# Multiplicity Results for Variable-Order Nonlinear Fractional Magnetic Schrödinger Equation with Variable Growth

Authors: Jianwen Zhou; Bianxiang Zhou; Yanning Wang
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we prove the multiplicity of nontrivial solutions for a class of fractional-order elliptic equation with magnetic field. Under appropriate assumptions, firstly, we prove that the system has at least two different solutions by applying the mountain pass theorem and Ekeland’s variational principle. Secondly, we prove that these two solutions converge to the two solutions of the limit problem. Finally, we prove the existence of infinitely many solutions for the system and its limit problems, respectively.


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## Body

## 1. Introduction

In this paper, we consider the multiplicity of nontrivial solutions of the following concave-convex elliptic equation involving variable-order nonlinear fractional magnetic Schrödinger equation:(1)−ΔAs·u+Vλxu=fxupx−2u+gxuqx−2uinΩ,u=0inℝN\Ω,whereN≥1; s·:ℝN×ℝN⟶0,1 is a continuous function; Ω is a bounded subset in ℝN with N>2sx,y for all x,y∈Ω×Ω; −ΔAs· is the variable-order fractional magnetic Laplace operator; the potential Vλx=λV+x−V−x with V±=max±V,0; λ>0 is a parameter; magnetic field A∈C0,αℝN,ℝN with α∈0,1; f,g>0 are two bounded nonnegative measurable function; p,q∈CΩ; and u:ℝN⟶ℂ. In [1], the fractional magnetic Laplacian has been defined as
(2)−ΔAsux=limr→0∫Brcxux−eix−y·Ax+y/2uyx−yN+2sdy,for x∈ℝN. In [2], the variable-order fractional magnetic Laplace −Δs· is defined as follows: for each x∈ℝN,
(3)−Δs·φx=2P.V∫ℝNφx−φyx−yN+2sx,ydy,along any φ∈C0∞Ω. Inspired by them, we define the variable-order fractional magnetic Laplacian −ΔAs· as follows: for each x∈ℝN,
(4)−ΔAs·ux=limr→0∫Brcxux−eix−y·Ax+y/2uyx−yN+2sx,ydy.Sinces· is a function, magnetic field A∈C0,αℝN,ℝN with α∈0,1, we see that operator −ΔAs· is variable order fractional magnetic Laplace operator. Especially, when s·≡constant, −ΔAs· reduce to the usual fractional magnetic Laplace operator. When s·≡constant,A=0,−ΔAs· reduce to the usual fractional Laplace operator. Very recently, for s·=1, px,qx≡constant, and A=0; in [3], under appropriate assumptions, the authors obtained the multiplicity and concentration of the positive solution of the following indefinite semilinear elliptic equations involving concave-convex nonlinearities by the variational method:
(5)−Δu+Vλxu=fxuq−2u+gxup−2uinℝN,u≥0inℝN.Fors·=α,px,qx≡constant, and A=0; in [4], the authors obtained the existence, multiplicity, and concentration of nontrivial solutions for the following indefinite fractional elliptic equation by using Nehari manifold decomposition:
(6)−Δαu+Vλxu=axuq−2u+bxup−2uinℝN,u≥0inℝN.WhenA=0,V−x=0, and fx,gx≡constant, the authors in [2] give some sufficient conditions to ensure the existence of two different weak solutions and use the variational method and the mountain pass theorem to obtain the two weak solutions of problem (12) which converge to two solutions of its limit problems and the existence of infinitely many solutions to its limit problem:
(7)−Δs·u+λVxu=αupx−2u+βuqx−2uinΩ,u=0inℝN\Ω.Fors·=s,px,qx≡constant, in [1], the authors study the existence of solutions for the following equation on the whole space by using the method of Nehari manifold decomposition,and obtain some sufficient conditions for the existence of nontrivial solutions of the following equation:
(8)−ΔAsu+Vλxu=fxuq−2u+gxup−2uinℝN.In recent years, with the continuous deepening of research, the fractional magnetic problem has attracted extensive attention of researchers. More and more researchers have studied the solvability of the fractional magnetic problem (see [5–8]). We know that the fractional magnetic Laplacian operator −ΔAs is introduced in literature [9]; −ΔAs comes from magnetic Laplacian ∇−iA2s. However, as far as we know, up to now, few papers have studied the existence and multiplicity of solutions for the variable-order fractional magnetic Schrödinger equation. Therefore, motivated by the above literature, we are interested in the existence and multiplicity of solutions to problem (1) with variable growth. As far as we know, this is the first time to study the existence and multiplicity of nontrivial solutions for the variable-order nonlinear fractional magnetic Schrödinger equation with variable exponents.It is worth noting that in this paper, we not only prove that there exist two different nontrivial solutions to problem (1), but we also show that the two nontrivial solutions of problem (1) converge to two solutions of the limit problem for problem (1). The novelty of this paper is that, compared with [1], we write the original fractional magnetic Schrödinger equation without variable growth to a variable-order fractional magnetic Schrödinger equation with variable growth. In addition, compared with [2], we write the variable order fractional Schrödinger equation to a variable order fractional magnetic Schrödinger equation.Inspired by the above works, we assume thats:ℝN×ℝN⟶0,1 and V are continuous functions satisfying the following:(S1): 0<s−≔minx,y∈ℝN×ℝNsx,y≤s+≔maxx,y∈ℝN×ℝNsx,y<1.(S2): s· is symmetric, that is, sx,y=sy,x for all x,y∈ℝN×ℝN.(V1): X=intV+−10⊂Ω is a nonempty bounded domain and X~=V+−10.(V2): there exists a nonempty open domain Ω0⊂X such that V+≡0, V−≡0 for all x∈Ω0¯.(V3): V+ is a continuous function on Ω and V−∈LN/2Ω.(V4): there exists a constant ϑ0>1 such that
(9)infu∈H0,As·Ω,ℂ\0∫ℝ2Nux−eix−y·Ax+y/2uy2/x−yN+2sx,ydxdy+λ∫ΩV+u2dx∫ΩV−u2dx≥ϑ0,or all λ>0, where H0,As·Ω,ℂ is the Hilbert space related to magnetic field A (see Section 2).For the variable exponentsp,q, we assume that p,q∈CΩ¯ and satisfy the following assumption:(H1): 2<px<2N/N−2sx,x for all x∈Ω¯.(H2): 1<qx<2 for all x∈Ω¯.In addition, we assume thatf, g satisfy the following assumption:(H3): f,g:ℝN⟶0,∞ are bounded nonnegative measurable function such that f>0, g>0 on open interval Ωf,Ωg⊂Ω and
(10)f∞=fL∞ℝN≤p−2−q+ϑ0−1maxCpp+,Cpp−2ϑ0p+−q+,g∞=gL∞ℝN≤q−p+−2ϑ0−1maxCqq+,Cqq−2ϑ0p+−q+2−q+ϑ0−12A1ϑ0p+−q+2−q+/p+−2.Based on the hypothesisS2, we can give the following definition of weak solutions for problem (1).Definition 1.
We say thatu∈Eλ is a weak solution of problem (1), if
(11)R∫ℝ2Nux−eix−y·Ax+y/2uyvx−eix−y·Ax+y/2vy¯x−yN+2sx,ydxdy+λR∫ΩV+uv¯dx−R∫ΩV−uv¯dx−R∫Ωfxupx−2u+gxuqx−2uv¯dx=0,for any v∈Eλ, where Eλ will be defined in Section 2.Now, we will describe the first main result as follows.Theorem 2.
Assume thatS1,S2,V1−V4,andH1−H3 hold. Let N>2s+. Then, the problem (1) allows at least two different solutions for all λ>0.Theorem 3.
Letuλ1 and uλ2 be two solutions obtained in Theorem 2. Then, uλ1⟶u1 and uλ2⟶u2 in H0,As·Ω,ℂ as λ⟶∞, where u1≠u2 are two nontrivial solutions of the following problem:
(12)−ΔAs·u−V−xu=fxupx−2u+gxuqx−2uinΩ0,u=0inℝN\Ω0.Remark 4.
In general, ifs·:ℝN×ℝN⟶0,1 is a continuous function, magnetic field A∈C0,αℝN,ℝN with α∈0,1, then the variable-order fractional magnetic Laplacian can be defined as follows: for each u∈C0∞Ω,ℂ,
(13)−ΔAs·u,v=R∫ℝ2Nux−eix−y·Ax+y/2uyvx−eix−y·Ax+y/2vy¯x−yN+2sx,ydxdy,along any v∈C0∞Ω,ℂ.


## 2. Preliminaries and Notations

In this section, we first give the definition of the variable exponential Lebesgue space. Secondly, we define variable-order fractional magnetic Sobolev spaces and prove the compact conditions between them. Finally, we give the variational setting for problem (1) and theorems that will be used later.In this paper, we useΩ to represent n-dimensional Lebesgue measure of a measurable set Ω⊂ℝN. In addition, for each a∈ℂ, we will use Ra to represent the real part of a and a¯ to represent the complex conjugate of a. Let N≥1 and Ω⊂ℝN be a nonempty set. A measurable function r:Ω¯⟶1,∞ is called a variable exponent, and we define r+=esssupx∈Ωrx,r−=essinfx∈Ωrx. If r+ is finite, then the exponent r is said to be bounded. The variable exponent Lebesgue space is
(14)LrxΩ,ℂ=u:Ω⟶ℂisameasurablefunction;∫Ωuxrxdx<∞with the Luxemburg norm
(15)uLrxΩ,ℂ=infμ>0:∫Ωuxμrxdx≤1,then LrxΩ,ℂ is a Banach space, and when r is bounded, we have the following relations
(16)minuLrxΩ,ℂr−,uLrxΩ,ℂr+≤∫Ωuxrxdx≤maxuLrxΩ,ℂr−,uLrxΩ,ℂr+.For bounded exponent, the dual spaceLrxΩ,ℂ′ can be identified with Lr′xΩ,ℂ, where the conjugate exponent r′ is defined by r′=r/r−1. If 1<r−≤r+<∞, then the variable exponent Lebesgue space LrxΩ,ℂ is a separable and reflexive. In particular,
(17)L2Ω,ℂ=u:Ω⟶ℂisameasurablefunction;∫Ωux2dx<∞with the scalar product
(18)u,vL2Ω,ℂ=R∫Ωuv¯dx.By Lemma 3.2.20 of [10] and ·LrxΩ,ℂ=·LrxΩ,ℝ, we know that in the variable exponent Lebesgue space, Hölder inequality is still valid. For all u∈LrxΩ,ℂ,v∈Lr′xΩ,ℂ with rx∈1,∞, the following inequality holds:
(19)∫Ωuvdx≤1r−+1r′−uLrxΩ,ℂvLr′xΩ,ℂ≤2uLrxΩ,ℂvLr′xΩ,ℂ.LetΩ be a nonempty open subset of ℝN, and let s·:ℝN×ℝN⟶0,1 be a measurable function, and there exist two constants 0<s0<s1<1 such that s0<sx,y<s1 for all x,y∈ℝN×ℝN. Set
(20)Hs·Ω,ℂ=u∈L2Ω,ℂ:∫Ω∫Ωux−uy2x−yN+2sx,ydxdy1/2<∞.EquipHs·Ω,ℂ with the norm
(21)uHs·Ω,ℂ2=uL2Ω,ℂ2+∫Ω∫Ωux−uy2x−yN+2sx,ydxdy.Especially, ifs·≡constant, then the space Hs·Ω,ℂ is the usual fractional Sobolev space HsΩ,ℂ.Lemma 8.
LetΩ be a smooth bounded subset of ℝN and let s·:ℝN×ℝN⟶0,1 satisfying S1 and r:Ω¯⟶1,∞ satisfying 1≤r≤2N/N−2sx,x. Then, there exists Cr~=CN,r+,s+,s−>0 such that uLrxΩ,ℂ≤Cr~uHs·Ω,ℂ, for any u∈Hs·Ω,ℂ. That is, the embedding Hs·Ω,ℂ°LrxΩ,ℂ is continuous. Moreover, Hs·Ω,ℂ°LrxΩ,ℂ is compact.Proof.
By Theorem 2.1 of [2], we know that Hs·Ω,ℝ°LrxΩ,ℝ is continuous and compact, there exists Cr~=CN,r+,s+,s−>0 such that uLrxΩ,ℝ≤Cr~uHs·Ω,ℝ. then, for any u∈Hs·Ω,ℂ, we have
(22)uLrxΩ,ℂ=uLrxΩ,ℝ≤Cr~uHs·Ω,ℝ=Cr~uL2Ω,ℝ2+∫Ω∫Ωux−uy2x−yN+2sx,ydxdy1/2≤Cr~uL2Ω,ℂ2+∫Ω∫Ωux−uy2x−yN+2sx,ydxdy1/2=Cr~uHs·Ω,ℂ.
Hence, the embeddingHs·Ω,ℂ°LrxΩ,ℂ is continuous and compact.LetA:ℝN⟶ℝN be a continuous function and A∈Lloc∞ℝN,ℝN. For a function u:ℝN⟶ℂ, define
(23)uHAs·ℝN,ℂ2≔∫ℝN∫ℝNux−eix−y·Ax+y/2uy2x−yN+2sx,ydxdyand the corresponding norm denoted by uHAs·ℝN,ℂ2=uL2ℝN,ℂ2+uHAs·ℝN,ℂ2. We consider the space H of measurable functions u:ℝN⟶ℂ such that uHAs·ℝN,ℂ<∞; then, H,·,·HAs·ℝN,ℂ is a Hilbert space. Define HAs·ℝN,ℂ as the closure of Cc∞ℝN,ℂ in H; then, HAs·ℝN,ℂ is a Hilbert space. Especially, if A=0, then the space HAs·ℝN,ℂ is the variable-order fractional Sobolev space Hs·ℝN,ℂ; if A=0 and s·≡constant, then the space HAs·ℝN,ℂ is the usual fractional Sobolev space HsℝN,ℂ.In order to define weak solutions of problem (1), we introduce the functional space
(24)H0,As·Ω,ℂ=u∈HAs·ℝN,ℂ:u=0a.e.inℝN\Ω,equipping H0,As·Ω,ℂ with the scalar product
(25)u,vH0,As·Ω,ℂ≔R∫ℝ2Nux−eix−y·Ax+y/2uyvx−eix−y·Ax+y/2vy¯x−yN+2sx,ydxdy,which induces the following norm uH0,As·Ω,ℂ≔u,uH0,As·Ω,ℂ1/2. Hence, H0,As·Ω,ℂ generalizes to the variable-order fractional Sobolev space (see [2]) and the magnetic framework the space introduced in [9]. Next, we state and prove some properties of space H0,As·Ω,ℂ, which will be useful in the sequel.Lemma 6.
There exists a constantC2>0, depending only on N,s1 and Ω, such that
(26)uH0,As·Ω,ℂ≤uHAs·ℝN,ℂ≤C2uH0,As·Ω,ℂ,for any u∈H0,As·Ω,ℂ. Thus, uH0,As·Ω,ℂ is an equivalent norm of HAs·ℝN,ℂ.Proof.
For anyu∈H0,As·Ω,ℂ, by Lemma 3.1 in [9], we have the pointwise diamagnetic inequality
(27)ux−eix−y·Ax+y/2uy≥ux−uy,fora.e.x,y∈ℝN,from which we immediately have
(28)∫ℝN∫ℝNux−eix−y·Ax+y/2uy2x−yN+2sx,ydxdy≥∫ℝN∫ℝNux−uy2x−yN+2sx,ydxdy≥∫ℝN\Ω∫Ωux−uy2x−yN+2sx,ydydx=∫ℝNΩ∫Ωuy2x−yN+2sx,ydydx.SinceΩ is bounded, there exists r>1/2 such that Ω⊂Br and Br\Ω>0; then, we have
(29)∫ℝN\Ω∫Ωuy2x−yN+2sx,ydydx≥∫BrΩ∫Ωuy22rN+2sx,ydydx≥∫Br\Ω∫Ωuy22rN+2s1dydx=12rN+2s1∫Br\Ω∫Ωuy2dydx=Br\Ω2rN+2s1uL2Ω,ℂ2.Thus, we obtain(30)uL2Ω,ℂ2≤2rN+2s1Br\Ω∫ℝ2Nux−eix−y·Ax+y/2uy2x−yN+2sx,ydxdy=C1uH0,As·Ω,ℂ2,uHAs·ℝN,ℂ2=uL2Ω,ℂ2+uHAs·ℝN,ℂ2≤C1uH0,As·Ω,ℂ2+uH0,As·Ω,ℂ2=C22uH0,As·Ω,ℂ2,where C1=2rN+2s1/Br\Ω and C22=C1+1.In addition,(31)uH0,As·Ω,ℂ2≤∫Ωux2dx+∫ℝ2Nux−eix−y·Ax+y/2uy2x−yN+2sx,ydxdy=uHAs·ℝN,ℂ2.Combining the above two aspects, we have(32)uH0,As·Ω,ℂ≤uHAs·ℝN,ℂ≤C2uH0,As·Ω,ℂ,which implies that uH0,As·Ω,ℂ is the equivalent norm of a norm uHAs·ℝN,ℂ.Lemma 2.3.
LetΩ be a bounded subset of ℝN. Assume that s·:ℝN×ℝN⟶0,1 is a continuous function satisfying S1 and r:Ω¯⟶1,∞ is a continuous function satisfying 1≤r≤2N/N−2sx,x. If u∈H0,As·Ω,ℂ, then
(33)H0,As·Ω,ℂ°Hs·Ω,ℂis continuous and
(34)H0,As·Ω,ℂ°LrxΩ,ℂis compact, that is, there exists Cr=CN,r+,s+,s−>0 such that
(35)uLrxΩ,ℂ≤CruH0,As·Ω,ℂ.Proof.
For anyu∈H0,As·Ω,ℂ, we have
(36)uHs·Ω,ℂ2=∫Ωux2dx+∫Ω∫Ωux−uy2x−yN+2sx,ydxdy≤∫Ωux2dx+2∫Ω∫Ωux−eix−y·Ax+y/2uy2x−yN+2sx,ydxdy+2∫Ω∫Ωuy2eix−y·Ax+y/2−12x−yN+2sx,ydxdy=2uHAs·Ω,ℂ2+2J,where
(37)J=∫Ω∫Ωuy2eix−y·Ax+y/2−12x−yN+2sx,ydxdy=∫Ωuy2∫Ω∩x−y>1eix−y·Ax+y/2−12x−yN+2sx,ydxdy+∫Ωuy2∫Ω∩x−y≤1eix−y·Ax+y/2−12x−yN+2sx,ydxdy=J1+J2.
Sinceeit−1≤2, we have
(38)J1≤4∫Ωuy2∫Ω∩x−y>11x−yN+2sx,ydxdy≤4∫Ωuy2∫Ω∩x−y>11x−yN+2s0dxdy=4∫Ωuy2∫Ω∩z>11zN+2s0dzdy≤C3∫Ωuy2dy=C3uL2Ω,ℂ2.In view ofΩ which is bounded, there exists a compact set K⊂ℝN such that Ω⊂K. By Lemma 2.2 of [11], we know that A is locally bounded and K⊂ℝNis compact, eix−y·Ax+y/2−12≤C4x−y2,forx−y≤1,x,y∈K. Thus, we obtain
(39)J2≤∫Kuy2∫K∩x−y≤1eix−y·Ax+y/2−12x−yN+2sx,ydxdy≤∫Kuy2∫K∩x−y≤1C4x−yN+2sx,y−2dxdy≤∫Kuy2∫K∩x−y≤1C4x−yN+2s1−2dxdy≤C4∫Kuy2∫K∩z≤11zN+2s1−2dzdy≤C5∫Kuy2dy=C5∫K\Ωuy2dy+C5∫Ωuy2dy=C5∫Ωuy2dy=C5uL2Ω,ℂ2.Equations (36)–(39) together with Lemma 6, we have
(40)uHs·Ω,ℂ2≤2uHAs·Ω,ℂ2+2C3uL2Ω,ℂ2+2C5uL2Ω,ℂ2≤C6uHAs·Ω,ℂ2≤C6uHAs·ℝN,ℂ2≤C7uH0,As·Ω,ℂ2,which implies that the embedding H0,As·Ω,ℂ↪Hs·Ω,ℂis continuous. In addition, by Lemma 5, we know that Hs·Ω,ℂ↪LrxΩ,ℂ is compact. Therefore, the embedding H0,As·Ω,ℂ↪LrxΩ,ℂ is compact.Next, we give the variational setting for problem (1). For λ>0, we need the following scalar product and norm:
(41)u,vλ≔R∫ℝ2Nux−eix−y·Ax+y/2uyvx−eix−y·Ax+y/2vy¯x−yN+2sx,ydxdy+λ·R∫ΩV+uv¯dx,uλ≔u,uλ1/2.LetE=u∈H0,As·Ω,ℂ:∫ΩV+u2dx<∞ be equipped with the norm uE=u1 (that is, λ=1 in uλ). Obviously, uE≤uλ for λ≥1. Set Eλ=E,·λ. Moreover, for rx∈1,2N/N−2sx,x, we can get
(42)∫Ωuxrxdx≤maxuLr·Ω,ℂr+,uLr·Ω,ℂr−≤maxCrr+uH0,As·Ω,ℂr+,Crr−uH0,As·Ω,ℂr−≤maxCrr+uλr+,Crr−uλr−.For simplicity, we letuλ,V2≔uH0,As·Ω,ℂ2+∫ΩVλu2dx. Therefore, by condition V4,
(43)uλ2≥uλ,V2≥ϑ0−1ϑ0uλ2,forallλ≥0.Associated with problem (1), we consider the energy functional Ψλ:Eλ⟶ℝ,(44)Ψλu=12uλ2−12∫ΩV−u2dx−∫Ωfxpxupx+gxqxuqxdx=12uλ,V2−∫Ωfxpxupx+gxqxuqxdx.In fact, one can verify thatΨλ is well-defined of class C1 in Eλ and
(45)R∫ℝ2Nux−eix−y·Ax+y/2uyvx−eix−y·Ax+y/2vy¯x−yN+2sx,ydxdy+λR∫ΩV+uv¯dx−R∫ΩV−uv¯dx−R∫Ωfxupx−2u+gxuqx−2uv¯dx=0,for all u,v∈Eλ. Therefore, if u∈Eλ is a critical point of Ψλ, then u is a solution of problem (1).Now we give the theorems that we need later.Theorem 8 (see [2, 12]).
LetX be a real infinite dimensional Banach space and I∈C′X a functional satisfying the PSc condition as well as the following three properties:
(1)
I0=0, and there exists two constants ρ,δ>0 such that Iu≥δ for all u∈X with u=ρ(2)
I is even(3)
For all finite dimensional subspacesY⊂X, there exists R=RY>0 such that Iu≤0 for all u∈X\BRY, where BRY=u∈Y:u≤R. Then, I poses an unbounded sequence of critical values characterized by a minimax argumentTheorem 9 (see [13]).
LetX be a real Banach space and I∈C1X,ℝ. Suppose I satisfies the (PS) condition, which is even and bounded from below, and I0=0. If for any k∈N, there exists a k-dimensional subspace Xk of X and ρk>0 such that supXk∩SρkI<0, where Sρk=u∈X:uX=ρk, then at least one of the following conclusions holds:
(i)
There exists a sequence of critical pointsuk satisfying Iuk<0 for all k and ukX⟶0 as k⟶∞(ii)
There existsr>0 such that for any 0<a<r, there exists a critical point u such that uX=a and Iu=0
It is easy to verify thatEλ is a separable Hilbert space. Let Xi=spanei, then Eλ=⊕i≥1Xi. we define
(46)Yk=⊕i=1kXi,Zk=⊕i=k+1∞Xi¯.Theorem 10 (see [14], fountain theorem).
Suppose thatI∈C1E,ℝ satisfying I−u=Iu. Assume that for every k∈N, there exist rk>γk>0 such that
(D1): ak=maxIu:u∈Yk,u=rk≤0.
(D2): bk=infIu:u∈Zk,u=γk⟶∞ as k⟶∞.
(D3): I satisfies PSc condition for every c>0.
Then,I has an unbounded sequence of critical values which have the form
(47)ck=infr∈Γkmaxu∈BkIηu,where Γk=η∈CBk,E:ηisequivariantandη∂Bk=id.Theorem 11 (see [15], dual fountain theorem).
Suppose thatI∈C1E,ℝ satisfying I−u=Iu. Assume that for every k≥k0, there exist rk>γk>0 such that
(D4): ak=infIu:u∈Zk,u=rk≥0.
(D5): bk=maxIu:u∈Yk,u=γk<0.
(D6): dk=maxIu:u∈Zk,u≤γk⟶0 as k⟶∞.
(D7): I satisfies PSc∗ condition for every c∈dk0,0.
Then,I has a sequence of negative critical values converging to 0.


## 3. Proof of Theorem 1

In this part, we first recall that definitions of functionalΨλ satisfies the PSc condition and PSc∗ condition in Eλ at the level c∈ℝ and use the usual mountain pass theorem (see [2]) to find a PSc sequence in Eλ. Second, we show that functional Ψλ satisfies the PSc∗ condition in Eλ at the level c<c0. Finally, we give the proof of problem (1).Definition 12 (see [2]).
LetI∈C1E,ℝ and c∈ℝ. The functional I satisfies the PSc condition if any sequence un⊂E such that Iun⟶candI′un⟶0asn⟶∞ admits a strongly convergent subsequence in E.Definition 13 (see [16]).
LetI∈C1E,ℝ and c∈ℝ. The functional I satisfies the PSc∗ condition (with respect to Yn) if any sequence un⊂E such that un∈Yn, Iun⟶candI′Ynun⟶0asn⟶∞ admits a strongly convergent subsequence in E.Remark 14.
From Remark 2.1 in [16], we get that the PSc∗ condition means the PSc condition.Theorem 15 (Theorem 3.1, [2]).
LetE be a real Banach space and I∈C1E,ℝ with I0=0. Suppose that
(i)
there existδ>0 and ρ>0 such that Iu≥δ for each u∈E subject to uE=ρ(ii)
there existse∈E with eE>ρ such that Ie<0
DefineΓ=γ∈C10,1,E:γ0=1,γ1=e. Then,
(48)c=infγ∈Γmax0≤t≤1Iγt≥δ,and there exists a PSc sequence unn⊂E.In order to obtain our main results by using the mountain pass theorem, we first prove thatΨλ satisfies the mountain pass geometry (i) and (ii).Lemma 16.
Assume that (S1), (V1)–(V4), and (H1)–(H3) hold. Then, for each λ>0, there exists ρ>0 and τ>0 such that
(49)Ψλu>τforallu∈Eλwithuλ=ρ.Proof.
In view of (42) and the fractional Sobolev inequality, for each u∈Eλ, one has
(50)∫Ωfxpxupx+gxqxuqxdx≤f∞p−∫Ωupxdx+g∞q−∫Ωuqxdx≤f∞p−maxCpp+uλp+,Cpp−uλp−+g∞q−maxCqq+uλq+,Cqq−uλq−,where Cp,Cq are two constants of embedding from variable-order fractional Sobolev space H0,As·Ω,ℂ to LpxΩ,ℂ and LqxΩ,ℂ, respectively. Making use of (43) and (50), we obtain that
(51)Ψλu=12uλ2−12∫ΩV−u2dx−∫Ωfxpxupx+gxqxuqxdx=12uλ,V2−∫Ωfxpxupx+gxqxuqxdx≥12ϑ0−1ϑ0uλ2−f∞maxCpp+,Cpp−p−uλp+−g∞maxCqq+,Cqq−q−uλq+=ϑ0−12ϑ0uλ2−A1uλp+−A2uλq+=uλq+ϑ0−12ϑ0uλ2−q+−A1uλp+−q+−A2,for each u∈Eλ with uλ≥1.Set(52)A1=f∞maxCpp+,Cpp−p−,A2=g∞maxCqq+,Cqq−q−.DefineΦ1t:0,∞⟶ℝ as follows
(53)Φ1t=Φ2ttq+forallt≥0,where
(54)Φ2t=ϑ0−12ϑ0t2−q+−A1tp+−q+−A2.As long as(55)A2<2−q+ϑ0−12A1ϑ0p+−q+2−q+/p+−2ϑ0−1p+−22ϑ0p+−q+,that is,
(56)A12−q+A2p+−2≤2−q+ϑ0−12ϑ0p+−q+2−q+ϑ0−1p+−22ϑ0p+−q+p+−2,that is,
(57)g∞≤q−p+−2ϑ0−1maxCqq+,Cqq−2ϑ0p+−q+2−q+ϑ0−12A1ϑ0p+−q+2−q+/p+−2,we can easily show that for t=t~=2−q+ϑ0−1/2A1ϑ0p+−q+1/p+−2, we have
(58)Φ2t~=maxt≥0Φ2t>0.By(59)f∞≤p−2−q+ϑ0−1maxCpp+,Cpp−2ϑ0p+−q+,it is easy to derive that
(60)t~=2−q+ϑ0−12A1ϑ0p+−q+1/p+−2≥1.By lettingρ=t~>0 and τ=Φt~>0, we can easily get Ψλu>τforallu∈Eλwithuλ=ρ.Lemma 17.
Suppose thatS1,V1−V4, and H1−H3 hold. Then, there exists e∈Eλwitheλ>ρ such that Ψλe<0forallλ>0, where ρ>0 is given by Lemma 16.Proof.
Notice thatf:ℝN⟶0,∞ is a bounded nonnegative measurable function such that f>0 on open interval Ωf⊂Ω; then, we can select w0∈Eλ such that
(61)w0λ=1and∫Ωfxw0xpxdx>0.
For allt≥1, combining (43) with (44), we have
(62)Ψλtw0=12tw0λ2−12∫ΩV−tw02dx−∫Ωfxpxtw0px+gxqxtw0qxdx=12tw0λ,V2−∫Ωfxpxtw0px+gxqxtw0qxdx≤12tw0λ2−∫Ωfxpxtw0px+gxqxtw0qxdx≤t22w0λ2−tp−p+∫Ωfxw0pxdx.Sincep−>2, then there exists t∗≥1 large enough such that t∗w0λ>ρ and Ψλt∗w0<0. By letting e=t∗w0, we can easily reach the conclusion.Define(63)cλ=infγ∈Γ1max0≤t≤1Ψλγt,cΩ0=infγ∈Γ2max0≤t≤1ΨλH0,As·Ω0,ℂγt,where ΨλH0,As·Ω0,ℂ is a restriction of Ψλ on H0,As·Ω0,ℂ and
(64)Γ1=γ∈C0,1,Eλ:γ0=0,γ1=e,Γ2=γ∈C0,1,H0,As·Ω0,ℂ:γ0=0,γ1=e.Observe that(65)ΨλH0,As·Ω0,ℂu=12uH0,As·Ω0,ℂ2−12∫Ω0V−u2dx−∫Ω0fxpxupx+gxqxuqxdx,for all u∈H0,As·Ω0,ℂ. Obviously, cΩ0 is independent of λ. From the proofs of Lemma 16 and Lemma 17, we can easily derive that ΨλH0,As·Ω0,ℂ satisfies the mountain pass geometry. Since H0,As·Ω0,ℂ⊂Eλ for all λ>0, we have 0<τ≤cλ≤cΩ0 for all λ>0. Evidently, for any t∈0,1,te∈Γ2. Consequently, there exists c0>0 such that
(66)cΩ0≤max0≤t≤1Ψλte≤c0<∞,being p−>2. Then,
(67)0<τ≤cλ≤cΩ0<c0,for all λ>0. In view of Lemma 16, Lemma 17, and Theorem 15, it is easy to get that for all λ>0, there exists un⊂Eλ such that
(68)Ψλun⟶cλ>0,Ψλ′un⟶0,asn⟶∞.Lemma 18.
Assume that (S1),(V1)–(V4) and (H1)–(H3) hold. Then, Ψλ satisfies the PSc∗ condition in Eλ for all c<c0 and λ>0.Proof.
Assume thatun be a PSc∗ sequence in Eλ with c<c0; then, un∈Yn, Ψλun⟶cλandΨλ′Ynun⟶0asn⟶∞. It follows from (42) and (43) and the Hölder inequality that
(69)cλ+o1unλ≥Ψλun−1p−Ψλ′un,un=12unλ,V2−∫Ωfxpxunpx+gxqxunqxdx−1p−unλ,V2+1p−∫Ωfxunpx+gxunqxdx=12−1p−unλ,V2−∫Ωfx1px−1p−unpxdx−∫Ωgx1qx−1p−unqxdx≥p−−22p−ϑ0−1ϑ0unλ2−1q−−1p−∫Ωgxunqxdx≥p−−22p−ϑ0−1ϑ0unλ2−1q−−1p−g∞maxCqq+unλq+,Cqq−unλq−.
On the contrary, we suppose thatun is not bounded in Eλ. Then, there exists a subsequence still denoted by un such that unλ⟶∞ as n⟶∞. Then, it follows from (69) that
(70)cλunλ2+o11unλ≥p−−22p−ϑ0−1ϑ0−1q−−1p−g∞maxCqq+unλq+−2,Cqq−unλq−−2,which leads to a contradiction. Hence, un is bounded in Eλ for all λ>0. Therefore, there exist a subsequence of un still denoted by un and u0 in Eλ such that
(71)un⇀u0inEλ,un⟶u0a.e.inΩ,unrx−2un⇀u0rx−2u0inLr′xΩ,ℂ,where r′x=rx/rx−1. The next step is to show that un⟶u0 in Eλ. By Lemma 7, we can get un⟶u0 in LrxΩ,ℂ. Thus,
(72)limn→∞∫Ωun−u0pxdx=0,(73)limn→∞∫Ωun−u0qxdx=0.Making use of Hölder inequality, we can obtain(74)∫Ωfxunpx−2un−u0px−2u0un−u0¯dx≤f∞∫Ωunpx−2un−u0px−2u0un−u0¯dx=f∞∫Ωunpx−2un−u0px−2u0un−u0¯dx=f∞∫Ωunpx−2un−u0px−2u0un−u0dx≤f∞∫Ωunpx−2un−u0px−2u0px/px−1dxpx−1/px∫Ωun−u0pxdx1/px.Combining (H3), (71), and (72), we have
(75)limn→∞∫Ωfxunpx−2un−u0px−2u0un−u0¯dx=0.Similarly, we have(76)limn→∞∫Ωgxunqx−2un−u0qx−2u0un−u0¯dx=0.We notice thatun⇀u0 in Eλ and Ψλ′un⟶0; then, we obtain
(77)limn→∞Ψλ′un−Ψλ′u0,un−u0=0.Therefore,(78)o1=Ψλ′un−Ψλ′u0,un−u0=un,un−u0λ,V−R∫Ωfxunpx−2un+gxunqx−2unun−u0¯dx−u0,un−u0λ,V+R∫Ωfxu0px−2u0+gxu0qx−2u0un−u0¯dx=un−u0,un−u0λ,V−R∫Ωfxunpx−2un−u0px−2u0un−u0¯dx−R∫Ωgxunqx−2un−u0qx−2u0un−u0¯dx,which means that
(79)limn→∞un−u0λ,V=0.It follows from (43) that limn→∞un−u0λ=0.Proof of Theorem 2.
In view of Lemma16, Lemma 17, and Theorem 15, we can easily infer that for all λ>0, there exists a PScλ sequence un for Ψλ on Eλ. It derives from Lemma 18 and 0<cλ<cΩ0<c0 for all λ>0 that there exists a subsequence of un still denoted by un and uλ1∈Eλ such that un⟶uλ1 in Eλ. Furthermore, Ψλun⟶cλ≥τ and uλ1 is a solution of problem (1).The next step is to prove that system (1) has another solution. Set
(80)c¯λ=infΨλu:u∈Bρ¯,where Bρ=u∈Eλ:uλ<ρ and ρ>0 is given by Lemma 16. Then, c¯λ<0 for all λ>0. For this purpose, we first prove there exists v0∈Eλ such that Ψλσv0<0 for all σ>0 sufficiently small. Let v0∈H0,As·Ω,ℂ such that ∫Ωgxv0qxdx>0. Making use of the hypothesis H2 and (43), we obtain that for σ∈0,1 small enough,
(81)Ψλσv0=σ22v0λ2−12∫ℝNV−σv02dx−∫Ωfxpxσv0px+gxqxσv0qxdx≤σ22v0λ,V2−σp+∫Ωfxpxv0pxdx−σq+∫Ωgxqxv0qxdx≤σ22v0λ2−σq+q+∫Ωgxv0qxdx<0.Consequently, there existsv0∈Eλ such that Ψλσv0<0 for all σ>0 sufficiently small.Applying Lemma16 and the Ekeland variational principle to Bρ¯, there exists a sequence un such that
(82)c¯λ≤Ψλun≤c¯λ+1n,Ψλv≥Ψλun−un−vλn,for all v∈Bρ¯. Now we prove that unλ<ρ for n enough large. On the contrary, we suppose that unλ=ρ for infinitely many n. Without loss of generality, we can suppose that unλ=ρ for n∈N. It follows from Lemma 16 that
(83)Ψλun≥τ>0.Combining with (82), we obtain c¯λ≥τ>0, which is contradictive with c¯λ<0. Next, we prove Ψλ′un⟶0 in Eλ∗ as n⟶∞. Let
(84)yn=un+σv,forallv∈B1≔u∈Eλ:uλ=1,where σ>0 small enough such that 2σρ+σ2≤ρ2−unλ2 for fixed n large. Then,
(85)ynλ2=unλ2+2σρun,vλ+σ2≤unλ2+2σρ+σ2=ρ2,which implies that yn∈Bρ¯. Hence, by using (50), we get
(86)Ψλyn≥Ψλun−un−ynλn,that is,
(87)Ψλun+σv−Ψλunσ≥−1n.Setσ⟶0+; we obtain Ψλ′un,v≥−1/n for each fixed n large. Similarly, choosing σ<0 and σ small enough and repeating the process above, we can easily get that
(88)Ψλ′un,v≤1n,for each fixed n large.In short, we have(89)limn→∞supv∈B1Ψλ′un,v=0,which immediately concludes that Ψλ′un⟶0 in Eλ∗ as n⟶∞. Therefore, un is a PSc¯λ sequence for the functional Ψλ. Making use of a similar proof as Lemma 18, there exists uλ2 such that un⟶uλ2 in Eλ. Therefore, we obtain a nontrivial solution uλ2 of problem (1) satisfying
(90)Ψλuλ2≤ξ<0,uλ2λ<ρ.Hence, it is easy to conclude that(91)Ψλuλ2=c¯λ≤ξ<0<τ<cλ=Ψλuλ1,forallλ>0,which completes the proof.


## 4. Proof of Theorem 3

In this section, we mainly give the proof of Theorem3. In addition, inspired by [2, 17], we obtain the method to prove Theorem 3.Proof of Theorem 3.
For each sequenceλn such that 1≤λn⟶∞ as n⟶∞, set uni to be the critical points of Ψλ obtained in Theorem 2, where i=1,2. Therefore, one has
(92)Ψλnun2≤ξ<0<τ<cλn=Ψλnun1<c0,Ψλn′un1=Ψλn′un2=0(93)Ψλnuni=12uniλn2−12∫ΩV−uni2dx−∫Ωfxpxunipxdx−∫Ωgxqxuniqxdx=12uniλn,V2−∫Ωfxpxunipxdx−∫Ωgxqxuniqxdx≥12ϑ0−1ϑ0uniλn2−∫Ωgxpxunipxdx−∫Ωgxqxuniqxdx≥p−−22p−ϑ0−1ϑ0uniλn2+1p−−1q−∫Ωgxuniqxdx≥p−−22p−ϑ0−1ϑ0uniλn2−1q−−1p−g∞maxCqq+uniλnq+,Cqq−uniλnq−.In view of (92) and (93), it gains
(94)uniλn≤c2,where c2>0 is independent of λn. So we can suppose that uni⇀ui in H0,As·Ω,ℂ and uni⟶ui in LrxΩ,ℂ. Making use of Fatou’s lemma, we can easily obtain
(95)∫ΩV+ui2dx≤liminfn→∞∫ΩV+uni2dx≤liminfn→∞uniλn2λn=0.Consequently,ui=0 a.e. in ℝN\V+−10 and ui∈H0,As·Ω0,ℂ.Next, we will prove thatuni⟶ui in H0,As·Ω,ℂ. Indeed, combining uni⟶ui in LrxΩ,ℂ and H3, one has
(96)∫Ωfxuni−uipxdx≤f∞∫Ωuni−uipxdx,∫Ωgxuni−uiqxdx≤g∞∫Ωuni−uiqxdx.Then,(97)limn→∞∫Ωfxuni−uipxdx=0,(98)limn→∞∫Ωgxuni−uiqxdx=0.We notice that(99)limn→∞Ψλn′uni,uni=limn→∞Ψλn′uni,ui=0.Therefore, we have(100)uniλn2=∫ΩV−uni2dx+∫Ωfxunipx+gxuniqxdx+o1,(101)uni,uiλn=R∫ΩV−uniui¯dx+R∫Ωfxunipx−2uniui¯dx+R∫Ωgxuniqx−2uniui¯dx+o1.By (97)–(101), we have
(102)limn→∞uniλn2=limn→∞uni,uiλn=uiH0,As·Ω,ℂ2.On the other hand, the weak lower semicontinuity of norm yields that(103)uiH0,As·Ω,ℂ2≤liminfn→∞uniH0,As·Ω,ℂ2≤limsupn→∞uniH0,As·Ω,ℂ2≤limn→∞uniλn2.To sum up, we can see that(104)limsupn→∞uniH0,As·Ω,ℂ≤uiH0,As·Ω,ℂ.By Proposition 3.32 of [18], we can obtain that uni⟶ui in H0,As·Ω,ℂ. We notice that limn→∞Ψλn′uni,v=0, for any v∈C0∞Ω0,ℂ. Hence,
(105)R∫ℝ2Nuix−eix−y·Ax+y/2uiyvx−eix−y·Ax+y/2vy¯x−yN+2sx,ydxdy−R∫Ω0V−uiv¯dx−R∫Ω0fxuipx−2ui+gxuiqx−2uiv¯dx=0.Since the density ofC0∞Ω0,ℂ in H0,As·Ω0,ℂ, we can obtain that ui is a weak solution of problem (12).Together with (92), ui=0 a.e. in ℝN\V+−10 and the constants ξ,τ are independent of λ; we have
(106)12u1H0,As·Ω0,ℂ2−12∫Ω0V−u12dx−∫Ω0fxpxu1px+gxqxu1qxdx≥τ>0,12u2H0,As·Ω0,ℂ2−12∫Ω0V−u22dx−∫Ω0fxpxu2px+gxqxu2qxdx≤ξ<0,which implies that ui≠0 and u1≠u2.Now we consider the case wherefx,gx≡constants; that is
(107)−ΔAs·u+Vλxu=aupx−2u+buqx−2uinΩ,u=0inℝN\Ω,where a,b are two nonnegative constants. Correspondingly, the energy functional Ψλ: Eλ⟶ℝ is
(108)Ψλu=12uλ2−12∫ΩV−u2dx−∫Ωapxupx+bqxuqxdx=12uλ,V2−∫Ωapxupx+bqxuqxdx.Next, we mainly prove the existence of infinitely many solutions to problem (107) by using four different methods.Theorem 19.
Assume that (S1), (S2), (V1)–(V4), and (H1)–(H3) hold. Let N>2s+. Then, problem (107) has infinitely many solutions.Proof.
Method 1: It is easy to verify that functionalΨλ is even and satisfies Ψλ0=0. Furthermore, Lemma 18 shows that functional Ψλ is bounded from below in Eλ and satisfies the PS condition. For any k∈N and ρk>0, let Sρk=u∈Eλ,uλ=ρk; then, for any u∈Sρk, one has
(109)Ψλu=12uλ,V2−∫Ωapxupxdx−∫Ωbqxuqxdx≤12uλ2−bq+∫Ωuqxdx≤12uλ2−bq+minuLqxΩ,ℂq+,uLqxΩ,ℂq−.We find that there existsCSρk>0 such that uLqxΩ≥CSρkuλ, since all norms are equivalent on finite dimensional Banach space. Then, by 1<qx<2, it gains
(110)supu∈Xk∩SρkΨλu≤12uλ2−bp+minCSρkq+,CSρkq−uλq+.Lettinguλ=ρk small enough, we can obtain supu∈Xk∩SρkΨλu<0=Ψλ0. Furthermore, we assert that (ii) of Theorem 9 does not work. In fact, (109) means Ψλu≠0 since Ψλtu<0 as t small enough with the given u∈Eλ. Thus, by Theorem 9, we get that problem (1) has a sequence of solutions uk with ukλ⟶0 as k⟶∞. In short, problem (1) has infinitely many solutions for all λ>0.Method 2.
To start with, we assert that for any finite dimensional subspaceX of Eλ, there exists r1=r1X such that Ψλu<0 for all u∈Eλ\Br1X, where Br1X=u∈Eλ:uλ<r1. Indeed, for each t≥1, we can easily get that
(111)Ψλtu=t22uλ2−∫Ωapxtupxdx−∫Ωbqxtuqxdx≤t22uλ2−atp−p+∫Ωupxdx≤t22uλ2−atp−p+minuLpxΩ,ℂp+,uLpxΩ,ℂp−.We observe that there existsCX>0 such that uLpxΩ≥CXuλ, since all norms are equivalent on finite dimensional Banach space X. Then, by p−>2, it gains
(112)Ψλtu≤t22uλ2−atp−p+minCXp+,CXp−uλp+⟶−∞ast⟶∞.Thus, there existsr1>0 large enough such that Ψλu<0 for all u∈Eλ, with uλ=r2 and r2≥r1. Consequently, the assertion is valid.From Lemma18, we know thatΨλ satisfies the PSc condition for any c∈ℝ. Obliviously, Ψλ0=0 and Ψλ is an even functional. In short, it follows from Theorem 8 that there exists an unbounded sequence of solutions of problem (1) for all λ>0.Lemma 19 (see Lemma 4.1, [10]).
Let1<rx<2N/N−2sx,x for all x∈Ω¯. For any k∈N, define
(113)βk≔uLrxΩ,ℂ:u∈Zk,uλ=1.
Then,βk⟶0, as k⟶∞.Method 3.
By Remark14 and Lemma 18, we know that Ψλ satisfies the PSc condition for any c∈ℝ. To start with, we will prove (D1) is satisfied. It follows from (36) and (43) that
(114)Ψλu=12uλ,V2−∫Ωapxupxdx−∫Ωbqxuqxdx≤12uλ2−ap+∫Ωupxdx≤12uλ2−ap+minuLpxΩ,ℂp+,uLpxΩ,ℂp−.We find that there existsCYk>0 such that uLpxΩ≥CYkuλ, since all norms are equivalent on finite dimensional Banach space Yk. Then,
(115)Ψλu≤12uλ2−ap+minCYkp+,CYkp−uλp−.Then, byp−>2, we can easily obtain that (D1) is satisfied for uλ=rk>1 big enough. Next, we will show (D2) is fulfilled. In view of (43) and Lemma 20, for u∈Eλ, one has
(116)Ψλu=12uλ2−12∫ℝNV−u2dx−∫Ωapxupx+bqxuqxdx=12uλ,V2−∫Ωapxupx+bqxuqxdx≥ϑ0−12ϑ0uλ2−ap−∫Ωupxdx−bq−∫Ωuqxdx≥ϑ0−12ϑ0uλ2−ap−βkp−uλp+−bq−βkq−uλq+=uλq+ϑ0−12ϑ0uλ2−q+−ap−βkp−uλp+−q+−bq−βkq−=uλq+uλ2−q+ϑ0−12ϑ0−ap−βkp−uλp+−2−bq−βkq−=γkq+γk2−q+ϑ0−12ϑ0−ap−βkp−γkp+−2−bq−βkq−.Choosingγk=p−ϑ0−1/8aϑ0βkp−1/p+−2. Combing with Lemma 20, we know that βk⟶0 as k⟶∞. Thus, one has γk⟶+∞ as k⟶∞ and
(117)Ψλu≥γkq+38ϑ0−1ϑ0γk2−q+−bq−βkq−⟶+∞,as k⟶∞. In conclusion, (D2) is fulfilled. It is easy to check that satisfying Ψλ−u=Ψλu. Thus, by Theorem 10, we can obtain that problem (1) has infinitely many solutions for all λ>0.Method 4. First, we will show that (D4) is fulfilled. By (42) and (43), one has
(118)Ψλu≥12ϑ0−1ϑ0uλ2−ap−maxCpp+uλp+,Cpp−uλp−−bq−maxuLqxΩ,ℂq+,uLqxΩ,ℂq−.Since1<q−≤q+<2<p−≤p+<2N/N−2sx,x, we can choose r3∈0,1 small enough such that for all u∈Eλ with uλ≤r3,
(119)18ϑ0−1ϑ0uλ2≥ap−maxCpp+uλp+,Cpp−uλp−hold. Then,
(120)Ψλu≥38ϑ0−1ϑ0uλ2−bq−βkq−uλq−.Choosing
(121)rk=8bϑ0βkq−3q−ϑ0−11/2−q−.Combing with Lemma 20, we know thatβk⟶0 as k⟶∞. Thus, one has rk⟶0 as k⟶∞. Hence, there exists k¯ such that rk≤r3 as k≥k¯. Consequently, for k≥k¯ and u∈Zk with uλ=rk, we can obtain that Ψλu≥0. Next, we will show (D5) is fulfilled. For any u∈Yk, uλ=γk with rk>γk>0, we get
(122)Ψλu=12uλ,V2−∫Ωapxupxdx−∫Ωbqxuqxdx≤12uλ2−ap+∫Ωupxdx−bq+∫Ωuqxdx≤12uλ2−ap+minuLpxΩ,ℂp+,uLpxΩ,ℂp−−bq+minuLqxΩ,ℂq+,uLqxΩ,ℂq−.We find that there existsCYk>0 such that uLqxΩ≥CYkuλ and uLpxΩ≥CYkuλ, since all norms are equivalent on finite dimensional Banach space Yk. Then,
(123)Ψλu≤12uλ2−ap+minCYkp+uλp+,CYkp−uλp−−bq+minCYkq+uλq+,CYkq−uλq−≤12uλ2−ap+minCYkp+,CYkp−uλp+−bq+minCYkq+,CYkq−uλq+<0,as γk>0 small enough. Now we check (D6) is fulfilled. It follows from (D4) that for k≥k¯ and u∈Zk with uλ≤rk,
(124)Ψλu≥38ϑ0−1ϑ0uλ2−bq−βkq−uλq−≥−bq−βkq−uλq−≥−bq−βkq−rkq−,thanks to βk⟶0 as k⟶∞. Thus, one has rk⟶0 as k⟶∞. Thus, (D6) is also satisfied. In conclusion, by Theorem 11, we can obtain that problem (1) has infinitely many solutions for all λ>0.



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