In this paper, we study the radially symmetry of solutions to the equation in the form of

$ \det D^2 u(x)= (1+|Du(x)|^2)^pK(x, u(x), Du(x)), $

(1.1)

where $p=\frac{n+2}{2}-\frac{1}{2\alpha}$ for some number $\alpha$. This equation comes from the well-known Gauss curvature flow, $\nu =[\frac{\hat K}{ K(x, u(x), Du(x))}]^{\alpha} $, where $\nu$ is the velocity along the normal direction of the moving hyper-surface in $R^n$ and $\hat K $ is its Gauss curvature. In fact, (1.1) is exactly the equation of translating solutions to the Gauss curvature flow. See [1, 2] for the details. When $K$ is a positive constant and $\alpha=\frac{1}{n+2}$, the classical results of Jorgens ($n=2$, [3]), Calabi ($n\le 5$, [4]) and Pogorelov ($n\ge 2$, [5]) asserted that any convex solution to (1.1) must be a quadratic polynomial. Obviously, it is radially symmetric under the following condition (1.2). When $K$ is a positive constant and $\alpha\in (0, \frac{1}{2})$, there exists a radially symmetric solution to (1.1) as well as infinitely many smooth, non-radially symmetric convex solutions to (1.1). See Theorem 6 in [2] and Theorem 1.2 in [1], respectively.

When $K(x, u(x), Du(x))=[(xDu(x)-u(x)]^{n+2}$ and $p=0$, (1.1) is the equation of the well-known affine hyperbolic sphere, and its Bernstein property, which may implies the symmetry under the normal condition (1.2), was studied in [6]. Basing a transform introduced by Jian and Wang in [6], the authors in [7] studied the symmetry of solutions of (1.1) with $K(x, Du(x), u(x))=[(xDu(x)-u(x)]^{\beta}F(u(x))$ in the entire space $R^n$. They proved that a $C^4$-strictly convex solution to (1.1) in $R^n$ is the radially symmetric about the lowest point, say for example the origin, provided that

$ u(0)=0, \ \ Du(0)=0, \ \ D^2u(0) =I \ \ {\hbox{(the unit matrix)}}, $

(1.2)

$ u_{x_ix_jx_k}(0)=0, \ \ \forall i, j, k= 1, 2, \cdots, n $

(1.3)

and

$ \lim\limits_{x\to \infty }Du(x)=\infty . $

(1.4)

Obviously, the strict convexity, (1.2) and (1.4) are natural assumptions, which implies that for any small $r>0$,

$ u(x)> 0, \ x\cdot Du(x)-u(x)>b, \ \ \forall x\in \partial B_r(0) $

(1.5)

for some $b>0$. But assumptions (1.3) and the $C^4$-smoothness for the solutions is too restrictive.

Recall that a $C^2$-function $h(x)$ is called strictly convex in $\Omega$ if its Hessian matrix $[D^2h(x)]$ is positive definite in $\Omega.$

In this paper, we study the symmetry of solutions of (1.1) in the domain outside a ball. Interestingly, in this case we can remove assumption (1.3) and the smoothness of the solutions needs only $C^2$. Our main result is the following theorem.

Theorem 1.1 Suppose that $p, \beta \in R$ and

$ K(x, Du(x), u(x))=[(xDu(x)-u(x)]^{\beta}F(u(x)) $

(1.6)

with $F$ satisfies

$ F\in C^1(0, \infty ) , \ \ F(q)>0, \ \ \forall q\in (0, \infty ) $

(1.7)

and

$ \beta \leq n+1-2p. $

(1.8)

If there exist positive constants $c, r, b$ such that $u\in C^2(R^n\backslash B_r(0)) $ satisfies (1.1) in $R^n\backslash \overline{B_r(0)}$, (1.4) and

$ u=c\ \ {\hbox{and}} \ \ x\cdot Du(x)-c>b \ \ {\hbox{on}} \ \ \partial B_r(0), $

(1.9)

and if $u$ is strictly convex, then $u$ is radially symmetric in $R^n\backslash B_r(0) $ about the origin.

By Theorem 1.1 and (1.5), it is easy to see that the main result in [7] still holds true if assumption (1.3) is replaced by the assumption that $u$ is radially symmetric about the origin in $B_r(0) $ for some small $r>0$.

The radial symmetry for the positive solutions of the equations $ \Delta u +f(u)=0$ for $x\in R^n $ was studied in [8] under the ground state conditions: $u(x)\to 0$ as $|x|\to \infty .$ Afterwards, a lot of works were published along this direction, dealing with various symmetry problems for nonlinear elliptic equations. See, for example, [9-13] and the references in [7]. However, all those symmetric results, up to authors' knowledge, need the a priori ground state condition, which can not be satisfied for solutions to Monge-Ampère equations.

The proof of Theorem 1.1 is done in Section 3; in Section 2, some lemmas necessary for the proof are given.

In [6], Jian and Wang introduced a transform which reduces the behavior of $u$ near $\infty $ to that of the new function near the origin and preserves the form of equation (1.1). This transform is

$ y=\frac{x}{u(x)}, \ \ v(y)=\frac{1}{u(x)}, $

(2.1)

where we assume $u\in C^2$ and $u>0$. Then

$ x=\frac{y}{v(y)}, \ \ u(x)=\frac{1}{v(y)}. $

By calculations, we have

$ D_xu(x)= \frac{D_yv(y)}{y \cdot D_yv(y)-v(y)}, \ \ x\cdot D_xu(x)-u(x)=[ y\cdot D_yv(y)-v(y)]^{-1}, $

(2.2)

$ u_{x_ix_j}=\frac{v}{y\cdot Dv-v}(\delta_{ik}-v_{y_i}b_k)(v_{y_ly_k})(\delta_{lj}-b_lv_{y_k}), $

(2.3)

where $ b_k=\frac{y_k}{y\cdot Dv-v}$, and

$ \det D_x^2u(x)= [\frac{v(y)}{y\cdot D_yv(y)-v(y)}]^{n+2}\det (D_y^2v(y)). $

(2.4)

See (2.6)-(2.8) in [6] or (2.3)-(2.5) in [7].

Lemma 2.1    Suppose that a strictly convex function $u\in C^2(R^n\backslash B_r(0))$ satisfies (1.1) in $R^n \backslash \overline{B_r(0)}$, (1.4), (1.6) and (1.9). Let $ y$ and $ v$ be given by (2.1). Then

$ \{ y=\frac{x}{u(x)}: x\in R^n\backslash B_r(0)\}= \overline{B_{\frac{r}{c}}(0)}\backslash \{0\}, \ \ \lim\limits_{y\to 0} v(y)=0 $

(2.5)

and $v \in C^2(\overline{B_{\frac{r}{c}}(0)}\backslash \{0\} )$ is a strictly convex and positive function, satisfying the equation

$ \det D^2 v= [\frac{y\cdot Dv-v}{v}]^{n+2-\beta}F(\frac{1}{v})[1+\frac{ |Dv|^2}{(y\cdot Dv-v)^2}]^p, \ \ \forall y\in B_{\frac{r}{c}}(0)\backslash \{0\} , $

(2.6)

and

$ v(y)>0, \ \ 0 < y\cdot Dv(y)-v(y) <\frac{1}{b}, \forall y\in \overline{B_{\frac{r}{c}}}(0)\backslash \{0\}. $

(2.7)

Proof    Obviously, the convexity implies that $u(x)\geq c>0$ for all $x\in R^n\backslash B_r(0)$. To prove (2.7), we have, by (1.9) and the continuity, that $u(x)-x\cdot Du(x) <-b $ for $x$ near $\partial B_r(0)$. This means the intersecting point, $(0, u(x)-x\cdot Du(x))$, between the supporting plane of the graph of $u$ at $(x, u(x))$ with the $x_{n+1}$-axis is below the point $(0, -b)$. Since the convexity preserves this property as $x$ is away from $\partial B_r(0)$, we obtain

$ u(x)-x\cdot Du(x) <-b , \ \ \forall x\in R^n\backslash B_r(0), $

which, together with (2.2), implies (2.7).

Since $-b < 0, $ the position of the supporting plane of the graph of $u$ shows the graph $\{(x, u(x)): x\, \, {\hbox{near}} \, \, \partial B_r(0)\}$ is above the cone $\{(x, c\frac{|x|}{r}): x\, \, {\hbox{near}} \, \, \partial B_r(0)\}$, so the whole graph $\{(x, u(x)): x\in R^n \backslash \overline{B_r(0)} \}$ is above the whole cone $\{(x, c\frac{|x|}{r}): x\in R^n \backslash \overline{B_r(0)} \}$, i.e.,

$ u(x)>c\frac{|x|}{r}, \ \ \forall x\in R^n \backslash \overline{B_r(0)}, $

which implies

$ |y|=\frac{|x|}{u(x)} <\frac{r}{c}, \ \ \forall x\in R^n \backslash \overline{B_r(0)}. $

Obviously,

$ |y|=\frac{|x|}{u(x)}=\frac{r}{c}, \ \ \forall x\in \partial B_r(0) $

and

$ \lim\limits_{x\to \infty } \frac{x}{u(x)}=0 $

by (1.4) and (1.9). Hence, we obtain (2.5).

Now, by (1.1), (1.6), (2.2) and (2.3), we see that $v \in C^2(\overline{B_{\frac{r}{c}}}(0)\backslash \{0\} )$ is a strictly convex and positive function, satisfying (2.6).

Lemma 2.2  Let $w \in C^2(\Omega ) $ be a nonnegative solution to

$ \sum\limits_{i, j=1}^n a^{ij}(x)\partial _{ij} w + \sum\limits_{i=1}^nb^{i}(x)\partial _{i}w+ C (x)w\leq 0, \forall x \in \Omega , $

(2.8)

where $\Omega $ is an open set in $R^n$, $a^{ij}, b^{i}, C (x)\in L^{\infty }(\Omega ')$ and the matrix $[a^{ij}(x)]$ is positive-definite in $\Omega '$ for any compact set $\Omega '\subset \bar{\Omega } $. Then either $w\equiv 0$ in $\Omega $ or $w(x)>0$ for all $x\in \Omega $. Moreover, if $w\in C^2(\Omega )\bigcap C^1(\bar\Omega )$ and $w(x_0)>0$ for some $x_0\in \Omega $ and $w(\bar x)=0$ for some $\bar x \in \partial \Omega $ which is smooth near $\bar x$, then $\frac{\partial w}{\partial \nu}(\bar x) <0$ where $\nu $ is the unit outer normal of $\partial \Omega $.

The proof of Lemma 2.2 can be found in [14].

Lemma 2.3  Suppose that diam $(\Omega ) \leq d, $ $a^{ij}, b^{i}, C (x)\in L^{\infty }(\Omega )$ and the matrix $[a^{ij}(x)]$ is positive in $\bar{\Omega } $. Let $w \in C^2(\Omega ) $ satisfies (2.8) and $\underline{\lim}_{x\to \partial \Omega } w(x)\geq 0.$ There exists a $\delta>0$ depending only on $n, d$ and the bound of the coefficients such that $w(x)\geq 0$ in $\Omega $ provided that the measure $|\Omega | <\delta .$

This lemma is exactly Proposition 1.1 in [9].

Assume $u\in C^2(R^n)\backslash B_r(0) $ is strictly convex, satisfies (1.1) in $R^n\backslash \overline{B_r(0)}$, (1.4) and (1.9) with $K$ satisfying (1.6)-(1.8). Replacing $u$ by $\frac{u}{c}$, we may assume $c=1$ in Theorem 1.1. Let $y$ and $v(y)$ be given by (2.1). It is enough to prove that $v$ is radially symmetric about the origin in $B_r(0)\backslash \{0\}$.

It follows Lemma 2.1 that Then $v \in C^2(\overline{B_r(0)}\backslash \{0\} )$ is a strictly convex function, satisfying

$ \det D^2v(x)= G(v)(x), \ \forall x\in B_r(0)\backslash \{0\}, $

(3.1)

where

$ \begin{eqnarray} G(v)(x):=[ I(v)(x) ]^{n+2-\beta}\frac{F( v^{-1}(x))}{v^{n+2}(x)}[1+ \frac{ |Dv(x)|^2}{(I(v)(x))^2}]^p, \\ I(v)(x):=x\cdot Dv(x) -v(x)\nonumber\end{eqnarray} $

(3.2)

and

$ v(x)>0, \ \ 0 < x\cdot Dv(x)-v(x) <\frac{1}{b}, \ \ \, \, \forall x\in \overline{B_r}(0)\backslash \{0\} $

(3.3)

and

$ v=1 \ \ {\hbox{on}} \ \ \partial B_r(0), \ \ \lim\limits_{x\to 0}v(x)=0. $

(3.4)

We will use the moving planes method to prove that $v$ is radially symmetric with respect to the origin. This needs to show that $v$ is symmetric in any direction with respect to the origin. Since equation (3.1) is invariant under orthonogal transforms, it is sufficient to do this in one direction. Without loss of generality, we will do it in $e_1$-direction. In a word, to show Theorem 1.1, it is enough to prove that

$ v(-x_1, x_2, \cdots , x_n)= v(x_1, x_2, \cdots , x_n), \ \ \forall x=(x_1, x_2, \cdots , x_n)\in B_r(0) \backslash\{0\}. $

(3.5)

Use $(x_1, x'):=(x_1, x_2, \cdots , x_n) $ to denote a point $x$ in $R^n.$ For any $\lambda \in R, $ denote

$ \begin{eqnarray*}&& B_r^{\lambda }:=\{ x=(x_1, x_2, \cdots , x_n) \in B_r(0) : x_1>\lambda \}, \\ && T_r^{\lambda }:=\{ x=(x_1, x_2, \cdots , x_n) \in \overline{B_r(0) } : x_1=\lambda \}, \\ && x^{\lambda }=(2\lambda -x_1, x'), \, \, \, v_{\lambda } (x)= v(x^{\lambda }), \\ && w_{\lambda }(x)=v(x)-v_{\lambda }(x)\end{eqnarray*} $

and

$ G_{\lambda }(v)(x):=[ I_{\lambda }(v)(x) ]^{n+2-\beta}\frac{F( v_{\lambda }^{-1}(x))}{v_{\lambda }^{n+2}(x)}[1+ \frac{ |Dv_{\lambda }(x)|^2}{(I(v_{\lambda })(x))^2}]^p, $

where

$ I_{\lambda }(v)(x):=\left\{ {x_1v_{x_1}(x)+\sum\limits_{i=2}^nx_i(v_{\lambda })_{x_i}(x)-v_{\lambda }(x), \ \ \quad\hbox{if}~~ v_{x_1}(x^{\lambda })\geq 0, \atop I(v_{\lambda })(x)=\sum\limits_{i=1}^nx_i(v_{\lambda })_{x_i}(x)-v_{\lambda }(x), \quad\hbox{if}~~ v_{x_1}(x^{\lambda }) <0 . }\right. $

As in [7], we introduce the following differential operator

$ L_{\lambda }(v)(x):= \det D^2v(x)-\det D^2v_{\lambda }(x) + G_{\lambda }(v)(x)-G(v)(x). $

By assumption (1.7) and a mean value theorem, we have the following obvious result.

Lemma 3.1  Let $r$ be any positive constant. If $v \in C^2(\overline{B_r(0)}\backslash \{0\})$ is bounded, positive and strictly convex, then for any compact $\Omega $ in $\overline{B_r(0)}\backslash \{0, 0^{\lambda } \}$, there exist a constant $C_1>0$ independent of $\lambda \in (0, \infty )$ (but depending on $\Omega $) and piecewise continuous functions $\{a_{\lambda }^{ij}(x)\}$, $\{b_{\lambda }^{i}(x)\}$, $C_{\lambda } (x)$ (all depending on the $v$ and its derivatives up to second order in $\Omega $), such that

$ L_{\lambda }(v)(x)=a_{\lambda }^{ij}(x)\partial _{ij} w_{\lambda } (x) + b_{\lambda }^{i}(x)\partial _{i}w_{\lambda } (x)+ C_{\lambda } (x)w_{\lambda } (x), \forall x \in \Omega \cap\{x_1>\lambda \} $

and

$ C_1^{-1}I \leq (a_{\lambda }^{ij}(x))\leq C_1I, \, \, | b_{\lambda }^{i}(x)|+ |C_{\lambda } (x)| \leq C_1, \ \ \forall x \in \Omega \cap\{x_1>\lambda \} . $

We will complete the proof of Theorem 1.1 after proving the following three claims.

Claim 1   Here exists a $\bar \lambda \in (\frac{r}{2}, r)$ such that

$ w_{\lambda } >0 \, \, {\hbox{in}} \, \, B_r^{\lambda }\, \, {\hbox{and}} \, \, \frac{\partial v}{\partial x_1}>0 \, \, {\hbox{on}} \, \, T_r^{\lambda } , \ \ \forall \lambda \in (\bar \lambda , r). $

To show it, we see, by (3.1), (1.7) and (3.3), that

$ \Delta v \geq n [\det (D^2v)]^{\frac{1}{n}}>0 \, \, {\hbox{in}} \, \, B_r(0)\backslash\{0\}. $

So, the usual strong maximum principle and Hopf's boundary point lemma in [4] implies

$ 0 <v <1 \, \, {\hbox{in}} \, \, B_r(0)\backslash\{0\} $

(3.6)

and

$ \frac{\partial v}{\partial x_1}>0 \, \, {\hbox{on}} \, \, \partial B_r(0) \bigcap \{x_1>0\}, $

(3.7)

where we have used the fact that the angel between the vector $(x_1, x')\in \partial B_r(0) $ and the out normal of $\partial B_r(0) $ at the point $x=(x_1, x')$ is less than $\frac{\partial i}{2}$ for $x_1>0.$

By continuity and (3.7), we see that that there is a $\lambda _1\in (\frac{r}{2}, r)$ such that

$ \frac{\partial v}{\partial x_1} > 0\ \ {\hbox{on}} \ \ T_r^{\lambda }, \ \ \forall \lambda \in (\lambda _1, r) . $

(3.8)

By (3.4) and (3.6), we see that

$ w_{\lambda }> 0 \ \ {\hbox{on}} \ \ \ \partial B_r^{\lambda }\backslash \{ (x_1, x')\in B_r: x_1=\lambda \} , \forall \lambda \in (0, r). $

(3.9)

To complete the proof, we notice, for $\lambda >0$ and $\lambda <x_1 <2\lambda , $ that

$ \begin{equation*} (2\lambda -x_1) v_{x_1}(x^{\lambda }) <\left\{ \begin{array}{cc} x_1v_{x_1}(x), &{\rm if}\, v_{x_1}(x^{\lambda })\geq 0 , \\ -x_1v_{x_1}(x^{\lambda })=x_1(v_{\lambda })_{x_1}(x), &{\rm if}\, v_{x_1}(x^{\lambda }) <0 , \end{array} \right. \end{equation*} $

where we have used the fact $ v_{x_1}(x^{\lambda }) <v_{x_1}(x)$ by the convexity. It follows that

$ \begin{eqnarray} I(v)(x^{\lambda}) &=& x^{\lambda }\cdot Dv(x^{\lambda })-v(x^{\lambda }) \\ &=& (2\lambda -x_1) v_{x_1} (x^{\lambda })+ \sum\limits_{i=2}^n x_iv_i(x^{\lambda })-v(x^{\lambda }) \\ &=& (2\lambda -x_1) v_{x_1} (x^{\lambda }) + \sum\limits_{i=2}^n x_i(v_{\lambda })_{x_i}(x)-v_{\lambda } (x) \\ &<& I_{\lambda }(v)(x) . \end{eqnarray} $

(3.10)

This, together with assumptions (1.8), implies

$ G(v)(x^{\lambda })> G_{\lambda }(v)(x), \ \ \forall \lambda >0, \ \ \forall x\in B_r^{\lambda } \backslash (B_r^{2\lambda }\cup \{0^{\lambda }\}) . $

(3.11)

Noting $ \det D^2v(x^{\lambda })= \det D^2v_{\lambda }(x)$ and using (3.11), we have, by equation (3.1), that

$ \begin{eqnarray} 0 &=& \det D^2v(x) - G(v)(x)- \det D^2v(x^{\lambda })+ G(v)(x^{\lambda }) \\ &>& L_{\lambda }(v)(x) , \ \ \forall \lambda >0, \ \ \forall x\in B_r^{\lambda } \backslash (B_r^{2\lambda }\cup\{0^{\lambda }\}). \end{eqnarray} $

(3.12)

Observing that the measure of $B_r^{\lambda } $ can be smaller than any positive constant if $r-\lambda _1 >0$ is small. Applying Lemmas 3.1 and 2.3 to (3.12), we obtain a constant ${ \bar \lambda } \in (\lambda _1, r)$ such that $w_{\lambda }\geq 0$ in $B_r^{\lambda } \backslash (B_r^{2\lambda }\bigcup\{0^{\lambda }\})$ for all $ \lambda \in [{\bar \lambda } , r).$ Since $\lambda $ is arbitrary in $[{\bar \lambda } , r)$, we have $w_{\lambda }\geq 0$ in $B_r^{\lambda }$. Furthermore, Lemma 2.2 implies

$ w_{\lambda }> 0\, \, {\hbox{ in}} \, \, B_r^{\lambda }, \ \ \forall \lambda \in [{\bar \lambda } , r) , $

which, together with (3.7) and continuity, complete the proof of Claim 1.

Let

$ Q =\{\lambda \in (0, r ): w_{\lambda }>0\, \, {\hbox{in}} \, \, B_r^{\lambda } \, \, {\hbox{and}} \, \, \frac{\partial v}{\partial x_1}>0 \, \, {\hbox{on}} \, \, T_r^{\lambda }\}. $

Then Claim 1 means that $ [{\bar \lambda }, r) \subset Q.$

Claim 2   The set $Q$ is open in $(0, r ).$ Suppose this Claim is false, i.e., there exist a $\lambda '\in Q$ and a number sequence $\lambda _k$ and a sequence of points $\{x^k\}\subset B_r^{\lambda }$ such that

$ \lim\limits_{k\to \infty } \lambda _k =\lambda '>0 \ \ {\hbox{and}} \ \ w_{\lambda _k}(x^k)\leq 0 , \ \ \ k=1, 2, \cdots . $

(3.13)

Otherwise, we have a sequence $y_k\in T_r^{\lambda _k }$ such that $\frac{\partial v}{\partial x_1}(y_k)\leq 0$, which implies immediately that $\frac{\partial v}{\partial x_1}(y)\leq 0$ for some $y\in T_r^{\lambda '}$, contradicting with $\lambda '\in Q$.

Next, we want to get a contradiction by (3.13). First, by the boundedness of $\{x_k\}$, we can choose a subsequence such that

$ x^k\to x^0 \ \ {\hbox{as}} \ \ k\to \infty . $

(3.14)

Then

$ w_{\lambda '}(x^0)\leq 0\ \ \text{and thus} \ \ x^0\in T_r^{\lambda '}. $

Because $\lambda '\in Q$, we have

$ \frac{\partial v}{\partial x_1}(x^0)>0. $

(3.15)

Sine (3.13) means that $v(x^k)\leq v((x^k)^{\lambda _k})$, and

$ (x_k)_1^{\lambda _k}=2\lambda _k-x_1^k <\lambda _k <x_1^k, $

we see that

$ \frac{\partial v}{\partial x_1}(\xi^k)\leq 0 $

for some $\xi^k $ in the segment connecting $x^k$ and $(x^k)^{\lambda _k}$ for $k=1, 2, \cdots .$ Moreover, $ \xi^k \to x^0$ by (3.14) and the fact $x^0\in T_r^{\lambda '}$. Then, $\frac{\partial v}{\partial x_1}(x^0)\leq 0$, contradicting (3.15). This proves Claim 2.

Claim 3 Let $(\lambda _0 , r )$ be the connected component of $Q$ in $ (0, r)$ containing $[\bar \lambda , r).$ Then $\lambda _0= 0$.

By the definition of $\lambda _0$, we have

$ w_{\lambda _0}\geq 0\, \, {\hbox{in}} \, \, B_r^{\lambda _0} $

(3.16)

and

$ w_{\lambda }> 0\, \, {\hbox{in}} \, \, \ \ B_r^{\lambda } , \ \ \forall \lambda \in (\lambda _0 , r). $

Observing $w_{\lambda }= 0 $ on $T_r^{\lambda }$, by Lemmas 3.1 and 2.2, we see that $\frac{\partial w_{\lambda }}{\partial x_1}>0$ and so $\frac{\partial v}{\partial x_1}>0$ on each $T_r^{\lambda }$ for all $\lambda \in (\lambda _0 , r)$. That is

$ \frac{\partial v}{\partial x_1}>0 \, \, {\hbox{in}} \, \, B_r^{\lambda _0} . $

(3.17)

Suppose the contrary $\lambda _0>0.$ We conclude that

$ w_{\lambda _0}> 0\, \, {\hbox{in}} \, \, B_r^{\lambda _0} . $

(3.18)

Where (3.18) false, $w_{\lambda _0}(\bar x) =0 $ for some $\bar x\in B_r^{\lambda _0} $, which is a minimum point of $w_{\lambda _0}$ in $ B_r^{\lambda _0}$ by (3.16). Then $ \frac{\partial w_{\lambda _0} }{\partial x_1}({\bar x})=0$, which, together with (3.17), implies

$ \frac{\partial v }{\partial x_1}({\bar x}^{\lambda _0}) = - \frac{\partial v }{\partial x_1}(\bar x) <0. $

Set

$ \Omega _0^{\lambda _0} =\{ x: x\in B_r(0) \backslash\{0\} : x_1 <\lambda _0\, \, {\hbox{and}} \, \, \frac{\partial v }{\partial x_1}(x) <0\}. $

Let $\Omega _0 $ be the symmetric set of $\Omega _0^{\lambda _0}$ with respect to the plane $x_1=\lambda _0.$ Then $\Omega _0$ is an open set and $\bar x$ is its interior point, and

$ \frac{\partial v }{\partial x_1}(x^{\lambda _0}) <0, \ \ \forall x\in \Omega _0. $

Recalling the definition of $I_{\lambda _0}(v)(x)$, we have, by the assumption $\lambda _0>0$, that

$ \begin{eqnarray*} I(v)(x^{\lambda _0})&=& x^{\lambda _0 }\cdot Dv(x^{\lambda _0})-v(x^{\lambda _0}) \\ &<& -x_1 v_{x_1} (x^{\lambda _0})+ \sum\limits_{i=2}^n x_iv_{x_i}(x^{\lambda _0 })-v(x^{\lambda _0}) \\ & =&x_1 (v_{\lambda _0})_{x_1} (x) + \sum\limits_ {i=2}^n x_i(v_{\lambda _0})_{x_i}(x)-v_{\lambda _0} (x) \\ & =& I_{\lambda _0}(v)(x)\end{eqnarray*} $

for all $x\in \Omega _0.$

Hence, (3.10)-(3.12) hold for $\lambda =\lambda _0$ and all $x\in \Omega _0, $ i.e.,

$ L_{\lambda _0}(v)(x) <0, \ \ \forall x\in \Omega _0. $

(3.19)

Using Lemmas 2.2 and 3.1, we obtain that $w_{\lambda _0}\equiv 0$ in a ball $B_\delta (\bar x)$ contained in $\Omega _0$. Therefore $ v\equiv v_{\lambda _0}$ and so $L_{\lambda _0}(v)\equiv 0$ in $B_\delta (\bar x), $ contradicting (3.19). This proves (3.18).

With (3.18) in hands, we use Lemmas 2.2 and 3.1 again to obtain

$ \frac{\partial w_{\lambda _0}}{\partial x_1}> 0\ \ {\hbox{on}} \ \ T_r^{\lambda _0}, $

which implies

$ \frac{\partial v }{\partial x_1}> 0\ \ {\hbox{on}} \ \ T_r^{\lambda _0}. $

This and (3.18) mean $\lambda _0\in Q$, contradicting the definition of $\lambda _0$. In this way, we have proved Claim 3.

Now we complete the proof of Theorem 1.1. Since $\lambda _0 =0 , $ by Claim 3 and (3.16), we have

$ v(x_1, x')\geq v(-x_1, x'), \forall x=(x_1 , x')\in B_r(0) \, , x_1 > 0. $

The opposite inequality is also true, because $V(x):=v(-x_1, x')$ is a solution to (3.1) in $ B_r(0)\setminus\{0\}$ and the same conditions as $v$ holds for $V$. This proves (3.5) and thus Theorem 1.1.