Abstract

Using the minimax methods in critical point theory, we study the multiplicity of solutions for a class of Neumann problems in the case near resonance. The results improve and generalize some of the corresponding existing results.

1. Introduction

The aim of this paper is to study the following semilinear Neumann problem: Here is a bounded domain with a boundary and denotes the outward unit normal on . Also with and it may change sign. The reaction is a Carathéodory function and satisfies the following assumptions:for every , there exists a function such that for all and a.e. ; uniformly for .

Recently, there have been many papers concerned with the Neumann problems; see [15] and the references therein. Or, more specifically, in Li [1] and Qian [2], the left hand side differential operator is , with , . In Motreanu et al. [3], Tang and Wu [4], and Motreanu et al. [5], the differential operator is (i.e., ). Semilinear Neumann problems with unbounded and indefinite potential, especially, were studied by Gasiński and Papageorgiou [6]. They obtained two multiplicity theorems. In addition, the same problems were studied by Papageorgiou and Rdulescu [7]. They dealt with equations in which the reaction exhibits an asymmetric behavior at and at (jumping nonlinearity) and they proved multiplicity theorems providing sign information for all the solutions.

On the other hand, for the perturbed problem, Mawhin and Schmitt [8] first considered the two-point boundary value problem Under the assumption that is bounded and satisfies a sign condition, if the parameter is sufficiently close to from left, problem (2) has at least three solutions; if , problem (2) has at least one solution, where , are the first and second eigenvalues of the corresponding linear problem. Ma et al. [9] considered the boundary value problem defined on a bounded open set , no matter whether the boundary conditions are Dirichlet or Neumann condition; as the parameter approaches from left, there exist three solutions. Moreover, existence of three solutions was obtained for the quasilinear problem in bounded domains as the parameter approaches from left. In [10, 11], these results were extended to the perturbed -Laplacian equation in . In [12], Ou and Tang extended above some results to some elliptic systems with the Dirichlet boundary conditions. de Paiva and Massa in [13], especially, studied the semilinear elliptic boundary value problem in any spatial dimension and using variational techniques; they showed that a suitable perturbation will turn the almost resonant situation ( near to , i.e., near resonance with a nonprincipal eigenvalue) in a situation where the solutions are at least two. In [14], those results were extended to the degenerate elliptic equations in the bounded domain.

Motivated by the above idea, we have the goal in this paper of extending these results in [6, 1214] to some elliptic equations with the Neumann boundary conditions. Here, it is worth pointing out that is weaker than in [13] (or in [14]). More to the point, there are functions satisfying the assumptions of our main results in Section 2 and not satisfying the assumptions in [13, 14]. For example, let . Then satisfies the assumptions for our Theorem 5 in Section 2 and does not satisfy in [13] (or in [14]).

The rest of our paper is organized as follows. In Section 2 we give some preliminary lemmas and our main results. Section 3 gives the detailed proofs of our main results based on several estimates, whose proof will be presented in Sections 4 and 5.

2. Preliminaries and Main Results

Let the Sobolev space . Denote to be the norm of in and the norm of in . The space is a Hilbert space. For a discussion about the space setting, we refer to [15] and the references therein.

Again, we recall the properties of the eigenvalue problem as follows (see [6]): The eigenvalue problems in (4) have a sequence of eigenvalues, such that There is also a corresponding sequence of eigenfunctions which form an orthonormal basis of and an orthogonal basis of . Moreover, we know that for some and all .

We denote by the eigenspace corresponding to an eigenvalue , and we can decompose . We set The eigenvalues admit the following variational characterizations in terms of the Rayleigh quotient for all : In (7), the infimum is realized on . Also, in (8), both the infimum and the supremum are realized on . All the eigenspaces have the so-called unique continuation property. The first eigenvalue is simple and it is clear from (7) that the corresponding eigenfunctions do not change sign. Namely, we can suppose that is strictly positive on . We mention that all the other eigenvalues have nodal eigenfunctions. For more properties to the eigenvalue problem (4), see [6, 7].

By the presence of function , weak solutions of (1) must be found in a suitable space. To this purpose, letting , we introduce a new inner product on by for and the associated norm

Lemma 1. Let with and it may change sign, if ; then is equivalent to the usual Sobolev norm .

Proof. By virtue of Hlder’s inequality, we have where is the constant of Sobolev imbedding from , , .
On the other hand, if , then there exists such that If not, it is clear from (7) that Exploiting the 2-homogeneity of we can find a sequence , such that Passing to a subsequence if necessary, we may assume that The sequential weak lower semicontinuity of and (15) imply that So and thus , with some .
If , then in , which contradicts the fact that for all . If , then, from (17), we have , which contradicts (7).
Combining (11) and (12), we have Namely, is equivalent to the usual Sobolev norm .

From now on we take as our working space, , . In view of Lemma 1 and the Rellich-Kondrachov Compactness theorem (see [15, Theorem  1]), we directly get the following lemma.

Lemma 2. Let with and it may change sign. Then the space is compactly embedded in for and continuously embedded in ; hence there exists such that

In addition, we also need the following lemmas.

Lemma 3 (from Lemma  4.6 of [13]). Let be a Hilbert space with orthonormal direct sum splitting . Moreover, let . For , set Then and link.

Lemma 4 (from Theorem  8.1 of [16]). Let be a Hilbert space where has finite dimension and satisfying the condition and such that, for given , where and represent the unit ball and the unit sphere in . Then there exists a critical point such that .

Next, let ; in order to state our main results, we introduce the following assumptions on the nonlinear term: uniformly with respect to ; uniformly with respect to ; uniformly with respect to ; uniformly with respect to .

Our main results are given by the following theorems.

Theorem 5. Let   () be an eigenvalue of multiplicity and with and it may change sign. Suppose that the conditions and hold and one of the sets of hypotheses and . Then there exists such that for problem (1) has at least two solutions.

Theorem 6. Let   () be an eigenvalue of multiplicity and with and it may change sign. Suppose that the conditions and hold and one of the sets of hypotheses and . Then there exists such that for problem (1) has at least two solutions.

Theorem 7. Let with and it may change sign. Suppose that the conditions and hold, and the nonlinearity satisfies . Then, for sufficiently close to , problem (1) has at least three solutions.

3. Proof of Theorems

The associated functional of problem (1) is for . Under the conditions and , it is easy to verify that, for every , and for . Moreover, critical points of are exactly weak solutions of problem (1).

It follows from and that for every there exist and such that for all and a.e. , which implies that for all and a.e. . From this and Hölder’s inequality, we have where and is the best embedding constant.

In addition, we will use several times the estimates below.

From (7) and (8), we have

Proposition 8. Assume that and hold, and suppose that for any . Then satisfies the condition.

Proof. For any sequence such that we need to prove that has a convergent subsequence. By the standard argument, it suffices to show that is bounded in . Suppose by contradiction that as . Let . Then , so we may suppose that there is such that as . Then, for any , from (24) and (27), we have which shows that By the arbitrariness of , one has Like in the proof of (34), from (26), it follows that Thus, for any , by (30), (31), and (34), passing to the limit in gives which implies that In addition, by (30), (31), and (35), passing to the limit in gives
If , then, from (40), we have , which is a contradiction.
If , then, from (40), we have . From this and (38) it follows that is an eigenvalue of operator , a contradiction; the proof is completed.

Hereafter, we set and we define and , , and , respectively, are their relative boundaries.

Theorems 5 and 6 will be a consequence of the geometry in Propositions 9 and 10 stated below, whose proofs will be postponed to Sections 4 and 5.

Proposition 9. If and hypotheses and are satisfied, then there exist constants and such that Moreover, if one of the sets of hypotheses and is satisfied, then there exists such that for there exist , such that, in addition to (43) and (44), (The values with index depend on ; the others may be fixed uniformly.)

Proposition 10. If and hypotheses and are satisfied, then there exist constants and such that Moreover, if one of the sets of hypotheses and is satisfied, then there exists such that for there exist , , and such that, in addition to (48) and (49), (The values with index depend on ; the others may be fixed uniformly.)

Proof of Theorem 5. Since the functional satisfies the condition, we can apply two times the saddle point theorem (see, e.g., [17]); let
The first solution, which we denote by and may be obtained for any with just hypotheses and , corresponds to a critical point at the level The criticality of this level is guaranteed by the estimates (43) and (44), since and link; that is, the image of any map in intersects .
The second solution, which we denote by , corresponds to a critical point at the critical level Actually, this is a critical level because of the estimates (45) and (46), since and link.
To conclude the proof, we need to show that these two solutions are distinct.
We observe first that by estimate (45) we have that , then we observe that we may build a map in such a way that its image is the union between the annulus and the image of a -dimensional ball in whose boundary is . By the estimates (46) and (47), we deduce that , and as a consequence , proving that the two solutions are distinct, for being at different critical levels.

Proof of Theorem 6. Since the functional satisfies the condition, we can apply the saddle point theorem and Lemma 4.
The first solution, which we denote by and may be obtained for any with just hypotheses and , is again obtained through the saddle point theorem and corresponds to a critical point at the critical level where now The criticality is guaranteed by estimates (48) and (49), since and link.
The second solution, which we denote by , comes from Lemma 3, where we set and ; actually we have the structure and then we have a critical point at the level .
Finally, in order to prove that these two solutions are distinct, we need a sharper estimate for than that given by (49). For this we use Lemma 3 to guarantee that, for any map , since , one has that the image of either intersects or has a point with . This implies that by estimates (51) and (52), and then proving that the two solutions are distinct, for being at different critical levels.

Proof of Theorem 7. The proof will be divided into four steps.
Step  1. For , from the definition of , (26), and (27), we get Letting , it follows that is coercive in .
Similarly, from (26) and (29), we obtain for all . Let ; hence is coercive in and is bounded from below on , and, moreover, there is a constant , independent of , such that .
Step  2. If is sufficiently close to , we have such that . In fact, for , we have For any fixed with , from , we get uniformly in . From Fatou’s lemma and in , it follows that and hence, taking large enough, we get For , combining (64) and (66) yields . A similar conclusion holds for some .
Step  3. If , let The functional satisfies the and condition for all .
In fact, let satisfy and as . From Proposition 8, there is such that strongly in . If , from the second conclusion of Step 1, we get , which is impossible. Hence, and satisfies the . Similarly we have that holds for all .
Step  4. Three solutions are obtained.
If is sufficiently close to , from Steps 1 and 2, we get , which implies that is bounded below in . Consequently, from Ekeland’s variational principle, there exists such that and as . Since satisfies for all , there is such that ; that is, the infimum is attained in . A similar conclusion holds in . So has two distinct critical points, denoted by , .
Suppose that . Letting we have , and since is the local minimum of , there are such that for . Since and satisfies the condition, from the mountain pass theorem, the number where , is a critical value of . From the definition of , we have and obtain a third critical point of . Hence, the proof is completed.

4. Proof of Estimates

In this section we will prove all the estimates in Propositions 9 and 10.

4.1. Estimates of the Saddle Geometry

Lemma 11. Under hypotheses and , one gets the following: for , there exists satisfying (43) and satisfying (45); for one has the following: there exists satisfying (48), for a given , there exists satisfying (53).

Proof. Let ; using estimates (26) and (29) we get
For , letting , it follows that is bounded below in ; that is, there exists a as in (45).
For , then the same estimate holds but the constant cannot be made independent of , giving (48).
In the same way, let and set ; we get Letting , it follows that is bounded below in ; that is, there exists a such that for all we have (43), where again the constant depends on , that is, on .
Finally, (71) with implies Then, no matter the value of , is bounded from below in any bounded subset of , giving (53) for a suitable value of .

Lemma 12. Under hypotheses and , one gets the following: for , given the constant , there exists satisfying (44); for one has the following: there exists satisfying (50), for a given , there exists satisfying (49), for a given , there exists satisfying (54).

Moreover, given the values , , one may always choose , as claimed in Propositions 9 and 10.

Proof. Let , by estimates (26) and (28); we get
For , letting , then one obtains (44) for suitably large .
For , letting , one obtains, for suitable and , (50) and (54).
Finally, letting and setting , we get Letting , it is clear that (once is fixed) this goes to and then we may find the claimed such that (49) holds.
Observe that and can be chosen uniformly for , while , will in fact depend on .

4.2. Estimating the Effect of the Nontrivial Perturbation

In this section we will prove the remaining inequalities in Propositions 9 and 10, those which rely on the hypothesis or or or , which, roughly speaking, say that the perturbation is nontrivial in such a way that a new solution arises when is sufficiently near to the eigenvalue . The proof is simpler for Theorem 5, since we need to estimate the functional in the compact set , while for Theorem 6 the same kind of estimate is required in the noncompact set .

4.2.1. Estimating in

For the next estimates, we will need the following lemma.

Lemma 13. Hypothesis implies that there exists a nondecreasing function such that

Proof. First we claim that there exists a constant such that the sets have measure , for all .
Actually, is a finite-dimensional subspace and the functions are smooth; they are uniformly bounded; that is, there exists such that for all . Suppose that for there exists such that .
On one hand, by (28), one obtains
On the other hand, That is a contradiction.
Now, for any , we will show that we can find a large enough so that for any and ; this means that
Actually, it follows from that for any there exists such that for . For , one has , and then one gets
For , by and , there exist and such that , for and a.e. .
Let ; one finally obtains It is elementary that is well defined and satisfies the claim.

Now we may prove the following.

Lemma 14. Consider Theorem 5 with one of the sets of hypotheses and . Given the constant , there exist , such that, for any , (46) and (47) hold.

Proof. We consider the two sets of hypotheses separately.
(i) In case , it follows from and that for any there exist and such that for and ; in particular we set . Let ; for being in a finite-dimensional subspace, all the norms are equivalent, so that (set and use estimates (28) and (82)) where .
(ii) In case , let be as in Lemma 13, for ; let with and : We assume that ; it is easy to see that for some constant ; we estimate Considering (83) and (85), we see that since by Lemma 13, we may fix so that (or for the case ) and then for one gets (46).
To obtain (47), we observe that (since ) if , that is, if , then in estimates (83) and (85) we may avoid the term so that (remember that is nondecreasing) for .

4.2.2. Estimating in

We consider the corresponding estimates of the previous lemma, for Theorem 6.

Lemma 15. Consider Theorem 6 with one of the sets of hypotheses and . Given the constant , there exist , such that, for any , (51) and (52) hold.

Proof. Letting , we see from (70) that property (52) will be satisfied provided that is large enough (say ); observe that this value can be made independent from once is small enough.
Now we consider the two sets of hypotheses separately.
(i) In case , suppose ; we can assume that , with and . Since is a finite dimension subspace, all the norms are equivalent, so that there exists such that for all we have . In addition, by and , there exist and such that uniformly in . So by (29) and (86), where , . Since supposing , (87) becomes since , so is bounded below for all ; that is, there exists such that ; by (89) one gets
(ii) In case , first we give some conclusions which are similar to Lemma 3 of [18]. Under the property of , there exists a const , and which is subadditive, that is, for all , and coercive, that is, as , and satisfies that for all , such that for all and .
In fact, since as uniformly for all , there exists a sequence of positive integers with for all positive integers such that for all and all . Let and define for , where .
By the definition of we have for all . By and , there exists such that It follows that where . In fact, when , for some , one has, by (96) and (98), for all . When , we have, by (98) and (99), for all .
It is obvious that is continuous and coercive. Moreover, one has for all . In fact, for every , there exists such that which implies that for all by (98) and the fact that for all integers .
Now we only need to prove the subadditivity of . Let and . Then we have Hence we obtain, by (98), which shows that is subadditive.
For , we assume that , with and ; letting , by (28), (92), (94), and (95), one gets where , . Since , is a finite-dimensional subspace, and is coercive, from the proof of (75), one can get That is, is coercive on . Since , so is coercive on , and and are bounded below, so it is obvious that for all .
Considering (91), (109), and (111), we can choose large enough such that for all one gets (or for the case ) and property (52) holds; then for and one gets ; that is, the property (51) holds.

5. Proof of the Geometry in Propositions 9 and 10

We finally give the proof of Propositions 9 and 10, which is nothing but a resume of the lemmata above, verifying that all the constants can be chosen sequentially without contradictions.

Proof of Proposition 9. Under hypotheses and , if we fix a value , then we obtain the constant from Lemma 11 and with this we get from Lemma 12. If we consider also one of the two sets of hypotheses and , then we proceed as follows. First of all, we determine (once forever) the constant from Lemma 11; with this we obtain from Lemma 14 the values and . Then, for any (now fixed) , we obtain from Lemma 11 the value . Finally, we can get from Lemma 12 the corresponding value of .

Proof of Proposition 10. Under hypotheses and , if we fix a value , then we obtain the constant from Lemma 11 and with this we get from Lemma 12. If we consider also one of the two sets of hypotheses and , then we proceed as follows. First of all, we determine (once forever) the constant from Lemma 12; with this we obtain from Lemma 15 the values and . Since we have , we can get from Lemma 11 the constant and with this obtain from Lemma 12.
Finally, for any (now fixed) , we obtain from Lemma 11 the constant and with this we get from Lemma 12 the corresponding value of .

Conflict of Interests

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

Acknowledgment

This work was supported by the Science and Technology Foundation of Guizhou Province (no. LKB[2012]19; no. [2013]2141).