# Nonlocal Cahn-Hilliard equation with degenerate mobility: Incompressible
limit and convergence to stationary states

Charles Elbar Sorbonne Université, CNRS, Université de Paris, Inria,
Laboratoire Jacques-Louis Lions, F-75005 Paris, France email:
<EMAIL_ADDRESS>Benoît Perthame11footnotemark: 1 email:
<EMAIL_ADDRESS>Andrea Poiatti Dipartimento di
Matematica, Politecnico di Milano, Milano 20133, Italyemail:
<EMAIL_ADDRESS>Jakub Skrzeczkowski Institute of Mathematics, Polish
Academy of Sciences and Faculty of Mathematics, Informatics and Mechanics,
University of Warsaw, Polandemail<EMAIL_ADDRESS>
###### Abstract

The link between compressible models of tissue growth and the Hele-Shaw free
boundary problem of fluid mechanics has recently attracted a lot of attention.
In most of these models, only repulsive forces and advection terms are taken
into account. In order to take into account long range interactions, we
include for the first time a surface tension effect by adding a nonlocal term
which leads to the degenerate nonlocal Cahn-Hilliard equation, and study the
incompressible limit of the system. The degeneracy and the source term are the
main difficulties. Our approach relies on a new $L^{\infty}$ estimate obtained
by De Giorgi iterations and on a uniform control of the energy despite the
source term. We also prove the long-term convergence to a single constant
stationary state of any weak solution using entropy methods, even when a
source term is present. Our result shows that the surface tension in the
nonlocal (and even local) Cahn-Hilliard equation will not prevent the tumor
from completely invading the domain.

Conflict of interest statement: The authors have no conflicts of interest to
declare that are relevant to the content of this article.

Data availability statement: Data availability is not applicable to this
article as no new data were created or analyzed in this study.

2010 Mathematics Subject Classification. 35B40; 35B45; 35G20 ; 35Q92
Keywords and phrases. Degenerate Cahn-Hilliard equation; Asymptotic analysis;
Convergence to equilibrium; Incompressible limit; Hele-Shaw equations.

##  1 Introduction

Nonlocal parabolic equations are commonly used to describe living tissues
because cells experience two types of forces: repulsive and attractive. The
repulsion arises at high volume fraction because, locally, cells occupy a non-
vanishing volume, while cell adhesion and chemotaxis create attraction at long
range, i.e., low densities [4]. These effects, as well as surface tension
effects, can be considered by using the Cahn-Hilliard equation (see, e.g.,
[44] for a review on these models). Our work is dedicated to the analysis of
the nonlocal Cahn-Hilliard equation for long range interactions with a
repulsive potential. More precisely we are interested in two results: the
incompressible limit connecting mechanistic and free boundary descriptions of
the tissue and the long-time asymptotics of equations. Concerning the first
result, the main difficulty is that we lose any maximum principle and we have
to rely on different arguments to obtain the same results concerning the
incompressible limit. Concerning the convergence to the stationary state, we
prove that it converges to a nonnegative constant which shows that the surface
tension effect (modeled by the Cahn-Hilliard equation) is not strong enough to
prevent the tumor from invading the entire domain.

###  1.1 Mathematical setting

Our settings is as follows: we let $\Omega$ be the $d=1,2,3$ dimensional flat
torus in $\mathbb{R}^{d}$, which is particularly useful when treating nonlocal
terms and we consider the equation

$\begin{split}\partial_{t}u-\text{div}(u\nabla\mu)=uG(p)\quad&\text{in
}\Omega\times(0,T),\\\ \mu=p+B_{\varepsilon}(u)\quad&\text{in
}\Omega\times(0,T),\end{split}$ (1.1)

with the initial condition $u(0)=u_{0}\geq 0\text{ in }\Omega$ and $u(x,t)\geq
0$ represents the cell density. Here, $B_{\varepsilon}$ denotes the nonlocal
operator defined as

$B_{\varepsilon}[u](x)=\frac{1}{\varepsilon^{2}}(u(x)-\omega_{\varepsilon}\ast
u(x))=\frac{1}{\varepsilon^{2}}\int_{\Omega}\omega_{\varepsilon}(y)(u(x)-u(x-y))\mathop{}\\!\mathrm{d}y$
(1.2)

for fixed $\varepsilon$ small enough (in order to be able to use [23, Lemma
C.1] and Lemma A.3) and $\omega_{\varepsilon}$ is a usual mollification kernel
$\omega_{\varepsilon}(x)=\frac{1}{\varepsilon^{d}}\omega(\frac{x}{\varepsilon})$
with $\omega$ compactly supported in the unit ball of $\mathbb{R}^{d}$
satisfying

$\int_{\mathbb{R}^{d}}\omega(y)\mathop{}\\!\mathrm{d}y=1,\quad\omega(\cdot)\text{
is radial}.$ (1.3)

The pressure and source term are defined, for $u\geq 0$, as

$p=u^{\gamma},\quad\gamma\geq 1,\qquad G(p)=p_{H}-p,$ (1.4)

with $p_{H}>0$ a constant called the homeostatic pressure, which is the
threshold where cells begin to die, assuming that pressure produces an
inhibitory effect on cell proliferation.

We comment the different terms appearing in the equation. First, $u$ is the
density of tumor cells and can be thought of as being between 0 and 1.
However, this fact is not easy to prove since the maximum principle does not
hold here. Using a De Giorgi iteration technique, we can prove however that
the bound holds with a small perturbation term which vanishes as
$\gamma\to\infty$ (see Lemma 3.1). From the Cahn-Hilliard terminology, we
refer to $\mu$ as the chemical potential, which is composed by two terms: one
is the pressure $p=u^{\gamma}$ and the other is $B_{\varepsilon}$, the
approximation of the Laplace operator, which takes into account surface
tension effects, see for instance [23]. Concerning the initial condition, we
distinguish two sets of assumptions.

######  Assumption 1.1 (Initial condition).

We assume:
(A) $0\leq u_{0}\leq p_{H}^{\frac{1}{\gamma}}$ for almost any $x\in\Omega$.

Note that the same assumption has already been considered, e.g., in [13] and
that it implies $u_{0}\in L^{q}(\Omega)$, for any $q\geq 1$, since $\Omega$ is
of finite Lebesgue measure. For the single Section 3, we need additionally:

######  Assumption 1.2 (Additional assumption for Section 3).

We assume:
(B) There is $\gamma_{0}>0$ and $C=C(\gamma_{0})>0$ such that
$\|\Delta(u_{0}^{1+\gamma})\|_{L^{1}(\Omega)}+\|\nabla
u_{0}\|_{L^{2}(\Omega)}+\|\Delta u_{0}\|_{L^{1}(\Omega)}\leq
C,\quad\forall\gamma\geq\gamma_{0}$.

System (1.1) is associated with the energy $\mathcal{E}$ and entropy $\Phi$,
respectively defined by

$\displaystyle\mathcal{E}(u):=\frac{1}{4\varepsilon^{2}}\int_{\Omega}\int_{\Omega}\omega_{\varepsilon}(y)|u(x)-u(x-y)|^{2}\mathop{}\\!\mathrm{d}y\mathop{}\\!\mathrm{d}x+\int_{\Omega}\dfrac{u^{\gamma+1}}{\gamma+1}\mathop{}\\!\mathrm{d}x\geq
0,$ (1.5)
$\displaystyle\Phi(u):=\int_{\Omega}\left(\frac{u}{p_{H}^{\frac{1}{\gamma}}}\log\left(\frac{u}{p_{H}^{\frac{1}{\gamma}}}\right)-\frac{u}{p_{H}^{\frac{1}{\gamma}}}+1\right)\mathop{}\\!\mathrm{d}x\geq
0.$ (1.6)

They formally satisfy the identities

$\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}t}\mathcal{E}(u)+\int_{\Omega}u|\nabla\mu|^{2}=\int_{\Omega}u\mu
G(p),$ (1.7)
$\displaystyle\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}t}\Phi(u)$
$\displaystyle+\frac{1}{2\varepsilon^{2}p_{H}^{\frac{1}{\gamma}}}\int_{\Omega}\int_{\Omega}\omega_{\varepsilon}(y)|\nabla
u(x)-\nabla u(x-y)|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}y$
$\displaystyle+\frac{1}{p_{H}^{\frac{1}{\gamma}}}\int_{\Omega}\dfrac{4\gamma}{(\gamma+1)^{2}}\left|\nabla|u|^{\frac{\gamma+1}{2}}\right|^{2}\mathop{}\\!\mathrm{d}x-\frac{1}{p_{H}^{\frac{1}{\gamma}}}\int_{\Omega}u\log\left(\frac{u}{p_{H}^{\frac{1}{\gamma}}}\right)G(p)\mathop{}\\!\mathrm{d}x=0.$
(1.8)

Moreover they provide us with direct a priori estimates, provided we can
control the integral related to the source term in (1.7), which may change
sign. Here, we assume that we have existence of solutions with regularity
typical of the Cahn-Hilliard equation. We do not include the proof, since most
of the a priori estimates are derived in Section 2. For a rigorous proof of
existence by means of an approximating scheme, we refer for instance to [23].

######  Lemma 1.3.

Let $u_{0}$ satisfy assumption (1.1). Then, for any $T>0$, there exist
constants $C_{0}(T,\mathcal{E}({u_{0}}),\Phi(u_{0}),\varepsilon)$ and
$C_{1}(T,\mathcal{E}({u_{0}}),\Phi(u_{0}),\varepsilon,\gamma)$ and a global
weak solution $u$ such that,

$\displaystyle u\geq 0\quad\text{a.e. in }\Omega\times(0,\infty),$ (1.9)
$\displaystyle u\in{C([0,\infty);(W^{1,r}(\Omega))^{\prime})}\cap
C_{weak}([0,\infty);L^{\gamma+1}(\Omega)),\quad
r=\frac{(\gamma+1)(2\gamma+1)}{\gamma^{2}},$ (1.10)
$\displaystyle\overline{u}(t):=\frac{1}{|\Omega|}\int_{\Omega}u(x,t)\mathop{}\\!\mathrm{d}x\leq
p_{H}^{\frac{1}{\gamma}}\quad\forall t\geq 0,$ (1.11)
$\displaystyle\|u\|_{L^{2}(0,T;H^{1}(\Omega))}+\frac{1}{\gamma+1}\|u\|^{\gamma+1}_{L^{\infty}(0,T;L^{\gamma+1}(\Omega))}+\|u\|_{L^{2\gamma+1}(\Omega\times(0,T))}\leq
C_{0},$ (1.12)
$\displaystyle\|\partial_{t}u\|_{L^{2}(0,T;(W^{1,r}(\Omega))^{\prime})}+\|\partial_{t}u\|_{L^{q^{\prime}}(0,T;(W^{1,q}(\Omega))^{\prime})}\leq
C_{1},$ (1.13)

where $\frac{1}{q}+\frac{1}{q^{\prime}}=1$, $q=\frac{2(2\gamma+1)}{\gamma}$.
Moreover, for any $v\in W^{1,r}(\Omega)$, $u$ satisfies

$\displaystyle\langle\partial_{t}u,v\rangle_{(W^{1,r}(\Omega))^{\prime},W^{1,r}(\Omega))}$
$\displaystyle+\int_{\Omega}u\nabla p\cdot\nabla
v\mathop{}\\!\mathrm{d}x+\frac{1}{\varepsilon^{2}}\int_{\Omega}u\nabla
u\cdot\nabla
v\mathop{}\\!\mathrm{d}x-\frac{1}{\varepsilon^{2}}\int_{\Omega}u(\nabla\omega_{\varepsilon}\ast
u)\cdot\nabla v\mathop{}\\!\mathrm{d}x$
$\displaystyle=\int_{\Omega}uG(p)v\mathop{}\\!\mathrm{d}x,\qquad\text{for
almost every }t>0,$ (1.14)

with $u(0)=u_{0}$ almost everywhere in $\Omega$. Here
$\langle\cdot,\cdot\rangle_{(W^{1,r}(\Omega))^{\prime},W^{1,r}(\Omega)}$
denotes the duality product between $(W^{1,r}(\Omega))^{\prime}$ and
$W^{1,r}(\Omega)$.

######  Remark 1.4.

The first continuity result follows from the bounds on $\partial_{t}u$ in
(1.13). Moreover, being $u\in L^{\infty}(0,T;L^{\gamma+1}(\Omega))$, we also
deduce that $u\in C_{weak}([0,T];L^{\gamma+1}(\Omega))$, see [7, Lemma
II.5.9]. This clearly also implies that
$\overline{u}=\frac{\int_{\Omega}u\mathop{}\\!\mathrm{d}x}{|\Omega|}\in
C([0,T])$.

###  1.2 The main results

Our first result establishes the incompressible limit $\gamma\to\infty$ of the
system (1.1) which links two descriptions of the tumor growth: mechanistic and
free-boundary. The main mathematical novelty here is the nonlocality which
makes it difficult to establish the uniform $L^{\infty}$ bound on
$u_{\gamma}$. To overcome this problem, we apply the De Giorgi iterations, see
Lemma 3.1, in the spirit of [28, 48].

######  Theorem 1.5 (Incompressible limit).

Let $u_{\gamma}$ be a weak solution to (1.1) as defined in Lemma 1.3 and
initial datum satisfying Assumptions 1.1-1.2. Then, as $\gamma\to\infty$, we
have for all $T>0$, up to a (not relabeled) subsequence

$\displaystyle u_{\gamma}\overset{\ast}{\rightharpoonup}u_{\infty}\text{ in
}L^{\infty}(\Omega\times(0,\infty)),$ $\displaystyle u_{\gamma}\to
u_{\infty}\text{ in }L^{q}(\Omega\times(0,T))\quad\forall q\in[2,+\infty),$
$\displaystyle u_{\gamma}\rightharpoonup u_{\infty}\text{ in
}L^{2}(0,T;H^{1}(\Omega)),$
$\displaystyle\partial_{t}u_{\gamma}\rightharpoonup\partial_{t}u_{\infty}\text{
in }L^{2}(0,T;(H^{1}(\Omega))^{\prime}),$ $\displaystyle
p_{\gamma}\rightharpoonup p_{\infty}\text{ in }L^{2}(0,T;H^{1}(\Omega)),$
$\displaystyle p_{\gamma}\to p_{\infty}\text{ in
}L^{r}(\Omega\times(0,T)),\quad\forall r\in[2,3),$

where $u_{\infty}$ and $p_{\infty}$ satisfy in
$\mathcal{D}^{\prime}(\Omega\times[0,\infty))$

$\displaystyle\partial_{t}u_{\infty}-\operatorname{div}(u_{\infty}\nabla(p_{\infty}+B_{\varepsilon}(u_{\infty})))=u_{\infty}G(p_{\infty}),$
(1.15) $\displaystyle
p_{\infty}\Big{(}\Delta{p}_{\infty}+\frac{1}{2\varepsilon^{2}}\Delta
u_{\infty}^{2}-\frac{1}{\varepsilon^{2}}\operatorname{div}(u_{\infty}(\nabla\omega_{\varepsilon}\ast
u_{\infty}))+u_{\infty}G(p_{\infty})\Big{)}=0,$ (1.16) $\displaystyle 0\leq
u_{\infty}\leq 1,\quad p_{\infty}\geq 0,\quad
p_{\infty}(1-u_{\infty})=0\quad\text{almost everywhere in
}\Omega\times(0,\infty),$ (1.17)

with $u_{\infty}(0)=u_{0}$. Furthermore it holds
$\langle\partial_{t}u_{\infty},p_{\infty}\rangle_{(H^{1}(\Omega))^{\prime},H^{1}(\Omega))}=0$
for almost any $t\in(0,\infty)$.

This theorem entails that in the limit $\gamma\to\infty$ we can consider the
measurable set $\Omega(t):=\\{x\in\Omega:\ p_{\infty}(t)>0\\}$, where
$u_{\infty}=1$ by the graph relation (1.17) so that it can be interpreted as
the ‘tumor zone’. Note that it must hold

$\displaystyle-\Delta{p}_{\infty}-\frac{1}{2\varepsilon^{2}}\Delta
u_{\infty}^{2}=-\frac{1}{\varepsilon^{2}}\operatorname{div}(u_{\infty}(\nabla\omega_{\varepsilon}\ast
u_{\infty}))+u_{\infty}G(p_{\infty})\quad\text{ in }{\cal I}nt(\Omega(t))$

which yields a Hele-Shaw type equation.

Our second result is concerned with the convergence to stationary states. We
distinguish two cases: when $G(p)=p_{H}-p$ and when $G(p)=0$. The main novelty
lies in the first case which is not conservative and its proof requires a
careful analysis of the entropy. We prove that as $t\to\infty$, the solution
converges to the constant $p_{H}^{\frac{1}{\gamma}}>0$, which shows that the
surface tension is not strong enough to prevent the tumor from invading the
entire domain. We have the

######  Theorem 1.6 (Long time behaviour).

Let $u$ be a solution to (1.1) with fixed $\gamma\geq 1$ in the sense of Lemma
1.3 and initial datum satisfying Assumption 1.1 Then, if $u_{0}\not\equiv 0$,
there are two cases:

  * •

for $G(p)=p_{H}-p$, we have

$\|u(t)-p_{H}^{\frac{1}{\gamma}}\|_{L^{q}(\Omega)}\to 0\text{ as
}t\to\infty,\quad\forall q\in[1,\gamma+1).$ (1.18)

  * •

For $G(p)=0$ and $\gamma\geq 1$, we have an exponential decay towards the mean
value: there exists a constant
$C=C(\Omega,\gamma,q,\Phi(u_{0}),\overline{u}_{0})$ such that

$\|u(t)-\overline{u}_{0}\|_{L^{q}(\Omega)}\lesssim e^{-Ct},\quad\forall
q\in[1,\gamma+1).$ (1.19)

There are two possibilities to prevent steady states to be constant. The first
one is to consider different potentials than just repulsive ones like
$u^{\gamma}$. The second possibility is to include an external force, which
can be taken into account either by a generic force that acts directly on the
cells like it was done in [22]. One can also include the effects of nutriments
and impose that the tumor cells die in the regions where there are no
nutriments.

######  Remark 1.7.

In the case of nonzero source term $G$, Theorem 1.6 implies that the constant
solution $u\equiv 0$ is unstable, whereas $p_{H}^{\frac{1}{\gamma}}$ is the
only asymptotically stable equilibrium and any weak solution $u$ (in the sense
of Lemma 1.3) departing from any nonzero initial datum converges to
$p_{H}^{\frac{1}{\gamma}}$.

###  1.3 Literature review

Incompressible limit for tumor model. The incompressible limit connects two
models of tumor growth: the compressible one studied in [47] and the free-
boundary one analysed in [42]. Many substantial contributions followed [47],
allowing nutrients [14], advection effects [15, 35, 34], additional
structuring variable [13], congesting flows [33], two species [19, 31] or
including additional surface tension effects via the degenerate Cahn-Hilliard
equation [21, 22].

One major difficulty for establishing the incompressible limit is proving
strong compactness of the gradient of the pressure. The main tool is the
celebrated Aronson-Benilan estimate [3, 12]. The estimate has been recently
readressed in [5] but the generalization available there are not applicable
for the pressure $p=u^{\gamma}+\frac{1}{\varepsilon^{2}}u$ as in our case cf.
[5, Theorem 4.1]. Another direct technique was developed in [38] which is
based on deducing strong convergence from a sort of energy equality. This is
the strategy we follow in our proof.

Nonlocal Cahn-Hilliard equation. The nonlocal Cahn-Hilliard is a variant of
the Cahn-Hilliard equation proposed to model dynamics of phase separation [9].
While originally introduced in the context of material science, it is
currently widely applied also in biology [40, 2, 46]. The nonlocal equation
was obtained for the first time by Giacomin and Lebowitz as the limit of
interacting particle systems [30, 29]. Their work can be considered as the
first derivation of the Cahn-Hilliard equation up to a delicate limit from the
nonlocal equation to the local one. The latter problem was in fact addressed
only recently, first for the case of the constant mobility [18, 16, 41, 17]
and finally for the case of degenerate mobility [23, 10]. Another derivation
as a hydrodynamic limit of the Vlasov equation was proposed recently in [20],
following [51]. In recent years nonlocal Cahn-Hilliard equation was also
studied in couplings with other hydrodynamic models, like Navier-Stokes
equations (see, e.g., [26, 24, 25, 27] and the references therein). Moreover,
it has been adopted in many optimal control problems, we just mention [49,
50].

Entropy dissipation methods and asymptotic analysis. For establishing
convergence to stationary states we use methods based on the entropy
dissipation. In the simplest scenario, it can be applied to PDEs equipped with
the entropy $\Phi$ which decreases with some dissipation

$\partial_{t}\Phi(t)+\mathcal{D}\Phi(t)\leq 0,\qquad\mathcal{D}\Phi(t)\geq 0.$

Then, one tries to prove that the dissipation is bounded from below by the
entropy $|\Phi(t)|^{\alpha}\leq\mathcal{D}\Phi(t)$ so that $\Phi(t)\to 0$ with
an explicit convergence rate (exponential if $\alpha=1$ and polynomial if
$\alpha>1$). Finally, by virtue of Csiszár-Kullback inequality, one deduces
convergence in $L^{1}$. The last step requires conservation of mass which is
not available when $G(p)\neq 0$ in (1.1). We present the method in detail for
the Cahn-Hilliard equation without the source term. Another method to obtain
convergence to equilibrium, applied in the context of the Cahn-Hilliard
equation, is via the Łojasiewicz-Simon inequality [11, 1, 39]. This method
cannot be applied here due to the degenerate mobility and the lack of
separation property from the degenerate case $u=0$.

##  2 Basic a priori estimates

The energy/entropy structure usually provides a priori estimates on the
solutions. However, in the case of a source term which may change sign, we
first need to control their dissipation. Before tackling this problem, we
first show a basic estimate which ensures the control of the mass of the
system, uniformly in time. This estimate is useful to obtain a first
$L^{\infty}_{t}L^{1}_{x}$ bound on the solution. Our proof of these estimates
is somehow formal but can be carried out rigorously with an approximation
scheme as, e.g., in [23]. These estimates are also fundamental to prove the
existence Lemma 1.3.

###  2.1 Control of the mass

We recall that the total mass of the system is defined in (1.11) and we prove
the corresponding bound.

######  Proposition 2.1 (Mass control).

For all $t\geq 0$ we have $\overline{u}(t)\leq p_{H}^{\frac{1}{\gamma}}$.

###### Proof.

Integrating equation (1.1), we obtain

$\displaystyle\frac{d}{dt}\overline{u}=\frac{1}{|\Omega|}\int_{\Omega}(u\,p_{H}-u^{\gamma+1})\mathop{}\\!\mathrm{d}x.$
(2.1)

By the Hölder inequality,

$\int_{\Omega}u\mathop{}\\!\mathrm{d}x\leq\left(\int_{\Omega}u^{\gamma+1}\mathop{}\\!\mathrm{d}x\right)^{\frac{1}{\gamma+1}}|\Omega|^{\frac{\gamma}{\gamma+1}}$

we deduce

$\displaystyle\frac{d}{dt}\overline{u}$
$\displaystyle\leq\frac{p_{H}}{|\Omega|}\int_{\Omega}{u}\mathop{}\\!\mathrm{d}x-\left(\int_{\Omega}{u}\mathop{}\\!\mathrm{d}x\right)^{\gamma+1}\frac{1}{|\Omega|^{{\gamma+1}}}=p_{H}\overline{u}-\overline{u}^{1+\gamma},$

so that, with the Assumption 1.1 on the initial condition, we can conclude
(1.11). ∎

###  2.2 Energy and entropy estimates

We recall that the energy, the entropy as well as their dissipation have been
defined in (1.5)–(1.8). We prove that, for a fixed time horizon $T$, we have
the following inequalities.

######  Proposition 2.2 (Control of the energy and entropy dissipation).

The inequalities hold

$\begin{split}&\sup_{t\geq
0}\Phi(u(t))+\frac{1}{2\varepsilon^{2}p_{H}^{\frac{1}{\gamma}}}\int_{0}^{\infty}\int_{\Omega}\int_{\Omega}\omega_{\varepsilon}(y)|\nabla
u(x)-\nabla
u(x-y)|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}y\mathop{}\\!\mathrm{d}t\\\
&+\frac{1}{p_{H}^{\frac{1}{\gamma}}}\dfrac{4\gamma}{(\gamma+1)^{2}}\int_{0}^{\infty}\int_{\Omega}\left|\nabla|u|^{\frac{\gamma+1}{2}}\right|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t+\int_{0}^{\infty}\int_{\Omega}u\log\left(\frac{u}{p_{H}^{\frac{1}{\gamma}}}\right)(p-p_{H})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t\leq\Phi(u_{0}),\end{split}$
(2.2)
$\begin{split}\frac{d}{dt}\mathcal{E}(u)&+\frac{1}{2\varepsilon^{2}}\int_{\Omega}\omega_{\varepsilon}(y)(u(x)-u(x-y))(u^{\gamma+1}(x)-u^{\gamma+1}(x-y))\mathop{}\\!\mathrm{d}x+\int_{\Omega}u|\nabla\mu|^{2}\mathop{}\\!\mathrm{d}x\\\
&+\frac{1}{2}\int_{\Omega}u^{2\gamma+1}\mathop{}\\!\mathrm{d}x\leq
C(\mathcal{E}(u)+1),\end{split}$ (2.3)

and thus there exists $C(\mathcal{E}(u_{0}),T,\varepsilon)>0$ such that

$\sup_{t\in[0,T]}\mathcal{E}(u(t))+\frac{1}{2}\int_{0}^{T}\int_{\Omega}u^{2\gamma+1}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t\leq
C(\mathcal{E}(u_{0}),T,\varepsilon).$

######  Remark 2.3.

The above estimate in the energy $\mathcal{E}$ depends exponentially on the
final time $T$. We improve this result to a global one in Proposition 2.5.

###### Proof.

Control of the entropy. Note that $\Phi(u)\geq 0$ by the inequality
$x\log\left(\frac{x}{y}\right)-x+y\geq(\sqrt{x}-\sqrt{y})^{2}$ for $x\geq 0$
and $y>0$ ($x=\frac{u}{p_{H}^{\frac{1}{\gamma}}}$, $y=1$ here). Then we have

$\displaystyle-\int_{\Omega}u\log\left(\frac{u}{p_{H}^{\frac{1}{\gamma}}}\right)G(p)\mathop{}\\!\mathrm{d}x=-\int_{\Omega}u\log\left(\frac{u}{p_{H}^{\frac{1}{\gamma}}}\right)(p_{H}-p)\mathop{}\\!\mathrm{d}x\geq
0,$

since $u\geq 0$ and $x\mapsto\log x$ is increasing for $x\geq 0$. Therefore,
all the terms in the dissipation of entropy in (1.8) are nonnegative so that
we can integrate in time and obtain (2.2), which clearly implies the control
of the entropy independently of $T$.

Energy control. Turning to the energy $\mathcal{E}$, departing from (1.7), we
observe that

$\displaystyle\int_{\Omega}uG(p)\mu\mathop{}\\!\mathrm{d}x=\int_{\Omega}u(p_{H}-p)p\mathop{}\\!\mathrm{d}x+\int_{\Omega}u\,B_{\varepsilon}(u)(p_{H}-p)\mathop{}\\!\mathrm{d}x.$

The first term can be written as

$\displaystyle\int_{\Omega}u(p_{H}-p)p\mathop{}\\!\mathrm{d}x$
$\displaystyle=-\int_{\Omega}up^{2}\mathop{}\\!\mathrm{d}x+\int_{\Omega}up_{H}p\mathop{}\\!\mathrm{d}x$
$\displaystyle\leq-\int_{\Omega}up^{2}\mathop{}\\!\mathrm{d}x+\frac{1}{2}\int_{\Omega}up^{2}\mathop{}\\!\mathrm{d}x+\frac{1}{2}p_{H}^{2}\int_{\Omega}u\mathop{}\\!\mathrm{d}x\leq-\frac{1}{2}\int_{\Omega}up^{2}\mathop{}\\!\mathrm{d}x+C,$

where we used the mass control (1.11). For the second term we have, by
symmetry of $\omega$,

$\displaystyle\int_{\Omega}u\,B_{\varepsilon}(u)(p_{H}-p)\mathop{}\\!\mathrm{d}x=$
$\displaystyle\frac{p_{H}}{2\varepsilon^{2}}\int_{\Omega}\int_{\Omega}\omega_{\varepsilon}(y)|u(x)-u(x-y)|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}y$
$\displaystyle-\frac{1}{2\varepsilon^{2}}\int_{\Omega}\int_{\Omega}\omega_{\varepsilon}(y)(u(x)-u(x-y))(u^{\gamma+1}(x)-u^{\gamma+1}(x-y))\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}y.$

All together, these inequalities give immediately (2.3) and by the Gronwall
lemma the energy control. ∎

######  Remark 2.4.

In the limit $\gamma\to\infty$ it holds that $u_{\infty}\leq 1$. This follows
from the bound obtained with the energy in Proposition 2.2 since

$\|u_{\infty}\|_{L^{\infty}}=\lim_{\gamma\to\infty}\|u_{\infty}\|_{L^{\gamma}}\quad\text{and}\quad\|u\|_{L^{\gamma+1}}\leq
C^{\frac{1}{\gamma+1}}(\gamma+1)^{\frac{1}{\gamma+1}}\to 1$

because of the weak convergence of $u^{\gamma}$. We refer for instance to [21,
52]. However in the next section we obtain a better control on $u$ by the De
Giorgi iteration method.

Now, we improve the local in time estimate on $\mathcal{E}$ to a global one,
which is nontrivial due to the presence of the source term. Since Proposition
2.2 gives the uniform control $\Phi(u(t))\leq\Phi(u_{0})$ for any $t\geq 0$,
our aim is to control in a uniform way the energy $\mathcal{E}$ as well.

######  Proposition 2.5 (Uniform in time estimates for the energy).

There exists a constant independent of time and $\gamma$ such that

$\displaystyle\mathcal{E}(t)\leq C,\quad\forall t\geq 0.$ (2.4)

###### Proof.

Firstly, we estimate separately the two terms defining the energy in (1.5)
using the entropy estimate (2.2). It immediately gives that, for a constant
$C>0$ independent of time and $\gamma$, for any sufficiently small
$\varepsilon$,

$\displaystyle\int_{t}^{t+1}$
$\displaystyle\frac{1}{4\varepsilon^{2}}\int_{\Omega}\int_{\Omega}\omega_{\varepsilon}(y)|u(x)-u(x-y)|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}y\mathop{}\\!\mathrm{d}s\leq
C(\Omega)\int_{t}^{t+1}\|u\|_{H^{1}(\Omega)}^{2}\mathop{}\\!\mathrm{d}s$
$\displaystyle\leq
C\int_{t}^{t+1}\frac{1}{4\varepsilon^{2}}\int_{\Omega}\int_{\Omega}\omega_{\varepsilon}(y)|\nabla
u(x)-\nabla
u(x-y)|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}y\mathop{}\\!\mathrm{d}s+C\int_{t}^{t+1}\|u\|_{L^{1}(\Omega)}^{2}\mathop{}\\!\mathrm{d}s$
(2.5) $\displaystyle\leq C+Cp_{H}^{\frac{2}{\gamma}},\quad\forall t\geq 0.$

where the second inequality follows from Equations (A.4) and (1.11).

Secondly, we control the remaining part of the energy $\mathcal{E}$, the one
related to
$\frac{1}{\gamma+1}\int_{\Omega}u^{\gamma+1}\mathop{}\\!\mathrm{d}x$. To this
aim, we integrate Equation (2.1) in time over $[t,t+1]$ and get

$\overline{u}(t+1)-\overline{u}(t)=\frac{1}{|\Omega|}\int_{t}^{t+1}\int_{\Omega}(up_{H}-u^{\gamma+1})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s,$

so that, rearranging the terms, we control the second term of the energy as

$\frac{1}{|\Omega|}\int_{t}^{t+1}\int_{\Omega}u^{\gamma+1}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s\leq\frac{1}{|\Omega|}\int_{t}^{t+1}\int_{\Omega}up_{H}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s+\overline{u}(t)=(p_{H}+1)\int_{t}^{t+1}\overline{u}(s)\mathop{}\\!\mathrm{d}s\leq
C$

because $\overline{u}(\cdot)\leq p_{H}^{\frac{1}{\gamma}}$ for any $t\geq 0$
thanks to (1.11). This, together with (2.5), implies that

$\displaystyle\int_{t}^{t+1}\mathcal{E}(s)\mathop{}\\!\mathrm{d}s\leq C,$
(2.6)

with $C>0$ independent of $t\geq 0$ and $\gamma$.

We may now conclude the energy estimate. By Proposition 2.2, we have

$\frac{d}{dt}\mathcal{E}(u)\leq C(\mathcal{E}(u)+1).$

Using the Gronwall lemma, we obtain for all $t\geq 0$ and all $0\leq s\leq 1$,

$\mathcal{E}(u(t+1))\leq C\,\mathcal{E}(u(t+s))+C.$

Integrating in $s$ and using the bound (2.6), we conclude the proof of
Proposition 2.5.

∎

###  2.3 A control on $\partial_{t}u$

Here we prove the estimate on time derivative, which appears also in Lemma 1.3
and is used in the proof of Theorem 1.6.

######  Proposition 2.6.

There exists $C=C(\gamma,T)>0$ such that the bounds hold

$\displaystyle\|\partial_{t}u\|_{L^{q^{\prime}}(0,T;(W^{1,q}(\Omega))^{\prime})}\leq
C,\quad
q={\dfrac{2(2\gamma+1)}{\gamma}},\quad\frac{1}{q}+\frac{1}{q^{\prime}}=1,$
(2.7)
$\displaystyle\|\partial_{t}u\|_{L^{2}(0,T;(W^{1,r}(\Omega))^{\prime})}\leq
C,\quad r=\frac{(\gamma+1)(2\gamma+1)}{\gamma^{2}}.$ (2.8)

###### Proof.

For $\gamma\geq 1$ fixed, any $T>0$, any $\varphi\in
L^{q}(0,T;W^{1,q}(\Omega))$, we have

$\displaystyle\Big{|}\int_{0}^{T}\langle\partial_{t}u,\varphi\rangle_{(W^{1,q}(\Omega))^{\prime},W^{1,q}(\Omega))}\mathop{}\\!\mathrm{d}t\Big{|}$
$\displaystyle\leq\left|\int_{0}^{T}\int_{\Omega}\gamma u^{\gamma}\nabla
u\cdot\nabla\varphi\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t\right|+\left|\int_{0}^{T}\frac{1}{\varepsilon^{2}}\int_{\Omega}u\nabla
u\cdot\nabla\varphi\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t\right|$
$\displaystyle\qquad\quad+\left|\int_{0}^{T}\frac{1}{\varepsilon^{2}}\int_{\Omega}u(\nabla\omega_{\varepsilon}\ast
u)\cdot\nabla\varphi\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t\right|+\left|\int_{0}^{T}\int_{\Omega}u(p_{H}-u^{\gamma})\varphi\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t\right|$
$\displaystyle\leq\gamma^{\frac{1}{2}}\|u\|^{\frac{\gamma+1}{2(2\gamma+1)}}_{L^{2\gamma+1}(\Omega\times(0,T))}\|\gamma^{\frac{1}{2}}u^{\frac{\gamma-1}{2}}\nabla
u\|_{L^{2}(\Omega\times(0,T))}\|\nabla\varphi\|_{L^{q}(\Omega\times(0,T))}$
$\displaystyle\quad+C(\varepsilon)\|u\|_{L^{2\gamma+1}(\Omega\times(0,T))}\|\nabla
u\|_{L^{2}(\Omega\times(0,T))}\|\nabla\varphi\|_{L^{\frac{2(2\gamma+1)}{2\gamma-1}}(\Omega\times(0,T))}$
$\displaystyle\quad+C(\varepsilon)\|u\|_{L^{2}(\Omega\times(0,T))}\|\nabla\omega_{\varepsilon}\ast
u\|_{L^{\infty}(\Omega\times(0,T))}\|\nabla\varphi\|_{L^{2}(\Omega\times(0,T))}$
$\displaystyle\quad+C\|u\|_{L^{2\gamma+1}(\Omega\times(0,T))}\|p_{H}-u^{\gamma}\|_{L^{2}(\Omega\times(0,T))}\|\varphi\|_{L^{\frac{2(2\gamma+1)}{2\gamma-1}}(\Omega\times(0,T))}$
$\displaystyle\leq
C(\varepsilon,T,\gamma)\|\varphi\|_{L^{q}(0,T;W^{1,q}(\Omega))}.$

More precisely, to estimate the first and second terms, we used the Hölder
inequality with exponents $\frac{2(2\gamma+1)}{\gamma+1}$, $2$, $q$ and
$2\gamma+1$, $2$ and $\frac{2(2\gamma+1)}{2\gamma-1}$, respectively. Then,
$\|u\|^{\frac{\gamma+1}{2(2\gamma+1)}}_{L^{2\gamma+1}(\Omega\times(0,T))}$ is
bounded due to (2.3), $\|\gamma^{\frac{1}{2}}u^{\frac{\gamma-1}{2}}\nabla
u\|_{L^{2}(\Omega\times(0,T))}$ is estimated by (2.2) while the bound on
$\nabla u$ follows from (2.2) and nonlocal Poincaré inequality (A.4). The
fourth term is bounded in the same spirit. Concerning the third one, we simply
estimate

$\|\nabla\omega_{\varepsilon}\ast
u\|_{L^{\infty}(\Omega\times(0,T))}\leq\|\nabla\omega_{\varepsilon}\|_{L^{\infty}(\Omega)}\|u\|_{L^{\infty}(0,T;L^{1}(\Omega))}\leq
C(\varepsilon)$

and use the estimate on the total mass (1.11). The final conclusion follows
from the inequalities $q>2$ and $q\geq\frac{2(2\gamma+1)}{2\gamma-1}$ for any
$\gamma\geq 1$.

Concerning (2.8), let $\varphi\in L^{2}(0,T;W^{1,r}(\Omega))$, with
$r=\frac{(\gamma+1)(2\gamma+1)}{\gamma^{2}}$. Then,

$\displaystyle\left|\int_{0}^{T}\langle\partial_{t}u,\varphi\rangle_{(W^{1,r}(\Omega))^{\prime},W^{1,r}(\Omega))}\mathop{}\\!\mathrm{d}t\right|$
$\displaystyle\leq\left|\int_{0}^{T}\int_{\Omega}u\nabla\mu\cdot\nabla\varphi\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t\right|+\left|\int_{0}^{T}\int_{\Omega}u(p_{H}-u^{\gamma})\varphi\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t\right|$
$\displaystyle\leq\|u^{\frac{1}{2}}\|_{L^{\infty}(0,T;L^{2(\gamma+1)})}\|u^{\frac{1}{2}}\nabla\mu\|_{L^{2}(\Omega\times(0,T))}\|\nabla\varphi\|_{L^{2}(0,T;L^{\frac{2(\gamma+1)}{\gamma}}(\Omega))}$
$\displaystyle\qquad+p_{H}\|u\|_{L^{\infty}(0,T;L^{\gamma+1}(\Omega))}\|\varphi\|_{L^{1}(0,T;L^{\frac{\gamma+1}{\gamma}}(\Omega))}$
$\displaystyle\qquad+\|u\|_{L^{\infty}(0,T;L^{\gamma+1}(\Omega))}\|u\|_{L^{2\gamma+1}(\Omega\times(0,T))}\|\varphi\|_{L^{\frac{2\gamma+1}{2\gamma}}(0,T;L^{r}(\Omega))}$
$\displaystyle\leq C\|\varphi\|_{L^{2}(0,T;W^{1,r}(\Omega))},$

exploiting Proposition 2.2, recalling that
$r\geq\frac{2(\gamma+1)}{\gamma}>\frac{\gamma+1}{\gamma}$ and
$\frac{2\gamma+1}{2\gamma}\leq 2$. This concludes the proof.

∎

##  3 Incompressible limit: proof of Theorem 1.5

We see the incompressible limit of (1.1) as the limit $\gamma\to\infty$ and
the resulting problem turns out to be a free boundary problem of Hele-Shaw
type. Concerning the techniques adopted here, we first show that, for any
fixed $T>0$, $u_{\gamma}$ is bounded in $L^{\infty}(\Omega\times(0,T))$ by a
quantity which converges to $1$ as $\gamma\to\infty$ (see (3.1)). Due to the
presence of the convolution term, which makes the equation nonlocal, we cannot
apply any classical maximum principle, so that we need to resort to De Giorgi
iterations, exploiting the fact that the equation is a second order
differential equation.

With uniform estimates in $L^{\infty}$ at hand, we may apply standard energy
estimates to gain sufficient regularity on the pressure $p_{\gamma}$, which we
bounded in $L^{3}(\Omega\times(0,T))\cap L^{2}(0,T;H^{1}(\Omega))$ uniformly
in $\gamma$. Then, one can obtain uniform controls in
$L^{\infty}(0,T;L^{1}(\Omega))$ for $\partial_{t}u_{\gamma}$ and in
$L^{1}(\Omega\times(0,T))$ for $\partial_{t}p_{\gamma}$, so as to deduce the
strong convergence of $u_{\gamma}$ and $p_{\gamma}$ in
$L^{2}(\Omega\times(0,T))$ to some $u_{\infty}$ and $p_{\infty}$. With the
help of these bounds, we are able to pass to the limit as $\gamma\to\infty$ in
Equation (1.1) and obtain Equations (1.15) and (1.17) for the limit
concentration $u_{\infty}$.

In order to obtain more information on $u_{\infty},p_{\infty}$, like
complementarity conditions (1.16), we need a stronger convergence for $\nabla
p_{\gamma}$. The standard technique uses some control on $\Delta p$ thanks to
the Aronson-Bénilan inequality (see, e.g., [15]) which does not apply here due
to the higher-order term. In particular, the (formal) CH equation can be
written as

$\partial_{t}u_{\gamma}-\frac{\gamma}{\gamma+1}\Delta
u^{\gamma+1}_{\gamma}-\frac{1}{2\varepsilon^{2}}\Delta
u^{2}_{\gamma}+\frac{1}{\varepsilon^{2}}\text{div}(u_{\gamma}(\nabla\omega_{\varepsilon}\ast
u)_{\gamma})=u_{\gamma}G(p_{\gamma}),$

and the extra term $\Delta u_{\gamma}^{2}$, independent of $\gamma$,
appearing, and this prevents us from obtaining the Aronson-Bénilan inequality.
Taking inspiration from [13, 38], in the second part of the present section,
we instead show the strong $L^{2}(\Omega\times(0,T))$ convergence of

$\nabla\left(\frac{\gamma}{\gamma+1}u_{\gamma}^{\gamma+1}+\frac{1}{2\varepsilon^{2}}u^{2}_{\gamma}\right),$

which is shown to be enough to guarantee the validity of the condition (1.16).

###  3.1 An $L^{\infty}$ bound on $u_{\gamma}$

######  Lemma 3.1.

Assume $0\leq u_{0}\leq p_{H}^{\frac{1}{\gamma}}$. For any $T>0$ there exists
$\overline{\gamma}(T,\varepsilon)>2$, explicitly computed as a function of
$T$, such that

$\displaystyle 0\leq u_{\gamma}\leq
p_{H}^{\frac{1}{\gamma}}+\frac{2}{\sqrt[3]{\gamma}},\quad\text{a.e. on
}\Omega\times(0,T),\quad\forall\gamma\geq\overline{\gamma}(T,\varepsilon).$
(3.1)

######  Remark 3.2.

Notice that the bound (3.1) is useless to control the pressure, since
$\left(p_{H}^{\frac{1}{\gamma}}+\frac{2}{\sqrt[3]{\gamma}}\right)^{\gamma}\to+\infty$
as $\gamma\to+\infty$.

###### Proof.

To simplify notations, we set $\xi:=\frac{1}{\sqrt[3]{\gamma}}$. The iterative
scheme is as follows. Let us consider the sequence

$k_{n}=p_{H}^{\frac{1}{\gamma}}+2\xi-\frac{\xi}{2^{n}},\quad n\geq 0,$

and note that ${p}_{H}^{\frac{1}{\gamma}}+\xi\leq
k_{n}<{p}_{H}^{\frac{1}{\gamma}}+2\xi$. Now we define the sequences

$A_{n}(t):=\\{x\in\Omega:\ u_{\gamma}\geq k_{n}\\},\quad
y_{n}:=\int_{0}^{T}\int_{A_{n}(s)}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s.$

By testing the equation against $u_{n,\gamma}:=(u_{\gamma}-k_{n})^{+}$, we
immediately infer that

$\displaystyle\frac{1}{2}\dfrac{d}{dt}\|(u_{\gamma}-k_{n})^{+}\|^{2}_{L^{2}(\Omega)}+$
$\displaystyle\int_{\Omega}\gamma
u_{\gamma}^{\gamma}|\nabla(u_{\gamma}-k_{n})^{+}|^{2}\mathop{}\\!\mathrm{d}x+\int_{\Omega}u_{\gamma}\nabla
B_{\varepsilon}(u_{\gamma})\cdot\nabla(u_{\gamma}-k_{n})^{+}\mathop{}\\!\mathrm{d}x$
$\displaystyle=\int_{\Omega}u_{\gamma}G(p_{\gamma})(u_{\gamma}-k_{n})^{+}\mathop{}\\!\mathrm{d}x.$

By the definition of $B_{\varepsilon}$, we get

$\int_{\Omega}u_{\gamma}\nabla
B_{\varepsilon}(u_{\gamma})\cdot\nabla(u_{\gamma}-k_{n})^{+}\mathop{}\\!\mathrm{d}x=\frac{1}{\varepsilon^{2}}\int_{\Omega}{u_{\gamma}}|\nabla
u_{n,\gamma}|^{2}\mathop{}\\!\mathrm{d}x-\frac{1}{\varepsilon^{2}}\int_{\Omega}u_{\gamma}(\nabla\omega_{\varepsilon}\ast
u_{\gamma})\cdot\nabla u_{n,\gamma}\mathop{}\\!\mathrm{d}x.$

The first term is nonnegative. For the second we use that $\gamma>2$,
$\|u_{\gamma}\|_{L^{\infty}(0,\infty;L^{1}(\Omega))}\leq{p}_{H}^{\frac{1}{\gamma}}|\Omega|$
thanks to (1.11) as well as $u_{\gamma}\geq k_{n}\geq p_{H}^{1/\gamma}+\xi$ on
$A_{n}$ to obtain

$\displaystyle\frac{1}{\varepsilon^{2}}\int_{\Omega}u_{\gamma}(\nabla\omega_{\varepsilon}\ast
u_{\gamma})\cdot\nabla
u_{n,\gamma}\mathop{}\\!\mathrm{d}x\leq\frac{1}{\varepsilon^{4}}\frac{1}{2\gamma}\int_{A_{n}}u^{2-\gamma}_{\gamma}|\nabla\omega_{\varepsilon}\ast
u_{\gamma}|^{2}\mathop{}\\!\mathrm{d}x+\frac{\gamma}{2}\int_{\Omega}u^{\gamma}_{\gamma}|\nabla
u_{n,\gamma}|^{2}\mathop{}\\!\mathrm{d}x$
$\displaystyle\leq\frac{1}{\varepsilon^{4}}\frac{1}{2\gamma\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{\gamma-2}}\|\nabla\omega_{\varepsilon}\ast
u_{\gamma}\|_{L^{\infty}(\Omega)}^{2}\int_{A_{n}}\mathop{}\\!\mathrm{d}x+\frac{\gamma}{2}\int_{\Omega}u_{\gamma}^{\gamma}|\nabla
u_{n,\gamma}|^{2}\mathop{}\\!\mathrm{d}x$
$\displaystyle\leq\frac{1}{\varepsilon^{4}}\frac{1}{2\gamma\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{{\gamma-2}}}\|\nabla\omega_{\varepsilon}\|_{L^{\infty}(\Omega)}^{2}\|u_{\gamma}\|_{L^{1}(\Omega)}^{2}\int_{A_{n}}\mathop{}\\!\mathrm{d}x+\frac{\gamma}{2}\int_{\Omega}u_{\gamma}^{\gamma}|\nabla
u_{n,\gamma}|^{2}\mathop{}\\!\mathrm{d}x$
$\displaystyle\leq\frac{C_{1}}{\gamma\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{{\gamma-2}}}\int_{A_{n}}\mathop{}\\!\mathrm{d}x+\frac{\gamma}{2}\int_{\Omega}u_{\gamma}^{\gamma}|\nabla
u_{n,\gamma}|^{2}\mathop{}\\!\mathrm{d}x,$

where $C_{1}>0$ is a constant that depends on $\varepsilon$ but not on $T$ and
$\gamma$. Moreover, since on $A_{n}$ we have
$u_{\gamma}\geq{p}_{H}^{\frac{1}{\gamma}}$ and thus $p_{\gamma}\geq p_{H}$, we
immediately infer that

$\displaystyle\int_{\Omega}u_{\gamma}G(p_{\gamma})(u_{\gamma}-k_{n})^{+}\mathop{}\\!\mathrm{d}x\leq
0.$

We can then sum up the results above to obtain

$\displaystyle\frac{1}{2}\dfrac{d}{dt}\|(u_{\gamma}-k_{n})^{+}\|^{2}_{L^{2}(\Omega)}+\frac{\gamma}{2}\int_{\Omega}u_{\gamma}^{\gamma}|\nabla(u_{\gamma}-k_{n})^{+}|^{2}\mathop{}\\!\mathrm{d}x\leq\frac{C_{1}}{\gamma\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{{\gamma-2}}}\int_{A_{n}}\mathop{}\\!\mathrm{d}x,$

which also implies, since on $A_{n}(t)$ we have
$u^{\gamma}_{\gamma}\geq\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{{\gamma}}$,

$\displaystyle\frac{1}{2}\dfrac{d}{dt}\|(u_{\gamma}-k_{n})^{+}\|^{2}_{L^{2}(\Omega)}+\frac{\gamma\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{{\gamma}}}{2}\int_{\Omega}|\nabla(u_{\gamma}-k_{n})^{+}|^{2}\mathop{}\\!\mathrm{d}x\leq\frac{C_{1}}{\gamma\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{{\gamma-2}}}\int_{A_{n}}\mathop{}\\!\mathrm{d}x.$

It is now clear that,

$\displaystyle\sup_{t\in[0,T]}\|u_{n,\gamma}(t)\|^{2}_{L^{2}(\Omega)}\leq\frac{2C_{1}}{\gamma\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{{\gamma-2}}}y_{n}=:Z_{n},$
(3.2)
$\displaystyle\gamma\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{{\gamma}}\int_{0}^{T}\int_{\Omega}|\nabla(u_{\gamma}-k_{n})^{+}(s)|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s\leq
Z_{n}.$ (3.3)

where we used, by the assumptions on the initial conditions,
$\|u_{n,\gamma}(0)\|_{L^{2}(\Omega)}=0$. Now for any $t$ and for almost any
$x\in A_{n+1}(t)$, we get

$\displaystyle
u_{n,\gamma}(x,t)=\underbrace{u_{\gamma}(x,t)-\left[{p}_{H}^{\frac{1}{\gamma}}+2\xi-\frac{\xi}{2^{n+1}}\right]}_{u_{n+1,\gamma}(x,t)\geq
0}+\xi\left[\frac{1}{2^{n}}-\frac{1}{2^{n+1}}\right]\geq\frac{\xi}{2^{n+1}}.$

Then we have

$\displaystyle\int_{0}^{T}\int_{\Omega}|u_{n,\gamma}|^{3}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s\geq\int_{0}^{T}\int_{A_{n+1}(s)}|u_{n,\gamma}|^{3}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s\geq\left(\frac{\xi}{2^{n+1}}\right)^{3}\int_{0}^{T}\int_{A_{n+1}(s)}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s=\left(\frac{\xi}{2^{n+1}}\right)^{3}y_{n+1}.$

Then we have

$\displaystyle\left(\frac{\xi}{2^{n+1}}\right)^{3}y_{n+1}\leq\left(\int_{0}^{T}\int_{\Omega}|u_{n,\gamma}|^{\frac{10}{3}}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s\right)^{\frac{9}{10}}\left(\int_{0}^{T}\int_{A_{n}(s)}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s\right)^{\frac{1}{10}}.$
(3.4)

For the sake of clarity we now present the argument in the case $d=3$, but it
can be easily adapted to any dimension $d\geq 1$. We recall that by a variant
of the three-dimensional Sobolev-Gagliardo-Nirenberg inequality (see, e.g.,[8,
Ch.9]) we get

$\displaystyle\|v-\overline{v}\|_{L^{\frac{10}{3}}(\Omega)}\leq{C_{G}}\,\|v\|_{L^{2}(\Omega)}^{\frac{2}{5}}\|\nabla
v\|_{L^{2}(\Omega)}^{\frac{3}{5}}\quad\forall v\in H^{1}(\Omega),$

with ${C_{G}}>0$ depending only on $\Omega$. Therefore,

$\displaystyle\int_{0}^{T}\int_{\Omega}|u_{n,\gamma}|^{\frac{10}{3}}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s\leq
2^{\frac{7}{3}}\int_{0}^{T}\int_{\Omega}|u_{n,\gamma}-\overline{u}_{n,\gamma}|^{\frac{10}{3}}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s+2^{\frac{7}{3}}\int_{0}^{T}\int_{\Omega}|\overline{u}_{n,\gamma}|^{\frac{10}{3}}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s$
$\displaystyle\leq C_{G}2^{\frac{7}{3}}\int_{0}^{T}\|\nabla
u_{n,\gamma}\|^{2}_{L^{2}(\Omega)}\|u_{n,\gamma}\|^{\frac{4}{3}}_{L^{2}(\Omega)}\mathop{}\\!\mathrm{d}s+\frac{2^{\frac{7}{3}}}{|\Omega|^{\frac{7}{3}}}\int_{0}^{T}\|u_{n,\gamma}\|_{L^{1}(\Omega)}^{\frac{10}{3}}$
$\displaystyle\leq C_{G}2^{\frac{7}{3}}\int_{0}^{T}\|\nabla
u_{n,\gamma}\|^{2}_{L^{2}(\Omega)}\|u_{n,\gamma}\|_{L^{2}(\Omega)}^{\frac{4}{3}}\mathop{}\\!\mathrm{d}s+\frac{2^{\frac{7}{3}}}{|\Omega|^{\frac{2}{3}}}\int_{0}^{T}\|u_{n,\gamma}\|_{L^{2}(\Omega)}^{\frac{10}{3}},$

so that by (3.2) and (3.3) we immediately infer

$\displaystyle\int_{0}^{T}\int_{\Omega}$
$\displaystyle|u_{n,\gamma}|^{\frac{10}{3}}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s$
$\displaystyle\leq\frac{C_{G}}{\gamma\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{{\gamma}}}\sup_{t\in[0,T]}\|u_{n,\gamma}(t)\|^{\frac{4}{3}}_{L^{2}(\Omega)}2^{\frac{7}{3}}\gamma\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{{\gamma}}\int_{0}^{T}\|\nabla
u_{n,\gamma}(s)\|^{2}_{L^{2}(\Omega)}\mathop{}\\!\mathrm{d}s$
$\displaystyle\qquad+\sup_{t\in[0,T]}\|u_{n,\gamma}(t)\|_{L^{2}(\Omega)}^{\frac{10}{3}}\frac{T2^{\frac{7}{3}}}{|\Omega|^{\frac{2}{3}}}$
$\displaystyle\leq\frac{2^{\frac{7}{3}}C_{G}}{\gamma\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{{\gamma}}}Z_{n}^{\frac{5}{3}}+\frac{T2^{\frac{7}{3}}}{|\Omega|^{\frac{2}{3}}}Z_{n}^{\frac{5}{3}}\leq\frac{T2^{\frac{10}{3}}}{|\Omega|^{\frac{2}{3}}}Z_{n}^{\frac{5}{3}}\leq\frac{C_{2}T}{\gamma^{\frac{5}{3}}\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{{\frac{5}{3}\gamma-\frac{10}{3}}}}y_{n}^{\frac{5}{3}},$

with $C_{2}=C_{2}(\varepsilon)>0$. Note that we have assumed $\gamma$
sufficiently large, say $\gamma\geq\gamma_{0}(T)>2$ so that

$\displaystyle\gamma\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{\gamma}\geq\dfrac{C_{G}|\Omega|^{\frac{2}{3}}}{T}.$
(3.5)

Coming back to (3.4), we get

$\displaystyle\left(\frac{\xi}{2^{n+1}}\right)^{3}y_{n+1}\leq\frac{C_{2}^{\frac{9}{10}}T^{\frac{9}{10}}}{\gamma^{\frac{3}{2}}\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{\frac{3}{2}\gamma-3}}y_{n}^{\frac{8}{5}},$

i.e., recalling the definition of $\xi$,

$\displaystyle
y_{n+1}\leq\frac{2^{3n+3}C_{2}^{\frac{9}{10}}T^{\frac{9}{10}}}{\gamma^{\frac{1}{2}}\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{\frac{3}{2}\gamma-3}}y_{n}^{\frac{8}{5}}.$

Due to Lemma A.1 with $b=2^{3}>1$,
$C=\frac{2^{3}C_{2}^{\frac{9}{10}}T^{\frac{9}{10}}}{\gamma^{\frac{1}{2}}\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{\frac{3}{2}\gamma-3}}>0$,
$\epsilon=\frac{3}{5}$, we get that ${y}_{n}\to 0$ if

$\displaystyle{y}_{0}\leq C^{-\frac{5}{3}}b^{-\frac{25}{9}}\iff y_{0}\leq
2^{-\frac{25}{3}}\frac{\gamma^{\frac{5}{6}}\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{\frac{5}{2}\gamma-5}}{2^{5}C_{2}^{\frac{3}{2}}T^{\frac{3}{2}}}$
(3.6)

As we have $y_{0}\leq T|\Omega|,$ it is enough to ask for $\gamma$
sufficiently large, say $\gamma\geq\gamma_{1}(T,\varepsilon)\geq\gamma_{0}(T)$
such that

$T^{\frac{5}{2}}|\Omega|\leq
2^{-\frac{25}{3}}\frac{\gamma^{\frac{5}{6}}\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{\frac{5}{2}\gamma-5}}{2^{5}C_{2}^{\frac{3}{2}}},$

i.e.,

$y_{0}\leq T|\Omega|\leq
2^{-\frac{25}{3}}\frac{\gamma^{\frac{5}{6}}\left({p}_{H}^{\frac{1}{\gamma}}+\xi\right)^{\frac{5}{2}\gamma-5}}{2^{5}C_{2}^{\frac{3}{2}}T^{\frac{3}{2}}}.$

This way $y_{n}\to 0$ as long as
$\overline{\gamma}(T,\varepsilon)\geq\gamma_{1}(T,\varepsilon)$ and any
$\gamma\geq\overline{\gamma}(T,\varepsilon)$. ∎

###  3.2 Higher-order regularity results, uniformly in $\gamma$.

######  Lemma 3.3.

For any $T>0$ there exists $C=C(T,\varepsilon)>0$ such that

$\displaystyle\|p_{\gamma}\|_{L^{2}(0,T;H^{1}(\Omega))}+\|p_{\gamma}\|_{L^{3}(\Omega\times(0,T))}\leq
C,\qquad\forall\gamma>1,$ (3.7)
$\displaystyle\|\partial_{t}u_{\gamma}\|_{L^{2}(0,T;(H^{1}(\Omega))^{\prime})}\leq
C,\quad\forall\gamma\geq\overline{\gamma}(T,\varepsilon)\qquad\text{(see
}\eqref{Linfty}),$ (3.8)
$\displaystyle\|\partial_{t}u_{\gamma}\|_{L^{\infty}(0,T;L^{1}(\Omega))}+\|\partial_{t}p_{\gamma}\|_{L^{1}(\Omega)\times(0,T)}\leq
C,\qquad\forall\gamma\geq 1.$ (3.9)

###### Proof.

The arguments of the proof are often written formally for simplicity, but can
be easily made rigorous in a suitable approximating scheme. Note that from
Proposition 2.2 we are able to deduce that

$\displaystyle p_{\gamma}=u_{\gamma}^{\gamma}\in
L^{2+\frac{1}{\gamma}}(\Omega\times(0,T))\hookrightarrow
L^{2}(\Omega\times(0,T))$ (3.10)

uniformly in $\gamma$. Thus, to prove the $H^{1}$ bound in (3.7), we only need
to find an estimate for the gradient of $p_{\gamma}$. Let us consider
$\int_{\Omega}u_{\gamma}^{\gamma}\mathop{}\\!\mathrm{d}x$ and compute its time
derivative: from (1.1) we infer

$\displaystyle\frac{d}{dt}\int_{\Omega}u_{\gamma}^{\gamma}\mathop{}\\!\mathrm{d}x=\int_{\Omega}\gamma
u_{\gamma}^{\gamma-1}\partial_{t}u_{\gamma}\mathop{}\\!\mathrm{d}x$
$\displaystyle=-\gamma\int_{\Omega}u_{\gamma}\nabla(u^{\gamma-1})\nabla(u_{\gamma}^{\gamma})\mathop{}\\!\mathrm{d}x-\gamma\int_{\Omega}u_{\gamma}\nabla(u_{\gamma}^{\gamma-1})\cdot\nabla
B_{\varepsilon}(u_{\gamma})\mathop{}\\!\mathrm{d}x+\gamma\int_{\Omega}u_{\gamma}(p_{H}-p_{\gamma})u_{\gamma}^{\gamma-1}\mathop{}\\!\mathrm{d}x$
$\displaystyle=-\gamma^{2}(\gamma-1)\int_{\Omega}u_{\gamma}^{2\gamma-2}|\nabla
u_{\gamma}|^{2}\mathop{}\\!\mathrm{d}x-\gamma\int_{\Omega}u_{\gamma}\nabla(u_{\gamma}^{\gamma-1})\cdot\nabla
B_{\varepsilon}(u_{\gamma})\mathop{}\\!\mathrm{d}x+\gamma\int_{\Omega}u_{\gamma}(p_{H}-p_{\gamma})u_{\gamma}^{\gamma-1}\mathop{}\\!\mathrm{d}x$
$\displaystyle=-(\gamma-1)\int_{\Omega}\left|\nabla(u_{\gamma}^{\gamma})\right|^{2}\mathop{}\\!\mathrm{d}x-\gamma(\gamma-1)\int_{\Omega}u_{\gamma}^{\gamma-1}\nabla
u_{\gamma}\cdot\nabla
B_{\varepsilon}(u_{\gamma})\mathop{}\\!\mathrm{d}x+\gamma\int_{\Omega}u_{\gamma}(p_{H}-p_{\gamma})u_{\gamma}^{\gamma-1}\mathop{}\\!\mathrm{d}x.$
(3.11)

Due to (1.2), we have

$\displaystyle\gamma(\gamma-1)\int_{\Omega}u_{\gamma}^{\gamma-1}\nabla
u_{\gamma}\cdot\nabla B_{\varepsilon}(u_{\gamma})\mathop{}\\!\mathrm{d}x$
$\displaystyle=\frac{1}{\varepsilon^{2}}{\gamma(\gamma-1)}\int_{\Omega}u_{\gamma}^{\gamma-1}|\nabla
u_{\gamma}|^{2}-\frac{1}{\varepsilon^{2}}{\gamma(\gamma-1)}\int_{\Omega}u_{\gamma}^{\gamma-1}\nabla
u_{\gamma}\cdot\nabla(\omega_{\varepsilon}\ast
u_{\gamma})\mathop{}\\!\mathrm{d}x$

By the Young inequality, recalling that
$\|u_{\gamma}\|_{L^{\infty}(0,T;L^{2}(\Omega))}\leq C$ (due to the uniform
bound on the energy and the non-local Poincaré inequality (A.3), see Lemma
1.3), we get

$\displaystyle\frac{1}{\varepsilon^{2}}{\gamma(\gamma-1)}\int_{\Omega}u_{\gamma}^{\gamma-1}\nabla
u_{\gamma}\cdot\nabla(\omega\ast u_{\gamma})\mathop{}\\!\mathrm{d}x$
$\displaystyle\leq\frac{\gamma^{2}(\gamma-1)}{2}\int_{\Omega}u_{\gamma}^{2\gamma-2}|\nabla
u_{\gamma}|^{2}\mathop{}\\!\mathrm{d}x+\frac{1}{\varepsilon^{4}}\frac{\gamma-1}{2}\int_{\Omega}|\nabla\omega_{\varepsilon}\ast
u_{\gamma}|^{2}\mathop{}\\!\mathrm{d}x$
$\displaystyle\leq\frac{\gamma^{2}(\gamma-1)}{2}\int_{\Omega}u_{\gamma}^{2\gamma-2}|\nabla
u_{\gamma}|^{2}\mathop{}\\!\mathrm{d}x+\frac{1}{\varepsilon^{4}}\frac{\gamma-1}{2}\|\nabla\omega_{\varepsilon}\|^{2}_{L^{1}(\Omega)}\|{u}_{\gamma}\|^{2}_{L^{2}(\Omega)}$
$\displaystyle\leq\frac{(\gamma-1)}{2}\int_{\Omega}|\nabla(u_{\gamma}^{\gamma})|^{2}\mathop{}\\!\mathrm{d}x+\frac{C(T,\varepsilon)(\gamma-1)}{2}.$

The last term in (3.11) can be controlled by

$\int_{\Omega}u_{\gamma}(p_{H}-p_{\gamma})\gamma
u_{\gamma}^{\gamma-1}\mathop{}\\!\mathrm{d}x\leq\int_{\Omega}(p_{H}-p_{\gamma})\gamma
u_{\gamma}^{\gamma}\mathop{}\\!\mathrm{d}x\leq\int_{p_{\gamma}\leq
p_{H}}(p_{H}-p_{\gamma})\gamma
u^{\gamma}_{\gamma}\mathop{}\\!\mathrm{d}x\leq\gamma|\Omega|p_{H}^{2}.$

Therefore, from (3.11) we get

$\displaystyle\frac{1}{\gamma-1}\frac{d}{dt}\int_{\Omega}u_{\gamma}^{\gamma}\mathop{}\\!\mathrm{d}x+\frac{1}{2}\int_{\Omega}\left|\nabla(u_{\gamma}^{\gamma})\right|^{2}\mathop{}\\!\mathrm{d}x\leq\frac{C}{2}+C\frac{\gamma}{\gamma-1}\leq
C(T,\varepsilon),$ (3.12)

showing that $\nabla u^{\gamma}\in L^{2}(\Omega\times(0,T))$ uniformly in
$\gamma$ so that $u_{\gamma}^{\gamma}$ is bounded uniformly in
$L^{2}(0,T;H^{1}(\Omega))$. We now need a similar estimate to show the $L^{3}$
bound in (3.7). We have

$\displaystyle\frac{1}{2\gamma}$
$\displaystyle\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}t}\int_{\Omega}u_{\gamma}^{2\gamma}\mathop{}\\!\mathrm{d}x=\int_{\Omega}u_{\gamma}^{2\gamma-1}\partial_{t}u_{\gamma}\mathop{}\\!\mathrm{d}x$
$\displaystyle=-\int_{\Omega}u_{\gamma}\nabla(u_{\gamma}^{2\gamma-1})\nabla(u_{\gamma}^{\gamma})\mathop{}\\!\mathrm{d}x-\int_{\Omega}u_{\gamma}\nabla(u_{\gamma}^{2\gamma-1})\cdot\nabla
B_{\varepsilon}(u_{\gamma})\mathop{}\\!\mathrm{d}x+\int_{\Omega}u_{\gamma}(p_{H}-p)u_{\gamma}^{2\gamma-1}\mathop{}\\!\mathrm{d}x$
$\displaystyle=-\gamma(2\gamma-1)\int_{\Omega}u_{\gamma}^{3\gamma-2}|\nabla
u_{\gamma}|^{2}\mathop{}\\!\mathrm{d}x-\frac{1}{\varepsilon^{2}}(2\gamma-1)\int_{\Omega}u_{\gamma}^{2\gamma-1}|\nabla
u_{\gamma}|^{2}\mathop{}\\!\mathrm{d}x$
$\displaystyle\quad+\frac{1}{\varepsilon^{2}}{(2\gamma-1)}\int_{\Omega}u_{\gamma}^{2\gamma-1}\nabla
u_{\gamma}\cdot(\nabla\omega_{\varepsilon}\ast
u_{\gamma})\mathop{}\\!\mathrm{d}x+\int_{\Omega}(p_{H}-p_{\gamma})u_{\gamma}^{2\gamma}\mathop{}\\!\mathrm{d}x.$
(3.13)

Notice now that, by Young’s inequality for convolutions and after integration
by parts,

$\displaystyle\frac{1}{\varepsilon^{2}}{(2\gamma-1)}$
$\displaystyle\int_{\Omega}u^{2\gamma-1}_{\gamma}\nabla
u_{\gamma}\cdot(\nabla\omega_{\varepsilon}\ast
u_{\gamma})\mathop{}\\!\mathrm{d}x$
$\displaystyle=\frac{1}{\varepsilon^{2}}\frac{2\gamma-1}{2\gamma}\int_{\Omega}\nabla
u^{2\gamma}_{\gamma}\cdot(\nabla\omega_{\varepsilon}\ast
u_{\gamma})\mathop{}\\!\mathrm{d}x=\frac{1}{\varepsilon^{2}}\frac{1-2\gamma}{2\gamma}\int_{\Omega}(\Delta\omega_{\varepsilon}\ast
u_{\gamma})u_{\gamma}^{2\gamma}\mathop{}\\!\mathrm{d}x$
$\displaystyle\leq\frac{1}{\varepsilon^{2}}\frac{2\gamma-1}{2\gamma}\|\Delta\omega_{\varepsilon}\|_{L^{\infty}(\Omega)}\|u_{\gamma}\|_{L^{1}(\Omega)}\int_{\Omega}p_{\gamma}^{2}\mathop{}\\!\mathrm{d}x\leq
C(\varepsilon)\frac{2\gamma-1}{2\gamma}\int_{\Omega}p_{\gamma}^{2}\mathop{}\\!\mathrm{d}x.$

We thus get, integrating (3.13) over $(0,T)$ and using the $L^{2}$ bound on
$p_{\gamma}$ in (3.10),

$\displaystyle\frac{1}{2\gamma}\int_{\Omega}p_{\gamma}^{2}(T)\mathop{}\\!\mathrm{d}x$
$\displaystyle+{\gamma(2\gamma-1)}\int_{0}^{T}\int_{\Omega}u_{\gamma}^{3\gamma-2}|\nabla
u_{\gamma}|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s+\frac{2\gamma-1}{\varepsilon^{2}}\int_{0}^{T}\int_{\Omega}u_{\gamma}^{2\gamma-1}|\nabla
u_{\gamma}|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s$
$\displaystyle+\int_{0}^{T}\int_{\Omega}p_{\gamma}^{3\gamma}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s\leq
C(T,\varepsilon)\left(\dfrac{2\gamma-1}{2\gamma}+1\right)\leq
C(T,\varepsilon),$

where we used that $\frac{2\gamma-1}{2\gamma}\to 1$ as $\gamma\to\infty$. From
this we deduce the uniform $L^{3}$ estimate in (3.7).

To prove (3.8), we compute for any $\varphi\in L^{2}(0,T;H^{1}(\Omega))$ and
$\gamma\geq\overline{\gamma}$, with $\overline{\gamma}(T)$ as in (3.1)

$\displaystyle\left|\int_{0}^{T}\langle\partial_{t}u_{\gamma},\varphi\rangle\mathop{}\\!\mathrm{d}t\right|\leq\left|\int_{0}^{T}\int_{\Omega}u_{\gamma}\nabla
p_{\gamma}\cdot\nabla\varphi\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t\right|+\left|\int_{0}^{T}\frac{1}{\varepsilon^{2}}\int_{\Omega}u_{\gamma}\nabla
u_{\gamma}\cdot\nabla\varphi\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t\right|$
$\displaystyle+\left|\int_{0}^{T}\frac{1}{\varepsilon^{2}}\int_{\Omega}u_{\gamma}(\nabla\omega_{\varepsilon}\ast
u_{\gamma})\cdot\nabla\varphi\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t\right|+\left|\int_{0}^{T}\int_{\Omega}u_{\gamma}(p_{H}-u_{\gamma}^{\gamma})\varphi\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t\right|.$

Note that by Young’s inequality for convolutions we get

$\displaystyle\|\nabla\omega_{\varepsilon}\ast
u_{\gamma}\|_{L^{2}(\Omega\times(0,T))}^{2}\leq\int_{0}^{T}\|\nabla\omega_{\varepsilon}\ast
u_{\gamma}\|_{L^{\infty}(\Omega)}^{2}|\Omega|\mathop{}\\!\mathrm{d}t\leq|\Omega|\|\nabla\omega_{\varepsilon}\|_{L^{\infty}(\Omega)}^{2}\int_{0}^{T}\|u_{\gamma}\|_{L^{1}(\Omega)}^{2}\mathop{}\\!\mathrm{d}t$
$\displaystyle\leq|\Omega|\|\nabla\omega_{\varepsilon}\|_{L^{\infty}(\Omega)}^{2}\int_{0}^{T}\left(\int_{\Omega}p_{H}^{\frac{1}{\gamma}}\mathop{}\\!\mathrm{d}x\right)^{2}\mathop{}\\!\mathrm{d}t\leq|\Omega|^{3}\|\nabla\omega_{\varepsilon}\|_{L^{\infty}(\Omega)}^{2}p_{H}^{\frac{2}{\gamma}}T\leq
C(\varepsilon,T),$

by (1.11). Therefore, by Lemma 1.3, (3.1) and (3.7)

$\displaystyle\left|\int_{0}^{T}\langle\partial_{t}u_{\gamma},\varphi\rangle\mathop{}\\!\mathrm{d}t\right|\leq\,$
$\displaystyle\|u_{\gamma}\|_{L^{\infty}(\Omega\times(0,T))}\|\nabla
p_{\gamma}\|_{L^{2}(0,T;L^{2}(\Omega))}\|\nabla\varphi\|_{L^{2}(0,T;L^{2}(\Omega))}$
$\displaystyle+C(\varepsilon)\|u_{\gamma}\|_{L^{\infty}(\Omega\times(0,T))}\|\nabla
u_{\gamma}\|_{L^{2}(\Omega\times(0,T))}\|\nabla\varphi\|_{L^{2}(\Omega\times(0,T))}$
$\displaystyle+C(\varepsilon)\|u_{\gamma}\|_{L^{\infty}(\Omega\times(0,T))}\|(\nabla\omega_{\varepsilon}\ast
u)_{\gamma}\|_{L^{2}(\Omega\times(0,T))}\|\nabla\varphi\|_{L^{2}(\Omega\times(0,T))}$
$\displaystyle+C\|u_{\gamma}\|_{L^{\infty}(\Omega\times(0,T))}\|p_{H}-u^{\gamma}_{\gamma}\|_{L^{2}(\Omega\times(0,T))}\|\varphi\|_{L^{2}(\Omega\times(0,T))}$
$\displaystyle\leq\,$ $\displaystyle
C(\varepsilon,T)\|\varphi\|_{L^{2}(0,T;H^{1}(\Omega))},$

Therefore, for any $\gamma\geq\overline{\gamma}$, we infer that
$\|\partial_{t}u_{\gamma}\|_{L^{2}(0,T;(H^{1}(\Omega)^{\prime})}\leq C(T)$,
thus showing (3.8).

It remains to prove (3.9). First note that, clearly, $\partial_{t}u_{\gamma}$
and $\partial_{t}p_{\gamma}$ share the same sign since $u_{\gamma}\geq 0$
almost everywhere in $\Omega\times(0,T)$. Then we differentiate in time (1.1)
and get

$\displaystyle\partial_{tt}u_{\gamma}-\dfrac{\gamma}{\gamma+1}\Delta\partial_{t}(u_{\gamma}^{\gamma+1})-$
$\displaystyle\frac{1}{2\varepsilon^{2}}\Delta\partial_{t}(u_{\gamma}^{2})+\frac{1}{\varepsilon^{2}}\operatorname{div}(\partial_{t}u_{\gamma}(\nabla\omega_{\varepsilon}\ast
u_{\gamma}))+\frac{1}{\varepsilon^{2}}\text{div}(u_{\gamma}(\nabla\omega_{\varepsilon}\ast\partial_{t}u_{\gamma}))$
$\displaystyle=\partial_{t}u_{\gamma}(p_{H}-p_{\gamma})-u_{\gamma}\partial_{t}p_{\gamma}.$

We test it against $\text{sign}(\partial_{t}u_{\gamma})$ and use Kato’s
inequality to obtain

$\displaystyle\partial_{t}|\partial_{t}u_{\gamma}|\leq$
$\displaystyle\frac{\gamma}{\gamma+1}\Delta(|\partial_{t}(u_{\gamma}^{\gamma+1})|)+\frac{1}{2\varepsilon^{2}}\Delta(|\partial_{t}(u_{\gamma}^{2})|)+\frac{1}{\varepsilon^{2}}\operatorname{div}(|\partial_{t}u_{\gamma}|(-\nabla\omega_{\varepsilon}\ast
u_{\gamma}))$
$\displaystyle-\frac{1}{\varepsilon^{2}}\text{div}(u_{\gamma}(\nabla\omega_{\varepsilon}\ast\partial_{t}u_{\gamma}))\,\text{sign}(\partial_{t}u_{\gamma})+|\partial_{t}u_{\gamma}|p_{H}-p_{\gamma}|\partial_{t}u_{\gamma}|-u_{\gamma}|\partial_{t}p_{\gamma}|.$

Now we rearrange the terms and integrate in space, deducing

$\displaystyle\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}t}\int_{\Omega}|\partial_{t}u_{\gamma}|\mathop{}\\!\mathrm{d}x+\int_{\Omega}u_{\gamma}|\partial_{t}p_{\gamma}|\mathop{}\\!\mathrm{d}x+\int_{\Omega}p_{\gamma}|\partial_{t}u_{\gamma}|\mathop{}\\!\mathrm{d}x$
$\displaystyle\leq-\frac{1}{\varepsilon^{2}}\int_{\Omega}\text{div}(u_{\gamma}(\nabla\omega_{\varepsilon}\ast\partial_{t}u_{\gamma}))\,\text{sign}(\partial_{t}u_{\gamma})\mathop{}\\!\mathrm{d}x+p_{H}\int_{\Omega}|\partial_{t}u_{\gamma}|\mathop{}\\!\mathrm{d}x.$

Then we have, by Young’s inequality for convolutions,

$\displaystyle-\frac{1}{\varepsilon^{2}}\int_{\Omega}$
$\displaystyle\text{div}(u_{\gamma}(\nabla\omega_{\varepsilon}\ast\partial_{t}u_{\gamma}))\,\text{sign}(\partial_{t}u_{\gamma})\mathop{}\\!\mathrm{d}x$
$\displaystyle=-\frac{1}{\varepsilon^{2}}\int_{\Omega}\nabla
u_{\gamma}\cdot(\nabla\omega_{\varepsilon}\ast\partial_{t}u_{\gamma})\,\text{sign}(\partial_{t}u_{\gamma})\mathop{}\\!\mathrm{d}x-\frac{1}{\varepsilon^{2}}\int_{\Omega}u_{\gamma}(\Delta\omega_{\varepsilon}\ast\partial_{t}u_{\gamma})\,\text{sign}(\partial_{t}u_{\gamma})\mathop{}\\!\mathrm{d}x$
$\displaystyle\leq\frac{1}{\varepsilon^{2}}\|\nabla
u_{\gamma}\|_{L^{1}(\Omega)}\|\nabla\omega_{\varepsilon}\ast\partial_{t}u_{\gamma}\|_{L^{\infty}(\Omega)}+\frac{1}{\varepsilon^{2}}\|u_{\gamma}\|_{L^{1}(\Omega)}\|\Delta\omega_{\varepsilon}\ast\partial_{t}u_{\gamma}\|_{L^{\infty}(\Omega)}$
$\displaystyle\leq C(\varepsilon)(1+\|\nabla
u_{\gamma}\|_{L^{1}(\Omega}))\int_{\Omega}|\partial_{t}u_{\gamma}|\mathop{}\\!\mathrm{d}x,$

since

$\|\nabla\omega_{\varepsilon}\ast\partial_{t}u_{\gamma}\|_{L^{\infty}(\Omega)}\leq\|\nabla\omega_{\varepsilon}\|_{L^{\infty}(\Omega)}\int_{\Omega}|\partial_{t}u_{\gamma}|\mathop{}\\!\mathrm{d}x\leq
C(\varepsilon)\int_{\Omega}|\partial_{t}u_{\gamma}|\mathop{}\\!\mathrm{d}x,$

and the same for
$\|\Delta\omega_{\varepsilon}\ast\partial_{t}u_{\gamma}\|_{L^{\infty}(\Omega)}$.
Therefore we end up with

$\displaystyle\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}t}\int_{\Omega}|\partial_{t}u_{\gamma}|\mathop{}\\!\mathrm{d}x+\int_{\Omega}u_{\gamma}|\partial_{t}p_{\gamma}|\mathop{}\\!\mathrm{d}x+\int_{\Omega}p_{\gamma}|\partial_{t}u_{\gamma}|\mathop{}\\!\mathrm{d}x\leq
C(\varepsilon)(1+\|\nabla
u_{\gamma}\|_{L^{1}(\Omega}))\int_{\Omega}|\partial_{t}u_{\gamma}|\mathop{}\\!\mathrm{d}x.$
(3.14)

To apply the Gronwall inequality, we need to estimate
$\|\partial_{t}u_{\gamma}(0)\|_{L^{1}(\Omega)}$. From (1.1) we have

$\displaystyle\|\partial_{t}u_{\gamma}(0)\|_{L^{1}(\Omega)}\leq\,$
$\displaystyle
C(\varepsilon)\Big{(}\frac{\gamma}{\gamma+1}\|\Delta(u_{0}^{\gamma+1})\|_{L^{1}(\Omega)}+\|\Delta(u_{0}^{2})\|_{L^{1}(\Omega)}+$
$\displaystyle+\|\text{div}(u_{0}(\nabla\omega_{\varepsilon}\ast
u_{0}))\|_{L^{1}(\Omega)}+\|u_{0}(p_{H}-p_{0})\|_{L^{1}(\Omega)}\Big{)}.$

Due to Assumption 1.2, the first term is bounded. Concerning next terms, we
have by Assumption 1.1 and 1.2

$\|\Delta(u_{0}^{2})\|_{L^{1}(\Omega)}\leq 2\|\nabla
u_{0}\|_{L^{2}(\Omega)}^{2}+2\|u_{0}\Delta u_{0}\|_{L^{1}(\Omega)}\leq
2\|\nabla u_{0}\|_{L^{2}(\Omega)}^{2}+2p_{H}^{\frac{1}{\gamma}}\|\Delta
u_{0}\|_{L^{1}(\Omega)}\leq C,$ $\displaystyle\|\text{div}(u_{0}$
$\displaystyle(\nabla\omega_{\varepsilon}\ast
u_{0}))\|_{L^{1}(\Omega)}\leq\|\nabla
u_{0}\cdot(\nabla\omega_{\varepsilon}\ast
u_{0})\|_{L^{1}(\Omega)}+\|u_{0}(\Delta\omega_{\varepsilon}\ast
u_{0})\|_{L^{1}(\Omega)}$ $\displaystyle\leq\|\nabla
u_{0}\|_{L^{1}(\Omega)}\|\nabla\omega_{\varepsilon}\|_{L^{\infty}(\Omega)}\|u_{0}\|_{L^{1}(\Omega)}+p_{H}^{\frac{1}{\gamma}}|\Omega|\|\Delta\omega_{\varepsilon}\|_{L^{\infty}(\Omega)}\|u_{0}\|_{L^{1}(\Omega)}\leq
C(\varepsilon).$

It follows that $\|\partial_{t}u_{\gamma}(0)\|_{L^{1}(\Omega)}\leq
C(\varepsilon)$. Since $u_{\gamma}\in L^{2}(0;T;H^{1}(\Omega))$ uniformly in
$\gamma$ (Lemma 1.3), we may apply the Gronwall Lemma in (3.14) and obtain

$\displaystyle\|\partial_{t}u_{\gamma}\|_{L^{\infty}(0,T;L^{1}(\Omega))}+\|u_{\gamma}\partial_{t}p_{\gamma}\|_{L^{1}(0,T;L^{1}(\Omega))}\leq
C(T,\varepsilon),$ (3.15)

with $C=C(T,\varepsilon)>0$ independent of $\gamma$. To conclude the argument
we notice that

$\displaystyle\int_{\Omega}|\partial_{t}p_{\gamma}|\mathop{}\\!\mathrm{d}x\leq\int_{u_{\gamma}\leq\frac{1}{2}}|\partial_{t}p_{\gamma}|\mathop{}\\!\mathrm{d}x+2\int_{u_{\gamma}>\frac{1}{2}}u_{\gamma}|\partial_{t}p_{\gamma}|\mathop{}\\!\mathrm{d}x\leq\frac{\gamma}{2^{\gamma-1}}\int_{\Omega}|\partial_{t}u_{\gamma}|\mathop{}\\!\mathrm{d}x+2\int_{\Omega}u_{\gamma}|\partial_{t}p_{\gamma}|\mathop{}\\!\mathrm{d}x,$

so that, being $\frac{\gamma}{2^{\gamma-1}}\leq 1$ for any $\gamma\geq 1$,
from (3.15) we deduce the second estimate in (3.9) and conclude the proof of
Lemma 3.3. ∎

###  3.3 The limit $\gamma\to\infty$

We complete the convergence as $\gamma\to\infty$ distinguishing two steps.

#### Step 1. Consequences of Lemmas 1.3, 3.1 and 3.3.

Exploiting those lemmas, we can obtain the following convergences, up to
subsequences, which are deduced by standard compactness arguments: for any
$T>0$, as $\gamma\to\infty$,

$\displaystyle u_{\gamma}\to u_{\infty}\text{ almost everywhere in
}\Omega\times(0,T),$ (3.16) $\displaystyle
u_{\gamma}\overset{\ast}{\rightharpoonup}u_{\infty}\text{ in
}L^{\infty}(\Omega\times(0,T)),$ (3.17) $\displaystyle u_{\gamma}\to
u_{\infty}\text{ in }L^{p}(\Omega\times(0,T))\quad\forall p\in[1,+\infty),$
(3.18) $\displaystyle u_{\gamma}\rightharpoonup u_{\infty}\text{ in
}L^{2}(0,T;H^{1}(\Omega)),$ (3.19)
$\displaystyle\partial_{t}u_{\gamma}\rightharpoonup\partial_{t}u_{\infty}\text{
in }L^{2}(0,T;(H^{1}(\Omega))^{\prime}),$ (3.20) $\displaystyle
p_{\gamma}\rightharpoonup p_{\infty}\text{ in
}L^{2}(0,T;H^{1}(\Omega)),\;L^{3}((0,T)\times\Omega).$ (3.21)

Moreover, by the Aubin-Lions-Simon Lemma,

$\displaystyle p_{\gamma}\to p_{\infty}\text{ in }L^{2}(0,T;L^{2}(\Omega)),$
(3.22)

which, thanks to (3.7) can be improved by interpolation, to

$\displaystyle p_{\gamma}\to p_{\infty}\text{ in
}L^{q}(0,T;L^{q}(\Omega)),\quad\forall q\in[2,3).$ (3.23)

In order to obtain the complementarity condition, we study the function
$v_{\gamma}:=u_{\gamma}^{\gamma+1}=u_{\gamma}p_{\gamma}$. First we have

$\|v_{\gamma}\|_{L^{2}(\Omega\times(0,T))}\leq\|u_{\gamma}\|_{L^{\infty}(\Omega\times(0,T))}\|p_{\gamma}\|_{L^{2}(\Omega\times(0,T))}\leq
C(T,\varepsilon),$

and

$\|\nabla
v_{\gamma}\|_{L^{2}(\Omega\times(0,T))}=\frac{\gamma+1}{\gamma}\|u_{\gamma}\nabla
p_{\gamma}\|_{L^{2}(\Omega\times(0,T))}\leq\frac{\gamma+1}{\gamma}\|u_{\gamma}\|_{L^{\infty}(\Omega\times(0,T))}\|\nabla
p_{\gamma}\|_{L^{2}(\Omega\times(0,T))}\leq C(T,\varepsilon),$

since $\frac{\gamma+1}{\gamma}\to 1$ as $\gamma\to\infty$. Therefore, up to
subsequences, we have

$v_{\gamma}=u_{\gamma}p_{\gamma}\rightharpoonup v_{\infty}\qquad\text{in
}\;L^{2}(0,T;H^{1}(\Omega)),$

for some $v_{\infty}\in L^{2}(0,T;H^{1}(\Omega))$. To identify $v_{\infty}$,
we observe that

$\displaystyle\|u_{\gamma}$ $\displaystyle
p_{\gamma}-u_{\infty}p_{\infty}\|_{L^{2}(\Omega\times(0,T))}\leq\|u_{\gamma}(p_{\gamma}-p_{\infty})\|_{L^{2}(\Omega\times(0,T))}+\|p_{\infty}(u_{\gamma}-u_{\infty})\|_{L^{2}(\Omega\times(0,T))}$
$\displaystyle\leq\|u_{\gamma}\|_{L^{\infty}(\Omega\times(0,T))}\|p_{\gamma}-p_{\infty}\|_{L^{2}(\Omega\times(0,T))}+\|u_{\gamma}-u_{\infty}\|_{L^{6}(\Omega\times(0,T))}\|p_{\infty}\|_{L^{3}(\Omega\times(0,T))}\to
0$

as $\gamma\to\infty$, thanks to the above results, in particular (3.1), (3.7),
(3.18) and (3.22). From this we clearly identify
$v_{\infty}=u_{\infty}p_{\infty}$ and obtain

$\displaystyle v_{\gamma}=u_{\gamma}p_{\gamma}\rightharpoonup
v_{\infty}=u_{\infty}p_{\infty}\text{ in }L^{2}(0,T;H^{1}(\Omega)).$ (3.24)
$\displaystyle v_{\gamma}\to p_{\infty}u_{\infty}\text{ in
}L^{2}(\Omega\times(0,T)).$ (3.25)

We are now able to pass to the limit in $\gamma$ to obtain (1.15) and

$\displaystyle p_{\infty}(1-u_{\infty})=0,\quad 0\leq u_{\infty}\leq 1,\quad
p_{\infty}\geq 0.$ (3.26)

Indeed, we can argue as in [37] to see that, for any $\epsilon>0$ there exists
$\gamma_{0}=\gamma_{0}(\epsilon)$ such that for any $y\geq 0$ and
$\gamma\geq\gamma_{0}$

$y^{\gamma+1}\geq y^{\gamma}-\epsilon.$

Applying this to $y=u_{\gamma}$, since we have that, up to subsequences,
$u^{\gamma}=p_{\gamma}\to p_{\infty}$ and $v_{\gamma}=u_{\gamma}^{\gamma+1}\to
p_{\infty}u_{\infty}$ almost everywhere in $\Omega\times(0,T)$, we can pass to
the limit and obtain,

$u_{\infty}p_{\infty}\geq p_{\infty}-\epsilon,\qquad\forall\epsilon.>0,$

which implies that $p_{\infty}u_{\infty}\geq p_{\infty}$. Since by (3.1) and
(3.16) we get $0\leq u_{\infty}\leq 1$ almost everywhere in
$\Omega\times(0,T)$, we get $p_{\infty}\geq p_{\infty}u_{\infty}\geq
p_{\infty}$, i.e., (3.26), since it holds for any $T>0$. In order to pass to
Step 2, we introduce the quantities

$\widetilde{v}_{\gamma}:=\frac{\gamma}{\gamma+1}v_{\gamma}+\frac{1}{2\varepsilon^{2}}u_{\gamma}^{2},\qquad\widetilde{v}_{\infty}=v_{\infty}+\frac{1}{2\varepsilon^{2}}u_{\infty}^{2},$
(3.27)

and study $\nabla\widetilde{v}_{\gamma}$ which is essential to obtain the
complementarity condition (1.16).

#### Step 2. Strong convergence of $\nabla\widetilde{v}_{\gamma}$.

######  Lemma 3.4.

Let $\widetilde{v}_{\gamma}$, $\widetilde{v}_{\infty}$ be as in (3.27). Then,
for any $T>0$,

$\widetilde{v}_{\gamma}\to\widetilde{v}_{\infty}\quad\text{in
}L^{2}(0,T;H^{1}(\Omega))\quad\text{ as }\gamma\to\infty.$

###### Proof.

Using (3.18) and (3.25) we obtain
$\widetilde{v}_{\gamma}\to\widetilde{v}_{\infty}$ in
$L^{2}(\Omega\times(0,T))$ so it is sufficient to prove
$\nabla\widetilde{v}_{\gamma}\to\nabla\widetilde{v}_{\infty}$ in
$L^{2}(\Omega\times(0,T))$. Of course, by (3.16)-(3.21) and (3.24), we have
weak convergence

$\displaystyle\nabla\widetilde{v}_{\gamma}\rightharpoonup\nabla\widetilde{v}_{\infty}\,\text{in
}L^{2}(0,T;L^{2}(\Omega)),\quad\gamma\to\infty.$ (3.28)

Let us first observe that (1.1) can be rewritten highlighting the presence of
$\widetilde{v}_{\gamma}$:

$\displaystyle\partial_{t}u_{\gamma}-\Delta\widetilde{v}_{\gamma}=u_{\gamma}G(p_{\gamma})-\frac{1}{\varepsilon^{2}}\text{div}(u_{\gamma}(\nabla\omega_{\varepsilon}\ast
u_{\gamma})).$ (3.29)

We multiply (3.29) by $\widetilde{v}_{\gamma}-\widetilde{v}_{\infty}$ and
integrate over $\Omega_{T}:=\Omega\times(0,T)$. Since

$\int_{\Omega}\partial_{t}u_{\gamma}v_{\gamma}\mathop{}\\!\mathrm{d}x=\frac{1}{\gamma+2}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}t}\int_{\Omega}u_{\gamma}^{\gamma+2}\mathop{}\\!\mathrm{d}x.$

we obtain

$\displaystyle\frac{\gamma}{(\gamma+2)(\gamma+1)}\int_{\Omega}u_{\gamma}^{\gamma+2}(T)\mathop{}\\!\mathrm{d}x+\int_{\Omega_{T}}\nabla\widetilde{v}_{\gamma}\cdot\nabla(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s$
$\displaystyle=\frac{\gamma}{(\gamma+2)(\gamma+1)}\int_{\Omega}u_{\gamma}^{\gamma+2}(0)\mathop{}\\!\mathrm{d}x-\frac{1}{2\varepsilon^{2}}\int_{\Omega_{T}}\partial_{t}u_{\gamma}(u_{\gamma}^{2}-u_{\infty}^{2})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t+\int_{\Omega_{T}}u_{\gamma}G(p_{\gamma})(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t$
$\displaystyle+\frac{1}{\varepsilon^{2}}\int_{\Omega_{T}}u_{\gamma}(\nabla\omega_{\varepsilon}\ast
u_{\gamma})\cdot\nabla(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t+\int_{0}^{T}\langle\partial_{t}u_{\gamma},v_{\infty}\rangle\mathop{}\\!\mathrm{d}t.$
(3.30)

The plan is to estimate
$\limsup_{\gamma\to\infty}\int_{\Omega_{T}}|\nabla\widetilde{v}_{\gamma}-\nabla\widetilde{v}_{\infty}|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t$
from (3.30). First, we rewrite the term
$\frac{1}{2\varepsilon^{2}}\int_{\Omega_{T}}\partial_{t}u_{\gamma}(u_{\gamma}^{2}-u_{\infty}^{2})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t$.
We have

$\displaystyle\frac{1}{2\varepsilon^{2}}\int_{\Omega}\partial_{t}u_{\gamma}(u^{2}_{\gamma}-u_{\infty}^{2})\mathop{}\\!\mathrm{d}x=\frac{1}{2\varepsilon^{2}}\langle\partial_{t}u_{\gamma}-\partial_{t}u_{\infty},u_{\gamma}^{2}-u_{\infty}^{2}\rangle+\frac{1}{2\varepsilon^{2}}\langle\partial_{t}u_{\infty},u^{2}_{\gamma}-u_{\infty}^{2}\rangle$
$\displaystyle=\frac{1}{4\varepsilon^{2}}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}t}\int_{\Omega}(u_{\gamma}-u_{\infty})^{2}(u_{\gamma}+u_{\infty})\mathop{}\\!\mathrm{d}x-\frac{1}{4\varepsilon^{2}}\langle\partial_{t}(u_{\gamma}+u_{\infty}),(u_{\gamma}-u_{\infty})^{2}\rangle+\frac{1}{2\varepsilon^{2}}\langle\partial_{t}u_{\infty},u_{\gamma}^{2}-u_{\infty}^{2}\rangle$
$\displaystyle=\frac{1}{4\varepsilon^{2}}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}t}\int_{\Omega}(u_{\gamma}-u_{\infty})^{2}(u_{\gamma}+u_{\infty})\mathop{}\\!\mathrm{d}x-\frac{1}{12\varepsilon^{2}}\frac{\mathop{}\\!\mathrm{d}}{\mathop{}\\!\mathrm{d}t}\int_{\Omega}(u_{\gamma}-u_{\infty})^{3}\mathop{}\\!\mathrm{d}x-\frac{1}{2\varepsilon^{2}}\langle\partial_{t}u_{\infty},(u_{\gamma}-u_{\infty})^{2}\rangle$
$\displaystyle\phantom{=}+\frac{1}{2\varepsilon^{2}}\langle\partial_{t}u_{\infty},u^{2}_{\gamma}-u_{\infty}^{2}\rangle,$

so that, integrating over $[0,T]$ and recalling that $u_{\infty}(0)\equiv
u_{\gamma}(0)\equiv u_{0}$, we get

$\displaystyle-\frac{1}{2\varepsilon^{2}}\int_{\Omega_{T}}\partial_{t}u_{\gamma}(u_{\gamma}^{2}-u_{\infty}^{2})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t=-\frac{1}{4\varepsilon^{2}}\int_{\Omega}(u_{\gamma}(T)-u_{\infty}(T))^{2}(u_{\gamma}(T)+u_{\infty}(T))\mathop{}\\!\mathrm{d}x$
$\displaystyle+\frac{1}{12\varepsilon^{2}}\int_{\Omega}(u_{\gamma}(T)-u_{\infty}(T))^{3}\mathop{}\\!\mathrm{d}x+\frac{1}{2\varepsilon^{2}}\int_{0}^{T}\langle\partial_{t}u_{\infty},(u_{\gamma}-u_{\infty})^{2}\rangle\mathop{}\\!\mathrm{d}t-\frac{1}{2\varepsilon^{2}}\int_{0}^{T}\langle\partial_{t}u_{\infty},u^{2}_{\gamma}-u_{\infty}^{2}\rangle\mathop{}\\!\mathrm{d}t.$
(3.31)

Note that

$\begin{split}\frac{1}{4\varepsilon^{2}}\int_{\Omega}&(u_{\gamma}(T)-u_{\infty}(T))^{2}(u_{\gamma}(T)+u_{\infty}(T))\mathop{}\\!\mathrm{d}x-\frac{1}{12\varepsilon^{2}}\int_{\Omega}(u_{\gamma}(T)-u_{\infty}(T))^{3}\mathop{}\\!\mathrm{d}x\\\
&=\frac{1}{4\varepsilon^{2}}\int_{\Omega}(u_{\gamma}(T)-u_{\infty}(T))^{2}\left(u_{\gamma}(T)+u_{\infty}(T)-\frac{1}{3}u_{\gamma}(T)+\frac{1}{3}u_{\infty}(T)\right)\mathop{}\\!\mathrm{d}x\geq
0.\end{split}$ (3.32)

Therefore, taking into account (3.3), (3.32) and $u_{\gamma}\geq 0$ we obtain
from (3.30)

$\displaystyle\int_{\Omega_{T}}\nabla\widetilde{v}_{\gamma}\cdot$
$\displaystyle\nabla(\widetilde{v}-\widetilde{v}_{\infty})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s\leq\frac{\gamma}{(\gamma+2)(\gamma+1)}\int_{\Omega}u_{\gamma}^{\gamma+2}(0)\mathop{}\\!\mathrm{d}x+\frac{1}{2\varepsilon^{2}}\int_{0}^{T}\langle\partial_{t}u_{\infty},(u_{\gamma}-u_{\infty})^{2}\rangle\mathop{}\\!\mathrm{d}t$
$\displaystyle-\int_{0}^{T}\langle\partial_{t}u_{\infty},\frac{1}{2\varepsilon^{2}}u_{\gamma}^{2}-\frac{1}{2\varepsilon^{2}}u_{\infty}^{2}\rangle\mathop{}\\!\mathrm{d}t+\int_{\Omega_{T}}u_{\gamma}G(p_{\gamma})(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t$
$\displaystyle+\frac{1}{\varepsilon^{2}}\int_{\Omega_{T}}u_{\gamma}(\nabla\omega_{\varepsilon}\ast
u_{\gamma})\cdot\nabla(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t+\int_{0}^{T}\langle\partial_{t}u_{\gamma},v_{\infty}\rangle\mathop{}\\!\mathrm{d}t.$
(3.33)

Observe that

$\displaystyle\int_{\Omega_{T}}\nabla\widetilde{v}_{\gamma}\cdot\nabla(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s=\int_{\Omega_{T}}|\nabla(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s+\int_{\Omega_{T}}\nabla\widetilde{v}_{\infty}\cdot\nabla(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s,$

but by the weak convergence in (3.28) we have

$\int_{\Omega_{T}}\nabla\widetilde{v}_{\infty}\cdot\nabla(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s\to
0,$

since $\nabla\widetilde{v}_{\infty}\in L^{2}(0,T;L^{2}(\Omega))$. Thus we
deduce that

$\displaystyle\limsup_{\gamma\to\infty}\int_{\Omega_{T}}|\nabla(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s=\limsup_{\gamma\to\infty}\int_{\Omega_{T}}\nabla\widetilde{v}_{\gamma}\cdot\nabla(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s.$

Then, recalling also that
$\partial_{t}u_{\gamma}\rightharpoonup\partial_{t}u_{\infty}$ in
$L^{2}(0,T;(H^{1}(\Omega))^{\prime})$ and that $v_{\infty}\in
L^{2}(0,T;H^{1}(\Omega))$, we deduce from (3.33)

$\displaystyle\limsup_{\gamma\to\infty}\int_{\Omega_{T}}|\nabla(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s\leq\limsup_{\gamma\to\infty}\left(\frac{\gamma}{(\gamma+2)(\gamma+1)}\int_{\Omega}u_{\gamma}^{\gamma+2}(0)\mathop{}\\!\mathrm{d}x\right.$
$\displaystyle+\frac{1}{2\varepsilon^{2}}\int_{0}^{T}\langle\partial_{t}u_{\infty},(u_{\gamma}-u_{\infty})^{2}\rangle\mathop{}\\!\mathrm{d}t\left.-\frac{1}{2\varepsilon^{2}}\int_{0}^{T}\langle\partial_{t}u_{\infty},u_{\gamma}^{2}-u_{\infty}^{2}\rangle\mathop{}\\!\mathrm{d}t+\int_{\Omega_{T}}u_{\gamma}G(p_{\gamma})(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t\right.$
$\displaystyle\left.+\frac{1}{\varepsilon^{2}}\int_{\Omega_{T}}u_{\gamma}\nabla\omega_{\varepsilon}\ast
u_{\gamma}\cdot\nabla(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t\right)+\int_{0}^{T}\langle\partial_{t}u_{\infty},v_{\infty}\rangle\mathop{}\\!\mathrm{d}t.$
(3.34)

The plan is to prove that all the terms on the (RHS) of (3.34) converge to 0.
First, we have that $0\leq u_{\gamma}(0)=u_{0}\leq p_{H}^{\frac{1}{\gamma}}$
for any $\gamma>0$ by assumption, so that in the end we deduce
$u_{\gamma}^{\gamma+1}(0)\leq p_{H}^{\frac{\gamma+2}{\gamma}}\leq C$ and thus,
as $\gamma\to\infty$,

$\frac{\gamma}{(\gamma+2)(\gamma+1)}\int_{\Omega}u_{\gamma}^{\gamma+2}(0)\mathop{}\\!\mathrm{d}x\to
0.$

Then, using (3.18) and the fact that $\partial_{t}u_{\infty}\in
L^{2}(0,T;(H^{1}(\Omega))^{\prime})$, we immediately deduce

$\frac{1}{2\varepsilon^{2}}\int_{0}^{T}\langle\partial_{t}u_{\infty},(u_{\gamma}-u_{\infty})^{2}\rangle\mathop{}\\!\mathrm{d}t-\frac{1}{2\varepsilon^{2}}\int_{0}^{T}\langle\partial_{t}u_{\infty},u_{\gamma}^{2}-u_{\infty}^{2}\rangle\mathop{}\\!\mathrm{d}t\to
0\quad\text{ as }\gamma\to\infty.$

Concerning the term
$\int_{\Omega_{T}}u_{\gamma}G(p_{\gamma})(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t$,
we simply use the fact that $\\{u_{\gamma}\\}$ is bounded in
$L^{\infty}(\Omega\times(0,T))$ (cf. (3.1)), $\\{G(p_{\gamma})\\}$ is bounded
in $L^{2}(\Omega\times(0,T))$ (cf. (3.7)) and
$\widetilde{v}_{\gamma}\to\widetilde{v}_{\infty}$ strongly in
$L^{2}(\Omega\times(0,T))$. Similarly,
$\int_{\Omega_{T}}u_{\gamma}\nabla\omega_{\varepsilon}\ast
u_{\gamma}\cdot\nabla(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}t\to
0$ because of weak convergence (3.28) and strong convergence
$u_{\gamma}\nabla\omega_{\varepsilon}\ast u_{\gamma}\to
u_{\infty}\nabla\omega_{\varepsilon}\ast u_{\infty}$ in
$L^{2}(\Omega\times(0,T))$ (which follows by (3.18) and simple properties of
convolutions).

We are left with the analysis of the last term, i.e.,
$\int_{0}^{T}\langle\partial_{t}u_{\infty},v_{\infty}\rangle\mathop{}\\!\mathrm{d}t$.
Our aim is to show that this term vanishes, exploiting Theorem A.2. We
introduce the following indicator function on $\mathbb{R}$:

$\displaystyle I_{S}(s):=\begin{cases}0\quad\text{ if }s\leq 1,\\\
+\infty\text{ if }s>1,\end{cases}$ (3.35)

and define $S=(-\infty,1]$, which is a closed, convex and nonempty set, so
that $I_{S}:\mathbb{R}\to(-\infty,+\infty]$ is proper, convex and lower
semicontinuous (see, e.g., [45, Appendix 1]) and it holds

$\partial I_{S}(x)=\\{y\in\mathbb{R}:y\cdot(x-s)\geq 0,\quad\forall s\leq
1\\}.$

We see that we have

  * •

$u_{\infty}\in L^{2}(0,T;H^{1}(\Omega))$ and $\partial_{t}u_{\infty}\in
L^{2}(0,T;(H^{1}(\Omega))^{\prime})$;

  * •

for almost any $(x,t)\in\Omega_{T}$,

$v_{\infty}(x,t)\in\partial I_{S}(u_{\infty}(x,t)),$

by (3.26) and being $v_{\infty}=p_{\infty}$. Indeed,

$v_{\infty}(x,t)(u_{\infty}(x,t)-s)\geq 0,\quad\forall s\leq 1,$

since, when $u_{\infty}(x,t)=1$, being $v_{\infty}(x,t)\geq 0$, the inequality
is always verified for any $s\leq 1$, whereas, when $u_{\infty}(x,t)<1$, it
holds $v_{\infty}(x,t)=0$ and thus the inequality is verified for any $s\leq
1$ as well.

  * •

$v_{\infty}\in L^{2}(0,T;H^{1}(\Omega))$.

Therefore, all the assumptions are verified and we can apply Theorem A.2 with
$h=I_{S}$, $f=u_{\infty}$, $g=v_{\infty}$, to infer, after an integration over
$[0,T]$,

$\displaystyle
0=\int_{\Omega}I_{S}(u_{\infty}(x,T))\mathop{}\\!\mathrm{d}x-\int_{\Omega}I_{S}(u_{\infty}(x,0))\mathop{}\\!\mathrm{d}x=\int_{0}^{T}\langle\partial_{t}u_{\infty},v_{\infty}\rangle\mathop{}\\!\mathrm{d}t,$
(3.36)

by the definition of $I_{S}$. Indeed, it holds
$\int_{\Omega}I_{S}(u_{\infty}(x,\cdot))\mathop{}\\!\mathrm{d}x\equiv 0$ on
$[0,T]$ for any $T>0$. To see this, first notice that, being $u_{\infty}\leq
1$ almost everywhere in $\Omega\times[0,\infty)$, we deduce that, for almost
any $t\in[0,\infty)$,

$u_{\infty}(x,t)\leq 1\quad\text{ for almost any }x\in\Omega.$

Therefore, $\int_{\Omega}I_{S}(u_{\infty}(x,\cdot))\mathop{}\\!\mathrm{d}x=0$,
for almost any $t\in[0,T]$ and for any $T>0$. Recall now by Theorem A.2 that
$\int_{\Omega}I_{S}(u_{\infty}(x,\cdot))\mathop{}\\!\mathrm{d}x\in AC([0,T])$
for any $T>0$, which ensures that
$\int_{\Omega}I_{S}(u_{\infty}(x,t))\mathop{}\\!\mathrm{d}x\equiv 0$ for any
$t\in[0,T]$ and for any $T>0$.

Having studied all the terms in the right-hand side of (3.34), in the end we
conclude that

$\limsup_{\gamma\to\infty}\int_{\Omega_{T}}|\nabla(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s\leq
0,$

implying that

$\lim_{\gamma\to\infty}\int_{\Omega_{T}}|\nabla(\widetilde{v}_{\gamma}-\widetilde{v}_{\infty})|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}s=0.$

∎

#### Step 3. The complementarity condition.

Now that we have all the necessary convergences, let us consider the following
equation in distributional sense (this equation comes from multiplying (1.1)
by $v_{\gamma}=u_{\gamma}^{\gamma+1}$):

$\displaystyle\frac{1}{\gamma+2}\partial_{t}u^{\gamma+2}_{\gamma}-v_{\gamma}\Delta\widetilde{v}_{\gamma}+\frac{1}{\varepsilon^{2}}v_{\gamma}\text{div}(u_{\gamma}(\nabla\omega_{\varepsilon}\ast
u_{\gamma}))=u_{\gamma}G(p_{\gamma})v_{\gamma}.$

Thanks to the results of Step 2. and Lemma 3.4, we can then pass the limit as
$\gamma\to\infty$ and obtain the complementarity condition:

$\displaystyle
v_{\infty}(\Delta\widetilde{v}_{\infty}-\frac{1}{\varepsilon^{2}}\text{div}(u_{\infty}(\nabla\omega\ast
u_{\infty}))+u_{\infty}G(p_{\infty}))=0\quad\text{ in
}\mathcal{D}^{\prime}(\Omega\times(0,T)),$ (3.37)

for any $T>0$, so that in the end, recalling $v_{\infty}=p_{\infty}$ and
$\widetilde{v}_{\infty}=p_{\infty}+\frac{1}{2\varepsilon^{2}}u_{\infty}^{2}$,

$\displaystyle p_{\infty}(\Delta{p}_{\infty}+\frac{1}{2\varepsilon^{2}}\Delta
u_{\infty}^{2}-\frac{1}{\varepsilon^{2}}\text{div}(u_{\infty}(\nabla\omega_{\varepsilon}\ast
u_{\infty}))+u_{\infty}G(p_{\infty}))=0\quad\text{ in
}\mathcal{D}^{\prime}(\Omega\times(0,\infty)),$ (3.38)

which is the complementarity condition (1.16). In conclusion, since
$v_{\infty}=p_{\infty}$, we can repeat the argument leading to (3.36), to
infer that, for any $t\in[0,\infty)$,

$\int_{0}^{t}\langle\partial_{t}u_{\infty},p_{\infty}\rangle\mathop{}\\!\mathrm{d}s=0,\qquad\langle\partial_{t}u_{\infty}(t),p_{\infty}(t)\rangle=0,$

thus concluding the proof of Theorem 1.5.

##  4 Convergence to equilibria: proof of Theorem 1.6

Here, we prove Theorem 1.6. Numerical simulations illustrating the result in
dimension $1$ with a source term are depicted in Figure 1.

###  4.1 Case $G(p)=p_{H}-p$

Proof of Theorem 1.6. We now divide the proof of Theorem 1.6 in different
steps.

Step 1: characterization of possible limits. We fix the initial datum
$u_{0}\neq 0$ as in the statement of Theorem 1.6. We show the following key
result:

######  Lemma 4.1.

From any divergent sequence $\\{t_{n}\\}_{n}\subset\mathbb{R}^{+}$ we can
extract a subsequence (not relabeled) such that, as $n\to\infty$,
$u_{n}(t):=u(t+t_{n})$ converges to the same limit $u_{\ast}$ strongly in
$L^{p}_{t,x}$ for all $1\leq p<2\gamma+1$. Moreover, either $u_{\ast}\equiv 0$
and $\Phi(u(t))\to|\Omega|$ as $t\to\infty$, otherwise $u_{\ast}\equiv
p_{H}^{1/\gamma}$ and $\Phi(u(t))\to 0$ as $t\to\infty$.

We fix $T>0$, we also consider $n$ large enough such that $t_{n}>T$. Observe
that $u_{n}$ solves the problem

$\begin{cases}\partial_{t}u_{n}-\text{div}(u_{n}\nabla\mu_{n})=u_{n}G(p_{n})\quad\text{in
}\Omega\times(-T,T),\\\ \mu_{n}=p_{n}+B_{\varepsilon}(u_{n}),\qquad\quad
p_{n}=u_{n}^{\gamma}.\end{cases}$ (4.1)

By Propositions 2.2 and 2.5, we have the following uniform-in-$n$ bounds :

$\displaystyle\mathcal{E}(u_{n}(t))\leq C\quad\forall t\in[-T,T],$ (4.2)
$\displaystyle\Phi(u_{n}(t))\leq\Phi(u_{0})\quad\forall
t\in(-T,T),\qquad\Phi(u_{n}(t))\text{ non-increasing}.$ (4.3)

From this and Propositions 2.2 and 2.6, we deduce the following uniform
bounds, for any $T>0$,

$\begin{split}&\|u_{n}\|_{L^{\infty}(-T,T;L^{\gamma+1}(\Omega))}+\|u_{n}\|_{L^{2}(-T,T;H^{1}(\Omega))}+\|u_{n}\|_{L^{2\gamma+1}(\Omega\times(-T,T))}\\\
&+\|\partial_{t}u_{n}\|_{L^{q^{\prime}}(-T,T;(W^{1,q}(\Omega))^{\prime})}+\|\partial_{t}u_{n}\|_{L^{2}(-T,T;(W^{1,r}(\Omega))^{\prime})}+\|\nabla
u_{n}^{\frac{\gamma+1}{2}}\|_{L^{2}(-T,T;L^{2}(\Omega))}\leq C(T),\end{split}$
(4.4)

with $q$ and $r$ as in Proposition 2.6 which implies, by standard arguments,
the following convergences (up to subsequences) as $n\to\infty$ to the same
function $u_{\ast}\geq 0$

$\displaystyle u_{n}\rightharpoonup u_{\ast}\quad\text{in
}L^{2}(-T,T;H^{1}(\Omega))\quad\text{and }L^{2\gamma+1}(\Omega\times(-T,T))$
$\displaystyle\partial_{t}u_{n}\rightharpoonup\partial_{t}u_{\ast}\quad\text{in
}L^{q^{\prime}}(-T,T;(W^{1,q}(\Omega))^{\prime})\quad\text{and in
}L^{2}(-T,T;(W^{1,r}(\Omega))^{\prime}),$ $\displaystyle u_{n}\to
u_{\ast}\quad\text{in }L^{p}(\Omega\times(0,T)),\quad
p\in[1,2\gamma+1),\qquad\text{and almost everywhere},$ $\displaystyle\nabla
u_{n}^{\frac{\gamma+1}{2}}\rightharpoonup\nabla
u_{\ast}^{\frac{\gamma+1}{2}}\quad\text{in }L^{2}(\Omega\times(0,T)).$

We want to characterize $u_{\ast}$. Notice that from (2.2), we have for all
$T>0$,

$\displaystyle\int_{-T}^{T}\int_{\Omega}\int_{\Omega}$
$\displaystyle\omega_{\varepsilon}(y)|\nabla u_{n}(x)-\nabla
u_{n}(x-y)|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}y\mathop{}\\!\mathrm{d}s$
$\displaystyle=\int_{t_{n}-T}^{t_{n}+T}\int_{\Omega}\int_{\Omega}\omega_{\varepsilon}(y)|\nabla
u(x)-\nabla
u(x-y)|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}y\mathop{}\\!\mathrm{d}s\to
0\qquad\text{as }n\to\infty$

by integrability on $(0,\infty)$. By weak-lower semicontinuity, in the limit

$\int_{-T}^{T}\int_{\Omega}\int_{\Omega}\omega_{\varepsilon}(y)|\nabla
u_{\ast}(x)-\nabla
u_{\ast}(x-y)|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}y\mathop{}\\!\mathrm{d}s=0$

so that $u_{\ast}(t)$ is constant in space for a.e. $t\in[0,T]$. In fact,
$u_{\ast}(t)$ is constant in space for all $t\in[0,T]$ because $u_{\ast}\in
C_{weak}([-T,T];L^{\gamma+1}(\Omega))$ (see Remark 1.4) so that for all
$\varphi\in C_{c}^{\infty}(\Omega)$, the function
$t\mapsto\int_{\Omega}u_{\ast}(t,x)\,\operatorname{div}\varphi(x)\mathop{}\\!\mathrm{d}x$
is continuous. Similarly, from (2.2) and the Fatou lemma,

$u_{\ast}\log\left(\frac{u_{\ast}}{p_{H}^{\frac{1}{\gamma}}}\right)(u_{\ast}^{\gamma}-p_{H})=0,\text{
for a.e. }t\in(-T,T),$

so that either $u_{\ast}(t)=0$ or $u_{\ast}(t)=p_{H}^{\frac{1}{\gamma}}$.
Since $u_{\ast}\in C_{weak}([-T,T];L^{\gamma+1}(\Omega))$, the average
$\overline{u}_{\ast}(\cdot)\in C([0,T])$ so that
$u_{\ast}\equiv\overline{u}_{\ast}$ can attain only one of the values for all
times.

Because $\Phi(u(t))$ is non-increasing in time, it has a limit as
$t\to\infty$, say $\Phi_{\ast}$. Then clearly
$\Phi_{\ast}=\lim_{n\to\infty}\Phi(u(t+t_{n}))=\Phi(u_{\ast})$, and thus
either $\Phi_{\ast}=\Phi(0)=|\Omega|$ if $u_{\ast}=0$ or
$\Phi_{\ast}=\Phi(p_{H}^{\frac{1}{\gamma}})=0$ if
$u_{\ast}=p_{H}^{\frac{1}{\gamma}}$. Clearly this also implies that, given
another sequence of times $\\{t_{m}\\}_{m}$, we can repeat the same argument
and extract a (non relabeled) subsequence $\\{u(t_{m})\\}_{m}$ converging to
the same constant $u_{\ast}$. This concludes the proof of Lemma 4.1.

Step 2. Stability of the equilibria. We complete Lemma 4.1 with the following

######  Lemma 4.2.

Under notation of Lemma 4.1, if $u_{*}\equiv 0$ then $u_{0}=0$ almost
everywhere in $\Omega$.

###### Proof.

Since $\Phi(u(t))$ is nonincreasing in time and, by Lemma 4.1,
$\Phi(u(t))\to|\Omega|$, we obtain

$\displaystyle|\Omega|=\lim_{t\to\infty}\Phi(u(t))\leq\Phi(u_{0})\leq|\Omega|.$

Thus we infer $\Phi(u_{0})=|\Omega|$. Being the entropy function
$g:x\mapsto\frac{x}{p_{H}^{\frac{1}{\gamma}}}\log\left(\frac{x}{p_{H}^{\frac{1}{\gamma}}}\right)-\frac{x}{p_{H}^{\frac{1}{\gamma}}}+1$
decreasing for $x\in[0,p_{H}^{\frac{1}{\gamma}})$, and since $u_{0}\leq
p_{H}^{\frac{1}{\gamma}}$, it follows that $u_{0}=0$ almost everywhere in
$\Omega$. ∎

Step 3. Existence of the $L^{q}(\Omega)$-limit as $t\to\infty$.

######  Lemma 4.3.

Assume that $u_{0}\not\equiv 0$. Then it holds

$\lim_{t\to\infty}u(t)=p_{H}^{\frac{1}{\gamma}}\quad\text{ in
}L^{q}(\Omega)\qquad\forall q\in[1,\gamma+1).$

###### Proof.

First note that, the pointwise values $u(t)$ as an
$L^{\gamma+1}(\Omega)$-function makes sense thanks to the weak continuity
obtained in Lemma 1.3. Now, we consider the decomposition of $\Phi$:

$\displaystyle\Phi(u)=\dfrac{1}{p_{H}^{\frac{1}{\gamma}}}\left(\int_{\Omega}u\log\left(\dfrac{u}{\overline{u}}\right)\mathop{}\\!\mathrm{d}x+\int_{\Omega}u\log\left(\dfrac{\overline{u}}{p_{H}^{\frac{1}{\gamma}}}\right)\mathop{}\\!\mathrm{d}x+\int_{\Omega}(p_{H}^{\frac{1}{\gamma}}-u)\mathop{}\\!\mathrm{d}x\right)$
(4.5)

and we study the limits of terms appearing in (4.5). Since $u_{0}\not\equiv
0$, from Lemma 4.1 we obtain
$\lim_{t\to\infty}\Phi(u(t))=\Phi(p_{H}^{1/\gamma})=0$.

By Proposition 2.1, for any sequence $\\{t_{n}\\}_{n}$ there exists a
(nonrelabeled) subsequence such that, for some $\kappa$,

$\overline{u}(t_{n})\to\kappa\in[0,p_{H}^{\frac{1}{\gamma}}].$ (4.6)

We prove that $\kappa=p_{H}^{\frac{1}{\gamma}}$. Indeed, the function
$g:x\mapsto\frac{x}{p_{H}^{\frac{1}{\gamma}}}\log\left(\frac{x}{p_{H}^{\frac{1}{\gamma}}}\right)-\frac{x}{p_{H}^{\frac{1}{\gamma}}}+1$
is convex and continuous. As
$\Phi(u(t))=\int_{\Omega}g(u(t))\mathop{}\\!\mathrm{d}x$, by Jensen’s
inequality we get

$0\leq g(\overline{u}(t))\leq\frac{1}{|\Omega|}\Phi(u(t))\to 0\text{ as
}t\to\infty$

so that $g(\kappa)=0$ and the claim follows. It follows that
$\overline{u}(t)\to p_{H}^{\frac{1}{\gamma}}$ as $t\to\infty$.

Hence, passing to the limit in (4.5)

$\displaystyle\int_{\Omega}u(t)\log\left(\dfrac{u(t)}{\overline{u}(t)}\right)\mathop{}\\!\mathrm{d}x\to
0\text{ when }k\to\infty.$ (4.7)

From Lemma A.4 and (4.7), together with the fact that $\overline{u}(t)\to
p_{H}^{\frac{1}{\gamma}}$ as $n\to\infty$, we then deduce

$\|u(t)-p_{H}^{1/\gamma}\|_{L^{1}(\Omega)}\to 0\quad\text{when $k\to\infty$}.$

Furthermore, by the bound in $L^{\infty}_{t}L^{\gamma+1}_{x}$ given by the
control of the energy $\mathcal{E}$ in Proposition 2.2, we can deduce the
convergence (1.18) by interpolation. The proof of Theorem 1.6 in the case of a
nonzero source term $G$ is thus concluded. ∎

(a) Initial condition

(b) Evolution at $t=0.03$

(c) Evolution at $t=0.14$

(d) Evolution at $t=0.25$

Figure 1: We provide here 1D numerical simulations illustrating this
theoretical result. We assume $\gamma=10$ and $p_{H}=0.7$. We thus have
$p_{H}^{1/\gamma}\approx 0,96$. The initial condition is taken as a double
gaussian as in Figure (a).

###  4.2 Case $G(p)=0$

When there is no source term, the solution converges to the mean value. The
argument is a simple consequence of the logarithmic Sobolev inequality and the
Csiszár–Kullback–Pinsker inequality. We refer to [43] for other systems where
it is applied. We consider the relative entropy between the solution $u$ and a
stationary state $u_{\ast}$, defined as

$\Phi(u|u_{\ast})=\int_{\Omega}\left(u\log\left(\frac{u}{u_{\ast}}\right)-u+u_{\ast}\right)\mathop{}\\!\mathrm{d}x.$

In our case $u_{\ast}=\overline{u}_{0}$ a.e. in $\Omega$. Notice that, by the
conservation of mass, we have

$\Phi(u|\overline{u})=\int_{\Omega}u\log\left(\frac{u}{\overline{u}}\right)\mathop{}\\!\mathrm{d}x.$

Since $\overline{u}(\cdot)\equiv\overline{u}_{0}$, we see that
$\Phi(u|\overline{u})$ satisfies the identity (or at least the inequality for
weak solutions): for almost any $t\geq 0$,

$\frac{d\Phi(u|\overline{u})}{dt}+\frac{1}{2\varepsilon^{2}}\int_{\Omega}\int_{\Omega}\omega_{\varepsilon}(y)|\nabla
u(x)-\nabla
u(x-y)|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}y+\int_{\Omega}\dfrac{4\gamma}{(\gamma+1)^{2}}\left|\nabla|u|^{\frac{\gamma+1}{2}}\right|^{2}\mathop{}\\!\mathrm{d}x=0.$

The generalized logarithmic Sobolev inequality [43, Section 3.1] provides us
the exponential decay of the relative entropy. Indeed we have

######  Lemma 4.4.

For any $m\geq(d-2)_{+}/d$, there exists $C(\Omega,m)$ such that

$\int_{\Omega}|\nabla(u^{\frac{m}{2}})|^{2}\mathop{}\\!\mathrm{d}x\geq
C(\Omega,m)\overline{u}^{m-1}\int_{\Omega}u\log\left(\frac{u}{\overline{u}}\right)\mathop{}\\!\mathrm{d}x.$

In our case $m=\gamma+1$ and $\overline{u}\equiv\overline{u}_{0}$ is bounded
from below by a positive constant since we consider an initial condition
$u_{0}\not\equiv 0$. Therefore, by the Gronwall Lemma we conclude that the
entropy $\Phi$ experiences an exponential decay as $t\to\infty$:

$\frac{d\Phi(u|\overline{u})}{dt}+C(\Omega,\gamma)\overline{u}_{0}^{\gamma}\Phi(u|\overline{u})\leq
0$

Therefore, we have, for some
$C=C(\Omega,\gamma,\Phi(u_{0}),\overline{u}_{0})$,

$\Phi(u|\overline{u})\leq Ce^{-Ct},\quad\forall t\geq 0.$

To prove that this implies the exponential decay of the solution we use the
Csiszár–Kullback–Pinsker inequality of Lemma A.4: there exists $C(\Omega)$
such that

$\Phi(u|\overline{u})\geq
C(\Omega)\|u-\overline{u}_{0}\|^{2}_{L^{1}(\Omega)}.$

The exponential decay in $L^{1}(\Omega)$ of $u$ towards $\overline{u}_{0}$
then easily follows. By the $L^{\infty}(0,T;L^{\gamma+1}(\Omega))$ bound given
by the control of the energy $\mathcal{E}$ (which is the same as in the case
with a source term $G$ given in Proposition 2.2), we can in conclusion deduce
the exponential convergence (1.19) by interpolation. This ends the proof of
Theorem 1.6.

###  4.3 Longtime behavior of the local Cahn-Hilliard equation

The nonlocal Cahn-Hilliard equation can be also seen as an approximation of
the local Cahn-Hilliard equation:

$\begin{cases}\partial_{t}u-\text{div}(u\nabla\mu)=uG(p)\quad\text{in
}\Omega\times(0,T),\\\ \mu=p-\Delta u,\quad p=u^{\gamma},\\\
u(0)=u_{0}\quad\text{ in }\Omega.\end{cases}$ (4.8)

This follows, at least formally, with a Taylor expansion, using the symmetry
of the kernel $\omega_{\varepsilon}$ in the operator $B_{\varepsilon}$ of
(1.1), and for a rigorous proof we refer, e.g., to [23]. Therefore, one may
wonder whether the previous results obtained for the nonlocal Cahn-Hilliard
equation also hold for the local one. It turns out that for the convergence to
the stationary states, the result is the same. Indeed, one mainly uses
arguments based on the entropy, so that the nonlocal term does not play a
role: concerning the entropy $\Phi$, we can consider again (1.6) and formally
get for (4.8)

$\displaystyle\frac{d}{dt}\Phi(u)+\frac{1}{p_{H}^{\frac{1}{\gamma}}}\int_{\Omega}|\Delta
u|^{2}\mathop{}\\!\mathrm{d}x+\frac{1}{p_{H}^{\frac{1}{\gamma}}}\int_{\Omega}\dfrac{4\gamma}{(\gamma+1)^{2}}\left|\nabla|u|^{\frac{\gamma+1}{2}}\right|^{2}\mathop{}\\!\mathrm{d}x-\int_{\Omega}u\log\left(\frac{u}{p_{H}^{\frac{1}{\gamma}}}\right)G(p)\mathop{}\\!\mathrm{d}x=0,$
(4.9)

which is very similar to the result in Proposition 2.2. Concerning the energy,
we set

$\mathcal{E}_{l}(u):=\frac{1}{2}\int_{\Omega}|\nabla
u|^{2}\mathop{}\\!\mathrm{d}x+\int_{\Omega}\dfrac{u^{1+\gamma}}{1+\gamma}\mathop{}\\!\mathrm{d}x$

and thus

$\displaystyle\frac{d}{dt}\mathcal{E}_{l}(u)+\int_{\Omega}u|\nabla\mu|^{2}\mathop{}\\!\mathrm{d}x=\int_{\Omega}uG(p)\mu\mathop{}\\!\mathrm{d}x.$

Now we observe that, integrating by parts,

$\displaystyle\int_{\Omega}uG(p)\mu\mathop{}\\!\mathrm{d}x$
$\displaystyle=\int_{\Omega}u(p_{H}-p)p\mathop{}\\!\mathrm{d}x-\int_{\Omega}u(p_{H}-p)\Delta
u\mathop{}\\!\mathrm{d}x$
$\displaystyle=\int_{\Omega}u(p_{H}-p)p\mathop{}\\!\mathrm{d}x+\int_{\Omega}|\nabla
u|^{2}(p_{H}-p)\mathop{}\\!\mathrm{d}x-\int_{\Omega}u\nabla p\cdot\nabla
u\mathop{}\\!\mathrm{d}x$
$\displaystyle=\int_{\Omega}u(p_{H}-p)p\mathop{}\\!\mathrm{d}x+\int_{\Omega}|\nabla
u|^{2}(p_{H}-p)\mathop{}\\!\mathrm{d}x-\gamma\int_{\Omega}p|\nabla
u|^{2}\mathop{}\\!\mathrm{d}x.$

Moreover, notice that, since the relation (1.11) still holds with the same
proof,

$p_{H}\int_{\Omega}up\mathop{}\\!\mathrm{d}x\leq\frac{1}{2}\int_{\Omega}up^{2}\mathop{}\\!\mathrm{d}x+\frac{p_{H}^{2}}{2}\int_{\Omega}u\mathop{}\\!\mathrm{d}x\leq\frac{1}{2}\int_{\Omega}up^{2}\mathop{}\\!\mathrm{d}x+C.$

Then we can rewrite the energy inequality as

$\displaystyle\frac{d}{dt}\mathcal{E}_{l}(u)+\int_{\Omega}u|\nabla\mu|^{2}\mathop{}\\!\mathrm{d}x+\frac{1}{2}\int_{\Omega}p^{2}u\mathop{}\\!\mathrm{d}x+\int_{\Omega}|\nabla
u|^{2}p\mathop{}\\!\mathrm{d}x+\gamma\int_{\Omega}p|\nabla
u|^{2}\mathop{}\\!\mathrm{d}x$ $\displaystyle\leq\int_{\Omega}|\nabla
u|^{2}\mathop{}\\!\mathrm{d}x+C\leq C(\mathcal{E}_{l}(u)+1),$

which is again very similar to the one obtained in Proposition 2.2 for the
nonlocal case. Therefore, with these estimates we can basically perform again
all the arguments of Section 4 and obtain again the same result as in Theorem
1.6. Note that in the local case, differently from the nonlocal one, we can
also repeat the same arguments in the case of a smooth bounded domain
$\Omega\subset\mathbb{R}^{d}$ with homogeneous Neumann boundary conditions
$u\nabla\mu\cdot\textbf{n}=0$ and $\nabla u\cdot\textbf{n}=0$ on
$\partial\Omega\times(0,\infty)$, where n is the outward unit normal.

######  Remark 4.5.

The incompressible limit, $\gamma\to\infty$, is very different and remains an
open question in the local Cahn-Hilliard case. Indeed, obtaining an equation
for the pressure $p$ from which to deduce a uniform-in-$\gamma$
$L^{2}(0,T;H^{1}(\Omega))$-control on $p$ seems still out of reach. Therefore
no analogous of Theorem 1.5 can be stated in this local case.

## Acknowledgements

J.S. was supported by the National Science Center grant 2017/26/M/ST1/00783.
A.P. has been partially funded by MIUR-PRIN research grant n. 2020F3NCPX and
is also member of Gruppo Nazionale per l’Analisi Matematica, la Probabilità e
le loro Applicazioni (GNAMPA), Istituto Nazionale di Alta Matematica (INdAM).

##  Appendix A Technical tools

Several tools have been used to carry out some proofs. First, we present a
lemma about geometric convergence of numerical sequences, whose proof can be
easily obtained by induction (see, e.g., [36, Ch.2, Lemma 5.6] ):

######  Lemma A.1.

Let $\\{y_{n}\\}_{n\in\mathbb{N}\cup\\{0\\}}\subset\mathbb{R}^{+}$ satisfy the
recursive inequality

$\displaystyle y_{n+1}\leq Cb^{n}y_{n}^{1+\epsilon},\quad\forall n\geq
0,\qquad\text{and}\qquad
y_{0}\leq\theta:=C^{-\frac{1}{\epsilon}}b^{-\frac{1}{\epsilon^{2}}},$ (A.1)

for some $C>0$, $b>1$ and $\epsilon>0$. Then, $y_{n}\to 0$ for $n\to\infty$
with geometric rate

$\displaystyle y_{n}\leq\theta b^{-\frac{n}{\epsilon}},\qquad\forall n\geq 0.$
(A.2)

Next, we state a theorem concerning the absolute continuity of some integrals
of convex functions in $\mathbb{R}$, whose proof can be found, e.g. in [32,
p.101]:

######  Theorem A.2.

Let $T>0$ and let $h:\mathbb{R}\to\mathbb{R}$ be a convex and lower
semicontinuous function. Assume that

  * •

$f\in L^{2}(0,T;H^{1}(\Omega))$ and $\partial_{t}f\in
L^{2}(0,T;(H^{1}(\Omega))^{\prime})$,

  * •

$g(x,t)\in\partial h(x,t)$ for almost every $(x,t)\in\Omega\times(0,T)$,

  * •

$g\in L^{2}(0,T;H^{1}(\Omega))$.

Then, the function $t\mapsto\int_{\Omega}h(f(x,t))$ is absolutely continuous
on $[0,T]$ and,

$\frac{d}{dt}\int_{\Omega}h(f)\mathop{}\\!\mathrm{d}x=\langle\partial_{t}f,g\rangle\quad\text{for
almost any }t\in(0,T).$

We then propose a control on the $H^{1}(\Omega)$-norm related to the use of
$\omega_{\varepsilon}$.

######  Lemma A.3.

There exists $\varepsilon_{0}>0$ and a constant $C$ such that for
$\varepsilon\in(0,\varepsilon_{0})$ and all $f\in L^{2}(\Omega)$ we have

$\displaystyle\|f-\overline{f}\|_{L^{2}(\Omega)}^{2}\leq\frac{C}{2\varepsilon^{2}}\int_{\Omega}\int_{\Omega}\omega_{\varepsilon}(y)|f(x)-f(x-y)|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}y,$
(A.3)

where $\overline{f}$ is the average of $f$ over $\Omega$. Similarly, for all
$\alpha$, there exists $\varepsilon_{0}(\alpha)>0$ and constant $C(\alpha)$
such that for all $\varepsilon\in(0,\varepsilon_{0})$ and all $f\in
H^{1}(\Omega)$ we have

$\displaystyle\|f\|_{H^{1}(\Omega)}^{2}\leq\frac{\alpha}{2\varepsilon^{2}}\int_{\Omega}\int_{\Omega}\omega_{\varepsilon}(y)|\nabla
f(x)-\nabla
f(x-y)|^{2}\mathop{}\\!\mathrm{d}x\mathop{}\\!\mathrm{d}y+C(\alpha)\|f\|_{L^{1}(\Omega)}^{2}.$
(A.4)

###### Proof.

The proof is identical to the one in [23, Lemma C.3] by substituting the norm
$\|\leavevmode\nobreak\ \cdot\leavevmode\nobreak\ \|_{L^{2}(\mathbb{T}^{d})}$
with the norm $\|\cdot\|_{L^{1}(\mathbb{T}^{d})}$. Indeed, with the notation
of the proof of that Lemma, also $n\|g_{n}\|_{L^{1}(\mathbb{T}^{d})}<1$
implies that the limit function $g=0$, exactly as in the case
$L^{2}(\mathbb{T}^{d})$. ∎

In conclusion, we recall the Csiszár–Kullback–Pinsker inequality (see, e.g.,
[6]), which is essential to study the asymptotic behavior of weak solutions

######  Lemma A.4.

For any non-negative $u\in L^{1}(\Omega)$

$4|\Omega|\overline{u}\int_{\Omega}u\log\left(\dfrac{u}{\overline{u}}\right)\mathop{}\\!\mathrm{d}x\geq\|u-\overline{u}\|^{2}_{L^{1}(\Omega)}.$

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