# Existence and Multiplicity of Solutions for a Robin Problem Involving thep(x)-Laplace Operator

Authors: Najib Tsouli; Omar Chakrone; Omar Darhouche; Mostafa Rahmani
Journal: Conference Papers in Mathematics
 (2013)

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## Abstract

We study the following nonlinear Robin boundary-value problem−Δp(x)u=λf(x,u) in Ω, |∇u|p(x)-2(∂u/∂v)+β(x)|u|p(x)−2u=0 on ∂Ω, where Ω⊂ℝN is a bounded domain with smooth boundary ∂Ω, ∂u/∂v is the outer unit normal derivative on ∂Ω, λ>0 is a real number, p is a continuous function on Ω¯ with infx∈Ω¯p(x)>1, β∈L∞(∂Ω) with β−:=infx∈∂Ωβ(x)>0, and f:Ω×ℝ→ℝ is a continuous function. Using the variational method, under appropriate assumptions on f, we obtain results on existence and multiplicity of solutions.


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## Body

## 1. Introduction

The purpose of this paper is to study the existence and multiplicity of solutions for the following Robin problem involving thep(x)-Laplacian:
(1)-Δp(x)u=λf(x,u)inΩ,|∇u|p(x)-2∂u∂ν+β(x)|u|p(x)-2u=0on∂Ω,
where Ω⊂ℝN is a bounded smooth domain, ∂u/∂ν is the outer unit normal derivative on ∂Ω,λ>0 is a real number, p is a continuous function on Ω¯ with p-:=infx∈Ω¯p(x)>1, and β∈L∞(∂Ω) with β-:=infx∈∂Ωβ(x)>0. The main interest in studying such problems arises from the presence of the p(x)-Laplace operator div(|∇u|p(x)-2∇u), which is a natural extension of the classical p-Laplace operator div(|∇u|p-2∇u) obtained in the case when p is a positive constant. However, such generalizations are not trivial since the p(x)-Laplace operator possesses a more complicated structure than p-Laplace operator; for example, it is inhomogeneous.In the recent years increasing attention has been paid to the study of differential and partial differential equations involving variable exponent conditions. The interest in studying such problems was stimulated by their applications in elastic mechanics, fluid dynamics, and calculus of variations; for information on modelling physical phenomena by equations involvingp(x)-growth condition we refer to [1–10]. In the past decades a vast amount of literature that deals  with the existence for type problems -Δp(x)u=f(x,u) with different boundary conditions (Dirichlet, Neumann, Robin, nonlinear, etc.) has  appeared. See, for instance, [11–16] and references therein.In [14], by applying the subsupersolution method and the variational method, under appropriate assumptions on f, the author proves that there exists λ*>0 such that problem (1) has at least two positive solutions if λ∈(0,λ*), has at least one positive solution if λ=λ*<+∞, and has no positive solution if λ=λ*. Recently in [11], the authors obtain the existence of at least two nontrivial solutions for problem (1) using a variational approach based on the nonsmooth critical point theory for locally Lipschitz functions.We make the following assumptions on the functionf:(H0)
f:Ω×ℝ→ℝ satisfies the Carathéodory condition and there exists a constant C≥0 such that
(2)|f(x,s)|≤C(1+|s|α(x)-1)∀(x,s)∈Ω×ℝ,
where α(x)∈C(Ω) and 1<α(x)<p*(x) for all x∈Ω.(H1)
There existR>0,  μ>p+:=supx∈Ω¯p(x) such that, for all |s|≥R and x∈Ω,
(3)0<μF(x,s)≤f(x,s)s.(H2)
One hasf(x,s)=o(|s|p+-1) as s→0 and uniformly for x∈Ω.(H3)
One hasf(x,-s)=-f(x,s),  x∈Ω,  s∈ℝ.(H4)
One hassupt∈ℝ∫0t‍f(x,s)ds>0, for all x∈Ω.(H5)
There existsq∈C(Ω) such that p+<q-:=infx∈Ωq(x)≤q(x)<q*(x) and
(4)limsupt→0supx∈Ω∫0t‍f(x,s)ds|t|q(x)<+∞.The main results of this paper are as follows.Theorem 1.
If (H0), (H1), and (H2) hold andα->p+, then, for any λ∈(0,+∞), (1) has at least a nontrivial weak solution.Theorem 2.
If (H0), (H1), and (H3) hold andα->p+, then, for any λ∈(0,+∞), (1) has infinite many pairs of weak solutions.Theorem 3.
If (H0), (H4), and (H5) hold andp->α+, then there exist an open interval Λ⊂(0,∞) and a positive real number ρ such that, for each λ∈Λ, (1) has at least three solutions whose norms are less than ρ.To prove our results, we will use a variational method and the theory of variable exponent Sobolev spaces. For the proof of Theorem1, we will use the Mountain Pass Theorem (see [17, 18]). For the proof of Theorem 2, we will use the Fountain Theorem (see [18, 19]). For the proof of Theorem 3, we will use Ricceri three critical points Theorem (see [20, 21]).This paper is organized as follows. First, we will introduce some basic preliminary results and lemmas in Section2. In Section 3, we will give the proofs of our main results.


## 2. Preliminaries

For completeness, we first recall some facts on the variable exponent spacesLp(x)(Ω) and W1,p(x)(Ω). Suppose that Ω is a bounded open domain of ℝN with smooth boundary ∂Ω and p∈C+(Ω¯), where
(5)C+(Ω¯)={p∈C(Ω¯),infx∈Ω¯p(x)>1}.
Denote p-:=infx∈Ω¯p(x) and p+:=supx∈Ω¯p(x). Define the variable exponent Lebesgue space Lp(x)(Ω) by
(6)Lp(x)(Ω)={u:Ω⟶ℝisameasurable,∫Ω‍|u|p(x)dx<+∞},
with the norm
(7)|u|Lp(x)(Ω)=|u|p(x)=inf{τ>0;∫Ω‍|uτ|p(x)dx≤1}.
Define the variable exponent Sobolev space W1,p(x)(Ω) by
(8)W1,p(x)(Ω)={u∈Lp(x)(Ω):|∇u|∈Lp(x)(Ω)},
with the norm
(9)∥u∥=inf{τ>0;∫Ω‍(|∇uτ|p(x)+|uτ|p(x))dx≤1},∥u∥=|∇u|p(x)+|u|p(x).
We refer the reader to [14, 22, 23] for the basic properties of the variable exponent Lebesgue and Sobolev spaces.Lemma 4 (see [23]).
Both(Lp(x)(Ω),|·|p(x)) and (W1,p(x)(Ω),∥·∥) are separable, reflexive, and uniformly convex Banach spaces.Lemma 5 (see [23]).
Hölder inequality holds, namely,(10)∫Ω‍|uv|dx≤2|u|p(x)|v|p′(x)∀u∈Lp(x)(Ω),v∈Lp′(x)(Ω),
where 1/p(x)+1/p′(x)=1.Now, we introduce a norm, which will be used later. Foru∈W1,p(x)(Ω), define
(11)∥u∥β=inf{λ>0:∫Ω‍|∇u(x)λ|p(x)dx=====+∫∂Ω‍β(x)|u(x)λ|p(x)dσx≤1}.
Then, by Theorem 2.1 in [14], ∥u∥β is also a norm on W1,p(x)(Ω) which is equivalent to ∥u∥.Hereafter, let(12)p*(x)={Np(x)N-p(x),ifp(x)<N,+∞,ifp(x)≥N,p∂(x)={(N-1)p(x)N-p(x),ifp(x)<N,+∞,ifp(x)≥N.Lemma 6 (see [12, 16, 23]).
(1) Ifq∈C+(Ω¯) and q(x)<p*(x) for any x∈Ω¯, then the imbedding from W1,p(x)(Ω) to Lq(x)(Ω) is compact and continuous.
(2) Ifq∈C+(Ω¯) and q(x)<p∂(x) for any x∈∂Ω, then the trace imbedding from W1,p(x)(Ω) to Lq(x)(∂Ω) is compact and continuous.An important role in manipulating the generalized Lebesgue-Sobolev spaces is played by the mapping defined by the following.Lemma 7 (see [14]).
Denoting(13)Iβ(u)=∫Ω‍|∇u|p(x)dx+∫∂Ω‍β(x)|u|p(x)dσx,==================∀u∈W1,p(x)(Ω),
with β->0, then(1)
∥u∥β≥1⇒∥u∥βp-≤Iβ(u)≤∥u∥βp+,(2)
∥u∥β≤1⇒∥u∥βp+≤Iβ(u)≤∥u∥βp-,(3)
∥u∥β→0 if and only if Iβ(u)→0 (as k→∞),(4)
|u(x)|p(x)→∞ if and only if Iβ(u)→∞ (as k→∞).Lemma 8 (see [20, 21, 24]).
LetX be a separable and reflexive real Banach space; ϕ:X→ℝ is a continuous Gâteaux differentiable and sequentially weakly lower semicontinuous functional whose Gâteaux derivative admits a continuous inverse on X*; ψ:X→ℝ is a continuous Gâteaux differentiable functional whose Gâteaux derivative is compact. Assume that(i)
lim∥u∥X→∞(ϕ(u)+λψ(u))=∞ for all λ>0,(ii)
there existr∈ℝ and u0,u1∈X such that ϕ(u0)<r<ϕ(u1),(iii)
(14)infu∈ϕ-1(-∞,r]ψ(u)>(ϕ(u1)-r)ψ(u0)+(r-ϕ(u0))ψ(u1)ϕ(u1)-ϕ(u0).Then there exist an open interval Λ⊂(0,∞) and a positive constant ρ>0 such that for any λ∈Λ the equation ϕ′(u)+λψ′(u)=0 has at least three solutions in X whose norms are less than ρ.Theorem 9.
LetX=W1,p(x)(Ω) and f:Ω×ℝ→ℝ be a continuous function with primitive F(x,u)=∫0u‍f(x,t)dt. If the following condition hold:
(15)|f(x,s)|≤C(1+|s|α(x)-1),∀(x,s)∈Ω¯×ℝ,
where C≥0 is a constant and α(x)∈C+(Ω¯) such that, for all x∈Ω¯,  α(x)<p*(x), then ψ(u)=-∫Ω‍F(x,u(x))dx∈C1(X,ℝ) and Dψ(u,φ)=〈ψ′(u),φ〉=-∫Ω‍f(x,u(x))φdx; moreover, the operator ψ′:X→X* is compact.Proof.
It is easily adapted from Theorem 2.1 in [13].LetX=W1,p(x)(Ω) and
(16)ϕ(u)=∫Ω‍1p(x)|∇u|p(x)dx+∫∂Ω‍β(x)p(x)|u|p(x)dσx,ψ(u)=-∫Ω‍F(x,u)dx,J(u)=ϕ(u)+λψ(u),
where F(x,t)=∫0t‍f(x,s)ds.Obviouslyϕ∈C1(X,ℝ) and
(17)(ϕ′(u),v)=∫Ω‍|∇u|p(x)-2∇u∇vdx+∫∂Ω‍β(x)|u|p(x)-2uvdσx,(ψ′(u),v)=-∫Ω‍f(x,u)vdx.
Moreover, we have the following.Proposition 10 (see [11]).
(
1
)  ϕ′:X→X* is a continuous, bounded, and strictly monotone operator.
(
2
)  ϕ′:X→X* is a mapping of type (S)+; that is, if un⇀u in X and limsupn→∞(ϕ′(un)-ϕ′(u),un-u)≤0, then un→u in X.
(
3
)  ϕ′:X→X* is a homeomorphism.Definition 11.
One says thatu∈W1,p(x)(Ω) is a weak solution of problem (1) if for all v∈W1,p(x)(Ω)(18)∫Ω‍|∇u|p(x)-2∇u∇vdx+∫∂Ω‍β(x)|u|p(x)-2uvdσx=λ∫Ω‍f(x,u)vdx,
where dσx is the measure on the boundary ∂Ω.


## 3. Proof of Main Results

To prove Theorem1, we have to check that the functional J satisfies the following compactness condition (PS).Definition 12.
One says that theC1-functional H:X→ℝ satisfies the Palais-Smale condition ((PS) condition for short) if any sequence (un)n∈ℕ⊆X for which (H(un))n∈ℕ⊆ℝ is bounded and H′(un)→0 as n→∞ has a convergent subsequence.Lemma 13.
If  (H0), (H1) hold, then for anyλ∈(0,+∞) the functional J satisfies the Palais-Smale condition (PS).Proof.
Suppose that(un)⊂X is a (PS) sequence; that is,
(19)sup|J(un)|≤M,J′(un)⟶0asn⟶∞.
We claim that (un) is bounded in X. Using hypothesis (H1), since J(un) is bounded, we have for n large enough
(20)M+1≥J(un)-1μ〈J′(un),un〉+1μ〈J′(un),un〉=∫Ω‍1p(x)|∇un|p(x)dx+∫∂Ω‍β(x)p(x)|un|p(x)dσx=-λ∫Ω‍F(x,un)dx=-1μ(Iβ(un)-λ∫Ω‍f(x,un)undx)=+1μ〈J′(un),un〉≥1p+Iβ(un)-λ∫Ω‍unμf(x,un)dx=-1μ(Iβ(un)-λ∫Ω‍f(x,un)undx)=+1μ〈J′(un),un〉≥(1p+-1μ)∥un∥βp--1μ∥J′(un)∥X*∥un∥-C1≥(1p+-1μ)C3∥un∥p--C2μ∥un∥-C1,
where C1,  C2, and C3 are three positive constants. Hence (un) is bounded in X since μ>p+. Without loss of generality, we assume that un⇀u, then ψ′(un)→ψ′(u) since ψ′:X→X* is completely continuous. The hypothesis J′(un)→0, that is,  ϕ′(un)+λψ′(un)→0 and the fact that ψ′(un)→ψ′(u) imply that ϕ′(un)→-λψ′(u). From Proposition 10, ϕ′ is a homeomorphism, then un→u, and so J satisfies the (PS) condition. The proof is complete.Lemma 14.
There existsr1,C′>0 such that J(u)≥C′ for all u∈X such that ∥u∥β=r1.Proof.
Conditions (H0) and (H2) assure that(21)|F(x,s)|≤ε|s|p++C(ε)|s|α(x)∀(x,s)∈Ω×ℝ.
For ∥u∥β small enough, we have
(22)J(u)≥1p+∥u∥βp+-λ∫Ω‍F(x,u)dx≥1p+∥u∥βp+-λ∫Ω‍(ε|u|p++C(ε)|u|α(x))dx.
Note that 1<p+<α-≤α(x)<p*(x), for all x∈Ω; then, by Lemma 6, we have X↪Lp+(Ω) and X↪Lα(x)(Ω) with a continuous and compact embedding. Furthermore, there exists C4, C5>0 such that
(23)|u|Lp+(Ω)≤C4∥u∥β,|u|Lα(x)(Ω)≤C5∥u∥β,∀u∈X.
Since ∥u∥β is small enough, we deduce
(24)∫Ω‍|u|α(x)dx≤max{|u|α(x)α-,|u|α(x)α+}≤C6∥u∥βα-.
Replacing in (22), it results that
(25)J(u)≥1p+∥u∥βp+-λεC4p+∥u∥βp+-λC(ε)C6∥u∥βα-.
Choosing ε>0 small enough such that 0<λεC4p+<1/2p+, then we obtain
(26)J(u)≥12p+∥u∥βp+-λC(ε)C6∥u∥βα-≥∥u∥βp+(12p+-λC(ε)C6∥u∥βα--p+).
Since p+<α-, the function t↦(1/2p+-λC(ε)C6tα--p+) is strictly positive in a neighborhood of zero. It follows that there exist r1>0 and C′>0 such that
(27)J(u)≥C′∀u∈X:∥u∥β=r1.
The proof is complete.Proof of Theorem1.
To apply the Mountain Pass Theorem [17, 18], we need to prove that J(tu)→-∞ as t→+∞, for a certain u∈X. From condition (H1), we obtain
(28)F(x,s)≥c|s|μ∀(x,s)∈Ω×ℝ.
Letting u∈X and t>1, we have
(29)J(tu)=∫Ω‍tp(x)p(x)|∇u|p(x)dx+∫∂Ω‍tp(x)β(x)p(x)|u|p(x)dσx-λ∫Ω‍F(x,tu)dx≤tp+(∫Ω‍1p(x)|∇u|p(x)dx+∫∂Ω‍β(x)p(x)|u|p(x)dσx)-tμλc∫Ω‍|u|μdx.
The fact μ>p+ implies for any λ∈(0,+∞),  J(tu)→-∞ as t→+∞.
It follows that there existse∈X such that ∥e∥β>r1 and J(e)<0. According to the Mountain Pass Theorem, J admits a critical value τ≥C′ which is characterized by
(30)τ=infh∈Γsupt∈[0,1]J(h(t)),
where
(31)Γ={h∈C([0,1],X):h(0)=0,h(1)=e}.
This completes the proof.SinceX is a separable and reflexive Banach space [22, 25], there exist {en}n=1∞⊂X and {fn}n=1∞⊂X* such that
(32)fn(em)=δn,m={1,ifn=m,0,ifn≠m.X=span¯{en:n=1,2,…},X*=span¯W*{fn:n=1,2,…}.
For k=1,2,… denote
(33)Xn=span{en},Yn=⨁j=1nXj,Zn=⨁j=n∞Xj¯.Lemma 15.
Forα(x)∈C+(Ω¯), α(x)<p*(x), and x∈Ω¯, let
(34)γk=sup{|u|Lα(x)(Ω):∥u∥β=1,u∈Zk}.
Then limk→∞γk=0.Proof.
It is clear that0<γk+1<γk, so γk converges to γ≥0. Let uk∈Zk such that
(35)∥uk∥β=1,0≤γk-|uk|Lα(x)(Ω)<1k.
Then, there exists a subsequence, noted also by (uk), such that uk⇀u in X and
(36)〈fi,u〉=limk→+∞〈fi,uk〉=0,∀i=1,2,….
Thus u=0 and so uk⇀0 in X. According to Lemma 6, there is a compact embedding of X into Lα(x)(Ω), which assure that uk→0 in Lα(x)(Ω). Hence we get γk→0 as k→+∞.Proof of Theorem2.
We will use the Fountain theorem [18, 19]. Obviously, J is an even functional and satisfies the (PS) condition according to (H1) and (H2). We will prove that if k is large enough, then there exist ρk>rk>0 such that(A1)
bk:=inf{J(u):u∈Zk,∥u∥β=rk}→+∞ as k→+∞,(A2)
ak:=max{J(u):u∈Yk,∥u∥β=ρk}≤0 as k→+∞.(A1) For u∈Zk such that ∥u∥β=rk>1, by condition (H0), we have
(37)J(u)=∫Ω‍1p(x)|∇u|p(x)dx+∫∂Ω‍β(x)p(x)|u|p(x)dσx-λ∫Ω‍F(x,u)dx≥1p+∥u∥βp--λ∫Ω‍C(1+|u|α(x))dx≥1p+∥u∥βp--λC∫Ω‍|u|α(x)dx-C7.
If |u|α(x)≤1 then ∫Ω‍|u|α(x)dx≤|u|α(x)α-≤1. And if |u|α(x)>1 then ∫Ω‍|u|α(x)dx≤|u|α(x)α+≤(γk∥u∥β)α+. So we obtain that
(38)J(u)≥{1p+∥u∥βp--C8(λ)-C7if|u|α(x)≤11p+∥u∥βp--C8(λ)(γk∥u∥β)α+-C7if|u|α(x)>1≥1p+∥u∥βp--C8(λ)(γk∥u∥β)α+-C9=rkp-(1p+-C8(λ)γkα+rkα+-p-)-C9.
If we take C8(λ)γkα+rkα+-p-=1/α+, that is, rk=(C8(λ)α+γkα+)1/(p--α+), we obtain
(39)J(u)≥rkp-(1p+-1α+)-C9.
Since γk→0 and p-≤p+<α-≤α+, we have rk→+∞ as k→+∞. Consequently,
(40)J(u)⟶+∞as∥u∥β⟶+∞,u∈Zk.
So (A1) holds.
(A2) Condition (H1) implies that there exist positive constantsC10,  C11 such that
(41)F(x,s)≥C10|s|μ-C11,∀(x,s)∈Ω×ℝ.
Let u∈Yk be such that ∥u∥β=ρk>rk>1. Then
(42)J(u)≤1p-∥u∥βp+-λ∫Ω‍(C10|u|μ-C11)dx≤1p-∥u∥βp+-λC10∫Ω‍|u|μdx+C12.
Note that the space Yk has finite dimension; then all norms are equivalents and we obtain
(43)J(u)≤1p-∥u∥βp+-λC13∥u∥βμ+C12.
Finally,
(44)J(u)⟶-∞as∥u∥β⟶+∞,u∈Yk
since μ>p+. The assertion (A2) is then satisfied and the proof of Theorem 2 is complete.Proof of Theorem3.
We will use Ricceri three critical points Theorem (see [20, 21]). Notice that ϕ is a continuous convex functional, so it is weakly lower semicontinuous and its inverse derivative is continuous. From Theorem 9 the precondition of Lemma 8 is satisfied. Now we only need to verify that conditions (i), (ii), and (iii) in Lemma 8 are fulfilled.
Firstly foru∈X such that ∥u∥β≥1, we have
(45)ψ(u)=-∫Ω‍F(x,u)dx=-∫Ω‍(∫0u(x)‍f(x,t)dt)dx≤C∫Ω‍(|u(x)|+1α(x)|u|α(x))dx≤C∫Ω‍|u(x)|dx+Cα-∫Ω‍|u(x)|α(x)dx.
Using the Hölder inequality and the Sobolev embedding theorem, we have for some positive constants C14 and C15(46)C∫Ω‍|u(x)|dx≤2C|1|α′(x)|u|α(x)≤C14∥u∥β,Cα-∫Ω‍|u|α(x)dx≤Cα-max{|u|α(x)α+,|u|α(x)α-}≤C15∥u∥βα+,
where 1/α(x)+1/α′(x)=1. Combining all together we obtain
(47)|ψ(u)|≤C14∥u∥β+C15∥u∥βα+.
On the other hand,
(48)ϕ(u)=∫Ω‍1p(x)|∇u|p(x)dx+∫∂Ω‍β(x)p(x)|u|p(x)dσx≥1p+∥u∥βp-.
Then for any λ>0,
(49)ϕ(u)+λψ(u)≥1p+∥u∥βp--λC14∥u∥β-λC15∥u∥βα+.
From p->α+ we obtain
(50)lim∥u∥β→∞(ϕ(u)+λψ(u))=∞,
then (i) of Lemma 8 is verified.
Secondly, lettingu0=0, then obviously we have ϕ(u0)=ψ(u0)=0.
We claim that there existr>0 and u1∈X such that ϕ(u0)>r and
(51)infu∈ϕ-1((-∞,r])ψ(u)>rψ(u1)ϕ(u1).
From (H5), there exitsη∈[0,1] and C16>0 such that
(52)F(x,t)≤C16|t|q(x)≤C16|t|q-,∀t∈[-η,η]uniformlyforx∈Ω.
In view of (H0), if we put(53)C17:=max{C16,supη≤|t|<1A(1+|t|α-)|t|q-,sup|t|≥1A(1+|t|α+)|t|q-},
where A is a positive constant, then we have
(54)F(x,t)≤C17|t|q-,∀t∈ℝuniformlyforx∈Ω.
Fix r such that 0<r<1. If (1/p+)∥u∥βp+≤r<1, then, by the Sobolev embedding theorem (X↪Lq-(Ω) is continuous), we have
(55)∫Ω‍F(x,u)dx≤C17∫Ω‍|u|q-dx≤C18∥u∥βq-≤C19rq-/p+,
where C18 and C19 are two positive constants.
Sinceq->p+, then we have
(56)limr→0+1rsup(1/p+)∥u∥βp+≤r∫Ω‍F(x,u)dx=0.
By (H4), we can choose a constant b∈X∖{0} such that ∫Ω‍F(x,b)dx>0.
Fixr0 such that r0<(1/p+)min{∥b∥βp+,∥b∥βp-,1}.
If∥b∥β>1, then we have
(57)1p+∥b∥βp-≤ϕ(b)≤1p-∥b∥βp+.
From (56) and (57), we deduce that, when 0<r<r0, then ϕ(b)>r and
(58)sup(1/p+)∥u∥βp+≤r∫Ω‍F(x,u)dx≤r2∫Ω‍F(x,u)dx(1/p-)∥b∥βp+.
Thus
(59)sup(1/p+)∥u∥βp+≤r∫Ω‍F(x,u)dx≤r∫Ω‍F(x,u)dx(1/p-)∥b∥βp+.
Since r<r0, we get
(60)ϕ-1((-∞,r])⊆{u∈X:1p+∥u∥βp+≤r}.
Then
(61)supϕ(u)≤r-ψ(u)<-rψ(u1)ϕ(u1),
with u1=b which implies that
(62)infu∈ϕ-1((-∞,r])ψ(u)>rψ(u1)ϕ(u1).
So we can find r>0,u1=b, and ϕ(b)>r satisfying (ii) and (iii) of Lemma 8.
If∥b∥β≤1, we have
(63)1p+∥b∥βp+≤ϕ(b)≤1p-∥b∥βp-.
From (56) and (63), we deduce that, when 0<r<r0, then ϕ(b)>r and
(64)sup(1/p+)∥u∥βp+≤r∫Ω‍F(x,u)dx≤r2∫Ω‍F(x,u)dx(1/p-)∥b∥βp-.
Thus
(65)sup(1/p+)∥u∥βp+≤r∫Ω‍F(x,u)dx≤r∫Ω‍F(x,u)dx(1/p-)∥b∥βp-.
Since r<r0, we get
(66)ϕ-1((-∞,r])⊆{u∈X:1p+∥u∥βp+≤r}.
Then
(67)supϕ(u)≤r-ψ(u)<-rψ(u1)ϕ(u1),
with u1=b.
Therefore(68)infu∈ϕ-1((-∞,r])ψ(u)>rψ(u1)ϕ(u1).
Then we obtain
(69)infu∈ϕ-1((-∞,r])ψ(u)>(ϕ(u1)-r)ψ(u0)+(r-ϕ(u0))ψ(u1)ϕ(u1)-ϕ(u0).
This means that condition (iii) in Lemma 8 is verified. Now since all the assumptions of Lemma 8 are verified, there exist an open interval Λ⊂(0,∞) and a positive constant ρ>0 such that for any λ∈Λ the equation ϕ′(u)+λψ′(u)=0 has at least three solutions in X whose norms are less than ρ.



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*Source: 231898-2013-08-01.xml*