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A Liouville-type theorem for an integral system on a half-space R + n

Journal of Inequalities and Applications20132013:37

https://doi.org/10.1186/1029-242X-2013-37

  • Received: 13 October 2012
  • Accepted: 17 January 2013
  • Published:

Abstract

Let R + n be an n-dimensional upper half Euclidean space, and let α be any real number satisfying 0 < α < n . In our previous paper (Cao and Dai in J. Math. Anal. Appl. 389:1365-1373, 2012), we considered the single equation

u ( x ) = R + n ( 1 | x y | n α 1 | x y | n α ) u r ( y ) d y ,
(0.1)

where x = ( x 1 , , x n 1 , x n ) is the reflection of the point x about the R + n . We obtained the monotonicity and nonexistence of positive solutions to equation (0.1) under some integrability conditions when r > n n α . In (Zhuo and Li in J. Math. Anal. Appl. 381:392-401, 2011), the authors discussed the following system of integral equations in R + n :

{ u ( x ) = R + n ( 1 | x y | n α 1 | x y | n α ) v q ( y ) d y , v ( x ) = R + n ( 1 | x y | n α 1 | x y | n α ) u p ( y ) d y
(0.2)

with 1 q + 1 + 1 p + 1 = n α n . They obtained rotational symmetry of positive solutions of (0.2) about some line parallel to x n -axis under the assumption u L p + 1 ( R + n ) and v L q + 1 ( R + n ) . In this paper, we derive nonexistence results of such positive solutions for (0.2). In particular, we present a simple and more general method for the study of symmetry and monotonicity which has been extensively used in various forms on a half-space.

AMS Subject Classification:35B05, 35B45.

Keywords

  • Liouville-type theorem
  • HLS inequality
  • systems of integral equations
  • monotonicity
  • moving planes method

1 Introduction

By a Liouville-type theorem, we here mean the statement of nonexistence of nontrivial (bounded or not) solutions on the whole space or on a half-space. In the last two decades, Liouville-type theorems have been widely used, in conjunction with rescaling arguments, to derive a priori estimates for solutions of boundary value problems.

Let R + n be the n-dimensional upper half Euclidean space
R + n = { x = ( x 1 , x 2 , , x n ) R n | x n > 0 } .
In our previous paper [1], we studied the integral equation in R + n :
u ( x ) = R + n ( 1 | x y | n α 1 | x y | n α ) u r ( y ) d y , u ( x ) > 0 , x R + n .
(1.1)

For r > n n α ( 0 < α < n ), we obtained the following Liouville-type theorem.

Theorem 1.1 [1]

Suppose r > n n α . If the solution u of (1.1) satisfies u L n ( r 1 ) α ( R + n ) and is nonnegative, then u 0 .

The result above motivates us to further study positive solutions of the systems of integral equations in R + n ,
{ u ( x ) = R + n ( 1 | x y | n α 1 | x y | n α ) v q ( y ) d y , v ( x ) = R + n ( 1 | x y | n α 1 | x y | n α ) u p ( y ) d y ,
(1.2)
where p and q satisfy
1 q + 1 + 1 p + 1 = n α n .

This is the so-called critical case.

In [2] Zhuo and Li discussed regularity and rotational symmetry of solutions for integral system (1.2).

Theorem 1.2 [2]

Let ( u , v ) be a pair of positive solutions of (1.2) with p , q 1 . Assume that u L p + 1 ( R + n ) and v L q + 1 ( R + n ) , then every positive solution ( u , v ) of (1.2) is rotationally symmetric about some line parallel to x n -axis.

They also showed close relationships between integral equation (1.2) and the following PDEs system:
{ ( ) α 2 u = v q , u > 0 ,  in  R + n ; ( ) α 2 v = u p , v > 0 ,  in  R + n ; u = ( ) u = = ( ) α 2 1 u = 0 , on  R + n ; v = ( ) v = = ( ) α 2 1 v = 0 , on  R + n ,
(1.3)

where α is an even number.

Theorem 1.3 [2]

Let ( u , v ) be a pair of solutions of (1.2) up to a constant, then ( u , v ) satisfies (1.3).

In this paper, we use a simple and more general method to derive that the solution pair ( u , v ) of (1.2) is strictly monotonically increasing with respect to the variable x n and further present the nonexistence of positive solutions of (1.2) under some integrability conditions.

Theorem 1.4 Let ( u , v ) be a pair of positive solutions of (1.2) with p , q 1 . Assume that u L p + 1 ( R + n ) and v L q + 1 ( R + n ) , then both u and v are strictly monotonically increasing with respect to the variable x n .

Theorem 1.5 Let ( u , v ) be a pair of positive solutions of (1.2) with p , q 1 . Assume that u L p + 1 ( R + n ) and v L q + 1 ( R + n ) are nonnegative, then u = v 0 .

2 Properties of the function G ( x , y )

In this section, we introduce some properties of the function G ( x , y ) which is defined on a half-space. By using the properties, one could find a simple and general method for the study of symmetry and monotonicity which has been used in various forms defined in a half-space. More precisely, for x , y R + n , define
G ( x , y ) = 1 | x y | n α 1 | x y | n α ,

where x = ( x 1 , , x n 1 , x n ) is a reflection of the point x about the R + n .

Let λ be a positive real number. Define
Σ λ = { x = ( x 1 , x 2 , , x n ) R n | 0 < x n < λ } , T λ = { x R + n | x n = λ }
and
Σ λ C = R + n Σ λ ,

the complement of Σ λ in R + n .

Let
x λ = ( x 1 , x 2 , , x n 1 , 2 λ x n )

be a reflection of the point x = ( x 1 , x 2 , , x n ) about the plane T λ .

To this end, for x , y R + n , define
d ( x , y ) = | x y | 2
and
θ ( x , y ) = { 4 x n y n , if  x , y R + n , 0 , x R + n  or  y R + n .
Then, for x , y R + n , x y , we have the following expression:
G ( x , y ) = H ( d ( x , y ) , θ ( x , y ) ) .
Here H : ( 0 , ) × [ 0 , ) R ,
H ( s , t ) = 1 s γ 1 ( s + t ) γ , γ = n α 2 .

The following lemma states some properties of the function G ( x , y ) . Here we present a proof.

Lemma 2.1
  1. (i)
    For any x , y Σ λ , x y , we have
    G ( x λ , y λ ) > max { G ( x λ , y ) , G ( x , y λ ) }
    (2.1)
     
and
G ( x λ , y λ ) G ( x , y ) > | G ( x λ , y ) G ( x , y λ ) | .
(2.2)
  1. (ii)
    For any x Σ λ , y Σ λ C , it holds
    G ( x λ , y ) > G ( x , y ) .
    (2.3)
     
Proof Since x , y Σ λ , it is easy to verify that
d ( x λ , y λ ) = d ( x , y ) < d ( x λ , y )
(2.4)
and
θ ( x λ , y λ ) > θ ( x λ , y ) > θ ( x , y ) .
(2.5)
In fact,
i.e.,
θ ( x λ , y λ ) max { θ ( x λ , y ) , θ ( x , y λ ) } min { θ ( x λ , y ) , θ ( x , y λ ) } θ ( x , y ) .
(2.6)
Consider
G ( x , y ) = H ( s , t ) = 1 s γ 1 ( s + t ) γ
with
s = d ( x , y ) and t = θ ( x , y ) .
Then, for s , t > 0 , we have
(2.7)
(2.8)
(2.9)
  1. (i)

    From (2.4), (2.5), (2.7) and (2.8), we obtain (2.1).

     
While by (2.6) and (2.9), we have
G ( x λ , y λ ) G ( x , y ) = θ ( x , y ) θ ( x λ , y λ ) H ( d ( x , y ) , t ) t d t > θ ( x , y ) θ ( x λ , y λ ) H ( d ( x λ , y ) , t ) t d t θ ( x , y λ ) θ ( x λ , y ) H ( d ( x λ , y ) , t ) t d t = | H ( d ( x λ , y ) , θ ( x λ , y ) ) H ( d ( x , y λ ) , θ ( x , y λ ) ) | = | G ( x λ , y ) G ( x , y λ ) | .
Here we have used the fact that d ( x λ , y ) = d ( x , y λ ) .
  1. (ii)
    Noticing that for x Σ λ and y Σ λ C , we have
    | x λ y | < | x y | and θ ( x λ , y ) > θ ( x , y ) .
     

Then (2.3) follows immediately from (2.7) and (2.8).

This completes the proof of Lemma 2.1. □

Remark The properties of the function G ( x , y ) defined on a half-space are very similar to the properties of Green’s function for a poly-harmonic operator on the ball with Dirichlet boundary conditions. One could find this interesting relation from [3, 4] and [5].

3 The proof of main theorems

In this section, by using the method of moving planes in integral forms, we derive the nonexistence of positive solutions to integral system (1.2) and obtain a new Liouville-type theorem on a half-space. To prove the theorems, we need several lemmas.

Let λ > 0 ,
Σ λ = { x R n | 0 < x n < λ } , Σ ˜ λ = { x λ | x Σ λ } .
Set
u λ ( x ) = u ( x λ ) and v λ ( x ) = v ( x λ ) .
Lemma 3.1 Let ( u , v ) be any pair of positive solutions of (1.2). For any x Σ λ , we have
(3.1)
(3.2)

Proof

Obviously, we have
Now, by properties (2.2) and (2.3) of the function G ( x , y ) and the pair of positive solutions of (1.2), we have
u ( x ) u λ ( x ) Σ λ [ G ( x λ , y λ ) G ( x , y λ ) ] ( v q ( y ) v λ q ( y ) ) d y + Σ λ C Σ ˜ λ [ G ( x , y ) G ( x λ , y ) ] v q ( y ) d y Σ λ [ G ( x λ , y λ ) G ( x , y λ ) ] ( v q ( y ) v λ q ( y ) ) d y .

Similarly, we could derive the second inequality in the lemma. This completes the proof of Lemma 3.1. □

Proof of Theorem 1.4 To prove Theorem 1.4, we compare ( u ( x ) , v ( x ) ) and ( u λ ( x ) , v λ ( x ) ) on Σ λ . The proof consists of two steps.

In the first step, we start from the very low end of our region R + n , i.e., x n = 0 . We will show that for λ sufficiently small,
u λ ( x ) u ( x ) and v λ ( x ) v ( x ) , x Σ λ .
(3.3)

In the second step, we will move our plane T λ toward the positive direction of x n -axis as long as inequality (3.3) holds.

Step 1. Define
Σ λ u = { x | x Σ λ , u ( x ) > u λ ( x ) } ,
and
Σ λ v = { x | x Σ λ , v ( x ) > v λ ( x ) } .
We show that for sufficiently small positive λ, Σ λ u and Σ λ v must both be measure zero. In fact, by Lemma 3.1, it is easy to verify that
u ( x ) u λ ( x ) Σ λ [ G ( x λ , y λ ) G ( x , y λ ) ] ( v p ( y ) v λ p ( y ) ) d y = Σ λ Σ λ v [ G ( x λ , y λ ) G ( x , y λ ) ] ( v p ( y ) v λ p ( y ) ) d y + Σ λ v [ G ( x λ , y λ ) G ( x , y λ ) ] ( v p ( y ) v λ p ( y ) ) d y Σ λ v [ G ( x λ , y λ ) G ( x , y λ ) ] ( v p ( y ) v λ p ( y ) ) d y Σ λ v G ( x λ , y λ ) [ v p ( y ) v λ p ( y ) ] d y p Σ λ v 1 | x y | n α ψ λ p 1 ( y ) [ v ( y ) v λ ( y ) ] d y p Σ λ v 1 | x y | n α v p 1 ( y ) [ v ( y ) v λ ( y ) ] d y ,
where ψ λ ( y ) is valued between v ( y ) and v λ ( y ) . Therefore, on Σ λ v we have
0 v λ ( y ) ψ λ ( y ) v ( y ) .
It follows from the Hardy-Littlewood-Sobolev inequality that
u λ u L p + 1 ( Σ λ u ) C v q 1 ( v λ v ) L ( q + 1 ) / q ( Σ λ v ) .
(3.4)
Then by the Hölder inequality,
u λ u L p + 1 ( Σ λ u ) C v L q + 1 ( Σ λ v ) q 1 v λ v L q + 1 ( Σ λ v ) .
(3.5)
Similarly, one can show that
v λ v L q + 1 ( Σ λ v ) C u L p + 1 ( Σ λ u ) p 1 u λ u L p + 1 ( Σ λ u ) .
(3.6)
Combining (3.5) and (3.6), we arrive at
u λ u L p + 1 ( Σ λ u ) C v L q + 1 ( Σ λ v ) q 1 u L p + 1 ( Σ λ u ) p 1 u λ u L p + 1 ( Σ λ u ) .
(3.7)
By the conditions that u L p + 1 ( R + n ) and v L q + 1 ( R + n ) , we can choose sufficiently small positive λ such that
C v L q + 1 ( Σ λ v ) q 1 u L p + 1 ( Σ λ u ) p 1 1 2 .

Now, inequality (3.7) implies u λ u L p + 1 ( Σ λ u ) = 0 , and therefore Σ λ u must be measure zero. Similarly, one can show that Σ λ v is measure zero. Therefore, (3.3) holds. This completes Step 1.

Step 2. (Move the plane to the limiting position to derive symmetry and monotonicity.)

Inequality (3.3) provides a starting point to move the plane T λ . Now, we start from the neighborhood of x n = 0 and move the plane up as long as (3.3) holds to the limiting position. We will show that the solution u ( x ) must be symmetric about the limiting plane and be strictly monotonically increasing with respect to the variable x n . More precisely, define
λ 0 = sup { λ | u ( x ) u μ ( x )  and  v ( x ) v μ ( x ) , x Σ μ , μ λ } .
Suppose that for such a λ 0 , we will show that both u ( x ) and v ( x ) must be symmetric about the plane T λ 0 by using a contradiction argument. Assume that on Σ λ 0 , we have
u ( x ) u λ 0 ( x ) and v ( x ) v λ 0 ( x ) , but u ( x ) u λ 0 ( x ) or v ( x ) v λ 0 ( x ) .
We show that the plane can be moved further up. More precisely, there exists an ϵ > 0 depending on n, α, and the solution ( u ( x ) , v ( x ) ) such that
u ( x ) u λ ( x ) and v ( x ) v λ ( x ) on  Σ λ  for all  λ  in  [ λ 0 , λ 0 + ϵ ) .
(3.8)
In the case
v ( x ) v λ 0 ( x ) on  Σ λ 0 ,
by Lemma 3.1, we have in fact u ( x ) < u λ 0 ( x ) in the interior of Σ λ 0 . Let
Σ λ 0 u ¯ = { x Σ λ 0 | u ( x ) u λ 0 ( x ) } and Σ λ 0 v ¯ = { x Σ λ 0 | v ( x ) v λ 0 ( x } .
Then, obviously, Σ λ 0 u ¯ has measure zero and lim λ λ 0 Σ λ u Σ λ 0 u ¯ . The same is true for that of v. From (3.5) and (3.6), we deduce
u λ u L p + 1 ( Σ λ u ) C v L q + 1 ( Σ λ v ) q 1 u L p + 1 ( Σ λ u ) p 1 u λ u L p + 1 ( Σ λ u ) .
(3.9)
Again, the conditions that u L p + 1 ( R + n ) and v L q + 1 ( R + n ) ensure that one can choose ϵ sufficiently small, so that for all λ in [ λ 0 , λ 0 + ϵ ) ,
C v L q + 1 ( Σ λ v ) q 1 u L p + 1 ( Σ λ u ) p 1 1 2 .

Now, by (3.9), we have u λ u L p + 1 ( Σ λ u ) = 0 , therefore Σ λ u must be measure zero. Similarly, Σ λ v must also be measure zero. This verifies (3.8), therefore both u ( x ) and v ( x ) are symmetric about the plane T λ 0 . Also, the monotonicity easily follows from the argument. This completes the proof of Theorem 1.4. □

Proof of Theorem 1.5 To prove the theorem, firstly we will show that the plane cannot stop at x n = λ 0 for some λ 0 < + , that is, we will prove that λ 0 = + .

Suppose that λ 0 < + , the process of Theorem 1.4 shows that the plane x n = 2 λ 0 is the symmetric points of the boundary R + n with respect to the plane T λ 0 , and we derive that u ( x ) = 0 and v ( x ) = 0 when x is on the plane x n = 2 λ 0 . This contradicts the pair of positive solutions ( u ( x ) , v ( x ) ) of (1.2), thus λ 0 = + .

Besides, we know that both u ( x ) and v ( x ) of positive solutions of (1.2) are strictly monotonically increasing in the positive direction of x n -axis, but u L p + 1 ( R + n ) and v L q + 1 ( R + n ) , so we come to the conclusion that the pair of positive solutions ( u ( x ) , v ( x ) ) of (1.2) does not exist.

This completes the proof of Theorem 1.5. □

Declarations

Acknowledgements

Most of this work was completed when the first author was visiting Yeshiva University, and she would like to thank the Department of Mathematics for the hospitality. Besides, the authors would like to express their gratitude to Professor Wenxiong Chen for his hospitality and many valuable discussions. This work is partially supported by the National Natural Science Foundation of China (No. 11171091; No. 11001076), NSF of Henan Provincial Education Committee (No. 2011A110008) and Foundation for University Key Teacher of Henan Province.

Authors’ Affiliations

(1)
College of Mathematics and Information Science, Henan Normal University, Xinxiang, Henan, 453007, P.R. China
(2)
Department of Computer Science, Henan Normal University, Xinxiang, Henan, 453007, P.R. China

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Copyright

© Cao and Dai; licensee Springer 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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