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\markboth{\hfil Gradient method in Sobolev spaces  \hfil EJDE--2000/51}
{EJDE--2000/51\hfil J. Kar\'atson \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol.~{\bf 2000}(2000), No.~51, pp.~1--17. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)}
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%
 Gradient method in Sobolev spaces for nonlocal boundary-value problems 
\thanks{ {\em Mathematics Subject Classifications:} 35J65, 46N20, 49M10.
\hfil\break\indent
{\em Key words:} nonlocal boundary-value problems, gradient method in
Sobolev space, \hfil\break\indent
infinite-dimensional  preconditioning. 
\hfil\break\indent
\copyright 2000 Southwest Texas State University  and University of
North Texas. \hfil\break\indent
Submitted November 29, 1999. Published June 30, 2000. \hfil\break\indent
 Supported by the Hungarian National Research Funds
AMFK under \hfil\break\indent
Magyary Zolt\'an Scholarship and OTKA under grant no. F022228.} }
\date{}
%
\author{ J. Kar\'atson }
\maketitle

\begin{abstract}
 An infinite-dimensional gradient method is proposed for the 
 numerical solution of nonlocal quasilinear boundary-value problems. 
 The iteration is executed for the boundary-value problem itself 
 (i.e. on the continuous level) in the corresponding Sobolev space,
 reducing the nonlinear boundary-value problem to auxiliary linear 
 problems. We extend earlier results concerning
 local (Dirichlet) boundary-value problems. We show linear 
 convergence of our method, and present a numerical example.
\end{abstract}      
 
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{proposition}{Proposition}[section]
\newtheorem{corollary}{Corollary}[section]

\section{Introduction}
                                
The object of this paper is to study the numerical solution to the 
nonlocal quasilinear boundary-value problem
$$                  
 \begin{array}{c}   
T(u)\equiv -\mathop{\rm div} f(x,\nabla u)+q(x,u)= g(x) 
        \quad {\rm in \ } \Omega \\[2mm]                 
\displaystyle Q(u)\equiv f(x,\nabla u)\cdot \nu + \int_{\partial
\Omega} \varphi(x,y)u(y)\, d\sigma(y) = 0
        \quad {\rm on \ } \partial\Omega 
\end{array}                 
$$
on a bounded domain $\Omega\subset{\mathbb R}^N$.   
The exact conditions on the domain $\Omega$
and the functions $f,q,g$ and $\varphi$ will be given in Section 2.


The nonlocal boundary condition allows the normal component of the
nonlinearity
$f(x,\nabla u)$ to depend on a nonlocal expression of $u$,
in contrast to a   function of $u(x)$ in the usual case of mixed
boundary conditions (or especially 0 in the case of Neumann
problems). 
This kind of boundary condition has been analysed in detail e.g. in
\cite{Li, Sam}. Most often the studied nonlocal expression depends on
a composite function of $u$, this boundary condition arises e.g. in
plasma physics. General theoretical 
results on existence and uniqueness of weak solutions to such problems
have been proved in \cite{Sku} and \cite{Simon} for linear and
nonlinear equations, respectively. In this paper we consider the case
when the nonlocal expression involves an integral for all the values
of $u_{\mid\partial\Omega}$ (cf. \cite{Li}).
(The weak formulation of our problem will also be given in Section 2.)

The usual way of the numerical solution of elliptic equations is
to discretize the problem and use an iterative method for the solution
of the arising nonlinear system of algebraic equations (see
e.g. \cite{Ke, O-R}).
However, the condition number of the Jacobians of these systems can be
arbitrarily large when discretization is refined. This phenomenon
would yield
very slow convergence of iterative methods, hence suitable nonlinear 
preconditioning technique has to be used \cite{Ax}.

Our approach is opposite to the above: the iteration can be executed
for the boundary-value problem itself (i.e. on the continuous level)
directly in the corresponding Sobolev space,  reducing the nonlinear
boundary-value problem to
auxiliary linear problems. Then discretization may be used
for these auxiliary problems. This approach can be regarded as
infinite-dimensional preconditioning, and yields automatically a fixed
ratio of convergence for
the iteration, namely, that which is explicitly obtained from the
coefficients
$f$, $q$ and $g$. Concerning this, we note that the method in question
is 
related to the Sobolev gradient technique, developed in \cite{Neu1,
Neu2, Neu3}.
Especially,  in \cite{Neu1} nonlocal boundary conditions are discussed
in connection with
Sobolev gradients.

 The theoretical background of this approach is the generalization of
the
gradient method to Hilbert spaces. This was first developed by
Kantorovich for linear equations
(see \cite{K-A}). For the numerous results so far, we refer e.g. to
\cite{Ax-Chr, Da, GGZ, Neu4, Vai}; the investigations of the author
have included non-differentiable operators \cite{JAA} and
non-uniformly monotone
operators \cite{Pu2}. The mentioned results 
focus on partial differential operators. Concerning
numerical realization to local (Dirichlet) boundary-value problems
relying on the Hilbert space gradient method,
we refer to \cite{GFEM, GGZ}.
        
This paper consists of three parts. The exact formulation of the
problem is given in Section 2.  The gradient method for the nonlocal
 boundary value problem is constructed and its linear convergence is 
proven in   Section 3. The numerical realization is
illustrated in   Section 4.     
                            
        \section{Formulation of the problem}

The exact formulation of the nonlocal boundary condition requires the
following notion. 
(Therein and throughout the paper $\sigma$ denotes Lebesgue
  measure on the boundary.)

\paragraph{Definition 2.1} Let $\Omega\subset{\mathbb R}^N$, $\partial\Omega\in C^1$. A
function 
$\varphi:\partial \Omega^2\to{\mathbb R}$ is called
\begin{enumerate}               
\item[(i)] a {\it positive kernel} if it fulfills
$$
\varphi(x,y)= \int_{\partial \Omega} \psi(x,z)\psi(z,y) \,
d\sigma(z) \quad (x,y\in\partial \Omega)
$$  
with some $\psi\in L^2(\partial \Omega^2)$ satisfying 
$\psi(x,y)=\psi(y,x)$ 
$(x,y\in\partial \Omega)$;
\item[(ii)] {\it regular} if the function \ 
$\displaystyle x \mapsto \int_{\partial \Omega} \varphi(x,z)\,
d\sigma(z)$ does not a.e. vanish on $\partial \Omega$.
\end{enumerate}
                                

\noindent The following properties are elementary to prove.
                
\begin{proposition} A positive kernel $\varphi$ fulfills 
$\varphi\in L^2(\partial \Omega^2)$ and 
$\varphi(x,y)=\varphi(y,x)$ 
$(x,y\in\partial \Omega)$.
\end{proposition}   

\begin{proposition} Consider the linear integral operator $A:L^2(\partial
\Omega)\to L^2(\partial \Omega)$,
\begin{equation}
(Au)(x)= \int_{\partial \Omega} \varphi(x,y)u(y) \, d\sigma(y)
\, .
\label{AU?}
\end{equation}
\begin{enumerate}               
\item[(i)] If $\varphi$ is a positive kernel then $A$ is a
positive operator, i.e.
$$
\int_{\partial \Omega} (Au)u \ge 0 \quad (u\in L^2(\partial \Omega).
$$
\item[(ii)] If $\varphi$ is regular then $A$ does not carry
constants to
the (a.e.) zero function.
\end{enumerate}
\end{proposition}

\paragraph{Definition 2.2} Let $\varphi$ be a regular positive kernel and
$m>0$. Then we define
\begin{equation}
\langle u,v \rangle  \equiv 
\int_\Omega \nabla u\cdot \nabla v + 
    {1 \over m} \int \hskip-2mm \int_{\partial \Omega^2}
\varphi(x,y)u(y)v(x)\, d\sigma(y)\, d\sigma(x)\, .
\label{SCP}
\end{equation}


\begin{proposition} Formula (\ref{SCP}) defines an inner product on
$H^1(\Omega)$.
\end{proposition}

The above inner product will be used in $H^1(\Omega)$ (with $m>0$ to
be defined
in condition (C3) below) throughout the paper, and the corresponding
norm will be denoted by $\|\, . \, \|$. We note that if $u\in
H^2(\Omega)$
and $\frac{\partial {u}}{\partial \nu}+A(u)=0$ on $\partial \Omega$,
then the divergence theorem yields
\begin{equation}
\langle u,v \rangle = \int_\Omega (-\Delta u)v
\label{Lauv}
\end{equation}
with $m=1$. (This is a special case of Remark 2.4 below with
$T=-\Delta$.)

We will use notation $\nu$ for the outward normal vector on $\partial
\Omega$, and dot product
to denote the inner product in ${\mathbb R}^N$.

Now the nonlocal boundary-value problem can be formulated.
\smallskip

We consider the problem
\begin{equation}
 \begin{array}{l}   
T(u)\equiv -\mathop{\rm div} f(x,\nabla u)+q(x,u)= g(x) 
        \quad {\rm in } \ \Omega 
\\[2mm]                             
\displaystyle Q(u)\equiv f(x,\nabla u)\cdot \nu + \int_{\partial
\Omega} \varphi(x,y)u(y) \, d\sigma(y) = 0
        \quad {\rm on} \ \partial\Omega 
\end{array}         
\label{BVP}
\end{equation}
with the following conditions:
\begin{enumerate}       
\item[\rm (C1)] $\Omega\subset{\mathbb R}^N$ is bounded,
$\partial\Omega\in C^1$;
$f\in C^1(\overline\Omega\times{\mathbb R}^N,{\mathbb R}^N)$, $q\in
C^1(\overline\Omega\times{\mathbb R}^N)$, $g\in L^2(\Omega)$;
\item[\rm (C2)] $\varphi$ is a regular positive kernel;

\item[\rm (C3)] there exist constants $m'\ge m>0$ such that 
for all $(x,\eta)\in  \overline\Omega\times{\mathbb R}^N$
the Jacobians        
$\displaystyle \frac{\partial f(x,\eta)}{\partial \eta} \in {\bf
R}^{N\times N}$ are symmetric
and their eigenvalues $\lambda$ fulfill
$$                  
m\le \lambda \le m'\, ;
$$
further, there exist constants $\kappa, \beta\ge 0$
such that for all $(x,u)\in  \overline\Omega\times{\mathbb R}$
$$                  
0 \le \frac{\partial q(x,u)}{\partial u} \le \kappa +\beta |u|^{p-2}
$$
where $2\le p$ if $N=2$ and 
$2\le p< {2N \over N-2}$ if $N>2$.
\end{enumerate}
                    
\paragraph{Remark 2.1} It is worth mentioning the following special cases of
$f$.
\begin{enumerate}       
\item[(a)] $f(x,\nabla u)=p(x,\nabla u)\nabla u$ where 
$p\in C^1(\overline\Omega\times{\mathbb R}^N)$. Then the boundary
condition takes the form
$$
p(x,\nabla u)\frac{\partial {u}}{\partial \nu} + \int_{\partial
\Omega} \varphi(x,y)u(y) \, d\sigma(y) = 0\, .
$$              
\item[(b)]  $f(x,\nabla u)=a(|\nabla u|)\nabla u$ where 
$a\in C^1[0,\infty)$ (a special case of (a)). The corresponding
type of operator $T$ arises e.g. in elasto-plasticity theory or in the
study of magnetic potential \cite{Kach, NK}.
\end{enumerate}

                    
\paragraph{Remark 2.2} The assumption $2\le p$ (if $N=2$), 
$2\le p< {2N \over N-2}$ (if $N>2$) in condition (C3) yields
  \cite{Ad}  that  there holds the  Sobolev embedding 
\begin{equation}
H^1(\Omega) \subset L^p(\Omega) \, .
\label{Sob}
\end{equation}

\paragraph{Remark 2.3} The condition that $\varphi$ is a regular kernel is
required to avoid the lack of
injectivity when $f(x,0)=0$ (e.g. in the cases of Remark 2.1). Namely,
there would otherwise hold $Q(c)=0$ on $\partial \Omega$ for constant
functions $c$ as in the case of Neumann boundary condition.


\begin{proposition} For any $u,v\in H^1(\Omega)$ let 
\begin{equation}
\langle F(u),v \rangle  \equiv 
\int_\Omega \bigl( f(x,\nabla u)\cdot \nabla v + q(x,u)v \bigr) + 
     \int \hskip-2mm \int_{\partial \Omega^2}
\varphi(x,y)u(y)v(x)\, d\sigma(y)\, d\sigma(x)\, .
\label{GDO}
\end{equation}
 Then formula  (\ref{GDO}) defines an operator $F:H^1(\Omega)\to
H^1(\Omega)$.
\end{proposition}
            
\paragraph{Proof} Condition (C3) implies that for all
$i,j=1,..,N$ and $(x,\eta)\in \overline\Omega \times {\mathbb R}^N $
 $$\left| \frac{\partial f_i}{\partial\eta_j}(x,\eta) \right| \le m' \,
.$$   
  Lagrange's inequality yields that for all
$(x,\eta) \in \overline\Omega \times {\mathbb R}^N $ we have
$$
|f_i(x,\eta)| \le | f_i(x,0)| + m'N^{1/2}|\eta|\, ,
\quad   
|q(x,u)| \le |q(x,0)| + \kappa |u| + \beta |u|^{p-1}\, .
$$
Consequently, the integral on $\Omega$ in (\ref{GDO}) can be estimated
by
\begin{eqnarray*}
\lefteqn{ \int\limits_\Omega \Big( 
 \sum\limits_{i=1}^N \big( | f_i(x,0)| + m'N^{1/2}|\nabla u| \big)
|\partial_i v| +
    (|q(x,0)| + \kappa |u|)|v| + \beta |u|^{p-1}|v|
\Big)}\\
&\le& \left( \|f(x,0)\|_{L^2(\Omega)^N} + m'N \|\nabla
u\|_{L^2(\Omega)^N} \right)
    \|\nabla v\|_{L^2(\Omega)^N} \\
&&+\left( \|q(x,0)\|_{L^2(\Omega)} + \kappa \|u\|_{L^2(\Omega)} \right)
\|v\|_{L^2(\Omega)}
     + \beta \|u\|_{L^p(\Omega)}^{p-1} \|v\|_{L^p(\Omega)}\, 
.\hspace{15mm}
\end{eqnarray*}
Using (\ref{SCP}) and (\ref{Sob}), we obtain the following estimate
for the
right side of (\ref{GDO}):
$$\displaylines{ 
\Big( \|f(x,0)\|_{L^2(\Omega)^N} + m'N \|\nabla u\|_{L^2(\Omega)^N} +
    K_{2,\Omega} \big( \|q(x,0)\|_{L^2(\Omega)} \cr
+ \kappa \|u\|_{L^2(\Omega)} \big)     
+ \beta K_{p,\Omega} \|u\|_{L^p(\Omega)}^{p-1} + \|u\| \Big)\|v\|\, ,
}$$                      
where $K_{p,\Omega}$ $(p\ge 2)$ is the 
embedding constant in the inequality
\begin{equation}
\|u\|_{L^p(\Omega)} \le  K_{p,\Omega} \|u\|  \quad (u\in H^1(\Omega)) 
\label{Sobin}       
\end{equation}
corresponding to (\ref{Sob}).
Hence for all fixed $u\in H^1(\Omega)$   Riesz's theorem ensures the
existence
of $F(u)\in H^1(\Omega)$. \hfill$\diamondsuit$
                        
\paragraph{Definition 2.2} A {\it weak solution} of problem (\ref{BVP}) is defined
in the usual way
as a function $u^\ast\in H^1(\Omega)$ satisfying
\begin{equation}
\langle F(u^\ast),v \rangle = \int_\Omega gv  \quad (v\in
H^1(\Omega)). 
\label{WS}  
\end{equation}
                        
\paragraph{Remark 2.4} For any $u\in H^2(\Omega)$ with $Q(u)=0$ on $\partial
\Omega$, we have
$$
\langle F(u),v \rangle = \int_\Omega T(u)v  \quad (v\in
H^1(\Omega)). 
$$                      
This follows from the divergence theorem:
$$ \int_\Omega T(u)v=
    \int_\Omega \bigl( f(x,\nabla u)\cdot \nabla v + q(x,u)v
\bigr) 
-\int_{\partial\Omega} \bigl( f(x,\nabla u)\cdot \nu \bigr) v\,
d\sigma \ . 
$$ 
Consequently (as usual), a solution of (\ref{BVP}) is a weak solution,
and
a weak solution  $u^\ast\in H^2(\Omega)$ with $Q(u^\ast)=0$  on
$\partial \Omega$ satisfies (\ref{BVP}).


\section{Construction and convergence of the gradient method
in Sobolev space}
                            

The construction of the gradient method relies on the following
property
of the generalized differential operator.


\begin{theorem} Let $F:H^1(\Omega)\to H^1(\Omega)$ be defined in (\ref{GDO}). 
Then $F$ is Gateaux differentiable and $F'$ satisfies 
\begin{equation}
m\| h \|^2 \le \langle F'(u)h,h \rangle \le M(\|u\|)\| h \|^2 
        \quad (u,h\in H^1(\Omega)),
\label{uell}
\end{equation}
where 
\begin{equation}
M(r) = m' + \kappa K_{2,\Omega}^2 + \beta K_{p,\Omega}^p r^{p-2}
\label{MR}
\end{equation}
with $K_{p,\Omega}$ defined in (\ref{Sobin}). \end{theorem} 


\paragraph{Proof} For any  $u\in H^1(\Omega)$ let $S(u):H^1(\Omega)\to
H^1(\Omega)$ be the bounded linear
operator defined by   
\begin{eqnarray} \label{SUHV}
 \langle S(u)h,v \rangle  &\equiv&   \int_\Omega \big( \frac{\partial
f}{\partial\eta}(x,\nabla u)\nabla h\cdot \nabla v 
        + \frac{\partial q}{\partial u}(x,u) hv \big) \\
&&+ \int \hskip-2mm \int_{\partial \Omega^2} \varphi(x,y)h(y)v(x)\,
d\sigma(y)\, d\sigma(x)\,,  \nonumber
\end{eqnarray} 
for all $u,h,v\in H^1(\Omega)$.
The existence of  $S(u)$  is provided by Riesz's theorem similarly as 
in Proposition 2.4,
now                            
using   the estimate    
$$ \left( m' +  \kappa K_{2,\Omega}^2
 + \beta K_{p,\Omega}^2 \|u\|_{L^p(\Omega)}^{p-2} \right) \|h\| \|v\|
$$                      
for the integral term on $\Omega$. We   will   prove that 
\begin{equation}
F'(u)=S(u)\quad (u\in H^1(\Omega)) 
\label{FVS}
\end{equation}
in  Gateaux sense. Therefore, let  $u,h \in  H^1(\Omega)$  and
${\cal E}:=
\left \{ v\in H^1(\Omega): \|v\|= 1 \right \}$.   Then
\begin{eqnarray*}
D_{u,h}(t)&\equiv& \frac{1}{t}\| F(u+th)-F(u)-tS(u)h\| \\
&=& \frac{1}{t} \sup_{v\in {\cal E}} \left\langle F(u+th)-
F(u)-tS(u)h,v\right\rangle   \\
&=& \frac{1}{t}\sup_{v\in {\cal E}} \int _\Omega 
\Big[ \big(
 f(x,\nabla u + t\nabla h) - f(x,\nabla u) - t\frac{\partial
f}{\partial\eta}(x,\nabla u)\nabla h
        \big)\cdot \nabla v \\
&& +\big( q(x, u + t h) - q(x, u) - t\frac{\partial q}{\partial
u}(x, u) h \big) v \Big] \\
&=& \sup_{v\in {\cal E}} \int _\Omega        
 \Big[ \big(
\frac{\partial f}{\partial\eta}(x,\nabla u + t\theta \nabla h)  
 -\frac{\partial f}{\partial\eta}(x,\nabla u) \big) \nabla h\cdot
\nabla v \\
&&+  \big( \frac{\partial q}{\partial u}(x, u + t\theta  h) 
     -\frac{\partial q}{\partial u}(x, u) \big) hv \Big] \\ 
&\le& \sup_{v\in {\cal E}} \Big[ \big\| 
\big(   \frac{\partial f}{\partial\eta}(x,\nabla u + t\theta\nabla h)
 -\frac{\partial f}{\partial\eta}(x,\nabla u) \big)\nabla
h\big\|_{L^2(\Omega)}  \|\nabla v \|_{L^2(\Omega)} \\
&&+\big\| \big( \frac{\partial q}{\partial u}(x,u + t\theta h)
 -\frac{\partial q}{\partial u}(x,u) \big) h \big\|_{L^q(\Omega)}
 \|v \|_{L^p(\Omega)} \Big]\,,       
\end{eqnarray*}      
where $p^{-1}+q^{-1}=1$.
Here $\|\nabla v\|_{L^2(\Omega )}\le \| v\| \le 1$ and
$\|v\|_{L^2(\Omega )}\le K_{2,\Omega} \| v\| \le K_{2,\Omega}$.
Further, $|t\theta\nabla h| \to 0 $ and $|t\theta h| \to 0 $
(as $ t\to 0)$  a.e. on $\Omega $,
hence the continuity of $\frac{\partial f}{\partial\eta}$ and
$\frac{\partial q}{\partial u}$
implies that  the integrands tend to  0  as  $t\to 0$. For $|t|\le
t_0$
the integrands are majorated   by $(2m'|\nabla h|)^2\in L^1(\Omega)$
and                     
$ (2 \kappa  + \beta (|u+t_0 h|^{p-2} + |u|^{p-2}) h)^q \le $
$ const.\cdot (2 \kappa  + \beta (|u+t_0 h|^{(p-2)q} + |u|^{(p-2)q})
h^q)
\in L^1(\Omega)$.
(The latter holds since $u,h\in L^p(\Omega)$ implies 
$u^{(p-2)q}\in L^{p \over (p-2)q}(\Omega)$ and $h^q\in L^{p \over
q}(\Omega)$,
and here ${(p-2)q \over p} + {q \over p} = 1$ from
$p^{-1}+q^{-1}=1$.)             
Hence Lebesgue's
theorem yields that the obtained expression tends to 0
(as $t \to 0$), thus
$$          
\lim_{t\rightarrow 0} D_{u,h}(t) =0\, .
$$
\quad Now the inequality (\ref{uell}) is left to prove. From
(\ref{FVS}) and
(\ref{SUHV}) we have for any  $u,h \in H^1(\Omega)$
\begin{eqnarray*}   
 \langle F'(u)h,h \rangle  &=& 
\int_\Omega \big( \frac{\partial
f}{\partial\eta}(x,\nabla u)\nabla h\cdot \nabla h  +
\frac{\partial q}{\partial u}(x,u)h^2 \big) \\
&&+\int \hskip-2mm \int_{\partial \Omega^2} \varphi(x,y)h(y)h(x)\,
d\sigma(y)\, d\sigma(x)\, .
 \end{eqnarray*}               
From condition (C3)  we have 
$$
m|\nabla h|^2 \le  \frac{\partial f}{\partial\eta}(x,\nabla u)\nabla
h\cdot \nabla h  \le
m'|\nabla h|^2 \, ,
$$ 
which, together with $\frac{\partial q}{\partial u}\ge 0$, implies
directly that
$$                  
m\|h\|^2 \le \langle F'(u)h,h \rangle \, .
$$      
Further,
\begin{eqnarray*}
 \langle F'(u)h,h \rangle
&\le& \int_\Omega \left[ m'|\nabla h|^2 + (\kappa  + \beta |u|^{p-2})h^2
\right] \\
&&+  \int \hskip-2mm \int_{\partial \Omega^2} \varphi(x,y)h(y)h(x)\,
d\sigma(y)\, d\sigma(x)\\
&\le& m'\|h\|^2 +     \kappa \|h\|_{L^2(\Omega)}^2 + \beta
\|u\|_{L^p(\Omega)}^{p-2} \|h\|^2_{L^p(\Omega)} \\
&\le& (m' + \kappa K_{2,\Omega}^2 + \beta K_{p,\Omega}^p \|u\|^{p-2} )
\|h\|^2 \, ,
\end{eqnarray*}
i.e. the right side of (\ref{uell}) is also satisfied.
\hfill$\diamondsuit$\medskip            

 
Now we quote an abstract result on the gradient method in Hilbert
space, which in this form
follows from \cite{Pu2} (Theorem 2 and Corollary 1).
                

\begin{theorem} Let $H$ be a real Hilbert space, $b\in H$ and
let $F :H\rightarrow H$ satisfy the following
properties:
\begin{enumerate}               
\item[(i)]  $F$ is Gateaux differentiable;
\item[(ii)] for any $u,k,w,h\in H$  the mapping $s,t\mapsto
F'(u+sk+tw)h$ is continuous from ${\mathbb R}^2$ to $H$;
\item[(iii)] for any $u\in H$ the operator $F'(u)$ is self-adjoint;
\item[(iv)]  there exists $m>0$ and an increasing function
$ M:[0,\infty)\rightarrow (0,\infty) $              
such that for all $u,h\in H$            
$$                              
m\| h \|^2 \le \langle F'(u)h,h \rangle \le M(\|u\|)\| h \|^2\, .
$$
\end{enumerate}
Then    
\begin{enumerate}               
\item[(1)]  the equation $F(u)=b$ has a unique solution $u^\ast \in
H$.
\item[(2)] Let $u_0\in H$, $M_0:= M\left( \|u_0\| +
         {1 \over m}\|F(u_0)-b\| \right)$.   
Then the sequence
$$                  
u_{n+1}= u_n - \frac{2}{M_0+m} (F(u_n)-b) \quad (n\in {\mathbb N})
$$                          
converges linearly to $u^\ast$, namely,
$$          
\| u_n - u^\ast \| \le 
\frac{1}{m}\left \| F(u_0)-b \right \| 
    \left ( \frac{M_0-m}{M_0+m}\right ) ^n  \quad (n\in {\mathbb N})\,. 
$$
\end{enumerate}                 
\end{theorem}

Now we are in  position for constructing the gradient method for
(\ref{BVP}) in $H^1(\Omega)$ and to verify its convergence.

\begin{theorem}
\begin{enumerate}               
\item[(1)]  Problem (\ref{BVP}) has a unique weak solution 
 $u^\ast \in H^1(\Omega)$.
\item[(2)] Let $b\in H^1(\Omega)$ such that
$$
\langle b,v \rangle = \int_\Omega gv  \quad (v\in H^1(\Omega)),
$$              
and let $F$ denote the generalized differential operator as in
(\ref{GDO}).        
 Let $u_0\in H^1(\Omega)$, $M_0:= M\left( \|u_0\| +
         {1 \over m}\|F(u_0)-b\| \right)$, where
$M(r) = m' + \kappa K_{2,\Omega}^2 + \beta K_{p,\Omega}^p r^{p-2}$. 
Then the sequence
\begin{equation}
u_{n+1}= u_n - \frac{2}{M_0+m} (F(u_n)-b) \quad (n\in {\mathbb N})
\label{iter}
\end{equation}
converges linearly to $u^\ast$, namely,
$$          
\| u_n - u^\ast \| \le 
\frac{1}{m}\left \| F(u_0)-b \right \| 
    \left ( \frac{M_0-m}{M_0+m}\right ) ^n  \quad (n\in {\mathbb N})\,. 
$$                  
\end{enumerate}                 
\end{theorem}

\paragraph{Proof} Our task is to verify   conditions (i)-(iv) of Theorem 3.2
for
(\ref{BVP}) in $H^1(\Omega)$. Conditions (i) and (iv) have been proved
in Theorem 3.1.
The hemicontinuity of $F'$ follows similarly to the differentiability
of $F$
if in the proof of Theorem 3.1 we examine 
$ \tilde D_{u,k,w,h}(t)\equiv  \| (F'(u+sk+tw)-F'(u)) h \|$
instead of $D_{u,h}(t)$. Finally, the symmetry   of  $F'(u)$
 follows immediately from (\ref{FVS}), (\ref{SUHV}) and the 
symmetry   of  $\varphi$ and of the Jacobians   $\frac{\partial
f}{\partial \eta}(x,\eta)$.
\hfill$\diamondsuit$
        
\paragraph{Remark 3.1} Assume that $u_n$ is constructed. Then
$$
u_{n+1}= u_n - \frac{2}{M_0+m} z_n\ ,
$$                          
where $z_n\in H^1(\Omega)$ satisfies 
$$
\langle z_n,v \rangle = \langle F(u_n),v \rangle - \int_\Omega gv
\quad (v\in H^1(\Omega)).
$$                          
That is, in order to find $z_{n}$ we need to solve the auxiliary
linear  variational problem
\begin{eqnarray} \label{ZN}
\lefteqn{\int_\Omega \nabla z_n\cdot \nabla v + 
   {1 \over m} \int \hskip-2mm \int_{\partial \Omega^2}
\varphi(x,y)z_n(y)v(x)\, d\sigma(y)\, d\sigma(x)}\\
&=& \langle F(u_n),v \rangle - \int_\Omega gv \quad (v\in
H^1(\Omega)).  \hspace{25mm}\nonumber  
\end{eqnarray}

        
\paragraph{Remark 3.2} If there hold the regularity properties $u_n\in
H^2(\Omega)$ and
$z_n\in H^2(\Omega)$, then the auxiliary problem (\ref{ZN}) can be
written in strong form
as follows. Using the divergence theorem, we obtain from (\ref{ZN})
that
$$\displaylines{
 \int_\Omega (-\Delta z_n)v + 
  \int_{\partial \Omega}  \left( \frac{\partial {z_n}}{\partial
\nu}(x) + {1 \over m}\int_{\partial \Omega}
    \varphi(x,y)z_n(y)\, d\sigma(y)\right) v(x) \,d\sigma(x)\cr
 =\int_\Omega (T(u_n)-g)v +              
  \int_{\partial \Omega}  \left( f(x,\nabla u_n)\cdot \nu +
\int_{\partial \Omega}
        \varphi(x,y)u_n(y)\, d\sigma(y)\right) v(x) \,d\sigma(x)
}$$                  
holds for all $v\in H^1(\Omega)$. If especially all $v\in
H^1_0(\Omega)$ are considered, then we obtain
$$
-\Delta z_n= T(u_n)-g\, .
$$
Hence for all $v\in H^1(\Omega)$ the boundary integral terms coincide,
which implies
that
\begin{eqnarray*}  
& \hskip-12mm \displaystyle \frac{\partial {z_n}}{\partial \nu} + {1 \over m}\int_{\partial
\Omega} \varphi(x,y)z_n(y)\, d\sigma(y)	&
\\
=& \displaystyle  f(x,\nabla u_n)\cdot \nu + \int_{\partial \Omega}   \varphi(x,y)u_n(y)\, d\sigma(y)        															%&
=& Q(u_n).
\end{eqnarray*}
Consequently, in this case $z_n$ is the solution of the linear
boundary-value problem
\begin{equation}
 \begin{array}{c}   
-\Delta z_n= T(u_n)-g\,,
\\[2mm]         
 \displaystyle \frac{\partial {z_n}}{\partial \nu} + {1 \over
m}\int_{\partial \Omega}
            \varphi(x,y)z_n(y)\, d\sigma(y)= Q(u_n).
\end{array} 
\label{LBVP}
\end{equation}
(In the general case -- without regularity of $z_n$ and $u_n$ --
(\ref{ZN})
is the weak formulation of (\ref{LBVP}).)

        
\paragraph{Remark 3.3} Consider the semilinear special case 
$T(u)\equiv -\Delta u + q(x,u)$ and assume that $u_0$ is chosen to
satisfy
$Q(u_0)=0$, further, that $z_n\in H^2(\Omega)$ for all $n\in{\bf
N}$. Then $m=1$ and 
the boundary condition 
in (\ref{LBVP}) is $Q(z_n)=Q(u_n)$. Hence by induction
$Q(z_n)=Q(u_n)=0$
$(n\in{\mathbb N})$, i.e. in each step homogeneous boundary condition
is imposed on the
auxiliary problem.


\paragraph{Remark 3.4} The construction of the method requires an estimate
for
the embedding constants $K_{p,\Omega}$. For this we can rely on the
exact constants
for the embedding of $H^1(\Omega)$ into $L^p(\Omega)$ obtained in
\cite{Bu}. When the 
lower order term of the equation   has at most linear growth (or is
not present at all),
 then only $K_{2,\Omega}$ is needed, which can be estimated, as usual,
using a
suitable Cauchy-Schwarz inequality. (The numerical example in the
following section
includes a direct estimation of the required constants.)

    \section{Numerical example}

The summary of the result in the previous section is as follows. The
Sobolev space 
gradient method reduces the solution of the nonlinear boundary value
problem (\ref{BVP}) to auxiliary linear 
problems given by (\ref{ZN}). The ratio of convergence of the
iteration is the number
$\frac{M_0-m}{M_0+m}$, which is determined by the original
coefficients $f$, $q$, 
$g$ and $\varphi$ and is independent of the numerical method used for
the solution 
of the auxiliary linear problems.

The numerical realization of the obtained gradient method is
established by choosing
a suitable numerical method for the solution of the auxiliary problems
(\ref{ZN}).
The latter method may be a finite difference or finite element
discretization.
In this case the advantage of having executed the iteration for the
original
problem (\ref{BVP}) in the Sobolev space lies in the fact that the
numerical questions
concerning discretization arise only for the linear problems
(\ref{ZN}) instead of the nonlinear one (\ref{BVP}), whereas the 
convergence of the iteration is guaranteed as mentioned in the 
preceding paragraph. This kind of coupling the Sobolev space gradient
 method with discretization of the auxiliary problems has been
developed for local (Dirichlet) boundary-value problems \cite{ GFEM,
GGZ}. It is plausible that this coupling may have a similarly 
effective realization for our nonlocal boundary-value problem 
(\ref{BVP}). Nevertheless, we prefer another situation for giving a 
numerical example, namely, when the auxiliary linear problems can be
solved directly  (without discretization).



\paragraph{The model problem.} Let $\Omega=[0,\pi]^2\subset{\mathbb R}^2$, 
and
$$                  
g(x,y)=\frac{2 \cos x \cos y}{\pi (2-0.249\cos 2x)(2-0.249\cos 2y)}\, .
$$
We consider the semilinear problem
\begin{equation}
  \begin{array}{c}      
 -\Delta u + u^3 = g(x,y) \quad {\rm in \  } \Omega
 \\[2mm]                    
 \displaystyle \frac{\partial {u}}{\partial \nu} + \int_{\partial
\Omega} u(y) \, d\sigma(y) = 0 \quad {\rm on} \ \partial\Omega
\, .
 \end{array}        
\label{MP}
\end{equation}
The calculations will be made up to accuracy  $10^{-4}$.
                                
The function $g(x,y)$ is approximated by its cosine Fourier partial
sum 
\begin{equation}
\tilde g(x,y)=            
     \sum\limits_{k,l \mbox{ are  odd}\atop k+l\le 6}
        a_{kl}\cos kx \cos ly\, , \qquad     
a_{kl}=2.9200\cdot 4^{-(k+l)} 
\label{GH}
\end{equation}
which yields $\|g-\tilde g \|_{L^2(\Omega)} \le 0.0001$. We
consider instead of (\ref{MP}) the equation 
$ -\Delta u + u^3 = \tilde g(x,y)$ with the
given boundary condition, and denote its solution by $\tilde u$.

The main idea of the numerical realization is the following. Let
$$          
 {\cal P}= \{ \sum\limits_{k,l \mbox{ are  odd} \atop k+l\le m}
 c_{kl} \cos kx \cos ly:
    \, m\in{\mathbb N}^+, \, c_{kl}\in{\mathbb R} \}.
$$          
Then $T$ is invariant on ${\cal P}$, i.e. $u\in{\cal P}$ implies
$T(u)\in{\cal P}$. Hence
also  $T(u)-\tilde g \in{\cal P}$. Further, any $u\in{\cal P}$ fulfills
the considered boundary condition 
(in fact, there even holds $ \frac{\partial {u}}{\partial \nu} =
\int_{\partial \Omega} u \, d\sigma =0$). Hence
for any $h\in{\cal P}$ the solution of the problem
$$
  \begin{array}{c}  
 -\Delta z = h \quad {\rm in \  } \Omega
 \\[2mm]                    
 \displaystyle \frac{\partial {z}}{\partial \nu} + \int_{\partial
\Omega} z \, d\sigma = 0 \quad {\rm on} \ \partial\Omega  
 \end{array}            
$$
fulfills $z\in{\cal P}$, namely, if
$$
 h(x,y)= \sum\limits_{k,l \mbox{ are  odd}\atop k+l\le m} c_{kl} 
                    \cos kx \cos ly
$$
 then
$$
z(x,y)= \sum\limits_{k,l \mbox{ are  odd} \atop k+l\le m}
{c_{kl}  \over k^2+l^2} \cos kx \cos ly \, .
$$
(That is, the inversion of the Laplacian is now elementary.) Summing
up: 
using Remark 3.3, we obtain that for any $u_0\in{\cal P}$  the GM
iteration 
\begin{equation}
  \begin{array}{c}      
  -\Delta z_n = T(u_n)-\tilde g \, , \quad  \frac{\partial
{z_n}}{\partial \nu} + \int_{\partial \Omega} z_n \, d\sigma = 0\, ;
 \\[2mm]                    
\displaystyle  u_{n+1}= u_n - \frac{2}{M_0+m} z_n
 \end{array}        
\label{ALG}
\end{equation}
fulfills $u_n\in{\cal P}$ for all $n\in{\mathbb N}^+$, and in each step
$u_{n+1}$ is elementary to obtain from $u_n$.

Now our remaining task is to choose an initial approximation
$u_0\in{\cal P}$
and to determine the corresponding ellipticity constants $M_0$ and
$m$.
For simplicity, we choose
$$
u_0 \equiv 0.
$$
Using the notations of conditions (C1)-(C3) in Section 2, the
coefficients are
$$
f(x,\eta)=\eta, \ \  q(x,u)=u^3 \ \ {\rm and} \ \varphi\equiv 1.
$$
Hence we have
$$
m=m'=1, \ \ \kappa=0, \ \ \beta =3 \ \ {\rm and} \ p=4.
$$
Thus Theorem 3.1 yields
\begin{equation}
M(r) = 1 +  3 K_{4,\Omega}^4 r^2,
\label{MR2} 
\end{equation}
and from Theorem 3.3 we obtain
\begin{equation}
M_0= M(\|b\|) = 1 +  3 K_{4,\Omega}^4 \|b\|^2
\label{M0}
\end{equation}
where $b\in H^1(\Omega)$ such that
$$
\langle b,v \rangle = \int_\Omega \tilde g v  \quad (v\in H^1(\Omega)).
$$

We recall that now, owing to $m=1$ and $\varphi\equiv 1$, the inner
product (\ref{SCP}) on
$H^1(\Omega)$ is 
\begin{equation}
\langle u,v \rangle  = \int_\Omega \nabla u\cdot \nabla v +
     \left( \int_{\partial \Omega} u \, d\sigma \right) \left(
\int_{\partial \Omega} v \, d\sigma \right) \, .
\label{ujscp}
\end{equation}

\begin{proposition} There holds 
$$  
b(x,y)= \sum\limits_{k,l \ are \ odd \atop k+l\le m}
{a_{kl}  \over k^2+l^2} \cos kx \cos ly \, ,
$$          
where (from (\ref{GH}))
$$
a_{kl}=2.92 \cdot 4^{-(k+l)}.
$$
\end{proposition}

\paragraph{Proof} {We have $-\Delta b = \tilde g$, hence  (\ref{Lauv})
yields
$$
\langle b,v \rangle = \int_\Omega (-\Delta b)v = \int_\Omega \tilde g
v \quad (v\in H^1(\Omega)).
$$

\begin{corollary} Since $\displaystyle \int_{\partial \Omega} b \, d\sigma
= 0$, therefore (\ref{ujscp}) yields
$$
\|b\|^2 = \int_\Omega |\nabla b|^2 = \left( {\pi \over 2} \right)^2
    \sum\limits_{k,l \ are \ odd \atop k+l\le m} {a_{kl}^2 \over
k^2+l^2}
        = 0.1014\, .
$$                              
\end{corollary}
        
\paragraph{Remark 4.1} In the same way as above, we have for all $u\in{\cal
P}$  
\begin{equation}
\|u\|^2 = \int_\Omega |\nabla u|^2 .
\label{unorm}
\end{equation}

            
In order to find now an estimate for
$K_{4,\Omega}$, we note that its value is only required
for the (closure of the) subspace ${\cal P}$ where $(u_n)$ runs.
That is, it suffices to determine $\tilde K_{4,\Omega}$ satisfying
$$
\|u\|_{L^4(\Omega)} \le  \tilde K_{4,\Omega} \|u\|  \quad (u\in{\cal
P}). 
$$

\begin{proposition} There holds $\tilde K_{4,\Omega}^4\le 10.3776$ .
\end{proposition}

The proof of this proposition consists of some calculations sketched in 
the Appendix.\medskip

Substituting in (\ref{M0}), we obtain $M_0$.

\begin{corollary} The ellipticity constants are
$m=1$ and $M_0 = 4.1569$. \end{corollary}

The corresponding stepsize and  convergence quotient are
$$ 
\frac{2}{M_0+m} = 0.3878, \quad \frac{M_0-m}{M_0+m} = 0.6122\,.
$$
The algorithm (\ref{ALG}) has been performed in MATLAB, which is
convenient
for the required elementary matrix operations determined by storing
the functions $u_n$ as matrices of coefficients.
 (In order to avoid the inconvenient growth of the matrix
sizes, the high-index almost zero coefficients were dropped within a
$10^{-4}$
error calculated from the square sum of the coefficients.)

                
The actual error $\|\tilde u-u_n\|$ was estimated using the residual
$$  
r_n = \|T(u_n)-\tilde g \|_{L^2(\Omega)}\, . 
$$          
The connection between $\|\tilde u-u_n\|$ and $r_n$ is based on the
following propositions.

\begin{proposition} For any $u\in{\cal P}$
$$
\|u\|_{L^2(\Omega)} \le 2^{-1/2} \|u\|.
$$
\end{proposition}
\paragraph{Proof} {Let 
$$  
 u(x,y)= \sum\limits_{k,l \ are \ odd \atop k+l\le m} c_{kl} 
                    \cos kx \cos ly\, .
$$
Then from (\ref{unorm})
\begin{eqnarray*} 
\|u\|^2 &=&  \int_\Omega |\nabla u|^2 =  \left( {\pi \over 2} \right)^2
    \sum\limits_{k,l \mbox{ are  odd} \atop k+l\le m} (k^2+l^2)c_{kl}^2 \\
&\ge&     2  \left( {\pi \over 2} \right)^2
\sum\limits_{k,l \mbox{ are  odd} \atop k+l\le m} c_{kl}^2 = 
        2 \|u\|_{L^2(\Omega)}^2\, . 
\end{eqnarray*}                             

                    
\begin{proposition} For all $u,v\in{\cal P}$ 
$$
\|u-v\| \le 2^{-1/2} \|T(u)-T(v)\|_{L^2(\Omega)}\, .
$$
\end{proposition}

\paragraph{Proof} The uniform ellipticity of $T$ implies
\begin{eqnarray*}                     
\|u-v\|^2 &\le& \int_\Omega (T(u)-T(v))(u-v) \\
&\le& \|T(u)-T(v)\|_{L^2(\Omega)} \|u-v\|_{L^2(\Omega)} \\
&\le& 2^{-1/2} \|T(u)-T(v)\|_{L^2(\Omega)} \|u-v\| \, .  
\end{eqnarray*}


\begin{corollary} Let
\begin{equation}
e_n = 2^{-1/2} r_n = 2^{-1/2} \|T(u_n)-\tilde g \|_{L^2(\Omega)} \quad
(n\in{\mathbb N}). 
\label{EN}
\end{equation}
Then, applying Proposition 4.4 to $u_n$ and $\tilde u$, we obtain
$$
\|\tilde u-u_n\|   \le e_n\, .
$$
\end{corollary}
                
Based on these, the error was measured by $e_n$ defined in (\ref{EN}).
(Since  $T(u_n)$ and $\tilde g$ are trigonometric polynomials, this
only requires
square summation of the coefficients.) 
                        
 The following table contains the error $e_n$ versus
the number of steps $n$.
        
\bigskip
    
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}\hline
step $n$ &    1  &  2   &  3   &  4   &  5  &   6  & 7   \\ \hline
error $e_n$ & 1.1107 & 0.6754 & 0.3992 & 0.2290 & 0.1288  & 0.0718 &
0.0402  \\ \hline
\end{tabular} 
\\[3mm]
\begin{tabular}{|c|c|c|c|c|c|c|c|}\hline  
step $n$ & 8  &   9  &   10   &   11   &   12   &   13   &   14   \\ \hline
error $e_n$& 0.0225 & 0.0127 & 0.0072 & 0.0042 & 0.0024 & 0.0014 & 0.0008 
\\ \hline   
\end{tabular}  
\\[3mm]
\begin{tabular}{|c|c|c|c|c|c|c|c|}\hline    
step $n$ & 15   &   16 &   17   &   18   &   19   &   20   &   21   \\ \hline  
error $e_n$ & 0.0005 & 0.0003 & 0.0003 & 0.0002 & 0.0002 & 0.0002 & 0.0001 \\
\hline  
\end{tabular} \\[3mm]
Table 1.  
\end{center} 
                            
\paragraph{Remark 4.2} We have determined above numerically, 
up to accuracy  $10^{-4}$, the solution $\tilde u$
of the approximated  problem with $\tilde g$ instead of $g$.
Since $\tilde u$ and $u^\ast$ are in $\overline{{\cal P}}$,
Proposition 4.4 yields
$$  
\|\tilde u-u^\ast\| \le 2^{-1/2} \|\tilde g-g\|_{L^2(\Omega)}  \le
2^{-1/2}\cdot 0.0001 \, .
$$                          


\section{Appendix}
\paragraph{Proof of Proposition 4.2.} 
The proof can be achieved through two lemmata.

\begin{lemma} For any $u\in{\cal P}$,
$$          
\int_\Omega u^4 \le {1 \over 8} 
    \left( \int_{\partial \Omega} u^2 \, d\sigma + 8^{1/2} \|u\|^2
\right)\, .
$$
\end{lemma}

\paragraph{Proof} It is proved in \cite{Lyons} that for any $u\in
H^1_0(\Omega)$
$$
\int_\Omega u^4 \le 4 \|u\|_{L^2(\Omega)}^2 \|\partial_1
u\|_{L^2(\Omega)}\|\partial_2 u\|_{L^2(\Omega)} \le
    2 \|u\|_{L^2(\Omega)}^2 \|\nabla u\|_{L^2(\Omega)}^2\, .
$$
Taking into account the boundary, we obtain in the same way that
for any $u\in H^1(\Omega)$
$$          
\int_\Omega u^4 \le 
2 \left( {1 \over 4} \int_{\partial \Omega} u^2 \, d\sigma +
\|u\|_{L^2(\Omega)} \|\nabla u\|_{L^2(\Omega)} \right)^2 .
$$                          
This yields the desired estimate for any $u\in{\cal P}$,
using Remark 4.1 and Proposition 4.3 for $\|u\|_{L^2(\Omega)}$ and
$\|\nabla u\|_{L^2(\Omega)}$. \hfill$\diamondsuit$
                            
        
\begin{lemma} For any $u\in{\cal P}$,
$$          
\int_{\partial \Omega} u^2 \, d\sigma \le 2\pi \|u\|^2 .
$$
\end{lemma}
    
\paragraph{Proof} Let   $\Gamma_1= [0,\pi ]\times \{ 0 \}$,  
$\Gamma_2=\{ \pi \}\times [0,\pi]$, \,
    $\Gamma_3= [0,\pi ]\times \{ \pi \}$, 
$\Gamma_4=\{ 0 \}\times [0,\pi]$.
Then $\partial \Omega = \cup \{ \Gamma_i:\, i=1,\dots ,4\}$. 
Now let $u\in{\cal P}$. For any $x,y\in [0,\pi]$ we have 
$$
u(x,\pi)-u(0,y) = \int_0^x \partial_1 u(s,y) ds +\int_y^\pi \partial_2
u(x,t)\, dt\, .
$$      
Raising to square and integrating over  $\Omega$, we obtain
\begin{eqnarray*}
\lefteqn{ \pi \left( \int_{\Gamma_3} u^2 \, d\sigma +\int_{\Gamma_4} u^2 \,
d\sigma \right)
    - 2 \left( \int_{\Gamma_3} u \, d\sigma  \right)
            \left(  \int_{\Gamma_4} u \, d\sigma \right) }\\
&\le &
2 \int_0^\pi \int_0^\pi 
\left[
     \left( \int_0^x \partial_1 u(s,y) ds  \right)^2 +
            \left(\int_y^\pi \partial_2 u(x,t) dt
\right)^2
                            \right] dx dy \\
&\le& \pi^2 \int_\Omega \left[ (\partial_1 u)^2 + (\partial_2 u)^2 \right]\,,
\end{eqnarray*}
where Cauchy-Schwarz inequality was used. We can repeat the same
argument
for the pairs of edges $(\Gamma_1,\Gamma_2)$, $(\Gamma_2,\Gamma_3)$
and $(\Gamma_1,\Gamma_4)$
in the place of $(\Gamma_3,\Gamma_4)$. Then, summing up and using
$\partial \Omega = \cup \{ \Gamma_i:\, i=1,\dots ,4\}$, we obtain
\begin{equation}
2\pi \int_{\partial \Omega} u^2 \, d\sigma          
    - 2 \left( \int_{\Gamma_1 \cup \Gamma_3} u \, d\sigma  \right)
            \left(  \int_{\Gamma_2 \cup \Gamma_4} u \,
d\sigma \right) \le 4\pi^2 \int_\Omega |\nabla u|^2 .
\label{unio}
\end{equation}
Using notations $\Gamma_x = \Gamma_1 \cup \Gamma_3$ and 
$\Gamma_y = \Gamma_2 \cup \Gamma_4$, there holds
\begin{eqnarray*}
2 \left( \int_{\Gamma_x} u \, d\sigma  \right)  
            \left(  \int_{\Gamma_y} u \, d\sigma \right) 
&=&\left( \int_{\Gamma_x \cup \Gamma_y} u \, d\sigma  \right)^2    
- \left( \int_{\Gamma_x} u \, d\sigma  \right)^2 - \left(
\int_{\Gamma_x} u \, d\sigma  \right)^2 \\
 &\le& \int_{\partial \Omega} u \, d\sigma = 0\, ,
\end{eqnarray*}
hence (\ref{unio}) yields
$$
2\pi \int_{\partial \Omega} u^2 \, d\sigma 
\le  4\pi^2 \int_\Omega |\nabla u|^2 =  4\pi^2 \|u\|^2 .  
$$  
                        
\paragraph{Proof of the proposition.} Lemmata 1 and 2 yield
$$
\|u\|_{L^4(\Omega)}^4 \le {1 \over 8} (2\pi + 8^{1/2}) \|u\|^4\, ,
$$                          
that is
$$      
\tilde K_{4,\Omega}^4  \le {1 \over 8} (2\pi + 8^{1/2}) = 10.3776
$$
up to accuracy $10^{-4}$.


 
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\noindent{\sc J\'anos Kar\'atson }\\
E\"otv\"os Lor\'and University \\
Dept. of Applied Analysis \\
H-1053 Budapest, Kecskem\'eti u. 10-12. \\
Hungary \\
email: {\tt karatson@cs.elte.hu}
                        
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