#  Nonlinear elliptic boundary value problems at resonance with nonlinear
Wentzell boundary conditions

Ciprian G. Gal  C. G. Gal, Department of Mathematics, University of
Missouri,Columbia, MO 65211 (USA<EMAIL_ADDRESS>and  Mahamadi Warma  M.
Warma, University of Puerto Rico, Department of Mathematics (Rio Piedras
Campus), PO Box 70377 San Juan PR 00936-8377 (USA<EMAIL_ADDRESS><EMAIL_ADDRESS>Dedicated to the 70th birthday of Jerome A. Goldstein

###### Abstract.

In the first part of the article, we give necessary and sufficient conditions
for the solvability of a class of nonlinear elliptic boundary value problems
with nonlinear boundary conditions involving the $q$-Laplace-Beltrami
operator. 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 $L^{\infty}$-setting.

###### Key words and phrases:

Nonlinear Wentzell-Robin boundary conditions, necessary and sufficient
conditions for existence of weak solutions, subdifferentials, a priori
estimates, boundary value problems at resonance.

###### 2010 Mathematics Subject Classification:

35J65, 35D30, 35B45

##  1\. Introduction

Let $\Omega\subset\mathbf{R}^{N},$ $N\geq 1,$ be a bounded domain with a
Lipschitz boundary $\partial\Omega$ and consider the following nonlinear
boundary value problem with nonlinear second order boundary conditions:

$\begin{cases}-\Delta_{p}u+\alpha_{1}\left(u\right)=f\left(x\right),&\text{ in
}\Omega,\\\ &\\\ b\left(x\right)\left|\nabla
u\right|^{p-2}\partial_{\mathbf{n}}u-\rho
b\left(x\right)\Delta_{q,\Gamma}u+\alpha_{2}\left(u\right)=g\left(x\right),&\text{
on }\partial\Omega,\end{cases}$ (1.1)

where $b\in L^{\infty}\left(\partial\Omega\right),$ $b(x)\geq b_{0}>0,$ for
some constant $b_{0}$, $\rho$ is either $0$ or $1,$ and $\alpha_{1},$
$\alpha_{2}\in C\left(\mathbb{R},\mathbb{R}\right)$ are monotone nondecreasing
functions such that $\alpha_{i}\left(0\right)=0$. Moreover,
$\Delta_{p}u=\mbox{div}(\left|\nabla u\right|^{p-2}\nabla u)$ is the
$p$-Laplace operator, $p\in\left(1,+\infty\right)$ and $f\in
L^{2}\left(\Omega,dx\right),$ $g\in L^{2}(\partial\Omega,\sigma)$ are given
real-valued functions. Here, $dx$ denotes the usual $N$-dimensional Lebesgue
measure in $\Omega$ and $\sigma$ denotes the restriction to $\partial\Omega$
of the $(N-1)$-dimensional Hausdorff measure. Recall that $\sigma$ coincides
with the usual Lebesgue surface measure since $\Omega$ has a Lipschitz
boundary, and $\partial_{\mathbf{n}}u$ denotes the normal derivative of $u$ in
direction of the outer normal vector $\overrightarrow{\mathbf{n}}$.
Furthermore, $\Delta_{q,\Gamma}$ is defined as the generalized $q$-Laplace-
Beltrami operator on $\partial\Omega,$ that is,
$\Delta_{q,\Gamma}u=\mbox{div}_{\Gamma}(\left|\nabla_{\Gamma}u\right|^{q-2}\nabla_{\Gamma}u),$
$q\in\left(1,+\infty\right)$. In particular, $\Delta_{2}=\Delta$ and
$\Delta_{2,\Gamma}=\Delta_{\Gamma}$ become the well-known Laplace and Laplace-
Beltrami operators on $\Omega$ and $\partial\Omega,$ respectively. Here, for
any real valued function $v,$

$\mbox{div}_{\Gamma}v=\sum_{i=1}^{N-1}\partial_{\tau_{i}}v,$

where $\partial_{\tau_{i}}v$ denotes the directional derivative of $v$ along
the tangential directions $\tau_{i}$ at each point on the boundary, whereas
$\nabla_{\Gamma}v=\left(\partial_{\tau_{1}}v,...,\partial_{\tau_{N-1}}v\right)$
denotes the tangential gradient at $\partial\Omega$. It is worth mentioning
again that when $\rho=0$ in (1.1), the boundary conditions are of lower order
than the order of the $p$ -Laplace operator, while for $\rho=1,$ we deal with
boundary conditions which have the same differential order as the operator
acting in the domain $\Omega.$ Such boundary conditions arise in many
applications, such as phase-transition phenomena (see, e.g., [13, 14] and the
references therein) and have been studied by several authors (see, e.g., [2,
12, 16, 24, 28]).

In a recent paper [12], the authors have formulated necessary and sufficient
conditions for the solvability of (1.1) 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
$\lambda_{1},$ $\lambda_{2}\in\mathbb{R}_{+}$ by

$\lambda_{1}=\int_{\Omega}dx,\text{
}\lambda_{2}=\int_{\partial\Omega}\frac{d\sigma}{b},$ (1.2)

this result reads that a necessary condition for the existence of a weak
solution of (1.1) is that

$\int_{\Omega}f\left(x\right)dx+\int_{\partial\Omega}g\left(x\right)\frac{d\sigma}{b\left(x\right)}\in\left(\lambda_{1}\mathcal{R}\left(\alpha_{1}\right)+\lambda_{2}\mathcal{R}\left(\alpha_{2}\right)\right),$
(1.3)

while a sufficient condition is

$\int_{\Omega}f\left(x\right)dx+\int_{\partial\Omega}g\left(x\right)\frac{d\sigma}{b\left(x\right)}\in\mbox{int}\left(\lambda_{1}\mathcal{R}\left(\alpha_{1}\right)+\lambda_{2}\mathcal{R}\left(\alpha_{2}\right)\right),$
(1.4)

where $\mathcal{R}(\alpha_{j})$ denotes the range of $\alpha_{j}$, $j=1,2$ and
$\mbox{int}(G)$ denotes the interior of the set $G$.

Relation (1.3) turns out to be both necessary and sufficient if either of the
sets $\mathcal{R}\left(\alpha_{1}\right)$ or
$\mathcal{R}\left(\alpha_{2}\right)$ is an open interval. This particular
result was established in [12, 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

$\left\\{\begin{array}[]{ll}-\Delta
u+\alpha_{1}\left(u\right)=f\left(x\right),&\text{in }\Omega,\\\
b\left(x\right)\partial_{n}u=g\left(x\right),&\text{on
}\partial\Omega,\end{array}\right.$ (1.5)

which is only a special case of (1.1) (i.e., $\rho=0,$ $\alpha_{2}\equiv 0$
and $p=2$). According to [12, Theorem 3] (see also (1.4)), this problem has a
weak solution if

$\int_{\Omega}f\left(x\right)dx+\int_{\partial\Omega}g\left(x\right)\frac{d\sigma}{b\left(x\right)}\in\mbox{int}\left(\lambda_{1}\mathcal{R}\left(\alpha_{1}\right)\right),$
(1.6)

which yields the result of Landesman and Lazer [17] for $g\equiv 0$. This last
condition is both necessary and sufficient when the interval
$\mathcal{R}\left(\alpha_{1}\right)$ is open. This was put into an abstract
context and significantly extended by Brezis and Haraux [8]. Their work was
much further extended by Brezis and Nirenberg [9]. The goal of the present
article is comparable to that of [12] since we want to establish similar
conditions to (1.4) and (1.6) for the existence of solutions to (1.1) when
$p,q\neq 2,$ with main emphasis on the generality of the boundary conditions.

Recall that $\lambda_{1}$ and $\lambda_{2}$ are given by (1.2). Let
$\mathbb{I}$ be the interval
$\lambda_{1}\mathcal{R}\left(\alpha_{1}\right)+\lambda_{2}\mathcal{R}\left(\alpha_{2}\right).$
Our first main result is as follows (see Section 4 also).

######  Theorem 1.1.

Let $\alpha_{j}:\mathbb{R}\rightarrow\mathbb{R}$ $(j=1,2)$ be odd, monotone
nondecreasing, continuous function such that $\alpha_{j}(0)=0$. Assume that
the functions $\Lambda_{j}(t):=\int_{0}^{|t|}\alpha_{j}(s)ds$ satisfy

$\Lambda_{j}(2t)\leq C_{j}\Lambda_{j}(t),\;\mbox{ for all }\;t\in\mathbb{R},$
(1.7)

for some constants $C_{j}>1$, $j=1,2$. If $u$ is a weak solution of (1.1) (in
the sense of Definition 4.10 below), then

$\int_{\Omega}f\left(x\right)dx+\int_{\partial\Omega}g\left(x\right)\frac{d\sigma}{b\left(x\right)}\in\mathbb{I}.$
(1.8)

Conversely, if

$\int_{\Omega}f\left(x\right)dx+\int_{\partial\Omega}g\left(x\right)\frac{d\sigma}{b\left(x\right)}\in
int\left(\mathbb{I}\right),$ (1.9)

then (1.1) has a weak solution.

Our second main result of the paper deals with a modified version of (1.1)
which is obtained by replacing the functions $\alpha_{1}\left(s\right),$
$\alpha_{2}\left(s\right)$ in (1.1) by
$\overline{\alpha}_{1}\left(s\right)+\left|s\right|^{p-2}s$ and
$\overline{\alpha}_{2}\left(s\right)+\rho b\left|u\right|^{q-2}u$,
respectively, and also allowing $\overline{\alpha}_{1},$
$\overline{\alpha}_{2}$ to depend on $x\in\overline{\Omega}$. Under additional
assumptions on $\overline{\alpha}_{1},\overline{\alpha}_{2}$ and under higher
integrability properties for the data $\left(f,g\right)$, 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 (1.1) with respect to the data $\left(f,g\right)$ can be also
established. In particular, we prove the following

######  Theorem 1.2.

Let all the assumptions of Theorem 1.1 be satisfied for the functions
$\overline{\alpha}_{1},$ $\overline{\alpha}_{2}$. Moreover, for each $j=1,2$,
assume that $\overline{\alpha}_{j}\left(t\right)/t\rightarrow 0,$ as
$t\rightarrow 0$ and $\overline{\alpha}_{j}\left(t\right)/t\rightarrow\infty,$
as $t\rightarrow\infty$, respectively.

(a) Then, for every $\left(f,g\right)\in L^{p_{1}}(\Omega)\times
L^{q_{1}}(\partial\Omega)$ with

$p_{1}>\max\left\\{1,\frac{N}{p}\right\\},\text{
}q_{1}>\left\\{\begin{array}[]{ll}\max\left\\{1,\frac{N-1}{p-1}\right\\},&\text{if
}\rho\in\left\\{0,1\right\\},\\\
\max\left\\{1,\frac{N-1}{p}\right\\},&\text{if }\rho=1\text{ and
}p=q,\end{array}\right.$

there exists a unique weak solution to problem (1.1) (in the sense of
Definition 5.3 below) which is bounded.

(b) Let $\overline{\alpha}_{j},$ $j=1,2,$ be such that

$c_{j}\left|\overline{\alpha}_{j}(\xi-\eta)\right|\leq\left|\overline{\alpha}_{j}(\xi)-\overline{\alpha}_{j}(\eta)\right|,\text{
}\mbox{ for all }\;\xi,\eta\in\mathbb{R},$

for some constants $c_{j}\in(0,1]$. Then, the weak (bounded) solution of
problem (1.1) depends continuously on the data $\left(f,g\right)$. Precisely,
let us indicate by $u_{F_{j}}$ the unique solution corresponding to the data
$F_{j}:=\left(f_{j},g_{j}\right)\in L^{p_{1}}(\Omega)\times
L^{q_{1}}(\partial\Omega),$ for each $j=1,2$. Then, the following estimate
holds:

$\|u_{F_{1}}-u_{F_{2}}\|_{L^{\infty}(\Omega)}+\|u_{F_{1}}-u_{F_{2}}\|_{L^{\infty}(\partial\Omega)}\leq
Q\left(\|f_{1}-f_{2}\|_{L^{p_{1}}(\Omega)},\|g_{1}-g_{2}\|_{L^{q_{1}}(\partial\Omega)}\right),$

for some nonnegative function $Q:\mathbb{R}_{+}^{2}\rightarrow\mathbb{R}_{+}$,
$Q\left(0,0\right)=0$, which can be computed explicitly.

We organize the paper as follows. In Section 2, 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 (1.1) satisfies a sort of ”quasilinear” version of the
Fredholm alternative (cf. Theorem 3.5), which is needed in order to obtain the
result in Theorem 1.1. Finally, in Sections 4 and 5, we provide detailed
proofs of Theorem 1.1 and Theorem 1.2. 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 [4, 7, 20, 21, 27]. We will
also introduce some notations.

###  2.1. Maximal monotone operators

Let $H$ be a real Hilbert space with scalar product $(\cdot,\cdot)_{H}$.

######  Definition 2.1.

Let $A:\;D(A)\subset H\to H$ be a closed (nonlinear) operator. The operator
$A$ is said to be:

  1. (i)

_monotone_ , if for all $u,v\in D(A)$ one has

$(Au-Av,u-v)_{H}\geq 0.$

  2. (ii)

_maximal monotone_ , if it is monotone and the operator $I+A$ is invertible.

Next, let $V$ be a real reflexive Banach space which is densely and
continuously embedded into the real Hilbert space $H$, and let $V^{\prime}$ be
its dual space such that $V\hookrightarrow H\hookrightarrow V^{\prime}$.

######  Definition 2.2.

Let $\mathcal{A}:\;V\times V\to\mathbb{R}$ be a continuous map.

  1. (a)

The map $\mathcal{A}:\;V\times V\to\mathbb{R}$ is called a _nonlinear form_ on
$H$ if for all $u\in V$ one has $\mathcal{A}(u,\cdot)\in V^{\prime}$, that is,
if $\mathcal{A}$ is linear and bounded in the second variable.

  2. (b)

The nonlinear form $\mathcal{A}:\;V\times V\to\mathbb{R}$ is said to be:

    1. (i)

_monotone_ if $\mathcal{A}(u,u-v)-\mathcal{A}(v,u-v)\geq 0\;\mbox{ for all
}\;u,v\in V$;

    2. (ii)

_hemicontinuous_ if $\displaystyle\lim_{t\downarrow
0}\mathcal{A}(u+tv,w)=\mathcal{A}(u,w),\;\;\forall\;u,v,w\in V$ ;

    3. (iii)

_coercive_ , if
$\displaystyle\lim_{\|v\|_{V}\to+\infty}\frac{\mathcal{A}(v,v)}{\|v\|_{V}}=+\infty$.

Now, let $\varphi:\;H\rightarrow(-\infty,+\infty]$ be a proper, convex, lower
semicontinuous functional with effective domain

$D(\varphi):=\\{u\in H:\;\varphi(u)<\infty\\}.$

The subdifferential $\partial\varphi$ of the functional $\varphi$ is defined
by

$\begin{cases}\displaystyle D(\partial\varphi)&:=\\{u\in
D(\varphi):\;\exists\;w\in H\;\forall\;v\in
D(\varphi):\;\varphi(v)-\varphi(u)\geq(w,v-u)_{H}\\};\\\ &\\\
\displaystyle\partial\varphi(u)&:=\\{w\in H:\;\forall\;v\in
D(\varphi):\;\varphi(v)-\varphi(u)\geq(w,v-u)_{H}\\}.\end{cases}$

By a classical result of Minty [20] (see also [7, 21]), $\partial\varphi$ is a
maximal monotone operator.

###  2.2. Functional setup

Let $\Omega\subset\mathbb{R}^{N}$ be a bounded domain with a Lipschitz
boundary $\partial\Omega$. For $1<p<\infty$, we let $W^{1,p}(\Omega)$ be the
first order Sobolev space, that is,

$W^{1,p}(\Omega)=\\{u\in L^{p}(\Omega):\;\nabla u\in(L^{p}(\Omega))^{N}\\}.$

Then $W^{1,p}(\Omega),$ endowed with the norm

$\|u\|_{W^{1,p}(\Omega)}:=\left(\|u\|_{\Omega,p}^{p}+\|\nabla
u\|_{\Omega,p}^{p}\right)^{1/p}$

is a Banach space, where we have set

$\|u\|_{\Omega,p}^{p}:=\int_{\Omega}|u|^{p}\;dx.$

Since $\Omega$ has a Lipschitz boundary, it is well-known that there exists a
constant $C>0$ such that

$\|u\|_{\Omega,p_{s}}\leq C\|u\|_{W^{1,p}(\Omega)},\;\text{for all }u\in
W^{1,p}(\Omega),$ (2.1)

where $p_{s}=\frac{pN}{N-p}$ if $p<N,$ and $1\leq p_{s}<\infty$ if $N=p$.
Moreover the trace operator $\mbox{Tr}(u):=u_{|_{\partial\Omega}}$ initially
defined for $u\in C^{1}(\bar{\Omega})$ has an extension to a bounded linear
operator from $W^{1,p}(\Omega)$ into $L^{q_{s}}(\partial\Omega)$ where
$q_{s}:=\frac{p(N-1)}{N-p}$ if $p<N$, and $1\leq q_{s}<\infty$ if $N=p$.
Hence, there is a constant $C>0$ such that

$\|u\|_{\partial\Omega,q_{s}}\leq C\|u\|_{W^{1,p}(\Omega)},\;\text{for
all}\;u\in W^{1,p}(\Omega).$ (2.2)

Throughout the remainder of this article, for $1<p<N$, we let

$p_{s}:=\frac{pN}{N-p}\;\mbox{ and }\;q_{s}:=\frac{p(N-1)}{N-p}.$ (2.3)

If $p>N$, one has that

$W^{1,p}(\Omega)\hookrightarrow C^{0,1-\frac{N}{p}}(\bar{\Omega}),$ (2.4)

that is, the space $W^{1,p}(\Omega)$ is continuously embedded into
$C^{0,1-\frac{N}{p}}(\bar{\Omega})$. For more details, we refer to [23,
Theorem 4.7] (see also [19, Chapter 4]).

For $1<q<\infty$, we define the Sobolev space $W^{1,q}(\partial\Omega)$ to be
the completion of the space $C^{1}(\partial\Omega)$ with respect to the norm

$\|u\|_{W^{1,q}(\partial\Omega)}:=\left(\int_{\partial\Omega}|u|^{q}\;d\sigma+\int_{\partial\Omega}|\nabla_{\Gamma}u|^{q}\;d\sigma\right)^{1/q},$

where we recall that $\nabla_{\Gamma}u$ denotes the tangential gradient of the
function $u$ at the boundary $\partial\Omega$. It is also well-known that
$W^{1,q}(\partial\Omega)$ is continuously embedded into
$L^{q_{t}}(\partial\Omega)$ where $q_{t}:=\frac{q(N-1)}{N-1-q}$ if $1<q<N-1,$
and $1\leq q_{t}<\infty$ if $q=N-1$. Hence, for $1<q\leq N-1$, there exists a
constant $C>0$ such that

$\|u\|_{q_{t},\partial\Omega}\leq C\|u\|_{W^{1,q}(\partial\Omega)},\text{ for
all }u\in W^{1,q}(\partial\Omega).$ (2.5)

Let $\lambda_{N}$ denote the $N$-dimensional Lebesgue measure and let the
measure $\mu:=\lambda_{N}|_{\Omega}\oplus\sigma$ on $\overline{\Omega}$ be
defined for every measurable set $A\subset\overline{\Omega}$ by

$\mu(A):=\lambda_{N}(\Omega\cap A)+\sigma(A\cap\partial\Omega).$

For $p,q\in[1,\infty],$ we define the Banach space

$X^{p,q}(\overline{\Omega},\mu):=\\{F=(f,g):\;f\in L^{p}(\Omega)\mbox{ and
}g\in L^{q}(\partial\Omega)\\}$

endowed with the norm

$\|F\|_{X^{p,q}(\overline{\Omega})}=\||F\||_{p,q}:=\|f\|_{\Omega,p}+\|g\|_{\partial\Omega,q},$

if $1\leq p,q<\infty,$ and

$\|F\|_{X^{\infty,\infty}(\overline{\Omega},\mu)}=\||F\||_{\infty}:=\max\\{\|f\|_{\Omega,\infty},\|g\|_{\partial\Omega,\infty}\\}.$

If $p=q$, we will simply denote $\||F\||_{p,p}=\||F\||_{p}$.

Identifying each function $u\in W^{1,p}(\Omega)$ with
$U=(u,u|_{\partial\Omega})$, we have that $W^{1,p}(\Omega)$ is a subspace of
$X^{p,p}(\overline{\Omega},\mu)$.

For $1<p,q<\infty,$ we endow

$\mathcal{V}_{1}:=\\{U:=(u,u|_{\partial\Omega}),u\in
W^{1,p}(\Omega),\;u|_{\partial\Omega}\in W^{1,q}(\partial\Omega)\\}$

with the norm

$\|U\|_{\mathcal{V}_{1}}:=\|u\|_{W^{1,p}(\Omega)}+\|u\|_{W^{1,q}(\partial\Omega)},$

while

$\mathcal{V}_{0}:=\\{U=(u,u|_{\partial\Omega}):\;u\in W^{1,p}(\Omega)\\}$

is endowed with the norm

$\|U\|_{\mathcal{V}_{0}}:=\|u\|_{W^{1,p}(\Omega)}.$

It follows from (2.1)-(2.2) that $\mathcal{V}_{0}$ is continuously embedded
into $X^{p_{s},q_{s}}(\overline{\Omega},\mu),$ with $p_{s}$ and $q_{s}$ given
by (2.3), for $1<p<N$. Moreover, by (2.1) and (2.5), $\mathcal{V}_{1}$ is
continuously embedded into $X^{p_{s},q_{t}}(\overline{\Omega},\mu)$.

###  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 Section 5).

######  Definition 2.3.

Let $(X,\Sigma,\nu)$ be a complete measure space. We call a function
$B:X\times\mathbb{R}\to[0,\infty]$ a _Musielak-Orlicz function_ on $X$ if

  1. (a)

$B(x,\cdot)$ is non-trivial, even, convex for $\nu$-a.e. $x\in X$;

  2. (b)

$B(x,\cdot)$ is vanishing and continuous at $0$ for $\nu$-a.e. $x\in X$;

  3. (c)

$B(x,\cdot)$ is left continuous on $[0,\infty);$

  4. (d)

$B(\cdot,t)$ is $\Sigma$-measurable for all $t\in[0,\infty)$;

  5. (e)

$\displaystyle\lim_{t\rightarrow\infty}\frac{B(x,t)}{t}=\infty$.

The _complementary Musielak-Orlicz function_ $\widetilde{B}$ is defined by

$\widetilde{B}(x,t):=\sup\\{s|t|-B(x,s):s>0\\}.$

It follows directly from the definition that for $t,s\geq 0$ (and hence for
all $t,s\in\mathbb{R}$)

$st\leq B(x,t)+\widetilde{B}(x,s).$

######  Definition 2.4.

We say that a Musielak-Orlicz function $B$ satisfies the
$(\triangle_{\alpha}^{0})$-condition $(\alpha>1)$ if there exists a set
$X_{0}$ of $\nu$-measure zero and a constant $C_{\alpha}>1$ such that

$B(x,\alpha t)\leq C_{\alpha}B(x,t),$

for all $t\in\mathbb{R}$ and every $x\in X\setminus X_{0}$.

We say that $B$ satisfies the $(\nabla_{2}^{0})$-condition if there is a set
$X_{0}$ of $\nu$-measure zero and a constant $c>1$ such that

$B(x,t)\leq\frac{1}{2c}B(x,ct),$

for all $t\in\mathbb{R}$ and all $x\in X\setminus X_{0}$.

######  Definition 2.5.

A function $\Phi:\mathbb{R}\rightarrow[0,\infty)$ is called an
${\mathcal{N}}$-function if

  * •

$\Phi$ is even, strictly increasing and convex;

  * •

$\Phi(t)=0$ if and only if $t=0$;

  * •

$\displaystyle\lim_{t\rightarrow 0}\frac{\Phi(t)}{t}=0$ and
$\displaystyle\lim_{t\rightarrow\infty}\frac{\Phi(t)}{t}=\infty$.

We say that an ${\mathcal{N}}$-function $\Phi$ satisfies the
$(\triangle_{2})$-condition if there exists a constant $C_{2}>1$ such that

$\Phi(2t)\leq C_{2}\Phi(t),\;\;\;\mbox{ for all }\;t\in\mathbb{R},$

and it satisfies the $(\nabla_{2})$-condition if there is a constant $c>1$
such that

$\Phi(t)\leq\Phi(ct)/(2c),\;\;\mbox{ for all }\;t\in\mathbb{R}.$

For more details on ${\mathcal{N}}$-functions, we refer to the monograph of
Adams [1, Chapter VIII] (see also [25, Chapter I], [26, Chapter I]).

######  Remark 2.6.

_For an ${\mathcal{N}}$-function $\Phi$, we let $\varphi$ be its left-sided
derivative. Then $\varphi$ is left continuous on $(0,\infty)$ and
nondecreasing. Let $\psi$ be given by_

$\psi\left(s\right):=\inf\left\\{t>0:\varphi\left(t\right)>s\right\\}.$

_Then_

$\Phi(t)=\int_{0}^{|t|}\varphi(s)\;ds;\qquad\Psi(t):=\int_{0}^{|t|}\psi(s)\;ds=\sup\\{|t|s-\Phi(s):s>0\\}.$

_As before for all $s,t\in\mathbb{R}$,_

$st\leq\Phi(t)+\Psi(s).$ (2.6)

_Moreover, if $s=\varphi(t)$ or $t=\psi(s)$ then we have equality, that is,_

$\Psi(\varphi(t))=t\varphi(t)-\Phi(t).$ (2.7)

_The function $\Psi$ is called the complementary ${\mathcal{N}}$-function of
$\Phi$. It is also known that an ${\mathcal{N}}$-function $\Phi$ satisfies the
$(\triangle_{2})$-condition if and only if_

$ct\varphi(t)\leq\Phi(t)\leq t\varphi(t),$ (2.8)

_for some constant $c\in(0,1]$ and for all $t\in\mathbb{R}$, where $\varphi$
is the left-sided derivative of $\Phi$._

######  Lemma 2.7.

Let $\Phi$ be an ${\mathcal{N}}$-function which satisfies the
$(\triangle_{2})$-condition with the constant $C_{2}>1$ and let $\Psi$ be its
complementary ${\mathcal{N}}$-function. Then $\Psi$ satisfies the
$(\nabla_{2})$-condition with the constant $c:=2^{C_{2}-1}$.

###### Proof.

We have

$t\varphi(t)\leq\int_{t}^{2t}\varphi(s)\;ds\leq\int_{0}^{2t}\varphi(s)\;ds=\Phi(2t)\leq
C_{2}\Phi(t).$

Since $\varphi(\psi(s))\geq s$ for all $s\geq 0$ and $s/\Psi(s)$ and $s/(s-1)$
are decreasing, we get for $t:=\psi(s),$ that

$\frac{s\psi(s)}{\Psi(s)}\geq\frac{\varphi(\psi(s))\psi(s)}{\Psi(\varphi(\psi(s)))}=\frac{t\varphi(t)}{\Psi(\varphi(t))}=\frac{t\varphi(t)}{t\varphi(t)-\Phi(t)}\geq\frac{C_{2}}{C_{2}-1}.$

Now let $c:=2^{C_{2}-1}$. Then for $t\geq 0,$

$\displaystyle\ln\left(\frac{\Psi(ct)}{\Psi(t)}\right)$
$\displaystyle=\int_{t}^{ct}\frac{\psi(s)}{\Psi(s)}\;ds\geq\int_{t}^{ct}\frac{C_{2}}{s(C_{2}-1)}\;ds$
$\displaystyle=\frac{C_{2}}{C_{2}-1}\ln(c)=C_{2}\log(2)=\ln(2\cdot
2^{C_{2}-1}).$

Hence, $\Psi(t)2c\leq\Psi(ct)$.

######  Corollary 2.8.

Let $B$ be a Musielak-Orlicz function such that $B(x,\cdot)$ is an
${\mathcal{N}}$-function for $\nu$-a.e. $x$. If $B$ satisfies the
$(\triangle_{2}^{0})$-condition, then $\widetilde{B}$ satisfies the
$(\nabla_{2}^{0})$-condition.

######  Definition 2.9.

Let $B$ be a Musielak-Orlicz function. Then the Musielak-Orlicz space
$L^{B}(X)$ associated with $B$ is defined by

$L^{B}(X):=\\{u:X\rightarrow\mathbb{R}\text{ measurable
}:\rho_{B}(u/\alpha)<\infty\text{ for some }\alpha>0\\},$

where

$\rho_{B}(v):=\int_{X}B(x,v(x))\;d\nu(x).$

On this space we consider the Luxemburg norm $\|\cdot\|_{X,B}$ defined by

$\|u\|_{X,B}:=\inf\\{\alpha>0:\rho_{B}(u/\alpha)\leq 1\\}.$

######  Proposition 2.10.

Let $B$ be a Musielak-Orlicz function which satisfies the $(\nabla_{2}^{0})$
-condition. Then

$\lim_{\|u\|_{X,B}\rightarrow+\infty}\frac{\rho_{B}(u)}{\|u\|_{X,B}}=+\infty.$

###### Proof.

If $B$ satisfies the $(\nabla_{2}^{0})$-condition, then there exists a set
$X_{0}\subset X$ of measure zero such that for every $\varepsilon>0$ there
exists $\alpha=\alpha(\varepsilon)>0$,

$B(x,\alpha t)\leq\alpha\varepsilon B(x,t),$ (2.9)

for all $t\in\mathbb{R}$ and all $x\in X\backslash X_{0}$. Let
$\lambda\in(0,\infty)$ be fixed. For $\varepsilon:=1/\lambda$ there exists
$\alpha>0$ satisfying the above inequality. We will show that
$\rho_{B}(u)\geq\lambda\|u\|_{X,B}$ whenever $\|u\|_{X,B}>1/\alpha$. Assume
that $\|u\|_{X,B}>1/\alpha$ and let $\delta>0$ be such that
$\alpha=(1+\delta)/\|u\|_{X,B}$. Then

$\displaystyle\rho_{B}(\alpha u)$
$\displaystyle=\int_{X}B(x,u(1+\delta)/\|u\|_{X,B})\;d\mu$
$\displaystyle\geq(1+\delta)^{1-1/n}\int_{X}B(x,u(1+\delta)^{1/n}/\|u\|_{X,B})\;d\mu\geq(1+\delta)^{1-1/n},$

for all $n\in\mathbb{N}$. If we assume that the last inequality does not hold,
then

$\|u\|_{X,B}/(1+\delta)\in\\{\alpha>0:\rho(u/\alpha)\leq 1\\},$

and this clearly contradicts the definition of $\|u\|_{X,B}$. Therefore, we
must have

$\rho_{B}(\alpha u)\geq 1+\delta=\alpha\|u\|_{X,B}.$ (2.10)

From (2.9), (2.10), we obtain

$\rho_{B}(u)=\int_{X}B(x,u(x))\;d\mu\geq\frac{\lambda}{\alpha}\int_{X}B(x,\alpha
u(x))\;d\mu=\frac{\lambda}{\alpha}\rho_{B}(\alpha u)\geq\lambda\|u\|_{X,B}.$

The proof is finished.

######  Corollary 2.11.

Let $B$ be a Musielak-Orlicz function such that $B(x,\cdot)$ is an
${\mathcal{N}}$-function for $\nu$-a.e. $x$. If its complementary
${\mathcal{N}}$-function $\widetilde{B}$ satisfies the
$(\triangle_{2}^{0})$-condition, then $B$ satisfies the
$(\nabla_{2}^{0})$-condition and

$\lim_{\|u\|_{X,B}\to+\infty}\frac{\rho_{B}(u)}{\|u\|_{X,B}}=+\infty.$

###  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 2.12.

Let $a,b\in\mathbb{R}^{N}$ and $p\in(1,\infty)$. Then, there exists a constant
$C_{p}>0$ such that

$\left(|a|^{p-2}a-|b|^{p-2}b\right)(a-b)\geq
C_{p}\left(|a|+|b|\right)^{p-2}|a-b|^{2}\geq 0.$ (2.11)

If $p\in[2,\infty)$, then there exists a constant $c_{p}\in(0,1]$ such that

$\left(|a|^{p-2}a-|b|^{p-2}b\right)(a-b)\geq c_{p}|a-b|^{p}.$ (2.12)

###### Proof.

The proof of (2.12) is included in [10, Lemma I.4.4]. In order to show (2.11),
one only needs to show that the left hand side is non-negative, 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 2.13.

Let $\psi:[k_{0},\infty)\rightarrow\mathbb{R}$ be a non-negative, non-
increasing function such that there are positive constants $c,\alpha$ and
$\delta$ ($\delta>1$) such that

$\psi(h)\leq c(h-k)^{-\alpha}\psi(k)^{\delta},\qquad\forall\;h>k\geq k_{0}.$

Then $\psi(k_{0}+d)=0$ with
$d=c^{1/\alpha}\psi(k_{0})^{(\delta-1)/\alpha}2^{\delta(\delta-1)}$.

##  3\. The Fredholm alternative

In what follows, we assume that $\Omega\subset\mathbb{R}^{N}$ is a bounded
domain with Lipschitz boundary $\partial\Omega$. Let $b\in
L^{\infty}(\partial\Omega)$ satisfy $b(x)\geq b_{0}>0$ for some constant
$b_{0}$. Let $\mathbb{X}_{2}$ be the real Hilbert space
$L^{2}\left(\Omega,dx\right)\oplus
L^{2}\left(\partial\Omega,\frac{d\sigma}{b}\right)$. Then, it is clear that
$\mathbb{X}_{2}$ is isomorphic to
$X^{2,2}(\overline{\Omega},\lambda_{N}\oplus\sigma)$ with equivalent norms.

Next, let $\rho\in\\{0,1\\}$ and $p,q\in(1,+\infty)$ be fixed. We define the
functional $\mathcal{J}_{\rho}:\;\mathbb{X}_{2}\rightarrow[0,+\infty]$ by
setting

$\mathcal{J}_{\rho}\left(U\right)=\begin{cases}\displaystyle\frac{1}{p}\int_{\Omega}\left|\nabla
u\right|^{p}dx+\frac{1}{q}\int_{\partial\Omega}\rho\left|\nabla_{\Gamma}u\right|^{q}\;d\sigma,&\text{
if }U=\left(u,u_{\mid\partial\Omega}\right)\in
D\left(\mathcal{J}_{\rho}\right),\\\ +\infty,&\text{ if
}U\in\mathbb{X}_{2}\diagdown D\left(\mathcal{J}_{\rho}\right),\end{cases}$
(3.1)

where the effective domain is given
$D(\mathcal{J}_{\rho})=\mathcal{V}_{\rho}\cap\mathbb{X}_{2}$.

Throughout the remainder of this section, we let
$\displaystyle\mu:=\lambda_{N}\oplus\frac{d\sigma}{b}$. The following result
can be obtained easily.

######  Proposition 3.1.

The functional $\mathcal{J}_{\rho}$ defined by (3.1) is proper, convex and
lower semicontinuous on $\mathbb{X}_{2}=X^{2,2}(\overline{\Omega},\mu)$.

The following result contains a computation of the subdifferential
$\partial\mathcal{J}_{\rho}$ for the functional $\mathcal{J}_{\rho}$.

######  Remark 3.2.

_Let $U=(u,u|_{\partial\Omega})\in D(\mathcal{J}_{\rho})$ and let
$F:=(f,g)\in\partial\mathcal{J}_{\rho}(U)$. Then, by definition,
$F\in\mathbb{X}_{2}$ and for all $V=(v,v|_{\partial\Omega})\in
D(\mathcal{J}_{\rho})$, we have_

$\int_{\overline{\Omega}}F(V-U)\;d\mu\leq\frac{1}{p}\int_{\Omega}\bigg{(}|\nabla
v|^{p}-|\nabla
u|^{p}\bigg{)}\;dx+\frac{1}{q}\rho\int_{\Omega}\bigg{(}|\nabla_{\Gamma}v|^{q}-|\nabla_{\Gamma}u|^{q}\bigg{)}\;d\sigma.$

_Let $W=(w,w|_{\partial\Omega})\in D(\mathcal{J}_{\rho})$, $0<t\leq 1$ and set
$V:=tW+U$ above. Dividing by $t$ and taking the limit as $t\downarrow 0$, we
obtain that_

$\int_{\overline{\Omega}}FW\;d\mu\leq\int_{\Omega}|\nabla u|^{p-2}\nabla
u\cdot\nabla
w\;dx+\rho\int_{\partial\Omega}|\nabla_{\Gamma}|^{q-2}\nabla_{\Gamma}u\cdot\nabla_{\Gamma}w\,d\sigma,$
(3.2)

_where we recall that_

$\int_{\overline{\Omega}}F\;d\mu=\int_{\Omega}f\;dx+\int_{\partial\Omega}g\;\frac{d\sigma}{b}.$

_Choosing $w=\pm\psi$ with $\psi\in\mathcal{D}(\Omega)$ (the space of test
functions) and integrating by parts in (3.2), we obtain_

$-\Delta_{p}u=f\;\;\mbox{ in }\;\mathcal{D}^{\prime}(\Omega)$

_and_

$g=b(x)\left|\nabla u\right|^{p-2}\partial_{n}u-\rho
b\left(x\right)\Delta_{q,\Gamma}u\;\mbox{ weakly on }\;\partial\Omega.$

_Therefore, the single valued operator $\partial\mathcal{J}_{\rho}$ is given
by_

$D(\partial\mathcal{J}_{\rho})=\\{U=(u,u_{\mid\partial\Omega})\in
D(\mathcal{J}_{\rho}),\;\left(-\Delta_{p}u,b(x)\left|\nabla
u\right|^{p-2}\partial_{n}u-\rho
b\left(x\right)\Delta_{q,\Gamma}u\right)\in\mathbb{X}_{2}\\},$

_and_

$\partial\mathcal{J}_{\rho}(U)=\left(-\Delta_{p}u,b(x)\left|\nabla
u\right|^{p-2}\partial_{n}u-\rho b\left(x\right)\Delta_{q,\Gamma}u\right).$
(3.3)

Since the functional $\mathcal{J}_{\rho}$ is proper, convex and lower
semicontinuous, it follows that its subdifferential
$\partial\mathcal{J}_{\rho}$ is a maximal monotone operator.

In the following two lemmas, we establish a relation between the null space of
the operator $A_{\rho}:=\partial\mathcal{J}_{\rho}$ and its range.

######  Lemma 3.3.

Let $\mathcal{N}\left(A_{\rho}\right)$ denote the null space of the operator
$A_{\rho}$. Then

$\mathcal{N}\left(A_{\rho}\right)=C\mathbf{1}=\left\\{C=(c,c):\;c\in\mathbb{R}\right\\},$

that is, $\mathcal{N}\left(A_{\rho}\right)$ consists of all the real constant
functions on $\overline{\Omega}$.

###### Proof.

We say that $U\in\mathcal{N}\left(A_{\rho}\right)$ if and only if (by
definition) $U=(u,u|_{\partial\Omega})$ is a weak solution of

$\begin{cases}-\Delta_{p}u=0,&\text{ in }\Omega,\\\
b\left(x\right)\left|\nabla u\right|^{p-2}\partial_{n}u-\rho
b\left(x\right)\Delta_{q,\Gamma}u=0,&\text{ on }\partial\Omega.\end{cases}$
(3.4)

A function $U=(u,u|_{\partial\Omega})\in\mathcal{V}_{\rho}\cap\mathbb{X}_{2}$
is said to be a weak solution of (3.4), if for every
$V=(v,v|_{\partial\Omega})\in\mathcal{V}_{\rho}\cap\mathbb{X}_{2}$, there
holds

$\mathcal{A}_{\rho}(U,V):=\int_{\Omega}\left|\nabla u\right|^{p-2}\nabla
u\cdot\nabla
v\;dx+\rho\int_{\partial\Omega}\left|\nabla_{\Gamma}u\right|^{q-2}\nabla_{\Gamma}u\cdot\nabla_{\Gamma}v\;d\sigma=0.$
(3.5)

Let $C:=(c,c)$ with $c\in\mathbb{R}$. Then it is clear that
$C\in\mathcal{N}\left(A_{\rho}\right)$.

Conversely, let
$U=(u,u|_{\partial\Omega})\in\mathcal{N}\left(A_{\rho}\right)$. Then, it
follows from (3.5) that

$\mathcal{A}_{\rho}(U,U):=\int_{\Omega}\left|\nabla
u\right|^{p}\;dx+\rho\int_{\partial\Omega}\left|\nabla_{\Gamma}u\right|^{q}\;d\sigma=0.$
(3.6)

Since $\Omega$ is bounded and connected, this implies that $u$ is equal to a
constant. Therefore, $U=C\mathbf{1}$ and this completes the proof.

######  Lemma 3.4.

The range of the operator $A_{\rho}$ is given by

$\mathcal{R}(A_{\rho})=\left\\{F:=(f,g)\in\mathbb{X}_{2}:\;\int_{\overline{\Omega}}F\;d\mu:=\int_{\Omega}f\;dx+\int_{\partial\Omega}g\;\frac{d\sigma}{b(x)}=0\right\\}.$

###### Proof.

Let $F\in\mathcal{R}(A_{\rho})\subset\mathbb{X}_{2}$. Then there exists
$U=(u,u|_{\partial\Omega})\in D(A_{\rho})$ such that $A_{\rho}(U)=F$. More
precisely, for every
$V=(v,v|_{\partial\Omega})\in\mathcal{V}_{\rho}\cap\mathbb{X}_{2},$ we have

$\displaystyle\mathcal{A}_{\rho}(U,V)$
$\displaystyle=\int_{\Omega}\left|\nabla u\right|^{p-2}\nabla u\cdot\nabla
v\;dx+\rho\int_{\partial\Omega}\left|\nabla_{\Gamma}u\right|^{q-2}\nabla_{\Gamma}u\cdot\nabla_{\Gamma}v\;d\sigma$
(3.7) $\displaystyle=\int_{\overline{\Omega}}FV\;d\mu.$

Taking $V=(1,1)\in\mathcal{V}_{\rho}\cap\mathbb{X}_{2}$, we obtain that
$\displaystyle\int_{\overline{\Omega}}F\;d\mu=0$. Hence,

$\displaystyle\mathcal{R}(A_{\rho})\subseteq\left\\{F\in\mathbb{X}_{2}:\;\int_{\overline{\Omega}}F\;d\mu=0\right\\}.$

Let us now prove the converse. To this end, let $F\in\mathbb{X}_{2}$ be such
that $\displaystyle\int_{\overline{\Omega}}F\;d\mu=0$. We have to show that
$F\in\mathcal{R}(A_{\rho})$, that is, there exists
$U\in\mathcal{V}_{\rho}\cap\mathbb{X}_{2}$ such that (3.7) holds, for every
$V\in\mathcal{V}_{\rho}\cap\mathbb{X}_{2}$. To this end, consider

$\mathcal{V}_{\rho,0}:=\left\\{U=(u,u|_{\partial\Omega})\in\mathcal{V}_{\rho}\cap\mathbb{X}_{2}:\;\int_{\overline{\Omega}}U\;d\mu:=\int_{\Omega}u\;dx+\int_{\partial\Omega}u\frac{d\sigma}{b}=0\right\\}.$

It is clear that $\mathcal{V}_{\rho,0}$ is a closed linear subspace of
$\mathcal{V}_{\rho}\cap\mathbb{X}_{2}\hookrightarrow\mathbb{X}_{2}$, and
therefore is a reflexive Banach space. Using [18, Section 1.1], we have that
the norm

$\|U\|_{\mathcal{V}_{\rho,0}}:=\|\nabla
u\|_{p,\Omega}+\rho\|\nabla_{\Gamma}u\|_{q,\partial\Omega}$

defines an equivalent norm on $\mathcal{V}_{\rho,0}$. Hence, there exists a
constant $C>0$ such that for every $U\in\mathcal{V}_{\rho,0}$,

$\||U\||_{2}\leq C\|U\|_{\mathcal{V}_{\rho,0}}:=\|\nabla
u\|_{p,\Omega}+\rho\|\nabla_{\Gamma}u\|_{q,\partial\Omega}.$ (3.8)

Define the functional
$\mathcal{F}_{\rho}:\;\mathcal{V}_{\rho,0}\rightarrow\mathbb{R}$ by

$\mathcal{F}_{\rho}(U)=\frac{1}{p}\int_{\Omega}|\nabla
u|^{p}\;dx+\frac{\rho}{q}\int_{\partial\Omega}|\nabla_{\Gamma}u|^{q}\;d\sigma-\int_{\overline{\Omega}}FU\;d\mu.$

It is easy to see that $\mathcal{F}_{\rho}$ is convex and lower-semicontinuous
on $\mathbb{X}_{2}$ (see Proposition 3.1). We show now that
$\mathcal{F}_{\rho}$ is coercive. By exploiting a classical Hölder inequality
and using (3.8), we have

$\displaystyle\left|\int_{\overline{\Omega}}FU\;d\mu\right|$
$\displaystyle\leq C\||F\||_{2}\||U\||_{2}\leq
C\||F\||_{2}\|U\|_{\mathcal{V}_{\rho,0}}$
$\displaystyle=C\||F\||_{2}\left(\|\nabla
u\|_{p,\Omega}+\rho\|\nabla_{\Gamma}u\|_{q,\partial\Omega}\right).$

Obviously, this estimate yields

$-\int_{\overline{\Omega}}FU\;d\mu\geq-C\||F\||_{2}\left(\|\nabla
u\|_{p,\Omega}+\rho\|\nabla_{\Gamma}u\|_{q,\partial\Omega}\right).$ (3.9)

Therefore, from (3.9), we immediately get

$\frac{\mathcal{F}_{\rho}(U)}{\|U\|_{\mathcal{V}_{\rho,0}}}\geq\frac{\frac{1}{p}\|\nabla
u\|_{p,\Omega}^{p}+\frac{\rho}{q}\|\nabla_{\Gamma}u\|_{q,\partial\Omega}^{q}}{{\|\nabla
u\|_{p,\Omega}+\rho\|\nabla_{\Gamma}u\|_{q,\partial\Omega}}}-C\||F\||_{2}.$

This inequality implies that

$\lim_{\|U\|_{\mathcal{V}_{\rho,0}}\rightarrow+\infty}\frac{\mathcal{F}_{\rho}(U)}{\|U\|_{\mathcal{V}_{\rho,0}}}=+\infty,$

and this shows that the functional $\mathcal{F}_{\rho}$ is coercive. Since
$\mathcal{F}_{\rho}$ is also convex, lower-semicontinuous, it follows from [3,
Theorem 3.3.4] that, there exists a function $U^{\ast}\in\mathcal{V}_{\rho,0}$
which minimizes $\mathcal{F}_{\rho}$. More precisely, for all
$V\in\mathcal{V}_{\rho,0}$,
$\mathcal{F}_{\rho}(U^{\ast})\leq\mathcal{F}_{\rho}(V);$ this implies that for
every $0<t\leq 1$ and every $V\in\mathcal{V}_{\rho,0}$,

$\mathcal{F}_{\rho}(U^{\ast}+tV)-\mathcal{F}_{\rho}(U^{\ast})\geq 0.$

Hence,

$\lim_{t\downarrow
0}\frac{\mathcal{F}_{\rho}(U^{\ast}+tV)-\mathcal{F}_{\rho}(U^{\ast})}{t}\geq
0.$

Using the Lebesgue Dominated Convergence, an easy computation shows that

$\displaystyle 0\leq\lim_{t\downarrow
0}\frac{\mathcal{F}_{\rho}(U^{\ast}+tV)-\mathcal{F}_{\rho}(U^{\ast})}{t}$
$\displaystyle=\int_{\Omega}|\nabla u^{\ast}|^{p-2}\nabla u^{\ast}\cdot\nabla
v\;dx$ (3.10)
$\displaystyle+\rho\int_{\partial\Omega}|\nabla_{\Gamma}u^{\ast}|^{q-2}\nabla_{\Gamma}u^{\ast}\cdot\nabla_{\Gamma}v\;d\sigma-\int_{\overline{\Omega}}FV\;d\mu.$

Changing $V$ to $-V$ into (3.10) gives that

$\int_{\Omega}|\nabla u^{\ast}|^{p-2}\nabla u^{\ast}\cdot\nabla
v\;dx+\rho\int_{\partial\Omega}|\nabla_{\Gamma}u^{\ast}|^{q-2}\nabla_{\Gamma}u^{\ast}\cdot\nabla_{\Gamma}v\;d\sigma=\int_{\overline{\Omega}}FV\;d\mu,$
(3.11)

for every $V\in\mathcal{V}_{\rho,0}$. Now, let
$V\in\mathcal{V}_{\rho}\cap\mathbb{X}_{2}$. Writing $V=V-C+C$ with $C=(c,c),$

$c:=\frac{1}{\left(\lambda_{1}+\lambda_{2}\right)}\left(\int_{\Omega}v\;dx+\int_{\partial\Omega}v\;\frac{d\sigma}{b}\right),$

and using the fact that $\displaystyle\int_{\overline{\Omega}}F\;d\mu=0$, we
obtain, for every $V\in\mathcal{V}_{\rho}\cap\mathbb{X}_{2}$, that

$\int_{\Omega}|\nabla u^{\ast}|^{p-2}\nabla u^{\ast}\cdot\nabla
v\;dx+\rho\int_{\partial\Omega}|\nabla_{\Gamma}u^{\ast}|^{q-2}\nabla_{\Gamma}u^{\ast}\cdot\nabla_{\Gamma}v\;d\sigma=\int_{\overline{\Omega}}FV\;d\mu.$

Therefore, $A_{\rho}(U)=F$. Hence, $F\in\mathcal{R}(A_{\rho})$ and this
completes the proof of the lemma.

The following result is a direct consequence of Lemmas 3.3, 3.4. This is the
main result of this section.

######  Theorem 3.5.

The operator $A_{\rho}=\partial\mathcal{J}_{\rho}$ satisfies the following
type of ”quasi-linear” Fredholm alternative:

$\mathcal{R}\left(A_{\rho}\right)=\mathcal{N}\left(A_{\rho}\right)^{\perp}=\left\\{F\in\mathbb{X}_{2}:\left\langle
F,\mathbf{1}\right\rangle_{\mathbb{X}_{2}}=0\right\\}.$

##  4\. Necessary and sufficient conditions for existence of solutions

In this section, we prove the first main result (cf. Theorem 1.1) for problem
(1.1). Before we do so, we will need the following results from maximal
monotone operators theory and convex analysis.

######  Definition 4.1.

Let $\mathcal{H}$ be a real Hilbert space. Two subsets $K_{1}$ and $K_{2}$ of
$\mathcal{H}$ are said to be almost equal, written, $K_{1}\simeq K_{2},$ if
$K_{1}$ and $K_{2}$ have the same closure and the same interior, that is,
$\overline{K_{1}}=\overline{K_{2}}$ and
$\mbox{int}\left(K_{1}\right)=\mbox{int}\left(K_{2}\right).$

The following abstract result is taken from [8, Theorem 3 and Generalization
in p.173–174].

######  Theorem 4.2 (Brezis-Haraux).

Let $A$ and $B$ be subdifferentials of proper convex lower semicontinuous
functionals $\varphi_{1}$ and $\varphi_{2}$, respectively, on a real Hilbert
space $\mathcal{H}$ with $D(\varphi_{1})\cap D(\varphi_{2})\neq\emptyset$, and
let $C$ be the subdifferential of the proper, convex lower semicontinuous
functional $\varphi_{1}+\varphi_{2}$, that is
$C=\partial(\varphi_{1}+\varphi_{2})$. Then

$\mathcal{R}(A)+\mathcal{R}(B)\subset\overline{\mathcal{R}(C)}\;\;\;\mbox{ and
}\;\;\;\mbox{Int}\left(\mathcal{R}(A)+\mathcal{R}(B)\right)\subset\mathcal{R}(C)$

In particular, if the operator $A+B$ is maximal monotone, then

$\mathcal{R}\left(A+B\right)\simeq\mathcal{R}\left(A\right)+\mathcal{R}\left(B\right),$

and this is the case if
$\partial(\varphi_{1}+\varphi_{2})=\partial\varphi_{1}+\partial\varphi_{2}$.

###  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:

$\begin{cases}-\Delta_{p}u+\alpha_{1}\left(u\right)=f,&\text{ in }\Omega,\\\
&\\\ b\left(x\right)\left|\nabla u\right|^{p-2}\partial_{n}u-\rho
b\left(x\right)\Delta_{q,\Gamma}u+\alpha_{2}\left(u\right)=g,&\text{ on
}\partial\Omega,\end{cases}$ (4.1)

where $p,q\in(1,+\infty)$ are fixed.We also assume that
$\alpha_{j}:\mathbb{R}\rightarrow\mathbb{R}$ ($j=1,2$) satisfy the following
assumptions.

######  Assumption 4.3.

The functions $\alpha_{j}:\;\mathbb{R}\rightarrow\mathbb{R}$ $(j=1,2)$ are
odd, monotone nondecreasing, continuous and satisfy $\alpha_{j}(0)=0$.

Let $\tilde{\alpha}_{j}$ be the inverse of $\alpha_{j}$. We define the
functions
$\Lambda_{j},\;\widetilde{\Lambda}_{j}:\;\mathbb{R}\rightarrow\mathbb{R}_{+}$
($j=1,2$) by

$\Lambda_{j}(t):=\int_{0}^{|t|}\alpha_{j}(s)ds\;\mbox{ and
}\;\widetilde{\Lambda}_{j}(t):=\int_{0}^{|t|}\tilde{\alpha}_{j}(s)ds.$ (4.2)

Then it is clear that $\Lambda_{j}$, $\widetilde{\Lambda}_{j}$ are even,
convex and monotone increasing on $\mathbb{R}_{+},$ with
$\Lambda_{j}(0)=\widetilde{\Lambda}_{j}(0),$ for each $j=1,2$. Moreover, since
$\alpha_{j}$ are odd, we have
$\Lambda_{j}^{{}^{\prime}}\left(t\right)=\alpha_{j}\left(t\right),$ for all
$t\in\mathbb{R}$ and $j=1,2$, with a similar relation holding for
$\widetilde{\Lambda}_{j}$ as well. The following result whose proof is
included in [25, Chap. I, Section 1.3, Theorem 3] holds.

######  Lemma 4.4.

The functions $\Lambda_{j}$ and $\widetilde{\Lambda}_{j}$ $(j=1,2)$ satisfy
(2.6) and (2.7). More precisely, for all $s,t\in\mathbb{R}$,

$st\leq\Lambda_{j}(s)+\widetilde{\Lambda}_{j}(t).$

If $s=\alpha_{j}(t)$ or $t=\tilde{\alpha}_{j}(s),$ then we also have equality,
that is,

$\widetilde{\Lambda}_{j}(\alpha_{j}(s))=s\alpha_{j}(s)-\Lambda_{j}(s),\text{
}j=1,2.$

We note that in [25], the statement of Lemma 4.4 assumed that $\Lambda_{j}$,
$\widetilde{\Lambda}_{j}$ are $\mathcal{N}$-functions in the sense of
Definition 2.5. However, the conclusion of that result holds under the weaker
hypotheses of Lemma 4.4.

Define the functional $\mathcal{J}_{2}:\;\mathbb{X}_{2}\rightarrow[0,+\infty]$
by

$\mathcal{J}_{2}(u,v):=\begin{cases}\displaystyle\int_{\Omega}\Lambda_{1}(u)\;dx+\int_{\partial\Omega}\Lambda_{2}(v)\;\frac{d\sigma}{b},\;\;\;&\mbox{
if }\;(u,v)\in D(\mathcal{J}_{2}),\\\ +\infty,&\mbox{ if
}(u,v)\in\mathbb{X}_{2}\setminus D(\mathcal{J}_{2}),\end{cases}$

with the effective domain

$D(\mathcal{J}_{2}):=\left\\{(u,v)\in\mathbb{X}_{2}:\;\int_{\Omega}\Lambda_{1}(u)\;dx+\int_{\partial\Omega}\Lambda_{2}(v)\;\frac{d\sigma}{b}<\infty\right\\}.$

######  Lemma 4.5.

Let $\alpha_{j}$ $(j=1,2)$ satisfy Assumption 4.3. Then the functional
$\mathcal{J}_{2}$ is proper, convex and lower semicontinuous on
$\mathbb{X}_{2}$.

###### Proof.

It is routine to check that $\mathcal{J}_{2}$ is convex and proper. This
follows easily from the convexity of $\Lambda_{j}$ and the fact that
$\Lambda_{j}(0)=0$. To show the lower semicontinuity on $\mathbb{X}_{2}$, let
$U_{n}=(u_{n},v_{n})\in D(\mathcal{J}_{2})$ be such that $U_{n}\rightarrow
U:=(u,v)$ in $\mathbb{X}_{2}$ and $\mathcal{J}_{2}(U_{n})\leq C$ for some
constant $C>0$. Since $U_{n}\rightarrow U$ in $\mathbb{X}_{2}$, then there is
a subsequence, which we also denote by $U_{n}=(u_{n},v_{n}),$ such that
$u_{n}\rightarrow u$ a.e. on $\Omega$ and $v_{n}\rightarrow v$ $\sigma$-a.e.
on $\Gamma$. Since $\Lambda_{j}(\cdot)$ are continuous (thus, lower-
semicontinuous), we have

$\Lambda_{1}(u)\leq\liminf_{n\rightarrow\infty}\Lambda_{1}(u_{n})\;\mbox{ and
}\;\Lambda_{2}(v)\leq\liminf_{n\rightarrow\infty}\Lambda_{2}(v_{n}).$

By Fatou’s Lemma, we obtain

$\int_{\Omega}\Lambda_{1}(u)dx\leq\int_{\Omega}\liminf_{n\rightarrow\infty}\Lambda_{1}(u_{n})dx\leq\liminf_{n\rightarrow\infty}\int_{\Omega}\Lambda_{1}(u_{n})dx$

and

$\int_{\partial\Omega}\Lambda_{2}(v)\frac{d\sigma}{b}\leq\int_{\partial\Omega}\liminf_{n\rightarrow\infty}\Lambda_{2}(v_{n})\frac{d\sigma}{b}\leq\liminf_{n\rightarrow\infty}\int_{\partial\Omega}\Lambda_{2}(v_{n})\frac{d\sigma}{b}.$

Hence, $\mathcal{J}_{2}$ is lower semicontinuous on $\mathbb{X}_{2}$.

We have the following result whose proof is contained in [25, Chap. III,
Section 3.1, Theorem 2].

######  Lemma 4.6.

Let $\alpha_{j}$ $(j=1,2)$ satisfy Assumption 4.3 and assume that there exist
constants $C_{j}>1$ $(j=1,2)$ such that

$\Lambda_{j}(2t)\leq C_{j}\Lambda_{j}(t),\;\mbox{ for all }\;t\in\mathbb{R}.$
(4.3)

Then $D(\mathcal{J}_{2})$ is a vector space.

Let the operator $B_{2}$ be defined by

$\begin{cases}D\left(B_{2}\right)=\left\\{U:=\left(u,v\right)\in\mathbb{X}_{2}:\left(\alpha_{1}\left(u\right),\alpha_{2}\left(v\right)\right)\in\mathbb{X}_{2}\right\\},\\\
B_{2}(U)=\left(\alpha_{1}\left(u\right),\alpha_{2}\left(v\right)\right).\end{cases}$
(4.4)

We have the following result.

######  Lemma 4.7.

Let the assumptions of Lemma 4.6 be satisfied. Then the subdifferential of
$\mathcal{J}_{2}$ and the operator $B_{2}$ coincide, that is, for all
$(u,v)\in D(B_{2})=D(\partial\mathcal{J}_{2})$,

$\partial\mathcal{J}_{2}(u,v)=B_{2}(u,v).$

###### Proof.

Let $U=(u,v)\in D(\mathcal{J}_{2})$ and
$F=(f,g)\in\partial\mathcal{J}_{2}(u,v)$. Then by definition,
$F\in\mathbb{X}_{2}$ and, for every $V=(u_{1},v_{1})\in D(\mathcal{J}_{2}),$
we get

$\int_{\overline{\Omega}}F(V-U)\;d\mu\leq\mathcal{J}_{2}(V)-\mathcal{J}_{2}(U).$

Let $V=U+tW,$ with $W=(u_{2},v_{2})\in D(\mathcal{J}_{2})$ and $0<t\leq 1$.
Then by Lemma 4.6, $V=U+tW\in D(\mathcal{J}_{2})$. Now, dividing by $t$ and
taking the limit as $t\downarrow 0$, we obtain

$\int_{\Omega}FW\;d\mu\leq\int_{\Omega}\alpha_{1}(u)u_{2}dx+\int_{\partial\Omega}\alpha_{2}(v)v_{2}\,\frac{d\sigma}{b}.$
(4.5)

Changing $W$ to $-W$ in (4.5) gives that

$\int_{\overline{\Omega}}FW\;d\mu=\int_{\Omega}\alpha_{1}(u)u_{2}dx+\int_{\partial\Omega}\alpha_{2}(v)v_{2}\,\frac{d\sigma}{b}.$

In particular, if $W=(u_{2},0)$ with $u_{2}\in\mathcal{D}(\Omega)$, we have

$\int_{\Omega}fu_{2}\;dx=\int_{\Omega}\alpha_{1}(u)u_{2}\;dx,$

and this shows that $\alpha_{1}(u)=f$. Similarly, one obtains that
$\alpha_{2}(v)=g$. We have shown that $U\in D(B_{2})$ and

$B_{2}(U):=B_{2}(u,v)=(\alpha_{1}(u),\alpha_{2}(v))=(f,g).$

Conversely, let $U=(u,v)\in D(B_{2})$ and set
$F=(f,g):=B_{2}(u,v)=(\alpha_{1}(u),\alpha_{2}(v))$. Since
$(\alpha_{1}(u),\alpha_{2}(v))\in\mathbb{X}_{2}$, from (4.2) and (4.3), it
follows that

$\int_{\Omega}\Lambda_{1}(u)dx+\int_{\partial\Omega}\Lambda_{2}(v)\frac{d\sigma}{b}<\infty.$

Hence, $U=(u,v)\in D(\mathcal{J}_{2})$. Let $V=(u_{1},v_{1})\in
D(\mathcal{J}_{2})$. Using Lemma 4.4, we obtain

$\displaystyle\alpha_{1}(u)(u_{1}-u)$
$\displaystyle=\alpha_{1}(u)u_{1}-\alpha_{1}(u)u$ (4.6)
$\displaystyle\leq\Lambda_{1}(u_{1})+\Lambda_{1}(\alpha_{1}(u))-\alpha_{1}(u)u$
$\displaystyle=\Lambda_{1}(u_{1})-\Lambda_{1}(u)$

and similarly,

$\alpha_{2}(v)(v_{1}-v)\leq\Lambda_{2}(v_{1})-\Lambda_{2}(v).$

Therefore,

$\displaystyle\int_{\overline{\Omega}}F(V-U)\;d\mu$
$\displaystyle=\int_{\Omega}\alpha_{1}(u)(u_{1}-u)dx+\int_{\partial\Omega}\alpha_{2}(v)(v_{1}-v)\frac{d\sigma}{b}$
$\displaystyle\leq\mathcal{J}_{2}(V)-\mathcal{J}_{2}(U).$

By definition, this shows that
$F=(\alpha_{1}(u),\alpha_{2}(v))=B_{2}(U)\in\partial\mathcal{J}_{2}(U)$. We
have shown that $U\in D(\partial\mathcal{J}_{2})$ and
$B_{2}(U)\in\partial\mathcal{J}_{2}(U)$. This completes the proof of the
lemma.

Next, we define the functional
$\mathcal{J}_{3,\rho}:\;\mathbb{X}_{2}\rightarrow[0,+\infty]$ by

$\mathcal{J}_{3,\rho}(U)=\begin{cases}\mathcal{J}_{\rho}(U)+\mathcal{J}_{2}(U)\;\;&\mbox{
if }\;U\in D(\mathcal{J}_{3,\rho}):=D(\mathcal{J}_{\rho})\cap
D(\mathcal{J}_{2}),\\\ +\infty&\mbox{ if }\;U\in\mathbb{X}_{2}\backslash
D(\mathcal{J}_{3,\rho}).\end{cases}$ (4.7)

Note that for $\rho=0$,

$D(\mathcal{J}_{3,0})=\\{U=(u,u|_{\partial\Omega})\in D(\mathcal{J}_{2}):u\in
W^{1,p}(\Omega)\cap L^{2}(\Omega),\;u|_{\partial\Omega}\in
L^{2}(\partial\Omega)\\},$ (4.8)

while for $\rho=1$,

$D(\mathcal{J}_{3,1})=\\{U=(u,u|_{\partial\Omega})\in D(\mathcal{J}_{2}):u\in
W^{1,p}(\Omega)\cap L^{2}(\Omega),\;u|_{\partial\Omega}\in
W^{1,q}(\partial\Omega)\cap L^{2}(\partial\Omega)\\}.$ (4.9)

We have the following result.

######  Lemma 4.8.

Let the assumptions of Lemma 4.6 be satisfied. Then the subdifferential of the
functional $\mathcal{J}_{3,\rho}$ is given by

$\displaystyle D(\partial\mathcal{J}_{3,\rho})$
$\displaystyle=\left\\{U=(u,u_{\mid\partial\Omega})\in
D(\mathcal{J}_{3,\rho}):-\Delta_{p}u+\alpha_{1}(u)\in L^{2}(\Omega)\right.$
$\displaystyle\left.\text{and }b(x)|\nabla
u|^{p-2}\partial_{n}u-b(x)\rho\Delta_{q,\Gamma}u+\alpha_{2}(u)\in
L^{2}(\partial\Omega,{d\sigma}/{b})\right\\}$

and

$\partial\mathcal{J}_{3,\rho}(U)=\bigg{(}-\Delta_{p}u+\alpha_{1}(u),b(x)|\nabla
u|^{p-2}\partial_{n}u-b(x)\rho\Delta_{q,\Gamma}u+\alpha_{2}(u)\bigg{)}.$
(4.10)

In particular, if for every $U=(u,u_{\mid\partial\Omega})\in
D(\mathcal{J}_{3,\rho}),$ the function
$\left(\alpha_{1}(u),\alpha_{2}(u)\right)\in\mathbb{X}_{2}$, then

$\partial\mathcal{J}_{3,\rho}:=\partial(\mathcal{J}_{\rho}+\mathcal{J}_{2})=\partial\mathcal{J}_{\rho}+\partial\mathcal{J}_{2}.$

###### Proof.

We calculate the subdifferential $\partial\mathcal{J}_{3,\rho}$. Let
$F=(f,g)\in\partial\mathcal{J}_{3,\rho}(U)$, that is, $F\in\mathbb{X}_{2}$,
$U\in D(\mathcal{J}_{3,\rho})=D(\mathcal{J}_{\rho})\cap D(\mathcal{J}_{2})$
and for every $V\in D(\mathcal{J}_{3,\rho})$, we have

$\int_{\overline{\Omega}}F(V-U)d\mu\leq\mathcal{J}_{3,\rho}(V)-\mathcal{J}_{3,\rho}(U).$

Proceeding as in Remark 3.2 and the proof of Lemma 4.7, we obtain that

$-\Delta_{p}u+\alpha_{1}(u)=f\;\;\mbox{ in }\;\mathcal{D}(\Omega)^{\prime},$

and

$b(x)|\nabla
u|^{p-2}\partial_{n}u-b(x)\rho\Delta_{q,\Gamma}u+\alpha_{2}(u)=g\;\mbox{
weakly on }\;\partial\Omega.$

Noting that $\partial\mathcal{J}_{3,\rho}$ is also a single-valued operator
(which follows from the assumptions on $\alpha_{j}$ and $\Lambda_{j}$), we
easily obtain (4.10), and this completes the proof of the first part.

To show the last part, note that it is clear that
$\partial\mathcal{J}_{\rho}+\partial\mathcal{J}_{2}\subset\partial\mathcal{J}_{3,\rho}$
always holds. To show the converse inclusion, let assume that for every
$U=(u,u_{\mid\partial\Omega})\in D(\mathcal{J}_{3,\rho}),$ the function
$(\alpha_{1}(u),\alpha_{2}(u))\in\mathbb{X}_{2}$. Then it follows from (3.3),
(4.4) (since $\partial\mathcal{J}_{2}=B_{2}$) and (4.10), that
$D(\partial\mathcal{J}_{3,\rho})=D(\partial\mathcal{J}_{\rho})\cap
D(\partial\mathcal{J}_{2})$ and

$\displaystyle\partial\mathcal{J}_{3,\rho}(U)$
$\displaystyle=\left(-\Delta_{p}u+\alpha_{1}(u),b(x)|\nabla
u|^{p-2}\partial_{n}u-b(x)\rho\Delta_{q,\Gamma}u+\alpha_{2}(u)\right)$
$\displaystyle=\left(-\Delta_{p}u,b(x)|\nabla
u|^{p-2}\partial_{n}u-b(x)\rho\Delta_{q,\Gamma}u\right)+\left(\alpha_{1}(u),\alpha_{2}(u)\right)$
$\displaystyle=\partial\mathcal{J}_{\rho}(U)+\partial\mathcal{J}_{2}(U).$

This completes the proof.

The following lemma is the main ingredient in the proof of Theorem 4.11 below.

######  Lemma 4.9.

Let $B_{1}:=A_{\rho}$ and set $B_{3}:=\partial\mathcal{J}_{3,\rho}$. Then

$\mathcal{R}\left(B_{1}\right)+\mathcal{R}\left(B_{2}\right)\subset\overline{\mathcal{R}(B_{3})}\;\mbox{
and
}\;\mbox{Int}(\mathcal{R}\left(B_{1}\right)+\mathcal{R}\left(B_{2}\right))\subset\mathcal{R}(B_{3}).$
(4.11)

In particular, if for every $U=(u,u_{\mid\partial\Omega})\in
D(\mathcal{J}_{3,\rho}),$ the function
$(\alpha_{1}(u),\alpha_{2}(u))\in\mathbb{X}_{2}$, then

$\mathcal{R}\left(B_{3}\right):=\mathcal{R}\left(B_{1}+B_{2}\right)\simeq\mathcal{R}\left(B_{1}\right)+\mathcal{R}\left(B_{2}\right).$
(4.12)

###### Proof.

By Remark 3.2 and Lemmas 4.7, 4.8, the operators $B_{1}$, $B_{2}$ and $B_{3}$
are subdifferentials of proper, convex and lower semicontinuous functionals
$\mathcal{J}_{\rho},\mathcal{J}_{2}$ and $\mathcal{J}_{\rho}+\mathcal{J}_{2}$,
respectively, on $\mathbb{X}_{2}$. Hence, $B_{1}$, $B_{2}$ and $B_{3}$ are
maximal monotone operators. In particular, if
$(\alpha_{1}(u),\alpha_{2}(u))\in\mathbb{X}_{2}$, for every
$U=(u,u_{\mid\partial\Omega})\in D(\mathcal{J}_{3,\rho}),$ then by Lemma 4.8,
one has $B_{3}=B_{1}+B_{2}$. Now, the lemma follows from the celebrated
Brezis-Haraux result in Theorem 4.2.

###  4.2. Statement and proof of the main result

Next, let $\mathcal{V}_{\rho}:=D(\mathcal{J}_{3,\rho})$ be given by (4.8) if
$\rho=0$ and by (4.9) if $\rho=1$.

######  Definition 4.10.

Let $F=(f,g)\in\mathbb{X}_{2}$. A function $u\in W^{1,p}(\Omega)$ is said to
be a weak solution of (4.1), if $\alpha_{1}(u)\in L^{1}(\Omega),$
$\alpha_{2}(u)\in L^{1}(\partial\Omega)$, $u|_{\partial\Omega}\in
W^{1,q}(\partial\Omega),$ if $\rho>0$ and

$\displaystyle\int_{\Omega}|\nabla u|^{p-2}\nabla u\cdot\nabla
vdx+\rho\int_{\partial\Omega}|\nabla_{\Gamma}u|^{q-2}\nabla_{\Gamma}u\cdot\nabla_{\Gamma}vd\sigma$
(4.13)
$\displaystyle+\int_{\Omega}\alpha_{1}(u)vdx+\int_{\partial\Omega}\alpha_{2}(u)v\frac{d\sigma}{b}=\int_{\Omega}fvdx+\int_{\partial\Omega}gv\frac{d\sigma}{b},$

for every $v\in W^{1,p}(\Omega)\cap C(\overline{\Omega})$ with
$v|_{\partial\Omega}\in W^{1,q}(\partial\Omega),$ if $\rho>0$.

Recall that $\lambda_{1}:=\int_{\Omega}dx$ and
$\displaystyle\lambda_{2}:=\int_{\partial\Omega}\frac{d\sigma}{b}$. We also
define the average $\left\langle F\right\rangle_{\overline{\Omega}}$ of
$F=\left(f,g\right)$ with respect to the measure $\mu,$ as follows:

$\left\langle
F\right\rangle_{\overline{\Omega}}:=\frac{1}{\mu\left(\overline{\Omega}\right)}\int_{\overline{\Omega}}Fd\mu=\frac{1}{\mu\left(\overline{\Omega}\right)}\left(\int_{\Omega}fdx+\int_{\partial\Omega}g\frac{d\sigma}{b}\right),$

where $\mu\left(\overline{\Omega}\right)=\lambda_{1}+\lambda_{2}$. Now, we are
ready to state the main result of this section.

######  Theorem 4.11.

Let $\alpha_{j}$ $(j=1,2)$ satisfy Assumption 4.3 and assume that the
functions $\Lambda_{j}$ $(j=1,2)$ satisfy (4.3). Let
$F=\left(f,g\right)\in\mathbb{X}_{2}$. The following hold:

  1. (a)

Suppose that the nonlinear elliptic problem (4.1) possesses a weak solution.
Then

$\left\langle
F\right\rangle_{\overline{\Omega}}\in\frac{\lambda_{1}\mathcal{R}\left(\alpha_{1}\right)+\lambda_{2}\mathcal{R}\left(\alpha_{2}\right)}{\lambda_{1}+\lambda_{2}}.$
(4.14)

  2. (b)

Assume that

$\left\langle
F\right\rangle_{\overline{\Omega}}\in\mbox{int}\left(\frac{\lambda_{1}\mathcal{R}\left(\alpha_{1}\right)+\lambda_{2}\mathcal{R}\left(\alpha_{2}\right)}{\lambda_{1}+\lambda_{2}}\right).$
(4.15)

Then the nonlinear elliptic problem (4.1) has at least one weak solution.

###### Proof.

We show that condition (4.14) is necessary. Let $F:=(f,g)\in\mathbb{X}_{2}$
and let $U=\left(u,u_{\mid\partial\Omega}\right)\in
D(B_{3})\subset\mathcal{V}_{\rho}$ be a weak solution of $B_{3}U=F$. Then, by
definition, for every $V=(v,v|_{\partial\Omega})\in\mathcal{V}_{\rho},$ (4.13)
holds. Taking $v\equiv 1$ in (4.13) yields

$\int_{\Omega}f\;dx+\int_{\partial\Omega}g\;\frac{d\sigma}{b}=\int_{\Omega}\alpha_{1}\left(u\right)dx+\int_{\partial\Omega}\alpha_{2}\left(u\right)\frac{d\sigma}{b}.$

Hence,

$\int_{\Omega}f\;dx+\int_{\partial\Omega}g\;\frac{d\sigma}{b}\in\left(\lambda_{1}\mathcal{R}\left(\alpha_{1}\right)+\lambda_{2}\mathcal{R}\left(\alpha_{2}\right)\right),$

and so (4.14) holds. This completes the proof of part (a).

We show that the condition (4.15) is sufficient.

(i) First, let $C\in\mathbf{C}$, where

$\mathbf{C:}=\left\\{C=(c_{1},c_{2}):(c_{1},c_{2})\in\mathcal{R}(\alpha_{1})\times\mathcal{R}(\alpha_{2})\right\\}.$

By definition, one has that $\mathbf{C}\subset\mathcal{R}\left(B_{2}\right)$
since $c_{1}=\alpha_{1}\left(d_{1}\right)$ for some constant function $d_{1}$
on $\Omega$ and $c_{2}=\alpha_{2}(d_{2})$ for some constant function $d_{2}$
on $\partial\Omega$. Let $F\in\mathbb{X}_{2}$ be such that (4.15) holds. We
must show $F\in\mathcal{R}\left(B_{3}\right)$. By (4.15), we may choose
$C=\left(c_{1},c_{2}\right)\in\mathbf{C}$ such that

$\left\langle
F\right\rangle_{\overline{\Omega}}=\frac{\lambda_{1}c_{1}+\lambda_{2}c_{2}}{\lambda_{1}+\lambda_{2}}\in\mbox{int}\left(\frac{\lambda_{1}\mathcal{R}\left(\alpha_{1}\right)+\lambda_{2}\mathcal{R}\left(\alpha_{2}\right)}{\lambda_{1}+\lambda_{2}}\right).$

Then, for $F\in\mathbb{X}_{2}$, we have $F=F_{1}+F_{2}$ with

$F_{1}:=F-C\mbox{ and }\;F_{2}=C.$

First,
$F_{1}\in\mathcal{R}\left(B_{1}\right)=\mathcal{N}\left(B_{1}\right)^{\perp}=\mathbf{1}^{\perp}$,
since

$\displaystyle\displaystyle\int_{\overline{\Omega}}F_{1}d\mu$ $\displaystyle=$
$\displaystyle\int_{\overline{\Omega}}\left(F-C\right)d\mu$ $\displaystyle=$
$\displaystyle\int_{\Omega}f\text{ }dx+\int_{\partial\Omega}g\text{
}\frac{d\sigma}{b}-\left(\lambda_{1}c_{1}+\lambda_{2}c_{2}\right)$
$\displaystyle=$
$\displaystyle\left(\lambda_{1}+\lambda_{2}\right)\left\langle
F\right\rangle_{\overline{\Omega}}-\left(\lambda_{1}c_{1}+\lambda_{2}c_{2}\right)=0.$

Obviously, $F_{2}=C\in\mathcal{R}\left(B_{2}\right)$. Hence, it is readily
seen that

$F\in(\mathcal{R}\left(B_{1}\right)+\mathcal{R}\left(B_{2}\right)).$

(ii) Next, denote by $\mathbb{B}_{\mathbb{R}}(x,r)$ the open ball in
$\mathbb{R}$ of center $x$ and radius $r>0$. Since

$\displaystyle\left\langle
F\right\rangle_{\overline{\Omega}}\in\mbox{int}\left(\frac{\lambda_{1}\mathcal{R}\left(\alpha_{1}\right)+\lambda_{2}\mathcal{R}\left(\alpha_{2}\right)}{\lambda_{1}+\lambda_{2}}\right),$

there exists $\delta>0$ such that the open ball

$\displaystyle\mathbb{B}_{\mathbb{R}}(\left\langle
F\right\rangle_{\overline{\Omega}},\delta)\subset\left(\frac{\lambda_{1}\mathcal{R}\left(\alpha_{1}\right)+\lambda_{2}\mathcal{R}\left(\alpha_{2}\right)}{\lambda_{1}+\lambda_{2}}\right).$

Since the mapping $F\mapsto\left\langle F\right\rangle_{\overline{\Omega}}$
from $\mathbb{X}_{2}$ into $\mathbb{R}$ is continuous, then there exists
$\varepsilon>0$ such that

$\left\langle
G\right\rangle_{\overline{\Omega}}\in\mathbb{B}_{\mathbb{R}}(\left\langle
F\right\rangle_{\overline{\Omega}},\delta)\subset\left(\frac{\lambda_{1}\mathcal{R}\left(\alpha_{1}\right)+\lambda_{2}\mathcal{R}\left(\alpha_{2}\right)}{\lambda_{1}+\lambda_{2}}\right),$

for all $G\in\mathbb{X}_{2}$ satisfying $\||F-G\||_{2}<\varepsilon$. It
finally follows from part (i) above that
$(\mathcal{R}\left(B_{1}\right)+\mathcal{R}\left(B_{2}\right))$ contains an
$\varepsilon$-ball in $\mathbb{X}_{2}$ centered at $F$. Therefore,

$F\in\mbox{int}(\mathcal{R}\left(B_{1}\right)+\mathcal{R}\left(B_{2}\right))\subset\mathcal{R}(B_{3}).$

Consequently, problem (4.1) is (weakly) solvable for every function
$F=(f,g)\in\mathbb{X}_{2},$ if (4.15) holds. This completes the proof of the
theorem.

######  Remark 4.12.

It is important to remark that in order to prove Theorem 4.11, we do not
require that $(\alpha_{1}(u),\alpha_{2}(u))$ should belong to
$\mathbb{X}_{2},$ for every $U=(u,u_{\mid\Gamma})\in D(\mathcal{J}_{3,\rho}).$
In particular, only the assumption (4.11) was needed. However, if this
happens, then we get the much stronger result in (4.12) which would require
that the nonlinearities $\alpha_{1},\alpha_{2}$ satisfy growth assumptions at
infinity.

We conclude this section with the following corollary and some examples.

######  Corollary 4.13.

Let the assumptions of Theorem 4.11 be satisfied. Let
$F=\left(f,g\right)\in\mathbb{X}_{2}$. Assume that at least one of the sets
$\mathcal{R}\left(\alpha_{1}\right)$, $\mathcal{R}\left(\alpha_{2}\right)$ is
open. Then the nonlinear elliptic problem (4.1) possesses a weak solution if
and only if (4.15) holds.

######  Remark 4.14.

Similar results to Theorem 4.11 and Corollary 4.13 were also obtained in [12,
Theorem 4.4], but only when $p=q=2.$

###  4.3. Examples

We will now give some examples as applications of Theorem 4.11.
Let$p,q\in(1,+\infty)$ be fixed.

######  Example 4.15.

Let $\alpha_{1}\left(s\right)$ or $\alpha_{2}\left(s\right)$ be equal to
$\alpha\left(s\right)=c\left|s\right|^{r-1}s,$ where $c,$ $r>0$. Note that
$\mathcal{R}\left(\alpha\right)=\mathbb{R}$. It is easy to check that $\alpha$
satisfies all the conditions of Assumption 4.3 and that the function
$\Lambda(t)=\int_{0}^{|t|}\alpha(s)ds$ satisfies (4.3). Then, it follows that
problem (4.1) is solvable for any $f\in L^{2}\left(\Omega\right),$ $g\in
L^{2}\left(\partial\Omega\right)$.

######  Example 4.16.

Consider the case when $\rho=\alpha_{2}\equiv 0$ in (4.1), that is, consider
the following boundary value problem:

$\left\\{\begin{array}[]{c}-\Delta_{p}u+\alpha_{1}\left(u\right)=f\text{ in
}\Omega,\\\ b\left(x\right)\left|\nabla u\right|^{p-2}\partial_{n}u=g\text{ on
}\Gamma.\end{array}\right.$

Then, by Theorem 4.11, this problem has a weak solution if

$\int_{\Omega}f\text{ }dx+\int_{\partial\Omega}g\text{
}\frac{d\sigma}{b}\in\lambda_{1}\mbox{int}(\mathcal{R}\left(\alpha_{1}\right)),$

which yields the classical Landesman-Lazer result (see (1.6)) for $g\equiv 0$
and $p=2$.

######  Example 4.17.

Let us now consider the case when $\alpha_{1}\equiv\alpha$ and
$\alpha_{2}\equiv 0,$ where $\alpha$ is a continuous, odd and nondecreasing
function on $\mathbb{R}$ such that $\alpha\left(0\right)=0$. The problem

$\left\\{\begin{array}[]{cc}-\Delta_{p}u+\alpha\left(u\right)=f,&\text{in
}\Omega,\\\ b\left(x\right)\left|\nabla u\right|^{p-2}\partial_{n}u-\rho
b\left(x\right)\Delta_{q,\Gamma}u=g,&\text{on
}\partial\Omega,\end{array}\right.$ (4.16)

has a weak solution if

$\int_{\Omega}f\text{ }dx+\int_{\partial\Omega}g\text{
}\frac{d\sigma}{b}\in\lambda_{2}\mbox{int}\bigg{(}\mathcal{R}\left(\alpha\right)\bigg{)}.$
(4.17)

Let us now choose $\alpha\left(s\right)=\arctan\left(s\right)$ in (4.16).
Then, it is easy to check that

$\Lambda(t):=\int_{0}^{|t|}\alpha(s)ds=\left|t\right|\arctan\left(\left|t\right|\right)-\frac{1}{2}\ln\left(1+t^{2}\right),\text{
}t\in\mathbb{R}$

is monotone increasing on $\mathbb{R}_{+}$ and that it satisfies
$\Lambda(2t)\leq C_{2}\Lambda(t),$ $\forall t\in\mathbb{R}$, for some constant
$C_{2}>1$. Therefore, (4.17) becomes the necessary and sufficient condition

$\left|\frac{1}{\lambda_{2}}\left(\int_{\Omega}f\text{
}dx+\int_{\partial\Omega}g\text{
}\frac{d\sigma}{b}\right)\right|<\frac{\pi}{2}.$ (4.18)

##  5\. A priori estimates

Let $\Omega\subset\mathbb{R}^{N}$ be a bounded Lipschitz domain with boundary
$\partial\Omega$. Recall that $1<p,q<\infty$, $\rho\in\\{0,1\\}$ and $b\in
L^{\infty}(\partial\Omega)$ with $b(x)\geq b_{0}>0,$ for some constant
$b_{0}$. We consider the nonlinear elliptic boundary value problem formally
given by

$\begin{cases}\displaystyle-\Delta_{p}u+\alpha_{1}(x,u)+|u|^{p-2}u=f,&\mbox{
in }\;\Omega\\\ &\\\ \displaystyle-\rho b(x)\Delta_{q,\Gamma}u+\rho
b(x)|u|^{q-2}u+b(x)|\nabla u|^{p-2}\partial_{n}u+\alpha_{2}(x,u)=g,&\mbox{ on
}\;\partial\Omega,\end{cases}$ (5.1)

where $f\in L^{p_{1}}(\Omega)$ and $g\in L^{q_{1}}(\partial\Omega)$ for some
$1\leq p_{1},q_{1}\leq\infty$. If $\rho=0$, then the boundary conditions in
(5.1) are of Robin type. Existence and regularity of weak solutions for this
case have been obtained in [5] for $p=2$ (see also [29] for the linear case)
and for general $p$ in [6]. Therefore, we will concentrate our attention to
the case $\rho=1$ only; in this case, the boundary condition in (5.1) is a
generalized Wentzell-Robin boundary condition. For the sake of simplicity,
from now on we will also take $b\equiv 1$.

###  5.1. General assumptions

Throughout this section, we assume that the functions
$\alpha_{1}:\Omega\times\mathbb{R}\rightarrow\mathbb{R}$ and
$\alpha_{2}:\partial\Omega\times\mathbb{R}\rightarrow\mathbb{R}$ satisfy the
following conditions:

######  Assumption 5.1.

$\begin{cases}\displaystyle\alpha_{j}(x,\cdot)\mbox{ is odd and strictly
increasing},\\\
\displaystyle\alpha_{j}(x,0)=0,\;\;\displaystyle\alpha_{j}(x,\cdot)\mbox{ is
continuous },\\\ \displaystyle\lim_{t\rightarrow
0}\frac{\alpha_{j}(x,t)}{t}=0,\;\;\displaystyle\lim_{t\rightarrow\infty}\frac{\alpha_{j}(x,t)}{t}=\infty,\end{cases}$

for $\lambda_{N}$-a.e. $x\in\Omega$ if $j=1$ and $\sigma$-a.e.
$x\in\partial\Omega$ if $j=2$.

Since $\alpha_{j}(x,\cdot)$ are strictly increasing for $\lambda_{N}$-a.e.
$x\in\Omega$ if $j=1$ and $\sigma$-a.e. $x\in\partial\Omega$ if $j=2$, then
they have inverses which we denote by $\widetilde{\alpha}_{j}(x,\cdot)$ (cf.
also Section 4). We define the functions $\Lambda_{1},$
$\widetilde{\Lambda}_{1}:\Omega\times\mathbb{R}\rightarrow[0,\infty)$ and
$\Lambda_{2},$
$\widetilde{\Lambda}_{2}:\partial\Omega\times\mathbb{R}\rightarrow[0,\infty)$
by

$\Lambda_{j}(x,t):=\int_{0}^{|t|}\alpha_{j}(x,s)\;ds\;\mbox{ and
}\;\widetilde{\Lambda}_{j}(x,t):=\int_{0}^{|t|}\widetilde{\alpha}_{j}(x,s)\;ds.$

Then, it is clear that, for $\lambda_{N}$-a.e. $x\in\Omega$ if $j=1$ and
$\sigma$-a.e. $x\in\partial\Omega$ if $j=2$, $\Lambda_{j}(x,\cdot)$ and
$\widetilde{\Lambda}_{j}(x,\cdot)$ are differentiable, monotone and convex
with $\Lambda_{j}(x,0)=\widetilde{\Lambda}_{j}(x,0)=0.$ Furthermore,
$\Lambda_{j}(x,\cdot)$ is an ${\mathcal{N}}$-function and
$\widetilde{\Lambda}_{j}(x,\cdot)$ is its complementary
${\mathcal{N}}$-function. The function $\widetilde{\Lambda}_{j}$ is then the
complementary Musielak-Orlick function of $\Lambda_{j}$ in the sense of Young
(see Definition 2.3).

######  Assumption 5.2.

We assume, for $\lambda_{N}$-a.e. $x\in\Omega$ if $j=1$ and $\sigma$-a.e.
$x\in\partial\Omega$ if $j=2$, that $\Lambda_{j}(x,\cdot)$ and
$\widetilde{\Lambda}_{j}(x,\cdot)$ satisfy the ($\triangle_{2}$)-condition in
the sense of Definition 2.5.

It follows from Assumption 5.2 that there exist two constants
$c_{1},c_{2}\in(0,1]$ such that for $\lambda_{N}$-a.e. $x\in\Omega$ if $j=1$
and $\sigma$-a.e. $x\in\partial\Omega$ if $j=2$ and for all $t\in\mathbb{R}$,

$c_{j}t\alpha_{j}(x,t)\leq\Lambda_{j}(x,t)\leq t\alpha_{j}(x,t).$ (5.2)

Next, let

$L_{\Lambda_{1}}(\Omega):=\left\\{u:\Omega\rightarrow\mathbb{R}\;\mbox{
measurable: }\;\int_{\Omega}\Lambda_{1}(x,u)dx<\infty\right\\}$

and

$L_{\Lambda_{2}}(\partial\Omega):=\left\\{u:\partial\Omega\rightarrow\mathbb{R}\;\mbox{
measurable: }\;\int_{\partial\Omega}\Lambda_{2}(x,u)d\sigma<\infty\right\\}.$

Since $\Lambda_{j}(x,\cdot)$ and $\widetilde{\Lambda}_{j}(x,\cdot)$ satisfy
the $(\triangle_{2})$-condition, it follows from [1, Theorem 8.19], that
$L_{\Lambda_{1}}(\Omega)$ and $L_{\Lambda_{2}}(\partial\Omega),$ endowed
respectively with the norms

$\|u\|_{\Lambda_{1},\Omega}:=\inf\left\\{k>0:\int_{\Omega}\Lambda_{1}\left(x,\frac{u(x)}{k}\right)dx\leq
1\right\\},$

and

$\|u\|_{\Lambda_{2},\partial\Omega}:=\inf\left\\{k>0:\int_{\partial\Omega}\Lambda_{2}\left(x,\frac{u(x)}{k}\right)d\sigma\leq
1\right\\},$

are reflexive Banach spaces. Moreover, by [1, Section 8.11, p.234], the
following generalized versions of Hölder’s inequality will also become useful
in the sequel,

$\left|\int_{\Omega}uvdx\right|\leq
2\|u\|_{\Lambda_{1},\Omega}\|v\|_{\widetilde{\Lambda}_{1},\Omega}$ (5.3)

and

$\left|\int_{\partial\Omega}uv\;d\sigma\right|\leq
2\|u\|_{\Lambda_{2},\partial\Omega}\|v\|_{\widetilde{\Lambda}_{2},\partial\Omega}.$
(5.4)

###  5.2. Existence and uniqueness of weak solutions of perturbed equations

Let

$\mathcal{V}:=\\{U:=(u,u|_{\partial\Omega}):\;u\in W^{1,p}(\Omega)\cap
L_{\Lambda_{1}}(\Omega),\;u_{\mid\partial\Omega}\in
W^{1,q}(\partial\Omega)\cap L_{\Lambda_{2}}(\partial\Omega)\\}.$

Then for every $1<p,q<\infty$, $\mathcal{V}$ endowed with the norm

$\|U\|_{\mathcal{V}}=\|u\|_{W^{1,p}(\Omega)}+\|u\|_{\Lambda_{1},\Omega}+\|u\|_{W^{1,q}(\partial\Omega)}+\|u\|_{\Lambda_{2},\partial\Omega}$

is a reflexive Banach space. Recall that $\rho=1$. Throughout the following,
we denote by $\mathcal{V}^{\prime}$ the dual of $\mathcal{V}$.

######  Definition 5.3.

A function $U=(u,u|_{\partial\Omega})\in\mathcal{V}$ is said to be a weak
solution of (5.1), if for every $V\in\mathcal{V}=(v,v|_{\partial\Omega}),$

$\mathcal{A}(U,V)=\int_{\Omega}fvdx+\int_{\partial\Omega}gvd\sigma,$ (5.5)

provided that the integrals on the right-hand side exist. Here,

$\displaystyle\mathcal{A}(U,V)$ $\displaystyle:=\int_{\Omega}|\nabla
u|^{p-2}\nabla u\cdot\nabla vdx+\int_{\Omega}|u|^{p-2}uvdx$
$\displaystyle+\int_{\Omega}\alpha_{1}(x,u)vdx+\int_{\partial\Omega}|\nabla_{\Gamma}u|^{q-2}\nabla_{\Gamma}u\cdot\nabla_{\Gamma}vd\sigma$
$\displaystyle+\int_{\partial\Omega}|u|^{q-2}uvd\sigma+\int_{\partial\Omega}\alpha_{2}(x,u)vd\sigma.$

######  Lemma 5.4.

Assume Assumptions 5.1 and 5.2. Let $1<p,q<\infty$ and $U\in\mathcal{V}$ be
fixed. Then the functional $V\mapsto\mathcal{A}(U,V)$ belongs to
$\mathcal{V}^{\prime}$. Moreover, $\mathcal{A}$ is strictly monotone,
hemicontinuous and coercive.

###### Proof.

Let $U=(u,u|_{\partial\Omega})\in\mathcal{V}$ be fixed. It is clear that
$\mathcal{A}(U,\cdot)$ is linear. Let
$V=(v,v|_{\partial\Omega})\in\mathcal{V}$. Then, exploiting (5.3) and (5.4),
we obtain

$\displaystyle\left|\mathcal{A}(U,V)\right|\leq\|u\|_{W^{1,p}(\Omega)}^{p-1}\|v\|_{W^{1,p}(\Omega)}+\|u\|_{W^{1,q}(\partial\Omega)}^{q-1}\|v\|_{W^{1,q}(\partial\Omega)}$
(5.6)
$\displaystyle+2\max\left\\{1,\int_{\Omega}\widetilde{\Lambda}_{1}(x,\alpha_{1}(x,u))\;dx\right\\}\|v\|_{\Lambda_{1},\Omega}$
$\displaystyle+2\max\left\\{1,\int_{\partial\Omega}\widetilde{\Lambda}_{2}(x,\alpha_{2}(x,u))\;d\sigma\right\\}\|v\|_{\Lambda_{2},\partial\Omega}$
$\displaystyle\leq K(U)\|V\|_{\mathcal{V}},$

where

$\displaystyle K(U):=$
$\displaystyle\|u\|_{W^{1,p}(\Omega)}^{p-1}+2\max\left\\{1,\int_{\Omega}\widetilde{\Lambda}_{1}(x,\alpha_{1}(x,u))\;dx\right\\}$
$\displaystyle+\|u\|_{W^{1,q}(\partial\Omega)}^{q-1}+2\max\left\\{1,\int_{\partial\Omega}\widetilde{\Lambda}_{2}(x,\alpha_{2}(x,u))\;d\sigma\right\\}.$

This shows $\mathcal{A}(U,\cdot)\in\mathcal{V}^{\prime},$ for every
$U\in\mathcal{V}$.

Next, let $U,V\in\mathcal{V}$. Then, using (2.11) and the fact that
$\alpha_{j}(x,\cdot)$ are monotone nondecreasing, that is,
$(\alpha_{j}(x,t)-\alpha_{j}(x,s))(t-s)\geq 0,$ for all $t,s\in\mathbb{R},$ we
obtain

$\displaystyle\mathcal{A}(U,U-V)-\mathcal{A}(V,U-V)$ (5.7)
$\displaystyle=\int_{\Omega}\left(|\nabla u|^{p-2}\nabla u-|\nabla
v|^{p-2}\nabla
v\right)\cdot\nabla(u-v)dx+\int_{\Omega}\left(|u|^{p-2}u-|v|^{p-2}v\right)(u-v)dx$
$\displaystyle+\int_{\Omega}\left(\alpha_{1}(x,u)-\alpha_{1}(x,v)\right)(u-v)dx+\int_{\partial\Omega}\left(|u|^{q-2}u-|v|^{q-2}v\right)(u-v)d\sigma$
$\displaystyle+\int_{\partial\Omega}\left(|\nabla_{\Gamma}u|^{q-2}\nabla_{\Gamma}u-|\nabla_{\Gamma}v|^{q-2}\nabla
v\right)\cdot\nabla_{\Gamma}(u-v)d\sigma$
$\displaystyle+\int_{\partial\Omega}\left(\alpha_{2}(x,u)-\alpha_{1}(x,v)\right)(u-v)d\sigma$
$\displaystyle\geq\int_{\Omega}\left(|\nabla u|+|\nabla
v|\right)^{p-2}|\nabla(u-v)|^{2}dx+\int_{\Omega}\left(|u|+|v|\right)^{p-2}|u-v|^{2}dx$
$\displaystyle+\int_{\partial\Omega}\left(|\nabla_{\Gamma}u|+|\nabla_{\Gamma}v|\right)^{p-2}|\nabla_{\Gamma}(u-v)|^{2}d\sigma+\int_{\partial\Omega}\left(|u|+|v|\right)^{p-2}|u-v|^{2}d\sigma$
$\displaystyle\geq 0.$

This shows that $\mathcal{A}$ is monotone. The estimate (5.7) also shows that

$\mathcal{A}(U,U-V)-\mathcal{A}(V,U-V)>0,$

for all $U,V\in V$ with $U\neq V$, that is, $u\neq v$ or
$u|_{\partial\Omega}\neq v|_{\partial\Omega}$. Thus, $\mathcal{A}$ is strictly
monotone.

The continuity of the norm function and the continuity of
$\alpha_{j}(x,\cdot),$ $j=1,2$ imply that $\mathcal{A}$ is hemicontinuous.

Finally, since $\Lambda_{j}$ and $\widetilde{\Lambda}_{j}$ satisfy the
$(\triangle_{2}^{0})$-condition, from Proposition 2.10 and Corollary 2.11, it
follows

$\displaystyle\lim_{\|u\|_{\Lambda_{1},\Omega}\rightarrow+\infty}\frac{\int_{\Omega}u\alpha_{1}(x,u)\;dx}{\|u\|_{\Lambda_{1},\Omega}}=+\infty,\mbox{
and
}\;\lim_{\|u\|_{\Lambda_{2},\partial\Omega}\rightarrow+\infty}\frac{\int_{\partial\Omega}u\alpha_{2}(x,u)\;d\sigma}{\|u\|_{\Lambda_{2},\partial\Omega}}$
$\displaystyle=+\infty.$

Consequently, we deduce

$\lim_{\|U\|_{\mathcal{V}}\rightarrow+\infty}\frac{\mathcal{A}(U,U)}{\|U\|_{\mathcal{V}}}=+\infty,$
(5.8)

which shows that $\mathcal{A}$ is coercive. The proof of the lemma is
finished.

The following result is concerned with the existence and uniqueness of weak
solutions to problem (5.1).

######  Theorem 5.5.

Assume Assumptions 5.1 and 5.2. Let $1<p,q<\infty$, $p_{1}\geq p\ast$ and
$q_{1}\geq q\ast$, where $p\ast:=p/(p-1)$ and $q\ast:=q/(q-1)$. Then for every
$(f,g)\in X^{p_{1},q_{1}}(\overline{\Omega},\mu)$, there exists a unique
function $U\in\mathcal{V}$ which is a weak solution to (5.1).

###### Proof.

Let $\langle\cdot,\cdot\rangle$ denote the duality between $\mathcal{V}$ and
$\mathcal{V}^{\prime}$. Then, from Lemma 5.4, it follows that for each
$U\in\mathcal{V}$, there exists $A(U)\in\mathcal{V}^{\prime}$ such that

$\mathcal{A}(U,V)=\langle A(U),V\rangle,$

for every $V\in\mathcal{V}$. Hence, this relation defines an operator
$A:\;\mathcal{V}\rightarrow\mathcal{V}^{\prime},$ which is bounded by (5.6).
Exploiting Lemma 5.4 once again, it is easy to see that $A$ is monotone and
coercive. It follows from Brodwer’s theorem (see, e.g., [11, Theorem 5.3.22]),
that $A(\mathcal{V})=\mathcal{V}^{\prime}$. Therefore, for every
$F\in\mathcal{V}^{\prime}$ there exists $U\in\mathcal{V}$ such that $A(U)=F$,
that is, for every $V\in\mathcal{V}$,

$\langle A(U),V\rangle=\mathcal{A}(U,V)=\langle V,F\rangle.$

Since $W^{1,p}(\Omega)\hookrightarrow L^{p}(\Omega)$ and
$W^{1,q}(\partial\Omega)\hookrightarrow L^{q}(\partial\Omega)$ with dense
injection, by duality, we have
$X^{p\ast,q\ast}(\overline{\Omega},\mu)\hookrightarrow\mathcal{V}^{\prime}$.
Since $\Omega$ is bounded and $\sigma(\partial\Omega)<\infty$, we obtain that

$X^{p_{1},q_{1}}(\overline{\Omega},\mu)\hookrightarrow
X^{p\ast,q\ast}(\overline{\Omega},\mu)\hookrightarrow\mathcal{V}^{\prime}.$

This shows the existence of weak solutions. The uniqueness follows from the
fact that $\mathcal{A}$ is strictly monotone (cf. Lemma 5.4). This completes
the proof of the theorem.

######  Corollary 5.6.

Let the assumptions of Theorem 5.5 be satisfied. Let

$p_{h}:=\frac{Np}{N(p-1)+p},\;\;q_{h}:=\frac{p(N-1)}{N(p-1)},\;\mbox{ and
}q_{k}:=\frac{q(N-1)}{N(q-1)+1}.$ (5.9)

  1. (a)

Let $1<p<N$, $1<q<p(N-1)/N$, $p_{1}\geq p_{h}$ and $q_{1}\geq q_{k}$. Then for
every $(f,g)\in X^{p_{1},q_{1}}(\Omega,\mu)$, there exists a function
$U\in\mathcal{V}$ which is the unique weak solution to (5.1).

  2. (b)

Let $1<q<N-1$, $1<p<Nq/(N-1)$, $p_{1}\geq p_{h}$ and $q_{1}\geq q_{h}$. Then
for every $(f,g)\in X^{p_{1},q_{1}}(\Omega,\mu)$, there exists a function
$U\in\mathcal{V}$ which is the unique weak solution to (5.1).

###### Proof.

We first prove (1). Let $1<p<N$ and $1<q<p(N-1)/N$ and let $p_{1}\geq p_{h}$
and $q_{1}\geq q_{k},$ where $p_{h}$ and $q_{k}$ are given by (5.9). Let
$p_{s}:=Np/(N-p)$ and $q_{t}:=(N-1)q/(N-1-q)$. Since
$W^{1,p}(\Omega)\hookrightarrow L^{p_{s}}(\Omega)$ and
$W^{1,q}(\partial\Omega)\hookrightarrow L^{q_{t}}(\partial\Omega)$ with dense
injection, then by duality,
$X^{p_{h},q_{k}}(\overline{\Omega},\mu)\hookrightarrow\mathcal{V}^{\prime}$,
where $\displaystyle 1/p_{s}+1/p_{h}=1$ and $\displaystyle 1/q_{t}+1/q_{k}=1$.
Since $\mu(\overline{\Omega})<\infty$, we have that

$X^{p_{1},q_{1}}(\overline{\Omega},\mu)\hookrightarrow
X^{p_{h},q_{h}}(\overline{\Omega},\mu)\hookrightarrow\mathcal{V}^{\prime}.$

Hence, for every $F:=(f,g)\in
X^{p_{1},q_{1}}(\overline{\Omega},\mu)\hookrightarrow\mathcal{V}^{\prime}$,
there exists $U\in\mathcal{V}$ such that for every $V\in\mathcal{V}$,

$\langle
A(U),V\rangle=\mathcal{A}(U,V)=\int_{\Omega}fv\;dx+\int_{\partial\Omega}gv\;d\sigma.$

The uniqueness of the weak solution follows again from the fact that
$\mathcal{A}$ is strictly monotone.

In order to prove the second part, we use the the embeddings
$W^{1,p}(\Omega)\hookrightarrow L^{p_{s}}(\Omega)$,
$W^{1,p}(\Omega)\hookrightarrow L^{q_{s}}(\partial\Omega)$ 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
operator $A$ to problem (5.1). We begin by assuming the following.

######  Assumption 5.7.

Suppose that $\alpha_{j},$ $j=1,2,$ satisfy the following conditions:

$\begin{cases}\displaystyle\mbox{there are constants }\;c_{j}\in(0,1]\;\mbox{
such that }\\\ \displaystyle
c_{j}\left|\alpha_{j}(x,\xi-\eta)\right|\leq\left|\alpha_{j}(x,\xi)-\alpha_{j}(x,\eta)\right|\;\mbox{
for all }\;\xi,\eta\in\mathbb{R}.\end{cases}$ (5.10)

######  Theorem 5.8.

Assume Assumptions 5.1, 5.2 and 5.7. Let $p,q\geq 2$ and let
$A:\;\mathcal{V}\rightarrow\mathcal{V}^{\prime}$ be the continuous and bounded
operator constructed in the proof of Theorem 5.5. Then $A$ is injective and
hence, invertible and its inverse $A^{-1}$ is also continuous and bounded.

###### Proof.

First, we remark that, since

$\left(\alpha_{j}(x,t)-\alpha_{j}(x,s)\right)(t-s)\geq 0\text{, for all
}t,s\in\mathbb{R},$

for $\lambda_{N}$-a.e.$x\in\Omega$ if $j=1$ and $\sigma$-a.e.
$x\in\partial\Omega$ if $j=2$, it follows from (5.10) that, for all
$t,s\in\mathbb{R}$,

$\left(\alpha_{j}(x,t)-\alpha_{j}(x,s)\right)(t-s)\geq
c_{j}\alpha_{j}(x,t-s)\cdot(t-s).$ (5.11)

Let $U,V\in\mathcal{V}$ and $p,q\in[2,\infty)$. Then, exploiting (2.12),
(5.11) and the ($\triangle_{2}$)-condition, we obtain

$\displaystyle\langle
A(U)-A(V),U-V\rangle=\mathcal{A}(U,U-V)-\mathcal{A}(V,U-V)$ (5.12)
$\displaystyle=\int_{\Omega}\left(|\nabla u|^{p-2}\nabla u-|\nabla
v|^{p-2}\nabla
v\right)\cdot\nabla(u-v)dx+\int_{\Omega}\left(|u|^{p-2}u-|v|^{p-2}v\right)(u-v)dx$
$\displaystyle+\int_{\Omega}\left(\alpha_{1}(x,u)-\alpha_{1}(x,v)\right)(u-v)dx+\int_{\partial\Omega}\left(|\nabla_{\Gamma}u|^{q-2}\nabla_{\Gamma}u-|\nabla_{\Gamma}v|^{q-2}\nabla_{\Gamma}v\right)\cdot\nabla_{\Gamma}(u-v)d\sigma$
$\displaystyle+\int_{\partial\Omega}\left(|u|^{q-2}u-|v|^{q-2}v\right)(u-v)d\sigma+\int_{\partial\Omega}\left(\alpha_{2}(x,u)-\alpha_{2}(x,v)\right)(u-v)d\sigma$
$\displaystyle\geq\|u-v\|_{W^{1,p}(\Omega)}^{p}+c_{1}\int_{\Omega}\Lambda_{1}(x,u-v)dx+\|u-v\|_{W^{1,q}(\partial\Omega)}^{q}+c_{2}\int_{\partial\Omega}\Lambda_{2}(x,u-v)d\sigma.$

This implies that $\langle A(U)-A(V),U-V\rangle>0,$ for all
$U,V\in\mathcal{V}$ with $U\neq V$ (that is, $u\neq v$, or
$u|_{\partial\Omega}\neq v|_{\partial\Omega}$). Therefore, the operator $A$ is
injective and hence, $A^{-1}$ exists. Since for every $U\in\mathcal{V}$,

$\mathcal{A}(U,U)=\langle
A(U),U\rangle\leq\|A(U)\|_{\mathcal{V}^{\prime}}\|U\|_{\mathcal{V}},$

from the coercivity of $\mathcal{A}$ (see (5.8)), it is not difficult to see
that

$\lim_{\|U\|_{\mathcal{V}}\rightarrow+\infty}\|A(U)\|_{\mathcal{V}^{\prime}}=+\infty.$
(5.13)

Thus, $A^{-1}:\;\mathcal{V}^{\prime}\rightarrow\mathcal{V}$ is bounded.

Next, we show that $A^{-1}:\;\mathcal{V}^{\prime}\rightarrow\mathcal{V}$ is
continuous. Assume that $A^{-1}$ is not continuous. Then there is a sequence
$F_{n}\in\mathcal{V}^{\prime}$ with $F_{n}\rightarrow F$ in
$\mathcal{V}^{\prime}$ and a constant $\delta>0$ such that

$\|A^{-1}(F_{n})-A^{-1}(F)\|_{\mathcal{V}}\geq\delta,$ (5.14)

for all $n\in\mathbb{N}$. Let $U_{n}:=A^{-1}(F_{n})$ and $U=A^{-1}(F)$. Since
$\left\\{F_{n}\right\\}$ is a bounded sequence and $A^{-1}$ is bounded, we
have that $\left\\{U_{n}\right\\}$ is bounded in $\mathcal{V}$. Thus, we can
select a subsequence, which we still denote by $\left\\{U_{n}\right\\},$ which
converges weakly to some function $V\in\mathcal{V}$. Since
$A(U_{n})-A(V)\rightarrow F-A(V)$ strongly in $\mathcal{V}$ and $U_{n}-V$
converges weakly to zero in $\mathcal{V}$, we deduce

$\lim_{n\rightarrow\infty}\langle A(U_{n})-A(V),U_{n}-V\rangle=0.$ (5.15)

From (5.12) and (5.15), it follows that

$\lim_{n\rightarrow\infty}\|u_{n}-v\|_{W^{1,p}(\Omega)}=0\;\mbox{and
}\;\lim_{n\rightarrow\infty}\int_{\Omega}\Lambda_{1}(x,u_{n}-v)dx=0,$

while

$\lim_{n\rightarrow\infty}\|u_{n}-v\|_{W^{1,q}(\partial\Omega)}=0\;\mbox{ and
}\;\lim_{n\rightarrow\infty}\int_{\partial\Omega}\Lambda_{2}(x,u_{n}-v)d\sigma=0.$

Therefore, $U_{n}\rightarrow V$ strongly in $\mathcal{V}$. Since $A$ is
continuous and

$F_{n}=A(U_{n})\rightarrow A(V)=F=A(U)$

it follows from the injectivity of $A,$ that $U=V$. This shows that

$\lim_{n\rightarrow\infty}\|A^{-1}(F_{n})-A^{-1}(F)\|_{\mathcal{V}}=\lim_{n\rightarrow\infty}\|U_{n}-U\|_{\mathcal{V}}=0,$

which contradicts (5.14). Hence,
$A^{-1}:\;\mathcal{V}^{\prime}\rightarrow\mathcal{V}$ is continuous. The proof
is finished.

######  Corollary 5.9.

Let the assumptions of Theorem 5.8 be satisfied. Let $p_{h},q_{h}$ and $q_{k}$
be as in (5.9) and let $A:\;\mathcal{V}\rightarrow\mathcal{V}^{\prime}$ be the
continuous and bounded operator constructed in the proof of Theorem 5.5.

  1. (a)

If $2\leq p<N$, $2\leq q<p(N-1)/N$, $p_{1}\geq p_{h}$ and $q_{1}\geq q_{k}$,
then $A^{-1}:\;X^{p_{1},q_{1}}(\overline{\Omega},\mu)\rightarrow
X^{p_{s},q_{t}}(\overline{\Omega},\mu)$ is continuous and bounded. Moreover,
$A^{-1}:\;X^{p_{1},q_{1}}(\overline{\Omega},\mu)\rightarrow\mathcal{V}\cap
X^{r,s}(\overline{\Omega},\mu)$ is compact for every $r\in(1,p_{s})$ and
$s\in(1,q_{s})$.

  2. (b)

If $2\leq q<N-1$, $2\leq p<qN/(N-1)$, $p_{1}\geq p_{h}$ and $q_{1}\geq q_{h}$,
then the operator $A^{-1}:\;X^{p_{1},q_{1}}(\overline{\Omega},\mu)\rightarrow
X^{p_{s},q_{s}}(\overline{\Omega},\mu)$ is continuous and bounded. Moreover,
$A^{-1}:\;X^{p_{1},q_{1}}(\overline{\Omega},\mu)\rightarrow\mathcal{V}\cap
X^{r,s}(\overline{\Omega},\mu)$ is compact for every $r\in(1,p_{s})$ and
$s\in(1,q_{s})$.

###### Proof.

We only prove the first part. The second part of the proof follows by analogy
and is left to the reader. Let $2\leq p<N$, $2\leq q<p(N-1)/N$, $p_{1}\geq
p_{h}$ and $q_{1}\geq q_{k}$ and let $F\in
X^{p_{1},q_{1}}(\overline{\Omega},\mu)$. Proceeding exactly as in the proof of
Theorem 5.8, we obtain

$\|A^{-1}(F)\|_{p_{s},q_{t}}\leq C_{1}\|A^{-1}(F)\|_{\mathcal{V}}\leq
C\|F\|_{{\mathcal{V}^{\prime}}}\leq C_{2}\|F\|_{p_{1},q_{1}}.$

Hence, the operator
$A^{-1}:\;X^{p_{1},q_{1}}(\overline{\Omega},\mu)\rightarrow
X^{p_{s},q_{t}}(\overline{\Omega},\mu)$ is bounded. Finally, using the facts
that
$X^{p_{1},q_{1}}(\overline{\Omega},\mu)\hookrightarrow\mathcal{V}^{\prime}$,
$A^{-1}:\;\mathcal{V}^{\prime}\rightarrow\mathcal{V}$ is continuous and
$\mathcal{V}\hookrightarrow X^{p_{s},q_{t}}(\overline{\Omega},\mu)$, we easily
deduce that $A^{-1}:\;X^{p_{1},q_{1}}(\overline{\Omega},\mu)\rightarrow
X^{p_{s},q_{t}}(\overline{\Omega},\mu)$ is continuous.

Now, let $1<r<p_{s}$ and $1<s<q_{s}$. Since the injection
$\mathcal{V}\hookrightarrow X^{r,s}(\overline{\Omega},\mu)$ is compact, then
by duality, the injection
$X^{r^{\prime},s^{\prime}}(\overline{\Omega},\mu)\hookrightarrow(\mathcal{V})^{*}$
is compact for every $r^{\prime}>p_{s}^{\prime}=p_{h}$ and
$s^{\prime}>q_{s}^{\prime}=q_{h}$. This, together with the fact that
$A^{-1}:\;(\mathcal{V})^{*}\to\mathcal{V}$ is continuous and bounded, imply
that $A^{-1}:\;X^{p_{1},q_{1}}(\overline{\Omega},\mu)\to\mathcal{V}$ is
compact for every $p_{1}>p_{h}$ and $q_{1}>q_{h}$.

It remains to show that $A^{-1}$ is also compact as a map into
$X^{r,s}(\overline{\Omega},\mu)$ for every $r\in(1,p_{s})$ and
$s\in(1,q_{s})$. Since $A^{-1}$ is bounded, we have to show that the image of
every bounded set $\mathcal{B}\subset\mathbb{X}^{p_{1},q_{1}}(\Omega,\mu)$ is
relatively compact in $X^{r,s}(\overline{\Omega},\mu)$ for every
$r\in(1,p_{s})$ and $s\in(1,q_{s})$. Let $U_{n}$ be a sequence in
$A^{-1}(\mathcal{B})$. Let $F_{n}=A(U_{n})\in\mathcal{B}$. Since $\mathcal{B}$
is bounded, then the sequence $F_{n}$ is bounded. Since $A^{-1}$ is compact as
a map into $\mathcal{V}$, it follows that there is a subsequence $F_{n_{k}}$
such that $A^{-1}(F_{n_{k}})\to U\in\mathcal{V}$. We may assume that
$U_{n}=A^{-1}(F_{n})\to U$ in $\mathcal{V}$ and hence, in
$X^{p,p}(\overline{\Omega},\mu)$. It remains to show that $U_{n}\to U$ in
$X^{r,s}(\overline{\Omega},\mu)$. Let $r\in[p,p_{s})$ and $s\in[p,q_{s})$.
Since $U_{n}:=(u_{n},u_{n}|_{\partial\Omega})$ is bounded in
$X^{p_{s},q_{s}}(\overline{\Omega},\mu)$, a standard interpolation inequality
shows that there exists $\tau\in(0,1)$ such that

$\||U_{n}-U_{m}\||_{r,s}\leq\||U_{n}-U_{m}\||_{p,p}^{\tau}\||U_{n}-U_{m}\||_{p_{s},q_{s}}^{1-\tau}\leq
C\||U_{n}-U_{m}\||_{p,p}^{\tau}.$

As $U_{n}$ converges in $X^{p,p}(\overline{\Omega},\mu)$, it follows from the
preceding inequality that $U_{n}$ is a Cauchy sequence in
$X^{r,s}(\overline{\Omega},\mu)$ and therefore converges in
$X^{r,s}(\overline{\Omega},\mu)$. Hence,
$A^{-1}:\;X^{p_{1},q_{1}}(\overline{\Omega},\mu)\to\mathcal{V}\cap
X^{r,s}(\overline{\Omega},\mu)$ is compact for every $r\in[p,p_{s})$ and
$s\in[p,q_{s})$. The case $r,s\in(1,p)$ follows from the fact that
$X^{p,p}(\overline{\Omega},\mu)\hookrightarrow X^{r,s}(\overline{\Omega},\mu)$
and the proof is finished

###  5.4. Statement and proof of the main result

We will now establish under what conditions the operator $A^{-1}$ maps
$X^{p_{1},q_{1}}(\overline{\Omega},\mu)$ boundedly and continuously into
$X^{\infty}(\overline{\Omega},\mu)$. The following is the main result of this
section.

######  Theorem 5.10.

Let the assumptions of Theorem 5.8 be satisfied.

  1. (a)

Suppose $2\leq p<N$ and $2\leq q<\infty$. Let

$p_{1}>\frac{p_{s}}{p_{s}-p}=\frac{N}{p}\;\mbox{ and
}\;q_{1}>\frac{q_{s}}{q_{s}-p}=\frac{N-1}{p-1}.$

Let $f\in L^{p_{1}}(\Omega),\;g\in L^{q_{1}}(\partial\Omega)$ and
$U,V\in\mathcal{V}$ be such that for every function
$\Phi=(\varphi,\varphi|_{\partial\Omega})\in\mathcal{V}$,

$\mathcal{A}(U,\Phi)-\mathcal{A}(V,\Phi)=\int_{\Omega}f\varphi\;dx+\int_{\partial\Omega}g\varphi\;d\sigma.$
(5.16)

Then there is a constant $C=C(N,p,q,\Omega)>0$ such that

$\||U-V\||_{\infty}^{p-1}\leq
C(\|f\|_{p_{1},\Omega}+\|g\|_{q_{1},\partial\Omega}).$

  2. (b)

Suppose $2\leq p=q<N-1$. Let

$p_{1}>\frac{p_{s}}{p_{s}-p}=\frac{N}{p}\;\mbox{ and
}\;q_{1}>\frac{p_{t}}{p_{t}-p}=\frac{N-1}{p}.$

Let $f\in L^{p_{1}}(\Omega),\;g\in L^{q_{1}}(\partial\Omega)$ and
$U,V\in\mathcal{V}$ satisfy (5.16). Then there is a constant
$C=C(N,p,q,\Omega)>0$ such that

$\||U-V\||_{\infty}^{p-1}\leq
C(\|f\|_{p_{1},\Omega}+\|g\|_{q_{1},\partial\Omega}).$

###### Proof.

Let $U,V\in\mathcal{V}$ satisfy (5.16). Let $k\geq 0$ be a real number and set

$w_{k}:=(|u-v|-k)^{+}\mathop{\mathrm{s}gn}(u-v)\;W_{k}:=(w_{k},w_{k}|_{\partial\Omega})\;\mbox{
and }\;w:=|u-v|.$

Let $A_{k}:=\\{x\in\overline{\Omega}:|w(x)|\geq k\\}$, and
$A_{k}^{+}:=\\{x\in\overline{\Omega}:w(x)\geq
k\\},\;\;A_{k}^{-}:=\\{x\in\overline{\Omega}:w(x)\leq-k\\}$. Clearly
$W_{k}\in\mathcal{V}$ and $A_{k}=A_{k}^{+}\cup A_{k}^{-}$. We claim that there
exists a constant $C>0$ such that

$C\mathcal{A}(W_{k},W_{k})\leq\mathcal{A}(U,W_{k})-\mathcal{A}(V,W_{k}),$
(5.17)

for all $U,V\in\mathcal{V}$. Using the definition of the form $\mathcal{A}$,
we have

$\displaystyle\mathcal{A}(U,W_{k})-\mathcal{A}(V,W_{k})$ (5.18)
$\displaystyle=\int_{\Omega}(|\nabla u|^{p-2}\nabla u-|\nabla v|^{p-2}\nabla
v)\cdot\nabla w_{k}dx+\int_{\Omega}(|u|^{p-2}u-|v|^{p-2}v)w_{k}dx$
$\displaystyle+\int_{\Omega}(\alpha_{1}(x,u)-\alpha_{2}(x,v))w_{k}dx+\int_{\partial\Omega}(|u|^{q-2}u-|v|^{q-2}v)w_{k}d\sigma$
$\displaystyle+\int_{\partial\Omega}(|\nabla_{\Gamma}u|^{p-2}\nabla_{\Gamma}u-|\nabla_{\Gamma}v|^{p-2}\nabla_{\Gamma}v)\cdot\nabla_{\Gamma}w_{k}d\sigma+\int_{\partial\Omega}(\alpha_{2}(x,u)-\alpha_{2}(x,v))w_{k}d\sigma.$

Since$\nabla w_{k}=\begin{cases}\nabla(u-v)&\mbox{ in }A(k),\\\ 0&\mbox{
otherwise, }\end{cases}$ we can rewrite (5.18) as follows:

$\displaystyle\mathcal{A}(U,W_{k})-\mathcal{A}(V,W_{k})=\int_{A(k)\cap\Omega}(|\nabla
u|^{p-2}\nabla u-|\nabla v|^{p-2}\nabla v)\cdot\nabla(u-v)dx$ (5.19)
$\displaystyle{}+\int_{A(k)\cap\partial\Omega}(|\nabla_{\Gamma}u|^{q-2}\nabla_{\Gamma}u-|\nabla_{\Gamma}v|^{q-2}\nabla_{\Gamma}v)\cdot\nabla_{\Gamma}(u-v)d\sigma$
$\displaystyle{}+\lambda\int_{A(k)\cap\Omega}(|u|^{p-2}u-|v|^{p-2}v)w_{k}dx+\int_{A(k)\cap\Omega}(\alpha_{1}(x,u)-\alpha_{1}(x,v))w_{k}dx$
$\displaystyle{}+\int_{A(k)\cap\partial\Omega}(\alpha_{2}(x,u)-\alpha_{2}(x,v))w_{k}d\sigma.$

Exploiting inequality (2.12), from (5.19) and (5.11), we deduce

$\displaystyle\mathcal{A}(U,W_{k})-\mathcal{A}(V,W_{k})$ (5.20)
$\displaystyle\geq\int_{A(k)\cap\Omega}\left(|\nabla
w_{k}|^{p}+|w_{k}|^{p}\right)dx+\int_{A(k)\cap\Omega}c_{1}\alpha_{1}(x,w_{k})w_{k}dx$
$\displaystyle+\int_{A(k)\cap\Omega}(|u|^{p-2}uw_{k}-|v|^{p-2}vw_{k}-|w_{k}|^{p})dx$
$\displaystyle+\int_{A(k)\cap\Omega}(\alpha_{1}(x,u)-\alpha_{1}(x,v)-c_{1}\alpha_{1}(x,w_{k}))w_{k}dx$
$\displaystyle+\int_{A(k)\cap\partial\Omega}\left(|\nabla_{\Gamma}w_{k}|^{q}+|w_{k}|^{q}\right)d\sigma+\int_{A(k)\cap\partial\Omega}c_{2}\alpha_{2}(x,w_{k})w_{k}d\sigma$
$\displaystyle+\int_{A(k)\cap\partial\Omega}(|u|^{q-2}uw_{k}-|v|^{q-2}vw_{k}-|w_{k}|^{q})d\sigma$
$\displaystyle+\int_{A(k)\cap\partial\Omega}(\alpha_{2}(x,u)-\alpha_{2}(x,v)-c_{2}\alpha_{2}(x,w_{k}))w_{k}d\sigma$
$\displaystyle\geq
C\mathcal{A}(W_{k},W_{k})+\int_{A(k)\cap\Omega}(|u|^{p-2}uw_{k}-|v|^{p-2}vw_{k}-|w_{k}|^{p})dx$
$\displaystyle{}+\int_{A(k)\cap\Omega}(\alpha_{1}(x,u)-\alpha_{1}(x,v)-c_{1}\alpha_{1}(x,w_{k}))w_{k}dx$
$\displaystyle{}+\int_{A(k)\cap\partial\Omega}(|u|^{q-2}uw_{k}-|v|^{q-2}vw_{k}-|w_{k}|^{q})d\sigma$
$\displaystyle{}+\int_{A(k)\cap\partial\Omega}(\alpha_{2}(x,u)-\alpha_{2}(x,v)-c_{2}\alpha_{2}(x,w_{k}))w_{k}d\sigma,$

where $c_{1},c_{2}$ are the constants from (5.11). Using (5.10) and the fact
that $\alpha_{j}(x,\cdot)$ are strictly increasing, for $x\in A_{k}^{+}$, we
have

$\displaystyle c_{j}\alpha_{j}(x,w_{k}(x))$
$\displaystyle=c_{j}\alpha_{j}(x,u(x)-v(x)-k)\leq
c_{j}\alpha_{j}(x,u(x)-v(x))$
$\displaystyle\leq\alpha_{j}(x,u(x))-\alpha_{j}(x,v(x)).$

Multiplying this inequality by $w_{k}(x)\geq 0,$ $x\in A_{k}^{+},$ yields

$(\alpha_{j}(x,u(x))-\alpha_{j}(x,v(x))-c_{j}\alpha_{j}(x,w_{k}(x)))w_{k}(x)\geq
0.$ (5.21)

Similarly, for $x\in A_{k}^{-},$

$\displaystyle c_{j}\alpha_{j}(x,w_{k}(x))$
$\displaystyle=c_{j}\alpha_{j}(x,u(x)-v(x)+k)\geq
c_{j}\alpha_{j}(x,u(x)-v(x))$
$\displaystyle\geq\alpha_{j}(x,u(x))-\alpha_{j}(x,v(x)).$

Hence, multiplying this inequality by $w_{k}(x)\leq 0$, we get

$(\alpha_{j}(x,u(x))-\alpha_{j}(x,v(x))-c_{j}\alpha_{j}(x,w_{k}(x)))w_{k}(x)\geq
0,$ (5.22)

for all $x\in A_{k}^{-}$. Hence, on account of (5.21) and (5.22), from (5.20)
we obtain the required estimate of (5.17).

(a) To prove this part, note that from Definition 5.3 it is clear that,

$\|w_{k}\|_{W^{1,p}(\Omega)}^{p}\leq\mathcal{A}(W_{k},W_{k}).$ (5.23)

Let $f\in L^{p_{1}}(\Omega)$ and $g\in L^{q_{1}}(\partial\Omega)$ with

$p_{1}>\frac{p_{s}}{p_{s}-p}=\frac{N}{p}\;\mbox{ and
}\;q_{1}>\frac{q_{s}}{q_{s}-p}=\frac{N-1}{p-1},$

and let $B\subset\overline{\Omega}$ be any $\mu$-measurable set. We claim that
there exists a constant $C\geq 0$ such that, for every $F\in
X^{p_{1},q_{1}}(\overline{\Omega},\mu)$ and $\varphi\in W^{1,p}(\Omega),$ we
have

$\||F\varphi 1_{B}\||_{1,1}\leq
C\||F\||_{p_{1},q_{1}}\|\varphi\|_{W^{1,p}(\Omega)}\||\chi_{B}\||_{p_{3},q_{3}},$
(5.24)

where $p_{3}$ and $q_{3}$ are such that $1/p_{3}+1/p_{1}+1/p_{s}=1$ and
$1/q_{3}+1/q_{1}+1/q_{s}=1$. In fact, note that if $n\in\mathbb{N}$ and
$p_{i},$ $q_{i}\in[1,\infty],\;(i=1,\dots,n)$ are such that

$\sum_{i=1}^{n}\frac{1}{p_{i}}=\sum_{i=1}^{n}\frac{1}{q_{i}}=1,$

and, if $F_{i}\in X^{p_{i},q_{i}}(\overline{\Omega},\mu),\;(i=1,\dots,n)$,
then by Hölder’s inequality,

$\||\prod_{i=1}^{n}F_{i}\||_{1,1}\leq\prod_{i=1}^{n}\||F_{i}\||_{p_{i},q_{i}}.$
(5.25)

Since $W^{1,p}(\Omega)\hookrightarrow X^{p_{s},q_{s}}(\overline{\Omega},\mu)$,
(5.24) follows immediately from (5.25) and the claim (5.24) is proved. Next,
it follows from (5.24), that

$\displaystyle\int_{\overline{\Omega}}FW_{k}d\mu=\||FW_{k}\||_{1,1}$
$\displaystyle=$ $\displaystyle\||FW_{k}\chi_{A_{k}}\||_{1,1}$
$\displaystyle\leq$
$\displaystyle\||F\||_{p_{1},q_{1}}\||w_{k}\||_{W^{1,p}(\Omega)}\||\chi_{A_{k}}\||_{p_{3},q_{3}},$

where we recall that
$1/p_{3}=\left(1-1/p_{s}-1/p_{1}\right)>\left(p-1\right)/p_{s}$ and
$q_{3}<q_{s}/(p-1)$. Therefore, for every $k\geq 0$,

$\mathcal{A}(U,W_{k})-\mathcal{A}(V,W_{k})\leq\||F\||_{p_{1},q_{1}}\|w_{k}\|_{W^{1,p}(\Omega)}\||\chi_{A_{k}}\||_{p_{3},q_{3}},$

which together with estimate (5.17) yields the desired inequality

$C\mathcal{A}(W_{k},W_{k})\leq\mathcal{A}(U,W_{k})-\mathcal{B}(V,W_{k})\leq\||F\||_{p_{1},q_{1}}\|w_{k}\|_{W^{1,p}(\Omega)}\||\chi_{A_{k}}\||_{p_{3},q_{3}},$
(5.26)

It follows from (5.23) and (5.26), that for every $k>0$,

$\displaystyle C\|w_{k}\|_{W^{1,p}(\Omega)}^{p}$
$\displaystyle\leq\mathcal{A}(W_{k},W_{k})\leq\mathcal{A}(U,W_{k})-\mathcal{A}(V,W_{k})$
$\displaystyle\leq\||F\||_{p_{1},q_{1}}\|w_{k}\|_{W^{1,p}(\Omega)}\||\chi_{A_{k}}|\|_{p_{3},q_{3}}.$

Hence, for every $k>0$, $\|w_{k}\|_{W^{1,p}(\Omega)}^{p-1}\leq
C_{1}\||\chi_{A_{k}}\||_{p_{3},q_{3}}$. Using the fact
$W^{1,p}(\Omega)\hookrightarrow X^{p_{s},q_{s}}(\overline{\Omega},\mu)$, we
obtain for every $k>0$, that

$\||w_{k}\||_{p_{s},q_{s}}^{p-1}\leq
C\||F\||_{p_{1},q_{1}}\||\chi_{A_{k}}\||_{p_{3},q_{3}}.$

Let $h>k$. Then $A_{h}\subset A_{k}$ and on $A_{h}$ the inequality
$|w_{k}|\geq\left(h-k\right)$ holds. Therefore,

$\||(h-k)\chi_{A_{h}}\||_{p_{s},q_{s}}^{p-1}\leq
C\||F\||_{p_{1},q_{1}}\||\chi_{A_{k}}\||_{p_{3},q_{3}},$

which shows that

$\||\chi_{A_{h}}\||_{p_{s},q_{s}}^{p-1}\leq
C\||F\||_{p_{1},q_{1}}(h-k)^{-(p-1)}\||\chi_{A_{k}}\||_{p_{3},q_{3}}.$ (5.27)

Let $C_{3}:=\||1_{\overline{\Omega}}\||_{p_{s},q_{s}}$, and

$\delta:=\min\left\\{\frac{p_{s}}{p_{3}},\frac{q_{s}}{p_{3}}\right\\}>p-1,\;\delta_{0}:=\frac{\delta}{p-1}>1.$

Then

$\displaystyle\|C_{3}^{-p_{s}/p_{3}}\chi_{A_{k}}\|_{\Omega,p_{3}}$
$\displaystyle=\|C_{3}^{-1}\chi_{A_{k}}\|_{\Omega,p_{s}}^{p_{s}/p_{3}}\leq\|C_{3}^{-1}\chi_{A_{k}}\|_{\Omega,p_{s}}^{\delta}$
(5.28)
$\displaystyle\leq\||\chi_{A_{k}}\||_{p_{s},q_{s}}^{\delta}C_{3}^{-\delta}$

and

$\displaystyle\|C_{3}^{-q_{s}/q_{3}}\chi_{A_{k}}\|_{\partial\Omega,q_{3}}$
$\displaystyle=\|C_{3}^{-1}\chi_{A_{k}}\|_{\partial\Omega,q_{s}}^{q_{s}/q_{3}}\leq\|C_{3}^{-1}\chi_{A_{k}}\|_{\partial\Omega,q_{s}}^{\delta}$
(5.29)
$\displaystyle\leq\||\chi_{A_{k}}\||_{p_{s},q_{s}}^{\delta}C_{3}^{-\delta}.$

Choosing $C_{\Omega}:=C_{3}^{p_{s}/p_{3}-\delta}+C_{3}^{q_{s}/q_{3}-\delta}$,
from (5.28)-(5.29) we have

$\||\chi_{A_{k}}\||_{p_{3},q_{3}}\leq
C_{\Omega}\||\chi_{A_{k}}\||_{p_{s},q_{s}}^{\delta}.$ (5.30)

Therefore, combining (5.27) with (5.30), we get

$\displaystyle\||\chi_{A_{h}}\||_{p_{s},q_{s}}^{p-1}$ $\displaystyle\leq
C\||F\||_{p_{1},q_{1}}(h-k)^{-(p-1)}\||\chi_{A_{k}}\||_{p_{s},q_{s}}^{\delta}$
(5.31)
$\displaystyle=C\||F\||_{p_{1},q_{1}}(h-k)^{-(p-1)}\left[\||\chi_{A_{k}}\||_{p_{s},q_{s}}^{p-1}\right]^{\delta_{0}}.$

Setting $\psi(h):=\||\chi_{A_{h}}\||_{p_{s},q_{s}}^{p-1}$ in Lemma 2.13, on
account of (5.31), we can find a constant $C_{2}$ (independent of $F$) such
that

$\||\chi_{A_{K}}\||_{p_{s},q_{s}}^{p-1}=0\;\text{ with
}\;K:=C_{2}\||F\||_{p_{1},q_{1}}^{1/(p-1)}.$

This shows that $\mu(A_{K})=0,$ where
$A_{K}=\\{x\in\overline{\Omega}:|(u-v)(x)|\geq K\\}$. Hence, we have
$|u-v|\leq K,$ $\mu$-a.e. on $\overline{\Omega}$ so that

$\||U-V\||_{\infty}^{p-1}\leq
C_{2}\||F\||_{p_{1},q_{1}}=C_{2}\left(\|f\|_{p_{1},\Omega}+\|g\|_{q_{1},\partial\Omega}\right),$

which completes the proof of part (a).

(b) To prove this part, instead of (5.23) and (5.24), one uses
$\|W_{k}\|_{\mathcal{V}_{1}}^{p}\leq\mathcal{A}(W_{k},W_{k})$ and $\||F\varphi
1_{B}\||_{1,1}\leq
C\||F\||_{p_{1},q_{1}}\|\varphi\|_{W^{1,p}(\Omega)}\||\chi_{B}\||_{p_{3},q_{3}}$,
(where $p_{3}$ and $q_{3}$ are such that $1/p_{3}+1/p_{1}+1/p_{s}=1$ and
$1/q_{3}+1/q_{1}+1/p_{t}=1$) and the embedding
$\mathcal{V}\hookrightarrow\mathcal{V}_{1}\hookrightarrow
X^{p_{s},p_{t}}(\overline{\Omega},\mu)$. The remainder of the proof follows as
in the proof of part (a).

We conclude this section with the following example.

######  Example 5.11.

Let $p\in[2,\infty)$, $b:\partial\Omega\rightarrow(0,\infty)$ be a strictly
positive and $\sigma$-measurable function and let

$\beta(x,\xi):=b(x)|\xi|^{p-2}\xi,\quad\xi\in\mathbb{R}.$

Then, it is easy to verify that $\beta$ satisfies Assumptions 5.1, 5.2 and 5.7
(see, e.g., [5, Example 4.17]).

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