On the Properties of Operators of the Stokes Problem with Corner Singularity in Nonsymmetric Variational Formulation

Abstract

The weighted finite element method makes it possible to find an approximate solution of a boundary value problem with corner singularity without loss of accuracy. The construction of this numerical method is based on the introduction of the concept of an $R_{\nu }$-generalized solution for a boundary value problem with a singularity. In this paper, special weighted sets based on the corresponding operators from the definition of the $R_{\nu }$-generalized solution of the Stokes problem in a nonsymmetric variational formulation are introduced. The properties and relationships of these weighted sets are established.

1. Introduction

Mathematical models of natural processes in domains with reentrant corners on the boundary play an important role in fracture mechanics. The presence of reentrant corner on the boundary causes a singularity in the solution of the problem. The solution of the Dirichlet problem for the Poisson equation in the domain $\Omega$ with one corner $\omega \in (\pi ,2\pi )$ with vertex at the origin can be written as

$$u(x_1,x_2)=r^{\frac{\pi }{\omega }}\chi (r,\phi )\Psi \left(r\right)+v(x_1,x_2),v\in W_2^2(\Omega ),$$

where r is a distance from point to $\mathcal{O}=(0,0),$$\chi (r,\phi )$ is a sufficiently smooth function, $(r,\phi )$ are polar coordinates at the point $(0,0)$ and $\Psi \left(r\right)$ is the cutoff function [1].

The generalized solution of this problem belongs to the space $W_2^{1+\alpha -\epsilon }(\Omega )$, where $0.25\le \alpha \le 0.63$ for $\frac{3\pi }{2}\le \omega \le 2\pi ,\epsilon$ is any positive number. The classical finite element method (FEM) or finite difference method lose accuracy in the process of finding an approximate solution This happens due to the presence of a singular component in the solution of a boundary value problem. According to the principle of coordinated estimates, the approximate solution converges to the exact one with a rate $\mathcal{O}\left(h^{\alpha }\right)(0.25\le \alpha \le 0.63)$ (see [2,3]). This leads to a significant increase in computer power and computation time to find a solution with a given accuracy. Traditionally, the acceptable rate of convergence of the approximate solution to the exact solution is $\mathcal{O}\left(h\right)$.

We note several approaches to find an approximate solution of hydrodynamic problems in domains with corner singularity that increase its convergence rate compared to classical approaches. The first one [4] is based on enrichment of the FE spaces by singular components. The second approach [5] use mesh geometrically refined in the neighborhood of a reentrant corner. The third method [6] relies on the definition of dual functions to the singular components and the extension of the variational problem statement using auxiliary equations. The fourth approach [7] is based on the approximation of stress coefficients for the singular components of the solution, knowing that we find the regular components of the solution. In the fifth method (see [8]), the original non-convex polygonal domain is divided into subdomains with simple geometry (without reentrant corners) and the Strang-Fix algorithm is implemented So that the dimension of the discrete space increases in the vicinity of the singularity point. The sixth one relies on the selection of several neighborhoods for the singularity point and introduction of auxiliary bilinear forms, the so-called energy-corrected FEM (see [9]).

We proposed to define the solution of a boundary value problem with a singularity as an $R_{\nu }$-generalized one [10]. The introduction of a weight function into the definition of a weak solution makes it possible to reduce the influence of the singularity on the accuracy in finding the approximate solution. We have proposed a weighted FEM, which allow us to find an approximate solution of boundary value problems with a strong and corner singularity, without loss of accuracy with a rate $\mathcal{O}\left(h\right)$. The method was constructed and investigated for the third boundary value problem for a second-order elliptic Equation [11], for the Maxwell’s Equations [12,13], for the Lamé system [14,15] and for an elasticity problem with a crack [16,17]. In [18,19,20], weighted FEM was developed for the Stokes and Oseen problems.

To create and research the convergence of numerical methods for problems with a corner singularity, it is necessary to study the existence, uniqueness and regularity of the solution. The early work on the study of the regularity theory in non-convex domains for hydrodynamics problems include a paper [21], which was developed in [22,23,24,25]. In particular, in [22,23] authors used and generalized the results for the Stokes and Navier–Stokes problems in the Sobolev spaces $W_{\beta }^k(\Omega )$ with Kondrat’ev-type weights, which were proposed in [1] for elliptic and parabolic problems. In recent years, it should be noted papers [26,27,28] which are devoted to study the differential properties of hydrodynamic problems solutions.

The study of the existence and uniqueness of the $R_{\nu }$-generalized solution of boundary value problems for the second-order elliptic equations and the Lamé system was held in [29,30,31]. To investigate the $R_{\nu }$-generalized solution of hydrodynamic problems with a corner singularity, it is required to examine the properties of their operators in a nonsymmetric variational formulation. This paper is devoted to the study of special weighted sets related to the operators of the Stokes problem.

The structure of the paper is as follows. In Section 2, we state a Stokes problem and introduce the necessary notation. Define the $R_{\nu }$-generalized solution in a nonsymmetric variation formulation in weighted sets. Section 3 is devoted to the study of properties of sets related to the operators of the Stokes problem. Finally, some concluding remarks are given in Section 4.

2. Problem Statement

The Stokes problem is to find the velocity field $\mathbf{w}=(w_1,w_2)$ and pressure p which satisfy the system of differential equations and boundary conditions

$$-▵\mathbf{w}+\nabla p=\mathbf{f},\mathrm{in}\Omega ,$$

(1)

$$div\mathbf{w}=0,\mathrm{in}\Omega ,$$

(2)

$$\mathbf{w}=\mathbf{0},\mathrm{on}\mathsf{\partial }\Omega .$$

(3)

Let $\Omega$ be a non-convex polygonal domain with one reentrant corner $\omega \in (\pi ,2\pi )$ with vertex at the origin $\mathcal{O}=(0,0).$ Let us briefly describe the behavior of the solution $(\mathbf{w},p)$ of Equations (1)–(3) in a neighborhood of the reentrant corner (for more details see, for example, [1,21]). The components of solution $(\mathbf{w},p)$ in polar coordinates $(r,\phi )$ are linear combinations of singular components and regular remainders. The singular ones of the function $w_i$ and p have an asymptotic $r^{{\lambda }_i}$ and $r^{{\lambda }_i-1}$, respectively, where ${\lambda }_i$ is an eigenvalue of the Stokes operator, satisfying, in the case homogeneous Dirichlet boundary conditions, the following equation:

$${\lambda }_i^2{\sin }^2\omega -{\sin }^2\left({\lambda }_i\omega \right)=0,{\lambda }_i\in \mathbb{C},{\lambda }_i\ne 0.$$

In particular, if $\omega$ is equal to $\frac{3\pi }{2},$ then the smallest positive eigenvalue, characterizing the behavior of the solution in the neighborhood of the reentrant corner is approximately equal to $0.544483.$

In order to determine the $R_{\nu }$-generalized solution of the Equtions (1)–(3) we introduce the spaces and sets of functions. Denote by ${\Omega }_{\delta }$ the intersection of the disk of radius $\delta ,\delta >0,$ centered at the origin $\mathcal{O}=(0,0)$ with a closure of $\Omega .$ Define the function $\rho \left(\mathbf{x}\right)$ in $\overline{\Omega },$ which we will call a weight function satisfying the conditions: $\rho \left(\mathbf{x}\right)=\sqrt{x_1^2+x_2^2},$ if $x\in {\Omega }_{\delta }$ and $\rho \left(\mathbf{x}\right)=\delta$ otherwise. Let $D^{l}z\left(\mathbf{x}\right)=\frac{{\partial }^{\left|l\right|}z\left(\mathbf{x}\right)}{\partial x_1^{l_1}\partial x_2^{l_2}}$, $l=(l_1,l_2),$$\left|l\right|=l_1+l_2,l_i$ are non-negative integers, $i\in \{1,2\}$, $d\mathbf{x}=d{x}_{1}d{x}_2$.

Denote by $L_{2,\beta }(\Omega ),W_{2,\beta }^k(\Omega )$ weighted spaces of functions $z\left(\mathbf{x}\right)$ with bounded norms

$${\parallel z\parallel }_{L_{2,\beta }(\Omega )}=(\underset{\Omega }{\int }{\rho }^{2\beta }\left(\mathbf{x}\right)z^2\left(\mathbf{x}\right)d\mathbf{x}{)}^{1/2},$$

(4)

$${\parallel z\parallel }_{W_{2,\beta }^k(\Omega )}=({\parallel z\parallel }_{L_{2,\beta }(\Omega )}^2+\sum _{1\le \left|l\right|\le k}\underset{\Omega }{\int }{\rho }^{2\beta }\left(\mathbf{x}\right){\left|D^{l}z\left(\mathbf{x}\right)\right|}^{2}d\mathbf{x}{)}^{1/2}$$

(5)

respectively. Let ${\left|z\right|}_{W_{2,\beta }^k(\Omega )}={\left({\displaystyle \sum _{\left|l\right|=k}}\underset{\Omega }{\int }{\rho }^{2\beta }\left(\mathbf{x}\right){\left|D^{l}z\left(\mathbf{x}\right)\right|}^{2}d\mathbf{x}\right)}^{1/2}$ be the seminorm of the second space. Denote by $W_{2,\beta }^{k,0}(\Omega )$ a closure relative to the norm (5) of the set of infinitely differentiable compactly supported functions in $\Omega$.

We define the following conditions for the functions $z\left(\mathbf{x}\right):$

$$0<C_1\le {\parallel z\parallel }_{L_{2,\beta }(\Omega \setminus {\Omega }_{\delta })},$$

(6)

$$\left|z\left(\mathbf{x}\right)\right|\le C_2{\delta }^{\beta -\tau }{\rho }^{\tau -\beta }\left(\mathbf{x}\right),\mathbf{x}\in {\Omega }_{\delta },$$

(7)

$$|D^{1}z\left(\mathbf{x}\right)|\le C_2{\delta }^{\beta -\tau }{\rho }^{\tau -\beta -1}\left(\mathbf{x}\right),\mathbf{x}\in {\Omega }_{\delta },$$

(8)

where $C_2$ is a positive constant, $\tau$ is a small positive parameter that does not depend on $\delta ,\beta ,l$ and $z\left(\mathbf{x}\right).$

Denote by $L_{2,\beta }(\Omega ,\delta )$ the set of functions $z\left(\mathbf{x}\right)$ from the space $L_{2,\beta }(\Omega ),$ satisfying (6) and (7) with bounded norm (4). Define the subset $L_{2,\beta }^0(\Omega ,\delta )=\{z\left(\mathbf{x}\right)\in L_{2,\beta }(\Omega ,\delta ):\parallel {\rho }^{\beta }z{\parallel }_{L_1(\Omega )}=0\}$ with bounded norm (4). Let $W_{2,\beta }^1(\Omega ,\delta )$ and $W_{2,\beta }^{1,0}(\Omega ,\delta )$ are sets of functions $z\left(\mathbf{x}\right)$ from the spaces $W_{2,\beta }^1(\Omega )$ and $W_{2,\beta }^{1,0}(\Omega )$, respectively, satisfying (6)–(8) with bounded norm (5). We will assume that a linear combination of functions from $L_{2,\beta }(\Omega ,\delta )$ ($W_{2,\beta }^{1,0}(\Omega ,\delta )$) also belongs to $L_{2,\beta }(\Omega ,\delta )$ ($W_{2,\beta }^{1,0}(\Omega ,\delta )$).

We will highlight space (set) of vector functions in bold style, that is, ${\mathbf{L}}_{2,\beta }(\Omega )=\{\mathbf{z}\left(\mathbf{x}\right)=(z_1\left(\mathbf{x}\right),z_2\left(\mathbf{x}\right)):z_i\left(\mathbf{x}\right)\in L_{2,\beta }(\Omega )\}$ (${\mathbf{L}}_{2,\beta }(\Omega ,\delta )=\{\mathbf{z}\left(\mathbf{x}\right)=(z_1\left(\mathbf{x}\right),z_2\left(\mathbf{x}\right)):z_i\left(\mathbf{x}\right)\in L_{2,\beta }(\Omega ,\delta )\}$) with bounded vector norm ${\parallel \mathbf{z}\parallel }_{{\mathbf{L}}_{2,\beta }(\Omega )}={\left(\parallel z_1{\parallel }_{L_{2,\beta }(\Omega )}^2+{\parallel z_2\parallel }_{L_{2,\beta }(\Omega )}^2\right)}^{1/2}.$ Analogically spaces (sets) of vector functions ${\mathbf{W}}_{2,\beta }^1(\Omega )$ and ${\mathbf{W}}_{2,\beta }^{1,0}(\Omega )$ (${\mathbf{W}}_{2,\beta }^1(\Omega ,\delta )$ and ${\mathbf{W}}_{2,\beta }^{1,0}(\Omega ,\delta )$) with a vector norm (5).

Define bilinear and linear forms

$$a(\mathbf{u},\mathbf{v})=\underset{\Omega }{\int }\left[\nabla \mathbf{u}:\nabla \left({\rho }^{2\nu }\mathbf{v}\right)\right]d\mathbf{x},$$

(9)

$$b_1(\mathbf{v},s)=-\underset{\Omega }{\int }s\mathrm{div}\left({\rho }^{2\nu }\mathbf{v}\right)d\mathbf{x},b_2(\mathbf{u},q)=-\underset{\Omega }{\int }\left({\rho }^{2\nu }q\right)\mathrm{div}\mathbf{u}d\mathbf{x},$$

$$l\left(\mathbf{v}\right)=\underset{\Omega }{\int }\mathbf{f}·\left({\rho }^{2\nu }\mathbf{v}\right)d\mathbf{x}.$$

Definition 1.

The pair$({\mathbf{w}}_{\nu },p_{\nu })\in {\mathbf{W}}_{2,\nu }^{1,0}(\Omega ,\delta )\times L_{2,\nu }^0(\Omega ,\delta )$is called an$R_{\nu }$-generalized solution of the problem (1)–(3),${\mathbf{w}}_{\nu }$satisfies a condition (3) almost everywhere on$\partial \Omega ,$such that for all pairs$(\mathbf{z},g)\in {\mathbf{W}}_{2,\nu }^{1,0}(\Omega ,\delta )\times L_{2,\nu }^0(\Omega ,\delta )$integral identities

$$a({\mathbf{w}}_{\nu },\mathbf{z})+b_1(\mathbf{z},p_{\nu })=l\left(\mathbf{z}\right),b_2({\mathbf{w}}_{\nu },g)=0$$

hold, where$\mathbf{f}\in {\mathbf{L}}_{2,\gamma }(\Omega ,\delta ),\gamma \le \nu .$

Let us introduce the notation

$$K_1=\{\mathbf{v}\in {\mathbf{W}}_{2,\nu }^{1,0}(\Omega ,\delta ):b_1(\mathbf{v},s)=-\underset{\Omega }{\int }s\mathrm{div}\left({\rho }^{2\nu }\mathbf{v}\right)d\mathbf{x}=0\forall s\in L_{2,\nu }^0(\Omega ,\delta )\}.$$

$$K_2=\{\mathbf{u}\in {\mathbf{W}}_{2,\nu }^{1,0}(\Omega ,\delta ):b_2(\mathbf{u},q)=-\underset{\Omega }{\int }\left({\rho }^{2\nu }q\right)\mathrm{div}\mathbf{u}d\mathbf{x}=0\forall q\in L_{2,\nu }^0(\Omega ,\delta )\}.$$

3. Properties of Functions from Sets ${\mathit{K}}_{\mathbf{1}}$ and ${\mathit{K}}_{\mathbf{2}}$

3.1. Construction of the Function $\mathbf{v}\in K_1$ Using the Function $\mathbf{u}\in K_2$ and Their Relationship

Let $\nu >0.$ For any function $\mathbf{u}\in K_2$ there exists a function $\psi$ (see [32]), such that $\mathbf{u}=\left(\frac{\partial \psi }{\partial x_2},-\frac{\partial \psi }{\partial x_1}\right).$ The function $\psi$ belongs to the space $W_{2,\nu }^{2,0}(\Omega )$ and satisfies the following conditions:

$$0<C_1\le {\parallel \psi \parallel }_{L_{2,\nu }(\Omega \setminus {\Omega }_{\delta })},$$

(10)

$$\left|\psi \left(\mathbf{x}\right)\right|\le C_2{\delta }^{\nu -\tau }{\rho }^{\tau -\nu +1}\left(\mathbf{x}\right),\mathbf{x}\in {\Omega }_{\delta },$$

(11)

$$|D^k\psi \left(\mathbf{x}\right)|\le C_2{\delta }^{\nu -\tau }{\rho }^{\tau -\nu -k+1}\left(\mathbf{x}\right),\mathbf{x}\in {\Omega }_{\delta },k=1,2.$$

(12)

We define a function $\mathbf{v}=\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_2},-{\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_1}\right)$ and prove the following lemma.

Lemma 1.

Components of the function$\mathbf{v}=\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_2},-{\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_1}\right)$satisfy a condition (6) and conditions (7) and (8) up to constants.

Proof. 

Without loss of generality, consider $v_1.$

Firstly, note that $v_1=u_1$ in $\Omega \setminus {\Omega }_{\delta }$, hence the condition (6) is satisfied.

Secondly,

$$v_1={\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_2}=\frac{\partial \psi }{\partial x_2}+{\rho }^{-2\nu }\left(\frac{\partial {\rho }^{2\nu }}{\partial x_2}\right)\psi .$$

For an arbitrary $\beta$, we have

$$\frac{\partial {\rho }^{\beta }}{\partial x_i}=\left\{\begin{array}{c}\beta {\rho }^{\beta -2}{x}_i,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.$$

(13)

Apply Equation (13), with $\beta =2\nu ,$ then

$${\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_2}=\left\{\begin{array}{c}2\nu {\rho }^{-2}{x}_2,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.$$

using the inequalities $x_2\le \rho$, (10) and (11) for $k=1$, we have

$$|v_1|\le C_2{\delta }^{\nu -\tau }{\rho }^{\tau -\nu }+2\nu C_2{\rho }^{-2}{\rho }^1{\delta }^{\nu -\tau }{\rho }^{\tau -\nu +1}=(1+2\nu )C_2{\delta }^{\nu -\tau }{\rho }^{\tau -\nu }.$$

(14)

The condition (7), up to a constant, is satisfied.

Thirdly,

$$\frac{\partial v_1}{\partial x_1}=\frac{{\partial }^2\psi }{\partial x_1\partial x_2}+\frac{\partial }{\partial x_1}\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_2}\right)\psi +\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_2}\right)\left(\frac{\partial \psi }{\partial x_1}\right).$$

Using (13), we conclude

$$\frac{\partial }{\partial x_1}\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_2}\right)=\left\{\begin{array}{c}-4\nu {\rho }^{-4}{x}_1{x}_2,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.$$

then, applying the inequalities $2x_1{x}_2\le x_1^2+x_2^2={\rho }^2$, (11) and (12), we derive

$$\left|\frac{\partial v_1}{\partial x_1}\right|\le C_2{\delta }^{\nu -\tau }{\rho }^{\tau -\nu -1}+4\nu C_2{\delta }^{\nu -\tau }{\rho }^2{\rho }^{-4}{\rho }^{\tau -\nu +1}+$$

$$+2\nu C_2{\delta }^{\nu -\tau }{\rho }^{-2}{\rho }^1{\rho }^{\tau -\nu }=(1+6\nu )C_2{\delta }^{\nu -\tau }{\rho }^{\tau -\nu -1}.$$

Further

$$\frac{\partial v_1}{\partial x_2}=\frac{{\partial }^2\psi }{\partial x_2^2}+\frac{\partial }{\partial x_2}\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_2}\right)\psi +\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_2}\right)\left(\frac{\partial \psi }{\partial x_2}\right),$$

$$\frac{\partial }{\partial x_2}\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_2}\right)=\left\{\begin{array}{c}2\nu {\rho }^{-2}-4\nu {\rho }^{-4}{x}_2^2,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.$$

then, applying the inequalities $x_2\le \rho$, (11) and (12), we get

$$\left|\frac{\partial v_1}{\partial x_2}\right|\le C_2{\delta }^{\nu -\tau }{\rho }^{\tau -\nu -1}+2\nu C_2{\delta }^{\nu -\tau }{\rho }^{-2}{\rho }^{\tau -\nu +1}+4\nu C_2{\delta }^{\nu -\tau }{\rho }^{-4}{\rho }^2{\rho }^{\tau -\nu +1}+$$

$$+2\nu C_2{\delta }^{\nu -\tau }{\rho }^{-2}{\rho }^1{\rho }^{\tau -\nu }=(1+8\nu )C_2{\delta }^{\nu -\tau }{\rho }^{\tau -\nu -1}.$$

The condition (8), up to a constant, is satisfied.

Lemma 1 is proved. □

Remark 1.

It will be proved in Theorem 1 that the function$\mathbf{v}\in {\mathbf{W}}_{2,\nu }^{1,0}(\Omega ).$In view of this fact, Lemma 1 and an equality$-\underset{\Omega }{\int }sdiv\left({\rho }^{2\nu }\mathbf{v}\right)d\mathbf{x}=0$$\forall s\in L_{2,\nu }^0(\Omega ,\delta ),$we conclude that$\mathbf{v}\in K_1.$

3.1.1. Auxiliary Statements

We need the following auxiliary statements.

Lemma 2.

(Friedrichs’s inequality). For any$z\in W_{2,0}^{1,0}(\Omega ,\delta )$the inequality

$${\parallel z\parallel }_{L_{2,0}(\Omega )}\le C_3{\parallel \nabla z\parallel }_{L_{2,0}(\Omega )}$$

(15)

holds, where$C_3$is a positive constant that does not depend on z.

Let us prove the statement connecting the integrals in ${\Omega }_{\delta }$ and $\Omega$.

Lemma 3.

For any$z\in L_{2,\nu }(\Omega ),$satisfying the conditions (6) and (7), the inequality

$$\underset{{\Omega }_{\delta }}{\int }{\rho }^{2(\nu -1)}{z}^{2}d\mathbf{x}\le C_4^2{\delta }^{2\nu }{\parallel z\parallel }_{L_{2,\nu }(\Omega )}^2$$

(16)

holds, where$C_4$is a positive constant equal to$\frac{C_2}{C_1}\sqrt{\frac{{\phi }_1-{\phi }_0}{2\tau }}$,${\phi }_1-{\phi }_0$is the value of of the corner ω change in polar coordinates.

Proof. 

Taking into account the condition (7), we conclude

$$\underset{{\Omega }_{\delta }}{\int }{\rho }^{2(\nu -1)}{z}^{2}d\mathbf{x}\le C_2^2{\delta }^{2\nu -2\tau }\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -2}{\rho }^{-2\nu +2\tau }d\mathbf{x}=C_2^2{\delta }^{2\nu -2\tau }\underset{{\phi }_0}{\overset{{\phi }_1}{\int }}\left(\underset{0}{\overset{\delta }{\int }}{\rho }^{2\tau -2}\rho d\rho \right)d\phi ,$$

$$\underset{{\Omega }_{\delta }}{\int }{\rho }^{2(\nu -1)}{z}^{2}d\mathbf{x}\le \frac{({\phi }_1-{\phi }_0)C_2^2{\delta }^{2\nu }}{2\tau }.$$

(17)

Using the condition (6), we have

$$C_1^2\le {\parallel z\parallel }_{L_{2,\nu }(\Omega \setminus {\Omega }_{\delta })}^2\le {\parallel z\parallel }_{L_{2,\nu }(\Omega )}^2.$$

(18)

We combine the inequalities (17) and (18), we get the estimate (16) with a constant $C_4=\frac{C_2}{C_1}\sqrt{\frac{{\phi }_1-{\phi }_0}{2\tau }}.$

Lemma 3 is proved. □

Lemma 3 implies the following corollaries.

Corollary 1.

For any$z\in L_{2,\nu }(\Omega )$satisfying the conditions (6) and (14), the inequality

$$\underset{{\Omega }_{\delta }}{\int }{\rho }^{2(\nu -1)}{z}^{2}d\mathbf{x}\le C_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel z\parallel }_{L_{2,\nu }(\Omega )}^2$$

(19)

holds, where$C_{\nu }=(1+2\nu ).$

Corollary 2.

For any$z\in L_{2,\nu }(\Omega )$satisfying the conditions (10) and (11), the inequality

$$\underset{{\Omega }_{\delta }}{\int }{\rho }^{2(\nu -2)}{z}^{2}d\mathbf{x}\le C_4^2{\delta }^{2\nu }{\parallel z\parallel }_{L_{2,\nu }(\Omega )}^2$$

(20)

holds.

Let us prove the connection between functions from the sets $W_{2,\nu }^{1,0}(\Omega ,\delta )$ and $W_{2,0}^{1,0}(\Omega ,\delta ).$

Lemma 4.

Function$z\in W_{2,\nu }^{1,0}(\Omega ,\delta )$if and only if${\rho }^{\nu }z\in W_{2,0}^{1,0}(\Omega ,\delta )$and the inequalities

$$\parallel \nabla \left({\rho }^{\nu }z\right){\parallel }_{L_{2,0}(\Omega )}^2\le {2\parallel \nabla z\parallel }_{L_{2,\nu }(\Omega )}^2+2{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel z\parallel }_{L_{2,\nu }(\Omega )}^2,$$

(21)

$${\left|z\right|}_{W_{2,\nu }^1(\Omega )}^2\le 2\parallel \nabla \left({\rho }^{\nu }z\right){\parallel }_{L_{2,0}(\Omega )}^2+2{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel {\rho }^{\nu }z\parallel }_{L_{2,0}(\Omega )}^2$$

(22)

hold.

Proof. 

1. Let $z\in W_{2,\nu }^{1,0}(\Omega ,\delta )$, then the function ${\rho }^{\nu }z\in L_{2,0}(\Omega ,\delta )$ and vanishes on $\partial \Omega$. Further

$$\parallel \nabla \left({\rho }^{\nu }z\right){\parallel }_{L_{2,0}(\Omega )}^2=\underset{\Omega }{\int }{\left(\frac{\partial \left({\rho }^{\nu }z\right)}{\partial x_1}\right)}^2+{\left(\frac{\partial \left({\rho }^{\nu }z\right)}{\partial x_2}\right)}^{2}d\mathbf{x}\le 2\underset{\Omega }{\int }{\rho }^{2\nu }\left[{\left(\frac{\partial z}{\partial x_1}\right)}^2+{\left(\frac{\partial z}{\partial x_2}\right)}^2\right]d\mathbf{x}+$$

$$+2\underset{\Omega }{\int }{z}^2\left[{\left(\frac{\partial {\rho }^{\nu }}{\partial x_1}\right)}^2+{\left(\frac{\partial {\rho }^{\nu }}{\partial x_2}\right)}^2\right]d\mathbf{x}.$$

Hence,

$$\parallel \nabla \left({\rho }^{\nu }z\right){\parallel }_{L_{2,0}(\Omega )}^2\le 2{\parallel \nabla z\parallel }_{L_{2,\nu }(\Omega )}^2+2\underset{\Omega }{\int }{z}^2\left[{\left(\frac{\partial {\rho }^{\nu }}{\partial x_1}\right)}^2+{\left(\frac{\partial {\rho }^{\nu }}{\partial x_2}\right)}^2\right]d\mathbf{x}.$$

(23)

Let us estimate the second term on the right-hand side (23). Using (13), we have

$${\left(\frac{\partial {\rho }^{\beta }}{\partial x_1}\right)}^2+{\left(\frac{\partial {\rho }^{\beta }}{\partial x_2}\right)}^2=\left\{\begin{array}{c}{\beta }^2{\rho }^{2\beta -2},\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }.\hfill \end{array}\right.$$

(24)

Let $\beta =\nu$ in (24), hence by Lemma 3, Equation (16):

$$\underset{\Omega }{\int }{z}^2\left[{\left(\frac{\partial {\rho }^{\nu }}{\partial x_1}\right)}^2+{\left(\frac{\partial {\rho }^{\nu }}{\partial x_2}\right)}^2\right]d\mathbf{x}={\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2(\nu -1)}{z}^{2}d\mathbf{x}\le {\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel z\parallel }_{L_{2,\nu }(\Omega )}^2.$$

(25)

Substituting (25) into (23), we obtain the estimate (21). Since $\parallel {\rho }^{\nu }{z\parallel }_{L_{2,0}(\Omega )}^2={\parallel z\parallel }_{L_{2,\nu }(\Omega )}^2,$ then combined with (21), we have a sequence of inequalities

$$\parallel {\rho }^{\nu }{z\parallel }_{W_{2,0}^1(\Omega )}^2=\parallel {\rho }^{\nu }{z\parallel }_{L_{2,0}(\Omega )}^2+{\parallel \nabla \left({\rho }^{\nu }z\right)\parallel }_{L_{2,0}(\Omega )}^2\le$$

$$\le {2\parallel \nabla z\parallel }_{L_{2,\nu }(\Omega )}^2+(1+2{\nu }^2{C}_4^2{\delta }^{2\nu }){\parallel z\parallel }_{L_{2,\nu }(\Omega )}^2\le max\{2,1+2{\nu }^2{C}_4^2{\delta }^{2\nu }\}{\parallel z\parallel }_{W_{2,\nu }^1(\Omega )}^2.$$

Thus, ${\rho }^{\nu }z\in W_{2,0}^{1,0}(\Omega ,\delta ).$

2. Let ${\rho }^{\nu }z\in W_{2,0}^{1,0}(\Omega ,\delta ),$ then the function $z\in L_{2,\nu }(\Omega ,\delta )$ and vanishes on $\partial \Omega$. Let us show that $z\in W_{2,\nu }^{1,0}(\Omega )$ under conditions (6)–(8), i.e., $z\in W_{2,\nu }^{1,0}(\Omega ,\delta ).$ We estimate the quantity ${\left|z\right|}_{W_{2,\nu }^1(\Omega )}^2$. We have

$$\frac{\partial \left({\rho }^{\nu }z\right)}{\partial x_i}={\rho }^{\nu }\frac{\partial z}{\partial x_i}+z\frac{\partial {\rho }^{\nu }}{\partial x_i}$$

and

$${\rho }^{2\nu }{\left(\frac{\partial z}{\partial x_i}\right)}^2={\left(\frac{\partial \left({\rho }^{\nu }z\right)}{\partial x_i}-z\frac{\partial {\rho }^{\nu }}{\partial x_i}\right)}^2\le 2{\left(\frac{\partial \left({\rho }^{\nu }z\right)}{\partial x_i}\right)}^2+2z^2{\left(\frac{\partial {\rho }^{\nu }}{\partial x_i}\right)}^2,$$

hence

$${\left|z\right|}_{W_{2,\nu }^1(\Omega )}^2=\underset{\Omega }{\int }{\rho }^{2\nu }\left[{\left(\frac{\partial z}{\partial x_1}\right)}^2+{\left(\frac{\partial z}{\partial x_2}\right)}^2\right]d\mathbf{x}\le$$

$$\le 2\underset{\Omega }{\int }\left[{\left(\frac{\partial \left({\rho }^{\nu }z\right)}{\partial x_1}\right)}^2+{\left(\frac{\partial \left({\rho }^{\nu }z\right)}{\partial x_2}\right)}^2\right]d\mathbf{x}+2\underset{\Omega }{\int }{z}^2\left[{\left(\frac{\partial {\rho }^{\nu }}{\partial x_1}\right)}^2+{\left(\frac{\partial {\rho }^{\nu }}{\partial x_2}\right)}^2\right]d\mathbf{x}.$$

Using the same reasoning as in the output (25), we conclude

$${\left|z\right|}_{W_{2,\nu }^1(\Omega )}^2\le 2\parallel \nabla \left({\rho }^{\nu }z\right){\parallel }_{L_{2,0}(\Omega )}^2+2{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel z\parallel }_{L_{2,\nu }(\Omega )}^2.$$

(26)

Since ${\parallel z\parallel }_{L_{2,\nu }(\Omega )}^2={\parallel {\rho }^{\nu }z\parallel }_{L_{2,0}(\Omega )}^2,$ applying (26), we get estimates (22) and

$${\parallel z\parallel }_{W_{2,\nu }^1(\Omega )}^2={\parallel z\parallel }_{L_{2,\nu }(\Omega )}^2+{\left|z\right|}_{W_{2,\nu }^1(\Omega )}^2\le 2\parallel \nabla \left({\rho }^{\nu }z\right){\parallel }_{L_{2,0}(\Omega )}^2+(1+2{\nu }^2{C}_4^2{\delta }^{2\nu }){\parallel {\rho }^{\nu }z\parallel }_{L_{2,0}(\Omega )}^2\le$$

$$\le max\{2,1+2{\nu }^2{C}_4^2{\delta }^{2\nu }\}{\parallel {\rho }^{\nu }z\parallel }_{W_{2,0}^1(\Omega )}^2.$$

Combined with the conditions (6)–(8) we conclude that $z\in W_{2,\nu }^{1,0}(\Omega ,\delta ).$

Lemma 4 is proved. □

Let us prove an analogue of the Friedrichs’s inequality in the set $W_{2,\nu }^{1,0}(\Omega ,\delta ).$

Lemma 5.

Let$\nu >0,$then there exists${\delta }_0={\delta }_0\left(\nu \right)>0,$such that for any$\delta <{\delta }_0$and any function$z\in W_{2,\nu }^{1,0}(\Omega ,\delta )$the inequality

$${\parallel z\parallel }_{L_{2,\nu }(\Omega )}\le C_5{\parallel \nabla z\parallel }_{L_{2,\nu }(\Omega )}$$

(27)

holds, where constant$C_5=2C_3.$

Proof. 

Consider an arbitrary function $z\in W_{2,\nu }^{1,0}(\Omega ,\delta ).$ By Lemma 4 ${\rho }^{\nu }z\in W_{2,0}^{1,0}(\Omega ,\delta ),$ then using the estimates (15) and (21), we conclude

$${\parallel z\parallel }_{L_{2,\nu }(\Omega )}^2=\parallel {\rho }^{\nu }{z\parallel }_{L_{2,0}(\Omega )}^2\le C_3^2\parallel \nabla \left({\rho }^{\nu }z\right){\parallel }_{L_{2,0}(\Omega )}^2\le 2C_3^2{\parallel \nabla z\parallel }_{L_{2,\nu }(\Omega )}^2+2{\nu }^2{C}_3^2{C}_4^2{\delta }^{2\nu }{\parallel z\parallel }_{L_{2,\nu }(\Omega )}^2,$$

that is

$$(1-2{\nu }^2{C}_3^2{C}_4^2{\delta }^{2\nu }){\parallel z\parallel }_{L_{2,\nu }(\Omega )}^2\le 2C_3^2{|\nabla z\parallel }_{L_{2,\nu }(\Omega )}^2.$$

For $\nu >0$ there exists ${\delta }_0={\delta }_0\left(\nu \right)>0$ such that ${\nu }^2{C}_3^2{C}_4^2{\delta }_0^{2\nu }=\frac{1}{4}$ and for any $\delta \in (0,{\delta }_0)$ sequence of inequalities

$$\frac{1}{2}{\parallel z\parallel }_{L_{2,\nu }(\Omega )}^2\le (1-2{\nu }^2{C}_3^2{C}_4^2{\delta }^{2\nu }){\parallel z\parallel }_{L_{2,\nu }(\Omega )}^2\le 2C_3^2{\parallel \nabla z\parallel }_{L_{2,\nu }(\Omega )}^2$$

holds. Therefore, the inequality (27) is true.

Lemma 5 is proved. □

3.1.2. Belonging of the Function $\mathbf{v}$ to the Set $K_1$

In Lemma 6 we estimate the norm of the function $\mathbf{v}$ by the norm of the function $\mathbf{u}$ of the space ${\mathbf{L}}_{2,\nu }(\Omega ),$ and in Lemma 7 we estimate their seminorms of the space ${\mathbf{W}}_{2,\nu }^1(\Omega ).$

Lemma 6.

Let$\nu >0,$then there exists${\delta }_0={\delta }_0\left(\nu \right)>0,$such that for any$\delta \in (0,{\delta }_0),$any function$\mathbf{u}=\left(\frac{\partial \psi }{\partial x_2},-\frac{\partial \psi }{\partial x_1}\right)\in K_2$and for$\mathbf{v},$represented as$\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_2},-{\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_1}\right),$an estimate

$${\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}\le C_6{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}$$

(28)

holds, where$C_6=\sqrt{2+4{\nu }^2{C}_4^2{C}_5^2{\delta }^{2\nu +2}{(1+\tau )}^{-1}}.$

Proof. 

By definition we have

$${\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2=\underset{\Omega }{\int }{\rho }^{2\nu }{\rho }^{-4\nu }\left[{\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_1}\right)}^2+{\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_2}\right)}^2\right]d\mathbf{x},$$

$${\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2=\underset{\Omega }{\int }{\rho }^{2\nu }\left[{\left(\frac{\partial \psi }{\partial x_1}\right)}^2+{\left(\frac{\partial \psi }{\partial x_2}\right)}^2\right]d\mathbf{x},$$

then

$${\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2=\underset{\Omega }{\int }{\rho }^{-2\nu }\left[\sum _{i=1}^2{\left({\rho }^{2\nu }\frac{\partial \psi }{\partial x_i}+\psi \frac{\partial {\rho }^{2\nu }}{\partial x_i}\right)}^2\right]d\mathbf{x}\le$$

$$\le 2\underset{\Omega }{\int }{\rho }^{2\nu }\left[\sum _{i=1}^2{\left(\frac{\partial \psi }{\partial x_i}\right)}^2\right]d\mathbf{x}+2\underset{\Omega }{\int }{\rho }^{-2\nu }{\psi }^2\left[\sum _{i=1}^2{\left(\frac{\partial {\rho }^{2\nu }}{\partial x_i}\right)}^2\right]d\mathbf{x},$$

that is

$${\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\le 2{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+2\underset{\Omega }{\int }{\rho }^{-2\nu }{\psi }^2\left[\sum _{i=1}^2{\left(\frac{\partial {\rho }^{2\nu }}{\partial x_i}\right)}^2\right]d\mathbf{x}.$$

(29)

Estimate the second term on the right-hand side (29), using (13) and (24) for $\beta =2\nu$. Applying (10) and (11) and the arguments of Lemma 3, we get

$$\underset{{\Omega }_{\delta }}{\int }{\rho }^{-2\nu }{\psi }^2{\left(2\nu \right)}^2{\rho }^{4\nu -2}d\mathbf{x}\le 2{\nu }^2{C}_4^2{(1+\tau )}^{-1}{\delta }^{2\nu +2}{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2.$$

(30)

Substituting (30) into (29), we conclude

$${\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\le {2\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+4{\nu }^2{C}_4^2{(1+\tau )}^{-1}{\delta }^{2\nu +2}{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2.$$

(31)

Let us apply Lemma 5 to the function $\psi$ and use the fact that ${\parallel \nabla \psi \parallel }_{L_{2,\nu }(\Omega )}^2={\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2$, then (31) will take the form

$${\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\le {2\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+4{\nu }^2{C}_4^2{C}_5^2{(1+\tau )}^{-1}{\delta }^{2\nu +2}{\parallel \nabla \psi \parallel }_{L_{2,\nu }(\Omega )}^2\le$$

$$\le (2+4{\nu }^2{C}_4^2{C}_5^2{(1+\tau )}^{-1}{\delta }^{2\nu +2}){\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

Lemma 6 is proved. □

Lemma 7.

Let$\nu >0,$then there exists${\delta }_0={\delta }_0\left(\nu \right)>0,$such that for any$\delta \in (0,{\delta }_0),$any function$\mathbf{u}=\left(\frac{\partial \psi }{\partial x_2},-\frac{\partial \psi }{\partial x_1}\right)\in K_2$and for$\mathbf{v},$represented as$\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_2},-{\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_1}\right),$an estimate

$${\left|\mathbf{v}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}\le C_7{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}$$

(32)

holds, where$C_7=\sqrt{3+12{\nu }^2{C}_4^2{C}_5^2{\delta }^{2\nu }(12C_5^2+1)}.$

Proof. 

By definition we have

$${\left|\mathbf{v}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2=\underset{\Omega }{\int }{\rho }^{2\nu }\left[\sum _{i,j=1}^2{\left(\frac{\partial }{\partial x_i}\left({\rho }^{-2\nu }\frac{\partial }{\partial x_j}\left({\rho }^{2\nu }\psi \right)\right)\right)}^2\right]d\mathbf{x},$$

$${\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2=\underset{\Omega }{\int }{\rho }^{2\nu }\left[\sum _{i,j=1}^2{\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right)}^2\right]d\mathbf{x}.$$

For arbitrary $i,j=1,2$:

$$\frac{\partial }{\partial x_i}\left({\rho }^{-2\nu }\frac{\partial }{\partial x_j}\left({\rho }^{2\nu }\psi \right)\right)=\frac{\partial }{\partial x_i}\left({\rho }^{-2\nu }{\rho }^{2\nu }\right(\frac{\partial \psi }{\partial x_j}\left)+\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\psi \right)=$$

$$=\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)+\frac{\partial }{\partial x_i}\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\psi +{\rho }^{-2\nu }\left(\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\left(\frac{\partial \psi }{\partial x_i}\right).$$

Hence

$${\left(\frac{\partial }{\partial x_i}\left({\rho }^{-2\nu }\frac{\partial }{\partial x_j}\left({\rho }^{2\nu }\psi \right)\right)\right)}^2\le 3{\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right)}^2+$$

$$+3{\left(\frac{\partial }{\partial x_i}\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\right)}^2{\psi }^2+3{\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)}^2{\left(\frac{\partial \psi }{\partial x_i}\right)}^2.$$

(33)

Summing up the inequalities (33) for all $i,j=1,2$:

$${\left|\mathbf{v}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2\le 3\underset{\Omega }{\int }{\rho }^{2\nu }\left[\sum _{i,j=1}^2{\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right)}^2\right]d\mathbf{x}+3\underset{\Omega }{\int }{\rho }^{2\nu }\left[\sum _{i,j=1}^2{\left(\frac{\partial }{\partial x_i}\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\right)}^2\right]{\psi }^{2}d\mathbf{x}+$$

$$+3\underset{\Omega }{\int }{\rho }^{2\nu }\left[\sum _{i=1}^2{\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_i}\right)}^2\right]{\left(\frac{\partial \psi }{\partial x_1}\right)}^{2}d\mathbf{x}+3\underset{\Omega }{\int }{\rho }^{2\nu }\left[\sum _{i=1}^2{\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_i}\right)}^2\right]{\left(\frac{\partial \psi }{\partial x_2}\right)}^{2}d\mathbf{x}=$$

$$={3\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+3A_1+3A_2+3A_3.$$

We get

$${\left|\mathbf{v}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2\le 3{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+3A_1+3A_2+3A_3.$$

(34)

Now we estimate $A_1,A_2$ and $A_3$ separately, using (13) and (24).

For $A_1:$

$${\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}=\left\{\begin{array}{c}2\nu {\rho }^{-2}{x}_j,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }.\hfill \end{array}\right.$$

(35)

Consider two cases

(1) if $i=j,$ then

$$\frac{\partial }{\partial x_j}\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)=\left\{\begin{array}{c}2\nu {\rho }^{-2}-4\nu {\rho }^{-4}{x}_j^2,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta };\hfill \end{array}\right.$$

(2) if $i\ne j,$ then

$$\frac{\partial }{\partial x_i}\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)=\left\{\begin{array}{c}-4\nu {\rho }^{-4}{x}_j{x}_i,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }.\hfill \end{array}\right.$$

Therefore

$$A_1=\underset{\Omega }{\int }{\rho }^{2\nu }{\left[\frac{\partial }{\partial x_1}\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_1}\right)\right]}^2{\psi }^{2}d\mathbf{x}+\underset{\Omega }{\int }{\rho }^{2\nu }{\left[\frac{\partial }{\partial x_1}\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_2}\right)\right]}^2{\psi }^{2}d\mathbf{x}+$$

$$+\underset{\Omega }{\int }{\rho }^{2\nu }{\left[\frac{\partial }{\partial x_2}\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_1}\right)\right]}^2{\psi }^{2}d\mathbf{x}+\underset{\Omega }{\int }{\rho }^{2\nu }{\left[\frac{\partial }{\partial x_2}\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_2}\right)\right]}^2{\psi }^{2}d\mathbf{x}=$$

$$=\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu }{\left[2\nu {\rho }^{-2}-4\nu {\rho }^{-4}{x}_1^2\right]}^2{\psi }^{2}d\mathbf{x}+16{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu }{\rho }^{-8}{x}_1^2{x}_2^2{\psi }^{2}d\mathbf{x}+16{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu }{\rho }^{-8}{x}_1^2{x}_2^2{\psi }^{2}d\mathbf{x}+$$

$$+\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu }{\left[2\nu {\rho }^{-2}-4\nu {\rho }^{-4}{x}_2^2\right]}^2{\psi }^{2}d\mathbf{x}\le \left[8{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{\psi }^{2}d\mathbf{x}+32{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}{x}_1^4{\psi }^{2}d\mathbf{x}\right]+$$

$$+32{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}{x}_1^2{x}_2^2{\psi }^{2}d\mathbf{x}+\left[8{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{\psi }^{2}d\mathbf{x}+32{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}{x}_2^4{\psi }^{2}d\mathbf{x}\right]=$$

$$=16{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{\psi }^{2}d\mathbf{x}+32{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}(x_1^4+x_1^2{x}_2^2+x_2^4){\psi }^{2}d\mathbf{x}.$$

Due to the fact that $x_1^4+x_1^2{x}_2^2+x_2^4\le {(x_1^2+x_2^2)}^2={\rho }^4,$ we conclude

$$A_1\le 48{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{\psi }^{2}d\mathbf{x}.$$

(36)

Applying Corollary 2, its estimate (20), to (36), we derive

$$A_1\le 48{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2.$$

(37)

For $A_2$ we use (35), so that

$$A_2=\underset{\Omega }{\int }{\rho }^{2\nu }\left[\sum _{i=1}^2{\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_i}\right)}^2\right]{\left(\frac{\partial \psi }{\partial x_1}\right)}^{2}d\mathbf{x}=4{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -2}{u}_2^{2}d\mathbf{x},$$

applying Lemma 3, its inequality (16), we obtain

$$A_2\le 4{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel u_2\parallel }_{L_{2,\nu }(\Omega )}^2.$$

(38)

Similarly to $A_2,$ we estimate $A_3:$

$$A_3\le 4{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel u_1\parallel }_{L_{2,\nu }(\Omega )}^2.$$

(39)

Substituting the right-hand sides of inequalities (37)–(39) instead of the left-hand ones in (34), we have

$${\left|\mathbf{v}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2\le {3\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+144{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2+12{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(40)

Applying Lemma 5 to the second term on the right-hand side (40), then taking into account ${\parallel \nabla \psi \parallel }_{L_{2,\nu }(\Omega )}={\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}$ and using Lemma 5 to the function $\mathbf{u}$ (obtained sum of the last two terms on the right-hand side), we conclude

$${\left|\mathbf{v}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2\le {3\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+144{\nu }^2{C}_4^2{C}_5^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+12{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\le 3{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+$$

$$+12{\nu }^2{C}_4^2{C}_5^2{\delta }^{2\nu }(12C_5^2+1){\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2=(3+12{\nu }^2{C}_4^2{C}_5^2{\delta }^{2\nu }(12C_5^2+1)){\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2.$$

Lemma 7 is proved. □

Let us prove that the function $\mathbf{v}$ belongs to the set $K_1.$ We obtain an estimate of its norm by the norm of the function $\mathbf{u}$ in the spaces ${\mathbf{W}}_{2,\nu }^1(\Omega ).$

Theorem 1.

Let$\nu >0,$then there exists${\delta }_0={\delta }_0\left(\nu \right)>0$, such that for any$\delta \in (0,{\delta }_0),$any function$\mathbf{u}\in K_2,$represented as$\left(\frac{\partial \psi }{\partial x_2},-\frac{\partial \psi }{\partial x_1}\right),$function$\mathbf{v}=\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_2},-{\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_1}\right)$belongs to the set$K_1$and an estimate

$${\parallel \mathbf{v}\parallel }_{{\mathbf{W}}_{2,\nu }^1(\Omega )}\le C_8{\parallel \mathbf{u}\parallel }_{{\mathbf{W}}_{2,\nu }^1(\Omega )}$$

(41)

holds, where$C_8=max\{C_6,C_7\}.$

Proof. 

Combining the results of Lemmas 6 and 7, their inequalities (28) and (32), respectively, we obtain

$${\parallel \mathbf{v}\parallel }_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2={\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+{\left|\mathbf{v}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2\le$$

$$\le \left[C_6^2{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+C_7^2{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2\right]\le max\{C_6^2,C_7^2\}{\parallel \mathbf{u}\parallel }_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2.$$

Considering Remark 1, we conclude that $\mathbf{v}\in K_1.$

Theorem 1 is proved. □

3.1.3. Connection between the Function $\nabla \left({\rho }^{\nu }\mathbf{u}\right)$ in the Norm of the Space ${\mathbf{L}}_{2,0}(\Omega )$ and the Bilinear Form $a(\mathbf{u},\mathbf{v})$

We introduce the notation

$$J=\underset{\Omega }{\int }\sum _{i,j=1}^2{\left[{\rho }^{-\nu }\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]}^{2}d\mathbf{x}.$$

(42)

Let us estimate J from (42) in the next lemma.

Lemma 8.

Let$\nu >0,$then there exists${\delta }_0={\delta }_0\left(\nu \right)>0,$such that for any$\delta \in (0,{\delta }_0),$any function$\mathbf{u}=\left(\frac{\partial \psi }{\partial x_2},-\frac{\partial \psi }{\partial x_1}\right)\in K_2$and for$\mathbf{v}\in K_1,$represented as$\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_2},-{\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_1}\right),$an equality

$$J\le {12\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+{\nu }^2{C}_4^2{\delta }^{2\nu }\left[{576\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2+{48\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+6C_{\nu }^2{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\right]$$

(43)

holds.

Proof. 

We have

$$\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2=\underset{\Omega }{\int }\sum _{i,j=1}^2{\left[\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]}^{2}d\mathbf{x}.$$

(44)

Due to the fact that

$$\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)=\left(\frac{\partial {\rho }^{-\nu }}{\partial x_i}\right)\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)+{\rho }^{-\nu }\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right),$$

we express

$${\rho }^{-\nu }\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)=\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)-\left(\frac{\partial {\rho }^{-\nu }}{\partial x_i}\right)\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right).$$

Using this, definitions (44) and a function $\mathbf{v}$ in terms of $\psi$, we estimate J presented by (42):

$$J=\underset{\Omega }{\int }\sum _{i,j=1}^2{\left[{\rho }^{-\nu }\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]}^{2}d\mathbf{x}\le 2\underset{\Omega }{\int }\sum _{i,j=1}^2{\left[\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]}^{2}d\mathbf{x}+$$

$$+2\underset{\Omega }{\int }\sum _{i,j=1}^2{\left(\frac{\partial {\rho }^{-\nu }}{\partial x_i}\right)}^2{\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)}^{2}d\mathbf{x}=2{\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right)\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2+$$

$$+2\underset{\Omega }{\int }{\rho }^{4\nu }\left[{\left(\frac{\partial {\rho }^{-\nu }}{\partial x_1}\right)}^2+{\left(\frac{\partial {\rho }^{-\nu }}{\partial x_2}\right)}^2\right]{\left(-{\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_1}\right)}^{2}d\mathbf{x}+$$

$$+2\underset{\Omega }{\int }{\rho }^{4\nu }\left[{\left(\frac{\partial {\rho }^{-\nu }}{\partial x_1}\right)}^2+{\left(\frac{\partial {\rho }^{-\nu }}{\partial x_2}\right)}^2\right]{\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_2}\right)}^{2}d\mathbf{x}=2{\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right)\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2+$$

$$+2\underset{\Omega }{\int }{\rho }^{4\nu }\left[{\left(\frac{\partial {\rho }^{-\nu }}{\partial x_1}\right)}^2+{\left(\frac{\partial {\rho }^{-\nu }}{\partial x_2}\right)}^2\right]v_2^{2}d\mathbf{x}+2\underset{\Omega }{\int }{\rho }^{4\nu }\left[{\left(\frac{\partial {\rho }^{-\nu }}{\partial x_1}\right)}^2+{\left(\frac{\partial {\rho }^{-\nu }}{\partial x_2}\right)}^2\right]v_1^{2}d\mathbf{x}.$$

Applying formula (24), for $\beta =-\nu ,$ we conclude

$${\left(\frac{\partial {\rho }^{-\nu }}{\partial x_1}\right)}^2+{\left(\frac{\partial {\rho }^{-\nu }}{\partial x_2}\right)}^2=\left\{\begin{array}{c}{\nu }^2{\rho }^{-2\nu -2},\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.$$

then, using this and Corollary 1, its inequality (19), we have

$$J\le 2\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2+2{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(45)

The first term on the right-hand side (45) is estimated by analogy with (21), applying inequalities (14) and (40). So that we get a sequence of inequalities

$$\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2\le {2\left|\mathbf{v}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+2{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\le 6{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+$$

$$+{\nu }^2{C}_4^2{\delta }^{2\nu }\left[{288\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2+{24\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+2C_{\nu }^2{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\right].$$

Due to this fact, the estimate (45) will take the form (43).

Lemma 8 is proved. □

Let us prove the main result of Section 3.1.3.

Theorem 2.

There exists${\epsilon }_1>0,$such that for$\nu >0$there exists${\delta }_1={\delta }_1({\epsilon }_1,\nu )>0,$such that for any$\delta \in (0,{\delta }_1),$any function$\mathbf{u}=\left(\frac{\partial \psi }{\partial x_2},-\frac{\partial \psi }{\partial x_1}\right)\in K_2$and for$\mathbf{v}\in K_1,$represented as$\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_2},-{\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_1}\right),$the estimate

$$\frac{1}{4}{\parallel \nabla \left({\rho }^{\nu }\mathbf{u}\right)\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2\le a(\mathbf{u},\mathbf{v})$$

(46)

holds.

Proof. 

We write out the bilinear form $a(\mathbf{u},\mathbf{v})$ (see (9)) for $\mathbf{u}$ and $\mathbf{v},$ represented by $\psi :$

$$a(\mathbf{u},\mathbf{v})=\underset{\Omega }{\int }[\frac{\partial }{\partial x_1}\left(\frac{\partial \psi }{\partial x_2}\right)\frac{\partial }{\partial x_1}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_2}\right)+\frac{\partial }{\partial x_2}\left(\frac{\partial \psi }{\partial x_2}\right)\frac{\partial }{\partial x_2}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_2}\right)+$$

$$+\frac{\partial }{\partial x_1}\left(\frac{\partial \psi }{\partial x_1}\right)\frac{\partial }{\partial x_1}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_1}\right)+\frac{\partial }{\partial x_2}\left(\frac{\partial \psi }{\partial x_1}\right)\frac{\partial }{\partial x_2}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_1}\right)]d\mathbf{x}.$$

(47)

In addition, we need the definition of the function $\nabla \left({\rho }^{\nu }\mathbf{u}\right)$ in the norm of the space ${\mathbf{L}}_{2,0}(\Omega ):$

$$\parallel \nabla \left({\rho }^{\nu }\mathbf{u}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2=\underset{\Omega }{\int }[{\left(\frac{\partial }{\partial x_1}\left({\rho }^{\nu }\frac{\partial \psi }{\partial x_2}\right)\right)}^2+{\left(\frac{\partial }{\partial x_2}\left({\rho }^{\nu }\frac{\partial \psi }{\partial x_2}\right)\right)}^2+$$

$$+{\left(\frac{\partial }{\partial x_1}\left({\rho }^{\nu }\frac{\partial \psi }{\partial x_1}\right)\right)}^2+{\left(\frac{\partial }{\partial x_2}\left({\rho }^{\nu }\frac{\partial \psi }{\partial x_1}\right)\right)}^2]d\mathbf{x}.$$

(48)

Insofar as

$$\frac{\partial }{\partial x_j}\left({\rho }^{2\nu }\psi \right)={\rho }^{2\nu }\left(\frac{\partial \psi }{\partial x_j}\right)+\left(\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\psi ,$$

then

$${\rho }^{\nu }\left(\frac{\partial \psi }{\partial x_j}\right)={\rho }^{-\nu }\frac{\partial }{\partial x_j}\left({\rho }^{2\nu }\psi \right)-{\rho }^{-\nu }\left(\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\psi .$$

(49)

Let us take the derivative in (49) with respect to the variable $x_i,$ we get

$$\frac{\partial }{\partial x_i}\left({\rho }^{\nu }\frac{\partial \psi }{\partial x_j}\right)={\rho }^{-\nu }\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)+\left(\frac{\partial {\rho }^{-\nu }}{\partial x_i}\right)\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)-$$

$$-\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\psi -\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\left(\frac{\partial \psi }{\partial x_i}\right).$$

(50)

Now we express $\frac{\partial }{\partial x_i}\left({\rho }^{\nu }\frac{\partial \psi }{\partial x_j}\right)$, as in (50), another way:

$$\frac{\partial }{\partial x_i}\left({\rho }^{\nu }\frac{\partial \psi }{\partial x_j}\right)={\rho }^{\nu }\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)+\left(\frac{\partial {\rho }^{\nu }}{\partial x_i}\right)\left(\frac{\partial \psi }{\partial x_j}\right).$$

(51)

Substitute the representations (51) and (50) into (48), then

$$\parallel \nabla \left({\rho }^{\nu }\mathbf{u}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2=\sum _{i,j=1}^2\underset{\Omega }{\int }\left[\frac{\partial }{\partial x_i}\left({\rho }^{\nu }\frac{\partial \psi }{\partial x_j}\right)\right]\left[\frac{\partial }{\partial x_i}\left({\rho }^{\nu }\frac{\partial \psi }{\partial x_j}\right)\right]d\mathbf{x}=$$

$$=\sum _{i,j=1}^2\underset{\Omega }{\int }\left[{\rho }^{\nu }\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right)+\left(\frac{\partial {\rho }^{\nu }}{\partial x_i}\right)\left(\frac{\partial \psi }{\partial x_j}\right)\right][{\rho }^{-\nu }\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)+\left(\frac{\partial {\rho }^{-\nu }}{\partial x_i}\right)(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j})-$$

$$-\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\psi -\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\left(\frac{\partial \psi }{\partial x_i}\right)]d\mathbf{x}=\sum _{i,j=1}^2\underset{\Omega }{\int }\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right)\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)d\mathbf{x}+$$

$$+\sum _{i,j=1}^2\underset{\Omega }{\int }\left({\rho }^{\nu }\frac{\partial {\rho }^{-\nu }}{\partial x_i}\right)\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right)\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)d\mathbf{x}-$$

$$-\sum _{i,j=1}^2\underset{\Omega }{\int }{\rho }^{\nu }\left(\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\right)\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right)\psi d\mathbf{x}-$$

$$-\sum _{i,j=1}^2\underset{\Omega }{\int }\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right)\left(\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\left(\frac{\partial \psi }{\partial x_i}\right)d\mathbf{x}+\sum _{i,j=1}^2\underset{\Omega }{\int }\left({\rho }^{-\nu }\frac{\partial {\rho }^{\nu }}{\partial x_i}\right)\left(\frac{\partial \psi }{\partial x_j}\right)\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)d\mathbf{x}+$$

$$+\sum _{i,j=1}^2\underset{\Omega }{\int }\left(\frac{\partial {\rho }^{\nu }}{\partial x_i}\right)\left(\frac{\partial {\rho }^{-\nu }}{\partial x_i}\right)\left(\frac{\partial \psi }{\partial x_j}\right)\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)d\mathbf{x}-$$

$$-\sum _{i,j=1}^2\underset{\Omega }{\int }\left(\frac{\partial {\rho }^{\nu }}{\partial x_i}\right)\left(\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\right)\left(\frac{\partial \psi }{\partial x_j}\right)\psi d\mathbf{x}-\sum _{i,j=1}^2\underset{\Omega }{\int }\left(\frac{\partial {\rho }^{\nu }}{\partial x_i}\right)\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\left(\frac{\partial \psi }{\partial x_j}\right)\left(\frac{\partial \psi }{\partial x_i}\right)d\mathbf{x}=$$

$$=\sum _{i,j=1}^2{I}_1^{i,j}+\sum _{i,j=1}^2{I}_2^{i,j}-\sum _{i,j=1}^2{I}_3^{i,j}-\sum _{i,j=1}^2{I}_4^{i,j}+\sum _{i,j=1}^2{I}_5^{i,j}+\sum _{i,j=1}^2{I}_6^{i,j}-\sum _{i,j=1}^2{I}_7^{i,j}-\sum _{i,j=1}^2{I}_8^{i,j}.$$

Thus

$$\parallel \nabla \left({\rho }^{\nu }\mathbf{u}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2=\sum _{i,j=1}^2{I}_1^{i,j}+\sum _{i,j=1}^2{I}_2^{i,j}-\sum _{i,j=1}^2{I}_3^{i,j}-$$

$$-\sum _{i,j=1}^2{I}_4^{i,j}+\sum _{i,j=1}^2{I}_5^{i,j}+\sum _{i,j=1}^2{I}_6^{i,j}-\sum _{i,j=1}^2{I}_7^{i,j}-\sum _{i,j=1}^2{I}_8^{i,j}.$$

(52)

Now we estimate ${\displaystyle \sum _{i,j=1}^2}{I}_k^{i,j},k=1,\dots ,8$ from (52) separately.

1. We have

$$\sum _{i,j=1}^2{I}_1^{i,j}=a(\mathbf{u},\mathbf{v}).$$

(53)

2. Using (13), with $\beta =-\nu$, we get

$$\frac{\partial {\rho }^{-\nu }}{\partial x_i}=\left\{\begin{array}{c}-\nu {\rho }^{-\nu -2}{x}_i,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }.\hfill \end{array}\right.$$

Applying the ${\epsilon }_1$ inequality, we conclude

$$|I_2^{i,j}|=\left|\underset{{\Omega }_{\delta }}{\int }\left[{\rho }^{\nu }\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right]\left[(-\nu ){\rho }^{\nu -2}{x}_i\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]d\mathbf{x}\right|\le$$

$$\le {\epsilon }_1\underset{\Omega }{\int }{\rho }^{2\nu }{\left[\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right]}^{2}d\mathbf{x}+\frac{{\nu }^2}{{\epsilon }_1}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_i^2{\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)}^{2}d\mathbf{x}.$$

(54)

Find the sum of terms from (54) over $i,j=1,2,$ using the representation $\mathbf{u}$ and $\mathbf{v}$ in terms of $\psi$, the definition ${\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )},$$x_1^2+x_2^2={\rho }^2,$ and Corollary 1 for $v_l,l=1,2$:

$$|\sum _{i,j=1}^2{I}_2^{i,j}|\le {\epsilon }_1\sum _{i,j=1}^2\underset{\Omega }{\int }{\rho }^{2\nu }{\left[\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right]}^{2}d\mathbf{x}+\frac{{\nu }^2}{{\epsilon }_1}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_1^2{v}_2^{2}d\mathbf{x}+\frac{{\nu }^2}{{\epsilon }_1}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_1^2{v}_1^{2}d\mathbf{x}+$$

$$+\frac{{\nu }^2}{{\epsilon }_1}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_2^2{v}_2^{2}d\mathbf{x}+\frac{{\nu }^2}{{\epsilon }_1}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_2^2{v}_1^{2}d\mathbf{x}={\epsilon }_1{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+\frac{{\nu }^2}{{\epsilon }_1}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -2}{v}_1^{2}d\mathbf{x}+$$

$$+\frac{{\nu }^2}{{\epsilon }_1}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -2}{v}_2^{2}d\mathbf{x}\le {\epsilon }_1{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+\frac{{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }}{{\epsilon }_1}{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2,$$

that is

$$|\sum _{i,j=1}^2{I}_2^{i,j}|\le {\epsilon }_1{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+\frac{{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }}{{\epsilon }_1}{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(55)

3. Using (13), with $\beta =2\nu$, we have

$${\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}=\left\{\begin{array}{c}2\nu {\rho }^{\nu -2}{x}_j,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }.\hfill \end{array}\right.$$

Consider two cases: (a) if $i=j,$ then

$${\rho }^{\nu }\frac{\partial }{\partial x_j}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)=\left\{\begin{array}{c}2\nu {\rho }^{2\nu -2}+2\nu (\nu -2){\rho }^{2\nu -4}{x}_j^2,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }\hfill \end{array}\right.$$

and

$$I_3^{j,j}=\underset{{\Omega }_{\delta }}{\int }\left[{\rho }^{\nu }\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right]\left[2\nu {\rho }^{\nu -2}\psi \right]d\mathbf{x}+\underset{{\Omega }_{\delta }}{\int }\left[{\rho }^{\nu }\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right]\left[2\nu (\nu -2){\rho }^{\nu -4}{x}_j^2\psi \right]d\mathbf{x}.$$

(56)

We use the $\frac{{\epsilon }_1}{2}$ inequality for both terms on the right-hand side (56):

$$|I_3^{j,j}|\le {\epsilon }_1\underset{\Omega }{\int }{\rho }^{2\nu }{\left[\frac{\partial }{\partial x_j}\left(\frac{\partial \psi }{\partial x_j}\right)\right]}^{2}d\mathbf{x}+\frac{8{\nu }^2}{{\epsilon }_1}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{\psi }^{2}d\mathbf{x}+\frac{8{\nu }^2{(\nu -2)}^2}{{\epsilon }_1}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}{x}_j^4{\psi }^{2}d\mathbf{x}.$$

(57)

(b) if $i\ne j,$ then

$${\rho }^{\nu }\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)=\left\{\begin{array}{c}2\nu (\nu -2){\rho }^{2\nu -4}{x}_i{x}_j,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }\hfill \end{array}\right.$$

and

$$I_3^{i,j}=\underset{{\Omega }_{\delta }}{\int }\left[{\rho }^{\nu }\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right]\left[2\nu (\nu -2){\rho }^{\nu -4}{x}_i{x}_j\psi \right]d\mathbf{x}.$$

(58)

We use the ${\epsilon }_1$ inequality for the right-hand side (58):

$$|I_3^{i,j}|\le {\epsilon }_1\underset{\Omega }{\int }{\rho }^{2\nu }{\left[\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right]}^{2}d\mathbf{x}+\frac{4{\nu }^2{(\nu -2)}^2}{{\epsilon }_1}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}{x}_i^2{x}_j^2{\psi }^{2}d\mathbf{x}.$$

(59)

Find the sum of the terms (57) and (59) over $i,j=1,2,$ using the representation of $\mathbf{u}$ and $\mathbf{v}$ by $\psi$, the definition ${\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )},$$x_1^4+x_1^2{x}_2^2+x_2^4\le {\rho }^4$ and Corollary 2, then

$$|\sum _{i,j=1}^2{I}_3^{i,j}|\le {\epsilon }_1\sum _{i,j=1}^2\underset{\Omega }{\int }{\rho }^{2\nu }{\left[\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right]}^{2}d\mathbf{x}+\frac{16{\nu }^2}{{\epsilon }_1}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{\psi }^{2}d\mathbf{x}+$$

$$+\frac{8{\nu }^2{(\nu -2)}^2}{{\epsilon }_1}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}(x_1^4+x_1^2{x}_2^2+x_2^4){\psi }^{2}d\mathbf{x}\le {\epsilon }_1{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+$$

$$+\frac{16{\nu }^2{C}_4^2{\delta }^{2\nu }}{{\epsilon }_1}{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2+\frac{8{\nu }^2{(\nu -2)}^2{C}_4^2{\delta }^{2\nu }}{{\epsilon }_1}{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2,$$

that is

$$|\sum _{i,j=1}^2{I}_3^{i,j}|\le {\epsilon }_1{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+\frac{8{\nu }^2{C}_4^2{\delta }^{2\nu }(2+{(\nu -2)}^2)}{{\epsilon }_1}{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2.$$

(60)

4. Using (13), with $\beta =2\nu$, we have

$$\frac{\partial {\rho }^{2\nu }}{\partial x_j}=\left\{\begin{array}{c}2\nu {\rho }^{2\nu -2}{x}_j,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.$$

then

$$I_4^{i,j}=\underset{{\Omega }_{\delta }}{\int }\left[{\rho }^{\nu }\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right]\left[2\nu {\rho }^{\nu -2}{x}_j\frac{\partial \psi }{\partial x_i}\right]d\mathbf{x}.$$

(61)

Let us use the ${\epsilon }_1$ inequality for the right-hand side (61), the definition of $\mathbf{u}$ in terms of $\psi$ and Lemma 3 for $u_i$:

$$|\sum _{i,j=1}^2{I}_4^{i,j}|\le {\epsilon }_1\sum _{i,j=1}^2\underset{\Omega }{\int }{\rho }^{2\nu }{\left[\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right]}^{2}d\mathbf{x}+\frac{4{\nu }^2}{{\epsilon }_1}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2)u_1^{2}d\mathbf{x}+$$

$$+\frac{4{\nu }^2}{{\epsilon }_1}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2)u_2^{2}d\mathbf{x}\le {\epsilon }_1{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+\frac{4{\nu }^2{C}_4^2{\delta }^{2\nu }}{{\epsilon }_1}{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2,$$

that is

$$|\sum _{i,j=1}^2{I}_4^{i,j}|\le {\epsilon }_1{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+\frac{4{\nu }^2{C}_4^2{\delta }^{2\nu }}{{\epsilon }_1}{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(62)

5. Using (13), with $\beta =\nu$, we have

$$\frac{\partial {\rho }^{\nu }}{\partial x_i}=\left\{\begin{array}{c}\nu {\rho }^{\nu -2}{x}_i,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.$$

then

$$|I_5^{i,j}|=\left|\underset{{\Omega }_{\delta }}{\int }\left[{\rho }^{-\nu }\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]\left[\nu {\rho }^{\nu -2}{x}_i\left(\frac{\partial \psi }{\partial x_j}\right)\right]d\mathbf{x}\right|.$$

(63)

Let us use the ${\epsilon }_1$ inequality for the right-hand side (63), the definition of $\mathbf{u}$ in terms of $\psi$ and Lemma 3 for $u_i$:

$$|\sum _{i,j=1}^2{I}_5^{i,j}|\le {\epsilon }_1\sum _{i,j=1}^2\underset{\Omega }{\int }{\rho }^{-2\nu }{\left[\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]}^{2}d\mathbf{x}+\frac{{\nu }^2}{{\epsilon }_1}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2)u_1^{2}d\mathbf{x}+$$

$$+\frac{{\nu }^2}{{\epsilon }_1}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2)u_2^{2}d\mathbf{x}\le {\epsilon }_1\sum _{i,j=1}^2\underset{\Omega }{\int }{\rho }^{-2\nu }{\left[\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]}^{2}d\mathbf{x}+\frac{{\nu }^2{C}_4^2{\delta }^{2\nu }}{{\epsilon }_1}{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2,$$

that is

$$|\sum _{i,j=1}^2{I}_5^{i,j}|\le {\epsilon }_1\sum _{i,j=1}^2\underset{\Omega }{\int }{\rho }^{-2\nu }{\left[\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]}^{2}d\mathbf{x}+\frac{{\nu }^2{C}_4^2{\delta }^{2\nu }}{{\epsilon }_1}{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(64)

6. Using (13), with $\beta =\nu$ and $\beta =-\nu$, we have

$$\frac{\partial {\rho }^{\nu }}{\partial x_i}=\left\{\begin{array}{c}\nu {\rho }^{\nu -2}{x}_i,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.\frac{\partial {\rho }^{-\nu }}{\partial x_i}=\left\{\begin{array}{c}-\nu {\rho }^{-\nu -2}{x}_i,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.$$

then

$$|I_6^{i,j}|=\left|\underset{{\Omega }_{\delta }}{\int }\left[-\nu {\rho }^{\nu -2}{x}_i\left(\frac{\partial \psi }{\partial x_j}\right)\right]\left[\nu {\rho }^{\nu -2}\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]d\mathbf{x}\right|.$$

(65)

Let us estimate the right-hand side (65) using the definitions of $\mathbf{u}$ and $\mathbf{v}$ in terms of $\psi$, Lemma 3 for $u_i$ and Corollary 1 for $v_i$:

$$|\sum _{i,j=1}^2{I}_6^{i,j}|\le \frac{{\nu }^2}{2}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2){\left(\frac{\partial \psi }{\partial x_1}\right)}^{2}d\mathbf{x}+\frac{{\nu }^2}{2}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2){\left(\frac{\partial \psi }{\partial x_2}\right)}^{2}d\mathbf{x}+$$

$$+\frac{{\nu }^2}{2}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2){\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_1}\right)}^{2}d\mathbf{x}+\frac{{\nu }^2}{2}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2){\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_2}\right)}^{2}d\mathbf{x}\le$$

$$\le \frac{{\nu }^2}{2}{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+\frac{{\nu }^2}{2}{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2,$$

that is

$$|\sum _{i,j=1}^2{I}_6^{i,j}|\le \frac{{\nu }^2}{2}{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+\frac{{\nu }^2}{2}{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(66)

7. Using (13), with $\beta =\nu$ and $\beta =2\nu$, we have

$$\frac{\partial {\rho }^{\nu }}{\partial x_j}=\left\{\begin{array}{c}\nu {\rho }^{\nu -2}{x}_j,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.{\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}=\left\{\begin{array}{c}2\nu {\rho }^{\nu -2}{x}_j,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }.\hfill \end{array}\right.$$

Similarly, as in the study of the $I_3^{i,j}$, we consider two cases:

(a) if $i=j,$ then

$$\frac{\partial }{\partial x_j}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)=\left\{\begin{array}{c}2\nu {\rho }^{\nu -2}+2\nu (\nu -2){\rho }^{\nu -4}{x}_j^2,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }\hfill \end{array}\right.$$

and

$$\left(\frac{\partial {\rho }^{\nu }}{\partial x_j}\right)\frac{\partial }{\partial x_j}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)=\left\{\begin{array}{c}2{\nu }^2{\rho }^{2\nu -4}{x}_j+2{\nu }^2(\nu -2){\rho }^{2\nu -6}{x}_j^3,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.$$

$$|I_7^{j,j}|=|\underset{{\Omega }_{\delta }}{\int }\left[\sqrt{2}\nu {\rho }^{\nu -2}\psi \right]\left[\sqrt{2}\nu {\rho }^{\nu -2}{x}_j\left(\frac{\partial \psi }{\partial x_j}\right)\right]d\mathbf{x}+$$

$$+\underset{{\Omega }_{\delta }}{\int }\left[\sqrt{2}\nu {\rho }^{\nu -4}{x}_j^2\psi \right]\left[\sqrt{2}\nu (\nu -2){\rho }^{\nu -2}{x}_j\left(\frac{\partial \psi }{\partial x_j}\right)\right]d\mathbf{x}|\le$$

$$\le {\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{\psi }^{2}d\mathbf{x}+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_j^2{\left(\frac{\partial \psi }{\partial x_j}\right)}^{2}d\mathbf{x}+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}{x}_j^4{\psi }^{2}d\mathbf{x}+$$

$$+{\nu }^2{(\nu -2)}^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_j^2{\left(\frac{\partial \psi }{\partial x_j}\right)}^{2}d\mathbf{x},$$

that is

$$|I_7^{j,j}|\le {\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{\psi }^{2}d\mathbf{x}+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_j^2{\left(\frac{\partial \psi }{\partial x_j}\right)}^{2}d\mathbf{x}+$$

$$+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}{x}_j^4{\psi }^{2}d\mathbf{x}+{\nu }^2{(\nu -2)}^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_j^2{\left(\frac{\partial \psi }{\partial x_j}\right)}^{2}d\mathbf{x};$$

(67)

(b) if $i\ne j,$ then

$$\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)=\left\{\begin{array}{c}2\nu (\nu -2){\rho }^{\nu -4}{x}_i{x}_j,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.$$

$$\left(\frac{\partial {\rho }^{\nu }}{\partial x_i}\right)\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)=\left\{\begin{array}{c}2{\nu }^2(\nu -2){\rho }^{2\nu -6}{x}_i^2{x}_j,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }\hfill \end{array}\right.$$

and

$$|I_7^{i,j}|=\left|\underset{{\Omega }_{\delta }}{\int }\left[\sqrt{2}\nu {\rho }^{\nu -4}{x}_i{x}_j\psi \right]\left[\sqrt{2}\nu (\nu -2){\rho }^{\nu -2}{x}_i\left(\frac{\partial \psi }{\partial x_j}\right)\right]d\mathbf{x}\right|\le {\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}{x}_i^2{x}_j^2{\psi }^{2}d\mathbf{x}+$$

$$+{\nu }^2{(\nu -2)}^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_i^2{\left(\frac{\partial \psi }{\partial x_j}\right)}^{2}d\mathbf{x},$$

that is

$$|I_7^{i,j}|\le {\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}{x}_i^2{x}_j^2{\psi }^{2}d\mathbf{x}+{\nu }^2{(\nu -2)}^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_i^2{\left(\frac{\partial \psi }{\partial x_j}\right)}^{2}d\mathbf{x}.$$

(68)

We estimate the sum of terms (67) and (68) $i,j=1,2,$ using Lemma 3 and Corollary 2:

$$|\sum _{i,j=1}^2{I}_7^{i,j}|\le 2{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{\psi }^{2}d\mathbf{x}+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2){\left(\frac{\partial \psi }{\partial x_1}\right)}^{2}d\mathbf{x}+$$

$$+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2){\left(\frac{\partial \psi }{\partial x_2}\right)}^{2}d\mathbf{x}+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}(x_1^4+2x_1^2{x}_2^2+x_2^4){\psi }^{2}d\mathbf{x}+$$

$$+{\nu }^2{(\nu -2)}^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2){\left(\frac{\partial \psi }{\partial x_1}\right)}^{2}d\mathbf{x}+{\nu }^2{(\nu -2)}^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2){\left(\frac{\partial \psi }{\partial x_2}\right)}^{2}d\mathbf{x}\le$$

$$\le 3{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2+(1+{(\nu -2)}^2){\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2,$$

that is

$$|\sum _{i,j=1}^2{I}_7^{i,j}|\le 3{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2+(1+{(\nu -2)}^2){\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(69)

8. Using (13), with $\beta =\nu$ and $\beta =2\nu$, we have

$$\frac{\partial {\rho }^{\nu }}{\partial x_i}=\left\{\begin{array}{c}\nu {\rho }^{\nu -2}{x}_i,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.{\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}=\left\{\begin{array}{c}2\nu {\rho }^{\nu -2}{x}_j,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }.\hfill \end{array}\right.$$

Consider two cases:

(a) if $i=j$, then

$$|I_8^{j,j}|=\underset{{\Omega }_{\delta }}{\int }2{\nu }^2{\rho }^{2\nu -4}{x}_j^2{\left(\frac{\partial \psi }{\partial x_j}\right)}^{2}d\mathbf{x};$$

(70)

(b) if $i\ne j,$ then

$$|I_8^{i,j}|=\left|\underset{{\Omega }_{\delta }}{\int }\left[\sqrt{2}\nu {\rho }^{\nu -2}{x}_i\left(\frac{\partial \psi }{\partial x_j}\right)\right]\left[\sqrt{2}\nu {\rho }^{\nu -2}{x}_j\left(\frac{\partial \psi }{\partial x_i}\right)\right]d\mathbf{x}\right|\le$$

$$\le {\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_i^2{\left(\frac{\partial \psi }{\partial x_j}\right)}^{2}d\mathbf{x}+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_j^2{\left(\frac{\partial \psi }{\partial x_i}\right)}^{2}d\mathbf{x},$$

that is

$$|I_8^{i,j}|\le {\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_i^2{\left(\frac{\partial \psi }{\partial x_j}\right)}^{2}d\mathbf{x}+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_j^2{\left(\frac{\partial \psi }{\partial x_i}\right)}^{2}d\mathbf{x}.$$

(71)

We estimate the sum of terms (70) and (71) $i,j=1,2,$ and using Lemma 3 for $u_i$:

$$|\sum _{i,j=1}^2{I}_8^{i,j}|\le 2{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2){\left(\frac{\partial \psi }{\partial x_1}\right)}^{2}d\mathbf{x}+$$

$$+2{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2){\left(\frac{\partial \psi }{\partial x_2}\right)}^{2}d\mathbf{x}\le 2{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2,$$

that is

$$|\sum _{i,j=1}^2{I}_8^{i,j}|\le 2{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(72)

Apply inequalities (53), (55), (60), (62), (64), (66), (69) and (72) to evaluate the right-hand side of (52):

$$\parallel \nabla \left({\rho }^{\nu }\mathbf{u}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2\le a(\mathbf{u},\mathbf{v})+\left[{\epsilon }_1{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+\frac{{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }}{{\epsilon }_1}{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\right]+[{\epsilon }_1{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+$$

$$+\frac{8{\nu }^2{C}_4^2{\delta }^{2\nu }(2+{(\nu -2)}^2)}{{\epsilon }_1}{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2]+\left[{\epsilon }_1{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+\frac{4{\nu }^2{C}_4^2{\delta }^{2\nu }}{{\epsilon }_1}{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\right]+$$

$$+\left[{\epsilon }_1\sum _{i,j=1}^2\underset{\Omega }{\int }{\rho }^{-2\nu }{\left[\frac{\partial }{\partial x_i}\left(\frac{\partial {\rho }^{2\nu }\psi }{\partial x_j}\right)\right]}^{2}d\mathbf{x}+\frac{{\nu }^2{C}_4^2{\delta }^{2\nu }}{{\epsilon }_1}{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\right]+[\frac{{\nu }^2}{2}{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+$$

$$+\frac{{\nu }^2}{2}{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2]+\left[3{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2+(1+{(\nu -2)}^2){\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\right]+$$

$$+\left[2{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\right],$$

then, using the definition (42) we have

$$\parallel \nabla \left({\rho }^{\nu }\mathbf{u}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2\le a(\mathbf{u},\mathbf{v})+3{\epsilon }_1{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+{\epsilon }_{1}J+{\nu }^2{C}_4^2{\delta }^{2\nu }[\left(\frac{5}{{\epsilon }_1}+3.5+{(\nu -2)}^2\right){\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+$$

$$+C_{\nu }^2\left(\frac{1}{{\epsilon }_1}+0.5\right){\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+\left(\frac{16+8{(\nu -2)}^2}{{\epsilon }_1}+3\right){\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2].$$

(73)

We apply the result of Lemma 8, the inequality (43), to estimate the third term on the right-hand side of (73):

$$\parallel \nabla \left({\rho }^{\nu }\mathbf{u}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2\le a(\mathbf{u},\mathbf{v})+15{\epsilon }_1{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+{\nu }^2{C}_4^2{\delta }^{2\nu }[\left(\frac{5}{{\epsilon }_1}+51.5+{(\nu -2)}^2\right){\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+$$

$$+C_{\nu }^2\left(\frac{1}{{\epsilon }_1}+6.5\right){\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+\left(\frac{16+8{(\nu -2)}^2}{{\epsilon }_1}+579\right){\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2].$$

(74)

Now we estimate the second, fourth and fifth terms on the right-hand side (74). For the second term we apply Lemma 4, the inequality (22) for vector functions, then

$${\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2\le 2\parallel \nabla \left({\rho }^{\nu }\mathbf{u}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2+2{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{L_{2,\nu }(\Omega )}^2.$$

We estimate the fourth term using the inequality (28) of Lemma 6:

$${\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\le C_6^2{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

For the fifth term we apply the inequality (27) of Lemma 5, then

$${\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2\le 4C_3^2{\parallel \nabla \psi \parallel }_{L_{2,\nu }(\Omega )}^2=4C_3^2{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

Substituting the relations obtained above into (74), we conclude

$$\parallel \nabla \left({\rho }^{\nu }\mathbf{u}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2\le a(\mathbf{u},\mathbf{v})+30{\epsilon }_1{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+{\nu }^2{C}_4^2{C}_0^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2,$$

(75)

where $C_0^2=\frac{5}{{\epsilon }_1}+51.5+{(\nu -2)}^2+C_{\nu }^2{C}_6^2\left(\frac{1}{{\epsilon }_1}+6.5\right)+4C_3^2\left(\frac{16+8{(\nu -2)}^2}{{\epsilon }_1}+579\right).$

It remains to apply Lemma 2 for the last term on the right-hand side (75):

$${\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2=\parallel {\rho }^{\nu }{\mathbf{u}\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2\le C_3^2{\parallel \nabla \left({\rho }^{\nu }\mathbf{u}\right)\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2.$$

Hence

$$\parallel \nabla \left({\rho }^{\nu }\mathbf{u}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2\le a(\mathbf{u},\mathbf{v})+30{\epsilon }_1\parallel \nabla \left({\rho }^{\nu }\mathbf{u}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2+C(\delta ,\nu ,{\epsilon }_1){\parallel \nabla \left({\rho }^{\nu }\mathbf{u}\right)\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2,$$

(76)

where $C(\delta ,\nu ,{\epsilon }_1)={\nu }^2{C}_4^2{C}_3^2{C}_0^2{\delta }^{2\nu }$.

If we choose in (76) ${\epsilon }_1=\frac{1}{60},$ then for $\nu >0$ there exists ${\delta }_1={\delta }_1({\epsilon }_1,\nu )>0$ for which $C({\delta }_1,\nu ,{\epsilon }_1)=\frac{1}{4}$ and for any $\delta \in (0,{\delta }_1)$ (${\delta }_1$ is less than ${\delta }_0$ of Lemma 5) the sequence of inequalities

$$\frac{1}{4}\parallel \nabla \left({\rho }^{\nu }\mathbf{u}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2\le \left(1-\frac{1}{2}-C(\delta ,\nu ,{\epsilon }_1)\right){\parallel \nabla \left({\rho }^{\nu }\mathbf{u}\right)\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2\le a(\mathbf{u},\mathbf{v})$$

holds.

Theorem 2 is proved. □

3.2. Construction of the Function $\mathbf{u}\in {\mathit{K}}_2$ Using the Function $\mathbf{v}\in {\mathit{K}}_1$ and Their Relationship

Let $\nu >0.$ Consider a function $\mathbf{v}\in K_1$ such that there exists a function $\theta ={\rho }^{2\nu }\psi$, where $\psi \in W_{2,\nu }^{2,0}(\Omega )$ ( $\psi$ satisfies the conditions (10)–(12)), which has the form

$$\mathbf{v}=\left({\rho }^{-2\nu }\frac{\partial \theta }{\partial x_2},-{\rho }^{-2\nu }\frac{\partial \theta }{\partial x_1}\right)=\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_2},-{\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_1}\right).$$

(77)

Next, we define the function

$$\mathbf{u}=\left(\frac{\partial \left({\rho }^{-2\nu }\theta \right)}{\partial x_2},-\frac{\partial \left({\rho }^{-2\nu }\theta \right)}{\partial x_1}\right)=\left(\frac{\partial \psi }{\partial x_2},-\frac{\partial \psi }{\partial x_1}\right).$$

(78)

Remark 2.

It will be proved in Theorem 3 that the function$\mathbf{u}$of the form (78) belongs to the space${\mathbf{W}}_{2,\nu }^{1,0}(\Omega ).$Due to this, the conditions (6)–(8) and equality—$\underset{\Omega }{\int }qdiv \mathbf{u}d\mathbf{x}=0$$\forall q\in L_{2,\nu }^0(\Omega ,\delta )$we conclude that$\mathbf{u}\in K_2.$

3.2.1. Belonging of the Function $\mathbf{u}$ to the Set $K_2$

In Lemma 9 we estimate the norm of the function $\mathbf{u}$ using the norm of the function $\mathbf{v}$ in the space ${\mathbf{L}}_{2,\nu }(\Omega ),$ and in Lemma 10 we estimate the seminorm $\mathbf{u}$ in the space ${\mathbf{W}}_{2,\nu }^1(\Omega )$ using the seminorm of $\mathbf{v}$ in the space ${\mathbf{W}}_{2,\nu }^1(\Omega )$ and norm $\mathbf{v}$ in the space ${\mathbf{L}}_{2,\nu }(\Omega ).$

Lemma 9.

Let$\nu >0,$then there exists${\delta }_2={\delta }_2\left(\nu \right)>0$, such that for any$\delta \in (0,{\delta }_2),$arbitrary function$\mathbf{v}\in K_1,$which has the form (77) and for$\mathbf{u}$represented as (78), the inequality

$${\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}\le 2{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}$$

(79)

holds.

Proof. 

We have $\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_i}={\rho }^{2\nu }\left(\frac{\partial \psi }{\partial x_i}\right)+\psi \left(\frac{\partial {\rho }^{2\nu }}{\partial x_i}\right),$ then ${\rho }^{\nu }\left(\frac{\partial \psi }{\partial x_i}\right)={\rho }^{-\nu }\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_i}\right)-\psi \left(\frac{\partial {\rho }^{2\nu }}{\partial x_i}\right){\rho }^{-\nu }$ and

$$\parallel {\rho }^{\nu }{\nabla \psi \parallel }_{L_{2,0}(\Omega )}^2=\underset{\Omega }{\int }{\rho }^{2\nu }[\sum _{i=1}^2{\left(\frac{\partial \psi }{\partial x_i}\right)}^2]d\mathbf{x}\le 2\underset{\Omega }{\int }{\rho }^{-2\nu }[\sum _{i=1}^2{\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_i}\right)}^2]d\mathbf{x}+$$

$$+2\underset{\Omega }{\int }{\rho }^{-2\nu }{\psi }^2[\sum _{i=1}^2{\left(\frac{\partial {\rho }^{2\nu }}{\partial x_i}\right)}^2]d\mathbf{x}=2{\parallel {\rho }^{-\nu }\nabla \left({\rho }^{2\nu }\psi \right)\parallel }_{L_{2,0}(\Omega )}^2+2\underset{\Omega }{\int }{\rho }^{-2\nu }{\psi }^2[\sum _{i=1}^2{\left(\frac{\partial {\rho }^{2\nu }}{\partial x_i}\right)}^2]d\mathbf{x},$$

that is

$$\parallel {\rho }^{\nu }{\nabla \psi \parallel }_{L_{2,0}(\Omega )}^2\le 2{\parallel {\rho }^{-\nu }\nabla \left({\rho }^{2\nu }\psi \right)\parallel }_{L_{2,0}(\Omega )}^2+2\underset{\Omega }{\int }{\rho }^{-2\nu }{\psi }^2[\sum _{i=1}^2{\left(\frac{\partial {\rho }^{2\nu }}{\partial x_i}\right)}^2]d\mathbf{x}.$$

(80)

Applying (24), for $\beta =2\nu$, estimates (27) and (30), we obtain a sequence of inequalities

$$\underset{\Omega }{\int }{\rho }^{-2\nu }{\psi }^2[\sum _{i=1}^2{\left(\frac{\partial {\rho }^{2\nu }}{\partial x_i}\right)}^2]d\mathbf{x}=\underset{{\Omega }_{\delta }}{\int }{\rho }^{-2\nu }{\psi }^2{\left(2\nu \right)}^2{\rho }^{4\nu -2}d\mathbf{x}\le 2{\nu }^2{C}_4^2{(1+\tau )}^{-1}{\delta }^{2\nu +2}{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2\le$$

$$\le 2{\nu }^2{C}_4^2{C}_5^2{(1+\tau )}^{-1}{\delta }^{2\nu +2}{\parallel \nabla \psi \parallel }_{L_{2,\nu }(\Omega )}^2=2{\nu }^2{C}_4^2{C}_5^2{(1+\tau )}^{-1}{\delta }^{2\nu +2}{\parallel {\rho }^{\nu }\nabla \psi \parallel }_{L_{2,0}(\Omega )}^2,$$

thus

$$\underset{\Omega }{\int }{\rho }^{-2\nu }{\psi }^2[\sum _{i=1}^2{\left(\frac{\partial {\rho }^{2\nu }}{\partial x_i}\right)}^2]d\mathbf{x}\le 8{\nu }^2{C}_4^2{C}_3^2{(1+\tau )}^{-1}{\delta }^{2\nu +2}{\parallel {\rho }^{\nu }\nabla \psi \parallel }_{L_{2,0}(\Omega )}^2.$$

(81)

We estimate the second term on the right-hand side (80) using the inequality (81), then

$$\parallel {\rho }^{\nu }{\nabla \psi \parallel }_{L_{2,0}(\Omega )}^2\le 2\parallel {\rho }^{-\nu }\nabla \left({\rho }^{2\nu }\psi \right){\parallel }_{L_{2,0}(\Omega )}^2+16{\nu }^2{C}_4^2{C}_3^2{(1+\tau )}^{-1}{\delta }^{2\nu +2}{\parallel {\rho }^{\nu }\nabla \psi \parallel }_{L_{2,0}(\Omega )}^2$$

and

$$(1-16{\nu }^2{C}_4^2{C}_3^2{(1+\tau )}^{-1}{\delta }^{2\nu +2})\parallel {\rho }^{\nu }{\nabla \psi \parallel }_{L_{2,0}(\Omega )}^2\le 2{\parallel {\rho }^{-\nu }\nabla \left({\rho }^{2\nu }\psi \right)\parallel }_{L_{2,0}(\Omega )}^2.$$

For $\nu >0$ there exists ${\delta }_2={\delta }_2\left(\nu \right)=min\{{\delta }_0\left(\nu \right),\sqrt{\frac{1+\tau }{8}}\}$, such that for any $\delta \in (0,{\delta }_2)$ the following sequence of inequalities

$$\frac{1}{2}\parallel {\rho }^{\nu }{\nabla \psi \parallel }_{L_{2,0}(\Omega )}^2\le (1-16{\nu }^2{C}_4^2{C}_3^2{(1+\tau )}^{-1}{\delta }^{2\nu +2})\parallel {\rho }^{\nu }{\nabla \psi \parallel }_{L_{2,0}(\Omega )}^2\le 2{\parallel {\rho }^{-\nu }\nabla \left({\rho }^{2\nu }\psi \right)\parallel }_{L_{2,0}(\Omega )}^2$$

holds, thus

$$\parallel {\rho }^{\nu }{\nabla \psi \parallel }_{L_{2,0}(\Omega )}\le 2{\parallel {\rho }^{-\nu }\nabla \left({\rho }^{2\nu }\psi \right)\parallel }_{L_{2,0}(\Omega )}.$$

(82)

Due to the fact that

$${\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}=\parallel {\rho }^{\nu }{\nabla \psi \parallel }_{L_{2,0}(\Omega )}{,\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}={\parallel {\rho }^{-\nu }\nabla \left({\rho }^{2\nu }\psi \right)\parallel }_{L_{2,0}(\Omega )},$$

the estimate (79) of the lemma follows directly from the relation (82).

Lemma 9 is proved. □

Lemma 10.

Let$\nu >0,$then there exists${\delta }_2={\delta }_2\left(\nu \right)>0$, such that for any$\delta \in (0,{\delta }_2),$arbitrary function$\mathbf{v}\in K_1,$which has the form (77) and for$\mathbf{u}$represented as (78), the inequality

$${\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2\le {3\left|\mathbf{v}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+C_9^2{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2,$$

(83)

holds, where a constant$C_9=4\sqrt{3}\nu C_4{\delta }^{\nu }\sqrt{1+12C_5^2}$.

Proof. 

By definition of the functions $\mathbf{u}$ and $\mathbf{v},$ we have

$${\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2=\underset{\Omega }{\int }{\rho }^{2\nu }\left[\sum _{i,j=1}^2{\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right)}^2\right]d\mathbf{x},$$

$${\left|\mathbf{v}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2=\underset{\Omega }{\int }{\rho }^{2\nu }\left[\sum _{i,j=1}^2{\left(\frac{\partial }{\partial x_i}\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)}^2\right]d\mathbf{x}.$$

For arbitrary $i,j=1,2,$ (see Lemma 7), we get

$$\frac{\partial }{\partial x_i}\left({\rho }^{-2\nu }\frac{\partial }{\partial x_j}\left({\rho }^{2\nu }\psi \right)\right)=\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right)+\frac{\partial }{\partial x_i}\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\psi +{\rho }^{-2\nu }\left(\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\left(\frac{\partial \psi }{\partial x_i}\right),$$

then

$${\rho }^{\nu }\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right)={\rho }^{\nu }\frac{\partial }{\partial x_i}\left({\rho }^{-2\nu }\frac{\partial }{\partial x_j}\left({\rho }^{2\nu }\psi \right)\right)-{\rho }^{\nu }\frac{\partial }{\partial x_i}\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\psi -{\rho }^{-\nu }\left(\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\left(\frac{\partial \psi }{\partial x_i}\right)$$

and

$${\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2=\underset{\Omega }{\int }{\rho }^{2\nu }\left[\sum _{i,j=1}^2{\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right)}^2\right]d\mathbf{x}\le 3\underset{\Omega }{\int }{\rho }^{2\nu }\left[\sum _{i,j=1}^2{\left(\frac{\partial }{\partial x_i}\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)}^2\right]d\mathbf{x}+$$

$$+3\underset{\Omega }{\int }{\rho }^{2\nu }\left[\sum _{i,j=1}^2{\left(\frac{\partial }{\partial x_i}\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\right)}^2\right]{\psi }^{2}d\mathbf{x}+3\underset{\Omega }{\int }{\rho }^{2\nu }[\sum _{i=1}^2{\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_i}\right)}^2]{\left(\frac{\partial \psi }{\partial x_1}\right)}^{2}d\mathbf{x}+$$

$$+3\underset{\Omega }{\int }{\rho }^{2\nu }[\sum _{i=1}^2{\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_i}\right)}^2]{\left(\frac{\partial \psi }{\partial x_2}\right)}^{2}d\mathbf{x}=3{\left|\mathbf{v}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+3L_1+3L_2+3L_3,$$

thus

$${\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2\le 3{\left|\mathbf{v}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+3L_1+3L_2+3L_3.$$

(84)

Note that $L_1,L_2$ and $L_3$ coincide with the corresponding $A_1,A_2$ and $A_3$ in the inequality (34) of Lemma 7. Hence, by analogy with the derivation of (40), we conclude

$${\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2\le {3\left|\mathbf{v}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+144{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2+12{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(85)

Applying Lemma 5, its estimate (27), to the second term on the right-hand side (85) and using the fact that ${\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}={\parallel \nabla \psi \parallel }_{L_{2,\nu }(\Omega )},$ we have

$${\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2\le {3\left|\mathbf{v}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+12{\nu }^2{C}_4^2{\delta }^{2\nu }(1+12C_5^2){\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(86)

Using the inequality (79) to estimate the second term on the right-hand side (86), we obtain the estimate (83).

Lemma 10 is proved. □

Let us prove the main result of Section 3.2.1.

Theorem 3.

Let$\nu >0,$then there exists${\delta }_2=min\{{\delta }_0\left(\nu \right),\sqrt{\frac{1+\tau }{8}}\}>0$, that for any$\delta \in (0,{\delta }_2),$arbitrary function$\mathbf{v}\in K_1,$represented as$\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_2},-{\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_1}\right),$function$\mathbf{u}=\left(\frac{\partial \psi }{\partial x_2},-\frac{\partial \psi }{\partial x_1}\right)$belongs to the set$K_2$and an estimate

$${\parallel \mathbf{u}\parallel }_{{\mathbf{W}}_{2,\nu }^1(\Omega )}\le C_{10}{\parallel \mathbf{v}\parallel }_{{\mathbf{W}}_{2,\nu }^1(\Omega )}$$

(87)

holds, where a constant$C_{10}$is equal to$\sqrt{4+C_9^2}.$

Proof. 

Let us use Lemmas 9 and 10, their estimates (79) and (83), respectively, then

$${\parallel \mathbf{u}\parallel }_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2={\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2\le \left[{4\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+{3\left|\mathbf{v}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+C_9^2{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\right]\le$$

$$\le (4+C_9^2){\parallel \mathbf{v}\parallel }_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2.$$

Taking into account Remark 2, we conclude that $\mathbf{u}\in K_2.$

Theorem 3 is proved. □

3.2.2. Connection between the Function $\nabla \left({\rho }^{\nu }\mathbf{v}\right)$ in the Norm of the Space ${\mathbf{L}}_{2,0}(\Omega )$ and the Bilinear Form $a(\mathbf{u},\mathbf{v})$

Let us prove the main result of Section 3.2.2.

Theorem 4.

There exists${\epsilon }_2>0,$such that for$\nu >0$there exists${\delta }_3={\delta }_3({\epsilon }_2,\nu )>0,$such that for arbitrary$\delta \in (0,min\{{\delta }_2,{\delta }_3\}),$an arbitrary function$\mathbf{v}\in K_1,$which has the form (77), and for the function$\mathbf{u}\in K_2$represented as (78), the inequality

$$\frac{1}{4}{\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right)\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2\le a(\mathbf{u},\mathbf{v})$$

(88)

holds.

Proof. 

For $j=1,2,$ we express ${\rho }^{-\nu }\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)$ from the equality (49), we have

$${\rho }^{-\nu }\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)={\rho }^{\nu }\left(\frac{\partial \psi }{\partial x_j}\right)+{\rho }^{-\nu }\left(\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\psi .$$

Then, we take the derivative with respect to the variable $x_i,i=1,2:$

$$\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)={\rho }^{\nu }\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right)+\left(\frac{\partial {\rho }^{\nu }}{\partial x_i}\right)\left(\frac{\partial \psi }{\partial x_j}\right)+$$

$$+\left(\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\right)\psi +\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\left(\frac{\partial \psi }{\partial x_i}\right).$$

(89)

Let us express $\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right),i,j=1,2,$ as in (89), in another way:

$$\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)={\rho }^{-\nu }\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)+\left(\frac{\partial {\rho }^{-\nu }}{\partial x_i}\right)\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right).$$

(90)

Using the representations (90) and (89), we have

$$\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2=\sum _{i,j=1}^2\underset{\Omega }{\int }\left[\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]\left[\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]d\mathbf{x}=$$

$$=\sum _{i,j=1}^2\underset{\Omega }{\int }\left[{\rho }^{-\nu }\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)+\left(\frac{\partial {\rho }^{-\nu }}{\partial x_i}\right)\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right][{\rho }^{\nu }\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right)+\left(\frac{\partial {\rho }^{\nu }}{\partial x_i}\right)\left(\frac{\partial \psi }{\partial x_j}\right)+$$

$$+\left(\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\right)\psi +\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\left(\frac{\partial \psi }{\partial x_i}\right)]d\mathbf{x}=\sum _{i,j=1}^2\underset{\Omega }{\int }\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right)\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)d\mathbf{x}+$$

$$+\sum _{i,j=1}^2\underset{\Omega }{\int }\left({\rho }^{-\nu }\frac{\partial {\rho }^{\nu }}{\partial x_i}\right)\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)\left(\frac{\partial \psi }{\partial x_j}\right)d\mathbf{x}+$$

$$+\sum _{i,j=1}^2\underset{\Omega }{\int }{\rho }^{-\nu }\left(\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\right)\psi \left(\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)d\mathbf{x}+$$

$$+\sum _{i,j=1}^2\underset{\Omega }{\int }\left({\rho }^{-2\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)\left(\frac{\partial \psi }{\partial x_i}\right)d\mathbf{x}+$$

$$+\sum _{i,j=1}^2\underset{\Omega }{\int }\left({\rho }^{\nu }\frac{\partial {\rho }^{-\nu }}{\partial x_i}\right)\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \psi }{\partial x_j}\right)\right)d\mathbf{x}+$$

$$+\sum _{i,j=1}^2\underset{\Omega }{\int }\left(\frac{\partial {\rho }^{-\nu }}{\partial x_i}\right)\left(\frac{\partial {\rho }^{\nu }}{\partial x_i}\right)\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\left(\frac{\partial \psi }{\partial x_j}\right)d\mathbf{x}+$$

$$+\sum _{i,j=1}^2\underset{\Omega }{\int }\left(\frac{\partial {\rho }^{-\nu }}{\partial x_i}\right)\left(\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\right)\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\psi d\mathbf{x}+$$

$$+\sum _{i,j=1}^2\underset{\Omega }{\int }\left(\frac{\partial {\rho }^{-\nu }}{\partial x_i}\right)\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\left(\frac{\partial \psi }{\partial x_i}\right)d\mathbf{x}=\sum _{i,j=1}^2{J}_1^{i,j}+\sum _{i,j=1}^2{J}_2^{i,j}+$$

$$+\sum _{i,j=1}^2{J}_3^{i,j}+\sum _{i,j=1}^2{J}_4^{i,j}+\sum _{i,j=1}^2{J}_5^{i,j}+\sum _{i,j=1}^2{J}_6^{i,j}+\sum _{i,j=1}^2{J}_7^{i,j}+\sum _{i,j=1}^2{J}_8^{i,j},$$

thus

$$\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2=\sum _{k=1}^8\sum _{i,j=1}^2{J}_k^{i,j}.$$

(91)

We will estimate each term ${\displaystyle \sum _{i,j=1}^2}{J}_k^{i,j},k=1,\dots ,8$ in (91) separately.

1. We have

$$\sum _{i,j=1}^2{J}_1^{i,j}=a(\mathbf{u},\mathbf{v}).$$

(92)

2. Due to the fact that $J_2^{i,j}=I_5^{i,j}$ (see Theorem 2), then using the ${\epsilon }_2$ inequality, by analogy with (64), we conclude

$$|\sum _{i,j=1}^2{J}_2^{i,j}|\le {\epsilon }_2\sum _{i,j=1}^2\underset{\Omega }{\int }{\rho }^{-2\nu }{\left[\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]}^{2}d\mathbf{x}+\frac{{\nu }^2{C}_4^2{\delta }^{2\nu }}{{\epsilon }_2}{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(93)

The first term on the right-hand side (93) has the form (42), then applying the inequality (45), we derive

$$|\sum _{i,j=1}^2{J}_2^{i,j}|\le 2{\epsilon }_2\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2+2{\epsilon }_2{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+\frac{{\nu }^2{C}_4^2{\delta }^{2\nu }}{{\epsilon }_2}{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(94)

3. To estimate $J_3^{i,j}$, we employ the ${\epsilon }_2$ inequality

$$J_3^{i,j}=\underset{\Omega }{\int }\left[{\rho }^{-\nu }\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)\right]\left[\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\psi \right]d\mathbf{x}\le$$

$$\le {\epsilon }_2\underset{\Omega }{\int }{\rho }^{-2\nu }{\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)}^{2}d\mathbf{x}+\frac{1}{{\epsilon }_2}\underset{\Omega }{\int }{\left(\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\right)}^2{\psi }^{2}d\mathbf{x}.$$

(95)

Using (13), for $\beta =2\nu ,$ we have

$${\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}=\left\{\begin{array}{c}2\nu {\rho }^{\nu -2}{x}_j,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }.\hfill \end{array}\right.$$

Consider two cases:

(a) if $i=j,$ then

$$\frac{\partial }{\partial x_j}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)=\left\{\begin{array}{c}2\nu {\rho }^{\nu -2}+2\nu (\nu -2){\rho }^{\nu -4}{x}_j^2,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }\hfill \end{array}\right.$$

and

$$\underset{\Omega }{\int }{\left(\frac{\partial }{\partial x_j}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\right)}^2{\psi }^{2}d\mathbf{x}\le 8{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{\psi }^{2}d\mathbf{x}+8{\nu }^2{(\nu -2)}^2\underset{{\Omega }_{\delta }}{\int }{x}_j^4{\rho }^{2\nu -8}{\psi }^{2}d\mathbf{x};$$

(96)

(b) if $i\ne j,$ then

$$\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)=\left\{\begin{array}{c}2\nu (\nu -2){\rho }^{\nu -4}{x}_i{x}_j,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }\hfill \end{array}\right.$$

and

$$\underset{\Omega }{\int }{\left(\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\right)}^2{\psi }^{2}d\mathbf{x}\le 4{\nu }^2{(\nu -2)}^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}{x}_i^2{x}_j^2{\psi }^{2}d\mathbf{x}.$$

(97)

Summing the inequalities (95) over all $i,j=1,2$, and applying the estimates (96) and (97), $x_1^4+x_1^2{x}_2^2+x_2^4\le {\rho }^4$ and Corollary 2, we conclude

$$\sum _{i,j=1}^2\underset{\Omega }{\int }{\left(\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)\right)}^2{\psi }^{2}d\mathbf{x}\le 8{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{\psi }^{2}d\mathbf{x}+$$

$$+8{\nu }^2{(\nu -2)}^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}(x_1^4+x_1^2{x}_2^2+x_2^4){\psi }^{2}d\mathbf{x}\le 8(1+{(\nu -2)}^2){\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2,$$

hence

$$|\sum _{i,j=1}^2{J}_3^{i,j}|\le {\epsilon }_2\underset{\Omega }{\int }{\rho }^{-2\nu }\sum _{i,j=1}^2{\left[\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]}^{2}d\mathbf{x}+\frac{8(1+{(\nu -2)}^2)}{{\epsilon }_2}{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2.$$

(98)

By analogy with (93), applying the estimate (45) to the first term on the right-hand side (98), we derive

$$|\sum _{i,j=1}^2{J}_3^{i,j}|\le 2{\epsilon }_2{\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right)\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2+$$

$$+2{\epsilon }_2{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+\frac{8(1+{(\nu -2)}^2)}{{\epsilon }_2}{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2.$$

(99)

4. Using (13), for $\beta =2\nu ,$ we have

$${\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}=\left\{\begin{array}{c}2\nu {\rho }^{\nu -2}{x}_j,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.{\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)}^2=\left\{\begin{array}{c}4{\nu }^2{\rho }^{2(\nu -2)}{x}_j^2,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.$$

then

$$J_4^{i,j}=\underset{\Omega }{\int }\left[{\rho }^{-\nu }\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]\left[{\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\left(\frac{\partial \psi }{\partial x_i}\right)\right]d\mathbf{x}\le {\epsilon }_2\underset{\Omega }{\int }{\left({\rho }^{-\nu }\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)}^{2}d\mathbf{x}+$$

$$+\frac{1}{{\epsilon }_2}\underset{\Omega }{\int }{\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)}^2{\left(\frac{\partial \psi }{\partial x_i}\right)}^{2}d\mathbf{x}={\epsilon }_2\underset{\Omega }{\int }{\rho }^{-2\nu }{\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)}^{2}d\mathbf{x}+\frac{4{\nu }^2}{{\epsilon }_2}\underset{{\Omega }_{\delta }}{\int }{\rho }^{2(\nu -2)}{x}_j^2{\left(\frac{\partial \psi }{\partial x_i}\right)}^{2}d\mathbf{x}.$$

Summing over $i,j=1,2,$ applying Corollary 2 and using the equality ${\parallel \nabla \psi \parallel }_{L_{2,\nu }(\Omega )}^2={\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2,$ we conclude

$$|\sum _{i,j=1}^2{J}_4^{i,j}|\le {\epsilon }_2\sum _{i,j=1}^2\underset{\Omega }{\int }{\rho }^{-2\nu }{\left(\frac{\partial }{\partial x_i}\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right)}^{2}d\mathbf{x}+\frac{4}{{\epsilon }_2}{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(100)

By analogy with (93), applying the estimate (45) to the first term on the right-hand side (100), we get

$$|\sum _{i,j=1}^2{J}_4^{i,j}|\le 2{\epsilon }_2\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2+2{\epsilon }_2{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+\frac{4}{{\epsilon }_2}{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(101)

5. Due to the fact that $J_5^{i,j}=I_2^{i,j}$ (see Theorem 2) and applying the ${\epsilon }_2$ inequality, by analogy with (55), we have

$$|\sum _{i,j=1}^2{J}_5^{i,j}|\le {\epsilon }_2{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+\frac{{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }}{{\epsilon }_2}{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(102)

6. Due to the fact that $J_6^{i,j}=I_6^{i,j}$ (see Theorem 2), by analogy with (66), we conclude

$$|\sum _{i,j=1}^2{J}_6^{i,j}|\le \frac{{\nu }^2}{2}{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+\frac{{\nu }^2}{2}{C}_{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(103)

7. Using (13), with $\beta =2\nu$ and $\beta =-\nu ,$ we have

$${\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_i}=\left\{\begin{array}{c}2\nu {\rho }^{\nu -2}{x}_i,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.\frac{\partial {\rho }^{-\nu }}{\partial x_j}=\left\{\begin{array}{c}-\nu {\rho }^{-\nu -2}{x}_j,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }.\hfill \end{array}\right.$$

Consider two cases:

(a) if $i=j,$ then

$$\frac{\partial }{\partial x_j}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)=\left\{\begin{array}{c}2\nu {\rho }^{\nu -2}+2\nu (\nu -2){\rho }^{\nu -4}{x}_j^2,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.$$

$$\left(\frac{\partial {\rho }^{-\nu }}{\partial x_j}\right)\frac{\partial }{\partial x_j}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)=\left\{\begin{array}{c}-2{\nu }^2{\rho }^{-4}{x}_j-2{\nu }^2(\nu -2){\rho }^{-6}{x}_j^3,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }\hfill \end{array}\right.$$

and

$$|J_7^{j,j}|=|-\underset{{\Omega }_{\delta }}{\int }\left(\sqrt{2}\nu {\rho }^{\nu -2}\psi \right)\left(\sqrt{2}\nu {\rho }^{\nu -2}{x}_j\left[{\rho }^{-2\nu }\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]\right))d\mathbf{x}-$$

$$-\underset{{\Omega }_{\delta }}{\int }\left(\sqrt{2}\nu {\rho }^{\nu -4}{x}_j^2\psi \right)\left(\sqrt{2}\nu (\nu -2){\rho }^{\nu -2}{x}_j\left[{\rho }^{-2\nu }\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]\right)d\mathbf{x}|\le$$

$$\le {\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{\psi }^{2}d\mathbf{x}+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_j^2{\left[{\rho }^{-2\nu }\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]}^{2}d\mathbf{x}+$$

$$+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}{x}_j^4{\psi }^{2}d\mathbf{x}+{\nu }^2{(\nu -2)}^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_j^2{\left[{\rho }^{-2\nu }\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]}^{2}d\mathbf{x},$$

therefore

$$|J_7^{j,j}|\le {\nu }^2[\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{\psi }^{2}d\mathbf{x}+\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}{x}_j^4{\psi }^{2}d\mathbf{x}+$$

$$+{\nu }^2(1+{(\nu -2)}^2)\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_j^2{\left[{\rho }^{-2\nu }\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]}^{2}d\mathbf{x}];$$

(104)

(b) if $i\ne j,$ then

$$\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)=\left\{\begin{array}{c}2\nu (\nu -2){\rho }^{\nu -4}{x}_i{x}_j,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.$$

$$\left(\frac{\partial {\rho }^{-\nu }}{\partial x_i}\right)\frac{\partial }{\partial x_i}\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)=\left\{\begin{array}{c}-2{\nu }^2(\nu -2){\rho }^{-6}{x}_i^2{x}_j,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }\hfill \end{array}\right.$$

and

$$|J_7^{i,j}|=\left|-\underset{{\Omega }_{\delta }}{\int }\left(\sqrt{2}\nu {\rho }^{\nu -4}{x}_i{x}_j\psi \right)\left(\sqrt{2}\nu (\nu -2){\rho }^{\nu -2}{x}_i\left[{\rho }^{-2\nu }\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]\right)d\mathbf{x}\right|\le$$

$$\le {\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}{x}_i^2{x}_j^2{\psi }^{2}d\mathbf{x}+{\nu }^2{(\nu -2)}^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_i^2{\left[{\rho }^{-2\nu }\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]}^{2}d\mathbf{x},$$

we get

$$|J_7^{i,j}|\le {\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}{x}_i^2{x}_j^2{\psi }^{2}d\mathbf{x}+{\nu }^2{(\nu -2)}^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_i^2{\left[{\rho }^{-2\nu }\left(\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right)\right]}^{2}d\mathbf{x}.$$

(105)

Combining the inequalities (104) and (105) for $i,j=1,2,$ and using the definition of the vector function $\mathbf{v}$, Corollaries 1 and 2 for the components $\mathbf{v}$ and the function $\psi ,$ estimates (19) and (20), respectively, we conclude

$$|\sum _{i,j=1}^2{J}_7^{i,j}|\le {\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -8}(x_1^4+2x_1^2{x}_2^2+x_2^4){\psi }^{2}d\mathbf{x}+$$

$$+{\nu }^2\left[1+{(\nu -2)}^2\right]\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2)v_1^{2}d\mathbf{x}+{\nu }^2{(\nu -2)}^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2)v_2^{2}d\mathbf{x}\le$$

$$\le {\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2+{\nu }^2\left[1+{(\nu -2)}^2\right]C_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2,$$

hence

$$|\sum _{i,j=1}^2{J}_7^{i,j}|\le {\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2+{\nu }^2\left[1+{(\nu -2)}^2\right]C_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(106)

8. Using (13), with $\beta =2\nu$ and $\beta =-\nu ,$ we have

$${\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}=\left\{\begin{array}{c}2\nu {\rho }^{\nu -2}{x}_j,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta },\hfill \end{array}\right.\frac{\partial {\rho }^{-\nu }}{\partial x_i}=\left\{\begin{array}{c}-\nu {\rho }^{-\nu -2}{x}_i,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }\hfill \end{array}\right.$$

and

$$\left(\frac{\partial {\rho }^{-\nu }}{\partial x_i}\right)\left({\rho }^{-\nu }\frac{\partial {\rho }^{2\nu }}{\partial x_j}\right)=\left\{\begin{array}{c}-2{\nu }^2{\rho }^{-4}{x}_i{x}_j,\mathbf{x}\in {\Omega }_{\delta },\hfill \\ 0,\mathbf{x}\in \overline{\Omega }\setminus {\Omega }_{\delta }.\hfill \end{array}\right.$$

Consider two cases:

(a) if $i=j,$ then

$$|J_8^{j,j}|=\left|\underset{{\Omega }_{\delta }}{\int }\left(-\sqrt{2}\nu {\rho }^{\nu -2}{x}_j\frac{\partial \psi }{\partial x_j}\right)\left(\sqrt{2}\nu {\rho }^{\nu -2}{x}_j\left[{\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right]\right)d\mathbf{x}\right|\le {\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_j^2{\left(\frac{\partial \psi }{\partial x_j}\right)}^{2}d\mathbf{x}+$$

$$+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_j^2{\left[{\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right]}^{2}d\mathbf{x},$$

that is

$$|J_8^{j,j}|\le {\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_j^2{\left(\frac{\partial \psi }{\partial x_j}\right)}^{2}d\mathbf{x}+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_j^2{\left[{\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right]}^{2}d\mathbf{x};$$

(107)

(b) if $i\ne j,$ then

$$|J_8^{i,j}|=\left|\underset{{\Omega }_{\delta }}{\int }\left(-\sqrt{2}\nu {\rho }^{\nu -2}{x}_j\frac{\partial \psi }{\partial x_i}\right)\left(\sqrt{2}\nu {\rho }^{\nu -2}{x}_i\left[{\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right]\right)d\mathbf{x}\right|\le {\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_j^2{\left(\frac{\partial \psi }{\partial x_i}\right)}^{2}d\mathbf{x}+$$

$$+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_i^2{\left[{\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right]}^{2}d\mathbf{x},$$

that is

$$|J_8^{i,j}|\le {\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_j^2{\left(\frac{\partial \psi }{\partial x_i}\right)}^{2}d\mathbf{x}+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}{x}_i^2{\left[{\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_j}\right]}^{2}d\mathbf{x}.$$

(108)

Combining the inequalities (107) and (108) for $i,j=1,2,$ using the definition functions $\mathbf{u}$ and $\mathbf{v}$, Lemma 3 and Corollary 1 for their components, estimates (16) and (19), respectively, we conclude

$$|\sum _{i,j=1}^2{J}_8^{i,j}|\le {\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2){\left(\frac{\partial \psi }{\partial x_1}\right)}^{2}d\mathbf{x}+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2){\left(\frac{\partial \psi }{\partial x_2}\right)}^{2}d\mathbf{x}+$$

$$+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2){\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_1}\right)}^{2}d\mathbf{x}+{\nu }^2\underset{{\Omega }_{\delta }}{\int }{\rho }^{2\nu -4}(x_1^2+x_2^2){\left({\rho }^{-2\nu }\frac{\partial \left({\rho }^{2\nu }\psi \right)}{\partial x_2}\right)}^{2}d\mathbf{x}\le$$

$$\le {\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2,$$

hence

$$|\sum _{i,j=1}^2{J}_8^{i,j}|\le {\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(109)

Substituting the obtained estimates (92), (94), (99), (101)–(103), (106), (109) to (91), we get

$$\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2\le a(\mathbf{u},\mathbf{v})+[2{\epsilon }_2\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2+2{\epsilon }_2{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+$$

$$+\frac{{\nu }^2{C}_4^2{\delta }^{2\nu }}{{\epsilon }_2}{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2]+[2{\epsilon }_2\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2+2{\epsilon }_2{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+$$

$$+\frac{8(1+{(\nu -2)}^2)}{{\epsilon }_2}{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2]+[2{\epsilon }_2\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2+2{\epsilon }_2{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+$$

$$+\frac{4}{{\epsilon }_2}{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2]+\left[{\epsilon }_2{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+\frac{{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }}{{\epsilon }_2}{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\right]+[\frac{{\nu }^2}{2}{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+$$

$$+\frac{{\nu }^2}{2}{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2]+\left[{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2+{\nu }^2\left[1+{(\nu -2)}^2\right]C_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\right]+$$

$$+\left[{\nu }^2{C}_4^2{\delta }^{2\nu }{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }{\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\right],$$

that is

$$\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2\le a(\mathbf{u},\mathbf{v})+6{\epsilon }_2\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2+{\epsilon }_2{\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+$$

$$+{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }\left(4{\epsilon }_2+\frac{1}{{\epsilon }_2}+2.5+{(\nu -2)}^2\right){\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+$$

$$+{\nu }^2{C}_4^2{\delta }^{2\nu }\left(\frac{5}{{\epsilon }_2}+1.5\right){\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+{\nu }^2{C}_4^2{\delta }^{2\nu }\left(1+\frac{8}{{\epsilon }_2}(1+{(\nu -2)}^2)\right){\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2.$$

(110)

Evaluating the third term on the right-hand side (110) using Lemma 10 (see (83)), by analogy with (22), taking into account (14), we conclude

$${\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2\le {3\left|\mathbf{v}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2+48{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }(1+12C_5^2){\parallel \mathbf{v}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2\le 6{\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right)\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2+$$

$$+6{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }\parallel {\rho }^{\nu }{\mathbf{v}\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2+48{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }(1+12C_5^2){\parallel {\rho }^{\nu }\mathbf{v}\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2,$$

hence

$${\left|\mathbf{u}\right|}_{{\mathbf{W}}_{2,\nu }^1(\Omega )}^2\le \left[6\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2+64{\nu }^2{C}_4^2{C}_{\nu }^2{\delta }^{2\nu }(1+9C_5^2){\parallel {\rho }^{\nu }\mathbf{v}\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2\right].$$

(111)

To estimate the norm of the last term on the right-hand side (110), we apply Lemma 5, its inequality (27), then

$${\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2\le 4C_3^2{\parallel \nabla \psi \parallel }_{L_{2,\nu }(\Omega )}^2=4C_3^2{\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2.$$

(112)

To evaluate the last two terms on the right-hand side (110), we apply (112) in combination with the inequality (79) of Lemma 9, then

$${\nu }^2{C}_4^2{\delta }^{2\nu }\left[\left(\frac{5}{{\epsilon }_2}+1.5\right){\parallel \mathbf{u}\parallel }_{{\mathbf{L}}_{2,\nu }(\Omega )}^2+\left(1+\frac{8}{{\epsilon }_2}\left(1+{(\nu -2)}^2\right)\right){\parallel \psi \parallel }_{L_{2,\nu }(\Omega )}^2\right]\le$$

$$\le 4{\nu }^2{C}_4^2{\delta }^{2\nu }\left[\frac{5}{{\epsilon }_2}+1.5+4C_3^2\left(1+\frac{8}{{\epsilon }_2}(1+{(\nu -2)}^2)\right)\right]{\parallel {\rho }^{\nu }\mathbf{v}\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2.$$

(113)

Due to the fulfillment of Lemma 2, taking into account the inequalities (111) and (113), the estimate (110) takes the following form:

$$\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2\le a(\mathbf{u},\mathbf{v})+12{\epsilon }_2\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2+D(\delta ,\nu ,{\epsilon }_2){\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right)\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2,$$

where $D(\delta ,\nu ,{\epsilon }_2)={\nu }^2{C}_3^2{C}_4^2{\delta }^{2\nu }{D}_0(\nu ,{\epsilon }_2)$ and $D_0(\nu ,{\epsilon }_2)=C_{\nu }^2\left[64{\epsilon }_2(1+9C_5^2)+4{\epsilon }_2+\frac{1}{{\epsilon }_2}+2.5+{(\nu -2)}^2\right]+\frac{20}{{\epsilon }_2}+6+16C_3^2\left[1+\frac{8}{{\epsilon }_2}(1+{(\nu -2)}^2)\right].$

If we choose ${\epsilon }_2=\frac{1}{24},$ then for $\nu >0,$ there exists ${\delta }_3={\delta }_3({\epsilon }_2,\nu )>0$, such that $D({\delta }_3,\nu ,{\epsilon }_2)=\frac{1}{4}$ and for any $\delta \in (0,min\{{\delta }_2,{\delta }_3\})$:

$$\frac{1}{4}\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right){\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2\le \left(1-\frac{1}{2}-D(\delta ,\nu ,{\epsilon }_2)\right){\parallel \nabla \left({\rho }^{\nu }\mathbf{v}\right)\parallel }_{{\mathbf{L}}_{2,0}(\Omega )}^2\le a(\mathbf{u},\mathbf{v}).$$

Theorem 4 is proved. □

4. Conclusions

In the present paper, an $R_{\nu }$-generalized solution of the Stokes problem with a corner singularity in a nonsymmetric variational formulation is defined. The properties of functions from special sets of the corresponding operators of the variational formulation are proved. The statements established in the paper will contribute to the study of the existence and uniqueness of the $R_{\nu }$-generalized solution in weighted sets. The results and methods of the paper are supposed to be generalized to other problems of hydrodynamics with a corner singularity in a nonsymmetric variational formulation. In particular, for solving problems with the mixed boundary conditions. In the case when the domain filled up with fluid. One part of the boundary is a fixed wall and the other one is both the input and output of the channel. For example, when the homogeneous Dirichlet condition is set on the first part expressing a no-slip behavior of the fluid on the fixed walls of the channel. On the second part a condition $p\mathbf{n}+\frac{\partial \mathbf{w}}{\partial \mathbf{n}}=\mathbf{0},$ where $\mathbf{n}$ is the outer normal vector expresses the “do nothing“ boundary condition.

Previously, we assumed that an $R_{\nu }$-generalized solution of hydrodynamic problems (see, for example, [18,20]) exists and is unique in the corresponding weighted sets. So certain decision makes it possible to create an efficient numerical approach—weighted FEM for finding an approximate solution of the problem with high accuracy. Using the weighted method, the ranges for choosing the optimal approach parameters, such as $\delta$, $\nu$ and ${\nu }^{*}$ (${\nu }^{*}$ is the exponent of the weight function in the FE basis) are experimentally established depending on the value of the reentrant corner to achieve convergence $\mathcal{O}\left(h\right)$. We have established that the order of convergence does not depend on the value of the reentrant corner for the Stokes [19], Oseen [20] and elasticity theory problems (see, for example, [15]) in the case when the Dirichlet conditions are set on the boundary. In addition, a numerical analysis of the elasticity problem [33] was carried out in the case when the Dirichlet boundary condition is set on one side of the reentrant corner, and the Neumann condition on the other one. As it is known [2], in this case the classical FEM loses its order of accuracy twice as compared to when the Dirichlet-Dirichlet or Neumann-Neumann conditions are given on the sides of the reentrant corner. If we apply a weighted FEM based on the concept of the $R_{\nu }$-generalized solution, then there is no loss of accuracy. Moreover, the approximate solution (see [33]) converges to the exact one with the first order with respect to the grid step, regardless of the reentrant corner. The weighted FEM is simple to implement and does not require mesh refinement in the vicinity of the singularity point.