# Nonlinear Elliptic Boundary Value Problems at Resonance with Nonlinear Wentzell Boundary Conditions

Authors: Ciprian G. Gal; Mahamadi Warma
Journal: Advances in Mathematical Physics
 (2017)

---

## Abstract

Given a bounded domainΩ⊂RN with a Lipschitz boundary ∂Ω and p,q∈(1,+∞), we consider the quasilinear elliptic equation -Δpu+α1u=f in Ω complemented with the generalized Wentzell-Robin type boundary conditions of the form bx∇up-2∂nu-ρbxΔq,Γu+α2u=g on ∂Ω. In the first part of the article, we give necessary and sufficient conditions in terms of the given functions f, g and the nonlinearities α1, α2, for the solvability of the above nonlinear elliptic boundary value problems with the nonlinear boundary conditions. In other words, we establish a sort of “nonlinear Fredholm alternative” for our problem which extends the corresponding Landesman and Lazer result for elliptic problems with linear homogeneous boundary conditions. In the second part, we give some additional results on existence and uniqueness and we study the regularity of the weak solutions for these classes of nonlinear problems. More precisely, we show some global a priori estimates for these weak solutions in an L∞-setting.


---

## Body

## 1. Introduction

LetΩ⊂RN, N≥1, be a bounded domain with a Lipschitz boundary ∂Ω and consider the following nonlinear boundary value problem with nonlinear second-order boundary conditions:(1)-Δpu+α1u=finΩ,bx∇up-2∂nu-ρbxΔq,Γu+α2u=gon∂Ω,where b∈L∞∂Ω, b(x)≥b0>0, for some constant b0, ρ is either 0 or 1, and α1, α2∈CR are monotone nondecreasing functions such that αi0=0. Moreover, Δpu=div⁡(∇up-2∇u) is the p-Laplace operator, p∈1,+∞, and f∈L2Ω,dx, g∈L2(∂Ω,σ) are given real-valued functions. Here, dx denotes the usual N-dimensional Lebesgue measure in Ω and σ denotes the restriction to ∂Ω of the (N-1)-dimensional Hausdorff measure. Recall that σ coincides with the usual Lebesgue surface measure since Ω has a Lipschitz boundary, and ∂nu denotes the normal derivative of u in direction of the outer normal vector n→. Furthermore, Δq,Γ is defined as the generalized q-Laplace-Beltrami operator on ∂Ω; that is, Δq,Γu=divΓ(∇Γuq-2∇Γu), q∈1,+∞. In particular, Δ2=Δ and Δ2,Γ=ΔΓ become the well-known Laplace and Laplace-Beltrami operators on Ω and ∂Ω, respectively. Here, for any real-valued function v,(2)divΓv=∑i=1N-1∂τiv,where ∂τiv denotes the directional derivative of v along the tangential directions τi at each point on the boundary, whereas ∇Γv=∂τ1v,…,∂τN-1v denotes the tangential gradient at ∂Ω. It is worth mentioning again that when ρ=0 in (1), the boundary conditions are of lower order than the order of the p-Laplace operator, while, for ρ=1, we deal with boundary conditions which have the same differential order as the operator acting in the domain Ω. Such boundary conditions arise in many applications, such as phase-transition phenomena (see, e.g., [1, 2] and the references therein), and have been studied by several authors (see, e.g., [3–7]).In [4], the authors have formulated necessary and sufficient conditions for the solvability of (1) when p=q=2, by establishing a sort of “nonlinear Fredholm alternative” for such elliptic boundary value problems. We shall now state their main result. Defining two real parameters λ1, λ2∈R+ by(3)λ1=∫Ωdx,λ2=∫∂Ωdσb,this result reads that a necessary condition for the existence of a weak solution of (1) is that(4)∫Ωfxdx+∫∂Ωgxdσbx∈λ1Rα1+λ2Rα2,while a sufficient condition is(5)∫Ωfxdx+∫∂Ωgxdσbx∈intλ1Rα1+λ2Rα2,where R(αj) denotes the range of αj, j=1,2, and int(G) denotes the interior of the set G.Relation (4) turns out to be both necessary and sufficient if either of the sets Rα1 or Rα2 is an open interval. This particular result was established in [4, Theorem  3], by employing methods from convex analysis involving subdifferentials of convex, lower semicontinuous functionals on suitable Hilbert spaces. As an application of our results, we can consider the following boundary value problem:(6)-Δu+α1u=finΩ,bx∂nu=gon∂Ω,which is only a special case of (1) (i.e., ρ=0, α2≡0, and p=2). According to [4, Theorem  3] (see also (5)), this problem has a weak solution if(7)∫Ωfxdx+∫∂Ωgxdσbx∈intλ1Rα1,which yields the result of Landesman and Lazer [8] for g≡0. This last condition is both necessary and sufficient when the interval Rα1 is open. This was put into an abstract context and significantly extended by Brézis and Haraux [9]. Their work was much further extended by Brézis and Nirenberg [10]. The goal of the present article is comparable to that of [4] since we want to establish similar conditions to (5) and (7) for the existence of solutions to (1) when p,q≠2, with main emphasis on the generality of the boundary conditions.Recall thatλ1 and λ2 are given by (3). Let I be the interval λ1Rα1+λ2Rα2. Our first main result is as follows (see Section 4 also).Theorem 1.
Letαj:R→R  (j=1,2) be odd, monotone nondecreasing, continuous function such that αj(0)=0. Assume that the functions Λj(t)≔∫0tαj(s)ds satisfy(8)Λj2t≤CjΛjt,∀t∈R,for some constants Cj>1, j=1,2. If u is a weak solution of (1) (in the sense of Definition 30 below), then(9)∫Ωfxdx+∫∂Ωgxdσbx∈I.Conversely, if(10)∫Ωfxdx+∫∂Ωgxdσbx∈intI,then (1) has a weak solution.Our second main result of the paper deals with a modified version of (1) which is obtained by replacing the functions α1s and α2s in (1) by α¯1s+sp-2s and α¯2s+ρbuq-2u, respectively, and also allowing α¯1, α¯2 to depend on x∈Ω¯. Under additional assumptions on α¯1, α¯2 and under higher integrability properties for the data f,g, the next theorem provides us with conditions for unique solvability results for solutions to such boundary value problems. Then, we obtain some regularity results for these solutions. In addition to these results, the continuous dependence of the solution to (1) with respect to the data f,g can be also established. In particular, we prove the following.Theorem 2.
Let all the assumptions of Theorem1 be satisfied for the functions α¯1, α¯2. Moreover, for each j=1,2, assume that α¯jt/t→0, as t→0, and α¯jt/t→∞, as t→∞, respectively.(a)
Then, for everyf,g∈Lp1(Ω)×Lq1(∂Ω) with (11)p1>max⁡1,Np,(12)q1>max⁡1,N-1p-1,ifρ∈0,1,max⁡1,N-1p,ifρ=1,p=q,there exists a unique weak solution to problem (1) (in the sense of Definition 40 below) which is bounded.(b)
Letα¯j, j=1,2, be such that (13)cjα¯jξ-η≤α¯jξ-α¯jη,∀ξ,η∈R,for some constantscj∈0,1. Then, the weak (bounded) solution of problem (1) depends continuously on the data f,g. Precisely, let us indicate by uFj the unique solution corresponding to the data Fj≔fj,gj∈Lp1Ω×Lq1∂Ω, for each j=1,2. Then, the following estimate holds: (14)uF1-uF2L∞Ω+uF1-uF2L∞∂Ω≤Qf1-f2Lp1Ω,g1-g2Lq1∂Ω,for some nonnegative functionQ:R+2→R+, Q0,0=0, which can be computed explicitly.We organize the paper as follows. In Section2, we introduce some notations and recall some well-known results about Sobolev spaces, maximal monotone operators, and Orlicz type spaces which will be needed throughout the article. In Section 3, we show that the subdifferential of a suitable functional associated with problem (1) satisfies a sort of “quasilinear” version of the Fredholm alternative (cf. Theorem 20), which is needed in order to obtain the result in Theorem 1. Finally, in Sections 4 and 5, we provide detailed proofs of Theorems 1 and 2. We also illustrate the application of these results with some examples.


## 2. Preliminaries and Notations

In this section we put together some well-known results on nonlinear forms, maximal monotone operators, and Sobolev spaces. For more details on maximal monotone operators, we refer to the monographs [11–15]. We will also introduce some notations.


### 2.1. Maximal Monotone Operators

LetH be a real Hilbert space with scalar product (·,·)H.Definition 3.
LetA:D(A)⊂H→H be a closed (nonlinear) operator. The operator A is said to be(i)
monotone if for allu,v∈D(A) one has (15)Au-Av,u-vH≥0,(ii)
maximal monotone if it is monotone and the operatorI+A is invertible.Next, letV be a real reflexive Banach space which is densely and continuously embedded into the real Hilbert space H, and let V′ be its dual space such that V↪H↪V′.Definition 4.
LetA:V×V→R be a continuous map.(a)
The mapA:V×V→R is called a nonlinear form on H if for all u∈V one has A(u,·)∈V′, that is, if A is linear and bounded in the second variable.(b)
The nonlinear formA:V×V→R is said to be
(i)
monotone ifA(u,u-v)-A(v,u-v)≥0forallu,v∈V;(ii)
hemicontinuous iflimt↓0A(u+tv,w)=A(u,w), ∀u,v,w∈V;(iii)
coercive, iflimvV→+∞(Av,v/vV)=+∞.Now, letφ:H→-∞,+∞ be a proper, convex, lower semicontinuous functional with effective domain (16)Dφ≔u∈H:φu<∞.The subdifferential ∂φ of the functional φ is defined by (17)D∂φ≔u∈Dφ:∃w∈H,∀v∈Dφ:φv-φu≥w,v-uH;∂φu≔w∈H:∀v∈Dφ:φv-φu≥w,v-uH.By a classical result of Minty [13] (see also [12, 14]), ∂φ is a maximal monotone operator.


### 2.2. Functional Setup

LetΩ⊂RN be a bounded domain with a Lipschitz boundary ∂Ω. For 1<p<∞, we let W1,p(Ω) be the first-order Sobolev space; that is,(18)W1,pΩ=u∈LpΩ:∇u∈LpΩN.Then W1,p(Ω), endowed with the norm (19)uW1,pΩ≔uΩ,pp+∇uΩ,pp1/p,is a Banach space, where we have set (20)uΩ,pp≔∫Ωupdx.Since Ω has a Lipschitz boundary, it is well-known that there exists a constant C>0 such that(21)uΩ,ps≤CuW1,pΩ,∀u∈W1,pΩ,where ps=pN/(N-p) if p<N and 1≤ps<∞ if N=p. Moreover the trace operator tr(u)≔u∣∂Ω initially defined for u∈C1(Ω¯) has an extension to a bounded linear operator from W1,p(Ω) into Lqs(∂Ω) where qs≔p(N-1)/(N-p) if p<N and 1≤qs<∞ if N=p. Hence, there is a constant C>0 such that(22)u∂Ω,qs≤CuW1,pΩ,∀u∈W1,pΩ.Throughout the remainder of this article, for 1<p<N, we let(23)ps≔pNN-p,qs≔pN-1N-p.Ifp>N, one has that(24)W1,pΩ↪C0,1-N/pΩ¯;that is, the space W1,p(Ω) is continuously embedded into C0,1-N/p(Ω¯). For more details, we refer to [16, Theorem  4.7] (see also [17, Chapter  4]).For1<q<∞, we define the Sobolev space W1,q(∂Ω) to be the completion of the space C1(∂Ω) with respect to the norm (25)uW1,q∂Ω≔∫∂Ωuqdσ+∫∂Ω∇Γuqdσ1/q,where we recall that ∇Γu denotes the tangential gradient of the function u at the boundary ∂Ω. It is also well-known that W1,q∂Ω is continuously embedded into Lqt(∂Ω) where qt≔q(N-1)/(N-1-q) if 1<q<N-1 and 1≤qt<∞ if q=N-1. Hence, for 1<q≤N-1, there exists a constant C>0 such that(26)uqt,∂Ω≤CuW1,q∂Ω,∀u∈W1,q∂Ω.LetλN denote the N-dimensional Lebesgue measure and let the measure μ≔λNΩ⊕σ on Ω¯ be defined for every measurable set A⊂Ω¯ by (27)μA≔λNΩ∩A+σA∩∂Ω.For p,q∈1,∞, we define the Banach space (28)Xp,qΩ¯,μ≔F=f,g:f∈LpΩ,g∈Lq∂Ω,endowed with the norm (29)FXp,qΩ¯=Fp,q≔fΩ,p+g∂Ω,q,if 1≤p,q<∞, and (30)FX∞,∞Ω¯,μ=F∞≔maxfΩ,∞,g∂Ω,∞.If p=q, we will simply denote Fp,p=Fp.Identifying each functionu∈W1,p(Ω) with U=(u,u∂Ω), we have that W1,p(Ω) is a subspace of Xp,p(Ω¯,μ).For1<p,q<∞, we endow (31)V1≔U≔u,u∂Ω,u∈W1,pΩ,u∂Ω∈W1,q∂Ω, with the norm (32)UV1≔uW1,pΩ+uW1,q∂Ω, while (33)V0≔U=u,u∂Ω:u∈W1,pΩ is endowed with the norm (34)UV0≔uW1,pΩ.It follows from (21) and (22) that V0 is continuously embedded into Xps,qs(Ω¯,μ), with ps and qs given by (23), for 1<p<N. Moreover, by (21) and (26), V1 is continuously embedded into Xps,qt(Ω¯,μ).


### 2.3. Musielak-Orlicz Type Spaces

For the convenience of the reader, we introduce the Orlicz and Musielak-Orlicz type spaces and prove some properties of these spaces which will be frequently used in the sequel (see Section5).Definition 5.
Let(X,Σ,ν) be a complete measure space. We call a function B:X×R→[0,∞] aMusielak-Orlicz function on X if(a)
B(x,·) is nontrivial, even, and convex for ν, a.e. x∈X;(b)
B(x,·) is vanishing and continuous at 0 for ν, a.e. x∈X;(c)
B(x,·) is left continuous on 0,∞;(d)
B(·,t) is Σ-measurable for all t∈0,∞;(e)
limt→∞(Bx,t/t)=∞.Thecomplementary Musielak-Orlicz functionB~ is defined by (35)B~x,t≔sup⁡st-Bx,s:s>0.It follows directly from the definition that for t,s≥0 (and hence for all t,s∈R) (36)st≤Bx,t+B~x,s.Definition 6.
We say that a Musielak-Orlicz functionB satisfies the (▵α0)-condition (α>1) if there exists a set X0 of ν-measure zero and a constant Cα>1 such that (37)Bx,αt≤CαBx,t,for all t∈R and every x∈X∖X0.
We say thatB satisfies the (∇20)-condition if there is a set X0 of ν-measure zero and a constant c>1 such that (38)Bx,t≤12cBx,ct,for all t∈R and all x∈X∖X0.Definition 7.
A functionΦ:R→0,∞ is called an N-function if
(i)
Φ is even, strictly increasing, and convex;
(ii)
Φ
(
t
)
=
0 if and only if t=0;
(iii)
l
i
m
t
→
0
(
Φ
t
/
t
)
=
0 and limt→∞(Φt/t)=∞.
We say that anN-function Φ satisfies the (▵2)-condition if there exists a constant C2>1 such that (39)Φ2t≤C2Φt,∀t∈R,and it satisfies the (∇2)-condition if there is a constant c>1 such that (40)Φt≤Φct2c,∀t∈R.For more details on N-functions, we refer to the monograph of Adams [18, Chapter  VIII] (see also [19, Chapter  I], [20, Chapter  I]).Remark 8.
For anN-function Φ, we let φ be its left-sided derivative. Then φ is left continuous on (0,∞) and nondecreasing. Let ψ be given by (41)ψs≔inf⁡t>0:φt>s.Then (42)Φt=∫0tφsds;Ψt≔∫0tψsds=sup⁡ts-Φs:s>0.As before for all s,t∈R(43)st≤Φt+Ψs.Moreover, if s=φ(t) or t=ψ(s) then we have equality; that is,(44)Ψφt=tφt-Φt.The function Ψ is called the complementary N-function of Φ. It is also known that an N-function Φ satisfies the (▵2)-condition if and only if(45)ctφt≤Φt≤tφt,for some constant c∈0,1 and for all t∈R, where φ is the left-sided derivative of Φ.Lemma 9.
LetΦ be an N-function which satisfies the (▵2)-condition with the constant C2>1 and let Ψ be its complementary N-function. Then Ψ satisfies the (∇2)-condition with the constant c≔2C2-1.Proof.
We have(46)tφt≤∫t2tφsds≤∫02tφsds=Φ2t≤C2Φt.Since φ(ψ(s))≥s for all s≥0 and s/Ψ(s) and s/(s-1) are decreasing, we get, for t≔ψ(s), that (47)sψsΨs≥φψsψsΨφψs=tφtΨφt=tφttφt-Φt≥C2C2-1.Now let c≔2C2-1. Then for t≥0(48)ln⁡ΨctΨt=∫tctψsΨsds≥∫tctC2sC2-1ds=C2C2-1ln⁡c=C2log⁡2=ln⁡2·2C2-1.Hence, Ψ(t)2c≤Ψ(ct).Corollary 10.
LetB be a Musielak-Orlicz function such that B(x,·) is an N-function for ν, a.e. on x. If B satisfies the (▵20)-condition, then B~ satisfies the (∇20)-condition.Definition 11.
LetB be a Musielak-Orlicz function. Then the Musielak-Orlicz space LB(X) associated with B is defined by (49)LBX≔u:X⟶Rmeasurable:ρBuα<∞for  someα>0,where (50)ρBv≔∫XBx,vxdνx. On this space we consider the Luxemburg norm ·X,B defined by (51)uX,B≔inf⁡α>0:ρBuα≤1.Proposition 12.
LetB be a Musielak-Orlicz function which satisfies the (∇20)-condition. Then (52)limuX,B→+∞⁡ρBuuX,B=+∞.Proof.
IfB satisfies the (∇20)-condition, then there exists a set X0⊂X of measure zero such that for every ε>0 there exists α=α(ε)>0(53)Bx,αt≤αεBx,t,for all t∈R and all x∈X∖X0. Let λ∈0,∞ be fixed. For ε≔1/λ there exists α>0 satisfying the above inequality. We will show that ρBu≥λuX,B whenever uX,B>1/α. Assume that uX,B>1/α and let δ>0 be such that α=(1+δ)/uX,B. Then (54)ρBαu=∫XBx,u1+δuX,Bdμ≥1+δ1-1/n∫XBx,u1+δ1/nuX,Bdμ≥1+δ1-1/n,for all n∈N. If we assume that the last inequality does not hold, then (55)uX,B1+δ∈α>0:ρuα≤1,and this clearly contradicts the definition of uX,B. Therefore, we must have(56)ρBαu≥1+δ=αuX,B.From (53), (56), we obtain (57)ρBu=∫XBx,uxdμ≥λα∫XBx,αuxdμ=λαρBαu≥λuX,B.
The proof is finished.Corollary 13.
LetB be a Musielak-Orlicz function such that B(x,·) is an N-function for ν, a.e. on x. If its complementary N-function B~ satisfies the (▵20)-condition, then B satisfies the (∇20)-condition and (58)limuX,B→+∞ρBuuX,B=+∞.


### 2.4. Some Tools

For the reader’s convenience, we report here below some useful inequalities which will be needed in the course of investigation.Lemma 14.
Leta,b∈RN and p∈(1,∞). Then, there exists a constant Cp>0 such that(59)ap-2a-bp-2ba-b≥Cpa+bp-2a-b2≥0.If p∈2,∞, then there exists a constant cp∈(0,1] such that(60)ap-2a-bp-2ba-b≥cpa-bp.Proof.
The proof of (60) is included in [21, Lemma  I.4.4]. In order to show (59), one only needs to show that the left hand side is nonnegative, which follows easily.The following result which is of analytic nature and whose proof can be found in [22, Lemma  3.11] will be useful in deriving some a priori estimates of weak solutions of elliptic equations.Lemma 15.
Letψ:k0,∞→R be a nonnegative, nonincreasing function such that there are positive constants c,α and δ (δ>1) such that (61)ψh≤ch-k-αψkδ,∀h>k≥k0.Then ψ(k0+d)=0 with d=c1/αψ(k0)(δ-1)/α2δ(δ-1).


## 2.1. Maximal Monotone Operators

LetH be a real Hilbert space with scalar product (·,·)H.Definition 3.
LetA:D(A)⊂H→H be a closed (nonlinear) operator. The operator A is said to be(i)
monotone if for allu,v∈D(A) one has (15)Au-Av,u-vH≥0,(ii)
maximal monotone if it is monotone and the operatorI+A is invertible.Next, letV be a real reflexive Banach space which is densely and continuously embedded into the real Hilbert space H, and let V′ be its dual space such that V↪H↪V′.Definition 4.
LetA:V×V→R be a continuous map.(a)
The mapA:V×V→R is called a nonlinear form on H if for all u∈V one has A(u,·)∈V′, that is, if A is linear and bounded in the second variable.(b)
The nonlinear formA:V×V→R is said to be
(i)
monotone ifA(u,u-v)-A(v,u-v)≥0forallu,v∈V;(ii)
hemicontinuous iflimt↓0A(u+tv,w)=A(u,w), ∀u,v,w∈V;(iii)
coercive, iflimvV→+∞(Av,v/vV)=+∞.Now, letφ:H→-∞,+∞ be a proper, convex, lower semicontinuous functional with effective domain (16)Dφ≔u∈H:φu<∞.The subdifferential ∂φ of the functional φ is defined by (17)D∂φ≔u∈Dφ:∃w∈H,∀v∈Dφ:φv-φu≥w,v-uH;∂φu≔w∈H:∀v∈Dφ:φv-φu≥w,v-uH.By a classical result of Minty [13] (see also [12, 14]), ∂φ is a maximal monotone operator.


## 2.2. Functional Setup

LetΩ⊂RN be a bounded domain with a Lipschitz boundary ∂Ω. For 1<p<∞, we let W1,p(Ω) be the first-order Sobolev space; that is,(18)W1,pΩ=u∈LpΩ:∇u∈LpΩN.Then W1,p(Ω), endowed with the norm (19)uW1,pΩ≔uΩ,pp+∇uΩ,pp1/p,is a Banach space, where we have set (20)uΩ,pp≔∫Ωupdx.Since Ω has a Lipschitz boundary, it is well-known that there exists a constant C>0 such that(21)uΩ,ps≤CuW1,pΩ,∀u∈W1,pΩ,where ps=pN/(N-p) if p<N and 1≤ps<∞ if N=p. Moreover the trace operator tr(u)≔u∣∂Ω initially defined for u∈C1(Ω¯) has an extension to a bounded linear operator from W1,p(Ω) into Lqs(∂Ω) where qs≔p(N-1)/(N-p) if p<N and 1≤qs<∞ if N=p. Hence, there is a constant C>0 such that(22)u∂Ω,qs≤CuW1,pΩ,∀u∈W1,pΩ.Throughout the remainder of this article, for 1<p<N, we let(23)ps≔pNN-p,qs≔pN-1N-p.Ifp>N, one has that(24)W1,pΩ↪C0,1-N/pΩ¯;that is, the space W1,p(Ω) is continuously embedded into C0,1-N/p(Ω¯). For more details, we refer to [16, Theorem  4.7] (see also [17, Chapter  4]).For1<q<∞, we define the Sobolev space W1,q(∂Ω) to be the completion of the space C1(∂Ω) with respect to the norm (25)uW1,q∂Ω≔∫∂Ωuqdσ+∫∂Ω∇Γuqdσ1/q,where we recall that ∇Γu denotes the tangential gradient of the function u at the boundary ∂Ω. It is also well-known that W1,q∂Ω is continuously embedded into Lqt(∂Ω) where qt≔q(N-1)/(N-1-q) if 1<q<N-1 and 1≤qt<∞ if q=N-1. Hence, for 1<q≤N-1, there exists a constant C>0 such that(26)uqt,∂Ω≤CuW1,q∂Ω,∀u∈W1,q∂Ω.LetλN denote the N-dimensional Lebesgue measure and let the measure μ≔λNΩ⊕σ on Ω¯ be defined for every measurable set A⊂Ω¯ by (27)μA≔λNΩ∩A+σA∩∂Ω.For p,q∈1,∞, we define the Banach space (28)Xp,qΩ¯,μ≔F=f,g:f∈LpΩ,g∈Lq∂Ω,endowed with the norm (29)FXp,qΩ¯=Fp,q≔fΩ,p+g∂Ω,q,if 1≤p,q<∞, and (30)FX∞,∞Ω¯,μ=F∞≔maxfΩ,∞,g∂Ω,∞.If p=q, we will simply denote Fp,p=Fp.Identifying each functionu∈W1,p(Ω) with U=(u,u∂Ω), we have that W1,p(Ω) is a subspace of Xp,p(Ω¯,μ).For1<p,q<∞, we endow (31)V1≔U≔u,u∂Ω,u∈W1,pΩ,u∂Ω∈W1,q∂Ω, with the norm (32)UV1≔uW1,pΩ+uW1,q∂Ω, while (33)V0≔U=u,u∂Ω:u∈W1,pΩ is endowed with the norm (34)UV0≔uW1,pΩ.It follows from (21) and (22) that V0 is continuously embedded into Xps,qs(Ω¯,μ), with ps and qs given by (23), for 1<p<N. Moreover, by (21) and (26), V1 is continuously embedded into Xps,qt(Ω¯,μ).


## 2.3. Musielak-Orlicz Type Spaces

For the convenience of the reader, we introduce the Orlicz and Musielak-Orlicz type spaces and prove some properties of these spaces which will be frequently used in the sequel (see Section5).Definition 5.
Let(X,Σ,ν) be a complete measure space. We call a function B:X×R→[0,∞] aMusielak-Orlicz function on X if(a)
B(x,·) is nontrivial, even, and convex for ν, a.e. x∈X;(b)
B(x,·) is vanishing and continuous at 0 for ν, a.e. x∈X;(c)
B(x,·) is left continuous on 0,∞;(d)
B(·,t) is Σ-measurable for all t∈0,∞;(e)
limt→∞(Bx,t/t)=∞.Thecomplementary Musielak-Orlicz functionB~ is defined by (35)B~x,t≔sup⁡st-Bx,s:s>0.It follows directly from the definition that for t,s≥0 (and hence for all t,s∈R) (36)st≤Bx,t+B~x,s.Definition 6.
We say that a Musielak-Orlicz functionB satisfies the (▵α0)-condition (α>1) if there exists a set X0 of ν-measure zero and a constant Cα>1 such that (37)Bx,αt≤CαBx,t,for all t∈R and every x∈X∖X0.
We say thatB satisfies the (∇20)-condition if there is a set X0 of ν-measure zero and a constant c>1 such that (38)Bx,t≤12cBx,ct,for all t∈R and all x∈X∖X0.Definition 7.
A functionΦ:R→0,∞ is called an N-function if
(i)
Φ is even, strictly increasing, and convex;
(ii)
Φ
(
t
)
=
0 if and only if t=0;
(iii)
l
i
m
t
→
0
(
Φ
t
/
t
)
=
0 and limt→∞(Φt/t)=∞.
We say that anN-function Φ satisfies the (▵2)-condition if there exists a constant C2>1 such that (39)Φ2t≤C2Φt,∀t∈R,and it satisfies the (∇2)-condition if there is a constant c>1 such that (40)Φt≤Φct2c,∀t∈R.For more details on N-functions, we refer to the monograph of Adams [18, Chapter  VIII] (see also [19, Chapter  I], [20, Chapter  I]).Remark 8.
For anN-function Φ, we let φ be its left-sided derivative. Then φ is left continuous on (0,∞) and nondecreasing. Let ψ be given by (41)ψs≔inf⁡t>0:φt>s.Then (42)Φt=∫0tφsds;Ψt≔∫0tψsds=sup⁡ts-Φs:s>0.As before for all s,t∈R(43)st≤Φt+Ψs.Moreover, if s=φ(t) or t=ψ(s) then we have equality; that is,(44)Ψφt=tφt-Φt.The function Ψ is called the complementary N-function of Φ. It is also known that an N-function Φ satisfies the (▵2)-condition if and only if(45)ctφt≤Φt≤tφt,for some constant c∈0,1 and for all t∈R, where φ is the left-sided derivative of Φ.Lemma 9.
LetΦ be an N-function which satisfies the (▵2)-condition with the constant C2>1 and let Ψ be its complementary N-function. Then Ψ satisfies the (∇2)-condition with the constant c≔2C2-1.Proof.
We have(46)tφt≤∫t2tφsds≤∫02tφsds=Φ2t≤C2Φt.Since φ(ψ(s))≥s for all s≥0 and s/Ψ(s) and s/(s-1) are decreasing, we get, for t≔ψ(s), that (47)sψsΨs≥φψsψsΨφψs=tφtΨφt=tφttφt-Φt≥C2C2-1.Now let c≔2C2-1. Then for t≥0(48)ln⁡ΨctΨt=∫tctψsΨsds≥∫tctC2sC2-1ds=C2C2-1ln⁡c=C2log⁡2=ln⁡2·2C2-1.Hence, Ψ(t)2c≤Ψ(ct).Corollary 10.
LetB be a Musielak-Orlicz function such that B(x,·) is an N-function for ν, a.e. on x. If B satisfies the (▵20)-condition, then B~ satisfies the (∇20)-condition.Definition 11.
LetB be a Musielak-Orlicz function. Then the Musielak-Orlicz space LB(X) associated with B is defined by (49)LBX≔u:X⟶Rmeasurable:ρBuα<∞for  someα>0,where (50)ρBv≔∫XBx,vxdνx. On this space we consider the Luxemburg norm ·X,B defined by (51)uX,B≔inf⁡α>0:ρBuα≤1.Proposition 12.
LetB be a Musielak-Orlicz function which satisfies the (∇20)-condition. Then (52)limuX,B→+∞⁡ρBuuX,B=+∞.Proof.
IfB satisfies the (∇20)-condition, then there exists a set X0⊂X of measure zero such that for every ε>0 there exists α=α(ε)>0(53)Bx,αt≤αεBx,t,for all t∈R and all x∈X∖X0. Let λ∈0,∞ be fixed. For ε≔1/λ there exists α>0 satisfying the above inequality. We will show that ρBu≥λuX,B whenever uX,B>1/α. Assume that uX,B>1/α and let δ>0 be such that α=(1+δ)/uX,B. Then (54)ρBαu=∫XBx,u1+δuX,Bdμ≥1+δ1-1/n∫XBx,u1+δ1/nuX,Bdμ≥1+δ1-1/n,for all n∈N. If we assume that the last inequality does not hold, then (55)uX,B1+δ∈α>0:ρuα≤1,and this clearly contradicts the definition of uX,B. Therefore, we must have(56)ρBαu≥1+δ=αuX,B.From (53), (56), we obtain (57)ρBu=∫XBx,uxdμ≥λα∫XBx,αuxdμ=λαρBαu≥λuX,B.
The proof is finished.Corollary 13.
LetB be a Musielak-Orlicz function such that B(x,·) is an N-function for ν, a.e. on x. If its complementary N-function B~ satisfies the (▵20)-condition, then B satisfies the (∇20)-condition and (58)limuX,B→+∞ρBuuX,B=+∞.


## 2.4. Some Tools

For the reader’s convenience, we report here below some useful inequalities which will be needed in the course of investigation.Lemma 14.
Leta,b∈RN and p∈(1,∞). Then, there exists a constant Cp>0 such that(59)ap-2a-bp-2ba-b≥Cpa+bp-2a-b2≥0.If p∈2,∞, then there exists a constant cp∈(0,1] such that(60)ap-2a-bp-2ba-b≥cpa-bp.Proof.
The proof of (60) is included in [21, Lemma  I.4.4]. In order to show (59), one only needs to show that the left hand side is nonnegative, which follows easily.The following result which is of analytic nature and whose proof can be found in [22, Lemma  3.11] will be useful in deriving some a priori estimates of weak solutions of elliptic equations.Lemma 15.
Letψ:k0,∞→R be a nonnegative, nonincreasing function such that there are positive constants c,α and δ (δ>1) such that (61)ψh≤ch-k-αψkδ,∀h>k≥k0.Then ψ(k0+d)=0 with d=c1/αψ(k0)(δ-1)/α2δ(δ-1).


## 3. The Fredholm Alternative

In what follows, we assume thatΩ⊂RN is a bounded domain with Lipschitz boundary ∂Ω. Let b∈L∞(∂Ω) satisfy b(x)≥b0>0 for some constant b0. Let X2 be the real Hilbert space L2Ω,dx⊕L2∂Ω,dσ/b. Then, it is clear that X2 is isomorphic to X2,2(Ω¯,λN⊕σ) with equivalent norms.Next, letρ∈{0,1} and p,q∈(1,+∞) be fixed. We define the functional Jρ:X2→0,+∞ by setting(62)JρU=1p∫Ω∇updx+1q∫∂Ωρ∇Γuqdσ,ifU=u,u∣∂Ω∈DJρ,+∞,ifU∈X2∖DJρ,where the effective domain is given by D(Jρ)=Vρ∩X2.Throughout the remainder of this section, we letμ≔λN⊕dσ/b. The following result can be obtained easily.Proposition 16.
The functionalJρ defined by (62) is proper, convex, and lower semicontinuous on X2=X2,2(Ω¯,μ).The following result contains a computation of the subdifferential∂Jρ for the functional Jρ.Remark 17.
LetU=(u,u∂Ω)∈D(Jρ) and let F≔(f,g)∈∂Jρ(U). Then, by definition, F∈X2 and, for all V=(v,v∂Ω)∈D(Jρ), we have (63)∫Ω¯FV-Udμ≤1p∫Ω∇vp-∇updx+1qρ∫Ω∇Γvq-∇Γuqdσ.Let W=(w,w∂Ω)∈D(Jρ), 0<t≤1, and set V≔tW+U above. Dividing by t and taking the limit as t↓0, we obtain that(64)∫Ω¯FWdμ≤∫Ω∇up-2∇u·∇wdx+ρ∫∂Ω∇Γq-2∇Γu·∇Γwdσ,where we recall that (65)∫Ω¯Fdμ=∫Ωfdx+∫∂Ωgdσb.Choosing w=±ψ with ψ∈D(Ω) (the space of test functions) and integrating by parts in (64), we obtain (66)-Δpu=finD′Ω,g=bx∇up-2∂nu-ρbxΔq,Γuweakly  on∂Ω.Therefore, the single-valued operator ∂Jρ is given by (67)D∂Jρ=U=u,u∣∂Ω∈DJρ,-Δpu,bx∇up-2∂nu-ρbxΔq,Γu∈X2,(68)∂JρU=-Δpu,bx∇up-2∂nu-ρbxΔq,Γu.Since the functionalJρ is proper, convex, and lower semicontinuous, it follows that its subdifferential ∂Jρ is a maximal monotone operator.In the following two lemmas, we establish a relation between the null space of the operatorAρ≔∂Jρ and its range.Lemma 18.
LetNAρ denote the null space of the operator Aρ. Then (69)NAρ=C1=C=c,c:c∈R;that is, NAρ consists of all the real constant functions on Ω¯.Proof.
We say thatU∈NAρ if and only if (by definition) U=(u,u∂Ω) is a weak solution of(70)-Δpu=0,inΩ,bx∇up-2∂nu-ρbxΔq,Γu=0,on∂Ω.A function U=(u,u∂Ω)∈Vρ∩X2 is said to be a weak solution of (70), if, for every V=(v,v∂Ω)∈Vρ∩X2, there holds(71)AρU,V≔∫Ω∇up-2∇u·∇vdx+ρ∫∂Ω∇Γuq-2∇Γu·∇Γvdσ=0.Let C≔(c,c) with c∈R. Then it is clear that C∈NAρ.
Conversely, letU=(u,u∂Ω)∈NAρ. Then, it follows from (71) that(72)AρU,U≔∫Ω∇updx+ρ∫∂Ω∇Γuqdσ=0. Since Ω is bounded and connected, this implies that u is equal to a constant. Therefore, U=C1 and this completes the proof.Lemma 19.
The range of the operatorAρ is given by (73)RAρ=F≔f,g∈X2:∫Ω¯Fdμ≔∫Ωfdx+∫∂Ωgdσbx=0.Proof.
LetF∈R(Aρ)⊂X2. Then there exists U=(u,u∂Ω)∈D(Aρ) such that Aρ(U)=F. More precisely, for every V=(v,v∂Ω)∈Vρ∩X2, we have(74)AρU,V=∫Ω∇up-2∇u·∇vdx⁡+ρ∫∂Ω∇Γuq-2∇Γu·∇Γvdσ=∫Ω¯FVdμ.Taking V=(1,1)∈Vρ∩X2, we obtain that ∫Ω¯Fdμ=0. Hence(75)RAρ⊆F∈X2:∫Ω¯Fdμ=0.
Let us now prove the converse. To this end, letF∈X2 be such that ∫Ω¯Fdμ=0. We have to show that F∈R(Aρ); that is, there exists U∈Vρ∩X2 such that (73) holds, for every V∈Vρ∩X2. To this end, consider (76)Vρ,0≔U=u,u∂Ω∈Vρ∩X2:∫Ω¯Udμ≔∫Ωudx+∫∂Ωudσb=0.It is clear that Vρ,0 is a closed linear subspace of Vρ∩X2↪X2 and therefore is a reflexive Banach space. Using [23, Section  1.1], we have that the norm (77)UVρ,0≔∇up,Ω+ρ∇Γuq,∂Ωdefines an equivalent norm on Vρ,0. Hence, there exists a constant C>0 such that for every U∈Vρ,0(78)U2≤CUVρ,0≔∇up,Ω+ρ∇Γuq,∂Ω.Define the functional Fρ:Vρ,0→R by (79)FρU=1p∫Ω∇updx+ρq∫∂Ω∇Γuqdσ-∫Ω¯FUdμ.It is easy to see that Fρ is convex and lower semicontinuous on X2 (see Proposition 16). We show now that Fρ is coercive. By exploiting a classical Hölder inequality and using (78), we have(80)∫Ω¯FUdμ≤CF2U2≤CF2UVρ,0=CF2∇up,Ω+ρ∇Γuq,∂Ω.Obviously, this estimate yields(81)-∫Ω¯FUdμ≥-CF2∇up,Ω+ρ∇Γuq,∂Ω.Therefore, from (81), we immediately get (82)FρUUVρ,0≥1/p∇up,Ωp+ρ/q∇Γuq,∂Ωq∇up,Ω+ρ∇Γuq,∂Ω-CF2.This inequality implies that (83)limUVρ,0→+∞⁡FρUUVρ,0=+∞,and this shows that the functional Fρ is coercive. Since Fρ is also convex and lower semicontinuous, it follows from [24, Theorem  3.3.4] that there exists a function U∗∈Vρ,0 which minimizes Fρ. More precisely, for all V∈Vρ,0, Fρ(U∗)≤Fρ(V); this implies that for every 0<t≤1 and every V∈Vρ,0(84)FρU∗+tV-FρU∗≥0.Hence(85)limt↓0FρU∗+tV-FρU∗t≥0.Using the Lebesgue Dominated Convergence, easy computation shows that(86)0≤limt↓0⁡FρU∗+tV-FρU∗t=∫Ω∇u∗p-2∇u∗·∇vdx+ρ∫∂Ω∇Γu∗q-2∇Γu∗·∇Γvdσ-∫Ω¯FVdμ.Changing V to -V in (86) gives that(87)∫Ω∇u∗p-2∇u∗·∇vdx+ρ∫∂Ω∇Γu∗q-2∇Γu∗·∇Γvdσ=∫Ω¯FVdμ,for every V∈Vρ,0. Now, let V∈Vρ∩X2. Writing V=V-C+C with C=(c,c)(88)c≔1λ1+λ2∫Ωvdx+∫∂Ωvdσb,and using the fact that ∫Ω¯Fdμ=0, we obtain, for every V∈Vρ∩X2, that (89)∫Ω∇u∗p-2∇u∗·∇vdx+ρ∫∂Ω∇Γu∗q-2∇Γu∗·∇Γvdσ=∫Ω¯FVdμ.Therefore, Aρ(U)=F. Hence, F∈R(Aρ) and this completes the proof of the lemma.The following result is a direct consequence of Lemmas18 and 19. This is the main result of this section.Theorem 20.
The operatorAρ=∂Jρ satisfies the following type of “quasilinear” Fredholm alternative: (90)RAρ=NAρ⊥=F∈X2:F,1X2=0.


## 4. Necessary and Sufficient Conditions for Existence of Solutions

In this section, we prove the first main result (cf. Theorem1) for problem (1). Before we do so, we will need the following results from maximal monotone operators theory and convex analysis.Definition 21.
LetH be a real Hilbert space. Two subsets K1 and K2 of H are said to be almost equal, written as K1≃K2, if K1 and K2 have the same closure and the same interior, that is, K1¯=K2¯ and intK1=intK2.The following abstract result is taken from [9, Theorem  3 and Generalization in p. 173-174].Theorem 22 (Brézis-Haraux).
LetA and B be subdifferentials of proper convex lower semicontinuous functionals φ1 and φ2, respectively, on a real Hilbert space H with D(φ1)∩D(φ2)≠∅, and let C be the subdifferential of the proper, convex lower semicontinuous functional φ1+φ2; that is, C=∂(φ1+φ2). Then (91)RA+RB⊂RC¯,intRA+RB⊂RC. In particular, if the operator A+B is maximal monotone, then (92)RA+B≃RA+RB, and this is the case if ∂(φ1+φ2)=∂φ1+∂φ2.


### 4.1. Assumptions and Intermediate Results

Let us recall that the aim of this section is to establish some necessary and sufficient conditions for the solvability of the following nonlinear elliptic problem:(93)-Δpu+α1u=f,inΩ,bx∇up-2∂nu-ρbxΔq,Γu+α2u=g,on∂Ω,where p,q∈(1,+∞) are fixed. We also assume that αj:R→R (j=1,2) satisfy the following assumptions.Assumption 23.
The functionsαj:R→R(j=1,2) are odd, monotone nondecreasing, and continuous and satisfy αj(0)=0.Letα~j be the inverse of αj. We define the functions Λj,Λ~j:R→R+(j=1,2) by(94)Λjt≔∫0tαjsds,Λ~jt≔∫0tα~jsds.Then it is clear that Λj, Λ~j are even, convex, and monotone increasing on R+, with Λj(0)=Λ~j(0), for each j=1,2. Moreover, since αj are odd, we have Λj′t=αjt, for all t∈R and j=1,2, with a similar relation holding for Λ~j as well. The following result whose proof is included in [19, Chap.  I, Section  1.3, Theorem  3] holds.Lemma 24.
The functionsΛj and Λ~j(j=1,2) satisfy (43) and (44). More precisely, for all s,t∈R(95)st≤Λjs+Λ~jt. If s=αj(t) or t=α~j(s), then we also have equality; that is, (96)Λ~jαjs=sαjs-Λjs,j=1,2.We note that, in [19], the statement of Lemma 24 assumed that Λj, Λ~j are N-functions in the sense of Definition 7. However, the conclusion of that result holds under the weaker hypotheses of Lemma 24.Define the functionalJ2:X2→0,+∞ by (97)J2u,v≔∫ΩΛ1udx+∫∂ΩΛ2vdσb,ifu,v∈DJ2,+∞,ifu,v∈X2∖DJ2,with the effective domain (98)DJ2≔u,v∈X2:∫ΩΛ1udx+∫∂ΩΛ2vdσb<∞.Lemma 25.
Letαj(j=1,2) satisfy Assumption 23. Then the functional J2 is proper, convex, and lower semicontinuous on X2.Proof.
It is routine to check thatJ2 is convex and proper. This follows easily from the convexity of Λj and the fact that Λj(0)=0. To show the lower semicontinuity on X2, let Un=(un,vn)∈D(J2) be such that Un→U≔(u,v) in X2 and J2(Un)≤C for some constant C>0. Since Un→U in X2, then there is a subsequence, which we also denote by Un=(un,vn), such that un→u a.e. on Ω and vn→vσ, a.e. on Γ. Since Λj(·) are continuous (thus, lower semicontinuous), we have (99)Λ1u≤lim⁡infn→∞⁡Λ1un,Λ2v≤lim⁡infn→∞⁡Λ2vn.By Fatou’s Lemma, we obtain (100)∫ΩΛ1udx≤∫Ωlim⁡infn→∞⁡Λ1undx≤lim⁡infn→∞⁡∫ΩΛ1undx,∫∂ΩΛ2vdσb≤∫∂Ωlim⁡infn→∞Λ2vndσb≤lim⁡infn→∞∫∂ΩΛ2vndσb.Hence, J2 is lower semicontinuous on X2.We have the following result whose proof is contained in [19, Chap.  III, Section  3.1, Theorem  2].Lemma 26.
Letαj(j=1,2) satisfy Assumption 23 and assume that there exist constants Cj>1(j=1,2) such that(101)Λj2t≤CjΛjt,∀t∈R.Then D(J2) is a vector space.Let the operatorB2 be defined by(102)DB2=U≔u,v∈X2:α1u,α2v∈X2,B2U=α1u,α2v.We have the following result.Lemma 27.
Let the assumptions of Lemma26 be satisfied. Then the subdifferential of J2 and the operator B2 coincide; that is, for all (u,v)∈D(B2)=D(∂J2)(103)∂J2u,v=B2u,v.Proof.
LetU=(u,v)∈D(J2) and F=(f,g)∈∂J2(u,v). Then, by definition, F∈X2 and, for every V=(u1,v1)∈D(J2), we get (104)∫Ω¯FV-Udμ≤J2V-J2U.Let V=U+tW, with W=(u2,v2)∈D(J2) and 0<t≤1. Then by Lemma 26, V=U+tW∈D(J2). Now, dividing by t and taking the limit as t↓0, we obtain(105)∫ΩFWdμ≤∫Ωα1uu2dx+∫∂Ωα2vv2dσb.Changing W to -W in (105) gives that (106)∫Ω¯FWdμ=∫Ωα1uu2dx+∫∂Ωα2vv2dσb.In particular, if W=(u2,0) with u2∈D(Ω), we have (107)∫Ωfu2dx=∫Ωα1uu2dx,and this shows that α1(u)=f. Similarly, one obtains that α2(v)=g. We have shown that U∈D(B2) and (108)B2U≔B2u,v=α1u,α2v=f,g.Conversely, let U=(u,v)∈D(B2) and set F=(f,g)≔B2(u,v)=(α1(u),α2(v)). Since (α1(u),α2(v))∈X2, from (94) and (101), it follows that (109)∫ΩΛ1udx+∫∂ΩΛ2vdσb<∞. Hence, U=(u,v)∈D(J2). Let V=(u1,v1)∈D(J2). Using Lemma 24, we obtain(110)α1uu1-u=α1uu1-α1uu≤Λ1u1+Λ1α1u-α1uu=Λ1u1-Λ1u,and similarly(111)α2vv1-v≤Λ2v1-Λ2v.Therefore(112)∫Ω¯FV-Udμ=∫Ωα1uu1-udx+∫∂Ωα2vv1-vdσb≤J2V-J2U.By definition, this shows that F=(α1(u),α2(v))=B2(U)∈∂J2(U). We have shown that U∈D(∂J2) and B2(U)∈∂J2(U). This completes the proof of the lemma.Next, we define the functionalJ3,ρ:X2→0,+∞ by(113)J3,ρU=JρU+J2UifU∈DJ3,ρ≔DJρ∩DJ2,+∞ifU∈X2∖DJ3,ρ. Note that for ρ=0(114)DJ3,0=U=u,u∂Ω∈DJ2:u∈W1,pΩ∩L2Ω,u∂Ω∈L2∂Ω,while for ρ=1(115)DJ3,1=U=u,u∂Ω∈DJ2:u∈W1,pΩ∩L2Ω,u∂Ω∈W1,q∂Ω∩L2∂Ω.We have the following result.Lemma 28.
Let the assumptions of Lemma26 be satisfied. Then the subdifferential of the functional J3,ρ is given by (116)D∂J3,ρ=U=u,u∣∂Ω∈DJ3,ρ:-Δpu+α1u∈L2Ω,bx∇up-2∂nu-bxρΔq,Γu+α2u∈L2∂Ω,dσb,∂J3,ρU=-Δpu+α1u,bx∇up-2∂nu-bxρΔq,Γu+α2u.In particular, if, for every U=(u,u∣∂Ω)∈D(J3,ρ), the function α1(u),α2(u)∈X2, then (117)∂J3,ρ≔∂Jρ+J2=∂Jρ+∂J2.Proof.
We calculate the subdifferential∂J3,ρ. Let F=(f,g)∈∂J3,ρ(U); that is, F∈X2, U∈D(J3,ρ)=D(Jρ)∩D(J2), and for every V∈D(J3,ρ), we have (118)∫Ω¯FV-Udμ≤J3,ρV-J3,ρU.Proceeding as in Remark 17 and the proof of Lemma 27, we obtain that (119)-Δpu+α1u=finDΩ′,bx∇up-2∂nu-bxρΔq,Γu+α2u=gweaklyon∂Ω.Noting that ∂J3,ρ is also a single-valued operator (which follows from the assumptions on αj and Λj), we easily obtain (116), and this completes the proof of the first part.
To show the last part, note that it is clear that∂Jρ+∂J2⊂∂J3,ρ always holds. To show the converse inclusion, let us assume that, for every U=(u,u∣∂Ω)∈D(J3,ρ), the function (α1(u),α2(u))∈X2. Then it follows from (68), (102) (since ∂J2=B2), and (116) that D(∂J3,ρ)=D(∂Jρ)∩D(∂J2) and (120)∂J3,ρU=-Δpu+α1u,bx∇up-2∂nu-bxρΔq,Γu+α2u=-Δpu,bx∇up-2∂nu-bxρΔq,Γu+α1u,α2u=∂JρU+∂J2U.This completes the proof.The following lemma is the main ingredient in the proof of Theorem31 below.Lemma 29.
LetB1≔Aρ and set B3≔∂J3,ρ. Then(121)RB1+RB2⊂RB3¯,intRB1+RB2⊂RB3.In particular, if, for every U=(u,u∣∂Ω)∈D(J3,ρ), the function (α1(u),α2(u))∈X2, then(122)RB3≔RB1+B2≃RB1+RB2.Proof.
By Remark17 and Lemmas 27 and 28, the operators B1, B2, and B3 are subdifferentials of proper, convex, and lower semicontinuous functionals Jρ, J2, and Jρ+J2, respectively, on X2. Hence, B1, B2, and B3 are maximal monotone operators. In particular, if (α1(u),α2(u))∈X2, for every U=(u,u∣∂Ω)∈D(J3,ρ), then, by Lemma 28, one has B3=B1+B2. Now, the lemma follows from the celebrated Brézis-Haraux result in Theorem 22.


### 4.2. Statement and Proof of the Main Result

Next, letVρ≔D(J3,ρ) be given by (114) if ρ=0 and by (115) if ρ=1.Definition 30.
LetF=(f,g)∈X2. A function u∈W1,p(Ω) is said to be a weak solution of (93), if α1(u)∈L1(Ω), α2(u)∈L1(∂Ω), u∂Ω∈W1,q(∂Ω), if ρ>0, and(123)∫Ω∇up-2∇u·∇vdx+ρ∫∂Ω∇Γuq-2∇Γu·∇Γvdσ+∫Ωα1uvdx+∫∂Ωα2uvdσb=∫Ωfvdx+∫∂Ωgvdσb,for every v∈W1,p(Ω)∩C(Ω¯) with v∂Ω∈W1,q(∂Ω), if ρ>0.Recall thatλ1≔∫Ωdx and λ2≔∫∂Ω(dσ/b). We also define the average FΩ¯ of F=f,g with respect to the measure μ, as follows: (124)FΩ¯≔1μΩ¯∫Ω¯Fdμ=1μΩ¯∫Ωfdx+∫∂Ωgdσb,where μΩ¯=λ1+λ2. Now, we are ready to state the main result of this section.Theorem 31.
Letαj(j=1,2) satisfy Assumption 23 and assume that the functions Λj(j=1,2) satisfy (101). Let F=f,g∈X2. The following hold:(a)
Suppose that the nonlinear elliptic problem (93) possesses a weak solution. Then(125)FΩ¯∈λ1Rα1+λ2Rα2λ1+λ2.(b)
Assume that(126)FΩ¯∈intλ1Rα1+λ2Rα2λ1+λ2.Then the nonlinear elliptic problem (93) has at least one weak solution.Proof.
We show that condition (125) is necessary. Let F≔(f,g)∈X2 and let U=u,u∣∂Ω∈D(B3)⊂Vρ be a weak solution of B3U=F. Then, by definition, for every V=(v,v∂Ω)∈Vρ, (123) holds. Taking v≡1 in (123) yields (127)∫Ωfdx+∫∂Ωgdσb=∫Ωα1udx+∫∂Ωα2udσb.Hence(128)∫Ωfdx+∫∂Ωgdσb∈λ1Rα1+λ2Rα2,and so (125) holds. This completes the proof of part (a).
We show that condition (126) is sufficient.
(i) First, letC∈C, where (129)C≔C=c1,c2:c1,c2∈Rα1×Rα2.By definition, one has that C⊂RB2 since c1=α1d1 for some constant function d1 on Ω and c2=α2(d2) for some constant function d2 on ∂Ω. Let F∈X2 be such that (126) holds. We must show F∈RB3. By (126), we may choose C=c1,c2∈C such that (130)FΩ¯=λ1c1+λ2c2λ1+λ2∈intλ1Rα1+λ2Rα2λ1+λ2.Then, for F∈X2, we have F=F1+F2 with (131)F1≔F-C,F2=C.First, F1∈RB1=NB1⊥=1⊥, since (132)∫Ω¯F1dμ=∫Ω¯F-Cdμ=∫Ωfdx+∫∂Ωgdσb-λ1c1+λ2c2=λ1+λ2FΩ¯-λ1c1+λ2c2=0.Obviously, F2=C∈RB2. Hence, it is readily seen that (133)F∈RB1+RB2.
(ii) Next, denote byBR(x,r) the open ball in R of center x and radius r>0. Since (134)FΩ¯∈intλ1Rα1+λ2Rα2λ1+λ2,there exists δ>0 such that the open ball (135)BRFΩ¯,δ⊂λ1Rα1+λ2Rα2λ1+λ2.Since the mapping F↦FΩ¯ from X2 into R is continuous, then there exists ε>0 such that (136)GΩ¯∈BRFΩ¯,δ⊂λ1Rα1+λ2Rα2λ1+λ2,for all G∈X2 satisfying F-G2<ε. It finally follows from part (i) above that (RB1+RB2) contains an ε-ball in X2 centered at F. Therefore(137)F∈intRB1+RB2⊂RB3.Consequently, problem (93) is (weakly) solvable for every function F=(f,g)∈X2, if (126) holds. This completes the proof of the theorem.Remark 32.
It is important to remark that, in order to prove Theorem31, we do not require that (α1(u),α2(u)) should belong to X2, for every U=(u,u∣Γ)∈D(J3,ρ). In particular, only the assumption (121) was needed. However, if this happens, then we get the much stronger result in (122) which would require that the nonlinearities α1,α2 satisfy growth assumptions at infinity.We conclude this section with the following corollary and some examples.Corollary 33.
Let the assumptions of Theorem31 be satisfied. Let F=f,g∈X2. Assume that at least one of the sets Rα1, Rα2 is open. Then the nonlinear elliptic problem (93) possesses a weak solution if and only if (126) holds.Remark 34.
Similar results to Theorem31 and Corollary 33 were also obtained in [4, Theorem  4.4], but only when p=q=2.


### 4.3. Examples

We will now give some examples as applications of Theorem31. Let p,q∈(1,+∞) be fixed.Example 35.
Letα1s or α2s be equal to αs=csr-1s, where , r>0. Note that Rα=R. It is easy to check that α satisfies all the conditions of Assumption 23 and that the function Λ(t)=∫0tα(s)ds satisfies (101). Then, it follows that problem (93) is solvable for any ∈L2Ω, g∈L2∂Ω.Example 36.
Consider the case whenρ=α2≡0 in (93); that is, consider the following boundary value problem: (138)-Δpu+α1u=finΩ,bx∇up-2∂nu=gonΓ.Then, by Theorem 31, this problem has a weak solution if (139)∫Ωfdx+∫∂Ωgdσb∈λ1intRα1,which yields the classical Landesman-Lazer result (see (7)) for g≡0 and p=2.Example 37.
Let us now consider the case whenα1≡α and α2≡0, where α is a continuous, odd, and nondecreasing function on R such that α0=0. The problem(140)-Δpu+αu=f,inΩ,bx∇up-2∂nu-ρbxΔq,Γu=g,on∂Ω, has a weak solution if(141)∫Ωfdx+∫∂Ωgdσb∈λ2intRα.Let us now choose αs=arctan⁡s in (140). Then, it is easy to check that (142)Λt≔∫0tαsds=tarctan⁡t-12ln⁡1+t2,t∈R,is monotone increasing on R+ and that it satisfies Λ(2t)≤C2Λ(t), ∀t∈R, for some constant C2>1. Therefore, (141) becomes the necessary and sufficient condition(143)1λ2∫Ωfdx+∫∂Ωgdσb<π2.


## 4.1. Assumptions and Intermediate Results

Let us recall that the aim of this section is to establish some necessary and sufficient conditions for the solvability of the following nonlinear elliptic problem:(93)-Δpu+α1u=f,inΩ,bx∇up-2∂nu-ρbxΔq,Γu+α2u=g,on∂Ω,where p,q∈(1,+∞) are fixed. We also assume that αj:R→R (j=1,2) satisfy the following assumptions.Assumption 23.
The functionsαj:R→R(j=1,2) are odd, monotone nondecreasing, and continuous and satisfy αj(0)=0.Letα~j be the inverse of αj. We define the functions Λj,Λ~j:R→R+(j=1,2) by(94)Λjt≔∫0tαjsds,Λ~jt≔∫0tα~jsds.Then it is clear that Λj, Λ~j are even, convex, and monotone increasing on R+, with Λj(0)=Λ~j(0), for each j=1,2. Moreover, since αj are odd, we have Λj′t=αjt, for all t∈R and j=1,2, with a similar relation holding for Λ~j as well. The following result whose proof is included in [19, Chap.  I, Section  1.3, Theorem  3] holds.Lemma 24.
The functionsΛj and Λ~j(j=1,2) satisfy (43) and (44). More precisely, for all s,t∈R(95)st≤Λjs+Λ~jt. If s=αj(t) or t=α~j(s), then we also have equality; that is, (96)Λ~jαjs=sαjs-Λjs,j=1,2.We note that, in [19], the statement of Lemma 24 assumed that Λj, Λ~j are N-functions in the sense of Definition 7. However, the conclusion of that result holds under the weaker hypotheses of Lemma 24.Define the functionalJ2:X2→0,+∞ by (97)J2u,v≔∫ΩΛ1udx+∫∂ΩΛ2vdσb,ifu,v∈DJ2,+∞,ifu,v∈X2∖DJ2,with the effective domain (98)DJ2≔u,v∈X2:∫ΩΛ1udx+∫∂ΩΛ2vdσb<∞.Lemma 25.
Letαj(j=1,2) satisfy Assumption 23. Then the functional J2 is proper, convex, and lower semicontinuous on X2.Proof.
It is routine to check thatJ2 is convex and proper. This follows easily from the convexity of Λj and the fact that Λj(0)=0. To show the lower semicontinuity on X2, let Un=(un,vn)∈D(J2) be such that Un→U≔(u,v) in X2 and J2(Un)≤C for some constant C>0. Since Un→U in X2, then there is a subsequence, which we also denote by Un=(un,vn), such that un→u a.e. on Ω and vn→vσ, a.e. on Γ. Since Λj(·) are continuous (thus, lower semicontinuous), we have (99)Λ1u≤lim⁡infn→∞⁡Λ1un,Λ2v≤lim⁡infn→∞⁡Λ2vn.By Fatou’s Lemma, we obtain (100)∫ΩΛ1udx≤∫Ωlim⁡infn→∞⁡Λ1undx≤lim⁡infn→∞⁡∫ΩΛ1undx,∫∂ΩΛ2vdσb≤∫∂Ωlim⁡infn→∞Λ2vndσb≤lim⁡infn→∞∫∂ΩΛ2vndσb.Hence, J2 is lower semicontinuous on X2.We have the following result whose proof is contained in [19, Chap.  III, Section  3.1, Theorem  2].Lemma 26.
Letαj(j=1,2) satisfy Assumption 23 and assume that there exist constants Cj>1(j=1,2) such that(101)Λj2t≤CjΛjt,∀t∈R.Then D(J2) is a vector space.Let the operatorB2 be defined by(102)DB2=U≔u,v∈X2:α1u,α2v∈X2,B2U=α1u,α2v.We have the following result.Lemma 27.
Let the assumptions of Lemma26 be satisfied. Then the subdifferential of J2 and the operator B2 coincide; that is, for all (u,v)∈D(B2)=D(∂J2)(103)∂J2u,v=B2u,v.Proof.
LetU=(u,v)∈D(J2) and F=(f,g)∈∂J2(u,v). Then, by definition, F∈X2 and, for every V=(u1,v1)∈D(J2), we get (104)∫Ω¯FV-Udμ≤J2V-J2U.Let V=U+tW, with W=(u2,v2)∈D(J2) and 0<t≤1. Then by Lemma 26, V=U+tW∈D(J2). Now, dividing by t and taking the limit as t↓0, we obtain(105)∫ΩFWdμ≤∫Ωα1uu2dx+∫∂Ωα2vv2dσb.Changing W to -W in (105) gives that (106)∫Ω¯FWdμ=∫Ωα1uu2dx+∫∂Ωα2vv2dσb.In particular, if W=(u2,0) with u2∈D(Ω), we have (107)∫Ωfu2dx=∫Ωα1uu2dx,and this shows that α1(u)=f. Similarly, one obtains that α2(v)=g. We have shown that U∈D(B2) and (108)B2U≔B2u,v=α1u,α2v=f,g.Conversely, let U=(u,v)∈D(B2) and set F=(f,g)≔B2(u,v)=(α1(u),α2(v)). Since (α1(u),α2(v))∈X2, from (94) and (101), it follows that (109)∫ΩΛ1udx+∫∂ΩΛ2vdσb<∞. Hence, U=(u,v)∈D(J2). Let V=(u1,v1)∈D(J2). Using Lemma 24, we obtain(110)α1uu1-u=α1uu1-α1uu≤Λ1u1+Λ1α1u-α1uu=Λ1u1-Λ1u,and similarly(111)α2vv1-v≤Λ2v1-Λ2v.Therefore(112)∫Ω¯FV-Udμ=∫Ωα1uu1-udx+∫∂Ωα2vv1-vdσb≤J2V-J2U.By definition, this shows that F=(α1(u),α2(v))=B2(U)∈∂J2(U). We have shown that U∈D(∂J2) and B2(U)∈∂J2(U). This completes the proof of the lemma.Next, we define the functionalJ3,ρ:X2→0,+∞ by(113)J3,ρU=JρU+J2UifU∈DJ3,ρ≔DJρ∩DJ2,+∞ifU∈X2∖DJ3,ρ. Note that for ρ=0(114)DJ3,0=U=u,u∂Ω∈DJ2:u∈W1,pΩ∩L2Ω,u∂Ω∈L2∂Ω,while for ρ=1(115)DJ3,1=U=u,u∂Ω∈DJ2:u∈W1,pΩ∩L2Ω,u∂Ω∈W1,q∂Ω∩L2∂Ω.We have the following result.Lemma 28.
Let the assumptions of Lemma26 be satisfied. Then the subdifferential of the functional J3,ρ is given by (116)D∂J3,ρ=U=u,u∣∂Ω∈DJ3,ρ:-Δpu+α1u∈L2Ω,bx∇up-2∂nu-bxρΔq,Γu+α2u∈L2∂Ω,dσb,∂J3,ρU=-Δpu+α1u,bx∇up-2∂nu-bxρΔq,Γu+α2u.In particular, if, for every U=(u,u∣∂Ω)∈D(J3,ρ), the function α1(u),α2(u)∈X2, then (117)∂J3,ρ≔∂Jρ+J2=∂Jρ+∂J2.Proof.
We calculate the subdifferential∂J3,ρ. Let F=(f,g)∈∂J3,ρ(U); that is, F∈X2, U∈D(J3,ρ)=D(Jρ)∩D(J2), and for every V∈D(J3,ρ), we have (118)∫Ω¯FV-Udμ≤J3,ρV-J3,ρU.Proceeding as in Remark 17 and the proof of Lemma 27, we obtain that (119)-Δpu+α1u=finDΩ′,bx∇up-2∂nu-bxρΔq,Γu+α2u=gweaklyon∂Ω.Noting that ∂J3,ρ is also a single-valued operator (which follows from the assumptions on αj and Λj), we easily obtain (116), and this completes the proof of the first part.
To show the last part, note that it is clear that∂Jρ+∂J2⊂∂J3,ρ always holds. To show the converse inclusion, let us assume that, for every U=(u,u∣∂Ω)∈D(J3,ρ), the function (α1(u),α2(u))∈X2. Then it follows from (68), (102) (since ∂J2=B2), and (116) that D(∂J3,ρ)=D(∂Jρ)∩D(∂J2) and (120)∂J3,ρU=-Δpu+α1u,bx∇up-2∂nu-bxρΔq,Γu+α2u=-Δpu,bx∇up-2∂nu-bxρΔq,Γu+α1u,α2u=∂JρU+∂J2U.This completes the proof.The following lemma is the main ingredient in the proof of Theorem31 below.Lemma 29.
LetB1≔Aρ and set B3≔∂J3,ρ. Then(121)RB1+RB2⊂RB3¯,intRB1+RB2⊂RB3.In particular, if, for every U=(u,u∣∂Ω)∈D(J3,ρ), the function (α1(u),α2(u))∈X2, then(122)RB3≔RB1+B2≃RB1+RB2.Proof.
By Remark17 and Lemmas 27 and 28, the operators B1, B2, and B3 are subdifferentials of proper, convex, and lower semicontinuous functionals Jρ, J2, and Jρ+J2, respectively, on X2. Hence, B1, B2, and B3 are maximal monotone operators. In particular, if (α1(u),α2(u))∈X2, for every U=(u,u∣∂Ω)∈D(J3,ρ), then, by Lemma 28, one has B3=B1+B2. Now, the lemma follows from the celebrated Brézis-Haraux result in Theorem 22.


## 4.2. Statement and Proof of the Main Result

Next, letVρ≔D(J3,ρ) be given by (114) if ρ=0 and by (115) if ρ=1.Definition 30.
LetF=(f,g)∈X2. A function u∈W1,p(Ω) is said to be a weak solution of (93), if α1(u)∈L1(Ω), α2(u)∈L1(∂Ω), u∂Ω∈W1,q(∂Ω), if ρ>0, and(123)∫Ω∇up-2∇u·∇vdx+ρ∫∂Ω∇Γuq-2∇Γu·∇Γvdσ+∫Ωα1uvdx+∫∂Ωα2uvdσb=∫Ωfvdx+∫∂Ωgvdσb,for every v∈W1,p(Ω)∩C(Ω¯) with v∂Ω∈W1,q(∂Ω), if ρ>0.Recall thatλ1≔∫Ωdx and λ2≔∫∂Ω(dσ/b). We also define the average FΩ¯ of F=f,g with respect to the measure μ, as follows: (124)FΩ¯≔1μΩ¯∫Ω¯Fdμ=1μΩ¯∫Ωfdx+∫∂Ωgdσb,where μΩ¯=λ1+λ2. Now, we are ready to state the main result of this section.Theorem 31.
Letαj(j=1,2) satisfy Assumption 23 and assume that the functions Λj(j=1,2) satisfy (101). Let F=f,g∈X2. The following hold:(a)
Suppose that the nonlinear elliptic problem (93) possesses a weak solution. Then(125)FΩ¯∈λ1Rα1+λ2Rα2λ1+λ2.(b)
Assume that(126)FΩ¯∈intλ1Rα1+λ2Rα2λ1+λ2.Then the nonlinear elliptic problem (93) has at least one weak solution.Proof.
We show that condition (125) is necessary. Let F≔(f,g)∈X2 and let U=u,u∣∂Ω∈D(B3)⊂Vρ be a weak solution of B3U=F. Then, by definition, for every V=(v,v∂Ω)∈Vρ, (123) holds. Taking v≡1 in (123) yields (127)∫Ωfdx+∫∂Ωgdσb=∫Ωα1udx+∫∂Ωα2udσb.Hence(128)∫Ωfdx+∫∂Ωgdσb∈λ1Rα1+λ2Rα2,and so (125) holds. This completes the proof of part (a).
We show that condition (126) is sufficient.
(i) First, letC∈C, where (129)C≔C=c1,c2:c1,c2∈Rα1×Rα2.By definition, one has that C⊂RB2 since c1=α1d1 for some constant function d1 on Ω and c2=α2(d2) for some constant function d2 on ∂Ω. Let F∈X2 be such that (126) holds. We must show F∈RB3. By (126), we may choose C=c1,c2∈C such that (130)FΩ¯=λ1c1+λ2c2λ1+λ2∈intλ1Rα1+λ2Rα2λ1+λ2.Then, for F∈X2, we have F=F1+F2 with (131)F1≔F-C,F2=C.First, F1∈RB1=NB1⊥=1⊥, since (132)∫Ω¯F1dμ=∫Ω¯F-Cdμ=∫Ωfdx+∫∂Ωgdσb-λ1c1+λ2c2=λ1+λ2FΩ¯-λ1c1+λ2c2=0.Obviously, F2=C∈RB2. Hence, it is readily seen that (133)F∈RB1+RB2.
(ii) Next, denote byBR(x,r) the open ball in R of center x and radius r>0. Since (134)FΩ¯∈intλ1Rα1+λ2Rα2λ1+λ2,there exists δ>0 such that the open ball (135)BRFΩ¯,δ⊂λ1Rα1+λ2Rα2λ1+λ2.Since the mapping F↦FΩ¯ from X2 into R is continuous, then there exists ε>0 such that (136)GΩ¯∈BRFΩ¯,δ⊂λ1Rα1+λ2Rα2λ1+λ2,for all G∈X2 satisfying F-G2<ε. It finally follows from part (i) above that (RB1+RB2) contains an ε-ball in X2 centered at F. Therefore(137)F∈intRB1+RB2⊂RB3.Consequently, problem (93) is (weakly) solvable for every function F=(f,g)∈X2, if (126) holds. This completes the proof of the theorem.Remark 32.
It is important to remark that, in order to prove Theorem31, we do not require that (α1(u),α2(u)) should belong to X2, for every U=(u,u∣Γ)∈D(J3,ρ). In particular, only the assumption (121) was needed. However, if this happens, then we get the much stronger result in (122) which would require that the nonlinearities α1,α2 satisfy growth assumptions at infinity.We conclude this section with the following corollary and some examples.Corollary 33.
Let the assumptions of Theorem31 be satisfied. Let F=f,g∈X2. Assume that at least one of the sets Rα1, Rα2 is open. Then the nonlinear elliptic problem (93) possesses a weak solution if and only if (126) holds.Remark 34.
Similar results to Theorem31 and Corollary 33 were also obtained in [4, Theorem  4.4], but only when p=q=2.


## 4.3. Examples

We will now give some examples as applications of Theorem31. Let p,q∈(1,+∞) be fixed.Example 35.
Letα1s or α2s be equal to αs=csr-1s, where , r>0. Note that Rα=R. It is easy to check that α satisfies all the conditions of Assumption 23 and that the function Λ(t)=∫0tα(s)ds satisfies (101). Then, it follows that problem (93) is solvable for any ∈L2Ω, g∈L2∂Ω.Example 36.
Consider the case whenρ=α2≡0 in (93); that is, consider the following boundary value problem: (138)-Δpu+α1u=finΩ,bx∇up-2∂nu=gonΓ.Then, by Theorem 31, this problem has a weak solution if (139)∫Ωfdx+∫∂Ωgdσb∈λ1intRα1,which yields the classical Landesman-Lazer result (see (7)) for g≡0 and p=2.Example 37.
Let us now consider the case whenα1≡α and α2≡0, where α is a continuous, odd, and nondecreasing function on R such that α0=0. The problem(140)-Δpu+αu=f,inΩ,bx∇up-2∂nu-ρbxΔq,Γu=g,on∂Ω, has a weak solution if(141)∫Ωfdx+∫∂Ωgdσb∈λ2intRα.Let us now choose αs=arctan⁡s in (140). Then, it is easy to check that (142)Λt≔∫0tαsds=tarctan⁡t-12ln⁡1+t2,t∈R,is monotone increasing on R+ and that it satisfies Λ(2t)≤C2Λ(t), ∀t∈R, for some constant C2>1. Therefore, (141) becomes the necessary and sufficient condition(143)1λ2∫Ωfdx+∫∂Ωgdσb<π2.


## 5. A Priori Estimates

LetΩ⊂RN be a bounded Lipschitz domain with boundary ∂Ω. Recall that 1<p,q<∞, ρ∈{0,1}, and b∈L∞(∂Ω) with b(x)≥b0>0, for some constant b0. We consider the nonlinear elliptic boundary value problem formally given by(144)-Δpu+α1x,u+up-2u=f,inΩ-ρbxΔq,Γu+ρbxuq-2u+bx∇up-2∂nu+α2x,u=g,on∂Ω,where f∈Lp1(Ω) and g∈Lq1(∂Ω) for some 1≤p1,q1≤∞. If ρ=0, then the boundary conditions in (144) are of Robin type. Existence and regularity of weak solutions for this case have been obtained in [25] for p=2 (see also [26] for the linear case) and for general p in [27]. Therefore, we will concentrate our attention to the case ρ=1 only; in this case, the boundary condition in (144) is a generalized Wentzell-Robin boundary condition. For the sake of simplicity, from now on we will also take b≡1.


### 5.1. General Assumptions

Throughout this section, we assume that the functionsα1:Ω×R→R and α2:∂Ω×R→R satisfy the following conditions.Assumption 38.
(145)
α
j
x
,
·
i
s
o
d
d
a
n
d
s
t
r
i
c
t
l
y
i
n
c
r
e
a
s
i
n
g
,
α
j
x
,
0
=
0
,
α
j
x
,
·
i
s
c
o
n
t
i
n
u
o
u
s
,
lim
t
→
0
⁡
α
j
x
,
t
t
=
0
,
lim
t
→
∞
⁡
α
j
x
,
t
t
=
∞
, for λN, a.e. x∈Ω if j=1, and σ, a.e. x∈∂Ω if j=2.Sinceαj(x,·) are strictly increasing for λN, a.e. x∈Ω if j=1, and σ, a.e. x∈∂Ω if j=2, then they have inverses which we denote by α~j(x,·) (cf. also Section 4). We define the functions Λ1, Λ~1:Ω×R→0,∞, and Λ2, Λ~2:∂Ω×R→0,∞ by (146)Λjx,t≔∫0tαjx,sds,Λ~jx,t≔∫0tα~jx,sds.Then, it is clear that, for λN, a.e. x∈Ω if j=1, and σ, a.e. x∈∂Ω if j=2, Λj(x,·) and Λ~j(x,·) are differentiable, monotone, and convex with Λj(x,0)=Λ~j(x,0)=0. Furthermore, Λj(x,·) is an N-function and Λ~j(x,·) is its complementary N-function. The function Λ~j is then the complementary Musielak-Orlick function of Λj in the sense of Young (see Definition 5).Assumption 39.
We assume, forλN, a.e. x∈Ω if j=1, and σ, a.e. x∈∂Ω if j=2, that Λj(x,·) and Λ~j(x,·) satisfy the (▵2)-condition in the sense of Definition 7.It follows from Assumption39 that there exist two constants c1,c2∈(0,1] such that for λN, a.e. x∈Ω if j=1, and σ, a.e. x∈∂Ω if j=2, and for all t∈R(147)cjtαjx,t≤Λjx,t≤tαjx,t.Next, let(148)LΛ1Ω≔u:Ω⟶Rmeasurable:∫ΩΛ1x,udx<∞,LΛ2∂Ω≔u:∂Ω⟶Rmeasurable:∫∂ΩΛ2x,udσ<∞.Since Λj(x,·) and Λ~j(x,·) satisfy the (▵2)-condition, it follows from [18, Theorem  8.19] that LΛ1(Ω) and LΛ2(∂Ω), endowed, respectively, with the norms (149)uΛ1,Ω≔inf⁡k>0:∫ΩΛ1x,uxkdx≤1,uΛ2,∂Ω≔inf⁡k>0:∫∂ΩΛ2x,uxkdσ≤1,are reflexive Banach spaces. Moreover, by [18, Section  8.11, p. 234], the following generalized versions of Hölder’s inequality will also become useful in the sequel(150)∫Ωuvdx≤2uΛ1,ΩvΛ~1,Ω,∫∂Ωuvdσ≤2uΛ2,∂ΩvΛ~2,∂Ω.


### 5.2. Existence and Uniqueness of Weak Solutions of Perturbed Equations

Let(151)V≔U≔u,u∂Ω:u∈W1,pΩ∩LΛ1Ω,u∣∂Ω∈W1,q∂Ω∩LΛ2∂Ω.Then, for every 1<p,q<∞, V endowed with the norm (152)UV=uW1,pΩ+uΛ1,Ω+uW1,q∂Ω+uΛ2,∂Ωis a reflexive Banach space. Recall that ρ=1. Throughout the following, we denote by V′ the dual of V.Definition 40.
A functionU=(u,u∂Ω)∈V is said to be a weak solution of (144), if for every V∈V=(v,v∂Ω)(153)AU,V=∫Ωfvdx+∫∂Ωgvdσ,provided that the integrals on the right-hand side exist. Here(154)AU,V≔∫Ω∇up-2∇u·∇vdx+∫Ωup-2uvdx+∫Ωα1x,uvdx+∫∂Ω∇Γuq-2∇Γu·∇Γvdσ+∫∂Ωuq-2uvdσ+∫∂Ωα2x,uvdσ.Lemma 41.
Assume Assumptions38 and 39. Let 1<p,q<∞ and U∈V be fixed. Then the functional V↦A(U,V) belongs to V′. Moreover, A is strictly monotone, hemicontinuous, and coercive.Proof.
LetU=(u,u∂Ω)∈V be fixed. It is clear that A(U,·) is linear. Let V=(v,v∂Ω)∈V. Then, exploiting (150), we obtain(155)AU,V≤uW1,pΩp-1vW1,pΩ+uW1,q∂Ωq-1vW1,q∂Ω+2max⁡1,∫ΩΛ~1x,α1x,udxvΛ1,Ω+2max⁡1,∫∂ΩΛ~2x,α2x,udσvΛ2,∂Ω≤KUVV,where (156)KU≔uW1,pΩp-1+2max⁡1,∫ΩΛ~1x,α1x,udx+uW1,q∂Ωq-1+2max⁡1,∫∂ΩΛ~2x,α2x,udσ.This shows A(U,·)∈V′, for every U∈V.
Next, letU,V∈V. Then, using (59) and the fact that αj(x,·) are monotone nondecreasing, that is, (αj(x,t)-αj(x,s))(t-s)≥0, for all t,s∈R, we obtain(157)AU,U-V-AV,U-V=∫Ω∇up-2∇u-∇vp-2∇v·∇u-vdx+∫Ωup-2u-vp-2vu-vdx+∫Ωα1x,u-α1x,vu-vdx+∫∂Ωuq-2u-vq-2vu-vdσ+∫∂Ω∇Γuq-2∇Γu-∇Γvq-2∇v·∇Γu-vdσ+∫∂Ωα2x,u-α1x,vu-vdσ≥∫Ω∇u+∇vp-2∇u-v2dx+∫Ωu+vp-2u-v2dx+∫∂Ω∇Γu+∇Γvp-2∇Γu-v2dσ+∫∂Ωu+vp-2u-v2dσ≥0.This shows that A is monotone. Estimate (157) also shows that (158)AU,U-V-AV,U-V>0,for all U,V∈V with U≠V, that is, u≠v or u∂Ω≠v∂Ω. Thus, A is strictly monotone.
The continuity of the norm function and the continuity ofαj(x,·), j=1,2 imply that A is hemicontinuous.
Finally, sinceΛj and Λ~j satisfy the (▵20)-condition, from Proposition 12 and Corollary 13, it follows that (159)limuΛ1,Ω→+∞⁡∫Ωuα1x,udxuΛ1,Ω=+∞,limuΛ2,∂Ω→+∞⁡∫∂Ωuα2x,udσuΛ2,∂Ω=+∞.Consequently, we deduce(160)limUV→+∞⁡AU,UUV=+∞,which shows that A is coercive. The proof of the lemma is finished.The following result is concerned with the existence and uniqueness of weak solutions to problem (144).Theorem 42.
Assume Assumptions38 and 39. Let 1<p,q<∞, p1≥p∗, and q1≥q∗, where p∗≔p/(p-1) and q∗≔q/(q-1). Then, for every (f,g)∈Xp1,q1(Ω¯,μ), there exists a unique function U∈V which is a weak solution to (144).Proof.
Let〈·,·〉 denote the duality between V and V′. Then, from Lemma 41, it follows that, for each U∈V, there exists A(U)∈V′ such that (161)AU,V=AU,V,for every V∈V. Hence, this relation defines an operator A:V→V′, which is bounded by (155). Exploiting Lemma 41 once again, it is easy to see that A is monotone and coercive. It follows from Browder’s theorem (see, e.g., [28, Theorem  5.3.22]) that A(V)=V′. Therefore, for every F∈V′ there exists U∈V such that A(U)=F; that is, for every V∈V(162)AU,V=AU,V=V,F.Since W1,p(Ω)↪Lp(Ω) and W1,q(∂Ω)↪Lq(∂Ω) with dense injection, by duality, we have Xp∗,q∗(Ω¯,μ)↪V′. Since Ω is bounded and σ(∂Ω)<∞, we obtain that (163)Xp1,q1Ω¯,μ↪Xp∗,q∗Ω¯,μ↪V′.This shows the existence of weak solutions. The uniqueness follows from the fact that A is strictly monotone (cf. Lemma 41). This completes the proof of the theorem.Corollary 43.
Let the assumptions of Theorem42 be satisfied. Let(164)ph≔NpNp-1+p,qh≔pN-1Np-1,qk≔qN-1Nq-1+1.(a)
Let1<p<N, 1<q<p(N-1)/N, p1≥ph, and q1≥qk. Then, for every (f,g)∈Xp1,q1(Ω,μ), there exists a function U∈V which is the unique weak solution to (144).(b)
Let1<q<N-1, 1<p<Nq/(N-1), p1≥ph, and q1≥qh. Then, for every (f,g)∈Xp1,q1(Ω,μ), there exists a function U∈V which is the unique weak solution to (144).Proof.
We first prove (a). Let1<p<N and 1<q<p(N-1)/N and let p1≥ph and q1≥qk, where ph and qk are given by (164). Let ps≔Np/(N-p) and qt≔(N-1)q/(N-1-q). Since W1,p(Ω)↪Lps(Ω) and W1,q(∂Ω)↪Lqt(∂Ω) with dense injection, then, by duality, Xph,qk(Ω¯,μ)↪V′, where 1/ps+1/ph=1 and 1/qt+1/qk=1. Since μ(Ω¯)<∞, we have that (165)Xp1,q1Ω¯,μ↪Xph,qhΩ¯,μ↪V′.Hence, for every F≔(f,g)∈Xp1,q1(Ω¯,μ)↪V′, there exists U∈V such that for every V∈V(166)AU,V=AU,V=∫Ωfvdx+∫∂Ωgvdσ.The uniqueness of the weak solution follows again from the fact that A is strictly monotone.
In order to prove the second part, we use the embeddingsW1,p(Ω)↪Lps(Ω), W1,p(Ω)↪Lqs(∂Ω) and proceed exactly as above. We omit the details.


### 5.3. Properties of the Solution Operator of the Perturbed Equation

In the sequel, we establish some interesting properties of the solution operatorA to problem (144). We begin by assuming the following.Assumption 44.
Suppose thatαj, j=1,2, satisfy the following conditions:(167)thereareconstantscj∈0,1suchthatcjαjx,ξ-η≤αjx,ξ-αjx,η∀ξ,η∈R.Theorem 45.
Assume Assumptions38, 39, and 44. Let p,q≥2 and let A:V→V′ be the continuous and bounded operator constructed in the proof of Theorem 42. Then A is injective and, hence, invertible and its inverse A-1 is also continuous and bounded.Proof.
First, we remark that since(168)αjx,t-αjx,st-s≥0,∀t,s∈R,for λN, a.e. x∈Ω if j=1, and σ, a.e. x∈∂Ω if j=2, it follows from (167) that, for all t,s∈R,(169)αjx,t-αjx,st-s≥cjαjx,t-s·t-s.Let U,V∈V and p,q∈2,∞. Then, exploiting (60), (169), and the (▵2)-condition, we obtain(170)AU-AV,U-V=AU,U-V-AV,U-V=∫Ω∇up-2∇u-∇vp-2∇v·∇u-vdx+∫Ωup-2u-vp-2vu-vdx+∫Ωα1x,u-α1x,vu-vdx+∫∂Ω∇Γuq-2∇Γu-∇Γvq-2∇Γv·∇Γu-vdσ+∫∂Ωuq-2u-vq-2vu-vdσ+∫∂Ωα2x,u-α2x,vu-vdσ≥u-vW1,pΩp+c1∫ΩΛ1x,u-vdx+u-vW1,q∂Ωq+c2∫∂ΩΛ2x,u-vdσ.This implies that 〈A(U)-A(V),U-V〉>0, for all U,V∈V with U≠V (that is, u≠v, or u∂Ω≠v∂Ω). Therefore, the operator A is injective and, hence, A-1 exists. Since for every U∈V(171)AU,U=AU,U≤AUV′UV,from the coercivity of A (see (160)), it is not difficult to see that(172)limUV→+∞⁡AUV′=+∞.Thus, A-1:V′→V is bounded.
Next, we show thatA-1:V′→V is continuous. Assume that A-1 is not continuous. Then there is a sequence Fn∈V′ with Fn→F in V′ and a constant δ>0 such that(173)A-1Fn-A-1FV≥δ,for all n∈N. Let Un≔A-1(Fn) and U=A-1(F). Since Fn is a bounded sequence and A-1 is bounded, we have that Un is bounded in V. Thus, we can select a subsequence, which we still denote by Un, which converges weakly to some function V∈V. Since A(Un)-A(V)→F-A(V) strongly in V and Un-V converges weakly to zero in V, we deduce(174)limn→∞⁡AUn-AV,Un-V=0.From (170) and (174), it follows that (175)limn→∞un-vW1,pΩ=0,limn→∞⁡∫ΩΛ1x,un-vdx=0,while (176)limn→∞⁡un-vW1,q∂Ω=0,limn→∞⁡∫∂ΩΛ2x,un-vdσ=0.Therefore, Un→V strongly in V. Since A is continuous and (177)Fn=AUn⟶AV=F=AU,it follows from the injectivity of A that U=V. This shows that (178)limn→∞A-1Fn-A-1FV=limn→∞Un-UV=0,which contradicts (173). Hence, A-1:V′→V is continuous. The proof is finished.Corollary 46.
Let the assumptions of Theorem45 be satisfied. Let ph,qh, and qk be as in (164) and let A:V→V′ be the continuous and bounded operator constructed in the proof of Theorem 42.(a)
If2≤p<N, 2≤q<p(N-1)/N, p1≥ph, and q1≥qk, then A-1:Xp1,q1(Ω¯,μ)→Xps,qt(Ω¯,μ) is continuous and bounded. Moreover, A-1:Xp1,q1(Ω¯,μ)→V∩Xr,s(Ω¯,μ) is compact for every r∈(1,ps) and s∈(1,qs).(b)
If2≤q<N-1, 2≤p<qN/(N-1), p1≥ph, and q1≥qh, then the operator A-1:Xp1,q1(Ω¯,μ)→Xps,qs(Ω¯,μ) is continuous and bounded. Moreover, A-1:Xp1,q1(Ω¯,μ)→V∩Xr,s(Ω¯,μ) is compact for every r∈(1,ps) and s∈(1,qs).Proof.
We only prove the first part. The second part of the proof follows by analogy and is left to the reader. Let2≤p<N, 2≤q<p(N-1)/N, p1≥ph, and q1≥qk and let F∈Xp1,q1(Ω¯,μ). Proceeding exactly as in the proof of Theorem 45, we obtain (179)A-1Fps,qt≤C1A-1FV≤CFV′≤C2Fp1,q1.Hence, the operator A-1:Xp1,q1(Ω¯,μ)→Xps,qt(Ω¯,μ) is bounded. Finally, using the facts that Xp1,q1(Ω¯,μ)↪V′, A-1:V′→V is continuous and V↪Xps,qt(Ω¯,μ), we easily deduce that A-1:Xp1,q1(Ω¯,μ)→Xps,qt(Ω¯,μ) is continuous.
Now, let1<r<ps and 1<s<qs. Since the injection V↪Xr,s(Ω¯,μ) is compact, then, by duality, the injection Xr′,s′(Ω¯,μ)↪V′ is compact for every r′>ps′=ph and s′>qs′=qh. This, together with the fact that A-1:V′→V is continuous and bounded, implies that A-1:Xp1,q1(Ω¯,μ)→V is compact for every p1>ph and q1>qh.
It remains to show thatA-1 is also compact as a map into Xr,s(Ω¯,μ) for every r∈(1,ps) and s∈(1,qs). Since A-1 is bounded, we have to show that the image of every bounded set B⊂Xp1,q1(Ω,μ) is relatively compact in Xr,s(Ω¯,μ) for every r∈(1,ps) and s∈(1,qs). Let Un be a sequence in A-1(B). Let Fn=A(Un)∈B. Since B is bounded, then the sequence Fn is bounded. Since A-1 is compact as a map into V, it follows that there is a subsequence Fnk such that A-1(Fnk)→U∈V. We may assume that Un=A-1(Fn)→U in V and, hence, in Xp,p(Ω¯,μ). It remains to show that Un→U in Xr,s(Ω¯,μ). Let r∈p,ps and s∈p,qs. Since Un≔(un,un∂Ω) is bounded in Xps,qs(Ω¯,μ), a standard interpolation inequality shows that there exists τ∈(0,1) such that (180)Un-Umr,s≤Un-Ump,pτUn-Umps,qs1-τ≤CUn-Ump,pτ.As Un converges in Xp,p(Ω¯,μ), it follows from the preceding inequality that Un is a Cauchy sequence in Xr,s(Ω¯,μ) and therefore converges in Xr,s(Ω¯,μ). Hence, A-1:Xp1,q1(Ω¯,μ)→V∩Xr,s(Ω¯,μ) is compact for every r∈p,ps and s∈p,qs. The case r,s∈(1,p) follows from the fact that Xp,p(Ω¯,μ)↪Xr,s(Ω¯,μ) and the proof is finished


### 5.4. Statement and Proof of the Main Result

We will now establish under what conditions the operatorA-1 maps Xp1,q1(Ω¯,μ) boundedly and continuously into X∞(Ω¯,μ). The following is the main result of this section.Theorem 47.
Let the assumptions of Theorem45 be satisfied.(a)
Suppose2≤p<N and 2≤q<∞. Let (181)p1>psps-p=Np,q1>qsqs-p=N-1p-1.Letf∈Lp1(Ω),g∈Lq1(∂Ω) and U,V∈V be such that for every function Φ=(φ,φ∂Ω)∈V(182)AU,Φ-AV,Φ=∫Ωfφdx+∫∂Ωgφdσ.Then there is a constantC=C(N,p,q,Ω)>0 such that (183)U-V∞p-1≤Cfp1,Ω+gq1,∂Ω.(b)
Suppose2≤p=q<N-1. Let (184)p1>psps-p=Np,q1>ptpt-p=N-1p.Letf∈Lp1(Ω),g∈Lq1(∂Ω) and U,V∈V satisfy (182). Then there is a constant C=C(N,p,q,Ω)>0 such that (185)U-V∞p-1≤Cfp1,Ω+gq1,∂Ω.Proof.
LetU,V∈V satisfy (182). Let k≥0 be a real number and set (186)wk≔u-v-k+sgn⁡u-vWk≔wk,wk∂Ω,w≔u-v.Let Ak≔{x∈Ω¯:wx≥k} and Ak+≔x∈Ω¯:wx≥k and Ak-≔x∈Ω¯:wx≤-k. Clearly Wk∈V and Ak=Ak+∪Ak-. We claim that there exists a constant C>0 such that(187)CAWk,Wk≤AU,Wk-AV,Wk,for all U,V∈V. Using the definition of the form A, we have(188)AU,Wk-AV,Wk=∫Ω∇up-2∇u-∇vp-2∇v·∇wkdx+∫Ωup-2u-vp-2vwkdx+∫Ωα1x,u-α2x,vwkdx+∫∂Ωuq-2u-vq-2vwkdσ+∫∂Ω∇Γup-2∇Γu-∇Γvp-2∇Γv·∇Γwkdσ+∫∂Ωα2x,u-α2x,vwkdσ.Since(189)∇wk=∇u-vinAk,0otherwise,we can rewrite (188) as follows:(190)AU,Wk-AV,Wk=∫Ak∩Ω∇up-2∇u-∇vp-2∇v·∇u-vdx+∫Ak∩∂Ω∇Γuq-2∇Γu-∇Γvq-2∇Γv·∇Γu-vdσ+λ∫Ak∩Ωup-2u-vp-2vwkdx+∫Ak∩Ωα1x,u-α1x,vwkdx+∫Ak∩∂Ωα2x,u-α2x,vwkdσ.Exploiting inequality (60), from (190) and (169), we deduce that(191)AU,Wk-AV,Wk≥∫Ak∩Ω∇wkp+wkpdx+∫Ak∩Ωc1α1x,wkwkdx+∫Ak∩Ωup-2uwk-vp-2vwk-wkpdx+∫Ak∩Ωα1x,u-α1x,v-c1α1x,wkwkdx+∫Ak∩∂Ω∇Γwkq+wkqdσ+∫Ak∩∂Ωc2α2x,wkwkdσ+∫Ak∩∂Ωuq-2uwk-vq-2vwk-wkqdσ+∫Ak∩∂Ωα2x,u-α2x,v-c2α2x,wkwkdσ≥CAWk,Wk+∫Ak∩Ωup-2uwk-vp-2vwk-wkpdx+∫Ak∩Ωα1x,u-α1x,v-c1α1x,wkwkdx+∫Ak∩∂Ωuq-2uwk-vq-2vwk-wkqdσ+∫Ak∩∂Ωα2x,u-α2x,v-c2α2x,wkwkdσ,where c1,c2 are the constants from (169). Using (167) and the fact that αj(x,·) are strictly increasing, for x∈Ak+, we have (192)cjαjx,wkx=cjαjx,ux-vx-k≤cjαjx,ux-vx≤αjx,ux-αjx,vx.Multiplying this inequality by wk(x)≥0, x∈Ak+, yields(193)αjx,ux-αjx,vx-cjαjx,wkxwkx≥0.Similarly, for x∈Ak-(194)cjαjx,wkx=cjαjx,ux-vx+k≥cjαjx,ux-vx≥αjx,ux-αjx,vx.Hence, multiplying this inequality by wk(x)≤0, we get(195)αjx,ux-αjx,vx-cjαjx,wkxwkx≥0,for all x∈Ak-. Hence, on account of (193) and (195), from (191) we obtain the required estimate of (187).
(a) To prove this part, note that from Definition40 it is clear that(196)wkW1,pΩp≤AWk,Wk.Let f∈Lp1(Ω) and g∈Lq1(∂Ω) with (197)p1>psps-p=Np,q1>qsqs-p=N-1p-1,and let B⊂Ω¯ be any μ-measurable set. We claim that there exists a constant C≥0 such that, for every F∈Xp1,q1(Ω¯,μ) and φ∈W1,p(Ω), we have(198)Fφ1B1,1≤CFp1,q1φW1,pΩχBp3,q3,where p3 and q3 are such that 1/p3+1/p1+1/ps=1 and 1/q3+1/q1+1/qs=1. In fact, note that if n∈N and pi, qi∈1,∞(i=1,…,n) are such that (199)∑i=1n1pi=∑i=1n1qi=1, and if Fi∈Xpi,qi(Ω¯,μ)(i=1,…,n), then by Hölder’s inequality(200)∏i=1nFi1,1≤∏i=1nFipi,qi.Since W1,p(Ω)↪Xps,qs(Ω¯,μ), (198) follows immediately from (200) and claim (198) is proved. Next, it follows from (198) that (201)∫Ω¯FWkdμ=FWk1,1=FWkχAk1,1≤Fp1,q1wkW1,pΩχAkp3,q3,where we recall that 1/p3=1-1/ps-1/p1>p-1/ps and q3<qs/(p-1). Therefore, for every k≥0(202)AU,Wk-AV,Wk≤Fp1,q1wkW1,pΩχAkp3,q3,which together with estimate (187) yields the desired inequality(203)CAWk,Wk≤AU,Wk-BV,Wk≤Fp1,q1wkW1,pΩχAkp3,q3.It follows from (196) and (203) that for every k>0(204)CwkW1,pΩp≤AWk,Wk≤AU,Wk-AV,Wk≤Fp1,q1wkW1,pΩχAkp3,q3.Hence, for every k>0, wkW1,p(Ω)p-1≤C1χAkp3,q3. Using the fact W1,p(Ω)↪Xps,qs(Ω¯,μ), we obtain, for every k>0, that (205)wkps,qsp-1≤CFp1,q1χAkp3,q3.Let h>k. Then Ah⊂Ak and on Ah the inequality wk≥h-k holds. Therefore(206)h-kχAhps,qsp-1≤CFp1,q1χAkp3,q3,which shows that(207)χAhps,qsp-1≤CFp1,q1h-k-p-1χAkp3,q3.Let C3≔1Ω¯ps,qs and (208)δ≔min⁡psp3,qsp3>p-1,δ0≔δp-1>1.Then(209)C3-ps/p3χAkΩ,p3=C3-1χAkΩ,psps/p3≤C3-1χAkΩ,psδ≤χAkps,qsδC3-δ,C3-qs/q3χAk∂Ω,q3=C3-1χAk∂Ω,qsqs/q3≤C3-1χAk∂Ω,qsδ≤χAkps,qsδC3-δ.Choosing CΩ≔C3ps/p3-δ+C3qs/q3-δ, from (209) we have(210)χAkp3,q3≤CΩχAkps,qsδ.Therefore, combining (207) with (210), we get (211)χAhps,qsp-1≤CFp1,q1h-k-p-1χAkps,qsδ=CFp1,q1h-k-p-1χAkps,qsp-1δ0.Setting ψ(h)≔χAhps,qsp-1 in Lemma 15, on account of (211), we can find a constant C2 (independent of F) such that (212)χAKps,qsp-1=0withK≔C2Fp1,q11/p-1.This shows that μ(AK)=0, where AK=x∈Ω¯:u-vx≥K. Hence, we have u-v≤K, μ, a.e. on Ω¯, so that (213)U-V∞p-1≤C2Fp1,q1=C2fp1,Ω+gq1,∂Ω,which completes the proof of part (a).
(b) To prove this part, instead of (196) and (198), one uses WkV1p≤A(Wk,Wk) and Fφ1B1,1≤CFp1,q1φW1,p(Ω)χBp3,q3, where p3 and q3 are such that 1/p3+1/p1+1/ps=1 and 1/q3+1/q1+1/pt=1, and the embedding V↪V1↪Xps,pt(Ω¯,μ). The remainder of the proof follows as in the proof of part (a).We conclude this section with the following example.Example 48.
Letp∈2,∞, b:∂Ω→(0,∞) be a strictly positive and σ-measurable function and let (214)βx,ξ≔bxξp-2ξ,ξ∈R.Then, it is easy to verify that β satisfies Assumptions 38, 39, and 44 (see, e.g., [25, Example  4.17]).


## 5.1. General Assumptions

Throughout this section, we assume that the functionsα1:Ω×R→R and α2:∂Ω×R→R satisfy the following conditions.Assumption 38.
(145)
α
j
x
,
·
i
s
o
d
d
a
n
d
s
t
r
i
c
t
l
y
i
n
c
r
e
a
s
i
n
g
,
α
j
x
,
0
=
0
,
α
j
x
,
·
i
s
c
o
n
t
i
n
u
o
u
s
,
lim
t
→
0
⁡
α
j
x
,
t
t
=
0
,
lim
t
→
∞
⁡
α
j
x
,
t
t
=
∞
, for λN, a.e. x∈Ω if j=1, and σ, a.e. x∈∂Ω if j=2.Sinceαj(x,·) are strictly increasing for λN, a.e. x∈Ω if j=1, and σ, a.e. x∈∂Ω if j=2, then they have inverses which we denote by α~j(x,·) (cf. also Section 4). We define the functions Λ1, Λ~1:Ω×R→0,∞, and Λ2, Λ~2:∂Ω×R→0,∞ by (146)Λjx,t≔∫0tαjx,sds,Λ~jx,t≔∫0tα~jx,sds.Then, it is clear that, for λN, a.e. x∈Ω if j=1, and σ, a.e. x∈∂Ω if j=2, Λj(x,·) and Λ~j(x,·) are differentiable, monotone, and convex with Λj(x,0)=Λ~j(x,0)=0. Furthermore, Λj(x,·) is an N-function and Λ~j(x,·) is its complementary N-function. The function Λ~j is then the complementary Musielak-Orlick function of Λj in the sense of Young (see Definition 5).Assumption 39.
We assume, forλN, a.e. x∈Ω if j=1, and σ, a.e. x∈∂Ω if j=2, that Λj(x,·) and Λ~j(x,·) satisfy the (▵2)-condition in the sense of Definition 7.It follows from Assumption39 that there exist two constants c1,c2∈(0,1] such that for λN, a.e. x∈Ω if j=1, and σ, a.e. x∈∂Ω if j=2, and for all t∈R(147)cjtαjx,t≤Λjx,t≤tαjx,t.Next, let(148)LΛ1Ω≔u:Ω⟶Rmeasurable:∫ΩΛ1x,udx<∞,LΛ2∂Ω≔u:∂Ω⟶Rmeasurable:∫∂ΩΛ2x,udσ<∞.Since Λj(x,·) and Λ~j(x,·) satisfy the (▵2)-condition, it follows from [18, Theorem  8.19] that LΛ1(Ω) and LΛ2(∂Ω), endowed, respectively, with the norms (149)uΛ1,Ω≔inf⁡k>0:∫ΩΛ1x,uxkdx≤1,uΛ2,∂Ω≔inf⁡k>0:∫∂ΩΛ2x,uxkdσ≤1,are reflexive Banach spaces. Moreover, by [18, Section  8.11, p. 234], the following generalized versions of Hölder’s inequality will also become useful in the sequel(150)∫Ωuvdx≤2uΛ1,ΩvΛ~1,Ω,∫∂Ωuvdσ≤2uΛ2,∂ΩvΛ~2,∂Ω.


## 5.2. Existence and Uniqueness of Weak Solutions of Perturbed Equations

Let(151)V≔U≔u,u∂Ω:u∈W1,pΩ∩LΛ1Ω,u∣∂Ω∈W1,q∂Ω∩LΛ2∂Ω.Then, for every 1<p,q<∞, V endowed with the norm (152)UV=uW1,pΩ+uΛ1,Ω+uW1,q∂Ω+uΛ2,∂Ωis a reflexive Banach space. Recall that ρ=1. Throughout the following, we denote by V′ the dual of V.Definition 40.
A functionU=(u,u∂Ω)∈V is said to be a weak solution of (144), if for every V∈V=(v,v∂Ω)(153)AU,V=∫Ωfvdx+∫∂Ωgvdσ,provided that the integrals on the right-hand side exist. Here(154)AU,V≔∫Ω∇up-2∇u·∇vdx+∫Ωup-2uvdx+∫Ωα1x,uvdx+∫∂Ω∇Γuq-2∇Γu·∇Γvdσ+∫∂Ωuq-2uvdσ+∫∂Ωα2x,uvdσ.Lemma 41.
Assume Assumptions38 and 39. Let 1<p,q<∞ and U∈V be fixed. Then the functional V↦A(U,V) belongs to V′. Moreover, A is strictly monotone, hemicontinuous, and coercive.Proof.
LetU=(u,u∂Ω)∈V be fixed. It is clear that A(U,·) is linear. Let V=(v,v∂Ω)∈V. Then, exploiting (150), we obtain(155)AU,V≤uW1,pΩp-1vW1,pΩ+uW1,q∂Ωq-1vW1,q∂Ω+2max⁡1,∫ΩΛ~1x,α1x,udxvΛ1,Ω+2max⁡1,∫∂ΩΛ~2x,α2x,udσvΛ2,∂Ω≤KUVV,where (156)KU≔uW1,pΩp-1+2max⁡1,∫ΩΛ~1x,α1x,udx+uW1,q∂Ωq-1+2max⁡1,∫∂ΩΛ~2x,α2x,udσ.This shows A(U,·)∈V′, for every U∈V.
Next, letU,V∈V. Then, using (59) and the fact that αj(x,·) are monotone nondecreasing, that is, (αj(x,t)-αj(x,s))(t-s)≥0, for all t,s∈R, we obtain(157)AU,U-V-AV,U-V=∫Ω∇up-2∇u-∇vp-2∇v·∇u-vdx+∫Ωup-2u-vp-2vu-vdx+∫Ωα1x,u-α1x,vu-vdx+∫∂Ωuq-2u-vq-2vu-vdσ+∫∂Ω∇Γuq-2∇Γu-∇Γvq-2∇v·∇Γu-vdσ+∫∂Ωα2x,u-α1x,vu-vdσ≥∫Ω∇u+∇vp-2∇u-v2dx+∫Ωu+vp-2u-v2dx+∫∂Ω∇Γu+∇Γvp-2∇Γu-v2dσ+∫∂Ωu+vp-2u-v2dσ≥0.This shows that A is monotone. Estimate (157) also shows that (158)AU,U-V-AV,U-V>0,for all U,V∈V with U≠V, that is, u≠v or u∂Ω≠v∂Ω. Thus, A is strictly monotone.
The continuity of the norm function and the continuity ofαj(x,·), j=1,2 imply that A is hemicontinuous.
Finally, sinceΛj and Λ~j satisfy the (▵20)-condition, from Proposition 12 and Corollary 13, it follows that (159)limuΛ1,Ω→+∞⁡∫Ωuα1x,udxuΛ1,Ω=+∞,limuΛ2,∂Ω→+∞⁡∫∂Ωuα2x,udσuΛ2,∂Ω=+∞.Consequently, we deduce(160)limUV→+∞⁡AU,UUV=+∞,which shows that A is coercive. The proof of the lemma is finished.The following result is concerned with the existence and uniqueness of weak solutions to problem (144).Theorem 42.
Assume Assumptions38 and 39. Let 1<p,q<∞, p1≥p∗, and q1≥q∗, where p∗≔p/(p-1) and q∗≔q/(q-1). Then, for every (f,g)∈Xp1,q1(Ω¯,μ), there exists a unique function U∈V which is a weak solution to (144).Proof.
Let〈·,·〉 denote the duality between V and V′. Then, from Lemma 41, it follows that, for each U∈V, there exists A(U)∈V′ such that (161)AU,V=AU,V,for every V∈V. Hence, this relation defines an operator A:V→V′, which is bounded by (155). Exploiting Lemma 41 once again, it is easy to see that A is monotone and coercive. It follows from Browder’s theorem (see, e.g., [28, Theorem  5.3.22]) that A(V)=V′. Therefore, for every F∈V′ there exists U∈V such that A(U)=F; that is, for every V∈V(162)AU,V=AU,V=V,F.Since W1,p(Ω)↪Lp(Ω) and W1,q(∂Ω)↪Lq(∂Ω) with dense injection, by duality, we have Xp∗,q∗(Ω¯,μ)↪V′. Since Ω is bounded and σ(∂Ω)<∞, we obtain that (163)Xp1,q1Ω¯,μ↪Xp∗,q∗Ω¯,μ↪V′.This shows the existence of weak solutions. The uniqueness follows from the fact that A is strictly monotone (cf. Lemma 41). This completes the proof of the theorem.Corollary 43.
Let the assumptions of Theorem42 be satisfied. Let(164)ph≔NpNp-1+p,qh≔pN-1Np-1,qk≔qN-1Nq-1+1.(a)
Let1<p<N, 1<q<p(N-1)/N, p1≥ph, and q1≥qk. Then, for every (f,g)∈Xp1,q1(Ω,μ), there exists a function U∈V which is the unique weak solution to (144).(b)
Let1<q<N-1, 1<p<Nq/(N-1), p1≥ph, and q1≥qh. Then, for every (f,g)∈Xp1,q1(Ω,μ), there exists a function U∈V which is the unique weak solution to (144).Proof.
We first prove (a). Let1<p<N and 1<q<p(N-1)/N and let p1≥ph and q1≥qk, where ph and qk are given by (164). Let ps≔Np/(N-p) and qt≔(N-1)q/(N-1-q). Since W1,p(Ω)↪Lps(Ω) and W1,q(∂Ω)↪Lqt(∂Ω) with dense injection, then, by duality, Xph,qk(Ω¯,μ)↪V′, where 1/ps+1/ph=1 and 1/qt+1/qk=1. Since μ(Ω¯)<∞, we have that (165)Xp1,q1Ω¯,μ↪Xph,qhΩ¯,μ↪V′.Hence, for every F≔(f,g)∈Xp1,q1(Ω¯,μ)↪V′, there exists U∈V such that for every V∈V(166)AU,V=AU,V=∫Ωfvdx+∫∂Ωgvdσ.The uniqueness of the weak solution follows again from the fact that A is strictly monotone.
In order to prove the second part, we use the embeddingsW1,p(Ω)↪Lps(Ω), W1,p(Ω)↪Lqs(∂Ω) and proceed exactly as above. We omit the details.


## 5.3. Properties of the Solution Operator of the Perturbed Equation

In the sequel, we establish some interesting properties of the solution operatorA to problem (144). We begin by assuming the following.Assumption 44.
Suppose thatαj, j=1,2, satisfy the following conditions:(167)thereareconstantscj∈0,1suchthatcjαjx,ξ-η≤αjx,ξ-αjx,η∀ξ,η∈R.Theorem 45.
Assume Assumptions38, 39, and 44. Let p,q≥2 and let A:V→V′ be the continuous and bounded operator constructed in the proof of Theorem 42. Then A is injective and, hence, invertible and its inverse A-1 is also continuous and bounded.Proof.
First, we remark that since(168)αjx,t-αjx,st-s≥0,∀t,s∈R,for λN, a.e. x∈Ω if j=1, and σ, a.e. x∈∂Ω if j=2, it follows from (167) that, for all t,s∈R,(169)αjx,t-αjx,st-s≥cjαjx,t-s·t-s.Let U,V∈V and p,q∈2,∞. Then, exploiting (60), (169), and the (▵2)-condition, we obtain(170)AU-AV,U-V=AU,U-V-AV,U-V=∫Ω∇up-2∇u-∇vp-2∇v·∇u-vdx+∫Ωup-2u-vp-2vu-vdx+∫Ωα1x,u-α1x,vu-vdx+∫∂Ω∇Γuq-2∇Γu-∇Γvq-2∇Γv·∇Γu-vdσ+∫∂Ωuq-2u-vq-2vu-vdσ+∫∂Ωα2x,u-α2x,vu-vdσ≥u-vW1,pΩp+c1∫ΩΛ1x,u-vdx+u-vW1,q∂Ωq+c2∫∂ΩΛ2x,u-vdσ.This implies that 〈A(U)-A(V),U-V〉>0, for all U,V∈V with U≠V (that is, u≠v, or u∂Ω≠v∂Ω). Therefore, the operator A is injective and, hence, A-1 exists. Since for every U∈V(171)AU,U=AU,U≤AUV′UV,from the coercivity of A (see (160)), it is not difficult to see that(172)limUV→+∞⁡AUV′=+∞.Thus, A-1:V′→V is bounded.
Next, we show thatA-1:V′→V is continuous. Assume that A-1 is not continuous. Then there is a sequence Fn∈V′ with Fn→F in V′ and a constant δ>0 such that(173)A-1Fn-A-1FV≥δ,for all n∈N. Let Un≔A-1(Fn) and U=A-1(F). Since Fn is a bounded sequence and A-1 is bounded, we have that Un is bounded in V. Thus, we can select a subsequence, which we still denote by Un, which converges weakly to some function V∈V. Since A(Un)-A(V)→F-A(V) strongly in V and Un-V converges weakly to zero in V, we deduce(174)limn→∞⁡AUn-AV,Un-V=0.From (170) and (174), it follows that (175)limn→∞un-vW1,pΩ=0,limn→∞⁡∫ΩΛ1x,un-vdx=0,while (176)limn→∞⁡un-vW1,q∂Ω=0,limn→∞⁡∫∂ΩΛ2x,un-vdσ=0.Therefore, Un→V strongly in V. Since A is continuous and (177)Fn=AUn⟶AV=F=AU,it follows from the injectivity of A that U=V. This shows that (178)limn→∞A-1Fn-A-1FV=limn→∞Un-UV=0,which contradicts (173). Hence, A-1:V′→V is continuous. The proof is finished.Corollary 46.
Let the assumptions of Theorem45 be satisfied. Let ph,qh, and qk be as in (164) and let A:V→V′ be the continuous and bounded operator constructed in the proof of Theorem 42.(a)
If2≤p<N, 2≤q<p(N-1)/N, p1≥ph, and q1≥qk, then A-1:Xp1,q1(Ω¯,μ)→Xps,qt(Ω¯,μ) is continuous and bounded. Moreover, A-1:Xp1,q1(Ω¯,μ)→V∩Xr,s(Ω¯,μ) is compact for every r∈(1,ps) and s∈(1,qs).(b)
If2≤q<N-1, 2≤p<qN/(N-1), p1≥ph, and q1≥qh, then the operator A-1:Xp1,q1(Ω¯,μ)→Xps,qs(Ω¯,μ) is continuous and bounded. Moreover, A-1:Xp1,q1(Ω¯,μ)→V∩Xr,s(Ω¯,μ) is compact for every r∈(1,ps) and s∈(1,qs).Proof.
We only prove the first part. The second part of the proof follows by analogy and is left to the reader. Let2≤p<N, 2≤q<p(N-1)/N, p1≥ph, and q1≥qk and let F∈Xp1,q1(Ω¯,μ). Proceeding exactly as in the proof of Theorem 45, we obtain (179)A-1Fps,qt≤C1A-1FV≤CFV′≤C2Fp1,q1.Hence, the operator A-1:Xp1,q1(Ω¯,μ)→Xps,qt(Ω¯,μ) is bounded. Finally, using the facts that Xp1,q1(Ω¯,μ)↪V′, A-1:V′→V is continuous and V↪Xps,qt(Ω¯,μ), we easily deduce that A-1:Xp1,q1(Ω¯,μ)→Xps,qt(Ω¯,μ) is continuous.
Now, let1<r<ps and 1<s<qs. Since the injection V↪Xr,s(Ω¯,μ) is compact, then, by duality, the injection Xr′,s′(Ω¯,μ)↪V′ is compact for every r′>ps′=ph and s′>qs′=qh. This, together with the fact that A-1:V′→V is continuous and bounded, implies that A-1:Xp1,q1(Ω¯,μ)→V is compact for every p1>ph and q1>qh.
It remains to show thatA-1 is also compact as a map into Xr,s(Ω¯,μ) for every r∈(1,ps) and s∈(1,qs). Since A-1 is bounded, we have to show that the image of every bounded set B⊂Xp1,q1(Ω,μ) is relatively compact in Xr,s(Ω¯,μ) for every r∈(1,ps) and s∈(1,qs). Let Un be a sequence in A-1(B). Let Fn=A(Un)∈B. Since B is bounded, then the sequence Fn is bounded. Since A-1 is compact as a map into V, it follows that there is a subsequence Fnk such that A-1(Fnk)→U∈V. We may assume that Un=A-1(Fn)→U in V and, hence, in Xp,p(Ω¯,μ). It remains to show that Un→U in Xr,s(Ω¯,μ). Let r∈p,ps and s∈p,qs. Since Un≔(un,un∂Ω) is bounded in Xps,qs(Ω¯,μ), a standard interpolation inequality shows that there exists τ∈(0,1) such that (180)Un-Umr,s≤Un-Ump,pτUn-Umps,qs1-τ≤CUn-Ump,pτ.As Un converges in Xp,p(Ω¯,μ), it follows from the preceding inequality that Un is a Cauchy sequence in Xr,s(Ω¯,μ) and therefore converges in Xr,s(Ω¯,μ). Hence, A-1:Xp1,q1(Ω¯,μ)→V∩Xr,s(Ω¯,μ) is compact for every r∈p,ps and s∈p,qs. The case r,s∈(1,p) follows from the fact that Xp,p(Ω¯,μ)↪Xr,s(Ω¯,μ) and the proof is finished


## 5.4. Statement and Proof of the Main Result

We will now establish under what conditions the operatorA-1 maps Xp1,q1(Ω¯,μ) boundedly and continuously into X∞(Ω¯,μ). The following is the main result of this section.Theorem 47.
Let the assumptions of Theorem45 be satisfied.(a)
Suppose2≤p<N and 2≤q<∞. Let (181)p1>psps-p=Np,q1>qsqs-p=N-1p-1.Letf∈Lp1(Ω),g∈Lq1(∂Ω) and U,V∈V be such that for every function Φ=(φ,φ∂Ω)∈V(182)AU,Φ-AV,Φ=∫Ωfφdx+∫∂Ωgφdσ.Then there is a constantC=C(N,p,q,Ω)>0 such that (183)U-V∞p-1≤Cfp1,Ω+gq1,∂Ω.(b)
Suppose2≤p=q<N-1. Let (184)p1>psps-p=Np,q1>ptpt-p=N-1p.Letf∈Lp1(Ω),g∈Lq1(∂Ω) and U,V∈V satisfy (182). Then there is a constant C=C(N,p,q,Ω)>0 such that (185)U-V∞p-1≤Cfp1,Ω+gq1,∂Ω.Proof.
LetU,V∈V satisfy (182). Let k≥0 be a real number and set (186)wk≔u-v-k+sgn⁡u-vWk≔wk,wk∂Ω,w≔u-v.Let Ak≔{x∈Ω¯:wx≥k} and Ak+≔x∈Ω¯:wx≥k and Ak-≔x∈Ω¯:wx≤-k. Clearly Wk∈V and Ak=Ak+∪Ak-. We claim that there exists a constant C>0 such that(187)CAWk,Wk≤AU,Wk-AV,Wk,for all U,V∈V. Using the definition of the form A, we have(188)AU,Wk-AV,Wk=∫Ω∇up-2∇u-∇vp-2∇v·∇wkdx+∫Ωup-2u-vp-2vwkdx+∫Ωα1x,u-α2x,vwkdx+∫∂Ωuq-2u-vq-2vwkdσ+∫∂Ω∇Γup-2∇Γu-∇Γvp-2∇Γv·∇Γwkdσ+∫∂Ωα2x,u-α2x,vwkdσ.Since(189)∇wk=∇u-vinAk,0otherwise,we can rewrite (188) as follows:(190)AU,Wk-AV,Wk=∫Ak∩Ω∇up-2∇u-∇vp-2∇v·∇u-vdx+∫Ak∩∂Ω∇Γuq-2∇Γu-∇Γvq-2∇Γv·∇Γu-vdσ+λ∫Ak∩Ωup-2u-vp-2vwkdx+∫Ak∩Ωα1x,u-α1x,vwkdx+∫Ak∩∂Ωα2x,u-α2x,vwkdσ.Exploiting inequality (60), from (190) and (169), we deduce that(191)AU,Wk-AV,Wk≥∫Ak∩Ω∇wkp+wkpdx+∫Ak∩Ωc1α1x,wkwkdx+∫Ak∩Ωup-2uwk-vp-2vwk-wkpdx+∫Ak∩Ωα1x,u-α1x,v-c1α1x,wkwkdx+∫Ak∩∂Ω∇Γwkq+wkqdσ+∫Ak∩∂Ωc2α2x,wkwkdσ+∫Ak∩∂Ωuq-2uwk-vq-2vwk-wkqdσ+∫Ak∩∂Ωα2x,u-α2x,v-c2α2x,wkwkdσ≥CAWk,Wk+∫Ak∩Ωup-2uwk-vp-2vwk-wkpdx+∫Ak∩Ωα1x,u-α1x,v-c1α1x,wkwkdx+∫Ak∩∂Ωuq-2uwk-vq-2vwk-wkqdσ+∫Ak∩∂Ωα2x,u-α2x,v-c2α2x,wkwkdσ,where c1,c2 are the constants from (169). Using (167) and the fact that αj(x,·) are strictly increasing, for x∈Ak+, we have (192)cjαjx,wkx=cjαjx,ux-vx-k≤cjαjx,ux-vx≤αjx,ux-αjx,vx.Multiplying this inequality by wk(x)≥0, x∈Ak+, yields(193)αjx,ux-αjx,vx-cjαjx,wkxwkx≥0.Similarly, for x∈Ak-(194)cjαjx,wkx=cjαjx,ux-vx+k≥cjαjx,ux-vx≥αjx,ux-αjx,vx.Hence, multiplying this inequality by wk(x)≤0, we get(195)αjx,ux-αjx,vx-cjαjx,wkxwkx≥0,for all x∈Ak-. Hence, on account of (193) and (195), from (191) we obtain the required estimate of (187).
(a) To prove this part, note that from Definition40 it is clear that(196)wkW1,pΩp≤AWk,Wk.Let f∈Lp1(Ω) and g∈Lq1(∂Ω) with (197)p1>psps-p=Np,q1>qsqs-p=N-1p-1,and let B⊂Ω¯ be any μ-measurable set. We claim that there exists a constant C≥0 such that, for every F∈Xp1,q1(Ω¯,μ) and φ∈W1,p(Ω), we have(198)Fφ1B1,1≤CFp1,q1φW1,pΩχBp3,q3,where p3 and q3 are such that 1/p3+1/p1+1/ps=1 and 1/q3+1/q1+1/qs=1. In fact, note that if n∈N and pi, qi∈1,∞(i=1,…,n) are such that (199)∑i=1n1pi=∑i=1n1qi=1, and if Fi∈Xpi,qi(Ω¯,μ)(i=1,…,n), then by Hölder’s inequality(200)∏i=1nFi1,1≤∏i=1nFipi,qi.Since W1,p(Ω)↪Xps,qs(Ω¯,μ), (198) follows immediately from (200) and claim (198) is proved. Next, it follows from (198) that (201)∫Ω¯FWkdμ=FWk1,1=FWkχAk1,1≤Fp1,q1wkW1,pΩχAkp3,q3,where we recall that 1/p3=1-1/ps-1/p1>p-1/ps and q3<qs/(p-1). Therefore, for every k≥0(202)AU,Wk-AV,Wk≤Fp1,q1wkW1,pΩχAkp3,q3,which together with estimate (187) yields the desired inequality(203)CAWk,Wk≤AU,Wk-BV,Wk≤Fp1,q1wkW1,pΩχAkp3,q3.It follows from (196) and (203) that for every k>0(204)CwkW1,pΩp≤AWk,Wk≤AU,Wk-AV,Wk≤Fp1,q1wkW1,pΩχAkp3,q3.Hence, for every k>0, wkW1,p(Ω)p-1≤C1χAkp3,q3. Using the fact W1,p(Ω)↪Xps,qs(Ω¯,μ), we obtain, for every k>0, that (205)wkps,qsp-1≤CFp1,q1χAkp3,q3.Let h>k. Then Ah⊂Ak and on Ah the inequality wk≥h-k holds. Therefore(206)h-kχAhps,qsp-1≤CFp1,q1χAkp3,q3,which shows that(207)χAhps,qsp-1≤CFp1,q1h-k-p-1χAkp3,q3.Let C3≔1Ω¯ps,qs and (208)δ≔min⁡psp3,qsp3>p-1,δ0≔δp-1>1.Then(209)C3-ps/p3χAkΩ,p3=C3-1χAkΩ,psps/p3≤C3-1χAkΩ,psδ≤χAkps,qsδC3-δ,C3-qs/q3χAk∂Ω,q3=C3-1χAk∂Ω,qsqs/q3≤C3-1χAk∂Ω,qsδ≤χAkps,qsδC3-δ.Choosing CΩ≔C3ps/p3-δ+C3qs/q3-δ, from (209) we have(210)χAkp3,q3≤CΩχAkps,qsδ.Therefore, combining (207) with (210), we get (211)χAhps,qsp-1≤CFp1,q1h-k-p-1χAkps,qsδ=CFp1,q1h-k-p-1χAkps,qsp-1δ0.Setting ψ(h)≔χAhps,qsp-1 in Lemma 15, on account of (211), we can find a constant C2 (independent of F) such that (212)χAKps,qsp-1=0withK≔C2Fp1,q11/p-1.This shows that μ(AK)=0, where AK=x∈Ω¯:u-vx≥K. Hence, we have u-v≤K, μ, a.e. on Ω¯, so that (213)U-V∞p-1≤C2Fp1,q1=C2fp1,Ω+gq1,∂Ω,which completes the proof of part (a).
(b) To prove this part, instead of (196) and (198), one uses WkV1p≤A(Wk,Wk) and Fφ1B1,1≤CFp1,q1φW1,p(Ω)χBp3,q3, where p3 and q3 are such that 1/p3+1/p1+1/ps=1 and 1/q3+1/q1+1/pt=1, and the embedding V↪V1↪Xps,pt(Ω¯,μ). The remainder of the proof follows as in the proof of part (a).We conclude this section with the following example.Example 48.
Letp∈2,∞, b:∂Ω→(0,∞) be a strictly positive and σ-measurable function and let (214)βx,ξ≔bxξp-2ξ,ξ∈R.Then, it is easy to verify that β satisfies Assumptions 38, 39, and 44 (see, e.g., [25, Example  4.17]).



---

*Source: 5196513-2017-06-27.xml*