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BY 4.0 license Open Access Published by De Gruyter Open Access December 31, 2022

Symmetric results of a Hénon-type elliptic system with coupled linear part

  • Zhenluo Lou EMAIL logo , Huimin Li and Ping Zhang
From the journal Open Mathematics

Abstract

In this article, we study the elliptic system:

Δ u + μ 1 u = x α u 3 + λ v , x Ω Δ v + μ 2 v = x α v 3 + λ u , x Ω u , v > 0 , x Ω , u = v = 0 , x Ω ,

where Ω R 3 is the unit ball. By the variational method, we prove that if α is sufficiently small, the ground state solutions of the system are radial symmetric, and if α > 0 is sufficiently large, the ground state solutions are nonradial; however, the solutions are Schwarz symmetry.

MSC 2010: 35A15; 35B06; 35B09; 35B38; 35J57

1 Introduction

In this article, we study the following elliptic system:

(1.1) Δ u + μ 1 u = x α u 3 + λ v , x Ω Δ v + μ 2 v = x α v 3 + λ u , x Ω u , v > 0 , x Ω , u = v = 0 , x Ω ,

where Ω R 3 is the unit ball, μ 1 , μ 2 > 0 , α > 0 . The parameter λ > 0 satisfies several conditions.

System (1.1) is the nonautonomous case of the following elliptic system:

(1.2) Δ u + μ 1 u = u 3 + λ v , x Ω Δ v + μ 2 v = v 3 + λ u , x Ω u , v > 0 , x Ω , u = v = 0 , x Ω ,

where Ω R N ( N = 2 , 3 ) is an open domain. (1.2) arises in a binary mixture of the Bose-Einstein condensates with a coupled linear part. In recent years, many researchers have been concerned with system (1.2). They have proved the existence, multiplicity, and limit problems of the solutions of system (1.2), and see [14] and the references therein.

Gidas et al. [5] studied the elliptic equation:

(1.3) Δ u = f ( x , u ) , u > 0 , x Ω , u = 0 , x Ω .

They proved that if Ω is a ball and f ( x , u ) = f ( u ) C 1 , then the solutions are radially symmetric. After that, many researchers studied the symmetric properties of elliptic equations, and see [69] and the references therein. In [7], they pointed out that if the domain cannot be convex or f ( x , u ) is increasing by x , one can have the nonradial positive solutions. Smets et al. [9] studied the Hénon equation:

(1.4) Δ u = x α u p , u > 0 , x Ω , u = 0 x Ω ,

where Ω is the unit ball and α > 0 , 1 < p < 2 1 . They proved that if α > 0 is large enough, the ground state solutions of (1.4) is nonradial symmetry, and Boheure et al. [6] studied the symmetry-breaking results of the Hénon-Lane-Emden system, and see also [7] and the references therein.

Following these ideas, we prove the following.

Theorem 1.1

Let λ > 0 be small. Then there exists α ¯ > 0 such that for all α > α ¯ the ground state solution is nonradial. However, the solutions are foliated Schwarz symmetric.

In [8,10,11], Kajikiya obtained the multiple nonradial solutions of equation (1.4) by the group critical theory. By the unique and nondegenerate results, they proved, if α is sufficiently small, the ground state solutions of (1.4) is radially symmetric. It is an interesting result; since for the Hénon equation, the moving plane method cannot be used, one also prove the radial results. In [12,13], the authors also considered the nonradial solutions. In this article, we first obtain a result of unicity, then we prove the following:

Theorem 1.2

For all λ > 0 be sufficiently small, the system (1.2) has a unique positive radial solution. If this solution is nondegenerate, then for α small, the ground state solution of system (1.1) should be radially symmetric.

This article is organized as follows. In Section 2, we present several preliminary results, and in Sections 3 and 4, we present the proof of the main results.

2 Some preliminaries and lemmas

Let H 0 1 ( Ω ) be the classical Sobolev space and = H 0 1 ( Ω ) × H 0 1 ( Ω ) with the norm:

( u , v ) = Ω u 2 + μ 1 u 2 + v 2 + μ 2 v 2 d x 1 2 .

We assume 0 < λ < min { μ 1 , μ 2 } .

The energy functionals of (1.1) and (1.2) are

I α ( u , v ) = 1 2 ( u , v ) 2 1 4 Ω x α ( u 4 + v 4 ) d x λ Ω u v d x

and

I ( u , v ) = 1 2 ( u , v ) 2 1 4 Ω ( u 4 + v 4 ) d x λ Ω u v d x .

The Nehari manifolds are

N α = { ( u , v ) \ { ( 0 , 0 ) } : I α ( u , v ) ( u , v ) = 0 }

and

N = { ( u , v ) \ { ( 0 , 0 ) } : I ( u , v ) ( u , v ) = 0 } .

Definition 2.1

( u ¯ , v ¯ ) is the ground state solution of (1.1), if it satisfies

c α = I α ( u ¯ , v ¯ ) = inf N α I α ( u , v ) .

If the infimum is obtained from the radial functions, we denote it as c α r .

By the moving plane method, one can prove that the solutions of system (1.2), denoted ( ω , h ) , are radial. If

(2.1) Δ ϕ + μ 1 ϕ = 3 ω 2 ϕ + λ ψ , Δ ψ + μ 2 ψ = 3 h 2 ψ + λ ϕ

has only trivial solutions, in this case ( ω , h ) is nondegenerate. One can check Damascelli and Pacella [14] and Troy [15] for the radially symmetric results, see also Dancer et al. [16], Sato and Wang [17], and Tavares and Weth [18]. For unique and nondegenerate results of the elliptic system, one can see [1921].

3 The Proof of Theorem 1.1

Lemma 3.1

For any ( u , v ) , there exists a unique t > 0 such that ( t u , t v ) N α .

Proof

For any ( u , v ) H , by calculating directly, we have

(3.1) I α ( t u , t v ) ( t u , t v ) = t 2 ( u , v ) 2 2 t 2 λ Ω u v d x t 4 Ω x α ( u 4 + v 4 ) d x ,

and thus,

t 2 = ( u , v ) 2 2 λ Ω u v d x Ω x α ( u 4 + v 4 ) d x .

We have completed the proof.□

Lemma 3.2

c α is achieved, and if α > 0 is large enough, we have that

c α C α ,

where C is a constant independent of α .

Proof

By the variational method, we can prove that c α and c α r are achieved, and c α c α r . Now we give the estimate of c α .

Let φ , ψ C 0 ( Ω ) and x 0 = 1 1 α , 0 , 0 , and

φ α ( x ) = φ ( α ( x x 0 ) ) , ψ α ( x ) = ψ ( α ( x x 0 ) ) .

When α is large, one can prove supp φ α Ω , supp ψ α Ω . Since C 0 ( Ω ) H 0 1 ( Ω ) , then

c α I α ( t φ , t ψ ) ,

where t satisfies ( t φ , t ψ ) N α . Then we have

I α ( t φ α , t ψ α ) = 1 4 t 4 Ω x α ( φ α 4 + ψ α 4 ) d x

and

t 2 = ( φ α , ψ α ) 2 2 λ Ω φ α ψ α d x Ω x 4 ( φ α 4 + ψ α 4 ) d x .

By combining these results, we have that

I α ( t φ α , t ψ α ) = 1 4 ( φ α , ψ α ) 2 2 λ Ω φ α ψ α d x 2 Ω x 4 ( φ α 4 + ψ α 4 ) d x .

Let y = α ( x x 0 ) , then d x = 1 α 3 d y . Because of supp φ α Ω , then we have y Ω , so x x α Ω 1 α , which is a ball with r = 1 α , and then we have that

(3.2) Ω x α ( φ α 4 + ψ α 4 ) d x 1 2 α α Ω ( φ α 4 + ψ α 4 ) d x = 1 2 α α Ω ( φ 4 + ψ 4 ) d x = 1 2 α α Ω 1 α 3 ( φ 4 + ψ 4 ) d y .

On the other hand, we have

(3.3) Ω ( φ α 2 + φ α 2 ) d x = Ω α 3 ( φ 2 + φ 2 ) d y = α 1 Ω ( α 2 φ 2 + α 2 φ 2 ) d y α 1 Ω ( φ 2 + α 2 φ 2 ) d y ,

and similarly, we have that

(3.4) Ω ( ψ α 2 + ψ α 2 ) d x α 1 Ω ( ψ 2 + α 2 ψ 2 ) d y .

Combining the previous inequalities (3.2)–(3.4), we have that

c α C α ,

where C is a positive constant independent of α .□

Lemma 3.3

Let ( u α , v α ) be a radial ground state solution of (1.1) and c α r is the energy value. Then we have that

c α r C α 3 ,

where C is a constant independent of α .

Proof

For convenience, we replace ( u , v ) of ( u α , v α ) . Define

Y ( x ) = ( u , x ) u + ( v , x ) v 1 2 ( u 2 + v 2 ) x + 1 4 x α ( u 4 + v 4 ) x + λ u v x 1 2 ( μ 1 u 2 + μ 2 v 2 ) x ,

where x R 3 . Let

r = x 1 2 + x 2 2 + x 3 2 ,

then we have

( u , x ) u = ( u r ) 2 ( x 1 , x 2 , x 3 ) , ( v , x ) v = ( v r ) 2 ( x 1 , x 2 , x 3 ) ,

and

(3.5) div ( u , x ) u 1 2 u 2 x + 1 4 x α u 4 x 1 2 u 2 x = 3 2 u 2 + 3 + α 4 x α u 4 3 2 u 2 + r u r ( u r + u 3 x α u ) ,

and

(3.6) div ( v , x ) v 1 2 v 2 x + 1 4 x α v 4 x 1 2 v 2 x = 3 2 v 2 + 3 + α 4 x α v 4 3 2 v 2 + r v r ( v r + v 3 x α v ) ,

and

(3.7) div ( λ u v x ) = 3 λ u v + λ r u v r + λ r v u r .

By combining the previous equalities, we obtain the following:

(3.8) div Y ( x ) = 3 2 ( u 2 + v 2 ) + 3 + α 4 x α ( u 4 + v 4 ) + 3 λ u v 3 2 ( μ 1 u 2 + μ 2 v 2 ) + r u r ( u r + u 3 x α μ 1 u + λ v ) + r v r ( v r + v 3 x α μ 2 v + λ u ) .

Since ( u , v ) is a radial solution,

u r = Δ u 2 u r r , v r = Δ v 2 v r r ,

r u r ( u r + u 3 x α μ 1 u + λ v ) = r u r Δ u 2 u r r + x α u 3 μ 1 u + λ v = r u r 2 u r r = 2 ( u r ) 2 = 2 u 2 ,

2 v 2 = r v r ( v r + v 3 x α μ 2 v + λ u ) .

Thus,

div Y ( x ) = 3 + α 4 x α ( u 4 + v 4 ) + 3 λ u v 1 2 ( u 2 + v 2 ) 3 2 ( μ 1 u 2 + μ 2 v 2 ) .

By divergence theorem, we obtain that , i.e.,

Ω div Y ( x ) d x = Ω Y ( x ) n d S

Ω ( u , x ) u n + ( v , x ) v n 1 2 ( u 2 x + v x ) n d S = Ω 3 + α 4 x α ( u 4 + v 4 ) + 3 λ ( u 2 + v 2 ) 3 2 ( μ 1 u 2 + μ 2 v 2 ) d x .

Again since ( u , v ) is a solution, we have that

Ω ( u , x ) u n + ( v , x ) v n 1 2 ( u 2 + v 2 ) ( x n ) d S = α 3 4 Ω x α ( u 4 + v 4 ) d x + Ω ( u 2 + v 2 ) d x α 3 4 Ω x α ( u 4 + v 4 ) d x .

Since u and v are radial functions, following pp. 72 of [6], we have

(3.9) ( u , x ) ( v , x ) = ( u , v ) x 2

and

Ω u n 2 d S = C Ω u n d S 2 , Ω v n 2 d S = C Ω v n d S 2 .

Since

c α = 1 4 Ω x α ( u 4 + v 4 ) d x

and

Ω ( u , x ) u n + ( v , x ) v n 1 2 ( u 2 + v 2 ) ( x n ) d S = 1 2 Ω u n 2 + v n 2 d S ,

and then we obtain

(3.10) ( α 3 ) c α 1 2 Ω u n 2 + v n 2 d S .

By the divergence theorem, we have Ω Δ u d x = Ω u n d S . Notice that

Δ u + μ 1 u = x α u 3 + λ v ,

and

(3.11) Ω x α u 3 d x = Ω x α 4 x 3 α 4 u 3 d x Ω x α d x 1 4 Ω x α u 4 d x 3 4 = C 0 1 r α + 2 d r 1 4 Ω x α u 4 d x 3 4 = C ( α + 3 ) 1 4 ( Ω x α u 4 d x ) 3 4 ,

and

(3.12) Ω λ v + μ 1 u d x C Ω u 2 d x 1 2 + Ω v 2 d x 1 2 2 C Ω u 2 + v 2 d x 1 2 .

Thus, we obtain that

(3.13) Ω Δ u d x 1 ( α + 3 ) 1 4 Ω x α u 4 d x 3 4 + 2 C Ω u 2 + v 2 d x 1 2 1 ( α + 3 ) 1 4 c α 3 4 + C c α 1 2

and

(3.14) Ω Δ v d x 1 ( α + 3 ) 1 4 c α 3 4 + C ( λ ) c α 1 2 .

We also obtain that

Ω u n 2 + v n 2 d S = C Ω Δ u d x 2 + Ω Δ v d x 2 ,

and by the previous estimations, one can obtain that

(3.15) Ω u n 2 + v n 2 d S C 1 ( α + 3 ) 1 4 c α 3 4 + C c α 1 2 2 .

From (3.10), we obtain

( α 3 ) c α 1 ( α + 3 ) 1 4 c α 3 4 + C c α 1 2 2 ,

and then we have

( α 3 ) 1 2 c α 1 2 C 1 ( α + 3 ) 1 4 c α 3 4 + C 2 c α 1 2

and

( α 3 ) 1 2 C 1 ( α + 3 ) 1 4 c α 1 4 + C 2 .

If α > 0 is large enough, we obtain α 3 4 C c α 1 4 , i.e.,

c α C α 3 ,

where C , C 1 , and C 2 are positive constants independently of α . In conclusion, we complete the proof.□

Now we list the proof of Theorem 1.1.

The proof of Theorem 1.1

By Lemmas 3.2 and 3.3, we can prove that if α > 0 is large enough, the ground state solutions cannot be radial. Let

f 1 ( x , u , v ) = x α u 3 + λ v , f 2 ( x , u , v ) = x α v 3 + λ u ,

then f 1 ( x , u , v ) and f 2 ( x , u , v ) are convex with respect to u and v . Since ( u , v ) is a ground state solution, then the Morse index M ( u , v ) 1 < N = 3 , and then by [14], we have that ( u , v ) is foliated Schwarz symmetry, and one can also follow the idea of Wang and Willem [23].

4 The Proof of Theorem 1.2

By using the idea of Wei and Yao [20], we obtain the following results.

Lemma 4.1

Let λ > 0 be small enough. Then the positive solution of system (1.2) is radially symmetric, and it is unique.

Proof

Following the idea of Gidas et al. [5], Damascelli and Pacella [14], or Troy [15], we prove that the positive solutions are radially symmetric.

Let ω i be the unique positive solution of the following scalar equation with the subcritical exponent:

Δ u + μ i u = u 3 x Ω , u = 0 x Ω ,

where i = 1 , 2 .

Let Φ ( λ , u , v ) = I α ( u , v ) , one can prove that Φ ( 0 , ω 1 , ω 2 ) = 0 , and we have

Φ ( 0 , ω 1 , ω 2 ) ( ϕ , ψ ) = Ω ϕ 2 + μ 1 ϕ 2 + ψ 2 + ψ 2 d x 3 Ω ω 1 2 ϕ 2 + ω 2 ψ 2 d x .

By the compact theorem, one can prove that Φ ( 0 , ω 1 , ω 2 ) is invertible, and then by implicity function theorem, if λ > 0 is small enough, the solution of system is uniqueness.□

Next we present the proof of Theorem 1.2.

By contradiction, the ground state solutions of (1.1) are nonradial. By the variational method, we prove that (1.1) has a nontrivial radial solution, one can see [22], denoting ( u ˜ α , v ˜ α ) , and another nonradial solution, denoting ( u α , v α ) . Now we give the L estimate of ( u α , v α ) and ( u ˜ α , v ˜ α ) , for convenience, we replace ( u α , v α ) by ( u , v ) . Since ( u , v ) is a solution of (1.1), let u 2 q 1 be a test function, then we have

Ω Δ u u 2 q 1 + μ 1 u 2 q d x = Ω x α u 2 q + 2 d x + λ Ω u 2 q 1 v d x ,

where 2 q 1 > 1 . By Green’s formula, we obtain that

LHS = Ω u ( u 2 q 1 ) + μ 1 u 2 q d x = ( 2 q 1 ) Ω u 2 u 2 q 2 d x + μ 1 Ω u 2 q d x = 2 q 1 q 2 Ω u q 2 d x + μ 1 Ω u q 2 d x .

By Hölder’s and Young’s inequality, we have that

Ω v u 2 q 1 d x Ω u 2 q + 2 d x 2 q 1 2 q + 2 Ω v 2 q + 2 3 d x 3 2 q + 2 2 q 1 2 q + 2 Ω u 2 q + 2 d x + 3 2 q + 2 Ω v 2 q + 2 3 d x 2 q 1 2 q + 2 Ω u 2 q + 2 d x + 3 2 q + 2 Ω v 2 q + 2 1 3 Ω 1 d x 2 3 Ω u 2 q + 2 d x + C C Ω u 2 q + 2 d x ,

where C is a constant independent of q . On the other hand,

Ω x α u 2 q + 2 d x Ω u 2 q + 2 d x .

Then we obtain that

(4.1) Ω u q 2 + u q 2 d x C q 2 2 q 1 Ω u 2 q + 2 d x .

Again by Hölder and Sobolev embedding inequality, we obtain

Ω u 2 q + 2 d x Ω u 3 q d x 2 3 ( u 6 d x ) 1 3 C Ω u 3 q d x 2 3

and

Ω u 4 q d x 1 4 Ω u q 2 + u q 2 d x 1 2 .

Thus, we obtain that

Ω u 4 q d x 1 4 q c q 2 2 q 1 1 2 q Ω u 3 q d x 1 3 q .

Now let q ( 1 ) = 1 and q ( n + 1 ) = 4 3 q ( n ) , and one can prove that q ( n ) + as n + . By the proceeding estimate, we have that

u 4 q ( n + 1 ) C i = 2 n + 1 q 2 ( i ) 2 q ( i ) 1 1 2 q ( i ) u 3 q ( 1 ) .

Notice that q ( n + 1 ) = 4 3 n , and thus, we can prove that

ln i = 2 n + 1 q 2 ( i ) 2 q ( i ) 1 1 2 q ( i ) = i = 2 n + 1 2 ln q ( i ) ln ( 2 q ( i ) 1 ) 2 q ( i ) < ,

and thus, i = 2 n + 1 q 2 ( i ) 2 q ( i ) 1 1 2 q ( i ) < , then we complete the L estimate.

Let λ > 0 satisfy the previous conditions, and { α n } satisfy α n 0 as n + . We denote that { ( u n , v n ) } is the ground state solutions of system (1) and { ( u ˜ n , v ˜ n ) } is the radial ground state solutions of system (1).

Following the idea of [8], by contradiction, u n u ˜ n , v n v ˜ n , let

ω n = u n u ˜ n u n u ˜ n , h n = v n v ˜ n v n v ˜ n L ,

then ω n 0 , h n 0 and ω n , h n W 2 , t ( Ω ) , which is classical Sobolev space, for any 1 < t < + . Let

f 1 ( u , v ) = x α u 3 + λ v and f 2 ( u , v ) = x α v 3 + λ u ,

then

(4.2) Δ ω n + μ 1 ω n = f 1 ( u n , v n ) f 1 ( u ˜ n , v ˜ n ) u n u ˜ n ω n , Δ h n + μ 2 h n = f 2 ( u n , v n ) f 2 ( u ˜ n , v ˜ n ) v n v ˜ n h n .

By the L estimate, we have ω n ω , h n h in C 1 ( Ω ) , and ω and h satisfy

(4.3) Δ ω + μ 1 ω = 3 u 0 2 ω + λ c 1 h , Δ h + μ 2 h = 3 v 0 2 h + λ c 2 ω ,

where c 1 = lim n + v n v ˜ n u n u ˜ n and c 2 = 1 c 1 are positive constants, and ( u o , v 0 ) is the uniqueness solution of (1.2).

Now let ω ¯ = ω c 1 and h ¯ = h , we have that

(4.4) Δ ω ¯ + μ 1 ω ¯ = 3 u 0 2 ω ¯ + λ h ¯ , Δ h ¯ + μ 2 h ¯ = 3 λ 2 v 0 2 h ¯ + λ ω ¯ .

It is a contradiction, then we complete the proof.

Acknowledgement

The authors thank the referees for valuable comments and suggestions, which improved the presentation of the manuscript.

  1. Funding information: Lou is supported by NSFC 11571339 and 12101192 and Key Scientific Research Projects of Higher Education Institutions in Henan Province 20B110004.

  2. Conflict of interest: The authors state no conflict of interest.

  3. Data availability statement: No data, models, or code are generated or used during the study.

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Received: 2022-02-13
Revised: 2022-11-28
Accepted: 2022-12-03
Published Online: 2022-12-31

© 2022 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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