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Symmetry of solutions to parabolic Monge-Ampère equations

Abstract

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

u t det ( D 2 u ) =f(t,u)in Ω×(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 R n ×(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 [57] and the study of parabolic Monge-Ampère equations; see the references [811]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 ( D 2 u ) =f(t,u),(x,t)Q,
(1.1)
u=0,(x,t)SQ,
(1.2)
u= u 0 (x),(x,t)BQ,
(1.3)

where D 2 u is the Hessian matrix of u in x, Q=Ω×(0,T], Ω is a bounded and convex open subset in R n , SQ=Ω×(0,T) denotes the side of Q, BQ= Ω ¯ ×{0} denotes the bottom of Q, and p Q=SQBQ 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 Ω= R n or Ω is a smooth bounded domain in R n , convex in x 1 and symmetric with respect to the hyperplane {x R n : x 1 =0}, then any positive solution of the Dirichlet problem

Δ u + f ( u ) = 0 , x Ω , u = 0 , x Ω

satisfies the following symmetry and monotonicity properties:

u( x 1 , x 2 ,, x n )=u( x 1 , x 2 ,, x n ),
(1.4)
u x 1 ( x 1 , x 2 ,, 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 R n with smooth boundary and symmetric with respect to the hyperplane {x R n : x 1 =0}, then each solution of the Dirichlet problem

det ( D 2 u ) = e u , x Ω , u = 0 , x Ω

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=Ω×J, J=(0,T], Gidas et al. [16] studied parabolic equations u t +Δu+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,), Hess and Poláčik [25] first studied the asymptotic symmetry results for classical, bounded, positive solutions of the problem

u t Δu=f(t,u),(x,t)Ω×J,
(1.6)
u=0,(x,t)Ω×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=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=R satisfying

inf t R u(x,t)>0(xΩ,tJ),

then u has the symmetry and monotonicity properties for each tR:

u ( x 1 , x , t ) = u ( x 1 , x , t ) ( x = ( x 1 , x ) Ω , t R ) , u x 1 ( x , t ) < 0 ( x Ω , x 1 > 0 , t R ) .

The result of J=(0,) is as follows.

Assume that u is a bounded positive solution of (1.6) and (1.7) with J=(0,) such that for some sequence t n ,

lim inf n u(x, t n )>0(xΩ).

Then u is asymptotically symmetric in the sense that

lim t ( u ( x 1 , x , t ) u ( x 1 , x , t ) ) = 0 ( x Ω ) , lim sup t u x 1 ( x , t ) 0 ( x Ω , x 1 > 0 ) .

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 R n , let a i j (x,t), b(x,t), c(x,t) be continuous functions in Q ¯ ,Q=Ω×(0,T]. Suppose that b(x,t)<0, c(x,t) is bounded and there exist positive constants λ 0 and Λ 0 such that

λ 0 | ξ | 2 a i j (x,t) ξ i ξ j Λ 0 | ξ | 2 ,ξ R n .

Here and in the sequel, we always denote

D i = x i , D i j = 2 x i x j .

We use the standard notation C 2 k , 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+2j2k.

Theorem 2.1 Let λ(x,t) be a bounded continuous function on Q ¯ , and let the positive function φ C 2 , 1 ( Q ¯ ) satisfy

b(x,t) φ t + a i j (x,t) D i j φλ(x,t)φ0.
(2.1)

Suppose that u C 2 , 1 (Q) C 0 ( Q ¯ ) satisfies

b(x,t) u t + a i j (x,t) D i j uc(x,t)u0,(x,t)Q,
(2.2)
u0,(x,t) p Q.
(2.3)

If

c(x,t)>λ(x,t),(x,t)Q,
(2.4)

then u0 in Q.

Proof We argue by contradiction. Suppose there exists ( x ¯ , t ¯ )Q such that u( x ¯ , t ¯ )<0. Let

v(x,t)= u ( x , t ) φ ( x , t ) ,(x,t)Q.

Then v( x ¯ , t ¯ )<0. Set v( x 0 , t 0 )= min Q ¯ v(x,t), then x 0 Ω and v( x 0 , t 0 )<0. Since v(, t 0 ) attains its minimum at x 0 , we have Dv( x 0 , t 0 )=0, D 2 v( x 0 , t 0 )0. In addition, we have v t ( x 0 , t 0 )0. A direct calculation gives

v t = u t φ u φ t φ 2 , D i j v = 1 φ D i j u u φ 2 D i j φ 1 φ D i v D j φ 1 φ D j v D i φ .

Taking into account u( x 0 , t 0 )<0, we have at ( x 0 , t 0 ),

0 φ a i j D i j v = a i j D i j u a i j D i j φ φ u a i j D i j u + u φ ( b φ t λ φ ) a i j D i j u + b φ u t φ λ u = a i j D i j u + b u t λ u < a i j D i j u + b u t c u 0 .

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=Ω×(0,T]. Besides the conditions of Theorem 2.1, we assume

lim inf | x | u(x,t)0.
(2.5)

Then u0 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,

Ω= { x R n | 0 < x 1 < l } ,

then we have the following narrow region principle.

Corollary 2.3 (Narrow region principle)

Suppose that u C 2 , 1 (Q) C 0 ( Q ¯ ) satisfies (2.2) and (2.3). Let the width l of Ω be sufficiently small. If on p Q, u0, then we have u0 in Q. If Ω is unbounded, and lim inf | x | u(x,t)0, then the conclusion is also true.

Proof Let 0<ε<l,

φ(x,t)=t+sin x 1 + ε l .

Then φ is positive and

φ t = 1 , a i j D i j φ = ( 1 l ) 2 a 11 φ .

Choose λ(x,t)= λ 0 / l 2 . In virtue of the boundedness of c(x,t), when l is sufficiently small, we have c(x,t)>λ(x,t), and thus

b φ t + a i j D i j φ λ φ = b ( 1 l ) 2 a 11 φ ( λ 0 l 2 ) φ = b ( 1 l ) 2 a 11 φ + λ 0 l 2 φ b < 0 .

From Theorem 2.1, we have u0. □

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):QR 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 (t,u)/f(t,u) is bounded in [0,T]×R.

  2. (B)

    u 0 / x 1 <0 and

    u 0 (x) u 0 ( x λ ) ,x Ω λ ,
    (3.1)

where x λ =(2λ x 1 , x 2 ,, x n ), Ω λ =Ω{xΩ: x 1 λ} (λ<0).

Theorem 3.1 Let Ω be a strictly convex domain in R n and symmetric with respect to the plane {xΩ: x 1 =0}, Q=Ω×(0,T]. Assume that conditions (A) and (B) hold and u C 2 , 1 (Q) C 0 ( Q ¯ ) is any parabolically convex solution of (1.1), (1.2) and (1.3). Then u( x 1 , x ,t)=u( x 1 , x ,t), where (x,t)=( x 1 , x ,t) R n + 1 , and when x 1 0, u(x,t)/ x 1 0.

Proof Let in Ω λ ×(0,T], u λ (x,t)=u( x λ ,t), that is,

u λ ( x 1 , x 2 ,, x n ,t)=u(2λ x 1 , x 2 ,, x n ,t),(x,t) Ω λ ×(0,T].

Then

D 2 u λ ( x 1 , x 2 ,, x n ,t)= P T D 2 u(2λ x 1 , x 2 ,, x n ,t)P,

where P=diag(1,1,,1). Therefore,

u t λ det ( D 2 u λ ) = u t ( 2 λ x 1 , x 2 , , x n , t ) det ( D 2 u ( 2 λ x 1 , x 2 , , x n , t ) ) = f ( t , u ( 2 λ x 1 , x 2 , , x n , t ) ) = f ( t , u λ ) .
(3.2)

We rewrite (3.2) in the form

log ( u t λ ) +log ( det ( D 2 u λ ) ) =logf ( t , u λ ) .
(3.3)

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

log( u t )+log ( det ( D 2 u ) ) =logf(t,u).
(3.4)

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

log( u t )log ( u t λ ) +log ( det ( D 2 u ) ) log ( det ( D 2 u λ ) ) =logf(t,u)logf ( t , u λ ) .

Therefore

0 1 d d s log ( s u t ( 1 s ) u t λ ) d s + 0 1 d d s log det ( s D 2 u + ( 1 s ) D 2 u λ ) d s = 0 1 d d s log f ( t , s u + ( 1 s ) u λ ) d s .

As a result, we have

b(x,t) ( u u λ ) t + a i j (x,t) ( u u λ ) i j c(x,t) ( u u λ ) =0,(x,t) Ω λ ×(0,T],
(3.5)

where

b ( x , t ) = 0 1 d s s u t + ( 1 s ) u t λ , a i j ( x , t ) = 0 1 g s i j d s , c ( x , t ) = 0 1 f u f ( t , s u + ( 1 s ) u λ ) d s ,

g s i j is the inverse matrix of s D 2 u+(1s) D 2 u λ . Then b(x,t)<0, c(x,t) is bounded and by the a priori estimate [9] we know there exist positive constants λ 0 and Λ 0 such that

λ 0 | ξ | 2 a i j ξ i ξ j Λ 0 | ξ | 2 ,ξ R n .

Let

w λ =u u λ ,

then from (3.5),

b(x,t) w t λ + a i j (x,t) w i j λ c(x,t) w λ =0,(x,t) Ω λ ×(0,T].
(3.6)

Clearly,

w λ (x,t)=0,x Ω λ { x 1 =λ},0<tT.
(3.7)

Because the image of Ω Ω λ about the plane { x 1 =λ} lies in Ω, according to the maximum principle of parabolic Monge-Ampère equations,

u λ (x,t)0,xΩ Ω λ .

Thus

w λ (x,t)=u u λ =0 u λ 0,xΩ Ω λ ,0<tT.
(3.8)

On the other hand, from (3.1),

w λ (x,0)= u 0 (x) u 0 ( x λ ) 0,x Ω λ .
(3.9)

From Corollary 2.3, when the width of Ω λ is sufficiently small, w λ (x,t)0, (x,t) Ω λ ×(0,T].

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

Λ=sup { λ < 0 | w λ ( x , t ) 0 , x Ω λ , 0 < t T } .

We claim that

Λ=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 Λ<0, then the image of Ω Ω Λ under the reflection about { x 1 =Λ} lies inside Ω. According to the strong maximum principle of parabolic Monge-Ampère equations, for xΩ, u Λ <0. Therefore, for x Ω Λ Ω, we have w Λ >0. On the other hand, by the definition of Λ, we have for x Ω Λ , w Λ 0. So, from the strong maximum principle [28] of linear parabolic equations and (3.6), we have for (x,t) Ω Λ ×(0,T],

w Λ (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 Λ+δ<0, δ d 0 /2Λ. We consider the function w Λ + δ (x,t) on the narrow region

Σ Λ + δ ×(0,T]= ( Ω Λ + δ { x 1 > Λ d 0 2 } ) ×(0,T].

Then w Λ + δ (x,t) satisfies

b(x,t) w t Λ + δ + a i j (x,t) D i j w Λ + δ c(x,t) w Λ + δ =0,(x,t) Σ Λ + δ ×(0,T].
(3.11)

Now we prove the boundary condition

w Λ + δ (x,t)0,(x,t) p ( Σ Λ + δ × ( 0 , T ] ) .
(3.12)

Similar to boundary conditions (3.7), (3.8) and (3.9), boundary condition (3.12) is satisfied for x Σ Λ + δ Ω, x Σ Λ + δ { x 1 =Λ+δ} and for t=0. In order to prove (3.12) is satisfied for x Σ Λ + δ { x 1 =Λ d 0 /2}, we apply the continuity argument. By (3.10) and the fact that (Λ d 0 /2, x 2 ,, x n ) is inside Ω Λ , there exists a positive constant c 0 such that

w Λ ( Λ d 0 2 , x 2 , , x n , t ) c 0 .

Because w λ is continuous in λ, then for small δ, we still have

w Λ + δ ( Λ d 0 2 , x 2 , , x n , t ) 0.

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

w Λ + δ (x,t)0,x Σ Λ + δ ,0<tT.
(3.13)

Combining (3.10) and the fact that w λ is continuous for λ, we know that w Λ + δ (x,t)0 for x Ω Λ when δ is small. Then from (3.13), we know that

w Λ + δ (x,t)0,x Ω Λ + δ ,0<tT.

This contradicts with the definition of Λ, and so Λ=0.

As a result, w 0 (x,t)0 for x Ω 0 , which means that as x 1 <0,

u( x 1 , x 2 ,, x n ,t)u( x 1 , x 2 ,, x n ,t).

Since Ω is symmetric about the plane { x 1 =0}, then for x 1 0, u( x 1 , x 2 ,, 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

u ( x , t ) / x 1 0 , x 1 0 , u ( x 1 , x 2 , , x n , t ) = u ( x 1 , x 2 , , x n , t ) .
(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=Ω×(0,T], then any parabolically convex solution u C 2 , 1 ( Q ¯ ) of (1.1), (1.2) and (1.3) is radially symmetric about the origin.

Remark 3.1 Solutions of (1.1) in R n ×(0,T] may not be radially symmetric. For example,

u t det ( D 2 u ) = e u ,(x,t) R n ×(0,T]
(3.15)

has a non-radially symmetric solution. In fact, we know that f(x)=2log(1+ e 2 x ) 2 xlog4 (x>0) satisfies f = e f in R 1 , and f(x)=f(x), x<0. Define

u(x,t)=log(Tt)+f( x 1 )+f( x 2 )++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 R n , and let radially symmetric function u 0 (x)= u 0 (r), r=|x| satisfy

u 0 ( r ) ( u 0 ( r ) ) n 1 u 0 ( r ) r n 1 =1,0<r<1,
(3.16)
u 0 (1)= u 0 (0)=0.
(3.17)

Example 3.1 Let u 0 satisfy (3.16) and (3.17). Then any solution of

u t det ( D 2 u ) =1,(x,t)B×(0,T],
(3.18)
u=0,(x,t)B×(0,T),
(3.19)
u= u 0 ,(x,t) B ¯ ×{0}
(3.20)

is of the form

u= [ ( n + 1 ) t + 1 ] 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

u i = u ( r , t ) r x i r , u i j = 2 u ( r , t ) r 2 x i x j r 2 + u ( r , t ) r ( δ i j r x i x j r 3 ) , det ( D 2 u ) = ( u / r r ) n 1 2 u r 2 .

Therefore (3.18) is

u t ( u / r r ) n 1 2 u 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 (t) ( u 0 ( r ) T ( t ) ) n 1 r n 1 u 0 (r)T(t)=1.

That is,

u 0 ( r ) ( u 0 ( r ) ) n 1 u 0 ( r ) r n 1 = 1 T ( t ) ( T ( t ) ) n .
(3.23)

Therefore

T (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 ] 1 n + 1 .

As a result,

u(r,t)= [ ( n + 1 ) t + 1 ] 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). □

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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).

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Dai, L. Symmetry of solutions to parabolic Monge-Ampère equations. Bound Value Probl 2013, 185 (2013). https://doi.org/10.1186/1687-2770-2013-185

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