# Existence of Solutions for Inclusion Problems in Musielak-Orlicz-Sobolev Space Setting

Authors: Ge Dong; Xiaochun Fang
Journal: Journal of Function Spaces
 (2023)

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## Abstract

In this paper, we mainly prove the existence of (weak) solutions of an inclusion problem with the Dirichlet boundary condition of the following form:L∈Ax,u,Du+Fx,u,Du,inΩ, and u=0,on∂Ω, in Musielak-Orlicz-Sobolev spaces W01LΦΩ by using the surjective theorem, where Ω⊂ℝN is a bounded Lipschitz domain, L belongs to the dual space W01LΦΩ∗ of W01LΦΩ, A is a multivalued maximal monotone operator, and F is a multivalued convection term. Some examples for A and F are given in the paper. Then, we give some properties of the solution set of the inclusion problem. We also show the existence of (weak) solutions of the inclusion problem with an obstacle effect.


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## Body

## 1. Introduction

In [1], Liu and Motreanu established in Sobolev spaces the existence and location of solutions for an elliptic inclusion problem driven by a p,q-Laplacian operator with Dirichlet boundary and a multivalued term that has gradient dependence (the so-called multivalued convection term). In [2], Zeng et al. proved the existence of positive solutions for an elliptic inclusion problem driven by an abstract nonhomogeneous operator with Dirichlet boundary and a multivalued convection term in Sobolev spaces. In [3], we proved the existence of positive solutions for an elliptic inclusion problem driven by a nonlinear operator with a Dirichlet boundary and a multivalued convection term in Orlicz-Sobolev spaces, where the nonlinear operator depends on the gradient of the solution. In [1–3], the conditions which ensured the existence of subsolutions and supersolutions were given, and the proof of the existence of solutions depended on the subsolutions and supersolutions.Precup and Rodríguez-López [4] proved the existence of solutions for an inclusion problem driven by a ϕ-Laplacian operator depending on the differential of the solution. Chen and Tang [5] discussed periodic solutions for a differential inclusion problem involving the pt-Laplacian. In [6], Zeng et al. proved the existence and boundedness of the weak solutions to inclusion problems driven by double-phase partial differential operators, obstacle effects, and multivalued convection terms in Sobolev spaces. In [7], Zeng et al. introduced a family of the approximating inclusion problems corresponding to an elliptic obstacle problem with double-phase phenomena and a multivalued reaction convection term in Sobolev spaces. Liu and Papageorgiou [8] proved the existence of a nontrivial bounded solution for a double-phase Dirichlet problem with unilateral constraints in Sobolev spaces. Crespo-Blanco et al. [9] showed the existence and uniqueness of a quasilinear elliptic equation driven by a double-phase operator with variable exponents. In the above results, the differential operators are single-valued functions.The differential inclusions governed by maximal monotone operators have many applications in heat equations, obstacle problems, mechanics, electricity, and management, for example, [10, 11]. Oppezzi and Rossi [12] proved the existence of solutions of a variational inequality with a maximal monotone operator in Sobolev spaces, where the maximal monotone operator depends on the solution and its gradient. Le [13] proved the existence of the solution of variational inequalities with maximal monotone operators by using a subsupersolution method in variable exponent Sobolev spaces, where the maximal monotone operator depends on the gradient of the solution. The conditions which ensure the existence of subsolutions or supersolutions are not given. Le [14] studied the inclusion problems containing maximal monotone and generalized pseudomonotone mappings in Sobolev spaces. Papageorgiou et al. [15] proved the existence of extremal solutions for nonlinear multivalued systems with maximal monotone terms in Sobolev spaces. Avci and Pankov [16] proved the existence of the (weak) solution of a Dirichlet boundary value problem driven by a maximal monotone differential operator in Musielak-Orlicz-Sobolev spaces, where the maximal monotone operator depends on the gradient of the solution.Variational inequalities and differential equations, as is known to all, correspond to different function spaces. The variable exponent Sobolev spaces and Orlicz-Sobolev spaces are two special kinds of Musielak-Orlicz-Sobolev spaces. They are distinct extensions of the classical Sobolev spaces. In recent years, there are many results of differential equations in Musielak-Orlicz-Sobolev spaces. For example, Fan [17] and we [18] proved the existence of weak solutions of a class of differential equations of divergence form by using a subsupersolution method in reflexive Musielak-Orlicz-Sobolev spaces and nonreflexive Musielak-Orlicz-Sobolev spaces, respectively; Li et al. [19] proved the existence and uniqueness of entropy solutions and the uniqueness of renormalized solutions to the nonlinear elliptic equations in Musielak-Orlicz-Sobolev spaces; we [20] proved the existence of barrier solutions of elliptic differential equations in Musielak-Orlicz-Sobolev spaces; and Baasandorj et al. [21] established optimal regularity estimates for the gradient of solutions to nonuniformly elliptic equations of the Orlicz double phase with variable exponent types. Musielak-Orlicz functions (described in Section 2) have many applications such as non-Newtonian fluids (see, e.g., [22]), thermistor problem (see, e.g., [23]), and image restoration (see, e.g., [24]).In this paper, we are interested to find a (weak) solutionu of the following inclusion problem with the Dirichlet boundary condition and a convection multivalued term
(1)L∈Ax,u,Du+Fx,u,Du,inΩu=0,on∂Ω,in Musielak-Orlicz-Sobolev spacesW01LΦΩ, where Ω⊂ℝN is a bounded Lipschitz domain, L belongs to the dual-space W01LΦΩ∗ of W01LΦΩ, A:Ω×ℝ×ℝN⟶2ℝN is maximal monotone with respect to the last variable, and F:Ω×ℝ×ℝN⟶2ℝ is a multivalued mapping.WhenA is maximal monotone with respect to the last variable, some particular cases of this problem in variable exponent Sobolev spaces and Musielak-Orlicz-Sobolev spaces were obtained, for example, by Le in [13] and by Avci and Pankov in [16], respectively. There are some other particular cases of problem (1). When A is an Orlicz double phase with variable exponents, a case of this problem was established by Baasandorj et al. in [21]. When L=0, F is a single-valued function in L1Ω and A is a multivalued and maximal monotone; a particular case of this problem in Sobolev spaces was developed, for example, by Oppezzi and Rossi in [12]. When L=0, F is a multivalued convection term and A is a single-valued elliptic differential operator; the existence results of this problem in Orlicz-Sobolev spaces were provided, for example, by us in [3]. When L=0, A is a double-phase operator and F is a multivalued convection term in Sobolev spaces; a particular case of this problem was developed, for example, by Zeng et al. in [6]. When L=0, A is a double-phase operator with variable exponents and F is a single-valued term in variable exponent Sobolev spaces; a particular case of this problem was obtained, for example, by Crespo-Blanco et al. in [9].The paper is organized as follows: Section2 contains some preliminaries and some technical lemmas which will be needed in Section 3. In Section 3, we first define the multivalued differential operator A defined on W01LΦΩ with values in W01LΦΩ∗ by using the multivalued function A. Then, we show that A is a pseudomonotone operator. Some examples for A are given. Next, we deal with the multivalued convection term. We also give an example for the multivalued convection term. Then, we prove the existence of solutions for problem (1) by using the surjective theorem. Then, some properties of the solution set of problem (1) are given. Finally, we show the existence of the solutions for the inclusion problem with an obstacle effect and give some properties of the solution set of the problem.In this paper, we always assume thatΩ⊂ℝN is a bounded domain with Lipschitz boundary and denote by L0Ω the set of all real measurable functions defined on Ω.


## 2. Preliminaries

Now, we first list briefly some definitions and facts about Musielak-Orlicz-Sobolev spaces; for more details, see Musielak [25] or Harjulehto and Hästö [26].Letℝ+=0,+∞. A real function Φ:Ω×ℝ+⟶ℝ+ will be called a Musielak-Orlicz function, denoted by Φ∈NΩ, if it satisfies the following conditions:
(i)
Φx,u is an N-function of the variable u≥0 for every x∈Ω; i.e., it is a convex, nondecreasing, and continuous function of u such that Φx,0=0, Φx,u>0 for u>0, and there hold the conditions(2)limu⟶0+supx∈ΩΦx,uu=0,limu⟶∞infx∈ΩΦx,uu=+∞,for every x∈Ω.
(ii)
Φx,u is a measurable function of x for all u≥0.Equivalently, for allx∈Ω and all u∈ℝ, Φ admits the representation Φx,u=∫0uφx,τdτ, where φx,t is the right-hand derivative of Φx,· at t, for a fixed x∈Ω and all t≥0. Then, for every x∈Ω, φx,τ is a right-continuous and nondecreasing function of τ≥0, φx,0=0, φx,τ>0 for τ>0, and limu⟶+∞infx∈Ωφx,τ=+∞.The complementary functionΦ¯ to a Musielak-Orlicz function Φ is defined by Φ¯x,v=supu≥0uv−Φx,uforv≥0,x∈Ω. Then, Φ¯ is a Musielak-Orlicz function and Φ is also the complementary function to Φ¯.Equivalently,Φ¯ admits the representation Φ¯x,v=∫0vϕx,σdσ, where ϕ is given by ϕx,σ=supτ:φx,τ≤σ, for all x∈Ω.ForΦ∈NΩ, the inequality uv≤Φx,u+Φ¯x,v for all u,v≥0, x∈Ω, is called the Young inequality.LetΦ∈NΩ. Φ is said to satisfy the Δ2 condition (Φ∈Δ2, for short), if, for ℓ>0, there exist a constant K>0 and h∈L1Ω with h≥0 such that
(3)Φx,ℓu≤KΦx,u+hx,for all u≥0 and a.e. x∈Ω.For eachx∈Ω, the inverse function of Φx,· is denoted by Φ−1x,·; i.e., Φ−1x,Φx,u=Φx,Φ−1x,u=u, for u≥0.LetΦ1,Φ2∈NΩ. Φ1≼Φ2 means that Φ1 is weaker than Φ2; that is, there exist positive constants K1,K2 and a nonnegative function h∈L1Ω such that Φ1x,u≤K1Φ2x,K2u+hx for all u≥0 and a.e. x∈Ω.Φ is called locally integrable, if ∫ΩΦx,udx<+∞ for every u>0.The following assumptions will be used.(Φ1) infx∈ΩΦx,1=c1>0.(Φ2) For every u0>0, there exists c=cu0>0 such that infx∈ΩΦx,u/u≥c and infx∈ΩΦ¯x,u/u≥c, for u≥u0.Surely, (Φ2) ⇒ (Φ1).Φ∈NΩ is said to satisfy the condition (Φ) if Φ,Φ¯∈Δ2, and both Φ and Φ¯ are locally integrable and satisfy Φ2.LetΦ∈NΩ. The Musielak-Orlicz space (i.e., the generalized Orlicz space) LΦΩ is defined by
(4)LΦΩ=u∈L0Ω:∫ΩΦx,uxλdx<+∞,forsomeλ>0,with the (Luxemburg) norm
(5)uΦ=infλ>0:∫ΩΦx,uxλdx≤1.Moreover, the setKΦΩ=u∈L0Ω:∫ΩΦx,uxdx<+∞ will be called the Musielak-Orlicz class (i.e., the generalized Orlicz class). A function u∈L0Ω is called a finite element of LΦΩ, if λu∈KΦΩ for every λ>0. The space of all finite elements of L0Ω is denoted by EΦΩ. Then, KΦΩ is a convex subset of LΦΩ, LΦΩ is the smallest vector subspace of L0Ω containing KΦΩ, and EΦΩ is the largest vector subspace of L0Ω contained in KΦΩ.IfΦ is locally integrable, then EΦΩ is a separable space, and EΦΩ=KΦΩ=LΦΩ if and only if Φ∈Δ2. Then, LΦΩ is reflexive if Φ∈NΩ satisfies the condition (Φ).The Musielak-Orlicz-Sobolev spaceW1LΦΩ is defined by
(6)W1LΦΩ=u∈LΦΩ:∀α≤1,Dαu∈LΦΩ,where α=α1,α2,⋯,αN is a multi-index with nonnegative integers αi, i=1,2,⋯,N, which has α=α1+α2+⋯+αN, and Dαu denotes the distributional derivatives.LetuΦ,Ω=infλ>0:∑α≤1∫ΩΦx,Dαux/λdx≤1 for u∈W1LΦΩ. Then, ·Φ,Ω is a norm on W1LΦΩ, and the pair W1LΦΩ,·Φ,Ω is a Banach space if Φ is locally integrable and satisfies (Φ1). Taking Φx,u=Φu, W1LΦΩ is an Orlicz-Sobolev space. Taking Φx,u=upx, W1LΦΩ is the variable exponent Sobolev space W1,p·Ω.It is easy to see that(7)W1LΦΩ=u∈LΦΩ:Du∈LΦΩ.DenoteDuΦ=DuΦ and u1,Φ=uΦ+DuΦ. Then, u1,Φ and uΦ,Ω are two equivalent norms.The spaceW1LΦΩ will always be identified with a subspace of the product ∏α≤1LΦΩ=ΠLΦ; this subspace is σΠLΦ,ΠEΦ¯ closed. Let W01LΦΩ be the σΠLΦ,ΠEΦ¯ closure of the Schwartz space DΩ in W1LΦΩ.LetΦ∈NΩ be locally integrable and satisfy (Φ1). Then, the space W1LΦΩ is reflexive if LΦΩ is reflexive.Next, we recall the following notations and lemmas which will be used later.We denote by2X all subsets of a set X.LetX be a Banach space and X∗ be its dual. We denote the dual-pairing X∗,X by ·,·. Given a multivalued mapping T:X⟶2X∗, let us fix some notations associated with it.
(i)
The domain ofT is the set DT=u∈X:Tu≠∅.(ii)
The graph ofT is the set GrT=u,u∗∈X×X∗:u∗∈Tu.(iii)
PfcX=E⊂X:Eisnonempty,closed,convex.Definition 1 (see [27], Definition 3.2.2).
LetX be a Banach space. Given a multivalued mapping T:X⟶2X∗, we say that
(i)
T is monotone if for any u,v∈DT and u∗∈Tu, v∗∈Tv, we have u∗−v∗,u−v≥0;(ii)
T is maximal monotone if it is monotone and, for u,u∗∈X×X∗, the inequalities u∗−v∗,u−v≥0 for all v,v∗∈GrT imply u,u∗∈GrT;(iii)
T is bounded if it maps bounded sets in X to bounded sets in X∗.Definition 2 (see [28], Definition 3.7).
LetX and Y be Hausdorff topological spaces and T:X⟶2Y be a multivalued operator. Then, T is called upper semicontinuous at u0∈X, if for every open subset V⊂Y with Tu0⊂V, there exists a neighborhood Uu0 of u0 such that TUu0⊂V. T is called upper semicontinuous in X, if T is upper semicontinuous at every u0∈X.Definition 3 (see [27], Definition 6.1.29).
LetX,d be a metric space and A,B⊂X. Set
(8)h∗A,B=supda,B:a∈A.
h∗A,B is called the excess of A over B.Definition 4 (see [27], Definition 6.1.33, or [29], Definition 2.60).
LetX be a Hausdorff topological space and Y,d a metric space. A multifunction F:X⟶2Y\∅ is said to be h-upper semicontinuous at x0∈X if x⟶h∗Fx,Fx0 is continuous at x0; i.e., for every ε>0, there is Uε∈Nx0 such that for all x∈Uε, h∗Fx,Fx0<ε, where Nx0 is the filter of neighborhoods of x0. If F is h-upper semicontinuous at every x0∈X, then we say that F is h-upper semicontinuous.Lemma 5 (see [27], Proposition 6.1.35).
IfF:X⟶2Y\∅ is upper semicontinuous, then F is h-upper semicontinuous.Lemma 6 (see [30], Theorem 1.1.1).
LetX be a reflexive Banach space, X∗ be its dual, and T be a multivalued map from X into X∗. Suppose that for each u∈X, Tu is a nonempty closed convex subset of X∗, and for each line interval u1,u2≔u∈X:u=λu1+1−λu2,λ∈0,1 in X, T is an upper semicontinuous map from the line interval into X∗ endowed with its weak topology. Then, T is maximal monotone.Definition 7 (see [28], Definition 3.57).
LetX be a reflexive Banach space and T:X⟶2X∗ be a multivalued operator. T is said to be pseudomonotone, if the following conditions hold:
(i)
The setTu is nonempty, bounded, closed, and convex for all u∈X;(ii)
T is upper semicontinuous from each finite-dimensional subspace of X to X∗ endowed with the weak topology;(iii)
Ifun⊂X with un⇀u weakly in X, and un∗∈Tun is such that
(9)limsupn⟶∞un∗,un−u≤0,then for every v∈X, there exists u∗v∈Tu such that
(10)liminfn⟶∞un∗,un−v≥u∗v,u−v.Lemma 8 (see [27], Proposition 3.2.55).
LetX be a real reflexive Banach space and assume that T:X⟶2X∗ satisfies the following conditions:
(i)
For eachu∈X, we have that Tu is a nonempty, closed, and convex subset of X∗;(ii)
T is bounded;(iii)
Ifun⇀u weakly in X and un∗⇀u∗ weakly in X∗ with un∗∈Tun and if limsupn⟶∞un∗,un−u≤0, then u∗∈Tu and un∗,un⟶u∗,u.
Then,T is pseudomonotone.Lemma 9 (see [28], Proposition 3.59).
LetX be a reflexive Banach space.
(i)
IfT:X⟶2X∗ is a maximal monotone operator with DT=X, then T is pseudomonotone.(ii)
IfT1,T2:X⟶2X∗ are pseudomonotone operators, then T1+T2 is pseudomonotone.Lemma 10 (see [28], Definition 3.61).
LetX be a Banach space and T:X⟶2X∗ be an operator. We say that T is coercive if either DT is bounded or DT is unbounded and
(11)limuX⟶+∞,u∈DTinfu∗,u:u∗∈TuuX=+∞.Lemma 11 (see [28], Theorem 3.61).
LetX be a reflexive Banach space and T:X⟶2X∗ be pseudomonotone and coercive. Then, T is surjective, i.e., RT=X∗.Lemma 12 (see [31], Lemma 2.1).
Suppose thatM is an N-function. Then,
(12)∫ΩMuxdx≤∫ΩMdDuxdx,for any u∈W01LMΩ, where d is twice the diameter of Ω.


## 3. Main Results

LetΩ⊂ℝN be a bounded domain with the Lipschitz boundary and Φ∈NΩ satisfy the condition (Φ). Then, LΦΩ, W01LΦΩ, and W1LΦΩ are all reflexive. Hence, the weak topology of W1LΦΩ (resp. W01LΦΩ) is equivalent to the weak-∗ topology of W1LΦΩ (resp. W01LΦΩ). We assume that there exists Ψ∈NΩ satisfying the condition (Φ) such that Φ≼Ψ and the embedding W01LΦΩ↪LΨΩ is compact.


### 3.1. The Maximal Monotone Operator

LetA:Ω×ℝ×ℝN⟶PfcℝN satisfy the following conditions:(A1)A is LΩ×Bℝ×BℝN-measurable, where LΩ is the family of Lebesgue measurable subsets of Ω, Bℝ is the σ-algebra of Borel sets in ℝ, and BℝN is the σ-algebra of Borel sets in ℝN; i.e., for every open set U⊂ℝN, A−1U≔x,s,ξ∈Ω×ℝ×ℝN:Ax,s,ξ∩U≠∅∈LΩ×Bℝ×BℝN.(A2) For a.e.x∈Ω, every s∈ℝ, Ax,s,·:ℝN⟶2ℝN is maximal monotone.(A3) Ifξ,u,un,η are given, being ξ∈ℝN, u∈W01LΦΩ, un in W01LΦΩ such that un⟶u a.e. in Ω, and η an LΩ-measurable selection of the map x∈Ω⟶Ax,ux,ξ⊂ℝN, then there exists a sequence ηn converging a.e. to η in Ω, such that, for every n∈ℕ, ηn is an LΩ-measurable selection of the map x∈Ω⟶Ax,unx,ξ⊂ℝN.(A4) There exista1∈LΦ¯Ω, a2∈L1Ω, and b1≥0, b2>0 such that
(13)ζ≤a1x+b1Φ¯−1x,Φx,s+b1Φ¯−1x,Φx,ξ,(14)ζξ≥a2x+b2Φx,ξ,for a.e. x∈Ω, all s∈ℝ, all ξ∈ℝN, and all ζ∈Ax,s,ξ.By Definition 6.2.1 (a) and Proposition 6.2.3 in [27], we can see that the condition (A1) is weaker than the condition Definition 3.3 (ii) in [16].Ifu:Ω⟶ℝ and w:Ω⟶ℝN are LΩ-measurable, then x∈Ω⟶A·,u,wx=Ax,ux,wx∈2ℝN turns out to be measurable as well (see [12], Remark 1.2). If η:Ω⟶2ℝN is a measurable selection of A·,u,Dv, u,v∈W01LΦΩ, then, by (13), η∈LΦ¯ΩN.LetA~:W01LΦΩ⟶2LΦ¯ΩN be defined by
(15)A~u=ζ∈LΦ¯ΩN:ζx∈Ax,ux,Duxfora.e.x∈Ω,andA:W01LΦΩ⟶2W01LΦΩ∗ by
(16)Au=−divζ:ζ∈A~u.It follows from (A1) and (13) that A~u≠∅, for any u∈W01LΦΩ. It is easy to see that DA=W01LΦΩ.Example 1.
LetΦx,t=tp, for t≥0, all x∈Ω¯, where 1<q<p<+∞ are given. For all x∈Ω¯, s∈ℝ, and ξ∈ℝN, we put Ax,s,ξ=Φx,ξ/ξ2ξ. Then, Au becomes the p-Laplace operator −divDup−2Du, e.g., [1].Example 2.
LetΦx,t=tp+μtq, for t≥0, all x∈Ω¯, where the numbers μ≥0 and 1<q<p<+∞ are given. For all x∈Ω¯, s∈ℝ, and ξ∈ℝN, we put Ax,s,ξ=Φx,ξ/ξ2ξ. Then, Au becomes the double-phase operator −divDup−2Du+μDuq−2Du, e.g., [6].Example 3.
LetΦx,t=tpx, for x∈Ω¯ and t≥0, where p∈CΩ¯ is given such that 1<p−≔infx∈Ω¯px≤px≤p+≔supx∈Ω¯px<+∞, for all x∈Ω¯. For all x∈Ω¯, s∈ℝ, and ξ∈ℝN, we put Ax,s,ξ=Φx,ξ/ξ2ξ. Then, Au becomes the px-Laplace differential operator −divDupx−2Du, e.g., [21].Example 4.
LetΦx,t=tpx+μxtqx, for x∈Ω¯ and t≥0, where μ∈L1Ω is a nonnegative function and p,q∈CΩ¯ are given such that 1<px<N, px<qx, for all x∈Ω¯. For all x∈Ω¯, s∈ℝ, and ξ∈ℝN, we put Ax,s,ξ=Φx,ξ/ξ2ξ. Then, Au becomes the double-phase operator with variable exponents −divDupx−2Du+μxDuqx−2Du, e.g., [9].Example 5.
Letφx,t=tpx−2tlog1+trx, for x∈Ω¯ and t≥0, where p∈L∞Ω with p−>1, and r∈L∞Ω with r−≥0, and let Φx,t=∫0tφx,τdτ for x∈Ω¯ and t∈ℝ. Then, Φ satisfies (Φ) (see [17]). For all x∈Ω, s∈ℝ, and ξ∈ℝN, we put A1x,s,ξ=a1x,sΦx,ξ/ξ2ξ and A2x,s,ξ=a2x,sΦx,ξ/ξ2ξ, where a1 and a2 are Carathéodory functions (i.e., for each s∈ℝ, x⟶a1x,s and x⟶a2x,s are measurable; for a.e. x∈Ω, s⟶a1x,s and s⟶a2x,s are continuous) on Ω×ℝ satisfying 0<α≤a1x,s≤a2x,s≤β for positive constants α and β, for a.e. x∈Ω. As in [12], we define Ax,s,ξ=λA1x,s,ξ+1−λA2x,s,ξ:λ∈0,1, for all x∈Ω, s∈ℝ, and ξ∈ℝN. Then, A satisfies (A1)–(A4).Lemma 13.
Under assumptions (A1)–(A4), we have that(i)
A is bounded;(ii)
Au∈PfcW01LΦΩ∗, ∀u∈X;(iii)
GrA is (sequentially) closed in W01LΦΩ×W01LΦΩ∗ with respect to the strong-weak topology (i.e., with respect to the product of the norm topology of W01LΦΩ and the weak topology of W01LΦΩ∗);(iv)
A is upper semicontinuous from W01LΦΩ with the strong topology to 2W01LΦΩ∗ with the weak topology on W01LΦΩ∗.Proof.
(i)
For anyu∈W01LΦΩ, by (13), we can obtain that(17)Φ¯x,1βζx≤Φ¯x,a1x+Φx,ux+Φx,Dux,
for a.e.x∈Ω, all ζ∈A~u, where β=max3,3b1. Therefore, A is bounded.
(ii)
A is closed and convex valued, so that the same is true for A.(iii)
Assume thatun and un∗ are sequences in W01LΦΩ and W01LΦΩ∗, respectively, such that un∗∈Aun, ∀n∈ℕ,(18)un⟶ustronglyinW01LΦΩ,(19)un∗⇀u∗weaklyinW01LΦΩ∗forσ∏LΦ,∏EΦ¯.
We will show thatu∗∈Au. From (16), for every n∈ℕ, there exists some ζn∈A~un such that un∗=−divζn. Hence,
(20)ζnx∈Ax,unx,Dunxfora.e.x∈Ω.
In view of (18), we see that un is bounded in W01LΦΩ and
(21)∫ΩΦx,2unx−uxdx⟶0,∫ΩΦx,Dunx−Duxdx⟶0,asn⟶∞.
Based on [32], taking to a subsequence if necessary, we have
(22)unx⟶ux,Dunx⟶Duxfora.e.x∈Ω,
and there exist nonnegative functionsh1,h2∈L1Ω such that Φx,2unx−ux≤h1x for a.e. x∈Ω. Since Φ is convex with respect to the second variable,
(23)Φx,unx≤12Φx,2unx−ux+12Φx,2ux≤12h1x+12Φx,2ux.
It follows from (13) that ζn is a bounded sequence in LΦ¯ΩN, and also, by passing to a subsequence if necessary, there exists ζ∈LΦ¯ΩN such that
(24)ζn⇀ζweaklyinLΦ¯ΩNforσLΦ¯ΩN,EΦΩN.
Lettingn⟶∞, from (24), we have
(25)un∗,v=−divζn,v=∫ΩζnDvdx⟶∫ΩζDvdx=−divζ,v,∀v∈W01LΦΩ.
Thanks to (19), we get −divζ=u∗.
By Definition 6.2.1 (a) and (b) and Proposition 6.2.10 in [27], A is graph measurable. Hence, based on Theorem 2.1.4 of [30], for every ξ∈ℝN, there exists a measurable selection η of A·,u,ξ. Based on (A3), there exists a sequence ηn such that, for every n∈ℕ, ηn is an LΩ-measurable selection of the map x∈Ω⟶Ax,unx,ξ and
(26)ηnx⟶ηx,fora.e.x∈Ω.
Now, we show that(27)ηn⟶ηstronglyinLΦ¯ΩN,asn⟶∞.
Indeed, suppose that there existsε0>0 such that ηn−ηΦ¯≥ε0, for any n∈ℕ, where ·Φ¯=·Φ¯ is the Luxemburg norm in LΦ¯ΩN. Then,
(28)∫ΩΦ¯x,ηnx−ηxε0dx≥1,foranyn∈ℕ.
On the one hand, based on (26), for a.e. x∈Ω, we have
(29)Φ¯x,ηnx−ηxε0⟶0,asn⟶∞.
On the other hand, via (13), Φ¯∈Δ2, and the convexity of Φ¯, we can deduce that
(30)Φ¯x,ηnx−ηxε0≤K~Φ¯x,a1x+Φx,unx+Φx,ux+Φx,ξx+hx,
for anyn∈ℕ and a.e. x∈Ω, where K~ is a positive constant and h∈L1Ω. Since Φ≼Ψ and Ψ∈Δ2, combing (26), (30), and (23), we obtain, by Lebesgue’s theorem, that
(31)∫ΩΦ¯x,ηnx−ηxε0dx⟶0,asn⟶∞.
It is a contradiction.
For every nonnegative functionν∈DΩ, in view of (A2), we have
(32)∫Ωζnx−ηnxDunx−ξνxdx≥0.
Lettingn⟶∞, and combining (18), (24), and (27), we have
(33)∫Ωζnx−ηnxDunx−ξνxdx⟶∫Ωζx−ηxDux−ξνxdx.
Thanks to (32), we get
(34)∫Ωζx−ηxDux−ξνxdx≥0.
Sinceν is arbitrary, we get ζx−ηxDux−ξ≥0 for a.e. x∈Ω. Based on (A1) and Theorem III.9 of [33], there exists a sequence of measurable selections σn of A·,u,ξ, such that σnx:n∈ℕ is dense in Ax,ux,ξ. It implies that ζx−ϑDux−ξ≥0 for a.e. x∈Ω and for every ϑ∈Ax,ux,ξ. Thanks to (A2), ζx∈Ax,ux,Dux for a.e. x∈Ω. Consequently, u∗∈Au.
(iv)
To prove (iv), we need to show that ifu∈W01LΦΩ, un⊂W01LΦΩ with(35)un⟶ustronglyinW01LΦΩ,asn⟶∞,and V⊂W01LΦΩ∗ is a weakly open set such that Au⊂V; then, we can find n0∈ℕ such that Aun⊂V for all n≥n0. If we suppose not, then there exists a subsequence uk of un such that Auk∩VC≠∅. Let wk∈Auk∩VC. Based on (35), un is bounded in W01LΦΩ. In view of (13), Aun is bounded in W01LΦΩ∗. It follows that wk is bounded in W01LΦΩ∗. Consequently, by passing to a subsequence if necessary, there exists w∈W01LΦΩ∗ such that wk⇀w weakly in W01LΦΩ∗ for σΠLΦ¯,ΠEΦ, as k⟶+∞. Based on (iii), w∈Au. Since W01LΦΩ is reflexive, w∈VC. It is a contradiction to the choice of V.Theorem 14.
Under assumptions (A1)–(A4), the mappingA is pseudomonotone.Proof.
Let sequencesun⊂W01LΦΩ and un∗⊂W01LΦΩ∗ satisfy
(36)un⇀u0weaklyinW01LΦΩforσ∏LΦ,∏EΦ¯,(37)un∗⇀u0∗weaklyinW01LΦΩ∗forσ∏LΦ¯,∏EΦ,(38)limsupn⟶∞un∗,un−u0≤0.
We will show thatu0∗∈Au0 and
(39)limn⟶∞un∗,un=u0∗,u0.
For eachu∈W01LΦΩ, we define the multivalued operator Bu:W01LΦΩ⟶2W01LΦΩ∗, by
(40)Buv≔−divη:ηx∈Ax,ux,Dvx,fora.e.x∈Ω,∀v∈W01LΦΩ.
Using the similar and more simple proof ((A3) is not needed in this case), we can deduce that Lemma13 holds if A is replaced by Bu. Moreover, based on (A2), Bu is monotone. Therefore, it is maximal monotone. Clearly, Buu=Au, for any u∈W01LΦΩ.
For eachn∈ℕ, there exists ζn∈A~u0 such that un∗=−divζn. From (36), un is bounded in W01LΦΩ. Then, by (13), ζn is bounded in LΦ¯ΩN. Therefore, there exists ζ∈LΦ¯ΩN such that, by extracting a subsequence if necessary, ζn⇀ζ in LΦ¯ΩN weakly for σLΦ¯ΩN,EΦΩN, as n⟶∞. Analogous with the proof in Lemma 13, we have −divζ=u0∗.
Letv∈W01LΦΩ and η∈LΦ¯ΩN be such that ηx∈Ax,u0x,Dvx for a.e. x∈Ω. Since W01LΦΩ↪LΨΩ is compact, in view of (36), there exists a subsequence of un still denoted by un such that
(41)un⟶u0stronglyinLΨΩ,asn⟶∞.
By passing to a further subsequence if necessary, we can suppose(42)un⟶u0a.e.inΩ,asn⟶∞,(43)Ψx,2unx−u0x≤h0x,fora.e.x∈Ω,
for someh0∈L1Ω (see [32]). Since Ψ∈Δ2 and Ψ is convex with respect to the second variable,
(44)Ψx,unx≤12Ψx,2unx−u0x+12Ψx,2u0x≤12h0x+12Ψx,2u0x,
for a.e.x∈Ω. According to (A3), there exists a sequence ηn such that
(45)ηn⟶ηa.e.inΩ,asn⟶∞,
andηn is an LΩ-measurable selection of x∈Ω⟶Ax,unx,Dvx⊂ℝN, for every n∈ℕ. Since Φ≼Ψ and Ψ∈Δ2, similar to the proof of (27), we have
(46)ηn⟶ηstronglyinLΦ¯ΩN,asn⟶∞.
For eachn∈ℕ, from the monotonicity of Ax,unx,· for a.e. x∈Ω, we obtain
(47)0≤limsupn⟶∞∫Ωζnx−ηnxDunx−Dvxdx=limsupn⟶∞un∗,un−u0+∫ΩζnxDu0x−Dvxdx−∫ΩηnxDunx−Dvxdx=limsupn⟶∞un∗,un−u0+∫Ωζx−ηxDu0x−Dvxdx≤−divζ−−divη,u0−v.
SinceBu0 is maximal monotone, we get that u0∗=−divζ∈Bu0u0=Au0.
Thanks to (A3), there existsνn such that νn⟶ζ a.e. in Ω, as n⟶∞, and νn is an LΩ-measurable selection of x∈Ω⟶Ax,unx,Dux⊂ℝN, for each n∈ℕ. Then, by using the monotonicity of Ax,unx,· for a.e. x∈Ω, we have
(48)un∗,un≥∫ΩηnxDunx−Du0xdx+∫ΩζnxDu0xdx.
It yields thatliminfn⟶∞un∗,un≥u0∗,u0. Consequently, based on (38), we can deduce that (39) holds. Based on Lemma 8, A is pseudomonotone.


### 3.2. The Multivalued Convection Term

We defineΦ−t≔essinfx∈ΩΦx,t, for any t≥0 (see [26]). By the condition (Φ1), we obtain that Φ−0,+∞≢0 and Φ−0,+∞≢+∞. According to Lemma 2.5.16 of [26], Φ− is an N-function. Hence, Φ−t>0 for t>0. Moreover, analogous to the proof of Lemma 2.5.24 of [26], we can obtain that Φ−∈Δ2 whenever Φ∈Δ2. Then, there exists a constant Kd>0 such that
(49)Φ−ds≤KdΦ−s,∀s≥s0,for some s0≥0, where d is twice the diameter of Ω.LetF:Ω×ℝ×ℝN⟶Pfcℝ satisfy the following conditions:(F1)F is graph measurable.(F2) For a.e.x∈Ω, the multivalued mapping Fx,·,·:ℝ×ℝN⟶2ℝ is upper semicontinuous.(F3) There exist a functiona3∈LΨ¯Ω and a constant b3>0 such that
(50)supf:f∈Fx,s,ξ≤a3x+b3Ψ¯−1x,Ψx,s+b3Ψ¯−1x,Φx,ξ,for a.e. x∈Ω, all s∈ℝ and all ξ∈ℝN.(F4) There exist a functionω∈L1Ω and two constants b4,b5≥0 such that
(51)fs≤ωx+b4Φ−s+b5Φx,ξ,for a.e. x∈Ω, all s∈ℝ and all ξ∈ℝN, and for all f∈Fx,s,ξ. Moreover,
(52)b2−λ~−1b4−b5>0,where
(53)λ~=infu∈Dℏ∫ΩΦ−Duxdx∫ΩΦ−uxdx,with Dℏ=u∈W01LΦ−Ω\0:∫ΩΦ−Duxdx≥ℏ for some ℏ>0.Remark 15.
WhenDux≢0 for a.e x∈Ω, we have ∫ΩΦ−Duxdx>0.Remark 16.
From Lemma12 and (49),
(54)∫ΩΦ−uxdx≤Kd∫ΩΦ−Duxdx+measΩ·Φ−ds0,which yields that
(55)∫ΩΦ−Duxdx∫ΩΦ−uxdx≥∫ΩΦ−DuxdxKd∫ΩΦ−Duxdx+measΩ·Φ−ds0.
Hence,λ~≥ℏ/Kdℏ+measΩ·Φ−ds0>0.Example 17.
Assume thata,b,c,d∈L∞Ω, e,k∈LΨ¯Ω, ax≤bx, cx≤dx, and ex≤kx, for a.e. x∈Ω, and
(56)b2−λ~−1maxaL∞Ω,bL∞Ω−maxcL∞Ω,dL∞Ω>0.
PutFx,s,ξ=−d1Φ−s/s2s+d2ς+υ:d1∈ax,bx,d2∈cx,dx,ς∈Φ¯−1x,Φx,ξ,Ψ¯−1x,Φx,ξ,υ∈ex,kx, for x∈Ω, s∈ℝ, and ξ∈ℝN. Then, F satisfies (F1)–(F4).For eachu∈W01LΦΩ, let
(57)F~u=f∈L0Ω:fx∈Fx,ux,Dux,fora.e.x∈Ω.It follows from (F1) thatF~u≠∅ whenever u∈W01LΦΩ. Moreover, from (F3), we can see that F~u⊂LΨ¯Ω whenever u∈W01LΦΩ.LetF:W01LΦΩ⟶2W01LΦΩ∗ be defined by Fu=f^∈W01LΦΩ∗:f∈F~u, where f^ is defined for each f∈LΨ¯Ω by
(58)f^,v=∫Ωfxvxdx,∀v∈W01LΦΩ.Analogous to the proof of Theorem 6.4.16 of [27], we have the following lemma, whose proof is omitted.Lemma 18.
Letw:Ω×ℝ⟶ℝ¯=ℝ∪±∞ be a LΩ×Bℝ-measurable function, T:Ω⟶2ℝ\∅ be graph-measurable, and the integral functional
(59)Iwv=∫Ωwx,vxdx,v∈LΦΩbe defined for all v∈SΦT=v∈LΦΩ:vx∈Txfora.e.x∈Ω, where Φ∈NΩ. If there exists v0∈LΦΩ such that Iwv0>−∞, then
(60)supv∈SΦTIwv=∫Ωsupη∈Txwx,ηdx.If there existsv0∈LΦΩ such that Iwv0<+∞, then x⟶infη∈Txwx,η is (Lebesgue) measurable and
(61)infv∈SΦTIwv=∫Ωinfη∈Txwx,ηdx.Now, we give the following lemma abouth-upper semicontinuous of multivalued mappings.Lemma 19.
Assume that (F1)–(F3) hold. Then,(i)
F~u∈PfcLΨ¯Ω for all u∈W01LΦΩ;(ii)
F~ is bounded;(iii)
F~:W01LΦΩ⟶PfcLΨ¯Ω is h-upper semicontinuous; that is, if un⟶u0 strongly in W01LΦΩ, then(62)h∗F~un,F~u0=supv′∈F~uninfv∈F~u0v′−vΨ¯⟶0,as n⟶∞.Proof.
(i)
The proof of the convexity and closedness ofF~u is a direct consequence of the fact that Fx,s,ξ is convex, closed, and bounded in ℝ for a.e. x∈Ω, all s∈ℝ, and all ξ∈ℝN.(ii)
The boundedness ofF~ follows directly from (F3).(iii)
The following proof is referred to Theorem 2.7 of [13]. Assume that un⊂W01LΦΩ is a sequence such that(63)un⟶u0stronglyinW01LΦΩ,asn⟶∞,
for someu0∈W01LΦΩ. We will show that
(64)supv′∈F~uninfv∈F~u0v′−vΨ¯⟶0,asn⟶∞.
SinceW01LΦΩ↪LΨΩ is compact,
(65)un⟶u0stronglyinLΨΩ,asn⟶∞.
It follows from (65) and (63) that there exists a subsequence of un still denoted by un and h1,h2∈L1Ω such that
(66)unx⟶u0x,Dunx⟶Du0x,fora.e.x∈Ω,asn⟶∞,
andΨx,2unx−u0x≤h1x, Φx,2Dunx−Du0x≤h2x, for a.e. x∈Ω. Hence,
(67)Ψx,unx≤12h1x+12Ψx,2u0x,Φx,Dunx≤12h2x+12Φx,2Du0x,
for a.e.x∈Ω. Based on (F2) and Lemma 5, the multivalued mapping Fx,·,·:ℝ×ℝN⟶2ℝ is h-upper semicontinuous, for a.e. x∈Ω. It follows that supξ∈Fx,unx,Dunxinfη∈Fx,u0x,Du0xξ−η⟶0 as n⟶∞, for a.e. x∈Ω, and thus,
(68)supξ∈Fx,unx,Dunxinfη∈Fx,u0x,Du0xΨ¯x,ξ−η⟶0asn⟶∞,for a.e. x∈Ω.
On the other hand, sinceΨ,Ψ¯∈Δ2, based on (F3) and (67), there exist some constants K1,K2>0 and a function h∈L1Ω such that
(69)Ψ¯x,ξ−η≤K12Ψ¯x,ξ+Ψ¯x,η+hx≤K1K262Ψ¯x,a3x+Ψx,unx+Φx,Dunx+Ψx,u0x+Φx,Du0x+K1+1hx≤K1K262Ψ¯x,a3x+12h1x+12Ψx,2u0x+12h2x+12Φx,2Du0x+Ψx,u0x+Φx,Du0x+K1+1hx≔Hx,for all n∈ℕ, a.e. x∈Ω, and all ξ∈Fx,unx,Dunx, η∈Fx,u0x,Du0x. Thus,
(70)supξ∈Fx,unx,Dunxinfη∈Fx,u0x,Du0xΨ¯x,ξ−η≤Hx,
for alln∈ℕ, a.e. x∈Ω. Evidently, the function H belongs to L1Ω. Using Lebesgue’s dominated convergence theorem, we deduce that
(71)∫Ωsupξ∈Fx,unx,Dunxinfη∈Fx,u0x,Du0xΨ¯x,ξ−ηdx⟶0asn⟶∞.
Letn∈ℕ. Based on (F1), F is graph measurable. For v′∈F~un, using Lemma 18, we can obtain that
(72)infv∈F~u0∫ΩΨ¯x,v′x−vxdx=∫Ωinfη∈Fx,u0x,Du0xΨ¯x,v′x−ηdx.
Using Lemma18 again, we get
(73)supv′∈F~un∫Ωinfη∈Fx,u0x,Du0xΨ¯x,v′x−ηdx=∫Ωsupξ∈Fx,unx,Dunxinfη∈Fx,u0x,Du0xΨ¯x,ξ−ηdx.
Lettingn⟶∞, and combining (72), (73), and (71), we have
(74)supv′∈F~uninfv∈F~u0∫ΩΨ¯x,v′x−vxdx⟶0.
SinceΨ¯∈Δ2, we can deduce (64), which completes our proof.Then, we have the following theorem.Theorem 20.
IfF satisfies (F1)–(F3), then
(i)
F is bounded;(ii)
GrF is (sequentially) closed in W01LΦΩ×W01LΦΩ∗ with respect to the strong-weak topology.Proof.
(i)
It is easy from (F3) to see thatF is bounded.(ii)
We show thatGrF is (sequentially) closed in W01LΦΩ×W01LΦΩ∗ with respect to the strong-weak topology. Assume that un and f^n are sequences in W01LΦΩ and W01LΦΩ∗, respectively, such that(75)un⟶ustronglyinW01LΦΩ,(76)f^n⇀f^weaklyinW01LΦΩ∗forσΠLΦ¯Ω,ΠEΦΩ,(77)f^n∈Fun,∀n∈ℕ.
Let us prove that(78)f^∈Fu.
Thanks to (77), there exists fn∈F~un such that, for every n∈ℕ,
(79)f^n,v=∫Ωfnxvxdx,∀v∈W01LΦΩ.
From (75) and Lemma 19 (iii), we obtain that h∗F~un,F~u⟶0, and thus,
(80)infw∈F~ufn−wΨ¯⟶0,asn⟶∞.
Consequently, there is a sequencewn⊂F~u such that
(81)fn−wnΨ¯⟶0,asn⟶∞.
SinceF~u is bounded in LΨ¯Ω, by passing to a subsequence if necessary, there exists w∈LΨ¯Ω such that
(82)wn⇀wweaklyinLΨ¯ΩforσLΨ¯Ω,EΨΩ,asn⟶∞.
According to Lemma19 (i), w∈F~u. Letting n⟶∞, and combing (81) and (82), we get that
(83)fn⇀wweaklyinLΨ¯ΩforσLΨ¯Ω,EΨΩ.
It implies thatf^,v=∫Ωwxvxdx,∀v∈W01LΦΩ. Therefore, f^∈Fu.


### 3.3. Existence Theorem for Inclusion Problems

A functionu∈W01LΦΩ is called a (weak) solution of problem (1), if there exist ζ∈LΦ¯ΩN and f∈LΨ¯Ω such that
(84)ζx∈Ax,ux,Dux,fx∈Fx,ux,Dux,fora.e.x∈Ω,∫ΩζxDvxdx+∫Ωfxvxdx=∫ΩLxvxdx,∀v∈W01LΦΩ.In this section, we are going to prove an existence result for our problem (1).Theorem 21.
LetA and f satisfy (A1)–(A4) and (F1)–(F4); then, there exists a solution for problem (1).Proof.
We define the multivalued mappingT:W01LΦΩ⟶2W1LΦΩ∗ by
(85)Tu=Au+Fu.
Then,T is well-defined.
Step 1. We are going to show thatT is pseudomonotone. Let sequences un⊂W01LΦΩ and ωn⊂W01LΦΩ∗ have the properties
(86)un⇀uweaklyinW01LΦΩforσ∏LΦ,∏EΦ¯,asn⟶∞,(87)ω^n⇀ω^weaklyinW01LΦΩ∗forσ∏LΦ¯,∏EΦ,asn⟶∞,with ω^n∈Tun and limsupn⟶∞ω^n,un−u≤0. From (85), there exist ζ^n∈Aun, f^n∈Fun such that ω^n=ζ^n+f^n, for every n∈ℕ, where ζ^n=−divζn with ζn∈A~un and f^n satisfies f^n,v=∫Ωfnxvxdx, ∀v∈W01LΦΩ, for some fn∈F~un.
Now, we prove that(88)un⟶ustronglyinW01LΦΩ,asn⟶∞.
To show (88), based on Proposition 1.9 of [17], it needs to be proven that
(89)Dun⟶DustronglyinLΦΩN,asn⟶∞.
We suppose that there exitsε0>0 such that
(90)Dun−DuΦ≥ε0,foranyn∈ℕ.
Then,(91)∫ΩΦx,Dunx−Duxε0dx≥1,foranyn∈ℕ.
From (86), the sequence un is bounded in W01LΦΩ, and by passing to a subsequence if necessary,
(92)un⟶ustronglyinLΨΩ,(93)un⟶ua.e.inΩ,
asn⟶∞. By passing to a further subsequence if necessary, there exists h0∈L1Ω such that
(94)Ψx,unx≤12h0x+12Ψx,2ux,
for a.e.x∈Ω. Thanks to (F3), fn is bounded in LΨ¯Ω. It follows that
(95)f^n,un−u=∫Ωfnxunx−uxdx≤2fnΨ¯un−uΨ⟶0,
asn⟶∞. Consequently, limsupn⟶∞ζ^n,un−u≤0. By (13), ζ^n is bounded in W01LΦΩ∗. Hence, by extracting a subsequence if necessary, there exists ζ^∈W01LΦΩ∗ such that
(96)ζ^n⇀ζ^weaklyinW01LΦΩ∗forσ∏LΦ¯,∏EΦ,asn⟶∞.
From Theorem14, A is pseudomonotone; then, we get ζ^∈Au and
(97)limn⟶∞ζ^n,un=ζ^,u.
Letη be a measurable selection of A·,u,Du. Based on (93) and (A3), there exists a sequence ηn converging a.e. to η in Ω, as n⟶∞, such that ηn is an LΩ-measurable selection of the map x∈Ω⟶Ax,unx,Dux∈2ℝN, for every n∈ℕ. Same as the proof of (46), we can obtain that
(98)ηn⟶ηstronglyinLΦ¯ΩN,asn⟶∞.
Denoteη^n=−divηn and η^=−divη.
On the one hand, combing (86), (96), (97), and (98), we can obtain that
(99)ζ^n−η^n,un−u=ζ^n,un−ζ^n,u−η^n−η^,un+η^n,u−η^,un⟶0,
asn⟶∞. On the other hand, for a.e. x∈Ω, dnx≔ζnx−ηnxDunx−Dux≥0. Hence, dn⟶0 in L1Ω. By passing to a subsequence if necessary,
(100)dnx⟶0fora.e.x∈Ω,asn⟶∞.
SinceΦ∈Δ2 and Φ≼Ψ, by using (13), (14), (94), (100), and the Young inequality, one has Φx,Dunx≤ℓx, for some function ℓ∈L1Ω, for all n∈ℕ, for a.e. x∈Ω. It implies that Dunx is bounded in ℝN, for a.e. x∈Ω. Taking a subsequence if necessary, there exists ϑ∈ℝN such that Dunx⟶ϑ as n⟶∞, for a.e. x∈Ω. Similar to Lemma 1.4 of [34], we can show that ϑ∈LΦΩN and Dun⟶ϑ for σLΦΩN,LΦ¯ΩN, as n⟶∞. Therefore, ϑ=Du, and thus, Dunx⟶Dux as n⟶∞, for a.e. x∈Ω. Based on Lebesgue’s Theorem, we can show that
(101)∫ΩΦx,Dunx−Duxε0dx⟶0,asn⟶∞.
It is a contradiction with (89).
Denotef^=ω^−ζ^. Combing (87) and (96),
(102)f^n⇀f^weaklyinW01LΦΩ∗,forσ∏LΦ¯,∏EΦ,asn⟶∞.
Invoking Theorem20, we obtain that f^∈Fu. Thus, ω^∈Tu. In view of (88), and (87), we get
(103)ω^n,un=ω^n,un−u+ω^n,u⟶ω^,u,asn⟶∞.
Step 2. We will see thatT is coercive. Let u∈W01LΦΩ and ω^∈Tu. Then, there exist ζ^∈Au and f^∈Fu such that ω^=ζ^+f^. From (A4), for any ζ^∈Au, we have that
(104)ζ^,u≥b2∫ΩΦx,Duxdx−a2L1Ω.
For anyf^∈Fu, by (51) and (53), we deduce that
(105)f^,u≤ωL1Ω+b4∫ΩΦ−uxdx+b5∫ΩΦx,Duxdx≤ωL1Ω+b4λ~∫ΩΦ−Duxdx+b5∫ΩΦx,Duxdx,for all u∈Dℏ.
Hence, we obtain that(106)ω^,u≥b2−b4λ~−b5∫ΩΦx,Duxdx−a2L1Ω−ωL1Ω,
for allu∈Dℏ. Based on Proposition 1.9 of [17], there exists a constant C∗>0 such that
(107)u1,Ω≤1+C∗DuΦ.
Then, we get that(108)ω^,uu1,Ω≥b2−b4λ~−b5∫ΩΦx,Duxdx1+C∗DuΦ−a2L1Ω+ωL1Ω1u1,Ω,for all u∈Dℏ.
Based on Proposition 3.1 of [18], there are functions h∈L1Ω with h≥0 and G:k,+∞⟶ℝ for some k>2 such that Gs⟶+∞ as s⟶+∞ and
(109)Φx,su≥sGsΦx,u−shx,foralls≥k,u≥0,a.e.x∈Ω.
It follows that(110)∫ΩΦx,Duxdx≥DuΦ−εGDuΦ−ε∫ΩΦx,DuxDuΦ−εdx−hL1Ω≥DuΦ−εGDuΦ−ε−hL1Ω,if DuΦ is large enough, for any ε∈0,1.
Combining (108) and (110), we can deduce that
(111)limu1,Ω⟶+∞,u∈W01LΦΩinfω^,u:ω^∈Tuu1,Ω=+∞.
Step 3. According to Lemma11, there exists u∈W01LΦΩ such that
(112)ω^,u=L,u,
for someω^∈Tu, where L,u=∫ΩLxuxdx. Therefore, there exist ζ^∈Au and f^∈Fu such that ω^=ζ^+f^. This completes the proof.The set of solutions of problem (1) is denoted by S. Then, we have the following results for set S.Theorem 22.
Assume (A1)–(A4) and (F1)–(F4) hold. Then,S is nonempty, bounded, and closed.Proof.
(1)
According to Theorem21, S is nonempty.(2)
We are going to prove thatS is bounded. Assume that S is unbounded. Then, there exists a sequence un in S such that un1,Ω⟶+∞, as n⟶∞. Since there exists a constant C>0 such that L,un≤CLW01LΦΩ∗un1,Ω, combining (108) and (110), we have that(113)limun1,Ω⟶+∞,un∈W01LΦΩinfω^n−L,un:ω^n∈Tunun1,Ω=+∞.
However, it follows from (112) that ω^n−L,un=0, where ω^n∈Tun, for any n∈ℕ. It is a contradiction.
(3)
We will show thatS is closed. Let un be a sequence in S such that(114)un⟶u0inW01LΦΩ,asn⟶∞,
for someu0∈W01LΦΩ. Then, there exist ζ^n∈Aun and f^n∈Fun such that ω^n=ζ^n+f^n∈Tun and
(115)ω^n,un=L,un,foranyn∈ℕ.
Based on (114), un is bounded in W01LΦΩ. In view of (13), and (F3), ζ^n and f^n are bounded in LΦ¯ΩN and LΨ¯Ω, respectively. Extracting a subsequence if necessary, there exist ζ^0,f^0∈W01LΦΩ∗ such that
(116)ζ^n⇀ζ^0weaklyinW01LΦΩ∗forσ∏LΦ¯,∏EΦ,(117)f^n⇀f^0weaklyinW01LΦΩ∗forσ∏LΦ¯,∏EΦ,as n⟶∞. Using Lemma 13 (iii), we have ζ^0∈Au0. Based on Theorem 20 (ii), we get f^0∈Fu0.
Based on (114),
(118)ζ^n,un−u0⟶0,f^n,un−u0⟶0,(119)L,un−u0⟶0,as n⟶∞.
Based on (115)–(119),
(120)ω^n−L,un⟶ζ^0+f^0−L,u0,asn⟶∞.
Denoteω^0=ζ^0+f^0. Then, ω^0∈Tu0. Consequently, ω^0,u0=L,u0. It implies that u0∈S.


### 3.4. Obstacle Problems

Now, we consider the following inclusion problem with an obstacle effect(121)L∈Ax,u,Du+Fx,u,Du,inΩ,ux≤υx,inΩ,u=0,on∂Ω,whereυ:Ω⟶ℝ+∪+∞ is a given function. Let K=u∈W01LΦΩ:u≤υa.e.inΩ. It is easy to see that K is a nonempty, closed, and convex subset of W01LΦΩ and contains 0. The indicator function IK:W01LΦΩ⟶ℝ∪+∞ of K is defined by
(122)IKu≔0,ifu∈K,+∞,otherwise.Let∂IK be the subdifferential of IK in the sense of convex analysis (see Definition 32.11 of [35]).A functionu∈K is called a (weak) solution of problem (121), if there exist ζ∈LΦ¯ΩN and f∈LΨ¯Ω such that
(123)ζx∈Ax,ux,Dux,fx∈Fx,ux,Dux,fora.e.x∈Ω,(124)∫ΩζxDvxdx+∫Ωfxvxdx=∫ΩLxvxdx,∀v∈K.From the definition above, it is not difficult to show that problem (121) is equivalent to the following variational inequality:Findu∈W01LΦΩ such that
(125)Au,v−u+Fu,v−u+IKv−IKu≥L,v−u,∀v∈W01LΦΩ,which is formulated as the following inclusion:Findu∈W01LΦΩ such that
(126)L∈Au+Fu+∂IKu.We have the following theorem.Theorem 23.
Assume that (A1)–(A4) and (F1)–(F4) hold. Then, there exists a solutionu of problem (121) such that u1,Ω<R for some R>0.Proof.
LetT be defined by (85). Then, based on the proof of Theorem 21, T is pseudomonotone.
According to Example 32.15 of [35], ∂IK is maximal monotone.
We are going to check that there exists a positive constantR such that
(127)ω^+η^−L,u>0,for all u∈W01LΦΩ with u1,Ω=R, for all ω^∈Tu and all η^∈∂IKu.
Letu∈W01LΦΩ with DuΦ>k, where k is the constant in (109). Based on Remark 15, there exists ℏ>0 such that u∈Dℏ. From (106) and (110), we obtain that
(128)ω^,u≥b2−b4λ~−b5DuΦ−εGDuΦ−ε−hL1Ω−a2L1Ω−ωL1Ω,for any ε∈0,1 and for all ω^∈Tu.
Similar to the proof of (2.50) in [36], there exists w∈W01LΦΩ∗ and αK∈ℝ such that
(129)IKu≥−C~wW01LΦΩ∗u1,Ω−αK,∀u∈W01LΦΩ,for some constant C~>0.
Combining (107), (128), and (129), and based on the arbitrariness of ε, we have
(130)ω^+η^−L,u≥b2−b4λ~−b5DuΦGDuΦ−b2−b4λ~−b5hL1Ω−a2L1Ω−ωL1Ω−C~wW01LΦΩ∗u1,Ω−αK−CLW01LΦΩ∗u1,Ω≥DuΦb2−b4λ~−b5GDuΦ−1+C∗C~wW01LΦΩ∗+CLW01LΦΩ∗−b2−b4λ~−b5hL1Ω−a2L1Ω−ωL1Ω−αK,
for allω^∈Tu and all η^∈∂IKu, where C is a positive constant. It is easy to see that we can take R0>k to be large enough such that
(131)R0b2−b4λ~−b5GR0−1+C∗C~wW01LΦΩ∗+CLW01LΦΩ∗−b2−b4λ~−b5hL1Ω−a2L1Ω−ωL1Ω−αK>0.
LetR≥1+C∗R0. Then, (127) holds for all u∈W01LΦΩ with u1,Ω=R, for all ω^∈Tu and all η^∈∂IKu. Using Theorem 2.2 of [14], inclusion (126) has a solution u such that u1,Ω<R.We denote the set of solutions of inclusion (126) based on SK. With the analogous proof of Theorem 22, we can deduce the following theorem.Theorem 24.
Assume that (A1)–(A4) and (F1)–(F4) hold. Then,SK is nonempty, bounded, and closed.


## 3.1. The Maximal Monotone Operator

LetA:Ω×ℝ×ℝN⟶PfcℝN satisfy the following conditions:(A1)A is LΩ×Bℝ×BℝN-measurable, where LΩ is the family of Lebesgue measurable subsets of Ω, Bℝ is the σ-algebra of Borel sets in ℝ, and BℝN is the σ-algebra of Borel sets in ℝN; i.e., for every open set U⊂ℝN, A−1U≔x,s,ξ∈Ω×ℝ×ℝN:Ax,s,ξ∩U≠∅∈LΩ×Bℝ×BℝN.(A2) For a.e.x∈Ω, every s∈ℝ, Ax,s,·:ℝN⟶2ℝN is maximal monotone.(A3) Ifξ,u,un,η are given, being ξ∈ℝN, u∈W01LΦΩ, un in W01LΦΩ such that un⟶u a.e. in Ω, and η an LΩ-measurable selection of the map x∈Ω⟶Ax,ux,ξ⊂ℝN, then there exists a sequence ηn converging a.e. to η in Ω, such that, for every n∈ℕ, ηn is an LΩ-measurable selection of the map x∈Ω⟶Ax,unx,ξ⊂ℝN.(A4) There exista1∈LΦ¯Ω, a2∈L1Ω, and b1≥0, b2>0 such that
(13)ζ≤a1x+b1Φ¯−1x,Φx,s+b1Φ¯−1x,Φx,ξ,(14)ζξ≥a2x+b2Φx,ξ,for a.e. x∈Ω, all s∈ℝ, all ξ∈ℝN, and all ζ∈Ax,s,ξ.By Definition 6.2.1 (a) and Proposition 6.2.3 in [27], we can see that the condition (A1) is weaker than the condition Definition 3.3 (ii) in [16].Ifu:Ω⟶ℝ and w:Ω⟶ℝN are LΩ-measurable, then x∈Ω⟶A·,u,wx=Ax,ux,wx∈2ℝN turns out to be measurable as well (see [12], Remark 1.2). If η:Ω⟶2ℝN is a measurable selection of A·,u,Dv, u,v∈W01LΦΩ, then, by (13), η∈LΦ¯ΩN.LetA~:W01LΦΩ⟶2LΦ¯ΩN be defined by
(15)A~u=ζ∈LΦ¯ΩN:ζx∈Ax,ux,Duxfora.e.x∈Ω,andA:W01LΦΩ⟶2W01LΦΩ∗ by
(16)Au=−divζ:ζ∈A~u.It follows from (A1) and (13) that A~u≠∅, for any u∈W01LΦΩ. It is easy to see that DA=W01LΦΩ.Example 1.
LetΦx,t=tp, for t≥0, all x∈Ω¯, where 1<q<p<+∞ are given. For all x∈Ω¯, s∈ℝ, and ξ∈ℝN, we put Ax,s,ξ=Φx,ξ/ξ2ξ. Then, Au becomes the p-Laplace operator −divDup−2Du, e.g., [1].Example 2.
LetΦx,t=tp+μtq, for t≥0, all x∈Ω¯, where the numbers μ≥0 and 1<q<p<+∞ are given. For all x∈Ω¯, s∈ℝ, and ξ∈ℝN, we put Ax,s,ξ=Φx,ξ/ξ2ξ. Then, Au becomes the double-phase operator −divDup−2Du+μDuq−2Du, e.g., [6].Example 3.
LetΦx,t=tpx, for x∈Ω¯ and t≥0, where p∈CΩ¯ is given such that 1<p−≔infx∈Ω¯px≤px≤p+≔supx∈Ω¯px<+∞, for all x∈Ω¯. For all x∈Ω¯, s∈ℝ, and ξ∈ℝN, we put Ax,s,ξ=Φx,ξ/ξ2ξ. Then, Au becomes the px-Laplace differential operator −divDupx−2Du, e.g., [21].Example 4.
LetΦx,t=tpx+μxtqx, for x∈Ω¯ and t≥0, where μ∈L1Ω is a nonnegative function and p,q∈CΩ¯ are given such that 1<px<N, px<qx, for all x∈Ω¯. For all x∈Ω¯, s∈ℝ, and ξ∈ℝN, we put Ax,s,ξ=Φx,ξ/ξ2ξ. Then, Au becomes the double-phase operator with variable exponents −divDupx−2Du+μxDuqx−2Du, e.g., [9].Example 5.
Letφx,t=tpx−2tlog1+trx, for x∈Ω¯ and t≥0, where p∈L∞Ω with p−>1, and r∈L∞Ω with r−≥0, and let Φx,t=∫0tφx,τdτ for x∈Ω¯ and t∈ℝ. Then, Φ satisfies (Φ) (see [17]). For all x∈Ω, s∈ℝ, and ξ∈ℝN, we put A1x,s,ξ=a1x,sΦx,ξ/ξ2ξ and A2x,s,ξ=a2x,sΦx,ξ/ξ2ξ, where a1 and a2 are Carathéodory functions (i.e., for each s∈ℝ, x⟶a1x,s and x⟶a2x,s are measurable; for a.e. x∈Ω, s⟶a1x,s and s⟶a2x,s are continuous) on Ω×ℝ satisfying 0<α≤a1x,s≤a2x,s≤β for positive constants α and β, for a.e. x∈Ω. As in [12], we define Ax,s,ξ=λA1x,s,ξ+1−λA2x,s,ξ:λ∈0,1, for all x∈Ω, s∈ℝ, and ξ∈ℝN. Then, A satisfies (A1)–(A4).Lemma 13.
Under assumptions (A1)–(A4), we have that(i)
A is bounded;(ii)
Au∈PfcW01LΦΩ∗, ∀u∈X;(iii)
GrA is (sequentially) closed in W01LΦΩ×W01LΦΩ∗ with respect to the strong-weak topology (i.e., with respect to the product of the norm topology of W01LΦΩ and the weak topology of W01LΦΩ∗);(iv)
A is upper semicontinuous from W01LΦΩ with the strong topology to 2W01LΦΩ∗ with the weak topology on W01LΦΩ∗.Proof.
(i)
For anyu∈W01LΦΩ, by (13), we can obtain that(17)Φ¯x,1βζx≤Φ¯x,a1x+Φx,ux+Φx,Dux,
for a.e.x∈Ω, all ζ∈A~u, where β=max3,3b1. Therefore, A is bounded.
(ii)
A is closed and convex valued, so that the same is true for A.(iii)
Assume thatun and un∗ are sequences in W01LΦΩ and W01LΦΩ∗, respectively, such that un∗∈Aun, ∀n∈ℕ,(18)un⟶ustronglyinW01LΦΩ,(19)un∗⇀u∗weaklyinW01LΦΩ∗forσ∏LΦ,∏EΦ¯.
We will show thatu∗∈Au. From (16), for every n∈ℕ, there exists some ζn∈A~un such that un∗=−divζn. Hence,
(20)ζnx∈Ax,unx,Dunxfora.e.x∈Ω.
In view of (18), we see that un is bounded in W01LΦΩ and
(21)∫ΩΦx,2unx−uxdx⟶0,∫ΩΦx,Dunx−Duxdx⟶0,asn⟶∞.
Based on [32], taking to a subsequence if necessary, we have
(22)unx⟶ux,Dunx⟶Duxfora.e.x∈Ω,
and there exist nonnegative functionsh1,h2∈L1Ω such that Φx,2unx−ux≤h1x for a.e. x∈Ω. Since Φ is convex with respect to the second variable,
(23)Φx,unx≤12Φx,2unx−ux+12Φx,2ux≤12h1x+12Φx,2ux.
It follows from (13) that ζn is a bounded sequence in LΦ¯ΩN, and also, by passing to a subsequence if necessary, there exists ζ∈LΦ¯ΩN such that
(24)ζn⇀ζweaklyinLΦ¯ΩNforσLΦ¯ΩN,EΦΩN.
Lettingn⟶∞, from (24), we have
(25)un∗,v=−divζn,v=∫ΩζnDvdx⟶∫ΩζDvdx=−divζ,v,∀v∈W01LΦΩ.
Thanks to (19), we get −divζ=u∗.
By Definition 6.2.1 (a) and (b) and Proposition 6.2.10 in [27], A is graph measurable. Hence, based on Theorem 2.1.4 of [30], for every ξ∈ℝN, there exists a measurable selection η of A·,u,ξ. Based on (A3), there exists a sequence ηn such that, for every n∈ℕ, ηn is an LΩ-measurable selection of the map x∈Ω⟶Ax,unx,ξ and
(26)ηnx⟶ηx,fora.e.x∈Ω.
Now, we show that(27)ηn⟶ηstronglyinLΦ¯ΩN,asn⟶∞.
Indeed, suppose that there existsε0>0 such that ηn−ηΦ¯≥ε0, for any n∈ℕ, where ·Φ¯=·Φ¯ is the Luxemburg norm in LΦ¯ΩN. Then,
(28)∫ΩΦ¯x,ηnx−ηxε0dx≥1,foranyn∈ℕ.
On the one hand, based on (26), for a.e. x∈Ω, we have
(29)Φ¯x,ηnx−ηxε0⟶0,asn⟶∞.
On the other hand, via (13), Φ¯∈Δ2, and the convexity of Φ¯, we can deduce that
(30)Φ¯x,ηnx−ηxε0≤K~Φ¯x,a1x+Φx,unx+Φx,ux+Φx,ξx+hx,
for anyn∈ℕ and a.e. x∈Ω, where K~ is a positive constant and h∈L1Ω. Since Φ≼Ψ and Ψ∈Δ2, combing (26), (30), and (23), we obtain, by Lebesgue’s theorem, that
(31)∫ΩΦ¯x,ηnx−ηxε0dx⟶0,asn⟶∞.
It is a contradiction.
For every nonnegative functionν∈DΩ, in view of (A2), we have
(32)∫Ωζnx−ηnxDunx−ξνxdx≥0.
Lettingn⟶∞, and combining (18), (24), and (27), we have
(33)∫Ωζnx−ηnxDunx−ξνxdx⟶∫Ωζx−ηxDux−ξνxdx.
Thanks to (32), we get
(34)∫Ωζx−ηxDux−ξνxdx≥0.
Sinceν is arbitrary, we get ζx−ηxDux−ξ≥0 for a.e. x∈Ω. Based on (A1) and Theorem III.9 of [33], there exists a sequence of measurable selections σn of A·,u,ξ, such that σnx:n∈ℕ is dense in Ax,ux,ξ. It implies that ζx−ϑDux−ξ≥0 for a.e. x∈Ω and for every ϑ∈Ax,ux,ξ. Thanks to (A2), ζx∈Ax,ux,Dux for a.e. x∈Ω. Consequently, u∗∈Au.
(iv)
To prove (iv), we need to show that ifu∈W01LΦΩ, un⊂W01LΦΩ with(35)un⟶ustronglyinW01LΦΩ,asn⟶∞,and V⊂W01LΦΩ∗ is a weakly open set such that Au⊂V; then, we can find n0∈ℕ such that Aun⊂V for all n≥n0. If we suppose not, then there exists a subsequence uk of un such that Auk∩VC≠∅. Let wk∈Auk∩VC. Based on (35), un is bounded in W01LΦΩ. In view of (13), Aun is bounded in W01LΦΩ∗. It follows that wk is bounded in W01LΦΩ∗. Consequently, by passing to a subsequence if necessary, there exists w∈W01LΦΩ∗ such that wk⇀w weakly in W01LΦΩ∗ for σΠLΦ¯,ΠEΦ, as k⟶+∞. Based on (iii), w∈Au. Since W01LΦΩ is reflexive, w∈VC. It is a contradiction to the choice of V.Theorem 14.
Under assumptions (A1)–(A4), the mappingA is pseudomonotone.Proof.
Let sequencesun⊂W01LΦΩ and un∗⊂W01LΦΩ∗ satisfy
(36)un⇀u0weaklyinW01LΦΩforσ∏LΦ,∏EΦ¯,(37)un∗⇀u0∗weaklyinW01LΦΩ∗forσ∏LΦ¯,∏EΦ,(38)limsupn⟶∞un∗,un−u0≤0.
We will show thatu0∗∈Au0 and
(39)limn⟶∞un∗,un=u0∗,u0.
For eachu∈W01LΦΩ, we define the multivalued operator Bu:W01LΦΩ⟶2W01LΦΩ∗, by
(40)Buv≔−divη:ηx∈Ax,ux,Dvx,fora.e.x∈Ω,∀v∈W01LΦΩ.
Using the similar and more simple proof ((A3) is not needed in this case), we can deduce that Lemma13 holds if A is replaced by Bu. Moreover, based on (A2), Bu is monotone. Therefore, it is maximal monotone. Clearly, Buu=Au, for any u∈W01LΦΩ.
For eachn∈ℕ, there exists ζn∈A~u0 such that un∗=−divζn. From (36), un is bounded in W01LΦΩ. Then, by (13), ζn is bounded in LΦ¯ΩN. Therefore, there exists ζ∈LΦ¯ΩN such that, by extracting a subsequence if necessary, ζn⇀ζ in LΦ¯ΩN weakly for σLΦ¯ΩN,EΦΩN, as n⟶∞. Analogous with the proof in Lemma 13, we have −divζ=u0∗.
Letv∈W01LΦΩ and η∈LΦ¯ΩN be such that ηx∈Ax,u0x,Dvx for a.e. x∈Ω. Since W01LΦΩ↪LΨΩ is compact, in view of (36), there exists a subsequence of un still denoted by un such that
(41)un⟶u0stronglyinLΨΩ,asn⟶∞.
By passing to a further subsequence if necessary, we can suppose(42)un⟶u0a.e.inΩ,asn⟶∞,(43)Ψx,2unx−u0x≤h0x,fora.e.x∈Ω,
for someh0∈L1Ω (see [32]). Since Ψ∈Δ2 and Ψ is convex with respect to the second variable,
(44)Ψx,unx≤12Ψx,2unx−u0x+12Ψx,2u0x≤12h0x+12Ψx,2u0x,
for a.e.x∈Ω. According to (A3), there exists a sequence ηn such that
(45)ηn⟶ηa.e.inΩ,asn⟶∞,
andηn is an LΩ-measurable selection of x∈Ω⟶Ax,unx,Dvx⊂ℝN, for every n∈ℕ. Since Φ≼Ψ and Ψ∈Δ2, similar to the proof of (27), we have
(46)ηn⟶ηstronglyinLΦ¯ΩN,asn⟶∞.
For eachn∈ℕ, from the monotonicity of Ax,unx,· for a.e. x∈Ω, we obtain
(47)0≤limsupn⟶∞∫Ωζnx−ηnxDunx−Dvxdx=limsupn⟶∞un∗,un−u0+∫ΩζnxDu0x−Dvxdx−∫ΩηnxDunx−Dvxdx=limsupn⟶∞un∗,un−u0+∫Ωζx−ηxDu0x−Dvxdx≤−divζ−−divη,u0−v.
SinceBu0 is maximal monotone, we get that u0∗=−divζ∈Bu0u0=Au0.
Thanks to (A3), there existsνn such that νn⟶ζ a.e. in Ω, as n⟶∞, and νn is an LΩ-measurable selection of x∈Ω⟶Ax,unx,Dux⊂ℝN, for each n∈ℕ. Then, by using the monotonicity of Ax,unx,· for a.e. x∈Ω, we have
(48)un∗,un≥∫ΩηnxDunx−Du0xdx+∫ΩζnxDu0xdx.
It yields thatliminfn⟶∞un∗,un≥u0∗,u0. Consequently, based on (38), we can deduce that (39) holds. Based on Lemma 8, A is pseudomonotone.


## 3.2. The Multivalued Convection Term

We defineΦ−t≔essinfx∈ΩΦx,t, for any t≥0 (see [26]). By the condition (Φ1), we obtain that Φ−0,+∞≢0 and Φ−0,+∞≢+∞. According to Lemma 2.5.16 of [26], Φ− is an N-function. Hence, Φ−t>0 for t>0. Moreover, analogous to the proof of Lemma 2.5.24 of [26], we can obtain that Φ−∈Δ2 whenever Φ∈Δ2. Then, there exists a constant Kd>0 such that
(49)Φ−ds≤KdΦ−s,∀s≥s0,for some s0≥0, where d is twice the diameter of Ω.LetF:Ω×ℝ×ℝN⟶Pfcℝ satisfy the following conditions:(F1)F is graph measurable.(F2) For a.e.x∈Ω, the multivalued mapping Fx,·,·:ℝ×ℝN⟶2ℝ is upper semicontinuous.(F3) There exist a functiona3∈LΨ¯Ω and a constant b3>0 such that
(50)supf:f∈Fx,s,ξ≤a3x+b3Ψ¯−1x,Ψx,s+b3Ψ¯−1x,Φx,ξ,for a.e. x∈Ω, all s∈ℝ and all ξ∈ℝN.(F4) There exist a functionω∈L1Ω and two constants b4,b5≥0 such that
(51)fs≤ωx+b4Φ−s+b5Φx,ξ,for a.e. x∈Ω, all s∈ℝ and all ξ∈ℝN, and for all f∈Fx,s,ξ. Moreover,
(52)b2−λ~−1b4−b5>0,where
(53)λ~=infu∈Dℏ∫ΩΦ−Duxdx∫ΩΦ−uxdx,with Dℏ=u∈W01LΦ−Ω\0:∫ΩΦ−Duxdx≥ℏ for some ℏ>0.Remark 15.
WhenDux≢0 for a.e x∈Ω, we have ∫ΩΦ−Duxdx>0.Remark 16.
From Lemma12 and (49),
(54)∫ΩΦ−uxdx≤Kd∫ΩΦ−Duxdx+measΩ·Φ−ds0,which yields that
(55)∫ΩΦ−Duxdx∫ΩΦ−uxdx≥∫ΩΦ−DuxdxKd∫ΩΦ−Duxdx+measΩ·Φ−ds0.
Hence,λ~≥ℏ/Kdℏ+measΩ·Φ−ds0>0.Example 17.
Assume thata,b,c,d∈L∞Ω, e,k∈LΨ¯Ω, ax≤bx, cx≤dx, and ex≤kx, for a.e. x∈Ω, and
(56)b2−λ~−1maxaL∞Ω,bL∞Ω−maxcL∞Ω,dL∞Ω>0.
PutFx,s,ξ=−d1Φ−s/s2s+d2ς+υ:d1∈ax,bx,d2∈cx,dx,ς∈Φ¯−1x,Φx,ξ,Ψ¯−1x,Φx,ξ,υ∈ex,kx, for x∈Ω, s∈ℝ, and ξ∈ℝN. Then, F satisfies (F1)–(F4).For eachu∈W01LΦΩ, let
(57)F~u=f∈L0Ω:fx∈Fx,ux,Dux,fora.e.x∈Ω.It follows from (F1) thatF~u≠∅ whenever u∈W01LΦΩ. Moreover, from (F3), we can see that F~u⊂LΨ¯Ω whenever u∈W01LΦΩ.LetF:W01LΦΩ⟶2W01LΦΩ∗ be defined by Fu=f^∈W01LΦΩ∗:f∈F~u, where f^ is defined for each f∈LΨ¯Ω by
(58)f^,v=∫Ωfxvxdx,∀v∈W01LΦΩ.Analogous to the proof of Theorem 6.4.16 of [27], we have the following lemma, whose proof is omitted.Lemma 18.
Letw:Ω×ℝ⟶ℝ¯=ℝ∪±∞ be a LΩ×Bℝ-measurable function, T:Ω⟶2ℝ\∅ be graph-measurable, and the integral functional
(59)Iwv=∫Ωwx,vxdx,v∈LΦΩbe defined for all v∈SΦT=v∈LΦΩ:vx∈Txfora.e.x∈Ω, where Φ∈NΩ. If there exists v0∈LΦΩ such that Iwv0>−∞, then
(60)supv∈SΦTIwv=∫Ωsupη∈Txwx,ηdx.If there existsv0∈LΦΩ such that Iwv0<+∞, then x⟶infη∈Txwx,η is (Lebesgue) measurable and
(61)infv∈SΦTIwv=∫Ωinfη∈Txwx,ηdx.Now, we give the following lemma abouth-upper semicontinuous of multivalued mappings.Lemma 19.
Assume that (F1)–(F3) hold. Then,(i)
F~u∈PfcLΨ¯Ω for all u∈W01LΦΩ;(ii)
F~ is bounded;(iii)
F~:W01LΦΩ⟶PfcLΨ¯Ω is h-upper semicontinuous; that is, if un⟶u0 strongly in W01LΦΩ, then(62)h∗F~un,F~u0=supv′∈F~uninfv∈F~u0v′−vΨ¯⟶0,as n⟶∞.Proof.
(i)
The proof of the convexity and closedness ofF~u is a direct consequence of the fact that Fx,s,ξ is convex, closed, and bounded in ℝ for a.e. x∈Ω, all s∈ℝ, and all ξ∈ℝN.(ii)
The boundedness ofF~ follows directly from (F3).(iii)
The following proof is referred to Theorem 2.7 of [13]. Assume that un⊂W01LΦΩ is a sequence such that(63)un⟶u0stronglyinW01LΦΩ,asn⟶∞,
for someu0∈W01LΦΩ. We will show that
(64)supv′∈F~uninfv∈F~u0v′−vΨ¯⟶0,asn⟶∞.
SinceW01LΦΩ↪LΨΩ is compact,
(65)un⟶u0stronglyinLΨΩ,asn⟶∞.
It follows from (65) and (63) that there exists a subsequence of un still denoted by un and h1,h2∈L1Ω such that
(66)unx⟶u0x,Dunx⟶Du0x,fora.e.x∈Ω,asn⟶∞,
andΨx,2unx−u0x≤h1x, Φx,2Dunx−Du0x≤h2x, for a.e. x∈Ω. Hence,
(67)Ψx,unx≤12h1x+12Ψx,2u0x,Φx,Dunx≤12h2x+12Φx,2Du0x,
for a.e.x∈Ω. Based on (F2) and Lemma 5, the multivalued mapping Fx,·,·:ℝ×ℝN⟶2ℝ is h-upper semicontinuous, for a.e. x∈Ω. It follows that supξ∈Fx,unx,Dunxinfη∈Fx,u0x,Du0xξ−η⟶0 as n⟶∞, for a.e. x∈Ω, and thus,
(68)supξ∈Fx,unx,Dunxinfη∈Fx,u0x,Du0xΨ¯x,ξ−η⟶0asn⟶∞,for a.e. x∈Ω.
On the other hand, sinceΨ,Ψ¯∈Δ2, based on (F3) and (67), there exist some constants K1,K2>0 and a function h∈L1Ω such that
(69)Ψ¯x,ξ−η≤K12Ψ¯x,ξ+Ψ¯x,η+hx≤K1K262Ψ¯x,a3x+Ψx,unx+Φx,Dunx+Ψx,u0x+Φx,Du0x+K1+1hx≤K1K262Ψ¯x,a3x+12h1x+12Ψx,2u0x+12h2x+12Φx,2Du0x+Ψx,u0x+Φx,Du0x+K1+1hx≔Hx,for all n∈ℕ, a.e. x∈Ω, and all ξ∈Fx,unx,Dunx, η∈Fx,u0x,Du0x. Thus,
(70)supξ∈Fx,unx,Dunxinfη∈Fx,u0x,Du0xΨ¯x,ξ−η≤Hx,
for alln∈ℕ, a.e. x∈Ω. Evidently, the function H belongs to L1Ω. Using Lebesgue’s dominated convergence theorem, we deduce that
(71)∫Ωsupξ∈Fx,unx,Dunxinfη∈Fx,u0x,Du0xΨ¯x,ξ−ηdx⟶0asn⟶∞.
Letn∈ℕ. Based on (F1), F is graph measurable. For v′∈F~un, using Lemma 18, we can obtain that
(72)infv∈F~u0∫ΩΨ¯x,v′x−vxdx=∫Ωinfη∈Fx,u0x,Du0xΨ¯x,v′x−ηdx.
Using Lemma18 again, we get
(73)supv′∈F~un∫Ωinfη∈Fx,u0x,Du0xΨ¯x,v′x−ηdx=∫Ωsupξ∈Fx,unx,Dunxinfη∈Fx,u0x,Du0xΨ¯x,ξ−ηdx.
Lettingn⟶∞, and combining (72), (73), and (71), we have
(74)supv′∈F~uninfv∈F~u0∫ΩΨ¯x,v′x−vxdx⟶0.
SinceΨ¯∈Δ2, we can deduce (64), which completes our proof.Then, we have the following theorem.Theorem 20.
IfF satisfies (F1)–(F3), then
(i)
F is bounded;(ii)
GrF is (sequentially) closed in W01LΦΩ×W01LΦΩ∗ with respect to the strong-weak topology.Proof.
(i)
It is easy from (F3) to see thatF is bounded.(ii)
We show thatGrF is (sequentially) closed in W01LΦΩ×W01LΦΩ∗ with respect to the strong-weak topology. Assume that un and f^n are sequences in W01LΦΩ and W01LΦΩ∗, respectively, such that(75)un⟶ustronglyinW01LΦΩ,(76)f^n⇀f^weaklyinW01LΦΩ∗forσΠLΦ¯Ω,ΠEΦΩ,(77)f^n∈Fun,∀n∈ℕ.
Let us prove that(78)f^∈Fu.
Thanks to (77), there exists fn∈F~un such that, for every n∈ℕ,
(79)f^n,v=∫Ωfnxvxdx,∀v∈W01LΦΩ.
From (75) and Lemma 19 (iii), we obtain that h∗F~un,F~u⟶0, and thus,
(80)infw∈F~ufn−wΨ¯⟶0,asn⟶∞.
Consequently, there is a sequencewn⊂F~u such that
(81)fn−wnΨ¯⟶0,asn⟶∞.
SinceF~u is bounded in LΨ¯Ω, by passing to a subsequence if necessary, there exists w∈LΨ¯Ω such that
(82)wn⇀wweaklyinLΨ¯ΩforσLΨ¯Ω,EΨΩ,asn⟶∞.
According to Lemma19 (i), w∈F~u. Letting n⟶∞, and combing (81) and (82), we get that
(83)fn⇀wweaklyinLΨ¯ΩforσLΨ¯Ω,EΨΩ.
It implies thatf^,v=∫Ωwxvxdx,∀v∈W01LΦΩ. Therefore, f^∈Fu.


## 3.3. Existence Theorem for Inclusion Problems

A functionu∈W01LΦΩ is called a (weak) solution of problem (1), if there exist ζ∈LΦ¯ΩN and f∈LΨ¯Ω such that
(84)ζx∈Ax,ux,Dux,fx∈Fx,ux,Dux,fora.e.x∈Ω,∫ΩζxDvxdx+∫Ωfxvxdx=∫ΩLxvxdx,∀v∈W01LΦΩ.In this section, we are going to prove an existence result for our problem (1).Theorem 21.
LetA and f satisfy (A1)–(A4) and (F1)–(F4); then, there exists a solution for problem (1).Proof.
We define the multivalued mappingT:W01LΦΩ⟶2W1LΦΩ∗ by
(85)Tu=Au+Fu.
Then,T is well-defined.
Step 1. We are going to show thatT is pseudomonotone. Let sequences un⊂W01LΦΩ and ωn⊂W01LΦΩ∗ have the properties
(86)un⇀uweaklyinW01LΦΩforσ∏LΦ,∏EΦ¯,asn⟶∞,(87)ω^n⇀ω^weaklyinW01LΦΩ∗forσ∏LΦ¯,∏EΦ,asn⟶∞,with ω^n∈Tun and limsupn⟶∞ω^n,un−u≤0. From (85), there exist ζ^n∈Aun, f^n∈Fun such that ω^n=ζ^n+f^n, for every n∈ℕ, where ζ^n=−divζn with ζn∈A~un and f^n satisfies f^n,v=∫Ωfnxvxdx, ∀v∈W01LΦΩ, for some fn∈F~un.
Now, we prove that(88)un⟶ustronglyinW01LΦΩ,asn⟶∞.
To show (88), based on Proposition 1.9 of [17], it needs to be proven that
(89)Dun⟶DustronglyinLΦΩN,asn⟶∞.
We suppose that there exitsε0>0 such that
(90)Dun−DuΦ≥ε0,foranyn∈ℕ.
Then,(91)∫ΩΦx,Dunx−Duxε0dx≥1,foranyn∈ℕ.
From (86), the sequence un is bounded in W01LΦΩ, and by passing to a subsequence if necessary,
(92)un⟶ustronglyinLΨΩ,(93)un⟶ua.e.inΩ,
asn⟶∞. By passing to a further subsequence if necessary, there exists h0∈L1Ω such that
(94)Ψx,unx≤12h0x+12Ψx,2ux,
for a.e.x∈Ω. Thanks to (F3), fn is bounded in LΨ¯Ω. It follows that
(95)f^n,un−u=∫Ωfnxunx−uxdx≤2fnΨ¯un−uΨ⟶0,
asn⟶∞. Consequently, limsupn⟶∞ζ^n,un−u≤0. By (13), ζ^n is bounded in W01LΦΩ∗. Hence, by extracting a subsequence if necessary, there exists ζ^∈W01LΦΩ∗ such that
(96)ζ^n⇀ζ^weaklyinW01LΦΩ∗forσ∏LΦ¯,∏EΦ,asn⟶∞.
From Theorem14, A is pseudomonotone; then, we get ζ^∈Au and
(97)limn⟶∞ζ^n,un=ζ^,u.
Letη be a measurable selection of A·,u,Du. Based on (93) and (A3), there exists a sequence ηn converging a.e. to η in Ω, as n⟶∞, such that ηn is an LΩ-measurable selection of the map x∈Ω⟶Ax,unx,Dux∈2ℝN, for every n∈ℕ. Same as the proof of (46), we can obtain that
(98)ηn⟶ηstronglyinLΦ¯ΩN,asn⟶∞.
Denoteη^n=−divηn and η^=−divη.
On the one hand, combing (86), (96), (97), and (98), we can obtain that
(99)ζ^n−η^n,un−u=ζ^n,un−ζ^n,u−η^n−η^,un+η^n,u−η^,un⟶0,
asn⟶∞. On the other hand, for a.e. x∈Ω, dnx≔ζnx−ηnxDunx−Dux≥0. Hence, dn⟶0 in L1Ω. By passing to a subsequence if necessary,
(100)dnx⟶0fora.e.x∈Ω,asn⟶∞.
SinceΦ∈Δ2 and Φ≼Ψ, by using (13), (14), (94), (100), and the Young inequality, one has Φx,Dunx≤ℓx, for some function ℓ∈L1Ω, for all n∈ℕ, for a.e. x∈Ω. It implies that Dunx is bounded in ℝN, for a.e. x∈Ω. Taking a subsequence if necessary, there exists ϑ∈ℝN such that Dunx⟶ϑ as n⟶∞, for a.e. x∈Ω. Similar to Lemma 1.4 of [34], we can show that ϑ∈LΦΩN and Dun⟶ϑ for σLΦΩN,LΦ¯ΩN, as n⟶∞. Therefore, ϑ=Du, and thus, Dunx⟶Dux as n⟶∞, for a.e. x∈Ω. Based on Lebesgue’s Theorem, we can show that
(101)∫ΩΦx,Dunx−Duxε0dx⟶0,asn⟶∞.
It is a contradiction with (89).
Denotef^=ω^−ζ^. Combing (87) and (96),
(102)f^n⇀f^weaklyinW01LΦΩ∗,forσ∏LΦ¯,∏EΦ,asn⟶∞.
Invoking Theorem20, we obtain that f^∈Fu. Thus, ω^∈Tu. In view of (88), and (87), we get
(103)ω^n,un=ω^n,un−u+ω^n,u⟶ω^,u,asn⟶∞.
Step 2. We will see thatT is coercive. Let u∈W01LΦΩ and ω^∈Tu. Then, there exist ζ^∈Au and f^∈Fu such that ω^=ζ^+f^. From (A4), for any ζ^∈Au, we have that
(104)ζ^,u≥b2∫ΩΦx,Duxdx−a2L1Ω.
For anyf^∈Fu, by (51) and (53), we deduce that
(105)f^,u≤ωL1Ω+b4∫ΩΦ−uxdx+b5∫ΩΦx,Duxdx≤ωL1Ω+b4λ~∫ΩΦ−Duxdx+b5∫ΩΦx,Duxdx,for all u∈Dℏ.
Hence, we obtain that(106)ω^,u≥b2−b4λ~−b5∫ΩΦx,Duxdx−a2L1Ω−ωL1Ω,
for allu∈Dℏ. Based on Proposition 1.9 of [17], there exists a constant C∗>0 such that
(107)u1,Ω≤1+C∗DuΦ.
Then, we get that(108)ω^,uu1,Ω≥b2−b4λ~−b5∫ΩΦx,Duxdx1+C∗DuΦ−a2L1Ω+ωL1Ω1u1,Ω,for all u∈Dℏ.
Based on Proposition 3.1 of [18], there are functions h∈L1Ω with h≥0 and G:k,+∞⟶ℝ for some k>2 such that Gs⟶+∞ as s⟶+∞ and
(109)Φx,su≥sGsΦx,u−shx,foralls≥k,u≥0,a.e.x∈Ω.
It follows that(110)∫ΩΦx,Duxdx≥DuΦ−εGDuΦ−ε∫ΩΦx,DuxDuΦ−εdx−hL1Ω≥DuΦ−εGDuΦ−ε−hL1Ω,if DuΦ is large enough, for any ε∈0,1.
Combining (108) and (110), we can deduce that
(111)limu1,Ω⟶+∞,u∈W01LΦΩinfω^,u:ω^∈Tuu1,Ω=+∞.
Step 3. According to Lemma11, there exists u∈W01LΦΩ such that
(112)ω^,u=L,u,
for someω^∈Tu, where L,u=∫ΩLxuxdx. Therefore, there exist ζ^∈Au and f^∈Fu such that ω^=ζ^+f^. This completes the proof.The set of solutions of problem (1) is denoted by S. Then, we have the following results for set S.Theorem 22.
Assume (A1)–(A4) and (F1)–(F4) hold. Then,S is nonempty, bounded, and closed.Proof.
(1)
According to Theorem21, S is nonempty.(2)
We are going to prove thatS is bounded. Assume that S is unbounded. Then, there exists a sequence un in S such that un1,Ω⟶+∞, as n⟶∞. Since there exists a constant C>0 such that L,un≤CLW01LΦΩ∗un1,Ω, combining (108) and (110), we have that(113)limun1,Ω⟶+∞,un∈W01LΦΩinfω^n−L,un:ω^n∈Tunun1,Ω=+∞.
However, it follows from (112) that ω^n−L,un=0, where ω^n∈Tun, for any n∈ℕ. It is a contradiction.
(3)
We will show thatS is closed. Let un be a sequence in S such that(114)un⟶u0inW01LΦΩ,asn⟶∞,
for someu0∈W01LΦΩ. Then, there exist ζ^n∈Aun and f^n∈Fun such that ω^n=ζ^n+f^n∈Tun and
(115)ω^n,un=L,un,foranyn∈ℕ.
Based on (114), un is bounded in W01LΦΩ. In view of (13), and (F3), ζ^n and f^n are bounded in LΦ¯ΩN and LΨ¯Ω, respectively. Extracting a subsequence if necessary, there exist ζ^0,f^0∈W01LΦΩ∗ such that
(116)ζ^n⇀ζ^0weaklyinW01LΦΩ∗forσ∏LΦ¯,∏EΦ,(117)f^n⇀f^0weaklyinW01LΦΩ∗forσ∏LΦ¯,∏EΦ,as n⟶∞. Using Lemma 13 (iii), we have ζ^0∈Au0. Based on Theorem 20 (ii), we get f^0∈Fu0.
Based on (114),
(118)ζ^n,un−u0⟶0,f^n,un−u0⟶0,(119)L,un−u0⟶0,as n⟶∞.
Based on (115)–(119),
(120)ω^n−L,un⟶ζ^0+f^0−L,u0,asn⟶∞.
Denoteω^0=ζ^0+f^0. Then, ω^0∈Tu0. Consequently, ω^0,u0=L,u0. It implies that u0∈S.


## 3.4. Obstacle Problems

Now, we consider the following inclusion problem with an obstacle effect(121)L∈Ax,u,Du+Fx,u,Du,inΩ,ux≤υx,inΩ,u=0,on∂Ω,whereυ:Ω⟶ℝ+∪+∞ is a given function. Let K=u∈W01LΦΩ:u≤υa.e.inΩ. It is easy to see that K is a nonempty, closed, and convex subset of W01LΦΩ and contains 0. The indicator function IK:W01LΦΩ⟶ℝ∪+∞ of K is defined by
(122)IKu≔0,ifu∈K,+∞,otherwise.Let∂IK be the subdifferential of IK in the sense of convex analysis (see Definition 32.11 of [35]).A functionu∈K is called a (weak) solution of problem (121), if there exist ζ∈LΦ¯ΩN and f∈LΨ¯Ω such that
(123)ζx∈Ax,ux,Dux,fx∈Fx,ux,Dux,fora.e.x∈Ω,(124)∫ΩζxDvxdx+∫Ωfxvxdx=∫ΩLxvxdx,∀v∈K.From the definition above, it is not difficult to show that problem (121) is equivalent to the following variational inequality:Findu∈W01LΦΩ such that
(125)Au,v−u+Fu,v−u+IKv−IKu≥L,v−u,∀v∈W01LΦΩ,which is formulated as the following inclusion:Findu∈W01LΦΩ such that
(126)L∈Au+Fu+∂IKu.We have the following theorem.Theorem 23.
Assume that (A1)–(A4) and (F1)–(F4) hold. Then, there exists a solutionu of problem (121) such that u1,Ω<R for some R>0.Proof.
LetT be defined by (85). Then, based on the proof of Theorem 21, T is pseudomonotone.
According to Example 32.15 of [35], ∂IK is maximal monotone.
We are going to check that there exists a positive constantR such that
(127)ω^+η^−L,u>0,for all u∈W01LΦΩ with u1,Ω=R, for all ω^∈Tu and all η^∈∂IKu.
Letu∈W01LΦΩ with DuΦ>k, where k is the constant in (109). Based on Remark 15, there exists ℏ>0 such that u∈Dℏ. From (106) and (110), we obtain that
(128)ω^,u≥b2−b4λ~−b5DuΦ−εGDuΦ−ε−hL1Ω−a2L1Ω−ωL1Ω,for any ε∈0,1 and for all ω^∈Tu.
Similar to the proof of (2.50) in [36], there exists w∈W01LΦΩ∗ and αK∈ℝ such that
(129)IKu≥−C~wW01LΦΩ∗u1,Ω−αK,∀u∈W01LΦΩ,for some constant C~>0.
Combining (107), (128), and (129), and based on the arbitrariness of ε, we have
(130)ω^+η^−L,u≥b2−b4λ~−b5DuΦGDuΦ−b2−b4λ~−b5hL1Ω−a2L1Ω−ωL1Ω−C~wW01LΦΩ∗u1,Ω−αK−CLW01LΦΩ∗u1,Ω≥DuΦb2−b4λ~−b5GDuΦ−1+C∗C~wW01LΦΩ∗+CLW01LΦΩ∗−b2−b4λ~−b5hL1Ω−a2L1Ω−ωL1Ω−αK,
for allω^∈Tu and all η^∈∂IKu, where C is a positive constant. It is easy to see that we can take R0>k to be large enough such that
(131)R0b2−b4λ~−b5GR0−1+C∗C~wW01LΦΩ∗+CLW01LΦΩ∗−b2−b4λ~−b5hL1Ω−a2L1Ω−ωL1Ω−αK>0.
LetR≥1+C∗R0. Then, (127) holds for all u∈W01LΦΩ with u1,Ω=R, for all ω^∈Tu and all η^∈∂IKu. Using Theorem 2.2 of [14], inclusion (126) has a solution u such that u1,Ω<R.We denote the set of solutions of inclusion (126) based on SK. With the analogous proof of Theorem 22, we can deduce the following theorem.Theorem 24.
Assume that (A1)–(A4) and (F1)–(F4) hold. Then,SK is nonempty, bounded, and closed.



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*Source: 8531992-2023-03-23.xml*