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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 308, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/308\hfil 
 H\"older continuity for $(p,q)$-Laplace equations]
{H\"older continuity for $(p,q)$-Laplace equations 
 that degenerate uniformly on part of the domain}

\author[S. T. Huseynov \hfil EJDE-2017/308\hfilneg]
{Sarvan T. Huseynov}

\address{Sarvan T. Huseynov \newline
Baku State University, Baku, Azerbaijan}
\email{sarvanhuseynov@rambler.ru}

\dedicatory{Communicated by Ludmila S. Pulkina}

\thanks{Submitted November 3, 2017. Published December 14, 2017.}
\subjclass[2010]{35J92, 35J65, 35J70, 35J62}
\keywords{$(p,q)$-Laplacian; elliptic equation; H\"older continuity}

\begin{abstract}
 In this article we consider $p(x)$-Laplace equations with two-phase
 degree $p(x)$, taking two values  $p$ and $q$, when the boundary of
 the phase interface is a hyperplane.
 Assuming that in the part of the domain where $q<p$ the equation
 degenerates uniformly for a small parameter, H\"older continuity
 of the solution is established.
\end{abstract}

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

\section{Formulation of results}

Consider in the domain  $D\subset\mathbb{R}^n$, $n\ge 2$ and the family
of the elliptic equations
\begin{equation}\label{eq}
 L_\varepsilon u= \operatorname{div}(\omega_\varepsilon(x)
|\nabla u|^{p(x)-2}\nabla u)=0
\end{equation}
with positive weight $\omega_\varepsilon(x)$ and degree $p(x)$, that
will be defined below. Suppose that the domain $D$ is divided by the
hyperplane $\Sigma=\{x:x_n=0\}$ into two parts
$D^{(1)}=D\cap\{x: x_n>0\}$ and $D^{(2)}=D\cap\{x: x_n<0\}$.
Also assume that for $\varepsilon\in (0,1]$,
\begin{equation} \label{w}
  \omega_\varepsilon(x)= \begin{cases}
  \varepsilon,& x\in D^{(1)} \\
  1,& x\in D^{(2)},
  \end{cases}
\end{equation}
and for $1<q<p$,
\begin{equation}\label{pdef}
p(x)= \begin{cases}
  q, & x\in D^{(1)} \\
  p, & x\in D^{(2)},
  \end{cases}
\end{equation}

To define the solution of equation \eqref{eq} we define a
class of functions related with the degree $p(x)$:
$$
W_{\rm loc}(D)=\{u: u\in W^{1,1}_{\rm loc}(D),\; |\nabla u|^{p(x)}\in
L^1_{\rm loc}(D)\},
$$
where $W^{1,1}_{\rm loc}(D)$  is a Sobolev space of the locally
integrable in $D$ functions together with their first order
generalized derivatives.

As a solution of equation  \eqref{eq} we take the function
 $u\in W_{\rm loc}(D)$, satisfying 
\begin{equation} \label{id} 
\int_D\omega_\varepsilon(x)|\nabla u|^{p(x)-2}\nabla
u\cdot\nabla\varphi\,dx=0
\end{equation}
for all test functions $\varphi\in C^\infty_0(D)$.

For the degree $p(\cdot)$, given by equality \eqref{pdef}, the
smooth functions are dense in  $W_{\rm loc}(D)$ (see [1]), and as a
result in integral identity \eqref{id} as test functions the finite
functions from $W_{\rm loc}(D)$ may be taken.

The $p$-Laplace type equation with variable nonlinearity degree
$p(x)$ and the variational problems with integrant satisfying the
non standard coercivity and growth conditions  usually arise in the
modeling of the composite materials, electroreological fluids (the
characteristics of which depend on the electromagnetic filed), in
the problems of image processing. In this paper the plane junction
of two different phases is considered as a model case. The case is
complicated by the presence of the uniform degeneracy over  $\varepsilon$ in
the domain $D^{(1)}$.

In each of the domains $D^{(i)}$, $i=1,2$ regularity of the solution
is described by the well developed theory   (see [2]). 
In [3] is
proved that for the degree $p$, given by equality \eqref{pdef}, any
solution of equation \eqref{eq} by each fixed value $\varepsilon\in (0,1]$
in the arbitrary subdomain  $D'\Subset D$ belongs to the space
$C^\alpha(D')$ of the H\"older in $D'$ functions. We
are interested in the problem of independence of the degree $\alpha$
on $\varepsilon$.

Consider the family $\{u^\varepsilon(x)\}$ of the solutions of the equation
$L_\varepsilon u^\varepsilon=0$, uniformly bounded over $\varepsilon$ in $L^\infty$
 on the compact subspaces of $D$. The main aim of this work is 
to prove the following statement.

\begin{theorem} \label{as} 
There exists a constant  $\alpha\in (0,1)$, not depending
on $\varepsilon$, such that the family $\{u^\varepsilon(x)\}$ is compact in
$C^\alpha(D')$ for arbitrary subdomain $D'\Subset D$.
\end{theorem}


Note that in the case  $p=q$ a similar result is obtained in 
[4,5].

Choice of the weight of type \eqref{w} and the degree $p$ from
\eqref{pdef} makes the case nonsymmetric with respect to the domains
 $D^{(1)}$ and $D^{(2)}$, and use of known results does not allow 
the one to prove the above statement.
 We will proceed from the modification of the Mozer's technique
[6], developed in [7,8], where the domains $D^{(1)}$ and
$D^{(2)}$ play different roles in the proof of the Theorem \ref{as}.

Statement of Theorem  \ref{as} remains true also for the solutions
of the equation
$$
\operatorname{div}\Big( \omega_{\varepsilon}(x) |\nabla u|^{p(x)-2}a\nabla u \Big)=0
$$
with measurable uniformly positive defined matrix $a$. Wherein
H\"older degree of the solutions will be additionally depend on
the ellipticity coefficients of this matrix.


\section{Auxiliary statements}


Here and below $u$ denotes the solution of equation \eqref{eq},
 $B_R\subset D$ are balls with centers in $\Sigma$, 
$B^{(i)}_R=B_R\cap D^{(i)}$ are semiballs ($i=1,2$), $|E|$ is
 $n$-dimensional Lebesgue measure of
the measurable set $E\subset\mathbb{R}^n$, and
$$
\hbox{--}\hskip-9pt\int_{E} fdx=\frac{1}{|E|}\int_{E}f\,dx.
$$

Below we  use for $i=1,2$ Sobolev's embedding theorem in the
semiballs:
\begin{equation}\label{sob}
\begin{gathered}
\Big(\hbox{--}\hskip-9pt\int_{B^{(i)}_R}|\varphi|^{kq}dx\Big)^{1/k}
\le CR^q\hbox{--}\hskip-9pt\int_{B^{(i)}_R} |\nabla \varphi|^q\,dx,\\
q\ge 1,\quad k=\frac{n}{n-1},\quad\varphi\in C_0^{\infty}(B_R),
\end{gathered}
\end{equation}
where $C=C(n,q)$.

Everywhere below $M=\sup_{B_{R_0}}|u(x)|$, where
$B_{R_0}\subset D$, $R_0\le 1/4$, and for $R\le R_0/6$ is taken
\begin{equation} \label{def}
 M_6=\sup_{B_{6R}}u,\quad
m_6=\inf_{B_{6R}}u,\quad
v(x)=\ln{\frac{M_6-m_6+2R}{M_6-u(x)+R}}.
\end{equation}
It is easy to see that
\begin{equation}\label{3} 
\frac{R}{4(M+1)}\le v(x)\le\frac{2(M+1)}{R}\quad x\in B_{6R}.
\end{equation}
The odd continuation of the function $f$ from $D^{(2)}$ in to $D^{(1)}$
with respect to the hyperplane $\Sigma$ is denoted as $\tilde f$.

\begin{lemma} \label{l2} 
For any $R\le\rho<r\le 3R$  the inequality
\begin{equation} \label{i1}
   \sup_{B_{\rho}}v\le C(n,p,q,M)\Big(\frac {r}{r-\rho}
   \Big)^a\Big(\hbox{--}\hskip-9pt\int_{B_r}v^p\,dx\Big)^{1/p}
\end{equation}
holds  with constant $a(n,p)>0$.
\end{lemma}

\begin{proof}
 Choosing in  \eqref{id} the test function as
$\varphi(x)=v^{\gamma+q-p}(x)(M_6-u(x)+R)^{1-p}\eta^p(x)$, where
\begin{equation}
\label{beta} \gamma\ge 1+p-q,
\end{equation}
$\eta\in {\it C}^\infty_0(B_{4R})$, $0\le\eta(x)\le 1$, we find that
\begin{align*}
&(\gamma+q-p)\int_{B_{4R}}\omega_\varepsilon(x)|\nabla
u|^{p(x)}(M_6-u+R)^{-p} v^{\gamma+q-p-1}\eta^p\,dx \\
& +(p-1)\int_{B_{4R}} \omega_\varepsilon(x)|\nabla
u|^{p(x)}(M_6-u+R)^{-p} v^{\gamma+q-p}\eta^p\,dx \\
&\le  p \int_{B_{4R}} \omega_\varepsilon(x)|\nabla u|^{p(x)-1}(M_6-u+R)^{1-p}
v^{\gamma+q-p}\eta^{p-1}|\nabla\eta|\,dx.
\end{align*}

Omitting the second term in the left hand side and applying to the
integrant in the right hand side the Young inequality with
corresponding $\varepsilon$ we arrive to the estimate
\begin{equation}\label{l22} 
\begin{aligned}
&\int_{B_{4R}}\omega_\varepsilon(x)|\nabla u|^{p(x)}(M_6-u+R)^{-p}
v^{\gamma+q-p-1}\eta^p\,dx\\
&\le C(p)\int_{B_{4R}}\omega_\varepsilon(x)(M_6-u+R)^{p(x)-p}
v^{\gamma+q-p+p(x)-1}|\nabla\eta|^{p(x)}\,dx.
\end{aligned}
\end{equation}

We narrow the integration domain in the left-hand side of
\eqref{l22}  to the 
 semiball $B^{(2)}_{4R}$. Then considering \eqref{w} and
\eqref{pdef} we can  write
\begin{equation}\label{yua} 
\begin{aligned}
&\int_{B^{(2)}_{4R}}|\nabla
u|^{p}(M_6-u+R)^{-p} v^{\gamma+q-p-1}\eta^p\,dx\\
&\le C(p)\Big(\int_{B^{(1)}_{4R}}(M_6-u+R)^{q-p}
v^{\gamma+2q-p-1}|\nabla\eta|^{q}\,dx \\
&\quad +\int_{B^{(2)}_{4R}}v^{\gamma+q-1}|\nabla\eta|^{p}\,dx\Big ).
\end{aligned}
\end{equation}

According to \eqref{3} and \eqref{pdef}, in $B^{(1)}_{4R}$ the
following inequalities are valid
$$
(M_6-u+R)^{q-p}\le R^{q-p},\quad 
v^{q}\le C(p,q,M)R^{q-p}v^{p}.
$$
Additionally
\begin{equation}
\label{t} |\nabla v|=\frac{ |\nabla u|}{
M_6-u+R},
\end{equation}
and from \eqref{yua} considering given above relations we obtain the
estimate
\begin{equation}\label{l27} 
\begin{aligned}
&\int_{B^{(2)}_{4R}}|\nabla
v|^{p}v^{\gamma+q-p-1}\eta^p\,dx \\
&\leq C(p,q,M)\Big(R^{q-p}\int_{B^{(1)}_{4R}}
v^{\gamma+q-1}|\nabla\eta|^{q}\,dx+
\int_{B^{(2)}_{4R}}v^{\gamma+q-1}|\nabla\eta|^{p}\,dx\Big).
\end{aligned}
\end{equation}

By the Soboloev's embedding theorem  \eqref{sob},
\begin{align*}
&\Big(\hbox{--}\hskip-9pt\int_
{B^{(2)}_{4R}}(v^{\gamma+q-1}\eta)^k\,dx\Big )^{1/k}\\
&\le C(n,p,q,M)(\gamma+q-1)^p\Big(R^{q}\hbox{--}\hskip-9pt\int_{B^{(1)}_{4R}}
v^{\gamma+q-1}|\nabla\eta|^{q}\,dx 
+ R^{p}\hbox{--}\hskip-9pt\int_{B^{(2)}_{4R}}v^{\gamma+q-1}
|\nabla\eta|^{p}\,dx\Big)
\end{align*}
 in the semiball $B^{(2)}_{4R}$.

Choosing here radial-symmetric with respect to the center of the
ball $B_R$, cutoff function $\eta=1$ in $B_\rho$, $|\nabla\eta|\le
Cr(R(r-\rho))^{-1}$, we have
\begin{equation}\label{l23} 
\Big(\hbox{--}\hskip-9pt\int_{B^{(2)}_\rho}v^{(\gamma+q-1)k}\,dx\Big)^{1/k} 
\le C(n,p,q,M)(\gamma+q-1)^p\big(\frac{r}{ r-\rho}\big)^p
\hbox{--}\hskip-9pt\int_{B_r}v^{\gamma+q-1}\,dx.
\end{equation}

Now we prove a similar estimate in the semiball $B^{(1)}_{4R}$. Let
\begin{equation} \label{g1}
 G_R=B^{(1)}_{4R}\cap\{x: v(x)>\tilde v(x)\}.
\end{equation}
In \eqref{id} use the test function
$$
  \varphi(x)=\begin{cases}
  (v^{\gamma}(x)-\tilde v^{\gamma}(x))
(M_6-u(x)+R)^{1-q}\eta^p(x) &\text{in } G_R \\
  0 &B_{4R}\setminus G_R,
  \end{cases}
$$
where $\gamma>1$ satisfies to condition \eqref{beta}, the
radial-symmetric function $\eta(x)$ has the same properties as
above. Considering \eqref{w}, and \eqref{pdef} we have
\begin{equation}\label{l24} 
\begin{aligned}
& \gamma\int_{G_R}|\nabla
u|^{q}v^{\gamma-1}(M_6-u+R)^{-q} \eta^p\, dx \\
& \le \gamma\int_{G_R}|\nabla u|^{q-1}|\nabla
\tilde v|\tilde v^{\gamma-1} (M_6-u+R)^{1-q}\eta^p\,dx\\
&\quad + p\int_{G_R}|\nabla u|^{q-1}(v^\gamma+\tilde
v^\gamma) (M_6-u+R)^{1-q}|\nabla \eta| \eta^{p-1}\,dx.
\end{aligned}
\end{equation}

Since $\tilde v(x)<v(x)$ by $x\in G_R$ and $0\le\eta\le 1$, applying
the Young inequality to the integrant in the right hand side of
\eqref{l24}, we obtain
\begin{align*}
&|\nabla u|^{q-1}|\nabla \tilde v|\tilde v^{\gamma-1}
(M_6-u+R)^{1-q}\eta^p \\
&\le \delta |\nabla u|^{q}v^{\gamma-1}(M_6-u+R)^{-q}\eta^p 
+C(\delta,q)|\nabla\tilde v|^{q}\tilde v^{\gamma-1} \eta^p,
\end{align*}
\begin{align*}
& |\nabla u|^{q-1}(v^\gamma+\tilde v^\gamma)
(M_6-u+R)^{1-q}|\nabla \eta| \eta^{q-1} \\
&\leq \delta |\nabla u|^{q}v^{\gamma-1} (M_6-u+R)^{-q}\eta^p+
C(\delta,q)v^{\gamma+q-1}|\nabla\eta|^{q}.
\end{align*}

From this and \eqref{l24} (see also \eqref{t}) after a corresponding
choice  of $\delta$ we find that
\begin{equation}\label{yua1} 
\begin{aligned}
&\int_{G_R}|\nabla v|^{q}v^{\gamma-1} \eta^p\, dx \\
&\le C(p,q)\Big(\int_{G_R}|\nabla \tilde v|^{q}\tilde
v^{\gamma-1} \eta^p\, dx +\int_{G_R} v^{\gamma+q-1}|\nabla
\eta|^{q}\, dx\Big ).
\end{aligned}
\end{equation}

Expanding the integrals in the right hand side of \eqref{yua1} to
larger set $B^{(1)}_{4R}$, we rewrite  \eqref{yua1} in the form
\begin{equation} \label{l25}
\begin{aligned}
&\int_{G_R}|\nabla v|^{q}v^{\gamma-1}\eta^p\, dx\\
&\leq C(p,q) \Big(\int_{B^{(1)}_{4R}}|\nabla \tilde v|^{q}\tilde
v^{\gamma-1} \eta^p\, dx
+\int_{B^{(1)}_{4R}}v^{\gamma+q-1}|\nabla \eta|^{q}\,
dx\Big ).
\end{aligned}
\end{equation}

We can not estimate the gradient $v(x)$ over the set
$B^{(1)}_{4R}\setminus G_R$. But this is not important. 
Consider in $D^{(1)}$ the auxiliary function
\begin{equation}\label{ag}
w(x)=\max{(v(x),\tilde v(x))}.
\end{equation}

Since $w(x)=v(x)$ for $x\in G_R$ and $w(x)=\tilde v(x)$ for
$x\in B^{(1)}_{4R}\setminus G_R$, we have
\[
\int_{B^{(1)}_{4R}}|\nabla
w|^{q}w^{\gamma-1}\eta^p\, dx 
 \le\int_{G_R}|\nabla v|^{q}v^{\gamma-1}\eta^p\, dx+
\int_{B^{(1)}_{4R}}|\nabla \tilde v|^{q}\tilde
v^{\gamma-1}\eta^p\, dx,
\]
and considering \eqref{l25},
\begin{equation}\label{l26}
\begin{aligned}
&\int_{B^{(1)}_{4R}}|\nabla
w|^{q}w^{\gamma-1}\eta^p\, dx\\
&  \le C(p,q) \Big(\int_{B^{(1)}_{4R}}|\nabla \tilde v|^{q}\tilde
v^{\gamma-1}\eta^p \, dx+
\int_{B^{(1)}_{4R}}v^{\gamma+q-1}|\nabla\eta|^{q}\, dx\Big).
\end{aligned}
\end{equation}

Now let us modify the first integrant in the right-hand side of
\eqref{l26}. Since $p>q$, according to Young's theorem
$$
|\nabla \tilde v|^{q}\tilde v^{\gamma-1}< R^{p-q}|\nabla \tilde v|
^{p}\tilde v^{\gamma+q-p-1}+R^{-q} \tilde v^{\gamma+q-1}.
$$
So from \eqref{l26} we obtain 
\begin{equation}\label{lo} 
\begin{aligned}
&\int_{B^{(1)}_{4R}}|\nabla w|^{q}w^{\gamma-1}\eta^p\, dx \\
&\le C(p,q)\Big(R^{p-q}\int_{B^{(1)}_{4R}} |\nabla\tilde v|^{p}\tilde
v^{\gamma+q-p-1}\eta^p\, dx \\
&\quad +\int_{B^{(1)}_{4R}}v^{\gamma+q-1} |\nabla \eta|^{q}\, dx
+R^{-q}\int_{B^{(1)}_{4R}}\tilde v^{\gamma+q-1}\eta^p\, dx
\Big ).
\end{aligned}
\end{equation}

Since $\tilde v(x)$ is an odd continuation of $v(x)$ from $D^{(2)}$
to $D^{(1)}$ and the cutoff function $\eta(x)$ is radial symmetric, 
it follows that
$$
\int_{B^{(1)}_{4R}}\tilde v^{\gamma+q-1}\eta^p\, dx
=\int_{B^{(2)}_{4R}}v^{\gamma+q-1}\eta^p\, dx
\le\int_{B_{4R}}v^{\gamma+q-1}\eta^p\, dx,
$$
and considering \eqref{l27},
\begin{align*}
& \int_{B^{(1)}_{4R}}|\nabla \tilde v|^{p}
\tilde v^{\gamma+q-p-1}\eta^p\,dx= \int_{B^{(2)}_{4R}}|\nabla
v|^{p}v^{\gamma+q-p-1}\eta^p\,dx \\
&\le C(p,q,M)\Big(R^{q-p}\int_{B^{(1)}_{4R}}
v^{\gamma+q-1}|\nabla\eta|^{q}\,dx+
\int_{B^{(2)}_{4R}}v^{\gamma+q-1}|\nabla\eta|^{p}\,dx\Big).
\end{align*}

Therefore from \eqref{lo} we arrive at the estimate
\begin{align*}
&\int_{B^{(1)}_{4R}}|\nabla w|^{q}w^{\gamma-1}\eta^p\, dx \\
&\le C(p,q,M) \Big(\int_{B^{(1)}_{4R}} v^{\gamma+q-1}|\nabla\eta|^{q}\,dx
+ R^{p-q}\int_{B^{(2)}_{4R}}
v^{\gamma+q-1}|\nabla\eta|^{p}\,dx \\
&\quad + R^{-q}\int_{B_{4R}}v^{\gamma+q-1}\eta^p\,dx\Big).
\end{align*}
It follows from the above inequality that
\begin{align*}
&\int_{B^{(1)}_{4R}}|\nabla
(w^{(\gamma+q-1)/q}\eta^{p/q})|^{q}\, dx\\
&\le C(p,q,M) (\gamma+q-1)^q\Big(\int_{B^{(1)}_{4R}}
v^{\gamma+q-1}|\nabla\eta|^{q}\,dx \\
&+R^{p-q}\int_{B^{(2)}_{4R}} v^{\gamma+q-1}|\nabla\eta|^{p}\,dx+
R^{-q}\int_{B_{4R}}v^{\gamma+q-1}\eta^p\,dx+\int_{B^{(1)}_{4R}}w^{\gamma+q-1}
|\nabla\eta|^q\,dx\Big ).
\end{align*}

From  definition \eqref{ag} of the function $w$ and the radial
symmetricity of the cutoff function $\eta$, we obtain
$$
\int_{B^{(1)}_{4R}}w^{\gamma+q-1} |\nabla\eta|^q\,dx
\le \int_{B_{4R}}v^{\gamma+q-1} |\nabla\eta|^q\,dx.
$$
From \eqref{beta} and \eqref{pdef}, we have
 $(\gamma+q-1)^q\le (\gamma+q-1)^p$. Therefore
\begin{align*}
&\int_{B^{(1)}_{4R}}|\nabla
(w^{(\gamma+q-1)/q}\eta^{p/q})|^{q}\, dx  \\
&\le C(p,q,M) (\gamma+q-1)^p\Big(\int_{B^{(1)}_{4R}}
v^{\gamma+q-1}|\nabla\eta|^{q}\,dx\\
&\quad +R^{p-q}\int_{B^{(2)}_{4R}}
v^{\gamma+q-1}|\nabla\eta|^{p}\,dx+
R^{-q}\int_{B_{4R}}v^{\gamma+q-1}\eta^p\,dx+\int_{B_{4R}}v^{\gamma+q-1}
|\nabla\eta|^q\,dx\Big ).
\end{align*}

From this following to the Sobolev's embedding theorem  \eqref{sob}
in the semiball $B^{(1)}_{4R}$ and from the choice of the cutoff
function $\eta$ we find that
$$
\Big(\hbox{--}\hskip-9pt\int_{B^{(1)}_\rho}w^{(\gamma+q-1)k}\,dx\Big)^{1/k} 
\le C(n,p,q,M)(\gamma+q-1)^p\big(\frac{r}{ r-\rho}\big)^p
\hbox{--}\hskip-9pt\int_{B_r}v^{\gamma+q-1}\,dx.
$$
Or, since $w\ge v$ on  $B^{(1)}_\rho$, it follows that
\begin{equation}\label{l28} 
\Big(\hbox{--}\hskip-9pt\int_{B^{(1)}_\rho}v^{(\gamma+q-1)k}\,dx\Big )^{1/k}
\le C(n,p,q,M)(\gamma+q-1)^p\big(\frac{r}{ r-\rho}\big)^p
\hbox{--}\hskip-9pt\int_{B_r}v^{\gamma+q-1}\,dx.
\end{equation}
Summing \eqref{l23} and \eqref{l28} one can get
\begin{equation}\label{l29} 
\Big(\hbox{--}\hskip-9pt\int_{B_\rho}v^{(\gamma+q-1)k}\,dx\Big )^{1/k} 
\le C(n,p,q,M)(\gamma+q-1)^p \big(\frac{ r}{
r-\rho}\big)^p \hbox{--}\hskip-9pt\int_{B_r}v^{\gamma+q-1}\,dx.
\end{equation}

Let us iterate this inequality. Let $j=0,1,\ldots$. Denote
${r_j=\rho+2^{-j}(r-\rho)}$, $\chi _j=pk^j$ and take $r=r_j$,
$\rho=r_{j+1}$, $\gamma=\chi _j+1-q$ in \eqref{l29}. Note that by
such choice of $\gamma$ the above assumption \eqref{beta} becomes
true. As a result for
$$
\Phi_j=\Big(\hbox{--}\hskip-9pt\int_{B_{r_j}}v^{\chi _j}\,dx\Big)^{1/\chi_j}
$$
we get the  recurrence relation
$$
 \Phi_{j+1}\le C^{1/{\chi _j}}(n,p,q,M)\left (2^j\left (1+\chi _j \right )
 \right )^{p/{\chi _j}}\Big(\frac {r}{r-\rho}\Big)^{p/{\chi _j}}\Phi_j ,
$$
from which follows (see [6]) the estimate \eqref{i1}. 
The proof is complete.
\end{proof}

The proof of the next statement uses the scheme given in [9].

\begin{lemma}\label{l3} 
If for any $R\le\rho<r\le 3R$, inequality (2.4) is valid
then the following inequity holds
\begin{equation}\label{l31}
 \sup_{B_R}v\le C(n,p,q,M)\hbox{--}\hskip-9pt\int_{B_{2R}}v\,dx.
\end{equation}
\end{lemma}

\begin{proof} 
Without loss of generality we assume that
\begin{equation}\label{l34} 
\hbox{--}\hskip-9pt\int_{B_{2R}}v\,dx=1.
\end{equation}
Denote by $B(t)$ the concentric with $B_R$ ball of radius $3Rt$ and
let
$$
J(t)=\Big(\hbox{--}\hskip-9pt\int_{B(t)}v^p\,dx\Big )^{1/p}.
$$
Taking $r=3Rt$, $\rho=3R\tau$, rewrite \eqref{i1} in the form
\begin{equation}\label{l33} 
\sup_{B(\tau)}v(x)\le C(n,p,q,M)(t-\tau)^{-a}J(t),\quad 
1/3\le \tau<t\le 1.
\end{equation}
In particular $\sup_{B_R}v\le C(n,p,q,M)J(1/2)$, and for the
proof of the lemma it is sufficient to set the estimate 
$J(1/2)\le C(n,p,q,M)$. Since, considering \eqref{l34}
  $$
J(\tau)\le C(n,p,q,M) \bigl (\sup_{B(\tau)}v\bigr
)^{\delta},\quad \delta=1-p^{-1},
  $$
then according to \eqref{l33},
\begin{gather*}
   J(\tau)\le C^{\delta}\left (t-\tau \right )^{-a\delta}J^{\delta}(t),
\quad 1/3\le \tau <t \le 1, \\
 \ln J(\tau) \le {\delta}\ln C+a{\delta}\ln{\frac {1}{t-\tau}}+
                    {\delta}\ln J(t).
\end{gather*}
Take here $\tau=t^b$, where $b>1$. It easy to see that
  $$
   \int_{{(1/2)}^{1/b}}^{1}\frac {\ln J(t^b)}{t}\,dt
   \le C(n,p,q,b,M)+{\delta}\int_{1/2}^{1}\frac {\ln J(t)}{t}\,dt .
  $$
Making substitution of the variables  $\xi=t^b$ one can get
  $$
 (1/b-\delta)\int_{1/2}^{1}\frac {\ln J(\xi)}{\xi}\,d\xi
                \le  C(n,p,q,b,M).
  $$
Let us choose here the constant $b>1$ satisfying the inequality
$1/b-\delta >0$. As ${J(\xi)\ge C(n,p)J(1/2)}$ by 
${\xi \in [1/2,1]}$. Then
  $$
    \ln \left (C(n,p)J(1/2)\right )\le \frac {C(n,p,q,b,M)}
    {\left (1/b-\delta\right )\ln2}
  $$
that leads us to the seeking estimate for $J(1/2)$.
This completes the proof.
\end{proof}

Inequality \eqref{l31} will be applied in some modified form. Denote
by $Q_r$, $r\ge 3R$ the balls with centers in $D^{(2)}$, obtained by
the parallel replacement of the ball $B_r$ with the center in $x_0$
along the normal to $\Sigma$ in the distance $R$. Suppose that
$Q^{(i)}_r=D^{(i)}\cap Q_r$, $i=1,2$. Let additionally
$w(x)=\max(v(x),\tilde v(x))$ by $x\in B^{(1)}_{4R}$ and $w(x)=v(x)$
by $x\in B^{(2)}_{4R}$. Expanding the integral in the right hand
side of \eqref{l31} up to larger set and replacing in the part
 $D^{(1)}$ the function
$v(x)$ by $w(x)$, we obtain
\begin{equation} \label{ii1}
  \sup_{B_R}v(x)\le C (n,p,M)
 \Big(\hbox{--}\hskip-9pt\int_{Q_{3R}} w\,dx+
 \hbox{--}\hskip-9pt\int_{ Q^{(2)}_{3R}} v\,dx \Big).
\end{equation}

\section{H\"older continuity of the solutions}

From the results in [2] it is known that the solutions of equation
\eqref{eq} are H\"older property inside of $D^{(1)}$ and
$D^{(2)}$. It remains to prove the H\"older property of the
solution on $\Sigma \cap D$, since the seeking holder property
inside of  $D$ may be obtained by elementary  "union" of the
H\"older property on $\Sigma\cap D$ and in $D^{(1)}$, $D^{(2)}$.

Let  $M$ means exact upper bound of the module of the solution in
the ball $B_{R_0}\subset D$ of radius $R_0\le 1/2$ and
$\operatorname{osc}\{u,B_r\}=\sup_{B_r}u(x)-\inf_{B_r}u(x)$, where
$B_r$ are balls with centers in $x_0\in \Sigma\cap D$. H\"older
continuity of the solutions in the point $x_0$ follows from the
following ``scattering lemma'':
\begin{equation}\label{o4} 
\operatorname{osc}\{u,B_{R}\}\le(1-\delta)\operatorname{osc}\{u,B_{6R}\}+R,\quad
\delta=\delta(n,p,q,M) >0, \quad R\le R_0/6.
\end{equation}
From this lemma  (see [10]) it follows the estimate
$$
\operatorname{osc}\{u,B_r\}\le C r^\alpha (R_0^{-\alpha}\operatorname{osc}
\{u,B_{R_0}\}+1), \quad r\le R_0
$$
with positive constants  $C=C(\delta)$ and $\alpha=\alpha(\delta)$.
In particular
$$
|u(x)-u(x_0)|\le C
|x-x_0|^\alpha(R_0^{-\alpha}\operatorname{osc}\{u,B_{R_0}\}+1), \quad |x-x_0|\le
R_0,
$$
that sets H\"older property of the solutions in the point  $x_0$.

Using denotation \eqref{def}, consider two sets:
\begin{gather}\label{f1} 
F=\{x\in Q_{3R}: u(x)\le (M_6+m_6)/2\}, \\
G=\{x\in Q_{3R}: M_6+m_6-u(x)\le (M_6+m_6)/2\}.
\end{gather}
One of the following inequalities is always true:
\begin{equation} \label{o1}
 |F|\ge \frac{1}{2}|Q_{3R}|
\end{equation}
or
\begin{equation}\label{o2}
 |G|\ge \frac{1}{2}|Q_{3R}|.
\end{equation}
If we show that from the condition \eqref{o1} for $u(x)$ follows
\begin{equation} \label{o3}
   \sup_{B_{R}}u(x)\le M_6-\delta \operatorname{osc}\{u,B_{6R}\}+R,\quad
\delta>0,
\end{equation}
then this result applied to the function $M_6+m_6-u(x)$ guarantees
under condition \eqref{o2} the estimate
$$
  \sup_{B_R}\left (M_6+m_6-u(x)\right )\le M_6-\delta
    \operatorname{osc}\{u,B_{6R}\}+R
$$
and in both cases we arrive to \eqref{o4}.

The following embedding fact will be used below.
\begin{equation}\label{fr} 
\int_{B_r}|\varphi|\,dx \le
C(n,\nu)r\int_{B_r}|\nabla\varphi|\, dx,
\end{equation}
for $\varphi\in C^{\infty}\left (\bar B_r\right )$,
$\varphi\vert_E=0$, $|E|\ge\nu |B_r|$, $\nu>0$.

Note that this embedding theorem holds also in the case of
truncated balls  $B_r\cap D^{(2)}$ with centers in $D^{(2)}$.

\begin{proof}[Proof of theorem \ref{as}]
 For the sake of simplicity assuming the fulfilment of condition \eqref{o1}, 
consider the function $v(x)$, introduced in \eqref{def}. Our aim is obtaining 
the estimate
\begin{equation}
\label{iv} \sup_{B_R}v(x)\le c_0(n,p,q,M).
\end{equation}
From this explicitly follows the scattering property \eqref{o3} 
($\delta=e^{-c_0}$), effecting H\"older property of the solution in
the point $x_0$. To result \eqref{iv} it needs to estimate the
integrals in the right hand side of \eqref{ii1}. Those estimations
are based on the following inequalities
\begin{gather}\label{pr2} 
\int_{Q^{(2)}_{3R}}|\nabla v|\,dx\le C(n,p,q,M)R^{n-1}, \\
\label{pr1} 
\int_{Q_{3R}}|\nabla w|\,dx\le C(n,p,q,M) R^{n-1}
\end{gather}
that will we set now.

Choosing in  \eqref{id} the test function
$$
  \varphi (x)=(M_6-u(x)+R)^{1-p}\eta^p(x),
$$
where $\eta\in {\it C}^{\infty}_0(Q_{4R})$ is a radial-symmetric
with respect to the center of the ball  $Q_{4R}$ cutoff function,
satisfying the condition $0\le\eta\le 1$, $\eta=1$ in $Q_{3R}$ and
$|\nabla\eta|\le CR^{-1}$, we obtain
\begin{align*}
&\int_{Q_{4R}}\omega_{\varepsilon}(x)|\nabla
u|^{p(x)}(M_6-u+R)^{-p}\eta^p\,dx  \\
&\le C(p)\int_{Q_{4R}}\omega_{\varepsilon}(x)|\nabla
u|^{p(x)-1}(M_6-u+R)^{1-p} \eta^{p-1}|\nabla\eta|\,dx.
\end{align*}
Appling Young's inequality  to the integrant in the right-hand side gives
\begin{align*}%\label{a1}
&\int_{Q_{4R}}\omega_{\varepsilon}(x)|\nabla
u|^{p(x)}(M_6-u+R)^{-p} \eta^p\,dx \\
&\le C(p)\int_{Q_{4R}}\omega_{\varepsilon}(x)(M_6-u+R)^{p(x)-p}
|\nabla\eta|^{p(x)}\,dx.
\end{align*}
and
\[
 \int_{Q_{4R}^{(2)}}|\nabla u|^{p}(M_6-u+R)^{-p} \eta^p\,dx\le
C(p)\int_{Q_{4R}}(M_6-u+R)^{p(x)-p} |\nabla\eta|^{p(x)}\,dx.
\]
Repeating now the considerations of lemma \ref{l2}, used in the
setting of estimate \eqref{l27} from relation \eqref{yua}, it can be 
obtained
$$
\int_{Q^{(2)}_{4R}}|\nabla v|^{p}\eta^p\,dx
\le C(p,q,M)\Big(R^{q-p}\int_{Q^{(1)}_{4R}}
|\nabla\eta|^{q}\,dx+\int_{Q^{(2)}_{4R}}|\nabla\eta|^{p}dx\Big).
$$
From this, considering the choice of the cutoff function, follows the
estimate
\begin{equation}
\label{a2} \int_{Q^{(2)}_{3R}}|\nabla v|^{p}\,dx
\le C(n,p,q,M)R^{n-p},
\end{equation}
that leads to \eqref{pr2}. In particular since $p>q$, following
Young's inequality
$$
|\nabla\tilde v|^{q}<R^{p-q}|\nabla \tilde v| ^{p}+R^{-q}
$$
in the domain $Q^{(1)}_{3R}$ \eqref{a2} and radial simmetricity of
$\eta$ we find that
\begin{equation}
\label{a3} \int_{Q^{(1)}_{4R}}|\nabla \tilde
v|^{q}\eta^p\,dx\le R^{p-q}\int_{Q^{(1)}_{4R}}|\nabla \tilde
v|^{p}\tilde \eta^p\,dx +R^{n-q}\le C(n,p,q,M)R^{n-q}.
\end{equation}

To proof estimate \eqref{pr1} we use more complicated test function.
Note that $u(x)>\tilde u(x)$ on the set $G_R\subset B^{(1)}_{4R}$
(see\eqref{g1}). Let us chose in integral identity \eqref{id}
$$
  \varphi (x)=\begin{cases}
     \left(( M_6-u(x)+R)^{1-q}-
  (M_6-\tilde u(x)+R)^{1-q}\right )
             \eta^p(x)&\text{in } G_R\\
  0 &\text{in }Q_{4R}\setminus G_R,
\end{cases}
 $$
where $\eta$ has the same sense as above. Then
\begin{align*}
&\int_{G_R}|\nabla u|^{q}(M_6-u+R)^{-q}\eta^p\,dx \\
& \le \int_{G_R}|\nabla u|^{q-1}|\nabla\tilde u| (M_6-\tilde u+R)^{-q}\eta^p\,dx \\
&\quad +\frac{p}{q-1}\int_{G_R}|\nabla u|^{q-1}(M_6-u+R)^{-q} 
|\nabla\eta|\eta^{p-1}\,dx.
\end{align*}

Applying Young's inequality to the integrant in the right-hade side,
and using the definition $G_R$, one  gets
\begin{align*}
&\int_{G_R}|\nabla u|^{q}(M_6-u+R)^{-q}\eta^p\,dx \\
&\le C(p,q)\Big(\int_{G_R}|\nabla \tilde u|^{q} (M_6-\tilde
u+R)^{-q}\eta^p\,dx +\int_{G_R}|\nabla\eta|^{q}\,dx\Big ).
\end{align*}
From this, by  relation  \eqref{t} and the choice the cutoff function
 we obtain 
$$
\int_{G_R}|\nabla v|^{q}\eta^p\,dx\le C(n,p,q)\Big(
\int_{G_R}|\nabla \tilde v|^{q}\eta^p\,dx+R^{n-q}\Big ).
$$
Thus,
\begin{align*}
\int_{Q^{(1)}_{4R}}|\nabla w|^{q}\eta^p\,dx
&= \int_{Q^{(1)}_{4R}\setminus G_R}|\nabla\tilde
v|^{q}\eta^p\,dx+ \int_{G_R}|\nabla v|^{q}\eta^p\,dx\\
&\le  C(n,p,q)\Big(\int_{B^{(1)}_{4R}}|\nabla \tilde
v|^{q}\eta^p\,dx+R^{n-q}\Big ).
\end{align*}
Considering \eqref{a3},
$$
\int_{Q^{(1)}_{3R}}|\nabla w|^{q}\,dx\le C(n,p,q,M)R^{n-q},
$$
which together with \eqref{pr2} gives \eqref{pr1}.

Now let us estimate the integrals in the right-hand side of
\eqref{ii1}. We use the assumption \eqref{o1} for the first one and
note that  $|F\cap Q^{(2)}_{3R}| \ge {\rm const.} |Q_{3R}|$ (see
\eqref{f1}). Since $v(x)\le\ln 2$ on $F$ and $w(x)=v(x)$ in
$Q^{(2)}_{3R}$, then for $E=\{x\in \ Q_{3R}: w(x)\le\ln 2\}$ we have
the estimate $|E|\ge {\rm const.}|Q_{3R}|$. Therefore by inequality
\eqref{fr} in the ball $Q_{3R}$,
$$
 \int_{Q_{3R}\setminus E}\left |w-\ln 2\right |\,dx\le C(n)R
 \int_{Q_{3R}} |\nabla w|\,dx
$$
and according to \eqref{pr1},
$$
 \int_{Q_{3R}} w\,dx\le C(n,p,q,M)R^n.
$$
The second integral in \eqref{ii1} may be estimated similarly.
Really, $|E\cap Q^{(2)}_{3R}|\ge {\rm const.}|Q_{3R}|$ and again by
the inequality  \eqref{fr} in the truncated ball $Q^{(2)}_{3R}$,
$$
 \int_{Q^{(2)}_{3R}\setminus E}\left |v-\ln 2\right |\,dx\le C(n)R
   \int_{Q^{(2)}_{3R}} |\nabla v|\,dx.
$$
The inequality \eqref{pr2} leads us to the estimate
$$
\int_{Q^{(2)}_{3R}} v\,dx\le C(n,p,q,M)R^n,
$$
that proves \eqref{o4}. 
\end{proof}

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\end{document}