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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 238, pp. 1--7.\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-2017/238\hfil Well-posedness of a porous medium flow]
{Well-posedness of a porous medium flow with fractional pressure
in Sobolev spaces}

\author[X. Zhou, W. Xiao \hfil EJDE-2017/238\hfilneg]
{Xuhuan Zhou, Weiliang Xiao}

\address{Xuhuan Zhou \newline
Department of Information Technology,
Nanjing Forest Police College, 210023 Nanjing, China}
\email{zhouxuhuan@163.com}

\address{Weiliang Xiao \newline
School of Applied Mathematics,
Nanjing University of Finance and Economics,
 210023 Nanjing, China}
\email{xwltc123@163.com}

\thanks{Submitted December 22, 2016. Published October 3, 2017.}
\subjclass[2010]{35K55, 35K65, 76S05}
\keywords{Fractional porous medium equation; Sobolev space;
\hfill\break\indent  degenerate diffusion transport equation}

\begin{abstract}
 We prove the existence of a non-negative solution for a linear degenerate
 diffusion transport equation from which we derive the existence and
 uniqueness of the solution for  the fractional porous medium equation in
 Sobolev spaces $H^\alpha$ with nonnegative initial data, 
 $\alpha>\frac d2+1$.  We also correct a mistake in our previous paper
 \cite{zhou01}. 
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\allowdisplaybreaks

\section{Introduction}

We  consider the porous medium type equation
\begin{equation}\label{e1.1}
\partial_t u=\nabla\cdot(u\nabla p),\quad
 p=(-\Delta)^{-s}u,\quad  0<s<1, \quad u(x,0) \geq 0,
\end{equation}
where $x\in \mathbb{R}^n$, $n\geq 2$, and $t>0$ and the fractional 
Laplacian $(-\triangle)^{s/2}: =\Lambda^{s}$ is given by the
psuedo differential operator with symbol  $|\xi|^{s}$, that is:
$$
(-\triangle)^{s/2}f=\Lambda^sf=\mathcal{F}^{-1}|\xi|^s\mathcal{F}f.
$$
Using the Riesz potential, one can also define this operator as
$$
(-\triangle)^{s/2}f(x)=\Lambda^sf(x)
=c_{n,s}\int_{\mathbb{R}^n}\frac{f(x)-f(y)}{|y|^{n+s}}{\rm d}y.
$$

This model is based on Darcy's law with pressure, $p$,  is given by an 
inverse fractional Laplacian operator. It was first introduced by
 Caffarelli and V\'azquez \cite{caffarelli02}, in which they proved 
the existence of a weak solution when $u_0$ is a bounded function with 
exponential decay at infinity. For $\alpha=\frac{n}{n+2-2s}$, 
Caffarelli, Soria and V\'azquez \cite{caffarelli01} proved that the
 bounded nonnegative solutions are $C^\alpha$ continuous in a strip of 
space-time for $s\neq1/2$. And same conclusion for the index $s=1/2$ 
was proved by Caffarelli and V\'azquez in \cite{caffarelli03}.
\cite{carrillo01,caffarelli04,vazquez01} give a detailed description of 
the large-time asymptotic behaviour of the solutions of \eqref{e1.1}.
\cite{biler01, stan01} consider degenerate cases and show the existence and 
properties of self-similar solutions.
Allen, Caffarelli and Vasseur \cite{allen01} study the equation with fractional
time derivative, and proved the H\"older continuity for its weak solutions.

In this paper, we study the existence and uniqueness of solutions of 
\eqref{e1.1} in Sobolev spaces.  Unlike
considering the existence of  weak solution in $L^{\infty}$ or constructing 
approximate solutions of linear transport systems, we solve equation \eqref{e1.1}
by constructing solutions to a linear degenerate diffusion transport systems. 
The well-posedness and properties of the  linear degenerate diffusion 
transport are interesting results by themselves and lead us to proving that 
for  $s\in [\frac12,1)$, $\alpha>\frac d2+1$, $u_0\in H^\alpha(\mathbb{R}^n)$ 
nonnegative,  and some $T_0>0$, the equation \eqref{e1.1} in $\mathbb{R}^n\times[0,T_0]$ has
a unique solutions. Besides, using the methods and results in this paper, 
we correct a mistake in our previous paper \cite{zhou01}.

\section{Preliminaries}

Define $\rho \in C_c^\infty(\mathbb{R}^n)$ by
\begin{equation*}
\rho(x)=
\begin{cases}
c_0\exp(-\frac 1{1-|x|^2}), |x|<1,\\
0, |x|\geq 1,
\end{cases}
\end{equation*}
where $c_0$ is selected such that $\int \rho(x)dx=1$.
Let
$J_\epsilon$ be defined by
\begin{equation*}
J_\epsilon u=\rho_\epsilon*u=\epsilon^{-n}\rho(\frac\cdot\epsilon)*u.
\end{equation*}
This operator satisfies the following properties.

\begin{proposition} \label{prop2.1}
\begin{itemize}
\item[(1)] $\Lambda^s J_\epsilon u= J_\epsilon \Lambda^s u$, $s\in \mathbb{R}$.
\item[(2)] For all $u\in L^p(\mathbb{R}^n)$, $v\in H^\alpha(\mathbb{R}^n)$, 
with $\frac 1p+\frac 1q=1$,
$\int (J_\epsilon f)g=\int f(J_\epsilon g)$.
\item[(3)] For all $u\in H^\alpha(\mathbb{R}^n)$,
\begin{equation*}
\lim_{\epsilon\rightarrow 0}\|J_\epsilon u-u\|_{H^\alpha}=0,\quad
 \lim_{\epsilon\rightarrow 0}\|J_\epsilon u-u\|_{H^{\alpha-1}}\leq C\|u\|_{H^\alpha}.
\end{equation*}
\item[(4)] For all $u\in H^\alpha(\mathbb{R}^n)$, $s\in \mathbb{R}$, 
$k\in \mathbb{Z}\cup\{0\}$, then
\begin{equation*}
\|J_\epsilon u\|_{H^{\alpha+k}}
 \leq \frac {C_{\alpha k}}{\epsilon^k}\|u\|_{H^\alpha},\ \
\|J_\epsilon D^ku\|_{L^\infty}
 \leq \frac {C_{k}}{\epsilon^{\frac n2+k}}\|u\|_{H^\alpha}
\end{equation*}
\end{itemize}
\end{proposition}

The following propositions can be found in \cite{cordoba01,ju01}.

\begin{proposition} \label{prop2.2}
Suppose that $s>0$ and $1<p<\infty$. If $f,g\in \mathcal {S}$, the 
Schwartz class, then we have
\begin{gather*}
\|\Lambda^s(fg)-f\Lambda^s g\|_{L^p}\leq c\|\nabla f\|_{L^{p_1}}
\|g\|_{\dot{H}^{s-1,p_2}}+c\|g\|_{L^{p_4}}\|f\|_{\dot{H}^{s,p_3}}, \\
\|\Lambda^s(fg)\|_{L^p}\leq c\|f\|_{L^{p_1}}\|g\|_{\dot{H}^{s,p_2}}
+c\|g\|_{L^{p_4}}\|f\|_{\dot{H}^{s,p_3}}
\end{gather*}
with $p_2,p_3\in(1,+\infty)$ such that 
$\frac 1p=\frac 1{p_1}+\frac 1{p_2}=\frac 1{p_3}+\frac 1{p_4}$.
\end{proposition}

\begin{proposition} \label{prop2.3}
If $0\leq s\leq 2$, $f\in \mathcal {S}(\mathbb{R}^n)$, then
\begin{equation*}
2f(x)\Lambda^s f(x)\geq \Lambda^s f^2(x) \quad
\text{for all }  x \in \mathbb{R}^n.
\end{equation*}
\end{proposition}

\begin{proposition} \label{prop2.4}
Let $\alpha_1$ and $\alpha_2$ be two real numbers such that 
$\alpha_1<\frac n2$, $\alpha_2<\frac n2$ and $\alpha_1+\alpha_2>0$. 
Then there exists a constant $C=C_{\alpha_1,\alpha_2}\geq 0$ such that
 for all $f\in \dot{H}^{\alpha_1}$ and $g\in \dot{H}^{\alpha_2}$,
\begin{equation*}
\|fg\|_{\dot{H}^{\alpha}}\leq C\|f\|_{\dot{H}^{\alpha_1}}
\|g\|_{\dot{H}^{\alpha_2}},
\end{equation*}
where $\alpha=\alpha_1+\alpha_2-\frac n2$.
\end{proposition}

\section{Main results}

\begin{theorem} \label{thm3.1}
If $s\in [1/2,1]$, $T>0$, $\alpha>\frac n2+1$, 
$u_{0}\in H^\alpha(\mathbb{R}^n)$, $v\geq 0$ and
$v\in C([0,T];H^\alpha(\mathbb{R}^n))$,
then the linear initial-value problem
\begin{equation}
\begin{gathered}
\partial_t{u}=\nabla u\cdot \nabla(-\Delta)^{-s}v-v(-\Delta)^{1-s}u,\\
u(x,0)=u_{0}.
\end{gathered}
\end{equation}
has a unique solution $u\in C^1([0,T];H^\alpha(\mathbb{R}^n))$.
If the initial data $u_0\geq 0$, then   $u\geq0, (x,t)\in \mathbb{R}^n\times[0,T]$.
\end{theorem}

\begin{proof}
For any $\epsilon>0$, we consider the  linear problem
\begin{equation} \label{e3.2}
\begin{gathered}
\partial_t{u^\epsilon}=F_\epsilon(u^\epsilon)
=J_\epsilon(\nabla J_\epsilon u^\epsilon\cdot \nabla(-\Delta)^{-s} v)-J_\epsilon (v(-\Delta)^{1-s}J_\epsilon u^\epsilon),\\
u^\epsilon(x,0)=u_{0}.
\end{gathered}
\end{equation}
By Propositions \ref{prop2.1} and \ref{prop2.2} and by $s\geq \frac 12$
 we can estimate
\begin{align*}
&\|F_\epsilon(u_1^\epsilon)-F_\epsilon(u_2^\epsilon)\|_{H^\alpha}\\
&=\|J_\epsilon(\nabla J_\epsilon (u_1^\epsilon-u_2^\epsilon)
 \cdot \nabla(-\Delta)^{-s} v)-J_\epsilon (v(-\Delta)^{1-s}
 J_\epsilon (u_1^\epsilon-u_2^\epsilon)\|_{H^\alpha}\\
&\leq C(\epsilon,\|v\|_{H^\alpha})\|u_1^\epsilon-u_2^\epsilon\|_{H^\alpha}.
\end{align*}
Using Picard iterations, for any $\alpha>\frac n2+1, \epsilon>0$, 
there exists a $T_\epsilon=T_\epsilon(u_*)>0$, problem \eqref{e3.2}
 has a unique solution $u^\epsilon\in C^1([0,T_\epsilon);H^\alpha)$. 
By Propositions \ref{prop2.1} and \ref{prop2.3},
\begin{align*}
\frac 12 \frac {d}{dt}\|u^\epsilon\|^2_{L^2}
&=\int \nabla J_\epsilon u^\epsilon \cdot\nabla(-\Delta)^{-s}vJ_\epsilon u^\epsilon-
\int v(-\Delta)^{1-s}J_\epsilon u^\epsilon J_\epsilon u^\epsilon\\
&\leq \frac 12\int\nabla |J_\epsilon u^\epsilon|^2\cdot\nabla(-\Delta)^{-s}v
 -\frac 12\int v(-\Delta)^{1-s}|J_\epsilon u^\epsilon|^2\\
&\leq  \frac 12\int|J_\epsilon u^\epsilon|^2(-\Delta)^{1-s}v
 -\frac 12\int|J_\epsilon u^\epsilon|^2(-\Delta)^{1-s}v=0.
\end{align*}
Moreover, for any $\alpha>0$,
\begin{align*}
&\frac 12 \frac {d}{dt}\|\Lambda^\alpha u^\epsilon\|^2_{L^2}\\
&=\int \Lambda^\alpha(\nabla J_\epsilon u^\epsilon 
 \cdot\nabla(-\Delta)^{-s}v)J_\epsilon\Lambda^\alpha u^\epsilon-
\int \Lambda^\alpha(v(-\Delta)^{1-s}J_\epsilon u^\epsilon) 
 \Lambda^\alpha J_\epsilon u^\epsilon\\
&\leq C\|[\Lambda^\alpha,\nabla(-\Delta)^{-s}v]
 \nabla J_\epsilon u^\epsilon\|_{L^2}\|\Lambda^\alpha u^\epsilon\|_{L^2}
+\int\nabla(-\Delta)^{-s}v\Lambda^\alpha \nabla J_\epsilon 
 u^\epsilon\Lambda^\alpha J_\epsilon u^\epsilon\\
&+ C\|[\Lambda^\alpha,v](-\Delta)^{1-s}J_\epsilon
  u^\epsilon\|_{L^2}\|\Lambda^\alpha u^\epsilon\|_{L^2}
 -\int v\Lambda^\alpha (-\Delta)^{1-s}J_\epsilon u^\epsilon 
 \Lambda^\alpha J_\epsilon u^\epsilon.
\end{align*}
By Proposition \ref{prop2.2} and Sobolev embeddings,
\begin{align*}
&\|[\Lambda^\alpha,\nabla(-\Delta)^{-s}v]\nabla 
 J_\epsilon u^\epsilon\|_{L^2}\\
&\leq C\|(-\Delta)^{1-s}v\|_{L^\infty}\|\Lambda^{\alpha-1} \nabla 
 J_\epsilon u^\epsilon\|_{L^2} 
 + \|\nabla(-\Delta)^{-s}v\|_{\dot{H}^\alpha}\|\nabla J_\epsilon 
 u^\epsilon\|_{L^\infty}\\
&\leq C\|v\|_{H^\alpha}\|u^\epsilon\|_{H^\alpha},
\end{align*}
\begin{align*}
&\|[\Lambda^\alpha,v](-\Delta)^{1-s}J_\epsilon u^\epsilon\|_{L^2}\\
&\leq C\|\nabla v\|_{L^\infty}\|(-\Delta)^{1-s}J_\epsilon 
 u^\epsilon\|_{\dot{H}^{\alpha-1}}
 +\|v\|_{\dot{H}^\alpha}\|(-\Delta)^{1-s}J_\epsilon u^\epsilon\|_{L^\infty}\\
&\leq C\|v\|_{H^\alpha}\|u^\epsilon\|_{H^\alpha}.
\end{align*}
By Proposition \ref{prop2.3},
\begin{align*}
&\int\nabla(-\Delta)^{-s}v\Lambda^\alpha \nabla J_\epsilon 
 u^\epsilon\Lambda^\alpha J_\epsilon u^\epsilon-\int v\Lambda^\alpha
 (-\Delta)^{1-s}J_\epsilon u^\epsilon \Lambda^\alpha J_\epsilon u^\epsilon\\
& \leq\frac12\int \nabla(-\Delta)^{-s}v\nabla(\Lambda^\alpha 
 J_\epsilon u^\epsilon)^2-\frac12\int v(-\Delta)^{1-s}
 (\Lambda^\alpha J_\epsilon u^\epsilon)^2\\
&\leq C\|v\|_{H^\alpha}\|u^\epsilon\|^2_{H^\alpha}.
\end{align*}
Combining the above estimates,
\begin{equation*}
\frac d {dt}\|u^\epsilon(\cdot,t)\|_{H^{\alpha}}
\leq C\|v\|_{H^\alpha}\|u^\epsilon\|_{H^\alpha}.
\end{equation*}
By Gronwall's inequality,
\begin{equation*}
\|u^\epsilon(\cdot,t)\|_{H^\alpha}
\leq \|u_0\|_{H^\alpha}\exp(C\sup_{0\leq t\leq T}\|v\|_{H^\alpha}).
\end{equation*}
Such the solution $u^\epsilon$ exists on $[0,T]$.
Similarly,
\begin{equation*}
\frac d {dt}\|u^\epsilon(\cdot,t)\|_{H^{\alpha-1}}
\leq C\|v\|_{H^\alpha}\|u^\epsilon\|_{H^\alpha}
\leq C(\|v\|_{H^\alpha}, \|u_0\|_{H^\alpha}, T).
\end{equation*}
By Aubin compactness theorem \cite{simon}, there is a subsequence of 
$\{u^{\frac 1n}\}_{n\geq1}$ that convergence strongly to $u$ in 
$C([0,T];H^\alpha)$. If $\alpha>\frac d2+1$,
$H^\alpha \hookrightarrow C^1$, so $u$ is a solution of \eqref{e3.1}.

If $u$ and $\tilde{u}$ are two solutions of problem \eqref{e3.1}, 
then $w=u-\tilde{u}$ satisfies
\begin{gather*}
\partial_t{w}=\nabla w\cdot \nabla(-\Delta)^{-s}v-v(-\Delta)^{1-s}w,\\
w(x,0)=0.
\end{gather*}
Similarly, we  get $\frac d{dt}\|w\|_{L^2}\leq 0$ and 
$\frac d{dt}\|w\|_{\dot{H}^\alpha}\leq \|v\|_{H^\alpha}\|w\|_{H^\alpha}$,
 i.e,
$\frac d{dt}\|w\|_{H^\alpha}\leq \|v\|_{H^\alpha}\|w\|_{H^\alpha}$. 
By Gronwall's inequality, $u(x,t)=0$, $(x,t)\in \mathbb{R}^n\times[0,T]$.

Since $u_0\geq 0$ then if there exists a first time $t_0$ where for some 
point $x_0$ we have $u(x_0,t_0)=0$, then $(x_0,t_0)$ will correspond 
to a minimum point
and therefore $\nabla u(x_0,t_0)=0$, and
\begin{equation*}
(-\Delta)^{1-s}u(x)=c\int \frac {u(x)-u(y)}{|y|^{n+2-2s}}{\rm d}y\leq0.
\end{equation*}
 Hence $ u_t|_{(x_0,t_0)}\geq 0$. So $u(x,t)\geq 0$ for all 
$(x,t)\in \mathbb{R}^n\times[0,T]$.
\end{proof}

\begin{theorem} \label{thm3.2}
Let $n\geq 2$, $s\in [\frac12,1)$, $\alpha>\frac d2+1$, 
$u_0\in H^\alpha(\mathbb{R}^n)$, and $u_0\geq 0$. Then there  the linear 
initial value problem
\begin{gather*}
\partial_t{u}=\nabla\cdot(u\nabla(-\Delta)^{-s}u),\\
u(x,0)=u_{0}.
\end{gather*}
has a unique solution $u\in C^1([0,T_0],H^\alpha(\mathbb{R}^n))$. 
If the initial data $u_0\geq 0$, then $u\geq0, (x,t)\in \mathbb{R}^n\times[0,T_0]$.
\end{theorem}

\begin{proof}
Set $u^1=u_0$. Note that  
$\partial_t{u}=\nabla\cdot(u\nabla(-\Delta)^{-s}u)
=\nabla u\cdot \nabla (-\Delta)^{-s}u-u(-\Delta)^{1-s}u$,
and let  $\{u^{n}\}$ be the sequence defined by
\begin{equation}\label{e3.1}
\begin{gathered}
\partial_t{u^{n+1}}=\nabla u^{n+1}\cdot \nabla(-\Delta)^{-s}u^{n}
-u^n(-\Delta)^{1-s}u^{(n+1)},\\
u^{n+1}(x,0)=u_{0}.
\end{gathered}
\end{equation}
By Theorem \ref{thm3.1}, $u^2\in C([0,T);H^\alpha)$, for all $T<\infty$, satisfies 
$u^2\geq 0$ and
\begin{equation*}
\sup_{0\leq t\leq T}\|u^2\|_{H^\alpha}
\leq \|u_0\|_{H^\alpha}\exp(C\|u^1\|_{H^\alpha}T).
\end{equation*}
If $\exp(2C\|u_1\|_{H^\alpha}T_0)\leq2$, for example 
$T_0=\frac {\ln 2}{2C(1+\|u_0\|_{H^\alpha})}$, we have 
\[
\sup_{0\leq t\leq T_0}\|u^2\|_{H^\alpha}\leq 2\|u_0\|_{H^\alpha}. 
\]
By the standard induction argument, if $u^{n}\in C([0,T_0];H^\alpha)$,
 $u^n\geq 0$ is a solution of \eqref{e3.1} with
 $\|u^n\|_{H^\alpha}\leq 2\|u_0\|_{H^\alpha}$. 
By Theorem \ref{thm3.1}  $u^{n+1}\in C([0,T_0];H^\alpha)$ , $u^{n+1}\geq0$ and
\begin{gather*}
\sup_{0\leq t\leq T_0}\|u^{n+1}\|_{H^\alpha}
\leq \|u_0\|_{H^\alpha}\exp(C\|u^{n}\|_{H^\alpha}T_0)\leq 2\|u_0\|_{H^\alpha}, \\
\frac d{dt}\|u^{n+1}\|_{H^{\alpha-1}}
\leq C\|u^n\|_{H^\alpha}\|u^{n+1}\|_{H^\alpha}\leq C\|u_0\|^2_{H^\alpha}.
\end{gather*}
By Aubin compactness theorem \cite{simon}, there is a subsequence of $u^{n}$ 
that convergence strongly to $u$ in $C([0,T];H^\alpha)$. 
If $u\geq 0,\tilde{u}\geq0$ are two solutions of problem \eqref{e3.1}, 
then $w=u-\tilde{u}$ satisfies 
\begin{gather*}
\partial_t{w}=\nabla\cdot(w \nabla(-\Delta)^{-s}u)
+\nabla \cdot(\tilde{u}\nabla(-\Delta)^{-s}w),\\
w(x,0)=0.
\end{gather*}
By Proposition \ref{prop2.2},
\begin{align*}
\frac12\frac d{dt}{\|w\|_{L^2}^2}
&=\int w\nabla\cdot(w \nabla(-\Delta)^{-s}u)
 +\int w\nabla \tilde{u}\cdot\nabla(-\Delta)^{-s}w
 -\int w \tilde{u}(-\Delta)^{1-s}w \\
& =: I_1+I_2+I_3.
\end{align*}
Note that
\begin{gather*}
\begin{aligned}
I_1&=\int w\nabla w\cdot \nabla(-\Delta)^{-s}u
 =\frac12\int\nabla w^2\cdot \nabla(-\Delta)^{-s}u\\
&=\frac12\int w^2(-\Delta)^{1-s}u
 \leq C\|u\|_{H^{\alpha}}\|w\|^2_{L^2},
\end{aligned}\\
I_3\leq-\frac 12\int \tilde{u}(-\Delta)^{1-s}w^2
 =\int-\frac 12(-\Delta)^{1-s}\tilde{u}\cdot w^2\leq C\|u\|_{H^{\alpha}}\|w\|^2_{L^2}.
\end{gather*}
When $s>1/2$,
\begin{align*}
I_2&\leq C\|w\|_{L^2}\|\nabla u\cdot\nabla(-\Delta)^{-s}w\|_{L^2} \\
   &\leq C\|w\|_{L^2}\|\nabla u\|_{\dot{H}^{\frac n2+1-2s}}
 \|\nabla(-\Delta)^{-s}w\|_{\dot{H}^{2s-1}}
   \leq C\|u\|_{H^{\alpha}}\|w\|^2_{L^2}.
\end{align*}
When $s=1/2$, the above estimates are still valid. 
 Combining the above estimates we have  
$\frac {d}{dt}\|w\|_{L^2}\leq C\|w\|_{L^2}\|u\|_{H^{\alpha}}$. 
By Gronwall's inequality we can deduce $w(x,t)=0$ on $[0,T_0]$.
\end{proof}

\section{Correction}
In  \cite{zhou01}, trying to establish the  well-posedness of 
\eqref{e1.1} in Besov spaces the authors incurred in a mistake in page 9
when estimating the term $J_4'$ in equation \cite[(4.5)]{zhou01}.
 To correct the mistake, we modify our proof the following way.
\medskip

\textbf{\cite[Theorem 1.1]{zhou01}}
Let $n\geq2,s\in[\frac12,1],\alpha>n+1$. If  the initial data 
$u_0\in B_{1,\infty}^{\alpha}$, then there exists 
$T=T(\|u_0\|_{B_{1,\infty}^{\alpha}})$ such that
\eqref{e1.1} has a unique solution in $[0,T]\times \mathbb{R}^n$.
Such a  solution belongs to 
$C^1([0,T];B_{1,\infty}^{\alpha+2s-2})\cap L^{\infty}([0,T];B_{1,\infty}^{\beta})$,
 with $\beta\in[\alpha+2s-2,\alpha]$.
\smallskip

\begin{proof}
First we construct the approximate equation
\begin{equation}\label{equation41}
\begin{gathered}
 u_t^{(n+1)}=\nabla u^{(n+1)} \cdot \nabla(-\triangle)^{-s}u_{\epsilon}^{(n)}
 -u_{\epsilon}^{(n)}(-\triangle)^{1-s}u^{(n+1)};   \\
 u^{(n+1)}(0)=\sigma_{\epsilon}*u_0,\quad u^{(1)}=\sigma_{\epsilon}*u_0.
\end{gathered}
\end{equation}
By the argument in section 2, there exists a sequence $u^{(n)}$ that solves 
the linear systems \eqref{equation41}. Assuming that  $u_0\geq0$, 
we prove that $u^{(n+1)}\geq0$.
Inspired by \cite{caffarelli02}, we assume that $x_0$ is a point of 
minimum of $u^{(n+1)}$ at time $t=t_0$. This indicates that
$\nabla u^{(n+1)}(x_0)=0$, and
\[
 (-\triangle)^{1-s} u^{(n+1)}(x_0)=c\int \frac{u(x_0)-u(y)}{|y|^{n+2(1-s)}}dy\leq0.
\]
Thus we deduce that $\frac{\partial}{\partial t}u^{(n+1)}\big|_{t=t_0}\geq0$, 
and by induction we have
$u^{(n+1)}\geq 0.$
Arguing as in   \cite{zhou01}, taking $\triangle_j$ on \eqref{equation41},
 we obtain
\begin{align*}
 \partial_t \triangle_j u^{(n+1)}
& =\sum[\triangle_j,\partial_i (-\triangle)^{-s} u_{\epsilon}^{(n)}]
  \partial_i u^{{n+1}}
   +\sum \partial_i(-\triangle)^{-s} u_{\epsilon}^{(n)} 
 \triangle_j (\partial_i u^{(n+1)})	\\
&\quad -[\triangle_j,u_{\epsilon}^{(n)}] (-\triangle)^{1-s} u^{(n+1)}
  -u_{\epsilon}^{(n)} \triangle_j (-\triangle)^{1-s}u^{(n+1)}.
\end{align*}
Multiplying both sides by $\frac{\triangle_j u^{(n+1)}}{|\triangle_j u^{(n+1)}|}$, 
and integrating over $\mathbb{R}^d$, then denote the corresponding part in 
the right side by $J_1', J_2', J_3', J_4'$, respectively. We obtain the estimates,
\begin{gather*}
 J_1' \leq C 2^{-j\alpha}\|u^{(n+1)}\|_{B_{1,\infty}^{\alpha}} 
 \|u^{(n)}\|_{B_{1,\infty}^{\alpha+1-2s}}	\\
 J_2' \leq C 2^{-j\alpha}\|u^{(n+1)}\|_{B_{1,\infty}^{\alpha}} 
 \|u^{(n)}\|_{B_{1,\infty}^{\alpha+1-2s}}	\\
 J_3' \leq C 2^{-j\alpha}\|u^{(n)}\|_{B_{1,\infty}^{\alpha}} 
 \|u^{(n+1)}\|_{B_{1,\infty}^{\alpha+1-2s}}.
\end{gather*}
The estimate for the term $J_4'$ is now replaced by
\begin{align*}
J_4' =
 &-\int u^{(n)}\triangle_j (-\triangle)^{1-s} u^{(n+1)} 
 \frac{\triangle_j u^{(n+1)}}{|\triangle_j u^{(n+1)}|} \\
 &\leq -\int u^{(n)} (-\triangle)^{1-s} |\triangle_j u^{(n+1)}|      \\
 &\leq -\int (-\triangle)^{1-s} u^{(n)}  |\triangle u^{(n+1)}|   \\
 &\leq 2^{-j\alpha}\|u^{n}\|_{B_{1,\infty}^{r+2-2s}}
 \|u^{(n+1)}\|_{B_{1,\infty}^{\alpha}}.
\end{align*}
Here $r>d$ is any real number.  The  first inequality uses the 
following pointwise estimate.

\begin{proposition}[\cite{miao01}]
If $0\leq \alpha \leq 2$, $p\geq1$, then
 $$
p|f(x)|^{p-2}f(x)\Lambda^{\alpha}f(x)\geq \Lambda^{\alpha}|f(x)|^p.
$$
for any $f\in\mathcal{S}(\mathbb{R}^d)$
\end{proposition}

Taking $r$ such that $r+2-2s<\alpha$, e.g. set $r=\alpha-1$, we conclude
\[
  \frac{d}{dt}\|u^{(n+1)}\|_{B_{1,,\infty}^\alpha}
\leq \|u^{(n)}\|_{B_{1,,\infty}^\alpha}\|u^{(n+1)}\|_{B_{1,,\infty}^\alpha}.
\]
The other parts of the proof need no modification.
\end{proof}

\subsection*{Acknowledgement}
We are very grateful to Ph. D. Mitia Duerinckx for pointing out our 
mistake and share some good views on this problem. This paper 
is supported by the NNSF of China under grants No. 11601223 and No.11626213.

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