THE PERTURBATION PROBLEM OF AN ELLIPTIC SYSTEM WITH SOBOLEV CRITICAL GROWTH?

Qi LI (李奇)

School of Mathematics and Statistics, Central China Normal University, Wuhan 430079, China

E-mail : qili@mails.ccnu.edu.cn

1 Introduction

In this paper, we consider the following coupled elliptic system:

Systems like (1.1) are general versions of nonlinear elliptic systems, such as the Bose-Einstein condensate, that arise in mathematical physics. We refer to [1–9, 21, 22]and references therein for more on systems (equations) with both critical and subcritical exponent. In particular, the case in which the coupling is nonlinear and critical has received a great deal of attention recently, and significant progress has been made in the last thirty years since the celebrated work of Brezis and Nirenberg [10].

Denote HRN= D1,2(RN)× D1,2(RN) with the normWe call a solution (u,v) positive if both u and v are positive, (u,v) nontrivial ifand (u,v) semi-trivial if (u,v) is of the form (u,0) or (0,v). It is well known that solutions of problem (1.1) can be attained by finding nontrivial critical points of the functional

Here u±:=max{±u, 0}.

We will mention some results related to problem (1.1). As we know (e.g. [11–13]), the radial function

solves the problem

and zμ,ξachieves the best constant for the embeddingwhere zμ,ξis defined by

We denote

and τ is a solution of equation

From[14],there exists at least one positive root to equation(1.7). More precisely,we have that

(A1) If 1< β <2, 1< α <2 andthen equation (1.7) has a unique positive root.

(A2) If 1< β <2 and α ≥ 2, then equation (1.7) has exactly two positive roots.

(A3) If β =2 and N =3,4,5, then equation (1.7) has a unique positive root.

(A4) If β > 2 and N = 4,5 or 4 < β < 5 and N = 3, then equation (1.7) has exactly two positive roots.

(A5) If 2< β ≤ 4 and N =3, then equation (1.7) has a unique positive root.

In particular, in this paper, we assume that α,β ≥ 2, N = 3, 4. Thus equation (1.7) has a unique positive root τ. Combining (1.5) with (1.6), we get that positive constants s, t and the manifold Z are uniquely determined.

In [15], Abdellaoui, Felli and Peral studied the following coupling system:

where λ1,λ2∈ (0,λN) and λNThey analyzed the behavior of (PS) sequence in order to recover compactness for some ranges of energy levels,and they proved the existence of a ground state solution. In [16], Chen and Zou studied following nonlinear Schrdinger system which is related to the Bose-Einstein condensate:

where ? ? RNis a smooth bounded domain. They proved the existence of positive least energy solutions when N ≥5. In [14], Peng,Peng and Wang considered the following coupling system with critical exponent:

where α, β > 1, α + β = 2?. They got a uniqueness result on the least energy solutions and showed that the manifold Z of the synchronized positive solutions is non-degenerate for some ranges of the parameters α,β,N.

A natural question is whether there are positive solutions to problem (1.1) which approximate (szμ,ξ, tzμ,ξ). The main difficulty is that the embeddingis not compact, so the (PS) sequence will fail to be compact. Motivated by [17, 18], we will adopt a perturbation argument and a finite dimensional reduction method to find positive solutions of problem (1.1).

More precisely, for the existence of positive solutions, we make the following assumptions:

(V2) h, l are continuous functions and have compact supports;

(V3) K, Q ≤ 0 and K(x), Q(x)→ 0 as |x|→ ∞;

(V4) There exists ξ0such that h(ξ0), l(ξ0) > 0 and K(ξ0) = Q(ξ0) = 0. Moreover, we assume that there are positive constants a, b such that as x ? ξ0→ 0, it holds that

where (N +2)/2 ≤a, b

Our main result in this paper can be stated as follows:

Theorem 1.1Suppose that N =3 or 4,α, β ≥ 2 and(V1)–(V4)hold. Then there exists ε0> 0, μ?> 0 and ξ?∈ RNsuch that for all |ε| < ε0, problem (1.1) has a positive solution(uε,vε) with (uε,vε) → (szμ?,ξ?,tzμ?,ξ?) as ε → 0.

In particular, if we suppose that K =Q ≡0, then we have

Corollary 1.2Suppose that N = 3 or 4, α, β ≥ 2, (V1) and (V2) hold. If there exists ξ0such that h and l have the same sign at ξ0, then there exists ε0> 0, μ?> 0 and ξ?∈RNsuch that for all |ε| < ε0, problem (1.1) has a positive solution (uε,vε) with (uε,vε) →(szμ?,ξ?,tzμ?,ξ?) as ε → 0.

2 Preliminary Results

In this section, we will give some lemmas which will be used to prove our main result.

Lemma 2.1Z is a non-degenerate critical manifold of I0, in the sense that

where TzZ is defined by the tangent space to Z at z.

ProofFor proof, the reader can refer to Theorem 1.4 in [14].

Lemma 2.2Let z =(szμ,ξ,tzμ,ξ). Thenwhere I is the identity operator and C is a compact operator.

ProofIt is easy to see that

where ? =(?1,?2),ψ =(ψ1,ψ2),

Suppose that {?n} is a bounded sequence in D1,2(RN). Then,

Thus, it is sufficient to prove that

Then we get

On the other hand, for any E ?RN, we have that

if |E| is small enough. Therefore, we apply the Vitali convergence theorem to get

Next we will use a perturbation argument, developed in [19, 20], which permits us to find critical points of the C2functional

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near a manifold Z of critical points of I0under suitable non-degeneracy conditions. Since 0 < p, q < 1, G fails to be C2on D1,2RN× D1,2RN. To overcome this lack of regularity,we have to modify the abstract approach a little.

First, it is convenient to work in the Banach space

with the norm

where zμ,ξhas been introduced in Section 1. R, μ1and μ2will be chosen later on. In any case,we shall take R in such a way that ω = ω1∪ ω2? BR(0), ω1, ω2are the supports of h(x) and l(x), respectively. Denote

where a depends on R, μ1and μ2.

For (u,v)=z+w ∈U and |x|

In particular, since ω ? BR, it holds that

then Φ ∈ L∞RN. Therefore, if we denote the solution Φ =JΨ, we have following results:

Lemma 2.3If (V1) and (V2) are satisfied, then (JAε,JBε)(U)? X.

ProofFrom (V1) and (V2), it is not hard to see that for any (u,v)∈U,

Lemma 2.4If (V1) and (V2) are satisfied, then (JAε,JBε)∈ C1(U,X) and

where (ζ1,ζ2) is the unique weak solution of the problem

ProofBy a direct calculation, it is not hard to verify that (JAε,JBε) is differentiable and that its Frechet derivative is given by (2.1)–(2.2). Let us now show that d(JAε,JBε) :U → L(X) is continuous. Indeed, let (u,v) ∈ U and (un, vn) ∈ U such that (un,vn) → (u,v)in X. Thus we have that

Direct calculation yields that

Since (un,vn)→(u,v) in X, we obtain that

Moreover, by the Sobolev embedding inequality, we deduce that

On the other hand, by the elliptic regularity, we have that

By (2.4) and (2.5), we have that

Thus the proof is complete.

Remark 2.1When α =2 or β =2, it is easier to check the above result.

Let TzZ =span{q1,··· ,qN+1} denote the tangent space to Z at z =(szμ,ξ,tzμ,ξ), where

We now have the following significant lemma:

Lemma 2.5Suppose that (V1) and (V2) hold. Then there exists ε0and a C1function

such that for any μ1< μ < μ2, ξ ∈ BR(0) and ε ∈ (?ε0,ε0), we have that

(i) (w1(μ,ξ,ε), (w2(μ,ξ,ε)) is orthogonal to TzZ;

(ii) z+w(μ,ξ,ε)? J(Aε,Bε)(szμ,ξ+w1(μ,ξ,ε),tzμ,,ξ+w2(μ,ξ,ε))∈ TzZ;

ProofLet

where z =(szμ,ξ,tzμ,ξ), w =(w1,w2). We defineby

where u=(u1,u2), v =(v1,v2).

From Lemma 2.4, it follows that H is of class C1and that its derivative with respect to variables (w,σ) is given by

where φ =(φ1,φ2)∈ HRN, d=(d1,··· ,dN+1)∈ RN+1. For any z ∈ Z0,

and its derivative at z ∈ Z0, ε=0, σ =0 and w =0 is as follows

From Lemma 2.2, we can see thatand that it is a Fredholm operator of index 0. On the other hand,is injective, since Z is a nondegenerate manifold of I0. Thus,is invertible. Finally, it is easy to get our result by using the Implicit Function Theorem; for details see [19].

From Lemma 2.5, it is natural to introduce the perturbed manifold

which is a natural constraint for Iε; namely, if u ∈ Zεand

In fact, if zεis a critical point of Iεconstrained on Zε, then we have that

On the other hand, from Lemma 2.5 (ii) we have that

Moreover, we have that

Consequently, if z is a critical point of the restriction G|Z, which is a proper local minimum or maximum point, then z+w(μ,ξ,ε) turns out to be a critical point of Iε. From the above analysis,the search for solutions to problem(1.1)is reduced to the search for the local minimum or maximum point of the finite dimensional functional

on (μ1,μ2)× BR(0). In order to simplify the functional, we denote

3 Proof of the Main Result

In this section, we will solve the finite dimensional problem. Before proving our result, we give some important lemmas.

Lemma 3.1Assume that K ∈L1(RN)∩L∞(RN) and K(x)→0 as |x|→∞. Then,

ProofIf μ → 0 and |ξ|→ ∞, then

Note that

Thus, by the Dominated Convergence Theorem, we have proven the result.

where R>0, I1and I2are defined by

By a direct calculation, we obtain that

Thus, for given δ > 0, by K ∈ L1(RN), we take R large enough such that |I2| < δ. On the other hand,

It is not hard to see that the last integral tends to 0. Therefore, we have proven the result.

Lemma 3.2Suppose that N = 3 or 4,and h is a continuous function with compact support. Then we have that

(i) Φ(μ,ξ)=O(μγ) as μ → 0+;

(iii) If h(ξ0)>0, then there exists a positive constant C such that

Proof(i) Let r >0 be such that ω1∈ Br(0). Assume first that|ξ|≥ 2r. If|x|

On the other hand, if |ξ|<2r, then we have that

where ωN?1is the surface area of the N ? 1 dimensional unit ball and BR(x) is the ball with center x, radius R>0. The last inequality is due to N ? 2γ ? 1> ?1 when N =3 or 4. Thus we have proved (i).

Therefore there exists a positive constant C such that

Lemma 3.3Suppose that N =3 or 4 and (V1)–(V4) hold. Then we have that

(iii) There exists μ0>0 such that

Proof(i) From (V1), (V2) and Lemma 3.2 (i), we have that

Thus the result directly follows from (V3) and (3.1).

(ii) From (V1)–(V3), Lemma 3.1 and Lemma 3.2 (ii), it is easy to get that

(iii) Since

there exists a positive constant C such that

Hence, from (V3) and (3.2), we get that

Similarly, there exists a positive constant C2such that

From Lemma 3.2 (iii), there exists a positive constant C3such that

Similarly, there exists a positive constant C4such that

Thus, from (3.3), (3.4), (3.5) and (3.6), we have that

On the other hand, since

Therefore, it is possible to choose μ0>0 small and a constant C5>0 such that

Proof of Theorem 1.1By (ii) of Lemma 3.3, there exist a large μ2> 0 and R2> R such that

On the other hand, from Lemma 3.3 (ii) we can find a small μ1>0 such that

In correspondence with μ1,μ2and R=R2, we fix Z0. From (3.7) and (3.8), it follows that

is achieved at some (μ?,ξ?), and the perturbation argument allows us to conclude that

is a critical point of Iε. Thus we have that

Then multiplying the first equation bythe second equation byand integrating by parts,we findHence (uε,vε) are nonnegative solutions of problem (1.1). Finally, the strong maximum principle implies that uε, vε>0.

Remark 3.1Our assumptions (V3) and (V4) may be replaced by the following:

where (N +2)/2 ≤a, b

In fact, by a similar method, we can get a negative minimum point ofwhich gives rise to a solution to problem (1.1).

AcknowledgementsThe author would like to thank Prof. Shuangjie Peng for stimulating discussions and helpful suggestions on the present paper.