1 Introduction

It has been known that the time evolution of a dilute Bose gas and the formation of Bose–Einstein condensation can be described by the Boltzmann equation for Bose–Einstein particles. Derivations of this important semiclassical model can be found in [35, 17, 19, 24]. The present paper is a continuation of our previous work [15] on the study of space-homogeneous solutions of the equation for the hard sphere model:

$$\begin{aligned} \frac{\partial }{\partial t}f= \int \limits _{{{{\mathbb {R}}^{3}}\times {\mathbb {S}}^{2}}}\frac{|(\mathbf{v}-\mathbf{v}_*)\cdot \omega |}{(4\pi )^{2}}[f'f_*'(1+f+f_*)-ff_*(1+f'+f_*')] \mathrm{d}\omega \mathrm{d}\mathbf{v_*}. \end{aligned}$$
(1.1)

Here the solution \(f=f(\mathbf{v},t)\ge 0\) is the number density of particles at time \(t\) with the velocity \(\mathbf{v}\); \(f_*=f(\mathbf{v_*},t), f'=f(\mathbf{v'},t),f_*'=f(\mathbf{v_*'},t)\), and \(\mathbf{v},\mathbf{v_*}\) and \(\mathbf{v'},\mathbf{v_*'}\) are velocities of two particles before and after their collision:

$$\begin{aligned} \mathbf{v}'=\mathbf{v}- ((\mathbf{v}-\mathbf{v}_*)\cdot \omega )\omega ,\quad \mathbf{v}_*'=\mathbf{v}_*+ ((\mathbf{v}-\mathbf{v}_*)\cdot \omega )\omega ,\quad \omega \in {{\mathbb {S}}^{2}} \end{aligned}$$

which conserves the kinetic energy

$$\begin{aligned} |\mathbf{v}'|^2+|\mathbf{v}_*'|^2=|\mathbf{v}|^2+|\mathbf{v}_*|^2. \end{aligned}$$
(1.2)

The main difficulty in investigating Eq.(1.1) comes from the cubic terms \(ff'f_*'\), \(f_*f'f_*'\), etc., because their collision integrals (even with a strong cutoff on the collision kernel) over \({\mathbb R}^3\times {{\mathbb {R}}^{3}}\times {{\mathbb {S}}^{2}}\) in the usual weak form (i.e. involving smooth test functions) are unbounded on a set of non-isotropic functions in \(\{0\le f\in C^{\infty }_c({{\mathbb {R}}^{3}})\,|\,\Vert f\Vert _{L^1({{\mathbb {R}}^{3}})}\le 1\}\) ([13]). So far the results about global existence, convergence to equilibrium, singular behavior, kinetics of Bose–Einstein condensation, numerical analysis, etc. of solutions of Eq.(1.1) and its modifications have been mainly concerned with isotropic solutions, see e.g. [1, 2, 68, 11, 13, 14, 18, 2023]. Recent works [9, 10, 15] also considered blow up and condensation in finite time for such solutions (see below).

By velocity translation one can assume that the mean velocity is zero so that the isotropic solution \(f(\mathbf{v},t)\) can be written as \(f(|\mathbf{v}|^2/2,t)\). Accordingly, let \(\epsilon , \epsilon _*,\epsilon ',\epsilon _*'\) stand for \(|\mathbf{v}|^2/2, |\mathbf{v}_*|^{2}/2, |\mathbf{v}'|^2/2, |\mathbf{v}_*'|^2/2\) respectively. Then (1.2) implies that \(\epsilon , \epsilon ',\epsilon _*'\) are independent variables and \(\epsilon _*=\epsilon '+\epsilon _*'-\varepsilon \), and this 3-degree of freedom is just enough for the total integration of the cubic isotropic term \(ff'f_*'\) over \({\mathbb R}_{+}^3\). As in the previous work [15], we are only interested in such solutions that have finite mass and energy, i.e.

$$\begin{aligned} \int \limits _{{\mathbb {R}}^{3}}(1+|\mathbf{v}|^2/2)f(|\mathbf{v}|^2/2,t)\mathrm{d}\mathbf{v}=4\pi \sqrt{2}\int \limits _{{\mathbb R}_{+}}(1+\varepsilon )f(\varepsilon ,t)\sqrt{\varepsilon }\mathrm{d}\varepsilon <\infty . \end{aligned}$$

It will be more convenient to write \(\epsilon , \epsilon _*,\epsilon ',\epsilon _*'\) as \(x,x_*,y,z\) respectively. Then introducing

$$\begin{aligned} w(x,y,z)=\frac{\min \{\sqrt{x},\sqrt{x_*}, \sqrt{y},\sqrt{z}\,\}}{\sqrt{x}},\quad x>0, y\ge 0, z\ge 0,\, x_*=(y+z-x)_{+}\nonumber \\ \end{aligned}$$
(1.3)

where \((u)_{+}=\max \{u, 0\}\), and using the formula (see e.g. [15])

$$\begin{aligned} \int \limits _{{{{\mathbb {R}}^{3}}\times {\mathbb {S}}^{2}}}\frac{|\mathbf{(v-v_*)\cdot \omega }|}{(4\pi )^2}\Phi \Big (|\mathbf{v}|^2/2, |\mathbf{v}'|^2/2,|\mathbf{v_*}'|^2/2\Big )\mathrm{d}\omega \mathrm{d}\mathbf{v_*}=\int \limits _{{\mathbb R}_{+}^2} w(x,y,z)\Phi (x, y,z)\mathrm{d}y \mathrm{d}z\nonumber \\ \end{aligned}$$
(1.4)

for \(\mathbf{v}\in {{\mathbb {R}}^{3}}{\setminus }\{0\}\), \(x=|\mathbf{v}|^2/2\), the equation (1.1) for isotropic solutions becomes

$$\begin{aligned} \frac{\partial }{ \partial t}f(x,t)=Q(f)(x,t) \end{aligned}$$
(1.5)

where \(Q(f)(x,t)=Q(f(\cdot ,t))(x)\) is the isotropic version of the collision integral in (1.1):

$$\begin{aligned} Q(f)(x)=\int \limits _{{\mathbb R}_{+}^2}w(x,y,z)[ f'f_*'(1+f+f_*))- ff_*(1+f'+f_*')]\mathrm{d}y\mathrm{d}z, \end{aligned}$$
(1.6)
$$\begin{aligned} f=f(x), \quad f_*=f(x_*),\quad f'=f(y),\quad f_*'=f(z). \end{aligned}$$
(1.7)

As usual, in order to study kinetic equations, the first step is to extend the concept of classical solutions to that of mild solutions:

Definition 1.1

Let \(f(x,t)\) be a nonnegative measurable function on \({\mathbb R}_{+}\times [0, T_{\infty })\,(0<T_{\infty }\le \infty )\). We say that \(f(\cdot ,t)\) is a mild solution of Eq.(1.5) on the time-interval \([0, T_{\infty })\) if \(f\) satisfies

  1. (i)

    \(\sup _{t\in [0,T]}\int \limits _{{\mathbb R}_{+}}(1+x)f(x,t)\sqrt{x}\mathrm{d}x<\infty \quad \forall \, 0<T<T_{\infty },\)

  2. (ii)

    there is a null set \(Z\subset {\mathbb R}_{+}\) (i.e. \(\mathrm{mes}(Z)=0\)) such that for all \(x\in {\mathbb R}_{+}{\setminus } Z\) and all \(t\in [0, T_{\infty })\), \(\int \nolimits _{0}^{t}\mathrm{d}\tau \int \nolimits _{{\mathbb R}_{+}^2}w(x,y,z)[ f'f_*'(1+f+f_*))+ ff_*(1+f'+f_*')]\mathrm{d}y\mathrm{d}z<\infty \) and

    $$\begin{aligned} f(x,t)=f_0(x) +\int \limits _{0}^{t}Q(f)(x,\tau )\mathrm{d}\tau . \end{aligned}$$

Here \(f_0=f(\cdot ,0)\) denotes the initial datum of \(f(\cdot ,t)\).

By definition, a mild solution describes only “regular” behavior of a Bose gas in a time-interval in which there is no condensation. In order to cover the whole time of evolution, in particular in order to describe the formation, nucleation, and growth of Bose–Einstein condensation, the class of mild solutions has to be extended, and any extension should include at least the low temperature equilibria \(\mathrm{d}F_\mathrm{be}(x)=f_{\mathrm{be}}(x)\sqrt{x}\mathrm{d}x+N_0\delta (x)\mathrm{d}x\), where \(f_\mathrm{be}(x)=(e^{x/\kappa }-1)^{-1}, \kappa >0, N_0\ge 0\), and \(\delta (x)\) is the Dirac delta function concentrating at \(x=0\). Some authors extend mild solution \(f(\cdot ,t)\) to the solution \((f(\cdot ,t),n(t))\) of an equation system where \(n(t)\ge 0\) is the coefficient of the delta function, i.e., \(n(t)\) describes the condensate, see e.g. [22, 23] on the kinetics and structural analysis of \((f(\cdot ,t),n(t))\) including nucleation of the condensate. See also [1, 2] for the global in time existence of such solutions for the case of low temperature and \(n(0)>0\). In this paper we (as before) extend mild solutions to distributional solutions ([13, 15]) and we will use the existence result of distributional solutions to study both mild solutions and condensation.

Multiplying by a bounded smooth function \(\varphi (x)\) to Eq. (1.5) and taking integration over \({\mathbb R}_{\ge 0}\) with respect to the weighted Lebesgue measure \(\sqrt{x}\mathrm{d}x\), and then replacing the density element \(f(x,t)\sqrt{x}\mathrm{d}x\) with a measure element \(\mathrm{d}F_t(x)\), we obtain a weak form (1.8) of Eq. (1.5) which leads to the definition of distributional solutions ([15]):

Definition 1.2

Let \(\{F_t\}_{t\ge 0}\subset \mathcal{B}_{1}^{+}({\mathbb R}_{\ge 0})\). We say that \(\{F_t\}_{t\ge 0}\), or simply \(F_t\), is a distributional solution of Eq. (1.5) on the time-interval \([0, \infty )\) if

  1. (i)

    \(\sup _{t\in [0,T]}\int _{{\mathbb R}_{\ge 0}}(1+x)\mathrm{d}F_t(x)<\infty \) for all \(T\in (0, \infty ),\)

  2. (ii)

    for every \(\varphi \in C^{1,1}_b({\mathbb R}_{\ge 0})\), the function \(t\mapsto \int _{{\mathbb R}_{\ge 0}}\varphi (x)\mathrm{d}F_t(x)\) belongs to \(C^1([0, \infty ))\),

  3. (iii)

    for every \(\varphi \in C^{1,1}_b({\mathbb R}_{\ge 0})\),

    $$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\int \limits _{{\mathbb R}_{\ge 0}}\varphi \mathrm{d}F_t= \int \limits _{{\mathbb R}_{\ge 0}^2}\mathcal{J}[\varphi ]\mathrm{d}^2F_t+ \int \limits _{{\mathbb R}_{\ge 0}^3}\mathcal{K}[\varphi ]\mathrm{d}^3F_t \qquad \forall \, t\in [0, \infty ). \end{aligned}$$
    (1.8)

    Here \(\mathrm{d}^2F=\mathrm{d}(F\otimes F)=\mathrm{d}F(y)\mathrm{d}F(z), \mathrm{d}^3F=\mathrm{d}(F\otimes F\otimes F)=\mathrm{d}F(x)\mathrm{d}F(y)\mathrm{d}F(z)\),

    $$\begin{aligned} \mathcal{J}[\varphi ](y,z)&= \frac{1}{2} \int \limits _{0}^{y+z}\mathcal{K}[\varphi ](x,y,z) \sqrt{x}\mathrm{d}x, \quad \mathcal{K}[\varphi ](x, y,z)=W(x,y,z)\Delta \varphi (x,y,z), \\ \Delta \varphi (x,y,z)&= \varphi (x)+\varphi (x_*)-\varphi (y)-\varphi (z), \quad x_*=(y+z-x)_{+},\\ W(x,y,z)&= \frac{\sqrt{x}w(x,y,z)}{\sqrt{xyz}}=\frac{\min \{\sqrt{x},\sqrt{x_*},\sqrt{y}, \sqrt{z}\,\}}{\sqrt{xyz}}\qquad \mathrm{if} \quad x,y,z>0,\\ W(x,y,z)&= \frac{1}{\max \{\sqrt{yz},\,\sqrt{xz},\,\sqrt{xy}\,\}}\quad \mathrm{if}\quad xyz=0,\, x_*\max \{yz,\,xz,\,xy\}>0,\\ W(x,y,z)&= 0\,\,\,\,\, \quad \quad \qquad \qquad \qquad \qquad \mathrm{others}. \end{aligned}$$

Remark 1.1

  1. (a)

    The spaces of test functions and Borel measures appeared above are taken as follows

    $$\begin{aligned}&C^{1,1}_b({\mathbb R}_{\ge 0})=\Big \{ \varphi \in C^1({\mathbb R}_{\ge 0})\, \big |\, \,\varphi \in C_b({\mathbb R}_{\ge 0}),\, \frac{\mathrm{d}}{\mathrm{d}x}\varphi \in C_b({\mathbb R}_{\ge 0})\cap \mathrm{Lip}({\mathbb R}_{\ge 0})\Big \},\\&\mathcal{B}_k({\mathbb R}_{\ge 0})= \Bigg \{ F\in \mathcal{B}({\mathbb R}_{\ge 0})\,\,\Big |\,\,\Vert F\Vert _k:= \int \limits _{{\mathbb R}_{\ge 0}}(1+x)^k\mathrm{d}|F|(x)<\infty \Bigg \},\\&\mathcal{B}_k^{+}({\mathbb R}_{\ge 0})= \Big \{ F\in \mathcal{B}_k({\mathbb R}_{\ge 0}) \,\,\big |\, \, F\ge 0\Big \},\quad k\ge 0;\quad \mathcal{B}^{+}({\mathbb R}_{\ge 0}):=\mathcal{B}^{+}_0({\mathbb R}_{\ge 0}), \end{aligned}$$

    where \(C_b({\mathbb R}_{\ge 0})\) is the class of bounded continuous on \({\mathbb R}_{\ge 0}\), \(\mathrm{Lip}({\mathbb R}_{\ge 0})\) is the class of functions satisfying Lipschitz condition on \({\mathbb R}_{\ge 0}\), and \(\mathcal{B}({\mathbb R}_{\ge 0})\equiv \mathcal{B}_0({\mathbb R}_{\ge 0})\) is the class of signed real Borel measures \(F\) on \({\mathbb R}_{\ge 0}\) satisfying \(\int \nolimits _{{\mathbb R}_{\ge 0}}\mathrm{d}|F|(x)<\infty \) where \(|F|\) is the total variation of \(F\).

  2. (b)

    Distributional solutions can be also defined on finite time-intervals by replacing \([0,\infty )\) with \([0, T]\) or \([0, T_{\infty })\). But this is not very necessary because it is different from mild solutions that the time-intervals of existence of distributional solutions can be always \([0,\infty )\), see Theorem 1.1 below.

  3. (c)

    Under the the condition (i) in Definition 1.2, the collision integrals \(t\mapsto \int _{{\mathbb R}_{\ge 0}^2}\mathcal{J}[\varphi ]\mathrm{d}^2F_t, t\mapsto \int _{{\mathbb R}_{\ge 0}^3}\mathcal{K}[\varphi ]\mathrm{d}^3F_t\) are well defined on \([0,\infty )\) for all \(\varphi \in C^{1,1}_b({\mathbb R}_{\ge 0})\) (see [15]), and the Corollary of Lemma 4 in [13] insures that if \(t\mapsto \int _{{\mathbb R}_{\ge 0}}\varphi (x)\mathrm{d}F_t(x)\) is continuous on \([0, \infty )\) for every \(\varphi \in C^{1,1}_b({\mathbb R}_{\ge 0})\), so are \(t\mapsto \int _{{\mathbb R}_{\ge 0}^2}\mathcal{J}[\varphi ]\mathrm{d}^2F_t,\, t\mapsto \int _{{\mathbb R}_{\ge 0}^3}\mathcal{K}[\varphi ]\mathrm{d}^3F_t\) on \([0,\infty )\). Therefore under the condition (i), the conditions (ii)–(iii) in Definition 1.2 are equivalent to the following (ii)\('\)–(iii)\('\): (ii)\('\) for every \(\varphi \in C^{1,1}_b({\mathbb R}_{\ge 0})\), the function \(t\mapsto \int _{{\mathbb R}_{\ge 0}}\varphi (x)\mathrm{d}F_t(x)\) is continuous on \([0, \infty )\); (iii)\('\) for every \(\varphi \in C^{1,1}_b({\mathbb R}_{\ge 0})\) and every \(t\in [0, \infty )\)

    $$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}}\varphi \mathrm{d}F_t=\int \limits _{{\mathbb R}_{\ge 0}}\varphi \mathrm{d}F_0+\int \limits _{0}^t\left( \int \limits _{{\mathbb R}_{\ge 0}^2}\mathcal{J}[\varphi ]\mathrm{d}^2F_\tau + \int \limits _{{\mathbb R}_{\ge 0}^3}\mathcal{K}[\varphi ]\mathrm{d}^3F_\tau \right) \mathrm{d}\tau . \end{aligned}$$
    (1.9)

Before introducing our main results and some recent relevant works, let us recall some terminologies.

1.1 Moments

Including negative orders, moments for a positive Borel measure \(F\) on \({\mathbb R}_{\ge 0}\) are defined by

$$\begin{aligned} M_{p}(F) =\int \limits _{{\mathbb R}_{\ge 0}} x^{p}\mathrm{d}F(x), \qquad p\in (-\infty ,\infty ). \end{aligned}$$
(1.10)

Here for the case \(p<0\) we adopt the convention \(0^{p}=(0^+)^p=\infty \), and we recall that \(\infty \cdot 0=0\). Then it should be noted that

$$\begin{aligned} M_{p}(F)<\infty \quad \mathrm{and}\quad p<0\quad \Longrightarrow \quad F(\{0\})=0. \end{aligned}$$
(1.11)

Moments of orders \(0, 1\) correspond to the mass and energy and are particularly denoted as

$$\begin{aligned} N(F)=M_0(F) , \quad E(F)=M_1(F). \end{aligned}$$
(1.12)

Moments for a nonnegative measurable function \(f\) on \({\mathbb R}_{+}\) are defined in consistence with the case of measures: \(M_p(f)=M_p(F)\) with \(\mathrm{d}F(x) =f(x)\sqrt{x}\mathrm{d}x\), i.e.

$$\begin{aligned} M_{p}(f)=\int \limits _{{\mathbb R}_{+}}x^pf(x)\sqrt{x}\mathrm{d}x,\qquad p\in (-\infty ,\infty ). \end{aligned}$$
(1.13)

We also denote \(N(f)=M_0(f), E(f)=M_1(f).\) And notice that \(M_{-1/2}(f)=\int \limits _{{\mathbb R}_{+}}f(x)\mathrm{d}x.\)

1.2 Weighted \(L^1\) Spaces

We denote by \(L^1({\mathbb R}_{+}, \varrho (x)\mathrm{d}x)\) the space of measurable functions \(f\) on \({\mathbb R}_{+}\) satisfying \(\int \nolimits _{{\mathbb R}_{+}}|f(x)|\varrho (x)\mathrm{d}x<\infty \), where \(\varrho (x)\) is a given nonnegative measurable function on \({\mathbb R}_{+}\). For the case \(\varrho (x)\equiv 1\) we denote \(L^1({\mathbb R}_{+})=L^1({\mathbb R}_{+}, \mathrm{d}x)\).

1.3 Kinetic Temperature

Let \(N=N(F_0), E=E(F_0)\). If \(m\) is the mass of one particle, then \(m4\pi \sqrt{2}N\), \(m 4\pi \sqrt{2}E\) are total mass and kinetic energy of the particle system per unite space volume. The kinetic temperature \(\overline{T}\) and the critical temperature \(\overline{T}_c\) are defined by (see e.g.[13] and references therein)

$$\begin{aligned} \overline{T}=\frac{2}{3k_\mathrm{B}}\cdot \frac{mE}{N} ,\qquad \frac{\overline{T}}{\overline{T}_c}=\frac{(2\pi )^{1/3}[\zeta (3/2)]^{5/3}}{3\zeta (5/2)} \cdot \frac{E}{N^{5/3}} \end{aligned}$$
(1.14)

where \(k_\mathrm{B}\) is the Boltzmann constant, \(\zeta (\cdot )\) is the Riemann zeta function.

1.4 Conservative Solutions

\(F_t\) is called a conservative distributional solution if it conserves both mass and energy, i.e., \(N(F_t)=N(F_0) ,E(F_t)=E(F_0)\) for all \(t\in [0, \infty )\). Similarly \(f\) is called conservative mild solution on the time-interval \([0, T_{\infty })\) if \(N(f(t))=N(f_0), E(f(t))=E(f_0)\) for all \(t\in [0, T_{\infty })\). Since, by definition of \(\mathcal{J}[\varphi ]\) and \(\mathcal{K}[\varphi ]\), \(\mathcal{J}[\varphi ]=\mathcal{K}[\varphi ]= 0\) for \(\varphi (x)\equiv 1\) and \(\varphi (x)\equiv x\), and since the constant \(\varphi (x)\equiv 1\) belongs to the test function space while the function \(\varphi (x)\equiv x\) does not, it follows from Definition 1.2 that every distributional solution automatically conserves the mass, but does not automatically conserve the energy. In general, for every distributional solution \(F_t\), the energy \(t\mapsto E(F_t)\) is non-decreasing on \([0, \infty )\) ([13]). On the other hand, in practice, most of distributional solutions obtained by weak convergence of conservative approximate solutions satisfy the energy inequality \(E(F_t)\le E(F_0)\) (\(\forall \, t\ge 0)\). These two together imply that the conservation of energy actually holds for such solutions. Throughout this paper we always assume that \(N(F_0)>0, N(f_0)>0.\) We also assume that \(E(F_0)>0\) when dealing with conservative distributional solutions. This is because if \(E(F_0)=0\), then the conservation of mass and energy imply that \(F_t\equiv F_0\) is a Dirac mass concentrating at \(x=0\), which is a trivial equilibrium. Basic result to be used in the paper on the existence and conservation law of distributional solutions is the following

Theorem 1.1

([13, 15])

  1. (a)

    Let \(F_0\in \mathcal{B}_1^{+}({\mathbb R}_{\ge 0})\). Then there exists a conservative distributional solution \(F_t\) of Eq. (1.5) on \([0,\infty )\) with the initial datum \(F_0\), and \(F_t\) has the moment production:

    $$\begin{aligned} M_{p}(F_t)\le C_{p}(1+1/t)^{2(p-1)},\quad \forall \, t>0,\quad \forall \, p\ge 1. \end{aligned}$$
    (1.15)

    Here \(0<C_p=C_{N,E,p}<\infty \) depends only on \(N=N(F_0), E=E(F_0)\) and \(p\).

  2. (b)

    For every distributional solution \(F_t\) of Eq. (1.5) on \([0,\infty )\), the energy \(t\mapsto E(F_t)\) is non-decreasing on \([0, \infty )\). Consequently if \(E(F_t)\le E(F_0)\) for all \(t\ge 0\), then \(E(F_t)\equiv E(F_0)\).

This theorem is in fact stated in Theorems 3–4 of [13] for distributional solutions \(\bar{F}_t\) defined in [13,  p. 1611], see also Definition 5.1 in [15]. It is proved in [15] that the two definitions of distributional solutions are equivalent and the corresponding solutions \(F_t\) and \(\bar{F}_t\) determine to each other through the functional representation:

$$\begin{aligned} 4\pi \sqrt{2}\int \limits _{{\mathbb R}_{\ge 0}}\varphi (x)\mathrm{d}F_t(x)=\int \limits _{{\mathbb R}_{\ge 0}}\varphi (r^2/2)\mathrm{d}\bar{F}_t(r) \end{aligned}$$
(1.16)

for all \(t\ge 0\) and all \(\varphi \in C_b({\mathbb R}_{\ge 0})\). By monotone convergence theorem, (1.16) holds also for all \(t\ge 0\) and all \(0\le \varphi \in C({\mathbb R}_{\ge 0})\). Correspondingly, \(N, E\) and \(\bar{N}=\int \nolimits _{{\mathbb R}_{\ge 0}}\mathrm{d}\bar{F}_0(r)\), \(\bar{E}=\frac{1}{2}\int \nolimits _{{\mathbb R}_{\ge 0}}r^2\mathrm{d}\bar{F}_0(r)\) have the relation \(\bar{N}=4\sqrt{2}\, N, \bar{E}=4\sqrt{2}\, E\). The conservative distributional solution \(\bar{F}_t\) obtained in [13] has the moment production: \(\int \nolimits _{{\mathbb R}_{\ge 0}}r^{2p}\mathrm{d}\bar{F}_t(r)\le \bar{C}_{p}(1+1/t)^{2(p-1)}\) for all \(t>0\), \(p\ge 1\), where \(\bar{C}_{p}=\bar{C}_{\bar{N},\bar{E},p}<\infty \) depends only on \(\bar{N},\bar{E},\) and \(p\). If we let the initial datum \(\bar{F}_0\) of \(\bar{F}_t\) be defined by \(F_0\) through (1.16) for \(t=0\), and let \(F_t\) be defined by \(\bar{F}_t\) through (1.16), then \(F_t\) is a conservative distributional solution of Eq. (1.5) on \([0,\infty )\) with the initial datum \(F_0\). Since \(M_p(F_t)=\frac{1}{4\sqrt{2}}\int \nolimits _{{\mathbb R}_{\ge 0}}(r^2/2)^p\mathrm{d}\bar{F}_t(r)\), it follows that \(F_t\) also satisfies (1.15).

1.5 Regular-Singular Decomposition

Let \(F_t\) be a distributional solution \(F_t\) of Eq. (1.5) on \([0,\infty )\). According to measure theory, \(F_t\) at every time \(t\) can be uniquely decomposed as the regular part \(f(\cdot ,t)\in L^1({\mathbb R}_{+},\sqrt{x}\mathrm{d}x)\) and the singular part \(\mu _t\in \mathcal{B}^{+}_1({\mathbb R}_{\ge 0})\) with respect to the Lebesgue measure: there is a Borel null set \(Z_t\subset {\mathbb R}_{\ge 0}\) (i.e. \(\mathrm{mes}(Z_t)=0\)) such that

$$\begin{aligned} \mathrm{d}F_t(x)=f(x,t)\sqrt{x}\mathrm{d}x+\mathrm{d}\mu _t(x),\quad \mu _t({\mathbb R}_{\ge 0}{\setminus } Z_t)=0. \end{aligned}$$
(1.17)

We say that \(F_t\) is regular at time \(t\) if \(\mathrm{d}F_t(x)=f(x,t)\sqrt{x}\mathrm{d}x\), i.e. \(\mu _t=0\). In this case, \(f(\cdot ,t)\) is called the density or density function of \(F_t\) at time \(t\). Similarly \(F_t\) is called regular on a time-interval \(I\) if \(\mathrm{d}F_t(x)=f(x,t)\sqrt{x}\mathrm{d}x\) for all \(t\in I\), i.e. \(\mu _t=0\) for all \(t\in I\). We say that \(F_t\) has condensation at time \(t\) if \(F_t(\{0\})>0\). It is generally believed that if \(F_0\in \mathcal{B}_1^{+}({\mathbb R}_{\ge 0})\) is regular, then there exists a conservative distributional solution \(F_t\) of Eq. (1.5) on \([0,\infty )\) with the initial datum \(F_0\) such that for any \(t\in [0,\infty )\) the singular part \(\mu _t\) (if not zero) will be only a condensate, i.e. \(\mathrm{d}\mu _t(x)=n(t)\delta (x)\mathrm{d}x\), where \(n(t)=F_t(\{0\})\) is the amount of condensation as mentioned above. But this has not been proven yet. The main difficulty is that while smooth test functions \(\varphi \) kill the singularity of the collision kernel \(W(x,y,z)\) so that the collision integrals are convergent, they also suppress the singular behavior of \(F_t\) near the origin and thus the direct capture of condensation becomes difficult. We then have to consider some indirect way: we carefully study regular behavior of \(F_t\) (this part has its own interests) and by assuming \(F_t\) in some cases has certain regularity in a time-interval we hope to find some contradiction so as to obtain some information about condensation.

1.6 Main Results

For convenience of statement, results introduced below are only concerned with conservative solutions. Details and other results will be given in the following sections.

1.6.1 Strong Solutions and Regularity

The definition of strong solutions will be given in Sect. 3. We observe that an important condition that implies the existence of strong solutions and regularity is \(M_{-1/2}(F_0)<\infty \). Recall that if \(F_t\) is regular and \(f(\cdot ,t)\) is its density, then \(M_{-1/2}(F_t)=\int \limits _{{\mathbb R}_{+}}f(x,t)\mathrm{d}x\). In view of the structure of \(Q(f)(x)\) and \(L^{\infty }\)-bounds \(0\le w(x,y,z)\le 1\), the integrability \(f(\cdot ,t)\in L^1({\mathbb R}_{+})\) is easily produced from a closed form of \(L^1\)-inequalities. For the case \(M_{-1/2}(F_0)=\infty \) and \(F_0\) is regular, if (for instance) the density \(f_0\) of \(F_0\) satisfies \(f_0(x){\sim }\mathrm{const}. x^{\alpha -3/2}\,(x\rightarrow 0+)\) with \(0<\alpha <1/2\), then the solution \(F_t\) cannot be regular on any time-interval \(I\subset (0, \infty )\) (see below), but for the critical case \(\alpha =1/2\), for instance if \(f_0(x)\le \mathrm{const}.x^{-1}\,(\forall \, x>0)\), then \(F_t\) can be regular on a time-interval. Main results on strong solutions and regularity are as follows:

  • Let \(F_t\) be a conservative distributional solution on \([0,\infty )\) with the initial datum \(F_0\) satisfying \(M_{-1/2}(F_0)<\infty \). Then there exists the largest \(0<T_{\max }\le \infty \) such that \(M_{-1/2}(F_t)<\infty \) for all \(t\in [0, T_{\max })\), and \(F_t\) is a unique strong solutions of Eq. (1.5) on \([0, T_{\max })\). Furthermore if \(M_{-1/2}(F_0)\) is small enough as compared with \([N(F_0)E(F_0)]^{1/4}\), for instance \(M_{-1/2}(F_0)\le \frac{1}{80} [N(F_0)E(F_0)]^{1/4}\), then \(T_{\max }=\infty \). In particular if \(F_0\) is regular and \(f_0\) is its density, then \(F_t\) is also regular on \([0, T_{\max })\) and its density \(f(\cdot ,t)\) is the unique conservative mild solution of Eq. (1.5) on \([0, T_{\max })\) satisfying \(f\in C([0,T_{\max }); L^1({\mathbb R}_{+}))\) and \(f(\cdot ,0)=f_0\). Finally if in addition \(f_0\in L^{\infty }({\mathbb R}_{+})\), then \(f(\cdot ,t)\in L^{\infty }({\mathbb R}_{+})\) for all \(t\in [0, T_{\max })\) including the case \(T_{\max }=\infty \).

  • Let \(F_0\in \mathcal{B}_1^{+}({\mathbb R}_{\ge 0})\) be regular with density \(f_0\) satisfying \(f_0(x)\le Kx^{-1}\,(\forall \, x>0)\) for some \(K>0\). Then there exist an explicit constant \(T_K>0\) and a conservative distributional solution \(F_t\) of Eq. (1.5) on \([0,\infty )\) with the initial datum \(F_0\) such that \(F_t\) is a regular strong solution on \([0, T_K]\) and its density \(f(x,t)\) satisfies \(f(x,t)\le 5K x^{-1}\) for all \( (x,t)\in {\mathbb R}_{+}\times [0, T_K].\)

As one will see in Sect. 3 that the relative smallness \(M_{-1/2}(F_0)\le \frac{1}{80} [N(F_0)E(F_0)]^{1/4}\) belongs to the case of very high temperature, \(\overline{T}/\overline{T}_c>>1\). For the case of low temperature, \(\overline{T}/\overline{T}_c<1\), we do not know whether the intervals of \(t\) satisfying \(M_{-1/2}(F_t)<\infty \) (or \(f(\cdot ,t)\in L^{1}({\mathbb R}_{+})\)) are only finite, but Escobedo and Velázquez proved in [9, 10] that this is the case for mild solutions with initial data \(f_0\in L^{\infty }({\mathbb R}_{+},(1+x)^{\gamma })(\gamma >3)\) if the blow up is taken in the \(L^{\infty }({\mathbb R}_{+})\) norm, see Theorem 1.2 below for details.

1.6.2 Condensation in Finite Time

The main result is based on the propagation of condensation

$$\begin{aligned} F_t(\{0\})\ge F_s(\{0\})e^{\int \limits _{s}^{t}[L_{\varphi }(F_\tau )-M_{1/2}(F_{\tau })]\mathrm{d}\tau }\qquad \forall \,0\le s<t<\infty \end{aligned}$$
(1.18)

(see [15], where \(t\mapsto L_{\varphi }(F_t)\) is a positive function) and the consideration for Lebesgue derivatives \(\underline{D}_{\alpha }(F_0), \overline{D}_{\alpha }(F_0)\) of the initial data \(F_0\) at the origin \(x=0\), where

$$\begin{aligned} \underline{D}_{\alpha }(F)=\liminf _{\varepsilon \rightarrow 0^+}\frac{F([0,\varepsilon ])}{\varepsilon ^{\alpha }},\quad \overline{D}_{\alpha }(F)=\limsup _{\varepsilon \rightarrow 0^+}\frac{F([0,\varepsilon ])}{\varepsilon ^{\alpha }},\quad \alpha >0 \end{aligned}$$
(1.19)

and in case \(\underline{D}_{\alpha }(F)=\overline{D}_{\alpha }(F)\) we define \(D_{\alpha }(F)=\underline{D}_{\alpha }(F)=\overline{D}_{\alpha }(F)\). Notice that if initially \(F_0(\{0\})>0\), then (1.18) implies \(F_t(\{0\})>0\) for all \(t\ge 0\). Thus the most interesting case for the occurrence of condensation is \(F_0(\{0\})=0\) as satisfied by all regular initial data.

For the case \(0<\alpha <1/2\), we have proved in [15] that if \(\underline{D}_{\alpha }(F_0)>0\), then \(F_t(\{0\})>0\) for all \(t>0\). A typical example is the regular measure \(\mathrm{d}F_0(x)=f_0(x)\sqrt{x}\mathrm{d}x\) whose density \(f_0\) satisfies \(f_0(x)\sim \mathrm{const}. x^{\alpha -3/2}\,(x\rightarrow 0+)\). This example shows also that \(F_t\) is not regular for every \(t>0\).

For the case \(\alpha >1/2\), it is easily seen that the bigness of \(\underline{D}_{\alpha }(F_0)\) does not destroy the smallness of \(M_{-1/2}(F_0)\); we will show in Example 7.1 that there is a large class of regular initial data \(F_0\) satisfying \(D_{\alpha }(F_0)=\infty \) such that \(F_t\) are regular for all \(t\in [0, \infty )\) hence there are no condensation at all.

For the critical case \(\alpha =1/2\), an important referential example is the equilibrium \(F_t\equiv F_\mathrm{be}\) at the critical temperature, i.e., \(F_\mathrm{be}\) is regular with density \(f_\mathrm{be}(x)=(e^{x/\kappa }-1)^{-1}\sim \kappa x^{-1}\,(x\rightarrow 0+)\) hence \(D_{1/2}(F_\mathrm{be})=2\kappa >0\). Of course there is no condensation. However by computing we find \(D_{1/2}(F_\mathrm{be})= 1.403 \ldots [N(F_\mathrm{be})E(F_\mathrm{be})]^{1/4}\) which indicates that \(D_{1/2}(F_\mathrm{be})\) is not large as compared with \([N(F_\mathrm{be})E(F_\mathrm{be})]^{1/4}\). This motives us to establish the following theorem:

  • Let \(N=N(F_0), E=E(F_0)\). If \([\underline{D}_{1/2}(F_0)\overline{D}_{1/2}(F_0)]^{1/2}\) is sufficiently large as compared with \((NE)^{1/4}\), then the condensation still occurs in finite time. For instance if

    $$\begin{aligned}{}[\underline{D}_{1/2}(F_0)\overline{D}_{1/2}(F_0)]^{1/2}> 213(NE)^{1/4} \end{aligned}$$
    (1.20)

    then \(F_t(\{0\})>0\) for all \(t\ge \frac{3}{4}(NE)^{-1/2}.\)

Notice that since \(\underline{D}_{1/2}(F_0),\overline{D}_{1/2}(F_0)\) are determined only by local behavior of \(F_0\) near the origin, they have almost no influence on macroscopic quantities such as mass, energy and temperature; we will give a large class of initial data \(F_0\) which satisfy (1.20) so that condensation occurs in finite time, but the corresponding temperature can be arbitrarily high/low (see Example 7.2).

1.6.3 Critical Time of Condensation

Suppose a distributional solution \(F_t\) has condensation in finite time and let \(t_c=\inf \{ t\ge 0\,|\, F_t(\{0\})>0\}\). It is easily seen from the condensate propagation (1.18) that the time-sets of non-condensation and condensation are separated by the single point \(t_c\), i.e.

$$\begin{aligned} F_t(\{0\})=0\quad \forall \, t\in [0, t_c);\quad F_t(\{0\})>0\quad \forall \, t\in (t_c, \infty ). \end{aligned}$$
(1.21)

We call \(t_c\) the critical time of condensation of \(F_t\). Notice that the case \(t_c=0\) is possible even if \(F_0(\{0\})=0\) as shown above. For that case, (1.21) is understood as just the second inequality appears. From (1.18) one sees also that the function \(t\mapsto F_t(\{0\})\exp \{\int \limits _{0}^{t}M_{1/2}(F_{\tau })\mathrm{d}\tau \}\) is monotone non-decreasing on \([0,\infty )\), which implies that \(t\mapsto F_t(\{0\})\) has both left and right limits at every \(t\in (0,\infty )\). We also proved in [15] that \(t\mapsto F_t(\{0\})\) is at least right-continuous on \([0,\infty )\). In particular it holds \(\lim _{t\rightarrow t_c+}F_t(\{0\})=F_{t_c}(\{0\})\) which means that the phase transition is at least right-continuous, i.e. there is no jump starting from \(t_c\) in the forward direction. Our second result on condensation is about the strict positivity of \(t_c\), which should be studied since in general the condensation of a dilute Bose gas takes place only after a shorter or longer time.

  • If a regular measure \(F_0\in \mathcal{B}_1^{+}({\mathbb R}_{\ge 0})\) satisfies (1.20) (with \(N=N(F_0), E=E(F_0)\)) and its density satisfies \(f_0(x)\le K x^{-1}\,\,(\forall \, x>0)\) for some constant \(K>0\), then there exist an explicit constant \(0<T_K< \frac{3}{4}(NE)^{-1/2}\) and a conservative distributional solution \(F_t\) of Eq. (1.5) on \([0,\infty )\) with the initial datum \(F_0\), such that \(t_c\in [T_K,\, \frac{3}{4}(NE)^{-1/2}]\).

In Example 7.3 we will show that the regular data \(F_0\) in the above result exist extensively.

We emphasize that the inequality (1.20) and similar ones given in Theorem 6.1 are only sufficient conditions for the occurrence of condensation in finite time. For the case where \(F_0\) does not satisfy (1.20), in particular for the case where \(F_0\) is regular and its density \(f_0\) is bounded (which implies \(\overline{D}_{1/2}(F_0)=0\)), one has to consider different methods, see Theorem 1.2 below.

1.7 Recent Progress

Escobedo and Velázquez recently obtained important results on the blow up and condensation in finite time ([9, 10]). We summarize them as follows using our notations and the definition of critical time \(t_c\) in (1.21).

Theorem 1.2

([9, 10])

  1. (I)

    (Local Well Posedness [9]) For any \(\gamma >3\) and any \(0\le f_0\in L^{\infty }({\mathbb R}_{+},(1+x)^{\gamma }):=\{ g\in L^{\infty }({\mathbb R}_{+})\, |\, \mathrm{ess}\sup _{x\in {\mathbb R}_+}|g(x)|(1+x)^{\gamma }<\infty \}\), there exist \(0<T_{\max }\le \infty \) and a unique conservative mild solution \(f\in L^{\infty }_{\mathrm{loc}}([0, T_{\max }); L^{\infty }({\mathbb R}_{+},(1+x)^{\gamma }))\) of Eq. (1.5), with the initial datum \(f_0\). And \(T_{\max }\) is the maximal existence time in the sense that

    $$\begin{aligned} T_{\max }<\infty \quad \Longrightarrow \quad \Vert f(t)\Vert _{L^{\infty }({\mathbb R}_{+})}\rightarrow \infty \quad as\quad t \nearrow T_{\max }. \end{aligned}$$
    (1.22)
  2. (II)

    (For Arbitrary Temperature [9]) There exists a universal constant \(\theta _*>0\) such that for all \(N>0, E>0, \nu >0\), there exist \(\rho _0=\rho _0(N,E,\nu )>0\), \(K^*=K^*(N,E,\nu )>0, T_0= T_0(N,E,\nu )>0\) depending only on \(N,E,\nu \) such that the following properties hold:

  3. (II.1)

    If \(0\le f_0\in L^{\infty }({\mathbb R}_{+},(1+x)^{\gamma })\) with \(\gamma >3\) satisfies \(N(f_0)=N, E(f_0)=E\) and

    $$\begin{aligned} \sup _{0<\rho \le \rho _0}\min \left\{ \inf _{0<\varepsilon \le \rho }\frac{1}{\nu \varepsilon ^{3/2}} \int \limits _{0}^{\varepsilon }f_0(x)\sqrt{x}\mathrm{d}x,\, \frac{1}{K^* \rho ^{\theta _*}}\int \limits _{0}^{\rho }f_0(x)\sqrt{x}\mathrm{d}x\right\} \ge 1, \end{aligned}$$
    (1.23)

    then the maximal existence time \(T_{\max }\) of \(f\) obtained in part (I) must be finite: \(T_{\max }<T_0\), hence \(f\) satisfies the \(L^{\infty }\)- blow up (1.22).

  4. (II.2)

    If the initial datum \(F_0\) of \(F_t\) satisfies \(N(F_0)=N, E(F_0)=E\), and

    $$\begin{aligned} \sup _{0<\rho \le \rho _0}\min \left\{ \inf _{0<\varepsilon \le \rho }\frac{F_0([0,\varepsilon ])}{\nu \varepsilon ^{3/2}}\,,\, \frac{F_0([0,\rho ])}{K^* \rho ^{\theta _*}}\right\} \ge 1, \end{aligned}$$
    (1.24)

    then \(F_t\) has condensation in finite time and \(t_c\in [0, T_0]\). Furthermore if in addition \(F_0\) is regular with density \(f_0\in L^{\infty }({\mathbb R}_{+},(1+x)^{\gamma }) (\gamma >3)\), then \(t_c\in [T_{\max }, T_0]\).   [Note: it is pointed out in [9] that the first condition in () does not imply the second since \(\theta _*\) might be small.]

  5. (III)

    (For Low Temperature [10]). Let \(N,E\) be the mass and energy of the conservative solutions appeared below and let \(\overline{T},\overline{T}_c\) be the kinetic temperature and critical temperature defined in (1.14). Then under only the low temperature condition \(\overline{T}/\overline{T}_c<1\), if \(t_0\) in the time-interval of existence is large enough, then the conditions (1.23), (1.24) are satisfied for \(\widetilde{f}_0(x)=f(x, t_0)\) and \(\widetilde{F}_0=F_{t_0}\) respectively and thus one has:

  6. (III.1)

    If \(f\) obtained in part (I) satisfies \(\overline{T}/\overline{T}_c<1\), then \(T_{\max }<\infty \) and \(f\) satisfies the \(L^{\infty }\)-blow up (1.22).

  7. (III.2)

    If \(F_0\) satisfies \(\overline{T}/\overline{T}_c<1\), then \(F_t\) has condensation in finite time and there exists \(T_0=T_0(N,E)>0\) depending only on \(N,E\) such that \(t_c\in [0, T_0]\). Furthermore if \(F_0\) is also regular with density \(f_0\in L^{\infty }({\mathbb R}_{+},(1+x)^{\gamma }))\,(\gamma >3)\), then \(t_c\in [T_{\max }, T_0].\)

Part (III.2) of the theorem is a fundamental result in low temperature kinetic theory; it demonstrates that the occurrence of condensation in finite time does not depend on any local information of a conservative solution if the temperature is low enough. An application of part (III.2) will be given in Example 7.4 for dealing with a class of initial data \(F_0\) which do not satisfy (1.20). Part (II.2) of Theorem 1.2 is the key result and is more closer to practical cases since it includes certain regular measures whose densities are bounded. Initial data satisfying all conditions of part (II.1) hence part (II.2) have been constructed in [9]. As for comparison between part (II.2) and our result, the main difference is that our condition (1.20) is purely local (near the origin) while () is not; and there is no implication relation between (1.20) and () except the subtle case \(\theta _*=1/2\). In fact, on the one hand, it is obvious that the condition () dose not imply (1.20) because the latter excludes all such regular measures whose densities are bounded near the origin. On the other hand, we will show in Example 7.5 that \(\theta _*\le 1/2\), and if \(\theta _*<1/2\), then the condition (1.20) does not imply (); if \(\theta _*=1/2\) and \(K^*>214(NE)^{1/4}\), the non-implication still holds. We stress however that both part (II.2) and our results on condensation in finite time are proved for arbitrary temperature, which may be helpful for the study of Bose–Einstein condensation at room temperature. Finally we note as pointed out at the end of Remark 2.2 of [15] that the practical lifetime of condensation \(F_t(\{0\})>0\) is “finite” since \(\lim _{t\rightarrow \infty }F_t(\{0\})=0\) for high temperature \(\overline{T}/\overline{T}_c\ge 1\), but it remains unclear whether \(\liminf _{t\rightarrow \infty }F_t(\{0\})>0\) holds true for low temperature \(\overline{T}/\overline{T}_c<1\).

The rest of the paper is organized as follows. In Sect. 2 we collect and prove some basic properties of collision integrals and propose suitable approximate solutions which will be used here to prove some special properties of solutions. In Sect. 3 we prove the local and global existence of strong solutions and establish stability estimates. In Sect. 4 we prove the regularity of distributional solutions and the global existence of mild solutions. Section 5 is devoted to the proof of regularity of certain distributional solutions whose initial data are regular measures with densities satisfying \(f_0(x)\le Kx^{-1}\). In Sect. 6 we study the condensation in finite time. By observing a necessary condition on the initial datum for the absence of condensation at some finite time, we propose some sufficient conditions as (1.20) for the occurrence of condensation in finite time, and this together with the result of Sect. 5 enables us to prove the existence and the strict positivity of the critical time \(t_c\) for a class of solutions. The examples mentioned above are given in Sect. 7.

Throughout this paper, \({\mathbb R}_{+}\) stands for either \({\mathbb R}_{\ge 0}:=[0,\infty )\) or \({\mathbb R}_{> 0}:=(0,\infty )\) and they are only used to denote domains of \(x\) in order to distinguish intervals of \(t\).

2 Basic Properties of Collision Integrals and Approximate Solutions

First of all we note that the space \(\mathcal{B}({\mathbb R}_{\ge 0}^n)\) of finite Borel measures on \({\mathbb R}_{\ge 0}^n\) can be treated as a subspace of \(\mathcal{B}({\mathbb R}^n)\) by zero-extension: \(\bar{\mu }(E)=\mu ({\mathbb R}_{\ge 0}^n\cap E)\) for all Borel sets \(E\subset {\mathbb R}^n\). Thus many properties that hold for \(\mathcal{B}({\mathbb R}^n)\) also hold for \(\mathcal{B}({\mathbb R}_{\ge 0}^n)\).

Basic properties of collision integrals of measures proved below will be used in Sect. 3 to study strong solutions. Roughly speaking, “strong” means that the smooth test functions can be replaced by bounded Borel functions, i.e., there is no need of the cancelation effect resulted from the symmetric difference \(\Delta \varphi (x,y,z)\) of smooth functions (see also (3.6) below). Thus we have to establish integrability conditions with respect to each single term in the decompositions

$$\begin{aligned} \mathcal{J}[\varphi ]=\mathcal{J}^{+}[\varphi ]- \mathcal{J}^{-}[\varphi ],\quad \mathcal{K}[\varphi ]=\mathcal{K}^{+}[\varphi ]- \mathcal{K}^{-}[\varphi ], \end{aligned}$$

where

$$\begin{aligned} \mathcal{J}^{+}[\varphi ](y,z)&= \frac{1}{2}\int \limits _{0}^{y+z}\mathcal{K}^{+}[\varphi ](x, y,z) \sqrt{x}\mathrm{d}x,\quad \mathcal{J}^{-}[\varphi ](y,z)=\frac{\varphi (y)+\varphi (z)}{2}\mathcal{J}^{+}[1](y,z),\quad \\ \mathcal{K}^{+}[\varphi ](x, y,z)&= W(x,y,z)[\varphi (x)+\varphi (x_*)],\quad \mathcal{K}^{-}[\varphi ](x, y,z)=W(x,y,z)[\varphi (y)+\varphi (z)]\quad \end{aligned}$$

and \(\varphi \) is a locally bounded Borel function on \({\mathbb R}_{\ge 0}\). To do this we consider measure spaces

$$\begin{aligned} \mathcal{B}_{p,1}({\mathbb R}_{\ge 0})= \{ F\in \mathcal{B}_{1}({\mathbb R}_{\ge 0})\,|\,\, M_{p}(|F|)<\infty \},\quad \mathcal{B}^{+}_{p,1}({\mathbb R}_{\ge 0})=\mathcal{B}_{p,1}({\mathbb R}_{\ge 0})\cap \mathcal{B}^{+}({\mathbb R}_{\ge 0}). \end{aligned}$$

It is easily seen that

$$\begin{aligned} p<q<0\quad \Longrightarrow \quad \mathcal{B}_{p,1}({\mathbb R}_{\ge 0}) \subset \mathcal{B}_{q,1}({\mathbb R}_{\ge 0}),\quad \mathcal{B}^{+}_{p,1}({\mathbb R}_{\ge 0})\subset \mathcal{B}^{+}_{q,1}({\mathbb R}_{\ge 0}). \end{aligned}$$

Let us define

$$\begin{aligned} M_{p,q}(|F|)=M_{p}(|F|)+M_{q}(|F|),\quad -\infty <p,q<\infty . \end{aligned}$$

As usual the notation \( F_1\otimes F_2\otimes \cdots \otimes F_n\) stands for the product measure of \(F_1,F_2,\ldots ,F_n\).

In this paper we denote by \(L^{\infty }_0({\mathbb R}_{\ge 0})\) the set of bounded Borel functions on \({\mathbb R}_{\ge 0}\). In general, for any \(k\ge 0\), we denote by \(L^{\infty }_{-k}({\mathbb R}_{\ge 0})\) the set of Borel functions \(\varphi \) on \({\mathbb R}_{\ge 0}\) satisfying

$$\begin{aligned} \Vert \varphi \Vert _{L^{\infty }_{-k}}:=\sup _{x\in {\mathbb R}_{\ge 0}}|\varphi (x)|(1+x)^{-k}<\infty . \end{aligned}$$

Lemma 2.1

  1. (a)

    Let \(F,G\in \mathcal{B}_{k+1/2}({\mathbb R}_{\ge 0})\) with \(k\in [0, 1]\). Then for any \(\varphi \in L^{\infty }_{-k}({\mathbb R}_{\ge 0})\) satisfying \(\Vert \varphi \Vert _{L^{\infty }_{-k}}\le 1\) we have

    $$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}^2}|\mathcal{J}^{\pm }[\varphi ]|\mathrm{d}(|F|\otimes |G|)\le \Vert F\Vert _{k+1/2}\Vert G\Vert _{k+1/2}. \end{aligned}$$
    (2.1)
  2. (b)

    Let \(F,G, H\in \mathcal{B}_{-1/3,1}({\mathbb R}_{\ge 0})\). Then for any \(\varphi \in L^{\infty }_0({\mathbb R}_{\ge 0})\) satisfying \(\Vert \varphi \Vert _{L^{\infty }_0}\le 1\) we have

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}^3}|\mathcal{K}^{\pm }[\varphi ]|\mathrm{d}(|F|\otimes |G|\otimes |H|) \le 2M_{-1/3}(|F|)M_{-1/3}(|G|)M_{-1/3}(|H|). \end{aligned}$$
(2.2)

In general, for any \(\varphi \in L^{\infty }_{-k}({\mathbb R}_{\ge 0})\) satisfying \(\Vert \varphi \Vert _{L^{\infty }_{-k}}\le 1\) with \(k\in [0,1]\), we have

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}^3}|\mathcal{K}^{\pm }[\varphi ]|\mathrm{d}(|F|\otimes |G|\otimes |H|) \le 2M_{-1/3}(|F|)M_{-1/3, k-1/3}(|G|)M_{-1/3,k-1/3}(|H|)\nonumber \\ \end{aligned}$$
(2.3)

and if assuming further that \(F,G, H\in \mathcal{B}_{-1/2,1}({\mathbb R}_{\ge 0})\), then

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}^3}|\mathcal{K}^{\pm }[\varphi ]|\mathrm{d}(|F|\otimes |G|\otimes |H|) \le 2a(F,G,H)\min \{\Vert F\Vert _k, \Vert G\Vert _k,\Vert H\Vert _k\} \end{aligned}$$
(2.4)

where

$$\begin{aligned} a(F,G,H):= [ M_{-1/2, 1/2}(|F|)+ M_{-1/2,1/2}(|G|)+M_{-1/2,1/2}(|H|)]^2. \end{aligned}$$
(2.5)

Proof

By definition of \(W(x,y,z)\) we compute

$$\begin{aligned} \mathcal{J}^{+}[1](y,z)=\int \limits _{0}^{y+z}W(x,y,z)\sqrt{x}\mathrm{d}x=\frac{y\wedge z}{3\sqrt{y\vee z}} \mathbf{1}_{\{y\vee z>0\}}+ \sqrt{y\vee z}. \end{aligned}$$
(2.6)

Here \(a\wedge b=\min \{a,b\}, a\vee b=\max \{a,b\}\); \(\mathbf{1}_{\{a>b\}}=1\) if \(a>b\); \(\mathbf{1}_{\{a>b\}}=0\) if \(a\le b\). From (2.6) we have

$$\begin{aligned} \mathcal{J}^{+}[1](y,z)\le \sqrt{y}+\sqrt{z}\le \sqrt{1+y}\sqrt{1+z}. \end{aligned}$$

In general, using \(x+x_*=y+z\) for \(W(x,y,z)>0\) and using \(1+y+z\le (1+y)(1+z)\) we see that if \(k\in [0,1]\) and \(|\varphi (x)|\le (1+x)^{k}\) (\(\forall \, x\in {\mathbb R}_{\ge 0}\)), then

$$\begin{aligned} |\mathcal{J}^{\pm }[\varphi ](y,z)|\le (1+y)^{k+1/2}(1+z)^{k+1/2}, \end{aligned}$$
(2.7)
$$\begin{aligned} |\mathcal{K}^{\pm }[\varphi ](x, y,z)|\le 2(1+y^k+z^k)W(x,y,z). \end{aligned}$$
(2.8)

The inequality (2.1) follows from (2.7).

Next we note that \(F,G, H\in \mathcal{B}_{-1/3,1}({\mathbb R}_{\ge 0})\) implies \((|F|\otimes |G|\otimes |H|)({\mathbb R}_{\ge 0}^3{\setminus } {\mathbb R}_{>0}^3)=0\) so that \(\int _{{\mathbb R}_{\ge 0}^3}{\{\cdots \}}\mathrm{d}(|F|\otimes |G|\otimes |H|)= \int _{{\mathbb R}_{>0}^3}\{\cdots \}\mathrm{d}(|F|\otimes |G|\otimes |H|).\) And by definition of \(W(x,y,z)\) we have

$$\begin{aligned} W(x, y,z) \le \min \Bigg \{\frac{1}{(xyz)^{1/3}}\,,\, \frac{1}{\sqrt{yz}}\,,\,\frac{1}{\sqrt{xz}}\,,\,\frac{1}{\sqrt{xy}} \Bigg \}\qquad \forall \, (x, y,z)\in {\mathbb R}_{>0}^3. \end{aligned}$$
(2.9)

From this and elementary calculations we deduce

$$\begin{aligned}&\int \limits _{{\mathbb R}_{\ge 0}^3}W(x,y,z)\mathrm{d}(|F|\otimes |G|\otimes |H|) \le M_{-1/3}(|F|)M_{-1/3}(|G|)M_{-1/3}(|H|), \end{aligned}$$
(2.10)
$$\begin{aligned}&\int \limits _{{\mathbb R}_{\ge 0}^3}(1\!+\!y^k\!+\!z^k)W\mathrm{d}(|F|\otimes |G|\otimes |H|) \!\le \! M_{-1/3}(|F|)M_{-1/3,k-1/3}(|G|)M_{-1/3, k-1/3}(|H|),\nonumber \\ \end{aligned}$$
(2.11)

and, assuming further that \(F,G, H\in \mathcal{B}_{-1/2,1}({\mathbb R}_{\ge 0})\),

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}^3}(1+y^k+z^k)W\mathrm{d}(|F|\otimes |G|\otimes |H|) \le a(F,G,H)\min \{\Vert F\Vert _k, \Vert G\Vert _k,\Vert H\Vert _k\}.\nonumber \\ \end{aligned}$$
(2.12)

Thus (2.2) follows from (2.10), while (2.3), (2.4) follow from (2.8) and (2.11)–(2.12). \(\square \)

According to Lemma 2.1 we can define Borel measures \(\mathcal{Q}_2^{\pm }(F,G)\in \mathcal{B}_k({\mathbb R}_{\ge 0})\) for \(F,G\in \mathcal{B}_{k+1/2}({\mathbb R}_{\ge 0})\,(k\in [0,1])\) and \(\mathcal{Q}_3^{\pm }(F,G,H)\in \mathcal{B}_1({\mathbb R}_{\ge 0})\) for \(F,G, H\in \mathcal{B}_{-1/3, 1}({\mathbb R}_{\ge 0})\) through Riesz representation theorem by

$$\begin{aligned}&\int \limits _{{\mathbb R}_{\ge 0}}\varphi (x)\mathrm{d} \mathcal{Q}_2^{\pm }(F,G)(x) =\int \limits _{{\mathbb R}_{\ge 0}^2}\mathcal{J}^{\pm }[\varphi ]\mathrm{d}(F\otimes G),\end{aligned}$$
(2.13)
$$\begin{aligned}&\int \limits _{{\mathbb R}_{\ge 0}}\varphi (x)\mathrm{d} \mathcal{Q}_3^{\pm }(F,G, H)(x) =\int \limits _{{\mathbb R}_{\ge 0}^3}\mathcal{K}^{\pm }[\varphi ]\mathrm{d}(F\otimes G\otimes H) \end{aligned}$$
(2.14)

for all \(\varphi \in C_b({\mathbb R}_{\ge 0})\). It is obvious that \((F,G)\mapsto \mathcal{Q}_2^{\pm }(F,G)\) and \((F,G, H)\mapsto \mathcal{Q}_3^{\pm }(F,G, H)\) are bounded bilinear and trilinear operators from \([\mathcal{B}_{k+1/2}({\mathbb R}_{\ge 0})]^2 \) to \(\mathcal{B}_{k}({\mathbb R}_{\ge 0})\) (\(k\in [0,1]\)) and from \( [\mathcal{B}_{-1/3, 1}({\mathbb R}_{\ge 0})]^3\) to \(\mathcal{B}_{1}({\mathbb R}_{\ge 0})\) respectively, and

$$\begin{aligned} \Vert \mathcal{Q}_2^{\pm }(F,G)\Vert _k&\le \Vert F\Vert _{k+1/2}\Vert G\Vert _{k+1/2}, \end{aligned}$$
(2.15)
$$\begin{aligned} \Vert \mathcal{Q}_3^{\pm }(F,G, H)\Vert _0&\le 2M_{-1/3}(|F|)M_{-1/3}(|G|)M_{-1/3}(|H|), \end{aligned}$$
(2.16)
$$\begin{aligned} \Vert \mathcal{Q}_3^{\pm }(F,G, H)\Vert _k&\le 2M_{-1/3}(|F|)M_{-1/3, k-1/3}(|G|)M_{-1/3,k-1/3}(|H|), \end{aligned}$$
(2.17)
$$\begin{aligned} \Vert \mathcal{Q}_3^{\pm }(F,G, H)\Vert _k&\le 2a(F,G,H) \min \{\Vert F\Vert _k, \Vert G\Vert _k,\Vert H\Vert _k\}. \end{aligned}$$
(2.18)

Here in the third inequality (2.18) we assume further that \(F,G, H\in \mathcal{B}_{-1/2,1}({\mathbb R}_{\ge 0})\) so that \(a(F,G,H)<\infty \). Notice that the equalities (2.13), (2.14) hold also for all \(\varphi \in L^{\infty }_0({\mathbb R}_{\ge 0})\).

In connecting with the equation Eq. (1.8) we define

$$\begin{aligned} \mathcal{Q}_2^{\pm }(F)&= \mathcal{Q}_2^{\pm }(F,F),\quad \mathcal{Q}_2(F)=\mathcal{Q}_2^{+}(F)-\mathcal{Q}_2^{-}(F),\\ \mathcal{Q}_3^{\pm }(F)&= \mathcal{Q}_3^{\pm }(F, F, F),\quad \mathcal{Q}_3(F)= \mathcal{Q}_3^{+}(F)-\mathcal{Q}_3^{-}(F),\\ \mathcal{Q}(F)&= \mathcal{Q}_2(F)+\mathcal{Q}_3(F). \end{aligned}$$

We then deduce from

$$\begin{aligned} F\otimes F-G\otimes G&= \frac{1}{2}(F-G)\otimes (F+G)+\frac{1}{2} (F+G)\otimes (F-G),\\ \mathcal{Q}_2^{\pm }(F)-\mathcal{Q}_2^{\pm }(G)&= \frac{1}{2}\mathcal{Q}_2^{\pm }(F-G,F+G) +\frac{1}{2}\mathcal{Q}_2^{\pm }(F+G,F-G), \end{aligned}$$

and (2.15) that for all \(F,G\in \mathcal{B}_{k+1/2}({\mathbb R}_{\ge 0})\) (with \(k\in [0,1]\))

$$\begin{aligned} \Vert \mathcal{Q}_2^{\pm }(F)-\mathcal{Q}_2^{\pm }(G)\Vert _k\le \Vert F+G\Vert _{k+1/2}\Vert F-G\Vert _{k+1/2}. \end{aligned}$$
(2.19)

Similarly we deduce from

$$\begin{aligned}&F\otimes F\otimes F-G\otimes G\otimes G =(F-G)\otimes F\otimes F +G\otimes (F-G)\otimes F +G\otimes G\otimes (F-G),\\&\Vert \mathcal{Q}_3^{\pm }(F)-\mathcal{Q}_3^{\pm }(G)\Vert _k \le \Vert \mathcal{Q}_3^{\pm }(F-G, F,F)\Vert _k+\Vert \mathcal{Q}_3^{\pm }(G, F-G,F)\Vert _k+\Vert \mathcal{Q}_3^{\pm }(G, G,F-G)\Vert _k \end{aligned}$$

and (2.16), (2.18) that

$$\begin{aligned}&\Vert \mathcal{Q}_3^{\pm }(F)-\mathcal{Q}_3^{\pm }(G)\Vert _0\le 2[M_{-1/3}(|F|)+M_{-1/3}(|G|)]^2M_{-1/3}(|F-G|) ,\end{aligned}$$
(2.20)
$$\begin{aligned}&\Vert \mathcal{Q}_3^{\pm }(F)-\mathcal{Q}_3^{\pm }(G)\Vert _k\le b(F,G)\Vert F-G\Vert _k,\quad k\in [0, 1] \end{aligned}$$
(2.21)

where for the inequality (2.21) we assume that \(F,G\in \mathcal{B}_{-1/2,1}^{+}({\mathbb R}_{\ge 0})\) so that

$$\begin{aligned} b(F,G):=36[M_{-1/2, 1/2}(|F|) +M_{-1/2,1/2}(|G|)]^2<\infty . \end{aligned}$$
(2.22)

Now we turn to the original form of collision integrals, say the case of functions. The following lemma collects some estimates for the study of mild solutions.

Lemma 2.2

Let \(w(x,y,z)\) be given by (1.3) and let \(f\) be a nonnegative measurable function on \({\mathbb R}_{+}\). Then for any \(x>0, y\ge 0, z\ge 0\) we have

$$\begin{aligned} \int \limits _{{\mathbb R}_{+}}w(x,y,z)\mathrm{d}x\le 2\sqrt{yz}, \end{aligned}$$
(2.23)
$$\begin{aligned} \int \limits _{{\mathbb R}_{+}}w(x,y,z) f(x_*) \left\{ \begin{array}{ll} \mathrm{d}x\\ \mathrm{d}y\\ \mathrm{d}z \end{array}\right\} \le M_{-1/2}(f), \end{aligned}$$
(2.24)
$$\begin{aligned} M_{1/2}(f)\le \int \limits _{{\mathbb R}_{+}^2}w(x,y,z) f(x_*)\mathrm{d}y\mathrm{d}z \le \sqrt{x} N(f)+M_{1/2}(f). \end{aligned}$$
(2.25)

Consequently (with the notation (1.7))

$$\begin{aligned} \int \limits _{{\mathbb R}_{+}^3}w(x,y,z)[ f'f_*'(1+f+f_*)+ ff_*(1+f'+f_*')]\mathrm{d}x\mathrm{d}y\mathrm{d}z&\le 4 M_{1/2}(f)M_{-1/2}(f) \nonumber \\&+\,4[M_{-1/2}(f)]^3 \end{aligned}$$
(2.26)

Proof

(2.23) is obvious by definition of \(w(x,y,z)\) and \(x_*=(y+z-x)_{+}\). (2.24) follows from the reflection and translation for the variable \(y+z-x\) with respect to \(x, y, z\) respectively. Similarly, to prove (2.25) we use translation to get

$$\begin{aligned} \int \limits _{{\mathbb R}_{+}^2}w(x,y,z)f(x_*)\mathrm{d}y\mathrm{d}z=\int \limits _{{\mathbb R}_{+}}I(x,y) f(y)\mathrm{d}y \end{aligned}$$

and the inner integral \(I(x,y)\) is calculated, with \(z_*=(x+y-z)_{+}\),

$$\begin{aligned} I(x,y)=\int \limits _{0}^{x+y}\frac{\min \{\sqrt{x},\sqrt{y},\sqrt{z_*}, \sqrt{z}\,\}}{\sqrt{x}}\mathrm{d}z =\frac{1}{\sqrt{x}}\Big (\frac{1}{3}(x\wedge y)^{3/2}+\sqrt{x\wedge y}(x \vee y)\Big ) \end{aligned}$$

from which we have \(y\le I(x,y)\le \sqrt{xy}+y\). This gives the lower and upper bound estimates in (2.25). The inequality (2.26) follows from (2.23)-(2.25), \(0\le w(x,y,z)\le 1\), and \([N(f)]^2\le M_{1/2}(f)M_{-1/2}(f)\). By the way, the lemma can be also proved by using classical argument: for instance using (1.4) and \(x_*=y+z-x\), \(|\mathbf{v}'|^2+|\mathbf{v}_*'|^2-|\mathbf{v}|^2=|\mathbf{v}_*|^2\) with \(\frac{1}{2}|\mathbf{v}|^2=x\) we compute

$$\begin{aligned}&\int \limits _{{\mathbb R}_{+}^2}w(x,y,z)f(x_*)\mathrm{d}y\mathrm{d}z =\frac{1}{(4\pi )^2}\int \limits _{{\mathbb R}^3\times {\mathbb S}^2} |\langle \mathbf{v}-\mathbf{v}_*, \omega \rangle |f(|\mathbf{v}_*|^2/2) \mathrm{d}\omega \mathrm{d}\mathbf{v}_*\\&=\frac{2\pi }{(4\pi )^2} \int \limits _{{\mathbb R}^3}\frac{ |\mathbf{v}_*-\mathbf{v}|+|\mathbf{v}_*+\mathbf{v}| }{2} f(|\mathbf{v}_*|^2/2)\mathrm{d}\mathbf{v}_* \ge \frac{2\pi }{(4\pi )^2} \int \limits _{{\mathbb R}^3}|\mathbf{v}_*|f(|\mathbf{v}_*|^2/2) \mathrm{d}\mathbf{v}_* \end{aligned}$$

which is equal to \(M_{1/2}(f)\). \(\square \)

We next observe a new positivity of the collision integrals, see Lemma 2.3 below, which enables us to establish Theorem 5.1. In order to prove Theorem 5.1 we have to use suitable approximate solutions \(\{f^n\}_{n=1}^{\infty }\). These are mild solutions of approximate equations of Eq. (1.5) defined by replacing \(Q(f)\) with \(Q_n(f)\), i.e., for each \(n\ge 1\),

$$\begin{aligned}&f^n(x,t)=f_0(x) +\int \limits _{0}^{t}Q_n(f^n)(x,\tau )\mathrm{d}\tau ,\end{aligned}$$
(2.27)
$$\begin{aligned}&Q_n(f)(x)=\int \limits _{{\mathbb R}_{+}^2}w_n(x,y,z)[ f'f_*'(1+f+f_*))- ff_*(1+f'+f_*')]\mathrm{d}y\mathrm{d}z \end{aligned}$$
(2.28)

where \(w_n(x,y,z)\) is given by \(\sqrt{x}w_n(x,y,z)=S_n(x,x_*; y,z)\) and \(S_n(\cdot , \cdot \,;\,\cdot ,\cdot )\) is a well-constructed cutoff for the original function \(\sqrt{x}w(x,y,z)=\min \{\sqrt{x}, \sqrt{x_*}, \sqrt{y}, \sqrt{z}\}\), satisfying the following properties (2.33)-(2.39). Two types of such cutoffs and the corresponding cutoffs \(W_n(x,y,z)\) of \(W(x,y,z)\) can be chosen as follows:

$$\begin{aligned}&S_n(x,x_*;y,z)=\min \{\sqrt{x}, \sqrt{x_*}, \sqrt{y}, \sqrt{z}\}\min \Big \{1, n\max \{x\wedge x_*, y\wedge z\}\Big \},\end{aligned}$$
(2.29)
$$\begin{aligned}&W_n(x,y,z)=W(x,y,z)\min \Big \{1, n\max \{x\wedge x_*, y\wedge z\}\Big \}; \end{aligned}$$
(2.30)
$$\begin{aligned}&S_n(x,x_*;y,z)=\min \{\sqrt{x}, \sqrt{x_*}, \sqrt{y}, \sqrt{z}\}\min \Big \{1, n\min \{x, x_*, y, z\}\Big \}, \end{aligned}$$
(2.31)
$$\begin{aligned}&W_n(x,y,z)=W(x,y,z)\min \Big \{1, n\min \{x, x_*, y, z\}\Big \}. . \end{aligned}$$
(2.32)

It is easily verified that both cutoffs (2.29)–(2.30) and (2.31)–(2.32) satisfy

$$\begin{aligned}&S_n(x,x_*;y,z)\le n\min \Big \{ \sqrt{xyz},\, \sqrt{x_*yz},\, \sqrt{xx_*y},\, \sqrt{xx_*z} \Big \},\quad (x,y,z)\in {\mathbb R}_{\ge 0}^3\end{aligned}$$
(2.33)
$$\begin{aligned}&0\le W(x,y,z)-W_n(x,y,z)\le \Big (\frac{1}{n}\Big )^{\frac{1}{2}-\alpha }\frac{(\min \{x, x_*,y,z\})^{\alpha }}{\sqrt{xyz}},\quad \, (x,y,z)\in {\mathbb R}_{>0}^3\nonumber \\ \end{aligned}$$
(2.34)

(for \(0\le \alpha <1/2\)) and preserve the collisional symmetries on \({\mathbb R}_{\ge 0}^4\):

$$\begin{aligned}&S(\varepsilon ,\varepsilon _*; \varepsilon ',\varepsilon _*')=S(\varepsilon _*,\varepsilon ; \varepsilon ',\varepsilon _*')=S(\varepsilon ',\varepsilon _*'; \varepsilon _*,\varepsilon )=S(\varepsilon ',\varepsilon _*'; \varepsilon ,\varepsilon _*),\end{aligned}$$
(2.35)
$$\begin{aligned}&S(\varepsilon ,\varepsilon _*; \varepsilon ',\varepsilon _*')=0\quad \mathrm{if}\quad (\varepsilon ,\varepsilon _*, \varepsilon ',\varepsilon _*')\in {\mathbb R}_{\ge 0}^4{\setminus } {\mathbb R}_{>0}^4. \end{aligned}$$
(2.36)

From (2.33) we have for all \(0\le f,g, h\in L^1({\mathbb R}_{+}, (1+x)\sqrt{x}\mathrm{d}x)\)

$$\begin{aligned}&\int \limits _{{\mathbb R}_{+}^3}S_n(x,x_*;y,z)\left\{ \begin{array}{ll} f(x)g(y)h(z)\\ f(x_*)g(y)h(z) \end{array}\right\} \mathrm{d}x\mathrm{d}y\mathrm{d}z \le n N(f)N(g)N(h),\end{aligned}$$
(2.37)
$$\begin{aligned}&\int \limits _{{\mathbb R}_{+}^3} x S_n(x,x_*;y,z) f(x)g(y)h(z) \mathrm{d}x\mathrm{d}y\mathrm{d}z \le n E(f)N(g)N(h) \end{aligned}$$
(2.38)
$$\begin{aligned}&\int \limits _{{\mathbb R}_{+}^3}xS_n(x,x_*;y,z) f(x_*)g(y)h(z) \mathrm{d}x\mathrm{d}y\mathrm{d}z \le n N(f)[E(g)N(h)+N(g)E(h)] \nonumber \\ \end{aligned}$$
(2.39)

where the last inequality is due to \(x\le y+z\) when \(S_n(x,x_*;y,z)\ne 0\).

In order to prove the existence of conservative mild solutions \(f^n\), one also needs a further cutoff, for instance \(S_{n,k}(x,x_*; y,z)\!=\!S_n(x,x_*;y,z)(1+\frac{1}{k}\max \{ \sqrt{x}, \sqrt{x_*}, \sqrt{y}, \sqrt{z}\})^{-1}\,(k\!\ge \! 1)\), which is only used for dealing with the quadratic terms in (2.28) so as to prove the finiteness and conservation of energy. Let \(w_{n,k}(x,y,z)\) be defined by \(\sqrt{x}w_{n,k}(x,y,z) =S_{n,k}(x,x_*; y,z)\) and let \(Q_{n,k}(f)\) correspond to the kernel \(w_{n,k}(x,y,z)\). Then for every \(n,k\ge 1\), using the same argument as in [12] it is easily proved that the equation \(\frac{\mathrm{d}}{\mathrm{d}t}f=Q_{n,k}(f)\) with the initial datum \(f_0\) has a conservative mild solution \(f^{n,k}(x,t)\) on \({\mathbb R}_{+}\times [0,\infty )\). Then, for every fixed \(n\ge 1\), following the same argument in [12] (proving the \(L^1\)-weak compactness of the sequence \(\{f^{n,k}(x, t)\sqrt{x}\}_{k=1}^{\infty }\), etc.) we obtain a conservative mild solution \(f^{n}(x,t)\) to the equation \(\frac{\mathrm{d}}{\mathrm{d}t}f=Q_{n}(f)\) on \({\mathbb R}_{+}\times [0,\infty )\) with the initial datum \(f_0\). Another method for proving the existence of \(f^n\) is to go back to the vector version \(\mathbf{v}\mapsto f^n(|\mathbf{v}|^2/2,t)\) and use the first or the second cutoffs:

\(B_n(\mathbf{v}, \mathbf{v}_*,\omega )=(4\pi )^{-2}|(\mathbf{v}-\mathbf{v}_*)\cdot \omega |\min \{1, n\max \{(|\mathbf{v}|^2/2)\wedge (|\mathbf{v}_*|^2/2), (|\mathbf{v}'|^2/2)\wedge (|\mathbf{v}_*'|^2/2)\}\}\) or \(B_n(\mathbf{v}, \mathbf{v}_*,\omega )=(4\pi )^{-2}|(\mathbf{v}-\mathbf{v}_*)\cdot \omega | \min \{1, n\min \{|\mathbf{v}|^2/2,\, |\mathbf{v}_*|^2/2,\, |\mathbf{v}'|^2/2, |\mathbf{v}_*'|^2/2\}\}\). Thanks to the identities (1.2),(1.4), and the inequalities (2.37)–(2.39), the cutoff kernels \(B_n(\mathbf{v}, \mathbf{v}_*,\omega )\) have all main properties presented in [12], and thus by just checking the proofs in [12], one obtains a conservative mild isotropic solution \(f^n(|\mathbf{v}|^2/2,t)\) on \({{\mathbb {R}}^{3}}\times [0,\infty )\) with the initial datum \(f_0(|\mathbf{v}|^2/2)\).

Now let \(\mathcal{J}_n[\varphi ], \mathcal{K}_n[\varphi ], \mathcal{J}_n^{\pm }[\varphi ], \mathcal{K}_n^{\pm }[\varphi ]\) be defined as \(\mathcal{J}[\varphi ], \mathcal{K}[\varphi ], \mathcal{J}^{\pm }[\varphi ], \mathcal{K}^{\pm }[\varphi ]\) by replacing \(W(x,y,z)\) with \(W_n(x,y,z)\). Then from (2.37)–(2.39) and the collisional symmetries (2.35)–(2.36) we see that all integrals appeared below are absolutely convergent and thus the weak form (1.9) for the approximate mild solutions \(f^n\) is rigorously written (denoting \(\mathrm{d}F^n_t(x)=f^n(x,t)\sqrt{x}\mathrm{d}x\) )

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}}\varphi \mathrm{d}F^n_t=\int \limits _{{\mathbb R}_{\ge 0}}\varphi \mathrm{d}F_0+\int \limits _{0}^{t}\mathrm{d}\tau \int \limits _{{\mathbb R}_{\ge 0}^2}\mathcal{J}_n[\varphi ]\mathrm{d}^2F^n_{\tau } +\int \limits _{0}^{t}\mathrm{d}\tau \int \limits _{{\mathbb R}_{\ge 0}^3}\mathcal{K}_n[\varphi ]\mathrm{d}^3 \mathrm{d}F^n_{\tau } \end{aligned}$$
(2.40)

for all \(t\ge 0\) and all \(\varphi \in L^{\infty }_0({\mathbb R}_{\ge 0})\). It should be noted that it is the first cutoff (2.29)–(2.30) that can be used to prove the existence of distributional solutions for general initial data. This is because the first cutoff (2.29)–(2.30) keeps the possibility of condensation at origin and insures the pointwise convergence on the whole \({\mathbb R}_{\ge 0}^3\):

$$\begin{aligned} \lim _{n\rightarrow \infty }| W(x,y,z)-W_n(x,y,z)|=0\qquad \forall \, (x,y,z)\in {\mathbb R}_{\ge 0}^3. \end{aligned}$$
(2.41)

And in fact we also have, for all \(\varphi \in C^{1,1}_b({\mathbb R}_{\ge 0})\) (denoting \(D^2\varphi (x)=\frac{\mathrm{d}^2\varphi }{\mathrm{d}x^2}(x)\)),

$$\begin{aligned} |\mathcal{K}[\varphi ](x,y,z)-\mathcal{K}_n[\varphi ](x,y,z) | \le \frac{\Vert D^2\varphi \Vert _{L^{\infty }}}{\sqrt{n}} \max \{\sqrt{y},\sqrt{z}\},\qquad \forall \, (x,y,z)\in {\mathbb R}_{\ge 0}^3 \end{aligned}$$

which is enough for proving the existence of distributional solutions \(F_t\) as did in [13].

The second cutoff (2.31)–(2.32) kills the possibility of condensation at the origin, and it does not satisfy (2.41): for instance

$$\begin{aligned} \lim _{n\rightarrow \infty }| W(0,y,z)-W_n(0,y,z)|=W(0,y,z)\ne 0\qquad \forall \, y>0, z>0. \end{aligned}$$

However if the initial datum \(F_0\) is regular and its density \(f_0\) has only weaker singularity at the origin (for instance \(f_0(x)\le Kx^{-1}\,(x>0)\)), the second cutoff (2.31)–(2.32) and the corresponding approximate solutions \(f^n\) are good ones for proving the existence of some expected solutions on a finite time-interval \([0, T]\) (see the proof of Theorem 5.1). This is because the second cutoff (2.31)–(2.32) as well as the original kernel has the following positivity:

Lemma 2.3

Let \(w(x,y,z)\) be given by (1.3) and let \(w_n(x,y,z)\) be the second cutoff of \(w(x,y,z)\):

$$\begin{aligned} w_n(x,y,z)=w(x,y,z)\min \Big \{1,\, n\min \{ x,x_*, y,z\}\Big \}. \end{aligned}$$

Then for any \(0\le f\in L^1({\mathbb R}_{+}, \sqrt{x}\mathrm{d}x)\) (i.e. \(N(f)<\infty \)) we have

$$\begin{aligned}&\int \limits _{{\mathbb R}_{+}^2}w(x,y,z)[ f(x_*)(f(y)+f(z))-f(y)f(z) ]\mathrm{d}y\mathrm{d}z\ge 0\qquad \forall \, x>0,\end{aligned}$$
(2.42)
$$\begin{aligned}&\int \limits _{{\mathbb R}_{+}^2}w_n(x,y,z)[ f(x_*)(f(y)+f(z))-f(y)f(z) ]\mathrm{d}y\mathrm{d}z\ge 0\qquad \forall \, x>0. \end{aligned}$$
(2.43)

Proof

First we note that the above integrals make sense since \((y,z)\mapsto w(x,y,z)f(y)f(z)\) is integrable, see Lemma 5.1 below. The proofs for (2.42) and (2.43) are the same. Let us prove (2.43). Recall the second cutoff (2.31), \(w_n(x,y,z)=\frac{1}{\sqrt{x}}S_n(x,x_*; y,z)\) with \(x_*=(y+z-x)_{+}\). Using translation of variables \(y+z-x\rightarrow z,\, y+z-x\rightarrow y\) for the integrals of \(S_n(x,x_*;y,z)f(x_*)f(y), S_n(x,x_*; y,z)f(x_*)f(z)\) respectively, we compute for any \(x>0\)

$$\begin{aligned}&\int \limits _{{\mathbb R}_{+}^2}w_n(x,y,z)[ f(x_*)(f(y)+f(z))-f(y)f(z) ]\mathrm{d}y\mathrm{d}z\\&\quad =\frac{1}{\sqrt{x}}\int \limits _{{\mathbb R}_{+}^2}[ S_n(x,z; y,y_*)+S_n(x,y;z,z_*)- S_n(x,x_*; y,z) ] f(y)f(z)\mathrm{d}y\mathrm{d}z \end{aligned}$$

where \(y_*=(x+z-y)_{+}, z_*=(x+y-z)_{+}\). Now by definition of the second cutoff \(S_n(\cdot , \cdot \,;\,\cdot ,\cdot )\) in (2.31) and by checking three cases \(0<x\le y\wedge z, y\wedge z\le x\le y\vee z,\) and \(y\vee z\le x\), we obtain

$$\begin{aligned}&S_n(x,z; y,y_*)+S_n(x,y;z,z_*)- S_n(x,x_*; y,z) \\&\ge \min \{\sqrt{y}(1\wedge (ny)),\sqrt{z} (1\wedge (nz)), \sqrt{y_*}(1\wedge (ny_*)), \sqrt{z_*}(1\wedge (nz_*))\}\ge 0. \end{aligned}$$

This proves the lemma. \(\square \)

3 Strong Solutions and Stability

On the basis of existence of distributional solutions (Theorem 1.1), our strong solutions is directly defined from the class of distributional solutions.

Definition 3.1

Let \(F_t\) be a distributional solution of Eq. (1.5) on \([0,\infty )\). Let \(0<T_{\infty }\le \infty \). We say that \(F_t\) is a strong solution of Eq. (1.5) on \([0, T_{\infty })\) if it satisfies the following (i)-(iii):

  1. (i)

    \(t\mapsto F_t\) belongs to \( C^1([0,T_{\infty });{\mathcal B}_{0}({\mathbb R}_{\ge 0}))\),

  2. (ii)

    \(t\mapsto \mathcal{Q}_2^{\pm }(F_t), t\mapsto \mathcal{Q}_3^{\pm }(F_t)\) belong to \( C([0,T_{\infty });{\mathcal B}_{0}({\mathbb R}_{\ge 0}))\), and

  3. (iii)
    $$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}F_t={\mathcal Q}(F_t)\quad in\quad (\mathcal{B}_{0}({\mathbb R}_{\ge 0}), \Vert \cdot \Vert _0 )\qquad \forall \,t\in [0, T_{\infty }). \end{aligned}$$
    (3.1)

    Besides, if \(F_t\) also conserves the energy on \([0,T_{\infty })\), then \(F_t\) is also called a conservative strong solution of Eq. (1.5) on \([0, T_{\infty })\). Strong solutions can be also defined on a finite closed time-interval by replacing \([0,T_{\infty })\) with \([0, T]\) for \(0<T<\infty \).

Remark 3.1

  1. (a)

    Since \(\frac{\mathrm{d} }{\mathrm{d}t}F_t\) and \(\int _{a}^{b}\mathcal{Q}(F_t)\mathrm{d}t\) as finite Borel measures are defined with the norm \(\Vert \cdot \Vert _0\) of \({\mathcal B}_0({\mathbb R}_{\ge 0})\), this gives explicit representations \((\frac{\mathrm{d} }{\mathrm{d}t}F_t)(E)=\frac{\mathrm{d}}{\mathrm{d}t}F_t(E), (\int _{a}^{b}\mathcal{Q}(F_t)\mathrm{d}t)(E)= \int _{a}^{b}\mathcal{Q}(F_t)(E) \mathrm{d}t\) for all Borel sets \(E\subset \mathbb {R}_{\ge 0}\). Here we recall that \(\lim _{n\rightarrow \infty }\Vert \mu _n-\mu \Vert _0=0\) is equivalent to the uniform convergence \(\lim _{n\rightarrow \infty }\sup _{E\subset {\mathbb R}_{\ge 0}} |\mu _n(E)-\mu (E)|=0.\)

  2. (b)

    Under the condition (ii), the conditions (i), (iii) are equivalent to the integral equation:

$$\begin{aligned} F_t=F_0+\int \limits _{0}^t\mathcal{Q}(F_\tau )\mathrm{d}\tau \qquad \forall \,t\in [0, T_{\infty }) \end{aligned}$$
(3.2)

where the integral is taken as the Riemann integral defined with the norm \(\Vert \cdot \Vert _0\). This then implies that, under the condition (ii), the integral equation (3.1) is equivalent to its dual form:

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}}\varphi \mathrm{d}F_t=\int \limits _{{\mathbb R}_{\ge 0}}\varphi \mathrm{d}F_0 +\int \limits _{0}^{t}\mathrm{d}\tau \int \limits _{{\mathbb R}_{\ge 0}}\varphi \mathrm{d}\mathcal{Q}(F_\tau )\qquad \forall \, \varphi \in L^{\infty }_0({\mathbb R}_{\ge 0}) \end{aligned}$$
(3.3)

for all \(t\in [0, T_{\infty })\). In fact, the equation in (3.3) holds for all \(\varphi \in C_b({\mathbb R}_{\ge 0})\) and thus it holds for all \(\varphi \in L^{\infty }_0({\mathbb R}_{\ge 0})\).

Systematic results on strong solutions can be obtained for the case \(M_{-1/2}(F_0)<\infty \). A special case of \(M_{-1/2}(F_0)=\infty , M_{-1/3}(F_0)<\infty \) is considered in Sect. 5. We now begin with the following

Lemma 3.1

Let \(F\in \mathcal{B}^{+}_{1/2}({\mathbb R}_{\ge 0})\) and let \(0\le \varphi \in C_b^{1,1}({\mathbb R}_{\ge 0})\) be convex on \({\mathbb R}_{\ge 0}\). Then

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}^2}\mathcal{J}[\varphi ]\mathrm{d}^2F\ge -M_{1/2}(F)\int \limits _{{\mathbb R}_{\ge 0}}\varphi \mathrm{d}F,\quad \int \limits _{{\mathbb R}_{\ge 0}^3}\mathcal{K}[\varphi ]\mathrm{d}^3F\ge 0. \end{aligned}$$
(3.4)

Proof

The second inequality in (3.4) has been proven in [15]. To prove the first one, it suffices to prove

$$\begin{aligned} \mathcal{J}[\varphi ](y,z)\ge -\frac{1}{2}(\, \varphi (y)\sqrt{z}+\varphi (z)\sqrt{y}\, ),\qquad \forall \, y,z\ge 0. \end{aligned}$$
(3.5)

First, it is easily seen from the assumption on \(\varphi \) that \(\varphi \) is non-increasing on \({\mathbb R}_{\ge 0}\). Next we have

$$\begin{aligned} \Delta \varphi (x,y,z) =(x-y)(x-z) \int \limits _{0}^1\int \limits _{0}^{1}(D^2\varphi ) (\xi _{s,t}) \mathrm{d}s \mathrm{d}t \end{aligned}$$
(3.6)

for all \(x,y,z\ge 0\) satisfying \(x\le y+z\) where \(\xi _{s,t}=y+z-x+s(x-z)+t(x-y)\), \(D^2\varphi (x)=\frac{\mathrm{d}^2\varphi }{\mathrm{d}x^2}(x)\). Now given any \(y,z\ge 0\). By symmetry \(\mathcal{J}[\varphi ](y,z)= \mathcal{J}[\varphi ](z,y)\) we may assume \(y\le z\). The convexity of \(\varphi \) and (3.6) imply \(\Delta \varphi (x,y,z)\ge 0\) for \(x\in [0, y]\cup [z, y+z]\) and so by definition of \(\mathcal{J}[\varphi ]\)

$$\begin{aligned} \mathcal{J}[\varphi ](y,z)\ge \frac{1}{2}\int \limits _{y}^{z}\sqrt{x}W(x,y,z)\Delta \varphi (x,y,z) \mathrm{d}x. \end{aligned}$$

From this and \(\varphi \ge 0\) we see that to prove (3.5) we can assume \(y<z\). By definition of \(W(x,y,z)\) and the non-increase of \(\varphi \) we deduce for all \(x\in (y,z)\) that \(W(x,y,z)=\frac{1}{\sqrt{xz}}\) and \(\varphi (x_*)\ge \varphi (z)\) and so \(\Delta \varphi (x,y,z) \ge -\varphi (y).\) Thus

$$\begin{aligned} \mathcal{J}[\varphi ](y,z)\ge \frac{1}{2\sqrt{z}}\int \limits _{y}^{z} \Delta \varphi (x,y,z) \mathrm{d}x\ge -\frac{1}{2\sqrt{z}} \varphi (y)(z-y) \ge -\frac{1}{2} \varphi (y)\sqrt{z} \end{aligned}$$

and so (3.5) holds true. \(\square \)

In order to estimate the moment \(M_{p}(F)\) of negative order \(p<0\) for \(F\in \mathcal{B}^{+}({\mathbb R}_{\ge 0})\) we will often use a smooth approximation:

$$\begin{aligned} M_{p}^{(\varepsilon )}(F):=\int \limits _{{\mathbb R}_{\ge 0}}(\varepsilon +x)^{p}\mathrm{d}F(x),\,\,\,\varepsilon >0;\quad \lim _{\varepsilon \rightarrow 0^+}M_{p}^{(\varepsilon )}(F)=M_{p}(F) \end{aligned}$$
(3.7)

where the limit is due to the monotone convergence theorem.

A fact that will be frequently used for distributional solutions \(F_t\) is that the function \(t\mapsto M_{1/2}(F_t)\) is continuous on \([0,\infty )\) and \(M_{1/2}(F_t)\le \sqrt{N(F_t)E(F_t)}\) for all \(t\in [0,\infty )\). Another fact to be used is the following “monotone non-decrease” of the moment \(t\mapsto M_p(F_t)\) for \(p<0\).

Lemma 3.2

Let \(F_t\) be a distributional solution of Eq. (1.5) on \([0, \infty )\). Then for any \(p<0\), the function \(t\mapsto M_{p}(F_t)e^{\int _{0}^{t}M_{1/2}(F_\tau )\mathrm{d}\tau } \) is monotone non-decreasing on \([0,\infty )\), i.e.,

$$\begin{aligned} M_{p}(F_t) \le M_{p}(F_T)e^{\int \limits _{t}^{T}M_{1/2}(F_\tau )\mathrm{d}\tau }\qquad \forall \, 0\le t\le T<\infty \end{aligned}$$
(3.8)

including the possible case \(M_{p}(F_t) =\infty \) for some (or all) \(t\in [0,\infty )\).

Proof

Let \(M_{p}^{(\varepsilon )}(F_t)\) be defined in (3.7). Applying Eq. (1.8) and Lemma 3.1 to the smooth convex function \(x\mapsto \varphi _{\varepsilon }(x)=(\varepsilon +x)^{p}\,\,(\varepsilon >0)\) we have

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}M_{p}^{(\varepsilon )}(F_{t})\ge \int \limits _{{\mathbb {R}}_{\ge 0}^2}\mathcal{J}[\varphi _{\varepsilon }]\mathrm{d}^2F_t \ge - M_{1/2}(F_t)M_{p}^{(\varepsilon )}(F_{t}), \quad t\ge 0. \end{aligned}$$

This implies that \(t\mapsto M_{p}^{(\varepsilon )}(F_t)e^{\int _{0}^{t}M_{1/2}(F_\tau )\mathrm{d}\tau } \) is monotone non-decreasing on \([0,\infty )\), i.e.

$$\begin{aligned} M_{p}^{(\varepsilon )}(F_t)\le M_{p}^{(\varepsilon )}(F_T)e^{\int \limits _{t}^{T}M_{1/2}(F_\tau )\mathrm{d}\tau }\qquad \forall \, 0\le t\le T<\infty . \end{aligned}$$

Letting \(\varepsilon \rightarrow 0^+\) and using the limit (3.7) we obtain (3.8). \(\square \)

Lemma 3.3

Let \(F_t\) be a distributional solution of Eq. (1.5) on \([0,\infty )\) and let \(M_{-1/2}^{(\varepsilon )}(F_t)\) be defined in (3.7) with \(p=-1/2\). Then

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}M_{-1/2}^{(\varepsilon )}(F_t)\le 2N^2-M_{1/2}(F_t)M_{-1/2}^{(\varepsilon )}(F_t)+ [M_{-1/2}^{(\varepsilon )}(F_t)]^3 \qquad \forall \, t\in [0,\infty ). \end{aligned}$$
(3.9)

Proof

Let \(\varphi (x)=(\varepsilon +x)^{-1/2}\,\, (\varepsilon >0).\) Applying the differential equation (1.8) we see that to prove (3.9) it suffices to prove that for any \(x,y,z\ge 0\)

$$\begin{aligned} \mathcal{J}[\varphi ](y,z)\le 2-\frac{1}{2}[\varphi (y)+\varphi (z)]\sqrt{y\vee z},\quad \mathcal{K}[\varphi ](x,y,z)\le \varphi (x)\varphi (y)\varphi (z). \end{aligned}$$
(3.10)

By definition of \(\mathcal{J}^{\pm }[\varphi ](y,z)\) and (2.6) it is easily deduced \(\mathcal{J}^{+}[\varphi ](y,z)\le 2\) and \(\mathcal{J}^{-}[\varphi ](y,z)\ge \frac{1}{2}[\varphi (y)+\varphi (z)]\sqrt{y\vee z}\). This gives the first inequality in (3.10). To prove the second one in (3.10) we can assume that \(\mathcal{K} [\varphi ](x,y,z)>0\) for the given \(x,y,z\ge 0\). In this case we have \(x_*=y+z-x>0\) and \((y-x)(z-x)>0\), the latter is due to the convexity of \(\varphi \) and (3.6). Let us denote \(a=\sqrt{\varepsilon +x}, b=\sqrt{\varepsilon +x_*}, c=\sqrt{\varepsilon +y}, d=\sqrt{\varepsilon +z}.\) Then

$$\begin{aligned} \Delta \varphi (x,y,z) =\frac{1}{a}+\frac{1}{b}-\frac{1}{c}-\frac{1}{d}=\varphi (x)\varphi (x_*)\varphi (y)\varphi (z) [(a+b)cd-(c+d)ab] \end{aligned}$$

and \(c^2d^2-a^2b^2=(y-x)(z-x)>0\) i.e. \(cd>ab\) which together with \(a^2+b^2=c^2+d^2\) implies \(a+b<c+d\) and \(cd+ab\le a^2+b^2\). Thus, by computing derivative with respect to \(\varepsilon \), we conclude that the function \(\varepsilon \mapsto (a+b)cd-(c+d)ab\) is decreasing on \([0,\infty )\) and so

$$\begin{aligned} \Delta \varphi (x,y,z) \le \varphi (x)\varphi (x_*)\varphi (y)\varphi (z)[(\sqrt{x}+\sqrt{x_*})\sqrt{y}\sqrt{z} -(\sqrt{y}+\sqrt{z})\sqrt{x}\sqrt{x_*}\,]. \end{aligned}$$

Then it is easily deduced from definition of \(W(x,y,z)\) that

$$\begin{aligned} W(x,y,z)\varphi (x_*)[(\sqrt{x}+\sqrt{x_*})\sqrt{y}\sqrt{z} -(\sqrt{y}+\sqrt{z})\sqrt{x}\sqrt{x_*}\,]\le 1. \end{aligned}$$

This gives the second inequality in (3.10) by definition of \(K[\varphi ](x,y,z)\). \(\square \)

Lemma 3.4

Let \(F_t\) be a distributional solution of Eq. (1.5) on \([0, \infty )\) with the initial datum \(F_0\) satisfying \(M_{-1/2}(F_0)<\infty \) and let

$$\begin{aligned} T_{F_{\cdot },\max }=\sup \{ t\in [0,\infty )\,|\, M_{-1/2}(F_t)<\infty \}. \end{aligned}$$
(3.11)

Then

$$\begin{aligned}&T_{F_{\cdot },\max }\ge T_{F_0}:=\frac{1}{2}[N^{2/3}+M_{-1/2}(F_0)]^{-2}>0,\quad N:=N(F_0), \end{aligned}$$
(3.12)
$$\begin{aligned}&\sup _{0\le t\le T}M_{-1/2}(F_t)\le M_{-1/2}(F_T)e^{\int \limits _{0}^T M_{1/2}(F_{\tau })\mathrm{d}\tau }<\infty \quad \forall \, T\in (0, T_{F_{\cdot },\max }), \end{aligned}$$
(3.13)
$$\begin{aligned}&M_{-1/2}(F_t)\le \frac{N^{2/3}+M_{-1/2}(F_0)}{\sqrt{1-t/T_{F_0}}}-N^{2/3} \qquad \forall \,t\in [0, T_{F_0}), \end{aligned}$$
(3.14)
$$\begin{aligned}&if\,\,\, t_0:=T_{F_{\cdot },\max }<\infty ,\,\,\,then \,\,\, \lim _{t\rightarrow t_0-}M_{-1/2}(F_t)=M_{-1/2}(F_{t_0})=\infty . \end{aligned}$$
(3.15)

Proof

Let \(M_{-1/2}^{(\varepsilon )}(F_t)\) be defined in (3.7) with \(\varepsilon \in (0, N^{2/3})\). By writing \(1=(\varepsilon +x)^{1/4}(\varepsilon +x)^{-1/4}\) and using Cauchy-Schwarz inequality we have \(N^2\le (\sqrt{\varepsilon }N+M_{1/2}(F_t))M_{-1/2}^{(\varepsilon )}(F_t)\) so that

$$\begin{aligned} 2N^2-M_{1/2}(F_t)M_{-1/2}^{(\varepsilon )}(F_t)+ [M_{-1/2}^{(\varepsilon )}(F_t)]^3 \le [N^{2/3}+ M_{-1/2}^{(\varepsilon )}(F_t)]^3. \end{aligned}$$

If we define \(u_{\varepsilon }(t)=N^{2/3}+M_{-1/2}^{(\varepsilon )}(F_t)\), then we deduce from (3.9) that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}u_{\varepsilon }(t)\le [u_{\varepsilon }(t)]^3,\quad t\ge 0. \end{aligned}$$

This gives \( (u_{\varepsilon }(t))^{-2}\ge (u_{\varepsilon }(0))^{-2}-2t\) for all \(t\ge 0.\) Since \(u_{\varepsilon }(0)\le N^{2/3}+M_{-1/2}(F_0)\), it follows from definition of \(T_{F_0}\) that

$$\begin{aligned} M_{-1/2}^{(\varepsilon )}(F_t)\le \frac{N^{2/3}+M_{-1/2}(F_0)}{\sqrt{1-t/T_{F_0}}}-N^{2/3} \qquad \forall \,t\in [0, T_{F_0}). \end{aligned}$$

Letting \(\varepsilon \rightarrow 0^+\) we conclude from the limit (3.7) that \(F_t\in \mathcal{B}^{+}_{-1/2, 1}({\mathbb R}_{\ge 0})\) for all \(t\in [0, T_{F_0})\) and (3.14) holds true. This also proves (3.12).

Next let \(\mathcal{T}(F_{\cdot })=\{ t\in [0,\infty )\,|\, M_{-1/2}(F_t)<\infty \}\). We will prove that \(\mathcal{T}(F_{\cdot })=[0, T_{F_{\cdot },\max })\). First, it is easily seen from the monotone inequality (3.8) and \(T_{F_{\cdot },\max }=\sup \mathcal{T}(F_{\cdot })\) that \(\mathcal{T}(F_{\cdot })\) is an interval and \([0, T_{F_{\cdot },\max })\subset \mathcal{T}(F_{\cdot })\) which also implies (3.13) by (3.8). Next we prove \(T_{F_{\cdot },\max } \not \in \mathcal{T}(F_{\cdot })\). Suppose to the contrary that \(t_0:=T_{F_{\cdot },\max }\in \mathcal{T}(F_{\cdot })\). Then \(t_0<\infty \) and \(M_{-1/2}(F_{t_0})<\infty \) and so applying the above result to the distributional solution \(t\mapsto \widetilde{F}_t:=F_{t+t_0}\) of Eq. (1.5) on \([0,\infty )\) with the initial datum \(\widetilde{F}_0=F_{t_0}\) we conclude that \(T_{\widetilde{F}_0}>0\) and \([0, T_{\widetilde{F}_0})\subset \mathcal{T}(\widetilde{F}_{\cdot })\). So \(M_{-1/2}(F_{t+t_0})=M_{-1/2}(\widetilde{F}_t)<\infty \) for all \(t\in [0, T_{\widetilde{F}_0})\), i.e. \(M_{-1/2}(F_t)<\infty \) for all \(t\in [t_0, t_0+T_{\widetilde{F}_0})\). This implies \([0, t_0+T_{\widetilde{F}_0})\subset \mathcal{T}(F_{\cdot })\) hence \(t_0+T_{\widetilde{F}_0}\le \sup \mathcal{T}(F_{\cdot })=t_0\) which contradicts \(T_{\widetilde{F}_0}>0\). Thus we must have \(T_{F_{\cdot },\max } \not \in \mathcal{T}(F_{\cdot })\). Since \(T_{F_{\cdot },\max }=\sup \mathcal{T}(F_{\cdot })\), it follows that \(\mathcal{T}(F_{\cdot })\subset [0, T_{F_{\cdot },\max })\) and thus \(\mathcal{T}(F_{\cdot })=[0, T_{F_{\cdot },\max })\).

To prove the blow up (3.15) we denote again \(t_0:= T_{F_{\cdot },\max }<\infty \). We have proved that \(t_0\not \in \mathcal{T}(F_{\cdot })\), i.e. \(M_{-1/2}(F_{t_0})=\infty \). Using the continuity of \(t\mapsto M^{(\varepsilon )}_{-1/2}(F_t)\) on \([0,\infty )\) we have

$$\begin{aligned} M^{(\varepsilon )}_{-1/2}(F_{t_0})=\liminf _{t\rightarrow t_0-} M^{(\varepsilon )}_{-1/2}(F_t) \le \liminf _{t\rightarrow t_0-} M_{-1/2}(F_t)\qquad \forall \, \varepsilon >0. \end{aligned}$$

Letting \(\varepsilon \rightarrow 0^+\) we conclude from the limit (3.7) that \(M_{-1/2}(F_{t_0}) \le \liminf _{t\rightarrow t_0-} M_{-1/2}(F_t)\) and so (3.15) holds true since \(M_{-1/2}(F_{t_0})=\infty \). \(\square \)

Proposition 3.1

Let \(F_t\) be a distributional solution of Eq. (1.5) on \([0,\infty )\) with the initial datum \(F_0\) satisfying \(M_{-1/2}(F_0)<\infty \). Then \(F_t\) is a strong solution of Eq. (1.5) on \([0, T_{F_{\cdot },\max })\).

Proof

Take any \(T\in (0, T_{F_{\cdot },\max })\). Using the estimates (2.15), (2.18) for \(k=1/2\) we have \(\Vert \mathcal{Q}_2^{\pm }(F_t)\Vert _{1/2},\, \Vert \mathcal{Q}_3^{\pm }(F_t)\Vert _{1/2}\le C_T\) for all \( t\in [0, T]\), where \(C_T<\infty \) depends only on \(\sup _{t\in [0,T]}M_{-1/2}(F_t)\) and \(\sup _{t\in [0,T]}\Vert F_t\Vert _1\). From this and the integral equation (1.9) which also reads

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}}\varphi \mathrm{d}(F_t-F_s)=\int \limits _{s}^{t}\mathrm{d}\tau \int \limits _{{\mathbb R}_{\ge 0}}\varphi \mathrm{d}\mathcal{Q}(F_{\tau })\qquad \forall \,\varphi \in C_b^{1,1}({\mathbb R}_{\ge 0}) \end{aligned}$$
(3.16)

we obtain \(\Vert F_t-F_s\Vert _0\le C_T|t-s|\) for all \(t,s\in [0, T].\) Since \(\Vert F_t-F_s\Vert _{1/2}\le \Vert F_t-F_s\Vert _1^{1/2}\Vert F_t-F_s\Vert _0^{1/2}\) by Cauchy-Schwarz inequality, it follows that \(t\mapsto F_t\) also belongs to \( C([0,T_{F_{\cdot },\max });{\mathcal B}_{1/2}({\mathbb R}_{\ge 0}))\) and thus we conclude from (2.19)–(2.21) with \(k=0\) that \(t\mapsto \mathcal{Q}_2^{\pm }(F_t), t\mapsto \mathcal{Q}_3^{\pm }(F_t)\) hence \(t\mapsto \mathcal{Q}(F_t)\) all belong to \(C([0, T_{F_{\cdot },\max });{\mathcal B}_{0}({\mathbb R}_{\ge 0}))\). Next for any \(T\in (0, T_{F_{\cdot },\max })\), using smooth approximation it is easily deduced that (3.16) with \(s=0\) and \(t\in [0, T]\) holds for all bounded Borel functions \(\varphi \) on \({\mathbb R}_{\ge 0}\), in particular it holds for all characteristic functions \(\varphi (x)=\mathbf{1}_{E}(x)\) of Borel sets \(E\subset {\mathbb R}_{\ge 0}\). Therefore \(F_t\) satisfies the integral equation (3.2) and so, according to the equivalent definition of strong solutions discussed in Remark 3.1, \(F_t\) is a strong solutions of Eq. (1.5) on \([0, T_{F_{\cdot },\max })\). \(\square \)

Now we are going to establish the stability estimate for conservative strong solutions. The method is similar to those for the space homogeneous measure-valued solutions of the classical Boltzmann equation (see e.g. [16]). Since we have here the cubic term \(\mathcal{Q}_3(F_t)\) which determines the Bose–Einstein model, we would like to present a complete proof.

Let \(F,G, H\in \mathcal{B}({\mathbb R}_{\ge 0}^n)\). Recall that the inequality \(F\le G\) means \(F(E)\le G(E)\) for all Borel sets \(E\subset {\mathbb R}_{\ge 0}^n\). Let \(h(x)\) be the sign function of \(H\), i.e., \(h\) is a real Borel function satisfying \([h(x)]^2\equiv 1\) and \( \mathrm{d}H(x)=h(x)\mathrm{d}|H|(x)\), which is equivalent to \(h(x)\mathrm{d}H(x)=\mathrm{d}|H|(x)\). If we define \(\kappa (x)=\frac{1}{2}(1+h(x))\) and \(H^+=\frac{1}{2}(H+|H|)\), then \(H^{+}\ge 0\) and \(\kappa (x)\mathrm{d}H(x)=\mathrm{d}H^{+}(x)\).

Lemma 3.5

Let \(F\in \mathcal{B}_{-1/2, 1}^+({\mathbb R}_{\ge 0})\cap \mathcal{B}_{3/2}^+({\mathbb R}_{\ge 0}), G\in \mathcal{B}_{-1/2, 1}^+({\mathbb R}_{\ge 0})\) and \(H=F-G\). Let \(\kappa : {\mathbb R}_{\ge 0}\mapsto \{0,1\}\) be the Borel function such that \(\kappa (\cdot )\mathrm{d}H=\mathrm{d}H^{+}.\) Then for all \(n\ge 1\)

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}^2}(1+ x\wedge n)\kappa (x)\mathrm{d}(\mathcal{Q}(F)- \mathcal{Q}(G))&\le \Vert F\Vert _{1/2}\Vert F\Vert _{3/2} +2\Vert F\Vert _{3/2}\Vert H\Vert _0\nonumber \\&+\, 2\Vert F\Vert _{1}\Vert H\Vert _{1/2}+2b(F,G)\Vert H\Vert _1,\qquad \end{aligned}$$
(3.17)
$$\begin{aligned} \limsup _{n\rightarrow \infty }\int \limits _{{\mathbb R}_{\ge 0}^2}(1+ x\wedge n)\kappa (x)\mathrm{d}(\mathcal{Q}(F)- \mathcal{Q}(G))&\le 2\Vert F\Vert _{3/2}\Vert H\Vert _0+ 2\Vert F\Vert _{1}\Vert H\Vert _{1/2}\nonumber \\&+\,2b(F,G)\Vert H\Vert _1 \end{aligned}$$
(3.18)

where \(b(F,G)\) is given in (2.22).

Proof

Let \(\varphi _n(x)=(1+ x\wedge n)\kappa (x)\). Since \(x\mapsto \varphi _n(x)\) is bounded, there is no problem of integrability in the proof. We have

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}^2}\varphi _n\mathrm{d}( \mathcal{Q}(F)- \mathcal{Q}(G) )\le \mathcal{Q}_{2,n}^{+}-\mathcal{Q}_{2,n}^{-}+2b(F,G)\Vert H\Vert _1, \end{aligned}$$
(3.19)
$$\begin{aligned} \mathcal{Q}_{2,n}^{+}:=\int \limits _{{\mathbb R}_{\ge 0}^2}\mathcal{J}^{+}[\varphi _n]\mathrm{d}( F\otimes F-G\otimes G),\quad \mathcal{Q}_{2,n}^{-}:=\int \limits _{{\mathbb R}_{\ge 0}^2}\varphi _n(y)\mathcal{J}^{+}[1]\mathrm{d}( F\otimes F-G\otimes G) \end{aligned}$$

where we have used (2.21) for \(k=1\) to get \(\Vert \mathcal{Q}_3(F)-\mathcal{Q}_3(G)\Vert _1\le 2b(F,G)\Vert H\Vert _1\). Recall that \(W(x,y,z)>0\) implies \(x+x_*=y+z\) so that \(\varphi _n(x)+\varphi _n(x_*) \le 2+x+x_*=2+y+z\). This gives \(\mathcal{J}^{+}[\varphi _n](y, z)\le (1+\frac{y+z}{2})\mathcal{J}^{+}[1](y, z) \) and so

$$\begin{aligned} \mathcal{Q}_{2,n}^{+} \!\!&\le \!\! \int \limits _{{\mathbb R}_{\ge 0}^2}\mathcal{J}^{+}[\varphi _n]\mathrm{d}( F\otimes F-G\otimes G)^{+}\nonumber \\ \!\!&\le \!\! \int \limits _{{\mathbb R}_{\ge 0}^2}\Big (1+\frac{y+z}{2}\Big )\mathcal{J}^{+}[1](y,z)\mathrm{d}( F\otimes F-G\otimes G)^+ \nonumber \\ \!\!&= \!\!\int \limits _{{\mathbb R}_{\ge 0}^2}(1+y)\mathcal{J}^{+}[1](y,z)\mathrm{d}( F\otimes F-G\otimes G)^+ \end{aligned}$$
(3.20)

where the last equality is due to the symmetry \(\mathcal{J}^{+}[1](y,z)=\mathcal{J}^{+}[1](z,y)\) and the equality

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}^2}\psi (y,z)\mathrm{d}( F\otimes F-G\otimes G)^+ =\int \limits _{{\mathbb R}_{\ge 0}^2}\psi (z,y)\mathrm{d}( F\otimes F-G\otimes G)^+ \end{aligned}$$

which holds at least for all bounded Borel functions \(\psi \) on \({\mathbb R}_{\ge 0}^2\) (see e.g. Lemma 5.2 of [16]) and thus it holds for all nonnegative Borel functions on \({\mathbb R}_{\ge 0}^2\) by monotone convergence. Since

$$\begin{aligned} (F\otimes F-G\otimes G)^+\le F\otimes H^+ + H^+\otimes G, \end{aligned}$$

\(1+y=(y-n)_{+}+1+y\wedge n\), and

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}^2}(y-n)_{+}\mathcal{J}^{+}[1]\mathrm{d}( F\otimes F-G\otimes G)^+\le \Vert F\Vert _{1/2}\int \limits _{{\mathbb R}_{\ge 0}} (y-n)_{+}\sqrt{1+y}\mathrm{d} F(y) =: e_n \end{aligned}$$

it follows from (3.20) that

$$\begin{aligned} \mathcal{Q}_{2,n}^{+}\le e_n+\int \limits _{{\mathbb R}_{\ge 0}^2}(1+y\wedge n)\mathcal{J}^{+}[1] \mathrm{d}(F\otimes H^+)+\int \limits _{{\mathbb R}_{\ge 0}^2}(1+y\wedge n)\mathcal{J}^{+}[1] \mathrm{d}(H^+\otimes G).\nonumber \\ \end{aligned}$$
(3.21)

For the negative part \(\mathcal{Q}_{2,n}^{-}\), recalling \(\varphi _n(y)=(1+ y\wedge n)\kappa (y)\) we have

$$\begin{aligned}&\varphi _n(y)\mathrm{d}( F\otimes F-G\otimes G) =\varphi _n(y)\mathrm{d}(F\otimes H)+(1+y\wedge n)\mathrm{d}(H^{+}\otimes G), \nonumber \\&\mathcal{Q}_{2,n}^{-}= \int \limits _{{\mathbb R}_{\ge 0}^2}\varphi _n(y)\mathcal{J}^{+}[1] \mathrm{d}(F\otimes H)+ \int \limits _{{\mathbb R}_{\ge 0}^2}(1+y\wedge n)\mathcal{J}^{+}[1]\mathrm{d}(H^+\otimes G). \end{aligned}$$
(3.22)

The common terms in the right hand sides of (3.21)–(3.22) cancel each other in \(\mathcal{Q}_{2,n}^{+}-\mathcal{Q}_{2,n}^{-}\) and thus using \(\mathcal{J}^{+}[1](y,z)\le \sqrt{y}+\sqrt{z}\) we obtain

$$\begin{aligned} \mathcal{Q}_{2,n}^{+}-\mathcal{Q}_{2,n}^{-}\le e_n+2\Vert F\Vert _{3/2}\Vert H\Vert _0+ 2\Vert F\Vert _{1}\Vert H\Vert _{1/2}. \end{aligned}$$
(3.23)

Finally by definition of \(e_n\) and the assumption on \(F\) we have \(e_n\le \Vert F\Vert _{1/2}\Vert F\Vert _{3/2}, n=1,2,3,\dots ;\) and \( \lim _{n\rightarrow \infty }e_n=0\) by dominated convergence theorem. The lemma then follows from these and (3.19), (3.23). \(\square \)

For any given \(F_0\in \mathcal{B}_{1}^{+}({\mathbb R}_{\ge 0})\) we define a function \(\Psi _{F_0}(\varepsilon )\) on \([0,\infty )\) by

$$\begin{aligned} \Psi _{F_0}(\varepsilon )=\varepsilon +\sqrt{\varepsilon }+\int \limits _{\frac{1}{\sqrt{\varepsilon }}}^{\infty }x \mathrm{d}F_0(x),\quad \varepsilon >0;\quad \Psi _{F_0}(0)=0. \end{aligned}$$
(3.24)

Here \(\int _{\frac{1}{\sqrt{\varepsilon }}}^{\infty }\) can be understood as either \(\int _{(\frac{1}{\sqrt{\varepsilon }}, \infty )}\) or \(\int _{[\frac{1}{\sqrt{\varepsilon }}, \infty )}\).

Theorem 3.1

Let \(F_t, G_t\) be conservative distributional solutions to Eq. (1.5) on \([0, \infty )\) with their initial data \(F_0, G_0\) satisfying \(M_{-1/2}(F_0)<\infty , M_{-1/2}(G_0)<\infty \). Let \(T_{F_{\cdot },\max }, T_{G_{\cdot },\max }\) be defined in (3.11) for \(F_t, G_t\) respectively. Then for any \(T\in (0, T_{F_{\cdot },\max }\wedge T_{G_{\cdot },\max })\)

$$\begin{aligned} \Vert F_t-G_t\Vert _1\le C\Psi _{F_0}(\Vert F_0-G_0\Vert _1) e^{ct}\qquad \forall \, t\in [0, T] \end{aligned}$$
(3.25)

where \(\Psi _{F_0}(\cdot )\) is defined in (3.24) and \(C=C_T, c=c_T\) are finite positive constants depending only on \(N(F_0), E(F_0)\), \(\sup _{0\le t\le T}M_{-1/2}(F_t)\), and \(\sup _{0\le t\le T}M_{-1/2}(G_t)\) with \(T\in (0, T_{F_{\cdot },\max }\wedge T_{G_{\cdot },\max }).\)

In particular if \(F_0=G_0\), then \(T_{F_{\cdot },\max }=T_{G_{\cdot },\max }:=T_{\max }\) and \(F_t=G_t\) for all \(t\in [0,T_{\max }).\)

Proof

The proof is divided into five steps. First of all, according to Proposition 3.1, \(F_t, G_t\) are strong solutions on \([0,T_{F_{\cdot },\max })\) and \([0, T_{G_{\cdot },\max })\) respectively. In Steps 1–3 we assume that one of the two solutions, e.g. \(F_t\), has the moment production (1.15) for all \(t\in (0, T_{F_{\cdot },\max })\). The existence of such \(F_t\) is assured by Theorem 1.1. Let us denote

$$\begin{aligned} H_t=F_t-G_t. \end{aligned}$$

By conservation of mass we have \(\Vert F_t\pm G_t\Vert _1\le \Vert F_0\Vert _1+\Vert G_0\Vert _1\) for all \(t\ge 0\). So if \(\Vert H_0\Vert _1\ge 1\), then \(\Vert H_t\Vert _1\le 2\Vert F_0\Vert _1+\Vert H_0\Vert _1\le (2\Vert F_0\Vert _1+1)\Vert H_0\Vert _1\) for all \(t\ge 0\). Therefore to prove (3.25) we can assume \(\Vert H_0\Vert _1<1\). Given any \(T\in (0, T_{F_{\cdot },\max }\wedge T_{G_{\cdot },\max })\) and \(s\in (0, T)\).

Step 1: We prove that

$$\begin{aligned}&\Vert H_t\Vert _0\le \Vert H_0\Vert _0 +C\int \limits _{0}^{t}\Vert H_\tau \Vert _1\mathrm{d}\tau ,\quad t\in [0, T],\end{aligned}$$
(3.26)
$$\begin{aligned}&\Vert H_t\Vert _1\le \Vert H_s\Vert _1+ C_0\int \limits _{s}^{t}(1+1/\tau )\Vert H_\tau \Vert _{0}\mathrm{d}\tau + C\int \limits _{s}^{t}\Vert H_\tau \Vert _1\mathrm{d}\tau , \quad t\in [s, T]. \qquad \end{aligned}$$
(3.27)

Here and below the constant \(0<C_0<\infty \) depends only on \(N(F_0)\) and \(E(F_0)\), while the constants \(0<C, c<\infty \) depend only on \(N(F_0)\), \(E(F_0)\), \(\sup _{0\le t\le T}M_{-1/2}(F_t)\), and \(\sup _{0\le t\le T}M_{-1/2}(G_t)\). And \(C_0, C\) may have different value in different places. Also recall that \(\Vert F_0\Vert _1=N(F_0)+E(F_0).\)

The inequality (3.26) follows from \(H_t=H_0+\int _{0}^{t}[\mathcal{Q}(F_\tau )-\mathcal{Q}(G_\tau )]\mathrm{d}\tau \) and the estimates (2.19), (2.21) for \(k=0\). To prove (3.27) we first use \(|H_t|=G_t-F_t+2(H_t)^{+}\) and the conservation of mass and energy to write

$$\begin{aligned} \Vert H_t\Vert _1=\Vert G_s\Vert _1-\Vert F_s\Vert _1+2\Vert (H_t)^{+}\Vert _1,\quad t\ge s. \end{aligned}$$
(3.28)

Let \( x\mapsto \kappa _t(x)\in \{0,1\}\) be the Borel function on \({\mathbb R}_{\ge 0}\) such that \(\kappa _t\mathrm{d}H_t=\mathrm{d}(H_t)^{+}\). Since \(t\mapsto \mathcal{Q}(F_t)-\mathcal{Q}(G_t)\) belongs to \(C([0, T_{F_{\cdot },\max }\wedge T_{G_{\cdot },\max }); \mathcal{B}_0({\mathbb R}_{\ge 0}))\), applying Lemma 5.1 of [16] to the measure equation \( H_t=H_s+\int _{s}^t ( \mathcal{Q}(F_\tau )-\mathcal{Q}(G_\tau ))\mathrm{d}\tau ,\, t\in [s, T_{F_{\cdot },\max }\wedge T_{G_{\cdot },\max }), \) we have

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}}\varphi (x)\mathrm{d}(H_t)^{+}= \int \limits _{{\mathbb R}_{\ge 0}}\varphi (x)\mathrm{d}(H_s) +\int \limits _{s}^t\mathrm{d}\tau \int \limits _{{\mathbb R}_{\ge 0}}\varphi (x) \kappa _\tau (x)\mathrm{d}( \mathcal{Q}(F_\tau )-\mathcal{Q}(G_\tau )) \end{aligned}$$

for all \(t\in [s, T]\) and all \(\varphi \in L^{\infty }_0({\mathbb R}_{\ge 0})\). In particular we have

$$\begin{aligned} \int \limits _{\mathbb {R}_{\ge 0}}(1+x \wedge n) \mathrm{d}(H_t)^{+} \le \Vert (H_s)^{+}\Vert _1+\int \limits _{s}^{t}\mathrm{d}\tau \int \limits _{\mathbb {R}_{\ge 0}}(1+x \wedge n)\kappa _\tau (x) \mathrm{d}( \mathcal{Q}(F_\tau )-\mathcal{Q}(G_\tau ) ). \end{aligned}$$

Next applying (1.15) with \(p=3/2\) we see that the function \(t\mapsto \Vert F_t\Vert _{3/2}\le C_{0}(1+1/t)\) is integrable on \([s, T]\) and thus we deduce from Lemma 3.5 and the reverse Fatou’s Lemma that

$$\begin{aligned}&\limsup _{n\rightarrow \infty }\int \limits _{s}^{t}\mathrm{d}\tau \int \limits _{{\mathbb R}_{\ge 0}}(1+x\wedge n)\kappa _\tau (x)\mathrm{d}(\mathcal{Q}_2(F_\tau )- \mathcal{Q}_2(G_\tau ))(x)\\&\le C_0 \int \limits _{s}^{t}(1+1/\tau )\Vert H_\tau \Vert _0\mathrm{d}\tau +C\int \limits _{s}^{t}\Vert H_\tau \Vert _1\mathrm{d}\tau . \end{aligned}$$

Here the second constant \(C\) is obtained by using the conservation of mass and energy which gives upper bounds \(M_{1/2}(F_t)+M_{1/2}(G_t) \le \frac{1}{2}\Vert F_0\Vert _1+\frac{1}{2}\Vert G_0\Vert _1 \le \Vert F_0\Vert _1+\frac{1}{2}, \) etc. Letting \(n\rightarrow \infty \) we conclude

$$\begin{aligned} \Vert (H_t)^{+}\Vert _1\le \Vert (H_s)^{+}\Vert _1+ C_0\int \limits _{s}^{t}(1+1/\tau )\Vert H_\tau \Vert _0\mathrm{d}\tau +C\int \limits _{s}^{t}\Vert H_\tau \Vert _1\mathrm{d}\tau . \end{aligned}$$

This together with (3.28) and \(\Vert G_s\Vert _1-\Vert F_s\Vert _1+2\Vert (H_s)^{+}\Vert _1 =\Vert H_s\Vert _1\) gives (3.27).

Step 2: We prove that for any \(R\ge 1\)

$$\begin{aligned} \Vert H_t\Vert _1\le 5R\Vert H_0\Vert _1+C Rt +2\int \limits _{x>R}x \mathrm{d}F_0(x),\quad t\in [0, T]. \end{aligned}$$
(3.29)

In fact using \(|H_t|=G_t-F_t+2(H_t)^{+}\) and conservation of mass and energy we have

$$\begin{aligned} \Vert H_t\Vert _1\le \Vert H_0\Vert _1+ 4R\Vert H_t\Vert _0 +2\int \limits _{x> R}x\mathrm{d}F_t(x) \end{aligned}$$
(3.30)

and applying (3.3) to the bounded function \(\psi (x)= \mathbf{1}_{\{x\le R\}}x\) we deduce

$$\begin{aligned} \int \limits _{x> R}x\mathrm{d}F_t =E(F_0)-\int \limits _{{\mathbb R}_{\ge 0}}\mathbf{1}_{\{x\le R\}}x\mathrm{d}F_t&= \int \limits _{x> R}x\mathrm{d}F_0 -\int \limits _{0}^{t}\mathrm{d}\tau \int \limits _{{\mathbb R}_{\ge 0}}\mathbf{1}_{\{x\le R\}}x \mathrm{d}\mathcal{Q}(F_\tau )\\&\le \int \limits _{x> R}x\mathrm{d}F_0+ R\int \limits _{0}^{t}\Vert \mathcal{Q}(F_\tau )\Vert _0\mathrm{d}\tau \\&\le \int \limits _{x> R}x\mathrm{d}F_0+C R t. \end{aligned}$$

This together with (3.30) and \(\Vert H_t\Vert _0\le \Vert H_0\Vert _0 +C t\) (by (3.26)) yields (3.29).

Step 3: If \(T\le \Vert H_0\Vert _1\), we take \(R=\frac{1}{\sqrt{\Vert H_0\Vert _1}}\) and use (3.29) to get

$$\begin{aligned} \Vert H_t\Vert _1 \le C\Big (\sqrt{\Vert H_0\Vert _1} +\int \limits _{x> \frac{1}{\sqrt{\Vert H_0\Vert _1}}}x\mathrm{d}F_0(x) \Big ) \le C\Psi _{F_0}(\Vert H_0\Vert _1),\quad t\in [0, T]. \end{aligned}$$

Suppose now \(\Vert H_0\Vert _1<T\) and let \(\varepsilon >0\) satisfy \(\Vert H_0\Vert _1\le \varepsilon <T\wedge 1.\) Taking \(R=\frac{1}{\sqrt{\varepsilon }}\) and using (3.29) we have

$$\begin{aligned} \Vert H_t\Vert _1\le C\sqrt{\varepsilon }\,+2\int \limits _{x>\frac{1}{\sqrt{\varepsilon }}}x \mathrm{d}F_0(x) \le C\Psi _{F_0}(\varepsilon ), \quad \forall \, t\in [0, \varepsilon ]. \end{aligned}$$
(3.31)

In particular this inequality holds for \(t=\varepsilon \). Thus using (3.27) for \(s=\varepsilon \) gives

$$\begin{aligned} \Vert H_t\Vert _1\!\le \! C\Psi _{F_0}(\varepsilon )\!+\!C_0\int \limits _{\varepsilon }^{t}(1+1/\tau )\Vert H_\tau \Vert _{0}\mathrm{d}\tau \!+\! C\int \limits _{\varepsilon }^{t}\Vert H_\tau \Vert _1\mathrm{d}\tau , \quad t\in [\varepsilon , T\wedge 1]\qquad \end{aligned}$$
(3.32)

Next, using (3.26) and \(\Vert H_0\Vert _0\le \varepsilon \le t\le 1\),

$$\begin{aligned} \int \limits _{\varepsilon }^{t}(1+1/\tau )\Vert H_\tau \Vert _{0}\mathrm{d}\tau \!\!&\le \!\! 2\varepsilon \log (1/\varepsilon ) +2C\int \limits _{\varepsilon }^{t}\frac{1}{\tau }\,\int \limits _{0}^{\tau }\Vert H_u\Vert _1\mathrm{d}u \mathrm{d}\tau \nonumber \\ \!\!&\le \!\! 2\sqrt{\varepsilon }+2C\int \limits _{0}^{t} \Vert H_u\Vert _1|\log u| \mathrm{d}u,\quad t\in [\varepsilon , T\wedge 1]. \end{aligned}$$
(3.33)

This together with (3.32) and (3.31) gives

$$\begin{aligned} \Vert H_t\Vert _1\le C\Psi _{F_0}(\varepsilon )+C \int \limits _{0}^{t}(1+|\log \tau |)\Vert H_\tau \Vert _1\mathrm{d}\tau ,\quad t\in [0, T\wedge 1]. \end{aligned}$$
(3.34)

By Gronwall lemma we then obtain

$$\begin{aligned} \Vert H_t\Vert _1\le C\Psi _{F_0}(\varepsilon )\exp \Big (C\int \limits _{0}^{t}(1+|\log \tau |)\mathrm{d}\tau \Big ) \le C\Psi _{F_0}(\varepsilon ),\quad t\in [0, T\wedge 1]. \end{aligned}$$
(3.35)

Now if \(T\le 1\), then (3.25) follows from (3.35). Suppose \(T>1\). Then (3.35) holds for all \(t\in [0,1]\). In particular \(\Vert H_1\Vert _1 \le C\Psi _{F_0}(\varepsilon )\). On the other hand from (3.27) with \(s=1\) we have \(\Vert H_t\Vert _1\le \Vert H_1\Vert _1+ c\int \limits _{1}^{t}\Vert H_\tau \Vert _1\mathrm{d}\tau \) for all \(t\in [1,T]\) and so \(\Vert H_t\Vert _1\le \Vert H_1\Vert _1 e^{c(t-1)}\le C\Psi _{F_0}(\varepsilon )e^{ct}\) for all \(t\in [1,T]\) by Gronwall Lemma. This together with the estimate for \(t\in [0,1]\) leads to

$$\begin{aligned} \Vert H_t\Vert _1\le C\Psi _{F_0}(\varepsilon )e^{ct},\quad t\in [0,T]. \end{aligned}$$
(3.36)

Finally if \(\Vert H_0\Vert _1>0\) then taking \(\varepsilon =\Vert H_0\Vert _1\) in (3.36) gives (3.25). If \(\Vert H_0\Vert _1=0\), then in (3.36) letting \(\varepsilon \rightarrow 0^+\) we conclude \(\Vert H_t\Vert _1=0\) for all \(t\in [0, T]\) and thus (3.25) still holds true. This proves (3.25) for the case where \(F_t\) has the moment production (1.15) for all \(t\in (0, T_{F_{\cdot },\max })\).

Step 4: We prove that if any two conservative distributional solutions \(F_t, G_t\) stated in the theorem satisfy \(F_t=G_t\) for all \(t\in [0, T_{F_{\cdot },\max }\wedge T_{G_{\cdot },\max })\), then \(T_{F_{\cdot },\max }=T_{G_{\cdot },\max }\).

Suppose this is not true and let for instance \(t_0:=T_{F_{\cdot },\max }<T_{G_{\cdot },\max }\). Then \(t_0<\infty \) and, by Lemma 3.5, \(M_{-1/2}(F_{t_0})=\infty \) but \(M_{-1/2}(G_{t_0})<\infty \). Let \(M_{-1/2}^{(\varepsilon )}(F_t)\), \(M_{-1/2}^{(\varepsilon )}(G_t)\) be defined in (3.7). By the assumption \(F_t=G_t\) for all \(t\in [0, t_0)\) we have \(M_{-1/2}^{(\varepsilon )}(F_t)=M_{-1/2}^{(\varepsilon )}(G_t)\) for all \(t\in [0, t_0)\). Since \(t\mapsto M_{-1/2}^{(\varepsilon )}(F_t), t\mapsto M_{-1/2}^{(\varepsilon )}(G_t) \) are continuous on \([0, \infty )\), letting \(t\nearrow t_0\) gives \(M_{-1/2}^{(\varepsilon )}(F_{t_0})=M_{-1/2}^{(\varepsilon )}(G_{t_0})\), and then letting \(\varepsilon \rightarrow 0^+\) and using the limit (3.7) leads to \(M_{-1/2}(F_{t_0})= M_{-1/2}(G_{t_0})<\infty \) which contradicts \(M_{-1/2}(F_{t_0})=\infty \). Thus we must have \(T_{F_{\cdot },\max }=T_{G_{\cdot },\max }\).

Step 5: Let \(G_t\) be any conservative distributional solution of Eq. (1.5) on \([0,\infty )\) having the same initial datum \(G_0=F_0\) of \(F_t\) where \(F_t\) is used in Steps 1–3. Using (3.25) we conclude \(G_t=F_t\) for all \(t\in [0, T_{F_{\cdot },\max }\wedge T_{G_{\cdot },\max })\) and thus Step 4 insures that \(T_{F_{\cdot },\max }=T_{G_{\cdot },\max }:=T_{\max }\) hence \(G_t=F_t\) for all \(t\in [0, T_{\max })\). In particular \(G_t\) also has the moment production (1.15) for all \(t\in (0, T_{\max })\). This proves that the moment condition added on \(F_t\) in Steps 1–3 is indeed satisfied for every conservative distributional solution satisfying the condition in the theorem and thus the stability estimate (3.25) holds true for any two conservative distributional solutions \(F_t, G_t\) stated in the theorem.

Finally let \(F_t, G_t\) be given in the theorem and suppose \(F_0=G_0\). Then (3.25) implies that \(F_t=G_t\) for all \(t\in [0, T_{F_{\cdot },\max }\wedge T_{G_{\cdot },\max })\) and Step 4 insures that \(T_{F_{\cdot },\max }=T_{G_{\cdot },\max }:=T_{\max }\) hence \(F_t=G_t\) for all \(t\in [0,T_{\max }).\) This finishes the proof of the theorem. \(\square \)

Now we are going to prove the global existence of strong solutions for a class of initial data. We will use the Hölder inequality of moments for \(F\in \mathcal{B}^{+}({\mathbb R}_{\ge 0})\) (with \(F({\mathbb R}_{> 0})>0\))

$$\begin{aligned} M_r(F)\le [M_p(F)]^{\frac{q-r}{q-p}}[M_q(F)]^{\frac{r-p}{q-p}},\qquad -\infty <p<r<q<\infty \end{aligned}$$
(3.37)

and the following lemma:

Lemma 3.6

Let \(F_t\) be a conservative distributional solution of Eq. (1.5) on \([0, \infty )\) satisfying the moment production (1.15), and let \(N=N(F_0),E=E(F_0)\). Then for all \(t>0\)

$$\begin{aligned} M_{1/2}(F_t)\ge \frac{1}{6}\left[ 1-\exp \big (-(\sqrt{NE}+ N^3/E)t \big )\right] \frac{NE}{\sqrt{NE}+ N^3/E}. \end{aligned}$$
(3.38)

Proof

By conservation of energy and (3.37) we have \(E\le [M_{1/2}(F_t)]^{2/3}[M_2(F_t)]^{1/3}\), i.e.

$$\begin{aligned} M_{1/2}(F_t)\ge \frac{E^{3/2}}{\sqrt{M_2(F_t)}}\quad \forall \,t>0. \end{aligned}$$
(3.39)

Thus to prove (3.38) we need only to prove that \(M_2(F_t)\) has the corresponding upper bound. To do this we must use the differential equation

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}M_{2}(F_t)=\int \limits _{{\mathbb R}_{\ge 0}^2} \mathcal{J}[\varphi ]\mathrm{d}^2F_{t}+\int \limits _{{\mathbb R}_{\ge 0}^3} \mathcal{K}[\varphi ]\mathrm{d}^3F_{t} \qquad \forall \, t>0 \end{aligned}$$
(3.40)

where \(\varphi (x)=x^2\). Of course this \(\varphi \) does not belong to the test function space \(C^{1,1}({\mathbb R}_{\ge 0})\), but thanks to the moment production (1.15) we are able to prove that (3.40) does hold rigourously. First, from the moment production (1.15) we have \(\sup _{t\ge s}\Vert F_t\Vert _p<\infty \) for all \(s>0\) and \(p\ge 1\) and thus using Corollary of Lemma 4 in [13] to \(\{F_t\}_{t\ge s}\) we conclude that the collision integrals \(t\mapsto \int \nolimits _{{\mathbb R}_{\ge 0}^2} \mathcal{J}[\varphi ]\mathrm{d}^2F_{t},\, t\mapsto \int \nolimits _{{\mathbb R}_{\ge 0}^3} \mathcal{K}[\varphi ]\mathrm{d}^3F_{t} \) are continuous on \([s, \infty )\,(\forall \, s>0)\). Also we note that the measure \(t\mapsto F_{t+s}\) is a distributional solution of Eq. (1.5) on \([0,\infty )\) with the initial datum \(F_s\) and thus it satisfies the integral equation (1.9). Next in order to use our test functions in \(C^{1,1}_b({\mathbb R}_{\ge 0})\) we consider smooth truncations \(\varphi _{n}(x)=x^2e^{-x/n}\). We compute \(\sup _{n\ge 1,x\ge 0}|D^2\varphi _{n}(x)|\le 4\) and, using (3.6),

$$\begin{aligned}&\sup _{n\ge 1}|\mathcal{K}[\varphi _{n}](x,y,z)|\le 4(1+y)(1+z),\quad \sup _{n\ge 1} |\mathcal{J}[\varphi _{n}](y,z)|\le 2(1+y)^{5/2}(1+z)^{5/2},\\&\lim _{n\rightarrow \infty }\mathcal{K}[\varphi _{n}](x,y,z)=\mathcal{K}[\varphi ](x,y,z), \,\,\quad \quad \lim _{n\rightarrow \infty }\mathcal{J}[\varphi _{n}](y,z)=\mathcal{J}[\varphi ](y,z) \end{aligned}$$

for all \(x,y,z\ge 0.\) Since \(\sup _{t\ge s}\Vert F_t\Vert _{5/2}<\infty \), it follows from dominated convergence theorem that the integral equation (1.9) of the solution \(t\mapsto F_{t+s}\) holds also for the function \(\varphi (x)=x^2\). Since \(s>0\) is arbitrary, this proves that the function \(t\mapsto M_2(F_t)\) belongs to \(C^1((0, \infty ))\) and satisfies the differential equation (3.40).

Now let us compute \(\mathcal{J}[\varphi ](y,z)\). By definition of \(\mathcal{J}[\varphi ]\) we have \(\mathcal{J}[\varphi ](y,z)=\mathcal{J}[\varphi ](z,y)\) and \(\mathcal{J}[\varphi ](0,0)=0\). Suppose for instance \(0\le y\le z\) and \(z>0\). By considering \(\mathcal{J}^{+}[\varphi ](y,z)=\int \nolimits _{0}^{y} +\int \nolimits _{y}^{z}+\int \nolimits _{z}^{y+z}\) we compute

$$\begin{aligned} \mathcal{J}[\varphi ](y,z)=\left( \frac{1}{2}+\frac{8}{15}\cdot \frac{y}{z}- \frac{13}{210}\Big (\frac{y}{z}\Big )^2\right) y{z}^{3/2}+\frac{{z}^{5/2}}{3}-\frac{1}{2}(y^2+{z}^2)\sqrt{z}. \end{aligned}$$

Since \(\frac{1}{2}+\frac{8}{15}\theta - \frac{13}{210}\theta ^2\le \frac{102}{105}<1\) for all \(0\le \theta \le 1\), it follows that

$$\begin{aligned} \mathcal{J}[\varphi ](y,z)\le y{z}^{3/2}\vee {y}^{3/2}z +\frac{1}{3}({y}^{5/2}\vee {z}^{5/2})-\frac{1}{2}({y}^2+{z}^2)\sqrt{y\vee z}\qquad \forall \,y,z\ge 0. \end{aligned}$$

From this we obtain

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}^2} \mathcal{J}[\varphi ]\mathrm{d}^2F_{t} \le 2EM_{3/2}(F_t)-\frac{1}{3}NM_{5/2}(F_t). \end{aligned}$$

Using the Hölder inequality (3.37) and the conservation of mass and energy \(M_0(F_t)= N, M_1(F_t)=E\) we have \(E\le \sqrt{N}\sqrt{M_2(F_t)}, M_{3/2}(F_t)\le \sqrt{E}\sqrt{M_2(F_t)}\) and \(M_{5/2}(F_t)\ge [M_2(F_t)]^{3/2}/\sqrt{E}. \) Thus

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}^2} \mathcal{J}[\varphi ]\mathrm{d}^2F_{t} \le 2\sqrt{NE}M_2(F_t)-\frac{N}{3\sqrt{E}}[M_{2}(F_t)]^{3/2}. \end{aligned}$$
(3.41)

Next we estimate the cubic term. Recalling that \(W(x,y,z)>0\) implies \(x_*=y+z-x>0\) and so \(\Delta \varphi (x,y,z)={x}^2+x_*^2-{y}^2-{z}^2 =2(yz-xx_*)\), it follows that

$$\begin{aligned} \mathcal{K}[\varphi ](x,y,z)= 2W(x,y,z)(yz-xx_*)\le 2W(x,y,z)yz\le 2\sqrt{yz} \end{aligned}$$

for all \(x,y,z\ge 0\) hence

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}^3} \mathcal{K}[\varphi ]\mathrm{d}^3F_{t} \le 2N[M_{1/2}(F_t)]^2\le 2N^2E\le 2\frac{N^3}{E}M_{2}(F_t). \end{aligned}$$
(3.42)

Here we used \(M_{1/2}(F_t)^2\le NE\) and \(E^2\le N M_2(F_t).\) Combining (3.41) and (3.42) with (3.40) we obtain

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}M_{2}(F_t)\le A M_{2}(F_t) -B [M_{2}(F_t)]^{3/2},\quad t>0 \end{aligned}$$
(3.43)

where

$$\begin{aligned} A=2\Big (\sqrt{NE}+ \frac{N^3}{E}\Big ),\quad B=\frac{N}{3\sqrt{E}}. \end{aligned}$$
(3.44)

By solving the differential inequality (3.43) we conclude

$$\begin{aligned} M_{2}(F_t)\le (A/B)^2(1-e^{-\frac{1}{2}A t})^{-2},\quad t>0. \end{aligned}$$
(3.45)

Therefore the inequality (3.38) follows from (3.39), (3.45) and (3.44). \(\square \)

Theorem 3.2

Let \(F_0\in \mathcal{B}^{+}_{-1/2,1}({\mathbb R}_{\ge 0})\) and suppose

$$\begin{aligned} M_{-1/2}(F_0)\le \frac{1}{80}[N(F_0)E(F_0)]^{1/4}. \end{aligned}$$
(3.46)

Then there exists a unique conservative distributional solution \(F_t\) of Eq. (1.5) on \([0, \infty )\) with the initial datum \(F_0\). Moreover \(F_t\) is also a strong solution on \([0,\infty )\) and satisfies

$$\begin{aligned} \sup _{t\ge 0}M_{-1/2}(F_t)\le \frac{1}{4}[N(F_0)E(F_0)]^{1/4}. \end{aligned}$$
(3.47)

In particular \(F_t\) has no condensation for all \(t\in [0, \infty )\).

Proof

Denote \(N=N(F_0), E=E(F_0)\) and let \(F_t\) be a conservative distributional solution of Eq. (1.5) on \([0, \infty )\) obtained by Theorem 1.1 with the initial datum \(F_0\), in particular \(F_t\) has the moment production (1.15). Then, by Lemma 3.6, \(F_t\) satisfies (3.38). To prove the theorem, we need only to prove that \(T_{F_{\cdot },\max }=\infty \) and \(F_t\) satisfies (3.47). In fact if \(T_{F_{\cdot },\max }=\infty \) holds true, then we conclude from Proposition 3.1 and Theorem 3.1 that this \(F_t\) is a strong solution on \([0,\infty )\) and \(F_t\) is the unique one in the class of conservative distributional solutions of Eq. (1.5) on \([0,\infty )\) having the same initial datum \(F_0\).

Let us introduce two numbers which will play important roles:

$$\begin{aligned} \beta =\frac{NE}{[M_{-1/2}(F_0)]^4},\quad \gamma =\frac{E}{N^{5/3}}. \end{aligned}$$
(3.48)

Using Hölder inequality (3.37) we have \(N\le (M_{-1/2}(F_0))^{2/3} E^{1/3}\) i.e.

$$\begin{aligned} M_{-1/2}(F_0)\ge \sqrt{N^3/E}. \end{aligned}$$
(3.49)

From (3.48), (3.49) we deduce

$$\begin{aligned} (\gamma /\beta )^{1/4}=N^{-2/3}M_{-1/2}(F_0),\quad \gamma \ge \beta ^{1/3}. \end{aligned}$$

Then, by Lemma 3.4,

$$\begin{aligned} M_{-1/2}(F_t)\le N^{2/3}\left( \frac{1+(\gamma /\beta )^{1/4}}{\sqrt{1-t/T_{F_0}}}-1\right) \qquad \forall \, t\in [0,T_{F_0}) \end{aligned}$$
(3.50)

where \(T_{F_0}\) is given in (3.12) which can be also rewritten in terms of \((\gamma /\beta )^{1/4}\) as

$$\begin{aligned} T_{F_0}=\frac{1}{ 2N^{4/3}(1+(\gamma /\beta )^{1/4})^2}. \end{aligned}$$

According to the differential inequality (3.9), in order to obtain an upper bound of \(M_{-1/2}(F_t)\) on \([t_0,\infty )\) for some \(0<t_0<T_{F_0}\), we need to have a lower bound for \(M_{1/2}(F_t)\) on \([t_0, \infty )\). The condition (3.46) which now reads \(\beta \ge (80)^4\) is just designed for this popups. Numerical computation suggests that a good choice for \(t_0\) is given by

$$\begin{aligned} 1-\exp (-(\sqrt{NE}+ N^3/E)t_0 ) =\frac{36}{37}\quad \mathrm{i.e.}\quad t_0=\frac{\log (37)}{\sqrt{NE}+ N^3/E}. \end{aligned}$$
(3.51)

We need to prove that this \(t_0\) satisfies

$$\begin{aligned} t_0\le \frac{1}{2}T_{F_0}\quad \mathrm{i.e.}\quad 4\log (37)\le \frac{\sqrt{NE}+ N^3/E}{N^{4/3} (1+(\gamma /\beta )^{1/4})^{2}}. \end{aligned}$$
(3.52)

To prove (3.52) we use \(\gamma =E/N^{5/3}\) to write

$$\begin{aligned} \sqrt{NE}+ N^3/E=N^{4/3}(\sqrt{\gamma }+\gamma ^{-1}) \end{aligned}$$
(3.53)

so that the inequality (3.52) is equivalent to

$$\begin{aligned} 4\log (37)\le \frac{\sqrt{\gamma }+\gamma ^{-1}}{(1+(\gamma /\beta )^{1/4})^{2}}. \end{aligned}$$
(3.54)

Using \(\gamma \ge \beta ^{1/3}\) and omitting \(\gamma ^{-1}\) we see that the right hand side of (3.54) is larger than \(\left( \gamma ^{1/4}/(1+(\gamma /\beta )^{1/4})\right) ^{2}\ge \beta ^{1/6}\left( 1+\beta ^{-1/6}\right) ^{-2}>4\log (37)\) where the last inequality is because \(\beta ^{1/6}\ge (80)^{2/3}\). Thus (3.54) holds true. By the way, from \(\gamma \ge \beta ^{1/3}\) we also have

$$\begin{aligned} \frac{E}{N^{5/3}}=\gamma \ge (80)^{4/3}. \end{aligned}$$
(3.55)

Now applying (3.38) with (3.51), (3.53) and \(E=N^{5/3}\gamma \) we obtain

$$\begin{aligned} M_{1/2}(F_t)\ge \frac{6}{37}\cdot \frac{NE}{\sqrt{NE}+ N^3/E}=3M_*^2\qquad \forall \, t\ge t_0 \end{aligned}$$
(3.56)

where

$$\begin{aligned} M_*=N^{2/3}\left( \frac{2\sqrt{\gamma }}{37(1+\gamma ^{-3/2})}\right) ^{1/2}. \end{aligned}$$
(3.57)

Inserting (3.56) into the differential inequality (3.9) leads to

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}M^{(\varepsilon )}_{-1/2}(F_t)\le 2N^2-3M_*^2 M^{(\varepsilon )}_{-1/2}(F_t)+ [M^{(\varepsilon )}_{-1/2}(F_t)]^3\qquad \forall \, t\ge t_0 \end{aligned}$$
(3.58)

from which we see that in order to get a global upper bound of \((t,\varepsilon )\mapsto M^{(\varepsilon )}_{-1/2}(F_t)\), it needs only to prove the following inequalities (the reason will be clear later):

$$\begin{aligned} N^2\le M_*^3,\quad M_{-1/2}(F_t)\le M_*\quad \forall \, t\in [0, t_0]. \end{aligned}$$
(3.59)

The first inequality in (3.59) is equivalent to \(\frac{\sqrt{\gamma } }{1+\gamma ^{-3/2}}\ge 37/2\) which is exactly satisfied since \(\gamma \ge (80)^{4/3}\). To prove the second one in (3.59) we first use (3.50) and \(t_0\le \frac{1}{2}T_{F_0}\) to get

$$\begin{aligned} M_{-1/2}(F_{t})\le N^{2/3}\Big (\sqrt{2}\,(1+(\gamma /\beta )^{1/4}) -1\Big )\qquad \forall \, t\in [0, t_0]. \end{aligned}$$

Then, by definition of \(M_*\) and \(\gamma ^{-3/2}\le (80)^{-2}\), we see that a sufficient condition for the second inequality in (3.59) to hold is

$$\begin{aligned} \sqrt{2}\,(1+(\gamma /\beta )^{1/4}) -1\le \Big (\frac{2\sqrt{\gamma }}{37(1+(80)^{-2})}\Big )^{1/2}. \end{aligned}$$

But this inequality does hold true because \(\beta ^{1/4}\ge 80\) and \(\gamma ^{1/4}\ge (80)^{1/3}\).

Now from (3.58), \(N^2\le M_*^3\), \(2M_*^3-3M_*^2M+M^3=(M-M_*)^2(M+2M_*)\), and denoting \(M_{\varepsilon }(t)=M^{(\varepsilon )}_{-1/2}(F_t)\) we obtain

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}M_{\varepsilon }(t)\le (M_{\varepsilon }(t)-M_*)^2(M_{\varepsilon }(t)+2M_*)\qquad \forall \, t\ge t_0. \end{aligned}$$

Since \(M_{\varepsilon }(t_0)\le M_{-1/2}(F_{t_0})\le M_*\), it follows that

$$\begin{aligned} (M_{\varepsilon }(t)-M_*)_{+}\le \int \limits _{t_0}^{t} [(M_{\varepsilon }(\tau )-M_*)_{+}]^2(M_{\varepsilon }(\tau )+2M_*)\mathrm{d}\tau ,\quad t\in [t_0,\infty ). \end{aligned}$$

Thus we conclude from Gronwall Lemma that \((M_{\varepsilon }(t)-M_*)_{+} \equiv 0\), i.e. \(M^{(\varepsilon )}_{-1/2}(F_t)=M_{\varepsilon }(t)\le M_*\) for all \(t\ge t_0\) and all \(\varepsilon >0\). Letting \(\varepsilon \rightarrow 0^+\) leads to \(M_{-1/2}(F_t)\le M_*\) for all \(t\ge t_0\). Combining this with the second inequality in (3.59) we conclude that

$$\begin{aligned} M_{-1/2}(F_{t})\le M_*\qquad \forall \, t\in [0,\infty ). \end{aligned}$$
(3.60)

This implies that \(T_{F_{\cdot },\max }=\infty \) by definition of \(T_{F_{\cdot },\max }\). Finally since \(N^{2/3}\gamma ^{1/4}=(NE)^{1/4}\), it follows from (3.57) that \(M_*\le (NE)^{1/4}\sqrt{2/37}\le \frac{1}{4}(NE)^{1/4}\) and thus we obtain (3.47) from (3.60). This completes the proof. \(\square \)

There are many \(F_0\) that satisfy the condition (3.46). For instance for any \(G_0\in \mathcal{B}_{-1/2,1}^{+}({\mathbb R}_{\ge 0})\) with \(M_{-1/2}(G_0)>0\), define \(F_0=\rho G_0\) with a constant \(\rho >0\). Then

$$\begin{aligned} \frac{M_{-1/2}(F_0)}{[N(F_0)E(F_0)]^{1/4}} =\sqrt{\rho }\, \frac{M_{-1/2}(G_0)}{[N(G_0)E(G_0)]^{1/4}} \end{aligned}$$
(3.61)

and so \(F_0\) satisfies (3.46) when \(\rho \) is small enough.

We have proved that the condition (3.46) implies (3.55) i.e. \(\frac{E}{N^{5/3}}\ge (80)^{4/3}\), from which and (1.14) one sees that the condition (3.46) belongs to the case of high temperature: \(\overline{T}/\overline{T}_c>783.\)

4 Regularity and Mild Solutions

In this section we use the above results to study regularity of distributional solutions and prove the existence and stability of mild solutions. Without risk of confusion we use short notations for the norms of \(L^1({\mathbb R})_{+}\) and \(L^{\infty }({\mathbb R})_{+}\):

$$\begin{aligned} \Vert f\Vert _{L^1}\equiv \Vert f\Vert _{L^1({\mathbb R}_{+})},\quad \Vert f\Vert _{L^{\infty }}\equiv \Vert f\Vert _{L^{\infty }({\mathbb R}_{+})}. \end{aligned}$$

As usual we denote \(f(t)=f(\cdot ,t)\) when the \(x\)-variable has been taken certain integration or norm, for instance \(N(f(t))=N(f(\cdot ,t)), \Vert f(t)\Vert _{L^1}=\Vert f(\cdot , t)\Vert _{L^1}\), etc. From Lemma 2.2 and the following propositions one will see that just as \(M_{-1/2}(F_t)\) plays the important role in the existence of strong solutions, \(\Vert f(t)\Vert _{L^1}=M_{-1/2}(f(t))\) also controls everything for the existence and stability of local and global (bounded or unbounded) mild solutions.

Proposition 4.1

Let \(F_t\) be a distributional solution of Eq. (1.5) on \([0, \infty )\) whose initial datum \(F_0\) is regular and satisfies \(M_{-1/2}(F_0)<\infty \), and let \(T_{F_{\cdot },\max }\) be defined in (3.11). Then \(F_t\) is regular for all \(t\in [0, T_{F_{\cdot },\max })\) and its density \(f(\cdot ,t)\) is a mild solution of Eq. (1.5) on \([0, T_{F_{\cdot },\max })\) satisfying \(f\in C([0, T_{F_{\cdot },\max }); L^1({\mathbb R}_{+}))\) and \(f(\cdot ,0)=f_0\), where \(f_0\) is the density of \(F_0\). In particular if \(F_t\) is conservative, so is \(f(\cdot ,t)\) on \([0, T_{F_{\cdot },\max })\).

Proof

Denote \(T_{\max }=T_{F_{\cdot },\max }\). By Proposition 3.1, \(F_t\) is a strong distributional solution on \([0, T_{\max })\), and from the relation (1.11) we have \(F_t(\{0\})=0\) for all \(t\in [0, T_{\max })\), which means that the origin \(x=0\) has no contribution with respect to the measure \(F_t\) and thus the integration domain \({\mathbb R}_{\ge 0}\) can be replaced by \({\mathbb R}_{+}={\mathbb R}_{>0}\). Let

$$\begin{aligned} \nu _t(E)=\int \limits _{E}\frac{1}{\sqrt{x}}\mathrm{d}F_t(x),\quad V_t(\delta )=\sup _{\mathrm{mes}(U)<\delta }\nu _t(U),\quad 0\le t<T_{\max } \end{aligned}$$

where \(E\subset {\mathbb R}_{+}\) is any Borel set, \(U\) is chosen from all open sets in \({\mathbb R}_{+}\), and \(\mathrm{mes}(\cdot )\) denotes the Lebeague measure. We are going to establish Gronwall inequality for \(V_t(\delta )\) on \(t\in [0, T_{\max })\).

Given any \(T\in (0,T_{\max })\) and take any open set \(U\subset {\mathbb R}_{+}\) satisfying \(\mathrm{mes}(U)<\delta \). Applying the integral equation (3.3) to a monotone sequence \(0\le \varphi _n\in C_b({\mathbb R}_{\ge 0})\) satisfying

$$\begin{aligned} \varphi _n(x) \nearrow \psi _U(x):=\frac{1}{\sqrt{x}}\mathbf{1}_{U}(x)\quad (n\rightarrow \infty )\quad \forall \,x\in {\mathbb R}_{+} \end{aligned}$$

(for instance one may take \( \varphi _n(x)=(x+\frac{1}{n})^{-1/2}(1-\exp (-n\mathrm{dist}(x,U^c)))\) where \(U^c={\mathbb R}{\setminus } U\)) and then omitting negative parts, we deduce from monotone convergence that

$$\begin{aligned} \nu _t(U) \le \nu _0(U) +\int \limits _{0}^{t}\mathrm{d}\tau \int \limits _{{\mathbb R}_{+}^2}\mathcal{J}^{+} [\psi _U]\mathrm{d}^2F_\tau + \int \limits _{0}^{t}\mathrm{d}\tau \int \limits _{{\mathbb R}_{+}^3}\mathcal{K}^{+}[\psi _U]\mathrm{d}^3F_\tau . \end{aligned}$$

Next we compute for all \(x,y,z>0\)

$$\begin{aligned}&\mathcal{J}^{+}[\psi _U](y,z) \le \frac{1}{\sqrt{yz}}\mathrm{mes}(U)\le \frac{1}{\sqrt{yz}}\delta \,,\\&\mathcal{K}^{+}[\psi _U](x,y,z)\le \frac{1}{\sqrt{xyz}}[ \mathbf{1}_{U}(x)+ \mathbf{1}_{U}(y+z-x) ] . \end{aligned}$$

Since \({\mathbb R}_{+}\cap (U+x-z)\) is open and \(\mathrm{mes}({\mathbb R}_{+}\cap (U+x-z))\le \mathrm{mes}(U)<\delta \), this gives

$$\begin{aligned} \int \limits _{{\mathbb R}_{+}} \mathbf{1}_{U}(y+z-x)\frac{1}{\sqrt{y}} \mathrm{d}F_\tau (y) =\int \limits _{{\mathbb R}_{+}\cap (U+x-z)}\frac{1}{\sqrt{y}} \mathrm{d}F_\tau (y) \le V_\tau (\delta ). \end{aligned}$$

It follows that

$$\begin{aligned} \nu _t(U) \le V_0(\delta )+\int \limits _{0}^{t}[M_{-1/2}(F_\tau )]^2\mathrm{d}\tau \,\delta + 2\int \limits _{0}^{t}[M_{-1/2}(F_\tau )]^2V_\tau (\delta )\mathrm{d}\tau . \end{aligned}$$

Taking \(\sup _{\mathrm{mes}(U)<\delta }\) leads to

$$\begin{aligned} V_t(\delta )\le V_0(\delta )+2\int \limits _{0}^{t}[M_{-1/2}(F_\tau )]^2\Big (\frac{\delta }{2}+ V_\tau (\delta )\Big )\mathrm{d}\tau ,\quad 0\le t\le T \end{aligned}$$

and so, by Gronwall Lemma,

$$\begin{aligned} \frac{\delta }{2}+V_t(\delta )\le \Big (\frac{\delta }{2}+V_0(\delta ) \Big )\exp \Big (2\int \limits _{0}^{t}[M_{-1/2}(F_\tau )]^2\mathrm{d}\tau \Big ),\quad 0\le t\le T. \end{aligned}$$

By assumption on \(F_0\) we have \(\lim _{\delta \rightarrow 0^+}V_0(\delta )=0\) and thus \(\lim _{\delta \rightarrow 0^+}V_t(\delta )=0\) for all \(t\in [0,T].\) Since \(T\in (0,T_{\max })\) is arbitrary, this proves that \(\nu _t\) is absolutely continuous with respect to the Lebesgue measure for every \(t\in [0,T_{\max })\), and thus there is a unique \(0\le f(\cdot ,t)\in L^1({\mathbb R}_{\ge 0})\) such that \(\frac{1}{\sqrt{x}}\mathrm{d}F_t(x)=\mathrm{d}\nu _t(x)=f(x,t)\mathrm{d}x.\) That is, we have proved that \(F_t\) is regular for all \(t\in [0, T_{\max })\) and its density \(f(\cdot ,t)\) belongs to \(L^1({\mathbb R}_{+})\) for all \(t\in [0, T_{\max })\).

Since \(\Vert f(t)\Vert _{L^1}= M_{-1/2}(F_t)<\infty \) for all \(t\in [0,T_{\max })\), it follows from (2.26) and (3.13) that

$$\begin{aligned} \sup _{0\le t\le T}&\int \limits _{{\mathbb R}_{+}^3}w(x,y,z)[ f'f_*'(1+f+f_*))+ ff_*(1+f'+f_*')]\mathrm{d}x\mathrm{d}y\mathrm{d}z\nonumber \\&\qquad \le 4\sup _{0\le t\le T}\Big ( M_{1/2}(f(t))\Vert f(t)\Vert _{L^1}+\Vert f(t)\Vert _{L^1}^3\Big )<\infty \quad \forall \, T\in (0,T_{\max }).\qquad \end{aligned}$$
(4.1)

From this and that \(F_t\) is a strong solution of Eq. (1.5) on \([0, T_{\max })\) we conclude that the equation

$$\begin{aligned}&\int \limits _{{\mathbb R}_{\ge 0}}\varphi (x)\Big ( f(x,t)-f_0(x)-\int \limits _{0}^{t}Q(f)(x,\tau )\mathrm{d}\tau \Big )\sqrt{x}\, \mathrm{d}x\\&\quad =\int \limits _{{\mathbb R}_{\ge 0}}\varphi (x)\mathrm{d}\Big (F_t -F_0 -\int \limits _{0}^{t}\mathcal{Q}(F_\tau )\mathrm{d}\tau \Big )\mathrm{d}x=0 \end{aligned}$$

holds for all \(t\in [0, T_{\max })\) and all \(\varphi \in L^{\infty }_0({\mathbb R}_{\ge 0})\). Thus for any \(t\in [0, T_{\max })\), there is a null set \(Z_t\subset {\mathbb R}_{\ge 0}\) such that

$$\begin{aligned} f(x,t)=f_0(x)+\int \limits _{0}^{t}Q(f)(x,\tau )\mathrm{d}\tau \qquad \forall \, x\in {\mathbb R}_{\ge 0}{\setminus } Z_t. \end{aligned}$$

Now we consider the nonnegative measurable function \(\widetilde{f}(x,t):=|f_0(x)+\int \limits _{0}^{t}Q(f)(x,\tau )\mathrm{d}\tau |\) on \({\mathbb R}_{+}\times [0, T_{\max })\). We have \(\widetilde{f}(x,0)\equiv f_0(x)\) and, by nonnegativity of \(f\), \(\widetilde{f}(x,t)=f(x,t)\) for all \(t\in [0, T_{\max })\) and all \(x\in {\mathbb R}_{+}{\setminus } Z_t\). But the advantage of \(\widetilde{f}\) is that there is a null set \(Z\) which is independent of \(t\) such that for every \(x\in {\mathbb R}_{+}{\setminus } Z\) the function \(t\mapsto \widetilde{f}(x,t)\) is continuous on \([0, T_{\max })\). Thus it follows from Fubini theorem that \(\widetilde{f}(\cdot ,t)\) is a mild solution of Eq. (1.5) on \([0, T_{\max })\). Again, since \(\widetilde{f}(x,t)=f(x,t)\) for all \(t\in [0, T_{\max })\) and all \(x\in {\mathbb R}_{+} {\setminus } Z_t\), it follows that \(\widetilde{f}(\cdot ,t)\) is also the same density of \(F_t\) for \(t\in [0, T_{\max })\). Thus by rewriting \(\widetilde{f}(\cdot , t)\) as \(f(\cdot ,t)\) we conclude that the density \(f(\cdot ,t)\) of \(F_t\) is a mild solution of Eq. (1.5) on \([0, T_{F_{\cdot },\max })\).

Finally for any \(T\in (0, T_{\max })\), let \(C_T\) be the left hand side of (4.1). Then we deduce from (4.1) and the definition of mild solutions that \(\Vert f(\tau )-f(t)\Vert _{L^1}\le 2C_T |s-t|\) for all \(s, t\in [0, T]\). Therefore \(f\in C([0, T_{F_{\cdot },\max }); L^1({\mathbb R}_{+}))\). \(\square \)

Proposition 4.2

Let \(0\le f_0\in L^1({\mathbb R}_{+})\) have finite mass and energy and let \(\mathcal{T}(f_0)= \{0<T\le \infty \,|\) Eq. (1.5) has a conservative mild solution \(f(\cdot ,t)\) on \([0, T)\) satisfying \(f\in C([0, T); L^1({\mathbb R}_{+}))\) and \(f(\cdot ,0)=f_0\}, T_{f_0,\max }=\sup \mathcal{T}(f_0)\). Then

  1. (a)

    \(\mathcal{T}(f_0)\) is non-empty, \(T_{f_0,\max }\in \mathcal{T}(f_0)\), and there exists a unique conservative mild solution \(f(\cdot ,t)\) of Eq.(1.5) on \([0, T_{f_0,\max })\) satisfying \(f\in C([0, T_{f_0,\max }); L^1({\mathbb R}_{+})) \) and \(f(\cdot ,0)=f_0\). Besides, if \(T_{f_0,\max }<\infty \) then \(\Vert f(t)\Vert _{L^1}\rightarrow \infty \) as \( t \nearrow T_{f_0,\max }\).

  2. (b)

    Let \(F_0\in \mathcal{B}_{-1/2, 1}^{+}({\mathbb R}_{\ge 0})\) be regular with the density \(f_0\). Then there exists a conservative distributional solution \(F_t\) of Eq. (1.5) on \([0,\infty )\) such that \(T_{F_{\cdot },\max } =T_{f_0,\max }\) and \(F_t\) is regular on \([0, T_{f_0,\max })\) with the density \(f(\cdot , t)\) obtained in part (a). And such an \(F_t\) is unique on \([0, T_{f_0,\max })\).

  3. (c)

    Let \(0\le g_0\in L^1({\mathbb R}_{+})\) have finite mass and energy and let \(g(\cdot ,t)\) be conservative mild solutions of Eq. (1.5) on \([0, T_{g_0,\max })\) satisfying \(g\in C([0, T_{g_0,\max }); L^1({\mathbb R}_{+})) \) and \(g(0,\cdot )=g_0\). Then for any \(T\in (0, T_{f_0,\max }\wedge T_{g_0,\max })\)

    $$\begin{aligned} \Vert f(t)-g(t)\Vert _{1}\le C\Psi _{f_0}(\Vert f_0-g_0\Vert _{1})e^{c t},\quad \forall \, t\in [0, T] \end{aligned}$$
    (4.2)

    where

    $$\begin{aligned} \Vert h\Vert _{1}:=\int \limits _{{\mathbb R}_{+}}(1+x)|h(x)| \sqrt{x}\mathrm{d}x, \end{aligned}$$

    \(\Psi _{f_0}(\varepsilon ):=\Psi _{F_0}(\varepsilon )\) is defined by (3.24) with \(F_0\) defined by \(\mathrm{d}F_0(x)=f_0(x)\sqrt{x}\mathrm{d}x\), and \(C=C_T, c=c_T\) are finite positive constants depending only on \(N(f_0), E(f_0), \sup _{0\le t\le T} \Vert f(t)\Vert _{L^1},\) and \(\sup _{0\le t\le T}\Vert g(t)\Vert _{L^1}\).

  4. (d)

    Suppose \(T_{f_0,\max }=\infty \) and let \(f(\cdot ,t)\) be the unique conservative mild solution of Eq. (1.5) on \([0, \infty )\) satisfying \(f\in C([0,\infty ); L^1({\mathbb R}_{+})) \) and \(f(\cdot ,0)=f_0\) obtained in part (a). Let \(F_0, F_t\in \mathcal{B}_{1}^{+}({\mathbb R}_{\ge 0})\) be regular measures with densities \(f_0, f(\cdot ,t)\) respectively. Then \(F_t\) is the unique conservative strong solution of Eq. (1.5) on \([0, \infty )\) with the initial datum \(F_0\).

Proof

(a)–(b): Let \(F_0\in \mathcal{B}_{-1/2, 1}^{+}({\mathbb R}_{\ge 0})\) be regular with the density \(f_0\), let \(F_t\) be a conservative distributional solution of Eq. (1.5) on \([0,\infty )\) with the initial datum \(F_0\), and let \(T_{F_{\cdot },\max }\) be defined in (3.11). By Proposition 4.1, \(F_t\) is regular on \([0, T_{F_{\cdot },\max })\) and its density \(f(\cdot ,t)\) is a conservative mild solution of Eq. (1.5) on \([0, T_{F_{\cdot },\max })\) satisfying \(f\in C([0, T_{F_{\cdot },\max }); L^1({\mathbb R}_{+}))\) and \(f(\cdot ,0)=f_0\). This implies \(T_{F_{\cdot },\max }\in \mathcal{T}(f_0)\). We claim \(T_{F_{\cdot },\max }=T_{f_0,\max }\). Otherwise, \(T_{F_{\cdot },\max }<T_{f_0,\max }\), then there exists \(T^*\in \mathcal{T}(f_0)\) such that \(T^* >T_{F_{\cdot },\max }\) and Eq. (1.5) has a conservative mild solution \(f^*(\cdot ,t)\) of Eq. (1.5) on \([0, T^*)\) satisfying \(f^*(\cdot ,0)=f_0\) and \(f^*\in C([0, T^*); L^1({\mathbb R}_{+}))\). Take \(t_0\in (T_{F_{\cdot },\max },\, T^*\)) and let \(F^*_t\) be defined by \(\mathrm{d}F^*_t(x)=f^*(x,t)\sqrt{x}\mathrm{d}x,\,t\in [0, t_0]\). Applying Theorem 1.1 to the initial datum \(F^*_{t_0}\), there exists a conservative distributional solution \(F^{**}_t\) of Eq. (1.5) on \([t_0,\infty )\) with the initial datum \(F^{**}_{t_0}=F^*_{t_0}\). Let \(\widetilde{F}_t=F^{*}_t\) for \(t\in [0, t_0],\widetilde{F}_t=F^{**}_t\) for \(t\in [t_0, \infty )\). Then \(\widetilde{F}_t\) satisfies (i), (ii)\('\), (iii)\('\) in the equivalent definition of distributional solutions Eq. (1.5) proved in Remark 1.1. So \(\widetilde{F}_t\) is a conservative distributional solution of Eq. (1.5) on \([0,\infty )\) with the initial datum \(\widetilde{F}_0=F_0\). By Theorem 3.1 we conclude \(T_{\widetilde{F}_{\cdot },\max }=T_{F_{\cdot },\max }\). On the other hand, from \(M_{-1/2}(\widetilde{F}_{t_0})=M_{-1/2}(F^*_{t_0})=\Vert f^*(t_0)\Vert _{L^1}<\infty \) we have \(T_{\widetilde{F}_{\cdot },\max }\ge t_0>T_{F_{\cdot },\max }\). This contradiction proves \(T_{F_{\cdot },\max }=T_{f_0,\max }\).

Next let \(\tilde{f}(\cdot ,t)\) be another conservative mild solution of Eq. (1.5) on \([0, T_{f_0,\max })\) satisfying \(\tilde{f}(\cdot ,0)=f_0\) and \(\tilde{f}\in C([0, T_{f_0,\max }); L^1({\mathbb R}_{+})) \). Take any \(T\in (0, T_{f_0,\max })\) and let \(F^{(T)}_t\) be defined by \(\mathrm{d}F^{(T)}_t(x)=\tilde{f}(x,t)\sqrt{x}\mathrm{d}x,\,t\in [0, T]\). As shown above, \(F^{(T)}_t\) can be extended as a conservative distributional solution \(\widetilde{F}_t\) of Eq. (1.5) on \([0,\infty )\) with the initial datum \(\widetilde{F}_0=F_0\). By Theorem 3.1 we conclude \(T_{\widetilde{F}_{\cdot },\max }=T_{F_{\cdot },\max }=T_{f_0,\max }\) and \(\widetilde{F}_t=F_t\) for all \(t\in [0, T_{f_0,\max })\). In particular, \(F_t^{(T)}=\widetilde{F}_t=F_t\) for all \(t\in [0, T]\) and so \(\tilde{f}(\cdot ,t)=f(\cdot ,t)\) for all \(t\in [0, T]\). Since \(T\) is arbitrary in \((0, T_{f_0,\max })\), we conclude \(\tilde{f}(\cdot ,t)=f(\cdot ,t)\) for all \(t\in [0, T_{f_0,\max })\). Thus \(f(\cdot ,t)\) is the unique conservative mild solution \(f(\cdot ,t)\) of Eq. (1.5) on \([0, T_{f_0,\max })\) satisfying \(f(\cdot ,0)=f_0\) and \(f\in C([0, T_{f_0,\max }); L^1({\mathbb R}_{+})) \). The uniqueness of \(F_t\) on \([0, T_{f_0,\max })\) follows from Theorem 3.1.

Now assume \(T_{f_0,\max }<\infty \). Then using Proposition 3.1 and \(T_{F_{\cdot },\max }= T_{f_0,\max }<\infty \) we have \(\Vert f(t)\Vert _{L^1}=M_{-1/2}(F_t)\rightarrow \infty \) as \(t \nearrow T_{f_0,\max }\).

(c) Let \(G_t\) be the conservative distributional solutions of Eq. (1.5) on \([0, \infty )\) obtained in part (b) corresponding to \(g(\cdot ,t)\), i.e. \(G_t\) is regular on \([0, T_{g_0,\max })\) with the density \(g(\cdot ,t)\) for all \(t\in [0, T_{g_0,\max })\). Since \(T_{F_{\cdot },\max }= T_{f_0,\max }, T_{G_{\cdot },\max }=T_{g_0,\max }\), \(\Vert f(t)\Vert _{L^1}=M_{-1/2}(F_t)\) for all \(t\in [0, T_{f_0,\max })\), \(\Vert g(t)\Vert _{L^1}=M_{-1/2}(G_t)\) for all \(t\in [0, T_{g_0,\max })\), and \(\Vert f(t)-g(t)\Vert _{1} =\Vert F_t-G_t\Vert _1\) for all \(t\in [0, T_{f_0,\max }\wedge T_{g_0,\max })\), the stability estimate (4.2) follows from Theorem 3.1.

(d) From \(\mathrm{d}F_t(x)=f(x,t)\sqrt{x}\mathrm{d}x\) and \(f\in C([0, \infty ); L^1({\mathbb R}_{+}))\), it is easily seen that \(t\mapsto \mathcal{Q}_2^{\pm }(F_t), t\mapsto \mathcal{Q}_3^{\pm }(F_t)\) belong to \( C([0,\infty );{\mathcal B}_{0}({\mathbb R}_{\ge 0}))\) and \(F_t\) satisfies the integral equation (3.3) for all \(t\ge 0\) and thus \(F_t\) is a conservative distributional solution with the initial datum \(F_0\). The conclusion of part (d) then follows from the equivalent definition of strong solutions (see Remark 3.1) and the uniqueness theorem (Theorem 3.1). \(\square \)

As did for the classical Boltzmann equation, the collision integral \(Q(f)\) can be decomposed as positive and negative parts:

$$\begin{aligned}&Q(f)(x)=Q^{+}(f)(x)-Q^{-}(f)(x),\end{aligned}$$
(4.3)
$$\begin{aligned}&Q^{+}(f)(x)=\int \limits _{{\mathbb R}_{+}^2}w(x,y,z)f(y)f(z)(1+f(x_*))\mathrm{d}y\mathrm{d}z, \end{aligned}$$
(4.4)
$$\begin{aligned}&Q^{-}(f)(x)=f(x)L(f)(x), \end{aligned}$$
(4.5)
$$\begin{aligned}&L(f)(x)=\int \limits _{{\mathbb R}_{+}^2}w(x,y,z)[f(x_*)(1+f(y)+f(z))-f(y)f(z)]\mathrm{d}y\mathrm{d}z. \end{aligned}$$
(4.6)

Notice that, according to Lemma 2.2 and Lemma 2.3, for any \(0\le f\in L^1({\mathbb R}_{+},\sqrt{x}\mathrm{d}x)\), the function \(x\mapsto L(f)(x)\) is well-defined, nonnegative on \({\mathbb R}_{+}\), and satisfies

$$\begin{aligned} M_{1/2}(f)\le L(f)(x)\le \sqrt{x} N(f)+M_{1/2}(f)+2[M_{-1/2}(f)]^2. \end{aligned}$$
(4.7)

The following proposition gives an exponential-positive representation (i.e. Duhamel’s formula) for a class of mild solutions.

Proposition 4.3

Let \(0\le f_0\in L^1({\mathbb R}_+)\) have finite mass and energy and let \(f\in C([0, T_{f_0,\max }); L^1({\mathbb R}_{+}))\) be the unique conservative mild solution of Eq. (1.5) on \([0, T_{f_0,\max })\) satisfying \(f(\cdot ,0)=f_0\). Then there is a null set \(Z\subset {\mathbb R}_{+}\) such that for all \(x\in {\mathbb R}_{+}{\setminus } Z\) and all \(t\in [0, T_{f_0,\max })\)

$$\begin{aligned} f(x,t)=f_0(x) e^{-\int \limits _{0}^{t}L(f)(x,\tau )\mathrm{d}\tau } +\int \limits _{0}^{t}Q^{+}(f)(x,\tau )e^{-\int \limits _{\tau }^{t}L(f)(x,s)\mathrm{d}s} \mathrm{d}\tau \end{aligned}$$
(4.8)

where \(Q^{+}(f), L(f)\) are defined in (4.4)–(4.6).

Proof

By definition of mild solutions and \(Q(f)=Q^{+}(f)-fL(f)\) there is a null set \(Z\subset {\mathbb R}_{+}\) which is independent of \(t\) such that for every \(x\in {\mathbb R}_{+}{\setminus } Z\)

$$\begin{aligned} \frac{\partial }{\partial t}f(x,t)=Q^{+}(f)(x,t)-f(x,t)L(f)(x,t)\quad \mathrm{a.e.}\quad t\in [0, T_{f_0,\max }). \end{aligned}$$
(4.9)

Notice that for every \(x>0\) the nonnegative function \(t\mapsto L(f)(x,t)\) is locally integrable on \([0, T_{f_0,\max })\). In fact applying (4.7) and \(f\in C([0, T_{f_0,\max }); L^1({\mathbb R}_{+}))\) we have

$$\begin{aligned} \sup _{t\in [0, T]}L(f)(x,t)\le \sup _{t\in [0, T]}( \sqrt{x}N(f(t))+M_{1/2}(f(t))+2\Vert f(t)\Vert _{L^1}^2 )<\infty \end{aligned}$$
(4.10)

for all \(T\in (0, T_{f_0,\max })\) and all \(x>0\). Therefore, for every \(x\in {\mathbb R}_{+}{\setminus } Z\), the function \(t\mapsto f(x,t)e^{\int \limits _{0}^{t}L(f)(x,\tau )\mathrm{d}\tau }\) is also absolutely continuous on \([0, T]\) for all \(T\in (0, T_{f_0,\max })\) and thus the Duhamel’s formula (4.8) follows from the differential equation (4.9). \(\square \)

For bounded mild solutions we have the following

Proposition 4.4

Let \(0\le f_0\in L^1({\mathbb R}_+)\) have finite mass and energy and let \(f\in C([0, T_{f_0,\max }); L^1({\mathbb R}_{+}))\) be the unique conservative mild solution of Eq. (1.5) on \([0, T_{f_0,\max })\) satisfying \(f(\cdot ,0)=f_0\). Suppose in addition \(f_0\in L^{\infty }({\mathbb R}_+)\). Then \(f(\cdot , t)\in L^{\infty }({\mathbb R}_{+})\) for all \(t\in [0, T_{f_0,\max })\) and there hold two estimates: for all \(t\in [0, T_{f_0,\max })\),

$$\begin{aligned} \Vert f(t)\Vert _{L^{\infty }} \!\! \le \!\! \max \{1\,,\,\Vert f_0\Vert _{L^{\infty }}\}\exp \left( 2\int \limits _{0}^{t}\Vert f(\tau )\Vert _{L^1}^2 \mathrm{d}\tau \right) , \end{aligned}$$
(4.11)
$$\begin{aligned} \Vert f(t)\Vert _{L^{\infty }}\!\!&\le \!\! \Vert f_0\Vert _{L^{\infty }} \exp \left( \int \limits _{0}^{t}[ \Vert f(\tau )\Vert _{L^1}^2-M_{1/2}(f(\tau )) ]\mathrm{d}\tau \right) \nonumber \\ \!\!&+ \!\! \int \limits _{0}^{t}\Vert f(\tau )\Vert _{L^1}^2 \exp \left( \int \limits _{\tau }^{t} [ \Vert f(s)\Vert _{L^1}^2-M_{1/2}(f(s)) ] \mathrm{d}s\right) \mathrm{d}\tau . \end{aligned}$$
(4.12)

Besides, if \(T_{f_0,\max }<\infty \) then \(\Vert f(t)\Vert _{L^{\infty }}\rightarrow \infty \) as \(t\nearrow T_{f_0,\max }\). \(\square \)

Proof

Let \(A(t)\) be the right hand side of (4.11) i.e.

$$\begin{aligned} A(t):=\max \{1,\,\Vert f_0\Vert _{L^{\infty }}\}e^{2\int \limits _{0}^{t}a(\tau )\mathrm{d}\tau }, \quad a(t):=\Vert f(t)\Vert _{L^1}^2,\quad t\in [0, T_{f_0,\max }). \end{aligned}$$

By definition of mild solutions and \(f(x,0)=f_0(x)\le A(0)\) for all \(x\in {\mathbb R}_{+}{\setminus } Z\) (here and below \(Z\subset {\mathbb R}_{+}\) denotes any null set which is independent of time variable) we have for all \(t\in [0,T_{f_0,\max })\)

$$\begin{aligned} (f(x,t)-A(t))_{+}=\int \limits _{0}^{t}( Q(f)(x,\tau )-2A(\tau )a(\tau ) )\mathbf{1}_{\{f(x,\tau )>A(\tau )\}}\mathrm{d}\tau . \end{aligned}$$

Taking integration with respect to \(x\in {\mathbb R}_{+}\) and omitting the negative part \(\mathcal{Q}^{-}(f)\ge 0\) gives

$$\begin{aligned} \int \limits _{{\mathbb R}_+}(f(x,t)-A(t))_{+} \mathrm{d}x \!\!\;\le&\!\! \int \limits _{0}^{t}\mathrm{d}\tau \int \limits _{{\mathbb R}_+}Q^{+}(f)(x,\tau )\mathbf{1}_{\{f(x,\tau )>A(\tau )\}} \mathrm{d}x \\ \!\!&-\;\!\!\int \limits _{0}^{t}\mathrm{d}\tau \int \limits _{{\mathbb R}_+} 2A(\tau )a(\tau )\mathbf{1}_{\{f(x,\tau )>A(\tau )\}} \mathrm{d}x. \end{aligned}$$

Next for the integrand of \(Q^{+}(f)(x,\tau )\), we have

$$\begin{aligned} f(y,\tau )f(z,\tau )(1+f(x_*,\tau ))\le f(y,\tau )f(z,\tau )(f(x_*,\tau )-A(\tau ))_{+}+2A(\tau )f(y,\tau )f(z,\tau ) \end{aligned}$$

where we used \(1\le A(\tau )\). Then applying the first inequality in (2.24) to \((f(x_*,\tau )-A(\tau ))_{+}\) and recalling \(a(\tau )=\Vert f(\tau )\Vert _{L^1}^2\) we deduce

$$\begin{aligned} \int \limits _{0}^{t}\mathrm{d}\tau \int \limits _{{\mathbb R}_{+}}Q^{+}(f)(x,\tau )\mathbf{1}_{\{f(x,\tau )>A(\tau )\}}\mathrm{d}x\!\! \;\le&\!\!\!\int \limits _{0}^{t}a(\tau )\mathrm{d}\tau \int \limits _{{\mathbb R}_{+}} ( f(x,\tau )-A(\tau ) )_{+} \mathrm{d}x\quad \\ \!\!&+\;\!\!\! \int \limits _{0}^{t}\mathrm{d}\tau \int \limits _{{\mathbb R}_{+}}2A(\tau )a(\tau )\mathbf{1}_{\{f(x,\tau )>A(\tau )\}}\mathrm{d}x.\quad \end{aligned}$$

Notice that the common terms \(\int _{0}^{t}\mathrm{d}\tau \int _{{\mathbb R}_{+}}2A(\tau )a(\tau )\mathbf{1}_{\{f(x,\tau )>A(\tau )\}}\mathrm{d}x\) in the right hand sides of the above successive inequalities cancel each other. It follows that for all \(t\in [0,T_{\max })\)

$$\begin{aligned} \int \limits _{{\mathbb R}_+}(f(x,t)-A(t))_{+} \mathrm{d}x\le \int \limits _{0}^{t}a(\tau )\mathrm{d}\tau \int \limits _{{\mathbb R}_{+}} (f(x,\tau )-A(\tau ))_{+} \mathrm{d}x. \end{aligned}$$

By Gronwall Lemma we conclude \(\int \limits _{{\mathbb R}_+}(f(x,t)-A(t))_{+} \mathrm{d}x=0\) for all \(t\in [0,T_{\max }).\) This implies \(f(\cdot ,t)\in L^{\infty }({\mathbb R}_{+})\) and \(\Vert f(t)\Vert _{L^{\infty }}\le A(t)\) for all \(t\in [0,T_{\max })\), i.e. (4.11) holds true.

To prove (4.12) we denote \(b(t)=M_{1/2}(f(t))\) and use the first inequality in (2.25) to get \(L(f)(x,t)\ge b(t)\) for all \(x>0\). Then we deduce from Duhamel’s formula (4.8) that

$$\begin{aligned} f(x,t)\le f_0(x) e^{-\int \limits _{0}^{t}b(\tau )\mathrm{d}\tau } +\int \limits _{0}^{t}a(\tau )(1+\Vert f(\tau )\Vert _{L^{\infty }}) e^{-\int \limits _{\tau }^{t}b(s)\mathrm{d}s}\mathrm{d}\tau \end{aligned}$$

for all \(x\in {\mathbb R}_{+}{\setminus } Z\) and all \(t\in [0, T_{f_0,\max })\). This gives

$$\begin{aligned} \Vert f(t)\Vert _{L^{\infty }}e^{\int \limits _{0}^{t}b(\tau )\mathrm{d}\tau } \le \Vert f_0\Vert _{L^{\infty }} +\int \limits _{0}^{t}a(\tau )(1+\Vert f(\tau )\Vert _{L^{\infty }}) e^{\int \limits _{0}^{\tau }b(s)\mathrm{d}s}\mathrm{d}\tau \end{aligned}$$

for all \(t\in [0, T_{f_0,\max })\) and thus (4.12) follows from Gronwall Lemma.

Finally by the conservation of mass we have \(\Vert f(t)\Vert _{L^1}\le 2[N(f_0)]^{2/3}[\Vert f(t)\Vert _{L^{\infty }}]^{1/3}\) for all \(t\in [0, T_{f_0,\max })\) and so the \(L^{\infty }\)-blow up follows from the \(L^1\)-blow up in part (a) of Proposition 4.2. \(\square \)

Now we turn to the global mild solutions. As explained at the end of Sect. 3, the following result belongs to the case of high temperature.

Theorem 4.1

Let \(0\le f_0\in L^1({\mathbb R}_+)\) have finite mass and energy and suppose

$$\begin{aligned} \Vert f_0\Vert _{L^1}\le \frac{1}{80}[N(f_0)E(f_0)]^{1/4}. \end{aligned}$$
(4.13)

Then there exists a unique conservative mild solution \(f(\cdot ,t)\) of Eq. (1.5) on \([0,\infty )\) satisfying \(f\in C([0,\infty ); L^1({\mathbb R}_{+}))\) and \(f(\cdot ,0)=f_0\). Moreover we have

$$\begin{aligned} \sup _{t\ge 0}\Vert f(t)\Vert _{L^1} \le \frac{1}{4}[N(f_0)E(f_0)]^{1/4}. \end{aligned}$$
(4.14)

Besides, if in addition \(f_0\in L^{\infty }({\mathbb R}_{+})\), then \(f(\cdot ,t) \in L^{\infty }({\mathbb R}_{+})\) for all \(t\ge 0\) and

$$\begin{aligned} \sup _{t\ge 0}\Vert f(t)\Vert _{L^{\infty }}\le 2\max \{1,\,\Vert f_0\Vert _{L^{\infty }}\}. \end{aligned}$$
(4.15)

Proof

Let \(F_0\) be defined by \(\mathrm{d}F_0(x)=f_0(x)\sqrt{x}\mathrm{d}x\). Then (4.13) says that \(F_0\) satisfies the condition (3.46) in Theorem 3.2 which insures that there exists a unique conservative distributional solution \(F_t\) of Eq. (1.5) on \([0,\infty )\) with the initial datum \(F_0\). Since \(F_t\) is also a strong solution on \([0,\infty )\), and \(T_{F_{\cdot },\max }=\infty \), by Proposition 4.1 and Proposition 4.2 we conclude that \(F_t\) is regular on \([0, \infty )\) and its density \(f(\cdot ,t)\) is the unique conservative mild solution of Eq. (1.5) on \([0,\infty )\) satisfying \(f(\cdot ,0)=f_0\) and \(f\in C([0,\infty ); L^1({\mathbb R}_{+}))\). Since \(\Vert f(t)\Vert _{L^1}=M_{-1/2}(F_t), N(F_0)=N(f_0), E(F_0)=E(f_0)\), the estimate (4.14) follows from (3.46).

Let us denote \(N=N(f_0), E=E(f_0)\). By Theorem 1.1 and the uniqueness of \(F_t\), \(F_t\) has the moment production (1.15). Thus from Lemma 3.6 and \(M_{1/2}(f(t))=M_{1/2}(F_t)\) we have for all \(t>0\)

$$\begin{aligned} M_{1/2}(f(t)) \ge \frac{1}{6}\left[ 1-\exp \big (-(\sqrt{NE}+ N^3/E)t \big )\right] \frac{NE}{\sqrt{NE}+ N^3/E}. \end{aligned}$$
(4.16)

Now assume further that \(f_0\in L^{\infty }({\mathbb R}_{+})\). Then, from inequalities (4.14), (4.11) and \(T_{f_0,\max }=\infty \) we have for all \(t\ge 0\)

$$\begin{aligned} \Vert f(t)\Vert _{L^1}^2\le \frac{1}{16}\sqrt{NE}=:a,\quad \Vert f(t)\Vert _{L^{\infty }}\le \max \{1\,,\, \Vert f_0\Vert _{L^{\infty }}\} e^{2at}. \end{aligned}$$
(4.17)

Let us choose \(t_0\) satisfying

$$\begin{aligned} 1-\exp \big (-(\sqrt{NE}+ N^3/E)t_0 \big ) =\frac{6}{7}\quad \mathrm{i.e.}\quad t_0=\frac{\log 7}{\sqrt{NE}+ N^3/E}. \end{aligned}$$

Then we deduce from (4.16) that

$$\begin{aligned} M_{1/2}(f(t))\ge \frac{1}{7}\cdot \frac{NE}{\sqrt{NE}+ N^3/E}=: b,\quad \forall \, t\ge t_0. \end{aligned}$$
(4.18)

Applying (4.12) to the mild solution \(t\mapsto f(\cdot ,t+t_0)\) on \([0, \infty )\) with the initial datum \(f(\cdot ,t_0)\) and using the first inequality in(4.17) and (4.18) we have

$$\begin{aligned} \Vert f(t)\Vert _{L^{\infty }}\le \Vert f(t_0)\Vert _{L^{\infty }} e^{(a-b)(t-t_0)} +a\int \limits _{t_0}^{t}e^{(a-b)(t-\tau )}\mathrm{d}\tau ,\quad t\ge t_0. \end{aligned}$$
(4.19)

This gives

$$\begin{aligned} \Vert f(t)\Vert _{L^{\infty }}\le \max \Bigg \{\Vert f(t_0)\Vert _{L^{\infty }}\,,\,\frac{a}{b-a}\Bigg \} ,\quad t\ge t_0 \end{aligned}$$
(4.20)

provided \(b>a\). But this is true because \(b>2a\). In fact by definition of \(a,b\) in (4.17), (4.18), it is easily checked that \(b>2a\) is equivalent to \(\frac{E^{3/2}}{N^{5/2}}>7\), and the latter is obvious since the assumption \(M_{-1/2}(F_0)=\Vert f_0\Vert _{L^1}\le \frac{1}{80}(NE)^{1/4}\) implies (3.55) and so \(\frac{E^{3/2}}{N^{5/2}}\ge (80)^2>7\). Thus (4.20) holds true. From (4.20), the second inequality in (4.17) and \(a/(b-a)<1\) we obtain

$$\begin{aligned} \Vert f(t)\Vert _{L^{\infty }}\le e^{2at_0}\max \{1\,,\, \Vert f_0\Vert _{L^{\infty }}\} ,\quad t\ge 0. \end{aligned}$$

Finally by definition of \(t_0\) and \(a\) we have \(e^{2at_0}\le 7^{\frac{1}{8}} <2.\) This proves (4.15) and completes the proof of the theorem. \(\square \)

5 Regularity for Solutions with \(f_0(x)\le K x^{-1}\)

In this section we study regularity of such a distributional solution \(F_t\) whose initial datum \(F_0\) is regular with density \(f_0\) satisfying \(f_0(x)\le Kx^{-1}\,(\forall \, x>0)\). It seems very difficult to prove any expected regularity without using suitable approximate solutions. This is why we introduced the second cutoff (2.31)–(2.32) and constructed approximate solutions. Theorem 5.1 below is based on the new positivity in Lemma 2.3 and the following two relevant lemmas.

Lemma 5.1

Let \(f\) be a nonnegative measurable function on \({\mathbb R}_{+}\) satisfying \(N(f)<\infty \). Then

$$\begin{aligned} \int \limits _{{\mathbb R}_{+}^2}w(x,y,z)f(y)f(z)\mathrm{d}y\mathrm{d}z\le 2\sqrt{2}[N(f)]^2x^{-1}\quad \forall \,x>0. \end{aligned}$$
(5.1)

Proof

Since \(w(x,y,z)\) is symmetric with respect to \((y,z)\) and that \(w(x,y,z)>0\) implies \(y+z>x\), it follows that

$$\begin{aligned} \int \limits _{{\mathbb R}_{+}^2}w(x,y,z)f(y)f(z)\mathrm{d}y\mathrm{d}z&\le 2\int \limits _{0<y\le z,\, z\ge x/2}\frac{\sqrt{y}}{\sqrt{x}}f(y)f(z)\mathrm{d}y\mathrm{d}z\\&\le \frac{2\sqrt{2}}{x} \int \limits _{0<y\le z,\, z\ge x/2}\sqrt{y} \sqrt{z}f(y)f(z)\mathrm{d}y\mathrm{d}z\le \frac{2\sqrt{2}}{x}[N(f)]^2. \end{aligned}$$

\(\square \)

Lemma 5.2

Let \(5/6<\alpha <5/4\). Then, for all \(x>0\) we have

$$\begin{aligned} I_{\alpha }(x):=\int \limits _{{\mathbb R}_{+}^2}w(x,y,z)x_*^{-\alpha }y^{-\alpha } z^{-\alpha } \mathrm{d}y\mathrm{d}z=I_{\alpha }(1)x^{2-3\alpha }<\infty . \end{aligned}$$
(5.2)

In particular for \(\alpha =1\) we have \(I_1(1)<19\) and

$$\begin{aligned} \int \limits _{{\mathbb R}_{+}^2}w(x,y,z)x_*^{-1}y^{-1}z^{-1} \mathrm{d}y\mathrm{d}z=I_{1}(1)x^{-1}. \end{aligned}$$
(5.3)

Proof

Fix \(x>0\). The equality (5.2) follows by changing variables \(y=xu, z=xv\) which gives the constant

$$\begin{aligned} I_{\alpha }(1) =2\int \limits _{u\ge v>0, u+v>1}\frac{\min \{1,\sqrt{u+v-1}, \sqrt{u},\sqrt{v}\}}{(u+v-1)^{\alpha }u^{\alpha } v^{\alpha }} \mathrm{d}u\mathrm{d}v. \end{aligned}$$

To prove \(I_{\alpha }(1)<\infty \) we use \(\min \{1,\sqrt{u+v-1}, \sqrt{u},\sqrt{v}\} \le (u+v-1)^{1/4}v^{1/4}\) and change variables \((u,v)=(r^2\cos ^2\theta , r^2\sin ^2\theta )\,(r>1, 0<\theta <\pi /4)\) to get

$$\begin{aligned} I_{\alpha }(1)\le 8\left( \int \limits _{1}^{\infty }\frac{\mathrm{d}r}{(r^2-1)^{\alpha -1/4}r^{4\alpha -7/2}}\right) \left( \int \limits _{0}^{\pi /4} \frac{\mathrm{d}\theta }{\cos ^{2\alpha -1}(\theta ) \sin ^{2\alpha -3/2}(\theta )} \right) . \end{aligned}$$

The two integrals are finite when \(5/6<\alpha <5/4\). For \(\alpha =1\), by direct and careful calculation we have \(I_1(1)=8\int \limits _{0}^{\pi /2}\theta \sin (\theta ) (1+\cos ^2(\theta ))^{-1/2}\mathrm{d}\theta +7\pi ^2/6<19.\) \(\square \)

The following simple lemma is also useful for proving Theorem 5.1.

Lemma 5.3

Let \(f_i, g_i\in [0, \infty ), i=1,2,\ldots , n\,(n\ge 2)\). Then

$$\begin{aligned} \prod _{i=1}^nf_i\le \prod _{i=1}^n f_i\wedge g_i +\sum _{i=1}^n\Big (\prod _{1\le j\le n, j\ne i}f_j\Big )(f_i-g_i)_+. \end{aligned}$$
(5.4)

Proof

Consider \(\prod _{i=1}^{n}f_i-\prod _{i=1}^{n} f_i\wedge g_i\) and use \(0\le f_i\wedge g_i\le f_i\) and \(f_i-f_i\wedge g_i=(f_i-g_i)_{+}\). The inequality follows easily by induction on the number \(n\). \(\square \)

Theorem 5.1

Let \(F_0\in \mathcal{B}_1^{+}({\mathbb R}_{\ge 0})\) be a regular measure with density \(f_0\) satisfying

$$\begin{aligned} f_0(x)\le Kx^{-1}\qquad \forall \, x\in {\mathbb R}_{+} \end{aligned}$$

for some \(0<K<\infty \). Let \(T_{K}=(6[N(F_0)]^2K^{-1}+40K^2)^{-1}\). Then there exists a conservative distributional solution \(F_t\) of Eq. (1.5) on \([0,\infty )\) with the initial datum \(F_0\) such that \(F_t|_{[0,T_{K}]}\) is a regular strong solution on \([0, T_{K}]\) and its density \(f(x,t)\) satisfies

$$\begin{aligned} f(x,t)\le 5K x^{-1}\qquad \forall \, (x,t)\in {\mathbb R}_{+}\times [0, T_{K}]. \end{aligned}$$
(5.5)

In particular \(F_t\) has no condensation for all \(t\in [0, T_{K}]\).

Proof

Step 1: Let \(f^n\) with \(f^n|_{t=0}=f_0\) be conservative mild solutions of the approximate equations constructed in Sect. 2 (see (2.27) (2.28)) with \(w_n(x,y,z)=\frac{1}{\sqrt{x}}S_n(x,x_*;y,z)\), where \(S_n(x,x_*;y,z)\) are taken as the second cutoff (2.31). Let \(C_1=I_1(1)\) be given in Lemma 5.2 and let \(C_0=2\sqrt{2}[N(F_0)]^2, \lambda =C_0K^{-1}+C_1K^2\), and

$$\begin{aligned} \Phi (x,t)=(1-2\lambda t)^{-1/2}K x^{-1},\quad t\in [0, T_{K}]. \end{aligned}$$

In this step we prove that \(f^n(x,t)\le \Phi (x,t)\) for all \((x,t)\in {\mathbb R}_{+}\times [0, T_{K}]\) and all \(n\ge 1\).

First of all by computing differences of coefficients \(6-4\sqrt{2}, 40-2I_1(1)(>2)\) etc., we have \(2\lambda T_{K}<1\) and \((1-2\lambda T_{K})^{-1/2}<5\). This implies \(\Phi (x,t)\le 5Kx^{-1}\) for all \(x>0, t\in [0, T_{K}]\). Next using the absolute continuity of \(t\mapsto f^n(x,t)\) and \(f^n(x,0)=f_0(x)\le \Phi (x,0)\) we have, for almost every \(x\in {\mathbb R}_+\) and for all \(t\in [0, T_{K}]\),

$$\begin{aligned} \big (f^n(x,t)-\Phi (x,t)\big )_{+} =\int \limits _{0}^{t}( Q_{n}(f^n)(x,\tau )-{\partial }_{\tau }\Phi (x,\tau ) )\mathbf{1}_{\{f^n(x,\tau )>\Phi (x,\tau )\}}\mathrm{d}\tau . \end{aligned}$$
(5.6)

Now consider decomposition \(Q_n(f)= Q_n^{+}(f)-fL_n(f)\) as given in (4.3)–(4.6) by replacing \(w(x,y,z)\) with \(w_n(x,y,z)\). Thanks to Lemma 2.3 we have \(L_n(f^n)(x,\tau )\ge 0\) hence \(Q_n(f^n)(x,\tau )\le Q_n^{+}(f^n)(x,\tau )\). Multiplying \(\sqrt{x}\) to both sides of (5.6) and taking integration over \({\mathbb R}_{+}\) we deduce

$$\begin{aligned} N((f^n(t)-\Phi (t))_{+}) \le \int \limits _{0}^{t}\mathrm{d}\tau \int \limits _{{\mathbb R}_{+}}\sqrt{x}( Q_{n}^{+}(f^n)(x,\tau )- {\partial }_{\tau }\Phi (x,\tau ) )\mathbf{1}_{\{f^n(x,\tau )>\Phi (x,\tau )\}}\mathrm{d}x \end{aligned}$$

for all \(t\in [0, T_{K}]\). Notice that there is no problem of integrability because, for all \(t\in [0, T_{K}]\),

$$\begin{aligned}&{\partial }_t\Phi (x,t)=\lambda (1-2\lambda t)^{-3/2}Kx^{-1},\\&Kx^{-1}\mathbf{1}_{\{f^n(x,t)>\Phi (x,t)\}} \le \Phi (x,t)\mathbf{1}_{\{f^n(x,t)>\Phi (x,t)\}}\le f^n(x,t), \end{aligned}$$

and \((1-2\lambda t)^{-3/2}\le (1-2\lambda T_{K})^{-3/2}<5^3.\) For convenience of derivation, let us use the notation (1.7), i.e. \(f^n=f^n(x,\tau ), f_*^n=f^n(x_*,\tau ), {f^n}'=f^n(y,\tau ), {f_*^n}'=f^n(z,\tau )\), etc. For the quadratic term of \(Q_{n}^{+}(f^n)(x,\tau )\) we use Lemma 5.1 and \(w_n(x,y,z)\le w(x,y,z)\) to get

$$\begin{aligned} \int \limits _{{\mathbb R}_{+}^2}w_n(x,y,z){f^n}' {f_*^n}'\mathrm{d}y\mathrm{d}z\le C_0x^{-1} \end{aligned}$$

where we have used the conservation of mass and \(N(f_0)=N(F_0)\). For the cubic term of \(Q_{n}^{+}(f^n)(x,\tau )\) we use Lemma 5.3 to get

$$\begin{aligned} \int \limits _{{\mathbb R}_{+}^2}w_n(x,y,z){f^n}'{f_*^n}'f_*^n\mathrm{d}y\mathrm{d}z\,\le&\int \limits _{{\mathbb R}_{+}^2}w_n(x,y,z) \Phi '\Phi _*'\Phi _* \mathrm{d}y\mathrm{d}z\\&+\;\int \limits _{{\mathbb R}_{+}^2}w_n(x,y,z)\big [ {f_*^n}'f_*^n({f^n}'-\Phi ')_{+}\\&+\; {f^n}'f_*^n({f_*^n}'-\Phi _*')_{+} +{f^n}'{f_*^n}'(f_*^n-\Phi _*)_{+}\big ]\mathrm{d}y\mathrm{d}z \end{aligned}$$

and using Lemma 5.2 with \(\alpha =1\) we have (since \(w_n(x,y,z)\le w(x,y,z)\))

$$\begin{aligned} \int \limits _{{\mathbb R}_{+}^2}w_n(x,y,z) \Phi '\Phi _*'\Phi _* \mathrm{d}y\mathrm{d}z\le (1-2\lambda \tau )^{-3/2}K^3C_1 x^{-1},\quad x>0. \end{aligned}$$

Combining these with the inequalities (2.37) for cubic terms we deduce

$$\begin{aligned}&\int \limits _{{\mathbb R}_{+}}\sqrt{x}(Q_{n}^{+}(f^n)(x,\tau ) -{\partial }_{\tau }\Phi (x,\tau ))\mathbf{1}_{\{f^n(x,\tau )>\Phi (x,\tau )\}}\mathrm{d}x \\&\quad \le \int \limits _{{\mathbb R}_{+}^3}S_n(x,x_*; y,z) [ {f_*^n}'f_*^n({f^n}'-\Phi ')_{+}\!+\! {f^n}'f_*^n({f_*^n}'\!-\!\Phi _*')_{+} \!+\!{f^n}'{f_*^n}'(f_*^n-\Phi _*)_{+}]\mathrm{d}x\mathrm{d}y\mathrm{d}z \\&\qquad +\int \limits _{{\mathbb R}_{+}}\sqrt{x}[ C_0x^{-1}+ C_1(1-2\lambda \tau )^{-3/2}K^3 x^{-1} -\lambda (1-2\lambda \tau )^{-3/2}Kx^{-1} ] \mathbf{1}_{\{f^n(x,\tau )>\Phi (x,\tau )\}}\mathrm{d}x \\&\quad \le a_n N((f^n(\tau )-\Phi (\tau ))_{+})+b(\tau ) \end{aligned}$$

where \(a_{n}=3n[N(F_0)]^2\),

$$\begin{aligned} b(\tau )=\int \limits _{{\mathbb R}_{+}}\sqrt{x}[ C_{0}K^{-1}-(\lambda -C_1K^2) (1-2\lambda \tau )^{-3/2} ] Kx^{-1}\mathbf{1}_{\{f^n(x,\tau )>\Phi (x,\tau )\}}\mathrm{d}x. \end{aligned}$$

Now by our choice for \(\lambda \) we have

$$\begin{aligned} C_{0}K^{-1}-(\lambda -C_1K^2) (1-2\lambda \tau )^{-3/2}= C_{0}K^{-1}-C_0K^{-1} (1-2\lambda \tau )^{-3/2}\le 0 \end{aligned}$$

and so \(b(\tau )\le 0\) for all \(\tau \in [0, T_{K}]\). Thus

$$\begin{aligned} N((f^n(t)-\Phi (t))_{+}) \le a_n\int \limits _{0}^{t} N((f^n(\tau )-\Phi (\tau ))_{+})\mathrm{d}\tau \qquad \forall \,t\in [0, T_{K}]. \end{aligned}$$

By Gronwall Lemma we conclude \(N((f^n(t)-\Phi (t))_{+}) =0\) for all \(t\in [0,T_{K}]\). Thus \(f^n(x,t)\le \Phi (x,t)\) for all \(t\in [0, T_{K}]\) and for a.e. \(x\in {\mathbb R}_{+}\) and so the function \(f^n\wedge \Phi \) is the same (up to a null set in \({\mathbb R}_{+}\)) mild solution restricted on \({\mathbb R}_{+}\times [0, T_{K}]\) with the initial datum \(f_0\). If we rewrite \(f^n\wedge \Phi \) as \(f^n\), then the mild solution \(f^n\) satisfies \(f^n\le \Phi \) on \({\mathbb R}_{+}\times [0, T_{K}]\). In particular we have

$$\begin{aligned} f^n(x,t)\le 5Kx^{-1}\quad \forall \, (x, t)\in {\mathbb R}_{+}\times [0, T_{K}],\quad \forall \,n\ge 1. \end{aligned}$$
(5.7)

Step 2: In this step we prove that a subsequence of \(\{f^n\}_{n=1}^{\infty }\) (restricted on \({\mathbb R}_{+}\times [0, T_{K}]\)) converges in \(L^1\)-weak topology to a density of a strong solution \(F_t\) on \([0, T_{K}]\) with the initial datum \(F_0\). To shorten notations we define \(F^n_t\in \mathcal{B}_1^{+}({\mathbb R}_{\ge 0})\) by

$$\begin{aligned} \mathrm{d}F^n_t(x)=f^n(x,t)\sqrt{x}\mathrm{d}x,\quad t\in [0, T_{K}] \end{aligned}$$

and recall that \(\mathrm{d}F_0(x)=f_0(x)\sqrt{x}\mathrm{d}x\). By conservation of mass and energy we have

$$\begin{aligned} \int \limits _{{\mathbb R}_{+}}(1+x)\mathrm{d}F^n_t(x)\equiv N(F_0)+E(F_0),\qquad t\in [0, T_{K}], \,\,\,n\ge 1. \end{aligned}$$
(5.8)

From (5.7) and \(\int _{E}x^{-1/2}\mathrm{d}x\le \int _{0}^{\mathrm{mes}(E)}x^{-1/2}\mathrm{d}x =2\sqrt{\mathrm{mes}(E)}\) for every measurable set \(E\subset {\mathbb R}_{+}\), we have

$$\begin{aligned} \sup _{n\ge 1, t\in [0, T_{K}]}\int \limits _{E} \mathrm{d}F^n_t(x)\le 10K\sqrt{\mathrm{mes}(E)}\rightarrow 0\quad \mathrm{as}\quad \mathrm{mes}(E)\rightarrow 0 . \end{aligned}$$
(5.9)

Also we have for all \(0\le \alpha <1/2\)

$$\begin{aligned} \sup _{n\ge 1, t\in [0, T_{K}]}\int \limits _{{\mathbb R}_{+}}x^{-\alpha }\mathrm{d}F^n_t(x) \le 5K\int \limits _{0}^{1}x^{-1/2-\alpha }\mathrm{d}x+N(F_0)<\infty . \end{aligned}$$

Taking \(\alpha =1/3\) and using \(W_n(x,y,z)\le W(x,y,z)\) and

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}}\sqrt{x}W(x,y,z) \mathrm{d}x \le \sqrt{y}+\sqrt{z}\,, \quad W(x,y,z)\le (xyz)^{-1/3} \end{aligned}$$
(5.10)

we obtain for all \(\psi \in L^{\infty }({\mathbb R}_{\ge 0})\)

$$\begin{aligned} \sup _{n\ge 1, t\in [0, T_{K}]}\Big ( \int \limits _{{\mathbb R}_{+}^2}|\mathcal{J}_n^{\pm }[\psi ]|\mathrm{d}^2F^n_t+\int \limits _{{\mathbb R}_{+}^3}|\mathcal{K}_n^{\pm }[\psi ]| \mathrm{d}^3F^n_t\Big )\le C\Vert \psi \Vert _{L^{\infty }} \end{aligned}$$
(5.11)

here and below \(C\) denotes any constant that depends only on \(N(F_0), E(F_0)\) and \(K\). Next using the weak formula (2.40) we have, for all \(0\le s<t\le T_{K}\) and all \(\psi \in L^{\infty }({\mathbb R}_{+})\),

$$\begin{aligned} \int \limits _{{\mathbb R}_{+}}\psi \mathrm{d}(F^n_t-F^n_s) =\int \limits _{s}^{t}\mathrm{d}\tau \int \limits _{{\mathbb R}_{+}^2}\mathcal{J}_n[\psi ]\mathrm{d}^2F^n_{\tau } +\int \limits _{s}^{t}\mathrm{d}\tau \int \limits _{{\mathbb R}_{+}^3}\mathcal{K}_n[\psi ]\mathrm{d}^3F^n_{\tau } \end{aligned}$$
(5.12)

which together with (5.11) gives the uniform strong continuity on \([0, T_{K}]\):

$$\begin{aligned} \sup _{n\ge 1}\Vert F^n_t-F^n_s\Vert _{0} \le C|t-s|\quad \forall \, t,s\in [0, T_{K}]. \end{aligned}$$
(5.13)

From (5.8), (5.9), (5.13) and the criterion of \(L^1\)-weakly relative compactness we conclude that there exist a subsequence of \(\{ F^n_t\}_{n=1}^{\infty }\), still denote it as \(\{ F^n_t\}_{n=1}^{\infty }\), and a positive regular Borel measure \(F_t\in C([0,T_{K}]; \mathcal{B}_0({\mathbb R}_{+}))\) such that for all \(t\in [0, T_{K}]\)

$$\begin{aligned} \lim _{n\rightarrow \infty }\int \limits _{{\mathbb R}_{+}}\psi \mathrm{d}F^n_t= \int \limits _{{\mathbb R}_{+}}\psi \mathrm{d}F_t \qquad \forall \,\psi \in L^{\infty }({\mathbb R}_{+}). \end{aligned}$$
(5.14)

Let \(f(\cdot ,t)\) be the density of \(F_t\). From (5.14), (5.7) we have \(f(x,t)\le \Phi (x,t)\) for all \(t\in [0, T_{K}]\) and for a.e. \(x\in {\mathbb R}_{+}\). Let \(f(x,t)\) be replaced by \(f(x,t)\wedge \Phi (x,t)\) which we still denote as \(f(x,t)\). Then \(f(\cdot ,t)\) is the same density of \(F_t\) for all \(t\in [0, T_{K}]\) and satisfies (5.5). From (5.14) and the conservation of mass and energy for \(F^n_t\) we also have

$$\begin{aligned} N(F_t)=N(F_0),\quad E(F_t)\le E(F_0),\quad t\in [0, T_{K}]. \end{aligned}$$
(5.15)

From these and \(f^n(x,t)\sqrt{x}\le 5Kx^{-1/2}, f(x,t)\sqrt{x}\le 5Kx^{-1/2}\) on \({\mathbb R}_{+}\times [0, T_{K}]\) and (5.14) we see that the convergence (5.14) can be extended as follows: for all \(t\in [0, T_{K}]\) and all \(\psi \in L^{\infty }({\mathbb R}_{+})\)

$$\begin{aligned}&\lim _{n\rightarrow \infty }\int \limits _{{\mathbb R}_{+}}\psi (x)(1+\sqrt{x})\mathrm{d}F^n_t(x)= \int \limits _{{\mathbb R}_{+}}\psi (x)(1+\sqrt{x})\mathrm{d}F_t(x),\end{aligned}$$
(5.16)
$$\begin{aligned}&\lim _{n\rightarrow \infty }\int \limits _{{\mathbb R}_{+}}\psi (x)x^{-1/3}\mathrm{d}F^n_t(x)= \int \limits _{{\mathbb R}_{+}}\psi (x)x^{-1/3}\mathrm{d}F_t(x). \end{aligned}$$
(5.17)

Also from the pointwise convergence (2.34) and (5.10) we have for all \((x,y,z)\in {\mathbb R}_{>0}^3\)

$$\begin{aligned}&\lim _{n\rightarrow \infty }\mathcal{J}_n^{\pm }[\psi ](y,z)=\mathcal{J}^{\pm }[\psi ](y,z),\quad \lim _{n\rightarrow \infty }\mathcal{K}_n^{\pm }[\psi ](x,y,z)=\mathcal{K}^{\pm }[\psi ](x,y,z);\\&\sup _{n\ge 1}|\mathcal{J}_n^{\pm }[\psi ](y,z)|, |\mathcal{J}^{\pm }[\psi ](y,z)|\le \Vert \psi \Vert _{L^{\infty }}(\sqrt{y}+\sqrt{z}),\\&\sup _{n\ge 1}|\mathcal{K}_n^{\pm }[\psi ](x,y,z)|, |\mathcal{K}^{\pm }[\psi ](x,y,z)|\le 2\Vert \psi \Vert _{L^{\infty }} (xyz)^{-1/3}. \end{aligned}$$

Thus we deduce from elementary convergence properties of integrals with product measures (see e.g. Lemma 4 in [12] and its application in the same paper) that

$$\begin{aligned} \lim _{n\rightarrow \infty } \int \limits _{{\mathbb R}_{+}^2}\mathcal{J}_n^{\pm }[\psi ]\mathrm{d}^2F^n_t =\int \limits _{{\mathbb R}_{+}^2}\mathcal{J}^{\pm }[\psi ]\mathrm{d}^2F_t,\quad \lim _{n\rightarrow \infty }\int \limits _{{\mathbb R}_{+}^3}\mathcal{K}_n^{\pm }[\psi ]\mathrm{d}^3F^n_t =\int \limits _{{\mathbb R}_{+}^3}\mathcal{K}^{\pm }[\psi ]\mathrm{d}^3F_t \end{aligned}$$

for all \(t\in [0, T_{K}]\). From these and the bound (5.11) which also holds for \(\int _{{\mathbb R}_{\ge 0}^2}|\mathcal{J}^{\pm }[\psi ]|\mathrm{d}^2F_t+ \int \limits _{{\mathbb R}_{\ge 0}^3}|\mathcal{K}^{\pm }[\psi ]|\mathrm{d}^3F_t\) we conclude that for all \(t\in [0, T_{K}]\) and all \(\psi \in L^{\infty }({\mathbb R}_{+})\)

$$\begin{aligned} \int \limits _{{\mathbb R}_{\ge 0}}\psi \mathrm{d}F_t= \int \limits _{{\mathbb R}_{\ge 0}}\psi \mathrm{d}F_0+ \int \limits _{0}^{t}\mathrm{d}\tau \int \limits _{{\mathbb R}_{\ge 0}^2}\mathcal{J}[\psi ]\mathrm{d}^2F_{\tau } +\int \limits _{0}^{t}\mathrm{d}\tau \int \limits _{{\mathbb R}_{\ge 0}^3}\mathcal{K}[\psi ]\mathrm{d}^3F_{\tau }. \end{aligned}$$
(5.18)

Thus as shown above we have \(\Vert F_t-F_s\Vert _0\le C|t-s|\) for all \(t,s\in [0, T_{K}]\). And since \(x^{-1/3}f(x,t)\sqrt{x}\le 5K x^{-5/6}\), it follows that \(\sup _{t\in [0, T_{K}]}M_{-1/3}(F_t)<\infty \) and

$$\begin{aligned} M_{-1/3}(|F_t-F_s|)\le C\Vert F_t-F_s\Vert _0^{1/3}\le C|t-s|^{1/3},\quad s,t\in [0, T_{K}] \end{aligned}$$

from which and the basic estimates of collision integrals (2.19) (with \(k=0\)) and (2.20) we see that \(t\mapsto \mathcal{Q}_2^{\pm }(F_t), t\mapsto \mathcal{Q}_3^{\pm }(F_t)\) belong to \( C([0,T_{K}]);{\mathcal B}_{0}({\mathbb R}_{\ge 0}))\). These together with (5.18) imply that the dual form (3.3) holds true. Thus we conclude from the equivalent definition of strong solutions showed in Remark 3.1 that \(F_t\) is a strong solution of Eq. (1.5) on \([0, T_{K}]\).

Step 3 (extension): Taking \(F_{T_{K}}\) as an initial datum, according to Theorem 1.1, there exists a conservative distributional \(F_{t}\) of Eq. (1.5) on \([T_{K}, \infty )\) such that \(F_t|_{t=T_{K}}=F_{T_{K}}\). As before, it is easily seen that the measure \(F_t\) defined for all \(t\in [0, \infty )\) in that way is a distributional solution of Eq. (1.5) on \([0, \infty )\). And from (5.15) we have \(E(F_t)\le E(F_0)\) for all \(t\in [0, \infty )\) and so it follows from Theorem 1.1(b) that \(F_t\) conserves also the energy. Thus \(F_t\) is a desired solution claimed in the theorem. \(\square \)

6 Condensation in Finite Time

As mentioned in the Introduction, our strategy for investigating the problem of condensation in finite time is to assume to the contrary that the distributional solution under consideration has no condensation at a finite time, then derive some necessary condition on the initial datum.

Let \(\underline{D}_{\alpha }(F), \overline{D}_{\alpha }(F)\) be defined in (1.19) for \(F\in \mathcal{B}^{+}({\mathbb R}_{\ge 0})\). As did in [15], in order to connect Eq. (1.8), we use a smooth version of \(\underline{D}_{\alpha }(F), \overline{D}_{\alpha }(F)\). For \(F\in \mathcal{B}^{+}({\mathbb R}_{\ge 0})\), \(\alpha \ge 0\), and \(\varepsilon >0\), let

$$\begin{aligned} N_{\alpha }(F,\varepsilon )&= \frac{1}{\varepsilon ^{\alpha }} N_{0}(F,\varepsilon ),\qquad \,\, N_{0}(F,\varepsilon )=\int \limits _{[0, \varepsilon ]}\Big (1-\frac{x}{\varepsilon }\Big )^2\mathrm{d}F(x),\\ {\underline{N}}_{\alpha }(F,\varepsilon )&= \mathop \mathrm{inf }\limits _{0<\delta \le \varepsilon }N_{\alpha }(F,\delta ), \quad {\overline{N}}_{\alpha }(F,\varepsilon )=\sup _{0<\delta \le \varepsilon }N_{\alpha }(F,\delta ),\\ \underline{N}_{\alpha }(F)&= \lim _{\varepsilon \rightarrow 0^+}\underline{N}_{\alpha }(F,\varepsilon ),\quad \quad {\overline{N}}_{\alpha }(F)=\lim _{\varepsilon \rightarrow 0^+}\overline{N}_{\alpha }(F,\varepsilon ) \,,\\ A_{\alpha }(F,\varepsilon )&= \frac{1}{\varepsilon ^{\alpha }}\int \limits _{[0,\varepsilon ]}\Big (\frac{x}{\varepsilon }\Big )^{2} \mathrm{d}F(x). \end{aligned}$$

It is easily seen that \(\underline{N}_{\alpha }(F), \overline{N}_{\alpha }(F)\) are equivalent to \(\underline{D}_{\alpha }(F),\overline{D}_{\alpha }(F)\) respectively. For instance for the most interesting case \(\alpha =1/2\) we have

$$\begin{aligned} \underline{D}_{1/2}(F)\le (5/4)^2\sqrt{5}\,\underline{N}_{1/2}(F),\quad \overline{D}_{1/2}(F)\le (5/4)^2\sqrt{5}\,\overline{N}_{1/2}(F). \end{aligned}$$
(6.1)

In fact for any \(r>1\) and any \(\varepsilon >0\)

$$\begin{aligned} N_{1/2}(F, r\varepsilon ) \ge \frac{1}{\sqrt{r\varepsilon }}\int \limits _{[0,\varepsilon ]}\Big (1-\frac{x}{r\varepsilon }\Big )^2\mathrm{d}F(x)\ge \frac{1}{\sqrt{r}}\Big (1-\frac{1}{r}\Big )^2 \frac{F([0,\varepsilon ])}{\sqrt{\varepsilon }}. \end{aligned}$$

Choosing \(r=5\) gives the best constant in (6.1).

Notice that the integrand \(\varphi _{\varepsilon }(x)=[(1-x/\varepsilon )_{+}]^2=(1-x/\varepsilon )^2\mathbf{1}_{[0,\varepsilon ]}(x)\) defining \(N_{\alpha }(F,\varepsilon )\) is convex and belongs to \(C^{1,1}({\mathbb R}_{\ge 0})\). In the following, \(\varphi _{\varepsilon }\) always stands for this special function.

Lemma 6.1

([15]) Let \(F\in \mathcal{B}^{+}({\mathbb R}_{\ge 0})\), \(\alpha \ge 0, \varepsilon >0\). Then

$$\begin{aligned} \underline{N}_{\alpha }(F,\varepsilon )[A_{\frac{1-\alpha }{2}}(F,\varepsilon )]^2 \le \int \limits _{{\mathbb R}_{\ge 0}^3}\mathcal{K}[\varphi _{\varepsilon }]\mathrm{d}^3F\quad for\quad 0\le \alpha \le 1/2 \end{aligned}$$
(6.2)

Also we have

$$\begin{aligned} F(\{0\})=0\quad \Longrightarrow \quad \sqrt{N_{\alpha }(F,\varepsilon )}\le \int \limits _{0}^{1} \theta ^{\alpha /2-1}\sqrt{A_{\alpha }(F,\varepsilon \theta )}\, \mathrm{d}\theta . \end{aligned}$$
(6.3)

Note To check (6.3) with [15] one may replace the closed interval \([0,\varepsilon ]\) by \((0,\varepsilon ]\) and define \(N_{\alpha }^{\circ }(F,\varepsilon ) =\frac{1}{\varepsilon ^{\alpha }}\int \limits _{(0,\varepsilon ]}(1-x/\varepsilon )^2\mathrm{d}F(x)\) as used in [15]. Then \(F(\{0\})=0 \Longrightarrow N_{\alpha }(F,\varepsilon )\equiv N_{\alpha }^{\circ }(F,\varepsilon )\).

The following proposition provides a necessary condition on the initial data for the absence of condensation at a finite time.

Proposition 6.1

Let \(F_t\) be a distributional solution of Eq. (1.5) with the initial datum \(F_0\) and suppose there is \(T\in (0,\infty )\) such that \(F_T(\{0\})=0\). Then

$$\begin{aligned} \sqrt{\underline{D}_{1/2}(F_0)\overline{D}_{1/2}(F_0)}\, \le \frac{50\sqrt{5}}{\sqrt{T}} \exp \Big (\frac{2}{3}\sqrt{N(F_0)E(F_T)}\,T\Big ). \end{aligned}$$
(6.4)

Proof

Applying (6.1) to \(F=F_0\) we see that to prove (6.4) it suffices to prove

$$\begin{aligned} \sqrt{\underline{N}_{1/2}(F_0)\overline{N}_{1/2}(F_0)}\, \le \frac{2^{5}}{\sqrt{T}} \exp \Big (\frac{2}{3}\sqrt{N(F_0)E(F_T)}\,T\Big ). \end{aligned}$$
(6.5)

Define for \(t\ge 0, \alpha \ge 0\) and \(\varepsilon >0\)

$$\begin{aligned}&m(t)=\exp \Big (\int \limits _{0}^{t}M_{1/2}(F_\tau )\mathrm{d}\tau \Big ),\qquad \qquad \qquad N_{\alpha }^{*}(t,\varepsilon )\!=\! m(t)N_{\alpha }(F_t,\varepsilon ) ,\\&\underline{N}_{\alpha }^{*}(t,\varepsilon ) = m(t)\underline{N}_{\alpha }(F_t,\varepsilon )=\inf _{0<\delta \le \varepsilon }N_{\alpha }^{*}(t,\delta ),\quad \overline{N}_{\alpha }^{*}(t,\varepsilon ) = m(t)\overline{N}_{\alpha }(F_t,\varepsilon )=\sup _{0<\delta \le \varepsilon }N_{\alpha }^{*}(t,\delta ). \end{aligned}$$

Notice that since all \(\overline{N}_{\alpha }^*(t,\varepsilon )\) are nonnegative, the following estimates involving these terms make sense even if possibly \(\overline{N}_{\alpha }^*(t,\varepsilon ) =\infty \) for large \(\alpha \).

Step 1: We prove that for any \(\varepsilon >0\) and any \(\alpha \ge 0\)

$$\begin{aligned} N_{\alpha }^*(t,\varepsilon )\ge N_{\alpha }^*(s,\varepsilon )+\frac{1}{\varepsilon ^{\alpha }}\int \limits _{s}^{t} m(\tau )\mathrm{d}\tau \int \limits _{{\mathbb R}_{\ge 0}^3}\mathcal{K}[\varphi _{\varepsilon }]\mathrm{d}^3F_{\tau } ,\quad 0\le s<t. \end{aligned}$$
(6.6)

Using Eq. (1.8), Lemma 3.1 and recalling \(\int _{{\mathbb R}_{\ge 0}}\varphi _{\varepsilon }\mathrm{d}F_t=N_{0}(F_t,\varepsilon )\) we have

$$\begin{aligned} \frac{\partial }{\partial t}N_{0}(F_t,\varepsilon )\ge -M_{1/2}(F_t) N_{0}(F_t,\varepsilon )+\int \limits _{{\mathbb R}_{\ge 0}^3} \mathcal{K}[\varphi _{\varepsilon }] \mathrm{d}^3F_t,\quad \forall \, t\ge 0 \end{aligned}$$

and so by definition of \( m(t)\) and \(N_{0}^*(t,\varepsilon )= m(t)N_{0}(F_t,\varepsilon )\),

$$\begin{aligned} \frac{\partial }{\partial t}N_{0}^*(t,\varepsilon ) \ge m(t)\int \limits _{{\mathbb R}_{\ge 0}^3} \mathcal{K}[\varphi _{\varepsilon }] \mathrm{d}^3F_t,\quad \forall \, t\ge 0. \end{aligned}$$

Since \(N_{\alpha }^*(t,\varepsilon )=\frac{1}{\varepsilon ^{\alpha }} N_{0}^*(t,\varepsilon )\), this gives (6.6).

By the way, applying (6.6) to \(\alpha =0\) and letting \(\varepsilon \rightarrow 0^+\) leads to a weak version of (1.18): \( m(t)F_{t}(\{0\}) \ge m(s)F_{s}(\{0\})\) for \(0\le s<t\). In particular we see that the assumption \(F_T(\{0\})=0\) implies \(F_t(\{0\})=0\) for all \(t\in [0, T]\).

Step 2: To prove (6.5) we can assume \(\underline{N}_{1/2}(F_0)>0\). Let us define for any \(\lambda \in (1/2,1)\)

$$\begin{aligned} D_{\lambda }=\lambda \underline{N}_{1/2}(F_0)\quad \mathrm{if}\quad \underline{N}_{1/2}(F_0)<\infty ;\quad D_{\lambda }=1\quad \mathrm{if}\quad \underline{N}_{1/2}(F_0)=\infty . \end{aligned}$$

Here the case \(``\underline{N}_{1/2}(F_0)=\infty ''\) is just considered in logic; if (6.5) is true, we must have \(\underline{N}_{1/2}(F_0)<\infty \). By definitions of \(\underline{N}_{1/2}(F_0,\varepsilon )\) and \(D_{\lambda }\), there is \(\varepsilon _{\lambda }>0\) such that

$$\begin{aligned} \underline{N}_{1/2}(F_0,\varepsilon )> D_{\lambda }\qquad \forall \, \varepsilon \in (0,\varepsilon _{\lambda }]. \end{aligned}$$

We now prove that for any \(\alpha \ge 0\) and any \(\varepsilon \in (0,\varepsilon _{\lambda }]\)

$$\begin{aligned} \overline{N}_{\frac{1}{4}+\frac{\alpha }{2}}^*(s,\varepsilon ) \le \frac{2^4}{\sqrt{ D_{\lambda }}}\cdot \frac{e^{c t}}{\sqrt{t-s}}\Big (\frac{1}{2}+\alpha \Big )^{-2}\sqrt{\overline{N}_{\alpha }^*(t,\varepsilon )} \quad \forall \, 0\le s<t\le T. \end{aligned}$$
(6.7)

where \(c=\sqrt{N(F_0)E(F_T)}.\)

Applying (6.6) to \(\alpha =1/2\) and \(s=0\) we have \(N_{1/2}^*(t,\varepsilon )\ge N_{1/2}^*(0,\varepsilon )\) for all \(t\ge 0\) and so

$$\begin{aligned} \underline{N}_{1/2}^*(t,\varepsilon )\ge \underline{N}_{1/2}^*(0,\varepsilon )=\underline{N}_{1/2}(F_0,\varepsilon )>D_{\lambda }\qquad \forall \, t\ge 0,\,\,\forall \,\varepsilon \in (0,\varepsilon _{\lambda }]. \end{aligned}$$
(6.8)

Given any \(\alpha \ge 0\) and \(0\le s<t\le T\). Using (6.6) we have ( omitting \(N_{\alpha }^*(s,\varepsilon )\) )

$$\begin{aligned} \frac{1}{\varepsilon ^{\alpha }}\int \limits _{s}^{t} m(\tau )\mathrm{d}\tau \int \limits _{{\mathbb R}_{\ge 0}^3}\mathcal{K}[\varphi _{\varepsilon }]\mathrm{d}^3F_{\tau }\le N_{\alpha }^*(t,\varepsilon )\qquad \forall \, \varepsilon \in (0,\varepsilon _{\lambda }]. \end{aligned}$$
(6.9)

While by Lemma 6.1 we have

$$\begin{aligned} {\underline{N}}_{1/2}(F_\tau ,\varepsilon ) [A_{\frac{1}{4}}(F_\tau ,\varepsilon )]^2\le \int \limits _{{\mathbb R}_{\ge 0}^3}\mathcal{K}[\varphi _{\varepsilon }]\mathrm{d}^3F_\tau . \end{aligned}$$
(6.10)

Multiplying \(\frac{1}{\varepsilon ^{\alpha }} m(\tau )\) to both sides of (6.10) and recalling definition of \({\underline{N}}^*_{\alpha }(t,\varepsilon )\) gives

$$\begin{aligned} {\underline{N}}_{1/2}^*(\tau ,\varepsilon ) \Big [A_{\frac{1}{4}+\frac{\alpha }{2}}(F_{\tau },\varepsilon )\Big ]^2\le \frac{1}{\varepsilon ^{\alpha }} m(\tau )\int \limits _{{\mathbb R}_{\ge 0}^3}\mathcal{K}[\varphi _{\varepsilon }]\mathrm{d}^3F_\tau ,\quad \tau \in [s, t]. \end{aligned}$$

Then taking integration \(\int \limits _{s}^{t}\) and using (6.9) and the lower bound (6.8) of \({\underline{N}}_{1/2}^*(\tau ,\varepsilon )\) we obtain

$$\begin{aligned} \int \limits _{s}^{t}[A_{\frac{1}{4}+\frac{\alpha }{2}}(F_{\tau },\varepsilon )]^2\mathrm{d}\tau \le \frac{1}{D_{\lambda }} N_{\alpha }^*(t,\varepsilon )\qquad \forall \, \varepsilon \in (0,\varepsilon _{\lambda }]. \end{aligned}$$
(6.11)

On the other hand, using \(F_t(\{0\})\equiv 0\) on \([0,T]\) and Lemma 6.1 we have

$$\begin{aligned} \sqrt{ N_{\frac{1}{4}+\frac{\alpha }{2}}(F_\tau ,\varepsilon )} \le \int \limits _{0}^{1}\theta ^{\frac{1}{8}+\frac{\alpha }{4}-1}\sqrt{ A_{\frac{1}{4}+\frac{\alpha }{2}}(F_\tau ,\varepsilon \theta )}\,\mathrm{d}\theta \qquad \forall \, \tau \in [s, t]. \end{aligned}$$

Multiplying \(\sqrt{ m(\tau )}\) to both sides and noticing that \( m(\tau )\le m(t)\) we have

$$\begin{aligned} \sqrt{ N_{\frac{1}{4}+\frac{\alpha }{2}}^*(\tau ,\varepsilon )} \le \sqrt{ m(t)}\int \limits _{0}^{1}\theta ^{\frac{1}{8}+\frac{\alpha }{4}-1}\sqrt{ A_{\frac{1}{4}+\frac{\alpha }{2}}(F_{\tau },\varepsilon \theta )}\,\mathrm{d}\theta \qquad \forall \,\tau \in [s,t]. \end{aligned}$$
(6.12)

By conservation of mass \(N(F_\tau )\equiv N(F_0)\) and the non-decrease of the energy \(\tau \mapsto E(F_\tau )\) on \([0,\infty )\) we have \(M_{1/2}(F_\tau )\le \sqrt{N(F_0)E(F_T)}=:c\) for all \(\tau \in [0, T]\) and so \( m(t)\le e^{ct}.\) By taking square of both sides of (6.12) and using Cauchy–Schwarz inequality twice together with (6.11) we deduce

$$\begin{aligned} \int \limits _{s}^{t} N_{\frac{1}{4}+\frac{\alpha }{2}}^*(\tau ,\varepsilon ) \mathrm{d}\tau&\le \frac{e^{ct}}{\frac{1}{8}+\frac{\alpha }{4}} \int \limits _{0}^{1}\theta ^{\frac{1}{8}+\frac{\alpha }{4}-1}\sqrt{t-s}\sqrt{\int \limits _{s}^{t} \Big [A_{\frac{1}{4}+\frac{\alpha }{2}}(F_{\tau },\varepsilon \theta )\Big ]^2\mathrm{d}\tau }\,\mathrm{d}\theta \nonumber \\&\le \frac{e^{ct}}{\sqrt{D_{\lambda }}}\sqrt{t-s}\Big (\frac{1}{8}+\frac{\alpha }{4}\Big )^{-2} \sqrt{\overline{N}_{\alpha }^*(t,\varepsilon )} \end{aligned}$$
(6.13)

where we used inequalities \(N_{\alpha }^*(t,\varepsilon \theta )\le \overline{N}_{\alpha }^*(t,\varepsilon \theta )\le \overline{N}_{\alpha }^*(t,\varepsilon ) \) for all \(\theta \in (0,1)\). On the other hand using (6.6) we have \(N_{\frac{1}{4}+\frac{\alpha }{2}}^*(s,\varepsilon )\le N_{\frac{1}{4}+\frac{\alpha }{2}}^*(\tau ,\varepsilon )\) for all \(\tau \in [s, t]\). This gives

$$\begin{aligned} N_{\frac{1}{4}+\frac{\alpha }{2}}^*(s,\varepsilon )\le \frac{2^4}{\sqrt{D_{\lambda }}}\frac{e^{ct}}{\sqrt{t-s}}\Big (\frac{1}{2}+\alpha \Big )^{-2} \sqrt{\overline{N}_{\alpha }^*(t,\varepsilon )} \qquad \forall \,\varepsilon \in (0,\varepsilon _{\lambda }]. \end{aligned}$$
(6.14)

Since \(\varepsilon \mapsto \overline{N}_{\alpha }^*(t,\varepsilon )\) is non-decreasing, (6.7) follows from (6.14).

Step 3: We now use an iteration process supplied by the inequality (6.7) to deduce (6.5). Let us fix any \(\varepsilon \in (0,\varepsilon _{\lambda }]\) and consider

$$\begin{aligned} \alpha _n=\frac{1}{2}-\frac{1}{2^n},\quad t_k=\Big (1-\frac{1}{2^k}\Big )T,\quad n=1,2,3,\ldots \,;\,\, k=0,1,2,\ldots \,. \end{aligned}$$

Applying (6.7) to \(\alpha =\alpha _{n-k}\), \(t=t_k, s=t_{k-1}\) for \(k=1,2,\ldots , n-1; n\ge 2,\) and noticing that \(\alpha _{n-(k-1)}=\frac{1}{4}+\frac{\alpha _{n-k}}{2}, \frac{1}{2}+\alpha _{n-k}=1-\frac{1}{2^{n-k}}\) and \(t_k-t_{k-1}=\frac{T}{2^k}\), we compute

$$\begin{aligned} \overline{N}_{\alpha _{n-(k-1)}}^*(t_{k-1},\varepsilon ) \le C2^{k/2}e^{-\frac{cT}{2^k}}\Big (1-\frac{1}{2^{n-k}}\Big )^{-2} \sqrt{\overline{N}_{\alpha _{n-k}}^*(t_k,\varepsilon )}. \end{aligned}$$
(6.15)

where \(C=\frac{2^4}{\sqrt{D_{\lambda }T}} e^{cT}.\) For any fixed \(n\ge 2\) let us denote

$$\begin{aligned} A_k:=\overline{N}_{\alpha _{n-k}}^*(t_k,\varepsilon ),\quad B_k:= C 2^{k/2} e^{-\frac{cT}{2^k}}\Big (1-\frac{1}{2^{n-k}}\Big )^{-2},\quad k=0,1,\ldots ,n-1. \end{aligned}$$

Then (6.15) reads

$$\begin{aligned} A_{k-1}\le B_k\sqrt{A_k},\quad k=1,2,\ldots ,n-1. \end{aligned}$$
(6.16)

Since \(\alpha _1=0\) and the conservation of mass imply that \(A_{n-1}=\overline{N}_{0}^*(t_{n-1},\varepsilon ) \le m(T)N(F_0)\), it follows from (6.16) that \(A_k\) are finite \((k=1,2,\ldots , n-1)\) and iterating (6.16) gives

$$\begin{aligned} A_0\le \Big (\prod _{k=1}^{n-1}(B_k)^{\frac{1}{2^{k-1}}}\Big )(A_{n-1})^{\frac{1}{2^{n-1}}}. \end{aligned}$$
(6.17)

Also since \(t_0=0\) hence \(A_0= \overline{N}_{\alpha _n}^*(0,\varepsilon )\), it follows from (6.17) and \(A_{n-1}\le m(T)N(F_0)\) that

$$\begin{aligned} \overline{N}_{\alpha _n}^*(0,\varepsilon ) \le P_n [ m(T)N(F_0) ]^{\frac{1}{2^{n-1}}},\quad n=2,3,4,\ldots \end{aligned}$$

where

$$\begin{aligned} P_n=\prod _{k=1}^{n-1}\left( C 2^{k/2}e^{-\frac{cT}{2^k}} \Big (1-\frac{1}{2^{n-k}}\Big )^{-2} \right) ^{\frac{1}{2^{k-1}}}. \end{aligned}$$

On the other hand by definition of \(\overline{N}_{\alpha }^*(t,\varepsilon )\) and \(\alpha _n=\frac{1}{2}-\frac{1}{2^n}\) we have

$$\begin{aligned} \overline{N}_{\alpha _n}^*(0,\varepsilon )=\overline{N}_{\alpha _n}(F_0,\varepsilon ) \ge N_{\alpha _n}(F_0,\varepsilon )=\varepsilon ^{\frac{1}{2^n}}N_{1/2}(F_0,\varepsilon ) \end{aligned}$$

and so

$$\begin{aligned} N_{1/2}(F_0,\varepsilon ) \le (1/\varepsilon )^{\frac{1}{2^n}}P_n [ m(T)N(F_0)]^{\frac{1}{2^{n-1}}},\quad n=2,3,4,\ldots \,. \end{aligned}$$

Letting \(n\rightarrow \infty \) and computing

$$\begin{aligned} \lim _{n\rightarrow \infty }P_n =\prod _{k=1}^{\infty }C^{\frac{2}{2^{k}}} \prod _{k=1}^{\infty } 2^{\frac{k}{2^{k}}}\prod _{k=1}^{\infty }e^{-\frac{2cT}{4^{k}}} \lim _{n\rightarrow \infty }\prod _{k=1}^{n-1}\Big (1-\frac{1}{2^{n-k}}\Big )^{\frac{-4}{2^{k}}} =C^2\cdot 2^2\cdot e^{-\frac{2cT}{3}}\cdot 1 \end{aligned}$$

we obtain

$$\begin{aligned} N_{1/2}(F_0,\varepsilon )\le (2Ce^{-\frac{cT}{3}})^2=\frac{2^{10}}{D_{\lambda }T} e^{\frac{4}{3}cT}. \end{aligned}$$

Since \(\varepsilon \in (0,\varepsilon _{\lambda }]\) is arbitrary, taking upper limit \(\limsup _{\varepsilon \rightarrow 0^+}\) leads to

$$\begin{aligned} \overline{N}_{1/2}(F_0) \le \frac{2^{10}}{D_{\lambda }T} e^{\frac{4}{3}cT}. \end{aligned}$$

This implies \(\overline{N}_{1/2}(F_0)<\infty \) hence \(\underline{N}_{1/2}(F_0)<\infty \) and so \(D_{\lambda }=\lambda \underline{N}_{1/2}(F_0)\). Thus

$$\begin{aligned} \underline{N}_{1/2}(F_0)\overline{N}_{1/2}(F_0) \le \frac{2^{10}}{\lambda T} e^{\frac{4}{3}cT}. \end{aligned}$$

Since \(\lambda \in (1/2,1)\) is arbitrary, letting \(\lambda \rightarrow 1^{-}\) gives (6.5) and finishes the proof. \(\square \)

As a consequence of Proposition 6.1 we obtain the following

Theorem 6.1

Let \(F_t\) be a distributional solution of Eq. (1.5) with the initial datum \(F_0\). Then

  1. (a)
    $$\begin{aligned} \underline{D}_{1/2}(F_0)\overline{D}_{1/2}(F_0)=\infty \,\,\,\Longrightarrow \,\,\,F_t(\{0\})>0\quad for \,\,\, all\,\,\,\, t>0. \end{aligned}$$

    In general, if

    $$\begin{aligned} \sqrt{\underline{D}_{1/2}(F_0)\overline{D}_{1/2}(F_0)}\,>\frac{50\sqrt{5}}{\sqrt{T}} \exp \Big (\frac{2}{3}\sqrt{N(F_0)E(F_T)}\,T\Big ) \quad for\,\,some\,\,\, T\in (0,\infty ) \end{aligned}$$

    then

    $$\begin{aligned} F_t(\{0\})>0\quad for \,\,\, all\,\,\,\, t\ge T. \end{aligned}$$
  2. (b)

    Suppose \(F_t\) also conserves the energy and

    $$\begin{aligned} \sqrt{\underline{D}_{1/2}(F_0)\overline{D}_{1/2}(F_0)}\,> 213[N(F_0)E(F_0)]^{1/4}. \end{aligned}$$

    Then

    $$\begin{aligned} F_t(\{0\})>0 \quad for\,\,\, all\,\,\,\, t\ge \frac{3}{4}(N(F_0)E(F_0))^{-1/2}. \end{aligned}$$

Proof

Part (a) follows directly from Proposition 6.1 and the propagation of condensation (1.18). Part (b) is a special case of part (a) with \(E(F_T)=E(F_0)\) and \(T=\frac{3}{4}(N(F_0)E(F_0))^{-1/2}\) which is the minimizer of \(\frac{50\sqrt{5}}{\sqrt{T}}\exp (\frac{2}{3}\sqrt{N(F_0)E(F_0)}\,T)\) with respect to \(T\in (0,\infty )\). The minimum is \(50\sqrt{20e/3}\,[N(F_0)E(F_0)]^{1/4}\) and \(213>50\sqrt{20e/3}.\) \(\square \)

As a corollary of Theorem 5.1 and Theorem 6.1 we have the following result concerning the strict positivity of the critical time of condensation.

Theorem 6.2

Let \(F_0\in \mathcal{B}_1^{+}({\mathbb R}_{\ge 0})\) be a regular measure with density \(f_0\) satisfying

$$\begin{aligned} \sqrt{\underline{D}_{1/2}(F_0)\overline{D}_{1/2}(F_0)}\,> 213[N(F_0)E(F_0)]^{1/4},\quad f_0(x)\le Kx^{-1}\quad \forall \, x\in {\mathbb R}_{+} \end{aligned}$$

for some \(0<K<\infty \). Then for the positive number \(T_{K}=(6[N(F_0)]^2K^{-1}+40K^2)^{-1}\), there exists a conservative distributional solution \(F_t\) of Eq. (1.5) with the initial datum \(F_0\), such that \(F_t\) is regular for all \(t\in [0, T_{K}]\) and thus the critical time \(t_c\) of condensation of \(F_t\) is strictly positive and \(t_c\in [T_{K},\,\frac{3}{4}(N(F_0)E(F_0))^{-1/2}]\).

Proof

By assumption we have \(2K\ge \overline{D}_{1/2}(F_0)\ge \sqrt{\underline{D}_{1/2}(F_0)\overline{D}_{1/2}(F_0)}\) and so \(4K^2>(213)^2[N(F_0)E(F_0)]^{1/2}\) which apparently implies that \(T_{K}<\frac{3}{4}[N(F_0)E(F_0)]^{-1/2}\). The conclusion of the theorem then follows from Theorem 5.1, Theorem 6.1, and the definition of the critical time \(t_c\) (see (1.21)). \(\square \)

7 Some Examples

As applications of main results of the paper we present here several examples mentioned in the Introduction.

Example 7.1

Here we show that if \(\alpha >1/2\), then there exist extensively conservative distributional solutions \(F_t\) of Eq. (1.5) on \([0,\infty )\) satisfying \(D_{\alpha }(F_t)=\infty \) for all \(t\in [0, \infty )\), but \(F_t\) have no condensation for all \(t\in [0,\infty )\). Given any regular measure \(G_0\in \mathcal{B}_{-1/2, 1}^{+}({\mathbb R}_{\ge 0})\) satisfying \(M_{-1/2}(G_0) <\frac{1}{85}[N(G_0)E(G_0)]^{1/4}.\) For any \(\alpha >1/2\), take \(1/2<\beta <\alpha \wedge 1\) and consider a local surgery \(F_0\) to \(G_0\) near the origin: \(\mathrm{d}F_0(x)=f_0(x)\sqrt{x}\mathrm{d}x\), \(f_0(x)= x^{\beta -3/2}\mathbf{1}_{\{0<x\le \eta \}}+ g_0(x)\mathbf{1}_{\{x>\eta \}}\) with \(\eta >0\), where \(g_0\) is the density of \(G_0\). We have \(\lim _{\eta \rightarrow 0^+}(M_{-1/2}(F_0), N(F_0), E(F_0)) =(M_{-1/2}(G_0), N(G_0), E(G_0))\). So we can fix a small \(\eta >0\) such that \(M_{-1/2}(F_0)<\frac{1}{80}[N(F_0)E(F_0)]^{1/4}\), i.e. \(\Vert f_0\Vert _{L^1}<\frac{1}{80}[N(f_0)E(f_0)]^{1/4}\). According to Theorem 3.2, Theorem 4.1 and part (d) of Proposition 4.2, there is a unique conservative distributional solution \(F_t\) of Eq. (1.5) on \([0,\infty )\) with the initial datum \(F_0\), and \(F_t\) is regular with density \(f(\cdot ,t)\) which is a unique conservative mild solution of Eq. (1.5) on \([0,\infty )\) satisfying \(f\in C([0,\infty ); L^1({\mathbb R}_{+}))\) and \(f(\cdot ,0)=f_0\). Of course \(F_t\) has no condensation for all \(t\ge 0\). On the other hand, using the Duhamel’s formula (4.8) and (4.10) we deduce

$$\begin{aligned} f(x,t)\ge f_0(x) e^{-c(x) t}\qquad \mathrm{a.e.}\quad x\in {\mathbb R}_{+}\quad \mathrm{for\,\,\,all}\,\,\, t\ge 0 \end{aligned}$$

where \(c(x)=\sqrt{x}N(f_0)+\frac{9}{8}\sqrt{N(f_0)E(f_0)}\) (by (4.7), (4.14)). This implies \(\underline{D}_{\beta }(F_t)>0\) and so \(D_{\alpha }(F_t)=\infty \) for all \(t\ge 0\) because \(\alpha >\beta \). This shows also that the unboundedness of a mild solution near the origin does not generally imply the condensation in finite time. \(\square \)

Example 7.2

In this example we show that the condensation can happen for arbitrary temperature. Given any \(G_0\in \mathcal{B}^{+}_{1}({\mathbb R}_{\ge 0})\). As did above we make a surgery to \(G_0\) near the origin: for any \(K>1\) we define \(F_0\in \mathcal{B}^{+}_1({\mathbb R}_{\ge 0})\) by \(\mathrm{d}F_0(x)=K^{1/4}x^{-1}\mathbf{1}_{\{0<x\le 1/K\}}\sqrt{x}\mathrm{d}x+\mathbf{1}_{\{x>1/K\}} \mathrm{d}G_0(x).\) We have \(\Vert F_0-G_0\Vert _1\rightarrow 0\) as \(K\rightarrow \infty \) and

$$\begin{aligned} D_{1/2}(F_0)=2K^{1/4},\,\, N(F_0)\le 2K^{-1/4}+N(G_0),\,\, E(F_0)\le \frac{2}{3}K^{-5/4}+E(G_0). \end{aligned}$$
(7.1)

So for any \(T\in (0,\infty )\) we can choose \(K\) large enough such that

$$\begin{aligned} D_{1/2}(F_0)> \frac{50\sqrt{5}}{\sqrt{T}}\exp \Big (\frac{2}{3}\sqrt{N(F_0)E(F_0)}\,T\Big ). \end{aligned}$$

Let \(F_t, G_t\) be conservative distributional solutions of Eq. (1.5) on \([0,\infty )\) with the initial data \(F_0, G_0\) respectively. By Theorem 6.1, \(F_t(\{0\})>0\) for all \(t\ge T.\) Notice that \(D_{1/2}(F_0)>0\) implies \(M_{-1/2}(F_0)=\infty \). So there is no contradiction to the stability theorem (Theorem 3.1) even if we can assume that the original initial datum \(G_0\) satisfies \(M_{-1/2}(G_0)\le \frac{1}{80}[N(G_0)E(G_0)]^{1/4}\) so that \(G_t\) has no condensation for all \(t\ge 0\). Notice also that \(F_0\) satisfies \(D_{\alpha }(F_0)=0\) for all \(0<\alpha <1/2\) and so our previous result Theorem 2.1(b) of [15] cannot apply to \(F_t\). Finally we note that if \(K\) is large enough, then \(F_0\) has almost the same temperature as \(G_0\) so that it can be arbitrarily high/low as that of \(G_0\). \(\square \)

Example 7.3

We show that the initial data \(F_0\) satisfying the conditions in Theorem 6.2 exist extensively. Take any regular measure \(G_0\in \mathcal{B}^{+}_{1}({\mathbb R}_{\ge 0})\) and let \(g_0\) be its density. As in Example 7.2, for any \(K>1\) we define \(F_0\in \mathcal{B}^{+}_{1}({\mathbb R}_{\ge 0})\) by \(\mathrm{d}F_0(x)=f_0(x)\sqrt{x}\mathrm{d}x\) with \(f_0(x)=K^{1/4}x^{-1}\mathbf{1}_{\{0<x\le 1/K\}}+\mathbf{1}_{\{x>1/K\}}\min \{ g_0(x), Kx^{-1}\},\,x\in {\mathbb R}_{+}.\) It is easily seen that \(f_0(x)\le Kx^{-1}\) for all \(x\in {\mathbb R}_{+}\) and (7.1) holds for all \(K>1\). Thus for \(K>1\) large enough, the condition \(D_{1/2}(F_0)>213[N(F_0)E(F_0)]^{1/4}\) is also satisfied. By Theorem 6.2, this example shows also that the critical time of condensation can be strictly positive even for a large class of initial data which are unbounded near the origin. \(\square \)

Example 7.4

Consider a family of regular initial data \(F_0\in \mathcal{B}^{+}_{1}({\mathbb R}_{\ge 0}), \mathrm{d}F_0(x)=f_0(x)\sqrt{x}\mathrm{d}x\), given by \(f_0(x)=a[x\log (1/x)]^{-1}\mathbf{1}_{\{0<x\le 1/2\}}+ bx^{-\gamma }\mathbf{1}_{\{x>1/2\}}\) with constants \(a>0,b>0,\gamma >5/2\). For such an \(F_0\) we have \(D_{1/2}(F_0)=0\) and \(M_{-1/2}(F_0)=\infty \) so that neither the condensation results in Sect. 6 nor the regularity results in Sect. 4 can be used, but \(f_0\) satisfies \(f_0(x)\le K x^{-1}\,(\forall \, x>0)\) with \(K=\max \{ a(\log 2)^{-1}, b2^{\gamma -1}\}\) and so by Theorem 5.1 there exists a conservative distributional solution \(F_t\) of Eq. (1.5) on \([0,\infty )\) with the initial datum \(F_0\) such that \(F_t\) is regular on \([0, T_{K}]\). To see whether \(F_t\) has condensation in finite time, we now have no other method but to check the low temperature condition. Let \(\overline{T}, \overline{T}_c\) be the kinetic temperature and the critical temperature defined in (1.14) with \(N=N(F_0), E=E(F_0)\). Then, for some explicit constants \(C_2\ge C_1>0\) depending only on \(\gamma \), we have \(C_1(a\vee b)^{-2/3}\le \overline{T}/\overline{T}_c\le C_2(a\vee b)^{-2/3}.\) If \(a\vee b\) is large enough such that \(\overline{T}/\overline{T}_c<1\), then according to part (III.2) of Theorem 1.2, \(F_t\) has condensation in finite time with the critical time \(t_c\in [T_{K}, \infty )\); while if \(a\vee b\) is so small that \(\overline{T}/\overline{T}_c\ge 1\), we do not know whether \(F_t\) has condensation in finite time since as mentioned above the results obtained so far do not work for this case of such \(F_0\). \(\square \)

Example 7.5

As the last example we discuss the size of the universal constant \(\theta _*\) and its influence on the working area of the condensation condition (1.24).

We first show that \(\theta _*\le 1/2\). Assume \(\theta _*>1/2\). Let \(1/2<\beta <\theta _*\wedge 1\) and let \(F_t\) be the conservative regular distributional solution of Eq. (1.5) on \([0,\infty )\) constructed in Example 7.1. In the condition (1.24) let us choose \(N=N(F_0), E=E(F_0)\). From \(D_{\beta }(F_0)>0\) and \(3/2>\beta , \theta _*>\beta \) we have

$$\begin{aligned} \lim _{\rho \rightarrow 0^+}\inf _{0<\varepsilon \le \rho }\frac{F_0([0,\varepsilon ])}{\varepsilon ^{3/2}} =D_{3/2}(F_0)=\infty ,\quad \lim _{\rho \rightarrow 0^+}\frac{F_0([0,\rho ])}{\rho ^{\theta _*}} =D_{\theta _*}(F_0)=\infty \end{aligned}$$

and so \(F_0\) satisfies (1.24) sufficiently: the left hand side of (1.24) is \(\infty \). But \(F_t\) has no condensation for all \(t\ge 0\). This proves \(\theta _*\le 1/2\). Another example for proving \(\theta _*\le 1/2\) is the regular equilibrium \(F_t\equiv F_\mathrm{be}\) with density \(f_\mathrm{be}(x)=(e^{x/\kappa }-1)^{-1}\).

Next we show that if \(\theta _*<1/2\), then for any \(N>0, E>0\), the condition (1.20) does not imply (1.24), and the same holds also true for \(\theta _*=1/2\) if \(K^*>214(NE)^{1/4}\).

Fix \(\gamma \ge 3\) and let \(a=107(NE)^{1/4}\). Consider a conservative distributional solution \(F_t\) of Eq. (1.5) on \([0,\infty )\) with the initial datum \(F_0\in \mathcal{B}_{1}^+({\mathbb R}_{\ge 0})\) given by

$$\begin{aligned} \mathrm{d}F_0(x)=(a x^{-1}\mathbf{1}_{\{0<x\le \eta \}}+b x^{-\gamma }\mathbf{1}_{\{x\ge R\}})\sqrt{x}\mathrm{d}x \end{aligned}$$

where

$$\begin{aligned} 0<\eta <\eta _0(N,E):=\min \left\{ \Bigg (\frac{N}{4a}\Bigg )^{2},\, \Bigg (\frac{3E}{4a}\Bigg )^{2/3},\, \frac{E}{6N}\right\} \end{aligned}$$

and \(b>0, R>0\) are determined by the given mass and energy: \( N(F_0)=N, E(F_0)=E\), i.e.,

$$\begin{aligned} b=(\gamma -3/2) R^{\gamma -3/2}N_{\eta },\quad R=\frac{\gamma -5/2}{\gamma -3/2}\cdot \frac{E_\eta }{N_{\eta }} \end{aligned}$$

where \(N_{\eta }=N-2a\eta ^{1/2},\, E_{\eta }=E-\frac{2}{3}a\eta ^{3/2}.\) Notice that our restriction on \(\eta \) and \(\gamma \ge 3\) imply \(N_{\eta }>N/2, E_{\eta }>E/2\), and \(R>E/(6N) >\eta \). By construction, \(F_0\) satisfies (1.20): \(D_{1/2}(F_0)=2a>213(NE)^{1/4}\) and so \(F_t\) has condensation in finite time. On the other hand, let \(\nu >0, \rho _0=\rho _0(N,E,\nu )>0, K^*=K^*(N,E,\nu )>0\) be given in part (II) of Theorem 1.2, and suppose first that \(\theta _*<1/2\). In this case we take any \(0<\delta <1\) and let \(\eta \) be further small:

$$\begin{aligned} 0<\eta <\min \left\{ \eta _0(N,E),\,\Bigg (\frac{\delta \nu }{2a}\Bigg )^{2}\Bigg (\frac{E}{6N}\Bigg )^3, \Bigg (\frac{\delta K^*}{2a}\Bigg )^{\frac{1}{\frac{1}{2}-\theta _*}} \right\} . \end{aligned}$$

We show that \(F_0\) does not satisfy (1.24):

$$\begin{aligned} \sup _{0<\rho \le \rho _0}\min \left\{ \inf _{0<\varepsilon \le \rho }\frac{F_0([0,\varepsilon ])}{\nu \varepsilon ^{3/2}}\,,\, \frac{F_0([0,\rho ])}{K^* \rho ^{\theta _*}}\right\} \le \delta . \end{aligned}$$
(7.2)

In fact, for any \(0<\rho \le \rho _0\), if \(\rho \ge R\), then \(\rho >E/(6N):=\varepsilon _0\) and so (since \(\eta <\varepsilon _0<R\))

$$\begin{aligned} \inf _{0<\varepsilon \le \rho }\frac{F_0([0,\varepsilon ])}{\nu \varepsilon ^{3/2}} \le \frac{F_0([0,\varepsilon _0])}{\nu \varepsilon _0^{3/2}} =\frac{2a}{\nu \varepsilon _0^{3/2}} {\eta }^{1/2} <\delta ; \end{aligned}$$

while if \(\rho <R\), then

$$\begin{aligned} \frac{F_0([0,\rho ])}{K^*\rho ^{\theta _*}} =\frac{2a}{K^*\rho ^{\theta _*}}(\eta \wedge \rho )^{1/2} \le \frac{2a}{K^*}\eta ^{\frac{1}{2}-\theta _*} <\delta . \end{aligned}$$

Thus for any \(0<\rho \le \rho _0\), one of the two is less than \(\delta \). This proves (7.2).

Finally suppose \(\theta _*=1/2\). In this case we assume that there is \(0<\delta <1\) such that \(K^*\ge \frac{214}{\delta }(NE)^{1/4}=\frac{2a}{\delta }\). Then for \(\eta \) satisfying

$$\begin{aligned} 0<\eta <\min \left\{ \eta _0(N,E),\,\Bigg (\frac{\delta \nu }{2a}\Bigg )^{2}\Bigg (\frac{E}{6N}\Bigg )^3\right\} \end{aligned}$$

the same argument applies and so (7.2) with \(\theta _*=1/2\) holds also for such \(\delta \). \(\square \)