# Multiple Solutions for ap (x)-Laplacian Problem: A Variational Approach

Authors: Mahnaz Khosravi Rashti; Mohsen Alimohammady; Sirous Ghobadi
Journal: Journal of Mathematics
 (2023)

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## Abstract

We focus on an anisotropicp (x)-Laplacian equation defined on a bounded domain with smooth boundary. Moreover, using variational methods to find the existence of at least three solutions due to Ricceri is discussed.


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## Body

## 1. Introduction

Here, the following is considered a Neumann boundary value problem:(1)−divb0x,∇ω+ωpx−2ω=λf0x,ω,inΩ,b0x,∇ω.∂ω∂v=μ0g0x,ω,on∂Ω,where λ and μ0 are non-negative and Ω⊂ℝNN≥2 is its smooth boundary. We assume that v is the outward normal vector on ∂Ω,f0:Ω×ℝ⟶ℝ,g0:∂Ω×ℝ⟶ℝ and b0:Ω¯×ℝN⟶ℝN are three Carathéodory functions. p∈CΩ¯ is a variable exponent and is regular enough. There are many articles, for example, [1–4] published in existence and multiplicity of solutions for different problems with different types of conditions.Recently, the study of fourth-order boundary value problems involving Laplacian equations andp (x)-growth conditions was taken into consideration for many physical motivations and they can be modeled similar to this kind of equations (for example, see [5–11]). We also referred the reader to [12–23], wherein the existence of one, two, three, and infinity solutions for px-Laplacian problems was studied by using variant variational methods and under some hypothesis on two nonlinear functions f0andg0 and for the parameters λ and μ0 which are in some intervals. For example, Fan and Deng in [8] studied the multiplicities of positive solutions for inhomogeneous Neumann problem of the kind px-Laplacian as follows:(2)−div∇ωpx−2∇ω+λωpx−2ω=f0x,ω,inΩ,∇ωpx−2∂ω∂v=ψ,on∂Ω.They considered the multiplicity of solutions for two casesλ=λ∗ in two cases λ∗>0orλ∗<0.Deng and Wang in [24] proved the existence and nonexistence of the following problem:(3)−Δpxω+λωpx−2ω=f0x,ω,inΩ,∇ωpx−2∇ω.v=g0x,ω,on∂Ω.Heidarkhani and Salari in [25] by using the critical point theory discussed the existence of two and three solutions for the following problem:(4)−div1+∇ωpx1+∇ω2px∇ωpx−2∇ω+ηxωpx−2ω=λf0x,ω,inΩ,∂ω∂v=0,on∂Ω.Finally, in [26], Papageorgiou and Winkert proved the multiplicity of solutions for the following equation:(5)−Δpxω+ξxωpx−2ω=f0x,ω,inΩ,∇ωpx−2∇ω.v+β0xωpx−2ω=0,on∂Ω.In this article, we would check the existence of at least three weak solutions for problem (1) when f0andg0 have a subcritical growth (conditions I1 and I2). Precisely, employing a three-critical point theorem due to Ricceri [27], we established the existing result for equations of px-Laplacian type. Moreover, an example for illustrating our main result is given.


## 2. Basic Definitions and Preliminary Results

Given the Banach spaceX, we denoted by WX the set of all maps Q0:X⟶ℝ which has the following properties.Weakly converging of a sequenceωn to ω in X in which lim¯n⟶∞Q0ωn≤Q0ω implies that ωn has a subsequence strong convergence to ω.Theorem 1.
Assuming thatX is a separable reflexive real Banach space, N:X⟶ℝ a coercive map in WX which is C1-map and sequentially weakly lower semicontinuous, bounded on each bounded subset of X, and N′ has a continuous inverse on X∗ (dual of X). Suppose that ℱ0:X⟶ℝ is a C1-map such that ℱ0′ is a compact function, N has a strict local minimum x0 such that Nx0=ℱ0x0=0.
Set(6)ρ=maxlim¯x⟶x0ℱ0xNx,lim¯x⟶+∞ℱ0xNx,0,σ=supx∈N−10,+∞ℱ0xNx.
In the case thatρ<σ, for any compact interval c,d⊆1/σ,1/ρ (with convention 1/∞=0), there is R0>0 such that for any λ∈c,d and any map G0:X⟶ℝ which G0′ is compact, there exists δ0>0 in which for any μ0∈0,δ0, the equation N′x−λℱ0′x−μ0G0′x=0 admits at least three weak solutions in X in BR00 (ball with center 0 and radius R0).
Considerp∈CΩ¯, then(7)1≤p−=infΩ¯p≤p+=supΩ¯p<+∞.
For(8)p∗x=+∞ifpx≥N,NpxN−pxifpx<N,p∂x=+∞ifpx≥N,N−1pxN−pxifpx<N,we set C+Ω¯≔=p∈CΩ¯:1<p−<p+<+∞ for p∈C+Ω¯, and we consider some conditions as follows:H0b0x,−t=−b0x,t for almost all x∈Ω¯ and all t∈ℝn.H1 There is a Carathéodory function B0:Ω¯×ℝN⟶ℝ which is continuously differentiable such that b0x,t=∇tB0x,t∀t∈ℝN and for almost all x∈Ω¯.H2B0x,0=0 for almost all x∈Ω¯.H3 There is c0>0 such that the growth condition b0x,t≤c0tpx−1 for almost all x∈Ω¯ and all t∈ℝN satisfy, where ⋅ is the usual Euclidean norm.H4 The monotonicity relation 0≤b0x,t1−b0x,t2t1−t2 holds for almost all x∈Ω¯ and each t1,t2∈ℝN. This is equality if and only if t1=t2.H5 The inequalities tpx≤b0x,tt≤pxB0x,t hold for almost all x∈Ω¯ and all t∈ℝN.I1 For s∈CΩ¯ in which p∗x>sx for all x∈Ω¯, then(9)supx,t∈Ω×ℝf0x,t1+tsx−1<+∞.I2 Given p2∈CΩ¯ such that p∂x>p2x for all x∈Ω¯ and p−>p2+, then(10)supx,t∈∂Ω×ℝg0x,t1+tp2x−1<+∞.The variable exponent Lebesgue space class is defined as follows:(11)LpxΩ=ω:ω:Ω⟶ℝis a measurable and∫Ωωpxdx<+∞,where LpxΩ is reflexive and separable Banach space endowed with the following Luxemburg norm (see [28]):(12)ωpx≔=infη>0:∫Ωωxηpxdx≤1.Moreover,(13)W1,pxΩ=ω∈LpxΩ:∇ω∈LpxΩ,is endowed with(14)ω=infη>0:∫Ωωxηpx+∇ωxηpxdx≤1;for anyω∈W1,pxΩ.We denotedW01,pxΩ as the closure of C0∞Ω in W1,pxΩ.Proposition 1 (see [28–31]).
(1)
W01,pxΩ is a reflexive and separable Banach space.(2)
r∈C+Ω¯ and p∗x>rx for all x∈Ω¯, and there is a continuous embedding W1,pxΩ⟶LrxΩ which is compact.(3)
r∈C+Ω¯ and rx<p∂x for any x∈Ω¯, and there is a continuous embedding W1,pxΩ⟶LrxΩ which is compact.(4)
(Poincaré)∃C0>0, a number such that(15)∇ωpx≥1C0ωpx,∀ω∈W01,pxΩ.Proposition 2.
There exists a positive constantk0 such that for all ω∈W1,pxΩ,(16)ωC0Ω≤k0ω.Proof.
For anys∈Ω,(17)ωs≤∫Ω|∇ωs|+ωsds,∀ω∈W1,pxΩ.
Using the Hölder inequality, there isk1>0 such that(18)ωL1Ω≤k1ωLpxΩ,∇ωL1Ω≤k1∇ωLpxΩ.
Hence,(19)1k1ωC0Ω≤∇ωLpxΩ+ωLpxΩ,where ω∈W1,pxΩ and s∈Ω. Moreover,(20)ωW1,pxΩ=∇ωLpxΩ+ωLpxΩon W1,pxΩ is an equivalent to the one in equation (14), so for a suitable positive constant k0,(21)ωC0Ω≤k0ω,∀ω∈W1,pxΩ.
Set(22)ρω=∫Ωωpx+∇ωpxdx,∀ω∈W1,pxΩ.Proposition 3 (see [32]).
Givenω,ωk∈W1,pxΩ,k=1,2,…, the following properties hold:(1)
ωLpx>1⇒ωp−≤ρω≤ωp+(2)
ωLpx<1⇒ωp−≥ρω≥ωp+(3)
ωk⟶0respectively,⟶+∞⇔ρωk⟶0respectively,⟶+∞.
Throughout this article, we assumed thatf0:Ω×ℝ⟶ℝ and g0:∂Ω×ℝ⟶ℝ are two L1-Carathéodory function, i.e.,(a)
x↦f0x,ξis measurable for anyξ∈ℝ.(b)
ξ↦f0x,ξis continuous for almost eachx∈Ω.(c)
Given anys>0,there is a functionls∈L1Ωin which(23)for almost allx∈Ω,supξ≤sf0x,ξ≤lsx,and similarly for the function g0 but with ∂Ω instead of Ω.
Corresponding to the functionsf0 and g0, we considered the maps F0:Ω×ℝ⟶ℝ and G0:∂Ω×ℝ⟶ℝ which are defined as follows:(24)F0x,s=∫0sf0x,ξdξ,x,s∈Ω×ℝ,G0x,s=∫0sg0x,ξdξ,x,s∈∂Ω×ℝ.Definition 1.
A weak solutionω for equation (1) means that(25)∫Ωb0x,∇ω∇vdx+∫Ωωpx−2ωvdx=λ∫Ωf0x,ωvdx+μ0∫∂Ωg0x,ωvdσ,∀v∈ω1,pxΩ,where dσ is the N−1 dimensional Hausdorff measure on the boundary.


## 3. Main Results

Forθ>0 and η0∈CΩ with 1<η0−, we set(26)θη0=maxθη0−,θη0+,θη0≔minθη0−,θη0+,λ1=inf∫Ω1/pxωpxdx+∫ΩB0x,∇ωdx∫ΩF0x,ωdx:∫ΩF0x,ωdx>0,ω∈X,and λ2≔1/max0,λ0,λ∞, where(27)λ0=lim¯ω⟶0∫ΩF0x,ωxdx∫Ω1/pxωxpxdx+∫ΩB0x,∇ωxdx,λ∞=lim¯ω⟶+∞∫ΩF0x,ωxdx∫Ω1/pxωxpxdx+∫ΩB0x,∇ωxdx.Define the functionalsN,ℱ0,G0:X=W1,pxΩ⟶ℝ by(28)Nω=∫Ω1pxωxpxdx+∫ΩB0x,∇ωxdx,ℱ0ω=∫ΩF0x,ωxdx,G0ω=∫∂ΩG0x,ωxdσ,for each ω∈X.N is well defined (see Theorem 3.1 in [23] and Proposition 4 in [15]), and it has continuous Gâteaux differentiable and it is sequentially weakly lower semicontinuous whose Gâteaux derivative N′:X⟶X∗ is as follows:(29)N′ωv=∫Ωωxpx−2ωxvxdx+∫Ωb0x,∇ωx∇vxdx∀v,ω∈X,which admits a continuous inverse on the dual space X∗≔W1,pxΩ∗. From I1 and I2, ℱ0 and G0 are continuously Gâteaux differentiable functions in which the embeddings W1,pxΩ⟶LpxΩ and W1,pxΩ⟶Lpx∂Ω (Proposition 1) are compact which implies the compactness of the following derivatives:(30)ℱ0′ωv=∫Ωf0x,ωxvxdx,G0′ωv=∫Ωg0x,ωxvxdσ,∀ω,v∈X,(for more details, see the proof of Theorem 1.1 in [23]).Lemma 1.
The functionalN is coercive and bounded on each bounded subset of X.Proof.
For checking the coercivity of it is enough to assume thatω≥1. From H5 and Proposition 3,(31)Nω=∫Ω1pxωxpxdx+∫ΩB0x,∇ωxdx≥∫Ω1pxωxpx+∇ωxpxdx≥1p+ρω≥1p+ωp−.
So, 𝒩 is coercive. Assume thatB⊆X is bounded, there is m1>0 such that ω≤m1∀ω∈B.
According toH3,(32)Nω=∫Ω1pxωxpxdx+∫ΩB0x,∇ωxdx≤1P−∫Ωωxpxdx+m2∫Ω1px∇ωxpxdx≤m3∫Ω∇ωxpx+ωxpxdx=m3ρω≤m3ωp≤m3m1p,for some suitable positive constants m2 and m3. Hence, on each bounded subset of X, N is bounded and(33)1p+ωp≤Nω≤m3ωp,∀ω∈X.
Moreover,N admits a strict local minimum 0 with N0=ℱ00=0.Lemma 2 (see Theorem 4.1 in [33]).
Suppose that the Carathéodory functionb0:Ω¯×ℝN⟶ℝN satisfies in H0−H5 and is of the type S+ that is ωn⇀ω in W1,pxΩasn⟶+∞ and limsupn⟶+∞∫Ωb0x,∇ωn∇ωn−∇ωdx≤0 implies ωn⟶ω strongly in W1,pxΩasn⟶+∞.Lemma 3.
N lies in WX.Proof.
Letωn⇀ω weakly in X and liminfn⟶+∞Nωn≤Nω. Since N is weakly lower semicontinuous, liminfn⟶+∞Nωn=Nω. From the compact embedding X⟶LpxΩ (see Lemma 2.5 in [34]), ∫Ω1/pxωnxpxdx converges to ∫Ω1/pxωxpxdx,asn⟶+∞. Now, from Lemma 2, we need only to show that(34)limn⟶+∞∫Ωb0x,∇ωn∇ωn−∇ωdx=0.
SinceNω=liminfn⟶+∞Nωn=liminfn⟶+∞∫ΩB0x,∇ωnxdx, for a subsequence of ωn which for simplicity we denoted it again by ωn,(35)limn⟶+∞∫ΩB0x,∇ωndx=∫ΩB0x,∇ωdx.
FromH3 and Lemmas 2.1 and 2.5 in [34],(36)0≤∫Ωb0x,∇ωn∇ωn−∇ωdx≤c0∫Ω∇ωn−∇ω∇ωnpx−1dx≤c01p−+1p′−∇ωn−∇ωLpxΩ∇ωnLp′xΩpx−1⟶0,asn⟶+∞.
Thus,limn⟶+∞∫Ωb0x,∇ωn∇ωn−∇ωdx=0. According to Lemma 2, ωn converges to ω strongly in X≔W1,pxΩ, so N∈WX.
Applying Lemmas1and 3, we like to show the multiplicity of solutions.Theorem 2.
Suppose thatf1 There exists ε>0 in which(37)maxlim¯s⟶+∞maxx∈Ω¯F0x,ssp−,lim¯s⟶0maxx∈Ω¯F0x,ssp−<ε.f2 There exists a function ω0∈X such that(38)Kω0=∫Ω1pxω0xpxdx+∫ΩB0x,∇ω0xdx≠0ε<∫ΩF0x,ω0xdxkp+p+Kω0measΩ.Then, for an intervalc,d⊂λ1,λ2, there exists R0>0 in which “for every λ∈c,d, there exists δ0>0 such that for each μ0∈0,δ0, problem (1) has at least three solutions in which whose norms in X are less than R0.”Proof.
Fromf1, there exist τ1 and τ2 with 0<τ1<τ2 such that(39)F0x,s≤εsp−,for each x∈Ω,s∈0,τ1∪τ2,+∞.F0x,s is continuous on Ω×ℝ, and it is bounded on Ω and s∈τ1,τ2, so choosing η0>0 and v>p+ leads to F0x,s≤ϵsp++η0sv for all x,s∈Ω×ℝ. Hence, from Proposition 2,(40)ℱ0ω=∫ΩF0x,ωxdx≤measΩεkp+ωp++η0kvωv,for all ω∈X. Hence, from equations (33) and (40),(41)lim¯ω⟶0ℱ0ωNω≤lim¯ω⟶0measΩεkp+ωp++η0kvωv1/p+ωp+=measΩp+kp+ε.
From equations (33) and (39) and Proposition 2, it follows that for each ω∈X/0,(42)ℱ0ωNω=∫ω>τ2F0x,ωdxNω+∫ω>τ2F0x,ωdxNω≤p+supx∈Ω,ω∈0,τ2F0x,ωmeasΩωp+kp+p+εωpmeasΩωp,lim¯ω⟶+∞ℱ0ωNω≤εkp+p+measΩ.
According to equations (41) and (42), the following relation is established:(43)ρ=max0,lim¯ω⟶+∞ℱ0ωNω,lim¯ω⟶0ℱ0ωNω≤εp+2p+measΩ.
Assumptionf2 in conjunction with equations (43) implies the following equation:(44)σ=supω∈N−10,+∞ℱ0ωNω=supX/0ℱ0ωNω≥∫Ωℱ0x,ω0xdxNω0x=∫Ωℱ0x,ω0xdxKω0>εp+kp+measΩ≥ρ.
Thus, taking into account Lemmas1 and 3, assumptions of Theorem 1 satisfy for λ1=1/σ and λ2=1/ρ. Using Theorem 1, for interval c,d⊂λ1,λ2, there exists R0>0 such that for any λ∈c,d, there exists δ0>0 and such that for each μ0∈0,δ0, problem (1) has at least three weak solutions whose norms in X are less than R0 (standard arguments show that I=:N−λF0−μ0G0 is a Gâteaux differentiable functional and ω∈X is a solution for problem (1) if and only if ω is a critical point of the function I).Theorem 3.
Let(45)maxlim¯s⟶+∞maxx∈Ω¯F0x,ssp−,lim¯s⟶0maxx∈Ω¯F0x,ssp−≤0,and(46)supω∈X∫ΩF0x,ωxdxp+kp+measΩ∫Ω1/pxωxpxdx+∫ΩB0x,∇ωxdx>0.Then, for each intervalc,d⊂λ1,+∞, there exists R0>0 in which for every λ∈c,d, there exists δ0>0 such that for each μ0∈0,δ0, problem (1) has at least three solutions whose norms are less than R0 in X.Proof.
From equation (45), there exists an arbitrary ϵ>0 and τ1 and τ2 with 0<τ1<τ2 in which F0x,s≤ϵsp− for any Ω and any s with s∈0,τ1∪τ2,+∞. Since F0x,s is continuous on Ω×ℝ, it is bounded on Ω and s∈τ1,τ2. We can choose η0>0 and v>p+ in a similar manner that F0x,s≤ϵsp++η0sv for all x,s∈Ω×ℝ. By the same argument as in proof of Theorem 2, we have equations (41) and (42). Since ε is arbitrary so from equations (41) and (42),(47)max0,lim¯ω⟶+∞ℱ0ωNω,lim¯ω⟶0ℱ0ωNω≤0.
Then, using notation of Theorem1, ρ=0, and equation (46) implies σ>0. In this case, λ1=1/σ and λ2=+∞. Finally, using Theorem 1, the proof is completed.
Now, we pay to the following example.Example 1.
LetΩ=x,y∈ℝ2|x2+y2<10−10,px,y=x2+y2+6,ω0x,y=105 and f0x,y,s=e−ss78−sx2+y2 for every x,y∈Ω,s∈ℝ. Thus, p−=6,p+=6+10−10,F0x,y,s=s8e−sx2+y2 for every x,y,s∈Ω×ℝ, and(48)∫ΩF0x,ω0dx=∫∫x2+y2<10−10F0x,y,105dxdy=π2e−10010.
Also, by settingI≔∫66+10−1010010s/sds, one has the following equation:(49)∫ΩF0x,ω0dxp+kp+measΩ∫∫Ω10010x2+y2+6/x2+y2+6dx,dy=1010e−100102p+kp+πI>0.
However,(50)limsups⟶+∞maxx∈Ω¯F0x,ss6=limsups⟶+∞10−10s8e−ss6=0,limsups⟶0maxx∈Ω¯F0x,ss6=limsups⟶0s21010es=0.
Hence, Theorem3 is applicable for the following system:(51)−divb0x,∇ω+ωx2+y2+4ω=λe−ωω78−ωx2+y2,inΩ,b0x,∇ω.∂ω∂v=μ0g0x,ω,on∂Ω.
For anys0∈ℝ,r0>0,ρ0∈0,1 and each x0∈ℝN, we put the following equation:(52)ϑ1=ϑ1s0,r0,ρ0,x0=c0p−∫BNx0,r0/BNx0,ρ0r0s0r01−ρ0pxdx,ϑ2=ϑ2s0,r0,ρ0,x0=1p−∫BNx0,ρ0r0s0pxdx+1p−∫BNx0,r0/BNx0,ρ0r0s0r01−ρ0pxr0−x−x0pxdx,L=ϑ1+ϑ2,where c0 is introduced in the condition H3, BNx0,r0=x∈Ω:x−x0≤r0⊆Ω. The next theorem provides sufficient conditions for applying Theorem 2 without satisfying ω0 in f2.Theorem 4.
Assume that the conditionf1 in Theorem 2 satifies andf3 there exist s0∈ℝ,x0∈ℝN, and r0>0 so small such that(53)F0x,s≥0,∫BNx0,r0F0x,s0dx>0,for almost allx∈BNx0,r0/BNx0,ρ0r0, with ρ0∈0,1 and 0≤s≤s0.f4L≔ϑ2+ϑ2≠0andε<∫BNx0,ρ0r0F0x,s0dx/p+kp+measΩL.
Then, for each intervalc,d⊂λ1,λ2, there exists R0>0 in which for every λ0∈c,d, there exists δ0>0 such that for each μ0∈0,δ, problem (1) has at least three weak solutions in X whose norms are less than R0.Proof.
Lets0≠0 be from f3 and fix ρ0∈0,1. We like to examine the validity of all assumptions of Theorem 2. We set the following equation:(54)ω0x≔0,ifx∈ΩBNx0,r0,s0,ifx∈BNx0,ρ0r0,s0r01−ρ0r0−x−x0,ifx∈BNx0,r0BNx0,ρ0r0.
Directly,0≤ω0x≤s0, for all x∈Ω and so ω0∈X. On the other hand, since for every x∈BNx0,r0/BNx0,ρ0r0,(55)∂ω0∂xi=−s0r01−ρ0xi−x0,ix−x0.
So, we obtain∇ω0x=s0/r01−ρ0 for every x∈BNx0,r0/BNx0,ρ0r0. Moreover, the existence of continuous embedding.
LpxΩ⟶Lp−Ω (see Theorem 2.8 of [28]) implies the following equation:(56)Kω0=∫Ω1pxω0xpxdx+∫ΩB0x,∇ω0xdx≥1p+ρω0≥1p+ω0p≥1p+∇ω0LpxΩp≥m4∫Ω∇ω0xp−dx=m4s0r01−ρ0p−∫BNx0,r0/BNx0,ρ0r0dx=m4s0r01−ρ0p−ωN∫ρ0r0r0rN−1dr=m5s0p−1−ρ0N1−ρ0p−r0N−p−ωN>0,for positive constants m4andm5, where ωN is the volume of BN0,1 (for more details, see Theorem 2.11 [34]). Thus, Kω0≠0. On the other hand, from f3, f4, and H3, one has the following equations:(57)∫ΩF0x,ω0xdx≥∫BNx0,ρ0r0F0x,s0dx,and(58)Kω0=∫Ω1pxω0xpxdx+∫ΩB0x,∇ω0xdx≤∫Ωc0p−∫Ω∇ω0xpxdx+1p−∫Ωω0xpxdx=ϑ1+ϑ2≔=L.
So, from equations (57) and (58) and f4, we obtain the following equation:(59)∫ΩF0x,ω0xdxkp+p+Kω0measΩ≥∫BNx0,ρ0r0F0x,s0dxLkp+p+measΩ>ε.
Therefore,f2 in Theorem 2 satisfies. Hence, Theorem 2 follows.
We studied the following problem to get results that the functionf0 has separated variables.(60)−divb0x,∇ω=λΥ0xf0ω−ωpx−2ω,inΩ,μ0g0x,ω=b0x,∇ω.∂ω∂v,on∂Ω,where Υ0:Ω⟶ℝ is a non-negative and nonzero function and Y0∈L1Ωandf0:ℝ⟶ℝ is a continuous function. Set F0x,s=Y0xF0s for every x,s∈Ω×ℝ, where F0s=∫0sf0ξdξ for all s∈ℝ.
The following existing results are consequences of Theorem4.Corollary 1.
Suppose thatf5 There is ε>0 in which(61)supx∈ΩY0xmaxlim¯s⟶+∞F0ssp−,lim¯s⟶0F0ssp−<ε.f6 There exist s0∈ℝ, x0∈ℝN, and r0>0 so small such that(62)∫0sf0ξdξ>0,for all 0≤s≤s0,L=ϑ1+ϑ2≠0 and(63)ε<F0s0∫BNx0,ρ0r0Y0xdxp+kp+measΩL.Then, for an intervalc,d⊂λ3,λ4 (where λ3 and λ4 are the same as λ1 and λ2 in which ∫ΩF0x,ωxdx is substituted by ∫ΩY0xF0ωxdx, respectively), there exists R0>0 in which for any λ∈c,d, there exists δ0>0 such that, for each μ0∈0,δ, problem (60) has at least three weak solutions in X whose norms are less than R0.Theorem 5.
Assume that there existss0∈ℝ, x0∈ℝN, and r0>0 so small such that(64)L=ϑ1+ϑ2>0,F0s0>0.
Moreover,(65)lim¯s⟶0f0ssp−−1=lim¯s⟶+∞f0ssp−−1=0.Then, for the given intervalc,d⊂λ3,+∞, where λ3 is the same as λ1 in which ∫ΩF0x,ωxdx is replaced by ∫ΩY0xF0ωxdx, there exists R0>0 such that for every λ∈c,d, there exists δ0>0 such that, for each μ0∈0,δ0, problem (60) has at least three weak solutions in X whose norms are less than R0.Proof.
From equation (65), f5 satisfies for any ϵ>0. Moreover, using equation (64) and ϵ>0 small enough, we can assume f6. Then, the conclusion follows from Corollary 1.Remark 1.
From cited results, we realized that nowhere in theorems asymptotic conditions on the functionsf0 and g0 are required and only the algebraic assumptions on f0 are supposed to give the existence of solutions.
The following example shows that nonlinearityf0 verifies the hypotheses of Theorem 5.Example 2.
Letpx=x2+6,Y0x=1 for all x∈Ω and(66)f0s=2s+sins2,ifs<π,2π2+tanhs−π,ifs≥π.
Thus,p−=6 and f0 is a continuous and non-negative function and by which choosing s0=1, we have the following equation: (67)F0s0=F01=2∫01s+sins2ds>0.
On the other hand, sinces0=1>0, one has ϑ1+ϑ2>0. Moreover,(68)lim¯s⟶0f0ssp−−1=lims⟶02s+sins2s5=0,lims⟶+∞f0ssp−−1=lims⟶+∞2π2+tanhs−πs5=0.
Hence, by applying Theorem5 for each compact interval c,d⊂0,+∞, there exists R0>0 with the following property: for every λ∈c,d, there exists δ0>0 in which for any μ0∈0,δ0, the problem(69)−divb0x,∇ω=λf0ω−ωx2+4ω,inΩ,μ0g0x,ω=b0x,∇ω.∂ω∂v,on∂Ω,has at least three weak solutions whose norms in Xbypx=x2+6 are less than R0.Remark 2.
Similar to the one in Proposition 2.13 in [34], we can consider and establish positive principal eigenvalues by following the px-Laplacian Neumann problem:(70)−div∇ωpx−2∇ω=ωpx−2ωλα0x−1,inΩ,∂ω∂n=0,on∂Ω,where α0∈L∞Ω and α0x>0 for almost all x∈Ω, and we discuss on number of solutions of problem (1) using problem (70) (precisely, to find out this the subject matter, see Theorem 2.14 in [34] and Theorem 4.4 in [35]).Remark 3.
In comparison with [23] when the authors worked with two parameters λ and μ0, we obtained a precise range for the parameter λ in all the results of existence and multiplicity. We also noticed that this article differs from [13, 15, 16, 18, 19, 22, 34] either in the operators, in the boundary conditions, or in the nonlinearities of f0 and g0. As noted, in the case of variable exponents, the question of multiplicity is recent, so our results contribute to these recent contributions by explaining the interval to the parameters λ and μ0.Conclusion 1.
In this article, by using a variational method and Ricceri variational principle, we studied thepx-Laplacian equation under the Neumann boundary condition. We proved the existence and multiplicity solutions for problem (1), where μ0>0, Ω⊂ℝN for N≥2, f0,g0, and b0 are Carathéodory functions and p∈CΩ¯.



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