# Weak solvability of a boundary value problem
for a parabolic equation with a global-in-time term
that contains a weighted integral

V. N. Starovoitov
Lavrentyev Institute of Hydrodynamics, Novosibirsk, Russian Federation
E-mail<EMAIL_ADDRESS>
###### Abstract

This paper deals with a parabolic partial differential equation that includes
a non-linear nonlocal in time term. This term is the product of a so-called
interaction potential and the solution of the problem. The interaction
potential depends on a weighted integral of the solution over the entire time
interval, where the problem is considered, and satisfies fairly general
conditions. Namely, it is assumed to be a continuous bounded from below
function that can behave arbitrarily at infinity. This fact implies that the
interaction term is not a lower order term in the equation. The weak
solvability of the initial boundary value problem for this equation is proven.
The proof does not use any continuity properties of the solution with respect
to time and is based on the energy estimate only.

Key words: nonlocal in time parabolic equation, weighted integral, initial
boundary value problem, solvability

2010 Mathematics Subject Classification: 35K58, 35Q92

##  1 Introduction

Let $\varOmega$ be a bounded domain in $\mathbb{R}^{n}$, $n\geq 2$, with a
Lipschitz boundary $\partial\varOmega$. In the space-time cylinder
$\varOmega_{T}=\varOmega\times(0,T)$, $T\in(0,\infty)$, we consider the
following differential equation:

$\partial_{t}u-\Delta
u+\varphi\Big{(}\int_{0}^{T}\alpha(\cdot,s)\,u(\cdot,s)\,ds\Big{)}\,u=f,$
(1.1)

where $u=u(x,t)$ is an unknown scalar function, $x=(x_{1},\ldots,x_{n})$ the
vector of the spatial variables in $\mathbb{R}^{n}$, $t$ the time variable on
$[0,T]$, $\varphi:\mathbb{R}\to\mathbb{R}$ a function that will be specified
below, $f:\varOmega_{T}\to\mathbb{R}$ a prescribed function. We suppose that
the following boundary and initial conditions are satisfied:

$u(x,t)=0\quad\text{for}\quad x\in\partial\varOmega,\quad t\in[0,T],$ (1.2)
$u(x,0)=u_{0}(x)\quad\text{for}\quad x\in\varOmega,$ (1.3)

where the function $u_{0}:\varOmega\to\mathbb{R}$ is prescribed.

Equation (1.1) contains a non-local in time $t$ term with the integral over
the whole interval $(0,T)$, where the problem is considered. For this reason,
we call this equation global-in-time. A problem of this type arose when
modeling the chaotic dynamics of a polymer molecule (polymer chain) in an
aqueous solution [1]. The function $u(x,t)$ corresponds to the density of
probability that the $t$-th segment of the chain is in a certain region of the
space. The role of time in the equation is played by the arc length parameter
along the chain whose total length is $T$. The equation contains a term
responsible for the interaction of chain’s segments and the function $\varphi$
is called the _interaction potential_. Since each segment interacts with all
others through the surrounding fluid, $\varphi$ depends on the integral of the
density of probability over the entire chain, i.e., over the entire time
interval $(0,T)$.

Equations with a global-in-time term appear also in other fields, for example,
in the filtration theory [2], in the population dynamics (see [3, 4] and the
references therein). Notice that the equation in the population dynamics is
ultra-parabolic and the age of individuals plays the role of the second time.
In the steady case, the equation becomes parabolic. The global terms contain
the integral with respect to the age and can be in the data of the problem and
in the equation as well. There are a lot of works, where a global-in-time term
stands in the data of the problem. These are problems with non-local boundary
or initial conditions. The latter means that the initial data depend on the
time integral of the solution. This topic is widely represented in the
literature, and there is no problem to find relevant publications. The
problems with global-in-time data are very different from ours.

The problem we study looks like parabolic one, however it has some unusual
features. First of all, since the potential $\varphi$ in equation (1.1) is
global-in-time, the state of the system depends not only on the past but also
on the future. That is, the causality principle is violated. Besides, from a
mathematical point of view, the solution of a nonlinear parabolic problem is
commonly being constructed locally in time and is extended afterwards. In our
case, this procedure is impossible. Finally, as a rule, the local-in-time
uniqueness of the solution implies the global one. We cannot prove the
uniqueness without restrictions on $T$.

There are several papers devoted to problems close to (1.1)–(1.3). In [5, 6],
the weak solvability of problem (1.1)–(1.3) with $\alpha\equiv 1$ and $f=0$ is
proven. Notice that the smoothness of the obtained in this papers weak
solution can be easily improved for better initial data. The strong
solvability of the same problem was proven in [7], where the semigroup
approach was employed. In [8], the authors consider the problem with the
Laplace operator replaced by a more complex non-local operator that can be
thought as the fractional Laplacian. The uniqueness of the solution is proven
for sufficiently small $T$ in all these works. The smallness condition means
that there is a restriction for $T$ which depends on the data of the problem.

In this paper, we consider a more general problem compared with the previous
works. At first, the global-in-time term contains a weighted integral and the
smoothness of the weight $\alpha$ is assumed to be not sufficient to apply the
technique from [5, 6]. Secondly, equation (1.1) contains the right-hand side
$f$. The point is that the proof of the main results of the paper employs only
the energy estimate and does not use any smoothness properties of the solution
with respect to time. The right-hand side $f$ prevents or makes it difficult
to obtain good time-estimates. Finally, the potential $\varphi$ is assumed to
be just a continuous bounded from below function. There are no conditions on
the growth of this function at infinity, no convexity, monotonicity, or
differentiability conditions. In [6], the assumption on the potential
$\varphi$ also does not contain the growth conditions, but $\varphi$ satisfies
other requirements that made it possible to obtain a higher degree of
integrability of $\varphi$ than in the presented paper. The absence of the
growth conditions at infinity leads to the fact that the interaction term is
not a lower order term in equation (1.1). Notice that in [7] this term has a
lower order.

Although the continuous function $\varphi$ is assumed to be just bounded from
below, by standard arguments the problem can be reduced to the problem with a
positive potential. Really, suppose that $\varphi(\xi)\geq-K$ for some
positive number $K$ and all $\xi\in\mathbb{R}$. If $u$ is a solution of
problem (1.1)–(1.3), then $\bar{u}(x,t)=e^{-Kt}\,u(x,t)$ is a solution of the
following problem

$\partial_{t}\bar{u}-\Delta\bar{u}+\bar{\varphi}(\bar{v})\,\bar{u}=\bar{f},\quad\bar{u}|_{\partial\varOmega}=0,\quad\bar{u}|_{t=0}=u_{0},$

where

$\bar{\varphi}=\varphi+K,\quad\bar{v}(x)=\int_{0}^{T}\bar{\alpha}(x,t)\,\bar{u}(x,t)\,dt,\quad\bar{\alpha}(x,t)=e^{Kt}\,\alpha(x,t),\quad\bar{f}(x,t)=e^{-Kt}\,f(x,t).$

Thus, we have the same problem but with the non-negative potential
$\bar{\varphi}$. In what follows, we will assume that $\varphi$ is a non-
negative function.

We will use the standard Lebesgue and Sobolev spaces $L^{p}(\varOmega)$,
$H^{1}_{0}(\varOmega)$, $L^{q}(0,T;L^{p}(\varOmega))$, and
$L^{q}(0,T;H^{1}_{0}(\varOmega))$, where $p,q\in[1,\infty]$. As usual,
$H^{-1}(\varOmega)$ is the dual space of $H^{1}_{0}(\varOmega)$ with respect
to the pivot space $L^{2}(\varOmega)$. The norm in $L^{2}(\varOmega)$ will be
denoted by $\|\cdot\|$.

######  Definition 1.1.

Let $\varphi:\mathbb{R}\to\mathbb{R}$ be a non-negative continuous function,
$T\in(0,\infty)$, $\alpha\in L^{1}(0,T;L^{2}(\varOmega))$, $f\in
L^{1}(0,T;L^{2}(\varOmega))$, and $u_{0}\in L^{2}(\varOmega)$. A function
$u:\varOmega_{T}\to\mathbb{R}$ is said to be a _weak solution of problem (
1.1)–(1.3)_, if

  1. 1.

$u\in L^{2}(0,T;H^{1}_{0}(\varOmega))$ and $\varphi(\zeta)\,u\in
L^{1}(\varOmega_{T})$, where
$\zeta=\int_{0}^{T}\alpha(\cdot,t)\,u(\cdot,t)\,dt$;

  2. 2.

the following integral identity

$\int_{0}^{T}\int_{\varOmega}\big{(}u\,\partial_{t}h-\nabla u\cdot\nabla
h-\varphi(v)\,u\,h+f\,h\big{)}\,dx\,dt+\int_{\varOmega}u_{0}h_{0}\,dx=0$

holds for an arbitrary smooth in the closure of $\varOmega_{T}$ function $h$
such that $h(x,t)=0$ for $x\in\partial\varOmega$ and for $t=T$. Here,
$h_{0}=h|_{t=0}$. •

The main result of the paper is Theorem 3.1 that states the existence of the
weak solution of problem (1.1)–(1.3). In the next section, we consider the
case where the potential is assumed to be bounded and prove Theorem 2.1 that
is an auxiliary result for Theorem 3.1. We emphasize once again that in the
proofs of these theorems we do not use any continuity properties of the
solution in time. This leads to certain difficulties, for example, in proving
the energy estimate (3.1) (see Step 4 of the proof of Theorem 3.1) which is
usually obtained automatically from the corresponding estimate for the
approximate solutions by using the weak semicontinuity of norms in Banach
spaces.

##  2 Problem with a bounded potential

In this section, we consider the case of the problem where the potential
$\varphi:\mathbb{R}\to\mathbb{R}$ is a continuous bounded function.

######  Theorem 2.1.

Assume that $T\in(0,\infty)$, $\alpha\in L^{1}(0,T;L^{2}(\varOmega))$, $f\in
L^{1}(0,T;L^{2}(\varOmega))$, $u_{0}\in L^{2}(\varOmega)$, and
$\varphi:\mathbb{R}\to\mathbb{R}$ is a continuous function such that
$0\leq\varphi(\xi)\leq K$ for all $\xi\in\mathbb{R}$, where $K$ is a positive
number. Then there exists a weak solution $u\in
L^{\infty}(0,T;L^{2}(\varOmega))\cap L^{2}(0,T;H^{1}_{0}(\varOmega))$ of
problem (1.1)–(1.3).

Proof. To prove the solvability of the problem, we employ the Schauder fixed
point theorem. Let $B_{R}$ be the ball of the radius $R$ centered at the zero
in the function space $L^{1}(\varOmega)$. The number $R$ will be determined
later. Let us define the mapping $\varPsi:B_{R}\to B_{R}$ that is referred to
in the Schauder theorem. For every $w\in B_{R}$, we define $u_{w}$ as the
solution of the following problem:

$\partial_{t}u_{w}-\Delta u_{w}+\varphi(w)\,u_{w}=f,\quad
u_{w}|_{\partial\varOmega}=0,\quad u_{w}|_{t=0}=u_{0}.$ (2.1)

As it follows from the classical theory of parabolic equations (see, e.g., [9,
Ch. 7], this problem has a unique weak solution. Now, we set

$\varPsi(w)=\int_{0}^{T}\alpha(\cdot,t)\,u_{w}(\cdot,t)\,dt.$

It necessary to prove that there exists $R>0$ such that $\varPsi(B_{R})\subset
B_{R}$ and $\varPsi:B_{R}\to L^{1}(\varOmega)$ is a compact continuous map.

The usual energy estimate for problem (2.1) looks as follows:

$\|u_{w}(\cdot,t)\|^{2}+\int_{0}^{t}\|\nabla
u_{w}(\cdot,s)\|^{2}\,ds\leq\Big{(}\|u_{0}\|+\int_{0}^{t}\|f(\cdot,s)\|\,ds\Big{)}^{2}+\frac{1}{2}\,\Big{(}\int_{0}^{t}\|f(\cdot,s)\|\,ds\Big{)}^{2}$

for almost all $t\in[0,T]$. We have used the positiveness of $\varphi$
although a similar estimate holds for $\varphi$ which is just bounded. Thus,

$\operatorname*{ess\,sup}_{t\in[0,T]}\|u_{w}(\cdot,t)\|^{2}+\int_{0}^{T}\|\nabla
u_{w}(\cdot,s)\|^{2}\,ds\leq C_{1},$ (2.2)

where
$C_{1}=\big{(}\|u_{0}\|+\int_{0}^{T}\|f(\cdot,s)\|\,ds\big{)}^{2}+\frac{1}{2}\,\big{(}\int_{0}^{T}\|f(\cdot,s)\|\,ds\big{)}^{2}$.
As a consequence of this inequality, we find that

$\|\varPsi(w)\|_{1}\leq C_{2}\sqrt{C_{1}},$

where $C_{2}=\int_{0}^{T}\|\alpha(\cdot,s)\|\,ds$. Therefore,
$\varPsi(B_{R})\subset B_{R}$ with $R=C_{2}\sqrt{C_{1}}$.

In order to prove the continuity of the mapping $\varPsi$, let us take an
arbitrary sequence $\\{w_{k}\\}$ in $B_{R}$ that converges in
$L^{1}(\varOmega)$ to some $w\in B_{R}$. Then $w_{k}\to w$ in measure on
$\varOmega$ and, due to the Lebesgue dominated convergence theorem,
$\varphi(w_{k})\to\varphi(w)$ in $L^{p}(\varOmega)$ for all $p\in[1,\infty)$.
The difference $u_{w_{k}}-u_{w}$ is a weak solution of the following problem:

$\displaystyle\partial_{t}(u_{w_{k}}-u_{w})-\Delta(u_{w_{k}}-u_{w})+\varphi(w)\,(u_{w_{k}}-u_{w})+\big{(}\varphi(w_{k})-\varphi(w)\big{)}\,u_{w_{k}}=0,$
$\displaystyle\quad(u_{w_{k}}-u_{w})|_{\partial\varOmega}=0,\quad(u_{w_{k}}-u_{w})|_{t=0}=0.$

Therefore, it satisfies estimate (2.2) with
$f=\big{(}\varphi(w_{k})-\varphi(w)\big{)}\,u_{w_{k}}$ and $u_{0}=0$. Thus,

$\operatorname*{ess\,sup}_{t\in[0,T]}\|(u_{w_{k}}-u_{w})(\cdot,t)\|^{2}+\int_{0}^{T}\|\nabla(u_{w_{k}}-u_{w})(\cdot,s)\|^{2}\,ds\leq
c_{k},$

where
$c_{k}=\frac{3}{2}\,\big{(}\int_{0}^{T}\|\big{(}\varphi(w_{k})-\varphi(w)\big{)}\,u_{w_{k}}(\cdot,s)\|\,ds\big{)}^{2}$.
Due to the Sobolev embedding theorems, estimate (2.2) implies that

$\int_{0}^{T}\|u_{w_{k}}(\cdot,s)\|_{L^{q}(\varOmega)}^{2}\,ds\leq C_{3},$

for all $k\in\mathbb{N}$ and some independent of $k$ constant $C_{3}$, where
$q=2n/(n-2)$, if $n>2$, and $q=\infty$ otherwise. As a consequence of the
Hölder inequality, we have

$\int_{0}^{T}\|\big{(}\varphi(w_{k})-\varphi(w)\big{)}\,u_{w_{k}}(\cdot,s)\|\,ds\leq\|(\varphi(w_{k})-\varphi(w)\|_{L^{p}(\varOmega)}\,\int_{0}^{T}\|u_{w_{k}}(\cdot,s)\|_{L^{q}(\varOmega)}\,ds,$

where $p=2q/(q-2)$. Thus, $c_{k}\to 0$ and
$\operatorname*{ess\,sup}_{t\in[0,T]}\|(u_{w_{k}}-u_{w})(\cdot,t)\|\to 0$ as
$k\to\infty$. This means that $\varPsi(w_{k})\to\varPsi(w)$ in
$L^{1}(\varOmega)$ as $k\to\infty$. Really,

$\|\varPsi(w_{k})-\varPsi(w)\|_{L^{1}(\varOmega)}\leq\int_{0}^{T}\|\alpha(\cdot,s)\|\,\|(u_{w_{k}}-u_{w})(\cdot,s)\|\,ds\\\
\leq\int_{0}^{T}\|\alpha(\cdot,s)\|\,ds\,\operatorname*{ess\,sup}_{t\in[0,T]}\|(u_{w_{k}}-u_{w})(\cdot,t)\|\to
0\quad\text{as}\quad k\to\infty.$

Thus, the mapping $\varPsi$ is continuous on $B_{R}$.

It remains to prove the compactness of $\varPsi$. It is possible to draw
arguments based on the available higher smoothness of the solution to problem
(2.1). In fact, $u_{w}$ satisfies a stronger estimate than (2.2). However, we
will show that this estimate is sufficient to prove the compactness of
$\varPsi$. Let $\\{w_{k}\\}$ be a bounded sequence in $B_{R}$. We need to show
that there exists a subsequence $\\{w_{k^{\prime}}\\}$ such that
$\varPsi(w_{k^{\prime}})$ converges in $L^{1}(\varOmega)$. According to
estimate (2.2), the sequence $\\{u_{w_{k}}\\}$ is bounded in
$L^{\infty}(0,T;L^{2}(\varOmega))\cap L^{2}(0,T;H^{1}_{0}(\varOmega))$.
Besides that, $\|\nabla\int_{0}^{T}h\,u_{w_{k}}\,dt\|\leq C_{4}$ for an
arbitrary smooth function $h:\varOmega_{T}\to\mathbb{R}$. The constant $C_{4}$
depends, of course, on $h$. Therefore, there exists a function $v\in
L^{\infty}(0,T;L^{2}(\varOmega))\cap L^{2}(0,T;H^{1}_{0}(\varOmega))$ and a
subsequence $\\{u_{w_{k^{\prime}}}\\}$ such that $u_{w_{k^{\prime}}}\to v$ as
$k^{\prime}\to\infty$ weakly in $L^{2}(0,T;H^{1}_{0}(\varOmega))$ and
$*$-weakly in $L^{\infty}(0,T;L^{2}(\varOmega))$, and

$\int_{0}^{T}h\,u_{w_{k^{\prime}}}\,dt\to\int_{0}^{T}h\,v\,dt\quad\text{in}\quad
L^{2}(\varOmega).$

Let us take an arbitrary $\varepsilon>0$. There exists a smooth function
$\alpha_{\varepsilon}:\varOmega_{T}\to\mathbb{R}$ such that
$\|\alpha-\alpha_{\varepsilon}\|_{L^{1}(0,T;L^{2}(\varOmega))}<\varepsilon$.
If we take $h=\alpha_{\varepsilon}$ in the previous relation, then we find
that

$\limsup_{k^{\prime}\to\infty}\Big{\|}\int_{0}^{T}\alpha\,(u_{w_{k^{\prime}}}-v)\,dt\Big{\|}_{L^{1}(\varOmega)}\leq\limsup_{k^{\prime}\to\infty}\Big{\|}\int_{0}^{T}(\alpha-\alpha_{\varepsilon})\,(u_{w_{k^{\prime}}}-v)\,dt\Big{\|}_{L^{1}(\varOmega)}\\\
+\limsup_{k^{\prime}\to\infty}\Big{\|}\int_{0}^{T}\alpha_{\varepsilon}(u_{w_{k^{\prime}}}-v)\,dt\Big{\|}_{L^{1}(\varOmega)}\leq
C_{5}\,\|\alpha-\alpha_{\varepsilon}\|_{L^{1}(0,T;L^{2}(\varOmega))}<C_{5}\,\varepsilon,$

where the constant $C_{5}$ is an upper bound for the norm of
$u_{w_{k^{\prime}}}-v$ in $L^{\infty}(0,T;L^{2}(\varOmega))$. Since
$\varepsilon$ is arbitrary,

$\lim_{k^{\prime}\to\infty}\Big{\|}\int_{0}^{T}\alpha\,(u_{w_{k^{\prime}}}-v)\,dt\Big{\|}_{L^{1}(\varOmega)}=0.$

Thus, the sequence $\\{\varPsi(w_{k^{\prime}})\\}$ converges in
$L^{1}(\varOmega)$.

So, all the conditions of the Schauder theorem is fulfilled and there exists
$w\in L^{1}(\varOmega)$ such that $\varPsi(w)=w$. Then $u=u_{w}$ is a weak
solution of problem (1.1)–(1.3) and the theorem is proven. $\square$

######  Remark 2.2.

In Theorem 2.1, we assumed that $f\in L^{1}(0,T;L^{2}(\varOmega))$. Without
changing the proof, we could suppose that $f\in L^{2}(0,T;H^{-1}(\varOmega))$.
Really, the proof is based on the energy estimate (2.2). This estimate is also
true with
$C_{1}=\|u_{0}\|^{2}+c^{2}\int_{0}^{T}\|f(\cdot,s)\|^{2}_{H^{-1}(\varOmega)}\,ds$,
where the constant $c$ is such that $\|u\|_{H^{1}_{0}(\varOmega)}\leq
c\,\|\nabla u\|$. $\bullet$

##  3 Problem with an unbounded potential

In this section, we consider a more general case where the potential $\varphi$
is a non-negative continuous function which is not necessarily bounded.

######  Theorem 3.1.

Assume that $T\in(0,\infty)$, $\alpha\in L^{\infty}(\varOmega;L^{2}(0,T))$,
$f\in L^{1}(0,T;L^{2}(\varOmega))$, $u_{0}\in L^{2}(\varOmega)$, and
$\varphi:\mathbb{R}\to\mathbb{R}$ is a non-negative continuous function. Then
there exists a weak solution $u\in L^{\infty}(0,T;L^{2}(\varOmega))\cap
L^{2}(0,T;H^{1}_{0}(\varOmega))$ of problem (1.1)–(1.3) such that

$\operatorname*{ess\,sup}_{t\in[0,T]}\|u(\cdot,t)\|^{2}+\int_{0}^{T}\|\nabla
u(\cdot,t)\|^{2}\,dt+\int_{0}^{T}\int_{\varOmega}\varphi(\zeta(x))\,u^{2}(x,t)\,dx\,dt\leq
C_{1},$ (3.1) $\text{$\varphi(\zeta)$, $\varphi(\zeta)\zeta$, and
$\varphi(\zeta)\zeta^{2}$ are in $L^{1}(\varOmega)$},$ (3.2)
$\text{$\varphi(\zeta)u$ and $\varphi(\zeta)u^{2}$ are in
$L^{1}(\varOmega_{T})$},$ (3.3)

where $\zeta=\int_{0}^{T}\alpha(\cdot,t)\,u(\cdot,t)\,dt\in L^{2}(\varOmega)$
and
$C_{1}=\big{(}\|u_{0}\|+\int_{0}^{T}\|f(\cdot,s)\|\,ds\big{)}^{2}+\frac{1}{2}\,\big{(}\int_{0}^{T}\|f(\cdot,s)\|\,ds\big{)}^{2}$.

Proof. For convenience, we split the proof into several steps.

Step 1. For every $k\in\mathbb{N}$, we define the following function:

$\varphi_{k}(\xi)=\begin{cases}\varphi(\xi),&\varphi(\xi)\leq k,\\\
k,&\text{otherwise}.\end{cases}$

Let us denote by $u_{k}$ the weak solution of problem (1.1)–(1.3) with the
potential $\varphi_{k}$. Since $\varphi_{k}$ is bounded and $\alpha\in
L^{\infty}(\varOmega;L^{2}(0,T))\subset L^{1}(0,T;L^{2}(\varOmega))$, the
existence of $u_{k}$ is proven in the previous section (Theorem 2.1). For
brevity, we introduce the notation:

$\zeta_{k}=\int_{0}^{T}\alpha(\cdot,t)\,u_{k}(\cdot,t)\,dt.$

According to Definition 1.1, $u_{k}$ satisfies the integral identity

$\int_{0}^{T}\int_{\varOmega}\big{(}u_{k}\,\partial_{t}h-\nabla
u_{k}\cdot\nabla
h-\varphi_{k}(\zeta_{k})\,u_{k}\,h+f\,h\big{)}\,dx\,dt+\int_{\varOmega}u_{0}\,h_{0}\,dx=0$
(3.4)

for an arbitrary smooth in the closure of $\varOmega_{T}$ function $h$ such
that $h(x,t)=0$ for $x\in\partial\varOmega$ and for $t=T$. The energy estimate
for $u_{k}$ looks as follows:

$\operatorname*{ess\,sup}_{t\in[0,T]}\|u_{k}(\cdot,t)\|^{2}+\int_{0}^{T}\|\nabla
u_{k}(\cdot,t)\|^{2}\,dt+\int_{0}^{T}\int_{\varOmega}\varphi_{k}(\zeta_{k}(x))\,u_{k}^{2}(x,t)\,dx\,dt\leq
C_{1},$ (3.5)

where the constant $C_{1}$ is the same as in (2.2) and (3.1). This estimate
implies that the sequence $\\{u_{k}\\}$ has a subsequence (denoted again by
$\\{u_{k}\\}$) which converges $*$-weakly in
$L^{\infty}(0,T;L^{2}(\varOmega))$ and weakly in
$L^{2}(0,T;H^{1}_{0}(\varOmega))$ to a function $u$. Our goal is to pass to
the limit in (3.4) as $k\to\infty$. Clearly,

$\lim_{k\to\infty}\int_{0}^{T}\int_{\varOmega}\big{(}u_{k}\,\partial_{t}h-\nabla
u_{k}\cdot\nabla
h\big{)}\,dx\,dt=\int_{0}^{T}\int_{\varOmega}\big{(}u\,\partial_{t}h-\nabla
u\cdot\nabla h\big{)}\,dx\,dt.$

The only problem is to prove that

$\lim_{k\to\infty}\int_{0}^{T}\int_{\varOmega}\varphi_{k}(\zeta_{k})\,u_{k}\,h\,dx\,dt=\int_{0}^{T}\int_{\varOmega}\varphi(\zeta)\,u\,h\,dx\,dt.$
(3.6)

Notice that we also need to justify that the right-hand side in the last
relation makes sense.

Step 2. The reasoning at the end of the proof of Theorem 2.1 implies that

$\zeta_{k}\to\zeta\quad\text{in $L^{1}(\varOmega)$ as $k\to\infty$.}$

Consequently, there exists a subsequence (denoted again by $\\{\zeta_{k}\\}$)
such that

$\zeta_{k}\to\zeta\quad\text{almost everywhere in $\varOmega$ as
$k\to\infty$}.$ (3.7)

Due to (3.5), the following estimate holds for all $k\in\mathbb{N}$:

$\int_{\varOmega}\varphi_{k}(\zeta_{k})\,\zeta_{k}^{2}\,dx=\int_{\varOmega}\varphi_{k}(\zeta_{k})\,\Big{(}\int_{0}^{T}\alpha(\cdot,t)\,u_{k}(\cdot,t)\,dt\Big{)}^{2}\,dx\\\
\leq\int_{\varOmega}\varphi_{k}(\zeta_{k})\Big{(}\int_{0}^{T}\alpha^{2}(\cdot,t)\,dt\int_{0}^{T}u_{k}^{2}(\cdot,t)\,dt\Big{)}dx\leq
C_{6},$ (3.8)

where $C_{6}=C_{1}\,\|\alpha\|^{2}_{L^{\infty}(\varOmega;L^{2}(0,T))}$ and
$C_{1}$ is the constant from (3.5). As a consequence of the Fatou lemma, this
estimate together with the continuity of $\varphi$ and (3.7) implies that

$\int_{\varOmega}\varphi(\zeta)\,\zeta^{2}\,dx\leq C_{6}$ (3.9)

which is a part of (3.2). The rest of the assertions in (3.2) are simple
consequence of (3.9).

We can also prove now that $\zeta\in L^{2}(\varOmega)$, and namely that

$\int_{\varOmega}\zeta^{2}\,dx\leq TC_{6},$ (3.10)

where $C_{6}$ is the constant from estimate (3.8). Indeed, exactly as in
(3.8), we obtain that

$\int_{\varOmega}\zeta_{k}^{2}\,dx\leq\int_{\varOmega}\int_{0}^{T}\alpha^{2}(\cdot,t)\,dt\int_{0}^{T}u_{k}^{2}(\cdot,t)\,dt\,dx\leq
T\,\|\alpha\|^{2}_{L^{\infty}(\varOmega;L^{2}(0,T))}\operatorname*{ess\,sup}_{t\in[0,T]}\|u_{k}(\cdot,t)\|^{2}\leq
TC_{6},$

and (3.10) follows from the Fatou lemma. Notice that we can put the constant
from the Poincare inequality instead of $T$ in (3.10), if we estimate
$\int_{\varOmega}u_{k}^{2}\,dx$ by $\int_{\varOmega}|\nabla u_{k}|^{2}\,dx$
and use again (3.5).

Step 3. In this step, we prove (3.6), which completes the proof of the weak
solvability of the problem. Let $h:\varOmega_{T}\to\mathbb{R}$ be an arbitrary
smooth bounded function and $R=\max_{(x,t)\in\varOmega_{T}}|h(x,t)|$. We will
prove that

$\varphi_{k}(\zeta_{k})\int_{0}^{T}u_{k}\,h\,dt\to\varphi(\zeta)\int_{0}^{T}u\,h\,dt\quad\text{as
$k\to\infty$ in $L^{1}(\varOmega)$},$ (3.11)

which clearly implies (3.6) and, as a consequence, the weak solvability of
problem (1.1)–(1.3). To do this, we will apply the Vitali convergence theorem
(see, e.g., [10, Sec. 4.8.7]).

Exactly as for the sequence $\\{\zeta_{k}\\}$ at the end of the proof of
Theorem 2.1, we can establish (selecting, if necessary, a subsequence) that
$\int_{0}^{T}u_{k}\,h\,dt\to\int_{0}^{T}u\,h\,dt$ in $L^{2}(\varOmega)$ and
almost everywhere in $\varOmega$ as $k\to\infty$. Besides that, due to (3.7),

$\varphi_{k}(\zeta_{k})\to\varphi(\zeta)\quad\text{as $k\to\infty$ almost
everywhere in $\varOmega$.}$ (3.12)

Really, let $E\subset\varOmega$ be the set such that $\zeta_{k}(x)\to\zeta(x)$
as $k\to\infty$ for all $x\in E$. As it follows from (3.7),
$\mu_{n}(\varOmega\setminus E)=0$, where $\mu_{n}$ is the $n$-dimensional
Lebesgue measure. For an arbitrary $x\in E$, there exists
$k_{x,1}\in\mathbb{N}$ such that $\varphi(\zeta(x))+1<k_{x,1}$. Since the
function $\varphi$ is continuous, there exists $k_{x,2}\in\mathbb{N}$ such
that $|\varphi(\zeta_{k}(x))-\varphi(\zeta(x))|<1$ for all $k>k_{x,2}$.
Therefore, $\varphi(\zeta_{k}(x))<k$ and, as a consequence,
$\varphi_{k}(\zeta_{k}(x))=\varphi(\zeta_{k}(x))$ whenever
$k>\max\\{k_{x,1},k_{x,2}\\}$. Due to the continuity of $\varphi$, we conclude
that $\varphi_{k}(\zeta_{k}(x))=\varphi(\zeta_{k}(x))\to\varphi(\zeta(x))$ as
$k\to\infty$ for all $x\in E$. Thus,

$\varphi_{k}(\zeta_{k})\int_{0}^{T}u_{k}\,h\,dt\to\varphi(\zeta)\int_{0}^{T}u\,h\,dt\quad\text{
almost everywhere in $\varOmega$ as $k\to\infty$}.$

According to the Vitali theorem, in order to prove (3.11), it is necessary to
establish that the sequence
$\\{\varphi_{k}(\zeta_{k})\int_{0}^{T}u_{k}\,h\,dt\\}$ is uniformly integrable
on $\varOmega$.

Let $E$ be an arbitrary measurable subset of $\varOmega$,
$E_{T}=E\times[0,T]$, and $G_{M}^{k}=\\{(x,t)\in E_{T}\;|\;|u_{k}(x,t)|\geq
M\\}$. Then

$\int_{G_{M}^{k}}\varphi_{k}(\zeta_{k})|u_{k}\,h|\,dx\,dt\leq\frac{R}{M}\int_{G_{M}^{k}}\varphi_{k}(\zeta_{k})\,u_{k}^{2}\,dx\,dt\leq\frac{R}{M}\int_{\varOmega_{T}}\varphi_{k}(\zeta_{k})\,u_{k}^{2}\,dx\,dt\leq\frac{C_{1}R}{M},$

where $C_{1}$ is the constant from (3.5). Besides that,

$\int_{E_{T}\setminus G_{M}^{k}}\varphi_{k}(\zeta_{k})|u_{k}\,h|\,dx\,dt\leq
MR\int_{E_{T}\setminus G_{M}^{k}}\varphi_{k}(\zeta_{k})\,dx\,dt\leq
MR\,T\int_{E}\varphi_{k}(\zeta_{k})\,dx.$

Therefore,

$\Big{|}\int_{E}\varphi_{k}(\zeta_{k})\int_{0}^{T}u_{k}\,h\,dt\,dx\Big{|}=\Big{|}\int_{E_{T}}\varphi_{k}(\zeta_{k})\,u_{k}\,h\,dx\,dt\Big{|}\leq\frac{C_{1}R}{M}+MR\,T\int_{E}\varphi_{k}(\zeta_{k})\,dx.$
(3.13)

Let us estimate $\int_{E}\varphi_{k}(\zeta_{k})\,dx$. If $E_{N}^{k}=\\{x\in
E\;|\;|\zeta_{k}(x)|>N\\}$, where $N\in(0,\infty)$, then (3.8) implies that

$\int_{E_{N}^{k}}\varphi_{k}(\zeta_{k})\,dx\leq\frac{1}{N^{2}}\int_{E_{N}^{k}}\varphi_{k}(\zeta_{k})\zeta_{k}^{2}\,dx\leq\frac{C_{6}}{N^{2}}.$

Since the function $\varphi$ is continuous, there exists a constant
$\gamma_{N}>0$ such that $\varphi(\xi)\leq\gamma_{N}$ for all $\xi\in[-N,N]$.
Therefore,

$\int_{E\setminus E_{N}^{k}}\varphi_{k}(\zeta_{k})\,dx\leq\int_{E\setminus
E_{N}^{k}}\varphi(\zeta_{k})\,dx\leq\gamma_{N}\mu_{n}(E).$

Thus,

$\int_{E}\varphi_{k}(\zeta_{k})\,dx\leq\frac{C_{6}}{N^{2}}+\gamma_{N}\mu_{n}(E)$

and, as it follows from (3.13),

$\Big{|}\int_{E}\varphi_{k}(\zeta_{k})\int_{0}^{T}u_{k}\,h\,dt\,dx\Big{|}\leq\frac{C_{1}R}{M}+\frac{MR\,TC_{6}}{N^{2}}+\gamma_{N}MR\,T\mu_{n}(E)$

For an arbitrary $\varepsilon>0$, we take $M$ and $N$ such that
$C_{1}R/M<\varepsilon/3$ and $MR\,TC_{6}/N^{2}<\varepsilon/3$. If
$\delta=\varepsilon/(3\gamma_{N}MR\,T)$, then, as a consequence of the last
inequality,

$\Big{|}\int_{E}\varphi_{k}(\zeta_{k})\int_{0}^{T}u_{k}\,h\,dt\,dx\Big{|}<\varepsilon\quad\text{whenever
$\mu_{n}(E)<\delta$},$

which implies the required uniform integrability of the sequence
$\\{\varphi_{k}(\zeta_{k})\int_{0}^{T}u_{k}\,h\,dt\\}$. Thus, (3.11) is
proven.

Step 4. In this step, we prove (3.1) and (3.3), which completes the proof of
the theorem. The fact that $\varphi(\zeta)u^{2}\in L^{1}(\varOmega_{T})$ is a
consequence of (3.1). In order to establish (3.1), we have to pass to the
limit as $k\to\infty$ in (3.5). The first two terms on the left-hand side of
(3.1) are obtained in the usual way due to the weak lower semicontinuity of
the norms. Let us obtain the third term.

Using the Hölder inequality, we get:

$\int_{\varOmega}\Big{|}\nabla\int_{0}^{T}u_{k}^{2}\,dt\Big{|}\,dx\leq
2\int_{\varOmega_{T}}|u_{k}|\,|\nabla u_{k}|\,dx\,dt\\\ \leq
2\Big{(}\int_{\varOmega_{T}}|u_{k}|^{2}\,dx\,dt\Big{)}^{1/2}\,\Big{(}\int_{\varOmega_{T}}|\nabla
u_{k}|^{2}\,dx\,dt\Big{)}^{1/2}\leq 2T^{1/2}C_{1},$

where $C_{1}$ is the constant from (3.5). This estimate implies that there
exists a function $v\in L^{1}(\varOmega)$ such that (up to a subsequence)

$\int_{0}^{T}u_{k}^{2}\,dt\to v\quad\text{as $k\to\infty$ in
$L^{1}(\varOmega)$ and almost everywhere in $\varOmega$.}$

Therefore, due to (3.12),
$\varphi(\zeta_{k})\int_{0}^{T}u_{k}^{2}\,dt\to\varphi(\zeta)\,v$ almost
everywhere in $\varOmega$. As it follows from (3.5) and the Fatou lemma,
$\varphi(\zeta)\,v\in L^{1}(\varOmega)$ and

$\int_{\varOmega}\varphi(\zeta)\,v\,dx\leq\liminf_{k\to\infty}\int_{0}^{T}\int_{\varOmega}\varphi_{k}(\zeta_{k})\,u_{k}^{2}\,dx\,dt.$

Thus, in order to establish (3.1), it suffices to show that

$\int_{0}^{T}\int_{\varOmega}\varphi(\zeta)\,u^{2}\,dx\,dt\leq\int_{\varOmega}\varphi(\zeta)\,v\,dx.$

Since $\varphi(\zeta)\geq 0$, it is enough to prove that

$\int_{0}^{T}u^{2}\,dt\leq v\quad\text{almost everywhere in $\varOmega$.}$

Let $G$ be an arbitrary measurable subset of $\varOmega$ and
$G_{T}=G\times[0,T]$. Since $u_{k}$ converges to $u$ weakly in $L_{2}(G_{T})$
and the norm in $L^{2}(G_{T})$ is weakly lower semicontinuous,

$\int_{G}\int_{0}^{T}u^{2}\,dt\,dx\leq\liminf_{k\to\infty}\int_{G}\int_{0}^{T}u_{k}^{2}\,dt\,dx=\int_{G}v\,dx.$

Due to the arbitrariness of $G$, we have the required relation.

Finally, the assertion that $\varphi(\zeta)u\in L^{1}(\varOmega_{T})$ follows
from the Hölder inequality and the facts that $\varphi(\zeta)\in
L^{1}(\varOmega)$ and $\varphi(\zeta)u^{2}\in L^{1}(\varOmega_{T})$.

The theorem is proven. $\square$

Notice that Remark 2.2 is also true for Theorem 3.1, since we used only the
energy estimate in its proof.

Acknowledgement
This work was supported by the Russian Science Foundation (Grant No.
19-11-00069).

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