On the Minimal Solutions of Variational Inequalities in Orlicz-Sobolev Spaces

Ge DONG

Abstract In this paper， the author studies the existence of the minimal nonnegative solutions of some elliptic variational inequalities in Orlicz-Sobolev spaces on bounded or unbounded domains.She gets some comparison results between different solutions as tools to pass to the limit in the problems and to show the existence of the minimal solutions of the variational inequalities on bounded domains or unbounded domains.In both cases，coercive and noncoercive operators are handled.The sufficient and necessary conditions for the existence of the minimal nonnegative solution of the noncoercive variational inequality on bounded domains are established.

Keywords Orlicz-Sobolev spaces，Elliptic variational inequalities， Minimal nonnegative solutions， Bounded domains， Unbounded domains

1 Introduction

In [10–12， 23]， the sub-supersolution method was used to prove the existence of solutions and extremal solutions， confined between their subsolutions and supersolutions， for elliptic variational inequalities and nonlinear elliptic equations in Sobolev spaces or Orlicz-Sobolev spaces defined on bounded domains.There are many existence results for elliptic equations in Orlicz-Sobolev spaces on bounded domains such as[2–4，7–9，13–14，16–19].Landes[22]proved the existence of a weak solution of a quasilinear elliptic equation without any lower order term on an unbounded domain.In [6]， a different method was used to establish the existence of the minimal solution to some elliptic variational inequalities in Sobolev spaces defined on bounded or unbounded domains.When trying to weaken the restriction on the operators in [6]， one is led to replace Sobolev spaces by Orlicz-Sobolev spaces.In this paper， we will take into account the minimal solutions of elliptic variational inequalities on bounded and unbounded domains by using a method in [6] different from the above.

In the present work，we will show the existence of the minimal nonnegative solutions for some variational inequalities with coercive operators or noncoercive operators on bounded domains or unbounded domains.Some comparison principles are investigated for the minimal solutions of variational inequalities on bounded domains or unbounded domains.In particular noncoercive one，the sufficient and necessary conditions for the existence of the minimal nonnegative solution are established on bounded domains.We will use the idea introduced in [6].

The paper is organized as follows.Section 2 contains some preliminaries which will be needed in the next two sections.In Section 3， we start with the coercive case and prove the existence theorems of nonnegative solutions for some elliptic variational inequalities on bounded domain.Some comparison results between different solutions are established as tools to pass to the limit in the problems and we show the existence of the minimal solutions for the variational inequalities with coercive operators or noncoercive operators defined on bounded domains.The sufficient and necessary conditions for the existence of the minimal nonnegative solution for the noncoercive one are established.Section 4 is devoted to showing the existence of the minimal solutions for variational inequalities on unbounded domains.In both cases， coercive and noncoercive operators are handled.

2 Preliminaries

For quick reference， we recall some basic results of Orlicz spaces.Good references are[1， 7， 18， 21， 25].

2.1 N-function

Let M : R+→R+be an N-function， i.e.， M is continuous， convex， with M(u) ＞0 for u ＞0，+∞as u →+∞.Equivalently， M admits the representation M(u) =φ(t)dt， where φ : R+→R+is a nondecreasing， right continuous function， with φ(0)=0， φ(t)＞0 for t ＞0， and φ(t)→+∞as t →+∞.

Clearly， M(u)≤uφ(u)≤M(2u) for all u ≥0.

φ，φ are called the right-hand derivatives of M，， respectively.

The N-function M is said to satisfy the Δ2condition near infinity (M ∈Δ2for short)， if for some k ＞1 and u0＞0， M(2u)≤kM(u)， ?u ≥u0.It is readily seen that this will be the case if and only if for every l ＞1 there exists a positive constant k =k(l) and＞0， such that

Moreover， one has the following Young inequality: uv ≤M(u)+(v)， ?u，v ≥0， and the equality holds if and only if v =φ(u) or u=φ(v).

We will extend these N-functions into even functions on all R.

Let P，Q be two N-functions.We say that P grows essentially less rapidly than Q near infinity， denoted as P ?Q， if for every ε ＞0，→0 as t →+∞.This is the case if and only if

For a measurable function u on Ω， its modular is defined by

The Sobolev conjugate N-function M*of M is defined by

2.2 Orlicz spaces

Let Ω be an open and bounded subset of RNand M be an N-function.The Orlicz class KM(Ω) (respectively， the Orlicz space LM(Ω)) is defined as the set of (equivalence classes of)real valued measurable functions u on Ω such that

LM(Ω) is a Banach space under the (Luxemburg) norm

and KM(Ω) is a convex subset of LM(Ω) but not necessarily a linear space.

The closure in LM(Ω) of the set of bounded measurable functions with compact support inis denoted by EM(Ω).

The equality EM(Ω)=LM(Ω) holds if and only if M ∈Δ2， moreover， LM(Ω) is separable.

LM(Ω) is reflexive if and only if M ∈Δ2and∈Δ2.

Convergences in norm and in modular are equivalent if and only if M ∈Δ2.

The dual space of EM(Ω)can be identified withby means of the pairingu(x)v(x)dx，and the dual norm ofis equivalent to

2.3 Orlicz-Sobolev spaces

We now turn to the Orlicz-Sobolev space: W1LM(Ω) (respectively， W1EM(Ω)) is the space of all functions u such that u and its distributional partial derivatives lie in LM(Ω)(respectively，EM(Ω)).It is a Banach space under the norm

Denote ‖Du‖(M)=‖|Du|‖(M)and ‖u‖1，M，Ω=‖u‖(M)+‖Du‖(M).Clearly， ‖u‖1，M，Ωis equivalent to ‖u‖Ω，M.

Thus W1LM(Ω) and W1EM(Ω) can be identified with subspaces of the product of N +1 copies of LM(Ω).Denoting this product by ΠLM，we will use the weak topologies σ(ΠLM，ΠEM)and σ(ΠLM，).

If M ∈Δ2， then W1LM(Ω) = W1EM(Ω).If M ∈Δ2and∈Δ2， then W1LM(Ω) =W1EM(Ω) are reflexive and the weak topologies σ(ΠLM，) and σ(ΠLM，) are equivalent.

The space W10EM(Ω) is defined as the (norm) closure of the Schwartz space D(Ω) in W1EM(Ω) and the space W10LM(Ω) as the σ(ΠLM，ΠEM) closure of D(Ω) in W1LM(Ω).If M ∈Δ2， then W10LM(Ω) = W10EM(Ω) and W10LM(Ω) is separable.If M ∈Δ2and∈Δ2，then W10LM(Ω) is reflexive.

3 Variational Inequalities in Bounded Domains

This section is devoted to studying the existence of nonnegative solutions and their minimal solutions for some quasilinear variational inequalities in bounded domains.We investigate variational inequalities with coercive operators in Subsection 3.1 and with noncoercive operators in Subsection 3.2， respectively.

3.1 Variational inequalities with coercive operator

Let Ω be a bounded domain in RN(N ≥1) with Lipschitz boundary， M be an N-function andbe the complementary function of M， and φ，φ are the right-hand derivatives of M，respectively.Assume that M，∈Δ2.

In what follows we denote by L0(Ω) the set of all (equivalence classes of) Lebesgue measurable functions from Ω to R.For u，v ∈L0(Ω)， U，V ?L0(Ω)， we use the standard notation: u ≤v ?u(x) ≤v(x) for a.e.x ∈Ω， u ∧v = min{u，v}，u ∨v = max{u，v}，U ∧V ={u ∧v :u ∈U，v ∈V}， U ∨V ={u ∨v :u ∈U，v ∈V}， u+:=u ∨0，u-:=-u ∧0.We consider the usual ordering relation (W10LM(Ω)，≤).

We denote by K a closed convex subset of W10LM(Ω) containing 0 such that the lattice condition

is satisfied.This type of lattice convex sets usually occurs in applications.For example，K =LM(Ω) for equations， and K = {u ∈W10LM(Ω) : u(x) ≥ψ(x) for a.e.x ∈Ω} for obstacle problems， where ψ :Ω →R is an obstacle function (see [8–9， 13， 19]).

Let a(x，ξ) = (ai(x，ξ))1≤i≤Nand a0(x，ξ) be a family of Carath′eodory’s functions defined on Ω×RN+1such that

where A is a nonlinear operator defined fromLM(Ω) into its dual by

From (3.5)， the operator A is well defined.For simplicity we set the dual pairing 〈·，·〉 =The above duality is equivalent to

Let ? ＞0 be a real number and Ωlbe the cylinder (-?，?)×Ω.The points in RN+1are denoted by (y，x) with x=(x1，··· ，xN)∈RNand the gradient operator defined over RN+1is also denoted by ?′=(?y，?) with ?=(?x1，?x2，··· ，?xN).We set

This is a closed convex subset ofLM(Ω?).

Let ψ : [0，+∞) →[0，+∞) be an increasing function.By [20， Lemma 8.2]，is monotone on Rn(n ≥1).However，we will show the strict monotonicity ofon Rnand give another method to proof the monotonicity of ψ(|ξ|).

Lemma 3.1Ifψ :[0，+∞)→[0，+∞)is increasing， thenψ(|ξ|)is monotone onRn(n ≥1).Moreover， ifψis strictly increasing， thenψ(|ξ|)is strictly monotone.

ProofLet ξ，η be two nonzero vectors in Rn， with angle θ ∈[0，π] between them at the origin.Then

The first term of the right-hand side of (3.8) is nonnegative since ψ is increasing.Since

the second term of the right-hand side of (3.8) is nonnegative.Consequently，

Moreover， suppose that there exist ξ1，ξ2∈RNwith ξ1/=ξ2such thatIt follows from (3.8)–(3.9) that

and cos θ1= 1， where θ1∈[0，π] is the angle between ξ1and ξ2at the origin.This yields that θ1= 0.Therefore， there exists λ ＞0 with λ /= 1 such that ξ1= λξ2.It implies that|ξ1| /= |ξ2|.Immediately， ψ(|ξ1|) /= ψ(|ξ2|) since ψ is strictly increasing.It contradicts (3.10).This completes the proof of the lemma.

We recall the following notation which will be used later.It can be referred to [24， p.25] or[5， Definition 1].

Definition 3.1(see [24])LetXbe a reflexive Banach space.The operatorT : X →X*is called pseudomonotone if

(i) Tis bounded;and

(ii)for any sequence{un}?Xsuch thatun?u0weakly inXand0， the inequality

holds.

Theorem 3.1Assume thatf ∈(Ω)is nonnegative and the assumptions(3.1)–(3.5)are satisfied.Then for every? ＞0， the following variational inequality

has at least one solutionu?withu?≥0.Moreover， ifφis strictly increasing， oraanda0are strictly monotone， i.e.， for a.e.x ∈Ωand allξ =(ξi)0≤i≤N，ξ′=(ξ′i)0≤i≤N∈RN+1，

then there exists a unique solutionu?of(3.11)andu?≥0.

ProofFor ? ＞0， we define the operator T?:LM(Ω?)→(LM(Ω?))*by

Denote z = (y，x).Thanks to (2.1)， (3.5) and the Young inequality， we can deduce that there exist positive constants C(β，N，K) and C(N，?，) such that

for all u，v ∈W10LM(Ω?).Therefore，T?is well defined.From(3.13)， it is easy to see that T?is bounded.In view of (3.4) and Lemma 3.1， we can show that T?is monotone.In a similar way as [10， Lemma 8]， we can check that T?is continuous.Thanks to [26， Propositon 27.6(a)]， T?is pseudomonotone.For u0∈K?， in a way similar to the proof of [10， Theorem 11]， we have

for some positive constants C1and C2.For u ∈W10LM(Ω?)， define

Therefore， 〈T?(u)-f，u-u0〉＞0 when ‖u‖1，M，Ω?is sufficiently large.From the above results，we can deduce that the conditions (i)–(iv) in [19] hold.According to [19， Proposition 1]， the variational inequality (3.11) has at least one solution u?.

Now， we prove that u?is nonnegative.Taking v =∈K?in (3.11) we have

In view of (3.4)， we have

Consequently， u?≥0.

Suppose that there exists another solution u′?of (3.11) with u?/=u′?.When we take v =u′?in (3.11) and v =u?in (3.11) written for u′?and add the two inequalities， it comes

If φ is strictly increasing or a and a0are strictly monotone， we can deduce， by Lemma 3.1 and(3.4)， or by (3.12)， u?=u′?.It is a contradiction.

Similar to the proof of Theorem 3.1， it is easy to show the following theorem.

Theorem 3.2Assume thatf ∈LM(Ω)is nonnegative and the assumptions(3.1)–(3.5)are satisfied.Then(3.6)has at least one solutionuwithu ≥0.Moreover， ifaanda0satisfy(3.12)， then there exists a unique solution of(3.6)andu ≥0.

Does there exist the minimal nonnegative solution of problem (3.6) when it has more than one solution? The following theorem will give the answer.

Theorem 3.3Assume thatf ∈LM(Ω)is nonnegative，φis strictly increasing and the assumptions(3.1)–(3.5)are satisfied.Then the pointwise limit of{u?}?is the minimal nonnegative solution of(3.6)， whereu?is the solution of(3.11)， for any? ＞0.

Moreover， ifu1andu2are the minimal nonnegative solutions of(3.6)obtained by replacingfwithf1andf2respectively， thenf1≤f2impliesu1≤u2.

ProofStep 1 The sequence{u?}?is nondecreasing and bounded above by any nonnegative solution of (3.6).

Let 0 ＜? ＜?′.Extending u?by 0 on Ω?′and since u?′is nonnegative， when we take v =u?-(u?-u?′)+∈K?in (3.11)and v =u?′+(u?-u?′)+∈K?′in (3.11)written for u?′and Ω?′and add the two inequalities， it comes

Thanks to the condition (3.4) we deduce

Using Lemma 3.1， we have u?(y，x) ≤u?′(y，x) for a.e.(y，x) ∈Ω?， which shows that the sequence {u?}?is nondecreasing.

Let ? ＞0 and u be a nonnegative solution of (3.6).Taking v =u+(u?(y，·)-u)+∈K as a test function in (3.6)， for a.e.y ∈(-?，?)， and integrating in y we derive

Taking v =u?-(u?-u)+∈K?as a test function in (3.11) we can deduce that

Adding the two inequalities (3.15)and (3.16)and using the fact that u is independent of y and the monotone condition (3.4) we obtain that

By Lemma 3.1，

Immediately， {u?}?is bounded above by any nonnegative solution of (3.6).

Step 2 The pointwise limit of {u?}?is independent of y.

It follows from Step 1 that u?has a pointwise limit we denote bysuch that

Let h ＞0.Denote Thu?(y，x)=u?(y+h，x).Then the functions Thu?(y，x)and(Thu?(y，x)-u?+h(y，x))+are supported in the closure of:= (-l-h，l-h)×Ω.Thanks to (3.11)， we have

where K?，h:={Thv |v ∈K?}={v ∈LM()|v(y，·)∈K a.e.in(-?-h，?-h)}.Choosing v = Thu?-(Thu?-u?+h)+∈K?，hin (3.19) and v = u?+h+(Thu?-u?+h)+∈K?+hin (3.11)written for u?+hand adding the two inequalities， we have

Using (3.4) we can obtain that

By Lemma 3.1，

Letting ? →+∞in (3.20)， we have

Similarly， we can show that (3.21) holds whenever h ＜0 for a.e.(y，x) ∈Ω?.Since h is arbitrary，(y，x)=(x) for a.e.(y，x)∈Ω?， that is，is independent of y.

Step 3 For all ?0＞0， there exists a constant C(?0) independent of ? such that

Let now ?0＞0.Clearly， it is need to consider the case ? ＞?0.Let ? ∈D(-2?0，2?0)such that 0 ≤? ≤1，? = 1 on (-?0，?0).Let u be a nonnegative solution of (3.6).Taking v =u?-?2(u?-u)∈K?as a test function in (3.11)， we derive that

By the fact ?=1 on (-?0，?0) and u is independent of y， it yields that

By (2.1)， (3.3)， (3.5) and the Young inequality， we can obtain that

where ε ∈(0，1) and the constant C ＞0 is independent of ?.Since M ∈Δ2， there exists a constant Kε＞0 and some tε＞0 such thatfor all t ＞tε.In view of (3.17)，we get

where ε ∈(0，1) and the positive constants C and C(ε) are independent of ?.Choosing ε small enough， we get

for some positive constant C(?0) independent of ?.For u ∈W1LM(Ω?0)， define

and

Then ‖u‖ρ，Ω?0is a norm of W1LM(Ω?0) equivalent to ‖u‖1，M，Ω?0(see [15]).It implies (3.22).

Let ?0＞0.In view of (3.22)， {u?}?is bounded in W10LM(Ω?0) for all ? ＞0.Consequently，and {ai(x，u?，?u?)}?(i=0，1，··· ，N) are bounded in(Ω?0).Hence， there exist d and diinsuch that

as ? →+∞，i=0，1，··· ，N.The two first convergences hold for the whole sequence since{u?}?is nondecreasing， which guarantees the uniqueness of the limit and the last two convergences hold up to a subsequence.

Let ω be a nonnegative function in D(-?0，?0).By (3.4)， (3.23) and (3.26)， it follows that

and thus

Combining (3.27) and (3.28)， we have

and

Let w ∈K and ω /≡0 be a nonnegative function inK?as a test function in (3.11)， we have

Thanks to (3.4)， one has

From(3.30)， it is clear that ?yu?→0 strongly inThis can imply that φ(|?yu?|)→0 strongly inPassing to the limit in the above inequality as ? →+∞yields

This implies

Let u be an arbitrary solution of the problem (3.6).Letting ? →+∞in (3.17)， we deduce≤u.This means thatis the minimal solution of the problem (3.6).

Step 6 u1≤u2.

Let u?，1and u?，2be the solutions of (3.11)obtained if we replace f by f1and f2，respectively.Then u?，1and u?，2converge to u1and u2， respectively， as ? →+∞.Taking v = u?，1-(u?，1-u?，2)+and v =u?，1+(u?，1-u?，2)+in (3.11) for f1and f2， respectively， we get

Using the condition (3.4) one has

This implies u?，1≤u?，2a.e.in Ω?.Letting ? →+∞， it follows that u1≤u2.

3.2 Noncoercive variational inequalities

We keep the notation and the assumptions of Subsection 3.1.Then we consider the following problem

where F :Ω×R →R is a nonnegative function satisfying

Clearly， (3.31) is the extension of (3.6).

Remark 3.1If F : Ω×R →R is a nonnegative Carath′eodory function such that for a.e.x ∈Ω and all t ∈R， |F(x，t)| ≤q(x)， where q(x) ∈Ω)， then F satisfies the conditions(3.32)–(3.34).

Lemma 3.2LetFbe a nonnegative function satisfying the hypotheses(3.32)–(3.34)，φbe strictly increasing， and suppose that the assumptions(3.1)–(3.5)are fulfilled.Define thatunis the minimal nonnegative solution of the variational inequality in the last line of the following problem:

?n ≥1.Then the sequence{un}n∈Nis well defined and nondecreasing.

ProofThe existence of unis insured by Theorem 3.3 since F(x，un-1)∈(Ω).In a way similar to the proof in[6]，the sequence of functions{un}n∈Nis nonnegative and nondecreasing.This completes the proof of Lemma 3.2.

Denote by u∞the pointwise nonnegative limit of {un}nwhich is not necessarily in LM(Ω)and may equal ∞.We also denotewhich may also be infinite on some subset.Assume that

Note that the above assumption is satisfied.For example，Then we have the following result.

Theorem 3.4LetFbe a nonnegative function satisfying the hypotheses(3.32)–(3.34)，φbe strictly increasing， and suppose that the assumptions(3.1)–(3.5)are fulfilled.Then we have the equivalence between the following assertions:

(1) (3.31)has at least one nonnegative solution，

(2) (3.31)has a minimal nonnegative solution，

(3)the hypothesis(3.36)holds.Moreover if the hypothesis(3.36)holds， thenu∞， the limit ofun， belongs toKand is the minimal solution of(3.31).

ProofClearly，(2)?(1).Suppose that(3.31)has a nonnegative solution w ∈K.According to Theorem 3.3， there exits a minimal nonnegative solution∈K of the following problem:

It is clearly that w is also a solution of (3.37).Then w ≥w.Since w is nonnegative and F is nondecreasing in the second variable， F(x，w) ≥F(x，0) for a.e.x ∈Ω.By Theorem 3.3，we have w ≥u1， where u1is the minimal nonnegative solution of the variational inequality in the last line of (3.35) for n = 1.Consequently， F(x，w) ≥F(x，w) ≥F(x，u1) for a.e.x ∈Ω.This implies that w ≥u2by Theorem 3.3， where u2is the minimal nonnegative solution of the variational inequality in the last line of (3.35) for n=2.By induction， we can obtain that

and

which yields (3.36).Hence， (1)?(3).

Let the hypothesis (3.36) hold.By Theorem 3.3， there exists the minimal solution u∞of the following problem:

Since F(·，un-1)≤F∞， ?n ≥1， and thanks to Theorem 3.3， we deduce that

It follows that there exists u∞such that

as n →∞.Therefore， u∞≤u∞， and F∞=F(·，u∞) a.e.in Ω.Consequently， u∞∈LM(Ω).

for some positive constants C and C(ε) independent of n， where ε ∈(0，1) and ?x0un= un.Choosing ε small enough， we have

for some constant C ＞0 independent of n.This yields that ‖un‖1，M，Ω≤C for some positive constant C independent of n.Combining the above results， it follows， as n →∞， that

Since K is a closed convex subset of W10LM(Ω)， it is also weakly closed， which yields that u∞∈K.

From (3.4) and (3.35)， it follows that

Taking w =u∞+t(v-u∞) with 0 ＜t ＜1 and v ∈K， and letting t →0， one has

that is， u∞is a solution of (3.31).Therefore， (3)?(1).

Suppose that the hypothesis (3.36) holds.Let w ∈K be a nonnegative solution of (3.31).By the above arguments， letting ? →+∞in (3.38)， we get w ≥u∞.Since w is an arbitrary solution of (3.31)， we have that u∞is the minimal solution of (3.31).Hence (3)?(2) and the proof is achieved.

4 Variational Inequalities in Unbounded Domains

This section is devoted to studying the existence of nonnegative solutions and their minimal solutions for some quasilinear variational inequalities in unbounded domains.We investigate variational inequalities with coercive operators in Subsection 4.1 and with noncoercive operators in Subsection 4.2， respectively.

4.1 Variational inequalities with coercive operator

Let G be a bounded domain in RN-1(N ≥2)with Lipschitz boundary，and KGbe a closed convex subset of W10LM(G) containing 0 such that the lattice condition

is satisfied.Let M be an N-function，be the complementary function of M， and φ，φ are the right-hand derivatives of M，， respectively.Assume that M，∈Δ2.

For x ∈R×G， denote x=(x1，X2) with X2=(x2，··· ，xN)， and

We set

and

Note that if ?x1a1(x，u，?u) ∈the above variational inequality can be written as

Since the domain is unbounded and f is not necessarily in the dual ofLM(R × G)，the existence of nonnegative solutions to problem (4.1) is not an ordinary issue.Once this is ensured， we can then look for the minimal nonnegative solution.Here， we will use the same approach as in Subsection 3.1 to prove these existence results.To this end， in addition to the hypotheses (3.1)–(3.5)， assume that

i = 0，1，··· ，N.That is to say if ξ1= 0 then the coefficients aifor i = 0，1，··· ，N are independent of x1.We also assume that there exists h ∈G) such that

For ? ＞0，let Ω?，G=(-?，?)2×G.For simplicity we denoteandWe set

Let ? ＞0.Consider the following variational inequalities:

and

Theorem 4.1Suppose that the assumptions(3.1)–(3.5)and(4.2)–(4.3)are satisfied， whereΩis replaced byR×Gin(3.1)–(3.5).Assume thatφis strictly increasing.Then the pointwise limit of{u?}?is the minimal nonnegative solution of(4.1)， whereu?is the solution of(4.5)，for every? ＞0.

Moreover， the following assertions hold:

(i)Letu1andu2be the minimum nonnegative solutions of(4.1)obtained by replacingfwithf1andf2， respectively.Iff1≤f2， thenu1≤u2.

(ii)Letbe the minimum nonnegative solution of(4.1)obtained by replacingfwithf1，andbe a nonnegative solution of(4.1)obtained by replacingfwithf2， wheref2does not necessarily satisfy(4.3).Iff1≤f2， then≤.

ProofStep 1 The sequence{u?}?is nondecreasing and bounded by any solution u of (4.4).

As the same arguments in the proof of Theorem 3.3， we can get {u?}?is a nondecreasing sequence.

Taking v =u+(u?(y，x1，·)-u)+∈KGas a test function in(4.4)and integrating on(-?，?)2，we obtain by (4.2) that

Taking v =u?-(u?-u)+∈as a test function in (4.5) and adding the resulting inequality with (4.6) one yields

In view of (3.4) and (4.3)， we derive

Since u is independent of y， this implies that

It follows that u?≤u.

Step 2 The pointwise limit of u?is independent of y.

Step 3 For all ?0＞0， there exists a constant C?0independent of ? such that Let ? ∈D((-2?0，2?0)2) such that

Let ?0＞0 and u be a nonnegative solution of (4.4).Taking v =u?-?2(u?-u)∈in (4.5)，then following the same arguments as in the proof of Theorem 3.3 we can deduce (4.8).

For ?0＞0， according to (4.7)–(4.8)， we can deduce that

and

as ? →+∞.

where diis the weak limit of ai(x，u?，?u?) in LM(Ω?0，G).(4.11)–(4.12)hold for a subsequence of {u?}?， still denoted by {u?}?.

Passing to the limit as ? →+∞and taking into account (4.9)–(4.12) we obtain

Let t ＞0 and ψ be a nonnegative function inThen it follows from the condition(3.4) that

Passing to the limit as ? →+∞， it follows from (3.5)， (4.9)–(4.11) that

which implies that

Combining (4.13) and (4.14)， we have

Since ?0is arbitrary， we get

Let u be an arbitrary nonnegative solution of (4.1).Then (u?-u)+are supported in the closure of Ω?，G.Choosing ? ∈D(R)such that ?=1 on(-?，?)，taking v =u+(u?(y，·)-u)+∈as a test function in (4.1)， and integrating on (-?，?)， we have

Taking v =u?-(u?-u)+∈as a test function in(4.5)and summing the produced inequality with (4.16)， we obtain

According to (3.4)， we get

since u is independent of y.It follows from Lemma 3.1 that u?≤u a.e.in Ω?.Letting ? →+∞in (4.7)， we have≤u a.e.in R×G.This means thatis the minimal nonnegative solution of (4.1).

Let u1?and u2?be the converging sequences defined above as solutions of (4.5) for f = f1and f = f2， respectively.By the same arguments as in the proof of Theorem 3.3，≤since f1≤f2.Letting ? →+∞， it follows that≤a.e.in R×G， whereandare the minimum nonnegative solutions of (4.1) obtained by replacing f with f1and f2， respectively.Hence， the assertion (i) holds.

Let u1?be the converging sequence defined above as the solution of (4.5)for f =f1，be the minimum nonnegative solution of (4.1)obtained by replacing f with f1，andbe a nonnegative solution of (4.1) obtained by replacing f with f2， where f2does not necessarily satisfy (4.3).Note that (u1?-)+are supported in the closure of Ω?，G.Takingas a test function in (4.5) for f = f1and v =+(u1?(y，·)-)+∈as a test function in(4.1) for f =f2， choosing ? ∈D(R) such that ?=1 on (-?，?) and integrating on (-?，?)， then summing the produced inequalities we have

Consider the following nonlinear elliptic problem defined on the infinite cylinder R×G，

A function u is called a (weak) solution of (4.17) if u ∈and

Then any solution of problem (4.1) for KG=LM(G) is a solution of (4.17) and vice versa.Thus the existence of nonnegative solutions of problem(4.17)is proved in Theorem 4.1.Indeed，let u ∈be a solution of (4.1).Choosing v =u±v′with v′∈D(R×G) in (4.1)and ?=1 on the support of v′， we can obtain (4.18).The converse is an immediate consequence of (4.18).

Therefore， we have the following result as an immediate consequence of Theorem 4.1.

Corollary 4.1Under the assumptions of Theorem4.1， there exists a minimal nonnegative solution of(4.17).Moreover， letandbe the minimal nonnegative solutions of(4.17)obtained by replacingfwithf1andf2respectively.Then iff1≤f2we have

4.2 Noncoercive variational inequalities

We consider the following nonlinear variational inequality defined on the infinite cylinder R×G，

where F is defined as in Subsection 3.2， replacing Ω by R×G.In addition， we assume that

satisfies

Lemma 4.1LetFandhbe nonnegative functions satisfying the hypotheses above.Suppose that the assumptions(3.1)–(3.5)and(4.2)hold， whereΩis replaced byR×Gin(3.1)–(3.5).Assume thatφis strictly increasing.Defineunandunare respectively the minimal nonnegative solutions of variational inequalities in the last line of the following problems:

and

respectively， for everyn ≥1.Then the sequences{un}n∈Nand{un}n∈Nare well defined and nondecreasing satisfying

ProofIt is clear that u0， u0satisfy (4.24).Suppose that {un-1} and {un-1} are defined and satisfy (4.24)， i.e.，

Thanks to (4.21)， one has h(X2，un-1)∈LM(G).Consequently， unexists by Theorem 3.3.In view of (4.25)， F(x，un-1)∈LM(G).Therefore， unexists by Theorem 4.1.Arguing as Step 1 in the proof of Theorem 4.1 one can deduce that un≤un.According to (3.33) and (4.20)， we have

i.e.， (4.24) holds.

Theorem 4.2LetFandhbe nonnegative functions satisfying the hypotheses above and suppose that the assumptions(3.1)–(3.5)and(4.2)hold， whereΩis replaced byR×Gin(3.1)–(3.5).Assume thatφis strictly increasing.Then there exists a minimal nonnegative solution of(4.19).

ProofBy (3.33)， F(·，un-1) ≤F∞， ?n ≥1 a.e.in R×G.It follows， by using Theorem 4.1， that {un}n∈Nis nondecreasing and

where u is the minimal solution of

Note that F∞≤h∞a.e.in R×G， and h∞is independent of x1.By Lemma 4.1， there existssuch that un→a.e.on R×G.

Let ?0＞0.Taking v =un-?(un-u) as a test function in (4.23)and choosing ? such that ?=1 on (-?0，?0)， we have

which implies that

Since M ∈Δ2，using the conditions(3.3)，(3.5)，(4.27)and the Young inequality，we can deduce

as Theorem 3.3， for some constant C =C(?0) independent of n， and consequently，

Therefore， one has

and

as n →∞.Since ?0is arbitrary，

By the same arguments as in the proof of Theorem 4.1， it is easy to see thatis a nonnegative solution of (4.19).

Let w be a nonnegative solution of (4.19).Then w is a solution of the following problem:

Since w ≥0， F(x，w) ≥F(x，0) = F(x，u0) for a.e.x ∈R×G.By Lemma 4.1 and Theorem 4.1(ii)， w ≥u1.Therefore，F(xiàn)(x，w)≥F(x，u1)for a.e.x ∈R×G.By Lemma 4.1 and Theorem 4.1(ii)， w ≥u2.By induction， we can obtain that w ≥un， ?n ∈N.Letting n →∞， w ≥，that is，is the minimal solution of (4.19).

AcknowledgementThe author is highly grateful for the referees’ careful reading and comments on this paper.