existence of multiple solutions with precise sign ... · annali di matematica (2009) 188:679–719...

Annali di Matematica (2009) 188:679–719DOI 10.1007/s10231-009-0096-7

Existence of multiple solutions with precise signinformation for superlinear Neumann problems

Sergiu Aizicovici · Nikolaos S. Papageorgiou ·Vasile Staicu

Received: 26 August 2008 / Accepted: 26 January 2009 / Published online: 26 February 2009© Fondazione Annali di Matematica Pura ed Applicata and Springer-Verlag 2009

Abstract We consider a nonlinear Neumann problem driven by the p-Laplacian differen-tial operator and having a p-superlinear nonlinearity. Using truncation techniques combinedwith the method of upper–lower solutions and variational arguments based on critical pointtheory, we prove the existence of five nontrivial smooth solutions, two positive, two negativeand one nodal. For the semilinear (i.e., p = 2) problem, using critical groups we produce asecond nodal solution.

Keywords Neumann problem · p-Laplacian · Constant sign solutions · Nodal solutions ·Second deformation theorem · Linking theorem · Critical groups

Mathematics Subject Classification (2000) 35J25 · 35J70

1 Introduction

Let Z ⊆ RN be a bounded domain with a C2 boundary ∂Z . In this paper we study the

following nonlinear Neumann problem:

This paper was completed while N.S. Papageorgiou was visiting the University of Aveiro as an invitedscientist. The hospitality and financial support of the host institution are gratefully acknowledged. V. Staicuacknowledges partial financial support from the Portuguese Foundation for Sciences and Technology (FCT)under the project POCI/MAT/55524/2004.

S. AizicoviciDepartment of Mathematics, Ohio University, Athens, OH 45701, USAe-mail: [email protected]

N. S. Papageorgiou (B)Department of Mathematics, National Technical University, Zografou Campus, 15780 Athens, Greecee-mail: [email protected]

V. StaicuDepartment of Mathematics, University of Aveiro, 3810-193 Aveiro, Portugale-mail: [email protected]

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680 S. Aizicovici et al.

{−div

(‖Dx(z)‖p−2 Dx(z)) + β |x(z)|p−2 x(z) = f (z, x(z)) a.e. on Z ,

∂x∂n = 0 on ∂Z , 2 ≤ p < ∞.

(1.1)

Here β > 0, f (z, x) is a Caratheodory nonlinearity, ‖Dx(z)‖ = ‖Dx(z)‖RN and n is the

outward unit normal on ∂Z . The goal of this paper is to establish the existence of multipleconstant sign solutions and of nodal (sign-changing) solutions for problem (1.1), when thenonlinearity f exhibits a (p −1) superlinear growth near infinity. This question was recentlyinvestigated in the framework of the Dirichlet boundary value problem by several authors.We mention the works of Bartsch and Liu [6], Garcia Azorero et al. [17], Motreanu et al. [25],Papageorgiou and Papageorgiou [27] for p-superlinear Dirichlet problems and by Carl andPerera [10], Zhang and Li [32], Zhang et al. [31] for p-linear Dirichlet problems. In contrast,the study of the Neumann case is lagging behind. There have been some multiplicity results,but under symmetry hypotheses on the nonlinearity and generally without any informationconcerning the sign of the solutions. We mention the works of Anello [3], Binding et al.[7], Bonnano and Candito [8], Filippakis et al. [16], Motreanu and Papageorgiou [26], Wuand Tan [30] and the very recent work by the authors [1]. Only Binding et al. [7] provideinformation on the sign of their solutions. They have a nonlinearity of the form

f (z, x, λ) = λa(z) |x |p−2 x + b(z) |x |q−2 x,

with a, b ∈ L∞(Z), λ ∈ R, 1 < p < N , 1 < q < p∗ where

p∗ ={

N pN−p if p < N

+∞ if p ≥ N(1.2)

and they prove the existence of one or two positive solutions. In this paper we provide preciseinformation about the sign of all solutions that we obtain. Our approach combines variationalmethods with truncation techniques and the use of ordered pairs of upper and lower solutions.The rest of the paper is organized as follows. Section 2 presents some background material,which will be used in the sequel. In Sect. 3 we prove the existence of multiple solutions ofconstant sign (two positive and two negative). In Sect. 4, we produce a nodal solution andfinally in Sect. 5, we examine the semilinear problem (i.e., p = 2).

2 Mathematical background

First, let us recall some basic facts about the spectrum of the negative p-Laplacian withNeumann boundary condition. For details, we refer to Lê [22] and Gasinski-Papageorgiou[19]. So, let m ∈ L∞(Z)+,m �= 0 and consider the following nonlinear weighted (withweight m) eigenvalue problem{−div

(‖Dx(z)‖p−2 Dx(z)) = λm(z) |x(z)|p−2 x(z) a.e. on Z ,

∂x∂n = 0, on ∂Z , 1 < p < ∞, λ ∈ R.

(2.1)

A λ ∈ R for which problem (2.1) has a nontrivial solution is said to be an eigenvalueof

(−p,W 1,p(Z),m)

(where pu = div(‖Du‖p−2 Du

)is the p-Laplacian differential

operator), and the nontrivial solution is an eigenfunction corresponding to the eigenvalue λ.It is easy to check that a necessary condition for λ to be an eigenvalue is λ ≥ 0. More-over , zero is an eigenvalue with corresponding eigenspace R (that is, the space of constant

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Multiple solutions with precise sign information 681

functions). This eigenvalue, denoted by λ0 (m) , is isolated and admits the following varia-tional characterization:

0 = λ0 (m) = inf

{‖Dx‖p

p∫Z m |x(z)|p dz

: x ∈ W 1,p(Z), x �= 0

}. (2.2)

Hereafter, by ‖.‖p we denote the norm of L p(Z) or L p(Z ,RN

), and by ‖.‖ the norm of the

Sobolev space W 1,p(Z) or the norm of RN , whenever no confusion arises.

Clearly, constant functions realize the infimum in (2.2). By virtue of the Lusternik–Schnirelmann theory, in addition to λ0 (m), we have a whole strictly increasing sequence{λk := λk (m)

}k≥0 ⊆ R+, such that λk → +∞ as k → ∞, and which are eigenvalues of(−p,W 1,p(Z),m

). These are the so-called LS-eigenvalues of

(−p,W 1,p(Z),m).

If p = 2 (linear eigenvalue problem), then the LS-eigenvalues are all the eigenvaluesof

(−,W 1,2(Z),m). If p �= 2 (nonlinear eigenvalue problem), we do not know if this is

the case. However, since λ0 (m) = 0 is isolated and the set σ (p,m) of all eigenvalues of(−p,W 1,p(Z),m)

is closed, then

0 < λ∗1 := inf

{λ : λ ∈ σ (p,m) , λ > 0

} ∈ σ (p,m)

and we have λ∗1 = λ1, i.e., the second eigenvalue and the second LS-eigenvalue coincide.

Of all the eigenvalues of(−p,W 1,p(Z),m

), only the principal eigenvalue λ0 has eigen-

functions which do not change sign. All the other eigenvalues have eigenfunctions whichare nodal. This suggests that it is reasonable to expect that most of the solutions of (1.1) arenodal. Nevertheless, as it will become evident in Sect. 4, to produce a nodal solution is ahighly nontrivial task.

If m ≡ 1, then we write λk = λk for all k ≥ 0 and (−p,W 1,p(Z))=(−p,W 1,p(Z), 1).In our study of problem (1.1) we will make use of the following two spaces:

W 1,pn (Z) =

{x ∈ W 1,p(Z) : x = lim

k→∞ xk in W 1,p(Z), xk ∈ C∞ (Z),∂xk

∂n= 0 on ∂Z

}

and

C1n

(Z) =

{x ∈ C1 ( Z

) : ∂x

∂n= 0 on ∂Z

}.

Both these spaces are ordered Banach spaces, with order cones given by

W+ ={

x ∈ W 1,pn (Z) : x(z) ≥ 0 a.e. on Z

}and

C+ = {x ∈ C1

n

(Z) : x(z) ≥ 0 for all z ∈ Z

}.

Moreover, int C+ �= ∅ and more precisely we have

intC+ = {x ∈ C1

n

(Z) : x(z) > 0 for all z ∈ Z

}.

Let

u0(z) = 1

|Z |1pN

for all z ∈ Z

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be the L p-normalized principal eigenfunction of(−p,W 1,p(Z)

)(hereafter, by |.|N we

denote the Lebesgue measure on RN ). Also let

∂BL p

1 = {x ∈ L p(Z) : ‖x‖p = 1

}.

In [1], we obtained the following alternative variational characterization of λ1 > 0 (for thecorresponding result for the Dirichlet eigenvalue problem, see Cuesta et al. [14]):

Proposition 1 If S = W 1,pn (Z) ∩ ∂BL p

1 and

� = {γ ∈ C ([−1, 1] , S) : γ (−1) = −u0, γ (1) = u0},then

λ1 = infγ∈� max

t∈[−1,1]‖Dγ (t)‖p

p .

The topological notion of linking sets is crucial in the minimax characterization of thecritical values of a C1-functional.

Definition Let Y be a Hausdorff topological space and E0, E, and D nonempty closedsubsets of Y with E0 ⊆ E . We say that the pair {E0, E} links with D in Y , if

(a) E0 ∩ D = ∅;(b) for any γ ∈ C (E, Y ) such that γ |E0= id |E0 , we have γ (E) ∩ D �= ∅.

Recall that, if X is a Banach space and ϕ ∈ C1 (X), then for c ∈ R, we say that ϕ satisfiesthe PSc-condition, if every sequence {xn}n≥1 ⊆ X such that

ϕ(xn) → c and ϕ′(xn) → 0 in X∗ as n → ∞,

has a strongly convergent subsequence. If this is true for every c ∈ R, then we say that ϕsatisfies the PS-condition. Using this notion, we have the following general minimax theoremfor the critical values of C1-functionals (see, for example, [19, p. 644]):

Theorem 1 If X is a Banach space, E0, E, and D are nonempty, closed subsets of X suchthat the pair {E0, E} links with D in X, ϕ ∈ C1 (X) is such that

supE0

ϕ < infDϕ,

� = {γ ∈ C (E, X) : γ |E0= id |E0

},

c = infγ∈� sup

v∈Eϕ (γ (v))

and ϕ satisfies the PSc condition, then c ≥ inf D ϕ and c is a critical value of ϕ, i.e., thereexists x0 ∈ X such that ϕ′ (x0) = 0 and ϕ (x0) = c.

By employing particular choices of linking sets, from the above theorem, we can pro-duce as corollaries the mountain pass theorem, the saddle point theorem and the generalizedmountain pass theorem. For future use, let us state here the mountain pass theorem.

Theorem 2 If X is a Banach space, ϕ ∈ C1 (X) , x0, x1 ∈ X and ρ > 0 satisfy

max {ϕ (x0) , ϕ (x1)} < inf {ϕ(x) : ‖x − x0‖ = ρ} = c0,

�0 = {γ ∈ C ([0, 1] , X) : γ (0) = x0, γ (1) = x1} ,c = inf

γ∈�0sup

t∈[0,1]ϕ (γ (t))

and ϕ satisfies the PSc condition, then c ≥ c0 and c is a critical value of ϕ.

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Remark It is easy to see that we can deduce Theorem 2 from Theorem 1, if we choose

E0 = {x0, x1} ,E = [x0, x1] = {x ∈ X : x = (1 − t) x0 + t x1, t ∈ [0, 1]}D = ∂Bρ (x0) = {x ∈ X : ‖x − x0‖ = ρ} .

In producing a nodal solution (see Sect. 4), we will need the so-called “second deformationtheorem” (see [12, p. 23], [19, p. 628]). For easy reference, we state this result here. First weneed to introduce some notation. So, let X be a Banach space, ϕ ∈ C1 (X) and c ∈ R. Wedefine the sublevel set of ϕ at c by

ϕc = {x ∈ X : ϕ(x) ≤ c}, (2.3)

the critical set of ϕ by

K = {x ∈ X : ϕ′(x) = 0

}, (2.4)

and the critical set of ϕ at the level c ∈ R by

Kc = {x ∈ K : ϕ(x) = c}. (2.5)

In the next theorem (the “second deformation theorem”), we allow b = +∞, in which caseϕb\Kb = X.

Theorem 3 If X is a Banach space, ϕ ∈ C1 (X), a ∈ R, a < b ≤ +∞, ϕ satisfies theP Sc− condition for every c ∈ [a, b), ϕ has no critical values in (a, b) and ϕ−1 (a) containsat most a finite number of critical points of ϕ, then there exists a continuous deformationh : [0, 1] × (

ϕb\Kb) → ϕb such that

• h(1, ϕb\Kb

) ⊆ ϕa

• h (t, x) = x for all t ∈ [0, 1] and all x ∈ ϕa,

• ϕ (h (t, x)) ≤ ϕ (h (s, x)) for all s, t ∈ [0, 1] , s ≤ t and all x ∈ ϕb\Kb.

Let now X be a reflexive Banach space and X∗ its topological dual. By 〈., .〉 we denotethe duality brackets for the pair (X∗, X). Also

w−→ denotes weak convergence in X.

Definition We say that a map A : X → X∗ is of type (S)+ , if for every sequence {xn}n≥1 ⊆ X

such that xnw−→ x in X and

lim supn→∞

〈A(xn), xn − x〉 ≤ 0,

one has

xn → x in X.

Let X = W 1,pn (Z), X∗ = W 1,p

n (Z)∗ and let A : X → X∗ be defined by

〈A(x), y〉 =∫Z

‖Dx‖p−2 (Dx, Dy)RN dz for all x , y ∈ X. (2.6)

Proposition 2 The map A : X → X∗ defined by (2.6) is of type (S)+.

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Proof Clearly A is continuous monotone, hence it is maximal monotone. Let xnw−→ x in

W 1,pn (Z) and suppose

lim supn→∞

〈A(xn), xn − x〉 ≤ 0. (2.7)

The map A being maximal monotone, it is generalized pseudomonotone (see [18, p. 330]).So, from (2.7) it follows that

‖Dxn‖pp = 〈A(xn), xn〉 → 〈A(x), x〉 = ‖Dx‖p

p .

Recall that Dxnw−→ Dx in L p

(Z ,RN

), which is a uniformly convex space. Therefore,

from the Kadec–Klee property we infer that Dxn → Dx in L p(Z ,RN

), hence xn → x in

W 1,pn (Z). ��The next result, due to Barletta–Papageorgiou [4] relates local C1

n

(Z)-minimizers with

local W 1,pn (Z)-minimizers for a large class of C1-functionals. It was first proved for p = 2

and Dirichlet boundary conditions by Brezis–Nirenberg [9]. It was extended to the case p �= 2and Dirichlet boundary conditions by Garcia Azorero et al. [17] (see also Guo and Zhang[20], and for a nonsmooth version, Kyritsi and Papageorgiou [21]).

We consider a function f0(z, x) satisfying the following hypotheses:

(H0) f0 : Z × R →R is a function such that:

(i) for all x ∈ R, z → f0(z, x) is measurable;(ii) for almost all z ∈ Z , x → f0(z, x) is continuous;

(iii) for almost all z ∈ Z and all x ∈ R

| f0(z, x)| ≤ a0(z)+ c0 |x |r−1 ,

with a0 ∈ L∞(Z)+, c0 > 0 and 1 < r < p∗, where p∗ is given by (1.2).

Let F0(z, x) = ∫ x0 f0(z, s)ds (the primitive of f0) and consider the functional ϕ0 :

W 1,pn (Z) → R defined by

ϕ0(x) = 1

p‖Dx‖p

p −∫Z

F0 (z, x(z)) dz for all x ∈ W 1,pn (Z).

Evidently ϕ0 ∈ C1(

W 1,pn (Z)

).

Proposition 3 If x0 ∈ W 1,pn (Z)(2 ≤ p < ∞) is a local C1

n

(Z)-minimizer of ϕ0, i.e., there

exists r0 > 0 such that

ϕ0 (x0) ≤ ϕ0 (x0 + h) for all h ∈ C1n

(Z), ‖h‖C1

n(

Z) ≤ r0,

then x0 ∈ C1n

(Z)

and it is a local W 1,pn (Z)-minimizer of ϕ0, i.e., there exists r1 > 0 such

that

ϕ0 (x0) ≤ ϕ0 (x0 + h) for all h ∈ W 1,pn (Z), ‖h‖ ≤ r1.

In Sect. 3, using truncation techniques to produce additional solutions of constant sign,we introduce a very mild nonsmoothness of the functional involved. We take care of this,using nonsmooth critical point theory, which is based on Clarke’s subdifferential for locallyLipschitz functions (see [13]).

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Briefly, we recall that, if X is a Banach space and θ : X → R is locally Lipschitz, thegeneralized directional derivative θ0(x; h) of θ at x ∈ X in the direction of h ∈ X , isdefined by

θ0(x; h) = lim supx ′→xλ↓0

θ(x ′ + λh)− θ(x ′)λ

.

It is easy to check that the function h �→ θ0(x; h) is sublinear continuous, and so it is thesupport function of a nonempty, convex and weakly star compact set, ∂θ(x) ⊆ X∗, defined by

∂θ(x) = {x∗ ∈ X∗ : 〈x∗, h〉 ≤ θ0(x; h) for all h ∈ X

}.

The multifunction x �→ ∂θ(x) is called the generalized subdifferential of θ . If θ : X → R

is continuous and convex, then it is locally Lipschitz, and the generalized subdifferential coin-cides with the subdifferential in the sense of convex analysis, defined by

∂cθ(x) = {x∗ ∈ X∗ : 〈x∗, h〉 ≤ θ(x + h)− θ(x) for all h ∈ X

}.

Also, if θ ∈ C1(X), then θ is locally Lipschitz and we have ∂θ(x) = {θ ′(x)}. For details seeClarke [13].

A point x ∈ X is a critical point of the locally Lipschitz function θ : X → R, if 0 ∈ ∂θ(x).Also we say that θ satisfies the nonsmooth Palais-Smale condition(the nonsmoothP S-con-dition for short), if every sequence {xn}n≥1 ⊆ X such that

|θ(xn)| ≤ c1 for some c1 > 0, all n ≥ 1, with

m(xn) = inf{||x∗||X∗ : x∗ ∈ ∂θ(xn)} → 0,

has a strongly convergent subsequence.Using this notion, we can have the nonsmooth analog of Theorem 2 (the nonsmooth

mountain pass theorem; see Chang [11], Gasinski and Papageorgiou [18]).In Sect. 5, for the semilinear problem (i.e., p = 2), we produce a second nodal solution

for a total of six nontrivial solutions using Morse theory and in particular critical groups. So,let us briefly recall some basic definitions and facts from this theory.

If X is a Hausdorff topological space and Y a subspace of it, for every integer n ≥ 0, wedenote by Hn (X, Y ) the nth-singular relative homology group with integer coefficients. LetH be a Hilbert space and ϕ ∈ C1 (H). The critical groups of ϕ at an isolated critical pointx0 ∈ H with ϕ (x0) = c are defined by

Cn (ϕ, x0) = Hn(ϕc ∩ U,

(ϕc ∩ U

) \ {x0}),

where U is a neighborhood of x0 such that K ∩ϕc ∩U = {x0} ,where ϕc is defined by (2.3).By the excision property of the singular homology theory, we see that the above definitionof critical groups is independent of U .

Suppose that −∞ < inf ϕ (K ) and choose c < inf ϕ (K ) . Then, the critical groups of ϕat infinity are defined by

Cn (ϕ,∞) = Hn(H, ϕc) for all n ≥ 0.

If K is finite, then the Morse type numbers of ϕ are defined by

Mn =∑x∈K

rankCn (ϕ, x) .

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The Betti-type numbers of ϕ are defined by

βn = rankCn (ϕ,∞) .

From Morse theory, we know that

m∑n=0

(−1)m−n Mn ≥m∑

n=0

(−1)m−nβn (2.8)

and ∑n≥0

(−1)n Mn =∑n≥0

(−1)nβn . (2.9)

From (2.8), we deduce that βn ≤ Mn for all n ≥ 0. Therefore, if βn �= 0 for some n ≥ 0,then ϕ must have a critical point x ∈ H and the critical group Cn (ϕ, x) is nontrivial. Rela-tion (2.9) is known as the “Poincare-Hopf formula”. Finally, if K = {x0} , then Ck (ϕ,∞) =Ck (ϕ, x0) for all k ≥ 0. For details we refer to the books of Chang [11] and Mawhin-Willem[24] and the paper of Bartsch-Li [5].

3 Solutions of constant sign

The hypotheses on the nonlinearity f (z, x) are the following:

H( f )1: The function f : Z × R →R is such that f (z, 0) = 0 a.e. on Z and

(i) for every x ∈ R, z → f (z, x) is measurable;(ii) for almost all z ∈ Z , x → f (z, x) is continuous;

(iii) for almost all z ∈ Z and all x ∈ R we have

| f (z, x)| ≤ a(z)+ c |x |r−1 ,

where a ∈ L∞(Z)+, c > 0 and 1 < r < p∗;(iv) there exist µ > max

{pβ, p

β

}and M > 0, such that for almost all z ∈ Z and

all |x | ≥ M

f (z, x)x ≥ µF(z, x) > 0 where F(z, x) =x∫

0

f (z, s) ds;

(v) there exist a− < 0 < a+, such that for almost all z ∈ Z

f (z, a−) = 0 = f (z, a+)

and {0 ≤ f (z, x) ≤ c+ (a+ − x)p−1 for all x ∈ [

0, a+]

−c− (x − a−)p−1 ≤ f (z, x) ≤ 0 for all x ∈ [a−, 0

]for some c+, c− > 0;

(vi) there exist δ > 0 and ξ > β such that for almost all z ∈ Z and all |x | ≤ δ,

ξ |x |p ≤ pF(z, x).

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Example 1 The following function satisfies hypotheses H( f )1. For simplicity, we drop thez-dependence:

f (x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

c2 (ex − 1) (1 + x)p−1 if x ∈ [−1,− 12

]c2

(1 − e− 1

2

)|x |p−2 x if x ∈ [− 1

2 , 0]

c2

(1 − e− 1

2

)x p−1 if x ∈ [

0, 12

]c2 (ex − 1) (1 − x)p−1 if x ∈ [ 1

2 , 1]

|x |q−2 x − x if |x | > 1

with

2 ≤ p < q < p∗, c2

(1 − e− 1

2

)> β.

We consider the Euler functional ϕ : W 1,pn (Z) → R for problem (1.1), defined by

ϕ(x) = 1

p‖Dx‖p

p + β

p‖x‖p

p −∫Z

F (z, x(z)) dz for all x ∈ W 1,pn (Z). (3.1)

Clearly, ϕ ∈ C1(

W 1,pn (Z)

). Also, we introduce the following truncations of the nonlinearity

f (z, x)

f+(z, x) =⎧⎨⎩

0 if x < 0f (z, x) if 0 ≤ x ≤ a+0 if a+ < x

and

f−(z, x) =⎧⎨⎩

0 if x < a−f (z, x) if a− ≤ x ≤ 00 if 0 < x .

Then, we introduce the corresponding primitive functions

F±(z, x) =x∫

0

f±(z, s)ds for all x ∈ R

and finally the functionals ϕ± : W 1,pn (Z) → R defined by

ϕ±(x) = 1

p‖Dx‖p

p + β

p‖x‖p

p −∫Z

F± (z, x(z)) dz for all x ∈ W 1,pn (Z).

Again we have ϕ± ∈ C1(

W 1,pn (Z)

).

In the next proposition, we produce the first two solutions of constant sign for problem(1.1).

Proposition 4 If hypotheses H( f )1 hold, then problem (1.1) has at least two solutionsx0 ∈ intC+ and v0 ∈ −intC+, which are local minimizers of ϕ.

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Proof First, we produce the positive solution. Clearly ϕ+ is coercive and exploiting thecompact embedding of W 1,p

n (Z) into L p(Z), we can easily verify that ϕ+ is weakly lowersemicontinuous. So, by the Weierstrass theorem, we can find x0 ∈ W 1,p

n (Z) such that

ϕ+ (x0) = inf{ϕ+(x) : x ∈ W 1,p

n (Z)}

= m+.

Let u0(z) = u0 > 0 for all z ∈ Z . Then for t > 0 small, we have

0 < tu0 ≤ min {δ, a+}where δ > 0 is as in hypothesis H( f )1 (vi) . Since F (z, tu0) = F+ (z, tu0), we have

ϕ+ (tu0) = β

pt pu p

0 |Z |N −∫Z

F (z, tu0) dz

≤ 1

p(β − ξ) (tu0)

p |Z |N < 0

(see hypothesis H( f )1 (vi)), hence

ϕ+ (x0) = m+ < 0 = ϕ+ (0);therefore

x0 �= 0.

Of course, x0 is a critical point of ϕ+ and so

ϕ′+ (x0) = 0;therefore

A (x0)+ βK p (x0) = N+ (x0), (3.2)

where, as before, K p(x)(.) = |x(.)|p−2 x(.) and N+(x)(.) = f+ (., x(.)) for all x ∈W 1,p

n (Z).Recall that for x ∈ Lr (Z) (1 ≤ r ≤ ∞), we define

x+ = max {x, 0} and x− = max {−x, 0}and we have

x = x+ − x− and |x | = x+ + x−.

On (3.2) we act with the test function −x−0 ∈ W 1,p

n (Z). Hence∥∥Dx−0

∥∥pp + β

∥∥x−0

∥∥pp = 0

(since f+ ≥ 0, see hypothesis H( f )1(v)); therefore∥∥x−0

∥∥ = 0, i.e., x0 ≥ 0, x0 �= 0.

Then, from (3.2) we have{− p x0(z)+ βx0(z)p−1 = f+ (z, x0(z)) a.e. on Z ,∂x0∂n = 0 on ∂Z .

(3.3)

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Also, the nonlinear regularity theory implies that x0 ∈ C+.Moreover, since f+ (z, x0(z)) ≥ 0a.e. on Z , we obtain from (3.3)

px0(z) ≤ βx0(z)p−1 a.e. on Z .

Invoking the nonlinear strong maximum principle of Vazquez [29], we infer that x0 ∈ intC+.Note that

A (x0 − a+) = A (x0) = N+ (x0)− βx p−10 . (3.4)

On (3.4) we act with (x0 − a+)+ ∈ W 1,pn (Z). Hence

‖D (x0 − a+)‖pp = −

∫{x0>a+}

βx p−10 (x0 − a+) dz,

therefore∥∥D (x0 − a+)+∥∥p

p + c3∥∥(x0 − a+)+

∥∥pp ≤ −

∫{x0>a+}

βa p−1+ (x0 − a+) dz, (3.5)

for some c3 > 0. Here we have used the fact that since p ≥ 2, we can find c3 > 0 such that

c3[(x0(z)− a+)+

]p ≤(

x0(z)p−1 − a p−1

+)(x0(z)− a+)+ for all z ∈ Z

(see, for example [19, p. 740]). From (3.5) it follows that

c3∥∥(x0 − a+)+

∥∥p ≤ 0;hence

x0 ≤ a+.

Then

A (a+ − x0) = −A (x0) = −N (x0)+ βK p (x0) ,

where N (x)(.) = f (., x(.)) for all x ∈ W 1,pn (Z). Hence, a.e. on Z ,

− p (a+ − x0) (z) = βx0(z)p−1 − f (z, x0(z))

≥ −c+ (a+ − x0) (z)p−1

(see H( f )1(v)), therefore,

p (a+ − x0) (z) ≤ c+ (a+ − x0) (z)p−1 a.e. on Z . (3.6)

Since x0 ∈ W 1,pn (Z) is a solution of (1.1) (see (3.3) and note that 0 ≤ x0 ≤ a+), we have

x0 �= a+ (cf. H( f )1(v)) and so from (3.6) and the nonlinear strong maximum principle ofVazquez [29], we obtain

x0(z) < a+ for all z ∈ Z

and, if for some z0 ∈ ∂Z we have a+ = x0 (z0) , then

∂

∂n(a+ − x0) (z0) < 0;

123


hence

∂x0

∂n(z0) > 0,

a contradiction to the fact that

∂x0

∂n(z) = 0 for all z ∈ ∂Z

(see (3.3) and recall that x0 ∈ intC+). Therefore,

a+ − x0 ∈ intC+. (3.7)

Because of (3.7) and the fact that x0 ∈ intC+, we can find r > 0 small such that, if

BC1

n(

Z)

r (x0) ={

x ∈ C1n

(Z) : ‖x − x0‖C1

n(

Z) ≤ r

},

then

ϕ+(x) = ϕ(x), for all x ∈ BC1

n(

Z)

r (x0) ,

hence

x0 is a local C1n

(Z)

-minimizer of ϕ;therefore (see Proposition 3)

x0 is a local W 1,pn (Z)-minimizer of ϕ.

In a similar manner, working this time with the functional ϕ−, we obtain v0 ∈ −intC+, asolution of (1.1), which too is a local minimizer of ϕ. ��

Next, using the solutions x0 ∈ intC+ and v0 ∈ −intC+ from Proposition 4 and a suitablemodification of the Euler functional ϕ, we will produce two additional smooth solutions ofconstant sign. For this purpose let

ξ+ =∫Z

F (z, x0(z)) dz and ξ− =∫Z

F (z, v0(z)) dz

and consider the functionals ψ± : W 1,pn (Z) → R defined by

ψ+(y) = 1

p‖D(x0 + y)‖p

p + β

p‖(x0 + y)‖p

p − 1

p‖Dx0‖p

p − β

p‖x0‖p

p

−∫Z

F(z,(x0 + y+) (z)) dz +

∫Z

f (z, x0) y−dz + ξ+ (3.8)

and

ψ−(y) = 1

p‖D(v0 + y)‖p

p + β

p‖(v0 + y)‖p

p − 1

p‖Dv0‖p

p − β

p‖v0‖p

p

−∫Z

F(z, (v0 − y−)(z))dz −∫Z

f (z, x0) y+dz + ξ−. (3.9)

Note that ψ± are Lipschitz continuous and in fact

ψ± ∈ C1(

W 1,pn (Z)\ {0}

).

123


Moreover, we have

‖D(x0 + y)‖pp = ∥∥D

(x0 + y+)∥∥p

p + ∥∥D(x0 − y−)∥∥p

p − ‖Dx0‖pp ,

‖D (v0 + y)‖pp = ∥∥D

(v0 + y+)∥∥p

p + ∥∥D(v0 − y−)∥∥p

p − ‖Dv0‖pp ,

‖x0 + y‖pp = ∥∥x0 + y+∥∥p

p + ∥∥x0 − y−∥∥pp − ‖x0‖p

p ,

‖v0 + y‖pp = ∥∥v0 + y+∥∥p

p + ∥∥v0 − y−∥∥pp − ‖v0‖p

p .

Therefore, we can rewrite ψ± (see (3.8) and (3.9)) as follows:

ψ+(y) = ϕ(x0 + y+) − ϕ (x0)+ 1

p

[∥∥D(x0 − y−)∥∥p

p − ‖Dx0‖pp

]

+ β

p

[∥∥x0 − y−∥∥pp − ‖x0‖p

p

]+∫Z

f (z, x0) y−dz (3.10)

and

ψ−(y) = ϕ(v0 − y−) − ϕ (v0)+ 1

p

[∥∥D(x0 + y+)∥∥p

p − ‖Dx0‖pp

]

+ β

p

[∥∥v0 + y+∥∥pp − ‖v0‖p

p

]−∫Z

f (z, v0) y−dz. (3.11)

Proposition 5 If hypotheses H( f )1 hold, then we can find ρ > 0 small such that

inf {ψ+(x) : ‖x‖ = ρ} = γ+ρ > 0

and

inf {ψ−(x) : ‖x‖ = ρ} = γ−ρ > 0.

Proof We do the proof for ψ+, the proof for ψ− being similar. From Proposition 4 we knowthat x0 ∈ intC+ is a local minimizer of ϕ.We can always assume that it is an isolated criticalpoint and local minimizer of ϕ, or otherwise, in the same way as above, we can generate awhole sequence of distinct positive solutions for problem (1.1). So, we can find ρ0 > 0 suchthat

ϕ (x0) < ϕ(y) and ϕ′(y) �= 0 for all y ∈ Bρ0 (x0) \ {x0} . (3.12)

Here Bρ0(x0) = {x ∈ W 1,pn (Z) : ‖x − x0‖ ≤ ρ0}.

Claim For all ρ ∈ (0, ρ0), we have

inf{ϕ(x) : x ∈ Bρ0 (x0) \Bρ(x0)

}> ϕ(x0), (3.13)

where

Bρ(x0) ={

x ∈ W 1,pn (Z) : ‖x − x0‖ < ρ

}.

We argue indirectly. So, suppose that the Claim is not true. Then, we can find ρ ∈ (0, ρ0)

and a sequence {xn}n≥1 ⊆ Bρ0 (x0) \Bρ (x0) such that

ϕ(xn) → ϕ (x0) , as n → ∞. (3.14)

123


The sequence {xn}n≥1 ⊆ W 1,pn (Z) is bounded and, so, by passing to a suitable subsequence

if necessary, we may assume that

xnw−→ x in W 1,p

n (Z) and xn → x in Lr (Z)

(recall that r < p∗). Then x ∈ Bρ0 (x0) , and since the functional ϕ is weakly lower semi-continuous, we have

ϕ (x) ≤ limn→∞ϕ(xn) = ϕ (x0)

(see (3.14)), hence

x = x0

(see (3.12)). From the mean value theorem, we have

ϕ(xn)− ϕ

(xn + x0

2

)=⟨x∗

n ,xn − x0

2

⟩,

with

x∗n = ϕ′

(tn xn + (1 − tn)

xn + x0

2

)and tn ∈ (0, 1) for all n ≥ 1.

Recall that

x∗n = A

(tn xn + (1 − tn)

xn + x0

2

)+ βK p

(tn xn + (1 − tn)

xn + x0

2

)

− N

(tn xn + (1 − tn)

xn + x0

2

).

Hence

ϕ(xn)− ϕ

(xn + x0

2

)=

⟨A

(tn xn + (1 − tn)

xn + x0

2

),

xn − x0

2

⟩

+ β

∫Z

∣∣∣∣tn xn + (1 − tn)xn + x0

2

∣∣∣∣p−2

×(

tn xn + (1 − tn)xn + x0

2

)xn − x0

2dz

−∫Z

f

(z, tn xn + (1 − tn)

xn + x0

2

)xn − x0

2dz. (3.15)

Since xn → x0 in Lr (Z) and in L p(Z), we have∫Z

∣∣∣∣tn xn + (1 − tn)xn + x0

2

∣∣∣∣p−2 [

tn xn + (1 − tn)xn + x0

2

]xn − x0

2dz → 0 (3.16)

and ∫Z

f

(z, tn xn + (1 − tn)

xn + x0

2

)xn − x0

2dz → 0 (3.17)

123


as n → ∞. Also, by virtue of the weak lower semicontinuity of ϕ, we have

ϕ (x0) ≤ lim infn→∞ ϕ

(xn + x0

2

). (3.18)

So, if in (3.15) we pass to the limit as n → ∞, then from (3.14), (3.16), (3.17) and (3.18)we obtain

lim supn→∞

⟨A

(tn xn + (1 − tn)

xn + x0

2

),

xn − x0

2

⟩≤ 0

hence

lim supn→∞

⟨A

(tn xn + (1 − tn)

xn + x0

2

), tn xn + (1 − tn)

xn + x0

2− x0

⟩≤ 0. (3.19)

We may assume that tn → t ∈ [0, 1] . Then

tn xn + (1 − tn)xn + x0

2w−→ x0 in W 1,p

n (Z). (3.20)

From (3.19), (3.20) and the fact that A is an operator of type (S)+ (see Proposition 2), weconclude that

tn xn + (1 − tn)xn + x0

2→ x0 in W 1,p

n (Z). (3.21)

But note that ∥∥∥∥tn xn + (1 − tn)xn + x0

2− x0

∥∥∥∥ = (1 + tn)

∥∥∥∥ xn − x0

2

∥∥∥∥ ≥ ρ

2. (3.22)

Comparing (3.21) and (3.22), we reach a contradiction. This proves the claim.Let ρ ∈ (0, ρ0) and consider ‖y‖ = ρ. Then

∥∥y+∥∥ ≥ ρ2 or

∥∥y−∥∥ ≥ ρ2 . First suppose that∥∥y+∥∥ ≥ ρ

2 . From (3.13) we see that we can find η1 = η1 (ρ) > 0 such that

ϕ(x0 + y+) − ϕ (x0) ≥ η1 > 0. (3.23)

Also from the convexity of the norm functionals, we have

1

p

[∥∥D(x0 − y−)∥∥p

p − ‖Dx0‖pp

]+∫Z

‖Dx0‖p−2 (Dx0, Dy−)RN dz ≥ 0 (3.24)

and

β

p

[∥∥x0 − y−∥∥pp − ‖x0‖p

p

]+ β

∫Z

|x0|p−2 x0 y−dz ≥ 0. (3.25)

Using (3.23)–(3.25), (3.10) and the fact that∫Z

‖Dx0‖p−2 (Dx0, Dy−)RN dz + β

∫Z

|x0|p−2 x0 y−dz =∫Z

f (z, x0(z))y−dz,

we conclude that

ψ+(y) ≥ η1 > 0 for all ‖y‖ = ρ,∥∥y+∥∥ ≥ ρ

2. (3.26)

123


Next, we assume that∥∥y−∥∥ ≥ ρ

2 . Recall that since p ≥ 2, for all ξ1, ξ2 ∈ Rk(k ≥ 1), we

have

‖ξ2‖pRk − ‖ξ1‖p

Rk ≥ p ‖ξ1‖p−2Rk (ξ1, ξ2 − ξ1)Rk + 1

2p−1 − 1‖ξ2 − ξ1‖p

Rk . (3.27)

(see [19, p. 740]). Using (3.27) twice, first with ξ1 = Dx0(z), ξ2 = D(x0 − y−) (z) and

then with ξ1 = x0(z), ξ2 = (x0 − y−) (z) and integrating over Z , from (3.10) we conclude

ψ+(y) ≥ η2 > 0 for all ‖y‖ = ρ,∥∥y−∥∥ ≥ ρ

2. (3.28)

From (3.26) and (3.28), we infer that

ψ+(y) ≥ γ+ρ = min {η1, η2} for all ‖y‖ = ρ.

The argument for ψ− is similar, based this time on (3.11). ��By considering the functionals ψ+ and ψ− we have introduced a mild nonsmoothness.

Namely, as already observed, both ψ+ and ψ− are nondifferentiable at 0. However, bothare Lipschitz continuous and so we can take care of the nonsmoothness at the origin, usingnonsmooth critical point theory.

Proposition 6 If hypotheses H( f )1 hold, then the functionals ψ+ and ψ− satisfy the non-smooth P S-condition.

Proof We do the proof for ψ+, the proof for ψ− being similar. So, we consider a sequence{yn}n≥1 ⊆ W 1,p

n (Z) such that

|ψ+(yn)| ≤ M1 for some M1 > 0, all n ≥ 1 and m(yn) → 0 as n → ∞.

Since ∂ψ+(yn) ⊆ W 1,pn (Z)∗ is w−compact and the norm functional in a Banach space is

weakly lower semicontinuous, from the Weierstrass theorem we can find y∗n ∈ ∂ψ+(yn) such

that m(yn) = ||y∗n ||, n ≥ 1.

Let V1, V2 : W 1,pn (Z) → R be the integral functionals defined by

V1(y) =∫Z

F(z, x0 + y+) (z)dz and V2(y) =

∫Z

f (z, x0(z))y−(z)dz.

Note that V1 ∈C1(W 1,pn (Z)) and V2 is Lipschitz continuous and belongs to C1(W 1,p

n (Z)\{0}).Let θ : R → R be the function defined by

θ(r) = r− = max{−r, 0}.Evidently, θ is Lipschitz continuous and convex. We have

y∗n = A (x0 + yn)+ βK p (x0 + yn)− N

(x0 + y+

n

)χ{yn≥0} + u∗

n

with

u∗n ∈ Lr ′

(Z)

(1

r+ 1

r ′ = 1

), u∗

n(z) ∈ f (z, x0(z)) ∂θ(y−

n (z))

a.e. on Z

(see [13, p. 47, 83]). We know that

u∗n(z) =

⎧⎨⎩

− f (z, x0(z)) for a.a. z ∈ {yn < 0}−sn f (z, x0(z)) for a.a. z ∈ {yn = 0} , sn ∈ [0, 1] .0 for a.a. z ∈ {yn > 0}

(3.29)

123


From the choice of the sequence {yn}n≥1 ⊆ W 1,pn (Z), we have∣∣⟨y∗

n , v⟩∣∣ ≤ εn ‖v‖ for all v ∈ W 1,p

n (Z) with εn ↓ 0. (3.30)

Note that

A (x0 + yn) = A(x0 + y+

n

) + A(x0 − y−

n

) − A (x0) (3.31)

and

K p (x0 + yn) = K p(x0 + y+

n

) + K p(x0 − y−

n

) − K p (x0). (3.32)

In (3.30), we use as a test function v = −y−n ∈ W 1,p

n (Z). Then, because of (3.31) and (3.32),we have ⟨

A(x0 + y+

n

),−y−

n

⟩ + ⟨A(x0 − y−

n

),−y−

n

⟩ − ⟨A (x0) ,−y−

n

⟩+ β

∫Z

∣∣x0 + y+n

∣∣p−2 (x0 + y+

n

) (−y−n

)dz

+ β

∫Z

∣∣x0 − y−n

∣∣p−2 (x0 − y−

n

) (−y−n

)dz − β

∫Z

|x0|p−2 x0(−y−

n

)dz

−∫Z

N(x0 + y+

n

)χ{yn≥0}

(−y−n

)dz +

∫Z

u∗n(z)

(−y−n

)dz ≤ εn

∥∥y−n

∥∥ . (3.33)

We examine each summand on the left hand side of (3.33) separately. First,⟨A(x0 + y+

n

),−y−

n

⟩ + ⟨A(x0 − y−

n

),−y−

n

⟩ + ⟨A (x0) , y−

n

⟩= − ⟨

A (x0) , y−n

⟩ + ⟨A(x0 − y−

n

),−y−

n

⟩ + ⟨A (x0) , y−

n

⟩= ⟨

A(x0 − y−

n

),−y−

n

⟩. (3.34)

Similarly,∫Z

∣∣x0 + y+n

∣∣p−2 (x0 + y+

n

) (−y−n

)dz +

∫Z

∣∣x0 − y−n

∣∣p−2 (x0 − y−

n

) (−y−n

)dz

+∫Z

|x0|p−2 x0 y−n dz =

∫Z

∣∣x0 − y−n

∣∣p−2 (x0 − y−

n

) (−y−n

)dz. (3.35)

Also, ∫Z

N(x0 + y+

n

)χ{yn≥0}

(−y−n

)dz = 0, (3.36)

and ∫Z

u∗n(z)

(−y−n

)dz =

∫Z

f (z, x0) y−n dz ≥ 0 (3.37)

(see (3.29)).We return to (3.33) and use (3.34)–(3.37). We obtain

⟨A(x0 − y−

n

),−y−

n

⟩ + β

∫Z

∣∣x0 − y−n

∣∣p−2 (x0 − y−

n

) (−y−n

)dz ≤ εn

∥∥y−n

∥∥;

123


hence

c4∥∥x0 − y−

n

∥∥p ≤ εn∥∥y−

n

∥∥ + c5∥∥x0 − y−

n

∥∥p−1 ‖x0‖for some c4, c5 > 0; therefore{

y−n

}n≥1 ⊆ W 1,p

n (Z) is bounded.

Recall that

|ψ+(yn)| ≤ M1 for all n ≥ 1,

hence (cf. (3.8))

µ

p‖D (x0 + yn)‖p

p + βµ

p‖x0 + yn‖p

p − µ

p‖Dx0‖p

p

−βµp

‖x0‖pp −

∫Z

µF(z, x0 + y+

n

)(z)dz +

∫Z

µ f (z, x0) y−n dz + µξ+

≤ µM1. (3.38)

Note that

‖D (x0 + yn)‖pp = ∥∥D

(x0 + y+

n

)∥∥pp + ∥∥D

(x0 − y−

n

)∥∥pp − ‖Dx0‖p

p (3.39)

and

‖x0 + yn‖pp = ∥∥x0 + y+

n

∥∥pp + ∥∥x0 − y−

n

∥∥pp − ‖x0‖p

p . (3.40)

In (3.38), we use (3.39) and (3.40) and the fact that{

y−n

}n≥1 ⊆ W 1,p

n (Z) is bounded. Then

c6µ

p

∥∥x0 + y+n

∥∥pp −

∫Z

µF(z, x0 + y+

n

)(z)dz ≤ M2 (3.41)

with

c6 = min {β, 1} > 0 and M2 > 0.

Also, if in (3.30) we use the test function x0 + y+n ∈ W 1,p

n (Z), then∣∣⟨y∗n , x0 + y+

n

⟩∣∣ ≤ εn∥∥x0 + y+

n

∥∥ for all n ≥ 1,

hence

− c7∥∥x0 + y+

n

∥∥p +∫

{yn≥0}N(x0 + y+

n

) (x0 + y+

n

)dz ≤ c8

∥∥x0 + y+n

∥∥ + M3 (3.42)

with

c7 = max{β, 1} > 0, c8 and M3 > 0.

We add (3.41) and (3.42) and obtain(c6µ

p− c7

)∥∥x0 + y+n

∥∥p +∫

{yn≥0}

(N(x0 + y+

n

) (x0 + y+

n

) − µF(z, x0 + y+

n

))dz

≤ c8∥∥x0 + y+

n

∥∥ + M4, (3.43)

123


with M4 > 0. Because of hypothesis H( f )1 (iv), from (3.43) we infer that∥∥x0 + y+n

∥∥p ≤ c8∥∥x0 + y+

n

∥∥ + M5 with M5 > 0,

hence {y+

n

}n≥1 ⊆ W 1,p

n (Z) is bounded,

and we conclude that

{yn}n≥1 ⊆ W 1,pn (Z) is bounded.

Therefore, we may assume that

ynw−→ y in W 1,p

n (Z), yn → y in L p(Z), yn(z) → y(z) a.e. on Z ,

and

|yn(z)| ≤ k(z) for a.a. z ∈ Z and all n ≥ 1, with k ∈ L p(Z)+.

From the choice of the sequence {yn}n≥1 ⊆ W 1,pn (Z), we have∣∣∣∣∣∣〈A (x0 + yn) , yn − y〉 + β

∫Z

|x0 + yn |p−2 (x0 + yn) (yn − y) dz

−∫Z

N(x0 + y+

n

)χ{yn≥0} (yn − y) dz +

∫Z

u∗n (yn − y) dz

∣∣∣∣∣∣ ≤ εn ‖yn − y‖ .

Passing to the limit as n → ∞, we obtain

〈A (x0 + yn) , yn − y〉 → ∞ as n → ∞.

But A is an (S)+-map (see Proposition 2). So, it follows that

yn → y in W 1,pn (Z),

hence ψ+ satisfies the nonsmooth PS-condition. ��In the next proposition, we produce two additional nontrivial smooth solutions of constant

sign.

Proposition 7 If hypotheses H( f )1 hold, then problem (1.1) has at least four solutionsx0, x ∈ intC+, x0 ≤ x, x0 �= x and v0, v ∈ −intC+, v ≤ v0, v0 �= v.

Proof It is well known that hypothesis H( f )1 (iv) implies that

F(z, x) ≥ c9 |x |µ − c10 for a.a. z ∈ Z and all x ∈ R (3.44)

with c9, c10 > 0. Let σ ∈ R, σ > 0. Then

ψ+ (σ ) = β

ρ‖x0 + σ‖p

p − β

ρ‖x0‖p

p −∫Z

F (z, x0 + σ) dz + ξ+

≤ β

ρ‖x0 + σ‖p

p − c9 ‖x0 + σ‖µµ + c11 for some c11 > 0

≤ c12 ‖x0 + σ‖pµ − c9 ‖x0 + σ‖µµ + c11 for some c12 > 0;

123


hence ψ+ (σ ) → −∞ as σ → +∞ (since µ > p). This fact combined with Propositions 5and 6, permits the use of the nonsmooth mountain pass theorem (see Chang [11], Gasinski andPapageorgiou [18, p. 183 and Theorem 2], with the PS-condition replaced by the nonsmoothPS-condition). So, we obtain y0 ∈ W 1,p

n (Z), y0 �= 0 such that

0 ∈ ∂ψ+ (y0);hence

A (x0 + y0)+ βK p (x0 + y0) = N(x0 + y+

0

)χ{y0≥0} − u∗ (3.45)

with

u∗ ∈ Lr ′(Z), u∗(z) ∈ f (z, x0(z)) ∂θ

(y−

0 (z))

a.e. on Z .

From (3.45), as before, we obtain⎧⎪⎨⎪⎩

− p (x0 + y0) (z)+ β |(x0 + y0) (z)|p−2 (x0 + y0) (z)

= f(z,(x0 + y+

0

)(z)

)χ{y0≥0}(z)− u∗(z) a.e. on Z ,

∂(x0+y0)∂n = 0 on Z .

(3.46)

The nonlinear regularity theory implies that

x0 + y0 ∈ C1n

(Z),

hence y0 ∈ C1n

(Z). So, in (3.46) the Neumann boundary condition is understood in the

pointwise sense and

∂y0

∂n(z) = 0 for all z ∈ ∂Z .

From (3.46), and since x0 ∈ intC+ is a solution of (1.1), we see that

u∗(z) = 0 a.e. on {y0 = 0} . (3.47)

On (3.45) we act with the test function −y−0 ∈ W 1,p

n (Z). We obtain

⟨A (x0 + y0) ,−y−

0

⟩ + β

∫Z

|x0 + y0|p−2 (x0 + y0)(−y−

0

)dz

=∫Z

f(z, x0 + y+

0

)χ{y0≥0}

(−y−0

)dz +

∫Z

f (z, x0)(−y−

0

)dz

(see (3.29) with yn replaced by y0); hence

⟨A(x0 − y−

0

),−y−

0

⟩ + β

∫Z

∣∣x0 − y−0

∣∣p−2 (x0 − y−

0

) (−y−0

)dz

= ⟨A (x0) ,−y−

0

⟩ + β

∫Z

|x0|p−2 x0(−y−

0

)dz

(since x0 ∈ intC+ is a solution of (1.1)). Therefore,⟨A(x0 − y−

0

) − A(x0),−y−0

⟩+ β

∫Z

(∣∣x0 − y−0

∣∣p−2 (x0 − y−

0

) − |x0|p−2 x0

) (−y−0

)dz = 0. (3.48)

123


But the map x → A(x) + βK p(x) is strongly monotone from W 1,pn (Z) into W 1,p

n (Z)∗.So, from (3.48) we infer that y−

0 = 0, hence y0 ≥ 0. Set x = x0 + y0. Then x ∈ intC+, x0 ≤x, x0 �= x and x is a solution of (1.1).

Similarly, working withψ−,we obtain v ∈ −intC+, v ≤ v0, v �= v0. Therefore, we haveproduced four nontrivial smooth solutions of constant sign, namely x0, x, v0, v. ��

4 Nodal solutions

In this section we establish the existence of a smooth nodal solution. To achieve this weimplement the following strategy. First, we show that problem (1.1) has a smallest positivesolution x+ ∈ intC+ and a biggest negative solution v− ∈ −intC+. Then, we focus onthe order interval

[v−, x+

]. Using suitable truncations of the nonlinearity and a variational

argument, we produce a solution y0 of (1.1) in the interval[v−, x+

], which is distinct from

v− and x+. Then, using Proposition 1 and Theorem 3, we can show that y0 is nontrivial,hence y0 must be nodal.

Working in the framework dictated by this strategy, first we prove some lattice-type prop-erties for the sets of upper and, respectively, lower solutions for problem (1.1). We start byrecalling the definitions of upper and lower solutions for problem (1.1).

Definition (a) A function x ∈ W 1,p(Z) is an “upper solution” for problem (1.1), if∫Z

‖Dx‖p−2 (Dx, Dh)RN dz + β

∫Z

|x |p−2 xhdz ≥∫Z

f (z, x) hdz

for all h ∈ W 1,pn (Z), h(z) ≥ 0 a.e. on Z . We say that an upper solution x is a “strict

upper solution” for problem (1.1), if it is not a solution of (1.1) .(b) A function x ∈ W 1,p(Z) is a “lower solution” for problem (1.1), if∫

Z

∥∥Dx∥∥p−2 (

Dx, Dh)RN dz + β

∫Z

∣∣x∣∣p−2xhdz ≤

∫Z

f(z, x

)hdz

for all h ∈ W 1,pn (Z), h(z) ≥ 0 a.e. on Z . We say that a lower solution x is a “strict

lower solution” for problem (1.1), if it is not a solution of (1.1).

Definition We say that a nonempty set S ⊆ W 1,p(Z) is downward (respectively upward)directed, if for every y1, y2 ∈ S, there exists y ∈ S, such that y ≤ y1 and y ≤ y2 (respectivelyy1 ≤ y and y2 ≤ y).

Lemma 1 The set of upper solutions for problem (1.1) is downward directed. More precisely,if y1, y2 are two upper solutions for problem (1.1), then min {y1, y2} is an upper solutiontoo.

Proof Let y1, y2 be two upper solutions for problem (1.1). Given ε > 0, we consider thetruncation function ξε : R → R defined by

ξε (s) =⎧⎨⎩

−ε if s ≤ −εs if s ∈ [−ε, ε]ε if s ≥ ε.

Clearly ξε is Lipschitz continuous. So, we have

ξε((y1 − y2)

−) ∈ W 1,p(Z)

123


and the chain rule implies

Dξε((y1 − y2)

−) = ξ ′ε

((y1 − y2)

−) D (y1 − y2)− (4.1)

(see [23]). Consider the test function ψ ∈ C1n

(Z), ψ ≥ 0. Then

ξε((y1 − y2)

−)ψ ∈ W 1,p(Z) ∩ L∞(Z)

and

D(ξε((y1 − y2)

−)ψ) = ψD(ξε((y1 − y2)

−)) + ξε((y1 − y2)

−) Dψ. (4.2)

Since y1 and y2 are upper solutions for problem (1.1), we have

⟨A (y1) , ξε

((y1 − y2)

−)ψ ⟩ + β

∫Z

K p (y1) ξε((y1 − y2)

−)ψdz

≥∫Z

f (z, y1) ξε((y1 − y2)

−)ψdz

and

⟨A(y2),

(ε − ξε

((y1 − y2)

−))ψ ⟩ + β

∫Z

K p(y2)(ε − ξε

((y1 − y2)

−))ψdz

≥∫Z

f (z, y2)(ε − ξε

((y1 − y2)

−))ψdz.

Adding these two inequalities, we obtain⟨A (y1) , ξε

((y1 − y2)

−)ψ ⟩ + ⟨A (y2) ,

(ε − ξε

((y1 − y2)

−))ψ ⟩+ β

∫Z

K p(y1)ξε((y1 − y2)

−)ψdz + β

∫Z

K p(y2)(ε − ξε

((y1 − y2)

−))ψdz

≥∫Z

f (z, y1)ξε((y1 − y2)

−)ψdz +∫Z

f (z, y2)(ε − ξε

((y1 − y2)

−))ψdz. (4.3)

Using (4.1) and (4.2), we have

⟨A (y1) , ξε

((y1 − y2)

−)ψ ⟩ =∫Z

‖Dy1‖p−2 (Dy1, D((y1−y2)

−))RN ξ

′ε

((y1−y2)

−)ψdz

+∫Z

‖Dy1‖p−2 (Dy1, Dψ)RN ξε((y1 − y2)

−) dz

= −∫

{−ε≤y1−y2≤0}‖Dy1‖p−2 (Dy1, D ((y1 − y2)))RN ψdz

+∫Z

‖Dy1‖p−2 (Dy1, Dψ)RN ξε((y1 − y2)

−)ψdz, (4.4)

123


and in a similar way

⟨A(y2),

(ε − ξε

((y1 − y2)

−))ψ ⟩=

∫{−ε≤y1−y2≤0}

‖Dy2‖p−2 (Dy2, D (y1 − y2))RN ψdz

+∫Z

‖Dy2‖p−2 (Dy2, Dψ)RN

(ε − ξε

((y1 − y2)

−)) dz. (4.5)

From (4.4) and (4.5), the fact that ψ ≥ 0 and the monotonicity of the homeomorphismσp : R

N → RN defined by

σp(y) = ‖y‖p−2 y,

we have

⟨A (y1) , ξε

((y1 − y2)

−)ψ ⟩ + ⟨A (y2) ,

(ε − ξε

((y1 − y2)

−))ψ ⟩≤∫Z

‖Dy1‖p−2 (Dy1, Dψ)RN ξε((y1 − y2)

−) dz

+∫Z


(ε − ξε

((y1 − y2)

−)) dz. (4.6)

We return to (4.3), use (4.6) and divide by ε > 0. Then

∫Z

‖Dy1‖p−2 (Dy1, Dψ)RN1

εξε((y1 − y2)

−) dz

+∫Z


(1 − 1

εξε((y1 − y2)

−)) dz

+ β

∫Z

K p (y1)1

εξε((y1 − y2)

−)ψdz + β

∫Z

K p (y2)

(1 − 1

εξε((y1 − y2)

−))ψdz

≥∫Z

f (z, y1)1

εξε((y1 − y2)

−)ψdz

+∫Z

f (z, y2)

(1 − 1

εξε((y1 − y2)

−))ψdz. (4.7)

Note that

1

εξε((y1 − y2)

− (z)) → χ{y1<y2}(z) a.e. on Z as ε ↓ 0,

and

χ{y1≥y2} = 1 − χ{y1<y2}.

123


So, if we pass to the limit as ε ↓ 0 in (4.7), we obtain∫{y1<y2}

‖Dy1‖p−2 (Dy1, Dψ)RN dz +∫

{y1≥y2}‖Dy2‖p−2 (Dy2, Dψ)RN dz

+ β

∫{y1<y2}

K p (y1) ψdz + β

∫{y1≥y2}

K p (y2) ψdz

≥∫

{y1<y2}f (z, y1) ψdz +

∫{y1≥y2}

f (z, y2) ψdz. (4.8)

We know that y = min {y1, y2} ∈ W 1,p(Z) and

Dy(z) ={

Dy1(z) for a.a. z ∈ {y1 < y2}Dy2(z) for a.a. z ∈ {y1 ≥ y2} .

Also note that

f (z, y(z)) = χ{y1<y2}(z) f (z, y1(z))+ χ{y1≥y2}(z) f (z, y2(z)).

Therefore from (4.8), we have

〈A(y), ψ〉 + β

∫Z

K p(y)ψdz ≥∫Z

f (z, y) ψdz. (4.9)

Because C1n

(Z)+ is dense in W 1,p

n (Z)+, from (4.9) we conclude that y = min {y1, y2} isan upper solution too. ��

In a similar manner, we can prove the following lemma:

Lemma 2 The set of lower solutions for problem (1.1) is upward directed. More precisely,if w1, w2 are two lower solutions for problem (1.1), then w = max {w1, w2} is a lowersolution too.

Now, that we have established these lattice-type properties for the sets of upper and lowersolutions, we can proceed and show that the problem (1.1) has a smallest positive solutionx+ ∈ intC+ and a biggest negative solution v− ∈ −intC+.To this end, we need to strengthenour hypotheses on the nonlinearity f (z, x). The new hypotheses on f are:

H( f )2: f : Z × R →R is a function such that f (z, 0) = 0 a.e. on Z and



| f (z, x)| ≤ a(z)+ c |x |r−1,


{pβ, p

β

}and M > 0, such that for almost all z ∈ Z

and all |x | ≥ M

f (z, x) x ≥ µF(z, x) > 0,

where F(z, x) = ∫ x0 f (z, s)ds;

123


(v) there exist a− < 0 < a+ such that for almost all z ∈ Z

f (z, a−) = 0 = f (z, a+)


0, a+]



(vi) there exist δ > 0 and η > ξ > β such that for almost all z ∈ Z

ξ x p−1 ≤ f (z, x) ≤ ηx p−1 for x ∈ [0, δ]

and

η |x |p−2 x ≤ f (z, x) ≤ ξ |x |p−2 x for x ∈ [−δ, 0] .

Remark It is easily seen that the function in Example 1 satisfies assumption H( f )2 (vi)with

ξ = c2

(1 − e− 1

2

), η = c2. In hypothesis H( f )2 (vi), we can always take δ < min {a+, a−} .

Note that

x = a+ ∈ int C+ is a strict upper solution for problem (1.1).

Also, if γ ∈ (0, δ] , then

x = γ ∈ int C+ is a strict lower solution for problem (1.1).

Similarly, in the negative half-axis

v = a− ∈ −int C+ is a strict lower solution for problem (1.1).

Also, if λ ∈ [−δ, 0) , then

v = λ ∈ −int C+ is a strict upper solution for problem (1.1).

Using the ordered upper–lower solution pairs{

x, x},{v, v

},we introduce the order intervals

[x, x

] ={

x ∈ W 1,pn (Z) : x ≤ x(z) ≤ x a.e. on Z

}and [

v, v] =

{v ∈ W 1,p

n (Z) : v ≤ v(z) ≤ v a.e. on Z}.

Proposition 8 If hypotheses H( f )2 hold, then problem (1.1) has a smallest solution in[x, x

]and a biggest solution in

[v, v

].

Proof Let S+ be the set of solutions of (1.1), which belong to the order interval J+ = [x, x

].

By choosing x small we ensure that S+ is nonempty (see the proof of Proposition 4). Weshow that the set S+ is downward directed. To this end, let x1, x2 ∈ S+. Then both x1 and x2

are upper solutions for problem (1.1). According to Lemma 1, x = min {x1, x2} ∈ W 1,pn (Z)

is an upper solution too. We consider the following order interval

J+ = [x, x

] ={

x ∈ W 1,pn (Z) : x ≤ x(z) ≤ x(z) a.e. on Z

}.

123


Define

f+(z, x) =⎧⎨⎩

f(z, x

)if x < x

f (z, x) if x ≤ x ≤ x(z)f (z, x(z)) if x(z) < x

.

Evidently, f+ is a Caratheodory function. We consider its primitive

F+(z, x) =x∫

0

f+(z, s)ds,

and then define the functional ϕ+ : W 1,pn (Z) → R by

ϕ+(x) = 1

p‖Dx‖p

p + β

p‖x‖p

p −∫Z

F+ (z, x(z)) dz for all x ∈ W 1,pn (Z).

Note that ϕ+ ∈ C1(

W 1,pn (Z)

). Moreover, it is clear that ϕ+ is coercive and weakly lower

semicontinuous. So, by the Weierstrass theorem, we can find x0 ∈ J+ such that

ϕ+ (x0) = infx∈ J+

ϕ+(x).

For any y ∈ J+, let

k0(t) = ϕ+ (t y + (1 − t )x0) , t ∈ [0, 1] .

We have k0(0) ≤ k0(t) for all t ∈ [0, 1], hence

0 ≤ 〈A (x0) , y − x0〉 + β

∫Z

|x0|p−2 x0 (y − x0) dz

−∫Z

f+ (z, x0) (y − x0) dz for all y ∈ J+. (4.10)

Given ε > 0 and h ∈ W 1,pn (Z), we define

y(z) =

⎧⎪⎨⎪⎩

x if z ∈ {x0 + εh ≤ x

}x0(z)+ εh(z) if z ∈ {

x < x0 + εh < x}

x(z) if z ∈ {x ≤ x0 + εh} .Clearly, y ∈ J+. So, we can use it as a test function in (4.10). Then

0 ≤ ε

∫Z

‖Dx0‖p−2 (Dx0, Dh)RN dz + εβ

∫Z

|x0|p−2 x0hdz − ε

∫Z

f+ (z, x0) hdz

−∫

{x≤x0+εh}‖Dx‖p−2 (Dx, D (x0 + εh − x))RN dz

−∫

{x≤x0+εh}

(β |x |p−2 x (x0 + εh − x)− f+ (z, x) (x0 + εh − x)

)dz

123


+∫

{x0+εh≤x}

(β∣∣x∣∣p−2

x(x − x0 − εh

) − f+(z, x

) (x − x0 − εh

))dz

+∫

{x≤x0+εh}

(f+ (z, x)− f+ (z, x0)

)(x − x0 − εh) dz

+∫

{x0+εh≤x}

(f+

(z, x

) − f+ (z, x0)) (

x − x0 − εh)

dz −∫

{x0+εh≤x}‖Dx0‖p dz

−β∫

{x0+εh≤x}

(|x0|p−2 x0 − ∣∣x∣∣p−2

x) (

x0 − x)

dz

− ε∫

{x0+εh≤x}‖Dx0‖p−2 (Dx0, Dh)RN dz

− εβ∫

{x0+εh≤x}

(|x0|p−2 x0 − ∣∣x∣∣p−2

x)

hdz

+∫

{x≤x0+εh}

(‖Dx‖p−2 Dx − ‖Dx0‖p−2 Dx0, D (x0 − x))RN dz

+β∫

{x≤x0+εh}

(|x |p−2 x − |x0|p−2 x0)(x0 − x) dz

+ ε∫

{x≤x0+εh}

(‖Dx‖p−2 Dx − ‖Dx0‖p−2 Dx0, Dh)RN dz

+ εβ∫

{x≤x0+εh}

(|x |p−2 x − |x0|p−2 x0)

hdz. (4.11)

Since x0 ∈ intC+ is an upper solution for problem (1.1), we have

−∫

{x≤x0+εh}‖Dx‖p−2 (Dx, D (x + εh − x0))RN dz

+∫

{x≤x0+εh}

(β |x |p−2 x (x0 + εh − x)− f+ (z, x) (x0 + εh − x)

)dz ≤ 0 (4.12)

Also, because x ∈ R+ is a lower solution for problem (1.1), we have

∫{x0+εh≤x}

(β∣∣x∣∣p−2

x(x − x0 − εh

) − f+(z, x

) (x − x0 − εh

))dz ≤ 0. (4.13)

123


Recall that the homeomorphism σp : Rk → R

k (k ≥ 1 ) defined by σp(y) = ‖y‖p−2Rk y is

monotone. Hence∫{x ≤ x0+εh}

(‖Dx‖p−2 Dx − ‖Dx0‖p−2 Dx0, Dx0 − Dx)Rk dz ≤ 0, (4.14)

∫{x ≤ x0+εh}

(|x |p−2 x − |x0|p−2 x0)(x0 − x) dz ≤ 0 (4.15)

and

−∫

{x0+εh≤x}

(|x0|p−2 x0 − ∣∣x∣∣p−2

x) (

x0 − x)

dz ≤ 0. (4.16)

Note that since x0 ≤ x , we have

h(z) ≥ 0 for a.a. z ∈ {x ≤ x0 + εh} .This combined with hypothesis H( f )2(iii) yields∫

{x≤x0+εh}

(f+ (z, x)− f+ (z, x0)

)(x − x0 − εh) dz ≤ c13

∫{x≤x0+εh}

(x0 − x + εh) dz

≤ εc13

∫{x0<x≤x0+εh}

hdz (4.17)

for some c13 > 0. Similarly, for some c∗13 > 0∫

{x0+εh≤x}

(f+

(z, x

) − f+ (z, x0)) (

x − x0 − εh)

dz

≤ c∗13

∫{x0+εh≤x}

(x − x0 − εh

)dz

≤ −εc∗13

∫{x0+εh≤x<x0}

hdz. (4.18)

We return to (4.11) and use (4.12)–(4.18). We obtain

0 ≤ ε

∫Z

(‖Dx0‖p−2 (Dx0, Dh)RN + β |x0|p−2 x0h − f+ (z, x0) h)

dz

+ εc13

∫{x0<x≤x0+εh}

hdz − εc∗13

∫{x0+εh≤x<x0}

hdz

− ε

∫{x0+εh≤x}

‖Dx0‖p−2 (Dx0, Dh)RN dz

− εβ

∫{x0+εh≤x}

(|x0|p−2 x0 − ∣∣x∣∣p−2

x)

hdz

123


+ ε

∫{x≤x0+εh}

(‖Dx‖p−2 Dx − ‖Dx0‖p−2 Dx0, Dh)RN

+ εβ

∫{x≤x0+εh}

(|x |p−2 x − |x0|p−2 x0)

hdz. (4.19)

From Stampacchia’s theorem (see, for example [19, pp. 195–196])

Dx0(z) = 0 a.e. on{

x0 = x}

and Dx0(z) = Dx(z) a.e. on {x0 = x} .So, if we divide (4.19) by ε > 0 and then let ε ↓ 0, we obtain

0 ≤∫Z

‖Dx0‖p−2 (Dx0, Dh)RN dz + β

∫Z

|x0|p−2 x0hdz −∫Z

f+ (z, x0) hdz. (4.20)

Since h ∈ W 1,pn (Z) is arbitrary, from (4.20) we infer that

A (x0)+ βK p (x0) = N+ (x0) , (4.21)

where N+(x)(.) = f+ (., x(.)) for all x ∈ W 1,pn (Z). From (4.21) it follows that{−p x0(z)+ β x0(z)p−1 = f (z, x0(z)) a.e. on Z ,

∂ x0∂n = 0 on ∂Z ,

since x0 ∈ J+. Therefore x0 ∈ intC+ (cf. nonlinear regularity theory) is a solution of (1.1).Since x0 ≤ x, it follows that S+ is downward directed.

Consider a chain D ⊆ S+ (i.e., D is a totally ordered subset of S+). From Dunford andSchwartz [15, Corollary 7, p. 336], we know that we can find {xn}n≥1 ⊆ D such that

infn≥1

xn = inf D.

Since S+ is downward directed, we may assume that {xn}n≥1 is decreasing. Also, becausethe xn’s are solutions of (1.1), we can find c14 > 0 such that

‖xn‖ ≤ c14 for all n ≥ 1.

Hence, we may assume that

xnw−→ u in W 1,p

n (Z) and xn → u in Lr (Z) as n → ∞.

We have

A(xn)+ βK p(xn) = N (xn). (4.22)

Acting with xn − u ∈ W 1,pn (Z) in (4.22) and then passing to the limit as n → ∞, we obtain

limn→∞ 〈A(xn), xn − u〉 = 0;

hence

xn → u in W 1,pn (Z)

(see Proposition 2). So, if we pass to the limit as n → ∞ in (4.22), we arrive at

A (u)+ βK p (u) = N (u) ;

123


hence

u ∈ S+ and u = inf D.

Invoking Zorn’s lemma, we obtain x∗ ∈ S+, a minimal element of S+. Since S+ is downwarddirected, we conclude that x∗ is the smallest solution of problem (1.1) in the order interval[x, x

].

Similarly, we obtain v∗ ∈ −intC+, the biggest solution of problem (1.1) in the interval[v, v

]. ��

Using this proposition, we can show that problem (1.1) has a smallest positive solutionand a biggest negative one.

Proposition 9 If hypotheses H( f )2 hold, then problem (1.1) has a smallest positive solutionx+ ∈ intC+ and a biggest negative solution v− ∈ −intC+.

Proof Let xn ∈ R+, xn ↓ 0 and J n+ = [xn, x

]. According to Proposition 8, problem (1.1)

has a smallest solution xn∗ in J n+. We know that{

xn∗}

n≥1 ⊆ W 1,pn (Z) is bounded and so we

may assume that

xn∗w−→ x+ in W 1,p

n (Z), and xn∗ → x+ in L p(Z) and Lr (Z) as n → ∞.

We know that

A(xn∗) + βK p

(xn∗) = N

(xn∗). (4.23)

As before, acting with xn∗ − x+ in (4.23), passing to the limit as n → ∞ and using the factthat A is an (S)+ −map (see Proposition 2), we obtain

xn∗ → x+ in W 1,pn (Z).

If x+ = 0, then∥∥xn∗

∥∥ → 0. We set yn = xn∗‖xn∗‖ , n ≥ 1. Then ‖yn‖ = 1 for all n ≥ 1 and we

may assume that

ynw−→ y in W 1,p

n (Z) and yn → y in L p(Z) as n → ∞.

We have

A(yn)+ βK p(yn) = N(xn∗)∥∥xn∗∥∥ (4.24)

(see (4.23)). From hypotheses H( f )2 (i i i) and (vi) , we have

| f (z, x)| ≤ c15(|x |p−1 + |x |r−1)

= c15(1 + |x |r−p) |x |p−1 for a.a. z ∈ Z and all x ∈ R

with c15 > 0, hence∣∣ f(z, xn∗ (z)

)∣∣ ≤ c16∣∣xn∗ (z)

∣∣p−1 for a.a. z ∈ Z , all n ≥ 1 and some c16 > 0;therefore {

hn = N(xn∗)

∥∥xn∗∥∥p−1

}n≥1

⊆ L p′(Z) is bounded

(1

p+ 1

p′ = 1

).

123


Hence, we may assume that

hnw−→ h in L p′

(Z) as n → ∞.

In (4.24), we act with yn − y and pass to the limit as n → ∞ to obtain

limn→∞ 〈A(yn), yn − y〉 = 0.

Because of Proposition 2, it follows that

yn → y in W 1,pn (Z), ‖y‖ = 1 and y ≥ 0.

By virtue of hypothesis H( f )2 (vi) and arguing as in Aizicovici et al. [2] (see the Proof ofProposition 31), we obtain

h = gy p−1 with g ∈ L∞(Z)+, ξ ≤ g(z) ≤ η a.e. on Z .

So, if in (4.24) we pass to the limit as n → ∞, then

A(y)+ βK p(y) = gK p(y);hence {−p y(z) = (g(z)− β) |y(z)|p−2 y(z) a.e. on Z ,

∂y∂n = 0 on ∂Z .

(4.25)

Since g(z) ≥ ξ > β a.e. on Z (see hypothesis H( f )2 (vi)), from (4.25) it follows that ymust change sign, a contradiction. So, x+ �= 0.

Passing to the limit as n → ∞ in (4.23), we obtain

A (x+)+ βK p (x+) = N (x+) ,

hence {−px+(z)+ βx+(z)p−1 = f (z, x+(z)) a.e. on Z ,∂x+∂n = 0 on ∂Z .

(4.26)

From the nonlinear regularity theory, x+ ∈ C+\ {0} and of course x+ is a solution of problem(1.1). From (4.26), by invoking the nonlinear strong maximum principle of Vazquez [29],we conclude that x+ ∈ intC+.

We claim that x+ is the smallest positive solution of (1.1). Indeed, let x ∈ C+\ {0} bea solution of (1.1). Since the set of upper solutions is downward directed, without any lossof generality, we may assume that x ≤ x . As before, via the nonlinear strong maximumprinciple of Vazquez [29], we have x ∈ intC+. For n ≥ 1 large, we will have xn ≤ x andthen xn∗ ≤ x, hence x+ ≤ x .

Similarly, we obtain v− ∈ −intC+, the biggest negative solution of (1.1). ��To obtain a nodal solution, we need to further strengthen our hypotheses on the nonlinearity

f (z, x). The new hypotheses are the following:



123



| f (z, x)| ≤ a(z)+ c |x |r−1 ,


{pβ, p

β


and all |x | ≥ M

f (z, x)x ≥ µF(z, x) > 0;(v) there exist a− < 0 < a+ such that for almost all z ∈ Z

f (z, a−) = 0 = f (z, a+)


0, a+]



(vi) there exist δ > 0 and η > ξ > λ1 + β such that for almost all z ∈ Z ,

ξ x p−1 ≤ f (z, x) ≤ ηx p−1, for 0 ≤ x ≤ δ

and

η |x |p−2 x ≤ f (z, x) ≤ ξ |x |p−2 x for − δ ≤ x ≤ 0.

The function in Example 1 satisfies condition H( f )3 (vi) provided that c2

(1 − e− 1

2

)>

λ1 + β. With these stronger hypotheses on f (z, x), we can produce a nodal solution andstate the full multiplicity result for problem (1.1).

Theorem 4 If hypotheses H( f )3 hold, then problem (1.1) has at least five nontrivial smoothsolutions, x0, x ∈ intC+, x0 ≤ x, x0 �= x, v0, v ∈ −intC+, v ≤ v0, v0 �= v and y0 ∈C1

n

(Z), a nodal solution.

Proof By Proposition 7, we already have four smooth constant sign solutions x0, x ∈intC+, x0 ≤ x, x0 �= x, v0, v ∈ −intC+, v ≤ v0, v0 �= v. We need to produce the nodalsolution y0 ∈ C1

n(Z).To this end, let x+ ∈ intC+, v− ∈ −intC+ be the two extremal constant sign solutions

of (1.1) obtained in Proposition 9. We consider the following truncations of the nonlinearityf (z, x) :

f+(z, x) =⎧⎨⎩

0 if x < 0f (z, x) if 0 ≤ x ≤ x+(z)f (z, x+(z)) if x+(z) < x,

f−(z, x) =⎧⎨⎩

f (z, v−(z)) if x < v−(z)f (z, x) if v−(z) ≤ x ≤ 00 if 0 < x,

and

f0(z, x) =⎧⎨⎩

f (z, v−(z)) if x < v−(z)f (z, x) if v−(z) ≤ x ≤ x+(z)f (z, x+(z)) if x+(z) < x .

123


We also introduce the corresponding primitive functions

F±(z, x) =x∫

0

f±(z, s)ds and F0(z, x) =x∫

0

f0(z, s)ds.

Finally, we define the functionals ϕ±, ϕ0 : W 1,pn (Z) → R by

ϕ±(x) = 1

p‖Dx‖p

p + β

p‖x‖p

p −∫Z

F± (z, x(z)) dz

and

ϕ0(x) = 1

p‖Dx‖p

p + β

p‖x‖p

p −∫Z

F0 (z, x(z)) dz

for all x ∈ W 1,pn (Z). Evidently, ϕ±, ϕ0 ∈ C1

(W 1,p

n (Z)).

We also consider the following three order intervals

T+ = [0, x+

], T− = [

v−, 0]

and T0 = [v−, x+

].

Claim 1 The critical points of ϕ+, ϕ−, ϕ0 are in T+, T−, T0,, respectively.

We prove Claim 1 for ϕ0, the proof for ϕ+ and ϕ− being similar. So, let x ∈ W 1,pn (Z) be

a critical point of ϕ0. Then ϕ′0(x) = 0, hence

A(x)+ βK p(x) = N0(x) (4.27)

with

N0(u)(.) = f0(., u(.)) for all u ∈ W 1,pn (Z).

On (4.27) we act with (x − x+)+ ∈ W 1,pn (Z). Then

⟨A(x), (x − x+)+

⟩ + β

∫Z

|x |p−2 x(x − x+)+dz

=∫Z

f0(z, x)(x − x+)+dz

=∫Z

f (z, x+)(x − x+)+dz

= ⟨A(x+), (x − x+)+

⟩ + β

∫Z

|x+|p−2 x+(x − x+)+dz,

hence⟨A(x)− A(x+), (x − x+)+

⟩ + β

∫Z

(|x |p−2 x − |x+|p−2 x+)(x − x+)+dz = 0. (4.28)

Due to the strong monotonicity of the operator x → A(x)+ βK p(x), from (4.28) it followsthat

(x − x+)+ = 0, i.e., x ≤ x+.

In a similar fashion, we show that v− ≤ x . Hence x ∈ T0.

123


Claim 2 The critical points of ϕ+ are {0, x+} and the critical points of ϕ− are {v−, 0}.We do the proof for ϕ+, the proof for ϕ− being similar. So, let x �= 0 be a critical point

of ϕ+. Then ϕ′+(x) = 0, hence

A(x)+ βK p(x) = N+(x)

with

N+(u)(.) = f+(., u(.)) for all u ∈ W 1,pn (Z).

We act with −x− ∈ W 1,pn (Z) in the above relation. Then∥∥Dx−∥∥p

p + β∥∥x−∥∥p

p = 0

(recall the definition of f+), hence

x− = 0, and so x ≥ 0, x �= 0.

Due to the extremality of x+ (see Proposition 9), we must have x = x+.

Claim 3 x+ and v− are local minimizers of ϕ0.

Let 0 < γ ≤ min{δ,minZ x+} (recall that x+ ∈ intC+). Then

ϕ+(γ ) = β

pγ p |Z |N −

∫Z

F+(z, γ )dz

≤ β

pγ p |Z |N − ξ

pγ p |Z |N < 0

(see hypotheses H( f )3(vi)), hence

inf ϕ+ < 0 = ϕ+(0). (4.29)

Note that the functional ϕ+ is coercive and sequentially weakly lower semicontinuous. So,by the Weierstrass theorem, we can find w ∈ W 1,p

n (Z) such that

inf ϕ+ = ϕ+(w),

hence w �= 0 (see (4.29)). Since w �= 0 is a critical point of ϕ+, from Claim 2 we infer thatw = x+ ∈ int C+. So, we can find r > 0 small such that

ϕ+ |B

C1n( Z )

r (x+)= ϕ0 |

BC1

n( Z )r (x+)

,

where

BC1

n(

Z)

r (x+) = {x ∈ C1

n

(Z) : ‖x − x+‖ ≤ r

}.

This means that x+ is a local C1n

(Z)-minimizer of ϕ0, and by Proposition 3 it is also a local

W 1,pn (Z)− minimizer of ϕ0.

Similarly, we show that v− ∈ −intC+ is a local W 1,pn (Z)−minimizer of ϕ0.

Without any loss of generality, we may assume that v− and x+ are isolated local minimiz-ers and critical points of ϕ0. Otherwise, we can find {xn}n≥1 ⊆ W 1,p

n (Z), xn /∈ {0, x+, v−}such that, for example, xn → x+ and

ϕ′0(xn) = 0.

123


Then

xn ∈ T0

(see Claim 1), hence xn is nodal. Therefore, we have produced a whole sequence of distinctnodal solutions and so we are done. Arguing as in the proof of Proposition 5, we can findr > 0 small, such that

ϕ0 (x+) < inf {ϕ0(x) : ‖x − x+‖ = r}and

ϕ0 (v−) < inf {ϕ0(v) : ‖v − v−‖ = r} .We may assume without any loss of generality that

ϕ0 (v−) ≤ ϕ0 (x+) .

If

E0 = {v−, x+} , E = T0

and

D = ∂Br (x+) ={

x ∈ W 1,pn (Z) : ‖x − x+‖ = r

},

then we can easily check (see, for example, [19, p. 642]) that the pair {E0, E} links withD in W 1,p

n (Z). The functional ϕ0 being coercive, it is readily verified that it satisfies thePS-condition. So, we can apply Theorem 1 and find y0 ∈ W 1,p

n (Z) such that

y0 �= v−, x+;ϕ′0 (y0) = 0 and ϕ0 (y0) = inf

γ0∈�0max

t∈[−1,1]ϕ0 (γ0 (t)) , (4.30)

where

�0 ={γ0 ∈ C

([−1, 1] ,W 1,p

n (Z))

: γ0(−1) = v−, γ0 (1) = x+}.

Since y0 is a critical point of ϕ0, it follows that y0 is a solution of problem (1.1) and y0 ∈C1

n

(Z)

(by the nonlinear regularity theory). Since y0 ∈ T0 (Claim 1), if we can show that y0

is nontrivial, then y0 must be nodal (recall that v−, x+ are extremal constant sign solutionsand y0 �= v−, x+; see (4.30)). In view of (4.30), to show the nontriviality of y0, it suffices toproduce a continuous path that joins v− and x+, along which ϕ0 is negative.

To this end, let S = W 1,pn (Z) ∩ ∂BL p

1 endowed with the relative W 1,pn (Z)-topology and

Sc = S ∩ C1n

(Z)

endowed with the relative C1n

(Z)-topology. Recall that

∂BL p

1 = {x ∈ L p(Z) : ‖x‖p = 1

}(see Sect. 2). Evidently, Sc is dense in S for the W 1,p

n (Z)− topology. It follows thatC ([−1, 1] , Sc) is dense in C ([−1, 1] , S) . So, if we set

� = {γ ∈ C ([−1, 1] , S) : γ (−1) = −u0, γ (1) = u0} .and

�c = {γ ∈ C ([−1, 1] , Sc) : γ (−1) = −u0, γ (1) = u0} ,then we see that �c is dense in �. Invoking Proposition 1, we can find γ ∈ �c such that

max{‖Dx‖p

p : x ∈ γ ([−1, 1])}< λ1 + δ0, (4.31)

123


with δ0 > 0 such that

2δ0 ≤ ξ − (λ1 + β) (4.32)

(see hypothesis H( f )3 (vi)). Since −v−, x+ ∈ intC+ and γ ∈ C ([−1, 1] , Sc) , we can findε > 0 small, such that

ε |x(z)| ≤ δ for all z ∈ Z and all x ∈ γ ([−1, 1])

and

εx ∈ [v−, x+

]for all x ∈ γ ([−1, 1]) . (4.33)

By virtue of hypothesis H( f )3 (vi) , we have

ξ

p|x |p ≤ F(z, x) for a.a. z ∈ Z and all |x | ≤ δ. (4.34)

If x ∈ γ ([−1, 1]) , then

ϕ0 (εx) = ε p

p‖Dx‖p

p + βε p

p‖x‖p

p −∫Z

F0 (z, εx) dz

≤ ε p

p(λ1 + δ0)+ βε p

p−∫Z

F (z, εx) dz (see (4.31), (4.32))

≤ ε p

p(λ1 + δ0 + β − ξ) (see (4.34))

< 0 (see (4.32)).

Therefore, if γε = εγ, then γε is a continuous path that joins −εu0 and εu0, and we have

ϕ0 |γε< 0. (4.35)

Next, we produce a continuous path that joins εu0 and x+, along which ϕ0 is negative. Forthis purpose, let

a = ϕ+ (x+) and b = 0 = ϕ+ (0) .

We know that a < b = 0.Also, from Claim 2, we know that if K +b is the set of critical points

of ϕ+ at the level b, then K +b = {0} . Since ϕ+ is coercive, it is easily seen that it satisfies

the PS-condition. Because of Claim 2, we can apply Theorem 3 and obtain a continuousdeformation h : [0, 1] × (

ϕb+\ {0}) → ϕb+ such that

h(

1, ϕb+\ {0})

⊆ ϕa+ = {x+} ,h (t, x+) = x+ for all t ∈ [0, 1]

(4.36)

and

h (t, x) ≤ h (s, x) for all t, s ∈ [0, 1] , s ≤ t, x ∈ ϕb+\ {0} . (4.37)

We set

γ+ (t) = h (t, εu0) for all t ∈ [0, 1] .

Evidently, γ+ : [0, 1] → W 1,pn (Z) is continuous and

γ+ (0) = h (0, εu0) = εu0

123


(since h is a continuous deformation),

γ+ (1) = h (1, εu0) = x+(see (4.36)) and

ϕ+ (γ+ (t)) ≤ ϕ+ (γ+ (0)) < 0

(see (4.37) and (4.35)). Therefore, we see that γ+ is a continuous path which joins εu0 andx+ and ϕ+ |γ+< 0. Taking into account that x+(z) ∈ (0, a+) and v−(z) ∈ (a−, 0) for allz ∈ Z we can invoke hypothesis H( f )3(v) to conclude that ϕ0 ≤ ϕ+. Hence

ϕ0 |γ+< 0. (4.38)

In a similar way, we produce a continuous path γ−, which joins −εu0 and v−, and such that

ϕ0 |γ−< 0. (4.39)

We concatenate the paths γ−, γε, γ+, to produce a continuous path γ ∗0 ∈ �0 such that

ϕ0 |γ ∗0< 0

(see (4.35), (4.38) and (4.39)). Then, from (4.30) we infer that

ϕ0 (y0) < 0 = ϕ0 (0) ,

hence y0 �= 0 and so y0 ∈ C1n

(Z)

(cf. the nonlinear regularity theory) is nodal. ��

5 The semilinear problem

In this section we examine the semilinear problem (i.e., the case when p = 2 in (1.1) ) So,the boundary value problem under consideration is the following{−x(z)+ βx(z) = f (z, x(z)) a.e. on Z ,

∂x∂n = 0 on ∂Z .

(5.1)

With additional regularity conditions on the nonlinearity x → f (z, x), we can show thatproblem (5.1) has two nodal solutions, hence a total of six nontrivial smooth solutions withprecise sign information. In what follows, {λk}k≥0 denotes the sequence of distinct eigen-

values of(−, H1

n (Z)), where H1

n (Z) = W 1,2n (Z).

The hypotheses on the nonlinearity f (z, x) are the following:


(i) for all x ∈ R, z → f (z, x) is measurable;(ii) for almost all z ∈ Z , x → f (z, x) is of class C1;

(iii) for almost all z ∈ Z and all x ∈ R we have∣∣ f ′x (z, x)

∣∣ ≤ a(z)+ c |x |r−2 ,

where a ∈ L∞(Z)+, c > 0 and 1 < r < 2∗;(iv) there exist µ > max

{2β, 2

β


and all |x | ≥ M

f (z, x)x ≥ µF(z, x) > 0;

123


(v) there exist a− < 0 < a+ such that for almost all z ∈ Z

f (z, a−) = 0 = f (z, a+)

and {0 ≤ f (z, x) for all x ∈ [

0, a+]

f (z, x) ≤ 0 for all x ∈ [a−, 0

].

(vi) there exists an integer m ≥ 1 such that for almost all z ∈ Z ,

λm ≤ f ′x (z, 0)− β ≤ λm+1

f ′x (z, 0)− β /∈ σ (−)

with strict inequalities on sets (not necessarily the same) of positive measureand if m = 1 then

f ′x (z, 0)− β ≥ σ > λ1.

Theorem 5 If hypotheses H( f )4 hold, then problem (5.1) has at least six nontrivial smoothsolutions

x0, x ∈ intC+, x0 ≤ x, x0 �= x,

v0, v ∈ −intC+, v ≤ v0, v0 �= v,

y0, w0 ∈ C1n

(Z), nodal solutions.

Proof From the nonlinear case we already have the five solutions x0, x, v0, v, y0 (see The-orem 4). Note that now hypothesis H( f )4(v) is weaker than H( f )3(v). The reason for thisis the following: hypothesis H( f )3(v) was used in the proof of Proposition 4 and permittedthe application of the nonlinear strong maximum principle of Vazquez [29]. In the presentsetting, due to the higher regularity of f (z, .)we can modify our reasoning and relax hypoth-esis H( f )3(v) to H( f )4(v). So, we can find k > 0 large enough such that x → f (z, x)+ kxincreases on

[a−, a+

]. Referring to the proof of Proposition 4, we have

− (a+ − x0) (z) = x0(z) = − f (z, x0(z))+ βx0(z) a.e. on Z . (5.2)

We know that x0 ∈ [0, a+

]. To show this, we only needed that f (z, x) ≥ 0 for a.a. z ∈ Z

and all x ∈ [0, a+

](see the proof of Proposition 4). Then

f (z, a+)− f (z, x0(z))+ k (a+ − x0(z)) ≥ 0 a.e. on Z , (5.3)

hence

− f (z, x0(z))+ k (a+ − x0(z)) ≥ 0 a.e. on Z

(see hypothesis H( f )4(v)). Then we can rewrite (5.2) as follows:

− (a+ − x0) (z) = − f (z, x0(z))+ k (a+ − x0(z))

+ (β + k) x0(z)− ka+ for a.a. z ∈ Z ,

hence

− (a+ − x0) (z) ≥ k (x0(z)− a+) for a.a. z ∈ Z

(see (5.3) and recall β > 0, x0 ≥ 0); therefore

(a+ − x0) (z) ≤ k (a+ − x0(z)) for a.a.z ∈ Z .

123


Invoking the Vazquez strong maximum principle (see [29]), we obtain

a+ − x0 ∈ int C+,

which is what we need in order for x0 to be a local minimizer of ϕ (see the proof ofProposition 4).

We use the functionalϕ0 introduced in the proof of Theorem 4. We know thatϕ0 is coerciveand v−, v+, y0 and 0 are critical points of ϕ0.More precisely, v− and x+ are local minimizersof ϕ0 (see Claim 3 in the proof of Theorem 4). So, we have

Ck (ϕ0, x+) = Ck (ϕ0, v−) = δk,0Z for all k ≥ 0 (5.4)

(see [12, p. 33], [24, p. 175]). Also, y0 ∈ C1n

(Z)

is a critical point of ϕ0 of mountain passtype (see the proof of Theorem 4). So,

Ck (ϕ0, y0) = δk,1Z for all k ≥ 0 (5.5)

(see [24, p. 195]). Because of hypothesis H( f )4(vi), the origin is a nondegenerate criticalpoint of ϕ0 with Morse index

τm = dim Hm = dimm⊕

k=0

E (λk) ,

where E(λk) denotes the eigenspace corresponding to λk . Hence

Ck(ϕ0, 0) = δk,τm Z for all k ≥ 0 (5.6)

(see [12, p. 34]). Next, we consider the one-parameter family of functionalsψt : H1n (Z) → R,

t ∈ [0, 1], defined by

ψt (x) = 1

2‖Dx‖2

2 + β

2‖x‖2

2 − t∫Z

F0(z, x(z))dz for all x ∈ H1n (Z).

Clearly, ψ ′t and ∂tψt are both locally Lipschitz. Also, we claim that there exists R > 0 such

that

inf{∥∥ψ ′

t (x)∥∥ : t ∈ [0, 1] , ‖x‖ ≥ R

}> 0. (5.7)

We argue indirectly. So, suppose we cannot find R > 0 for which (5.7) holds. This meansthat there exist {tn}n≥1 ⊆ [0, 1] and {xn}n≥1 ⊆ H1

n (Z) such that

tn → t, ‖xn‖ → ∞ and ψ ′tn (xn) → 0 in H1

n (Z)∗ as n → ∞.

Set yn = xn‖xn‖ . Then ‖yn‖ = 1 for all n ≥ 1, and so we can assume that

ynw−→ y in H1

n (Z) and yn → y in L2(Z) as n → ∞.

We have ⟨ψ ′

tn (xn), v⟩ ≤ εn ‖v‖ for all v ∈ H1

n (Z) with εn ↓ 0,

hence

|〈A(xn)+ βxn − tn N0(xn), v〉| ≤ εn ‖v‖with

N0 (u) (.) = f0 (., u(.)) for all u ∈ H1n (Z).

123


Therefore, ∣∣∣∣⟨

A(yn)+ βyn − tnN0(xn)

‖xn‖ , v

⟩∣∣∣∣ ≤ ε′n ‖v‖ with ε′n = εn

‖xn‖ .

If v = yn − y, then 〈A(yn), yn − y0〉 → 0; hence yn → y in H1n (Z), and so

‖y‖ = 1.

Note that N0(xn)‖xn‖ → 0 in L2(Z). So, in the limit as n → ∞, we obtain

A(y) = −βy

hence y = 0, a contradiction to the fact that ‖y‖ = 1. Therefore, we can find R > 0 forwhich (5.7) holds. Moreover, it is clear that

inf {‖ψt (x)‖ : t ∈ [0, 1] , ‖x‖ ≤ R} > −∞.

Hence, we can apply Lemma 2.4 of Perera and Schechter [28], and obtain

Ck (ψ0,∞) = Ck (ψ1,∞) for all k ≥ 0. (5.8)

Note that

ψ0(x) = 1

2‖Dx‖2

2 + β

2‖x‖2

2 and ψ1 = ϕ0. (5.9)

But ψ0 has only one critical point x = 0, which is a global minimizer of ψ0. Hence

Ck (ψ1,∞) = δk,0Z for all k ≥ 0. (5.10)

From (5.8), (5.9) and (5.10), it follows that

Ck (ϕ0,∞) = δk,0Z. (5.11)

Suppose that v−, x+, y0 and 0 are the only critical points ofϕ0.Then, from the Poincare–Hopfformula (see (2.9) ) and from (5.4), (5.5) (5.6) and (5.10), we have

(−1)0 = (−1)0 + (−1)0 + (−1)1 + (−1)τm ,

and hence (−1)τm = 0, a contradiction. This means that ϕ0 has one additional critical pointw0 ∈ H1

n (Z), which is distinct from v−, x+, y0 and 0. From the proof of Theorem 4 (seeClaim 3 there), we have w0 ∈ T0 and so w0 is a solution of (5.1), which clearly is nodal.Finally, by regularity theory, w0 ∈ C1

n(Z). ��

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