# Entropy solutions to the fully non-local diffusion equations

Ying Li, Chao Zhang∗ Ying Li
School of Mathematics, Harbin Institute of Technology, Harbin 150001, China
<EMAIL_ADDRESS>Chao Zhang
School of Mathematics and Institute for Advanced Study in Mathematics, Harbin
Institute of Technology, Harbin 150001, China
<EMAIL_ADDRESS>
###### Abstract.

We consider the fully non-local diffusion equations with non-negative
$L^{1}$-data. Based on the approximation and energy methods, we prove the
existence and uniqueness of non-negative entropy solutions for such problems.
In particular, our results are valid for the time-space fractional Laplacian
equations.

∗ Corresponding author.

Keywords: Non-local diffusion; fractional partial differential equations;
entropy solution; existence; $L^{1}$-data.

MSC 2020: 35D99, 35R11, 45K05.

Acknowledgment: This work was supported by the National Natural Science
Foundation of China (No. 12071098).

##  1\. Introduction

Let $\Omega$ be a bounded domain of ${\mathbb{R}^{N}}(N>2)$. The present work
is concerned with the existence of solutions to the following integro-
differential equation

$\left\\{\begin{array}[]{cl}\partial_{t}(k*(u-u_{0}))+(-\Delta)^{s}u=f&\quad\text{in
}\Omega_{T}\equiv\Omega\times(0,T),\\\ u=0&\quad\text{in
}({\mathbb{R}^{N}}\setminus\Omega)\times(0,T),\\\
u(x,0)=u_{0}(x)&\quad\text{in }\Omega.\end{array}\right.$ (1.1)

Here, $k\in L^{1}_{loc}(\mathbb{R^{+}})$ is a non-negative kernel that belongs
to a certain kernel class, and $k*v$ denotes the convolution on the positive
half-line w.r.t the time variable, that is
$(k*v)(t)=\int^{t}_{0}k(t-\tau)v(\tau)\,d\tau$ with $t\geq 0$. The fractional
Laplacian operator $(-\Delta)^{s}$ with $0<s<1$ is defined as

$\displaystyle(-\Delta)^{s}u(x,t):=$ $\displaystyle\quad{\rm
P.V.}\int_{{\mathbb{R}^{N}}}\frac{u(x,t)-u(y,t)}{|x-y|^{N+2s}}\,dy$
$\displaystyle=$ $\displaystyle\quad\lim_{\epsilon\to
0}\int_{{\mathbb{R}^{N}}\setminus
B_{\epsilon}(x)}\frac{u(x,t)-u(y,t)}{|x-y|^{N+2s}}\,dy,$

where $(x,t)\in{\mathbb{R}^{N}}\times\mathbb{R^{+}}$ and P.V. is a commonly
used abbreviation in the principal value sense. We consider the above problem
with non-negative $L^{1}$-data, i.e.,

$0\leq f\in L^{1}(\Omega_{T})\quad\text{and}\quad 0\leq u_{0}\in
L^{1}(\Omega).$

The kernel $k$ satisfies the following condition:

  * (Hk)

$k$ is non-negative, non-increasing, and there exists a kernel $l\in
L^{p}(0,T)$ with $p>1$ such that $k*l=1$ in $(0,\infty)$.

Let $k_{\lambda}$ be the kernels which arised in characterisation (2.2) of the
Yosida approximation. To be able to show the existence of entropy solutions to
(1.1), we further assume that $k$ satisfies the following conditions:

  * (K1)

There exist constants $C_{1},C_{2}>0$ such that

$0\leq k_{\lambda}(t)\leq C_{1}k(t)+C_{2},~{}\lambda>0,~{}t\in(0,T);$

  * (K2)

$k\in{\rm AC}_{loc}((0,T])$ and there exist constants $C_{1},C_{2}>0$ such
that

$0\leq-k^{\prime}_{\lambda}(t)\leq-
C_{1}k^{\prime}(t)+C_{2},~{}\lambda>0,~{}t\in(0,T),$

where $k^{\prime}$ represents the differentiation of $k$. Moreover, the
convergence $k^{\prime}_{\lambda}(t)\to k^{\prime}(t)$ as $\lambda\to 0$ holds
for almost every $t\in(0,T)$.

Note that kernels $k$ satisfying $\rm(Hk)$ are in particular kernels of type
$\mathcal{PC}$, see Definition 2.1. The kernels of type $\mathcal{PC}$ was
introduced by Zacher in [45]. $\mathcal{PC}$-kernels cover most of the
relevant integro-differential operators w.r.t. time that appear in physics
applications in the context of subdiffusion processes, which is a special case
of anomalous diffusive behaviour. We refer to [34, 35] and references therein
for the physical background. There have been wide research activities [25, 36,
42, 46, 44] on the kernels of $\mathcal{PC}$ type.

Next, we shall give some examples of $\mathcal{PC}$-kernels that satisfy the
conditions (Hk), (K1) and (K2).

######  Example 1.1.

The most typical example is given by

$k(t)=g_{1-\alpha}(t),~{}l(t)=g_{\alpha}(t),\quad t>0,$

where $\alpha\in(0,1)$ and $g_{\beta}$ denotes the Riemann-Liouville kernel

$g_{\beta}=\frac{t^{\beta-1}}{\Gamma(\beta)},\quad t>0,\quad\beta>0.$

In this case, the term $\partial_{t}(k*\cdot)$ denotes the fractional time
derivative of order $\alpha$ in the sense of Riemann-Liouville, and
$k*\partial_{t}(\cdot)$ the Caputo derivative if $v$ is sufficient smooth, see
[39, Example 2.7].

######  Example 1.2.

Another example is the time-fractional case with exponential weight, i.e.,

$k(t)=g_{1-\alpha}(t)e^{-\mu t},\quad l(t)=g_{\alpha}(t)e^{-\mu
t}+\mu(1*[g_{\alpha}(\cdot)e^{-\mu\cdot}])(t),\quad t>0,$

where $\mu>0$ and $\alpha\in(0,1)$. We refer to [39, Example 2.8] for more
information.

Our focus in this manuscript is to establish the existence and uniqueness of
solutions for problem (1.1). As we consider problem with data of low
integrability, it is reasonable to work with entropy solutions, which need
less regularity of the data than usual weak solutions. The notion of entropy
solutions has been proposed by Bénilan et al.[6] for the study of the
nonlinear elliptic problems. There already exists a vast literature that
concerns about the entropy solutions for elliptic and parabolic problems with
$L^{1}$-data, see [5, 7, 15, 32, 47, 48].

In recent years, the study of the fractional Laplacian operators and non-local
operators have received significant attention. The main reason lies in its
wide range of applications, such as continuum mechanics, phase transition
phenomena, population dynamics, image process, game theory and so on. We refer
to [8, 9, 10, 35] and the references therein for more details. We point out
that the linear elliptic problems with non-local operator $(-\Delta)^{s}_{p}$
in the case $p=2$ were considered in [3, 24, 31]. The corresponding parabolic
problem was also discussed in [31]. It is worth mentioning that in [3],
Alibaud et al. established the existence and uniqueness of solution for the
problem

$\beta(u)+(-\Delta)^{s}u\ni f\quad\text{in}~{}\mathbb{R}^{N},$

where $f\in L^{1}(\mathbb{R}^{N})$ and $\beta$ is a maximal monotone graph in
$\mathbb{R}$. Moreover, there are a large number of papers devoted to the
study of elliptic and parabolic equations with non-local operators
$(-\Delta)^{s}_{p}$ in the case $p\neq 2$. The existence of a unique entropy
positive solution to fractional elliptic $p$-Laplacian equation with weight
and general datum was developed by Abdellaoui et al. [2], see also [1] for the
parabolic framework. For more recent works concern about the non-local
operator $(-\Delta)^{s}_{p}$, we refer the reader to [28, 29, 33, 40].

As to the problems with integro-differential operators, we mention that
Jakubowski and Wittbold [23] generalized the concept of entropy solutions for
parabolic equations with $L^{1}$-data and considered a class of nonlinear
history-dependent degenerated elliptic-parabolic equation. The existence of
entropy solutions for the doubly nonlinear history-dependent problems of the
form

$\partial_{t}[k*(b(u)-b(u_{0}))]-\operatorname{div}a(x,\nabla u)=f$

was obtained by Scholtes and Wittbold in [39], where $k$ is assumed to be of
type $\mathcal{PC}$, $b$ is strictly increasing and continuous satisfying
$b(0)=0$, and $a$ is a Carathédory function satisfying the Leary-Lions
condition. Sapountzoglou [37] generalized this result to the case that $b$ is
not strictly increasing. In addition, [38, 43] were devoted to the study of
existence of solutions for the time-fractional porous medium type equations,
which were equipped with bounded measurable coefficients that may explicitly
depend on time. Moreover, we point out that an abstract evolutionary integro-
differential equation in Hilbert spaces with a kernel $k$ of type
$\mathcal{PC}$ was investigated in [45]. Vergara and Zacher [42] have studied
the decay estimates of time-fractional equations via energy methods, see also
[25] on this issue.

The equation which is non-local both in space and time is called the fully
non-local diffusion equation. Fully non-local equations play a crucial role in
model situations with long-range interactions and memory effects and have been
proposed to describe plasma transport, see [16, 17]. The study of fully non-
local equations has attracted considerable interest during recent years.
Allen, Caffarelli and Vasseur [4] have discussed the regularity of weak
solutions to a parabolic problem with fractional diffusion in space and a
fractional time derivative. Kim and Lim [27] have considered the behavior of
fundamental solutions to time-space fractional differential equations. The
decay estimates for time-space fractional equations were investigated in [26,
20]. For other recent developments on the issues of the fully non-local
problems, let us refer to [21, 18, 14].

To our best knowledge, there is no result concerns about the fully non-local
problems with the right-hand side is merely integrable. Inspired by the papers
mentioned above, we aim to establish the existence and uniqueness results of
entropy solution for the fully non-local equation (1.1). We point out that we
cannot use the method of Steklov average in time to obtain the appropriate
time regularization since the Steklov average operators and convolution do not
commute. We shall introduce a regularization in time which adapts to the non-
local nature of the problem (1.1), see Definition 2.2. The method is a
modification of the regularization method proposed by Landes [30] and has been
used in [39, 37]. It is important to note that the authors in [39, 37]
obtained the existence and uniqueness of entropy solutions for the time-
fractional problems by using the theory of accretive operators. They showed
that the generalized solution of the associated abstract Volterra equation is
an entropy solution. Unlike the proofs of [39, 37], we first construct an
approximate problem of (1.1) and obtain the existence and uniqueness of weak
solutions for the approximate problem by using the results in [45]. Next, we
shall establish some a priori estimates for the approximate solution sequence.
Then we draw a subsequence to obtain a limit function and prove that the limit
function is the entropy solution for problem (1.1). The uniqueness of the
entropy solution is obtained by a comparison principle. In addition, we use
the method of Yosida approximation to regularize the kernel $k$ which will be
discussed in Section 2. This method has already been used in [37, 46, 45, 41].

Before giving the main results of this manuscript, we first introduce some
functional settings that will be used below. We refer to [19, 31, 40] for more
details.

Let $s\in(0,1)$ and $p>1$. The fractional Sobolev space

$W^{s,p}(\mathbb{R}^{N})\equiv\bigg{\\{}u\in
L^{p}(\mathbb{R}^{N}):\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}\frac{|u(x)-u(y)|^{p}}{|x-y|^{N+ps}}\,dxdy<\infty\bigg{\\}}$

is a Banach space endowed with the norm

$\|u\|_{W^{s,p}(\mathbb{R}^{N})}=\|u\|_{L^{p}(\mathbb{R}^{N})}+\bigg{(}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}\frac{|u(x)-u(y)|^{p}}{|x-y|^{N+ps}}\,dxdy\bigg{)}^{\frac{1}{p}}.$

For $\omega\in W^{s,p}(\mathbb{R}^{N})$, we define the fractional
$p$-Laplacian as

$(-\Delta)^{s}_{p}\omega(x)={\rm
P.V.}\int_{\mathbb{R}^{N}}\frac{|\omega(x)-\omega(y)|^{p-2}(\omega(x)-\omega(y))}{|x-y|^{N+ps}}\,dy.$

Note that for all $\omega,v\in W^{s,p}(\mathbb{R}^{N})$, we have

$\langle(-\Delta)^{s}_{p}\omega(x),v\rangle=\frac{1}{2}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}|\omega(x)-\omega(y)|^{p-2}(\omega(x)-\omega(y))(v(x)-v(y))\,dv.$

Denote
$\mathcal{D}_{\Omega}=(\mathbb{R}^{N}\times\mathbb{R}^{N})\setminus(\mathcal{C}\Omega\times\mathcal{C}\Omega)$,
where $\mathcal{C}\Omega=\mathbb{R}^{N}\setminus\Omega$. $X^{s,p}$ denotes the
linear space of Lebesuge measurable function $u:\mathbb{R}^{N}\to\mathbb{R}$
such that the quantity

$X^{s,p}(\Omega)=\bigg{(}\int_{\Omega}|u|^{p}\,dx+\int_{\mathcal{D}_{\Omega}}\frac{|u(x)-u(y)|^{p}}{|x-y|^{N+sp}}\,dxdy\bigg{)}$

is finite. $X_{0}^{s,p}$ denotes the space of functions $u\in X^{s,p}$ that
vanish a.e. in $\mathcal{C}\Omega$. As it is explained in [19, 40], we know
that there exists a positive constant $C$ such that for any $u\in
X^{s,p}_{0}(\Omega)$,

$\int_{\mathcal{D}_{\Omega}}|u(x)-u(y)|^{p}\,dv\leq\|u\|^{p}_{W^{s,p}(\mathbb{R}^{N})}\leq
C\int_{\mathcal{D}_{\Omega}}|u(x)-u(y)|^{p}\,dv,$

where

$dv=\frac{dxdy}{|x-y|^{N+sp}}.$

Thus, we can endow $X^{s,p}_{0}(\Omega)$ with the equivalent norm

$\|u\|_{X^{s,p}_{0}(\Omega)}=\bigg{(}\int_{\mathcal{D}_{\Omega}}|u(x)-u(y)|^{p}\,dv\bigg{)}^{\frac{1}{p}}.$

It is obvious that for all $\omega,v\in X_{0}^{s,p}(\Omega)$,

$\langle(-\Delta)^{s}_{p}\omega(x),v\rangle=\frac{1}{2}\int_{\mathcal{D}_{\Omega}}|\omega(x)-\omega(y)|^{p-2}(\omega(x)-\omega(y))(v(x)-v(y))\,dv.$

Moreover, we notice that
$(-\Delta)^{s}_{p}:X^{s,p}_{0}(\Omega)\hookrightarrow(X^{s,p}_{0}(\Omega))^{\prime}$,
where $(X^{s,p}_{0}(\Omega))^{\prime}$ denotes the dual space of
$X^{s,p}_{0}(\Omega)$. In addition, we claim that the space
$L^{p}(0,T;X^{s,p}_{0}(\Omega))$ is defined as the set of function $u$ such
that $u\in L^{p}(\Omega_{T})$ with
$\|u\|_{L^{p}(0,T;X^{s,p}_{0}(\Omega))}<\infty$, where

$\|u\|_{L^{p}(0,T;X^{s,p}_{0}(\Omega))}=\bigg{(}\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}|u(x,t)-u(y,t)|^{p}\,dvdt\bigg{)}^{\frac{1}{p}}.$

We define an entropy solution to problem (1.1) based on the definition in
[23]. We set

$\mathcal{P}:=\\{S\in C^{1}{(\mathbb{R})}:0\leq S^{\prime}\leq
1,\quad\text{supp}\,S^{\prime}~{}\text{compact},\quad S(0)=0\\}.$

In the following definition and throughout the paper, $T_{K}$ denotes the
truncation function at level $K>0$:

$T_{K}(r):=\max\\{\min(r,K),-K\\}.$

Moreover, we denote $T_{K,M}(r):=T_{M}(r)-T_{K}(r)$ for $M>K>0$.

######  Definition 1.1.

A measurable function $u\in L^{1}(\Omega_{T})$ with $T_{k}(u)\in
L^{2}(0,T;X^{s,2}_{0}(\Omega))$ is called an entropy solution to (1.1) if

$\displaystyle-\int^{T}_{0}\int_{\Omega}\zeta_{t}\bigg{[}k_{1}*\int^{u}_{u_{0}}S(\sigma-\phi)\,d\sigma\bigg{]}\,dxdt+\int^{T}_{0}\int_{\Omega}\zeta\partial_{t}[k_{2}*(u-u_{0})]S(u-\phi)\,dxdt$
(1.2)
$\displaystyle\quad+\frac{1}{2}\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}(u(x)-u(y))\bigg{(}S(u(x,t)-\phi(x,t))-S(u(y,t)-\phi(y,t))\bigg{)}\zeta\,dvdt$
$\displaystyle\quad\leq\int^{T}_{0}\int_{\Omega}\zeta fS(u-\phi)\,dxdt$

for all $\phi\in X^{s,2}_{0}(\Omega)\cap L^{\infty}(\Omega)$, $\zeta\in
C^{\infty}_{0}([0,T))$, $\zeta\geq 0$, $S\in\mathcal{P}$, and $k_{1},k_{2}\in
L^{1}(0,T)$ non-increasing and non-negative with $k=k_{1}+k_{2}$ and
$k_{2}(0^{+})<\infty$.

Our main results read as follows.

######  Theorem 1.2.

Let $0\leq f\in L^{1}(\Omega_{T})$, $0\leq u_{0}\in L^{1}(\Omega)$ and let $k$
satisfy the conditions (Hk), (K1) and (K2). Then there exists an entropy
solution for problem (1.1).

######  Proposition 1.3 ( Comparison principle).

Suppose that $u^{i}(i=1,2)$ are entropy solutions to problem (1.1) with
$f=f^{i}\in L^{1}(\Omega_{T})$ and $u_{0}=u^{i}_{0}\in L^{1}(\Omega)$. Then we
have

$\int^{T}_{0}\int_{\Omega}(u^{1}-u^{2})^{+}\,dxdt\leq
T\int_{\Omega}(u_{0}^{1}-u^{2}_{0})^{+}\,dx+\|l\|_{L^{1}(0,T)}\int^{T}_{0}\int_{\Omega}(f^{1}-f^{2})^{+}\,dxdt$
(1.3)

and

$\|u^{1}-u^{2}\|_{L^{1}(\Omega_{T})}\leq
T\|u_{0}^{1}-u^{2}_{0}\|_{L^{1}(\Omega)}+\|l\|_{L^{1}(0,T)}\|f^{1}-f^{2}\|_{L^{1}(\Omega_{T})}.$
(1.4)

######  Remark 1.4.

Proposition 1.3 provides that the entropy solution of problem (1.1) is unique
and non-negative.

##  2\. Preliminaries

In this section, we shall give a short introduction to kernels of type
$\mathcal{PC}$. For detailed discussion we refer the reader to [11, 41, 45].
We begin with the definition of the $\mathcal{PC}$-kernels.

######  Definition 2.1.

A kernel $k\in L^{1}_{loc}([0,\infty))$ is called to be of type $\mathcal{PC}$
if it is non-negative, non-increasing, and there exists a kernel $l\in
L^{1}_{loc}([0,\infty))$ such that

$(k*l)(t)=1,\quad\forall t>0.$

We write it as $(k,l)\in\mathcal{PC}$.

We next introduce an important method of approximation kernels of type
$\mathcal{PC}$. For $1\leq p<\infty$, $T>0$, and a real Banach space $X$ we
consider the operator $B$ defined by

$Bu=\partial_{t}(k*u),\quad D(B)=\\{u\in
L^{p}(0,T;X):k*u\in{}_{0}W^{1,p}(0,T;X)\\},$ (2.1)

where the zero means vanishing at $t=0$. It is known that this operator is
$m$-accretive in $L^{p}(0,T;X)$, see [12, 13]. Its Yosida approximation
$B_{\lambda}$, defined by $B_{\lambda}=\frac{B}{I+\lambda B}$, $\lambda>0$,
enjoy the property that for any $u\in D(B)$,

$B_{\lambda}u\to Bu\quad\text{in}~{}L^{p}(0,T;X)\text{ as}~{}\lambda\to 0.$

Denote $J^{B}_{\lambda}=(I+\lambda B)^{-1}$, $\lambda>0$, it had been proved
in [41, Theorem 2.1] that the Yosida approximation
$B_{\lambda}=BJ^{B}_{\lambda}$ could be wirtten in the form

$B_{\lambda}u=\frac{B}{I+\lambda
B}u=BJ^{B}_{\lambda}u=\partial_{t}(k_{\lambda}*u),$ (2.2)

where $k_{\lambda}=\frac{s_{\lambda}}{\lambda}=k*r_{\lambda}$, and
$s_{\lambda},r_{\lambda}\in L^{1}_{loc}({\mathbb{R}}^{+})$ are defined by the
scalar Volterra equations

$\begin{split}&s_{\lambda}(t)+\lambda^{-1}(l*s_{\lambda})(t)=1,\qquad t>0,\\\
&r_{\lambda}(t)+\lambda^{-1}(l*r_{\lambda})(t)=\lambda^{-1}l(t),\qquad
t>0.\end{split}$

Moreover, we have

$J^{B}_{\lambda}u=\frac{u}{(I+\lambda B)}=r_{\lambda}*u.$

In addition, according to the formula for the variation of the constants of
the Volterra integral equation [22, Theorem 3.5], it follows that

$s_{\lambda}(t)=1-\int^{t}_{0}r_{\lambda}(\tau)d\tau,~{}t>0,\lambda>0.$

Thus, we know that $s_{\lambda}\in W^{1,1}(0,T)$. We point out that both
$s_{\lambda}$ and $r_{\lambda}$ are non-negative for all $\lambda>0$. This is
a consequence of $l$ is completely positive as $(k,l)\in\mathcal{PC}$, see
[11, Theorem 2.2]. It is further known that the kernels $k_{\lambda}$ are also
non-negative and non-increasing and belong to $W^{1,1}(0,T)$. We note that for
any $v\in L^{p}(0,T;X)$,

$J^{B}_{\lambda}v=r_{\lambda}*v\to v\quad\text{in
}L^{p}(0,T;X)~{}\text{as}~{}\lambda\to 0.$ (2.3)

In particular, we have

$k_{\lambda}=r_{\lambda}*k\to
k\quad\text{in}~{}L^{1}(0,T)~{}\text{as}~{}\lambda\to 0.$ (2.4)

In fact, for any $v\in L^{p}(0,T;X)$, it is easy to check that

$k*l*v=1*v=\int^{t}_{0}v(\tau)\,d\tau\in{}_{0}W^{1,p}(0,T;X),$

which implies that

$l*v\in D(B)=\\{\omega\in L^{p}(0,T;X):k*\omega\in{}_{0}W^{1,p}(0,T;X)\\}.$

Hence, we deduce that

$r_{\lambda}*v=\partial_{t}(k*r_{\lambda}*l*v)=\partial_{t}(k_{\lambda}*l*v)=B_{\lambda}(l*v)\to
B(l*v)=v$

in $L^{p}(0,T;X)$ as $\lambda\to 0$.

Next, we will give a regularization in time which adapts to the nonlocal
nature of the problem (1.1). We point out that the method is a modification of
the regularzation method proposed by Landes [30].

######  Definition 2.2.

Let $X$ be a Banach space, $X^{\prime}$ its dual and $1\leq
p^{\prime}<\infty$. For $v\in L^{p^{\prime}}(0,T;X^{\prime})$, let $v_{\mu}\in
L^{p^{\prime}}(0,T;X^{\prime})$ be defined by

$v_{\mu}(t)=\int^{T}_{t}r_{\mu}(\tau-t)v(\tau)\,d\tau,~{}t\in(0,T),~{}\mu>0.$
(2.5)

Note that $v_{\mu}=J^{B^{*}}_{\mu}v$ for $p^{\prime}=\frac{p}{p-1}$, where
$B^{*}:D(B^{*})\subset L^{p^{\prime}}(0,T;X^{\prime})\to
L^{p^{\prime}}(0,T;X^{\prime})$ is the adjoint of the operator $B$ defined in
(2.1). Thus, we have $v_{\mu}\to v$ in $L^{p^{\prime}}(0,T;X^{\prime})$ for
any $v\in L^{p^{\prime}}(0,T;X^{\prime})$ as $\mu\to 0$.

######  Lemma 2.3.

[39, Lemma 2.3] Let $\mu>0$, and assume that $v\in L^{\infty}(\Omega_{T})$.
Then $v_{\mu}\in L^{\infty}(\Omega_{T})$ and

$\|v_{\mu}\|_{L^{\infty}(\Omega_{T})}\leq\|v\|_{L^{\infty}(\Omega_{T})}.$

Then, we give a fundamental identify for integro-differential operator of the
form $\partial_{t}(k*u)$.

######  Lemma 2.4.

Let $k\in W^{1,1}(0,T)$, $H\in C^{1}(\mathbb{R})$ and $u\in L^{1}(0,T)$ with
$u(t)\in\mathbb{R}$ for almost all $t\in(0,T)$. Suppose that the functions
$H(u)$, $H^{\prime}(u)u$ and $H^{\prime}(u)(k^{\prime}*u)$ belong to
$L^{1}(0,T)$. Then we have for almost all $t\in(0,T)$,

$\begin{split}&H^{\prime}(u(t))\partial_{t}(k*u)(t)\\\
&\quad=\partial_{t}(k*H(u))(t)+(H^{\prime}(u(t))u(t)-H(u(t)))k(t)\\\
&\qquad+\int^{t}_{0}\bigg{(}H(u(t-s))-H(u(t))-H^{\prime}(u(t))[u(t-s)-u(t)]\bigg{)}[-k^{\prime}(s)]\,ds.\end{split}$
(2.6)

The assertion of Lemma 2.4 follows from a straightforward computation, see
also [46, 44]. We note that a more general version of (2.6) in integrated form
can be found in [22, Lemma 18.4.1]. The following two lemmas are immediate
consequence of (2.6).

######  Lemma 2.5.

[25, Corollary 6.1] Let $u_{0}\in\mathbb{R}$, $k$, $H$ and $u$ be as in Lemma
2.4. And assume in addition that $k$ is non-negative and non-increasing and
$H$ is convex. Then

$H^{\prime}(u(t))\partial_{t}(k*(u-u_{0}))(t)\geq\partial_{t}(k*[H(u)-H(u_{0})])(t)~{}\text{a.a.}~{}t\in(0,T).$
(2.7)

######  Lemma 2.6.

[39, Lemma 2.5] Let $u\in L^{1}(0,T)$, $k$ a $\mathcal{PC}$-kernel, and for
$\lambda>0$, let $k_{\lambda}$ be the kernel of the Yosida approximation of
the operator defined in (2.2). Then we have for all $K>0$ and almost every
$t>0$,

$\begin{split}&\partial_{t}[k_{\lambda}*u](t)T_{K}(u(t))\\\
&\quad=\partial_{t}[k_{\lambda}*\int^{u}_{0}T_{K}(\sigma)\,d\sigma](t)+\bigg{[}T_{K}(u(t))u-\int^{u}_{0}T_{K}(\sigma)\,d\sigma\bigg{]}k_{\lambda}(t)\\\
&\qquad+\int^{t}_{0}\bigg{[}\int^{u(t-s)}_{u(t)}T_{K}(\sigma)\,d\sigma-
T_{K}(u(t))(u(t-s)-u(t))\bigg{]}[-k^{\prime}_{\lambda}(s)]\,ds\\\
&\quad\geq\partial_{t}\left[k_{\lambda}*\int^{u}_{0}T_{K}(\sigma)\,d\sigma\right](t).\end{split}$
(2.8)

##  3\. Weak solutions

In this section, we will introduce an approximate problem. For
$m\in\mathbb{N}$ we define $f^{m}=T_{m}(f)$ and $u^{m}_{0}=T_{m}(u_{0})$, then
we know that $f^{m}$ and $u^{m}_{0}$ are non-negative, $(f^{m},u^{m}_{0})\in
L^{\infty}(\Omega_{T})\times L^{\infty}(\Omega)$ and
$(f^{m},u^{m}_{0})\nearrow(f,u_{0})$ strongly in $L^{1}(\Omega_{T})\times
L^{1}(\Omega)$ such that

$\|f^{m}\|_{L^{1}(\Omega_{T})}\leq\|f\|_{L^{1}(\Omega_{T})},\quad\|u^{m}_{0}\|_{L^{1}(\Omega)}\leq\|u_{0}\|_{L^{1}(\Omega)}.$
(3.1)

We set

$W(u_{0},X^{s,2}_{0}(\Omega),L^{2}(\Omega)):=\\{\omega\in
L^{2}(0,T;X_{0}^{s,2}(\Omega)):k*(\omega-
u_{0})\in{}_{0}W^{1,2}(0,T;(X_{0}^{s,2}(\Omega))^{\prime})\\}.$

Then we consider the approximate problem of (1.1).

######  Lemma 3.1.

Let (Hk) be satisfied. Then the following problem

$\left\\{\begin{array}[]{cl}\partial_{t}(k*(u^{m}-u^{m}_{0}))+(-\Delta)^{s}u^{m}=f^{m}&\quad\text{in
}\Omega_{T}\equiv\Omega\times(0,T),\\\ u^{m}=0&\quad\text{in
}({\mathbb{R}^{N}}\setminus\Omega)\times(0,T),\\\
u^{m}(x,0)=u^{m}_{0}(x)&\quad\text{in }\Omega\end{array}\right.$ (3.2)

admits a unique weak solution $u^{m}\in
W(u_{0},X^{s,2}_{0}(\Omega),L^{2}(\Omega))$ with $k*u^{m}\in
C([0,T];L^{2}(\Omega))$ such that for any $\phi\in
L^{2}(0,T;X^{s,2}_{0}(\Omega))\cap L^{\infty}(\Omega_{T})$, $t_{1}\in(0,T)$,

$\begin{split}&\int^{t_{1}}_{0}\int_{\Omega}\phi\partial_{t}[k*(u^{m}-u^{m}_{0})]\,dxdt+\frac{1}{2}\int^{t_{1}}_{0}\int_{\mathcal{D}_{\Omega}}(u^{m}(x)-u^{m}(y))(\phi(x)-\phi(y))\,dvdt\\\
&\quad=\int_{0}^{t_{1}}\int_{\Omega}f^{m}\phi\,dxdt\end{split}$ (3.3)

holds.

###### Proof.

Arguing as the arguments of [45, Theorem 3.1], we can find a unique weak
solution $u^{m}\in W(u^{m}_{0},X^{s,2}_{0}(\Omega),L^{2}(\Omega))$ for problem
(3.2). ∎

######  Lemma 3.2 (Comparison principle).

Suppose that $u^{m}_{i}(i=1,2)$ are weak solutions to problem (1.1) with
$u^{m}_{0}=u^{m}_{0i}\in L^{\infty}(\Omega)$ and $f=f^{m}_{i}\in
L^{\infty}(\Omega_{T})$. Then, we have

$\displaystyle\int^{T}_{0}\int_{\Omega}(u^{m}_{1}-u^{m}_{2})^{+}\,dxdt\leq
T\int_{\Omega}(u^{m}_{01}-u^{m}_{02})^{+}\,dx+\|l\|_{L^{1}(0,T)}\int^{T}_{0}\int_{\Omega}(f^{m}_{1}-f^{m}_{2})^{+}\,dx$
(3.4)

and

$\displaystyle\|u^{m}_{1}-u^{m}_{2}\|_{L^{1}(\Omega_{T})}\leq
T\|u^{m}_{01}-u^{m}_{02}\|_{L^{1}(\Omega)}+\|l\|_{L^{1}(0,T)}\|f^{m}_{1}-f^{m}_{2}\|_{L^{1}(\Omega_{T})}.$
(3.5)

###### Proof.

We shall give the proof of (3.4) and the proof of (3.5) is similar to (3.4).

Set

$v=u^{m}_{1}-u^{m}_{2},\quad v_{0}=u^{m}_{01}-u^{m}_{02}\quad and\quad
F=f^{m}_{1}-f^{m}_{2}.$

For $\epsilon>0$, we define a convex function

$H_{\epsilon}(y)=\sqrt{y^{2}+\epsilon^{2}}-\epsilon,\quad y\in{\mathbb{R}}.$
(3.6)

After computation, we can get

$H^{\prime}_{\epsilon}(y)=\frac{y}{\sqrt{y^{2}+\epsilon^{2}}},\quad
H^{\prime\prime}_{\epsilon}(y)=\frac{\epsilon^{2}}{(y^{2}+\epsilon^{2})^{\frac{3}{2}}},\quad
y\in{\mathbb{R}}.$

Taking $H^{\prime}_{\epsilon}(v^{+})\in L^{2}(0,T;X^{s,2}_{0}(\Omega))\cap
L^{\infty}(\Omega_{T})$ as a test function in the weak formulation of the
problem for both $u^{m}_{1}$ and $u^{m}_{2}$, we deduce that

$\begin{split}&\int^{t_{1}}_{0}\int_{\Omega}H^{\prime}_{\epsilon}(v^{+})\partial_{t}[k*(v-v_{0})]\,dxdt+\frac{1}{2}\int^{t_{1}}_{0}\int_{\mathcal{D}_{\Omega}}(v(x)-v(y))(H^{\prime}_{\epsilon}(v^{+}(x))-H^{\prime}_{\epsilon}(v^{+}(y)))\,dvdt\\\
&\quad=\int_{0}^{t_{1}}\int_{\Omega}FH^{\prime}_{\epsilon}(v^{+})\,dxdt.\end{split}$

It is not difficult to see that

$\begin{split}&\int^{t_{1}}_{0}\int_{\mathcal{D}_{\Omega}}(v(x)-v(y))(H^{\prime}_{\epsilon}(v^{+}(x))-H^{\prime}_{\epsilon}(v^{+}(y)))\,dvdt\\\
&\quad=\int^{t_{1}}_{0}\int_{\mathcal{D}_{\Omega}}(v(x)-v(y))(v^{+}(x)-v^{+}(y))H_{\epsilon}^{\prime\prime}(\xi)\,dvdt\geq
0,\end{split}$

where $\xi$ is between $v^{+}(x)$ and $v^{+}(y)$. Combining with the fact that
$0<H^{\prime}_{\epsilon}(v^{+})\leq 1$, we deduce that

$\int^{t_{1}}_{0}\int_{\Omega}H^{\prime}_{\epsilon}(v^{+})\partial_{t}[k*(v-v_{0})]\,dxdt\leq\int_{0}^{t}\int_{\Omega}FH^{\prime}_{\epsilon}(v^{+})\,dxdt\leq\int^{t_{1}}_{0}\int_{\Omega}F^{+}\,dxdt.$
(3.7)

For $\lambda>0$, let $k_{\lambda}$ be the kernel associated to the Yosida
approximation of the operator

$\displaystyle B_{1}\omega$ $\displaystyle:=$
$\displaystyle\partial_{t}(k*\omega)$ $\displaystyle D(B_{1})$
$\displaystyle:=$ $\displaystyle\\{\omega\in
L^{2}(0,T;(X_{0}^{s,2}(\Omega))^{\prime}):k*\omega\in{}_{0}W^{1,2}(0,T;(X_{0}^{s,2}(\Omega))^{\prime})\\}.$

Note that

$\begin{split}\int^{t_{1}}_{0}\int_{\Omega}H^{\prime}_{\epsilon}(v^{+})\partial_{t}[k*(v-v_{0})]\,dxdt&=\int^{t_{1}}_{0}\int_{\Omega}H^{\prime}_{\epsilon}(v^{+})\partial_{t}[k_{\lambda}*(v-v_{0})]\,dxdt\\\
&\quad-\int^{t_{1}}_{0}\int_{\Omega}H^{\prime}_{\epsilon}(v^{+})\partial_{t}[(k_{\lambda}-k)*(v-v_{0})]\,dxdt.\end{split}$

Since $H^{\prime}_{\epsilon}(v^{+})$ is a convex function, it follows from
Lemma 2.5 that

$\begin{split}\int^{t_{1}}_{0}\int_{\Omega}H^{\prime}_{\epsilon}(v^{+})\partial_{t}[k_{\lambda}*(v-v_{0})]\,dxdt&\geq\int^{t_{1}}_{0}\int_{\Omega}\partial_{t}[k_{\lambda}*(H_{\epsilon}(v^{+})-H_{\epsilon}(v^{+}_{0})]\,dxdt\\\
&=\int_{\Omega}k_{\lambda}*(H_{\epsilon}(v^{+})-H_{\epsilon}(v^{+}_{0}))(t_{1},x)\,dx\end{split}$

for $t_{1}\in(0,T)$. Thus, we infer that

$\int_{\Omega}k_{\lambda}*(H_{\epsilon}(v^{+})-H_{\epsilon}(v^{+}_{0}))(t_{1},x)\,dx\leq\int^{t_{1}}_{0}\int_{\Omega}F^{+}\,dxdt+\int^{t_{1}}_{0}\int_{\Omega}H^{\prime}_{\epsilon}(v^{+})\partial_{t}[(k_{\lambda}-k)*(v-v_{0})]\,dxdt.$

According to the fact that $k_{\lambda}\to k$ in $L^{1}(0,T)$ as $\lambda\to
0$, it follows from the Young’s inequality that

$\int_{\Omega}k_{\lambda}*(H_{\epsilon}(v^{+})-H_{\epsilon}(v^{+}_{0}))(t_{1},x)\,dx\to\int_{\Omega}k*(H_{\epsilon}(u^{+})-H_{\epsilon}(v^{+}_{0}))(t_{1},x)\,dx$

in $L^{1}(0,T)$, and selecting a subsequence if necessary, a.e. on $(0,T)$. In
addition, we notice that
$k*(v-v_{0})\in{}_{0}W^{1,2}(0,T;(X_{0}^{s,2}(\Omega))^{\prime})$. Thus, we
deduce that

$\partial_{t}(k_{\lambda}*(v-v_{0}))\to\partial_{t}(k*(v-v_{0}))\quad\text{in}~{}L^{2}(0,T;(X_{0}^{s,2}(\Omega))^{\prime}),$

which implies

$\lim_{\lambda\to
0}\int^{t_{1}}_{0}\int_{\Omega}H^{\prime}_{\epsilon}(v^{+})\partial_{t}[(k_{\lambda}-k)*(v-v_{0})]\,dxdt=0.$

Then, we conclude that

$\int_{\Omega}k*(H_{\epsilon}(v^{+})-H_{\epsilon}(v^{+}_{0}))(t_{1},x)\,dx\leq\int^{t_{1}}_{0}\int_{\Omega}F^{+}\,dxdt,~{}\text{for
a.e.}~{}t_{1}\in(0,T).$ (3.8)

Convolving (3.8) with the kernel $l$ and evaluating at $t_{1}=T$ gives

$\int^{T}_{0}\int_{\Omega}H_{\epsilon}(v^{+})\,dxdt\leq
T\int_{\Omega}H_{\epsilon}(v^{+}_{0})\,dx+\|l\|_{L^{1}(0,T)}\int^{T}_{0}\int_{\Omega}F^{+}\,dxdt.$
(3.9)

Since $H_{\epsilon}(y^{+})\to y^{+}$ as $\epsilon\to 0$ and
$|H^{\prime}_{\epsilon}(y^{+})|\leq 1$, it follows from the Lebesgue’s
dominated convergence theorem that

$\int^{T}_{0}\int_{\Omega}v^{+}\,dxdt\leq
T\int_{\Omega}v^{+}_{0}\,dx+\|l\|_{L^{1}(0,T)}\int^{T}_{0}\int_{\Omega}F^{+}\,dxdt.$

This yields the assertion. ∎

##  4\. The proof of main results

In this section we provide proofs of the main goals–existence and uniqueness
of entropy solutions for problems (1.1). Some of the reasoning is based on the
ideas developed in [38, 39, 40, 43, 45, 48]. According to Lemma 3.1 and Lemma
3.2, we can find a unique non-negative weak solution $u^{m}\in
W(u_{0},X^{s,2}_{0}(\Omega),L^{2}(\Omega))$ for the approximate problem (3.2).
Our aim is to prove that a subsequence of these solutions $\\{u^{m}\\}$
converges to a measurable function $u$, which is an entropy solution of
problem (1.1). The uniqueness of entropy solution is obtained by a comparison
principle. Although some of the arguments are not new, we present a self-
contained proof for the sake of clarity and readability. We will divide the
proof into several steps.

Proof of Theorem 1.2. Step 1. Prove the convergence of $u^{m}$ in
$L^{1}(\Omega_{T})$ and find its subsequence which is almost everywhere
convergent in $\Omega_{T}$.

It follows from Lemma 3.2 that

$\begin{split}\sup_{m\in\mathbb{N}}\|u^{m}\|_{L^{1}(\Omega_{T})}&\leq\sup_{m\in\mathbb{N}}(T\|u^{m}_{0}\|_{L^{1}(\Omega)}+\|l\|_{L^{1}(0,T)}\|f^{m}\|_{L^{1}(\Omega_{T})})\\\
&\leq
T\|u_{0}\|_{L^{1}(\Omega)}+\|l\|_{L^{1}(0,T)}\|f\|_{L^{1}(\Omega_{T})}.\end{split}$

Moreover, according to (3.4), we know that

$u^{m}\leq u^{m+1}\quad\text{for all}~{}m\in\mathbb{N},$

which implies that $u^{m}$ is a non-negative increasing sequence. Therefore,
we know that there exists an element $u$ such that

$u^{m}\to u\quad\text{a.e. in}~{}\Omega_{T}~{}\text{for}~{}m\to\infty.$

In particular, we have $u^{m}\leq u$ a.e. in $\Omega_{T}$. Using the
Lebesgue’s dominated theorem, we deduce that

$u^{m}\to u\quad\text{in}~{}L^{1}(\Omega_{T})~{}\text{for}~{}m\to\infty.$

Step 2. Prove $T_{K}(u^{m})$ strongly converges to $T_{K}(u)$ in
$L^{2}(0,T;X^{s,2}_{0}(\Omega))$ for every $K>0$.

Choosing $T_{K}(u^{m})$ as a test function in (3.3), we have

$\displaystyle\int^{t_{1}}_{0}\int_{\Omega}T_{K}(u^{m})\partial_{t}[k*(u^{m}-u^{m}_{0})]\,dxdt$
$\displaystyle\quad+\frac{1}{2}\int^{t_{1}}_{0}\int_{\mathcal{D}_{\Omega}}(u^{m}(x,t)-u^{m}(y,t))\bigg{(}T_{K}(u^{m}(x,t))-T_{K}(u^{m}(y,t))\bigg{)}\,dvdt$
$\displaystyle\quad=\int^{t_{1}}_{0}\int_{\Omega}f^{m}T_{K}(u^{m})\,dxdt,$

for all $t_{1}\in(0,T)$.

Due to the following elementary algebraic inequality

$\displaystyle(a-b)(T_{K}(a)-T_{K}(b))\geq|T_{K}(a)-T_{K}(b)|^{2},\quad
a,b\in{\mathbb{R}},$

we deduce that

$\displaystyle\int^{t_{1}}_{0}\int_{\Omega}T_{K}(u^{m})\partial_{t}[k_{\lambda}*(u^{m}-u^{m}_{0})]\,dxdt$
(4.1)
$\displaystyle\quad+\frac{1}{2}\int^{t_{1}}_{0}\int_{\mathcal{D}_{\Omega}}\bigg{|}T_{K}(u^{m}(x,t))-T_{K}(u^{m}(y,t))\bigg{|}^{2}\,dvdt$
$\displaystyle\quad\leq\int^{t_{1}}_{0}\int_{\Omega}f^{m}T_{K}(u^{m})\,dxdt+\int^{t_{1}}_{0}\int_{\Omega}T_{K}(u^{m})\partial_{t}[(k_{\lambda}-k)*(u^{m}-u^{m}_{0})]\,dxdt.$

Combining with Lemma 2.6, we obtain

$\begin{split}&\int_{\Omega}\left[k_{\lambda}*\int^{u^{m}}_{0}T_{K}(\sigma)\,d\sigma\right](t_{1})\,dx+\int^{t_{1}}_{0}\int_{\Omega}\bigg{[}T_{K}(u^{m}(t))u^{m}-\int^{u^{m}}_{0}T_{K}(\sigma)\,d\sigma\bigg{]}k_{\lambda}(t)\,dxdt\\\
&\quad+\int^{t_{1}}_{0}\int_{\Omega}\int^{t}_{0}\bigg{[}\int^{u^{m}(t-s)}_{u^{m}(t)}T_{K}(\sigma)\,d\sigma-
T_{K}(u^{m}(t))(u^{m}(t-s)-u^{m}(t))\bigg{]}[-k^{\prime}_{\lambda}(s)]\,dsdxdt\\\
&\quad+\frac{1}{2}\int^{t_{1}}_{0}\int_{\mathcal{D}_{\Omega}}\bigg{|}T_{K}(u^{m}(x,t))-T_{K}(u^{m}(y,t))\bigg{|}^{2}\,dvdt\\\
&\quad\leq\int^{t_{1}}_{0}\int_{\Omega}f^{m}T_{K}(u^{m})\,dxdt+\int^{t_{1}}_{0}\int_{\Omega}T_{K}(u^{m})\partial_{t}[(k_{\lambda}-k)*(u^{m}-u^{m}_{0})]\,dxdt\\\
&\qquad+\int^{t_{1}}_{0}\int_{\Omega}T_{K}(u^{m})k_{\lambda}(t)u^{m}_{0}\,dxdt.\end{split}$
(4.2)

Since $(u^{m}-u^{m}_{0})\in D(B_{1})$ the second term in the right-hand side
converges to zero as $\lambda\to 0$. Recalling the fact that $k_{\lambda}\to
k$ in $L^{1}(0,T)$ as $\lambda\to 0$, we obtain

$\displaystyle\int_{\Omega}\left[k*\int^{u^{m}}_{0}T_{K}(\sigma)\,d\sigma\right](t_{1})\,dx+\frac{1}{2}\int^{t_{1}}_{0}\int_{\mathcal{D}_{\Omega}}\bigg{|}T_{K}(u^{m}(x,t))-T_{K}(u^{m}(y,t))\bigg{|}^{2}\,dvdt$
$\displaystyle\quad\leq\int^{t_{1}}_{0}\int_{\Omega}f^{m}T_{K}(u^{m})\,dxdt+\int^{t_{1}}_{0}\int_{\Omega}T_{K}(u^{m})k(t)u^{m}_{0}\,dxdt,$

which implies that

$\displaystyle\frac{1}{2}\int^{t_{1}}_{0}\int_{\mathcal{D}_{\Omega}}\bigg{|}T_{K}(u^{m}(x,t))-T_{K}(u^{m}(y,t))\bigg{|}^{2}\,dvdt$
$\displaystyle\quad\leq\int^{t_{1}}_{0}\int_{\Omega}f^{m}T_{K}(u^{m})\,dxdt+K\int^{t_{1}}_{0}\int_{\Omega}k(t)u^{m}_{0}\,dxdt$
$\displaystyle\quad\leq
K\|f\|_{L^{1}(\Omega_{T})}+K\|k\|_{L^{1}(0,T)}\|u_{0}\|_{L^{1}(\Omega)}\leq
C,$

since $k$ is non-negative. Then, up to a subsequence, we deduce that

$\displaystyle T_{K}(u^{m})$
$\displaystyle\overset{m\to\infty}{\rightharpoonup}T_{K}(u)\quad\text{weakly
in}~{}L^{2}(0,T;X^{s,2}_{0}(\Omega)).$

Next, we aim to prove that

$\begin{split}&\limsup_{m\to\infty}\int^{t_{1}}_{0}\int_{\mathcal{D}_{\Omega}}\bigg{|}T_{K}(u^{m}(x))-T_{K}(u^{m}(y))\bigg{|}^{2}\,dvdt\\\
&\quad\leq\int^{t_{1}}_{0}\int_{\mathcal{D}_{\Omega}}\bigg{|}T_{K}(u(x))-T_{K}(u(y))\bigg{|}^{2}\,dvdt.\end{split}$

For $l>0$ we denote by $h_{l}$ the function defined by

$h_{l}(u)=\min\\{(l+1-|u|)^{+},1\\}.$

We take $\bigg{(}T_{K}(u^{m})-h_{l}(u^{m})(T_{K}(u))_{\mu}\bigg{)}$ as a test
function in equation (3.3), where $(T_{K}(u))_{\mu}$ is defined by (2.5).
Thus, we obtain

$J^{\lambda,m,\mu,l}_{1}+J^{\lambda,m,\mu,l}_{2}=J^{\lambda,m,\mu,l}_{3}+J^{\lambda,m,\mu,l}_{4},$
(4.3)

where

$\begin{split}&J^{\lambda,m,\mu,l}_{1}=\int^{t_{1}}_{0}\int_{\Omega}\bigg{(}T_{K}(u^{m})-h_{l}(u^{m})(T_{K}(u))_{\mu}\bigg{)}\partial_{t}[k_{\lambda}*(u^{m}-u^{m}_{0})]\,dxdt,\\\
&J^{\lambda,m,\mu,l}_{2}=\int^{t_{1}}_{0}\langle(-\Delta)^{s}_{2}u^{m},T_{K}(u^{m})-h_{l}(u^{m})(T_{K}(u))_{\mu}\rangle\,dt,\\\
&J^{\lambda,m,\mu,l}_{3}=\int^{t_{1}}_{0}\int_{\Omega}f^{m}\bigg{(}T_{K}(u^{m})-h_{l}(u^{m})(T_{K}(u))_{\mu}\bigg{)}\,dxdt,\\\
&J^{\lambda,m,\mu,l}_{4}=\int^{t_{1}}_{0}\int_{\Omega}(T_{K}(u^{m})-h_{l}(u^{m})(T_{K}(u))_{\mu})\partial_{t}[(k_{\lambda}-k)*(u^{m}-u^{m}_{0})]\,dxdt.\end{split}$

We are going to pass the limit with $\lambda\to 0$, $m\to\infty$, then $\mu\to
0$, and finally with $l\to\infty$. Roughly speaking, we show that the limits
of $J^{\lambda,m,\mu,l}_{3}$ and $J^{\lambda,m,\mu,l}_{4}$ are zero, and the
limit of $J^{\lambda,m,\mu,l}_{1}$ is non-negative. Then the limit of
$J^{\lambda,m,\mu,l}_{2}$ is nonpositive.

Limit of $J^{\lambda,m,\mu,l}_{3}$. To deal with the limit with $m\to\infty$,
we apply the Lebesgue’s dominated convergence theorem due to the continuity of
the integrand and the fact that $u^{m}\to u$ a.e. in $\Omega_{T}$. Moreover,
we know that $(T_{K}(u))_{\mu}$ strongly convergence to $T_{K}(u)$ in
$L^{2}(0,T;X_{0}^{s,2}(\Omega))$ and a.e. in $\Omega_{T}$ as $\mu\to\infty$,
and $h_{l}(u)\to 1$ a.e. in $\Omega_{T}$ as $l\to\infty$. Therefore, it is
obvious that

$\begin{split}\lim_{l\to\infty}\lim_{\mu\to
0}\lim_{m\to\infty}\int^{t_{1}}_{0}\int_{\Omega}f^{m}\bigg{(}T_{K}(u^{m})-h_{l}(u^{m})(T_{K}(u))_{\mu}\bigg{)}\,dxdt=0.\end{split}$
(4.4)

Limit of $J^{\lambda,m,\mu,l}_{4}$. As an immediate consequence of
$(u^{m}-u^{m}_{0})\in D(B_{1})$, we obtain that

$\begin{split}\lim_{\lambda\to
0}\int^{t_{1}}_{0}\int_{\Omega}(T_{K}(u^{m})-h_{l}(u^{m})(T_{K}(u))_{\mu})\partial_{t}[(k_{\lambda}-k)*(u^{m}-u^{m}_{0})]\,dxdt=0.\end{split}$
(4.5)

Limit of $J^{\lambda,m,\mu,l}_{1}$. We are aim to prove that

$\begin{split}&\liminf_{l\to\infty}\liminf_{\mu\to
0}\liminf_{m\to\infty}\liminf_{\lambda\to
0}\int^{t_{1}}_{0}\int_{\Omega}\partial_{t}[k_{\lambda}*(u^{m}-u^{m}_{0})]\\\
&\qquad\qquad\qquad\times\bigg{(}T_{K}(u^{m})-h_{l}(u^{m})(T_{K}(u))_{\mu}\bigg{)}\,dxdt\geq
0\end{split}$ (4.6)

for almost every $t_{1}\in(0,T)$. Let us consider a decomposition

$\begin{split}&\int^{t_{1}}_{0}\int_{\Omega}\partial_{t}[k_{\lambda}*(u^{m}-u^{m}_{0})]\times\bigg{(}T_{K}(u^{m})-h_{l}(u^{m})(T_{K}(u))_{\mu}\bigg{)}\,dxdt\\\
&\quad=\int^{t_{1}}_{0}\int_{\Omega}\partial_{t}[k_{\lambda}*u^{m}]\times
T_{K}(u^{m})\,dxdt-\int^{t_{1}}_{0}\int_{\Omega}\partial_{t}[k_{\lambda}*u^{m}]\times
h_{l}(u^{m})(T_{K}(u))_{\mu}\,dxdt\\\
&\qquad-\int^{t_{1}}_{0}\int_{\Omega}k_{\lambda}u^{m}_{0}\times\bigg{(}T_{K}(u^{m})-h_{l}(u^{m})(T_{K}(u))_{\mu}\bigg{)}\,dxdt\\\
&\quad=:J^{\lambda,m,\mu,l}_{11}+J^{\lambda,m,\mu,l}_{12}+J^{\lambda,m,\mu,l}_{13}.\end{split}$
(4.7)

Next, we shall give the estimates of $J^{\lambda,m,\mu,l}_{1i}(i=1,2,3)$ one
by one.

_Limit of $J^{\lambda,m,\mu,l}_{13}$._ As $k_{\lambda}\xrightarrow{\lambda\to
0}k$ in $L^{1}(0,T)$ and $(T_{K}(u^{m})-h_{l}(u^{m})(T_{K}(u))_{\mu})$ is
uniformly bounded by $2K$, combining with the fact that
$u^{m}\xrightarrow{m\to\infty}u$ a.e. in $\Omega_{T}$, we conclude from
Lebesgue’s dominated theorem that

$\displaystyle\int^{t_{1}}_{0}\int_{\Omega}k_{\lambda}u^{m}_{0}[T_{K}(u^{m})-h_{l}(u^{m})(T_{K}(u))_{\mu}]\,dxdt$
$\displaystyle\quad\xrightarrow[m\to\infty]{\lambda\to
0}\int^{t_{1}}_{0}\int_{\Omega}k(t)u_{0}[T_{K}(u)-h_{l}(u)(T_{K}(u))_{\mu}]\,dxdt.$

Since $(T_{K}(u))_{\mu}\xrightarrow{\mu\to 0}T_{K}(u)$ a.e in $\Omega_{T}$,
and $h_{l}(u)\xrightarrow{l\to\infty}1$ for every $u\in\mathbb{R}$, using the
Lebesgue’s dominated theorem again, we have

$\displaystyle\lim_{l\to\infty}\lim_{\mu\to
0}\int^{t_{1}}_{0}\int_{\Omega}k(t)u_{0}[T_{K}(u)-h_{l}(u)(T_{K}(u))_{\mu}]\,dxdt=0.$

Thus, we conclude that

$\begin{split}&\liminf_{l\to\infty}\liminf_{\mu\to
0}\liminf_{m\to\infty}\liminf_{\lambda\to 0}J^{\lambda,m,\mu,l}_{13}\\\
&\quad=\liminf_{l\to\infty}\liminf_{\mu\to
0}\liminf_{m\to\infty}\liminf_{\lambda\to
0}-\int^{t_{1}}_{0}\int_{\Omega}k_{\lambda}u^{m}_{0}\times\bigg{(}T_{K}(u^{m})-h_{l}(u^{m})(T_{K}(u))_{\mu}\bigg{)}\,dxdt\\\
&\quad=0.\end{split}$ (4.8)

_Limit of $J^{\lambda,m,\mu,l}_{11}$._ According to Lemma 2.6, we obtain

$\begin{split}\int^{t_{1}}_{0}\int_{\Omega}\partial_{t}(k_{\lambda}*u^{m})T_{K}(u^{m})\,dxdt&=\int_{\Omega}\bigg{[}k_{\lambda}*\int^{u^{m}}_{0}T_{K}(\sigma)\,d\sigma\bigg{]}(t_{1})\,dx\\\
&\qquad+\int^{t_{1}}_{0}\int_{\Omega}\bigg{[}T_{K}(u^{m}(t))u^{m}-\int^{u^{m}}_{0}T_{K}(\sigma)\,d\sigma\bigg{]}k_{\lambda}(t)\,dxdt\\\
&\qquad+\int^{t_{1}}_{0}\int_{\Omega}\int^{t}_{0}\bigg{[}\int^{u^{m}(t-s)}_{u^{m}(t)}T_{K}(\sigma)\,d\sigma\\\
&\qquad\qquad-
T_{K}(u^{m}(t))(u^{m}(t-s)-u^{m}(t))\bigg{]}[k^{\prime}_{\lambda}(s)]\,dsdxdt.\end{split}$

Since $u^{m}\to u$ a.e. in $\Omega_{T}$, it follows that
$\int^{u^{m}}_{0}T_{K}(\sigma)\,d\sigma\to\int^{u}_{0}T_{K}(\sigma)\,d\sigma$
a.e. in $\Omega_{T}$. Combining with the fact that
$\left|\int^{u^{m}}_{0}T_{K}(\sigma)\,d\sigma\right|\leq K|u^{m}|\leq K|u|$,
we deduce by the Lebesgue’s dominated theorem that

$\int^{u^{m}}_{0}T_{K}(\sigma)\,d\sigma\xrightarrow{m\to\infty}\int^{u}_{0}T_{K}(\sigma)\,d\sigma\quad\text{in
}L^{1}(\Omega_{T})$ (4.9)

and

$\int_{\Omega}\int^{u^{m}}_{0}T_{K}(\sigma)\,d\sigma\,dx\xrightarrow{m\to\infty}\int_{\Omega}\int^{u}_{0}T_{K}(\sigma)\,d\sigma
dx\quad\text{in }L^{1}(0,T).$

Then, we can find an a.e. convergent subsequence, still denoted the same way,
such that

$\int_{\Omega}\int^{u^{m}}_{0}T_{K}(\sigma)\,d\sigma
dx\xrightarrow{m\to\infty}\int_{\Omega}\int^{u}_{0}T_{K}(\sigma)\,d\sigma
dx\quad\text{a.e. in }(0,T).$

As $k_{\lambda}\to k$ in $L^{1}(0,T)$, we conclude by Young’s inequality that

$\int_{\Omega}\left[k_{\lambda}*\int^{u^{m}}_{0}T_{K}(\sigma)\,d\sigma\right](\cdot)\,dx\xrightarrow[m\to\infty]{\lambda\to
0}\int_{\Omega}\left[k*\int^{u}_{0}T_{K}(\sigma)\,d\sigma\right](\cdot)\,dx$

in $L^{1}(0,T)$ and a.e. in $(0,T)$ for subsequence.

According to

$\partial_{t}k_{\lambda}*(u^{m}-u^{m}_{0})\to\partial_{t}k*(u^{m}-u^{m}_{0})\quad\text{in
}L^{2}(0,T;(X_{0}^{s,2}(\Omega))^{\prime})$

as $\lambda\to 0$, we deduce that

$\lim_{\lambda\to\infty}\int^{t_{1}}_{0}\int_{\Omega}T_{K}(u^{m})\partial_{t}(k_{\lambda}-k)*(u^{m}-u^{m}_{0})\,dxdt=0,$

which implies that there exists a constant $C$ which is independent of
$\lambda$ such that

$\int^{t_{1}}_{0}\int_{\Omega}T_{K}(u^{m})\partial_{t}(k_{\lambda}-k)*(u^{m}-u^{m}_{0})\,dxdt\leq
C.$

Thus, combining with the fact that the kernel $k$ satisfies $\rm(K1)$ and
$\rm(K2)$, we deduce from (4.2) that for all $t_{1}\in(0,T)$,

$\displaystyle\int^{t_{1}}_{0}\int_{\Omega}\bigg{[}T_{K}(u^{m}(t))u^{m}-\int^{u^{m}}_{0}T_{K}(\sigma)\,d\sigma\bigg{]}k_{\lambda}(t)\,dxdt$
$\displaystyle\quad+\int^{t_{1}}_{0}\int_{\Omega}\int^{t}_{0}\bigg{[}\int^{u^{m}(t-s)}_{u^{m}(t)}T_{K}(\sigma)\,d\sigma-
T_{K}(u^{m}(t))(u^{m}(t-s)-u^{m}(t))\bigg{]}[-k^{\prime}_{\lambda}(s)]\,dsdxdt$
$\displaystyle\quad\leq C,$

where $C=C(K,m)$ is independent of $\lambda$. Then, the Fatou’s lemma yields
that

$[T_{K}(u^{m})u^{m}-\int^{u^{m}}_{0}T_{K}(\sigma)\,d\sigma]k(t)\in
L^{1}(\Omega_{T}),$ (4.10)

and

$\int^{t}_{0}\bigg{[}\int^{u^{m}(t-s)}_{u^{m}(t)}T_{K}(\sigma)\,d\sigma-
T_{K}(u^{m}(t))(u^{m}(t-s)-u^{m}(t))\bigg{]}[-k^{\prime}(s)]\,ds\in
L^{1}(\Omega_{T}).$ (4.11)

Combining with (4.2), we know that

$\begin{split}&\liminf_{m\to\infty}\int^{t_{1}}_{0}\int_{\Omega}\bigg{[}T_{K}(u^{m}(t))u^{m}-\int^{u^{m}}_{0}T_{K}(\sigma)\,d\sigma\bigg{]}k(t)\,dxdt\\\
&\quad+\liminf_{m\to\infty}\int^{t_{1}}_{0}\int_{\Omega}\int^{t}_{0}\bigg{[}\int^{u^{m}(t-s)}_{u^{m}(t)}T_{K}(\sigma)\,d\sigma-
T_{K}(u^{m}(t))(u^{m}(t-s)-u^{m}(t))\bigg{]}[-k^{\prime}(s)]\,dsdxdt\\\
&\quad\leq\limsup_{m\to\infty}\int^{t_{1}}_{0}\int_{\Omega}f^{m}T_{K}(u^{m})\,dxdt+\limsup_{m\to\infty}\int^{t_{1}}_{0}\int_{\Omega}T_{K}(u^{m})k(t)u^{m}_{0}\,dxdt\\\
&\quad\leq
K\|f\|_{L^{1}(\Omega_{T})}+K\|k(t)\|_{L^{1}(0,T)}\|u_{0}\|_{L^{1}(\Omega)}.\end{split}$

Therefore, using the Fatou’s lemma again we conclude that

$\begin{split}&\liminf_{l\to\infty}\liminf_{\mu\to
0}\liminf_{m\to\infty}\liminf_{\lambda\to 0}J^{\lambda,m,\mu,l}_{11}\\\
&\quad=\liminf_{l\to\infty}\liminf_{\mu\to
0}\liminf_{m\to\infty}\liminf_{\lambda\to
0}\int^{t_{1}}_{0}\int_{\Omega}\partial_{t}(k_{\lambda}*u^{m})T_{K}(u^{m})\,dxdt\\\
&\quad\geq\int_{\Omega}[k*\int^{u}_{0}T_{K}(\sigma)\,d\sigma](t_{1})\,dx+\int^{t_{1}}_{0}\int_{\Omega}\bigg{[}T_{K}(u(t))u-\int^{u}_{0}T_{K}(\sigma)\,d\sigma\bigg{]}k(t)\,dxdt\\\
&\qquad+\int^{t_{1}}_{0}\int_{\Omega}\int^{t}_{0}\bigg{[}\int^{u(t-s)}_{u(t)}T_{K}(\sigma)\,d\sigma\\\
&\qquad\qquad\qquad\qquad-
T_{K}(u(t))(u(t-s)-u(t))\bigg{]}[-k^{\prime}(s)]\,dsdxdt.\end{split}$ (4.12)

_Limit of $J^{\lambda,m,\mu,l}_{12}$._ Thanks to Lemma 2.4, we get

$\begin{split}J^{\lambda,m,\mu,l}_{12}&=-\int^{t_{1}}_{0}\int_{\Omega}\partial_{t}(k_{\lambda}*u^{m})h_{l}(u^{m})(T_{K}(u^{m}))_{\mu}\,dxdt\\\
&=-\int^{t_{1}}_{0}\int_{\Omega}\partial_{t}(k_{\lambda}*\int^{u^{m}}_{0}h_{l}(\sigma)\,d\sigma)(T_{K}(u^{m}))_{\mu}\,dxdt\\\
&\quad-\int^{t_{1}}_{0}\int_{\Omega}[h_{l}(u^{m})u^{m}-\int^{u^{m}}_{0}h_{l}(\sigma)\,d\sigma]k_{\lambda}(t)(T_{K}(u^{m}))_{\mu}\,dxdt\\\
&\quad-\int^{t_{1}}_{0}\int_{\Omega}\int^{t}_{0}\bigg{[}\int^{u^{m}(t-s)}_{u^{m}(t)}h_{l}(\sigma)d\sigma\\\
&\qquad\qquad\qquad-
h_{l}(u^{m}(t))(u^{m}(t-s)-u^{m}(t))\bigg{]}[-k^{\prime}_{\lambda}(s)]\,ds(T_{K}(u^{m}))_{\mu}\,dxdt\\\
&:=-I^{1}_{\lambda,m,\mu,l}-I^{2}_{\lambda,m,\mu,l}-I^{3}_{\lambda,m,\mu,l}.\end{split}$
(4.13)

First, we are going to prove that

$\limsup_{l\to\infty}\limsup_{\mu\to 0}\limsup_{m\to\infty}\limsup_{\lambda\to
0}(I^{2}_{\lambda,m,\mu,l}+I^{3}_{\lambda,m,\mu,l})=0.$

Define $T^{+}_{l,l+1}(r):=T_{l,l+1}(\max(r,0))$ and
$T^{-}_{l,l+1}(r):=T_{l,l+1}(-\max(-r,0))$ for $l\in\mathbb{R}$. Since
$h_{l}(\sigma)=T^{-}_{l,l+1}(\sigma)-T^{+}_{l,l+1}(\sigma)+1$, it follows

$\begin{split}I^{2}_{\lambda,m,\mu,l}&=\int^{t_{1}}_{0}\int_{\Omega}[T^{-}_{l,l+1}(u^{m})u^{m}-\int^{u^{m}}_{0}T^{-}_{l,l+1}(\sigma)d\sigma]k_{\lambda}(t)(T_{K}(u^{m}))_{\mu}\,dxdt\\\
&\quad-\int^{t_{1}}_{0}\int_{\Omega}[T^{+}_{l,l+1}(u^{m})u^{m}-\int^{u^{m}}_{0}T^{+}_{l,l+1}(\sigma)d\sigma]k_{\lambda}(t)(T_{K}(u^{m}))_{\mu}\,dxdt\end{split}$

and

$\displaystyle I^{3}_{\lambda,m,\mu,l}$
$\displaystyle=\int^{t_{1}}_{0}\int_{\Omega}\int^{t}_{0}\bigg{[}\int^{u^{m}(t-s)}_{u^{m}(t)}T^{-}_{l,l+1}(\sigma)d\sigma$
$\displaystyle\qquad\qquad\qquad-T^{-}_{l,l+1}(u^{m})(u^{m}(t-s)-u^{m}(t))\bigg{]}[-k^{\prime}_{\lambda}(s)]\,ds(T_{K}(u^{m}))_{\mu}\,dxdt$
$\displaystyle\quad-\int^{t_{1}}_{0}\int_{\Omega}\int^{t}_{0}\bigg{[}\int^{u^{m}(t-s)}_{u^{m}(t)}T^{+}_{l,l+1}(\sigma)d\sigma$
$\displaystyle\qquad\qquad\qquad-T^{+}_{l,l+1}(u^{m})(u^{m}(t-s)-u^{m}(t))\bigg{]}[-k^{\prime}_{\lambda}(s)]\,ds(T_{K}(u^{m}))_{\mu}\,dxdt.$

Choosing $T_{l,l+1}(u^{m})=T_{1}(u^{m}-T_{l}(u^{m}))$ ($l>0$) as a test
function for problem (3.2), we obtain

$\begin{split}&\frac{1}{2}\int^{t_{1}}_{0}\int_{\\{(u(x,t),u(y,t))\in
R_{l}\\}}|u^{m}(x,t)-u^{m}(y,t)|\,dvdt+\int_{\Omega}[k_{\lambda}*\int^{u^{m}}_{0}T_{K}(\sigma)\,d\sigma](t_{1})\,dx\\\
&\quad+\int^{t_{1}}_{0}\int_{\Omega}[T_{l,l+1}(u^{m})u^{m}-\int^{u^{m}}_{0}T_{l,l+1}(\sigma)d\sigma]k_{\lambda}(t)\,dxdt\\\
&\quad+\int^{t_{1}}_{0}\int_{\Omega}\int^{t}_{0}\bigg{[}\int^{u^{m}(t-s)}_{u^{m}(t)}T_{l,l+1}(\sigma)d\sigma\\\
&\qquad\qquad\qquad\qquad-
T_{l,l+1}(u^{m})(u^{m}(t-s)-u^{m}(t))\bigg{]}[-k^{\prime}_{\lambda}(s)]\,ds\,dxdt\\\
&\quad\leq\int_{\Omega_{T}\cap\\{|u^{m}|>l\\}}f^{m}\,dxdt+\int_{\Omega_{T}\cap\\{|u^{m}|>l\\}}k_{\lambda}u^{m}_{0}\,dxdt\\\
&\quad\quad+\int_{\Omega_{T}\cap\\{|u^{m}|>l\\}}T_{l,l+1}(u^{m})|\partial_{t}[(k_{\lambda}-k)*(u^{m}-u^{m}_{0})]|\,dxdt\end{split}$
(4.14)

for every $t_{1}\in(0,T)$, where

$R_{l}=\\{(w,v)\in\mathbb{R}^{2}:l+1\leq\max{\\{|w|,|v|\\}}~{}and~{}\min{\\{|w|,|v|\\}}\leq
l~{}or~{}wv<0\\}.$

Since $k_{\lambda}\to k$ in $L^{1}(0,T)$, $u^{m}\to u$ a.e. in $\Omega_{T}$
and $|u|<\infty$ a.e. in $\Omega_{T}$ it follows that

$\limsup_{l\to\infty}\limsup_{m\to\infty}\limsup_{\lambda\to
0}\bigg{(}\int_{\Omega_{T}\cap\\{|u^{m}|>l\\}}f^{m}\,dxdt+\int_{\Omega_{T}\cap\\{|u^{m}|>l\\}}k_{\lambda}u^{m}_{0}\,dxdt\bigg{)}=0.$

Moreover, since $(u_{m}-u_{0})\in D(B_{1})$, one immediately obtain that the
last term in (4.14) tends to 0 as $\lambda\to 0$. Since all terms on the left-
hand side in (4.14) are non-negative, we conclude that, for every
$t_{1}\in(0,T)$,

$\begin{split}&\limsup_{l\to\infty}\limsup_{m\to\infty}\limsup_{\lambda\to
0}\int^{t_{1}}_{0}\int_{\Omega}[T_{l,l+1}(u^{m})u^{m}-\int^{u^{m}}_{0}T_{l,l+1}(\sigma)d\sigma]k_{\lambda}\,dxdt=0,\\\
&\limsup_{l\to\infty}\limsup_{m\to\infty}\limsup_{\lambda\to
0}\int^{t_{1}}_{0}\int_{\Omega}\int^{t}_{0}\bigg{[}\int^{u^{m}(t-s)}_{u^{m}(t)}T_{l,l+1}(\sigma)d\sigma\\\
&\qquad\qquad\qquad\qquad\qquad\qquad-
T_{l,l+1}(u^{m}(t))\bigg{(}u^{m}(t-s)-u^{m}\bigg{)}\bigg{]}[-k^{\prime}_{\lambda}(s)]\,dsdxdt=0.\end{split}$
(4.15)

It is easy to see that the corresponding result holds in the case $T_{l,l+1}$
is replaced by $T^{\pm}_{l,l+1}$. Thus, we obtain

$\limsup_{l\to\infty}\limsup_{\mu\to 0}\limsup_{m\to\infty}\limsup_{\lambda\to
0}(I^{2}_{\lambda,m,\mu,l}+I^{3}_{\lambda,m,\mu,l})=0.$

Next, we will focus on the term $I^{1}_{\lambda,m,\mu,l}$. It is easy to check
that $\int^{u^{m}}_{0}h_{l}(\sigma)\,d\sigma$ is uniformly bounded. Thus, it
follows from the Lebesgue’s dominated theorem that

$\displaystyle\int^{u^{m}}_{0}h_{l}(\sigma)\,d\sigma\xrightarrow{m\to\infty}\int^{u}_{0}h_{l}(\sigma)\,d\sigma\quad\text{in
}L^{2}(0,T;L^{2}(\Omega)).$

Moreover, we know that

$\displaystyle
B_{\lambda}\int^{u^{m}}_{0}h_{l}(\sigma)\,d\sigma=\partial_{t}(k_{\lambda}*\int^{u^{m}}_{0}h_{l}(\sigma)\,d\sigma)\in
L^{2}(0,T;(X_{0}^{s,2}(\Omega))^{\prime}).$

Using the fact that

$J^{B}_{\mu}B_{\lambda}v=\frac{B}{(I+\mu B)(I+\lambda
B)}v=B_{\mu}J^{B}_{\lambda}v\quad\text{for any }v\in D(B),$

we obtain

$\begin{split}I^{1}_{\lambda,m,\mu,l}&=\int^{t_{1}}_{0}\int_{\Omega}\partial_{t}(k_{\lambda}*\int^{u^{m}}_{0}h_{l}(\sigma)\,d\sigma)(T_{K}(u))_{\mu}\,dxdt\\\
&=\bigg{\langle}B_{\lambda}\int^{u^{m}}_{0}h_{l}(\sigma)\,d\sigma,J^{B^{*}}_{\mu}T_{K}(u)\bigg{\rangle}_{L^{2}(0,t_{1};(X_{0}^{s,2}(\Omega))^{\prime})\times
L^{2}(0,t_{1};X_{0}^{s,2}(\Omega))}\\\
&=\bigg{\langle}B_{\mu}J^{B}_{\lambda}\int^{u^{m}}_{0}h_{l}(\sigma)\,d\sigma,T_{K}(u)\bigg{\rangle}_{L^{2}(0,t_{1};(X_{0}^{s,2}(\Omega))^{\prime})\times
L^{2}(0,t_{1};X_{0}^{s,2}(\Omega))}\end{split}$ (4.16)

for any $t_{1}\in(0,T)$. Since $J^{B}_{\lambda}$ is bounded operator
satisfying $J^{B}_{\lambda}v\to v$ in
$L^{2}(0,T;(X_{0}^{s,2}(\Omega))^{\prime})$ for all $v\in
L^{2}(0,T;(X_{0}^{s,2}(\Omega))^{\prime})$ as $\lambda\to 0$, the continuous
embedding $L^{2}(\Omega)\hookrightarrow(X_{0}^{s,2}(\Omega))^{\prime}$ yields

$\displaystyle
J_{\lambda}^{B}\int^{u^{m}}_{0}h_{l}(\sigma)\,d\sigma\xrightarrow{\lambda\to
0}\int^{u^{m}}_{0}h_{l}(\sigma)\,d\sigma\xrightarrow{m\to\infty}\int^{u}_{0}h_{l}(\sigma)\,d\sigma\quad\text{in}~{}~{}L^{2}(0,T;(X_{0}^{s,2}(\Omega))^{\prime}).$

Thus, the continuity of the Yosida approximation $B_{\mu}$ implies

$\begin{split}I^{1}_{\lambda,m,\mu,l}&\xrightarrow[m\to\infty]{\lambda\to
0}\langle B_{\mu}\int^{u}_{0}h_{l}(\sigma)\,d\sigma,T_{K}(u)\rangle\\\
&=\int^{t_{1}}_{0}\int_{\Omega}\partial_{t}(k_{\mu}*\int^{u}_{0}h_{l}(\sigma)\,d\sigma)T_{K}(u)\,dxdt.\end{split}$

Define $g_{l}(u):=\int^{u}_{0}h_{l}(\sigma)\,d\sigma$, $u\in\mathbb{R}$ and

$H(u):=\int^{u}_{0}T_{K}\circ g^{-}_{l}(\sigma)\,d\sigma,\quad u\in
ran(g_{l}).$

Using the transformation $\sigma\mapsto g_{l}(\sigma)$, we see that

$H(g_{l}(u(t)))=\int^{u(t)}_{0}T_{K}(\sigma)\,dg_{l}(\sigma).$

By the definition of $g_{l}$, the measure $dg_{l}$ is absolutely continuous
with respect to $d\sigma$ and the Radon-Nikodym derivative is given by
$h_{l}$. According to Lemma 2.6, we get

$\begin{split}&-\int^{t_{1}}_{0}\int_{\Omega}\partial_{t}\bigg{(}k_{\mu}*\int^{u}_{0}h_{l}(\sigma)\,d\sigma\bigg{)}T_{K}(u)\,dxdt\\\
&=-\int_{\Omega}\bigg{[}k_{\mu}*\int^{u}_{0}T_{K}(\sigma)h_{l}(\sigma)\,d\sigma\bigg{]}(t_{1})\,dx\\\
&\quad-\int^{t_{1}}_{0}\int_{\Omega}\bigg{[}T_{K}(u(t))\int^{u(t)}_{0}h_{l}(\sigma)\,d\sigma-\int^{u(t)}_{0}T_{K}(\sigma)h_{l}(\sigma)\,d\sigma\bigg{]}k_{\mu}(t)\,dxdt\\\
&\quad-\int^{t_{1}}_{0}\int_{\Omega}\int^{t}_{0}\bigg{[}\int^{u(t-s)}_{u(t)}T_{K}(\sigma)h_{l}(\sigma)\,d\sigma-
T_{K}(u(t))\int^{u(t-s)}_{u^{(}t)}h_{l}(\sigma)\,d\sigma\bigg{]}[-k^{\prime}_{\mu}(s)]\,dsdxdt.\end{split}$
(4.17)

According to the Young’s inequality and $k_{\mu}\to k$ in $L^{1}(0,T)$, we
have

$\int_{\Omega}\bigg{[}k_{\mu}*\int^{u}_{0}T_{K}(\sigma)h_{l}(\sigma)\,d\sigma\bigg{]}dx\xrightarrow{\mu\to
0}\int_{\Omega}\bigg{[}k*\int^{u}_{0}T_{K}(\sigma)h_{l}(\sigma)\,d\sigma\bigg{]}dx$

in $L^{1}(0,T)$ and a.e. in $(0,T)$ for subsequence. In addition, we note that

$\int^{u}_{0}T_{K}(\sigma)h_{l}(\sigma)\,d\sigma\leq\int^{u}_{0}T_{K}(\sigma)\,d\sigma\quad\text{a.e.
in }\Omega_{T},$

then the Lebesgue’s dominated theorem yields that

$\int_{\Omega}\bigg{[}k*\int^{u}_{0}T_{K}(\sigma)h_{l}(\sigma)\,d\sigma\bigg{]}\,dx\xrightarrow{l\to\infty}\int_{\Omega}\bigg{[}k*\int^{u}_{0}T_{K}(\sigma)\,d\sigma\bigg{]}\,dx\quad\text{in
}L^{1}(0,T)$

and a.e. in $(0,T)$ for subsequence.

Note that $0\leq h_{l}\leq 1$ also implies

$\begin{split}0&\leq
T_{K}(u(t))\int^{u}_{0}h_{l}(\sigma)\,d\sigma-\int^{u}_{0}T_{K}(\sigma)h_{l}(\sigma)\,d\sigma\\\
&\leq T_{K}(u(t))u(t)-\int^{u}_{0}T_{K}(\sigma)\,d\sigma\end{split}$

a.e. on $\Omega_{T}$, and

$\begin{split}0&\leq\int^{u(t-s)}_{u(t)}T_{K}(\sigma)h_{l}(\sigma)\,d\sigma-
T_{K}(u(t))\int^{u(t-s)}_{u(t)}h_{l}(\sigma)\,d\sigma\\\
&\leq\int^{u(t-s)}_{u(t)}T_{K}(\sigma)\,d\sigma-
T_{K}(u(t))(u(t-s)-u(t))\end{split}$

a.e. on $\Omega_{T}$. Due to $k_{\mu}$ is non-negative, non-increasing and
satisfies (K1) and (K2), we deduce from the Lebesgue’s dominated theorem that

$\begin{split}&\liminf_{l\to\infty}\liminf_{\mu\to
0}\int^{t_{1}}_{0}\int_{\Omega}\bigg{[}T_{K}(u(t))\int^{u}_{0}h_{l}(\sigma)\,d\sigma-\int^{u}_{0}T_{K}(\sigma)h_{l}(\sigma)\,d\sigma\bigg{]}k_{\mu}(t)\,dxdt\\\
&\quad=\int^{t_{1}}_{0}\int_{\Omega}\bigg{[}T_{K}(u(t))u(t)-\int^{u}_{0}T_{K}(\sigma)\,d\sigma\bigg{]}k(t)\,dxdt\end{split}$

and

$\begin{split}&\liminf_{l\to\infty}\liminf_{\mu\to
0}\int^{t_{1}}_{0}\int_{\Omega}\int^{t}_{0}\bigg{[}\int^{u(t-s)}_{u(t)}T_{K}(\sigma)h_{l}(\sigma)\,d\sigma-
T_{K}(u(t))\int^{u(t-s)}_{u^{(}t)}h_{l}(\sigma)\,d\sigma\bigg{]}[-k^{\prime}_{\mu}(s)]\,dsdxdt\\\
&\quad=\int^{t_{1}}_{0}\int_{\Omega}\int^{t}_{0}\bigg{[}\int^{u(t-s)}_{u(t)}T_{K}(\sigma)\,d\sigma-
T_{K}(u(t))(u(t-s)-u(t))\bigg{]}[-k^{\prime}(s)]\,dsdxdt.\end{split}$

Thus, passing to the limit in (4.16) yields

$\begin{split}&\liminf_{l\to\infty}\liminf_{\mu\to
0}\liminf_{m\to\infty}\liminf_{\lambda\to 0}J_{12}^{\lambda,m,\mu,l}\\\
&\quad=-\liminf_{l\to\infty}\liminf_{\mu\to
0}\liminf_{m\to\infty}\liminf_{\lambda\to 0}I^{1}_{\lambda,m,\mu,l}\\\
&\quad=-\int_{\Omega}[k*\int^{u}_{0}T_{K}(\sigma)\,d\sigma](t_{1})\,dx\\\
&\quad\quad-\int^{t_{1}}_{0}\int_{\Omega}\bigg{[}T_{K}(u(t))u-\int^{u}_{0}T_{K}(\sigma)\,d\sigma\bigg{]}k(t)\,dxdt\\\
&\quad\quad-\int^{t_{1}}_{0}\int_{\Omega}\int^{t}_{0}\bigg{[}\int^{u(t-s)}_{u(t)}T_{K}(\sigma)\,d\sigma-
T_{K}(u(t))(u(t-s)-u(t))\bigg{]}[-k^{\prime}(s)]\,dsdxdt.\end{split}$

Combining with (4.7) and (4.12), we obtain

$\begin{split}&\liminf_{l\to\infty}\liminf_{\mu\to
0}\liminf_{m\to\infty}\liminf_{\lambda\to
0}\int^{t_{1}}_{0}\int_{\Omega}J_{1}^{\lambda,m,n,\mu,l}\\\
&\quad=\liminf_{l\to\infty}\liminf_{\mu\to
0}\liminf_{m\to\infty}\liminf_{\lambda\to
0}\int^{t_{1}}_{0}\int_{\Omega}J_{11}^{\lambda,m,n,\mu,l}+J_{12}^{\lambda,m,n,\mu,l}\geq
0,\end{split}$

which end the proof of (4.6).

Limit of $J^{\lambda,m,\mu,l}_{2}$ Therefore, we conclude that for
$t_{1}\in(0,T)$,

$\begin{split}&\limsup_{l\to\infty}\limsup_{\mu\to
0}\limsup_{m\to\infty}\limsup_{\lambda\to 0}J^{\lambda,m,\mu,l}_{2}\\\
&\quad=\limsup_{l\to\infty}\limsup_{\mu\to
0}\limsup_{m\to\infty}\int^{t_{1}}_{0}\langle(-\Delta)^{s}_{2}u^{m},T_{K}(u^{m})-h_{l}(u^{m})(T_{K}(u))_{\mu}\rangle
dt\\\ &\quad\leq 0.\end{split}$ (4.18)

Then, it is easy to deduce that

$\begin{split}&\limsup_{m\to\infty}\int^{T}_{0}\langle(-\Delta)^{s}_{2}u^{m},T_{K}(u^{m})-T_{K}(u)\rangle
dt\\\ &\quad\leq\limsup_{l\to\infty}\limsup_{\mu\to
0}\limsup_{m\to\infty}\int^{t_{1}}_{0}\langle(-\Delta)^{s}_{2}u^{m},h_{l}(u^{m})(T_{K}(u))_{\mu}-T_{K}(u)\rangle
dt.\end{split}$

Denote that

$\begin{split}&H^{m,\mu,l}:=(u^{m}(x)-u^{m}(y))\bigg{(}h_{l}(u^{m}(x))(T_{K}(u(x)))_{\mu}-T_{K}(u(x))\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad-(h_{l}(u^{m}(y))(T_{K}(u(y)))_{\mu}-T_{K}(u(y)))\bigg{)}.\end{split}$

We have

$\begin{split}&\int^{T}_{0}\langle(-\Delta)^{s}_{2}u^{m},h_{l}(u^{m})(T_{K}(u))_{\mu}-T_{K}(u)\rangle
dt=\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}H^{m,\mu,l}\,dvdt.\end{split}$

Taking $l\geq K$, we set

$\begin{split}&D_{1}=\\{(x,y,t)\in\mathcal{D}\times(0,T):u^{m}(x,t)\leq
l,\qquad u^{m}(y,t)\leq l\\},\\\
&D_{2}=\\{(x,y,t)\in\mathcal{D}\times(0,T):u^{m}(x,t)\leq l,\qquad l+1\leq
u^{m}(y,t)\\},\\\ &D_{3}=\\{(x,y,t)\in\mathcal{D}\times(0,T):u^{m}(x,t)\leq
l,\qquad l\leq u^{m}(y,t)\leq l+1\\},\\\
&D_{4}=\\{(x,y,t)\in\mathcal{D}\times(0,T):l\leq u^{m}(x,t)\leq l+1,\qquad
u^{m}(y,t)\leq l\\},\\\ &D_{5}=\\{(x,y,t)\in\mathcal{D}\times(0,T):l\leq
u^{m}(x,t)\leq l+1,\qquad l+1\leq u^{m}(y,t)\\},\\\
&D_{6}=\\{(x,y,t)\in\mathcal{D}\times(0,T):l\leq u^{m}(x,t)\leq l+1,\qquad
l\leq u^{m}(y,t)\leq l+1\\},\\\
&D_{7}=\\{(x,y,t)\in\mathcal{D}\times(0,T):l+1\leq u^{m}(x,t),\qquad
u^{m}(y,t)\leq l\\},\\\ &D_{8}=\\{(x,y,t)\in\mathcal{D}\times(0,T):l+1\leq
u^{m}(x,t),\qquad l+1\leq u^{m}(y,t)\\},\\\
&D_{9}=\\{(x,y,t)\in\mathcal{D}\times(0,T):l+1\leq u^{m}(x,t),\qquad l\leq
u^{m}(y,t)\leq l+1\\}.\end{split}$

Then

$\mathcal{D}\times(0,T)=D_{1}\cup D_{2}\cup D_{3}\cup D_{4}\cup D_{5}\cup
D_{6}\cup D_{7}\cup D_{8}\cup D_{9}.$

In $D_{1}$, we have $h_{l}(u^{m}(x))=h_{l}(u^{m}(y))=1$. Therefore, we obtain

$\begin{split}&\limsup_{\mu\to
0}\limsup_{m\to\infty}\int_{D_{1}}H^{m,\mu,l}\,dvdt\\\ &\quad=\limsup_{\mu\to
0}\limsup_{m\to\infty}\int_{D_{1}}(u^{m}(x)-u^{m}(y))\bigg{(}(T_{K}(u(x)))_{\mu}-T_{K}(u(x))\\\
&\qquad\qquad\qquad\qquad-((T_{K}(u(y)))_{\mu}-T_{K}(u(y)))\bigg{)}\,dvdt\\\
&\quad\leq\limsup_{\mu\to
0}\limsup_{m\to\infty}\int_{D}|T_{l}(u^{m}(x))-T_{l}(u^{m}(y))|\bigg{|}(T_{K}(u(x)))_{\mu}-T_{K}(u(x))\\\
&\qquad\qquad\qquad\qquad-((T_{K}(u(y)))_{\mu}-T_{K}(u(y)))\bigg{|}\,dvdt.\end{split}$

Since $(T_{K}(u))_{\mu}$ strongly convergence to $T_{K}(u)$ in
$L^{2}(0,T;X_{0}^{s,2}(\Omega))$ and a.e. in $\Omega_{T}$ as $\mu\to 0$, it
follows that

$\limsup_{\mu\to 0}\limsup_{m\to\infty}\int_{D_{1}}H^{m,\mu,l}\,dvdt\leq 0.$

In $D_{2}$, we have

$\begin{split}\int_{D_{2}}H^{m,\mu,l}\,dvdt=\int_{D_{2}}(u^{m}(x)-u^{m}(y))\bigg{(}(T_{K}(u(x)))_{\mu}-T_{K}(u(x))+T_{K}(u(y))\bigg{)}\,dvdt.\end{split}$

Combining with the fact that $K\leq l+1\leq u^{m}(y)\leq u(y)$, we know that

$T_{K}(u(y))-T_{K}(u(x))\geq 0.$

Thus, we conclude that

$\limsup_{\mu\to 0}\limsup_{m\to\infty}\int_{D_{2}}H^{m,\mu,l}\,dvdt\leq 0.$

We point out out that the estimates in $D_{7}$ are similarly to the case in
$D_{2}$.

In $D_{3}$, we have

$\begin{split}&\limsup_{\mu\to
0}\limsup_{m\to\infty}\int_{D_{3}}H^{m,\mu,l}\,dvdt\\\ &\quad=\limsup_{\mu\to
0}\limsup_{m\to\infty}\int_{D_{3}}(u^{m}(x)-u^{m}(y))\bigg{(}(T_{K}(u(x)))_{\mu}-T_{K}(u(x))\\\
&\qquad\qquad\qquad\qquad-(h_{l}(u^{m}(y))(T_{K}(u(y)))_{\mu}-T_{K}(u(y)))\bigg{)}\,dvdt\\\
&\quad\leq\limsup_{\mu\to
0}\limsup_{m\to\infty}\int_{D}|T_{l+1}(u^{m}(x))-T_{l+1}(u^{m}(y))|\bigg{|}(T_{K}(u(x)))_{\mu}-T_{K}(u(x))\\\
&\qquad\qquad\qquad\qquad-((T_{K}(u(y)))_{\mu}-T_{K}(u(y)))\bigg{|}\,dvdt.\end{split}$

Similarly to the estimates in $D_{1}$, we have

$\limsup_{\mu\to 0}\limsup_{m\to\infty}\int_{D_{3}}H^{m,\mu,l}\,dvdt\leq 0.$

In $D_{4}$, it can be done similarly to the estimate in $D_{3}$.

In $D_{5}$, we have

$\begin{split}\int_{D_{5}}H^{m,\mu,l}\,dvdt=\int_{D_{5}}(u^{m}(x)-u^{m}(y))\bigg{(}h_{l}(u^{m}(x))(T_{K}(u(x)))_{\mu}-T_{K}(u(x))+T_{K}(u(y))\bigg{)}\,dvdt.\end{split}$

Similarly to estimate in $D_{2}$, we have

$\limsup_{\mu\to 0}\limsup_{m\to\infty}\int_{D_{5}}H^{m,\mu,l}\,dvdt\leq 0.$

It is easy to check that the estimates in $D_{9}$ are similarly to the case in
$D_{5}$.

In $D_{6}$, since $K\leq l\leq u^{m}(x)\leq u(x)$, $K\leq l\leq u^{m}(y)\leq
u(y)$, we have

$\begin{split}\int_{D_{6}}H^{m,\mu,l}\,dvdt&=\int_{D_{6}}(u^{m}(x)-u^{m}(y))\bigg{(}h_{l}(u^{m}(x))-h_{l}(u^{m}(y))\bigg{)}K_{\mu}\,dvdt,\\\
&\leq K\int_{D_{6}}(u^{m}(x)-u^{m}(y))^{2}h^{\prime}_{l}(\xi)\,dvdt\\\ &\leq
0,\end{split}$

where $\xi$ is between $u^{m}(x)$ and $u^{m}(y)$.

In $D_{8}$, since $h_{l}(u^{m}(x))=h_{l}(u^{m}(y))=0$,
$T_{K}(u(x))=T_{K}(u(y))=K$, we immediately obtain

$\int_{D_{9}}H^{m,\mu,l}\,dvdt=0.$

Combining all the above estimates, we get

$\begin{split}&\limsup_{m\to\infty}\int^{T}_{0}\langle(-\Delta)^{s}u^{m},T_{K}(u^{m})-T_{K}(u)\rangle
dt\\\ &\quad\leq\limsup_{l\to\infty}\limsup_{\mu\to
0}\limsup_{m\to\infty}\int^{T}_{0}\langle(-\Delta)^{s}u^{m},h_{l}(u^{m})(T_{K}(u))_{\mu}-T_{K}(u)\rangle
dt\\\ &\quad\leq 0,\end{split}$

which implies that

$\begin{split}&\limsup_{m\to\infty}\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}(u^{m}(x)-u^{m}(y))(T_{K}(u^{m}(x))-T_{K}(u^{m}(y)))\,dvdt\\\
&\quad\leq\limsup_{m\to\infty}\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}(u^{m}(x)-u^{m}(y))(T_{K}(u(x))-T_{K}(u(y)))\,dvdt.\end{split}$
(4.19)

Then following the arguments of the proof of [2, Lemma 3.6], we have

$\begin{split}&\limsup_{m\to\infty}\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}\bigg{|}T_{K}(u^{m}(x))-T_{K}(u^{m}(y))\bigg{|}^{2}\,dvdt\\\
&\quad\leq\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}\bigg{|}T_{K}(u(x))-T_{K}(u(y))\bigg{|}^{2}\,dvdt.\end{split}$
(4.20)

Since $T_{K}(u^{m})\overset{m\to\infty}{\rightharpoonup}T_{K}(u)$ weakly in
$L^{2}(0,T;X^{s,2}_{0}(\Omega))$, we obtain
$T_{K}(u^{m})\xrightarrow{m\to\infty}T_{K}(u)$ strongly in
$L^{2}(0,T;X^{s,2}_{0}(\Omega))$.

Step 3. Show that $u$ is an entropy solution.

Let $S\in\mathcal{P}$, $\phi\in X^{s,2}_{0}(\Omega)\cap L^{\infty}(\Omega)$,
$\zeta\in C^{\infty}_{0}([0,T))$, $\zeta\geq 0$. Taking $S(u^{m}-\phi)\zeta$
as a test function in (3.3), we obtain

$\begin{split}&\int^{T}_{0}\int_{\Omega}\zeta
S(u^{m}-\phi)\partial_{t}[k*(u^{m}-u^{m}_{0})]\,dxdt\\\
&\quad+\frac{1}{2}\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}\zeta(u^{m}(x)-u^{m}(y))(S(u^{m}-\phi)(x)-S(u^{m}-\phi)(y))\,dvdt\\\
&\quad=\int_{0}^{T}\int_{\Omega}f^{m}S(u^{m}-\phi)\zeta\,dxdt.\end{split}$

Choosing arbitrary $k_{1},k_{2}\in L^{1}(0,T)$ non-increasing and non-negative
with $k_{2}(0^{+})<\infty$ such that $k=k_{1}+k_{2}$. For $\lambda>0$ we
define $k_{1,\lambda}$ by the kernel associated to the Yosida-approximation of
the operator $B^{1}:=\partial_{t}(k_{1}*\cdot)$, $D(B^{1}):=\\{\omega\in
L^{2}(0,T;(X^{s,2}_{0}(\Omega))^{\prime});k_{1}*\omega\in{}_{0}W^{1,2}(0,T;(X^{s,2}_{0}(\Omega))^{\prime})\\}$.
Then there holds

$\begin{split}&\int^{T}_{0}\int_{\Omega}\zeta
S(u^{m}-\phi)\partial_{t}[k_{1,\lambda}*(u^{m}-u^{m}_{0})]\,dxdt+\int^{T}_{0}\int_{\Omega}\zeta
S(u^{m}-\phi)\partial_{t}[k_{2}*(u^{m}-u^{m}_{0})]\,dxdt\\\
&\quad+\frac{1}{2}\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}\zeta(u^{m}(x)-u^{m}(y))(S(u^{m}-\phi)(x,t)-S(u^{m}-\phi)(y,t))\,dvdt\\\
&\quad=\int_{0}^{T}\int_{\Omega}f^{m}S(u^{m}-\phi)\zeta\,dxdt+\int^{T}_{0}\int_{\Omega}\zeta
S(u^{m}-\phi)\partial_{t}[(k_{1,\lambda}-k_{1})*(u^{m}-u^{m}_{0})]\,dxdt.\end{split}$

According to Lemma 2.5, we obtain

$\begin{split}&-\int^{T}_{0}\int_{\Omega}\zeta_{t}(k_{1,\lambda}*\int^{u^{m}}_{u^{m}_{0}}S(\sigma-\phi)\,d\sigma
dxdt+\int^{T}_{0}\int_{\Omega}\zeta
S(u^{m}-\phi)\partial_{t}[k_{2}*(u^{m}-u^{m}_{0})]\,dxdt\\\
&\quad+\frac{1}{2}\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}\zeta(u^{m}(x)-u^{m}(y))(S(u^{m}-\phi)(x)-S(u^{m}-\phi)(y))\,dvdt\\\
&\quad=\int_{0}^{T}\int_{\Omega}f^{m}S(u^{m}-\phi)\zeta\,dxdt+\int^{T}_{0}\int_{\Omega}\zeta
S(u^{m}-\phi)\partial_{t}[(k_{1,\lambda}-k_{1})*(u^{m}-u^{m}_{0})]\,dxdt.\end{split}$

Recalling the fact that $k_{1,\lambda}\to k_{1}$ in $L^{1}(0,T)$ and
$u^{m}-u^{m}_{0}\in D(B^{1})$, we obtain
$\partial_{t}(k_{1,\lambda}*(u^{m}-u^{m}_{0}))\to\partial_{t}(k_{1}*(u^{m}-u^{m}_{0}))$
in $L^{2}(0,T;(X^{s,2}_{0}(\Omega))^{\prime})$. Passing to the limit as
$\lambda\to 0$ in the above equation, we have

$\begin{split}&-\int^{T}_{0}\int_{\Omega}\zeta_{t}\bigg{(}k_{1}*\int^{u^{m}}_{u^{m}_{0}}S(\sigma-\phi)\,d\sigma\bigg{)}\,dxdt+\int^{T}_{0}\int_{\Omega}\zeta\partial_{t}[k_{2}*(u^{m}-u^{m}_{0})]S(u^{m}-\phi)\,dxdt\\\
&\quad+\frac{1}{2}\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}(u^{m}(x)-u^{m}(y))\bigg{(}S(u^{m}(x,t)-\phi(x,t))-S(u^{m}(y,t)-\phi(y,t))\bigg{)}\zeta\,dvdt\\\
&\leq\int^{T}_{0}\int_{\Omega}\zeta f^{m}S(u^{m}-\phi)\,dxdt.\end{split}$
(4.21)

We first consider the term in the right-hand side of (4.21). Since $S$ is
bounded and continuous, we immediately see that

$\lim_{m\to\infty}\int^{T}_{0}\int_{\Omega}\zeta
f^{m}S(u^{m}-\phi)\,dxdt=\int^{T}_{0}\int_{\Omega}\zeta fS(u-\phi)\,dxdt.$

We now deal with the first term of (4.21). Since
$u^{m}\xrightarrow[]{m\to\infty}u$ a.e. in $\Omega_{T}$, it follows that

$\int^{u^{m}}_{u^{m}_{0}}S(\sigma-\phi)\,d\sigma\to\int^{u}_{u_{0}}S(\sigma-\phi)\,d\sigma\quad\text{a.e
in }\Omega_{T}.$

Since $|u^{m}|\leq|u|$, it follows that

$\bigg{|}\int^{u^{m}}_{u^{m}_{0}}S(\sigma-\phi)\,d\sigma\bigg{|}\leq C|u|\in
L^{1}(\Omega_{T}).$

Hence, we deduced from the Lebesgue’s dominated theorem that

$\lim_{m\to\infty}\int^{T}_{0}\int_{\Omega}\zeta_{t}\bigg{(}k_{1}*\int^{u^{m}}_{u^{m}_{0}}S(\sigma-\phi)\,d\sigma\bigg{)}\,dxdt=\int^{T}_{0}\int_{\Omega}\zeta_{t}\bigg{(}k_{1}*\int^{u}_{u_{0}}S(\sigma-\phi)\,d\sigma\bigg{)}\,dxdt.$

We now deal with the second term of (4.21). Note that

$\begin{split}&\partial_{t}[k_{2}*(u^{m}-u^{m}_{0})](t)\\\
&\quad=k_{2}(0^{+})(u^{m}(t)-u^{m}_{0})+\int^{t}_{0}u^{m}(t-s)-u^{m}_{0}\,dk_{2}(s)\\\
&\overset{m\to\infty}{\to}k_{2}(0^{+})(u(t)-u_{0})+\int^{t}_{0}u(t-s)-u_{0}\,dk_{2}(s)\end{split}$

in $L^{1}(\Omega_{T})$. Consequently, we have

$\lim_{m\to\infty}\int^{T}_{0}\int_{\Omega}\zeta\partial_{t}[k_{2}*(u^{m}-u^{m}_{0})]S(u^{m}-\phi)\,dxdt=\int^{T}_{0}\int_{\Omega}\zeta\partial_{t}[k_{2}*(u-u_{0})]S(u-\phi)\,dxdt.$

Next, we focus on the third term in the left-hand side of equation (4.21). We
set

$w^{m}=u^{m}-\phi,\quad U^{m}(x,y,t)=u^{m}(x,t)-u^{m}(y,t)\quad\text{and}\quad
W^{m}(x,y,t)=w^{m}(x,t)-w^{m}(y,t),$

then

$\begin{split}&U^{m}(x,y,t)[S(w^{m}(x,t))-S(w^{m}(y,t))]\zeta\\\
&\quad=:M^{1,m}(x,y,t)\zeta+M^{2,m}(x,y,t)\zeta,\end{split}$

where

$M^{1,m}(x,y,t)=W^{m}(x,y,t)[S(w^{m}(x,t))-S(w^{m}(y,t))],$

and

$M^{2,m}(x,y,t)=[U^{m}(x,y,t)-W^{m}(x,y,t)][S(w^{m}(x,t))-S(w^{m}(y,t))].$

Since $\zeta\in\mathcal{D}([0,T))$, $\zeta\geq 0$,

$\begin{split}M^{1,m}(x,y,t)&=W^{m}(x,y,t)[S(w^{m}(x,t))-S(w^{m}(y,t))]\\\
&=S^{\prime}(\omega_{m}(\cdot))[w^{m}(x,t)-w^{m}(y,t)]^{2}\geq 0,\end{split}$

and

$M^{1,m}(x,y,t)\xrightarrow[]{m\to\infty}W(x,y,t)[S(w(x,t))-S(w(y,t))]\quad\text{a.e.
in }\Omega_{T},$

where $w=u-\phi$ and $W(x,y,t)=w(x,t)-w(y,t)$, it follows from the Fatou’s
Lemma that

$\begin{split}\lim_{m\to\infty}\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}M^{1,m}(x,y,t)\zeta\,dvdt\geq\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}W(x,y,t)[S(w(x,t))-S(w(y,t))]\zeta\,dvdt.\end{split}$

In addition, it is easy to check that

$\begin{split}|M^{2,m}(x,y,t)|\leq|\phi(x,t)-\phi(y,t)||S(w^{m}(x,t))-S(w^{m}(y,t))|.\end{split}$

Since $S\in\mathcal{P}$, we claim that there exists a constant $K$ large
enough such that $S(v)=S(T_{K}(v))$ for any $v\in{\mathbb{R}}$. Recalling the
fact that $T_{K}(u^{m})\xrightarrow{m\to\infty}T_{K}(u)$ strongly in
$L^{2}(0,T;X^{s,2}_{0}(\Omega))$, we get

$S(w^{m})\xrightarrow{m\to\infty}S(w)\quad\text{ strongly in
}L^{2}(0,T;X^{s,2}_{0}(\Omega)),$

which implies that

$\frac{S(w^{m}(x,t))-S(w^{m}(y,t))}{|x-y|^{\frac{N+2s}{2}}}\to\frac{S(w(x,t))-S(w(y,t))}{|x-y|^{\frac{N+2s}{2}}}$
(4.22)

strongly in $L^{2}(0,T;\mathcal{D}_{\Omega})$. Since $\phi\in
X^{s,2}_{0}(\Omega)\cap L^{\infty}(\Omega)$, we deduce by duality argument
that

$\begin{split}&\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}\zeta(\phi(x,t)-\phi(y,t))[S(w^{m}(x,t))-S(w^{m}(y,t))]\,dvdt\\\
&\quad\xrightarrow{m\to\infty}\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}\zeta(\phi(x,t)-\phi(y,t))[S((w(x,t))-S(w(y,t))]\,dvdt.\end{split}$

Using the Lebesgue’s dominated theorem, we get

$\begin{split}&\lim_{m\to\infty}\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}\zeta
M^{2,m}(x,y,t)\,dvdt\\\
&\quad=\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}\zeta[U(x,y,t)-W(x,y,t)][S(w(x,t))-S(w(y,t))]\,dvdt.\end{split}$

Therefore, we conclude that

$\begin{split}&\lim_{m\to\infty}\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}\zeta
U^{m}(x,y,t)[S(w^{m}(x,t))-S(w^{m}(y,t))]\,dvdt\\\
&\quad\geq\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}\zeta
U(x,y,t)[S(w(x,t))-S(w(y,t))]\,dvdt.\end{split}$

Combining all the above estimates we conclude that

$\displaystyle-\int^{T}_{0}\int_{\Omega}\zeta_{t}\bigg{[}k_{1}*\int^{u}_{u_{0}}S(\sigma-\phi)\,d\sigma\bigg{]}+\int^{T}_{0}\int_{\Omega}\zeta\partial_{t}[k_{2}*(u-u_{0})]S(u-\phi)\,dxdt$
$\displaystyle\quad+\frac{1}{2}\int^{T}_{0}\int_{\mathcal{D}_{\Omega}}(u(x)-u(y))\bigg{(}S(u(x,t)-\phi(x,t))-S(u(y,t)-\phi(y,t))\bigg{)}\zeta\,dvdt$
$\displaystyle\quad\leq\int^{T}_{0}\int_{\Omega}\zeta fS(u-\phi)\,dxdt$

for all $\phi\in X^{s,2}_{0}(\Omega)\cap L^{\infty}(\Omega)$, $\zeta\in
C^{\infty}_{0}([0,T))$, $\zeta\geq 0$, $S\in\mathcal{P}$, and $k_{1},k_{2}\in
L^{1}(0,T)$ non-increasing and non-negative with $k=k_{1}+k_{2}$ and
$k_{2}(0^{+})<\infty$. This completes the proof of Theorem 1.2. $\Box$

Proof of Proposition 1.3. Suppose that $u^{im}(i=1,2)$ are weak solutions to
problem (3.2) with $u_{0}=u^{im}_{0}$, $f=f^{im}$ respectively. We take the
test function as $H^{\prime}_{\epsilon}((u^{1m}-u^{2m})^{+})$ in the weak
formulation of the approximate problem for both $u^{1m}$ and $u^{2m}$. We
point out that the function $H_{\epsilon}$ is defined as (3.6). Arguing as the
proof of Lemma 3.2, we obtain

$\displaystyle\int^{T}_{0}\int_{\Omega}(u^{1m}-u^{2m})^{+}\,dxdt\leq
T\int_{\Omega}(u_{0}^{1m}-u_{0}^{2m})^{+}\,dx+\|l\|_{L^{1}(0,T)}\int^{T}_{0}\int_{\Omega}(f^{1m}-f^{2m})^{+}\,dx.$

Passing to the limit with $m\to\infty$, it is easy to deduce that

$\displaystyle\int^{T}_{0}\int_{\Omega}(u^{1}-u^{2})^{+}\,dxdt\leq
T\int_{\Omega}(u_{0}^{1}-u_{0}^{2})^{+}\,dx+\|l\|_{L^{1}(0,T)}\int^{T}_{0}\int_{\Omega}(f^{1}-f^{2})^{+}\,dx.$

So, we obtain the desired result. The proof of (1.4) is analogous to (1.3)
with the only difference that in the case of (1.4), we take the test function
as $H^{\prime}_{\epsilon}(u^{1m}-u^{2m})$. $\Box$

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