Symmetry of solutions to parabolic Monge-Ampère equations

Abstract

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

$$-u_{t}det\left(D^{2}u\right)=f(t,u)\text{in }\mathrm{\Omega }\times (0,T].$$

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 ${\mathbb{R}}^n\times (0,T]$ and an example in a cylinder to illustrate the results.

MSC:35K96, 35B06.

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 [5–7] and the study of parabolic Monge-Ampère equations; see the references [8–11]etc. The parabolic Monge-Ampère equation $-u_{t}det(D^{2}u)=f$ 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 $-u_{t}det(D^{2}u)=f$ is the most appropriate parabolic version of the elliptic Monge-Ampère equation $det(D^{2}u)=f$ 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

$$-u_{t}det\left(D^{2}u\right)=f(t,u),(x,t)\in Q,$$

(1.1)

$$u=0,(x,t)\in SQ,$$

(1.2)

$$u=u_0(x),(x,t)\in BQ,$$

(1.3)

where $D^{2}u$ is the Hessian matrix of u in x, $Q=\mathrm{\Omega }\times (0,T]$, Ω is a bounded and convex open subset in ${\mathbb{R}}^n$, $SQ=\partial \mathrm{\Omega }\times (0,T)$ denotes the side of Q, $BQ=\overline{\mathrm{\Omega }}\times \{0\}$ 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 $\{x\in {\mathbb{R}}^n:x_1=0\}$, then any positive solution of the Dirichlet problem

$$\begin{array}{r}\mathrm{\Delta }u+f(u)=0,x\in \mathrm{\Omega },\\ u=0,x\in \partial \mathrm{\Omega }\end{array}$$

satisfies the following symmetry and monotonicity properties:

$$u(-x_1,x_2,\dots ,x_n)=u(x_1,x_2,\dots ,x_n),$$

(1.4)

$$u_{x_1}(x_1,x_2,\dots ,x_n)<0(x_1>0).$$

(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 $det(D^{2}u)=e^{-u}$ 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 $\{x\in {\mathbb{R}}^n:x_1=0\}$, then each solution of the Dirichlet problem

$$\begin{array}{r}det\left(D^{2}u\right)=e^{-u},x\in \mathrm{\Omega },\\ u=0,x\in \partial \mathrm{\Omega }\end{array}$$

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 $Q=\mathrm{\Omega }\times J$, $J=(0,T]$, Gidas et al. [16] studied parabolic equations $-u_t+\mathrm{\Delta }u+f(t,r,u)=0$ and $-u_t+F(t,x,u,Du,D^{2}u)=0$, and they proved that parabolic equations possessed the same symmetry as the above elliptic equations. When $J=(0,\mathrm{\infty })$, Hess and Poláčik [25] first studied the asymptotic symmetry results for classical, bounded, positive solutions of the problem

$$u_t-\mathrm{\Delta }u=f(t,u),(x,t)\in \mathrm{\Omega }\times J,$$

(1.6)

$$u=0,(x,t)\in \partial \mathrm{\Omega }\times J.$$

(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

$$\underset{t\in \mathbb{R}}{inf}u(x,t)>0(x\in \mathrm{\Omega },t\in J),$$

then u has the symmetry and monotonicity properties for each $t\in \mathbb{R}$:

$$\begin{array}{c}u(-x_1,x^{\mathrm{\prime }},t)=u(x_1,x^{\mathrm{\prime }},t)(x=(x_1,x^{\mathrm{\prime }})\in \mathrm{\Omega },t\in \mathbb{R}),\hfill \\ u_{x_1}(x,t)<0(x\in \mathrm{\Omega },x_1>0,t\in \mathbb{R}).\hfill \end{array}$$

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

Assume that u is a bounded positive solution of (1.6) and (1.7) with $J=(0,\mathrm{\infty })$ such that for some sequence $t_n\to \mathrm{\infty }$,

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}u(x,t_n)>0(x\in \mathrm{\Omega }).$$

Then u is asymptotically symmetric in the sense that

$$\begin{array}{c}\underset{t\to \mathrm{\infty }}{lim}(u(-x_1,x^{\mathrm{\prime }},t)-u(x_1,x^{\mathrm{\prime }},t))=0(x\in \mathrm{\Omega }),\hfill \\ \underset{t\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{u}_{x_1}(x,t)\le 0(x\in \mathrm{\Omega },x_1>0).\hfill \end{array}$$

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 $a^{ij}(x,t)$, $b(x,t)$, $c(x,t)$ be continuous functions in $\overline{Q},Q=\mathrm{\Omega }\times (0,T]$. Suppose that $b(x,t)<0$, $c(x,t)$ is bounded and there exist positive constants ${\lambda }_0$ and ${\mathrm{\Lambda }}_0$ such that

$${\lambda }_0{|\xi |}^2\le a^{ij}(x,t){\xi }_i{\xi }_j\le {\mathrm{\Lambda }}_0{|\xi |}^2,\mathrm{\forall }\xi \in {\mathbb{R}}^n.$$

Here and in the sequel, we always denote

$$D_i=\frac{\partial }{\partial x_i},D_{ij}=\frac{{\partial }^2}{\partial x_i\partial x_j}.$$

We use the standard notation $C^{2k,k}(Q)$ 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 $\lambda (x,t)$ be a bounded continuous function on $\overline{Q}$, and let the positive function $\phi \in C^{2,1}(\overline{Q})$ satisfy

$$b(x,t){\phi }_t+a^{ij}(x,t)D_{ij}\phi -\lambda (x,t)\phi \le 0.$$

(2.1)

Suppose that $u\in C^{2,1}(Q)\cap C^0(\overline{Q})$ satisfies

$$b(x,t)u_t+a^{ij}(x,t)D_{ij}u-c(x,t)u\le 0,(x,t)\in Q,$$

(2.2)

$$u\ge 0,(x,t)\in {\partial }_{p}Q.$$

(2.3)

If

$$c(x,t)>\lambda (x,t),(x,t)\in Q,$$

(2.4)

then $u\ge 0$ in Q.

Proof We argue by contradiction. Suppose there exists $(\overline{x},\overline{t})\in Q$ such that $u(\overline{x},\overline{t})<0$. Let

$$v(x,t)=\frac{u(x,t)}{\phi (x,t)},(x,t)\in Q.$$

Then $v(\overline{x},\overline{t})<0$. Set $v(x_0,t_0)={min}_{\overline{Q}}v(x,t)$, then $x_0\in \mathrm{\Omega }$ and $v(x_0,t_0)<0$. Since $v(\cdot ,t_0)$ attains its minimum at $x_0$, we have $Dv(x_0,t_0)=0$, $D^{2}v(x_0,t_0)\ge 0$. In addition, we have $v_t(x_0,t_0)\le 0$. 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 $u(x_0,t_0)<0$, we have at $(x_0,t_0)$,

$$\begin{array}{rcl}0& \le & \phi a^{ij}{D}_{ij}v=a^{ij}{D}_{ij}u-\frac{a^{ij}{D}_{ij}\phi }{\phi }u\\ \le & a^{ij}{D}_{ij}u+\frac{u}{\phi }(b{\phi }_t-\lambda \phi )\\ \le & a^{ij}{D}_{ij}u+\frac{b}{\phi }{u}_t\phi -\lambda u\\ =& a^{ij}{D}_{ij}u+b{u}_t-\lambda u\\ <& a^{ij}{D}_{ij}u+b{u}_t-cu\\ \le & 0.\end{array}$$

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, $Q=\mathrm{\Omega }\times (0,T]$. Besides the conditions of Theorem 2.1, we assume

$$\underset{|x|\to \mathrm{\infty }}{lim\hspace{0.17em}inf}u(x,t)\ge 0.$$

(2.5)

Then $u\ge 0$ in Q.

Proof Still consider $v(x,t)$ in the proof of Theorem 2.1. Condition (2.5) shows that the minimum of $v(x,t)$ 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,

$$\mathrm{\Omega }=\{x\in {\mathbb{R}}^n|0<x_1<l\},$$

then we have the following narrow region principle.

Corollary 2.3 (Narrow region principle)

Suppose that $u\in C^{2,1}(Q)\cap C^0(\overline{Q})$ 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 ${lim\hspace{0.17em}inf}_{|x|\to \mathrm{\infty }}u(x,t)\ge 0$, then the conclusion is also true.

Proof Let $0<\epsilon <l$,

$$\phi (x,t)=t+sin\frac{x_1+\epsilon }{l}.$$

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 $\lambda (x,t)=-{\lambda }_0/l^2$. In virtue of the boundedness of $c(x,t)$, when l is sufficiently small, we have $c(x,t)>\lambda (x,t)$, and thus

$$\begin{array}{r}b{\phi }_t+a^{ij}{D}_{ij}\phi -\lambda \phi \\ =b-{\left(\frac{1}{l}\right)}^2{a}^{11}\phi -(-\frac{{\lambda }_0}{l^2})\phi \\ =b-{\left(\frac{1}{l}\right)}^2{a}^{11}\phi +\frac{{\lambda }_0}{l^2}\phi \\ \le b<0.\end{array}$$

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

Suppose that the following conditions hold.

• (A) $f_u(t,u)/f(t,u)$ is bounded in $[0,T]\times \mathbb{R}$.

• (B) $\partial u_0/\partial x_1<0$ and

  $$u_0(x)\le u_0\left(x^{\lambda }\right),x\in {\mathrm{\Omega }}^{\lambda },$$

  (3.1)

where $x^{\lambda }=(2\lambda -x_1,x_2,\dots ,x_n)$, ${\mathrm{\Omega }}^{\lambda }=\mathrm{\Omega }\cap \{x\in \mathrm{\Omega }:x_1\le \lambda \}$ ($\lambda <0$).

Theorem 3.1 Let Ω be a strictly convex domain in ${\mathbb{R}}^n$ and symmetric with respect to the plane $\{x\in \mathrm{\Omega }:x_1=0\}$, $Q=\mathrm{\Omega }\times (0,T]$. Assume that conditions (A) and (B) hold and $u\in C^{2,1}(Q)\cap C^0(\overline{Q})$ is any parabolically convex solution of (1.1), (1.2) and (1.3). Then $u(x_1,x^{\mathrm{\prime }},t)=u(-x_1,x^{\mathrm{\prime }},t)$, where $(x,t)=(x_1,x^{\mathrm{\prime }},t)\in {\mathbb{R}}^{n+1}$, and when $x_1\ge 0$, $\partial u(x,t)/\partial x_1\le 0$.

Proof Let in ${\mathrm{\Omega }}^{\lambda }\times (0,T]$, $u^{\lambda }(x,t)=u(x^{\lambda },t)$, that is,

$$u^{\lambda }(x_1,x_2,\dots ,x_n,t)=u(2\lambda -x_1,x_2,\dots ,x_n,t),(x,t)\in {\mathrm{\Omega }}^{\lambda }\times (0,T].$$

Then

$$D^2{u}^{\lambda }(x_1,x_2,\dots ,x_n,t)=P^T{D}^{2}u(2\lambda -x_1,x_2,\dots ,x_n,t)P,$$

where $P=diag(-1,1,\dots ,1)$. Therefore,

$$\begin{array}{rcl}-u_t^{\lambda }det\left(D^2{u}^{\lambda }\right)& =& -u_t(2\lambda -x_1,x_2,\dots ,x_n,t)det(D^{2}u(2\lambda -x_1,x_2,\dots ,x_n,t))\\ =& f(t,u(2\lambda -x_1,x_2,\dots ,x_n,t))\\ =& f(t,u^{\lambda }).\end{array}$$

(3.2)

We rewrite (3.2) in the form

$$log(-u_t^{\lambda })+log(det\left(D^2{u}^{\lambda }\right))=logf(t,u^{\lambda }).$$

(3.3)

On the other hand, from (1.1), we have

$$log(-u_t)+log(det\left(D^{2}u\right))=logf(t,u).$$

(3.4)

According to (3.3) and (3.4), we have

$$log(-u_t)-log(-u_t^{\lambda })+log(det\left(D^{2}u\right))-log(det\left(D^2{u}^{\lambda }\right))=logf(t,u)-logf(t,u^{\lambda }).$$

Therefore

$$\begin{array}{r}{\int }_0^1\frac{d}{ds}log(-s{u}_t-(1-s)u_t^{\lambda })ds+{\int }_0^1\frac{d}{ds}logdet(s{D}^{2}u+(1-s)D^2{u}^{\lambda })ds\\ ={\int }_0^1\frac{d}{ds}logf(t,su+(1-s)u^{\lambda })ds.\end{array}$$

As a result, we have

$$b(x,t){(u-u^{\lambda })}_t+a^{ij}(x,t){(u-u^{\lambda })}_{ij}-c(x,t)(u-u^{\lambda })=0,(x,t)\in {\mathrm{\Omega }}^{\lambda }\times (0,T],$$

(3.5)

where

$$\begin{array}{c}b(x,t)={\int }_0^1\frac{ds}{s{u}_t+(1-s)u_t^{\lambda }},\hfill \\ a^{ij}(x,t)={\int }_0^1{g}_s^{ij}ds,\hfill \\ c(x,t)={\int }_0^1\frac{f_u}{f}(t,su+(1-s)u^{\lambda })ds,\hfill \end{array}$$

$g_s^{ij}$ is the inverse matrix of $s{D}^{2}u+(1-s)D^2{u}^{\lambda }$. Then $b(x,t)<0$, $c(x,t)$ is bounded and by the a priori estimate [9] we know there exist positive constants ${\lambda }_0$ and ${\mathrm{\Lambda }}_0$ such that

$${\lambda }_0{|\xi |}^2\le a^{ij}{\xi }_i{\xi }_j\le {\mathrm{\Lambda }}_0{|\xi |}^2,\mathrm{\forall }\xi \in {\mathbb{R}}^n.$$

Let

$$w^{\lambda }=u-u^{\lambda },$$

then from (3.5),

$$b(x,t)w_t^{\lambda }+a^{ij}(x,t)w_{ij}^{\lambda }-c(x,t)w^{\lambda }=0,(x,t)\in {\mathrm{\Omega }}^{\lambda }\times (0,T].$$

(3.6)

Clearly,

$$w^{\lambda }(x,t)=0,x\in \partial {\mathrm{\Omega }}^{\lambda }\cap \{x_1=\lambda \},0<t\le T.$$

(3.7)

Because the image of $\partial \mathrm{\Omega }\cap \partial {\mathrm{\Omega }}^{\lambda }$ about the plane $\{x_1=\lambda \}$ lies in Ω, according to the maximum principle of parabolic Monge-Ampère equations,

$$u^{\lambda }(x,t)\le 0,\mathrm{\forall }x\in \partial \mathrm{\Omega }\cap \partial {\mathrm{\Omega }}^{\lambda }.$$

Thus

$$w^{\lambda }(x,t)=u-u^{\lambda }=0-u^{\lambda }\ge 0,x\in \partial \mathrm{\Omega }\cap \partial {\mathrm{\Omega }}^{\lambda },0<t\le T.$$

(3.8)

On the other hand, from (3.1),

$$w^{\lambda }(x,0)=u_0(x)-u_0\left(x^{\lambda }\right)\ge 0,x\in {\mathrm{\Omega }}^{\lambda }.$$

(3.9)

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

Now we start to move the plane to its right limit. Define

$$\mathrm{\Lambda }=sup\{\lambda <0|w^{\lambda }(x,t)\ge 0,x\in {\mathrm{\Omega }}^{\lambda },0<t\le T\}.$$

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 $\{x_1=\mathrm{\Lambda }\}$ 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 $(x,t)\in {\mathrm{\Omega }}^{\mathrm{\Lambda }}\times (0,T]$,

$$w^{\mathrm{\Lambda }}(x,t)>0.$$

(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 $w^{\mathrm{\Lambda }+\delta }(x,t)$ on the narrow region

$${\mathrm{\Sigma }}^{\mathrm{\Lambda }+\delta }\times (0,T]=({\mathrm{\Omega }}^{\mathrm{\Lambda }+\delta }\cap \{x_1>\mathrm{\Lambda }-\frac{d_0}{2}\})\times (0,T].$$

Then $w^{\mathrm{\Lambda }+\delta }(x,t)$ satisfies

$$b(x,t)w_t^{\mathrm{\Lambda }+\delta }+a^{ij}(x,t)D_{ij}{w}^{\mathrm{\Lambda }+\delta }-c(x,t)w^{\mathrm{\Lambda }+\delta }=0,(x,t)\in {\mathrm{\Sigma }}^{\mathrm{\Lambda }+\delta }\times (0,T].$$

(3.11)

Now we prove the boundary condition

$$w^{\mathrm{\Lambda }+\delta }(x,t)\ge 0,(x,t)\in {\partial }_p({\mathrm{\Sigma }}^{\mathrm{\Lambda }+\delta }\times (0,T]).$$

(3.12)

Similar to boundary conditions (3.7), (3.8) and (3.9), boundary condition (3.12) is satisfied for $x\in \partial {\mathrm{\Sigma }}^{\mathrm{\Lambda }+\delta }\cap \partial \mathrm{\Omega }$, $x\in \partial {\mathrm{\Sigma }}^{\mathrm{\Lambda }+\delta }\cap \{x_1=\mathrm{\Lambda }+\delta \}$ and for $t=0$. In order to prove (3.12) is satisfied for $x\in \partial {\mathrm{\Sigma }}^{\mathrm{\Lambda }+\delta }\cap \{x_1=\mathrm{\Lambda }-d_0/2\}$, we apply the continuity argument. By (3.10) and the fact that $(\mathrm{\Lambda }-d_0/2,x_2,\dots ,x_n)$ is inside ${\mathrm{\Omega }}^{\mathrm{\Lambda }}$, there exists a positive constant $c_0$ such that

$$w^{\mathrm{\Lambda }}(\mathrm{\Lambda }-\frac{d_0}{2},x_2,\dots ,x_n,t)\ge c_0.$$

Because $w^{\lambda }$ is continuous in λ, then for small δ, we still have

$$w^{\mathrm{\Lambda }+\delta }(\mathrm{\Lambda }-\frac{d_0}{2},x_2,\dots ,x_n,t)\ge 0.$$

Therefore boundary condition (3.12) holds for small δ. From Corollary 2.3, we have

$$w^{\mathrm{\Lambda }+\delta }(x,t)\ge 0,x\in {\mathrm{\Sigma }}^{\mathrm{\Lambda }+\delta },0<t\le T.$$

(3.13)

Combining (3.10) and the fact that $w^{\lambda }$ is continuous for λ, we know that $w^{\mathrm{\Lambda }+\delta }(x,t)\ge 0$ for $x\in {\mathrm{\Omega }}^{\mathrm{\Lambda }}$ when δ is small. Then from (3.13), we know that

$$w^{\mathrm{\Lambda }+\delta }(x,t)\ge 0,x\in {\mathrm{\Omega }}^{\mathrm{\Lambda }+\delta },0<t\le T.$$

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

As a result, $w^0(x,t)\ge 0$ for $x\in {\mathrm{\Omega }}^0$, which means that as $x_1<0$,

$$u(x_1,x_2,\dots ,x_n,t)\ge u(-x_1,x_2,\dots ,x_n,t).$$

Since Ω is symmetric about the plane $\{x_1=0\}$, then for $x_1\ge 0$, $u(-x_1,x_2,\dots ,x_n,t)$ also satisfies (1.1). Thus we can move the plane from the right towards the left and get the reverse inequality. Therefore

$$\begin{array}{r}\partial u(x,t)/\partial x_1\le 0,x_1\ge 0,\\ u(x_1,x_2,\dots ,x_n,t)=u(-x_1,x_2,\dots ,x_n,t).\end{array}$$

(3.14)

Equation (3.14) means that u is symmetric about the plane $\{x_1=0\}$. 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, $Q=\mathrm{\Omega }\times (0,T]$, then any parabolically convex solution $u\in C^{2,1}(\overline{Q})$ of (1.1), (1.2) and (1.3) is radially symmetric about the origin.

Remark 3.1 Solutions of (1.1) in ${\mathbb{R}}^n\times (0,T]$ may not be radially symmetric. For example,

$$-u_{t}det\left(D^{2}u\right)=e^{-u},(x,t)\in {\mathbb{R}}^n\times (0,T]$$

(3.15)

has a non-radially symmetric solution. In fact, we know that $f(x)=2log(1+e^{\sqrt{2}x})-\sqrt{2}x-log4$ ($x>0$) satisfies $f^{\mathrm{\prime }\mathrm{\prime }}=e^{-f}$ in ${\mathbb{R}}^1$, and $f(x)=f(-x)$, $x<0$. Define

$$u(x,t)=log(T-t)+f(x_1)+f(x_2)+\cdots +f(x_n),$$

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 $B=B_1(0)$ be the unit ball in ${\mathbb{R}}^n$, and let radially symmetric function $u_0(x)=u_0(r)$, $r=|x|$ satisfy

$$\frac{u_0(r){(u_0^{\mathrm{\prime }}(r))}^{n-1}{u}_0^{\mathrm{\prime }\mathrm{\prime }}(r)}{r^{n-1}}=-1,0<r<1,$$

(3.16)

$$u_0(1)=u_0^{\mathrm{\prime }}(0)=0.$$

(3.17)

Example 3.1 Let $u_0$ satisfy (3.16) and (3.17). Then any solution of

$$-u_{t}det\left(D^{2}u\right)=1,(x,t)\in B\times (0,T],$$

(3.18)

$$u=0,(x,t)\in \partial B\times (0,T),$$

(3.19)

$$u=u_0,(x,t)\in \overline{B}\times \{0\}$$

(3.20)

is of the form

$$u=-{[(n+1)t+1]}^{\frac{1}{n+1}}{u}_0(r),$$

(3.21)

where $r=|x|$.

Proof According to Corollary 3.2, the solution is symmetric. Let

$$u(x,t)=u(r,t),r=|x|.$$

Then

$$\begin{array}{c}{u}_i=\frac{\partial u(r,t)}{\partial r}\frac{x_i}{r},\hfill \\ u_{ij}=\frac{{\partial }^{2}u(r,t)}{\partial r^2}\frac{x_i{x}_j}{r^2}+\frac{\partial u(r,t)}{\partial r}(\frac{{\delta }_{ij}}{r}-\frac{x_i{x}_j}{r^3}),\hfill \\ det\left(D^{2}u\right)={\left(\frac{\partial u/\partial r}{r}\right)}^{n-1}\frac{{\partial }^{2}u}{\partial r^2}.\hfill \end{array}$$

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

$$u(r,t)=T(t)u_0(r).$$

Then

$$-u_0(r)T^{\mathrm{\prime }}(t)\frac{{(u_0^{\mathrm{\prime }}(r)T(t))}^{n-1}}{r^{n-1}}{u}_0^{\mathrm{\prime }\mathrm{\prime }}(r)T(t)=1.$$

That is,

$$\frac{u_0(r){(u_0^{\mathrm{\prime }}(r))}^{n-1}{u}_0^{\mathrm{\prime }\mathrm{\prime }}(r)}{r^{n-1}}=-\frac{1}{T^{\mathrm{\prime }}(t){(T(t))}^n}.$$

(3.23)

Therefore

$$T^{\mathrm{\prime }}(t){(T(t))}^n=1.$$

(3.24)

By (3.20), we know that

$$T(0)=1.$$

(3.25)

From (3.24) and (3.25), we have

$$T(t)={[(n+1)t+1]}^{\frac{1}{n+1}}.$$

As a result,

$$u(r,t)=-{[(n+1)t+1]}^{\frac{1}{n+1}}{u}_0(r).$$

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). □