#  $q$-Moment Estimates for the Singular $p$-Laplace Equation and Applications

Samuel Drapeau  Shanghai Jiao Tong University, China Academy of Financial
Research (SAIF) and School of Mathematical Sciences, Shanghai, China
<EMAIL_ADDRESS>http://www.samuel-drapeau.info and  Liming Yin
Shanghai Jiao Tong University, School of Mathematical Sciences, Shanghai,
China<EMAIL_ADDRESS>
(Date: August 27, 2024)

###### Abstract.

We provide $q$-moment estimates on annuli for weak solutions of the singular
$p$-Laplace equation where $p$ and $q$ are conjugates. We derive $q$-uniform
integrability for some critical parameter range. As a application, we derive a
mass conservation as well as a weak convergence result for a larger critical
parameter range. Concerning the latter point, we further provide a rate of
convergence of order $t^{q-1}$ of the solution in the $q$-Wasserstein
distance.
Keywords: Singular $p$-Laplace Equation; Moment Estimates; Mass Conservation;
Convergence Rate in Wasserstein Distance.

National Science Foundation of China, Grants Numbers: 11971310 and 11671257
are gratefully acknowledged.

##  1\. Introduction

We consider the Cauchy problem for the parabolic $p$-Laplace equation posed in
the whole Euclidean space $\mathbb{R}^{N}$

$\begin{cases}\displaystyle\partial_{t}u=\mathrm{div}(|\nabla u|^{p-2}\nabla
u),\\\ u(0)=\mu,\end{cases}$ (1.1)

with a positive Radon measure $\mu$ as initial data and $p_{c}<p<2$ where the
critical value $p_{c}$ is defined as

$p_{c}:=\frac{2N}{N+1}.$

Solutions are understood in the weak sense on the whole space and finite time
horizon $S_{T}:=\mathbb{R}^{N}\times(0,T]$ where $N\geq 2$.

For $1<p<2$, such Cauchy problem is know as the fast diffusion or singular
equation. Existence and uniqueness and regularity of weak solutions with
various conditions for initial data are studied by Di Benedetto and Herrero
[6], Junning [10], Chen and Di Benedetto [5]. For further insights about
parabolic $p$-Laplace equation, we refer to [3, 15]. While for a Dirac measure
$\delta_{0}$ as initial data, this Cauchy problem does not admit a positive
solution for any $1<p\leq p_{c}$, see [10], it is known that for $p_{c}<p<2$,
there exists a fundamental solution called Barenblatt source-type solution,
see [15].

In the context of optimal matching problem, Ambrosio et al. [1] draw a link
between $2$-Wasserstein distance and Poisson equation through linearization of
Monge-Ampère equation and regularize empirical measures with heat semigroup.
They suggest that such a linearization for the $q$-Wasserstein distance leads
to $p$-Laplacian where $p$ and $q$ are conjugates. Such an approach has been
used in a different context by Evans and Gangbo [7].

Based on these ideas, it seems natural to regularize empirical measures with
$p$-Laplacian semigroup. And further, to connect the Cauchy problem (1.1) to
flows in the $q$-Wasserstein space, as done in Kell [12]. A key point in this
connection concerns the behavior of $q$-moment of weak solutions $u(x,t)$ of
Cauchy problem (1.1), namely,

$\int_{\mathbb{R}^{N}}|x|^{q}u(x,t)dx.$

For the degenerate case where $p>2$, that is, $1<q<2$, such $q$-moments are
easily estimated, at least for initial data with finite mass and compact
support, due to the finite propagation property, see [11]. However, for the
singular case $p_{c}<p<2$, solutions diffuse everywhere even with a Dirac
measure as initial data. Hence, the focus on the parameter range $p_{c}<p<2$.

Following similar methods as in [6, 10], our main result are the following
$q$-moment estimate for weak solutions of (1.1) on annuli

$\sup_{0<\tau\leq t}\int_{r\leq|x|\leq R}\left|x\right|^{q}u(x,\tau)dx\leq
C\int_{\frac{1}{2}r\leq|x|\leq 2R}|x|^{q}d\mu\\\
+Ct^{\frac{1}{2-p}}\left(r^{-\frac{p(N)}{(p-1)(2-p)}}+R^{-\frac{p(N)}{(p-1)(2-p)}}\right).$
(1.2)

for any $p_{c}<p<2$, where $C$ is a constant depending only on $N$ and $p$ and
$p(N)$ is the polynomial

$p(N)\colon=(N+2)p^{2}-(3N+3)p+2N.$

This estimate shows in particular that with a finite $q$-moment Radon measure
as initial data, there exists a weak solution which has uniformly bounded
$q$-moments on all compact time interval for any $p_{N}<p<2$, where the
critical value $p_{c}<p_{N}<2$ is the largest root of the polynomial $p(N)$.
This critical range $(p_{N},2)$ is sharp in the sense that the Barenblatt
source-type solution has finite $q$-moment if and only if $p_{N}<p<2$.

As applications, we show that for such a weak solution with a finite
$q$-moment Radon measure as initial data, the mass conservation as well as
weak convergence at $t=0$ holds for any $p_{c}<p<2$. Finally, we obtain a
$q$-Wasserstein convergence rate for weak solutions $u$ with a finite
$q$-moment Radon measure $\mu$ as initial data for any $p_{N}<p<2$, namely,

$W^{q}_{q}(\mu,u(x,t)dx)\leq Ct^{q-1},$ (1.3)

where $C$ is a constant depending only on $N$, $p$, $\mu$ and $T$. In
particular, when $p$ converges to $2$, the convergence rate coincides with the
well-known heat flow case in [1].

To the best of our knowledge, there does not exists $q$-moment estimates for
singular $p$-Laplacian equation. However, similar results in terms of mass
conservation and connection to $q$-Wasserstein space has been done. The mass
conservation of singular parabolic $p$-Laplace equation for $p_{c}<p<2$ is
proved by Fino et al. [8] for any weak solution where the initial data are
Radon measures with compact support. Our result only requires finite
$q$-moment instead of compact support for initial data, but for any weak
solution constructed through mollification. As for the connection to the flow
in the $q$-Wasserstein space, the work of Kell [12] is the closest to the
present one. There, the author shows that a smooth solution of Cauchy problem
(1.1) solves a generalized gradient flow problem of Rényi entropy functional
in the $q$-Wasserstein space based on gradient flow and functional analysis
methods. The author further provides a condition for the mass conservation of
the flow with an integrable function as initial data for $3/2<p<\infty$ in a
general metric space. In contrast, our $q$-moment estimate (1.2) is a local
estimate based on pure PDE and analysis method, and, while holding for
$\mathbb{R}^{N}$, our result shows that mass conservation holds for a
different range $p_{c}<p<2$.

The paper is organized as follows: in Section 2, we introduce notations and
definitions and auxiliary Lemmas as well as the critical parameters $p_{c}$,
$p(N)$ and $p_{N}$. Section 3, is dedicated to the main theorem for the local
$q$-moment estimate (1.2). Section 4 addresses the mass conservation and the
weak convergence. Finally, in Section 5 we prove the Wasserstein convergence
rate.

##  2\. Notation and preliminary results

On $\mathbb{R}^{N}$ equipped with the Euclidean norm $|\cdot|$, we denote by
$B_{\rho}(x)=\\{y\colon|y-x|\leq\rho\\}$ the closed ball centered at $x$ with
radius $\rho$, and set $B_{\rho}:=B_{\rho}(0)$. We denote by
$A_{r}^{R}=\\{x\colon r\leq|x|\leq R\\}$ be the closed annulus centered at $0$
with radius $r$ and $R$. Let further $\Omega\subseteq\mathbb{R}^{N}$ be a
measurable set. For $1\leq p\leq\infty$, we denote by
$L^{p}\left(\Omega\right)$ the space of $p$-integrable functions with respect
to the Lebesgue measure and set $L^{p}(\mathbb{R}^{N})=L^{p}$. We denote by
$W^{1,p}(\Omega)$ the space of first-order Sobolev space and set
$W^{1,p}(\mathbb{R}^{N})=W^{1,p}$. For a function $u$ in $W^{1,p}(\Omega)$, we
denote by $\nabla u$ the weak derivative of $u$. For $T>0$ and
$S_{T}=\mathbb{R}^{N}\times(0,T]$, we denote by
$L^{\infty}_{\mathrm{loc}}(S_{T})$ the space of functions which belong to
$L^{\infty}(K\times[s,t])$ for all intervals $[s,t]\subset(0,T]$ and compact
subsets $K$ of $\mathbb{R}^{N}$. For $X=L^{p}(\Omega)$ or $X=W^{1,p}(\Omega)$,
we denote by $C([0,T];X),C((0,T);X)$ and $L^{p}(0,T;X)$ in a Bochner sense. We
further denote by $\mathcal{M}^{+}$ the set of positive Radon measures on
$\mathbb{R}^{N}$.

######  Definition 2.1.

We say that a measurable function $u:S_{T}\to\mathbb{R}$ is a weak solution of
the Cauchy problem (1.1) with $L^{1}$-initial data, that is

$u_{0}=\mu\in L^{1},$ (2.1)

if for every bounded open set $\Omega\subset\mathbb{R}^{N}$, it holds

$\begin{cases}\displaystyle u\in
C\left((0,T);L^{1}\left(\Omega\right)\right)\cap
L^{p-1}\left(0,T;W^{1,p-1}\left(\Omega\right)\right)\cap
L^{\infty}_{loc}\left(S_{T}\right),\\\ |\nabla u|\in
L^{\infty}_{loc}\left(S_{T}\right),\end{cases}$ (2.2)

and for all $0<s<t<T$ and $\phi\in C^{1}(\mathbb{R}^{N}\times[0,T])$ such that
$\mathrm{supp}(\phi(\cdot,\tau))\subset B_{\rho}$ for all $\tau\in[0,T]$ for
some $\rho>0$,

$\int_{\mathbb{B_{\rho}}}(\phi
u)(x,t)dx+\int_{s}^{t}\int_{B_{\rho}}\left(-u\partial_{t}\phi+|\nabla
u|^{p-2}\nabla u\cdot\nabla\phi\right)dxd\tau=\int_{B_{\rho}}(\phi u)(x,s)dx,$
(2.3)

and for all $R>0$,

$\lim_{\tau\searrow 0}\int_{B_{R}}\left|u(x,t)-u_{0}(x)\right|dx=0.$ (2.4)

The measurable function $u:S_{T}\to\mathbb{R}^{N}$ is a weak solution of the
Cauchy problem (1.1) with positive Radon measure as initial data, that is,

$u_{0}=\mu\in\mathcal{M}^{+},$ (2.5)

if condition (2.4) is replaced by

$\lim_{\tau\searrow
0}\int_{\mathbb{R}^{N}}\psi(x)u(x,\tau)dx=\int_{\mathbb{R}^{N}}\psi(x)\mu(dx)$
(2.6)

for all $\psi\in C_{c}(\mathbb{R}^{N})$.

For the existence and uniqueness of weak solution of (1.1) with initial data
$u_{0}$ in $L^{1}$, we refer readers to [6]. In particular, if the initial
data $u_{0}$ is in $C^{\infty}_{c}(\mathbb{R}^{N})$, then the weak solution
$u$ is unique and satisfies that $u\in C([0,T];L^{1}(\mathbb{R}^{N}))$,
$\partial_{t}u\in L^{2}(0,T;L^{2}(\mathbb{R}^{N}))$ and $|\nabla u|\in
L^{p}(0,T;L^{p}(\mathbb{R}^{N}))$ and $u\in L^{\infty}(S_{T})$. In this case,
the condition (2.3) is equivalent to the following condition:

$\int_{\mathbb{R}^{N}}(\phi
u)(x,t)dx+\int_{0}^{t}\int_{\mathbb{R}^{N}}\left(-u\partial_{\tau}\phi+|\nabla
u|^{p-2}\nabla
u\cdot\nabla\phi\right)dxd\tau=\int_{\mathbb{R}^{N}}\phi(x,0)u_{0}dx.$ (2.7)

Furthermore, $u$ is locally $\alpha$-Hölder continuous on
$\Omega^{\varepsilon}_{T}:=\Omega\times[\varepsilon,T]$ for all bounded
subsets $\Omega\subset\mathbb{R}^{N}$ and $\varepsilon>0$, with
$\alpha\in(0,1)$ depending only on $N,p$ and supremum norm
$\|u\|_{\infty,\Omega^{\varepsilon}_{T}}$ on $\Omega^{\varepsilon}_{T}$, see
[5]. Moreover, if initial data $u_{0}\geq 0$, then the weak solution
$u(x,t)\geq 0$ for all $(x,t)\in S_{T}$.

For the existence of weak solution of (1.1) with a positive Radon measure as
initial data, we refer readers to [10, Theorem 1] and [6, Theorem III.8.1].
However, to our knowledge, in this case the uniqueness of weak solution is
unknown and depends on the choice of the definition of the solution
(viscosity, entropy, distributional, etc.).

Recall that a sequence $(\mu_{n})$ of positive Radon measures converges
vaguely, or weakly, to $\mu$ in $\mathcal{M}^{+}$ if
$\int_{\mathbb{R}^{N}}\phi d\mu_{n}\to\int_{R^{N}}\phi d\mu$ for every $\phi$
in $C_{c}(\mathbb{R}^{N}$, or for every $\phi$ in $C_{b}(\mathbb{R}^{N})$,
respectively.

We briefly recall two important Lemmas for the proofs that are formulated in
our notations.

######  Lemma 2.2.

(A priori estimate, Junning [10]). Let $k=N(p-2)+p$ and $u$ be the weak
solution of (1.1) with $u_{0}\in C^{\infty}_{c}(\mathbb{R}^{N})$ as initial
data. Then there exists $C_{1}:=C_{1}(N,p)$ such that for all $R>0$, it holds

$\sup_{0<\tau\leq t}\int_{B_{R}}\left|u(x,\tau)\right|dx\leq
C_{1}\int_{B_{2R}}\left|u_{0}\right|dx+C_{1}R^{-\frac{k}{2-p}}t^{\frac{1}{2-p}},$
(2.8)

and for any $R_{0}$, there exists $C_{2}:=C_{2}(N,p,R_{0})$ such that for all
$R>R_{0}$, it holds

$\sup_{x\in B_{R}}\left|u(x,t)\right|\leq
C_{2}\left(t^{-\frac{N}{k}}\left(\int_{B_{4R}}|u_{0}|dx\right)^{\frac{p}{k}}+R^{-\frac{p}{2-p}}t^{\frac{1}{2-p}}\right).$
(2.9)

######  Lemma 2.3.

([9, Lemma 3.1]) Let $f\colon[r_{0},r_{1}]\to[0,\infty)$ be a bounded function
with $0\leq r_{0}<r_{1}$. Suppose that for any $\sigma,\sigma^{\prime}$ with
$r_{0}\leq\sigma<\sigma^{\prime}\leq r_{1}$, we have

$f(\sigma)\leq\left[A\left(\sigma^{\prime}-\sigma\right)^{-a}+B\right]+\theta
f\left(\sigma^{\prime}\right),$

where $A,B,a,\theta$ are nonnegative constants with $0\leq\theta<1$. Then for
any $\lambda,\lambda^{\prime}$ with $r_{0}\leq\lambda<\lambda^{\prime}\leq
r_{1}$, it holds

$f(\lambda)\leq
C\left(a,\theta\right)\left[A\left(\lambda^{\prime}-\lambda\right)^{-a}+B\right],$

for some positive constant $C(a,\theta)$ depending only on $a$ and $\theta$.

Finally, recall the following critical values

$\begin{split}p_{c}&:=\frac{2N}{N+1},\\\ p(N)&:=(N+2)p^{2}-(3N+3)p+2N,\\\
p_{N}&:=\text{largest root of }p(N).\end{split}$ (2.10)

Note that since $N\geq 2$, it holds

$\frac{4}{3}\leq p_{c}<p_{N}<2.$

##  3\. Moments estimates

In this section, we derive $q$-moment estimates for a weak solution of (1.1)
with finite positive Radon measure as initial data. We denote by $q$ the
Hölder conjugate number of $p$, that is, $\frac{1}{p}+\frac{1}{q}=1$, and
denote by $k$ the constant $N(p-2)+p$. Given a $\mu\in\mathcal{M}^{+}$, we say
that $\mu$ has finite $q$-moment if $\int_{\mathbb{R}^{N}}|x|^{q}d\mu<\infty$.
We begin to state our main result in this section.

######  Theorem 3.1.

($q$-moment estimates.) Let $p_{c}<p<2$ and $\mu$ be in $\mathcal{M}^{+}$.
Then for any $0<r<R$ and $0<t\leq T$, there exists a weak solution $u$ of
(1.1) with $\mu$ as initial data satisfying

$\sup_{0<\tau\leq t}\int_{r\leq|x|\leq R}\left|x\right|^{q}u(x,\tau)dx\leq
C\int_{\frac{1}{2}r\leq|x|\leq 2R}|x|^{q}d\mu\\\
+Ct^{\frac{1}{2-p}}\left(r^{-\frac{p(N)}{(p-1)(2-p)}}+R^{-\frac{p(N)}{(p-1)(2-p)}}\right),$
(3.1)

for some constant $C:=C(N,p)$ depending only on $N$ and $p$. Furthermore, if
$\mu$ in $\mathcal{M}^{+}$ has finite total mass and finite $q$-moment, and
$p(N)>0$, then for any $0<t\leq T$, the family $\\{u(x,\tau)dx\colon
0\leq\tau\leq t\\}$ has uniformly bounded $q$-moment on $[0,t]$ and that

$\lim_{r\rightarrow\infty}\sup_{0<\tau\leq t}\int_{|x|\geq
r}|x|^{q}u(x,\tau)dx=0.$ (3.2)

######  Remark 3.2.

Let $\\{X_{\tau}\colon 0\leq\tau\leq t\\}$ be a family of
$\mathbb{R}^{N}$-valued random variables such that the law of $X_{\tau}$ is
$u(x,\tau)dx$. Then equality (3.2) is equivalent to say that
$\\{|X_{\tau}|^{q}\colon 0\leq\tau\leq t\\}$ is uniformly integrable for any
$0<t\leq T$.

######  Remark 3.3.

Note that $(p_{N},2)$ is a sharp range for $p$ in the following sense: let $U$
be the Barenblatt source-type solution, that is,

$U(x,t)=t^{-\frac{N}{k}}F(t^{-\frac{1}{k}}x),$ (3.3)

where $F(\xi)=(C+\gamma|\xi|^{\frac{p}{p-1}})^{\frac{p-1}{p-2}}$ for
$\gamma=(p-1)k/(p(2-p))$ and some positive constant $C$. It is well-known that
$w$ is a weak solution of (1.1) with initial data $\mu=M\delta_{0}$ where
$M=\|w(t)\|_{L^{1}}$. Computations show that $U$ has finite $q$-moment if and
only if $p_{N}<p<2$.

To show Theorem 3.1, we first establish the following auxiliary lemma.

######  Lemma 3.4.

Let $p\in(p_{c},2)$ and $\xi\in C^{1}_{c}(\mathbb{R}^{N})$ be a smooth
function such that $\mathrm{supp}(\xi)\subset\Omega$ for some bounded domain
$\Omega$. Let $u$ be the positive weak solution of (1.1) with initial data
$u_{0}\in C^{\infty}_{c}(\mathbb{R}^{N}),u_{0}\geq 0$. Then there exists a
positive constant $C:=C(N,p)$ such that for any $\varepsilon>0$,

$\int_{0}^{t}\int_{\Omega}\xi^{p}\tau^{\beta q}|\nabla
u|^{p}(u+\varepsilon)^{-\alpha q}dxd\tau\leq
C\int_{0}^{t}\int_{\Omega}\tau^{\beta q}(u+\varepsilon)^{p-\alpha
q}|\nabla\xi|^{p}dxd\tau\\\ +C\int_{\Omega}t^{\beta
q}(u+\varepsilon)^{2-\alpha q}\xi^{p}dx,$

where $\beta=\alpha/2$ and $\alpha=1/p$ if $p>1/(p-1)$ and $\alpha=p-1$ if
$p\leq 1/(p-1)$.

###### Proof.

Let $\varepsilon>0$ be fixed and $u_{\varepsilon}=u+\varepsilon$ and
$\psi(x,\tau)=\tau^{\beta q}u_{\varepsilon}^{1-\alpha q}\xi^{p}$. Note that
$u_{\varepsilon}^{1-\alpha p}$ is bounded since $-1<1-\alpha p<0$.

On the one hand, by the regularity of $u$ on $(0,T)$ it follows that

$\partial_{\tau}\left(\tau^{\beta q}u_{\varepsilon}^{2-\alpha
q}\xi^{p}\right)=\left(2-\alpha q\right)\psi\partial_{\tau}u+\beta
q\tau^{\beta q-1}u_{\varepsilon}^{2-\alpha q}\xi^{p}\geq\left(2-\alpha
q\right)\psi\partial_{\tau}u.$

Since $0<2-\alpha q<1$, with $\partial_{\tau}u$ in
$L^{2}(0,T;L^{2}(\mathbb{R}^{N}))$ and $u$ in $C([0,T];L^{1}(\mathbb{R}^{N}))$
and boundedness of $\psi$, we can integrate over $\Omega\times(0,t)$ on both
sides and get

$\int_{0}^{t}\int_{\Omega}\psi\partial_{\tau}udxd\tau\leq\frac{1}{2-\alpha
q}\int_{\Omega}t^{\beta q}u_{\varepsilon}^{2-\alpha q}(x,t)\xi^{p}dx.$ (3.4)

By the regularity of $u$ it follows that

$|\nabla u|^{p-2}\nabla u\nabla\psi=p\xi^{p-1}|\nabla u|^{p-2}\tau^{\beta
q}u_{\varepsilon}^{1-\alpha q}\nabla u\nabla\xi-\left(\alpha q-1\right)|\nabla
u|^{p}\xi^{p}\tau^{\beta q}u_{\varepsilon}^{-\alpha q}.$ (3.5)

For the first term on the right hand side of (3.5), applying Young’s
inequality $a\cdot
b\leq\varepsilon|a|^{q}+\varepsilon^{-\frac{1}{q-1}}|b|^{p}$ for
$\varepsilon=(\alpha q-1)/(2p)$ and

$a=\xi^{p-1}|\nabla u|^{p-2}\tau^{\beta}u_{\varepsilon}^{-\alpha}\nabla
u\quad\text{and}\quad
b=\tau^{\beta(q-1)}u_{\varepsilon}^{1-\alpha(q-1)}\nabla\xi.$

Equation (3.5) yields

$|\nabla u|^{p-2}\nabla u\nabla\psi\leq-\frac{1}{2}\left(\alpha
q-1\right)\xi^{p}|\nabla u|^{p}\tau^{\beta q}u_{\varepsilon}^{-\alpha q}\\\
+p\left(\frac{2p}{\alpha q-1}\right)^{\frac{1}{q-1}}\tau^{\beta
q}u_{\varepsilon}^{p-\alpha q}|\nabla\xi|^{p}.$ (3.6)

Since $|\nabla u|$ is in $L^{p}(0,T;L^{p}(\mathbb{R}^{N}))$ and
$u_{\varepsilon}^{-\alpha q}$ is bounded and $0<p-\alpha q<1$, it follows that

$\int_{0}^{t}\int_{\Omega}\xi^{p}|\nabla u|^{p}\tau^{\beta
q}u_{\varepsilon}^{-\alpha
q}dxd\tau<\infty\quad\text{and}\quad\int_{0}^{t}\int_{\Omega}\tau^{\beta
q}u_{\varepsilon}^{p-\alpha q}|\nabla\xi|^{p}dxd\tau<\infty.$

Hence, integrating (3.6) over $\Omega\times(0,t)$ yields

$\int_{0}^{t}\int_{\Omega}|\nabla u|^{p-2}\nabla u\nabla\psi
dxd\tau\leq-\frac{1}{2}\left(\alpha
q-1\right)\int_{0}^{t}\int_{\Omega}\xi^{p}|\nabla u|^{p}\tau^{\beta
q}u_{\varepsilon}^{-\alpha q}dxd\tau\\\ +p\left(\frac{2p}{\alpha
q-1}\right)^{\frac{1}{q-1}}\int_{0}^{t}\int_{\Omega}\tau^{\beta
q}u_{\varepsilon}^{p-\alpha q}|\nabla\xi|^{p}dxd\tau.$ (3.7)

On the other hand, multiplying (1.1) by $\psi$ and integrating by part, it
follows that

$\int_{0}^{t}\int_{\Omega}\psi\partial_{\tau}udxd\tau+\int_{0}^{t}\int_{\Omega}|\nabla
u|^{p-2}\nabla u\nabla\psi dxd\tau=0.$

Hence, adding (3.4) and (3.7) together yields

$\int_{0}^{t}\int_{\Omega}\xi^{p}|\nabla u|^{p}\tau^{\beta
q}u_{\varepsilon}^{-\alpha q}dxd\tau\leq\left(\frac{2p}{\alpha
q-1}\right)^{p}\int_{0}^{t}\int_{\Omega}\tau^{\beta
q}u_{\varepsilon}^{p-\alpha q}|\nabla\xi|^{p}dxd\tau\\\ +\frac{2}{\left(\alpha
q-1\right)\left(2-\alpha q\right)}\int_{\Omega}t^{\beta
q}u_{\varepsilon}^{2-\alpha q}(x,t)\xi^{p}dxd\tau.$

Taking

$C(N,p)=\max\left\\{\left(\frac{2p}{\alpha
q-1}\right)^{-p},\frac{2}{\left(\alpha q-1\right)\left(2-\alpha
q\right)}\right\\}$

yields the result. ∎

We address now the proof of Theorem 3.1. In the following proof, we denote by
$C$ a generic positive constant depending only on $p$ and $N$.

###### Proof of Theorem 3.1.

In a first step, we show inequality (3.1) in the case where $u$ is the
positive weak solution of (1.1) with initial data $u_{0}$ in
$C^{\infty}_{c}(\mathbb{R}^{N})$ and $u_{0}\geq 0$. Let $R>r>0$ and
$1\leq\sigma<\sigma^{\prime}\leq 2$. Let $A_{\sigma},A_{\sigma^{\prime}}$
denote closed annulus $A_{r/\sigma}^{R\sigma}$ and
$A_{r/\sigma^{\prime}}^{R\sigma^{\prime}}$ respectively. Let $\xi$ be a smooth
cut-off function such that $\xi=1$ in $A_{\sigma}$ and $\xi=0$ in
$A^{c}_{\sigma^{\prime}}$ and
$|\nabla\xi|\leq\frac{\sigma^{\prime}\sigma}{(\sigma^{\prime}-\sigma)r}$ in
$A^{r}:=A_{r/\sigma^{\prime}}^{r/\sigma}$ and
$|\nabla\xi|\leq\frac{1}{(\sigma^{\prime}-\sigma)R}$ in
$A^{R}:=A_{R\sigma}^{R\sigma^{\prime}}$. Let
$\phi(x,\tau)=|x|^{q}\xi^{p}\chi_{[0,T]}(\tau)$ be the test function where
$\chi_{[0,T]}$ is indicator function. Then for any $0<t\leq T$, by (2.7), it
follows that

$\int_{A_{\sigma}}|x|^{q}u(x,t)dx\leq\int_{A_{\sigma^{\prime}}}|x|^{q}u_{0}dx+p\int_{0}^{t}\int_{A_{\sigma^{\prime}}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{q}|\nabla\xi|dxd\tau\\\
+q\int_{0}^{t}\int_{A_{\sigma^{\prime}}}|\nabla
u|^{p-1}\xi^{p}|x|^{\frac{1}{p-1}}dxd\tau.$ (3.8)

For the second term on the right hand side of (3.8), since $|\nabla\xi|=0$ in
$A_{\sigma}$ and
$|\nabla\xi|\leq\frac{\sigma^{\prime}\sigma}{(\sigma^{\prime}-\sigma)r}$ and
$|\nabla\xi|\leq\frac{1}{(\sigma^{\prime}-\sigma)R}$ in $A^{r}$ and $A^{R}$
respectively, and note that $\sigma,\sigma^{\prime}\leq 2$, it follows that

$\int_{0}^{t}\int_{A_{\sigma^{\prime}}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{q}|\nabla\xi|dxd\tau=\int_{0}^{t}\int_{A^{r}\cup
A^{R}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{\frac{1}{p-1}}\left(|x||\nabla\xi|\right)dxd\tau\\\
\leq\frac{4}{\sigma^{\prime}-\sigma}\int_{0}^{t}\int_{A_{\sigma^{\prime}}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{\frac{1}{p-1}}dxd\tau.$ (3.9)

Plugging (3.9) into (3.8) and use the fact that $\xi^{p}\leq\xi^{p-1}$ and
$1\leq 1/(\sigma^{\prime}-\sigma)$ yields

$\int_{A_{\sigma}}|x|^{q}u(x,t)dx\leq\int_{A_{\sigma^{\prime}}}|x|^{q}u_{0}dx+\frac{C}{\sigma^{\prime}-\sigma}\int_{0}^{t}\int_{A_{\sigma^{\prime}}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{\frac{1}{p-1}}dxd\tau.$ (3.10)

Let $u_{\varepsilon}:=u+\varepsilon$ for $\varepsilon>0$. We consider the
second term on the right hand side of (3.10) for different cases of $p$:

  1. Case 1:

If $p>1/(p-1)$, then by Hölder’s inequality, it follows that

$\int_{0}^{t}\int_{A_{\sigma^{\prime}}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{\frac{1}{p-1}}dxd\tau\\\
\leq\left(\int_{0}^{t}\int_{A_{\sigma^{\prime}}}\tau^{\beta q}|\nabla
u|^{p}\xi^{p}u_{\varepsilon}^{-\alpha
q}dxd\tau\right)^{\frac{1}{q}}\left(\int_{0}^{t}\int_{A_{\sigma^{\prime}}}\tau^{-\beta
p}|x|^{q}u_{\varepsilon}^{\alpha p}dxd\tau\right)^{\frac{1}{p}},$ (3.11)

where $\alpha=1/p$ and $\beta=1/(2p)$.

For the first term on the right hand of (3.11), by Lemma 3.4 and upper bounds
of $|\nabla\xi|$ in $A^{r}$ and $A^{R}$, it follows that

$\int_{0}^{t}\int_{A_{\sigma^{\prime}}}\tau^{\beta q}|\nabla
u|^{p}\xi^{p}u_{\varepsilon}^{-\alpha
q}dxd\tau\leq\frac{C}{[(\sigma^{\prime}-\sigma)r]^{p}}\int_{0}^{t}\int_{A^{r}}\tau^{\beta
q}u_{\varepsilon}^{p-\alpha q}dxd\tau\\\
+\frac{C}{[(\sigma^{\prime}-\sigma)R]^{p}}\int_{0}^{t}\int_{A^{R}}\tau^{\beta
q}u_{\varepsilon}^{p-\alpha q}dxd\tau+C\int_{A_{\sigma^{\prime}}}t^{\beta
q}u_{\varepsilon}^{2-\alpha q}dx.$ (3.12)

For the first term on the right hand side of (3.12), since $0<p-1/(p-1)<1$,
Hölder’s inequality yields

$\int_{0}^{t}\int_{A^{r}}\tau^{\frac{q}{2p}}u_{\varepsilon}^{p-\frac{1}{p-1}}dxd\tau=\int_{0}^{t}\int_{A^{r}}\left(\tau^{\frac{p}{2}}|x|^{-q\left(p-\frac{1}{p-1}\right)}\right)\left(|x|^{q}\tau^{-\frac{1}{2}}u_{\varepsilon}\right)^{p-\frac{1}{p-1}}dxd\tau\\\
\leq\left(\int_{0}^{t}\int_{A^{r}}\tau^{\frac{p}{2(q-p)}}|x|^{-\frac{q}{q-p}\left(p-\frac{1}{p-1}\right)}dxd\tau\right)^{1-p+\frac{1}{p-1}}\left(\int_{0}^{t}\int_{A^{r}}\tau^{-\frac{1}{2}}|x|^{q}u_{\varepsilon}dxd\tau\right)^{p-\frac{1}{p-1}}\\\
\leq
Ct^{q-\frac{q}{2p}}r^{N(q-p)-q\left(p-\frac{1}{p-1}\right)}\left(\sup_{0<\tau\leq
t}\int_{A_{\sigma^{\prime}}}|x|^{q}u_{\varepsilon}dx\right)^{p-\frac{1}{p-1}}.$
(3.13)

The second term on the right hand side of (3.12) is the same as the first
term, replacing $r$ by $R$. As for the third term on the right hand side of
(3.12), Hölder’s inequality yields

$\int_{A_{\sigma^{\prime}}}t^{\frac{q}{2p}}u_{\varepsilon}^{2-\frac{1}{p-1}}dx\leq
t^{\frac{q}{2p}}\left(\int_{A_{\sigma^{\prime}}}|x|^{-\frac{p(2p-3)}{(p-1)(2-p)}}dx\right)^{\frac{1}{p-1}-1}\left(\int_{A_{\sigma^{\prime}}}|x|^{q}u_{\varepsilon}dx\right)^{2-\frac{1}{p-1}}\\\
\leq
Ct^{\frac{q}{2p}}\left(\max\left\\{r^{-\frac{p(N)}{(p-1)(2-p)}},R^{-\frac{p(N)}{(p-1)(2-p)}}\right\\}\right)^{\frac{2-p}{p-1}}\left(\sup_{0<\tau\leq
t}\int_{A_{\sigma^{\prime}}}|x|^{q}u_{\varepsilon}dx\right)^{\frac{2p-3}{p-1}}.$
(3.14)

We now turn to the second term on right hand side of (3.11) for which holds

$\left(\int_{0}^{t}\int_{A_{\sigma^{\prime}}}\tau^{-\frac{1}{2}}|x|^{q}u_{\varepsilon}dxd\tau\right)^{\frac{1}{p}}\leq
Ct^{\frac{1}{2p}}\left(\sup_{0<\tau\leq
t}\int_{A_{\sigma^{\prime}}}|x|^{q}u_{\varepsilon}dx\right)^{\frac{1}{p}}.$
(3.15)

Plugging (3.13), (3.14) into (3.12), and then applying inequality
$(a+b)^{1/q}\leq a^{1/q}+b^{1/q}$ for $a,b>0$ to (3.12), and plugging it into
(3.11) together with (3.15) and taking $\varepsilon\searrow 0$, it follows
that

$\frac{C}{\sigma^{\prime}-\sigma}\int_{0}^{t}\int_{A_{\sigma^{\prime}}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{\frac{1}{p-1}}dxd\tau\\\ \leq
C(\sigma^{\prime}-\sigma)^{-p}t\left(r^{1-k-\left(p-\frac{1}{p-1}\right)}+R^{1-k-\left(p-\frac{1}{p-1}\right)}\right)M_{q}(\sigma^{\prime})^{p-1}\\\
+C(\sigma^{\prime}-\sigma)^{-1}t^{\frac{1}{p}}\left(\max\left\\{r^{-\frac{p(N)}{(p-1)(2-p)}},R^{-\frac{p(N)}{(p-1)(2-p)}}\right\\}\right)^{\frac{2-p}{p}}M_{q}(\sigma^{\prime})^{\frac{2(p-1)}{p}},$
(3.16)

where $M_{q}(\sigma^{\prime})=\sup_{0<\tau\leq
t}\int_{A_{\sigma^{\prime}}}|x|^{q}udx$. Applying Young’s inequaliy
$ab\leq\varepsilon b^{1/(p-1)}+C_{\varepsilon}a^{1/(2-p)}$ and
$ab\leq\varepsilon b^{p/(2p-2)}+C_{\varepsilon}a^{p/(2-p)}$ to the first and
second terms of the right hand side of (3.16), respectively, whereby
$\varepsilon=1/4$, and using identity $1-k-p+1/(p-1)=-p(N)/(p-1)$, it follows
that

$\frac{C}{\sigma^{\prime}-\sigma}\int_{0}^{t}\int_{A_{\sigma^{\prime}}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{\frac{1}{p-1}}dxd\tau\leq\frac{1}{2}\sup_{0<\tau\leq
t}\int_{A_{\sigma^{\prime}}}|x|^{q}udx\\\
+C(\sigma^{\prime}-\sigma)^{-\frac{p}{2-p}}t^{\frac{1}{2-p}}\left(r^{-\frac{p(N)}{(p-1)(2-p)}}+R^{-\frac{p(N)}{(p-1)(2-p)}}\right).$
(3.17)

  2. Case 2:

If $p\leq 1/(p-1)$, by Hölder’s inequality, it follows that

$\int_{0}^{t}\int_{A_{\sigma^{\prime}}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{\frac{1}{p-1}}dxd\tau\\\
\leq\left(\int_{0}^{t}\int_{A_{\sigma^{\prime}}}\tau^{\beta q}|\nabla
u|^{p}u_{\varepsilon}^{-\alpha
q}\xi^{p}|x|^{\left(\frac{1}{p-1}-p\right)q}dxd\tau\right)^{\frac{1}{q}}\left(\int_{0}^{t}\int_{A_{\sigma^{\prime}}}\tau^{-\beta
p}|x|^{p^{2}}u_{\varepsilon}^{\alpha p}dxd\tau\right)^{\frac{1}{p}},$ (3.18)

where $\alpha=p-1$ and $\beta=(p-1)/2$.

For the first term on the right hand side of (3.18), taking
$\tilde{\xi}=\xi|x|^{(1/(p-1)-p)/(p-1)}$ and applying Lemma 3.4, yields

$\int_{0}^{t}\int_{A_{\sigma^{\prime}}}\tau^{\beta q}|\nabla
u|^{p}u_{\varepsilon}^{-\alpha
q}\left(\xi|x|^{\left(\frac{1}{p-1}-p\right)\frac{1}{p-1}}\right)^{p}dxd\tau\leq
C\int_{0}^{t}\int_{A_{\sigma^{\prime}}}\tau^{\beta q}u_{\varepsilon}^{p-\alpha
q}|\nabla\tilde{\xi}|^{p}dxd\tau\\\ +C\int_{A_{\sigma^{\prime}}}t^{\beta
q}u_{\varepsilon}^{2-\alpha q}\tilde{\xi}^{p}dx.$ (3.19)

For the first term on the right hand side of (3.19), note that we have

$\nabla\tilde{\xi}=|x|^{\left(\frac{1}{p-1}-p\right)\frac{1}{p-1}}\nabla\xi+\xi\left(\frac{1}{p-1}-p\right)\frac{1}{p-1}|x|^{\left(\frac{1}{p-1}-p\right)\frac{1}{p-1}-2}x.$

Hence, applying inequality $|a+b|^{p}\leq 2^{p-1}(|a|^{p}+|b|^{p})$ and upper
bounds of $|\nabla\xi|$ in $A^{r}$ and $A^{R}$ and $|\xi|\leq 1$, it follows
that

$\int_{0}^{t}\int_{A_{\sigma^{\prime}}}\tau^{\frac{p}{2}}|\nabla\tilde{\xi}|^{p}dxd\tau\leq
Ct^{\frac{p}{2}+1}\left\\{\int_{A_{\sigma^{\prime}}}|x|^{\left(\frac{1}{p-1}-p\right)q}|\nabla\xi|^{p}+|x|^{\left(\frac{1}{p-1}-p\right)q-p}dx\right\\}\\\
\leq
Ct^{\frac{p}{2}+1}\left(\sigma^{\prime}-\sigma\right)^{-p}\left(r^{-p}\int_{A^{r}}|x|^{\left(\frac{1}{p-1}-p\right)q}dx+R^{-p}\int_{A^{R}}|x|^{\left(\frac{1}{p-1}-p\right)q}dx\right)\\\
+Ct^{\frac{p}{2}+1}\int_{A_{\sigma^{\prime}}}|x|^{\left(\frac{1}{p-1}-p\right)q-p}dx\\\
\leq
Ct^{\frac{p}{2}+1}\left(\sigma^{\prime}-\sigma\right)^{-p}\left(r^{N+\left(\frac{1}{p-1}-p\right)q-p}+R^{N+\left(\frac{1}{p-1}-p\right)q-p}\right)\\\
+Ct^{\frac{p}{2}+1}\max\left\\{r^{N+\left(\frac{1}{p-1}-p\right)q-p},R^{N+\left(\frac{1}{p-1}-p\right)q-p}\right\\}.$
(3.20)

By the fact that $N+(1/(p-1)-p)q-p\geq 0$ when $N\geq 2$ and $r\leq R$ and
$1\leq(\sigma^{\prime}-\sigma)^{-p}$, it follows from (3.20) that

$\int_{0}^{t}\int_{A_{\sigma^{\prime}}}\tau^{\frac{p}{2}}|\nabla\tilde{\xi}|^{p}dxd\tau\leq
Ct^{\frac{p}{2}+1}\left(\sigma^{\prime}-\sigma\right)^{-p}R^{N+\left(\frac{1}{p-1}-p\right)q-p}.$
(3.21)

As for the second term on the right hand side of (3.19), note that
$|\tilde{\xi}|^{p}\leq|x|^{(1/(p-1)-p)q}$, which by Hölder’s inequality yields

$\int_{A_{\sigma^{\prime}}}t^{\frac{p}{2}}u_{\varepsilon}^{2-p}\tilde{\xi}^{p}dx\leq
t^{\frac{p}{2}}\left(\int_{A_{\sigma^{\prime}}}|x|^{\left(\frac{1}{p-1}-p\right)q-(2-p)q}u_{\varepsilon}^{2-p}|x|^{(2-p)q}dx\right)\\\
\leq
t^{\frac{p}{2}}\left(\int_{A_{\sigma^{\prime}}}|x|^{\left(\frac{1}{p-1}-2\right)\frac{q}{p-1}}dx\right)^{p-1}\left(\int_{A_{\sigma^{\prime}}}|x|^{q}u_{\varepsilon}dx\right)^{2-p}\\\
\leq
Ct^{\frac{p}{2}}\max\left\\{r^{N+\frac{p(3-2p)}{(p-1)^{3}}},R^{N+\frac{p(3-2p)}{(p-1)^{3}}}\right\\}^{p-1}\left(\sup_{0<\tau\leq
t}\int_{A_{\sigma^{\prime}}}|x|^{q}u_{\varepsilon}dx\right)^{2-p}.$ (3.22)

Note that if $p\leq 3/2$, it is obvious that $N+p(3-2p)/(p-1)^{3}\geq 0$. If
$p>3/2$, since $N\geq 2$, then it holds that

$N(p-1)^{3}+p(3-2p)\geq\frac{2}{2}(p-1)^{2}+p(3-2p)=-p^{2}+p+1\geq 0.$

Taking $\varepsilon\searrow 0$ on the right hand side of (3.22), yields

$\int_{A_{\sigma^{\prime}}}t^{\frac{p}{2}}u_{\varepsilon}^{2-p}\tilde{\xi}^{p}dx\leq
Ct^{\frac{p}{2}}R^{\left(N+\frac{p(3-2p)}{(p-1)^{3}}\right)(p-1)}M_{q}(\sigma^{\prime})^{2-p}.$
(3.23)

Plugging (3.21) and (3.23) into (3.19) and applying inequality
$(a+b)^{1/q}\leq a^{1/q}+b^{1/q}$ yields

$\left(\int_{0}^{t}\int_{A_{\sigma^{\prime}}}\tau^{\beta q}|\nabla
u|^{p}u_{\varepsilon}^{-\alpha
q}\xi^{p}|x|^{\left(\frac{1}{p-1}-p\right)q}dxd\tau\right)^{\frac{1}{q}}\\\
\leq
Ct^{\frac{p+2}{2q}}\left(\sigma^{\prime}-\sigma\right)^{1-p}R^{\frac{N}{q}+\frac{1}{p-1}-p-\frac{p}{q}}+Ct^{\frac{p}{2q}}R^{\left(N+\frac{p(3-2p)}{(p-1)^{3}}\right)\frac{(p-1)}{q}}M_{q}(\sigma^{\prime})^{\frac{2-p}{q}}.$
(3.24)

As for the second term on the right hand side of (3.18), Hölder’s inequality
and then letting $\varepsilon\searrow 0$, yields

$\left(\int_{0}^{t}\int_{A_{\sigma^{\prime}}}\tau^{-\frac{p(p-1)}{2}}|x|^{p^{2}}u_{\varepsilon}^{p(p-1)}dxd\tau\right)^{\frac{1}{p}}\\\
\leq\left\\{\left(\int_{0}^{t}\int_{A_{\sigma^{\prime}}}dxd\tau\right)^{1-p^{2}+p}\left(\int_{0}^{t}\int_{A_{\sigma^{\prime}}}\tau^{-\frac{1}{2}}|x|^{q}u_{\varepsilon}dxdx\right)^{p^{2}-p}\right\\}^{\frac{1}{p}}\\\
\leq
Ct^{\frac{1}{p}-\frac{p-1}{2}}R^{\frac{N(-p^{2}+p+1)}{p}}M_{q}(\sigma^{\prime})^{p-1}.$
(3.25)

Plugging (3.24) and (3.25) into (3.18), it follows that

$\frac{C}{\sigma^{\prime}-\sigma}\int_{0}^{t}\int_{A_{\sigma^{\prime}}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{\frac{1}{p-1}}dxd\tau\leq
C\left(\sigma^{\prime}-\sigma\right)^{-p}tR^{N(2-p)-\frac{p(2p-3)}{p-1}}M_{q}(\sigma^{\prime})^{p-1}\\\
+C\left(\sigma^{\prime}-\sigma\right)^{-1}t^{\frac{1}{p}}R^{\frac{N(2-p)}{p}+\frac{3-2p}{p-1}}M_{q}(\sigma^{\prime})^{\frac{2(p-1)}{p}}.$
(3.26)

Applying Young’s inequality $ab\leq\varepsilon
b^{1/(p-1)}+C_{\varepsilon}a^{1/(2-p)}$ as well as $ab\leq\varepsilon
b^{p/(2p-2)}+C_{\varepsilon}a^{p/(2-p)}$ to the first and second terms on the
right hand side of (3.26) respectively, where $\varepsilon=1/4$, it follows
that

$\frac{C}{\sigma^{\prime}-\sigma}\int_{0}^{t}\int_{A_{\sigma^{\prime}}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{\frac{1}{p-1}}dxd\tau\leq\frac{1}{2}\sup_{0<\tau\leq
t}\int_{A_{\sigma^{\prime}}}|x|^{q}udx\\\
+C\left(\sigma^{\prime}-\sigma\right)^{\frac{p}{2-p}}t^{\frac{1}{2-p}}R^{-\frac{p(N)}{(p-1)(2-p)}}.$
(3.27)

Together with (3.17) and (3.27), we deduce from (3.10) that

$\int_{A_{\sigma}}|x|^{q}u(x,t)dx\leq\int_{A_{\sigma^{\prime}}}|x|^{q}u_{0}dx+\frac{1}{2}\sup_{0<\tau\leq
t}\int_{A_{\sigma^{\prime}}}|x|^{q}udx\\\
+C\left(\sigma^{\prime}-\sigma\right)^{-\frac{p}{2-p}}t^{\frac{1}{2-p}}\left(r^{-\frac{p(N)}{(p-1)(2-p)}}+R^{-\frac{p(N)}{(p-1)(2-p)}}\right).$
(3.28)

Inequality (3.28) holds by replacing $t$ by any $\tau$ in $(0,t]$ implying
that

$\sup_{0<\tau\leq
t}\int_{A_{\sigma}}|x|^{q}udx\leq\int_{A_{\sigma^{\prime}}}|x|^{q}u_{0}dx+\frac{1}{2}\sup_{0<\tau\leq
t}\int_{A_{\sigma^{\prime}}}|x|^{q}udx\\\
+C\left(\sigma^{\prime}-\sigma\right)^{-\frac{p}{2-p}}t^{\frac{1}{2-p}}\left(r^{-\frac{p(N)}{(p-1)(2-p)}}+R^{-\frac{p(N)}{(p-1)(2-p)}}\right).$
(3.29)

By Lemma 2.2, it holds that

$\sup_{\sigma\in[1,2]}M_{q}(\sigma)\leq(2R)^{q}\sup_{0<\tau\leq
t}\int_{B_{2R}}udx\leq
C(2R)^{q}\left(\int_{B_{2R}}u_{0}dx+R^{-\frac{k}{2-p}}t^{\frac{1}{2-p}}\right)<\infty.$

Hence $M_{q}(\sigma)$ has bounded value on $\sigma\in[1,2]$. Applying Lemma
2.3 for $\lambda=1$ and $\lambda^{\prime}=2$, it follows that

$\sup_{0<\tau\leq t}\int_{r\leq|x|\leq R}|x|^{q}udx\leq
C\int_{\frac{1}{2}r\leq|x|\leq 2R}|x|^{q}u_{0}dx\\\
+Ct^{\frac{1}{2-p}}\left(r^{-\frac{p(N)}{(p-1)(2-p)}}+R^{-\frac{p(N)}{(p-1)(2-p)}},\right)$
(3.30)

for some $C:=C(N,p)$. Thus we prove inequality (3.1) for the case where the
initial data $u_{0}$ is in $C^{\infty}_{c}(\mathbb{R}^{N})$.

Let us finally address the inequality (3.1) in the general case where
$\mu\in\mathcal{M}^{+}$ is taken as initial data. Let $m\in
C^{\infty}_{c}(\mathbb{R}^{N})$ be a mollifier function with
$\mathrm{supp}(m)\subset B_{1},m\geq 0$ and $\int_{\mathbb{R}^{N}}mdx=1$ and
$m_{n}(x)=n^{N}m(nx)$. Let $\xi_{n}\in C^{\infty}_{c}(\mathbb{R}^{N})$ be a
smooth cut-off function such that $\xi_{n}=1$ in $B_{n}$ and $\xi=0$ in
$B^{c}_{2n}$. Let $u_{0n}\in C^{\infty}_{c}(\mathbb{R}^{N})$ be defined as
follows:

$u_{0n}(x)=\xi_{n}(x)\left(m_{n}*\mu\right)(x)=\xi_{n}(x)\int_{\mathbb{R}^{N}}m_{n}(x-y)d\mu(y).$

Obviously, $u_{0n}\geq 0$. Moreover, for any $\rho>0$, since
$\mathrm{supp}(m_{n}(x-\cdot))\subset B_{1/n}(x)$, it follows that

$\int_{B_{\rho}}u_{0n}dx\leq\int_{y\in B_{\rho+1/n}}\int_{B_{1/n}(y)\cap
B_{\rho}}m_{n}(x-y)dxd\mu(y)\leq\mu(B_{\rho+1}).$ (3.31)

Furthermore, for any $\phi\in C^{\infty}_{c}(\mathbb{R}^{N})$, take $n$ large
enough such that $\mathrm{supp}(\phi)_{1}\subset B_{n}$, where
$\mathrm{supp}(\phi)_{1}:=\\{x\colon d(x,\mathrm{supp}(\phi))\leq 1\\}$. Then
it follows that

$\int_{\mathbb{R}^{N}}\phi
u_{0n}dx=\int_{\mathrm{supp}(\phi)_{1}}\int_{B_{1/n}(y)}\xi_{n}(x)\phi(x)m_{n}(x-y)dxd\mu(y)=\int_{\mathrm{supp}(\phi)_{1}}(\phi*\tilde{m}_{n})(y)d\mu(y),$

where $\tilde{m}_{n}(x)=m_{n}(-x)$. By the uniform convergence of
$\phi*\tilde{m}_{n}$ to $\phi$ on compact sets, see [4, Proposition 4.21], it
follows that

$\lim_{n\rightarrow\infty}\int_{\mathbb{R}^{N}}\phi
u_{0n}dx=\int_{\mathbb{R}^{N}}\phi d\mu.$ (3.32)

Let $u_{n}$ be the unique positive weak solution of (1.1) and (2.1) with
initial data $u_{0n}$ for $n\geq 1$. Then by inequality (2.9) of Lemma 2.2 and
(3.31), it follows that $(u_{n})_{n}$ is locally equibounded in $S_{T}$, that
is, for any $\varepsilon>0$ and bounded domain $\Omega\subset\mathbb{R}^{N}$,
$(u_{n})_{n}$ is equibounded in
$\Omega_{\varepsilon}:=\Omega\times[\varepsilon,T]$. Then by [5, Theorem 1],
it follows that $(u_{n})_{n}$ is uniformly equicontinuous in
$\Omega_{\varepsilon}$ for all $\varepsilon>0$ and all bounded domains
$\Omega\subset\mathbb{R}^{N}$. By diagonalization procedure, we can find a
subsequence of $(u_{n})$, which is relabelled by $n$, such that

$u_{n},\nabla u_{n}\rightarrow u,\nabla u\quad\text{uniformly on every compact
subset of }S_{T},$ (3.33)

and $u$ is a positive weak solution of (1.1) and (2.5) with initial data
$\mu$, see [6, Theorem III.8.1].

Now for any $R>r>0$ and $0<t\leq T$, by inequality (3.30), it follows that for
any $\tau\in(0,t]$,

$\int_{r\leq|x|\leq R}|x|^{q}u_{n}(x,\tau)dx\leq
C\int_{\frac{1}{2}r\leq|x|\leq 2R}|x|^{q}u_{0n}dx\\\
+Ct^{\frac{1}{2-p}}\left(r^{-\frac{p(N)}{(p-1)(2-p)}}+R^{-\frac{p(N)}{(p-1)(2-p)}}\right).$
(3.34)

For the left hand side of (3.34), by dominated convergence and (3.33), it
follows that

$\lim_{n\rightarrow\infty}\int_{r\leq|x|\leq
R}|x|^{q}u_{n}(x,\tau)dx=\int_{r\leq|x|\leq R}|x|^{q}u(x,\tau)dx.$ (3.35)

For the first term on the right hand side of (3.34), let $\varepsilon>0$ and
$\phi\in C_{c}(\mathbb{R}^{N})$ such that $\phi=|x|^{q}$ on
$\\{\frac{1}{2}r\leq|x|\leq 2R\\}$ and $\phi=0$ on
$\\{\frac{1}{2}r-\varepsilon\leq|x|\leq 2R+\varepsilon\\}^{c}$. Then taking
$n\rightarrow\infty$ and by (3.32), it follows that

$\limsup_{n\rightarrow\infty}\int_{\frac{1}{2}r\leq|x|\leq
2R}|x|^{q}u_{0n}dx\leq\int_{\mathbb{R}^{N}}\phi
d\mu\leq\int_{\frac{1}{2}r-\varepsilon\leq|x|\leq 2R+\varepsilon}|x|^{q}d\mu.$
(3.36)

For $\varepsilon\searrow 0$ together with (3.35) and (3.36), it follows that
for any $\tau\in(0,t]$,

$\int_{r\leq|x|\leq R}|x|^{q}u(x,\tau)dx\leq C\int_{\frac{1}{2}r\leq|x|\leq
2R}|x|^{q}d\mu\\\
+Ct^{\frac{1}{2-p}}\left(r^{-\frac{p(N)}{(p-1)(2-p)}}+R^{-\frac{p(N)}{(p-1)(2-p)}}\right).$

Taking the supremum over $\tau\in(0,t]$ on the left hand side yields the
desired result.

We are left to show uniform boundedness of $q$-moment and inequality (3.2).
Assume that $\mu$ is in $\mathcal{M}^{+}$ with finite total mass and finite
$q$-moment, and $p\in(p_{c},2)$ is such that $p(N)>0$. By inequality (2.8) in
Lemma 2.2 and similar arguments, see also [10, Theorem 1], it follows that for
any $t\in(0,T]$ and $r>0$,

$\sup_{0<\tau\leq t}\int_{B_{r}}u(x,\tau)dx\leq
C\int_{B_{2r}}d\mu+Cr^{-\frac{k}{2-p}}t^{\frac{1}{2-p}}.$ (3.37)

Taking $r=1$ and together with (3.1), it follows that for any $R>1$

$\sup_{0<\tau\leq t}\int_{B_{R}}|x|^{q}u(x,\tau)dx\leq\sup_{0<\tau\leq
t}\int_{|x|<1}|x|^{q}u(x,\tau)dx+\sup_{0<\tau\leq t}\int_{1\leq|x|\leq
R}|x|^{q}u(x,\tau)dx\\\ \leq C\int_{\frac{1}{2}\leq|x|\leq
2R}|x|^{q}d\mu+C\int_{|x|\leq
2}d\mu+Ct^{\frac{1}{2-p}}\left(1+R^{-\frac{p(N)}{(p-1)(2-p)}}\right).$

Since $p(N)>0$, taking $R\rightarrow\infty$ and it follows that

$\sup_{0<\tau\leq t}\int_{\mathbb{R}^{N}}|x|^{q}u(x,\tau)dx\leq
C\int_{|x|\geq\frac{1}{2}}|x|^{q}d\mu+C\int_{|x|\leq
2}d\mu+Ct^{\frac{1}{2-p}}.$

From this we obtain that $\\{\mu_{\tau}=u(x,\tau)dx\colon\tau\in[0,t]\\}$ has
uniformly bounded $q$-moment for any $t>0$. For inequality (3.2), taking
$R\rightarrow 0$ for both side of (3.1) and it follows that

$\sup_{0<\tau\leq t}\int_{|x|\geq r}|x|^{q}u(x,\tau)dx\leq
C\int_{|x|\geq\frac{1}{2}r}|x|^{q}d\mu+Ct^{\frac{1}{2-p}}r^{-\frac{p(N)}{(p-1)(2-p)}}.$

Since $\mu$ has finite $q$-moment, taking $r\rightarrow\infty$ and it follows
that

$\lim_{r\rightarrow\infty}\sup_{0<\tau\leq t}\int_{|x|\geq
r}|x|^{q}u(x,\tau)dx=0.$

∎

##  4\. Mass conservation and weak convergence

In [8], the authors show that mass conservation of any positive weak solution
of (1.1) and (2.5) with initial data $\mu$ in $\mathcal{M}^{+}_{f}$ with
compact support. In this section, we will show that the weak solution
constructed in Theorem 3.1 with initial data $\mu$ in $\mathcal{M}^{+}$ with
finite total mass and finite $q$-moment, preserves mass. As a by-product,
measure $\mu_{t}:=u(x,t)dx$ converges to $\mu$ weakly.

######  Theorem 4.1 (Mass Conservation).

Let $p\in(p_{c},2)$ and $\mu$ be a positive finite Radon measure in
$\mathbb{R}^{N}$ and $u$ is the positive weak solution in Theorem 3.1. If
$\mu$ has $q$-finite moment, then $u$ preserves mass, that is,

$\int_{\mathbb{R}^{N}}u(x,t)dx=\int_{\mathbb{R}^{N}}d\mu,\quad\text{for
all}\quad 0<t\leq T.$ (4.1)

######  Corollary 4.2 (Weak Convergence).

Let $\mu$ be a positive finite Radon measure in $\mathbb{R}^{N}$ with finite
$q$-moment and $u$ be the positive weak solution in Theorem 3.1. Then the
measure $\mu_{t}:=u(x,t)dx$ converges weakly to the initial data $\mu$ as $t$
goes to $0$.

We start with a lemma providing gradient estimate of the weak solution of
(1.1) with respect to $q$-moment of smooth and compact-supported initial data
on the annulus.

######  Lemma 4.3.

Let $u$ be the positive weak solution of (1.1) with initial data $u_{0}$ in
$C^{\infty}_{c}(\mathbb{R}^{N})$ and $u_{0}\geq 0$. Then there exists a
positive constant $C:=C(N,p)$ such that for any $0<t\leq T$ and $R>r>0$, it
holds that

$\int_{0}^{t}\int_{r\leq|x|\leq R}|\nabla u|^{p-1}dxd\tau\\\ \leq
Ct^{\frac{1}{p}}r^{-\frac{1}{p-1}}\left(r^{-\frac{p(N)}{p(p-1)}}+R^{-\frac{p(N)}{p(p-1)}}\right)\left(M_{q}\left(u_{0},A_{r/4}^{4R}\right)\right)^{\frac{2(p-1)}{p}}\\\
+Ct^{\frac{1}{2-p}}r^{-\frac{1}{p-1}}\left(r^{-\frac{2p(N)}{p(2-p)}}+R^{-\frac{2p(N)}{p(2-p)}}\right)\left(r^{-\frac{p(N)}{p(p-1)}}+R^{-\frac{p(N)}{p(p-1)}}\right),$
(4.2)

where $M_{q}(u_{0},A):=\int_{A}|x|^{q}u_{0}dx$.

###### Proof.

Let $R>r>0$ and denote by $A_{\sigma}$ the annulus $A_{r/\sigma}^{R\sigma}$
for $\sigma\geq 1$. Let $\xi$ be a smooth cut-off functions such that $\xi=1$
on $A_{1}$ and $\xi=0$ on $A^{c}_{2}$, and $|\nabla\xi|\leq 2r^{-1}$ on
$A_{r/2}^{r}$ and $|\nabla\xi|\leq R^{-1}$ on $A_{R}^{2R}$. Since $|x|\geq r$
on $A_{1}$, it follows that

$\int_{0}^{t}\int_{A_{1}}|\nabla u|^{p-1}dxd\tau\leq
r^{-\frac{1}{p-1}}\int_{0}^{t}\int_{A_{2}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{\frac{1}{p-1}}dxd\tau.$ (4.3)

We consider the right hand side of (4.3) for different cases of $p$:

  1. Case 1:

If $p>1/(p-1)$, then by inequality (3.16) in which we choose $\sigma=1$ and
$\sigma^{\prime}=2$, it follows that

$\int_{0}^{t}\int_{A_{2}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{\frac{1}{p-1}}dxd\tau\leq
Ct\left(r^{1-k-p+\frac{1}{p-1}}+R^{1-k-p+\frac{1}{p-1}}\right)M_{q}(2)^{p-1}\\\
+Ct^{\frac{1}{p}}\left(\max\left\\{r^{-\frac{p(N)}{(p-1)(2-p)}},R^{-\frac{p(N)}{(p-1)(2-p)}}\right\\}\right)^{\frac{2-p}{p}}M_{q}(2)^{\frac{2(p-1)}{p}},$
(4.4)

where $M_{q}(2)=\sup_{0<\tau\leq t}\int_{A_{2}}|x|^{q}udx$. Since that
$(p-1)\leq\frac{2(p-1)}{p}$, applying Young’s inequality $ab\leq
a^{\frac{2}{2-p}}+b^{\frac{2}{p}}$ to the first term on the right hand of
(4.4), yields

$\displaystyle tr^{1-k-p+\frac{1}{p-1}}M_{q}(2)^{p-1}$
$\displaystyle=\left(t^{\frac{1}{2}}r^{1-k-p+\frac{1}{p-1}+\frac{p(N)}{2(p-1)}}\right)\left(t^{\frac{1}{2}}r^{-\frac{p(N)}{2(p-1)}}M_{q}(2)^{p-1}\right)$
$\displaystyle\leq
t^{\frac{1}{2-p}}r^{-\frac{p(N)}{(p-1)(2-p)}}+t^{\frac{1}{p}}r^{-\frac{p(N)}{p(p-1)}}M_{q}(2)^{\frac{2(p-1)}{p}},$

which plugged into (4.4) yields

$\int_{0}^{t}\int_{A_{2}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{\frac{1}{p-1}}dxd\tau\leq
Ct^{\frac{1}{2-p}}\left(r^{-\frac{p(N)}{(p-1)(2-p)}}+R^{-\frac{p(N)}{(p-1)(2-p)}}\right)\\\
+Ct^{\frac{1}{p}}\left(r^{-\frac{p(N)}{p(p-1)}}+R^{-\frac{p(N)}{p(p-1)}}\right)M_{q}(2)^{\frac{2(p-1)}{p}}.$
(4.5)

  2. Case 2:

If $p\leq 1/(p-1)$, then by inequality (3.26) in which we choose $\sigma=1$
and $\sigma^{\prime}=2$, it follows that

$\int_{0}^{t}\int_{A_{2}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{\frac{1}{p-1}}dxd\tau\\\ \leq
CtR^{-\frac{p(N)}{p-1}}M_{q}(2)^{p-1}+Ct^{\frac{1}{p}}R^{-\frac{p(N)}{p(p-1)}}M_{q}(2)^{\frac{2(p-1)}{p}}.$
(4.6)

Applying the same Young’s inequality as in the previous step to the first term
on the right hand side, yields

$\displaystyle tR^{-\frac{p(N)}{p-1}}M_{q}(2)^{p-1}$
$\displaystyle=\left(t^{\frac{1}{2}}R^{-\frac{p(N)}{2(p-1)}}\right)\left(t^{\frac{1}{2}}R^{-\frac{p(N)}{2(p-1)}}M_{q}(2)^{p-1}\right)$
$\displaystyle\leq
t^{\frac{1}{2-p}}R^{-\frac{p(N)}{(p-1)(2-p)}}+t^{\frac{1}{p}}R^{-\frac{p(N)}{p(p-1)}}M_{q}(2)^{\frac{2(p-1)}{p}},$

which plugged into (4.6), implies

$\int_{0}^{t}\int_{A_{2}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{\frac{1}{p-1}}dxd\tau\leq
Ct^{\frac{1}{2-p}}R^{-\frac{p(N)}{(p-1)(2-p)}}+Ct^{\frac{1}{p}}R^{-\frac{p(N)}{p(p-1)}}M_{q}(2)^{\frac{2(p-1)}{p}}.$
(4.7)

Hence, from (4.5) and (4.7), we obtain that for any $p\in(p_{c},2)$, it holds
that

$\int_{0}^{t}\int_{A_{2}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{\frac{1}{p-1}}dxd\tau\leq
Ct^{\frac{1}{2-p}}\left(r^{-\frac{p(N)}{(p-1)(2-p)}}+R^{-\frac{p(N)}{(p-1)(2-p)}}\right)\\\
+Ct^{\frac{1}{p}}\left(r^{-\frac{p(N)}{p(p-1)}}+R^{-\frac{p(N)}{p(p-1)}}\right)M_{q}(2)^{\frac{2(p-1)}{p}}.$
(4.8)

Let $\alpha_{1}=-p(N)/[(p-1)(2-p)]$ and $\alpha_{2}=-p(N)/[p(p-1)]$ and
$\alpha_{3}=-2p(N)/[p(2-p)]$ and denote by $F_{i}$ the terms
$r^{\alpha_{i}}+R^{\alpha_{i}}$ for $i=1,2,3$. By applying inequality (3.1) in
Theorem 3.1 to $M_{q}(2)$ and inequality $(a+b)^{\gamma}\leq
a^{\gamma}+b^{\gamma}$ for $\gamma=2(p-1)/p$, it follows that

$\int_{0}^{t}\int_{A_{2}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{\frac{1}{p-1}}dxd\tau\leq
Ct^{\frac{1}{2-p}}F_{1}+Ct^{\frac{1}{p}}F_{2}\left(\int_{A_{4}}|x|^{q}u_{0}dx+t^{\frac{1}{2-p}}F_{1}\right)^{\frac{2(p-1)}{p}}\\\
\leq
Ct^{\frac{1}{2-p}}\left(F_{1}+F_{2}\left(F_{1}\right)^{\frac{2(p-1)}{p}}\right)+Ct^{\frac{1}{p}}F_{2}\left(\int_{A_{4}}|x|^{q}u_{0}dx\right)^{\frac{2(p-1)}{p}}.$
(4.9)

By using inequality $(a+b)^{\gamma}\leq a^{\gamma}+b^{\gamma}$ for
$\gamma=(2-p)/p$ and $\gamma=2(p-1)/p$ respectively, we have
$(F_{1})^{(2-p)/p}\leq F_{2}$ and $(F_{1})^{2(p-1)/p}\leq F_{3}$. Then for the
first two terms on the right hand side in (4.9), it follows that

$F_{1}+F_{2}\left(F_{1}\right)^{\frac{2(p-1)}{p}}=\left(F_{1}\right)^{\frac{2-p}{p}+\frac{2(p-1)}{p}}+F_{2}\left(F_{1}\right)^{\frac{2(p-1)}{p}}\leq
2F_{2}F_{3}.$

Plugging it into (4.9), we obtain

$\int_{0}^{t}\int_{A_{2}}|\nabla
u|^{p-1}\xi^{p-1}|x|^{\frac{1}{p-1}}dxd\tau\leq
Ct^{\frac{1}{2-p}}F_{2}F_{3}+Ct^{\frac{1}{p}}F_{2}\left(\int_{A_{4}}|x|^{q}u_{0}dx\right)^{\frac{2(p-1)}{p}}.$

which together with (4.3) yields the result. ∎

###### Proof of Theorem 4.1.

Let $\mu\in\mathcal{M}^{+}$ with finite total mass be fixed and $(u_{0n})$ be
the sequence of $C^{\infty}_{c}$ functions constructed in the proof of Theorem
3.1, and $u_{n}$ be the positive weak solution of (1.1) with $u_{0n}$ as
initial data, and $u$ be the positive weak solution constructed in the proof
of Theorem 3.1. Let $\rho>0$ and $\xi$ be a smooth cut-off function such that
$\xi=1$ on $B_{\rho}$ and $\xi=0$ on $B^{c}_{2\rho}$ and
$|\nabla\xi|\leq\rho^{-1}$ on $A_{\rho}^{2\rho}$. By using
$\phi(x,\tau)=\xi(x)\chi_{[0,T]}(\tau)$ as the test function in (2.7), it
follows that for any $0<t\leq T$ and $n\geq 1$,

$\left|\int_{B_{2\rho}}\xi(x)u_{n}(x,t)dx-\int_{B_{2\rho}}\xi(x)u_{0n}(x)dx\right|\leq\int_{0}^{t}\int_{B_{2\rho}}|\nabla
u_{n}|^{p-1}|\nabla\xi|dxd\tau\\\
\leq\rho^{-1}\int_{0}^{t}\int_{\rho\leq|x|\leq 2\rho}|\nabla
u_{n}|^{p-1}dxd\tau.$ (4.10)

By Lemma 4.3 with $r=\rho$ and $R=2\rho$, it follows that

$\left|\int_{B_{2\rho}}\xi(x)u_{n}(x,t)dx-\int_{B_{2\rho}}\xi(x)u_{0n}(x)dx\right|\leq\\\
Ct^{\frac{1}{p}}\rho^{-1-\frac{1}{p-1}-\frac{p(N)}{p(p-1)}}\left\\{\int_{\frac{1}{4}\rho\leq|x|\leq
8\rho}|x|^{q}u_{0n}dx\right\\}^{\frac{2(p-1)}{p}}+Ct^{\frac{1}{2-p}}\rho^{-1-\frac{1}{p-1}-\frac{2p(N)}{p(2-p)}-\frac{p(N)}{p(p-1)}}.$
(4.11)

By (3.33) in the proof of Theorem 3.1, it holds that $u_{n}$ converges to $u$,
up to a subsequence, uniformly on each compact subset of $S_{T}$. Together
with equality (3.32), taking $n\rightarrow\infty$ on both side of (4.11) and
it follows that

$\left|\int_{B_{2\rho}}\xi(x)u(x,t)dx-\int_{B_{2\rho}}\xi(x)d\mu\right|\leq
Ct^{\frac{1}{p}}\rho^{-1-\frac{1}{p-1}-\frac{p(N)}{p(p-1)}}\left\\{\int_{\frac{1}{4}\rho\leq|x|\leq
8\rho}|x|^{q}d\mu\right\\}^{\frac{2(p-1)}{p}}\\\
+Ct^{\frac{1}{2-p}}\rho^{-1-\frac{1}{p-1}-\frac{2p(N)}{p(2-p)}-\frac{p(N)}{p(p-1)}}.$
(4.12)

For the left hand side of (4.12), by letting $r\rightarrow\infty$ on both side
of (3.37) in the proof of Theorem 3.1, it follows that
$\int_{\mathbb{R}^{N}}u(x,t)dx<\infty$. Together with $|\xi(x)u(x,t)|\leq
u(x,t)$, by dominated convergence theorem, it follows that

$\lim_{\rho\rightarrow\infty}\left|\int_{B_{2\rho}}\xi(x)u(x,t)dx-\int_{B_{2\rho}}\xi(x)d\mu\right|=\left|\int_{\mathbb{R}^{N}}u(x,t)dx-\mu(\mathbb{R}^{N})\right|.$
(4.13)

For the right hand side of (4.12), it is easy to check that for
$p\in(p_{c},2)$,

$\displaystyle-1-\frac{1}{p-1}-\frac{p(N)}{p(p-1)}=-\frac{(N+3)p-2N}{p}<0,$
$\displaystyle-1-\frac{1}{p-1}-\frac{2p(N)}{p(2-p)}-\frac{p(N)}{p(p-1)}=-\frac{(N+1)p-2N}{2-p}<0.$

Since $\mu$ has finite $q$-moment, the term on the right hand side of (4.12)
converges to $0$ as $\rho\rightarrow\infty$. So we obtain that for any
$0<t\leq T$,

$\int_{\mathbb{R}^{N}}u(x,t)dx=\mu(\mathbb{R}^{N}).$

∎

###### Proof of Corollary 4.2.

By Theorem 3.1 and definition of weak solution of (1.1) and (2.5), it follows
that measure $\mu_{t}=u(x,t)dx$ converges vaguely to initial data $\mu$. Since
$\mu_{t}(\mathbb{R}^{N})=\mu(\mathbb{R}^{N})$ for all $0<t<T$, by the
classical result(for example, [13, Theorem 13.16], it follows that $\mu_{t}$
converges to $\mu$ weakly as $t$ goes to $0$. ∎

##  5\. Convergence rate of Wasserstein distance

We address now the convergence rate of the constructed weak solution
$u(t,\cdot)dx$ to $\mu$ in the $q$-Wasserstein distance. We recall that the
$q$-Wasserstein distance $W_{q}(\mu,\nu)$ between finite Borel measures
$\mu,\nu$ in $\mathbb{R}^{N}$ with equal mass is defined as

$W^{q}_{q}(\mu,\nu)=\min\left\\{\int_{\mathbb{R}^{N}\times\mathbb{R}^{N}}|x-y|^{q}d\pi(x,y)\colon\pi\in\Pi(\mu,\nu)\right\\},$
(5.1)

where $\Pi(\mu,\nu)$ is the family of all Borel measures on
$\mathbb{R}^{N}\times\mathbb{R}^{N}$ having $\mu$ and $\nu$ as their first and
second marginal measures respectively.

######  Theorem 5.1.

Let $\mu$ be a finite Radon measure on $\mathbb{R}^{N}$ and $u$ be the weak
solution of (1.1) constructed in Theorem 3.1 with $\mu$ as initial data. If
$p\in(p_{N},2)$ and $\mu$ has finite $q$-moment, then there exist constant
$C:=C(N,p,\mu,T)$ such that for all $t\in[0,T]$,

$W^{q}_{q}(\mu_{t},\mu)\leq Ct^{q-1}.$ (5.2)

To address the proof of Theorem 5.1, we first establish an auxiliary lemma,
prove the theorem in the case where the initial data is in
$C^{\infty}_{c}(\mathbb{R}^{N}$), and then show the general case by an
approximation procedure. The key ingredients of the proof are to use Brenier-
Benamou formulation for $q$-Wasserstein distance and write the parabolic
$p$-Laplace equation as a law of mass conservation
$\partial_{t}u=\mathrm{div}(\mathbf{v}u)$ where $\mathbf{v}=|\nabla
u|^{p-2}(-\nabla u)/u$.

######  Lemma 5.2.

Let $p\in(p_{N},2)$ and $u$ be the positive weak solution of Cauchy problem
(1.1) with $u_{0}\in C_{c}^{\infty}(\mathbb{R}^{N})$ and $u_{0}\geq 0$ as
initial data. Then for any $t>0$, it holds that

$\int_{0}^{t}\int_{\mathbb{R}^{N}}|\nabla u|^{p}u^{-\frac{1}{p-1}}dxd\tau\leq
C\left(\int_{|x|\leq 2}u_{0}dx\right)^{2-\frac{1}{p-1}}+C\left(\int_{|x|\geq
1/2}|x|^{q}u_{0}dx\right)^{2-\frac{1}{p-1}}\\\ +Ct^{\frac{2p-3}{(p-1)(2-p)}}$
(5.3)

for some $C:=C(N,p)$.

###### Proof.

Let $\varepsilon>0$ and $R>0$. Let $\xi$ be a smooth cut-off function such
that $\xi=1$ on $B_{R}$ and $\xi=0$ on $B^{c}_{2R}$ and $|\nabla\xi|\leq
R^{-1}$ on $A_{R}^{2R}$. Let $u_{\varepsilon}:=u+\varepsilon$ and
$u_{0\varepsilon}:=u_{0}+\varepsilon$ and
$\psi:=u_{\varepsilon}^{1-1/(p-1)}\xi$. Multiplying $\psi$ on (1.1) and
integrating it over $B_{2R}\times[0,t]$, by integral by part and similar
argument in Lemma 3.4, it follows that

$\displaystyle\int_{0}^{t}\int_{B_{2R}}|\nabla
u|^{p}u_{\varepsilon}^{-\frac{1}{p-1}}\xi dxd\tau$ $\displaystyle\leq
C\int_{0}^{t}\int_{B_{2R}}u_{\varepsilon}^{p-\frac{1}{p-1}}|\nabla\xi|^{p}dxd\tau$
$\displaystyle\quad\quad+C\left(\int_{B_{2R}}u_{\varepsilon}^{2-\frac{1}{p-1}}\xi
dx-\int_{B_{2R}}u_{0\varepsilon}^{2-\frac{1}{p-1}}\xi dx\right)$
$\displaystyle\leq
C\int_{0}^{t}\int_{B_{2R}}u_{\varepsilon}^{p-\frac{1}{p-1}}|\nabla\xi|^{p}dxd\tau+C\int_{B_{2R}}u_{\varepsilon}^{2-\frac{1}{p-1}}dx$
(5.4)

Note that $p\in(p_{N},2)$ implies that $p-1/(p-1)>1$. Hence, for the first
term on the right hand side of (5), by using the similar argument as in the
proof of Theorem 3.1, it follows that

$\displaystyle\int_{0}^{t}\int_{B_{2R}}u_{\varepsilon}^{p-\frac{1}{p-1}}|\nabla\xi|^{p}dxd\tau$
$\displaystyle\leq
R^{-p}\int_{0}^{t}\int_{A_{R}^{2R}}|x|^{-q\left(p-\frac{1}{p-1}\right)}|x|^{q\left(p-\frac{1}{p-1}\right)}u_{\varepsilon}^{p-\frac{1}{p-1}}dxd\tau$
$\displaystyle\leq
CtR^{q\left(1-k-p+\frac{1}{p-1}\right)}\left(\sup_{0<\tau\leq
t}\int_{A_{R}^{2R}}|x|^{q}u_{\varepsilon}dx\right)^{p-\frac{1}{p-1}}.$ (5.5)

As for the second term on the right hand side of (5), we take
$\phi_{1}(x)=1_{\\{|x|<1\\}}+|x|^{q(2-1/(p-1))}1_{A_{1}^{2R}}$ and
$\phi_{2}(x)=1_{\\{|x|<1\\}}+|x|^{-q(2-1/(p-1))}1_{A_{1}^{2R}}$. By Hölder’s
inequality and $(a+b)^{\gamma}\leq a^{\gamma}+b^{\gamma}$ for
$\gamma=2-1/(p-1)$, it follows that

$\displaystyle\int_{B_{2R}}u_{\varepsilon}^{2-\frac{1}{p-1}}dx$
$\displaystyle\leq\left(\int_{B_{2R}}\phi_{2}^{\frac{p-1}{2-p}}dx\right)^{\frac{1}{p-1}-1}\left(\int_{B_{2R}}\phi_{1}^{\frac{p-1}{2p-3}}u_{\varepsilon}dx\right)^{2-\frac{1}{p-1}}$
$\displaystyle\leq
C\left(\int_{B_{1}}u_{\varepsilon}dx\right)^{2-\frac{1}{p-1}}+C\left(\int_{A_{1}^{2R}}|x|^{q}u_{\varepsilon}dx\right)^{2-\frac{1}{p-1}}.$
(5.6)

Plugging (5) and (5) into (5), taking $\varepsilon\rightarrow 0$ and using
monotone convergence theorem as well as identity $1-k-p+1/(p-1)=-p(N)/(p-1)$
yields

$\displaystyle\int_{0}^{t}\int_{B_{2R}}|\nabla u|^{p}u^{-\frac{1}{p-1}}\xi
dxd\tau$ $\displaystyle\leq
CtR^{-\frac{p(N)q}{p-1}}\left(M_{q}(t,A_{r}^{2R})\right)^{p-\frac{1}{p-1}}+C\left(M_{0}(t,B_{1})\right)^{2-\frac{1}{p-1}}$
$\displaystyle+C\left(M_{q}(t,A_{1}^{2R})\right)^{2-\frac{1}{p-1}},$ (5.7)

where $M_{q}(t,A):=\sup_{0<\tau\leq t}\int_{A}|x|^{q}u(x,\tau)dx$. Using
Inequality (3.1) from Theorem 3.1 and (2.8) from 2.2 in the right hand side of
(5), it follows that

$\int_{0}^{t}\int_{B_{R}}|\nabla
u|^{p}u^{-\frac{1}{p-1}}dxd\tau\leq\int_{0}^{t}\int_{B_{2R}}|\nabla
u|^{p}u^{-\frac{1}{p-1}}\xi dxd\tau\\\ \leq
C\left(M_{0}(0,B_{2})\right)^{2-\frac{1}{p-1}}+CtR^{-\frac{p(N)q}{p-1}}\left(M_{q}(0,A_{R/2}^{4R})\right)^{p-\frac{1}{p-1}}\\\
+C\left(M_{q}(0,A_{1/2}^{4R})dx\right)^{2-\frac{1}{p-1}}+Ct^{\frac{2p-3}{(p-1)(2-p)}}\left(1+R^{-\frac{p(N)}{(p-1)(2-p)}}\right),$
(5.8)

where $M_{q}(0,A):=\int_{A}|x|^{q}u_{0}dx$. Since $p(N)>0$, taking
$R\rightarrow\infty$ on both side of (5.8), monotone convergence theorem and
$q$-finite moment of $u_{0}$ yields the result. ∎

###### Proof of Theorem 5.1.

We first show that inequality (5.2) holds in the case where $u(x,\tau)$ is the
positive weak solution of (1.1) with $u_{0}$ in
$C^{\infty}_{c}(\mathbb{R}^{N})$ and $u_{0}\geq 0$ as initial data. Let
$\mu=u_{0}dx$ and $\mu_{t}=u(x,t)dx$. By Benamou-Brenier formulation for
$q$-Wasserstein, see [14, Theorem 5.28] and [2], it follows that

$W^{q}_{q}(\mu,\mu_{t})=\inf\left\\{\int_{0}^{1}\int_{\mathbb{R}^{N}}|v_{\tau}|^{q}\rho_{\tau}dxd\tau\colon\partial_{\tau}\rho_{\tau}+\mathrm{div}(\rho_{\tau}v_{\tau})=0,\rho_{0}=u_{0},\rho_{1}=u(t)\right\\}.$

Rescaling as $\tilde{\rho}(x,\tau)=\rho(x,t^{-1}\tau)$ and
$\tilde{v}(x,\tau)=t^{-1}v(x,t^{-1}\tau)$ and changing variable yields

$W^{q}_{q}(\mu,\mu_{t})=\\\
\inf\left\\{t^{q-1}\int_{0}^{t}\int_{\mathbb{R}^{N}}|\tilde{v}_{\tau}|^{q}\tilde{\rho}_{\tau}dxd\tau\colon\partial_{\tau}\tilde{\rho}_{\tau}+\mathrm{div}(\tilde{\rho}_{\tau}\tilde{v}_{\tau})=0,\tilde{\rho}_{0}=u_{0},\tilde{\rho}_{t}=u(t)\right\\}.$

Note that by the property of singular $p$-Laplace equation, $u(x,\tau)>0$ for
all $\tau\in(0,t]$. So choosing $\tilde{v}:=-|\nabla u|^{p-2}\frac{\nabla
u}{u}$ and $\tilde{\rho}=u$ for $\tau\in(0,t]$, it follows that

$W^{q}_{q}(\mu,\mu_{t})\leq t^{q-1}\int_{0}^{t}\int_{\mathbb{R}^{N}}|\nabla
u|^{p}u^{-\frac{1}{p-1}}dxd\tau.$ (5.9)

By Lemma 5.2, the inequality (5.2) follows.

We are left to show that inequality (5.2) holds in the case where $u$ is the
positive weak solution of the Cauchy problem (1.1) with finite Radon measure
with finite $q$-moment $\mu$ as initial data, constructed in the proof of
Theorem 3.1. Let $(u_{0n})=((m_{n}*\mu)\xi_{n})$ be the sequence of smooth
initial data in $C^{\infty}_{c}(\mathbb{R}^{N})$ defined in the proof of
Theorem 3.1 and $u_{n}$ be the corresponding positive weak solution, and
$(u_{n_{k}})_{k},(u_{0n_{k}})$ be the subsequence of $(u_{n}),(u_{0n})$ such
that $u_{nk}$ and $\nabla u_{nk}$ converges to $u$ and $\nabla u$ uniformly on
all compact subset of $S_{T}$. Let $\varepsilon>0$ be fixed. For
$W_{q}(\mu,\mu_{t})$, by triangle inequality, it follows that

$2^{1-q}W^{q}_{q}(\mu,\mu_{t})\leq
W^{q}_{q}(\mu,\mu^{0n_{k}})+W^{q}_{q}(\mu^{0n_{k}},\mu^{n_{k}}_{t})+W^{q}_{q}(\mu^{n_{k}}_{t}dx,\mu_{t}),$
(5.10)

where $d\mu^{0n}=u_{0n}dx$ and $d\mu^{n}_{t}=u_{n}(x,t)dx$.

As for the first term on the right hand side of (5.10), we claim that
$\mu^{0n}$ converges to $\mu$ weakly and that $\sup_{n}\int_{|x|\geq
R}|x|^{p}u_{0n}dx\rightarrow 0$ as $R\rightarrow\infty$. Indeed, for all
$n\in\mathbb{N}$ and $R>0$, by construction of $u_{n}$ and Fubini’s theorem,
it follows that

$\displaystyle\int_{|x|\geq R}|x|^{q}u_{0n}dx$ $\displaystyle=\int_{|x|\geq
R}|x|^{q}\xi_{n}(x)\int_{y\in B_{1/n}(x)}m_{n}(x-y)d\mu(y)dx$
$\displaystyle=\int_{|y|\geq(R-1)^{+}}\int_{x\in
B_{1/n}(y)}m_{n}(x-y)\xi_{n}(x)|x-y+y|^{q}dxd\mu(y)$ $\displaystyle\leq
2^{q-1}\int_{|y|\geq(R-1)^{+}}|y|^{q}d\mu(y)+2^{q-1}n^{-q}\int_{|y|\geq(R-1)^{+}}d\mu(y).$
(5.11)

Hence, $\sup_{n}\int_{|x|\geq R}|x|^{q}u_{0n}dx\rightarrow 0$ as
$R\rightarrow\infty$. Choose $R>0$ large enough such that
$\sup_{n}\int_{|x|\geq R}|x|^{q}u_{0n}dx<\varepsilon$ and
$\mu(B^{c}_{R})<\varepsilon$ and let $\xi$ in $C_{c}(\mathbb{R}^{N})$ such
that $\xi=1$ on $B_{R}$ and $\xi=0$ on $B^{c}_{2R}$. Since $\mu_{0n}$
converges to $\mu$ vaguely by (3.32), choose $n$ large enough such that

$\left|\int_{\mathbb{R}^{N}}u_{0n}dx-\mu(\mathbb{R}^{N})\right|\leq\left|\int_{\mathbb{R}^{N}}\xi
u_{0n}dx-\int_{\mathbb{R}^{N}}\xi d\mu\right|+\int_{|x|\geq
R}u_{0n}dx+\mu(B^{c}_{R})\leq 3\varepsilon.$ (5.12)

Hence, $\int_{\mathbb{R}^{N}}u_{0n}dx\rightarrow\mu(\mathbb{R}^{N})$ as
$n\rightarrow\infty$. By classical result [13, Theorem 13.16], $u_{0n}dx$
converges to $\mu$ weakly. Together with (5), by [16, Theorem 7.12], it
follows that $W_{q}(\mu^{0n},\mu)\rightarrow 0$ as $n\rightarrow\infty$.

For the third term on the right hand side of (5.10), we claim that
$\mu^{n_{k}}_{t}$ converges to $\mu_{t}$ weakly and that
$\sup_{k}\int_{|x|\geq R}|x|^{p}u_{n_{k}}dx\rightarrow 0$ as
$R\rightarrow\infty$. Indeed, by inequality (3.1) of Theorem 3.1, it follows
that

$\sup_{k}\int_{|x|\geq R}|x|^{q}u_{n_{k}}(x,t)dx\leq C\sup_{k}\int_{|x|\geq
R/2}|x|^{q}u_{0n_{k}}dx+Ct^{\frac{1}{2-p}}R^{-\frac{p(N)}{(p-1)(2-p)}}.$
(5.13)

Applied to (5), it follows that $\sup_{k}\int_{|x|\geq
R}|x|^{q}u_{n_{k}}dx\rightarrow 0$ as $R\rightarrow\infty$. Furthermore, using
inequality (5.13) as well as (3.2), choose $R>0$ large enough such that
$\sup_{k}\int_{|x|\geq R}|x|^{q}u_{n_{k}}(x,t)dx<\varepsilon$ and
$\int_{|x|\geq R}|x|^{q}u(x,t)dx<\varepsilon$. Then since $u_{n_{k}}$
converges to $u$ uniformly on all compact subsets on $S_{T}$, choose
$k\in\mathbb{N}$ large enough, such that

$\left|\int_{\mathbb{R}^{N}}u_{n_{k}}(x,t)dx-\int_{\mathbb{R}^{N}}u(x,t)dx\right|\leq\left|\int_{B_{R}}u_{n_{k}}(x,t)dx-\int_{B_{R}}u(x,t)dx\right|\\\
+\int_{|x|\geq R}|x|^{q}u_{n_{k}}(x,t)dx+\int_{|x|\geq R}|x|^{q}u(x,t)dx\leq
3\varepsilon.$

By the same argument as previously, it follows that
$W_{q}(\mu^{n_{k}}_{t},\mu_{t})\rightarrow 0$ as $k\rightarrow\infty$.

As for the second term on the right hand side of (5.10), by result from the
first step for smooth initial data and inequality (5), it follows that for any
$t\in[0,T]$ it holds

$W^{q}_{q}(\mu^{0n_{k}},\mu^{n_{k}}_{t})\\\ \leq
C(N,p)\left\\{\left(\int_{B_{2}}u_{0n_{k}}dx\right)^{2-\frac{1}{p-1}}+\left(\int_{|x|\geq
1/2}|x|^{q}u_{0n_{k}}dx\right)^{2-\frac{1}{p-1}}+T^{\frac{2p-3}{(p-1)(2-p)}}\right\\}t^{q-1}\\\
\leq
C(N,p)\left\\{\left(\mu(\mathbb{R}^{N})\right)^{2-\frac{1}{p-1}}+\left(\int_{\mathbb{R}^{N}}|x|^{q}d\mu\right)^{2-\frac{1}{p-1}}+T^{\frac{2p-3}{(p-1)(2-p)}}\right\\}t^{q-1}.$
(5.14)

Thus, taking $k\rightarrow\infty$ on both side of (5.10) and together with
inequality (5.14), the result follows. ∎

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