# Barrier Solutions of Elliptic Differential Equations in Musielak-Orlicz-Sobolev Spaces

Authors: Ge Dong; Xiaochun Fang
Journal: Journal of Function Spaces
 (2021)

---

## Abstract

In this paper, we study the solution set of the following Dirichlet boundary equation:−diva1x,u,Du+a0x,u=fx,u,Du in Musielak-Orlicz-Sobolev spaces, where a1:Ω×ℝ×ℝN⟶ℝN, a0:Ω×ℝ⟶ℝ, and f:Ω×ℝ×ℝN⟶ℝ are all Carathéodory functions. Both a1 and f depend on the solution u and its gradient Du. By using a linear functional analysis method, we provide sufficient conditions which ensure that the solution set of the equation is nonempty, and it possesses a greatest element and a smallest element with respect to the ordering “≤,” which are called barrier solutions.


---

## Body

## 1. Introduction

LetΩ⊂ℝN be a bounded domain with Lipschitz boundary. Fan [1] established a sub-supersolution method for the following differential equations of divergence form:
(1)−diva1x,Du+a0x,u=fx,u,in Ω coupled with Neumann or Dirichlet boundary condition in reflexive Musielak-Orlicz-Sobolev spaces W01LΦΩ and obtained the existence of greatest and smallest solutions within the order interval u¯,u¯, where u¯≔maxu¯1,u¯2,⋯,u¯k, u¯≔minu¯1,u¯2,⋯,u¯m, and u¯1,u¯2,⋯,u¯k and u¯1,u¯2,⋯,u¯m are subsolutions and supersolutions, respectively. Dong and Fang [2] studied the existence of weak solutions for the following equations:
(2)−diva1x,Du+a0x,u=fx,u,Du,in Ω coupled with Neumann or Dirichlet boundary condition in separable Musielak-Orlicz-Sobolev spaces and obtained the existence of greatest and smallest solutions within the order interval u¯,u¯. Carl and Le [3] established the enclosure of solutions, i.e., u¯,u¯, for multivalued elliptic variational inequalities by using the sub-supersolution method in Sobolev spaces. However, Carl [4] pointed out that there may exist some more solutions of some elliptic equations which are not within the corresponding interval u¯,u¯. For example, the solution set S of the following Dirichlet boundary value problem
(3)−Δu−λu=0,inΩ,u=0,on∂Ω,is given by S=u=cφ1:c∈ℝ, where φ1 is the eigenfunction belonging to λ1, and λ1>0 is the first eigenvalue of the negative Laplacian −Δ (see [4]). Then, S≠∅ neither is norm bounded nor has the extreme solutions. Therefore, Carl [4] provided some sufficient conditions which ensure that the solution set of an elliptic variational inequality possesses a greatest element u∗ and a smallest element u∗ with respect to the natural partial ordering of functions in Sobolev spaces. Here, u∗ and u∗ are called barrier solutions (see [4]). In [5], Chipot et al. established the existence of the minimal solution to some elliptic variational inequalities in Sobolev spaces.It is well known that different classes of differential equations correspond to different function space settings. The Musielak-Orlicz-Sobolev spaces are the extensions of Sobolev spaces, variable exponent Sobolev spaces, and Orlicz-Sobolev spaces. Gwiazda et al. [6] proved the existence of renormalized solutions to the following nonlinear elliptic equations:
(4)−divAx,Du=fin Ω coupled with Dirichlet boundary condition in Musielak-Orlicz-Sobolev spaces, where A:Ω×ℝN⟶ℝN is a Carathéodory function and f∈L1Ω. The growth of A is controlled by Musielak-Orlicz function Φ:
(5)Ax,ξξ≥cAΦx,ξ+Φ¯x,Ax,ξ,for all x∈Ω and ξ∈ℝN, where Φ¯ is a complementary function to Φ, the constant cA∈0,1. Although they did not require Φ∈Δ2 nor Φ¯∈Δ2, condition (5) is stronger than the growth condition (3.1) and the coercive condition (3.2) in [2]. In [7], Chlebicka and Karppinen studied removable sets for elliptic equations of the form
(6)−divAx,Du=0in Ω, where A satisfies the nonstandard growth condition and the coercivity condition expressed by means of a Musielak-Orlicz function.The purpose of this paper is to provide sufficient conditions which ensure that the solution setS of the following Dirichlet boundary problem
(7)−diva1x,u,Du+a0x,u=fx,u,Du,inΩ,u=0,on∂Ω,is nonempty and possesses barrier solutions in Musielak-Orlicz-Sobolev spaces, where a1:Ω×ℝ×ℝN⟶ℝN, a0:Ω×ℝ⟶ℝ, and f:Ω×ℝ×ℝN⟶ℝ are Carathéodory functions. Both a1 and f depend on the solution u and its gradient Du.This paper is organized as follows: Section2 contains some preliminaries and some technical lemmas which will be needed. In Section 3, we first give the sufficient conditions which ensure the existence of weak solutions for (7) in Musielak-Orlicz-Sobolev spaces. Some examples are given for the application of our results. We point that (5) is strong. Then, we provide some sufficient conditions which ensure that the solution set of (7) is bounded, compact, and direct. Finally, we provide some sufficient conditions which ensure that the solution set of (7) possesses barrier solutions in Musielak-Orlicz-Sobolev spaces.We refer to some results in variable exponent Sobolev or Orlicz-Sobolev spaces [8–22].


## 2. Preliminaries

Now, we list briefly some definitions and facts about Musielak-Orlicz-Sobolev spaces; for more details, see [1, 2, 23–25].Φ:Ω×ℝ+=0,+∞⟶ℝ+ is called a generalized N-function (i.e., 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(8)limu⟶0+supx∈ΩΦx,uu=0,limu⟶+∞infx∈ΩΦx,uu=+∞(ii)
Φx,u is a measurable function of x for all u≥0Equivalently, for allx∈Ω and all u≥0, 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,t is a right-continuous and nondecreasing function of t≥0, φx,0=0, φx,t>0 for t>0, and φx,t⟶+∞ as t⟶+∞ for every x∈Ω. The complementary function Φ¯ to Φ is defined as follows:
(9)Φ¯x,v=supu≥0uv−Φx,uforallv≥0,x∈Ω.Then,Φ¯∈NΩ 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.Moreover, one has the Young inequality:uv≤Φx,u+Φ¯x,v, for all u,v≥0,x∈Ω.Φ is said to satisfy the Δ2 condition (Φ∈Δ2, for short), if there exists a positive constant K>1 and a nonnegative function h∈L1Ω such that Φx,2u≤KΦx,u+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 t0>0, there exists a c=ct0>0 such that
(10)Φx,tt≥c,(11)Φ¯x,tt≥c,for all t≥t0, and all x∈Ω.Clearly, (10) implies (Φ1).LetΦ∈NΩ. We denote by L0Ω the set of all (equivalence classes of) Lebesgue measurable functions from Ω to ℝ. The Musielak-Orlicz space (i.e., the generalized Orlicz space) LΦΩ is defined by
(12)LΦΩ=u∈L0Ω:ρΦuλ≔∫ΩΦx,uxλdx<∞,forsomeλ>0,with the (Luxemburg) norm
(13)uΦ=infλ>0:ρΦuλ≤1.Moreover, the set(14)KΦΩ=u∈L0Ω:ρΦu<+∞will be called the Musielak-Orlicz class (i.e., the generalized Orlicz class). A function u∈L0Ω will be called a finite element of LΦΩ, if λu∈KΦΩ for every λ>0. The space of all finite elements of L0Ω will be 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.IfΦ is locally integrable and satisfy (10), then EΦΩ∗=LΦ¯Ω. Moreover, if Φ¯ is also locally integrable satisfying (11), and Φ,Φ¯∈Δ2, then LΦΩ is reflexive.The Musielak-Orlicz-Sobolev spaceW1LΦΩ is defined by
(15)W1LΦΩ=u∈LΦΩ:∀α≤1,Dαu∈LΦΩ,where α=α1,⋯,αN with nonnegative integers αi, i=1,⋯,N, α=α1+α2+⋯+αN and Dαu denote the distributional derivatives.Let(16)ρΦu=∑α≤1∫ΩΦx,Dαuxdx,uΦ,Ω=infλ>0:ρΦuλ≤1,for u∈W1LΦΩ. Then, uΦ,Ω is a norm on W1LΦΩ. The pair W1LΦΩ,uΦ,Ω is a Banach space if Φ is locally integrable and satisfies (Φ1).IfΦx,u=Φu is independent of x, then W1LΦΩ becomes the Orlicz-Sobolev space. Let p∈C+Ω¯, where C+Ω¯=p:Ω¯⟶ℝ is continuous, px>1,∀x∈Ω¯. If Φx,u=upx, then W1LΦΩ becomes the variable exponent Sobolev space
(17)W1,pxΩ=u∈LpxΩ∣Du∈LpxΩN,where LpxΩ=u∈L0Ω∫Ωuxpxdx<+∞. LpxΩ is equipped with the Luxemburg norm
(18)uLpxΩ=infλ>0:∫Ωuxλpxdx≤1u∈LpxΩ,and W1LΦΩ is equipped with the norm
(19)uW1,pxΩ=uLpxΩ+DuLpxΩu∈W1,pxΩ.It is easy to see that(20)W1LΦΩ=u∈LΦΩ:Du∈LΦΩ.DenoteDuΦ=DuΦ and u1,Φ=uΦ+DuΦ. Clearly, u1,Φ is equivalent to uΦ,Ω.The spaceW1LΦΩ can be identified with a subspace of the product of N+1 copies of LΦΩ. Denote the product by ∏LΦ. The subspace is σ∏LΦ,∏EΦ¯ closed. Let W01LΦΩ be the σ∏LΦ,∏EΦ¯ closure of the Schwartz space DΩ in W1LΦΩ.The spaceW1EΦΩ can be identified with a subspace of the product of N+1 copies of EΦΩ. Denote the product by ∏EΦ. Let W01EΦΩ be the (norm) closure of DΩ in W1LΦΩ.Lemma 1 (see [23], Lemma 14.4).
LetΦ∈NΩ, Φ¯ be a complementary function to Φ, and Φ−1,Φ¯−1 be the inverse functions to Φ,Φ¯, respectively. Then,
(21)u≤Φ−1x,uΦ¯−1x,u≤2u,foreveryu≥0,x∈Ω.By Lemma1, we can deduce the following lemma.Lemma 2.
LetΦ∈NΩ, Φ¯ be a complementary function to Φ and Φ−1,Φ¯−1 be the inverse functions to Φ,Φ¯, respectively. Then,
(22)Φx,uu≤Φ¯−1x,Φx,u≤2Φx,uu,foreveryu>0,x∈Ω.Lemma 3 (see [26], Lemma 2.1).
Ifu∈W1LΦΩ, then u+, u−∈W1LΦΩ and
(23)Du+=Du,ifu>0,0,ifu≤0,Du−=0,ifu≥0,−Du,ifu<0.Here,u+=maxu,0, u−=−minu,0.


## 3. Existence Theorems

LetΩ⊂ℝN be a bounded domain with Lipschitz boundary. Let Φ∈NΩ, Φ¯ be a complementary function to Φ with Φ¯∈Δ2, Φ and Φ¯ be locally integrable. Suppose that the embedding W01LΦΩ↪LΦΩ is compact.Foru, v∈L0Ω, we use the standard notations: u∧v=minu,v, u∨v=maxu,v, u+≔u∨0, u−≔−u∧0, u≤v⇔ux≤vx for a.e. x∈Ω. According to [1] (Remark 3.1), u∨v,u∧v∈W1LΦΩ, for any u,v∈W1LΦΩ. Clearly, u∨v,u∧v∈W01LΦΩ, if u,v∈W01LΦΩ.Leta1:Ω×ℝ×ℝℕ⟶ℝℕ be a Carathéodory function satisfying the following conditions:(A1) For a.e. x∈Ω, all s∈ℝ and all ξ,η∈ℝN,
(24)a1x,s,ξ≤g1x+b1Φ¯−1x,Φx,k1s+b2Φ¯−1x,Φx,k2ξ,(25)a1x,s,ξξ≥b3Φx,ξ−g2x,(26)a1x,s,ξ−a1x,s,ηξ−η>0,ξ≠η,where b1,b2,b3,k1,k2>0, g1∈EΦ¯Ω, g1≥0, g2∈L1Ω, and g2≥0.Leta0:Ω×ℝ⟶ℝ be a Carathéodory function satisfying the following conditions:(A0) For a.e. x∈Ω and all s∈ℝ,
(27)a0x,s≤g1x+b1Φ¯−1x,Φx,k1s,(28)a0x,ss≥b3Φx,s−g2x,where b1,b3,k1>0, g1∈EΦ¯Ω, g1≥0, g2∈L1Ω, and g2≥0.Letf:Ω×ℝ×ℝN⟶ℝ be a Carathéodory function satisfying the following growth condition:(F) For a.e. x∈Ω, all s∈ℝ, and all ξ∈ℝN,
(29)fx,s,ξ≤qx+b4Φ¯−1x,Φx,s+b5Φ¯−1x,Φx,ξ,where q∈EΦ¯Ω, b4,b5>0. Denote by F the Nemytskii operator associated to f, that is,
(30)Fux=fx,ux,Dux,forx∈Ω.Consider the following Dirichlet boundary value problem:(31)−diva1x,u,Du+a0x,u=fx,u,Du,inΩ,u=0,on∂Ω.A functionu is called a weak solution of (31) if u∈W01LΦΩ, Fu∈LΦ¯Ω, and u satisfies the equation
(32)∫Ωa1x,u,DuDvdx+∫Ωa0x,uvdx=∫Ωfx,u,Duvdx,forallv∈W01LΦΩ.A functionu is called a subsolution (resp., supersolution) of (31) if u∈W01LΦΩ, Fu∈LΦ¯Ω, and (32) holds with “=” replaced by “≤” (resp., “≥”) for every nonnegative functions v in W01LΦΩ (see [1]).Theorem 4.
LetA1, A0, and F hold. If
(33)2b4+b5<b3,then there exists at least one weak solution of (31).Proof.
DenoteV=W01EΦΩ. Define Γ:V⟶V∗,
(34)Γu,v≔∫Ωa1x,u,DuDvdx+∫Ωa0x,uvdx−∫ΩFuvdx,
∀v∈V. Then, Γ is well defined.
SinceV is separable, there is a sequence wnn=1∞⊂DΩ such that wnn=1∞ is dense in V. Let Vm=spanw1,⋯,wm and consider ΓVm.
Similar to the proof of Theorem 3.1 in [27], u⟶ΓVmu:Vm⟶Vm∗ is continuous.In view of (29), for every u∈V, we have
(35)∫ΩFuudx≤C1∗qΦ¯u1,Φ+2b4+b5∫ΩΦx,udx+b5∫ΩΦx,Dudx,for some constant C1∗>0.Combining (25), (28), (33), and (35), we obtain
(36)Γu,u≥b3−b5∫ΩΦx,Dudx+b3−2b4−b5∫ΩΦx,udx−C−C1∗qΦ¯u1,Φ≥b3−b5∫ΩΦx,Dudx−C−C1∗qΦ¯u1,Φ,for all u∈V, and some constant C>0 independent of u.Consequently,(37)Γu,uu1,Φ≥b3−b5∫ΩΦx,Dudxu1,Φ−Cu1,Φ−C1∗qΦ¯≥b3−b5∫ΩΦx,DudxC2∗DuL1Ω−Cu1,Φ−C1∗qΦ¯,for all u∈Vm, and some constant C2∗>0. Using ([2], Lemma 1), we have
(38)Γu,uu1,Φ⟶+∞,asu1,Φ⟶+∞.By [27] (Remark 2.1), there exists a Galerkin solution um∈Vm for every m∈ℕ such that
(39)Γum,v=0,∀v∈Vm.Using the density ofwn, we deduce that
(40)Γum,v=0,∀v∈V.Therefore, the sequenceum is bounded in V. Consequently, there exists u0∈V, and we can extract a subsequence un of um such that
(41)un⇀u0weaklyinVforσ∏LΦ,∏EΦ¯,(42)un⟶u0stronglyinLΦΩ,(43)unx⟶u0xfora.e.x∈Ω,as n⟶∞. By (24), a1x,un,Dun is bounded in LΦ¯ΩN, and a0x,un and fx,un,Dun are bounded in LΦ¯Ω. Thanks to (42),
(44)∫Ωa0x,unun−u0dx⟶0,∫Ωfx,un,Dunun−u0dx⟶0,as n⟶∞. It follows that
(45)∫Ωa1x,un,DunDun−Du0dx⟶0,as n⟶∞.Therefore, similar to [27], we can deduce that there exists a subsequence of un still denoted by un such that
(46)Dunx⟶Du0xfora.e.x∈Ω,as n⟶∞. Combining (43) and (46), we obtain that
(47)a1x,un,Dun⟶a1x,u0,Du0a.e.inΩ,a0x,un⟶a0x,u0a.e.inΩ,fx,un,Dun⟶fx,u0,Du0a.e.inΩ,as n⟶∞.Using [25] (Lemma 1) and letting n⟶∞, passing to a subsequence if necessary, we can deduce from (40) that
(48)Γu0,v=0,∀v∈V.By [28] (Lemma 2.5) and [25] (Lemma 7),
(49)Γu0,v=0,∀v∈W01LΦΩ.Thus, there exists at least one weak solutionu0 of (31).Example 1.
LetΦx,t=epxt−pxt−1, for x∈Ω and t∈ℝ, where p:Ω⟶ℝ is a measurable function such that 1<p−≔infx∈Ωpx≤px≤p+≔supx∈Ωpx<+∞. Then, Φ∉Δ2.Example 2.
Puta1x,s,ξ=ax,sφx,ξξ/ξ if ξ∈ℝN\0 and a1x,s,ξ=0 if ξ=0, for a.e. x∈Ω, all s∈ℝ, where a is a Carathéodory function on Ω×ℝ satisfying α≤ax,s≤β with α>0 and β>0 for all s∈ℝ, φx,s is the right-hand derivative of Φx,· at s, and 0 is the zero vector in ℝN. Let a0x,s=αφx,ss/s, and fx,s,ξ=bx,sφx,ξs, for a.e. x∈Ω, all s∈ℝ and all ξ∈ℝN, where b is a Carathéodory function on Ω×ℝ satisfying bx,s≤γ with 0<γ<α/3. Then, a1, a0, and f satisfy conditions (A1), (A0), and (29), respectively.Remark 5.
Clearly, condition (5) implies that
(50)Ax,ξξ≥cAΦx,ξ,Ax,ξξ≥cAΦ¯x,Ax,ξ,for all x∈Ω and ξ∈ℝN. By Lemma 2, we have
(51)ξcA≥Φ¯x,Ax,ξAx,ξ≥12Φ¯−1x,Φ¯x,Ax,ξ.It implies that(52)Φ¯x,Ax,ξ≤Φx,2cAξ.Consequently,(53)Ax,ξ≤Φ¯−1x,Φx,2cAξ.Hence, condition (5) is stronger than not only the growth condition (3.1) but also the coercive condition (3.2) in [2].LetS be the set of all weak solutions of (31). Under the assumptions of Theorem 4, the solution set S of (31) is nonempty.Theorem 6.
Assume that the assumptions of Theorem4 hold. Then S is bounded and closed in W01LΦΩ. Moreover, S is compact in W01LΦΩ.Proof.
DenoteV=W01LΦΩ. For u∈V, define ρu=∫ΩΦx,u+Φx,Dudx, and uρ=infλ>0:ρu/λ≤1, then uρ is a norm of V equivalent to u1,Φ (see [1]). Taking α0=minb3−b5,b3−2b4−b5 in the first inequality of (36) and by [2] (Proposition 3.1), there exists h∈L1Ω, such that, for every ε>0,
(54)Γu,u≥α0uρ−εGuρ−ε−α0uρ−ε∫Ωhxdx−C1−C2u1,Φ,for all ∀u∈V, as u1,Φ is large enough, where the constants C1 and C2 are independent of u. Consequently, we can deduce that S is bounded in W01LΦΩ.
Next, we prove thatS is closed. Indeed, assume unn=1∞⊂S, and un⟶u0 in W01LΦΩ as n⟶∞. Since S is bounded, unn=1∞ is bounded. Therefore, by passing to a subsequence if necessary, we can assume without loss of generality that
(55)un⇀u0weaklyinVforσΠLΦ,ΠEΦ¯,as n⟶∞. Similar to the proof of Theorem 4, we can obtain that u0 is a weak solution of (31), i.e., u0∈S. Consequently, S is closed in V.For any subsequenceunn=1∞ in S, unn=1∞ is bounded. Repeating the proof of Theorem 4, by passing to a subsequence if necessary, we can obtain that unn=1∞ is convergent to u0, and u0 is a weak solution of (31), i.e., u0∈S. Consequently, S is sequentially compact. Since S is closed, S is compact.Definition 7 ([4], Definition 2.1).
Let (P, ≤) be a partially ordered set. A subset C of P is said to be upward directed if for each pair x,y∈C there is a z∈C such that x≤z and y≤z. Similarly, C is downward directed if for each pair x,y∈C there is w∈C such that w≤x and w≤y. If C is both upward and downward directed, it is called directed.If there existu∗,u∗∈S such that u∗≤u∗, and u∗≤u≤u∗, for any u∈S, then u∗ and u∗ are called barrier solutions of (31) (see [4]). Apparently, u∗ and u∗ are the least element and the greatest element with respect to the ordering “≤,” respectively.To show thatS is directed with respect to the ordering “≤,” the following modulus of continuity condition on the coefficient a1:Ω×ℝ×ℝN⟶ℝN is required:(A1′) For a.e. x∈Ω, all s,s′∈ℝ and all ξ∈ℝN,
(56)a1x,s,ξ−a1x,s′,ξ≤g3x+Φ¯−1x,Φx,s+Φ¯−1x,Φx,s′+Φ¯−1x,Φx,ξωs−s′,where g3∈EΦ¯Ω, g3≥0, and ω:ℝ+⟶ℝ+ is a continuous function such that
(57)∫0ε1ωrdr=+∞,for every ε>0.Theorem 8.
Assume that the assumptions of Theorem4 and (A′1) hold. Let u1,u2∈S. Then u1∨u2 and u1∧u2 are the subsolution and supersolution of (31), respectively.Proof.
DenoteV=W01LΦΩ. Let u1,u2∈S. In view of assumption (A′1), for any ε>0, there exists δε∈0,ε such that ∫δεε1/ωrdr=1. Define the function θε:ℝ⟶ℝ+, which is defined by
(58)θεs=0,ifs<δε,∫δεs1ωrdr,ifδε≤s≤ε,1,ifs>ε.
Then, for everyε>0, the function θε is continuous, piecewise differentiable, and the derivative is nonnegative and bounded. Consequently, θε is Lipschitz continuous and nondecreasing, and it satisfies θε⟶χs>0 as ε⟶0. Moreover, θε′s=1/ωs if δε≤s≤ε, and θε′s=0 if s∉δε,ε (see [29]).Letψ∈V. Replacing u with u1,u2 and taking the test function v=ψ1−θεu2−u1,v=ψθεu2−u1 in (32), respectively, and adding the resulting equations, we obtain
(59)∫Ωa1x,u1,Du1Dψdx+∫Ωa1x,u2,Du2−a1x,u1,Du1Dψθεu2−u1dx+∫Ωa0x,u1ψdx+∫Ωa0x,u2−a0x,u1ψθεu2−u1dx=∫Ωfx,u1,Du1ψdx+∫Ωfx,u2,Du2−fx,u1,Du1ψθεu2−u1dx.By applying (26), (A′1), and the properties of θε, we have
(60)∫Ωa1x,u2,Du2−a1x,u1,Du1Dψθεu2−u1dx=∫Ωa1x,u2,Du2−a1x,u1,Du1θεu2−u1Dψdx+∫Ωa1x,u2,Du2−a1x,u2,Du1ψθε′u2−u1Du2−u1dx+∫Ωa1x,u2,Du1−a1x,u1,Du1ψθε′u2−u1Du2−u1dx≥∫Ωa1x,u2,Du2−a1x,u1,Du1θεu2−u1Dψdx−∫δε<u2−u1<εg3x+Φ¯−1x,Φx,u1+Φ¯−1x,Φx,u2+Φ¯−1x,Φx,Du1ψDu2−u1dx.The last term on the right-hand side of (60) tends to 0, as ε⟶0. Using Lebesgue’s dominated convergence theorem, we obtain that
(61)∫Ωa1x,u2,Du2−a1x,u1,Du1θεu2−u1Dψdx⟶∫Ωa1x,u2,Du2−a1x,u1,Du1χu2>u1Dψdx,(62)∫Ωa0x,u2−a0x,u1ψθεu2−u1dx⟶∫Ωa0x,u2−a0x,u1χu2>u1ψdx,(63)∫Ωfx,u2,Du2−fx,u1,Du1ψθεu2−u1dx⟶∫Ωfx,u2,Du2−fx,u1,Du1χu2>u1ψdx,as ε⟶0.Combining (59)–(63), we get that
(64)∫Ωa1x,u1,Du1Dψdx+∫Ωa1x,u2,Du2−a1x,u1,Du1χu2>u1Dψdx+∫Ωa0x,u1ψdx+∫Ωa0x,u2−a0x,u1χu2>u1ψdx≤∫Ωfx,u1,Du1ψdx+∫Ωfx,u2,Du2−fx,u1,Du1χu2>u1ψdx,which yields that u1∨u2 is a subsolution of (31), i.e., u1∨u2∈S. Analogously, we can deduce that u1∧u2 is a supersolution of (31).Theorem 9.
Assume that the assumptions of Theorem4 and (A′1) hold. Then, S is directed downward and upward; that is, for all u1,u2∈S, there exist u,v∈S such that v≤u1∧u2 and u1∨u2≤u.Proof.
DenoteV=W01LΦΩ. For u1,u2∈S, let u0=u1∨u2. We will prove that there exists u~∈S such that u0≤u~.
Define the cutoff functionb:Ω×ℝ⟶ℝ by
(65)bx,s=0,ifs>u0x,−Φ¯−1x,Φx,u0x−s,ifs≤u0x,for x∈Ω, s∈ℝ. Then for any u∈LΦΩ, bx,u∈LΦ¯Ω.Denoteu≤u0≔x∈Ω:ux≤u0x. For any u∈V, by Young inequality, one has
(66)∫Ωbx,uudx≥∫u≤u0Φx,u0−udx−∫u≤u012Φx,u0−u+Φx,2u0dx≥−12∫ΩΦx,2u0dx.DefineTu=u∨u0. By Remark 3.1 in [1], T:V⟶V is continuous. According to the definition of T, T is bounded.Thanks to (29), and by Young inequality, we have
(67)∫ΩFTuudx≤2b4+b5∫u>u0Φx,udx+b5∫ΩΦx,Dudx+C1u1,Φ,for all u∈V, and some constant C1>0 independent of u. From (27) and (28), we can deduce that
(68)∫Ωa0x,Tuudx≥b3∫u>u0Φx,udx−∫u>u0g2xdx−C2u1,Φ,for all u∈V, and some constant C2>0 independent of u.Let us consider the following auxiliary equation of findingu∈V such that
(69)∫Ωa1x,Tu,DuDvdx+∫Ωa0x,Tuvdx+∫Ωbx,uvdx=∫ΩFTuvdx.DefineΓT:V⟶V∗,
(70)ΓTu,v≔∫Ωa1x,Tu,DuDvdx+∫Ωa0x,Tuvdx+∫Ωbx,uvdx−∫ΩFTuvdx,∀v∈V. Then, ΓT is well defined.Combining (25), (28), (29), (33), (66), (67), and (68), we have
(71)ΓTu,u≥b2−b4∫ΩΦx,Dudx−C3u1,Φ−C4,∀u∈V, where C3,C4>0 are two constants independent of u.Similar to the proof of Theorem4, we can deduce that there exists u~∈V such that
(72)ΓTu~,v=0,∀v∈V,i.e., u~ is a weak solution of (69).Denoteu0>u~=x∈Ω:u0x>u~x. Replacing u with u~ in (69), and taking v=u0−u~+∈V, we can obtain that
(73)∫Ωa1x,Tu~,Du~Du0−u~+dx+∫Ωa0x,Tu~u0−u~+dx+∫Ωbx,u~u0−u~+dx=∫ΩFTu~u0−u~+dx.By Theorem8, we know
(74)∫Ωa1x,u0,Du0Dvdx+∫Ωa0x,u0vdx≤∫ΩFu0vdx,for any v∈V with v≥0.Takingv=u0−u~+∈V in (74), we have
(75)∫Ωa1x,u0,Du0Du0−u~+dx+∫Ωa0x,u0u0−u~+dx≤∫ΩFu0u0−u~+dx.Subtracting (73) from (75), we obtain that
(76)∫Ωa1x,u0,Du0−a1x,Tu~,Du~Du0−u~+dx+∫Ωa0x,u0−a0x,Tu~u0−u~+dx−∫Ωbx,u~u0−u~+dx≤∫ΩFu0−FTu~u0−u~+dx.Thanks to (26), the first term of the left-hand side of (76) is nonnegative. Clearly, both the second term of the left-hand side and the right-hand side of (76) equal 0. Indeed,
(77)∫Ωa0x,u0−a0x,Tu~u0−u~+dx=∫u0>u~a0x,u0−a0x,u0u0−u~dx=0,∫ΩFu0−FTu~u0−u~+dx=∫u0>u~Fu0−Fu0u0−u~dx=0.Therefore,(78)∫Ωbx,u~u0−u~+dx≥0.On the other hand,(79)∫Ωbx,u~u0−u~+dx=−∫u0>u~Φ¯−1x,Φx,u0−u~u0−u~dx≤−∫u0>u~Φx,u0−u~dx≤0.It follows thatu0≤u~.From the definitions ofl, T, we deduce that
(80)a1x,Tu~x,Du~x=a1x,u~x,Du~x,a0x,Tu~x=a0x,u~x,bx,u~x=0,fx,Tu~x,DTu~x=fx,u~x,Du~x,for a.e. x∈Ω. We note that (69) reduces to (32). Hence, u~ is a weak solution of (31). Consequently, u~∈S, and u1∨u2≤u~.Similarly, there existsv∈S, such that v≤u1∧u2. Therefore, S is a directed set.Similar to step (c) of the proof of Theorem 3.1 in [4], we have the following result.Theorem 10.
Assume that the assumptions of Theorem4 and (A′1) hold. Let Φ∈Δ2. Then (31) admits barrier solutions, i.e., there exist u∗,u∗∈S such that u∗≤u≤u∗, ∀u∈S.



---

*Source: 9927898-2021-10-07.xml*