Abstract and Applied Analysis

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

doi:10.1155/2012/760854

Research Article

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

Luisa Toscano and Speranza Toscano

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 (i₁₄) also holds for any .

Let us pass to () and suppose

() there exist and such thatand for any .

Proposition 2.5. If (i₂₂) 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

Let us consider the system:

Let us introduce the conditions:

Then (Propositions 2.2, 2.3 and 2.5)

Since ([1], Theorems 2.2 and 4.2; Remarks 2.3 and 4.4), we get the following proposition.

Proposition 2.27. Under assumption (2.95) we have:

(i)When (2.97) holds ((2.97) and (2.98) hold, resp.), set as in (2.99) (resp. (2.100)) system (2.96) has at least two weak solutions and with as ((ii)When (2.98) holds, set as in (2.101) system (2.96) has at least two weak solutions and with as .

Consequently, when (2.97) and (2.98) hold, with and system (2.96) has at least four different weak solutions.

We remark that (Proposition A.3, [5]) as :

Moreover, since is a weak supersolution of the equation:

where , we have [6]

Proposition 2.28. It results in

Proof. Since (2.104), we must show that

About (2.106), it is sufficient (Remark 1.1) to prove that

Let with . Let be a compact set such that

From Proposition A.1, there exists , with , such that

Then, with , we have

Let us prove (2.107). We recall that ([1], Theorem 4.2):

where as in Application 2.22. Reasoning by contradiction, let . Then, for some and consequently from (2.104) .

Let , with , such that . Let us consider the function:

Since

we have

We note that . In fact, if and, being , is bounded (else (within a subsequence) . Then (within a subsequence) with , from which .

We add that belongs to , and its derivative has the form:

Hence, set , it results in

As in Proposition 2.24, we introduce the open ball with centre included in and the functionals and belonging to . Chosen such that , we have

and consequently . Then, taking into account (2.112), with such that , we get the contradiction:

Proposition 2.29. If , then

Proof. In fact,

Application 2.30. Let for each :

where

Let as . Let . Set as and , let us consider the system:

under at least one of the following conditions

Evidently, about the validity of () we choose as in Application 2.26.

Proposition 2.31 (see [1], Theorem 3.2). Under assumptions (2.123), (2.125) ((2.125) and (2.126), resp.), if and is sufficiently small, for as in (2.99) (resp. (2.100)) system (2.124) has at least one weak solution , .

Let us note that

Application 2.32. Let , and for each :

under one of the following assumptions:

Set as in Application 2.30. Let us consider the system:

Let us verify that

Let with, for example, . Let and such that . Let us suppose and set . Then,

Proposition 2.33. Under assumption (2.129) (resp. (2.130)), system (2.131) with has at least two weak solutions and , and we have as :

Consequently,

Proof. The statement is due to ([1], Theorem 2.2, Remark 2.3), [5], Proposition A.3, [6].

Proposition 2.34 (see [1], Theorems 3.1, 3.2). Under assumption (2.129) (resp. (2.130)), system (2.131) with and sufficiently small has at least two different weak solutions ), and we have even if .

Remark 2.35. If (within a set with measure equal to zero), with the same reasoning used about (2.132), we get that

hence, even if .

3. Neumann Problems

Let be an open, bounded, and connected set. Let and as in Section 2, the measure on the outward unit normal to if if . Let us assume

We note that for each we set where is the trace operator from into . Morever we consider the functionals (as in ()) such that

It is easy to verify the following.

Proposition 3.1. Let as . Then,

Let us set and for each

Let us introduce the conditions:

()there exists ;()there exist and and .

Proposition 3.2. Let () holds with . Let . Let as . Then with and as : () holds if as .

Proof. Reasoning by contradiction, for each there exist , with , and such that

then, set , we have

Since , there exists such that (within a subsequence)

Consequently, from (3.6), passing to limit as , we get

from which , and then the contradiction .

Proposition 3.3. Let () holds with . Let . Then,

The proof as in Proposition 3.2.

Proposition 3.4. Let () holds with . Let as . Then,

Moreover, if as , we have

with and as holds if and for some .

Proof. The first statement is evident. Let us prove the second one. Reasoning by contradiction, for each there exist , with and for some , and a sequence such that

Let be a strictly increasing sequence such that as . Let . Then, and

moreover, there exists such that (within a subsequence)

Consequently,

then , and the contradiction .

Proposition 3.5. Let () holds with . Let as . Then,

The proof as in Proposition 3.4.

Remark 3.6. It is suitable to make some clarifications.

(i)The assumption “” (see Propositions 3.1, 3.2, and 3.4) can be replaced by “ do not change sign.” In this case we can choose and such that and .(ii)The assumption “” (see Propositions 3.4 and 3.5) can be replaced by “”. In this case, we can choose and such that and + for some , with instead of .(iii)When for each do not change sign, then the conclusion of the Proposition 3.2 [resp. Proposition 3.3] holds even if and as (resp. as ).

In order to simplify the presentation of the applications, we suppose in the next and, while the additional assumptions on and the assumptions on (the same of Propositions 3.1, 3.2, 3.4, and 3.5) will be pointed out case by case.

Passing to the applications (with ), we recall that in [3] Pohozaev and Véron in the case have studied the Neumann problem:

The existence theorems proved by these authors can be got by using some results of ([1], Theorems 2.1, 2.2, 4.1, and 4.2; Remarks 2.1, 2.3, 4.1, and 4.4), Propositions 3.3 and 3.5.

Application 3.7. Let for each :

where

Let us consider the system:

Let us introduce the conditions:

Evidently (3.20) . Moreover (3.21) (Proposition A.2). Hence (Propositions 3.3 and 3.5)

Proposition 3.8 (see ([1], Theorems 2.1 and 4.1; Remarks 2.1 and 4.1); Proposition A.4; [5, 6]). Under assumption (3.18), we have:

(i)When (3.20) and (3.21) hold, with as in (3.23) system (3.19) has at least two weak solutions and , and it results in

(ii)When (3.20) and (3.22) hold, with as in (3.24) system (3.19) has at least two weak solutions and , and it results in

Consequently, when (3.20)–(3.22) hold, with as in (3.24) and instead of system (3.19) has at least four different weak solutions.

Proposition 3.9. If , then as.

Proof. It is sufficient (Remark 1.1) to verify that

Let . Let, for example, . Since , there exists such that

Let a compact set and an open set such that

Since Propositions A.1 and A.2, there exist a compact set , with , and such that

where is the characteristic function of . Let us choose such that

and we set . Then,

Proposition 3.10. If

then .

Proof. We recall that ([1], Theorem 4.1)

Reasoning by contradiction let, for example, . As and for all , we have and since (3.33) . Then, it is possible to choose such that . Let us add that there exist and such that .

Let now and . Since is necessarily bounded. Then (within a subsequence) . Consequently, from the inequality:

as and from (3.34), we get the contradiction:

Remark 3.11. Let us note that the conditions (3.20), (3.21), and (3.33) are compatible.

Application 3.12. Let for each :

where

Let us consider the system:

Pointing out that , we advance the conditions

Taking into account that

we have (Propositions 3.2 and 3.4)

Proposition 3.13 . (see ([1], Theorems 2.1 and 4.1; Remark 2.1); Proposition A.4; [5, 6]). Under assumption (3.39), we have

(i)When (3.41) and (3.43) hold, with as in (3.45) system (3.40) has at least one weak solution (), and it results in

(ii)When (3.41)–(3.43) hold, with as in (3.46) system (3.40) has at least one weak solution , and it results in as .

Consequently, when (3.41)–(3.43) hold, with as in (3.46) and instead of system (3.40) has at least two different weak solutions.

About the properties of and expressed by Proposition 3.13, it is necessary to remark that if (3.40), then as . In fact,

Application 3.14. Let and for any :

where

Let us consider the system:

Let us introduce the conditions:

we have (Propositions 3.3 and 3.5)

Proposition 3.15 (see ([1], Theorems 2.2 and 4.2; Remarks 2.3 and 4.4); Proposition A.4; [5]). Under assumption (3.50), we have

(i)When (3.52) and (3.53) hold, with as in (3.55) system (3.51) has at least two weak solutions and , and it results in

(ii)When (3.53) and (3.54) hold, with as in (3.56) system (3.51) has at least two weak solutions and , and it results in

Consequently, when (3.52)–(3.54) hold, with as in (3.56), and instead of system (3.51) has at least four different weak solutions.

Proposition 3.16. Under the assumption and as , we have

(i)if for some , then as ;(ii)if for some , thenas.

Proof. First of all is a weak supersolution to the equation:

Also, has a similar property. Then [6] it is sufficient to verify that

About (3.60), let us prove (Remark 1.1) that

Let . Let, for example, . Let

Since Propositions A.1 and A.2, there exists , with and , such that

Then with , we have

Passing to (3.61), let us introduce the function , and let us remember that ([1], Theorem 4.2)

where .

Reasoning by contradiction, let us set, for example, and set . Since

as in Proposition 2.24, we get the contradiction:

Application 3.17. Let and set for each :

where

Let us consider the system:

Let us make the assumptions:

About Neumann’s problem (3.71), we have an existence result similar to the one of Proposition 3.15 related to system (3.51). About the positive sign of the components of the weak solutions and to system (3.71), as in Proposition 3.16, we show.

Proposition 3.18. Under the assumption as and as with , we have

(i)if either or for all for some , then as ;(ii)if for all for some , then as .

The following remark deals also with Application 3.14.

Remark 3.19. Making in (3.50) (resp. (3.70)) the change

system (3.51) (resp. (3.71)) has at least the two weak solutions and ([1], Theorem 4.2; Remark 4.4). The components of keep the properties that Propositions 3.15 and 3.16 (Proposition 3.15 and Proposition 3.18 resp.) underline.

Application 3.20. Let for each :

where

Let us consider the system:

Let us introduce the conditions:

We have (Propositions 3.2 and 3.4)

Proposition 3.21 (see ([1], Theorems 2.2 and 4.2; Remarks 2.3 and 4.4); Proposition A.4; [5, 6]). Under assumption (3.75), we have

(i)When (3.77)–(3.79) hold, with as in (3.81), system (3.76) has at least two weak solutions and ), and it results in , . If , then

(ii)When (3.78)–(3.80) hold, with as in (3.82), system (3.76) has at least two weak solutions and ), and it results in , . If , then

Consequently, when (3.77)–(3.80) hold, with as in (3.82), and instead of system (3.76) has at least four different weak solutions. Obviously, and if .

The following proposition gives a sufficient condition to

Proposition 3.22. Let . If and as , then (3.85) and (3.86) hold.

Proof. Since

using Propositions A.1 and A.2, we can verify that

from which (Remark 1.1) we get (3.85).

Let us prove (3.86). Reasoning by contradiction, let us set, for example, . If , we have

Then as in Proposition 3.16, we get a contradiction.

Remark 3.23. Making in (3.75) the change:

system (3.76) has at least the two weak solutions and ([1], Theorem 4.2; Remark 4.4). The components of , all bounded, are locally Hölderian with their first derivatives. If as , then (3.86) holds.

Application 3.24. Let for each :

where

Let us consider the system:

We advance the condition:

and we note that (Proposition 3.1)

Proposition 3.25. Under conditions (3.92) and (3.94), with as in (3.95), system (3.93) has at least two weak solutions and ), and it results in

Proof. We recall that ([1], Section 2), set , we have

We introduce the functional which is in . We still remember that ([1], Theorem 2.3; Remark 2.5)

The property is due to Proposition A.4. Let us verify that as . Reasoning by contradiction, let us set, for example, and . As , we have

Then, since (there exists such that (, from which the contradiction:

Application 3.26. Let for each :

where

Let as , where and (dual space of ). Let . Let us consider the system:

Let us introduce the conditions:

and let us note that (Proposition 3.3)

Proposition 3.27. Under assumptions (3.102) and (3.104), if and is sufficiently small, then with as in (3.105), system (3.103) has at least one weak solution . When , it results in

Proof. The existence of is due to ([1], Theorem 3.2). About (3.106), it is sufficiently (Remark 1.1) to verify that

Let with, for example, . Let . Let be a compact set having positive measure such that

Proposition A.1 lets us choose satisfying the following conditions:

Then with , we have

Now we replace conditions (3.104) with the following:

Proposition 3.28. Under assumptions (3.102) and (3.111), if and is sufficiently small, then with and as system (3.103) has at least two different weak solution and (. When , it results in

Proof. The existence of and is due to ([1], Theorems 3.1, 3.2, and 3.3; Remark 3.1). Relation (3.112) is proved as in Proposition 3.27.

Appendix

In this appendix, we present some results used previously. The first one is trivial. The second one is easy to prove. It is possible to show the third one and the fourth one with the technique developed by Drabek in ([7, Lemma 3.2]). The symbols are the same introduced in Section 3.

Proposition A.1. Let be an open set of . Let be a compact set with . If is an open set such that , then there exists a family of functions such that

where is the characteristic function of .

Proposition A.2. Let be an open, bounded, connected and set. Let be an open neighborhood of . If is a subset of with , then there exist a compact set with and a family of functions such that

where is the characteristic function of .

Let be an open, bounded, connected and set. Let as be a Carathèodory function into defined for for and for such that

where , .

Proposition A.3. Let . If there exist and with such that

then .

Proposition A.4. Let . If there exist with , with such that

then .

Remark A.5. If , we can suppose .

Acknowledgment

This paper is supported by the Second University of Naples.

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