The Neumann Problem for a Degenerate Elliptic System Near Resonance

An, Yu-Cheng; Suo, Hong-Min

doi:https://doi.org/10.1155/2017/2925065

Advances in Mathematical Physics

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Research Article | Open Access

Volume 2017 | Article ID 2925065 | https://doi.org/10.1155/2017/2925065

The Neumann Problem for a Degenerate Elliptic System Near Resonance

Yu-Cheng An¹and Hong-Min Suo²

Academic Editor: Pietro d’Avenia

Received27 Aug 2016

Revised01 Dec 2016

Accepted06 Dec 2016

Published11 Jan 2017

Abstract

This paper studies the following system of degenerate equations , , , , , Here is a bounded domain, and is the exterior normal vector on . The coefficient function may vanish in , with . We show that the eigenvalues of the operator are discrete. Secondly, when the linear part is near resonance, we prove the existence of at least two different solutions for the above degenerate system, under suitable conditions on , and .

1. Introduction

In recent decades, many kinds of perturbed problems were studied by many scholars, such as [1–11]. Here, we want to say that the authors in [5] studied the following Dirichlet boundary problem:When the parameter is close to an eigenvalue of the operator , they proved that problem (1) has two different solutions. Moreover, this result was extended to some equations and systems; see [6–10]. In particular, Massa and Rossato [11] studied a nondegenerate elliptic system and two solutions were obtained by using Galerkin techniques. On the other hand, we also mention that many scholars studied some elliptic equations with the Neumann or Robin boundary; see [12–17] and the references therein. Inspired by the above results, we study the following system of degenerate equations:where is a bounded domain, is the exterior normal vector on , , and . The coefficient may vanish in , with ; that is, problem (2) may be degenerate; see [18]. As in [11], we will use the critical point theory and Galerkin techniques to obtain the existence of two different solutions for the above degenerate system. Now, we introduce the function set which consists of functions such thatThroughout the paper, we always assume that there exist and such thatWe also assume that belongs to with , and is a Carathéodory mapping and satisfies the following conditions: For every , there exists such that, for all , uniformly in , .Although the conditions and were introduced in [10], it is weaker than in [5] (or (1.2) in [11]). In fact, let ; it is easy to see that satisfies , but the function does not satisfy the condition in [5] (or (1.2) in [11]).

In Section 2, we give some preliminary lemmas and our main results. Meanwhile, we show that the eigenvalues of the operator are discrete under Neumann boundary condition. In Section 3, we prove our main results through Galerkin techniques and saddle point theorem.

2. Preliminaries and Main Results

In this section, we first collect some basic facts and then give the properties of the eigenvalues of the operator . Secondly, we define a new norm and prove it is equivalent to the usual Sobolev norm. At the end of this section, we give the main results of this paper.

Firstly, let denote the completion of with respect to the norm The inner product in is denoted by From (4), we know that the spaces and are equivalent; see [18]. Let ; from with , one hasHence, by the Sobolev embedding theorem of [18], we know that is compactly embedded in . Moreover, it follows from Hölder’s inequality that

Now, we use a similar argument to that of Gasiński and Papageorgiou (see [15]). Let us study the following eigenvalue problem:Firstly, from (4) and the Sobolev embedding theorem of [18], we know thatand the first embedding is compact. Then, for any , we havefor some positive constant ; see [19].

Let us define It follows from (9) and (12) thatChoosing small enough, then from (14) one getsfor some positive constants . Hence, by Corollary 7.8 in [20], we conclude that there exists an eigenvalue sequence satisfyingas andLet be the corresponding eigenfunction sequence; then is complete in and for some ; see [21].

Now, let ; since the coefficient may vanish in , we need to define a new norm:and the corresponding inner product

Lemma 1. Let with ; then the norms and are equivalent.

Proof. Firstly, it follows from (17) thatWe prove that there exists such thatIn fact, if (21) is false, by mean of the 2-homogeneity of , there exists such that for all and as . Without loss of generality, we may assume thatBy the sequential weak lower semicontinuity of and the choice of , we know thatBy (17), one gets as well as for some constant . If , then in , which contradicts , ; if , by (23), one has ; this is a contradiction. Hence, (21) is true.
On the flip side, from (9), we haveHere is a positive constant. Then, by (21) and (24), one getsThis proved the norms and are equivalent.

From now on, we always assume .

Lemma 2. Under the hypotheses of Lemma 1, the embedding is continuous for , compact for .

Proof. By Lemma 1 and the Compactness Theorem in [18], we directly conclude Lemma 2.
In addition, from Lemma 2, there exists such that . For simplicity, we will assume that ; that is,

Now, let and . For fixed , suppose that is an eigenvalue of multiplicity and denote by the eigenspace associated with the eigenvalue , . The main results are as follows.

Theorem 3. Let be the first value above and suppose that conditions and hold. Also,uniformly in , andThen for any , there exists such that ; if , then problem (2) has at least two different solutions.

Theorem 4. If we replace condition (27) of Theorem 3 withuniformly in , then for any , there exists such that ; if , then problem (2) has at least two different solutions.

Theorem 5. In addition to conditions , , and (29), suppose that is the first value above andThen for any , there exists such that ; if , then problem (2) has at least two different solutions.

Theorem 6. Let be the first eigenvalue above and conditions , , (27), and (30) hold. Then for any , there exists such that ; if , then problem (2) has at least two different solutions.

3. Proof of Main Results

In this section, we firstly prove some preliminary lemmas, and then we prove our main results through variational methods and Galerkin techniques.

For the sake of simplicity, let and , with the norms the inner products and , respectively. In addition, we will always use the notation , unless otherwise specified.

Define and For every , we claim that there exist positive constants and such thatIn fact, by means of , , and (26), the arguments of (33) and (34) are quite similar to that of Lemma 3.1 in [11] and so is omitted.

Now, we define the functional , where , given bywhere . By means of and , one has and which implies that the critical points of are exactly weak solutions of problem (2).

Next, we need to consider the eigenvalue problem: , for ; that is,Firstly, for any , one has and for some . Now, by and using and in (38), then a straightforward calculation shows that (38) is equivalent toObviously, if and only if . Hence, we obtain two sequences of eigenvaluesand the corresponding eigenfunctionsare the corresponding eigenfunctions.

Let , for ; a simple calculation yieldswhere denotes the Kronecker symbol. Moreover, if , thenIn addition, for every , if , from (40) one getsLet us fix and defineMeanwhile, we denote by , , , and the unitary closed balls, with respect to the norm , in the spaces , , , and , respectively, and by , , , and their relative boundaries.

Lemma 7. Suppose that satisfies , , . For fixed , let and be the first eigenvalue above and , respectively.
If , then we havefor some constants and .
Further, if condition (27) also is satisfied, then there exists a positive constant such that, for , there exist , , such that (46) hold andHere, a value with index represents that depend on , and other cases are similar.

Proof. Firstly, if , then ; if , then .
For , if , then the sequence is nondecreasing, which impliesIf , then the sequence is nonincreasing, which impliesSimilarly, for , that is, , if , then the sequence is also nondecreasing, which impliesIf , then the sequence is nonincreasing, which implies for every .
In a word, for fixed , there exists such thatSecondly, because of the fact that is the first eigenvalue above and , thus , if ; , if . Proceeding as in the proof of the first step, we can also conclude that there exists such thatLet ; we have by (53) and (54)Hence, as in the proof of Lemma 4.1 in [11], we know thatFrom (33) and (56), we getFrom (33) and (57), we getBy (58) and (59) and choosing , we conclude that there exist and satisfying (46).
In addition, if is near enough to , in particular, if and , we claim that there exists such thatIn fact, if and , for , we haveAnd, for , we haveHence, for fixed , there exists such thatFrom this we easily get the estimates (60) and (61).
Next, we prove (47), (48), and (49). Let and . If , then .
If , it follows from (33) and (61) thatBy choosing , then there exists satisfying (47).
If , by (33) and (60), we obtainLet us choose small enough; then there exists satisfying (49) for .
Now, we prove the estimate (48). If (48) is not true, then, for any sequences and , there exist and such thatHere (or ) denotes the form (or the functional ) with and . And, denotes the eigenvalues of the bilinear form .
Let , and suppose and as . By (44), one has , which implies . So, by (60), we getfor all positive integers ; we get by (68)It follows from (33) and (69) thatWe note that ; then from (70) we obtainwhere . Note that Then, we get by (71)Let , where , . Then, by (73), there exists with , such thatLet . Then, we have From this and (74), without loss of generality, we assumeBy , one has for some positive constant . Besides, from (26) and the second inequality of (76), we obtain . Thus, by choosing small enough, there exists such thatFrom this and (74), there exist and such thatBy and (27), proceeding as in the proof of Proposition 4.6 in [11], we haveFurther, because of the fact that is bounded from below, we getIn addition, for fixed and , from (54) and (64), we can choose small enough such that ; then we get . Hence, by (28), it is easy to see that for some positive constant . Moreover, it follows from (69) that Recall that and as , which contradicts with (80). Hence, there exist and which satisfy (48).
Finally, by the process of the above proof, we easily know that all the constants of the estimates above are not contradictory; then we finished the proof of Lemma 7.

Remark 8. In fact, from (46), we can get a solution of problem (2). And, from (47), (48), and (49), we can also get a solution of problem (2). But we do not know whether they are different. So we will use Galerkin techniques to show the existence of two different solutions.
Now, we assume that It follows from this that . As before, let , , , , , and represent their relative boundary. In addition, we know that all the estimates of Lemma 7 are true when the conditions of Theorem 3 are satisfied. Further, if and , we can also check that the similar estimates of Lemma 7, with respect to the functional , hold on the spaces , , and .

Lemma 9. Fix , assume that , , and and satisfy and . In addition, let satisfywhere as . Then, there exists a subsequence such that converges in .

Proof. If , then . Hence, we may divide the space in the two orthogonal componentsFor , let . In addition, because of as , then we may set , and thenSetting , by testing (83) with we haveFrom this, (34), (42), and (85), we getIn the same way, by testing (83) with we haveFrom (88) and (89), we haveLet us choose ; then is bounded in . By , then there exists such that converges in .

From the proof of Lemma 9, we can also get Lemma 10 below.

Lemma 10. Assume that , , and and satisfy and . In addition, let satisfywhere as . Then is bounded in .
Clearly, by using the Saddle Point Theorem, Lemma 9, (46), (47), and (48), there exist and in such that , , and , , where

Lemma 11. Assume that the conditions are the same as Lemma 7, , and . Then, for all , there exists such that and .

Proof. Firstly, by (92), one gets and . DefineSo it follows from (48) and (49) that which implies . In addition, by , one gets .
For , it follows from thatFrom this and (33), we get for any Hence, there exists such that . This finishes the proof of Lemma 11.

Now, we start to prove our main theorems.

Proof of Theorem 3. Firstly, by Lemma 11 and without loss of generality, there exist and such that and as .
Next, we prove that there exists such that and . In fact, for all , we haveThen, it follows from Lemma 10 that is bounded in . Hence, without loss of generality, there exists such thatFor , let and in (98), respectively. From a direct calculation, we have for all Let ; we get for any . It follows from that . That is, is a critical point of the functional .
Next, we prove . Firstly, let us define the orthogonal projection which implies , in . ThenIt follows from (102) thatLet in (98); one getsFrom this, (103), we obtainwhich impliesBy and (106), one has . From the uniform convexity of , we have as , which implies that .
In the same way, we can also prove that there exists such that , , and . Note that Hence, we get , which finished the proof of Theorem 3.

Proof of Theorem 4. Let , and we use the same approach as before; we also get eigenvalues of :and the same eigenfunctionsIn the same way, we can also define the subspaces , , and . If for some , thenNext, we prove that (56), (57), (60), and (61), with respect to , hold on the new subspaces , , and . Firstly, we prove the following claim.
Assume that and are the first eigenvalue above and , respectively. If and , then there exists such thatBesides, if is close to , or , thenfor some positive constant .
Actually, from (108) one can prove that an estimate like (55) holds for these new eigenvalues , and then we get (111). Moreover, when is close to , we can also check that, as in the proof of (60) and (61), there exists the positive constant satisfying (112).
Similarly, we can also choose small enough such that . By (111) and (112), we can prove a similar result of Lemma 7. In other words, if the conditions of Theorem 4 are satisfied, we can conclude that there exist positive constants and such that the functional satisfies (46) on the new subspaces , , and . Further, there exists such that if , then the functional satisfies (46), (47), (48), and (49). Hence, we can also obtain two critical point sequences and of , at critical levels , , similar to (92). Next, the remainder of the argument is similar to the proof of Theorem 3.

Proof of Theorem 5 (or Theorem 6). In order to take advantage of the conclusion of Theorem 3 (or Theorem 4), we consider the following degenerate system:where , , , and . Obviously, under the hypotheses of Theorem 5 (or Theorem 6), the similar hypotheses of Theorem 3 (or Theorem 4) are satisfied for problem (113). Meanwhile, if is a solution of (113), then is a solution of problem (2). Hence, by the proof of Theorem 3 (or Theorem 4), we know that Theorem 5 (or Theorem 6) is true.

Competing Interests

The authors declare that they have no competing interests.

Acknowledgments

This work was supported by National Natural Science Foundation of China (no. 11661021) and Innovation Group Major Program of Guizhou Province (no. KY(2016)029; no. KY(2013)405).

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Copyright © 2017 Yu-Cheng An and Hong-Min Suo. 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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