International Journal of Differential Equations
Volume 2016 (2016), Article ID 1683759, 18 pages
http://dx.doi.org/10.1155/2016/1683759

Research Article

A New Result Concerning the Solvability of a Class of General Systems of Variational Equations with Nonmonotone Operators: Applications to Dirichlet and Neumann Nonlinear Problems

Luisa Toscano¹ and Speranza Toscano²

¹Department of Mathematics and Applications “R. Caccioppoli”, University of Naples “Federico II”, Via Cintia, Monte S. Angelo, I-80126 Naples, Italy
²University of Naples, Via Roma 29, 81031 Aversa, Italy

Received 30 July 2015; Accepted 28 September 2015

Academic Editor: Jianshe Yu

Copyright © 2016 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.

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 set

It is easy to control that, for each , the equation

has only one positive root and it results in

The implicit function theory assures the functional belongs to . Then, the functional belongs to and it results in

It is important to remark that (9) can also be written as follows:

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

Proof. Let us set

where is the positive root of equation . Pointing out that and

let us prove that

Let such that ; that is,

We note that

moreover,

since holds in and holds in .
Inequality (18) implies that exists such that (within a subsequence) weakly in . Then,

Let us verify that

In , 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 obtain

Let us add that

In , 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 that

In fact, taking into account that holds in and holds in , we obtain

On the other hand, since   , we have

Then, if , we get the contradiction

Relations (14), (15), and (26) allow (11).
From (11), a Lagrange multiplier exists such that

Setting 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 have

Consequently, [].

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 ,

where

Let 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. If

then 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 that

then,

About the function , we have

In 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 in

Then, with , we have

and, 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 have

Let us consider the function . Since

we have the following:

the function belongs to , , and

Then, set ; we get

in particular, .

Application 2. Let and, for each ,

where

Let us consider the system:

Evidently, and . Then, set ; we have (Propositions 7 and 8)

Proposition 12 (see Theorems 1 and 4, Remark 6). Under conditions (56), one has the following:

 With as in (58), system (57) has at least two weak solutions and , and one has as . When , with as in (59), system (57) has at least two weak solutions and , and one has as .

Consequently, when , with as ,   , and as some , system (57) has at least four different weak solutions.

Proposition 13. Let as and with . Let as . Then,

Proof. It is sufficient (Propositions 2 and 5) to prove that

About (61), let . Let be a compact set such that and in . Let [11, Proposition ] with such that

where is the characteristic function of . We choose such that

and let us set . With , we have

It is easy to verify that

Then, .
About (62), let . Set

we have

Relations (69) imply that

and the function belongs to and it results in

Then, with , we have

moreover,

from which since as some .
Finally, let us verify (63), for example, when . Let . Since

we can suppose . Set ; we have

we add that

from which .

4. Neumann Problems

We assume ( is an open, bounded, connected and set, ), with , and with ( is the measure on , , where is the trace operator).

Let us consider the functional (as in ) such that (). Moreover, and have been settled as in Section 3, is the outward orthogonal unitary vector to , if , and if .

About the validity of assumptions , we set , as  some    and we introduce the following assumptions:

exists: for every . exists: and for every .

Proposition 14. When and as , then holds if , , and as . When holds with , and as , , then with , and as holds if . When holds with and , then holds if .

Proof. We reason by contradiction as in [11, Propositions and ].

Proposition 15. When holds with , and as , and as , then with , , and as holds if and for some . When holds with , and as , then holds if and for some .

Proof. See [11, Propositions and ].

Passing to the applications, we state in advance that the results of [8] for problem

can be obtained by using Theorem 1 and Proposition 14, Theorems and of [10], and Propositions and of [11] (taking into account [17] too).

Application 3. Let and, for each .

We set

where

Let us consider the system:

Proposition 16. Let be a weak solution of system (80). Then,

Proof. Relation (81) is evident. About (82), it is easy to verify that

where are positive constants and . Consequently [11, Proposition ], from which, since in , we deduce that [17] and if [18].

Let us introduce the conditions:

Then (Propositions 14 and 15),

Proposition 17 (Theorems 1 and 4, Remark 6, Proposition 16). Under conditions (79), one has the following:

 When (84)–(86) hold, with as in (88), system (80) has at least two weak solutions and , and one has , as if . When and (85)–(87) hold, with as in (89), system (80) has at least two weak solutions and , and one has , as if .

Consequently, when and (84)–(87) hold, with as in (89) and instead of , system (80) has at least four different weak solutions.

Proposition 18. One gets the following:

Proof. Relation (90) is evident. About (91), it is sufficient (Proposition 2) to verify that

holds with

Let us check, for example, the case . Let ; then, exists such that and on . Let be a compact set with positive measure and an open set such that , . Thanks to [11, Propositions , ], a compact set with , , and exist such that

where is the characteristic function of . Set ; let us fix such that

Then, with , we have

Proposition 19. Let for some either or . Let , , and . Then, on .

Proof. Since in Proposition 18  , it is sufficient (Proposition 2) to prove that

holds with .

Let us suppose, for example, . Let and . Set =   ; we have

then, belonging to exists such that , if , , and . Consequently, set ; it results in

Proposition 20. If either or , it results in

Proof. About (97) [resp., (98)], it is sufficient (Proposition 5) to prove that

holds with

Let and . Let =    and . We note that

Then, belonging to exists such that

from which with [resp.,] we get

Evidently, on if .

Proposition 21. When and for some , , , then

Proof. Let us prove (102) as . Reasoning by contradiction, let on ; that is, on . Let . As =   , we have

Thanks to (103), (104), for each , it is possible to choose such that . Then, set ; it results in

moreover, since

exists such that

Relations (105), (107) imply that (Theorem 4)

Since , we have

from which

where .
From (108), (110), we get

We add that since

then and exist such that

from which

thanks to (103). Relations (111), (114) imply that

from which, as , we get the contradiction

Application 4. Let and, for each , we set

where

Let us consider the system:

Proposition 22. Let either or . If is a weak solution of system (119) with all nonnegative components, then .

Proof. The statement is true in virtue of [11, Proposition , Remark ], and of relation

where are positive constants.

Let us introduce the conditions:

We have (Propositions 14 and 15)

Proposition 23 (see Theorems 1 and 4, Remark 6). Under conditions (118), one has the following:

 When (121) and (122) hold, with , as in (124), system (119) has at least two weak solutions and , and one has as . When and (122), (123) hold, with as in (125), system (119) has at least two weak solutions and (), and one has as .

Consequently, when and (121)–(123) hold, with as in (125) and instead of , system (119) has at least four different weak solutions.

From Propositions 22 and 23, we deduce the following.

Proposition 24. We have as . If either or , then as .

Proposition 25. Let If and , then as .

Proof. It is sufficient (Proposition 2) to show that

 as holds  with .

Let us consider, for example, the case . Let . Firstly, let us prove that exists with such that

Let and    Since , then exists such that . In fact,     .
Firstly, let us suppose . Then, a compact set exists such that and in . Let [11, Proposition ] with such that

where is the characteristic function of . Set and ; we can consider two cases:

In case (130), thanks to (129), it is possible to choose such that (126)–(128) hold with . In case (131), also from (129), we can find such that satisfying (126)–(128), where , , and in .
Let us suppose now . Since , we have

Let us add that

Let such that , strongly in and as , where is the characteristic function of . Evidently, satisfies (126)–(128) with suitable and , , and in .
Let us introduce the function . Equation (126) implies that

Consequently, since and , and belonging to such that , and we have

and then, by (127).
In conclusion, set and, taking into account (128), (135), it results in

Proposition 26. If for some , then as .

Proof. It is sufficient (Proposition 2) to show that

 as holds  with .

Let us suppose and let . Let us verify that exists with such that (126) and (127) hold. At this aim, we note that

where . Let be a compact set with positive measure. Let such that , strongly in and as ( = characteristic function of ). With and as in Proposition 25, it is possible to find such that [resp., ] satisfying (126) and (127) in case (130) [resp., (131)]. Then, if and are as in Proposition 25, we have

Proposition 27. If for some , then as .

Proof. It is sufficient (Proposition 5) to show that

 as holds with , if .

Let us suppose, for example, and let us set . Let be a compact set such that and in for some . Let with and in . Set ; it results in

Then, belonging to such that , and we have

Consequently, with , we get

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

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