# Harnack inequality for doubly nonlinear mixed local and nonlocal parabolic
equations

Vicenţiu D. Rădulescu, Bin Shang, Chao Zhang∗ Vicen$\c{t}$iu D. Rădulescu
Faculty of Applied Mathematics, AGH University of Kraków, 30-059, Kraków,
Poland & Department of Mathematics, University of Craiova, Street A.I. Cuza
13, 200585 Craiova, Romania<EMAIL_ADDRESS>Bin Shang School of
Mathematics, Harbin Institute of Technology, Harbin 150001, P.R. China &
Department of Mathematics, University of Craiova, Street A.I. Cuza 13, 200585
Craiova, Romania<EMAIL_ADDRESS>Chao Zhang School of Mathematics and
Institute for Advanced Study in Mathematics, Harbin Institute of Technology,
Harbin 150001, P.R. China<EMAIL_ADDRESS>
###### Abstract.

In this paper, we establish the Harnack inequality of nonnegative weak
solutions to the doubly nonlinear mixed local and nonlocal parabolic
equations. This result is obtained by combining a related comparison
principle, a local boundedness estimate, and an integral Harnack-type
inequality. Our proof is based on the expansion of positivity together with a
comparison argument.

###### Key words and phrases:

Expansion of positivity; Harnack inequality; Doubly nonlinear parabolic
equations

###### 2020 Mathematics Subject Classification:

35K67, 35B45, 35B65, 35K65, 35K92

ORCID: 0000-0003-4615-5537 (Vicenţiu D. Rădulescu), 0000-0003-2702-2050 (Chao
Zhang)

∗Corresponding author.

##  1\. Introduction

Let $E_{T}:=E\times(0,T)$ be an open bounded set $E\subset\mathbb{R}^{N}$ and
$T>0$. In this paper, we discuss the Harnack-type estimate for nonnegative
weak solutions to the following doubly nonlinear parabolic equation

$\displaystyle\partial_{t}\left(|u|^{q-1}u\right)-\mathrm{div}(|\nabla
u|^{p-2}\nabla u)+\mathcal{L}u=0\quad\text{in }E_{T},$ (1.1)

where $p>1$, $q>0$ and the operator $\mathcal{L}$ is given by

$\displaystyle\mathcal{L}u(x,t)=\mathrm{P.V.}\int_{\mathbb{R}^{N}}K(x,y,t)|u(x,t)-u(y,t)|^{p-2}(u(x,t)-u(y,t))\,dy.$
(1.2)

Here, P.V. means the Cauchy principal value and
$K(x,y,t):\mathbb{R}^{N}\times\mathbb{R}^{N}\times(0,T]\rightarrow[0,\infty)$
is the symmetric kernel function satisfying

$\displaystyle\frac{\Lambda^{-1}}{|x-y|^{N+sp}}\leq K(x,y,t)\equiv
K(y,x,t)\leq\frac{\Lambda}{|x-y|^{N+sp}}\quad\mathrm{a.e.}\
x,y\in\mathbb{R}^{N}$

for some $\Lambda\geq 1$ and $s\in(0,1)$.

To state the definition of weak solutions, we denote by $W_{0}^{1,p}(E)$ the
Sobolev space with zero boundary values, namely

$\displaystyle W_{0}^{1,p}(E):=\left\\{u\in W^{1,p}(E):u=0\text{ in
}\mathbb{R}^{N}\backslash E\right\\}.$

Moreover, we introduce the tail space

$\displaystyle L_{\alpha}^{m}(\mathbb{R}^{N}):=\left\\{v\in
L_{\rm{loc}}^{m}(\mathbb{R}^{N}):\int_{\mathbb{R}^{N}}\frac{|v(x)|^{m}}{1+|x|^{N+\alpha}}\,dx<+\infty\right\\},\quad
m>0\text{ and }\alpha>0.$

The parabolic nonlocal tail is of the form

$\displaystyle\mathrm{Tail}_{\infty}(v;x_{0},R;t_{0}-S,t_{0}):=\operatorname*{ess\,\sup}_{t_{0}-S<t<t_{0}}\left(R^{p}\int_{\mathbb{R}^{N}\backslash
B_{R}(x_{0})}\frac{|v(x,t)|^{p-1}}{|x-x_{0}|^{N+sp}}\,dx\right)^{\frac{1}{p-1}}.$

Note that $\mathrm{Tail}_{\infty}(v;x_{0},R;t_{0}-S,t_{0})$ is well-defined
for any $v\in L^{\infty}(t_{0}-S,t_{0};L_{sp}^{p-1}(\mathbb{R}^{N}))$.

Eq. (1.1) can be seen as a mixed version of the doubly nonlinear parabolic
equation

$\displaystyle\partial_{t}\left(|u|^{q-1}u\right)-\mathrm{div}(|\nabla
u|^{p-2}\nabla u)=0\quad\text{in }E_{T},$ (1.3)

which attracts lots of interest both in light of its mathematical structure
and its significance to describe many physical phenomena including shallow
water flows [3] and glacier dynamics [24], see also [36, 21] for other
classical applications. It is well-known that Eq. (1.3) covers Trudinger’s
equation $(q=p-1)$, the evolutionary $p$-Laplace equation $(q=1)$, and the
porous medium equation $(p=2)$. There have been fruitful achievements on Eq.
(1.3), such as the existence of solutions [8, 26], Hölder regularity [9, 39],
higher integrability [7, 28], and Harnack type estimate [38, 6, 11, 40]. We
refer the readers to [16, 19, 27, 25] and reference therein for more related
results.

For the doubly nonlinear nonlocal parabolic equation

$\displaystyle\partial_{t}\left(|u|^{q-1}u\right)+(-\Delta)_{p}^{s}u=0\quad\text{in
}E_{T},$ (1.4)

the pointwise behavior of weak solutions was shown in [4] with $p>1$ and
$q>0$. For what concerns the nonlocal Trudinger equation, Banerjee-Garain-
Kinnunen [5] investigated the local boundedness of weak subsolutions by De
Giorgi’s method. Meanwhile, they also provide a crucial algebraic inequality
to deal with the nonlocal term in obtaining a reverse Hölder inequality for
positive weak supersolutions under the assumption that $p>2$. In particular, a
weak Harnack inequality for globally bounded positive weak solutions to the
nonlocal Trudinger equation was derived by Prasad [32]. Taking into account
the fractional $p$-Laplace parabolic equation, the local boundedness $(p>1)$
and the Hölder regularity $(p>2)$ for local weak solutions were explored in
[14]. Moreover, [22] generalized the regularity results to the region
$1<p<\infty$ by means of the intrinsic scaling method, but avoids using any
comparison principle. Further results can be found in [35, 37, 12].

Regarding the homogeneous scenario of (1.1), i.e., $q=p-1$, the local
boundedness of sign-changing weak solutions for $p\geq 2$ was achieved by
Nakamura [30]. Later, the author considered Harnack’s inequality for globally
bounded positive weak solutions through some quantitative estimates in [29].
When $q=1$, Eq. (1.1) becomes the mixed local and nonlocal parabolic
$p$-Laplace equation

$\displaystyle\partial_{t}u-\mathrm{div}(|\nabla u|^{p-2}\nabla
u)+\mathcal{L}u=0\quad\text{in }E_{T},$ (1.5)

some regularity properties of sign-changing weak solutions involving the local
boundedness, the semicontinuity, and the pointwise behavior were discussed in
[15]. By employing the expansion of positivity, Shang-Zhang studied Harnack’s
estimate and the Hölder continuity for weak solutions to (1.5) in [33] and
[34], respectively. Very recently, Adimurthi-Prasad-Tewary [2] developed the
$C^{1,\alpha}$ regularity of weak solutions to (1.5) by suitable comparison
estimates.

To the best of our knowledge, there are no results concerning Eq. (1.1) for
the general case $q\neq p-1$. Motivated by the works [6, 29], our purpose in
this paper is to study the Harnack inequality of globally bounded nonnegative
weak solutions to problem (1.1) under the assumptions that $0<p-1<q<p^{2}-1$
and $p>N$. Different from the homogeneous case, the solutions to Eq. (1.1) in
the fast diffusion range can not be scaled by multiplying any scale factor.
Exactly for this, we adopt the approach called the expansion of positivity and
choose suitable geometries to overcome the non-homogeneity. In addition, we
remark that the assumptions proposed on $p$ and $q$ address another difficulty
stemming from the nonlocal feature. To investigate the Harnack inequality, we
also establish the local boundedness, a comparison principle, and an integral-
type Harnack inequality as byproducts.

We now state the notion of weak solutions to problem (1.1).

######  Definition 1.1.

We identify a function $u$ is a weak subsolution (super-) to the doubly
nonlinear mixed local and nonlocal parabolic equation (1.1) if

$\displaystyle u\in C(0,T;L^{q+1}(E))\cap L^{p}(0,T;W^{1,p}(E))\cap
L^{\infty}(0,T;L_{sp}^{p-1}(\mathbb{R}^{N}))$

such that there holds the integral inequality

$\displaystyle\iint_{E_{T}}-|u|^{q-1}u\partial_{t}\varphi+|\nabla
u|^{p-2}\nabla
u\cdot\nabla\varphi\,dxdt+\int_{0}^{T}\mathcal{E}(u,\varphi,t)\,dt\leq(\geq)0$
(1.6)

for all nonnegative test functions $\varphi\in
W_{0}^{1,q+1}(0,T;L^{q+1}(E))\cap L^{p}(0,T;W_{0}^{1,p}(E))$, where

$\displaystyle\mathcal{E}(u,\varphi,t):=\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}K(x,y,t)|u(x,t)-u(y,t)|^{p-2}(u(x,t)-u(y,t))(\varphi(x,t)-\varphi(y,t))\,dydx.$

A function $u$ is called a weak solution to (1.1) if it is both a weak
subsolution and a weak supersolution.

At this stage, we present our results as follows. The first one is about the
local boundedness of weak subsolutions.

######  Theorem 1.2.

Let $0<p-1<q$ and let $r\geq 1$ such that

$\displaystyle\lambda_{r}:=N(p-q-1)+rp>0.$ (1.7)

Suppose that $u$ is a nonnegative, weak subsolution to (1.1). For
$r>m:=p\frac{N+q+1}{N}$, we further assume that $u$ is qualitatively locally
bounded. Let $Q_{\rho,s}=K_{\rho}(x_{0})\times(t_{0}-s,t_{0}]\subset\subset
E_{T}$. Then there holds

$\displaystyle\underset{Q_{\frac{1}{2}\rho,\frac{1}{2}s}}{\operatorname{ess}\sup}\
u\leq$
$\displaystyle\gamma\left(\frac{\rho^{p}}{s}\right)^{\frac{N}{\lambda_{r}}}\left({\int-}\\!\\!{\int-}_{Q_{\rho,s}}u^{r}\,dxdt\right)^{\frac{p}{\lambda_{r}}}+\gamma\left(\frac{s}{\rho^{p}}\right)^{\frac{1}{q+1-p}}$
$\displaystyle+\gamma\left(\frac{s}{\rho^{p}}[\mathrm{Tail}_{\infty}(u;x_{0},\rho/2;t_{0}-s,t_{0})]^{p-1}\right)^{\frac{1}{q}},$

where $\gamma$ is a constant depending only on $N,p,s,q,\Lambda$.

######  Remark 1.3.

We will discuss the local boundedness result in three cases. The first two
cases care about the exponent $r\leq m$, in which $u\in
L_{\rm{loc}}^{m}(E_{T})$ can be ensured by Sobolev embedding in Lemma 2.7. For
$r>m$, we further assume that the weak subsolutions of (1.1) are locally
bounded because Lemma 2.7 is generally not true. Notice that there exists some
$r\geq 1$ satisfying $r\leq m$ and $\lambda_{r}>0$, if and only if
$\lambda_{m}>0$. Owing to

$\lambda_{m}=(N+p)(m-q-1)=\frac{N+p}{N}\lambda_{q+1},$

$\lambda_{m}>0$ asking for $m>q+1$ or $\lambda_{q+1}>0$, which is equal to

$\displaystyle q<\frac{N(p-1)+p}{(N-p)_{+}}.$

Consequently, we can apply Theorem 1.2 for $r=q+1$ without assuming a prior
that $u$ is locally bounded in this case.

From Remark 1.3, we get a corollary of Theorem 1.2.

######  Corollary 1.4.

Let $0<p-1<q<\frac{N(p-1)+p}{(N-p)_{+}}$. Then we know that every nonnegative
weak subsolution to (1.1) is locally bounded.

Based on the boundedness result, we can derive the Harnack inequality in an
integral-type, which is a key ingredient to obtain the Harnack inequality.

######  Theorem 1.5.

Let $0<p-1<q<\min\left\\{p^{2}-1,\frac{N(p-1)}{(N-p)_{+}}\right\\}$ and
$\lambda_{q}:=N(p-1-q)+qp>0$. Suppose that $u$ is a nonnegative, weak solution
to (1.1), and the cylinder
$Q_{\rho,s}=K_{\rho}(x_{0})\times(t_{0}-s,t_{0}]\subset E_{T}$ with
$\rho\in(0,1]$. Then there holds

$\displaystyle\operatorname*{ess\sup}_{Q_{\frac{1}{2}\rho,\frac{1}{2}s}}u\leq$
$\displaystyle\gamma\left(\frac{\rho^{p}}{s}\right)^{\frac{N}{\lambda
q}}\left(\inf_{t\in\left[t_{0}-s,t_{0}\right]}{\int-}_{K_{\rho}(x_{0})\times\\{t\\}}u^{q}\,dx\right)^{\frac{p}{\lambda_{q}}}+\gamma\left(\frac{s}{\rho^{p}}\right)^{\frac{1}{q+1-p}}$
$\displaystyle+\gamma\left(\frac{s}{\rho^{p}}[\mathrm{Tail}_{\infty}(u;x_{0},\rho/2;t_{0}-s,t_{0})]^{p-1}\right)^{\frac{1}{q}}$
$\displaystyle+\gamma\left(\frac{s}{\rho^{p}}\right)^{\frac{p-N}{\lambda_{q}}}\left[\mathrm{Tail}_{\infty}(u;x_{0},\rho/2;t_{0}-s,t_{0})\right]^{\frac{p(p-1)}{\lambda_{q}}}$

with constant $\gamma$ depending only on $N,p,s,q,\Lambda$.

The last theorem is our main result regarding the Harnack inequality of
nonnegative weak solutions while we additionally assume that the weak
solutions are globally bounded.

######  Theorem 1.6.

Let $p>N$, $0<p-1<q<p^{2}-1$ and let $\rho\in(0,1]$. Suppose that $u\in
L^{\infty}(\mathbb{R}^{N}\times(0,T))$ is a nonnegative, continuous, weak
solution to (1.1), and $u(x_{0},t_{0})>0$. There exist constants $\gamma>1$
and $\sigma\in(0,1)$ depending only on
$N,p,s,q,\Lambda,\|u\|_{L^{\infty}(\mathbb{R}^{N}\times(0,T))}$, such that for
any

$\displaystyle(x,t)\in
K_{\rho}\left(x_{0}\right)\times\left(t_{0}-\sigma\left[u\left(x_{0},t_{0}\right)\right]^{q+1-p}\rho^{p},t_{0}+\sigma\left[u\left(x_{0},t_{0}\right)\right]^{q+1-p}\rho^{p}\right),$

we have

$\displaystyle\gamma^{-1}u\left(x_{0},t_{0}\right)\leq u(x,t)\leq\gamma
u\left(x_{0},t_{0}\right),$

provided

$\displaystyle
K_{8\rho}\left(x_{0}\right)\times\left(t_{0}-\gamma[u(x_{0},t_{0})]^{q+1-p}(8\rho)^{p},t_{0}+\gamma[u(x_{0},t_{0})]^{q+1-p}(8\rho)^{p}\right)\subset
E_{T}.$

Figure 1.

The range where Theorem 1.6 holds is shown in Figure 1. The Harnack inequality
we established above distinguishes from the normal parabolic Harnack
inequality in two aspects. First, we obtain the pointwise information in an
intrinsic cylinder since we perform a specific scaling of the equation.
Second, the effect of time in this Harnack inequality is weakened, so that it
presents a “elliptic” feature.

######  Remark 1.7.

In the statement of Theorem 1.6, we assume that $u$ is a continuous function
for giving a clear sense to $u(x_{0},t_{0})$. In fact, this property can be
proved by using De Giorgi-type Lemma 6.2 in the forthcoming context together
with Theorem 2.1 in [23], and the detailed proof can be found in [4].

This paper is organized as follows. In Section 2, we display some basic
notations and give several preliminary materials. Section 3 is devoted to
deriving a comparison principle. In Section 4, we will provide the
Caccioppoli-type inequality first, and then discuss the local boundedness
result Theorem 1.2. The integral-type Harnack inequality Theorem 1.5 will be
proved in Section 5. In Section 6, we will develop the expansion of positivity
of weak solutions. Finally, we complete the proof of main result (Theorem 1.6)
in Section 7.

##  2\. Preliminaries

###  2.1. Notation

First, we collect some notations used throughout the paper. We shall denote
$K_{\rho}(x_{0})$ is a cube centered at $x_{0}\in\mathbb{R}^{N}$, whose side
length $2\rho>0$, and whose faces are parallel to the coordinate planes in
$\mathbb{R}^{N}$. As is customary, we write the general backward parabolic
cylinders as

$\displaystyle(x_{0},t_{0})+Q_{R,S}:=K_{R}(x_{0})\times\left(t_{0}-S,t_{0}\right].$

We will omit $(x_{0},t_{0})$ if the context is clear or $(x_{0},t_{0})=(0,0)$.

For fixed $k\in\mathbb{R}$, define

$\displaystyle(u-k)_{+}=\max\\{u-k,0\\}\quad\textmd{and}\quad(u-k)_{-}=\max\\{-(u-k),0\\}.$

For a function $u$ defined in $E$ and a real number $l$, we denote

$[u>l]=\\{x\in E:u(x)>l\\}.$

We also use the shorthand notations

$\displaystyle d\mu=d\mu(x,y,t)=K(x,y,t)\,dxdy$

and

$\displaystyle U(x,y,t):=|u(x,t)-u(y,t)|^{p-2}(u(x,t)-u(y,t)).$

We denote by $\gamma$ some generic constants, that may vary from each other
even in the same line.

###  2.2. Technical lemmas

We give an algebraic inequality, see Lemma 2.2 in [1] for $0<\alpha<1$ and
inequality (2.4) in [17] for $\alpha>1$.

######  Lemma 2.1.

For every $\alpha>0$, there is a constant $\gamma$ depending only on $\alpha$
such that

$\displaystyle\frac{1}{\gamma}\left||b|^{\alpha-1}b-|a|^{\alpha-1}a\right|\leq(|a|+|b|)^{\alpha-1}|b-a|\leq\gamma\left||b|^{\alpha-1}b-|a|^{\alpha-1}a\right|$

for all $a,b\in\mathbb{R}$.

To work with the term involving the time derivative, we shall use auxiliary
functions $\bm{h}_{\pm}$ defined as

$\displaystyle\bm{h}_{\pm}(w,k):=\pm q\int_{k}^{w}|s|^{q-1}(s-k)_{\pm}\,ds,$
(2.1)

for $k,w\in\mathbb{R}$ and $q>0$. It is easy to check that
$\bm{h}_{\pm}(w,k)\geq 0$. We also write

$\displaystyle\bm{h}(w,k):=q\int_{k}^{w}|s|^{q-1}(s-k)\,ds.$

The next lemma can be deduced with the help of Lemma 2.1.

######  Lemma 2.2 (Lemma 2.2, [9]).

Let $q>0$. There exists a constant $\gamma=\gamma(q)>0$ such that for all
$a,b\in\mathbb{R}$, there holds

$\displaystyle\frac{1}{\gamma}(|a|+|b|)^{q-1}|a-b|^{2}\leq\bm{h}(a,b)\leq\gamma(|a|+|b|)^{q-1}|a-b|^{2}$

and

$\displaystyle\frac{1}{\gamma}(|a|+|b|)^{q-1}(a-b)_{\pm}^{2}\leq\bm{h}_{\pm}(a,b)\leq\gamma(|a|+|b|)^{q-1}(a-b)_{\pm}^{2}.$

What follows is a necessary tool for dealing with the nonlocal term to discuss
the integral-type Harnack inequality.

######  Lemma 2.3 (Lemma 2.9, [5]).

Let $a,b>0,\tau_{1},\tau_{2}\geq 0$ and $p>1$. Then there exists a constant
$\gamma=\gamma(p)>1$ such that

$\displaystyle|b-a|^{p-2}(b-a)\left(\tau_{1}^{p}a^{-\varepsilon}-\tau_{2}^{p}b^{-\varepsilon}\right)\geq$
$\displaystyle\gamma\xi(\varepsilon)\left|\tau_{2}b^{\frac{\alpha}{p}}-\tau_{1}a^{\frac{\alpha}{p}}\right|^{p}$
$\displaystyle-\left(\xi(\varepsilon)+1+\varepsilon^{-(p-1)}\right)|\tau_{2}-\tau_{1}|^{p}\left(b^{\alpha}+a^{\alpha}\right),$

where $\varepsilon\in(0,p-1)$, $\alpha:=p-1-\varepsilon$,
$\xi(\varepsilon)=\frac{\varepsilon p^{p}}{\alpha}$ if $0<\alpha<1$, and
$\xi(\varepsilon)=\varepsilon(\frac{p}{\alpha})^{p}$ otherwise.

We give a fast geometric convergence lemma from [13, Lemma 4.1].

######  Lemma 2.4.

Let $\\{Y_{j}\\}_{j=0}^{\infty}$ be a sequence of positive numbers satisfying

$\displaystyle Y_{j+1}\leq Kb^{j}Y_{j}^{1+\delta},\quad j=0,1,2,\ldots$

for some constants $K$, $b>1$ and $\delta>0$. If

$\displaystyle Y_{0}\leq K^{-\frac{1}{\delta}}b^{-\frac{1}{\delta^{2}}},$

then we have $Y_{j}\rightarrow 0$ as $j\rightarrow 0$.

The following iteration lemma is displayed in [18, Lemma 6.1].

######  Lemma 2.5.

Let constants $A,B,C\geq 0$, $\alpha>\beta\geq 0$ and $\theta\in(0,1)$.
Suppose that $f:[r,\rho]\rightarrow[0,\infty)$ is a bounded function that
satisfies

$\displaystyle f(R_{1})\leq\theta
f(R_{2})+\frac{A}{(R_{2}-R_{1})^{\alpha}}+\frac{B}{(R_{2}-R_{1})^{\beta}}+C$

for all $r<R_{1}<R_{2}<\rho$. Then we have

$\displaystyle
f(r)\leq\gamma(\alpha,\theta)\left[\frac{A}{(\rho-r)^{\alpha}}+\frac{B}{(\rho-r)^{\beta}}+C\right].$

We now present a Poincaré-type inequality from[13, Chapter I, Proposition
2.1].

######  Lemma 2.6.

Suppose that $\Omega\subset\mathbb{R}^{N}$ is a bounded convex set. Let
$\varphi\in C(\overline{\Omega})$ satisfy $0\leq\varphi\leq 1$, and the sets
$[\varphi>k]$ are convex for any $k\in(0,1)$. Let $v\in W^{1,p}(\Omega)$, and
the set

$\mathcal{E}:=[v=0]\cap[\varphi=1]$

has positive measure. Then we have

$\displaystyle\left(\int_{\Omega}\varphi|v|^{p}\,dx\right)^{\frac{1}{p}}\leq\gamma\frac{(\operatorname{diam}\Omega)^{N}}{|\mathcal{E}|^{\frac{N-1}{N}}}\left(\int_{\Omega}\varphi|\nabla
v|^{p}\,dx\right)^{\frac{1}{p}},$

where $\gamma>0$ depends only on $N$ and $p$, but independent of $v$ and
$\varphi$.

We finally state a parabolic Sobolev embedding lemma.

######  Lemma 2.7 (Chapter I, Proposition 3.1, [13]).

Suppose that $E\subset\mathbb{R}^{N}$ is a bounded domain. Let $m,p>1$ and
$q=p\frac{N+m}{N}$. Then for every

$u\in L^{\infty}(0,T;L^{m}(E))\cap L^{p}(0,T;W_{0}^{1,p}(E)),$

we have

$\displaystyle\iint_{E_{T}}|u|^{q}\,dxdt\leq\gamma\iint_{E_{T}}|\nabla
u|^{p}\,dxdt\left(\operatorname*{ess\sup}_{t\in[0,T]}\int_{E\times\\{t\\}}|u|^{m}\,dx\right)^{\frac{p}{N}},$

where $\gamma>0$ only depends on $N,p$ and $q$.

###  2.3. Time mollification

The exponential mollification in time will be introduced in this subsection to
solve the difficulty that weak solutions do not have a time derivation
generally. This technique is extracted from [20]. For any $v\in L^{1}(E_{T})$
and $h>0$, define

$\displaystyle[\\![v]\\!]_{h}(x,t):=\frac{1}{h}\int_{0}^{t}e^{\frac{s-t}{h}}v(x,s)\,ds$

and

$\displaystyle[\\![v]\\!]_{\bar{h}}(x,t):=\frac{1}{h}\int_{t}^{T}e^{\frac{t-s}{h}}v(x,s)\,ds.$

The following fundamental properties of mollified functions are given in [10,
Appendix].

######  Lemma 2.8.

Let $r\geq 1$. Then the following conclusions hold.

  * (i)

If $v\in L^{r}(E_{T})$, then $[\\![v]\\!]_{h}\in L^{r}(E_{T})$ and
$\|[\\![v]\\!]_{h}\|_{L^{r}(E_{T})}\leq\|v\|_{L^{r}(E_{T})}$. In addition, we
have $[\\![v]\\!]_{h}\rightarrow v$ strongly in $L^{r}(E_{T})$ and almost
everywhere on $E_{T}$ as $h\rightarrow 0$.

  * (ii)

There almost everywhere on $E_{T}$ holds

$\displaystyle\partial_{t}[\\![v]\\!]_{h}=\frac{1}{h}\left(v-[\\![v]\\!]_{h}\right),\quad\partial_{t}[\\![v]\\!]_{\bar{h}}=\frac{1}{h}\left([\\![v]\\!]_{\bar{h}}-v\right).$

  * (iii)

If $v\in L^{r}(E_{T})$, then $[\\![v]\\!]_{h}$ and $[\\![v]\\!]_{\bar{h}}$
belong to $C\left([0,T];L^{r}(E)\right)$.

  * (iv)

If $Du\in L^{r}(E_{T})$, then $D[\\![v]\\!]_{h}=[\\![Dv]\\!]_{h}\rightarrow
Dv$ strongly in $L^{r}(E_{T})$ and almost everywhere on $E_{T}$ as
$h\rightarrow 0$.

  * (v)

If $v\in C([0,T];L^{r}(E))$, then $[\\![v]\\!]_{h}(\cdot,t)\rightarrow
v(\cdot,t)$ strongly in $L^{r}(E)$ and almost everywhere on $E$ for any
$t\in(0,T]$ as $h\rightarrow 0$.

The same statements of $(\rm{{i}}),(\rm{{iv}})$ and $(\rm{{v}})$ also apply to
$[\\![v]\\!]_{\bar{h}}$.

##  3\. Comparison principles

In this part, we will consider the continuity with respect to the time
variable of weak solutions and establish a comparison principle. We introduce
the Cauchy-Dirichlet problem

$\displaystyle\left\\{\begin{array}[]{cl}\partial_{t}(|u|^{q-1}u)-\mathrm{div}(|\nabla
u|^{p-2}\nabla u)+\mathcal{L}u=0&\text{ in }E_{T},\\\ u=0&\text{ in
}\mathbb{R}^{N}\backslash E\times(0,T],\\\ u(\cdot,0)=u_{0}&\text{ in
}E,\end{array}\right.$ (3.4)

where $p>1,q>0$, $\mathcal{L}$ is defined as in (1.2), and $u_{0}\in
L^{q+1}(E)$.

We give the definition of weak solutions to (3.4) as below.

######  Definition 3.1.

We say a function

$\displaystyle u\in C([0,T];L^{q+1})\cap L^{p}(0,T;W_{0}^{1,p}(E))\cap
L^{\infty}(0,T;L_{sp}^{p-1}(\mathbb{R}^{N}))$

is a weak solution to (3.4), if there holds

$\displaystyle\iint_{E_{T}}(|u_{0}|^{q-1}u_{0}-|u|^{q-1}u)\partial_{t}\zeta+|\nabla
u|^{p-2}\nabla
u\cdot\nabla\zeta\,dxdt+\int_{0}^{T}\mathcal{E}(u,\zeta,t)\,dt=0$ (3.5)

for any test function $\zeta\in W^{1,q+1}(0,T;L^{q+1}(E))\cap
L^{p}(0,T;W_{0}^{1,p}(E))$ with $\zeta(\cdot,T)=0$, where

$\displaystyle\mathcal{E}(u,\zeta,t):=\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}K(x,y,t)|u(x,t)-u(y,t)|^{p-2}(u(x,t)-u(y,t))(\zeta(x,t)-\zeta(y,t))\,dydx.$

###  3.1. Parabolicity

In what follows, we show Eq. (1.1) is parabolicity, which describes the
property that if $u$ is a weak subsolution (super-) to (1.1), then the
corresponding truncation function is also a weak subsolution (super-) to
(1.1).

######  Proposition 3.2.

Let $p>1$ and $q>0$. If $u$ is a weak subsolution (super-) to (1.1), then the
truncation function $u_{k}=k+(u-k)_{+}$, $u_{k}=k-(u-k)_{-}$ with
$k\in\mathbb{R}$ is a weak subsolution (super-) to (1.1).

###### Proof.

We verify this result holds for subsolutions, while the claim for
supersolutions can proceed in the same way. For $\sigma>0$, test the weak
formulation (1.6) with function

$\displaystyle\varphi_{h}:=\frac{\zeta([\\![u]\\!]_{\bar{h}}-k)_{+}}{([\\![u]\\!]_{\bar{h}}-k)_{+}+\sigma},$

where $\zeta\in W_{0}^{1,q+1}(0,T;L^{q+1}(E))\cap L^{p}(0,T;W_{0}^{1,p}(E).$
The term including the time derivative, and the local term can be dealt with
the same as in [6, Proposition 4.7], we only concentrate on the nonlocal term.
Letting $h\rightarrow 0$ and using Lemma 2.8 $(\rm{{i}})$, we have

$\displaystyle\lim_{h\rightarrow
0}\int_{0}^{T}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}U(x,y,t)(\varphi_{h}(x,t)-\varphi_{h}(y,t))\,d\mu
dt$
$\displaystyle=\int_{0}^{T}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}U(x,y,t)\left[\frac{\zeta(u-k)_{+}(x,t)}{(u-k)_{+}(x,t)+\sigma}-\frac{\zeta(u-k)_{+}(y,t)}{(u-k)_{+}(y,t)+\sigma}\right]\,d\mu
dt.$

We shall send $\sigma\rightarrow 0$, and get

$\displaystyle-\iint_{E_{T}}|u_{k}|^{q-1}u_{k}\partial_{t}\zeta\,dxdt+\iint_{E_{T}}|\nabla
u|^{p-2}\nabla u\cdot\nabla\zeta\chi_{[u>k]}\,dxdt$ $\displaystyle\ \ \
+\int_{0}^{T}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}U(x,y,t)\left[\zeta(x,t)\chi_{[u(x,t)>k]}-\zeta(y,t)\chi_{[u(y,t)>k]}\right]\,d\mu
dt\leq 0.$

Moreover, one can check that

$\displaystyle\int_{0}^{T}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}|u_{k}(x,t)-u_{k}(y,t)|^{p-2}(u_{k}(x,t)-u_{k}(y,t))(\zeta(x,t)-\zeta(y,t))\,d\mu
dt$ $\displaystyle\leq$
$\displaystyle\int_{0}^{T}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}U(x,y,t)\left[\zeta(x,t)\chi_{[u(x,t)>k]}-\zeta(y,t)\chi_{[u(y,t)>k]}\right]\,d\mu
dt.$

Thus, we conclude that

$\displaystyle-\iint_{E_{T}}|u_{k}|^{q-1}u_{k}\partial_{t}\zeta\,dxdt+\iint_{E_{T}}|\nabla
u_{k}|^{p-2}\nabla u_{k}\cdot\nabla\zeta\,dxdt$ $\displaystyle\ \ \
+\int_{0}^{T}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}|u_{k}(x,t)-u_{k}(y,t)|^{p-2}(u_{k}(x,t)-u_{k}(y,t))(\zeta(x,t)-\zeta(y,t))\,d\mu
dt\leq 0$

for every nonnegative $\zeta\in W^{1,q+1}_{0}(0,T;L^{q+1}(E))\cap
L^{p}(0,T;W_{0}^{1,p}(E))$. ∎

###  3.2. Time continuity of weak solutions

The forthcoming proposition tells the weak solutions of (1.1) belong to
$C\left([0,T];L^{q+1}(E)\right)$. We refer the readers to [30] for details.

######  Proposition 3.3.

Let $p>1$, $q>0$ and let $u\in L^{\infty}(0,T;L^{q+1}(E))\cap
L^{p}(0,T;W_{0}^{1,p}(E))\cap L^{\infty}(0,T;L_{sp}^{p-1}(\mathbb{R}^{N}))$
satisfy the weak formula (3.5) for some $u_{0}\in L^{q+1}(E)$. Then there is a
representative $u\in C\left([0,T];L^{q+1}(E)\right)$, indicating that

$\displaystyle\lim_{t\rightarrow 0}\int_{E}|u(\cdot,t)-u_{0}|^{q+1}\,dx=0.$

###  3.3. Comparison principle

Let us denote the Lipschitz function $H_{\delta}(s)$ by

$\displaystyle H_{\delta}(s):=\left\\{\begin{array}[]{cl}1,&\text{ for
}s\geq\delta,\\\ \frac{1}{\delta}s,&\text{ for }0<s<\delta,\\\ 0,&\text{ for
}s\leq 0.\end{array}\right.$

Define the functions

$\displaystyle
h_{\delta}(z,z_{0}):=\int_{z_{0}}^{z}H_{\delta}(s-z_{0})qs^{q-1}\,ds\ \ \text{
for }z,z_{0}\in\mathbb{R}_{\geq 0},$
$\displaystyle\widehat{h}_{\delta}(z,z_{0}):=\int_{z_{0}}^{z}\widehat{H}_{\delta}(s-z_{0})qs^{q-1}\,ds\
\ \text{ for }z,z_{0}\in\mathbb{R}_{\geq 0},$

where $\widehat{H}_{\delta}(s):=-H_{\delta}(-s)$ is the odd reflection of
$H_{\delta}$. In the rest of this section, we write

$\displaystyle V(x,y,t)=|v(x,t)-v(y,t)|^{p-2}(v(x,t)-v(y,t)),$ $\displaystyle
W(x,y,t)=|w(x,t)-w(y,t)|^{p-2}(w(x,t)-w(y,t)).$

Now, we provide a comparison principle for nonnegative weak solutions to (1.1)
and (3.4). The comparison of two weak solutions in $\mathbb{R}^{N}\backslash
E\times(0,T)$ can be directly checked through the boundary value of (3.4) and
the non-negativity of weak solutions to (1.1). Hence, we just need to give the
assumption on the initial value to develop the comparison principle on
cylinders.

######  Proposition 3.4.

Let $p>1,q>0$, and let $w$ be a nonnegative weak solution to $\eqref{1.1}$ and
$v$ be a nonnegative weak solution to (3.4) with $v_{0}\geq 0$. If $v_{0}\leq
w(\cdot,0)$ a.e. in $E$, then $v\leq w$ a.e. in $E_{T}$.

For proving Proposition 3.4, we discuss the result for the function $v$ first.

######  Lemma 3.5.

Let $p>1$, $q>0$, and let $v$ be a nonnegative weak solution to (3.4) with
$v_{0}\geq 0$. Suppose that $\widetilde{v}\in L^{q+1}(E)\cap W_{0}^{1,p}(E)$
is a nonnegative function. For any $\psi\in
C_{0}^{\infty}(\mathbb{R}^{N}\times(0,T))$, there holds that

$\displaystyle\iint_{E_{T}}-h_{\delta}(v,\widetilde{v})\partial_{t}\psi+|\nabla
v|^{p-2}\nabla v\cdot\nabla\left[H_{\delta}(v-\widetilde{v})\psi\right]\,dxdt$
$\displaystyle+\int_{0}^{T}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}V(x,y,t)[H_{\delta}(v(x,t)-\widetilde{v}(x))\psi(x,t)-H_{\delta}(v(y,t)-\widetilde{v}(y))\psi(y,t)]\,d\mu
dt=0.$

###### Proof.

According to $v=0$ in $\mathbb{R}^{N}\backslash E\times(0,T)$ along with
$\widetilde{v}=0$ in $\mathbb{R}^{N}\backslash E$, we get

$\displaystyle H_{\delta}(v(\cdot,t)-\widetilde{v}(\cdot))=0\quad\text{ in
}\mathbb{R}^{N}\backslash E$

for a.e. $t\in(0,T)$, which implies
$\zeta=H_{\delta}\left([\\![v]\\!]_{h}-\widetilde{v}\right)\psi$ is a
admissible test function in (3.5). Due to $\zeta(\cdot,0)=0$, the term
including $v_{0}$ vanishes. In light of Lemma 2.8 $(\rm{{i}})$ and
$(\rm{{iv}})$, we have

$\displaystyle H_{\delta}([\\![v]\\!]_{h}-\widetilde{v})\psi\rightarrow
H_{\delta}(v-\widetilde{v})\psi\quad\text{as }h\rightarrow 0$

and

$\displaystyle\nabla[H_{\delta}([\\![v]\\!]_{h}-\widetilde{v})\psi]\rightarrow\nabla[H_{\delta}(v-\widetilde{v})\psi]\quad\text{as
}h\rightarrow 0.$

Then we obtain

$\displaystyle\lim_{h\rightarrow 0}\iint_{E_{T}}|\nabla v|^{p-2}\nabla
v\cdot\nabla\zeta\,dxdt=\iint_{E_{T}}|\nabla v|^{p-2}\nabla
v\cdot\nabla[H_{\delta}(v-\widetilde{v})\psi]\,dxdt$

and

$\displaystyle\lim_{h\rightarrow
0}\int_{0}^{T}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}V(x,y,t)(\zeta(x,t)-\zeta(y,t))\,d\mu
dt$ $\displaystyle=$
$\displaystyle\int_{0}^{T}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}V(x,y,t)[H_{\delta}(v(x,t)-\widetilde{v}(x))\psi(x,t)-H_{\delta}(v(y,t)-\widetilde{v}(y))\psi(y,t)]\,d\mu
dt.$

Thanks to Lemma 2.8 $(\rm{{ii}})$, we estimate the time part as

$\displaystyle\iint_{E_{T}}$
$\displaystyle-v^{q}\partial_{t}\zeta\,dxdt=\iint_{E_{T}}(-[\\![v]\\!]_{h}^{q}+[\\![v]\\!]_{h}^{q}-v^{q})\partial_{t}\zeta\,dxdt$
$\displaystyle=$
$\displaystyle-\iint_{E_{T}}[\\![v]\\!]_{h}^{q}\partial_{t}\zeta\,dxdt+\iint_{E_{T}}([\\![v]\\!]_{h}^{q}-v^{q})H_{\delta}([\\![v]\\!]_{h}-\widetilde{v})\partial_{t}\psi\,dxdt$
$\displaystyle+\iint_{E_{T}}([\\![v]\\!]_{h}^{q}-v^{q})H_{\delta}^{\prime}([\\![v]\\!]_{h}-\widetilde{v})\frac{1}{h}(v-[\\![v]\\!]_{h})\psi\,dxdt$
$\displaystyle\leq$
$\displaystyle\iint_{E_{T}}\partial_{t}[\\![v]\\!]_{h}^{q}\zeta\,dxdt+\iint_{E_{T}}\left([\\![v]\\!]_{h}^{q}-v^{q}\right)H_{\delta}([\\![v]\\!]_{h}-\widetilde{v})\partial_{t}\psi\,dxdt$
$\displaystyle=$
$\displaystyle-\iint_{E_{T}}h_{\delta}([\\![v]\\!]_{h},\widetilde{v})\partial_{t}\psi\,dxdt+\iint_{E_{T}}([\\![v]\\!]_{h}^{q}-v^{q})H_{\delta}([\\![v]\\!]_{h}-\widetilde{v})\partial_{t}\psi\,dxdt.$
(3.6)

Observe that the second term on the right-hand side of (3.3) vanishes as
$h\rightarrow 0$ by Lemma 2.8 $(\rm{{i}})$. In addition, we can find

$\displaystyle\left|h_{\delta}([\\![v]\\!]_{h},\widetilde{v})-h_{\delta}(v,\widetilde{v})\right|=\left|\int_{v}^{[\\![v]\\!]_{h}}H_{\delta}(s-\widetilde{v})qs^{q-1}\,ds\right|\leq\left|[\\![v]\\!]_{h}^{q}-v^{q}\right|,$

which combines with Lemma 2.8 $(\rm{{i}})$ leads to

$\displaystyle-\lim_{h\rightarrow
0}\iint_{E_{T}}h_{\delta}([\\![v]\\!]_{h},\widetilde{v})\partial_{t}\psi\,dxdt=-\iint_{E_{T}}h_{\delta}(v,\widetilde{v})\partial_{t}\psi\,dxdt.$

So far, we get the desired result with “$\geq$ ” by collecting the above
estimates. Analogously, we can prove the reverse inequality by choosing
$\zeta=H_{\delta}([\\![v]\\!]_{\bar{h}}-\widetilde{v})\psi$ as a test
function. ∎

Indeed, if taking
$\zeta=\widehat{H}_{\delta}([\\![w]\\!]_{h}-\widetilde{w})\psi$ as a test
function in (3.5) and running similarly, we can obtain the result for the
function $w$.

######  Lemma 3.6.

Let $p>1,q>0$, and let $w$ be a nonnegative weak solution to $\eqref{1.1}$.
Suppose that $\widetilde{w}\in L^{q+1}(E)\cap W_{0}^{1,p}(E)$ is a nonnegative
function. For any $\psi\in
C_{0}^{\infty}\left(\mathbb{R}^{N}\times(0,T)\right)$, there holds that

$\displaystyle\iint_{E_{T}}-\widehat{h}_{\delta}(w,\widetilde{w})\partial_{t}\psi+|\nabla
w|^{p-2}\nabla
w\cdot\nabla\left(\widehat{H}_{\delta}(w-\widetilde{w})\psi\right)\,dxdt$
$\displaystyle+\int_{0}^{T}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}W(x,y,t)\left(\widehat{H}_{\delta}(w(x,t)-\widetilde{w}(x))\psi(x,t)-\widehat{H}_{\delta}(w(y,t)-\widetilde{w}(y))\psi(y,t)\right)\,d\mu
dt=0.$

Now we are ready to prove Proposition 3.4 by using the argument proposed in
[31].

###### Proof of Proposition 3.4.

We set

$\displaystyle(x,t_{1},t_{2})\in\widetilde{Q}:=E\times(0,T)^{2}.$

Let $0\leq\psi\in C_{0}^{\infty}((0,T)^{2})$, and extend functions $v$ and $w$
to $\widetilde{Q}$ with

$\displaystyle v(x,t_{1},t_{2}):=v(x,t_{1}),\quad
w(x,t_{1},t_{2}):=w(x,t_{2}).$

For fixed $\delta>0$ and a.e. $t_{2}\in(0,T)$, we note that $H_{\delta}$,
$\widetilde{v}(x)=w(x,t_{2})=:w_{t_{2}}(x)$,
$\widetilde{v}(y)=w(y,t_{2})=:w_{t_{2}}(y)$, and
$\psi_{t_{2}}(t_{1}):=\psi(t_{1},t_{2})$ for $t_{1}\in(0,T)$ are allowed in
Lemma 3.5 based on the fact $v=0$ in $\mathbb{R}^{N}\backslash E\times(0,T]$.
Thus, it holds that

$\displaystyle\iint_{E_{T}}-h_{\delta}(v,w_{t_{2}})\partial_{t_{1}}\psi_{t_{2}}+|\nabla
v|^{p-2}\nabla
v\cdot\nabla\left[H_{\delta}(v(x,t_{1})-w_{t_{2}})\psi_{t_{2}}\right]\,dxdt_{1}$
$\displaystyle+\int_{0}^{T}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}V(x,y,t_{1})[H_{\delta}(v(x,t_{1})-w_{t_{2}})-H_{\delta}(v(y,t_{1})-w_{t_{2}})]\psi_{t_{2}}\,d\mu
dt_{1}=0.$ (3.7)

Likewise, for fixed $\delta>0$ and a.e. $t_{1}\in(0,T)$, we note that
$\widehat{\mathcal{H}}_{\delta}$, $\widetilde{w}(x)=v(x,t_{1})=:v_{t_{1}}(x)$,
$\widetilde{w}(y)=v(y,t_{1})=:v_{t_{1}}(y)$, and
$\psi_{t_{1}}(t_{2}):=\psi(t_{1},t_{2})$ for $t_{2}\in(0,T)$ are permitted in
Lemma 3.6 due to $v=0$ in $\mathbb{R}^{N}\backslash E\times(0,T]$.
Subsequently, we get

$\displaystyle\iint_{E_{T}}-\widehat{h}_{\delta}(w,v_{t_{1}})\partial_{t_{2}}\psi_{t_{1}}+|\nabla
w|^{p-2}\nabla
w\cdot\nabla\left(\widehat{H}_{\delta}(w(x,t_{2})-v_{t_{1}})\psi_{t_{1}}\right)\,dxdt_{2}$
$\displaystyle+\int_{0}^{T}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}W(x,y,t_{2})\left(\widehat{H}_{\delta}(w(x,t_{2})-v_{t_{1}})-\widehat{H}_{\delta}(w(y,t_{2})-v_{t_{1}})\right)\psi_{t_{1}}\,d\mu
dt_{2}=0.$ (3.8)

Integrating (3.3) over $t_{2}\in(0,T)$, (3.3) over $t_{1}\in(0,T)$, and adding
two integral equalities yields that

$\displaystyle\iiint_{\widetilde{Q}}-\left(h_{\delta}(v,w)\partial_{t_{1}}\psi+\widehat{h}_{\delta}(w,v)\partial_{t_{2}}\psi\right)\,dxdt_{1}dt_{2}$
$\displaystyle+\iiint_{\widetilde{Q}}\left(|\nabla v|^{p-2}\nabla v-|\nabla
w|^{p-2}\nabla
w\right)\cdot\nabla\left[H_{\delta}(v-w)\right]\psi\,dxdt_{1}dt_{2}$
$\displaystyle+\int_{0}^{T}\\!\int_{0}^{T}\\!\int_{\mathbb{R}^{N}}\\!\int_{\mathbb{R}^{N}}\left[V(x,y,t_{1},t_{2})-W(x,y,t_{1},t_{2})\right]$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\times\left[H_{\delta}(v-w)(x,t_{1},t_{2})-H_{\delta}(v-w)(y,t_{1},t_{2})\right]\psi\,d\mu
dt_{1}dt_{2}=0,$ (3.9)

where we employed the property $H_{\delta}(z)=-\widehat{H}_{\delta}(-z)$.
Before sending $\delta\rightarrow 0$, we treat the local and nonlocal terms
that do not possess a regular limit. For the second term in (3.3), recalling
the algebraic inequality

$\displaystyle\left(|\eta|^{p-2}\eta-|\zeta|^{p-2}\zeta\right)(\eta-\zeta)\geq
0$ (3.10)

holds for all $\eta,\zeta\in\mathbb{R}$, thus we have

$\displaystyle(|\nabla v|^{p-2}\nabla v-|\nabla w|^{p-2}\nabla
w)\cdot\nabla\left[H_{\delta}(v-w)\right]$ $\displaystyle=$ $\displaystyle
H_{\delta}^{\prime}(v-w)(\nabla v-\nabla w)(|\nabla v|^{p-2}\nabla v-|\nabla
w|^{p-2}\nabla w)\geq 0.$

For the nonlocal term, it infers from the Mean Value Theorem that there exists
a function $\xi$ which lies between $(v-w)(x,t_{1},t_{2})$ and
$(v-w)(y,t_{1},t_{2})$ such that

$\displaystyle\left(V(x,y,t_{1},t_{2})-W(x,y,t_{1},t_{2})\right)\left[H_{\delta}(v-w)(x,t_{1},t_{2})-H_{\delta}(v-w)(y,t_{1},t_{2})\right]$
$\displaystyle=$
$\displaystyle[|v(x,t_{1},t_{2})-v(y,t_{1},t_{2})|^{p-2}(v(x,t_{1},t_{2})-v(y,t_{1},t_{2}))$
$\displaystyle-|w(x,t_{1},t_{2})-w(y,t_{1},t_{2})|^{p-2}(w(x,t_{1},t_{2})-w(y,t_{1},t_{2}))]$
$\displaystyle\times
H^{\prime}_{\delta}(\xi)[(v(x,t_{1},t_{2})-v(y,t_{1},t_{2}))-(w(x,t_{1},t_{2})-w(y,t_{1},t_{2}))]\geq
0,$

where we used (3.10) again. Dropping the nonnegative terms in (3.3) and
letting $\delta\rightarrow 0$, we derive

$\displaystyle\limsup_{\delta\rightarrow
0}\iiint_{\widetilde{Q}}-\left(h_{\delta}(v,w)\partial_{t_{1}}\psi+\widehat{h}_{\delta}(w,v)\partial_{t_{2}}\psi\right)\,dxdt_{1}dt_{2}\leq
0.$ (3.11)

Once we obtain the integral inequality (3.11), Proposition 3.4 follows from
the argument in [6, Proposition 4.16]. ∎

##  4\. Energy estimate and local boundedness

###  4.1. Energy estimate

In this section, we first present the Caccioppoli-type inequality which can
proceed similarly as in [4].

######  Lemma 4.1.

Let $p>1$ and $q>0$. Let $u$ be a nonnegative weak subsolution to (1.1) and
let $Q_{\rho,s}=K_{\rho}(x_{0})\times(t_{0}-s,t_{0}]\subset\subset E_{T}$. For
any nonnegative, piecewise smooth cutoff function $\psi$ vanishing on
$\partial K_{\rho}(x_{0})\times(t_{0}-s,t_{0})$, there exists a constant
$\gamma(N,p,s,q,\Lambda)>0$ such that

$\displaystyle\operatorname*{ess\sup}_{t_{0}-s<t<t_{0}}\int_{K_{\rho}(x_{0})\times\\{t\\}}\bm{h}_{+}(u,k)\psi^{p}(x,t)\,dx+\iint_{Q_{\rho,s}}|\nabla(u-k)_{+}|^{p}\psi^{p}(x,t)\,dxdt$
$\displaystyle+\int_{t_{0}-s}^{t_{0}}\int_{K_{\rho}(x_{0})}\int_{K_{\rho}(x_{0})}\left|(u-k)_{+}(x,t)\psi(x,t)-(u-k)_{+}(y,t)\psi(y,t)\right|^{p}\,d\mu
dt$ $\displaystyle\leq$
$\displaystyle\gamma\iint_{Q_{\rho,s}}\bm{h}_{+}(u,k)|\partial_{t}\psi|+(u-k)_{+}^{p}|\nabla\psi|^{p}\,dxdt$
$\displaystyle+\gamma\int_{t_{0}-s}^{t_{0}}\int_{K_{\rho}(x_{0})}\int_{K_{\rho}(x_{0})}\max\\{(u-k)_{+}(x,t),(u-k)_{+}(y,t)\\}^{p}|\psi(x,t)-\psi(y,t)|^{p}\,d\mu
dt$ $\displaystyle+\gamma\mathop{\mathrm{ess}\,\sup}_{\stackrel{{\scriptstyle
t_{0}-s<t<t_{0}}}{{x\in\mathrm{supp}\,\psi(\cdot,t)}}}\int_{\mathbb{R}^{N}\backslash
K_{\rho}(x_{0})}\frac{(u-k)_{+}^{p-1}(y,t)}{|x-y|^{N+sp}}\,dy\iint_{Q_{\rho,s}}(u-k)_{+}\psi^{p}(x,t)\,dxdt$
$\displaystyle+\int_{K_{\rho}(x_{0})\times\\{t_{0}-s\\}}\bm{h}_{+}(u,k)\psi^{p}(x,t)\,dx,$
(4.1)

where $k\in\mathbb{R}$ and the function $\bm{h}_{+}$ is defined as in (2.1).

###  4.2. Local boundedness

In the following, we devote to showing the quantitative $L^{\infty}$ bound of
weak solutions to (1.1) presented in Theorem 1.2. To start with, we deduce an
iterative inequality.

####  4.2.1. An iterative inequality for $r\geq q+1$

Let $(x_{0},t_{0})=(0,0)$. For $\sigma\in(0,1)$, define decreasing sequences

$\displaystyle\rho_{0}=\rho,\ \ \rho_{j}=\sigma\rho+2^{-j}(1-\sigma)\rho,\ \
\widetilde{\rho}_{j}=\frac{\rho_{j}+\rho_{j+1}}{2},\ \
\widehat{\rho}_{j}=\frac{3\rho_{j}+\rho_{j+1}}{4},\ \ j=0,1,2\ldots,$
$\displaystyle s_{0}=s,\ \ s_{j}=\sigma s+2^{-j}(1-\sigma)s,\ \
\widetilde{s}_{j}=\frac{s_{j}+s_{j+1}}{2},\ \
\widehat{s}_{j}=\frac{3s_{j}+s_{j+1}}{4},\ \ j=0,1,2\ldots.$

Set the domains

$\displaystyle K_{j}=K_{\rho_{j}},\ \
\widetilde{K}_{j}=K_{\widetilde{\rho}_{j}},\ \
\widehat{K}_{j}=K_{\widehat{\rho}_{j}},\ \ j=0,1,2\ldots,$ $\displaystyle
Q_{j}=K_{j}\times(-s_{j},0],\ \
\widetilde{Q}_{j}=\widetilde{K}_{j}\times(-\widetilde{s}_{j},0],\ \
\widehat{Q}_{j}=\widehat{K}_{j}\times(-\widehat{s}_{j},0],\ \ j=0,1,2\ldots.$

Take increasing sequences

$\displaystyle k_{j}=k-\frac{k}{2^{j}},\ \
\widetilde{k}_{j}=\frac{k_{j}+k_{j+1}}{2},\ \ j=0,1,2\ldots$

with level $k>0$ will be specified later. Consider the cutoff function
$\zeta\in C^{\infty}_{0}(Q_{j})$ vanishing outside $\widehat{Q}_{j}$,
satisfying

$\displaystyle 0\leq\zeta\leq 1,\ \
|\nabla\zeta|\leq\frac{2^{j+2}}{(1-\sigma)\rho},\ \
|\partial_{t}\zeta|\leq\frac{2^{j+2}}{(1-\sigma)s},\ \ \zeta\equiv
1\quad\text{in }\widetilde{Q}_{j}.$

Applying the energy estimate (4.1) in this framework yields that

$\displaystyle\operatorname*{ess\sup}_{-\widetilde{s}_{j}<t<0}\int_{\widetilde{K}_{j}\times\\{t\\}}\bm{h}_{+}(u,\widetilde{k}_{j})\,dx+\iint_{\widetilde{Q}_{j}}|\nabla(u-\widetilde{k}_{j})_{+}|^{p}\,dxdt$
$\displaystyle\leq$ $\displaystyle\frac{\gamma
2^{j}}{(1-\sigma)s}\iint_{Q_{j}}\bm{h}_{+}(u,\widetilde{k}_{j})\,dxdt+\frac{\gamma
2^{pj}}{(1-\sigma)^{p}\rho^{p}}\iint_{Q_{j}}(u-\widetilde{k}_{j})_{+}^{p}\,dxdt$
$\displaystyle+\frac{\gamma}{(1-\sigma)^{p}\rho^{p}}\int_{-s_{j}}^{0}\int_{K_{j}}\int_{K_{j}}\frac{\max\\{(u-\widetilde{k}_{j})_{+}^{p}(x,t),(u-\widetilde{k}_{j})_{+}^{p}(y,t)\\}}{|x-y|^{N-(1-s)p}}\,dxdydt$
$\displaystyle+\gamma\mathop{\mathrm{ess}\,\sup}_{\stackrel{{\scriptstyle-
s_{j}<t<0}}{{x\in\mathrm{supp}\,\zeta(\cdot,t)}}}\int_{\mathbb{R}^{N}\backslash
K_{j}}\frac{(u(y,t)-\widetilde{k}_{j})_{+}^{p-1}}{|x-y|^{N+sp}}\,dy\iint_{Q_{j}}(u(x,t)-\widetilde{k}_{j})_{+}\zeta^{p}(x,t)\,dxdt,$
(4.2)

where we drop the nonnegative term on the left-hand side, and the constant
$\gamma$ depends only on $N,p,s,q,\Lambda$. Next, we estimate the first term
on the right-hand side of (4.2.1). On the set $[u>k_{j+1}]$, we have

$\displaystyle
1\leq\frac{u+\widetilde{k}_{j}}{u-\widetilde{k}_{j}}\leq\frac{2u}{u-\widetilde{k}_{j}}\leq\frac{2k_{j+1}}{k_{j+1}-\widetilde{k}_{j}}\leq
2^{j+3},$

which in conjunction with Lemma 2.2 gives that

$\displaystyle\bm{h}_{+}(u,\widetilde{k}_{j})$
$\displaystyle\geq\frac{1}{\gamma}(u+\widetilde{k}_{j})^{q-1}(u-\widetilde{k}_{j})_{+}^{2}$
$\displaystyle\geq\frac{1}{\gamma}2^{-(j+3)(1-q)_{+}}(u-k_{j+1})_{+}^{q+1},$
(4.3)

where $\gamma>0$ depends only on $q$. Besides, on the set
$[u>\widetilde{k}_{j}]$,

$\displaystyle
1\leq\frac{u+\widetilde{k}_{j}}{u-k_{j}}\leq\frac{2u}{u-k_{j}}\leq\frac{2\widetilde{k}_{j}}{\widetilde{k}_{j}-k_{j}}\leq
2^{j+3}.$

Therefore, we have by Lemma 2.2 that

$\displaystyle\bm{h}_{+}(u,\widetilde{k}_{j})$
$\displaystyle\leq\gamma(u+\widetilde{k}_{j})^{q-1}(u-\widetilde{k}_{j})_{+}^{2}$
$\displaystyle\leq\gamma
2^{(j+3)(q-1)_{+}}(u-k_{j})_{+}^{q+1}\chi_{[u>\widetilde{k}_{j}]}$ (4.4)

with $\gamma$ only depending on $q$. For the last term of (4.2.1), it is easy
to find

$\displaystyle\frac{|y|}{|y-x|}\leq 1+\frac{|x|}{|y-x|}\leq
1+\frac{\widehat{\rho}_{j}}{\rho_{j}-\widehat{\rho}_{j}}\leq\frac{2^{j+4}}{1-\sigma}$

for every $|x|\leq\widehat{\rho}_{j}\text{ and }|y|\geq\rho_{j}$. Thus we
obtain

$\displaystyle\mathop{\mathrm{ess}\,\sup}_{\stackrel{{\scriptstyle-
s_{j}<t<0}}{{x\in\mathrm{supp}\,\zeta(\cdot,t)}}}\int_{\mathbb{R}^{N}\backslash
K_{j}}\frac{(u(y,t)-\widetilde{k}_{j})_{+}^{p-1}}{|x-y|^{N+sp}}\,dy\iint_{Q_{j}}(u(x,t)-\widetilde{k}_{j})_{+}\zeta^{p}(x,t)\,dxdt$
$\displaystyle\leq$ $\displaystyle\frac{\gamma
2^{(N+sp)j}}{(1-\sigma)^{N+sp}}\operatorname*{ess\sup}_{-s_{j}<t<0}\int_{\mathbb{R}^{N}\backslash
K_{\sigma\rho}}\frac{(u(y,t)-k_{0})_{+}^{p-1}}{|y|^{N+sp}}\,dy\iint_{Q_{j}}(u(x,t)-\widetilde{k}_{j})_{+}\,dxdt$
$\displaystyle\leq$ $\displaystyle\frac{\gamma
2^{(N+sp)j}}{(1-\sigma)^{N+sp}(\sigma\rho)^{p}}[\mathrm{Tail}_{\infty}(u;0,\sigma\rho;-s_{j},0)]^{p-1}\iint_{Q_{j}}(u(x,t)-\widetilde{k}_{j})_{+}\,dxdt.$
(4.5)

Substituting (4.2.1)–(4.2.1) into (4.2.1), we arrive at

$\displaystyle
2^{-j(1-q)_{+}}\operatorname*{ess\sup}_{-\widetilde{s}_{j}<t<0}\int_{\widetilde{K}_{j}\times\\{t\\}}(u-k_{j+1})_{+}^{q+1}\,dx+\iint_{\widetilde{Q}_{j}}|\nabla(u-\widetilde{k}_{j})_{+}|^{p}\,dxdt$
$\displaystyle\leq$ $\displaystyle\frac{\gamma
2^{j[1+(q-1)_{+}]}}{(1-\sigma)s}\iint_{\widetilde{A}_{j}}(u-k_{j})_{+}^{q+1}\,dxdt+\frac{\gamma
2^{pj}}{(1-\sigma)^{p}\rho^{p}}\iint_{Q_{j}}(u-\widetilde{k}_{j})_{+}^{p}\,dxdt$
$\displaystyle+\frac{\gamma
2^{(N+sp)j}}{(1-\sigma)^{N+sp}(\sigma\rho)^{p}}[\mathrm{Tail}_{\infty}(u;0,\sigma\rho;-s_{j},0)]^{p-1}\iint_{Q_{j}}(u-\widetilde{k}_{j})_{+}\,dxdt,$
(4.6)

where $\widetilde{A}_{j}:=[u>\widetilde{k}_{j}]\cap Q_{j}$, and $\gamma$
depends only on $N,p,s,q,\Lambda$. Note that

$\displaystyle|\widetilde{A}_{j}|=|[u>\widetilde{k}_{j}]\cap
Q_{j}|\leq\frac{2^{(j+2)r}}{k^{r}}\iint_{Q_{j}}(u-k_{j})_{+}^{r}\,dxdt,$ (4.7)

along with the assumption $r\geq q+1\geq p$ allows us to apply Hölder’s
inequality to get

$\displaystyle\iint_{\widetilde{A}_{j}}(u-k_{j})_{+}^{q+1}\,dxdt$
$\displaystyle\leq\left(\iint_{Q_{j}}(u-k_{j})_{+}^{r}\,dxdt\right)^{\frac{q+1}{r}}|\widetilde{A}_{j}|^{1-\frac{q+1}{r}}$
$\displaystyle\leq\frac{\gamma
2^{(r-q-1)j}}{k^{r-q-1}}\iint_{Q_{j}}(u-k_{j})_{+}^{r}\,dxdt.$ (4.8)

Similarly, we can derive

$\displaystyle\iint_{Q_{j}}(u-\widetilde{k}_{j})_{+}^{p}\,dxdt$
$\displaystyle\leq\frac{\gamma
2^{(r-p)j}}{k^{r-p}}\iint_{Q_{j}}(u-k_{j})_{+}^{r}\,dxdt,$ (4.9)

and

$\displaystyle\iint_{Q_{j}}(u-\widetilde{k}_{j})_{+}\,dxdt$
$\displaystyle\leq\frac{\gamma
2^{(r-1)j}}{k^{r-1}}\iint_{Q_{j}}(u-k_{j})_{+}^{r}\,dxdt.$ (4.10)

Merging estimates (4.2.1), (4.2.1)–(4.10) results to

$\displaystyle
2^{-j(1-q)_{+}}\operatorname*{ess\sup}_{-\widetilde{s}_{j}<t<0}\int_{\widetilde{K}_{j}\times\\{t\\}}(u-k_{j+1})_{+}^{q+1}\,dx+\iint_{\widetilde{Q}_{j}}|\nabla(u-\widetilde{k}_{j})_{+}|^{p}\,dxdt$
$\displaystyle\leq$ $\displaystyle\frac{\gamma
2^{j(N+p+r)}}{(1-\sigma)^{N+p}\sigma^{p}s}\frac{1}{k^{r-q-1}}\left(1+\frac{s}{\rho^{p}k^{q+1-p}}+\frac{s}{\rho^{p}k^{q}}[\text{Tail}_{\infty}(u;0,\sigma\rho;-s,0)]^{p-1}\right)$
$\displaystyle\times\iint_{Q_{j}}(u-k_{j})_{+}^{r}\,dxdt,$

where we utilized the fact $\sigma\in(0,1)$. Now, we choose $k$ to satisfy

$\displaystyle
k\geq\left(\frac{s}{\rho^{p}}\right)^{\frac{1}{q+1-p}}+\left(\frac{s}{\rho^{p}}[\text{Tail}_{\infty}(u;0,\sigma\rho;-s,0)]^{p-1}\right)^{\frac{1}{q}},$
(4.11)

so that

$\displaystyle
2^{-j(1-q)_{+}}\operatorname*{ess\sup}_{-\widetilde{s}_{j}<t<0}\int_{\widetilde{K}_{j}\times\\{t\\}}(u-k_{j+1})_{+}^{q+1}\,dx+\iint_{\widetilde{Q}_{j}}|\nabla(u-\widetilde{k}_{j})_{+}|^{p}\,dxdt$
$\displaystyle\leq$ $\displaystyle\frac{\gamma
2^{j(N+p+r)}}{(1-\sigma)^{N+p}\sigma^{p}s}\frac{1}{k^{r-q-1}}\iint_{Q_{j}}(u-k_{j})_{+}^{r}\,dxdt.$
(4.12)

Since $m=p\frac{N+q+1}{N}$, we deduce from the Sobolev embedding Lemma 2.7
that

$\displaystyle\iint_{Q_{j+1}}$
$\displaystyle(u-k_{j+1})_{+}^{m}\,dxdt\leq\iint_{\widetilde{Q}_{j}}(u-k_{j+1})_{+}^{m}\,dxdt$
$\displaystyle\leq$
$\displaystyle\gamma\iint_{\widetilde{Q}_{j}}|\nabla(u-k_{j+1})_{+}|^{p}\,dxdt\left(\operatorname*{ess\sup}_{-\widetilde{s}_{j}<t<0}\int_{\widetilde{K}_{j}\times\\{t\\}}(u-k_{j+1})_{+}^{q+1}\,dx\right)^{\frac{p}{N}}$
$\displaystyle\leq$ $\displaystyle\frac{\gamma
2^{j\left[N+p+r+(1-q)_{+}\right]\frac{N+p}{N}}}{(1-\sigma)^{\frac{(N+p)^{2}}{N}}\sigma^{\frac{p(N+p)}{N}}k^{(r-q-1)\frac{N+p}{N}}s^{\frac{N+p}{N}}}\left(\iint_{Q_{j}}(u-k_{j})_{+}^{r}\,dxdt\right)^{\frac{N+p}{N}},$

where we used (4.2.1) to get the last line. The above estimate implies

$\displaystyle{\int-}\\!\\!{\int-}_{Q_{j+1}}\left(u-k_{j+1}\right)_{+}^{m}\,dxdt$
$\displaystyle\leq$ $\displaystyle\frac{\gamma
2^{j\left[N+p+r+(1-q)_{+}\right]\frac{N+p}{N}}}{(1-\sigma)^{\frac{(N+p)^{2}}{N}}\sigma^{\frac{p(N+p)}{N}}k^{(r-q-1)\frac{N+p}{N}}}\frac{\rho^{p}}{s}\left({\int-}\\!\\!{\int-}_{Q_{j}}(u-k_{j})_{+}^{r}\,dxdt\right)^{\frac{N+p}{N}},$
(4.13)

where $\gamma$ depends only on $N,p,s,q,\Lambda$.

Note that inequality (4.2.1) holds for exponent $r\geq q+1$. Hereafter, we
distinguish three cases: $q+1\leq r\leq m$; $r\leq m$ and $r<q+1$; $r>m$ to
discuss the local boundedness of weak subsolutions to (1.1).

####  4.2.2. The case $q+1\leq r\leq m$

Denote

$\displaystyle Y_{j}={\int-}\\!\\!{\int-}_{Q_{j}}(u-k_{j})_{+}^{r}\,dxdt.$

By Hölder’s inequality and (4.7), there exists a constant $\gamma$ depending
only on $N,p,s,q,\Lambda$ such that

$\displaystyle Y_{j+1}$
$\displaystyle\leq\frac{1}{|Q_{j+1}|}\left(\iint_{Q_{j+1}}(u-k_{j+1})_{+}^{m}\,dxdt\right)^{\frac{r}{m}}|\widetilde{A}_{j}|^{1-\frac{r}{m}}$
$\displaystyle\leq\gamma\left({\int-}\\!\\!{\int-}_{Q_{j+1}}(u-k_{j+1})_{+}^{m}\,dxdt\right)^{\frac{r}{m}}\left(\frac{2^{(j+2)r}}{k^{r}}{\int-}\\!\\!{\int-}_{Q_{j}}(u-k_{j})_{+}^{r}\,dxdt\right)^{1-\frac{r}{m}}.$

Based on the definition $\lambda_{r}=N(p-q-1)+rp$, it reads from (4.2.1) that

$\displaystyle Y_{j+1}\leq\frac{\gamma
b^{j}}{(1-\sigma)^{\frac{r(N+p)^{2}}{Nm}}\sigma^{\frac{rp(N+p)}{Nm}}}\left(\frac{\rho^{p}}{s}\right)^{\frac{r}{m}}k^{-\frac{r\lambda_{r}}{Nm}}Y_{j}^{1+\frac{rp}{Nm}},$

where $b:=2^{\frac{(N+p)r}{Nm}[N+p+r+(1-q)_{+}]+r}$ and $\gamma$ depends only
on $N,p,s,q,\Lambda$. Moreover, Lemma 2.4 guarantees $Y_{j}\rightarrow 0$ as
$j\rightarrow\infty$ if

$\displaystyle
Y_{0}={\int-}\\!\\!{\int-}_{Q_{0}}u^{r}\,dxdt\leq\gamma^{-1}[\sigma(1-\sigma)]^{\frac{(N+p)^{2}}{p}}\left(\frac{s}{\rho^{p}}\right)^{\frac{N}{p}}k^{\frac{\lambda_{r}}{p}},$

which asks for enforcing

$\displaystyle
k\geq\frac{\gamma}{[(1-\sigma)\sigma]^{\frac{(p+N)^{2}}{\lambda_{r}}}}\left(\frac{\rho^{p}}{s}\right)^{\frac{N}{\lambda_{r}}}\left({\int-}\\!\\!{\int-}_{Q_{0}}u^{r}\,dxdt\right)^{\frac{p}{\lambda_{r}}}.$
(4.14)

Combining the choices of $k$ in (4.11) and (4.14), it follows form
$Y_{j}\rightarrow 0$ that

$\displaystyle\operatorname*{ess\sup}_{Q_{\sigma\rho,\sigma s}}\ u\leq$
$\displaystyle\frac{\gamma}{[(1-\sigma)\sigma]^{\frac{(p+N)^{2}}{\lambda_{r}}}}\left(\frac{\rho^{p}}{s}\right)^{\frac{N}{\lambda_{r}}}\left({\int-}\\!\\!{\int-}_{Q_{\rho,s}}u^{r}\,dxdt\right)^{\frac{p}{\lambda_{r}}}+\left(\frac{s}{\rho^{p}}\right)^{\frac{1}{q+1-p}}$
$\displaystyle+\left(\frac{s}{\rho^{p}}[\text{Tail}_{\infty}(u;0,\sigma\rho;-s,0)]^{p-1}\right)^{\frac{1}{q}},$
(4.15)

where $\gamma$ depends only on $N,p,s,q,\Lambda$. With taking
$\sigma=\frac{1}{2}$, we get the desired boundedness result for $q+1\leq r\leq
m$.

####  4.2.3. The case $r\leq m\text{ and }r<q+1$

Since $r<q+1$, it is not hard to find $0<\lambda_{r}<\lambda_{q+1}$, from
which we can deduce that $q+1<m$. Hence, we can exploit estimate (4.2.2) in
the first case directly by replacing $q+1$ with $r$. For any $\sigma\in(0,1)$,
we have

$\displaystyle\operatorname*{ess\sup}_{Q_{\sigma\rho,\sigma s}}\ u\leq$
$\displaystyle\frac{\gamma}{[(1-\sigma)\sigma]^{\frac{(p+N)^{2}}{\lambda_{q+1}}}}\left(\frac{\rho^{p}}{s}\right)^{\frac{N}{\lambda_{q+1}}}\left({\int-}\\!\\!{\int-}_{Q_{\rho,s}}u^{q+1}\,dxdt\right)^{\frac{p}{\lambda_{q+1}}}+\left(\frac{s}{\rho^{p}}\right)^{\frac{1}{q+1-p}}$
$\displaystyle+\left(\frac{s}{\rho^{p}}[\text{Tail}_{\infty}(u;0,\sigma\rho;-s,0)]^{p-1}\right)^{\frac{1}{q}}.$
(4.16)

Set

$\displaystyle M_{\sigma}=\operatorname*{ess\sup}_{Q_{\sigma\rho,\sigma s}}\
u\quad\text{ and }\quad M_{1}=\operatorname*{ess\sup}_{Q_{\rho,s}}\ u.$

Recalling the definition of $\lambda_{q+1}$, $\lambda_{r}$, we deduce from
(4.2.3) that

$\displaystyle M_{\sigma}\leq$ $\displaystyle
M_{1}^{1-\frac{\lambda_{r}}{\lambda_{q+1}}}\frac{\gamma}{[(1-\sigma)\sigma]^{\frac{(p+N)^{2}}{\lambda_{q+1}}}}\left(\frac{\rho^{p}}{s}\right)^{\frac{N}{\lambda_{q+1}}}\left({\int-}\\!\\!{\int-}_{Q_{\rho,s}}u^{r}\,dxdt\right)^{\frac{p}{\lambda_{q+1}}}+\left(\frac{s}{\rho^{p}}\right)^{\frac{1}{q+1-p}}$
$\displaystyle+\left(\frac{s}{\rho^{p}}[\text{Tail}_{\infty}(u;0,\sigma\rho;-s,0)]^{p-1}\right)^{\frac{1}{q}}.$

By Young’s inequality, we have

$\displaystyle M_{\sigma}\leq$
$\displaystyle\frac{1}{2}M_{1}+\frac{\gamma}{[(1-\sigma)\sigma]^{\frac{(p+N)^{2}}{\lambda_{r}}}}\left(\frac{\rho^{p}}{s}\right)^{\frac{N}{\lambda_{r}}}\left({\int-}\\!\\!{\int-}_{Q_{\rho,s}}u^{r}\,dxdt\right)^{\frac{p}{\lambda_{r}}}+\left(\frac{s}{\rho^{p}}\right)^{\frac{1}{q+1-p}}$
$\displaystyle+\left(\frac{s}{\rho^{p}}[\text{Tail}_{\infty}(u;0,\sigma\rho;-s,0)]^{p-1}\right)^{\frac{1}{q}}.$

Consider above estimate on the cylinders $Q_{\sigma_{2}\rho,\sigma_{2}s}$ and
$Q_{\sigma_{1}\rho,\sigma_{1}s}$ with
$\frac{1}{2}\leq\sigma_{1}\leq\sigma_{2}\leq 1$, it shows that

$\displaystyle M_{\sigma_{1}}\leq$
$\displaystyle\frac{1}{2}M_{\sigma_{2}}+\frac{\gamma}{(\sigma_{2}-\sigma_{1})^{\frac{(p+N)^{2}}{\lambda
r}}}\left(\frac{\rho^{p}}{s}\right)^{\frac{N}{\lambda_{r}}}\left({\int-}\\!\\!{\int-}_{Q_{\sigma_{1}\rho,\sigma_{1}s}}u^{r}\,dxdt\right)^{\frac{p}{\lambda_{r}}}+\left(\frac{s}{\rho^{p}}\right)^{\frac{1}{q+1-p}}$
$\displaystyle+\left(\frac{s}{\rho^{p}}[\text{Tail}_{\infty}(u;0,\rho/2;-s,0)]^{p-1}\right)^{\frac{1}{q}}.$

This enables us to use iteration Lemma 2.5 to arrive at the claim in this
case.

####  4.2.4. The case $r>m$

According to the assumption $\lambda_{r}>0$, we can see $r>q+1$ from $r>m$. If
not, there will hold $0<\lambda_{r}\leq\lambda_{q+1}=N(m-q-1)$, which implies
that $q+1<m$, and this would yield a contradiction. In this case, we assume a
prior that $u\in L_{\rm{loc}}^{\infty}(E_{T})$ due to the embedding Lemma 2.7
generally not true. By iterative inequality (4.2.1) and the definition of
$Y_{j}$, we have

$\displaystyle Y_{j+1}=$
$\displaystyle{\int-}\\!\\!{\int-}_{Q_{j+1}}(u-k_{j+1})_{+}^{r}\,dxdt$
$\displaystyle\leq$
$\displaystyle\|u\|_{\infty,Q_{0}}^{r-m}{\int-}\\!\\!{\int-}_{Q_{j+1}}(u-k_{j+1})_{+}^{m}\,dxdt$
$\displaystyle\leq$
$\displaystyle\gamma\|u\|_{\infty,Q_{0}}^{r-m}\frac{\rho^{p}}{s}\frac{b^{j}}{[(1-\sigma)\sigma]^{\frac{(N+p)^{2}}{N}}k^{(r-q-1)\frac{N+p}{N}}}Y_{j}^{1+\frac{p}{N}},$

where $b=2^{\frac{(N+p)^{2}}{N}+r\frac{N+p}{N}+\frac{N+p}{N}(1-q)_{+}}$.
Utilizing Lemma 2.4, it holds that $Y_{j}\rightarrow 0$ as
$j\rightarrow\infty$ if

$\displaystyle
Y_{0}={\int-}\\!\\!{\int-}_{Q_{0}}u^{r}\,dxdt\leq\gamma^{-1}\|u\|_{\infty,Q_{0}}^{-(r-m)\frac{N}{p}}[(1-\sigma)\sigma]^{\frac{(N+p)^{2}}{p}}\left(\frac{s}{\rho^{p}}\right)^{\frac{N}{p}}k^{\frac{(r-q-1)(N+p)}{p}},$

which requires us to choose $k$ fulling

$\displaystyle
k\geq\frac{\gamma\|u\|_{\infty,Q_{0}}^{\frac{N(r-m)}{(N+p)(r-q-1)}}}{[(1-\sigma)\sigma]^{\frac{N+p}{r-q-1}}}\left(\frac{\rho^{p}}{s}\right)^{\frac{N}{(N+p)(r-q-1)}}\left({\int-}\\!\\!{\int-}_{Q_{0}}u^{r}\,dxdt\right)^{\frac{p}{(N+p)(r-q-1)}}.$
(4.17)

Taking the choices of $k$ in (4.11) and (4.17) into account, we deduce from
$Y_{j}\rightarrow 0$ that

$\displaystyle\operatorname*{ess\sup}_{Q_{\sigma\rho,\sigma s}}\ u\leq$
$\displaystyle\frac{\gamma\|u\|_{\infty,Q_{0}}^{\frac{N(r-m)}{(N+p)(r-q-1)}}}{[(1-\sigma)\sigma]^{\frac{N+p}{r-q-1}}}\left(\frac{\rho^{p}}{s}\right)^{\frac{N}{(N+p)(r-q-1)}}\left({\int-}\\!\\!{\int-}_{Q_{\rho,s}}u^{r}\,dxdt\right)^{\frac{p}{(N+p)(r-q-1)}}$
$\displaystyle+\left(\frac{s}{\rho^{p}}\right)^{\frac{1}{q+1-p}}+\left(\frac{s}{\rho^{p}}[\text{Tail}_{\infty}(u;0,\sigma\rho;-s,0)]^{p-1}\right)^{\frac{1}{q}},$

where $\gamma$ depends only on $N,p,s,q,\Lambda$. We complete the proof by
using the analogous argument as the second case $r\leq m,r<q+1$.

##  5\. Integral-type Harnack inequality

In this section, we will derive Theorem 1.5 which is a direct result from
Theorem 1.2 and the following integral-type Harnack inequality

######  Proposition 5.1.

Let $0<p-1<q<p^{2}-1$, $\rho\in(0,1]$ and
$K_{2\rho}(\bar{y})\times[s,\tau]\subset\subset E_{T}$. Assume that $u$ is a
nonnegative weak solution to (1.1). Then there exists a constant $\gamma>0$
depending only on $N,p,s,q,\Lambda$ such that

$\displaystyle\sup_{t\in[s,\tau]}\int_{K_{\rho}(\bar{y})\times\\{t\\}}u^{q}\,dx\leq$
$\displaystyle\gamma\inf_{t\in[s,\tau]}\int_{K_{2\rho}(\bar{y})\times\\{t\\}}u^{q}\,dx+\gamma\left(\frac{\tau-s}{\rho^{\lambda}}\right)^{\frac{q}{q+1-p}}$
$\displaystyle+\gamma\frac{\tau-s}{\rho^{p-N}}\left[\mathrm{Tail}_{\infty}(u;\bar{y},\rho;s,\tau)\right]^{p-1},$

where

$\displaystyle\lambda:=\frac{\lambda_{q}}{q}=\frac{N}{q}(p-q-1)+p.$

######  Remark 5.2.

If $u$ is only a nonnegative weak supersolution to (1.1), there will hold

$\displaystyle\sup_{t\in[s,\tau]}\int_{K_{\rho}(\bar{y})\times\\{t\\}}u^{q}\,dx\leq$
$\displaystyle\gamma\int_{K_{2\rho}(\bar{y})\times\\{\tau\\}}u^{q}\,dx+\gamma\left(\frac{\tau-s}{\rho^{\lambda}}\right)^{\frac{q}{q+1-p}}$
$\displaystyle+\gamma\frac{\tau-s}{\rho^{p-N}}[\mathrm{Tail}_{\infty}(u;\bar{y},\rho;s,\tau)]^{p-1}.$

Now, we are in a position to show a valuable estimate for proving Proposition
5.1.

######  Lemma 5.3.

Let $0<p-1<q<p^{2}-1$, $\rho\in(0,1]$ and
$K_{\rho}(\bar{y})\times[s,\tau]\subset\subset E_{T}$. Let $u$ be a
nonnegative weak supersolution to (1.1). For any $\sigma\in(0,1)$, there
exists a constant $\gamma>0$ depending only on $N,p,s,q,\Lambda$ such that

$\displaystyle\int_{s}^{\tau}\int_{K_{\sigma\rho}(\bar{y})}(t-s)^{\frac{1}{p}}(u+\kappa)^{-\frac{1+q}{p}}|\nabla
u|^{p}\,dxdt$
$\displaystyle+\\!\int_{s}^{\tau}\\!\int_{K_{\sigma\rho}(\bar{y})}\\!\int_{K_{\sigma\rho}(\bar{y})}\\!(t-s)^{\frac{1}{p}}(|u(x,t)+\kappa|+|u(y,t)+\kappa|)^{-\frac{1+q}{p}}\frac{|u(x,t)-u(y,t)|^{p}}{|x-y|^{N+sp}}\,dxdydt$
$\displaystyle\leq$
$\displaystyle\frac{\gamma\rho}{(1-\sigma)^{N+p}}\left(\frac{\tau-s}{\rho^{\lambda}}\right)^{\frac{1}{p}}\left(\sup_{t\in[s,\tau]}\int_{K_{\rho}(\bar{y})\times\\{t\\}}u^{q}\,dx+\kappa^{q}\rho^{N}\right)^{\frac{(p-1)(q+1)}{pq}},$

where

$\displaystyle\lambda=\frac{N}{q}(p-q-1)+p\quad\textmd{and}\quad\kappa=\left(\frac{\tau-s}{\rho^{p}}\right)^{\frac{1}{q+1-p}}.$
(5.1)

###### Proof.

For simplicity, we may assume $(\bar{y},s)=(0,0)$. Choose

$\displaystyle\varphi_{h}(x,t)=t^{\frac{1}{p}}\left([\\![u]\\!]_{\bar{h}}+\kappa\right)^{-\frac{q+1-p}{p}}\zeta^{p}(x)\psi_{\varepsilon}(t)$

as a test function in (1.6), where $\zeta\in
C_{0}^{1}\left(K_{\frac{\rho(1+\sigma)}{2}};[0,1]\right)$ satisfies $\zeta=1$
in $K_{\sigma\rho}$ and $|\nabla\zeta|\leq\frac{2}{(1-\sigma)\rho}$, for
$\varepsilon>0$, $\psi_{\varepsilon}$ is a Lipschitz function such that
$\psi_{\varepsilon}=1$ in
$(\varepsilon,\tau-\varepsilon),\psi_{\varepsilon}=0$ outside $(0,\tau)$, and
it is linearly interpolated otherwise. Then we yield that

$\displaystyle\iint_{E_{T}}-u^{q}\partial_{t}\varphi_{h}\,dxdt+\iint_{E_{T}}|\nabla
u|^{p-2}\nabla u\cdot\nabla\varphi_{h}\,dxdt$
$\displaystyle+\int_{0}^{T}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}|u(x,t)-u(y,t)|^{p-2}(u(x,t)-u(y,t))(\varphi_{h}(x,t)-\varphi_{h}(y,t))\,d\mu
dt\geq 0.$ (5.2)

The treatment of the first and second terms in (5) is similar to that of [6,
Lemma 7.3], thus we have

$\displaystyle-\iint_{E_{T}}u^{q}\partial_{t}\varphi_{h}\,dxdt\leq$
$\displaystyle
q\tau^{\frac{1}{p}}\int_{K_{\rho}\times\\{\tau\\}}\zeta^{p}(x)\int_{0}^{u}s^{q-1}(s+\kappa)^{-\frac{q+1-p}{p}}\,dsdx$

and

$\displaystyle\iint_{E_{T}}|\nabla u|^{p-2}\nabla
u\cdot\nabla\varphi_{h}\,dxdt\leq$
$\displaystyle-\frac{q+1-p}{2p}\int_{0}^{\tau}\int_{K_{\rho}}|\nabla
u|^{p}(u+\kappa)^{-\frac{q+1}{p}}t^{\frac{1}{p}}\zeta^{p}(x)\,dxdt$
$\displaystyle+\gamma\int_{0}^{\tau}\int_{K_{\rho}}(u+\kappa)^{\frac{p^{2}-1-q}{p}}|\nabla\zeta|^{p}t^{\frac{1}{p}}\,dxdt.$

For the nonlocal term, we pass to the limit $h\rightarrow 0$ first and utilize
Lemma 2.8 $(\rm{{i}})$, and then send $\varepsilon\rightarrow 0$ leads to

$\displaystyle\int_{0}^{T}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}|u(x,t)-u(y,t)|^{p-2}(u(x,t)-u(y,t))(\varphi_{h}(x,t)-\varphi_{h}(y,t))\,d\mu
dt$ $\displaystyle\rightarrow$
$\displaystyle\int_{0}^{\tau}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}U(x,y,t)t^{\frac{1}{p}}\left[(u(x,t)+\kappa)^{-\frac{q+1-p}{p}}\zeta^{p}(x)-(u(y,t)+\kappa)^{-\frac{q+1-p}{p}}\zeta^{p}(y)\right]\,d\mu
dt$ $\displaystyle=$
$\displaystyle\int_{0}^{\tau}\int_{K_{\rho}}\int_{K_{\rho}}U(x,y,t)t^{\frac{1}{p}}\left[(u(x,t)+\kappa)^{-\frac{q+1-p}{p}}\zeta^{p}(x)-(u(y,t)+\kappa)^{-\frac{q+1-p}{p}}\zeta^{p}(y)\right]\,d\mu
dt$
$\displaystyle+2\int_{0}^{\tau}\\!\\!\int_{K_{\rho}}\\!\int_{\mathbb{R}^{N}\backslash
K_{\rho}}\\!U(x,y,t)t^{\frac{1}{p}}\\!\left[(u(x,t)+\kappa)^{-\frac{q+1-p}{p}}\zeta^{p}(x)\\!-\\!(u(y,t)+\kappa)^{-\frac{q+1-p}{p}}\zeta^{p}(y)\right]d\mu
dt.$

Since $p-1<q<p^{2}-1$, it holds that $0<\frac{q+1-p}{p}<p-1$, thus we can
employ Lemma 2.3 with $a=u(y,t)+\kappa$, $b=u(x,t)+\kappa$,
$\tau_{1}=\zeta(y)$, $\tau_{2}=\zeta(x)$, and $\varepsilon=\frac{q+1-p}{p}$ to
deduce

$\displaystyle
U(x,y,t)\left[(u(x,t)+\kappa)^{-\frac{q+1-p}{p}}\zeta^{p}(x)-(u(y,t)+\kappa)^{-\frac{q+1-p}{p}}\zeta^{p}(y)\right]$
$\displaystyle\leq$
$\displaystyle-\gamma(p)\xi(\varepsilon)\left|\zeta(x)(u(x,t)+\kappa)^{\frac{p^{2}-q-1}{p^{2}}}-\zeta(y)(u(y,t)+\kappa)^{\frac{p^{2}-q-1}{p^{2}}}\right|^{p}$
$\displaystyle+\left(\xi(\varepsilon)+1+\varepsilon^{-(p-1)}\right)|\zeta(x)-\zeta(y)|^{p}\left[(u(x,t)+\kappa)^{\frac{p^{2}-q-1}{p}}+(u(y,t)+\kappa)^{\frac{p^{2}-q-1}{p}}\right].$

Subsequently,

$\displaystyle\int_{0}^{\tau}\int_{K_{\rho}}\int_{K_{\rho}}U(x,y,t)t^{\frac{1}{p}}\left[(u(x,t)+\kappa)^{-\frac{q+1-p}{p}}\zeta^{p}(x)-(u(y,t)+\kappa)^{-\frac{q+1-p}{p}}\zeta^{p}(y)\right]\,d\mu
dt$ $\displaystyle\leq$
$\displaystyle-\gamma\int_{0}^{\tau}\int_{K_{\rho}}\int_{K_{\rho}}t^{\frac{1}{p}}\frac{\left|\zeta(x)(u(x,t)+\kappa)^{\frac{p^{2}-q-1}{p^{2}}}-\zeta(y)(u(y,t)+\kappa)^{\frac{p^{2}-q-1}{p^{2}}}\right|^{p}}{|x-y|^{N+sp}}\,dxdydt$
$\displaystyle+\gamma\int_{0}^{\tau}\int_{K_{\rho}}\int_{K_{\rho}}t^{\frac{1}{p}}\left[(u(x,t)+\kappa)^{\frac{p^{2}-q-1}{p}}+(u(y,t)+\kappa)^{\frac{p^{2}-q-1}{p}}\right]\frac{|\zeta(x)-\zeta(y)|^{p}}{|x-y|^{N+sp}}\,dxdydt,$

where $\gamma$ depends only on $N,p,s,q,\Lambda$. By the non-negativity of
weak solutions, we estimate

$\displaystyle\int_{0}^{\tau}\int_{K_{\rho}}\int_{\mathbb{R}^{N}\backslash
K_{\rho}}U(x,y,t)t^{\frac{1}{p}}\left[(u(x,t)+\kappa)^{-\frac{q+1-p}{p}}\zeta^{p}(x)-(u(y,t)+\kappa)^{-\frac{q+1-p}{p}}\zeta^{p}(y)\right]\,d\mu
dt$ $\displaystyle\leq$
$\displaystyle\gamma\int_{0}^{\tau}\int_{K_{\rho}}\int_{\mathbb{R}^{N}\backslash
K_{\rho}\cap[u(x,t)\geq
u(y,t)]}\frac{U(x,y,t)}{|x-y|^{N+sp}}t^{\frac{1}{p}}\zeta^{p}(x)(u(x,t)+\kappa)^{-\frac{q+1-p}{p}}\,dxdydt$
$\displaystyle\leq$
$\displaystyle\gamma\left(\sup_{x\in\rm{supp}\,\zeta(\cdot)}\int_{\mathbb{R}^{N}\backslash
K_{\rho}}\frac{dy}{|x-y|^{N+sp}}\right)\int_{0}^{\tau}\int_{K_{\rho}}u(x,t)^{p-1}t^{\frac{1}{p}}\zeta^{p}(x)(u(x,t)+\kappa)^{-\frac{q+1-p}{p}}\,dxdt.$

Gathering the above estimates, we arrive at

$\displaystyle\int_{0}^{\tau}\int_{K_{\rho}}|\nabla
u|^{p}(u+\kappa)^{-\frac{q+1}{p}}t^{\frac{1}{p}}\zeta^{p}(x)\,dxdt$
$\displaystyle+\int_{0}^{\tau}\int_{K_{\rho}}\int_{K_{\rho}}t^{\frac{1}{p}}\frac{\left|\zeta(x)(u(x,t)+\kappa)^{\frac{p^{2}-q-1}{p^{2}}}-\zeta(y)(u(y,t)+\kappa)^{\frac{p^{2}-q-1}{p^{2}}}\right|^{p}}{|x-y|^{N+sp}}\,dxdydt$
$\displaystyle\leq$
$\displaystyle\gamma\tau^{\frac{1}{p}}\\!\\!\int_{K_{\rho}\times\\{\tau\\}}\zeta^{p}(x)\\!\int_{0}^{u}s^{q-1}(s+\kappa)^{-\frac{q+1-p}{p}}\,dsdx\\!+\\!\frac{\gamma}{(1-\sigma)^{p}\rho^{p}}\int_{0}^{\tau}\\!\\!\int_{K_{\rho}}t^{\frac{1}{p}}(u+\kappa)^{\frac{p^{2}-1-q}{p}}\,dxdt$
$\displaystyle+\int_{0}^{\tau}\int_{K_{\rho}}\int_{K_{\rho}}t^{\frac{1}{p}}\left[(u(x,t)+\kappa)^{\frac{p^{2}-q-1}{p}}+(u(y,t)+\kappa)^{\frac{p^{2}-q-1}{p}}\right]\frac{|\zeta(x)-\zeta(y)|^{p}}{|x-y|^{N+sp}}\,dxdydt$
$\displaystyle+\gamma\left(\sup_{x\in\rm{supp}\,\zeta(\cdot)}\int_{\mathbb{R}^{N}\backslash
K_{\rho}}\frac{dy}{|x-y|^{N+sp}}\right)\int_{0}^{\tau}\int_{K_{\rho}}u(x,t)^{p-1}t^{\frac{1}{p}}\zeta^{p}(x)(u(x,t)+\kappa)^{-\frac{q+1-p}{p}}\,dxdt$
$\displaystyle=:$ $\displaystyle I_{1}+I_{2}+I_{3}+I_{4}.$ (5.3)

Estimate of $I_{1}$: From the hypothesis $0<p-1<q$, one can easily get

$\displaystyle\frac{(p-1)(q+1)}{pq}\in(0,1)\quad\text{ and
}\quad\int_{0}^{u}s^{q-1}(s+\kappa)^{-\frac{q+1-p}{p}}\,ds\leq\gamma
u^{\frac{(p-1)(q+1)}{p}}.$

Thus we apply Hölder’s inequality to obtain

$\displaystyle
I_{1}\leq\gamma\tau^{\frac{1}{p}}\int_{K_{\rho}\times\\{\tau\\}}u^{\frac{(p-1)(q+1)}{p}}\,dx\leq\gamma\rho\left(\frac{\tau}{\rho^{\lambda}}\right)^{\frac{1}{p}}\left(\sup_{t\in[0,\tau]}\int_{K_{\rho}\times\\{t\\}}u^{q}\,dx\right)^{\frac{(p-1)(q+1)}{pq}}.$
(5.4)

Estimate of $I_{2}$: Again using Hölder’s inequality and the definition of
$\kappa$, it follows that

$\displaystyle I_{2}$
$\displaystyle\leq\frac{\gamma\tau^{\frac{1}{p}}}{(1-\sigma)^{p}}\frac{\tau}{\rho^{p}}\kappa^{-(q+1-p)}\sup_{t\in[0,\tau]}\int_{K_{\rho}\times\\{t\\}}(u+\kappa)^{\frac{(p-1)(q+1)}{p}}\,dx$
$\displaystyle\leq\frac{\gamma\rho}{(1-\sigma)^{p}}\left(\frac{\tau}{\rho^{\lambda}}\right)^{\frac{1}{p}}\left(\sup_{t\in[0,\tau]}\int_{K_{\rho}\times\\{t\\}}u^{q}\,dx+\kappa^{q}\rho^{N}\right)^{\frac{(p-1)(q+1)}{pq}}.$
(5.5)

Estimate of $I_{3}$: By exchanging the role of $x$ and $y$ and by exploiting
$\rho\in(0,1]$, we have

$\displaystyle I_{3}$
$\displaystyle\leq\frac{\gamma\rho^{p-sp}}{(1-\sigma)^{p}\rho^{p}}\int_{0}^{\tau}\int_{K_{\rho}}t^{\frac{1}{p}}(u+\kappa)^{\frac{p^{2}-1-q}{p}}\,dxdt$
$\displaystyle\leq\frac{\gamma\rho}{(1-\sigma)^{p}}\left(\frac{\tau}{\rho^{\lambda}}\right)^{\frac{1}{p}}\left(\sup_{t\in[0,\tau]}\int_{K_{\rho}\times\\{t\\}}u^{q}\,dx+\kappa^{q}\rho^{N}\right)^{\frac{(p-1)(q+1)}{pq}}.$
(5.6)

Estimate of $I_{4}$: Notice that

$\displaystyle\frac{|y|}{|y-x|}\leq 1+\frac{|x|}{|y-x|}\leq
1+\frac{1+\sigma}{1-\sigma}\leq\frac{2}{1-\sigma}$

for any $|x|\leq\frac{(1+\sigma)\rho}{2}$ and $|y|\geq\rho$, thus there holds

$\displaystyle I_{4}$ $\displaystyle\leq\frac{\gamma
2^{N+sp}}{(1-\sigma)^{N+sp}\rho^{sp}}\int_{0}^{\tau}\int_{K_{\rho}}t^{\frac{1}{p}}(u(x,t)+\kappa)^{\frac{p^{2}-1-q}{p}}\,dxdt$
$\displaystyle\leq\frac{\gamma
2^{N+sp}\rho}{(1-\sigma)^{N+sp}}\left(\frac{\tau}{\rho^{\lambda}}\right)^{\frac{1}{p}}\left(\sup_{t\in[0,\tau]}\int_{K_{\rho}\times\\{t\\}}u^{q}\,dx+\kappa^{q}\rho^{N}\right)^{\frac{(p-1)(q+1)}{pq}}.$
(5.7)

On the other hand, the left-hand side of (5) can be estimated by utilizing
Lemma 2.1 with $\alpha=\frac{p^{2}-q-1}{p^{2}}>0$ and the property $\zeta=1$
in $K_{\sigma\rho}$, it gives

$\displaystyle\int_{0}^{\tau}\int_{K_{\rho}}\int_{K_{\rho}}t^{\frac{1}{p}}\frac{\left|\zeta(x)(u(x,t)+\kappa)^{\frac{p^{2}-q-1}{p^{2}}}-\zeta(y)(u(y,t)+\kappa)^{\frac{p^{2}-q-1}{p^{2}}}\right|^{p}}{|x-y|^{N+sp}}\,dxdydt$
$\displaystyle\geq$
$\displaystyle\int_{0}^{\tau}\int_{K_{\sigma\rho}}\int_{K_{\sigma\rho}}t^{\frac{1}{p}}\frac{\left|(u(x,t)+\kappa)^{\frac{p^{2}-q-1}{p^{2}}}-(u(y,t)+\kappa)^{\frac{p^{2}-q-1}{p^{2}}}\right|^{p}}{|x-y|^{N+sp}}\,dxdydt$
$\displaystyle\geq$
$\displaystyle\int_{0}^{\tau}\int_{K_{\sigma\rho}}\int_{K_{\sigma\rho}}t^{\frac{1}{p}}\left(|u(x,t)+\kappa|+|u(y,t)+\kappa|\right)^{-\frac{q+1}{p}}\frac{|u(x,t)-u(y,t)|^{p}}{|x-y|^{N+sp}}\,dxdydt.$
(5.8)

Combining estimates (5)–(5), we get the conclusion. ∎

Given Lemma 5.3, we can explore the following result.

######  Lemma 5.4.

Let $0<p-1<q<p^{2}-1$, $\rho\in(0,1]$ and
$K_{\rho}(\bar{y})\times[s,\tau]\subset\subset E_{T}$. Assume that $u$ is a
nonnegative weak supersolution to (1.1). For all $\delta,\sigma\in(0,1)$,
there exists a constant $\gamma>0$ depending only on $N,p,s,q,\Lambda$ such
that

$\displaystyle\frac{1}{\rho}\int_{s}^{\tau}\int_{K_{\sigma\rho}(\bar{y})}|\nabla
u|^{p-1}\,dxdt+\frac{1}{\rho}\int_{s}^{\tau}\int_{K_{\sigma\rho}(\bar{y})}\int_{K_{\sigma\rho}(\bar{y})}\frac{|u(x,t)-u(y,t)|^{p-1}}{|x-y|^{N+sp-1}}\,dxdydt$
$\displaystyle\quad\leq\delta\sup_{t\in[s,\tau]}\int_{K_{\rho}(\bar{y})\times\\{t\\}}u^{q}\,dx+\frac{\gamma}{\left[\delta^{q+1}(1-\sigma)^{pq}\right]^{\frac{N+p}{q+1-p}}}\left(\frac{\tau-s}{\rho^{\lambda}}\right)^{\frac{q}{q+1-p}},$

where $\lambda$ and $\kappa$ defined in (5.1).

###### Proof.

Let $(\bar{y},s)=(0,0)$. We derive from Hölder’s inequality that

$\displaystyle\int_{0}^{\tau}\int_{K_{\sigma\rho}}|\nabla u|^{p-1}\,dxdt\leq$
$\displaystyle\left(\int_{0}^{\tau}\int_{K_{\sigma\rho}}|\nabla
u|^{p}(u+\kappa)^{-\frac{q+1}{p}}t^{\frac{1}{p}}\,dxdt\right)^{\frac{p-1}{p}}$
$\displaystyle\times\left(\int_{0}^{\tau}\int_{K_{\sigma\rho}}(u+\kappa)^{\frac{(p-1)(q+1)}{p}}t^{\frac{1-p}{p}}\,dxdt\right)^{\frac{1}{p}}.$
(5.9)

The estimate of the first integral on the right-hand side of (5) yields from
Lemma 5.3. Coming to estimate the second integral, there holds by using
Hölder’s inequality that

$\displaystyle\int_{0}^{\tau}\int_{K_{\sigma\rho}}(u+\kappa)^{\frac{(p-1)(q+1)}{p}}t^{\frac{1-p}{p}}\,dxdt$
$\displaystyle\leq\int_{0}^{\tau}t^{\frac{1-p}{p}}\,dt\times\sup_{t\in[0,\tau]}\int_{K_{\sigma\rho}\times\\{t\\}}(u+\kappa)^{\frac{(p-1)(q+1)}{p}}\,dx$
$\displaystyle\leq\gamma\rho\left(\frac{\tau}{\rho^{\lambda}}\right)^{\frac{1}{p}}\left(\sup_{t\in[0,\tau]}\int_{K_{\sigma\rho}\times\\{t\\}}u^{q}\,dx+\kappa^{q}\rho^{N}\right)^{\frac{(p-1)(q+1)}{pq}}.$
(5.10)

Still by Hölder’s inequality, we compute

$\displaystyle\int_{0}^{\tau}\int_{K_{\sigma\rho}}\int_{K_{\sigma\rho}}\frac{|u(x,t)-u(y,t)|^{p-1}}{|x-y|^{N+sp-1}}\,dxdydt$
$\displaystyle\leq$
$\displaystyle\left(\int_{0}^{\tau}\int_{K_{\sigma\rho}}\int_{K_{\sigma\rho}}\frac{|u(x,t)-u(y,t)|^{p}}{|x-y|^{N+sp}}(u(x,t)+\kappa)^{-\frac{q+1}{p}}t^{\frac{1}{p}}\,dxdydt\right)^{\frac{p-1}{p}}$
$\displaystyle\times\left(\int_{0}^{\tau}\int_{K_{\sigma\rho}}\int_{K_{\sigma\rho}}\frac{(u(x,t)+\kappa)^{\frac{(p-1)(q+1)}{p}}}{|x-y|^{N+sp-p}}t^{\frac{1-p}{p}}\,dxdydt\right)^{\frac{1}{p}}.$
(5.11)

The first integral on the right-hand side of (5) has been provided in Lemma
5.3. We deal with the second integral as in (5) and obtain

$\displaystyle\int_{0}^{\tau}\int_{K_{\sigma\rho}}\int_{K_{\sigma\rho}}\frac{(u(x,t)+\kappa)^{\frac{(p-1)(q+1)}{p}}}{|x-y|^{N+sp-p}}t^{\frac{1-p}{p}}\,dxdydt$
$\displaystyle\leq$
$\displaystyle\gamma\rho\left(\frac{\tau}{\rho^{\lambda}}\right)^{\frac{1}{p}}\left(\sup_{t\in[0,\tau]}\int_{K_{\sigma\rho}\times\\{t\\}}u^{q}\,dx+\kappa^{q}\rho^{N}\right)^{\frac{(p-1)(q+1)}{pq}},$

where we used the assumption $\sigma\in(0,1)$, and the radius $\rho\in(0,1]$.
Putting the above estimates together, employing Young’s inequality as well as
the definition of $\kappa$ in (5.1), we infer that

$\displaystyle\int_{0}^{\tau}\int_{K_{\sigma\rho}}|\nabla
u|^{p-1}\,dxdt+\int_{0}^{\tau}\int_{K_{\sigma\rho}}\int_{K_{\sigma\rho}}\frac{|u(x,t)-u(y,t)|^{p-1}}{|x-y|^{N+sp-1}}\,dxdydt$
$\displaystyle\leq\frac{\gamma\rho}{(1-\sigma)^{N+p}}\left(\frac{\tau}{\rho^{\lambda}}\right)^{\frac{1}{p}}\left(\sup_{t\in[0,\tau]}\int_{K_{\rho}\times\\{t\\}}u^{q}\,dx+\kappa^{q}\rho^{N}\right)^{\frac{(p-1)(q+1)}{pq}}$
$\displaystyle\leq\delta\rho\sup_{t\in[0,\tau]}\int_{K_{\rho}\times\\{t\\}}u^{q}\,dx+\frac{\gamma\rho}{\left[\delta^{q+1}(1-\sigma)^{pq}\right]^{\frac{N+p}{q+1-p}}}\left(\frac{\tau}{\rho^{\lambda}}\right)^{\frac{q}{q+1-p}},$

where the constant $\delta\in(0,1)$ depends on $p,q$. Diving both sides by
$\rho$, we arrive at the claim. ∎

Hereafter, we give the proof of Proposition 5.1.

###### Proof of Proposition 5.1.

Assume that $(\bar{y},s)=(0,0)$. Denote

$\displaystyle\rho_{j}=\sum_{n=0}^{j}\frac{\rho}{2^{n}},\quad\widetilde{\rho}_{j}=\frac{\rho_{j}+\rho_{j+1}}{2},\quad\widehat{\rho}_{j}=\frac{3\rho_{j}+\rho_{j+1}}{4},\
\ j=0,1,2\ldots$

and

$\displaystyle\quad
K_{j}=K_{\rho_{j}},\quad\widetilde{K}_{j}=K_{\widetilde{\rho}_{j}},\quad\widehat{K}_{j}=K_{\widehat{\rho}_{j}},\
\ j=0,1,2\ldots.$

Consider the function $\zeta\in C_{0}^{1}(\widetilde{K}_{j};[0,1])$ that
vanishes outside $\widehat{K}_{j}$, equals to $1$ in $K_{j}$ such that
$|\nabla\zeta|\leq 2^{j+3}/\rho$. Testing (1.6) with $\zeta$, there holds for
any $t_{1},t_{2}\in[0,\tau]$ that

$\displaystyle\int_{\widetilde{K}_{j}\times\\{t_{1}\\}}u^{q}\zeta\,dx\leq$
$\displaystyle\int_{\widetilde{K}_{j}\times\\{t_{2}\\}}u^{q}\zeta\,dx+\frac{2^{j+3}}{\rho}\int_{t_{1}}^{t_{2}}\int_{\widetilde{K}_{j}}|\nabla
u|^{p-1}\,dxdt$
$\displaystyle+\frac{2^{j+3}}{\rho}\int_{t_{1}}^{t_{2}}\int_{\widetilde{K}_{j}}\int_{\widetilde{K}_{j}}\frac{|u(x,t)-u(y,t)|^{p-1}}{|x-y|^{N+sp-1}}\,dxdydt$
$\displaystyle+2\int_{t_{1}}^{t_{2}}\int_{\widetilde{K}_{j}}\int_{\mathbb{R}^{N}\backslash\widetilde{K}_{j}}\frac{|u(x,t)-u(y,t)|^{p-1}}{|x-y|^{N+sp}}\zeta(x)\,dxdydt.$
(5.12)

Through direct computation,

$\displaystyle\frac{|y|}{|y-x|}\leq 1+\frac{|x|}{|y-x|}\leq
1+\frac{\widehat{\rho}_{j}}{\widetilde{\rho}_{j}-\widehat{\rho}_{j}}\leq
2^{j+3}$

for any $|x|\leq\widehat{\rho}_{j}$ and $|y|\geq\widetilde{\rho}_{j}$, thus
the last term in (5) can be estimated as

$\displaystyle\int_{t_{1}}^{t_{2}}\int_{\widetilde{K}_{j}}\int_{\mathbb{R}^{N}\backslash\widetilde{K}_{j}}\frac{|u(x,t)-u(y,t)|^{p-1}}{|x-y|^{N+sp}}\zeta(x)\,dxdydt$
$\displaystyle\leq$ $\displaystyle\frac{\tau
2^{(j+3)(N+sp)}}{\rho^{p-N}}[\text{Tail}_{\infty}(u;0,\rho;0,\tau)]^{p-1}.$
(5.13)

For $t_{1}>t_{2}$, we may test (1.6) by $-\zeta$ instead of $\zeta$. To
proceed, we set

$\displaystyle\int_{K_{2\rho}\times\\{t_{2}\\}}u^{q}\,dx=\inf_{t\in[0,\tau]}\int_{K_{2\rho}\times\\{t\\}}u^{q}\,dx=:A.$

Define

$\displaystyle S_{j}:=\sup_{t\in[0,\tau]}\int_{K_{j}\times\\{t\\}}u^{q}\,dx.$

Since $t_{1}\in[0,\tau]$ is arbitrary, it tells from (5) and (5) that

$\displaystyle S_{j}\leq$ $\displaystyle
A+\frac{2^{j+3}}{\rho}\int_{0}^{\tau}\int_{\widetilde{K}_{j}}|\nabla
u|^{p-1}\,dxdt+\frac{2^{j+3}}{\rho}\int_{0}^{\tau}\int_{\widetilde{K}_{j}}\int_{\widetilde{K}_{j}}\frac{|u(x,t)-u(y,t)|^{p-1}}{|x-y|^{N+sp-1}}\,dxdydt$
$\displaystyle+\frac{\tau
2^{(j+4)(N+sp)}}{\rho^{p-N}}[\text{Tail}_{\infty}(u;0,\rho;0,\tau)]^{p-1}.$
(5.14)

From the definition of $\rho_{j}$, we can get $\rho\leq\rho_{j}<2\rho$.
Selecting $\sigma$ small enough such that $1-\sigma\geq 2^{-(j+4)}$, and we
deduce from Lemma 5.4 that

$\displaystyle\frac{1}{2\rho}\int_{0}^{\tau}\int_{\widetilde{K}_{j}}|\nabla
u|^{p-1}\,dxdt+\frac{1}{2\rho}\int_{0}^{\tau}\int_{\widetilde{K}_{j}}\int_{\widetilde{K}_{j}}\frac{|u(x,t)-u(y,t)|^{p-1}}{|x-y|^{N+sp-1}}\,dxdydt$
$\displaystyle\leq$
$\displaystyle\frac{1}{\rho_{j+1}}\int_{0}^{\tau}\int_{\widetilde{K}_{j}}|\nabla
u|^{p-1}\,dxdt+\frac{1}{\rho_{j+1}}\int_{0}^{\tau}\int_{\widetilde{K}_{j}}\int_{\widetilde{K}_{j}}\frac{|u(x,t)-u(y,t)|^{p-1}}{|x-y|^{N+sp-1}}\,dxdydt$
$\displaystyle\leq$
$\displaystyle\delta\sup_{t\in[0,\tau]}\int_{K_{j+1}\times\\{t\\}}u^{q}\,dx+\frac{\gamma}{\left[\delta^{q+1}(1-\sigma)^{pq}\right]^{\frac{N+p}{q+1-p}}}\left(\frac{\tau}{\rho_{j+1}^{\lambda}}\right)^{\frac{q}{q+1-p}}$
$\displaystyle\leq$ $\displaystyle\delta S_{j+1}+\frac{\gamma
2^{j\frac{pq(N+p)}{q+1-p}}}{\delta^{\frac{(q+1)(N+p)}{q+1-p}}}\left(\frac{\tau}{\rho^{\lambda}}\right)^{\frac{q}{q+1-p}}.$

Multiplying both sides of the above inequality by $2^{j+4}$ leads to

$\displaystyle\frac{2^{j+3}}{\rho}\int_{0}^{\tau}\int_{\widetilde{K}_{j}}|\nabla
u|^{p-1}\,dxdt+\frac{2^{j+3}}{\rho}\int_{0}^{\tau}\int_{\widetilde{K}_{j}}\int_{\widetilde{K}_{j}}\frac{|u(x,t)-u(y,t)|^{p-1}}{|x-y|^{N+sp-1}}\,dxdydt$
$\displaystyle\leq$ $\displaystyle\delta 2^{j+4}S_{j+1}+\frac{\gamma
2^{j\left[1+\frac{pq(N+p)}{q+1-p}\right]}}{\delta^{\frac{(q+1)(N+p)}{q+1-p}}}\left(\frac{\tau}{\rho^{\lambda}}\right)^{\frac{q}{q+1-p}}.$

For some $\varepsilon\in(0,1)$, letting $\delta=\varepsilon/2^{j+4}$ to get

$\displaystyle\frac{2^{j+3}}{\rho}\int_{0}^{\tau}\int_{\widetilde{K}_{j}}|\nabla
u|^{p-1}\,dxdt+\frac{2^{j+3}}{\rho}\int_{0}^{\tau}\int_{\widetilde{K}_{j}}\int_{\widetilde{K}_{j}}\frac{|u(x,t)-u(y,t)|^{p-1}}{|x-y|^{N+sp-1}}\,dxdydt$
$\displaystyle\leq$ $\displaystyle\varepsilon
S_{j+1}+\gamma(\varepsilon)b^{j}\left(\frac{\tau}{\rho^{\lambda}}\right)^{\frac{q}{q+1-p}}$
(5.15)

with $b=b(p,q,N)>1$. By virtue of (5) and (5), we have for any
$j\in\mathbb{N}\cup\\{0\\}$ that

$\displaystyle S_{j}\leq\varepsilon
S_{j+1}+\gamma(\varepsilon)b^{j}\left(A+\left(\frac{\tau}{\rho^{\lambda}}\right)^{\frac{q}{q+1-p}}+\tau\rho^{N-p}[\text{Tail}_{\infty}(u;0,\rho;0,\tau)]^{p-1}\right).$
(5.16)

Iterating inequality (5.16) gives that

$\displaystyle
S_{0}\leq\varepsilon^{j}S_{j}+\gamma(\varepsilon)\left(A+\left(\frac{\tau}{\rho^{\lambda}}\right)^{\frac{q}{q+1-p}}+\tau\rho^{N-p}[\text{Tail}_{\infty}(u;0,\rho;0,\tau)]^{p-1}\right)\sum_{i=0}^{j-1}(\varepsilon
b)^{i}.$

Then, by letting $\varepsilon=1/2b$, we get $\sum_{i=0}^{j-1}(\varepsilon
b)^{i}\leq 2$, and meanwhile $\varepsilon^{j}S_{j}\rightarrow 0$ as
$j\rightarrow\infty$. The proof is completed by recalling the definitions of
$S_{j}$ and $A$. ∎

Finally, Theorem 1.5 follows from a combination of Theorem 1.2 and Proposition
5.1.

##  6\. Expansion of positivity

This section aims to obtain the following result regarding the expansion of
positivity, which derives the pointwise estimate for weak supersolutions from
the measure theoretical condition.

######  Proposition 6.1.

Let $0<p-1\leq q$. Assume that $u\in L^{\infty}(\mathbb{R}^{N}\times(0,T))$ is
a nonnegative, weak supersolution to (1.1). If for some constants $M>0$,
$\alpha\in(0,1)$ we have

$\displaystyle|[u(\cdot,t_{0})\geq M]\cap
K_{\rho}(x_{0})|\geq\alpha|K_{\rho}|,$ (6.1)

then there exist parameters $\delta,\eta\in(0,1)$ depending only on
$N,p,s,q,\Lambda,\alpha$ such that

$\displaystyle u\geq\eta M\quad\text{ a.e. in
}K_{2\rho}(x_{0})\times\left(t_{0}+\tfrac{1}{2}\delta
M^{q+1-p}\rho^{p},t_{0}+\delta M^{q+1-p}\rho^{p}\right],$

provided

$\displaystyle K_{8\rho}(x_{0})\times(t_{0},t_{0}+\delta
M^{q+1-p}\rho^{p}]\subset E_{T}.$

Before proving Proposition 6.1, we give several preparatory estimates. The
first one is a De Giorgi-type lemma.

######  Lemma 6.2.

Let $q>0$, $p>1$, and let $u\in L^{\infty}(\mathbb{R}^{N}\times(0,T))$ be a
nonnegative, weak supersolution to (1.1). For some constants $M>0$ and
$\delta\in(0,1)$, assume that $\rho\in(0,1]$ and
$(x_{0},t_{0})+Q_{\rho}(\theta)\subset E_{T}$, where $\theta=\delta
M^{q+1-p}$. If there is a constant $\nu\in(0,1)$ only depending on
$N,p,s,q,\Lambda$ and $\delta$ such that

$\displaystyle\left|[u\leq
M]\cap(x_{0},t_{0})+Q_{\rho}(\theta)\right|\leq\nu|Q_{\rho}(\theta)|,$

then we have

$\displaystyle u\geq\frac{1}{2}M\quad\text{a.e. in
}(x_{0},t_{0})+Q_{\frac{1}{2}\rho}(\theta).$

###### Proof.

Let $(x_{0},t_{0})=(0,0)$. Take decreasing sequences

$\displaystyle k_{j}=\frac{M}{2}+\frac{M}{2^{j+1}},\ \ j=0,1,2\ldots.$

Denote

$\displaystyle\rho_{j}=\frac{\rho}{2}+\frac{\rho}{2^{j+1}},\quad\widetilde{\rho}_{j}=\frac{\rho_{j}+\rho_{j+1}}{2},\quad\widehat{\rho}_{j}=\frac{3\rho_{j}+\rho_{j+1}}{4},\
\ j=0,1,2\ldots.$

Set the domains

$\displaystyle
K_{j}=K_{\rho_{j}},\quad\widetilde{K}_{j}=K_{\widetilde{\rho}_{j}},\quad\widehat{K}_{j}=K_{\widehat{\rho}_{j}},\
\ j=0,1,2\ldots,$

and

$\displaystyle
Q_{j}=K_{j}\times\left(-\theta\rho_{j}^{p},0\right],\quad\widetilde{Q}_{j}=\widetilde{K}_{j}\times\left(-\theta\widetilde{\rho}_{j}^{p},0\right],\quad\widehat{Q}_{j}=\widehat{K}_{j}\times\left(-\theta\widehat{\rho}_{j}^{p},0\right],\
\ j=0,1,2\ldots.$

Consider the cutoff function $0\leq\zeta\leq 1$ in $Q_{j}$ vanishing outside
$\widehat{Q}_{j}$, and equals to $1$ in $\widetilde{Q}_{j}$, such that

$\displaystyle|\nabla\zeta|\leq\frac{2^{j+4}}{\rho}\quad\text{ and
}\quad|\partial_{t}\zeta|\leq\frac{2^{p(j+4)}}{\theta\rho^{p}}.$

An application of the energy estimate (4.1) and Lemma 2.2 in this setting
leads to

$\displaystyle\operatorname*{ess\sup}_{-\theta\rho_{j}^{p}<t<0}\int_{K_{j}}\zeta^{p}(u+k_{j})^{q-1}(u-k_{j})_{-}^{2}\,dx+\iint_{Q_{j}}\zeta^{p}|\nabla(u-k_{j})_{-}|^{p}\,dxdt$
$\displaystyle\leq$
$\displaystyle\gamma\iint_{Q_{j}}(u+k_{j})^{q-1}(u-k_{j})_{-}^{2}|\partial_{t}\zeta|\,dxdt+\gamma\iint_{Q_{j}}(u-k_{j})_{-}^{p}|\nabla\zeta|^{p}\,dxdt$
$\displaystyle+\gamma\int_{-\theta\rho_{j}^{p}}^{0}\int_{K_{j}}\int_{K_{j}}\max\\{(u-k_{j})_{-}(x,t),(u-k_{j})_{-}(y,t)\\}^{p}|\zeta(x,t)-\zeta(y,t)|^{p}\,d\mu
dt$
$\displaystyle+\gamma\mathop{\mathrm{ess}\,\sup}_{\stackrel{{\scriptstyle-\theta\rho_{j}^{p}<t<0}}{{x\in\mathrm{supp}\,\zeta(\cdot,t)}}}\int_{\mathbb{R}^{N}\backslash
K_{j}}\frac{(u-k_{j})_{-}^{p-1}(y,t)}{|x-y|^{N+sp}}\,dy\iint_{Q_{j}}(u-k_{j})_{-}\zeta^{p}(x,t)\,dxdt$
$\displaystyle=:$ $\displaystyle G_{1}+G_{2}+G_{3}+G_{4}.$ (6.2)

For the first term $G_{1}$, observe that when $u<k_{j}$, there holds

$\displaystyle\frac{1}{2}M\leq k_{j}\leq u+k_{j}\leq 2M.$ (6.3)

Thus we have

$\displaystyle G_{1}$
$\displaystyle\leq\gamma\frac{2^{pj}}{\theta\rho^{p}}\iint_{Q_{j}}(u+k_{j})^{q-1}(u-k_{j})_{-}^{2}\,dxdt$
$\displaystyle\leq\gamma\frac{2^{pj}}{\theta\rho^{p}}M^{q+1}|A_{j}|,$

where $A_{j}=[u<k_{j}]\cap Q_{j}$. With the properties of function $\zeta$, we
get the estimates of $G_{2}$ and $G_{3}$ as below

$\displaystyle G_{2},G_{3}\leq\gamma\frac{2^{pj}}{\rho^{p}}M^{p}|A_{j}|.$

Besides, we apparently see

$\displaystyle\frac{|y|}{|y-x|}\leq 1+\frac{|x|}{|y-x|}\leq
1+\frac{\widehat{\rho}_{j}}{\rho_{j}-\widehat{\rho}_{j}}\leq 2^{j+4}$

for any $|x|\leq\widehat{\rho}_{j}$ and $|y|\geq\rho_{j}$. Hence, it holds
that

$\displaystyle G_{4}$ $\displaystyle\leq\gamma
2^{(j+4)(N+sp)}M^{p}|A_{j}|\left(\frac{\gamma
2^{spj}}{\rho^{sp}}+\operatorname*{ess\sup}_{-\theta\rho_{j}^{sp}<t<0}\int_{\mathbb{R}^{N}\backslash
K_{\rho}}\frac{1}{|y|^{N+sp}}\,dy\right)$ $\displaystyle\leq\gamma
2^{(j+4)(N+sp)}\frac{M^{p}}{\rho^{p}}|A_{j}|.$

Now, let us turn to estimate the left-hand side of (6). By using (6.3) one can
get

$\displaystyle\operatorname*{ess\sup}_{-\theta\rho_{j}^{p}<t<0}\int_{K_{j}}\zeta^{p}(u+k_{j})^{q-1}(u-k_{j})_{-}^{2}\,dx\geq\frac{M^{q-1}}{2^{|q-1|}}\operatorname*{ess\sup}_{-\theta\rho_{j}^{p}<t<0}\int_{K_{j}}\zeta^{p}(u-k_{j})_{-}^{2}\,dx.$

We conclude from the above estimates that

$\displaystyle\frac{M^{q-1}}{2^{|q-1|}}\operatorname*{ess\sup}_{-\theta\widetilde{\rho}_{j}^{p}<t<0}\int_{\widetilde{K}_{j}}(u-k_{j})_{-}^{2}\,dx+\iint_{\widetilde{Q}_{j}}|\nabla(u-k_{j})_{-}|^{p}\,dxdt$
$\displaystyle\leq$
$\displaystyle\gamma\frac{2^{(N+p)j}}{\rho^{p}}M^{p}\left(1+\frac{M^{q+1-p}}{\theta}\right)|A_{j}|.$
(6.4)

Hölder’s inequality in conjunction with Sobolev embedding Lemma 2.7 indicates
that

$\displaystyle\frac{M}{2^{j+4}}$
$\displaystyle|A_{j+1}|\leq\iint_{\widetilde{Q}_{j}}(u-k_{j})_{-}\,dxdt$
$\displaystyle\leq$
$\displaystyle\left(\iint_{\widetilde{Q}_{j}}(u-k_{j})_{-}^{p\frac{N+2}{N}}\,dxdt\right)^{\frac{N}{p(N+2)}}|A_{j}|^{1-\frac{N}{p(N+2)}}$
$\displaystyle\leq$
$\displaystyle\gamma\left(\iint_{\widetilde{Q}_{j}}\left|\nabla(u-k_{j})_{-}\right|^{p}\,dxdt\right)^{\frac{N}{p(N+2)}}\left(\operatorname*{ess\sup}_{-\theta\widetilde{\rho}_{j}^{p}<t<0}\int_{\widetilde{K}_{j}}(u-k_{j})_{-}^{2}\,dx\right)^{\frac{1}{N+2}}|A_{j}|^{1-\frac{N}{p(N+2)}}$
$\displaystyle\leq$
$\displaystyle\gamma\left[\frac{2^{(N+p)j}}{\rho^{p}}M^{p}\left(1+\frac{M^{q+1-p}}{\theta}\right)\right]^{\frac{N+p}{p(N+2)}}\left(\frac{2^{|q-1|}}{M^{q-1}}\right)^{\frac{1}{N+2}}|A_{j}|^{1+\frac{1}{N+2}},$
(6.5)

where we used (6) in the last step. Denote $Y_{j}=|A_{j}|/|Q_{j}|$, we infer
from (6) that

$\displaystyle Y_{j+1}\leq\gamma
b^{j}\left(1+\frac{M^{q+1-p}}{\theta}\right)^{\frac{N+p}{p(N+2)}}\left(\frac{\theta}{M^{q+1-p}}\right)^{\frac{1}{N+2}}Y_{j}^{1+\frac{1}{N+2}},$

where $b=b(N,s,p)>1$, and $\gamma>0$ depends only on $N,p,s,q,\Lambda$. With
the help of Lemma 2.4, we get $Y_{j}\rightarrow 0$ as $j\rightarrow\infty$ if
$Y_{0}\leq\nu$. Recalling the definition of $\theta$, we find $\nu$ only
depends on $N,p,s,q,\Lambda$ and $\delta$. ∎

The following result is about extending the measure information of positivity
forward in time direction.

######  Lemma 6.3.

Let $q>0,p>1$ and $\rho\in(0,1]$. Suppose that $u$ is a nonnegative weak
solution to (1.1). If for some $M>0$ and $\alpha\in(0,1)$ there holds

$\displaystyle|[u(\cdot,t_{0})\geq M]\cap
K_{\rho}(x_{0})|\geq\alpha|K_{\rho}|,$

then there exist constants $\delta$ and $\varepsilon$ in $(0,1)$ depending on
$N,p,s,q,\Lambda$ and $\alpha$ such that

$\displaystyle|[u(\cdot,t)]\geq\varepsilon M]\cap
K_{\rho}(x_{0})|\geq\frac{1}{2}\alpha|K_{\rho}|\quad\text{ for all
}t\in\left(t_{0},t_{0}+\delta M^{q+1-p}\rho^{p}\right],$

provided

$\displaystyle Q:=K_{\rho}\times(t_{0},t_{0}+\delta M^{q+1-p}\rho^{p}]\subset
E_{T}.$

###### Proof.

Let $(x_{0},t_{0})=(0,0)$. Employing the energy estimate (4.1) in $Q$ with
$k=M$. Choose the test function $\zeta(x,t)=\zeta(x)$ equals to $1$ in
$K_{(1-\sigma)\rho}$, vanishing outside $K_{\frac{2-\sigma}{2}\rho}$ such that
$|\nabla\zeta|\leq(\sigma\rho)^{-1}$, where $\sigma\in(0,1)$ will be chosen
later. Then we have for any $0<t\leq\delta M^{q+1-p}\rho^{p}$ that

$\displaystyle\int_{K_{\rho}\times\\{t\\}}\int_{u}^{M}s^{q-1}(s-M)_{-}\,ds\zeta^{p}\,dx$
$\displaystyle\leq$
$\displaystyle\int_{K_{\rho}\times\\{0\\}}\int_{u}^{M}s^{q-1}(s-M)_{-}\,ds\zeta^{p}\,dx+\gamma\iint_{Q}(u-M)_{-}^{p}|\nabla\zeta|^{p}\,dxdt$
$\displaystyle+\int_{0}^{\delta
M^{q+1-p}\rho^{p}}\\!\\!\int_{K_{\rho}}\int_{K_{\rho}}\max\\{(u-M)_{-}(x,t),(u-M)_{-}(y,t)\\}^{p}|\zeta(x,t)-\zeta(y,t)|^{p}\,d\mu
dt$ $\displaystyle+\gamma\mathop{\mathrm{ess}\,\sup}_{\stackrel{{\scriptstyle
0<t<\delta
M^{q+1-p}\rho^{p}}}{{x\in\mathrm{supp}\,\zeta(\cdot,t)}}}\int_{\mathbb{R}^{N}\backslash
K_{\rho}}\frac{(u-M)_{-}^{p-1}(y,t)}{|x-y|^{N+sp}}\,dy\iint_{Q}(u-M)_{-}\zeta^{p}(x,t)\,dxdt$
$\displaystyle\leq$
$\displaystyle\int_{K_{\rho}\times\\{0\\}}\int_{u}^{M}s^{q-1}(s-M)_{-}\,ds\zeta^{p}\,dx+\gamma\frac{\delta
M^{q+1}}{\sigma^{N+p}}|K_{\rho}|,$

where $\gamma$ depends only on $N,p,s,q,\Lambda$. The rest of the proof is
standard, and we refer the readers to Lemma 6.2 in [6] for details. ∎

Next, we introduce a measure shrinking lemma.

######  Lemma 6.4.

Let $0<p-1\leq q$, constants $\delta,\alpha\in(0,1)$ and $M>0$. Suppose that
$u$ is a nonnegative, weak supersolution to (1.1). Let $\rho\in(0,1)$ and
$K_{2\rho}(x_{0})\times(t_{0},t_{0}+\delta M^{q+1-p}\rho^{p}]\subset E_{T}$.
For any $\nu\in(0,1)$, if

$\displaystyle|[u(\cdot,t)\geq M]\cap
K_{\rho}(x_{0})|\geq\alpha|K_{\rho}|\quad\text{ for all
}t\in\left(t_{0},t_{0}+\delta M^{q+1-p}\rho^{p}\right],$ (6.6)

then there exists $\xi\in(0,1)$ depending on $N,p,s,q,\Lambda,\delta,\nu$ and
$\alpha$ such that

$\displaystyle|[u(\cdot,t)\leq\xi M]\cap K_{\rho}(x_{o})|\leq\nu|K_{\rho}|$

for all time

$\displaystyle t\in\left(t_{0}+\tfrac{1}{2}\delta
M^{q+1-p}\rho^{p},t_{0}+\delta M^{q+1-p}\rho^{p}\right].$

At this point, some preparations are needed. Let $(x_{0},t_{0})=(0,0)$. Set

$\displaystyle I=(0,\delta M^{q+1-p}\rho^{p}],\quad\lambda
I=\left((1-\lambda)\delta M^{q+1-p}\rho^{p},\delta M^{q+1-p}\rho^{p}\right],$

and

$\displaystyle Q=K_{2\rho}\times I,\quad\lambda Q=K_{\lambda
2\rho}\times\lambda I,$

where $\lambda\in(0,1)$. For some $c\in(0,1)$ that will be selected later,
taking the sequence

$\displaystyle k_{j}:=c^{j}M,\ \ j=0,1,2\ldots.$ (6.7)

Denote

$\displaystyle Y_{j}:=\sup_{t\in
I}{\int-}_{K_{2\rho}\times\\{t\\}}\zeta^{p}\chi_{[u<k_{j}]}\,dx.$ (6.8)

Here, the cutoff function $\zeta(x,t)=\zeta_{1}(x)\zeta_{2}(t)$ is piecewise
smooth in $Q$, such that

$\displaystyle\left\\{\begin{array}[]{c}0\leq\zeta\leq 1\text{ in
}Q,\quad\zeta=1\text{ in }\frac{1}{2}Q,\quad\zeta=0\text{ in
}Q\backslash\frac{3}{4}Q,\\\ |\nabla\zeta_{1}|\leq\frac{2}{\rho},\quad
0\leq\partial_{t}\zeta_{2}\leq\frac{4}{\delta M^{q+1-p}\rho^{p}},\\\ \text{
the sets }[x\in K_{2\rho}:\zeta_{1}(x)>a]\text{ are convex for all
}a\in(0,1).\end{array}\right.$ (6.12)

In the next step, we give the estimate of $Y_{j}$ as a crucial tool to prove
Lemma 6.4.

######  Lemma 6.5.

Suppose the conditions in Lemma 6.4 hold. Then there exist $\sigma,c\in(0,1)$
depending on $N,p,s,q,\Lambda,\delta,\alpha$ and $\nu$, such that for every
$\nu\in(0,1)$ there holds either

$\displaystyle Y_{j}\leq\nu$

or

$\displaystyle Y_{j+1}\leq\max\left\\{\nu,\sigma Y_{j}\right\\}$

with $j\in\mathbb{N}\cup\\{0\\}$.

###### Proof.

We prove this lemma starts by showing an integral inequality under the
assumption

$\displaystyle\partial_{t}u^{q}\in C(I;L^{1}(K_{2\rho})),$ (6.13)

and drops it later. As a consequence of Proposition 3.2, $u_{k}:=k-(k-u)_{+}$
with $k\in(0,M)$ is a nonnegative, weak supersolution to (1.1) in $Q$, reads
that

$\displaystyle\partial_{t}u_{k}^{q}-\operatorname{div}(|\nabla
u_{k}|^{p-2}\nabla u_{k})$ $\displaystyle\ \ \ \ \ \
+\mathrm{P.V.}\int_{\mathbb{R}^{N}}|u_{k}(x,t)-u_{k}(y,t)|^{p-2}(u_{k}(x,t)-u_{k}(y,t))\,dy\geq
0\quad\text{ weakly in }Q.$ (6.14)

Testing (6) with the function

$\displaystyle\frac{\zeta^{p}}{\left[k-(k-u)_{+}+ck\right]^{p-1}},$ (6.15)

where $\zeta$ is given in (6.12), and constants $c\in(0,1),k\in(0,M)$ will be
determined later. Then for a.e. $t\in I$, we have

$\displaystyle\partial_{t}\int_{K_{2\rho}\times\\{t\\}}\zeta^{p}\Phi_{k}(u)\,dx+\int_{K_{2\rho}\times\\{t\\}}\zeta^{p}\left|\nabla\Psi_{k}(u)\right|^{p}\,dx$
$\displaystyle\leq$
$\displaystyle\int_{K_{2\rho}\times\\{t\\}}\left|\nabla\Psi_{k}(u)\right|^{p-1}\zeta^{p-1}|\nabla\zeta|\,dx+\int_{K_{2\rho}\times\\{t\\}}\Phi_{k}(u)\partial_{t}\zeta^{p}\,dx$
$\displaystyle+\int_{K_{2\rho}\times\\{t\\}}\int_{K_{2\rho}}\frac{U_{k}(x,y,t)}{|x-y|^{N+sp}}\left(\frac{\zeta^{p}(x,t)}{[u_{k}(x,t)+ck]^{p-1}}-\frac{\zeta^{p}(y,t)}{[u_{k}(y,t)+ck]^{p-1}}\right)\,dxdy$
$\displaystyle+2\int_{K_{2\rho}\times\\{t\\}}\int_{\mathbb{R}^{N}\backslash
K_{2\rho}}\frac{\zeta^{p}(x,t)U_{k}(x,y,t)}{[u_{k}(x,t)+ck]^{p-1}|x-y|^{N+sp}}\,dxdy$
$\displaystyle=:$ $\displaystyle J_{1}+J_{2}+J_{3}+J_{4},$ (6.16)

where

$\displaystyle
U_{k}(x,y,t):=|u_{k}(x,t)-u_{k}(y,t)|^{p-2}(u_{k}(x,t)-u_{k}(y,t)),$
$\displaystyle\Phi_{k}(u):=\int_{0}^{(k-u)_{+}}\frac{q(k-s)^{q-1}}{(k-s+ck)^{p-1}}\,ds,$
$\displaystyle\Psi_{k}(u):=\ln\left(\frac{k(1+c)}{k(1+c)-(k-u)_{+}}\right).$

The existence of the term containing time derivative is ensured by assumption
(6.13).

Next, we estimate $J_{1}$–$J_{4}$ in (6), separately.

Estimate of $J_{1}$: By Young’s inequality, it holds that

$\displaystyle
J_{1}\leq\frac{1}{2}\int_{K_{2\rho}\times\\{t\\}}\left|\nabla\Psi_{k}(u)\right|^{p}\zeta^{p}\,dx+\gamma(p)\int_{K_{2\rho}}|\nabla\zeta|^{p}\,dx.$

Estimate of $J_{2}$: Since $0<p-1\leq q$, we compute

$\displaystyle\Phi_{k}(u)$
$\displaystyle\leq\int_{0}^{k}\frac{q(k-s)^{q-1}}{(k-s+ck)^{p-1}}\,ds$
$\displaystyle=k^{q+1-p}\int_{0}^{1}\frac{qs^{q-1}}{(s+c)^{p-1}}\,ds$
$\displaystyle\leq\gamma(p,q)k^{q+1-p}\ln\left(\frac{1+c}{c}\right).$ (6.17)

By using (6), (6.12) and $k\in(0,M)$, it follows that

$\displaystyle J_{2}\leq\gamma\frac{k^{q+1-p}}{\delta
M^{q+1-p}\rho^{p}}\ln\left(\frac{1+c}{c}\right)|K_{2\rho}|\leq\frac{4\gamma(p,q)}{\delta\rho^{p}}\ln\left(\frac{1+c}{c}\right)|K_{2\rho}|.$

Estimate of $J_{3}$: $J_{3}$ can be estimated in virtue of (6.12) as

$\displaystyle J_{3}\leq$
$\displaystyle\int_{K_{2\rho}\times\\{t\\}}\int_{K_{2\rho}}\frac{|u_{k}(x,t)-u_{k}(y,t)|^{p-2}(u_{k}(x,t)-u_{k}(y,t))}{|x-y|^{N+sp}}\times\frac{\zeta^{p}(x,t)-\zeta^{p}(y,t)}{[u_{k}(x,t)+ck]^{p-1}}\,dxdy$
$\displaystyle\leq$
$\displaystyle\gamma(N,p,s,q,\Lambda)\frac{1}{\rho^{p}}|K_{2\rho}|,$

where we drop the non-positive term

$\displaystyle\int_{K_{2\rho}\times\\{t\\}}\int_{K_{2\rho}}\frac{U_{k}(x,y,t)}{|x-y|^{N+sp}}\left(\frac{\zeta^{p}(y,t)}{[u_{k}(x,t)+ck]^{p-1}}-\frac{\zeta^{p}(y,t)}{[u_{k}(y,t)+ck]^{p-1}}\right)\,dxdy.$

Estimate of $J_{4}$: Note that

$\displaystyle\frac{|y|}{|y-x|}\leq 1+\frac{|x|}{|y-x|}\leq
1+\frac{3/2\rho}{1/2\rho}\leq 4$

for any $|x|\leq\frac{3}{2}\rho$ and $|y|\geq 2\rho$. Thus we evaluate

$\displaystyle J_{4}$ $\displaystyle\leq\gamma\int_{\mathbb{R}^{N}\backslash
K_{2\rho}}\frac{1}{|y|^{N+sp}}\,dy\int_{K_{2\rho}\times\\{t\\}}\frac{u_{k}(x,t)^{p-1}\zeta^{p}(x,t)}{[u_{k}(x,t)+ck]^{p-1}}\,dx$
$\displaystyle\leq\gamma(N,p,s,q,\Lambda)\frac{1}{\rho^{p}}|K_{2\rho}|.$

Inserting estimates $J_{1}$–$J_{4}$ into (6), we derive for a.e. $t\in I$ that

$\displaystyle\partial_{t}\int_{K_{2\rho}\times\\{t\\}}\zeta^{p}\Phi_{k}(u)\,dx+\int_{K_{2\rho}\times\\{t\\}}\zeta^{p}\left|\nabla\Psi_{k}(u)\right|^{p}\,dx\leq\frac{\gamma}{\delta\rho^{p}}\ln\left(\frac{1+c}{c}\right)\left|K_{2\rho}\right|$
(6.18)

with $\gamma$ depending only on $N,p,s,q,\Lambda$. Here, we can chose
$c\in(0,\frac{1}{3})$ such that $\ln\left(\frac{1+c}{c}\right)\geq 1$. From
the measure information (6.6) and $k<M$, we have

$\displaystyle\left|[\Psi_{k}(u)=0]\cap K_{\rho}\right|\geq\alpha
2^{-N}|K_{2\rho}|\quad\text{ for all }t\in I.$

Employing Poincaré inequality in Lemma 2.6, it gives that

$\displaystyle\int_{K_{2\rho}\times\\{t\\}}\zeta^{p}\Psi_{k}^{p}(u)\,dx\leq\frac{\gamma_{*}\rho^{p}}{\alpha^{p}}\int_{K_{2\rho}\times\\{t\\}}\zeta^{p}\left|\nabla\Psi_{k}(u)\right|^{p}\,dx\quad\text{
for a.e. }t\in I,$ (6.19)

where $\gamma_{*}$ is the Sobolev constant depending only on $p,N$. Taking
(6.18) and (6.19) into account, we arrive at

$\displaystyle\partial_{t}\int_{K_{2\rho}\times\\{t\\}}\zeta^{p}\Phi_{k}(u)\,dx+\frac{\alpha^{p}}{\gamma_{*}\rho^{p}}\int_{K_{2\rho}\times\\{t\\}}\zeta^{p}\Psi_{k}^{p}(u)\,dx\leq\frac{\gamma}{\delta\rho^{p}}\ln\left(\frac{1+c}{c}\right)|K_{2\rho}|,$
(6.20)

which is the desired integral inequality.

Since the integral inequality (6.20) has been obtained, we can proceed as in
[6, §6.3.1.2-§6.3.1.4] to prove there holds $Y_{j+1}\leq\max\left\\{\nu,\sigma
Y_{j}\right\\}$ under the assumption (6.13). Then, we can remove (6.13) by
following the argument in [6, §6.3.1.5]. Moreover, the convergence of the
nonlocal term can be verified by Lemma 2.8 $(\rm{{i}})$. We omit the proof for
short. ∎

###### Proof of Lemma 6.4.

Iterating Lemma 6.3 yields that

$\displaystyle
Y_{j_{0}}\leq\max\left\\{\nu,\sigma^{j_{0}}Y_{0}\right\\}\quad\text{for}\
j_{0}\in\mathbb{N}.$

From the definition of $Y_{j}$ in (6.8), we know $Y_{0}\leq 1$. Now, choosing
$j_{0}$ such that $\sigma^{j_{0}}\leq\nu$, further to obtain
$Y_{j_{0}}\leq\nu$. In view of (6.7), (6.8) and (6.12), we have

$\displaystyle\frac{1}{|K_{2\rho}|}\left|[u(\cdot,t)\leq c^{j_{0}}M]\cap
K_{\rho}\right|\leq Y_{j_{0}}\leq\nu\quad\text{ for all }t\in\frac{1}{2}I.$

Let $\xi=c^{j_{0}}$, we conclude the proof by replacing $\nu$ by $2^{-N}\nu$
and adjusting some constants. ∎

We now proceeding in proving Proposition 6.1.

###### Proof of Proposition 6.1.

The measure theoretical hypothesis (6.1) implies that

$\displaystyle\left|[u(\cdot,t_{0})\geq M]\cap K_{4\rho}(x_{0})\right|\geq
4^{-N}\alpha\left|K_{4\rho}\right|.$

This along with Lemma 6.3 gives there exist $\delta$ and $\varepsilon$ in
$(0,1)$ depending only on $N,p,s,q,\Lambda,\alpha$ such that

$\displaystyle\left|[u(\cdot,t)\geq\varepsilon M]\cap
K_{4\rho}(x_{0})\right|\geq\frac{1}{2}4^{-N}\alpha|K_{4\rho}|$ (6.21)

for any

$\displaystyle t\in(t_{0},t_{0}+\delta M^{q+1-p}(4\rho)^{p}].$

With the help of (6.21), we can employ Lemma 6.4 to derive that for given
$\nu\in(0,1)$, there exists $\xi\in(0,1)$ depending only on
$N,p,s,q,\Lambda,\nu,\delta,\alpha$ such that

$\displaystyle\left|[u(\cdot,t)\leq\xi\varepsilon M]\cap
K_{4\rho}(x_{0})\right|\leq\nu\left|K_{4\rho}\right|$

for

$\displaystyle t\in\left(t_{0}+\tfrac{1}{2}\delta(\varepsilon
M)^{q+1-p}(4\rho)^{p},t_{0}+\delta(\varepsilon M)^{q+1-p}(4\rho)^{p}\right].$

Notice for every time

$\displaystyle\bar{t}\in\left(t_{0}+\tfrac{3}{4}\delta(\varepsilon
M)^{q+1-p}(4\rho)^{p},t_{0}+\delta(\varepsilon M)^{q+1-p}(4\rho)^{p}\right],$

we have

$\displaystyle\left(x_{0},\bar{t}\right)+Q_{4\rho}(\theta)\subset
K_{4\rho}(x_{0})\times\left(t_{0}+\tfrac{1}{2}\delta(\varepsilon
M)^{q+1-p}(4\rho)^{p},t_{0}+\delta(\varepsilon M)^{q+1-p}(4\rho)^{p}\right]$

with $\theta=\frac{1}{4}\delta(\xi\varepsilon M)^{q-1+p}$. Thus, we deduce for
any $\bar{t}$ that

$\displaystyle\left|[u\leq\xi\varepsilon
M]\cap(x_{0},\bar{t})+Q_{4\rho}(\theta)\right|\leq\nu|Q_{4\rho}(\theta)|.$
(6.22)

Thanks to the measure information (6.22), it permits us to exploit the De
Giorgi-type Lemma 6.2 in $(x_{0},\bar{t})+Q_{4\rho}(\theta))$ with $M$
replaced by $\xi\varepsilon M$, we also choose $\nu$ from Lemma 6.2, then we
deduce that

$\displaystyle u\geq\frac{1}{2}\xi\varepsilon M\quad\text{ a.e.
in}(x_{0},\bar{t})+Q_{2\rho}(\theta).$

Due to the arbitrariness of $\bar{t}$, we get the desired results. ∎

##  7\. Harnack inequality

In this section, we devote to establishing the Harnack inequality in Theorem
1.6. The first step is going to scale functions.

###  7.1. Scaling functions

Change the variables

$\displaystyle z\rightarrow\frac{x-x_{0}}{\rho},\quad
z^{\prime}\rightarrow\frac{y-x_{0}}{\rho},\quad\tau\rightarrow[u(x_{0},t_{0})]^{p-q-1}\frac{t-t_{0}}{\rho^{p}}.$

Consider the following rescaled function

$\displaystyle v(z,\tau):=\frac{u\left(x_{0}+\rho
z,t_{0}+[u(x_{0},t_{0})]^{q+1-p}\rho^{p}\tau\right)}{u(x_{0},t_{0})}\quad\text{
in }\widetilde{Q}_{8}=K_{8}\times(-8^{p},8^{p}).$ (7.1)

Via formula calculation, we can check that $v(0,0)=1$ and $v$ is a bounded
continuous nonnegative solution to

$\displaystyle\partial_{\tau}v^{q}-\mathrm{div}(|\nabla v|^{p-2}\nabla v)$
$\displaystyle\ \ \ \ \
+\mathrm{P.V.}\int_{\mathbb{R}^{N}}\widetilde{K}(z,z^{\prime},\tau)|v(z,\tau)-v(z^{\prime},\tau)|^{p-2}(v(z,\tau)-v(z^{\prime},\tau))\,dz^{\prime}=0\quad\text{
in }\widetilde{Q}_{8}$ (7.2)

with

$\displaystyle\widetilde{K}(z,z^{\prime},\tau)=\rho^{N+p}K(x_{0}+\rho
z,x_{0}+\rho z^{\prime},t_{0}+[u(x_{0},t_{0})]^{q+1-p}\rho^{p}\tau),$

satisfying

$\displaystyle\frac{\rho^{p-sp}\Lambda^{-1}}{|z-z^{\prime}|^{N+sp}}\leq\widetilde{K}(z,z^{\prime},\tau)\leq\frac{\rho^{p-sp}\Lambda}{|z-z^{\prime}|^{N+sp}}.$

Now, our intention turns to demonstrate there exist $\gamma>1$ and
$\sigma\in(0,1)$ depending only on $N,p,s,q,\Lambda$ and
$\|u\|_{L^{\infty}(\mathbb{R}^{N}\times(0,T))}$ such that

$\displaystyle v\geq\gamma^{-1}\quad\text{ in }K_{1}\times(-\sigma,\sigma).$
(7.3)

Notice the Harnack inequality on the left-hand side is a straightforward
consequence of (7.3).

###  7.2. The supremum of function $v$ in $K_{1}$.

For $\tau\in(0,1)$, define

$\displaystyle M_{\tau}:=\sup_{K_{\tau}}v(\cdot,0),\quad
N_{\tau}:=(1-\tau)^{-\beta},\quad\text{ where }\beta=\frac{p}{q+1-p}.$

Observe that $M_{0}=N_{0}=1$, $N_{\tau}\rightarrow\infty$ as $\tau\rightarrow
1$, and $M_{\tau}$ is bounded due to $v$ is a bounded function. Therefore, due
to $M_{\tau}=N_{\tau}$ must have roots, we denote the largest one by
$\tau_{*}$, specifically,

$\displaystyle M_{\tau_{*}}=N_{\tau_{*}}\quad\text{ and }\quad M_{\tau}\leq
N_{\tau}\quad\text{ for all }\tau\geq\tau_{*}.$

Pick $\bar{\tau}\in\left(\tau_{*},1\right)$ such that

$\displaystyle
N_{\bar{\tau}}=(1-\bar{\tau})^{-\beta}=4(1-\tau_{*})^{-\beta},\quad\text{
i.e., }\quad\bar{\tau}=1-4^{-\frac{1}{\beta}}\left(1-\tau_{*}\right),$

and let

$\displaystyle 2r:=\bar{\tau}-\tau_{*}=(1-4^{-\frac{1}{\beta}})(1-\tau_{*}).$

Clearly, $M_{\tau_{*}}$ can be achieved at some $\bar{x}\in K_{\tau_{*}}$
because $v$ is a continuous function. Hence, there holds
$K_{2r}(\bar{x})\subset K_{\bar{\tau}},M_{\bar{\tau}}\leq N_{\bar{\tau}}$, and

$\displaystyle\sup_{K_{\tau_{*}}}v(\cdot,0)$
$\displaystyle=M_{\tau_{*}}=v(\bar{x},0)\leq\sup_{K_{2r}(\bar{x})}v(\cdot,0)$
$\displaystyle\leq\sup_{K_{\bar{\tau}}}v(\cdot,0)=M_{\bar{\tau}}\leq
N_{\bar{\tau}}=4(1-\tau_{*})^{-\beta}.$ (7.4)

###  7.3. Expanding of positivity of $v$.

Set the cylinder

$\displaystyle\widetilde{Q}_{r}(\theta_{*}):=K_{r}(\bar{x})\times(-\theta_{*}r^{p},\theta_{*}r^{p}),\quad\theta_{*}:=(1-\tau_{*})^{-\beta(q+1-p)}$

with center $(\bar{x},0)$. By the choices of $\beta$ and $r$, we compute

$\displaystyle\theta_{*}r^{p}=(1-\tau_{*})^{-p}r^{p}=2^{-p}\left(1-4^{-\frac{1}{\beta}}\right)^{p}=:c,$
(7.5)

which implies that

$\widetilde{Q}_{2r}\left(\theta_{*}\right)=K_{2r}(\bar{x})\times\left(-2^{p}c,2^{p}c\right).$

Obviously,

$\displaystyle r=c^{\frac{1}{p}}\left(1-\tau_{*}\right).$ (7.6)

We next discuss the estimate of the supremum of $v$ in
$\widetilde{Q}_{r}\left(\theta_{*}\right)$.

######  Lemma 7.1.

Let $v$ be defined as in (7.1) and
$0<p-1<q<\min\left\\{p^{2}-1,\frac{N(p-1)}{(N-p)_{+}}\right\\}$. There exists
a constant $\gamma$ depending only on
$N,p,s,q,\Lambda,\|u\|_{L^{\infty}(\mathbb{R}^{N}\times(0,T))}$ such that

$\displaystyle\sup_{\widetilde{Q}_{r}\left(\theta_{*}\right)}v\leq\gamma(1-\tau_{*})^{-\beta}.$

###### Proof.

An application of Theorem 1.5 to $v$ over the cylinder
$\widetilde{Q}_{r}\left(\theta_{*}\right)\subset\widetilde{Q}_{2r}\left(\theta_{*}\right)$
leads to

$\displaystyle\sup_{\widetilde{Q}_{r}(\theta_{*})}v^{q}\leq$
$\displaystyle\frac{\gamma}{c^{\frac{N}{\lambda}}}\left(\int_{K_{2r}(\bar{x})}v^{q}(x,0)\,dx\right)^{\frac{p}{\lambda}}+\gamma\left(\frac{c}{r^{p}}\right)^{\frac{q}{q+1-p}}$
$\displaystyle+\gamma\frac{c}{r^{p}}\left[\mathrm{Tail}_{\infty}(u;\bar{x},r;-2^{p}c,2^{p}c)\right]^{p-1}\\!+\\!\gamma\left(\frac{c}{r^{p}}\right)^{\frac{p-N}{\lambda}}[\mathrm{Tail}_{\infty}(u;\bar{x},r;-2^{p}c,2^{p}c)]^{\frac{p(p-1)}{\lambda}}$
$\displaystyle\leq$
$\displaystyle\frac{\gamma}{c^{\frac{N}{\lambda}}}\left(\int_{K_{2r}(\bar{x})}v^{q}(x,0)\,dx\right)^{\frac{p}{\lambda}}+\gamma\left(\frac{c}{r^{p}}\right)^{\frac{q}{q+1-p}}+\gamma\frac{c}{r^{p}}+\gamma\left(\frac{c}{r^{p}}\right)^{\frac{p-N}{\lambda}}$
$\displaystyle\leq$
$\displaystyle\gamma\left[(1-\tau_{*})^{-\frac{pq}{q+1-p}}+(1-\tau_{*})^{-p}+(1-\tau_{*})^{\frac{-p(p-N)}{\lambda}}\right],$
(7.7)

where $\gamma$ depends only on
$N,p,s,q,\Lambda,\|u\|_{L^{\infty}(\mathbb{R}^{N}\times(0,T))}$. In the last
step of (7.3), we used (7.2) and (7.6).

By the definitions of $\beta,\lambda$ and the assumption $0<p-1<q$, we compute

$\displaystyle-p\geq\frac{-pq}{q+1-p}=-\beta q$ (7.8)

and

$\displaystyle\frac{-p(p-N)}{\lambda}=\frac{pq(p-N)}{N(q+1-p)-pq}\geq\frac{-pqN}{N(q+1-p)}=-\beta
q.$ (7.9)

Since $\tau_{*}\in(0,1)$, the claim follows from (7.3)–(7.9). ∎

With Lemma 7.1, we present the following result which provides the condition
of the expansion of positivity.

######  Lemma 7.2.

Let $v$ be defined as in (7.1). Let $p>N$ and $0<p-1<q<p^{2}-1$. There exist
constants $\delta,\bar{c},\alpha\in(0,1)$ just depending on
$N,p,s,q,\Lambda,\|u\|_{L^{\infty}(\mathbb{R}^{N}\times(0,T))}$ such that

$\displaystyle\left|\left[v(\cdot,t)\geq\bar{c}(1-\tau_{*})^{-\beta}\right]\cap
K_{r}(\bar{x})\right|\geq\alpha|K_{r}|\quad\text{ for all
}t\in\left[-\delta\theta_{*}r^{p},\delta\theta_{*}r^{p}\right].$

###### Proof.

By using Theorem 1.5 to function $v$ over the cylinder
$\widetilde{Q}_{\frac{1}{2}r}(\delta\theta_{*})\subset\widetilde{Q}_{r}(\delta\theta_{*})$
yields

$\displaystyle(1-\tau_{*})^{-\beta q}=$ $\displaystyle
v^{q}(\bar{x},0)\leq\sup_{K_{\frac{1}{2}r}(\bar{x})}v^{q}(\cdot,0)$
$\displaystyle\leq$
$\displaystyle\frac{\gamma}{\left(\delta\theta_{*}r^{p}\right)^{\frac{N}{\lambda}}}\left(\int_{K_{r}(\bar{x})}v^{q}(x,t)\,dx\right)^{\frac{p}{\lambda}}+\gamma(\delta\theta_{*})^{\frac{q}{q+1-p}}+\gamma\delta\theta_{*}+\gamma(\delta\theta_{*})^{\frac{p-N}{\lambda}}$
$\displaystyle\leq$
$\displaystyle\frac{\gamma}{\left(\delta\theta_{*}r^{p}\right)^{\frac{N}{\lambda}}}\left(\int_{K_{r}(\bar{x})}v^{q}(x,t)\,dx\right)^{\frac{p}{\lambda}}+\gamma\delta^{\frac{q}{q+1-p}}\left(1-\tau_{*}\right)^{-\beta
q}$
$\displaystyle+\gamma\delta(1-\tau_{*})^{-\beta(q+1-p)}+\gamma\delta^{\frac{p-N}{\lambda}}(1-\tau_{*})^{-\beta(q+1-p)\frac{p-N}{\lambda}}$
$\displaystyle\leq$
$\displaystyle\frac{\gamma}{\left(\delta\theta_{*}r^{p}\right)^{\frac{N}{\lambda}}}\left(\int_{K_{r}(\bar{x})}v^{q}(x,t)\,dx\right)^{\frac{p}{\lambda}}+\gamma\delta^{\frac{p-N}{\lambda}}(1-\tau_{*})^{-\beta
q}$

for all $t\in\left[-\delta\theta_{*}r^{p},\delta\theta_{*}r^{p}\right]$. In
the above display, we also employed the definition of $\theta_{*}$ and the
fact $0<(p-N)/\lambda<1$. With taking $\delta$ such that
$\delta^{\frac{p-N}{\lambda}}\leq\frac{1}{2}$, we have

$\displaystyle(1-\tau_{*})^{-\beta
q}\leq\gamma\left(\int_{K_{r}(\bar{x})}v^{q}(x,t)\,dx\right)^{\frac{p}{\lambda}},$

where $\gamma$ depends only on $N,p,s,q,\Lambda$ and
$\|u\|_{L^{\infty}(\mathbb{R}^{N}\times(0,T))}$, since $\theta_{*}r^{p}=c$
depends only on $p$ and $q$. In addition, we deduce from Lemma 7.1 that

$\displaystyle\int_{K_{r}(\bar{x})}v^{q}(x,t)\,dx=$
$\displaystyle\int_{K_{r}(\bar{x})}v^{q}(x,t)\chi_{\left[v<\bar{c}(1-\tau_{*})^{-\beta}\right]}\,dx+\int_{K_{r}(\bar{x})}v^{q}(x,t)\chi_{\left[v\geq\bar{c}(1-\tau_{*})^{-\beta}\right]}\,dx$
$\displaystyle\leq$ $\displaystyle\bar{c}^{q}(1-\tau_{*})^{-\beta
q}(2r)^{N}+\gamma^{q}(1-\tau_{*})^{-\beta
q}|[v\geq\bar{c}\left(1-\tau_{*}\right)^{-\beta}]\cap K_{r}(\bar{x})|$

with $\bar{c}\in(0,1)$ to be specified. In light of the convexity of
$s^{\frac{p}{\lambda}}$, we can derive

$\displaystyle(1-\tau_{*})^{-\beta q}\leq$
$\displaystyle\gamma\left[\bar{c}^{q}(1-\tau_{*})^{-\beta
q}(2r)^{N}\right]^{\frac{p}{\lambda}}+\gamma(1-\tau_{*})^{-\beta
q\frac{p}{\lambda}}\left|\left[v\geq\bar{c}(1-\tau_{*})^{-\beta}\right]\cap
K_{r}(\bar{x})\right|^{\frac{p}{\lambda}}$ $\displaystyle\leq$
$\displaystyle\gamma\bar{c}^{q\frac{p}{\lambda}}(1-\tau_{*})^{-\beta
q}+\gamma(1-\tau_{*})^{-\beta
q\frac{p}{\lambda}}\left|\left[v\geq\bar{c}(1-\tau_{*})^{-\beta}\right]\cap
K_{r}(\bar{x})\right|^{\frac{p}{\lambda}}.$

Fixing $\bar{c}$ such that $\gamma\bar{c}^{q\frac{p}{\lambda}}=\frac{1}{2}$,
we obtain

$\displaystyle(1-\tau_{*})^{-\beta q}\leq\gamma(1-\tau_{*})^{-\beta
q\frac{p}{\lambda}}\left|\left[v\geq\bar{c}(1-\tau_{*})^{-\beta}\right]\cap
K_{r}(\bar{x})\right|^{\frac{p}{\lambda}}$

with $\gamma$ depending only on $N,p,s,q,\Lambda$ and
$\|u\|_{L^{\infty}(\mathbb{R}^{N}\times(0,T))}$. Thus, we complete the proof
based on the definitions of $r,\beta$ and $\lambda$. ∎

The next pointwise estimate is a direct consequence of Proposition 6.1 and
Lemma 7.2.

######  Lemma 7.3.

Let $v$ be defined as in (7.1). Let $p>N$ and $0<p-1<q<p^{2}-1$. There exist
constants $\eta,\delta\in(0,1)$ only depending on
$N,p,s,q,\Lambda,\|u\|_{L^{\infty}(\mathbb{R}^{N}\times(0,T))}$, such that

$\displaystyle v\geq\eta(1-\tau_{*})^{-\beta}\quad\text{ in
}K_{2r}(\bar{x})\times\left[-\tfrac{1}{2}\delta\theta_{*}r^{p},\delta\theta_{*}r^{p}\right].$

In the next step, we expand the pointwise positivity of $v$ defined as in
(7.1) to $K_{2}(\bar{x})$, and thus justify (7.3). This proof proceeds by a
comparison argument.

###  7.4. A comparison argument

We give the following initial-boundary value problem

$\displaystyle\left\\{\begin{array}[]{cl}\partial_{t}w^{q}-\mathrm{div}(|\nabla
w|^{p-2}\nabla w)+\mathcal{L}w=0&\text{ in
}K_{4}(\bar{x})\times(-\sigma,1],\\\ w=0&\text{ in }\mathbb{R}^{N}\backslash
K_{4}(\bar{x})\times(-\sigma,1],\\\
w^{q}(\cdot,-\sigma)=\eta(1-\tau_{*})^{-N}\chi_{K_{2r}(\bar{x})}(\cdot)&\text{
in }K_{4}(\bar{x}),\end{array}\right.$ (7.13)

where $\mathcal{L}$ is given by (1.2), $\eta$ is defined as in Lemma 7.3,
$\sigma\in\left(0,\frac{1}{2}\delta c\right)$ to be determined and $c$ is
defined as in (7.5).

######  Lemma 7.4.

Let $0<p-1<q$ and $p>N$. Let $v$ defined as in (7.1) be a nonnegative weak
solution to (7.1) and $w$ be a nonnegative weak solution to (7.13). Then we
have $w\leq v$ a.e. in $K_{4}(\bar{x})\times[-\sigma,1]$.

###### Proof.

Since $v$ is a nonnegative solution to (7.1), we know that $w\leq v$ in
$\mathbb{R}^{N}\backslash K_{4}(\bar{x})\times(-\sigma,1]$. Thanks to the
assumptions on $p,q$, we have $\beta q>N$, this combines with Lemma 7.3 gives
$w(\cdot,-\sigma)\leq v(\cdot,-\sigma)$ a.e. in $K_{4}(\bar{x})$, and thus we
derive from Proposition 3.4 that $w\leq v$ a.e. in
$K_{4}(\bar{x})\times[-\sigma,1]$. ∎

Finally, we end this section with proving Theorem 1.6.

###### Proof of Theorem 1.6.

In view of Lemma 7.4, we only need to obtain the pointwise positivity of $w$
as below,

$\displaystyle w\geq\gamma^{-1}\quad\text{ in
}K_{2}(\bar{x})\times\left[-\tfrac{1}{4}\sigma,\tfrac{1}{4}\sigma\right]$
(7.14)

with $\gamma>1$ and $\sigma\in\left(0,\frac{1}{2}\delta c\right)$ depending
only on $N,p,s,q,\Lambda,\|u\|_{L^{\infty}(\mathbb{R}^{N}\times(0,T))}$. In
order to get (7.14), we utilize Proposition 5.1 over the cylinder
$K_{2}(\bar{x})\times[-\sigma,\sigma]$. Then for all $t\in[-\sigma,\sigma]$,
there holds

$\displaystyle\int_{K_{1}(\bar{x})}w^{q}(x,-\sigma)\,dx$
$\displaystyle\leq\gamma\int_{K_{2}(\bar{x})}w^{q}(x,t)\,dx+\gamma\sigma^{\frac{q}{q+1-p}}+\gamma\sigma\left[\mathrm{Tail}_{\infty}(w;\bar{x},1;-\sigma,\sigma)\right]^{p-1}$
$\displaystyle\leq\gamma\int_{K_{2}(\bar{x})}w^{q}(x,t)\,dx+\gamma\sigma,$
(7.15)

where we used $\sigma\in(0,1)$ and $p>1$. Then, by virtue of the initial data
in (7.13), we evaluate the left-hand side of (7.4) as

$\displaystyle\int_{K_{1}(\bar{x})}w^{q}(x,-\sigma)\,dx=\eta(1-\tau_{*})^{-N}(4r)^{N}=2^{N}\eta\left(1-4^{-\frac{1}{\beta}}\right)^{N}=:\eta
c_{0}.$ (7.16)

Now, we pick $\sigma$ such that $\gamma\sigma=\frac{1}{2}\eta c_{0}$, a
combination of (7.4) and (7.16) leads to

$\displaystyle\frac{1}{2}\eta
c_{0}\leq\gamma\int_{K_{2}(\bar{x})}w^{q}(x,t)\,dx\quad\text{ for all
}t\in[-\sigma,\sigma].$

Applying Theorem 1.5 in
$K_{2}(\bar{x})\times\left[-\frac{1}{4}\sigma,\frac{1}{4}\sigma\right]\subset
K_{4}(\bar{x})\times\left[-\sigma,\frac{1}{2}\sigma\right]$ along with the
initial data in (7.13) gives

$\displaystyle\sup_{K_{2}(\bar{x})\times\left[-\frac{1}{4}\sigma,\frac{1}{4}\sigma\right]}w^{q}$
$\displaystyle\leq\frac{\gamma}{\sigma^{\frac{N}{\lambda}}}\left(\eta
c_{0}\right)^{\frac{p}{\lambda}}+\gamma\sigma^{\frac{q}{q+1-p}}+\gamma\sigma+\gamma\sigma^{\frac{p-N}{\lambda}}$
$\displaystyle\leq\frac{\gamma}{\sigma^{\frac{N}{\lambda}}}\left(\eta
c_{0}\right)^{\frac{p}{\lambda}}+\gamma\sigma^{\frac{p-N}{\lambda}}=\gamma\sigma\times\sigma^{\frac{p-N}{\lambda}-1}\leq\gamma_{1}\sigma,$
(7.17)

where $\gamma_{1}$ depends only on
$N,p,s,q,\Lambda,\|u\|_{L^{\infty}(\mathbb{R}^{N}\times(0,T))}$ due to the
choice of $\sigma$. In light of (7.4), we derive for all
$t\in\left[-\frac{1}{4}\sigma,\frac{1}{4}\sigma\right]$ that

$\displaystyle\frac{1}{2}\eta c_{0}$
$\displaystyle\leq\gamma\int_{K_{2}(\bar{x})}w^{q}(x,t)\,dx$
$\displaystyle=\gamma\int_{K_{2}(\bar{x})\cap[w<b]}w^{q}(x,t)\,dx+\gamma\int_{K_{2}(\bar{x})\cap[w\geq
b]}w^{q}(x,t)\,dx$ $\displaystyle\leq\gamma b^{q}|K_{2}|+\gamma\gamma_{1}\eta
c_{0}\big{|}[w(\cdot,t)\geq b]\cap K_{2}(\bar{x})\big{|}$ (7.18)

with some $b>0$. In fact, we can choose $b$ such that $\gamma
b^{q}|K_{2}|=\tfrac{1}{4}\eta c_{0}$, which together with (7.4) indicates

$\displaystyle|[w(\cdot,t)\geq b]\cap
K_{2}(\bar{x})|\geq\frac{1}{4\gamma_{2}}\quad\text{ for all
}t\in\left[-\tfrac{1}{4}\sigma,\tfrac{1}{4}\sigma\right]$

with $\gamma_{2}=\gamma\gamma_{1}>1$ only depending on $N,p,s,q,\Lambda$ and
$\|u\|_{L^{\infty}(\mathbb{R}^{N}\times(0,T))}$. With the above measure
information at hand, the pointwise estimate of the function $w$ follows from
Proposition 6.1, consequently, the estimate of $v$. Finally, we establish the
left-hand side Harnack inequality in Theorem 1.6 by the definition of $v$, and
by choosing $\gamma,\sigma$ properly.

The argument to obtain Harnack inequality on the right-hand side is standard
by using the continuity of weak solutions together with the Harnack inequality
on the left-hand side, we refer the readers to [6, pages $85$–$86$]. Up to
now, we have finished the proof. ∎

### Conflict of interest

The authors declare that there is no conflict of interest.

### Data availability

No data was used for the research described in the article.

### Acknowledgments

This work was supported by the National Natural Science Foundation of China
(No. 12071098) and the Fundamental Research Funds for the Central Universities
(No. 2022FRFK060022). The paper was done when the second author visited
Department of Mathematics, University of Craiova. She would like to thank its
hospitality during her stay and thank the China Scholarship Council (No.
202306120202). The research of V.D. Rădulescu was supported by the grant
“Nonlinear Differential Systems in Applied Sciences” of the Romanian Ministry
of Research, Innovation and Digitization, within PNRR-III-C9-2022-I8/22.

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