The Choquard Equation with Weighted Terms and Sobolev-Hardy Exponent

Sang, Yanbin; Luo, Xiaorong; Wang, Yongqing

doi:https://doi.org/10.1155/2018/7210680

Journal of Function Spaces

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Abstract Introduction Preliminaries Data Availability Conflicts of Interest Authors’ Contributions Acknowledgments References Copyright Related Articles

Research Article | Open Access

Volume 2018 | Article ID 7210680 | https://doi.org/10.1155/2018/7210680

The Choquard Equation with Weighted Terms and Sobolev-Hardy Exponent

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

Academic Editor: Gennaro Infante

Received24 Mar 2018

Accepted27 May 2018

Published08 Jul 2018

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 whereFor 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 haveso 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 hasTherefore, 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 thatTherefore 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 thatWe 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 obtainwhere 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 thatBy and Lemma 19, we know that and Let . According to Brézis-Lieb Lemma and the Vitai’s theorem, one has ThusTherefore, 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).

DefineLet 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 SoWe may assume that , which leads to That is Combining (167), (169), one getsFrom (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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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.

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