Open Access

Symmetry of solutions to parabolic Monge-Ampère equations

Boundary Value Problems20132013:185

DOI: 10.1186/1687-2770-2013-185

Received: 3 April 2013

Accepted: 5 August 2013

Published: 20 August 2013

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.

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 ( 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 ) S Q ,
(1.2)
u = u 0 ( x ) , ( x , t ) B Q ,
(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 , S Q = Ω × ( 0 , T ) denotes the side of Q, B Q = Ω ¯ × { 0 } denotes the bottom of Q, and p Q = S Q B Q 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 + Δ u + f ( t , r , u ) = 0 and u t + F ( t , x , u , D u , 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 Δ u = 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 ( x Ω , t J ) ,
then u has the symmetry and monotonicity properties for each t R :
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 ( x Ω ) .
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 + 2 j 2 k .

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 φ λ ( x , t ) φ 0 .
(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 u c ( x , t ) u 0 , ( x , t ) Q ,
(2.2)
u 0 , ( x , t ) p Q .
(2.3)
If
c ( x , t ) > λ ( x , t ) , ( x , t ) Q ,
(2.4)

then u 0 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 D v ( 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 u 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,
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 , u 0 , then we have u 0 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 u 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 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 ( 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 = ( 2 λ x 1 , x 2 , , x n ) , Ω λ = Ω { x Ω : 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 { x Ω : 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 ( 2 λ 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 ( 2 λ 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 ) ) = log f ( t , u λ ) .
(3.3)
On the other hand, from (1.1), we have
log ( u t ) + log ( det ( D 2 u ) ) = log f ( 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 ) ) = log f ( t , u ) log f ( 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 + ( 1 s ) 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 < t T .
(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 , x Ω Ω λ .
Thus
w λ ( x , t ) = u u λ = 0 u λ 0 , x Ω Ω λ , 0 < t T .
(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 x Ω , 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 / 2 Λ . 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 < t T .
(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 < t T .

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 ) = 2 log ( 1 + e 2 x ) 2 x log 4 ( x > 0 ) satisfies f = e f in R 1 , and f ( x ) = f ( x ) , x < 0 . Define
u ( x , t ) = log ( T t ) + 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). □

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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