# On a Class of Schrödinger System Problem in Orlicz–Sobolev Spaces

Authors: H. El-Houari; L. S. Chadli; H. Moussa
Journal: Journal of Function Spaces
 (2022)

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## Abstract

Using the mountain pass theorem, we obtain the existence of a nontrivial and nonnegative weak solution of a quasi-linear Schrödinger system driven by theω⋅-Laplacian operator in Orlicz–Sobolev spaces.


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## Body

## 1. Introduction

For a long time, the theory of function spaces has been more or less strongly attached with the investigation of partial differential equations, between 1935 and 1938. S. L. Sobolev introduced the classical Sobolev spacesWq,p, p∈1,+∞, q∈ℕ, as follows:(1)Wq,p♌=u∈Lp♌:Dαu∈Lp♌,for0≤α≤q,where ♌⊂ℝN is an open set and Dα is the weak (or distributional) partial derivative. The study of diverse mathematical problems with homogeneous operators satisfies nonpolynomial growth conditions rather than the classical p-structure using the classical model of monotonic operators which are mainly based on the analysis of Sobolev spaces W1,p♌ for finding the weak solutions. However, the problems do not have any equivalence in the traditional Lebesgue and Sobolev spaces. This has led to the employment of a general class of spaces of functions called Orlicz–Sobolev spaces. Those spaces are composed of functions that have weak derivatives and satisfy specific integrability conditions. We can quote here [1–7] and reference therein.In recent years, much of the work focused on the study of pulse propagation in nonlinear optical fiber. In a single-mode optical fiber, when nonlinear third-order effects are involved, the pulse propagation is qualified by the nonlinear Schrödinger equation. This equation was widely studied by several authors (see [8–11] and reference therein), and it has been proved that the existence and uniqueness of a soliton solution in the ground state, i.e., a positive, radial, symmetrical solution with the lowest energy level among all possible solutions, is a significant advantage. A mono-mode optical fiber is not precisely mono-mode; however, it is dual-mode due to the occurrence of birefringence. Birefringence tends to split a pulse in half in both polarization directions, but nonlinear effects can trap them together to prevent them from splitting. Menyuk in [11] has shown that both polarization components in a birefringent fiber are regulated by a nonlinear coupled Schrödinger system:(2)−ϕ2+bψ2ϕ=iϕt+ϕyy,−ψ2+bϕ2ψ=iψt+ψyy,where b>0 which depends on the fiber anisotropy. In the search for standing wave solutions of (2), i.e., solutions of the form:(3)ϕy,t=expiω12tuy,ψy,t=expiω22tvy,and by rescaling the variables, we obtain that u and v satisfy the following system:(4)−uyy+u=u2u+bv2u,inℝ,−vyy+ω2v=v2v+bu2v,inℝ.In this work, we deal with the existence of a weak solution to the following Schrödinger problem system type:(5)−Δω1⋅u+ϱ1yω1uu=Fuy,u,v,in♌,−Δω2⋅v+ϱ2yω2vv=Fvy,u,v,in♌,u=v=0,on∂♌,where −Δωi⋅ defined by(6)−Δωi⋅u=−divωi∇u∇u,i=1;2,and ϱi are two continuous functions satisfied the following:(i)
A1: ϱi∈CℝN,ℝ and infy∈ℝNϱiy≥ϱ0>0(ii)
A2: measy∈ℝN:ϱiy≤H<∞, for all H>0, where meas⋅ denotes the Lebesgue measure in ℝNRecently, problem (5) has attracted the interest of researchers. It was studied by L. Wang et al, in [12]. We weakened the hypotheses used to solve this problem in [12] by defining a new uniformly convex Banach space E, which is injected into the Orlicz–Sobolev space W1,Λ♌, where Λ:ℝ+⟶ℝ+ is an N-function, i.e., Λ is a convex even function and is represented as follows:(7)Λt=∫0tωsds,where ω:ℝ+⟶ℝ+ satisfies(8)ωt>0fort>0,limt⟶∞ωt=∞;ω0=0,and is right continuous function and nondecreasing function. We present some examples of an N-functions as follows:(1)
ωt=tp−1, t>0, 1<p+1<N(2)
ωt=2p1+t2p−1,t>0,1≤p<N/2(3)
ωt=log1+t/t,t∈ℝ∗We note byN the set of all N-functions. The relationship between Λ¯∈N the complementary function of Λ is given by(9)Λ¯t=supr≥0tr−Λr.TheN-function Λ∈Δ2, if for a certain constant 2<k,(10)kΛt≥Λ2t,∀t>0.This article proceeds in this way. In the second section, we remind some of the properties on Orlicz and Orlicz–Sobolev spaces. The third section discusses the data assumption specification. In section four, we explain the existence of a result and its proof based on the mountain pass theorem (see [13]).


## 2. Some Preliminary Results and Hypotheses

In this section, we mention a few basic properties of the Orlicz–Sobolev space. We refer the reader to [14–16] for some proofs of the results of this section.Let♌ be an open subset of ℝN, N≥2. The Orlicz space LΛ♌ is defined as the set of real-valued measurable functions u on ♌ such that(11)∫♌Λiuyλdy<∞,for someΛ>0.Recall thatLΛ♌ is a Banach space under the norm(12)uΛ=infλ>0∫♌Λuy/λdy≤1.The closure inLΛ♌ of all bounded measurable functions with compact support in ♌¯ is denoted by EΛ♌. The Space LΛ♌ is reflexive if and only if Λ,Λ¯∈Δ2 (see Adams [14]). In LΛ♌, the Orlicz norm uΛ is defined by(13)uΛ≔sup∫♌uyvydy,for all v∈EΛ¯♌ such that vΛ¯≤1. One major inequality in LΛ♌ is(14)∫♌ΛuyuΛdy≤1,for allu∈LΛ♌∖0.We have the following inequality:(15)uΛ≤∫♌Λuydy+1,for allu∈LΛ♌.Moreover, the Hölder inequality holds(16)uΛvΛ¯≥∫♌uyvydy,∀u∈LΛ♌∀v∈LΛ¯♌.For Orlicz spaces, Young inequality reads as follows:(17)Λ¯t+Λs≥st,∀t,s≥0.Also, in [14], for all s,t≥0 and p∈1,∞, we have that(18)s+tp≤2p−1tp+sp.LetΛ1,Λ2∈N. The notation Λ1≪Λ2 means that for each ε>0,(19)Λ1tΛ2εt⟶0ast⟶∞.We now focus on the Orlicz–Sobolev space defined by(20)W1LΛ♌≔u∈LΛ♌:∂u∂xi∈LΛ♌,i=1,…,N,equipped with the norm(21)u1,Λ=∇uΛ+uΛ.Thus,W1LΛ♌,u1,Λ is a Banach space. Denote(22)W01LΛ♌=C∞¯♌⋅1,Λ.Definition 1.
Letuk⊂LΛ♌ and u⊂LΛ♌. uk⟶modu in LΛ♌ if for certain μ>0, we have that ∫♌Λuk−u/μdy⟶0.
The reason for this is thatΛ∈Δ2 implies that(23)∫♌Λuk−udy⟶0⇔uk⟶moduinLΛ♌.Theorem 1 (Theorem 3.2 in [17]).
Let♌⊂ℝN be an open, bounded set and satisfies a cone condition. Then,(1)
If∫1+∞Λ−1t/t1+1/Ndt=+∞, W1LΛ♌21AALΛ∗♌ with continuous injection(2)
If∫1+∞Λ−1t/t1+1/Ndt<+∞, W1LΛ♌21AAC♌∩L∞♌ with continuous injectionTheorem 2.
LetΛ∈N. Then, there is μ>0 such that(24)uΛ≤μu0,Λ,∀u∈W01LΛ♌.
In this work, we noteW01LΛ♌ by W01,Λ♌ and W1LΛ♌ by W1,Λ♌.After that, we list few inequalities to be used in our proofs (see [14,18]).Lemma 1.
Letχ0t=mintl,tn and χ1t=maxtl,tn for t≥0 and Λ∈N, then these are equivalent:(1)
(25)1<l≔inft>0tωtΛt≤supt>0tωtΛt≔n<∞.(2)
χ0tΛs≤Λst≤χ1tΛsfor allt,s≥0.(3)
Λ∈Δ2.Lemma 2.
If (25) holds, then(26)χ0uΛ≤∫♌Λudy≤χ1uΛ,∀u∈LΛ♌.Lemma 3.
LetΛ¯ be the complement of Λ and χ2t=mintl¯,tn¯, χ3t=maxtl¯,tn¯ for t≥0 where l¯=l/l−1 and n¯=n/n−1. If Λ is an N-function and (25) holds with l>1, then Λ¯ satisfies the following:(1)
n¯=inft>0tΛ¯′t/Λ¯t≤supt>0tΛ¯′t/Λ¯t=l¯(2)
χ2tΛ¯ρ≤Λ¯ρt≤χ3tΛ¯ρ,∀t,ρ≥0(3)
χ2uΛ¯≤∫♌Λ¯udy≤χ3uΛ¯,∀u∈LΛ¯♌Remark 1.
We can see that1/l¯+1/l=1 and 1/n¯+1/n=1.We assume that(27)A0∫01Λ−1tt1+1/Ndt<∞,A∞∫1+∞Λ−1tt1+1/Ndt=+∞.Under the assumptionsA0 and A∞, we may introduce Λ∗∈N, say the conjugate given its inverse in ℝ+:(28)Λi∗−1s=∫0sΛi−1rrN+1/Ndrfors≥0,Λi∗s=Λi∗−sfors≤0.Lemma 4.
Letχ4t=mintl∗,tn∗, χ5t=maxtl∗,tn∗ for t≥0 where l∗=lN/N−l, and n∗=nN/N−n. If Λ∈N function γ satisfies the conditions A0, A∞. If (25) in Lemma 1 holds with l,n∈1,N, then Λ∗ satisfies the following:(1)
l∗=inft>0tΛ∗′t/Λ∗t≤supt>0tΛ∗′t/Λ∗t=n∗(2)
χ4tΛ∗s≤Λ∗st≤χ5tΛ∗s∀t,s≥0(3)
χ4uΛ∗≤∫♌Λ∗udy≤χ5uΛ∗∀u∈LΛ∗♌Lemma 5.
Λ≪Λ∗t near infinity, i.e., limt⟶∞Λkt/Λ∗t=0∀k>0.


### 2.1. Working SpaceE

This subsection is dedicated to the definition of the subsite of Orlicz–Sobolev spacesE.Due to nature ofω⋅-Laplacian operator defined in (6), we need to consider the Orlicz–Sobolev framework and we will examine some specific techniques to Orlicz and the Orlicz–Sobolev spaces. For that we define(29)Ei=1;2=u∈W1,ΛiℝN:∫♌ϱiyΛiudy<∞;u=0a.eℝN∖♌,equipped with the following norm uEi,Λi=∇uΛi+uϱi,Λi, where(30)uϱi,Λi=infλ>0:∫♌ϱiyΛiuyλdy≤1=infλ>0pu,Λi≤1,with(31)pu,Λi=∫♌ϱiyΛiuyλdy.Remark 2.
IfΛt∈N is a convex function, then we can see that Ei is uniformly convex Banach space.Remark 3.
It can be noticed by means of Fatou’s lemma that(32)puuϱi,Λi≤1,∀u∈∖0.For anyu∈Ei, we may assume u≠0; by the definition of ⋅ϱi,Λi, there exists λn↘uϱi,Λi such that(33)puλn≤1.By Fatou’s lemma,(34)puuϱi,Λi≤limn⟶+∞puλn≤1.Now, we check that⋅Ei,Λi is a norm in E; for that, it is sufficient to verify that ⋅ϱi,Λi is a norm in Ei. Let us prove that ⋅ϱi,Λi is a norm in Ei.(i)
Ifuϱi,Λi=0, then ∫♌ϱiyΛiuy/Λdy=0. Since Λi:ℝ+⟶ℝ+ and ϱi:ℝℕ⟶ℝ+ are tow continuous and positive functions, then ϱiyΛiuy/Λ=0, and by the condition A1, we infer that ϱiy>0, and then Λiuy/Λ=0. Or Λi0=0 then uy/Λ=0, which implies that uy=0.(ii)
Forα∈ℝ and u∈Ei, we have(35)αuϱi,Λi=infΛ>0,∫♌ϱiyΛiαuyΛdy≤1=infpαu/k,Λi≤1K.(i)
SetK1=K/α with α≠0, then K=αK1, which gives that(36)αui,Λi=infpαu/k,Λi≤1K=αinfpu/K1,Λi≤1K1=αui,Λi.(iii)
For the triangle inequality, letu,v∈Ei2. Using Remark 3, then(37)pu+vuϱi,Λi+vϱi,Λi=puϱi,Λiuϱi,Λi+vϱi,Λiuuϱi,Λi+vϱi,Λiuϱi,Λi+vϱi,Λivvϱi,Λi≤uϱi,Λiuϱi,Λi+vϱi,Λipuuϱi,Λi+vϱi,Λiuϱi,Λi+vϱi,Λipvvϱi,Λi≤1.Thus,u+vϱi,Λi≤uϱi,Λi+vϱi,Λi.Now, we recall some basic results which are needed to our study.


## 2.1. Working SpaceE

This subsection is dedicated to the definition of the subsite of Orlicz–Sobolev spacesE.Due to nature ofω⋅-Laplacian operator defined in (6), we need to consider the Orlicz–Sobolev framework and we will examine some specific techniques to Orlicz and the Orlicz–Sobolev spaces. For that we define(29)Ei=1;2=u∈W1,ΛiℝN:∫♌ϱiyΛiudy<∞;u=0a.eℝN∖♌,equipped with the following norm uEi,Λi=∇uΛi+uϱi,Λi, where(30)uϱi,Λi=infλ>0:∫♌ϱiyΛiuyλdy≤1=infλ>0pu,Λi≤1,with(31)pu,Λi=∫♌ϱiyΛiuyλdy.Remark 2.
IfΛt∈N is a convex function, then we can see that Ei is uniformly convex Banach space.Remark 3.
It can be noticed by means of Fatou’s lemma that(32)puuϱi,Λi≤1,∀u∈∖0.For anyu∈Ei, we may assume u≠0; by the definition of ⋅ϱi,Λi, there exists λn↘uϱi,Λi such that(33)puλn≤1.By Fatou’s lemma,(34)puuϱi,Λi≤limn⟶+∞puλn≤1.Now, we check that⋅Ei,Λi is a norm in E; for that, it is sufficient to verify that ⋅ϱi,Λi is a norm in Ei. Let us prove that ⋅ϱi,Λi is a norm in Ei.(i)
Ifuϱi,Λi=0, then ∫♌ϱiyΛiuy/Λdy=0. Since Λi:ℝ+⟶ℝ+ and ϱi:ℝℕ⟶ℝ+ are tow continuous and positive functions, then ϱiyΛiuy/Λ=0, and by the condition A1, we infer that ϱiy>0, and then Λiuy/Λ=0. Or Λi0=0 then uy/Λ=0, which implies that uy=0.(ii)
Forα∈ℝ and u∈Ei, we have(35)αuϱi,Λi=infΛ>0,∫♌ϱiyΛiαuyΛdy≤1=infpαu/k,Λi≤1K.(i)
SetK1=K/α with α≠0, then K=αK1, which gives that(36)αui,Λi=infpαu/k,Λi≤1K=αinfpu/K1,Λi≤1K1=αui,Λi.(iii)
For the triangle inequality, letu,v∈Ei2. Using Remark 3, then(37)pu+vuϱi,Λi+vϱi,Λi=puϱi,Λiuϱi,Λi+vϱi,Λiuuϱi,Λi+vϱi,Λiuϱi,Λi+vϱi,Λivvϱi,Λi≤uϱi,Λiuϱi,Λi+vϱi,Λipuuϱi,Λi+vϱi,Λiuϱi,Λi+vϱi,Λipvvϱi,Λi≤1.Thus,u+vϱi,Λi≤uϱi,Λi+vϱi,Λi.Now, we recall some basic results which are needed to our study.


## 3. Hypotheses

We are going to show that the representation given by(38)Λi=1;2t≔∫0trωirdrfor allt∈ℝ,where ωi=1;2⋅ verified (8) exists and it is an N-function (see [16]). We suppose through our paper that Λi∈Δ2. Then, by Lemma 1, we have for all t>0 that(39)1<li≔inft>0t2ωitΛit≤supt>0t2ωitΛit≔ni<∞.Related to the functionF, our hypotheses are the following.F satisfies the following:F1: F:♌×ℝ×ℝ⟶ℝ is a C1 function such that Fy,0,0=0 for all y∈♌ and there exist Ψi=1,2∈N such that Ψi≪Λi=1,2 and satisfy that(40)ni<lΨi≔inft>0tψitΨit≤supt>0tψitΨit≔nΨi<∞,where(41)Ψit≔∫0tψirdr,∀t∈ℝ.Moreover,(42)Fuy,u,v≤c11+ψ1u+Ψ¯1−1Ψ2v,Fvy,u,v≤c11+ψ2v+Ψ¯2−1Ψ1u,for every y,u,v∈♌×ℝ×ℝ, where c1>0 and Ψ¯ denote the complements of Ψi=1,2, respectively.F2:(43)limu,v⟶0Fy,u,vΛ1u+Λ2v=0,uniformly∀y∈♌,(44)limu,v⟶+∞Fy,u,vΛ1u+Λ2v=∞,uniformly∀y∈♌.F3: γ:0,∞⟶ℝ is a continuous function, such that(45)1<lΓ≔inft>0tγtΓt≤supt>0tγtΓt≔nΓ<+∞,where(46)Γt≔∫0tγrdr,for allt∈ℝ,is in N and functions Mit≔tlilΓ/lΓ−1≪Λi∗ and satisfies(47)ΓFy,u,vul1+vl2≤c2F¯y,u,v,x∈ℝ,u,v≥r,where c2,r>0 and(48)F¯y,u,v≔1n1Fuy,u,vu+1n2Fvy,u,vv−Fy,u,v,∀y,u,v∈♌×ℝ×ℝ.Now, the functionalℱi:Ei⟶ℝ is defined by(49)ℱiu=∫♌ϱiyΛiuydy.Lemma 6.
These properties are true:(1)
ℱiu/uϱi,Λi≤1,for allu∈Ei∖0(2)
χ0uϱi,Λi≤ℱiu≤χ1uϱi,Λi,∀u∈EiProof.
(1)
See Remark3.(2)
Letu∈Ei, by choosing β=uϱi,Λi in Lemma 6, we obtain(50)Λiu≤χ1uϱi⋅ΛiΛiuuϱi⋅Λi,and then(51)ϱiyΛiu≤χ1uϱi⋅ΛiaiyΛuua⋅Λ,for ally∈♌.
We deduce that from the definition of the norm (30),(52)∫♌ϱiyΛiudy≤χ1uϱi⋅Λi∫♌ϱiyΛiuuϱi⋅Λidy≤χ1uϱi,Λi.
Furthermore, letβ>0 and β=uϱi,Λi−ϵ in Lemma 6, we get(53)χ0uϱi,Λi−ϵϱiyΛiuuϱi⋅Λi−ϵ≤ϱiyΛiu,and then(54)∫♌ϱiyΛiudy≥χ0uϱi,Λi−ϵ∫♌ϱiyΛiuuϱi⋅Λi−ϵdy≥χ0uϱi,Λi−ϵ.
Forϵ⟶0 in the above inequality, we obtain that(55)χ0uϱi,Λi≤∫♌ϱiyΛiudy.
This completes the proof.Theorem 3 (see [19]).
LetΛi,Φi∈N. If A1 and A2 and (25) hold then Ei21AALΛi♌ is continuous, and Ei21AALΦi♌ is compact for Φi≪Λi.


## 4. Main Results

In the following, we show the existence of a weak solution to problem (5).Theorem 4.
Let us suppose thatA1, A2, F1,F2, and F3 hold. Then, system (5) has a nontrivial weak solution.Remark 4.
Under assumption (39) and Lemma 1, Λ¯∈N satisfies Δ2.Remark 5.
Under assumptionsF1 and F3, by Theorem 3, the following embeddings Ei⟶LΨi♌, Ei⟶Llil¯Γ♌, and Ei⟶Llin¯Γ♌ are compact where l¯=lΓ/lΓ−1 and n¯=nΓ/nΓ−1.Remark 6.
By (2) in Lemma1, (44) and assumptions F3 show(56)limu,v⟶+∞F¯y,u,v⟶+∞,∀y∈♌.Remark 7.
By Young’s inequality (17),(57)Fy,u,v=∫0uFsy,s,0ds+∫0vFty,0,tdt+Fy,0,0,∀y,u,v∈♌×ℝ×ℝ.
By (42) and (2) in Lemma 1, it is shown that there is c3>0 such that(58)Fy,u,v≤c3u+v+Ψ1u+Ψ2v,∀y,u,v∈♌×ℝ×ℝ,which together with (25) and (40) show that there is c4>0 such that(59)Fy,u,v≤c41+Ψ1u+Ψ2v,∀y,u,v∈♌×ℝ×ℝ.We now have the tools to investigate problem (5). For this purpose, we define our working space W≔E1×E2 under the following norm:(60)u,v≔uE1,Λ1+vE2,Λ2.Notice that, according to Remark2, W is a separable and reflexive Banach space. We observe that the energy function I on W corresponding to system (5) is(61)Iu,v≔∫♌Λ1∇udy+∫♌Λ1∇vdy+∫♌ϱ1yΛ1udy+∫♌ϱ2yΛ2vdy−∫♌Fy,u,vdy,for allu,v∈W. Label by Ii=1,2: W⟶ℝ, the functionals(62)I1u,v≔∫♌Λ1∇udy+∫♌Λ2∇vdy+∫♌ϱ1yΛ1udy+∫♌ϱ2yΛ2vdy,I2u,v=∫♌Fy,u,vdy.Then,Iu,v=I1u,v−I2u,v.Theω⋅-Laplacian operator defined in (6) is well-defined from Ei to its dual space Ei∗. By (39) and Theorem 3, by means of similar arguments in [12], we can show that I1 is well-defined and C1W,ℝ and(63)I′u,v,u¯,v¯=∫♌ω1∇u∇u∇u¯dy+∫♌ω2∇v∇v∇v¯dy+∫♌ϱ1yω1uuu¯dy+∫♌ϱ2yω2vvv¯dy−∫♌Fuy,u,vu¯dy−∫♌Fvy,u,vv¯dy,for all u¯,v¯∈W. Then, the critical points of I on W are weak solutions of system (5).Lemma 7.
The functionalℱi is weak lower semicontinuous.Proof.
By Corollary III.8 in [20], it is suffice to show that ℱi is lower semicontinuous.
Letuk⊂Ei such that uk⇀u in Ei. By Theorem 3, if Φit=tμ, then Ei is compactly embedded in LμRN so it follows that uk⟶u in LμRN. As far as a subsequence,(64)uky⟶uya.e∈ℝN.
With Fatou’s lemma, we obtainℱiu≤liminfk⟶∞ℱiuk. This ends the proof.Definition 2.
LetJ∈C1Ei,ℝ. A sequence uk in Ei is a Cerami sequence (abbreviated to Cc) at the level c∈ℝ for the functional J when(65)Juk⟶c,1+ukEiJ′ukEi∗⟶0.Remark 8.
By means of arguments in [21], it appears thatPSc due to Palais–Smale may be replaced by Cc due to Cerami sequence in the mountain pass theorem.Lemma 8 (Theorem 2.2 in [13]).
ConsiderA as a real Banach space and I∈C1A,R fulfilling PSc. Assume that I0=0 and(i)
I1 We have I|∂Bβ≥α for certain β,α>0(ii)
I2 We have Ie≤0 for some e∈A\Bβ
Then,I has a critical value c≥α.Lemma 9.
By (39), F1, and (43), there is β,α>0 such that I|∂Bβ≥α.Proof.
By (43), we have that for each ϵ>0, there is η>0 such that(66)Fy,u,vΛ1u+Λ2v<ϵ,for eachu,v<η.
Hence,(67)Fy,u,v<ϵΛ1u+Λ2vforu,v<η.
AsΨi∈N, it also means that(68)limt⟶∞Ψitt=∞.
Therefore,K>0 exists so that(69)Ψitt≥1,t>K.
WhenK>η, it results from the continuity of t↦Ψit/t in η≤t≤K that a>0 exists such that Ψit/t≥a. Therefore, using (59), we infer that(70)Fy,u,v<CΨ1u+Ψ2vforu,v>η.
Under these conditions, it follows from (67) and (70) and continuity of F that for ϵ=ϱ0>0 there is C>0 in such a way that(71)Fy,u,v≤ϱ0Λ1u+Λ2u+CΨ1u+Ψ2v∀y,u,v∈♌×ℝ×ℝ.
Whenu,v≤1, for a given ε∈0,1 by monotony Λi, (24), and (2) in Lemma 6, we infer that(72)Iu,v=∫♌Λ1∇udy+∫♌Λ2∇vdy+∫♌ϱ1yΛ1udy+∫♌ϱ2yΛ2vdy−∫♌Fy,u,vdy>ε∫♌Λ1∇udy+ε∫♌Λ2∇vdy+∫♌ϱ1yΛ1udy+∫♌ϱ2yΛ2vdy−∫♌Fy,u,vdy≥ε∫♌Λ1∇udy+ε∫♌Λ2∇vdy+ϱ0∫♌Λ1udy+ϱ0∫♌Λ2vdy−ϱ0∫♌Λ1u+Λ2vdy−C∫♌Ψ1udy−C∫♌Ψ2vdy≥ε∇uΛ1n1+ε∇vΛ2n2−CuΨ1lΨ1−CvΨ2lΨ2≥ε∇uΛ1n1+ε∇vΛ2n2−C′uΛ1lΨ1−C′vΛ2lΨ2≥∇uΛ1n1ε−C′uE,Λ1lΨ1−n1+∇vΛ2n2ε−C′vE,Λ2lΨ2−n2,where C′>0 is obtained by Poincaré’s Inequality and compact embedding. From 1<ni<lΨi, we may choose β,α>0 sufficiently small that Iu,v>α for all u,v∈W with u,v=β.Lemma 10.
There is a pointu,v∈W∖Bβ such that Iu,v<0.Proof.
By (44) and since F is continuous, for every given constant K>0, there is a constant CK>0 such that(73)Fy,u,v≥KΛ1u+Λ2v−CK∀y,u,v∈♌×ℝ×ℝ.
Now, chooseu0∈Cc1♌∖0 with 0≤u0y≤1. Then, u0,0∈W, and by (73) and (2) in Lemma 1, when t>0, we have(74)Itu0,0=∫♌Λ1t∇u0dy+∫♌ϱ1yΛ1tu0dy−∫♌Fy,tu0,0dy≤Λ1t∫♌max∇u0l1;∇u0n1dy+Λ1t∫♌ϱ1ymaxu0l1;u0n1dy−KΛ1t∫♌×♌minu0l1;u0n1dy+CΛ♌≤Λ1t∇u0Ll1♌l1+∇u0Ln1♌n1+ϱ1∞u0Ll1♌l1−Ku0Ln1♌n1+CK♌.
SinceK>0 is arbitrary and limt⟶∞Λ1t=+∞, we can choose(75)K>∇u0Ll1♌l1+∇u0Ln1♌n1+ϱ1∞u0Ll1♌l1u0Ln1♌n1,and large t such that Itu0,0≤0 and tu0,0>ρ.Lemma 11.
AnyCc-sequence in W is bounded.Proof.
Letuk,vk⊂W be a Cc-sequence of I on W, then for large enough k by (39) and (65), we obtain(76)c+1≥Iuk,vk−I′uk,vk;1n1uk,1n2vk=∫♌Λ1∇ukdy+∫♌Λ2∇vkdy+∫♌ϱ1yΛ1ukdy+∫♌ϱ2yΛ2vkdy−∫♌Fy,uk,vkdy−1n1∫♌ω1∇uk∇uk2dy−1n2∫♌ω2∇vk∇vk2dy−1n1∫♌ϱ1yω1ukuk2dy−1n1∫♌ϱ2yω2vkvk2dy+1n1∫♌Fuy,uk,vkukdy+1n2∫♌Fvy,uk,vkvkdy=∫♌Λ1∇uk−1n1ω1∇uk∇uk2dy+∫♌Λ2∇vk−1n2ω2∇vk∇vk2dy+∫♌ϱ1yΛ1uk−1n1ω1ukuk2dy+∫♌ϱ2yΛ1vk−1n2ω2vkvk2dy+∫♌1n1Fuy,uk,vkukdy+1n2Fvy,uk,vkvk−Fy,uk,vkdy≥∫♌F¯y,uk,vkdy.
As a contradiction, we demonstrate the boundedness of the sequenceuk,vk. Consider that there exists a subsequence of uk,vk, which always referred as uk,vk, so that(77)uk,vk=ukE1,Λ1+vkE2,Λ2⟶+∞.
Next, we discuss the problem in two cases.Case 1.
Suppose thatukE1,Λ1⟶+∞ and vkE2,Λ2⟶+∞. Let u¯k=uk/ukE1,Λ1 and v¯k=vk/vkE2,Λ2 Then, u¯k,v¯k is bounded in a separable and reflexive space W. By switching to a subsequence u¯k,v¯k by Remark 5, there is a point u¯,v¯∈W so that(a)
u¯k⇀u¯ in E1, u¯k⇀u¯ in Ll1l¯Γ♌, and Ll2n¯Γ♌u¯k⇀u¯ in a.e in ♌(b)
v¯k⇀v¯ in E2, v¯k⇀v¯ in Ll1l¯Γ♌, and Ll2n¯Γ♌, v¯k⇀v¯ in a.e in ♌
First of all, we suppose thatu¯≠0≔y∈♌:u¯y≠0 or v¯≠0≔y∈♌:v¯y≠0 has a nonzero Lebesgue measure. It is clear that(78)uk=u¯kukE1,Λ1⟶+∞∈u≠0,vk=v¯kvkE2,Λ1⟶+∞∈v≠0.
Then, by Fatou’s lemma, (76), and Remark 6, we have(79)c+1≥∫♌F¯y,uk,vkdy⟶+∞,which is a contradiction. Secondly, we assume that the two u¯≠0 and v¯≠0 have zero Lebesgue measure, that is, u¯=0 in E1 and v¯=0 in E2. By Lemma 2, we have(80)min∇ukΛ1l1,∇ukΛ1n1+min∇vΛ2l2,∇vkΛ2n2+minukϱ1,Λ1l1,ukϱ1,Λ1n1+minvkϱ2,Λ2l2,vkϱ2,Λ2n2≤∫♌Λ1∇ukdy+∫♌Λ2∇vkdy+∫♌ϱ1yΛ1ukdy+∫♌ϱ2yΛ2vkdy=Iuk,vk+∫♌Fy,uk,vkdy.
For a largek, the inequality (80) becomes as follows:(81)∇ukΛ1l1+∇vkΛ2l2+ukϱ1,Λ1l1+vkϱ2,Λ2l2≤Iuk,vk+∫♌Fy,uk,vkdy.
By (18), we infer that(82)12l1−1uE1,Λ1l1+12l2−1vE2,Λ2l2≤∇ukΛ1l1+∇vkΛ2l2+ukϱ1,Λ1l1+vkϱ2,Λ2l2,then(83)12l1−1uE1,Λ1l1+12l2−1vE2,Λ2l1≤Iuk,vk+∫♌Fy,uk,vkdy,which is equivalent to(84)1≤Iuk,vk1/2l1−1uE1,Λ1l1+1/2l2−1vE2,Λ2l2+∫uk,vk≤R∫uk,vk>R+Fy,uk,vk1/2l1−1uE1,Λ1l1+1/2l2−1vE2,Λ2l2dy=ok1+∫uk,vk≤RFy,uk,vk1/2l1−1uE1,Λ1l1+1/2l2−1vE2,Λ2l2dy+∫uk,vk>RFy,uk,vk1/2l1−1uE1,Λ1l1+1/2l2−1vE2,Λ2l2dy,where R is given in F3 and(85)Fy,u,v≥0,x∈♌u,v>Rby48.
By (43), there is CR>0 so that(86)Fy,u,v<CR,∀x∈♌,u,v≤R,then(87)∫uk,vk≤RFy,uk,vk1/2l1−1uE1,Λ1l1+1/2l2−1vE2,Λ2l2dy≤CR♌1/2l1−1uE1,Λ1l1+1/2l2−1vE2,Λ2l2=ok1.
Furthermore, it is clear from Hölder’s inequality that(88)∫uk,vk>RFy,uk,vk1/2l1−1uE1,Λ1l1+1/2l2−1vE2,Λ2l2dy=∫uk,vk>RFy,uk,vk1/2l1−1ukl1/u¯kl1+1/2l2−1vkl2/u¯kl2dy≤∫uk,vk>RFy,uk,vkukl1+vkl212l1−1u¯kl1+12l2−1v¯kl2dy≤2max12l1−1;12l2−1Fy,uk,vkukl1+vkl2χuk,vk>RΓ×u¯nl1+v¯nl2χuk,vk>RΓ¯,where χ denotes the characteristic function which satisfies χuky,vky>R=1,forx∈y∈♌:uky,vky>R,0,forx∈y∈♌:uky,vky≤R. For k large enough, by (47) and (76) and whether it is true that F¯ is continuous, we have(89)∫♌ΓFy,uk,vkukl1+vkl2χuk,vk>Rdy≤c2∫♌F¯y,uk,vkdy+C≤c3c+1+C.
So, fork being sufficiently large, by Lemma 2, it exists S>0 such that(90)Fy,uk,vkukl1+vkl2χuk,vk>RΓ≤S.
Moreover, it is clear that(91)u¯kl1+v¯kl2χuk,vk>RΓ¯≤u¯kl1+v¯kl2Γ¯≤u¯kl1Γ¯+v¯kl2Γ¯.
By Lemmas2 and 3, F3 implies that Γ¯∈Δ2. Then, by (23), uΓ¯⟶0 as ∫♌Γ¯udy⟶0. It results from Lemma 3a and b that(92)∫♌Γ¯u¯kl1dy+∫♌Γ¯v¯kl2dy≤Γ¯1∫♌maxu¯kl1l¯Γ,u¯kl1m¯Γdy+Γ¯1∫♌maxv¯kl2l¯Γ,v¯kl2n¯Γdy≤Γ¯1∫♌u¯kl1l¯Γdy+∫♌u¯kl1n¯Γdy+∫♌v¯kl2l¯Γdy+∫♌v¯kl2n¯Γdy=ok1,which implies(93)u¯kl1+v¯kl2χuk,vk>RΓ¯≤u¯kl1Γ¯+v¯kl2Γ¯=ok1.
By combining (67), (88), (90), and (93) with (84), we obtain a contradiction.Case 2.
Assume thatukE1,Λ1≤C or vkE2,Λ2≤C for certain C>0 and all k∈ℕ. In order to be general, we suppose that ukE1,Λ1⟶+∞ and vkE2,Λ2≤C for some C>0 and all k∈ℕ. Let u¯k=uk/ukE1,Λ1 and v¯k=vk/ukE1,Λ1, then v¯kE2,Λ2⟶0 and u¯kE1,Λ1=1 by Remark 5, and there is a point u¯,v¯∈W so that(c)
u¯k⇀u¯ in E1, u¯k⇀u¯ in Ll1l¯Γ♌, and Ll2n¯Γ♌u¯k⇀u¯ in a.e in ♌(d)
v¯k⇀v¯ in E2, v¯k⇀v¯ in Ll1l¯Γ♌ and Ll2n¯Γ♌, v¯k⇀v¯ in a.e in ♌
Likewise, we first assume thatu¯≠0 has a nonzero Lebesgue measure. We may see that(94)uk=u¯kukE1,Λ1⟶+∞∈u≠0.
Then, by Remark6, (76), and Fatou’s lemma, we get a contradiction by(95)c+1≥∫♌F¯y,uk,vkdy⟶+∞.
Secondly, let us assume thatu¯≠0 has a null Lebesgue measure, i.e., u¯=0 in E1. By Lemma 3(c) and d, we have(96)minvkl2Γ¯l¯Γ,vkl2Γ¯n¯Γ≤∫♌Γ¯vkl2dy≤Γ¯1∫♌maxvkl2l¯Γ,vkl2n¯Γdy≤Γ¯1∫♌vkl2l¯Γdy+∫♌vkl2n¯Γdy⟶C.
This shows that a constantL>0 exists such that(97)vkl2Γ¯≤L,∀k∈ℕ.
Ifk is large enough, (83) turns into(98)12l1−1ukE1,Λ1l1+K≤Iuk,vk+∫♌Fy,uk,vk+K,where K>0 with K>4SL (see (90) and (97)). Then, by (86), (90), (93), (97), and Hölder’s inequality, the estimate above means(99)1≤Iuk,vk+K1/2l1−1ukE1,Λ1l1+K+∫♌Fy,uk,vk1/2l1−1ukE1,Λ1l1+Kdy=Iuk,vk+K1/2l1−1ukE1,Λ1l1+K+∫uk,vk≤R|Fy,uk,vk1/2l1−1ukE1,Λ1l1+Kdy+∫uk,vk≤R|Fy,uk,vk1/2l1−1ukE1,Λ1l1+Kdy≤ok1+∫uk,vk≤R|Fy,uk,vk1/2l1−1ukE1,Λ1l1+Kdy≤ok1+∫uk,vk≤R|Fy,uk,vkukl1+vkl212l1−1u¯kl1+1Kv¯kl2dy≤ok1+Fy,uk,vkukl1+vkl2χuk,vk>RΓ12l1−1u¯kl1+1Kv¯kl2χuk,vk>RΓ¯≤ok1+2S12l1−1ukl1Γ¯+1Kvkl2Γ¯≤ok1+2Sukl1Γ¯+1Kvkl2Γ¯≤ok1+2Sok1+LK=ok1+2SLK<ok1+12,which is a contradiction.Lemma 12.
The energy functionI satisfies Cc-condition.Proof.
Consideruk,vk as any Cc sequence of I in W. Lemma 8 proves that uk,vk is bounded. By switching to a subsequence uk,vk, by Remark 5, there is a point u,v∈W such that(e)
uk⇀u in E1, uk⇀u in LΨ1♌, uk⇀u a.e ♌(f)
vk⇀v in E2, vk⇀v in LΨ2♌, vk⇀v a.e ♌
Now, define the operatorsℋi:Ei⟶Ei∗ by(100)ℋiu≔∫♌ωi∇u∇u∇u¯dy+∫♌ϱiyωiuuu¯dy.
Then, we have(101)ℋ1uk,uk−u=∫♌ω1∇uk∇uk∇uk−udy+∫♌ϱ1yω1ukuk−uukdy=I′uk,vk,uk−u,0+∫♌Fuy,uk,vkuk−vkdy.
Equation (65) shows that(102)I′uk,vk,uk−u,0≤I′uk,vkW∗uk−uE1⟶0.
FromF1 and Hölder’s inequality, we get(103)∫♌Fuy,uk,vkuk−udy≤c1∫♌1+ψ1uk+Ψ¯1−1Ψ2vkuk−udy≤2c11+ψ1uk+Ψ¯1−1Ψ2vkΨ¯1uk−uΨ1.
In fact inequality (44) the functions Ψ1,Ψ¯1∈N and Ψ1,Ψ¯1∈Δ2, which with the convexity of N-function, Lemma 2, Remark 5, and the boundedness of uk,vk implies that(104)∫♌Ψ¯11+ψ1uk+Ψ¯1−1Ψ2vkdy≤C∫♌1+Ψ1uk+Ψ2vkdy≤C,which with Lemma 2 again shows that(105)1+ψ1uk+Ψ¯1−1Ψ2vkΨ¯1≤C,for some C>0. Furthermore, e and f give evidence that(106)uk−uΨ1⟶0.
Then, combining (101)–(103), (105), and (106), we obtain(107)ℋ1uk,uk−u⟶0ask⟶∞.
According to the following lemma, we obtain the result.Lemma 13 (see [22]).
Suppose thatuk⇀u in E1 and(108)limn⟶∞supℋ1uk,uk−u≤0.
Then,uk⟶u in E1. Similarly, we can obtain that, vk⟶v in E2. Therefore, uk,vk⟶u,v in W.Proof of Theorem4.
By Lemmas9, 11, and 12 and the evident fact that I0=0, all conditions of Lemma 8 hold. Then, the functional I possesses a critical point u,v∈W which is a weak solution of system (5) satisfying Iu,v=infWI. Lemma 10 implies that u,v≠0. Thus, system (5) possesses at least one nontrivial weak solution.


## 5. Conclusion

In this paper, we established a Schrödinger system involvingω⋅-Laplacian operator. Our results are extensions and generalizations of some existing results in the literature.



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*Source: 2486542-2022-03-30.xml*