International Journal of Mathematics and Mathematical SciencesVolume 2011 (2011), Article ID 419341, 11 pagesdoi:10.1155/2011/419341

Research Article

A Multiplicity Result for Quasilinear Problems with Nonlinear Boundary Conditions in Bounded Domains

S. Khademloo

Department of Mathematics, Faculty of Basic Sciences, Babol Noshirvani University of Technology, 47148-71167 Babol, Iran

Received 28 September 2011; Accepted 2 November 2011

Academic Editor: Enrico Obrecht

Copyright © 2011 S. Khademloo. 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.

Abstract.

We study the following quasilinear problem with nonlinear boundary condition , in and on , where is a connected bounded domain with smooth boundary , the outward unit normal to which is denoted by . is the -Laplcian operator defined by , the functions and are sign changing continuous functions in , , where if and otherwise. The properties of the first eigenvalue and the associated eigenvector of the related eigenvalue problem have been studied in (Khademloo, In press). In this paper, it is shown that if , the original problem admits at least one positive solution, while if , for a positive constant , it admits at least two distinct positive solutions. Our approach is variational in character and our results extend those of Afrouzi and Khademloo (2007) in two aspects: the main part of our differential equation is the -Laplacian, and the boundary condition in this paper also is nonlinear.

1. Introduction and Results

In this paper, we consider the problem where is a connected bounded domain with a smooth boundary , the outward unit normal to which is denoted by . is the -Laplcianian operator defined by . The functions and are assumed to be sign changing in . Here, we say a function changes sign if the measure of the sets and are both positive. is a real parameter and exponent is assumed to satisfy the condition , where if and otherwise.

A host of literature exists for this type of problem when . For the works concerning with problems similar to (1.1) in the case , we refer to [1–3] and references therein.

The growing attention in the study of the -Laplace operator is motivated by the fact that it arises in various applications, for example, non-Newtonian fluids, reaction diffusion problems, flow through porus media, glacial sliding, theory of superconductors, biology, and so forth (see [4, 5] and the references therein).

In this paper, we obtain new existence results by using a variational method based on the properties of eigencurves, that is, properties of the map , where denotes the principal eigenvalue of the problem Similar to [6] our method works provided that the eigenvalue problem has principal eigenvalues and it can be shown that this occurs on an interval where . Thus, we are able to obtain existence results for problem (1.2) even in the case of nonlinear Neumann boundary conditions where is small and negative. Our method depends on using eigencurves to produce an equivalent norm on ; such an equivalent norm is also introduced in [2]. The results that we obtain in this paper are generalization of the previous results obtained by Pohozaev and Veron [7].

It can be shown that has the variational characterization: from whence it follows that(i) is a principal eigenvalue of (1.2) if and only if ,(ii) is an increasing function,(iii) is a concave function with a unique maximum such that as [6].

If , then , and so has exactly one negative zero and one positive zero . Thus, and are principal eigenvalues for (1.2).

If , then . If , then is decreasing, and if , then is increasing. Assume now that changes sign in : if , there exists a unique such that and for . If , then and for . If , then there exists a unique such that and for .

Suppose now that and that is small and negative. Then, since is increasing, it follows that there still exist principal eigenvalues of (2.5), but now both of them are positive.

It can be shown that there exists such that the above is true for all , but for , for all so that principal eigenvalues no longer exist.

Similar considerations show that when , there exists such that there are principal eigenvalues for but when , there are no principal eigenvalues for (see [6]).

It is easy to see that if and exist, for all .

Thus, we assume that the following conditions hold:, (), , , ,

where is the positive principal eigenfunction corresponding to .

With these constructions we have the following.

Proposition 1.1. Assume , then for every , defines a norm in which is equivalent to the usual norm of , that is,

Proof. See [2].

Now we can state our main results.

Theorem 1.2. Assume , , , and . Then, for every , problem (1.2) admits at least one positive solution .

Theorem 1.3. Assume , , , , and . Then, problem (1.2) has at least one positive solution for .

Theorem 1.4. Assume , , , , and . Then, there exists such that problem (1.2) admits at least two distinct positive weak solutions in , whenever .

When and , we have the following.

Corollary 1.5. Assume . Then, the problem has a positive solution.

Throughout this paper, c denotes a positive constant. We will use fibrering method in a similar way to those in [8]. A brief description of the method and the proof of Theorem 1.2 are presented in Section 2. We then study the cases and in Sections 3 and 4.

2. The Case When

In this section, motivated by Pohozaev [8], we will introduce the fibrering map as our framework for the study of problem (1.2).

For a (weak) solution of problem (1.2), we mean a function such that for every , there holds

Now let us define the variational functional corresponding to problem (1.2). We set as

It is easy to see that , and for , there holds

Since , we know that critical points of are weak solutions of the problem (1.2). Thus, to prove our main theorems, it suffices to show that admits critical point. We will do this in this paper. Our main tool is the theory of the fibrering maps. First, we will introduce this map for .

Let . We refer to such maps as fibrering maps. It is clear that if is a local minimizer of , then has a local minimum at .

Lemma 2.1. Let and . Then, if and only if .

Proof. The result is an immediate consequence of the fact that

Thus critical points of correspond to stationary points of the maps that can be given by where Hence, Thus, if and have the same sign, has exactly one turning point at provided that , and if and have opposite signs, has no turning points. Now substituting (2.8) into (2.5), we get

Lemma 2.2. Suppose that is a critical point of , where and have the same sign. Then, .

Proof. Let , , then for all . This completes the proof.

The following lemma shows that the critical points of the functional can be found by using the conditional variational problem associated with .

Lemma 2.3. Suppose that is a well-defined functional on and is a minimizer of on for some . Then,.

Proof. If is a local minimizer of on , then is a solution of the optimization problem: where . Hence, by the theory of the Lagrange multipliers, there exists such that . Thus, Note that the functional is 0-homogeneous and the Gateaux derivative of at the point , , in direction , is zero, that is, . It then follows that due to . Hence, the proof is complete.

The following scheme for the investigation of the solvability of (1.2) is based on previous lemmas. First, we will prove the existence of nonzero critical points of under the constraint given by suitable functional . This will be an actual critical point of and it will generate critical point of the Euler functional which will coincide with the weak solution of problem (1.2).

Proof of Theorem 1.2. Suppose that , , , satisfy. It follows from variational characterization of that for all and . Moreover, we have Hence, for using Lemma 2.3, it is sufficient to consider the case , for example . In this case, we have From the necessary condition for the existence of , we have . It follows that we must consider a critical point with .The functional is nonnegative and so bounded below, hence we can look for positive local minimizer for on .Let us consider variational problem: Note that for , this set is not empty, and from Lemma 2.3, the solution of this problem is a minimizer of on .Suppose is the maximizing sequence of this problem. It follows from the equivalent property in Proposition 1.1, is bounded and so we may assume that in . Since may be compactly embedded in , , and , we have in , and . Hence, Moreover, we have Here, the weak lower semicontinuity of the equivalent norm was used.Assume that . By using the map: we have and and so for some , that is, . So we derive which is a contradiction. Hence, and so is a maximizer. As Lemma 2.2, the result would follow by considering . Then, is a weak solution of (1.2) and in . Now, following the bootstrap argument (used, e.g., in [9]), we prove . Then, we can apply the Harnack inequality due to Trudinger [5] in order to get in (cf. [9]).

3. The Case When

Proof of Theorem 1.3. If , It is easy to see that the set is unbounded and we are forced to require an additional assumption . In this case, again we are looking for a maximizing of the problem: Suppose is a maximizing sequence of this problem. First, we investigate the case when is unbounded. Then, we may assume without loss of generality that . So, we obtain where . Thus, we have Since , we may assume that in . Again, since may be compactly embedded in , and , we have So, And, therefore, . Returning to , we have also Due to simplicity of , there exists such that . Therefore, we have that implies , which contradicts . Therefore, is bounded and so in and in , and . Hence, , and so .Moreover, we obtain . Here, the variational characteristic of and the weak lower semicontinuity of the norm were used. Let us now consider the case . Again, we obtain for some , and so which contradicts .Now we prove that . Suppose otherwise, then by using in (2.20), we obtain some such that . By direct calculation, we get a contradiction like (4.1). This yields that is a nonnegative solution of problem (1.2). Using the same ideas as the proof of Theorem 1.2, we have proved Theorem 1.3.

4. The Case When

As it is proved in [10], we will show that for but close enough to , we have two distinct positive solutions for problem (1.2). The existence of one of them is obtained by using the following lemma.

Lemma 4.1. Under the assumption of Theorem 1.4, there exists a maximizer of the problem: whenever , for some . Moreover, is a positive weak solution of problem (1.2).

Proof. First note that using in (2.20), it is easy to see that is a maximizer of problem (4.2) if and only if is a maximizer of the problem: Now suppose that the result is false. Then, there exists a sequence such that and problem (4.1) has no solution for . For simplicity, we use . Let be a maximizing sequence of this problem, that is, We prove that if be bounded or unbounded, we arrive at a contradiction and so the lemma is proved.The first case: is bounded. Thus, in for some and in , , and . Hence, and the weak lower semicontinuity of the norm gives . Therefore, is a solution of problem (4.5), which is a contradiction.The case is unbounded. Then we may assume that . Let . Then, in for some and in , and . This implies that Therefore, . Furthermore, and so . Thus, we arrive at which implies It is a direct consequent of the comactly embedding of in , and . Now, we pass to the limit for . Then, and since is a bounded sequence, we may assume that in for some and in , , and .It follows from (4.10) that and from the variational characteristic of and (4.6) that which contradicts .Hence, for some , problem (4.5) has at least one nonnegative solution for any .

In order to find the second positive solution, we consider the minimizing problem:

Note that this set is empty for (because of the variational characterization of ). Hence, this second problem does not have a solution for .

Lemma 4.2. Under the assumption of Theorem 1.4, there exists such that for , problem (1.1) has a nonnegative solution satisfying .

Proof. First note that using the auxiliary function and the assumption , it is easy to see that for some . Thus, the set is nonempty. Also, for this , we have for . Again, we assume that the result is not true. Then, there exists such that for , problem (1.1) has no solution.Let be a minimizing sequence of this problem, that is,Assume that is bounded. Using similar argument as in the proof of Lemma 4.1, we find a solution of problem (1.1) which is a contradiction. Let us assume is bounded. Again, we consider is bounded, and so in for some . Thus,.Letting , we arrive at . It follows easily that , and so . As might be expected, we arrive at a contradiction with assumption , by using similar steps as in the proof of Theorem 1.3. This completes the proof.

Proof of Theorem 1.4. Let , and and , where is defined by (2.16).It easy to see that and , and so . Thus, and are two distinct non negative weak solutions to (1.2). Other properties of and follow in the same way as in Section 2.

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