\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 278, pp. 1--11.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2016 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2016/278\hfil Harnack type inequality] {Harnack type inequality for non-negative solutions of second-order degenerate parabolic equations in divergent form} \author[S. T. Huseynov \hfil EJDE-2016/278\hfilneg] {Sarvan T. Huseynov } \address{Sarvan T. Huseynov \newline Baku State University, Baku, AZ1148, Azerbaijan} \email{sarvanhuseynov@rambler.ru} \thanks{Submitted May 30, 2016. Published October 18, 2016.} \subjclass[2010]{35K65, 35K10} \keywords{Degenerate equations; weighted Sobolev space; weak solution; \hfill\break\indent Poincare inequality; Harnack inequality} \begin{abstract} We study a class of second-order degenerate parabolic equations in divergent form. We prove two analogues of the Harnack inequality, one for non-negative weak solutions, an another for non-negative solutions. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \allowdisplaybreaks \section{Introduction} Let $\mathbb{R}^n$ be a Euclidean space of the points $x=(x_1 ,x_2 ,\dots ,x_n )$ and $D$ be a bounded domain in $\mathbb{R}^{n+1}$ with the parabolic boundary $\Gamma(D)$, $(0,0)\in D$. Consider the parabolic equation \begin{equation} \label{s1} Lu=\frac{\partial u}{\partial t} -\sum _{i,j=1}^n\frac{\partial }{\partial x_i } \Big(a_{ij} (x,t)\frac{\partial u}{\partial x_{j} } \Big) =0,\quad (x,t)\in D, \end{equation} and assume that $\{a_{ij} (x,t)\}$ is a real symmetric matrix with measurable elements and for all $(x,t)\in D$ and $\xi \in \mathbb{R}^n$ the following condition is fulfilled: \begin{equation} \label{s2} \gamma \sum _{i=1}^n\lambda _i (x,t)\xi _i^2 \le \sum _{i,j=1}^na_{ij} (x,t)\xi _i \xi _{j} \le \gamma ^{-1} \sum _{i=1}^n\lambda _i (x,t)\xi _i^2 , \end{equation} where $\gamma \in (0,1]$ is a constant, \begin{gather*} \lambda _i (x,t)=g_i\big(\rho (x)+\sqrt{|t|} \big), \\ \rho(x)=\sum_{i=1}^nw_i (|x_i|),\quad g_i(z)=\frac{(w_i^{-1}(z))^2 }{z^2 } ,\quad i=1,2,\dots ,n. \end{gather*} We assume that the functions $w_i (t)$ increase strictly monotonically, $w_i (0)=0$, $w_i^{-1} (t)$ is the function inverse to $w_i (t)$ and for $i=1,2,\dots ,n$, \begin{gather} \label{s3} w_i (2t)\le 2w_i (t), \\ \label{s4} \Big(\frac{w_i (t)}{t} \Big)^{q-1} \int_0^{w_i^{-1} (t)}(\frac{w_i(z)}{z} )^{q} dz\le c_1t \end{gather} with some constant $q>n$ and positive constant $c_1 $ independent of $t$. A simple example of function $w_i$ is $w_i(t)=t^{\alpha_i}$ where \[ \alpha_i\ge \frac{-1+\sqrt{1+4q(q-1)}}{2(q-1)}. \] The principal result of this article is the Harnack type inequality for non-negative weak solutions of equation \eqref{s1}. For uniformly second-order parabolic equations of divergent structure, with discontinuous coefficients the Harnack inequality was obtained in the well known paper by Nash \cite{n1}. Moser \cite{m1} obtained another proof of this fact. For parabolic equations of divergent structure with uniform degeneration we refer to \cite{c1,f1}. When $w_i (t)$ are power functions, the Harnack type inequality was obtained in \cite{g1}. Now we introduce some notation: let $D$ be a cylindrical domain $\Omega \times [T_0,T]$, where $\Omega$ is a bounded domain in $\mathbb{R}^n$ and $-\infty 1$ will be defined later (it is the exponent of the imbedding theorem corresponding to the weights $\lambda_i$). Now we state a Sobolev-type embedding theorem with weights, whose proof can be found in \cite{g1}. We set $$ -\hspace{-3.5mm}\int _{\Pi_R}f\,dx=\frac{1}{|\Pi_R|}\int_{\Pi_R}f\,dx,\quad -\hspace{-3.5mm}\int_{S_\rho}g\,dx\,dt=\frac{1}{|S(\rho)|}\int_{S(\rho)}g\,dx\,dt. $$ \begin{theorem}[Sobolev theorem with weights] \label{thm1} For any function $\varphi\in W_{2,\Lambda }^{1,0}(S(\rho))$ with zero trace on the lateral boundary of $S(\rho)$, and any $R\le R_0$ it holds \begin{equation}\label{sob} \begin{aligned} \Bigl (-\hspace{-3.5mm}\int_{S_\rho}|\varphi|^{2\sigma}\,dx\,dt\Bigr)^{1/\sigma} &\le c\Big( \sup_{t\in (-(\frac{1}{3} +\rho )R^2 ,-(\frac{3}{4}-\rho )R^2)} -\hspace{-3.5mm}\int_{\Pi(R)}\varphi^2\,dx \\ &\quad +R^2-\hspace{-3.5mm}\int_{S_\rho}\sum _{i=1}^n\lambda _i (x,t)\varphi_{x_i }^2\, dx\,dt\Bigr ), \end{aligned} \end{equation} for $\sigma>1$, where the constant $c$ does not depend on $\varphi$, $R$ and $\rho$. \end{theorem} \begin{theorem} \label{thm2.1} Let $u(x,t)$ be a non-negative weak solution of \eqref{s1} with coefficients that satisfy \eqref{s2}--\eqref{s4}. For any $r>0$ and $1/3\le \rho '<\rho \le 1/2$ it holds \begin{equation} \label{s5} \sup_{S(\rho ')} u \le c (\rho -\rho ')^{-\xi} \Bigl(-\hspace{-3.5mm}\int_{S(\rho )}u^{r}\, dx\,dt \Bigr)^{-1/r}, \end{equation} where positive constants $c$ and $\xi$ depend only on $q$, $c_1$, $n$, $r$ and $\gamma$. \end{theorem} \begin{proof} First, we prove the statement of this theorem for $r=2$. The case $r>2$ follows then by the H\"older inequality. To treat the case $r\in (0,2)$, we use an additional iteration. Take a function $\eta$ such that $\eta (x,t)=1$ in $S(\rho ')$, $\eta (x,t)=0 $ outside of $S(\rho )$, $0\le \eta (x,t)\le 1$, and there exists a constant $c(n)$ such that \begin{equation}\label{tes} |\eta _{x_i} |\le \frac{c}{(\rho -\rho ')w_i^{-1} (R)} ,\quad i=1,2,\dots ,n;\quad |\eta _{t} |\le \frac{c}{(\rho -\rho ')R^2 } . \end{equation} In \eqref{IntId} choose a test function $\psi=u^\beta \eta^2$, where $\beta>0$. We obtain \begin{align*} &\sup_{t\in (-(\frac{1}{3} +\rho )R^2,-(\frac{3}{4} -\rho )R^2 )} \frac{1}{\beta+1}\int_{\Pi(\rho)} u^{\beta+1} \eta^2 \, dx + \beta \int_{S(\rho)} u^{\beta-1}\eta^2 a_{ij} u_{x_i}u_{x_j}\, dx\,dt\\ &=\frac{2}{\beta+1}\int_{S(\rho)} u^{\beta+1}\eta\eta_t \, dx\,dt -2\int_{S(\rho)}u^\beta \eta a_{ij} u_{x_i} \eta_{x_i}\, dx\,dt. \end{align*} Let $v=u^{(\beta+1)/2}$. Using \eqref{s2} and the Young inequality, we arrive at \begin{align*} &\sup_{t\in (-(\frac{1}{3}+\rho )R^2,-(\frac{3}{4}-\rho )R^2)} \int_{\Pi(\rho)} v^2 \eta^2 \, dx+ \frac{4\beta}{\beta+1}\int_{S(\rho)} \eta^2 v_{x_i}^2 \lambda_i(x,t)\, dx\,dt\\ &\leq C \int_{S(\rho)} v^2\eta|\eta_t| \, dx\,dt+C(\beta+1) \beta^{-1} \int_{S(\rho)}v^2 \eta_{x_i}^2 \lambda_i(x,t)\,dx\,dt \end{align*} The above integral is taken over the set $S(\rho )\setminus S(\rho ')$, since $\eta _{x_i } =0$ in $S(\rho ')$. But in this set $$ \rho (x)\le cR, \quad \sqrt{|t|} \le R, \quad \lambda _i (x,t)\le c \frac{(w_i^{-1}(R))^2 }{R^2 }; $$ therefore, \begin{equation*} \int_{S(\rho )}\eta ^2 \sum _{i=1}^n\lambda _i (x,t)v_{x_i }^2dx\,dt \le \frac{c (\beta +1)^2 }{\beta ^2 (\rho -\rho ')^2 R^2 } \int_{S(\rho )}v^2 dx\,dt . \end{equation*} On the other hand, we have \begin{equation*} \sup_{t\in (-(\frac{1}{3} +\rho )R^2 ,-(\frac{3}{4} -\rho )R^2 )}\int_{\Pi_{\rho R} }v^2 \eta ^2\,dx \le C \frac{\beta+1}{\beta}(\rho -\rho ')^{-2} R^{-2} \int_{S(\rho )}v^2\, dx\,dt. \end{equation*} For $\beta\geq 1$ these estimates take the form \begin{gather} \label{s6} \sup_{t\in (-(\frac{1}{3} +\rho )R^2 ,-(\frac{3}{4} -\rho )R^2 )} -\hspace{-3.5mm}\int_{\Pi_{\rho R} }\eta ^2 v^2\, dx \le c (\rho -\rho ')^{-2} -\hspace{-3.5mm}\int_{S(\rho )}v^2\, dx\,dt, \\ \label{s7} -\hspace{-3.5mm}\int_{S(\rho)}\sum _{i=1}^n\lambda _i (x,t) v_{x_i }^2 \eta ^2\, dx\,dt \le c (\rho -\rho ')^{-2} R^{-2} -\hspace{-3.5mm}\int_{S(\rho )}v^2\, dx\,dt . \end{gather} Further more we assume that $\beta \geq 1$. Applying \eqref{s6},\eqref{s7} and the embedding theorem \eqref{sob} we obtain \begin{equation} \label{s10} \begin{aligned} &\Bigl(-\hspace{-3.5mm}\int_{S(\rho ')}v^{2\sigma }\, dx\,dt \Bigr)^{1/\sigma}\\ &\le c \Bigl(-\hspace{-3.5mm}\int_{S(\rho)}v^{2\sigma } \eta ^{2\sigma }\, dx\,dt \Bigr)^{1/\sigma} \\ &\le c \Bigl(\sup_{t\in (-(\frac{1}{3} +\rho )R^2 ,-(\frac{3}{4} -\rho )R^2 )} -\hspace{-3.5mm}\int_{\Pi _{\rho R} }\eta ^2 v ^2\, dx+R^2 -\hspace{-3.5mm}\int_{S(\rho )}\sum _{i=1}^n\lambda _i (x,t)(v \eta )_{x_i }^2 \, dx\,dt\Bigr)\\ &\le c (\rho -\rho ')^2 -\hspace{-3.5mm}\int_{S(\rho )}v^2 \,dx\,dt . \end{aligned} \end{equation} We define the sequences \begin{gather*} \rho '_m =\rho '+\frac{\rho -\rho '}{2^{m+1} },\quad \rho _m =\rho '+\frac{\rho -\rho '}{2^{m} } ,\\ \beta _m=2\sigma^m-1,\quad v_m =u^{\frac{\beta _m +1}{2} } . \end{gather*} Then from \eqref{s10} we deduce \begin{align*} \phi_{m+1}:= &\Bigl (-\hspace{-3.5mm}\int_{S(\rho _{m+1} )}u^{2\sigma^{m+1}} \, dx\,dt \Bigr)^\frac{1}{2\sigma ^{m+1}} \\ =&\Bigl (-\hspace{-3.5mm}\int_{S(\rho _{m+1} )}v_{m+1}^2 \, dx\,dt \Bigr )^\frac{1}{2\sigma ^{m+1}}\\ =&\Bigl(-\hspace{-3.5mm}\int_{S(\rho_m')} v_m^{2\sigma} \, dx\,dt \Bigr)^\frac{1}{2\sigma ^{m+1}} \\ \le& \Bigl (c (\rho_m-\rho_m' )^{-2} -\hspace{-3.5mm}\int_{S(\rho_m )}v_m^2\, dx\,dt \Bigr)^\frac{1}{2\sigma^m} \\ \leq& (c 2^m (\rho-\rho')^{-2})^\frac{1}{2\sigma^m}\phi_m. \end{align*} It easily follows that $$ \phi_{m+1}\leq C (\rho-\rho')^{\sigma/(1-\sigma)} \phi_0,\quad m\geq 0. $$ Thus, \begin{equation*} \limsup_{m\to \infty } \Bigl(-\hspace{-3.5mm}\int_{S(\rho _m )}u^{2\sigma^m} \, dx\,dt \Bigr )^\frac{1}{2\sigma ^m} \leq C (\rho-\rho')^{\sigma/(1-\sigma)} \Bigl (-\hspace{-3.5mm}\int_{S(\rho_0 )}u^2 \, dx\,dt \Bigr )^{1/2} \end{equation*} The statement of the theorem for $r=2$ easily follows, in view of the well-known property $$ \operatorname{ess\,sup} \{u; A\} =\limsup_{q\to\infty} \Bigl(~\int_A u^q\,dx\,dt \Bigr )^{1/q}. $$ The statement of the theorem for $r>2$ follows by a direct application of the H\"older inequality $$ \Bigl(-\hspace{-3.5mm}\int_{S(\rho )}u^2\, dx\,dt \Bigr )^{\frac{1}{2}} \le \Bigl(-\hspace{-3.5mm}\int_{S(\rho )}u^{r}\, dx\,dt \Bigr )^{-1/r}, \quad r>2. $$ Now, we treat the case $r\in (0,2)$. Here we need an additional iteration. In the integral identity \eqref{IntId} we choose the test function $\psi=u^\beta \eta^2$, where $\beta=-1+r$, and the cut-off function $\eta$ has the same meaning as in \eqref{tes}. We arrive at the estimate \eqref{s10} with the constant $c$, which depends on $r$. Iterating this relation as above, by a finite number of steps we obtain $$ -\hspace{-3.5mm}\int_{S(\rho^\prime)}u^2\, dx\,dt \le c (\rho -\rho ')^{-\xi_0}\Bigl(-\hspace{-3.5mm}\int_{S(\rho)}u^{r}\, dx\,dt \Bigr)^{-1/r}, $$ where positive constants $c$ and $\xi_0$ depend only on $q$, $c_1$, $n$, $r$ and $\gamma$. Combining this inequality with the estimate \eqref{s5} obtained earlier for $r\ge 2$, and using that $\rho'$ can be taken arbitrarily, we arrive at the desired statement. \end{proof} Now let \begin{equation*} Q(\rho )=\Pi_{\rho R} \times (-\rho ^2 R^2 ,0);\quad \rho \in (0,1). \end{equation*} The following statement is proved as in the previous theorem. The only difference is that the value of $\beta$ in the proof is taken to be less than $-1$. \begin{lemma} \label{lem2.1} Let $r>0$ and $u(x,t)$ be a weak non-negative solution of \eqref{s1}. Then the following estimate holds \begin{equation*} \inf_{Q(\rho ')} u \ge c(\rho -\rho^\prime)^{-\xi} \Bigl(-\hspace{-3.5mm}\int_{Q(\rho )}u^{-r}\, dx\,dt \Bigr)^{-1/r} \end{equation*} where $1/3 \le \rho'<\rho \le 1/2$. \end{lemma} The next Lemma is a variant of Theorem \ref{thm2.1} with a slightly different choice of the outer and inner cylinders. \begin{lemma} \label{lem2.2} Let the conditions of the previous lemma be fulfilled. Then the following estimate is valid \begin{equation*} \sup _{Q(1/3 )} u\le c \Bigl(-\hspace{-3.5mm}\int_{Q(1/2)}u^2\, dx\,dt \Bigr )^{1/2}. \end{equation*} \end{lemma} \section{Harnack type inequality} The technique of this section is based on ideas from \cite{m1}. \begin{theorem} \label{thm3.1} Let $u(x,t)$ be a non-negative weak solution of equation \eqref{s1}. Then there exist the constants $a_1 (\Lambda ,n)$ and $a_2 (\Lambda ,n)$ such that for any $s>0$, \begin{gather*} |\{(x,t)\in D_1 ,\, \ln u(x,t)>s+a_1 \}|\le c \frac{R^2|\Pi_{R}|}{s}, \\ |\{(x,t)\in D_2 ,\, \ln u(x,t)<-s+a_1 \}|\le c \frac{R^2|\Pi_{R}|}{s}, \end{gather*} where \begin{equation*} D_1 =\Pi_{R/2} \times (-R^2,-\frac{R^2 }{2}),\quad D_2=\Pi_{R/2} \times (-\frac{R^2}{2} ,0). \end{equation*} \end{theorem} \begin{proof} Assume $\upsilon (x,t)=-\ln u(x,t)$ and let $\eta (x,t)=\xi (x)w(t)$, where $w(t)=1$ for $t\le -\tau _1R^2$, $w(t)=0$ for $t\ge -\frac{\tau _1 }{2} R^2$, $0\le w(t)\le 1$, $|w_{t} |\le \frac{c }{\tau _1 R^2} $; and $\xi (x)=1$ for $x\in \Pi_{R/2}$, $\xi (x)=0$ for $x\notin\Pi _{\frac{5R}{2} }$, $0\le \xi(x)\le 1$, $|\xi _{x_i } |\le \frac{c}{w_i^{-1} (R)}$; $i=1,2,\dots ,n$ moreover for $0<\tau _1 <1$, and the function $\xi(x)$ such that for an arbitrary $C$ the set $\{x:\xi(x)\geq C\}$ is convex. From Theorem \ref{thm2.1} we have (if only $\tau_1<\tau_2\leq1$) \begin{equation}\label{s11} \int_{\Pi_\frac{5R}{6} } v\xi ^2\, dx \Bigl|_{-\tau_1 R^2}^{-\tau _2 R^2} +\frac{\gamma }{2} \int_{-\tau _2 R^2 }^{-\tau _1 R^2}\,dt \int_{{\Pi_{\frac{5R}{6} } } }\xi ^2 \sum_{i=1}^n\lambda _i (x,t)v _{x_i }^2\, dx \leq c (\tau _2 -\tau _1 ) |\Pi_{R}|. \end{equation} Indeed, since $\eta _{t} =0$ for $t\in (-\tau _2 R^2 ,-\tau_1 R^2 )$, according to Theorem \ref{thm2.1}, the left-hand side of \eqref{s11} is estimated by \begin{equation*} J=c (\gamma )\int_{-\tau _2 R^2 }^{-\tau _1 R^2 }dt \int_{\Pi_{R} \backslash \Pi_{R\mathrm{/2}} } \sum_{i=1}^n\lambda _i (x,t)\xi _{x_i }^2\, dx. \end{equation*} (since $\xi _{x_i } \equiv 0$ in $\Pi_{R\mathrm{/2}} $). Note that, for $x\in \Pi_{R} \backslash \Pi_{R/2}$, $w_i(|x_i |)\le cR$. Thus, $\rho (x)+\sqrt{|t|} \le cR$, i.e. $\lambda _i (x,t)\le c \frac{(w_i^{-1}(R))^2 }{R^2 }$. Hence we deduce that \begin{equation*} \sum _{i=1}^n\lambda _i (x,t) \xi _{x_i }^2 \le c \frac{(w_i^{-1} (R))^2 }{R^2 } \frac{1}{(w_i^{-1} (R))^2 } =\frac{c}{R^2 } . \end{equation*} So, \begin{equation*} J\le \frac{c}{R^2 } (\tau _2 -\tau _1 )R^2 =c(\tau _2 -\tau _1 ) \end{equation*} and \eqref{s11} is proved. \end{proof} Now consider the functions \begin{equation*} V(t)=\frac{\int_{\Pi_{R} }v(x,t)\xi ^2 (x)\,dx} {\int_{\Pi_{R} }\xi ^2 (x)\,dx}, \quad D(t)=\frac{\int_{\Pi_{R}}\big(v(x,t)-V(t)\big)^2{\xi ^2 (x)\,dx}} {\int_{\Pi_{R} }\xi ^2(x)\,dx}. \end{equation*} By the Poincare inequality \cite{g1}, we have \begin{equation*} D(t)\Big(\int_{\Pi_{R} }\xi ^2 (x)\,dx\Big)^2 \le c R^2 |\Pi_{R}| \int_{\Pi_{R} }\xi ^2 (x)\sum _{i=1}^n\lambda _i (x,t)v_{x_i }^2\, dx, \end{equation*} that together with \eqref{s11} gives \begin{equation*} V(-\tau _1 \cdot R^2 )-V(-\tau _2 \cdot R^2 )+\frac{c} {R^2 |\Pi_{R}|} \int_{-\tau _2 R^2 }^{-\tau _1 R^2 }dt \int_{\Pi_{R/2} }(v-V)^2\, dx \le c (\tau _2 -\tau _1 ). \end{equation*} When let $\tau _2 $ to $\tau _1 $ and assume $t=-\tau _1 R^2$. Then it follows from the above inequality that \begin{equation} \label{s12} R^2 \frac{dV}{dt} +\frac{c}{|\Pi_{R}|} \int_{\Pi_{R/2} }(v-V)^2\, dx\le c. \end{equation} Now consider the functions \begin{gather*} \omega (x,t)=v(x,t)+\frac{c}{R^2 } (-\frac{R^2}{2} -t), \\ W(t)=V(t)+\frac{c}{R^2 } (-\frac{R^2 }{2}-t). \end{gather*} Then from \eqref{s12} we deduce \begin{equation} \label{s13} R^2 \frac{dW}{dt} +\frac{c}{|\Pi_{R}|} \int_{\Pi_{R/2}} (\omega -W)^2\, dx\le 0\,. \end{equation} From \eqref{s13} it follows that the function $W(t)$ does not increase with respect to $t$, therefore for all $t\in (-R^2 ,-\frac{R^2 }{2})$, we have \begin{equation*} W(t)\ge W(-\frac{R^2 }{2} )=V(-\frac{R^2 }{2} ). \end{equation*} By the same reason, for $t\in (-\frac{R^2 }{2} ,R^2 )$, we have \[ W(t)\le W(-\frac{R^2 }{2} )=V(-\frac{R^2 }{2} ). \] Assume that $s_1 0$, \begin{equation*} \operatorname{meas}\big\{(x,t)\in D_1:\omega (x,t)<-s+V(-\frac{R^2 }{2} )\big\} \le c \frac{R^2 |\Pi_{R}|}{s}, \end{equation*} and \begin{equation}\label{s14} \operatorname{meas}\big\{(x,t)\in D_1:\ln u(x,t)>s-V(-\frac{R^2}{2}) +\frac{c}{R^2}(-\frac{R^2}{2}-t)\big\} \le c \frac{R^2|\Pi_{R}|}{s}. \end{equation} Now it suffices to take into account that $t\in (-R^2,-\frac{R^2 }{2} )$, and from \eqref{s14} it follows that for $a_1 =-V(-\frac{R^2 }{2} )+\frac{c}{2}$, \begin{align*} &\operatorname{meas}\big\{(x,t)\in D_1:\ln u(x,t)>s+a_1 \big\}\\ &\le \operatorname{meas}\big\{(x,t)\in D_1:\ln u(x,t)>s-V(-\frac{R^2}{2} ) +c (-\frac{R^2 }{2} -t)\big\}\\ &\le c \frac{R^2|\Pi_{R}|}{s} \end{align*} and the right side of the statement of the lemma is proved. Its second part is proved in the same way. Indeed, it suffices to obtain $s_2 >V(-\frac{R^2 }{2} )$ and \begin{equation*} m_2(s_2 )=|\{(x,t)\in D_2 :\, \, \omega (x,t)>s_2\}|. \end{equation*} Then \begin{equation*} m_2 (s_2 )\le c \frac{R^2|\Pi_{R}|}{(s_2-V(-\frac{R^2 }{2} ))} , \end{equation*} i.e. for any $s>0$ and \[ a_2 =-V(-\frac{R^2 }{2} )-\frac{c}{2} \] we have \begin{equation*} |\{(x,t)\in D_2:\ln u(x,t)<-s+a_2 \}|\le c \frac{R^2|\Pi_{R}|}{s}\,. \end{equation*} The proof is complete. It is easy to see that \begin{equation*} a_1 -a_2 =c. \end{equation*} Now consider the functions $\omega _1 (x,t)=u(x,t)e^{-a_1 } $ and $\omega _2 (x,t)=(u(x,t))^{-1} e^{a_2 } $, where $u(x,t)$ is a non-negative weak solution of equation \eqref{s1}. Let $\frac{1}{3} \le \rho '<\rho \le \frac{1}{2}$, $r_{\nu } =\sigma ^{-\nu } (1+\sigma )^{-1}$, $\nu=0,1,2,\dots$; $s_1 (\rho )=S(\rho )$, $s_2 (\rho )=Q(\rho)$. In fact, from Theorem \ref{thm3.1} it follows that \begin{gather*} \sup_{s_{j}(\rho ')}\omega_{j}^{r_{\nu}} \le c (\rho -\rho ')^{-(n+1)} \Bigl(-\hspace{-3.5mm}\int_{s_j(\rho)}\omega_j^2\,dx\Bigr )^{1/2}, \\ \big|\big\{(x,t)\in s_{j} (\frac{1}{2} ), \ln \omega_{j} >s\big\}\big|\le c \frac{R^2 |\Pi_{R}|}{s}, \end{gather*} where $j=1,2$. \begin{lemma} \label{lem3.1} If the conditions of the previous theorem are fulfilled, then the following estimates hold: \begin{equation*} \sup_{s_{j}(\frac{1}{3})}\omega _{j} \le c,\quad j=1,2. \end{equation*} \end{lemma} \begin{proof} It is obvious that it suffices to prove the lemma for $j=1$. Consider the function $\varphi (\rho )=\sup _{s(\rho )} \ln \omega _1 (x,t)$, and let $\kappa =\max \{c , 1\}$. Then $\varphi (\rho )$ does not decrease with respect to $\rho $. If $\varphi (1/3)\le 3\kappa $, then the lemma is proved with $c =e^{3\kappa } $. Now let $\varphi (1/3)>3\kappa $. Then for $\rho \in [1/3 ,1/2]$, \begin{equation*} \varphi (\rho )>3\kappa \end{equation*} We show that for $\rho '$ and $\rho $ satisfying \begin{equation*} \frac{1}{3} \le \rho '<\rho \le \frac{1}{2} , \end{equation*} the it holds \begin{equation} \label{s15} \varphi (\rho ')<\frac{3}{4} \varphi (\rho )+c (\rho -\rho ')^{-8(n+1)}. \end{equation} Let $s(\rho )=s^{1} (\rho )+s^2 (\rho )$, where \begin{gather*} s^{1}(\rho )=\{(x,t)\in s(\rho ):\frac{1}{2}\varphi(\rho )<\ln \omega_1 (x,t)\le \varphi (\rho)\}, \\ s^2 (\rho )=\{(x,t)\in s(\rho ):\frac{1}{2} \varphi (\rho )\ge \ln \omega _1 (x,t)\}. \end{gather*} We have \begin{align*} -\hspace{-3.5mm}\int_{s(\rho)} \omega _1^{2r_{\nu}}\,dx\,dt &=\frac{1}{R^2|\Pi_{\rho}|}\Bigl (-\hspace{-3.5mm}\int_{s^1(\rho)}\omega _1^{2r_{\nu}}\,dx\,dt+-\hspace{-3.5mm}\int_{s^2(\rho)}\omega _1^{2r_{\nu}}\,dx\,dt\Bigr )\\ &\le \frac{1}{R^2|\Pi_{\rho}|}\Bigl(c \frac{R^2|\Pi_{R}| }{\frac{1}{2} \varphi (\rho )} e^{2r_{\nu } \varphi (\rho )}+R^2 |\Pi_{\rho}| e^{r_{\nu } \varphi (\rho )}\Big)\\ &\le \frac{\kappa }{\varphi (\rho )} e^{2r_{\nu } \varphi (\rho )} +e^{r_{\nu } \varphi (\rho )}. \end{align*} Since $\frac{\kappa }{\varphi (\rho )} <1/3 $, then for any $\rho \in [\frac{1}{3} ,\frac{1}{2} ]$ there exists $r_{\nu } $ such that \begin{equation*} \frac{\kappa }{\varphi (\rho )} e^{2r_{\nu } \varphi (\rho )} \le e^{r_{\nu } \varphi (\rho )} \end{equation*} and we can choose the non-negative integer $\nu $ so large that \begin{equation*} r_{\nu } =\sigma ^{-\nu } (1+\sigma )^{-1} \le \frac{1}{\varphi (\rho )} \ln \frac{\varphi (\rho )}{\kappa } , \end{equation*} and furthermore for any $\rho \in [\frac{1}{3} ,\frac{1}{2} ]$ \begin{equation*} r_{\nu } \sigma =\frac{\sigma }{\sigma ^{\nu } (1+\sigma )} >\frac{1}{\varphi (\rho )} \ln \frac{\varphi (\rho )}{\kappa } , \end{equation*} since $\sigma >1$ and $\kappa \ge 1$, $\frac{\varphi (\rho )}{\kappa } >3$; therefore, \begin{equation*} \frac{1}{\varphi (\rho )} \ln \frac{\varphi (\rho )}{\kappa } =\frac{1}{\kappa }\cdot \frac{\ln \frac{\varphi (\rho )}{\kappa } } {\frac{\varphi (\rho )}{\kappa } } \le \frac{\ln 3}{3} <\frac{1}{2} . \end{equation*} We have taken into account that for $x\ge 3$ the function $\frac{\ln \chi }{\chi } $ decreases. Thus, we obtain \begin{align*} \varphi (\rho ') &=\sup_{s(\rho ')}\ln \omega _1 (x,t) =\frac{1}{2r_{\nu } \varphi (\rho )} \ln \sup _{s(\rho ')} \omega _1^{2r_{\nu }}\\ &\le \frac{1}{2r_{\nu } } \ln \big(c (\rho -\rho ')^{-2(n+1)} \big) +\frac{\varphi (\rho )}{2}. \end{align*} Then we have \begin{equation*} \varphi (\rho ')\le \frac{1}{2} \varphi (\rho )\big(\frac{\sigma}{\ln{\frac{\varphi (\rho )}{\kappa}}} \ln{(c (\rho -\rho ')^{-2(n+1)})} +1\big). \end{equation*} From the above estimate it follows \eqref{s15}. Indeed, if the first term of the right-hand side is no greater than $1/2$, then $\varphi (\rho ')\le \frac{3}{4} \varphi (\rho)$. But if \begin{equation*} \frac{\sigma }{\ln \frac{\varphi (\rho )}{\kappa }}\ln \Big( c (\rho -\rho ')^{-2(n+1)} \Big)>\frac{1}{2}, \end{equation*} then \begin{equation*} \ln \frac{\varphi (\rho )}{\kappa }<2\sigma \ln \big(c (\rho -\rho ')^{-2(n+1)} \big) \le 4\ln\big(c (\rho -\rho ')^{-2(n+1)}\big). \end{equation*} Hence it follows that \begin{equation*} \varphi (\rho ')\le \varphi (\rho )\le c (\rho -\rho ')^{-8(n+1)} , \end{equation*} and \eqref{s15} is proved. Now consider the sequence \begin{equation*} \rho _{j} =\frac{1}{2} -\frac{1}{\sigma (1+j)} ,\quad j=0,1,2,\dots . \end{equation*} and using \eqref{s15} we obtain \begin{align*} \varphi (\frac{1}{3} ) &=\varphi (\rho _0 )<\frac{3}{4} \varphi (\rho _1 )+\frac{c}{(\rho _1-\rho _0 )^{8(n+1)} } \\ &<(\frac{3}{4} )^2 \varphi (\rho _2 )+c \Big((\rho _1 -\rho _0 )^{-8(n+1)} +\frac{3}{4} (\rho _2-\rho _1 )^{-8(n+1)} \Big)\\ &<\dots< (\frac{3}{4} )^{m} \varphi (\rho _m )+c \sum _{j=0}^{m-1}(\frac{3}{4} )^{j} (\rho _{j+1} -\rho _{j})^{-8(n+1)} \\ &=(\frac{3}{4} )^{m} \varphi (\rho _m )+c \sum _{j=0}^{m-1}(\frac{3}{4} )^{j} \Big(\sigma (j+1)(2+j)\Big). \end{align*} From the continuity of the function $\omega _1$ it follows $\varphi (\frac{1}{2} )<\infty $, thus \begin{equation*} \varphi (\frac{1}{3} )\le 1+c \sum _{j=0}^{\infty }(\frac{3}{4} )^{j}\left (\sigma (j+1)(2+j)\right )\le c <\infty , \end{equation*} and the proof is complete. \end{proof} \begin{theorem} \label{thm3.2} Let $u(x,t)$ be a non-negative weak solution of \eqref{s1} whose coefficients satisfy conditions \eqref{s2}-\eqref{s4}. Then there exists a constant $c=c(\gamma, n, q, c_1)$ such that \begin{equation*} \sup _{S(1/3)} u\le c \mathop{\inf }_{Q(1/3)} u. \end{equation*} \end{theorem} \begin{proof} From Lemma \ref{lem3.1}, we have \begin{equation*} \sup _{S(1/3 )} \omega _1 (x,t)\sup _{Q(1/3)} \omega _2(x,t) =e^{-a_1 +a_2 } \sup _{S(1/3)} u(x,t)\sup _{Q(1/3)}(u(x,t))^{-1} \le c. \end{equation*} Thus, \begin{equation*} \sup _{S(1/3)} u(x,t)\le c\mathop{\inf }_{Q(1/3)} u(x,t), \end{equation*} and the proof is complete. \end{proof} \subsection*{Acknowledgements} The author wants to express his gratitude to the referees for their very valuable comments and suggestions. \begin{thebibliography}{00} \bibitem{c1} Chiarenza, F. M.; Serapioni, R. P.; \emph{A Harnack inequality for degenerate parabolic equations}. Comm. Part. Differ. Eq., 1984, vol. 9, pp. 719-749. \bibitem{f1} Fernandes, J.; \emph{Mean value and Harnack's inequalities for a certain class of degenerate parabolic equations}. Revista Math. 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