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1 Introduction and main results

Let and . are the critical Hardy–Sobolev–Maz’ya exponents, is an open bounded domain in We study the following equation:

where is a positive function, It is well known that solutions of (1) are critical points of the corresponding functional

given by

By using the following Hardy–Sobolev–Maz’ya inequality (Lemma 1), we know that is well defined and functional on for any open subset of

Since (1) involves the double critical Hardy–Sobolev–Maz’ya exponents, we can use the pioneering idea of Brézis and Nirenberg [5], or the concentration compactness principle of Lions [16, 17], or the global compactness of Struwe [23] to show that (2) has a critical point, then get a positive solution to (1).

When and (1) is related to the well-known Brézis– Nirenberg problem [5]

where is the critical Sobolev exponent. Since the pioneering work of [5], there are some important results on this problem. See, e.g., [6, 8, 9, 11, 25]. Here we would like to point out [10]. In this paper, Devillanova and Solimini proved that when , (3) has infinitely many solutions for each . Let us now briefly recall the main results concerning the sign-changing solutions of (3) obtained before. If and is a ball, then for any , (3) has infinitely many nodal solutions, which are built by using particular symmetries of the domain (see [12]). In [22], Solimini proved that if is a ball and , for each , (3) has infinitely many sign-changing radial solutions. When is a ball and, there is a such that (3) has no radial solutions, which change sign if (see [2]). In [12, 22], the symmetry of the ball plays an essential role, hence their methods are invalid for general domains.

When , (1) is becoming Hardy–Sobolev–Maz’ya equation

By using the idea of [10], the authors of [26] obtained infinitely many solutions for Hardy– Sobolev–Maz’ya equation. Ganguly [13] and Wang [29] used different methods to get infinitely many sign-changing solutions. For the existence of infinitely many solutions or infinitely many sign-changing solutions for the related equations, see [14, 24, 30, 32] and the references therein. Very recently, Wang and Yang [27] proved the existence of infinitely many sign-changing solutions for (1).

Theorem 1. Suppose that and is a bounded domain. Ifin a neighborhood of , where is the outward normal of . If when and if when, then (1)has infinitely many sign-changing solutions.

Wang and Yang also considered the following nonexistence theorem.

Theorem 2. (See [27].) Suppose thatandfor every . Then (1)does not have nontrivial solution in a domain, which is star shaped domain with respect to the origin.

Remark 1. Let be the first eigenvalue of

Since and is strictly positive, system (4) has infinitely many eigenvaluessuch that . It is characterized by the following variational principle:

Let be the orthonormal eigenfunction corresponding to and . Denote

Then and It is easy to know that if , equation (1) has infinitely many sign-changing solutions. Indeed, by multiplying the first eigenfunction and integrating both sides, then we can check that if , any nontrivial solution of (1) has to change sign. Therefore, by the result of [28], to prove Theorem 1 it suffices to consider the case of .

Remark 2. When and , Cao and Peng [6] considered the following system:

They obtained a pair of sign-changing solutions to (6). In [8,32], the authors get infinitely many sign-changing solutions for (6). They only considered the case In another case, , the mean curvature of at plays an important role in the existence of mountain pass solutions, see [3,6,14]. As it is pointed in [4,31], there are some differences between the case and . When, solutions of (6) have a singularity at , and the authors of [8, 32] impose the condition. If , no such condition is needed. So the estimates for the case and the case are very different. Therefore we have generalize the results in [32] to the case .

Remark 3. In order to prove the results, Wang and Yang [27] first used an abstract theorem, which is introduced by Schechter and Zou [21]. Then by combining with the uniform bounded theorem due to [28], the authors of [27] obtained infinitely many sign-changing solutions. The methods introduced in [4, 8, 13, 21, 31] sometimes are limited because, by general minimax procedure to get the Morse indices of sign-changing critical points, sometimes are not clear. Another limited condition is that the corresponding functional is also needed to be .

Before giving our main results, we give some notations first. We will always denote Let be endowed with the standard scalar and norm

The norm on with is given by with the norm , where denote the Lebesgue measure in . Denote and

We will use the usual Ljusternik–Schnirelman-type minimax method and invariant set method to prove Theorem 1. Our method is much simpler than the proof of [27]. In fact, our approach also works for the Brézis–Nirenberg problem involving subcritical perturbation term , which is not . However, the techniques developed by Wang and Yang [27] or Schechter and Zou [21] cannot be applied directly. Let us outline the proof of Theorem 1 and explain the difficulties we will encounter.

In general, by using the combination of invariant sets method and minimax method to obtain infinitely many nodal critical points, we need the energy functional satisfies the Palais–Smale condition in all energy level. This fact prevents us from using the variational methods directly to prove the existence of infinitely many sign-changing solutions for (1) because does not satisfy the Palais–Smale condition for large energy level due to the double critical Hardy–Sobolev–Maz’ya exponents and .

In order to overcome the difficulty, we will adopt the idea in [10, 24] and [4, 31]. We first study the following perturbed problem:

where is a small constant. The corresponding energy functional is

By the following lemmas, we will know is a function on and satisfies the Palais–Smale condition. It follows from [1, 20] that has infinitely many critical points. More precisely, there are positive numbers, with as . Moreover, a critical point for satisfies

Next, we will show that for any fixed are uniformly bounded with respect to , then we can apply the following compactness result Proposition 1 (see [28, Thm. 1.3]), which essentially follows from the uniform bounded theorem due to Devillanova and Solimini [10], to show that converges strongly toin as .

Therefore it is easy to prove that is a solution of (1) with

Proposition 1. (See [28].) Suppose that and satisfies the conditions in Theorem 1. If when and when , then for any sequence , which is a solution of (7) with satisfying for some constant independent of has a sequence, which converges strongly in .

In the end, we will distinguish two cases to prove that has infinitely many sign- changing critical points.

Case I. There are satisfying

Case II. There is a positive integer such that for all .

The central task in this procedure is to deal with case II. In fact, we can prove that the usual Krasnoselskii genus of is denoted in Section 2) is at least two, where Then our result is obtained.

Throughout this paper, the letterswill be used to denote various positive constants, which may vary from line to line and are not essential to the problem. The closure and the boundary of set are denoted by and , respectively. We denote weak convergence and by strong convergence. Also if we take a subsequence of a sequence , we shall denote it again .

The paper is organized as follows. In Section 2, we introduce some notations and Hardy–Sobolev–Maz’ya inequality. In Section 3, we give an auxiliary operator and construct the invariant sets. We give the proof of Theorem 1 in Section 4.

2 Preliminaries

Now we give some integrals inequalities, for details we refer to [19].

Lemma 1 [Hardy–Sobolev–Maz’ya inequality]. Let then there exist a positive constant such that

for all

Lemma 2. (See [13].) If is a bounded subset of then

with the inclusion being continuous whenever

Remark 4. If for then with

For each and, we define

Lemma 3. (See [13].) Let and , then the embedding is compact.

By Lemmas 2, 3 and Hardy–Sobolev–Maz’ya inequality, we know that the singular term and are finite and where isindependent of . Therefore is a function on By Lemma 3, satisfies the Palais–Smale condition. In order to prove Theorem 1, it is enough to obtain sign-changing critical points for the functional .

Fix In the following, we will always assume that . In order to construct the minimax values for the perturbed functional , the following two technique lemmas are needed.

Lemma 4. Assume Then there exists such that for all

where

Proof. Since is finite dimensional, by Lemma 2, we know that is defined as the norm on There is a constant such that for all Therefore

Since and , we have that The proof is complete.

Lemma 5. For any , there exists such that

Proof.

Since and , there exists such that The proof is complete.

Lemma 5 implies that 0 is a strict local minimum critical point. Then we can construct invariant sets containing all the positive and negative solutions of (1) for the gradient flow of . Therefore nodal solutions can be found outside of these sets.

3 Auxiliary operator and invariant subsets of descending flow

For any , let be given by

for . Then the gradient of has the form

Note that the set of fixed points of is the same as the set of critical points of , which is . It is easy to check that is locally Lipschitz continuous.

We consider the negative gradient flow of defined by

Here and in the sequel, for , denote , the convex cones

For we define

In the following, we will show that there exists such that is an invariant set under the descending flow for all . Note that contains only signchanging functions, where

since contains only signchanging functions. By a version of the symmetric mountain pass theorem, which provides the minimax critical values on , we can prove that (6) has infinitely many sign-changing solutions.

For any and denotes the open neighborhood of N , i.e.,

whose closure and boundary are denoted by and . By the following result, we can know that a neighborhood of is an invariant set. We can use similar way as Lemma 2 in [9] and Lemma 3.1 in [3] to get the following lemma.

Lemma 6. There exists such that for any there holds

and

Moreover, every nontrivial solutions and of (5) are positive and negative, respectively.

By using the combination of invariant sets method and minimax method, we can construct a nodal solution first, then to prove our main result. We need a deformation lemma in the presence of invariant sets.

Definition 1. A subset is an invariant set with respect to if, for any , for all

From Lemma 6 we may choose an sufficiently small such that are invariant set. Set . Note that int and only contains sign-changing functions.

Since satisfies the Palais–Smale condition, we have the following deformation lemma, which follows from Lemma 5.1 in [18] (also see Lemma 2.4 in [15]).

Define where Let be such that where dist

We can use the similar method to the proof of Lemma 5.1 [18] and Lemma 2.4 [15] to prove the following lemma.

Lemma 7. Assume that satisfies Palais–Smale condition, then there exists an such that for any, there exists satisfying:

• 

• 

• is odd and a homeomorphism of

• is nonincreasing

• for any

4 The proof of Theorem 1

In the following, we assume that . For any small, we define the minimax value for the perturbed functional with We now define a family of sets for the minimax procedure here. We essentially follow [3], also see [18] and [20]. Define

where is given by Lemma 4. Note that since . Set

for . From [20] possess the following properties:

• and for all .

• If is odd and on , then for all .

• If is open and and then

Now, for , we can define the minimax value by

Lemma 8. For any and then is well defined, and where is given by Lemma 5.

Proof. Consider the attracting domain of in :

Note that is open since is a local minimum of and by the continuous dependence of ODE on initial data. Moreover, is an invariant set, and In particular, the following holds

for every (see [3, Lemma 3.4]). Now we claim that for any with , it holds

If this is true, then we have and because and sup by Lemma 5.

To prove (10), let

with and Define

Then is a bounded open symmetric set with and Thus, it follows from the Borsuk–Ulam theorem that and, by the continuity of , . As a consequence,

and therefore

by the “monotone, subadditive and supervariant” property of the genus [23, Prop. 5.4]. Since

Thus for , we conclude that

which proves (10).

Thus is well defined for all and

The proof is complete.

Lemma 9.

Proof. If not, we assume that

By Lemma 7, for the functional , there exist and a map such that is odd, and

By the definition of , there exists such that . Let It follows from (14) that

On the other hand, it is easy to show that by Lemma 4 and the property (ii) of above. As a result, This contradicts with . The proof is complete.

Lemma 9 implies that there exists a sign-changing critical point such that

As a consequence of Lemma 8, we have that is well defined for all and Now we can show the following lemma.

Lemma 10.

Proof. Here we deduce by a negation. Suppose Since satisfies Palais–Smale condition, it follows that and is compact. Moreover, we have

Indeed, assume is a sequence of sign-changing solutions to (6) with , and we have

By using the variantional principle of (5), we obtain

It follows that, by Sobolev embedding theorem, where is a constant independent of This implies that the limit of the subsequence of is still sign-changing.

Assume Since and is compact, by the “continuous” property of the genus [23, Prop. 5.4], there exists an open neighborhood in with such that Now using Lemma 7 for the functional , there exist and a map such that is odd, and

Since we can choose sufficiently large such that Clearly, By the definition of we can find a set that is, where such that

for any which implies It follows from (13) that

Let Then is symmetric and open, and

Then it is easy to check by (ii) and (iii) above. As a result, by (14),

This is a contradiction to . The proof is complete.

Lemma 11. For any fixed is uniformly bounded with respect to , and then converges strongly to ul in as .

Proof. Indeed, by using the same above, we can also define the minimax value for the following auxiliary function:

Here we choose sufficiently large if necessary such that Lemma 4 also holds for . Then by a version of the mountain pass theorem [20, Thm. 9.2], for each , is well defined, and because

where

Therefore, for any fixed , is uniformly bounded for , that is, there is independent on such that uniformly for because is a nodal solution of (6) and By the definition of , we can obtain the following:

where and Therefore is uniformly with respect to . So we can apply Proposition 1 and obtain a subsequence such that strongly in for some and also . Thus is a solution of (5), and . Moreover, since is sign-changing, similar to Lemma 10, by Sobolev embedding theorem, we can prove that is still sign-changing. The proof is complete.

Proof. Proof of Theorem 1 Noting that is nondecreasing with respect to , we have the following two cases:

Case I. There are satisfying Obviously, in this case, equation (1) has infinitely many sign solutions such that

Case II. There is a positive integer such that for all .

From now on we assume that there exists a such that has no sign-changing critical point with

Otherwise, we are done. In this case, we claim that , where and Then as a consequence, has infinitely many sign-changing critical points.

Now we adopt a technique in the proof of Theorem 1.1 in [7]. Suppose, on the contrary, that (note that ). Moreover, we assume contains only finitely many critical points, otherwise, we are done. Then it follows that is compact. Obviously,. Then there exists a open neighborhood in with such that .

Define

We now claim that if small, has no sign-changing critical point Indeed, arguing indirectly, suppose that there exist and satisfying with and

Then, by Proposition 1, up to a subsequence, converges strongly to in . Therefore ,

and .

This is a contradiction to our assumption and the fact that u is still sign-changing. The following proof is similar to that of Lemma 9. By using Lemma 7, for the functional , there exist and a map such that is odd, for and

Now fix Since we can find an small such that By the definition of , we can find a set that is,

where such that

for any , which implies . Then by (15), we have

Let . Then is symmetric and open, and

Then it is easy to check by (ii) and (iii) above. As a result, by (16),

This contradicts to. Then the proof for case II is finished. The proof is complete.

Acknowledgments

The first authors like to thank Chern Institute of Mathematics for visiting scholar programme and the authors are grateful to the anonymous referees for their useful suggestions, which improve the contents of this article.

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