\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 59, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/59\hfil Magnetohydrodynamic flows in pipes]
{Existence and uniqueness for magnetohydrodynamic flows in pipes with
  viscosity dependent on the temperature}

\author[G. Cimatti\hfil EJDE-2009/59\hfilneg]
{Giovanni Cimatti}

\address{Giovanni Cimatti \newline
Dipartimento di Matematica, Universita di Pisa,
Largo Bruno Pontecorvo 5, Pisa, Italy}
\email{cimatti@dm.unipi.it}

\thanks{Submitted December 15, 2008. Published April 29, 2009.}
\subjclass[2000]{76W05, 35J55}
\keywords{Elliptic system; magnetohydrodynamic flow;\hfill\break\indent
temperature-dependent viscosity}

\begin{abstract}
 The steady motion of a viscous fluid in pipes of arbitrary cross-sections
 under a transverse magnetic field is studied, assuming that the
 viscosity and the electric and thermal conductivity are given functions
 of the temperature. Theorems of existence and uniqueness for the
 nonlinear elliptic system governing the problem are presented.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

In this paper we study a class of steady, incompressible rectilinear
flows of viscous electrically and thermally conducting fluids along
cylindrical channels of arbitrary cross-section in the framework of
the equations of magnetohydrodynamics. The open and bounded subset
$\Omega$ of $\mathbb{R}^2$ representing the cross-section of the pipe 
is referred to the orthogonal frame $Oxy$ with unit vectors $\mathbf{i}$ 
and $\mathbf{j}$. $Oz$ is the axis of the channel with $\mathbf{k}$ as
unit vector. The magnetic field $\mathbf{H}$ is assumed of the form
\begin{equation}
\mathbf{H}=\mathbf{M}+h(x,y)\mathbf{k},
\notag
\end{equation}
where  $\mathbf{M}$ is a vector constant and parallel to $\Omega$.
Rotating the $Oxy$ frame we can write
\begin{equation}
\mathbf{H}=M\mathbf{i}+h(x,y)\mathbf{k}.
\label{1.2}
\end{equation}
Since the flow is laminar and rectilinear,
\begin{equation}
\mathbf{v}=v(x,y)\mathbf{k}
\label{2.2}
\end{equation}
is the velocity of the fluid and
$p=p(z)$ the pressure. In the steady state, Maxwell's equations reads
\begin{gather}
\nabla\times \mathbf{E}=0 , \label{4.2}\\
\nabla\times \mathbf{H}=\mathbf{J},
\label{5.2}
\end{gather}
where $\mathbf{E}$ is the electric field and $\mathbf{J}$ the current density. Taking the
curl of Ohm's law
\begin{equation*}
\rho\mathbf{J}=\mathbf{E}+\mathbf{v}\times\mathbf{H}
%\label{1.3}
\end{equation*}
where $\rho$ is the resistivity and recalling (\ref{1.2}), (\ref{2.2}),
(\ref{4.2}) and (\ref{5.2}), we have
\begin{equation}
\nabla\cdot(\rho\nabla h)+M \frac{\partial v}{\partial x} =0.
\label{2.3}
\end{equation}
Moreover, in view of (\ref{1.2}) and (\ref{2.2}) the equation of motion
reduces to
\begin{equation}
\nabla\cdot(\eta\nabla v)+M\frac{\partial h}{\partial x}=-k
\label{1.4}
\end{equation}
where $k$ is the constant pressure gradient and $\eta$ the viscosity.
The system of partial differential equations (\ref{2.3}), (\ref{1.4})
has been studied in  \cite{JS1} and \cite{JS2} for its relevance
in applications, as e.g. in electromagnetic
flow-measurements \cite{JS4}. Crucial in this treatment is the
hypothesis of a constant viscosity and resistivity. In this paper we
study a non-linear version of the system (\ref{2.3}), (\ref{1.4})
in which viscosity and resistivity are given functions of the
temperature $\theta$. In practical cases this dependence can be quite strong.
Thus we need to add the energy equation
\begin{equation}
-\nabla\cdot(\kappa\nabla\theta)=\eta|\nabla v|^2+\rho|\nabla h|^2 \label{1.5}
\end{equation}
to the system. In (\ref{1.5}) $\kappa$ is the thermal conductivity, also a
function of the temperature. The first term on the right hand side of
(\ref{1.5}) reflects the viscous attrition and the second the Joule
heating. Let $\Gamma$ be the boundary of $\Omega$. Assuming the walls of the pipe to
be a perfect electrical insulant we have
\begin{equation}
\mathbf{J}\cdot {\bf n}=0\quad \hbox{on } \Gamma.
\label{1.6}
\end{equation}
Moreover, (\ref{5.2}) reads
\begin{equation}
J_x=\frac{\partial h}{\partial y},\quad J_y=-\frac{\partial h}{\partial x},
\label{1.7}
\end{equation}
thus (\ref{1.6}) and (\ref{1.7}) imply that $h$ is constant on $\Gamma$ (with
possibly different values if $\Omega$ is not simply connected). We shall study two different boundary value problems for the system
(\ref{2.3}), (\ref{1.4}), (\ref{1.5}), more precisely a ``Poiseuille''
case $Pb_P$
\begin{gather}
\begin{gathered}
\nabla\cdot(\eta(\theta)\nabla v)+M\frac{\partial h}{\partial
  x}=-k\quad\text{in }\quad \Omega\quad
v=0\quad\text{on }\Gamma,
\end{gathered}\label{1.8}
\\
\begin{gathered}
\nabla\cdot(\rho(\theta)\nabla h)+M\frac{\partial v}{\partial x}=0\quad\text{in }\quad\Omega\quad
h=0\quad\text{on }\Gamma,
\end{gathered}\label{3.8}
\\
\begin{gathered}
-\nabla\cdot(\kappa(\theta)\nabla\theta)=\eta(\theta)|\nabla v|^2+\rho(\theta)|\nabla h|^2\quad\text{in }\Omega \\
\theta=\Theta_b\quad\text{on }\Gamma,
\end{gathered}\label{1.9}
\end{gather}
and a ``Couette'' case in which $\Omega$ is doubly-connected with boundary
consisting of two curves $\Gamma_1$ and $\Gamma_2$ with
$\Gamma=\Gamma_1\cup\Gamma_2$, and $\Gamma_1\cap\Gamma_2=\emptyset$. The external wall
of the pipe, of cross-section $\Gamma_2$, moves with respect to the internal one
with cross-section $\Gamma_1$
with constant velocity $V$ and with absence of pressure gradient. This implies
$k=0$ in (\ref{1.4}). In this way we obtain problem $Pb_C$
\begin{gather*}
\nabla\cdot(\eta(\theta)\nabla v)+M\frac{\partial h}{\partial x}=0\quad\text{in }\Omega \\%\label{1.11}
v=0\quad\text{on }\Gamma_1,\quad v=V\quad\text{on }\Gamma_2,\\ %\label{2.11}
\nabla\cdot(\rho(\theta)\nabla h)+M\frac{\partial v}{\partial x}=0\quad\text{in }\Omega \\ %\label{3.11}
h=0\quad\text{on }\Gamma_1,\quad h=H\quad\text{on }\Gamma_2, \\ %\label{4.11}
-\nabla\cdot(\kappa(\theta)\nabla\theta)=\eta(\theta)|\nabla v|^2+\rho(\theta)|\nabla h|^2\quad\text{in }\Omega \\ %\label{5.11}
\theta=\Theta_b\quad\text{on }\Gamma, %\label{6_11}
\end{gather*}
where $H$ is a given constant. The boundary value of the temperature is
supposed to be the trace of a function $\Theta\in H^2(\Omega)$ harmonic in $\Omega$. Moreover we
assume $\Gamma$ to be of class $\mathcal{C}^2$. In Section 2 we prove, using an elliptic
regularization, that problems $Pb_P$
and $Pb_C$ have at least one weak solution. A result of uniqueness for
 problem $Pb_C$ is presented in Section 3.

\section{Existence and Uniqueness of Weak Solutions}

 We assume $\Gamma$ to be regular (e.g. $C^2$). For later use we recall the following results.

\begin{lemma} \label{lem1}
Let  $a(\mathbf{x})$, $ b(\mathbf{x})\in L^\infty(\Omega)$, $\mathbf{x}=(x,y)$ and
\begin{equation}
a(\mathbf{x})\geq a_0>0,\quad b(\mathbf{x})\geq b_0>0.
\label{1.17}
\end{equation}
Then the system
\begin{gather}
v\in H^1_0(\Omega),\quad \int_\Omega\big[a(\mathbf{x})\nabla v\cdot\nabla\varphi-Mh_x\varphi\big]dX=k\int_\Omega\varphi
dX,\quad \forall \varphi\in H_0^1(\Omega), \label{2.17} \\
h\in H^1_0(\Omega),\quad \int_\Omega\big[b(\mathbf{x})\nabla h\cdot\nabla\psi-Mv_x\psi\big]dX=0,\quad
 \forall \psi\in H^1_0(\Omega),
\label{3.17}
\end{gather}
has one and only one solution. Moreover,
\begin{gather}
\| v\|_{H^1_0(\Omega)}+\| h\|_{H^1_0(\Omega)}\leq C, \label{1.18} \\
\max_\Omega|v|+\max_\Omega|h|\leq C,
\label{2.18}
\end{gather}
where the constant $C$ depends only on $a_0$, $b_0$, $k$, $M$ and $\Omega$.
\end{lemma}

\begin{proof}
The bilinear form
\begin{equation*}
a((v,h),(\varphi,\psi))=\int_\Omega\Big[a(\mathbf{x})\nabla v\cdot\nabla\varphi+b(\mathbf{x})\nabla
 h\cdot\nabla\psi-M\Big(\frac{\partial h}{\partial x}\varphi
 +\frac{\partial v}{\partial  x}\psi\Big)\Big] dX
\end{equation*}
is bounded and coercive in $H^1_0(\Omega)\times H^1_0(\Omega)$, as easily
verified. Therefore, by the Lax-Milgram lemma, the system
(\ref{2.17}), (\ref{3.17}) has
one and only one solution which satisfies (\ref{1.18}).
By standard elliptic
regularity (see \cite{La}), (\ref{2.18}) follows.
\end{proof}

The main difficulty in problem $Pb_P$ lies in the quadratic growth in the
right hand side of equation (\ref{1.9}). However,
from (\ref{1.8}) and (\ref{3.8}) we have
\begin{equation}
\eta(\theta)|\nabla v|^2+\rho(\theta)|\nabla h|^2=\nabla\cdot(h\rho(\theta)\nabla h)+\nabla\cdot(v\eta(\theta)\nabla
v)+Mh\frac{\partial v}{\partial x}+Mv\frac{\partial h}{\partial x}+kv.
\label{1.21}
\end{equation}
This suggests the following weak formulation:
\begin{gather}
v\in H^1_0(\Omega),\quad \int_\Omega\Big[\eta(\theta)\nabla
v\cdot\nabla\varphi -M\frac{\partial h}{\partial x}\varphi\Big]dX=\int_\Omega
k\varphi dX,\quad \forall  \varphi\in H^1_0(\Omega),
\label{1.22}
\\
h\in H^1_0(\Omega),\quad \int_\Omega\Big[\rho(\theta)\nabla h\cdot\nabla\zeta 
-M\frac{\partial v}{\partial  x}\zeta\Big]dX=0,\quad \forall  \zeta\in H^1_0(\Omega),
\label{1.23}
\\
\begin{aligned}
&\theta-\Theta\in H^1_0(\Omega),\quad \int_\Omega\kappa(\theta)\nabla\theta\cdot\nabla\xi dX\\
&=-\int_\Omega h\rho(\theta)\nabla h\cdot\nabla\xi dX
+M\int_\Omega h\frac{\partial v}{\partial x}\xi dX-\int_\Omega v\eta(\theta)\nabla v\cdot\nabla\xi dX\\
&\quad + \int_\Omega v\Big(M\frac{\partial h}{\partial x}+k\Big)\xi dX,\quad
\forall  \xi\in H^1_0(\Omega),
\end{aligned}\label{2.23}
\end{gather}
where we assume $\eta(u)$, $\rho(u)$, $\kappa(u)$ to be continuous and to satisfy
\begin{equation}
\eta_1\geq\eta(u)\geq\eta_0>0,\quad
\rho_1\geq\rho(u)\geq\rho_0>0,\quad
\kappa_1\geq \kappa(u)\geq \kappa_0>0.
\label{ipotesi}
\end{equation}
To prove existence of a weak solution we consider the following sequence of
approximating problems $Pb_\epsilon$:
\begin{gather}
v_\epsilon\in\ H^1_0(\Omega),\quad \int_\Omega\eta(\theta_\epsilon)\nabla v_\epsilon\cdot\nabla\varphi dX-M\int_\Omega\frac{\partial h_\epsilon}{\partial x}\varphi
dX=k\int_\Omega\varphi dX,\quad \forall \varphi\in H^1_0(\Omega)
\label{1.25}
\\
h_\epsilon\in H^1_0(\Omega),\quad \int_\Omega\rho(\theta_\epsilon)\nabla h_\epsilon\cdot\nabla\zeta dX-M\int_\Omega\frac{\partial v_\epsilon}{\partial x}\zeta
dX=0,\forall \zeta\in H^1_0(\Omega)
\label{1.26}
\\
\label{2.26}
\begin{aligned}
&\theta_\epsilon-\Theta\in H^2_0(\Omega),\quad \epsilon\int_\Omega\Delta \theta_\epsilon\Delta\xi dX
+\int_\Omega\kappa(\theta_\epsilon)\nabla \theta_\epsilon\cdot\nabla\xi dX \\
&=-\int_\Omega h_\epsilon\rho(\theta_\epsilon)\nabla h_\epsilon\cdot\nabla\xi dX+
M\int_\Omega h_\epsilon\frac{\partial v_\epsilon}{\partial x}\xi dX-\int_\Omega v_\epsilon\eta(\theta_\epsilon)\nabla v_\epsilon\cdot\nabla\xi dX\\
&\quad +\int_\Omega v_\epsilon\Big(M\frac{\partial h_\epsilon}{\partial x}+k\Big)\xi dX,\quad\forall \xi\in H^2_0(\Omega).
\end{aligned}
\end{gather}

\begin{lemma} \label{lem2}
Let $(v_\epsilon,h_\epsilon,\theta_\epsilon)$ be a solution to $Pb_\epsilon$.
Then the following ``a priori'' estimates hold:
\begin{gather}
\| v_\epsilon\|_{H^1_0(\Omega)}\leq C, \label{1.28} \\
\| h_\epsilon\|_{H^1_0(\Omega)}\leq C, \label{2.28} \\
\| \theta_\epsilon-\Theta|_{H^1_0(\Omega)}\leq C, \label{3.28} \\
\epsilon\|\Delta \theta_\epsilon\|^2_{L^2(\Omega)}\leq C, \label{4.28} \\
\max_\Omega|v_\epsilon|\leq C,\quad  \max_\Omega|h_\epsilon|\leq C
\label{5.28}
\end{gather}
where the C's denote constants, generally different, depending on $\eta_0$,
$\rho_0$, $\kappa_0$, $\Omega$, $M$ and $k$, but not on $\epsilon$
\end{lemma}

\begin{proof}
Setting $\varphi=v_\epsilon$ in (\ref{1.25}) and $\zeta=h_\epsilon$ in (\ref{1.26}), we have
\begin{gather*}
\int_\Omega\eta(\theta_\epsilon)|\nabla v_\epsilon|^2 dX-M\int_\Omega\frac{\partial h_\epsilon}{\partial x} v_\epsilon dX=k\int_\Omega v_\epsilon dX,\\
\int_\Omega\rho(\theta_\epsilon)|\nabla h_\epsilon|^2 dX+M\int_\Omega\frac{\partial h_\epsilon}{\partial x} v_\epsilon dX=0.
\end{gather*}
Adding and using the Poincar\`e inequality we obtain (\ref{1.28})
and (\ref{2.28}) by (\ref{ipotesi}). Applying Lemma \ref{lem1}
 we get (\ref{5.28}).
Choosing $\xi=\theta_\epsilon-\Theta$ in (\ref{2.26}) we have
\begin{align*}
&\epsilon\int_\Omega\Delta \theta_\epsilon\Delta(\theta_\epsilon-\Theta)dX+\int_\Omega\kappa(\theta_\epsilon)\nabla \theta_\epsilon\cdot\nabla(\theta_\epsilon-\Theta)dX\\
&=-\int_\Omega h_\epsilon\rho(\theta_\epsilon)\nabla h_\epsilon\cdot\nabla(\theta_\epsilon-\Theta)dX+
M\int_\Omega h_\epsilon\frac{\partial v_\epsilon}{\partial x}(\theta_\epsilon-\Theta)dX\\
&-\int_\Omega v_\epsilon\eta(\theta_\epsilon)\nabla v_\epsilon\cdot\nabla(\theta_\epsilon-\Theta)dX
+\int_\Omega v_\epsilon\Big(M\frac{\partial h_\epsilon}{\partial x}+k\Big)(\theta_\epsilon-\Theta)dX.
\end{align*}
Using repeatedly the H\"older inequality and (\ref{ipotesi}), ( \ref{1.28}),
(\ref{2.28}) we obtain
\begin{equation}
\epsilon\|\Delta \theta_\epsilon\|^2_{L^2(\Omega)}+\|\nabla \theta_\epsilon\|^2_{L^2(\Omega)}
\leq C\bigl(\|\Delta \theta_\epsilon\|_{L^2(\Omega)}+\|\nabla \theta_\epsilon\|_{L^2(\Omega)}+1\bigl)
\label{1.32}
\end{equation}
from which (\ref{3.28}) and (\ref{4.28}) follow.
\end{proof}

We recall the classical Leray-Schauder fixed point theorem.

\begin{theorem} \label{thm1}
Let $\mathcal{B}$ be a Banach space and $\mathcal{T}$ a continuous and compact mapping
from $\mathcal{B}\times[0,1]$ into $\mathcal{B}$ such that
$\mathcal{T}(w;0)=\bar u$ for all $w\in\mathcal{B}$.
If all solutions of the equation
\begin{equation*}
w=\mathcal{T}(w;\lambda),\quad w\in\mathcal{B},\; \lambda\in [0,1]
\end{equation*}
are bounded in $\mathcal{B}$ by a constant not depending on $\lambda$,
then $\mathcal{T}(w,1)$ has a fixed point in $\mathcal{B}$.
\end{theorem}

\begin{lemma} \label{lem3}
For every $\epsilon>0$ there exists at least one solution to $Pb_\epsilon$.
\end{lemma}

\begin{proof}
We omit the dependence of $\epsilon$ in $v_\epsilon$, $h_\epsilon$ and $\theta_\epsilon$.
Let $\mathcal{B}=H^1_0(\Omega)$ and define
\begin{equation*}
\theta=\mathcal{T}(w,\lambda),\quad \mathcal{T}:\mathcal{B}\times[0,1]\to\mathcal{B}
\end{equation*}
via the linear problem
\begin{gather}
v\in H^1_0(\Omega),\quad \nabla\cdot(\eta(\lambda w)\nabla v)+M\frac{\partial h}{\partial x}=-k, \label{1.35} \\
h\in H^1_0(\Omega),\quad \nabla\cdot(\rho(\lambda w)\nabla h)+M\frac{\partial v}{\partial x}=0, \label{2.35} \\
\begin{aligned}
&\theta-\Theta\in H^2_0(\Omega),\quad \epsilon\Delta\Delta\theta+\nabla\cdot(\kappa(\lambda w)\nabla\theta)\\
&=\nabla\cdot(h\rho(\lambda w)\nabla h)+
Mh\frac{\partial v}{\partial x}+\nabla\cdot(v\eta(\lambda w)\nabla v)+kv+Mv\frac{\partial h}{\partial x}.
\end{aligned} \label{3.35}
\end{gather}

Given $w\in\mathcal{B}$ the system (\ref{1.35}), (\ref{2.35}) is solvable by
Lemma \ref{lem1}. Moreover, the right hand side of (\ref{3.35}) defines a
bounded linear functional in $H^2(\Omega)$. Hence (\ref{3.35})
is solvable with respect to $\theta$ and
the mapping $(w,\lambda)\to\theta$ is well-defined. Let $(\bar v,\bar h,\bar
\theta)$ solve
\begin{gather*}
\bar v\in H^1_0(\Omega),\quad \nabla\cdot(\eta(0)\nabla\bar v)
+M\frac{\partial\bar h}{\partial x}=-k,\\
\bar h\in H^1_0(\Omega),\quad \nabla\cdot(\rho(0)\nabla\bar h)
+M\frac{\partial\bar v}{\partial x}=0,\\
\begin{aligned}
&\bar\theta-\Theta\in H^2_0(\Omega),\quad \epsilon\Delta\Delta\bar \theta+\nabla\cdot(\kappa(0)\nabla\bar \theta)\\
&=\nabla\cdot(\bar h\rho(0)\nabla\bar h)+M\bar h
\frac{\partial\bar v}{\partial x}
+\nabla\cdot(\bar v\eta(0)\nabla\bar v)+k\bar v+M\bar v
\frac{\partial\bar h}{\partial x}.
\end{aligned}
\end{gather*}
We have $\mathcal{T}(w,0)=\bar \theta$ for all $w\in\mathcal{B}$. To prove the continuity of
$\mathcal{T}(w,\lambda)$, suppose $(w_i,\lambda_i)\to(w^*,\lambda^*)$ in
$\mathcal{B}\times[0,1]$ and
\begin{equation*}
\theta_i=\mathcal{T}(w_i,\lambda_i),\quad \theta^*=\mathcal{T}(w^*,\lambda^*).
\end{equation*}
Let $(v_i,h_i)\in H^1_0(\Omega)\times H^1_0(\Omega)$ be solution of the system
\begin{gather}
\int_\Omega\eta(\lambda_iw_i)\nabla v_i\cdot\nabla\varphi dX-M\int_\Omega
\frac{\partial h_i}{\partial x}\varphi
dX=k\int_\Omega\varphi dX,\quad \hbox{for all } \varphi\in H^1_0(\Omega),
\label{1.37} \\
\int_\Omega\rho(\lambda_iw_i)\nabla h_i\cdot\nabla\zeta dX-M\int_\Omega
\frac{\partial v_i}{\partial x}\zeta
dX=0,\quad \hbox{for all } \zeta\in H^1_0(\Omega),
\label{1.38}
\end{gather}
and $(v^*,h^*)$ of the system
\begin{gather}
\int_\Omega\eta(\lambda^*w^*)\nabla v^*\cdot\nabla\varphi dX-M\int_\Omega
\frac{\partial h^*}{\partial x}\varphi
dX=k\int_\Omega\varphi dX,\quad \hbox{for all } \varphi\in H^1_0(\Omega),
\label{2.38} \\
\int_\Omega\rho(\lambda^*w^*)\nabla h_i\cdot\nabla\zeta dX-M\int_\Omega
\frac{\partial v^*}{\partial x}\zeta
dX=0,\quad \hbox{for all } \zeta\in H^1_0(\Omega).
\label{3.38}
\end{gather}
Choosing $\varphi=v_i-v^*$ in (\ref{1.37}) and (\ref{2.38}) and $\zeta=h_i-h^*$ in
(\ref{1.38}) and (\ref{3.38}) we have, after simple calculations,
\begin{equation}
\begin{aligned}
&\| v_i-v^*\|_{H^1_0(\Omega)}^2+\| h_i-h^*\|_{H^1_0(\Omega)}^2\\
&\leq C\Big[\|\eta(w_i\lambda_i)-\eta(\lambda^*w^*)\|_{L^\infty(\Omega)}\| v_i\|_{H^1_0(\Omega)}\|
v_i-v^*\|_{H^1_0(\Omega)}\\
&\quad +\|\rho(w_i\lambda_i)-\rho(\lambda^*w^*)\|_{L^\infty(\Omega)}\| h_i\|_{H^1_0(\Omega)}
\| h_i-h^*\|_{H^1_0(\Omega)}\Big].
\end{aligned}\label{1.40}
\end{equation}
Hence
\begin{equation}
v_i\to v^*,\quad h_i\to h^*\quad\text{in }H^1_0(\Omega).
\label{2.40}
\end{equation}
Let $\theta_i$ and $\theta^*$ be given respectively by
\begin{gather}
\begin{aligned}
&\theta_i-\Theta\in H^2_0(\Omega),\quad \epsilon\int_\Omega\Delta\theta_i\Delta\xi dX
+\int_\Omega\kappa(\theta_i)\nabla\theta_i\cdot\nabla\xi dX\\
&=-\int_\Omega h_i\rho(\lambda_iw_i)\nabla h_i\cdot\nabla\xi dX
+k\int_\Omega v_i\xi dX+M\int_\Omega h_i
\frac{\partial v_i}{\partial x}\xi dX\\
&+M\int_\Omega v_i\frac{\partial h_i}{\partial x}\xi dX
-\int_\Omega v_i\eta(\lambda_iw_i)\nabla v_i\cdot\nabla\xi dX,\quad
  \hbox{for all } \xi\in H^2_0(\Omega),
\end{aligned}\label{1.41}
\\
\begin{aligned}
&\theta^*-\Theta\in H^2_0(\Omega),\quad \epsilon\int_\Omega\Delta\theta^*\Delta\xi dX
+\int_\Omega\kappa(\theta^*)\nabla\theta^*\cdot\nabla\xi dX\\
&=-\int_\Omega h^*\rho(\lambda^*w^*)\nabla h^*\cdot\nabla\xi dX
+k\int_\Omega v^*\xi dX+M\int_\Omega h^*
\frac{\partial v^*}{\partial x}\xi dX\\
&\quad +M\int_\Omega v^*\frac{\partial   h^*}{ \partial x}\xi dX
 -\int_\Omega v^*\eta(\lambda^*w^*)\nabla v^*\cdot\nabla\xi dX,\quad
  \hbox{for all } \xi\in H^2_0(\Omega).
\end{aligned}\label{1.42}
\end{gather}
By difference from (\ref{1.41}) and (\ref{1.42}),  setting
$\xi=\theta_i-\theta^*$ in the resulting equation, using the Poincare inequality and
the Sobolev imbedding theorem we have
\begin{equation}
\begin{aligned}
&\epsilon\|\Delta(\theta_i-\theta^*)\|_{L^2(\Omega)}^2+\|\nabla(\theta_i-\theta^*)\|_{L^2(\Omega)}^2\\
&\leq C\Big[\| h_i-h^*\|_{H^1_0(\Omega)}+\| v_i-v^*\|_{H^1_0(\Omega)}
+\|\rho(\lambda_i w_i)-\rho(\lambda^*w^*)\|_{L^\infty(\Omega)}\\
&\quad +\|\eta(\lambda_i w_i)-\eta(\lambda^*w^*)\|_{L^\infty(\Omega)}\Big]
\|\Delta(\theta_i-\theta^*)\|_{L^2(\Omega)}.
\end{aligned}\label{1.44}
\end{equation}
 From (\ref{1.44}) the continuity of $\mathcal{T}(w,\lambda)$ easily follows.
The mapping $\mathcal{T}(w,\lambda)$ is also compact, since in dimension $2$
bounded subsets of $H^2_0(\Omega)$ are compact in $H^1_0(\Omega)$.
Finally, repeating with minor changes the proof of
 Lemma \ref{lem2}, we can prove that all solutions of the equation
\begin{equation*}
\theta=\mathcal{T}(\theta,\lambda)
\end{equation*}
are bounded in the $\mathcal{B}$-norm by a constant not depending on $\lambda$. Hence
problem $Pb_\epsilon$ has at least one solution by the Leray-Schauder principle.
\end{proof}

\begin{theorem} \label{thm2}
There exists at least one weak solution to problem $Pb_P$.
\end{theorem}

\begin{proof}
By (\ref{1.28}), (\ref{2.28}) and (\ref{3.28}) we can extract from
$\{v_\epsilon\}$, $\{h_\epsilon\}$ and $\{\theta_\epsilon\}$ subsequences (not relabelled)
such that
\begin{equation}
\begin{gathered}
v_\epsilon\to v\quad \hbox{weakly in } H^1_0(\Omega),\quad
h_\epsilon\to h\quad \hbox{weakly  in } H^1_0(\Omega),\\
\theta_\epsilon\to \theta\quad \hbox{weakly in } H^1(\Omega)
\end{gathered}\label{1.46}
\end{equation}
and
\begin{equation}
\begin{gathered}
\theta_\epsilon\to\theta\quad\text{in }L^2(\Omega),\quad
\kappa(\theta_\epsilon)\to\kappa(\theta)\quad\text{in }L^p(\Omega),\\
 \rho(\theta_\epsilon)\to\rho(\theta)\quad\text{in }L^p(\Omega),\quad
 \eta(\theta_\epsilon)\to\eta(\theta)\quad\text{in }L^p(\Omega),\quad 1\leq p<\infty.
\end{gathered}\label{2.46}
\end{equation}
Letting $\epsilon\to 0$ in (\ref{1.25}) and (\ref{1.26}), we have (\ref{1.22})
and (\ref{1.23}). It remains to pass to the limit for $\epsilon\to 0$ in
(\ref{2.26}). By (\ref{4.28}), the first term in the left hand side of
(\ref{2.26}) vanishes when $\epsilon\to 0$. Moreover, by (\ref{1.46}),
(\ref{2.46}),
\begin{equation*}
\int_\Omega\kappa(\theta_\epsilon)\nabla \theta_\epsilon\cdot\nabla\xi dX\to\int_\Omega\kappa(\theta)\nabla\theta\cdot\nabla\xi dX.
\end{equation*}
To pass to the limit in the first term in the right hand side
 of (\ref{2.26}) we write
\begin{equation*}
\int_\Omega\Big[h_\epsilon\rho(\theta_\epsilon)\nabla h_\epsilon-h\rho(\theta)\nabla h\Big]\cdot\nabla\xi dX=I_1+I_2+I_3
\end{equation*}
where
\begin{gather*}
 I_1=\int_\Omega\bigl[(h_\epsilon-h)\rho(\theta_\epsilon)\nabla h_\epsilon\bigl]\cdot\nabla\xi dX, \quad
I_2=\int_\Omega\Big[h\Big(\rho(\theta_\epsilon)-\rho(\theta)\Big)\nabla h_\epsilon\Big]\cdot\nabla\xi dX,\\
I_3=\int_\Omega h\rho(\theta)(\nabla h_\epsilon-\nabla h)\cdot\nabla\xi dX.
\end{gather*}
We have
\begin{gather*}
|I_1|\leq \rho_1\|h_\epsilon-h\|_{L^2(\Omega)}\|\nabla h_\epsilon\|_{L^2(\Omega)}
\|\nabla\xi\|_{L^\infty(\Omega)},\\
|I_2|\leq\|
 h\|_{L^p(\Omega)}\|\rho(\theta_\epsilon)-\rho(\theta)\|_{L^p(\Omega)}\|\nabla h_\epsilon
\|_{L^2(\Omega)}\|\nabla\xi\|_{L^\infty(\Omega)},\quad  1/p+1/q+1/2=1.
\end{gather*}
Moreover, by (\ref{1.46}) and (\ref{2.46}),
\begin{equation*}
I_3=\int_\Omega h\rho(\theta)\bigl(\nabla h_\epsilon-\nabla h)\cdot\nabla\xi dX\to 0,\quad \hbox{as }
\epsilon\to 0.
\end{equation*}
The remaining terms in the right hand side of (\ref{2.26}) can be dealt with
similarly. In the end we obtain (\ref{2.23}). Thus problem
$Pb_P$ has a weak solution.
\end{proof}

\section{The Couette Case}

Only minor changes are needed to prove that also
problem $Pb_C$ has a solution. Uniqueness seems to be, in general, an open
question for both problems $Pb_P$ and $Pb_C$. However, in special cases
existence, non-existence and uniqueness can be proved for problem $Pb_C$, even
suppressing the hypothesis of ellipticity (\ref{ipotesi}).

\begin{theorem} \label{thm3}
Suppose that in problem $Pb_C$:
$M=0$, $\Theta_b=0$, $\kappa(\theta)>0$, $\rho(\theta)>0$, $\eta(\theta)>0$,
$\rho(\theta)=\gamma\eta(\theta)$, $\gamma>0$.
Assume that
\begin{equation}
\int_0^\infty\frac{\kappa(t)}{\rho(t)}dt={\it l}<\infty.
\label{2.51}
\end{equation}
Then, if $1+\frac{\gamma V^2}{H^2}<{\it l}$  problem $Pb_C$ has
one and only one solution. If ${\it l}\leq 1+\frac{\gamma V^2}{ H^2}
$ problem $Pb_C$ has no solution. If, on the contrary,
\begin{equation*}
\int_0^\infty\frac{\kappa(t)}{\rho(t)}dt=\infty,
\end{equation*}
then problem $Pb_C$ has   one and only one solution.
\end{theorem}

 The proof is based on the  transformation
\begin{equation}
\Psi=\frac12 h^2+\frac{\gamma}{2}v^2+\int_0^\theta\frac{\kappa(t)}{\rho(t)}dt,
\label{2.52}
\end{equation}
which gives the equations
\begin{gather}
\nabla\cdot(\rho(\theta)\nabla\Psi)=0, \label{3.52} \\
\nabla\cdot(\rho(\theta)\nabla v)=0, \label{4.52} \\
\nabla\cdot(\rho(\theta)\nabla h)=0,\label{5.52}
\end{gather}
and the boundary conditions
\begin{gather*}
\Psi=0\quad\text{on }\Gamma_1,\quad
\Psi=\frac12 H^2+\frac{\gamma}{2}V^2\quad\text{on }\Gamma_2,
\\
v=0\quad\text{on }\Gamma_1,\quad
v=V\quad\text{on }\Gamma_2, %\label{2.53}
\\
h=0\quad\text{on }\Gamma_1,\quad
h=H\quad\text{on }\Gamma_2. %\label{2.53}
\end{gather*}
The system of the three equations (\ref{3.52}), (\ref{4.52}), (\ref{5.52}),
together with the functional relation (\ref{2.52}), can be reduced,
quite surprisingly, to the linear Dirichlet problem
\begin{equation*}
\Delta\Phi=0\quad\text{in }\Omega,\quad \Phi=0\quad\text{on }\Gamma_1,\quad
 \Phi=1\quad\text{on }\Gamma_2.
%\label{4.53}
\end{equation*}
For more details, we refer the reader to \cite{GC}.

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\bibitem{La} O. A. Ladyzhenskaya, N. N. Ural`tseva;
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Academic Press, New York (1968)

\bibitem{JS1} J. A. Shercliff;
\emph{Steady motion of conducting fluids in pipes under
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\bibitem{JS2} J. A. Shercliff;
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\bibitem{JS4} J. A. Shercliff;
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\end{thebibliography}

\end{document}