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\begin{document}

\title{A fixed point approach to the semi-linear Stokes problem}
\author{David BRUMAR}
\email{david.brumar@ubbcluj.ro} \subjclass{35Q30, 35J25, 35J61,
35Q35} \keywords{Stokes system, semi-linear problem, operator
method, fixed point theorem, Sobolev space}

\begin{abstract}
The aim of this paper is to study the Dirichlet problem for semi-linear
Stokes equations. The approach of this study is based on the operator
method, using abstract results of nonlinear functional analysis. We first
study the problem using Schauder's fixed point theorem and we prove the
existence of a solution in case that the nonlinear term has a linear growth.
Next we establish whether the existence of solutions can still be obtained
without this linear growth restriction. Such a result is obtained by applying the
Leray-Schauder fixed point theorem.
\end{abstract}


\address{Babe\c{s}-Bolyai University, \\ Faculty of Mathematics and Computer Sciences\\
1, Kog\u{a}lniceanu Street,\\
400084 Cluj-Napoca,\\
Romania}

\maketitle

\section{Introduction}

The field of fluid dynamics does not only engage the attention of
mathematicians and physicists but also of astrophysicists, oceanographers
and many others and this is due to the fact that it addresses real-world
natural phenomena and tries to come up with mathematical models that help us
to understand them.

An inertial fluid flow that is Newtonian, incompressible and homogeneous
follows the Navier-Stokes equations, which are essentially derived from
Newton's second law of motion applied to the fluid and the law of mass
conservation in the context of constant density flow. If the velocity field
is not time-dependent, then the flow is called steady, and it means that the
fluid particles follow the streamlines, which do not change in time.
Neglecting the nonlinear term in the Navier-Stokes system we get the Stokes
system, which is in fact, the one that we are here interested in.

The aim of this paper is to study the existence of solutions of the
semi-linear Dirichlet problem for the steady Stokes system
\begin{equation}
\begin{cases}
-\mu\Delta u+\nabla p= f(x,u(x)) & \text{ in }\Omega \\
\text{div }u=0 & \text{ in }\Omega \\
u=0 & \text{ on }\partial\Omega.%
\end{cases}
\label{semiStokes1}
\end{equation}

\section{Preliminaries}

In this section we briefly recall without proof, some important
results from functional analysis and some basic results regarding
the Stokes system that are used in the forthcoming material. For
additional details, we refer the reader to the following works
\cite{mbm,fabrie,brezis,jebelean,precup3,abdul,herma,temam}.

\subsection{The Nemytskii operator}

First we recall some properties of the Nemytskii superposition operator
(see, e.g., \cite{precup1}).

\begin{definition}[Nemytskii operator]
Let $\Omega\subset \mathbb{R}^N, N\geq 1,$ be an open set and let $%
f:\Omega\times\mathbb{R}^n\to\mathbb{R}^m, n,m\geq 1$. By the Nemystkii
operator associated to $f$ we understand the operator $N_f$ which, to each
function $u:\Omega\to\mathbb{R}^n$, assigns $f\circ u$, that is
\begin{align*}
N_fu(x)=(f\circ u)(x)=f(x,u(x)), \text{ for } x\in\Omega.
\end{align*}
\end{definition}

\begin{definition}[Carath\'{e}odory function]
Let $\Omega \subset \mathbb{R}^{N},N\geq 1,$ be an open set. We say that $%
f:\Omega \times \mathbb{R}^{n}\rightarrow \mathbb{R}^{m},n,m\geq 1$ is a
Carath\'{e}odory function if it satisfies the following conditions:

\begin{itemize}
\item[(i)] $x\mapsto f(x,y)$ is measurable in $\Omega$ for every $y\in%
\mathbb{R}^n;$

\item[(ii)] $y\mapsto f(x,y)$ is continuous on $\mathbb{R}^n$ for a.e. $%
x\in\Omega$.
\end{itemize}
\end{definition}

\begin{proposition}[{see \protect\cite[Proposition 9.1]{precup3}}]
If $f$ is a Carath\'{e}odory function, then the Nemystkii operator
associated to the function $f$ maps measurable functions into measurable
functions. \label{propNemit}
\end{proposition}

\begin{theorem}[{see \protect\cite[Theorem 9.1]{precup3}}]
\label{th.nem}Let $\Omega \subset \mathbb{R}^{N}$ be an open set, $f:\Omega
\times \mathbb{R}^{n}\rightarrow \mathbb{R}^{m}$ and $1\leq p,q<+\infty $.
If $f$ satisfies the Carath\'{e}odory conditions and there exists $a\in
\mathbb{R}_{+}$ and $h\in L^{q}(\Omega ;\mathbb{R}_{+})$ such that
\begin{equation*}
||f(x,y)||\leq a||y||^{\frac{p}{q}}+h(x)
\end{equation*}%
for every $y\in \mathbb{R}^{n}$ and a.e. $x\in \Omega $, then the operator
\begin{equation*}
N_{f}:L^{p}(\Omega ;\mathbb{R}^{n})\rightarrow L^{q}(\Omega ;\mathbb{R}%
^{m})\ \text{ given by\  }N_{f}(u)=f(\cdot ,u)
\end{equation*}%
is well defined, continuous and bounded. Moreover, the following inequality
holds:
\begin{equation*}
||N_{f}(u)||_{L^{q}}\leq a||u||_{L^{p}}^{\frac{p}{q}}+||h||_{L^{q}}\ \ \text{
for all \ }u\in L^{p}(\Omega ;\mathbb{R}^{n}).
\end{equation*}%
\label{Nmt_th}
\end{theorem}

\subsection{Embedding results}

\noindent The purpose of our work, namely the study of the existence of
solutions for a semi-linear boundary valued problem, is achieved by looking
for a weak solution which lead us to use continuous or compact embeddings of
function spaces. In particular, we use the following embedding results due
to Sobolev and Rellich-Kondrachov regarding the continuous and compact
embeddings of Sobolev spaces into Lebesgues spaces.

Let $1\leq q\leq +\infty $. Then the critical exponent associated to $q$ is
denoted by $q^{\ast }$ and is defined by
\begin{equation*}
\begin{cases}
\cfrac{1}{q^*}=\cfrac{1}{q}-\cfrac{1}{n}, & q<n \\
q^{\ast }=+\infty , & q\geq n,%
\end{cases}%
\end{equation*}%
where by $n$ is denoted the dimension of the space.

\begin{theorem}[Sobolev]
Let $\Omega \subset \mathbb{R}^{n}$ be an open set of class $C^{1}$ (or $%
\Omega =\mathbb{R}^{n}$). Then the following continuous embeddings hold:

\begin{enumerate}
\item[a)] $H^{1}(\Omega )\subset L^{q}(\Omega )\ $ for every $q\in \lbrack
2,2^{\ast }]$, where $n\geq 3$.

\item[b)] $H^{1}(\Omega )\subset L^{q}(\Omega )\ $ for every $q\in \lbrack
2,+\infty )$, if $n=2$.
\end{enumerate}
\end{theorem}

\begin{theorem}[Rellich-Kondrachov]
\label{th.RK} Let $\Omega\subset\mathbb{R}^n$ be a bounded open set of class
$C^1$.

\begin{enumerate}
\item[a)] If $n\geq 3$, then the embedding $H^1(\Omega)\subset L^q(\Omega)$
is compact for $q\in [1,2^*)$, where $2^*:=2n/(n-2)$.

\item[b)] If $n=2$, then the embedding $H^1(\Omega)\subset L^q(\Omega)$ is
compact for every $q\in [1,+\infty)$.
\end{enumerate}
\end{theorem}

\noindent We recall that a real number $\lambda$ is said to be an eigenvalue
of the Dirichlet problem for $-\Delta$ if the problem
\begin{equation*}
\begin{cases}
-\Delta u=\lambda u & \text{in }\Omega \\
u=0 & \text{on }\partial\Omega%
\end{cases}%
\end{equation*}
has nonzero weak solutions.

\begin{theorem}[Poincar\'{e}'s inequality]
Let $\Omega $ be a bounded open set of $\mathbb{R}^{n}$. Then there exists a
constant $C$ that depends on $\Omega $ such that
\begin{equation*}
||u||_{L^{2}(\Omega )}\leq C||\nabla u||_{L^{2}},\ \text{ for every }u\in
H_{0}^{1}(\Omega ).
\end{equation*}
\end{theorem}

\noindent Due to this result, the Sobolev space $H_{0}^{1}(\Omega )$ can be
endowed with an equivalent norm
\begin{equation*}
||u||_{H_{0}^{1}}:=||\nabla u||_{L^{2}}=\left( \int_{\Omega }||\nabla
u||^{2}\right) ^{1/2}
\end{equation*}%
that comes from the scalar product in $H_{0}^{1}(\Omega )$
\begin{equation*}
(u,v)_{H_{0}^{1}}=(\nabla u,\nabla v)_{L^{2}}=\int_{\Omega }\nabla u\cdot
\nabla v.
\end{equation*}%
Hence in terms of the new norm, Poincar\'{e}'s inequality can be written as
\begin{equation*}
||u||_{L^{2}}\leq C||u||_{H_{0}^{1}},\ \ u\in H_{0}^{1}(\Omega ).
\end{equation*}%
Since the first eigenvalue of the Dirichlet problem for $-\Delta $ is
\begin{equation*}
\lambda _{1}=\inf\limits_{u\in H_{0}^{1}(\Omega )\setminus \{0\}}%
\cfrac{||u||^2_{H_0^1}}{||u||^2_{L^2}},
\end{equation*}%
it follows that the smallest constant $C$ for which the Poincar\'{e}'s
inequality holds, is in fact $\cfrac{1}{\sqrt{\lambda_1}}.$ Therefore,
\begin{equation*}
||u||_{L^{2}}\leq \cfrac{1}{\sqrt{\lambda_1}}||u||_{H_{0}^{1}},\ \text{ for
all }u\in H_{0}^{1}(\Omega ).
\end{equation*}%
Moreover, the Poincar\'{e}'s inequality also holds for the embedding $%
L^{2}(\Omega )\subset H^{-1}(\Omega )$ with the same constant, namely
\begin{equation*}
||u||_{H^{-1}}\leq \cfrac{1}{\sqrt{\lambda_1}}||u||_{L^{2}},\ \text{ for all
}u\in L^{2}(\Omega ).
\end{equation*}%
For a more detailed exposition of these results we refer the reader to \cite[%
Chapter 3]{precup3}.

\begin{remark}
Since we are concerned with $n$-dimensional vector-valued functions, we
shall use the notations
\begin{equation*}
L^{p}(\Omega ):=\left( L^{p}(\Omega )\right) ^{n},\hspace{1cm}H^{m}(\Omega
):=\left( H^{m}(\Omega )\right) ^{n},\hspace{1cm}H_{0}^{m}(\Omega ):=\left(
H_{0}^{m}(\Omega )\right) ^{n}.
\end{equation*}
\end{remark}

\subsection{The variational form of the Stokes system}

Let us consider the Dirichlet problem for the steady non-homogeneous Stokes
system
\begin{equation}
\begin{cases}
-\mu \Delta u+\nabla p=f & \text{ in }\Omega  \\
\text{div }u=0 & \text{ in }\Omega  \\
u=0 & \text{ on }\partial \Omega .%
\end{cases}
\label{stokes_symp}
\end{equation}%
Here $\Omega \subset \mathbb{R}^{n}$ is a bounded open set, $\mu >0$ is a
constant representing the kinematic viscosity, $u:\Omega \rightarrow \mathbb{%
R}^{n}$ is the velocity field, $p$ is the pressure and $f\in L^{2}(\Omega )$
is the external force. In this subsection we give the variational
formulation of problem (\ref{stokes_symp}). For a very detailed way of
getting to the variational form of the Stokes equation we refer the reader
to \cite{temam}.

We define the Hilbert space
\begin{equation*}
V:=\{v\in H_{0}^{1}(\Omega ):\ \text{div }v=0\},
\end{equation*}%
endowed with the scalar product
\begin{equation*}
(u,v)_{V}=\int_{\Omega }\nabla u\cdot \nabla v,\ \text{ for }u,v\in V
\end{equation*}%
and the corresponding norm
\begin{equation*}
\Vert u\Vert _{V}=\left( \int_{\Omega }|\nabla u|^{2}\right) ^{\frac{1}{2}}.
\end{equation*}%
We can now state the variational formulation of problem (\ref{stokes_symp}):%
\newline
\textit{Given $f\in L^{2}(\Omega )$ find $u\in V$ such that}
\begin{equation}
\mu (u,v)_{V}=(f,v)_{L^{2}},\ \text{ for all }v\in V.
\label{variational_form}
\end{equation}

\begin{definition}[Weak solution]
Let $f\in L^2(\Omega)$. By the \textit{weak solution} of the Stokes problem (%
\ref{stokes_symp}) we mean a function $u_f\in V$ that satisfies (\ref%
{variational_form}).
\end{definition}

\noindent One has the following embeddings:
\begin{equation*}
V\subset H_{0}^{1}(\Omega )\subset L^{2}(\Omega )\subset H^{-1}(\Omega
)\subset V^{\prime }.
\end{equation*}%
Then, by the Riesz's representation theorem, we can extend (\ref%
{variational_form}) so that for any $f\in V^{\prime }$ there exist a unique $%
u_{f}\in V$ such that
\begin{equation}
\label{duality_Riesz}
\mu (u_{f},v)_{V}=(f,v),\ \text{ for all }v\in V.
\end{equation}%
Notice that the notation $(f,v),$ for $f\in V^{\prime }$ and $v\in V,$
stands for the value at $v$ of the linear functional $f$.

\begin{definition}[Solution operator]
The operator $S:V^{\prime }\rightarrow V$ defined by $Sf:=u_{f}$ for any $%
f\in V^{\prime }$ is called the \textit{solution operator}.
\end{definition}

\noindent If in (\ref{duality_Riesz}) we take in particular $v:=u_{f}$ we
obtain
\begin{equation*}
\Vert u_{f}\Vert _{V}^{2}=\cfrac{1}{\mu}(f,u_{f})\leq \cfrac{1}{\mu}\Vert
f\Vert _{V^{\prime }}\Vert u_{f}\Vert _{V}.
\end{equation*}%
Hence we have $\Vert u_{f}\Vert _{V}\leq \mu ^{-1}\Vert f\Vert _{V^{\prime
}},$ that is
\begin{equation*}
\Vert Sf\Vert _{V}\leq \cfrac{1}{\mu}\Vert f\Vert _{V^{\prime }}.
\end{equation*}%
Thus the linear operator $S$ is continuous from $V^{\prime }$ to $V.$

\begin{remark}
Note that the existence of the pressure $p$ is guaranteed as a consequence
of De Rham's Lemma.
\end{remark}

\section{Main results}

Let us now turn back to the semi-linear problem (\ref{semiStokes1}), where $%
\Omega \subset \mathbb{R}^{n},n\geq 2$ is a bounded open set, $f:\Omega
\times \mathbb{R}^{n}\rightarrow \mathbb{R}^{n}$, $p:\Omega \rightarrow
\mathbb{R}$. \newline
We seek weak solutions, i.e., functions $u\in V$ such that
\begin{equation*}
f(\cdot ,u(\cdot ))\in H^{-1}(\Omega )
\end{equation*}%
and
\begin{equation*}
\mu (u,v)_{V}=(f(\cdot ,u),v)\ \ \text{for all }v\in V.
\end{equation*}%
The system (\ref{semiStokes1}) can be written as an equivalent fixed point
equation
\begin{equation*}
u=T(u),\ \ \ u\in V,
\end{equation*}%
where
\begin{equation*}
T:=S\circ F,
\end{equation*}%
where $F:V\rightarrow V^{\prime },$ $F\left( u\right) =f\left( \cdot
,u\left( \cdot \right) \right) .$

\subsection{Application of Schauder's fixed point theorem}

In this section we find sufficient conditions that assure the existence of a
solution of problem (\ref{semiStokes1}), having in mind Schauder's fixed
point theorem on the space $V$.

First we show that $T$\textit{\ is a completely continuous
operator}. In order to do so we would like to have the
representation of $\ F=I\circ N_{f}\circ P,$ where

\begin{itemize}
\item $P:V\to L^2(\Omega), \ Pu= u$;

\item $N_{f}:L^{2}(\Omega )\rightarrow L^{2}(\Omega ),\ N_{f}(w)=f(\cdot
,w(\cdot ))$;

\item $I:L^2(\Omega)\to V^{\prime }, \ I(v)=(v,\cdot)_{L^2}$.
\end{itemize}

Let us observe that by the Theorem \ref{th.RK}, the embedding $%
H_{0}^{1}(\Omega )\subset L^{2}(\Omega )$ is continuous. Then it follows
that $P$ is a continuous linear operator, hence bounded. Also, since the
embedding $L^{2}(\Omega )\subset H^{-1}(\Omega )$ is compact it follows that
operator $I$ is a completely continuous linear operator. It remains to see
whether the operator $N_{f}:L^{2}(\Omega )\rightarrow L^{2}(\Omega )$ is
well-defined. For this purpose let us assume that $f$ is a Carath\'{e}odory
function. Hence, for any $w\in L^{2}(\Omega ),\ N_{f}(w)$ is also
measurable. We impose a linear growth condition on $f$, that is
\begin{equation}
\label{growth_cond}
||f(x,u)||\leq a||u||+k(x),\ \text{ for all }u\in \mathbb{R}^{n}\text{ and
a.e. }x\in \Omega ,
\end{equation}%
for some $k\in L^{2}(\Omega ,\mathbb{R}_{+})$ and $a\in \mathbb{R}_{+}$.
Then, we have
\begin{equation*}
||N_{f}(w)(x)||\leq a||w(x)||+k(x),\ \text{ for a.e. }x\in \Omega .
\end{equation*}%
Hence, by these assumptions over $f$, it follows that $N_{f}$ is
well-defined, continuous and bounded.

Due to the boundedness of the operators $P$ and $N_f$, it follows that $%
N_f\circ P$ is bounded too. Therefore, since $I$ is completely continuous,
it follows that the operator $F$ is completely continuous from $V$ to $%
V^{\prime }$. Next, by the linearity and continuity of the solution operator
$S$ we have that $T=S\circ F$ is completely continuous from $V$ to itself.

Secondly, we show that $T$\textit{\ is a self-map of a closed ball of }$V$.
To this purpose, let $u\in V$. Notice that for every $h\in H^{-1}(\Omega )$
one has
\begin{equation*}
||h||_{V^{\prime }}\leq ||h||_{H^{-1}}.
\end{equation*}%
Indeed, since $V\subset H_{0}^{1}(\Omega )$ we have that
\begin{equation*}
||h||_{V^{\prime }}=\sup\limits_{v\in V}\cfrac{|(h,v)|}{||v||_{V}}\leq
\sup\limits_{v\in H_{0}^{1}(\Omega )}\cfrac{|(h,v)|}{||v||_{H_0^1}}%
=||h||_{H^{-1}}.
\end{equation*}%
Then, since the operator $S$ is linear and continuous and also by the Poincar%
\'{e}%
%TCIMACRO{\U{b4}}%
%BeginExpansion
\'{}%
%EndExpansion
s inequality we have
\begin{align*}
\Vert T(u)\Vert _{V}& =\Vert S\circ F(u)\Vert _{V}\leq \cfrac{1}{\mu}\Vert
F(u)\Vert _{V^{\prime }}\leq \cfrac{1}{\mu}||F(u)||_{H^{-1}} \\
& \leq \cfrac{1}{\mu\sqrt{\lambda_1}}\Vert F(u)\Vert _{L^{2}}=%
\cfrac{1}{\sqrt{\mu\lambda_1}}\Vert f(\cdot ,u(\cdot ))\Vert _{L^{2}}.
\end{align*}%
By the growth condition (\ref{growth_cond}), we deduce that
\begin{equation*}
||T(u)||_{V}\leq \cfrac{a}{\mu\sqrt{\lambda_1}}||u||_{L^{2}}+%
\cfrac{1}{\mu\sqrt{\lambda_1}}\cdot ||k||_{L^{2}}.
\end{equation*}%
Since $u\in V$, we can apply again the Poincar\'{e} inequality and we obtain
that
\begin{equation*}
||T(u)||_{V}\leq \cfrac{a}{\mu\lambda_1}||u||_{V}+\cfrac{1}{\mu\sqrt{%
\lambda_1}}\cdot ||k||_{L^{2}}.
\end{equation*}%
In the end, we assume that $\cfrac{a}{\mu\lambda_1}<1$ so that there exists
a radius $r>0$ such that if $||u||_{V}\leq r$ then $||T(u)||_{V}\leq r$.
Indeed, from $\cfrac{a}{\mu\lambda_1}<1$ and $||u||_{V}\leq r$ we have that
\begin{equation*}
||T(u)||_{V}\leq \cfrac{a}{\mu\lambda_1}r+\cfrac{1}{\mu\lambda_1}%
||k||_{L^{2}}\leq r\text{ \ for }r>0\text{ large enough}.
\end{equation*}%
Hence, $||T(u)||_{V}\leq r$.

Therefore, based on Schauder's fixed point theorem we can state the
following result:

\begin{theorem}
Let $\Omega\subset\mathbb{R}^n$ be a bounded open set and $f:\Omega\times%
\mathbb{R}^n\to\mathbb{R}^n$ be a mapping such that:

\begin{enumerate}
\item[(a)] $f$ is a Carath\'eodory function;

\item[(b)] there is a positive constant $a$ and $k\in L^{2}(\Omega ,\mathbb{R%
}_{+})$ such that
\begin{equation*}
||f(x,u)||\leq a||u||+k(x)\ \text{ for all }u\in \mathbb{R}^{n}\text{ and
a.e. }x\in \Omega .
\end{equation*}
\end{enumerate}

Also, assume that $\cfrac{a}{\mu\lambda_1}<1$. Then the semi-linear Stokes
problem (\ref{semiStokes1}) has at least one solution $(u,p)$ with $u\in V.$
\end{theorem}

\subsection{Application of Lerray-Schauder's fixed point theorem}

In this section we consider more generally that the right hand side of the
problem (\ref{semiStokes1}) is of the form $\ f_{0}+f_{1}(\cdot ,u(\cdot )),$
where $\ f_{0}\in H^{-1}(\Omega )$ and $\ f_{1}:\Omega \times \mathbb{R}%
^{n}\rightarrow \mathbb{R}^{n}$. As before, the problem can be written as an
equivalent fixed point equation
\begin{equation*}
u=T(u),
\end{equation*}%
where this time
\begin{equation*}
T=S\circ (F+f_{0}),
\end{equation*}%
with
\begin{equation*}
F:V\rightarrow V^{\prime },\ \ F(u)=f_{1}(\cdot ,u(\cdot )).
\end{equation*}%
We are now interested if one can still obtain the existence of the solution
of the new problem without a linear growth restriction. We shall see this is
possible due to the Lerray-Schauder's fixed point theorem (see \cite{precup2}%
).

We first guarantee the \textit{complete continuity} of the operator
$T$. The idea we follow here is similar to the one in the previous
section: since the operator $S$ is linear and bounded, in order for
$T=S\circ F$ to be completely continuous, we need that the operator
$F$ is completely continuous. To this end, we write the operator $F$
as $F=I\circ N_{f_{1}}\circ P$, where

\begin{itemize}
\item $P:V\rightarrow L^{2^{\ast }}(\Omega ),$ $\ Pu=u$;

\item $N_{f_{1}}:L^{2^{\ast }}(\Omega )\rightarrow L^{q}(\Omega ),$ \ $%
N_{f_{1}}(\omega )=f_{1}(\cdot ,\omega )$;

\item $I:L^{q}(\Omega )\rightarrow V^{\prime },$\ $\ I(v)=(v,\cdot
)_{L^{2}},\ $ for some $\ q\in ((2^{\ast })^{\prime },+\infty )$.
\end{itemize}
Due to Theorem \ref{th.RK} it follows that the embedding $\ \left(
V\subset \right) H_{0}^{1}(\Omega )\subset L^{2^{\ast }}(\Omega )$
is continuous. Therefore the operator $P$ is a continuous linear
operator, hence bounded. Since $H_{0}^{1}(\Omega )\subset
L^{p}(\Omega )\text{ is compact for }p\in \lbrack 1,2^{\ast })$,
passing to duals, we get that $L^{q}(\Omega )\subset
H^{-1}(\Omega )$ is also compact for $q\in ((2^{\ast })^{\prime },+\infty )$%
, where $(2^{\ast })^{\prime }=2n/(n+2)$ is the conjugate of $2^{\ast }$.
Therefore, if $q>(2^{\ast })^{\prime }$, the inclusion operator $I$ is
completely continuous. We now show the operator $\ N_{f_{1}}$ is
well-defined. For this purpose we will make use of Theorem \ref{th.nem}. In
view of this result, we need to impose a growth condition on the function $%
f_{1}$, namely, for some $a\in \mathbb{R}_{+}$ and $\overline{h}\in
L^{q}(\Omega )$ to have
\begin{equation*}
||f_{1}(x,\omega )||\leq a||\omega ||^{\frac{2^{\ast }}{q}}+\overline{h}(x),
\end{equation*}%
for every $\omega \in \mathbb{R}^{n}$ and a.e. $x\in \Omega $. To this aim,
it sufficies to have
\begin{equation}
||f_{1}(x,\omega )||\leq a||\omega ||^{\alpha }+h(x).  \label{cond_LS}
\end{equation}%
for some $\alpha \in \lbrack 1,2^{\ast }/q]$ and $h\in L^{q}(\Omega )$. Note
that from $\alpha \leq 2^{\ast }/q$ it follows that $q\leq 2^{\ast }/\alpha $%
. Together with the condition $q>(2^{\ast })^{\prime },$ this shows that
\begin{equation*}
(2^{\ast })^{\prime }<q\leq \cfrac{2^*}{\alpha},
\end{equation*}%
and so, such a $q$ exists if
\begin{equation*}
\alpha <\frac{2^{\ast }}{\left( 2^{\ast }\right) ^{\prime }}=\cfrac{n+2}{n-2}%
.
\end{equation*}%
Thus, the condition (\ref{cond_LS}) holds for $\alpha \in \left[
1,(n+2)/(n-2)\right) $; hence, we can let $h\in L^{2^{\ast }/\alpha }(\Omega
)$.

Then from Theorem \ref{Nmt_th} it follows that the Nemytskii operator $%
N_{f_{1}}$ is well defined, continuous and bounded. Therefore, the operator $%
F$ is well-defined and completely continuous. Hence the operator $T$ is
completely continuous.

Finally, we carry on with the \textit{a priori bounds} of solutions, that is
to show there is a positive constant $R>0$ such that $||u||_{V}<R$ for any
solution $u\in V$ of the equation $\lambda T(u)=u$ and any $\lambda \in (0,1)
$. Let $u\in V$ be any solution of the equation $\lambda T(u)=u$ for some $%
\lambda \in (0,1)$. Thus, $u$ is a weak solution of the problem
\begin{equation}
\begin{cases}
-\mu \Delta u=-\nabla p+\lambda f_{0}(x)+\lambda f_{1}(x,u) & \text{ in }%
\Omega  \\
\text{div }u=0 & \text{ in }\Omega  \\
u=0 & \text{ on }\partial \Omega .%
\end{cases}
\label{stokes_symp4}
\end{equation}%
Therefore
\begin{equation*}
(u,v)_{V}=\cfrac{\lambda}{\mu}(f_{0}(\cdot )+f_{1}(\cdot ,u(\cdot )),v),\ \
\text{for any }v\in V.
\end{equation*}%
If we take in particular $v=u$ we obtain
\begin{equation*}
||u||_{V}^{2}=\cfrac{\lambda}{\mu}(f_{0}+f_{1}(\cdot ,u),u)=%
\cfrac{\lambda}{\mu}(f_{0},u)+\cfrac{\lambda}{\mu}(f_{1}(\cdot ,u),u).
\end{equation*}%
Note that since $f_{1}(\cdot ,u(\cdot ))\in L^{(2^{\ast })^{\prime }}(\Omega
),$ one has
\begin{equation*}
(f_{1}(\cdot ,u),u)=\int_{\Omega }u(x)\cdot f_{1}(x,u(x)).
\end{equation*}%
Let us now assume that there exists a positive constant $k$ such that
\begin{equation*}
y\cdot f_{1}(x,y)\leq k||y||^{2},\ \text{ for all }y\in \mathbb{R}^{n}\text{
and a.e. }x\in \Omega .
\end{equation*}%
\noindent Then $\ (f_{1}(\cdot ,u),u)\leq k||u||_{L^{2}}^{2}$ and using
Poincar\'{e}'s inequality we obtain
\begin{align*}
||u||_{V}^{2}& =\cfrac{\lambda}{\mu}(f_{0},u)+\cfrac{\lambda}{\mu}%
\int_{\Omega }u(x)\cdot f_{1}(x,u(x)) \\
& \leq \cfrac{\lambda}{\mu}\left(
||f_{0}||_{H^{-1}}||u||_{V}+k||u||_{L^{2}}^{2}\right)  \\
& <\cfrac{1}{\mu}||f_{0}||_{H^{-1}}||u||_{V}+\cfrac{k}{\mu}||u||_{L^{2}}^{2}
\\
& \leq \cfrac{1}{\mu}||f_{0}||_{H^{-1}}||u||_{V}+\cfrac{k}{\mu\lambda_1}%
||u||_{V}^{2}.
\end{align*}%
Hence, we have
\begin{equation*}
||u||_{V}\left( 1-\cfrac{k}{\mu \lambda_1}\right) \leq \cfrac{1}{\mu}%
||f_{0}||_{H^{-1}}.
\end{equation*}%
Assuming that $k<\mu \lambda _{1}$ it follows that
\begin{equation*}
||u||_{V}<\cfrac{\lambda_1}{\mu\lambda_1-k} ||f_{0}||_{H^{-1}}:=R.
\end{equation*}%
Therefore, we can state the following result:

\begin{theorem}
Let $\Omega \subset \mathbb{R}^{n}$ be a bounded open set, $f=f_{0}+f_{1}$
with $f_{0}\in H^{-1}(\Omega )$ and $f_{1}$ a function such that

\begin{enumerate}
\item[(a)] $f_{1}$ is a Carath\'{e}odory function;

\item[(b)] there is a positive constant $a$ and $\alpha \in \left[
1,(n+2)/(n-2)\right) $ and a function $h\in L^{2^{\ast }/\alpha }(\Omega )$
such that
\begin{equation*}
||f_{1}(x,u)||\leq a||u||^{\alpha }+h(x),
\end{equation*}%
for any $u\in \mathbb{R}^{n}$ and a.e. $x\in \Omega $;

\item[(c)] there is a positive constant $k<\mu\lambda_1$ such that the
condition
\begin{equation*}
y\cdot f_1(x,y)\leq k||y||^2
\end{equation*}
holds for any $y\in\mathbb{R}^n$ and a.e. $x\in\Omega$.
\end{enumerate}

Then the problem (\ref{stokes_symp4}) has at least one solution $(u,p)$ with
$u\in V$ and
\begin{equation*}
||u||_{V}\leq \cfrac{\lambda_1}{\mu\lambda_1-k} ||f_{0}||_{H^{-1}}.
\end{equation*}
\end{theorem}

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\end{document}
