Research Article Dirichlet Problems with an Indefinite and Unbounded Potential and Concave-Convex Nonlinearities

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1 Abstract and Applied Analysis Volume 2012, Article ID , 36 pages doi: /2012/ Research Article Dirichlet Problems with an Indefinite and Unbounded Potential and Concave-Convex Nonlinearities Leszek Gasiński 1 and Nikolaos S. Papageorgiou 2 1 Faculty of Mathematics and Computer Science, Jagiellonian University, ulica Łojasiewicza 6, Kraków, Poland 2 Department of Mathematics, National Technical University, Zografou Campus, Athens, Greece Correspondence should be addressed to Leszek Gasiński, leszek.gasinski@ii.uj.edu.pl Received 5 March 2012; Accepted 18 April 2012 Academic Editor: Yuriy Rogovchenko Copyright q 2012 L. Gasiński and N. S. Papageorgiou. 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. We consider a parametric semilinear Dirichlet problem with an unbounded and indefinite potential. In the reaction we have the competing effects of a sublinear concave term and of a superlinear convex term. Using variational methods coupled with suitable truncation techniques, we prove two multiplicity theorems for small values of the parameter. Both theorems produce five nontrivial smooth lutions, and in the second theorem we provide precise sign information for all the lutions. 1. Introduction Let R N be a bounded domain with a C 2 -boundary. In this paper we study the following parametric nonlinear Dirichlet problem: Δu z β z u z g z, u z f z, u z in, u 0, > 0. P Here β L s with s > N/2 N 2 is a potential function which may change sign indefinite potential. Al>0 is a parameter, and g,f : R R are Carathéodory functions i.e., for all ζ R, functions z g z, ζ and z f z, ζ are measurable and for almost all z, functions ζ g z, ζ and ζ f z, ζ are continuous. We assume that for almost all z, the function ζ g z, ζ is strictly sublinear near ±, while the function ζ f z, ζ is superlinear near ±. So, problem P exhibits competing nonlinearities of

2 2 Abstract and Applied Analysis the concave-convex type. This situation was first studied with β 0 and the right hand side nonlinearity being ζ q 2 ζ ζ r 2 ζ, 1.1 with 2N 1 <q<2 <r<2 if N>2, N 2 if N 2, 1.2 by Ambrosetti et al. 1. In 1, the authors focus on positive lutions and proved certain bifurcation-type phenomena as >0 varies. Further results in this direction can be found in the works of I lyav 2, Lietal. 3, Lubyshev 4, andrădulescu and Repovš 5. Inall the aforementioned works β 0. We should al mention the recent work of Motreanu et al. 6, where the authors consider equations driven by the p-laplacian, with concave term of the form ζ q 2 ζ where 1 <q<p and a perturbation exhibiting an asymmetric behaviour at and at p 1 -superlinear near and p 1 -sublinear near. They prove multiplicity results producing four nontrivial lutions with sign information. In this work, we prove two multiplicity results for problem P when the parameter >0 is small. In both results we produce five nontrivial smooth lutions, and in the second we provide precise sign information for all the lutions. For the superlinear convex nonlinearity f z,, we do not employ the usual in such cases Ambrosetti-Rabinowitz condition. Instead, we use a more general condition, which incorporates in our framework al superlinear perturbations with slow growth near ±, which do not satisfy the Ambrosetti-Rabinowitz condition. We should point out that none of the works mentioned earlier provide sign information for all the lutions in particular, none of them produced a nodal sign changing lution and all use the Ambrosetti-Rabinowitz condition to express the superlinearity of the convex contribution in the reaction. Our approach is variational based on the critical point theory which is combined with suitable truncation techniques. In the next section we recall the main mathematical tools we will use in the analysis of problem P. We al introduce the hypotheses on the terms g and f. 2. Mathematical Background and Hypotheses Let X be a Banach space and let X be its topological dual. By, we denote the duality brackets for the pair X,X.Letϕ C 1 X. We say that ϕ satisfies the Cerami condition,ifthe following is true. Every sequence {x n } n 1 X, such that {ϕ x n } n 1 R is bounded and 1 x n ϕ x n 0 in X, 2.1 admits a strongly convergent subsequence. This compactness-type condition is in general weaker than the usual Palais-Smale condition. However, the Cerami condition suffices to prove a deformation theorem and from

3 Abstract and Applied Analysis 3 it we derive the minimax theory for certain critical values of ϕ C 1 X see Gasiński and Papageorgiou 7 and Motreanu and Rădulescu 8. In particular, we can have the following theorem, known in the literature as the mountain pass theorem. Theorem 2.1. If X is a Banach space, ϕ C 1 X satisfies the Cerami condition, x 0,x 1 X, ϱ> 0 x 1 x 0 >ϱ: max { ϕ x 0,ϕ x 1 } < inf { ϕ x : x x 0 ϱ } η ϱ, c inf maxϕ( γ t, γ Γ t 0,1 2.2 where Γ { γ C 0, 1 ; X : γ 0 x 0,γ 1 x 1 }. 2.3 Then c η ϱ and c are a critical value of ϕ. In the analysis of problem P, in addition to the Sobolev space H 1 0, we will al use the Banach space C 1 0 ( { u C 1( } : u This is an ordered Banach space with positive cone: C { ( } u C 1 0 : u z 0 z. 2.5 This cone has a nonempty interior given by { int C u C : u z > 0 z, u n } z < 0 z, 2.6 where n is the outward unit normal on. In the proof of the second multiplicity theorem and in order to produce a nodal sign changing lution, we will al use critical groups. So, let us recall their definition. Let ϕ C 1 X and let c R. We introduce the following sets: ϕ c { x X : ϕ x c }, K ϕ { x X : ϕ x 0 }. 2.7 Al, if Y 1,Y 2 is a topological pair with Y 2 Y 1 X and k 0 is an integer, by H k Y 1,Y 2 we denote the kth relative singular homology group for the pair Y 2,Y 1 with integer coefficients. The critical groups of ϕ at an ilated critical point x 0 X of ϕ with ϕ x 0 c are defined by C k ( ϕ, x0 Hk ( ϕ c U, ϕ c U \ {x 0 } k 0, 2.8

4 4 Abstract and Applied Analysis where U is a neighbourhood of x 0, such that K ϕ ϕ c U {x 0 }. The excision property of singular homology implies that this definition is independent of the particular choice of the neighbourhood U. Using the spectral theorem for compact self-adjoint operators, we can show that the differential operator H 1 0 u Δu βu 2.9 has a sequence of distinct eigenvalues { k } k 1, such that k as k The first eigenvalue is simple and admits the following variational characterization: { } σ u 1 inf : u H 1 u 2 0,u / 0, where σ u u 2 2 β z u z 2 dz u H Moreover, the corresponding eigenfunction û C 1 0 does not change sign, and in fact we can take û z > 0 for all z see Gasiński and Papageorgiou 9. Using 2.11 and this property of the principal eigenfunction, we can have the following lemma see Gasiński and Papageorgiou 9, Lemma 2.1. Lemma 2.2. If η L s, η z 1 for almost all z, η / 1, then there exists ĉ>0, such that σ u ηu 2 dz ĉ u 2 u H that Al, from Gasiński and Papageorgiou 9 we know that there exist ĉ 0, ĉ 1 > 0, such u 2 ĉ 0 (σ u ĉ 1 u 2 2 u H 1 0, 2.14 with u u 2, the norm of the Sobolev space H 1 0. Next we state the hypotheses on the two components g and f of the reaction in problem P. Let G z, ζ ζ 0 g z, s ds, F z, ζ ζ 0 f z, s ds. 2.15

5 Abstract and Applied Analysis 5 H g : g : R R is a Carathéodory function, such that g z, 0 0 for almost all z. i For every ϱ>0, there exists a function a ϱ L, such that g z, ζ a ϱ z for almost all z, all ζ ϱ ii We have g z, ζ lim 0 uniformly for almost all z ζ ± ζ iii There exist constants c 1,c 2 > 0, 1 <q<μ<2andδ 0 > 0, such that c 1 ζ q g z, ζ ζ for almost all z, all ζ R, μg z, ζ g z, ζ ζ c 2 ζ q for almost all z, all ζ δ iv For every ϱ>0, we can find γ ϱ > 0, such that for almost all z the map [ ϱ, ϱ ] ζ g z, ζ γ ϱ ζ q 2 ζ is nondecreasing Remark 2.3. Hypothesis H g ii implies that for almost all z, the function g z, is strictly sublinear near ±. Hence g z, is the concave component in the reaction of P the terminology concave and convex nonlinearities is due to Ambrosetti 1. Notethat hypothesis H g iii implies that for almost all z, the function g z, has a similar growth near 0 that is, we have a concave term near zero. Hypothesis H g iv is weaker than assuming the monotonicity of g z, for almost all z. H f : f : R R is a Carathéodory function, such that f z, 0 0 for almost all z. i There exist a L, c>0andr 2, 2, such that f z, ζ a z c ζ r 1 for almost all z, all ζ R ii We have F z, ζ lim uniformly for almost all z ζ ± ζ 2 iii There exist functions η 0, η 0 L, such that η 0 z 1 for almost all z, η 0 / 1 and f z, ζ η 0 z lim inf ζ 0 ζ f z, ζ lim sup η 0 z ζ 0 ζ 2.22 uniformly for almost all z.

6 6 Abstract and Applied Analysis iv For every ϱ>0, we can find γ ϱ > 0, such that for almost all z, we have the map [ ϱ, ϱ ] ζ f z, ζ γ ϱ ζ is nondecreasing Remark 2.4. Hypothesis H f ii implies that for almost all z, the function F z, is superquadratic near ±. Evidently, this is satisfied if the function f z, is superlinear near ±, that is, when f z, ζ lim uniformly for almost all z ζ ± ζ So, f z, ζ is the convex component of the reaction which competes with the concave component g z, ζ. Note that in H f, we did not include the Ambrosetti-Rabinowitz condition to characterize the superlinearity of f z,. We recall that the Ambrosetti-Rabinowitz condition says that there exist τ>2andm>0, such that 0 <τf z, ζ f z, ζ ζ uniformly for almost all z, all ζ M, 2.25 ess inf F,M > Integrating 2.25 and using 2.26, we obtain the weaker condition c 3 ζ τ F z, ζ uniformly for almost all z, all ζ M Therefore, for almost all z, the function ζ G z, ζ F z, ζ 2.28 is superquadratic with at least τ-growth near ±. Hence the Ambrosetti-Rabinowitz condition excludes superlinear perturbations with slower growth near ±. For this rean, here we employ a weaker condition. So, let ξ z, ζ ( g z, ζ f z, ζ ζ 2 G z, ζ F z, ζ We employ the following hypothesis. H 0 : For every >0, there exists a function ϑ L1, such that ( ξ z, ζ ξ z, y ϑ z for almost all z, all 0 ζ y, ( ξ z, ζ ξ z, y ϑ z for almost all z, all y ζ

7 Abstract and Applied Analysis 7 Remark 2.5. Hypothesis H 0 is a generalized version of a condition first introduced by Li and Yang 10, where the reader can find other possible extensions of the Ambrosetti-Rabinowitz condition and comparins between them. Hypothesis H 0 is a quasimonotonicity condition on ξ z,, and it is satisfied if there exists M>0, such that for almost all z, the function ζ g z, ζ f z, ζ ζ 2.31 is increasing on M, and is decreasing on, M see Li and Yang 10. Example 2.6. The following pairs of functions satisfy hypotheses H g, H f,andh 0 for the sake of simplicity we drop the z-dependence : g 1 ζ ζ q 2 ζ, f 1 ζ ζ p 2 ζ η 0 ζ, g 2 ζ ζ q 2 ζ, f 2 ζ ζ p 2 ζ ln 1 ζ η 0 ζ, g 3 ζ ζ q 2 ζ ζ τ 2 η 0 (ζ ζ r 2 ζ if ζ 1, ζ, f 3 ζ ζ p 2 ζ ln ζ if ζ > 1, 2.32 with 1 <q<τ<2 <p<2,2<r, η 0 < 1. Note that f 2 and f 3 do not satisfy the Ambrosetti-Rabinowitz condition see For every u H 1 0,weset u u by virtue of the Poincaré inequality. We mention that the notation will be al used to denote the R N -norm. It will always be clear from the context which norm is used. For ζ R, letζ ± max{±ζ, 0}. Then for u W 1,p 0, wesetu ± u ±. We have u ± H 1 0, u u u,andu u u. For a given measurable function h : R R e.g., a Carathéodory function, weset N h u h,u u W 1,p Finally, let A L H 1 0,H 1 be the operator, defined by ( A u, y u, y dz u, y H 1 R N For the properties of the operator A we refer to Gasiński and Papageorgiou 11, Proposition 3.1, page 852.

8 8 Abstract and Applied Analysis 3. Solutions of Constant Sign In this section, for >0 small, we generate four nontrivial smooth lutions of constant sign two positive and two negative. To this end, we introduce the following modifications of the nonlinearities g z, and f z, : g ± z, ζ g ( z, ±ζ ±, f± z, ζ f ( z, ±ζ ± ĉ 1 ( ±ζ ±, 3.1 with ĉ 1 > 0asin These modifications are Carathéodory functions. We set G ± z, ζ ζ 0 g ± z, s ds, F ± z, ζ ζ 0 f ± z, s ds 3.2 and consider the C 1 -functionals ϕ ± : H1 0 R, defined by ϕ ± u 1 ĉ1 σ u 2 2 u 2 2 G ± z, u z F ± z, u z dz u H Proposition 3.1. If hypotheses H g, H f, and H 0 hold and >0, then the functionals ϕ ± satisfy the Cerami condition. Proof. We do the proof for ϕ, the proof for ϕ being similar. So, let {u n } n 1 H 1 0 be a sequence, such that ϕ u n M1 n 1, 3.4 for me M 1 > 0and 1 u n ( ϕ un 0 in H From 3.5, we have A u n,h ε n h 1 u n ( β ĉ1 un hdz h H 1 0, g z, u n hdz f z, u n hdz 3.6 with ε n 0. In 3.6 we choose h u n H 1 0. Then σ ( u n ĉ1 u n 2 ε n n 1, ĉ 0 u n 2 εn n 1 3.8

9 Abstract and Applied Analysis 9 see 2.14, and hence u n 0 in H From 3.4 and 3.9, we have σ u n 2G z, u n dz 2F z, u n dz M 2 n 1, 3.10 for me M 2 > 0. Al, if in 3.6 we choose h u n H 1 0, then σ u n g z, u n u ndz f z, u n u n dz ε n n Adding 3.10 and 3.11, weobtain for me M 3 > 0 see 2.29 for the definition of ξ. ξ z, u n dz M 3 n 1, 3.12 Claim 1. The sequence {u n} n 1 H 1 0 is bounded. Arguing by contradiction, suppose that the claim is not true. Then by passing to a subsequence if necessary, we may assume that u n Let y n u n u n n Then yn 1 n And, passing to a subsequence if necessary, we may assume that y n y weakly in H 1 0, 3.16 y n y in L r If y / 0, then u n z for almost all z { y>0 } 3.18

10 10 Abstract and Applied Analysis recall that y 0. So, by virtue of hypotheses H g ii and H f ii, for almost all z,we have G z, u lim n z G z, u lim n z y n u n 2 n u n z 2 n z 2 0, F z, u lim n z F z, u lim n z y n u n 2 n u n z 2 n z Then Fatou s lemma implies that lim n ( G z, u n u n 2 dz F z, u n dz. u n But from 3.4 and 3.9, we have G z, u n dz F z, u n dz M 4 σ u n n 1, 3.21 for me M 4 > 0, G z, u n u n 2 F z, u dz n dz M u n 2 5 n 1, 3.22 for me M 5 > 0 since the sequence {σ y n } n 1 is bounded in R. Comparing 3.20 and 3.22, we reach a contradiction. So, we have y 0. We fix μ>0andset v n ( 2μ 1/2 yn n Evidently v n 0inL r 3.24 see Hence by Krasnoselskii s theorem see Gasiński and Papageorgiou 12, Proposition , page 87 and hypotheses H g i and H f i, we have G z, v n z dz 0, F z, v n z dz Since u n, we can find n 0 1, such that 0 < ( 2μ 1/2 1 u n < 1 n n

11 Abstract and Applied Analysis 11 Let t n 0, 1 be such that ϕ t nu n max 0 t 1 ϕ tu n By virtue of 3.26, we have ϕ t nu n ϕ v n 2μσ ( y n G z, v n dz F z, v n dz 2μ 2μ βyn 2 dz G z, v n dz F z, v n dz Note that βy 2 n dz This fact together with 3.25 and 3.28 implies that ϕ t nu n μ n n 1, 3.30 for me n 1 n 0. Since μ>0 is arbitrary, we infer that ϕ t nu n Note that ϕ 0 0, ϕ u n M 6 n 1, 3.32 for me M 6 > 0 see 3.4 and 3.9. Hence 3.31 implies that there exists n 2 n 1, such that t n 0, 1 n n And from the choice of t n, we have d dt ϕ tu n t tn 0 n n 2, 3.34 ( ϕ tn u n,u n 0 n n 2, 3.35

12 12 Abstract and Applied Analysis thus ( ϕ tn u n,t n u n 0 n n 2, 3.36 and hence σ t n u n g z, t n u n t n u n dz f z, t n u n t n u n dz Hypothesis H 0 implies that ξ z, t n u n dz ξ z, u n dz ϑ 1 n 1, ϕ t nu n ξ z, u n dz ϑ 1 n see 3.37,andthus 2 ϕ t nu n M 3 ϑ 1 n see Comparing 3.31 and 3.40, we reach a contradiction. This proves the claim. By virtue of the claim and 3.9, we have that the sequence {u n } n 1 bounded. So, we may assume that H 1 0 is u n u weakly in H 1 0, 3.41 u n u in L r In 3.6 we choose h u n u, pass to the limit as n,anduse Then lim n A u n,u n u 0, 3.43 u n u n 2 u 2 u 3.44 see Gasiński and Papageorgiou 11, Proposition 3.1, page 852, andthus u n u in H

13 Abstract and Applied Analysis 13 by the Kadec-Klee property of Hilbert spaces. This proves that ϕ condition. Similarly we show that ϕ al satisfies the Cerami condition. satisfies the Cerami Our aim is to apply Theorem 2.1 the mountain pass theorem to two functionals ϕ and ϕ. We have checked that both functionals satisfy the Cerami condition. So, it remains to show that they satisfy the mountain pass geometry as it is described in Theorem 2.1. The next proposition is a crucial step in satisfying the mountain pass geometry for the two functionals ϕ and ϕ. Proposition 3.2. If hypotheses H g and H f hold, then there exist ± > 0, such that for every 0, ±, we can find ϱ ± > 0, such that inf { ϕ ± u : u } ϱ± η ± > Proof. Hypotheses H g i and ii imply that for a given ε>0, we can find c 4 c 4 ε > 0, such that g z, ζ εζ c 4 ζ q 1 for almost all z, all ζ 0, 3.47 G z, ζ ε 2 ζ2 c 4 q ζq for almost all z, all ζ Similarly hypotheses H f i and iii imply that for a given ε>0, we can find c 5 c 5 ε > 0, such that f z, ζ ( η 0 z ε ζ c 5 ζ r 1 for almost all z, all ζ 0, 3.49 F z, ζ 1 ( η0 z ε ζ 2 c 5 2 r ζr for almost all z, all ζ Then, for u H 1 0, we have ϕ u 1 ĉ1 σ u 2 2 u 2 2 G z, u dz F z, u dz 1 2 σ u 1 η 0 u 2 dz ε 2 2 u 2 2 c 5 r u r r ε 2 u 2 2 c 4 q u q q 1 2 σ( u ĉ1 u ( ε 1 ĉ u 2 1 u 2 ( c 6 u r u q, 2 1 2ĉ for me c 6 > 0 see 3.48, 3.50, 2.11, 2.14,andLemma 2.2.

14 14 Abstract and Applied Analysis Choosing ε 0, 1 ĉ/ 1, we have ϕ u c 7 u 2 ( c 6 u r u q ( (c 7 c 6 u r 2 u q 2 u 2, 3.52 for me c 7 > 0. Let μ t t r 2 t q 2 t> Since q<2 <r, we have lim t 0 μ t lim μ t. r 3.54 Al μ is continuous in 0,. Therefore, we can find t 0 0,, such that μ t 0 inf t>0 μ t, 3.55 μ t 0 0, 3.56 and thus ( ( 1/ r q 2 q t 0 t r 2 Evidently μ t 0 0 as Hence we can find > 0, such that μ t 0 < c 7 c 6 0,, 3.59 inf { ϕ u : u ϱ t 0 } η > see 3.52.

15 Abstract and Applied Analysis 15 Similarly, for ϕ, we can find > 0, such that for all 0, there exists ϱ > 0, such that inf { ϕ u : u } ϱ η > With the next proposition we complete the mountain pass geometry for problem P. Proposition 3.3. If hypotheses H g and H f hold, >0 and ũ int C with ũ 2 1, then ϕ ± tũ as t ± Proof. By virtue of hypotheses H g i and ii, foragivenε>0, we can find c 8 c 8 ε > 0, such that G z, ζ ε 2 ζ2 c 8, for almost all z, all ζ R Similarly, hypotheses H f i and ii imply that for any given ξ>0, we can find c 9 c 9 ξ > 0, such that F z, ζ ξ 2 ζ2 c 9, for almost all z, all ζ R Then, we have ( ϕ tũ t2 ε ξ 2 σ ũ 2 t2 2 t 2 c 10 ( ũ 2 β s ũ 2s ε ξ c for me c 10 > 0 see 3.63, 3.64 and recall that ũ int C, ũ 2 1and1/s 1/s 1. Since ξ>0 is arbitrary, choosing ξ>ε ũ 2 β s ũ 2s, 3.66 from 3.65, we infer that ϕ tũ as t ± Now we are ready to produce the first two nontrivial smooth lutions of constant sign. In what follows, we set min{, }. 3.68

16 16 Abstract and Applied Analysis Proposition 3.4. If hypotheses H g, H f, and H 0 hold and 0,, then problem P has at least two nontrivial smooth lutions of constant sign: u 0,v 0 C 1 0 (, with v 0 z < 0 <u 0 z z Proof. Propositions 3.1, 3.2,and3.3 permit the application of the mountain pass theorem see Theorem 2.1 for the functional ϕ, and we obtain u 0 H 1 0, such that ϕ 0 0 < η ϕ u 0, 3.70 From 3.70,weseethatu 0 / 0. From 3.71, we have ( ϕ u A u 0 βu 0 ĉ 1 u 0 N g u 0 N f u On 3.72 we act with u 0 H1 0 and obtain σ ( u 0 ĉ1 u , u ĉ see 2.14 ; hence u 0 0, u 0 / 0. Therefore 3.72 becomes A u 0 βu 0 N g u 0 N f u 0, 3.75 Δu 0 z β z u 0 z g z, u 0 z f z, u 0 z in, u From the regularity theory for Dirichlet problems see Struwe 13, pp , we have that u 0 C 1 0. Moreover, invoking the weak Harnack inequality of Pucci and Serrin 14, page 154, we have that u 0 z > 0 for all z. Similarly working with ϕ, this time we obtain a nontrivial smooth negative lution v 0 C 1 0 with v 0 z < 0 for all z. We can improve the conclusion of this proposition by strengthening the condition on the potential β: H β : β L s with s>n/2andβ L.

17 Abstract and Applied Analysis 17 Remark 3.5. So, the potential function is bounded from above but in general can be unbounded from below. Proposition 3.6. If hypotheses H g, H f, H 0, and H β hold and 0,, then problem P has at least two nontrivial smooth lutions of constant sign: u 0 int C, v 0 int C Proof. From Proposition 3.4, we already have two lutions: u 0,v 0 C 1 0 (, with v 0 z < 0 <u 0 z z We have Δu 0 z β z u 0 z g z, u 0 z f z, u 0 z f z, u 0 z for almost all z 3.79 see hypothesis H g iii. Letϱ u 0 and let γ ϱ > 0 be as postulated by hypothesis H f iv. Then from 3.79, we have Δu 0 z ( β z γ ϱ u0 z f z, u 0 z γ ϱ u 0 z, for almost all z, 3.80 Δu 0 z ( β γ ϱ u0 z for almost all z and thus u 0 int C see Vázquez 15 and Pucci and Serrin 14, page 120. Similarly for the negative lution v 0. To continue and produce additional nontrivial smooth lutions of constant sign, we need to keep hypotheses H β. Proposition 3.7. If hypotheses H g, H f, H 0 and H β hold and 0,, then problem P has at least four nontrivial smooth lutions of constant sign: u 0, û int C, û u 0 int C, v 0, v int C, v 0 v int C Proof. From Proposition 3.6, we already have two lutions: u 0 int C, v 0 int C. 3.83

18 18 Abstract and Applied Analysis We introduce the following truncation perturbation of the reaction of the problem P : h z, ζ g z, u 0 z f z, u 0 z ĉ 1 u 0 z g z, ζ f z, ζ ĉ 1 ζ if ζ u 0 z, if u 0 z <ζ This is a Carathéodory function. We set ζ H z, ζ h z, s ds and consider the C 1 -functional ψ : H1 0 R, defined by ψ u 1 ĉ1 σ u 2 2 u 2 2 H z, u z dz u H Claim 2. We have K ψ u 0, where u 0 { } u H 1 0 : u 0 z u z for almost all z Let ũ K ψ. Then A ũ ( β ĉ 1 ũ Nh ũ On 3.88 we act with u 0 ũ H 1 0. Then A ũ, u0 ũ ( β ĉ1 ũ u0 ũ dz h z, ũ u 0 ũ dz ( g z, u0 f z, u 0 ĉ 1 u 0 u0 ũ dz A u 0, u 0 û ( β ĉ1 u0 u 0 ũ dz 3.89 see 3.84, A u0 ũ, u 0 ũ β u 0 ũ u 0 ũ dz ĉ u0 1 ũ 2 0, thus σ ( u 0 ũ ĉ u0 1 ũ 2 0,

19 Abstract and Applied Analysis 19 hence 1 ĉ 0 u0 ũ see 2.14, and finally u 0 ũ. This proves Claim 2. Claim 3. We may assume that u 0 is a local minimizer of ψ. Let μ,, and consider problem P μ. As we did in the proof of Proposition 3.4, via the mountain pass theorem, we obtain a nontrivial smooth positive lution u μ,and by virtue of the strong maximum principle, we have u μ int C see the proof of Proposition 3.7. Then Δu μ z β z u μ z μg ( z, u μ z f ( z, u μ z g ( z, u μ z f ( z, u μ z for almost all z 3.93 see H g iii and recall that <μ. We consider the following truncation perturbation of the reaction of problem P : γ z, ζ 0 if ζ<0, g z, ζ f z, ζ ĉ 1 ζ if 0 ζ u μ z, g ( z, u μ z f ( z, u μ z ĉ 1 u μ z if u μ z <ζ This is a Carathéodory function. We set Γ z, ζ ζ 0 γ z, s ds 3.95 and consider the C 1 -functional ψ : H1 0 R, defined by ψ u 1 ĉ1 σ u 2 2 u 2 2 Γ z, u z dz u H From 3.94, it is clear that ψ is coercive. Al, using the Sobolev embedding theorem, we check that ψ is sequentially weakly lower semicontinuous. So, by the Weierstrass theorem, we can find u H 1 0, such that ψ u inf u H 1 0 ψ u By virtue of hypothesis H f iii, we can find c 11 > 0and δ 0 > 0, such that F z, ζ c 11 2 ζ2 for almost all z, all ζ δ

20 20 Abstract and Applied Analysis Let ũ int C and let t 0, 1 be small, such that tũ u μ, tũ z δ 0 z 3.99 recall that u μ int C. Then, we have ψ tũ t2 2 σ ũ t2 ĉ 1 ũ t2 ( σ ũ ĉ 1 c 11 ũ G z, tũ dz F z, tũ dz tq c 1 q ũ q q see 3.94, hypothesis H g iii, and Since q<2, by choosing t 0, 1 even smaller if necessary, we have ψ tũ < 0, ψ u < 0 ψ see 3.97 ; hence u / 0. From 3.97, we have ( ψ u 0, A u ( β ĉ 1 u N γ u Acting on with u H1 0, weobtainthatu 0, u / 0. Al, acting on with u u μ H 1 0, we have A u, ( ( ( dz u u μ β ĉ1 u u u μ γ z, u ( dz u u μ ( ( ( ( dz g z, uμ f z, uμ ĉ1 u μ u u μ A ( ( ( ( dz u μ, u u μ β ĉ1 uμ u u μ see 3.94 and 3.93, A ( ( ( ( u u μ, u u μ β ĉ1 u u μ u u μ dz 0, 3.106

21 Abstract and Applied Analysis 21 thus hence ( (u ( 2 σ u μ ĉ 1 u u μ 0, ĉ 0 ( u u μ see 2.14 andfinally u u μ So, we have proved that u [ ] } 0,u μ {u H 1 0 :0 u z u μ z for almost all z Hence becomes A u βu N g u N f u see 3.94, Δu z β z u z g z, u z f z, u z in, u 0, thus u int C as in the proof of Proposition 3.6 and it is a lution of P. If u / u 0, then this is the desired second nontrivial positive smooth lution of P. So, we may assume that u u 0 and that there is no other lution of P in the order interval u 0,u { } u H 1 0 : u 0 z u z u μ z for almost all z We introduce the following truncation of γ z, : γ z, ζ γ z, u 0 z if ζ<u 0 z, γ z, ζ if u 0 z ζ This is a Carathéodory function. We set Γ z, ζ ζ 0 γ z, s ds 3.115

22 22 Abstract and Applied Analysis and consider the C 1 -functional ψ : H1 0 R, defined by ψ u 1 ĉ1 σ u 2 2 u 2 2 Γ z, u z dz u H From 3.94 and 3.114, it follows that ψ is coercive. Al, it is sequentially weakly lower semicontinuous. Hence, we can find ũ H 1 0, such that ψ ũ inf u H 1 0 ψ u, ( ψ ũ 0, and thus A ũ ( β ĉ 1 ũ N γ ũ As before, acting on with u 0 ũ H 1 0 and with ũ u μ H 1 0, weshow that ũ u 0,u μ. Then, from 3.94 and 3.114, it follows that A ũ βũ N g ũ N f ũ, ũ int C is a lution of P in u 0,u μ, hence ũ u 0. Let ϱ u 0 and let γ ϱ > 0and γ ϱ > 0 be as postulated by hypotheses H g iv and H f iv, respectively. We have Δu 0 z ( β z γ ϱ u0 z γ ϱ u 0 z q 1 g z, u 0 z γ ϱ u 0 z q 1 f z, u 0 z γ ϱ u 0 z g ( z, u μ z γ ϱ u μ z q 1 f ( z, u μ z γ ϱ u μ z μg ( z, u μ z μγ ϱ u μ z q 1 f ( z, u μ z γ ϱ u μ z Δu μ z ( β z γ ϱ uμ z μγ ϱ u μ z q 1 for almost all z see H g iv and iii, H f iv and recall that u 0 u μ and <μ, there exists c 12 > 0, such that Δ ( u μ u 0 z ( β γ ϱ ( uμ u 0 z μγϱ c 12 ( uμ u 0 z, for almost all z recall that the function ζ ζ q 1 is locally Lipschitz ; thus u μ u 0 int C 3.123

23 Abstract and Applied Analysis 23 see Struwe 13 and Pucci and Serrin 14, page 120. So, we have that u u 0 int C 1 0 [ 0,uμ ] Note that ψ 0,u μ ψ 0,u μ see 3.94 and 3.114, u 0 is alocal C 1 0 ( -minimizer of ψ see 3.124, andthus u 0 is alocal H 1 0 -minimizer of ψ Brézis and Nirenberg 16. This proves Claim 3. By virtue of Claim 3, asingasiński and Papageorgiou 17, proof of Theorem 3.4, we can find ϱ 0, 1 small, such that ψ u 0 < inf { ψ u : u u 0 ϱ } η As in Proposition 3.4,forũ int C with ũ 2 1, we have ψ tũ as t Note that ϕ ψ ξ with ξ R.HencebyvirtueofProposition 3.1, ψ satisfies the Cerami condition. This fact together with and permits the use of the mountain pass theorem see Theorem 2.1. So, we can find û H 1 0, such that ψ u 0 <η ψ û, ( ψ û From we have û / u 0.From and Claim 2, we have that u 0 û Hence A û βû N g û N f û see 3.84,andû int C lves problem P.

24 24 Abstract and Applied Analysis Moreover, as before, using the strong maximum principle see Vázquez 15 and Pucci and Serrin 14, page 120, we have û u 0 int C In a similar way, using v 0 int C, we define { g z, ζ f z, ζ h z, ζ ĉ1 ζ if ζ<v 0 z, g z, v 0 z f z, v 0 z ĉ 1 v 0 z if v 0 z ζ This is a Carathéodory function. We set ζ H z, ζ h z, s ds and consider the C 1 -functional ψ : H1 0 R, defined by ψ u 1 ĉ1 σ u 2 2 u 2 2 H z, u z dz u H Reaning as above, using this time ψ, we obtain a second negative smooth lution v int C of problem P, such that v 0 v int C Five Solutions In this section, we prove two multiplicity theorems, establishing five nontrivial smooth lutions when 0,. In the second multiplicity theorem, we provide sign information forallthelutions i.e., we show that the fifth lution is actually nodal. Theorem 4.1. If hypotheses H g,h f,h 0, and H β hold and 0,, then problem P has at least five nontrivial smooth lutions: u 0, û int C, û u 0 int C, v 0, v int C, v 0 v int C ( y 0 C 1 0, u 0 y 0 int C, y 0 v 0 int C. 4.1 Proof. From Proposition 3.7, we already have four nontrivial smooth lutions of constant sign: u 0, û int C, û u 0 int C,v 0, v int C, v 0 v int C. 4.2

25 Abstract and Applied Analysis 25 We consider the following truncation perturbation of the reaction of problem P : h z, ζ g z, v 0 z f z, v 0 z ĉ 1 v 0 z g z, ζ f z, ζ g z, u 0 z f z, u 0 z ĉ 1 u 0 z if ζ<v 0 z, if v 0 z ζ u 0 z, if u 0 z <ζ. 4.3 This is a Carathéodory function. We set ζ H z, ζ h z, s ds and consider the C 1 -functional ψ : H1 0 R, defined by ψ u 1 ĉ1 σ u 2 2 u 2 2 H z, u z dz u H From 4.3, it follows that ψ is coercive. Al, it is sequentially weakly lower semicontinuous. So, we can find y 0 H 1 0, such that ψ ( y0 inf u H 1 0 ψ u. 4.6 As in the proof of Proposition 3.7, using 3.98, we show that ψ ( y0 < 0 ψ 0, 4.7 hence y 0 / 0. From 4.6, we have ( ψ ( y0 0, 4.8 A ( ( y 0 β ĉ1 y0 N h ( y On 4.9 we act with y 0 u 0 H 1 0. Then A ( ( ( ( dz y 0, y0 u 0 β ĉ1 y0 y0 u 0 ( ( dz g z, u0 f z, u 0 ĉ 1 u 0 y0 u 0 A u 0, ( y 0 u ( ( dz β ĉ u0 y0 u 0

26 26 Abstract and Applied Analysis see 4.3, A ( ( ( ( ( dz y 0 u 0, y0 u 0 β ĉ1 y0 u 0 y0 u 0 0, 4.11 thus ( (y0 ( 2 σ u 0 ĉ 1 y0 u 0 0, and hence 1 ĉ 0 ( y0 u see 2.14 ; hence y 0 u 0. Similarly, acting on 4.9 with v 0 y 0 H 1 0, we show that v 0 y 0. Therefore, y 0 v 0,u 0 { } u H 1 0 : v 0 z u z u 0 z for almost all z, 4.14 and 4.9 becomes A ( y 0 βy0 N g ( y0 Nf ( y see 4.3 ; thus y 0 C 1 0 ( 4.16 regularity theory; see Struwe 13, and it lves problem P. Moreover, as in the proof of Proposition 3.7, using hypotheses H g iv and H f iv, we al show that u 0 y 0 int C, y 0 v 0 int C Next we will improve the conclusion of Theorem 4.1 and show that the fifth lution y 0 is nodal sign changing. To do this we need to strengthen a little bit the hypotheses on f z,. The new hypotheses of f are the following: H f : f : R R is a Carathéodory function, such that f z, 0 0for almost all z, i, iii, iv are the same as the corresponding hypotheses H f i, iii, iv,

27 Abstract and Applied Analysis 27 ii we have f z, ζ lim uniformly for almost all z ζ ± ζ Remark 4.2. Hypothesis H f is a slight restricted version of hypothesis H f.notethatif ξ z, ζ f z, ζ ζ 2F z, ζ 4.19 and there exists a function ϑ L 1, such that ξ z, ζ ξ ( z, y ϑ z for almost all z, all 0 ζ y or y ζ 0, 4.20 then 4.20 and H f ii imply H f ii see Li and Yang 10. Al, note that hypotheses H f i, ii,and iii imply that there exists c > 0, such that f z, ζ ζ c ζ 2 0 for almost all z, all ζ R We consider the following auxiliary Dirichlet problem: Δu z β z u z c 1 u z q 2 u z c u z in, u 0. Q Here c 1 > 0andq 1, 2 are as in hypothesis H g iii and c > 0isasin Proposition 4.3. For every >0, problem Q has a unique nontrivial positive lution u int C, and by oddness, we have that u v int C is the unique negative lution of Q. Proof. Let k : R R be the Carathéodory function, defined by { 0 if ζ 0, k z, ζ c 1 ζ q 1 ĉ 1 c ζ if ζ> Clearly, we can always assume that c ĉ 1 see 4.21.Let K z, ζ ζ 0 k z, s ds, 4.23 and consider the C 1 -functional θ : H1 0 R, defined by θ u 1 ĉ1 σ u 2 2 u 2 2 K z, u z dz u H

28 28 Abstract and Applied Analysis Using 2.14 and 4.22, we have θ 1 u u 2 c ĉ 1 2ĉ 0 2 u 2 2 c 1 q u q q Since q<2, from 4.25, we infer that θ is coercive. Al, it is sequentially weakly lower semicontinuous recall that c ĉ 1. So, by the Weierstrass theorem, we can find u H 1 0, such that θ ( u inf u H 1 0 θ u As before see the proof of Proposition 3.7, sinceq<2, we have θ ( u < 0 θ 0, 4.27 hence u / 0. From 4.26, we have ( θ ( u 0, 4.28 A ( u ( β ĉ1 u N k ( u Acting on 4.29 with u becomes H1 0, weshowthatu 0, u / 0 see Then 4.29 A ( u ( β c u c 1 u q 1, 4.30 Δu z ( β z c u z 0 for almost all z, 4.31 thus Δu z ( β c u z for almost all z 4.32 see hypothesis H β, and hence u int C 4.33 see Vázquez 15 and Pucci and Serrin 14, page 120.

29 Abstract and Applied Analysis 29 We claim that this lution is the unique nontrivial positive lution of Q.Tothis end, let u, v int C be two positive lutions of Q. We have c 1 u q 1 c u( u 2 v 2 dz u ( ( c 1 u q 1 c u u v2 dz u ( Δu u v2 ( dz β u 2 v 2 dz u ( ( v 2 ( u, u dz β u 2 v 2 dz u R N ( u 2 2 u, 2v ( v2 v u u u dz β u 2 v 2 dz 2 R N u 2 2 2v u u, v R dz N v 2 ( u 2 u 2 dz β u 2 v 2 dz Interchanging the roles of u and v in the above argument, we al have c 1 v q 1 c v ( v 2 u 2 dz v v 2 2 2u v v, u u 2 ( RN dz v 2 v 2 dz β v 2 u 2 dz Adding 4.34 and 4.35, weobtain ( c1 u q 1 c u c 1v q 1 c v ( u 2 v 2 dz u u u v v v 2 v v 2 2 u u Since q<2, the function ζ c 1 ζ q 1 c ζ /ζ is strictly decreasing on 0,. Hence, from 4.36, we infer that u v. This proves the uniqueness of the nontrivial positive lution u int C of problem Q. The oddness of problem Q implies that v u int C is the unique nontrivial negative lution of Q. Using Proposition 4.3, we can show that problem P for 0, has extremal constant sign lutions; that is, it has a smallest nontrivial positive lution and a biggest nontrivial negative lution.

30 30 Abstract and Applied Analysis Proposition 4.4. If hypotheses H g, H f, H 0, and H β hold and 0,, then problem P has a smallest nontrivial positive lution u int C and a biggest nontrivial negative lution v int C. Proof. Let u be a nontrivial positive lution of P. From the proof of Proposition 3.6, we know that u int C. We have Δu z β z u z g z, u z f z, u z c 1 u z q 1 c u z, 4.37 for almost all z see hypothesis H g iii and We consider the following Carathéodory function: k z, ζ 0 if ζ 0, c 1 ζ q 1 ĉ 1 c ζ if 0 ζ u, c 1 u z q 1 ĉ 1 c u z if u<ζ Let K z, ζ ζ 0 k z, s ds, 4.39 and consider the C 1 -functional ξ : H1 0 R, defined by ξ u 1 ĉ1 σ u 2 2 u 2 2 K z, u z dz u H From 4.38 and 2.14, it is clear that ξ is coercive. Al ξ semicontinuous. Thus we can find w 0 H 1 0, such that ξ w 0 inf u H 1 0 ξ u. is sequentially weakly lower 4.41 As before, the presence of the concave term c 1 ζ q 1 implies that ξ w 0 < 0 ξ 0, 4.42 that is, w 0 / 0. From 4.41, we have ( ξ w0 0, 4.43 A w 0 ( β ĉ 1 w0 w N k

31 Abstract and Applied Analysis 31 On 4.44 we act with w 0 H1 0 and obtain σ ( w 0 ĉ1 w see 4.38, 1 w ĉ see 2.14, and hence w 0 0, w 0 / 0. Al, on 4.44 we act with w 0 u H 1 0. Then A w 0, w 0 u ( β ĉ1 w0 w 0 u dz k z, w 0 w 0 u dz ( c 1 u q 1 c u w 0 u dz A u, w 0 u β ( w 0 u dz ĉ w0 1 u see 4.38 and 4.37, σ ( w 0 u ĉ w0 1 u 2 0, thus 1 ĉ 0 w0 u see 2.14, and hence w 0 u So, we have proved that w 0 0, u {u H 1 0 :0 u z u z for almost all z } This means that 4.44 becomes A w 0 βw 0 c 1 w q 1 0 c w 0, 4.52

32 32 Abstract and Applied Analysis w 0 int C regularity theory of Struwe 13 and strong maximum principle due to Vázquez 15 and Pucci Serrin 14, page 120 and it lves problem Q.Thus w 0 u 4.53 see Proposition 4.3,and u u This shows that every nontrivial positive lution u of P satisfies u u Similarly, we show that every nontrivial negative lution v of problem P satisfies v v u Let S resp., S be the set of nontrivial positive resp., negative lutions of problem P.LetC S be a chain i.e., a totally ordered subset of S. From Dunford and Schwartz 18, page 336, we can find a sequence {u n } n 1 C, such that inf C inf n 1 u n Lemma 1.5 of Heikkilä and Lakshmikantham 19, page 15 implies that we can have the sequence {u n } n 1 C to be decreasing. Then we have A u n βu n N g u n N f u n, u u n u 1 n 1, 4.58 the sequnece {u n } n 1 H 1 0 is bounded Hence by passing to a suitable subsequence if necessary, we may assume that u n u weakly in H 1 0, u n u in L 2s and in L r So, passing to the limit as n in 4.58 and using 4.61, weobtain A u βu N g u N f u, u u, 4.62

33 Abstract and Applied Analysis 33 inf C u S Since C was an arbitrary chain, invoking the Kuratowski-Zorn lemma, we infer that S has a minimal element u int C.FromGasiński and Papageorgiou 17, Lemma 4.2, page 5763, we know that S is downward directed i.e., if u 1,u 2 S ; then we can find u S, such that u u 1 and u u 2. So, it follows that u is the smallest nontrivial positive lution of problem P. Similarly, we introduce the biggest nontrivial negative lution v int C of problem P.NotethatS is upward directed i.e., if v 1,v 2 S, then we can find v S, such that v 1 v and v 2 v;seegasiński and Papageorgiou 17, Lemma 4.3, page Now that we have these extremal constant sign lutions, we can produce a nodal lution of problem P with 0,. Theorem 4.5. If hypotheses H g, H f, H 0, and H β hold and 0,, then problem P has at least five nontrivial smooth lutions: u 0, û int C, û u 0 int C, v 0, v int C, v 0 v int C ( y 0 C 1 0 nodal with u 0 y 0 int C, y 0 v 0 int C Moreover, problem P has a smallest nontrivial positive lution and a biggest negative lution. Proof. The existence of extremal nontrivial constant sing lutions is guaranteed by Proposition 4.4. Letu int C and v int C be these two extremal lutions. We introduce the following truncation perturbation of the reaction of problem P : ( ( g z, v z f z, v z ĉ 1 v z γ z, ζ g z, ζ f z, ζ ĉ 1 ζ g ( z, u z f ( z, u z ĉ 1 u z if ζ<v z, if v z ζ u z, if u z <ζ This is a Carathéodory function. We set Γ z, ζ ζ 0 γ z, s ds 4.66 and consider the C 1 -functional χ : H 1 0 R, defined by χ u 1 ĉ1 σ u 2 2 u 2 2 Γ z, u z dz u H

34 34 Abstract and Applied Analysis Al, we introduce γ ± z, ζ γ ( z, ±ζ ± ζ, Γ ± z, ζ 0 γ ± z, s ds 4.68 and consider the C 1 -functionals χ ± : H1 0 R, defined by χ ± u 1 ĉ1 σ u 2 2 u 2 2 Γ ± z, u z dz u H As in the proof of Theorem 4.1, we show that K χ [ ] ] [ ] v,u, K χ [0,u, K χ v, see The extremality of the lutions v and u implies that K χ [ ] } { } v,u, K χ {0,u, K χ v, Claim 4. Solutions u and v are both local minimizers of χ. Evidently χ is coercive see 4.65 and sequentially weakly lower semicontinuous. So, we can find u H 1 0, such that χ u inf u H 1 0 χ u As before see the proof of Proposition 3.7, the presence of the concave term implies that χ u < 0 χ 0, 4.73 hence u / 0, and u u see Since χ C χ C, 4.74 it follows that u u int C is local C 1 0 -minimizers of χ ; hence by Brézis and Nirenberg 16, it is al local H 1 0 -minimizers of χ. Similarly this is for v using this time the functional χ. This proves the claim. Without any loss of generality, we may assume that χ (v χ ( u. 4.75

35 Abstract and Applied Analysis 35 The analysis is similar, if the opposite inequality holds. Because of the claim, we can find ϱ 0, 1 small, such that ( { } χ (v χ u < inf χ u : u u ϱ η 4.76 see Gasiński and Papageorgiou 17, proof of Theorem 3.4. Since χ is coercive see 4.65, it satisfies the Cerami condition. This fact and 4.76 permit the use of the mountain pass theorem see Theorem 2.1. So, we can find y 0 K χ v,u see 4.71, such that η χ ( y0, 4.77 y 0 / { } v,u 4.78 see Since y 0 is a critical point of χ of mountain pass type, we have C 1 ( χ,y 0 / see e.g., Chang 20. On the other hand, hypothesis H f iii implies that we can find ξ 1 > 0, such that f z, ζ ζ μf z, ζ ξ 1 ζ 2 for almost all z, all ζ δ 1, 4.80 for me δ 1 δ 0. This combined with hypothesis H g iii implies that μg z, ζ μf z, ζ g z, ζ ζ f z, ζ ζ > 0 for almost all z, all ζ δ 2, 4.81 for me δ 2 δ 1 and ess sup G,δ 2 F,δ 2 > Hence invoking Proposition 2.1 of Jiu and Su 21, we infer that C k ( χ, 0 0 k Combining 4.79 and 4.83, we have that y 0 / 0. Since y 0 v,u, the extremality of v and u implies that y 0 must be a nodal lution of problem P, and the regularity theory see Struwe 13 implies that y 0 C 1 0.

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