Symmetry Results for classical solutions of Monge-Ampere system in the plane

TL;DR

This paper proves that classical solutions to a Monge-Ampere system in the plane are symmetric about a point using the moving planes method, linking the results to a differential geometry problem.

Contribution

It establishes symmetry of solutions for a Monge-Ampere system in the plane, a novel application of the moving planes method in this context.

Findings

Solutions are symmetric about some point in the plane.

The symmetry result applies to systems arising from differential geometry.

Method demonstrates the effectiveness of moving planes for Monge-Ampere systems.

Abstract

In this paper, by the method of moving planes, we prove the symmetry result which says that classical solutions of Monge-Ampere system in the whole plane are symmetric about some point. Our system under consideration comes from the differential geometry problem.

Equations

$$K_1(x,u,v)=\frac{det{D}^{2}u(x)}{(1+|Du(x){|}^2{)}^2},in{\mathbf{R}}^2$$

$$K_2(x,u,v)=\frac{det{D}^{2}v(x)}{(1+|Dv(x){|}^2{)}^2},in{\mathbf{R}}^2.$$

$$\displaystyle\left\{\begin{array}[]{l@{\quad\quad}l}det(D^{2}u)+g(u,v,\nabla u)=0,&in\ \mathbf{R}^{2},\\ det(D^{2}v)+f(u,v,\nabla v)=0,&in\ \mathbf{R}^{2},\\ \ (D^{2}u)>0,\ (D^{2}v)>0,&in\ \mathbf{R}^{2},\\ \lim_{\{|x|\to\infty;x_{1}<\lambda\}}(u(x)-u(x^{\lambda}))\geq 0,&for\ \lambda<0,\\ \lim_{\{|x|\to\infty;x_{1}<\lambda\}}(v(x)-v(x^{\lambda}))\geq 0,&for\ \lambda<0,\\ \lim_{\{|x|\to\infty;x_{1}>\lambda\}}(u(x)-u(x^{\lambda}))\geq 0,&for\ \lambda>0,\\ \lim_{\{|x|\to\infty;x_{1}>\lambda\}}(v(x)-v(x^{\lambda}))\geq 0,&for\ \lambda>0,\\ \lim_{|x|\to\infty}u(x)=\lim_{|x|\to\infty}v(x)=+\infty,\\ \lim_{|x|\to\infty}|\nabla u(x)|=\lim_{|x|\to\infty}|\nabla v(x)|=\infty,\end{array}\right.$$

$$(i)$$

$$(ii)$$

$$(iii)$$

$$(iv)$$

$${\Sigma }_{\lambda }:=\{x\in {\mathbf{R}}^2|x_1<\lambda \},T_{\lambda }:=\partial {\Sigma }_{\lambda }=\{x\in {\mathbf{R}}^2|x_1=\lambda \}.$$

$$u_{\lambda }(x):=u(x^{\lambda }),v_{\lambda }(x):=v(x^{\lambda }),$$

$$U_{\lambda }(x):=u_{\lambda }(x)-u(x),V_{\lambda }(x):=v_{\lambda }(x)-v(x).$$

$$det(D^2{u}_{\lambda })-det(D^{2}u)=a_{ij}{U}_{\lambda ,ij},$$

$$(a_{ij}(x))=\frac{1}{2}(det(D^2{u}_{\lambda })(D^2{u}_{\lambda }{)}^{-1}+det(D^{2}u)(D^{2}u{)}^{-1}).$$

$$(a_{ij}(x))>0,x\in {\Sigma }_{\lambda }.$$

$$det(D^2{u}_{\lambda })(x)$$

$$a_{ij}{U}_{\lambda ,ij}+\frac{\partial g}{\partial p_1}(u_{\lambda },v_{\lambda },{\theta }_1(x,\lambda ),u_{\lambda ,2})U_{\lambda ,1}+\frac{\partial g}{\partial p_2}(u_{\lambda },v_{\lambda },u_{\lambda ,1},{\theta }_2(x,\lambda ))U_{\lambda ,2}$$

$$+\frac{\partial g}{\partial u}({\xi }_1(x,\lambda ),v,\nabla u)U_{\lambda }+\frac{\partial g}{\partial v}(u_{\lambda },{\eta }_1(x,\lambda ),\nabla u)V_{\lambda }=0,$$

$$b_{ij}{V}_{\lambda ,ij}+\frac{\partial f}{\partial p_1}(u_{\lambda },v_{\lambda },{\tau }_1(x,\lambda ),v_{\lambda ,2})V_{\lambda ,1}+\frac{\partial f}{\partial p_2}(u_{\lambda },v_{\lambda },v_{\lambda ,1},{\tau }_2(x,\lambda ))V_{\lambda ,2}$$

$$+\frac{\partial f}{\partial u}({\xi }_2(x,\lambda ),v_{\lambda },\nabla v)U_{\lambda }+\frac{\partial f}{\partial v}(u,{\eta }_2(x,\lambda ),\nabla v)V_{\lambda }=0,$$

$${\xi }_i(x,\lambda )$$

$${\theta }_i(x,\lambda )\in (\overline{u_i(x),u_{\lambda ,i}(x)}),$$

$${\tau }_i(x,\lambda )\in (\overline{v_i(x),v_{\lambda ,i}(x)}).$$

$${\Sigma }_{\lambda }^{U^{-}}:=\{x\in {\Sigma }_{\lambda }|U_{\lambda }(x)>0\},{\Sigma }_{\lambda }^{V^{-}}:=\{x\in {\Sigma }_{\lambda }|V_{\lambda }(x)>0\}.$$

$$\forall x\in \{{\mathbf{R}}^2\backslash B_{R_1}(0)\}\cap {\Sigma }_{\lambda }^{U^{-}};\forall y\in \{{\mathbf{R}}^2\backslash B_{R_1}(0)\}\cap {\Sigma }_{\lambda }^{V^{-}}$$

$$\frac{\partial g}{\partial u}(z,v(x),\nabla u(x))<0,\frac{\partial f}{\partial v}(u(y),w,\nabla v(y))<0,$$

$$\forall y_1,y_2\in \{{\mathbf{R}}^2\backslash B_{R_2}(0)\}\cap {\Sigma }_{\lambda }^{U^{-}}\cap {\Sigma }_{\lambda }^{V^{-}},$$

$$\frac{\partial g}{\partial u}(z_1,v(y_1),\nabla u(y_1))\frac{\partial f}{\partial v}(u(y_2),w_2,\nabla v(y_2))$$

$$-\frac{\partial g}{\partial v}(u_{\lambda }(y_1),w_1,\nabla u(y_1))\frac{\partial f}{\partial u}(z_2,v_{\lambda }(y_2),\nabla v(y_2))>0,$$

$$\frac{\partial g}{\partial u}(z_1,v(y_1),\nabla u(y_1))\frac{\partial f}{\partial v}(u(y_2),w_2,\nabla v(y_2))$$

$$-\frac{\partial g}{\partial v}(u_{\lambda }(y_1),w_1,\nabla u(y_1))\frac{\partial f}{\partial u}(z_2,v_{\lambda }(y_2),\nabla v(y_2))$$

$$\ge \frac{1}{2}\{\frac{\partial g}{\partial u}(z_1,v(y_1),\nabla u(y_1))\frac{\partial f}{\partial v}(u(y_1),v(y_1),\nabla v(y_1))$$

$$-\frac{\partial g}{\partial v}(u(y_1),v(y_1),\nabla u(y_1))\frac{\partial f}{\partial u}(u(y_1),v(y_1),\nabla v(y_1))\}$$

$$\to \underset{|u{|}^{-1}+|v{|}^{-1}+|(p_1,p_2){|}^{-1}\to 0}{\lim }\frac{1}{2}(\frac{\partial g}{\partial u}\frac{\partial f}{\partial v}-\frac{\partial g}{\partial v}\frac{\partial f}{\partial u}){|}_{(u,v,p_1,p_2)}>0(asR\to +\infty ),$$

$$U_{\lambda }(x)\equiv V_{\lambda }(x)\equiv 0,\forall x\in {\Sigma }_{\lambda }.$$

$$U_{\lambda }(y_1)=\underset{y\in {\bar{\Sigma }}_{\lambda }}{\max }{U}_{\lambda }(y)>0.$$

$$\frac{\partial g}{\partial u}({\xi }_1(y_1,\lambda ),v(y_1),\nabla u(y_1))U_{\lambda }(y_1)+\frac{\partial g}{\partial v}(u_{\lambda }(y_1),{\eta }_1(y_1,\lambda ),\nabla u(y_1))V_{\lambda }(y_1)\ge 0.$$

Full text

Keyword: Moving plane, positive solutions, radial symmetric, Monge-Ampere system

Mathematics Subject Classification: 35J60, 53C21, 58J05

∗ The research is partially supported by the National Natural Science Foundation of China 10631020 and SRFDP 20060003002.

1. Introduction

It is an interesting question to consider the relation between two convex surfaces in space. There are at least two famous theorems for convex surfaces in 3-space. One is the Cohn-Vossen theorem, which says that if two closed convex surfaces differ by an isometry, then they are the same by a translation. The other is the Minkowski theorem, which says that if two closed convex surfaces share the same Gauss curvature, then they are differ only by a translation. For a simple and beautiful proof of the latter, one may see the paper of S.S.Chern [4]. We study the relation between two non-compact convex surfaces in space, which leads us to consider a symmetry result for the Monge-Ampere system in the plane $\mathbf{R}^{2}$. Assume that we are given two functions $K_{1}(x,u,v)$ and $K_{2}(x,u,v)$ for $(x,u,v)\in\mathbf{R}^{2}\times\mathbf{R}^{2}$. We are looking for a pair of function $u=u(x)$ and $v=v(x)$ with their graphs $\Gamma(x,u(x))$ and $\Gamma(x,v(x))$ such that $K_{1}(x,u(x),v(x))$ and $K_{2}(x,u(x),v(x))$ are Gaussian curvatures of the graphs $\Gamma(x,u(x))$ and $\Gamma(x,v(x))$ respectively. Then we are lead to solve the Monge-Ampere system

[TABLE]

and

[TABLE]

Single Monge-Ampere equation (with Dirichlet boundary condition) has been studied by many authors, one may see [3] and [11] for more references.

We study here a little more generalized version of the Monege-Ampere system above. We study the symmetry result for classical solutions of the Monege-Ampere system:

[TABLE]

for $x^{\lambda}=(2\lambda-x_{1},x_{2})$ is the reflection of the point $x$ with respect to any line $T_{\lambda}=\{x_{1}=\lambda\}$, where $f,g\in C^{1}(\mathbf{R}^{2}\times\mathbf{R}^{2},\mathbf{R})$.

We suppose that:

[TABLE]

where $C_{i}$ are positive constants.

Our result is the following.

Theorem 1**.**

Let $(u,v)$ be a classical solution of system (10). Suppose (i)-(iv) hold, then the solution $(u,v)$ is symmetric in $x_{1}$ direction, i.e., there is some $\lambda$ such that $u(x_{1},x_{2})=u(2\lambda-x_{1},x_{2})$ and $v(x_{1},x_{2})=v(2\lambda-x_{1},x_{2})$.

Notation: We shall use $(\overline{A,B})$ to denote the open interval in the line from $A$ to $B$.

The method to prove Theorem 1 is motivated by [2], [8], [10], and [9]. We use the method of moving planes, which has been used in [5], [1],[6], and [7].

2. Monge-Ampere System in $\mathbf{R}^{2}$

We prove our theorem in this section.

In what follows, we shall use the method of moving plane. To proceed, we start by considering lines parallel to $x_{1}=0$, coming from $-\infty$. For each $\lambda\in\mathbf{R}$, we define

[TABLE]

For any point $x=(x_{1},x_{2})\in\Sigma_{\lambda}$, let $x^{\lambda}=(2\lambda-x^{1},x_{2})$ be the reflected point with respect to the line $T_{\lambda}$. We define the reflected functions by

[TABLE]

and introduce the functions

[TABLE]

By the definition of $u_{\lambda}$, $det(D^{2}u_{\lambda})(x)=det(D^{2}u)(x^{\lambda})$. By using the integral form of the theorem of the mean, we obtain:

[TABLE]

where

[TABLE]

Noticing that the solution $(u,v)$ are convex everywhere, we have

[TABLE]

On the other hand, by using (i), we have

[TABLE]

Similarly we have the elliptic equation for $V_{\lambda}$ with the elliptic coefficient matrix $(b_{ij})=(b_{ij}(x))$.

Therefore, we find that $(U_{\lambda},V_{\lambda})$ satisfies an elliptic linear system: in $\Sigma_{\lambda}$

[TABLE]

[TABLE]

where for $i=1,2,$

[TABLE]

[TABLE]

[TABLE]

We define

[TABLE]

For simplicity, we first give two lemmata.

Lemma 2**.**

Assume (i)-(iv) hold. Then there exists a constant $R_{1}>0$, such that

[TABLE]

the following hold

[TABLE]

where $z\in(\overline{u_{\lambda}(x),u(x)})$ and $w\in(\overline{v_{\lambda}(y),v(y)})$ arbitrarily.

Proof.

Using (iii), we choose $\epsilon>0$ such that $\frac{\partial g}{\partial u}(u,v,p_{1},p_{2})<0$, if $|u|^{-1}+|v|^{-1}+|(p_{1},p_{2})|^{-1}<\epsilon$. For this particular $\epsilon$, since $\lim_{|x|\to+\infty}(|u(x)|^{-1}+|\nabla u|^{-1})=0$ and $\lim_{|x|\to+\infty}|v(x)|^{-1}=0$, there exists a constant $R_{1}>0$, such that $|u(x)|^{-1}+|v(x)|^{-1}+|\nabla u|^{-1}(x)<\epsilon$, if $|x|>R_{1}$. For all $x\in\{\mathbf{R}^{2}\backslash B_{R_{1}}(0)\}\cap\Sigma_{\lambda}^{U^{-}}$, $|u(x)|^{-1}+|v(x)|^{-1}+|\nabla u|^{-1}(x)<\epsilon$. Notice that in this case $z\in(\overline{u_{\lambda}(x),u(x)})=(u(x),u_{\lambda}(x))$. Thus, $|z|^{-1}+|v(x)|^{-1}+|\nabla u|^{-1}(x)<\epsilon$, $\forall z\in(\overline{u_{\lambda}(x),u(x)})$. Therefore, $\frac{\partial g}{\partial u}(z,v(x),\nabla u(x))<0$, $\forall z\in(\overline{u_{\lambda}(x),u(x)})$. Similarly, when $R_{1}$ is sufficiently large, $\frac{\partial f}{\partial v}(u(y),w,\nabla v(y))<0$, if $y\in\{\mathbf{R}^{2}\backslash B_{R_{1}}(0)\}\cap\Sigma_{\lambda}^{V^{-}}$. ∎

Lemma 3**.**

Assume (i)-(iv) hold. Then there exists a constant $R_{2}>0$, such that

[TABLE]

the following holds:

[TABLE]

*where $z_{i}\in(\overline{u_{\lambda}(y_{i}),u(y_{i})})$, $w_{i}\in(\overline{v_{\lambda}(y_{i}),v(y_{i})})$ $(i=1,2)$ arbitrarily.

Proof.

Since $f,g\in C^{1}(\mathbf{R}^{2}\times\mathbf{R}^{2},\mathbf{R})$ and the fact that the solutions growing to infinity at infinity, for all $R\geq 0$, $\forall\ y_{1},y_{2}\in\{\mathbf{R}^{2}\backslash B_{R}(0)\}\cap\Sigma_{\lambda}^{U^{-}}\cap\Sigma_{\lambda}^{V^{-}}$, $\frac{\partial g}{\partial u}(z_{1},v(y_{1}),\nabla u(y_{1}))$ and $\frac{\partial g}{\partial v}(z_{1},v(y_{1}),\nabla u(y_{1}))$ are bounded. Using the continuity of $\frac{\partial f}{\partial u}$ and $\frac{\partial f}{\partial v}$, we have

[TABLE]

which proves the lemma. ∎

We now give our Theorem 1, which will be proved by the use of the maximum principle and the method of moving planes as in [2].

Proof.

We just have to show that for $\lambda=0$ there hold

[TABLE]

In order to show this, we will apply the moving planes methods in three steps.

Step 1: There exists $\lambda^{*}<0$ such that $U_{\lambda}\leq 0$ and $V_{\lambda}\leq 0$ in $\Sigma_{\lambda}$, for all $\lambda\leq\lambda^{*}$.

Fix $\lambda^{*}<-\max\{R_{1},R_{2}\}$, then for all $\lambda<\lambda^{*}$, we have $\forall x\in\Sigma_{\lambda}$, $|x|>\max\{R_{1},R_{2}\}$. Assume for contradiction that there is a $\lambda<\lambda^{*}$ and a point $y_{0}\in\Sigma_{\lambda}$, such that $U_{\lambda}(y_{0})>0$.

Since $\lim_{|x|\to+\infty}U_{\lambda}(x)=0$ and $U_{\lambda}(x)\equiv 0$, $x\in T_{\lambda}=\partial\Sigma_{\lambda}$, we may take $y_{1}\in\Sigma_{\lambda}$, such that

[TABLE]

At point $y_{1}$, we have $(U_{\lambda,ij})(y_{1})\leq 0$ and $\nabla U_{\lambda}(y_{1})=0$. Thus, (16) turns to be

[TABLE]

Notice that $y_{1}\in\{\mathbf{R}^{2}\backslash B_{R_{1}}(0)\}\cap\Sigma_{\lambda}^{U^{-}}$, by using Lemma 2, we have

[TABLE]

Combining this with the assumption (ii), we have $V_{\lambda}(y_{1})>0$, i.e. $y_{1}\in\{\mathbf{R}^{2}\backslash B_{R_{2}}(0)\}\cap\Sigma_{\lambda}^{U^{-}}\cap\Sigma_{\lambda}^{V^{-}}$. On the other hand, we also have

[TABLE]

hence, we can take $y_{2}\in\Sigma_{\lambda}$ such that

[TABLE]

We can repeat the above argument for (17) and show that $U_{\lambda}(y_{2})>0$ i.e. $y_{2}\in\{\mathbf{R}^{2}\backslash B_{R_{2}}(0)\}\cap\Sigma_{\lambda}^{U^{-}}\cap\Sigma_{\lambda}^{V^{-}}$ and

[TABLE]

Let us put

[TABLE]

[TABLE]

[TABLE]

[TABLE]

By using (22) and (23), we obtain

[TABLE]

which implies

[TABLE]

However, as we have shown $y_{1},y_{2}\in\{\mathbf{R}^{2}\backslash B_{R_{2}}(0)\}\cap\Sigma_{\lambda}^{U^{-}}\cap\Sigma_{\lambda}^{V^{-}}$, by the result of Lemma 3,

[TABLE]

which leads to a contradiction. **Step 1 is completed.

**

We now move the plane $T_{\lambda}$ toward right, i.e. increase the value of $\lambda$, as long as both $U_{\lambda}(x)\leq 0$ and $V_{\lambda}(x)\leq 0$ hold. Define:

[TABLE]

Step 1 implies that $\lambda_{0}>-\infty$. On the other hand, as $u(x)>u(0)$, $\forall|x|>R$, for some sufficiently large $R$, $\lambda_{0}<+\infty$. We want to show that $\lambda_{0}\geq 0$. Assume not, i.e., $\lambda_{0}<0$.

Since all objects we consider are continuous with respect to $\lambda$, we know that

[TABLE]

Then it follows from (16) (17) and (ii) that in $\Sigma_{\lambda_{0}}$:

[TABLE]

[TABLE]

where $\theta_{i},\tau_{i},\eta_{i}$ are the same as in (16) and (17).

Step 2. Either $U_{\lambda_{0}}\equiv 0$ or $V_{\lambda_{0}}\equiv 0$ hold in $\Sigma_{\lambda_{0}}$.

Suppose $U_{\lambda_{0}}\nequiv 0$ and $V_{\lambda_{0}}\nequiv 0$, in $\Sigma_{\lambda_{0}}$. Thus we can choose $R_{3}>\max\{R_{1},R_{2}\}$ s.t. $U_{\lambda_{0}}\nequiv 0$ and $V_{\lambda_{0}}\nequiv 0$ in $\Sigma_{\lambda_{0}}\cap B_{R_{3}}(0)$.

The definition of $\lambda_{0}$ implies that there exist sequences $\{\lambda_{k}\}_{k=1}^{\infty}\subset\mathbf{R}$ and $\{y_{k}\}_{k=1}^{\infty}\subset\mathbf{R}^{2}$ such that $\lambda_{k}>\lambda_{0}$, $\lim_{k\to\infty}\lambda_{k}=\lambda_{0}$, $y_{k}\in\Sigma_{\lambda_{k}}$, and either $U_{\lambda_{k}}(y_{k})>0$, or $V_{\lambda_{k}}(y_{k})>0$. Taking $U_{\lambda_{k}}(y_{k})>0$ (up to a subsequence) as an example, we can rename $y_{k}$ to be the maximum points, i.e.

[TABLE]

Then there are two cases could happen.

Case 1. The sequence $\{y_{k}\}_{k=1}^{\infty}$ contains a bounded subsequence.

Without loss of generality, we assume $\{y_{k}\}_{k=1}^{\infty}\subset\{\Sigma_{\lambda_{0}}\cap B_{R_{3}-1}(0)\}.$

[TABLE]

Noticing that since $(a_{ij})$ is strictly convex in $\Sigma_{\lambda_{0}}\cap B_{R_{3}}(0)$, we can apply the Maximum Principle and the Hopf’s lemma to (LABEL:eq:u) in $\{\Sigma_{\lambda_{0}}\cap B_{R_{3}}(0)\}$.

From (27), we have $U_{\lambda_{k}}(y_{k})>0$, $D^{2}U_{\lambda_{k}}(y_{k})\geq 0$, and $\nabla U_{\lambda_{k}}(y_{k})=0$, which imply

[TABLE]

Since $U_{\lambda_{0}}(x)\leq 0$ $\forall x\in\Sigma_{\lambda_{0}}$, we have $U(y_{0})=0$. Using the Maximum Principle, we have either (a) $U_{\lambda_{0}}\equiv 0$ in $\{\Sigma_{\lambda_{0}}\cap B_{R_{3}}(0)\}$, or (b) $U_{\lambda_{0}}<0$ $\forall x\in\{\Sigma_{\lambda_{0}}\cap B_{R_{3}}(0)\}$. Since we haven chosen $R_{3}$ to be a large constant such that (a) does not happen, (b) mush hold. Therefore, $y_{0}\in\partial\{\Sigma_{\lambda_{0}}\cap B_{R_{3}}(0)\}$. By using Hopf’s lemma, we have $\frac{\partial U_{\lambda_{0}}}{\partial\nu}(y_{0})>0$, ($\nu$ is the outward normal vector on $\partial\{\Sigma_{\lambda_{0}}\cap B_{R_{3}}(0)\}$), which contradicts the second equation in (28).

Case 2. $\lim_{k\to\infty}|y_{k}|=+\infty$.

In this case, we can choose $k^{*}>0$ such that if $k>k^{*}$, $|y_{k}|>\max\{R_{1},R_{2}\}$, where $R_{1},R_{2}$ as in Lemma 2 and Lemma 3. For any fixed $k>k^{*}$, exactly as in Step 1, we can show $V_{\lambda_{k}}(y_{k})>0$. After choosing

[TABLE]

we also have $U_{\lambda_{k}}(x_{k})>0$. Therefore,

[TABLE]

Similarly defining

[TABLE]

[TABLE]

[TABLE]

[TABLE]

same as in Step 1, we can show $J_{11}(\lambda_{k})J_{22}(\lambda_{k})-J_{12}(\lambda_{k})J_{21}(\lambda_{k})\leq 0,$ which contradicts Lemma 3. **Step 2 is completed.

Step 3.** Both $U_{\lambda_{0}}\equiv 0$ and $V_{\lambda_{0}}\equiv 0$ hold in $\Sigma_{\lambda_{0}}$.

From Step 2, we let $U_{\lambda_{0}}\equiv 0$ for example, i.e. $u(x)$ is symmetry about $T_{\lambda_{0}}$. Putting $U_{\lambda_{0}}$ into (16) and using the assumption (ii), we obtain $V_{\lambda_{0}}\equiv 0$. Assume that $\lambda_{0}\geq 0$. Doing the same for planes from the right, we know that there is some $\lambda_{1}$ such that both $U_{\lambda_{1}}\equiv 0$ and $V_{\lambda_{1}}\equiv 0$ hold in $\Sigma_{\lambda_{1}}^{c}$.Step 3 is completed.

∎