1 Introduction

Correlations between the superfluid component and the normal fluid in a uniform condensed Bose gas, at temperature below but close to the condensation temperature, and for a small number density of condensed atoms, may be described by the equation

$$\begin{aligned} \frac{\partial n}{\partial t}(t, p)&=n_c(t)I_3(n(t))(p)\qquad t>0,\; p\in \mathbb {R}^3, \end{aligned}$$
(1.1)
$$\begin{aligned} I_3(n)(p)&=\!\!\iint _{(\mathbb {R}^3)^2}\!\!\big [R(p, p_1, p_2)\!-\!R(p_1, p, p_2)\!-\!R(p_2, p_1, p) \big ]\hbox {d}p_1\hbox {d}p_2, \end{aligned}$$
(1.2)
$$\begin{aligned} R(p, p_1, p_2)&= \left[ \delta ( |p|^2-|p_1|^2- |p_2|^2) \delta (p-p_1-p_2)\right] \nonumber \\&\quad \times \left[ n_1n_2(1+n)-(1+n_1)(1+n_2)n \right] , \end{aligned}$$
(1.3)

where n(tp) represents the density of particles in the normal gas that at time \(t>0\) have momentum p and \(n_c(t)\) is the density of the condensate at time t, that satisfies

$$\begin{aligned} \frac{\hbox {d}n_c}{\hbox {d}t} (t)=-n_c(t)\int _{ \mathbb {R}^3 } I_3(n(t))(p))\hbox {d}p \qquad t>0. \end{aligned}$$
(1.4)

The collision integral \(I_3\) in (1.1) was first derived in [9] and [20] and their treatment was afterwards extended to a trapped Bose gas. By including Hartree–Fock corrections to the energy of the excitations the so called ZNG system was obtained (cf. [18]). On the interest of system (1.1), (1.4) for the description of condensed Bose gases see also [19, 24, 26].

Of course, even in presence of the superfluid component, interactions between particles in the normal gas continue to take place. In order to take them into account the Nordheim–Boltzmann collision operator should be added at the right hand side of the equation (1.1) as it is done in the references above. Such an operator is generally written as

$$\begin{aligned}&I_4(n)(p)=\int _{D(\omega _1)}\!\!\!\! f_3f_4(1+f_1)(1+f_2) -f_1f_2(1+ f_3)(1+f_4)\, W\, \hbox {d}\omega _3 \hbox {d} \omega _4, \end{aligned}$$
(1.5)
$$\begin{aligned}&f_i = f(\omega _i ),\,\,\,\omega _i=|p_i|^2,\,\,\,i=1, 2, 3, 4;\,\,\,\,\, \omega _2 = \omega _3+\omega _4-\omega _1, \end{aligned}$$
(1.6)
$$\begin{aligned}&W= {\min \left( \sqrt{\omega _1},\sqrt{\omega _2}, \sqrt{ \omega _3},\sqrt{ \omega _4}\right) \over \sqrt{\omega _1}}, \end{aligned}$$
(1.7)
$$\begin{aligned}&D\left( \omega _1\right) = \left\{ \left( \omega _3 , \omega _4\right) : \omega _3>0, \omega _4>0, \omega _3+ \omega _4\ge \omega _1>0\right\} . \end{aligned}$$
(1.8)

It may be of some interest to understand the respective effects of each of the two terms \(I_3\) and \(I_4\) at the different stages of the evolution of the condensate and its thermal cloud. It is known that the term \(I_4\) provokes by itself the finite time blow up of some solutions of Nordheim equation (cf. [15]). The finite time blow up phenomena does not seem to be expected in presence of \(I_3\) alone. On the contrary, as proved below, the Cauchy problem for the linearization of (1.11.4) around an equilibria is globally well posed in a suitable functional space; its solutions conserve the natural physical quantities and converge to a suitable equilibrium. Some regularizing effects are also observed as shown in Theorem 1.9 and Theorem 3.2, below. Such effects are not observed for \(I_4\) (cf. [12]). However, a detailed understanding of the respective mathematical properties of \(I_3\) and \(I_4\) is still missing, and that makes difficult to precisely foresee what may be expected in presence of both terms. Several of these questions are considered in [3] where in particular some regularizing property of the term \(I_3\) is proved in a different functional setting.

Other theoretical models do exist to describe Bose gases in presence of a condensate (cf. [23]) but ZNG system, and (1.1), (1.4) in particular, are very appealing by their simplicity and are well suited for analytical PDE methods.

The two functions of time,

$$\begin{aligned} {\mathcal {N}}(t)=\int _{ \mathbb {R}^3 }n(t, p)\hbox {d}p,\,\,\,\,\,\,{\mathcal {E}}(t)=\int _{ \mathbb {R}^3 }n(t, p)|p|^2\hbox {d}p, \end{aligned}$$
(1.9)

give, respectively, the total number of particles and the total energy of the normal fluid part in the gas, with density function n(tp). The total number of particles at time t in the system condensate-normal fluid is \(n_c(t)+{\mathcal {N}}(t)\) and its total energy is \({\mathcal {E}}(t)\). It formally follows from (1.1), (1.4) that these two quantities are constant in time: \(\mathcal E(t)={\mathcal {E}}(0)\) and \(n_c(t)+{\mathcal {N}}(t)=n_c(0)+{\mathcal {N}}(0)\) for all \(t>0\). This corresponds to the conservation of the total mass and energy property that is satisfied by the particle system in the physical description (cf. for example [9]). It is also well known that equation (1.1) has a family of non trivial equilibria,

$$\begin{aligned}&n_0(p)=\left( e^{\beta |p|^2}-1\right) ^{-1} , \end{aligned}$$
(1.10)

where the mass of the particles is taken to be \(m=1/2\) and \(\beta \) is a positive constant related to the temperature of the gas whose particle’s density is at the equilibrium \(n_0\). It is easily checked that \(R(p, p_k, p_\ell )\equiv 0\) in (1.3) for \(n=n_0\).

Our purpose is to prove the existence of classical solutions to the Cauchy problem for the “radially symmetric linearization” of (1.1)–(1.4) around an equilibrium \(n_0\), and describe some of their properties. Such linearization is deduced through the change of variables (cf. [10, 13])

$$\begin{aligned} n(t, p)&=n_0(p)+n_0(p)(1+n_0(p))\Omega (t, |p|)=n_0(p)+\frac{\Omega (t, |p|)}{4 \, \sinh ^2 \left( \frac{\beta |p|^2}{2}\right) }, \end{aligned}$$
(1.11)
$$\begin{aligned} x&=\frac{\sqrt{\beta }\,|p|}{\sqrt{2}}=\frac{\sqrt{\beta }\, k }{\sqrt{2}}, \quad u(t, x)=\frac{\Omega (t, |p|)}{k^{2}}, \end{aligned}$$
(1.12)

and keeping only linear terms with respect to u in (1.1). Since equation (1.4) is linear with respect to \(n_c\) its linearization (1.15) follows by just keeping the terms of \(I_3(n_0(p)+n_0(p)(1+n_0(p))|p|^2u (t, |p|))\) that are linear with respect to u. This finally reads, for dimensionless variables in units which minimize the number of prefactors, as

$$\begin{aligned} \frac{\partial u}{\partial t }(t, x)&=p_c(t){\mathcal {L}}(u(t)) \end{aligned}$$
(1.13)
$$\begin{aligned} {\mathcal {L}}(u(t))&=\int _0^\infty (u(t , y)-u(t , x)) M(x, y) \hbox {d}y \end{aligned}$$
(1.14)
$$\begin{aligned} \frac{\hbox {d}p_c}{\hbox {d}t}(t)&=-p_c(t) \int _0^\infty \int _0^\infty W (x, y)(u(t, y)-u(t, x))y^4x^2\hbox {d}y\hbox {d}x, \end{aligned}$$
(1.15)

where, for all \(x>0\), \(y>0\), \(x\not =y\),

$$\begin{aligned} M(x, y)&= \left( \frac{1}{\sinh |x^2-y^2|}-\frac{1}{\sinh (x^2+y^2) } \right) \frac{y^3\sinh x^2}{x^3\sinh y^2}, \end{aligned}$$
(1.16)
$$\begin{aligned} W(x, y)&= \frac{M(x, y)}{(\sinh x^2)^2}\frac{x^2}{y^4} \end{aligned}$$
(1.17)
$$\begin{aligned}&= \left( \frac{1}{\sinh |x^2-y^2|}-\frac{1}{\sinh (x^2+y^2) } \right) \frac{1}{xy\sinh x^2\sinh y^2}. \end{aligned}$$
(1.18)

With some abuse of notation the function \(\nu _0(x)=n_0(p)\) is still denoted \(n_0(x)\).

For uniform condensed Bose gases at very low temperatures and large number density of condensed atoms, the limit of the ZNG system deduced in references [9] and [20] is slightly different. In that limit, the interactions involving only particles in the thermal cloud are neglected, ZNG system is reduced to (1.1), (1.4), up to lower order terms but with a different expression for \(R(p, p_1, p_2)\) in (1.3) since the dispersion relation and scattering amplitude are different. Related works in the mathematical literature for the isotropic case may be found in [1, 2, 27]. The non isotropic linearized system around an equilibrium is treated in [11].

1.1 The Isotropic Linearization of (1.1), (1.4).

The linearization of (1.1), detailed in [13] and recalled in [10], is briefly presented here for the sake of completeness. When \(R(p, p_1, p_2)\!-\!R(p_1, p, p_2)\!-\!R(p_2, p_1, p)\) is written in terms of the function \(\Omega \) defined in (1.11) and only linear terms in \(\Omega \) are kept, the result is

$$\begin{aligned} n_0(1+n_0)\frac{\partial \Omega (t)}{\partial t}=&\, n_c(t)L _{ I_3 }(\Omega (t)), \end{aligned}$$
(1.19)
$$\begin{aligned} L _{ I_3 }(\Omega (t))=&\int _0^\infty \left( {\mathscr {U}}(k, k')\Omega (t, k')-{\mathscr {V}}(k, k')\Omega (t, k)\right) k'^2\hbox {d}k', \end{aligned}$$
(1.20)
$$\begin{aligned} {\mathscr {U}}(k, k')&=\frac{16 n_\textrm{c} a^2 }{kk'} \Big [ \theta (k-k') \nonumber \\&\quad \times n_0(\omega (k))[1+n_0(\omega (k'))][1+n_0(\omega (k)-\omega (k'))] + (k \leftrightarrow k') \Big ] \nonumber \\&\quad - \, n_0(\omega (k)+\omega (k'))[1+n_0(\omega (k))][1+n_0(\omega (k'))] , \end{aligned}$$
(1.21)
$$\begin{aligned} {\mathscr {V}}(k, k') =&\frac{16 n_\textrm{c} a^2}{kk'}\Big [ \theta (k-k') \nonumber \\&\quad \times n_0(\omega (k))[1+n_0(\omega (k'))][1+n_0(\omega (k)-\omega (k'))]+ (k \leftrightarrow k')\Big ], \end{aligned}$$
(1.22)

where a is the s-wave scattering length, \(k=|p|\) and \(k'=|p'|\). The functions \({\mathscr {U}}(k, k')\) and \({\mathscr {V}}(k, k')\) have a non integrable singularity along the diagonal \(k=k'\). However, these singularities cancel each other when the two terms are combined as in (1.20) as far as it is assumed that, for all \(t>0\), \(\Omega (t)\in C^\alpha (0, \infty )\) for some \(\alpha >0\). However, the integrand \(\left( {\mathscr {U}}(k, k')\Omega (t, k')-\mathscr {V}(k, k')\Omega (t, k)\right) \) cannot be split as for example in the linearization of Boltzmann equations for classical particles. However an explicit calculation shows that, for all \(k>0\),

$$\begin{aligned} L _{ I_3 }(\omega )(k)=\int _0^\infty \left( \mathscr {U}(k, k') k'^2-{\mathscr {V}}(k, k')k^2\right) k'^2\hbox {d}k'=0 \end{aligned}$$
(1.23)

from which we deduce, for all \(k>0\),

$$\begin{aligned} \int _0^\infty&\left( {\mathscr {U}}(k, k') \frac{k'^2}{k^2}\Omega (t, k)-{\mathscr {V}}(k, k')\Omega (t, k)\right) k'^2\hbox {d}k' =\frac{\Omega (t, k) }{k}L _{ I_3 }(\omega )(k)=0. \end{aligned}$$

We may then write,

$$\begin{aligned} L _{ I_3 }(\Omega (t))&=\int _0^\infty \left( {\mathscr {U}}(k, k')\Omega (t, k')-{\mathscr {V}}(k, k')\Omega (t, k)\right) k'^2\hbox {d}k'\\&=\int _0^\infty {\mathscr {U}}(k, k')\left( \frac{\Omega (t, k')}{k'^2}- \frac{\Omega (t, k)}{k^2}\right) k'^4\hbox {d}k'. \end{aligned}$$

Since equation (1.4) is linear with respect to \(n_c\), its linearization (1.15) follows by just keeping the terms of \(I_3(n_0(p)+n_0(p)(1+n_0(p))|p|^2u (t, |p|))\) that are linear with respect to u. The linearized system then reads as

$$\begin{aligned}&n_0(1+n_0)\frac{\partial \Omega (t, k)}{\partial t}=p_c(t)\int _0^\infty {\mathscr {U}}(k, k')\left( \frac{\Omega (t, k')}{k'^2}- \frac{\Omega (t, k)}{k^2}\right) k'^4\hbox {d}k', \end{aligned}$$
(1.24)
$$\begin{aligned}&p_c'(t)=-p_c(t)\int _0^\infty \int _0^\infty {\mathscr {U}}(k, k')\left( \frac{\Omega (t, k')}{k'^2}- \frac{\Omega (t, k)}{k^2}\right) k'^4k^2\hbox {d}k'\hbox {d}k, \end{aligned}$$
(1.25)

or, in terms of \({\tilde{\Omega }} (t, k)=\Omega (t, k)/k^2\),

$$\begin{aligned} \frac{\partial {\tilde{\Omega }}(t, k)}{\partial t}=&p_c(t) \int _0^\infty \frac{ {\mathscr {U}}(k, k')}{n_0(k)(1+n_0(k))k^2}\left( {\tilde{\Omega }} (t, k')- {\tilde{\Omega }} (t, k)\right) k'^4\hbox {d}k' \\ p_c'(t)=&-p_c(t)\int _0^\infty \int _0^\infty {\mathscr {U}}(k, k')\left( {\tilde{\Omega }} (t, k')- {\tilde{\Omega }} (t, k)\right) k'^4k^2\hbox {d}k'\hbox {d}k. \end{aligned}$$

Since \(\left( k^2n_0(k)(1+n_0(k))\right) ^{-1}=4k^{-2}\sinh ^2\left( \frac{\beta k^2}{2}\right) \),

$$\begin{aligned} \frac{\partial {\tilde{\Omega }}(t, k)}{\partial t}=4p_c(t) \int _0^\infty \left[ {\mathscr {U}}(k, k')\sinh ^2\left( \frac{\beta k^2}{2}\right) \frac{k'^4}{k^2}\right] \left( {\tilde{\Omega }} (t, k')- {\tilde{\Omega }} (t, k)\right) \hbox {d}k'. \end{aligned}$$

Use of the change of variables (1.111.12) in (1.24) yields system (1.13), (1.15) for \((u, p_c)\), after scaling the time variable to get rid of some positive numerical constants.

1.2 A Nonlinear Approximation

Another approximation of the system (1.1), (1.4) is possible where, in the equation (1.4), the function n is replaced by \(n_0+n_0(1+n_0)x^2 u\) in the nonlinear collision term \(I_3\) given by (1.2) to obtain the system

$$\begin{aligned}&\frac{\partial v}{\partial t }(t, x)={\tilde{p}}_c(t)\int _0^\infty (v(t , y)-v(t , x)) M(x, y) \hbox {d}y \end{aligned}$$
(1.26)
$$\begin{aligned}&\frac{\partial {\tilde{p}}_c}{\partial t}(t)=-{\tilde{p}}_c(t)\int _{0}^\infty I_3\left( n_0+n_0(1+n_0)x^2 v(t, x)\right) x^2\hbox {d}x, \end{aligned}$$
(1.27)

instead of (1.13), (1.15). In that way the non linearity of \(I_3\) in the equation for \({\tilde{p}}_c\) is kept, but the conservation in time of \({\tilde{p}}_c(t)+N(t)\) does not hold, and so an important global property of the original system (1.1)–(1.4) is lost. As a consequence the time existence of the solutions to system (1.26), (1.27) cannot be proved to be \((0, \infty )\). Then, system (1.26), (1.27) is not too satisfactory to describe global properties of the particle’s system. But it may be a better approximation of the local properties of the solutions to the nonlinear system of equations (1.1), (1.4). In order to avoid any confusion, system (1.26), (1.27) is considered in the Appendix.

1.3 Further Motivation

It is known that for all non negative measure \(n _{ in }\) with a finite first moment, and for every constant \(\rho >0\), system (1.1)–(1.4) has a weak solution \((n(t), n_c(t))\) with initial data \((n _{ in }, \rho )\) that satisfies the conservation of mass and energy (cf. [7]). For all \(t>0\), n(t) is a non negative measure that does not charge the origin, with finite first moment, and \(n_c(t)>0\). However, one basic aspect of the non equilibrium behavior of the system condensate–normal fluid is the growth of the condensate after its formation (cf. [4, 18, 23] and references therein). In the kinetic formulation (1.1)–(1.4), this behavior is driven by the integral of \(I_3(n)\) in the right hand side of equation (1.4). A was shown in [25], the behaviour of that term crucially depends on the behavior of n(tp) as \(|p|\rightarrow 0\) (this was discussed also in [7, 21, 26]). If, for example, the measure n(t) is a radially symmetric, bounded function near the origin then, from a simple use of Fubini’s Theorem,

$$\begin{aligned} \int _{ \mathbb {R}^3 }I_3(n(t))(p)\hbox {d}p=C\int _0^\infty x^3 n(t, x ) \hbox {d}x \end{aligned}$$

for some constant \(C>0\) independent of n, and this would give a monotone decreasing behavior of \(n_c(t)\). On the contrary, as it is shown in [25], if the measure n(t) is a function such that

$$\begin{aligned} n(t, p) \underset{p\rightarrow 0 }{\sim }\ a(t)|p|^{-2} \end{aligned}$$
(1.28)

for some \(a(t)>0\), and satisfies some Hölder regularity property with respect to p in a neighborhood of the origin, then for some other constant \(C_1>0\) independent of n,

$$\begin{aligned} \int _{ \mathbb {R}^3 }I_3(n(t))(p)\hbox {d}p=-C_1a^2(t)+C\int _0^\infty x^3 n(t, p )\hbox {d}x. \end{aligned}$$
(1.29)

On the other hand, it was proved in [7] that if the measure \(|p|^2 n(t, p)\) has no atomic part and has an algebraic behavior as \(|p|\rightarrow 0\) then it satisfies (1.28). Both results in [25] and [7] assume some regularity of the solution n with respect to p, although no regular solutions to (1.1) are known yet. The existence of regular classical solutions to (1.1)–(1.4) satisfying (1.28) is one of the motivations of our present work.

Since (1.28) is the behavior of the equilibrium \(n_0\) (with \(a(t)\equiv \beta \)), it is natural to first consider the existence of such regular solutions for the linearization of (1.1) around \(n_0\). Because of the singular behavior (1.28) of \(n_0\) near the origin, the linear operator \({\mathcal {L}}\) in (1.14) has regularizing effects. Similar regularizing effects may be expected also in the non linear equation (1.1).

1.4 Basic Arguments and Main Results

The function \(p_c(t)\) in the right hand side of (1.13) may be absorbed by the change of variables,

$$\begin{aligned} \tau =\int _0^tp_c(s)\hbox {d}s,\,\,f(\tau , x)=u(t, x), \end{aligned}$$
(1.30)

to obtain

$$\begin{aligned} \frac{\partial f(\tau , x)}{\partial \tau }={\mathcal {L}}(f(\tau ))(x). \end{aligned}$$
(1.31)

This equation may be written as

$$\begin{aligned}&\frac{\partial f}{\partial \tau }(\tau , x)=L(f(\tau ))(x)+F(f(\tau ))(x) \end{aligned}$$
(1.32)
$$\begin{aligned}&L(f)(x)=\int _0^\infty (f (y)-f (x) )\left( \frac{1}{|x^2-y^2|}-\frac{1}{x^2+y^2} \right) \frac{y}{x}\hbox {d}y , \end{aligned}$$
(1.33)
$$\begin{aligned}&F(f(\tau ))(x)=({\mathcal {L}}-L)(f(\tau ))(x) \end{aligned}$$
(1.34)

where, from (1.14), (1.33) and (1.35), the operator F is given by,

$$\begin{aligned} F(f)(x)&=-f(x) \int _0^\infty T(x, y) \hbox {d}y+ \int _0^\infty T(x, y) f (y)\hbox {d}y \end{aligned}$$
(1.35)
$$\begin{aligned} T(x, y)&=\frac{y^3\sinh x^2}{x^3\sinh y^2} \left( \frac{1}{\sinh |x^2-y^2|}-\frac{1}{\sinh (x^2+y^2) } \right) \nonumber \\&\quad -\frac{y}{x}\left( \frac{1}{|x^2-y^2|}-\ \frac{1}{x^2+y^2} \right) . \end{aligned}$$
(1.36)

The equation (1.32) is then solved as a perturbation of

$$\begin{aligned} \frac{\partial f}{\partial \tau }(\tau , x)=L(f(\tau ))(x) \end{aligned}$$
(1.37)

with a forcing term in (1.35).

Equation (1.37) is of interest by itself and has been considered in [10]. For example, one may consider what is seen somehow as the “classical approximation” of the nonlinear equation (1.1)–(1.3), where the factor of the Dirac’s measure in (1.3) is replaced by \(n_1n_2-n(n_2+n_1)\). The resulting equation also appears in the theory of wave turbulence for nonlinear optic waves (cf. for example [8, 28]), related to the Schrödinger equation. The function \(|p|^{-2}\) is an equilibrium solution of that equation, and (1.37) is its isotropic linearization, around that equilibrium.

It is proved in [10] that equation (1.37) has a fundamental solution \(\Lambda \in C((0,\infty ); L^1(0, \infty ))\) that satisfies (1.37) in \({\mathscr {D}}'((0, \infty )\times (0, \infty ))\) and almost every \(t>0\), \(x>0\). For all initial data \(f_0\in L^1\), there exists a weak solution of (1.37), denoted \(S(t)f_0\),

$$\begin{aligned} S(t)f_0(x)=\int _0^\infty \Lambda \left( \frac{t}{y}, \frac{x}{y} \right) f_0(y)\frac{\hbox {d}y}{y},\,\,\forall t>0,\,\,\forall x>0, \end{aligned}$$
(1.38)

such that \(S(\cdot )f_0\in C([0, \infty ); L^1(0, \infty ))\), \(S(t)f_0\in C([0, \infty ))\) for all \(t>0\) and (1.37) is satisfied in \({\mathscr {D}}'((0, \infty )\times (0, \infty ))\). It was also proved that if \(f_0\in L^1(0, \infty )\cap L^\infty _{ loc }(0, \infty )\) then \(L(u)\in L^\infty ((0, \infty )\times (0, \infty ))\), \(u_t\in L^\infty ((0, \infty )\times (0, \infty ))\) and (1.37) is satisfied pointwise, for \(t>0\) and \(x>0\). (cf. Appendix for more detailed statements).

Once the Cauchy problem for equation (1.32) is solved using the semigroup S(t), the change of time variable in (1.30) is inverted to obtain the function u(t), and deduce \(p_c(t)\) using the conservation of mass of system and equation (1.15). Our first result, then, is as follows:

Theorem 1.1

Suppose that \(u_0\in L^1(0, \infty )\) satisfies

$$\begin{aligned} |||u_0|||_\theta \equiv \sup _{ 0<x<1 }x^\theta |u_0(x)|+\sup _{ x>1 }|u_0(x)|<\infty \end{aligned}$$
(1.39)

for some \(\theta \in (0, 1)\). Then, there exists a pair \((u, p_c)\),

$$\begin{aligned}&u\in C([0, \infty ); L^1 (0, \infty ))\cap L^\infty ((\delta , \infty ); L^\infty (0, \infty )), \forall \delta >0, \end{aligned}$$
(1.40)
$$\begin{aligned}&p_c\in C([0, \infty )), \end{aligned}$$
(1.41)

such that, for each \(t>0\), u(t) is locally Lipschitz on \((0, \infty )\) and, for all almost every \(t>0\) and \(x>0\),

$$\begin{aligned}&\frac{\partial u(t, x)}{\partial t}=p_c(t){\mathcal {L}}(u(t))(x), \end{aligned}$$
(1.42)
$$\begin{aligned}&\frac{\textrm{d}p_c(t)}{\textrm{d}t}=-p_c(t)\int _0^\infty {\mathcal {L}}(u(t))(x) n_0(x^2)(1+n_0(x^2))x^4\textrm{d}x. \end{aligned}$$
(1.43)

Moreover,

$$\begin{aligned} \frac{\partial u}{\partial t},\, {\mathcal {L}}(u) \in L^\infty _{ loc }((0, \infty )\times (0, \infty ))\cap L^1((0, T);L^1(0, \infty )),\,\,\forall T>0, \end{aligned}$$
(1.44)

and there exists a function \(H\in L^\infty ((\delta , 0)\times (0, \delta ))\) for all \(\delta >0\), defined in (5.12), such that

$$\begin{aligned}&\left| \frac{\partial u(t, x)}{\partial t} \right| + |{\mathcal {L}}(u)(t, x)|\le C\left( \sup _{ 0<y<1 }y^{\theta }|f_0(y)| +||u_0|| _{ L^\infty (1, \infty ) }+ ||u_0||_1\right) H(t, x), \end{aligned}$$
(1.45)
$$\begin{aligned}&\forall T>0, \exists C_T>0;\,\, ||u(t)||_1\le C_T||u_0||_1,\,\,\forall t\in (0, T), \end{aligned}$$
(1.46)
$$\begin{aligned}&\forall \delta >0,\,\,||u(t)||_\infty \le C(\theta ) \left( ||u_0|| _{ L^\infty (1, \infty ) }+ \delta ^{-\theta }\sup _{ 0<y<1 }y^\theta |u_0(y)|+ ||u_0||_1\right) , \,\forall t\ge \delta . \end{aligned}$$
(1.47)

For all \(\varphi \in C^1_0(0, \infty )\), the map \( t\mapsto \displaystyle {\int _0^\infty \varphi (x)u(t, x)\textrm{d}x} \) belongs to \(W _{ loc }^{1,1}(0, \infty )\) and for almost every \(t>0\),

$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}t}\int _0^\infty \varphi (x)u(t, x)\textrm{d}x=\int _0^\infty L(u(t))(x)\varphi (x)\textrm{d}x+\int _0^\infty F(u(t, x))\varphi (x)\textrm{d}x. \end{aligned}$$
(1.48)

Theorem 1.1, shows that the possible singular behavior as \(x^{-\theta }\) for some \(\theta \in (0, 1)\) of the initial data \(u_0\) at the origin is instantaneously regularized to \(u(t)\in L^\infty (0, \infty )\) for all \(t>0\). This is a direct consequence of the same property of the equation (1.37), where L may be seen as a pseudo differential operator with a symbol whose logarithmic growth at infinity induces a logarithmic regularizing effect (cf. Remark 6.6 below). The equation (1.37) enjoys a “local \(C^1\)” regularizing property (cf. Lemma (6.4) below), that cannot be extended to the equation (1.42), due precisely to its local character. However, the equation (1.42) regularizes the initial data \(u_0=\delta _1\) and gives a weak solution \(u \in C((0, \infty ); L^1(0, \infty ))\) (cf. Theorem 3.2). No regularizing effects have been observed for the isotropic linearization of the “classical approximation” of the Boltzmann Nordheim collision integral (1.5) around its stationary solutions ( [8, 16]).

In view of (1.11), (1.12) and (1.30), if u is a solution of (1.32) given by Theorem 1.1, the pair of functions

$$\begin{aligned} n(t, p)&=n_0(p)+n_0(p)(1+n_0(p))|p|^2 u (t , |p|) \end{aligned}$$
(1.49)
$$\begin{aligned} p_c(t)&=\exp \left( \int _0^t\int _0^\infty \int _0^\infty \frac{M(x, y)}{(\sinh x^2)^2}(u(t, y)-u(t, x))\frac{x^2}{y^4}\hbox {d}y\hbox {d}x\hbox {d}s\right) \end{aligned}$$
(1.50)
$$\begin{aligned}&\equiv \exp \left( \int _0^t\int _0^\infty {\mathcal {L}}(u(t))(x) \frac{x^2\hbox {d}x \hbox {d}s}{(\sinh x^2)^2}\right) \end{aligned}$$
(1.51)

may be seen as an approximated solution of (1.1), (1.4), as far as \(n_0(p)(1+n_0(p))|p|^2 u (t, |p|)\) remains small compared to \(n_0\). In view of (1.9) it is natural to look at the quantities

$$\begin{aligned} N(t )&=\int _0^\infty n_0(x)(1+n_0(x)u(t , x)x^4\hbox {d}x \end{aligned}$$
(1.52)
$$\begin{aligned} E(t )&=\int _0^\infty n_0(x)(1+n_0(x))u(t , x)x^6\hbox {d}x. \end{aligned}$$
(1.53)

These represent, respectively, the variation of the total number of particles and of energy caused by the initial perturbation \(n_0(p)(1+n_0(p))|p|^2 u(0)\) of the equlibrium \(n_0\). Let us also define,

$$\begin{aligned}&N_0=\int _0^\infty n_0(x)(1+n_0(x)x^4\hbox {d}x \end{aligned}$$
(1.54)
$$\begin{aligned}&E_0=\int _0^\infty n_0(x)(1+n_0(x))x^6\hbox {d}x. \end{aligned}$$
(1.55)

The two following properties then hold true:

Corollary 1.2

Let \(u_0\) and u be as in Theorem 1.1 and n defined by (1.49). Then,

$$\begin{aligned}&E(t )=E(0),\,\,\,\forall t >0, \end{aligned}$$
(1.56)
$$\begin{aligned}&p_c(t)+N(t)=p_c(0)+N(0),\,\,\forall t>0. \end{aligned}$$
(1.57)

Corollary 1.3

Let \(u_0\) and u be as in Theorem 1.1. Then,

$$\begin{aligned}&\lim _{ t \rightarrow \infty } \int _0^\infty |u(t, x)-C_*|^2n_0(x)(1+n_0(x))x^6\textrm{d}x=0, \end{aligned}$$
(1.58)
$$\begin{aligned}&\lim _{ t \rightarrow \infty } \int _0^\infty |u(t, x)-C_*|\,n_0(x)(1+n_0(x))x^4\textrm{d}x=0, \end{aligned}$$
(1.59)
$$\begin{aligned}&\hbox {where}\,\,\,\,C_*=\frac{E(0)}{E_0}. \end{aligned}$$
(1.60)

It follows from Corollary 1.3 that the mass and the energy variations due to the perturbation \(n_0(1+n_0)|p|^2u (t)\) tend to the mass and energy of \( C_*n_0(1+n_0)|p|^2\), and this however small the perturbation is at infinity, even if, for example, \(u_0\) is compactly supported. This kind of energy flux towards infinity could be expected, since it is well known to happen in the nonlinear homogeneous version of wave turbulence type of the system (1.1), (1.4) and is called direct energy cascade ( [8, 28] and [16, 27]).

Corollary 1.4

Let \(u_0\) and u be as in Theorem 1.1. Then, the function \(p_c\in C[0, \infty )\) is bounded on \([0, \infty )\) and

$$\begin{aligned}&\lim _{ t\rightarrow \infty }p_c(t)=p_c(0)\exp \left( -{\mathcal {M}}_ \infty \right) \end{aligned}$$
(1.61)
$$\begin{aligned}&{\mathcal {M}} _ \infty =C_*\int _0^\infty n_0(1+n_0)x^4\textrm{d}x - \int _0^\infty n_0(1+n_0)u_0(x)x^4\textrm{d}x. \end{aligned}$$
(1.62)

1.5 Some Remarks

Several remarks follow from the previous results.

1.5.1 On the Formal Approximation

The approximation of (1.1), (1.4) by (1.13), (1.15) may be expected to be reasonable only as long as the perturbation remains small with respect to \(n_0\),

$$\begin{aligned} n_0(1+n_0)|\Omega (t)|<<n_0, \end{aligned}$$
(1.63)

and this requires \(x^2|f(t, x)|\) small for \(x\rightarrow \infty \). However, although it could be proved that (1.63) holds for small values of time if it holds at \(t=0\), it follows from (1.58) that it can not be true for all \(t>0\). Notice indeed that, for all \(R\ge R_0>0\),

$$\begin{aligned}&\left| \int _R^\infty x^6n_0(1+n_0)|u(t, x)|\hbox {d}x-\int _R^\infty x^6n_0(1+n_0)|C_*| \right| \nonumber \\&\quad \le \int _R^\infty x^6n_0(1+n_0)|u(t, x)-C_*|\hbox {d}x\nonumber \\&\quad \le \left( \int _R^\infty x^6n_0(1+n_0)|u(t, x)-C_*|^2\hbox {d}x\right) ^{1/2} \left( \int _R^\infty x^6n_0(1+n_0)\hbox {d}x\right) ^{1/2} \end{aligned}$$
(1.64)

and the right hand side tends to zero as \(t\rightarrow \infty \). If, on the other hand, we had \(x^2|u(t, x)|\le C\), for some \(C>0\), \(R_0>0\) and \(t_0>0\) for all \(x>R_0\) and all \(t>t_0\),

$$\begin{aligned} \int _R^\infty x^6n_0(1+n_0)|u(t, x)|\hbox {d}x\le \frac{C}{R^2}\int _R^\infty x^6n_0(1+n_0)\hbox {d}x,\,\,\forall R>R_0, t>t_0, \end{aligned}$$

and then, for \(R>C/|C_*|\) and all \(t>t_0\),

$$\begin{aligned}&\left| \int _R^\infty x^6n_0(1+n_0)|C_*|\hbox {d}x-\int _R^\infty x^6n_0(1+n_0)|u(t, x)| \right| \\&\quad \ge \left( |C_*|-\frac{C}{R^2}\right) \int _R^\infty x^6n_0(1+n_0)\hbox {d}x>0,\,\,\forall t>t_0. \end{aligned}$$

and this would contradict (1.64). System (1.13), (1.15) may then be considered “close to” (1.1), (1.4) only for small values of t. Of course, u could be such that, for some C(t) that tends to \(\infty \) with t, \(x^2|u(t, x)|\le C(t)\) for all \(x>0\).

1.5.2 The Behavior of the Perturbation as \(\varvec{|p|\rightarrow 0}\)

For all \(t>0\), the perturbation \(n_0(1+n_0)\Omega (t, p)\) of \(n_0\) satisfies (1.28), for any \(f_0\) as in the hypothesis of Theorem 1.1, where a(t) is given in Proposition 4.16. The behavior \(|p|^{-2}\) at the origin (that of the equilibria of (1.1), (1.4)) is then instantaneously fixed, whatever the behavior at the origin of \(f_0\) may be, as far as the hypothesis of Theorem 1.1 are satisfied.

1.5.3 The Function \(p_c(t)\)

In view of Corollary 1.4, if the initial data \(u_0\) is such that

$$\begin{aligned} \frac{E(0)}{E_0}N_0<N(0) \end{aligned}$$
(1.65)

or equivalently,

$$\begin{aligned} {\mathcal {M}}_\infty =C_*\int _0^\infty n_0(1+n_0)x^4\hbox {d}x - \int _0^\infty n_0(1+n_0)u_0(x)x^4\hbox {d}x<0, \end{aligned}$$

then \(\lim _{ t\rightarrow \infty }p_c(t)>p_c(0)\), and conversely.

Condition (1.65) and its converse are both compatibles with \(n_0(1+n_0)x^2u_0\) being a small perturbation of \(n_0\). For example

$$\begin{aligned}&n_0(1+n_0)x^2u_0(x)=\left( \frac{1.02}{1+x^2} -1\right) n_0,\nonumber \\&N(0)\approx -0.344949,\,\,\,E(0)\approx -0.523546 \nonumber \\&\frac{N(0)}{E(0)}\approx 0.658872< \frac{N_0}{E_0}\approx 0.778949 \Longleftrightarrow \frac{E(0)}{E_0}N_0<N(0) \end{aligned}$$
(1.66)

and

$$\begin{aligned}&n_0(1+n_0)x^2u_0(x)=0.2 \hbox {Arctg}\left( \frac{x}{10}\right) n_0,\nonumber \\&N(0)\approx 0.0163705,\,\,\,E(0)\approx 0.0238295 \nonumber \\&\frac{E(0)}{N(0)}\approx 1.45564> \frac{E_0}{N_0}\approx 1.28378 \Longleftrightarrow \frac{E(0)}{E_0}N_0>N(0). \end{aligned}$$
(1.67)

1.6 Very Low Temperature and Large \(n_c\)

The linearization of system (1.1), (1.4) for large number density of condensed atoms and very low temperature may be performed following similar arguments as to those above (cf. [6] and [11]). No regularizing effects have been observed and the existence of a first positive eigenvalue and spectral gap for a suitable integrable operator ( [6] and [17]) provide a convergence rate to the equilibrium for a large set of initial data (cf. [11], Theorem 2.2) A necessary and sufficient condition on \(p_c(0)\) to have a global solution.

2 The Operator F

Equation (1.32) may be treated as a perturbation of (1.37) only whenever the term F(f) in (1.35) is bounded in spaces where the properties of the solutions of (1.37) may be used. The purpose of this section is to establish that this is the case.

Proposition 1.5

(i) For all \(g\in L^\infty (0, \infty )\), \(F(g)\in L^\infty (0, \infty )\) and

$$\begin{aligned} ||F(g)|| _{ \infty }\le 2M||g|| _{ \infty }. \end{aligned}$$

(ii) For all \(g\in L^1(0, \infty )\),

$$\begin{aligned} ||F(g)||_1\le C_F ||g||_1\,\,\, C_F=(M+{\widetilde{M}}), \end{aligned}$$

where

$$\begin{aligned} M=\sup _{ x>0 }\int _0^\infty |T(x, y)|\textrm{d}y,\,\,\,{\widetilde{M}}=\sup _{ y>0 }\int _0^\infty |T(x, y)|\textrm{d}x. \end{aligned}$$

(iii) For all \(\theta \in [0, 1)\) there exists a positive constant \(C(\theta )\) depending on \(\theta \), such that, if \(g\in L^\infty _{ loc }(0, \infty )\) satisfies \( |||g|||_\theta <\infty \) then \(|||F(g)|||_\theta \le C(\theta )|||g|||_\theta .\)

(iv) For all \(g\in L^1(0, \infty )\cap L^\infty _{ loc }(0, \infty )\), \(F(g)\in L^1(0, \infty )\cap L^\infty _{ loc }(0, \infty )\).

(v) For all \(R>0\) there exists a constant \(C=C(R)>0\) such that:

$$\begin{aligned} ||F(g)|| _{ \infty }\le C\left( ||g|| _{ L^1(0, R) }+||g|| _{ L^\infty (R, \infty ) }\right) ,\,\,\forall g\in&L^1(0, \infty )\cap L^\infty (R, \infty ). \end{aligned}$$

Proposition 2.1 follows from estimates of the kernel T defined in (1.36), that we split as follows:

$$\begin{aligned}&T(x, y)=T_1(x, y)+T_2(x, y)\\&T_1(x, y)=\frac{y}{x}\left( \frac{1}{\sinh |x^2-y^2|}-\frac{1}{|x^2-y^2|}-\frac{1}{\sinh (x^2+y)^2} +\frac{1}{|x^2-y^2|}\right) \\&T_2(x, y)=\frac{y^3}{x^3}\left( \frac{\sinh x^2}{\sinh y^2} -\frac{x^2}{y^2}\right) \left( \frac{1}{\sinh |x^2-y^2|}-\frac{1}{\sinh (x^2+y)^2 } \right) . \end{aligned}$$

The kernels \(T_1\) and \(T_2\) are estimated in the two next Propositions.

Proposition 1.6

$$\begin{aligned}&\forall R>0, \exists C_R>0;\,\,| T_1(x, y)|\le C_Rxy,\,\,\,\forall y\in (0, R),\,\forall x\in (0, R), \end{aligned}$$
(2.1)
$$\begin{aligned}&| T_1(x, y)|\le \frac{Cy}{x }\left( \min (x^2 , y^2 )+{\mathcal {O}}(x^2 +x '^2)^3\right) ,0\le x\le 1/2,\,\,0\le y\le 1/2, \end{aligned}$$
(2.2)
$$\begin{aligned}&|T_1(x, y)|\le C, \hbox {if}\,\,\, x+y>1, |y-x|\le 1/8, \end{aligned}$$
(2.3)
$$\begin{aligned}&\forall x\in (0, 1),\,y>\min (2, 3x/2),\,\,\,| T_1(x, y)|\le \frac{Cx}{y^3}. \end{aligned}$$
(2.4)
$$\begin{aligned}&\hbox {If}\,\, x+y>1, \,\,1/8<|x-y|<x/2,\nonumber \\&| T_1(x, y)|\le \frac{Cy}{x}\left( \frac{2\min (x,y)^2}{(x^2-y^2)(x^2+y^2)}+\left| \frac{1}{\sinh |x^2-y^2|}-\frac{1}{\sinh (x^2 +y^2)}\right| \right) , \end{aligned}$$
(2.5)
$$\begin{aligned}&\forall |x-y|>x/2,\nonumber \\&|T_1(x, y)|\le \frac{Cy}{x }\left( \frac{e^{-C_1\max (x,y)^2}}{\left( 1-e^{-\beta (x^2+y^2)}\right) \left( 1-e^{-C_1\max (y, x)^2} \right) }+\frac{\min (y,x)^2}{\max (y, x)^4}\right) , \end{aligned}$$
(2.6)
$$\begin{aligned}&|T_1(x, y)|\le \frac{Cy}{x}\left( \frac{1}{\sinh \frac{3x^2}{4}}+\frac{\min (y,x)^2}{\max (y, x)^4}\right) \,\,\,\hbox {if}\,\,\,0<y<\frac{x}{2}. \end{aligned}$$
(2.7)

Proof

0.- Proof of (2.1). When \(y\in (0, R)\) and \(x\in (0, R)\). We may use the series expansion of the function \(1/\sinh x\) to obtain

$$\begin{aligned}&h(z)=\frac{1}{z}-\frac{1}{\sinh z}=\sum _{ n=1 }^\infty \frac{2(2^{2n-1}-1)B_n}{(2n)!}z^{2n-1},\\&B_0=1,\, B_n=\sum _{ \ell =1 }^{n-1}\left( {\begin{array}{c}n\\ \ell \end{array}}\right) \frac{B_\ell }{n+1-\ell },\,\,n\ge 1,\,\,(\hbox {Bernoulli's numbers}),\\&h'(z)=\sum _{ n=1 }^\infty (2n-1)\frac{2(2^{2n-1}-1)B_n}{(2n)!}z^{2(n-1)}. \end{aligned}$$

For \(y\in (0, R)\) and \(x\in (0, R)\),

$$\begin{aligned}&h(x^2+y^2)-h(|x^2-y^2|)=2x^2h'(\xi ),\,\,\,\xi \in (y^2-x^2, y^2+x^2)\\&|h(x^2+y^2)-h(|x^2-y^2|)|\le 2x^2\sup _{ \xi \in (0, 2R^2) } |h'(\xi )| \end{aligned}$$

and

$$\begin{aligned} |T_1(x, y)|\le 2xy\sup _{ \xi \in (0, 2R^2) } |h'(\xi )|,\,\,\forall y\in (0, R), x\in (0, R). \end{aligned}$$

1.- Proof of (2.2). Consider first the set where \(y <1/2\) and \(x <1/2\), and use the Taylor’s expansion of \(1/\sinh z \) around \(z =0\),

$$\begin{aligned}&\frac{1}{\sinh |y^2-x^2 | }=\frac{2}{\beta |x^2-y^2|}-\frac{\beta |x^2-y^2|}{12}+{\mathcal {O}}(|x^2-y^2|)^3,\,\,|x-y|\rightarrow 0\\&\frac{1}{ \sinh (y^2+x^2 )}=\frac{2}{\beta (x^2+x^2)}-\frac{\beta (x^2+y^2)}{12}+{\mathcal {O}}((x^2+y^2))^3,\,\,|x+y|\rightarrow 0. \end{aligned}$$

Then

$$\begin{aligned}&\left( \frac{1}{\sinh |y^2-x^2 | } -\frac{1}{\beta |x^2-y^2|}-\frac{1}{\sinh (y^2+x^2 ) } +\frac{1}{\beta (x^2+y^2)}\right) \\&\quad =\frac{\min (x^2 ,y^2)}{3}+{\mathcal {O}}(x^2 +y^2)^3,\\&\qquad |T_1(x, y)|\le \frac{Cy}{x }\left( \min (x^2 , x '^2)+\mathcal {O}(x^2 +y^2)^3\right) , \end{aligned}$$

and this proves (2.2).

2.- Proof of (2.4). When \(x\in (0, 1)\) and \(y>x\), in the identity,

$$\begin{aligned} \frac{1}{\sinh |x^2-y^2|} - \frac{1}{\sinh (x^2 +y^2)}=\frac{\sin (y^2+x^2 )-\sinh (y^2-x^2 )}{\sin (y^2+x^2 )\sinh (y^2-x^2 )}. \end{aligned}$$

we use the mean value Theorem to obtain

$$\begin{aligned}&\sin (y^2+x^2 )-\sinh (y^2-x^2 )=-2x^2 \cosh \xi \\&\xi \equiv \xi (y^2-x^2 , y^2+x^2 )\in (y^2-x^2, y^2+x^2)\subset (y^2-1, y^2+1) \end{aligned}$$

and then,

$$\begin{aligned} \cosh \xi \le C\cosh (y^2). \end{aligned}$$

We deduce,

$$\begin{aligned} \left| \frac{1}{\sinh |x^2-y^2|} - \frac{1}{\sinh (x^2+y^2)}\right| \le C\frac{x^2\cosh y^2}{\sin (y^2+x^2)\sinh (y^2-x^2)} \le C\frac{x^2}{\sinh y^2}. \end{aligned}$$

On the other hand, since \(y>\min (2, 3x/2)\), \(y^2-x^2>Cy^2\) and

$$\begin{aligned} \frac{1}{|x^2-y^2|}-\frac{1}{|x^2+y^2|}=\frac{2x^2}{(y^2-x^2)(x^2+y^2)}\le C\frac{x^2}{y^4}, \end{aligned}$$

it follows that, \( |T_1(x, y)|\le C\frac{x^2}{y^4}\frac{y}{x}\le C\frac{x}{y^3} \) and that proves (2.4).

3.- Proof of (2.3) Suppose now that \(x+y>1\), and \(|x-y|\le 1/8\). We may still use the Taylor’s expansion of \(1/\sinh (|x^2-y^2|) \) around \(|x^2-y^2|=0\),

$$\begin{aligned} \frac{1}{\sinh |x^2-y^2|}-\frac{1}{|x^2-y^2|}=-\frac{ |x^2-y^2|}{6}+{\mathcal {O}}(|x^2-y^2|)^3,\,\,|x-y|\rightarrow 0. \end{aligned}$$

Then, if \(x+y>1\), \(|y-x|\le 1/8\),

$$\begin{aligned} |T_1(x, y)|&\le \frac{Cy}{x}\left( |x^2-y^2|-\frac{1}{\sinh |y^2+x^2| } +\frac{1}{ |x^2+y^2|}\right) \\&\le \frac{Cy}{x}\left( 1+\frac{1}{ (x^2+y^2)}\right) \le \frac{Cy}{x}, \end{aligned}$$

and this proves (2.3).

4.- Proof of (2.5). Consider now the cases where \(x+y>1\), \(|y-x|>1/8\) and \(|x-y|<x/2\). Since

$$\begin{aligned} \frac{1}{|x^2-y^2|}-\frac{1}{|x^2+y^2|}=\frac{2\min (x, y)^2}{(y^2-x^2)(x^2+y^2)}\le \frac{16\min (x, y)^2}{(y+x)(x^2+y^2)}, \end{aligned}$$

if \(x+y>1\), \(|y-x|>1/8\) and \(|x-y|<x/2\),

$$\begin{aligned} |T_1(x, y)|&\le \frac{Cy}{k}\left( \frac{2\min (x, y)^2}{(x^2-y^2)(x^2+y^2)}+\left| \frac{1}{\sinh |x^2-y^2|}-\frac{1}{\sinh (x^2+y^2)}\right| \right) \end{aligned}$$
(2.8)
$$\begin{aligned}&\le \frac{Cy}{x}\left( \frac{2\min (x, y)^2}{(y+x)(x^2+y^2)}+\frac{1}{\sinh \frac{ |x+x '|}{8}}+\frac{1}{\sinh (x^2+y^2)}\right) \nonumber \\&\le \frac{Cy}{x}\left( \frac{2\min (x, y)^2}{(y+x)(x^2+y^2)}+\frac{1}{\sinh \frac{ |x+x '|}{8}}\right) . \end{aligned}$$
(2.9)

The estimate (2.8) is nothing but (2.5).

5.- Proof of (2.6) On the other hand, if \(x>1\) and \(|y-x|>x/2\), we may still use hat,

$$\begin{aligned} \frac{1}{|x^2-y^2|}-\frac{1}{|x^2+y^2|}=\frac{2\min (x, y)^2}{(y^2-x^2)(x^2+y^2)}&\le {\left\{ \begin{array}{ll} \frac{Cx^2}{y^4},\,\,\forall y>3x/2>0\\ \frac{Cy^2}{x^4},\,\,\,\forall y<x/2. \end{array}\right. } \end{aligned}$$

If \(y>x\),

$$\begin{aligned}&\left( \frac{1}{\sinh |y^2-x^2| }- \frac{1}{\sinh |y^2+x^2| }\right) = \frac{e^{x^2+y^2}\left( 1-e^{-2(x^2+y^2)}-e^{-2x^2}+e^{-2y^2}\right) }{e^{x^{2}+y^2}\left( 1-e^{-2(x^2+y^2)}\right) \left( e^{y^2-x^2}-e^{-y^2+x^2}\right) }\\&\quad =\frac{1-e^{-2(x^2+y^2)}-e^{-2x^2}+e^{-2y^2}}{\left( 1-e^{-2(x^2+y^2)}\right) e^{y^2-x^2}\left( 1-e^{-2(y^2-x^2)} \right) } \le \frac{e^{-(y^2-x^2)}}{\left( 1-e^{-2(x^2+y^2)}\right) \left( 1-e^{-2(y^2-x^2)} \right) }. \end{aligned}$$

Then, if \(y>3x/2\),

$$\begin{aligned} \left( \frac{1}{\sinh |y^2-x^2| }- \frac{1}{\sinh (y^2+x^2)}\right)&\le \frac{e^{-(y^2-x^2)}}{\left( 1-e^{-2(x^2+y^2)}\right) \left( 1-e^{-2 (y^2-x^2)} \right) }\\&\le \frac{e^{-\frac{5 }{9}y^2}}{\left( 1-e^{-2 (x^2+y^2)}\right) \left( 1-e^{-\frac{5 }{9}y^2} \right) }. \end{aligned}$$

If \(x>2y\),

$$\begin{aligned} \left( \frac{1}{\sinh |y^2-x^2|}- \frac{1}{\sinh |y^2+x^2| }\right)&\le \frac{e^{-(x^2-y^2)}}{\left( 1-e^{-2 (x^2+y^2)}\right) \left( 1-e^{-2 (x^2-y^2)} \right) }\\&\le \frac{e^{-\frac{3}{4}x^2}}{\left( 1-e^{-2 (x^2+y^2)}\right) \left( 1-e^{-\frac{3 }{4}x^2}\right) }, \end{aligned}$$

which proves (2.6). However, we also have, in the case \(x>2y\), \(x^2-y^2>3x^2/4\), and so

$$\begin{aligned} \frac{1}{\sinh |y^2-x^2| }\le \frac{1}{\sinh \frac{3 x^2 }{4}};\,\,\,\, \frac{1}{\sinh (y^2+x^2)}\le \frac{1}{\sinh x^2}, \end{aligned}$$

which proves (2.7). \( \quad \square \)

Proposition 1.7

$$\begin{aligned}&\forall R>0,\exists C_R>0;\,\,\, |T_2(x, y)|\le C_R,\,\,\forall x\in (0, R),\,\,\forall y\in (0,R) \end{aligned}$$
(2.10)
$$\begin{aligned}&\forall \delta>0,\,\,\exists C_\delta>0;\,\,\,\forall x\in (0, 1),\,\forall y>x+\delta :\,\,\,\,\,\, |T_2(x, y)|\le C_\delta \frac{x y}{\sinh y^2} \end{aligned}$$
(2.11)
$$\begin{aligned}&\forall \,\,\,x+y>1, |x-y|<1/8:\,\,|T_2(x, y)|\le C \end{aligned}$$
(2.12)
$$\begin{aligned}&\forall \,\,\,x+y>1,\frac{1}{8}< |x-y|<x/2: \nonumber \\&|T_2x, y)|\le C\left| \frac{1}{\sinh |x^2 -y^2|}-\frac{1}{\sinh (x^2 +y^2)}\right| . \end{aligned}$$
(2.13)

If \(x+y>1\) and \(|x-y|>x/2\),

$$\begin{aligned}&|T_2(x, y)|\le \frac{C}{\sinh \frac{ 3 x^2 }{4} },\,\,y<x/2 \end{aligned}$$
(2.14)
$$\begin{aligned}&|T_2(x, y)|\le Ce^{-(y^2 -x^2 )}\left( e^{-(y^2 -x^2)}+\frac{x^2 }{y^2} \right) \frac{y^3}{x^3}.\,\,\,y\ge 3x/2 \end{aligned}$$
(2.15)

Proof

1.- Proof of (2.10). Let us write first,

$$\begin{aligned}&\left| \frac{\sinh x^2 }{\sinh y^2}-\frac{x^2}{y^2}\right| = \left| \frac{y^2\sinh x- x^2 \sinh y^2}{y^2\sinh y^2}\right| \\&\quad =\left| \frac{y^2\sinh x^2-x^2\sinh x^2+x^2\sinh x^2- x^2 \sinh y^2}{y^2\sinh y^2}\right| \end{aligned}$$

and

$$\begin{aligned} y^2\sinh x^2 -x^2 \sinh y^2=y^2(\sinh x^2 -\sinh y^2)+(y^2-x^2 )\sinh y^2. \end{aligned}$$

Using Taylor’s expansion we have, for some \(\xi (x^2, y^2)\) between \(x^2 \) and \(y^2\),

$$\begin{aligned} y^2&(\sinh x^2 -\sinh y^2)\\&=y^2\left( (x^2 -y^2)\cosh y^2+\frac{1}{2}(x^2 -y^2)^2+\frac{1}{6}(x^2 -y^2)^3\cosh \xi (x^2, y^2)\right) \end{aligned}$$

and

$$\begin{aligned}&y^2(\sinh x^2 -\sinh y^2)+(y^2-x^2 )\sinh y^2\\&\quad =y^2\left( (x^2 -y^2)\cosh y^2+\frac{1}{2}(x^2 -y^2)^2+\frac{1}{6}(x^2 -y^2)^3\cosh \xi (x^2 , x)^2\right) \\&\qquad +(y^2-x^2 )\sinh y^2\\&\quad =(x^2 -y^2)\left( y^2 \cosh y^2 -\sinh y^2\right) +\frac{y^2}{2}(x^2 -y^2)^2\\&\qquad +\frac{y^2}{6}(x^2 -y^2)^3\cosh \xi (x^2 , y^2). \end{aligned}$$

We deduce that, for all \(x\in (0, R)\) and \(y\in (0, R)\),

$$\begin{aligned}&|y^2(\sinh x^2 -\sinh y^2)+(y^2-x^2 )\sinh y^2| \le C_R|x^2 -y^2 \\&\quad \times |( y^6+y^2|x^2 -y^2|+|x^2 -y^2|^2y^2) \le C'_R|x^2 -y^2|\left( y^6+y^2x^2\right) . \end{aligned}$$

Similarly, using the change \(x\leftrightarrow y\),

$$\begin{aligned} |y^2(\sinh x^2 -\sinh y^2)+(y^2-x^2 )\sinh y^2|\le C_R|x^2 -y^2|\left( x^6+y^2x^2\right) . \end{aligned}$$

It follows that, for all \(x\in (0, R)\), \(y\in (0, R)\),

$$\begin{aligned} |y^2(\sinh x^2 -\sinh y^2)+(y^2-x^2 )\sinh y^2|\le C_R|x^2 -y^2| \\ \times \min \{y^2(x^2+y^4), x^2(x^4+y^2)\}\le C_R|x^2 -y^2| x^2y^2 \end{aligned}$$

and

$$\begin{aligned}&\left| \frac{\sinh x^2 }{\sinh y^2}-\frac{x^2}{y^2}\right| \le C_R\frac{|x^2 -y^2| x^2\, y^2 }{y^4}. \end{aligned}$$

On the other hand, since \(\sinh \) is locally Lipschitz,

$$\begin{aligned} \left| \frac{1}{\sinh |y^2-x^2|}- \frac{1}{\sinh (x^2+y^2)}\right| \le \frac{C_R\min (x^2, y^2)}{|y^2-x^2 |(y^2+x^2 )} \end{aligned}$$

for some constant \(C_R>0\). We deduce that

$$\begin{aligned} \left| \frac{\sinh x^2 }{\sinh y^2}-\frac{x^2}{y^2}\right| \left| \frac{1}{\sinh |y^2-x^2|}- \frac{1}{\sinh (x^2+y^2)}\right|&\le C_R\frac{x^2 y^2\min (x^2, y^2) }{y^4(y^2+x^2 )}\\&\le C_R\frac{x^2\min (x^2, y^2) }{y^2(x^2 +y^2)} \end{aligned}$$

and

$$\begin{aligned} |T_2(x, y)|\le C_R\frac{y \min (x^2, y^2)}{x(x^2+y^2)}\le C_R,\,\,\forall x\in (0, R),\, y\in (0, R). \end{aligned}$$

This proves (2.10).

2.- Proof of (2.11). Suppose now that \(x\in (0, 1)\) and \(y>x+\delta \) for any \(\delta >0\). Then, as we have ween in the proof of Proposition 2.2,

$$\begin{aligned} \left| \frac{1}{\sinh |x^2-y^2|} - \frac{1}{\sinh (x^2 +y^2)}\right| \le C\frac{x^2 \cosh y^2}{\sin (y^2+x^2 )\sinh (y^2-x^2 )} \le C_\delta \frac{x^2 }{\sinh y^2}. \end{aligned}$$

We have also, since \(\sinh x^2 \le \sinh 1\),

$$\begin{aligned} \left| \frac{\sinh x^2 }{\sinh y^2}-\frac{x^2}{y^2}\right|&=\left| \frac{y^2\sinh x^2 -x^2 \sinh y^2}{y^2\sinh y^2}\right| \\&\le C\frac{x^2 (y^2+\sinh y^2)}{y^2\sinh y^2}=Cx^2 \left( \frac{1}{\sinh y^2}+\frac{1}{y^2}\right) \le \frac{Cx^2 }{y^2}. \end{aligned}$$

Then,

$$\begin{aligned} |T_2(x, y)|\le C_\delta \frac{x^2 }{\sinh y^2}\frac{x^2 }{y^2}\frac{y^3}{x^3}=C_\delta \frac{x\ y}{\sinh y^2}, \end{aligned}$$

and this proves (2.11).

3.- Proof of (2.12) We consider now the case where \(x+y>1\) and \(|x-y|<1/8\). Suppose again that \(0<y^2<x \). We may still use the Taylor’s expansion around \(x^2 =y^2\), and write that

$$\begin{aligned} |y^2\sinh x^2 -x^2 \sinh y^2|&\le y^2|\sinh x^2 -\sinh y^2|+|x^2 -y^2 |\sinh y^2\\&\le y^2 |y^2-x^2 |\cosh \xi (x^2 , y^2)+|x^2 -y^2 |\sinh y^2 \end{aligned}$$

and

$$\begin{aligned}&\left| \frac{\sinh x^2}{\sinh y^2}-\frac{x^2}{y^2}\right| \le \frac{y^2 |y^2-x^2 |\cosh \xi (x^2 , y^2)+|x^2 -y^2 |\sinh y^2}{y^2\sinh y^2}\\&\quad =|x^2 -y^2|\left( \frac{\cosh \xi (x^2 , y^2)}{\sinh y^2}+\frac{1}{y^2}\right) . \end{aligned}$$

If \(|x-y|<1/8\) and \(x+y>1\) then, for some \(x_0>0\), \(y\ge x_0\) and \(x\ge x_0\). Then, there exists \(C>0\) such that

$$\begin{aligned} \frac{\cosh \xi (x^2, y^2)}{\sinh y^2}\le C,\,\,\forall x,\,\,\forall x; x+y>1, |x-y|<1/8 \end{aligned}$$

and,

$$\begin{aligned}&\left| \frac{\sinh x^2}{\sinh y^2}-\frac{x^2}{y^2}\right| \le C|x^2 -y^2|\,\,\forall x,\,\,\forall y; x+y>1, |x-y|<1/8. \end{aligned}$$

On the other hand, if \(|x+y|>1\) and \(|x-y|<1/8\),

$$\begin{aligned} \left| \frac{1}{\sinh |x^2-y^2|}- \frac{1}{\sinh (x^2+y^2)}\right| \le \frac{C}{|y^2-x^2 |}+\frac{1}{\sinh (x^2+y^2)} \end{aligned}$$

for some positive C. It follows that

$$\begin{aligned}&\left| \frac{\sinh x^2}{\sinh y^2}-\frac{x^2}{y^2}\right| \left| \frac{1}{\sinh |x^2-y^2|}- \frac{1}{\sinh (x^2+y^2)}\right| \le C; \,\, k+k'>1, |k-k'|<1/8 \end{aligned}$$

and

$$\begin{aligned} |T_2(x, y)|\le C \frac{y^3}{x^3}\le C\frac{(x+1/8)^3}{x^3}\le C',\,\,x+y>1, |x-y|<1/8, \end{aligned}$$

and this shows (2.12).

4.- Proof of (2.13) Suppose now that \(x+y>1\), \(|x-y|>1/8\) and \(|x-y|\le x/2\). Then

$$\begin{aligned}&\left| \frac{\sinh x^2 }{\sinh y^2}-\frac{x^2}{y^2}\right| \le C \end{aligned}$$

and

$$\begin{aligned} \left| \frac{\sinh x^2 }{\sinh y^2}-\frac{x^2}{y^2}\right| \left| \frac{1}{\sinh |x^2-y^2|}- \frac{1}{\sinh (x^2+y^2)}\right| \le C\left| \frac{1}{\sinh |x^2-y^2|}- \frac{1}{\sinh (x^2+y^2)}\right| , \end{aligned}$$

and this shows (2.13).

5.- Proof of (2.14) and (2.15). Suppose now \(x+y>1\), \(|x-y|>x/2\). As we have seen in the proof of Proposition 2.2, if \(y>x\),

$$\begin{aligned} \left| \frac{1}{\sinh |x^2-y^2|}- \frac{1}{\sinh (x^2+y^2)}\right| \le \frac{e^{-(y^2-x^2)}}{\left( 1-e^{-2(x^2+y^2)}\right) \left( 1-e^{-2(y^2-x^2)} \right) }\le Ce^{-(y^2-x^2 )}, \end{aligned}$$

and

$$\begin{aligned} \left| \frac{\sinh x^2 }{\sinh y^2}-\frac{x^2}{y^2}\right| \le Ce^{-(y^2 -x^2)}+\frac{x^2 }{y^2}. \end{aligned}$$

Then,

$$\begin{aligned} \left| \frac{\sinh x^2 }{\sinh y^2}-\frac{x^2}{y^2}\right| \left| \frac{1}{\sinh |x^2-y^2|}- \frac{1}{\sinh (x^2+y^2)}\right| \le Ce^{-(y^2-x^2 )}\left( e^{-(y^2 -x^2)}+\frac{x^2 }{y^2} \right) . \end{aligned}$$

If, on the other hand, \(y<x/2\), \(x^2-y^2>3x^2/4\), we then have that

$$\begin{aligned} \left| \frac{1}{\sinh |x^2-y^2|}- \frac{1}{\sinh (x^2+y^2)}\right| \le \frac{1}{\sinh \frac{3 x^2 }{4}}+\frac{1}{\sinh x^2}\le \frac{2}{\sinh \frac{3x^2 }{4} }. \end{aligned}$$

We deduce that

$$\begin{aligned}&|T_2(x, y)|\le \frac{C}{\sinh \frac{3 x^2 }{4} }\frac{y^3}{x^3}\le \frac{C}{\sinh \frac{ 3x^2}{4} },\,\,y<x/2\\&|T_2(x, y)|\le Ce^{-(y^2-x^2 )}\left( e^{-(y^2 -x^2)}+\frac{x^2 }{y^2} \right) \frac{y^3}{x^3},\,\,\,y\ge 3x/2 \end{aligned}$$

and this proves (2.14) and (2.15). \( \quad \square \)

Two Corollaries follow from Proposition (2.2) and Proposition (2.3). We first have

Corollary 1.8

$$\begin{aligned} M=\sup _{ x>0 }m(x)=\sup _{ x>0 }\int _0^\infty |T(x, y)|\textrm{d}y<\infty . \end{aligned}$$

Proof

By definition, for all \(x>0\),

$$\begin{aligned} m(x)=m_1(x)+m_2(x);\,\,\, |m_1(x)|\le \int _0^\infty |T_1(x, y)|\hbox {d}y;\,\,\, |m_2(x)|\le \int _0^\infty |T_2(x, y)|\hbox {d}y. \end{aligned}$$

Suppose first that \(x\in (0, 1)\). Then, by (2.2) and (2.4),

$$\begin{aligned} m_1(x)&\le \int _0^2 |T_1(x, y)|\hbox {d}y+\int _2^\infty |T_1(x, y)|\hbox {d}y\nonumber \\&\le Cx\int _0^1y\hbox {d}y+x \int _1^\infty \frac{\hbox {d}y}{y^3}=Cx,\,\forall x\in (0, 1). \end{aligned}$$
(2.16)

With a similar argument, using (2.10) and (2.11) instead,

$$\begin{aligned} m_2(x)&\le \int _0^2 |T_2(x, y)|\hbox {d}y+\int _2^\infty |T_2(x, y)|\hbox {d}y\nonumber \\&\le C+Cx \int _2^\infty \frac{y\hbox {d}y}{\sinh y^2}\le C,\,\forall x\in (0, 1). \end{aligned}$$
(2.17)

Suppose now that \(x>1\) and write, for \(\ell =1, 2\), that

$$\begin{aligned} m_\ell (x)=\int _0^{x/2} \hbox {d}y+\int _{ x/2 }^{x-\frac{1}{8}}\hbox {d}y+\int _{ x-\frac{1}{8} }^{x+\frac{1}{8}}\hbox {d}y+\int _{x+\frac{1}{8}}^{3x/2}\hbox {d}y+\int _{ 3x/2 }^\infty \hbox {d}y. \end{aligned}$$
(2.18)

When \(y\in (0, x/2)\) we may use (2.7) to obtain that

$$\begin{aligned} \int _0^{\frac{x}{2}}|T_1(x, y)|&\le \frac{C}{x} \int _0^{\frac{x}{2}}y\left( \frac{1}{\sinh \frac{3x^2}{4}}+\frac{y^2}{ x^4}\right) \le C\left( \frac{x}{\sinh \frac{3x^2}{4}}+\frac{1}{x}\right) \le \frac{C}{x},\,\,\,\,\forall x>1. \end{aligned}$$
(2.19)

If we use (2.14) in the same region we obtain

$$\begin{aligned}&|T_2(x, y)|\le \frac{C}{\sinh \frac{ 3 x^2 }{4} },\nonumber \\ \quad \hbox {and then}\quad&\int _0^{\frac{x}{2}}|T_2(x, y)| \le \frac{Cx}{\sinh \frac{ 3x^2 }{4} },\,\,\,\,\forall x>1. \end{aligned}$$
(2.20)

When \(y\in (x/2, x-1/8)\), by (2.9),

$$\begin{aligned} |T_1(x, y)|\le&\frac{Cy}{x}\left( \frac{2\min (x, y)^2}{(y+x)(x^2+y^2)}+\frac{1}{\sinh \frac{ |x+x '|}{8}}\right) \le C\frac{y}{x^2}, \end{aligned}$$
(2.21)
$$\begin{aligned} \quad \hbox {and then}\quad&\int _{x/2}^{x-\frac{1}{8}}|T_1(x, y)|\hbox {d}y\le C,\,\,\forall x>1. \end{aligned}$$
(2.22)

We now use (2.13) in the same region to get that

$$\begin{aligned}&|T_2(x, y)|\le C\left| \frac{1}{\sinh |x^2 -y^2|}-\frac{1}{\sinh (x^2 +y^2)}\right| \le C \left( \frac{1}{\sinh \frac{x}{8}}+\frac{1}{\sinh x^2} \right) ,\nonumber \\&\hbox {from which} \quad \int _{x/2}^{x-\frac{1}{8}} |T_2(x, y)|\hbox {d}y\le C\frac{x}{\sinh \frac{x}{8}},\,\,\forall x>1. \end{aligned}$$
(2.23)

Suppose now that \(y\in (x-1/8, x+1/8)\), by (2.3) and (2.12) \(T_1(x, y)\) and \(T_2(x, y)\) are both bounded on \([x-1/8, x+1/8]\), and so

$$\begin{aligned} \int _{ x-\frac{1}{8} }^{x+\frac{1}{8}} (|T_1(x, y)|+|T_2(x, y)|)\hbox {d}y\le C,\,\,\forall x>1. \end{aligned}$$
(2.24)

In the region \(y\in (x+1/8, 3x/2)\), we may use (2.9) again to obtain, as in (2.21), that

$$\begin{aligned} |T_1(x, y)|\le&\frac{Cy}{x}\left( \frac{2\min (s, y)^2}{(y+x)(x^2+y^2)}+\frac{1}{\sinh \frac{ |x+x '|}{8}}\right) \le \frac{Cy}{x^2}\nonumber \\ \quad \hbox {from which}\quad&\int _{x-\frac{1}{8}}^{\frac{3x}{2}}|T_1(x, y)|\hbox {d}y\le C,\,\,\forall x>1. \end{aligned}$$
(2.25)

By (2.13) we have, in the same region,

$$\begin{aligned} |T_2(x, y)|\le&C\left| \frac{1}{\sinh |x^2 -y^2|}-\frac{1}{\sinh (x^2 +y^2)}\right| \le C \left( \frac{1}{\sinh \frac{x}{8}}+\frac{1}{\sinh x^2} \right) ,\end{aligned}$$
(2.26)
$$\begin{aligned} \quad \hbox {and then}\quad&\int _{x-\frac{1}{8}}^{\frac{3x}{2}} |T_2(x, y)|\hbox {d}y\le C\frac{x}{\sinh \frac{x}{8}},\,\,\forall x>1. \end{aligned}$$
(2.27)

If \(y>3x/2\), by (2.6),

$$\begin{aligned} |T_1(x, y)|\le \frac{Cy}{x }\left( e^{-C_1y^2}+\frac{x^2}{y^4}\right) \end{aligned}$$

and we may write that

$$\begin{aligned}&e^{-C_1y^2}\le e^{-\frac{C_1x^2}{2}}e^{-\frac{C_1y^2}{2}}\,\,\hbox {and}\,\,\,\,\,\frac{x}{y^3}\le \frac{1}{y^2 } \end{aligned}$$

from which

$$\begin{aligned} \int _{ \frac{3x}{2} }^\infty |T_1(x, y)|\hbox {d}y\le Ce^{-\frac{C_1x^2}{2}} +\int _{ \frac{3x}{2} }^\infty \frac{\hbox {d}y}{y^2} \le C\left( e^{-\frac{C_1x^2}{2}}+\frac{1}{x}\right) ,\,\,\forall x>1. \end{aligned}$$
(2.28)

By (2.15),

$$\begin{aligned} |T_2(x, y)|&\le {\left\{ \begin{array}{ll} Ce^{-(y^2-x^2 )}\left( e^{-(y^2 -x^2)}+\frac{x^2 }{y^2} \right) \frac{y^3}{x^3},\,\,\,\,\forall x>\frac{3x}{2}\\ Ce^{-\frac{5y^2}{9}}\left( e^{-\frac{5y^2}{9}}+1\right) \frac{y^3}{x^3},\,\,\,\,\forall x>\frac{3x}{2}, \end{array}\right. }\\ \int _{ \frac{3x}{2} }^\infty&|T_2(x, y)|\hbox {d}y\le Ce^{-\frac{5x^2 }{9}},\,\,\forall x>1. \end{aligned}$$

\( \square \)

Similar arguments show the second Corollary,

Corollary 1.9

$$\begin{aligned} {\widetilde{M}}=\sup _{ y>0 }\mu (y)=\sup _{ y>0 }\int _0^\infty |T(x, y)|\hbox {d}x<\infty . \end{aligned}$$

Proof

As before,

$$\begin{aligned}&\mu (y)\le \mu _1(y)+\mu _2(y)\\&\mu _\ell (y)=\int _0^\infty |T_\ell (x, y)|\hbox {d}x,\,\ell =1, 2. \end{aligned}$$

Suppose first that \(y\in (0, 2)\). We have then,

$$\begin{aligned} \mu _1(y)\le \int _0^5|T_1(x, y)|\hbox {d}x+\int _5^\infty |T_1(x, y)|\hbox {d}x. \end{aligned}$$
(2.29)

We use (2.1) in the first integral of the right hand side of (2.29). Since \(x>5>2y\) in the second integral of the right hand side of (2.29), we may use (2.7) and deduce that \(\sup _{ y\in (0, 2) }\mu _1 (y)<\infty \). Similarly,

$$\begin{aligned} \mu _2(y)\le \int _0^5|T_2(x, y)|\hbox {d}x+\int _5^\infty |T_2(x, y)|\hbox {d}x \end{aligned}$$

where (2.10) and (2.14) yield \(\sup _{ y\in (0, 2) }\mu _2(y)<\infty \).

Suppose now that \(y\in (2, 15/2)\). We then write

$$\begin{aligned} \int _5^\infty |T_1(x, y)|\hbox {d}x=\int _{ 5 }^{2y}|T_1(x, y)|\hbox {d}x+\int _{2y}^\infty |T_1(x, y)|\hbox {d}x. \end{aligned}$$
(2.30)

The first integral in the right hand side of (2.30) is estimated using (2.1):

$$\begin{aligned} \int _{ 5 }^{2y}|T_1(x, y)|\hbox {d}x\le \int _{ 5 }^{15}|T_1(x, y)|\hbox {d}x\le C. \end{aligned}$$

The second integral in the right hand side of (2.30) may be estimated using (2.7) to obtain that

$$\begin{aligned} \int _{2y}^\infty |T_1(x, y)|\hbox {d}x\le Cy \int _{ 2y }^\infty \frac{1}{x}\left( \frac{1}{\sinh \frac{3x^2}{4}}+\frac{y^2}{x^4}\right) \hbox {d}x \le C\int _0^\infty e^{-\frac{3x^2}{8}}\hbox {d}x+Cy^{-1}. \end{aligned}$$

and it follows that \(\sup _{ y\in (2, 15/2 )}\mu _1(y)<\infty \). A similar argument using (2.10) and (2.14) gives \(\sup _{ y\in (2, 15/2 )}\mu _2(y)<\infty \).

Suppose now that \(y>15/2\). Then,

$$\begin{aligned} \int _5^\infty |T_1(x, y)|\hbox {d}x=\int _{ 5 }^{\frac{2y}{3}}|T_1(x, y)|\hbox {d}x+\int _{ \frac{2y}{3} }^{2y}|T_1(x, y)|\hbox {d}x+\int _{2y}^\infty |T_1(x, y)|\hbox {d}x. \end{aligned}$$
(2.31)

In the first and third integrals of the righthand side of (2.31) we use (2.6) to get that

$$\begin{aligned} \int _{ 5 }^{\frac{2y}{3}}|T_1(x, y)|\hbox {d}x\le Cy\int _5^{\frac{2y}{3}} \frac{1}{x }\left( e^{-C_1y^2} +\frac{x^2}{y^4}\right) \hbox {d}x<C\\ \int _{2y}^\infty |T_1(x, y)|\hbox {d}x\le Cy\int _{2y} ^\infty \frac{1}{x }\left( e^{-C_1x^2} +\frac{y^2}{x^4}\right) \hbox {d}x<C. \end{aligned}$$

Since \(|x-y|<x/2\) in the second integral at the right hand side of (2.31) we write,

$$\begin{aligned} \int _{ \frac{2y}{3} }^{2y}|T_1(x, y)|\hbox {d}x=\int _{ |y-x|<\frac{1}{8} } |T_1(x, y)|\hbox {d}x+\int _{ \frac{1}{8}<|y-x|<x/2 } |T_1(x, y)|\hbox {d}x. \end{aligned}$$
(2.32)

Using (2.3)

$$\begin{aligned} \int _{ |y-x|<\frac{1}{8} } |T_1(x, y)|\hbox {d}x\le C \end{aligned}$$

We use now (2.5) in the second integral on the right hand side of (2.32):

$$\begin{aligned}&\int _{ \frac{1}{8}<|y-x|<x/2 } |T_1(x, y)|\hbox {d}x \\&\quad \le Cy \int _{\frac{1}{8}<|x-y|<x/2 }\frac{1}{x}\left( \frac{2\min (y,x)^2}{(x^2-y^2)(x^2+y^2)}+\left| \frac{1}{\sinh |x^2-y^2|}-\frac{1}{\sinh (x^2 +y^2)}\right| \right) \hbox {d}x\\&\quad \le Cy \int _{\frac{1}{8}<|x-y|<x/2 }\frac{1}{x}\left( \frac{\min (y,x)^2}{(x+y)(x^2+y^2)}+\frac{1}{\sinh |(x+y)/8|}+\frac{1}{\sinh (x^2 +y^2)}\right) \hbox {d}x<C, \end{aligned}$$

from where \(\sup _{ y>15/2 }\mu _1(y)<\infty \). A similar argument shows that \(\sup _{ y>15/2 }\mu _2(y)<\infty \), using and (2.13), (2.12) instead of (2.3) and (2.15), (2.14) instead of (2.5). \(\quad \square \)

Proof

The proofs of (i) and (ii) are now straightforward:

$$\begin{aligned} |F(g(x))|&\le |g(x)|\int _0^\infty |T(x, y)|\text{ d }y+\int _0^\infty |g(y)|T(x, y)|\text{ d }y \nonumber \le 2M||g||_\infty \nonumber \\ ||F(g)||_1&\le \int _0^\infty |g(x)|\int _0^\infty |T(x, y)|\text{ d }y\text{ d }x+\int _0^\infty \int _0^\infty |g(y)| |T(x, y)|\text{ d }y\text{ d }x \nonumber \\ {}&\le {\widetilde{M}}||g||_1+M||g||_1. \end{aligned}$$
(2.33)

Proof of (iii). Consider again the right hand side of (2.33) and notice that, by (2.1) and (2.10), for \(x\in (0, 1)\),

$$\begin{aligned}&\int _0^1 |g(y)||T(x, y)|\hbox {d}y\le \int _0^1 |g(y)|\Big (|T_1(x, y)|+|T_2(x, y)|\Big )\hbox {d}y\\&\quad \le C\int _0^1 |g(y)|\Big (1+xy\Big )\hbox {d}y\le C\sup _{ 0\le y\le 1 } y^\theta |g(y)| \int _0^1y^{-\theta }\Big (1+xy \Big )\hbox {d}y\\&\quad \le C|||f(s)|||_\theta . \end{aligned}$$

By Corollary 2.4,

$$\begin{aligned}&\int _1^\infty |g(y)||T(x, y)|\hbox {d}y\le ||g|| _{ L^\infty (1, \infty ) }\int _0^\infty |T(x, y)|\hbox {d}y\le M ||g|| _{ L^\infty (1, \infty ) } \end{aligned}$$

and then,

$$\begin{aligned} \sup _{ 0\le y\le 1 } y^\theta |F(g)(y)|\le M \sup _{ 0\le y\le 1 } y^\theta |g(y)|+C|||g||| _{ \theta }. \end{aligned}$$

By (2.7) and (2.14), for \(x>2\),

$$\begin{aligned}&\int _0^1 |g(y)| \Big (|T_1(x, y)|+|T_2(x, y)|\Big )\hbox {d}y\le \sup _{ 0\le y\le 1 } y^\theta |g(y)| \\&\quad \times \int _0^1 y^{-\theta } \Big (|T_1(x, y)|+|T_2(x, y)|\Big )\hbox {d}y\le C \sup _{ 0\le x\le 1 } y^\theta |g(y)| \\&\quad \int _0^1 y^{1-\theta } \Big (1+y^2\Big )\hbox {d}y= C \sup _{ 0\le y\le 1 } y^\theta |g(y)|, \end{aligned}$$

and by, Corollary 2.4,

$$\begin{aligned}&\int _1^\infty |g(y)||T(x, y)|\hbox {d}y\le ||g|| _{ L^\infty (1, \infty ) }\int _0^\infty |T(x, y)|\hbox {d}y\le M ||g|| _{ L^\infty (1, \infty ) }, \end{aligned}$$

and \( |||F(g)||| _{ \theta }\le C|||g|||_\theta \).

Proof of (iv). By (ii) only \(F(g)\in L^\infty _{ loc }(0, \infty )\) remains to be proved. For \(K=[a, b] \subset (0, \infty )\), \([a, b] \subset (A, B )\) and all \(x\in [a, b]\),

$$\begin{aligned} \int _0^\infty T(x, y) g (y)\hbox {d}y= \int _{ y\in (A, B ) } T(x, y) g (y)\hbox {d}y+ \int _{ y\in (A, B )^c }T(x, y) g (y)\hbox {d}y=I_1+I_2, \end{aligned}$$

where,

$$\begin{aligned} |I_1(x)|=\left| \int _A ^B T(x, y) g (y)\hbox {d}x\right|&\le ||g|| _{ L^\infty (A , B ) } \int _A ^B |T(x, y)|\hbox {d}y\\&\le m(x)||g|| _{ L^\infty (A , B ) } \le M||g|| _{ L^\infty (A , B ) } . \end{aligned}$$

and

$$\begin{aligned} |I_2(x)|\le \int _0^A |T(x, y)| |g (y)|\hbox {d}y +\int _B ^\infty |T(x, y)| |g (y)|\hbox {d}y. \end{aligned}$$

By (2.1) and (2.10),

$$\begin{aligned} \int _0^A |T(x, y)| |g (y)|\hbox {d}y&\le \sup \{|T(x, y)|;x\in [a, b],\,y<A \}\int _0^\infty |g(y)|\hbox {d}y \\&\le C_\beta (1+bA )\int _0^\infty |g(y)|\hbox {d}y. \end{aligned}$$

A simple inspection of the expression of T(xy) given by (1.36) shows that \(C(K)=\sup \{|T(x, y)|;x\in [a, b],\,y>B \}<\infty \). Then,

$$\begin{aligned} \int _B ^\infty |T(x, y)| |g (y)|\hbox {d}y\le C(K)\int _0^\infty |g(y)|\hbox {d}y \end{aligned}$$

and

$$\begin{aligned} \sup _{ x\in K }|F(g)(x)|\le ||g|| _{ L^\infty ((A , B )) } M+( C_B +C(K)) (1+bA )||g||_1. \end{aligned}$$

Proof of (v). For all \(x>0\),

$$\begin{aligned} \left| \int _0^\infty T(x, y)g(y)\hbox {d}y\right| \le \int _0^R T(x, y)|g(y)|\hbox {d}y+\int _R^\infty T(x, y)|g(y)|\hbox {d}y\\ \le \int _0^R T(x, y)|g(y)|\hbox {d}y+M||g|| _{ L^\infty (R, \infty ) }. \end{aligned}$$

If \(x\le 2R\), by (2.1) and (2.10),

$$\begin{aligned} \int _0^R T(x, y)|g(y)|\hbox {d}y\le C _{ 2R }x\int _0^Ry |g(y)\hbox {d}y\le 2C_{ 2R }R^2||g||_1. \end{aligned}$$

For \(x>2R\), we use the original expression of T(xy) in (1.36):

$$\begin{aligned}&0\le \frac{y^3\sinh x^2}{x^3\sinh y^2} \left( \frac{1}{\sinh |x^2-y^2|}-\frac{1}{\sinh (x^2+y)^2 } \right) \le C\frac{e^{x^2}}{x^3} \left( e^{-x^2+y^2} \right) \le \frac{C_R}{x^3}\\&0\le \frac{y}{x}\left( \frac{1}{|x^2-y^2|}-\ \frac{1}{x^2+y^2} \right) \le \frac{R}{x^3}. \end{aligned}$$

Then

$$\begin{aligned} |T(x, y)|\le \frac{C_R}{x^3},\,\,\forall x>2R,\,y\in (0, R) \end{aligned}$$

and (iv) follow, since,

$$\begin{aligned} \int _0^R T(x, y)|g(y)|\hbox {d}y\le \frac{C _{ R }}{x^3}||g||_1,\,\,\forall x>2R. \end{aligned}$$

\( \square \)

3 Existence of Global Solution f

Using the properties of the operator L, Proposition  2.1 and a fixed point argument, classical solutions \(f\in C([0, \infty ); L^1(0, \infty ))\) of the Cauchy problem for (1.32) with initial data \(f_0\in L^1(0, \infty )\) are obtained. If, moreover, \(f_0\in L^1(0, \infty )\cap L^\infty (0, \infty )\) then \(f\in C([0, \infty ); L^1(0, \infty ))\cap L^\infty ((0, \infty ; L^\infty )\). However it is interesting to consider initial data slightly more general than in \(f_0\in L^1(0, \infty )\cap L^\infty (0, \infty )\) but whose solutions are more regular than just integrable with respect to x in \((0, \infty )\).

Theorem 1.10

Suppose that \(f_0\in L^1(0, \infty )\) satisfies

$$\begin{aligned} |||f_0|||_\theta \equiv \sup _{ 0<x<1 }x^\theta |f_0(x)|+\sup _{ x>1 }|f_0(x)|<\infty \end{aligned}$$
(3.1)

for some \(\theta \in (0, 1)\). Then, there exists a function

$$\begin{aligned} f\in C([0, \infty ); L^1 (0, \infty ))\cap L^\infty _{ loc }((0, \infty ); L^\infty (0, \infty )) \end{aligned}$$
(3.2)

satisfying that

$$\begin{aligned} f(t)=S(t)f_0+\int _0^tS(t-s)F(f(s))\textrm{d}s. \end{aligned}$$
(3.3)

The function f also satisfies that

$$\begin{aligned}&f(t)\in C(0, \infty )\,\,\,\forall t>0 \end{aligned}$$
(3.4)
$$\begin{aligned}&\forall T>0,\,\,\exists C=C(T, \theta );\nonumber \\&\quad ||f(t)||_1\le C_T||f_0||_1, \end{aligned}$$
(3.5)
$$\begin{aligned}&\quad ||f(t)||_1+|||f(t)||| _{ \theta }\le C(T)(||f_0||_1+|||f_0||| _{ \theta }),\,\forall t\in (0, T) \end{aligned}$$
(3.6)
$$\begin{aligned}&\quad ||f(t)||_\infty \le C(T, \theta ) \left( ||f_0|| _{ L^\infty (1, \infty ) }+ t^{-\theta }\sup _{ 0<y<1 }y^\theta |f_0(y)|+ ||f_0||_1\right) . \end{aligned}$$
(3.7)

Proof

Given \(f_0\) fixed and satisfying the hypothesis, consider the operator

$$\begin{aligned} {\mathscr {L}} (f)(t, x)=S(t)f_0(x)+\int _0^t S(t-s)(F(f(s)))\hbox {d}s \end{aligned}$$

on the space

$$\begin{aligned}&Z_T=C\left( [0, T); L^1 (0, \infty )\right) \cap C\left( (0, T);L^\infty (0, \infty )\right) , \\&||f|| _{ Z_T }=\sup _{ s\in (0, T) } \left( ||f(s)|| _{ 1 }+||f(s)|| _{ L^\infty (1, \infty ) }+s^{\theta }\sup _{ 0<x<1 }|f(s, x)|\right) \end{aligned}$$

By (ii) in Proposition (2.1),

$$\begin{aligned} ||{\mathscr {L}} (f(t))|| _{ 1 }\le C _{ s }||f_0|| _{ 1 }+C_sC_F\int _0^t||f(s)|| _{ 1 }\hbox {d}s. \end{aligned}$$
(3.8)

On the other hand,

$$\begin{aligned} |{\mathscr {L}} (f(t))(x)|\le |S(t)f_0(x)|+\int _0^t |S(t-s)F(f(s))(x)|\hbox {d}s. \end{aligned}$$

By (iv) in Proposition (2.1) and Proposition 6.3 in the Appendix, for all \(t>0\), \(s\in (0, t)\) and \(x\in (0, 2)\),

$$\begin{aligned}&|S(t-s)F(f(s))(x)|\le C||F(s)|| _{ L^\infty (1, \infty ) }+C(t-s)^{-\theta }\sup _{ 0<x<1 }x^{\theta }|F(f(s))(x)|\\&\quad \le C(1+(t-s)^{-\theta })||F(f)||_\infty \le C(1+(t-s)^{-\theta })(||f(t) || _{ L^\infty (1, \infty ) }+||f(t)||_1). \end{aligned}$$

Since we also have, for \(x\in (0, 2)\), that

$$\begin{aligned} |S(t)f_0(x)|\le C\left( ||f_0|| _{ L^\infty (1, \infty ) }+t^{-\theta }\sup _{ 0<x<1 }x^\theta |f_0(x)|\right) , \end{aligned}$$
(3.9)

we deduce, for \(t>0\), \(s\in (0, t)\) and \(x\in (0, 2)\),

$$\begin{aligned} |{\mathscr {L}} (f(t))(x)|\le C\left( ||f_0|| _{ L^\infty (1, \infty ) }+t^{-\theta }\sup _{ 0<x<1 }x^\theta |f_0(x)|\right) \\ +C\int _0^t (1+(t-s)^{-\theta })\left( ||f(s)||_1+||f(s)|| _{L^\infty (1, \infty )}\right) \hbox {d}s, \end{aligned}$$

and then

$$\begin{aligned} \sup _{ 0<x<2 }t^\theta&|{\mathscr {L}} (f(t))(x)| \le C\left( t^{\theta }||f_0|| _{ L^\infty (1, \infty ) }+\sup _{ 0<x<1 }x^\theta |f_0(x)|\right) \nonumber \\&+Ct^\theta \int _0^t (1+(t-s)^{-\theta })\left( ||f(s)||_1+||f(s)|| _{L^\infty (1, \infty )}\right) \hbox {d}s \end{aligned}$$
(3.10)

If \(x>2\),

$$\begin{aligned} |S(t-s)F(f(s))(x)|\le&C||F(s)|| _{ L^\infty (1, \infty ) }+C||F(f(s))||_1\\ \le&C\left( ||f(s)||_1+||f(s)|| _{ L^\infty (1, \infty ) }\right) \\ |S(t)f_0(x)|\le&C\left( ||f_0|| _{ L^\infty (1, \infty ) }+||f_0||_1\right) , \end{aligned}$$

from which,

$$\begin{aligned} \sup _{ x>2 }|{\mathscr {L}} (f(t))(x)|&\le C\left( ||f_0|| _{ L^\infty (1, \infty ) }+||f_0||_1\right) +\nonumber \\&\quad +C\int _0^t\left( ||f(s)||_1+||f(s)|| _{ L^\infty (1, \infty ) }\right) \hbox {d}s. \end{aligned}$$
(3.11)

Adding (3.8), (3.10) and (3.11),

$$\begin{aligned}&||{\mathscr {L}} (f(t))|| _{ 1 }+t^\theta \left| \sup _{ 0<x<1 }x^\theta {\mathscr {L}} (f(t))(x)\right| +\left| \sup _{ x>1 } {\mathscr {L}} (f(t))(x)\right| \\&\quad \le ||{\mathscr {L}} (f(t))|| _{ 1 }+ t^\theta \left| \sup _{ 0<x<1 } {\mathscr {L}} (f(t))(x)\right| +\left| \sup _{ x>1 } {\mathscr {L}} (f(t))(x)\right| \\&\quad \le C\left( ||f_0|| _{ L^\infty (R, \infty ) }+||f_0||_1\right) +C\left( t^{\theta }||f_0|| _{ L^\infty (1, \infty ) }+ \sup _{ 0<x<1 }x^\theta |f_0(x)|\right) \\&\qquad +C t^\theta \int _0^t (1+(t-s)^{-\theta })\left( ||f(s)||_1+||f(s)|| _{L^\infty (1, \infty )}\right) \hbox {d}s \\&\qquad +C\int _0^t\left( ||f(s)||_1+||f(s)|| _{ L^\infty (1, \infty ) }\right) \hbox {d}s, \end{aligned}$$

and we deduce

$$\begin{aligned}&||{\mathscr {L}} (f(t))|| _{ 1 }+t^\theta \left| \sup _{ 0<x<1 }x^\theta {\mathscr {L}} (f(t))(x)\right| +\left| \sup _{ x>1 } {\mathscr {L}} (f(t))(x)\right| \nonumber \\&\quad \le C(1+ t^\theta ) ||f_0|| _{ L^\infty (1, \infty ) }+||f_0||_1+C \sup _{ 0<x<1 }x^\theta |f_0(x)| \nonumber \\&\qquad +C t^\theta \sup _{ 0<s<t } \left( ||f(s)||_1+||f(s)|| _{L^\infty (R, \infty )}\right) (t+t^{1-\theta }) +T \!\!\! \sup _ {\begin{array}{c} 0<x<1\\ 0<s<T \end{array}} s^\theta |f(t, x)| \nonumber \\&\quad \le C_1\left( (1+T^\theta ) ||f_0|| _{ L^\infty (1, \infty ) }+||f_0||_1+ \sup _{ 0<x<1 }x^\theta |f_0(x)|\right) +C_2 T ||f|| _{ Z_T }. \end{aligned}$$
(3.12)

If we denote that

$$\begin{aligned} \gamma _0=C_1(2 ||f_0|| _{ L^\infty (1, \infty ) }+||f_0||_1+ \sup _{ 0<x<1 }x^\theta |f_0(x)|) \end{aligned}$$
(3.13)

we have then proved that

$$\begin{aligned} ||{\mathscr {L}}(f)|| _{ Z_T }\le \gamma _0+C _2T ||f|| _{ Z_T },\,\,\forall T\in (0, 1). \end{aligned}$$
(3.14)

Let \(\rho >0\) and \(T>0\) be such that

$$\begin{aligned}&T<\min \left( 1, \frac{1}{2C_2} \right) \end{aligned}$$
(3.15)
$$\begin{aligned}&\rho >2\gamma _0. \end{aligned}$$
(3.16)

Then, for all \(f\in Z_T\) such that \(||f|| _{ Z_T }\le \rho \),

$$\begin{aligned} ||{\mathscr {L}}(f)|| _{ Z_T }\le \gamma _0+C _2T\rho \le \rho , \end{aligned}$$
(3.17)

and then

$$\begin{aligned}&{\mathscr {L}}: B _{ Z_T }(0, \rho )\mapsto B _{ Z_T }(0, \rho ), \end{aligned}$$
(3.18)
$$\begin{aligned}&B _{ Z_T }(0, \rho )=\left\{ f\in Z_T; ||f|| _{ Z_T }\le \rho \right\} . \end{aligned}$$
(3.19)

On the other hand,

$$\begin{aligned} {\mathscr {L}} (f(t))-{\mathscr {L}} (g(t))=\int _0^t S(t-s)F(f(s)-g(s))\hbox {d}s, \end{aligned}$$

and arguing as before,

$$\begin{aligned} ||{\mathscr {L}} (f)-{\mathscr {L}} (g)|| _{ Z_T }\le C T||f-g|| _{ Z_T }. \end{aligned}$$

The map \({\mathscr {L}}\) is then a contraction form \(B _{ Z_T }(0, \rho )\) into itself if T is small enough, and has a fixed point \(u\in B _{ Z_T }(0, \rho )\) that satisfies

$$\begin{aligned} f(t)=S(t)f_0+\int _0^tS(t-s)F(f(s))\hbox {d}s \end{aligned}$$
(3.20)

in \(Z_T\). Property (3.5) follows from and Gönwall’s Lemma on (0, T) and by (3.10) and (3.11):

$$\begin{aligned}&||f(t)||_1+|||f(t)||| _{ \theta }\le C(||f_0||_1+|||f_0||| _{ \theta })+\end{aligned}$$
(3.21)
$$\begin{aligned}&\quad +C\left( 1+t^\theta \right) \int _0^t\Big (1+(t-s)^{-\theta }\Big )\Big (||f(s) || _{ L^\infty (1, \infty ) }+||f(s)||_1\Big ),\,\,\forall t\in (0, T). \end{aligned}$$
(3.22)

Then, there exists a constant \(C=C(T)>0\) such that, (3.31) holds true.

On the other hand, since \(f_0\in L^1(0, \infty )\) and \(|||f_0|||_\theta <\infty \), by Proposition 6.3 in the Appendix, \(S(t)f_0\in L^\infty (0, \infty )\). Moreover, by Proposition (6.3) and Proposition (2.1), for \(t\in (0, T)\) and \(x\in (0, 2)\),

$$\begin{aligned} |S(t-s)F(f(s))(x)|&\le C(t-s)^{-\theta } \sup _{0<x<1 }x^\theta |F(f(s)) (x)|+||F(f(s)|| _{ L^\infty (1, \infty ) } \nonumber \\&\le C (1+ (t-s)^{-\theta } ||F(f(s))||_\infty )\nonumber \\&\le C(1+ (t-s)^{-\theta })(||f(s)||_1+||f(s)|| _{ L^\infty (1, \infty ) })\nonumber \\&\le C(1+ (t-s)^{-\theta })\rho . \end{aligned}$$
(3.23)

It immediately follows that, for all \(t\in (0, T)\) and \(x\in (0, 2)\),

$$\begin{aligned} |f(t, x)|\le C||f_0|| _{ L^\infty (1, \infty ) }+Ct^{-\theta }\sup _{ 0<y<1 }y^\theta |f_0(y)|+C\rho \int _0^t (1+ (t-s)^{-\theta })\hbox {d}s \end{aligned}$$

and then \(f(t)\in L^\infty (0, \infty )\) for all \(t\in (0, T)\). We wish to extend now this function f for all \(t>0\). We notice, to this end that, for all \(x>1\),

$$\begin{aligned} |f(t, x)|\le&C||f_0|| _{ L^\infty (1, \infty ) }+Ct^{-\theta }\sup _{ 0<y<1 }y^\theta |f_0(y)|+ \int _0^t |S(t-s)F(f(s))(x)|\hbox {d}s\\ \le&C||f_0|| _{ L^\infty (1, \infty ) }+Ct^{-\theta }\sup _{ 0<y<1 }y^\theta |f_0(y)|\\&\quad +C\int _0^t(1+ (t-s)^{-\theta })(||f(s)||_1+||f(s)|| _{ L^\infty (1, \infty ) }). \end{aligned}$$

Since, by Proposition 2.1,

$$\begin{aligned} ||f(t)||_1\le C||f_0||_1+C\int _0^t||f(s)||_1\hbox {d}s, \end{aligned}$$
(3.24)

we obtain

$$\begin{aligned} \sup _{ x>1 } |f(t, x)|+||f(t)||_1\le C\left( t^{-\theta }\sup _{ x\in (0, 1)}x^\theta |f_0(x)|+||f_0||_1\right) \\ +C\int _0^t(1+ (t-s)^{-\theta })(||f(s)||_1+||f(s)|| _{ L^\infty (1, \infty ) }). \end{aligned}$$

It follows by Gönwall’s Lemma, that, for some constant C depending on T and \(\theta \),

$$\begin{aligned} \sup _{ x>1 } |f(t, x)|+||f(t)||_1\le C(T, \theta )\left( ||f_0|| _{ L^\infty (1, \infty ) }+t^{-\theta }\sup _{ 0<x<1 }x^\theta |f_0(x)| +||f_0||_1\right) . \end{aligned}$$
(3.25)

On the other hand, for \(x\in (0, 2)\), using (3.23),

$$\begin{aligned} |f(t, x)|&\le C||f_0|| _{ L^\infty (1, \infty ) }+Ct^{-\theta } \sup _{ 0<y<1 }y^\theta |f_0(y)|\\&\quad +C \int _0^t (1+ (t-s)^{-\theta })(||f(s)||_1+||f(s)|| _{ L^\infty (1, \infty ) })\hbox {d}s, \end{aligned}$$

and by (3.25), for all \(x\in (0, 2)\),

$$\begin{aligned} |f(t, x)|&\le C||f_0|| _{ L^\infty (1, \infty ) }+Ct^{-\theta }\sup _{ 0<y<1 }y^\theta |f_0(y)|\nonumber \\&\qquad +C(T, \theta ) \int _0^t (1+ (t-s)^{-\theta }) \left( ||f_0|| _{ L^\infty (1, \infty ) }+s^{-\theta }\sup _{ 0<x<1 }x^\theta |f_0(x)| +||f_0||_1\right) \hbox {d}s\nonumber \\&\quad \le C(T, \theta ) \left( ||f_0|| _{ L^\infty (1, \infty ) }+ t^{-\theta }\sup _{ 0<y<1 }y^\theta |f_0(y)|+ ||f_0||_1\right) . \end{aligned}$$
(3.26)

We deduce from (3.24), (3.25) and (3.26) for all \(t\in (0, T)\) that

$$\begin{aligned} ||f(t)||_1+||f(t)|| _\infty \le C(T, \theta ) \left( ||f_0|| _{ L^\infty (1, \infty ) }+ t^{-\theta }\sup _{ 0<y<1 }y^\theta |f_0(y)|+ ||f_0||_1\right) .\qquad \end{aligned}$$
(3.27)

By a classical argument it follows that the function f may be extended to a function, still denoted f, for all \(t>0\) such that \(f\in Z_t\) for all \(t>0\) and satisfies (3.20) for all \(t>0\).

The same arguments used to prove the estimates (3.5), (3.31) and (3.7) on the interval of time given by (3.15) may now be applied to obtain (3.5), (3.31), (3.7) on all finite interval (0, T) for all \(T>0\).

Since \(f_0\in L^1(0, \infty )\) and \(|||f_0|||_\theta <\infty \) it follows from Proposition 6.2 that \(S(t)f_0\in C(0, \infty )\) for every \(t>0\). Since \(f(s)\in L^\infty (0, \infty )\) for all \(s>0\) it follows by Proposition 2.1 that \(F(f(s))\in L^\infty (0, \infty )\) too and therefore, again by 6.2, \(S(t-s)F(f(s))\in C(0, \infty )\) for all \(t>0\) and \(s\in (0, t)\). This shows that \(f(t)\in C(0, \infty )\) for all \(t>0\). \(\quad \square \)

The \(L^1-L^\infty \) regularizing effect of the equation (1.30) observed in Theorem 3.1 follows from a similar property as to that of the equation (1.37) proved in [10]. Our next result shows how equation (1.30) regularizes the Dirac’s delta given as initial data. This is a consequence of the same property of (1.37) proved Theorem 1.2 of [10] (cf. Section  6.1.1 in Appendix below). It also holds for example for the linearized coagulation equation around its equilibria ([14]), but not for the linearization of the classical approximation (or wave turbulence “version”) of the Nordheim equation [12].

Theorem 1.11

There exists a function

$$\begin{aligned} f\in C((0, \infty ); L^1 (0, \infty )) \end{aligned}$$
(3.28)

satisfying

$$\begin{aligned} f(t)=S(t)\delta _1+\int _0^tS(t-s)F(f(s))\textrm{d}s, \end{aligned}$$
(3.29)

and such that

$$\begin{aligned}&f\in C((0, \infty ); L^1(0, \infty )) \end{aligned}$$
(3.30)
$$\begin{aligned}&||f(t)||_1 \le \frac{C_0}{1+t^2}+\int _0^t\frac{e^{C(t-s)}\textrm{d}s}{1+s^2},\,\,\,\forall t>0 \end{aligned}$$
(3.31)
$$\begin{aligned}&\forall \varphi \in {\mathscr {D}}(0, \infty ),\,\,\forall t>0: \nonumber \\&\frac{\textrm{d}}{\textrm{d}t}\int _0^\infty f(t, x)\varphi (x)\textrm{d}x= \langle L(f(t)), \varphi \rangle +\int _0^\infty F(f(s))(x)\varphi (x)\textrm{d}x, \end{aligned}$$
(3.32)

where the notation \( \langle L(f(t)), \varphi \rangle \) is defined in Section 6.1.1.

Proof

By definition, \(S(t)\delta _1\) with \(t>0\) is nothing but the fundamental solution \(\Lambda (t)\) of the equation (1.37). Consider the following operator T, defined on functions \(f\in C((0, T); L^1(0, \infty )\):

$$\begin{aligned} T(f)(t, x)=\Lambda (t, x)+\int _0^tS(t-s)F(f(s))\hbox {d}s. \end{aligned}$$

By (6.5) in the Appendix below, there exists a constant \(C_0>0\) such that,

$$\begin{aligned} ||\Lambda (t)||_1\le C_0,\,\,\,\forall t>0, \end{aligned}$$

and then, by (6.13) in Section 6.1.2 and Proposition (2.1), for all \(f\in C((0, T); L^1(0, \infty ))\) such that \(||f(t)||_1\le R\) for all \(t\in [0, T]\),

$$\begin{aligned} ||T(f)(t)||_1&\le C_0+\left| \left| \int _0^tS(t-s)F(f(s))\hbox {d}s \right| \right| _1\le C_0+C_F\int _0^t||F(f(s))||_1\hbox {d}s\\&\le \frac{C_0}{1+t^2}+C\int _0^t||f(s)||_1\hbox {d}s \le C_0+ C t R,\\ ||T(f)-T(g)||_1&\le C t \sup _{ 0\le t\le T }||f(t)-g(t)||_1. \end{aligned}$$

Therefore, if

$$\begin{aligned} C_0+CTR<R\,\,\,\text {and}\,\,CT<1, \end{aligned}$$

the operator T has a fixed point \(f\in C((0, T);L^1(0, \infty ))\) such that \(||f(t)||_1\le R\) for all \(t\in (0, T)\). It easilly follows using Gönwal’s Lemma that the solution f may be extended for all \(t>0\) and satisfies, for all \(t>0\),

$$\begin{aligned} ||f(t)||_1\le \frac{C_0}{1+t^2}+C\int _0^t ||f(s)||_1\hbox {d}s\Longrightarrow ||f(t)||_1\le \frac{C_0}{1+t^2}+\int _0^t\frac{e^{C(t-s)}\hbox {d}s}{1+s^2}. \end{aligned}$$

The proof of (3.32) follows now by classical arguments. Multiplication of (3.29) by \(\varphi \in {\mathscr {D}}(0, \infty )\) yields

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\int _0^\infty \!\!\! f(t, x)\varphi (x)\hbox {d}x&=\frac{\hbox {d}}{\hbox {d}t}\int _0^\infty \!\!\!\Lambda (t, x)\varphi (x)\hbox {d}x\\ {}&\quad +\frac{\hbox {d}}{\hbox {d}t}\int _0^\infty \int _0^t [S(t-s)F(f(s))](x)\hbox {d}s\varphi (x)\hbox {d}x, \end{aligned}$$

where in the first term (cf. Section  6.1.1 in the Appendix below),

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t} \int _0^\infty \Lambda (t, x) \varphi (x)\hbox {d}x = \left\langle L[\Lambda (t)], \varphi \right\rangle ,\,\,\,\forall \varphi \in {\mathscr {D}}(0, \infty ). \end{aligned}$$

The integral in the second term is denoted as

$$\begin{aligned} v(t)=\int _0^\infty \int _0^t [S(t-s)F(f(s))](x)\hbox {d}s\varphi (x)\hbox {d}x\equiv \int _0^t \left\langle [S(t-s)F(f(s))], \varphi \right\rangle \hbox {d}s, \end{aligned}$$

and, in order to obtain the derivative of v, consider the quotient,

$$\begin{aligned} \frac{v(t+h)-v(t)}{h}=&\int _0^t \left\langle [S(t-s)\frac{S(h)-I}{h}F(f(s))], \varphi \right\rangle \hbox {d}s\\&+\frac{1}{h}\int _t^{t+h} \left\langle [S(t+h-s)F(f(s))], \varphi \right\rangle \hbox {d}s. \end{aligned}$$

Since the function \(t\rightarrow \left\langle [S(t+h-s)F(f(s))], \varphi \right\rangle \) is continuous on \((0, \infty )\),

$$\begin{aligned} \lim _{ h\rightarrow 0 }\frac{1}{h}\int _t^{t+h} \left\langle [S(t+h-s)F(f(s))], \varphi \right\rangle \hbox {d}s=\left\langle F(f(t)), \varphi \right\rangle ,\,\,\forall t>0. \end{aligned}$$

On the other hand, since, for all \(g\in L^1(0, \infty )\),

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t} \int _0^\infty (S(t) g)(x) \varphi (x)\hbox {d}x= \left\langle L\left[ S(t)g \right] ,\varphi \right\rangle , \end{aligned}$$

integration from zero to \(h>0\) gives

$$\begin{aligned} \int _0^\infty [S(h) g](x) \varphi (x)\hbox {d}x= \int _0^\infty g(x) \varphi (x)\hbox {d}x+ \int _0^h\left\langle L\left[ S(t)g \right] ,\varphi \right\rangle \hbox {d}t. \end{aligned}$$
(3.33)

Use of (3.33) with \(g=S(t-s)F(f(s))\) yields, for all \(s\in (0, t)\), \(h>0\),

$$\begin{aligned} \langle [S(t-s)S(h)F(f(s))], \varphi \rangle =\langle S(t-s)F(f(s)), \varphi \rangle \\ +\int _0^h \langle L[S(\sigma )S(t-s)F(f(s))], \varphi \rangle \hbox {d}\sigma . \end{aligned}$$

Then,

$$\begin{aligned} \left\langle [S(t-s)\frac{S(h)-I}{h}F(f(s))], \varphi \right\rangle =\frac{1}{h}\int _0^h \langle L [S(\sigma )S(t-s)F(f(s))], \varphi \rangle \hbox {d}\sigma \end{aligned}$$

and by the continuity of the function \(\sigma \rightarrow \langle L [S(\sigma )S(t-s)F(f(s))], \varphi \rangle \),

$$\begin{aligned}&\lim _{ h\rightarrow 0 }\frac{1}{h}\int _0^h \langle L[S(\sigma )S(t-s)F(f(s))], \varphi \rangle \hbox {d}\sigma =\langle L[S(t-s) F(f(s))], \varphi \rangle \\&\lim _{ h\rightarrow 0 }\int _0^t \left\langle [S(t-s)\frac{S(h)-I}{h}F(f(s))], \varphi \right\rangle \hbox {d}s= \int _0^t \langle L[S(t-s) F(f(s))], \varphi \rangle \hbox {d}s. \end{aligned}$$

It follows that

$$\begin{aligned}&\frac{\hbox {d}}{\hbox {d}t}\int _0^\infty \!\!\! f(t, x)\varphi (x)\hbox {d}x=\frac{\hbox {d}}{\hbox {d}t}\int _0^\infty \!\!\!\Lambda (t, x)\varphi (x)\hbox {d}x+\frac{\hbox {d}}{\hbox {d}t}\int _0^\infty \int _0^t [S(t-s)F(f(s))](x)\hbox {d}s\varphi (x)\hbox {d}x\\&\quad =\left\langle L(\Lambda ), \varphi \right\rangle +\int _0^t \langle L[S(t-s) F(f(s))], \varphi \rangle \hbox {d}s+\left\langle F(f(t)), \varphi \right\rangle \\&\quad =\left\langle L\left[ \Lambda (t)+\int _0^tS(t-s)F(f(s))\hbox {d}s \right] +F(f(t)), \varphi \right\rangle =\langle L (f(t))+F(f(t)), \varphi \rangle . \end{aligned}$$

\( \square \)

A stronger regularizing effect of the equation (1.37) takes place for \(t>0\) and \(x\in (0, t)\), and is given below in Lemma 6.4 of the Appendix. However, because of its “local” feature, this property does not extends to the equation (1.30).

In the next Theorem are presented some additional properties of the function f obtained in Theorem 3.1.

Theorem 1.12

Suppose that \(f_0\in L^1(0, \infty )\) satisfies (3.1) and f is the function given by Theorem 3.1. Then,

$$\begin{aligned} \frac{\partial f}{\partial t},\, {\mathcal {L}}(f) \in L^\infty _{ loc }((0, \infty )\times (0, \infty ))\cap L^1((0, T)\times (0, \infty )),\,\forall T >0,\qquad \end{aligned}$$
(3.34)

and, for all almost every \(t>0,\, x>0\),

$$\begin{aligned} \frac{\partial f(t, x)}{\partial t}=\mathcal {L}(f(t))(x). \end{aligned}$$
(3.35)

There exists a function \({\tilde{H}}\in L^\infty ((\delta , \infty )\times (\delta , \infty ))\) for all \(\delta >0\) such that

$$\begin{aligned} \left| \frac{\partial f(t, x)}{\partial t} \right|&+ |\mathcal L(f)(t, x)|\le C\left( \sup _{ 0<y<1 }y^{\theta }|f_0(y)|+||f_0|| _{ L^\infty (1, \infty ) }+ ||f_0||_1\right) {\tilde{H}}(t, x) \end{aligned}$$
(3.36)

(\({\tilde{H}}\) is defined in (3.45) below). For all \(\varphi \in C^1_0([0, \infty ))\), the map \( t\mapsto \displaystyle {\int _0^\infty \varphi (x)f(t, x)}\textrm{d}x \) belongs to \(W _{ loc }^{1,1}(0, \infty )\) and, for almost every \(t>0\),

$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}t}\int _0^\infty \varphi (x)f(t, x)\textrm{d}x=\int _0^\infty {\mathcal {L}}(f(t))(x)\varphi (x)\textrm{d}x. \end{aligned}$$
(3.37)

Proof

We begin proving (3.34), (3.35), and (3.36). Since \(f_0\in L^1(0, \infty )\) and \(|||f_0|||_\theta <\infty \), by Proposition 6.3 in the Appendix \(S(t)u_0\in L^\infty (0, \infty )\) and Theorem 1.2 in [10],

$$\begin{aligned}&\frac{\partial }{\partial t}((S(t)f_0)\in L^\infty (0, \infty );\,\,\,\, L(S(t)f_0)(x)\in L^\infty (0, \infty ),\,\,\forall t>0\\&\frac{\partial }{\partial t}((S(t)f_0)(x))=L(S(t)f_0)(x),\,\, \hbox { for a. e.}\,\, t>0, x>0. \end{aligned}$$

Since for all \(s>0\), \(F(f(s))\in L^\infty (0, \infty )\cap L^1(0, \infty )\), for almost every \(x>0\), \(t\in (0, T)\) and \(s\in (0, t)\), by the same argument,

$$\begin{aligned} \frac{\partial }{\partial t}[(S(t-s)F(f(s)))(x)]=L(S(t-s)F(f(s)))(x), \end{aligned}$$
(3.38)

where both terms belong to \(L^\infty (0, \infty )\). Let us define

$$\begin{aligned}&\xi (t, x)=\frac{t^3}{\max (t^4, x^4) } \end{aligned}$$
(3.39)
$$\begin{aligned}&\zeta _\theta (t, x)=\frac{\min (t, t^{2-\theta })}{x^3}\mathbb {1}_{ t<2x/3 }+\frac{x\mathbb {1}_{ 2x< t}}{\max (t^{2+\theta }, t^3)} +\frac{\mathbb {1}_{ 2x/3<t<2x }}{\max (x^2, x^{1+\theta })}. \end{aligned}$$
(3.40)

By (6.27), and (iv) of Proposition (2.1), for all \(t\in (0, T)\), \(s\in (0, t)\) and \(x>0\),

$$\begin{aligned} \left| \frac{\partial }{\partial t}(S(t-s)F(f(s)))(x)\right|&\le C||F(f(s)||_\infty \left( 1+\xi (t-s, x)\right) +\zeta _0 (t-s, x)||F(f(s)|| _{ 1, 0 } \nonumber \\&\le C\left( ||f(s)|| _{ L^\infty (R, \infty ) } +||f(s)||_1\right) \left( 1+\xi (t-s, x)\right) \nonumber \\&\quad +\zeta _0 (t-s, x)\left( ||f(s)||_1+||f(s)|| _{L^ \infty (R, \infty ) } \right) \nonumber \\&\le C(1+\rho )\left( 1+\xi (t-s, x)+\zeta _0 (t-s, x) \right) . \end{aligned}$$
(3.41)

The right hand side of (3.41) may now be estimated for all \(s\in (0, t)\) using

$$\begin{aligned} \zeta _0 (t-s, x)+\xi (t-s, x)\le {\left\{ \begin{array}{ll} \displaystyle {\frac{C}{x}+ \frac{\mathbb {1}_{ {0<s<t-x} }}{x}+\frac{\mathbb {1}_{ t-x<s<t }t^3}{x^4}},\,\,\hbox {when}\,\,x\in (0, t)\\ \displaystyle {\,\,\,\,\,\frac{C}{x}+ \frac{1}{x^4}},\,\,\hbox {when}\,\,x>t. \end{array}\right. } \end{aligned}$$

We deduce, for all \(t>0\), \(x>0\),

$$\begin{aligned} \frac{\partial }{\partial t}\left( \int _0^tS(t-s)F(f(s))(x)\hbox {d}s \right) =F(f(t))(x)+\int _0^t\frac{\partial }{\partial t}\left( S(t-s)F(f(s))(x)\right) \hbox {d}s \end{aligned}$$
(3.42)

and

$$\begin{aligned}&\left| \frac{\partial }{\partial t}(S(t-s)F(f(s)))(x)\right| \le C||F(f(s)||_\infty \left( 1+\xi (t-s, x)\right) +\zeta (t-s, x)||F(f(s)||_1 \nonumber \\&\quad \le C\left( ||f(s)|| _{ L^\infty (R, \infty ) } +||f(s)||_1\right) \left( 1+\xi (t-s, x)+\zeta _0 (t-s, x)\right) . \end{aligned}$$
(3.43)

Then,

$$\begin{aligned}&\left| \frac{\partial f(t, x)}{\partial t} \right| \le \left| \frac{\partial }{\partial t}(S(t)f_0)(x)\right| +|F(f(t)(x)|+ \left| \int _0^t \frac{\partial }{\partial t}\left( S(t-s)F(f(s))\right) (x)\hbox {d}s \right| \\&\quad \le C||S(t)u_0)||_\infty \left( 1+\xi (t, x)\right) +\zeta _0 (t, x)||S(t)f_0|| _{ 1, 0 }+||F(f(t)|| _{ \infty } \\&\qquad +C\int _0^t \left( ||f(s)|| _{ L^\infty (1, \infty ) } +||f(s)||_1\right) \left( 1+\xi (t-s, x)+\zeta _0(t-s, x)\right) \hbox {d}s. \end{aligned}$$

Then, for all \(T>0\) and \(t\in (0, T)\), there exists \(C=C(T, \theta )\) such that

$$\begin{aligned}&\left| \frac{\partial f(t, x)}{\partial t} \right| \le C\left( ||f_0|| _{ L^\infty (1, \infty ) }+ t^{-\theta }\sup _{ 0<y<1 }y^\theta |f_0(y)|+ ||f_0||_1\right) (1+\xi (t, x)+\zeta _0(t, x))\\&\quad +C\int _0^t\left( ||f_0|| _{ L^\infty (1, \infty ) }+ s^{-\theta }\sup _{ 0<y<1 }y^\theta |f_0(y)|+ ||f_0||_1\right) \left( 1+\xi (t-s, x)+\zeta _0(t-s, x)\right) \hbox {d}s. \end{aligned}$$

Using Proposition 6.8 in the Appendix it follows that

$$\begin{aligned}&\int _0^t\left( ||f_0|| _{ L^\infty (1, \infty ) }+ s^{-\theta }\sup _{ 0<y<1 }y^\theta |f_0(y)|+ ||f_0||_1\right) \left( 1+\xi (t-s, x)+\zeta _0(t-s, x)\right) \hbox {d}s \\&\quad \le C(||f_0|| _{ L^\infty (1, \infty ) }+||f_0||_1)\left( \Xi _1(t, x)+ \omega _{ 1, 0 }(t, x)\right) +\sup _{ 0<y<1 }y^\theta |f_0(y)|\left( \Xi _2(t, x)+\omega _2(x, t)\right) . \end{aligned}$$

We deduce that

$$\begin{aligned}&\left| \frac{\partial f(t, x)}{\partial t} \right| \le C\left( ||f_0|| _{ L^\infty (1, \infty ) }+ t^{-\theta }\sup _{ 0<y<1 }y^\theta |f_0(y)|+ ||f_0||_1\right) (1+\xi (t, x)+\zeta _0(t, x))\nonumber \\&\qquad + C(||f_0|| _{ L^\infty (1, \infty ) }+||f_0||_1)\left( \Xi _1(t, x)+ \omega _{ 1, 0 }(t, x)\right) +\sup _{ 0<y<1 }y^\theta |f_0(y)|\left( \Xi _2(t, x)+\omega _{ 2 }(x, t)\right) \nonumber \\&\quad \le C\left( ||f_0|| _{ L^\infty (1, \infty ) }+ t^{-\theta }\sup _{ 0<y<1 }y^\theta |f_0(y)|+ ||f_0||_1\right) {\tilde{H}}(t, x) \end{aligned}$$
(3.44)

with

$$\begin{aligned} {\tilde{H}}(t, x)=(1+\xi (t, x)+\zeta _0(t, x))+\Xi _1(t, x)+ \omega _{ 1, 0 }(t, x)+ \Xi _2(t, x)+\omega _{ 2 }(x, t). \end{aligned}$$
(3.45)

Estimates (3.34) and (3.35) for \(\frac{\partial f}{\partial t}\) follow using again Proposition 6.8. Moreover, by (3.42), (3.43) and (3.38),

$$\begin{aligned} \frac{\partial }{\partial t}\left( \int _0^tS(t-s)F(f(s))\hbox {d}s \right)&=F(f(t))+\int _0^t\frac{\partial }{\partial t}\left( S(t-s)F(f(s))\right) \hbox {d}s.\\&=F(f(t))+\int _0^t L(S(t-s)F(f(s))) \hbox {d}s, \end{aligned}$$

and for almost every \(t>0\) and \(x>0\),

$$\begin{aligned} \frac{\partial f(t, x)}{\partial t}&=\frac{\partial }{\partial t}((S(t)f_0)(x))+\frac{\partial }{\partial t}\left( \int _0^tS(t-s)F(f(s))(x)\hbox {d}s\right) \\&=L(S(t)f_0)(x)+F(f(t))(x)+\int _0^t L(S(t-s)F(f(s)))(x) \hbox {d}s\\&=L\left( S(t)f_0+ \int _0^t S(t-s)F(f(s)) \hbox {d}s\right) (x)+F(f(t))(x)\\&=L(f(t))(x)+F(f(t))(x). \end{aligned}$$

By Proposition 2.1 this shows \({\mathcal {L}}(f)\in L^\infty _{ loc }((0,\infty ); L^\infty (\delta , \infty )\cap L^1(0, \infty ))\) for all \(\delta >0\). This ends the proof of (3.34) and (3.35), and proves (3.36).

On the other hand, if we multiply both sides of (3.20) by \(\varphi \in C^1_0([0, \infty ))\) and integrate,

$$\begin{aligned} \int _0^\infty f(t, x) \varphi (x)\hbox {d}x=\int _0^\infty \varphi (x)S(t)f_0(x)\hbox {d}x+\int _0^\infty \varphi (x)\int _0^tS(t-s)F(f(s))(x)\hbox {d}s\hbox {d}x. \end{aligned}$$

In order to derive this expression with respect to t, we use (3.38) and (3.43) again to obtain

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\int _0^\infty f(t, x) \varphi (x)\hbox {d}x&= \int _0^\infty \varphi (x)L(S(t)f_0)(x)\hbox {d}x+\int _0^\infty \varphi (x)F(f(t, x)\hbox {d}x\nonumber \\&\quad +\int _0^\infty \varphi (x)\int _0^t L(S(t-s)F(f(s))(x))\hbox {d}s\hbox {d}x, \end{aligned}$$
(3.46)

and, for all \(t\in (0, T)\),

$$\begin{aligned}&\left| \frac{\hbox {d}}{\hbox {d}t}\int _0^\infty f(t, x) \varphi (x)\hbox {d}x\right| \le C||\varphi ||_1||f_0||_\infty +C(1+T)||\varphi ||_1||f|| _\infty , \end{aligned}$$

which shows that

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\int _0^\infty f(t, x) \varphi (x)\hbox {d}x\in L^\infty _{ loc }([0, \infty ). \end{aligned}$$

Identity (3.37) follows now, since

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\int _0^\infty f(t, x) \varphi (x)\hbox {d}x&=\int _0^\infty \varphi (x)L\left( (S(t)f_0)(x)+\int _0^t S(t-s)F(f(s))(x)\hbox {d}s\right) \hbox {d}x\\&\quad +\int _0^\infty \varphi (x)F(f(t, x)\hbox {d}x=\int _0^\infty \varphi (x)L\left( f\right) \hbox {d}x+\int _0^\infty \varphi (x)F(f(t, x)\hbox {d}x. \end{aligned}$$

\( \square \)

4 Further Properties of the Solution \(\varvec{f}\)

We describe in this section some further properties of the solutions f given by Theorem 3.1. We first consider what are the variations of mass and energy induced by the initial perturbation \(n_0(1+n_0)x^2f(\tau )\) of the equilibrium \(n_0\) introduced in (1.11), (1.12). Then we prove that for all \(\delta >0\), \(f\in L^\infty ((\delta , \infty )\times (0, \infty ))\) and that for every \(t>0\) the function f(t) has a limit as \(x\rightarrow 0\).

4.1 Mass and Energy

It will be sometimes denoted in what follows that

$$\begin{aligned}&N_0(x)=n_0(x)(1+n_0(x))x^6;\,\,\,\,\,\,\hbox {d}\mu (x)=N_0(x)\hbox {d}x,\\&\int _0^\infty \int _0^\infty W(x, y)|f(\tau , y)-f(\tau , x)|^2y^4x^4\hbox {d}y\hbox {d}x=D(f(\tau )). \end{aligned}$$

With some abuse of notation the function \(\nu _0(x)=n_0(p)\) is still denoted \(n_0(x)\). A first basic property is the following:

Proposition 1.13

Let \(f_0\) and f be as in Theorem 3.3. Then, for all \(p>1\),

$$\begin{aligned}&\frac{\textrm{d}}{\textrm{d}t}\int _0^\infty |f(t, x)|^p\textrm{d}\mu (x) \le 0,\,\,\forall t>0. \end{aligned}$$
(4.1)
$$\begin{aligned}&||f(t_2)|| _{ L^\infty (\textrm{d}\mu ) }\le ||f(t_1)|| _{ L^\infty (\textrm{d}\mu ) },\,\,\forall t_2>t_1\ge 0. \end{aligned}$$
(4.2)

Proof

Since f satisfies (3.34 ), (3.36), and \(x^6n_0(1+n_0)|f|^{p-2}f \in L^1(0, \infty )\), multiplication of both sides of (3.36) and integration over \((0, \infty )\) gives, using (1.31) and the symmetry of W(xy) as follows:

$$\begin{aligned}&\frac{\hbox {d}}{\hbox {d}t}\int _0^\infty n_0(x)(1+n_0(x))|f(t, x)|^px^6\hbox {d}x \nonumber \\&\quad =\int _0^\infty \int _0^\infty W(x, y)(f(\tau , y)-f(\tau , x)) |f(x)|^{p-2}f(x)y^4x^4\hbox {d}y\hbox {d}x\nonumber \\&\quad =-\frac{1}{2}\int _0^\infty \int _0^\infty W(x, y)(f(\tau , y)-f(\tau , x)) \nonumber \\&\qquad \times \left( |f(y)|^{p-2}f(y)-|f(x)|^{p-2}f(x)\right) y^4x^4\hbox {d}y\hbox {d}x\le 0. \end{aligned}$$
(4.3)

Since \(\hbox {d}\mu \) is a non negative finite measure on \((0, \infty )\),

$$\begin{aligned} ||f(t_2)|| _{ L^\infty (\hbox {d}\mu ) }=\lim _{ p\rightarrow \infty }||f(t_2)|| _{ L^p(\hbox {d}\mu ) }\le \lim _{ p\rightarrow \infty }||f(t_1)|| _{ L^p(\hbox {d}\mu ) }=||f(t_1)|| _{ L^\infty (\hbox {d}\mu ) }. \end{aligned}$$

\( \square \)

Lemma 1.14

For all \(\theta \in [0, 3)\),

$$\begin{aligned}&(i)\qquad ||f||_{L^\infty (\textrm{d}\mu )}\le \max \left( ||f|| _{ L^\infty (1, \infty )}, \sup _{ x\in (0, 1) }x^\theta |f(x)|\right) \end{aligned}$$
(4.4)
$$\begin{aligned}&(ii)\qquad ||x^6n_0(1+n_0)f|| _{ \infty }\le ||f||_{L^\infty (\textrm{d}\mu )},\,\,\forall f\in L^\infty (\textrm{d}\mu )\cap L^\infty (0, \infty ) \end{aligned}$$
(4.5)

Proof

(i)  For all \(C>0\) denote,

$$\begin{aligned} A_C=\{x\in (0, \infty ); |f(x)|>C\}. \end{aligned}$$

Then, if \(C>||f|| _{ L^\infty (1, \infty ) }\), the Lebesgue measure of \(A_c\cap (1, \infty )\) is zero and

$$\begin{aligned} \hbox {d}\mu (A_c)&=\int _{ A_C\cap (0, 1) } n_0(1+n_0)x^6\hbox {d}x+\int _{ A_C\cap (1, \infty ) } n_0(1+n_0)x^6\hbox {d}x\\&=\int _{ A_C\cap (0, 1) } n_0(1+n_0)x^6\hbox {d}x. \end{aligned}$$

On the other hand,

$$\begin{aligned} \int _{ A_C\cap (0, 1) } n_0(1+n_0)x^6\hbox {d}x&\le \sup _{ x>0 }\left( n_0(1+n_0)x^4 \right) \int _{ A_C\cap (0, 1) }x^2\hbox {d}x\\&\le \sup _{ x>0 }\left( n_0(1+n_0)x^4 \right) \frac{1}{C}\int _{ A_C\cap (0, 1) }|f(x)|x^2\hbox {d}x\\&\le \sup _{ x>0 }\left( n_0(1+n_0)x^4 \right) \frac{1}{C} \int _{ A_C \cap (0, 1) }x^\theta |f(x)|\hbox {d}x. \end{aligned}$$

Then, since \(x^\theta f\in L^1(0, 1)\), if \(C>\sup _{ x\in (0, 1) }x^\theta |f(x)|\),

$$\begin{aligned} \int _{ A_C \cap (0, 1) }x^\theta |f(x)|\hbox {d}x=0. \end{aligned}$$

It follows that, for all \(C>\max (||f|| _{ L\infty (1, \infty ) }, \sup _{ x\in (0, 1) }x^\theta |f(x)|)\),

$$\begin{aligned} |f(x)|\le C,\,\,\,\hbox {except in a set of zero measure}, \end{aligned}$$

and this proves (4.4).

(ii) Let us denote now that

$$\begin{aligned} B_C=\{x>0;\, x^6n_0(1+n_0)|f(x)|\ge C\} \end{aligned}$$

and suppose that \(C>||f|| _{ L^\infty (\hbox {d}\mu ) }\). Then

$$\begin{aligned} m(B_C)&=\int _{ B_C }\hbox {d}x\le \frac{1}{C}\int _{ B_C } x^6n_0(1+n_0)|f(x)|\hbox {d}x\\&\le \frac{||f||_\infty }{C}\int _{ B_C } x^6n_0(1+n_0)\hbox {d}x= \frac{||f||_\infty }{C}\hbox {d}\mu (B_C)=0. \end{aligned}$$

\(\square \)

We how easily have the following:

Corollary 1.15

Let \(f_0\) and f be as in Theorem 3.3. Then,

$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}t} \int _0^\infty n_0(x)(1+n_0(x)f(t , x)x^6\textrm{d}x=0,\,\,\forall t>0. \end{aligned}$$

Proof

Since \(n_0(1+n_0)x^6\in C_0^1([0, \infty ))\), identity (3.37) with \(\varphi =n_0(1+n_0)x^6\) gives, by the definition of \({\mathcal {L}}\) (cf. (1.31)),

$$\begin{aligned}&\frac{\hbox {d}}{\hbox {d}t}\int _0^\infty n_0(1+n_0)x^4f(\tau , x)x^2\hbox {d}x=\int _0^\infty n_0(1+n_0)x^4{\mathcal {L}}(f(\tau ))(x)\hbox {d}x\\&\quad =\int _0^\infty \int _0^\infty W(x, y)(f(\tau , y)-f(\tau , x))y^4\hbox {d}yx^4\hbox {d}x. \end{aligned}$$

The result follows from the symmetry of the kernel W(xy). \( \quad \square \)

The following property will show the boundedness of the variation of the mass:

Proposition 1.16

Let \(f_0\) and f be as in Theorem 3.3. Then for all \(t>0\) and all \(p>3\),

$$\begin{aligned}&\int _0^\infty n_0(1+n_0)x^4|f(t, x)|\textrm{d}x\le C_p\max \left( ||f_0|| _{ L^\infty (1, \infty )}, \sup _{ x\in (0, 1) }x^\theta |f_0(x)|\right) ^{\frac{p-1}{p}}\!\!\! ||f_0||^{1/p} _{ L^1(\textrm{d}\mu ) }\\&C_p=\left( \int _0^\infty n_0(1+n_0) x^{\left( \frac{4p}{p-1}-\frac{6}{p-1} \right) }\textrm{d}x\right) ^{\frac{p-1}{p}} \end{aligned}$$

Proof

By Hölder’s inequality,

$$\begin{aligned}&\int _0^\infty n_0(1+n_0)x^4|f(t, x)|\hbox {d}x\le \left( \int _0^\infty n_0(1+n_0)x^6|f(t, x)|^p\hbox {d}x\right) ^{\frac{1}{p}}\\&\qquad \times \left( \int _0^\infty n_0(1+n_0) x^{\left( 4-\frac{6}{p} \right) \frac{p}{p-1}}\hbox {d}x \right) ^{\frac{p-1}{p}}\\&\quad =||f(t)|| _{ L^p(\hbox {d}\mu ) }\left( \int _0^\infty n_0(1+n_0) x^{\left( \frac{4p}{p-1}-\frac{6}{p-1} \right) }\hbox {d}x \right) ^{\frac{p-1}{p}}. \end{aligned}$$

If \(p>3\),

$$\begin{aligned}&\frac{4p}{p-1}-\frac{6}{p-1}>3,\,\,\,\hbox {and then},\\&\quad \int _0^\infty n_0(1+n_0) x^{\left( \frac{4p}{p-1}-\frac{6}{p-1} \right) }\hbox {d}x=C^{\frac{p}{p-1}}_p<\infty ,\\&\quad \lim _{ p\rightarrow 3 }C_p=\infty . \end{aligned}$$

It follows by Proposition 4.1, for all \(t> 0\),

$$\begin{aligned} \int _0^\infty n_0(1+n_0)x^4|f(t, x)|\hbox {d}x\le C_p ||f(t)|| _{ L^p(\hbox {d}\mu ) }\le C_p ||f_0|| _{ L^p(\hbox {d}\mu ) } \end{aligned}$$

Since,

$$\begin{aligned} ||f_0|| _{ L^p(\hbox {d}\mu ) } \le ||f_0|| _{L^ \infty (\hbox {d}\mu ) }^{\frac{p-1}{p}} ||f_0||^{1/p} _{ L^1(\hbox {d}\mu ) } \end{aligned}$$

it follows from Lemma 4.2 that

$$\begin{aligned} ||f_0|| _{ L^p(\hbox {d}\mu ) }&\le \max \left( ||f_0|| _{ L^\infty (1, \infty )}, \sup _{ x\in (0, 1) }x^\theta |f_0(x)|\right) ^{\frac{p-1}{p}} ||f_0||^{1/p} _{ L^1(\hbox {d}\mu ) }, \end{aligned}$$

and the Proposition follows. \(\quad \square \)

From Proposition 4.4 we immediately have

Corollary 1.17

Let \(f_0\) and f be as in Theorem 3.3. Then, for \(N(\tau )\) defined in (1.54), and all \(p>3\),

$$\begin{aligned} N(\tau )\le C_p\max \left( ||f_0|| _{ L^\infty (1, \infty )}, \sup _{ x\in (0, 1) }x^\theta |f_0(x)|\right) ^{\frac{p-1}{p}}\!\!\! ||f_0||^{1/p} _{ L^1(\textrm{d}\mu ) },\,\,\,\forall \tau >0, \end{aligned}$$

where \(C_p\) is given in Proposition 4.4.

The following property also follows from similar arguments:

Proposition 1.18

Suppose \(f_0\) and f are as in Theorem 3.3, \(C>0\) is a constant and \(f_0\le C\). Then \(f(t)\le C\) for all \(t>0\).

Proof

Since \({\mathcal {L}} (C)\equiv 0\), it follows that

$$\begin{aligned} \frac{\partial (f(t, x)-C)}{\partial t}=\mathcal {L}\Big (f(t)-C\Big )(x). \end{aligned}$$

If we multiply the equation by \(n_0(1+n_0)x^6f_C(t)^+\), with \(f_C(t)=(f(t)-C)\),

$$\begin{aligned}&\frac{\hbox {d}}{\hbox {d}t}\int _0^\infty \!\!\! n_0(1+n_0)|f_C^+(t, x)|^2x^6\hbox {d}x\\ {}&\quad =\int _0^\infty \int _0^\infty W(x, y)(f_C(t, y)-f_C(t, x))f_C^+(t, x)x^4y^4\hbox {d}y\hbox {d}x\\&\quad =-\int _0^\infty \int _0^\infty W(x, y)(f_C(t, y)-f_C(t, x))f^+(t, y)x^4y^4\hbox {d}y\hbox {d}x\\&\quad =-\frac{1}{2}\int _0^\infty \int _0^\infty W(x, y)(f_C(t, y)-f_C(t, x))(f_C^+(t, y)-f_C^+(t, x))x^4y^4\hbox {d}y\hbox {d}x. \end{aligned}$$

If \(f_C(t, y)>f_C(t, x)\) and \(f_C^+(t, y)\le f_C^+(t, x)\) then \(f_C(t, y)\le 0\) because if \(f_C(t, y)>0\) we would have \(f_C^+(t, y)=f_C(t, y)>f(t, x)=f_C^+(t, x)\), which is a contradiction. However this implies \(f_C^+(t, x)=f_C^+(t, y)=0\). On the other hand, if \(f_C(t, y)<f_C(t, x)\) and \(f_C^+(t, y)\ge f_C^+(t, x)\) then, \(f_C(t, x)<0\) because if \(f_C(t, x)>0\) then \(f_C^+(t, x)=f_C(t, x)>0\) then, as before we would have, \(f_C^+(t, y)\ge f_C^+(t, x)>0\) and so \(f_C(t, y)=f_C^+(t, y)\) but this would give \(_Cf^+(t, y)=f_C(t, y)<f_C(t, x)=f_C^+(t, x)\), which is a contradiction. Thus, in this case, \(f_C^+(t, x)=f_C^+(t, y)=0\). We deduce

$$\begin{aligned}&\frac{\hbox {d}}{\hbox {d}t}\int _0^\infty n_0(1+n_0)|f_C^+(t, x)|^2x^6\hbox {d}x\le 0\\&\quad \int _0^\infty n_0(1+n_0)|f_C^+(t, x)|^2x^6\hbox {d}x= 0, \end{aligned}$$

from which \(f_C^+(t)(x)=(f(t, x)-C)^+=0\,\, a. e.\) and the Proposition follows. \( \quad \square \)

We have the following Corollary:

Corollary 1.19

Suppose that f and g are two solutions of (3.35) given by Theorem 3.3 with initial data \(f_0\) and \(g_0\) satisfying the hypothesis of Theorem 3.3 and such that \(f_0\le g_0\). Then, \(f(t)\le g(t)\) for all \(t>0\). In particular, there exists a unique function f satisfying (3.2), (3.34) and (3.35).

Let let us also deduce the following:

Corollary 1.20

Suppose that \(f_0\) and f are as in Theorem 3.1. Then, for all \(\delta >0\) and all \(t\ge \delta \),

$$\begin{aligned} ||f(t)||_\infty \le C(\theta )\Big (||f_0|| _{ L^\infty (1, \infty ) }+ \delta ^{-\theta } \sup _{ 0\le y\le 1} y^\theta |f_0(y)|+||f_0||_1\Big ). \end{aligned}$$

Proof

By Theorem 3.1, for all\(\delta >0\), \(f(\delta )\in L^\infty (0, \infty )\cap L^1(0, \infty )\) and

$$\begin{aligned} ||f(\delta )||_\infty \le C(\theta )\Big (||f_0|| _{ L^\infty (1, \infty ) }+ \delta ^{-\theta } \sup _{ 0\le y\le 1} y^\theta |f_0(y)|+||f_0||_1\Big ). \end{aligned}$$

Since the constant \(||f(\delta )||_\infty \) is a solution of (3.36), the result follows, by Corollary 4.7. \( \quad \square \)

Our next result concerns the long time behavior of the solution f. The \(L^2(\hbox {d}\mu )\) norm and \(D(f(\tau ))\) play here the usual roles of entropy and entropy’s dissipation through the identity (cf. (4.3)),

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\int _0^\infty |f(t, x)|^2\hbox {d}\mu (x)=-D(f(\tau )). \end{aligned}$$
(4.6)

For every \(n\in \mathbb {N}\setminus \{0\}\), define the regularized kernel

$$\begin{aligned} W_1(x, y)=\left( \frac{\sinh (x^2+y^2)-\sinh |x^2-y^2 |}{\sinh (x^2+y^2)(\sinh |x^2-y^2 | + 1+\sinh (x^2+y^2))}\right) \frac{1}{xy (\sinh x^2)\,(\sinh y^2)}, \end{aligned}$$

so that, \(\hbox {d}\sigma _1(x, y)=W_1(x, y) x^4y^4\hbox {d}y\hbox {d}x\) is now a bounded measure on \(\mathbb {R}^2\).

Now let j be the function

$$\begin{aligned} j(u, v)= {\left\{ \begin{array}{ll} |u-v|^2,\,\,\hbox {if}\,\,u\in \mathbb {R}, v\in \mathbb {R},\,\\ \infty \,\,\,\hbox {elsewhere}, \end{array}\right. } \end{aligned}$$

consider, for any pair of functions fg defined on \((0, \infty )\) the function U defined on \((0, \infty )^2\) as

$$\begin{aligned} U(x, y)=(f(x), g(y))\in \mathbb {R}^2 \end{aligned}$$

and denote that

$$\begin{aligned}&J_1(U)= {\left\{ \begin{array}{ll} \displaystyle {\,\,\,\iint \limits _{ (0, \infty )^2 } j(f(x), g(y)) \hbox {d}\sigma _1(x, y),\,\,\,\hbox {if}\,\, j(f, g)\in L^1(\hbox {d}\sigma _n)}\\ \infty ,\,\,\hbox {elsewhere}, \end{array}\right. }\\&D_1(f)=J_1(f, f). \end{aligned}$$

Lemma 1.21

The function \(J_1\) is weakly l.s.c. on \(L^2(\textrm{d}\mu )\).

Proof

Let us show that \(J_1\) is convex and continuous on \(L^2(\hbox {d}\mu )\times L^2(\hbox {d}\mu )\). Since the Hessian of j is positive semi definite on \(\mathbb {R}^2\), the function j is convex on \(\mathbb {R}^2\) and then so is \(J_1\). On the other hand, the estimate of \(W_1(x, y)x^4y^4\) for \(x>0\) and \(y>0\) easily follows from this definition:

$$\begin{aligned} 0\le W_1(x, y)x^4y^4\le {\left\{ \begin{array}{ll} \displaystyle {\frac{y^3\cosh y^2}{\sinh y^2(1+\sinh y^2)}\frac{x^5}{\sinh x^2(1+\sinh x^2)}},\,\,\forall y>0,\,\forall x\in (0, y)\\ \displaystyle {\frac{y^2 }{\sinh y^2 }\frac{x^4}{\sinh x^2(1+\sinh x^2)}},\,\,\forall y>0,\,\forall x >y. \end{array}\right. } \end{aligned}$$

Then, for \(f\in L^2(\hbox {d}\mu )\),

$$\begin{aligned}&\int _0^\infty \int _0^\infty |f(x)|^2W_1(x, y)x^4y^4\hbox {d}x\hbox {d}y=\int _0^\infty \int _0^y |f(x)|^2W_1(x, y)x^4y^4\hbox {d}x\hbox {d}y\\&\quad +\int _0^\infty \int _y^\infty |f(x)|^2W_n(x, y)x^4y^4\hbox {d}x\hbox {d}y=I_1+I_2 \end{aligned}$$

and the terms \(I_i\), \(i=1, 2\) are bounded as

$$\begin{aligned}&I_1\le \int _0^\infty \frac{y^3\cosh y^2}{\sinh y^2(1+\sinh y^2)}\int _0^y\frac{|f(x)|^2x^5}{\sinh x^2(1+\sinh x^2)}\hbox {d}x\hbox {d}y\\&\quad \le C\int _0^\infty \frac{|f(x)|^2x^5}{\sinh x^2(1+\sinh x^2)}\hbox {d}x\le C||f||^2 _{ L^2(\hbox {d}\mu ) } \end{aligned}$$

and

$$\begin{aligned}&I_2\le \int _0^\infty \frac{y^2 }{\sinh y^2 }\int _0^y \frac{|f(x)|^2x^4}{\sinh x^2(1+\sinh x^2)}\hbox {d}x\hbox {d}y\\&\quad \le C\int _0^\infty \frac{|f(x)|^2x^4}{\sinh x^2(1+\sinh x^2)}\hbox {d}x\le C||f||^2 _{ L^2(\hbox {d}\mu ) }. \end{aligned}$$

Therefore, if \(\{f_n\} _{ n\in \mathbb {N}}\) and \(\{g_n\} _{ n\in \mathbb {N}}\) are two sequences in \(L^2(\hbox {d}\mu )\) converging, respectively, to \(f\in L^2(\hbox {d}\mu )\) and \(g\in L^2(\hbox {d}\mu )\),

$$\begin{aligned}&|J_1(f_n, g_n)-J_1(f, g)|\le \int _0^\infty \int _0^\infty |(f_n(x)-g_n(y))^2-(f(x)-g(y))^2|\hbox {d}\sigma _n(x, y)\\&\quad = \int _0^\infty \int _0^\infty |(f_n(x)-g_n(y)-(f(x)-g(y))||(f_n(x)-g_n(y)+(f(x)-g(y))|\hbox {d}\sigma _n(x, y)\\&\quad \le \left( \int _0^\infty \int _0^\infty |(f_n(x)-g_n(y)-(f(x)-g(y))|^2\hbox {d}\sigma _n(x, y)\right) ^{1/2} \\&\qquad \times \left( \int _0^\infty \int _0^\infty |(f_n(x)-g_n(y)+(f(x)-g(y))|^2\hbox {d}\sigma _n(x, y)\right) ^{1/2}. \end{aligned}$$

Using the previous estimate, we deduce in the second integral as

$$\begin{aligned}&\int _0^\infty \int _0^\infty |(f_n(x)-g_n(y)+(f(x)-g(y))|^2\hbox {d}\sigma _n(x, y)\le \int _0^\infty \int _0^\infty |f_n(x)|^2\hbox {d}\sigma _n(x, y)\\&\qquad +\int _0^\infty \int _0^\infty | g_n(y)|^2\hbox {d}\sigma _n(x, y)+\int _0^\infty \int _0^\infty |f(x)|^2\hbox {d}\sigma _n(x, y)+ \int _0^\infty \int _0^\infty | g(y)|^2\hbox {d}\sigma _n(x, y)\\&\quad \le C\left( ||f_n||^2 _{ L^2(\hbox {d}\mu ) }+||g_n||^2 _{ L^2(\hbox {d}\mu ) }+||f||^2 _{ L^2(\hbox {d}\mu ) }+||g||^2 _{ L^2(\hbox {d}\mu ) }\right) . \end{aligned}$$

In the first,

$$\begin{aligned}&\int _0^\infty \int _0^\infty |(f_n(x)-g_n(y)-(f(x)-g(x))|^2\hbox {d}\sigma _n(x, y) \\&\quad \le \int _0^\infty \int _0^\infty |f_n(x)-f(x)|^2\hbox {d}\sigma _n(x, y) +\int _0^\infty \int _0^\infty |g_n(y)-g(x)|^2\hbox {d}\sigma _n(x, y)\\&\quad \le C\left( ||f_n-f||^2 _{ L^2(\hbox {d}\mu ) }+||g_n-g||^2 _{ L^2(\hbox {d}\mu ) }\right) . \end{aligned}$$

Since the sequences \(\{f_n\} _{ n\in \mathbb {N}}\) and \(\{g_n\} _{ n\in \mathbb {N}}\) are bounded in \(L^2(\hbox {d}\mu )\) it follows that \(J_1(f_n, g_n)\rightarrow J_1(f, g)\) as \(n\rightarrow \infty \). The function \(J_1\) is then convex and continuous on \(L^2(\hbox {d}\mu )\times L^2(\hbox {d}\mu )\) and the weakly l.s.c. on \(L^2(\hbox {d}\mu )\times L^2(\hbox {d}\mu )\), (cf. for example [5], Corollary 3.9). \( \quad \square \)

Proposition 1.22

Let \(f_0\) and f be as in Theorem 3.1. Then, for all \(\varphi \in L^2(\textrm{d}\mu )\),

$$\begin{aligned} \lim _{ \tau \rightarrow \infty } \int _0^\infty f(\tau , x )\varphi (x)\textrm{d}\mu (x)=\frac{E(0)}{\int _0^\infty \textrm{d}\mu (x)}\int _0^\infty \varphi (x)\textrm{d}\mu (x). \end{aligned}$$

Remark 4.11

Notice that, since \(\hbox {d}\mu \) is a non negative bounded measure on \((0, \infty )\), \(L^2(\hbox {d}\mu )\subset L^1(\hbox {d}\mu )\). Corollary 4.10 shows the weak convergence,

$$\begin{aligned} n_0(1+n_0)x^6\,f(\tau )\underset{\tau \rightarrow \infty }{\rightharpoonup } \frac{E(0)}{\int _0^\infty \hbox {d}\mu (x)},\,\,\hbox {weakly in}\,L^2(0, \infty ). \end{aligned}$$

Proof

Consider any sequence \(\{\tau _k\} _{ k\in \mathbb {N}}\) where \(\tau _k\rightarrow \infty \) as \(k\rightarrow \infty \) and define

$$\begin{aligned} f_k(\tau )=f(\tau +\tau _k). \end{aligned}$$
(4.7)

By (4.6), for all \(T>0\),

$$\begin{aligned} \frac{1}{2}\int _0^TD(f(s))\hbox {d}t&=-\int _0^\infty |f(T, x)|^2\hbox {d}\mu (x)+\int _0^\infty |f(0, x)|\hbox {d}\mu (x)\\ {}&\le \int _0^\infty |f(0, x)|^2\hbox {d}\mu (x), \end{aligned}$$

from which we deduce that \(D(f(s))\in L^1(0, \infty )\) and

$$\begin{aligned} \int _0^T D(f_k(t))\hbox {d}t= \int _{ t_k }^{t_k+T} |Df(t)|\hbox {d}t\rightarrow 0,\,\,\hbox {as}\,\,k\rightarrow \infty . \end{aligned}$$
(4.8)

Since \(0\le D_1(f_k(t))\le D(f_k(t))|\) for all \(k\ge 1\),

$$\begin{aligned} \lim _{ k\rightarrow \infty }\int _0^T D_1(f_k(t))\hbox {d}t=0,\,\,\,\forall T>0, \end{aligned}$$
(4.9)

and since \(D_1(f_k(t))\ge 0\) for all \(t>0\),

$$\begin{aligned} \lim _{ k\rightarrow \infty } D_1(f_k(t))=0,\,\,\,\hbox {for a. e. }t\in (0, T). \end{aligned}$$
(4.10)

On the other hand, by (4.6) again, the sequence \(\{f_k\} _{ k\in \mathbb {N}}\) is bounded in \(L^\infty ((0, \infty ); L^2(\hbox {d}\mu ))\) and there exists a subsequence still denoted \(\{t_k\} _{ k\in \mathbb {N}}\), such that \(t_k\rightarrow \infty \) as \(k\rightarrow \infty \) and \(g\in L^\infty ((0, \infty ); L^2(\hbox {d}\mu ))\) satisfying

$$\begin{aligned} f_k \underset{k\rightarrow \infty }{\rightharpoonup }\ g,\,\,\,\hbox {in the weak}^*\hbox { topology of } \,\,L^\infty ((0, \infty ); L^2(\hbox {d}\mu )). \end{aligned}$$
(4.11)

It then follows by the weak lower semicontinuity of \(D_1\) in \(L^2(\hbox {d}\mu )\) that

$$\begin{aligned} \int _0^T D_n(g(s))\hbox {d}s\le \liminf _{ k\rightarrow \infty }\int _0^T D_n(f_k(s))\hbox {d}s=0. \end{aligned}$$

Then,

$$\begin{aligned} D_1(g(s)=\int _0^\infty \int _0^\infty W_1(x, y)|g(s, y)-g(s, x)|^2y^4x^4\hbox {d}y\hbox {d}x=0, \end{aligned}$$

and, for almost every \(s \in (0, T)\) the measure \(|g(s, y)-g(s, x)|^2y^4x^4\hbox {d}y\hbox {d}x\) is concentrated on the diagonal \(\{(x, y)\in (0, \infty )^2;\,y=x\}\). Since \(g(s )\in L^2(\hbox {d}\mu )\) for almost every \(s\in (0, T)\), it follows that \(g=C_*\) for some constant \(C_*\in \mathbb {R}\). Since \(\mathbb {1}_{ (0, 1) }\in L^1(0, \infty ; L^2(\hbox {d}\mu ))\), it follows that

$$\begin{aligned} C\int _0^\infty \hbox {d}\mu (x)&=\lim _{ k\rightarrow \infty } \int _0^1\int _0^\infty f_k(t, x)\hbox {d}\mu (x)\hbox {d}t\\&=\lim _{ k\rightarrow \infty }\int _0^1\int _0^\infty f_k(t, x)n_0(x)(1+n_0(x)x^6\hbox {d}x\hbox {d}t\\&=\lim _{ k\rightarrow \infty }\int _0^1 E(t+t_k)\hbox {d}t=E(0), \end{aligned}$$

and then

$$\begin{aligned} C_*=\frac{E(0)}{\int _0^\infty \hbox {d}\mu (x)}. \end{aligned}$$

For all \(\varphi \in C_c^1(0, \infty )\), the function

$$\begin{aligned} t\rightarrow \int _0^\infty f(t, x)\varphi (x)n_0(x)(1+n_0(x))x^6\hbox {d}x\in C(0, \infty ; \mathbb {R})\cap L^\infty (0, \infty ). \end{aligned}$$

and,

$$\begin{aligned}&\left| \frac{\hbox {d}}{\hbox {d}t}\int _0^\infty f(t, x)\varphi (x)n_0(x)(1+n_0(x))x^6\hbox {d}x\right| \\&\quad \le \int _0^\infty \left| \int _0^\infty W(x, y)(f(t, y)-f(t, x))y^4\hbox {d}y\right| \varphi (x)\hbox {d}x\\&\quad \le ||{\mathcal {L}}(f)(t)|| _{ \infty }||\varphi ||_1\in L^\infty _{ loc }(0, \infty ). \end{aligned}$$

There exists then a sequence of \(\{t_j\} _{ k\in \mathbb {N}}\) and \(C(\varphi )\) such that

$$\begin{aligned} \lim _{ j\rightarrow \infty }\int _0^\infty f(t_j)\varphi (x)\hbox {d}\mu (x)=C(\varphi ). \end{aligned}$$

Since \(\mathbb {1}_{ (0, 1) }(t)\varphi (x) \in L^1(0, \infty ; L^2(\hbox {d}\mu ))\), by (4.11),

$$\begin{aligned} \lim _{ j\rightarrow \infty }\int _0^1\int _0^\infty f(t_j)\varphi (x)\hbox {d}\mu (x)\hbox {d}t=C_*\int _0^1 \int _0^\infty \varphi (x)\hbox {d}\mu (x)\hbox {d}t=C_*\int _0^\infty \varphi (x)\hbox {d}\mu \end{aligned}$$

and

$$\begin{aligned} C(\varphi )=C_*\int _0^\infty \varphi (x)\hbox {d}\mu . \end{aligned}$$

\(\square \)

The following auxiliary result is used in the proof of the next Proposition:

Lemma 1.24

For all \(\varepsilon >0\), there exists a constant \(C_\varepsilon >0\) such that

$$\begin{aligned}&\forall (x, y)\in \left( \varepsilon , \frac{1}{\varepsilon }\right) \times \left( \varepsilon , \frac{1}{\varepsilon }\right) ,\\&W(x, y)\ge \left( \frac{C _{\varepsilon }}{\sinh (x^2+y^2)\sinh |x^2-y^2 | }\right) \frac{1}{xy (\sinh x^2)\,(\sinh y^2)} \end{aligned}$$

Proof

There certainly exists \(C _{1, \varepsilon } >0\) such that

$$\begin{aligned} x^2+y^2-|x^2-y^2|\ge C' _{ \varepsilon },\,\,\,\forall (x, y)\in \left( \varepsilon , \frac{1}{\varepsilon }\right) \times \left( \varepsilon , \frac{1}{\varepsilon }\right) . \end{aligned}$$

Then, by continuity,

$$\begin{aligned} \sinh (x^2+y^2)-\sinh |x^2-y^2 |\ge C _{ \varepsilon },\,\, \forall (x, y)\in \left( \varepsilon , \frac{1}{\varepsilon }\right) \times \left( \varepsilon , \frac{1}{\varepsilon }\right) , \end{aligned}$$

and the result follows. \( \quad \square \)

Proposition 1.25

Let \(f_0\) and f be as in Theorem 3.1. Then,

$$\begin{aligned} \lim _{ \tau \rightarrow \infty } \int _0^\infty |f(t, x)-C^*|^2\textrm{d}\mu (x)=0. \end{aligned}$$

Proof

For all \(\varepsilon >0\),

$$\begin{aligned}&\int _0^T\int _0^\varepsilon n_0(1+n_0)x^6 |f_k(t, x)-C_*|^2\hbox {d}x=\int _0^T\int _0^\varepsilon |f_k(t, x)-C_*|^2\hbox {d}\mu (x) \nonumber \\&\quad \le T ||f_k(t)-C_*|| _{ L^2(\hbox {d}\mu ) }\int _0^\varepsilon \hbox {d}\mu (x)\le T \left( C_*+ ||f_0|| _{ L^2(\hbox {d}\mu ) }\right) \int _0^\varepsilon \hbox {d}\mu (x) \end{aligned}$$
(4.12)

and similarly,

$$\begin{aligned} \int _0^T\int _{1/\varepsilon } ^\infty n_0(1+n_0)x^6 |f_k(t, x)-C_*|^2\hbox {d}x \le T \left( C_*+ ||f_0|| _{ L^2(\hbox {d}\mu ) }\right) \int _{1/\varepsilon }^\infty \hbox {d}\mu (x). \end{aligned}$$
(4.13)

Then, for \(\delta >0\) given, let \(\varepsilon \) be fixed such that

$$\begin{aligned} T \left( C_*+ ||f_0|| _{ L^2(\hbox {d}\mu ) }\right) \left( \int _0^\varepsilon \hbox {d}\mu (x)+\int _{1/\varepsilon }^\infty \hbox {d}\mu (x) \right) <\delta /2. \end{aligned}$$
(4.14)

On the other hand, it follows from Lemma 4.12, for all \(\varepsilon >0\), for all \(x\in (\varepsilon /2, 2/\varepsilon )\) and \(y\in (\varepsilon /2, 2/\varepsilon )\),

$$\begin{aligned}&x^4 y^4W_1(x, y)\ge \left( \frac{C _{\varepsilon }}{\sinh (x^2+y^2)(\sinh |x^2-y^2 |+1+\sinh (x^2+y^2)) }\right) \frac{x^3y^3}{ (\sinh x^2)\,(\sinh y^2)}\\&\quad =\frac{x^6}{(\sinh x^2)^2}\left\{ \left( \frac{C _{\varepsilon }}{\sinh (x^2+y^2)(\sinh |x^2-y^2 |+1+ \sinh (x^2+y^2))}\right) \frac{y^3(\sinh x^2)}{ x^3\,(\sinh y^2)}\right\} \\&\quad \ge \frac{x^6}{(\sinh x^2)^2}\left\{ \left( \frac{C _{\varepsilon }}{\sinh (x^2+y^2)(1+2\sinh (x^2+y^2)) }\right) \frac{y^3(\sinh x^2)}{ x^3\,(\sinh y^2)}\right\} \\&\quad \ge \frac{x^6}{(\sinh x^2)^2}\left\{ \left( \frac{C _{\varepsilon }}{\sinh (2\varepsilon ^{-2})(1+2 \sinh (2\varepsilon ^{-2})) }\right) \frac{\varepsilon ^6(\sinh \varepsilon ^2)}{(\sinh \varepsilon ^{-2})}\right\} =\gamma _{ \varepsilon } \frac{x^6}{(\sinh x^2)^2}. \end{aligned}$$

Let then be \(\theta _\varepsilon \in C_0(0, \infty )\), \(\theta _\varepsilon (x)=1\) for \(x\in (\varepsilon , 1/\varepsilon )\) and \(\theta _\varepsilon (x)=0\) if \(x\in (0, \varepsilon /2)\) or \(x>2/\varepsilon )\) and consider \(\{f_k\} _{ k\in \mathbb {N}}\) the sequence constructed in (4.7).

$$\begin{aligned} \gamma _{ \varepsilon }&\int _0^T\int _0^\infty \theta _\varepsilon (x) n_0(1+n_0)x^6 |f_k(t, x)-C_*|^2\hbox {d}x \nonumber \\&\le \int _0^T\int _0^\infty \int _0^\infty \theta _\varepsilon (x) \theta _\varepsilon (y)W_n(x, y)\left( |f_k(t, x)-C_*|^2+ |f_k(t, y)-C_*|^2\right) x^4y^4\hbox {d}x\hbox {d}y\hbox {d}t \nonumber \\&\le \int _0^TD(f_k(t))\hbox {d}t+2\int _0^\infty \int _0^\infty \theta _\varepsilon (x) \theta _\varepsilon (y) W_n(x, y) \nonumber \\&\quad \times \left( f_k(t, x)f_k(t, y)+C_*^2-C_*f_k(t, x)-C_*f_k(t, y)\right) x^4y^4\hbox {d}x\hbox {d}y\hbox {d}t. \end{aligned}$$
(4.15)

It was been proven in (4.8) that the first term in the right hand side of (4.15) tends to zero as \(k\rightarrow \infty \). In order to prove that the last term in the right hand side tends to zero the following Lemma is needed, whose proof is delayed after the end of the proof of Proposition 4.13.

Lemma 1.26

For all \(\varphi \in C_c(0, \infty )\) and all \(T>0\),

$$\begin{aligned} \lim _{ k \rightarrow \infty } \int _0^T \left| \int _0^\infty f_k(t, x)\varphi (x) \textrm{d}\mu (x)-C_* \int _0^\infty \varphi (x) \textrm{d}\mu (x) \right| \textrm{d}t=0. \end{aligned}$$

The second term in the right hand side of (4.15) may be split as follows:

$$\begin{aligned}&2\int _0^\infty \int _0^\infty \theta _\varepsilon (x) \theta _\varepsilon (y){\mathcal {U}}_n(x, y) \\&\qquad \times \left( f_k(t, x)f_k(t, y)+C_*^2-C_*f_k(t, x)-C_*f_k(t, y)\right) x^4y^4\hbox {d}x\hbox {d}y\hbox {d}t =J_1+J_2,\\&\quad J_1=\int _0^T \int _0^\infty \int _0^\infty \theta _\varepsilon (x) \theta _\varepsilon (y){\mathcal {U}}_n(x, y) \frac{f_k(t, x)(f_k(t, y)-C_*)x^4y^4\hbox {d}x\hbox {d}y\hbox {d}t}{xy (\sinh x^2)\,(\sinh y^2)}\\&\quad J_2=\int _0^T \int _0^\infty \int _0^\infty \theta _\varepsilon (x) \theta _\varepsilon (y){\mathcal {U}}_n(x, y) \frac{C_*(C_*-f_k(t, y))x^4y^4\hbox {d}x\hbox {d}y\hbox {d}t}{xy (\sinh x^2)\,(\sinh y^2)}. \end{aligned}$$

The first term may be written as

$$\begin{aligned}&J_1=\int _0^T\int _0^\infty \theta _\varepsilon (y)\frac{y^3(f_k(t, y)-C_*)\hbox {d}y}{\sinh y^2}\\&\qquad \times \int _0^\infty \theta _\varepsilon (x) \frac{f_k(t, x) x^3}{(\sinh x^2)}\left( \frac{\sinh (x^2+y^2)-\sinh |x^2-y^2 |}{\sinh (x^2+y^2)(\sinh |x^2-y^2 | +1+ \sinh (x^2+y^2))}\right) \hbox {d}x\hbox {d}t. \end{aligned}$$

Since,

$$\begin{aligned}&\int _0^\infty \theta _\varepsilon (x) \frac{|f_k(t, x)| x^3}{(\sinh x^2)}\left( \frac{\sinh (x^2+y^2)-\sinh |x^2-y^2 |}{\sinh (x^2+y^2)(\sinh |x^2-y^2 | +1+\sinh (x^2+y^2))}\right) \hbox {d}x \\&\quad \le n\int _0^\infty \theta _\varepsilon (x) \frac{|f_k(t, x)| x^3}{\sinh x^2}\\&\quad \le \frac{n}{\varepsilon ^2}\left( \int _0^\infty |f_k(t, x)|^2\hbox {d}\mu \right) ^{1/2}\le \frac{n}{\varepsilon ^2}\left( \int _0^\infty |f_0(x)|^2\hbox {d}\mu \right) ^{1/2} \end{aligned}$$

we deduce that

$$\begin{aligned} |J_1|&\le \frac{n}{\varepsilon ^2}||f_0|| _{ L^2(\hbox {d}\mu ) }\int _0^T \left| \int _0^\infty \theta _\varepsilon (y) \frac{y^3(f_k(t, y)-C_*)\hbox {d}y}{\sinh y^2}\right| \hbox {d}t\\&\quad = \frac{n}{\varepsilon ^2}||f_0|| _{ L^2(\hbox {d}\mu ) }\int _0^T \left| \int _0^\infty (f_k(t, y)-C_*) \theta _\varepsilon (y)\frac{\sinh y^2}{y^3}\hbox {d}\mu (y) \right| \hbox {d}t \underset{k\rightarrow \infty }{\rightarrow }\ 0, \end{aligned}$$

by Lemma 4.14 with \(\varphi (x)=\theta _\varepsilon (y)\frac{\sinh y^2}{y^3}\). The same argument shows that \(J_2\) tends to zero too as k goes to \(\infty \) and this shows that, for all \(\varepsilon >0\) fixed,

$$\begin{aligned} \lim _{ k\rightarrow \infty }\int _0^T\int _0^\infty \theta _\varepsilon (x) n_0(1+n_0)x^6 |f_k(t, x)-C_*|^2\hbox {d}x=0. \end{aligned}$$
(4.16)

We deduce from (4.14) and (4.16) that

$$\begin{aligned} \lim _{ k\rightarrow \infty }\int _0^T \int _0^\infty |f(t, x)-C^*|^2\hbox {d}\mu (x)\hbox {d}t=0, \end{aligned}$$

since, by Proposition 4.1,

$$\begin{aligned} \int _0^\infty |f(T, x)-C^*|^2\hbox {d}\mu (x)\le \int _0^\infty |f(t, x)-C^*|^2\hbox {d}\mu (x),\,\,\forall t\in (0, T). \end{aligned}$$

Proposition 4.13 follows. \( \quad \square \)

Proof

For all \(\varphi \in C_0(0, \infty )\),

$$\begin{aligned}&\int _0^\infty \int _0^\infty {\mathcal {U}}(x, y)(f(y)-f(x))\varphi (x)x^4y^4\hbox {d}y\hbox {d}x\\&\quad =-\int _0^\infty \int _0^\infty {\mathcal {U}}(x, y)(f(y)-f(x))\varphi (y)x^4y^4\hbox {d}y\hbox {d}x\\&\quad =\frac{1}{2}\int _0^\infty \int _0^\infty {\mathcal {U}}(x, y)(f(y)-f(x))(\varphi (x) -\varphi (y)) x^4y^4\hbox {d}y\hbox {d}x\\&\quad =\frac{1}{2}\int _0^\infty \int _0^\infty {\mathcal {U}}(x, y)f(y)(\varphi (x) -\varphi (y)) x^4y^4\hbox {d}y\hbox {d}x\\&\qquad -\frac{1}{2}\int _0^\infty \int _0^\infty {\mathcal {U}}(x, y)f(x)(\varphi (x) -\varphi (y)) x^4y^4\hbox {d}y\hbox {d}x\\&\quad =\frac{1}{2}\int _0^\infty \int _0^\infty {\mathcal {U}}(x, y)f(y)(\varphi (x) -\varphi (y)) x^4y^4\hbox {d}y\hbox {d}x\\&\qquad -\frac{1}{2}\int _0^\infty \int _0^\infty {\mathcal {U}}(x, y)f(y)(\varphi (y) -\varphi (x)) x^4y^4\hbox {d}y\hbox {d}x\\&\quad =\int _0^\infty \int _0^\infty {\mathcal {U}}(x, y)f(y)(\varphi (x)-\varphi (y)) x^4y^4\hbox {d}y\hbox {d}x \end{aligned}$$

and

$$\begin{aligned} I_k(t )=&\left| \frac{\hbox {d}}{\hbox {d}t}\int _0^\infty n_0(1+n_0)x^4f_k(t, x)\varphi (x)\hbox {d}x\right| \\&\quad \le \int _0^\infty \int _0^\infty {\mathcal {U}}(x, y) \left| \varphi (y)-\varphi (x) \right| |f_k(t, x)|x^4y^4\hbox {d}y\hbox {d}x. \end{aligned}$$

If \(\rho >0\) and \(R>0\) are such that \(\textrm{supp } \varphi \subset (\rho , R)\), then \(\left| \varphi (y)-\varphi (x) \right| =0\) for \(x\in (0, \rho )\) and \(y\in (0, \rho )\). If, on the other hand, \(x\ge \rho \) or \(y\ge \rho \), that

$$\begin{aligned} \sinh |x^2-y^2|=\sinh \left( (x+y)|x-y| \right) \ge \sinh (2\rho |x-y|)\\ \frac{|\varphi (y)-\varphi (x)|}{\sinh |x^2-y^2|}\le C_\varphi \frac{|\varphi (y)-\varphi (x)|}{\sinh (2\rho |x-y|)}. \end{aligned}$$

If \(x>R\) and \(y>R\), then \(\left| \varphi (y)-\varphi (x) \right| =0\) again, and therefore,

$$\begin{aligned}&I_k(t )\le \int _0^R \left( \int _0^\infty \left( \frac{\sinh (x^2+y^2)-\sinh |x^2-y^2 |}{\sinh (x^2+y^2)}\right) \right. \\&\qquad \left. \times \frac{|\varphi (y)-\varphi (x)|}{\sinh |x^2-y^2 |} \frac{ |f_k(t, x)|x^3y^3}{ (\sinh x^2)\,(\sinh y^2)}\hbox {d}y\right) \hbox {d}x\\&\qquad + \int _0^\infty \left( \int _0^R \left( \frac{\sinh (x^2+y^2)-\sinh |x^2-y^2 |}{\sinh (x^2+y^2)}\right) \right. \\&\qquad \left. \frac{|\varphi (y)-\varphi (x)|}{\sinh |x^2-y^2 |} \frac{ |f_k(t, x)|x^3y^3}{ (\sinh x^2)\,(\sinh y^2)}\hbox {d}y\right) \hbox {d}x\\&\quad =I _{ k, 1 }(t)+I _{ k,2 }(t). \end{aligned}$$

The term \(I _{ k, 1 }\) is easily estimated as

$$\begin{aligned} I _{ k,1 }(t )&\le \int _0^R\left( \int _0^\infty \left( \frac{\sinh (x^2+y^2)-\sinh |x^2-y^2 |}{\sinh (x^2+y^2)}\right) \right. \\&\qquad \left. \times \frac{|\varphi (y)-\varphi (x)|}{\sinh |x^2-y^2 |} \frac{ |f_k(t, x)|x^3y^3}{ (\sinh x^2)\,(\sinh y^2)}\hbox {d}y\right) \hbox {d}x\\&\le C_\varphi \int _0^R \left( \int _0^\infty \frac{ |f_k(t, x)|x^3}{ \sinh x^2} \frac{y^3}{\sinh y^2}\hbox {d}y\right) \hbox {d}x\\&= C_\varphi \int _0^R \frac{ |f_k(t, x)|x^3\hbox {d}x}{ \sinh x^2} \int _0^\infty \frac{y^3\hbox {d}y}{\sinh y^2}. \end{aligned}$$

Using Hölder’s inequality and Proposition 4.1,

$$\begin{aligned} I _{ k,1 }(t)&\le C_\varphi \sqrt{R} \left( \int _0^R \frac{|f(t, x)|^2x^6\hbox {d}x}{(\sinh x^2)^2}\right) ^{1/2}\int _0^\infty \frac{y^3\hbox {d}y}{\sinh y^2}\\&\le C'_\varphi \left( \int _0^\infty \frac{|f_0(x)|^2x^6\hbox {d}x}{(\sinh x^2)^2}\right) ^{1/2}. \end{aligned}$$

The integral \(I _{ k, 2 }\) may be split again as

$$\begin{aligned} I _{ k, 2 }=I _{ k, 2 , 1}+I _{ k, 2 , 2}, \end{aligned}$$

where,

$$\begin{aligned} I _{ k, 2 , 1}\le \int _0^{2R}\left( \int _0^R \frac{|\varphi (y)-\varphi (x)|}{\sinh |x^2-y^2 |} \frac{ |f_k(t, x)|x^3y^3}{ (\sinh x^2)\,(\sinh y^2)}\hbox {d}y\right) \hbox {d}x, \end{aligned}$$

and \(I _{ k, 2, 1}\) is then estimated as \(I _{ k,1 }\). The estimate of the term \(I _{ k, 2, 2}\) uses that when \(y\le R<x/2\) then \(\sinh |x^2-y^2|\ge \sinh (3x^2/4)\) as

$$\begin{aligned} I _{ k, 2 , 1}\le \int _{2R} ^\infty \left( \int _0^R\frac{|\varphi (y)-\varphi (x)|}{\sinh |x^2-y^2 |} \frac{ |f_k(t, x)|x^3y^3}{ (\sinh x^2)\,(\sinh y^2)}\hbox {d}y\right) \hbox {d}x\\ \le C_\varphi \int _{2R} ^\infty \int _0^R \frac{ |f_k(t, x)|x^3y^3\hbox {d}y\hbox {d}x}{\sinh (3x^2/4) (\sinh x^2)\,(\sinh y^2)}\\ =C_\varphi \int _{2R} ^\infty \frac{ |f_k(t, x)|x^3 \hbox {d}x}{\sinh (3x^2/4) (\sinh x^2)} \int _0^R \frac{y^3\hbox {d}y}{\sinh y^2}, \end{aligned}$$

and using Hölder’s inequality again,

$$\begin{aligned} I _{ k, 2 , 1} \le&C_\varphi \left( \int _{2R} ^\infty \frac{|f_k(t, x)|^2x^6\hbox {d}x}{(\sinh x^2)^2}\right) ^{1/2}\left( \int _{2R} ^\infty \frac{\hbox {d}x}{(\sinh (3x^2/4)^2}\right) ^{1/2} \int _0^R \frac{y^3\hbox {d}y}{\sinh y^2}. \end{aligned}$$

We deduce that

$$\begin{aligned} \int _0^T\left| \frac{\hbox {d}}{\hbox {d}t}\int _0^\infty f_k(t, x)\varphi (x)\hbox {d}\mu (x) \right| \hbox {d}t\le C' _{ \varphi } T \left( \int _0^\infty \frac{|f_0(x)|^2x^6\hbox {d}x}{(\sinh x^2)^2}\right) ^{1/2}. \end{aligned}$$

We have then, for all \(T>0\), that

$$\begin{aligned} \int _0^\infty f_k(t, x)\varphi (x)\hbox {d}\mu (x) \in W^{1, 1}(0, T),\,\,\, \end{aligned}$$

and by the compactness of the injection \(W^{1, 1}(0, T)\subset L^1(0, T)\), there exists a sequence \(t_j \underset{j\rightarrow \infty }{\rightarrow }~\infty \) and \(h\in L^1(0, T)\) satisfying

$$\begin{aligned} \lim _{ k\rightarrow \infty }\int _0^T\left| \int _0^\infty f_k(t, x)\varphi (x)\hbox {d}\mu (x) -h(t)\right| \hbox {d}t=0. \end{aligned}$$

However, since, by Corollary 4.10,

$$\begin{aligned} \lim _{ k\rightarrow \infty }\int _0^T \int _0^\infty f_k(t, x)\varphi (x)\hbox {d}\mu (x)\hbox {d}t=C_* T\int _0^\infty \varphi (x)\hbox {d}\mu (x), \end{aligned}$$

we deduce that

$$\begin{aligned} \int _0^Th(t)\hbox {d}t=C_* T\int _0^\infty \varphi (x)\hbox {d}\mu (x) ,\,\,\forall T>0 \end{aligned}$$

and by the fundamental Theorem of Calculus,

$$\begin{aligned} C_* \int _0^\infty \varphi (x)\hbox {d}\mu (x) =\frac{\hbox {d}}{\hbox {d}T}\int _0^Th(t)\hbox {d}t=h(T),\,\,\forall T>0. \end{aligned}$$

It follows that h is the constant given by \(C_*\int _0^\infty \varphi (x)\hbox {d}\mu (x)\) and this ends the proof of Lemma 4.14. \( \quad \square \)

Corollary 1.27

Let \(f_0\) and f be as in Theorem 3.1. Then,

$$\begin{aligned} \lim _{ \tau \rightarrow \infty } \int _0^\infty |f(\tau , x)-C^*|x^4n_0(x)(1+n_0(x))\textrm{d}x=0. \end{aligned}$$

Proof

Arguing as in Proposition 4.4, for all \(\varepsilon >0\) and \(t>1\),

$$\begin{aligned}&\int _0^\varepsilon n_0(1+n_0)x^4|f(t, x)-C_*|\hbox {d}x\le \left( \int _0^\varepsilon n_0(1+n_0)x^6(|f(t, x)|^p+C_*^p)\hbox {d}x\right) ^{\frac{1}{p}}\\&\qquad \times \left( \int _0^\varepsilon n_0(1+n_0) x^{\left( 4-\frac{6}{p} \right) \frac{p}{p-1}}\hbox {d}x \right) ^{\frac{p-1}{p}}\\&\quad \le \left( C+ ||f(1)|| _{ L^p(\hbox {d}\mu ) }\right) \left( \int _0^\varepsilon n_0(1+n_0) x^{\left( \frac{4p}{p-1}-\frac{6}{p-1} \right) }\hbox {d}x \right) ^{\frac{p-1}{p}}. \end{aligned}$$

For \(p>3\), \(n_0(1+n_0) x^{\left( \frac{4p}{p-1}-\frac{6}{p-1} \right) }\in L^\infty (0, \infty )\), and then

$$\begin{aligned} \lim _{ \varepsilon \rightarrow 0 }\int _0^\varepsilon n_0(1+n_0)x^4|f(t, x)-C_*|\hbox {d}x=0. \end{aligned}$$

On the other hand,

$$\begin{aligned} \int _\varepsilon ^\infty n_0(1+n_0)x^4|f(t, x)-C_*|\hbox {d}x&\le \left( \int _\varepsilon ^\infty x^6 n_0(x)(1+n_0(x)) |f(t, x)-C_*|^2\hbox {d}x \right) ^{1/2}\\&\quad \times \left( \int _\varepsilon ^\infty n_0(1+n_0)x^2 \hbox {d}x\right) \\&\le C_\varepsilon \left( \int _0 ^\infty |f(t, x)-C_*|^2\hbox {d}\mu (x) \right) ^{1/2}, \end{aligned}$$

and the Corollary 4.15 follows from Corollary 4.13. \(\quad \square \)

4.2 The Limit of f(tx) as \(x\rightarrow 0\) for \(t>0\)

We show now the existence of the limit

$$\begin{aligned} b(t)=\lim _{ x\rightarrow 0 }f(t, x)\,\,\,\forall t>0 \end{aligned}$$
(4.17)

for all \(f_0\) and f as in Theorem 1.1, and describe its time evolution. We use the property, proven in Proposition 1.3 of [10],

$$\begin{aligned}&\forall f_0\in L^1(0, \infty ),\,\forall t>0,\,\,\,\,\lim _{ x \rightarrow 0 }S(t)f_0(x)=\ell (f_0; t) \in (-\infty , \infty ) \end{aligned}$$
(4.18)
$$\begin{aligned}&\ell (f_0; t)=A_1t^{-3}\int _0^tf_0(y)y^2\hbox {d}y+A_2t^{-4}\int _0^t f_0(y)y^3\hbox {d}y +\int _0^t f_0(y)\,b_1\!\!\left( \frac{t}{y} \right) \frac{\hbox {d}y}{y}, \end{aligned}$$
(4.19)

for some constants \(A_1\), \(A_2\) and a function \(b_1\) such that \(b _1(t)={\mathcal {O}}( t^{-8})\) for \(t>1\). Slightly more precise information may be obtained and is shown here, since it is of further interest.

Proposition 1.28

Suppose that \(f_0\) and f are as in Theorem 3.1. Then for all \(t>0\), and \(\delta >0\) as small as desired,

$$\begin{aligned} f(t, x)=&b(t)+\left( ||f_0|| _{ L^\infty (1, \infty ) }+||f_0||_1\right) {\mathcal {O}}\left( t^\delta x^{1-\delta } \right) \nonumber \\&+\sup _{ 0<x<1 }x^\theta |f_0(x)|{\mathcal {O}}\left( t^{\delta -\theta } x^{1-\delta } \right) ,\,\,x\rightarrow 0, \end{aligned}$$
(4.20)

where,

$$\begin{aligned}&b (t)=\ell (f_0; t)+\int _0^t \ell (F(f(s)); t-s)\textrm{d}s \end{aligned}$$
(4.21)
$$\begin{aligned}&\ell (F(f(s)); t-s)=\frac{A_1}{(t-s)^{3}}\int _0^{t-s}\!\!\!\!\!\!F(f(s, y))y^2\textrm{d}y+\frac{A_2}{(t-s)^{4}}\int _0^{t-s} \!\!\!\!\!\!F(f(s, y))y^3\textrm{d}y\nonumber \\&\quad +\int _0^{t-s} \!\!\!\!\!\! F(f(s, y))\,b_1\!\!\left( \frac{t-s}{y} \right) \frac{\textrm{d}y}{y} \end{aligned}$$
(4.22)

and there exists a constant \(C>0\) such that

$$\begin{aligned} |b(t)|\le C|||f_0|||_\theta \left( t^{-\theta }+t \right) ,\,\,\forall t>0. \end{aligned}$$
(4.23)

Proof

By construction, for all \(t>0\) and \(x>0\),

$$\begin{aligned} f(t, x)=S(t)f_0(x)+\int _0^t S(t-s)(F(f(s))(x)\hbox {d}s. \end{aligned}$$

By Proposition 1.5 in [10],

$$\begin{aligned}&(S(t)f_0)(x)=\ell (f_0; t)+\left( t^{-2+\delta }\int _0^t|f_0(y)|\hbox {d}y+t^{5+\delta }\int _t^\infty \frac{|f_0(y)|\hbox {d}y}{y^7}\right) {\mathcal {O}}_\delta (x^{1-\delta })\\&S(t-s)F(f(s))(x)=\ell (F(f(s)); t-s)+\Bigg ( (t-s)^{-2+\delta }\int _0^{t-s}|F(f(s, y))|\hbox {d}y\\&\quad +(t-s)^{5+\delta }\int _{t-s}^\infty \frac{|F(f(s, y))|\hbox {d}y}{y^7}\Bigg ){\mathcal {O}}_\delta (x^{1-\delta }). \end{aligned}$$

By (iv) in Proposition (2.1),

$$\begin{aligned}&(t-s)^{-2+\delta }\int _0^{t-s}|F(f(s, y))|\hbox {d}y\le (t-s)^{-2+\delta }\int _0^{t-s}(||f(s)||_1+||f(s)|| _{ L^\infty (1, \infty ) }))\hbox {d}y\\&\quad \le (t-s)^{-1+\delta }(||f(s)||_1+||f(s)|| _{ L^\infty (1, \infty ) }))\\&\qquad (t-s)^{-1+\delta }\int _{t-s}^\infty \frac{|F(f(s, y))|\hbox {d}y}{y^7}\le (t-s)^5(||f(s)||_1+||f(s)|| _{ L^\infty (1, \infty ) })\int _{ t }^\infty \frac{\hbox {d}y}{y^7}\\&\quad = \frac{(t-s)^{-1+\delta }}{6}(||f(s)||_1+||f(s)|| _{ L^\infty (1, \infty ) }) \end{aligned}$$

and, by (3.25) for \(T>0\),

$$\begin{aligned} \sup _{ x>1 } |f(t, x)|+||f(t)||_1\le C(T, \theta )\left( ||f_0|| _{ L^\infty (1, \infty ) }+t^{-\theta }\sup _{ 0<x<1 }x^\theta |f_0(x)| +||f_0||_1\right) . \end{aligned}$$

On the other hand,

$$\begin{aligned}&S(t-s)F(f(s))(x)\\ {}&\quad =\ell (F(f(s)); t-s)+(||f(s)||_1+||f(s)|| _{ L^\infty (1, \infty ) }){\mathcal {O}}\left( (t-s)^{-1+\delta }x^{1-\delta } \right) \\&\quad = \ell (F(f(s)); t-s)+\left( ||f_0|| _{ L^\infty (1, \infty ) }+s^{-\theta }\sup _{ 0<x<1 }x^\theta |f_0(x)| +||f_0||_1\right) \\&\qquad \times {\mathcal {O}}_T\left( (t-s)^{-1+\delta }x^{1-\delta } \right) \end{aligned}$$

and then,

$$\begin{aligned}&\int _0^t S(t-s)F(f(s))(x)= \int _0^t \ell (F(f(s)); t-s)\hbox {d}s\\&\quad +(||f_0|| _{ L^\infty (1, \infty ) }+||f_0||_1)\mathcal {O}\left( t^\delta x^{1-\delta } \right) +\sup _{ 0<x<1 }x^\theta |f_0(x)|{\mathcal {O}}\left( t^{\delta -\theta } x^{1-\delta } \right) . \end{aligned}$$

This shows (4.20), (4.21), (4.22). On the other hand, by property (iv) in Proposition 2.1, there exists a constant C that depends on \(\theta \) such that

$$\begin{aligned} |||F(f(s))||| _{ \theta }\le C|||f(s)|||_\theta ,\,\,\,\forall s>0. \end{aligned}$$

Then, for \(t\in (0, 1)\) and \(s\in (0, t)\),

$$\begin{aligned} \int _0^t (t-s)^{-n}\int _0^{t-s}F(f(s, y))y^{n-1}\hbox {d}y\hbox {d}s&\le C|||f_0|||_\theta \int _0^t (t-s)^{-n}\int _0^{t-s}y^{n-1-\theta }\hbox {d}y\hbox {d}s\\&=C|||f_0|||_\theta t^{-\theta }. \end{aligned}$$

For \(t>1\),

$$\begin{aligned}&\int _0^t (t-s)^{-n}\int _0^{t-s}F(f(s, y))y^{n-1}\hbox {d}y\hbox {d}s=J_1+J_2+J_3\\&\quad J_1=\int _0^{t-1} (t-s)^{-n}\int _0^1F(f(s, y))y^{n-1}\hbox {d}y\hbox {d}s\\&\quad J_2=\int _0^{t-1} (t-s)^{-n}\int _1^{t-s}F(f(s, y))y^{n-1}\hbox {d}y\hbox {d}s\\&\quad J_3=\int _{t-1}^t (t-s)^{-n}\int _0^{t-s}F(f(s, y))y^{n-1}\hbox {d}y\hbox {d}s, \end{aligned}$$

with,

$$\begin{aligned} |J_1|&\le C|||f_0|||_\theta \int _0^{t-1} (t-s)^{-n}\int _0^1y^{n-1-\theta }\hbox {d}y\hbox {d}s\\&=C(n, \theta )\int _0^{t-1}(t-s)^{-n}\hbox {d}s\le C|||f_0|||_\theta ,\\ |J_2|&\le C|||f_0|||_\theta \int _0^{t-1} (t-s)^{-n}\int _1^{t-s}y^{n-1}\hbox {d}y\hbox {d}s\\&=C|||f_0|||_\theta \int _0^{t-1} \left( 1-(t-s)^{-n} \right) \hbox {d}s=\le C|||f_0|||_\theta \, t \end{aligned}$$

and

$$\begin{aligned} |J_3|&\le C|||f_0|||_\theta \int _{t-1}^t (t-s)^{-n}\int _0^{t-s}y^{n-1-\theta }\hbox {d}y\hbox {d}s\\&=C|||f_0|||_\theta \int _{t-1}^t (t-s)^{-\theta }\hbox {d}s\le C|||f_0|||_\theta . \end{aligned}$$

It follows, for \(t>1\), that

$$\begin{aligned} \left| \int _0^t (t-s)^{-n}\int _0^{t-s}F(f(s, y))y^{n-1}\hbox {d}y\hbox {d}s\right| \le C|||f_0|||_\theta \, t, \end{aligned}$$

and

$$\begin{aligned} |b(t)-\ell (f_0; t)|\le C|||f_0|||_\theta \left( t^{-\theta }+t \right) ,\,\,\forall t>0. \end{aligned}$$

The same arguments show that \(|\ell (f_0; t)|\le C|||f_0|||_\theta (1+t^{-\theta })\) for all \(t>0\) and (4.23) follows. \( \quad \square \)

5 The Functions \(\varvec{u(t)}\) and \(\varvec{p_c(t)}\)

We now return to the notation of the time variable as in sub Section 1.4. Then, given the function \(f(\tau , x)\) obtained in Theorem 3.1, \(t=t(\tau )\) and \(p_c(t)\) must be determined in order to define

$$\begin{aligned} u(t, x)=f(\tau , x),\,\,\forall t>0, \,\forall x>0. \end{aligned}$$

The functions \(t, \tau \) and \(p_c(t)\) are related by the change of time variable (1.30), i.e.,

$$\begin{aligned} \tau =\int _0^{t}p_c(s)\hbox {d}s. \end{aligned}$$
(5.1)

Proposition 1.29

For all \(\tau >0\),

$$\begin{aligned} \int _0^\tau \int _0^\infty \left| {\mathcal {L}}(f(\sigma ))(x)\right| n_0(x^2)(1+n_0(x^2))x^4\textrm{d}x\textrm{d}\sigma <\infty , \end{aligned}$$
(5.2)

and, if

$$\begin{aligned}&m(\tau )=\int _0^\infty {\mathcal {L}}(f(\tau ))(x) n_0(x^2)(1+n_0(x^2))x^4\textrm{d}x,\,\,\forall \tau >0 \end{aligned}$$
(5.3)
$$\begin{aligned}&|m(\tau )|<\infty ,\,\,\forall \tau >0. \end{aligned}$$
(5.4)

Proof

The proposition is a direct and straightforward consequence of the integrability property (3.34) of \({\mathcal {L}}(f)\). \( \quad \square \)

Let us denote that

$$\begin{aligned} {\mathcal {M}}(r )=\int _0^r m(\rho )d\rho ,\,\,\forall r>0. \end{aligned}$$
(5.5)
$$\begin{aligned} q_c(\tau )=q_c(0)e^{-{\mathcal {M}}(\tau )},\,\,\forall \tau >0. \end{aligned}$$
(5.6)

Proposition 1.30

For all \(t>0\) there exists a unique \(\tau >0\) such that

$$\begin{aligned} t=\int _0^\tau \frac{\textrm{d}\sigma }{q_c(\sigma )},\,\,\forall \tau >0. \end{aligned}$$
(5.7)

Proof

By Proposition 5.1, \(|{\mathcal {M}}(\tau )|<\infty \) for all \(\tau >0\) and then \(q_c(\tau )\in (0, \infty )\) for all \(\tau >0\) and the integral in the right hand side of (5.7) is well defined and convergent. Since \(q_c(t)>0\) this integral is a monotone increasing function of \(\tau \). It only remains to check that its range is \([0, \infty )\).

By Corollary 4.15, for \(\varepsilon >0\) and \(\tau _\varepsilon \) large enough,

$$\begin{aligned} \int _0^\infty f(\tau , x)n_0(1+n_0)x^4\hbox {d}x>C_*\int _0^\infty n_0(1+n_0)x^4\hbox {d}x-\varepsilon ,\,\,\forall \tau \ge \tau _\varepsilon . \end{aligned}$$

Since, on the other hand,

$$\begin{aligned} {\mathcal {M}}(\tau )&=\int _0 ^\tau \int _0^\infty \int _0^\infty W(x, y)(f(\sigma , y)-f(\sigma , x)) y^4x^2\hbox {d}y\hbox {d}x\hbox {d}\sigma \nonumber \\&=\int _0^\tau \frac{d}{\hbox {d}\sigma }\int _0^\infty n_0(1+n_0) f(\sigma , x)x^4\hbox {d}x\hbox {d}\sigma \nonumber \\&=\int _0^\infty n_0(1+n_0)f(\tau , x)x^4\hbox {d}x - \int _0^\infty n_0(1+n_0)f_0(x)x^4\hbox {d}x, \end{aligned}$$
(5.8)

it follows, for \(\tau >\tau _\varepsilon \), that

$$\begin{aligned} {\mathcal {M}}(\tau )\ge C_*\int _0^\infty n_0(1+n_0)x^4\hbox {d}x-\varepsilon - \int _0^\infty n_0(1+n_0)f_0(x)x^4\hbox {d}x. \end{aligned}$$

Therefore, the function \(e^{{\mathcal {M}}(\sigma )}\) is not integrable at infinity, and

$$\begin{aligned} \lim _{ \tau \rightarrow \infty }\int _0^\tau \frac{\hbox {d}\sigma }{q_c(\sigma )}=\infty , \end{aligned}$$

and, for all \(t>0\), there exists a unique \(\tau >0\) satisfying (5.7). \( \quad \square \)

Proof

For all \(t>0\), let \(\tau >0\) be given by Proposition 5.2 and define that

$$\begin{aligned}&u(t, x)=f(\tau , x),\,\,\,\forall x>0, \end{aligned}$$
(5.9)
$$\begin{aligned}&p_c(t)=q_c(\tau ), \end{aligned}$$
(5.10)

where f is obtained by Theorem 3.1 with initial data \(f_0=u_0\). From the definition of \(p_c\),

$$\begin{aligned} \frac{d t}{d \tau }=\frac{1}{q_c(\tau )}\Longrightarrow \frac{d \tau }{d t}=q_c(\tau )=p_c(t), \end{aligned}$$

and (5.1) is satisfied. On the other hand,

$$\begin{aligned} \frac{{\hbox {d}}p_c(t)}{\hbox {d}t}=\frac{\hbox {d}q_c(\tau )}{\hbox {d}\tau }\frac{\hbox {d}\tau }{\hbox {d}t} =-q_c(\tau )m(\tau )\frac{\hbox {d}\tau }{\hbox {d}t}=-p_c(t)m(\tau )\frac{\hbox {d}\tau }{\hbox {d}t}. \end{aligned}$$
(5.11)

By (5.3) and (5.9),

$$\begin{aligned} m(\tau )&=\int _0^\infty {\mathcal {L}}(f(\tau )) n_0(x^2)(1+n_0(x^2))x^4\hbox {d}x\\&=\int _0^\infty {\mathcal {L}}(u(t)) n_0(x^2)(1+n_0(x^2))x^4\hbox {d}x=\mu (t). \end{aligned}$$

It follows from (5.6) that

$$\begin{aligned} \frac{\hbox {d}p_c(t)}{\hbox {d}t}&=-p_c(t)m(\tau )\frac{\hbox {d}\tau }{\hbox {d}t}=-p_c(t)\frac{\hbox {d}\mu (t)}{\hbox {d}t}\\&= -p_c(t)\frac{\hbox {d}}{\hbox {d}t} \int _0^\infty \mathcal {L}(u(s))n_0(x^2)(1+n_0(x^2))x^4\hbox {d}x\hbox {d}s, \end{aligned}$$

and then \(p_c\in C^1(0, \infty )\) and satisfies (1.15). On the other hand, by (5.1) and Theorem 3.3,

$$\begin{aligned}&\frac{\partial u}{\partial t}(t, x)=\frac{\hbox {d}\tau }{\hbox {d}t}\frac{\partial f}{\partial \tau }(\tau , x)=-p_c(t)\frac{\partial f}{\partial \tau }(\tau , x). \end{aligned}$$

Then, Theorem 1.1 follows from Theorems 3.1 and 3.3, where the function H in (1.45) is given by

$$\begin{aligned}&H(t, x)={\tilde{H}}(\tau , x), \,\,\,a. e. \,\,(0, \infty )\times (0, \infty ), \end{aligned}$$
(5.12)

with \({\tilde{H}}\) given in (3.45). \( \quad \square \)

Proof

It follows from (1.42), and Corollary 4.3 that

$$\begin{aligned} \frac{\hbox {d}E(t)}{\hbox {d}t }&=p_c(t) \int _0^\infty n_0(1+n_0)x^4{\mathcal {L}}(u(t ))(x)\hbox {d}x\\&=p_c(t) \int _0^\infty n_0(x)(1+n_0(x){\mathcal {L}} f(\tau , x)x^4\hbox {d}x \\&=p_c(t)\frac{\hbox {d}}{\hbox {d}\tau }\int _0^\infty n_0(1+n_0)x^4f(\tau , x)x^2\hbox {d}x=0, \end{aligned}$$

and this yields property (1.56) in Corollary 1.2. Property (1.57) of Corollary 1.2 follows from (1.13) and (1.15), since by (1.44), integration of (1.13) on \((0, \infty )\) yields

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\int _0^\infty n_0(1+n_0)u(t, x)x^4\hbox {d}x=-\frac{\hbox {d}p_c(t)}{\hbox {d}t}. \end{aligned}$$

From Proposition 4.13 property (1.58) in Corollary 1.3 follows, and property (1.59) is deduced from Corollary 4.15.

Since \(f\in C([0, \infty ); L^1(0, \infty ))\), by the identity (5.8) and (5.6), \(q_c\in C([0, \infty ))\). It follows that \(p_c\in C([0, \infty ))\) and it is bounded on any compact subset of \([0, \infty )\). Passing to the limit in (5.8) as \(t\rightarrow \infty \) and using Corollary 4.15 property (1.61) is obtained. Then, it also follows that \(p_c\) is bounded on \((0, \infty )\). \( \quad \square \)

Proposition 1.31

Suppose that \(u_0\) and u are as in Theorem 1.1. Then for all \(t>0\), and \(\delta >0\) as small as desired,

$$\begin{aligned} u(t, x)&=a(t )+\left( ||f_0|| _{ L^\infty (1, \infty ) }+||f_0||_1\right) {\mathcal {O}}\left( \tau ^\delta x^{1-\delta } \right) \nonumber \\&\quad +\sup _{ 0<x<1 }x^\theta |f_0(x)|{\mathcal {O}}\left( \tau ^{\delta -\theta } x^{1-\delta } \right) ,\,\,x\rightarrow 0,\end{aligned}$$
(5.13)
$$\begin{aligned} \hbox {where,}\quad a(t)&=b\left( \int _0^{t}p_c(s)\textrm{d}s\right) \end{aligned}$$
(5.14)

satisfies, for some constant \(C>0\),

$$\begin{aligned} |a(t)|\le C|||u_0|||_\theta \left( \left( \int _0^{t}p_c(s)\textrm{d}s \right) ^{-\theta }+\int _0^{t}p_c(s)\textrm{d}s\right) ,\,\,\forall t>0. \end{aligned}$$
(5.15)

Proof

By construction, \(u(t, x)=f(\tau , x)\) were \(\tau \) is given in terms of t by (5.1) and therefore, (5.13), (5.14) follow from (4.20) and (5.15) follows from (4.23). \( \quad \square \)