# Non-stationary Navier-Stokes Equations with Mixed Boundary Conditions

Tujin Kim

Institute of Mathematics,
Academy of Sciences,
Pyongyang, DPR Korea
e-mail<EMAIL_ADDRESS>
Daomin Cao

Institute of Applied Mathematics,
AMSS, Chinese Academy of Sciences,
Beijing 100190, P. R. China
e-mail<EMAIL_ADDRESS>Partially supported by AMSS in Chinese Academy of
Sciences.Partially supported by NSFC grants (No. 11271354 and No. 11331010)
and Beijing Center for Mathematics and Information Interdisciplinary Sciences.

###### Abstract

In this paper we are concerned with the initial boundary value problem of the
2, 3-D Navier-Stokes equations with mixed boundary conditions including
conditions for velocity, static pressure, stress, rotation and Navier slip
condition together. Under a compatibility condition at the initial instance it
is proved that for the small data there exists a unique solution on the given
interval of time. Also, it is proved that if a solution is given, then there
exists a unique solution for small perturbed data satisfying the compatibility
condition. Our smoothness condition for initial functions in the compatibility
condition is weaker than one in such a previous result.

Mathematics Subject Classification: 35Q30, 35A02, 35A15, 76D03, 76D05

Keywords: Navier-Stokes equations, Pressure, Stress, Rotation, Slip, Mixed
boundary conditions, Mixed problem, Unique existence

##  1 Introduction

For the Navier-Stokes equations

$-\nu\Delta v+(v\cdot\nabla)v+\nabla p=f,\,\,\,\nabla\cdot
v=0\quad\mbox{in}\,\,\Omega\subset R^{l},l=2.3;$

and

$\frac{\partial v}{\partial t}-\nu\Delta v+(v\cdot\nabla)v+\nabla
p=f,\,\,\,\nabla\cdot v=0\quad\mbox{in}\,\,\Omega$

different natural and artificial boundary conditions are considered. For
example on solid walls, homogeneous Dirichlet condition $v=0$ is often used.
On a free surface a Neumann condition $2\nu\varepsilon(v)n-pn=0$ may be
useful. Here and in what follows $\varepsilon(v)$ denotes the so-called strain
tensor with the components
$\varepsilon_{ij}(v)=\frac{1}{2}(\partial_{x_{i}}v_{j}+\partial_{x_{j}}v_{i})$
and $n$ is the outward normal unit vector. For simulations of flows in the
presence of rough boundaries, the Navier slip-with-friction boundary
conditions $v\cdot n=0,\,\,(\nu\varepsilon_{n\tau}(v)+\alpha
v_{\tau})|_{\Gamma_{5}}=0$ is also used, where $\varepsilon_{n\tau}(v)$ and
$v_{\tau}$ are, respectively, the tangent components of $\varepsilon(v)n$ and
$v$. Combination of the condition $v_{n}=0$ and the tangential component of
the friction (slip condition for uncovered fluid surfaces) or the condition
$v_{\tau}=0$ and the normal component of the friction (condition for in/out-
stream surfaces) are frequently used. At the outlet of a channel “do nothing”
condition $\nu\frac{\partial v}{\partial n}-pn=0$, i.e. the outlet boundary
condition, is also used. Rotation boundary condition has been fairly
extensively studied over the past several years. Also, for inlet or outlet of
flow one deals with the static pressure $p$ or total pressure (Bernoulli’s
pressure) $\frac{1}{2}|v|^{2}+p$. For papers dealing with the problems
mentioned above one can refer to Introduction of [13].

In practice we deal with mixture of some kind of boundary conditions. For a
channel flow a mixture of Dirichlet condition $v=0$ on the wall and “do
nothing” condition on the outlet is used. But for a channel flow with a rough
boundary surface a mixture of Dirichlet condition, the Navier slip-with-
friction boundary condition and “do nothing” condition may be used. For a flow
in a vessel with in/outlet a mixture of Dirichlet condition $v=0$ on the wall
and pressure conditions on the inlet/outlet is used. But for the flow in a
vessel with in/outlet and a free surface a mixture of Dirichlet condition, a
Neumann condition $2\nu\varepsilon(v)n-pn=0$ and pressure conditions may be
used.

There are vast literatures for the Stokes and Navier-Stokes problems with
mixed boundary conditions and several variational formulations are used for
them, but two possible examples above are not considered except [13]. With
exception [13] mixtures of boundary conditions for Navier-Stokes equations may
be divided into three groups according to what bilinear form is used for a
variational formulation (for more concrete one can refer to Introduction of
[13]).

To include more different boundary conditions together, in [13] the relations
among strain, rotation, normal derivative of vector field and shape of
boundary surface are obtained and applied to the stationary and non-stationary
Navier-Stokes problems with mixture of seven kinds of boundary conditions.
However, for the non-stationary Navier-Stokes problems we only were concerned
with a mixed boundary condition including total pressure (not static
pressure), total stress (not stress) and so on. Thus, in this paper we will
study the non-stationary problems with a mixed boundary condition including
static pressure (not total pressure), stress (not total stress) and so on.

On the other hand, when one of static pressure (instead of total pressure),
stress (instead of total stress) or the outlet boundary condition is given on
a portion of boundary, for the initial boundary value problems of the Navier-
Stokes equations existence of a unique local-in-time solution and a unique
solution on a given interval for small given data (in what follows we call it
a solution for small data) are proved. In the mathematical point of view the
main difficulty of such problems consists in the fact that in priori
estimation the inertial term is not canceled, and in the mechanical point of
view it is in the fact that the kinetic energy of fluid is not controlled by
the data and uncontrolled “backward flow” can take place at the portion of
boundary(cf. preface in [2]).

The Navier-Stokes equations with mixture of Dirichlet condition and stress
condition are studied. In [15] under smoothness condition and a compatibility
condition of data at the initial instance existence of a unique local-in-time
solution to the 3-D Navier-Stokes equations is studied. In [2] for the Navier-
Stokes equations on the polyhedral domain with mixture of Dirichlet condition,
Navier slip condition and stress condition a local-in-time solution is
studied. Here smoothness of solutions to the corresponding steady Stokes
problem is used essentially.

The Navier-Stokes equations and the Boussinesq equations with mixture of
Dirichlet condition and the outlet boundary condition are studied. For 2-D
Navier-Stokes equations a local-in-time solution in [7] and a solution for
small data in [8] are studied. Here also smoothness of solutions to the
corresponding 2-D steady linear problem is important. For the Boussinesq
systems a local-in-time strong solutions in [3] on 2-D channel and in [6] on
3-D channel are studied. Here smoothness of solutions to the corresponding
steady linear problems, respectively, in [5] and [1] is the key. In [14] it is
proved that if under a compatibility condition at initial instance there
exists a unique solution, then so does for small perturbed data. This result
shows that under the compatibility condition there exists a unique solution
for small data. In [20] for the Boussinesq equations it is proved that under a
compatibility condition there exists a unique local-in-time solution, which is
similar to the result in [15]. Smoothness of initial function in the
compatibility condition of [14] is stronger than one in [15] and [20].

The 2-D Navier-Stokes equations with mixture of Dirichlet condition and
pressure is studied. In [16] existence of a unique solution for small data is
proved.

The Navier-Stokes equations with mixture of Dirichlet condition, outlet
condition and tangent stress condition is studied. In [4] existence of a
unique solution for small perturbation data of the given solution is studied.
Here also smoothness of solutions to the corresponding steady linear problem
is the key.

In the present paper as a continuation of [13], we are concerned with the non-
steady Navier-Stokes equations with mixed boundary conditions involving
conditions for Dirichlet, static pressure, rotation, stress and normal
derivative of velocity together. Owing to the relations among strain,
rotation, normal derivative of velocity and shape of boundary surface obtained
in [13] (Theorems 2.1, 2.2), we can consider all these boundary conditions
together.

In general, the solution of the Stokes problem with mixed boundary conditions
has singularities on the intersections of surfaces for different boundary
conditions and the leading singular exponent of the solution is a function of
the intersection angle (cf. [19]). For the problem with Dirichlet condition
and “do nothing” condition if the intersection angle is $\pi/2,$ then under
some conditions for data the solution belongs to $H^{2}(\Omega)$ (cf. [5]).
For the problem with Dirichlet condition and stress conditions, for similar
results refer to subsection 5.5 of [17] and section 10.3 of [18]. In our case
the boundary conditions are more complicated than others, and there is no
result for smoothness of solutions to the corresponding steady linear
problems. Thus, we prove existence of a unique weak solution for small data
under a compatibility condition at initial instance. We also prove that if a
solution smooth as in [14] is given, then under the compatibility condition
for the small perturbed data there exists a unique solution.

We are concerned with two problems distinguished according to boundary
conditions. Using relations among strain, rotation, normal derivative of
vector field and shape of boundary surface, which are obtained in [13], we
reflect all these boundary conditions into variational formulations for
problems.

This paper consists of 5 sections. In the end of this section the method in
this paper is compared with another one.

In Section 2, the formulations of problems and some results for definitions of
weak solutions are given. According to bilinear forms used for variational
formulations for problems, the involved boundary conditions are slightly
different. Thus, difference between our two problems is explained (Remark
2.1).

In Section 3, first, for the Navier-Stokes problems with seven kinds of
boundary conditions a variational formulation, which is based on the bilinear
form

(1.1)
$a(v,u)=2\Sigma_{i,j}(\varepsilon_{ij}(v),\varepsilon_{ij}(u))_{L_{2}(\Omega)}\quad\mbox{for}\,\,v,u\in\textbf{H}^{1}(\Omega),$

is given. Next, by a transformation of the unknown function, the problem is
reduced to an equivalent problem in which the linear main operator is positive
definite. Then, studying properties of linear operator differential equations
and using a local diffeomorphism theorem of nonlinear operator, we prove that
under a compatibility condition similar to one in [14], [15], [20] there
exists a unique solution for small data (Theorem 3.8).

In Section 4 for the Navier-Stokes problems with six kinds of boundary
conditions, which is a little different from one in Section 3, a variational
formulation based on the bilinear form

(1.2) $a(v,u)=(\nabla v,\nabla
u)_{\textbf{L}_{2}(\Omega)}\quad\mbox{for}\,\,v,u\in\textbf{H}^{1}(\Omega)$

is given. Also, by a transformation of the unknown function, the problem is
reduced to another equivalent problem in which the linear main operator is
positive definite. The result similar to one in Section 3 is obtained (Theorem
4.2).

Section 5 is considered in comparison with [14] rather than practical models.
Existence of a unique solution for the small data perturbed from a given
solutions is proved under a compatibility condition (Theorem 5.7).

The compatibility conditions in Sections 3, 4 and 5 are similar to one in
[15], [20] and [14]. In point of view of smoothness of the initial functions,
the conditions are the same with one in [15], [20] concerning with local-in-
time solutions (cf. Remark 3.4), but weaker than one in [14] concerning with
solutions for small data as our case (cf. Remarks 4.2, 5.2). In [14] the main
results for the nonlinear problem as perturbation of a linear problem is
obtained by a local diffeomorphism theorem relying on the properties of the
corresponding linear problem, and so is it in our paper.

Then, let us consider why smoothness of the initial functions in our
compatibility conditions is weaker than one in [14].

Let $\textbf{H}^{k}=(W^{k}_{2}(\Omega))^{l}$ be a Sobolev spaces on $\Omega$
with dimension $l,$ $V$ be a divergence-free subspace of $\textbf{H}^{1}$
satisfying appropriate boundary conditions, $H$ \- the closure of $V$ in
$(L^{2}(\Omega))^{l}$, $V^{*}$\- the adjoint space of $V$,
$V^{r_{0}}(\Omega)=V\cap\textbf{H}^{r_{0}}(\Omega),$ where $r_{0}>l/2,$
$\mathscr{X}=\\{w\in L_{2}(0,T;V);w^{\prime}\in
L_{2}(0,T;V),w^{\prime\prime}\in L_{2}(0,T;V^{*})\\},\mathscr{Y}=\\{w\in
L_{2}(0,T;V^{*});w^{\prime}\in L_{2}(0,T;V^{*})\\}$ and $A:V\rightarrow
V^{*}$\- the Stokes operator.

Considering a linear problem

$\left\\{\begin{aligned} &u^{\prime}(t)+Au(t)=f(t),\\\
&u(0)=\varphi,\end{aligned}\right.$

in [14] the author proved the fact that a map $u\rightarrow\\{u(0),Lu\equiv
u^{\prime}+Au\\}$ is linear continuous one-to-one from
${\mathcal{X}}=\\{u\in\mathscr{X}:u(0)\in V^{r_{0}}(\Omega)\\}$ onto
${\mathcal{Y}}=\\{[\varphi,h]:\varphi\in
V^{r_{0}}(\Omega),h\in\mathscr{Y},h(0)-A\varphi\in H\\}$ (Theorem 3.1 in
[14]). Then, starting from this fact, the author studied a nonlinear problem

(1.3) $\left\\{\begin{aligned} &u^{\prime}(t)+(A+B)u(t)=f(t),\\\
&u(0)=\varphi,\end{aligned}\right.$

where $B:V\rightarrow V^{*}$ is defined by
$\left<Bu,v\right>=\left<(u\cdot\nabla)u,v\right>$ for $u,v\in V$. To this
end, it was proved that the inverse of a nonlinear map
$u\rightarrow\\{u(0),\tilde{L}u\equiv u^{\prime}+(A+B)u\\}$ is one-to-one from
a neighborhood of $0_{\mathcal{Y}}$ onto a neighborhood of $0_{\mathcal{X}}.$
From this fact the author obtained that under the compatibility condition
$f(0)-A\varphi\in H,\varphi\in V^{r_{0}}(\Omega)$ and smallness of data, there
exists a unique solution to (1.3).

However, we prove that for a modified operator $A$ a map
$u\rightarrow\\{u^{\prime}(0),Lu\equiv u^{\prime}+Au\\}$ is linear continuous
one-to-one from $\mathscr{X}$ onto $H\times\mathscr{Y}$ (Lemmas 3.3, 4.3).
Then using this fact, we prove that for a modified operator $B(t)$ the inverse
of a nonlinear map $u\rightarrow\\{u^{\prime}(0),\tilde{L}u\equiv
u^{\prime}+(A+B(t))u\\}$ is one-to-one from a neighborhood of
$0_{H\times\mathscr{Y}}$ onto a neighborhood of $0_{\mathscr{X}}.$ By this
under the compatibility condition $f(0)-A\varphi-B(0)\varphi\in H$ without
$\varphi\in V^{r_{0}}(\Omega)$ and smallness of data, we get existence of
unique solution to (1.3) (Theorems 3.8, 4.2, 5.7, 5.8). Since
$B(0)\varphi=(\varphi\cdot\nabla)\varphi,$ for $\varphi\in V^{l/2}(\Omega)$ we
get $B(0)\varphi\in H$, and so our condition is weaker than one in [14].

##  2 Problems and preliminaries

Throughout this paper we will use the following notation.

Let $\Omega$ be a connected bounded open subset of $R^{l},\,l=2,3.$
$\partial\Omega\in C^{0,1}$,
$\partial\Omega=\cup_{i=1}^{N}\overline{\Gamma}_{i}$,
$\Gamma_{i}\cap\Gamma_{j}=\varnothing$ for $i\neq j$, $\Gamma_{i}\in C^{2}$
for $i=2,3$. For Problems I and II stated below we assume, respectively,
$\Gamma_{7}\in C^{2}$ and $\Gamma_{5}\in C^{2}$. Let $n(x)$ and $\tau(x)$ be,
respectively, outward normal and tangent unit vectors at $x$ in
$\partial\Omega$. When $X$ is a Banach space, $\mathbf{X}=X^{l}$ and
$\mathbf{X}^{*}$ is the dual of $\mathbf{X}$. Let $W^{k}_{\alpha}(\Omega)$ be
Sobolev spaces, $H^{k}(\Omega)=W^{k}_{2}(\Omega)$, and so
$\mathbf{H}^{1}(\Omega)=\\{H^{1}(\Omega)\\}^{l}$. $Q=\Omega\times(0,T)$,
$\Sigma_{i}=\Gamma_{i}\times(0,T)$, $0<T<\infty$.

An inner product and a norm in the space $\mathbf{L}_{2}(\Omega)$ are denoted,
respectively, by $(\cdot\,,\cdot)$ and $\|\cdot\|$; and
$\langle\cdot\,,\cdot\rangle$ means the duality pairing between a Sobolev
space $X$ and its dual one. Also, $(\cdot\,,\cdot)_{\Gamma_{i}}$ is an inner
product in the $\mathbf{L}_{2}(\Gamma_{i})$ or $L_{2}(\Gamma_{i})$; and
$\langle\cdot\,,\cdot\rangle_{\Gamma_{i}}$ means the duality pairing between
$\mathbf{H}^{\frac{1}{2}}(\Gamma_{i})$ and
$\mathbf{H}^{-\frac{1}{2}}(\Gamma_{i})$ or between
$H^{\frac{1}{2}}(\Gamma_{i})$ and $H^{-\frac{1}{2}}(\Gamma_{i})$. The inner
product and norms in $R^{l}$, respectively, are denoted by
$(\cdot\,,\cdot)_{R^{l}}$ and $|\cdot|$. Sometimes $a\cdot b$ is used for
inner product in $R^{l}$ between $a$ and $b$. When $\mathbf{X}$ is a Banach
space, the zero element of $\mathbf{X}$ is denoted by $0_{\mathbf{X}}$ and
$\mathscr{O}_{M}(0_{\mathbf{X}})$ means $M$-neighborhood of $0_{\mathbf{X}}$.

In this paper for the Navier-Stokes problem

(2.1) $\left\\{\begin{aligned} &\frac{\partial v}{\partial t}-\nu\Delta
v+(v\cdot\nabla)v+\nabla p=f,\,\,\,\quad\mbox{in}\,\,Q,\\\ &\nabla\cdot
v=0,\\\ &v(0)=v_{0}\end{aligned}\right.$

we are concerned with the problems I and II, which are distinguished according
to boundary conditions. Problem I is one with the boundary conditions

(2.2) $\displaystyle(1)\quad v|_{\Gamma_{1}}=h_{1},$ $\displaystyle(2)\quad
v_{\tau}|_{\Gamma_{2}}=0,\,\,-p|_{\Gamma_{2}}=\phi_{2},$
$\displaystyle(3)\quad v_{n}|_{\Gamma_{3}}=0,\,\,\mbox{rot}\,v\times
n|_{\Gamma_{3}}=\phi_{3}/\nu,$ $\displaystyle(4)\quad
v_{\tau}|_{\Gamma_{4}}=h_{4},\,\,(-p+2\nu\varepsilon_{nn}(v))|_{\Gamma_{4}}=\phi_{4},$
$\displaystyle(5)\quad
v_{n}|_{\Gamma_{5}}=h_{5},\,\,2(\nu\varepsilon_{n\tau}(v)+\alpha
v_{\tau})|_{\Gamma_{5}}=\phi_{5},$
$\displaystyle(6)\quad(-pn+2\nu\varepsilon_{n}(v))|_{\Gamma_{6}}=\phi_{6},$
$\displaystyle(7)\quad v_{\tau}|_{\Gamma_{7}}=0,\,(-p+\nu\frac{\partial
v}{\partial n}\cdot n)|_{\Gamma_{7}}=\phi_{7},$

and Problem II is one with the conditions

(2.3) $\displaystyle(1)\quad v|_{\Gamma_{1}}=h_{1},$ $\displaystyle(2)\quad
v_{\tau}|_{\Gamma_{2}}=0,\,\,-p|_{\Gamma_{2}}=\phi_{2},$
$\displaystyle(3)\quad v_{n}|_{\Gamma_{3}}=0,\,\,\mbox{rot}\,v\times
n|_{\Gamma_{3}}=\phi_{3}/\nu,$ $\displaystyle(4)\quad
v_{\tau}|_{\Gamma_{4}}=h_{4},\,\,(-p+2\nu\varepsilon_{nn}(v))|_{\Gamma_{4}}=\phi_{4},$
$\displaystyle(5)\quad
v_{n}|_{\Gamma_{5}}=h_{5},\,\,2(\nu\varepsilon_{n\tau}(v)+\alpha
v_{\tau})|_{\Gamma_{5}}=\phi_{5},$
$\displaystyle(6)\quad(-pn+\nu\frac{\partial v}{\partial
n})|_{\Gamma_{7}}=\phi_{7},$

together $\Gamma_{6}=\varnothing$. Here and in what follows $u_{n}=u\cdot n$,
$u_{\tau}=u-(u\cdot n)n$, $\varepsilon(v)=\\{\varepsilon_{ij}(v)\\},$
$\varepsilon_{n}(v)=\varepsilon(v)n$,
$\varepsilon_{nn}(v)=(\varepsilon(v)n,n)_{R^{l}}$,
$\varepsilon_{n\tau}(v)=\varepsilon(v)n-\varepsilon_{nn}(v)n$, and
$h_{i},\,\phi_{i}$ are functions or vector functions of $x,t$ defined on
$\Gamma_{i}\times(0,T)$.

###### Remark 2.1

The condition (6) of (2.3) (with $\phi_{7}=0$) is “do nothing” condition, but
(7) of (2.2) (with $\phi_{7}=0$) is rather different from “do nothing”
condition and we can not unify two problems.

First, let us consider why (7) of (2.2) is not changed with (6) of (2.3). In
Section 3 relying on the bilinear form (1.1) and integrating by parts
$(-\nu\Delta v+\nabla p,u)$, we get boundary integral
$(-2\nu(\varepsilon(v)n,u)_{\partial\Omega}+(p,u\cdot n)_{\partial\Omega}.$
Then, in order to reflect the boundary conditions into Formulation 3.1, using
$v_{\tau}=0$ or $v_{n}=0$ and applying Theorems 2.1 or 2.2, we transform the
boundary integrals on $\Gamma_{i},i=2,3,7.$ (cf. (3.1)-(3.3)). Concretely,
under conditions $v_{\tau}|_{\Gamma_{7}}=0$ we have

(2.4) $\displaystyle\big{<}-pn+\nu\frac{\partial v}{\partial
n},u\big{>}_{\Gamma_{7}}\quad\forall u\,\,\mbox{with}\,\,u_{\tau}=0.$

Usually, $v_{\tau}=0$ does not imply $\frac{\partial v}{\partial
n}\cdot\tau=0,$ but by virtue of the conditions $u_{\tau}=0$ and (7) of (2.2)
we have

(2.5) $\displaystyle\big{<}-pn+\nu\frac{\partial v}{\partial
n},u\big{>}_{\Gamma_{7}}=\big{<}-p+\nu\frac{\partial v}{\partial
n}n,u_{n}\big{>}_{\Gamma_{7}}=\left<\phi_{7},u_{n}\right>_{\Gamma_{7}}\quad\forall
u\,\,\mbox{with}\,\,u_{\tau}=0.$

Thus, substituting $\big{<}-pn+\nu\frac{\partial v}{\partial
n},u\big{>}_{\Gamma_{7}}$ with $\left<\phi_{7},u_{n}\right>_{\Gamma_{7}}$, we
reflect the boundary condition (7) of (2.2) into Formulation 3.1.

Changing (7) of (2.2) by $(-pn+\nu\frac{\partial v}{\partial
n})|_{\Gamma_{7}}=\phi_{7}$ with a vector $\phi_{7}$ and substituting
$\big{<}-pn+\nu\frac{\partial v}{\partial n},u\big{>}_{\Gamma_{7}}$ with
$\left<\phi_{7},u\right>_{\Gamma_{7}}$, we can come to a formal variational
formulation. But when a solution $v$ is smooth enough, trying to convert from
the formal variational formulation to the original problem, we come to

(2.6) $(-pn+\nu\frac{\partial v}{\partial
n},u)_{\Gamma_{7}}=\left<\phi_{7},u\right>_{\Gamma_{7}}\quad\forall
u,u_{\tau}=0$

on $\Gamma_{8}.$ If we have (2.6) without $u_{\tau}=0$, then from (2.6) we can
get $-pn+\nu\frac{\partial v}{\partial n}=\phi_{7}$ on $\Gamma_{8}.$ But owing
to $u_{\tau}=0$ we only get $(-pn+\nu\frac{\partial v}{\partial
n},n)_{\Gamma_{7}}=<\phi_{7},n>_{\Gamma_{7}}$. This shows that the formal
variational formulation is not equivalent to the original condition on
$\Gamma_{7}$ and equivalent to $(-p+\nu\frac{\partial v}{\partial
n}n)|_{\Gamma_{7}}=\phi_{7}\cdot n.$ (Thus, (7) of (3.3) in [13] was
corrected. See Erratum to: [13].)

Similarly, relying on the form (1.2), we can reflect “do nothing” condition
into Formulation 4.1, but can not do (6) of (2.2). (cf. (4.1)-(4.3)).
Therefore, two problems are not unified.

“Do nothing” boundary condition results from variational formulation based on
(1.2) and does not have a real physical meaning, but is rather used in
truncating large physical domains to smaller computational domains by assuming
parallel flow. If the flow is parallel in a near the boundary, then (7) of
(2.2) is same with “do nothing” condition.

For variational formulations of Problems I, II we need the following.

Let $\Gamma$ be a surface (curve for $l=2$) of $C^{2}$ and $v$ be a vector
field of $C^{2}$ on a domain of $R^{l}$ near $\Gamma$. In what follows the
surface is a piece of boundary of 3-D or 2-D bounded connected domains, and so
we can assume the surface is oriented.

###### Theorem 2.1

(Theorem 2.1 in [13]) Suppose that $v\cdot n|_{\Gamma}=0$. Then, on the
surface $\Gamma$ the followings hold.

(2.7)
$\left(\varepsilon(v)n,\tau\right)_{R^{l}}=\frac{1}{2}(\mbox{rot}\,v\times
n,\tau)_{R^{l}}-(S\tilde{v},\tilde{\tau})_{R^{l-1}},$ (2.8)
$(\mbox{rot}\,v\times n,\tau)_{R^{l}}=\left(\frac{\partial v}{\partial
n},\tau\right)_{R^{l}}+(S\tilde{v},\tilde{\tau})_{R^{l-1}},$ (2.9)
$\left(\varepsilon(v)n,\tau\right)_{R^{l}}=\frac{1}{2}\left(\frac{\partial
v}{\partial
n},\tau\right)_{R^{l}}-\frac{1}{2}(S\tilde{v},\tilde{\tau})_{R^{l-1}},$

where $S$ is the shape operator of the surface $\Gamma$ for $l=3$, i.e.

$\displaystyle\hskip 113.81102ptS=\begin{pmatrix}L&K\\\ M&N\end{pmatrix},$
$\displaystyle L=\left(e_{1},\frac{\partial n}{\partial
e_{1}}\right)_{R^{l}},\,\,K=\left(e_{2},\frac{\partial n}{\partial
e_{1}}\right)_{R^{l}},\,\,M=\left(e_{1},\frac{\partial n}{\partial
e_{2}}\right)_{R^{l}},\,\,N=\left(e_{2},\frac{\partial n}{\partial
e_{2}}\right)_{R^{l}},$

and the curvature of $\Gamma$ for $l=2$. Here $e_{i}$ are the unit vectors in
a local curvilinear coordinates on $\Gamma$ and $\tilde{v},\tilde{\tau}$ are
expressions of the vectors $v,\tau$ in the coordinate system.

###### Theorem 2.2

(Theorem 2.2 in [13]) If $v_{\tau}|_{\Gamma}=0$ and $\text{div}\,v=0$, then on
the surface $\Gamma$ the following holds.

$\left(\varepsilon(v)n,n\right)_{R^{l}}=\left(\frac{\partial v}{\partial
n},n\right)_{R^{l}}=-(k(x)v,n)_{R^{l}}$

where $k(x)=\mbox{div}\,n(x)$.

###### Remark 2.2

(cf. [13]) $k(x)=\mbox{div}\,n(x)=Tr(S(x))=2\times\mbox{mean curvature}.$

If $\Gamma$ is a piece of $\partial\Omega$, then since $\partial\Omega\in
C^{0,1}$ and $\Gamma\in C^{2}$, elements of $S$ belong to $C(\bar{\Gamma})$
and so does $k(x)$.

##  3 Existence of a unique solution to problem I

We use the following notation.

$\mathbf{V}=\\{u\in\mathbf{H}^{1}(\Omega):\mbox{div}\,u=0,\,u|_{\Gamma_{1}}=0,\,u_{\tau}|_{\Gamma_{2}\cup\Gamma_{4}\cup\Gamma_{7}}=0,\,u_{n}|_{\Gamma_{3}\cup\Gamma_{5}}=0\\}$
and $\mathbf{V}_{\Gamma
237}(\Omega)=\\{u\in\mathbf{H}^{1}(\Omega):\mbox{div}\,u=0,\,u_{\tau}|_{\Gamma_{2}\cup\Gamma_{7}}=0,\,u_{n}|_{\Gamma_{3}}=0\\}.$
Denote by $H$ the completion of $\mathbf{V}$ in the space
$\textbf{L}_{2}(\Omega)$. Through this paper
$\widetilde{\textbf{V}}=\\{u\in\mathbf{H}^{1}(\Omega):\mbox{div}\,u=0\\}$.

By Theorems 2.1 and 2.2 we have that for
$v\in\mathbf{H}^{2}(\Omega)\cap\mathbf{V}_{\Gamma 237}(\Omega)$ and
$u\in\mathbf{V}$

(3.1) $\displaystyle-(\Delta v,u)$
$\displaystyle=2(\varepsilon(v),\varepsilon(u))-2(\varepsilon(v)n,u)_{\cup_{i=2}^{7}\Gamma_{i}}$
$\displaystyle=2(\varepsilon(v),\varepsilon(u))+2(k(x)v,u)_{\Gamma_{2}}-(\mbox{rot}\,v\times
n,u)_{\Gamma_{3}}+2(S\tilde{v},\tilde{u})_{\Gamma_{3}}$
$\displaystyle\quad-2(\varepsilon_{n}(v),u)_{\cup_{i=4}^{7}\Gamma_{i}}$
$\displaystyle=2(\varepsilon(v),\varepsilon(u))+2(k(x)v,u)_{\Gamma_{2}}-(\mbox{rot}\,v\times
u,u)_{\Gamma_{3}}+2(S\tilde{v},\tilde{u})_{\Gamma_{3}}$
$\displaystyle\quad-2(\varepsilon_{nn}(v),u\cdot
n)_{\Gamma_{4}}-2(\varepsilon_{n\tau}(v),u)_{\Gamma_{5}}-2(\varepsilon_{n}(v),u)_{\Gamma_{6}}$
$\displaystyle\quad-\left(\frac{\partial v}{\partial
n},u\right)_{\Gamma_{7}}+(k(x)v,u)_{\Gamma_{7}}.$

Also, for $p\in H^{1}(\Omega)$ and $u\in\mathbf{V}$ we have

(3.2) $\displaystyle(\nabla p,u)$ $\displaystyle=(p,u\cdot
n)_{\cup_{i=2}^{7}\Gamma_{i}}=(p,u\cdot n)_{\Gamma_{2}}+(p,u\cdot
n)_{\Gamma_{4}}+(pn,u)_{\Gamma_{6}\cup\Gamma_{7}},$

where the fact that $u_{n}|_{\Gamma_{3}\cup\Gamma_{5}}=0$ was used.

Let

$\displaystyle\mathscr{X}=\\{w\in L_{2}(0,T;\textbf{V});w^{\prime}\in
L_{2}(0,T;\textbf{V}),w^{\prime\prime}\in L_{2}(0,T;\textbf{V}^{*})\\},$
$\displaystyle\hskip
28.45274pt\|w\|_{\mathscr{X}}=\|w\|_{L_{2}(0,T;\textbf{V})}+\|w^{\prime}\|_{L_{2}(0,T;\textbf{V})}+\|w^{\prime\prime}\|_{L_{2}(0,T;\textbf{V}^{*})},$
$\displaystyle\mathscr{Y}=\\{w\in L_{2}(0,T;\textbf{V}^{*});w^{\prime}\in
L_{2}(0,T;\textbf{V}^{*})\\},$ $\displaystyle\hskip
28.45274pt\|w\|_{\mathscr{Y}}=\|w\|_{L_{2}(0,T;\textbf{V}^{*})}+\|w^{\prime}\|_{L_{2}(0,T;\textbf{V}^{*})},$
$\displaystyle\mathscr{W}=\\{w\in
L_{2}(0,T;\widetilde{\textbf{V}});w^{\prime}\in
L_{2}(0,T;\widetilde{\textbf{V}}),w^{\prime\prime}\in
L_{2}(0,T;{\widetilde{\textbf{V}}}^{*})\\},$ $\displaystyle\hskip
28.45274pt\|w\|_{\mathscr{W}}=\|w\|_{L_{2}(0,T;\widetilde{\textbf{V}})}+\|w^{\prime}\|_{L_{2}(0,T;\widetilde{\textbf{V}})}+\|w^{\prime\prime}\|_{L_{2}(0,T;{\widetilde{\textbf{V}}}^{*})}.$

Here and in what follows $w^{\prime}$ means the derivative of $w(t)$ with
respect to $t$.

For Problem I, we use the following assumptions.

###### Assumption 3.1

$f,\,f^{\prime}\in
L_{2}(0,T;\mathbf{V}^{*}),\,\,\phi_{i},\,\phi_{i}^{\prime}\in
L_{2}(0,T;H^{-\frac{1}{2}}(\Gamma_{i})),\,i=2,4,7,$
$\phi_{i},\,\phi_{i}^{\prime}\in L_{2}(0,T;$
$\mathbf{H}^{-\frac{1}{2}}(\Gamma_{i})),i=3,5,6,\,\,\alpha_{ij}\in
L_{\infty}(\Gamma_{5})$, where $\alpha_{ij}$ are components of the matrix
$\alpha$, and $\Gamma_{1}\neq\emptyset$.

###### Assumption 3.2

There exists a function $U\in\mathscr{W}$ such that

$\displaystyle\mbox{div}\,U=0,\,U|_{\Gamma_{1}}=h_{1},\,U_{\tau}|_{\Gamma_{2}\cup\Gamma_{7}}=0,\,U_{n}|_{\Gamma_{3}}=0,\,U_{\tau}|_{\Gamma_{4}}=h_{4},\,U_{n}|_{\Gamma_{5}}=h_{5}.$

Also, $U(0,x)-v_{0}\in\textbf{V}$.

###### Remark 3.1

In practical situations $h_{4},h_{5}=0,$ and in the cases if for every fixed
$t$ $h_{1}(t,x)\in H^{\frac{1}{2}}_{00}(\Gamma_{1}),$
$\int_{\Gamma_{1}}h_{1}(t,x)\cdot n\,dx=0$ and
$\|h_{1}(t,x)\|_{H^{\frac{1}{2}}(\Gamma_{1})}$ is smooth enough with respect
to $t$, then there exists such a function $U$.

Taking (3.1) and (3.2) into account, we get the following variational
formulation for Problem I:

###### Formulation 3.1

Find $v$ such that

(3.3) $\displaystyle v-U\in L_{2}(0,T;\mathbf{V}),$ $\displaystyle
v(0)=v_{0},$ $\displaystyle\langle
v^{\prime},u\rangle+2\nu(\varepsilon(v),\varepsilon(u))+\langle(v\cdot\nabla)v,u\rangle+2\nu(k(x)v,u)_{\Gamma_{2}}$
$\displaystyle\hskip
28.45274pt+2\nu(S\tilde{v},\tilde{u})_{\Gamma_{3}}+2(\alpha(x)v,u)_{\Gamma_{5}}+\nu(k(x)v,u)_{\Gamma_{7}}$
$\displaystyle\hskip 28.45274pt=\langle
f,u\rangle+\sum_{i=2,4,7}\langle\phi_{i},u_{n}\rangle_{\Gamma_{i}}+\sum_{i=3,5,6}\langle\phi_{i},u\rangle_{\Gamma_{i}}\quad\mbox{for
all}\,\,u\in\textbf{V}.$

Taking Assumption 3.2 into account, put $v=\overline{z}+U$. Then, we have the
following problem equivalent to Formulation 3.1:

Find $\overline{z}$ such that

(3.4) $\displaystyle\overline{z}\in L_{2}(0,T;\mathbf{V}),$
$\displaystyle\overline{z}(0)=\overline{z}_{0}\equiv v_{0}-U(0)\in\mathbf{V},$
$\displaystyle\langle\overline{z}^{\prime},u\rangle+2\nu(\varepsilon(\overline{z}),\varepsilon(u))+\langle(\overline{z}\cdot\nabla)\overline{z},u\rangle+\langle(U\cdot\nabla)\overline{z},u\rangle+\langle(\overline{z}\cdot\nabla)U,u\rangle$
$\displaystyle\hskip
28.45274pt+2\nu(k(x)\overline{z},u)_{\Gamma_{2}}+2\nu(S\tilde{\overline{z}},\tilde{u})_{\Gamma_{3}}+2(\alpha(x)\overline{z},u)_{\Gamma_{5}}+\nu(k(x)\overline{z},u)_{\Gamma_{7}}$
$\displaystyle=-(U^{\prime},u)-2\nu(\varepsilon(U),\varepsilon(u))-\langle(U\cdot\nabla)U,u\rangle-2\nu(k(x)U,u)_{\Gamma_{2}}$
$\displaystyle\hskip
28.45274pt-2\nu(S\tilde{U},\tilde{u})_{\Gamma_{3}}-2(\alpha(x)U,u)_{\Gamma_{5}}-\nu(k(x)U,u)_{\Gamma_{7}}+\langle
f,u\rangle$ $\displaystyle\hskip
28.45274pt+\sum_{i=2,4,7}\langle\phi_{i},u_{n}\rangle_{\Gamma_{i}}+\sum_{i=3,5,6}\langle\phi_{i},u\rangle_{\Gamma_{i}}\qquad\mbox{for
all}\,\,u\in\mathbf{V}.$

Now, define an operator $A_{0}:\textbf{V}\rightarrow\textbf{V}^{*}$ by

(3.5) $\displaystyle\langle A_{0}y,u\rangle=$ $\displaystyle
2\nu(\varepsilon(y),\varepsilon(u))+2\nu(k(x)y,u)_{\Gamma_{2}}+2\nu(S\tilde{y},\tilde{u})_{\Gamma_{3}}$
$\displaystyle+2(\alpha(x)y,u)_{\Gamma_{5}}+\nu(k(x)y,u)_{\Gamma_{7}}\quad\mbox{for
all}\,\,y,u\in\mathbf{V}.$

###### Lemma 3.1

$\exists\delta>0$, $\exists k_{0}\geq 0$; $\langle
A_{0}u,u\rangle\geq\delta\|u\|_{\textbf{{V}}}^{2}-k_{0}\|u\|_{H}^{2}\quad\mbox{for
all}\,\,u\in\mathbf{V}.$

_Proof_ By Korn’s inequality

(3.6)
$2\nu(\varepsilon(u),\varepsilon(u))\geq\beta\|u\|^{2}_{\mathbf{V}}\quad\exists\beta>0,\mbox{for
all}\,\,u\in\mathbf{V}.$

By Remark 2.2 and Assumption 3.1, there exists a constant $M$ such that

$\displaystyle\|S(x)\|_{\infty},\,\|k(x)\|_{\infty},\|\alpha(x)\|_{\infty}\leq
M,$

and so there exists a constant $c_{0}$ (depending on $\beta$) such that

(3.7)
$\displaystyle\left|2\nu(k(x)z,z)_{\Gamma_{2}}+2\nu(S\tilde{z},\tilde{z})_{\Gamma_{3}}+\nu(k(x)z,z)_{\Gamma_{7}}+2(\alpha(x)y,u)_{\Gamma_{5}}\right|$
$\displaystyle\hskip
56.9055pt\leq\frac{\beta}{2}\|z\|^{2}_{\mathbf{H}^{1}(\Omega)}+c_{0}\|z\|_{H}^{2}\,dt\quad\mbox{for
all}\,\,z\in\mathbf{V}$

((cf. Theorem 1.6.6 in [9] or (1), p. 258 in [11])). Put
$\delta=\frac{\beta}{2}$, $k_{0}=c_{0}$. Then, by (3.6), (3.7) we come to the
asserted conclusion. $\square$

###### Remark 3.2

In process of proof of Lemma 3.1, we see that if
$\Gamma_{i}=\emptyset,\,i=2,3,7,$ or these are unions of pieces of planes
(segments in case of 2-D)and $\Gamma_{5}=\emptyset$ or $\alpha(x)=0$, then we
can take $k_{0}=0$.

When $k_{0}>0$, if $k_{0}$ is not small enough, then the operator defined by
(3.5) is not positive, and so let us transform the unknown function to get a
positive operator $A$ in (3.9) bellow. Now, let $k_{0}$ be the constant in
Lemma 3.1 and put $z=e^{-k_{0}t}\overline{z}$. Then, since
$e^{-k_{0}t}\overline{z}^{\prime}=z^{\prime}+k_{0}z$, we get the following
problem equivalent to problem (3.4):

Find $z$ such that

(3.8) $\displaystyle z\in L_{2}(0,T;\mathbf{V}),$ $\displaystyle
z(0)=v_{0}-U(0)\in\mathbf{V},$ $\displaystyle\langle
z^{\prime}(t),u\rangle+2\nu(\varepsilon(z(t)),\varepsilon(u))+e^{k_{0}t}\langle(z(t)\cdot\nabla)z(t),u\rangle+\langle(U(t)\cdot\nabla)z(t),u\rangle$
$\displaystyle\hskip
28.45274pt+\langle(z(t)\cdot\nabla)U(t),u\rangle+k_{0}(z(t),u)+2\nu(k(x)z(t),u)_{\Gamma_{2}}$
$\displaystyle\hskip
28.45274pt+2\nu(S\tilde{z}(t),\tilde{u})_{\Gamma_{3}}+2(\alpha(x)z(t),u)_{\Gamma_{5}}+\nu(k(x)z(t),u)_{\Gamma_{7}}$
$\displaystyle=e^{-k_{0}t}\Big{[}-(U^{\prime}(t),u)-2\nu(\varepsilon(U(t)),\varepsilon(u))-\langle(U(t)\cdot\nabla)U(t),u\rangle$
$\displaystyle\hskip
28.45274pt-2\nu(k(x)U(t),u)_{\Gamma_{2}}-2\nu(S\tilde{U}(t),\tilde{u})_{\Gamma_{3}}-2(\alpha(x)U(t),u)_{\Gamma_{5}}$
$\displaystyle\hskip 28.45274pt-\nu(k(x)U(t),u)_{\Gamma_{7}}+\langle
f(t),u\rangle+\sum_{i=2,4,7}\langle\phi_{i}(t),u_{n}\rangle_{\Gamma_{i}}$
$\displaystyle\hskip
28.45274pt+\sum_{i=3,5,6}\langle\phi_{i}(t),u\rangle_{\Gamma_{i}}\Big{]}\quad\mbox{for
all}\,\,u\in\mathbf{V}.$

Define operators $A$, $A_{U}(t):\textbf{V}\rightarrow\textbf{V}^{*}$ by

(3.9) $\langle Av,u\rangle=\langle A_{0}v,u\rangle+(k_{0}v,u)\quad\mbox{for
all}\,\,v,u\in\mathbf{V},$ (3.10) $\displaystyle\langle
A_{U}(t)v,u\rangle=\langle(U(t,x)\cdot\nabla)v,u\rangle+\langle(v\cdot\nabla)U(t,x),u\rangle\quad\mbox{for
all}\,\,v,u\in\mathbf{V},$

where $A_{0}$ is the operator by (3.5) and $k_{0}$ is one in Lemma 3.1. Since
$U\in\mathscr{W}$, we have $U\in C\left([0,T];\textbf{H}^{1}(\Omega)\right)$
and so such a definition is well. Then, the operator $A$ is positive definite,
and this fact is used in future.

Define an operator $B(t):\textbf{V}\rightarrow\textbf{V}^{*}$ and $F(t)\in
V^{*}$ by

(3.11) $\langle
B(t)v,u\rangle=e^{k_{0}t}\langle(v\cdot\nabla)v,u\rangle\quad\mbox{for
all}\,\,v,u\in\mathbf{V},$ (3.12) $\displaystyle\langle F(t),u\rangle$
$\displaystyle=e^{-k_{0}t}\Big{[}-(U^{\prime}(t),u)-2\nu(\varepsilon(U)(t),\varepsilon(u))-\langle(U(t)\cdot\nabla)U(t),u\rangle$
$\displaystyle-2\nu(k(x)U(t),u)_{\Gamma_{2}}-2\nu(S\tilde{U}(t),\tilde{u})_{\Gamma_{3}}-2(\alpha(x)U(t),u)_{\Gamma_{5}}$
$\displaystyle-\nu(k(x)U(t),u)_{\Gamma_{7}}+\langle
f(t),u\rangle+\sum_{i=2,4,7}\langle\phi_{i}(t),u_{n}\rangle_{\Gamma_{i}}$
$\displaystyle+\sum_{i=3,5,6}\langle\phi_{i}(t),u\rangle_{\Gamma_{i}}\Big{]}\quad\mbox{for
all}\,\,u\in\mathbf{V}.$

Then, (3.8) is written by

(3.13) $\displaystyle z\in L_{2}(0,T;\mathbf{V}),$ $\displaystyle
z(0)=v_{0}-U(0)\in\mathbf{V},$ $\displaystyle
z^{\prime}(t)+\left(A+A_{U}(t)+B(t)\right)z(t)=F(t).$

Now, define operators
$L,{\widetilde{A}}_{U},L_{U},{\widetilde{B}}:\mathscr{X}\rightarrow\mathscr{Y}$,
$C:\mathscr{X}\times\mathscr{X}\rightarrow\mathscr{Y}$ and $F\in\mathscr{Y}$
by

(3.14) $\displaystyle\langle(Lz)(t),u\rangle=\langle
z^{\prime}(t),u\rangle+\langle Az(t),u\rangle\quad\mbox{for
all}\,\,z\in\mathscr{X},\mbox{for all}\,\,u\in\mathbf{V},$
$\displaystyle\langle({\widetilde{A}}_{U}z)(t),u\rangle=\langle
A_{U}(t)z(t),u\rangle\quad\mbox{for all}\,\,z\in\mathscr{X},\mbox{for
all}\,\,u\in\mathbf{V},$ $\displaystyle\langle(L_{U}z)(t),u\rangle=\langle
z^{\prime}(t),u\rangle+\langle(A+A_{U}(t))z(t),u\rangle\quad\mbox{for
all}\,\,z\in\mathscr{X},\mbox{for all}\,\,u\in\mathbf{V},$
$\displaystyle\langle({\widetilde{B}}z)(t),u\rangle=\langle
B(t)z(t),u\rangle\quad\mbox{for all}\,\,z\in\mathscr{X},\mbox{for
all}\,\,u\in\mathbf{V},$ $\displaystyle\langle
C(w,z)(t),u\rangle=e^{k_{0}t}\langle(w(t)\cdot\nabla)z(t),u\rangle+e^{k_{0}t}\langle(z(t)\cdot\nabla)w(t),u\rangle$
$\displaystyle\hskip 85.35826pt\quad\mbox{for
all}\,\,w,z\in\mathscr{X},\mbox{for all}\,\,u\in\mathbf{V},$
$\displaystyle(F)(t)=F(t).$

###### Lemma 3.2

$C$ is a bilinear continuous operator such that
$\mathscr{X}\times\mathscr{X}\rightarrow\mathscr{Y}$. Under Assumptions 3.1
and 3.2, ${\widetilde{A}}_{U}$ is a linear continuous operator such that
$\mathscr{X}\rightarrow\mathscr{Y}$ and $F\in\mathscr{Y}$.

_Proof_ Obviously, $C$ is bilinear. When $w\in\mathscr{X}$,

$w\in L_{\infty}(0,T;\textbf{V}),\quad\|w\|_{L_{\infty}(0,T;\textbf{V})}\leq
c\big{[}\|w\|_{L_{2}(0,T;\textbf{V})}+\|w^{\prime}\|_{L_{2}(0,T;\textbf{V})}\big{]}$

and by virtue of Hölder inequality and the imbedding theorem

$\displaystyle\big{|}e^{k_{0}t}\langle(w\cdot\nabla)z,u\rangle+e^{k_{0}t}\langle(z\cdot\nabla)w,u\rangle\big{|}$
$\displaystyle\hskip 28.45274pt\leq c(\|w\|_{\textbf{L}_{3}}\|\nabla
z\|_{\textbf{L}_{2}}\|u\|_{\textbf{L}_{6}}+\|z\|_{\textbf{L}_{3}}\|\nabla
w\|_{\textbf{L}_{2}}\|u\|_{\textbf{L}_{6}})\leq
c\|w\|_{\textbf{V}}\|z\|_{\textbf{V}}\|u\|_{\textbf{V}}$ $\displaystyle\hskip
113.81102pt\mbox{for all}\,\,w,z,u\in\mathbf{V}.$

Thus,

(3.15) $\|C(w,z)\|_{L_{2}(0,T;\textbf{V}^{*})}\leq
c\|w\|_{L_{\infty}(0,T;\textbf{V})}\|z\|_{L_{2}(0,T;\textbf{V})}\leq
c\|w\|_{\mathscr{X}}\cdot\|z\|_{\mathscr{X}}.$

Also, since

$\displaystyle|\langle C(w,z)^{\prime}(t),u\rangle|$
$\displaystyle=e^{k_{0}t}\big{|}k_{0}\langle(w\cdot\nabla)z,u\rangle+k_{0}\langle(z\cdot\nabla)w,u\rangle+\langle(w^{\prime}\cdot\nabla)z,u\rangle$
$\displaystyle+\langle(w\cdot\nabla)z^{\prime},u\rangle+\langle(z^{\prime}\cdot\nabla)w,u\rangle+\langle(z\cdot\nabla)w^{\prime},u\rangle\big{|},$

taking (3.15) into account we have

(3.16) $\displaystyle\|C(w,z)^{\prime}\|$
$\displaystyle{}_{L_{2}(0,T;\textbf{V}^{*})}\leq
c\|C(w,z)\|_{L_{2}(0,T;\textbf{V}^{*})}$
$\displaystyle+c\big{[}\big{(}\|w^{\prime}\|_{L_{2}(0,T;\textbf{V})}+\|w\|_{L_{\infty}(0,T;\textbf{V})})(\|z^{\prime}\|_{L_{2}(0,T;\textbf{V})}+\|z\|_{L_{\infty}(0,T;\textbf{V})}\big{)}\big{]}$
$\displaystyle\leq c\|w\|_{\mathscr{X}}\cdot\|z\|_{\mathscr{X}}.$

(3.15) and (3.16) imply

(3.17) $\displaystyle\|C(w,z)\|_{\mathscr{Y}}\leq
c\|w\|_{\mathscr{X}}\cdot\|z\|_{\mathscr{X}}.$

By the same argument above, we have

(3.18) $\displaystyle\|{\widetilde{A}}_{U}z\|_{\mathscr{Y}}\leq
c\|U\|_{\mathscr{W}}\cdot\|z\|_{\mathscr{X}}.$

By Assumption 3.1, Remark 2.2 and the trace theorem, we can see that
$F\in\mathscr{Y}$. $\square$

###### Lemma 3.3

The operator $\overline{L}$ defined by $\overline{L}z=(z^{\prime}(0),Lz)$ for
$z\in\mathscr{X}$ is a linear continuous one-to-one operator from
$\mathscr{X}$ onto $H\times\mathscr{Y}$.

_Proof_ The linearity of $\overline{L}$ is obvious. The fact $z\in\mathscr{X}$
implies that $z^{\prime}\in C([0,T];H)$, $\|z^{\prime}\|_{C([0,T];H)}\leq
c\|z\|_{\mathscr{X}}$, and so we see that a map $z\in\mathscr{X}\rightarrow
z^{\prime}(0)\in H$ is continuous.

Clearly, $\|z^{\prime}\|_{\mathscr{Y}}\leq c\|z\|_{\mathscr{X}}$. Also, by
Assumption 3.1, Remark 2.2 and the trace theorem,

(3.19) $|\langle Av,u\rangle|\leq
c\|v\|_{\mathbf{V}}\cdot\|u\|_{\mathbf{V}}\quad\mbox{for
all}\,\,v,u\in\mathbf{V}.$

Formula (3.19) implies that the mapping $z\in\mathscr{X}\rightarrow
Az\in\mathscr{Y}$ is continuous. Therefore, $\overline{L}$ is continuous.

Let us show that $\overline{L}$ is a one-to-one and surjective operator from
$\mathscr{X}$ onto $H\times\mathscr{Y}$.

First, let us prove that this operator is injective. For this, it is enough to
prove that the inverse image of $(0_{H},0_{\mathscr{Y}})\in
H\times\mathscr{Y}$ by the operator $\overline{L}$ is $0_{\mathscr{X}}$. By
Lemma 3.1 and (3.9), we get

(3.20) $\langle
Av,v\rangle\geq\delta\|v\|^{2}_{\mathbf{V}}\quad\exists\delta>0,\mbox{for
all}\,\,v\in\mathbf{V}.$

By (3.19), (3.20) for any $q\in\textbf{V}^{*}$ there exists a unique solution
$y\in V$ to the following problem

(3.21) $Ay=q.$

Let $z\in\mathscr{X}$ be the inverse image of $(0_{H},0_{\mathscr{Y}})\in
H\times\mathscr{Y}$ by $\overline{L}$. Then since $z^{\prime}(0)=0_{H}$,
putting $t=0$ in the first equation of (3.14) we get

$\langle Az(0),u\rangle=0\quad\mbox{for all}\,\,u\in\mathbf{V},$

where $z(0)=z(0,x)$. This means that $z(0)$ is a unique solution to (3.21) for
$q=0_{\textbf{V}^{*}}$, i.e. $z(0)=0_{\textbf{V}}$. Putting $w=z^{\prime}$, we
get $w(0)=z^{\prime}(0)=0_{H}$. Taking $Lz=0$ into account and differentiating
the first equation of (3.14), we have

(3.22) $\displaystyle\langle w^{\prime}(t),u\rangle+\langle
Aw(t),u\rangle=0\quad\mbox{for all}\,\,u\in\mathbf{V}.$

The operator $A$ in (3.22) satisfies all conditions of Theorem 1.1, ch. 6 in
[12]. Thus, for problem (3.22) with an initial condition $w(0)\in H$ there
exists a unique solution $w$ such that $w\in
L_{2}(0,T;\textbf{V}),\,\,w^{\prime}\in L_{2}(0,T;\textbf{V}^{*})$. Since
$w(0)=0_{H}$, we have $w=0$, which means $z=0_{\mathscr{X}}$ since
$z(0)=0_{\textbf{V}}$.

Let us prove that $\overline{L}$ is surjective. Let $(w_{0},g)\in
H\times\mathscr{Y}$. Since $g\in\mathscr{Y}$, we have $g(0)\in\textbf{V}^{*}$.
Then, by (3.19) and (3.20), there exists a unique solution
$z_{0}\in\textbf{V}$ to problem

(3.23) $Az_{0}=g(0)-w_{0}.$

Let us consider problem

(3.24) $\left\\{\begin{aligned} &w^{\prime}+Aw=g^{\prime},\\\
&w(0)=w_{0}.\end{aligned}\right.$

There exists a unique solution $w$ such that $w\in
L_{2}(0,T;\textbf{V}),\,\,w^{\prime}\in L_{2}(0,T;\textbf{V}^{*})$ to problem
(3.24) (cf. Theorem 1.3 of ch. 6 in [12]). Now, put

(3.25) $z=z_{0}+\int_{0}^{t}w(s)\,ds,$

where $z_{0}$ is the solution to (3.23). Then, $z^{\prime}=w$ and
$z\in\mathscr{X}$. Integrating two sides of the first one in (3.24) from $0$
to $t$ and using (3.25), we have

(3.26) $\displaystyle\langle w(t),u\rangle+\left\langle
Az(t),u\right\rangle-\left[\langle w_{0},u\rangle+\langle
Az_{0},u\rangle\right]=\langle g(t),u\rangle-\langle
g(0),u\rangle\quad\mbox{for all}\,\,u\in\mathbf{V}.$

Taking (3.23), (3.25) into account, from (3.26) we get

(3.27) $\langle z^{\prime}(t),u\rangle+\left\langle
Az(t),u\right\rangle=\langle g(t),u\rangle\quad\mbox{for
all}\,\,u\in\mathbf{V}.$

This means that $z\in\mathscr{X}$ defined by (3.25) is the inverse image of
$(w_{0},g)\in H\times\mathscr{Y}$ by the operator $\overline{L}$, i.e.
$\overline{L}$ is surjective. Therefore, $\overline{L}$ is an epimorphism.
$\square$

###### Lemma 3.4

Under Assumption 3.2, let $\|U(0,x)\|_{\widetilde{\textbf{V}}}$ be small
enough. The operator $\overline{L}_{U}$ defined by
$\overline{L}_{U}z=(z^{\prime}(0),L_{U}z)$ for $z\in\mathscr{X}$ is a linear
continuous one-to-one operator from $\mathscr{X}$ onto $H\times\mathscr{Y}$.

_Proof_ When $z\in\mathscr{X}$, $z\in C([0,T];\textbf{V})$ and

$\|z\|_{C([0,T];\textbf{V})}\leq
c\big{[}\|z\|_{L_{2}(0,T;\textbf{V})}+\|z^{\prime}\|_{L_{2}(0,T;\textbf{V})}\big{]}.$

By virtue of this fact and Lemma 3.2, the operator
$\overline{A}_{U}\in(\mathscr{X}\rightarrow H\times\mathscr{Y})$ defined by
$\overline{A}_{U}z=(0_{H},{\widetilde{A}}_{U}z)$ is continuous. Thus, the
operator $\overline{L}_{U}$ defined on $\mathscr{X}$ is linear continuous.

As in Lemma 3.5 of [14] it is proved that the operator
${\widetilde{A}}_{U}\in(\mathscr{X}\rightarrow\mathscr{Y})$ is compact. Thus,
$\overline{A}_{U}\in(\mathscr{X}\rightarrow H\times\mathscr{Y})$ is also
compact. Since $\overline{L}_{U}=\overline{L}+\overline{A}_{U}$, by virtue of
Theorem 3.4 in [14] and Lemma 3.3 we know that in order to prove that the
operator $\overline{L}_{U}$ is one-to-one from $\mathscr{X}$ onto
$H\times\mathscr{Y}$ it is enough to prove that $\overline{L}_{U}$ is one-to-
one from $\mathscr{X}$ into $H\times\mathscr{Y}$. To prove the last fact it is
enough to show that the inverse image of $(0_{H},0_{\mathscr{Y}})$ by
$\overline{L}_{U}$ is $0_{\mathscr{X}}$. By Hölder inequality and imbedding
theorem

(3.28)
$\displaystyle\big{|}\big{\langle}(U(t,x)\cdot\nabla)v,v\big{\rangle}+\big{\langle}(v\cdot\nabla)U(t,x),v\big{\rangle}\big{|}\leq
K_{0}\|v\|_{\textbf{V}}\|U(t,x)\|_{\textbf{H}^{1}}\|v\|_{\textbf{V}}.$

Thus, if $\|U(0,x)\|_{\widetilde{V}}$ is so small that
$\|U(0,x)\|_{\textbf{H}^{1}}\leq\frac{\delta}{2K_{0}}$, then (3.19), (3.20)
and (3.28) imply

(3.29)
$\big{|}\big{\langle}\big{(}A+A_{U}(0)\big{)}v,u\big{\rangle}\big{|}\leq
c\|v\|_{\mathbf{V}}\cdot\|u\|_{\mathbf{V}},\quad\big{\langle}\big{(}A+A_{U}(0)\big{)}v,v\big{\rangle}\geq\frac{\delta}{2}\|v\|^{2}_{\mathbf{V}}\quad\mbox{for
all}\,\,v,u\in\mathbf{V}.$

By (3.29) for any $q\in\textbf{V}^{*}$ there exists a unique solution $y\in V$
to

(3.30) $(A+A_{U}(0))y=q.$

Let $z\in\mathscr{X}$ be the inverse image of $(0_{H},0_{\mathscr{Y}})$ by
$\overline{L}$. Then $z^{\prime}(0)=0_{H}$, and putting $t=0$ from the third
one in (3.14) we get

$\big{\langle}\big{(}A+A_{U}(0)\big{)}z(0),u\big{\rangle}=0\quad\mbox{for
all}\,\,u\in\mathbf{V},$

where $z(0)=z(0,x)$. This means that $z(0)$ is the unique solution to (3.30)
with $q=0_{\textbf{V}^{*}}$, i.e. $z(0)=0_{\textbf{V}}$. Therefore,
$z\in\mathscr{X}$ satisfies

(3.31) $\left\\{\begin{aligned}
&z^{\prime}(t)+\big{(}A+A_{U}(t)\big{)}z(t)=0,\\\
&z(0)=0_{\textbf{V}}.\end{aligned}\right.$

Now, making duality pairing with $z(t)$ on two sides of

$z^{\prime}(t)+Az(t)=-A_{U}(t)z(t)$

and taking (3.20) into account and using Gronwall’s inequality, we can prove
$z=0_{\mathscr{X}}$ as in Lemma 3.8 of [14]. It is finished to prove the
Lemma. $\square$

###### Lemma 3.5

Under Assumption 3.2 the operator $T$ defined by
$Tz=\big{(}z^{\prime}(0),(L_{U}$ $+{\widetilde{B}})z\big{)}$ for
$z\in\mathscr{X}$ is continuously differentiable,
$T(0_{\mathscr{X}})=\left(0_{H},0_{\mathscr{Y}}\right)$ and the Frechet
derivative of $T$ at $0_{\mathscr{X}}$ is $\overline{L}_{U}$.

_Proof_ It is easy to verify that
$T(0_{\mathscr{X}})=\left(0_{H},0_{\mathscr{Y}}\right)$. Since the operator
$L_{U}$ is linear, its Frechet derivative is the same with itself. Therefore,
if ${\widetilde{B}}$ is continuously differentiable, then so is $T$.

For any $w,z\in\mathscr{X}$,

$\big{(}{\widetilde{B}}(w+z)-{\widetilde{B}}w\big{)}(t)=e^{k_{0}t}\big{(}w(t)\cdot\nabla\big{)}z(t)+e^{k_{0}t}\big{(}z(t)\cdot\nabla\big{)}w(t)+({\widetilde{B}}z)(t).$

By (3.17), we get

$\lim_{\|z\|_{\mathscr{X}}\rightarrow
0}\frac{\|{\widetilde{B}}z\|_{\mathscr{Y}}}{\|z\|_{\mathscr{X}}}\leq\lim_{\|z\|_{\mathscr{X}}\rightarrow
0}\frac{c\|z\|^{2}_{\mathscr{X}}}{\|z\|_{\mathscr{X}}}=0.$

Then, put

$C(w,z)(t)\equiv
e^{k_{0}t}\big{(}w(t)\cdot\nabla\big{)}z(t)+e^{k_{0}t}\big{(}z(t)\cdot\nabla\big{)}w(t)=({\widetilde{B}}^{\prime}_{w}z)(t).$

By Lemma 3.2
${\widetilde{B}}^{\prime}_{w}\in(\mathscr{X}\rightarrow\mathscr{Y})$ is
continuous, and it is the Frechet derivative of $\widetilde{B}$ at $w$ and
also continuous with respect to $w$. Thus, $T$ is continuously differentiable.
Also from the formula above we can see that the Frechet derivative of
$\widetilde{B}$ at $0_{\mathscr{X}}$ is zero. Therefore, the Frechet
derivative of $T$ at $0_{\mathscr{X}}$ is $\overline{L}_{U}$. $\square$

Let us consider problem

(3.32) $\big{(}A+A_{U}(0)+B(0)\big{)}u=q.$

###### Lemma 3.6

Assume that $\|U(0,x)\|_{\widetilde{\textbf{V}}}$ is small enough. If the norm
of $q\in V^{*}$ is small enough, then there exists a unique solution to (3.32)
in some $\mathscr{O}_{M}(0_{\mathbf{V}})$.

_Proof_ Since $\|U(0,x)\|_{\widetilde{\textbf{V}}}$ is small enough, by
(3.29), for any fixed $z\in\mathbf{V}$ there exists a unique solution to
problem

(3.33) $\big{(}A+A_{U}(0)\big{)}w=q-B(0)z.$

On the other hand,

(3.34) $\displaystyle|\langle B(0)w_{1}-B(0)w_{2},u\rangle|\leq$
$\displaystyle
KM\|w_{1}-w_{2}\|_{\mathbf{V}}\cdot\|u\|_{\mathbf{V}}\quad\mbox{for
all}\,\,w_{i}\in\mathscr{O}_{M}(0_{\mathbf{V}}),\,\,\mbox{for
all}\,\,u\in\mathbf{V}.$

Owing to (3.29) the solution $w$ to (3.33) is estimated as follows

$\|w\|_{\mathbf{V}}\leq\frac{2}{\delta}\big{(}\|q\|_{\mathbf{V}^{*}}+\|B(0)z\|_{\mathbf{V}^{*}}\big{)}\leq\frac{2}{\delta}\big{(}\|q\|_{\mathbf{V}^{*}}+KM^{2}\big{)}.$

Thus, if $\|q\|_{\mathbf{V}^{*}}$ and $M$ are small enough, then the operator
$(z\rightarrow w)$ maps $\mathscr{O}_{M}(0_{\mathbf{V}})$ into itself and by
(3.34) this operator is strictly contract. Therefore, in
$\mathscr{O}_{M}(0_{\mathbf{V}})$ there exists a unique solution to (3.33).
Thus, we come to the asserted conclusion. $\square$

For proof of unique existence of a solution to Problem I, we use the following

###### Proposition 3.7

(cf. Theorem 4.1.1 in [10]) Let $X,Y$ be Banach spaces, $\cal G$ an open set
in $X$, $f:X\rightarrow Y$ continuously differentiable on $\cal G$. Let the
derivative $f^{\prime}(a)$ be an isomorphism of $X$ onto $Y$ for $a\in\cal G$.
Then there exist neighborhoods $\cal U$ of $a$, $\cal V$ of $f(a)$such that
$f$ is injective on $\cal U$, $f(\cal U)=\cal V$.

One of main results of this paper is the following

###### Theorem 3.8

Suppose that Assumptions 3.1 and 3.2 hold. Assume that $\|U\|_{\mathscr{W}}$
and the norms of $f,f^{\prime},\phi_{i},\phi_{i}^{\prime}$ in the spaces where
they belong to are small enough.

If

(3.35) $\displaystyle w_{0}\equiv F(0)-(A+A_{U}(0)+B(0))z_{0}\in H,$

where $z_{0}=v_{0}-U(0,\cdot),$ and $\|w_{0}\|_{H}$ is small enough, then
there exists a unique solution to (3.3) in the space $\mathscr{W}.$

_Proof_ First, let us prove existence of a solution.
If $\|U\|_{\mathscr{W}}$ and the norms of
$f,f^{\prime},\phi_{i},\phi_{i}^{\prime}$ in the spaces they belong to are
small enough, then $\|F\|_{\mathscr{Y}}$ is also small enough. By virtue of
Lemmas 3.4, 3.5 and Proposition 3.7, for any $R_{1}>0$ small enough if
$\|F\|_{\mathscr{Y}}$, $R$ are small enough and
$w_{1}\in\mathscr{O}_{R}(0_{H})$, there exists a unique
$z\in\mathscr{O}_{R_{1}}(0_{\mathscr{X}})$ such that

(3.36) $\displaystyle
z^{\prime}(t)+\big{(}A+A_{U}(t)+B(t)\big{)}z(t)=F(t),\quad
z^{\prime}(0)=w_{1}\in\mathscr{O}_{R}(0_{H}).$

Putting $t=0$ in (3.36), we get

$F(0)-\big{(}A+A_{U}(0)+B(0)\big{)}z(0)=w_{1}\in\mathscr{O}_{R}(0_{H}).$

On the other hand, if $\|U\|_{\mathscr{W}}$ is small enough, then so is
$\|U(0,x)\|_{\widetilde{\textbf{V}}}$. Thus, when $\|F(0)-w_{1}\|_{V^{*}}$ is
small enough, by Lemma 3.6 there exists a unique solution
$z_{0}\in\mathscr{O}_{R_{2}}(0_{V})$ for some $R_{2}>0$ to

(3.37) $\big{(}A+A_{U}(0)+B(0)\big{)}z_{0}=F(0)-w_{1}.$

Since $\|z(0)\|_{V}\leq c\|z\|_{\mathscr{X}},$ we can choose $R_{1}$ such that
$z(0)\in\mathscr{O}_{R_{2}}(0_{V}),$ and we have $z(0)=z_{0}$. Therefore, if
$\|F\|_{\mathscr{Y}}$ is small enough,
$F(0)-\big{(}A+A_{U}(0)+B(0)\big{)}z_{0}$ belongs to $H$ and its norm is small
enough, then $z\in\mathscr{X},$ the solution to (3.36), is a solution to
problem

(3.38) $\left\\{\begin{aligned}
&z^{\prime}(t)+\big{(}A+A_{U}(t)+B(t)\big{)}z(t)=F(t),\\\
&z(0)=z_{0}.\end{aligned}\right.$

By definitions of $A,A_{U}(t),B(t),F,$ the solution $z$ of (3.38) is also a
solution to (3.8) which is equivalent to (3.3). Thus,
$e^{k_{0}t}z+U\in\mathscr{W}$ is a solution to (3.3).

Second, let us prove uniqueness.
Let $v_{1},v_{2}$ be two solutions to (3.3) corresponding to the same data.
Putting $\overline{w}=v_{1}-v_{2}$, we have

(3.39) $\displaystyle\overline{w}\in L_{2}(0,T;\mathbf{V}),$
$\displaystyle\overline{w}(0)=0,$
$\displaystyle\langle\overline{w}^{\prime},u\rangle+2\nu(\varepsilon(\overline{w}),\varepsilon(u))+\langle(v_{1}\cdot\nabla)\overline{w},u\rangle+\langle(\overline{w}\cdot\nabla)v_{2},u\rangle+2\nu(k(x)\overline{w},u)_{\Gamma_{2}}$
$\displaystyle\hskip
28.45274pt+2\nu(S\tilde{\overline{w}},\tilde{u})_{\Gamma_{3}}+2(\alpha(x)\overline{w},u)_{\Gamma_{5}}+\nu(k(x)\overline{w},u)_{\Gamma_{7}}=0\quad\mbox{for
all}\,\,u\in\textbf{V}.$

Putting $w=e^{-k_{0}t}\overline{w}$, where $k_{0}$ is the constant in Lemma
3.1, we get $e^{-k_{0}t}\overline{w}^{\prime}=w^{\prime}+k_{0}w$. Then, we
have

(3.40) $\displaystyle w\in L_{2}(0,T;\mathbf{V}),$ $\displaystyle w(0)=0,$
$\displaystyle\langle
w^{\prime},u\rangle+2\nu(\varepsilon(w),\varepsilon(u))+\langle(v_{1}\cdot\nabla)w,u\rangle+\langle(w\cdot\nabla)v_{2},u\rangle+k_{0}(w,u)$
$\displaystyle\hskip
28.45274pt+2\nu(k(x)w,u)_{\Gamma_{2}}+2\nu(S\tilde{w},\tilde{u})_{\Gamma_{3}}+2(\alpha(x)w,u)_{\Gamma_{5}}+\nu(k(x)w,u)_{\Gamma_{7}}=0$
$\displaystyle\hskip 142.26378pt\mbox{for all}\,\,u\in\mathbf{V},$

which is equivalent to (3.39). By Lemma 3.1,

$\displaystyle 2\nu(\varepsilon(w),\varepsilon(w))$
$\displaystyle+k_{0}(w,w)+2\nu(k(x)w,w)_{\Gamma_{2}}+2\nu(S\tilde{w},\tilde{w})_{\Gamma_{3}}$
$\displaystyle+2(\alpha(x)w,w)_{\Gamma_{5}}+\nu(k(x)w,w)_{\Gamma_{7}}\geq\delta\|w\|_{\textbf{V}}^{2}.$

Taking it into account, we can prove $w=0_{\mathscr{X}}$ as in Lemma 3.8 of
[14]. Thus, uniqueness of a solution is proved, and we finished proof of the
theorem. $\square$

###### Remark 3.3

Let us consider more precisely the condition that
$F(0)-\big{(}A+A_{U}(0)+B(0)\big{)}z_{0}$ belongs to $H$ and its norm is small
enough. By (3.9)$\sim$(3.14) we have

(3.41)
$\displaystyle\big{\langle}F(0)-\big{(}A+A_{U}(0)+B(0)\big{)}z_{0},u\big{\rangle}=$
$\displaystyle\Big{[}-(U^{\prime}(0,x),u)-2\nu(\varepsilon(U(0,x)),\varepsilon(u))-\langle(U(0,x)\cdot\nabla)U(0,x),u\rangle$
$\displaystyle\hskip
14.22636pt-2\nu(k(x)U(0,x),u)_{\Gamma_{2}}-2\nu(S\tilde{U}(0),\tilde{u})_{\Gamma_{3}}-2(\alpha(x)U(0,x),u)_{\Gamma_{5}}$
$\displaystyle\hskip 14.22636pt-\nu(k(x)U(0,x),u)_{\Gamma_{7}}+\langle
f(0),u\rangle+\sum_{i=2,4,7}\langle\phi_{i}(t),u_{n}\rangle_{\Gamma_{i}}+\sum_{i=3,5,6}\langle\phi_{i}(t),u\rangle_{\Gamma_{i}}\Big{]}$
$\displaystyle-\big{[}2\nu(\varepsilon(z_{0}),\varepsilon(u))+2\nu(k(x)z_{0},u)_{\Gamma_{2}}+2\nu(S\tilde{z_{0}},\tilde{u})_{\Gamma_{3}}$
$\displaystyle\hskip
113.81102pt+2(\alpha(x)z_{0},u)_{\Gamma_{5}}+\nu(k(x)z_{0},u)_{\Gamma_{7}}\big{]}$
$\displaystyle-\big{[}\langle(U(0,x)\cdot\nabla)z_{0},u\rangle+\langle(z_{0}\cdot\nabla)U(0,x),u\rangle\big{]}-\langle(z_{0},\nabla)z_{0},u\rangle\quad\mbox{for
all}\,\,u\in\textbf{ V}.$

Taking into account the fact that $U(0,x)+z_{0}=v_{0}$,
$U^{\prime}(0,x)\in\textbf{L}_{2}(\Omega)$ and its norm is small enough, from
(3.41) we can see that the condition mentioned above is equivalent to the
condition $\overline{w}_{0}\in\mathscr{O}_{R}(0_{H})$ for $R>0$ small enough,
where $\overline{w}_{0}$ is defined by

(3.42) $\displaystyle\langle\overline{w}_{0},u\rangle$
$\displaystyle\equiv\langle
f(0),u\rangle+\sum_{i=2,4,7}\langle\phi_{i}(0,x),u_{n}\rangle_{\Gamma_{i}}+\sum_{i=3,5,6}\langle\phi_{i}(0,x),u\rangle_{\Gamma_{i}}$
$\displaystyle-\big{[}2\nu(\varepsilon(v_{0}),\varepsilon(u))+2\nu(k(x)v_{0},u)_{\Gamma_{2}}+2\nu(S\tilde{v}_{0},\tilde{u})_{\Gamma_{3}}+2(\alpha(x)v_{0},u)_{\Gamma_{5}}$
$\displaystyle+\nu(k(x)v_{0},u)_{\Gamma_{7}}+\langle(v_{0}\cdot\nabla)v_{0},u\rangle+k_{0}(v_{0},u)\big{]}\quad\mbox{for
all}\,\,u\in\mathbf{V}.$

###### Remark 3.4

If $\Gamma_{i}=\varnothing,i=2\sim 5,7,$ then the problem is reduced to one in
[15] where a local-in-time solution was studied. In this case $k_{0}=0$ (cf.
Remark 3.2), and the condition (3.35) is the same with (25) in [15]. And our
condition for $U$ is also the same with one in [15].

##  4 Existence of a unique solution to problem II

Let
$\mathbf{V}_{1}=\\{u\in\mathbf{H}^{1}(\Omega):\mbox{div}\,u=0,\,u|_{\Gamma_{1}}=0,\,u_{\tau}|_{(\Gamma_{2}\cup\Gamma_{4})}=0,\,u_{n}|_{(\Gamma_{3}\cup\Gamma_{5})}=0\\}$
and $\mathbf{V}_{\Gamma
2-5}(\Omega)=\\{u\in\mathbf{H}^{1}(\Omega):\mbox{div}\,u=0,\,u_{\tau}|_{(\Gamma_{2}\cup\Gamma_{4})}=0,\,u_{n}|_{(\Gamma_{3}\cup\Gamma_{5})}=0\\}$.
Denote by $H_{1}$ the completion of $\mathbf{V}_{1}$ in the space
$\textbf{L}_{2}(\Omega)$.

By Theorems 2.1 and 2.2, for $v\in\mathbf{H}^{2}(\Omega)\cap\mathbf{V}_{\Gamma
2-5}(\Omega)$ and $u\in\mathbf{V}_{1}$ we have that

(4.1) $\displaystyle-(\Delta v,u)$ $\displaystyle=(\nabla v,\nabla
u)-\left(\frac{\partial v}{\partial n},u\right)_{\partial\Omega}$
$\displaystyle=(\nabla v,\nabla u)+(k(x)v,u)_{\Gamma_{2}}-(\mbox{rot}\,v\times
n,u)_{\Gamma_{3}}+(S\tilde{v},\tilde{u})_{\Gamma_{3}}$
$\displaystyle\qquad-(\varepsilon_{n}(v),u)_{\Gamma_{4}}-2(\varepsilon_{n}(v),u)_{\Gamma_{5}}-(S\tilde{v},\tilde{u})_{\Gamma_{5}}-\left(\frac{\partial
v}{\partial n},u\right)_{\Gamma_{7}}$ $\displaystyle=(\nabla v,\nabla
u)+(k(x)v,u)_{\Gamma_{2}}-(\mbox{rot}\,v\times
n,u)_{\Gamma_{3}}+(S\tilde{v},\tilde{u})_{\Gamma_{3}}$
$\displaystyle\qquad-(\varepsilon_{nn}(v),u\cdot
n)_{\Gamma_{4}}-2(\varepsilon_{n\tau}(v),u)_{\Gamma_{5}}-(S\tilde{v},\tilde{u})|_{\Gamma_{5}}-\left(\frac{\partial
v}{\partial n},u\right)_{\Gamma_{7}}.$

Also, for $p\in H^{1}(\Omega)$ and $u\in\mathbf{V}_{1}$ we get

(4.2) $\displaystyle(\nabla p,u)$ $\displaystyle=(p,u\cdot
n)_{\cup_{i=2}^{7}\Gamma_{i}}=(p,u\cdot n)_{\Gamma_{2}}+(p,u\cdot
n)_{\Gamma_{4}}+(pn,u)_{\Gamma_{7}},$

where the fact that $u\cdot n|_{\Gamma_{3}\cup\Gamma_{5}}=0$ was used.

Let

$\displaystyle\mathscr{X}_{1}=\\{w\in L_{2}(0,T;\textbf{V}_{1});w^{\prime}\in
L_{2}(0,T;\textbf{V}_{1}),w^{\prime\prime}\in
L_{2}(0,T;\textbf{V}_{1}^{*})\\},$ $\displaystyle\hskip
28.45274pt\|w\|_{\mathscr{X}_{1}}=\|w\|_{L_{2}(0,T;\textbf{V}_{1})}+\|w^{\prime}\|_{L_{2}(0,T;\textbf{V}_{1})}+\|w^{\prime\prime}\|_{L_{2}(0,T;\textbf{V}_{1}^{*})},$
$\displaystyle\mathscr{Y}_{1}=\\{w\in
L_{2}(0,T;\textbf{V}_{1}^{*});w^{\prime}\in L_{2}(0,T;\textbf{V}_{1}^{*})\\},$
$\displaystyle\hskip
28.45274pt\|w\|_{\mathscr{Y}_{1}}=\|w\|_{L_{2}(0,T;\textbf{V}_{1}^{*})}+\|w^{\prime}\|_{L_{2}(0,T;\textbf{V}_{1}^{*})}.$

Unlike problem I, for problem II we do not require the condition
$v_{\tau}|_{\Gamma_{7}}=0$, and so instead of Assumptions 3.1 and 3.2, we use
the following assumptions.

###### Assumption 4.1

Assumption 3.1 holds with $\phi_{7},\phi_{7}^{\prime}\in
L_{2}(0,T;\textbf{H}^{-\frac{1}{2}}(\Gamma_{7}))$ instead of
$\phi_{7},\phi_{7}^{\prime}\in L_{2}(0,T;H^{-\frac{1}{2}}(\Gamma_{7}))$

###### Assumption 4.2

There exists a function $U\in\mathscr{W}$ such that

$\mbox{div}\,U=0,\,U|_{\Gamma_{1}}=h_{1},\,U_{\tau}|_{\Gamma_{2}}=0,\,U_{n}|_{\Gamma_{3}}=0,\,U_{\tau}|_{\Gamma_{4}}=h_{4},\,U_{n}|_{\Gamma_{5}}=h_{5},$

where $\mathscr{W}$ is the same as in the previous section. Also,
$U(0,x)-v_{0}\in\textbf{V}_{1}$.

Taking into account (4.1), (4.2), we get the following variational formulation
for Problem II:

###### Formulation 4.1

Find $v$ such that

(4.3) $\displaystyle v-U\in L_{2}(0,T;\mathbf{V}_{1}),$ $\displaystyle
v(0)=v_{0},$ $\displaystyle\langle v^{\prime},u\rangle+\nu(\nabla v,\nabla
u)+\langle(v\cdot\nabla)v,u\rangle+\nu(k(x)v,u)_{\Gamma_{2}}$
$\displaystyle\hskip
56.9055pt+\nu(S\tilde{v},\tilde{u})_{\Gamma_{3}}+2(\alpha(x)v,u)_{\Gamma_{5}}-\nu(S\tilde{v},\tilde{u})_{\Gamma_{5}}$
$\displaystyle\hskip 28.45274pt=\langle
f,u\rangle+\sum_{i=2,4}\langle\phi_{i},u_{n}\rangle_{\Gamma_{i}}+\sum_{i=3,5,7}\langle\phi_{i},u\rangle_{\Gamma_{i}}\quad\mbox{for
all}\,\,u\in\mathbf{V}_{1}.$

Taking into account Assumption 4.2 and putting $v=\overline{z}+U$, we get the
following problem which is equivalent to Formulation 4.1:

Find $\overline{z}$ such that

(4.4) $\displaystyle\overline{z}\in L_{2}(0,T;\mathbf{V}_{1}),$
$\displaystyle\overline{z}(0)\equiv v_{0}-U(0)\in\mathbf{V}_{1},$
$\displaystyle\langle\overline{z}^{\prime},u\rangle+\nu(\nabla\overline{z},\nabla
u)+\langle(\overline{z}\cdot\nabla)\overline{z},u\rangle+\langle(U\cdot\nabla)\overline{z},u\rangle+\langle(\overline{z}\cdot\nabla)U,u\rangle$
$\displaystyle\hskip
28.45274pt+\nu(k(x)\overline{z},u)_{\Gamma_{2}}+\nu(S\tilde{\overline{z}},\tilde{u})_{\Gamma_{3}}+2(\alpha(x)\overline{z},u)_{\Gamma_{5}}-\nu(S\tilde{\overline{z}},\tilde{u})_{\Gamma_{5}}$
$\displaystyle\hskip 14.22636pt=-\langle U^{\prime},u\rangle-\nu(\nabla
U,\nabla
u)-\langle(U\cdot\nabla)U,u\rangle-\nu(k(x)U,u)_{\Gamma_{2}}-\nu(S\tilde{U},\tilde{u})_{\Gamma_{3}}$
$\displaystyle\hskip
28.45274pt-2(\alpha(x)U,u)_{\Gamma_{5}}+\nu(S\tilde{U},\tilde{u})_{\Gamma_{5}}+\langle
f,u\rangle+\sum_{i=2,4}\langle\phi_{i},u_{n}\rangle_{\Gamma_{i}}+\sum_{i=3,5,7}\langle\phi_{i},u\rangle_{\Gamma_{i}}$
$\displaystyle\hskip 170.71652pt\mbox{for all}\,\,u\in\mathbf{V}_{1}.$

Define an operator $A_{01}:\textbf{V}_{1}\rightarrow\textbf{V}_{1}^{*}$ by

(4.5) $\displaystyle\langle A_{01}y,u\rangle=$ $\displaystyle\hskip
14.22636pt\nu(\nabla y,\nabla
u)+\nu(k(x)y,u)_{\Gamma_{2}}+\nu(S\tilde{y},\tilde{u})_{\Gamma_{3}}+2(\alpha(x)y,u)_{\Gamma_{5}}-\nu(S\tilde{y},\tilde{u})_{\Gamma_{5}}$
$\displaystyle\hskip 142.26378pt\mbox{for all}\,\,y,u\in\mathbf{V}_{1}.$

By virtue of the same argument used to prove Lemma 3.1 we get

###### Lemma 4.1

$\exists\delta>0$, $\exists k_{1}\geq 0$:

$\langle
A_{01}u,u\rangle\geq\delta\|u\|_{\textbf{V}_{1}}^{2}-k_{1}\|u\|_{H_{1}}^{2}\quad\mbox{for
all}\,\,u\in\mathbf{V}_{1}.$

Putting $z=e^{-k_{1}t}\overline{z}$, where $k_{1}$ is the constant in Lemma
4.1, and using the fact that
$e^{-k_{1}t}\overline{z}^{\prime}=z^{\prime}+k_{1}z$, we get the following
problem which is equivalent to (4.4):

Find $z$ such that

(4.6) $\displaystyle z\in L_{2}(0,T;\mathbf{V}_{1}),$ $\displaystyle
z(0)=z_{0}\equiv v_{0}-U(0)\in\mathbf{V}_{1},$ $\displaystyle\langle
z^{\prime}(t),u\rangle+\nu(\nabla z,\nabla
u)+e^{k_{1}t}\langle(z(t)\cdot\nabla)z(t),u\rangle+\langle(U(t)\cdot\nabla)z(t),u\rangle$
$\displaystyle\hskip
28.45274pt+\langle(z(t)\cdot\nabla)U(t),u\rangle+k_{1}(z(t),u)+\nu(k(x)z(t),u)_{\Gamma_{2}}+\nu(S\tilde{z}(t),\tilde{u})_{\Gamma_{3}}$
$\displaystyle\hskip
28.45274pt+2(\alpha(x)z(t),u)_{\Gamma_{5}}-\nu(S\tilde{z}(t),\tilde{u})_{\Gamma_{5}}$
$\displaystyle=e^{-k_{1}t}\Big{[}-(U^{\prime}(t),u)-\nu(\nabla U,\nabla
u)-\langle(U(t)\cdot\nabla)U(t),u\rangle$ $\displaystyle\hskip
28.45274pt-\nu(k(x)U(t),u)_{\Gamma_{2}}-\nu(S\tilde{U}(t),\tilde{u})_{\Gamma_{3}}-2(\alpha(x)U(t),u)_{\Gamma_{5}}-\nu(S\tilde{U}(t),\tilde{u})_{\Gamma_{5}}$
$\displaystyle\hskip 28.45274pt+\langle
f(t),u\rangle+\sum_{i=2,4}\langle\phi_{i}(t),u_{n}\rangle_{\Gamma_{i}}+\sum_{i=3,5,7}\langle\phi_{i}(t),u\rangle_{\Gamma_{i}}\Big{]}\quad\mbox{for
all}\,\,u\in\mathbf{V}_{1}.$

Define operators $A_{1}$, $A_{1U}(t)$ by

(4.7) $\langle A_{1}v,u\rangle=\langle
A_{01}v,u\rangle+(k_{1}v,u)\quad\mbox{for all}\,\,v,u\in\mathbf{V}_{1},$ (4.8)
$\displaystyle\langle
A_{1U}(t)v,u\rangle=\langle(U(t,x)\cdot\nabla)v,u\rangle+\langle(v\cdot\nabla)U(t,x),u\rangle\quad\mbox{for
all}\,\,v,u\in\mathbf{V}_{1},$

where $A_{01}$ is one defined in (4.5). $U\in\mathscr{W}$ implies $U\in
C\left([0,T];\textbf{H}^{1}(\Omega)\right)$, and such definitions have
meaning. Also, define an operator $B_{1}(t):V_{1}\rightarrow V_{1}^{*}$ by

(4.9) $\langle
B_{1}(t)v,u\rangle=e^{k_{1}t}\langle(v\cdot\nabla)v,u\rangle\quad\mbox{for
all}\,\,v,u\in\mathbf{V}_{1}.$

Define an element $F_{1}\in\mathscr{Y}_{1}$ by

(4.10) $\displaystyle\langle F_{1}(t),u\rangle=e^{-k_{1}t}\Big{[}-\langle
U^{\prime}(t),u\rangle-\nu(\nabla U(t),\nabla
u)-\langle(U(t)\cdot\nabla)U(t),u\rangle$ $\displaystyle\hskip
19.91684pt-\nu(k(x)U(t),u)_{\Gamma_{2}}-\nu(S\tilde{U}(t),\tilde{u})_{\Gamma_{3}}-2(\alpha(x)U(t),u)_{\Gamma_{5}}+\nu(S\tilde{U}(t),\tilde{u})_{\Gamma_{5}}$
$\displaystyle\hskip 19.91684pt+\langle
f,u\rangle+\sum_{i=2,4}\langle\phi_{i},u_{n}\rangle_{\Gamma_{i}}+\sum_{i=3,5,7}\langle\phi_{i},u\rangle_{\Gamma_{i}}\Big{]}\quad\mbox{for
all}\,\,u\in\mathbf{V}_{1}.$

Now, in the same way as Theorem 3.8 we can prove the following theorem which
is one of the main results of this paper.

###### Theorem 4.2

Suppose that Assumptions 4.1 and 4.2 hold. Assume that $\|U\|_{\mathscr{W}}$
and the norms of $f,f^{\prime},\phi_{i},\phi_{i}^{\prime}$ in the spaces they
belong to are small enough.

If

(4.11) $\displaystyle w_{1}\equiv F_{1}(0)-(A_{1}+A_{1U}(0)+B_{1}(0))z_{0}\in
H_{1},$

where $z_{0}=v_{0}-U(0,\cdot),$ and $\|w_{1}\|_{H_{1}}$ is small enough, then
in the space $\mathscr{W}$ there exists a unique solution to (4.3).

###### Remark 4.1

By the same argument as Remark 3.3, we can see that the condition (4.11) is
equivalent to the condition $\overline{w}_{0}\in H_{1},$ where
$\overline{w}_{0}\in\mathbf{V}_{1}^{*}$ is defined by

(4.12) $\displaystyle\langle\overline{w}_{1},u\rangle$ $\displaystyle=\langle
f(0),u\rangle+\sum_{i=2,4}\langle\phi_{i}(0,x),u_{n}\rangle_{\Gamma_{i}}+\sum_{i=3,5,7}\langle\phi_{i}(0,x),u\rangle_{\Gamma_{i}}$
$\displaystyle-\big{[}\nu(\nabla v_{0},\nabla
u)+\nu(k(x)v_{0},u)_{\Gamma_{2}}+\nu(S\tilde{v}_{0},\tilde{u})_{\Gamma_{3}}+2(\alpha(x)v_{0},u)_{\Gamma_{5}}$
$\displaystyle-\nu(S\tilde{v}_{0},\tilde{u})_{\Gamma_{5}}+\langle(v_{0}\cdot\nabla)v_{0},u\rangle+k_{1}(v_{0},u)\big{]}\quad\mbox{for
all}\,\,u\in\mathbf{V}_{1},$

with $k_{1}$ in Lemma 4.1.

###### Remark 4.2

If $U\equiv 0$ and $\Gamma_{i}=\varnothing,i=2\sim 5,$ then problem (4.3) is
reduced to one in [14]. In this case $k_{1}=0.$ (cf. Remark 3.2). If
$v_{0}\in\textbf{H}^{l/2}(\Omega)$, then
$(v_{0}\cdot\nabla)v_{0}\in\textbf{L}_{2}(\Omega)$. Thus, the condition above
for $\overline{w}_{1}$ being in $H_{1}$ is the same with one of conditions of
Theorems 3.5$\sim$3.8 in [14], but we do not demand
$v_{0}\in\textbf{H}^{r_{0}}(\Omega),\,r_{0}>\frac{l}{2}$.

##  5 Existence of a unique solution for perturbed data

In [14] it is proved that if a solution satisfying smoothness and a
compatibility condition is given, then there exists a unique solution for
small perturbed data satisfying the compatibility condition. In this section
we get such results for the Problems I and II. In our results the conditions
for a given solution is essentially the same with one in [14], but the
smoothness condition for the initial functions in the compatibility condition
for small perturbed data is weaker than one in [14](cf. Remark 5.2).

Let
$\widetilde{\textbf{V}}^{r_{0}}=\\{u\in\mathbf{H}^{r_{0}}(\Omega):\mbox{div}\,u=0\\}$,
$r_{0}>l/2$, and

$\displaystyle\overline{\mathscr{W}}=\left\\{w\in
L_{2}(0,T;\widetilde{\textbf{V}});w^{\prime}\in
L_{2}(0,T;\widetilde{\textbf{V}}),w^{\prime\prime}\in
L_{2}(0,T;{\widetilde{\textbf{V}}}^{*}),w(0)\in\widetilde{\textbf{V}}^{r_{0}}\right\\},$
$\displaystyle\hskip
28.45274pt\|w\|_{\mathscr{W}}=\|w\|_{L_{2}(0,T;\widetilde{\textbf{V}})}+\|w^{\prime}\|_{L_{2}(0,T;\widetilde{\textbf{V}})}+\|w^{\prime\prime}\|_{L_{2}(0,T;{\widetilde{\textbf{V}}}^{*})}+\|w(0)\|_{\widetilde{\textbf{V}}^{r_{0}}}.$

Let us consider Problem I.

Let $W(x,t)\in\overline{\mathscr{W}}$ be a given solution to Problem I. Let
$v$ be the solution for the data perturbed except $h_{i}$ and put
$v=\overline{z}+W$. Then, we get a problem for $\overline{z}$:

Find $\overline{z}$ such that

(5.1) $\displaystyle\overline{z}\in L_{2}(0,T;\mathbf{V}),$
$\displaystyle\overline{z}(0)=z_{0}\equiv v_{0}-W(0,x)\in\mathbf{V},$
$\displaystyle\langle\overline{z}^{\prime},u\rangle+2\nu(\varepsilon(\overline{z}),\varepsilon(u))+\langle(\overline{z}\cdot\nabla)\overline{z},u\rangle+\langle(W\cdot\nabla)\overline{z},u\rangle+\langle(\overline{z}\cdot\nabla)W,u\rangle$
$\displaystyle\hskip
28.45274pt+2\nu(k(x)\overline{z},u)_{\Gamma_{2}}+2\nu(S\tilde{\overline{z}},\tilde{u})_{\Gamma_{3}}+2(\alpha(x)\overline{z},u)_{\Gamma_{5}}+\nu(k(x)\overline{z},u)_{\Gamma_{7}}$
$\displaystyle=\langle
f,u\rangle+\sum_{i=2,4,7}\langle\phi_{i},u_{n}\rangle_{\Gamma_{i}}+\sum_{i=3,5,6}\langle\phi_{i},u\rangle_{\Gamma_{i}}\quad\mbox{for
all}\,\,u\in\mathbf{V},$

where $z_{0},f,\phi_{i}$ are perturbations of corresponding data.

###### Remark 5.1

Proofs of this section are similar to one in Section 3. Main difference is
that we do not assume smallness of $W(0,x)$ unlike $U(0,x)$ in Section 3.

Define an operator $A_{02}:\textbf{V}\rightarrow\textbf{V}^{*}$ by

(5.2) $\displaystyle\langle A_{02}y,u\rangle=$ $\displaystyle
2\nu(\varepsilon(y),\varepsilon(u))+2\nu(k(x)y,u)_{\Gamma_{2}}+2\nu(S\tilde{y},\tilde{u})_{\Gamma_{3}}$
$\displaystyle+2(\alpha(x)y,u)_{\Gamma_{5}}+\nu(k(x)y,u)_{\Gamma_{7}}+\langle(W(0,x)\cdot\nabla)y,u\rangle$
$\displaystyle+\langle(y\cdot\nabla)W(0,x),u\rangle\qquad\mbox{for
all}\,\,y,u\in\mathbf{V}.$

###### Lemma 5.1

There exists $\delta>0$ and $k_{2}\geq 0$ such that

$\langle
A_{02}u,u\rangle\geq\delta\|u\|_{\textbf{{V}}}^{2}-k_{2}\|u\|_{H}^{2}\quad\mbox{for
all}\,\,u\in\mathbf{V}.$

_Proof_ By Korn’s inequality

(5.3)
$2\nu(\varepsilon(u),\varepsilon(u))\geq\beta\|u\|^{2}_{\mathbf{V}}\quad\exists\beta>0,\mbox{for
all}\,\,u\in\mathbf{V}.$

By Remark 2.2, there exists a constant $M$ such that

$\displaystyle\|S(x)\|_{\infty},\,\|k(x)\|_{\infty},\|\alpha(x)\|_{\infty}\leq
M.$

Then, there exists a constant $c_{0}$ (depend on $\beta$) such that

(5.4)
$\displaystyle\left|2\nu(k(x)u,u)_{\Gamma_{2}}+2\nu(S\tilde{u},\tilde{u})_{\Gamma_{3}}+\nu(k(x)u,u)_{\Gamma_{7}}+2(\alpha(x)u,u)_{\Gamma_{5}}\right|$
$\displaystyle\hskip
56.9055pt\leq\frac{\beta}{4}\|u\|^{2}_{\mathbf{H}^{1}(\Omega)}+c_{0}\|u\|_{H}^{2}\quad\mbox{for
all}\,\,u\in\mathbf{V}$

(cf. Theorem 1.6.6 in [9]).

Let us estimate
$\langle(W(0,x)\cdot\nabla)u,u\rangle+\langle(u\cdot\nabla)W(0,x),u\rangle$.
Since $W(0,x)\in C(\overline{\Omega})$,

(5.5)
$\big{|}\big{\langle}(W(0,x)\cdot\nabla)u,u\big{\rangle}\big{|}\leq\frac{\beta}{8}\|u\|^{2}_{\mathbf{H}^{1}(\Omega)}+c_{1}\|u\|_{H}^{2}.$

Taking $\text{div}\,u=0$ into account, we get

$\displaystyle\big{\langle}(u\cdot\nabla)W(0,x),u\big{\rangle}$
$\displaystyle=\sum_{j}\int_{\Omega}u_{j}\frac{\partial W(0,x)}{\partial
x_{j}}u\,dx$ $\displaystyle=\int_{\partial\Omega}(u\cdot n)(W(0,x)\cdot
u)\,d\Gamma-\sum_{j}\int_{\Omega}u_{j}\frac{\partial u}{\partial
x_{j}}W(0,x)\,dx.$

On the right hand side of the formula above estimating the first term as in
(5.4) and applying Hölder inequality in the second term, we have

(5.6)
$\big{|}\big{\langle}(u\cdot\nabla)W(0,x),u\big{\rangle}\big{|}\leq\frac{\beta}{8}\|u\|^{2}_{\mathbf{H}^{1}(\Omega)}+c_{2}\|u\|_{H}^{2}.$

Putting $\delta=\frac{\beta}{2}$, $k_{2}=c_{0}+c_{1}+c_{2}$, from (5.3)-(5.6)
we get the asserted conclusion. $\square$

Put $z=e^{-k_{2}t}\overline{z}$, where $k_{2}$ is a constant in Lemma 5.1.
Then, $e^{-k_{2}t}\overline{z}^{\prime}=z^{\prime}+k_{2}z$ and we have the
following problem which is equivalent to (5.1).

Find $z$ such that

(5.7) $\displaystyle z\in L_{2}(0,T;\mathbf{V}),$ $\displaystyle
z(0)=z_{0}=v_{0}-W(0)\in\mathbf{V},$ $\displaystyle\langle
z^{\prime}(t),u\rangle+2\nu(\varepsilon(z)(t),\varepsilon(u))+e^{k_{2}t}\langle(z(t)\cdot\nabla)z(t),u\rangle+\langle(W(t)\cdot\nabla)z(t),u\rangle$
$\displaystyle\hskip
28.45274pt+\langle(z(t)\cdot\nabla)W(t),u\rangle+k_{2}(z(t),u)+2\nu(k(x)z(t),u)_{\Gamma_{2}}+2\nu(S\tilde{z}(t),\tilde{u})_{\Gamma_{3}}$
$\displaystyle\hskip
28.45274pt+2(\alpha(x)z(t),u)_{\Gamma_{5}}+\nu(k(x)z(t),u)_{\Gamma_{7}}$
$\displaystyle=e^{-k_{2}t}\Big{[}\langle
f(t),u\rangle+\sum_{i=2,4,7}\langle\phi_{i}(t),u_{n}\rangle_{\Gamma_{i}}+\sum_{i=3,5,6}\langle\phi_{i}(t),u\rangle_{\Gamma_{i}}\Big{]}\quad\mbox{for
all}\,\,u\in\mathbf{V}.$

Define operators $A_{2},A_{W}(t):\textbf{V}\rightarrow\textbf{V}^{*}$ by

(5.8) $\displaystyle\langle A_{2}y,u\rangle=$ $\displaystyle
2\nu(\varepsilon(y),\varepsilon(u))+2\nu(k(x)y,u)_{\Gamma_{2}}+2\nu(S\tilde{y},\tilde{u})_{\Gamma_{3}}$
$\displaystyle+2(\alpha(x)y,u)_{\Gamma_{5}}+\nu(k(x)y,u)_{\Gamma_{7}}+k_{2}(y,u)\quad\mbox{for
all}\,\,y,u\in\mathbf{V},$ (5.9) $\displaystyle\langle
A_{W}(t)v,u\rangle=\langle(W(t,x)\cdot\nabla)v,u\rangle+\langle(v\cdot\nabla)W(t,x),u\rangle\quad\mbox{for
all}\,\,v,u\in\mathbf{V},$

where $k_{2}$ is a constant in Lemma 5.1. $W\in\overline{\mathscr{W}}$ implies
$W\in C\left([0,T];\textbf{H}^{1}(\Omega)\right)$, and such definitions are
well.

In proof of Lemma 5.1 it is clear that

(5.10) $\left\langle
A_{2}u,u\right\rangle\geq\frac{3\beta}{4}\|u\|^{2}_{\textbf{V}}.$

Also, by Lemma 5.1

(5.11)
$\left\langle\big{(}A_{2}+A_{W}(0)\big{)}u,u\right\rangle\geq\frac{\beta}{4}\|u\|^{2}_{\textbf{V}}.$

Define an operator $B_{2}(t):V\rightarrow V^{*}$ by

(5.12) $\langle
B_{2}(t)v,u\rangle=e^{k_{2}t}\langle(v\cdot\nabla)v,u\rangle\quad\mbox{for
all}\,\,v,u\in\mathbf{V}.$

Define operators
$L_{2},{\widetilde{A}}_{W},L_{2W},{\widetilde{B}}_{2}:\mathscr{X}\rightarrow\mathscr{Y}$,
$C_{2}:\mathscr{X}\times\mathscr{X}\rightarrow\mathscr{Y}$ and an element
$F_{2}\in\mathscr{Y}$ by

(5.13) $\displaystyle\langle(L_{2}z)(t),u\rangle=\langle
z^{\prime}(t),u\rangle+\langle A_{2}z(t),u\rangle\quad\mbox{for
all}\,\,z\in\mathscr{X},\mbox{for all}\,\,u\in\mathbf{V},$
$\displaystyle\langle({\widetilde{A}}_{W}z)(t),u\rangle=\langle
A_{W}(t)z(t),u\rangle\quad\mbox{for all}\,\,z\in\mathscr{X},\mbox{for
all}\,\,u\in\mathbf{V},$ $\displaystyle\langle(L_{2W}z)(t),u\rangle=\langle
z^{\prime}(t),u\rangle+\big{\langle}\big{(}A_{2}+A_{W}(t)\big{)}z(t),u\big{\rangle}\quad\mbox{for
all}\,\,z\in\mathscr{X},\mbox{for all}\,\,u\in\mathbf{V},$
$\displaystyle\langle({\widetilde{B}}_{2}z)(t),u\rangle=\langle
B_{2}(t)z(t),u\rangle\quad\mbox{for all}\,\,z\in\mathscr{X},\mbox{for
all}\,\,u\in\mathbf{V},$ $\displaystyle\langle
C_{2}(w,z)(t),u\rangle=e^{k_{2}t}\langle(w\cdot\nabla)z,u\rangle+e^{k_{2}t}\langle(z\cdot\nabla)w,u\rangle\quad\mbox{for
all}\,\,w,z\in\mathscr{X},\mbox{for all}\,\,u\in\mathbf{V},$
$\displaystyle\langle(F_{2})(t),u\rangle=e^{-k_{2}t}\Big{[}\langle
f(t),u\rangle+\sum_{i=2,4,7}\langle\phi_{i}(t),u_{n}\rangle_{\Gamma_{i}}+\sum_{i=3,5,6}\langle\phi_{i}(t),u\rangle_{\Gamma_{i}}\Big{]}\quad\mbox{for
all}\,\,u\in\mathbf{V}.$

By the argument as Lemma 3.2 we get

###### Lemma 5.2

$C_{2}$ is a bilinear continuous operator such that
$\mathscr{X}\times\mathscr{X}\rightarrow\mathscr{Y}$. Under Assumption 3.1
${\widetilde{A}}_{W}$ is a linear continuous operator such that
$\mathscr{X}\rightarrow\mathscr{Y}$ and $F_{2}\in\mathscr{Y}$.

Using (5.10) instead of (3.20), as Lemma 3.3 we get

###### Lemma 5.3

The operator $\overline{L}_{2}$ defined by
$\overline{L}_{2}z=(z^{\prime}(0),L_{2}z)$ for $z\in\mathscr{X}$ is a linear
continuous one-to-one operator from $\mathscr{X}$ onto $H\times\mathscr{Y}$.

Now, using (5.11) without assuming the fact that
$\|W(0,x)\|_{\widetilde{\textbf{V}}}$ is small enough, as Lemma 3.4 we prove
the following

###### Lemma 5.4

The operator $\overline{L}_{2W}$ defined by
$\overline{L}_{2W}z=(z^{\prime}(0),L_{2W}z)$ for $z\in\mathscr{X}$ is a linear
continuous one-to-one operator from $\mathscr{X}$ onto $H\times\mathscr{Y}$.

_Proof_ As Lemma 3.5 in [14] it is proved that the operator
${\widetilde{A}}_{W}\in(\mathscr{X}\rightarrow\mathscr{Y})$ is compact. Thus,
$\overline{A}_{W}\in(\mathscr{X}\rightarrow H\times\mathscr{Y})$ defined by
$\overline{A}_{W}z=\\{0_{H},\widetilde{A}_{W}z\\}$ is also compact. Since
$\overline{L}_{2W}=\overline{L}_{2}+\overline{A}_{W}$, in order to get the
asserted conclusion by virtue of Theorem 3.4 in [14] it is enough to prove
that $\overline{L}_{2W}$ is one-to-one from $\mathscr{X}$ into
$H\times\mathscr{Y}$.

To prove the last fact it is enough to show that the inverse image of
$(0_{H},0_{\mathscr{Y}})$ by $\overline{L}_{2W}$ is $0_{\mathscr{X}}$. It is
easy to verify that

(5.14)
$\big{|}\big{\langle}\big{(}A_{2}+A_{W}(0)\big{)}v,u\big{\rangle}\big{|}\leq
c\|v\|_{\mathbf{V}}\cdot\|u\|_{\mathbf{V}}\quad\mbox{for
all}\,\,v,u\in\mathbf{V}.$

By (5.11), (5.14) for any $q\in\textbf{V}^{*}$ there exists a unique solution
$y\in V$ to

(5.15) $(A_{2}+A_{W}(0))y=q.$

Let $z\in\mathscr{X}$ be the inverse image of $(0_{H},0_{\mathscr{Y}})$ by
$\overline{L}$. Then, $z^{\prime}(0)=0_{H}$, and putting $t=0$ from the third
one in (5.13) we get

$\big{\langle}\big{(}A_{2}+A_{W}(0)\big{)}z(0),u\big{\rangle}=0\quad\mbox{for
all}\,\,u\in\mathbf{V},$

where $z(0)=z(0,x)$. This means that $z(0)$ is the unique solution to (5.15)
with $q=0_{\textbf{V}^{*}}$, i.e., $z(0)=0_{\textbf{V}}$. Therefore,
$z\in\mathscr{X}$ satisfies

(5.16) $\left\\{\begin{aligned}
&z^{\prime}(t)+\big{(}A_{2}+A_{W}(t)\big{)}z(t)=0,\\\
&z(0)=0_{\textbf{V}}.\end{aligned}\right.$

Now, using (5.16) and Gronwall’s inequality, as in Lemma 3.8 of [14] we can
prove $z=0_{\mathscr{X}}$. It is finished to prove the Lemma. $\square$

By the argument as Lemma 3.5 we get

###### Lemma 5.5

The operator $T_{2}$ defined by
$T_{2}z=\left(z^{\prime}(0),(L_{2W}+{\widetilde{B}}_{2})z\right)$ for
$z\in\mathscr{X}$ is continuously differentiable,
$T_{2}0_{\mathscr{X}}=\left(0_{H},0_{\mathscr{Y}}\right)$ and the Frechet
derivative of $T_{2}$ at $0_{\mathscr{X}}$ is $\overline{L}_{2W}$.

Let us consider the following problem

(5.17) $\big{(}A_{2}+A_{W}(0)+B_{2}(0)\big{)}u=q.$

Now, using (5.11) without assuming the fact that
$\|W(0,x)\|_{\widetilde{\textbf{V}}}$ is small enough, as Lemma 3.6 we can
prove

###### Lemma 5.6

If the norm of $q\in V^{*}$ is small enough, then there exists a unique
solution to (5.17) in some $\mathscr{O}_{M}(0_{\mathbf{V}})$.

Using Lemmas 5.2$\sim$5.5, Proposition 3.7, in the same way as Theorem 3.8 we
get

###### Theorem 5.7

Suppose that Assumptions 3.1 holds and the norms of $f,f^{\prime},\phi_{i},$
$\phi_{i}^{\prime}$ in the spaces they belong to are small enough.

If

(5.18) $\displaystyle w_{2}\equiv F_{2}(0)-(A_{2}+A_{2W}(0)+B_{2}(0))z_{0}\in
H,$

where $z_{0}=v_{0}-U(0,\cdot),$ and $\|w_{2}\|_{H}$ is small enough, then
there exists a unique solution to (5.1) in the space $\mathscr{W}.$

###### Remark 5.2

By the same argument as Remark 3.3, we can see that the condition (5.18) is
equivalent to the condition $\overline{w}_{2}\in H_{1},$ where
$\overline{w}_{2}\in\mathbf{V}_{1}^{*}$ is defined by

(5.19) $\displaystyle\langle\overline{w}_{2},u\rangle$ $\displaystyle=\langle
f(0),u\rangle+\sum_{i=2,4}\langle\phi_{i}(0,x),u_{n}\rangle_{\Gamma_{i}}+\sum_{i=3,5,7}\langle\phi_{i}(0,x),u\rangle_{\Gamma_{i}}$
$\displaystyle-\big{[}2\nu(\varepsilon(z_{0}),\varepsilon(u))+2\nu(k(x)z_{0},u)_{\Gamma_{2}}+2\nu(S\tilde{z}_{0},\tilde{u})_{\Gamma_{3}}$
$\displaystyle+2(\alpha(x)z_{0},u)_{\Gamma_{5}}+\nu(k(x)z_{0},u)_{\Gamma_{7}}+\langle(W(0,x)\cdot\nabla)z_{0},u\rangle$
$\displaystyle+\langle(z_{0}\cdot\nabla)W(0,x),u\rangle+k_{2}(z_{0},u)+\langle(z_{0}\cdot\nabla)z_{0},u\rangle\big{]}\quad\mbox{for
all}\,\,u\in\mathbf{V}$

with $k_{2}$ in Lemma 5.1.

Let us consider Problem II.

Let $W(x,t)\in\overline{\mathscr{W}}$ be given solution to Problem II. Let $v$
be the solution for the data perturbed except $h_{i}$ and put
$v=\overline{z}+W$. Then, we get a problem for $\overline{z}$:

Find $\overline{z}$ such that

(5.20) $\displaystyle\overline{z}\in L_{2}(0,T;\mathbf{V}_{1}),$
$\displaystyle\overline{z}(0)=z_{0}\equiv v_{0}-W(0,x)\in\mathbf{V}_{1},$
$\displaystyle\langle\overline{z}^{\prime},u\rangle+\nu(\nabla\overline{z},\nabla
u)+\langle(\overline{z}\cdot\nabla)\overline{z},u\rangle+\langle(W\cdot\nabla)\overline{z},u\rangle+\langle(\overline{z}\cdot\nabla)W,u\rangle$
$\displaystyle\hskip
28.45274pt+\nu(k(x)\overline{z},u)_{\Gamma_{2}}+\nu(S\tilde{\overline{z}},\tilde{u})_{\Gamma_{3}}+2(\alpha(x)\overline{z},u)_{\Gamma_{5}}-\nu(S\tilde{\overline{z}},\tilde{u})_{\Gamma_{5}}$
$\displaystyle\hskip 14.22636pt=\langle
f,u\rangle+\sum_{i=2,4}\langle\phi_{i},u_{n}\rangle_{\Gamma_{i}}+\sum_{i=3,5,7}\langle\phi_{i},u\rangle_{\Gamma_{i}}\quad\mbox{for
all}\,\,u\in\mathbf{V}_{1},$

where $z_{0},f,\phi_{i}$ are perturbations of corresponding data.

By the same argument as Theorem 5.7 we have

###### Theorem 5.8

Suppose that Assumptions 3.1 holds and the norms of $f,f^{\prime},$
$\phi_{i},\phi_{i}^{\prime}$ in the spaces they belong to are small enough.
Define an element $w_{3}\in\mathbf{V}_{1}^{*}$ by

(5.21) $\displaystyle\langle w_{3},u\rangle$ $\displaystyle=\langle
f(0),u\rangle+\sum_{i=2,4}\langle\phi_{i}(0,x),u_{n}\rangle_{\Gamma_{i}}+\sum_{i=3,5,7}\langle\phi_{i}(0,x),u\rangle_{\Gamma_{i}}$
$\displaystyle-\big{[}\nu(\nabla z_{0},\nabla
u)+\nu(k(x)z_{0},u)_{\Gamma_{2}}+\nu(S\tilde{z}_{0},\tilde{u})_{\Gamma_{3}}+2(\alpha(x)z_{0},u)_{\Gamma_{5}}$
$\displaystyle-\nu(S\tilde{z}_{0},\tilde{u})_{\Gamma_{5}}+\langle(W(0,x)\cdot\nabla)z_{0},u\rangle+\langle(z_{0}\cdot\nabla)W(0,x),u\rangle$
$\displaystyle+k_{3}(z_{0},u)+\langle(z_{0}\cdot\nabla)z_{0},u\rangle\big{]}\hskip
11.38092pt\mbox{for all}\,\,u\in\mathbf{V}_{1},$

where $k_{3}$ is a constant determined as in Lemma 5.1.

If $w_{3}\in\mathscr{O}_{R}(0_{H_{1}})$ for $R>0$ small enough, then there
exists a unique solution to (5.20) in the space $\mathscr{W}$.

###### Remark 5.3

If $\Gamma_{i}=\varnothing,i=2\sim 5,$ then problem (5.20) is reduced to one
in [14]. If $z_{0}\in\textbf{H}^{l/2}(\Omega)$, then
$(z_{0}\cdot\nabla)z_{0},(W(0,x)\cdot\nabla)z_{0},(z_{0}\cdot\nabla)W(0,x)\in\textbf{L}_{2}(\Omega)$
and $k_{3}z_{0}\in\textbf{L}_{2}(\Omega)$. Thus, the last 4 terms in the right
hand side of (5.21) do not give any effect to the condition for $w_{3}$ being
in $H_{1}$, and so the conditions in the Theorem 5.8 is the same with one of
conditions of Theorems 3.5$\sim$3.8 in [14]. Thus, Theorem 5.8 guarantees
existence of a unique solution under a condition weaker that one in [14].

Note that putting $W(t,x)\equiv 0$ in Theorems 5.7 and 5.8, we can not get
Theorems 3.8 and 4.2, since there $h_{i}\neq 0$.

Acknowledgment: The authors are grateful to the anonymous referee for his or
her valuable comments.

## References

  * [1] Beneš, M., Solutions to the mixed problem of viscous incompressible flows in a channel, Arch. Math., 93 (2009), 287-297
  * [2] Beneš, M., Mixed initial-boundary value problem for the three-dimentional Navier-Stokes equations in polyhedral domains, Discrete Contin. Dyn. Syst., suppl, 02 (2011), 135-144.
  * [3] Beneš, M., Strong solutions to non-stationary channel flows of heat-conducting viscous incompressible fluids with dissipative heating, Acta Appl. Math., 116 (2011), 237-254.
  * [4] Beneš, M., Solutions of the Navier-Stokes equations with various types of boundary conditions, Arch. Math., 98 (2012), 487-497.
  * [5] Beneš, M., A note on regularity and uniqueness of natural convection with effects of viscous dissipation in 3D open channels, Z. Angew. Math. Phys., 65 (2014), 961-975.
  * [6] Beneš, M., The “do nothing” problem for Boussinesq fluids, Appl. Math. Lett., 31 (2014), 25-28.
  * [7] Beneš, M. and Kučera, P., Non-steady Navier-Stokes equations with homogeneous mixed boundary conditions and arbitrarily large initial condition, Carpathian J. Math., 23 1-2 (2007), 32-40.
  * [8] Beneš, M. and Kučera, P., Solutions to the Navier-Stokes equations with mixed boundary conditions in two-dimensional bounded domains, Math. Nachr., 289 2-3 (2016), 194-212.
  * [9] Brenner, S. C. and Scott, L. R., The mathematical theory of finite element methods, Springer, 2008.
  * [10] Drábek, P. and Milota, J., Methods of Nonlinear Analysis, Birkh${\ddot{\mbox{a}}}$ser, 2007.
  * [11] Evans, L. C., Partial differential equations, American Mathematical Society, 1998.
  * [12] Gajewski, H., Gr${\ddot{\mbox{o}}}$ger, K. and Zacharias, K., Nichtlineare operatorgleichungen und operatordifferentialgleichungen, Academic-Verlag, 1974 (Russian 1978).
  * [13] Kim, T. and Cao, D., Some properties on the surfaces of vector fields and its application to the Stokes and Navier-Stokes problems with mixed boundary conditions, Nonlin. Anal. TMA, 113 (2015), 94-114.
  * [14] Kučera, P., Basic properties of solution of the non-steady Navier-Stokes equations with mixed boundary conditions in a bounded domain, Ann. Univ. Ferrara, 55 (2009), 289-308.
  * [15] Kučera, P. and Skalák, Z., Local solutions to the Navier-Stokes equations with mixed boundary conditions, Acta Appl. Math., 54 (1998), 275-288.
  * [16] Marušić, S., On the Navier-Stokes system with pressure boundary condition, Ann. Univ. Ferrara, 53 (2007), 319-311.
  * [17] Mazya, V. G. and Rossmann, J., $L^{p}$ estimates of solutions to mixed boundary value problems for the Stokes system in polyhedral domains, Math. Nachr., 280 (2007), 751-793.
  * [18] Mazya, V.G. and Rossmann, J., Elliptic Equations in Polyhedral Domains, Mathematical Surveys and Monographs, vol. 162, American Mathematical Society, 2010.
  * [19] Orlt, M. and Sändig, A.-M., Regularity of viscous Navier-Stokes fows in nonsmooth domains, Boundary value problems and integral equations in nonsmooth domains (Luminy, 1993), 185-201, Lecture Notes in Pure and Appl. Math., 167, Dekker, New York, 1995.
  * [20] Skalák, Z. and Kučera, P., An existence theorem for the Boussinesq equations with non-Dirichlet boundary conditions, Appl. Math., 45(2) (2000), 81-98.