THE DAMPED FOCUSING CUBIC WAVE EQUATION ON A BOUNDED DOMAIN

Thomas Perrin

Abstract. For the focusing cubic wave equation on a compact Riemannian
manifold of dimension $3$, the dichotomy between global existence and blow-up
for solutions starting below the energy of the ground state is known since the
work of Payne and Sattinger. In the case of a damped equation, we prove that
the dichotomy between global existence and blow-up still holds. In particular,
the damping does not prevent blow-up. Assuming that the damping satisfies the
geometric control condition, we then prove that any global bounded solution
converges to a stationary solution along a time sequence, and that global
solutions below the energy of the ground state can be stabilised, adapting the
proof of a similar result in the defocusing case.

## Introduction

Let $M$ be a compact Riemannian manifold of dimension $3$, with or without
boundary, and set $\Omega=M\backslash\partial M$, where $\partial M$ is the
boundary of $M$. Let $\beta\in\mathbb{R}$ be such that the Poincaré inequality

$\int_{\Omega}\left(\left|\nabla
u\right|^{2}+\beta|u|^{2}\right)\mathrm{d}x\gtrsim\int_{\Omega}|u|^{2}\mathrm{d}x$

is satisfied, for all $u\in H_{0}^{1}(\Omega)$. This specifically requires
$\beta>0$ if $\partial\Omega$ is empty. For real-valued initial data
$\left(u^{0},u^{1}\right)\in H_{0}^{1}(\Omega)\times L^{2}(\Omega)$, consider
the equation

$\left\\{\begin{array}[]{rcccl}\square u+\beta u&=&u^{3}&&\text{in
}\mathbb{R}\times\Omega,\\\
(u(0),\partial_{t}u(0))&=&\left(u^{0},u^{1}\right)&&\text{in }\Omega,\\\
u&=&0&&\text{on }\mathbb{R}\times\partial\Omega.\end{array}\right.$

If $\beta=0$, then (Introduction) is the focusing cubic wave equation, and if
$\beta>0$, it is the focusing cubic Klein-Gordon equation. For
$\left(u^{0},u^{1}\right)\in H_{0}^{1}(\Omega)\times L^{2}(\Omega)$, write
$\left\|u^{0}\right\|_{H_{0}^{1}}^{2}=\int_{\Omega}\left(\left|\nabla
u^{0}\right|^{2}+\beta|u^{0}|^{2}\right)\mathrm{d}x$ and set

$J(u^{0})=\frac{1}{2}\left\|u^{0}\right\|_{H_{0}^{1}}^{2}-\frac{1}{4}\left\|u^{0}\right\|_{L^{4}}^{4}\quad\text{
and }\quad
E\left(u^{0},u^{1}\right)=J(u^{0})+\frac{1}{2}\left\|u^{1}\right\|_{L^{2}}^{2}.$

The functionals $J$ and $E$ are respectively called _static energy_ and
_energy_. The energy of a solution of (Introduction) at time $t\in\mathbb{R}$
is defined by $E\left(u(t),\partial_{t}u(t)\right)$, and is conserved. Let $Q$
be the ground state of (Introduction), that is, the unique positive stationary
solution of (Introduction) of minimal energy

$E(Q,0)=J(Q)=d>0.$

For $u^{0}\in H_{0}^{1}(\Omega)$, write

$K(u^{0})=\left\|u^{0}\right\|_{H_{0}^{1}}^{2}-\left\|u^{0}\right\|_{L^{4}}^{4},$
(1)

and set

$\mathscr{K}^{+}=\left\\{\left(u^{0},u^{1}\right)\in H_{0}^{1}(\Omega)\times
L^{2}(\Omega),E\left(u^{0},u^{1}\right)<d,K(u^{0})\geq 0\right\\},$

and

$\mathscr{K}^{-}=\left\\{\left(u^{0},u^{1}\right)\in H_{0}^{1}(\Omega)\times
L^{2}(\Omega),E\left(u^{0},u^{1}\right)<d,K(u^{0})<0\right\\}.$

The following result is due to Payne and Sattinger [14].

######  Theorem 1.

The spaces $\mathscr{K}^{+}$ and $\mathscr{K}^{-}$ are stable under the flow
of (Introduction). A solution starting from $\mathscr{K}^{+}$ is defined on
$\mathbb{R}$, and a solution starting from $\mathscr{K}^{-}$ blows up in
finite positive and negative times.

For $\gamma\in L^{\infty}(\Omega,\mathbb{R}_{+})$, our main focus is the
damped equation

$\left\\{\begin{array}[]{rcccl}\square u+\gamma\partial_{t}u+\beta
u&=&u^{3}&&\text{in }\mathbb{R}_{+}\times\Omega,\\\
(u(0),\partial_{t}u(0))&=&\left(u^{0},u^{1}\right)&&\text{in }\Omega,\\\
u&=&0&&\text{on }\mathbb{R}_{+}\times\partial\Omega,\end{array}\right.$

with $\left(u^{0},u^{1}\right)\in H_{0}^{1}(\Omega)\times L^{2}(\Omega)$. For
$t\geq 0$, the energy equality is

$E\left(u(t),\partial_{t}u(t)\right)=E\left(u^{0},u^{1}\right)-\int_{0}^{t}\int_{\Omega}\gamma(x)\left|\partial_{t}u(s,x)\right|^{2}\mathrm{d}x\mathrm{d}s.$

In particular, the energy of a solution is nonincreasing. Our first result is
the following theorem.

######  Theorem 2.

Fix $\gamma\in L^{\infty}(\Omega,\mathbb{R}_{+})$. The spaces
$\mathscr{K}^{+}$ and $\mathscr{K}^{-}$ are stable under the forward flow of
(Introduction). A solution of (Introduction) initiated in $\mathscr{K}^{+}$ is
defined on $\mathbb{R}_{+}$, and a solution of (Introduction) initiated in
$\mathscr{K}^{-}$ blows up in finite time $t>0$.

In particular, below the energy of the ground state, a blow-up solution of
(Introduction) cannot be stabilised by the addition of a damping term of the
form $\gamma\partial_{t}u$.

In the defocusing case, that is, with the non-linearity $u^{3}$ changed into
$-u^{3}$, the energy of a solution is positive and governs its size. All
corresponding solutions are defined on $\mathbb{R}$, and the only stationary
solution is zero. In this case, Joly and Laurent [11] proved a stabilization
result under the geometric control condition of Bardos-Lebeau-Rauch [3], on
bounded or unbounded domains. In [10], the same authors consider the case of a
non-linearity that is asymptotically defocusing and establish a semi-global
controllability result. For this type of non-linearity, non-zero stationary
solutions exist, but no blow-up phenomenon occurs.

In the focusing case we consider here, few stabilization or controllability
results exist in the literature. For a focusing equation on the entire space,
or outside a star-shaped obstacle, Alaoui, Ibrahim, and Nakanishi [2]
established stabilization for solutions starting from $\mathcal{K}^{+}$ when
the damping is greater than a positive constant outside a ball. Note that
assuming that the obstacle is star-shaped is stronger than the geometric
control condition. Burq, Raugel, and Schlag [5] proved a dichotomy property
for radial solutions of the focusing damped Klein-Gordon equation on the
entire space with a constant positive damping: such a solution either blows up
in a finite positive time or exists globally in positive time and converges to
a stationary solution. In [4], the same authors established a similar
dichotomy property in the non-radial case, with a damping independent of $x$
and converging to zero as $t$ tends to infinity. In the case of the cubic
focusing radial Klein-Gordon equation on a closed ball of $\mathbb{R}^{3}$,
Krieger and Xiang [12] studied the stability of the system near the static
solution $u=1$, under different dissipative boundary conditions.

We will use the notion of generalized geodesic, for which we refer to [13].

######  Definition 3.

For $\omega\subset\Omega$, we say that $\omega$ satisfies the _geometric
control condition_ (in short, GCC) if there exists $L>0$ such that any
generalized geodesic of $\Omega$ of length $L$ meets the set $\omega$.

Our second result is the following theorem, which can be seen as an extension
of the stabilisation property proved in [11] to the focusing case.

######  Theorem 4.

Suppose $\gamma(x)\geq\alpha$ for almost all $x\in\omega$, with $\alpha>0$ a
constant and $\omega\subset\Omega$ an open set fulfilling the GCC. Then for
any $E_{0}\in[0,d)$, there exist $C>0$ and $\lambda>0$ such that for all
$\left(u^{0},u^{1}\right)\in\mathscr{K}^{+}$, the solution $u$ of
(Introduction) satisfies

$\left\|u(t)\right\|_{H_{0}^{1}}+\left\|\partial_{t}u(t)\right\|_{L^{2}}\leq
Ce^{-\lambda t},\quad t\geq 0,$

if $E\left(u^{0},u^{1}\right)\leq E_{0}$.

For a solution starting from $\mathscr{K}^{+}$, we will see that the energy
can be used almost as in the defocusing case, allowing us to adapt most of the
arguments of [11]. Note that $C$ and $\lambda$ depend on $E_{0}$ (see Remark
29).

The proof of Theorem 4 will relies on an asymptotic compactness result, which
roughly states that for all uniformly bounded sequence of global solutions of
(Introduction), one can find a subsequence which converges to a stationary
solution of (Introduction) along a time-sequence ; see Proposition 22 for a
precise statement. It implies the following theorem.

######  Theorem 5.

Fix $\gamma\in L^{\infty}(\Omega,\mathbb{R}_{+})$ satisfying assumption of
Theorem 4. Let $\left(u^{0},u^{1}\right)\in H_{0}^{1}(\Omega)\times
L^{2}(\Omega)$ be such that the associated solution $u$ of (Introduction)
exists on $\mathbb{R}_{+}$. Assume that there exists $C>0$ such that

$\left\|u(t)\right\|_{H_{0}^{1}}+\left\|\partial_{t}u(t)\right\|_{L^{2}}\leq
C$

for all $t\geq 0$. Then, we have the following properties.

  1. 1.

For all sequence $(T_{n})_{n\in\mathbb{N}}$ such that
$T_{n}\rightarrow+\infty$, there exist an increasing function
$\phi:\mathbb{N}\rightarrow\mathbb{N}$ and a stationary solution $w$ of
(Introduction) such that

$\left(u(T_{\phi(n)}+\cdot),\partial_{t}u(T_{\phi(n)}+\cdot)\right)\underset{n\rightarrow+\infty}{\longrightarrow}(w,0)$

in $L^{\infty}_{\mathrm{loc}}(\mathbb{R},H_{0}^{1}(\Omega)\times
L^{2}(\Omega))$.

  2. 2.

Assume that there exists an at most countable number of stationary solution
$w$ of (Introduction) such that

$J(w)=\liminf_{t\rightarrow+\infty}E\left(u(t),\partial_{t}u(t)\right).$ (1)

Then there exists a stationary solution $w$ of (Introduction) such that

$\left(u(t+\cdot),\partial_{t}u(t+\cdot)\right)\underset{t\rightarrow+\infty}{\longrightarrow}(w,0)$

in $L^{\infty}_{\mathrm{loc}}(\mathbb{R},H_{0}^{1}(\Omega)\times
L^{2}(\Omega))$.

######  Remark 6.

For example, if $\liminf E\left(u(t),\partial_{t}u(t)\right)\leq J(Q)$ then
there are at most 3 stationary solutions satisfying (1), which are $0$, $Q$
and $-Q$.

On the one hand, observe that Theorem 5 applies to any bounded solution, while
Theorem 4 only applies to solutions with energy below the energy of the ground
state. On the other hand, Theorem 4 provides an exponential convergence rate,
whereas Theorem 5 provides no convergence rate.

In [5], the authors proved that any solution defined on $\mathbb{R}_{+}$
converges to an asymptotic solution, when considering $\Omega=\mathbb{R}^{d}$.
In their case, the boundedness of the solution arises as a consequence of
their theorem rather than being an assumed condition, and they make no
assumption regarding the number of stationary solutions at a fixed energy
level. On the other hand, their damping is a positive constant, whereas
Theorem 5 holds true for a damping satisfying the geometric control condition.

######  Remark 7.

The choice of considering only a cubic non-linearity on a manifold of
dimension 3 has been made for simplicity, but the arguments of the present
article could be adapted to a more general focusing subcritical non-linearity.

The outline of this article is the following. In Section 1, we recall some
basic facts about (Introduction) and the stabilisation of the linear version
of (Introduction), and we prove some technical results related to the ground
state $Q$. In Section 2, we prove Theorem 2. In Section 3, we establish the
asymptotic compactness result (Proposition 22), and we show that it implies
Theorem 5. In Section 4, we first prove an easy stabilisation property in the
case of a positive damping, and then we prove Theorem 4, by adapting the proof
of [11].

#### Acknowledgements.

I warmly thank Thomas Duyckaerts and Jérôme Le Rousseau for their constant
support and guidance.

Keywords: damped wave equations, nonlinear wave equation, focusing
nonlinearity, exponential stabilization, blow-up, convergence to equilibrium.
MSC2020: 35B40, 35B44, 35L71, 93D20.

##  1 Preliminaries

###  1.1 The Cauchy problem and some properties of the linear equation

In this article, we only consider real-valued solutions of wave equations. For
$s\in[0,1]$, let $H_{0}^{s}(\Omega)$ denote the complex interpolation space
between $L^{2}(\Omega)$ and $H_{0}^{1}(\Omega)$. Define

$X^{s}=\left(H^{1+s}(\Omega)\cap H_{0}^{1}(\Omega)\right)\times
H_{0}^{s}(\Omega).$ (2)

For example, one has $X^{0}=H_{0}^{1}(\Omega)\times L^{2}(\Omega)$ and
$X^{1}=\left(H^{2}(\Omega)\cap H_{0}^{1}(\Omega)\right)\times
H_{0}^{1}(\Omega)$. For $s\geq 0$, we write $D\left(\Delta^{s}\right)$ for the
iterated domain of the Dirichlet Laplacian (defined by interpolation for
$s\notin\mathbb{N}$). Then for $s\in[0,1]$, we have

$X^{s}=D\left(\Delta^{\frac{1+s}{2}}\right)\times
D\left(\Delta^{\frac{s}{2}}\right)$ (3)

so the norm of $X^{s}$ can be represented by a Fourier series. Consider the
unbounded operator $A:X^{0}\rightarrow X^{0}$ with domain $D(A)=X^{1}$ defined
by

$A=\left(\begin{array}[]{cc}0&\mathrm{Id}\\\
\Delta-\beta&-\gamma\end{array}\right).$

Write $e^{tA}$ for the associated semi-group. The following result about the
Cauchy problem corresponding to the linear version of (Introduction) is well-
known.

######  Theorem 8.

For $s\in[0,1]$, $\left(u^{0},u^{1}\right)\in X^{s}$, $\gamma\in
L^{\infty}(\Omega,\mathbb{R}_{+})$, and $g\in
L_{\mathrm{loc}}^{1}(\mathbb{R},H_{0}^{s}(\Omega))$, there exists a unique
solution

$u\in\mathscr{C}^{0}(\mathbb{R},H^{1+s}(\Omega)\cap H_{0}^{1}(\Omega))\ \cap\
\mathscr{C}^{1}(\mathbb{R},H_{0}^{s}(\Omega))$

of the linear damped wave equation

$\left\\{\begin{array}[]{rcccl}\square u+\gamma\partial_{t}u+\beta
u&=&g&&\text{in }\mathbb{R}\times\Omega,\\\
(u(0),\partial_{t}u(0))&=&\left(u^{0},u^{1}\right)&&\text{in }\Omega,\\\
u&=&0&&\text{on }\mathbb{R}\times\partial\Omega.\end{array}\right.$

In addition, for $T>0$ and $s\in[0,1]$, there exists $C_{s,T}>0$ such that for
all $\left(u^{0},u^{1}\right)\in X^{s}$, $\gamma\in
L^{\infty}(\Omega,\mathbb{R}_{+})$, and $g\in
L_{\mathrm{loc}}^{1}(\mathbb{R},H_{0}^{s}(\Omega))$, one has

$\left\|u\right\|_{L^{\infty}((0,T),H^{1+s}\cap
H_{0}^{1})}+\left\|\partial_{t}u\right\|_{L^{\infty}((0,T),H_{0}^{s})}\leq
C_{s,T}\left(\left\|\left(u^{0},u^{1}\right)\right\|_{X^{s}}+\left\|g\right\|_{L^{1}((0,T),H_{0}^{s})}\right).$

######  Remark 9.

By the energy estimate, we can assume that $C_{0,T}\lesssim C_{0,1}$ for
$0<T\leq 1$.

To construct the solution of (Introduction), one uses Strichartz estimates
(see for example Theorem 2.1 of [11]).

######  Theorem 10 (Strichartz estimates).

Let $T>0$. There exists a constant $C>0$ such that for all $(q,r)$ satisfying

$\frac{1}{q}+\frac{3}{r}=\frac{1}{2}\quad\text{and}\quad
q\in\left[\frac{7}{2},+\infty\right]$ (3)

and for all $\left(u^{0},u^{1}\right)\in X^{0}$, $g\in
L^{1}([0,T],L^{2}(\Omega))$, the unique solution $u$ of (8) associated with
$\left(u^{0},u^{1}\right)$ and $g$ satisfies

$\|u\|_{L^{q}([0,T],L^{r})}\leq
C\left(\|u^{0}\|_{H_{0}^{1}}+\|u^{1}\|_{L^{2}}+\|g\|_{L^{1}([0,T],L^{2})}\right).$

Using those estimates, the solution of the Cauchy problem (Introduction) can
be constructed locally (see [8]). Unlike the defocusing case, it cannot be
proven that the solution is global in time.

######  Theorem 11.

Consider $\gamma\in L^{\infty}(\Omega,\mathbb{R}_{+})$. For any (real-valued)
initial data $\left(u^{0},u^{1}\right)\in X^{0}$, there exist a maximal time
of existence $T\in(0,+\infty]$ and a unique solution $u$ of (Introduction) in
$\mathscr{C}^{0}([0,T),H_{0}^{1}(\Omega))\cap\mathscr{C}^{1}([0,T),L^{2}(\Omega))$.
If $T<+\infty$, then

$\left\|u(t)\right\|_{H_{0}^{1}}\underset{t\rightarrow
T^{-}}{\longrightarrow}+\infty.$ (4)

###### Proof.

We only prove (4). If $T<+\infty$, then the Cauchy theory gives

$\left\|u(t)\right\|_{H_{0}^{1}}+\left\|\partial_{t}u(t)\right\|_{L^{2}}\underset{t\rightarrow
T^{-}}{\longrightarrow}+\infty.$

As the energy of a solution is nonincreasing, it implies

$\left\|u(t)\right\|_{L^{4}}\underset{t\rightarrow
T^{-}}{\longrightarrow}+\infty,$

yielding (4) by the Sobolev embedding $H^{1}(\Omega)\hookrightarrow
L^{4}(\Omega)$. ∎

In what follows, we refer to $u$ as "the solution of (Introduction) with
initial data $\left(u^{0},u^{1}\right)$ (and damping $\gamma$)". Finally, we
will need the fact that the geometric control condition implies the
stabilisation of the linear equation. More precisely, we will use the
following result, which is due to Bardos, Lebeau and Rauch [3].

######  Theorem 12.

Assume that $\gamma\in L^{\infty}(\Omega,\mathbb{R}_{+})$ satisfies
$\gamma(x)\geq\alpha$ for almost all $x\in\omega$, with $\alpha>0$ a constant
and $\omega\subset\Omega$ an open set fulfilling the GCC. There exist $C>0$
and $\lambda>0$ such that for all $s\in[0,1]$, and all
$\left(u^{0},u^{1}\right)\in X^{s}$, if $u=e^{tA}\left(u^{0},u^{1}\right)$ is
the solution of (8) with $g=0$, then

$\left\|\left(u(t),\partial_{t}u(t)\right)\right\|_{X^{s}}\leq Ce^{-\lambda
t}\left\|\left(u^{0},u^{1}\right)\right\|_{X^{s}},\quad t\geq 0.$

###  1.2 Ground state and some related properties

Consider $u\in H_{0}^{1}(\Omega)$, $u\neq 0$. For $\lambda\in\mathbb{R}$, set
$j(\lambda)=J(\lambda u)$, that is

$j(\lambda)=\frac{\lambda^{2}}{2}\left\|u\right\|^{2}_{H_{0}^{1}}-\frac{\lambda^{4}}{4}\left\|u\right\|^{4}_{L^{4}}.$

Write $\lambda^{\ast}=\lambda^{\ast}(u)>0$ for the positive argument of the
maximum of $j$. Using $j^{\prime}(\lambda^{\ast})=0$, one finds
$K\left(\lambda^{\ast}u\right)=0$, where $K$ is defined by (1). Set

$\displaystyle d$
$\displaystyle=\inf\left\\{J\left(\lambda^{\ast}(u)u\right),u\in
H_{0}^{1}(\Omega),u\neq 0\right\\}$ $\displaystyle=\inf\left\\{J(w),w\in
H_{0}^{1}(\Omega),w\neq 0,K(w)=0\right\\}.$ (5)

The fact that equality (5) holds is clear, because for $w\in
H_{0}^{1}(\Omega)$, $w\neq 0$, there exists a unique $\lambda>0$ such that

$\lambda\left\|w\right\|^{2}_{H_{0}^{1}}-\lambda^{3}\left\|w\right\|^{4}_{L^{4}}=0$

so that if $K(w)=0$ then $\lambda^{\ast}(w)=1$. Note that if $u\in
H_{0}^{1}(\Omega)$ satisfies $K(u)<0$, then $\lambda^{\ast}(u)\in(0,1)$. Note
also that we have

$\left|\int_{\Omega}\left(w^{4}-v^{4}\right)\mathrm{d}x\right|\lesssim\left\|w-v\right\|_{L^{2}}\left(\left\|w\right\|_{L^{6}(\Omega)}^{3}+\left\|v\right\|_{L^{6}(\Omega)}^{3}\right)$

so that $K$ and $J$ are continuous by the Sobolev embedding
$H^{1}(\Omega)\hookrightarrow L^{6}(\Omega)$. The following result is due to
Payne and Sattinger [14].

######  Theorem 13.

One has $d>0$, and there exists $w\in H_{0}^{1}(\Omega)$ such that $w\neq 0$,
$K(w)=0$, and $J(w)=d$. Such a function $w$ is a critical point of $J$, does
not change sign, satisfies $w\in H^{2}(\Omega)$, and $-\Delta w+\beta
w=w^{3}$. In addition, $J$ has exactly two critical points : $w$ and $-w$.

In what follows, we will write $Q$ for the unique nonnegative critical point
of $J$, which is called the ground state of (Introduction). We prove some
technical results for later use. The first one states that a function $u$ such
that $K(u)<0$ is far from the zero function.

######  Lemma 14.

Let $(u_{n})$ be a sequence of elements of $H_{0}^{1}(\Omega)$ such that
$K(u_{n})<0$ for all $n\in\mathbb{N}$, and
$K(u_{n})\underset{n\rightarrow\infty}{\longrightarrow}0$. Then, $\liminf
J(u_{n})\geq d$.

###### Proof.

For $n\in\mathbb{N}$, one has
$J(u_{n})=\frac{1}{4}K(u_{n})+\frac{1}{4}\left\|u_{n}\right\|_{H_{0}^{1}}^{2}$,
implying

$\liminf J(u_{n})=\frac{1}{4}\liminf\left\|u_{n}\right\|_{H_{0}^{1}}^{2}.$

If $\liminf J(u_{n})=+\infty$, the result holds. Otherwise, up to a
subsequence, one can assume that $\left(J(u_{n})\right)_{n}$ and
$\left(\left\|u_{n}\right\|_{H_{0}^{1}}\right)_{n}$ converge. In particular,
$(u_{n})$ is bounded and we can assume that (up to a subsequence) it converges
weakly in $H_{0}^{1}(\Omega)$ and strongly in $L^{4}(\Omega)$. Denote by $u\in
H_{0}^{1}(\Omega)\cap L^{4}(\Omega)$ the limit. For $n\in\mathbb{N}$, using
the Sobolev embedding and the fact that $K(u_{n})\leq 0$, we find

$\left\|u_{n}\right\|_{L^{4}}^{4}\lesssim\left\|u_{n}\right\|_{H_{0}^{1}}^{4}\leq\left\|u_{n}\right\|_{L^{4}}^{8}.$

As $u_{n}\neq 0$, this gives $\|u\|_{L^{4}}\gtrsim 1$, and $u\neq 0$ as a
result. One also has

$\|u\|_{H_{0}^{1}}^{2}\leq\lim\left\|u_{n}\right\|_{H_{0}^{1}}^{2}\leq\|u\|_{L^{4}}^{4},$

yielding $K(u)\leq 0$. We split the end of the proof into two cases.

First, assume that $K(u)=0$. Then
$\|u\|_{H_{0}^{1}}=\lim\left\|u_{n}\right\|_{H_{0}^{1}}$, and hence,
$(u_{n})_{n}$ converges to $u$ strongly in $H_{0}^{1}(\Omega)$. In particular,
by definition of $d$, we have

$\lim J(u_{n})=J(u)\geq d$

so the proof is complete in that case.

Second, assume that $K(u)<0$, that is,

$\|u\|_{H_{0}^{1}}<\lim\left\|u_{n}\right\|_{H_{0}^{1}}.$

As above, there exists $\lambda^{\ast}\in(0,1)$ such that
$K\left(\lambda^{\ast}u\right)=0$, and $d\leq J\left(\lambda^{\ast}u\right)$.
The fact that $K\left(\lambda^{\ast}u\right)=0$ gives

$J\left(\lambda^{\ast}u\right)=\frac{1}{4}\left\|\lambda^{\ast}u\right\|_{L^{4}}^{4}.$

Writing
$J(u_{n})=\frac{1}{2}K(u_{n})+\frac{1}{4}\left\|u_{n}\right\|_{L^{4}}^{4}$,
one finds $\lim J\left(u_{n}\right)=\frac{1}{4}\left\|u\right\|_{L^{4}}^{4}$.
This gives

$d\leq J\left(\lambda^{\ast}u\right)=\left(\lambda^{\ast}\right)^{4}\lim
J\left(u_{n}\right)<\lim J\left(u_{n}\right),$

implying the result in the second case. ∎

The second result will allow us to improve inequalities of the form $K(u)\geq
0$ or $K(u)<0$ for functions such that $J(u)$ is strictly below $d$.

######  Lemma 15.

Fix $\delta>0$. There exists a constant $c>0$ such that for all $u\in
H_{0}^{1}(\Omega)$ with $J(u)\leq d-\delta$, one has :

  1. (i)

if $K(u)\geq 0$ then $K(u)\geq c\|u\|_{H_{0}^{1}}^{2}$.

  2. (ii)

if $K(u)<0$ then $K(u)\leq-c\|u\|_{H_{0}^{1}}^{2}$ and $K(u)\leq-c$.

######  Remark 16.

The proof can be carried out by contradiction, but it does not provide any
information regarding the size of the constant. Our proof gives the following
explicit estimates (which will be of no use in the rest of the article) : if
$u$ is such that $J(u)\leq d-\delta$ then $K(u)\geq 0$ implies
$K(u)\geq\sqrt{\frac{\delta}{d}}\|u\|_{H_{0}^{1}}^{2}$, and $K(u)<0$ implies

$K(u)\leq-4\delta-4\sqrt{d\delta}\quad\text{ and }\quad
K(u)\leq-\frac{\delta+\sqrt{d\delta}}{d+\sqrt{d\delta}}\|u\|_{H_{0}^{1}}^{2}.$

###### Proof.

We split the proof into 3 steps.

#### Step 1 : an explicit Sobolev embedding.

We prove that for $u\in H_{0}^{1}(\Omega)$, one has

$\|u\|_{L^{4}}\|Q\|_{L^{4}}\leq\|u\|_{H_{0}^{1}}.$ (6)

Fix $u\in H_{0}^{1}(\Omega),u\neq 0$. As above, write $j(\lambda)=J(\lambda
u)$ for $\lambda\in\mathbb{R}$. Then

$\lambda^{\ast}=\frac{\|u\|_{H_{0}^{1}}}{\|u\|_{L^{4}}^{2}}$ (7)

is the positive argument of the maximum of $j$, and one has

$J\left(\lambda^{\ast}u\right)\geq J(Q)=d.$ (8)

Note that $K\left(\lambda^{\ast}u\right)=K(Q)=0$ implies

$J\left(\lambda^{\ast}u\right)=\frac{\left\|\lambda^{\ast}u\right\|_{L^{4}}^{4}}{4}\quad\text{
and }\quad J(Q)=\frac{\|Q\|_{L^{4}}^{4}}{4}.$

Together with (7) and (8), this gives (6).

#### Step 2 : estimation of $\|u\|_{H_{0}^{1}}$.

Here, we find $x^{+}$ and $x^{-}$ such that
$0<x^{-}<\|Q\|_{H_{0}^{1}}^{2}<x^{+}$, satisfying the following property : if
$u\in H_{0}^{1}(\Omega)$ is such that $J(u)\leq d-\delta$, then $K(u)\geq 0$
implies $\|u\|_{H_{0}^{1}}\leq x^{-}$ and $K(u)<0$ implies
$\|u\|_{H_{0}^{1}}\geq x^{+}$.

Fix $u\in H_{0}^{1}(\Omega)$. For $x\geq 0$, we put

$\alpha(x)=\frac{x^{2}}{2}-\frac{x^{4}}{4\|Q\|_{L^{4}}^{4}}$

so that (6) gives $\alpha\left(\|u\|_{H_{0}^{1}}\right)\leq J(u)\leq
d-\delta$. The graph of $\alpha$ is given in the following figure. Note that
the maximum of $\alpha$ is $d$, and is reached at $x=\|Q\|_{L^{4}}^{2}$.

$\bullet$$\bullet$$\bullet$$d$$d-\delta$$\|Q\|_{L^{4}}^{2}$$x^{+}$$x^{-}$

Note that there exists a unique $x^{+}>\|Q\|_{L^{4}}^{2}$ such that
$\alpha(x^{+})=d-\delta$, and if $\delta\leq d$ then there exists a unique
$x^{-}<\|Q\|_{L^{4}}^{2}$ such that $\alpha(x^{-})=d-\delta$. Explicitly, one
has

$x^{\pm}=2\sqrt{d\pm\sqrt{d\delta}}.$

First, assume that $K(u)\geq 0$. One has

$\frac{\|u\|_{H_{0}^{1}}^{2}}{4}\leq\frac{\|u\|_{H_{0}^{1}}^{2}}{4}+\frac{K(u)}{4}=J(u)\leq
d-\delta,$

implying $\|u\|_{H_{0}^{1}}^{2}\leq 4d=\|Q\|_{L^{4}}^{4}$. As
$\alpha\left(\|u\|_{H_{0}^{1}}\right)\leq d-\delta$, this gives
$\|u\|_{H_{0}^{1}}\leq x^{-}$.

Second, assume that $K(u)<0$. Then, using (6), we find

$\|u\|_{H_{0}^{1}}^{2}<\|u\|_{L^{4}}^{4}\leq\left(\frac{\|u\|_{H_{0}^{1}}}{\|Q\|_{L^{4}}}\right)^{4},$

yielding $\|u\|_{H_{0}^{1}}\geq\|Q\|_{L^{4}}^{2}$. As
$\alpha\left(\|u\|_{H_{0}^{1}}\right)\leq d-\delta$, this gives
$\|u\|_{H_{0}^{1}}\geq x^{+}$.

#### Step 3 : end of the proof.

First, assume that $K(u)\geq 0$. Then using (6) and step 2, we obtain (i) as
follows

$K(u)\geq\|u\|_{H_{0}^{1}}^{2}-\frac{\|u\|_{H_{0}^{1}}^{4}}{\|Q\|_{L^{4}}^{4}}\geq\|u\|_{H_{0}^{1}}^{2}\left(1-\frac{(x^{-})^{2}}{\|Q\|_{L^{4}}^{4}}\right)\geq\sqrt{\frac{\delta}{d}}\|u\|_{H_{0}^{1}}^{2}.$

Second, assume that $K(u)<0$. Using step 2 and $J(u)\leq d-\delta$, we find

$K(u)=4J(u)-\|u\|_{H_{0}^{1}}^{2}\leq
4(d-\delta)-(x^{+})^{2}=-4\delta-4\sqrt{d\delta}.$

That gives the second inequality of (ii). For the first inequality of (ii), it
suffices to find a constant $C>0$ such that

$K(u)\leq 4(d-\delta)-\|u\|_{H_{0}^{1}}^{2}\leq-C\|u\|_{H_{0}^{1}}^{2},$

using $\|u\|_{H_{0}^{1}}\geq x^{+}$. It holds if one chooses

$C\leq
1-\frac{4(d-\delta)}{(x^{+})^{2}}=\frac{\delta+\sqrt{d\delta}}{d+\sqrt{d\delta}}.$

The proof is complete. ∎

##  2 The damped equation

Here, we prove Theorem 2. To begin with, we check that the spaces
$\mathscr{K}^{+}$ and $\mathscr{K}^{-}$ are invariant under the flow of the
damped equation.

######  Lemma 17.

Suppose $\left(u^{0},u^{1}\right)\in\mathscr{K}^{\pm}$, $\gamma\in
L^{\infty}(\Omega,\mathbb{R}_{+})$, and $u$ is the solution of (Introduction).
If $t\geq 0$ is such that $u$ exists on $[0,t]$, then
$\left(u(t),\partial_{t}u(t)\right)\in\mathscr{K}^{\pm}$.

###### Proof.

Assume that $\left(u^{0},u^{1}\right)\in\mathscr{K}^{+}$, and that there
exists $0<t_{0}\leq t$ such that
$\left(u(t_{0}),\partial_{t}u(t_{0})\right)\notin\mathscr{K}^{+}$. As
$E\left(u(t_{0}),\partial_{t}u(t_{0})\right)\leq E\left(u^{0},u^{1}\right)<d$,
there exists a sequence $(t_{n})_{n}$ such that $0<t_{n}<t$, $K(u(t_{n}))<0$,
and $K(u(t_{n}))\underset{n\rightarrow\infty}{\longrightarrow}0$. Lemma 14
gives $\liminf J(u(t_{n}))\geq d$, a contradiction with

$J(u(t_{n}))\leq E\left(u(t_{n}),\partial_{t}u(t_{n})\right)\leq
E\left(u^{0},u^{1}\right)<d,\quad n\in\mathbb{N}.$

The proof for $\left(u^{0},u^{1}\right)\in\mathscr{K}^{-}$ is the same. ∎

###  2.1 Global solutions

The following result is straightforward, as in the case of the undamped
equation.

######  Theorem 18.

Suppose $\left(u^{0},u^{1}\right)\in\mathscr{K}^{+}$, $\gamma\in
L^{\infty}(\Omega,\mathbb{R}_{+})$, and $u$ is the solution of (Introduction).
Then $u$ is defined on $\mathbb{R}_{+}$.

###### Proof.

Using Lemma 17 and the fact that the energy is nonincreasing, we find

$\displaystyle E\left(u^{0},u^{1}\right)\geq
E\left(u(t),\partial_{t}u(t)\right)$
$\displaystyle=J(u(t))+\frac{1}{2}\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}$
$\displaystyle=\frac{1}{4}K(u(t))+\frac{1}{4}\left\|u(t)\right\|_{H_{0}^{1}}^{2}+\frac{1}{2}\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}$
$\displaystyle\geq\frac{1}{4}\left\|u(t)\right\|_{H_{0}^{1}}^{2}+\frac{1}{2}\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}.$

Hence, there is no blow-up, and $u$ is defined on $\mathbb{R}_{+}$ by the
Cauchy theory (see Theorem 11). ∎

The previous proof also gives the following lemma.

######  Lemma 19 (Equivalence between the energy and the square of the
$H_{0}^{1}\times L^{2}$ norm for solutions in $\mathscr{K}^{+}$).

For all $\left(u^{0},u^{1}\right)\in\mathscr{K}^{+}$ and $\gamma\in
L^{\infty}(\Omega,\mathbb{R}_{+})$, one has

$2E\left(u(t),\partial_{t}u(t)\right)\leq\left\|u(t)\right\|_{H_{0}^{1}}^{2}+\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}\leq
4E\left(u(t),\partial_{t}u(t)\right),\quad t\geq 0,$

where $u$ is the solution of (Introduction) with initial data
$\left(u^{0},u^{1}\right)$.

We will use this to prove the stabilisation of a solution starting from
$\mathscr{K}^{+}$. We have the following source-to-solution continuity result.

######  Lemma 20.

Fix $T>0$, $C_{0}>0$, and $\gamma\in L^{\infty}(\Omega,\mathbb{R}_{+})$. There
exists a constant $C>0$ such that for all
$\left(u^{0},u^{1}\right),\left(v^{0},v^{1}\right)\in H_{0}^{1}(\Omega)\times
L^{2}(\Omega)$, if the solutions $u$ and $v$ of (Introduction) with initial
data $\left(u^{0},u^{1}\right)$ and $\left(v^{0},v^{1}\right)$ are defined on
$[0,T]$ and satisfy

$\sup_{t\in[0,T]}\left(\left\|\left(u(t),\partial_{t}u(t)\right)\right\|_{H_{0}^{1}\times
L^{2}}+\left\|\left(v(t),\partial_{t}v(t)\right)\right\|_{H_{0}^{1}\times
L^{2}}\right)\leq C_{0},$ (9)

then one has

$\sup_{t\in[0,T]}\left\|\left(u(t),\partial_{t}u(t)\right)-\left(v(t),\partial_{t}v(t)\right)\right\|_{H_{0}^{1}\times
L^{2}}\leq
C\left\|\left(u(t_{0}),\partial_{t}u(t_{0})\right)-\left(v(t_{0}),\partial_{t}v(t_{0})\right)\right\|_{H_{0}^{1}\times
L^{2}}$

for $t_{0}\in\left\\{0,\frac{T}{2},T\right\\}$.

###### Proof.

It suffices to prove the result for $t_{0}=0$ and for $t_{0}=T$. In both
cases, we can assume that $T$ is arbitrary small : the result for large $T$
follows by iteration. We assume that $t_{0}=0$, the other case is similar.

Set $w=u-v$, solution of

$\left\\{\begin{array}[]{rcccl}\square w+\gamma\partial_{t}w+\beta
w&=&u^{3}-v^{3}&&\text{in }\mathbb{R}_{+}\times\Omega,\\\
(w(0),\partial_{t}w(0))&=&\left(u^{0},u^{1}\right)-\left(v^{0},v^{1}\right)&&\text{in
}\Omega,\\\ w&=&0&&\text{on
}\mathbb{R}_{+}\times\partial\Omega.\end{array}\right.$

Assume that $T<1$. Then, by Theorem 8 and Remark 9, there exists a constant
independent of $T$, $\left(u^{0},u^{1}\right)$ and $\left(v^{0},v^{1}\right)$
such that

$\sup_{t\in[0,T]}\left(\left\|w(t)\right\|_{H_{0}^{1}}+\left\|\partial_{t}w(t)\right\|_{L^{2}}\right)\lesssim\left\|\left(u^{0},u^{1}\right)-\left(v^{0},v^{1}\right)\right\|_{H_{0}^{1}\times
L^{2}}+\left\|u^{3}-v^{3}\right\|_{L^{1}((0,T),L^{2})}.$

Hölder’s inequality gives

$\left\|u^{3}-v^{3}\right\|_{L^{1}((0,T),L^{2})}\lesssim
T\left\|u-v\right\|_{L^{\infty}((0,T),L^{6})}\left(\left\|u\right\|_{L^{\infty}((0,T),L^{6})}^{2}+\left\|v\right\|_{L^{\infty}((0,T),L^{6})}^{2}\right).$

Using (9) and the Sobolev embedding $H^{1}(\Omega)\hookrightarrow
L^{6}(\Omega)$, one finds

$\displaystyle\left\|u^{3}-v^{3}\right\|_{L^{1}((0,T),L^{2})}$
$\displaystyle\lesssim
T\left\|w\right\|_{L^{\infty}((0,T),H_{0}^{1})}\left(\left\|u\right\|_{L^{\infty}((0,T),H_{0}^{1})}^{2}+\left\|v\right\|_{L^{\infty}((0,T),H_{0}^{1})}^{2}\right)$
$\displaystyle\lesssim T\left\|w\right\|_{L^{\infty}((0,T),H_{0}^{1})}.$

Hence, for $T$ sufficiently small, one obtains

$\sup_{t\in[0,T]}\left(\left\|w(t)\right\|_{H_{0}^{1}}+\left\|\partial_{t}w(t)\right\|_{L^{2}}\right)\lesssim\left\|\left(u^{0},u^{1}\right)-\left(v^{0},v^{1}\right)\right\|_{H_{0}^{1}\times
L^{2}}$

and this completes the proof. ∎

###  2.2 Blow-up solutions

Here, we prove that a solution initiated in $\mathscr{K}^{-}$ blows up in
finite time.

######  Theorem 21.

For $\gamma\in L^{\infty}(\Omega,\mathbb{R}_{+})$, if $u$ is the solution of
(Introduction) with initial data $\left(u^{0},u^{1}\right)\in\mathscr{K}^{-}$,
then the maximal time of existence of $u$ is finite.

###### Proof.

Take $\left(u^{0},u^{1}\right)\in\mathscr{K}^{-}$ and assume by contradiction
that $u(t)$ exists for all $t\geq 0$. As in the original proof of Payne and
Sattinger [14], set $M(t)=\left\|u(t)\right\|_{L^{2}}^{2}$. We prove that
$M\in\mathscr{C}^{2}(\mathbb{R}_{+},\mathbb{R})$, with

$M^{\prime}(t)=2\int_{\Omega}u(t)\partial_{t}u(t)\mathrm{d}x.$ (10)

and

$M^{\prime\prime}(t)=-2\left\|u(t)\right\|_{H_{0}^{1}}^{2}+2\left\|u(t)\right\|_{L^{4}}^{4}+2\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}-2\int_{\Omega}\gamma
u(t)\partial_{t}u(t)\mathrm{d}x$ (11)

As
$u\in\mathscr{C}^{0}(\mathbb{R}_{+},H_{0}^{1}(\Omega))\cap\mathscr{C}^{1}(\mathbb{R}_{+},L^{2}(\Omega))$,
one finds $M\in\mathscr{C}^{1}(\mathbb{R}_{+},\mathbb{R})$, and (10) holds
true. We show that (11) holds true in
$\mathcal{D}^{\prime}((0,+\infty),\mathbb{R})$. Take
$\phi\in\mathscr{C}_{\mathrm{c}}^{\infty}((0,+\infty),\mathbb{R})$, and write

$\int_{0}^{+\infty}M^{\prime}\phi^{\prime}\mathrm{d}t=2\int_{0}^{+\infty}\int_{\Omega}\left(\partial_{t}u\partial_{t}\left(u\phi\right)-\left(\partial_{t}u\right)^{2}\phi\right)\mathrm{d}x\mathrm{d}t.$

One has $u\phi\in H_{0}^{1}((0,+\infty)\times\Omega)$. Consider a sequence
$\left(u_{n}\right)_{n}$ of elements of
$\mathscr{C}_{\mathrm{c}}^{\infty}((0,+\infty)\times\Omega,\mathbb{R})$ that
converges to $u\phi$ in $H^{1}((0,+\infty)\times\Omega)$. For
$n\in\mathbb{N}$, using the equation satisfied by $u$, we find

$\int_{0}^{+\infty}\int_{\Omega}\partial_{t}u\partial_{t}u_{n}\mathrm{d}x\mathrm{d}t=\int_{0}^{+\infty}\int_{\Omega}\left(\nabla
u\nabla u_{n}+\gamma\partial_{t}uu_{n}+\beta
uu_{n}-u^{3}u_{n}\right)\mathrm{d}x\mathrm{d}t.$

In the limit $n\rightarrow+\infty$, one obtains

$\int_{0}^{+\infty}\int_{\Omega}\partial_{t}u\partial_{t}\left(u\phi\right)\mathrm{d}x\mathrm{d}t=\int_{0}^{+\infty}\int_{\Omega}\left(\nabla
u\nabla\left(u\phi\right)+\gamma u\partial_{t}u\phi+\beta
u^{2}\phi-u^{4}\phi\right)\mathrm{d}x\mathrm{d}t$

and as $\phi$ is independent of $x$, this gives

$\int_{0}^{+\infty}M^{\prime}\phi^{\prime}\mathrm{d}t=2\int_{0}^{+\infty}\int_{\Omega}\left(\left|\nabla
u\right|^{2}+\gamma u\partial_{t}u+\beta
u^{2}-u^{4}-\left(\partial_{t}u\right)^{2}\right)\phi\mathrm{d}x\mathrm{d}t.$

Hence, the weak derivative of $M^{\prime}$ exists and is given by (11). Note
that the right-hand side of (11) depends continuously on $t\geq 0$, so that in
fact, $M\in\mathscr{C}^{2}(\mathbb{R}_{+},\mathbb{R})$, and (11) holds true in
a strong sense.

One has

$M^{\prime\prime}(t)=-2K(u(t))+2\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}-2\int_{\Omega}\gamma
u(t)\partial_{t}u(t)\mathrm{d}x.$

Write $E(t)=E\left(u(t),\partial_{t}u(t)\right)$ for the energy. Recall that
the energy equality is

$E^{\prime}(t)=-\int_{\Omega}\gamma\left|\partial_{t}u(t)\right|^{2}\mathrm{d}x.$

As the energy is nonincreasing, either it is bounded or it tends to $-\infty$
as $t$ tends to infinity. We treat these two cases separately.

#### Case 1 : bounded energy.

For all $\varepsilon>0$, as $\gamma\in L^{\infty}(\mathbb{R}_{+})$, one has

$\left|2\int_{\Omega}\gamma
u(t)\partial_{t}u(t)\mathrm{d}x\right|\lesssim\varepsilon\left\|u(t)\right\|_{H_{0}^{1}}^{2}+\frac{1}{\varepsilon}\int_{\Omega}\gamma\left|\partial_{t}u(t)\right|^{2}\mathrm{d}x=\varepsilon\left\|u(t)\right\|_{H_{0}^{1}}^{2}-\frac{1}{\varepsilon}E^{\prime}(t).$

This gives

$M^{\prime\prime}(t)\geq-2K(u(t))-C_{1}\varepsilon\left\|u(t)\right\|_{H_{0}^{1}}^{2}+2\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}+\frac{C_{1}}{\varepsilon}E^{\prime}(t)$
(12)

for some $C_{1}>0$. Inequality (12) has two consequences.

First, we use Lemma 15 _(ii)_ : for $t\geq 0$, we have

$J(u(t))\leq E\left(u(t),\partial_{t}u(t)\right)\leq
E\left(u^{0},u^{1}\right)<d$

so there exists $C_{2}>0$ such that

$K\left(u(t)\right)\leq-C_{2}\left\|u(t)\right\|_{H_{0}^{1}}^{2}\quad\text{
and }\quad K\left(u(t)\right)\leq-C_{2},\quad t\geq 0.$ (13)

Using this in (12) and taking $\varepsilon$ sufficiently small, we find

$M^{\prime\prime}(t)-\frac{C_{1}}{\varepsilon}E^{\prime}(t)\gtrsim\left\|u(t)\right\|_{H_{0}^{1}}^{2}+\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}\geq\left\|u(t)\right\|_{H_{0}^{1}}^{2}.$

Note that by (13), one has $\left\|u(t)\right\|_{H_{0}^{1}}\gtrsim 1$ for
$t\geq 0$. Hence, we have
$M^{\prime\prime}(t)-\frac{C_{1}}{\varepsilon}E^{\prime}(t)\gtrsim 1$,
implying

$M^{\prime}(t)-\frac{C_{1}}{\varepsilon}E(t)-M^{\prime}(0)+\frac{C_{1}}{\varepsilon}E(0)\gtrsim
t$

for all $t\geq 0$. As the energy is bounded, this gives

$M^{\prime}(t)\underset{t\rightarrow+\infty}{\longrightarrow}+\infty\quad\text{
and }\quad M(t)\underset{t\rightarrow+\infty}{\longrightarrow}+\infty.$

Second, we use the energy equality. By definition, we have

$K(u(t))=4E(t)-2\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}-\left\|u(t)\right\|_{H_{0}^{1}}^{2}$

for all $t\geq 0$. Using this in (12), one obtains

$M^{\prime\prime}(t)-\frac{C_{1}}{\varepsilon}E^{\prime}(t)\geq
6\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}+(2-C_{1}\varepsilon)\left\|u(t)\right\|_{H_{0}^{1}}^{2}-8E(t).$

Set $C=\frac{C_{1}}{\varepsilon}$. As the energy is bounded and as $M(t)$
tends to infinity, for $\varepsilon$ chosen sufficiently small and $t$
sufficiently large, we find

$M^{\prime\prime}(t)-CE^{\prime}(t)\geq
6\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}.$

Together with the Cauchy-Schwarz inequality, this gives

$\left|M^{\prime}(t)\right|^{2}\leq
4\left(\int_{\Omega}\left|u(t)\right|^{2}\mathrm{d}x\right)\left(\int_{\Omega}\left|\partial_{t}u(t)\right|^{2}\mathrm{d}x\right)\leq\frac{2}{3}M(t)\left(M^{\prime\prime}(t)-CE^{\prime}(t)\right),\quad
t\geq 0\text{ large}.$

Dividing by $M(t)M^{\prime}(t)$, and using $M^{\prime}(t)\rightarrow+\infty$
and $E^{\prime}(t)\leq 0$, one writes

$\frac{M^{\prime}(t)}{M(t)}\leq\frac{2}{3}\left(\frac{M^{\prime\prime}(t)}{M^{\prime}(t)}-C\frac{E^{\prime}(t)}{M^{\prime}(t)}\right)\leq\frac{2}{3}\left(\frac{M^{\prime\prime}(t)}{M^{\prime}(t)}-CE^{\prime}(t)\right)$

for $t$ large. Fix $T>0$ such that the previous inequality holds for all
$t\geq T$. Integrating, we find

$\ln\left(M(t)\right)-\ln\left(M(T)\right)\leq\frac{2}{3}\left(\ln\left(M^{\prime}(t)\right)-\ln\left(M^{\prime}(T)\right)\right)-\frac{2C}{3}\left(E(t)-E(T)\right),\quad
t\geq T.$

As the energy is bounded and $M(t)\rightarrow+\infty$, this gives
$\ln\left(M(t)\right)\leq\frac{3}{4}\ln\left(M^{\prime}(t)\right)$ for $t$
large, yielding

$\frac{M^{\prime}(t)}{M(t)^{\frac{4}{3}}}\geq 1,\quad t\geq T^{\prime},\quad
T^{\prime}\text{ large}.$

Integrating between $t_{1}$ and $t_{2}$ with $T^{\prime}\leq t_{1}\leq t_{2}$,
one finds

$-\frac{3}{M(t_{2})^{\frac{1}{3}}}+\frac{3}{M(t_{1})^{\frac{1}{3}}}\geq
t_{2}-t_{1}.$

Letting $t_{2}$ tend to infinity gives a contradiction.

#### Case 2 : the energy tends to $-\infty$.

In that case, by definition of the energy, we have

$\left\|u(t)\right\|_{L^{4}}\underset{t\rightarrow+\infty}{\longrightarrow}+\infty.$

For $\varepsilon>0$, as $\gamma\in L^{\infty}(\mathbb{R}_{+})$ and as $\Omega$
is bounded, we write

$\left|2\int_{\Omega}\gamma
u(t)\partial_{t}u(t)\mathrm{d}x\right|\lesssim\frac{1}{\varepsilon}\left\|u(t)\right\|_{L^{2}}^{2}+\varepsilon\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}\lesssim\frac{1}{\varepsilon}\left\|u(t)\right\|_{L^{4}}^{2}+\varepsilon\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2},$

and with (11), this gives

$M^{\prime\prime}(t)\geq-2\left\|u(t)\right\|_{H_{0}^{1}}^{2}+(2-C\varepsilon)\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}+2\left\|u(t)\right\|_{L^{4}}^{4}-\frac{C}{\varepsilon}\left\|u(t)\right\|_{L^{4}}^{2}$

for some $C>0$. Fix $\varepsilon>0$ such that $2-C\varepsilon\geq\frac{3}{2}$.
For $t$ sufficiently large, one has

$2\left\|u(t)\right\|_{L^{4}}^{4}-\frac{C}{\varepsilon}\left\|u(t)\right\|_{L^{4}}^{2}\geq\frac{3}{2}\left\|u(t)\right\|_{L^{4}}^{4},$

and one obtains

$M^{\prime\prime}(t)\geq-2\left\|u(t)\right\|_{H_{0}^{1}}^{2}+\frac{3}{2}\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}+\frac{3}{2}\left\|u(t)\right\|_{L^{4}}^{4},\quad
t\geq 0\text{ large}.$

By definition of the energy, one has

$\left\|u(t)\right\|^{4}_{L^{4}}=2\left\|\partial_{t}u(t)\right\|^{2}_{L^{2}}+2\left\|u(t)\right\|^{2}_{H_{0}^{1}}-4E(t)$

and for $t$ sufficiently large, this gives

$M^{\prime\prime}(t)\geq\frac{9}{2}\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}+\left\|u(t)\right\|_{H_{0}^{1}}^{2}-6E(t).$

In particular, one has $M^{\prime\prime}(t)\rightarrow+\infty$, and
consequently $M(t)\rightarrow+\infty$. For $t$ sufficiently large, we also
have
$M^{\prime\prime}(t)\geq\frac{9}{2}\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}$.
Using (10) and the Cauchy-Schwarz inequality, we get

$M^{\prime}(t)^{2}\leq\frac{8}{9}M(t)M^{\prime\prime}(t),\quad t\geq 0\text{
large}.$

For $\alpha>0$, set $\tilde{M}(t)=M(t)^{-\alpha}$. For $t$ sufficiently large,
we have

$\displaystyle\tilde{M}^{\prime\prime}(t)$ $\displaystyle=\alpha
M(t)^{-\alpha-2}\left((\alpha+1)M^{\prime}(t)^{2}-M(t)M^{\prime\prime}(t)\right)$
$\displaystyle\leq\alpha\left(\frac{8}{9}(\alpha+1)-1\right)M(t)^{-\alpha-1}M^{\prime\prime}(t)$

Hence, $\tilde{M}$ is a concave function for $t$ sufficiently large if
$\alpha$ is chosen sufficiently small. As $\tilde{M}>0$ and
$\tilde{M}(t)\rightarrow 0$, this gives a contradiction. ∎

##  3 Convergence towards a stationary solution

###  3.1 Convergence of a bounded sequence along a sequence of time

Here, we show the following proposition.

######  Proposition 22.

Fix $\gamma\in L^{\infty}(\Omega,\mathbb{R}_{+})$ satisfying the geometric
control condition. Let $\left(u^{0}_{n},u^{1}_{n}\right)_{n\in\mathbb{N}}$ be
a sequence of elements of $H_{0}^{1}(\Omega)\times L^{2}(\Omega)$, and for
$n\in\mathbb{N}$, write $u_{n}$ for the solution of (Introduction) with
initial data $\left(u^{0}_{n},u^{1}_{n}\right)$. Let
$(T_{n})_{n\in\mathbb{N}}$ be a time-sequence satisfying
$T_{n}\rightarrow+\infty$. Assume that each $u_{n}$ exists on
$\mathbb{R}_{+}$, that there exists $C>0$ such that

$\left\|u_{n}(t)\right\|_{H_{0}^{1}}+\left\|\partial_{t}u_{n}(t)\right\|_{L^{2}}\leq
C,\quad t\geq 0,\quad n\in\mathbb{N},$ (14)

and that for all $T>0$,

$\int_{T_{n}-T}^{T_{n}+T}\int_{\Omega}\gamma\left|\partial_{t}u_{n}\right|^{2}\mathrm{d}x\mathrm{d}t\underset{n\rightarrow+\infty}{\longrightarrow}0.$
(15)

Then there exist an increasing function $\phi:\mathbb{N}\rightarrow\mathbb{N}$
and a stationary solution $w$ of (Introduction) such that for all $T>0$,

$\sup_{t\in[-T,T]}\left(\left\|u_{\phi(n)}(T_{\phi(n)}+t)-w\right\|_{H_{0}^{1}}+\left\|\partial_{t}u_{\phi(n)}(T_{\phi(n)}+t)\right\|_{L^{2}}\right)\underset{n\rightarrow+\infty}{\longrightarrow}0.$

###### Proof.

We divide the proof into three steps.

#### Step 1 : asymptotic compactness.

We use Corollary 4.2 of [11] (which relies on a result from [6]), which we
copy here for convenience.

######  Lemma 23 (Corollary 4.2 of [11]).

Take $f\in\mathscr{C}^{1}(\mathbb{R},\mathbb{R})$ such that

$f(0)=0,\quad xf(x)\geq
0,\quad|f(x)|\lesssim(1+|x|)^{p},\quad|f^{\prime}(x)|\lesssim(1+|x|)^{p-1}$

with $1\leq p<5$. Consider $R>0$, $T>0$, $0\leq s<1$, and set
$\varepsilon=\min\left(1-s,\frac{5-p}{2},\frac{17-3p}{14}\right)>0$. There
exist $C>0$ and $(q,r)$ satisfying

$\frac{1}{q}+\frac{3}{r}=\frac{1}{2},\quad
q\in\left[\frac{7}{2},+\infty\right]$

such that the following property holds : if $u\in
L^{\infty}([0,T],H^{1+s}(\Omega)\cap H_{0}^{1}(\Omega))$ satisfies

$\|u\|_{L^{q}([0,T],L^{r})}\leq R,$

then $f(u)\in L^{1}([0,T],H_{0}^{s+\varepsilon}(\Omega))$, with

$\left\|f(u)\right\|_{L^{1}([0,T],H^{s+\varepsilon})}\leq
C\left\|u\right\|_{L^{\infty}([0,T],H^{1+s})}.$

We will use this lemma with $f(x)=x^{3}$. Recall that $X^{s}$ is defined by
(2). We prove the following corollary.

######  Corollary 24.

Consider $T>0$, $0\leq s<1$, $C_{0}>0$, and set
$\varepsilon=\min\left(1-s,\frac{4}{7}\right)$. There exists $C>0$ such that
for all $\left(u^{0},u^{1}\right)\in X^{s}$, if the solution $u$ of
(Introduction) with initial data $\left(u^{0},u^{1}\right)$ exists on $[0,T]$
and satisfies $u\in L^{\infty}([0,T],H^{1+s}(\Omega)\cap H_{0}^{1}(\Omega))$,
with

$\|u\|_{L^{\infty}([0,T],H_{0}^{1})}+\left\|\partial_{t}u\right\|_{L^{\infty}([0,T],L^{2})}\leq
C_{0},$

then one has $u^{3}\in L^{1}([0,T],H^{s+\varepsilon}_{0}(\Omega))$, with

$\left\|u^{3}\right\|_{L^{1}([0,T],H_{0}^{s+\varepsilon})}\leq
C\left\|u\right\|_{L^{\infty}([0,T],H^{1+s})}.$

###### Proof.

The function $f(x)=x^{3}$ clearly satisfies the assumption of the previous
theorem. By Strichartz estimates (Theorem 10), one has

$\|u\|_{L^{q}([0,T],L^{r})}\lesssim\|u^{3}\|_{L^{1}([0,T],L^{2})}+\|u^{0}\|_{H_{0}^{1}}+\|u^{1}\|_{L^{2}}.$

for all $(q,r)$ satisfying $\frac{1}{q}+\frac{3}{r}=\frac{1}{2}$ and
$q\in\left[\frac{7}{2},+\infty\right]$. Using the Sobolev embedding
$H^{1}(\Omega)\hookrightarrow L^{6}(\Omega)$, one obtains

$\|u\|_{L^{q}([0,T],L^{r})}\lesssim\|u\|_{L^{\infty}([0,T],H_{0}^{1})}^{3}+\|u^{0}\|_{H_{0}^{1}}+\|u^{1}\|_{L^{2}}\leq
C_{0}^{3}+C_{0}.$

Hence, we can apply Corollary 4.2 of [11] : it gives

$\left\|u^{3}\right\|_{L^{1}([0,T],H_{0}^{s+\varepsilon})}\lesssim\left\|u\right\|_{L^{\infty}([0,T],H^{1+s})}$

with a constant depending only on $C_{0}$. This completes the proof of
Corollary 24. ∎

We will use this result to find a subsequence of
$\left(u_{n}(T_{n}),\partial_{t}u_{n}(T_{n})\right)$ which converges in
$H_{0}^{1}(\Omega)\times L^{2}(\Omega)$. Recall that

$A=\begin{pmatrix}0&\mathrm{Id}\\\ \Delta-\beta&-\gamma\end{pmatrix}$

is the infinitesimal generator of the linear part of (Introduction), that
$e^{tA}$ is the associated semi-group, and put

$U_{n}=\left(u_{n},\partial_{t}u_{n}\right)\quad\text{ and }\quad
F_{n}=\left(0,u_{n}^{3}\right)$

We will use the well-know fact that the geometric control condition implies
the stabilisation of the linear version of (Introduction), as stated in
Theorem 12. Using the Duhamel formula, we can write

$\displaystyle U_{n}(T_{n})$
$\displaystyle=e^{T_{n}A}U_{n}(0)+\sum_{k=0}^{\lfloor
T_{n}\rfloor-1}e^{kA}\int_{0}^{1}e^{sA}F_{n}(T_{n}-k-s)\mathrm{d}s+\int_{\lfloor
T_{n}\rfloor}^{T_{n}}e^{sA}F_{n}(T_{n}-s)\mathrm{d}s$
$\displaystyle=e^{T_{n}A}U_{n}(0)+\sum_{k=0}^{\lfloor
T_{n}\rfloor-1}e^{kA}I_{n,k}+I_{n}.$

We show that the Duhamel term is bounded in $X^{\varepsilon}$, with
$\varepsilon=\frac{4}{7}$. To do so, write

$\displaystyle\|I_{n,k}\|_{X^{\varepsilon}}$
$\displaystyle\leq\int_{0}^{1}\left\|e^{sA}F_{n}(T_{n}-k-s)\right\|_{X^{\varepsilon}}\mathrm{d}s$
$\displaystyle\lesssim\int_{0}^{1}\left\|F_{n}(T_{n}-k-s)\right\|_{X^{\varepsilon}}\mathrm{d}s$
$\displaystyle=\int_{0}^{1}\left\|u_{n}(T_{n}-k-s)^{3}\right\|_{H_{0}^{\varepsilon}}\mathrm{d}s.$

By (14), one has

$\left\|u_{n}(T_{n}-k-\cdot)\right\|_{L^{\infty}((0,1),H_{0}^{1})}\leq\left\|u_{n}\right\|_{L^{\infty}((0,\infty),H_{0}^{1})}\lesssim
1.$

Hence, we can apply Corollary 24 (with $s=0$) to find
$\|I_{n,k}\|_{X^{\varepsilon}}\lesssim 1$. Similarly, one shows
$\|I_{n}\|_{X^{\varepsilon}}\lesssim 1$. Using the linear stabilisation
(Theorem 12), we see that there exists $\lambda>0$ such that

$\left\|\sum_{k=0}^{\lfloor
T_{n}\rfloor-1}e^{kA}I_{n,k}+I_{n}\right\|_{X^{\varepsilon}}\lesssim\sum_{k=0}^{\lfloor
T_{n}\rfloor-1}e^{-\lambda k}+1\lesssim 1.$

We have proved that the sequence
$\left(U_{n}(T_{n})-e^{T_{n}A}U_{n}(0)\right)_{n}$ is bounded in
$X^{\varepsilon}$. By Rellich’s theorem, there exists $U_{\infty}(0)\in
H_{0}^{1}(\Omega)\times L^{2}(\Omega)$ such that up to a subsequence, one has

$U_{n}(T_{n})-e^{T_{n}A}U_{n}(0)\underset{n\rightarrow\infty}{\longrightarrow}U_{\infty}(0)$

in $H_{0}^{1}(\Omega)\times L^{2}(\Omega)$. Using linear stabilisation again,
one finds
$U_{n}(T_{n})\underset{n\rightarrow\infty}{\longrightarrow}U_{\infty}(0)$ in
$H_{0}^{1}(\Omega)\times L^{2}(\Omega)$.

Let $u_{\infty}$ be the solution of (Introduction) with initial data
$U_{\infty}(0)$. We show that $u_{\infty}$ is defined on $\mathbb{R}$. Fix
$T>0$. For $n$ and $m$ sufficiently large, Lemma 20 gives

$\displaystyle\sup_{t\in[-T,T]}\left(\left\|u_{n}(T_{n}+t)-u_{m}(T_{m}+t)\right\|_{H_{0}^{1}}^{2}+\left\|\partial_{t}u_{n}(T_{n}+t)-\partial_{t}u_{m}(T_{m}+t)\right\|_{L^{2}}^{2}\right)$
$\displaystyle\lesssim$
$\displaystyle\left\|u_{n}(T_{n})-u_{m}(T_{m})\right\|_{H_{0}^{1}}^{2}+\left\|\partial_{t}u_{n}(T_{n})-\partial_{t}u_{m}(T_{m})\right\|_{L^{2}}^{2}$

so that $\left(u_{n}(T_{n}+\cdot)\right)_{n}$ is a Cauchy sequence in
$\mathscr{C}^{0}([-T,T],H_{0}^{1}(\Omega))\cap\mathscr{C}^{1}([-T,T],L^{2}(\Omega))$.
The limit is a solution of (Introduction) on $[-T,T]$, and coincides with
$u_{\infty}$ near $0$. Hence, $u_{\infty}$ is defined on $\mathbb{R}$. Note
that (14) gives

$\left\|u_{\infty}(t)\right\|_{H_{0}^{1}}+\left\|\partial_{t}u_{\infty}(t)\right\|_{L^{2}}\leq
C,\quad t\in\mathbb{R},$ (16)

and (15) implies

$\int_{\mathbb{R}}\int_{\Omega}\gamma\left|\partial_{t}u_{\infty}\right|^{2}\mathrm{d}x\mathrm{d}t=0.$

Hence, one has
$E\left(u_{\infty}(t),\partial_{t}u_{\infty}(t)\right)=E\left(u_{\infty}(0),\partial_{t}u_{\infty}(0)\right)$
for all $t\in\mathbb{R}$, and $\partial_{t}u_{\infty}(t,x)=0$ for all
$t\in\mathbb{R}$ and almost all $x\in\omega$.

#### Step 2 : regularity of the limit.

To ease notations, we write $u=u_{\infty}$. In this step, we will use
Corollary 24 again to show that

$\left(u(0),\partial_{t}u(0)\right)\in H^{2}(\Omega)\times H_{0}^{1}(\Omega)$
(17)

and we will use a result of [9] to prove that for all
$\alpha\in\left(\frac{1}{2},1\right)$,

$u:\mathbb{R}\longrightarrow H^{1+\alpha}(\Omega)\cap H_{0}^{1}(\Omega)\text{
is analytic.}$ (18)

Finally, we will see that it implies

$u\in\mathscr{C}^{\infty}(\mathbb{R}\times\overline{\Omega}).$ (19)

Proof of (17). As above, set $U=\left(u,\partial_{t}u\right)$ and
$F=\left(0,u^{3}\right)$. By Duhamel’s formula, one has

$\displaystyle U(t)$
$\displaystyle=e^{nA}U(t-n)+\int_{0}^{n}e^{sA}F(t-s)\mathrm{d}s$
$\displaystyle=e^{nA}U(t-n)+\sum_{k=0}^{n-1}e^{kA}\int_{0}^{1}e^{sA}F(t-k-s)\mathrm{d}s$

for $t\in\mathbb{R}$ and $n\in\mathbb{N}$. By (16), linear stabilization gives

$e^{nA}U(t-n)\underset{n\rightarrow\infty}{\longrightarrow}0,\quad
t\in\mathbb{R},$

in $X^{0}$. Furthermore, one has

$\displaystyle\left\|\int_{0}^{1}e^{sA}F(t-k-s)\mathrm{d}s\right\|_{X^{0}}$
$\displaystyle\lesssim\int_{0}^{1}\left\|F(t-k-s)\right\|_{X^{0}}\mathrm{d}s$
$\displaystyle=\left\|u^{3}\right\|_{L^{1}([t-k-1,t-k],L^{2})}$

implying

$\left\|\int_{0}^{1}e^{sA}F(t-k-s)\mathrm{d}s\right\|_{X^{0}}\lesssim\left\|u\right\|_{L^{\infty}([t-k-1,t-k],H_{0}^{1})}^{3}\lesssim
1$

by the Sobolev embedding $H^{1}(\Omega)\hookrightarrow L^{6}(\Omega)$. In
particular, using linear stabilization again, one finds

$\left\|e^{kA}\int_{0}^{1}e^{sA}F(t-k-s)\mathrm{d}s\right\|_{X^{0}}\leq
Ce^{-\lambda k}$

for some $C>0$ and $\lambda>0$. This gives

$U(t)=\sum_{k=0}^{\infty}e^{kA}\int_{0}^{1}e^{sA}F(t-k-s)\mathrm{d}s,\quad
t\in\mathbb{R},$ (20)

in $X^{0}$.

Next, we prove

$U\in L^{\infty}(\mathbb{R},X^{1})$ (21)

by using Corollary 24 two times. First, by (16), Corollary 24 with $T=1$ and
$s=0$ gives

$\left\|u(t-k-\cdot)^{3}\right\|_{L^{1}((0,1),H_{0}^{\varepsilon})}\lesssim
1,\quad t\in\mathbb{R},\quad k\in\mathbb{N},$

where $\varepsilon=\frac{4}{7}$. In particular, using linear stabilisation in
$X^{\varepsilon}$, one has

$\left\|e^{kA}\int_{0}^{1}e^{sA}F(t-k-s)\mathrm{d}s\right\|_{X^{\varepsilon}}\lesssim
e^{-\lambda
k}\left\|u(t-k-\cdot)^{3}\right\|_{L^{1}((0,1),H_{0}^{\varepsilon})}\lesssim
e^{-\lambda k},$

implying that equality (20) holds in $X^{\varepsilon}$, and $U\in
L^{\infty}(\mathbb{R},X^{\varepsilon})$. Second, Corollary 24 with $T=1$ and
$s=\frac{4}{7}$ gives

$\left\|u(t-k-\cdot)^{3}\right\|_{L^{1}((0,1),H_{0}^{s+\varepsilon^{\prime}})}\lesssim
1,\quad t\in\mathbb{R},\quad k\in\mathbb{N}$

where $\varepsilon^{\prime}=\frac{3}{7}$. As above, this proves that equality
(20) holds in $X^{1}$, and that (21) holds true. In particular, (17) is true.
Note that using the Sobolev embedding
$H^{2}(\Omega)\hookrightarrow\mathscr{C}^{0}(\overline{\Omega})$ (see for
example [1], 4.12 Part II with $n=3$, $m=p=2$, $j=0$), one finds $u\in
L^{\infty}(\mathbb{R}\times\overline{\Omega})$.

Proof of (18). Following [11], we use Theorem 2.20 of [9] (applied with
hypotheses (H3mod) and (H5)), which we copy here for convenience.

######  Theorem 25.

Let $Y$ be a Banach space. Let $P_{n}\in\mathscr{L}(Y)$ be a sequence of
continuous linear maps and let $Q_{n}=\mathrm{Id}-P_{n}$. Let
$A:D(A)\rightarrow Y$ be the generator of a continuous semi-group $e^{tA}$ and
let $G\in\mathscr{C}^{1}(Y,Y)$. Let $U$ be a global solution in $Y$ of

$\partial_{t}U(t)=AU(t)+G(U(t)),\quad t\in\mathbb{R}.$

We further assume that :

  1. (i)

$\\{U(t),t\in\mathbb{R}\\}$ is contained in a compact set $K$ of $Y$.

  2. (ii)

For any $y\in Y$, $P_{n}y$ converges to $y$ when $n$ goes to infinity and
$(P_{n})$ and $(Q_{n})$ are sequences of $\mathscr{L}(Y)$ bounded by a
constant $C_{0}$.

  3. (iii)

The operator $A$ splits as $A=A_{1}+B_{1}$, where $B_{1}$ is bounded and
$A_{1}$ commutes with $P_{n}$.

  4. (iv)

There exist $M$ and $\lambda>0$ such that
$\left\|e^{tA}\right\|_{\mathscr{L}(Y)}\leq Me^{-\lambda t}$ for all $t\geq
0$.

  5. (v)

$G$ is analytic in the ball $B_{Y}(0,r)$, where $r$ is such that $r\geq
4C_{0}\sup_{t\in\mathbb{R}}\|U(t)\|_{Y}$. More precisely, there exists
$\rho>0$ such that $G$ can be extended to an holomorphic function of
$B_{Y}(0,r)+iB_{Y}(0,\rho)$.

  6. (vi)

$\\{DG(U(t))V,t\in\mathbb{R},\|V\|_{Y}\leq 1\\}$ is a relatively compact set
of $Y$.

Then the solution $U:\mathbb{R}\rightarrow Y$ is analytic.

Fix $\alpha\in\left(\frac{1}{2},1\right)$. We write
$\left(\lambda_{k}\right)_{k\in\mathbb{N}}$ for the eigenvalues of the
Dirichlet Laplacian, and $\left(\varphi_{k}\right)_{k\in\mathbb{N}}$ for a
basis of normalized eigenvectors. We will apply this theorem with
$Y=X^{\alpha}$. For $n\in\mathbb{N}^{\ast}$, let $P_{n}$ be the restriction to
$Y$ of the $L^{2}(M)\times L^{2}(M)$ orthogonal projection on the vector space
generated by the vectors

$(\varphi_{1},0),\cdots,(\varphi_{n},0),(0,\varphi_{1}),\cdots,\text{ and
}(0,\varphi_{n}).$

By (3), we can equip $Y$ with a norm represented by a Fourier series, using
the sequence $\left(\varphi_{k}\right)_{k\in\mathbb{N}}$. This implies
$P_{n}\in\mathscr{L}(Y)$ and $Q_{n}=\mathrm{Id}-P_{n}\in\mathscr{L}(Y)$, with

$\left\|P_{n}\right\|_{\mathscr{L}(Y)}\leq
C_{0}\quad\text{and}\quad\left\|Q_{n}\right\|_{\mathscr{L}(Y)}\leq C_{0}$ (22)

for some $C_{0}>0$.

Let $G:Y\rightarrow Y$ be given by
$G\left(u^{0},u^{1}\right)=\left(0,(u^{0})^{3}\right)$. Note that $G$ is well-
defined as $H^{1+\alpha}(\Omega)\cap H_{0}^{1}(\Omega)$ is an algebra, as
$\alpha>\frac{1}{2}$. We need to use a smooth version of the damping $\gamma$.
Recall that $\omega$ is an open subset of $\Omega$ satisfying the geometric
control condition, and that $\gamma(x)>\alpha>0$ for almost all $x\in\omega$.
Recall also that $u$ satisfies $\gamma\partial_{t}u=0$ and $\square u+\beta
u=u^{3}$. By continuity of generalized geodesics (see for example Theorem 3.34
of [13]), there exists an open subset $\tilde{\omega}$, compactly included in
$\omega$, such that the geometric control condition holds for
$\tilde{\omega}$. Fix
$\tilde{\gamma}\in\mathscr{C}_{\mathrm{c}}^{\infty}(\omega,[0,1])$ such that
$\tilde{\gamma}=1$ on $\tilde{\omega}$. We have
$\tilde{\gamma}\partial_{t}u=0$, and so

$\square u+\tilde{\gamma}\partial_{t}u+\beta u=u^{3}.$

Let $\tilde{A}$ be the infinitesimal generator of the linear part of this
equation, that is

$A=\begin{pmatrix}0&-1\\\ -\Delta+\beta&0\end{pmatrix}+\begin{pmatrix}0&0\\\
0&\tilde{\gamma}\end{pmatrix}=A_{1}+B_{1}.$

Now, we check that the assumptions of the previous theorem are satisfied. One
has $U\in L^{\infty}(\mathbb{R},X^{1})$, and as $\alpha<1$ the embedding
$X^{1}\rightarrow X^{\alpha}$ is compact by the Rellich theorem, so _(i)_ is
clear. The fact that for all $y\in Y$, $P_{n}y$ converges to $y$ when $n$ goes
to infinity is obvious when the norm of $Y$ is expressed as a Fourier series.
Together with (22), this gives _(ii)_. As $\tilde{\gamma}$ is smooth, _(iii)_
is clear. We have already used the fact that _(iv)_ is true (see Theorem 12).
For $\left(u^{0},u^{1}\right)\in Y$ and $(v_{0},v_{1})\in Y$, the function

$z\in\mathbb{C}\longmapsto
G\left(\left(u^{0},u^{1}\right)+z(v_{0},v_{1})\right)=\left(0,(u_{0}+zv_{0})^{3}\right)$

is well-defined and analytic, as $H^{1+\alpha}(\Omega)\cap H_{0}^{1}(\Omega)$
is an algebra, so _(v)_ is true.

Finally, we check that _(vi)_ holds. We have

$\left\\{DG(U(t))V,t\in\mathbb{R},\|V\|_{Y}\leq
1\right\\}=\left\\{\left(0,u(t)^{2}v_{0}\right),t\in\mathbb{R},\|v_{0}\|_{H^{1+\alpha}\cap
H_{0}^{1}}\leq 1\right\\}.$

Take $t\in\mathbb{R}$ and $v_{0}\in H^{1+\alpha}(\Omega)\cap
H_{0}^{1}(\Omega)$ such that $\|v_{0}\|_{H^{1+\alpha}\cap H_{0}^{1}}\leq 1$.
For $\varepsilon>0$ sufficiently small, using again the fact that
$H^{1+\alpha}(\Omega)\cap H_{0}^{1}(\Omega)$ is an algebra, we can write

$\left\|u(t)^{2}v_{0}\right\|_{H_{0}^{\alpha+\varepsilon}}\leq\left\|u(t)^{2}v_{0}\right\|_{H^{1+\alpha}\cap
H_{0}^{1}}\lesssim\left\|u(t)\right\|_{H^{1+\alpha}\cap
H_{0}^{1}}^{2}\left\|v_{0}\right\|_{H^{1+\alpha}\cap
H_{0}^{1}}\lesssim\left\|u\right\|_{L^{\infty}(\mathbb{R},X^{1})}^{2}.$

The embedding $H_{0}^{\alpha+\varepsilon}(\Omega)\rightarrow
H_{0}^{\alpha}(\Omega)$ is compact by the Rellich theorem, so _(vi)_ is true.
The previous theorem gives (18).

Proof of (19). We have $\partial_{t}^{k}u(t)\in H^{1+\alpha}(\Omega)\cap
H_{0}^{1}(\Omega)$ for all $k\in\mathbb{N}$ and all
$\alpha\in\left(\frac{1}{2},1\right)$. This will give (19) by standard
elliptic regularity properties and a bootstrap argument. More precisely, we
use the following result, which can be found (for example) in [7], Theorem
9.19. Note that for $\ell\in\mathbb{N}$ and $\lambda\in(0,1)$, the norm of
$\mathscr{C}^{\ell,\lambda}(\overline{\Omega})$ is given by

$\|v\|_{\mathscr{C}^{\ell,\lambda}}=\max_{|\beta|\leq\ell}\left(\|\partial_{x}^{\beta}v\|_{L^{\infty}(\overline{\Omega})}+\sup_{x,y\in\overline{\Omega},x\neq
y}\frac{\left|\partial_{x}^{\beta}v(x)-\partial_{x}^{\beta}v(y)\right|}{\left|x-y\right|^{\lambda}}\right),\quad
v\in\mathscr{C}^{\ell,\lambda}(\overline{\Omega}).$

######  Theorem 26.

Let $\mathcal{O}$ be a smooth Riemannian manifold, with or without boundary.
Let $u\in H^{2}(\mathcal{O})$ be such that $\Delta
u=f\in\mathscr{C}^{\ell,\lambda}(\overline{\mathcal{O}})$, with
$\ell\in\mathbb{N}$ and $\lambda\in(0,1)$. Then
$u\in\mathscr{C}^{\ell+2,\lambda}\left(\overline{\mathcal{O}}\right)$.

Set $\alpha=\frac{3}{4}$. We will use the fact that there exists
$\lambda\in(0,1)$ such that the Sobolev embedding
$H^{1+\alpha}(\Omega)\hookrightarrow\mathscr{C}^{0,\lambda}(\overline{\Omega})$
holds true (see for example [1], 4.12, Part II with $n=3$, $p=2$ and $j=1$).
We prove by induction on $\ell\in\mathbb{N}$ that for all
$\phi\in\mathscr{C}_{\mathrm{c}}^{\infty}(\mathbb{R})$ and all
$k\in\mathbb{N}$,

$\phi\partial_{t}^{k}u\in\mathscr{C}^{2\ell,\lambda}\left(\mathbb{R}\times\overline{\Omega}\right).$
(23)

We start with $\ell=0$. Consider
$\phi\in\mathscr{C}_{\mathrm{c}}^{\infty}(\mathbb{R})$, $k\in\mathbb{N}$, and
let $I\subset\mathbb{R}$ be a compact interval such that
$\operatorname{supp}\phi\subset I$. One has $\partial_{t}^{k}u$,
$\partial_{t}^{k+1}u\in\mathscr{C}^{0}(I,H^{1+\alpha}(\Omega))$, implying

$\partial_{t}^{k}u,\partial_{t}^{k+1}u\in
L^{\infty}(I,\mathscr{C}^{0,\lambda}(\overline{\Omega}))$

by the Sobolev embedding mentioned above. For $x,y\in\overline{\Omega}$,
$t,t^{\prime}\in I$, one has

$\displaystyle\left|\phi(t)\partial_{t}^{k}u(t,x)-\phi(t^{\prime})\partial_{t}^{k}u(t^{\prime},y)\right|$
$\displaystyle\leq\ $
$\displaystyle\left|\phi(t)\right|\left|\partial_{t}^{k}u(t,x)-\partial_{t}^{k}u(t,y)\right|+\left|\int_{t^{\prime}}^{t}\partial_{t}\left(\phi\partial_{t}^{k}u\right)(s,y)\mathrm{d}s\right|$
$\displaystyle\lesssim\ $
$\displaystyle\left\|\partial_{t}^{k}u\right\|_{L^{\infty}(I,\mathscr{C}^{0,\lambda})}\left|x-y\right|^{\lambda}+\left(\left\|\partial_{t}^{k}u\right\|_{L^{\infty}(I,\mathscr{C}^{0,\lambda})}+\left\|\partial_{t}^{k+1}u\right\|_{L^{\infty}(I,\mathscr{C}^{0,\lambda})}\right)\left|t-t^{\prime}\right|$

yielding (23) for $\ell=0$.

Now, let $\ell\in\mathbb{N}$ be such that (23) holds true. As above, consider
$\phi\in\mathscr{C}_{\mathrm{c}}^{\infty}(\mathbb{R})$, $k\in\mathbb{N}$, and
$I\subset\mathbb{R}$ a compact interval such that
$\operatorname{supp}\phi\subset I$. One has

$\left(\partial_{t}^{2}+\Delta\right)\left(\phi\partial_{t}^{k}u\right)=\partial_{t}^{2}\left(\phi\partial_{t}^{k}u\right)+\phi\partial_{t}^{k}\left(\partial_{t}^{2}u+\beta
u-u^{3}\right).$

As $\mathscr{C}^{2\ell,\lambda}(\mathbb{R}\times\overline{\Omega})$ is an
algebra, (23) gives
$\left(\partial_{t}^{2}+\Delta\right)\left(\phi\partial_{t}^{k}u\right)\in\mathscr{C}^{2\ell,\lambda}(\mathbb{R}\times\overline{\Omega})$.
One also has

$\left(\partial_{t}^{2}+\Delta\right)\left(\phi\partial_{t}^{k}u\right)\in
L^{2}(I\times\Omega),$

yielding $\phi\partial_{t}^{k}u\in H^{2}(I\times\Omega)$. By Theorem 26,
applied in a smooth open subset of $I\times\Omega$ containing
$\operatorname{supp}\phi\times\Omega$, one obtains (23) for $\ell+1$. This
proves (19).

#### Step 3 : identification of the limit.

Here, we complete the proof of Proposition 22 by showing that $u$ is a
stationary solution of (Introduction). We use Corollary 3.2 of [11] (which is
a consequence of Theorem A of [15]), which we copy here for convenience.

######  Theorem 27 (Corollary 3.2 of [11]).

Let $T\in(0,+\infty]$ and let $b$, $(c_{i})_{i=1,2,3}$ and $d$ be coefficients
in $\mathscr{C}^{\infty}(\Omega\times[0,T),\mathbb{R})$. Assume moreover that
$b$, $c$ and $d$ are analytic in time and that $u$ is a strong solution of

$\partial_{t}^{2}u=\Delta u+b\partial_{t}u+c\cdot\nabla
u+du,\quad(t,x)\in(-T,T)\times\Omega.$

Let $\mathcal{O}$ be a nonempty open subset of $\Omega$ such that $u(x,t)=0$
in $\mathcal{O}\times(-T,T)$. Then $u(x,0)=0$ in

$\mathcal{O}_{T}=\left\\{x\in\Omega,\mathrm{d}(x,\mathcal{O})<T\right\\}$

where the distance $\mathrm{d}(x,\mathcal{O})$ is defined as the infimum of
the lengths of the $\mathscr{C}^{1}-$paths between $x$ and a point of
$\mathcal{O}$.

We apply this theorem with $\mathcal{O}=\omega$ and $T=+\infty$ : the
geometric control condition implies that $\mathcal{O}_{T}=\Omega$ is that
case. As $u\in\mathscr{C}^{\infty}(\mathbb{R}\times\Omega)$, if
$v=\partial_{t}u$ then we have

$\partial_{t}^{2}v=\Delta v-\beta v+3u^{2}v.$

As $u$ is smooth and analytic in time, and as $v=0$ on $\omega$, we find
$v=0$. Hence, $u$ is a stationary solution of (Introduction), and this
completes the proof of Proposition 22. ∎

###  3.2 Convergence of a bounded solution

Here, we prove Theorem 5. Let $\left(u^{0},u^{1}\right)\in
H_{0}^{1}(\Omega)\times L^{2}(\Omega)$ be such that the solution $u$ of
(Introduction) with initial data $\left(u^{0},u^{1}\right)$ exists on
$\mathbb{R}_{+}$. Assume that there exists $C>0$ such that

$\left\|u(t)\right\|_{H_{0}^{1}}+\left\|\partial_{t}u(t)\right\|_{L^{2}}\leq
C$ (24)

for all $t\geq 0$. By (24), the energy of $u$ is bounded. Hence, as the energy
is nonincreasing, there exists $E_{\infty}\leq E\left(u^{0},u^{1}\right)$ such
that

$E\left(u(t),\partial_{t}u(t)\right)\underset{t\rightarrow+\infty}{\longrightarrow}E_{\infty}.$

The energy equality gives

$E\left(u^{0},u^{1}\right)-E_{\infty}=\int_{0}^{\infty}\int_{\Omega}\gamma\left|\partial_{t}u\right|^{2}\mathrm{d}x\mathrm{d}t<\infty$

and this allows us to use Proposition 22. More precisely, let
$(T_{n})_{n\in\mathbb{N}}$ be a time-sequence satisfying
$T_{n}\rightarrow+\infty$. For all $T>0$, we have

$\int_{T_{n}-T}^{T_{n}+T}\int_{\Omega}\gamma\left|\partial_{t}u\right|^{2}\mathrm{d}x\mathrm{d}t\underset{n\rightarrow+\infty}{\longrightarrow}0.$

Using also (24), we see that all the assumptions of Proposition 22 are
satisfied by the constant sequence
$\left(u^{0}_{n},u^{1}_{n}\right)_{n\in\mathbb{N}}=\left(u^{0},u^{1}\right)_{n\in\mathbb{N}}$.
Hence, there exist an increasing function
$\phi:\mathbb{N}\rightarrow\mathbb{N}$ and a stationary solution $w$ of
(Introduction), both depending on the sequence $(T_{n})$, such that for all
$T>0$,

$\sup_{t\in[-T,T]}\left(\left\|u(T_{\phi(n)}+t)-w\right\|_{H_{0}^{1}}+\left\|\partial_{t}u(T_{\phi(n)}+t)\right\|_{L^{2}}\right)\underset{n\rightarrow+\infty}{\longrightarrow}0.$

In particular, this implies

$J(w)=\lim_{n\rightarrow\infty}E\left(u(T_{\phi(n)}),\partial_{t}u(T_{\phi(n)})\right)=E_{\infty}.$

For example, if $E_{\infty}=0$ then we must have $w=0$, and if $E_{\infty}=d$
then $w=\pm Q$.

Now, we assume that

$\text{there is an at most countable number of stationary solution }w\text{
such that }J(w)=E_{\infty}.$ (25)

We show that this implies that $w$ is, in fact, independent of $(T_{n})$.
Assume by contradiction that there exist two sequences $(T_{n})$ and
$(T_{n}^{\prime})$ such that for all $T>0$, we have

$\sup_{t\in[-T,T]}\left(\left\|u(T_{n}+t)-w\right\|_{H_{0}^{1}}+\left\|\partial_{t}u(T_{n}+t)\right\|_{L^{2}}\right)\underset{n\rightarrow+\infty}{\longrightarrow}0$

and

$\sup_{t\in[-T,T]}\left(\left\|u(T_{n}^{\prime}+t)-w^{\prime}\right\|_{H_{0}^{1}}+\left\|\partial_{t}u(T_{n}^{\prime}+t)\right\|_{L^{2}}\right)\underset{n\rightarrow+\infty}{\longrightarrow}0$

where $w$ and $w^{\prime}$ are two distinct stationary solutions of
(Introduction). Fix $\varphi\in L^{2}(\Omega)$ such that

$\int_{\Omega}w\varphi\mathrm{d}x<\int_{\Omega}w^{\prime}\varphi\mathrm{d}x.$

For $t\geq 0$, set $\alpha(t)=\int_{\Omega}u(t,x)\varphi(x)\mathrm{d}x$. The
function $\alpha$ is real and continuous. One has

$\lim_{n\rightarrow\infty}\alpha(T_{n})=\int_{\Omega}w\varphi\mathrm{d}x<\int_{\Omega}w^{\prime}\varphi\mathrm{d}x=\lim_{n\rightarrow\infty}\alpha(T_{n}^{\prime})$

so by the intermediate value theorem, for all
$\ell\in\left[\int_{\Omega}w\varphi\mathrm{d}x,\int_{\Omega}w^{\prime}\varphi\mathrm{d}x\right]$,
there exists a sequence $(t_{n})_{n}$ such that $t_{n}\rightarrow+\infty$ and

$\alpha(t_{n})\underset{n\rightarrow+\infty}{\longrightarrow}\ell.$

Hence, for all such $\ell$, there exists a stationary solution
$w^{\prime\prime}$ such that $J(w^{\prime\prime})=E_{\infty}$ and

$\ell=\int_{\Omega}w^{\prime\prime}\varphi\mathrm{d}x.$

This is a contradiction with (25).

Summarizing, we have proved that there exists a stationary solution $w$ such
that for all sequence $(T_{n})_{n}$ such that $T_{n}\rightarrow+\infty$, there
exists a subsequence $(T_{\phi(n)})_{n}$ such that for all $T>0$, we have

$\sup_{t\in[-T,T]}\left(\left\|u(T_{\phi(n)}+t)-w\right\|_{H_{0}^{1}}+\left\|\partial_{t}u(T_{\phi(n)}+t)\right\|_{L^{2}}\right)\underset{n\rightarrow+\infty}{\longrightarrow}0.$

A basic contradiction argument ends the proof of Theorem 5.

##  4 Stabilisation of global solutions of the damped equation

###  4.1 Case of a positive damping

In the case of a positive damping, it is possible to make a short proof of the
stabilization of solutions of (Introduction) initiated in $\mathscr{K}^{+}$.
That proof is based on the arguments of the proof of Proposition 2.5 of [11]
and on Lemma 15.

######  Theorem 28.

Assume that $\gamma(x)\geq\alpha>0$ on $\Omega$. Then for any $E_{0}\in[0,d)$,
there exists $C>0$ and $\lambda>0$ such that for all
$\left(u^{0},u^{1}\right)\in\mathscr{K}^{+}$ such that
$E\left(u^{0},u^{1}\right)\in[0,E_{0}]$, if $u$ is the solution of
(Introduction) with initial data $\left(u^{0},u^{1}\right)$, then

$\left\|u(t)\right\|_{H_{0}^{1}}+\left\|\partial_{t}u(t)\right\|_{L^{2}}\leq
Ce^{-\lambda t},\quad t\geq 0.$

######  Remark 29.

Note that $C$ and $\lambda$ depend on $E_{0}$. The following result is false :
there exist $C>0$ and $\lambda>0$ such that for
$\left(u^{0},u^{1}\right)\in\mathscr{K}^{+}$ and $t\geq 0$, we have

$\left\|u(t)\right\|_{H_{0}^{1}}+\left\|\partial_{t}u(t)\right\|_{L^{2}}\leq
Ce^{-\lambda t}.$ (26)

Indeed, for $n\in\mathbb{N}^{\ast}$, consider
$\left(u_{n}^{0},u_{n}^{1}\right)=\left(\left(1-\frac{1}{n}\right)Q,0\right)$
and write $u_{n}$ for the solution with initial data
$\left(u_{n}^{0},u_{n}^{1}\right)$. Recall that if $j(\lambda)=J(\lambda Q)$,
then we have $j^{\prime}(\lambda)>0$ for $\lambda\in(0,1)$ and
$j^{\prime}(\lambda)<0$ for $\lambda>1$. In particular, we have
$K(u_{n}^{0})>0$ and $\left(u_{n}^{0},u_{n}^{1}\right)\in\mathscr{K}^{+}$.
Hence, if (26) is true, then there exists $T>0$ such that for all
$n\in\mathbb{N}^{\ast}$, we have

$\left\|u_{n}(T)\right\|_{H_{0}^{1}}\leq\frac{1}{2}\left\|Q\right\|_{H_{0}^{1}}.$

By continuity of the source-to-solution operator (see Lemma 20), we have

$\left\|u_{n}(T)-Q\right\|_{H_{0}^{1}}\underset{n\rightarrow\infty}{\longrightarrow}0$

and this is a contradiction.

###### Proof.

Fix $E_{0}\in[0,d)$, and $\left(u^{0},u^{1}\right)\in\mathscr{K}^{+}$ such
that $E\left(u^{0},u^{1}\right)\leq E_{0}$. For $\varepsilon>0$, we define

$E_{\varepsilon}(t)=E\left(u(t),\partial_{t}u(t)\right)+\varepsilon\int_{\Omega}u(t,x)\partial_{t}u(t,x)\mathrm{d}x.$

Recall that as $K(u(t))\geq 0$, one has

$E\left(u(t),\partial_{t}u(t)\right)\geq\frac{1}{4}\left\|u(t)\right\|_{H_{0}^{1}}^{2}+\frac{1}{2}\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2},\quad
t\geq 0.$

For $\varepsilon>0$ sufficiently small, one finds

$\varepsilon\left|\int_{\Omega}u(t,x)\partial_{t}u(t,x)\mathrm{d}x\right|\leq\frac{1}{2}E\left(u(t),\partial_{t}u(t)\right)$

implying

$\frac{1}{2}E\left(u(t),\partial_{t}u(t)\right)\leq
E_{\varepsilon}(t)\leq\frac{3}{2}E\left(u(t),\partial_{t}u(t)\right),\quad
t\geq 0.$ (27)

Using the energy equality, for $\varepsilon>0$ and $t\geq 0$, one obtains

$E_{\varepsilon}^{\prime}(t)=\int_{\Omega}(\varepsilon-\gamma)\left|\partial_{t}u(t)\right|^{2}\mathrm{d}x-\varepsilon\left\|u(t)\right\|_{H_{0}^{1}}^{2}+\varepsilon\left\|u(t)\right\|_{L^{4}}^{4}-\varepsilon\int_{\Omega}\gamma
u(t)\partial_{t}u(t)\mathrm{d}x.$

As $\gamma\geq\alpha$, this gives

$E_{\varepsilon}^{\prime}(t)\leq(\varepsilon-\alpha)\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}-\varepsilon\left\|u(t)\right\|_{H_{0}^{1}}^{2}+\varepsilon\left\|u(t)\right\|_{L^{4}}^{4}-\varepsilon\int_{\Omega}\gamma
u(t)\partial_{t}u(t)\mathrm{d}x.$

Note that for $t\geq 0$, we have

$J(u(t))\leq E\left(u(t),\partial_{t}u(t)\right)\leq
E\left(u^{0},u^{1}\right)\leq E_{0}<d$

so by Lemma 15 _(i)_ , there exists $C>0$ such that $K(u(t))\geq
C\left\|u(t)\right\|_{H_{0}^{1}}^{2}$ for all $t\geq 0$. This gives

$E_{\varepsilon}^{\prime}(t)\leq(\varepsilon-\alpha)\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}-\varepsilon
C\left\|u(t)\right\|_{H_{0}^{1}}^{2}-\varepsilon\int_{\Omega}\gamma
u(t)\partial_{t}u(t)\mathrm{d}x.$

Writing

$\varepsilon\left|\int_{\Omega}u(t,x)\partial_{t}u(t,x)\mathrm{d}x\right|\lesssim\varepsilon^{\frac{3}{2}}\left\|u(t)\right\|_{H_{0}^{1}}^{2}+\varepsilon^{\frac{1}{2}}\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}$

one finds

$E_{\varepsilon}^{\prime}(t)\leq(\varepsilon-\alpha+C^{\prime}\sqrt{\varepsilon})\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}+\varepsilon\left(C^{\prime}\sqrt{\varepsilon}-C\right)\left\|u(t)\right\|_{H_{0}^{1}}^{2}$

for some $C^{\prime}>0$. Fix $T>0$. For $\varepsilon>0$ sufficiently small,
one has

$E_{\varepsilon}(T)-E_{\varepsilon}(0)\lesssim-\int_{0}^{T}\left(\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}+\left\|u(t)\right\|_{H_{0}^{1}}^{2}\right)\mathrm{d}t,$

and using (27), one finds

$E\left(u(T),\partial_{t}u(T)\right)-3E\left(u(0),\partial_{t}u(0)\right)\lesssim-\int_{0}^{T}E\left(u(t),\partial_{t}u(t)\right)\mathrm{d}t.$

As the energy is decreasing, this gives

$E\left(u(T),\partial_{t}u(T)\right)-3E\left(u(0),\partial_{t}u(0)\right)\lesssim-
TE\left(u(T),\partial_{t}u(T)\right).$

Choosing $T$ sufficiently large, one obtains

$E\left(u(T),\partial_{t}u(T)\right)\leq\mu
E\left(u(0),\partial_{t}u(0)\right)$

for some $\mu\in(0,1)$. As
$\left(u(t),\partial_{t}u(t)\right)\in\mathscr{K}^{+}$ for all $t\geq 0$, we
can iterate this process to get

$E\left(u(nT),\partial_{t}u(nT)\right)\leq\mu^{n}E\left(u^{0},u^{1}\right),\quad
n\in\mathbb{N}.$

This completes the proof. ∎

###  4.2 Case of a damping satisfying the geometric control condition

Here, we prove Theorem 4, using Proposition 22. We spit the proof into two
steps.

#### Step 1 : observability estimate and application of Proposition 22.

To prove Theorem 4, it suffices to show the following observability inequality
: there exist $C>0$ and $T>0$ such that for all
$\left(u^{0},u^{1}\right)\in\mathscr{K}^{+}$ with
$E\left(u^{0},u^{1}\right)\leq E_{0}$, the solution $u$ of (Introduction) with
initial data $\left(u^{0},u^{1}\right)$ satisfies

$E\left(u^{0},u^{1}\right)\leq
C\int_{0}^{T}\int_{\Omega}\gamma(x)\left|\partial_{t}u(t,x)\right|^{2}\mathrm{d}x\mathrm{d}t.$

Indeed, if that observability inequality holds, then using the energy
equality, we find

$E\left(u(T),\partial_{t}u(T)\right)\leq\left(1-\frac{1}{C}\right)E\left(u^{0},u^{1}\right).$

As $\left(u(t),\partial_{t}u(t)\right)\in\mathscr{K}^{+}$ for all $t\geq 0$,
we can iterate this process to get

$E\left(u(nT),\partial_{t}u(nT)\right)\leq\left(1-\frac{1}{C}\right)^{n}E\left(u^{0},u^{1}\right),\quad
n\in\mathbb{N}.$

This proves that $t\mapsto E\left(u(t),\partial_{t}u(t)\right)$ decays
exponentially. As $K(u(t))\geq 0$, one has

$E\left(u(t),\partial_{t}u(t)\right)\geq\frac{1}{4}\left\|u(t)\right\|_{H_{0}^{1}}^{2}+\frac{1}{2}\left\|\partial_{t}u(t)\right\|_{L^{2}}^{2}$

and this gives the conclusion.

We prove the observability inequality by contradiction. We assume that there
exists a sequence $(T_{n})_{n\geq 1}$, satisfying $T_{n}\rightarrow+\infty$,
such that for all $n\in\mathbb{N}$, $n\geq 1$, there exists
$\left(u^{0}_{n},u^{1}_{n}\right)\in\mathscr{K}^{+}$ such that
$E\left(u^{0}_{n},u^{1}_{n}\right)\leq E_{0}$ and

$\int_{0}^{T_{n}}\int_{\Omega}\gamma(x)\left|\partial_{t}u_{n}(t,x)\right|^{2}\mathrm{d}x\mathrm{d}t<\frac{1}{n}E\left(u_{n}^{0},u_{n}^{1}\right).$
(28)

For $n\geq 1$ and $t\geq 0$, as $K(u_{n}(t))\geq 0$, one has

$\left\|u_{n}(t)\right\|_{H_{0}^{1}}^{2}+\left\|\partial_{t}u_{n}(t)\right\|_{L^{2}}^{2}\lesssim
E\left(u_{n}(t),\partial_{t}u_{n}(t)\right)\leq
E\left(u_{n}^{0},u_{n}^{1}\right)\leq E_{0}.$

Write $T_{n}^{\prime}=\frac{T_{n}}{2}$. For all $T>0$, if $n\in\mathbb{N}$ is
sufficiently large, then

$\int_{T_{n}^{\prime}-T}^{T_{n}^{\prime}+T}\int_{\Omega}\gamma\left|\partial_{t}u_{n}\right|^{2}\mathrm{d}x\mathrm{d}t\leq\int_{0}^{T_{n}}\int_{\Omega}\gamma\left|\partial_{t}u_{n}\right|^{2}\mathrm{d}x\mathrm{d}t.$

This gives

$\int_{T_{n}^{\prime}-T}^{T_{n}^{\prime}+T}\int_{\Omega}\gamma\left|\partial_{t}u_{n}\right|^{2}\mathrm{d}x\mathrm{d}t\underset{n\rightarrow+\infty}{\longrightarrow}0,\quad
T>0.$

Hence, we can apply Proposition 22 : there exist an increasing function
$\phi:\mathbb{N}\rightarrow\mathbb{N}$ and a stationary solution $w$ of
(Introduction) such that

$\sup_{t\in[-T,T]}\left(\left\|u_{\phi(n)}(T_{\phi(n)}^{\prime}+t)-w\right\|_{H_{0}^{1}}+\left\|\partial_{t}u_{\phi(n)}(T_{\phi(n)}^{\prime}+t)\right\|_{L^{2}}\right)\underset{n\rightarrow+\infty}{\longrightarrow}0,\quad
T>0.$

As the energy is nondecreasing, one has

$J(w)=\lim_{n\rightarrow\infty}E\left(u_{\phi(n)}(T_{\phi(n)}^{\prime}),\partial_{t}u_{\phi(n)}(T_{\phi(n)}^{\prime})\right)\leq
E_{0}<d.$

Hence, $w$ is a stationary solution satisfying $J(w)\in[0,d)$ : this gives
$w=0$. Using the energy equality, we have

$E\left(u_{n}(T_{n}^{\prime}),\partial_{t}u_{n}(T_{n}^{\prime})\right)\geq
E\left(u_{n}(T_{n}),\partial_{t}u_{n}(T_{n})\right)=E\left(u_{n}^{0},u_{n}^{1}\right)-\int_{0}^{T_{n}}\int_{\Omega}\gamma(x)\left|\partial_{t}u_{n}(t,x)\right|^{2}\mathrm{d}x\mathrm{d}t$

so that (28) gives

$E\left(u_{n}(T_{n}^{\prime}),\partial_{t}u_{n}(T_{n}^{\prime})\right)\geq\left(1-\frac{1}{n}\right)E\left(u_{n}^{0},u_{n}^{1}\right).$

In particular, one obtains

$E\left(u_{\phi(n)}^{0},u_{\phi(n)}^{1}\right)\underset{n\rightarrow+\infty}{\longrightarrow}0.$

For $n\in\mathbb{N}^{\ast}$, set
$\alpha_{n}=\sqrt{E\left(u_{n}^{0},u_{n}^{1}\right)}$. In the next step, we
will consider the equation scaled by $\alpha_{n}$ to get a contradiction.

#### Step 2 : the scaled equation.

For $n\in\mathbb{N}$, set $w_{n}=\frac{u_{n}}{\alpha_{n}}$. For
$n\in\mathbb{N}^{\ast}$, $w_{n}$ is the solution of

$\square w_{n}+\beta w_{n}=\alpha_{n}^{2}w_{n}^{3}$

and one has

$\left\|w_{n}(T_{n})\right\|_{H_{0}^{1}}^{2}+\left\|\partial_{t}w_{n}(T_{n})\right\|_{L^{2}}^{2}=\frac{1}{\alpha_{n}^{2}}\left(\left\|u_{n}(T_{n})\right\|_{H_{0}^{1}}^{2}+\left\|\partial_{t}u_{n}(T_{n})\right\|_{L^{2}}^{2}\right)\gtrsim\frac{E\left(u_{n}(T_{n}),\partial_{t}u_{n}(T_{n})\right)}{\alpha_{n}^{2}}.$

Using the energy equality and (28), one finds

$\left\|w_{n}(T_{n})\right\|_{H_{0}^{1}}^{2}+\left\|\partial_{t}w_{n}(T_{n})\right\|_{L^{2}}^{2}\gtrsim\frac{1}{\alpha_{n}^{2}}\left(E\left(u_{n}^{0},u_{n}^{1}\right)-\frac{1}{n}E\left(u_{n}^{0},u_{n}^{1}\right)\right)=1-\frac{1}{n}\geq\frac{1}{2},\quad
n\geq 2.$ (29)

Recall that $A$ denotes the infinitesimal generator of the linear part of
(Introduction), that $e^{tA}$ is the associated semi-group, and set

$W_{n}=\left(w_{n},\partial_{t}w_{n}\right)\quad\text{ and }\quad
F_{n}=\left(0,\alpha_{n}^{2}w_{n}^{3}\right).$

Using the Duhamel formula, we can write

$\displaystyle W_{n}(T_{n})$
$\displaystyle=e^{T_{n}A}W_{n}(0)+\int_{0}^{T_{n}}e^{(T_{n}-s)A}F_{n}(s)\mathrm{d}s$
$\displaystyle=e^{T_{n}A}W_{n}(0)+\sum_{k=0}^{\lfloor
T_{n}\rfloor-1}e^{(T_{n}-k)A}\int_{0}^{1}e^{-sA}F_{n}(k+s)\mathrm{d}s+\int_{\lfloor
T_{n}\rfloor}^{T_{n}}e^{(T_{n}-s)A}F_{n}(s)\mathrm{d}s.$ (30)

As $K(u_{n}(0))\geq 0$, we have

$\left\|w_{n}(0)\right\|_{H_{0}^{1}}^{2}+\left\|\partial_{t}w_{n}(0)\right\|_{L^{2}}^{2}=\frac{1}{\alpha_{n}^{2}}\left(\left\|u_{n}(0)\right\|_{H_{0}^{1}}^{2}+\left\|\partial_{t}u_{n}(0)\right\|_{L^{2}}^{2}\right)\lesssim\frac{E\left(u_{n}(0),\partial_{t}u_{n}(0)\right)}{\alpha_{n}^{2}}=1.$

Hence, the sequence $\left(W_{n}(0)\right)$ is bounded, and linear
stabilisation (Theorem 12) gives

$e^{T_{n}A}W_{n}(0)\underset{n\rightarrow\infty}{\longrightarrow}0$

in $H_{0}^{1}(\Omega)\times L^{2}(\Omega)$. Next, we write

$\left\|\int_{0}^{1}e^{-sA}F_{n}(k+s)\mathrm{d}s\right\|_{H_{0}^{1}\times
L^{2}}\lesssim\left\|F_{n}(k+\cdot)\right\|_{L^{1}((0,1),H_{0}^{1}\times
L^{2})}=\left\|\alpha_{n}^{2}w_{n}^{3}\right\|_{L^{1}((k,k+1),L^{2})}.$

A Sobolev embedding gives

$\left\|\int_{0}^{1}e^{-sA}F_{n}(k+s)\mathrm{d}s\right\|_{H_{0}^{1}\times
L^{2}}\lesssim\frac{1}{\alpha_{n}}\left\|u_{n}\right\|_{L^{3}((k,k+1),L^{6})}^{3}\lesssim\frac{1}{\alpha_{n}}\left\|\left(u_{n},\partial_{t}u_{n}\right)\right\|_{L^{\infty}((k,k+1),H_{0}^{1}\times
L^{2})}^{3}.$

Using $K(u_{n})\geq 0$ and the fact that the energy is nonincreasing, we find

$\left\|\int_{0}^{1}e^{-sA}F_{n}(k+s)\mathrm{d}s\right\|_{H_{0}^{1}\times
L^{2}}\lesssim\frac{E\left(u_{n}^{0},u_{n}^{1}\right)^{\frac{3}{2}}}{\alpha_{n}}=\sqrt{\alpha_{n}}.$

Hence linear stabilisation gives

$\left\|e^{(T_{n}-k)A}\int_{0}^{1}e^{-sA}F_{n}(k+s)\mathrm{d}s\right\|_{H_{0}^{1}\times
L^{2}}\lesssim e^{\lambda(k-T_{n})}\sqrt{\alpha_{n}}$

where $\lambda>0$ is a constant, implying

$\left\|\sum_{k=0}^{\lfloor
T_{n}\rfloor-1}e^{(T_{n}-k)A}\int_{0}^{1}e^{-sA}F_{n}(k+s)\mathrm{d}s\right\|_{H_{0}^{1}\times
L^{2}}\lesssim\sum_{k=0}^{\lfloor
T_{n}\rfloor-1}e^{\lambda(k-T_{n})}\sqrt{\alpha_{n}}\lesssim\sqrt{\alpha_{n}}.$

Similarly, one has

$\left\|\int_{\lfloor
T_{n}\rfloor}^{T_{n}}e^{(T_{n}-s)A}F_{n}(s)\mathrm{d}s\right\|_{H_{0}^{1}\times
L^{2}}\lesssim\sqrt{\alpha_{n}}.$

As $\alpha_{\phi(n)}\rightarrow 0$, coming back to (30), one finds

$W_{\phi(n)}(T_{\phi(n)})\underset{n\rightarrow\infty}{\longrightarrow}0$

in $H_{0}^{1}(\Omega)\times L^{2}(\Omega)$, a contradiction with (29). This
completes the proof of Theorem 4.

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Perrin Thomas<EMAIL_ADDRESS>
Laboratoire Analyse Géométrie et Application, Institut Galilée - UMR 7539,
CNRS/Université Sorbonne Paris Nord, 99 avenue J.B. Clément, 93430
Villetaneuse, France