# Instability of singular equilibria of a wave kinetic equation

Miguel Escobedo

Departamento de Matemáticas,

Universidad del País Vasco,

Apartado 644, E–48080 Bilbao, Spain.
E-mail<EMAIL_ADDRESS>
Angeliki Menegaki

Department of Mathematics,

Imperial College London,
Exhibition Road, London, SW7 2AZ, UK
E-mail<EMAIL_ADDRESS>

Abstract: We consider the singular Rayleigh-Jeans equilibrium of the $4$-waves
kinetic turbulence equation for the three dimensional Schrödinger equation. We
first show the formation in finite time of a Dirac measure at zero frequency
in the solution of the wave kinetic equation when the initial data has the
form of Rayleigh-Jeans, truncated at large values of the energy. The initial
value problem for the linearization around the singular Rayleigh Jeans
equilibria is then solved in several functional spaces. Then, long time
convergence to a Dirac measure at the origin is described in detail for some
of the solutions. This determines a basin of attraction of the Dirac measure.

##  1 Introduction

The description, within the framework of wave turbulence theory, of large
systems of weakly interacting waves gives rise to the study of nonlinear
kinetic equations, sometimes referred to as kinetic wave equations (KWE).
These equations are considered to describe the evolution of the density
function $n(t,\bf p)$ of waves with wave number $\bf p$ at time $t$. These
equations have a family of equilibrium solutions called Rayleigh Jeans (RJ)
thermodynamic equilibria. They can also have other stationary, non-equilibrium
solutions known as Kolmogorov-Zakharov (KZ) spectra that cancel the nonlinear
collision integral without the integrand being identically zero. These spectra
correspond to non-zero wave action and/or energy fluxes (cf. [26], [12],
[24]).

General stability properties of these stationary solutions may be found in
[26] (Chapter 4) or [2] (Chapter 3), although mainly for the KZ spectra, with
only some short remarks about the RJ equilibria (Section 4.2.3 of [26] or
Section 4.1 of [2]).

###  1.1 The “Schrödinger WKE”.

In what follows, we only consider the WKE for a system of waves governed by
the nonlinear Schrödinger equation in $\mathbb{R}^{3}$, which will sometimes
be referred to as Schrödinger WKE. This is the only case where the kinetic
equation has been rigorously deduced from the corresponding dispersive
equation for waves, but only for a spatially homogeneous and isotropic system.
First at equilibrium in [20], and more recently at non equilibrium, to
arbitrarily large time scales, up to the time where the kinetic equation is
well-posed, in [10], (cf. also [3, 5, 6, 8] for previous results). No such
deduction is known for a non isotropic system of waves, although the
corresponding WKE has been widely considered (cf. [26, 12]).

The $L^{2}$ stability property of RJ equilibria is proved in [23] for the
spatially homogeneous Schrödinger WKE with frequency cut off. There the
linearized operator around a RJ equilibria with $\mu\leq 0$ is shown to be
coercive in $L^{2}$. Local well-posedness in weighted $L^{2}$ and $L^{\infty}$
for more general initial data for wave kinetic equations similar to the
Schrödinger WKE without cut off and with slightly more general dispersion
relations is proved in [18].

On the other hand, for the spatially inhomogeneous WKE, global in time
existence, uniqueness and stability of mild solutions for small initial data
that depend on the space variable and not necessarily isotropic in frequency
are proved in [1]. Their result shows that the equilibrium $n\equiv 0$ is
stable under sufficiently small perturbations with uniform algebraic decay at
infinity both in space and frequencies.

Let us now consider the Schrödinger WKE and examine the stability of RJ
equilibria in the simplified situation of spatially homogeneous and radially
symmetric distributions $n(t,p)$, where $p=|{\bf p}|$. With a slight abuse of
notation, we write $n({\bf p})=n(p)$). The equation in this case reads

$\displaystyle\partial_{t}n(t,p)=\iint_{D(p)}\frac{\min\\{p,p_{1},p_{2},p_{3}\\}}{p}\left(n_{3}n_{4}(n_{1}+n_{2})-n_{1}n_{2}(n_{3}+n_{4})\right)p_{2}p_{3}\,{\rm
d}p_{2}\,{\rm d}p_{3}$ (1.1) $\displaystyle
D(p)=\left\\{(p_{2},p_{3});\,\,p_{2}>0,p_{3}>0,p_{2}+p_{3}>p\right\\},\,\,p_{1}=p_{2}+p_{3}-p.$

Or also, in terms of the new variables $\omega=p^{2}$, $f(\omega)=n(\bf p)$,

$\displaystyle\partial_{t}f=Q(f(t))$ (1.2) $\displaystyle
Q(f(t))=\iint_{D(\omega_{1})}W\left[\left(f_{1}+f_{2}\right)f_{3}f_{4}-\left(f_{3}+f_{4}\right)f_{1}f_{2}\right]\,{\rm
d}\omega_{3}\,{\rm d}\omega_{4},\ \ t>0$ (1.3) $\displaystyle
W=\frac{\min\left\\{\sqrt{\omega_{1}},\sqrt{\omega_{2}},\sqrt{\omega_{3}},\sqrt{\omega_{4}}\right\\}}{\sqrt{\omega_{1}}}\
\ ,\ \ \ \omega_{2}=\omega_{3}+\omega_{4}-\omega_{1}.$ (1.4)

Due to the change to radial variables, the functions $n$ and $f$ do not
describe any more a density of waves and the conservation of action wave and
energy are respectively:

$\displaystyle\int_{0}^{\infty}\\!\\!\\!n(t,p)p^{2}\,{\rm
d}p=\int_{0}^{\infty}\\!\\!\\!n(0,p)p^{2}\,{\rm
d}p,\,\,\,\int_{0}^{\infty}\\!\\!\\!f(t,\omega)\omega^{1/2}\,{\rm
d}\omega=\int_{0}^{\infty}\\!\\!\\!f(0,\omega)\omega^{1/2}\,{\rm
d}\omega,\,\,\forall t>0$ (1.5)
$\displaystyle\int_{0}^{\infty}\\!\\!\\!n(t,p)p^{4}\,{\rm
d}p=\int_{0}^{\infty}\\!\\!\\!n(t,p)p^{4}\,{\rm
d}p,\,\,\,\int_{0}^{\infty}\\!\\!\\!f(t,\omega)\omega^{3/2}\,{\rm
d}\omega=\int_{0}^{\infty}\\!\\!\\!f(0,\omega)\omega^{3/2}\,{\rm
d}\omega,\,\forall t>0.$ (1.6)

The KZ spectra are $Q^{1/3}p^{-3}$ and $P^{1/3}p^{-7/3}$ with $Q$ and $P$ real
constants corresponding to wave action and energy fluxes respectively (cf.
[26], [12]). The radially symmetric RJ equilibria of (1.1) or (1.2) are

$\displaystyle n(p)=\frac{1}{\omega(p)-\mu},\,\,\mu\leq 0$ (1.7)

for which the integrand is identically zero. In all that follows only the case
of (1.7) for $\mu=0$, that we sometimes call singular RJ, is considered
because our arguments based on homogeneity do not apply when $\mu<0$.

###  1.2 Set up and Main results

Our first result shows that singular RJ distributions are unstable equilibria
for the non linear equation (1.2) under some very simple perturbations. For
that purpose the following definition of weak solution of the nonlinear
equation is used. Consider the sub space of non-negative measures on
$[0,\infty)$:

$\displaystyle\mathcal{M}^{\rho}_{+}\left(\left[0,\infty\right)\right)=\left\\{\mu\in\mathcal{M}_{+}([0,\infty);\|\mu\|_{\rho}<\infty\right\\},\,\,\rho\in\mathbb{R},$
$\displaystyle\text{where},\,\|\mu\|_{\rho}=\sup_{R>1}\frac{1}{(1+R)^{\rho}}\frac{1}{R}\int_{\frac{R}{2}}^{R}\mu\left(\,{\rm
d}\omega\right)+\int_{0}^{1}\mu(\,{\rm d}\omega).$

and let $f_{0}$ be a non negative measure such that $g_{0}$, the measure
defined by $g_{0}(\omega)=\sqrt{\omega}f_{0}(\omega)$, satisfies
$g_{0}\in\mathcal{M}^{\rho}_{+}\left(\left[0,\infty\right)\right)$ for some
$\rho<-1$.

We say that a family of measures $f(t,\cdot)$ for $t>0$ is a weak solution of
(1.2) with initial data $f_{0}$ if $g(t,\cdot)$ defined as
$g(t,\omega)=\sqrt{\omega}f(t,\omega)$ satisfies:
1.- $g\in
C([0,\infty);\mathcal{M}^{\rho}_{+}\left(\left[0,\infty\right)\right))$
2.-for all $\varphi\in
C_{0}^{2}\left(\left[0,\infty\right)\times\left[0,\infty\right)\right)$ and
all $t^{*}>0$:

$\displaystyle\int_{\left[0,\infty\right)}g\left(t_{\ast},\omega\right)\varphi\left(t_{\ast},\omega\right)\,{\rm
d}\omega-\int_{\left[0,\infty\right)}g_{0}\varphi\left(0,\omega\right)\,{\rm
d}\omega=\int_{0}^{t_{\ast}}\int_{\left[0,\infty\right)}g\partial_{t}\varphi\,{\rm
d}\omega\,{\rm d}t$
$\displaystyle+\int_{0}^{t_{\ast}}\iiint_{\left[0,\infty\right)^{3}}\frac{g_{1}g_{2}g_{3}\Phi}{\sqrt{\omega_{1}\omega_{2}\omega_{3}}}\times$
$\displaystyle\hskip
56.9055pt\times\left[\varphi\left(\omega_{1}+\omega_{2}-\omega_{3}\right)+\varphi\left(\omega_{3}\right)-\varphi\left(\omega_{1}\right)-\varphi\left(\omega_{2}\right)\right]\,{\rm
d}\omega_{1}\,{\rm d}\omega_{2}\,{\rm d}\omega_{3}\,{\rm d}t$ (1.8)

where
$\Phi=\min\\{\sqrt{\omega_{1}},\sqrt{\omega_{2}},\sqrt{\omega_{3}},\sqrt{\omega_{4}}\\}$,
3.- for all $t>0$,

$\displaystyle\int_{[0,\infty)}g(t,\omega)\,{\rm
d}\omega=\int_{[0,\infty)}g_{0}(\omega)\,{\rm d}\omega,$
$\displaystyle\int_{[0,\infty)}g(t,\omega)\omega\,{\rm
d}\omega=\int_{[0,\infty)}g_{0}(\omega)\omega\,{\rm d}\omega$

and we say that $f$ conserves the mass and the energy.

It was proved in Theorem 2.19 and Proposition 2.28 of [14] that for all
initial data $f_{0}$ as above there exists at least one non negative weak
solution $f$ in the sense of the above definition. Our first result is then,

######  Theorem 1.1.

There exists a numerical constant $C>0$ such that, for all $A>0$, $B>0$,
$\rho<-1$, if
$g_{0}\in\mathcal{M}^{\rho}_{+}\left(\left[0,\infty\right)\right)$,

$f_{0}(\omega)=A\omega^{-1}\mathbbm{1}_{0<\omega<B}+\min\left(A\omega^{-1},\frac{g_{0}(\omega)}{\sqrt{\omega}}\right)\mathbbm{1}_{\omega>B},$

and $f$ is one weak non negative solution of (1.2), then

$g(t,\\{0\\})>0,\,\,\forall t>CA^{-2}.$

where $g$ is defined as $g(t,\omega)=\sqrt{\omega}f(t,\omega)$.

######  Remark 1.2.

Theorem 1.1 shows that for initial data that behave like a singular RJ close
to the origin, a Dirac’s measure is formed after some finite time
$t^{*}=CA^{-2}$ and this time may be as small as desired, depending on the
constant $A$. Local-well posedness of the equation up to time $t_{*}$ in some
space of locally bounded functions is natural to be conjectured and is under
current investigation, following techniques in [17, 16].

######  Remark 1.3.

Theorem 1.1 is very natural in view of part b) of Theorem 6.1 in [21] for the
Nordheim equation, and its proof follows along similar arguments, since for
non-negative density functions the collision integral of the Nordheim’s
equation is bounded from below by the collisional operator of the wave
turbulence equation.

After Theorem 1.1, our motivation to consider the linearization of the
equation (1.2) around the RJ equilibrium $n^{0}(p)=p^{-2}$ is twofold.

The instability of singular RJ equilibrium shown in Theorem 1.1 comes from the
concentration of waves at zero frequency to form a Dirac’s delta at
$\omega(p)=0$ in finite time. Precise information is lacking about how this
condensation phenomena happens (cf. for example [19]). A concentration is
still present in the linearized equation, although it takes there an infinite
time. Indeed, some of the solutions $A$ of the linear equation converge to a
Dirac measure asymptotically as $t\to\infty$, although $A(t)$ remains a
locally bounded function on $(0,\infty)$ for all time. In the linear equation
it is possible to describe this concentration mechanism in detail, to show in
particular that it happens with an explicit exponential rate and to describe
the pointwise behavior of some of the solutions.

On the other hand, the linearized equation around stationary solutions have
proved to be useful in order to solve initial value problems for the nonlinear
equation with a prescribed behavior as $p\to 0$. That is the case still for
equation (1.2) and its linearization around the KZ spectrum $\omega^{-7/6}$ as
used in [17, 16]. It could then be useful in order to obtain local-well
posedness for the equation (1.1) or (1.2).

####  1.2.1 Linearized equation

The linearization of equation (1.1) around the RJ equilibrium
$n^{0}(p)=|p|^{-2}$ has been considered in [13], where the perturbation is
written as follows,

$\displaystyle n(t,p)=n^{0}(p)(1+A(t,p)).$ (1.9)

If $q(n)=n_{3}n_{4}(n_{1}+n_{2})-n_{1}n_{2}(n_{3}+n_{4})$, a simple
computation yields,

$\displaystyle
q(n^{0}(1+A))=A_{1}\left(n^{0}_{3}n^{0}_{4}-n^{0}_{2}n^{0}_{3}-n^{0}_{2}n^{0}_{4}\right)-A_{2}n^{0}_{3}n^{0}_{4}+A_{3}n^{0}_{2}n^{0}_{4}+A_{4}n^{0}_{2}n^{0}_{3}$

where $n^{0}_{2}$ denotes $n^{0}(p_{4}+p_{3}-p_{1})$. If the perturbation $A$
is such that

$\displaystyle\iint_{D(p_{1})}\frac{\min\\{p_{1},p_{2},p_{3},p_{4}\\}}{p_{1}}A(p_{3})n^{0}(p_{4}+p_{3}-p_{1})n^{0}(p_{4})\,{\rm
d}p_{3}\,{\rm d}p_{4}<\infty,$

then by Fubini’s Theorem,

$\displaystyle\iint_{D(p_{1})}$
$\displaystyle\frac{\min\\{p_{1},p_{2},p_{3},p_{4}\\}}{p_{1}}q(n^{0}(1+A_{1}))\,{\rm
d}p_{3}\,{\rm d}p_{4}=$ $\displaystyle=$ $\displaystyle
A_{1}\iint_{D(p_{1})}\frac{\min\\{p_{1},p_{2},p_{3},p_{4}\\}}{p_{1}}\left(n^{0}_{3}n^{0}_{4}-n^{0}_{2}n^{0}_{3}-n^{0}_{2}n^{0}_{4}\right)\,{\rm
d}p_{3}\,{\rm d}p_{4}$
$\displaystyle-\iint_{D(p_{1})}\frac{\min\\{p_{1},p_{2},p_{3},p_{4}\\}}{p_{1}}A_{2}n^{0}_{3}n^{0}_{4}\,{\rm
d}p_{3}\,{\rm d}p_{4}$
$\displaystyle+2\iint_{D(p_{1})}\frac{\min\\{p_{1},p_{2},p_{3},p_{4}\\}}{p_{1}}A_{4}n^{0}_{2}n^{0}_{3}\,{\rm
d}p_{3}\,{\rm d}p_{4}.$ (1.10)

In this case it was shown in [13] that the linearized equation for $A$ may be
written,

$\displaystyle\frac{\partial A(t,p)}{\partial
t}=\gamma_{N}A(t,p)+\int_{0}^{\infty}\mathcal{V}\left(\frac{p}{p^{\prime}}\right)A(t,p^{\prime})\frac{\,{\rm
d}p^{\prime}}{p^{\prime}}$ (1.11)

where $\mathcal{V}$ is an explicit function and $\gamma_{N}$ is an explicit
negative constant, given in the Appendix below.

The findings in [13] regarding the linear equation (1.11) are described in
some detail in the Appendix, Section 4, but can be briefly summarised here as
follows. First, the explicit expressions of the function $\mathcal{V}$ and its
Mellin transform $W_{\mathcal{V}}$ were obtained in terms of special
functions. Solutions of the equation (1.11) were obtained using the Mellin
transform and for some of them, the asymptotic behavior as $p\to 0$ and as
$p\to\infty$, was described by means of the steepest descent method. It was
also observed that for some initial data, all the waves tend to concentrate
exponentially fast at the value $p=0$.

Equation (1.11) is considered here in more strictly mathematical terms, aiming
in particular at establishing the well-posedness of the equation, define and
estimate the linear semigroup defined by the equation, and determine the basin
of attraction of the Dirac measure at the origin. These aspects could not be
addressed in detail in [13] are of interest not only in their own right but
also for considering the nonlinear equation.

Our main results on the linearised equation (1.11) are summarised as follows.
Define the spaces:

$\displaystyle L_{\theta}^{q}\left(0,\infty\right)=\left\\{A\,\text{ is
measurable on}\,(0,\infty);\,\int_{0}^{\infty}|A(p)|^{q}p^{2\theta-1}\,{\rm
d}p<\infty\right\\}$ (1.12) $\displaystyle\mathscr{Y}_{A,B}(0,\infty):=\\{g\in
L^{\infty}_{\text{loc}}(0,\infty);||g||_{A,B}<\infty\\},\,\,\text{where},$
(1.13)
$\displaystyle||g||_{A,B}:=\sup_{0<p<1}(p^{2A}|g(p)|)+\sup_{p>1}(p^{2B}|g(p)|).$

######  Theorem 1.4.

The linear equation (1.11) has the following properties:

  * •

_(Well-Posedness)_ (i) For all non negative $A_{0}\in
L_{\theta}^{q}\left(0,\infty\right)$ with $q\geq 1$ and $\theta\in(-1,5/2)$
there exists a unique $A\in C([0,\infty);L^{q}_{\theta}(0,\infty))\cap
C^{1}((0,\infty);L_{\theta}^{q}(0,\infty))$ such that $A(t)\geq 0$ for all
$t>0$ and satisfying equation (1.11) in $L^{q}_{\theta}(0,\infty)$. Moreover,
for all $T>0$ there exists a constant $C_{T}>0$ satisfying,

$\displaystyle\|A(t)\|_{L^{q}_{\theta}}\leq
C_{T}\|A_{0}\|_{L_{\theta}^{q}},\quad\forall t\in[0,T].$

When $\theta=0$,

$\displaystyle\|A(t)\|_{L^{q}_{0}}\leq e^{\gamma
t}\left(1+\|\mathcal{V}\|_{L^{q}_{0}}\left(e^{\|\mathcal{V}\|_{L^{q}_{0}}t}-1\right)\right)\|A_{0}\|_{L^{q}_{0}},\,\forall
t\geq 0.$

(ii) For all non negative $A_{0}\in\mathscr{Y}_{A,B}$ with
$-\frac{5}{2}<B<-\frac{3}{4}<\frac{1}{2}<A<1$ there exists a unique $A\in
C([0,\infty);\mathscr{Y}_{A,B}\cap C^{1}((0,\infty);\mathscr{Y}_{A,B})$ such
that $A(t)\geq 0$ for all $t>0$ and satisfying equation (1.11) in
$L^{q}_{\theta}(0,\infty)$. For all $T>0$ there exists a constant $C_{T}>0$
satisfying,

$\displaystyle\|A(t)\|_{A,B}\leq C_{T}\|A_{0}\|_{A,B},\quad\forall t\in[0,T].$

  * •

_(Conservation of wave action)_ The solutions conserve the wave action,

$\displaystyle A_{0}\in
L^{1}(0,\infty)\Longrightarrow\int_{0}^{\infty}A(t,p)\,{\rm
d}p=\int_{0}^{\infty}A_{0}(p)\,{\rm d}p,\,\forall t>0.$ (1.14)

  * •

_(Non-Conservation of energy)_ There is growth of the energy along the
solutions: there exists $\kappa>0$ such that

$\displaystyle A_{0}\in L^{1}((0,\infty);p^{2}\,{\rm
d}p)\Longrightarrow\int_{0}^{\infty}A(t,p)p^{2}\,{\rm d}p=e^{\kappa
t}\int_{0}^{\infty}A_{0}(p)p^{2}\,{\rm d}p.$ (1.15)

  * •

_(Long time behaviour)_ If the initial data $A_{0}\in\mathscr{Y}_{A,B}$,
$A_{0}\geq 0$ and $A,B$ satisfy (• ‣ 1.4) then, for all $\varphi\in
C([0,\infty))\cap L^{\infty}((0,\infty))$,

$\displaystyle\lim_{t\to\infty}\int_{0}^{\infty}A(t,p)\varphi(p)\,{\rm
d}p=\varphi(0)\int_{0}^{\infty}A_{0}(p)\,{\rm d}p.$

The last property above shows that all the mass of the perturbation
$n^{0}(p)A(t,p)$ concentrates at the origin as $t\to\infty$.

###  1.3 Organisation of the article

The content of the article is as follows. Section 2 contains the proof of
finite-time condensation as stated in Theorem 1.1. Next, in Section 3 we
proceed with the set-up of proving Theorem 1.4. In particular, Subsection 3.1
contains the description of some properties of functions containing key
information on the linearised problem. In Subsection 3.2 the Cauchy problem
for equation (3.5) is proved to be well-posed in several functional spaces and
conservation of the wave action (cf.(3.7)) is proved to hold for some of the
solutions. Results on the long time behaviour of the solutions for some
initial data are presented in Subsection 3.3.

##  2 Nonlinear instability of the RJ equilibrium

We prove in this Section Theorem 1 and to this end follow closely [21]. Let us
then denote, for $g$ a non negative measure on $[0,\infty)$,

$\displaystyle
N_{0}(g,\varepsilon)=\int_{[0,\varepsilon]}\left(1-\frac{x}{\varepsilon}\right)^{2}\,{\rm
d}g(x),\,\,N_{\alpha}(g,\varepsilon)=\frac{1}{\varepsilon^{\alpha}}N_{0}(g,\varepsilon),$

and,

$\displaystyle\underline{N}_{\alpha}(g,\varepsilon)=\inf_{0<\delta\leq\varepsilon}N_{\alpha}(g,\delta),\,\,\overline{N}_{\alpha}(g,\varepsilon)=\sup_{0<\delta\leq\varepsilon}N_{\alpha}(g,\delta),$
(2.1) $\displaystyle\underline{N}_{\alpha}(g)=\lim_{\varepsilon\to
0^{+}}\underline{N}_{\alpha}(g,\varepsilon),\,\,\overline{N}_{\alpha}(g)=\lim_{\varepsilon\to
0^{+}}\overline{N}_{\alpha}(g,\varepsilon).$ (2.2)

Also we define

$\displaystyle
A_{\alpha}(g,\varepsilon)=\varepsilon^{-\alpha}\int_{[0,\varepsilon]}\left(\frac{x}{\varepsilon}\right)^{2}\,{\rm
d}g(x).$

######  Lemma 2.1.

Suppose that $f$ is a solution of (1.2) in the weak sense (1.8) and $g$ is the
measure such that $g(t,\omega)=\sqrt{\omega}f(t,\omega)$ Then,

  * (i)

For all $0\leq s<t$, $\varepsilon>0$,

$\displaystyle
N_{\alpha}(g(t),\varepsilon)-N_{\alpha}(g(s),\varepsilon)=\varepsilon^{-\alpha}\int_{s}^{t}\mathscr{K}(\varphi_{\varepsilon},g(\tau))\,{\rm
d}\tau,$
$\displaystyle\mathscr{K}(\varphi_{\varepsilon},g(\tau))=\iiint_{\left[0,\infty\right)^{3}}\frac{g_{1}(\tau)g_{2}(\tau)g_{3}(\tau)\Phi}{\sqrt{\omega_{1}\omega_{2}\omega_{3}}}\times$
$\displaystyle\hskip
56.9055pt\times\left[\varphi_{\varepsilon}\left(\omega_{1}+\omega_{2}-\omega_{3}\right)+\varphi_{\varepsilon}\left(\omega_{3}\right)-\varphi_{\varepsilon}\left(\omega_{1}\right)-\varphi_{\varepsilon}\left(\omega_{2}\right)\right]\,{\rm
d}\omega_{1}\,{\rm d}\omega_{2}\,{\rm d}\omega_{3}$
$\displaystyle\quad\varphi_{\varepsilon}(x):=1_{[0,\varepsilon]}(x)\left(1-\frac{x}{\varepsilon}\right)^{2},$

where
$\Phi=\min\\{\sqrt{\omega_{1}},\sqrt{\omega_{2}},\sqrt{\omega_{3}},\sqrt{\omega_{4}}\\}.$

  * (ii)

The functional $\bar{N}_{\alpha}(g,\sigma)$ is non-decreasing in the second
variable:

$\sigma\geq\sigma^{\prime}\Longrightarrow\overline{N}_{\alpha}(g,\sigma)\geq\overline{N}_{\alpha}(g,\sigma^{\prime})$

  * (iii)

For any $g$ a non-negative measure on $[0,\infty)$, for all $\alpha\in[0,1/2]$
and $\varepsilon>0$,

$\underline{N}_{\alpha}(g,\varepsilon)\left[A_{\frac{1-\alpha}{2}}(g,\varepsilon)\right]^{2}\leq\mathscr{K}(\varphi_{\varepsilon},g)$

  * (iv)

For any $g$ a non-negative measure on $[0,\infty)$ with $g(\\{0\\})=0$, for
all $\alpha\geq 0$ and $\varepsilon>0$,

$\sqrt{N_{\alpha}(g,\varepsilon)}\leq\int_{0}^{1}\sigma^{\alpha/2-1}[A_{\alpha}(g,\varepsilon\sigma)]^{1/2}\,{\rm
d}\sigma.$

###### Proof.

The proofs of these properties are contained in [21, 22]. We include them here
for the sake of completeness. The first part (i) follows directly by computing
$\frac{\,{\rm d}}{\,{\rm d}t}N_{\alpha}(g(t),\varepsilon)$ and applying the
definition of a distributional solution of our equation for the test function
$\varphi_{\varepsilon}$. Item (ii) is obvious from the definition of the
functionals. Regarding item (iv): we may neglect the point $x=0$ as
$g(\\{0\\})=0$ and rather work with the functional
$N_{0}(g,\varepsilon)=\int_{(0,\varepsilon]}\left(1-\frac{x}{\varepsilon}\right)^{2}dg(x)$.
We keep the same notation to avoid unnecessary complications. We write

$\displaystyle\frac{\partial}{\partial\varepsilon}N_{0}(g,\varepsilon)$
$\displaystyle=2\int_{0}^{\infty}\frac{\left(1-\frac{x}{\varepsilon}\right)_{+}}{\varepsilon}\frac{x}{\varepsilon}\,{\rm
d}g(x)\leq\frac{2}{\varepsilon}\sqrt{N_{0}(g,\varepsilon)}\sqrt{A_{0}(g,\varepsilon)}$

due to Cauchy-Schwarz. So for all $\kappa>0$,

$\frac{\partial}{\partial\varepsilon}\sqrt{N_{0}(g,\varepsilon)+\kappa}\lesssim\varepsilon^{-1}\sqrt{A_{0}(g,\varepsilon)}.$

Now integrating over $[\kappa,\varepsilon_{0}]$ we get

$\sqrt{N_{0}(g,\varepsilon_{0})+\kappa}-\sqrt{N_{0}(g,\kappa)+\kappa}\lesssim\int_{\kappa}^{\varepsilon_{0}}\varepsilon^{-1}\sqrt{A_{0}(g,\varepsilon)}\,{\rm
d}\varepsilon,$

and sending $\kappa\to 0^{+}$, we get

$\sqrt{N_{0}(g,\varepsilon_{0})}\lesssim\int_{0}^{\varepsilon_{0}}\varepsilon^{-1}\sqrt{A_{0}(g,\varepsilon)}\,{\rm
d}\varepsilon=\int_{0}^{1}\sigma^{-1}\sqrt{A_{0}(g,\varepsilon\sigma)}\,{\rm
d}\sigma.$

Multiply by $\varepsilon^{-\alpha/2}$ for $\alpha>0$ and use that
$A_{0}(g,\varepsilon\sigma)=(\varepsilon\sigma)^{\alpha}A_{\alpha}(g,\varepsilon\sigma)$
to prove the claim. For the item (iii), we first note that the collision
operator may be rewritten as follows by decomposing the frequency space into
$4$ regions:

$\displaystyle\mathscr{K}(\varphi_{\varepsilon},g)=\iiint_{\left[0,\infty\right)^{3}}\frac{g_{1}g_{2}g_{3}\Phi}{\sqrt{\omega_{1}\omega_{2}\omega_{3}}}\left[\varphi_{\varepsilon}\left(\omega_{1}+\omega_{2}-\omega_{3}\right)+\varphi_{\varepsilon}\left(\omega_{3}\right)-\varphi_{\varepsilon}\left(\omega_{1}\right)-\varphi_{\varepsilon}\left(\omega_{2}\right)\right]\,{\rm
d}\omega_{1}\,{\rm d}\omega_{2}\,{\rm d}\omega_{3}$
$\displaystyle=2\int_{0\leq\omega_{1}<\omega_{2}<\omega_{3}}\cdots\
+2\int_{0\leq\omega_{2}<\omega_{1}<\omega_{3}}\cdots\
+\int_{0\leq\omega_{1}<\omega_{3}=\omega_{2}}\cdots\
+\int_{\begin{subarray}{c}\omega_{2}\geq 0\\\
\omega_{3}<\omega_{1}\end{subarray}}\cdots$
$\displaystyle=2\int_{0\leq\omega_{1}<\omega_{2}<\omega_{3}}\frac{\Delta_{\text{sym}}\varphi_{\varepsilon}}{\sqrt{\omega_{2}}\sqrt{\omega_{3}}}\,{\rm
d}g(\omega_{1})\,{\rm d}g(\omega_{2})\,{\rm
d}g(\omega_{3})+\int_{0\leq\omega_{1}<\omega_{3}=\omega_{2}}\cdots\
+\int_{\begin{subarray}{c}\omega_{2}\geq 0\\\
\omega_{3}<\omega_{1}\end{subarray}}\cdots$
$\displaystyle=2\int_{0\leq\omega_{1}<\omega_{2}\leq\omega_{3}}\chi_{\omega_{2}\omega_{3}}\frac{\Delta_{\text{sym}}\varphi_{\varepsilon}}{\sqrt{\omega_{2}\omega_{3}}}\,{\rm
d}g(\omega_{1})\,{\rm d}g(\omega_{2})\,{\rm d}g(\omega_{3})+$ (2.3)
$\displaystyle\hskip 113.81102pt+\int_{\begin{subarray}{c}\omega_{2}\geq 0\\\
\omega_{3}<\omega_{1}\end{subarray}}\sqrt{\frac{(\omega_{2}+\omega_{3}-\omega_{1})_{+}}{\omega_{1}\omega_{2}\omega_{3}}}\Delta\varphi_{\varepsilon}\,{\rm
d}g(\omega_{1})\,{\rm d}g(\omega_{2})\,{\rm d}g(\omega_{3})$ (2.4)

where in the third line we used the symmetries between $\omega_{1}$ and
$\omega_{2}$, $\chi_{\omega_{2}\omega_{3}}$ is defined by
$\chi_{\omega_{2}\omega_{3}}=21_{\omega_{2}<\omega_{3}}+1_{\omega_{2}=\omega_{3}}$
and
$\Delta_{\text{sym}}\varphi_{\varepsilon}=\varphi_{\varepsilon}(\omega_{3}+\omega_{2}-\omega_{1})+\varphi_{\varepsilon}(\omega_{3}+\omega_{1}-\omega_{1})-2\varphi_{\varepsilon}(\omega_{3})$
while
$\Delta\varphi_{\varepsilon}=\varphi_{\varepsilon}(\omega_{1})+\varphi_{\varepsilon}((\omega_{2}+\omega_{3}-\omega_{1})_{+})-\varphi_{\varepsilon}(\omega_{3})-\varphi_{\varepsilon}(\omega_{2})$.
We also notice that

$\Delta_{\text{sym}}\varphi_{\varepsilon}=(\omega_{2}-\omega_{1})^{2}\iint_{[0,1]^{2}}\varphi_{\varepsilon}^{{}^{\prime\prime}}(\omega_{3}+(s-t)(\omega_{2}-\omega_{1}))\,{\rm
d}s\,{\rm d}t,\ \forall\ 0<\omega_{1}\leq\omega_{2}\leq\omega_{3},$

and since
$\varphi_{\varepsilon}^{{}^{\prime\prime}}(x)=2\varepsilon^{-2}1_{[0,1]}(\varepsilon^{-1}x)$
for all $0\leq x<\varepsilon$, we have the lower bound

$\Delta_{\text{sym}}\varphi_{\varepsilon}\geq(\omega_{2}-\omega_{1})^{2}\varepsilon^{-2},\
\forall\ 0<\omega_{1}\leq\omega_{2}\leq\omega_{3}\leq\varepsilon.$

We use these to lower bound the operator in (2.3) and (2.4). Of course for
both of them we have a positive sign for free, due to the convexity of the
test function $\varphi_{\varepsilon}$. We use the positive sign to bound
(2.4). While for (2.3) we write

$\displaystyle\eqref{eq: K_1}$
$\displaystyle\geq\varepsilon^{-2}\int_{0\leq\omega_{1}<\omega_{2}\leq\omega_{3}<\varepsilon}\chi_{\omega_{2}\omega_{3}}\frac{(\omega_{2}-\omega_{1})^{2}}{\sqrt{\omega_{2}\omega_{3}}}\,{\rm
d}g(\omega_{1})\,{\rm d}g(\omega_{2})\,{\rm d}g(\omega_{3})$
$\displaystyle=\varepsilon^{-2}\int_{0<\omega_{2}\leq\omega_{3}<\varepsilon}\chi_{\omega_{2}\omega_{3}}\frac{\omega_{2}^{\frac{3}{2}+\alpha}}{\sqrt{\omega_{3}}}\left[\omega_{2}^{-\alpha}\int_{0}^{\omega_{2}}\left(1-\frac{\omega_{1}}{\omega_{2}}\right)^{2}\,{\rm
d}g(\omega_{1})\right]\,{\rm d}g(\omega_{2})\,{\rm d}g(\omega_{3})$
$\displaystyle\geq\varepsilon^{-2}\underline{N}_{\alpha}(g,\varepsilon)\int_{0<\omega_{2}\leq\omega_{3}<\varepsilon}\chi_{\omega_{2}\omega_{3}}\frac{\omega_{2}^{\frac{3}{2}+\alpha}}{\sqrt{\omega_{3}}}\,{\rm
d}g(\omega_{2})\,{\rm d}g(\omega_{3})$
$\displaystyle\geq\underline{N}_{\alpha}(g,\varepsilon)\left[\varepsilon^{\frac{\alpha-1}{2}}\int_{0}^{\varepsilon}\left(\frac{\omega_{2}}{\varepsilon}\right)^{\frac{3}{2}+\alpha}\,{\rm
d}g(\omega_{2})\right]^{2}\geq\underline{N}_{\alpha}(g,\varepsilon)\left[A_{\frac{1-\alpha}{2}}(g,\varepsilon)\right]^{2},$

where in the last line we employed the more general inequality (due to the
definition of $\chi_{\omega_{2}\omega_{3}}$):

$\left[\int_{0}^{\varepsilon}\omega_{2}^{\gamma}\,{\rm
d}g(\omega_{2})\right]^{2}\leq\varepsilon^{\gamma+\beta}\iint_{0<\omega_{2}\leq\omega_{3}<\varepsilon}\chi_{\omega_{2}\omega_{3}}\omega_{2}^{\gamma}\omega_{3}^{-\beta}\,{\rm
d}g(\omega_{2})\,{\rm d}g(\omega_{3})$

for positive $\varepsilon,\gamma,\beta$, here applied to $\beta=1/2$ and
$\gamma=\alpha+3/2$. For the last bound, we also used that $\alpha\leq 1/2$.
In total we have

$\mathscr{K}(\varphi_{\varepsilon},g)\geq\underline{N}_{\alpha}(g,\varepsilon)\left[A_{\frac{1-\alpha}{2}}(g,\varepsilon)\right]^{2},$

as required. ∎

######  Proposition 2.2.

Suppose that $f$ is a non negative solution of (1.2) in the weak sense (1.8)
that conserves the mass and energy and $g$ is the measure such that
$g(t,\omega)=\sqrt{\omega}f(t,\omega)$, with initial data $f_{0}$. Then, there
exists a numerical constant $C>0$ such that

$\displaystyle g(t,\\{0\\})>0,\,\,\,\forall
t>t_{*}:=C\left(\underline{N}_{1/2}(g_{0})\overline{N}_{1/2}(g_{0})\right)^{-1}.$

###### Proof.

We are going to show that there exists a numerical constant $C>0$ such that if
$g(t,\\{0\\})=0$ for all $t\in[0,T]$, then

$\displaystyle\underline{N}_{1/2}(g_{0})\overline{N}_{1/2}(g_{0})\leq
CT^{-1}.$ (2.5)

We first observe that by definition for any $\lambda\in(1/2,1)$, there exists
$\varepsilon_{\lambda}>0$ so that
$\underline{N}_{1/2}(g_{0},\varepsilon)>\lambda\underline{N}_{1/2}(g_{0})$ for
all $\varepsilon\in(0,\varepsilon_{\lambda}].$ Thus combined with the item
(ii) in the Lemma 2.1,

$\underline{N}_{1/2}(g(t),\varepsilon)\geq\underline{N}_{1/2}(g_{0},\varepsilon)>\lambda\underline{N}_{1/2}(g_{0}).$

Moreover from item (i) and (iii) of the same Lemma 2.1,

$\displaystyle N_{\alpha}(g(t),\varepsilon)$
$\displaystyle\geq\varepsilon^{-\alpha}\int_{s}^{t}\mathscr{K}(\varphi_{\varepsilon},g(\tau))\,{\rm
d}\tau\geq\int_{s}^{t}\underline{N}_{1/2}(g(\tau),\varepsilon)\left[A_{\frac{1}{4}+\frac{\alpha}{2}}(g(\tau),\varepsilon)\right]^{2}\,{\rm
d}\tau.$
$\displaystyle\geq\underline{N}_{1/2}(g_{0},\varepsilon)\int_{s}^{t}\left[A_{\frac{1}{4}+\frac{\alpha}{2}}(g(\tau),\varepsilon)\right]^{2}d\tau>\lambda\underline{N}_{1/2}(g_{0})\int_{s}^{t}\left[A_{\frac{1}{4}+\frac{\alpha}{2}}(g(\tau),\varepsilon)\right]^{2}\,{\rm
d}\tau.$

In other words, for all $\varepsilon\in(0,\varepsilon_{\lambda}]$:

$\displaystyle\int_{s}^{t}\left[A_{\frac{1}{2}}(g(\tau),\varepsilon)\right]^{2}\,{\rm
d}\tau<\frac{N_{\alpha}(g(t),\varepsilon)}{\lambda\underline{N}_{1/2}(g_{0})}.$
(2.6)

Now we employ item (iv) of the Lemma 2.1 to get that for all $\tau\in[s,t]$

$\displaystyle\sqrt{N_{\frac{1}{4}+\frac{\alpha}{2}}(g(\tau),\varepsilon)}\leq\int_{0}^{1}\sigma^{\frac{1}{8}+\frac{\alpha}{4}-1}\left[A_{\frac{1}{4}+\frac{\alpha}{2}}(g(\tau),\varepsilon\sigma)\right]^{1/2}\,{\rm
d}\sigma.$

This implies

$\displaystyle\int_{s}^{t}N_{\frac{1}{4}+\frac{\alpha}{2}}(g(\tau),\varepsilon)\,{\rm
d}\tau\leq\int_{s}^{t}\left(\int_{0}^{1}\sigma^{\frac{1}{8}+\frac{\alpha}{4}-1}\left[A_{\frac{1}{4}+\frac{\alpha}{2}}(g(\tau),\varepsilon\sigma)\right]^{1/2}\,{\rm
d}\sigma\right)^{2}\,{\rm d}\tau$
$\displaystyle\lesssim\int_{s}^{t}\int_{0}^{1}\sigma^{\frac{1}{8}+\frac{\alpha}{4}-1}A_{\frac{1}{4}+\frac{\alpha}{2}}(g(\tau),\varepsilon\sigma)\,{\rm
d}\sigma\,{\rm
d}\tau=\int_{0}^{1}\sigma^{\frac{1}{8}+\frac{\alpha}{4}-1}\int_{s}^{t}A_{\frac{1}{4}+\frac{\alpha}{2}}(g(\tau),\varepsilon\sigma)\,{\rm
d}\tau\,{\rm d}\sigma$
$\displaystyle\lesssim\left(\frac{1}{2}+\alpha\right)^{-1}\int_{0}^{1}\sigma^{\frac{1}{8}+\frac{\alpha}{4}-1}\sqrt{t-s}\left(\int_{s}^{t}A_{\frac{1}{4}+\frac{\alpha}{2}}(g(\tau),\varepsilon\sigma)^{2}\,{\rm
d}\tau\right)^{1/2}\,{\rm d}\sigma$
$\displaystyle\lesssim\left(\frac{1}{2}+\alpha\right)^{-1}\int_{0}^{1}\sigma^{\frac{1}{8}+\frac{\alpha}{4}-1}\sqrt{t-s}\sqrt{\frac{N_{\alpha}(g(t),\varepsilon)}{\lambda\underline{N}_{1/2}(g_{0})}}\,{\rm
d}\sigma\lesssim\left(\frac{1}{2}+\alpha\right)^{-2}\sqrt{t-s}\sqrt{\frac{\overline{N}_{\alpha}(g(t),\varepsilon)}{\lambda\underline{N}_{1/2}(g_{0})}},$

where we applied twice Cauchy-Schwarz, Fubini, the estimate (2.6) and also
utilised item (ii). Due to the monotonicity properties, this yields for all
$0\leq s<t\leq T$:

$\displaystyle
N_{\frac{1}{4}+\frac{\alpha}{2}}(g(s),\varepsilon)\leq\frac{C}{\sqrt{t-s}}\sqrt{\frac{\overline{N}_{\alpha}(g(t),\varepsilon)}{\lambda\underline{N}_{1/2}(g_{0})}}\left(\frac{1}{2}+\alpha\right)^{-2},\text{
for all }\varepsilon\in(0,\varepsilon_{\lambda}],$

which due to the monotonicity of $\bar{N}_{\alpha}$ in the second variable,
gives for all $0\leq s<t\leq T$:

$\displaystyle\bar{N}_{\frac{1}{4}+\frac{\alpha}{2}}(g(s),\varepsilon)\leq\frac{C}{\sqrt{t-s}}\sqrt{\frac{\overline{N}_{\alpha}(g(t),\varepsilon)}{\lambda\underline{N}_{1/2}(g_{0})}}\left(\frac{1}{2}+\alpha\right)^{-2},\text{
for all }\varepsilon\in(0,\varepsilon_{\lambda}]$ (2.7)

for some numerical constant $C$ independent of $\alpha,\varepsilon$ and the
time variables. Now since this holds for any $\alpha\in[0,1/2]$ we proceed by
applying the estimate (2.7) iteratively where each step of iteration consists
of $\alpha_{i}$ converging to $1/2$ from below and time step converging to
$T$. To be concrete, we fix $\varepsilon\in(0,\varepsilon_{\lambda}]$, and
consider $\alpha_{n}:=\frac{1}{2}-\frac{1}{2^{n}}$ and
$t_{k}=\left(1-2^{-k}\right)T$, $k\in\mathbb{N},n\in\mathbb{N}^{*}$. The
iteration process is to apply the estimate (2.7) for
$\alpha=\alpha_{n-k},s=t_{k-1}<t_{k}=t$ for $k\in[n-1]$, $n\geq 2$. We then
have for each $k\in[n-1]$,

$\displaystyle
a_{k-1}:=\overline{N}_{\alpha_{n-(k-1)}}(g(t_{k-1}),\varepsilon)$
$\displaystyle\leq\frac{C_{1}2^{k/2}}{\sqrt{T}}\sqrt{\frac{\overline{N}_{\alpha_{n-k}}(g(t_{k}),\varepsilon)}{\lambda\underline{N}_{1/2}(g_{0})}}\left(1-2^{k-n}\right)^{-2}$
$\displaystyle=:b_{k}\sqrt{a_{k}}$

for some numerical constant $C_{1}$ independent of
$\alpha,\varepsilon,\lambda$ and the time variables. We define here
$b_{k}:=\frac{C_{1}2^{k/2}}{\sqrt{T}}\frac{\left(1-2^{k-n}\right)^{-2}}{\sqrt{\lambda\underline{N}_{1/2}(g_{0})}}$.
First thing to observe is that since the mass is conserved, we have
$a_{n-1}\leq m_{0}$, $m_{0}=\int_{[0,\infty)}\,{\rm d}g_{0}$, and thus
$a_{i}<\infty$ for all $i\in[n-1]$. Moreover we have

$\displaystyle
a_{0}=\bar{N}_{\alpha_{n}}(g_{0},\varepsilon)\leq\left[\prod_{k=1}^{n-1}b_{k}^{1/(2^{k-1})}\right]a_{n-1}^{1/(2^{n-1})}\leq\left[\prod_{k=1}^{n-1}b_{k}^{1/(2^{k-1})}\right]m_{0}^{1/(2^{n-1})},$
(2.8)

while the left-hand side is lower bounded by

$N_{\alpha_{n}}(g_{0},\varepsilon)=N_{\frac{1}{2}-\frac{1}{2^{n}}}(g_{0},\varepsilon)=\varepsilon^{2^{-n}}N_{1/2}(g_{0},\varepsilon).$

Inserting this into (2.8) above, we get

$\displaystyle
N_{1/2}(g_{0},\varepsilon)\leq\varepsilon^{-2^{-n}}m_{0}^{1/(2^{n-1})}\left[\prod_{k=1}^{n-1}\left(C_{1}2^{k/2}\frac{\left(1-2^{k-n}\right)^{-2}}{\sqrt{T\lambda\underline{N}_{1/2}(g_{0})}}\right)^{1/(2^{k-1})}\right].$
(2.9)

Finally we pass to the limit as $n\to\infty$, the product in the right-hand
side of (2.9) is converging to

$\prod_{k\geq
1}C_{2}^{2/2^{k}}2^{k/2^{k}}\lim_{n}\prod_{k=1}^{n-1}\left(1-2^{k-n}\right)^{-2^{2-k}}\sim(2C_{2})^{2},$

where $C_{2}=C_{1}(T\lambda\underline{N}_{1/2}(g_{0}))^{-1/2}$. Therefore,

$N_{1/2}(g_{0},\varepsilon)\leq\frac{C_{3}}{T\lambda\underline{N}_{1/2}(g_{0})},$

which implies after taking the $\limsup$ in $\varepsilon$ in the left-hand
side that

$\bar{N}_{1/2}(g_{0})\leq\frac{C_{3}}{T\lambda\underline{N}_{1/2}(g_{0})}$

or since all these quantities are finite, that

$\bar{N}_{1/2}(g_{0})\underline{N}_{1/2}(g_{0})\leq\frac{C_{3}}{T\lambda},$

which implies (2.5) since $\lambda$ is arbitrary in $(1/2,1)$. ∎

######  Corollary 2.3.

There exists a numerical constant $C>0$ such that, for all $A>0$, $B>0$, and
$g_{0}\in\mathcal{M}^{\rho}_{+}\left(\left[0,\infty\right)\right)$ for some
$\rho<-1$, if

$f_{0}(\omega)=A\omega^{-1}\mathbbm{1}_{0<\omega<B}+\min(A\omega^{-1},g_{0}(\omega))\mathbbm{1}_{\omega>B},$

there exists a non negative solution $f$ of (1.2) in the weak sense (1.8) that
conserves the mass and energy and such that,

$g(t,\\{0\\})>0,\,\,\forall t>CA^{-2}.$

######  Remark 2.4.

It is not known yet if the solutions $f$ described in Corollary (2.3) are such
that $g(t,\\{0\\})>0$ for all $t>0$ or if some of them remain a regular
measure for some time. In that case it would also be interesting to know the
behavior of $f(t,\omega)$ as $\omega\to 0$ and compare with [14, Theorem 2.33
] for perturbations of the KZ spectrum $\omega^{-7/6}$.

##  3 The linearized equation in Fourier variables

It is useful to formulate the problem in the new variables:

$\displaystyle\omega=p^{2}=e^{x},\,\,p=e^{\frac{x}{2}};\,\,x=2\log p,$ (3.1)

which then implies $2\,{\rm d}p=e^{\frac{x}{2}}\,{\rm d}x=p\,{\rm d}x$ or that
$\frac{\,{\rm d}p}{p}=\frac{\,{\rm d}x}{2}.$ The new unknown in the Fourier
variables is denoted by

$\displaystyle A(t,p)=f(t,x).$ (3.2)

Moreover, we define

$\displaystyle
K(x)=\mathcal{V}(p)=\mathcal{V}\left(e^{\frac{x}{2}}\right)\Longrightarrow\mathcal{V}\left(\frac{p}{p^{\prime}}\right)=\mathcal{V}\left(e^{\frac{x}{2}-\frac{x^{\prime}}{2}}\right)=\mathcal{V}\left(e^{\frac{x-x^{\prime}}{2}}\right)=K(x-x^{\prime}).$
(3.3)

In the new variables the equation (1.11) reads,

$\displaystyle\frac{\partial f(t,x)}{\partial
t}=\gamma_{N}f(t,x)+\frac{1}{2}\int_{\mathbb{R}}K(x-x^{\prime})f(t,x^{\prime})\,{\rm
d}x^{\prime}.$ (3.4)

With a new time variable $\tau=t/2$, and if, with some abuse of notation,
$f(2\tau,x)$ is still denoted $f(t,x)$ equation (3.4) reads,

$\displaystyle\frac{\partial f(t,x)}{\partial t}=\gamma
f(t,x)+\int_{\mathbb{R}}K(x-x^{\prime})f(t,x^{\prime})\,{\rm d}x^{\prime},$
(3.5) $\displaystyle\text{where},\,\,\gamma=2\gamma_{N}.$ (3.6)

In these new variables, the following quantities may have a physical
interpretation:
1.- Wave action of the perturbation $n_{0}(p)A(t,p)$,

$\displaystyle\int_{0}^{\infty}n_{0}(p)A(t,p)p^{2}\,{\rm
d}p=\int_{0}^{\infty}A(t,p)\,{\rm
d}p=\frac{1}{2}\int_{\mathbb{R}}f(t,x)e^{\frac{x}{2}}\,{\rm d}x$ (3.7)

2.- Total energy of the perturbation

$\displaystyle\int_{0}^{\infty}n_{0}(p)A(t,p)p^{4}\,{\rm
d}p=\int_{0}^{\infty}A(t,p)p^{2}\,{\rm
d}p=\frac{1}{2}\int_{\mathbb{R}}f(t,x)e^{\frac{3x}{2}}\,{\rm d}x.$ (3.8)

Equation 1.11 was studied in [13] using the Mellin transform of $\mathcal{V}$,
which is defined as follows

$\displaystyle
W_{\mathcal{V}}(s):=\mathcal{M}(\mathcal{V})(s)=\int_{0}^{\infty}\mathcal{V}(p)p^{s}\frac{\,{\rm
d}p}{p}.$ (3.9)

It was also proved that the function $W(s)=\gamma+W_{\mathcal{V}}(s)$ was
analytic in the strip $\Re e(s)\in(-2,5)$ with double. poles as $s=-2$ and
$s=5$. In the new variables,

$\displaystyle
W_{\mathcal{V}}(s)=\frac{1}{2}\int_{\mathbb{R}}K(x)e^{\frac{sx}{2}}\,{\rm
d}x=\frac{1}{2}\widehat{K}\left(\frac{is}{2}\right),$ (3.10)

where, $\widehat{K}(\xi)$ is the Fourier transform of $K$, defined as

$\displaystyle\widehat{K}(\xi)=\int_{\mathbb{R}}e^{-ix\xi}K(x)\,{\rm d}x.$
(3.11)

We have then the following relations between Mellin and Fourier transforms

$\displaystyle\widehat{K}(\xi)=2W_{\mathcal{V}}\left(-2i\xi\right),\,\,\,W_{\mathcal{V}}(s)=\frac{1}{2}\widehat{K}\left(\frac{is}{2}\right).$
(3.12)

In a similar way as in [13], if Fourier transform may be applied to both sides
of the equation (3.5), it follows

$\displaystyle\frac{\partial\widehat{f}(t,\xi)}{\partial
t}=\gamma\widehat{f}(t,\xi)+\widehat{K}(\xi)\widehat{f}(t,\xi).$ (3.13)

This yields

$\displaystyle\widehat{f}(t,\xi)$
$\displaystyle=\widehat{f}_{0}(\xi)e^{t\Omega(\xi)},\text{ where}$ (3.14)
$\displaystyle\Omega(\xi)$ $\displaystyle=\gamma+\widehat{K}(\xi)$ (3.15)

from where also,

$\displaystyle\Omega(\xi)=2W(-2i\xi),\,\,$ (3.17)

and the function $\Omega$ is well-defined and analytic in the strip $\Im
m(\xi)\in(-1,5/2)$. We denote, for all $t>0$, the following functions

$\displaystyle G(t):=\mathscr{F}^{-1}\left(e^{t\Omega}\right)$ (3.18)
$\displaystyle H(t):=\mathscr{F}^{-1}\left(e^{\widehat{K}t}-1\right),$ (3.19)

where $\mathscr{F}^{-1}$ denotes the inverse Fourier transform defined as,

$\displaystyle\mathscr{F}^{-1}(g)(x)=\frac{1}{2\pi}\int_{\mathbb{R}}e^{ix\xi}g(\xi)d\xi.$

As observed in [13], equation (3.5) has two trivial stationary solutions,

######  Proposition 3.1.

The functions $e^{-x}$ and $e^{-x/2}$ are stationary solutions of equation
(3.5).

###### Proof.

If $A_{k}(p)=p^{-k}$ then $\mathcal{M}(A)(s)=2\pi\delta(i(s-k))$. Since
$W(1)=W(2)=0$ it follows that $\mathcal{M}(A_{k})(s)W(s)=0$ for all $s$. Then
the Mellin image of the solution $\mathcal{M}(A_{k})(s)$ satisfies
$\mathcal{M}(A_{k})_{t}=\mathcal{M}(A_{k})(s)W(s)=0$. Taking inverse Mellin
transform, $A_{k}$ is a stationary solution of equation (1.11) for $k=1,2$. It
follows that $e^{-\frac{x}{2}}$ and $e^{-x}$ are solutions of (3.5). ∎

###  3.1 Properties of $K$, $\widehat{K}$, $H$ and $G$.

As with the Mellin transform for equation (1.11), to solve the Cauchy problem
for (3.5) with general initial data, use of the inverse Fourier transform
applied to equation (3.15) appears to be the most direct approach. This method
relies on certain properties of $K$, $\widehat{K}$ and $H$. Some of them
directly follow from those of $\mathcal{V}$ and $W_{\mathcal{V}}$ obtained in
[13], others are demonstrated in what follows in this subsection.

Even though the necessary details for this article are presented in the first
part of the Appendix 4, let us summarise here briefly the basic properties.
The function $\mathcal{V}\in C^{\infty}((0,1)\cup(1,\infty))$ has a
singularity at $p=1$ of order $|p-1|^{-1/2}$. It follows from the known
detailed behaviours of $\mathcal{V}(p)$ as $p\to 0$ and $p\to\infty$ that
$\mathcal{V}\in L^{p}(0,\infty)$ for $p\in[1,2)$.

The function $W=\gamma+W_{\mathcal{V}}$, where $W_{\mathcal{V}}$ is the Mellin
transform of $\mathcal{V}$, is meromorphic with two double poles at $s=-2$,
$s=5$, and analytic on the strip $\Re e(s)\in(-2,5)$. Its behaviours as
$s\to-2$ and $s\to 5$ are also known in detail (cf. (4.15), (4.16)).

The corresponding expressions and properties for the functions $K$ follow from
the change of variables $p=e^{x/2}$. The behaviour of $K,K^{\prime}$ and
$K^{\prime\prime}$ as $|x|\to\infty$ and $x\to 0$ are given also in the
Appendix 4. In particular $K\in L^{p}(\mathbb{R})\cap
L^{p}\left(e^{x/2}\right)$ for all $p\in[1,2)$ with
$\|K\|_{L^{p}\left(e^{x/2}\right)}=2\|\mathcal{V}\|_{p}$.

The following property of $H$ easily follows from the integrability properties
of $K$.

######  Proposition 3.2.

For all $t>0$,

$\displaystyle H(t)=\mathscr{F}^{-1}\left(e^{\widehat{K}t}-1\right)\in
L^{p}(\mathbb{R}),\quad\forall\,p\in[1,2)$ and
$\displaystyle\|H(t)\|_{p}\leq\|K\|_{p}\left(e^{\|K\|_{1}t}-1\right).$

###### Proof.

Since $K\in L^{1}(\mathbb{R})$, by the Riemann-Lebesgue Lemma,
$e^{\widehat{K}(\xi)t}-1\to 0$ as $|\xi|\to\infty$. It follows that for each
$t>0$, $e^{\widehat{K}(\cdot)t}-1\in
L^{\infty}(\mathbb{R})\subset\mathscr{S}^{\prime}(\mathbb{R})$ and tends to
zero as $|\xi|\to\infty$. Its inverse Fourier transform is then well defined
and belongs to $\mathscr{S}^{\prime}(\mathbb{R})$. Moreover,

$\displaystyle
e^{\widehat{K}(\xi)t}-1=\sum_{n=1}^{\infty}\frac{\widehat{K}(\xi)^{n}t^{n}}{n!}$

while

$\displaystyle H(t,x)$
$\displaystyle=\mathscr{F}^{-1}\left(e^{\widehat{K}t}-1\right)=\sum_{n=1}^{\infty}\mathscr{F}^{-1}\left(\widehat{K}^{n}\right)\frac{t^{n}}{n!}$
$\displaystyle=\sum_{n=1}^{\infty}(\underbrace{K\ast K\ast K\ast\cdots\ast
K}_{n\,\text{times}})(x)\frac{t^{n}}{n!}.$ (3.20)

Since $K\in L^{p}$ for $p\in[1,2)$, then $K\ast K\in L^{p}(\mathbb{R})$ too
and

$\displaystyle\|K\ast K\|_{p}\leq\|K\|_{1}\|K\|_{p}$

for all $p\in[1,2)$. Then inductively

$\displaystyle\underbrace{K\ast K\ast K\ast\cdots\ast K}_{n\,\text{times}}\in
L^{p}(\mathbb{R}),\,\quad\forall\,p\in[1,2)$

and

$\displaystyle\|\underbrace{K\ast K\ast K\ast\cdots\ast
K}_{n\,\text{times}}\|_{p}\leq\|K\|_{1}^{n}\|K\|_{p}.$

Consequently from (3.20),

$\displaystyle\left\|H(t)\right\|_{p}\leq\|K\|_{p}\sum_{n=1}^{\infty}\frac{\|K\|_{1}^{n}t^{n}}{n!}=\|K\|_{p}\left(e^{\|K\|_{1}t}-1\right).$

This concludes the stated estimate on $H(t)$. ∎

######  Proposition 3.3.

For all compact interval $I\subset(-1,5/2)$ and $\varepsilon>0$, there exist
constants $C_{I,1}$ and $C_{I,2}$ depending on $I$ and $\varepsilon$ such
that, for all $\xi\in\mathbb{C}$ with $\Im m\xi\in I$,

$\displaystyle\left|\widehat{K}(\xi)-\pi\sqrt{2\pi}\xi^{-1/2}+\left(K(\varepsilon)ie^{-i\varepsilon\xi}+K(-\varepsilon)ie^{i\varepsilon\xi}\right)\xi^{-1}\right|\leq
C_{I,1}|\xi|^{-3/2},$ (3.21)

and

$\displaystyle\left|\widehat{K}(\xi)\right|\leq
C_{I,2}(1+\xi)^{-1/2},\,\,\forall\,\xi.$ (3.22)

###### Proof.

Since $K\in L^{1}(\mathbb{R})$,

$\displaystyle\widehat{K}(\xi)=\int_{\mathbb{R}}K(x)e^{-ix\xi}\,{\rm
d}x=\int_{|x|>2}K(x)e^{-ix\xi}\,{\rm d}x+\int_{|x|<2}K(x)e^{-ix\xi}\,{\rm
d}x,$

and for the first term we write

$\displaystyle\int_{2}^{\infty}K(x)e^{-ix\xi}\,{\rm d}x$
$\displaystyle=\frac{i}{\xi}\int_{2}^{\infty}K(x)\left(e^{-ix\xi}\right)_{x}\,{\rm
d}x$
$\displaystyle=-\frac{i}{\xi}\int_{2}^{\infty}K^{\prime}(x)e^{-ix\xi}\,{\rm
d}x+\lim_{x\to\infty}K(x)\frac{ie^{-ix\xi}}{\xi}-K(2)\frac{ie^{-i2\xi}}{\xi}.$

Using integration by parts again gives the existence of a constant $C>0$ such
that for $\Im m\xi\in(-1,5/2)$,

$\displaystyle-\frac{i}{\xi}\int_{2}^{\infty}K^{\prime}(x)e^{-ix\xi}\,{\rm
d}x=-\frac{1}{\xi^{2}}\left(\int_{2}^{\infty}K^{\prime\prime}(x)e^{-ix\xi}\,{\rm
d}x-K^{\prime}(2)e^{-i2\xi}\right)$

and so

$\displaystyle\left|\int_{2}^{\infty}K^{\prime\prime}(x)e^{-ix\xi}\,{\rm
d}x-K^{\prime}(2)e^{-ix\xi}\right|\leq
C\int_{2}^{\infty}|K^{\prime\prime}(x)|e^{x\Im m\xi}\,{\rm
d}x+|K^{\prime}(2)|e^{2\Im m\xi}<C.$

The last bound follows from the behaviour of $K^{\prime\prime}(x)$ at infinity
given in Appendix 4 and the fact that $\Im m\xi<5/2$. By (4.21), when
$x\to-\infty$,

$\displaystyle\left|K(x)\frac{ie^{-ix\xi}}{\xi}\right|\leq
C|\xi|^{-1}|x|e^{x(1+\Im m\xi)},$ (3.23)

and by (4.22) when $x\to\infty$,

$\displaystyle\left|K(x)\frac{ie^{-ix\xi}}{\xi}\right|\leq
C|\xi|^{-1}|x|e^{x(-\frac{5}{2}+\Im m\xi)}.$ (3.24)

Then, for $\xi$ such that $\Im m\xi\in(-1,5/2)$,

$\displaystyle\lim_{x\to\infty}K(x)\frac{ie^{-ix\xi}}{\xi}=\lim_{x\to-\infty}K(x)\frac{ie^{-ix\xi}}{\xi}=0.$
(3.25)

On the other hand, by the behaviour of $K$ near $x=0$ given in (4.23) for
$x\in(0,2)$, $K(x)=\frac{\pi}{\sqrt{|x|}}+\mathcal{O}(x)^{1/2}$, we have that

$\displaystyle\int_{0}^{2}K(x)e^{-ix\xi}dx$
$\displaystyle=\pi\int_{0}^{2}\frac{e^{-ix\xi}\,{\rm
d}x}{\sqrt{x}}+\mathcal{O}\left(\int_{0}^{2}x^{1/2}e^{-ix\xi}\,{\rm
d}x\right),\,|\xi|\to\infty$ $\displaystyle\sim
A\pi\xi^{-\frac{1}{2}}+\mathcal{O}(\xi^{-3/2}),\,\,|\xi|\to\infty$

where

$\displaystyle A:=\int_{0}^{\infty}z^{-\frac{1}{2}}e^{-iz}\,{\rm
d}z=e^{-\frac{\pi i}{4}}\Gamma\left(\frac{1}{2}\right).$

It follows that for each $2>0$, there exists a constant $C>0$ such that for
$\Im m\xi\in(-1,5/2)$,

$\displaystyle\left|\int_{0}^{\infty}K(x)e^{-ix\xi}\,{\rm
d}x-A\pi\xi^{-\frac{1}{2}}+K(2)\frac{ie^{-i2\xi}}{\xi}\right|\leq
C|\xi|^{-3/2},\,\,|\xi|\to\infty.$

After using a similar argument on $(-\infty,0)$:

$\displaystyle\int_{-\infty}^{-\varepsilon}K(x)e^{-ix\xi}\,{\rm
d}x=\frac{i}{\xi}\int_{-\infty}^{-\varepsilon}K(x)\left(e^{-ix\xi}\right)_{x}\,{\rm
d}x$
$\displaystyle=-\frac{i}{\xi}\int_{-\infty}^{-\varepsilon}K^{\prime}(x)e^{-ix\xi}\,{\rm
d}x+K(-2)\frac{ie^{i2\xi}}{\xi}-\lim_{x\to\infty}K(x)\frac{ie^{-ix\xi}}{\xi}$
$\displaystyle=-\frac{i}{\xi}\int_{-\infty}^{-\varepsilon}K^{\prime}(x)e^{-ix\xi}\,{\rm
d}x+K(-2)\frac{ie^{i2\xi}}{\xi},$

where for $\Im m\xi\in(-1,5/2)$,

$\displaystyle-\frac{i}{\xi}\int_{-\infty}^{-2}K^{\prime}(x)e^{-ix\xi}\,{\rm
d}x=-\frac{1}{\xi^{2}}\left(\int_{-\infty}^{-\varepsilon}K^{\prime\prime}(x)e^{-ix\xi}\,{\rm
d}x+K^{\prime}(-2)e^{i2\xi}\right)$

and where

$\displaystyle\left|\int_{-\infty}^{-2}K^{\prime\prime}(x)e^{-ix\xi}\,{\rm
d}x+K^{\prime}(-2)e^{i2\xi}\right|\leq
C\int_{-\infty}^{-2}|K^{\prime\prime}(x)|e^{x\Im m\xi}\,{\rm
d}x+|K^{\prime}(-2)|e^{\varepsilon\Im m\xi}<C.$

One then arrives at the existence of a constant $C>0$ such that for $\Im
m\xi\in(-1,5/2)$,

$\displaystyle\left|\int_{-\infty}^{\infty}K(x)e^{-ix\xi}dx-(A+A^{\prime})\pi\xi^{-\frac{1}{2}}+K(2)\frac{ie^{-i2\xi}}{\xi}+K(-2)\frac{ie^{i2\xi}}{\xi}\right|\leq
C|\xi|^{-3/2},\,|\xi|\to\infty,$

where

$\displaystyle A^{\prime}=\int_{0}^{\infty}z^{-1/2}e^{iz}\,{\rm
d}z=e^{\frac{\pi i}{4}}\Gamma\left(\frac{1}{2}\right).$

It follows since $A+A^{\prime}=\sqrt{2\pi}$, that

$\displaystyle\left|\widehat{K}(\xi)-\pi\sqrt{2\pi}\xi^{-1/2}+K(2)\frac{ie^{-i2\xi}}{\xi}+K(-2)\frac{ie^{i\varepsilon\xi}}{\xi}\right|\leq
C|\xi|^{-3/2}$ (3.26)

which concludes (3.21). Since $\widehat{K}\in L^{\infty}(\mathbb{R})$, (3.22)
follows. ∎

Let us remark now that from the analyticity properties of $W$ in [13] It
follows that $\widehat{K}$ is then analytic in the strip $\Im
m\xi\in(-1,5/2)$, it has poles $\xi=-i$ and $\xi=5i/2$ and

$\displaystyle\widehat{K}(\xi)$
$\displaystyle=2W\left(-2i\xi\right)\sim\frac{2}{(-i\xi+1)^{2}},\,\,\xi\to-i$
$\displaystyle\widehat{K}(\xi)$
$\displaystyle=2W\left(-2i\xi\right)\sim\frac{8}{(2i\xi+5)^{2}},\,\,\xi\to\frac{5i}{2}.$

The behaviour of $H(t)$ for large values of $|x|$ and $t$ in bounded intervals
of $[0,\infty)$ is given in the following result.

######  Corollary 3.4.

Let $H$ be the distribution defined in (3.19). Then for all $t>0$,

$\displaystyle H(t)=tK+\frac{t^{2}(K\ast K)}{2}+\text{continuous and bounded
function on }\,\mathbb{R}.$

Moreover, for any $\beta\in(-3,-1)$, $\alpha\in\left(5/2,9/2\right)$ and
$t>0$,

$\displaystyle H(t,x)-tK(x)+\frac{t^{2}(K\ast
K)(x)}{2}=\begin{cases}\ell_{1}(t)e^{x}+\mathcal{O}\left(t^{3}e^{-\beta
x}\right),\,\,x\to-\infty\\\
\ell_{2}(t)e^{-\frac{5x}{2}}+\mathcal{O}\left(t^{3}e^{-\alpha
x}\right),\,\,x\to\infty\end{cases}$ (3.27)

where

$\displaystyle\ell_{1}(t)=\operatorname{Res}\left(P(t),\xi=-i\right),\,\,\,\ell_{2}(t)=\operatorname{Res}\left(P(t),\xi=-\frac{5i}{2}\right)$

with

$\displaystyle
P(t)=e^{t\widehat{K}}-1-t\widehat{K}-\frac{t^{2}\widehat{K}^{2}}{2}.$

There exist positive constants, $C_{H},D_{H}$, such that for all $t>0$

$\displaystyle|H(t,x)|\leq\begin{cases}C_{H}te^{D_{H}t}e^{x}(1+|x|^{3}),\,\,\forall\
x<0,,\,\,\forall\ t>0.\\\
C_{H}te^{D_{H}t}e^{-\frac{5x}{2}}(1+|x|^{3}),\,\,\forall\ x>0,\,\,\forall\
t>0\end{cases}$ (3.28)

###### Proof.

It follows from (3.22) that for all $t\in I$,

$\displaystyle
P(t,\xi)=e^{t\widehat{K}(\xi)}-1-t\widehat{K}(\xi)-\frac{t^{2}\widehat{K}^{2}(\xi)}{2}=\mathcal{O}\left(t^{3}|\xi|^{-3/2}\right),\,|\xi|\to\infty$

so that $P(t)\in L^{1}(\mathbb{R})\cap L^{\infty}(\mathbb{R})$ and it inverse
Fourier transform is a bounded continuous function. On the other hand

$\displaystyle\mathscr{F}^{-1}\left(t\widehat{K}+\frac{t^{2}}{2}\widehat{K}^{2}\right)=tK+\frac{t^{2}}{2}K\ast
K,$

and by definition of $H$,

$\displaystyle H(t,x)=tK(x)+\frac{t^{2}}{2}(K\ast K)(x)+\text{continuous and
bounded function.}$

For the next part of the Proposition, the behaviour of
$\mathscr{F}^{-1}(P(t))(x)$ as $x\to\pm\infty$ are given by classical
arguments. Indeed,

$\displaystyle\mathscr{F}^{-1}\left(P(t)\right)(x)$
$\displaystyle=\frac{1}{2\pi}\int_{\mathbb{R}}P(t,\xi)e^{ix\xi}\,{\rm d}\xi$

where the integral is absolutely continuous. The function $P(t,\cdot)$ is
analytic in the strip $\Im m\xi\in(-1,5/2)$ with two double poles at $\xi=-i$
and $\xi=5i/2$. It is also integrable since by Proposition 3.3,
$|P(t,\xi)|\leq Ct^{3}(1+|\xi|)^{-3/2}$. Therefore lassical contour
deformation argument and location of the poles of the function $\widehat{K}$
give,

$\displaystyle\int_{\mathbb{R}}P(t,\xi)e^{ix\xi}\,{\rm d}\xi$
$\displaystyle=2\pi i\operatorname{Res}\left(P(t),\xi=-i\right)e^{x}+\int_{\Im
m\xi=\beta}P(t,\xi)e^{ix\xi}\,{\rm d}\xi,\,\,x\to-\infty$

for all $\beta\in(-3,-1)$, and

$\displaystyle\int_{\mathbb{R}}P(t,\xi)e^{ix\xi}\,{\rm d}\xi$
$\displaystyle=2\pi
i\operatorname{Res}\left(P(t),\xi=\frac{5i}{2}\right)e^{-\frac{5x}{2}}+\int_{\Im
m\xi=\alpha}P(t,\xi)e^{ix\xi}\,{\rm d}\xi,\,\,x\to\infty,$

for all $\alpha\in\left(\frac{5}{2},9/2\right)$. Then

$\displaystyle\left|\int_{\Im m\xi=\beta}P(t,\xi)e^{ix\xi}\,{\rm
d}\xi\right|\leq Ce^{-\beta x}\int_{\Im m\xi=\beta}|P(t,\xi)|\,{\rm d}\xi\leq
Ce^{-\beta x}t^{3}$ $\displaystyle\left|\int_{\Im
m\xi=\alpha}P(t,\xi)e^{ix\xi}\,{\rm d}\xi\right|\leq Ce^{-\alpha x}\int_{\Im
m\xi=\alpha}|P(t,\xi)|\,{\rm d}\xi\leq Ce^{-\alpha x}t^{3}$

and (3.27) follows. By (3.12) and (4.15),

$\displaystyle\widehat{K}(\xi)\sim\left(\frac{2}{(1-i\xi)^{2}}+\frac{8}{3(1-i\xi)}+\mathscr{A}(\xi)\right),\,\xi\to-i$
$\displaystyle\mathscr{A}(\xi):\text{analytic
near}\,\,\xi=-i,\,\,\,\mathscr{A}(\xi)=\mathcal{O}(1),\,\,\xi\to-i.$

It follows that for all integers $n>1$,

$\displaystyle\widehat{K}(\xi)^{n}$
$\displaystyle=\left(\frac{2}{(1-i\xi)^{2}}+\frac{8}{3(1-i\xi)}+\mathscr{A}(\xi)\right)^{n},\,\xi\to-i$

and we then compute

$\displaystyle\operatorname{Res}\left(\widehat{K}(\xi)^{n},\xi=-i\right)=\frac{8in}{3}\,\mathscr{A}(-i)^{n-1}.$

Then,

$\displaystyle\sum_{n=3}^{\infty}\frac{t^{n}}{n!}\operatorname{Res}\left(\widehat{K}(\xi)^{n},\xi=-i\right)$
$\displaystyle=\frac{8i}{3}\sum_{n=3}^{\infty}\frac{t^{n}\mathscr{A}(-i)^{n-1}}{(n-1)!}$
$\displaystyle=\frac{8it}{3}\left(e^{\mathscr{A}(-i)t}-1-\mathscr{A}(-i)t\right)$

and, for some constants $C>0$, $D>0$,

$\displaystyle\left|\sum_{n=3}^{\infty}\frac{t^{n}}{n!}\text{Res}\left(\widehat{K}(\xi)^{n},\xi=-i\right)\right|\leq
Ct^{2}e^{Dt},\,\,\forall\ t>0.$ (3.29)

A similar argument analogously gives

$\displaystyle\sum_{n=3}^{\infty}\frac{t^{n}}{n!}\operatorname{Res}\left(P(t),\xi=\frac{5i}{2}\right)$
$\displaystyle=\frac{8it}{3}\sum_{n=3}^{\infty}\frac{t^{n}\mathscr{A}(5i/2)^{n-1}}{(n-1)!}$
$\displaystyle=\frac{8it}{3}\left(e^{\mathscr{A}(5i/2)t}-1-\mathscr{A}(5i/2)t\right)$

and, for some constants $C^{\prime}>0$, $D^{\prime}>0$,

$\displaystyle\left|\sum_{n=3}^{\infty}\frac{t^{n}}{n!}\operatorname{Res}\left(P(t),\xi=\frac{5i}{2}\right)\right|\leq
C^{\prime}t^{2}e^{D^{\prime}t},\,\,\forall\ t>0.$ (3.30)

On the other hand, $K\ast K$ is easily estimated as follows. For all $x>0$,

$\displaystyle|(K\ast K)(x)|\leq$ $\displaystyle
Ce^{-\frac{5x}{2}}\int_{-\infty}^{0}e^{\frac{7y}{2}}(x-y)y\,{\rm
d}y+Ce^{-\frac{5x}{2}}\int_{0}^{x}(x-y)y\,{\rm d}y$
$\displaystyle+Ce^{x}\int_{x}^{\infty}e^{-\frac{7y}{2}}(x-y)y\,{\rm d}y\leq
Ce^{-\frac{5x}{2}}(1+x)+Ce^{-\frac{5x}{2}}x^{3}.$

For $x<0$,

$\displaystyle|(K\ast K)(x)|\leq
Ce^{-\frac{5x}{2}}\int_{-\infty}^{x}e^{\frac{7y}{2}}(x-y)y\,{\rm
d}y+Ce^{x}\int_{x}^{0}(x-y)y\,{\rm d}y+$
$\displaystyle+Ce^{x}\int_{0}^{\infty}e^{-\frac{7y}{2}}(x-y)y\,{\rm d}y\leq
Ce^{x}(1+|x|)+Ce^{x}|x|^{3}.$

This ends the proof of (3.28), and of Corollary 3.4. ∎

Before we conclude this subsection, let us state and prove one more result
that provides a detailed description of the behaviour of $W(t)$ as $|x|\to 0$
and $|x|\to\infty$. While this fact is not necessary for the purposes of this
article \- namely, well-posedness and the long-term behaviour of the linear
problem - we include it here as we believe it will be valuable for future
studies on the nonlinear problem.

######  Corollary 3.5.

The function $G(t)$ defined in (3.18) admits the following decompositions:

  * (i)

For all $t>0$ in a bounded set,

$\displaystyle G(t)=J_{1}(t)+Q_{1}(t)$

where

$\displaystyle J_{1}(t,x)=e^{\gamma
t}\delta_{0}(x)+\alpha(t)\operatorname{sign}(x)+\beta(t)\left(\operatorname{sign}(x-2)+\operatorname{sign}(x+2)\right)$
(3.31)

and where $Q_{1}(t)$ is a Hölder continuous function in a neighbourhood of
$x=0$. The functions $\alpha,\beta$ and the Hölder constants of $Q_{1}(t)$ are
uniformly bounded in bounded intervals of $t$.

  * (ii)

There exist constants $\rho_{0}>0$, $\rho_{1}\in(0,1)$ as close to $1$ as
desired, $\rho_{2}\in(0,5/2)$ as close to $5/2$ as desired and a function
$Q_{2}$ Hölder continuous in a neighbourhood of $x=0$ such that, for some
constant $C>0$,

$\displaystyle G(t)=J_{2}(t)+Q_{2}(t)$ (3.32)

with

$\displaystyle\left|J_{2}(t,x)-e^{-\gamma t}\delta_{0}(x)\right|\leq
Ce^{-\rho_{0}t}\varphi(x),$ (3.33)
$\displaystyle|\varphi(x)|\leq\begin{cases}e^{\rho_{1}x},\,&x<-1\\\
e^{-\rho_{2}x},\,&x>1\\\ |x|^{-\frac{1}{2}},\,&|x|<1.\end{cases}$ (3.34)

###### Proof.

For all $t$, by Taylor expanding the exponential function and using the
Proposition 3.3 we have,

$\displaystyle e^{(\gamma+\widehat{K}(\xi))t}=e^{\gamma
t}\left(e^{\widehat{K}(\xi)t}\right)=e^{\gamma
t}\sum_{n=0}^{\infty}\frac{\widehat{K}(\xi)^{n}t^{n}}{n!}$

with

$\displaystyle\sum_{n=0}^{\infty}\frac{\widehat{K}(\xi)^{n}t^{n}}{n!}=1+\widehat{K}(\xi)t+\frac{\widehat{K}(\xi)^{2}t^{2}}{2}+\mathcal{O}\left(\widehat{K}(\xi)t\right)^{3}=g_{1}(t,\xi)+h_{1}(t,\xi)$

and where

$\displaystyle
g_{1}(t,\xi)=1+\pi\sqrt{2\pi}\,t|\xi|^{-1/2}+2\pi^{3}\frac{t^{2}}{\xi}-t^{2}\left(K(\varepsilon)ie^{-i\varepsilon\xi}+K(-\varepsilon)ie^{i\varepsilon\xi}\right)\xi^{-1}$
$\displaystyle
h_{1}(t,\xi)=\mathcal{O}\left(\widehat{K}(\xi)t\right)^{3}=\mathcal{O}\left(|\xi|^{-1/2}t\right)^{3},\,\,|\xi|\to\infty.$

Then,

$\displaystyle
G(t,x)=\mathscr{F}^{-1}\left(e^{(\gamma+\widehat{K})t}\right)(x)=e^{\gamma
t}\delta_{0}(x)+\mathscr{F}^{-1}\left(g(t)+h(t)\right)(x)$

from where (3.31) follows. On the other hand, consider a regular cut-off
function $\chi$ such that $\chi(s)=1$ for $|s|\leq 1$ and $\chi(s)=0$ for
$|s|>2$. Then,

$\displaystyle\mathscr{F}^{-1}\left(e^{(\gamma+\widehat{K})t}\right)=e^{\gamma
t}\Bigg{[}\mathscr{F}^{-1}(1)+$
$\displaystyle\mathscr{F}^{-1}\left(e^{\widehat{K}t}\left[1-\chi\left(\frac{|\xi|}{t^{2}}\right)\right]-1\right)$
$\displaystyle\hskip
85.35826pt+\mathscr{F}^{-1}\left(e^{\widehat{K}t}\chi\left(\frac{|\xi|}{t^{2}}\right)\right)\Bigg{]}.$
(3.35)

When $|\xi|<t^{2}$ the first and second terms in the right-hand side of (3.35)
cancel. When $|\xi|>2t^{2}$, in the second term in the right-hand side of
(3.35) is

$\displaystyle
e^{\widehat{K}(\xi)t}\left[1-\chi\left(\frac{|\xi|}{t^{2}}\right)\right]-1=\left(e^{\widehat{K}(\xi)t}-1\right)$
$\displaystyle\text{with}\,\,|\widehat{K}(\xi)|t\leq C\frac{t}{|\xi|^{1/2}}.$

Then, for $t\gg 1$ and $|\xi|>2t^{2}$,

$\displaystyle\left(e^{\widehat{K}(\xi)t}-1\right)=\sum_{n=1}^{\infty}\frac{\widehat{K}(\xi)^{n}t^{n}}{n!}$
$\displaystyle=\widehat{K}(\xi)t+\widehat{K}(\xi)^{2}t^{2}+\mathcal{O}\left(|\xi|^{-1/2}\,t\right)^{3}$
$\displaystyle=g_{2}(t,\xi)+h_{2}(t,\xi)$

where

$\displaystyle
g_{2}(t,\xi)=\pi\sqrt{2\pi}\frac{t}{|\xi|^{1/2}}+2\pi^{3}\frac{t^{2}}{\xi}-\frac{it^{2}}{\xi}\left(K(2)e^{-i2\xi}+K(-2)e^{i2\xi}\right)$
$\displaystyle h_{2}(t,\xi)=\mathcal{O}\left(|\xi|^{-1/2}t\right)^{3}.$

Then we compute

$\displaystyle\mathscr{F}^{-1}\left(e^{\widehat{K}(\xi)t}-1\right)=\pi\sqrt{2\pi}\frac{t}{|x|^{1/2}}+\sqrt{2}\,\pi^{7/2}t^{2}\operatorname{sign}(x)$
$\displaystyle\hskip
113.81102pt-\sqrt{\frac{\pi}{2}}t^{2}\left(K(2)\operatorname{sign}(x-2)-K(-2)\operatorname{sign}(x+2)\right)+H_{2}(t,x).$

Since $\widehat{K}$ is analytic in the strip $\Im m\xi\in(-1,5/2)$, the third
term in the right-hand side of (3.35) is estimated using contour deformation.
When $x\to-\infty$, for all $\rho>-1$ as close to $-1$ as desired,

$\displaystyle\int_{\mathbb{R}}e^{ix\xi}e^{\widehat{K}(\xi)t}\chi\left(\frac{|\xi|}{t^{2}}\right)d\xi=\int_{\Im
m\xi=\rho}e^{ix\xi}e^{\widehat{K}(\xi)t}\chi\left(\frac{|\xi|}{t^{2}}\right)\,{\rm
d}\xi$

and where

$\displaystyle\left|\int_{\Im
m\xi=\rho}e^{ix\xi}e^{\widehat{K}(\xi)t}\chi\left(\frac{|\xi|}{t^{2}}\right)\,{\rm
d}\xi\right|\leq Ce^{-\rho x}\int_{\begin{subarray}{c}\Im m\xi=\rho,\\\
|\xi|<2t^{2}\end{subarray}}e^{|\widehat{K}(\xi)|t}\,{\rm d}\xi$
$\displaystyle\leq Ce^{-\rho x}\int_{\begin{subarray}{c}\Im m\xi=\rho,\\\
|\xi|<2t^{2}\end{subarray}}e^{t(1+|\xi|)^{-1/2}}\,{\rm d}\xi=Ce^{-\rho
x}\int_{|u|<2t^{2}}e^{t(1+|u+i\rho|)^{-1/2}}\,{\rm d}u.$

Now if $t^{2}(1+u)^{-1}=v$, $1+u=t^{2}v^{-1}$ and $\,{\rm
d}u=-t^{2}v^{-2}\,{\rm d}v$. Thus,

$\displaystyle e^{\gamma
t}\int_{\mathbb{R}}e^{ix\xi}e^{(\widehat{K}(\xi)-\gamma)t}d\xi$
$\displaystyle\leq 2Ce^{\gamma t}e^{-\rho
x}\int_{0}^{2t^{2}}e^{t(1+u)^{-1/2}}\,{\rm d}u$ $\displaystyle=2Ce^{\gamma
t}e^{-\rho x}\
t^{2}\int_{\frac{t^{2}}{1+2t^{2}}}^{t^{2}}e^{\sqrt{v}}\frac{\,{\rm
d}v}{v^{2}}.$

These integrals are found to be

$\displaystyle\tau^{2}\int_{\frac{\tau^{2}}{1+2\tau^{2}}}^{t\tau^{2}}e^{\sqrt{v}}\frac{\,{\rm
d}v}{v^{2}}=-e^{t}\tau(1+\tau)+e^{\frac{t\tau}{(1+2\tau^{2}}^{1/2}}(1+2\tau^{2}+\tau(1+2\tau^{2})^{1/2}+$
$\displaystyle+\tau^{2}\left(\text{ExpIntegralEi}(\tau)-\text{ExpIntegralEi}\left(\frac{\tau}{(1+2\tau^{2})^{1/2}}\right)\right)\leq
Ce^{\tau}$

and since $\gamma>1$, there exists $\delta_{1}>0$ such that

$\displaystyle e^{\gamma
t}\int_{\mathbb{R}}e^{ix\xi}e^{(\widehat{K}(\xi)-\gamma)t}d\xi\leq
Ce^{-\delta_{1}\kappa t}e^{-\rho x}.$

A similar argument gives, for $x\to\infty$ the existence of $\delta_{2}>0$ and
$\rho^{\prime}<5/2$ as close to $5/2$ as desired, such that

$\displaystyle e^{\gamma
t}\int_{\mathbb{R}}e^{ix\xi}e^{(\widehat{K}(\xi)-\gamma)t}\,{\rm d}\xi$
$\displaystyle\leq Ce^{-\delta_{2}\kappa t}e^{-\rho^{\prime}x}$

and part (ii) of Corollary 3.5 follows. ∎

###  3.2 Existence, uniqueness and properties of solutions to (3.5).

Using the information obtained in the previous subsection, it is now easy to
obtain solutions to (3.5).

######  Proposition 3.6.

For all $f_{0}\in\mathscr{S}^{\prime}(\mathbb{R})$ and $t>0$, $e^{\Omega
t}\,\widehat{f}_{0}\in\mathscr{S}^{\prime}(\mathbb{R})$ and
$\widehat{K}\widehat{f}_{0}e^{\Omega t}\in\mathscr{S}^{\prime}(\mathbb{R})$.
Then, the distribution

$\displaystyle
f(t)=\mathscr{F}^{-1}\left(e^{t\Omega}\,\widehat{f}_{0}\right)=G(t)\ast
f_{0}\in\mathscr{S}^{\prime}(\mathbb{R})$ (3.36)

satisfies

$\displaystyle\frac{\partial f(t)}{\partial t}=\gamma f(t)+K\ast
f(t)\in\mathscr{S}^{\prime}(\mathbb{R})\quad\text{for all }t>0.$ (3.37)

###### Proof.

The idea is to write,

$\displaystyle e^{(\gamma+\widehat{K}(\xi))t}=e^{\gamma
t}e^{\widehat{K}(\xi))t}=e^{\gamma
t}\left(1+\left(e^{\widehat{K}(\xi)t}-1\right)\right)$

with $\mathscr{F}^{-1}\left(e^{\gamma t}\right)=e^{\gamma t}\delta_{0}$. Since
$\widehat{K}\in L^{\infty}$, $e^{(\gamma+\widehat{K})t}\in
L^{\infty}(\mathbb{R})$. Also, since $\widehat{K}$ is analytic in the strip
$\Im m(\xi)\in(-1,5/2)$ it follows that
$\widehat{f}_{0}e^{(\gamma+\widehat{K})t},\widehat{f}_{0}\widehat{K}e^{(\gamma+\widehat{K})t}\in\mathscr{S}^{\prime}(\mathbb{R})$
for all $f_{0}\in\mathscr{S}^{\prime}(\mathbb{R})$. Their inverse Fourier
transforms are then well-defined and

$\displaystyle f(t)$
$\displaystyle=\mathscr{F}^{-1}\left(\widehat{e}^{(\gamma+\widehat{K})t\,f_{0}}\right)\in\mathscr{S}^{\prime}(\mathbb{R}),\,\forall\
t>0$

while

$\displaystyle\frac{\partial f(t)}{\partial t}$
$\displaystyle=\mathscr{F}^{-1}\left(e^{(\gamma+\widehat{K})t}\widehat{f}_{0}(\gamma+\widehat{K})\right)\in\mathscr{S}^{\prime}(\mathbb{R}),\,\forall\
t>0,$

and equation (3.37) is satisfied by taking the inverse Fourier transform in
both sides of (3.13). ∎

######  Corollary 3.7.

Suppose that $f_{0}\in L^{p}$ for some $p\geq 1$ and let $f$ be the
distribution in (3.36). Then, $f\in C([0,\infty);L^{p}(\mathbb{R}))\cap
C^{1}((0,\infty);L^{p}(\mathbb{R}))$ and satisfies

$\displaystyle\frac{\partial f(t)}{\partial t}=\gamma f(t)+K\ast
f(t),\quad\forall t>0$ (3.38)

in $L^{p}(\mathbb{R})$. Moreover, for all $t>0$,

$\displaystyle\|f(t)\|_{p}\leq e^{\gamma
t}\|f_{0}\|_{p}\left(1+\|K\|_{1}\left(e^{\kappa\|K\|_{1}t}-1\right)\right).$
(3.39)

###### Proof.

Since (3.36) may also be written as

$\displaystyle f(t,x)=e^{\gamma t}\left(f_{0}(x)+(H(t)\ast f_{0})(x)\right)$

and $H(t)\in L^{1}(\mathbb{R})$, it follows that $f(t)\in L^{p}(\mathbb{R})$
if $f_{0}\in L^{p}(\mathbb{R})$. Then, $f(t)\ast K\in L^{p}(\mathbb{R})$,
since $K\in L^{1}(\mathbb{R})$. The right-hand side of (3.37) is then in
$L^{p}(\mathbb{R})$ and the Corollary follows using Proposition 3.2 with
$p=1$. ∎

######  Proposition 3.8.

For all $f_{0}\in L^{p}(\mathbb{R}),\,p\geq 1$ the function $f$ given in
(3.36) is the unique solution of (3.38) in $C^{1}((0,T);L^{p}(\mathbb{R})\cap
C([0,T);L^{p}(\mathbb{R}))$.

###### Proof.

Suppose that $f_{0}=0$, and multiply the equation by $pf^{p-1}$,

$\displaystyle\frac{\,{\rm d}}{\,{\rm d}t}\|f(t)\|^{p}_{p}$
$\displaystyle=2\gamma
p\|f(t)\|^{p}_{p}+p\int_{\mathbb{R}}f^{p-1}(t,x)(f(t)\ast K)(x)\,{\rm d}x$
$\displaystyle\leq 2\gamma
p\|f(t)\|^{p}_{p}+p\|f^{p-1}\|_{p^{\prime}}\|f(t)\ast K\|_{p}$
$\displaystyle\leq 2\gamma
p\|f(t)\|^{p}_{p}+p\left(\int_{\mathbb{R}}|f(t,x)|^{p}dx\right)^{\frac{p-1}{p}}\|K\|_{1}\|f(t)\|_{p}$
$\displaystyle=2\gamma p\|f(t)\|^{p}_{p}+p\|f(t)\|_{p}^{p},$

the fact that $K\in L^{1}(\mathbb{R})$. Then,
$\|f(t)\|_{p}^{p}\leq\|f(0)\|_{p}^{p}e^{(2\gamma+p)t}$ and
$\|f(t)\|_{p}^{p}=0$. ∎

######  Corollary 3.9.

If $f_{0}\in L^{p}(\mathbb{R})$ for some $p\geq 1$ is such that $f_{0}\geq 0$,
the solution of (3.38) given in (3.36) is such that $f(t)\geq 0$ for all
$t>0$.

###### Proof.

The solution $f$ may also be obtained from a solution of

$\displaystyle\frac{\partial g(t)}{\partial t}=K\ast
g(t),\,\,\,g(t,x)=e^{-2\gamma t}f(t,x).$

A solution of that equation may be obtained as a fixed point of

$\displaystyle\mathscr{T}(g)(t,x)=g_{0}(x)+\int_{0}^{t}(K\ast g(s))(x)\,{\rm
d}s$

for $t$ small enough. Since $K\geq 0$, if $f_{0}\geq 0$ and $g(s)\geq 0$ for
all $s\in[0,t]$ and thus the same is true for $\mathscr{T}(g(t))$. ∎

The next result solves the Cauchy problem for equation (3.5) in weighted
$L^{p}$ spaces.

######  Proposition 3.10.

Suppose that $\theta\in(-1,5/2)$. Then, for all $p\geq 1$ and for all bounded
intervals $I\subset[0,\infty)$, there exists a constant $C_{H,1}>0$ such that

$\displaystyle\|f_{0}\ast H(t)\|_{L^{p}(e^{\theta x})}\leq
C_{H,1}\ell(t)\|f_{0}\|_{L^{p}(e^{\theta x})},\quad\forall t\in I.$

###### Proof.

The proof follows as in [4, page 104], by exploiting properties of
convolutions. We first remind that from Corollary 3.4, for all
$\theta\in(-1,5/2)$

$\displaystyle\int_{\mathbb{R}}|H(t,x)|e^{\theta x}\,{\rm
d}x\leq\ell(t)\left(\int_{-\infty}^{0}e^{(\theta+1)x}(1+|x|^{3})\,{\rm
d}x+\int_{0}^{\infty}e^{(\theta-\frac{5}{2})x}(1+x^{3})\,{\rm
d}x\right)=C_{H,1}\ell(t),$

from where we deduce that $H(t)\in L^{1}\left(e^{\theta x}\right)$. Then
observe that if $f_{0}\in L^{1}\left(e^{\theta x}\right)$,

$\displaystyle\int_{\mathbb{R}}e^{\theta x}|(f_{0}\ast
H(t))(x)|dx=\|f_{0}\|_{L^{1}\left(e^{\theta
x}\right)}\|H(t)\|_{L^{1}\left(e^{\theta x}\right)}.$

Also, if $f_{0}\in L^{p}(e^{\theta x})$ then by the previous case the function
$y\mapsto H(t,x-y)f^{p}_{0}(y)$ belongs to $L^{1}\left(e^{\theta x}\right)$,
and since $H(t,y)^{1/p^{\prime}}\in L^{p^{\prime}}\left(e^{\theta x}\right)$,
it follows from Hölder’s inequality that

$\displaystyle|(H(t)\ast f_{0})(x)|^{p}\leq\|H(t)\|_{L^{1}(e^{\theta
x})}^{p/p^{\prime}}\left(|H(t)|\ast|f_{0}|^{p}\right)(x).$

Finally, $|f_{0}|^{p}\in L^{1}\left(e^{\theta x}\right)$, and so we conclude
as

$\displaystyle\|H(t)\ast f_{0}\|^{p}_{L^{p}\left(e^{\theta
x}\right)}\leq\|H(t)\|_{L^{1}\left(e^{\theta
x}\right)}^{p/p^{\prime}}\|H(t)\|_{L^{1}(e^{\theta
x})}\|f_{0}\|^{p}_{L^{p}\left(e^{\theta x}\right)}.$

∎

The Cauchy problem for equation (3.5) is also solved in the following spaces

$\displaystyle X_{A,B}=\\{f\in
L^{\infty}_{\text{loc}}(\mathbb{R})\mid\|f\|_{X_{A,B}}<\infty\\},$ (3.40)
$\displaystyle\|f\|_{X_{A,B}}=\sup_{x<0}e^{-Ax}|f(x)|+\sup_{x>0}e^{-Bx}|f(x)|.$
(3.41)

######  Corollary 3.11.

Suppose that $f_{0}\in L^{p}_{\text{loc}}(\mathbb{R})\cap X_{A,B}$ for some
$p\geq 1$.
If $-\frac{5}{2}<A,B<1$, $H(t)\ast f_{0}\in X_{A,B}$ and there exists a
constant $C=C(A,B)$ such that

$\displaystyle\|H(t)\ast f_{0}\|_{X_{A,B}}\leq C\ell(t).$ (3.42)

If $B<-\frac{5}{2}$ and $A>1$, then there exists a constant $C>0$ such that

$\displaystyle\sup_{x<0}|H(t)\ast
f_{0}(x)|(1+|x|^{3})e^{-x}+\sup_{x>0}|H(t)\ast
f_{0}(x)|(1+|x|^{3})e^{\frac{5x}{2}}<C\ell(t).$ (3.43)

If moreover $B<-\frac{1}{2}<A$, then

$\displaystyle\int_{\mathbb{R}}|(H(t)\ast f_{0})(x)|e^{\frac{x}{2}}\,\,{\rm
d}x<\infty,\quad\forall t>0.$ (3.44)

###### Proof.

By hypothesis $f_{0}\in L^{p}(\mathbb{R})$ and then, by Proposition 3.8, the
function $H(t)\ast f_{0}$ is the unique solution of (3.38) in
$C^{1}((0,T);L^{p}(\mathbb{R})\cap C([0,T);L^{p}(\mathbb{R}))$.

We start with the case when $x>0$. As long as $\frac{5}{2}+A>0$, $B-1<0$,

$\displaystyle|(H(t)\ast f_{0})(x)|\leq
C_{0}\ell(t)e^{-\frac{5}{2}x}\int_{-\infty}^{0}(1+(x-y)^{3})e^{\left(\frac{5}{2}+A\right)y}\,{\rm
d}y$ $\displaystyle\hskip
85.35826pt+C_{0}\ell(t)e^{-\frac{5}{2}x}\int_{0}^{x}e^{\left(\frac{5}{2}+B\right)y}(1+(x-y)^{3})\,{\rm
d}y$ $\displaystyle\hskip
99.58464pt+C_{0}\ell(t)e^{x}\int_{x}^{\infty}e^{(B-1)y}(1+(x-y)^{3})\,{\rm
d}y$ $\displaystyle\hskip 76.82234pt\leq
C^{\prime}\ell(t)\left(e^{-\frac{5}{2}x}\left(1+x^{3}\right)+e^{Bx}+e^{Bx}\right).$
(3.45)

Similarly for $x<0$, as long as $\frac{5}{2}+A>0$, $B-1<0$,

$\displaystyle|(H(t)\ast f_{0})(x)|\leq
C_{0}\ell(t)e^{-\frac{5}{2}x}\int_{-\infty}^{x}e^{\left(\frac{5}{2}+A\right)y}(1+(x-y)^{3})\,{\rm
d}y$ $\displaystyle\hskip
85.35826pt+C_{0}\ell(t)e^{x}\int_{x}^{0}e^{\left(A-1\right)y}(1+(x-y)^{3})\,{\rm
d}y$ $\displaystyle\hskip
99.58464pt+C_{0}\ell(t)e^{x}\int_{0}^{\infty}e^{(B-1)y}(1+(x-y)^{3})\,{\rm
d}y$ $\displaystyle\hskip 76.82234pt\leq
C^{\prime}\ell(t)\left(e^{Ax}+e^{Ax}+e^{x}(1+|x|^{3})\right).$ (3.46)

As $B<-5/2$, the term $e^{Bx}$ wins among all the terms for $x>0$ while as
$A<1$, $e^{Ax}$ wins among all the terms for $x<0$. This concludes the first
part of the Corollary. Property (3.43) follows as well from (3.2) and (3.2)
depending on the hypothesis on $A$ and $B$. Property (3.44) follows from,

$\displaystyle\int_{\mathbb{R}}|(H(t)\ast f_{0})(x)|e^{\frac{x}{2}}dx\leq
C_{0}\ell(t)\int_{-\infty}^{0}e^{\left(A+\frac{1}{2}\right)y}\,{\rm
d}y+C_{0}\ell(t)\int_{0}^{\infty}e^{\left(B+\frac{1}{2}\right)y}\,{\rm d}y$

where the two integrals in the right-hand side converge since
$A+\frac{1}{2}>0$ and $B+\frac{1}{2}<0$. ∎

######  Remark 3.12.

The map $S(t):f_{0}\to H(t)\ast f_{0}$ belongs to
$\mathscr{L}\left(L^{p}(\mathbb{R})\right)\cap\mathscr{L}\left(L^{p}(e^{\theta
x})\right)$ for $p\geq 1$ and $S(t)\in\mathscr{L}(X_{A,B})$ too when $A$ and
$B$ satisfy $-5/2<A,B<1$. For all $t>0$,

$\displaystyle\|S(t)\|_{\mathscr{L}\left(L^{p}(\mathbb{R})\right)}\leq
e^{\gamma t}\left(1+\|K\|_{1}\left(e^{\kappa\|K\|_{1}t}-1\right)\right)$
$\displaystyle\|S(t)\|_{\mathscr{L}\left(L^{p}(e^{\theta x})\right)}\leq
e^{\gamma t}\Big{(}1+C_{H,1}\ell(t)\Big{)}$
$\displaystyle\|S(t)\|_{\mathscr{L}(X_{A,B})}\leq e^{\gamma
t}\Big{(}1+C(A,B)\ell(t)\Big{)}.$

Let us show for further comparison, some properties of the maps $S(t)$ defined
in Remark 3.12 acting on spaces $X_{A,B}$.

######  Corollary 3.13.

Suppose that, for some $A\in\left(-\frac{5}{2},1\right)$,

$f_{0}(x)=e^{Ax}\mathbbm{1}_{x<0}.$ (3.47)

Then, for all $a^{\prime}\in(-1,-A)$, there exists a constant $C>0$ such that

$\displaystyle\left|f(t,x)-e^{t\Omega(-iA)}e^{Ax}\right|\leq
Cte^{t\gamma}e^{-a^{\prime}x},\quad\forall x<0,$ (3.48)

and for all $a^{\prime\prime}\in\left(-A,\frac{5}{2}\right)$, there exists a
constant $C^{\prime}>0$ such that

$\displaystyle\left|f(t,x)-e^{t\gamma}e^{Ax}\right|\leq
Cte^{t\gamma}e^{-a^{\prime\prime}x},\quad\forall x>0.$ (3.49)

If, on the other hand, $f_{0}(x)=x^{B}\mathbbm{1}_{x>0}$ for some
$B\in\left(-\frac{5}{2},1\right)$, then for all
$b^{\prime}\in\left(-B,\frac{5}{2}\right)$, there exists a constant $C>0$ such
that

$\displaystyle\left|f(t,x)-e^{t\Omega(-iB)}e^{Bx}\right|\leq
Cte^{t\gamma}e^{-b^{\prime}x},\quad\forall x>0,$ (3.50)

and for all $b^{\prime\prime}\in(-1,-B)$, there exists a constant
$C^{\prime}>0$ such that

$\displaystyle\left|f(t,x)-e^{t\gamma}e^{Bx}\right|\leq
Cte^{t\gamma}e^{-b^{\prime\prime}x},\quad\forall x>0.$ (3.51)

###### Proof.

Suppose first that $f_{0}(x)=e^{Ax}\mathbbm{1}_{x<0}$ with $A\in(-5/2,1)$.
Then, for all $\xi\in\mathbb{C}$ such that $\Im m(\xi)>-A$,

$\displaystyle\widehat{f}_{0}(\xi)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{0}e^{-ix\xi}e^{Ax}\,{\rm
d}\xi=\frac{1}{\sqrt{2\pi}(A-i\xi)}$ (3.52)

and then, if $\mathscr{C}_{a}=\\{\xi\in\mathbb{C};\,\,\Im m(\xi)=a\\}$ with
$a\in(-A,5/2)$, using the fact that the solution is given by the inverse
Fourier transform of $\widehat{f}_{0}(\xi)e^{t\Omega(\xi)}$ and inserting
(3.52), we get

$\displaystyle f(t,x)$
$\displaystyle=\frac{1}{2\pi}\int_{\mathscr{C}_{a}}\frac{e^{ix\xi}e^{t\Omega(\xi)}}{A-i\xi}\,{\rm
d}\xi=\frac{1}{2\pi}\int_{\mathscr{C}_{a}}\frac{e^{ix\xi}e^{t(\gamma+\widehat{K}(\xi))}}{A-i\xi}\,{\rm
d}\xi$
$\displaystyle=e^{t\gamma}e^{Ax}+\frac{e^{t\gamma}}{2\pi}\int_{\mathscr{C}_{a}}\left(e^{t\widehat{K}(\xi)}-1\right)\frac{e^{ix\xi}}{A-i\xi}\,{\rm
d}\xi.$ (3.53)

A contour deformation argument yields then for $a^{\prime}\in(-1,-A)$,

$\displaystyle f(t,x)$
$\displaystyle=e^{t\gamma}e^{Ax}+e^{t\gamma}\left(e^{t\widehat{K}(-iA)}-1\right)e^{Ax}+\frac{e^{t\gamma}}{2\pi}\int_{\mathscr{C}_{a^{\prime}}}\left(e^{t\widehat{K}(\xi)}-1\right)\frac{e^{ix\xi}}{A-i\xi}\,{\rm
d}\xi$

from where we get,

$\displaystyle\left|f(t,x)-e^{t\Omega(-iA)}e^{Ax}\right|\leq
Ce^{t\gamma}e^{-a^{\prime}x}\int_{\mathbb{R}}\frac{\left|e^{t\widehat{K}(u+ia^{\prime})}-1\right|\,{\rm
d}u}{((A+a^{\prime})^{2}+u^{2})^{1/2}}$

with $A+a^{\prime}<0$. It follows from (3.21) in Proposition 3.3,

$\displaystyle\int_{\mathbb{R}}\frac{\left|e^{t\widehat{K}(u+ia^{\prime})}-1\right|\,{\rm
d}u}{((A+a^{\prime})^{2}+u^{2})^{1/2}}\leq
Ct\int_{\mathbb{R}}\frac{\left(u^{2}+a^{\prime 2}\right)^{-1/2}\,{\rm
d}u}{((A+a^{\prime})^{2}+u^{2})^{1/2}}$

and (3.48) follows. If now $x>0$, a contour deformation in (3.53) gives for
$a^{\prime\prime}\in(-A,5/2)$,

$\displaystyle
f(t,x)-e^{t\gamma}e^{Ax}=\frac{e^{t\gamma}}{2\pi}\int_{\mathscr{C}_{a^{\prime\prime}}}\left(e^{t\widehat{K}(\xi)}-1\right)\frac{e^{ix\xi}}{A-i\xi}\,{\rm
d}\xi$

with,

$\displaystyle\left|\int_{\mathscr{C}_{a^{\prime\prime}}}\left(e^{t\widehat{K}(\xi)}-1\right)\frac{e^{ix\xi}}{A-i\xi}\,{\rm
d}\xi\right|\leq
Cte^{-a^{\prime\prime}x}\int_{\mathbb{R}}\frac{\left(u^{2}+a^{\prime\prime
2}\right)^{-1/2}\,{\rm d}u}{((A+a^{\prime\prime})^{2}+u^{2})^{1/2}}$

and (3.49) follows. Similarly, if $f_{0}(x)=e^{Bx}\mathbbm{1}_{x>0}$ with
$B\in(-5/2,1)$ then for all $\xi\in\mathbb{C}$ such that $\Im m(\xi)<-B$,

$\displaystyle\widehat{f}_{0}(\xi)=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-ix\xi}e^{Bx}\,{\rm
d}\xi=-\frac{1}{\sqrt{2\pi}(B-i\xi)}$

and if $b^{\prime\prime}\in(-1,-B)$,

$\displaystyle f(t,x)$
$\displaystyle=-\frac{1}{2\pi}\int_{\mathscr{C}_{b^{{}^{\prime\prime}}}}\frac{e^{ix\xi}e^{t\Omega(\xi)}}{B-i\xi}\,{\rm
d}\xi=-\frac{1}{2\pi}\int_{\mathscr{C}_{b^{{}^{\prime\prime}}}}\frac{e^{ix\xi}e^{t(\gamma+\widehat{K}(\xi))}}{B-i\xi}\,{\rm
d}\xi$
$\displaystyle=e^{t\gamma}e^{Bx}-\frac{e^{t\gamma}}{2\pi}\int_{\mathscr{C}_{b^{{}^{\prime\prime}}}}\left(e^{t\widehat{K}(\xi)}-1\right)\frac{e^{ix\xi}}{B-i\xi}\,{\rm
d}\xi.$

Then,

$\displaystyle f(t,x)$
$\displaystyle-e^{t\gamma}e^{Bx}=-\frac{e^{t\gamma}}{2\pi}\int_{\mathscr{C}_{b^{{}^{\prime\prime}}}}\left(e^{t\widehat{K}(\xi)}-1\right)\frac{e^{ix\xi}}{B-i\xi}\,{\rm
d}\xi$

and (3.51) follows in the same way as (3.49). If $x>0$, then arguing as for
(3.48),

$\displaystyle f(t,x)$
$\displaystyle=e^{t\gamma}e^{Bx}+e^{t\gamma}\left(e^{t\widehat{K}(-iB)}-1\right)e^{Bx}+\frac{e^{t\gamma}}{2\pi}\int_{\mathscr{C}_{b^{\prime}}}\left(e^{t\widehat{K}(\xi)}-1\right)\frac{e^{ix\xi}}{B-i\xi}\,{\rm
d}\xi$

for any $b^{\prime}\in(-B,5/2)$ and (3.50) follows. ∎

######  Remark 3.14.

It is remarkable how different the properties of the map $S(t)$ obtained in
Corollary 3.13 are with respect to the similar map $T(t)$ for the
linearization of equation (1.2) around the KZ solution $\omega^{-7/6}$. As
proved indeed in [16, Lemma 3.9], these have the following regularizing effect
at $-\infty$: for all initial data (3.47) with $A\in\left(-3/2,-7/6\right]$,
there is $\lambda(t)\in\mathbb{R}$ such that
$T(t)f_{0}(x)\sim\lambda(t)e^{-7x/6}$ as $x\to-\infty$ . This is in contrast
with Corollary 3.13, where the behavior $e^{Ax}$ of the initial as
$x\to-\infty$ is observed at all times.

As proved in the next Proposition, for some solutions of equation (3.5) the
quantity (3.7) (wave action) is constant in time. This property follows from
the fact that $W(1)=0$. In general, the zeros of the function $W$ correspond
to the conserved quantities. As seen in the first section, Proposition 3.1,
$W(2)=0$ makes $p^{-2}$ to be a stationary solution of (1.11).

######  Corollary 3.15.

Suppose that $f_{0}\in L^{1}(e^{rx})$ for some $r\in(-1,\frac{5}{2})$. Then,

$\displaystyle\int_{\mathbb{R}}f(t,x)e^{rx}\,\,{\rm
d}x=e^{\lambda_{r}t}\int_{\mathbb{R}}f_{0}(x)e^{rx}\,\,{\rm d}x,$

where $\lambda_{r}>0$ if $r\in(-1,\frac{1}{2})$ or $r\in(1,\frac{5}{2})$,
$\lambda_{r}<0$ if $r\in(\frac{1}{2},1)$, and $\lambda_{r}=0$ if
$r=\frac{1}{2}$ or $r=1$.

###### Proof.

For all $r\in(-1,5/2)$ and $f_{0}\in L^{1}(e^{rx})$, by Proposition 3.10,
$f(t)\in L^{1}(e^{rx})$. Then we write

$\displaystyle\int_{\mathbb{R}}f(t,x)e^{rx}\,{\rm d}x$
$\displaystyle=\widehat{f}(t,ir)=\widehat{f}_{0}(ir)\widehat{G}(t,ir)=\widehat{f}_{0}(ir)e^{(\gamma+\widehat{K}(ir))t}$
$\displaystyle=\widehat{f}_{0}(ir)e^{(\gamma+W(2r))t}=\int_{\mathbb{R}}f_{0}(x)e^{rx}\,{\rm
d}x\,e^{(\gamma+W(2r))t}.$

The result follows with $\lambda_{r}=\gamma+W(2r)$ and the properties of the
function $W$. ∎

###  3.3 Long time behavior of some solutions to (3.5).

Our next result shows that for some initial data at last, a concentration of
waves towards zero frequency. Unlike in the non linear equation, this process
takes infinite time, at an exponential rate that is easy to obtain explicitly.
More precisely,

######  Theorem 3.16.

Suppose that $f_{0}\in X_{A,B}$, where $A$ and $B$ satisfy (• ‣ 1.4). Then,
for $\sigma>\frac{1}{2}$ close to $1/2$,

$\displaystyle\int_{-\infty}^{L}f(t,x)e^{\frac{x}{2}}\,{\rm
d}x=\int_{\mathbb{R}}f(t,x)e^{\frac{x}{2}}\,{\rm d}x+\int_{\Im
m\xi=\sigma}\widehat{f}_{0}(\xi)e^{(\gamma+\widehat{K}(\xi))t}\frac{e^{L\left(i\xi+\frac{1}{2}\right)}}{i\xi+\frac{1}{2}}\,{\rm
d}\xi,$ (3.54)

where all the integrals are absolutely convergent and for $t\to\infty$,

$\displaystyle\left|\int_{\Im
m\xi=\sigma}\widehat{f}_{0}(\xi)e^{(\gamma+\widehat{K}(\xi))t}\frac{e^{L\left(i\xi+\frac{1}{2}\right)}}{i\xi+\frac{1}{2}}\,{\rm
d}\xi\right|\leq Ce^{-\nu t}\int_{\mathbb{R}}|f_{0}(x)|e^{\frac{3x}{4}}\,{\rm
d}x$ (3.55)

where $-\nu=(\gamma+\widehat{K}(3i/4))<0$.

###### Proof.

Since $f_{0}\in X_{A,B}$ with $-5/2<A,B<1$, $f(t)\in X_{A,B}$ with
$-5/2<A,B<1$ for all $t>0$ and the function $g(t)$ defined as
$g(t,x)=f(t,x)e^{x/2}$ belongs to $L^{1}(\mathbb{R})$ for all $t>0$. Since
$f_{0}\in L^{2}$ then $f(t)\in L^{2}(\mathbb{R})$,

$f(t,x)=\int_{\mathbb{R}}e^{ix\xi}\widehat{f}(t,\xi)\,{\rm
d}\xi=\lim_{R\to\infty}\int_{\mathbb{R}}e^{ix\xi}\widehat{f}(t,\xi)\,{\rm
d}\xi,\,\,\text{in}\,\,L^{2}(\mathbb{R}),$

and using that the function $x\to e^{\frac{x}{2}}$ belongs to
$L^{2}(-\infty,L)$ for all finite $L\in\mathbb{R}$:

$\displaystyle\int_{-\infty}^{L}f(t,x)e^{\frac{x}{2}}\,{\rm d}x$
$\displaystyle=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{L}\int_{\mathbb{R}}e^{ix\xi}\widehat{f}_{0}(\xi)e^{(\gamma+\widehat{K}(\xi))t}d\xi
e^{\frac{x}{2}}\,{\rm d}x$
$\displaystyle=\lim_{R\to\infty}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{L}\int_{-R}^{R}e^{ix\xi}\widehat{f}_{0}(\xi)e^{(\gamma+\widehat{K}(\xi))t}\,{\rm
d}\xi e^{\frac{x}{2}}\,{\rm d}x.$

Fubini Theorem then gives,

$\displaystyle\int_{-\infty}^{L}f(t,x)e^{\frac{x}{2}}\,{\rm d}x$
$\displaystyle=\lim_{R\to\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\widehat{f}_{0}(\xi)e^{(\gamma+\widehat{K}(\xi))t}\int_{-\infty}^{L}e^{ix\xi}e^{\frac{x}{2}}\,{\rm
d}x\,{\rm d}\xi$
$\displaystyle=\lim_{R\to\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\widehat{f}_{0}(\xi)e^{(\gamma+\widehat{K}(\xi))t}\frac{e^{L\left(i\xi+\frac{1}{2}\right)}}{i\xi+\frac{1}{2}}\,{\rm
d}\xi.$

Since $\widehat{f}_{0}$ and the function $\xi\to(i\xi+\frac{1}{2})^{-1}$
belong to $L^{2}(\mathbb{R})$,

$\displaystyle\lim_{R\to\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\widehat{f}_{0}(\xi)e^{(\gamma+\widehat{K}(\xi))t}\frac{e^{L\left(i\xi+\frac{1}{2}\right)}}{i\xi+\frac{1}{2}}\,{\rm
d}\xi=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\widehat{f}_{0}(\xi)e^{(\gamma+\widehat{K}(\xi))t}\frac{e^{L\left(i\xi+\frac{1}{2}\right)}}{i\xi+\frac{1}{2}}\,{\rm
d}\xi,$ (3.56)

where the integral in the right hand side of (3.56) is absolutely convergent.

Since $f(t)$ satisfies (• ‣ 1.4) it follows that $\widehat{f}(t)$ is analytic
on the strip $\Im m(\xi)\in(B,A)$ and then the function under the integral
sign at the right hand side of (3.56) is analytic in the strip $\Im
m(\xi)\in(B,\frac{1}{2})$ and has a pole at $\xi=i/2$. Moreover, for all
$\xi=u+i\sigma$ such that $u\in\mathbb{R}$,
$\sigma\in(-\varepsilon,\frac{1}{2}+\varepsilon)$ with $\varepsilon>0$ small,

$\displaystyle\left|\widehat{f}_{0}(\xi)e^{(\gamma+\widehat{K}(\xi))t}\frac{e^{L(i\xi+\frac{1}{2})}}{i\xi+\frac{1}{2}}\right|$
$\displaystyle\leq
Ce^{C^{\prime}t}|\widehat{f}_{0}(u+i\sigma)|\frac{e^{-L\sigma}}{|-\sigma+\frac{1}{2}+iu|}$
$\displaystyle\leq
Ce^{C^{\prime}t}|\widehat{g}_{0}(u)|\frac{e^{-L\sigma}}{|-\sigma+\frac{1}{2}+iu|},$

where $g_{0}(u)=e^{\sigma u}f_{0}(u)$ is such that,

$\displaystyle\int_{\mathbb{R}}|g_{0}(u)|\,{\rm d}u$
$\displaystyle=\int_{\mathbb{R}}|f_{0}(u)|e^{\sigma u}\,{\rm d}u$
$\displaystyle\leq C\int_{-\infty}^{0}e^{(A+\sigma)u}\,{\rm
d}u+C\int_{0}^{\infty}e^{(B+\sigma)u}\,{\rm d}u<\infty$

since $A+\sigma>A-\varepsilon>0$ and $B+\sigma<B+\frac{1}{2}+\varepsilon<0$.
Therefore,

$\displaystyle\left|\widehat{f}_{0}(\xi)e^{(\gamma+\widehat{K}(\xi))t}\frac{e^{L(i\xi+\frac{1}{2})}}{i\xi+\frac{1}{2}}\right|\leq
Ce^{C^{\prime}t}\|g_{0}\|_{1}|\frac{e^{-L\sigma}}{|-\sigma+\frac{1}{2}+iu|}\to
0,\,\,|u|\to\infty.$

Classical deformation of integration contour gives then,

$\displaystyle\int_{\mathbb{R}}\widehat{f}_{0}(\xi)e^{(\gamma+\widehat{K}(\xi))t}\frac{e^{L\left(i\xi+\frac{1}{2}\right)}}{i\xi+\frac{1}{2}}\,{\rm
d}\xi$ $\displaystyle=(-i)2\pi
i\widehat{f}_{0}(i/2)e^{\left(\gamma+\widehat{K}(i/2)\right)t}$
$\displaystyle\hskip 28.45274pt+\int_{\Im
m(\xi)=\sigma}\widehat{f}_{0}(\xi)e^{(\gamma+\widehat{K}(\xi))t}\frac{e^{L\left(i\xi+\frac{1}{2}\right)}}{i\xi+\frac{1}{2}}\,{\rm
d}\xi.$

Since $\gamma+\widehat{K}(i/2)=0$,

$\displaystyle\int_{-\infty}^{L}f(t,x)e^{\frac{x}{2}}dx=\sqrt{2\pi}\widehat{f}_{0}(i/2)+\frac{1}{\sqrt{2\pi}}\int_{\Im
m\xi=\sigma}\widehat{f}_{0}(\xi)e^{(\gamma+\widehat{K}(\xi))t}\frac{e^{L\left(i\xi+\frac{1}{2}\right)}}{i\xi+\frac{1}{2}}\,{\rm
d}\xi$ (3.57)

and (3.54) follows.

The function under the integral sign at the right hand side of (3.57) is now
analytic on the domain $\Im m(\xi)\in(i/2,5i/2)$. Moreover, since $f_{0}\in
X_{A,B}$, $-5/2<A,B<1$, the function $h_{\sigma}$ defined as
$h_{\sigma}(x)=e^{-\sigma x}f_{0}(x)$ is such that $h_{\sigma}\in
L^{2}(\mathbb{R})$, and then $\hat{h}_{\sigma}\in L^{2}(\mathbb{R})$, for all
$\sigma\in(B,A)$. Since for all $u\in\mathbb{R},\sigma\in\mathbb{R}$,

$\mathscr{F}\left(h_{\sigma}\right)(u)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f_{0}(x)e^{-\sigma
x}e^{ixu}\,{\rm
d}x=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f_{0}(x)e^{ix(u+i\sigma)}\,{\rm
d}x=\hat{f}_{0}(u+i\sigma)$

it follows, for all $\sigma\in(1/2,A)$,

$\displaystyle\int_{\Im
m\xi=\sigma}\left|\widehat{f}_{0}(\xi)e^{(\gamma+\widehat{K}(\xi))t}\frac{e^{L\left(i\xi+\frac{1}{2}\right)}}{i\xi+\frac{1}{2}}\right|\,{\rm
d}|\xi|\leq\int_{\mathbb{R}}|\hat{h}_{\sigma}(u)|\frac{\,{\rm
d}u}{|iu-\sigma+\frac{1}{2}|}\leq C\|h_{\sigma}\|_{2}$

and the integral

$\displaystyle\frac{1}{\sqrt{2\pi}}\int_{\Im
m\xi=\sigma}\widehat{f}_{0}(\xi)e^{(\gamma+\widehat{K}(\xi))t}\frac{e^{L\left(i\xi+\frac{1}{2}\right)}}{i\xi+\frac{1}{2}}\,{\rm
d}\xi$

is absolutely convergent for all $\sigma\in(1/2,A)$.

The same argument shows that the contour deformation argument may still be
applied to the second integral at the right hand side of (3.57) as long as the
contour does not leave the domain where $\Im m(\xi)\in(1/2,A)$.

The critical point of $\Omega(\xi)=(\gamma+\widehat{K}(\xi))$ is $\xi=3i/4$.
Since $\Omega$ is analytic on the strip $\Im m\xi\in(-1,5/2)$,

$\displaystyle\Omega(\xi)=\Omega(3i/4)+\frac{(\xi-3i/4)^{2}}{2}\Omega^{\prime\prime}(3i/4)+\mathcal{O}\left(\xi-3i/4\right)^{3},\,\,\xi\to
3i/4,$

where, use of Mathematica gives
$\Omega^{\prime\prime}(3i/4)=-4W^{\prime\prime}(3/2)<0$. Since the function
$\Omega$ is analytic in the strip $\Im m\xi\in(-1,5/2)$ the contour of
integration $\Im m(\xi)=\sigma$ may now be deformed to the line

$\displaystyle\mathcal{C}^{\prime}=\left\\{\xi=u+\frac{3i}{4},u\in\mathbb{R}\right\\}$

along which the following hold:

  * (i)

$\Omega(\xi)=\Omega(3i/4)+\frac{1}{2}(\xi-3i/4)^{2}\Omega^{\prime\prime}(3i/4)+\mathcal{O}((\xi-3i/4)^{3}),\
\xi\to 3i/4$

  * (ii)

$\Omega(u+3i/4)\in\mathbb{R}$ for all $u\in\mathbb{R},$ and

$\displaystyle\forall\rho>0,\,\exists\mu>0;\
\Omega(u+3i/4)\leq\Omega(3i/4)-\mu,\,\,\forall|u|\geq\rho\,\,\text{(cf.\
Figure 1)}.$

Figure 1: The function $W\left(\frac{3}{2}+iv\right)$ for $v\in(-70,70)$.

Then from (i), for $R_{0}$ small enough

$\displaystyle\Omega(\xi)=\Omega(3i/4)+\frac{u^{2}}{2}\Omega^{\prime\prime}(3i/4)+\mathcal{O}(u)^{3},\,\forall\xi\in\mathcal{C}^{\prime},\,|u|\leq
R_{0}.$

while from (ii), there exists $\mu>0$ such that

$\displaystyle\Omega(\xi)\leq\Omega(3i/4)-\mu,\,\,\forall\xi\in\mathcal{C}^{\prime},\,\,|u|>R_{0}.$

The curve $\mathcal{C}^{\prime}$ may then be decomposed into
$\mathcal{C}^{\prime}=\mathcal{C}_{1}^{\prime}+\mathcal{C}_{2}^{\prime}$,
where

$\displaystyle\mathcal{C}^{\prime}_{1}=\\{\xi\in\mathcal{C}^{\prime};\,|u|>R_{0}\\},\quad\mathcal{C}^{\prime}_{2}=\\{\xi\in\mathcal{C}^{\prime};\,|u|<R_{0}\\},$

so that

$\displaystyle\int_{\mathcal{C}^{\prime}}\widehat{f}_{0}(\xi)e^{t\Omega(\xi)}\frac{e^{L\left(i\xi+\frac{1}{2}\right)}}{i\xi+\frac{1}{2}}\,{\rm
d}\xi$
$\displaystyle=\int_{\mathcal{C}^{\prime}_{1}}\widehat{f}_{0}(\xi)e^{t\Omega(\xi)}\frac{e^{L\left(i\xi+\frac{1}{2}\right)}}{i\xi+\frac{1}{2}}\,{\rm
d}\xi+\int_{\mathcal{C}^{\prime}_{2}}\widehat{f}_{0}(\xi)e^{t\Omega(\xi)}\frac{e^{L\left(i\xi+\frac{1}{2}\right)}}{i\xi+\frac{1}{2}}\,{\rm
d}\xi$ $\displaystyle=:I_{1}+I_{2}.$

In the first integral, $I_{1}$

$\displaystyle\left|e^{t\Omega(\xi)}\frac{e^{L\left(i\xi+\frac{1}{2}\right)}}{i\xi+\frac{1}{2}}\right|\leq\frac{Ce^{(\Omega(3i/4)-\mu)t}}{\left|i\xi+\frac{1}{2}\right|}$

using the conclusion from point (ii). In total,

$\displaystyle\left|I_{1}\right|\leq
Ce^{(\Omega(3i/4)-\mu)t}\int_{\mathcal{C}^{\prime}}\left|\hat{f}_{0}(\xi)\right|\frac{d\xi}{\left|i\xi+\frac{1}{2}\right|}\leq
Ce^{(\Omega(3i/4)-\mu)t}\|h_{\sigma}\|_{2}.$

after C-S in the last inequality. In the second integral, for $R_{0}$ small,

$\displaystyle
I_{2}=e^{t\Omega(3i/4)+\mathcal{O}(R_{0})^{3}}\int_{|u|<R_{0}}\widehat{f}_{0}(u+3i/4)e^{\frac{t}{2}u^{2}\Omega^{\prime\prime}(3i/4)}\frac{e^{L\left(-\frac{1}{4}+iu\right)}}{-\frac{1}{4}+iu}\,{\rm
d}u,$

we perform the change of variables, $u\sqrt{-t\Omega^{\prime\prime}(3i/4)}=2z$
which gives,

$\displaystyle\int_{|u|<R_{0}}\widehat{f}_{0}(u+3i/4)e^{\frac{t}{2}u^{2}\Omega^{\prime\prime}(3i/4)}\frac{e^{L\left(-\frac{1}{4}+iu\right)}}{-\frac{1}{4}+iu}\,{\rm
d}u=$
$\displaystyle\frac{2}{\sqrt{-t\Omega^{\prime\prime}(3i/4)}}\int\limits_{2|z|\leq
R_{0}\sqrt{-t\Omega^{\prime\prime}(3i/4)}}\widehat{f}_{0}\left(\frac{2z}{\sqrt{-t\Omega^{\prime\prime}(3i/4)}}+\frac{3i}{4}\right)e^{-z^{2}}j(z)\,{\rm
d}z$

where

$\displaystyle j(z)$
$\displaystyle=\left(\frac{2iz}{\sqrt{-t\Omega^{\prime\prime}(3i/4)}}-\frac{1}{4}\right)^{-1}\exp\left(\frac{2iLz}{\sqrt{-t\Omega^{\prime\prime}(3i/4)}}-\frac{L}{4}\right)$
$\displaystyle=-4e^{-\frac{L}{4}}\left(1+\mathcal{O}\left(R_{0}\right)\right),\,R_{0}\to
0.$

By continuity of $f_{0}$,

$\displaystyle\widehat{f}_{0}\left(\frac{2z}{\sqrt{-t\Omega^{\prime\prime}(3i/4)}}+\frac{3i}{4}\right)=\widehat{f}_{0}\left(\frac{3i}{4}\right)(1+o(1)),\,\,t\to\infty$

and then, as $t\to\infty$ and $R_{0}\to 0$

$\displaystyle
I_{2}=\frac{-2e^{t\Omega(3i/4)}}{\sqrt{-t\Omega^{\prime\prime}(3i/4)}}\left(1+\mathcal{O}(R_{0})^{3}\right)\widehat{f}_{0}\left(\frac{3i}{4}\right)(1+o(1))4e^{-\frac{L}{4}}\left(1+\mathcal{O}\left(R_{0}\right)\right)$

from where (3.55) follows. ∎

The second and final result of this section pertains to initial data $f_{0}$
whose Fourier transform exhibits good properties of analyticity and
integrability. Under these conditions, the pointwise behaviour of $f(t,x)$ as
$|x|/t\to\infty$ and $|x|/t\to 0$ may be described. It is then possible to
observe the growth or decay of the solution for $x\lessgtr(4/3)\Omega(3i/4)t$,
where $\Omega(3i/4)\approx-0.1572$.

######  Proposition 3.17.

Suppose that $f_{0}$ is such that $\hat{f}_{0}\in X_{A,B}$ with $-5/2<A,B<1$
and the function $g_{0}$ defined as $g_{0}(x)=f_{0}(x)e^{\frac{xs_{*}}{2}}$
for $x\in\mathbb{R}$ satisfies $\widehat{g}_{0}\in L^{1}(\mathbb{R})$. Then,

$\displaystyle
f(t,x)\sim\begin{cases}\displaystyle{\frac{\widehat{f}_{0}(-i)}{\sqrt{3\pi}}\frac{e^{x}t^{\frac{1}{6}}}{|x|^{\frac{2}{3}}}e^{3t^{\frac{1}{3}}|x|^{\frac{2}{3}}},\,\frac{x}{t}\to-\infty,}\\\
\displaystyle{\frac{\widehat{f}_{0}(5i/2)}{\sqrt{3\pi}}\frac{e^{-\frac{5x}{2}}t^{\frac{1}{6}}}{|x|^{\frac{2}{3}}}e^{3t^{\frac{1}{3}}|x|^{\frac{2}{3}}},\,\frac{x}{t}\to\infty.}\end{cases}$
(3.58)

and,

$\displaystyle
f(t,x)\sim\frac{\widehat{f}_{0}(3i/4)e^{-3x/4}e^{\Omega(3i/4)t}}{\sqrt{2\pi\Omega^{\prime\prime}(3i/4)t}},\,\,\frac{|x|}{t}\to
0.$ (3.59)

###### Proof.

As in the Theorem 3.16, since $f_{0}\in X_{A,B}$ with $-5/2<A,B<1$, $f(t)\in
X_{A,B}$ with $-5/2<A,B<1$ for all $t>0$.The proof follows now from steepest
descent method applied to the integral

$\displaystyle
f(t,x)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{ix\xi+(\gamma+\widehat{K}(\xi))t}\widehat{f}_{0}(\xi)\,{\rm
d}\xi.$

However, the proof is simpler to write in the variables $A,p$, using then the
Mellin transform and its inverse. Write instead (notice the difference of
$p=e^{x/2},\,x=2\log p$ with the notation in [13]),

$\displaystyle f(t,x)=\frac{1}{2\pi
i}\int_{\sigma-i\infty}^{\sigma+i\infty}\Phi(s)e^{-\frac{xs}{2}+tW(s)}\,{\rm
d}s$

where $\Phi(s)$ is the Mellin transform of the initial data $A_{0}$. Notice
first that the unique critical point $s_{*}$ of the function
$\psi(s;t,x)=-\frac{xs}{2}+tW(s)$ is such that,

$\displaystyle W^{\prime}(s_{*})=-\frac{x}{2t}.$

Since $W$ is analytic for $\Re e(s)\in(-2,5)$ with only two poles at
$s=-2,s=5$ in $\Re e(s)\in[-2,5]$ and satisfies,

$\displaystyle W^{\prime}(s)$
$\displaystyle=-\frac{8}{(s+2)^{3}}-\frac{8}{3(s+2)^{2}}+\mathcal{O}(1),\,\,s\in(-2,5),\,s\to-2$
$\displaystyle W^{\prime}(s)$
$\displaystyle=-\frac{8}{(s-5)^{3}}+\frac{8}{3(s-5)^{2}}+\mathcal{O}(1),\,\,s\in(-2,5),\,\,s\to
5$

two cases may arise as $xt^{-1}\to\pm\infty$. Suppose first $x\to-\infty$, and
then $W^{\prime}(s_{*})\to\infty$. If $\Re e(s_{*})\in(-2,5)$, then

$\displaystyle-\frac{x}{2t}=W^{\prime}(s_{*})=-\frac{8}{(s_{*}+2)^{3}}+\mathcal{O}\left(s_{*}+2\right)^{-2}\to\infty,\,\,\,\,\frac{x}{t}\to-\infty,$

and since, for $|x/t|$ large enough,

$\displaystyle\frac{x}{2t}=\left|-\frac{8}{(s_{*}+2)^{3}}+\mathcal{O}\left(s_{*}+2\right)^{-2}\right|\leq
C\left|\frac{8}{(s_{*}+2)^{3}}\right|$

it follows that $s_{*}+2\to 0$ as $x/t\to-\infty$ and,

$\displaystyle
s_{*}+2=\mathcal{O}\left(\frac{t}{x}\right)^{1/3},\,\,\,\frac{x}{t}\to-\infty.$

Then,

$\displaystyle\frac{1}{-\frac{8}{(s_{*}+2)^{3}}+\mathcal{O}\left(s_{*}+2\right)^{-2}}=-\frac{2t}{x}$

meaning that

$\displaystyle(s_{*}+2)^{3}=\frac{16t}{x}+\mathcal{O}(s_{*}+2)^{4}=\frac{16t}{x}+\mathcal{O}\left(\frac{t}{x}\right)^{4/3}$

and so

$\displaystyle
s_{*}=-2+2\left(\frac{2t}{x}\right)^{1/3}+\mathcal{O}\left(\frac{t}{x}\right)^{4/9}$

and the point $s_{*}$ is then in the domain $\Re e(s)\in(-2,5)$ for
$x/t\to-\infty$. The integration contour $\Re e(s)=\sigma$ must now be
deformed to the contour
$\mathscr{C}=\left\\{s=s_{*}+iv,\,v\in\mathbb{R}\right\\}$, along which the
following hold,

  * (i)

$\psi(s;t,x)=\psi(s_{*};t,x)+\frac{1}{2}(s-s_{*})^{2}tW^{\prime\prime}(s_{*})+\mathcal{O}(s-s_{*})^{3},\,\,\frac{x}{t}\to-\infty$

  * (ii)

$\forall\rho>0,\,\exists\mu>0;\,\,\Re e(W(s_{*}+iv))\leq\Re
e(W(s_{*}))-\mu,\,\,\forall|v|\geq\rho.$

Figure 2: The function $Re[W\left(u+iv\right)]$ for $u\in(-2,-1),v\in(0,1)$.

The curve $\mathscr{C}$ may then be written as
$\mathscr{C}=\mathscr{C}_{1}\cup\mathscr{C}_{2}$,

$\displaystyle\mathscr{C}_{1}=\left\\{s=s_{*}+iv,\,|v|>\rho\right\\}$
$\displaystyle\mathscr{C}_{2}=\left\\{s=s_{*}+iv,\,|v|<\rho\right\\}$

and

$\displaystyle\int_{\mathscr{C}}\Phi(s)e^{\psi(s;t,x)}\,{\rm
d}s=\int_{\mathscr{C}_{1}}\Phi(s)e^{\psi(s;t,x)}\,{\rm
d}s+\int_{\mathscr{C}_{2}}\Phi(s)e^{\psi(s;t,x)}\,{\rm d}s=:I_{1}+I_{2}.$

In the term $I_{1}$, point (ii) gives $\Re e(W(s_{*}+iv))\leq\Re
e(W(s_{*}))-\mu$, and then,

$\displaystyle\left|\Phi(s)e^{\psi(s;t,x)}\right|=\left|\Phi(s)\right|e^{\Re
e(\psi(s;t,x))}\leq\left|\Phi(s)\right|e^{-\frac{xs_{*}}{2}+t(\Re
e(W(s_{*}))-\mu)}$

so that

$\displaystyle\left|\int_{\mathscr{C}_{1}}\Phi(s)e^{\psi(s;t,x)}\,{\rm
d}s\right|$ $\displaystyle\leq
C\int_{|v|>\rho}\left|\Phi(s_{*}+iv)\right|e^{-\frac{xs_{*}}{2}+t(\Re
e(W(s_{*}))-\mu)}$ $\displaystyle\leq Ce^{-\frac{xs_{*}}{2}+t(\Re
e(W(s_{*}))-\mu)}\int_{|v|>\rho}\left|\Phi(s_{*}+iv)\right|\,{\rm d}v.$

By hypothesis,

$\displaystyle\Phi(s_{*}+iv)$
$\displaystyle=\int_{0}^{\infty}p^{s_{*}+iv}A(p)\frac{\,{\rm
d}p}{p}=\int_{\mathbb{R}}e^{\frac{x(s_{*}+iv))}{2}}f_{0}(x)\,{\rm d}x$
$\displaystyle=\int_{\mathbb{R}}e^{\frac{xiv}{2}}e^{\frac{xs_{*}}{2}}f_{0}(x)\,{\rm
d}x=\sqrt{2\pi}\,\widehat{g}_{0}(v/2)\in L^{1}(\mathbb{R}).$

where $g_{0}(x)=f_{0}(x)e^{\frac{xs_{*}}{2}}$. On the other hand, if
$s=s_{*}+iv\in\mathscr{C}$ then $s-s_{*}=iv$ for some $v\in(-\rho,\rho)$ and
then

$\displaystyle I_{2}$
$\displaystyle=e^{\psi(s_{*};t,x)}\int_{|v|<\rho}\Phi(s_{*}+iv)e^{-\frac{1}{2}v^{2}tW^{\prime\prime}(s_{*})+\mathcal{O}(v)^{3}}\,{\rm
d}v$
$\displaystyle=e^{\psi(s_{*};t,x)}\int_{|v|<\rho}\Phi(s_{*}+iv)e^{-\frac{1}{2}v^{2}tW^{\prime\prime}(s_{*})}\left(1+\left(e^{\mathcal{O}(v)^{3}}-1\right)\right)\,{\rm
d}v$
$\displaystyle=e^{\psi(s_{*};t,x)}\int_{|v|<\rho}\Phi(s_{*}+iv)e^{-\frac{1}{2}v^{2}tW^{\prime\prime}(s_{*})}\left(1+\mathcal{O}(\rho)^{3}\right)\,{\rm
d}v$

where $W^{\prime\prime}(s_{*})>0$. Under the change of variables
$z=\sqrt{tW^{\prime\prime}(s_{*})}\,v$,

$\displaystyle I_{2}$
$\displaystyle=\frac{e^{\psi(s_{*};t,x)}}{\sqrt{tW^{\prime\prime}(s_{*})}}\int\limits_{|z|<\rho\sqrt{tW^{\prime\prime}(s_{*})}}\Phi\left(s_{*}+\mathcal{O}(\rho)\right)e^{-\frac{1}{2}z^{2}}\left(1+\mathcal{O}(\rho)\right)\,{\rm
d}z$
$\displaystyle=\frac{e^{\psi(s_{*};t,x)}}{\sqrt{tW^{\prime\prime}(s_{*})}}\Phi\left(s_{*}+\mathcal{O}(\rho)\right)\left(1+\mathcal{O}(\rho)\right)\int\limits_{|z|<\rho\sqrt{tW^{\prime\prime}(s_{*})}}e^{-\frac{1}{2}z^{2}}\,{\rm
d}z$
$\displaystyle=\sqrt{\frac{\pi}{2}}\frac{e^{\psi(s_{*};t,x)}}{\sqrt{tW^{\prime\prime}(s_{*})}}\Phi\left(s_{*}+\mathcal{O}(\rho)\right)\left(1+\mathcal{O}(\rho)\right)\left(1+\mathcal{O}\left(e^{-t\rho^{2}W^{\prime\prime}(s_{*})}\right)\right).$

Since

$\displaystyle
s_{*}=-2+2\left(\frac{2t}{x}\right)^{1/3}+\mathcal{O}\left(\frac{t}{x}\right)^{4/9}$

we compute

$\displaystyle\psi(s_{*};t,x)$
$\displaystyle=-\frac{xs_{*}}{2}+tW(s_{*})=x\left(1-\left(\frac{2\,t}{x}\right)^{1/3}+\mathcal{O}\left(\frac{t}{x}\right)^{4/9}\right)$
$\displaystyle\hskip
199.16928pt+\frac{4t}{(s_{*}+2)^{2}}+t\mathcal{O}\left(s_{*}+2\right)^{-1}$
$\displaystyle=x\left(1-\left(\frac{2t}{x}\right)^{1/3}+\mathcal{O}\left(\frac{t}{x}\right)^{4/9}\right)+4t\left(\frac{x}{16\,t}\right)^{2/3}\left(1+\mathcal{O}\left(\frac{t}{x}\right)^{1/9}\right)+t\,\mathcal{O}\left(\frac{x}{t}\right)^{1/3}$
$\displaystyle=x-(2t)^{1/3}x^{\frac{2}{3}}+t^{\frac{1}{3}}\left(\frac{x}{2}\right)^{\frac{2}{3}}+x\mathcal{O}\left(\frac{t}{x}\right)^{4/9}+t\mathcal{O}\left(\frac{x}{t}\right)^{\frac{5}{9}}+t\,\mathcal{O}\left(\frac{x}{t}\right)^{1/3}$
$\displaystyle=x-\left(2^{1/3}-2^{-2/3}\right)t^{\frac{1}{3}}x^{2/3}+t\mathcal{O}\left(\frac{x}{t}\right)^{\frac{5}{9}}.$

Also,

$\displaystyle
W^{\prime\prime}(s_{*})=\frac{24}{(s_{*}+2)^{4}}+\mathcal{O}\left(s_{*}+2\right)^{-3}$

which gives

$\displaystyle\frac{1}{\sqrt{tW^{\prime\prime}(s_{*})}}$
$\displaystyle=\frac{(s_{*}+2)^{2}}{2\sqrt{6t}}\left(1+\mathcal{O}\left(s_{*}+2\right)\right)$
$\displaystyle=\frac{1}{2\sqrt{6t}}\left(\frac{16t}{x}\right)^{2/3}\left(1+\mathcal{O}\left(\frac{t}{x}\right)^{1/9}\right)+\frac{1}{\sqrt{t}}\mathcal{O}\left(\frac{t}{x}\right)^{1/3}$
$\displaystyle=\frac{2\times
2^{1/6}}{\sqrt{3}}t^{1/6}x^{-2/3}+\frac{1}{\sqrt{t}}\mathcal{O}\left(\frac{t}{x}\right)^{1/3}.$

Inserting these into $I_{2}$, gives the stated behaviour of $f$ as
$x\to-\infty$. Analogously we conclude the behaviour at $x\to\infty$, as
$s_{*}\to 5$. ∎

######  Remark 3.18.

If the initial data $f_{0}$ satisfies $\widehat{g}_{0}\in L^{1}(\mathbb{R})$
for $g_{0}(x)=f_{0}(x)e^{\frac{xs_{*}}{2}}$, then $g$ defined as
$g(t,x)=f(t,x)e^{\frac{xs_{*}}{2}}$ for $t>0,x\in\mathbb{R}$ is such that
$\widehat{g}(t)\in L^{1}(\mathbb{R})$ for all $t>0$ as it immediately follows
from the following observation:

$\displaystyle\widehat{g}(t,\xi)$
$\displaystyle=\widehat{f}\left(t,\xi+\frac{s_{*}}{2}\right)=\widehat{G}\left(t,\xi+\frac{s_{*}}{2}\right)\widehat{f}_{0}\left(\xi+\frac{s_{*}}{2}\right)$
$\displaystyle=\widehat{G}\left(t,\xi+\frac{s_{*}}{2}\right)\widehat{g}_{0}\left(\xi\right)\in
L^{1}(\mathbb{R}),\,\,\text{since}\,\,\widehat{G}(t)\in
L^{\infty}(\mathbb{R}),\,\forall t>0.$

###  3.4 Back to the variables $A$ and $p>0$.

In this subsection, we translate our obtained results in Fourier variables
into the original variables $A$ and $p$. They can be summarized as follows.

  * 1.

Well-posedness: Equation (1.11) is well-posed in the spaces:

$\displaystyle(i)\quad L^{q}\left((0,\infty);\frac{\,{\rm
d}p}{p}\right),\quad\forall q\geq 1,$ (3.60) $\displaystyle(ii)\quad
L^{q}\left((0,\infty);p^{2\theta-1}\,{\rm d}p\right),\quad q\geq
1,\quad\theta\in(-1,5/2),$ (3.61)
$\displaystyle(iii)\quad\mathscr{Y}_{A,B}=\\{g\in
L^{\infty}_{\text{loc}}(0,\infty);\,||g||_{\mathscr{Y}}<\infty\\},$ (3.62)
$\displaystyle\qquad||g||_{\mathscr{Y}}=\sup_{0<p<1}(p^{2A}|g(p)|)+\sup_{p>1}(p^{2B}|g(p)|),$

where $A,B$ satisfy (• ‣ 1.4).

  * 2.

Non-negative solutions:

$A_{0}\in L^{q}\left((0,\infty);\frac{\,{\rm d}p}{p}\right)\ \&\ A_{0}\geq
0\Longrightarrow A(t)\geq 0,\quad\forall t>0.$ (3.63)

  * 3.

Conservation of the wave action:

$\displaystyle A_{0}\in
L^{1}(0,\infty)\Longrightarrow\int_{0}^{\infty}A(t,p)\,{\rm
d}p=\int_{0}^{\infty}A_{0}(p)\,{\rm d}p,$ (3.64)

and growth of the energy:

$\displaystyle A_{0}\in L^{1}((0,\infty);p^{2}\,{\rm
d}p)\Longrightarrow\int_{0}^{\infty}A(t,p)p^{2}\,{\rm
d}p=e^{W(3)t}\int_{0}^{\infty}A_{0}(p)p^{2}\,{\rm d}p.$ (3.65)

  * 4.

Long time behaviour: If $A_{0}\in\mathscr{Y}_{A,B}$ and $A,B$ satisfy (• ‣
1.4), then

$\displaystyle\int_{0}^{R}A(t,p)\,{\rm d}p=\int_{0}^{\infty}A_{0}(p)\,{\rm
d}p+Ce^{-\nu t}\int_{0}^{\infty}|A_{0}(p)|p^{-1/4}\,{\rm d}p.$ (3.66)

######  Remark 3.19.

If $A_{0}\in\mathscr{Y}_{A,B}$ and $A,B$ satisfy (• ‣ 1.4) it follows from
properties 3) and 4),

$\displaystyle\lim_{t\to\infty}\int_{R}^{\infty}A(t,p)\,{\rm d}p=0.$

If in addition $A_{0}\geq 0$, then for all $\varphi\in C([0,\infty))\cap
L^{\infty}((0,\infty))$,

$\displaystyle\lim_{t\to\infty}\int_{0}^{\infty}A(t,p)\varphi(p)\,{\rm
d}p=\varphi(0)m(A),\quad m(A)=\int_{0}^{\infty}A_{0}(p)\,{\rm d}p.$

Indeed first, $A(t)\geq 0$ for all $t>0$. By the continuity of $\varphi$, for
all $\varepsilon>0$ and $\rho$ small enough

$\displaystyle\int_{0}^{\infty}A(t,p)\varphi(p)\,{\rm
d}p-\int_{0}^{\infty}A_{0}(p)\,{\rm d}p\,\varphi(0)=$
$\displaystyle=\int_{0}^{\rho}A(t,p)\varphi(p)\,{\rm
d}p+\int_{\rho}^{\infty}A(t,p)\varphi(p)\,{\rm
d}p-\int_{0}^{\infty}A(t,p)\varphi(0)\,{\rm d}p$
$\displaystyle=\int_{0}^{\rho}A(t,p)(\varphi(p)-\varphi(0))\,{\rm
d}p+\int_{\rho}^{\infty}A(t,p)\varphi(p)\,{\rm
d}p-\int_{\rho}^{\infty}A(t,p)\varphi(0)\,{\rm d}p.$

We have

$\displaystyle\left|\int_{0}^{\rho}A(t,p)(\varphi(p)-\varphi(0))\,{\rm
d}p\right|\leq\varepsilon\|A_{0}\|_{1},$
$\displaystyle\left|\int_{\rho}^{\infty}A(t,p)\varphi(p)\,{\rm
d}p\right|\leq\|\varphi\|_{\infty}\int_{\rho}^{\infty}A(t,p)\,{\rm d}p\to
0,\,\,t\to\infty,$
$\displaystyle\left|\int_{\rho}^{\infty}A(t,p)\varphi(0)\,{\rm
d}p\right|\leq|\varphi(0)|\int_{\rho}^{\infty}A(t,p)\,{\rm d}p\to
0,\,\,t\to\infty.$

######  Remark 3.20.

By the previous Remark and (ii) of Proposition (3.15), for all $R>0$,

$\displaystyle\lim_{t\to\infty}\int_{0}^{R}A(t,p)p^{2}dp=0$

and then, for $\varepsilon>0$ as small as desired there exists $T$ large
enough, depending on $R$ and $\varepsilon$, such that

$\displaystyle\int_{R}^{\infty}A(t,p)p^{2}\,{\rm
d}p\geq\frac{1}{1+\varepsilon}\int_{0}^{\infty}A(t,p)p^{2}\,{\rm
d}p,\,\,\forall t\geq T.$

##  4 Appendix

###  4.1 Results of [13]

We record here some facts on the function $\mathcal{V}$ and its Mellin
transform, that have been proved in [13].

$\displaystyle\mathcal{V}(p)=R(p)+S(p)$ (4.1) $\displaystyle
R(p)=\left(\frac{2\log
p}{1+p^{2}}-\frac{2}{p\sqrt{1+p^{2}}}\text{arctanh}\left(\frac{p}{\sqrt{1+p^{2}}}\right)\right)\Theta(1-p)-$
$\displaystyle-\left(\frac{2\log
p}{p(1+p^{2})}+\frac{2}{p\sqrt{1+p^{2}}}\text{arctanh}\left(\frac{1}{\sqrt{1+p^{2}}}\right)\right)\Theta(p-1)$
(4.2)

and

$\displaystyle S(p)=\left(-\frac{2\log
p}{1-p^{2}}+\frac{2}{p\sqrt{1-p^{2}}}\text{arctanh}\left(\frac{p}{\sqrt{1-p^{2}}}\right)\right)\Theta(1-p)+$
$\displaystyle+\left(\frac{2\log
p}{p(p^{2}-1)}+\frac{2}{p\sqrt{p^{2}-1}}\text{arctanh}\left(\frac{1}{\sqrt{p^{2}-1}}\right)\right)\Theta(p-1)$
(4.3)

where $\Theta$ is the Heaviside’s function. The function $\mathcal{V}\in
C^{\infty}((0,1)\cup(1,\infty))$ is such that

$\displaystyle\mathcal{V}(p)\sim-4p^{2}(\log p-\frac{2}{3}+\cdots),\,\,p\to 0$
(4.4) $\displaystyle\mathcal{V}(p)\sim-\frac{4}{p^{5}}(\log
p+\frac{2}{3}+\cdots),\,\,p\to\infty$ (4.5)
$\displaystyle\mathcal{V}(p)\sim\frac{\pi}{\sqrt{2}\sqrt{|p-1|}}-1-\sqrt{2}\,\text{arcoth}(\sqrt{2})+\cdots,\,p\to
1.$ (4.6)

The function $\mathcal{V}$ has a singularity at $p=1$ of order $|p-1|^{-1/2}$.
Then $\mathcal{V}\in L^{p}(0,1)$ for $p\in[1,2)$. Moreover, we record the
following behaviour of the first derivative

$\displaystyle\mathcal{V}^{\prime}(p)\sim-\frac{4}{3}(-1+6\log
p)p+\mathcal{O}(p)^{3},\,\,p\to 0$ (4.7)
$\displaystyle\mathcal{V}^{\prime}(p)\sim-\frac{4(7+15\log
p)}{3p^{6}}+\mathcal{O}(p)^{-8},\,\,p\to\infty$ (4.8)
$\displaystyle\mathcal{V}^{\prime}(p)\sim\frac{\pi}{2\sqrt{2}\,(1-p)^{3/2}}-\frac{5\pi}{8\sqrt{2}\,\sqrt{1-p}}+\cdots,\,p\to
1^{-}$ (4.9)
$\displaystyle\mathcal{V}^{\prime}(p)\sim-\frac{\pi}{2\sqrt{2}\,(p-1)^{3/2}}-\frac{5\pi}{8\sqrt{2}\,\sqrt{p-1}}+\cdots,\,p\to
1^{+},$ (4.10)

while for the second derivative we have

$\displaystyle\mathcal{V}^{\prime\prime}(p)\sim-\frac{4}{3}(5+6\log
p)+\mathcal{O}(p),\,\,p\to 0$ (4.11)
$\displaystyle\mathcal{V}^{\prime\prime}(p)\sim-\frac{12(3+10\log
p)}{p^{7}}+\mathcal{O}(p)^{-8},\,\,p\to\infty$ (4.12)
$\displaystyle\mathcal{V}^{\prime\prime}(p)\sim\frac{3\pi}{4\sqrt{2}\,(1-p)^{5/2}}+\frac{5\pi}{16\sqrt{2}\,(1-p)^{3/2}}+\cdots,\,p\to
1^{-}$ (4.13)
$\displaystyle\mathcal{V}^{\prime\prime}(p)\sim\frac{3\pi}{4\sqrt{2}\,(p-1)^{5/2}}+\frac{5\pi}{16\sqrt{2}\,(p-1)^{3/2}}+\cdots,\,p\to
1^{+}.$ (4.14)

The Mellin transform of $\mathcal{V}$ was also obtained in [13]. It is a
meromorphic function $W_{\mathcal{V}}$ such that $\gamma+W_{\mathcal{V}}$ is
analytic on the domain $D$ where $\Re e(s)\in(-2,5)$, has two double poles at
$s=-2$ and $s=5$ and

$\displaystyle
W(s)=\frac{4}{(s+2)^{2}}+\frac{8}{3(s+2)}+\mathcal{O}(1),\,s\to-2,\,s\in D$
(4.15) $\displaystyle
W(s)=\frac{4}{(s-5)^{2}}-\frac{8}{3(s-5)}+\mathcal{O}(1),\,s\to 5,\,s\in D.$
(4.16)

Moreover,

$\displaystyle
W^{\prime}(s)=-\frac{8}{(s+2)^{3}}-\frac{8}{3(s+2)^{2}}+\mathcal{O}(1),\,s\to-2,\,s\in
D$ (4.17) $\displaystyle
W^{\prime}(s)=-\frac{8}{(s-5)^{3}}+\frac{8}{3(s-5)^{2}}+\mathcal{O}(1),\,s\to
5,\,s\in D,$ (4.18)

while

$\displaystyle
W^{\prime\prime}(s)=\frac{24}{(s+2)^{4}}+\frac{16}{3(s+2)^{3}}+\mathcal{O}(1),\,s\to-2,\,s\in
D$ (4.19) $\displaystyle
W^{\prime\prime}(s)=\frac{24}{(s-5)^{3}}-\frac{16}{3(s-5)^{2}}+\mathcal{O}(1),\,s\to
5,\,s\in D.$ (4.20)

By definition of $K$ and properties (4.4)-(4.6),

$\displaystyle K(x)$ $\displaystyle=\mathcal{V}(p)\sim-4p^{2}(\log
p-\frac{2}{3}+\cdots)=-4e^{x}\left(\frac{x}{2}-\frac{2}{3}+\cdots\right),\,\,x\to-\infty$
(4.21) $\displaystyle K(x)$
$\displaystyle=\mathcal{V}(p)\sim-\frac{4}{p^{5}}\left(\log
p+\frac{2}{3}+\cdots\right)=4e^{-\frac{5x}{2}}\left(\frac{x}{2}+\frac{2}{3}+\cdots\right),\,\,x\to\infty,$
(4.22) $\displaystyle K(x)$
$\displaystyle=\mathcal{V}(p)\sim\frac{\pi}{\sqrt{2}\sqrt{|e^{\frac{x}{2}}-1|}}-1-\sqrt{2}\,\text{arcoth}(\sqrt{2})+\cdots=\frac{\pi}{\sqrt{|x|}}-\frac{\pi\sqrt{|x|}}{8}+\cdots,\,x\to
0.$ (4.23)

Since,

$\displaystyle K^{\prime}(x)=\mathcal{V}^{\prime}(p)\frac{\,{\rm d}p}{\,{\rm
d}x}=\mathcal{V}^{\prime}(p)\frac{p}{2}$

it follows from (4.7)-(4.10),

$\displaystyle K^{\prime}(x)$
$\displaystyle\sim-\frac{4}{6}(-1+3x)e^{x}+\mathcal{O}(e^{\frac{x}{2}})^{4},\,x\to-\infty,$
(4.24) $\displaystyle K^{\prime}(x)$
$\displaystyle\sim-\frac{4}{6}\left(7+\frac{15}{2}x\right)e^{-\frac{5x}{2}}+\mathcal{O}(e^{\frac{x}{2}})^{-7},\,x\to\infty,$
(4.25) $\displaystyle K^{\prime}(x)$ $\displaystyle\sim\frac{\pi
e^{\frac{x}{2}}}{4\sqrt{2}\,(1-e^{\frac{x}{2}})^{3/2}}-\frac{5\pi
e^{\frac{x}{2}}}{16\sqrt{2}\,\sqrt{1-e^{\frac{x}{2}}}}+\cdots,\,x\to 0^{-}$
$\displaystyle\sim\frac{\pi}{2|x|^{3/2}}-\frac{3\pi}{8|x|^{1/2}}+\mathcal{O}(|x|)^{1/2},\,x\to
0^{-}$ (4.26) $\displaystyle K^{\prime}(x)$ $\displaystyle\sim-\frac{\pi
e^{\frac{x}{2}}}{4\sqrt{2}\,(e^{\frac{x}{2}}-1)^{3/2}}-\frac{5\pi
e^{\frac{x}{2}}}{16\sqrt{2}\,\sqrt{e^{\frac{x}{2}}-1}}+\cdots,\,x\to 0^{+}$
$\displaystyle\sim-\frac{\pi}{2|x|^{3/2}}-\frac{3\pi}{8|x|^{1/2}}+\mathcal{O}(|x|)^{1/2},\,x\to
0^{+}.$ (4.27)

Similarly, from (4.11) - (4.14), and
$K^{\prime\prime}(x)=\mathcal{V}^{\prime\prime}(p)\left(\frac{\,{\rm
d}p}{dx}\right)^{2}+\mathcal{V}^{\prime}(p)\frac{\,{\rm d}^{2}p}{\,{\rm
d}x^{2}}$, $\,{\rm d}p/\,{\rm d}x=e^{x/2}/2$,

$\displaystyle|K^{\prime\prime}(x)|\leq C|x|e^{x},\,\,x\to-\infty$ (4.28)
$\displaystyle|K^{\prime\prime}(x)|\leq C|x|e^{-\frac{5x}{2}},\,\,x\to\infty$
(4.29) $\displaystyle
K^{\prime\prime}(x)\sim\frac{3\pi}{4|x|^{5/2}}+\mathcal{O}\left(|x|^{3/2}\right),\,x\to
0.$ (4.30)

######  Proposition 4.1.

It holds that

$K\ast
K(x)=\begin{cases}\frac{8x^{3}}{3}e^{-\frac{5x}{2}}+\mathcal{O}(xe^{-\frac{5x}{2}}),\,x\to\infty\\\
-\frac{8x^{3}}{3}e^{x}+\mathcal{O}(xe^{x}),\,x\to-\infty.\end{cases}$ (4.31)

###### Proof.

Suppose $x\to\infty$ and write,

$\displaystyle K\ast K(x)=\int_{-\infty}^{0}K(x-y)K(y)\,{\rm
d}y+\int_{0}^{x}K(x-y)K(y)\,{\rm d}y+$
$\displaystyle+\int_{x}^{\infty}K(x-y)K(y)\,{\rm d}y.$ (4.32)

By (4.22),

$\displaystyle\int_{x}^{\infty}K(x-y)K(y)\,{\rm d}y\sim
4\int_{x}^{\infty}ye^{-\frac{5y}{2}}K(x-y)\,{\rm d}y$
$\displaystyle=4\int_{-\infty}^{0}(x-z)e^{-\frac{5(x-z)}{2}}K(z)\,{\rm d}z$
$\displaystyle=4xe^{-\frac{5x}{2}}\int_{-\infty}^{0}e^{\frac{5z}{2}}K(z)\,{\rm
d}z-4e^{-\frac{5x}{2}}\int_{-\infty}^{0}ze^{\frac{5z}{2}}K(z)\,{\rm d}z.$

where, by (4.21),

$\displaystyle\int_{-\infty}^{0}e^{\frac{5z}{2}}K(z)\,{\rm
d}z<\infty\,\,\text{ and}\,\,\,\int_{-\infty}^{0}ze^{\frac{5z}{2}}K(z)\,{\rm
d}z<\infty.$

This implies

$\displaystyle\int_{x}^{\infty}K(x-y)K(y)\,{\rm
d}y=4xe^{-\frac{5x}{2}}\int_{-\infty}^{0}e^{\frac{5z}{2}}K(z)\,{\rm
d}z+\mathcal{O}\left(e^{-\frac{5x}{2}}\right),\,x\to\infty.$ (4.33)

On the other hand, from (4.21), (4.22)

$\displaystyle\int_{-\infty}^{0}K(x-y)K(y)\,{\rm d}y\leq
C\int_{-\infty}^{0}(x-y)e^{-\frac{5(x-y)}{2}}ye^{y}\,{\rm d}y$
$\displaystyle\leq
Cxe^{-\frac{5x}{2}}\int_{-\infty}^{0}e^{\frac{7y}{2}}y\,{\rm
d}y+Ce^{-\frac{5x}{2}}\int_{-\infty}^{0}e^{\frac{7y}{2}}y^{2}\,{\rm
d}y=\mathcal{O}\left(xe^{-\frac{5x}{2}}\right).$ (4.34)

The second integral in the right hand side of (4.1) gives first,

$\displaystyle\int_{0}^{x}K(x-y)K(y)\,{\rm d}y=2\int_{0}^{x/2}K(x-y)K(y)\,{\rm
d}y.$

As $x\to\infty$, $x-y>x/2\to\infty$ and so by (4.22),

$\displaystyle\int_{0}^{x/2}K(x-y)K(y)\,{\rm d}y\sim
4\int_{0}^{x/2}(x-y)e^{-\frac{5(x-y)}{2}}K(y)\,{\rm d}y$
$\displaystyle=4xe^{-\frac{5x}{2}}\int_{0}^{x/2}e^{\frac{5y}{2}}K(y)\,{\rm
d}y-4e^{-\frac{5x}{2}}\int_{0}^{x/2}ye^{\frac{5y}{2}}K(y)\,{\rm d}y.$

For $R\in(0,x/2)$ large ,

$\displaystyle\int_{0}^{x/2}e^{\frac{5y}{2}}K(y)\,{\rm
d}y=\int_{0}^{R}e^{\frac{5y}{2}}K(y)\,{\rm
d}y+\int_{R}^{x/2}e^{\frac{5y}{2}}K(y)\,{\rm d}y$

For $y>R$,
$K(y)=ye^{-\frac{5y}{2}}+\mathcal{O}\left(e^{-\frac{5y}{2}}\right)$, then

$\displaystyle\int_{R}^{x/2}e^{\frac{5y}{2}}K(y)\,{\rm
d}y=\int_{R}^{x/2}e^{\frac{5y}{2}}\left(4ye^{-\frac{5y}{2}}+\mathcal{O}\left(e^{-\frac{5y}{2}}\right)\right)\,{\rm
d}y$ $\displaystyle=4\int_{R}^{x/2}y\,{\rm
d}y+\int_{R}^{x/2}e^{\frac{5y}{2}}\mathcal{O}\left(e^{-\frac{5y}{2}}\right)\,{\rm
d}y=2\left(\frac{x^{2}}{4}-R^{2}\right)+\mathcal{O}\left(x\right)$
$\displaystyle=\frac{x^{2}}{2}+\mathcal{O}(R^{2})+\mathcal{O}\left(x\right),\,x\to\infty,$

and then,

$\displaystyle\int_{0}^{x/2}e^{\frac{5y}{2}}K(y)\,{\rm
d}y=\frac{x^{2}}{2}+\mathcal{O}(R^{2})+\mathcal{O}\left(x\right),\,x\to\infty.$

A similar argument in the integral

$\displaystyle\int_{0}^{x/2}e^{\frac{5y}{2}}yK(y)\,{\rm
d}y=\int_{0}^{R}ye^{\frac{5y}{2}}K(y)\,{\rm
d}y+\int_{R}^{x/2}ye^{\frac{5y}{2}}K(y)\,{\rm d}y$

yields,

$\displaystyle\int_{R}^{x/2}ye^{\frac{5y}{2}}K(y)\,{\rm
d}y=\int_{R}^{x/2}e^{\frac{5y}{2}}y\left(4ye^{-\frac{5y}{2}}+\mathcal{O}\left(e^{-\frac{5y}{2}}\right)\right)dy$
$\displaystyle=4\int_{R}^{x/2}y^{2}\,{\rm
d}y+\int_{R}^{x/2}e^{\frac{5y}{2}}y\mathcal{O}\left(e^{-\frac{5y}{2}}\right)dy=\frac{4}{3}\left(\frac{x^{3}}{8}-R^{3}\right)+\mathcal{O}\left(x^{2}\right)$
$\displaystyle=\frac{x^{3}}{6}+\mathcal{O}(R^{3})+\mathcal{O}\left(x^{2}\right),\,x\to\infty.$

Then,

$\displaystyle\int_{0}^{x/2}e^{\frac{5y}{2}}yK(y)dy=\frac{x^{3}}{6}+\mathcal{O}(R^{3})+\mathcal{O}\left(x^{2}\right),\,x\to\infty.$

It follows,

$\displaystyle\int_{0}^{x/2}K(x-y)K(y)dy=4xe^{-\frac{5x}{2}}\Bigg{(}\frac{x^{2}}{2}+\mathcal{O}(R^{2})+\mathcal{O}\left(x\right)\Bigg{)}-$
$\displaystyle-4e^{-\frac{5x}{2}}\Bigg{(}\frac{x^{3}}{6}+\mathcal{O}(R^{3})+\mathcal{O}\left(x^{2}\right)\Bigg{)}=\frac{4x^{3}}{3}e^{-\frac{5x}{2}}+\mathcal{O}\left(x^{2}e^{-\frac{5x}{2}}\right),\,\,x\to\infty$

and

$\displaystyle\int_{0}^{x}K(x-y)K(y)dy=\frac{8x^{3}}{3}e^{-\frac{5x}{2}}+\mathcal{O}\left(x^{2}e^{-\frac{5x}{2}}\right),\,\,x\to\infty.$
(4.35)

Property (4.31) for $x\to\infty$ follows from (4.33), (4.1) and (4.35). A
similar argument shows (4.31) for $x\to-\infty$. ∎

Acknowledgments. The research of the first author is supported by grant
PID2020-112617GB-C21 of MCIN, grant IT1247-19 of the Basque Government and
grant RED2022-134784-T funded by MCIN/AEI/10.13039/501100011033. The second
author acknowledges support from the Chapman Fellowship at Imperial College
London.

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