Recent Developments and Applications on Qualitative Theory of Fractional Equations and Related TopicsView this Special Issue
Multiple Solutions to Elliptic Equations on with Combined Nonlinearities
In this paper, we are concerned with the multiplicity of nontrivial radial solutions for the following elliptic equations : ; as , where , , and are radial positive functions, which can be vanishing or coercive at infinity, and is asymptotically linear at infinity.
1. Introduction and Main Results
In this paper, we deal with the multiplicity of nontrivial radial solutions for the following elliptic equations: where , , , , and are radial positive functions, which can be vanishing or coercive at infinity.
When is a smooth bounded domain in , the problem where , , and , has been widely studied in the literature and plays a central role in modern mathematical sciences, in the theory of heat conduction in electrically conduction materials and in the study of non-Newtonian fluids. However, it is not possible to give here a complete bibliography. Here we just list some representative results. In the case where is superlinear near infinity, problem is the famous concave-convex problem; after the celebrated work [1, 2], this kind of problem has drawn much attention. In the case where is linear in , the authors in  have proved that there exist at least two nonnegative solutions for a more general question: where satisfies some additional conditions. For problem , in the special case , where , one nonnegative solution for any and was found in  via Mountain Pass Theorem. In the last years, several papers have also been devoted to the study of nonlinearities with indefinite sign, for example, [5, 6] and the references therein.
When , there are a large number of papers devoted to the following equation: So far, in almost all the results concerning , the nonlinear function is assumed to be globally superlinear, that is, and there exists such that for all , where . The case in which , , and is globally superlinear was first studied by Rabinowitz in . The assumptions in  ensure that the associated functional of the equation satisfies the Palais-Smale condition; this fact was observed in [8, 9] where the results in  were generalized. For a radially symmetric Schrödinger equation with an asymptotically linear term, one radial solution has been obtained in [10, 11] by Stuart and Zhou and their results were generalized to more general situations in [12–15].
Since the class Sobolev embedding is , we cannot study the sublinear problems in via variation method. In order to overcome this obstacle, a regular way is to add some restrictions on potentials and . For example, in , the authors obtained the existence of infinitely many nodal solutions for problem , where , ,, and the nonlinearity is symmetric in the sense of being odd in and may involve a combination of concave and convex terms. There are also some other results about concave and convex problem on , such as [17–19] and the references therein. However, as we have known, there are few results about problem with both sublinear terms and asymptotically linear terms.
Recently, in , the authors established a weighted Sobolev type embedding of radially symmetric functions which provides a basic tool to study quasilinear elliptic equations with sublinear nonlinearities. Motivated by the works of , we consider with more general potentials and combined nonlinearities. In our paper, we assume the following.) is radially symmetric and there exists such that () is radially symmetric and there exists such that It is clear that the indexes and describe the behavior of and near infinity. On , we assume the following:(), ;(), ;(), ;(), ;(), . According to the indexes , we define the bottom index :
Let denote the collection of smooth functions with compact support and Denote by the completion of under the norm Define which is a Hilbert space [21, 22] equipped with the norm Let which is a Banach space equipped with the norm
Following Theorem 1.2 in , under the assumptions , , and , , it holds that the embedding is compact for . We remark that the index by , , so it is possible to study with sublinear nonlinearities. We make the following assumptions on :();().
Since under the assumptions , , and , , it holds that the embedding is compact, the eigenvalue problem has the eigenvalue sequence Similar to the eigenvalue problem on bounded domain, is simple and isolated and has an associated eigenfunction which is positive in .
Our main results are the following.
Theorem 1. Under the assumptions (), (), and (), , if satisfies (), () with , moreover(), ;()there exist and small, such that , ,then there exists such that, for any , has at least four nontrivial solutions.
Theorem 2. Under the assumptions (), (), and (), , if satisfies (), () with for some , moreover, () is not an eigenvalue, , , , for some , then there exists such that, for , has at least one nontrivial solution.
Remark 3. In Theorem 1, may be assumed as superlinear near zero; we can get four nontrivial solutions by Mountain Pass Theorem and Ekeland’s variational principle and truncation technique. In Theorem 2, under the assumptions on near zero, the functional associated to problem enjoys linking structure, and has a linking solution.
Remark 4. In Theorem 1, may be an eigenvalue of problem ; then problem may be resonant near infinity.
Remark 5. As we have known, there are few results about problem on with both sublinear and asymptotically linear nonlinearities at the same time.
In this section we give some preliminaries that will be used to prove the main results of the paper. We begin with a special case of results on Sobolev type embedding which is due to .
Lemma 6 (see ). Let (), (), and (), , be satisfied; the space is compactly embedded in , for any such that .
For , we denote where , ; then, under conditions and , and .
Recall that a sequence is a sequence for the functional , if A sequence is a sequence for the functional , if
Definition 7. Assume is a Banach space, ; one says that satisfies the condition, if every sequence has a convergent subsequence. satisfies condition if satisfies at any .
Definition 8. Assume is a Banach space, ; one says that satisfies the condition, if every sequence has a convergent subsequence. satisfies condition if satisfies at any .
Lemma 9 (Ekeland’s variational principle, ). Let be a complete metric space and let be lower semicontinuous, bounded from below. For any , there is some point with
Lemma 10 (Mountain Pass Theorem, Ambrosetti-Rabinowitz, 1973, ). Let be a Banach space, . Let and be such that and If satisfies the condition with then is a critical value of .
Lemma 11 (Linking Theorem, Rabinowitz, 1978, ). Let be a Banach space with . Let and be such that . Define , , . Let be such that If satisfies the condition with then is a critical value of .
It is well known that the above two minimax theorems are still valid under condition. In our paper, we denote ; is denoted to be various positive constants.
Lemma 12. Under the assumptions (), (), and (), , if satisfies (), (), and () with , then there exists such that, for , one has the following.(i)There exist , such that (ii)There exists with such that .
Proof . We only prove the above results for .
(i) By , , and , there exists and , such that . Then Set where and . By , we have Then there exists such that . Thus, there exists such that, for , . Furthermore, set ; we have
(ii) Let be a -eigenfunction; for we have By , and ; then there exists large enough such that So, we can choose ; then (ii) is proved.
Lemma 13. Under the assumptions (), (), and (), , if satisfies ()–() with and (), then(i)for any given and , we have (ii)there exists satisfying the fact that for there exist two positive constants and such that for all , one has (iii)there exists such that, for any given and , and , we have .
Proof. (i) Let ; by , , , then
(ii) Let ; by and with , and , we have that there exist , , , such that . Then The rest of the proof is similar to the proof of (i) of Lemma 12.
(iii) For any , set ; by , we have , as , where . Then Since , for every , , , Arguing by contradiction, we find a sequence , satisfying , , where , , , such that Dividing in both sides of the above equality, there holds where , . Since , after passing to a subsequence , in , in . Let ; by (32), there exists a bounded domain , such that As , it follows from (34) that Clearly, , for some and , as . Since , in , in , then , in . It is easy to see from the Lebesgue dominated converge theorem that Hence ; this is impossible.
3. Proof of Main Results
Proof of Theorem 1. Firstly, we will prove that, for any fixed , the functionals have a local minimizer, respectively; then problem has two nontrivial solutions: one is nonnegative; the other one is nonpositive.
Similar to , for given by Lemma 12(i), define and is a complete metric space with the distance By Lemma 12, we have that Clearly, ; hence is lower semicontinuous and bounded from below on . Let By the definition of , we can easily claim that . Indeed, since if is small enough, By Lemma 9, for any , there exists a such that Then, for large enough. Otherwise, if for infinitely many , without loss of generality, we may assume that for all , and it follows from (40) that Let and combine (43); we can get that . This is a contradiction.
We prove now that , as . In fact, for any with , let and, for any fixed , we have if is small enough. So it follows from (43) that That is, Let ; we see that , and this gives So, , as , and, by (43), , as . Then, for any given , is a bounded sequence of . By the compactness of Sobolev embedding Lemma 6 and a standard procedure, we see that there exists such that . Since this implies that . That is, is a nontrivial solution of problem . For the case , by the same argument, we can get that problem has another nontrivial solution, which is nonpositive.
Define where .
Lemma 14. Under the assumptions , , and , , if satisfies , with , , and then, for any fixed , the functional satisfies the condition.
Proof. Here, we only prove the case for .
For every sequence , We claim that the sequence is bounded in . Seeking a contradiction, we suppose that . Let ; up to a subsequence, we get that in , in , , a.e. .We claim that . Otherwise, , since by (51) Dividing in both sides of (52), we get that , , and with imply that there exists , such that Combining (53) and (54), we have Letting , we get a contradiction. Thus, in .
Set From , for all . Dividing in both sides of the above equality, there holds By (54), for . Then we have On the other hand, since for a.e., we have for a.e. , which implies that , for a.e.. Besides , for a.e. . Using the Lebesgue’s Dominated Convergence theorem, we obtain that By (59) and (60), Combining (58) and (61), letting , there holds We claim that . Otherwise ; taking in (62), we have , which is impossible. Taking in (62), then we can get . Moreover, by the Hopf’ Lemma, we also can get in . Taking in (62), we obtain Since is the eigenfunction associated to , and , we have This is impossible, since . Then is bounded in . Since the embedding from into is compact, there exists , such that strongly in , and , .
Finally, since , then We have , i.e., . Thus, is a nonnegative solution for problem . Similarly, for we can also get a nonpositive solution for problem .
Thus, problem has at least four nontrivial solutions. The proof of Theorem 1 is complete.
Proof of Theorem 2. In order to prove Theorem 2, we firstly verify that the functional enjoys the linking structure. This can be easily got form Lemma 13. In fact, for , , with , , and or and , . Lemma 13 implies that there exists , such that, for , Define Next, we prove that the functional satisfies the condition.
Lemma 15. Under the assumptions (), (), and (), , if satisfies the assumptions of Theorem 2, then, for any given , the functional satisfies the condition.
Proof. For every sequence ,
Here we just prove that is bounded. Seeking a contradiction we suppose that . Letting , up to a subsequence, we get that in , in , , a.e. .Now, we consider the two possible cases.
Case 1 ( in ). From , we have Dividing in both sides of the above equality, we get that Since , , and with imply that combing (71) and (72), we have Let ; we get a contradiction.
Case 2 ( in ). From , for all . Dividing in both sides of the above equality, there holds By (72), for . Then we have