Journal of Function Spaces
Volume 2018 (2018), Article ID 7210680, 22 pages
https://doi.org/10.1155/2018/7210680

Research Article

The Choquard Equation with Weighted Terms and Sobolev-Hardy Exponent

Yanbin Sang,1,2 Xiaorong Luo,¹ and Yongqing Wang²

¹Department of Mathematics, School of Science, North University of China, Taiyuan, Shanxi 030051, China
²School of Mathematical Sciences, Qufu Normal University, Qufu, Shandong 273165, China

Correspondence should be addressed to Yanbin Sang; syb6662004@163.com

Received 24 March 2018; Accepted 27 May 2018; Published 8 July 2018

Academic Editor: Gennaro Infante

Copyright © 2018 Yanbin Sang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

We study a nonlinear Choquard equation with weighted terms and critical Sobolev-Hardy exponent. We apply variational methods and Lusternik-Schnirelmann category to prove the multiple positive solutions for this problem.

1. Introduction

The main goal of this paper is to consider the multiplicity of positive solution for the following nonlinear Choquard equation:

where is a bounded open set with smooth boundary. , , , , and is a parameter. There are two continuous weight functions, satisfying the following conditions:

 .  and .  there exist two positive constants such that and for every . , , there exists such that

In recent years, much attention has been paid to nonlinear Choquard equation involving the Hardy-Littlewood-Sobolev inequality, which generalizes and complements the classical elliptic boundary value problems and Schrödinger-Poisson system [1–21]. In particularly, more and more authors have studied the critical problems [22–36]. In the case that , Gao and Yang [22] considered the existence and multiplicity of (1) with upper critical exponent . The authors focus on the case that how the sublinear perturbation term has influence on the multiplicity of (1). When , Xiang [33] showed the uniqueness and nondegeneracy solutions for ground states of Choquard equation. Furthermore, by means of the tool of Nehari manifold, Zhang et al. [36] established the existence theorem of ground states for generalized Choquard equation when the nonlinear term is concave-convex. On the other hand, there are a great deal of results on the existence of elliptic boundary value problems with sign-changing weights [37–45]. We should point out that Hsu and Lin [38] showed the existence of positive solutions for elliptic equations with concave-convex nonlinearities and sign-changing weights. What is more, Wu [39] obtained three positive solutions discussed in [38] by using of the method of Nehari manifold combining with the Lusternik-Schnirelmann category. In view of the same method, Chen and Wu [41] obtained the existence of positive solutions for a class of critical semilinear problem. Chen et al. [42] established multiplicity theorems for Kirchhoff type equation with sign-changing weight functions. Goyal and Sreenadh [45] used fibering map approach to study -fractional Laplacian equation with sign-changing weight function.

Motivated by above results, we use variational approach to analyze the existence and multiplicity of positive solutions for (1). To the best of our knowledge, there is no result studying Choquard equation with upper critical exponent and two sign-changing weight functions.

Set

Our main results are the following theorems.

Theorem 1. Assume that hold, for every , the problem (1) has at least one positive solution in .

Theorem 2. Assume that hold; there exists , such that, for every , the problem (1) has at least two positive solutions in .

Furthermore, we will utilize the following condition.

(G) There are a nonempty closed set and such that

Remark 3. Let for . According to (G), we can assume that there exist two positive solutions and such that

Theorem 4. Assume that holds; for every , there exists such that for , (1) has at least positive solutions.

Definition 5. is called a positive weak solution of (1), if

The energy functional associated with the problem (1) is given by

where is the norm in .

2. Preliminaries

Proposition 6 (Hardy-Littlewood-Sobolev inequality [22]). Let and with and . There exists a constant independent of , such that

If , then

For , by Proposition 6, we have

The best constant is defined as

Proposition 7 (see [22]). The constant is attained if and only if

where ; therefore

where

For

is a solution of the problem

with = .

Definition 8. (i) For , if strongly in as , then the sequence is a sequence in ;
(ii) satisfies the condition if every sequence in for has a convergent subsequence.
Though is not bounded below on , we can construct the following Nehari manifold:

Define the fiber map by .

Lemma 9. is coercive and bounded below on .

Proof. If , by (14) we have

so is coercive and bounded below on . This lemma is completed.

Lemma 10. Assume that is a local minimizer for on and . Then . Moreover, if is a nontrivial nonnegative function in , then is a positive solution of (1).

Proof. Since is a local minimizer for on , is a solution of the optimization problem

where

Hence, using Lagrange multipliers, there exists , such that , which implies

Since and , one has

Hence, if , combining (21) , then in . Using Harnack inequality, thus, is a positive solution of (1) in . The lemma is completed.

From , we obtain that

for , we get

Obviously, ; we divide into the following three sets

For any , one has

Therefore, we can derive

Lemma 11. (i) For every , then .
(ii) For every , then .

Lemma 12. For , then .

Proof. Assuming the contrary, there is such that , then for , by (26) and Proposition 6, which implies

so

Similarly, coupling with (26) and (14), which infers that

hence

From the above inequalities, we deduce that

which is a contradiction.

Letting , one has

Let

Assume if and only if So

Clearly, , then . Hence, (or ) provided (or . can attain its maximum at duo to , where

where

Lemma 13. For every , it follows that
(i)if , there exists a unique such that , and is increasing on , decreasing on .
Moreover

(ii)if , then there exist unique such that and is increasing on , decreasing on . Moreover

Proof. (i)Assume that . There exists a solution , with , then , and . Consequently, has a unique critical point at and . Since . Hence, (36) holds and .
(ii) Assume , then . There are two solutions , satisfying , such that , which means that these two solutions of depend on , where . So according to has critical points at . Thus, is increasing on , and , (37) holds.
This lemma is completed.

Furthermore, we have

Define

Theorem 14. (i) for all .
(ii) If , then there exists a positive number , such that .
Especially, for all .

Proof. (i) Supposing that from (26), one has

by Lemma 9, we derive

Thus, .
(ii) Suppose that from (11), (26), and , we get

Moreover

Hence, if , such that

where depends on
This completes the proof.

Lemma 15. Every sequence that satisfies with and = has a convergent subsequence.

Proof. For , there exists such that

Therefore

So, by Theorem 14, , we know that

Thus

which implies

Furthermore

Lemma 16. For every , then has a minimizer in , satisfying that

(i);(ii) is a positive solution of (1);(iii) as .

Proof. (i)From Lemma 15, there exists a minimizing sequence for , such that

Noting that is coercive and bounded on in . Going if necessary to a subsequence, we can assume that there exists such that

We show that is a solution of (1). , which derives that

thus

By Theorem 14, (51), (52), (54), one has

combining Fatou’s Lemma, which implies that

Clearly,

(ii) Let . Using Brézis-Lieb Lemma, such that

which infers that in and . On the contrary, if , from (26), (55), which gets and . Therefore, by Lemma 13, there are unique and such that . Moreover, . By simple calculation, one has

So there exists such that . Coupling with Lemma 13 which indicates that

which is a contradiction. Furthermore, and . From Lemma 10, we claim that is a positive solution of (1). Combining the standard elliptic regularity argument and , we can get . By Harnack inequality, we have that in ; that is is a positive solution of (1).
(iii) By (26), we get

Moreover

Obviously, as . This completes the proof.

Lemma 17. For every , there exist and a differentiable function with and such that

Proof. ; define a function :

, and

Applying the implicit function theorem, we obtain

which infers that

so .

3. Proof of Theorem 1

According to Lemma 16, we can derive that the problem (1) has a positive solution.

4. Proof of Theorem 2

Proposition 18. (i) For every , there exists sequence in for ;
(ii)For every , there exists sequence in for .

Proof. By Lemma 9 and Ekeland variational principle there exists a minimizing sequence such that

is large enough, from Theorem 14(i), which yields that

which derives that

Hence by (70). From Lemma 15, we have

We claim that . Using Lemma 17, then there exists a differentiable function for some small enough number , such that . Take , and let , , . Coupling with (68), for , we obtain

where

Dividing by in (73) and passing to the limit as , which infers that

Applying (71), (72), we have

Noting that

By Lemma 17, for some constant , we have

Now we prove that

Arguing by contradiction, assume that there exists a subsequence such that

From Lemma 17 and (72), there exists a constant such that

Set

Moreover

On the other hand, combining (82) with (83), we conclude that

which is a contradiction; thus (80) is true.
Using the same methods to prove (ii). The proof is completed.

Lemma 19. Assume that and hold, if is a sequence, for with in , then and there exists a constant , such that .

Proof. Since is a sequence for with in . Obviously,

Thus

Therefore

where
The proof is completed.

Lemma 20. Assume that hold. For , satisfies the condition.

Proof. Let be a sequence, which implies

By Lemma 9, is bounded in , so there exists a subsequence still denoted by , we can suppose that there exists such that

By and Lemma 19, we know that and

Let . According to Brézis-Lieb Lemma and the Vitai’s theorem, one has

Thus

Therefore, we suppose that

By the Sobolev inequality, one has

Applying (94), we deduce that

and, obviously, , or . In fact, . Arguing by contradiction, suppose
Coupling with (91)-(94), we have

which is a contradiction, that is, in .
This completes the proof.

Lemma 21. Assume that hold, then there exist and , such that, for ,

In particular, for every , we have .

Proof. Set

Based on , we get that there exists , for every such that

For , let be a cut function, such that

Let

where

Step 1. We will show that .
At first, we prove the following estimates :

From [22] and definition of , let , which derives that

Furthermore

where

Hence

( is the volume of the unit ball in .)
Thus

moreover

which implies that

Taking sufficient small such that , by (114), we have

so

that is,

From (106),(107), we conclude that

moreover

Let

in fact , and , so

which leads to

From (119), we have

Step 2. Let , there exists , for all , and we have

We denote satisfying

By , we obtain that

which implies that there exists such that

Moreover, set , which leads to

Furthermore

Combining with (128) and (129), for , we have

It is easy to find such that

Hence, set and , if , then

Step 3. For every then
By and , we get

Using Lemma 13 and the definition of , we have that for every there exists such that and

The proof of (73) is valid.

Theorem 22. Suppose that hold, then there exists such that for , has a minimizer and
(i) ;
(ii) is a positive solution of (1) in .

Proof. According to Proposition 18 (ii), for any , there exists a sequence in for . In terms of Theorem 14 (ii) and Lemmas 20 and 21, for all , satisfies and condition. From Lemma 9, is coercive on , so is bounded. Moreover, there exists a subsequence still denoted by , and such that in , for , . At last, we apply the same arguments as those of Lemma 16 to prove fore , is a positive solution of (1).
The proof is completed.

Proof of Theorem 2. According to Lemma 16 and Theorem 22, we admit that and are two solutions of (1). It is easy to verify that is different because .
The proof is completed.

5. Proof of Theorem 4

In this part, we will show the proof of Theorem 4. At first, we consider the following problem:

and the associated energy functional of (135) in is

Recalling that

where

Lemma 23.

Moreover, (135) has no positive solution satisfying .

Proof. By Lemma 13, for any , we can obtain that and ,

Furthermore

Using (140)-(143), we have

Consequently

Applying Lemma (18)(i), for every , . In addition, there exists a unique such that . Hence

Furthermore,

so

Next, we assume that (135) has a solution satisfying . By contradiction, suppose there exists such that . In fact, is a positive solution owing to and , from Lemmas 10 and 13, we conclude that , and there exists a unique with such that

which indicates that That is , which contradicts with
This proof is completed.

Lemma 24. Assume that is a minimizing sequence for in , then (i) .
(ii)
In addition, for , is a sequence in .

Proof. For any , there exists such that , which implies that

Form Lemma 13(i)

By Lemma 23, we can get

Indeed, , there is such that . Arguing with contradiction, assume that as . Noticing that

By Lemma 9, we know that is bounded, thus or which is a contradiction with , which derives

and so

Using the same method as those of (i), we can prove (ii).

Define

Let be a barycentre map:

Lemma 25. For , there exists such that

Proof. According to Lemma 24, there exist a sequence in and . Since is bounded and using concentration compactness principle, there exist two sequences and with , and is a positive solution of (16), such that and as satisfying

Thus

In fact, . We know that is a minimizing sequence for , and, combining with Lemma 24 (ii), we have

which deduce that and

and this contradicts with assumption. This completes the proof.

Set

Lemma 26. For every , there exists , such that, for every, we have

Proof. For , combining (26) with Lemma 13(i), there exists a unique , such that . Based on Sobolev and Höder inequalities, one has

We assume that there exists a positive constant such that . Using (26) and Sobolev inequality, we deduce that

So

We may assume that , which leads to

That is

Combining (167), (169), one gets

From (165) and (170),

Using Lemma 9, if . For every , there exists such that for every .
Thus

Taking , for , there exists , and we have

Based on Lemma 25, we derive that and

Let , and

so, . At the same time, we conclude that

Lemma 27 (see [41]). Let closed sets with and , and suppose is homotopically equivalent to identity mapping id in , then .

Theorem 28. For every , has at least critical points on .

Proof. By Lemma 15, for every and

From Lemma 27, , which implies that . Combined with Lemma 16, (1) has at least solution.
This proof is completed.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Authors’ Contributions

All authors contributed equally to this work. All authors read and approved the final manuscript.

Acknowledgments

This project is supported by the Natural Science Foundation of Shanxi Province of China (201601D011003) and the Natural Science Foundation of Shandong Province of China (ZR2017MA036).

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