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Abstract and Applied Analysis

Volume 2012 (2012), Article ID 760854, 44 pages

http://dx.doi.org/10.1155/2012/760854

## Dirichlet and Neumann Problems Related to Nonlinear Elliptic Systems: Solvability, Multiple Solutions, Solutions with Positive Components

Department of Mathematics and Applications, “R. Caccioppoli.,” University of Naples “Federico II”, Via Claudio 21, 80125 Naples, Italy

Received 1 February 2012; Accepted 2 April 2012

Academic Editor: D. O'Regan

Copyright © 2012 Luisa Toscano and Speranza Toscano. 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 solvability of Dirichlet and Neumann problems for different classes of nonlinear elliptic systems depending on parameters and with nonmonotone operators, using existence theorems related to a general system of variational equations in a reflexive Banach space. We also point out some regularity properties and the sign of the found solutions components. We often prove the existence of at least two different solutions with positive components.

#### 1. Introduction

In this paper, we present some significant applications of the results got in [1] to Dirichlet problems (Section 2) of the type: and to Neumann problems (Section 3) of the type: where are real parameters, is a bounded connected open set of with regular boundary , and is the outward orthogonal unitary vector to .

The study deals with the solvability of the problems, the existence of multiple solutions with all the components not identically equal to zero and, in the homogeneous case, the existence of solutions with positive components, bounded and locally Hölderian with their first derivatives. It is suitable to recall the problem studied in [1] with some notations and hypotheses.

Let real reflexive Banach spaces . Let be the product space . Let be the norm on , the norm on (dual space of ), and () the duality between (dual space of ) and (resp. and ). Let us denote by “” Fréchet differential operator and by “” Fréchet differential operator with respect to . Let and be real functionals defined in and real functionals defined in satisfying the conditions:() is lower weakly semicontinuous in and , are weakly continuous in and,and and and;( is weakly continuous in and and .

Letwith and; let us consider the following problem.

*Problem (). *Find such that
Obviously Problem (P) means to find the critical points of the Euler functional:
where .

Let us set About Problem (P), using Lagrange multipliers and the “fibering method,” different existence theorems have been proved in [1]. They base on one of the following hypotheses: (; (; ( is not empty and bounded in W.

*Remark 1.1. * In this paper, we use some existence theorems ([1], Theorems 2.1, 2.2, 3.1, and 3.2), in which as , in relation to a set , we suppose for each with , there exist and the real functions such that and , and as , for all , for all , for some .

The condition () assures that for the solutions of Problem (P), found with the method used in the recalled theorems, we have if .

Before showing Dirichlet problems (including the problem studied in [2] by Drábek and Pohozaev when and ) we give Propositions 2.2–2.6 which show some cases in which hypotheses ()−() hold. These propositions are based on the comparison between the parameters with suitable eigenvalues connected to -Laplacian. About Neumann problems (including the one studied in [3] by Pohozaev and Véron when ) the same question is solved by Propositions 3.1–3.5 in which the parameters and have compared with zero. Finally, the results in Appendix are very useful: Propositions A.1 and A.2 in order to get condition (), Propositions A.3 and A.4 to get qualitative properties of the solutions and the positive sign of the components of the found solutions.

#### 2. Dirichlet Problems

Let be an open, bounded, connected and set with . Let the Lebesgue measure on otherwise.

Let us assume Moreover we consider the functionals (as in ()) such that Let us use the notation instead of .

As let , respectively, the first eigenvalue and the first eigenfunction of the problem: Let us remember that [4] with in ;; is simple, that is, each eigenfunction of (2.3) related to is of the type with ; is isolate, that is, there exists such that is the only eigenvalue of (2.3) belonging to .

*Remark 2.1. *About the results related to problem (2.3), it is sufficient to suppose and as . This holds also for the results of this section if we limit to consider only the parameters nonnegative.

Let us start by presenting some sufficient conditions such that , and hold.

Using the variational characterization of it is easy to verify the following proposition.

Proposition 2.2. *If , then () holds. Consequently, () holds when .*

When for some , it is possible to fulfil () with an additional condition on . Let . For any let and let us suppose There exists .

Proposition 2.3. *Let () holds with . Let . If we fix the parameters set with , then there exists such that () also holds for any *

*Proof. * Arguing by contradiction, for any there exist and such that
Set , we have
moreover, since , there exists such that (within a subsequence)
Taking into account that is weakly continuous in , from (2.6) as we get
Since
from (2.9), we deduce that
Let us add that for some , since if we have the contradiction . Then , and consequently from (). This last inequality contradicts (2.8).

In the same way the following propositions can be proved.

Proposition 2.4. *Let () holds with . Let . Then, there exists such that (i _{14}) also holds for any .*

Let us pass to () and suppose() there exist and such thatand for any .

Proposition 2.5. * If (i _{22}) holds with, then
*

*Moreover, if we fix the parameters set with , then there exists such that*

*Proof. * Let us prove (2.12). Let with if , then . Let , we have
Let us prove (2.13). Arguing by contradiction, for any there exist with and , where if , such that
Relation (2.15) implies that there exists strictly increasing such that
Let , we have
Then, as we get
From (2.18), we get that . Then since () inequality holds, which contradicts (2.19).

Proposition 2.6. *If () holds with , then
**The proof as in Proposition 2.5.*

*Remark 2.7. *The applications we now show, except the first one, deal with systems with equations. We consider the functionals with , and we suppose .

*Application 2.8. * Let . Let us consider the problem
where
Evidently
Let us advance the conditions:
Let us note that (Propositions 2.2, 2.4, and 2.6)

Proposition 2.9 (see [1], Theorems 2.1, 2.2, 4.1, and 4.2; Remarks 2.1, 2.3, 4.1, and 4.4; Proposition A.3; [5, 6]). *Under assumptions (2.22) we have:*(i)*When (2.24) holds, with [resp. (2.24) and (2.25) hold, with ] problem (2.21) has at least two weak solutions and), and it results in ;*(ii)*When (2.25) holds, with problem (2.21) has at least two weak solutions ), and it results in .**Consequently, when (2.24) and (2.25) hold, with problem (2.21) has at least four different weak solutions.*

*Remark 2.10. *Our results include the ones of Drábek and Pohozaev [2] when .

*Application 2.11. *Let us consider the system:
where
System (2.27) is included among Problem (P) with:
Let us advance the conditions (compatible):
there exist and a constant such that and
Then (Propositions 2.2, 2.3, and 2.5)
and set

Taking into account that and , from ([1], Theorem 2.1, Remark 2.1, and Theorem 4.1) we get the following proposition.

Proposition 2.12. *Under assumptions (2.28) we have:*(i)*When (2.31) holds, ((2.30) and (2.31) hold resp.), choosing as in (2.32) (resp. (2.33)) system (2.27) has at least two weak solutions and with as ; *(ii)*When (2.30) holds, choosing as in (2.34) system (2.27) has at least two weak solutions and .**Consequently, when (2.30) and (2.31) hold, with and system (2.27) has at least four different weak solutions.*

The following proposition is obvious.

Proposition 2.13. *The following relations hold:
*

Proposition 2.14. *If , then as :
*

*Proof. * It is easy to prove that
where . Then (Proposition A.3) and consequently [5] .

Let us note that is a weak supersolution to the equation:
Then, since (2.35), it must be [6] .

Let us continue the analysis of system (2.27) under the condition: then Hence (Proposition 2.5) if and :

Proposition 2.15. *Under assumptions (2.28) and (2.39), choosing as in (2.41) system (2.27) has at least two weak solutions and with as .*

*Proof. *Thanks to ([1], Theorem 4.1), there exists such that
where .

Reasoning by contradiction, let, for example, . Since and from (2.39) , setting we have
then . This implies that ([1], see the proof of Theorem 4.1) is a weak solution of system (2.27). Then from which too as .

Condition (2.39) holds in particular when

Proposition 2.16. *Replacing in Proposition 2.15 (2.39) with (2.44), it is right to say that and as . Consequently, if *

*Proof. *Set , as in Proposition 2.15 is a weak solution to system (2.27).

Let us add that since (2.44), there exists (Proposition 2.6) such that
Then the existence of is assured also choosing as in (2.46), and the conclusions of Proposition 2.16 hold.

*Application 2.17. * Let us set
where
Let us consider the system:
We advance the conditions
Therefore,
Then ([1], Theorems 2.1 and 4.1, and Remarks 2.1 and 4.1).

Proposition 2.18. * Under assumption (2.48), we have:*(i)*When (2.50) holds, ((2.50) and (2.51) hold resp.), if system (2.49) has at least two weak solutions and with as ;*(ii)*When (2.51) holdssystem (2.49) has at least two weak solutions and with as .**Consequently, when (2.50) and (2.51) hold, with system (2.49) has at least four different weak solutions.*

In order to establish some properties of and it is useful to recall that ([1], Theorems 2.1 and 4.1)

Proposition 2.19. *When , we have
**
besides
*

*Proof. *The relation comes from Proposition A.3. Then [5] .

About (2.56), it is sufficiently (Remark 1.1) to prove that

Let with . Since
let (Proposition A.1) with such that
where is the characteristic function of . Set such that
with it results in

Proposition 2.20. *When , we have
*

*Proof. * We can get (2.62) from Proposition A.3 and [5].

About (2.63), it is sufficiently [6] to prove that as . Reasoning by contradiction, let, for example, . We note that
Let us suppose and set . Then
Set such that and taking into account (2.54), we get the contradiction:

Proposition 2.21. *When , we allow that as :
*

*Proof. *The assumption implies that
Let, for example, and . Set and as, with , we have
If , set as , with, it results in
This method let us to find .

Then, if for some , with as in (2.68) we have from (2.53) (resp. (2.54)) . Consequently ([1], see the proof of Theorem 2.1 (resp. Theorem 4.1)) is a weak solution of system (2.49). Therefore [6] as .

*Application 2.22. * Let us assume , and as in Application 2.17,
where
Let us consider the system:
under almost one of the conditions:
By using some results ([1], Theorems 2.2 and 4.2, and Remarks 2.3 and 4.4), we can advance a proposition similar to Proposition 2.18 replacing in particular with and with.

Thanks to Proposition A.3 and a result of [5], for the solutions and to system (2.73), we have
We continue to analyze the properties of and . To this aim we recall that ([1], Theorems 2.2 and 4.2), set for each *,* we have:
Besides with , it results in

Proposition 2.23. *When as , then
*

*Proof. * It is sufficiently (Remark 1.1) to prove that
Let with . As in Proposition 2.19, it is possible to find such that with , it results in

Proposition 2.24. *When as , then
*

*Proof. * It is sufficiently [6] to prove that . Reasoning by contradiction, let, for example, and such that
Since
there exist an open ball of with centre included in and a unique functional belongs to such that
Then, the functional
belongs to *,* and we have
Then, for (2.78)
Now, let us remark that with , it results in
Then, since
we have. Consequently,
from which we get the contradiction:

Proposition 2.25. *When , we allow that
*

*Proof. *We reason as in Proposition 2.21, taking into account (2.77) and (2.78) ([1], see proofs of Theorems 2.2 and 4.2).

*Application 2.26. *Let for each :
where