Abstract

The existence, nonexistence, and multiplicity of vector solutions of the linearly coupled Choquard type equations are proved, where , , are positive functions, and denotes the Riesz potential.

1. Introduction

We deal with the linearly coupled Choquard type equations: where , , and are positive functions, is the lower critical exponent with respect to a Hardy-Littlewood-Sobolev inequality (see ([1], Theorem 3.1) or ([2], Theorem 4.3)), and denotes the Riesz potential defined on by

The single equation appears in various physical contexts (see [36]). Mathematically, equations of this type have received considerable attention due to the appearance of the nonlocal term , which makes the problem challenging and interesting. The readers can refer to [4, 718] and references therein for research on related problems.

Recently, Chen and Liu [19] established the existence and asymptotic behavior of the vector ground state of the linearly coupled system: where , . Xu, Ma and Xing [20] extended the results in [19] to (4) in the case that and are replaced with general subcritical nonlinearities and , respectively. Yang et al. [21] obtained the existence of the vector ground state of (4) in the following three cases:

They also proved that (4) has no nontrivial solutions if or .

As we know, when , the local system which has application in a large number of physical problems such as in nonlinear optics, can be regarded as a limiting system of (4). Systems of this type have received great attention in recent years (see [2228] for instance). However, linearly coupled systems with nonlocal nonlinearities have been less studied.

In this paper, we are interested in the existence, nonexistence, and multiplicity of solutions of system (1) with positive nonconstant potentials. We assume that

(H1) and

(H2) ,

(H3)

For simplicity, the integral is denoted by . According to (H1), the norm in can be defined by where

Then, a solution of system (1) can be found as a critical point of the energy functional defined by

Set

We first show that is attained.

Theorem 1. Assume that (H1), (H2), and (H3) hold. Then, there exists a vector ground state of system (1). Additionally, if is a sequence satisfying as , then up to a subsequence, either or in as , where is a ground state of and is a ground state of

Remark 2. We call a solution of system (1) a nontrivial solution if and a vector solution if and . A nontrivial solution satisfying for any nontrivial solutions of system (1) is called a ground state.

Remark 3. Under assumptions (H1) and (H2), the existence of ground states of equations (11) and (12) has been proved by Moroz and Van Schaftingen ([17], Theorem 3 and Theorem 6).

To prove Theorem 1, it is crucial to give an estimate of the upper bound of the least energy due to the lack of compactness. In our case, the estimate is quite involved, since we are dealing with a coupled system, which is more complex than a single equation. The method we follow can be sketched as follows. We first study the minimizing problem which can be considered an extension of the classical problem

By the results that is attained if and only if where is a fixed constant, , and (see ([1], Theorem 3.1) or ([2], Theorem 4.3)), and studying the minimum point of a function defined on by we show that is attained at if and at if (see Theorem 7 in Section 2), which combined with the existence of ground states for equations (11) and (12) enables us to obtain the precise upper bound of .

Our second goal is to show the existence of a higher energy vector solution of (1).

Theorem 4. Assume that (H1) and (H2) hold. Then, for some , there exists a vector solution of system (1) if . Additionally, if is a sequence satisfying as , then up to a subsequence, in , where is a ground state of (11) and is a ground state of (12).

Remark 5. For sufficiently small, it is trivial to see that the solutions obtained in Theorem 1 and Theorem 4 are different, which implies that there exists at least two vector solutions of system (1) if is small enough.

Finally, we prove the nonexistence of the nontrivial solution of system (1) by establishing the Pohozaev type identity.

Theorem 6. Assume that (H3) holds. If and then, system (1) has no nontrivial solutions in

This paper is structured as follows. Some preliminary results are provided in Section 2. The proofs of Theorems 1 and 4 are presented in Section 3 and Section 4, respectively. In Section 5, we show the nonexistence of nontrivial solutions.

2. Preliminary Results

In this section, we show the sharp constant defined in (13) is attained and give an estimate of the upper bound of

Theorem 7. If , then is attained. Moreover, (or ) is a solution of (13) for (or ), where is a minimum point of defined on by

Proof. First, we show that there exists such that Calculating directly, we have Set It can be easily seen that as , and as Then, there is such that and
In the next step, we prove where is defined in (14). We employ the idea in ([29], Theorem 5) to prove (21). For the case , taking gives Let be a minimizing sequence for Set , where Then, Collecting (23) and (24) leads to Then, (21) follows from (22) and (25). For the case , the conclusions follow by replacing with and repeating the proof previously.

Lemma 8. Assume that (H1) and (H3) holds, then for any , there exists such that and

Proof. This result is standard and the proof can be found in ([30], Lemma 12). We omit it.

For equations (11) and (12), we set and where

Then, according to ([17], Theorem 3 and Theorem 6), we have and is achieved, where is defined in (14). By Theorem 7 and ([17], Theorem 3 and Theorem 6), we are able to get the following estimate.

Lemma 9. Assume that (H1), (H2), and (H3) hold. Then,

Proof. We first show the positivity of . By (H3), we have for some , which suggests that there exists such that Thus, we obtain Second, we show From the assumptions (H1)–(H3), we see that , and so Theorem 7 holds. For the case , by Lemma 8, there is such that ; then, we have The last inequality in (34) follows from Theorem 7 and direct calculation. Denote To prove (33), it is enough to show for some . Since we have Then, by a transformation , we get Taking the assumption (H2) into consideration, we see that that (36) holds. Then, (33) follows from (34).
Now, it remains to show Denote ground states of (11) and (12) by and , respectively. Since and , we have If , then we see that at least one of and is a solution of system (1), which is impossible since , so (40) holds.

3. Proof of Theorem 11

Lemma 10. Assume that (H1), (H2), and (H3) hold. Then, there exists a vector ground state of system (1).

Proof. According to Ekeland’s variational principle, there exists such that For simplicity, we denote . First, we prove . Indeed, for some and sufficiently large . Particularly, From the proof of Lemma 9, we observe that there exists such that . Then, we have Taking (43) into consideration, we obtain that as . Then, from (42), we get
We may assume that Then, To complete the proof, it is sufficient to prove that Actually, if By Fatou’s lemma, we get Furthermore, since and , we see from (1) that and , that is, is a vector ground state of (1).
Suppose the assertion is false, that is, On the one hand, we know from (H1) that as . Then, it follows as . Using Theorem 7, we obtain On the other hand, Collecting (49) and (50) yields which contradicts Lemma 9. Thus,

Proof of Theorem 11. By Lemma 10, we need only show the asymptotic behavior of the vector ground state when First, we claim that decreases strictly monotonically with respect to Indeed, fix with Denoting a vector ground state of system (1) when by and letting be the constant such that , we obtain Then, by , we deduce that , which gives The claim is proved.
Now, choose satisfying as and denote a vector ground state of system (1) when by Then we have By and (29), we obtain Repeating an argument as in Lemma 10, we deduce that or in , where and are ground states of (11) and (12), respectively.

4. Proof of Theorem 18

In this section, we study the existence of a higher energy vector solution of system (1) for sufficiently small. We suppose that without loss of generality. Let , be ground states of (11) and (12), respectively. Then, we may assume that and are positive since and are also ground states of (11) and (12), respectively. Now, we set where

Then . Moreover, by a similar argument as that in ([28], Lemma 12), we obtain the following.

Lemma 12. Assume that (H1) and (H2) hold. Then, is compact, and there exist such that

Proof. The proof can be found in ([28], Lemma 12) and will be omitted here.

By the definition of and , we know that for some , , , and satisfying and . Denote and define by

Then, for some Define where and is defined in Lemma 12. Obviously, Moreover, we havethe following.

Lemma 13. Assume that (H1), (H2), and hold. Then,

Proof. By , we see that and Observing that since , we deduce Now, for , define a function on by where are given by Noting that are continuous and , we deduce Moreover, we know from (60) that for any which implies is well defined and Therefore, there exists satisfying , that is, Recalling the definition of we have . Then, it follows Thus, . Taking account of (66), we obtain Now, it remains to prove If (12) is not true, there exists a sequence , and satisfying For the given above, the definition of leads to for some sufficiently large. Then, it follows which contradicts (70), implying that (71) holds.

Set

Then, we show the compactness of the PS sequence.

Lemma 14. Assume that (H1) and (H2) hold. Denote and let If satisfies and as and is a sequence with then, there exists such that in

Proof. Observing that is bounded by the choice of and Lemma 12, we assume in . Then, by a similar argument as in ([28], Lemma 14), we obtain that Moreover, using the definition of again, we get
We now prove . Actually, for , we have Then, it holds Note that from (78), Then, combining Lemma 13 with (79) and (80), we have and in .

Next, we will construct a PS sequence using a perturbation approach.

Lemma 15. Assume that (H1) and (H2) hold. Then, for a where was defined in Lemma 14, there are and such that

Proof. We prove indirectly. Suppose that there exists satisfying and with as . Then, we see immediately that from Lemma 13, and in for some by Lemma 14. Thus, for sufficiently large, which is in contradiction with , so the conclusion holds.

In the sequel, we assume that be fixed such that Lemma 15 holds.

Lemma 16. Assume that (H1) and (H2) hold. Then, there exist and such that

Proof. Arguing indirectly, we suppose that there exist and satisfying We may suppose . Then, from Lemma 13 and (83), we deduce Recalling the definition of , we have , and so which is in contradiction with

For and given in Lemma 16, we define

Then, for some .

Lemma 17. Assume that (H1) and (H2) hold. For fixed , there is with

Proof. For , suppose contradictorily that for all and some Then, there is a pseudogradient vector field for on neighborhood of such that Define a function on satisfying , on and on , and a function on with , if , and if . Then both and are Lipschitz continuous. Set Then, the initial problem has a global solution on with the properties: (i) if or or (ii)(iii)Now, the proof can be divided into two steps.
Step 1. We show that there exists such that for and defined in (86).
Arguing indirectly, we suppose for some Noting and applying Lemma 16, we get By property (iii) and the fact that we have Then, If for all , then Thus, which contradicts (93). Hence, for some . Observing that we deduce that for some with , it holds , , and for all Then, according to Lemma 15, we get Moreover, we deduce from property (ii) that which yields . Then, we obtain which contradicts (93), and the proof of this step is complete.
By step 1, we define and . Obviously, for all
Step 2. We prove Noting that for all we have and
It remains to show , , and is a continuous function of . For any , if , then and so If , then and So we get for Then, we can prove that Indeed, if not, by similar arguments as in step 1, we know that for some satisfying , which is in contradiction with the definition of . Thus, and for some . Hence, from Lemma 12, Now, we prove that is continuous. For fixed , if , then and Since is continuous, we have for some , which implies Thus, the continuity of at is proved. If then we see from the proof previously that , so and for any . By the continuity of , we obtain , for some . Therefore, and If , then If , then for any with . Then, since is continuous, we deduce Combining with (110), we obtain the continuity of at Consequently, (102) holds.
By Step 1 and Step 2, we have showed that and which is in contradiction with the definition of Thus, the conclusion holds.

Proof of Theorem 18. Denote . From Lemma 17, we obtain that there exists such that for fixed where . Then, by Lemma 14, in for some and . Moreover, recalling the definition of , we have , that is, is a vector solution of system (1.1).
Now, choosing such that as , by a repeat of the proof in Lemma 14, we obtain in , with and being ground states of (11) and (12), respectively, which completes the proof.

5. Proof of Theorem 20

Lemma 19. Let , and be a solution of system (1). If then satisfies the Pohozaev identity

Proof. The lemma can be proved by a similar argument as that in ([17], Proposition 11).

Proof of Theorem 20. Let be a solution of system (1). By Lemma 19, we have Then, the conclusion follows from a classical Hardy inequality (see ([31], Theorem 6.4.10))

Data Availability

No data were used to support this study.

Conflicts of Interest

The author declares that she has no conflicts of interest in relation to this article.

Acknowledgments

This research is supported by the Scientific Research Foundation of Minjiang University (Nos. MJY18014 and MYK19020) and the Foundation of Educational Department of Fujian Province (No. JAT190614).