#### Abstract

A new result of solvability for a wide class of systems of variational equations depending on parameters and governed by nonmonotone operators is found in a Banach real and reflexive space with applications to Dirichlet and Neumann problems related to nonlinear elliptic systems.

#### 1. Introduction

Many researchers (e.g., [1–9]) have devoted (and are still devoting) their studies to the solvability and the investigation of multiple and positive solutions of nonlinear elliptic problems.

A class of general systems of variational equations has been studied in [10]. The Dirichlet and Neumann problems investigated in [11] belong to this class.

In this paper, we prove a further existence theorem related to the problem of [10] in the homogeneous case, by using the Lagrange multipliers and the “algebraic” approach which is based on the fibering method [12]. This theorem and the ones of [10] include the results of [8] and some results of [13–15]. Now, let us recall the problem studied in [10].

Let be Banach reflexive and real spaces. Let with . Let [] be the duality between dual space of [] and []. Let us denote by “” Fréchet differential operator and by “” Fréchet differential operator with respect to . Let and () be real functionals defined in and let and () be real functionals defined in satisfying the following conditions: is lower weakly semicontinuous in and , and are weakly continuous in and , and , and and . is weakly continuous in and , , and , if .Let us set the following:Let us consider the following problem.

*Problem (P)*. Find such that

In Section 2, we present new cases in which Problem (P) is solvable. In these cases, we introduce one of the following hypotheses: . ().Theorem 1 assures the existence of at least one solution. It is possible to get the existence of a second solution from a result of [10, Theorem ]. This result is based in particular on the following assumption: is not empty and bounded in .

The applications to Dirichlet problems in Section 3 [resp., Neumann problems in Section 4] (whose variational form is included in Problem (P)), for the sake of brevity, deal with the first case of solvability, since in this case we can use at the same time Theorems 1 and 4. When , thanks to Propositions 2 and 5, we have got sufficient conditions so that the components of the found solutions are not identically equal to zero.

#### 2. Solvability of Problem (P)

Let us consider the following cases:, , , , and holds., , , if , and holds., , , , changes sign, and holds., , changes sign, , , and holds.Let us introduce the open set of the space :and let us setIt is easy to control that, for each , the equationhas only one positive root and it results inThe implicit function theory assures the functional belongs to . Then, the functional belongs to and it results inIt is important to remark that (9) can also be written as follows:

Theorem 1. *Under assumptions , , in cases , one has*

*Proof. *Let us setwhere is the positive root of equation . Pointing out that andlet us prove thatLet such that ; that is,We note thatmoreover,since holds in and holds in .

Inequality (18) implies that exists such that (within a subsequence) weakly in . Then,Let us verify thatIn , let such that and . We set , , in , and in , in .

Let us denote by the positive root of equation . Since and , we have .

In , we note that from (17) ; then,from which (21) follows taking into account (16).

Relation (21) assures that exists such that (within a subsequence) .

Consequently, from (16), (17), and (20), we obtainLet us add thatIn , since from (7) , (23) implies that and ; then, in , while (24) ⇒ in and . In case (where from (8) ), since from (17) , we have from which we obtain (25).

Evidently, ((24), (25)) ⇒ ; then, (23) can be written in the form ; that is, . Then, (15) holds.

Let us prove thatIn fact, taking into account that holds in and holds in , we obtainOn the other hand, since , we haveThen, if , we get the contradictionRelations (14), (15), and (26) allow (11).

From (11), a Lagrange multiplier exists such thatSetting in (30) , we get , since , and from (9) − − . Then, (30), taking into account (10), implies ; then, (12) holds.

Proposition 2. *Let . Let as in Theorem 1 and . Let us suppose the following:** and a function belonging to () [resp., ()] exist such that , [resp., ] and for some [] [resp., ] [resp., ].**Then, .*

*Proof. *In fact, with and as in [] exists such that and [resp., ] [resp., ]; then, [resp., ].

*Remark 3. *Condition with includes condition introduced in [10] and it implies the conclusion of Theorem of [10].

In the case, holds, , and ,for each , (6) has only one positive root and we have .

Set , in [10, Theorem ], the following result has been proved.

Theorem 4. *Under conditions , , in case , one has the following:*

We add the following proposition.

Proposition 5. *Let . Let as in Theorem 4 and . Let one suppose the following:** and a function belonging to () [resp., ()] exist such that , [resp., ] and for some [resp., ] [resp., ] [resp., ].**Then, .*

*Proof. *Let and be as in . Since , and ; then, an open ball exists with center included in and a unique functional belonging to such that and . Evidently, . Set [resp., ] such that [resp., ]; we haveConsequently, [].

*Remark 6. *Let be a vector lattice. Let [resp., ] be as in Theorem 1 [resp., Theorem 4]. If and and , then , , [resp., ].

Then, reasoning as in Theorem 1 proof’s final step [resp., Theorem of [10]], we see that [] are solutions of Problem (P). Consequently, we can assume that ; that is, [resp., ].

#### 3. Dirichlet Problems

We assume () with , , and , where , is open, bounded, connected and set , with .

Let us consider the functional (as in ) such that .

Let us use the following notations: if , else; is the Lebesgue measure in ; , , are instead of , , ; is the first eigenvalue and is the first eigenfunction of the problem in [16].

We present some results about the validity of assumptions . To this aim, let us set and, for each , , let us introduce the following hypothesis: exists: for every . exists such that and for every

Proposition 7. *If for each , then holds. When and holds with , then, with as , holds if as . When and holds with , then holds if as .*

*Proof. *The first statement is obvious. We can prove the second and the third ones as in [11, Propositions and ].

Proposition 8. *When holds with , then with as holds if and as some . When holds with , then holds if and as some .*

*Proof. *See [11, Propositions and ].

Let us now investigate Problem (P) in two concrete cases where .

* Application 1*. Let and, for each ,whereLet us consider the following system:Let us introduce the following conditions:Then (Propositions 7 and 8),

Proposition 9 (Theorems 1 and 4, Remark 6,). *Under conditions (34), one has the following:* *When (36) holds [resp., (36) and (37) hold], with as in (38) [resp., (39)], system (35) has at least two weak solutions and , and one has as , .* *When and (37) holds, with as in (40), system (35) has at least two weak solutions and , and one has as , .**Consequently, when and (36), (37) hold, with as and , system (35) has at least four different weak solutions.*

Proposition 10. *Ifthen as .*

*Proof. *It is sufficient (Proposition 2) to verify the following.

As holds with .

Let, for example, . Let . First of all,Let us introduce the function . Let us note thatthen,About the function , we haveIn fact, with and , since , is necessarily bounded. Then (within a subsequence), with , from which . Then, belongs to .

Let us add that belongs to and it results inThen, with , we haveand, moreover, taking into account the first one of (43),from which .

Proposition 11. *If and as , then as .*

*Proof. *It is sufficient (Proposition 5) to prove the following.

As holds with .

Let, for example, . Let . Firstly, we haveLet us consider the function . Sincewe have the following: the function belongs to , , andThen, set ; we getin particular, .

*Application 2*. Let and, for each ,where