1 Introduction

Magnetohydrodynamics (MHD) describes the dynamics of conducting fluids, such as gases, under the in influence of an electromagnetic field. If a conducting fluid moves in a magnetic field, electric fields are induced and an electric current flow is developed. The magnetic field exerts forces on these currents which considerably modify the hydrodynamic motion of the fluid. There is a complex interaction between the magnetic field and fluid dynamic phenomena, and both hydrodynamic and electrodynamic effects have to be considered. The applications of magnetohydrodynamics cover a very wide range of physical areas from liquid metals to cosmic plasmas, for example, the intensely heated and ionized fluids in an electromagnetic field in astrophysics, geophysics, high-speed aerodynamics and plasma physics. The set of equations which are obtained by neglecting the motion of electrons and consider only heavy ions, are a combination of the compressible Navier–Stokes equations of fluid dynamics and Maxwell’s equations of electromagnetism, connecting the magnetic field, plasma velocity, pressure and density.

This paper is concerned with the steady compressible magnetohydrodynamic system with the gravitational force on \( \Omega = \mathbb {R}^2\times ]0,h[\)

$$\begin{aligned}{} & {} \hspace{4pc} \nabla \cdot (\varrho v) = 0, \end{aligned}$$
(1.1)
$$\begin{aligned}{} & {} \varrho (v\cdot \nabla )v-\mu \Delta v - (\mu +\lambda )\nabla \nabla \cdot v = -\nabla p - g\varrho e_3 + (\nabla \times B)\times B, \end{aligned}$$
(1.2)
$$\begin{aligned}{} & {} \nu \nabla \times (\nabla \times B) - \nabla \times (v\times B) = 0,\quad \nabla \cdot B = 0 \end{aligned}$$
(1.3)

with the following boundary conditions

$$\begin{aligned}{} & {} v(x',0)= v(x',h)= 0, \quad B(x',0)= {\bar{B}}, \ \ B(x',h) = {\bar{B}} + \delta (x')e_3, \end{aligned}$$
(1.4)
$$\begin{aligned}{} & {} v\rightarrow 0, \quad B\rightarrow B_\infty ,\quad \varrho \rightarrow \varrho _\infty \quad \hbox {as} \ \ |x'|\rightarrow +\infty ,\end{aligned}$$
(1.5)

where \({\bar{B}}\in \mathbb {R}^3\) is a given constant field while \(\delta (x')\in \mathbb {R}\) is small perturbation of magnetic field \({\bar{B}}\) on the upper plane \(x_3 = h\) and \(x'= (x_1,x_2)\) is a generic point in \(\mathbb {R}^2\) so that \(x=(x',x_3)\in \Omega = \mathbb {R}^2\times ]0,h[\). As to \(B_{\infty }(x_3)\in \mathbb {R}^3\) and \(\varrho _{\infty }(x_3)>0\) will be defined below (see (1.10)). We will assume that \(\varrho _{\infty }\in L^1([0,h])\) and

$$\begin{aligned} \int _0^h\varrho _\infty (x_3) dx_3 = M,\end{aligned}$$
(1.6)

where \(M>0\) is given.

In the system (1.1)–(1.3), the unknowns \(v,B:\Omega \rightarrow \mathbb {R}^3\) and \(\varrho :\Omega \rightarrow (0,\infty )\) stand for the velocity field, magnetic field and density respectively. The physical constants \(\mu \) and \(\lambda \) are the shear viscosity and bulk viscosity of the fluid respectively and satisfy \(\mu >0\) and \(2\mu +3\lambda >0\), while the other physical constant \(\nu > 0\) is the magnetic diffusivity coefficient, g the gravitational acceleration and \(e_3 =(0,0,1)\) is the horizontal axis.

We assume that the pressure p is given by the law of viscous barotropic fluid

$$\begin{aligned} p = \vartheta \varrho ^\gamma \ \ \text{ with } \ \ \gamma > 1 \ \ \text{ and } \ \ \vartheta \ \ \text{ is } \text{ a } \text{ positive } \text{ constant }.\end{aligned}$$
(1.7)

We look for a stationary solution in a neighborhood close to the equilibrium state \((0,B_{eq},\varrho _{eq})\) corresponding to the perturbation \(\delta \) in (1.4) identically zero. As can easily be seen, this rest state is given by the following density and magnetic field distributions

$$\begin{aligned} B_{eq}(x_3) = {\bar{B}},\quad \varrho _{eq}(x_3) = \Big (C_M - \displaystyle {\gamma -1\over \vartheta \gamma }gx_3\Big )^{1\over \gamma -1},\ \ \gamma >1\end{aligned}$$
(1.8)

where the only positive constant \(C_M\) is determined by the conditions

$$\begin{aligned} \int _0^h\varrho _{eq}(x_3)dx_3 = M,\ \ \displaystyle \inf _{0\le x_3\le h}{\varrho }_{eq}(x_3) = \Big (C_M - \displaystyle {\gamma -1\over \vartheta \gamma }gh\Big )^{1\over \gamma -1}>0.\end{aligned}$$
(1.9)

This being, we can now specify the boundary conditions (1.5) for magnetic field and density. We then set (see (1.8))

$$\begin{aligned}&B_{\infty }(x_3) = B_{eq}(x_3),\quad \varrho _{\infty }(x_3) = \varrho _{eq}(x_3). \end{aligned}$$
(1.10)

From now on we make the assumptions

$$\begin{aligned} M\ge M_0,\quad |{\bar{B}}|\ge B_0.\end{aligned}$$
(1.11)

where \(B_0>0\) and \(M_0>0\) are given large enough.

As to the small perturbation \(\delta \) of the given magnetic field \({\bar{B}}\) on the upper plane \(x_3=h\), we assume that

$$\begin{aligned} \delta \in H^3(\mathbb {R}^2)\cap L^1(\mathbb {R}^2),\quad \Vert \delta \Vert _{H^3(\mathbb {R}^2)}\le \delta _0,\quad \displaystyle \int _{\mathbb {R}^2}\delta (x')\ dx'=0,\end{aligned}$$
(1.12)

where \(\delta _0>0\) is given small enough. According to (1.9) and (1.8), we have

$$\begin{aligned} M = \int _0^h\Big (C_M - \displaystyle {\gamma -1\over \vartheta \gamma }gx_3\Big )^{1\over \gamma -1}dx_3 \le hC_M^{1\over \gamma -1}. \end{aligned}$$

Hence, (see (1.11))

$$\begin{aligned} C_M\ge \Big [\displaystyle {M_0\over h}\Big ]^{\gamma -1},\end{aligned}$$
(1.13)

where \(M_0\) is large enough so that we definitely get

$$\begin{aligned} \displaystyle {1\over 2}C_M\le \displaystyle \inf _{{\bar{\Omega }}}\varrho _{eq}^{\gamma -1}\le \varrho _{eq}^{\gamma -1}\le C_M.\end{aligned}$$
(1.14)

For the equations (1.1)–(1.3), Zhou [48] proved the existence of a spatially periodic weak solution to the steady compressible isentropic MHD equation with \(p(\varrho ) = a\varrho ^\gamma \) (\(a>0\)) in \(\mathbb {R}^3\) for any specific heat ratio \(\gamma >1\) and a periodic external force \(f\in L^\infty (\mathbb {R}^3)\). To the best our knowledge, this is the first result concerning the steady MHD equations of compressible flows. For the incompressible case, we refer the interested reader to the articles [1,2,3,4, 26, 34] for the existence of strong and weak solutions. In [23], Gerbeau et al. also considered several kinds of unsteady problems and gave some numerical analysis (see also [41]). Because of its physical importance, complexity and mathematical challenges, there have been numerous studies on unsteady MHD by physicists and mathematicians in the recent years (see e.g., [16, 18,19,20,21, 31, 32, 45] and the references cited therein). In particular, the one-dimensional problem has been studied in many papers, see, for examples, [6, 10, 15, 33]. On the other hand, there has been considerable interest in developing accurate and reliable numerical methods for the study of MHD system of equations (see e.g., [5, 12, 24, 29, 35, 40, 42]).

Almost all the literature mentioned above is concerned with the Cauchy problem for (1.1)–(1.3) or the initial boundary value problem for compressible MHD equations, with the homoegenous Dirichlet condition on the magnetic field. In contrast with the extensive researches on unsteady MHD flow, we find that there are only few results concerning the steady flow. In [46], Yang et al. have established the existence and uniqueness of a strong solution to the steady magnetohydrodynamic equations for the compressible barotropic fluids in a bounded smooth domain with a perfectly conducting boundary, under the assumption that the external force field is small. This solution is obtained in a neighborhood close to a constant equilibrium state.

In [7] the authors have investigated the question of a stationary motion in an horizontal layer of a viscous compressible fluid in the presence of a magnetic field subjected to gravitational force. By imposing periodic conditions in the horizontal directions, they showed that a non-homogeneous distribution of the magnetic field given around a large constant on the upper plane can generate a stationary motion in a state close to the equilibrium state \((0,B_{eq},\varrho _{eq})\) (see (1.8)).

In this work, we consider in an infinite horizontal layer the stationary motion of a compressible viscous fluid in a magnetic field subjected to the gravitational force with boundary conditions on the magnetic field close to a arbitrarily large constant. As will be seen later, the mathematical analysis of the equations describing this motion in question presents serious challenges.

Remark 1.1

If the perturbation \(\delta \) of magnetic field \({\bar{B}}\) on the upper plane \(x_3 = h\) (see (1.4)) is not identically zero, and if there exists a solution \((v,B,\varrho )\) of system equations (1.1)–(1.5), then necessarily v is not identically equal to zero (see Remark 7.1 section 5). This is obtained from the main result (Theorem 2.1 and see also Remark 2.1) in the next section.

2 Main Result and Preliminary of Proof

We recall that we look for a solution \((v,B,\varrho )\) in a small neighborhood of the stationary profile \((0,\widehat{B},\widehat{\varrho })\) close to the equilibrium state \((0,B_{eq},\varrho _{eq})\) given in (1.8). The stationary profile is defined by

$$\begin{aligned} \widehat{B}(x',x_3) = B_{eq}(x_3) + \displaystyle {x_3\over h}\delta e_3 + \widehat{b}(x',x_3),\quad \widehat{\varrho }(x_3) =\varrho _{eq}(x_3).\end{aligned}$$
(2.1)

where \(\widehat{b}\) is a solution of the boundary value problem

$$\begin{aligned}{} & {} \nabla \cdot \widehat{b} = -\displaystyle {1\over h}\delta \ \ \text{ in } \ \ \Omega = \mathbb {R}^2\times ]0,h[,\nonumber \\{} & {} \widehat{b}(x',0)= \widehat{b}(x',h)=0,\quad \widehat{b}\rightarrow 0\ \ \text{ if } \ \ |x'|\rightarrow \infty . \end{aligned}$$
(2.2)

According to (1.12)\(_3\), there exists (see e.g. [11, 22]) at least one solution \(\widehat{b}\in H^1_0(\Omega )\). Moreover we have

$$\begin{aligned} \Vert \widehat{b} \Vert _{H^k(\Omega )}\le C_\Omega \Vert \delta \Vert _{H^{k-1}(\Omega )}.\end{aligned}$$
(2.3)

Note that, considering (2.2) and (2.1), we have

$$\begin{aligned} \nabla \cdot \widehat{B} = 0.\end{aligned}$$
(2.4)

This being so, we set then

$$\begin{aligned} v = v,\quad b = B - \widehat{B},\quad \sigma = \varrho - \widehat{\varrho }.\end{aligned}$$
(2.5)

and, since as we seek \(\varrho \) close to \(\widehat{\varrho }\), we can assume that

$$\begin{aligned} \Vert \varrho - \widehat{\varrho }\Vert _{L^\infty }\le \displaystyle {1\over 2}\inf _{{\bar{\Omega }}} \widehat{\varrho }.\end{aligned}$$
(2.6)

By considering the set of identities:

$$\begin{aligned}{} & {} 2(\nabla \times B)\times B = 2(B\cdot \nabla )B - \nabla |B|^2, \\{} & {} \nabla \times (v\times B) = v\nabla \cdot B- B\nabla \cdot v + (B\cdot \nabla ) v - (v\cdot \nabla ) B,\\{} & {} \nabla \times (\nabla \times B) = \nabla \nabla \cdot B - \Delta B,\end{aligned}$$

and (2.5) (see also (2.4)), we rewrite the problem (1.1)–(1.5) with unknowns v, B and \(\varrho \) into a new problem with unknowns v, b and \(\sigma \) as follows:

$$\begin{aligned}{} & {} -\mu \Delta v - (\mu +\lambda )\nabla \nabla \cdot v = - \gamma \vartheta \nabla (\widehat{\varrho }^{\gamma -1}\sigma ) - g\sigma e_3 +(\widehat{B}\cdot \nabla )b - \nabla (b\cdot \widehat{B}) + F, \end{aligned}$$
(2.7)
$$\begin{aligned}{} & {} \hspace{4pc}\nabla \cdot (\sigma v) = - \nabla \cdot (\widehat{\varrho }v),\end{aligned}$$
(2.8)
$$\begin{aligned}{} & {} - \nu \Delta b = -\widehat{B}\nabla \cdot v + (\widehat{B}\cdot \nabla )v + v\nabla \cdot \widehat{B} + G, \ \ \nabla \cdot b = 0, \end{aligned}$$
(2.9)

with the boundary conditions

$$\begin{aligned}{} & {} v(x',0) = v(x',h)= 0, \quad b(x',0)= b(x',h) = 0,\end{aligned}$$
(2.10)
$$\begin{aligned}{} & {} v\rightarrow 0, \ \ b\rightarrow 0, \ \ \sigma \rightarrow 0 \ \ \text{ if } \ \ |x'|\rightarrow \infty ,\end{aligned}$$
(2.11)

where the functions \(F = F(v,b,\sigma )\) and \(G=G(v,b,\sigma )\) are given by

$$\begin{aligned} F= & {} (b\cdot \nabla )\widehat{B} - (\widehat{\varrho } + \sigma )(v\cdot \nabla )v\end{aligned}$$
(2.12)
$$\begin{aligned}{} & {} \quad - \ \displaystyle {1\over 2}\vartheta \gamma (\gamma -1)\nabla ((\widehat{\varrho }+\xi \sigma )^{\gamma -2}\sigma ^2) + (\nabla \times b)\times b + (\nabla \times \widehat{B})\times \widehat{B}\nonumber \\ G= & {} \nu \Delta \widehat{B} -(v\cdot \nabla )\widehat{B} + \ v\nabla \cdot b - b\nabla \cdot v + (b\cdot \nabla )v - (v\cdot \nabla )b, \end{aligned}$$
(2.13)

where \(0<\xi <1\) is a regular function depending on \(\widehat{\varrho }\) and \(\widehat{\varrho } + \sigma \).

Notice that the Eq. (2.9) is obtained from (1.3) which is written simply

$$\begin{aligned} - \nu \Delta B - \nabla \times (v \times B) = 0,\quad B = b+\widehat{B}, \end{aligned}$$

because \(\nabla \times (\nabla \times B) = \nabla \nabla \cdot B - \Delta B\), \(\nabla \cdot B = 0\), \(\nabla \cdot \widehat{B} = 0\) (see (2.4)) and so \(\nabla \cdot b= 0\). Thus the problem (1.1)–(1.5)) is reduced to find

$$\begin{aligned} (v, b,\sigma ) = (v,B - \widehat{B},\varrho -\widehat{\varrho }) \end{aligned}$$

satisfying the problem (2.7)–(2.11). Our main result is the following theorem.

Theorem 2.1

Under (1.11), the problem (1.1)–(1.5) admits at least one solution \((v,B,\varrho )\) such that

$$\begin{aligned} v\in H^3(\Omega ),\ \ B\in L^\infty (\Omega )\cap L^2(\Omega '_h), \ \ \nabla B\in H^1(\Omega ), \ \ \varrho \in L^\infty (\Omega )\cap H^2(\Omega '_h)\end{aligned}$$
(2.14)

where \(\Omega '_h = \Omega '\times ]0,h[\) and \(\Omega '\) is an arbitrary large bounded domain of \(\mathbb {R}^2\).

Remark 2.1

This result means that the non homogeneous repartition of the magnetic field around the constant field \({\bar{B}}\) in the upper plane \(x_3 = h\) can generate a stationary motion close to the equilibrium state \((0,B_{eq},\varrho _{eq})\) given by (1.8).

Remark 2.2

We look for a solution

$$\begin{aligned} (v,B,\varrho ) = (v,b + \widehat{B},\sigma + \widehat{\varrho }) \end{aligned}$$

of the system of equations (1.1)–(1.3) with boundary conditions (1.4) and (1.5)), where

$$\begin{aligned} (v,b,\sigma )\in H^3(\Omega )\times H^2(\Omega )\times H^2(\Omega ) \end{aligned}$$

is the appropriate solution of problem (2.7)–(2.11).

By recalling here (see (2.1), (1.8) and (2.2)) the expressions of \( \widehat{\rho }\) and \( \widehat{B}\) namely,

$$\begin{aligned} \widehat{\varrho }(x_3) = \Big (C_M - \displaystyle {\gamma -1\over \vartheta \gamma }gx_3\Big )^{1\over \gamma -1} \text{ and } \,\,\widehat{B}(x',x_3) = \bar{B} + \displaystyle {x_3\over h}\delta e_3 + \widehat{b}(x',x_3), \end{aligned}$$

we see that \(\widehat{\varrho }\) is only in \(H^2_{\text{ loc }}(\Omega )\) because it only depends on \(x_3\) and \(\Omega =\mathbb {R}^2\times ]0,h[\). We also have \( \widehat{B}\notin L^2(\Omega )\) otherwise, we would get that the constant field \(\bar{B}=\widehat{B}-\displaystyle {x_3\over h}\delta e_3 - \widehat{b}\in L^2(\Omega )\) (with \(\Omega \) being unbounded). However, we have \(\nabla \widehat{B}\in H^1(\Omega )\). This justifies why we only have local regularity for B and \(\varrho \) in the theorem 2.1.

The solution \((v,b,\sigma )\) will be obtained as limit of a sequence fixed points of some operators built starting from a suitable linearization of the system of equations (2.7)–(2.9).

3 Linearized Equations

Let D be an open set of \(\Omega = \mathbb {R}^2\times ]0,h[\). We put

$$\begin{aligned}{} & {} \widehat{H}^2(D) = H^2(D)\cap \widehat{H}^1_0(\Omega ), \ \ \widehat{H}^1_0(\Omega )=\{b\in H^1_0(\Omega ): \nabla \cdot b = 0\},\end{aligned}$$
(3.1)
$$\begin{aligned}{} & {} \mathbb {H}(D) = H^3(D)\times \widehat{H}^2(D)\times H^2(D),\quad {\mathbb {H}}_0(D) = H^2(D)\times H^1(D)\times H^1(D)\end{aligned}$$
(3.2)
$$\begin{aligned}{} & {} \Vert \cdot \Vert _{\mathbb {H}(D)}^2 = \Vert \cdot \Vert _{H^3(D)}^2 + \Vert \cdot \Vert _{H^2(D)}^2 + \Vert \cdot \Vert _{H^2(D)}^2,\end{aligned}$$
(3.3)
$$\begin{aligned}{} & {} \Vert \cdot \Vert _{{\mathbb {H}}_0(D)}^2 = \Vert \cdot \Vert _{H^2(D)}^2 + \Vert \cdot \Vert _{H^1(D)}^2 + \Vert \cdot \Vert _{H^1(D)}^2.\end{aligned}$$
(3.4)

Furthermore, we set

$$\begin{aligned}{} & {} V = \{u=(v,b,\sigma )\in \mathbb {H}(\Omega ): \ u \ \hbox {satisfy} \ (2.10)--(2.11)\},\ \ V_0 = (V, \Vert \cdot \Vert _{V_0})\end{aligned}$$
(3.5)
$$\begin{aligned}{} & {} \Vert \cdot \Vert _{V}= \Vert \cdot \Vert _{{\mathbb {H}}(\Omega )},\quad \Vert \cdot \Vert _{V_0}= \Vert \cdot \Vert _{{\mathbb {H}}_0(\Omega )}.\end{aligned}$$
(3.6)

Note that if D is bounded the embedding \(\mathbb {H}(D)\hookrightarrow \mathbb {H}_0(D)\) is compact.

Now, let \(u'=(v',\vartheta ',\sigma ')\in V\) and \(k>0\). We consider the equations system

$$\begin{aligned}{} & {} -\mu \Delta v - (\mu +\lambda )\nabla \nabla \cdot v =(\widehat{B}\cdot \nabla )b - \nabla (b\cdot \widehat{B}) - \gamma \vartheta \nabla (\widehat{\varrho }^{\gamma -1}\sigma ') - g\sigma ' e_3 + F' \end{aligned}$$
(3.7)
$$\begin{aligned}{} & {} - \nu \Delta b = v\nabla \cdot \widehat{B}+ (\widehat{B}\cdot \nabla )v -\widehat{B}\nabla \cdot v + G', \ \ \nabla \cdot b = 0 \end{aligned}$$
(3.8)
$$\begin{aligned}{} & {} k(\sigma -\sigma ')+\nabla \cdot (\sigma v) = - \nabla \cdot (\widehat{\varrho }v) \end{aligned}$$
(3.9)

with the conditions (2.10) and (2.11) where (see (2.12) and (2.13)),

$$\begin{aligned} G' = G(u'),\quad F'= F(u').\end{aligned}$$
(3.10)

We obtain in the following lemma, an existence result for the system (3.7)–(3.9) with the boundary conditions (2.10) and (2.11). In the sequel we note \(\left\Vert {\cdot } \right\Vert _{H^k}\) and \(\left\Vert {\cdot } \right\Vert _{L^2}\) the norm of \(H^k(\Omega )\) (\(k\ge 1\)) and \(L^2(\Omega )\). As for the constant \(c_\Omega \), it designates any constant depending only on \(\Omega \), \(\widehat{B}\) and \(\widehat{\varrho }\).

Lemma 3.1

Let \(u' = (v',b',\sigma ')\in V\). If \(k>0\) is large enough, the system (3.7)–(3.9) with the conditions (2.10) and (2.11) has a unique solution \(u=(v,b,\sigma )\in V\) such that

$$\begin{aligned}{} & {} \left\Vert {b} \right\Vert _{H^2}\le c_\Omega (\Vert \sigma '\Vert _{H^1}^2 + \Vert F'\Vert _{L^2}^2 + \Vert G'\Vert _{L^2}^2), \end{aligned}$$
(3.11)
$$\begin{aligned}{} & {} \left\Vert {v} \right\Vert _{H^3}\le c_\Omega (\Vert \sigma '\Vert _{H^2}^2 + \Vert F'\Vert _{H^1}^2 + \Vert G'\Vert _{L^2}^2),\end{aligned}$$
(3.12)
$$\begin{aligned}{} & {} k\left\Vert {\sigma } \right\Vert _{H^2}^2\le c_\Omega \left\Vert {v} \right\Vert _{H^3}\left\Vert {\sigma } \right\Vert _{H^2}^2 + k\left\Vert {\sigma '} \right\Vert _{H^2}^2 + c_\Omega \left\Vert {v} \right\Vert _{H^3}^2.\end{aligned}$$
(3.13)

Proof

Let us start by showing that the system of Eqs. (3.7) and (3.8) with the boundary conditions (2.10) and (2.11) has a unique weak solution

$$\begin{aligned} (v,b)\in H^1_0(\Omega )\times \widehat{H}^1_0(\Omega ) \end{aligned}$$

(see (3.1)) such that

$$\begin{aligned} a((v,b),(\varphi ,\psi )):= L(\varphi ,\psi )\quad \forall \ (\varphi ,\psi )\in H^1_0(\Omega )\times \widehat{H}^1_0(\Omega ),\end{aligned}$$
(3.14)

where

$$\begin{aligned} a((v,b),(\varphi ,\psi ))=&{} \displaystyle \int _\Omega (\mu \nabla v\cdot \nabla \varphi + (\mu +\lambda )\nabla \cdot v\nabla \cdot \varphi )dx + \nu \displaystyle \int _\Omega \nabla b\cdot \nabla \psi dx\\{}&{} \quad + \displaystyle \int _\Omega \big ((\widehat{B}\cdot \nabla )b - \nabla (b\cdot \widehat{B})\big )\cdot \varphi dx +\displaystyle \int _\Omega \big (v\nabla \cdot \widehat{B} + (\widehat{B}\cdot \nabla )v -\widehat{B}\nabla \cdot v\big )\cdot \psi dx,\\ L(\varphi ,\psi )=&{} \displaystyle \int _\Omega \big (- \gamma \vartheta \nabla (\widehat{\varrho }^{\gamma -1}\sigma ') - g\sigma ' e_3 + F'\big )\cdot \varphi dx + \displaystyle \int _\Omega G'\cdot \psi dx.\end{aligned}$$

Indeed, by endowing \(H^1_0(\Omega )\times \widehat{H}^1_0(\Omega )\) with the norm \(\Vert (v,b)\Vert = (\Vert \nabla v\Vert _{L^2}^2 + \Vert \nabla b\Vert _{L^2}^2)^{1\over 2}\) which makes it a Hilbert space, we can easily see that the bilinear form a is continuous on \({H}^1_0(\Omega )\times \widehat{H}^1_0(\Omega )\). Moreover, since

$$\begin{aligned} \displaystyle \int _\Omega \big [(\widehat{B}\cdot \nabla )b - \nabla (b\cdot \widehat{B})\big ]v \ dx + \displaystyle \int _\Omega \big [v\nabla \cdot \widehat{B} + (\widehat{B}\cdot \nabla )v -\widehat{B}\nabla \cdot v\big ]b \ dx=0\,, \end{aligned}$$

we have

$$\begin{aligned} a((v,b),(v,b)) = \mu \Vert \nabla v\Vert _{L^2}^2 + (\mu + \lambda ) \Vert \nabla \cdot v\Vert _{L^2}^2 + \nu \Vert \nabla b\Vert _{L^2}^2\ge \inf (\mu ,\mu + \lambda ,\nu )\Vert (v,b)\Vert ^2. \end{aligned}$$

So, the bilinear form a is coercive and, since the linear form L is obviously continuous over \(H^1_0(\Omega )\times \widehat{H}^1_0(\Omega )\), according to the Lax-Milgram theorem, the problem (3.14) has a unique solution \((v,b)\in H^1_0(\Omega )\times \widehat{H}^1_0(\Omega )\) such that

$$\begin{aligned} \Vert \nabla v\Vert _{L^2}^2 + \Vert \nabla b\Vert _{L^2}^2\le c_\Omega (\Vert \sigma '\Vert _{H^1}^2 + \Vert F'\Vert _{L^2}^2 + \Vert G'\Vert _{L^2}^2).\end{aligned}$$
(3.15)

Now, by standard \(L^2\)-regularity theory of linear elliptic systems, according to (3.15) we have (3.11) and (3.12). As for the solution \(\sigma \in {H}^2(\Omega )\) of equation (3.9) and its estimate (3.13) we refer the interested reader to [6, 37] for instance. \(\square \)

According to the lemma 3.1, we then define the nonlinear operator S as follows

$$\begin{aligned} S: V\rightarrow V,\quad u' = (v',b',\sigma '), \ \ S(u') = u,\end{aligned}$$
(3.16)

where \(u= (v,b,\sigma )\in V\) is the unique solution of the system (3.7)–(3.9) with the boundary conditions (2.10) and (2.11).

Lemma 3.2

The nonlinear operator \(S: V_0\rightarrow V_0\) is continuous. More precisely, for any bounded subset D of V, there exists \(c_D>0\) such that

$$\begin{aligned} \left\Vert {S(u'_{1})-S(u'_{2})} \right\Vert _{V_0} \le c_{D}\left\Vert {u'_1-u'_2} \right\Vert _{V_0}\qquad \forall u_1,\,u_2\in D.\end{aligned}$$
(3.17)

Proof

Let us note firstly that S transforms a bounded set of V into a bounded set of V. Indeed, taking into account (3.10) (see also (2.6), (2.3), (2.1) and (1.12)\(_2\)), it is easy to see that, for every \(u'=(v',\vartheta ',\sigma ')\in V\), we have

$$\begin{aligned} \left\Vert {F'} \right\Vert _{H^1}\le & {} c_\Omega \left\Vert {v'} \right\Vert _{H^2}^2( 1 + \left\Vert {\sigma '} \right\Vert _{H^2}) + c_\Omega (\left\Vert {\sigma '} \right\Vert _{H^2}^2 + \left\Vert {b'} \right\Vert _{H^2}^2 ) + c_\Omega (1+\left\Vert {b'} \right\Vert _{H^2})\left\Vert {\delta } \right\Vert _{H^2(\mathbb {R}^2)}, \\ \left\Vert {G'} \right\Vert _{L^2}\le & {} c_\Omega (\left\Vert {v'} \right\Vert _{H^2}^2 + \left\Vert {b'} \right\Vert _{H^2}^2) + c_\Omega (1+\left\Vert {v'} \right\Vert _{H^1})\left\Vert {\delta } \right\Vert _{H^2(\mathbb {R}^2)}.\end{aligned}$$

Hence, by recalling the V norm (see (3.6))

$$\begin{aligned} \left\| {G'} \right\| _{L^2}\le&{} c_\Omega \left\| {u'} \right\| _{V}^2 + c_\Omega (1+\left\| {u'} \right\| _{V})\left\| {\delta } \right\| _{H^2(\mathbb {R}^2)}.\end{aligned}$$
(3.18)

Let D be a V-bounded set. From (3.11)–(3.13) and the previous inequalities, if k is large enough, one sees easily that

$$\begin{aligned} \left\Vert {S(u')} \right\Vert _{V}^2 = \left\Vert {v} \right\Vert _{H^3}^2 + \left\Vert {b} \right\Vert _{H^2}^2 + \left\Vert {\sigma } \right\Vert _{H^2}^2\le c_{D}\quad \forall u'\in D,\end{aligned}$$
(3.19)

this means that S(D) is a bounded set of V. Now, let

$$\begin{aligned} u'_{i} = (v'_i,b'_i, \sigma '_ i)\in D \ \ \text{ and } \ \ S(u'_i) = u_i = (v_i,b_i, \sigma _ i), \ \ i=1,2.\end{aligned}$$
(3.20)

We write

$$\begin{aligned} u'_{1} - u'_{2} = (v',b',\sigma '),\quad S(u'_1) - S(u'_2) = u_{1} - u_{2} = (v,b,\sigma ).\end{aligned}$$
(3.21)

By the definition (3.16) of S, we have

$$\begin{aligned}{} & {} - \mu \Delta v - (\mu +\lambda )\nabla \nabla \cdot v = (\widehat{B}\cdot \nabla )b - \nabla (b\cdot \widehat{B}) - \gamma \vartheta \nabla (\widehat{\varrho }^{\gamma -1}\sigma ') - g\sigma ' e_3 + F(u'_{1}) - F(u'_{2}) \end{aligned}$$
(3.22)
$$\begin{aligned}{} & {} - \nu \Delta b = v\nabla \cdot \widehat{B}+ (\widehat{B}\cdot \nabla )v -\widehat{B}\nabla \cdot v + G(u'_{1}) - G(u'_{2}), \ \ \nabla \cdot b=0 \end{aligned}$$
(3.23)
$$\begin{aligned}{} & {} k(\sigma -\sigma ') + \nabla \cdot (\sigma v_1) = - \nabla \cdot ((\widehat{\varrho } + \sigma _{2}) v)). \end{aligned}$$
(3.24)

From (3.10) (see also (2.6), (2.12) and (2.13)), it is easy to see that

$$\begin{aligned} \left\Vert {G(u'_{1})-G(u'_{2})} \right\Vert _{H^{-1}}^2 + \left\Vert {F(u'_{1})-F(u'_{2})} \right\Vert _{L^2}^2\le P(\left\Vert {u'_1} \right\Vert _{V},\left\Vert {u'_2} \right\Vert _{V})(\left\Vert {v'} \right\Vert _{H^2}^2 + \left\Vert {b'} \right\Vert _{H^1}^2 + \left\Vert {\sigma '} \right\Vert _{H^1}^2) \end{aligned}$$

where P is polynomial function of two variables. With the same arguments as those of the Lemma 3.1, we can see that the system of Eqs. (3.22) and (3.23) has a unique weak solution (vb) such that (see (3.21) and (3.6))

$$\begin{aligned} \left\Vert {b} \right\Vert _{H^1}^2 + \left\Vert {v} \right\Vert _{H^2}^2\le P(\left\Vert {u'_1} \right\Vert _{V},\left\Vert {u'_2} \right\Vert _{V})\left\Vert {u'_1 - u'_2} \right\Vert _{V_0}^2.\end{aligned}$$
(3.25)

It remains us to estimate in \(H ^1\) norm the solution \(\sigma \) of Eq. (3.24). We first multiply that equation by \(\sigma \) and we integrate over \(\Omega \). Using

$$\begin{aligned} \displaystyle \int _\Omega \nabla \cdot (\sigma v_1)\sigma \,dx = \displaystyle {1\over 2}\int _\Omega |\sigma |^2\nabla \cdot v_1\,dx\le c_\Omega \left\| {v_1} \right\| _{H^3}\left\| {\sigma } \right\| _{L^2}^2, \end{aligned}$$

we easily obtain

$$\begin{aligned} k\left\Vert {\sigma } \right\Vert _{L^2}^2\le c_\Omega \left\Vert {v_1} \right\Vert _{H^3}\left\Vert {\sigma } \right\Vert _{L^2}^2 + k\left\Vert {\sigma '} \right\Vert _{L^2}^2 + c_\Omega (1 + \left\Vert {\sigma _2} \right\Vert _{H^1}^2)\left\Vert {v} \right\Vert _{H^1}^2.\end{aligned}$$
(3.26)

We next apply the nabla operator \(\nabla \) to the Eq. (3.24), multiply the new equation by \(\nabla \sigma \) and integrate over \(\Omega \). Integration by parts and the use of the inequality

$$\begin{aligned}\displaystyle \int _\Omega \nabla \sigma \cdot \nabla (\nabla \cdot (\sigma v_1))dx= & {} \displaystyle {1\over 2}\int _\Omega |\nabla \sigma |^2(\nabla \cdot v_1) dx + \displaystyle \int _\Omega (\nabla \sigma \cdot \nabla )v_1\cdot \nabla \sigma dx\\{} & {} \quad + \ \displaystyle \int _\Omega \sigma (\nabla \nabla \cdot v_1)\cdot \nabla \sigma dx\le c_\Omega \left\Vert {v_1} \right\Vert _{H^3}\left\Vert {\sigma } \right\Vert _{H^1}^2, \end{aligned}$$

give us

$$\begin{aligned} k\left\Vert {\nabla \sigma } \right\Vert _{L^2}^2\le c_\Omega \left\Vert {v_1} \right\Vert _{H^3}\left\Vert {\sigma } \right\Vert _{H^1}^2 + k\left\Vert {\nabla \sigma '} \right\Vert _{L^2}^2 + c_\Omega (1 + \left\Vert {\sigma _2} \right\Vert _{H^2}^2)\left\Vert {v} \right\Vert _{H^2}^2.\end{aligned}$$
(3.27)

Adding (3.26) and (3.27) we obtain according to (3.25) (see also (3.20) and (3.21)),

$$\begin{aligned}{} & {} k \left\Vert {\sigma } \right\Vert _{H^1}^2 \le c_\Omega \left\Vert {S(u'_1)} \right\Vert _{V}\left\Vert {\sigma } \right\Vert _{H^1}^2 + k\left\Vert {\sigma '} \right\Vert _{H^1}^2 + c_\Omega (1 + \left\Vert {S(u'_2)} \right\Vert _{V}^2)\left\Vert {v} \right\Vert _{H^2}^2\nonumber \\{} & {} \quad \le c_\Omega \left\Vert {S(u'_1)} \right\Vert _{V}\left\Vert {\sigma } \right\Vert _{H^1}^2 + k\left\Vert {u'_1-u'_2} \right\Vert _{V_0}^2\nonumber \\{} & {} \qquad + \ c_\Omega (1 + \left\Vert {S(u'_2)} \right\Vert _{V}^2)P(\left\Vert {u'_1} \right\Vert _{V},\left\Vert {u'_2} \right\Vert _{V})\left\Vert {u'_1-u'_2} \right\Vert _{V_0}^2. \end{aligned}$$
(3.28)

According now to (3.28) and (3.25), we obtain

$$\begin{aligned}{} & {} \left\Vert {v} \right\Vert _{H^2}^2 + \left\Vert {b} \right\Vert _{H^1}^2 + (k - c_\Omega \left\Vert {S(u'_1)} \right\Vert _{V})\left\Vert {\sigma } \right\Vert _{H^1}^2 \nonumber \\{} & {} \quad \le c_\Omega (1 + \left\Vert {S(u'_2)} \right\Vert _{V}^2)P(\left\Vert {u'_1} \right\Vert _{V},\left\Vert {u'_2} \right\Vert _{V})\left\Vert {u'_1-u'_2} \right\Vert _{V_0}^2. \end{aligned}$$
(3.29)

If k is large enough so that (see (3.19)) \(k - c_\Omega \left\Vert {S(u'_1)} \right\Vert _{V}\ge k - c_\Omega c_{D}\ge 1\) from the previous inequality it follows

$$\begin{aligned} \left\Vert {v} \right\Vert _{H^2}^2 + \left\Vert {\vartheta } \right\Vert _{H^1}^2 + \left\Vert {\sigma } \right\Vert _{H^1}^2 \le c_{D}\left\Vert {u'_1-u'_2} \right\Vert _{V_0}^2 \end{aligned}$$

from which (see (3.21)) follows (3.17) and this completes the proof of the Lemma 3.2. \(\square \)

Remark 3.3

The \(V_0\)-Continuous operator S (i.e continuous in the norm of \(V_0\)) will be viewed as realizing the hypotheses of the Schauder fixed point theorem. Following the work done in [7, 8], we can show that for a sufficiently small, we have that \(S(B^a_V)\subseteq B^a_V\) where \(B^a_V = \overline{B(0,a)}\) is the closed ball of V with radius a. However, \(\Omega = \mathbb {R}^2\times ]0,h[\) being unbounded, the closed ball \(B^a_V\) is not \(V_0\)-compact (because the embedding \(H^k(\Omega )\hookrightarrow H^{k-1}(\Omega )\)) is not compact when \(\Omega \) is unbounded). In order to overcome this lack of compactness, our strategy here largely inspired by the work of [8], is to construct a sequence \(V_n\) of closed subspaces of V, closed convex subsets \(B^a_n\) of \(V_n\) which are \(V_0\)-compact and also operators \(S_n\) such that \(S_n: B^a_n\rightarrow B^a_n\) is \(V_0\)-continuous. We then apply the Schauder fixed point theorem to each operator operator \(S_n\) and get that \(S_n\) has a fixed point \(u_n\) in \(B^a_n\). Then, we will show that the sequence \((u_n)\) of fixed points converges to some \(u = (v,b,\sigma )\in V\) which solves the boundary problem (2.7)–(2.11).

The following section is devoted to the construction of the subspaces \(V_n\), the bounded closed convex subsets of \(V_n\) relatively \(V_0\)-compact and finally the operators \(S_n\).

4 Operators \(\mathbf{S_\textbf{n}}\) and their Properties

Let \(\tau \in C^\infty (\mathbb {R}^2)\) such that for all \(x'=(x_1,x_2)\in \mathbb {R}^2\),

$$\begin{aligned}{} & {} 0\le \tau (x')\le 1,\quad \tau (x') = 1 \ \ \text { if } \ \ |x'|\le {1\over 2},\quad \tau (x') = 0 \ \ \text { if } \ \ |x'|\ge 1\,, \end{aligned}$$
(4.1)
$$\begin{aligned}{} & {} \tau _n(x') = \tau \big (\displaystyle {x'\over n}\big ). \end{aligned}$$
(4.2)

We recall that \(\Omega =\mathbb {R}^2\times ]0,h[\) and we set

$$\begin{aligned} \Omega _n = \Omega '_n\times ]0,h[,\quad \Omega '_n = \{x'\in \mathbb {R}^2: \ |x'| < n\}, \ \ ^c\Omega _n = \Omega \setminus \Omega _n = {^c\Omega '}_n\times ]0,h[\end{aligned}$$
(4.3)

and (see (3.2))

$$\begin{aligned}{} & {} {\mathbb {H}}_n = \{w\in {\mathbb {H}}(\Omega _n),\ w\ \text{ satisfies } \ (2.10)\},\quad \Vert \cdot \Vert _{{\mathbb {H}}_n} = \Vert \cdot \Vert _{{\mathbb {H}}(\Omega _n)}\end{aligned}$$
(4.4)
$$\begin{aligned}{} & {} V_n = \{ u = \tau _n w, \ w\in {\mathbb {H}}_n, \ u =0 \ \text{ in } \ ^c\Omega _n\} . \end{aligned}$$
(4.5)

Lemma 4.1

For any \(n\ge 1\) integer, \(V_n\) is a closed subspace of V.

Proof

It is clear that \(V_n\) is a subspace of V (see (3.5)). Let now \((u_p)\) be a sequence of \(V_n\) strongly convergent in V to some u. It is clear that \(u= 0\) outside \(\Omega _n\). It remains to show that \(u = \tau _n w\) with \(w\in {\mathbb {H}}_n\). Indeed, since \(u_p = \tau _n w_p\) in \(\Omega _n\), where \(w_p=(v_p,b_p,\sigma _p)\in {\mathbb {H}}_n \), we have (see (3.3) and (3.5))

$$\begin{aligned} \Vert u_p - u\Vert _V = \Vert u_p - u\Vert _{{\mathbb {H}}(\Omega )}=\Vert u_p - u\Vert _{{\mathbb {H}}(\Omega _n)} \end{aligned}$$

so the sequence \((u_p)\) strongly converges in \({\mathbb {H}}(\Omega _n)\). On the other hand, since

$$\begin{aligned} {\mathbb {H}}(\Omega _n)\hookrightarrow C^1({\bar{\Omega }}_n)\times C^0({\bar{\Omega }}_n)\times C^0({\bar{\Omega }}_n) \end{aligned}$$
(4.6)

because \(H^k(\Omega _n)\hookrightarrow C^{k-2}({\bar{\Omega }}_n)\) (\( k = 2,3,\cdots \)), it follows that each component of the sequence \((u_p)\) converges in \(C^0({\bar{\Omega }}_n)\) and hence, \((w_p)\) converges pointwise on \(\Omega _n\) to some \(w = (v,b,\sigma )\). Therefore \(u = \tau _n w\). To complete the proof, let us show that \(w\in {\mathbb {H}}_n\). In fact, notice first that \(w\in {\mathbb {H}}(\Omega _n)\) since \(u=\tau _n w\in {\mathbb {H}}(\Omega _n)\) (being in V). On the other hand, since \(u_p =\tau _n w_p\) satisfies the conditions (2.10) (because \(w_p\in {\mathbb {H}}_n\)), taking into account the continuous embedding (4.6) and the strong convergence of the sequence \((u_p)\) in \({\mathbb {H}}(\Omega _n)\), we deduce that u verifies (2.10). So \(w\in {\mathbb {H}}_n\) and therefore \(u\in V_n\). \(\square \)

Lemma 4.2

Let

$$\begin{aligned} B_{V}^a = \{w=(v,b,\sigma )\in V: \ \Vert w\Vert _V\le a\},\quad B_n^a=\tau _n B_{V}^a. \end{aligned}$$
(4.7)

Then \(B_n^a\) is a closed bounded convex set of \(V_n\).

Proof

Since (see (4.1))

$$\begin{aligned} \displaystyle \sup _{\mathbb {R}^2}|\tau _n(x')|\le 1,\quad \displaystyle \sup _{\mathbb {R}^2}|D^k\tau _n(x')|\le \displaystyle {1\over n^k}\displaystyle \sup _{|\zeta |\le 1}|D^k\tau (\zeta )|, \end{aligned}$$
(4.8)

notice that

$$\begin{aligned} \left\Vert {\tau _n\varphi } \right\Vert _{H^k}^2\le (1+\nu ^k_n)\left\Vert {\varphi } \right\Vert _{H^k}^2\quad \forall \varphi \in H^k (\Omega ),\quad k = 0,1,\ldots \end{aligned}$$
(4.9)

where \( \nu ^0_n = 0\) and for all \(k\ge 1\) fixed integer \(\nu ^k_n\downarrow 0\). Hence (see (3.5)),

$$\begin{aligned} \left\Vert {\tau _n w} \right\Vert _{V}^2= \left\Vert {\tau _nv} \right\Vert _{ H^3}^2 + \left\Vert {\tau _n\vartheta } \right\Vert _{ H^2}^2 +\left\Vert {\tau _n\sigma } \right\Vert _{ H^2}^2\le (1+\nu _n)\left\Vert {w} \right\Vert _{V}^2,\ \forall \ w=(v,\vartheta ,\sigma )\in V\end{aligned}$$
(4.10)

where \( \nu _n\downarrow 0\). Since \(V\subset {\mathbb {H}_n}\) it follows that \(B_n^a=\tau _nB_V^a\subset V_n\) and is convex and bounded (see (4.10)).

Let us now show that \(B_n^a\) is closed in V. To this aim, it sufficient to show that it is weakly closed in V. So, let \((u_p)\) with \(u_p = \tau _n w_p\), be a sequence of \(B_n^a\) which converges weakly in V to some \(u\in V\). Hence, we get that

$$\begin{aligned} u = \tau _n w,\qquad w\in B_V^a.\end{aligned}$$
(4.11)

Indeed, from the fact that \(w_p\in B_V^a \) we can extract a subsequence \((w_{p'})\) which converges weakly in V to \(w\in B_V^a\) and hence, \((u_{p'})\) converges weakly to u in V since \((u_{p'})\) with \(u_{p'}= \tau _n w_{p'}\), is a subsequence of \((u_p) \). On the other hand, as the linear mapping \( L: V \rightarrow V\) defined by \(L (w): = \tau _n w\) is (see (4.10)) strongly continuous, it follows that L is also weakly continuous (see e.g., [13, Theorem 3.10 page 61]). Therefore, \(L(w_{p'})\rightharpoonup L(w)\) i.e. \(u_{p'}\rightharpoonup \tau _n w = u\) and hence, (4.11) holds. \(\square \)

Lemma 4.3

\(B_n^a\) is \(V_0\)-compact.

Proof

Let \((u_p)\) with \( u_p =\tau _nw_p \), be a sequence of \(B_n^a = \tau _nB_V^a\). Given (4.10) (see also (3.5)), we have

$$\begin{aligned} \left\| {\tau _n w_p} \right\| _{\mathbb {H}(\Omega _n)} = \left\| {\tau _n w_p} \right\| _{\mathbb {H}(\Omega )} = \left\| {\tau _n w_p} \right\| _{V}\le a\sqrt{1+\nu _n}\,, \end{aligned}$$

and hence, \(\tau _n w_p \) is bounded in \({\mathbb {H}}(\Omega _n)\). So from the compact embedding \({\mathbb {H}}(\Omega _n) \hookrightarrow {\mathbb {H}}_0(\Omega _n)\), there exists a subsequence \(u_{p'} = \tau _n w_ {p'}\) of \(u_p \) which converges strongly to some u in \({\mathbb {H}}_0(\Omega _n) \) and weakly in V. Since \( w_{p '} \in B_V^a \), passing if necessary to a further subsequence, we can assume that \((w_{p'}) \) converges weakly in V to some \(w\in B_V ^a\). Arguing as (4.11), we find that \( u = \tau _n w\). Since (see (3.6) and (3.4))

$$\begin{aligned} \left\Vert {\tau _n w_{p'}- \tau _n w} \right\Vert _{V_0}= \left\Vert {\tau _n w_{p'}- \tau _n w} \right\Vert _{\mathbb {H}_0(\Omega )} = \left\Vert {\tau _n w_{p'}-\tau _n w} \right\Vert _{\mathbb {H}_0(\Omega _n)}, \end{aligned}$$

recalling that \(\tau _n w_{p'}\) converges strongly in \({\ H_0(\Omega _n)}\) to \(\tau _n w\), we see that from any sequence \(\tau _n w_{p}\in \tau _n B_V^a\) on can extract subsequence \(\tau _n w_{p'} \) strongly convergent in \(V_0\) so \(\tau _nB_V^a\) is \(V_0 \)-compact. \(\square \)

Now we are in a position to define the operators \(S_n\)

Lemma 4.4

Set \(S_n:= \tau _n S\). Then, \(S_n: V_n\rightarrow V_n\) is \(V_0\)-continuous.

Proof

Notice first that similarly to (4.10), we have

$$\begin{aligned} \left\Vert {\tau _n u} \right\Vert _{V_0}^2= \left\Vert {\tau _nv} \right\Vert _{ H^2(\Omega )}^2 + \left\Vert {\tau _n\vartheta } \right\Vert _{ H^1(\Omega )}^2 +\left\Vert {\tau _n\sigma } \right\Vert _{ H^1(\Omega )}^2\le (1+{\bar{\nu }}_n)\left\Vert {u} \right\Vert _{V_0}^2.\end{aligned}$$
(4.12)

Since \(S:V\rightarrow V\) is \(V_0\)-continuous (from Lemma 3.2), its restriction \(S:V_n\rightarrow V\) to the closed subspace \(V_n\) is also continuous. Moreover, taking into account (4.12), it is clear that \(\tau _n S: V_n\rightarrow \tau _nV\) is \(V_0\)-continuous and hence, it follows from \(\tau _nS(V_n)\subset \tau _nV\subset \tau _n{\mathbb {H}}_n = V_n,\) that \(S_n:V_n\rightarrow V_n\) is \(V_0\)-continuous. \(\square \)

Remark 4.5

From the Lemmas 4.2, 4.3 and 4.4, to show that \(S_n\) possess a fixed point via the Schauder theorem, it remains to prove that

$$\begin{aligned} S_n(B_n^a)\subseteq B_n^a,\quad B_n^a=\tau _nB^a_{V},\end{aligned}$$
(4.13)

which will be established in the proof of Lemma 6.1.

Lemma 4.6

Let \( a > 0\) be small enough. If \( S_n \) has a fixed point in \(\tau _nB^a_V\) then the boundary value problem (2.7)–(2.11) has at least one solution \(u\in V\).

Proof

Indeed, if (4.13) is satisfied, according to Remark 4.5, \(S_n\) has a fixed point \(\tau _nu'_n\in \tau _nB^a_{V}\), i.e,

$$\begin{aligned} S_n(\tau _nu'_n) = \tau _nu'_n,\quad u'_n\in B_{V}^a. \end{aligned}$$

Recalling the definition of \(S_n\) in Lemma 4.4, we have that \(\tau _nS(\tau _n u'_n)=\tau _n u'_n\) and hence,

$$\begin{aligned} S(\tau _n u'_n) = u'_n\quad \text{ in } \ \ \mathbb {H}(\Omega _n).\end{aligned}$$
(4.14)

Set

$$\begin{aligned} S(\tau _nu'_n) = u_n \ \ \text{ with } \ \ u_n= (v_n,b_n,\sigma _n) \ \ \text{ and } \ \ u'_n = (v'_n,b'_n,\sigma '_n)\in B_{V}^a.\end{aligned}$$
(4.15)

According to the definition (see (3.16)) of the operator S, we get that \(u_n\in V\) and satisfies in \(\Omega \) the system of equations

$$\begin{aligned}{} & {} -\mu \Delta v_n - (\mu +\lambda )\nabla \nabla \cdot v_n =(\widehat{B}\cdot \nabla )b_n - \nabla (b_n\cdot \widehat{B}) - \gamma \vartheta \nabla (\widehat{\varrho }^{\gamma -1}\tau _n\sigma '_n) - g\tau _n\sigma '_n e_3 + F'_n, \end{aligned}$$
(4.16)
$$\begin{aligned}{} & {} - \nu \Delta b_n = v_n\nabla \cdot \widehat{B}+ (\widehat{B}\cdot \nabla )v_n -\widehat{B}\nabla \cdot v_n + G'_n, \ \ \nabla \cdot b_n = 0, \end{aligned}$$
(4.17)
$$\begin{aligned}{} & {} k(\sigma _n - \tau _n\sigma '_n) + \nabla \cdot (\sigma _n v_n) = - \nabla \cdot (\widehat{\varrho }v_n), \end{aligned}$$
(4.18)

where

$$\begin{aligned}G'_n =G(\tau _nu'_n),\quad F'_n = F(\tau _nu'_n).\end{aligned}$$

As (see (4.15) and (4.14)) \( u_n = u'_n\) in \({\mathbb {H}}(\Omega _n)\), then \(u_n=(v_n,b_n,\sigma _n)\) verifies in \(\Omega _n\) the following system

$$\begin{aligned}{} & {} -\mu \Delta v_n - (\mu +\lambda )\nabla \nabla \cdot v_n =(\widehat{B}\cdot \nabla )b_n - \nabla (b_n\cdot \widehat{B}) - \gamma \vartheta \nabla (\widehat{\varrho }^{\gamma -1}\tau _n\sigma _n) - g\tau _n\sigma _n e_3 + F(\tau _nu_n), \\{} & {} - \nu \Delta b_n = v_n\nabla \cdot \widehat{B}+ (\widehat{B}\cdot \nabla )v_n -\widehat{B}\nabla \cdot v_n + G(\tau _nu_n), \ \ \nabla \cdot b_n = 0, \\{} & {} k(\sigma _n - \tau _n\sigma _n) + \nabla \cdot (\sigma _n v_n) = - \nabla \cdot (\widehat{\varrho }v_n). \end{aligned}$$

Let now fix N be large enough. Since for any \( n> 2 N \) (see (4.2)) \(\tau _n =1\) in \(\Omega _N\), from the previous system of equations, it follows that \(u_n=(v_n,b_n,\sigma _n)\) verifies in \(\Omega _N\) the following system

$$\begin{aligned}{} & {} -\mu \Delta v_n - (\mu +\lambda )\nabla \nabla \cdot v_n =(\widehat{B}\cdot \nabla )b_n - \nabla (b_n\cdot \widehat{B}) - \gamma \vartheta \nabla (\widehat{\varrho }^{\gamma -1}\sigma _n) - g\sigma _n e_3 + F(u_n), \end{aligned}$$
(4.19)
$$\begin{aligned}{} & {} - \nu \Delta b_n = v_n\nabla \cdot \widehat{B}+ (\widehat{B}\cdot \nabla )v_n -\widehat{B}\nabla \cdot v_n + G(u_n), \ \ \nabla \cdot b_n = 0, \end{aligned}$$
(4.20)
$$\begin{aligned}{} & {} \nabla \cdot (\sigma _n v_n) = - \nabla \cdot (\widehat{\varrho }v_n). \end{aligned}$$
(4.21)

Since \(u_n = S(\tau _nu'_n)\) then (see (3.19) and (4.10)), \(u_n\) is bounded in V so in \(\mathbb {H}(\Omega _N)\), passing if necessary to a subsequence, we can assume that \(u_n\) converges weakly in V and \({\mathbb {H}}(\Omega _N)\) so strongly in \( {\mathbb {H}}_0 (\Omega _N)\). Let u be the weak limit of \(u_n\) in V and \(u_N\) is weak limit in \({\mathbb {H}}(\Omega _N)\) of \(u_n|_{\Omega _N}\). Therefore

$$\begin{aligned} u_N = u|_{\Omega _N},\quad u\in V.\end{aligned}$$
(4.22)

Indeed, as the linear mapping \( R: V \rightarrow {\mathbb {H}}(\Omega _N)\) defined by \(R(w): = w|_{\Omega _N}\) is of course strongly continuous, it follows that R is also weakly continuous (see e.g., [13, Theorem 3.10 page 61]). Therefore, \(R(u_n)\rightharpoonup R(u)\) i.e. \(u_n|_{\Omega _N}\rightharpoonup u|_{\Omega _N}\) and hence, (4.22) holds.

On the other hand, considering the expressions (see (2.12) and (2.13)) of F and G, one can easily see that

$$\begin{aligned}{} & {} \left\Vert {G(u_n) - G(u)} \right\Vert _{L^2(\Omega _N)} + \left\Vert {F(u_n)-F(u)} \right\Vert _{L^2(\Omega _N)}\\{} & {} \quad \le C_{\Omega _N}(1+ \left\Vert {u_n} \right\Vert _{V}^2+ \left\Vert {u} \right\Vert _{V}^2) (\left\Vert {v_n - v} \right\Vert _{H^2(\Omega _N)} + \left\Vert {b_n - b} \right\Vert _{H^1(\Omega _N)} + \left\Vert {\sigma _n - \sigma } \right\Vert _{H^1(\Omega _N)}) \end{aligned}$$

and

$$\begin{aligned} \left\Vert {\nabla \cdot (v_n\sigma _n) - \nabla \cdot (v\sigma )} \right\Vert _{L^2(\Omega _N)}\le (\left\Vert {u_n} \right\Vert _{V}+ \left\Vert {u} \right\Vert _{V})(\left\Vert {v_n - v} \right\Vert _{H^2(\Omega _N)} + \left\Vert {\sigma _n - \sigma } \right\Vert _{H^1(\Omega _N)}).\end{aligned}$$

Hence, by recalling the \(V_0\)-norm (see (3.6) and (3.4)), we get

$$\begin{aligned}{} & {} \left\Vert {\nabla \cdot (v_n\sigma _n) - \nabla \cdot (v\sigma )} \right\Vert _{L^2(\Omega _N)}\le C'_{\Omega _N}(1 + \left\Vert {u} \right\Vert _{V})\left\Vert {u_n - u} \right\Vert _{{\mathbb {H}}_0(\Omega _N)},\nonumber \\{} & {} \quad \left\Vert {G(u_n) - G(u)} \right\Vert _{L^2(\Omega _N)} + \left\Vert {F(u_n)-F(u)} \right\Vert _{L^2(\Omega _N)} \le C'_{\Omega _N}(1+ \left\Vert {u} \right\Vert _{V}^2)\left\Vert {u_n - u} \right\Vert _{{\mathbb {H}}_0(\Omega _N)}. \end{aligned}$$
(4.23)

This being, in passing to \(L^2(\Omega _N)\)-weak limit in the Eqs. (4.19)–(4.21), given the inequalities (4.23), the strong convergence in \({\mathbb {H}}_0(\Omega _N)\) and weak convergence in \({\mathbb {H}}(\Omega _N)\) of the sequence \( (u_n) \) to \( u=(v,b,\sigma )\), we obtain

$$\begin{aligned}{} & {} -\mu \Delta v - (\mu +\lambda )\nabla \nabla \cdot v =(\widehat{B}\cdot \nabla )b - \nabla (b\cdot \widehat{B}) - \gamma \vartheta \nabla (\widehat{\varrho }^{\gamma -1}\sigma ) - g\sigma e_3 + F(v,b,\sigma ), \\{} & {} - \nu \Delta b = v\nabla \cdot \widehat{B}+ (\widehat{B}\cdot \nabla )v -\widehat{B}\nabla \cdot v + G(v,b,\sigma ), \ \ \nabla \cdot b = 0, \\{} & {} \nabla \cdot (\sigma v) = - \nabla \cdot (\widehat{\varrho }v) \end{aligned}$$

in \(\Omega _N\) for any N large enough and so, also in \(\Omega \). Therefore \(u\in V\) as a weak limit of \(u_n\) satisfies the Eqs. (2.7)–(2.9). \(\square \)

Now, to conclude on the existence of a solution of the Eqs. (2.7)–(2.9) with the boundary conditions (2.10) and (2.11), it remains, considering the Lemma 4.6, to show the crucial point (4.13). For this, we need suitable estimates of \(S(\tau _nu') = u_n\) (see (3.16)).

From now on, for any generic function \(\varphi \), we denote

$$\begin{aligned} \varphi ^{(n)}(x',x_3) = \tau _n(x')\varphi (x',x_3),\quad (x',x_3)\in \Omega =\mathbb {R}^2\times ]0,h[. \end{aligned}$$
(4.24)

Let us bear in mind that

$$\begin{aligned} S(u'^{(n)}) = u_n \ \ \text{ with } \ \ u_n=(v_n, b_n, \sigma _n)\in V\end{aligned}$$
(4.25)

is solution of the system

$$\begin{aligned}{} & {} -\mu \Delta v_n - (\mu +\lambda )\nabla \nabla \cdot v_n =(\widehat{B}\cdot \nabla )b_n - \nabla (b_n\cdot \widehat{B}) - \gamma \vartheta \nabla (\widehat{\varrho }^{\gamma -1}\sigma '^{(n)}) - g\sigma '^{(n)} e_3 + F'_n, \end{aligned}$$
(4.26)
$$\begin{aligned}{} & {} - \nu \Delta b_n = v_n\nabla \cdot \widehat{B}+ (\widehat{B}\cdot \nabla )v_n -\widehat{B}\nabla \cdot v_n + G'_n, \ \ \nabla \cdot b_n = 0, \end{aligned}$$
(4.27)
$$\begin{aligned}{} & {} k(\sigma _n - \sigma '^{(n)}) + \nabla \cdot (\sigma _n v_n) = - \nabla \cdot (\widehat{\varrho }v_n), \end{aligned}$$
(4.28)

with the boundary conditions (2.10)–(2.11) and (see (2.12) and (2.13))

$$\begin{aligned} G'_n =G(u'^{(n)}),\quad F'_n = F(u'^{(n)}).\end{aligned}$$
(4.29)

We recall that our goal is to establish the estimates of the solutions (4.25) from which we will obtain (4.13) in Remark 4.5. These estimates will require a more elaborate treatment and hence will be discussed in the next technical section. We will need some estimates of the nonlinear terms

$$\begin{aligned} F'_n=F(u'^{(n)}),\quad G'_n=G(u'^{(n)}),\quad v_n\sigma _n = v(u'^{(n)})\sigma (u'^{(n)}) \end{aligned}$$

which appear in the Eqs. (4.26)–(4.28). These estimates in terms of the norm \( \left\Vert {u'^{(n)}} \right\Vert _V \) of \(u'^{(n)}\in \tau _nB_V^a\) and therefore in terms of a, will be very useful in the proof of (4.13).

Indeed, let \( u'^{(n)} = (v'^{(n)},b'^{(n)},\sigma '^{(n)})\in \tau _n B_{V}^a\). If a is small enough then

$$\begin{aligned} \left\Vert {G'_n} \right\Vert _{L^2}^2 + \left\Vert {F'_n} \right\Vert _{H^1}^2 + \left\Vert {v_n} \right\Vert _{H^3}\left\Vert {\sigma _n} \right\Vert _{H^2}^2 \le c_\Omega a^3 \end{aligned}$$
(4.30)

In fact, notice first that, from (3.19) and (3.18) we have

$$\begin{aligned} \left\Vert {G'_n} \right\Vert _{L^2} + \left\Vert {F'_n} \right\Vert _{H^1} \le c_\Omega (1 + \Vert u'^{(n)}\Vert _{V})\left\Vert {\delta } \right\Vert _{H^2} + c_\Omega \Vert u'^{(n)}\Vert _{V}^{2}(1 + \Vert u'^{(n)}\Vert _{V}).\end{aligned}$$
(4.31)

Now by considering (3.12) and (3.13), we have

$$\begin{aligned}{} & {} \left\Vert {v_n} \right\Vert _{H^3}\le c_\Omega ({\big |\hspace{-1.42271pt}\big |\sigma '^{(n)}\big |\hspace{-1.42271pt}\big |}_{H^2} + \left\Vert {G'_n} \right\Vert _{L^2} + \left\Vert {F'_n} \right\Vert _{H^1}), \end{aligned}$$
(4.32)
$$\begin{aligned}{} & {} k\left\Vert {\sigma _n} \right\Vert _{H^2}^2\le c_\Omega \left\Vert {v_n} \right\Vert _{H^3}\left\Vert {\sigma _n} \right\Vert _{H^2}^2 + k\big |\hspace{-1.42271pt}\big | {\sigma '^{(n)}}\big |\hspace{-1.42271pt}\big |_{H^2}^2 + c_\Omega \left\Vert {v_n} \right\Vert _{H^3}^2.\end{aligned}$$
(4.33)

From (4.31) (see also (4.10)) it follows that

$$\begin{aligned} \big |\hspace{-1.42271pt}\big | {\sigma '^{(n)}}\big |\hspace{-1.42271pt}\big |_{H^2} + \left\Vert {G'_n} \right\Vert _{L^2} + \left\Vert {F'_n} \right\Vert _{H^1}\le c_\Omega [a + (1 + a)(\left\Vert {\delta } \right\Vert _{H^2} + a^2)]\le c_a.\end{aligned}$$
(4.34)

If k is large enough so that

$$\begin{aligned} k- c_\Omega (\big |\hspace{-1.42271pt}\big | {\sigma '^{(n)}}\big |\hspace{-1.42271pt}\big |_{H^2} + \left\Vert {G'_n} \right\Vert _{L^2} + \left\Vert {F'_n} \right\Vert _{H^1})\le k- c_{a}>0, \end{aligned}$$

from (4.32) and (4.33), it follows that

$$\begin{aligned} \left\Vert {\sigma _n} \right\Vert _{H^2}^2\le c_\Omega (\big |\hspace{-1.42271pt}\big | {\sigma '^{(n)}}\big |\hspace{-1.42271pt}\big |_{H^2}^2 + \left\Vert {G'_n} \right\Vert _{L^2}^2 + \left\Vert {F'_n} \right\Vert _{H^1}^2),\end{aligned}$$
(4.35)

whence, given (4.32)

$$\begin{aligned} \left\Vert {v_n} \right\Vert _{H^3}\left\Vert {\sigma _n} \right\Vert _{H^2}^2\le c_\Omega (\big |\hspace{-1.42271pt}\big | {\sigma '^{(n)}}\big |\hspace{-1.42271pt}\big |_{H^2} + \left\Vert {G'_n} \right\Vert _{L^2} + \left\Vert {F'_n} \right\Vert _{H^1})^3\end{aligned}$$
(4.36)

where (see (4.31))

$$\begin{aligned} \left\Vert {G'_n} \right\Vert _{L^2} + \left\Vert {F'_n} \right\Vert _{H^1}\le c_\Omega (1 + a)(\left\Vert {\delta } \right\Vert _{H^2(\mathbb {R}^2)} + a^2)\quad \text{ for } \text{ all } \ \ u'^{(n)}\in \tau _nB_{V}^a.\end{aligned}$$
(4.37)

From (4.37), (4.36), and (4.34) we get

$$\begin{aligned}{} & {} \left\Vert {G'_n} \right\Vert _{L^2}^2 + \left\Vert {F'_n} \right\Vert _{H^1}^2 + \left\Vert {v_n} \right\Vert _{H^3}\left\Vert {\sigma _n} \right\Vert _{H^2}^2\\{} & {} \quad \le c_\Omega (1 + a^2)(\left\Vert {\delta } \right\Vert _{H^2(\mathbb {R}^2)}^2 + a^4) + c_\Omega \big [a + (1+a)(\left\Vert {\delta } \right\Vert _{H^2(\mathbb {R}^2)} + a^2)\big ]^3.\end{aligned}$$

By choosing (see (1.12)) \(\delta _0\) so that \(\left\Vert {\delta } \right\Vert _{H^2(\mathbb {R}^2)}\le a^2\), it is easy to see that for all \(a\in ]0,1[\), we have

$$\begin{aligned} c_\Omega (1 + a^2)(\left\Vert {\delta } \right\Vert _{H^2(\mathbb {R}^2)}^2 + a^4) + c_\Omega \big [a + (1+a)(\left\Vert {\delta } \right\Vert _{H^2(\mathbb {R}^2)} + a^2)\big ]^3\le c_\Omega a^3 \end{aligned}$$

whence follows (4.30). \(\square \)

5 Estimates of the Solutions

This section is devoted to solutions \(u_n=(v_n,b_n,\sigma _n)\) the system of equations (4.26)–(4.28) with the boundary conditions (2.10) and (2.11). These estimate which will be obtained in the following lemmas are based largely on the ideas developed in [8]. We remind here that these lemmas will be proven under the assumption (1.11). As for the positive number k which appears in Eq. (4.28) it can, as in Lemma 3.1, be chosen arbitrarily large. It will be convenient to set

$$\begin{aligned} k = {{\bar{k}} k_1\over 2},\quad k_1 = {\gamma \vartheta \over 2\mu + \lambda } \Big (C_M - {\gamma -1\over \gamma \vartheta }gh\Big )^{\gamma \over \gamma -1},\end{aligned}$$
(5.1)

where \(\bar{k}\) is large positive number satisfying in particular

$$\begin{aligned} {\bar{k}}\ge 8\Big (1 - {\gamma -1\over \gamma \vartheta C_M}gh\Big )^{-{\gamma \over \gamma -1}}.\end{aligned}$$
(5.2)

We will denote in the statements of the following lemmas, by \(C'_i\) \( (i= 1, \cdots 10) \) the constants which depend on \(\Omega \) and \({\bar{B}}\) but not on M (possibly on its lower bound \(M_0\) see (1.11)) and by \({\tilde{C}}_i\) \( (i = 1,2,3) \) the constants which depend on \(\Omega \), \({\bar{B}}\) and M. In addition, in the proof of each lemma, if it is not necessary to specify them, one will denote by \(C_\Omega \) the constants independent of M and by \( \widehat{C}\) those which depend on M.

Lemma 5.1

Let \(u'=(v',b',\sigma ')\in V\) and \(u_n=S(u'^{(n)}) = (v_n,b_n,\sigma _n)\in V\) the solution of the system (4.26)–(4.28) with the boundary conditions (2.10) and (2.11) guaranteed by Lemma 3.1. Let k be given by (5.1) and suppose that the hypothesis (1.11)–(1.12) holds, then we have

$$\begin{aligned} {}&{} \mu \left\| {\nabla v_n} \right\| _{L^2}^2 + (\mu +\lambda )\left\| {\nabla \cdot v_n} \right\| _{L^2}^2 + \nu \left\| {\nabla b_n} \right\| _{L^2}^2 +k\Vert \widehat{\varrho }^{\gamma -2\over 2}\sigma _n\Vert _{L^2}^2 + C'_1C_M\Vert \sigma '^{(n)}\Vert _{L^2}^2 \nonumber \\{}&{} \quad \le k\Vert \widehat{\varrho }^{\gamma -2\over 2}\sigma '^{(n)}\Vert _{L^2}^2 + C'_1\big ( \left\| {F'_n} \right\| _{L^2}^2+ \left\| {G'_n} \right\| _{L^2}^2+\left\| {v_n} \right\| _{H^3}\left\| {\sigma _n} \right\| _{L^2}^2\big ) C_\Omega \left\| {\varepsilon } \right\| _{L^2(^c\Omega '_n)}^2.\end{aligned}$$
(5.3)

Proof

Let us first observe that

$$\begin{aligned}{} & {} -\displaystyle \int _\Omega [v_n\nabla (\widehat{\varrho }^{\gamma -1}\sigma '^{(n)}) +\widehat{\varrho }^{\gamma -2}\sigma _n\nabla \cdot (\widehat{\varrho } v_n)]dx \\{} & {} \quad = \ -\displaystyle \int _\Omega \widehat{\varrho }^{\gamma -2}(\sigma _n - \sigma '^{(n)})\nabla \cdot (\widehat{\varrho } v)dx - \displaystyle \int _\Omega \widehat{\varrho }^{\gamma -1}(\nabla \log \widehat{\varrho }\cdot v_n)\sigma '^{(n)} dx,\\{} & {} \qquad \displaystyle \int _\Omega \big [(\widehat{B}\cdot \nabla )b_n - \nabla (b_n\cdot \widehat{B})\big ]v_n dx + \displaystyle \int _\Omega b_n\cdot (v_n\nabla \cdot \widehat{B} + (\widehat{B}\cdot \nabla )v_n -\widehat{B}\nabla \cdot v_n) dx=0. \end{aligned}$$

Multiply (4.26) by \((\gamma \vartheta )^{-1}v_n\), (4.27) by \((\gamma \vartheta )^{-1}b_n\) and (4.28) by \( \widehat{\varrho }^{\gamma -2}\sigma _n\) and integrate the resulting equations over \(\Omega \). By using the above identities, we find after integration by parts that

$$\begin{aligned}{} & {} \displaystyle {\nu \over \gamma \vartheta }\left\Vert {\nabla b_n} \right\Vert _{L^2}^2 +\displaystyle {\mu \over \gamma \vartheta }\left\Vert {\nabla v_n} \right\Vert _{L^2}^2 + {\mu + \lambda \over \gamma \vartheta } \left\Vert {\nabla \cdot v_n} \right\Vert _{L^2}^2 + {k\over 2}\Vert \widehat{\varrho }^{\gamma -2\over 2}(\sigma _n - \sigma '^{(n)}\Vert _{L^2}^2 \nonumber \\{} & {} \quad + \ \displaystyle {k\over 2}\Vert \widehat{\varrho }^{\gamma -2\over 2}\sigma _n\Vert _{L^2}^2 =\displaystyle {k\over 2}\displaystyle \Vert \widehat{\varrho }^{\gamma -2\over 2}\sigma '^{(n)}\Vert _{L^2}^2 + \displaystyle \sum _{i=1}^4I_i. \end{aligned}$$
(5.4)

where

$$\begin{aligned}I_1= & {} -\displaystyle \int _\Omega \widehat{\varrho }^{\gamma -2}(\sigma _n - \sigma '^{(n)})\nabla \cdot (\widehat{\varrho } v_n)dx,\\ I_2= & {} - \displaystyle {g\over \gamma \vartheta }\int _\Omega (e_3\cdot v_n)\sigma '^{(n)} dx - \displaystyle \int _\Omega \widehat{\varrho }^{\gamma -1}(\nabla \log \widehat{\varrho }\cdot v_n)\sigma '^{(n)} dx,\\ I_3= & {} \displaystyle {1\over \gamma \vartheta }\int _\Omega (F'_n\cdot v_n + G'_n\cdot b_n )dx, \ \ I_4\, =\, - \displaystyle \int _\Omega \widehat{\varrho }^{\gamma -2}\sigma _n\nabla (v_n\sigma _n)dx.\end{aligned}$$

Recalling the expression of \(\widehat{\varrho }\) (see (2.1) and (1.8)), thanks to (1.11)–(1.14), one can easily see that \( \left\Vert {\nabla \log \widehat{\varrho }} \right\Vert _{L^\infty }\) is small enough so that

$$\begin{aligned} \left\Vert {v_n\cdot \nabla \log \widehat{\varrho }} \right\Vert _{L^2}^2\le \displaystyle {\mu \over 2\mu + \lambda }\left\Vert {\nabla v_n} \right\Vert _{L^2}^2\end{aligned}$$
(5.5)

thus, by using (5.1) and (5.2), we obtain

$$\begin{aligned}I_1\le & {} \displaystyle {k\over 2}\Vert \widehat{\varrho }^{\gamma -2\over 2}(\sigma _n - \sigma '^{(n)})\Vert _{L^2}^2 + \displaystyle {2\over {\bar{k}} k_1}\left\Vert {\widehat{\varrho }^\gamma } \right\Vert _{L^\infty }(\left\Vert {\nabla \cdot v_n} \right\Vert _{L^2}^2 + \left\Vert {v_n\cdot \nabla \log \widehat{\varrho }} \right\Vert _{L^2}^2)\\\le & {} \displaystyle {k\over 2}\Vert \widehat{\varrho }^{\gamma -2\over 2}(\sigma _n - \sigma '^{(n)})\Vert _{L^2}^2 + \displaystyle {\mu +\lambda \over 8\gamma \vartheta }\left\Vert {\nabla \cdot v_n} \right\Vert _{L^2}^2 + \displaystyle {\mu \over 4\gamma \vartheta }\left\Vert {\nabla v_n} \right\Vert _{L^2}^2,\\ I_2\le & {} \displaystyle {\mu \over 4\gamma \vartheta }\left\Vert {\nabla v_n} \right\Vert _{L^2}^2 + C_\Omega \Vert \sigma '^{(n)}\Vert _{L^2}^2,\\ I_3\le & {} \displaystyle {\mu \over 4\gamma \vartheta }\left\Vert {\nabla v_n} \right\Vert _{L^2}^2 + \displaystyle {3\nu \over 4\gamma \vartheta }\left\Vert {\nabla b_n} \right\Vert _{L^2}^2 + C_\Omega (\left\Vert {F'_n} \right\Vert _{L^2}^2 +\left\Vert {G'_n} \right\Vert _{L^2}^2).\end{aligned}$$

As for the last term, we have

$$\begin{aligned} I_4 = -{1\over 2}\int _\Omega \widehat{\varrho }^{\gamma -2}|\sigma _n|^2\nabla \cdot v_n dx + {1\over 2}\int _\Omega |\sigma _n|^2 v_n\cdot \nabla \widehat{\varrho }^{\gamma -2}dx\le C_\Omega \left\Vert {v_n} \right\Vert _{H^3}\left\Vert {\sigma _n} \right\Vert _{L^2}^2. \end{aligned}$$

Combining these estimates and using (5.4), we get

$$\begin{aligned}{} & {} \displaystyle {3\mu \over 4\gamma \vartheta }\left\Vert {\nabla v_n} \right\Vert _{L^2}^2 + {3\over 4\gamma \vartheta }(\mu +\lambda ) \left\Vert {\nabla \cdot v_n} \right\Vert _{L^2}^2 + \displaystyle {3\nu \over 4\gamma \vartheta }\left\Vert {\nabla b_n} \right\Vert _{L^2}^2 + \displaystyle {k\over 2}\Vert \widehat{\varrho }^{\gamma -2\over 2}\sigma _n\Vert _{L^2}^2 \nonumber \\{} & {} \quad \le \displaystyle {k\over 2}\displaystyle \Vert \widehat{\varrho }^{\gamma -2\over 2}\sigma '^{(n)}\Vert _{L^2}^2 + C_\Omega (\left\Vert {F'_n} \right\Vert _{L^2}^2 + \left\Vert {v_n} \right\Vert _{H^3}\left\Vert {\sigma _n} \right\Vert _{L^2}^2 + \Vert \sigma '^{(n)}\Vert _{L^2}^2). \end{aligned}$$
(5.6)

Next, let us establish the \(L^2\)-estimate of \(\sigma '^{(n)}\) which appears in the right-hand side of (5.6). To this aim, we introduce the following boundary value problem

$$\begin{aligned}&\nabla \cdot \varphi = a\quad \text{ in } \ \ \Omega =\mathbb {R}^2\times ]0,h[, \end{aligned}$$
(5.7)
$$\begin{aligned}&\varphi |_{x_3=0} = \varphi |_{x_3=h} =0,\quad \varphi \rightarrow 0 \ \ \text{ if } \ \ |x'|\rightarrow \infty \end{aligned}$$
(5.8)

where (see (4.3))

$$\begin{aligned} a = \sigma '^{(n)} \ \ \text{ in } \ \ \Omega _n,\quad a = \varepsilon \ \ \text{ in } \ \ ^c\Omega _n,\end{aligned}$$
(5.9)

with \(\varepsilon =\varepsilon (x')\) being a given positive function in \(L^2(\mathbb {R}^2)\setminus L^1(\mathbb {R}^2)\). There exists (see e.g. [11, 22, 39]) at least one solution \(\varphi \in H^1(\Omega )\) of the problem (5.7)–(5.8) such that

$$\begin{aligned} \left\Vert { \varphi } \right\Vert _{H^1}\le c_\Omega \left\Vert {a} \right\Vert _{L^2}.\end{aligned}$$
(5.10)

Now, we rewrite the Eq. (4.26) as

$$\begin{aligned} - \gamma \vartheta \nabla (\widehat{\varrho }^{\gamma -1}\sigma '^{(n)}) = - \mu \Delta v_n - (\lambda +\mu )\nabla \nabla \cdot v_n + g\sigma '^{(n)}e_3 - (\widehat{B}\cdot \nabla )b_n + \nabla (b_n\cdot \widehat{B})- F'_n. \end{aligned}$$

We multily this new equation by \((\gamma \vartheta )^{-1}\varphi \) and integrate over \(\Omega \). By integrating by parts, we obtain taking into account (5.9) and (5.7)

$$\begin{aligned} \Vert \widehat{\varrho }^{\gamma -1\over 2}\sigma '^{(n)}\Vert _{L^2}^2= & {} \displaystyle {g\over \gamma \vartheta }\int _{\Omega }\sigma '^{(n)}e_3\cdot \varphi dx + \displaystyle {1\over \gamma \vartheta }\int _{\Omega } (b_n\cdot \widehat{B}\nabla \cdot \varphi - (\widehat{B}\cdot \nabla )b_n\cdot \varphi ) dx \nonumber \\{} & {} \quad + \displaystyle {1\over \gamma \vartheta }\displaystyle \int _{\Omega }[\mu \nabla v_n\cdot \nabla \varphi + (\mu +\lambda )(\nabla \cdot v_n)(\nabla \cdot \varphi ) - F'_n\cdot \varphi ]dx. \end{aligned}$$
(5.11)

Given (5.10) and (5.9) (see also (2.1) and (1.14)), we get from (5.11) that

$$\begin{aligned} C_M\left\Vert {\sigma '^{(n)}} \right\Vert _{L^2}^2\le & {} C_\Omega \left\Vert {\varepsilon } \right\Vert _{L^2(^c\Omega '_n)}^2\\{} & {} \quad + C_\Omega (\displaystyle {\mu \over \gamma \vartheta }\left\Vert {\nabla v_n} \right\Vert _{L^2}^2 + \displaystyle {1\over \gamma \vartheta } (\mu +\lambda )\left\Vert {\nabla \cdot v_n} \right\Vert _{L^2}^2 + \displaystyle {\nu \over \gamma \vartheta }\left\Vert {\nabla b_n} \right\Vert _{L^2}^2 + \left\Vert {F'_n} \right\Vert _{L^2}^2). \end{aligned}$$

If we multiply now the above inequality by \((4C_\Omega )^{-1}\) and add it to (5.6) we obtain the estimate (5.3) and this completes the proof the Lemma 5.1. \(\square \)

Lemma 5.2

Under the same the hypotheses of Lemma 5.1, we have

$$\begin{aligned}{} & {} \displaystyle \sum _{i=1}^2\Big [{\mu \over \gamma \vartheta } \left\Vert {\nabla \partial _{x_i} v_n} \right\Vert _{L^2}^2 + {\mu +\lambda \over \gamma \vartheta } \left\Vert {\nabla \cdot \partial _{x_i}v_n} \right\Vert _{L^2}^2 + {\nu \over \gamma \vartheta } \left\Vert {\nabla \partial _{x_i} b_n} \right\Vert _{L^2}^2 + k\Vert \widehat{\varrho }^{\gamma -2\over 2}\partial _{x_i}\sigma _n\Vert _{L^2}^2\Big ]\nonumber \\{} & {} \quad \le k\displaystyle \sum _{i=1}^2\Vert \widehat{\varrho }^{\gamma -2\over 2}\partial _{x_i}\sigma '^{(n)}\Vert _{L^2}^2 + C'_2\bigl (\Vert \sigma '^{(n)}\Vert _{L^2}^2 + \left\Vert {\nabla v_n} \right\Vert _{L^2}^2 + \left\Vert {\nabla b_n} \right\Vert _{L^2}^2\bigr )\nonumber \\{} & {} \qquad +\,\,C'_2\bigl (\left\Vert {F'_n} \right\Vert _{L^2}^2 + \left\Vert {G'_n} \right\Vert _{L^2}^2 + \left\Vert {v_n} \right\Vert _{H^3}\left\Vert {\sigma _n} \right\Vert _{H^1}^2\bigr ). \end{aligned}$$
(5.12)

Proof

Let us remark that, since \(\partial _{x_i}v_n\) and \(\partial _{x_i}b_n\) (\(i=1,2\)) satisfy the same boundary conditions (2.10), then we have

$$\begin{aligned}{} & {} -\displaystyle \int _\Omega \big [(\partial _{x_i} v_n)\nabla \cdot (\partial _{x_i}(\widehat{\varrho }^{\gamma -1}\sigma '^{(n)}) + \widehat{\varrho }^{\gamma -2}\partial _{x_i}\nabla \cdot (\widehat{\varrho } v_n)\partial _{x_i}\sigma _n\big ]dx \\{} & {} \quad = \ - \displaystyle \int _\Omega \widehat{\varrho }^{\gamma -2}(\partial _{x_i}\nabla \cdot (\widehat{\varrho } v_n))(\partial _{x_i}\sigma _n - \partial _{x_i}\sigma '^{(n)})dx - \displaystyle \int _\Omega \widehat{\varrho }^{\gamma -2}(\partial _{x_i}v_n\cdot \nabla \widehat{\varrho }) \partial _{x_i}\sigma '^{(n)} dx,\\{} & {} \qquad \displaystyle \int _\Omega \big [(\widehat{B}\cdot \nabla )\partial _{x_i}b_n - \nabla (\partial _{x_i}b_n\cdot \widehat{B})\big ]\partial _{x_i}v_n dx \\{} & {} \qquad + \ \displaystyle \int _\Omega \partial _{x_i}b_n\cdot (\partial _{x_i}v_n\nabla \cdot \widehat{B} + (\widehat{B}\cdot \nabla )\partial _{x_i}v_n - \widehat{B}\nabla \cdot \partial _{x_i}v_n) dx=0. \end{aligned}$$

Now, we apply the operator \(\partial _{x_i}\) to both sides of each Eqs. (4.26)–(4.28), then we multiply the resulting equations by \( (\gamma \vartheta )^{-1}\partial _{x_i}v_n\), \((\gamma \vartheta )^{-1}\partial _{x_i}b_n\) and \(\widehat{\varrho }^{\gamma -2}\partial _{x_i}\sigma _n\) respectively and we integrate over \(\Omega \). The integration by parts gives us

$$\begin{aligned}{} & {} \displaystyle {\mu \over \gamma \vartheta } \left\Vert {\nabla \partial _{x_i} v_n} \right\Vert _{L^2}^2 + {\mu +\lambda \over \gamma \vartheta }\left\Vert {\nabla \cdot \partial _{x_i}v_n} \right\Vert _{L^2}^2 + \displaystyle {\nu \over \gamma \vartheta } \left\Vert {\nabla \partial _{x_i} b_n} \right\Vert _{L^2}^2 + \displaystyle {k\over 2}\Vert \widehat{\varrho }^{\gamma -2\over 2}\partial _{x_i}\sigma _n\Vert _{L^2}^2 \nonumber \\{} & {} \quad + \displaystyle {k\over 2}\Vert \widehat{\varrho }^{\gamma -2\over 2}(\partial _{x_i}\sigma _n -\partial _{x_i}\sigma '^{(n)})\Vert _{L^2}^2 = \displaystyle {k\over 2}\Vert \widehat{\varrho }^{\gamma -2\over 2}\partial _{x_i}\sigma '^{(n)}\Vert _{L^2}^2 + \sum _{i=1}^6I_i. \end{aligned}$$
(5.13)

where

$$\begin{aligned}I_1=&{} -\displaystyle \int _\Omega \widehat{\varrho }^{\gamma -2}(\partial _{x_i}\sigma _n - \partial _{x_i}\sigma '^{(n)})\partial _{x_i}\nabla \cdot ({\bar{\varrho }} v_n)\,dx,\\ I_2=&{} \displaystyle \int _\Omega \widehat{\varrho }^{\gamma -1}\sigma '^{(n)}(\partial _{x_i}\partial _{x_i}v_n)\cdot \nabla \log \widehat{\varrho }\,dx + \displaystyle {g\over \gamma \vartheta }\displaystyle \int _\Omega \sigma '^{(n)} e_3\cdot \partial _{x_i}\partial _{x_i}v_n \ dx,\\ I_3=&{} \displaystyle \int _\Omega \big [(\partial _{x_i}\widehat{B}\cdot \nabla )b_n - \nabla (b_n\cdot \partial _{x_i}\widehat{B})\big ]\partial _{x_i}v_n\, dx, \\ I_4=&{} \displaystyle \int _\Omega \partial _{x_i}b_n\cdot (v_n\nabla \cdot \partial _{x_i}\widehat{B} + (\partial _{x_i}\widehat{B}\cdot \nabla )v_n - \partial _{x_i}\widehat{B}\nabla \cdot v_n)\, dx,\\ I_5=&{} - \displaystyle {1\over \gamma \vartheta }\int _\Omega (F'_n\cdot \partial _{x_i}\partial _{x_i}v_n + G'_n\cdot \partial _{x_i}\partial _{x_i}v_n)\,dx, \\ I_6=&{} -\displaystyle \int _\Omega \widehat{\varrho }^{\gamma -2}(\partial _{x_i}\sigma _n)\partial _{x_i}\nabla \cdot ( \sigma _n v_n)\,dx. \end{aligned}$$

By the similar arguments used in the proof of the Lemma 5.1 (particulary for the term \(I_1\)), we obtain

$$\begin{aligned} I_1\le & {} \displaystyle {k\over 2}\Vert \big (\widehat{\varrho }^{\gamma -2\over 2}(\partial _{x_i}\sigma _n - \partial _{x_i}\sigma '^{(n)})\Vert _{L^2}^2 + \displaystyle {\mu +\lambda \over 8\gamma \vartheta }\left\Vert {\nabla \cdot ( \partial _{x_i}v_n)} \right\Vert _{L^2}^2 + \displaystyle {\mu \over 4\gamma \vartheta }\left\Vert {\nabla \partial _{x_i}v_n} \right\Vert _{L^2}^2,\\ I_2\le & {} \displaystyle {\mu \over 12\gamma \vartheta }\Vert \nabla \partial _{x_i}v_n\Vert _{L^2}^2 + C_\Omega \Vert \sigma '^{(n)}\Vert _{L^2}^2, \\ I_3+I_4\le & {} \displaystyle {\mu \over 12\gamma \vartheta }\left\Vert {\nabla \partial _{x_i}v_n} \right\Vert _{L^2}^2 + \displaystyle {\nu \over 4\gamma \vartheta }\left\Vert {\nabla \partial _{x_i}b_n} \right\Vert _{L^2}^2 + C_\Omega (\left\Vert {\nabla v_n} \right\Vert _{L^2}^2 + \left\Vert {\nabla b_n} \right\Vert _{L^2}^2),\\ I_5\le & {} \displaystyle {\mu \over 12\gamma \vartheta }\left\Vert {\nabla \partial _{x_i}v_n} \right\Vert _{L^2}^2 + \displaystyle {\nu \over 4\gamma \vartheta }\left\Vert {\nabla \partial _{x_i}b_n} \right\Vert _{L^2}^2 + C_\Omega (\left\Vert {F'_n} \right\Vert _{L^2}^2+ \left\Vert {G'_n} \right\Vert _{L^2}^2). \end{aligned}$$

As for the term \(I_6\), we have

$$\begin{aligned} I_6= & {} -\displaystyle {1\over 2}\int _\Omega \widehat{\varrho }^{\gamma -2}|\partial _{x_i}\sigma _n|^2(\nabla \cdot v_n)dx + \displaystyle {1\over 2}\int _\Omega |\partial _{x_i}\sigma _n|^2 \big (v_n\cdot \nabla \widehat{\varrho }^{\gamma -2}\big )dx\nonumber \\{} & {} \quad - \displaystyle \int _\Omega \widehat{\varrho }^{\gamma -2}(\partial _{x_i}\sigma _n)\nabla \cdot (\sigma _n\partial _{x_i} v_n)dx\le C_\Omega \left\Vert {v_n} \right\Vert _{H^3}\left\Vert {\sigma _n} \right\Vert _{H^1}^2.\end{aligned}$$
(5.14)

By adding these estimates to (5.13) and summing on \(i=1,2\), we get the estimate (5.12) and this completes the proof the lemma 5.2. \(\square \)

Lemma 5.3

Under the same hypotheses of Lemma 5.1, we have

$$\begin{aligned}{} & {} ({\bar{k}} + 1)C'_MC_M^2\left\Vert {\partial _{x_3}\sigma _n} \right\Vert _{L^2}^2\le ({\bar{k}} - 1)C'_MC_M^2\left\Vert {\partial _{x_3}\sigma '^{(n)}} \right\Vert _{L^2}^2 + C'_3C_M^{-2}\left\Vert {\nabla v_n} \right\Vert _{L^2}^2 \nonumber \\{} & {} \quad + \ C'_3\big (\displaystyle \sum _{i=1}^2\left\Vert {\nabla \partial _{x_i}v_n} \right\Vert _{L^2}^2 +\Vert \sigma '^{(n)}\Vert _{L^2}^2 + \left\Vert {\nabla b_n} \right\Vert _{L^2}^2\big ) + \tilde{C}_1\left\Vert {F'_n} \right\Vert _{L^2}^2 + C'_3\left\Vert {v_n} \right\Vert _{H^3}\left\Vert {\sigma _n} \right\Vert _{H^1}^2, \end{aligned}$$
(5.15)

where

$$\begin{aligned} C'_M = {\gamma \vartheta \over 2\mu + \lambda }\Big (1 - {\gamma -1\over \gamma \vartheta C_M}gh\Big )^{\gamma \over \gamma -1}.\end{aligned}$$
(5.16)

Proof

Using the identity

$$\begin{aligned} \Delta v_{n,3} = \partial _{x_3}\nabla \cdot v_n + \partial _{x_1}(\partial _{x_1}v_{n,3} - \partial _{x_3}v_{n,1}) + \partial _{x_2}(\partial _{x_2}v_{n,3} - \partial _{x_3}v_{n,2}), \end{aligned}$$

from the Eq. (4.26) it follows that

$$\begin{aligned}{} & {} \partial _{x_3}\nabla \cdot v_n = -\displaystyle {\mu \over 2\mu +\lambda }\bigl [\partial _{x_1}(\partial _{x_1}v_{n,3} - \partial _{x_3}v_{n,1}) + \partial _{x_2}(\partial _{x_2}v_{n,3} - \partial _{x_3}v_{n,2})\bigr ] \nonumber \\{} & {} \quad + \ \displaystyle {1\over 2\mu +\lambda }\bigl [\gamma \vartheta \widehat{\varrho }^{\gamma -1}\partial _{x_3}\sigma '^{(n)} + (\gamma \vartheta \partial _{x_3}\widehat{\varrho }^{\gamma -1} + g)\sigma '^{(n)} + \partial _{x_3}(b_n\cdot \widehat{B}) - (\widehat{B}\cdot \nabla )b_{n,3} - F'_{n,3}\bigr ]. \end{aligned}$$
(5.17)

Next, we apply the differential operator \(\partial _ {x_3}\) to both sides of (4.28), multiply the resulting equation by \(\partial _{x_3}\sigma _n\) and integrate over \(\Omega \). Integrating by parts and taking into account of (5.17), we get

$$\begin{aligned} \displaystyle \int _\Omega [k(\partial _{x_3}\sigma _n - \partial _{x_3}\sigma '^{(n)})\partial _{x_3}\sigma _n + {\gamma \vartheta \over 2\mu +\lambda }\widehat{\varrho }^\gamma (\partial _{x_3}\sigma _n)(\partial _{x_3}\sigma '^{(n)})]dx = \displaystyle \sum _{i=1}^6I_i,\end{aligned}$$
(5.18)

where

$$\begin{aligned}I_1= & {} \displaystyle {1\over 2\mu +\lambda }\int _\Omega \widehat{\varrho }(\gamma \vartheta \partial _{x_3}\widehat{\varrho }^{\gamma -1} + g)\sigma '^{(n)}\partial _{x_3}\sigma _n dx, \quad I_2 = \displaystyle {1\over 2\mu +\lambda } \int _\Omega \widehat{\varrho }F'_{n,3}\partial _{x_3}\sigma _n dx\\ I_3= & {} \displaystyle {\mu \over 2\mu +\lambda } \int _\Omega \widehat{\varrho }(\partial _{x_3}\sigma _n)(\partial _{x_1} (\partial _{x_1}v_{n,3} - \partial _{x_3}v_{n,1}) + \partial _{x_2}(\partial _{x_2}v_{n,3} - \partial _{x_3}v_{n,2}))dx, \\ I_4= & {} -\,\displaystyle \int _\Omega (\partial _{x_3}(v_n\cdot \nabla \widehat{\varrho }) + (\partial _{x_3}\widehat{\varrho })\nabla \cdot v_n)\partial _{x_3}\sigma _n dx,\\ I_5= & {} \displaystyle {1\over 2\mu +\lambda } \int _\Omega \widehat{\varrho }(\partial _{x_3}\sigma _n)(\partial _{x_3}(\widehat{B}\cdot b_n) - (\widehat{B}\cdot \nabla )b_{n,3})dx_3, \\ I_6= & {} -\,\displaystyle \int _\Omega (\partial _{x_3}\sigma _n)\nabla \cdot (\partial _{x_3}(v_n\sigma _n)) dx.\end{aligned}$$

Using the identity

$$\begin{aligned} (X-Y)X + aXY = {1 + a\over 2} X^2 + {1 - a\over 2}(X - Y)^2 - {1 - a\over 2} Y^2 \end{aligned}$$

with

$$\begin{aligned} a = {\gamma \vartheta \over k(2\mu +\lambda )}\widehat{\varrho }^\gamma \end{aligned}$$

and taking into account the expressions of \(\widehat{\varrho }\) and \(k_1\) (see (2.1), (1.8) and (5.1)), we obtain

$$\begin{aligned}{} & {} \displaystyle \int _\Omega k\big [(\partial _{x_3}\sigma _n - \partial _{x_3}\sigma ')\partial _{x_3}\sigma _n + {\gamma \vartheta \over k(2\mu +\lambda )}\widehat{\varrho }^\gamma (\partial _{x_3}\sigma )(\partial _{x_3}\sigma '^{(n)})\big ]dx \nonumber \\{} & {} \quad \ge \displaystyle {k +k_1\over 2}\left\Vert {\partial _{x_3}\sigma _n} \right\Vert _{L^2}^2 - \displaystyle {k - k_1\over 2}\Vert \partial _{x_3}\sigma '^{(n)}\Vert _{L^2}^2 + \displaystyle {k - k'_1\over 2}\Vert \partial _{x_3}\sigma _n - \partial _{x_3}\sigma '^{(n)}\Vert _{L^2}^2,\end{aligned}$$
(5.19)

where

$$\begin{aligned} \displaystyle k'_1 = {\gamma \vartheta \over 2\mu + \lambda }C_M^{\gamma \over \gamma -1}. \end{aligned}$$

Moreover, given (2.1) (see also, (1.8) and (1.14), we have

$$\begin{aligned} I_1\le & {} \displaystyle {k - k'_1\over 6} \left\Vert {\partial _{x_3}\sigma _n - \partial _{x_3}\sigma '^{(n)}} \right\Vert _{L^2}^2 + \displaystyle {k_1\over 12} \left\Vert {\partial _{x_3}\sigma '^{(n)}} \right\Vert _{L^2}^2 + C_\Omega C_M^{{2-\gamma \over \gamma -1}} \Vert \sigma '^{(n)}\Vert _{L^2}^2, \\ I_2\le & {} \widehat{C}\left\Vert {F'_n} \right\Vert _{L^2}^2 + \displaystyle {k_1\over 8} \left\Vert {\partial _{x_3}\sigma _n} \right\Vert _{L^2}^2,\\ I_3\le & {} \displaystyle {k - k'_1\over 6} \left\Vert {\partial _{x_3}\sigma _n - \partial _{x_3}\sigma '^{(n)}} \right\Vert _{L^2}^2 + \displaystyle {k_1\over 12} \left\Vert {\partial _{x_3}\sigma '^{(n)}} \right\Vert _{L^2}^2 + C_\Omega C_M^{{2-\gamma \over \gamma -1}} \sum _{i=1}^2\left\Vert {\nabla \partial _{x_i}v_n} \right\Vert _{L^2}^2,\\ I_4\le & {} \displaystyle {k - k'_1\over 6} \left\Vert {\partial _{x_3}\sigma _n - \partial _{x_3}\sigma '^{(n)}} \right\Vert _{L^2}^2 + \displaystyle {k_1\over 12} \left\Vert {\partial _{x_3}\sigma '^{(n)}} \right\Vert _{L^2}^2 + C_\Omega C_M^{{4-3\gamma \over \gamma -1}} \left\Vert {\nabla v_n} \right\Vert _{L^2}^2,\\ I_5\le & {} C_\Omega C_M^{{2-\gamma \over \gamma -1}}\left\Vert {\nabla b_n} \right\Vert _{L^2}^2 + \displaystyle {k_1\over 8} \left\Vert {\partial _{x_3}\sigma _n} \right\Vert _{L^2}^2..\end{aligned}$$

As for the last \(I_6\) term, one has

$$\begin{aligned}I_6= & {} -\displaystyle {1\over 2}\int _\Omega |\partial _{x_3}\sigma _n|^2(\nabla \cdot v_n)dx - \displaystyle \int _\Omega (\partial _{x_3}\sigma _n)(\nabla \sigma \cdot v_n)dx + \displaystyle \int _\Omega (\partial _{x_3}\sigma _n)(\nabla \cdot (\partial _{x_3}v_n))\sigma _n\\\le & {} C_\Omega \left\Vert {v_n} \right\Vert _{H^3}\left\Vert {\sigma _n} \right\Vert _{H^1}^2.\end{aligned}$$

Combining these estimates with (5.18) and taking into account of (5.19), we obtain

$$\begin{aligned}{} & {} \displaystyle \Big (k +{k_1\over 2}\Big ) \left\Vert {\partial _{x_3}\sigma _n} \right\Vert _{L^2}^2 \le \displaystyle \Big (k - {k_1\over 2}\Big ) \left\Vert {\partial _{x_3}\sigma '^{(n)}} \right\Vert _{L^2}^2 + C_\Omega C_M^{{2-\gamma \over \gamma -1}}\big (\Vert \sigma '^{(n)}\Vert _{L^2}^2 + \left\Vert {\nabla b_n} \right\Vert _{L^2}^2\big )\\{} & {} \quad + \,\, C_\Omega C_M^{{2-\gamma \over \gamma -1}} \displaystyle \sum _{i=1}^2\left\Vert {\nabla \partial _{x_i}v_n} \right\Vert _{L^2}^2 + C_\Omega C_M^{{4-3\gamma \over \gamma -1}} \left\Vert {\nabla v_n} \right\Vert _{L ^2}^2 + \,\,\widehat{C}\left\Vert {F'_n} \right\Vert _{L^2}^2 + C_\Omega \left\Vert {v_n} \right\Vert _{H^3}\left\Vert {\sigma _n} \right\Vert _{H^1}^2. \end{aligned}$$

By multiplying now this inequality by

$$\begin{aligned} C_M^ {-2+\gamma \over \gamma -1} = C_M^2C_M^{-\gamma \over \gamma -1}, \end{aligned}$$

we get the estimate (5.15) using (5.1), and this completes of the Lemma 5.3. \(\square \)

Lemma 5.4

Under the assumptions of the Lemma 5.1, we have

$$\begin{aligned}{} & {} \left\Vert {v_n} \right\Vert _{H^2}^2 + C_M^2\Vert \nabla \sigma '^{(n)}\Vert _{L^2}^2\le C'_4\Vert \sigma '^{(n)}\Vert _{L^2}^2 + C'_4\displaystyle \sum _{i=1}^2\left\Vert {\nabla \partial _{x_i}v_n} \right\Vert _{L^2}^2 \nonumber \\ {}{} & {} \quad + \, C'_4\big (C_M^2\Vert \partial _{x_3}\sigma '^{(n)}\Vert _{L^2}^2 + \left\Vert {\nabla b_n} \right\Vert _{L^2}^2 + \left\Vert {\nabla v_n} \right\Vert _{L^2}^2 + \left\Vert {F'_n} \right\Vert _{L^2}^2\big ).\end{aligned}$$
(5.20)

Proof

We first rewrite Eq. (4.26) as a Stokes problem in \(\Omega \)

$$\begin{aligned}{} & {} -\mu \Delta v_n + \gamma \vartheta \nabla (\widehat{\varrho }^{\gamma -1}\sigma '^{(n)}) = (\mu + \lambda )\nabla \nabla \cdot v_n - g\sigma '^{(n)} e_3 + (\widehat{B}\cdot \nabla )b_n - \nabla (b_n\cdot \widehat{B}) + F'_n,\end{aligned}$$
(5.21)
$$\begin{aligned}{} & {} \nabla \cdot v_n = \nabla \cdot v_n,\end{aligned}$$
(5.22)

with the boundary conditions (2.10)\(_1\). From the classical estimates (see [10]) of the Stokes problem, one has

$$\begin{aligned} \left\Vert {v_n} \right\Vert _{H^2}^2 + \Vert \nabla (\widehat{\varrho }^{\gamma -1}\sigma '^{(n)}) \Vert _{L^2}^2 \le C_\Omega (\left\Vert {\nabla \cdot v_n} \right\Vert _{H^1}^2 + \left\Vert {F'_n} \right\Vert _{L^2}^2 + \Vert \sigma '^{(n)} \Vert _{L^2}^2 + \left\Vert {\nabla b_n} \right\Vert _{L^2}^2).\end{aligned}$$
(5.23)

Since

$$\begin{aligned}\left\Vert {\nabla \cdot v_n} \right\Vert _{H^1}^2\le \left\Vert {\nabla \cdot v_n} \right\Vert _{L^2}^2 + \Vert \partial _{x_3}\nabla \cdot v_n \Vert _{L^2}^2 + \displaystyle \sum _{i=1}^2\left\Vert {\nabla \partial _{x_i}v_n} \right\Vert _{L^2}^2,\end{aligned}$$

using (5.17), we have

$$\begin{aligned} \left\Vert {\nabla \cdot v_n} \right\Vert _{H^1}^2\le & {} C_\Omega \displaystyle \sum _{i=1}^2\left\Vert {\nabla \partial _{x_i}v_n} \right\Vert _{L^2}^2\nonumber \\{} & {} \quad + \ C_\Omega \bigl ( \Vert \sigma '^{(n)}\Vert _{L^2}^2 + C_M^2 \Vert \partial _{x_3}\sigma '^{(n)} \Vert _{L^2}^2 + \left\Vert {\nabla b_n} \right\Vert _{L^2}^2 + \left\Vert {\nabla \cdot v_n} \right\Vert _{L^2}^2 + \left\Vert {F'_n} \right\Vert _{L^2}^2\bigr ).\end{aligned}$$
(5.24)

Moreover, thanks to (1.14) and (1.11), we have

$$\begin{aligned} \left\Vert {\nabla (\widehat{\varrho }^{\gamma -1}\sigma '^{(n)})} \right\Vert _{L^2}^2 \ge C_\Omega (C_M^2\Vert \nabla \sigma '^{(n)}\Vert _{L^2}^2 - \Vert \sigma '^{(n)}\Vert _{L^2}^2).\end{aligned}$$
(5.25)

We then obtain the estimate (5.20) from (5.23) using (1.7) and (5.24), which achieves the proof of the lemma 5.4. \(\square \)

Lemma 5.5

We have

$$\begin{aligned}&\left\Vert {b_n} \right\Vert _{H^2}^2\le C'_5(\left\Vert {\nabla v_n} \right\Vert _{L^2}^2 + \left\Vert {G'_n} \right\Vert _{L^2}^2).\end{aligned}$$
(5.26)

Proof

Since \(\left\Vert {b_n} \right\Vert _{H^2}\le C_\Omega \left\Vert {\Delta b_n} \right\Vert _{L^2}\) for all \(b_n\in H^2(\Omega )\cap H_0^1(\Omega )\), from the Eq. (3.8) and the boundary condition (1.4) we easily get (5.26). \(\square \)

Lemma 5.6

Under the same assumptions of the Lemma 5.1, we have.

$$\begin{aligned}{} & {} \displaystyle \sum _{i,j=1,2}^2\Big [{\mu \over \gamma \vartheta } \left\Vert {\nabla \partial _{x_i}\partial _{x_j} v_n} \right\Vert _{L^2}^2 + {\lambda +\mu \over \gamma \vartheta }\left\Vert {\partial _{x_i}\partial _{x_j}\nabla \cdot v_n} \right\Vert _{L^2}^2+ k\Vert \widehat{\varrho }^{\gamma -2\over 2}\partial _{x_i}\partial _{x_j}\sigma _n\Vert _{L^2}^2\Big ] \nonumber \\ {}{} & {} \quad \ \le k\displaystyle \sum _{i,j=1,2}^2\Vert \widehat{\varrho }^{\gamma -2\over 2} \partial _{x_i}\partial _{x_j}\sigma '^{(n)}\Vert _{L^2}^2 + \ C'_6 \bigl (\Vert \nabla \sigma '^{(n)}\Vert _{L^2}^2 + \left\Vert {v_n} \right\Vert _{H^2}^2 + \left\Vert {b_n} \right\Vert _{H^2}^2\bigr )\nonumber \\{} & {} \qquad + \, C'_6\bigl (\left\Vert {F'_n} \right\Vert _{H^1}^2 + \left\Vert {v_n} \right\Vert _{H^3}\left\Vert {\sigma _n} \right\Vert _{H^2}^2\bigr ). \end{aligned}$$
(5.27)

Proof

As in the Lemma 5.2, since \(\partial _{x_i}\partial _{x_j}v_n\) (\(i,j=1,2\)) satisfies the same boundary conditions (2.10), then we have

$$\begin{aligned}{} & {} -\displaystyle \int _\Omega \big [(\partial _{x_i}\partial _{x_j} v_n)\nabla \cdot (\partial _{x_i}\partial _{x_j}(\widehat{\varrho }^{\gamma -1}\sigma '^{(n)}) + \widehat{\varrho }^{\gamma -2}\partial _{x_i}\partial _{x_j}\nabla \cdot (\widehat{\varrho } v_n)\partial _{x_i}\partial _{x_j}\sigma _n\big ]dx\\{} & {} \quad = - \displaystyle \int _\Omega \widehat{\varrho }^{\gamma -2}(\partial _{x_i}\partial _{x_j}\nabla \cdot (\widehat{\varrho } v_n))(\partial _{x_i}\partial _{x_j}\sigma _n - \partial _{x_i}\partial _{x_j}\sigma '^{(n)})dx - \displaystyle \int _\Omega \widehat{\varrho }^{\gamma -2}(\partial _{x_i}\partial _{x_j}v_n\cdot \nabla \widehat{\varrho }) \partial _{x_i}\partial _{x_j}\sigma '^{(n)} dx.\end{aligned}$$

Now if we apply the differential operator \(\partial _{x_i}\partial _{x_j}\), \((i, j=1,2)\) to both sides of the Eqs. (4.26) and (4.28) and we multiply the resulting equations by \((\gamma \vartheta )^{-1}\partial _{x_i}\partial _ {x_j}v_n\) and \(\widehat{\varrho }^{\gamma -2}\partial _{x_i}\partial _{x_j}\sigma _n\) respectively, then we integrate over \(\Omega \). We then obtain through integratin by parts

$$\begin{aligned}{} & {} \displaystyle {\mu \over \gamma \vartheta } \Vert \nabla \partial _{x_i}\partial _{x_j} v_n\Vert _{L^2}^2 + \displaystyle {\lambda +\mu \over \gamma \vartheta }\Vert \partial _{x_i}\partial _{x_j}\nabla \cdot v_n\Vert _{L^2}^2 + \displaystyle {k\over 2}\Vert \widehat{\varrho }^{\gamma -2\over 2}\partial _{x_i} \partial _{x_j}\sigma _n\Vert _{L^2}^2\nonumber \\{} & {} \quad + \ \displaystyle {k\over 2}\Vert \widehat{\varrho }^{\gamma -2\over 2}(\partial _{x_i}\partial _{x_j}\sigma _n -\partial _{x_i}\partial _{x_j}\sigma '^{(n)})\Vert _{L^2}^2 = \displaystyle {k\over 2}\Vert \widehat{\varrho }^{\gamma -2\over 2}\partial _{x_i}\partial _{x_j}\sigma '^{(n)}\Vert _{L^2}^2 + \displaystyle \sum _{i=1}^7I_i. \end{aligned}$$
(5.28)

where

$$\begin{aligned} I_1= & {} -\displaystyle \int _\Omega \widehat{\varrho }^{\gamma -2}(\partial _{x_i}\partial _{x_j}\sigma _n - \partial _{x_i}\partial _{x_j}\sigma '^{(n)})\partial _{x_j} \partial _{x_i}\nabla \cdot (\widehat{\varrho } v_n)dx,\\ I_2= & {} -\displaystyle {g\over \gamma \vartheta }\displaystyle \int _\Omega (\partial _{x_i}\sigma '^{(n)})\partial _{x_i}\partial _{x_i} \partial _{x_j}v_{n,3} dx + \displaystyle \int _\Omega \widehat{\varrho }^{\gamma -2}\partial _{x_i}\partial _{x_i}\partial _{x_j} (v_n\cdot \nabla \widehat{\varrho }) (\partial _{x_j}\sigma '^{(n)}) dx,\\ I_3= & {} -\displaystyle {1\over \gamma \vartheta }\int _\Omega (\partial _{x_i}\partial _{x_i} \partial _{x_j}v_n)\cdot (\partial _{x_j}F'_n)dx,\ \ I_4 = -\displaystyle \int _\Omega \widehat{\varrho }^{\gamma -2}(\partial _{x_i}\partial _{x_j}\sigma _n) \partial _{x_i}\partial _{x_j}\nabla \cdot (v_n\sigma _n)dx,\\ I_5= & {} \displaystyle {1\over \gamma \vartheta }\int _\Omega (\partial _{x_i} \partial _{x_j}v_n)\cdot [\partial _{x_i} \partial _{x_j}\widehat{B}\cdot \nabla )b_n + ( \partial _{x_j}\widehat{B}\cdot \nabla )\partial _{x_i}b_n + ( \partial _{x_i}\widehat{B}\cdot \nabla )\partial _{x_j}b_n]dx, \\ I_6= & {} - \displaystyle {1\over \gamma \vartheta }\int _\Omega \big [\partial _{x_i} \partial _{x_j}(\widehat{B}\cdot b_n)\nabla \cdot (\partial _{x_i}\partial _{x_j}v_n) + (\nabla \cdot \widehat{B})(\partial _{x_i}\partial _{x_j}b_n)\cdot (\partial _{x_i}\partial _{x_j}v_n)\big ]dx, \\ I_7= & {} - \displaystyle {1\over \gamma \vartheta } \int _\Omega (\widehat{B}\cdot \nabla )\partial _{x_i}\partial _{x_j}v_n\cdot \partial _{x_i}\partial _{x_j}b_n dx. \end{aligned}$$

By estimating the terms \(I_i\) \((i = 1, \ldots 4)\) as in the proof of Lemma 5.2, we obtain

$$\begin{aligned}{} & {} I_1+\cdots +I_4\le \displaystyle {k\over 2}\Vert \big (\widehat{\varrho }^{\gamma -2\over 2}(\partial _{x_i}\partial _{x_j}\sigma _n - \partial _{x_i}\partial _{x_j}\sigma '^{(n)})\Vert _{L^2}^2 + \displaystyle {\mu +\lambda \over 4\gamma \vartheta }\left\Vert {\nabla \cdot ( \partial _{x_i}\partial _{x_j}v_n)} \right\Vert _{L^2}^2 \\{} & {} \quad + \ \displaystyle {\mu \over 4\gamma \vartheta }\left\Vert {\nabla \partial _{x_i}\partial _{x_j}v_n} \right\Vert _{L^2}^2 + C_\Omega (\Vert \nabla \sigma '^{(n)}\Vert _{L^2}^2 + \left\Vert {v_n} \right\Vert _{H^2}^2 + \left\Vert {F'_n} \right\Vert _{H^1}^2 + \left\Vert {v} \right\Vert _{H^3}\left\Vert {\sigma } \right\Vert _{H^2}^2). \end{aligned}$$

As for the terms \(I_i\) \((i = 5,\ldots 7)\), we have

$$\begin{aligned} I_5+I_6+I_7\le \displaystyle {\mu +\lambda \over 4\gamma \vartheta }\left\Vert {\nabla \cdot ( \partial _{x_i}\partial _{x_j}v_n)} \right\Vert _{L^2}^2 + \displaystyle {\mu \over 4\gamma \vartheta }\left\Vert {\nabla \partial _{x_i}\partial _{x_j}v_n} \right\Vert _{L^2}^2 + C_\Omega (\left\Vert {b_n} \right\Vert _{H^2}^2 + \left\Vert {v_n} \right\Vert _{H^2}^2). \end{aligned}$$

By adding now the estimates of \( I_i \) \( (i = 1, \ldots 7)\) to (5.28) and summing over \( i = 1,2 \), we obtain (5.27), and the lemma is proved. \(\square \)

Lemma 5.7

Under the assumptions of the Lemma 5.1, we have

$$\begin{aligned}{} & {} ({\bar{k}} + 1)C'_MC_M^2\displaystyle \sum _{i=1}^2\big |\hspace{-1.42271pt}\big | {\partial _{x_3}\partial _{x_i}\sigma _n}\big |\hspace{-1.42271pt}\big |_{L^2}^2 \le ({\bar{k}} - 1)C'_MC_M^2\displaystyle \sum _{i=1}^2\big |\hspace{-1.42271pt}\big | {\partial _{x_3}\partial _{x_i}\sigma '^{(n)}}\big |\hspace{-1.42271pt}\big |_{L^2}^2 \nonumber \\{} & {} \quad + \ C'_7\Big (\displaystyle \sum _{i,j=1}^2\left\Vert {\nabla \partial _{x_i}\partial _{x_j} v_n} \right\Vert _{L^2}^2 + C_M^{-2}\left\Vert {v_n} \right\Vert _{H^2}^2 + \left\Vert {b_n} \right\Vert _{H^2}^2 + \Vert \nabla \sigma '^{(n)}\Vert _{L^2}^2\Big ) \nonumber \\ {}{} & {} \quad + \ \tilde{C}_2\big (\left\Vert {F'_n} \right\Vert _{H^1}^2 + \left\Vert {v_n} \right\Vert _{H^3}\left\Vert {\sigma _n} \right\Vert _{H^2}^2\big ), \end{aligned}$$
(5.29)

where \(C'_M\) is given by (5.16).

Proof

We apply the differential operator \(\partial _{x_3}\partial _{x_i}\), \((i=1,2)\) to both sides of the Eq. (4.28) and we multiply the resulting equations by \(\partial _ {x_3}\partial _{x_i}\sigma _n\) and integrate over \(\Omega \). We then obtain through integration by parts taking into account (5.17)

$$\begin{aligned}{} & {} \displaystyle \int _\Omega k\Big [(\partial _{x_3}\partial _{x_i}\sigma _n - \partial _{x_3}\partial _{x_i}\sigma '^{(n)})\partial _{x_i}\partial _{x_3}\sigma _n \nonumber \\{} & {} \quad + {\gamma \vartheta \over 2\mu +\lambda }\widehat{\varrho }^\gamma (\partial _{x_3}\partial _{x_i}\sigma )(\partial _{x_3}\partial _{x_i}\sigma '^{(n)})\Big ]dx = \displaystyle \sum _{i=1}^5I_i, \end{aligned}$$
(5.30)

where

$$\begin{aligned} {}&{} \displaystyle \int _\Omega k\Big [(\partial _{x_3}\partial _{x_i}\sigma _n - \partial _{x_3}\partial _{x_i}\sigma '^{(n)})\partial _{x_i}\partial _{x_3}\sigma _n \nonumber \\{}&{} \quad + {\gamma \vartheta \over 2\mu +\lambda }\widehat{\varrho }^\gamma (\partial _{x_3}\partial _{x_i}\sigma )(\partial _{x_3}\partial _{x_i}\sigma '^{(n)})\Big ]dx = \displaystyle \sum _{i=1}^5I_i, \end{aligned}$$

Just like in the proof of Lemma 5.3 we use the inequality

$$\begin{aligned}{} & {} \displaystyle \int _\Omega k\big [(\partial _{x_i}\partial _{x_3}\sigma _n- \partial _{x_i}\partial _{x_3}\sigma '^{(n)})\partial _{x_3}\sigma _n + {\gamma \vartheta \over 2\mu +\lambda }\widehat{\varrho }^\gamma (\partial _{x_i}\partial _{x_3}\sigma _n)(\partial _{x_i}\partial _{x_3}\sigma '^{(n)})\big ]dx \\{} & {} \quad \ge \displaystyle {k +k_1\over 2}\left\Vert {\partial _{x_i}\partial _{x_3}\sigma _n} \right\Vert _{L^2}^2 - \displaystyle {k - k_1\over 2}\big |\hspace{-1.42271pt}\big | {\partial _{x_i}\partial _{x_3}\sigma '^{(n)}}\big |\hspace{-1.42271pt}\big |_{L^2}^2 + \displaystyle {k - k'_1\over 2}\big |\hspace{-1.42271pt}\big | {\partial _{x_i}\partial _{x_3}\sigma _n - \partial _{x_i}\partial _{x_3}\sigma '^{(n)}}\big |\hspace{-1.42271pt}\big |_{L^2}^2 \end{aligned}$$

and we estimate similarly the terms \(I_i\) \((I = 1, \ldots , 5)\). By combining these estimates with (5.18), we obtain

$$\begin{aligned}{}&{} \displaystyle \Big (k +{k_1\over 2}\Big ) \big |\hspace{-1.42271pt}\big | {\partial _{x_i}\partial _{x_3}\sigma _n}\big |\hspace{-1.42271pt}\big |_{L^2}^2\le \displaystyle \Big (k - {k_1\over 2}\Big ) \big |\hspace{-1.42271pt}\big | {\partial _{x_i}\partial _{x_3}\sigma '^{(n)}}\big |\hspace{-1.42271pt}\big |_{L^2}^2 \\{}&{}+\, C_\Omega \big (C_M^{{2-\gamma \over \gamma -1}}\sum _{j=1}^2 \left\| {\nabla \partial _{x_i}\partial _{x_j}v_n} \right\| _{L^2}^2 + C_M^{{4-3\gamma \over \gamma -1}} \left\| {v_n} \right\| _{H^2}^2 + C_M^{{2-\gamma \over \gamma -1}}(\Vert \nabla \sigma '^{(n)}\Vert _{L^2}^2 + \left\| {b_n} \right\| _{H^2}^2\big )\\ {}{}&{} \quad + \,\widehat{C}\left\| {F'_n} \right\| _{H^1}^2 + C_\Omega \left\| {v_n} \right\| _{H^3}\left\| {\sigma _n} \right\| _{H^2}^2. \end{aligned}$$

Multiplying now this inequality by

$$\begin{aligned} C_M^{-2+\gamma \over \gamma -1} = C_M^2C_M^ {-\gamma \over \gamma -1} \end{aligned}$$

and taking into account of (5.1), we obtain by summing on \(i=1,2\), the inequality (5.29). \(\square \)

Lemma 5.8

Under the assumptions of Lemma 5.1, we have

$$\begin{aligned}{} & {} \displaystyle \sum _{i=1}^2\bigl (\left\Vert {\partial _{x_i}v_n} \right\Vert _{H^2}^2 + C^2_M\Vert \nabla \partial _{x_i}\sigma '^{(n)}\Vert _{L^2}^2\bigr )\le C'_8\displaystyle \sum _{i,j=1}^2 \left\Vert {\nabla \partial _{x_i}\partial _{x_j}v_n} \right\Vert _{L^2}^2 \nonumber \\{} & {} \quad + \ C'_8\bigl (\left\Vert {v_n} \right\Vert _{H^2}^2 + \left\Vert {b_n} \right\Vert _{H^2}^2 + \Vert \nabla \sigma '^{(n)} \Vert _{L^2}^2 + C_M^2\sum _{i=1}^2 \Vert \partial _{x_3}\partial _{x_i}\sigma '^{(n)} \Vert _{L^2}^2\big ) + C'_8\left\Vert {F'_n} \right\Vert _{H^1}^2. \end{aligned}$$
(5.31)

Proof

We apply to Eq. (4.26) the differential operator \(\partial _{x_i}\) \( (i = 1,2) \) and we rewrite the obtained equation as Stokes problem in \( \Omega \)

$$\begin{aligned}{} & {} \mu \Delta (\partial _{x_i}v_n) + \gamma \vartheta \nabla (\partial _{x_i}(\widehat{\varrho }^{\gamma -1}\sigma '^{(n)})) \\{} & {} \quad = - (\mu +\lambda )\partial _{x_i}\nabla \nabla \cdot v - g\partial _{x_i}\sigma '^{(n)}e_3 + \partial _{x_i}(\nabla (\widehat{B}\cdot b_n) - (\widehat{B}\cdot \nabla )b_n) + \partial _{x_i}F'_n,\\{} & {} \qquad \nabla \cdot \partial _{x_i}v_n = \nabla \cdot \partial _{x_i}v_n,.\end{aligned}$$

with the same boundary conditions on \(\partial _{x_i}v\) as (2.10)\(_1\). We obtain by the same arguments of proof of Lemma 5.4 that

$$\begin{aligned}{} & {} \left\Vert {\partial _{x_i}v_n} \right\Vert _{H^2}^2 + \Vert \nabla \partial _{x_i}(\widehat{\varrho }^{\gamma -1}\sigma '^{(n)})\Vert _{L^2}^2 \nonumber \\{} & {} \quad \le C_\Omega \big (\left\Vert {\nabla \cdot \partial _{x_i} v_n} \right\Vert _{H^1}^2 + \left\Vert {F'_n} \right\Vert _{H^1}^2 + \Vert \nabla \sigma '^{(n)} \Vert _{L^2}^2 + \left\Vert {b_n} \right\Vert _{H^2}^2\big ). \end{aligned}$$
(5.32)

In addition, let us note that

$$\begin{aligned} \left\Vert {\partial _{x_i}\nabla \cdot v_n} \right\Vert _{H^1}^2\le C_\Omega \Bigl (\displaystyle \sum _{j=1}^2\left\Vert {\partial _{x_i}\partial _{x_j}\nabla \cdot v_n} \right\Vert _{L^2}^2 + \left\Vert {\partial _{x_i}\partial _{x_3}\nabla \cdot v_n} \right\Vert _{L^2}^2 + \left\Vert {v_n} \right\Vert _{H^2}^2\Bigr ) \end{aligned}$$
(5.33)

and, in applying \(\partial _{x_i}\) \((i=1,2) \) to (5.17), it follows that

$$\begin{aligned}{} & {} \left\Vert {\partial _{x_i}\partial _{x_3}\nabla \cdot v_n} \right\Vert _{L^2}^2\nonumber \\{} & {} \quad \le C_\Omega \Big (\displaystyle \sum _{j=1}^2 \left\Vert {\nabla \partial _{x_i}\partial _{x_j}v_n} \right\Vert _{L^2}^2 + C_M^2\Vert \partial _{x_3}\partial _{x_i}\sigma '^{(n)}\Vert _{L^2}^2 + \left\Vert {F'_n} \right\Vert _{H^1}^2 + \Vert \nabla \sigma '^{(n)}\Vert _{L^2}^2 + \left\Vert {b_n} \right\Vert _{H^2}^2\Big ). \end{aligned}$$
(5.34)

By substituting (5.33) and (5.34) into (5.32) and taking into account the estimate

$$\begin{aligned} \Vert \nabla \partial _{x_i}(\widehat{\varrho }^{\gamma -1}\sigma '^{(n)}) \Vert _{L^2}^2\ge C_\Omega \big (C_M^2 \Vert \nabla \partial _{x_i}\sigma '^{(n)} \Vert _{L^2}^2 - \Vert \nabla \sigma '^{(n)} \Vert _{L^2}^2\big ), \end{aligned}$$

(obtained similarly to (5.25) through (1.14)), we get the estimate (5.31). This completes the proof of the lemma. \(\square \)

Lemma 5.9

Under the assumptions of the Lemma 5.1 one has

$$\begin{aligned}{} & {} ({\bar{k}} + 1)C'_MC_M^2\left\Vert {\Delta \sigma _n} \right\Vert _{L^2}^2 \le ({\bar{k}} - 1)C'_MC_M^2\left\Vert {\Delta \sigma '^{(n)}} \right\Vert _{L^2}^2 \nonumber \\{} & {} \quad + \ C'_9(C_M^{-2}\left\Vert {v_n} \right\Vert _{H^2}^2 + \left\Vert {b} \right\Vert _{H^2}^2 + \left\Vert {\nabla \sigma '^{(n)}} \right\Vert _{L^2}^2) + \tilde{C}_3\left\Vert {F'_n} \right\Vert _{H^1}^2 + C'_9\left\Vert {v_n} \right\Vert _{H^3}\left\Vert {\sigma _n} \right\Vert _{H^2}^2. \end{aligned}$$
(5.35)

Proof

We apply the Laplacian operator to Eq. (3.9), multiply the resulting equation by \(\Delta \sigma _n\) and integrate over \(\Omega \). Following integration by parts, on account of (5.17) we get

$$\begin{aligned} \displaystyle \int _\Omega k\Bigl [(\Delta \sigma _n- \Delta \sigma '^{(n)})\Delta \sigma _n + {\gamma \vartheta \over 2\mu +\lambda }\widehat{\varrho }^\gamma \Delta \sigma _n\Delta \sigma '^{(n)}\Bigr ]dx =I_1 + I_2 + I_3, \end{aligned}$$
(5.36)

where

$$\begin{aligned} I _1=&{} -\displaystyle {1\over 2\mu +\lambda }\int _\Omega (2(\gamma -1)\vartheta \nabla \widehat{\varrho }^\gamma \cdot \nabla \sigma '^{(n)} + g\widehat{\varrho }\partial _{x_3}\sigma '^{(n)})\Delta \sigma _n dx,\\ I_2=&{} \displaystyle \int _\Omega (\Delta (v_n\cdot \nabla \widehat{\varrho }) + \Delta \widehat{\varrho }\nabla \cdot v_n + 2(\nabla \widehat{\varrho })\cdot \nabla (\nabla \cdot v_n))\Delta \sigma _n \,dx,\\ I_3=&{} \displaystyle \int _\Omega \widehat{\varrho }\Delta \sigma _n(\nabla \cdot (\nabla (\widehat{B}\cdot b_n) - (\widehat{B}\cdot \nabla )b_n))\,dx,\\ I_4=&{} -\displaystyle \int _\Omega (\Delta \nabla \cdot (\sigma _n v_n))\Delta \sigma _n\, dx + \displaystyle {1\over (2\mu +\lambda )}\int _\Omega \widehat{\varrho }(\nabla \cdot F'_n)\Delta \sigma _n\, dx\,. \end{aligned}$$

In a similar way to the proof of the lemmas 5.3 and 5.7, we have

$$\begin{aligned}{} & {} \displaystyle \int _\Omega k\big [(\Delta \sigma _n - \Delta \sigma '^{(n)})\Delta \sigma + {\gamma \vartheta \over (2\mu +\lambda )}\widehat{\varrho }^{\gamma }(\Delta \sigma _n)(\Delta \sigma '^{(n)})\big ]dx \\{} & {} \quad \ge \displaystyle {k +k_1\over 2}\left\Vert {\Delta \sigma _n} \right\Vert _{L^2}^2 - \displaystyle {k - k_1\over 2}\left\Vert {\Delta \sigma '^{(n)}} \right\Vert _{L^2}^2 + \displaystyle {k - k'_1\over 2}\left\Vert {\Delta \sigma _n - \Delta \sigma '^{(n)}} \right\Vert _{L^2}^2 \end{aligned}$$

and by estimating similarly the terms \(I_i\) (\(i=1,\ldots ,4)\), we obtain

$$\begin{aligned}{} & {} \displaystyle \big (k +{k_1\over 2}\big ) \left\Vert {\Delta \sigma _n} \right\Vert _{L^2}^2\le \displaystyle \big (k - {k_1\over 2}\big ) \left\Vert {\Delta \sigma '^{(n)}} \right\Vert _{L^2}^2 \\{} & {} \quad +\,\,C_\Omega (C_M^{{4-3\gamma \over \gamma -1}} \left\Vert {v_n} \right\Vert _{H^2}^2 + C_M^{{2-\gamma \over \gamma -1}}(\left\Vert {b_n} \right\Vert _{H^2}^2 + \left\Vert {\nabla \sigma '^{(n)}} \right\Vert _{L^2}^2)) + \widehat{C}\left\Vert {F'} \right\Vert _{H^1}^2 + C_\Omega \left\Vert {v_n} \right\Vert _{H^3}\left\Vert {\sigma _n} \right\Vert _{H^2}^2. \end{aligned}$$

Multiplying now this inequality by

$$\begin{aligned} C_M^{-2+\gamma \over \gamma -1} = C_M^2C_M^ {-\gamma \over \gamma -1} \end{aligned}$$

and taking into account (5.1), we obtain the estimate (5.35). \(\square \)

Lemma 5.10

Under the same assumptions of Lemma 5.1 one has

$$\begin{aligned}{} & {} \left\Vert {v_n} \right\Vert _{H^3}^2 + C_M^2\big |\hspace{-1.42271pt}\big | {\nabla \sigma '^{(n)}}\big |\hspace{-1.42271pt}\big |_{H^1}^2\le C'_{10}C_M^2\Big [\big |\hspace{-1.42271pt}\big | {\Delta \sigma '^{(n)}}\big |\hspace{-1.42271pt}\big |_{L^2}^2 + \displaystyle \sum _{i=1}^2 \big |\hspace{-1.42271pt}\big | {\nabla \partial _{x_i}\sigma '^{(n)}}\big |\hspace{-1.42271pt}\big |_{L^2}^2\Big ]\nonumber \\{} & {} \quad + \ C'_{10}\Big (\displaystyle \sum _{i=1}^2 \big |\hspace{-1.42271pt}\big | {\partial _{x_i}v_n}\big |\hspace{-1.42271pt}\big |_{H^2}^2 + \left\Vert {v_n} \right\Vert _{H^1}^2 +\Vert \sigma '^{(n)}\Vert _{H^1}^2 + \left\Vert {b_n} \right\Vert _{H^2}^2 + \big |\hspace{-1.42271pt}\big | {F'_n}\big |\hspace{-1.42271pt}\big |_{H^1}^2\Big ). \end{aligned}$$
(5.37)

Proof

As in Lemma 5.4, according the well-known theory estimates of the stokes problem, one deduced from (5.21) and (5.22) with the boundary conditions (2.10)

$$\begin{aligned} \left\Vert {v_n} \right\Vert _{H^3}^2 + \gamma \vartheta \left\Vert {\nabla (\widehat{\varrho }\sigma '_n)} \right\Vert _{H^1}^2 \le C_\Omega \bigl (\left\Vert {b_n} \right\Vert _{H^2}^2 + \left\Vert {\nabla \cdot v_n} \right\Vert _{H^2}^2 + \Vert \sigma '^{(n)}\Vert _{H^1}^2 + \left\Vert {F'_n} \right\Vert _{H^1}^2\bigr ).\end{aligned}$$
(5.38)

Let us first notice that

$$\begin{aligned} \left\Vert {\nabla \cdot v_n} \right\Vert _{H^2}^2\le \left\Vert {\partial _{x_3}\nabla \cdot v_n} \right\Vert _{H^1}^2 + \displaystyle \sum _{i=1}^2 \left\Vert {\partial _{x_i}v_n} \right\Vert _{H^2}^2 + \left\Vert {v_n} \right\Vert _{H^1}^2. \end{aligned}$$

In addition, taking into account (5.17), we have

$$\begin{aligned} \left\Vert {\partial _{x_3}\nabla \cdot v_n} \right\Vert _{H^1}^2\le C_\Omega \Big (\displaystyle \sum _{i=1}^2\left\Vert {\partial _{x_i}v_n} \right\Vert _{H^2}^2 + C_M^2\Vert \nabla \partial _{x_3}\sigma '^{(n)}\Vert _{L^2}^2 + \left\Vert {b_n} \right\Vert _{H^2}^2 + \left\Vert {F'_n} \right\Vert _{H^1}^2 + \Vert \nabla \sigma '^{(n)}\Vert _{L^2}^2\Big ). \end{aligned}$$

Since (see (2.1)\(_1\) and (1.14))

$$\begin{aligned} \big |\hspace{-1.42271pt}\big | {\nabla (\widehat{\varrho }^{\gamma -1}\sigma '^{(n)})}\big |\hspace{-1.42271pt}\big |_{H^1}^2\ge&{} C_\Omega \bigl (C_M^2\Vert \nabla \sigma '^{(n)}\Vert _{H^1}^2 - \Vert \nabla \sigma '^{(n)} \Vert _{L^2}^2 - \Vert \sigma '^{(n)}\Vert _{L^2}^2\bigr )\,,\\ \Vert \nabla \partial _{x_3}\sigma '^{(n)}\Vert _{L^2}^2\le&{} \Vert \Delta \sigma '^{(n)}\Vert _{L^2}^2 + \displaystyle \sum _{i=1}^2\Vert \nabla \partial _{x_i}\sigma '^{(n)}\Vert _{L^2}^2,\end{aligned}$$

by adding these inequalities to (5.38), one obtains (5.37). \(\square \)

6 Fixed Point of the Operator \(\textbf{S}_n\)

Having proved the lemmas 5.15.10 one is now in a position to establish that, for every n, the operator \(S_n\) has a fixed point. Following Remark 4.5 and Lemma 4.6, it remains for us to establish the crucial point (4.13), which is the goal of the following lemma.

Lemma 6.1

There is a norm \(|\cdot |_V\) equivalent (see (3.6) and (3.3)) to \(\Vert \cdot \Vert _V\) such that

$$\begin{aligned} S_n(\tau _nB_{V}^a)\subseteq \tau _nB_{V}^a,\quad \text{ with }\quad B_{V}^a= \{u=(v,b,\sigma )\in V: |u|_V\le a\}.\end{aligned}$$
(6.1)

Proof

Notice that, it suffices to prove that

$$\begin{aligned} S(\tau _nB_{V}^a)\subseteq B_{V}^a.\end{aligned}$$
(6.2)

Let then \(\lambda _1, \lambda _2, \ldots , \lambda _9\) be positive numbers which will be suitably chosen later. We set

$$\begin{aligned} \begin{array}{lll} \nu _1:= C'_M({\bar{k}} +1)\lambda _9C_M^2, &{} \nu _2: = C'_M({\bar{k}} +1)\lambda _7C_M^2, &{} \nu _3:= k\lambda _6,\\ \nu _4:= C'_M({\bar{k}} +1)\lambda _3C_M^2, &{} \nu _5:=: k\lambda _2,&{}\nu _6:= k\lambda _1, \end{array} \end{aligned}$$

where \(C'_M\) is given in (5.16) and \({\bar{k}}\) in (5.2). For any \(\varphi \in H^2\), we set

$$\begin{aligned}{} & {} |\varphi |_{2,\Omega }^2: = \ \nu _1\left\Vert {\Delta \varphi } \right\Vert _{L^2}^2 + \nu _2\displaystyle \sum _{i=1}^2 \Vert \partial _{x_i}\partial _{x_3}\varphi \Vert _{L^2}^2 + \nu _3\displaystyle \sum _{i,j=1}^2 \Vert \widehat{\varrho }^{\gamma -2\over 2}\partial _{x_i}\partial _{x_j} \varphi \Vert _{L^2}^2\nonumber \\{} & {} \quad + \ \nu _4\left\Vert {\partial _{x_3}\varphi } \right\Vert _{L^2}^2 + \nu _5\displaystyle \sum _{i=1}^2 \Vert \widehat{\varrho }^{\gamma -2\over 2}\partial _{x_i}\varphi \Vert _{L^2}^2 + \nu _6\displaystyle \Vert \widehat{\varrho }^{\gamma -2\over 2}\varphi \Vert _{L^2}^2. \end{aligned}$$
(6.3)

It is clear that \( |\cdot |_{2,\Omega }\) is equivalent to the \(H^2\)-norm. Moreover (see (4.8)), an easy computation shows that

$$\begin{aligned} |\varphi ^{(n)}|_{2,\Omega } = |\tau _n\varphi |_{2,\Omega }\le (1+\nu '_n)|\varphi |_{2,\Omega }.\end{aligned}$$
(6.4)

If we set

$$\begin{aligned} |u|_V^2: = \left\Vert {v} \right\Vert _{H^3}^2 + \left\Vert {b} \right\Vert _{H^2}^2 + |\sigma |_{2,\Omega }^2,\quad u=(v,b,\sigma )\in V,\end{aligned}$$
(6.5)

we obtain (see (3.5)) a equivalent norm to \(\Vert \cdot \Vert _V\). In addition, given (6.4) and (4.9), we can see that

$$\begin{aligned} |u^{(n)}|_V^2 = \Vert v^{(n)}\Vert _{H^3}^2 + \Vert b^{(n)}\Vert _{H^2}^2 + |\sigma ^{(n)}|_{2,\Omega }^2 \le (1+ \tilde{\nu }_n)|u|_V^2,\end{aligned}$$
(6.6)

where \((\tilde{\nu }_n)\) is a decreasing sequence of positive numbers converging to zero. Let now

$$\begin{aligned} S(u'^{(n)}) = u_n = (v_n,b_n,\sigma _n),\quad u'^{(n)}=\tau _n u\in \tau _nB_{V}^a\,.\end{aligned}$$
(6.7)

If we multiply the estimates on solutions \(u_n\) obtained in the lemmas 5.15.10 by the positive numbers \(\lambda _1, \lambda _2, \ldots , \lambda _9\) and \(\lambda _{10}= 1\) respectively and adding them, we obtain

$$\begin{aligned}{} & {} \left\Vert {v_n} \right\Vert _{H^3}^2 + |\sigma _n|_{2,\Omega }^2 +\Lambda _1\big (\displaystyle \sum _{i=1}^2\left\Vert {\partial _{x_i}v_n} \right\Vert _{H^2}^2 + C_M^2\displaystyle \sum _{i=1}^2\Vert \nabla \partial _{x_i}\sigma '^{(n)}\Vert _{L^2}^2\big ) \nonumber \\{} & {} \qquad +\, \Lambda _2\displaystyle \sum _{i,j=1}^2 \left\Vert {\nabla \partial _{x_i}\partial _{x_j}v_n} \right\Vert _{L^2}^2 + \Lambda _3\left\Vert {v_n} \right\Vert _{H^2}^2+ \Lambda _4\left\Vert {b_n} \right\Vert _{H^2}^2 + \Lambda _5\displaystyle \sum _{i=1}^2 \left\Vert {\nabla \partial _{x_i}v_n} \right\Vert _{L^2}^2\nonumber \\{} & {} \qquad + \, \Lambda _6\big (\left\Vert {\nabla v_n} \right\Vert _{L^2}^2 + \left\Vert {\nabla b_n} \right\Vert _{L^2}^2\big ) + \Lambda _7C_M\Vert \sigma '^{(n)}\Vert _{L^2}^2 + \Lambda _8C_M^2\Vert \nabla \sigma '^{(n)}\Vert _{L^2}^2\nonumber \\{} & {} \quad \le N(\sigma '^{(n)}) + {\tilde{C}}\big (\left\Vert {F'_n} \right\Vert _{H^1}^2 + \left\Vert {G'_n} \right\Vert _{L^2}^2 + \left\Vert {v_n} \right\Vert _{H^3}\left\Vert {\sigma _n} \right\Vert _{H^2}^2\big ), \end{aligned}$$
(6.8)

with some positive constant \({\tilde{C}}\). As for \(N(\sigma '^{(n)})\) it is given by

$$\begin{aligned}{} & {} N(\sigma '^{(n)}):= (C'_{10} + ({\bar{k}} - 1) C'_M\lambda _9)C_M^2\left\Vert {\Delta \sigma '^{(n)}} \right\Vert _{L^2}^2 \nonumber \\{} & {} \quad + \,\, (C'_8\lambda _8 + C'_M({\bar{k}} - 1)\lambda _7)C_M^2\displaystyle \sum _{i=1}^2 \Vert \partial _{x_i}\partial _{x_3}\sigma '^{(n)}\Vert _{L^2}^2 + k\lambda _6\displaystyle \sum _{i,j=1}^2 \Vert \widehat{\varrho }^{\gamma -2\over 2}\partial _{x_i}\partial _{x_j} \sigma '^{(n)}\Vert _{L^2}^2 \nonumber \\{} & {} \quad +\, \, k\lambda _2\displaystyle \sum _{i=1}^2 \Vert \widehat{\varrho }^{\gamma -2\over 2}\partial _{x_i}\sigma '^{(n)}\Vert _{L^2}^2 + (C'_4\lambda _4 + C'_M({\bar{k}} -1)\lambda _3)C_M^2\big |\hspace{-1.42271pt}\big | {\partial _{x_3}\sigma '^{(n)}}\big |\hspace{-1.42271pt}\big |_{L^2}^2 \nonumber \\{} & {} \quad + \,\, k\lambda _1\displaystyle \Vert \widehat{\varrho }^{\gamma -2\over 2} \sigma '^{(n)}\Vert _{L^2}^2 + k\lambda _1\displaystyle \Vert \widehat{\varrho }^{\gamma -2\over 2} \sigma '^{(n)}\Vert _{L^2}^2, \end{aligned}$$
(6.9)

and the numbers \(\Lambda _i\) (\(i=1,2, \ldots ,8)\) are given by

$$\begin{aligned} \begin{array}{l} \Lambda _1:= \lambda _8 - C'_{10}, \quad \Lambda _2: = \displaystyle {\mu \over \gamma \vartheta }\lambda _6 - C'_7\lambda _7 - C'_8\lambda _8,\\ \Lambda _3:= \lambda _4 - C'_6\lambda _6 - C'_7C_M^{-2}\lambda _7 - C'_8\lambda _8 - C'_9C_M^{-2}\lambda _9 - C'_{10}\,,\\ \Lambda _4:= \lambda _5 - C'_6\lambda _6 - C'_7\lambda _7 - C'_8\lambda _8 - C'_9\lambda _9 - C'_{10},\quad \Lambda _5: = \displaystyle {\mu \over \gamma \vartheta }\lambda _2 - C'_3\lambda _3 - C'_4\lambda _4, \\ \Lambda _6:= \displaystyle {1\over \gamma \vartheta }\inf (\mu ,\nu )\lambda _1 - C'_2\lambda _2 - C'_3\lambda _3- C'_4\lambda _4 - C'_5\lambda _5, \\ \Lambda _7:=C'_1C_M\lambda _1 - C'_2\lambda _2 - C'_3\lambda _3 - C'_4\lambda _4 - C'_{10},\\ \ \Lambda _8:= C_M^2\lambda _4 - C'_3\lambda _3 - C'_6\lambda _6 - C'_7\lambda _7 - C'_8\lambda _8 - C'_9\lambda _9 - C'_{10}. \end{array} \end{aligned}$$

Since (see (1.13) and (1.11)) \(C_M\) is large enough, we can see easily that it is possible to choose

$$\begin{aligned} \lambda _i = \lambda _i(M_0),\quad i=9,8,\ldots ,1 \end{aligned}$$

so that

$$\begin{aligned} \Lambda _i > 0 \quad \forall \ i=1,\ldots ,6,\ \ \text{ and } \quad N(\sigma ')\le (1-t)|\sigma '^{(n)}|_{2}^2\ \ \text{ for } \text{ some } \ \ t\in ]0,1[.\end{aligned}$$
(6.10)

Indeed for \(j = i+1,\ldots ,9\) and \(i=8,\ldots ,1\) given, as can be seen easily, considering the constraints (6.10) we can choose a \(\lambda _i\) large enough so that the inequalities containing only \(\lambda _i\) and \(\lambda _j\) are verified (and so we can proceed to the choice of \(\lambda _i\) starting from \(\lambda _9\) and then choosing successively \(\lambda _i\) for \(i=8,\ldots ,1\)). Such a choices of the numbers \(\lambda _i= \lambda _i(M_0)\) (\(i=1, \ldots , 9)\) imply in particular that \(\Lambda _7 > \)0 and \(\Lambda _8 > 0\), because \(C_M\) is large enough.

If we recall now the inequality (6.8), considering (6.10), we obtain

$$\begin{aligned} \left\Vert {v_n} \right\Vert _{H^3}^2 + \left\Vert {b_n} \right\Vert _{H^2}^2 + |\sigma _n|_{2,\Omega }^2\le & {} (1-t)|\sigma '^{(n)}|_{2,\Omega }^2\nonumber \\{} & {} \quad + \ {\tilde{C}}\big (\left\Vert {F'_n} \right\Vert _{H^1}^2 + \left\Vert {G'_n} \right\Vert _{L^2}^2 + \left\Vert {v_n} \right\Vert _{H^3}\left\Vert {\sigma _n} \right\Vert _{H^2}^2\big ), \end{aligned}$$
(6.11)

and given (4.30) (see also (6.5)–(6.7)), from (6.11) it follows that for all \(u'^{(n)}\in \tau _nB_{V}^a\)

$$\begin{aligned} |S(u'^{(n)})|_V^2\le & {} (1-{\bar{t}})|u'^{(n)}|_{V}^2 + \left\Vert {\varepsilon } \right\Vert _{L^2(^c\Omega '_n)}^2 + c_\Omega a^3\nonumber \\\le & {} (1-{\bar{t}})(1+\tilde{\nu }_n)a^2 + \left\Vert {\varepsilon } \right\Vert _{L^2(^c\Omega '_n)}^2 + c_\Omega a^3.\end{aligned}$$
(6.12)

Let n be large enough and a be small enough so that

$$\begin{aligned} 0<\tilde{\nu }_n\le {\bar{t}}<1,\qquad \left\Vert {\varepsilon } \right\Vert _{L^2(^c\Omega '_n)}^2\le {\bar{t}^2\over 4}a^2,\qquad a\le \displaystyle {{\bar{t}}^2\over 4c_\Omega }.\end{aligned}$$

From (6.12) it follows that

$$\begin{aligned} |S(u'^{(n)})|_{V}^2 \le \Big (1-\displaystyle {\bar{t}^2\over 2}\Big )a^2\quad \forall u'^{(n)}\in \tau _n B_{V}^a, \end{aligned}$$

which means that \(S(u'^{(n)})\in B_{V}^a\) for every \(u'^{(n)}\in \tau _nB_{V}^a\). Hence, we get (6.2) and this completes the proof of the lemma 6.1. \(\square \)

7 The Proof of the Theorem 2.1

After explaining the strategy to follow in order to obtain the proof of our existence result, we were able to establish some crucial non trivial estimates in the previous lemmas 5.15.10. We are now in a position to prove our main result from the estimates in those lemmas.

In fact, recalling Remark 4.5, we obtain from Lemma 6.1 together with Lemma 4.6 that the system of Eqs. (2.7)–(2.11) has a solution \(u=(v,b,\sigma )\). More precisely, the solution \(u = (v,b,\sigma )\) belongs to V and verifies in \(\Omega \) the system of equations

$$\begin{aligned}{} & {} -\mu \Delta v - (\mu +\lambda )\nabla \nabla \cdot v = - \gamma \vartheta \nabla (\widehat{\varrho }^{\gamma -1}\sigma ) - g\sigma e_3 +(\widehat{B}\cdot \nabla )b - \nabla (b\cdot \widehat{B}) + F, \\{} & {} \nabla \cdot (\sigma v) = - \nabla \cdot (\widehat{\varrho }v),\\{} & {} - \nu \Delta b = -\widehat{B}\nabla \cdot v + (\widehat{B}\cdot \nabla )v + v\nabla \cdot \widehat{B} + G, \ \ \nabla \cdot b = 0,\end{aligned}$$

with the boundary conditions (2.10) and (2.11). Now, if we set

$$\begin{aligned} B:= \widehat{B} + b\quad \text{ and } \quad \varrho : = \widehat{\varrho } + \sigma \end{aligned}$$

(see (2.5)), then the triple \((v,B,\varrho )\) solves the system of Eqs. (1.1)–(1.3) with the boundary conditions (1.4)–(1.5) and this completes the proof of Theorem 2.1. \(\square \)

Remark 7.1

It is clear that if \(\delta (x') = 0\) then \((v,B,\varrho ) = (0,B_{eq},\varrho _{eq})\) where (see (1.8))

$$\begin{aligned} B_{eq}(x_3) = {\bar{B}},\quad \varrho _{eq}(x_3)= \Big (C_M - \displaystyle {\gamma -1\over \vartheta \gamma }gx_3\Big )^{1\over \gamma -1} \end{aligned}$$

is the unique stationary solution to the boundary value problem (1.1)–(1.5). However, it is important to establish whether this system can still have a stationary solution \((v,B,\varrho ) = (0,B_{st},\varrho _{st})\) when \(\delta (x') \ne 0\).

Indeed suppose that

$$\begin{aligned} \delta \ \ \text{ is } \text{ not } \text{ identically } \text{ zero } \text{ in } \ \mathbb {R}^2,\end{aligned}$$
(7.1)

and that there exists a stationary solution \((0,B_{st},\varrho _{st})\) to the boundary value problem (1.1)–(1.5). Hence, the pair \((B_{st},\varrho _{st})\) solves in the domain \(\Omega =\mathbb {R}^2\times ]0,h[,\) the system of equations

$$\begin{aligned}{} & {} \vartheta \nabla (\varrho _{st}^{\gamma }) + g\varrho _{st}e_3 = - (\nabla \times B_{st})\times B_{st}, \end{aligned}$$
(7.2)
$$\begin{aligned}{} & {} \nu \nabla \nabla \cdot B_{st} - \nu \Delta B_{st}=0,\quad \nabla \cdot B_{st} = 0, \end{aligned}$$
(7.3)

with the boundary conditions

$$\begin{aligned} B_{st}(x',0) = {\bar{B}},\quad B_{st}(x',h) = {\bar{B}} + \delta (x')e_3,\ \ \forall x'\in \mathbb {R}^2,\ \ B_{st}\rightarrow {\bar{B}} \ \ \text{ as } |x'|\rightarrow \infty \end{aligned}$$
(7.4)

where, we recall here that \(e_3=(0,0,1)\), \({\bar{B}}\in \mathbb {R}^3\) is a given constant field and \(\delta (x')\) is a small perturbation of magnetic field \({\bar{B}}\) on the upper plane \(x_3=h\). It is obvious that

$$\begin{aligned} B_{st}(x',x_3) = {\bar{B}} + \displaystyle {x_3\over h}\delta (x')e_3 + A(x',x_3),\quad A= (A_1,A_2,A_3)\end{aligned}$$
(7.5)

is the unique solution of the boundary value problem (7.3)–(7.4), where A is the unique solution in \(\Omega \) of the system of equations

$$\begin{aligned} \Delta A = de_3,\quad d = -\displaystyle {1\over h}\Delta (x_3\delta (x')),\ \ \nabla \cdot A = -\displaystyle {1\over h}\nabla \cdot (x_3\delta (x')e_3) \end{aligned}$$
(7.6)

with the boundary conditions

$$\begin{aligned} A(x',0) = A(x',h)= 0,\ \ A\rightarrow 0 \ \ |x'|\rightarrow \infty .\end{aligned}$$
(7.7)

As can be easily seen, the only solution A of (7.6)\(_1\)–(7.7) is given by

$$\begin{aligned} A= ae_3\quad \text{ with } \quad a(x',x_3) = \int _{\mathbb {R}^2}\widehat{a}(\xi ,x_3)e^{2i\pi \xi \cdot x'} \ d\xi ,\end{aligned}$$
(7.8)

where \(\widehat{a}\) is such that for every \(\xi \in \mathbb {R}^2\), the function \(a(\xi ,\cdot )\) solves the second order boundary value problem

$$\begin{aligned} \partial _{x_3}^2\widehat{a}(\xi ,x_3) - |\xi |^2\widehat{a}(\xi ,x_3) = \widehat{d}(\xi ,x_3), \quad \widehat{a}(\xi ,0) = \widehat{a}(\xi ,h) = 0\end{aligned}$$
(7.9)

and

$$\begin{aligned} \widehat{d}(\xi ,x_3) = \int _{\mathbb {R}^2}d(x')e^{-2i\pi \xi \cdot x'}dx' \end{aligned}$$

which, considering (7.6)\(_2\), is

$$\begin{aligned} \widehat{d}(\xi ,x_3) = \displaystyle {|\xi |^2x_3\over h}\int _{\mathbb {R}^2}\delta (x')e^{-2i\pi \xi \cdot x'}dx' = \displaystyle {|\xi |^2x_3\over h}\widehat{\delta }(\xi ).\end{aligned}$$

Thus, as we can see now, the only solution of (7.9) is given by

$$\begin{aligned} \widehat{a}(\xi ,x_3) = \widehat{\delta }(\xi )\Big (\displaystyle {\text{ sh }(|\xi |x_3)\over \text{ sh }(|\xi |h)}-{x_3\over h}\Big ).\end{aligned}$$
(7.10)

As for the Eq. (7.6)\(_3\), it is satisfied by A (see (7.10 and (7.8)) if and only if

$$\begin{aligned} \partial _{x_3}a(x',x_3) = \displaystyle {1\over h}\delta (x') \ \ \text{ so } \text{ that } \ \ \partial _{x_3}\widehat{a}(\xi ,x_3) = \displaystyle {1\over h}\widehat{\delta }(\xi ).\end{aligned}$$
(7.11)

By considering the boundary conditions (7.9), it follows from (7.11)\(_2\) that

$$\begin{aligned} \widehat{\delta }(\xi ) = 0 \ \quad \forall \ \xi \in \mathbb {R}^2 \end{aligned}$$

and therefore,

$$\begin{aligned} \delta (x') = \int _{\mathbb {R}^2}\widehat{\delta }(\xi )e^{2i\pi \xi \cdot x'}d\xi =0 \quad \forall \ x'\in \mathbb {R}^2, \end{aligned}$$

which contradicts (7.1). Thus the system of Eqs. (1.1)–(1.5) does not have a stationary solution \((v,B,\varrho )=(0,B_{st},\varrho _{st})\) when the function \(\delta \) is not identically equal to 0.\(\square \)

8 Conclusion and Remarks

In our recent contribution [8], we developed a strategy through which we studied the stationary convective motion in an infinite layer of a viscous compressible and heat-conducive fluid and an existence result for a solution close to the hydrostatic state was obtained in Sobolev spaces. Subsequently, we studied in [7] the stationary motion of a viscous compressible fluid in a magnetic field under the action of a large external force field, and with non-homogeneous and rather large Dirichlet boundary conditions on the magnetic field. Two questions naturally arose: On the one hand, the question about the extension of the study in [7] in the case of the motion in an infinite layer, through the approach developed in [8]. The current work is precisely an answer to that question. On the other hand, we are currently investigating on the stability of the stationary solution with a large potential force field and large enough non-homogeneous Dirichlet boundary conditions on the magnetic field. In the current state, there are no such results in the literature. Let us mention that a stability result was obtained for instance in [30], for the stationary solution of the viscous compressible MHD equations for a barotropic fluid with a large potential force and rather a homogeneous Dirichlet condition on the magnetic field.