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., [19]) 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 [1315]. 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 ,whereLet 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 thatAbout (61), let . Let be a compact set such that and in . Let [11, Proposition ] with such thatwhere is the characteristic function of . We choose such thatand let us set . With , we haveIt is easy to verify thatThen, .
About (62), let . Setwe haveRelations (69) imply thatand the function belongs to and it results inThen, with , we havemoreover,from which since as some .
Finally, let us verify (63), for example, when . Let . Sincewe can suppose . Set ; we havewe add thatfrom 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 problemcan 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 setwhereLet 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 thatwhere 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 thatholds 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 thatwhere is the characteristic function of . Set ; let us fix such thatThen, 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 thatholds with .Let us suppose, for example, . Let and . Set =   ; we havethen, 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 thatholds with Let and . Let =    and . We note thatThen, belonging to exists such thatfrom 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 haveThanks to (103), (104), for each , it is possible to choose such that . Then, set ; it results inmoreover, since exists such thatRelations (105), (107) imply that (Theorem 4)Since , we havefrom whichwhere .
From (108), (110), we getWe add that sincethen and exist such thatfrom whichthanks to (103). Relations (111), (114) imply thatfrom which, as , we get the contradiction

Application 4. Let and, for each , we setwhereLet 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 relationwhere 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 thatas holds  with .Let us consider, for example, the case . Let . Firstly, let us prove that exists with such thatLet 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 thatwhere 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 haveLet us add thatLet 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 thatConsequently, since and , and belonging to such that , and we haveand 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 thatas holds  with .Let us suppose and let . Let us verify that exists with such that (126) and (127) hold. At this aim, we note thatwhere . 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 thatas 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 inThen, belonging to such that , and we haveConsequently, with , we get

Conflict of Interests

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