Global dynamics for a charged and colliding plasma in presence of a massive scalar field on the Robertson–Walker spacetime

Abstract

We consider the coupled Einstein–Maxwell–Boltzmann system with cosmological constant in presence of a massive scalar field. The background metric is that of Friedman–Lemaître–Robertson–Walker space time in the spatially homogeneous case where the unknown functions only depend on time and not on the space variables $(x^i)$, $i=1,2,3$. By combining the energy estimates method with that of characteristics we derive under suitable conditions on the collision kernel [see (2.20)], a local (in time) solution of the coupled system. Further, under the hypotheses that the data are small in some appropriate norms and that the cosmological constant satisfies $\Lambda > -4\pi m^2\Phi _0^2$, we derive a unique global (in time) solution (Theorem 6.1).

Appendices

An Energy estimate for a first order hyperbolic partial differential equation

Let us consider the first order PDE in $u=u(t,x)$:

$$\begin{aligned} u_t+\sum _{i=1}^na_i(x,t)u_{x_i}+b(x,t)u=f(x,t)\quad in\quad \mathbb {R}^n\times \mathbb {R}; \end{aligned}$$

(A.1)

with initial data:

$$\begin{aligned} u(x,0)=u_0(x)\quad in\quad \mathbb {R}^n. \end{aligned}$$

(A.2)

Here b and f are functions defined on $\mathbb {R}^{n+1}$ and $a=(a_1,\ldots ,a_n)$ a family of functions such that:

$$\begin{aligned} \sum _{i=1}^n\big (\sup _{(x,t)}|a_i(x,t)|\big )=:|a|\le \frac{1}{\kappa }, \end{aligned}$$

(A.3)

where $\kappa$ is a positive constant.

Remark A.1

We would like to the point out the fact that the notations in this subsection are independent of those of the other sections of the paper. For examples, the letter a here is used for a collection of real valued functions and has nothing to do with the expansion factor of the previous sections, f here is the source term of the PDE we are dealing with and must not be confused with the distribution function. We hope that this clash of notations will not confuse the reader.

Proposition A.1

Let $a=(a_1,\ldots ,a_n)$ be a family of class $C^1$ functions with bounded partial dérivatives with respect to $x_i$ defined on $\mathbb {R}^{n+1}$ and satisfying (A.3). Let b be a bounded function defined on $\mathbb {R}^{n+1}$, and u a solution of the initial value problem (A.1)–(A.2). Then, for every $T>0$, if $f\in C\big ([0,T);L^2(\mathbb {R}^n)\big )$ and $u_0\in L^2(\mathbb {R}^n)$, we have:

$$\begin{aligned} \int _{\mathbb {R}^n}e^{-\alpha t}u^2dx\le \int _{\mathbb {R}^n} u^2_0 dx+\int _0^te^{-\alpha s}\Vert f(s,\cdot )\Vert ^2_{L^2(\mathbb {R}^n)}ds ,\qquad t\le T; \end{aligned}$$

(A.4)

where $\alpha$ is a positive constant.

Proof

Let $\overline{t}>0$ be given. Define $D=D_{\;\kappa ,t,\overline{t}\;},0\le t<T<\overline{t}$ by:

$$\begin{aligned} D=\{(x,s)\in \mathbb {R}^{n+1};\;\kappa |x|<\overline{t}-s,\;0<s<t\} . \end{aligned}$$

(A.5)

Let us denote by $\Sigma _{t,\overline{t}}$, $\Sigma _{0,\overline{t}}$ and $S_{t,\overline{t}}$ respectively, the upper boundary, the lower boundary and the side of D; ie:

$$\begin{aligned} \left\{ \begin{array}{ll} \Sigma _{t,\overline{t}}= \{(x,t)\in \mathbb {R}^{n+1};\;\kappa |x|<\overline{t}-t\}&{} \\ \Sigma _{0,\overline{t}}= \{(x,0)\in \mathbb {R}^{n+1};\;\kappa |x|<\overline{t}\} &{}\\ S_{t,\overline{t}}= \{(x,s)\in \mathbb {R}^{n+1};\;\kappa |x|=\overline{t}-s,\;0<s<t\}&{} \end{array} \right. . \end{aligned}$$

(A.6)

For $\alpha >0$, we multiply Eq. A.1 by $2e^{-\alpha t}u$, and obtain an equation which can be written as:

$$\begin{aligned} \big (e^{-\alpha t}u^2\big )_t+\sum _{i=1}^n\big (e^{-\alpha t}a_iu^2\big )_{x_i}+e^{-\alpha t}\big (\alpha +2b-\sum _{i=1}^na_{i,x_i}\big )u^2=2e^{-\alpha t}u\,f . \end{aligned}$$

(A.7)

The first two terms of (A.7) can be written as a divergence. If we set $X = X^{\mu }\partial _{\mu }:= e^{-\alpha t}u^2 a^{\mu } \partial _\mu ,$ then

$$\begin{aligned}&div X=\sum _{\mu =0}^n\big (e^{-\alpha t}a^{\mu }u^2\big )_{x_{\mu }}\\&\quad = \big (e^{-\alpha t}u^2\big )_t+\sum _{i=1}^n\big (e^{-\alpha t}a_iu^2\big )_{x_i} ; \end{aligned}$$

with $a^{\mu }=(1, a_1, \ldots , a_n)$ and $x_0=t$. We integrate (A.7) on D. We have

$$\begin{aligned}&\int _D\left( \big (e^{-\alpha t}u^2\big )_t+\sum _{i=1}^n\big (e^{-\alpha t}a_iu^2\big )_{x_i}+e^{-\alpha t}\big (\alpha +2b-\sum _{i=1}^na_{i,x_i}\big )u^2\right) dt\,dx\nonumber \\&\quad =2\int _D e^{-\alpha t}u\, f dt\,dx . \end{aligned}$$

(A.8)

By the Stokes theorem, we have:

$$\begin{aligned}&\int _{D}\Big [\big (e^{-\alpha t}u^2\big )_t+\sum _{i=1}^n\big (e^{-\alpha t}a_iu^2\big )_{x_i}\Big ]dxdt\\&\quad = \int _{\partial D}X\cdot \eta dS= \int _{\Sigma _{0,\overline{t}}}X\cdot \eta dS+\int _{\Sigma _{t,\overline{t}}}X\cdot \eta dS+\int _{S_{t,\overline{t}}}X\cdot \eta dS , \end{aligned}$$

where $\eta$ is the outward unit normal vector to $\partial D$ and dS the surface element on $\partial D$. But,

• on $S_{t,\overline{t}}, \; \eta =(\eta _t,\eta _1,\ldots ,\eta _n)=\frac{1}{\sqrt{1+\kappa ^2}}\left( 1,\kappa \frac{x_1}{|x|}, \ldots ,\kappa \frac{x_n}{|x|} \right)$ from where we have

  $$\begin{aligned}\int _{S_{t,\overline{t}}}X\cdot \eta dS=\int _{S_{t,\overline{t}}}e^{-\alpha t}\big (\sum _{i=1}^n\eta _ia_i+\eta _t\big )u^2dS . \end{aligned}$$

• on $\Sigma _{t,\overline{t}}$ and $\Sigma _{0,\overline{t}}$ the outward unit normals are respectively $\eta =(1,\ldots ,0,0)$ and $\eta =(-1,\ldots ,0,0)$, and then

  $$\begin{aligned} \int _{\Sigma _{t,\overline{t}}}X\cdot \eta dS=\int _{\Sigma _{t,\overline{t}}}e^{-\alpha t}u^2dx \qquad \text{ and }\qquad \int _{\Sigma _{0,\overline{t}}}X\cdot \eta dS=-\int _{\Sigma _{0,\overline{t}}}u_0^2dx . \end{aligned}$$

Let us observe that, by (A.3) we have $\sum _{i=1}^n\eta _ia_i+\eta _t\ge 0$ which leads to the following inequality:

$$\begin{aligned} \int _{D}\Big [\big (e^{-\alpha t}u^2\big )_t+\sum _{i=1}^n\big (e^{-\alpha t}a_iu^2\big )_{x_i}\Big ]dxdt\ge \int _{\Sigma _{t,\overline{t}}}e^{-\alpha t}u^2dx-\int _{\Sigma _{0,\overline{t}}}u_0^2dx . \end{aligned}$$

(A.9)

Now since b and the partial derivatives of $a_i$ with respect to $x_i$ are bounded, we can choose $\alpha$ such that:

$$\begin{aligned} \alpha +2b-\sum _{i=1}^na_{i,x_i}\ge 1\quad \text{ in }\quad D . \end{aligned}$$

(A.10)

From now, we suppose that $\alpha$ is chosen such that (A.10) holds. We then have

$$\begin{aligned} \int _{D}e^{-\alpha t}\big (\alpha +2b-\sum _{i=1}^na_{i,x_i}\big )u^2 dtdx\ge \int _{D}e^{-\alpha t}u^2 dtdx . \end{aligned}$$

(A.11)

Now, recall the trivial inequality

$$\begin{aligned} \int _{D}2e^{-\alpha t}ufdxdt\le \int _{D}e^{-\alpha t}u^2dxdt+\int _{D}e^{-\alpha t}f^2dxdt . \end{aligned}$$

(A.12)

Adding (A.9) and (A.11) and using (A.7), (A.12) give:

$$\begin{aligned} \int _{\Sigma _{t,\overline{t}}}e^{-\alpha t}u^2dx\le \int _{\Sigma _{0,\overline{t}}}u_0^2dx+\int ^t_0ds\int _{\Sigma _{s,\overline{t}}}e^{-\alpha s}f^2(s,\cdot )dx ; \end{aligned}$$

(A.13)

where in (A.13), $\Sigma _{s,\overline{t}}= \{(x,s)\in \mathbb {R}^{n+1};\;\kappa |x|<\overline{t}-s\,\}$ for fixed s. Now, we let $\overline{t}$ tends to $\infty$, then $\Sigma _{t,\overline{t}}$, $\Sigma _{s,\overline{t}}$ et $\Sigma _{0,\overline{t}}$ tend respectively to $\{t\}\times \mathbb {R}^n$, $\{s\}\times \mathbb {R}^n$ and $\{0\}\times \mathbb {R}^n$. Therefore (A.13) gives:

$$\begin{aligned} \int _{\mathbb {R}^n}e^{-\alpha t}u^2dx\le \int _{\mathbb {R}^n}u_0^2dx+\int ^t_0ds\int _{\mathbb {R}^n}e^{-\alpha s}f^2(s,\cdot )dx ; \end{aligned}$$

(A.14)

which is the desired inequality. $\square$

Remark A.2

For $0<t\le T$, inequality (A.14) reads

$$\begin{aligned} e^{-\alpha t}\Vert u(t,\cdot )\Vert ^2_{L^2(\mathbb {R}^n)}\le \Vert u_0\Vert ^2_{L^2(\mathbb {R}^n)}+\int _0^t e^{-\alpha s}\Vert f(s,\cdot )\big \Vert ^2_{L^2(\mathbb {R}^n)}ds . \end{aligned}$$

(A.15)

The $H^k$-version of the previous Proposition reads:

Proposition A.2

Let $a=(a_1,\ldots ,a_n), b$ be a family of smooth functions defined on $\mathbb {R}^{n+1}$ such that the collection of functions a satisfies (A.3). Suppose that the partial derivatives up to order $k\in \mathbb {N}^{*}$ of a and b with respect to the space variables x are bounded. Assume that u is a $C^1$ solution of the initial value problem (A.1)–(A.2). Then, for every $T>0$, if $f\in C\big ([0,T);H^k(\mathbb {R}^n)\big )$ and $u_0\in H^k(\mathbb {R}^n)$, then for all $t\in [0,T], \; u(t, \cdot )\in H^k(\mathbb {R}^n)$ and

$$\begin{aligned} e^{-\delta _0 t}\Vert u(t,\cdot )\Vert ^2_{H^k(\mathbb {R}^n)}\le \Vert u_0\Vert ^2_{H^k(\mathbb {R}^n)}+ C_0\int _0^t\Vert e^{-\delta _0 s}f(s,\cdot )\big \Vert ^2_{H^k(\mathbb {R}^n)}ds ; \end{aligned}$$

(A.16)

where $\delta _0$ and $C_0$ are positive constants.

Proof

Note that (A.15) gives (A.16) for $k=0$. Consider the commutator $[L,\,\partial ^{\alpha }]$ defined by: $[L,\,\partial ^{\alpha }]u=L\partial ^{\alpha }u-\partial ^{\alpha }Lu$, where L is the differential operator associated to (A.1) (here, $\partial ^{\alpha }\equiv \partial ^{\alpha }_x)$. We have, since (A.1):

$$\begin{aligned} L\partial ^{\alpha }u=\partial ^{\alpha }f+[L,\,\partial ^{\alpha }]u . \end{aligned}$$

(A.17)

Applying Proposition A.1 to (A.17) with u replaced by $\partial ^{\alpha }u$ and f replaced by $\partial ^{\alpha }f+[L,\,\partial ^{\alpha }]u$ shows that there exists $\delta >0$ such that:

$$\begin{aligned} e^{-\delta t}\Vert (\partial ^{\alpha }u)(t,\cdot )\Vert ^2_{L^2(\mathbb {R}^n)}\le & {} \Vert (\partial ^{\alpha }u)(0,\cdot )\Vert ^2_{L^2(\mathbb {R}^n)}\nonumber \\&+ \int _0^te^{-\delta s}\big \Vert (\partial ^{\alpha }f+[L,\,\partial ^{\alpha }]u)(s,\cdot )\big \Vert ^2_{L^2(\mathbb {R}^n)}ds \nonumber \\\le & {} \Vert (\partial ^{\alpha }u)(0,\cdot )\Vert ^2_{L^2(\mathbb {R}^n)}+ 2\int _0^te^{-\delta s}\big \Vert \partial ^{\alpha }f\big \Vert ^2_{L^2(\mathbb {R}^n)}ds\nonumber \\&+2\int _0^te^{-\delta s}\Vert [L,\,\partial ^{\alpha }]u(s,\cdot )\big \Vert ^2_{L^2(\mathbb {R}^n)}ds ; \nonumber \\ \end{aligned}$$

(A.18)

where $|\alpha |\le k$. On the order hand,

$$\begin{aligned}{}[L,\,\partial ^{\alpha }]u= & {} L\partial ^{\alpha }u-\partial ^{\alpha }Lu \nonumber \\= & {} (\partial ^{\alpha }u)_t+\sum _{i=1}^na_i(\partial ^{\alpha }u)_{x_i}+b\partial ^{\alpha }u-\partial ^{\alpha } u_t-\partial ^{\alpha }\big (\sum _{i=1}^na_iu_{x_i}\big )-\partial ^{\alpha }(bu) \nonumber \\= & {} \sum _{i=1}^na_i(\partial ^{\alpha }u)_{x_i}+b\partial ^{\alpha }u -\sum _{i=1}^n\sum _{0\le \beta \le \alpha }\complement ^{\beta }_{\alpha }\partial ^{\beta } a_i\partial ^{\alpha -\beta }u_{x_i}-\sum _{0\le \beta \le \alpha } \complement ^{\beta }_{\alpha }\partial ^{\beta }b\partial ^{\alpha -\beta }u \nonumber \\= & {} -\sum _{i=1}^n\sum _{0<\beta \le \alpha }\complement ^{\beta }_{\alpha }\partial ^{\beta }a_i\partial ^{\alpha -\beta }u_{x_i} -\sum _{0<\beta \le \alpha }\complement ^{\beta }_{\alpha }\partial ^{\beta }b\partial ^{\alpha -\beta }u. \end{aligned}$$

(A.19)

Since $\partial ^{\beta }a_i$ and $\partial ^{\beta }b$ are bounded, the last equality gives the following estimate

$$\begin{aligned} \Vert [L,\,\partial ^{\alpha }]u\Vert _{L^2(\mathbb {R}^n)}\le & {} \Big \Vert \sum _{i=1}^n\sum _{0<\beta \le \alpha }\complement ^{\beta }_{\alpha } \partial ^{\beta }a_i\partial ^{\alpha -\beta }u_{x_i}\Big \Vert _{L^2(\mathbb {R}^n)}+ \Big \Vert \sum _{0<\beta \le \alpha }\complement ^{\beta }_{\alpha }\partial ^{\beta } b\partial ^{\alpha -\beta }u\Big \Vert _{L^2(\mathbb {R}^n)} \\\le & {} C\sum _{|\alpha |\le k}\Big \Vert \partial ^{\alpha }u\Big \Vert _{L^2(\mathbb {R}^n)}+C\sum _{|\alpha |\le k-1}\Big \Vert \partial ^{\alpha }u\Big \Vert _{L^2(\mathbb {R}^n)} \\\le & {} C\Vert u\Vert _{H^k(\mathbb {R}^n)}. \end{aligned}$$

From (A.18) we deduce that; $\forall \alpha$, $|\alpha |\le k$,

$$\begin{aligned} e^{-\delta t}\Vert (\partial ^{\alpha }u)(t,\cdot )\Vert ^2_{L^2(\mathbb {R}^n)}\le & {} \Vert (\partial ^{\alpha } u)(0,\cdot )\Vert ^2_{L^2(\mathbb {R}^n)}\\&+ 2\int _0^te^{-\delta s}\big \Vert (\partial ^{\alpha }f(s,\cdot )\big \Vert ^2_{L^2(\mathbb {R}^n)}ds\\&+2C\int _0^te^{-\delta s}\Vert u(s,\cdot )\Vert ^2_{H^k(\mathbb {R}^n)}ds . \end{aligned}$$

Let us take the sum over $|\alpha |\le k$; we obtain:

$$\begin{aligned} e^{-\delta t}\Vert u(t,\cdot )\Vert ^2_{H^k(\mathbb {R}^n)}\le & {} \Vert u(0,\cdot )\Vert ^2_{H^k(\mathbb {R}^n)}+ 2\int _0^te^{-\delta s}\big \Vert f(s,\cdot )\big \Vert ^2_{H^k(\mathbb {R}^n)}ds\\&+2C\int _0^te^{-\delta s}\Vert u(s,\cdot )\Vert ^2_{H^k(\mathbb {R}^n)}ds ; \end{aligned}$$

and using Gronwall’s Lemma, we have:

$$\begin{aligned} e^{-\delta t}\Vert u(t,\cdot )\Vert ^2_{H^k(\mathbb {R}^n)}\le \Big (\Vert u(0,\cdot )\Vert ^2_{H^k(\mathbb {R}^n)}+2\int _0^te^{-\delta s}\big \Vert f(s,\cdot )\big \Vert ^2_{H^k(\mathbb {R}^n)}ds\Big )e^{2Ct}ds . \end{aligned}$$

Note that $\delta$ and C are two positive constants which depend only on the bounds of the $C^k$-norms of a and b. Now setting $\delta _0=\delta +2C$ and $C_0=2e^{2CT}$ gives

$$\begin{aligned} e^{-\delta _0 t}\Vert u(t,\cdot )\Vert ^2_{H^k(\mathbb {R}^n)}\le \Vert u_0\Vert ^2_{H^k(\mathbb {R}^n)}+C_0\int _0^te^{-\delta _0s}\Vert f(s,\cdot )\big \Vert ^2_{H^k(\mathbb {R}^n)}ds ; \end{aligned}$$

which is the desired inequality. $\square$

The weighted version of Proposition A.2 can be stated as follows:

Corollary A.1

Let d be a non negative real number and k an integer. Under the hypotheses of Proposition A.2 assume further that the collection of functions $a=\left( a_1, \ldots , a_n\right)$ does not depend on the space variables ($a\equiv a(t)$) and that $b\equiv 0$. If u is a $C^1$-solution of the Cauchy problem (A.1)–(A.2), then for every $T>0$, if $f\in C\big ([0,T];H^k_d(\mathbb {R}^n)\big )$, $u_0\in H^k_d(\mathbb {R}^n)$ then for all $t\in [0,T], \; u(t, \cdot )\in H^k_d(\mathbb {R}^n)$ and

$$\begin{aligned} e^{-\delta _1t}\Vert u(t,\cdot )\Vert ^2_{H^k_d(\mathbb {R}^n)}\le \Vert u_0\Vert ^2_{H^k_d(\mathbb {R}^n)}+C_1\int _0^te^{-\delta _1s}\Vert f(s,\cdot )\big \Vert ^2_{H^k_d(\mathbb {R}^n)}ds \end{aligned}$$

(A.20)

where $\delta _1$ and $C_1$ are positive constants which depend only on $\kappa$.

Proof

Let $\beta \in \mathbb {N}^n$ be given such that $|\beta |\le k$. If we differentiate Eq. (A.1) with $\partial ^{\beta }$ (recall $a\equiv a(t)$ and $b\equiv 0$) and then multiply the differentiated equation by the weight $(1+|x|)^{d+|\beta |}$, then we obtain

$$\begin{aligned}&\big ((1+|x|)^{d+|\beta |}\partial ^{\beta }u\big )_t+\sum _{i=1}^na_i \big ((1+|x|)^{d+|\beta |}\partial ^{\beta }u\big )_{x_i}\\&\quad = (1+|x|)^{d+|\beta |}\partial ^{\beta }f +\sum _{i=1}^n(d+|\beta |)\frac{x_i}{|x|}(1+|x|)^{d+|\beta |-1} a_i\partial ^{\beta }u . \end{aligned}$$

But the l.h.s is defined by L; so we have:

$$\begin{aligned}&L\big [(1{+}|x|)^{d+|\beta |}\partial ^{\beta }u\big ]=(1+|x|)^{d+|\beta |}\partial ^{\beta }f +\sum _{i=1}^n(d+|\beta |)\frac{x_ia_i}{|x|(1+|x|)}(1+|x|)^{d+|\beta |}\partial ^{\beta }u .\qquad \qquad \end{aligned}$$

(A.21)

Since the function a is bounded and $\frac{|x_i|}{|x|(1+|x|)}\le 1$, we have:

$$\begin{aligned}&\bigg \Vert \sum _{i=1}^n(d+|\beta |)\frac{x_ia_i}{|x|(1+|x|)} (1+|x|)^{d+|\beta |}\partial ^{\beta }u\bigg \Vert _{L^2(\mathbb {R}^n)}\nonumber \\&\quad \le C\Vert (1+|x|)^{d+|\beta |}\partial ^{\beta }u\Vert _{L^2(\mathbb {R}^n)}\le C\Vert u\Vert _{H^k_d(\mathbb {R}^n)} . \end{aligned}$$

(A.22)

Now, applying Proposition A.1 to Eq. (A.21) shows that there exists a constant $\delta >0$ such that:

$$\begin{aligned}&e^{-\delta t}\Vert (1+|x|)^{d+|\beta |}\partial ^{\beta }u(t,\cdot )\Vert ^2_{L^2(\mathbb {R}^n)}\nonumber \\&\quad \le \Vert (1+|x|)^{d+|\beta |}\partial ^{\beta }u_0\Vert ^2_{L^2(\mathbb {R}^n)}+2\int _0^te^{-\delta s}\Vert f(s,\cdot )\big \Vert ^2_{H^k_d(\mathbb {R}^n)}ds\nonumber \\&\qquad +2C\int _0^te^{-\delta s}\Vert u(s,\cdot )\big \Vert ^2_{H^k_d(\mathbb {R}^n)}ds ; \end{aligned}$$

(A.23)

where we have used (A.22). Summing over $|\beta |\le k$, we obtain:

$$\begin{aligned} e^{-\delta t}\Vert u(t,\cdot )\Vert ^2_{H^k_d(\mathbb {R}^n)}\le & {} \Vert u_0\Vert ^2_{H^k_d(\mathbb {R}^n)}+2\int _0^te^{-\delta s}\Vert f(s,\cdot )\big \Vert ^2_{H^k_d(\mathbb {R}^n)}ds\nonumber \\&+2C\int _0^te^{-\delta s}\Vert u(s,\cdot )\big \Vert ^2_{H^k_d(\mathbb {R}^n)}ds ; \end{aligned}$$

(A.24)

from where using Gronwall’s Lemma, we get:

$$\begin{aligned}e^{-\delta t}\Vert u(t,\cdot )\Vert ^2_{H^k_d(\mathbb {R}^n)}\le \Big (\Vert u_0\Vert ^2_{H^k_d(\mathbb {R}^n)}+2\int _0^te^{-\delta s}\Vert f(s,\cdot )\big \Vert ^2_{H^k_d(\mathbb {R}^n)}ds\Big )e^{2Ct} . \end{aligned}$$

Note that $\delta$ and C only depends on the $L^{\infty }$ norm of a and therefore there exists two positive constants $\delta _1$ and $C_1$ which depends only on $\kappa$ such that

$$\begin{aligned} e^{-\delta _1t}\Vert u(t,\cdot )\Vert ^2_{H^k_d(\mathbb {R}^n)}\le \Vert u_0\Vert ^2_{H^k_d(\mathbb {R}^n)}+C_1\int _0^te^{-\delta _1s}\Vert f(s,\cdot )\big \Vert ^2_{H^k_d(\mathbb {R}^n)}ds . \end{aligned}$$

$\square$

B Proof of Proposition 5.1

Set

$$\begin{aligned}\Vert X^n(t)\Vert =|E^n(t)|+|U^n(t)|+|W^n(t)|+|Z^n(t)|+|\Phi ^n(t)|+|\psi ^n(t)|+\Vert f^n(t,\cdot )\Vert _{H^3_d(\mathbb {R}^3)}\end{aligned}$$

and

$$\begin{aligned}C_0= |E^0|+|U^0|+|W^0|+|Z^0|+|\Phi ^0|+|\psi ^0|+r . \end{aligned}$$

We will prove by induction that, there exists $T_0>0$ such that $\Vert X^n(t)\Vert \le 2C_0, \quad \forall n\in \mathbb {N}, \quad \forall t\in [0,\,T_0]$.

$\bullet$ For $n=0$, we have $\Vert X^0\Vert \le C_0\le 2C_0$.

$\bullet$ Let $n\in \mathbb {N}$ and suppose that $\forall k\,\le n$, we have $\Vert X^k(t)\Vert \le 2C_0$. We want to show that $\Vert X^{n+1}(t)\Vert \le 2C_0$ for all $t\in [0,T_0]$, the choice of $T_0$ will be given shortly.

• Integrate equation (4.35) on $[0,\,t]$, we have:

  $$\begin{aligned} E^{n+1}(t)=E_0-\int _0^tE^n(s)U^n(s)ds ; \end{aligned}$$

  then

  $$\begin{aligned} |E^{n+1}(t)|\le |E_0|+A_1t; \end{aligned}$$

  (a)

  where $A_1>0$ is a constant which only depends on $C_0$.

• Integrate equation (4.36) on $[0,\,t]$, we have

  $$\begin{aligned} U^{n+1}(t)= & {} U_0+ \int _0^t\Big [-\frac{3}{2}(U^n)^2+\frac{\Lambda }{2}-4\pi (E^n)^5\int _{\mathbb {R}^3}\frac{(v^1)^2}{v^0_n}f^n(s,\overline{v})d\overline{v}\nonumber \\&-2\pi \frac{(Z^n)^2}{(E^n)^2}-2\pi (2\psi ^n-m^2(\phi ^n)^2)\Big ]ds . \end{aligned}$$

  (B.1)

  We had, by (4.35): $\dot{E}^n=-U^{n-1}E^{n-1}$ which implies: $|\dot{E}^n|\le 4C_0^2$ i.e. $-4C_0^2\le \dot{E}^n\le 4C_0^2$. Integrating this last inequality gives $E_0-4C_0^2t\le E^n(t)$. But $E_0=\frac{1}{a_0}>0$. So if we take t such that $0<4C_0^2t<\frac{E_0}{2}$, then $E_0-4C_0^2t\ge \frac{E_0}{2}$; then $\frac{E_0}{2}\le E^n(t)$ i.e. $\frac{1}{E^n(t)}\le \frac{2}{E_0}$. This proves that one can find two constants $t_1>0$ and $A_2>0$ such that, identity (B.1) gives the following:

  $$\begin{aligned} |U^{n+1}(t)|\le |U_0|+A_2t, \forall \, 0\le t\le t_1. \end{aligned}$$

  (b)

• Integrate equation (4.37) on $[0,\,t]$, we have: $W^{n+1}(t)=W_0-\int _0^t(3U^n(s)W^n(s)+\rho ^2)ds$. Then there exists a constant $A_3>0$ such that:

  $$\begin{aligned} |W^{n+1}(t)|\le |W_0|+A_3t. \end{aligned}$$

  (c)

• Integrate equation (4.38) on $[0,\,t]$, we have: $Z^{n+1}(t)=Z_0-3\int _0^tU^n(s)Z^n(s)ds$. Then there exists a constant $A_4>0$ such that:

  $$\begin{aligned} |Z^{n+1}(t)|\le |Z_0|+A_4t . \end{aligned}$$

  (d)

• Integrate equation (4.39) on $[0,\,t]$, we have: $\Phi ^{n+1}(t)=\Phi _0+\int _0^t\sqrt{2\psi ^n(s)}ds .$ This shows that there exists a constant $A_5>0$ such that:

  $$\begin{aligned} |\Phi ^{n+1}(t)|\le |\Phi _0|+A_5t . \end{aligned}$$

  (e)

• Integrate equation (4.40) on $[0,\,t]$, we have:

  $$\begin{aligned} \psi ^{n+1}(t)=\psi _0-\int _0^t\big (6U^n(s)\psi ^n(s)+m^2\Phi ^n(s)\sqrt{2\psi ^n(s)}+\rho ^2\big )ds .\end{aligned}$$

  This shows that there exists a constant $A_6>0$ such that:

  $$\begin{aligned} |\psi ^{n+1}(t)|\le |\psi _0|+A_6t . \end{aligned}$$

  (f)

• Now we want to use Corollary A.1 to obtain a bound for $\Vert f^{n+1}(t,\cdot )\Vert ^2_{H^3_d(\mathbb {R}^3)}$. Observe that Eq. (4.41) is of the form (A.1) with $a= a(t)= E^nZ^n\int _{\mathbb {R}^3}f^n(t,\overline{v})d\overline{v}$, $\,b\equiv 0$ and since by induction hypothesis

  $$\begin{aligned} \Big | E^nZ^n\int _{\mathbb {R}^3}f^n(t,\overline{v})d\overline{v} \Big |\le 4C_0^2\Vert f^n(t, \cdot )\Vert _{L^1(\mathbb {R}^3)}\le C(C_0)\Vert f^n(t, \cdot )\Vert _{H^3_d(\mathbb {R}^3)}\le C(C_0) ; \end{aligned}$$

  one can use inequality (A.20) of Corollary A.1 (with $\frac{1}{\kappa }=C(C_0), \; k=n=3$) to obtain

  $$\begin{aligned} \Vert f^{n+1}(t,\cdot )\Vert ^2_{H^3_d(\mathbb {R}^3)}\le \left( \Vert f_0\Vert ^2_{H^3_d(\mathbb {R}^3)}+C(C_0)\int _0^t\Vert f^n(s,\cdot )\big \Vert ^2_{H^3_d(\mathbb {R}^3)}ds\right) e^{C(C_0)t} . \end{aligned}$$

  This proves that:

  $$\begin{aligned} \Vert f^{n+1}(t,\cdot )\Vert ^2_{H^3_d(\mathbb {R}^3)}\le e^{C(C_0)t}\left( \Vert f_0\Vert ^2_{H^3_d(\mathbb {R}^3)}+ C(C_0)t\right) . \end{aligned}$$

  We take $t_2>0$ such that $e^{C(C_0)t}\le 1, \,\forall t\in [0,\,t_2]$. Then, there exists a constant $A_7>0$, such that for $t\in [0,\,t_2]$:

  $$\begin{aligned} \Vert f^{n+1}(t)\Vert _{H^3_d(\mathbb {R}^3)}\le \Vert f_0\Vert _{H^3_d(\mathbb {R}^3)}+A_7\sqrt{t} . \end{aligned}$$

  (g)

Now we add inequalities (a), (b), (c), (d), (e), (f), (g) to obtain, for $t\le \min (t_1,t_2)$:

$$\begin{aligned} \Vert X^{n+1}(t,\cdot )\Vert \le C_0+\left( \sum _{i=1}^7A_i\right) (t+\sqrt{t}) . \end{aligned}$$

Now choose $t_3>0$ such that for $\forall \, 0\le t\le t_3$ we have $\big (\sum _{i=1}^7A_i\big )(t+\sqrt{t})\le C_0$. Finally, by setting $T_0=\min (t_1,t_2,t_3)$ we obtain that

$$\begin{aligned} \Vert X^{n+1}(t,\cdot )\Vert \le 2C_0\qquad \text{ for } \text{ all }\quad 0\le t\le T_0; \end{aligned}$$

and the proof is complete. $\square$

C Proof of Lemma

First, we prove (5.1). We have:

$$\begin{aligned}&\left| |E^n|^5\int _{\mathbb {R}^3}\frac{|v^1|^2}{v^0_n}f^n(\overline{v})d\overline{v}- |E^{n-1}|^5\int _{\mathbb {R}^3}\frac{|v^1|^2}{v^0_{n-1}}f^{n-1}(\overline{v})d\overline{v} \right| \\&\quad =\left| \left( |E^n|^5-|E^{n-1}|^5\right) \int _{\mathbb {R}^3}\frac{|v^1|^2}{v^0_n}f^n(\overline{v})d\overline{v}\right. \\&\qquad \left. + |E^{n-1}|^5\left( \int _{\mathbb {R}^3}\frac{|v^1|^2}{v^0_n}f^n(\overline{v})-\frac{|v^1|^2}{v^0_{n-1}}f^{n-1}(\overline{v})\right) d\overline{v}\right| \\&\quad \le C(C_0)|E^n-E^{n-1}|+|E^{n-1}|^5\left| \int _{\mathbb {R}^3}\Big (\frac{1}{v^0_{n}}-\frac{1}{v^0_{n-1}} \Big )|v^1|^2f^n(\overline{v})d\overline{v}\right. \\&\qquad \left. +\int _{\mathbb {R}^3}\frac{|v^1|^2}{v^0_{n-1}}\big (f^n-f^{n-1}\big )(\overline{v})d\overline{v}\right| \\&\quad \le C(C_0)\left( |E^n-E^{n-1}|+\Vert f^n-f^{n-1}\Vert _{H^2_d(\mathbb {R}^3)}\right) . \end{aligned}$$

Secondly we prove (5.2). From now, we choose to write $\tilde{Q}(E^n,f^{n},f^{n})$ for $\tilde{Q}_n$ to take advantage of the presence of $E^n$ in $\tilde{Q}_n$. We have

$$\begin{aligned} \frac{1}{u^0_n}\tilde{Q}_n-\frac{1}{u^0_{n-1}}\tilde{Q}_{n-1}= & {} \frac{1}{u^0_{n}} \left( \tilde{Q}(E^n,f^n,f^n)-\tilde{Q}(E^{n-1},f^n,f^n)\right) \nonumber \\&+ \left( \frac{1}{u^0_n}-\frac{1}{u^0_{n-1}}\right) \tilde{Q}(E^{n-1},f^n,f^n)\nonumber \\&+\frac{1}{u^0_{n-1}}\left( \tilde{Q}(E^{n-1},f^n,f^n)-\tilde{Q}\left( E^{n-1},f^{n-1},f^{n-1}\right) \right) .\nonumber \\ \end{aligned}$$

(C.1)

As far as the first term of (C.1) is concerned, we have:

$$\begin{aligned}&\frac{1}{u^0_{n}}\left( \tilde{Q}(E^n,f^{n},f^{n})-\tilde{Q}(E^{n-1},f^{n},f^{n})\right) \\&\quad = \frac{1}{u^0_{n}}\left[ \tilde{Q}^+(E^n,f^{n},f^{n})-\tilde{Q}^+(E^{n-1},f^{n},f^{n})\right] \\&\qquad - \frac{1}{u^0_{n}}\left[ \tilde{Q}^-(E^n,f^{n},f^{n})-\tilde{Q}^-(E^{n-1},f^{n},f^{n})\right] \\&\quad =: (I)+(II) . \end{aligned}$$

Let us estimate the first term. We have

$$\begin{aligned} (I)= & {} \frac{1}{u^0_{n}}\bigg [\int _{\mathbb {R}^{3}}\frac{(E^n)^3}{v_n^{0}}d\overline{v}\int _{S^2}f^n(t,\overline{u}') f^n(t,\overline{v}')\tilde{B}(E^n,\overline{u},\overline{v},\overline{u}',\overline{v}')d\omega \\&- \int _{\mathbb {R}^3}\frac{(E^{n-1})^3}{v_{n-1}^0}d\overline{v}\int _{S^2}f^n(t,\overline{u}')f^n(t,\overline{v}') \tilde{B}(E^{n-1},\overline{u},\overline{v},\overline{u}',\overline{v}') d\omega \bigg ] \\= & {} \frac{1}{u^0_{n}}\bigg [\int _{\mathbb {R}^{3}}\frac{(E^n)^3}{v_n^{0}}d\overline{v}\int _{S^2}f^n(t,\overline{u}') f^n(t,\overline{v}')\big (\tilde{B}(E^n,\overline{u},\overline{v},\overline{u}',\overline{v}')\\&- \tilde{B}(E^{n-1},\overline{u},\overline{v},\overline{u}',\overline{v}')\big )d\omega \\&+ \int _{\mathbb {R}^3}\left( \frac{(E^n)^3}{v_n^0}-\frac{(E^{n-1})^3}{v_{n-1}^0}\right) d\overline{v}\int _{S^2}f^n(t,\overline{u}')f^n(t,\overline{v}') \tilde{B}(E^{n-1},\overline{u},\overline{v},\overline{u}',\overline{v}') d\omega \bigg ] \\= & {} (I_1)+ (I_2) . \end{aligned}$$

Note that

$$\begin{aligned}&\frac{(E^n)^3}{v_n^0}-\frac{(E^{n-1})^3}{v_{n-1}^0}\\&\quad = \frac{(E^n)^3\big (v^0_{n-1}-v^0_n\big )+v^0_n\big ((E^{n})^3-(E^{n-1})^3\big )}{v^0_nv^0_{n-1}} \\&\quad = \frac{(E^{n})^3-(E^{n-1})^3}{v^0_{n-1}}+(E^n)^3\frac{\sqrt{1+(E^{n-1})^2|\overline{v}|^2}-\sqrt{1+(E^n)^2|\overline{v}|^2}}{v^0_nv^0_{n-1}} \\&\quad =\frac{E^n-E^{n-1}}{v^0_{n-1}}\underbrace{\left[ (E^n)^2+E^nE^{n-1}+(E^{n-1})^2+\frac{(E^n)^3(E^n+E^{n-1}) |\overline{v}|^2}{v^0_n(v^0_n+v^0_{n-1})}\right] }_{\;:\,=\, \xi ^n} ; \end{aligned}$$

thus

$$\begin{aligned} I_2= \left( E^n-E^{n-1}\right) \frac{u^0_{n-1}}{u^0_n}\, \frac{1}{u^0_{n-1}}\int _{\mathbb {R}^3}\frac{\xi ^n \,d\bar{v}}{v^0_{n-1}}\int _{S^2}f^n(t,\overline{u}')f^n(t,\overline{v}') \tilde{B}(E^{n-1},\overline{u},\overline{v},\overline{u}',\overline{v}') d\omega .\nonumber \\ \end{aligned}$$

(C.2)

Before continuing, we point out the following Lemma.

Lemma C.1

If f and g are functions such that the partial derivatives of f up to order k are uniformly bounded then

$$\begin{aligned} \Vert fg\Vert _{H^k_{d}(\mathbb {R}^3)}\le C \Vert g\Vert _{H^k_{d}(\mathbb {R}^3)} \end{aligned}$$

where the positive constant C only depends on the bounds of f and its derivatives.

This Lemma shows that in order to control $H^2_{d}$-norm of $(I_2)$, we then need to show that the function $\bar{u}\mapsto \frac{u^0_{n-1}}{u^0_n}$ and its derivatives up to order two are uniformly bounded. Note that for all $n\in \mathbb {N}, \;0< \frac{E_0}{2}\le E^n\le 2C_0$ thus we have

$$\begin{aligned} \frac{u^0_{n-1}}{u^0_n}= & {} \frac{\sqrt{1+(E^{n-1})^2|\bar{u}|^2}}{\sqrt{1+(E^{n})^2|\bar{u}|^2}}\le C(C_0,E_0) ; \\ \left| \partial _i\left( \frac{u^0_{n-1}}{u^0_n}\right) \right|= & {} \left| \frac{(E^{n-1})^2 u^i}{u^0_{n-1}u^0_{n}}-\frac{(E^{n})^2 u^iu^0_{n-1}}{(u^0_{n})^3}\right| \le C(C_0,E_0) ; \end{aligned}$$

and

$$\begin{aligned}&\left| \partial ^2_{ij}\left( \frac{u^0_{n-1}}{u^0_n}\right) \right| \nonumber \\&\quad =\left| \frac{(E^{n-1})^2\delta _j^i}{u^0_nu^0_{n-1}} -2\frac{(E^n)^2(E^{n-1})^2u^iu^j}{(u^0_n)^3u^0_{n-1}}-\frac{(E^{n-1})^4u^iu^j}{u^0_n(u^0_{n-1})^3} \right. \nonumber \\&\qquad \left. -\frac{\delta _j^i(E^n)^2u^0_{n-1}}{(u^0_n)^3} +3\frac{(E^n)^4u^iu^ju^0_{n-1}}{(u^0_n)^5}\right| \nonumber \\&\quad \le C(C_0,E_0) . \end{aligned}$$

(C.3)

Lemma C.1 implies that

$$\begin{aligned}&\Vert (I_2)\Vert _{H^2_d(\mathbb {R}^3)}\le C(C_0)|E^n\\&\quad -E^{n-1}|\left\| \frac{1}{u^0_{n-1}}\int _{\mathbb {R}^3}\frac{\xi ^n \,d\bar{v}}{v^0_{n-1}}\int _{S^2}f^n(t,\overline{u}')f^n(t,\overline{v}') \tilde{B}(E^{n-1},\overline{u},\overline{v},\overline{u}',\overline{v}') d\omega \right\| _{H^2_d(\mathbb {R}^3)} . \end{aligned}$$

Since $\xi ^n$ is bounded and does not depend on $\bar{u}$, the $H^2_d(\mathbb {R}^3)$-norm at the r.h.s. of the previous inequality will give an estimate similar to (4.7). More precisely, we have

$$\begin{aligned} \Vert (I_2)\Vert _{H^2_d(\mathbb {R}^3)}\le C|E^n-E^{n-1}|\,\Vert f^n\Vert ^2_{H^2_d(\mathbb {R}^3)}\le C|E^n-E^{n-1}| . \end{aligned}$$

(C.4)

In order to obtain an estimate for the $H^2_d(\mathbb {R}^3)$-norm of the term $(I_1)$, we proceed exactly as in the proof of Proposition 3.6 of [12], page 88. The term B in that reference is replaced by the difference $\big (\tilde{B}(E^n,\overline{u},\overline{v},\overline{u}',\overline{v}')- \tilde{B}(E^{n-1},\overline{u},\overline{v},\overline{u}',\overline{v}')\big )$ and we use instead the property that B and its derivatives are Lipschitz continuous. This leads to an estimate of the form

$$\begin{aligned} \Vert (I_1)\Vert _{H^2_d(\mathbb {R}^3)}\le C|E^n-E^{n-1}|\,\Vert f^n\Vert ^2_{H^2_d(\mathbb {R}^3)}\le C|E^n-E^{n-1}| ; \end{aligned}$$

(C.5)

and then

$$\begin{aligned} \left\| \frac{1}{u^0_{n}}\left[ \tilde{Q}^+(E^n,f^{n},f^{n})-\tilde{Q}^+(E^{n-1},f^{n},f^{n})\right] \right\| _{H^2_d(\mathbb {R}^3)}\le C|E^n-E^{n-1}| . \end{aligned}$$

Similarly we have:

$$\begin{aligned} \left\| (II)\right\| _{H^2_d(\mathbb {R}^3)}= \left\| \frac{1}{u^0_{n}}\left[ \tilde{Q}^-(E^n,f^{n},f^{n})-\tilde{Q}^-(E^{n-1},f^{n},f^{n})\right] \right\| _{H^2_d(\mathbb {R}^3)}\le C|E^n-E^{n-1}| ; \end{aligned}$$

and we deduce that:

$$\begin{aligned} \left\| \frac{1}{u^0_{n}}\left[ \tilde{Q}(E^n,f^{n},f^{n})-\tilde{Q}(E^{n-1},f^{n},f^{n})\right] \right\| _{H^2_d(\mathbb {R}^3)}\le C|E^n-E^{n-1}| . \end{aligned}$$

(C.6)

As far as the second term of (C.1) is concerned, we have:

$$\begin{aligned}&\left\| \left( \frac{1}{u^0_n}-\frac{1}{u^0_{n-1}}\right) \tilde{Q}(E^{n-1},f^n,f^n)\right\| _{H^2_d(\mathbb {R}^3)}\nonumber \\&\quad = \left\| \frac{u^0_{n-1}-u^0_n}{u^0_n}\,\frac{1}{u^0_{n-1}}\tilde{Q}(E^{n-1},f^n,f^n)\right\| _{H^2_d(\mathbb {R}^3)} \nonumber \\&\quad = \big |(E^{n-1})^2-(E^{n})^2\big |\left\| \frac{|\bar{u}|^2}{(u^0_{n-1}+u^0_n)u^0_n}\,\frac{1}{u^0_{n-1}}\tilde{Q}(E^{n-1},f^n,f^n) \right\| _{H^2_d(\mathbb {R}^3)} \nonumber \\&\quad \le C\big |E^{n-1}-E^{n}\big |\left\| \frac{1}{u^0_{n-1}}\tilde{Q}(E^{n-1},f^n,f^n) \right\| _{H^2_d(\mathbb {R}^3)} \nonumber \\&\quad \le C\big |E^{n-1}-E^{n}\big |\left\| f^n \right\| ^2_{H^2_d(\mathbb {R}^3)} \nonumber \\&\quad \le C\big |E^{n-1}-E^{n}\big | ; \end{aligned}$$

(C.7)

where in the first inequality we have used the fact that the function $\bar{u}\mapsto \frac{|\bar{u}|^2}{(u^0_{n-1}+u^0_n)u^0_n}$ and its derivatives up to order two are bounded (the details of computations can be found in [13]). For the third term of (C.1), applying directly (4.8) gives

$$\begin{aligned} \left\| \frac{1}{u^0_{n-1}}\left[ \tilde{Q}(E^{n-1},f^n,f^n)-\tilde{Q}(E^{n-1},f^{n-1},f^{n-1})\right] \right\| _{H^2_d(\mathbb {R}^3)}\le C\Vert f^n-f^{n-1}\Vert _{H^2_d(\mathbb {R}^3)} . \end{aligned}$$

(C.8)

Finally adding (C.6)–(C.8) gives (5.2) and the proof is complete. $\square$

D Proof of Proposition 5.2

In what follows, the constant C only depends on $C_0$ and $T_0$ and may be different from line to line.

• We first prove that $(X^n)$ is a Cauchy sequence in $\big (C^0([0,T_0];\mathbb {R})\big )^6\times C^0\big ([0,T_0];H^2_d(\mathbb {R}^3)\big )$.

  • We integrate Eq. (4.35) and obtain:

    $$\begin{aligned} |E^{n+1}(t)-E^n(t)|^2\le C\int _0^t\big (|E^n(s)-E^{n-1}(s)|^2+|U^n(s)-U^{n-1}(s)|^2\big )ds .\nonumber \\ \end{aligned}$$

    (D.1)

  • From Eq. (4.36) we have:

    $$\begin{aligned} U^{n+1}(t)-U^n(t)= & {} -\displaystyle \int _0^t\bigg [\frac{3}{2}\big ((U^n(s))^2-(U^{n-1}(s))^2\big )\\&+ 2\pi \Big (\frac{(Z^n)^2(s)}{(E^n)^2(s)}-\frac{(Z^{n-1})^2(s) }{(E^{n-1})^2(s)}\Big )\\&+\; 4\pi \Big ((E^{n})^5(s)\int _{\mathbb {R}^3}\frac{|v^1|^2}{v^0_n}f^n(s,\overline{v})d\overline{v}\\&- (E^{n-1})^5(s)\int _{\mathbb {R}^3}\frac{|v^1|^2}{v^0_{n-1}}f^{n-1}(s,\overline{v})d\overline{v}\Big ) \\&+ \;4\pi \big (\psi ^n(s)-\psi ^{n-1}(s)\big )\\&-2\pi m^2\big ((\Phi ^n(s))^2-(\Phi ^{n-1}(s))^2\big )\bigg ]ds; \end{aligned}$$

    which implies (since (5.1)) that

    $$\begin{aligned}&|U^{n+1}(t)-U^n(t)|\le C \int _0^t\Big [|E^n(s)-E^{n-1}(s)|\\&\quad +|U^n(s)-U^{n-1}(s)|+|Z^n(s)-Z^{n-1}(s)|\\&\quad +|\Phi ^n(s)-\Phi ^{n-1}(s)|+|\psi ^n(s)-\psi ^{n-1}(s)| +\Vert f^n(s,\cdot )-f^{n-1}(s,\cdot )\Vert _{H^2_d(\mathbb {R}^3)}\Big ]ds . \end{aligned}$$

    Thus,

    $$\begin{aligned}&|U^{n+1}(t)-U^n(t)|^2\le C\int _0^t\Big [|E^n(s)-E^{n-1}(s)|^2\nonumber \\&\quad +|U^n(s)-U^{n-1}(s)|^2+|Z^n(s)-Z^{n-1}(s)|^2\nonumber \\&\quad +|\Phi ^n(s)-\Phi ^{n-1}(s)|^2+|\psi ^n(s)-\psi ^{n-1}(s)|^2 +\Vert f^n(s,\cdot )-f^{n-1}(s,\cdot )\Vert ^2_{H^2_d(\mathbb {R}^3)}\Big ]ds.\nonumber \\ \end{aligned}$$

    (D.2)

  • Integrating Eq. (4.37) gives:

    $$\begin{aligned} |W^{n+1}(t)-W^n(t)|^2\le C\int _0^t\big (|U^n(s)-U^{n-1}(s)|^2+|W^n(s)-W^{n-1}(s)|^2\big )ds .\nonumber \\ \end{aligned}$$

    (D.3)

  • Similarly, Eq. (4.38) gives:

    $$\begin{aligned} |Z^{n+1}(t)-Z^n(t)|^2\le C \int _0^t\big (|U^n(s)-U^{n-1}(s)|^2+|Z^n(s)-Z^{n-1}(s)|^2\big )ds . \end{aligned}$$

    (D.4)

  • Now, Eq. (4.39) gives:

    $$\begin{aligned} \Phi ^{n+1}(t)-\Phi ^n(t)=\int _0^t\big (\sqrt{2\psi ^n(s)}-\sqrt{2\psi ^{n-1}(s)}\big )ds=2\int _0^t\frac{\psi ^n(s)-\psi ^{n-1}(s)}{\sqrt{2\psi ^n(s)}+\sqrt{2\psi ^{n-1}(s)}}ds . \end{aligned}$$

    In order to get rid of the denominator of the right hand side, we use Eq. (4.40) which we recall is $\dot{\psi }^{n+1}=-6U^n\psi ^n-m^2\Phi ^n\sqrt{2\psi ^n}-\rho ^2$. Since the sequence $(X^n)$ is uniformly bounded, on $[0,T_0]$, we have $|-6U^n\psi ^n-m^2\Phi ^n\sqrt{2\psi ^n}-\rho ^2|\le C_1\;$ and thus,

    $$\begin{aligned} \frac{d\psi ^{n+1}}{dt}\ge -C_1 . \end{aligned}$$

    The last inequality implies that $\psi ^{n+1}(t)\ge \psi _0-C_1t .$ Since $\psi _0>0$, we assume that $T_0$ is sufficiently small so that: $0\le C_1t\le \frac{\psi _0}{2}, \, 0\le t\le T_0$ and obtain that $\psi ^{n+1}(t)\ge \frac{\psi _0}{2}$ and $\frac{1}{\sqrt{2\psi ^{n+1}(t)}}\le \frac{1}{\sqrt{\psi _0}}$. From there, we deduce that there exists a constant $C>0$ such that:

    $$\begin{aligned} \forall t\in [0,T_0], \quad |\Phi ^{n+1}(t)-\Phi ^n(t)|^2\le C(C_0)\int _0^t|\psi ^n(s)-\psi ^{n-1}(s)|^2ds .\nonumber \\ \end{aligned}$$

    (D.5)

  • From Eq. (4.40) we have:

    $$\begin{aligned} \psi ^{n+1}(t)-\psi ^n(t)= & {} \int _0^t\Big [6\big (U^n(s)\psi ^n(s)-U^{n-1}(s)\psi ^{n-1}(s)\big )\\&+m^2\big (\Phi ^n(s)\sqrt{2\psi ^n(s)}-\Phi ^{n-1}(s)\sqrt{\psi ^{n-1}(s)}\big )\Big ]ds . \end{aligned}$$

    Since $\frac{1}{\sqrt{2\psi ^{n+1}(t)}}\le \frac{1}{\sqrt{\psi _0}}$, we deduce that there exists a constant $C>0$ such that:

    $$\begin{aligned}&|\psi ^{n+1}(t)-\psi ^n(t)|^2\le C\int _0^t\Big (|U^n(s)-U^{n-1}(s)|^2+|\Phi ^n(s)-\Phi ^{n-1}(s)|^2\nonumber \\&\quad +|\psi ^n(s)-\psi ^{n-1}(s)|^2\Big )ds . \end{aligned}$$

    (D.6)

  • Finally, from Eq. (4.41) we get:

    $$\begin{aligned}&\frac{\partial (f^{n+1}-f^n)}{\partial t}+\left( E^nZ^n\int _{\mathbb {R}^3}f^n(t,\overline{v})d\overline{v}\right) \sum _{i=1}^3\frac{\partial ( f^{n+1}-f^n)}{\partial u^i}\nonumber \\&\quad =\left( E^nZ^n\int _{\mathbb {R}^3}f^n(t,\overline{v})d\overline{v}-E^{n-1}Z^{n-1}\int _{\mathbb {R}^3}f^{n-1}(t,\overline{v}) d\overline{v}\right) \sum _{i=1}^3\frac{\partial f^{n}}{\partial u^i}\nonumber \\&\qquad +\frac{1}{u^0_n}\tilde{Q}_n-\frac{1}{u^0_{n-1}}\tilde{Q}_{n-1}.\nonumber \\ \end{aligned}$$

    (D.7)

    But since the sequence $(X^n)$ is bounded, we have:

    $$\begin{aligned}&\Big |E^n(t)Z^n(t)\int _{\mathbb {R}^3}f^n(t,\overline{v})d\overline{v}-E^{n-1}(t)Z^{n-1}(t)\int _{\mathbb {R}^3}f^{n-1}(t,\overline{v})d\overline{v}\Big | \\&\quad =\Big |(E^n(t)-E^{n-1}(t))Z^n(t)\int _{\mathbb {R}^3}f^n(t,\overline{v})d\overline{v}+(Z^n(t)-Z^{n-1}(t))E^{n-1}\int _{\mathbb {R}^3}f^n(t,\overline{v})d\overline{v}\\&\qquad +E^{n-1}(t)Z^{n-1}(t)\int _{\mathbb {R}^3}(f^n-f^{n-1})(t,\overline{v})d\overline{v}\Big |\\&\quad \le C\Big (|E^n(t)-E^{n-1}(t)|+|Z^n(t)-Z^{n-1}(t)|+\Vert f^n(t,\cdot )-f^{n-1}(t,\cdot )\Vert _{H^2_d(\mathbb {R}^3)}\Big ) . \end{aligned}$$

    Since $f^n\in H^3_{d}(\mathbb {R}^3)$ the second term in the r.h.s. of Eq. (D.7) is an element of $H^2_{d}(\mathbb {R}^3)$ and then,

    $$\begin{aligned}&\Big \Vert \Big (E^n(t)Z^n(t)\int _{\mathbb {R}^3}f^n(t,\overline{v})d\overline{v}\nonumber \\&\qquad -E^{n-1}(t)Z^{n-1}(t) \int _{\mathbb {R}^3}f^{n-1}(t,\overline{v})d\overline{v}\Big ) \sum _{i=1}^3\frac{\partial f^{n}}{\partial u^i}(t,\cdot )\Big \Vert _{H^2_d(\mathbb {R}^3)}\nonumber \\&\quad \le C \Big (|E^n(t)-E^{n-1}(t)|+|Z^n(t)-Z^{n-1}(t)|+\Vert f^n(t,\cdot )-f^{n-1}(t,\cdot )\Vert _{H^2_d(\mathbb {R}^3)}\Big ) . \end{aligned}$$

    (D.8)

    Further (see inequality (5.2) of Lemma 5.1),

    $$\begin{aligned} \Big \Vert \frac{1}{u_n^0}\tilde{Q}_n-\frac{1}{u_{n-1}^0}\tilde{Q}_{n-1}\Big \Vert _{H^2_d(\mathbb {R}^3)}\le C\left( |E^n-E^{n-1}|+\Vert f^n-f^{n-1}\Vert _{H^2_d(\mathbb {R}^3)}\right) . \end{aligned}$$

    Equation (D.7) has a form to which Corollary A.1 applies. Thus, for $k=2$ and $u= f^{n+1}-f^{n}$ in Inequality (A.20) and using the last two estimates, we have:

    $$\begin{aligned}&\Vert f^{n+1}(s,\cdot )-f^n(s,\cdot )\Vert ^2_{H^2_d(\mathbb {R}^3)}\nonumber \\&\quad \le C\int _0^t\Big (|E^n(s)-E^{n-1}(s)|^2+|Z^n(s)-Z^{n-1}(s)|^2\nonumber \\&\qquad +\Vert f^n(s,\cdot )-f^{n-1}(s,\cdot )\Vert ^2_{H^2_d(\mathbb {R}^3)}\Big )ds . \end{aligned}$$

    (D.9)

  Consider the space

  $$\begin{aligned} \Sigma _{T_0}:=\Big (\mathcal {C}\big ([0,T_0]),\mathbb {R}\big )\Big )^6\times \mathcal {C}\big ([0,T_0],H^2_d(\mathbb {R}^3)\big ) . \end{aligned}$$

  Endowed with the norm

  $$\begin{aligned} |\Vert X|\Vert :=\sum _{i=1}^6\sup _{0\le t\le T_0}|X_i(t)|+\sup _{0\le t\le T_0}\Vert X_7(t)\Vert _{H^2_d(\mathbb {R}^3)} \end{aligned}$$

  where $X=(X_i)_{1\le i\le 7}\in \Sigma _{T_0}$, $\Sigma _{T_0}$ is a Banach space. We want to show that there exists a constant $0<\alpha <1$ which depends only upon $C_0$ and $T_0$ such that $|\Vert X^{n+1}-X^n|\Vert \le \alpha |\Vert X^{n}-X^{n-1}|\Vert$ if $T_0$ is small enough. Summing up inequalities (D.1)–(D.6) and (D.9) gives:

  $$\begin{aligned}&|E^{n+1}(t)-E^{n}(t)|^2+|U^{n+1}(t)-U^{n}(t)|^2+|W^{n+1}(t)-W^{n}(t)|^2+|Z^{n+1}(t)-Z^{n}(t)|^2\nonumber \\&\qquad +|\Phi ^{n+1}(t)-\Phi ^{n}(t)|^2 +|\psi ^{n+1}(t)-\psi ^{n}(t)|^2 +\Vert f^{n+1}(t,\cdot )-f^{n}(t,\cdot )\Vert ^2_{H^2_d(\mathbb {R}^3)}\nonumber \\&\quad \le C \int _0^t\Big ( |E^{n}(s)-E^{n-1}(s)|^2+|U^{n}(s)-U^{n-1}(s)|^2+|W^{n}(s)-W^{n-1}(s)|^2\nonumber \\&\qquad +|Z^{n}(s)-Z^{n-1}(s)|^2+|\Phi ^{n}(s)-\Phi ^{n-1}(s)|^2 +|\psi ^{n}(s)-\psi ^{n-1}(s)|^2\nonumber \\&\qquad +\Vert f^{n}(s,\cdot )-f^{n-1}(s,\cdot )\Vert ^2_{H^2_d(\mathbb {R}^3)}\Big )ds . \end{aligned}$$

  (D.10)

  This last inequality implies that

  $$\begin{aligned} |\Vert X^{n+1}-X^n|\Vert \le \sqrt{C(C_0)T_0}|\Vert X^{n}-X^{n-1}|\Vert . \end{aligned}$$

  (D.11)

  Now, we assume that $T_0$ is small enough such that $C(C_0)T_0<1$, an obtain from (D.11) that the sequence $(X^n)$ is a Cauchy sequence in the Banach space $\Sigma _{T_0}$.

• Next, we show that $(\frac{dY^n}{dt})$ is a Cauchy sequence in $\big (C^0([0,T_0];\mathbb {R})\big )^6$. Since the sequence $(X^n)$ is bounded, from equations (4.35)–(4.40) we deduce that there exists a constant $C >0$ which only depends upon $C_0$ and T such that:

  $$\begin{aligned} \Big \Vert \Big |\frac{dY^{n+1}}{dt}-\frac{dY^n}{dt}\Big \Vert \Big |\le C|\Vert Y^n-Y^{n-1}|\Vert . \end{aligned}$$

  (D.12)

  Note that $|\Vert Y^n-Y^{n-1}|\Vert \le |\Vert X^n-X^{n-1}|\Vert$, thus inequality (D.11) shows that:

  $$\begin{aligned} \Big \Vert \Big |\frac{dY^{n+1}}{dt}-\frac{dY^n}{dt}\Big \Vert \Big |\le \big (C_2\sqrt{C(C_0)T_0}\big )^{n-1}\Vert |X^1-X^0|\Vert . \end{aligned}$$

  (D.13)

  As we did before, the constant $C_2\sqrt{C(C_0)T_0}$ can be assumed to be less that one even if it means to shrink $T_0$ again. This shows that $(\frac{dY^n}{dt})$ is a Cauchy sequence and thus, $(Y^n)$ is a Cauchy sequence in $\big (C^1([0,T_0],\mathbb {R})\big )^6$.

• Let us now prove that $(\frac{\partial f^n}{\partial t})$ is a Cauchy sequence in $C^0\big ([0,T_0];H^1_d(\mathbb {R}^3)\big )$. From Eq. (D.7) we have:

  $$\begin{aligned} \Big \Vert \frac{\partial (f^{n+1}-f^n)}{\partial t}\Big \Vert _{H^1_d(\mathbb {R}^3)}\le & {} C\Big \{\Vert f^{n+1}-f^n\Vert _{H^2_d(\mathbb {R}^3)}+ \Vert f^{n}-f^{n-1}\Vert _{H^1_d(\mathbb {R}^3)}\\&\quad +\;|E^n-E^{n-1}|+|Z^n-Z^{n-1}|\Big \} \\\le & {} C\big \{|\Vert X^{n+1}-X^n|\Vert +|\Vert X^{n}-X^{n-1}|\Vert \big \} .\nonumber \\ \end{aligned}$$

  Thus,

  $$\begin{aligned} \Big \Vert \frac{\partial (f^{n+1}-f^n)}{\partial t}\Big \Vert _{C^0([0,T_0];H^1_d(\mathbb {R}^3))}\le & {} C\big \{|\Vert X^{n+1}-X^n|\Vert +|\Vert X^{n}-X^{n-1}|\Vert \big \}\nonumber \\ \end{aligned}$$

  (D.14)
  $$\begin{aligned}\le & {} \;C \alpha ^{n}|\Vert X^{1}-X^{0}|\Vert . \end{aligned}$$

  (D.15)

  which shows $(\frac{\partial f^n}{\partial t})$ is a Cauchy sequence in $C^0\big ([0,T_0];H^1_d(\mathbb {R}^3)\big )$ and ends the proof.

$\square$

E Proof of Theorem 5.1

(1) Existence As the first step towards existence of solution we prove that the sequence $(X^n)$ converges in the space $\big (\mathcal {C}^1_b([0,T_0])\big )^6\times \mathcal {C}^1\big ([0,T_0]\times \mathbb {R}^3\big ).$ From Proposition 5.2 we know that the sequence $(Y^n)_n$ is a Cauchy sequence in the Banach space $(C^1([0,T_0];\mathbb {R}))^6$ thus there exists a set of functions $Y= (E,U,W,Z,\Phi ,\psi )$ such that $(Y^n)$ converges towards Y in $\big (\mathcal {C}^1_b([0,T_0])\big )^6$. Secondly, Proposition 5.2 also tells us that $(f^n)_n$ is a Cauchy sequence in the Banach space $C^0\big ([0,T_0];H^2_d(\mathbb {R}^3)\big )$. It follows that there exists a function f such that $(f^n)_n$ converges to f in the space $C^0\big ([0,T_0];H^2_d(\mathbb {R}^3)\big )$. But the space $C^0\big ([0,T_0];H^2_d(\mathbb {R}^3)\big )$ embeds continuously in $C^0\big ([0,T_0];H^2(\mathbb {R}^3)\big )$, therefore $(f^n)$ is a Cauchy sequence in $C^0\big ([0,T_0];H^2(\mathbb {R}^3)\big ).$ Now from interpolation inequality (5.3), for any real number $2<s<3$ we have:

$$\begin{aligned} \Vert f^n(t,\cdot )-f^p(t,\cdot )\Vert _{(s)}\le \Vert f^n(t,\cdot )-f^p(t,\cdot )\Vert ^{s-2}_{H^3(\mathbb {R}^3)} \Vert f^n(t,\cdot )-f^p(t,\cdot )\Vert ^{3-s}_{H^2(\mathbb {R}^3)} .\nonumber \\ \end{aligned}$$

(E.1)

Since $\big (f^n(t, \cdot )\big )_n$ is a uniformly bounded in $H^3_d(\mathbb {R}^3)$ and then in $H^3(\mathbb {R}^3)$, inequality (E.1) shows that $\big (f^n\big )_n$ is a Cauchy sequence in $C^0\big ([0,T_0];H^{(s)}(\mathbb {R}^3)\big )$ for any $2<s<3$.

Similarly, since the sequence $\left( \frac{\partial f^n}{\partial t}\right)$ is uniformly bounded (see Remark 5.2) and is a Cauchy sequence in $C^0\big ([0,T_0];H^1_d(\mathbb {R}^3)\big )$, the interpolation inequality shows that it is a Cauchy sequence in $C^0\big ([0,T_0];H^{(s)}(\mathbb {R}^3)\big )$ for any $1<s<2$. We then obtain that

$$\begin{aligned} (f^n) \; \text{ is } \text{ a } \text{ Cauchy } \text{ sequence } \text{ in }\; C^0\big ([0,T_0];H^{(s+1)}(\mathbb {R}^3)\big )\cap C^1\big ([0,T_0];H^{(s)}(\mathbb {R}^3)\big ); \quad 1<s<2 .\nonumber \\ \end{aligned}$$

(E.2)

Sobolev embedding inequality we know that

$$\begin{aligned} C^0\big ([0,T_0];H^{(s+1)}(\mathbb {R}^3)\big )\cap C^1\big ([0,T_0];H^{(s)}(\mathbb {R}^3)\big )\hookrightarrow C^1_b([0,T_0]\times \mathbb {R}^3)\quad \text{ for } \text{ any } \quad s>\frac{3}{2} . \end{aligned}$$

(E.3)

$\frac{3}{2}<s<2$ shows that $(f^n)$ is a Cauchy in $C^1_b([0,T_0]\times \mathbb {R}^3)$ and thus converges towards a function $\tilde{f}$ in $C^1_b([0,T_0]\times \mathbb {R}^3)$ and the embedding $C^0([0,T_0];H^2_d(\mathbb {R}^3))\hookrightarrow C^0([0,T_0]\times \mathbb {R}^3)$ shows that $f=\tilde{f}$. This shows that the collection of functions $X=(E,U,W,Z,\Phi ,\psi ,f)$ is the limit of the sequence $(X^n)$ in the space $\big (\mathcal {C}^1_b([0,T_0])\big )^6\times \mathcal {C}^1\big ([0,T_0]\times \mathbb {R}^3\big )$.

As the second step towards existence, we now prove that X is indeed a solution of (4.18)–(4.24). Since $(Y^n)_n$ converges towards $Y= (E,U,W,Z,\Phi ,\psi )$ in $(C^1([0,T_0];\mathbb {R}))^6$ taking the limit pointwise in Eqs. (4.35), (4.37)–(4.40) shows that $E,\,U, \,W,\, Z,\,\Phi$ and $\psi$ satisfy (4.18), (4.20)–(4.23). It remains to show that integrals $\displaystyle \int _{\mathbb {R}^3}f^n(t,\overline{v})d\overline{v}$ and $\displaystyle \int _{\mathbb {R}^3}\frac{|v|^2}{v_n^0}f^n(t,\overline{v})d\overline{v}$ converge respectively to $\displaystyle \int _{\mathbb {R}^3}f(t,\overline{v})d\overline{v}$ and $\displaystyle \int _{\mathbb {R}^3}\frac{|v|^2}{v^0}f(t,\overline{v})d\overline{v}$, $\forall t\in [0,T_0]$ as n goes to infinity. We notice that these last two integrals are convergent since $\forall t\in [0,T_0], \; f(t,\cdot )\in H^2_d(\mathbb {R}^3)$. We have

$$\begin{aligned} \Bigg |\int _{\mathbb {R}^3}f^n(t,\overline{v})d\overline{v}-\int _{\mathbb {R}^3}f(t,\overline{v})d\overline{v}\Bigg |\le & {} \int _{\mathbb {R}^3}|f^n(t,\overline{v})-f(t,\overline{v})|d\overline{v} \\\le & {} \Vert f^n(t,\cdot )-f(t,\cdot )\Vert _{H^2_d(\mathbb {R}^3)} . \end{aligned}$$

This shows that $\displaystyle \int _{\mathbb {R}^3}f^n(t,\overline{v})d\overline{v}\longrightarrow \int _{\mathbb {R}^3}f(t,\overline{v})d\overline{v}$, $\forall t\in [0,T_0]$. Similarly, from (5.1) we have

$$\begin{aligned}&\Bigg |\int _{\mathbb {R}^3}\frac{|v|^2}{v_n^0}f^n(t,\overline{v})d\overline{v}- \int _{\mathbb {R}^3}\frac{|v|^2}{v^0}f(t,\overline{v})d\overline{v}\Bigg |\\&\quad \le C\big (|E^n-E|+ \Vert f^n(t,\cdot )-f(t,\cdot )\Vert _{H^2_d(\mathbb {R}^3)}\big )\longrightarrow 0 . \end{aligned}$$

Finally let us prove that $\frac{1}{u_n^0}\tilde{Q}(f^n,f^n)\longrightarrow \frac{1}{u^0}\tilde{Q}(f,f)$. As we did in the proof of (5.2), we have

$$\begin{aligned}&\big \Vert \frac{1}{u_n^0}\tilde{Q}(f^n,f^n)-\frac{1}{u^0}\tilde{Q}(f,f)\big \Vert _{H^2_d(\mathbb {R}^3)}\\&\quad \le C(T_0) \left( \Vert f^n\Vert _{H^2_d(\mathbb {R}^3)}+\Vert f\Vert _{H^2_d(\mathbb {R}^3)}\right) \big (|E^n-E|+\Vert f^n-f\Vert _{H^2_d(\mathbb {R}^3)}\big ) \\&\quad \le C(T_0)\left( C_0+\Vert f\Vert _{H^2_d(\mathbb {R}^3)}\right) \big (|E^n-E|+\Vert f^n-f\Vert _{H^2_d(\mathbb {R}^3)}\big ) . \end{aligned}$$

Thus, $\frac{1}{u_n^0}\tilde{Q}_n$ converges towards $\frac{1}{u^0}\tilde{Q}(f,f)$ in the space $H^2_d(\mathbb {R}^3)$ and since $H^2_d(\mathbb {R}^3)\hookrightarrow \mathcal {C}^0_b(\mathbb {R}^3)$ it follows that $\frac{1}{u_n^0}\tilde{Q}_n$ converges towards $\frac{1}{u^0}\tilde{Q}(f,f)$ in $\mathcal {C}^0_b(\mathbb {R}^3)$.

We have thus proved that the limit can also be taken pointwise in the remaining Eqs. (4.36) and (4.41) to obtain that $X=(E,U,W,Z,\Phi ,\psi ,f)$ also satisfies 4.19 and (4.24). Therefore X is a local solution of (4.18)–(4.24).

(2) Uniqueness Suppose that there exists two sets of functions $X_i=(E_i,U_i,W_i,Z_i,\Phi _i,\psi _i,f_i),\; i=1, 2$ which solve the system (4.18)–(4.24) with the same Cauchy data. We proceed exactly as we did in the proof of Proposition 5.2 and obtain an estimate of the form (D.10) where $X^{n+1}-X^{n}$ and $X^{n}-X^{n-1}$ are replaced by $X_2-X_1$. Applying Gronwall’s inequality to the obtained estimate proves that $X_1=X_2.$

(3) Now, we prove that $\forall t\in [0,T], \; f(t,\cdot )\in H_d^3(\mathbb {R}^3)$. Recall, the sequence $\left( f^n(t,\cdot )\right)$ is uniformly bounded in the Hilbert space $H^3_d(\mathbb {R}^3)$. But it is well known that any bounded sequence in a Hilbert space has a weakly convergent subsequence (see for example [6], Theorem 5.4.2 page 151). Therefore there exists a subsequence $\left( f^{n_p}(t,\cdot )\right)$ and a function $g(t, \cdot )\in H^3_d(\mathbb {R}^3)$ such that $\left( f^{n_p}(t,\cdot )\right)$ converges to $g(t, \cdot )$ in $H^3_d(\mathbb {R}^3)$ endowed with its weak topology which continuously embeds in $H^2_d(\mathbb {R}^3)$ endowed with its weak topology. Moreover, recall again, $(f^n(t,\cdot ))_n$ converges to $f(t,\cdot )$ in the space $H^2_d(\mathbb {R}^3)$ thus this convergence also holds in $H^2_d(\mathbb {R}^3)$ endowed with its weak topology. Since the weak topology is Hausdorff, we thus conclude that $f(t, \cdot )=g(t, \cdot )$ and then $f(t, \cdot )\in H^3_d(\mathbb {R}^3)$. Note that

$$\begin{aligned} \Vert f(t,\cdot )\Vert _{H^3_d(\mathbb {R}^3)}\le \liminf \Vert f^{n_p}(t,\cdot )\Vert _{H^3_d(\mathbb {R}^3)} \le C,\quad \text{ uniformly } \text{ in }\; t .\end{aligned}$$

(E.4)

(4) Let us show that (5.5) holds. We proceed as in [22] by first proving the weak continuity and then the strong continuity. The following Lemma will be needed, its proof is given at the end of the paper.

Lemma E.1

The space $\mathcal {C}^{\infty }_c(\mathbb {R}^3)$ of Compactly supported smooth functions defined on $\mathbb {R}^3$ is dense in the space $H^3_d(\mathbb {R}^3)$.

Weak continuity First, let us prove that the solution is weakly continuous. Let F be in the dual of $H^3_d(\mathbb {R}^3)$. Then, due to the Riesz representation theorem there exists $\varphi _F\in H^3_d(\mathbb {R}^3)$ such that for all $h\in H^3_d(\mathbb {R}^3)$,

$$\begin{aligned} F(h)=\langle h,\varphi \rangle _{H^3_d(\mathbb {R}^3)}=\sum _{|\alpha |\le 3}\int _{\mathbb {R}^3}(1+|\bar{v}|)^{d+|\alpha |}\partial ^{\alpha }h(\bar{v})\cdot (1+|\bar{v}|)^{d+|\alpha |}\partial ^{\alpha }\varphi _F(\bar{v})d\bar{v}. \end{aligned}$$

We then have,

$$\begin{aligned}&F(f^n(t,\cdot ))-F(f(t,\cdot ))=\sum _{|\alpha |\le 3}\int _{\mathbb {R}^3}(1+|\bar{v}|)^{d+|\alpha |}\partial ^{\alpha }f^n\cdot (1+|\bar{v}|)^{d+|\alpha |}\partial ^{\alpha }\varphi _F d\bar{v}\\&\quad - \sum _{|\alpha |\le 3}\int _{\mathbb {R}^3}(1+|\bar{v}|)^{d+|\alpha |}\partial ^{\alpha }f\cdot (1+|\bar{v}|)^{d+|\alpha |}\partial ^{\alpha }\varphi _F d\bar{v} . \end{aligned}$$

Consequently, if $(\varphi _j)$ is a sequence of compactly supported smooth functions converging to $\varphi _F$ in $H^3_d(\mathbb {R}^3)$, we obtain:

$$\begin{aligned}&F(f^n(t,\cdot ))-F(f(t,\cdot ))\\&\quad = \sum _{|\alpha |\le 3}\int _{\mathbb {R}^3}(1+|\bar{v}|)^{d+|\alpha |}\partial ^{\alpha }(f^n-f)(t,\,\bar{v})\cdot (1+|\bar{v}|)^{d+|\alpha |}\partial ^{\alpha } (\varphi _F-\varphi _j)(\bar{v})d\bar{v} \\&\qquad +\; \sum _{|\alpha |\le 3}\int _{\mathbb {R}^3}(1+|\bar{v}|)^{d+|\alpha |}\partial ^{\alpha }(f^n-f)(t,\,\bar{v})\cdot (1+|\bar{v}|)^{d+|\alpha |}\partial ^{\alpha } \varphi _j(\bar{v})d\bar{v} . \end{aligned}$$

We then deduce since, the sequence $(f^n)$ is bounded in $H^3_d(\mathbb {R}^3)$ that:

$$\begin{aligned}&\big |F(f^n(t,\cdot ))-F(f(t,\cdot ))\big |\le C\Vert \varphi _F-\varphi _j\Vert _{H^3_d(\mathbb {R}^3)}\\&\quad +\bigg | \sum _{|\alpha |\le 3}\int _{\mathbb {R}^3}(1+|\bar{v}|)^{d+|\alpha |}\partial ^{\alpha }(f^n-f)(t,\,\bar{v})\cdot (1+|\bar{v}|)^{d+|\alpha |}\partial ^{\alpha } \varphi _j(\bar{v})d\bar{v}\bigg | \end{aligned}$$

Letting j be large enough that the first term on the right-hind side is less than or equal to $\frac{\varepsilon }{2}$, and then choosing n large enough, depending on j, so that the second term is less than $\frac{\varepsilon }{2}$, we conclude that the right-hand side is less than $\varepsilon$. We conclude that $F(f^n(t,\cdot ))$ converges uniformly to $F(f(t,\cdot ))$ which proves that the solution f is weakly continuous.

Strong continuity Let $t_0\in [0, T_0)$. We want to prove that $f: [0,T_0)\longrightarrow H_d^{3}(\mathbb {R}^3)$ is continuous at $t_0$ i.e.

$$\begin{aligned} \lim _{t\rightarrow t_0}\Vert f(t,\cdot )-f(t_0,\cdot )\Vert _{H_d^{3}(\mathbb {R}^3)}=0. \end{aligned}$$

Using the inner product $\langle \, ,\,\rangle$ on $H_d^{3}(\mathbb {R}^3)$, we can write for $t\in [0,T_0)$:

$$\begin{aligned}&\langle f(t,\cdot )-f(t_0,\cdot ),f(t,\cdot )-f(t_0,\cdot )\rangle \nonumber \\&\quad =\langle f(t,\cdot ),f(t,\cdot )\rangle -2\langle f(t,\cdot ),f(t_0,\cdot )\rangle +\langle f(t_0,\cdot ),f(t_0,\cdot )\rangle . \end{aligned}$$

(E.5)

Note that the last term on the right-hand side is $\Vert f(t_0,\cdot )\Vert ^2_{H_d^{3}(\mathbb {R}^3)}$. Due to the weak continuity of f, the limit as t goes to $t_0$ of second term on the right-hand side is $-2\Vert f(t_0,\cdot )\Vert ^2_{H_d^{3}(\mathbb {R}^3)}$.

For the first term on the right-hand side, we suppose that $t>t_0$ and use the fact that there exists $\delta >0$ such that [see estimate (A.20)]

$$\begin{aligned} e^{-\delta (t-t_0)}\Vert f(t,\cdot )\Vert ^2_{H^3_d(\mathbb {R}^3)}\le \Vert f(t_0,\cdot )\Vert ^2_{H^3_d(\mathbb {R}^3)}+ C\int _{t_0}^te^{-\delta s}\Vert f(s,\cdot )\big \Vert ^4_{H^3_d(\mathbb {R}^3)}ds; \end{aligned}$$

and the fact that $f(t,\cdot )$ is uniformly bounded [see (E.4)], to have $\lim _{t\rightarrow t_0^+} \langle f(t,\cdot ),f(t,\cdot )\rangle \le \Vert f(t_0,\cdot )\Vert ^2_{H_d^{3}(\mathbb {R}^3)}.$ Combine these observations with (E.5), to have $\lim _{t\rightarrow t_0^+} \langle f(t,\cdot )-f(t_0,\cdot ),f(t,\cdot )-f(t_0,\cdot )\rangle \le 0,$ and conclude, since $\langle f(t,\cdot )-f(t_0,\cdot ),f(t,\cdot )-f(t_0,\cdot )\rangle \ge 0,\;\forall t\in [0,T_0)$ that $\; \lim _{t\rightarrow t_0^+} \langle f(t,\cdot )-f(t_0,\cdot ),f(t,\cdot )-f(t_0,\cdot )\rangle = 0 ;$ i.e. $f: [0,T_0)\longrightarrow H_d^{3}(\mathbb {R}^3)$ is right continuous on $[0,T_0)$. By time reversal one obtains left continuity and thus continuity of f.

(5) Finally, since the Einstein–Maxwell–Boltzmann-massive scalar field system is equivalent to the system of first order partial differential equations (4.18)–(4.24), we have therefore proved that the Einstein–Maxwell–Boltzmann-massive scalar field equations have a unique local (in time) solution. $\square$

F Proof of Lemma E.1

Lemma F.1

The space $\mathcal {C}^{\infty }_c(\mathbb {R}^3)$ of Compactly supported smooth functions defined on $\mathbb {R}^3$ is dense in the space $H^3_d(\mathbb {R}^3)$.

Proof

The proof follows closely that of the density of $\mathcal {C}^{\infty }_c(\mathbb {R}^3)$ in the usual Sobolev space $H^k(\mathbb {R}^3)$ and will be done in two steps.

1.1 First step

Set $\mathcal {T}=\{f\in H^3_d(\mathbb {R}^3);\; supp ($f$) \; compact \}$ and let us show that $\mathcal {T}$ is dense in $H^3_d(\mathbb {R}^3)$. Let $\varphi \in \mathcal {C}^{\infty }_c(\mathbb {R}^3)$ such that $\varphi (x)=1$ for $|x|\le 1;\; 0\le \varphi \le 1$ and $supp(\varphi )\subset B_{\mathbb {R}^3}(0, 2)$. For every integer $j\ge 1$ set $\varphi _j(x)= \varphi \left( \frac{x}{j}\right) .$ Then,

$$\begin{aligned} \varphi _j\in \mathcal {C}^{\infty }_c(\mathbb {R}^3), \;0\le \varphi _j \le 1,\; supp \,\varphi _j\subset B(0,2j) . \end{aligned}$$

Moreover, $\,\forall \alpha \in \mathbb {N}^3, \; D^{\alpha }\varphi _j$ is uniformly bounded; more precisely:

$$\begin{aligned} \forall j\in \mathbb {N}^*,\;|D^{\alpha }\varphi _j|\le & {} \frac{C_{\alpha }}{j^{|\alpha |}}\\\le & {} C_{\alpha }\;\text{ where }\; C_{\alpha }=\sup _{x\in \mathbb {R}^3}|D^{\alpha }\varphi (x)|. \end{aligned}$$

Since $\varphi _j(x)=1$ for $|x|\le j$, the sequence $(\varphi _j)_j$ converges pointwise to 1 as j goes to $\infty$. Now let $f\in H^3_d(\mathbb {R}^3)$ and set $f_j = \varphi _jf$. Since $\varphi _j\in \mathcal {C}^{\infty }_c(\mathbb {R}^3),\; f_j\in \mathcal {T}$. Let us prove that $(f_j)$ converge towards f in $H^3_d(\mathbb {R}^3)$.

We have

$$\begin{aligned} \Vert f-f_j\Vert _{H^3_d(\mathbb {R}^3)}=\Vert f-\varphi _jf\Vert _{H^3_d(\mathbb {R}^3)}=\Bigg (\sum _{|\alpha |\le 3}\int _{\mathbb {R}^3}(1+|x|)^{2d+2|\alpha |}\big |D^{\alpha }\big (f(1-\varphi _j)\big )\big |^2dx\Bigg )^{\frac{1}{2}} .\nonumber \\ \end{aligned}$$

(F.1)

From Leibnitz’s formula and the convexity of the function $t\mapsto t^2$,

$$\begin{aligned}&\big |D^{\alpha }\big (f(1-\varphi _j)\big )\big |\le \sum _{|\beta |\le |\alpha |} \complement ^{\beta }_{\alpha } |D^{\alpha -\beta }f|\,| D^{\beta }(1-\varphi _j)| \\&\quad \le \Bigg (\sum _{|\beta |\le |\alpha |}\big (\complement ^{\beta }_{\alpha }\big )^2\Bigg )^{\frac{1}{2}} \Bigg (\sum _{|\beta |\le |\alpha |}|D^{\alpha -\beta }f|^2|D^{\beta }(1-\varphi _j)|^2\Bigg )^{\frac{1}{2}} ; \end{aligned}$$

thus,

$$\begin{aligned} (1+|x|)^{2d+2|\alpha |}\big |D^{\alpha }\big (f(1-\varphi _j)\big )\big |^2\le C_{\alpha }\sum _{|\beta |\le |\alpha |}(1+|x|)^{2d+2|\alpha |}|D^{\alpha -\beta }f|^2|D^{\beta }(1-\varphi _j)|^2 . \end{aligned}$$

(F.2)

Note that

$$\begin{aligned}&\sum _{|\beta |\le |\alpha |}(1+|x|)^{2d+2|\alpha |}|D^{\alpha -\beta }f|^2|D^{\beta }(1-\varphi _j)|^2\\&\quad =(1+|x|)^{2d+2|\alpha |}|D^{\alpha }f|^2|1-\varphi _j|^2\\&\qquad +\sum _{1\le |\beta |\le |\alpha |}(1+|x|)^{2d+2|\alpha |}|D^{\alpha -\beta }f|^2|D^{\beta }(1-\varphi _j)|^2 . \end{aligned}$$

(i) $\varphi _j\longrightarrow 1$ and $|1-\varphi _j|\le 1+|\varphi _j|\le 2$, thus,

$$\begin{aligned} \left\{ \begin{array}{ll} (1+|x|)^{2d+2|\alpha |}|D^{\alpha }f|^2|1-\varphi _j|^2\longrightarrow 0,\;\text{ as }\;j\longrightarrow +\infty \;\text{ pointwise }\\ \text{ and } \\ (1+|x|)^{2d+2|\alpha |}|D^{\alpha }f|^2|1-\varphi _j|^2\le 4(1+|x|)^{2d+2|\alpha |}|D^{\alpha }f|^2 \end{array} \right. ; \end{aligned}$$

but $f\in H^3_d(\mathbb {R}^3)$, thus $(1+|x|)^{2d+2|\alpha |}|D^{\alpha }f|^2$ is integrable and we can thus apply the Lebesgue dominated convergence theorem to obtain

$$\begin{aligned} \int _{\mathbb {R}^3}(1+|x|)^{2d+2|\alpha |}|D^{\alpha }f|^2|1-\varphi _j|^2dx\longrightarrow 0 \;\text{ as }\;j\longrightarrow +\infty . \end{aligned}$$

(ii) Recall that $|D^{\beta }\varphi _j|\le \frac{C_{\beta }}{j^{|\beta |}}, \,\forall |\beta |\ge 1$ and $supp \varphi _j\subset B(0,2j)$, thus,

$$\begin{aligned} (1+|x|)^{2|\beta |}|D^{\beta }\varphi _j|^2\le C^2_{\beta }\frac{(1+2j)^{2|\beta |}}{j^{2|\beta |}}\le C_{\beta }\frac{(1+j^{2|\beta |})}{j^{2|\beta |}}\le 2C_{\beta } \end{aligned}$$

and then,

$$\begin{aligned} \left\{ \begin{array}{ll} (1+|x|)^{2d+2|\alpha |}|D^{\alpha -\beta }f|^2|D^{\beta }\varphi _j|^2 \le (1+|x|)^{2d+2|\alpha |}|D^{\alpha -\beta }f|^2\frac{C_{\beta }}{j} \longrightarrow 0,\;\text{ as }\;j\longrightarrow +\infty \\ \text{ and } \\ (1+|x|)^{2d+2|\alpha |}|D^{\alpha -\beta }f|^2|D^{\beta }\varphi _j|^2\le 2C_{\beta }(1+|x|)^{2d+2|\alpha -\beta |}|D^{\alpha -\beta }f|^2. \end{array} \right. \end{aligned}$$

Again, $f\in H^3_d(\mathbb {R}^3)$, and then $(1+|x|)^{2d+2|\alpha -\beta |}|D^{\alpha -\beta }f|^2|$ is integrable. We use once more the dominated convergence theorem and obtain that $\forall \,\beta$ such that $1\le |\beta |\le |\alpha |$,

$$\begin{aligned} \int _{\mathbb {R}^3}(1+|x|)^{2d+2|\alpha |}|D^{\alpha -\beta }f|^2|D^{\beta }\varphi _j|^2dx\longrightarrow 0, \;\text{ as }\;j\longrightarrow +\infty . \end{aligned}$$

and thus

$$\begin{aligned} \sum _{1\le |\beta |\le |\alpha |}\int _{\mathbb {R}^3}(1+|x|)^{2d+2|\alpha |}|D^{\alpha -\beta }f|^2|D^{\beta }\varphi _j|^2dx\longrightarrow 0 ,\;\text{ as }\;j\longrightarrow +\infty . \end{aligned}$$

Finally we obtain from (F.2) that

$$\begin{aligned} \forall \alpha ,\;|\alpha |\le 3,\;\int _{\mathbb {R}^3}(1+|x|)^{2d+2|\alpha |}\big |D^{\alpha }\big (f(1-\varphi _j)\big )\big |^2dx\longrightarrow 0 \;\text{ as }\;j\longrightarrow +\infty \end{aligned}$$

and conclude from (F.1), that $f_j\longrightarrow f$ in the topology of $H^3_d(\mathbb {R}^3)$. Thus, $\mathcal {T}$ is dense in $H^3_d(\mathbb {R}^3)$.

1.2 Second step

We prove that $\mathcal {C}^{\infty }_c(\mathbb {R}^3)$ is dense in $\mathcal {T}$ endowed with the topology of $H^3_d(\mathbb {R}^3)$. Let $\theta$ be a standard mollifier, which means that $\theta$ is a positive $C^{\infty }$ function in $\mathbb {R}^3$ supported in the unit ball and such that $\int _{\mathbb {R}^3}\theta (x)dx=1$. For every $j\in \mathbb {N}^*$ , we define $\theta _j$ by

$$\begin{aligned} \theta _j(x)= j^3\theta (jx) . \end{aligned}$$

Then, the sequence $(\theta _j)_{j\in \mathbb {N}^{*}}$ has the following properties

$$\begin{aligned} \theta _j\ge 0;\quad \theta _j\in \mathcal {C}^{\infty }_c(\mathbb {R}^3);\quad \int _{\mathbb {R}^3}\theta _j(x)dx=1;\quad supp\,\theta _j\subset \bar{B}(0,\frac{1}{j}),\;j\in \mathbb {N}^* . \end{aligned}$$

Now let $f\in \mathcal {T}$, and let R be a positive real number such that $supp(f)\subset B(0,R)$. Consider the following convolution product $f_j=f\star \theta _j$, then $supp(f_j)$ is compact since f and $\theta _j$ are compactly supported, more precisely, $\theta _j\in \mathcal {C}^{\infty }_c(\mathbb {R}^3).$ We shall prove that $(f_j)$ converges towards f in $H^3_d(\mathbb {R}^3)$. For this purpose, it will be sufficient to prove that

$$\begin{aligned} \forall \alpha ,\; |\alpha |\le 3,\;(1+|x|)^{d+|\alpha |}D^{\alpha }f_j\longrightarrow (1+|x|)^{d+|\alpha |}D^{\alpha }f\;\text{ in }\;L^2(\mathbb {R}^3) . \end{aligned}$$

Let $\alpha ,\; |\alpha |\le 3$, we have: $D^{\alpha }f_j=D^{\alpha }f\star \theta _j\;\text{ and }\;D^{\alpha }f-D^{\alpha }f_j=D^{\alpha }f-D^{\alpha }f\star \theta _j$. Now

$$\begin{aligned} D^{\alpha }f\star \theta _j(x)=\int _{\mathbb {R}^3}D^{\alpha }f(y)\theta _j(x-y)dy \end{aligned}$$

and

$$\begin{aligned} D^{\alpha }f(x)=D^{\alpha }f(x)\int _{\mathbb {R}^3}\theta _j(x-y)dy=\int _{\mathbb {R}^3}D^{\alpha }f(x)\theta _j(x-y)dy \end{aligned}$$

thus

$$\begin{aligned}&(1+|x|)^{2d+2|\alpha |}\left| D^{\alpha }f(x)-D^{\alpha }f_j(x)\right| ^2\nonumber \\&\quad \le (1+|x|)^{2d+2|\alpha |}\left( \int _{\mathbb {R}^3} \big |D^{\alpha }f(x)-D^{\alpha }f(y)\big |\theta _j(x-y)dy\right) ^2 . \end{aligned}$$

(F.3)

Applying Holder’s inequality gives the following estimates

$$\begin{aligned}&(1+|x|)^{2d+2|\alpha |}\big |D^{\alpha }f(x)- D^{\alpha }f_j(x)\big |^2\\&\quad \le (1+|x|)^{2d+2|\alpha |}\int _{\mathbb {R}^3}\big |D^{\alpha }f(x)-D^{\alpha }f(y)\big |^2\theta _j(x-y)dy. \end{aligned}$$

which after integration on $\mathbb {R}^3$, gives:

$$\begin{aligned}&\int _{\mathbb {R}^3}(1+|x|)^{2d+2|\alpha |}\big |D^{\alpha }f(x)-D^{\alpha }f_j(x)\big |^2dx\nonumber \\&\quad \le \int _{\mathbb {R}^3}\int _{\mathbb {R}^3}(1+|x|)^{2d+2|\alpha |}\big |D^{\alpha }f(x)-D^{\alpha }f(y)\big |^2\theta _j(x-y)dxdy. \end{aligned}$$

(F.4)

Consider on $\mathbb {R}^{2n}$ the change of variables $u= x - y , \;v = y$, we have:

$$\begin{aligned}&\int _{\mathbb {R}^3}(1+|x|)^{2d+2|\alpha |}\big |D^{\alpha }f(x)-D^{\alpha }f_j(x)\big |^2dx\\&\quad \le \int _{|u|\le \frac{1}{j}}\int _{\mathbb {R}^3}(1+|v+u|)^{2d+2|\alpha |}\big |D^{\alpha }f(v+u)-D^{\alpha }f(v)\big |^2\theta _j(u)dudv, \end{aligned}$$

which can be written as

$$\begin{aligned}&\int _{\mathbb {R}^3}(1+|x|)^{2d+2|\alpha |}\big |D^{\alpha }f(x)-D^{\alpha }f_j(x)\big |^2dx\nonumber \\&\quad \le \int _{|u|\le \frac{1}{j}}\theta _j(u)du\int _{|v|<R+1}(1+|u+v|)^{2d+2|\alpha |}\big |D^{\alpha }f(u+v)-D^{\alpha }f(v)\big |^2dv.\nonumber \\ \end{aligned}$$

(F.5)

then, for $|u|\le \frac{1}{j}<1$, we have:

$$\begin{aligned}&\int _{\mathbb {R}^3}(1+|u+v|)^{2d+2|\alpha |}\big |D^{\alpha }f(u+v)-D^{\alpha }f(v)\big |^2dv\nonumber \\&\quad \le C_{\alpha }\int _{\mathbb {R}^3}\big |D^{\alpha }f(u+v)-D^{\alpha }f(v)\big |^2dv \end{aligned}$$

(F.6)

where $C_{\alpha }=(3+R|)^{2d+2|\alpha |}.$ Recall that $f\in \mathcal {T}\subset H^3_d(\mathbb {R}^3)\subset H^3(\mathbb {R}^3)$ thus $D^{\alpha }f\in L^2(\mathbb {R}^3)$ and by the continuity of the $L^2$-norm, one has:

$$\begin{aligned} \int _{\mathbb {R}^3}\big |D^{\alpha }f(y+z)-D^{\alpha }f(y)\big |^2dy\longrightarrow 0\quad \text{ as }\quad z\longrightarrow 0 \end{aligned}$$

and it follows from (F.6) that:

$$\begin{aligned} \int _{\mathbb {R}^3}(1+|y+z|)^{2d+2|\alpha |}\big |D^{\alpha }f(y+z)-D^{\alpha }f(y)\big |^2dy\longrightarrow 0\quad \text{ as }\quad z\longrightarrow 0 . \end{aligned}$$

Now let $\varepsilon >0$, there exists $\exists \delta >0$ such that

$$\begin{aligned} \forall z\in \mathbb {R}^3, \;|z|\le \delta \Longrightarrow \int _{\mathbb {R}^3}(1+|y+z|)^{2d+2|\alpha |}\big |D^{\alpha }f(y+z)-D^{\alpha }f(y)\big |^2dy<\varepsilon . \end{aligned}$$

Since $\lim _{j\rightarrow +\infty }\frac{1}{j}=0$, there exists $j_0>0$ such that $\forall j\in \mathbb {N}^*$, $j>j_0\Longrightarrow \frac{1}{j}<\delta$ thus, $j>j_0$ implies $|z|<\frac{1}{j}<\delta$ and then

$$\begin{aligned} \int _{\mathbb {R}^3}(1+|y+z|)^{2d+2|\alpha |}\big |D^{\alpha }f(y+z)-D^{\alpha }f(y)\big |^2dy<\varepsilon ; \end{aligned}$$

from where we obtain

$$\begin{aligned} j>j_0\Longrightarrow \int _{\mathbb {R}^3}(1+|x|)^{2d+2|\alpha |}\big |D^{\alpha }f(x)-D^{\alpha }f_j(x)\big |^2dx \le \varepsilon \int _{|z|\le \frac{1}{j}}\theta _j(z)dz<\varepsilon ; \end{aligned}$$

thus,

$$\begin{aligned} \forall \alpha ,\; |\alpha |\le 3,\;(1+|x|)^{d+|\alpha |}D^{\alpha }f_j\longrightarrow (1+|x|)^{d+|\alpha |}D^{\alpha }f\;\text{ in }\;L^2(\mathbb {R}^3) ; \end{aligned}$$

and consequently, $f_j\longrightarrow f$ in $H^3_d(\mathbb {R}^3)$ which proves that $\mathcal {C}^{\infty }_c(\mathbb {R}^3)$ is dense in $\mathcal {T}$ endowed with the topology of $H^3_d(\mathbb {R}^3)$. $\square$