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This article is cited in 1 scientific paper (total in 1 paper) Bojarski–Meyers estimate for a solution to the Zaremba problem for Poisson's equations with drift Yu. A. Alkhutova, G. A. Chechkinbcd a Vladimir State University, Vladimir, Russia b Faculty of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia c Institute of Mathematics with Computing Centre, Ufa Federal Research Centre of the Russian Academy of Sciences, Ufa, Russia d Institute of Mathematics and Mathematical Modeling, Almaty, Kazakhstan
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  § 1. IntroductionThis paper was written in the memory of the remarkable scientist Arlen Mikhailovich Il’in, on the occasion of the 90th anniversary of his birth. The problem of the increased integrability of the gradient of solutions to elliptic equations is classical and goes back to [1], where the case of linear divergent uniformly elliptic equations of the second order with measurable coefficients was considered. Subsequently, in the multidimensional case of equations of the same type, the increased integrability of the gradient of a solution to the Dirichlet problem in a domain with sufficiently regular boundary was established in [2]. Since then, estimates for the increased integrability of the gradient of solutions are generally called Meyers-type estimates. A Meyers-type estimate for a solution to the Dirichlet problem in a domain with Lipschitz boundary for the $p$-Laplace equation with variable exponent $p$ possessing a logarithmic modulus of continuity was first obtained in [3]. Then in [4] and [5] this result was refined and extended to systems of elliptic equations with a variable exponent of integrability. Note that in [3] the study of Meyers’ estimates was prompted by the thermistor problem, in which the electric field potential and temperature are simultaneously described (see [3], [6], and [7]). Systems of the same type also arise in the hydromechanics of quasi-Newtonian fluids. For Laplace’s equation the mixed Zaremba problem (proposed by Wirtinger) in a three-dimensional bounded domain with smooth boundary and inhomogeneous Dirichlet and Neumann conditions was first considered in [8]. The classical solvability of this problem was established by methods of potential theory, under the assumption that the boundary of the open set on which the Neumann data are prescribed also has a certain smoothness. The study of the properties of solutions of the Zaremba problem for second-order elliptic equations with variable regular coefficients goes back to [9]. In particular, it was established there that solutions lose their smoothness at the interface between the Dirichlet and Neumann data. For divergent uniformly elliptic second-order equations with measurable coefficients, integral and pointwise bounds for solutions of the Zaremba problem under fairly general assumptions about the boundary of the domain were presented in [10]. The problem of estimates for the increased integrability of the gradient of the solution to Zaremba’s problem was studied in [11]–[14], where an estimate for the increased integrability of the gradient of the solution to Zaremba’s problem was obtained for a linear elliptic equation in divergent form in domains with Lipschitz boundary and frequent alternation of Dirichlet and Neumann boundary conditions, with an increased exponent of integrability independent of the frequency of this alternation. Estimates of this type are important in the theory of homogenization of problems with frequent alternation of boundary conditions; they make it possible to improve the rate of convergence of pre-limiting solutions to the solution of the homogenized problem (see a similar problem in a domain perforated along the boundary in [15]). Similar estimates for the $p$-Laplacian were obtained in [16]. Our paper is devoted to estimates for solutions of the Zaremba problem for Poisson’s equation with lower-order terms defined in a bounded domain $D\in \mathbb{R}^n$ with Lipschitz boundary, where $n>1$. The increased integrability of the gradient of the solution to the Zaremba problem in the presence of lower-order terms has not been considered yet. Before stating the Zaremba problem, we introduce the Sobolev space of functions $W^1_2(D,F)$, where $F\subset \partial D$ is a closed set, as the completion of the set of infinitely differentiable functions in the closure of $D$ that are equal to zero in a neighbourhood of $F$, with respect to the norm
$$
\begin{equation*}
\| u\|_{W^{1}_2(D,F)}=\biggl (\int_{D} v^2\,dx+\int_{D}|\nabla v|^2\,dx\biggr)^{1/2}.
\end{equation*}
\notag
$$
A priori, functions $v\in W^1_2(D,F)$ are assumed to satisfy Friedrichs’s inequality: the validity of which is discussed below, depending on the conditions on the compact set $F$. Setting $G=\partial D\setminus F$ we consider the Zaremba problem
$$
\begin{equation}
\mathcal{L} u:=\Delta u +b\cdot\nabla u= l \quad\text{in } D, \qquad u=0 \quad\text{on } F, \qquad \frac{\partial u}{\partial \nu}=0 \quad\text{on } G,
\end{equation}
\tag{1.2}
$$
where ${\partial u}/{\partial \nu}$ denotes the outward normal derivative of $u$, and $l$ is a linear functional in the dual space of $W^1_2(D,F)$. The vector function $b(x)\mkern-2mu=\mkern-2mu(b_1(x), \dots, b_n(x))$ in (1.2) satisfies the following conditions:
$$
\begin{equation}
b_j(x)\in L_p(D), \quad p=n, \quad\text{if } n>2, \quad j=1,\dots, n,
\end{equation}
\tag{1.3}
$$
$$
\begin{equation}
b_j(x)\in L_p(D), \quad p>2, \quad \text{if } n=2, \quad j=1,2.
\end{equation}
\tag{1.4}
$$
By a solution of problem (1.2) we mean a function $u\in W^1_2(D,F)$ for which the integral identity
$$
\begin{equation}
\int_D \nabla u\cdot\nabla\varphi\,dx-\int_D (b\cdot\nabla u)\varphi\,dx =-l(\varphi)
\end{equation}
\tag{1.5}
$$
holds for all test functions $\varphi\in W^1_2(D,F)$. By Friedrichs’s inequality (1.1) the space $W^1_2(D,F)$ can be equipped with a norm involving the gradient alone. Then to every element of the Sobolev space one can assign, by a one-to-one isometric correspondence, its gradient, that is, an element of $(L_2(D))^n$. Using Riesz’ theorem on the representation of a linear continuous functional in a Hilbert space, it is easy to show that the functional $l$ can be written in the form
$$
\begin{equation}
l(\varphi)= -\sum_{i=1}^n\int_D f_i\varphi_{x_i}\,dx,
\end{equation}
\tag{1.6}
$$
where $f_i\in L_2(D)$. Therefore, by virtue of (1.5), for any particular functional a solution of (1.2) can be treated in the sense of the integral relation
$$
\begin{equation}
\int_D \nabla u\cdot\nabla\varphi\,dx-\int_D (b\cdot\nabla u)\varphi\,dx=\int_D f\cdot\nabla\varphi\,dx
\end{equation}
\tag{1.7}
$$
for all test functions $\varphi\in W^1_2(D,F)$, where components of the vector-valued function $f=(f_1,\dots,f_n)$ are functions in $L_2(D)$. For $n>2$ we will use the following representation for the lower-order coefficients $b\in (L_n(D))^n$ of the equations under consideration:
$$
\begin{equation}
b=\breve{b}+\widehat{b}, \quad\text{where }\breve{b}\in (L_\infty (D))^n, \quad \widehat{b}\in (L_n(D))^n, \quad\| \widehat{b} \|_{L_n(D)}\leqslant\sigma,
\end{equation}
\tag{1.8}
$$
where $\sigma\in (0,1)$ is a sufficiently small constant to be specified below. In this paper we assume that the Zaremba problem (1.2) is uniquely solvable. A theorem on the existence and uniqueness of solutions was proved in [17]. § 2. Main resultsWe are interested in the increased integrability of the gradients of solutions to (1.2). A condition concerning the structure of the support set of the Dirichlet data $F$ plays a key role here. To formulate the result we need the concept of capacity. Given a compact set $K\subset \mathbb{R}^n$, we define the capacity $C_q(K)$, where $1<q<n$, by the equality
$$
\begin{equation}
C_q(K)=\inf\biggl \{\int_{\mathbb{R}^n}|\nabla\varphi|^q\,dx\colon\varphi\in C^\infty_0 (\mathbb{R}^n),\, \varphi\geqslant 1\text{ on }K\biggr \}.
\end{equation}
\tag{2.1}
$$
Below $B^{x_0}_r$ denotes an open $n$-ball of radius $r$ with centre $x_0$. For $n>2$ the following condition is assumed to hold: for an arbitrary point $x_0\in F$ and $r\leqslant r_0$, the inequality
$$
\begin{equation}
C_q(F\cap \overline B^{x_0}_r)\geqslant c_0 r^{n-q}, \qquad q=\frac{2n}{n+2},
\end{equation}
\tag{2.2}
$$
holds, and for $n=2$ and $p>2$ we assume that
$$
\begin{equation}
C_q(F\cap \overline B^{x_0}_r)\geqslant c_0 r^{2-q_1}, \qquad q_1=\frac{p}{p-1},
\end{equation}
\tag{2.3}
$$
where the positive constant $c_0$ does not depend on $x_0$ and $r$. Note that, if condition (2.2) or (2.3) is satisfied for $v\in W^1_2(D,F)$, then Friedrichs’s inequality (1.1) holds, which follows from results due to Maz’ya (see [18], § 14.1.2) and is discussed in detail below. We dwell on this point, denoting by $\mathcal{Q}_d$ an open cube with edge length $d$ and faces parallel to the coordinate axes, assuming that the domain $D$ with Lipschitz boundary has diameter $d$ and $D\subset \mathcal{Q}_d$. Next, we need the notion of the capacity $C_q(K,\mathcal{Q}_{2d})$ of a compact set $K\subset\overline {\mathcal{Q}}_d$ with respect to the cube $\mathcal{Q}_{2d}$, which is defined by the equality
$$
\begin{equation}
C_q(K,\mathcal{Q}_{2d})=\inf\biggl \{\int_{\mathcal{Q}_{2d}}|\nabla\varphi|^q\,dx\colon \varphi\in C^\infty_0 (\mathcal{Q}_{2d}),\, \varphi\geqslant 1\text{ on }K\biggr \}.
\end{equation}
\tag{2.4}
$$
From the theorem in [18], § 14.1.2, and the comments to it, it follows, in particular, that for functions $v\in W^1_2(D,F)$ the inequality
$$
\begin{equation}
\int_D v^2\,dx\leqslant \frac{C(n,D)d^n}{C_2(F,\mathcal{Q}_{2d})}\int_D |\nabla v|^2\,dx
\end{equation}
\tag{2.5}
$$
holds, which is called Maz’ya’s inequality. Here $C(n,D)$ is a constant. Below we use condition (2.2) (the reasoning for (2.3) is similar). First we note that if $1<q<2$, then by the definition of the capacity $C_q(K,\mathcal{Q}_{2d})$ and Hölder’s inequality we have
$$
\begin{equation}
C_q(K,\mathcal{Q}_{2d})\leqslant |\mathcal{Q}_{2d}|^{(2-q)/2}C_2^{q/2}(K,\mathcal{Q}_{2d}),
\end{equation}
\tag{2.6}
$$
where $|\mathcal{Q}_{2d}|$ denotes the $n$-dimensional measure of the cube $\mathcal{Q}_{2d}$. Now we use the fact that for $1<q<n$ (see [19], Proposition 4) there exists a constant $\gamma(n,q)\geqslant 1$ such that
$$
\begin{equation}
C_q(K)\leqslant C_q(K,\mathcal{Q}_{2d})\leqslant \gamma\, C_q(K).
\end{equation}
\tag{2.7}
$$
It follows from (2.2) and the monotonicity of the capacity with respect to the domain that $C_q(F)>0$. Therefore, $C_2(F,\mathcal{Q}_{2d})>0$ too, by (2.6) and (2.7). Now, using (2.5) we arrive at Friedrichs’s inequality (1.1). To formulate our result and explain the scheme of the proof of the main assertion, we need a more detailed explanation of how we define a domain $D$ with Lipschitz boundary. Let $Q$ be a cube with centre $x_0\in \partial D$. We introduce a Cartesian coordinate system with origin at $x_0$ in which edges of the cube are parallel to coordinate axes and have length $2R_0$. We say that the boundary of the domain $D$ is strictly Lipschitz if, for every point $x_0\in\partial D$, the set $Q\cap\partial D$ is the graph of a Lipschitz function $x_n=g(x')$, where $x'=(x_1, \dots, x_{n-1})$, with Lipschitz constant $L$. Here we assume that the edge length of $Q$ and the Lipschitz constant $L$ do not depend on $x_0$. For definiteness we assume that the set $Q\cap D$ lies over the graph of $g$. Now we formulate the main assertions, assuming that the constant $r_0$ in (2.2) or (2.3) does not exceed the constant $R_0$ involved in the definition of the domain with Lipschitz boundary. Theorem 1. If $n>2$ and condition (1.3) is satisfied, and if $f\in \big(L_{2+\delta_0}(D)\big)^n$, where $\delta_0>0$, then there exist positive constants $\delta<\delta_0$ and $C$ such that the following bound holds for the solution of problem (1.2): Theorem 2. If $n=2$, condition (1.4) is satisfied, and $f\in \big(L_{2+\delta_0}(D)\big)^2$, where $\delta_0>0$, then there exist positive constants $\delta<\delta_0$ and $C$ such that the bound Remark 2. The theorem remains valid if, instead of the Laplace operator, we consider a second-order uniformly elliptic linear operator of the form where ${A}(x)$ is a uniformly elliptic measurable symmetric matrix. In this case the constants $C$ and $\delta$ in (2.8) and (2.9) depend also on the ellipticity constants of $A$. The proof of Theorems 1 and 2 uses interior and boundary estimates for the increased integrability of the gradient of a solutions of problem (1.2). First, a bound for the increased integrability of the gradient of a solution of (1.2) is established in a neighbourhood of the boundary of $D$. The technique of local straightening of the boundary $\partial D$ is used here. § 3. Auxiliary constructionsAssuming that $Q_{R_0}=\{x\colon |x_i|<R_0,\, i=1,\dots,n\}$ and using the definition of a domain with Lipschitz boundary, for an arbitrary point $x_0\in\partial D$ we consider a local Cartesian coordinate system with origin at $x_0$ such that the part of the boundary $\partial D$ that occurs in the cube $Q_{R_0}$ is defined in this coordinate system by the equation $x_n=g(x')$, where $g$ is a Lipschitz function with Lipschitz constant $L$. It is assumed that the domain $D_{R_0}=Q_{R_0}\cap D$ lies in the set of the points where $x_n>g(x')$. Next, we go over to a new coordinate system in $Q_{R_0}$ by performing the nondegenerate transformation of variables It is clear that the part $Q_{R_0}\cap\partial D$ of the boundary is transformed into a piece of a hyperplane:
$$
\begin{equation*}
P_{R_0}=\{y\colon|y_i|<R_0,\, i=1,\dots,n-1,\, y_n=0\}.
\end{equation*}
\notag
$$
For what follows we note that the domain $\widetilde Q_{R_0}$ (the image of $Q_{R_0}$ under this coordinate transformation) contains the cube
$$
\begin{equation}
K_{R_0}=\bigl\{y\colon|y_i|<\bigl(1+\sqrt{n-1} L\bigr)^{-1}R_0,\, i=1,\dots,n\bigr\}.
\end{equation}
\tag{3.2}
$$
In this case, in the half-cube $K^+_{R_0}=K_{R_0}\cap \{y\colon y_n>0\}$, which is contained in the image of $D\cap Q_{R_0}$, the problem (1.2), for whose solution we keep the original notation, acquires the form
$$
\begin{equation}
\widetilde{\mathcal{L}}u=\widetilde l \quad \text{in } K^+_{R_0}, \qquad u=0 \quad\text{on } \widetilde F_{R_0}, \qquad \frac{\partial u}{\partial \widetilde\nu}=0 \quad\text{on } \widetilde G_{R_0}.
\end{equation}
\tag{3.3}
$$
Here
$$
\begin{equation}
\widetilde{\mathcal{L}}u:=\operatorname{div} (a(y)\nabla u)+\widetilde b\cdot \nabla u
\end{equation}
\tag{3.4}
$$
is a uniformly elliptic operator with symmetric matrix $a(y)=\{a_{ij}(y)\}$ whose ellipticity constant depends on the Lipschitz constant $L$ of $g$, and the vector-valued function $f$ participating in the notation for the functional (1.6) is transformed into a function $\widetilde f$ whose components are defined by the equalities
$$
\begin{equation}
\begin{gathered} \, \widetilde f(y)=(\widetilde f_1(y),\dots, \widetilde f_n(y)), \quad \text{where } \widetilde f_i(y)= f_i(y',y_n+g(y')) \quad\text{for } i=1,\dots,n-1, \\ \widetilde f_n(y)=\sum_{i=1}^{n-1}\frac{\partial g(y')}{\partial y_i} f_i(y',y_n+g(y'))+f_n(y',y_n+g(y')). \end{gathered}
\end{equation}
\tag{3.5}
$$
The lower-order coefficients of the operator (3.4) are written out similarly:
$$
\begin{equation}
\begin{gathered} \, \widetilde b(y)=(\widetilde b_1(y),\dots, \widetilde b_n(y)), \quad\text{where }\ \widetilde b_i(y)= b_i(y',y_n+g(y')) \quad\text{for } i=1,\dots,n-1, \\ \widetilde b_n(y)=\sum_{i=1}^{n-1}\frac{\partial g(y')}{\partial y_i}b_i(y',y_n+g(y'))+b_n(y',y_n+g(y')). \end{gathered}
\end{equation}
\tag{3.6}
$$
The sets $\widetilde F_{R_0}$ and $\widetilde G_{R_0}$ in (3.3) have the form $\widetilde F_{R_0}=\widetilde F\cap P_{R_0}\cap K_{R_0}$ and $\widetilde G_{R_0}=\widetilde G\cap P_{R_0}\cap K_{R_0}$, where $\widetilde F$ and $\widetilde G$ are the images of the sets $F\cap Q_{R_0}$ and $G\cap Q_{R_0}$, respectively, and ${\partial u}/{\partial \widetilde\nu}=\sum_{i,j=1}^{n} a_{ij} ({\partial u}/{\partial y_j})\nu_j$ denotes the outward conormal derivative of $u$ associated with the operator (3.4). We extend the function $u$ satisfying (3.3) to an even function with respect to the hyperplane $\{y\colon y_n=0\}$. The extended function, for which we also keep the previous notation, satisfies the relation
$$
\begin{equation}
\widetilde{\mathcal{L}}_1 u=\operatorname{div} (c(y)\nabla u)+k\cdot\nabla u=l_h \quad \text{in } K_{R_0}\setminus \widetilde F_{R_0}, \qquad u=0 \quad\text{on } \widetilde F_{R_0}.
\end{equation}
\tag{3.7}
$$
Here the positive definite matrix $c=\{c_{ij}(y)\}$ consists of the elements $c_{jn}(y)=c_{jn}(y)$ for $j\ne n$ that are the odd extension of the elements of the matrix $a_{jn}(y)$ in (3.4), and all other elements $c_{ij}(y)$ are the even extensions of the $a_{ij}(y)$. The components of the vector-valued function $h=( h_1,\dots,h_n)$ in (3.7), involved in the representation for the functional $l_h$, are defined by similar equalities: its components $h_i(y)$ for $i=1,\dots,n-1$ are the even continuations of the components $\widetilde f_i(y)$ in (3.3), and $h_n(y)$ is the odd continuation of $\widetilde f_n(y)$. Similarly, the components of the vector function $k=(k_1,\dots,k_n)$ in (3.7) are defined by the following equalities: for $i=1,\dots,n-1$ its components $k_i(y)$ are the even extensions of the $\widetilde b_i(y)$ in (3.6), and $k_n(y)$ is the odd extension of $\widetilde b_n(y)$. It is easy to see that after this extension the matrix $c$ is uniformly elliptic and satisfies the same bounds as the matrix $a$, namely,
$$
\begin{equation}
\alpha^{-1}|\xi|^2\leqslant \sum_{i,j=1}^nc_{ij}(y)\xi_i\xi_j\leqslant \alpha |\xi|^2 \quad\text{for almost all } y\in K_{R_0} \text{ and for all } \xi\in \mathbb{R}^n.
\end{equation}
\tag{3.8}
$$
It is clear that the function $u\in W^1_2(K_{R_0})$ in (3.7) satisfies the integral relation (see (1.7))
$$
\begin{equation}
\int_{K_{R_0}} c(y)\nabla u\cdot\nabla\varphi\,dy-\int_{K_{R_0}} (k\cdot\nabla u)\varphi\,dy=\int_ {K_{R_0}}h\cdot\nabla\varphi\,dy
\end{equation}
\tag{3.9}
$$
for all functions $\varphi\in W^1_2(K_{R_0},\widetilde F_{R_0}\cup \partial K_{R_0})$ that lie in the closure of the set of functions infinitely differentiable in the closure of $K_{R_0}$ (and vanish in a neighbourhood of $\partial K_{R_0}$ and $\widetilde F_{R_0}$) with respect to the norm
$$
\begin{equation*}
\| u\|_{W^{1}_2(K_{R_0},F_{R_0})} =\biggl (\int_{K_{R_0}}u^2\, dx+\int_{K_{R_0}}|\nabla u|^2\,dx\biggr)^{1/2}.
\end{equation*}
\notag
$$
We denote by $Q_R^{y_0}$ the open cube with centre $y_0$ and edges of length $2R$ that are parallel to coordinate axes. Below we assume that
$$
\begin{equation*}
y_0\in K_{R_0/2}\setminus\partial K_{R_0/2}\quad\text{and} \quad R\leqslant \frac{1}{2}\operatorname{dist}(y_0,\partial K_{R_0}),
\end{equation*}
\notag
$$
and also that
$$
\begin{equation*}
\,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_R}f\, dx=\frac{1}{|Q^{y_0}_R|}\int_{Q^{y_0}_R}f\,dx,
\end{equation*}
\notag
$$
where $|Q^{y_0}_R|$ denotes the $n$-dimensional measure of $Q^{y_0}_R$. First, we consider the case when $Q^{y_0}_{3R/2}\subset K_{R_0}\setminus \widetilde F_{R_0} $ and choose the test function $\varphi=(u-\lambda)\eta^2$ in the integral identity (3.9), where
$$
\begin{equation*}
\lambda= \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{3R/2}} u\,dy,
\end{equation*}
\notag
$$
and the cutoff function $\eta\in C_0^{\infty}(Q^{y_0}_{3R/2})$ is such that $0<\eta\leqslant 1$, $\eta=1$ in $Q^{y_0}_{R}$ and $|\nabla \eta|\leqslant CR^{-1}$. Then (3.9) is transformed into
$$
\begin{equation}
\begin{aligned} \, \notag \int_{{Q^{y_0}_{3R/2}}} (c(y)\nabla u\cdot \nabla u) \eta^2\,dy &=\int_{{Q^{y_0}_{3R/2}}} (k\cdot\nabla u)(u-\lambda)\eta^2\,dy \\ \notag &\qquad -2\int_{{Q^{y_0}_{3R/2}}} c(y)\eta(u-\lambda)\nabla u\cdot\nabla \eta\,dy \\ &\qquad +2\int_ {{Q^{y_0}_{3R/2}}}h(u-\lambda)\eta\cdot\nabla\eta\,dy +\int_ {{Q^{y_0}_{3R/2}}}\eta^2 h\cdot\nabla u\,dy. \end{aligned}
\end{equation}
\tag{3.10}
$$
§ 4. Proof of the main assertions Proof of Theorem 1. First of all, we estimate the first term on the right-hand side of identity (3.10), bearing in mind that $n>2$. It follows from the representation (1.8) for the lower coefficients and (3.6) that the lower-order coefficients $k$ in the transformed equation (3.7) have the form
Therefore,
$$
\begin{equation}
\begin{aligned} \, \notag &\int_{Q^{y_0}_{3R/2}} (k\cdot\nabla u)(u-\lambda)\eta^2\,dy =\int_{Q^{y_0}_{3R/2}}(\breve{k}\cdot\nabla u)(u-\lambda)\eta^2\,dy \\ &\qquad\qquad +\int_{Q^{y_0}_{3R/2}}(\widehat{k}\cdot\nabla u)(u-\lambda)\eta^2\,dy. \end{aligned}
\end{equation}
\tag{4.2}
$$
For the first integral on the right-hand side of (4.2) we have
$$
\begin{equation*}
\biggl|\int_{Q^{y_0}_{3R/2}}(\breve{k}\cdot\nabla u)(u-\lambda)\eta^2\,dy\biggr| \leqslant R \| \breve{k}\|_{L_\infty(D)} \int_{Q^{y_0}_{3R/2}}|\nabla u|\, \biggl|\frac{u-\lambda}{R}\biggr |\eta^2\,dy.
\end{equation*}
\notag
$$
Since $R\leqslant R_0$ and $0\leqslant\eta\leqslant 1$, using Cauchy’s inequality for $\varepsilon>0$ we obtain
$$
\begin{equation}
\begin{aligned} \, \notag &\biggl|\int_{Q^{y_0}_{3R/2}}(\breve{k}\cdot\nabla u)(u-\lambda)\eta^2\,dy\biggr| \\ &\qquad \leqslant \varepsilon \int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\,dy+C(\varepsilon)R_0^2\| \breve{k}\|_{L_\infty(D)}^2R^{-2}\int_{Q^{y_0}_{2R}} (u-\lambda)^2\,dy. \end{aligned}
\end{equation}
\tag{4.3}
$$
We estimate the other integral on the right-hand side of (4.2) using Hölder’s inequality, which shows that
$$
\begin{equation}
\begin{aligned} \, \notag &\biggl|\int_{Q^{y_0}_{3R/2}} (\widehat{k}\cdot\nabla u)(u-\lambda)\eta^2\,dy\biggr| \\ \notag &\quad \leqslant R \biggl (\int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\, dy\biggr)^{1/2}\biggl (\ \int_{Q^{y_0}_{3R/2}}|\widehat{k}|^2\eta^2\left (\frac{u-\lambda}{R}\right)^2\,dy\biggr)^{1/2} \\ &\quad \leqslant R\| \widehat{k} \|_{L_n(D)} \biggl (\int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\, dy\biggr)^{1/2} \biggl (\int_{Q^{y_0}_{3R/2}} \biggl|\frac{(u-\lambda)\eta}{R}\biggr|^{2n/(n-2)}\,dy\biggr)^{(n-2)/(2n)}. \end{aligned}
\end{equation}
\tag{4.4}
$$
By Sobolev’s inequality
$$
\begin{equation}
\begin{aligned} \, \notag &\biggl (\int_{Q^{y_0}_{3R/2}} \biggl|\frac{(u-\lambda)\eta}{R}\biggr|^{2n/(n-2)}\,dy\biggr)^{(n-2)/(2n)} \\ &\qquad \leqslant C(n) \biggl (\biggl(\int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\,dy\biggr)^{1/2} +\biggl ( \int_{Q^{y_0}_{3R/2}}(u-\lambda)^2|\nabla \eta|^2\,dy\biggr)^{1/2}\biggr). \end{aligned}
\end{equation}
\tag{4.5}
$$
Since $R\leqslant R_0$, we see from (4.4), (4.5) and representation (4.1) that
$$
\begin{equation*}
\begin{aligned} \, &\biggl|\int_{Q^{y_0}_{3R/2}} (\widehat{k}\cdot\nabla u)(u-\lambda)\eta^2\,dy\biggr| \\ &\qquad \leqslant C(L,n,R_0)\sigma \biggl (\int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\, dy\biggr)^{1/2}\biggl(\biggl (\int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\,dy\biggr)^{1/2} \\ &\qquad\qquad + \biggl (\int_{Q^{y_0}_{3R/2}}(u-\lambda)^2|\nabla \eta|^2\,dy\biggr)^{1/2}\biggr), \end{aligned}
\end{equation*}
\notag
$$
which yields
$$
\begin{equation*}
\begin{aligned} \, &\biggl|\int_{Q^{y_0}_{3R/2}} (\widehat{k}\cdot\nabla u)(u-\lambda)\eta^2\,dy\biggr| \leqslant C(L,n,R_0)\sigma \int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\, dy \\ &\qquad\qquad +C(L,n,R_0)\sigma\biggl(\int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\,dy\biggr)^{1/2} \biggl ( \int_{Q^{y_0}_{3R/2}}(u-\lambda)^2|\nabla \eta|^2\,dy\biggr)^{1/2}. \end{aligned}
\end{equation*}
\notag
$$
From this, by Cauchy’s inequality we obtain
$$
\begin{equation}
\begin{aligned} \, \notag &\biggl|\int_{Q^{y_0}_{3R/2}} (\widehat{k}\cdot\nabla u)(u-\lambda)\eta^2\,dy\biggr| \\ &\qquad \leqslant C(L,n,R_0)\sigma \biggl (\int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\, dy +\int_{Q^{y_0}_{3R/2}}(u-\lambda)^2|\nabla \eta|^2\,dy \biggr). \end{aligned}
\end{equation}
\tag{4.6}
$$
As a result, from (4.3) and (4.6), by virtue of (4.2), taking the choice of the cutoff $\eta$ into account, we obtain
$$
\begin{equation}
\begin{aligned} \, \notag &\biggl |\int_{Q^{y_0}_{3R/2}} (k\cdot\nabla u)(u-\lambda)\eta^2\,dy\biggr |\leqslant (\varepsilon+C(L,n,R_0)\sigma) \int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\, dy \\ &\qquad\qquad +(C(\varepsilon)\| \breve{k}\|_{L_\infty(D)}^2+ C(L,n,R_0)\sigma) R^{-2} \int_{Q^{y_0}_{3R/2}}(u-\lambda)^2\,dy. \end{aligned}
\end{equation}
\tag{4.7}
$$
Let us now estimate the remaining integrals on the right-hand side of (3.10). For the second term we have
$$
\begin{equation}
\begin{aligned} \, \notag &\biggl|\int_{Q^{y_0}_{3R/2}} c(y)\eta(u-\lambda)\nabla u\cdot\nabla \eta\,dy\biggr| \\ &\qquad \leqslant \varepsilon \int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\,dy +\frac{C(\varepsilon)}{R^2} \int_{Q^{y_0}_{3R/2}} (u-\lambda)^2\,dy. \end{aligned}
\end{equation}
\tag{4.8}
$$
The third term is estimated as follows:
$$
\begin{equation}
\biggl|\int_{Q^{y_0}_{3R/2}}h(u-\lambda)\eta\cdot\nabla\eta\,dy\biggr| \leqslant \int_{Q^{y_0}_{3R/2}}|h|^2\, dy +\frac{C}{R^2}\int_{Q^{y_0}_{3R/2}} (u-\lambda)^2\, dy.
\end{equation}
\tag{4.9}
$$
For the fourth we derive that
$$
\begin{equation}
\biggl|\int_{Q^{y_0}_{3R/2}}\eta^2 h\cdot\nabla u\,dy\biggr| \leqslant \varepsilon\int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\, dy +C(\varepsilon)\int_{Q^{y_0}_{3R/2}} |h|^2\, dy.
\end{equation}
\tag{4.10}
$$
As a result, using (3.10), taking the last bounds (4.7)–(4.10) and (3.8) into account, for an appropriate choice of $\varepsilon$ and $\sigma$ we arrive at the inequality
$$
\begin{equation}
\int_{Q^{y_0}_R}|\nabla u|^2\, dy \leqslant C(n,k,\alpha,L,R_0)\biggl (\frac{1}{R^{2}} \int_{Q^{y_0}_{3R/2}}(u-\lambda)^2\, dy+\int_{Q^{y_0}_{3R/2}}|h|^2\, dy\biggr).
\end{equation}
\tag{4.11}
$$
Further, we find from the Poincaré–Sobolev inequality
$$
\begin{equation*}
\biggl ( \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{3R/2}}(u-\lambda)^2\,dx\biggr)^{1/2}\leqslant C(n)R \biggl ( \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{3R/2}}|\nabla u|^q\,dx\biggr)^{1/q}, \qquad q=\frac{2n}{n+2},
\end{equation*}
\notag
$$
and from (4.11) that
$$
\begin{equation}
\begin{aligned} \, \notag &\biggl( \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{R}}|\nabla u|^2\,dy\biggr)^{1/2} \\ &\qquad \leqslant C(n,k,\alpha,L,R_0)\biggl (\biggl ( \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{2R}}|\nabla u|^q\,dy\biggr)^{1/q} +\biggl( \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{2R}}|h|^2\,dy\biggr)^{1/2}\biggr). \end{aligned}
\end{equation}
\tag{4.12}
$$
Note that on the right-hand side the domain of integration is larger.
Now consider the case when $Q^{y_0}_{3R/2}\cap \widetilde F_{R_0}\ne \varnothing$. Choosing the test function $\varphi=u\eta^2$ for the same cutoff function $\eta$ in the integral identity (3.9) we obtain
$$
\begin{equation}
\begin{aligned} \, \notag &\int_{Q^{y_0}_{3R/2}} (c(y)\nabla u\cdot\nabla u) \eta^2\,dy =\int_{Q^{y_0}_{3R/2}} k\cdot\nabla u u\eta^2\,dy \\ \notag &\qquad\qquad -2\int_{Q^{y_0}_{3R/2}} c(y)\eta u\nabla u\cdot\nabla \eta\,dy \\ &\qquad\qquad +2\int_ {Q^{y_0}_{3R/2}}h u\eta\cdot\nabla\eta\,dy +\int_ {Q^{y_0}_{3R/2}}\eta^2 h\cdot\nabla u\,dy. \end{aligned}
\end{equation}
\tag{4.13}
$$
All integrals on the right-hand side of (4.13) are estimated in precisely the same way as above, up to (4.11). As a result, we arrive at the bound (4.11) for $\lambda=0$. We rewrite this bound in the form
$$
\begin{equation}
\int_{Q^{y_0}_R}|\nabla u|^2\, dy\leqslant C(n,k,\alpha,L,R_0) \biggl (\frac{1}{R^{2}}\int_{Q^{y_0}_{2R}}u^2\, dy+\int_{Q^{y_0}_{2R}}|h|^2\, dy\biggr).
\end{equation}
\tag{4.14}
$$
Here we use Maz’ya’s inequality (see [18], § 14.1.2) in estimates for the first integral on the right-hand side of (4.14). To formulate this inequality we denote by $W^1_q(Q^{y_0}_{2R},K)$, where $K\subset \overline Q^{y_0}_{2R}$ is a compact set, the closure of the set of functions in $W^1_q(Q^{y_0}_{2R})$ with support outside a neighborhood of $K$ in the norm of the Sobolev space $W^1_q$. According to Maz’ya’s inequality, if $v\in W^1_q(Q^{y_0}_{2R},K)$, $q<n$, and $s\in [1,nq/(n-q)]$, then
$$
\begin{equation}
\| v\|_{L_s(Q^{y_0}_{2R})}\leqslant\frac{C(n)R^{n/s}}{C_q^{1/q}(K,Q^{y_0}_{4R})}\| \nabla v\|_{L_q(Q^{y_0}_{2R})},
\end{equation}
\tag{4.15}
$$
where $C_q(K,Q^{y_0}_{4R})$ is the capacity of $K$ (defined in (2.4)).
To use this inequality, note that, since $Q^{y_0}_{3R/2}\cap \widetilde F_{R_0}\ne \varnothing$, there is a point $z_0\in Q^{y_0}_{3R/2}\cap \widetilde F_{R_0}$ such that $\overline {Q^{z_0}_{R/2}}\subset \overline {Q^{y_0}_{2R}}$. Let $z\in F\cap Q_{R_0}$ denote the preimage of $z_0$ under the transformation (3.1). It is easy to see that the preimage of the closed cube $\overline Q^{z_0}_{R/2}$ contains a closed ball $\overline B^z_{cR}$, where $c=c(L,n)>0$ is independent of $z$. By (2.2) the inequality $C_q(F\cap \overline B^{z}_{cR})\geqslant C(L,n,c_0)R^{n-q}$ holds, where $q=2n/(n+2)$. It follows from this and the definition of capacity in (2.1) that $C_q(\widetilde F_{R_0}\cap \overline Q^{y_0}_{2R})\geqslant C(L,n,c_0)R^{n-q}$. Since $C_q(\widetilde F_{R_0}\cap \overline Q^{y_0}_{2R}) \leqslant C_q(\widetilde F_{R_0}\cap \overline Q^{y_0}_{2R}, Q^{y_0}_{4R})$, it follows from inequality (4.15) for $s=2$, $q=2n/(n+2)$ and $K=\widetilde F_{R_0}\cap \overline Q^{y_0}_{2R}$, as applied to the first integral on the right-hand side of (4.14), that
$$
\begin{equation*}
\biggl( \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{2R}}u^2\,dy\biggr)^{1/2}\leqslant C(L,n,c_0)R \biggl( \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{2R}}|\nabla u|^{q}\,dy\biggr)^{1/q}.
\end{equation*}
\notag
$$
From this and (4.14) we arrive at (4.12) again, with a constant $C$ depending additionally on $c_0$. It follows from the bound (4.12), which holds for all cubes $Q^{y_0}_{R}$ under consideration, and from the generalized Gehring lemma (see [20], [21], and also Ch. VII in [22]), taking the edge size of $K_{R_0}$ into account (see (3.2)), that, under the assumption $h\in L_{2+\delta_0}(K_{R_0})$, where $\delta_0>0$, the bound
$$
\begin{equation}
\begin{aligned} \, \notag &\|\nabla u\|_{L_{2+\delta}(K_{R_0/4})} \\ &\qquad \leqslant C(n,\alpha,k,\delta_0,c_0,L, R_0) (\|\nabla u\|_{L_{2}(K_{R_0/2})}+\|h\|_{L_{2+\delta}(K_{R_0/2})}) \end{aligned}
\end{equation}
\tag{4.16}
$$
holds with positive constant $\delta=\delta(n,\delta_0)$. Since $u$ is even with respect to the hyperplane $\{y\colon y_n=0\}$, this bound can be rewritten in the form
$$
\begin{equation}
\begin{aligned} \, \notag &\|\nabla u\|_{L_{2+\delta}(K^+_{R_0/4})} \\ &\qquad \leqslant C(n,\alpha,\widetilde b,\delta_0,c_0,L, R_0) (\|\nabla u\|_{L_{2}(K^+_{R_0/2})}+\|\widetilde f\|_{L_{2+\delta}(K^+_{R_0/2})}) \end{aligned}
\end{equation}
\tag{4.17}
$$
(see (3.3)).
Let us now make the transformation inverse to (3.1). It is easy to see that the preimage of the half-cube $K^+_{R_0/2}$ is contained in $D_{R_0}$, and the preimage of $K^+_{R_0/4}$ contains the set $D_{\theta R_0}$, where $\theta=\theta(n,L)>0$. Taking (3.5) into account, by (4.17) we have
$$
\begin{equation*}
\|\nabla u\|_{L_{2+\delta}(D_{\theta R_0})} \leqslant C(n,\alpha,b,\delta_0,c_0,L, R_0)(\|\nabla u\|_{L_{2}(D_{R_0})}+\|f\|_{L_{2+\delta}(D_{R_0})}).
\end{equation*}
\notag
$$
Going over to the coordinate system with the origin at the point $x_0\in\partial D$ from which we began our reasoning, we obtain
$$
\begin{equation*}
\|\nabla u\|_{L_{2+\delta}(D\cap Q^{x_0}_{\theta R_0})} \leqslant C(n,\alpha,b,\delta_0,c_0,L, R_0)(\|\nabla u\|_{L_{2}(D\cap Q^{x_0}_{R_0})}+\|f\|_{L_{2+\delta}(D\cap Q^{x_0}_{R_0})}).
\end{equation*}
\notag
$$
Since $x_0\in\partial D$ is an arbitrary boundary point and the boundary $\partial D$ is compact, we can find a finite cover of $\partial D$ such that the closed set
$$
\begin{equation*}
\mathcal{D}_{\theta_1 R_0}=\{x\in D\colon \operatorname{dist}(x,\partial D)\leqslant \theta_1 R_0\}, \qquad \theta_1=\theta_1(n,L)>0,
\end{equation*}
\notag
$$
is covered by the union of the sets $D\cap Q^{t_i}_{\theta R_0}$, where $t_i\in\partial D$. Therefore, summing the inequalities
$$
\begin{equation*}
\|\nabla u\|_{L_{2+\delta}(D\cap Q^{t_i}_{\theta R_0})} \leqslant C(n,\alpha,b,\delta_0,c_0,L, R_0)(\|\nabla u\|_{L_{2}(D\cap Q^{t_i}_{R_0})}+\|f\|_{L_{2+\delta}(D\cap Q^{t_i}_{R_0})}),
\end{equation*}
\notag
$$
we arrive at the bound
$$
\begin{equation*}
\|\nabla u\|_{L_{2+\delta}(\mathcal{D}_{\theta_1 R_0})} \leqslant C(n,\alpha, b,\delta_0,c_0,L, R_0)(\|\nabla u\|_{L_{2}(D)}+\|f\|_{L_{2+\delta}(D)}).
\end{equation*}
\notag
$$
The interior bound
$$
\begin{equation*}
\|\nabla u\|_{L_{2+\delta}(D\setminus \mathcal{D}_{\theta_1 R_0})} \leqslant C(n,\alpha,b,\delta_0,c_0,L, R_0)(\|\nabla u\|_{L_{2}(D)}+\|f\|_{L_{2+\delta}(D)})
\end{equation*}
\notag
$$
is much easier. Its proof does not require the technique of straightening the boundary of $D$ locally or going over to new local coordinates. It is sufficient to obtain a bound of the form (4.12) in cubes interior to $D$, which has, in fact, already been established. After this the interior bound follows, for example, from [2].
As a result, combining the last two inequalities we arrive at
$$
\begin{equation}
\|\nabla u\|_{L_{2+\delta}(D)} \leqslant C(n,\alpha,b,\delta_0,c_0,L, R_0)(\|\nabla u\|_{L_{2}(D)}+\|f\|_{L_{2+\delta}(D)}).
\end{equation}
\tag{4.18}
$$
By [17], the first term on the right-hand side of (4.18) satisfies
$$
\begin{equation}
\|\nabla u\|_{L_2(D)}\leqslant C \|f\|_{L_2(D)}\leqslant C \|f\|_{L_{2+\delta}(D)}
\end{equation}
\tag{4.19}
$$
with constant $C$ depending only on the coefficients of the operator $\mathcal{L}$, the domain $D$ and the dimension of the space. Now, combining (4.18) and (4.19) we arrive at the required bound (2.8).
Theorem 1 is proved. Proof of Theorem 2. The line of reasoning remains the same as in the proof of Theorem 1. Therefore, we present the distinctive features of the flat case. The only difference is the way of estimating the first integrals on the right-hand sides of (3.10) and (4.13). First assume that $Q^{y_0}_{3R/2}\subset K_{R_0}\setminus \widetilde F_{R_0}$. The first integral on the right-hand side of (3.10) is estimated by Hölder’s inequality, which shows that
$$
\begin{equation}
\begin{aligned} \, \notag &\biggl|\int_{Q^{y_0}_{3R/2}} (k\cdot\nabla u)(u-\lambda)\eta^2\,dy\biggr| \\ \notag &\qquad \leqslant R \biggl (\int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\, dy\biggr)^{1/2} \biggl (\int_{Q^{y_0}_{3R/2}}|k|^2\eta^2\biggl (\frac{u-\lambda}{R}\biggr)^2\,dy\biggr)^{1/2} \\ &\qquad \leqslant R \biggl (\int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\, dy\biggr)^{1/2} \biggl (\int_{Q^{y_0}_{3R/2}}|k|^p\eta^p\,dy \biggr)^{1/p} \nonumber \\ &\qquad\qquad\times \biggl (\int_{Q^{y_0}_{3R/2}} \biggl|\frac{u-\lambda}{R}\biggr|^{2p/(p-2)}\,dy\biggr)^{(p-2)/(2p)}. \end{aligned}
\end{equation}
\tag{4.20}
$$
Further, by the Poincaré–Sobolev inequality we have
$$
\begin{equation}
\biggl (\int_{Q^{y_0}_{3R/2}}\biggl|\frac{u-\lambda}{R} \biggr|^{2p/(p-2)}\,dy\biggr)^{(p-2)/(2p)} \leqslant C(p)\biggl (\int_{Q^{y_0}_{3R/2}}|\nabla u|^{q_1}\,dy\biggr)^{1/q_1},
\end{equation}
\tag{4.21}
$$
where $q_1=p/(p-1)$. From (4.21) and (4.20), taking into account that $0\leqslant \eta\leqslant 1$ we obtain
$$
\begin{equation*}
\begin{aligned} \, &\biggl|\int_{Q^{y_0}_{3R/2}} (k\cdot\nabla u)(u-\lambda)\eta^2\,dy\biggr| \\ &\qquad \leqslant C(p)R \biggl (\int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\, dy\biggr)^{1/2} \| k \|_{L_p(Q^{y_0}_{3R/2})} \biggl (\int_{Q^{y_0}_{3R/2}}|\nabla u|^{q_1}\,dy\biggr)^{1/q_1}. \end{aligned}
\end{equation*}
\notag
$$
By Cauchy’s inequality for $\varepsilon>0$,
$$
\begin{equation}
\begin{aligned} \, \notag &\biggl|\int_{Q^{y_0}_{3R/2}} (k\cdot\nabla u)(u-\lambda)\eta^2\,dy\biggr| \leqslant \varepsilon \int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\, dy \\ &\qquad\qquad + \frac{C(p)}{4\varepsilon}R^2\| k \|_{L_p(Q^{y_0}_{3R/2})}^2 \biggl ( \int_{Q^{y_0}_{3R/2}}|\nabla u|^{q_1}\,dy\biggr)^{2/q_1}. \end{aligned}
\end{equation}
\tag{4.22}
$$
The remaining integrals on the right-hand side of (3.10) are estimated in just the same way as in (4.8)–(4.10). As a result, since $R\leqslant R_0$, using (3.10) and taking into account the last bounds and (3.8), after an appropriate choice of $\varepsilon$ we arrive at the inequality
$$
\begin{equation}
\begin{aligned} \, \notag & \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_R}|\nabla u|^2\, dy\leqslant C(k,p,\alpha,L, D)\biggl (\frac{1}{R^{2}} \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{3R/2}}(u-\lambda)^2\, dy \\ &\qquad\qquad + \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{3R/2}}|h|^2\, dy +\biggl ( \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{3R/2}}|\nabla u|^{q_1}\,dy\biggr)^{2/q_1}\biggr). \end{aligned}
\end{equation}
\tag{4.23}
$$
By the Poincaré–Sobolev inequality we have
$$
\begin{equation*}
\biggl ( \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{3R/2}}(u-\lambda)^2\,dx\biggr)^{1/2}\leqslant C(p)R \biggl ( \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{3R/2}}|\nabla u|^{q_1}\,dx\biggr)^{1/q_1}, \qquad q_1=\frac{p}{p-1},
\end{equation*}
\notag
$$
and from (4.23) we see that
$$
\begin{equation}
\begin{aligned} \, \notag &\biggl( \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{R}}|\nabla u|^2\,dy\biggr)^{1/2} \\ &\qquad \leqslant C(k,p,\alpha,L,D)\biggl (\biggl ( \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{2R}}|\nabla u|^{q_1}\,dy\biggr)^{1/q_1} +\biggl( \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{2R}}|h|^2\,dy\biggr)^{1/2}\biggr). \end{aligned}
\end{equation}
\tag{4.24}
$$
If $Q^{y_0}_{3R/2}\cap \widetilde F_{R_0}\ne \varnothing$, then the first integral on the right-hand side of (4.13) is estimated by Hölder’s inequality in the same way as in (4.20). As a result, we obtain
$$
\begin{equation*}
\begin{aligned} \, &\biggl|\int_{Q^{y_0}_{3R/2}} (k\cdot\nabla u)u\eta^2\,dy\biggr| \\ &\qquad \leqslant \biggl (\int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\, dy\biggr)^{1/2}\| k \|_{L_p(Q^{y_0}_{3R/2})} \biggl (\int_{Q^{y_0}_{3R/2}}|u|^{2p/(p-2)}\,dy\biggr)^{(p-2)/(2p)}. \end{aligned}
\end{equation*}
\notag
$$
Hence by Cauchy’s inequality, for $\varepsilon>0$ we derive that
$$
\begin{equation}
\begin{aligned} \, \notag &\biggl|\int_{Q^{y_0}_{3R/2}} (k\cdot\nabla u)u\eta^2\,dy\biggr| \\ &\qquad \leqslant \varepsilon \int_{Q^{y_0}_{3R/2}}|\nabla u|^2\eta^2\, dy+ \frac{C(p,k)}{4\varepsilon}\biggl (\int_{Q^{y_0}_{3R/2}}|u|^{2p/(p-2)}\,dy\biggr)^{(p-2)/p}. \end{aligned}
\end{equation}
\tag{4.25}
$$
The remaining integrals on the right-hand side of (4.13) are estimated in the same way as for $n>2$. As a result, using (3.8) and (4.25), after an appropriate choice of $\varepsilon$ we arrive at the bound
$$
\begin{equation*}
\begin{aligned} \, \int_{Q^{y_0}_R}|\nabla u|^2\, dy &\leqslant C(k,p,\alpha,L,D)\biggl (\biggl (\int_{Q^{y_0}_{3R/2}}|u|^{2p/(p-2)}\,dy\biggr)^{(p-2)/p} \\ &\qquad+\frac{1}{R^{2}}\int_{Q^{y_0}_{3R/2}}u^2\, dy+\int_{Q^{y_0}_{3R/2}}|h|^2\, dy\biggr). \end{aligned}
\end{equation*}
\notag
$$
Since $R\leqslant R_0$, applying Hölder’s inequality to the second integral on the right-hand side we obtain
$$
\begin{equation}
\begin{aligned} \, \notag &\int_{Q^{y_0}_R}|\nabla u|^2\, dy \\ &\qquad \leqslant C(k,p,\alpha,L,D)\biggl (R^{(4-2p)/p} \biggl (\int_{Q^{y_0}_{2R}}|u|^{2p/(p-2)}\,dy\biggr)^{(p-2)/p} +\int_{Q^{y_0}_{2R}}|h|^2\, dy\biggr). \end{aligned}
\end{equation}
\tag{4.26}
$$
The first integral on the right-hand side of (4.26) is estimated by Maz’ya’s inequality (4.15). Since $Q^{y_0}_{3R/2}\cap \widetilde F_{R_0}\ne \varnothing$, just as in the case of $n>2$, by condition (2.3) we have $C_{q_1}(\widetilde F_{R_0}\cap \overline Q^{y_0}_{2R})\geqslant C(L,c_0)R^{2-q_1}$. It is clear that $C_q(\widetilde F_{R_0}\cap \overline Q^{y_0}_{2R}) \leqslant C_q(\widetilde F_{R_0}\cap \overline Q^{y_0}_{2R}, Q^{y_0}_{4R})$, and from inequality (4.15) for $s=2p/(p-2)$, $q=q_1=p/(p-1)$ and $K=\widetilde F_{R_0}\cap \overline Q^{y_0}_{2R}$ we obtain
$$
\begin{equation*}
\biggl( \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{2R}}|u|^{2p/(p-2)}\,dy\biggr)^{(p-2)/(2p)}\leqslant C(L,p,c_0)R\biggl ( \,\rlap{-}\kern-1.5mm \int _{Q^{y_0}_{2R}}|\nabla u|^{q_1}\,dy\biggr)^{1/q_1}.
\end{equation*}
\notag
$$
Hence by (4.26) we arrive again at the bound (4.24) in which the constant $C$ also depends on $c_0$. Thus, the (4.24) is valid for all cubes under consideration. The further reasoning, based on the use of Gehring’s lemma mentioned before, does not differ from the above argument. From (4.24) and the generalized Gehring lemma we arrive at the bound (4.16) with constants $C$ and $\delta$ depending additionally on $p$. All other arguments in the proof of Theorem 1, starting from (4.17) and up to (4.19), are repeated verbatim. As a result, we arrive at the required bound (2.9). Theorem 2 is proved. AcknowledgementsThe authors express their gratitude to M. D. Surnachev for his useful comments on this paper. The authors are also very grateful to anonymous referees: the text has significantly been improved thanks to their painstaking work. The authors express their special gratitude to A. I. Nazarov, who drew our attention to the ‘critical’ case of the exponent in the embedding theorem. 
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\Bibitem{AlkChe25} |
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