Generalized Solutions for Nonlocal Elliptic Equations and Systems with Nonlinear Singularities

Abstract

We use the topological degree method to study the existence of solutions for nonlocal elliptic equations (systems) with a strong singular nonlinearity.

1. Introduction and Main Results

Given , an integer , and a bounded open set of with Lipschitz boundary, let be a function satisfying the following properties:(i) with .(ii)There exists such that for any .(iii) for any .

The so-called nonlocal elliptic operator is defined by In particular, when , is equal to the fractional Laplace operator (up to normalization factors).

For a Carathéodory function , the following problem and its special case have been widely studied under various contexts; see a recent survey [1] for details.

1.1. Previous Work

Motivated by the work of Caffarelli and Silvestre [2], several authors have considered an equivalent problem of (3) by means of an auxiliary variable; see [2–6]. Precisely, let denote the points in and . Take and as the completion of with respect to the norm where is a normalization constant. For , let and consider the problem An energy solution to this problem is a function such that Such an energy solution yields a function in the sense of traces, which belongs to the space and is a weak solution of (3). The converse is also true. The reader may refer to [2–6] for dealing with (3) with this method. In particular, Stinga and Torrea [6] generalized the arguments and results in [2] to the fractional powers , , of a linear second order partial differential operator that is nonnegative, densely defined, and self-adjoint in with a positive measure on .

Servadei and Valdinoci developed a variational framework to study the problem (2) in a series of papers [7–11]. They introduced the following Hilbert space in [7, 8]. Let , where , and let be the space of all Lebesgue measurable functions such that and that the map is a Banach space endowed with the so-called Gagliardo norm and is contained in . Consider the subspace of : It was proved in [12, Theorem 6] that this space is the closure of in . Clearly, the space depends on . In fact, when ,   ([11, Lemma 7-b]).   can be endowed with a Hilbert space structure given by the inner product ([8, Lemma 7]) and contains ([7, Lemma 11]). Denote by the induced norm of the inner product in (10). This norm is equivalent to the restriction of to . Call a weak solution of the problem (2) if satisfies for all . Define for the above Carathéodory function . Suppose that there exist , and , , such that Then, the functional defined by is of class and the critical points of are exactly the weak solutions of (2). Condition (12) is always assumed in the proofs of present several existence results on (2) via variational methods [7–11, 13–15]. Except for [13] is also assumed to be a solution in all other works. For studies of (3), there is a great deal of literature; see [2, 4, 16] and references therein.

However, all previous results cannot include the following case: and , where such that for and that for .

1.2. Main Results of This Paper

We will use the topological degree theory developed by [17] to study generalized solutions of problem (2).

Our result can apply to the example just mentioned. Without special statements, we write (the latter plays the role of a critical Sobolev exponent).

Theorem 1. For an integer let satisfy (i)–(iii) and let be a Carathéodory function verifying the following conditions: there exist positive numbers , , , and functions and such that Then, problem (2) has at least one generalized solution in ; that is, it satisfies In particular, it must have a nontrivial generalized solution if is not identically zero.

Corollary 2. Under the assumptions of Theorem 1, let , and let . If either or belongs to with , then has at least one nontrivial solution in provided is not identically zero.

Corollary 3. Under the assumptions of Theorem 1, let and with and . If is such that , then has at least one nontrivial solution in provided is not identically zero.

In particular, this corollary includes the example at the end of Section 1.2. See Example 1 for more general cases.

Our methods can also be used to study the case of nonlocal elliptic operator systems. Let be functions satisfying conditions (i)–(iii). Given two Carathéodory functions , , consider the following problem: Call a generalized solution of system (21) if for every . Here is the second main result.

Theorem 4. Under the above assumptions, suppose also that there exist positive numbers , , , , and functions and such that Then, problem (21) has at least one generalized solution in . In particular, it must have a nontrivial generalized solution if one of and is not identically zero.

Similarly, consider the following problem: or Assume(A) are Carathéodory functions, ;(B)there exist constants  and measurable functions , , such that (C)there exist measurable functions and constants such that, for any ,  Combing the proof of [17] and that of Theorem 4, we can prove the following.

Theorem 5. Under the conditions (A), (B), and (C), if or is not zero, then the equation systems (26) and (27) have at least a nontrivial generalized solution .

Finally, let us point out that the corresponding results of Theorems 1 and 4 can be also proved if the operator is replaced by in [6, (1.10)]. They will be given in other places.

The arrangements of this paper are as follows. In Section 2, we give some necessary preliminaries. The proof of Theorem 1 will be completed in Section 3. In Section 4, we will prove Corollaries 2 and 3 and give an example. Theorem 4 will be proved in Section 5.

2. Preliminaries

Firstly, we review the topological degree theory for mappings of class developed in [17]. Let be a Hilbert space with inner product and let be a strictly increasing sequence of finite dimensional subspace of such that is dense in . Denote by the orthogonal projection from onto for every integer . Let be an open bounded set in and let be a mapping from , the closure of in , into . Put Since and induce equivalent topologies on all finite dimensional spaces the subset has the same closure in , , and , denoted by .

Definition 6. Under the above assumptions, is said to be of class on if and only if the following conditions are satisfied: (a) is a continuous mapping for each .(b)There is not any sequence in such that the sequence is weakly convergent in and for all and in .

Lemma 7 (see [17, Lemma 2.3]). Let and be as in Definition 6. Assume that is of class on . Then, there exists an integer such that the Leray-Schauder degree is defined and

It follows that is defined. It was the topological degree of on at   in [17]. The corresponding versions with usual properties of the Leray-Schauder degree were given in [17, Theorem 2.1]. In particular, the identity map is of class (), and if . Moreover, the following proposition is key for the proof of our main results.

Proposition 8 (see [17, Corollary 2.1]). Let and be as in Definition 6. Let be a mapping from into such that is continuous on for any . Suppose that contains 0 and Then there is a weakly Cauchy sequence in such that

Next, we need the following results on the space .

Lemma 9.    and are continuously embedded in and , respectively ([8, Lemma 5]).
If is a bounded open subset with continuous boundary, the embedding is compact for any   ([8, Lemma 8] and [11, Lemma 9-a]).
The embedding is continuous for ([11, Lemma 9-b]).
The embedding is continuous for any ([18, Theorem 6.5]).
If is an open set in of class with bounded boundary, then there exist continuous embeddings and for any and ([18, Proposition 2.2]).

Lemma 10. Let be a bounded open set in with boundary of class . Then, the space is separable. Furthermore, there exists a sequence in such that is a maximal orthogonal set of .

Proof. By Proposition 9(f) of [9], there exists a Hilbert basis of , which implies separability of . So is a dense countable subset in . Let denote this countable set. Since is dense in by [12, Theorem 6], for each we can take such that . Then, is also dense in . Let be a maximal subset of such that any finite elements in are linearly independent. Then, is dense in . Making the Hilbert-Schmidt orthogonalization procedure for , we obtain an orthogonal set , which is also a maximal in .

3. Proof of Theorem 1

Take an increasing sequence of open subsets of , , such that each of them has -boundary and that By Lemma 10, we may choose a sequence in such that is a maximal orthogonal set of . Then, we can find a sequence in such that is a maximal orthogonal set of . By the mathematical induction, it is easy to find the set in such that is a maximal orthogonal set of for every . Let us rewrite the countable set as a sequence . Let be the vector subspace of spanned by , and . For conveniences we set and denote by the orthogonal projection from onto .

Lemma 11.    is dense in .
For each , there are and a sequence in such that the supports of all are contained in and that in as .
For every and for every given , there exists a unique in such that Moreover, if , then Suppose that a sequence weakly converges to in . Then, weakly converges to in for .
For every given (the support of must be contained in some by the construction of ), there exists a unique such that    is continuous for every .
There exists a constant such that

Proof. (a) Since is a maximal orthogonal set of , is dense in .
(b) For a given , by the choices of , there exists such that the support of is contained in . Let be the maximal orthogonal set of as constructed above. Then, is dense in . Hence, we can find a sequence in such that as .
(c) By (16), we can write . Then, , and (15) implies Note that and . Moreover, since we get For given , by [19, Theorem 3.2.4] (see also [20, page 30]) we have a continuous mapping from into , where For , we have and by Lemma 9(b) and (c). Thus, and Using Lemma 9(c) again, there exists a constant such that for all . It follows that Hence, Riesz representative theorem yields a unique such that If , for each integer we deduce (d) Let be a sequence weakly converging to in . Since from Lemma 9(b), we deduce that converges to in . Then, the continuity of the map implies that converges to in . Moreover, for , we have . Recall that and . We deduce that and hence by (46).
(e) Since , and , by (46) we deduce that This shows that is a continuous linear functional. Using the Riesz representative theorem again we obtain a unique such that Clearly, for all .
(f) By the construction of , we have an integer such that each has a support contained in . Let converge to . By (d), weakly converges to in for every . Then, (e) implies that as . In particular, since satisfies , we have This shows that weakly converges to in . However, the strong converge and the weak ones on finitely dimensional space are equivalent. Hence, as .
(g) Since , by Lemma 9(b) and (c) there is a constant such that . It follows from this and (17) that This leads to (g).

Proof of Theorem 1. Let , , and be as above. Since , we have such that Let . Define by Let us prove that is of class on . Note that has the same closure in , , and , denoted by . Let be defined by for each . Suppose that a sequence converges to . Lemma 11(f) implies in . Hence, is continuous. By the proof of Lemma 11(g) and (54) we deduce that This implies that Definition 6(b) is satisfied. Hence, is of class on . Moreover, it also shows that satisfies the conditions of Proposition 8. Thus, we have a weakly Cauchy sequence such that Let be the weak limit of in . For a given , the support of it is contained in some , and thus by Lemma 11(c). For each fixed , there exists such that . Hence, Lemma 11(e) yields Taking in both sides of and using (57) and Lemma 11(d), we deduce For any given , by Lemma 11(b), we have an integer and a sequence such that for any and that in . So (60) leads to that is, . Letting , we get which shows that is a generalized solution. Note that might be zero! But is not a solution if takes nonzero values on a nonzero measure set. The proof is completed.

4. Proofs of Corollaries and Examples

Proof of Corollary 2. Let . It suffices to check that satisfies (15)–(17). Clearly, we can assume . Let . Since , we have and thus Then, and By Young’s inequality, we obtain Note that and imply . Let , which sits in . From these and (15), it follows that for some constants and , where Young’s inequality is used again. So satisfies (15). Moreover, ; that is, satisfies (16).
Finally, let us check that (17) holds for . If , then For another case, observe that Since , we may choose a real number in . Let . By Young’s inequality, we have Note that since . Moreover, and Let , which belongs to . Using Young’s inequality, we can derive for some constants and . Hence, for a.e. and all , we have where belongs to as above. This shows that (17) is true for . The desired conclusion follows from Theorem 1 immediately.

Proof of Corollary 3. Let . Since , we see that satisfies (17) from the above proof. It remains to prove that satisfies (15)-(16). Let . Note that By Young’s inequality, we have Now, implies and because . We obtain . As in the proof of Corollary 2, using Young’s inequality, we may derive from this and (73)-(74) that satisfies (15) and (16).

Example 1. Let and be as above. Consider where belongs to with , is absolutely continuous, , , and , a.e., , . Then, (75) has a nontrivial generalized solution.
In fact, taking in Corollary 3, we should require . Since , there is sufficiently small such that This means that we can take . Moreover, , and Hence, (15)–(17) are satisfied for .

5. Proof of Theorem 4

Consider the product Hilbert space equipped with inner product for . The induced norm is

Let and be as in Section 3. For every integer let and . Denote by the orthogonal projection from onto . Corresponding to Lemma 11, we have the following.

Lemma 13.    is dense in .
For each , there are and a sequence in such that the supports of all are contained in and that in as .
For every , and for every given , there exists a unique in such that for any . Moreover, if , then .
Suppose that a sequence weakly converges to in . Then, weakly converges to in for .
For every given (the support of must be contained in some by the construction of ), there exists a unique such that    is continuous for every .
There exists a constant such that for all

Proof. (a) and (b) follow from Lemma 11(a) and (b) immediately.
(c) By (24), we can write as in the proof of Lemma 11(c); then, , and Moreover, for every given and , we have continuous mappings where for , . Recall that . For , by Lemma 9(b) and (c), we have Thus, , , and Using Lemma 9(c) again, there exist constants such that for all , . It follows that By the Riesz representative theorem, for each , we have a unique such that Setting , we obtain Another claim can be proved as that of Lemma 11(c).
(d) Let for each . Then, weakly converges to in and weakly converges to in . For each , by Lemma 11(d), weakly converges to in and weakly converges to in . Thus, weakly converges to in for .
(e) Since , , and , for any integer and , we derive from (89) that These show that are two continuous linear functionals. Using the Riesz representative theorem again we obtain a unique such that for all . Set ; then, Clearly, for all .
(f) follows the above (e) and Lemma 11(f) directly.
(g) Since , by Lemma 9(b) and (c) there is a constant such that It follows from this and (25) that Here, . This leads to (g).

Proof of Theorem 4. We replace the space in the proof of Theorem 1 by . Since for , as in (54), we have such that Then repeating the proof of Theorem 1, we get a such that for any ; namely, (22) holds.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors would like to thank the anonymous referee for the valuable comments and suggestions on the paper. This work is partially supported by the NNSF 10971014 and 11271044, of China, and PCSIRT.