Abstract

Questions on the existence of positive solutions for the following class of elliptic problems are studied: , in , , on , where is a bounded smooth domain, and are given functions.

1. Introduction

In this paper, we are concerned with the elliptic problemwhere is a bounded smooth domain, and are given functions, is the -Laplacian: and is the usual normin the Sobolev space .

Such a problem, which will be named -Kirchhoff problem, is a generalization of the classical stationary Kirchhoff equationwhere is the usual norm in .

As it is well known, problem (1.4) is the stationary counterpart of the hyperbolic Kirchhoff equationthat appeared at the first time in the work of Kirchhoff [1], in 1883. The equation in (1.5) is called Kirchhoff equation and it extends the classical D'Alembert wave equation, by considering the effects of the changes in the length of the strings during the vibrations.

The interest of the mathematicians on the so-called nonlocal problems like (1.1) (nonlocal because of the presence of the term , which implies that equations in (1.1) and (1.4) are no longer pointwise equalities) has increased because they represent a variety of relevant physical and engineering situations and requires a nontrivial apparatus to solve them.

Particularly, problem (1.1) presents some combinations that, at least to our knowledge, seem to be new. Indeed, in problem (1.1) appears the nonlocal term motivated, among other things, by the above physical situations. Furthermore, we have the presence of the -Laplacian operator that appears in several areas of the science such as astronomy, glaciology, climatology, nonnewtonian fluids, petroleum extraction. Problems that involve these two terms, and , present several difficulties such as uniqueness, regularity, degeneracy, as we will see throughout this paper.

Beside these considerations, we also consider a case with the presence of a singular term which poses an additional difficulty in our study. Singular elliptic problems arise in chemical heterogeneous catalysts, nonnewtonian fluids, nonlinear heat conduction, among other phenomena.

In case , problem (1.1) has been studied by several authors. See [27], and the references therein. Particularly, this work was motivated by [24, 6].

We will establish existence results for problem (1.1) by considering several classes of functions and .

An outline of this work is as follows:

In Section 2, we recall some properties of the -Laplacian. In Section 3, we study the case in which depends only on . This is the -Linear case. In Section 4, we attack problem (1.1) when is sublinear, that is, , for some .

In both Sections 3 and 4, we suitably adapt ideas developed in [2, 3, 6].

In Section 5, we analyze the case in which possesses a singular term. More precisely, is of the following form:for and , with . In this section, we have to use some arguments different from those in [4].

To end this introduction we recall that is a weak solution of problem (1.1) if

2. Preliminaries on The -Laplacian

We will briefly expose some properties of the -Laplacian operator defined by

First, we consider the problemwhere , and the boundary condition will be understood as .

The following results holds as a simple consequence of a minimization of a suitable functional.

Theorem 2.1. If , then problem (2.2) has only a solution in the weak sense, namely,

So, we have defined an operator , the inverse of , which satisfies the following.

(a) is uniformly continuous on bounded sets, where such operator is defined aswhereHere, we denote by the duality pairing between and .

(b) If , then the weak solution of (2.2) belongs to , for some , and the mapping is compact.

Theorem 2.2 (weak comparison principle). Let satisfyThen, a.e. in .

Theorem 2.3 (a Hopf-type maximum principle). If and verifiesthen on , where is the outward normal to .

Theorem 2.4 (a strong maximum principle). Assume that is a nonnegative number and is a bounded domain . Suppose that satisfies and in . Then, in . The conclusion is still true for all when .

We now consider the eigenvalue problemwhere is a parameter. We say that is an eigenvalue of (2.9) if there exists a function , satisfying (2.9) in the weak sense. Such a function is called an eigenfunction of (2.9) associated to the eigenvalue .

There exists the first positive eigenvalue of problem (2.9) which is characterized as the minimum of the Rayleigh quotientMoreover, is simple (i.e., all the associated first eigenfunctions are merely constant multiples of each other) and isolated (i.e., there are no eigenvalues less than and no eigenvalues in some right reduced neighborhood of ). There is a positive (in ) eigenfunction corresponding to .

For more informations on the -Laplacian the reader may consult [811].

3. The -Linear Case

This section is devoted to the study of the so-called -Linear problemwhere is the conjugate exponent of , that is, and is the topological dual of .

The next result is an adaptation of some ideas contained in [12, 13] (which were used for another nonlocal problem) for problem (3.1).

Theorem 3.1. For each , problem (3.1) possesses as many solutions as the folloing equation: where is the only solution of problem (2.2).

Proof. First, let us suppose that is a solution of (3.1), where is fixed. Henceand so is the solution ofand observe that from which we conclude that is a solution of (3.2).
Conversely, let be the solution of (3.4) and suppose that is a solution of (3.2).
Defineand so . A straightforward calculation shows thatand, because is a solution of , one has . Hence,which concludes the proof of the theorem.

Remark 3.2. Let us suppose that is a continuous function. So,

Assume, in addition, that there are such that for . In this case,Because of the intermediate value theorem, for , there is such thatIn particular, if is an increasing function for , problem (3.1) possesses only a solution for each . As an example, we take for . Thus, problem (3.1) possesses as many solutions as the following equation:

A simple exercise shows that the function attains its maximum in and . Consequently, if , problem (3.1) does not possess any solution; if , problem (3.1) possesses exactly one solution; and if , problem (3.1) possesses exactly two solutions.

As we see, the presence of the term produces great difference between problems (2.2) and (3.1).

Remark 3.3. Let us consider the case , that is,

If for all , problem (3.12) possesses only the null solution. If , for some , problem (3.12) possesses infinitely many solutions. Indeed, if , we have thatis a solution of (3.12).

4. on A Sublinear Problem

In this section, we are going to study the sublinear problemwhere satisfies with .

Such a kind of problem belongs to a class of problems known as sublinear whose prototype iswith (note that, in this case, ), which has been vastly studied. See [1417]. For the nonlocal problem, with , we cite [2, 3, 6, 7], and the references therein.

In particular, Díaz and Saá [16] study the problem, and show that it possesses only a positive solution . Indeed, such a solution also belongs to . Our next result describes what happens with the nonlocal sublinear problem (4.1).

Theorem 4.1. Suppose that is a continuous function satisfying for all . Then problem (4.1) has at least as many solutions as the equationwhere is the solution of (4.1).

Proof. We adapt for problem (4.1) the ideas developed in [3, 6]. Let us suppose that is a solution of the (4.4) and set which implies . So, a simple calculation shows that
Define and let us show that such a function is a solution of problem (4.1). Indeed, this follows from the calculation below:Hence, is a solution of problem (4.1).

Remark 4.2. Remark 3.2, mutatis mutandis, remains valid for problem (4.1).

5. A Singular Problem Via The Galerkin Method

In this section, we will study problemwhich is a singular perturbation of problem (4.1), where is a bounded smooth domain, and is a suitable function defined in .

We will attack problem (5.1) by using the Galerkin method which rests heavily on the following result, which is a variant of the well-known Brouwer fixed point theorem, whose proof may be found in Lions [18].

Proposition 5.1. Suppose is a continuous function such that with , for some , where denotes the usual norm in . Then, there is such that .

We will suppose that satisfies

there are and such that if ;

If for all , condition is vacuous.

Theorem 5.2. Let be a continuous function with on and is a continuous function satisfying . Then problem (5.1) possesses a positive solution.

We will split the proof of this theorem in several lemmas. First, for each fixed , we will consider the following problem:

In what follows, throughout Section 5, we are always supposing that and enjoy assumptions of Theorem 5.2

Lemma 5.3. For each fixed , problem (5.2) possesses a solution .

Proof. Let us consider the problemwhere is given by
We now consider a Schauder's basis for (we recall that a Schauder's basis for a Banach space is a sequence such that to each there exists a unique sequence of scalars for which the partial sums of converge to in the norm of ). For more informations on Schauder's basis see [19] and the references therein.
For each , let be the finite dimensional vector space spanned by the functions . So, each is written as . We will use on the norm
We note that and are isomorphic through the following map:with where .
Since, for each is a finite dimensional vector space the norms and , induced from on , are equivalent to each other and so, there exist positive constants and such that
We now consider, for each , the following applicationwhere
A simple calculation leads us toWe now note thatand because , we have , and soHence,and from , for , we have . Thus,
We now fix and for and noting that , we obtainSo, if is large enough, we havewhich implieswhere , because and are equivalent in .
From Proposition 5.1 there is , with , such that and note that for some is large enough.
Through the isometric identification of with , we find , that is, , such thatfor each , which gives usfor all . Since ,
Because we obtain
Let us show that the sequence is bounded. Indeed, suppose, on the contrary, that is not bounded. So, up to a subsequence, we may suppose that . From condition and inequality (5.22) we obtainwhich give usSince we arrive inwhich is impossible. Thus, is bounded. Consequently, up to a subsequence,And in view of continuity of
We now fix and so . For we havefor all .
We now remark thatBy the Lebesgue dominated convergence theoremFurthermore, because , we haveand because is a bounded domain and we have and soConsequently,and there is such thatThus,and using again the Lebesgue dominated convergence theorem
We will now consider the following claim, whose proof will be postponed to Section 6.
Claim 1. The following convergence holdsfor all .
In the inequality (5.28) we make and we use (5.27), (5.30), and (5.36) to obtainfor all .
Since is arbitrary, the above inequality holds true for all . Because of this and so , which impliesfor all .
Taking as a test function in the inequality (5.39), we get
We now take in (5.28) to obtain
Sinceand because , it follows that a.e. in and there is such that a.e. in for all . Consequently,
Reasoning as before, we obtain
Taking limits on both sides of the equality (5.41) and using , (5.27), (5.44), we obtain
Comparing this last equality with the one in (5.40) we obtain because . Then, from (5.39)for all .
This shows that is a weak solution of the auxiliary problem (5.2) which is positive by virtue of the maximum principle. This proves the lemma.

In what follows, for each , we set and where is the solution obtained in the last lemma.

Lemma 5.4. There exists such that , for all .

Proof. We will reason by contradiction. Suppose thatIf it is the case, the sequence is bounded because, on the contrary, we would have , perhaps for a subsequence, and soand this would implywhich is impossible in view of (5.47). Therefore, up to subsequences,Since is continuous, we have
We now note thatfor all and , because the function , attains a positive minimum.
SinceTaking in , as a test function in the last equality, we obtainand so . Taking limit on the above expression we getwhich is impossible. This completes the proof of the lemma.

Lemma 5.5. The sequence is bounded, where is as above.

Proof. First of all we note thatwhere and are positive constants independent of . Consequently,Since , we conclude that is bounded.

As a consequence of the preceding lemma we have

Lemma 5.6. The sequence , obtained in the last lemma, converges to a solution of problem (5.1).

Proof. As is a bounded sequence in we have, up to subsequence, thatFrom the continuity of ,
Let be an eigenfunction of associated to first eigenvalue and satisfyingwhere is the constant obtained in (5.54). Consequently,Sinceit follows thatand as on . By the comparison principle,and from (5.58)From which we conclude that for each .
Next, we will use the following.

Hardy-Sobolev Inequality. If and , then , where , andwhere is a constant and is an eigenfunction of associated to the first eigenvalue .

Whose proof may be found in [20].
Asand using (5.66) it followsFrom the Hardy-Sobolev inequalityFurthermore,and so, by the Lebesgue dominated convergence theorem
Since is bounded and , we have and soSince is bounded and , we have and soand there exists such thatHence,and using again the Lebesgue dominated convergence theorem, we obtainfor all .
As is solution of the auxiliary problemfor all , taking limits on both sides of the above expression, using the same reasoning as in the proof of the claim and the convergences in (5.60), (5.72), and (5.77) we obtainfor all . Using as a test function in the last expressionTaking in the expression (5.78), it follows thatNoticing that in , and we have and soand there exists such thatThus,By the Lebesgue dominated convergence theoremand because , it follows that We now take limits on both sides of (5.81), by using (5.60), (5.85), and (5.86), to obtainthat is,
Comparing (5.80), (5.88) and using , we get . Therefore, from (5.79)for all , which shows that is a weak solution of (5.1).

6. Proof of The Claim

We recall that we have to prove thatfor all .

First of all, we claim that is a bounded sequence. Indeed,and because , we have and . Applying the Hölder inequality, by using the exponents and , it follows thatwhich implieswhere is a constant, because the boundedness of was proved in Lemma 5.5.

Since is a separable Banach space and is a bounded sequence in then, up to a subsequence,that is,which is equivalent to

Sinceit follows from (6.7) thatfor all . Taking limits on both sides of (5.28) and using the convergences (5.30) and (5.36), we obtainfor all . So, from (6.9), (6.10) and the uniqueness of the limitfor all .

Using the monotonicity of the operator , that is,we haveTaking and in the last expression, we obtainand soIn (5.28) we take to obtainConsequently,using the definition of taking limits as and using (5.44) we haveWe note thatfor all , and soTherefore,that is,Settingwe obtaintaking limits on both sides of the above expression as and using the continuity of the operator So,and we obtainHence,which implies , because is arbitrary. This means that the above equality is valid for all . From (6.10) and (6.11) we conclude thatfor all . Hence,for all . This concludes the proof of the claim.

Acknowledgment

The author is partially supported by CNP-q-Brazil-301603/2007-3 and FADESP.