)| ≤ 0.
That is φ(t) = for all |t| < . So we get, for all |t| < , and |ω| = 1, we have
u(x∗ + tω) = u(x∗ ) = a.
In particular, we know
B (x∗ ) ⊂ Γa (u).
So Γa (u) is open. So Γa (u) = Rn . Therefore, we know
u(x) ≡ u(x0 ),
for all x ∈ Rn .
References
[1] David Gilbarg and Neil S. Trudinger. Elliptic partial differential equations of second order. Springer, 2001.
[2] Changfeng Gui. Allen-Cahn equation and its generalizations. Notes, 2009.
[3] Luciano Modica. A gradient bound and a liouville theorem for nonlinear poisson equations. Communications on Pure and Applied
Mathematics, 38:679–684, 1985.
[4] Mingfeng Zhao. The Gronwall’s inequality. Personal Notes on Mathematical Problems or Theorems, 2011.
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