Abstract

We study the existence of positive solutions for the homogeneous Dirichlet boundary value problem of -Laplacian systems with a singular weight which may not be in .

1. Introduction

In this paper, we study nonlinear differential systems of the form where with an odd increasing homeomorphism, with , on any subinterval in , and with ; here we denote , , and the Hadamard product of and in . Thus problem can be rewritten as

We first give assumptions on and .(A)There exist an increasing homeomorphism of onto and a function of into such that (H) is locally integrable satisfying  for .

For convenience, we introduce a new class of weight functions. For a bijection , define as a subset of given by By the notation, condition means .

The case of -Laplace operator, namely, , , , satisfies condition () with . We give one more example of and satisfying conditions () and ().

Example 1. Define as an odd function with Then is obviously an increasing homeomorphism. Define functions and given as Then and is an increasing homeomorphism. This implies that satisfies condition (). Moreover, for , we can easily calculate to see .

We note that given in the example above is not integrable near a boundary ; that is, , and, in this paper, we focus on studying generalized Laplacian systems of condition with singular weights which may not be in . We now give assumptions on .(F) is continuous, .

Problems of -Laplacian or more generalized ones like problem appear in various applications which describe reaction-diffusion systems, nonlinear elasticity, glaciology, population biology, combustion theory, and non-Newtonian fluids (see [1–4]). Recently there is a vast literature related to existence, multiplicity, or nonexistence of positive solutions of problem for either -Laplacian or more generalized Laplacian problems (see [5–11] and the references therein). Specially, for generalized Laplacian problems, one may refer to works of Agarwal et al. (see [12–14]). Let us denote where for all and .

Among the variety of works mentioned above, we are interested in the following result.

Res A. Problem has at least one positive solution if either , or , .

Wang [10] proved Res A when each is continuous and satisfies that there exist two increasing homeomorphisms and of onto such that Do Ó et al. [7] also proved Res A when and each .

The aim of this paper is to prove Res A when satisfies condition () and each . More precisely, we state our main theorem as follows.

Theorem 2. Assume (), (), and () hold. Then problem satisfies Res A.

Extension of results in [10] or [7] to Theorem 2 is not obvious mainly due to the singularity of in comparison with Wang and lack of homogeneity of the general operator in comparison with Do et al.

For proofs, we introduce a newly developed solution operator for motivated by Sim and Lee [15]. And then we make use of the fixed point theorem of a cone for the existence of positive solutions.

This paper is organized as follows. In Section 2, we introduce a solution operator for problem and prove the compactness of the operator. In Section 3, we prove our main theorem.

2. A Solution Operator

Let us consider a simple scalar problem of the form where satisfies () and with .

Since may not be in as we see the example in the introduction section, in this case, the solution of + may not be in . So by a solution to this problem, we understand a function with absolutely continuous which satisfies .

We first give some remarks for calculations later on.

Remark 3. From condition (), we get

Remark 4. Let . Then for any fixed , we know . Applying and in Remark 3, we get This implies .

Remark 5. If , then, for any fixed ,

We need a lemma which guarantees concavity of solutions. The proof is similar to Lemma 2.3 in Wang [10].

Lemma 6. Let satisfy on . Then is concave on and , where is the supremum norm of .

Let be a solution of + .

Then integrating both sides of on the interval for and for , respectively, we find that + is equivalent to where . We show that . Indeed, by Lemma 6, solution has a unique maximal point. That is, there exists a unique such that . Since , we see from (13) that Since is an odd homeomorphism, , and by Remark 5, we get Similar argument shows that . Now we integrate both sides of (13) on the interval for and on the interval for , respectively. Then we get

Let us check . For , define Then the function is well-defined. If has a unique zero, then . For this, we give the following lemma. The proof generally follows the lines of proof of Lemma 2.2 in Sim and Lee [15].

Lemma 7. For given , the function defined in (17) has a unique zero in .

Consequently, if satisfies () and , then the solution of + can be represented by where uniquely satisfies

On the other hand, it is not hard to see that a function defined in (18) satisfies , and is absolutely continuous on and is in turn a solution of + .

Now we come back to our main problem We finally introduce the corresponding solution operator for and prove compactness of the operator. For this purpose, we need a preliminary lemma.

Lemma 8. If , then, for given , .

Proof. Let and be given. Then applying Remark 3 with , and using the fact , we get Similarly, we can prove

This lemma should be more natural if it is valid under assumption . Even though it is true for the case , the -Laplace operator, it seems not easy to prove in general mainly caused by lack of homogeneity of .

To set up the solution operator for , let us define as the Banach space with norm and define a cone by taking Let and , ; then and by Lemma 8, . Let us apply the solution representation for + replacing with ; then we get where is a unique zero of

Now for , let us define Then by Lemma 6, and we see that is a positive solution of if and only if on .

We finally prove the solution operator is completely continuous. For this, we need a couple of lemmas about the properties of . Since and are fixed, we regard as a function of . The proofs of the following two lemmas are mainly induced by the monotonicity of and similar to proofs of Lemmas 3.1 and 3.2 in Sim and Lee [15].

Lemma 9. sends bounded sets in into bounded sets in for .

Lemma 10. is continuous for .

Lemma 11. is completely continuous.

Proof. Continuity of can be done by using the Lebesgue Dominated Convergence Theorem with aid of the continuity of . Let be a bounded subset of . Then it is enough to prove is uniformly bounded and equicontinuous. We first prove that is uniformly bounded. Indeed, take , , and denote simply . We compute the bound on the interval ; the bound on the interval can be obtained by the similar way. Consider

Case1 ()

Case2 (). Let ; then implies that is continuous on , for and . Thus we may choose satisfying If , then On the other hand, if , then Applying Remark 3 with and , we get By the fact , all bounds above are finite and independent on and . Thus is uniformly bounded.