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INTERIOR GRADIENT ESTIMATES
FOR MEAN CURVATURE EQUATIONS
Xu-Jia Wang
School of Mathematical Sciences
Australian National University
Abstract. In this paper we give a simple proof for the interior gradient estimate for
curvature and Hessian equations. We also derive a Liouville type result for these equations.
§0. Introduction
The interior gradient estimate for the prescribed mean curvature equation has been
extensively studied, see [9] and the references therein. For high order mean curvature
equations it has also been obtained in [11, 18]. In most articals such estimates were
obtained by carrying out analysis on the graphs of solutions and so the arguments depend
on the invariance of equations under rigid motion.
From the view point of partial differential equations such estimate should hold for
equations with similar structural conditions. In [13, 6] the gradient estimate was obtained
for certain fully nonlinear elliptic equations under various conditions. Different proofs
for the gradient estimate for mean curvature equations have been given in [1, 3].
In this note we show that for curvature and Hessian equations the interior gradient
estimate can be obtained very easily. Our proof is elementary which is based on suitable
choice of auxiliary functions and avoids geometric computations on the graph of solutions.
The technique in this note has actually been widely used in literature, see, e.g., [9].
§1. Mean curvature equation.
Let us consider the mean curvature equation
aij uij =: (δij −
p
ui uj
)u
=
f
(x)
=:
H(x)
1 + |∇u|2
ij
1 + |∇u|2
(1.1)
1991 Mathematics Subject Classification. 35B45.
Key words and phrases. Gradient estimate, curvature equations, Hessian equations.
Typeset by AMS-TEX
Typeset by AMS-TEX
1
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where H ∈ C 0,1 (Br (0)), uij = uxi xj , and summation convention is used. Suppose
u ∈ C 3 (Br (0)) is a nonnegative solution of (1.1). We want to give an upper bound for
|∇u(0)|.
u
Let G(x, ξ) = g(x)ϕ(u) log uξ (x), where g(x) = 1 − |x|2 /r2 , ϕ(u) = 1 + M
, M =
n−1
supΩ u(x). Suppose the supremum sup{G(x, ξ), x ∈ Br (0), ξ ∈ S
} is attained at
some point x0 and in the x1 direction. Then at x0 ,
0 = (log G)i =
gi
ϕ0
u1i
+ ui +
,
g
ϕ
u1 log u1
(1.2)
and the matrix {(log G)ij } ≤ 0, where
(log G)ij =
2
¡ gij
gi gj ¢ ¡ ϕ00
ϕ0 ¢
ϕ0
− 2 +
− 2 ui uj + uij
g
g
ϕ
ϕ
ϕ
¡
u1ij
1 ¢ u1i u1j
+
− 1+
.
u1 log u1
log u1 u21 log u1
By (1.2),
2
u1i u1j
ϕ0
ϕ0
gi gj
+
u
u
+
(gi uj + gj ui ).
=
i
j
g2
ϕ2
ϕg
u21 log2 u1
Hence we have
gij
ϕ00
ϕ0
ϕ0
(log G)ij =
+
ui uj + uij +
(gi uj + gj ui )
g
ϕ
ϕ
ϕg
¡
u1ij
2 ¢ u1i u1j
+
− 1+
u1 log u1
log u1 u21 log u1
Since ui (x0 ) = 0 for i ≥ 2, we have at x0 , a11 =
i 6= j. Differentiating the equation (1.1) gives
aii uii1 = f1 +
1
,
1+u21
aii = 1 for i ≥ 2, and aij = 0 for
2u1 2
2u1
2
u
+
Σ
u
i≥2
11
2
(1 + u1 )2
1 + u21 1i
at x0 .
Hence if u1 (x0 ) is suitably large, we have
µ
aii uii1 − (1 +
Note that ϕ(u) = 1 +
aii
u
M,
2
u2
) 1i
log u1 u1
¶
≥ f1 +
(1.3)
u1 u211
u1 u21i
+
Σ
.
i≥2
2(1 + u21 )2
2(1 + u21 )
we have
© gii
ª ϕ0
ϕ00 2 ϕ0
2ϕ0
2n
4ϕ0 u1
+
ui + uii +
gi ui ≥ f − 2 −
.
g
ϕ
ϕ
ϕg
ϕ
r g rgϕ 1 + u21
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We may suppose G(x0 ) is suitably large so that gu1 ≥ 1. Then by (1.3) we obtain
ϕ0
f1
2n
C
u211
0 ≥ aii (log G)ii ≥
+ f− 2 −
+
.
u1 log u1
ϕ
r g M r 2(1 + u21 )2 log u1
p
Note that f (x) = H(x) 1 + |∇u|2 , by (1.2) we have,
f1
ϕ0
(1 + u21 )1/2
ϕ0 H
g1 Hu1
+ f = H1
+
−
2
1/2
u1 log u1
ϕ
u1 log u1
ϕ(1 + u1 )
g(1 + u21 )1/2
≥ −C1 − C2 /gr.
If G(x0 ) is large enough so that | gg1 | ≤
ϕ0
2ϕ u1
(1.4)
at x0 , then by (1.2),
2
u211
ϕ0 2 u41 log2 u1
ϕ0
≥
(
)
≥
log2 u1 .
(1 + u21 )2
2ϕ (1 + u21 )2
8ϕ2
It follows
0 ≥ aii (log G)ii ≥ −C1 −
Hence
C2
log u1
+
.
gr2
64M 2
M2
g log u1 ≤ C1 + C2 2 .
r
(1.5)
We have thus proved
Theorem 1.1.
Let u ∈ C 3 (Br (0)) be a nonnegative solution of (1.1).
|H(x)| ≤ C0 and |∇H(x)| ≤ C0 . Then
¡
M2 ¢
|∇u(0)| ≤ exp C1 + C2 2 ,
r
Suppose
(1.6)
where M = supΩ u(x), C1 depends only on n, M , and C0 ; C2 depends only on n and C0 .
Remarks
(1) By approximation as in [18] one sees that Theorem 1 holds for viscosity solutions.
(2) The same argument can also be applied to mean curvature flow equations.
(3) From (1.4) it is easy to see that Theorem 1.1 can be extended to the case H =
H(x, u, γ) with Hu ≥ 0, where γ = (∇u, −1)/(1 + |∇u|2 )1/2 .
(4) The estimate (1.6) is not sharp, see [9]. From (1.5) we also have |∇u(x)| ≤
¡
¢
exp C(1 + d−1
x ) , where C depends on M and r, dx = dist(x, ∂Br ).
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§2. High Order Mean Curvature Equations.
The above proof can be extended to high order mean curvature equations. For u ∈
C 2 (Br (0)), the downwards directed normal of the graph of u is (ν, νn+1 ) = (∇u, −1)/w,
p
where w = 1 + |∇u|2 . The principal curvatures of the graph of u are the eigenvalues
of the Jacobian matrix Dν. But the matrix Dν is not symmetric, so instead we make
use of the matrix
ui uk ukj
uj uk uki
ui uj uk ul ukl
aij = uij −
−
+ 2
.
w(1 + w) w(1 + w)
w (1 + w)2
Let F (aij ) = σk (λ), where λ ∈ IRn denotes the eigenvalue vector of the matrix {aij } and
σk (λ) denotes the k th elementary symmetric function of λ. Then the k th -mean curvature
H(x) of the graph of u is given by H(x) = F (aij )/wk , see [5]. Hence we get the equation
F (aij ) = f (x) =: H(x)(1 + |∇u|2 )k/2 .
(2.1)
Let Γk denote the component of {λ ∈ IRn , σk > 0} which contains the point (1, · · · , 1).
A function u ∈ C 2 (Br (0)) is said to be admissible if the eigenvalue vector of the matrix
(aij ) lies in Γk . If u is admissible, F is elliptic. Moreover we have
Σni,j=1 aij
∂
F = kF ;
∂aij
Σni=1
∂
F 1/k (aij ) ≥ Cn,k > 0.
∂aii
(2.2)
Theorem 2.1.
Let u ∈ C 3 (Br (0)) be an nonnegative admissible solution of (2.1).
Suppose H(x) ≥ 0 and |∇H 1/k (x)| ≤ C0 . Then
|∇u(x0 )| ≤ exp(C1 + C2
M2
),
r2
where C1 depends only on n, C0 , and M = supx∈Br u(x), C2 depends only on n and C0 .
Proof. Let G(x) = G(x, ξ) = g(x)ϕ(x) log uξ (x), where g(x) and ϕ(u) are as in section
1. Suppose G attains its maximum at x0 and in the direction e1 . Then at x0 , uj = 0 for
j ≥ 2,

2
if i = j = 1,

 1/w
aij = αij uij , αij = 1/w
if i = 1 or j = 1, and i + j ≥ 3,


1
otherwise;
and (1.2) and (1.3) hold. Multiplying (1.3) by Feij and taking summation, we obtain
¾
½
ϕ00
ϕ0
ϕ0
gij
e
e
+
ui uj + uij +
(gi uj + gj ui )
0 ≥ Fij (log G)ij =Fij
g
ϕ
ϕ
ϕg
½
¾
¡
¢ u1i u1j
u
2
1ij
+ Feij
− 1+
=: B + A,
u1 log u1
log u1 u21 log u1
(2.3)
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where Fij = ∂F/∂aij , Feij = Fij αij .
To estimate B we suppose at x0 , uij = 0 for i 6= j and i, j ≥ 2; and u22 ≥ · · · ≥ unn .
Then aij = 0 for i 6= j and i, j ≥ 2 and we have F1j = −a1j σk−2 ({a22 , · · · , ann }\ajj ),
where σk−2 ({· · · }\ajj ) ≥ 0. Hence by (1.2) we have, for i ≥ 2,
gi F1i = −gi
u1i
g 2 u1 log u1
σk−2 ({a22 , · · · , ann }\aii ) = i
σk−2 (· · · ) ≥ 0
w
gw
We obtain, by (2.2),
B≥−
2F
4u1 ϕ0 F11
ϕ0
−
+
kf
,
gr2
rgϕw2
ϕ
(2.4)
where F = Σni=1 Fii .
To estimate A we differentiate the equation (2.1) to obtain
Feij uij1 = f1 +
where



bij =
We have


2u1
Fij bij
w(1 + w)
at x0 ,
u211
u211
u21k
w2 + w3 + Σk>1 w
u11 u1j
u u1j
1
+ 11
w
2w2 + 2 u1j ujj
i = j = 1,
i = 1, j ≥ 2,
i, j ≥ 2.
u1i u1j
1
Fij bij ≥ [1 + O( )]Fij αij u1i u1j + Σj>1 F1j u1j ujj .
w
Hence
A log u1 ≥
f1
1
1
2F1j u1j ujj
+
[1 − O(
)]Fij αij u1i u1j + Σj>1
.
u1
w(w + 1)
log u1
w(w + 1)
0
0
ϕ 2
Suppose G(x0 ) is suitably large so that in (1.2), | gg1 | < 21 ϕϕ . Then u11 ≤ − 2ϕ
u1 log u1 < 0
and
2
gj2 2
u211
ϕ0 2
2
2
≥
u
log
u
,
u
=
u log2 u1 j ≥ 2.
(2.5)
1
1j
w2
4ϕ2 1
g2 1
Observe that F1j u1j ≤ 0 for j ≥ 2 and Fij = 0 for i 6= j and i, j ≥ 2, we have
A log u1 ≥
f1
1
2F1j u1j ujj
+
F11 α11 u211 + Σj∈J
u1
2w(w + 1)
w(1 + w)
(2.6)
provided u1 is suitably large, where J = {2 ≤ j ≤ n, ujj ≥ 0}. Recall that u11 < 0, we
have F11 ≥ δF and F ≥ δa22 · · · akk for some δ > 0 depending only on n, k (see [16]),
where a22 ≥ · · · ≥ ann as above. Hence for j ∈ J (note that ujj = ajj for j ≥ 2),
F1j u1j ujj = −a1j u1j ujj σk−2 ({a22 , · · · , ann }\ajj )
≥ −Ca1j u1j a22 · · · akk ≥ −Ca1j u1j F11 = −Cu21j F11 /w.
(2.7)
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By (2.5) we obtain,
2
A log u1 ≥
ϕ0
f1
C|∇g|2
+[
]F11 log2 u1 .
−
u1
16ϕ2
g2 w
(2.8)
2
02
C|∇g|
ϕ
Suppose G(x0 ) is large enough so that 32ϕ
2 ≥
g 2 w . Insert (2.8) and (2.4) into (2.3)
we obtain
2
f1
ϕ0
F
ϕ0
ϕ0 F11
0≥
+
+
+ kf
F
log
u
−
C(
).
(2.9)
11
1
u1 log u1
ϕ
32ϕ2
rgϕw
gr2
Similar to (1.4) we have
f1
ϕ0
w k H1
1
+ kf
≥
− CH(1 + )wk−1 .
u1 log u1
ϕ
u1 log u1
gr
By (2.2),
C ≤ Σi
∂
1
1
[F (aij )]1/k = F (1−k)/k F = H (1−k)/k w1−k F,
∂aii
k
k
we have F ≥ CH (k−1)/k wk−1 . By assumption, |H (1−k)/k Hi | ≤ C0 , we obtain F ≥
C|Hi |wk−1 . Hence
ϕ0
1
f1
+ kf
≥ −C(1 + )F.
u1 log u1
ϕ
gr
Dividing (2.9) by F and noticing that F11 ≥ δF, we obtain
2
ϕ0
C2
0≥
log u1 − C1 − 2 .
2
32ϕ
gr
2
Hence g log u1 ≤ C1 + C2 M
r 2 . This completes the proof.
We remark that the above argument can also be applied to Hessian equations. For
Hessian equations one may choose the auxiliary function G(x, ξ) = g(x)ϕ(u)uξ , where
g(x) = 1 − |x|2 /r2 , ϕ(u) = (1 − u/M )−α , M = 4 sup{|u(x)|, x ∈ Br (0)}, and α ∈ (0, 1)
2
is chosen small so that ϕ00 − Cϕ0 /ϕ ≥ 0.
§3. Liouville type result.
Slight modification of the above proof yields the following Liouville type result.
Theorem 3.1.
Let u ∈ C 3 (IRn ) be an admissible solution of F (aij ) = 0. Suppose
|u(x)| = o(1 + |x|)
as
|x| → ∞.
Then u ≡ constant, where the notation F, aij etc is as in §2.
(3.1)
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Proof.
By Theorem 2.1 we have
|∇u(x)| ≤ C
∀ x ∈ IRn .
(3.2)
To prove Theorem 3.1 we suppose by contradiction that |∇u(0)| ≥ δ > 0. Let G(x) =
G(x, ξ) = g(x)ϕ(u)uξ , where g(x) = 1 − |x|2 /r2 , ϕ(u) = (1 − u/M )−α , M = Mr =
4 sup{|u(x)|, x ∈ Br (0)}, and α ∈ (0, 1) will be determined below.
Suppose G attains its maximum at x0 and in the direction e1 . By (3.2) we have
g(x0 ) ≥ δ1
and
u1 (x0 ) ≥ δ1
(3.3)
for some δ1 > 0 depending only on δ. At x0 we have
u1i
gi
ϕ0
(3.4)
+ + ui
u1
g
ϕ
2
¡ ϕ00
¡ gij
ϕ0 ¢
ϕ0
gi gj ¢ ϕ0
+
− 2 2 ui uj + uij +
−2 2 −
(gi uj + gj ui )
ϕ
ϕ
ϕ
g
g
gϕ
0 = (log G)i =
(log G)ij =
u1ij
u1
Hence
0 ≥ Feij (log G)ij ≥
2
2Fij bij
ϕ00
ϕ0
gij
gi gj
+[
− C 2 ]Feij ui uj + [
− C 2 ]Feij .
w(1 + w)
ϕ
ϕ
g
g
By (3.3) we obtain
gij
gi gj
CF
Feij [
−C 2 ]≥− 2 .
g
g
r
u1
By (3.1) we have for r large enough, u11 = − gϕ [gϕ0 u1 + g1 ϕ] < 0. Noticing that
F1j u1j ≤ 0 for j ≥ 2. We have
Fij bij ≥Fij αij u1i u1j + Σj≥2
1
F1j u11 u1j + Σj≥2 F1j u1j ujj
w2
1
F1j u11 u1j + Σj≥2 F1j u1j ujj ≥ Σj∈J F1j u1j ujj ,
w2
where J = {2 ≤ j ≤ n, ujj > 0}. For j ∈ J, we have by (2.7) and (3.4),
≥Σj≥2
F1j u1j ujj ≥ −
C 2
C
u1j F11 ≥ − 2 F11 .
w
r
Hence Fij bij ≥ − rC2 F11 . We thus obtain
2
ϕ0
u2
C
[ϕ − C
]F11 12 ≤ 2 F.
ϕ
w
r
00
Let α ∈ (0, 1) be small enough so that
2
ϕ00
ϕ0
α(1 − Cα)
α
−C 2 ≥
≥
.
2
ϕ
ϕ
M
2M 2
u2
2
Noticing that F11 ≥ δF, we obtain w12 ≤ C M
r 2 . Let r → ∞, by (3.1) we conclude
u1 (x0 ) = 0, which is in contradiction with (3.3). Hence ∇u(0) = 0.
Page 8
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