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Constant sign and nodal solutions for

nonhomogeneous Robin boundary value problems with asymmetric reactions

Antonio Iannizzotto

B1

, Monica Marras

1

and Nikolaos S. Papageorgiou

2

1Department of Mathematics and Computer Science, University of Cagliari, Viale L. Merello 92, 09123 Cagliari, Italy

2Department of Mathematics, National Technical University, Zografou Campus, Athens 15780, Greece

Received 28 January 2018, appeared 25 August 2018 Communicated by Patrizia Pucci

Abstract. We study a nonlinear, nonhomogeneous elliptic equation with an asymmetric reaction term depending on a positive parameter, coupled with Robin boundary con- ditions. Under appropriate hypotheses on both the leading differential operator and the reaction, we prove that, if the parameter is small enough, the problem admits at least four nontrivial solutions: two of such solutions are positive, one is negative, and one is sign-changing. Our approach is variational, based on critical point theory, Morse theory, and truncation techniques.

Keywords: non-linear elliptic equations, asymmetric reactions, variational methods.

2010 Mathematics Subject Classification: 35J20, 35J60, 58E05.

1 Introduction

We study the following nonlinear, nonhomogeneous Robin problem:

−diva(∇u) +ξ(x)|u|p2u =λg(x,u) + f(x,u) in Ω

∂u

∂na +β(x)|u|p2u=0 on ∂Ω. (1.1)

HereΩ⊂RN(N>1) is a bounded domain with aC2-boundary∂Ω,p>1, anda:RNRN is a continuous, monotone mapping (hence maximal monotone too) which satisfies certain growth and regularity conditions (see hypotheses Ha below). These conditions are mild enough to include in our framework many non-linear operators of interest, such as the p- Laplacian, the (p,q)-Laplacian, and the generalized mean curvature operator. The potential functionξ ∈ L()is indefinite (i.e., sign-changing, see hypothesisHξ).

BCorresponding author. Email: antonio.iannizzotto@unica.it

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On the right-hand side, λ > 0 is a parameter and g,f : Ω×RR are Carathéodory functions. We assume that for a.a. x ∈ the mapping g(x,·) is strictly (p−1)-sublinear at ± (concave nonlinearity), while f(x,·) exhibits an asymmetric behavior, being (p−1)- superlinear at+and asymptotically(p−1)-linear at−(see hypothesesHg,Hf below). So, in the positive semiaxis we have a competition phenomenon between a concave and a convex nonlinearity, while in the negative semiaxis and in the particular case of the p-Laplacian the equation may be resonant with respect to the first eigenvalue.

In the boundary condition,∂u/∂na denotes the generalized normal derivative correspond- ing to the mappinga, namely the extension of

u 7→ ha(∇u),ni, u∈C1()

toW1,p()(ndenotes the outward unit normal to∂Ω). The boundary coefficientβ∈C0,α(∂Ω) is non-negative, and the special caseβ=0 corresponds to the Neumann problem (see hypoth- esisHβ below).

In this paper, using variational methods based on the critical point theory, together with suitable truncation/perturbation techniques and Morse theory, we prove that, forλ>0 small enough, problem (1.1) has at least four nontrivial solutions: two positive, one negative, and one nodal (see Theorem 2.4 below and the ensuing discussion for a short account on our method).

Recently, elliptic boundary value problems with asymmetric reactions were studied in [15,28,29] (semilinear Dirichlet problems with zero potential), [18] (semilinear Neumann problem with indefinite potential), [3,14,23,24] (semilinear Robin problems with indefinite potential). For nonlinear elliptic equations we mention [13,16] (Dirichlet problems driven by thep-Laplacian), [20,26] (Dirichlet problems driven by the(p, 2)-Laplacian).

Compared with the existing literature, our result is novel in a twofold sense: unlike most of the aforementioned works, our result proves existence of four nontrivial solutions with precise sign information; and it holds for a very general problem, incorporating Robin and Neumann boundary conditions and several nonlinear leading differential operators as special cases (the only exception is represented by [16], which provides four solutions but only for Dirichlet conditions and the p-Laplace operator).

For the sake of completeness, we mention some more results on nonlinear Robin problems (with symmetric reactions) contained in [19,21,25].

The paper has the following structure: in Section2we introduce our hypotheses and main result, and we also establish some preliminary results and notations; in Section3we deal with constant sign solutions; and in Section4we investigate extremal constant sign solutions and nodal solutions.

2 Hypotheses and main result

We start this section by introducing and commenting the precise hypotheses on all features of problem (1.1). We begin with the mapping a:

Ha a : RNRN is defined by a(y) =a0(|y|)yfor all y ∈ RN with a0 : R+R+, and we set for allt>0

H0(t) =

Z t

0 a0(τ)τdτ and for ally∈ RN H(y) =H0(|y|). Moreover:

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(i) a0 ∈ C1(0,+), a0(t)>0 for all t >0, t 7→ a0(t)t is strictly increasing on(0,+), and

tlim0+a0(t)t=0, lim

t0+

a00(t)t a0(t) >−1;

(ii) there existsθ ∈C1(0,+)s.t. for allt >0 c06 θ

0(t)t

θ(t) 6c1, c2tp16θ(t)6c3(tσ1+tp1) (c0,c1,c2,c3 >0, 16σ< p), and for ally∈RN\ {0}

|∇a(y)|6c4

θ(|y|)

|y| (c4 >0); (iii) for ally,z∈RN,y6=0

h∇a(y)z,zi> θ(|y|)|z|2

|y| ; (iv) there existsr∈ (1,p]s.t.t 7→ H0(t1r)is convex,

lim sup

t0+

rH0(t)

tr 6c5 (c5 >0), and for allt>0

pH0(t)−a0(t)t2>−c6 (c6 >0).

Hypotheses Ha (i)–(iii) are dictated by the nonlinear regularity theory of [12] and the nonlinear maximum principle of [27]. Hypothesis Ha (iv) serves the needs of our problem but is general enough to include several cases of interest (see Example2.2below). As a whole, Ha implies that H0 is strictly convex and increasing onR+, and His convex with H(0) =0,

∇H(0) =0, and ∇H(y) = a(y)for ally ∈ RN\ {0}, i.e., His the primitive of a. This, along with convexity, clearly implies for all y∈RN

H(y)6ha(y),yi. (2.1)

HypothesesHa (i)–(iii)and (2.1) lead to the following properties ofaand H.

Lemma 2.1. IfHa (i)–(iii)hold, then

(i) a:RNRN is continuous and monotone (hence maximal monotone);

(ii) |a(y)|6c7(1+|y|p1)for all y∈ RN (c7 >0);

(iii) ha(y),yi> cp2|y|1p for all yRN;

(iv) pc(2p|y|1p) 6 H(y)6c8(1+|y|p)for all y∈RN (c8 >0).

In what follows we shall denoteA :W1,p()→W1,p() the nonlinear differential oper- ator defined for allu,v ∈W1,p()by

hA(u),vi=

Z

ha(∇u),∇vidx,

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which is well defined by virtue ofHa (ii). Such operator enjoys the(S)+-property, i.e., when- ever(un)is a sequence inW1,p()s.t. un*uinW1,p()and

lim sup

n

hA(un),un−ui60,

thenun →uinW1,p()(see [17, p. 405]). Here follow some examples.

Example 2.2. The following mapsa:RNRN satisfyHa: (a) a(y) =|y|p2y, corresponding to the p-Laplace operator

pu=div(|∇u|p2∇u);

(b) a(y) =|y|p2y+|y|q2y(1< q< p< +), corresponding to the(p,q)-Laplace operator

pu+qu=div |∇u|p2∇u+|∇u|q2∇u .

Such operators arise in problems of mathematical physics, see [2] (reaction-diffusion equa- tions), [4] (elementary particles), [30] (plasma physics). Further:

(c) a(y) = (1+|y|2)p22y, corresponding to the generalized p-mean curvature operator div (1+|∇u|2)p22∇u

;

(d) a(y) = 2 ln(1+|y|p) + (1+|y|p)1y, corresponding to the operator div

2 ln(1+|∇u|p)∇u+ ∇u 1+|∇u|p

;

(e) a(y) =|y|p2y+|y|p2(1+|y|p)1y, corresponding to the operator

pu+div

|∇u|p2∇u 1+|∇u|p

.

Such operators arise in problems of nonlinear elasticity [7] and plasticity.

The other ingredients of (1.1) are subject to the following hypotheses:

Hξ ξ ∈ L();

Hβ β∈C0,α(∂Ω)for someα∈ (0, 1),β(x)>0 for allx∂Ω.

We note that the potential ξ may change sign, and that for β = 0 we recover the Neumann problem. Finally we introduce our hypotheses on the reactions, starting withg:

Hg g:Ω×RRis a Carathéodory function, for all(x,t)∈×Rwe set G(x,t) =

Z t

0 g(x,τ)dτ.

Moreover:

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(i) for allρ >0 there existsaρ ∈ L()+s.t. for a.a.x ∈Ω, all|t|6ρ

|g(x,t)|6aρ(x); (ii) lim|t|→+ g(x,t)

|t|p2t =0 uniformly for a.a. x∈Ω;

(iii) there existsq∈(1,r)s.t. for a.a. x∈ Ω, allt∈R

g(x,t)t >c9|t|q (c9>0); (iv) lim supt0|gt(|qx,t2)t 6c10 uniformly for a.a. x∈(c10 >0);

(v) there existsδ0 >0 s.t. for a.a.x ∈Ω, all|t|6δ0

g(x,t)t6qG(x,t).

We set ξ0 = (p−1)ξ/c2, β0 = (p−1)β/c2 (c2 > 0 as in Ha (ii)) and we denote by ˆλ1 = λˆ1(p,ξ0,β0)>0 the first eigenvalue of the auxiliary problem





pu+ξ0(x)|u|p2u=λ|u|p2u in Ω

∂u

∂np +β0(x)|u|p2u=0 on ∂Ω. (2.2) where

∂u

∂np =h|∇u|p2∇u,ni

(nbeing as usual the outward unit normal toΩ). Now we consider the asymmetric term f: Hf f :Ω×RRis a Carathéodory function, for all(x,t)∈×Rwe set

F(x,t) =

Z t

0 f(x,τ)dτ.

Moreover:

(i) for allρ >0 there existsbρ ∈ L()+ s.t. for a.a.x ∈Ω, all|t|6ρ

|f(x,t)|6bρ(x); (ii) limt→+ f(x,t)

tp∗ −1 =0 uniformly for a.a. x∈ ; (iii) limt→+ f(x,t)

tp1 = +uniformly for a.a.x∈ Ω;

(iv) f(x,t)6c11(tp1+tr1)−c12tp1for a.a. x∈Ω, allt >0 (c11,c12>0);

(v) uniformly for a.a.x∈

−c136lim inf

t→−

f(x,t)

|t|p2t 6lim sup

t→−

f(x,t)

|t|p2t 6 c2λˆ1

p−1 (c13>0); (vi) limt0 f(x,t)

|t|p2t =0 uniformly for a.a. x∈Ω;

(vii) there existsδ1 >0 s.t. for a.a.x ∈Ω, all|t|6δ1

f(x,t)t >0.

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Finally, we set for allλ>0 and all(x,t)∈×R eλ(x,t) =λg(x,t)t+ f(x,t)t−p

λG(x,t) +F(x,t) and we assume the following condition:

He for allλ>0

(i) there existsηλ ∈ L1()+s.t. for a.a. x∈ Ω, all 06t6t0 eλ(x,t)6eλ(x,t0) +ηλ(x); (ii) lim

t→−eλ(x,t) = +uniformly for a.a.x∈Ω.

We will writeHto mean all hypothesesHa,Hξ,Hβ,Hg,Hf, andHe.

By Hg (ii) g(x,·) is strictly (p−1)-sublinear at ±, so it gives a ’concave’ contribution to the reaction of (1.1). Hypotheses Hf (iii), (v) imply that f(x,·) has an asymmetric be- havior at ±∞. More precisely, Hf (iii) means that f(x,·) is strictly (p−1)-superlinear at +, so onR+ it represents a ’convex’ contribution to the reaction, leading to a competition phenomenon (concave-convex nonlinearities). We point out that the (p−1)-superlinearity of f(x,·) is not coupled with the usual Ambrosetti–Rabinowitz (AR) condition. Instead we use the less restrictive quasimonotonicity conditionHe (i), which includes in our framework (p−1)-superlinear reactions with a slower growth at+∞, that fail to satisfy(AR). Note that He (i)holds whenever we can find ρ>0 s.t. the mapping

t7→ λg(x,t) + f(x,t) tp1

is nondecreasing in [ρ,+) for a.a. x ∈ [11]. On the negative semiaxis R, by Hf (v) the mapping f(x,·) is asymptotically (p−1)-linear at −∞, and in the special case of the p-Laplacian (Example 2.2 (a) with c2 = p−1) resonance with the principal eigenvalue is allowed. Resonance occurs from the left, so byHe(ii)problem (1.1) is coercive on the negative direction, which permits the use of the direct method of the calculus of variations. Finally we remark that byHf (ii) f(x,·)does not satisfy the usual subcritical growth. Instead we have

’almost-critical’ growth, namely for allε >0 we can findcε >0 s.t. for a.a. x∈ Ω, allt∈R

|f(x,t)|6ε|t|p1+cε.

This kind of growth is a source of technical difficulties, sinceW1,p()is not compactly em- bedded intoLp(). We shall overcome such difficulties by using Vitali’s theorem.

Example 2.3. The following functions (of the typeλg+ f,λ > 0) satisfy hypothesesHg, Hf, andHe:

(a) t7→ λ|t|q2t+

(λˆ1|t|p2t ift 60

ts1+tr1 ift >0 (q<r < p<s< p); (b) t7→ λ|t|q2t+

(λˆ1|t|p2t ift 60

tp1ln(1+t)−tr1 ift >0 (q<r < p);

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(c) t 7→λ|t|q2t+





λˆ1|t|p2t ift60 tp1

ln(1+tp)− pt

p+p1

p(1+tp)ln(1+tp)2 −tr1 ift >0 (q<r< p); Note that (a)satisfies (AR)while(b)does not, and that (c)has an almost critical growth at +∞.

Our main result is the following.

Theorem 2.4. IfHhold, then there existsλ >0s.t. for allλ∈ (0,λ)problem(1.1)admits at least four nontrivial solutions u+,v+,u, ˜u∈ C1()with u+, v+positive in Ω, unegative inΩ, andu˜ nodal.

Our approach is entirely variational, based on critical point theory. For allλ>0, we define an energy functional ϕλ for problem (1.1), which always admits 0 as a critical point. Then we introduce two truncated/perturbed functionals ˆϕ±λ, by replacing the indefinite potential ξ with a positive one and truncating the reactions at 0: thus, nontrivial critical points of ˆϕ+λ (resp. ˆϕλ) are positive (resp. negative) solutions of (1.1). Then we study separately the critical sets of such functionals: forλ>0 small enough, ˆϕ+λ turns out to admit at least two nontrivial critical points, namely, a local minimizer v+ and another critical point u+ produced by the mountain pass theorem; while ˆϕλ contributes a negative global minimizeru.

Then we go further, proving existence of a smallest positive solution u+ and a biggest negative solution u of (1.1), and we truncate again the reactions. The resulting functional

˜

ϕλ selects solutions lying in the interval [u, u+], and admits a critical point ˜u of mountain pass type. By computing the critical groups at 0 and at ˜u, we see that ˜u6= 0, hence it must be nodal.

2.1 Notation

We establish some notation: we set R+ = [0,+), R = (−∞, 0]; c0,c1, . . . denote positive constants; for allt ∈Rwe set

t±=max{0,±t}.

In any Banach space X, * denotes weak convergence and → strong convergence; if X is a function space on the domainD, then we denote the positive order cone by

X+={u∈ X: u(x)>0 for a.a. x∈ D}.

We will say that a functional ϕ ∈ C1(X)satisfies the Cerami condition (C), if any sequence (un)s.t. (ϕ(un))is bounded inR and (1+kunk)ϕ0(un) → 0 in X, admits a (strongly) con- vergent subsequence. We will denote the set of critical points of ϕby

K(ϕ) ={u∈X: ϕ0(u) =0}.

We also recall the basic notion from Morse theory: let ϕ∈ C1(X)andu∈K(ϕ)be an isolated critical point, namely there exists a neighborhood U ⊂ X of u s.t. K(ϕ)∩U = {u}, and ϕ(u) =c. Then, for allk∈N, thek-th critical group of ϕatuis defined by

Ck(ϕ,u) =Hk {v∈U: ϕ(v)6c},{v ∈U: ϕ(v)6c, v6=u}, where Hk,·)denotes thek-th singular homology group of a topological pair.

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We shall use the function spaces(W1,p(),k · k)and(C1(),k · kC1()), endowed with the usual norms. Brackets h·,·i denote both the inner product of RN and the duality between W1,p() andW1,p(), with no possible confusing arising. We shall also use the Lebesgue spaces (Lν(),k · kν) for all ν ∈ [1,+], and the trace space (Lp(∂Ω),k · kLp(∂Ω)) (any u ∈ W1,p()will be identified with its trace on ∂Ω). We set

D+ ={u∈C1(): u(x)>0 for allx ∈}, noting thatD+⊆int(C1()+).

3 Constant sign solutions

For allλ>0, u∈W1,p()we set ϕλ(u) =

Z

H(∇u)dx+ 1 p

Z

ξ(x)|u|pdx+ 1 p

Z

∂Ωβ(x)|u|pdσ−

Z

λG(x,u) +F(x,u)dx

(the integral onΩis computed with respect to the(N−1)-dimensional Hausdorff measure).

By Lemma2.1 (iv),Hξ,Hβ,Hg (i) (ii),Hf (i) (ii) (v), we have ϕλ ∈ C1(W1,p()). Moreover, ϕλ is the energy functional for problem (1.1). Indeed, for allu∈K(ϕλ),v∈W1,p()we have

hA(u),vi+

Z

ξ(x)|u|p2uv dx+

Z

∂Ωβ(x)|u|p2uv dσ=

Z

[λg(x,u) + f(x,u)]v dx, (3.1) i.e.,u is a (weak) solution of (1.1). Besides, let

µ>max

1, p−1 c2

kξk (3.2)

and for all(x,t)∈×Rset

kλ(x,t) =λg(x,t) + f(x,t) +µ|t|p2t, k±λ(x,t) =kλ(x,±t±),

and the primitives

K(±)λ (x,t) =

Z t

0 k(±)λ (x,τ)dτ.

Now we define two truncated/perturbed functionals by setting for allu∈W1,p() ˆ

ϕ±λ(u) =

Z

H(∇u)dx+ 1 p

Z

(ξ(x) +µ)|u|pdx+ 1 p

Z

β(x)|u|pdσ−

Z

K±λ(x,u)dx (note thatξ+µis positive by (3.2)). We shall study separately the properties of ˆϕ+λ and ˆϕλ, which are different by the asymmetry of f.

Lemma 3.1. IfHhold, then for allλ>0 ˆϕ+λ ∈C1(W1,p())satisfies(C).

Proof. Clearlyk+λ :Ω×RRis a Carathéodory function, with a growth defined byHg (ii), Hf (ii) (v), so ˆϕ+λ ∈C1(W1,p()).

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Let (un) be a sequence in W1,p() s.t. |ϕˆ+λ(un)| 6 c14 for all n ∈ N (c14 > 0) and (1+ kunk)(ϕˆ+λ)0(un) → 0 inW1,p(). We can find a sequence (εn)in R s.t. εn → 0+ and for all n∈N,v∈W1,p()

hA(un),vi+

Z

(ξ(x) +µ)|un|p2unv dx+

Z

∂Ωβ(x)|un|p2unv dσ−

Z

k+λ(x,un)v dx 6 εnkvk

1+kunk. (3.3)

Choosing v=−un in (3.3) and using Lemma2.1(iii)we get for alln∈ N c2

p−1k∇unkpp+

Z

(ξ(x) +µ)(un)pdx6εn.

Passing to the limit we see that un → 0 inW1,p(). Now we deal with u+n. By definition of K+λ we have for all n∈N

pc14> p

Z

H(∇un)dx+

Z

(ξ(x) +µ)|un|pdx+

Z

∂Ωβ(x)|un|pdσ−p Z

K+λ(x,un)dx

> p

Z

H(∇u+n)dx+

Z

ξ(x)(u+n)pdx+

Z

∂Ωβ(x)(u+n)pdσ−p Z

λG(x,u+n) +F(x,u+n)dx, while (3.3) with v=−u+n yields

−hA(u+n),u+ni −

Z

ξ(x)(u+n)pdx−

Z

∂Ωβ(x)(u+n)pdσ+

Z

[λg(x,u+n) + f(x,u+n)]u+n dx6εn. Adding up we get

Z

pH(∇u+n)− ha(∇u+n),∇u+nidx+

Z

eλ(x,u+n)dx6c15 (c15>0), which byHa (iv)implies

Z

eλ(x,u+n)dx6c16 (c16>0). (3.4) We claim that (u+n)is bounded in W1,p(). Arguing by contradiction, we may assume that (passing if necessary to a subsequence) ku+nk → +∞. Then we set for all n ∈ N wn = u+nku+nk1, so wn∈W1,p()withkwnk=1. Passing again to a subsequence we havewn *w in W1,p() and wn → w both in Lp() and in Lp(Ω) (due to the compact embeddings W1,p(),→ Lp(), Lp(∂Ω)). Clearlyw∈W1,p()+. Two cases may occur:

(a) First we assumew6=0. Let

+= {x∈: w(x)>0},

then |+|> 0 and for a.a. x ∈ + we haveu+n(x)→ +∞. ByHf (iii)we have for a.a.

x∈+

F(x,u+n(x))

ku+nkp = F(x,u+n(x))

u+n(x)p wn(x)p →+ By Fatou’s lemma we have

limn

Z

+

F(x,u+n)

ku+nkp dx= +∞.

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ByHf (i) (iii)we have for a.a.x∈ Ω, allt>0

F(x,t)>tp−c17 (c17>0), so we have

Z

\+

F(x,u+n) ku+nkp dx>

Z

\+

wnpdx− c17|| ku+nkp, and the latter is bounded from below. Summarizing,

limn

Z

F(x,u+n)

ku+nkp dx= +. (3.5)

Besides,Hg (ii)implies, as above, limn

Z

+

G(x,u+n)

ku+nkp dx=0.

ByHg (i) (ii), for anyε>0 we can findc18=c18(ε)>0 s.t. for a.a. x∈Ω, allt>0 G(x,t)6 ε

ptp+c18. So we have

lim sup

n

Z

\+

G(x,u+n)

ku+nkp dx6lim sup

n

Z

\+

ε

pwnp+ c18 ku+nkp

dx6 ε pkwkpp. sinceε>0 is arbitrary, adding the two integrals we get

limn

Z

G(x,u+n)

ku+nkp dx =0. (3.6)

Now (3.5), (3.6) imply limn

Z

λG(x,u+n) +F(x,u+n)

ku+nkp dx = +∞.

But again from boundedness of (ϕˆ+λ(un)), and recalling that un → 0 in W1,p(), we have for alln∈N

Z

H(∇u+n)dx+ 1 p Z

ξ(x)(u+n)pdx+1 p Z

∂Ωβ(x)(u+n)pdσ−

Z

λG(x,u+n) +F(x,u+n)dx

>−c19 (c19>0),

which, along with Lemma2.1(iv), implies Z

λG(x,u+n) +F(x,u+n) ku+nkp dx 6 c19

ku+nkp +

Z

c8(1+|∇u+n|p)

ku+nkp dx+ 1 p

Z

ξ(x)wnpdx+ 1 p

Z

∂Ωβ(x)wnpdσ 6c20(1+kwnkp) (c20>0),

and the latter is bounded from above. Thus we reach a contradiction.

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(b) Now we assume w = 0. Fix M > 0 and set ˆwn = (Mp)1pwn for all n ∈ N, so ˆwn * 0 inW1,p()and ˆwn0 in Lp()and Lp(Ω). ByHg (i) (ii)we have for a.a.x∈ Ω, all t ∈R

|G(x,t)|6c21(1+|t|p) (c21 >0). So we get

limn

Z

G(x, ˆwn)dx=0. (3.7)

Clearly(wˆn)is bounded inLp(), so set K0 =sup

nN

kwˆnkpp.

ByHf (i) (ii), for anyε>0 we can findc21=c22(ε)>0 s.t. for a.a. x∈Ω, allt∈R

|F(x,t)|6 ε

2K0|t|p+c22.

So, the sequence (F(·, ˆwn)) is bounded in L1(). Furthermore, for any measurable set B⊂with |B|6ε(2c22)1 we have for alln∈N

Z

B

|F(x, ˆwn)|dx6 ε

2K0kwˆnkpp+c22|B|6ε.

So the sequence(F(·, ˆwn))is uniformly integrable inΩ(see [8, Problem 1.6]). Passing to a subsequence, we have F(x, ˆwn(x)) →0 asn → ∞, for a.a. x ∈ Ω. By Vitali’s theorem [8, p. 5] we have

limn

Z

F(x, ˆwn)dx=0. (3.8)

Sinceku+nk →+∞, forn∈Nbig enough we have 0< (Mp)1p

ku+nk 61. (3.9)

Let ˆψ+λ ∈C1(W1,p())be defined for allu ∈W1,p()by ψˆ+λ(u) = c2

p(p−1)k∇ukpp+ 1 p

Z

(ξ(x) +µ)|u|pdx+ 1 p

Z

∂Ωβ(x)|u|pdσ−

Z

K+λ(x,u)dx.

For alln∈Nthere existstn∈[0, 1]s.t.

ψˆλ+(tnu+n) = max

t∈[0,1]

ψˆ+λ(tu+n). In particular, by (3.9) we have forn∈Nbig enough

ψˆ+λ(tnu+n)>ψˆ+λ(wˆn)

> c2M

p−1k∇wnkpp+M Z

(ξ(x) +µ)wpndx−

Z

λG(x, ˆwn) +F(x, ˆwn) +Mµwnp dx

>M(c23µkwnkpp)−

Z

λG(x, ˆwn) +F(x, ˆwn)dx (c23>0)

(recall thatµ> kξk andkwnk=1). Now by (3.7), (3.8) we have forn∈Neven bigger ψˆ+λ(tnu+n)> Mc24 (c24 >0)

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which by arbitrarity of M > 0 implies ˆψ+λ(tnu+n)→ +as n → ∞. By Lemma 2.1 (iv) we have ˆϕ+λ(u)>ψˆ+λ(u)for allu ∈W1,p(), hence the sequence (ψˆλ+(u+n))is bounded from above. Besides, clearly ˆψλ+(0) = 0. So, for all n ∈ N big enough we must have tn∈(0, 1). By definition oftn, then,

d

dt ψˆλ+(tu+n)

t=tn =h(ψˆ+λ)0(tnu+n),u+ni=0.

Multiplying bytnwe get c2

p−1k∇(tnu+n)kpp+

Z

ξ(x)(tnu+n)pdx+

Z

β(x)(tnu+n)p

=

Z

[λg(x,tnu+n) + f(x,tnu+n)]tnu+n dx.

ByHe (i), tn <1, and (3.4) we have Z

eλ(x,tnu+n)dx6

Z

eλ(x,u+n)dx+kηλk1< c25 (c25 >0), which implies

Z

[λg(x,tnu+n) + f(x,tnu+n)]tnu+n dx6 p

Z

λG(x,tnu+n) +F(x,tnu+n)dx+c25. Thus, for alln∈Nbig enough we have

pψˆ+λ(tnu+n) =

Z

[λg(x,tnu+n) + f(x,tnu+n)]tnu+n dx−p Z

λG(x,tnu+n) +F(x,tnu+n)dx 6c25,

a contradiction.

By the claim above and un → 0, we see that (un) is bounded in W1,p(). Passing to a subsequence, we may assumeun * u inW1,p() andun → uin Lp()and Lp(∂Ω). ByHg (i) (ii)we have for a.a.x ∈, allt ∈R

|g(x,t)|6c26(1+|t|p1) (c26 >0), hence by Hölder’s inequality

limn

Z

g(x,u+n)(un−u)dx=0. (3.10) Besides,(un)is bounded inLp(), so we set

K1 =sup

nN

kunkp+kukp.

ByHf (i) (ii), for anyε>0 we can findc27= c27(ε)>0 s.t. for a.a.x∈ Ω, allt>0

|f(x,t)|6 ε

3K1ptp1+c27.

Passing if necessary to a subsequence, we have f(x,u+n(x))(un(x)−u(x))→0 as n→ ∞, for a.a.x ∈Ω. Moreover, for any measurableB⊂ with

|B|6 ε

6K1c27 (p)0

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we have by Hölder’s inequality

Z

B f(x,u+n)(un−u)dx 6 ε

3K1p Z

B

(u+n)p1|un−u|dx+c27kun−uk1

6 ε

3K1pku+nkpp1kun−ukp+c27|B|(p1)0kun−ukp 6

3 + ε 3 =ε.

So, the sequence (f(·,u+n)(un−u))is uniformly integrable inΩ. By Vitali’s theorem we get limn

Z

f(x,u+n)(un−u)dx=0. (3.11) If we choose v=un−uin (3.3), pass to the limit asn→∞, and use (3.10) and (3.11), we now get

lim sup

n

hA(un),un−ui=0.

By the(S)+-property of Awe haveun→u inW1,p(), which concludes the proof.

The following lemmas show that forλ >0 small enough ˆϕ+λ exhibits the ‘mountain pass’

geometry.

Lemma 3.2. IfHhold, then there existsλ >0s.t. for allλ∈(0,λ)there existsρλ >0s.t.

kuinfk=ρλ

ˆ

ϕ+λ(u) =mˆ+λ >0.

Proof. ByHg(ii) (iv), for allε>0 we can findc28 =c28(ε)>0 s.t. for a.a.x ∈Ω, allt >0 G(x,t)6 ε

ptp+c28tq

(recall thatq< p). ByHf (iv)we have as well for a.a.x∈Ω, allt>0 F(x,t)6 c11

ptp+ c11

r trc12 p tp.

Recalling that q<r < p< p and choosing ε<c12/λwe get for a.a.x∈ Ω, allt>0 λG(x,t) +F(x,t)6λc28tq+c29tpc30

p tp,

where, taking c28,c29 > 0 big enough, we may assume c30 > kξk. So, recalling Hβ and µ>kξk, for allu∈W1,p()we have

ˆ

ϕ+λ(u)> c2

p(p−1)k∇ukpp+ 1 p

Z

(ξ(x) +µ)(u)pdx + c2

p(p−1)k∇u+kpp+ 1 p

Z

ξ(x)(u+)pdx−

Z

λG(x,u+) +F(x,u+)dx

>c31kukp+ c2

p(p−1)k∇u+kpp+ 1 p

Z

(ξ(x) +c30)(u+)pdx−λc28ku+kqq−c29ku+kpp

>c31kukp+c32ku+kpλc33kukq−c34kukp

>c35kukpλc33kukq−c34kukp (c31, . . . ,c35>0).

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Summarizing, we have

ˆ

ϕ+λ(u)>(c35−jλ(kuk))kukp, (3.12) where we have set for alltρ >0

jλ(ρ) =λc33ρqp+c34ρp

p. Sinceq< p< p, we have for allλ>0

lim

ρ0+jλ(ρ) = lim

ρ→+jλ(ρ) = +∞, so there existsρλ >0 s.t.

jλ(ρλ) = inf

ρ>0jλ(ρ). In particular we have

0= j0λ(ρλ) =λc33(q−p)ρqλp1+c34(p−p)ρp

p1

λ ,

which yields

ρλ =

λc33(p−q) c34(p−p)

p∗ −1q . We are interested in the mappingλ7→ jλ(ρλ), which amounts to

jλ(ρλ) =c36λ

p∗ −p

p∗ −q (c36>0 independent ofλ),

and the latter tends to 0 as λ → 0+. So there exists λ > 0 s.t. for all λ ∈ (0,λ) we have jλ(ρλ)<c35. Thus, by (3.12) we have for allu ∈W1,p()with kuk=ρλ

ˆ

ϕ+λ(u)>(c35−jλ(ρλ))ρpλ =: ˆm+λ >0, which concludes the proof.

Let ˆu1 = uˆ1(p,ξ0,β0) ∈ D+ be the positive, Lp()-normalized first eigenfunction of the eigenvalue problem (2.2).

Lemma 3.3. IfHhold, then for allλ>0

t→+limϕˆ+λ(tuˆ1) =−∞.

Proof. Fix λ > 0. By Hg (ii), Hf (iii), for all η > 0 we can find c37 = c37(η) > 0 s.t. for a.a.

x∈ Ω, allt>c37

λG(x,t) +F(x,t)> ηµ p tp.

Since ˆu1 ∈D+, for allt>0 big enough we havetuˆ1(x)>c37for allx∈Ω. Then by Lemma2.1 (iv)we have

ˆ ϕ+λ(tuˆ1)

6c8

Z

(1+tp|∇uˆ1|p)dx+ 1 p

Z

(ξ(x) +µ)(tuˆ1)pdx+ 1 p

Z

∂Ωβ(x)(tuˆ1)pdσ−

Z

η

p(tuˆ1)pdx 6c38+

c39kuˆ1kpη pkuˆ1kpp

tp (c38,c39 >0).

Choosing η > 0 big enough, the latter tends to − as soon as t → +∞, concluding the proof.

Hivatkozások

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