Properties of Translating Solutions to Mean Curvature Flow

TL;DR

This paper investigates the geometric and asymptotic properties of translating solutions to mean curvature flow and related nonlinear flows, focusing on convexity, gradient estimates, and behavior at infinity.

Contribution

It provides new insights into the convexity, gradient bounds, and asymptotic analysis of translating solutions for mean curvature flow and nonlinear curvature flows.

Findings

Convexity of translating solutions established.

Interior gradient estimates derived.

Asymptotic behavior at infinity characterized.

Abstract

In this paper, we study the convexity, interior gradient estimate, Liouville type theorem and asymptotic behavior at infinity of translating solutions to mean curvature flow as well as the nonlinear flow by powers of the mean curvature.

Equations

$$a_{ij}{u}_{ij}:=({\delta }_{ij}-\frac{u_i{u}_j}{1+|\nabla u{|}^2})u_{ij}=(\frac{1}{\sqrt{1+|\nabla u{|}^2}}{)}^{\alpha -1},\forall x\in R^n$$

$$div(\frac{Du}{\sqrt{1+|Du{|}^2}})=\frac{1}{\sqrt{1+|Du{|}^2}}in{R}^n,$$

$$r^{\prime\prime}(t)>0,\ \ and\ \ \frac{t}{n}<r^{\prime}(t)\bigl{(}1+(r^{\prime}(t))^{2}\bigr{)}^{\frac{\alpha-1}{2}}<\frac{t}{n-1}$$

$$r(t)=\frac{t^2}{2(n-1)}-\ln t+C_1-\frac{(n-1)(n-4)}{2}{t}^{-2}+o(t^{-2})if\alpha =1$$

$$r(t)=\frac{\alpha}{\alpha+1}\bigl{(}\frac{1}{n-1}\bigr{)}^{1/\alpha}t^{1+\frac{1}{\alpha}}-C(\alpha,n)t^{1-\frac{1}{\alpha}}+o(t^{1-\frac{1}{\alpha}}),\quad\hbox{if }\,\alpha\not=1$$

$$C(\alpha,n)=\frac{1}{\alpha-1}(n-1)^{1/\alpha}\bigl{(}\frac{1}{\alpha(n-1)}+\frac{\alpha-1}{2}\bigr{)}.$$

$$\underset{|x|\to \infty }{\lim }\frac{|u(x)|}{|x|}=0.$$

$$|\nabla u(0)|\le \exp \{C_1+C_2\frac{M^2}{r^2}\},$$

$$\frac{r^{\prime\prime}}{1+(r^{\prime})^{2}}+\frac{n-1}{t}r^{\prime}=\bigl{(}1+(r^{\prime})^{2}\bigr{)}^{\frac{1-\alpha}{2}},\forall t\in(0,\infty)$$

$$r(0)=r^{\prime }(0)=0.$$

$$y^{\prime }+[(n-1)y(1+y^2{)}^{\frac{\alpha -1}{2}}-e^s](1+y^2{)}^{\frac{3-\alpha }{2}}=0,\forall s\in (-\infty ,\infty )$$

$$\underset{s\to -\infty }{\lim }\frac{y(s)(1+y^2{)}^{\frac{\alpha -1}{2}}}{e^s}=\frac{1}{n},$$

$$r^{\prime }(0):=\underset{t\to 0^{+}}{\lim }{r}^{\prime }(t)=\underset{s\to -\infty }{\lim }y(s)=0=r(0)$$

$$r^{\prime \prime }(0):=\underset{t\to 0^{+}}{\lim }{r}^{\prime \prime }(t)=\underset{s\to -\infty }{\lim }\frac{y^{\prime }(s)}{e^s}=\frac{1}{n}$$

$$y^{\prime }(s)>0and\frac{e^s}{n}<y(s)(1+y^2{)}^{\frac{\alpha -1}{2}}<\frac{e^s}{n-1},\forall s\in (-\infty ,\infty ).$$

$$\underset{s\to -\infty }{\lim }z(s)=-\frac{1}{n},\underset{s\to \infty }{\lim }z(s)=0and{z}^{\prime }(s)>0,\forall s\in (-\infty ,\infty ).$$

$$\underset{s\to -\infty }{\lim }y(s)=0.$$

$$y^{\prime }(s)\ge 0andy(s)(1+y^2{)}^{\frac{\alpha -1}{2}}\le \frac{e^s}{n-1},\forall s\in (-\infty ,T).$$

$$g(y)=y(1+y^2{)}^{\frac{\alpha -1}{2}}.$$

$$g^{\prime }(y)=y^{\prime }(s)(1+y^2{)}^{\frac{\alpha -3}{2}}(1+\alpha y^2)>0.$$

$$y^{\prime }(s)<0,andg(y(s))>\frac{e^s}{n-1},\forall s\in (-\infty ,s_0).$$

$$y^{\prime }(s_1)=0,g(y(s_1))=\frac{e^{s_1}}{n-1}and{y}^{\prime }(s)<0,\forall s\in (s_1,s_0).$$

$$z^{\prime }+nz+1+\alpha (n-1)z{y}^2=0andz\le 0,\forall s\in (-\infty ,\infty ).$$

$$z(s)\ge -\frac{1}{n},\forall s\in (-\infty ,\infty ).$$

$$z^{\prime \prime }+n{z}^{\prime }+\alpha (n-1)z^{\prime }{y}^2+2\alpha (n-1)zy{y}^{\prime }=0,\forall s\in (-\infty ,\infty )$$

$$\underset{s\to \infty }{\lim }{z}^{\prime }(s)=0.$$

$$g(y(s))=\frac{e^s}{n-1}(1+o(1))andy(s)=(\frac{1}{n-1}{)}^{1/\alpha }{e}^{s/\alpha }(1+o(1))$$

$$2\alpha (n-1)zy{y}^{\prime }=\frac{2\alpha (n-1)z{e}^s[z^{\prime }+(z+1)]y}{(1+y^2{)}^{\frac{\alpha -3}{2}}(1+\alpha y^2)}=2(n-1{)}^{2-2/\alpha }z[z^{\prime }+(z+1)]e^{2s/\alpha }(1+o(1)).$$

$$z(s)=-\frac{1}{\alpha (n-1)}{y}^{-2}(1+o(1))=-\frac{1}{\alpha }(n-1{)}^{\frac{2}{\alpha }-1}{e}^{-\frac{2s}{\alpha }}(1+o(1)),ass\to \infty .$$

$$g(y)=\frac{1}{n-1}e^{s}\bigl{(}1-\frac{1}{\alpha}(n-1)^{\frac{2}{\alpha}-1}e^{-\frac{2s}{\alpha}}\bigr{)}(1+o(1)),\ \ as\ \ s\to\infty.$$

Full text

Properties of Translating Solutions to Mean Curvature

Flow ††thanks: Supported by Natural Science Foundation of China (No. 10631020 and 10871061) and the Grant for Ph.D Program of Ministry of Education of China.

Changfeng Gui

Department of Mathematics

University of Connecticut, Storrs, CT 06269-3009,USA

(e-mail: gui@math.uconn.edu)

Huaiyu Jian and Hongjie Ju

Department of Mathematics

Tsinghua University, Beijing 100084, P.R.China

(e-mail: hjian@math.tsinghua.edu.cn)

Abstract. In this paper, we study the convexity, interior gradient estimate, Liouville type theorem and asymptotic behavior at infinity of translating solutions to mean curvature flow as well as the nonlinear flow by powers of the mean curvature.

Key Words. elliptic equation, mean curvature flow, asymptotic behavior, convex solution, gradient estimate.

1991 Mathematical Subject Classification. 35J60, 52C44.

1. Introduction and Main Results

In this paper, we study the convexity, interior gradient estimate, Liouville type theorem and asymptotic behavior at infinity of the solutions to equation

[TABLE]

where constants $\alpha>0$ and we have used the notation $u_{i}=\frac{\partial u}{\partial x_{i}}$ and the convention for summing. It is called the translating soliton equation of the nonlinear evolution flow of hypersurfaces by powers ($\frac{1}{\alpha}$) of the mean curvature. This nonlinear flow was studied in [10, 11] and has important applications in minimal surfaces [2] and isoperimetric inequalities [11]. When $\alpha=1,$ (1.1) is reduced to

[TABLE]

which plays a key role in classifying the type II-singularity of mean curvature flows [4,5,15]. Scaling the space and time variables in a proper way near type II-singularity points on the surfaces evolved by mean curvature vector with a mean convex initial surface, Huisken-Sinestrari [4,5] and White [15] proved that the limit flow can be represented as $M_{t}=\{(x,u(x)+t)\in R^{n+1}:x\in R^{n},t\in R\}$ and is also a solution (called a translating solutions or solitons) to mean curvature flow. Equivalently, $u$ is a solution to equation (1.2). Therefore, the classification of type II-singularity of mean curvature flow is reduced to the classification of solutions of equation (1.2). There was a well known conjecture among the geometric flows researchers which asserts that any complete strictly convex solution of (1.2) is radially symmetric [15]. A few years ago, Wang [13] proved this conjecture for $n=2$ and found a non-radially symmetric solution of (1.2) for $n>2.$ Sheng and Wang [12] used a direct argument to study the Singularity profile in mean curvature flow.

One natural question is that how are about the asymptotic behavior at infinity of the solutions to (1.2), or more generally to equation (1.1). Obviously, the first step is to make clear the asymptotic behavior of radially symmetric solutions of (1.1). This leads us to prove the following theorem 1.1 in Section 2.

Theorem 1.1 Equation (1.1) has a unique solution of the form $u(x)=r(|x|)$ up to a translation in $R^{n+1}.$ Moreover, the function $r\in C^{2}[0,\infty)$ satisfies that

[TABLE]

for all $t>0,$ and

[TABLE]

[TABLE]

as $t\to\infty,$ where $C_{1}$ is a constant depending on $r(1)$ and

[TABLE]

We should mention that when $\alpha=1$, (1.3) was proved in [8] and a asymptotic result similar to (1.4) was proven in [1]. Our method for general $\alpha>0$ is different and yield more properties of the solutions. See Section 2 for details.

Another natural question, formulated explicitly as an open problem in [13], is whether any solution of (1.2) is strictly convex. We will prove the following theorem 1.2 which is related to this question in Section 3.

Theorem 1.2 Let $u\in C^{2}(R^{n})$ be a convex solution of equation (1.1). If $u$ is strictly convex in some nonempty set, then $u$ is is strictly convex in $R^{n}.$ Particularly, $u$ is is strictly convex in $R^{n}$ if $u(x)\to\infty$ as $|x|\to\infty;$ and when $\alpha=1$ (i.e., $u$ is a solution to (1.2)) and $u(x)\to\infty$ as $|x|\to\infty,$ then after a rotation of coordinate system, $\lim_{h\to\infty}h^{-2}u(hx)\to\sum_{i=1}^{k}x_{i}^{2}$ in $R^{n}$ for some $k\geq 2,$ and $u$ is radially symmetric if $n=2.$

This theorem generalizes the main results in [6] which asserts that the Hessian $(D^{2}u(x))$ has constant rank for all $x\in R^{n}$ if $(D^{2}u(x))$ is positive semi-definite and $\Delta u=f(u,\nabla u)$ in $R^{n},$ where $f\in C^{2,\alpha}$ is strictly positive and convex in $u.$ In the case of Minkowski space [9], similar convexity result was proved by the second author in [7].

Theorem 1.1 tells us that the radially symmetric solution of (1.1) is of the $1+\frac{1}{\alpha}$ order growth at infinity. This order tends to 1 as $\alpha$ goes to $\infty$. Motivated by this, we have the following Liouville result.

Theorem 1.3 There is no nonnegative solution $u\in C^{3}(R^{n})$ of (1.1) such that

[TABLE]

We will prove this theorem in section 5. For this purpose, in section 4 we will want to use the gradient estimate techniques by Xu-Jia Wang in [14] and the methods in [3] to obtain the following interior gradient result for equation (1.1).

Theorem 1.4 Suppose $u\in C^{3}(B_{r}(0))$ is a nonnegative solution of (1.1), then

[TABLE]

where $M=\sup_{x\in B_{r}(0)}u(x),$ $C_{i}$(i=1,2) are constants depending only on $n$ and $\alpha.$

2. Asymptotic Behavior - Proof of Theorem 1.1

We will use a few lemmas to prove theorem 1.1. The main difficulty is to prove asymptotic expansion (1.4) and (1.5).

Lemma 2.1 Suppose that $u(x)=r(|x-x_{0}|)+u(x_{0})$ for some $x_{0}\in R^{n}$ and for all $x\in R^{n}.$ Then $u\in C^{2}(R^{n})$ is a solution of (1.1) if and only if $r\in C^{2}(0,\infty)$ satisfies

[TABLE]

and

[TABLE]

Proof. We may assume $x_{0}=0.$ Let $e_{i}$ be the unit vector in positive $x_{i}$-axis. Since $r(t)=u(te_{i})-u(0)=u(-te_{i})-u(0)$ for all $t\geq 0,$ then $r\in C^{2}[0,\infty)$ and it satisfies (2.2) if and only if $u\in C^{2}(R^{n}).$ Writing (1.1) in $r,$ we see that (1.1) is equivalent to (2.1).

Lemma 2.2 If $y\in C^{1}(-\infty,\infty)$ satisfies

[TABLE]

and

[TABLE]

then $r(t)=\int_{0}^{t}y(\ln s)ds\in C^{2}[0,\infty)$ satisfies (2.1) and (2.2).

Proof. (2.1) can be verified directly by (2.3). Note that (2.4) implies

[TABLE]

and

[TABLE]

by equation (2.3) as well as equation (2.1).

Lemma 2.3 There exists a $y\in C^{1}(-\infty,\infty)$ solving (2.3) and (2.4) such that

[TABLE]

Moreover, the function $z(s):=(n-1)e^{-s}y(s)(1+y^{2})^{\frac{\alpha-1}{2}}-1$ satisfies

[TABLE]

Proof. By local existence we have a function $r\in C^{2}[0,\varepsilon)$ which solves (2.1) and (2.2) in $(0,\varepsilon)$ for some $\varepsilon>0$ (see [8]). Then $y(s)=r^{\prime}(t),t=e^{s}$ satisfies (2.3) in $(-\infty,\ln\varepsilon)$ and

[TABLE]

Let $(-\infty,T)$ be the maximal interval for which $y$ solves (2.3). First we prove $T=\infty.$ In fact, we shall show that

[TABLE]

For convenience, we may define

[TABLE]

Note that $\alpha>0$ and

[TABLE]

Suppose that there exists $s_{0}\in(-\infty,T)$ such that $y^{\prime}(s_{0})<0$ and then $g(y(s_{0}))>\frac{e^{s_{0}}}{n-1}$ by (2.3). We claim that

[TABLE]

Otherwise, there exists a $s_{1}<s_{0}$ such that

[TABLE]

Then $\frac{e^{s_{1}}}{n-1}=g(y(s_{1}))>g(y(s_{0}))>\frac{e^{s_{0}}}{n-1}.$ This contradiction implies that (2.9) holds, and hence $\lim_{s\to-\infty}g(y(s))>\frac{e^{s_{0}}}{n-1},$ which contradicts (2.7). Therefore we have proven (2.8), and then $T=\infty.$ follows by the standard existence theory of ordinary differential equations.

Next, we shall prove (2.6). Since $y$ satisfies (2.3) and (2.8) in $(-\infty,\infty),$ the function $z(s):=(n-1)e^{-s}g(y(s))-1$ satisfies

[TABLE]

Thus, $(ze^{ns})^{\prime}\geq-e^{ns}$ for all $s\in(-\infty,\infty).$ Integrating this inequality over $(-\infty,s)$ and then using (2.7) we obtain

[TABLE]

Now we make two observations. The first one is that $z$ has no local maxima. Indeed, if $z^{\prime}(s_{0})=0$ for some $s_{0}$, we can obtain $z^{\prime\prime}(s_{0})>0$ since $z$ satisfies

[TABLE]

.

The second observation is that $z^{\prime}(s)>0$ for all $s\in(-\infty,\infty).$ Otherwise, if $z^{\prime}(s_{0})\leq 0$ for some $s_{0}.$ Then $z^{\prime}(s)\leq 0$ for all $s\leq s_{0},$ since $z$ has no local maxima. This, together with (2.11), implies that $\lim_{k\to-\infty}z^{\prime}(s_{k})=0$ for some sequence $s_{k}\to-\infty.$ Thus, $\lim_{k\to-\infty}z(s_{k})=-\frac{1}{n}$ by (2.10). Since (2.10) and (2.11) imply $z^{\prime}(s)<0$ in $(-\infty,s_{0}),$ we see that $z(s)<-\frac{1}{n}$ for $s\in(-\infty,s_{0}),$ which contradicts (2.11).

Now that $z^{\prime}(s)>0$ for all $s\in(-\infty,\infty),$ by (2.10) and (2.11) we have $\lim_{s\to-\infty}z(s)=-\frac{1}{n}$ and $\lim_{s\to\infty}z(s)=0.$ This proves (2.6), which, together with equation (2.3), implies (2.5). The proof of Lemma 2.3 has been completed.

Lemma 2.4 Let $y$ be a function as in Lemma 2.3. Then $r(t)=\int_{1}^{t}y(\ln s)ds$ satisfies (1.4) and (1.5) .

Proof. It follows from (2.6) that $\lim_{k\to\infty}z^{\prime}(s_{k})=0$ for some sequence $s_{k}\to\infty.$ By this, we claim that

[TABLE]

Assume that this is not true, then there exists a sequence of local maxima $\theta_{k}\to\infty$ of $z^{\prime}(s)$. Note that we have $z^{\prime\prime}(\theta_{k})=0.$ From (2.6), we can derive

[TABLE]

as $s\to\infty.$ By (2.10) and (2.6), we have $\lim_{s\to\infty}z^{\prime}(s)e^{-2s/\alpha}=0.$ By the definition of $z$, we have

[TABLE]

This, together with (2.12) and (2.6) again, implies $\lim_{k\to\infty}z^{\prime}(\theta_{k})=0$. The claim is hence proven.

Then it follows from (2.13), (2.10) and (2.6) that

[TABLE]

and hence

[TABLE]

Therefore, by straightforward computation we obtain

[TABLE]

where

[TABLE]

This implies (1.5).

In particular, when $\alpha=1$, we have

[TABLE]

In theory, we can repeat the above procedure to obtain higher order expansion. Take the simple case $\alpha=1$ as example, we can let $w(s)=-\frac{e^{2s}}{n-1}z(s)-1.$ Then

[TABLE]

and equation (2.10) read as

[TABLE]

Thus,

[TABLE]

Since (2.15) means $\lim_{k\to\infty}w^{\prime}(s_{k})=0$ for some sequence $s_{k}\to\infty,$ we claim that

[TABLE]

If not, then there exists a $\delta>0$ and a sequence $\theta_{k}\to\infty$ such that $w^{\prime\prime}(\theta_{k})=0$ and $w^{\prime}(\theta_{k})>\delta$ (or $w^{\prime}(\theta_{k})<-\delta$.) But it follows from (2.15) and (2.16) that

[TABLE]

Taking $s=\theta_{k}$ in (2.17) and using (2.15) and (2.19), we have

[TABLE]

which implies $\lim_{k\to\infty}w^{\prime}(\theta_{k})=0,$ a contradiction.

Thus, using (2.15) and (2.18) we can rewrite (2.16) as

[TABLE]

By ( 2.14), we have

[TABLE]

which implies

[TABLE]

Therefore, $r(t)=\int_{1}^{t}y(\ln s)ds$ satisfies (1.4).

Proof of Theorem 1.1: the existence follows from Lemmas 2.1-2.4; while the uniqueness from the well-known comparison principle.

3. Convexity - Proof of Theorem 1.2

Lemma 3.1 Let $u\in C^{2}(R^{n}).$ Suppose that there is a constant $c$ such that the set $\Omega_{c}=\{x\in R^{n}:u(x)<c\}$ is nonempty and bounded. If $u=c$ on $\partial\Omega_{c},$ then the Hessian matrix $(u_{ij}(x_{0}))>0$ (positive definite) for some $x_{0}\in\Omega_{c}.$

Proof. By the assumption we see that $u=c$ on $\partial\Omega_{c}$ and

[TABLE]

for some $x_{1}\in\Omega_{c}.$ This implies that the function

[TABLE]

must attain interior minimum in $\Omega_{c}.$ Consequently, $(u_{ij}(x_{0}))>0$ for some $x_{0}\in\Omega_{c},$ which implies the desired result.

Lemma 3.2 Let $u\in C^{2}(R^{n})$ be a convex solution of equation (1.4). If the set $\Omega_{0}=\{x\in R^{n}:(u_{ij}(x))>0\}$ is nonempty, then $\Omega_{0}=R^{n}.$

Proof. We follow the arguments of theorem 1.3 by the second author in [7]. Suppose the contrary that there exists a $x_{1}\in R^{n}\backslash\Omega_{0}.$ We will derive a contradiction. We may assume $\Omega_{0}$ is nonempty and connected. (Otherwise, we replace it by one of its connected components ). Then there exists a short segment $l\subset\Omega_{0}$ such that $\bar{l}\cap\partial\Omega_{0}=\{x_{1}\}.$ Take $x_{2}\in l$ and $\varepsilon>0$ such that $\overline{B_{\varepsilon}(x_{2})}\subset\Omega_{0}.$ Translating the ball $B_{\varepsilon}(x_{2})$ along the line $l$ toward $x_{1}$ we come to a point $\bar{x}$ where the ball and $\partial\Omega_{0}$ are touched at the first time. It follows that

[TABLE]

for some $x_{0}\in\Omega_{0}.$ Moreover, the minimum eigenvalue $\lambda(x)$ of the Hessian $(u_{ij}(x))$ satisfies $\lambda(\bar{x})=0.$ By a coordinate translation and rotation we may arrange that

[TABLE]

Thus, the origin $0\in\partial B_{\varepsilon}(x_{0})$ and

[TABLE]

Rewrite equation (1.4) as

[TABLE]

where $A(t)=\frac{1}{t+1}$ and $B(t)=(1+t)^{\frac{1-\alpha}{2}}$ are both analytic for $t>-1.$ Differentiating (3.4) twice with respect to $\frac{\partial}{\partial x_{1}},$ we have

[TABLE]

Since $u$ is analytic in $R^{n},$ we expand $u_{11}$ at $x=0$ as a power series to obtain $u_{11}(x)=P_{k}(x)+R(x)$ for all $x\in B_{\delta}(0)$ for some $\delta>0$ such that $\overline{B_{\varepsilon}(x_{0})}\subset B_{\delta}(0)$ (one can choose a smaller $\varepsilon$ in advance if necessary), where $P_{k}(x)$ is the lowest order term, which, by (3.2) and (3.3), is a nonzero homogeneous polynomial of degree $k$, and $R(x)$ is the rest. Note that $k\geq 2$ by the convexity of $u.$ It follows from (3.3) that $u_{ii}u_{11}-(u_{i1})^{2}>0$ in $B_{\varepsilon}(x_{0}).$ Summing over $i$ we have

[TABLE]

for each $i=1,2,\cdots,n.$

We claim that each $u_{i1}$ is of order at least $\frac{k}{2}.$ Otherwise, we expand $u_{i1}$ at $x=0$ as a power series so that the lowest order term $h(x)$ must be a a nonzero homogeneous polynomial. Choose

[TABLE]

so that the segment

[TABLE]

Now restricting (3.6) on $L,$ multiplying the both sides by $t^{-k}$ and then letting $t\to 0^{+},$ we see the limit of the left-hand side of (3.6) is a nonzero constant multiplied by $\Delta u(0)$ which equals to $1$ by (3.4), but the limit of the right-hand side is positive infinite. This is a contradiction.

Therefore, each $u_{i1}$ is of order at least $\frac{k}{2}.$ Hence $u_{ij1},$ $u_{11i}$ and $u_{11ij}$ are of order at least $\frac{k}{2}-1,$ $k-1$ and $k-2$ respectively. Also note that each $u_{i}$ is of order at least 1 by (3.2). With these facts one can check that the right-hand side of equation (3.5) is of order at least of $k;$ while the left-hand side, $\Delta u_{11},$ is either of order $k-2,$ or $\Delta P_{k}=0$ for all $x\in B_{\varepsilon}(x_{0}).$ Since the first case is impossible by comparing the orders of the two sides, we obtain that $P_{k}$ is a harmonic polynomial in $B_{\varepsilon}(x_{0}).$

We claim that $P_{k}\geq 0$ for all $x\in B_{\varepsilon}(x_{0}).$ Otherwise, there exists $a=(a_{1},a_{2},\cdots,a_{n})\in B_{\varepsilon}(x_{0})$ such that $P_{k}(a)<0.$ Then

[TABLE]

which implies $\lim_{t\to 0^{+}}\frac{u_{11}(ta)}{t^{k}}=P_{k}(a)<0$ contradicting the fact that $u_{11}>0$ in $B_{\varepsilon}(x_{0})$ (see (3.3)).

Now we use the strong maximal principle to see that $P_{k}>0$ for all $x\in B_{\varepsilon}(x_{0}).$ But $P_{k}(0)=0,$ and it follows from Hopf’s lemma that $\frac{\partial P_{k}}{\partial\nu}(0)<0,$ where $\nu$ is the unit outward normal to the sphere $\partial B_{\varepsilon}(x_{0}).$ This means that the degree of $P_{k}$ is only one, contradicting the fact $k\geq 2.$ This contradiction proves the lemma.

Proof of Theorem 1.2: it is direct from Lemmas 3.1 and 3.2. Note that if $u(x)\to\infty$ as $|x|\to\infty,$ then $u$ is strictly convex by Lemmas 3.1 and 3.2. Therefore, when $\alpha=1$, we use the results in [13] to know that after a rotation of coordinate system, $\lim_{h\to\infty}h^{-2}u(hx)\to\sum_{i=1}^{k}x_{i}^{2}$ in $R^{n}$ for some $k\geq 2$ by [13; Theorem 1.3] and $u$ is radially symmetric if $n=2$ by [13; Theorem 1.1].

4. Interior Gradient Estimate - Proof of Theorem 1.4

Let $G(x,\xi)=g(x)\varphi(u)\log u_{\xi}(x),$ where $u$ satisfies the hypothesis of theorem 1.4, $g(x)=1-\frac{|x|^{2}}{r^{2}},\,\varphi(u)=1+\frac{u}{M},\,M=\sup_{x\in B_{r(0)}}u(x).$ Suppose that $\sup\{G(x,\xi),\,x\in B_{r}(0),\xi\in S^{n-1}\}$ is attained at point $x_{0}$ and in the direction $e_{1}$. Then at $x_{0},$ $u_{i}(x_{0})=0$ for $i\geq 2$ since directive derivatives attain the maximum along the gradient direction, and so $a_{11}=\frac{1}{1+u_{1}^{2}},\,a_{ii}=1$ for $i\geq 2,$ $a_{ij}=0$ for $i\neq j.$ As the arguments from (1.2) to (1.4) in [14], at $x_{0}$ we have

[TABLE]

and

[TABLE]

Where

[TABLE]

In particular,

[TABLE]

Suppose $G(x_{0},e_{1})$ is large enough so that $logu_{1}>1$ and $|\frac{g^{\prime}}{g}|\leq\frac{\varphi^{\prime}}{2\varphi}u_{1}$ at $x_{0},$ then by (4.1) we can obtain $u_{11}\leq-\frac{\varphi^{\prime}}{2\varphi}u_{1}^{2}\log u_{1}<0.$ Therefore in the case of $\alpha\geq 1$,

[TABLE]

and

[TABLE]

If $0<\alpha<1,$ then there exists a positive integer $m$ such that $\alpha m>1,$ we may suppose that $G(x_{0},e_{1})$ is still suitably large so that $logu_{1}>1$ and $|\frac{g^{\prime}}{g}|\leq\frac{\varphi^{\prime}}{m\varphi}u_{1}$ at $x_{0}$, then by (4.1) we have

[TABLE]

Obviously, in this case (4.2) becomes

[TABLE]

where the constant $C$ depends only on $\alpha.$

To sum up the above two cases we see that at $x_{0}$ and for any $\alpha>0,$

[TABLE]

Recall we may have assumed $\log u_{1}(x_{0})>1.$ Then we obtain

[TABLE]

where $C_{3}$ and $C_{4}$ depend only on $n$ and $\alpha$.

Since $1\leq\varphi\leq 2,$ we use Cauchy inequality to get

[TABLE]

Therefore, we have proved that for any $\xi\in S^{n-1},$

[TABLE]

which implies

[TABLE]

Noting that $\xi$ can be any vector in $S^{n-1},$ we have

[TABLE]

This proves theorem 1.4.

5. Liouville Type Theorem - Proof of Theorem 1.3

Suppose $u\in C^{3}(R^{n})$ is a nonnegative solution of (1.4),and

[TABLE]

We will prove $u\equiv constant$. Since any constant is not a solution to (1.4), the theorem will be proved.

By Theorem 1.4 and (5.1) we have

[TABLE]

We claim that

[TABLE]

Otherwise, there is a $y$ such that $\nabla u(y)\neq 0$ for some $y.$ We may assume $y=0$ and thus

[TABLE]

for some positive $\delta.$ We will induce a contradiction.

Let

[TABLE]

where

[TABLE]

and $\beta\in(0,1)$ is to be determined. Suppose $\sup\{G(x,\xi),\,x\in B_{r}(0),\xi\in S^{n-1}\}$ is attained at point $x_{0}$ and in the direction $e_{1}$.

By (5.3) we have

[TABLE]

for some $\delta_{1}>0$ depending only on $\delta$.

Then at $x_{0}$,

[TABLE]

and

[TABLE]

where we have used (5.5). Note that at $x_{0}$, $u_{i}=0$ for $i\geq 2,$ $a_{11}=\frac{1}{1+u_{1}^{2}},\ a_{ii}=1$ for $i\geq 2,$ and $a_{ij}=0$ for $i\neq j.$ Therefore at $x_{0},$

[TABLE]

By (5.4) we obtain

[TABLE]

and

[TABLE]

where constants $C_{1},C_{2}$ depend only on $n,$ $\delta_{1}$ and $\beta$.

Differentiating equation (1.4) with respect to $x_{1}$, we have

[TABLE]

which implies

[TABLE]

at $x_{0}$. Thus, we obtain that at $x_{0},$

[TABLE]

Observing that (5.5) implies $u_{11}=-\frac{u_{1}^{2}\varphi^{\prime}}{\varphi}-\frac{u_{1}g_{1}}{g}$ at $x_{0}$, we see that

[TABLE]

Since $\delta_{1}<u_{1}\leq C$ and $\lim_{r\to\infty}\frac{r}{M-u}=+\infty$ by (5.1), we have for $r$ large enough,

[TABLE]

Inserting (5.7),(5.8) and (5.9)into (5.6), we obtain that at $x_{0}$ and for large $r$,

[TABLE]

Now chose $\beta\in(0,1)$ so that

[TABLE]

Then (5.10) implies

[TABLE]

at $x_{0}$ and for large $r$. Let $r\rightarrow\infty,$ by (5.1) we obtain $u_{1}(x_{0})=0,$ contradicting (5.4). In this way, we have proved theorem 1.3.