Boundary Value Problems

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Symmetry of solutions to parabolic Monge-Ampère equations

Boundary Value Problems20132013:185

https://doi.org/10.1186/1687-2770-2013-185

Accepted: 5 August 2013

Published: 20 August 2013

Abstract

In this paper, we study the parabolic Monge-Ampère equation

Using the method of moving planes, we show that any parabolically convex solution is symmetric with respect to some hyperplane. We also give a counterexample in ${\mathbb{R}}^{n}×\left(0,T\right]$ and an example in a cylinder to illustrate the results.

MSC:35K96, 35B06.

Keywords

parabolic Monge-Ampère equations symmetry method of moving planes

1 Introduction

The Monge-Ampère equation has been of much importance in geometry, optics, stochastic theory, mass transfer problem, mathematical economics and mathematical finance theory. In optics, the reflector antenna system satisfies a partial differential equation of Monge-Ampère type. In [1, 2], Wang showed that the reflector antenna design problem was equivalent to an optimal transfer problem. An optimal transportation problem can be interpreted as providing a weak or generalized solution to the Monge-Ampère mapping problem [3]. More applications of the Monge-Ampère equation and the optimal transportation can be found in [3, 4]. In the meantime, the Monge-Ampère equation turned out to be the prototype for a class of questions arising in differential geometry.

For the study of elliptic Monge-Ampère equations, we can refer to the classical papers [57] and the study of parabolic Monge-Ampère equations; see the references [811]etc. The parabolic Monge-Ampère equation $-{u}_{t}det\left({D}^{2}u\right)=f$ was first introduced by Krylov [12] together with the other parabolic versions of elliptic Monge-Ampère equations; see [8] for a complete description and related results. It is also relevant in the study of deformation of surfaces by Gauss-Kronecker curvature [13, 14] and in a maximum principle for parabolic equations [15]. Tso [15] pointed out that the parabolic equation $-{u}_{t}det\left({D}^{2}u\right)=f$ is the most appropriate parabolic version of the elliptic Monge-Ampère equation $det\left({D}^{2}u\right)=f$ in the proof of Aleksandrov-Bakelman maximum principle of second-order parabolic equations. In this paper, we study the symmetry of solutions to the parabolic Monge-Ampère equation
$-{u}_{t}det\left({D}^{2}u\right)=f\left(t,u\right),\phantom{\rule{1em}{0ex}}\left(x,t\right)\in Q,$
(1.1)
$u=0,\phantom{\rule{1em}{0ex}}\left(x,t\right)\in SQ,$
(1.2)
$u={u}_{0}\left(x\right),\phantom{\rule{1em}{0ex}}\left(x,t\right)\in BQ,$
(1.3)

where ${D}^{2}u$ is the Hessian matrix of u in x, $Q=\mathrm{\Omega }×\left(0,T\right]$, Ω is a bounded and convex open subset in ${\mathbb{R}}^{n}$, $SQ=\partial \mathrm{\Omega }×\left(0,T\right)$ denotes the side of Q, $BQ=\overline{\mathrm{\Omega }}×\left\{0\right\}$ denotes the bottom of Q, and ${\partial }_{p}Q=SQ\cup BQ$ denotes the parabolic boundary of Q, f and ${u}_{0}$ are given functions.

There is vast literature on symmetry and monotonicity of positive solutions of elliptic equations. In 1979, Gidas et al. [16] first studied the symmetry of elliptic equations, and they proved that if $\mathrm{\Omega }={\mathbb{R}}^{n}$ or Ω is a smooth bounded domain in ${\mathbb{R}}^{n}$, convex in ${x}_{1}$ and symmetric with respect to the hyperplane $\left\{x\in {\mathbb{R}}^{n}:{x}_{1}=0\right\}$, then any positive solution of the Dirichlet problem
$\begin{array}{r}\mathrm{\Delta }u+f\left(u\right)=0,\phantom{\rule{1em}{0ex}}x\in \mathrm{\Omega },\\ u=0,\phantom{\rule{1em}{0ex}}x\in \partial \mathrm{\Omega }\end{array}$
satisfies the following symmetry and monotonicity properties:
$u\left(-{x}_{1},{x}_{2},\dots ,{x}_{n}\right)=u\left({x}_{1},{x}_{2},\dots ,{x}_{n}\right),$
(1.4)
${u}_{{x}_{1}}\left({x}_{1},{x}_{2},\dots ,{x}_{n}\right)<0\phantom{\rule{1em}{0ex}}\left({x}_{1}>0\right).$
(1.5)

The basic technique they applied is the method of moving planes first introduced by Alexandrov [17] and then developed by Serrin [18]. Later the symmetry results of elliptic equations have been generalized and extended by many authors. Especially, Li [19] considered fully nonlinear elliptic equations on smooth domains, and Berestycki and Nirenberg [20] found a way to deal with general equations with nonsmooth domains using the maximum principles on domains with small measure. Recently, Zhang and Wang [21] investigated the symmetry of the elliptic Monge-Ampère equation $det\left({D}^{2}u\right)={e}^{-u}$ and they got the following results.

Let Ω be a bounded convex domain in ${\mathbb{R}}^{n}$ with smooth boundary and symmetric with respect to the hyperplane $\left\{x\in {\mathbb{R}}^{n}:{x}_{1}=0\right\}$, then each solution of the Dirichlet problem
$\begin{array}{r}det\left({D}^{2}u\right)={e}^{-u},\phantom{\rule{1em}{0ex}}x\in \mathrm{\Omega },\\ u=0,\phantom{\rule{1em}{0ex}}x\in \partial \mathrm{\Omega }\end{array}$

has the above symmetry and monotonicity properties (1.4) and (1.5). Extensions in various directions including degenerate problems [22] or elliptic systems of equations [23] were studied by many authors.

For the symmetry results of parabolic equations on bounded and unbounded domains, the reader can be referred to [16, 24, 25] and the references therein. In particular, when $Q=\mathrm{\Omega }×J$, $J=\left(0,T\right]$, Gidas et al. [16] studied parabolic equations $-{u}_{t}+\mathrm{\Delta }u+f\left(t,r,u\right)=0$ and $-{u}_{t}+F\left(t,x,u,Du,{D}^{2}u\right)=0$, and they proved that parabolic equations possessed the same symmetry as the above elliptic equations. When $J=\left(0,\mathrm{\infty }\right)$, Hess and Poláčik [25] first studied the asymptotic symmetry results for classical, bounded, positive solutions of the problem
${u}_{t}-\mathrm{\Delta }u=f\left(t,u\right),\phantom{\rule{1em}{0ex}}\left(x,t\right)\in \mathrm{\Omega }×J,$
(1.6)
$u=0,\phantom{\rule{1em}{0ex}}\left(x,t\right)\in \partial \mathrm{\Omega }×J.$
(1.7)

The symmetry of general positive solutions of parabolic equations was investigated in [24, 26, 27] and the references therein. A typical theorem of $J=\mathbb{R}$ is as follows.

Let Ω be convex and symmetric in ${x}_{1}$. If u is a bounded positive solution of (1.6) and (1.7) with $J=\mathbb{R}$ satisfying
$\underset{t\in \mathbb{R}}{inf}u\left(x,t\right)>0\phantom{\rule{1em}{0ex}}\left(x\in \mathrm{\Omega },t\in J\right),$
then u has the symmetry and monotonicity properties for each $t\in \mathbb{R}$:
$\begin{array}{c}u\left(-{x}_{1},{x}^{\mathrm{\prime }},t\right)=u\left({x}_{1},{x}^{\mathrm{\prime }},t\right)\phantom{\rule{1em}{0ex}}\left(x=\left({x}_{1},{x}^{\mathrm{\prime }}\right)\in \mathrm{\Omega },t\in \mathbb{R}\right),\hfill \\ {u}_{{x}_{1}}\left(x,t\right)<0\phantom{\rule{1em}{0ex}}\left(x\in \mathrm{\Omega },{x}_{1}>0,t\in \mathbb{R}\right).\hfill \end{array}$

The result of $J=\left(0,\mathrm{\infty }\right)$ is as follows.

Assume that u is a bounded positive solution of (1.6) and (1.7) with $J=\left(0,\mathrm{\infty }\right)$ such that for some sequence ${t}_{n}\to \mathrm{\infty }$,
$\underset{n\to \mathrm{\infty }}{lim inf}u\left(x,{t}_{n}\right)>0\phantom{\rule{1em}{0ex}}\left(x\in \mathrm{\Omega }\right).$
Then u is asymptotically symmetric in the sense that
$\begin{array}{c}\underset{t\to \mathrm{\infty }}{lim}\left(u\left(-{x}_{1},{x}^{\mathrm{\prime }},t\right)-u\left({x}_{1},{x}^{\mathrm{\prime }},t\right)\right)=0\phantom{\rule{1em}{0ex}}\left(x\in \mathrm{\Omega }\right),\hfill \\ \underset{t\to \mathrm{\infty }}{lim sup}{u}_{{x}_{1}}\left(x,t\right)\le 0\phantom{\rule{1em}{0ex}}\left(x\in \mathrm{\Omega },{x}_{1}>0\right).\hfill \end{array}$

In this paper, using the method of moving planes, we obtain the same symmetry of solutions to problem (1.1), (1.2) and (1.3) as elliptic equations.

2 Maximum principles

In this section, we prove some maximum principles. Let Ω be a bounded domain in ${\mathbb{R}}^{n}$, let ${a}^{ij}\left(x,t\right)$, $b\left(x,t\right)$, $c\left(x,t\right)$ be continuous functions in $\overline{Q},Q=\mathrm{\Omega }×\left(0,T\right]$. Suppose that $b\left(x,t\right)<0$, $c\left(x,t\right)$ is bounded and there exist positive constants ${\lambda }_{0}$ and ${\mathrm{\Lambda }}_{0}$ such that
${\lambda }_{0}{|\xi |}^{2}\le {a}^{ij}\left(x,t\right){\xi }_{i}{\xi }_{j}\le {\mathrm{\Lambda }}_{0}{|\xi |}^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }\xi \in {\mathbb{R}}^{n}.$
Here and in the sequel, we always denote
${D}_{i}=\frac{\partial }{\partial {x}_{i}},\phantom{\rule{2em}{0ex}}{D}_{ij}=\frac{{\partial }^{2}}{\partial {x}_{i}\phantom{\rule{0.2em}{0ex}}\partial {x}_{j}}.$

We use the standard notation ${C}^{2k,k}\left(Q\right)$ to denote the class of functions u such that the derivatives ${D}_{x}^{i}{D}_{t}^{j}u$ are continuous in Q for $i+2j\le 2k$.

Theorem 2.1 Let $\lambda \left(x,t\right)$ be a bounded continuous function on $\overline{Q}$, and let the positive function $\phi \in {C}^{2,1}\left(\overline{Q}\right)$ satisfy
$b\left(x,t\right){\phi }_{t}+{a}^{ij}\left(x,t\right){D}_{ij}\phi -\lambda \left(x,t\right)\phi \le 0.$
(2.1)
Suppose that $u\in {C}^{2,1}\left(Q\right)\cap {C}^{0}\left(\overline{Q}\right)$ satisfies
$b\left(x,t\right){u}_{t}+{a}^{ij}\left(x,t\right){D}_{ij}u-c\left(x,t\right)u\le 0,\phantom{\rule{1em}{0ex}}\left(x,t\right)\in Q,$
(2.2)
$u\ge 0,\phantom{\rule{1em}{0ex}}\left(x,t\right)\in {\partial }_{p}Q.$
(2.3)
If
$c\left(x,t\right)>\lambda \left(x,t\right),\phantom{\rule{1em}{0ex}}\left(x,t\right)\in Q,$
(2.4)

then $u\ge 0$ in Q.

Proof We argue by contradiction. Suppose there exists $\left(\overline{x},\overline{t}\right)\in Q$ such that $u\left(\overline{x},\overline{t}\right)<0$. Let
$v\left(x,t\right)=\frac{u\left(x,t\right)}{\phi \left(x,t\right)},\phantom{\rule{1em}{0ex}}\left(x,t\right)\in Q.$
Then $v\left(\overline{x},\overline{t}\right)<0$. Set $v\left({x}_{0},{t}_{0}\right)={min}_{\overline{Q}}v\left(x,t\right)$, then ${x}_{0}\in \mathrm{\Omega }$ and $v\left({x}_{0},{t}_{0}\right)<0$. Since $v\left(\cdot ,{t}_{0}\right)$ attains its minimum at ${x}_{0}$, we have $Dv\left({x}_{0},{t}_{0}\right)=0$, ${D}^{2}v\left({x}_{0},{t}_{0}\right)\ge 0$. In addition, we have ${v}_{t}\left({x}_{0},{t}_{0}\right)\le 0$. A direct calculation gives
$\begin{array}{c}{v}_{t}=\frac{{u}_{t}\phi -u{\phi }_{t}}{{\phi }^{2}},\hfill \\ {D}_{ij}v=\frac{1}{\phi }{D}_{ij}u-\frac{u}{{\phi }^{2}}{D}_{ij}\phi -\frac{1}{\phi }{D}_{i}v{D}_{j}\phi -\frac{1}{\phi }{D}_{j}v{D}_{i}\phi .\hfill \end{array}$
Taking into account $u\left({x}_{0},{t}_{0}\right)<0$, we have at $\left({x}_{0},{t}_{0}\right)$,
$\begin{array}{rcl}0& \le & \phi {a}^{ij}{D}_{ij}v={a}^{ij}{D}_{ij}u-\frac{{a}^{ij}{D}_{ij}\phi }{\phi }u\\ \le & {a}^{ij}{D}_{ij}u+\frac{u}{\phi }\left(b{\phi }_{t}-\lambda \phi \right)\\ \le & {a}^{ij}{D}_{ij}u+\frac{b}{\phi }{u}_{t}\phi -\lambda u\\ =& {a}^{ij}{D}_{ij}u+b{u}_{t}-\lambda u\\ <& {a}^{ij}{D}_{ij}u+b{u}_{t}-cu\\ \le & 0.\end{array}$

This is a contradiction and thus completes the proof of Theorem 2.1. □

Theorem 2.1 is also valid in unbounded domains if u is nonnegative at infinity. Thus we have the following corollary.

Corollary 2.2 Suppose that Ω is unbounded, $Q=\mathrm{\Omega }×\left(0,T\right]$. Besides the conditions of Theorem 2.1, we assume
$\underset{|x|\to \mathrm{\infty }}{lim inf}u\left(x,t\right)\ge 0.$
(2.5)

Then $u\ge 0$ in Q.

Proof Still consider $v\left(x,t\right)$ in the proof of Theorem 2.1. Condition (2.5) shows that the minimum of $v\left(x,t\right)$ cannot be achieved at infinity. The rest of the proof is the same as the proof of Theorem 2.1. □

If Ω is a narrow region with width l,
$\mathrm{\Omega }=\left\{x\in {\mathbb{R}}^{n}|0<{x}_{1}

then we have the following narrow region principle.

Corollary 2.3 (Narrow region principle)

Suppose that $u\in {C}^{2,1}\left(Q\right)\cap {C}^{0}\left(\overline{Q}\right)$ satisfies (2.2) and (2.3). Let the width l of Ω be sufficiently small. If on ${\partial }_{p}Q$, $u\ge 0$, then we have $u\ge 0$ in Q. If Ω is unbounded, and ${lim inf}_{|x|\to \mathrm{\infty }}u\left(x,t\right)\ge 0$, then the conclusion is also true.

Proof Let $0<\epsilon ,
$\phi \left(x,t\right)=t+sin\frac{{x}_{1}+\epsilon }{l}.$
Then φ is positive and
$\begin{array}{c}{\phi }_{t}=1,\hfill \\ {a}^{ij}{D}_{ij}\phi =-{\left(\frac{1}{l}\right)}^{2}{a}^{11}\phi .\hfill \end{array}$
Choose $\lambda \left(x,t\right)=-{\lambda }_{0}/{l}^{2}$. In virtue of the boundedness of $c\left(x,t\right)$, when l is sufficiently small, we have $c\left(x,t\right)>\lambda \left(x,t\right)$, and thus
$\begin{array}{r}b{\phi }_{t}+{a}^{ij}{D}_{ij}\phi -\lambda \phi \\ \phantom{\rule{1em}{0ex}}=b-{\left(\frac{1}{l}\right)}^{2}{a}^{11}\phi -\left(-\frac{{\lambda }_{0}}{{l}^{2}}\right)\phi \\ \phantom{\rule{1em}{0ex}}=b-{\left(\frac{1}{l}\right)}^{2}{a}^{11}\phi +\frac{{\lambda }_{0}}{{l}^{2}}\phi \\ \phantom{\rule{1em}{0ex}}\le b<0.\end{array}$

From Theorem 2.1, we have $u\ge 0$. □

3 Main results

In this section, we prove that the solutions of (1.1), (1.2) and (1.3) are symmetric by the method of moving planes.

Definition 3.1 A function $u\left(x,t\right):Q\to \mathbb{R}$ is called parabolically convex if it is continuous, convex in x and decreasing in t.

Suppose that the following conditions hold.
1. (A)

${f}_{u}\left(t,u\right)/f\left(t,u\right)$ is bounded in $\left[0,T\right]×\mathbb{R}$.

2. (B)
$\partial {u}_{0}/\partial {x}_{1}<0$ and
${u}_{0}\left(x\right)\le {u}_{0}\left({x}^{\lambda }\right),\phantom{\rule{1em}{0ex}}x\in {\mathrm{\Omega }}^{\lambda },$
(3.1)

where ${x}^{\lambda }=\left(2\lambda -{x}_{1},{x}_{2},\dots ,{x}_{n}\right)$, ${\mathrm{\Omega }}^{\lambda }=\mathrm{\Omega }\cap \left\{x\in \mathrm{\Omega }:{x}_{1}\le \lambda \right\}$ ($\lambda <0$).

Theorem 3.1 Let Ω be a strictly convex domain in ${\mathbb{R}}^{n}$ and symmetric with respect to the plane $\left\{x\in \mathrm{\Omega }:{x}_{1}=0\right\}$, $Q=\mathrm{\Omega }×\left(0,T\right]$. Assume that conditions (A) and (B) hold and $u\in {C}^{2,1}\left(Q\right)\cap {C}^{0}\left(\overline{Q}\right)$ is any parabolically convex solution of (1.1), (1.2) and (1.3). Then $u\left({x}_{1},{x}^{\mathrm{\prime }},t\right)=u\left(-{x}_{1},{x}^{\mathrm{\prime }},t\right)$, where $\left(x,t\right)=\left({x}_{1},{x}^{\mathrm{\prime }},t\right)\in {\mathbb{R}}^{n+1}$, and when ${x}_{1}\ge 0$, $\partial u\left(x,t\right)/\partial {x}_{1}\le 0$.

Proof Let in ${\mathrm{\Omega }}^{\lambda }×\left(0,T\right]$, ${u}^{\lambda }\left(x,t\right)=u\left({x}^{\lambda },t\right)$, that is,
${u}^{\lambda }\left({x}_{1},{x}_{2},\dots ,{x}_{n},t\right)=u\left(2\lambda -{x}_{1},{x}_{2},\dots ,{x}_{n},t\right),\phantom{\rule{1em}{0ex}}\left(x,t\right)\in {\mathrm{\Omega }}^{\lambda }×\left(0,T\right].$
Then
${D}^{2}{u}^{\lambda }\left({x}_{1},{x}_{2},\dots ,{x}_{n},t\right)={P}^{T}{D}^{2}u\left(2\lambda -{x}_{1},{x}_{2},\dots ,{x}_{n},t\right)P,$
where $P=diag\left(-1,1,\dots ,1\right)$. Therefore,
$\begin{array}{rcl}-{u}_{t}^{\lambda }det\left({D}^{2}{u}^{\lambda }\right)& =& -{u}_{t}\left(2\lambda -{x}_{1},{x}_{2},\dots ,{x}_{n},t\right)det\left({D}^{2}u\left(2\lambda -{x}_{1},{x}_{2},\dots ,{x}_{n},t\right)\right)\\ =& f\left(t,u\left(2\lambda -{x}_{1},{x}_{2},\dots ,{x}_{n},t\right)\right)\\ =& f\left(t,{u}^{\lambda }\right).\end{array}$
(3.2)
We rewrite (3.2) in the form
$log\left(-{u}_{t}^{\lambda }\right)+log\left(det\left({D}^{2}{u}^{\lambda }\right)\right)=logf\left(t,{u}^{\lambda }\right).$
(3.3)
On the other hand, from (1.1), we have
$log\left(-{u}_{t}\right)+log\left(det\left({D}^{2}u\right)\right)=logf\left(t,u\right).$
(3.4)
According to (3.3) and (3.4), we have
$log\left(-{u}_{t}\right)-log\left(-{u}_{t}^{\lambda }\right)+log\left(det\left({D}^{2}u\right)\right)-log\left(det\left({D}^{2}{u}^{\lambda }\right)\right)=logf\left(t,u\right)-logf\left(t,{u}^{\lambda }\right).$
Therefore
$\begin{array}{r}{\int }_{0}^{1}\frac{d}{ds}log\left(-s{u}_{t}-\left(1-s\right){u}_{t}^{\lambda }\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{0}^{1}\frac{d}{ds}logdet\left(s{D}^{2}u+\left(1-s\right){D}^{2}{u}^{\lambda }\right)\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}={\int }_{0}^{1}\frac{d}{ds}logf\left(t,su+\left(1-s\right){u}^{\lambda }\right)\phantom{\rule{0.2em}{0ex}}ds.\end{array}$
As a result, we have
$b\left(x,t\right){\left(u-{u}^{\lambda }\right)}_{t}+{a}^{ij}\left(x,t\right){\left(u-{u}^{\lambda }\right)}_{ij}-c\left(x,t\right)\left(u-{u}^{\lambda }\right)=0,\phantom{\rule{1em}{0ex}}\left(x,t\right)\in {\mathrm{\Omega }}^{\lambda }×\left(0,T\right],$
(3.5)
where
$\begin{array}{c}b\left(x,t\right)={\int }_{0}^{1}\frac{ds}{s{u}_{t}+\left(1-s\right){u}_{t}^{\lambda }},\hfill \\ {a}^{ij}\left(x,t\right)={\int }_{0}^{1}{g}_{s}^{ij}\phantom{\rule{0.2em}{0ex}}ds,\hfill \\ c\left(x,t\right)={\int }_{0}^{1}\frac{{f}_{u}}{f}\left(t,su+\left(1-s\right){u}^{\lambda }\right)\phantom{\rule{0.2em}{0ex}}ds,\hfill \end{array}$
${g}_{s}^{ij}$ is the inverse matrix of $s{D}^{2}u+\left(1-s\right){D}^{2}{u}^{\lambda }$. Then $b\left(x,t\right)<0$, $c\left(x,t\right)$ is bounded and by the a priori estimate [9] we know there exist positive constants ${\lambda }_{0}$ and ${\mathrm{\Lambda }}_{0}$ such that
${\lambda }_{0}{|\xi |}^{2}\le {a}^{ij}{\xi }_{i}{\xi }_{j}\le {\mathrm{\Lambda }}_{0}{|\xi |}^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }\xi \in {\mathbb{R}}^{n}.$
Let
${w}^{\lambda }=u-{u}^{\lambda },$
then from (3.5),
$b\left(x,t\right){w}_{t}^{\lambda }+{a}^{ij}\left(x,t\right){w}_{ij}^{\lambda }-c\left(x,t\right){w}^{\lambda }=0,\phantom{\rule{1em}{0ex}}\left(x,t\right)\in {\mathrm{\Omega }}^{\lambda }×\left(0,T\right].$
(3.6)
Clearly,
${w}^{\lambda }\left(x,t\right)=0,\phantom{\rule{1em}{0ex}}x\in \partial {\mathrm{\Omega }}^{\lambda }\cap \left\{{x}_{1}=\lambda \right\},0
(3.7)
Because the image of $\partial \mathrm{\Omega }\cap \partial {\mathrm{\Omega }}^{\lambda }$ about the plane $\left\{{x}_{1}=\lambda \right\}$ lies in Ω, according to the maximum principle of parabolic Monge-Ampère equations,
${u}^{\lambda }\left(x,t\right)\le 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in \partial \mathrm{\Omega }\cap \partial {\mathrm{\Omega }}^{\lambda }.$
Thus
${w}^{\lambda }\left(x,t\right)=u-{u}^{\lambda }=0-{u}^{\lambda }\ge 0,\phantom{\rule{1em}{0ex}}x\in \partial \mathrm{\Omega }\cap \partial {\mathrm{\Omega }}^{\lambda },0
(3.8)
On the other hand, from (3.1),
${w}^{\lambda }\left(x,0\right)={u}_{0}\left(x\right)-{u}_{0}\left({x}^{\lambda }\right)\ge 0,\phantom{\rule{1em}{0ex}}x\in {\mathrm{\Omega }}^{\lambda }.$
(3.9)

From Corollary 2.3, when the width of ${\mathrm{\Omega }}^{\lambda }$ is sufficiently small, ${w}^{\lambda }\left(x,t\right)\ge 0$, $\left(x,t\right)\in {\mathrm{\Omega }}^{\lambda }×\left(0,T\right]$.

Now we start to move the plane to its right limit. Define
$\mathrm{\Lambda }=sup\left\{\lambda <0|{w}^{\lambda }\left(x,t\right)\ge 0,x\in {\mathrm{\Omega }}^{\lambda },0
We claim that
$\mathrm{\Lambda }=0.$

Otherwise, we will show that the plane can be further moved to the right by a small distance, and this would contradict with the definition of Λ.

In fact, if $\mathrm{\Lambda }<0$, then the image of $\partial \mathrm{\Omega }\cap \partial {\mathrm{\Omega }}^{\mathrm{\Lambda }}$ under the reflection about $\left\{{x}_{1}=\mathrm{\Lambda }\right\}$ lies inside Ω. According to the strong maximum principle of parabolic Monge-Ampère equations, for $x\in \mathrm{\Omega }$, ${u}^{\mathrm{\Lambda }}<0$. Therefore, for $x\in \partial {\mathrm{\Omega }}^{\mathrm{\Lambda }}\cap \partial \mathrm{\Omega }$, we have ${w}^{\mathrm{\Lambda }}>0$. On the other hand, by the definition of Λ, we have for $x\in {\mathrm{\Omega }}^{\mathrm{\Lambda }}$, ${w}^{\mathrm{\Lambda }}\ge 0$. So, from the strong maximum principle [28] of linear parabolic equations and (3.6), we have for $\left(x,t\right)\in {\mathrm{\Omega }}^{\mathrm{\Lambda }}×\left(0,T\right]$,
${w}^{\mathrm{\Lambda }}\left(x,t\right)>0.$
(3.10)
Let ${d}_{0}$ be the maximum width of narrow regions so that we can apply the narrow region principle. Choose a small positive constant δ such that $\mathrm{\Lambda }+\delta <0$, $\delta \le {d}_{0}/2-\mathrm{\Lambda }$. We consider the function ${w}^{\mathrm{\Lambda }+\delta }\left(x,t\right)$ on the narrow region
${\mathrm{\Sigma }}^{\mathrm{\Lambda }+\delta }×\left(0,T\right]=\left({\mathrm{\Omega }}^{\mathrm{\Lambda }+\delta }\cap \left\{{x}_{1}>\mathrm{\Lambda }-\frac{{d}_{0}}{2}\right\}\right)×\left(0,T\right].$
Then ${w}^{\mathrm{\Lambda }+\delta }\left(x,t\right)$ satisfies
$b\left(x,t\right){w}_{t}^{\mathrm{\Lambda }+\delta }+{a}^{ij}\left(x,t\right){D}_{ij}{w}^{\mathrm{\Lambda }+\delta }-c\left(x,t\right){w}^{\mathrm{\Lambda }+\delta }=0,\phantom{\rule{1em}{0ex}}\left(x,t\right)\in {\mathrm{\Sigma }}^{\mathrm{\Lambda }+\delta }×\left(0,T\right].$
(3.11)
Now we prove the boundary condition
${w}^{\mathrm{\Lambda }+\delta }\left(x,t\right)\ge 0,\phantom{\rule{1em}{0ex}}\left(x,t\right)\in {\partial }_{p}\left({\mathrm{\Sigma }}^{\mathrm{\Lambda }+\delta }×\left(0,T\right]\right).$
(3.12)
Similar to boundary conditions (3.7), (3.8) and (3.9), boundary condition (3.12) is satisfied for $x\in \partial {\mathrm{\Sigma }}^{\mathrm{\Lambda }+\delta }\cap \partial \mathrm{\Omega }$, $x\in \partial {\mathrm{\Sigma }}^{\mathrm{\Lambda }+\delta }\cap \left\{{x}_{1}=\mathrm{\Lambda }+\delta \right\}$ and for $t=0$. In order to prove (3.12) is satisfied for $x\in \partial {\mathrm{\Sigma }}^{\mathrm{\Lambda }+\delta }\cap \left\{{x}_{1}=\mathrm{\Lambda }-{d}_{0}/2\right\}$, we apply the continuity argument. By (3.10) and the fact that $\left(\mathrm{\Lambda }-{d}_{0}/2,{x}_{2},\dots ,{x}_{n}\right)$ is inside ${\mathrm{\Omega }}^{\mathrm{\Lambda }}$, there exists a positive constant ${c}_{0}$ such that
${w}^{\mathrm{\Lambda }}\left(\mathrm{\Lambda }-\frac{{d}_{0}}{2},{x}_{2},\dots ,{x}_{n},t\right)\ge {c}_{0}.$
Because ${w}^{\lambda }$ is continuous in λ, then for small δ, we still have
${w}^{\mathrm{\Lambda }+\delta }\left(\mathrm{\Lambda }-\frac{{d}_{0}}{2},{x}_{2},\dots ,{x}_{n},t\right)\ge 0.$
Therefore boundary condition (3.12) holds for small δ. From Corollary 2.3, we have
${w}^{\mathrm{\Lambda }+\delta }\left(x,t\right)\ge 0,\phantom{\rule{1em}{0ex}}x\in {\mathrm{\Sigma }}^{\mathrm{\Lambda }+\delta },0
(3.13)
Combining (3.10) and the fact that ${w}^{\lambda }$ is continuous for λ, we know that ${w}^{\mathrm{\Lambda }+\delta }\left(x,t\right)\ge 0$ for $x\in {\mathrm{\Omega }}^{\mathrm{\Lambda }}$ when δ is small. Then from (3.13), we know that
${w}^{\mathrm{\Lambda }+\delta }\left(x,t\right)\ge 0,\phantom{\rule{1em}{0ex}}x\in {\mathrm{\Omega }}^{\mathrm{\Lambda }+\delta },0

This contradicts with the definition of Λ, and so $\mathrm{\Lambda }=0$.

As a result, ${w}^{0}\left(x,t\right)\ge 0$ for $x\in {\mathrm{\Omega }}^{0}$, which means that as ${x}_{1}<0$,
$u\left({x}_{1},{x}_{2},\dots ,{x}_{n},t\right)\ge u\left(-{x}_{1},{x}_{2},\dots ,{x}_{n},t\right).$
Since Ω is symmetric about the plane $\left\{{x}_{1}=0\right\}$, then for ${x}_{1}\ge 0$, $u\left(-{x}_{1},{x}_{2},\dots ,{x}_{n},t\right)$ also satisfies (1.1). Thus we can move the plane from the right towards the left and get the reverse inequality. Therefore
$\begin{array}{r}\partial u\left(x,t\right)/\partial {x}_{1}\le 0,\phantom{\rule{1em}{0ex}}{x}_{1}\ge 0,\\ u\left({x}_{1},{x}_{2},\dots ,{x}_{n},t\right)=u\left(-{x}_{1},{x}_{2},\dots ,{x}_{n},t\right).\end{array}$
(3.14)

Equation (3.14) means that u is symmetric about the plane $\left\{{x}_{1}=0\right\}$. Theorem 3.1 is proved. □

If we put the ${x}_{1}$ axis in any direction, from Theorem 3.1, we have the following.

Corollary 3.2 If Ω is a ball, $Q=\mathrm{\Omega }×\left(0,T\right]$, then any parabolically convex solution $u\in {C}^{2,1}\left(\overline{Q}\right)$ of (1.1), (1.2) and (1.3) is radially symmetric about the origin.

Remark 3.1 Solutions of (1.1) in ${\mathbb{R}}^{n}×\left(0,T\right]$ may not be radially symmetric. For example,
$-{u}_{t}det\left({D}^{2}u\right)={e}^{-u},\phantom{\rule{1em}{0ex}}\left(x,t\right)\in {\mathbb{R}}^{n}×\left(0,T\right]$
(3.15)
has a non-radially symmetric solution. In fact, we know that $f\left(x\right)=2log\left(1+{e}^{\sqrt{2}x}\right)-\sqrt{2}x-log4$ ($x>0$) satisfies ${f}^{\mathrm{\prime }\mathrm{\prime }}={e}^{-f}$ in ${\mathbb{R}}^{1}$, and $f\left(x\right)=f\left(-x\right)$, $x<0$. Define
$u\left(x,t\right)=log\left(T-t\right)+f\left({x}_{1}\right)+f\left({x}_{2}\right)+\cdots +f\left({x}_{n}\right),$

then u is a solution of (3.15) but not radially symmetric.

We conclude this paper with a brief examination of Theorem 3.1. Let $B={B}_{1}\left(0\right)$ be the unit ball in ${\mathbb{R}}^{n}$, and let radially symmetric function ${u}_{0}\left(x\right)={u}_{0}\left(r\right)$, $r=|x|$ satisfy
$\frac{{u}_{0}\left(r\right){\left({u}_{0}^{\mathrm{\prime }}\left(r\right)\right)}^{n-1}{u}_{0}^{\mathrm{\prime }\mathrm{\prime }}\left(r\right)}{{r}^{n-1}}=-1,\phantom{\rule{1em}{0ex}}0
(3.16)
${u}_{0}\left(1\right)={u}_{0}^{\mathrm{\prime }}\left(0\right)=0.$
(3.17)
Example 3.1 Let ${u}_{0}$ satisfy (3.16) and (3.17). Then any solution of
$-{u}_{t}det\left({D}^{2}u\right)=1,\phantom{\rule{1em}{0ex}}\left(x,t\right)\in B×\left(0,T\right],$
(3.18)
$u=0,\phantom{\rule{1em}{0ex}}\left(x,t\right)\in \partial B×\left(0,T\right),$
(3.19)
$u={u}_{0},\phantom{\rule{1em}{0ex}}\left(x,t\right)\in \overline{B}×\left\{0\right\}$
(3.20)
is of the form
$u=-{\left[\left(n+1\right)t+1\right]}^{\frac{1}{n+1}}{u}_{0}\left(r\right),$
(3.21)

where $r=|x|$.

Proof According to Corollary 3.2, the solution is symmetric. Let
$u\left(x,t\right)=u\left(r,t\right),\phantom{\rule{2em}{0ex}}r=|x|.$
Then
$\begin{array}{c}{u}_{i}=\frac{\partial u\left(r,t\right)}{\partial r}\frac{{x}_{i}}{r},\hfill \\ {u}_{ij}=\frac{{\partial }^{2}u\left(r,t\right)}{\partial {r}^{2}}\frac{{x}_{i}{x}_{j}}{{r}^{2}}+\frac{\partial u\left(r,t\right)}{\partial r}\left(\frac{{\delta }_{ij}}{r}-\frac{{x}_{i}{x}_{j}}{{r}^{3}}\right),\hfill \\ det\left({D}^{2}u\right)={\left(\frac{\partial u/\partial r}{r}\right)}^{n-1}\frac{{\partial }^{2}u}{\partial {r}^{2}}.\hfill \end{array}$
Therefore (3.18) is
$-\frac{\partial u}{\partial t}{\left(\frac{\partial u/\partial r}{r}\right)}^{n-1}\frac{{\partial }^{2}u}{\partial {r}^{2}}=1.$
(3.22)
We seek the solution of the form
$u\left(r,t\right)=T\left(t\right){u}_{0}\left(r\right).$
Then
$-{u}_{0}\left(r\right){T}^{\mathrm{\prime }}\left(t\right)\frac{{\left({u}_{0}^{\mathrm{\prime }}\left(r\right)T\left(t\right)\right)}^{n-1}}{{r}^{n-1}}{u}_{0}^{\mathrm{\prime }\mathrm{\prime }}\left(r\right)T\left(t\right)=1.$
That is,
$\frac{{u}_{0}\left(r\right){\left({u}_{0}^{\mathrm{\prime }}\left(r\right)\right)}^{n-1}{u}_{0}^{\mathrm{\prime }\mathrm{\prime }}\left(r\right)}{{r}^{n-1}}=-\frac{1}{{T}^{\mathrm{\prime }}\left(t\right){\left(T\left(t\right)\right)}^{n}}.$
(3.23)
Therefore
${T}^{\mathrm{\prime }}\left(t\right){\left(T\left(t\right)\right)}^{n}=1.$
(3.24)
By (3.20), we know that
$T\left(0\right)=1.$
(3.25)
From (3.24) and (3.25), we have
$T\left(t\right)={\left[\left(n+1\right)t+1\right]}^{\frac{1}{n+1}}.$
As a result,
$u\left(r,t\right)=-{\left[\left(n+1\right)t+1\right]}^{\frac{1}{n+1}}{u}_{0}\left(r\right).$

From the maximum principle, we know that the solution of (3.18)-(3.20) is unique. Thus any solution of (3.18), (3.19) and (3.20) is of the form of (3.21). □

Declarations

Acknowledgements

The research was supported by NNSFC (11201343), Shandong Province Young and Middle-Aged Scientists Research Awards Fund (BS2011SF025), Shandong Province Science and Technology Development Project (2011YD16002).

Authors’ Affiliations

(1)
School of Mathematics and Information Science, Weifang University

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