Discontinuous Galerkin Methods with Generalized Numerical Fluxes for the Vlasov-Viscous Burgers’ System

Abstract

In this paper, semi-discrete numerical scheme for the approximation of the periodic Vlasov-viscous Burgers’ system is developed and analyzed. The scheme is based on the coupling of discontinuous Galerkin approximations for the Vlasov equation and local discontinuous Galerkin approximations for the viscous Burgers’ equation. Both these methods use generalized numerical fluxes. The proposed scheme is both mass and momentum conservative. Based on generalized Gauss–Radau projections, optimal rates of convergence in the case of smooth compactly supported initial data are derived. Finally, computational results confirm our theoretical findings.

Appendix A.

The following lemma yields estimate for $u_x$ in $L^\infty$-norm which helps us to prove the regularity result for viscous Burgers’ equation.

Lemma A.1

For a periodic function u with $u \in C^2(I)$, there holds

$$\begin{aligned} \Vert u_x(t)\Vert _{L^\infty (I)} \le \sqrt{2}\,\Vert u_x(t)\Vert ^\frac{1}{2}_{L^2(I)}\Vert u_{xx}(t)\Vert _{L^2(I)}^\frac{1}{2}. \end{aligned}$$

Proof

As u is periodic. Hence, $\int _I u_x\,{\textrm{d}}x = 0$ and there is some $x_0 \in I$ such that $u_x(x_0) = 0$

$$\begin{aligned} \begin{aligned} |u_x|^2 = \left| \int _{x_0}^{x}\partial _x(u_x)^2\,{\textrm{d}}x\right| = 2\left| \int _{x_0}^{x}u_x\,u_{xx}\,{\textrm{d}}x\right| \le 2\,\Vert u_x\Vert _{L^2(I)}\,\Vert u_{xx}\Vert _{L^2(I)}. \end{aligned} \end{aligned}$$

Here, in first step use fundamental theorem of Calculus and in last step use the Cauchy-Schwarz inequality. After taking square root on both side and supremum over $x \in I$ completes the proof. $\square$

The following lemma shows the regularity result for the solution of viscous Burgers’ equation.

Lemma A.2

Let $\int _{{\mathbb R}}\int _I\,|v|^pf_0\,{\textrm{d}}x\,{\textrm{d}}v < \infty$ for $0 \le p \le 3$ and $u_0 \in H^1(I)$. Then

$$\begin{aligned} u \in L^2(0,T;H^2(I)) \cap L^\infty (0,T;H^1(I)) \cap H^1(0,T;L^2(I)). \end{aligned}$$

(A.1)

Proof

Multiplying Eq. (1.2) by $- u_{xx}$ and integrate with respect to x to obtain

$$\begin{aligned} \frac{1}{2}\partial _t\Vert u_x\Vert ^2_{L^2(I)} + \epsilon \,\Vert u_{xx}\Vert ^2_{L^2(I)} = -\left( \rho V, u_{xx}\right) + \left( \rho u, u_{xx}\right) + \left( uu_x,u_{xx}\right) . \end{aligned}$$

(A.2)

A use of the Hölder inequality with Lemma A.1 and the Young’s inequality yields

$$\begin{aligned} \begin{aligned} \left( uu_x,u_{xx}\right)&\le \Vert u\Vert _{L^2(I)}\Vert u_x\Vert _{L^\infty (I)}\Vert u_{xx}\Vert _{L^2(I)}\\&\le \sqrt{2}\,\Vert u\Vert _{L^2(I)}\Vert u_x\Vert _{L^2(I)}^\frac{1}{2}\Vert u_{xx}\Vert ^\frac{3}{2}_{L^2(I)}\\&\le \epsilon ^{-3}\,\Vert u\Vert ^4_{L^2(I)}\Vert u_x\Vert ^2_{L^2(I)} + \frac{3\epsilon }{4}\Vert u_{xx}\Vert ^2_{L^2(I)}. \end{aligned} \end{aligned}$$

(A.3)

From Eq. (A.2) after a use of the Hölder inequality, the Young’s inequality and Eq. (A.3) with kickback argument, we obtain

$$\begin{aligned} \begin{aligned} \partial _t\Vert u_x\Vert ^2_{L^2(I)} + \frac{2\epsilon }{5}\,\Vert u_{xx}\Vert ^2_{L^2(I)} \lesssim \epsilon ^{-3}\left( \Vert \rho V\Vert ^2_{L^2(I)} + \Vert \rho \Vert ^2_{L^2(I)}\Vert u\Vert ^2_{L^\infty (I)} + \Vert u\Vert ^4_{L^2(I)}\Vert u_x\Vert ^2_{L^2(I)}\right) . \end{aligned} \end{aligned}$$

A use of Sobolev inequality and an integration in time from 0 to t leads to

$$\begin{aligned} \begin{aligned} \Vert u_x\Vert ^2_{L^2(I)} + \frac{2\epsilon }{5}\int _{0}^{t}\Vert u_{xx}\Vert ^2_{L^2(I)}\,{\textrm{d}}s&\lesssim \epsilon ^{-3}\,\Vert \rho V\Vert ^2_{L^2([0,T]\times I)}\\&\quad + \epsilon ^{-3}\,\int _{0}^{t}\left( \Vert \rho \Vert ^2_{L^2(I)}+ \Vert u\Vert ^4_{L^2(I)}\right) \Vert u_x\Vert ^2_{L^2(I)}\,{\textrm{d}}s \\&\lesssim \epsilon ^{-3}\,\Vert \rho V\Vert ^2_{L^2([0,T]\times I)} + \epsilon ^{-3}\,\left( \Vert \rho \Vert ^2_{L^\infty (0,T;L^2(I))} \right. \\&\quad \left. + \Vert u\Vert ^4_{L^\infty (0,T;L^2(I))}\right) \Vert u_x\Vert ^2_{L^2([0,T]\times I)}. \end{aligned} \end{aligned}$$

A use of (2.2) and (2.5) shows

$$\begin{aligned} \begin{aligned} \Vert u_x\Vert ^2_{L^2(I)} + \frac{2\epsilon }{5}\int _{0}^{t}\Vert u_{xx}\Vert ^2_{L^2(I)}\,{\textrm{d}}s \le C\,\epsilon ^{-3}. \end{aligned} \end{aligned}$$

After taking supremum over $t \in [0,T]$, we obtain the first result.

Now, multiply Eq. (1.2) by $u_t$ and integration with respect to x follows

$$\begin{aligned} \Vert u_t\Vert ^2_{L^2(I)} + \frac{\epsilon }{2}\partial _t\Vert u_x\Vert ^2_{L^2(I)} = \left( \rho V,u_t\right) + \left( \rho u, u_t\right) - \left( u\,u_x, u_t\right) . \end{aligned}$$

A use of the Hölder inequality with the Young’s inequality, kickback argument and an integration in time completes the rest of the proof. $\square$

As a consequence of above lemma, it follows that

$$\begin{aligned} \begin{aligned} u \in L^2(0,T;H^2(I)) \subset L^2(0,T;W^{1,\infty }(I)) \subset L^1(0,T;W^{1,\infty }(I)). \end{aligned} \end{aligned}$$

Here, in second step use Sobolev inequality and in third step use the Hölder inequality.

The following result is on the propagation of velocity moments which is crucial for the proof of existence and uniqueness of strong solution.

Lemma A.3

Let $u \in L^1(0,T;W^{1,\infty }(I))$ and let $f_0 \ge 0$ be such that

$$\begin{aligned} \int _{{\mathbb R}}\int _I\,\vert v\vert ^k\left\{ f_0 + \vert \partial _xf_0\vert ^2 + \vert \partial _vf_0\vert ^2\right\} \,{\textrm{d}}x\,{\textrm{d}}v \le C, \end{aligned}$$

for $k \ge 0$. Then, the solution f of the Vlasov equation satisfies

$$\begin{aligned} \int _{{\mathbb R}}\int _I\,\vert v\vert ^k\left\{ f + \vert \partial _xf\vert ^2 + \vert \partial _vf\vert ^2\right\} \,{\textrm{d}}x\,{\textrm{d}}v \le C, \end{aligned}$$

for $k \ge 0$ and for all $t>0$.

Proof

Consider the equation for $\partial _xf$

$$\begin{aligned} \partial _t\partial _xf + v\partial _x^2f + \partial _v\left( \partial _xu\, f\right) + \partial _v\left( \left( u - v\right) \partial _xf\right) = 0. \end{aligned}$$

Multiplying the above equation by $\left( 1 + |v|^k\right) \partial _xf$ and integrating with respect to x, v yields

$$\begin{aligned} \frac{1}{2}\frac{{\textrm{d}}}{{\textrm{d}}t}\int _{{\mathbb R}}\int _I \left( 1 + \vert v\vert ^k\right) \left| \partial _xf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v = I_1 + I_2 + I_3 \end{aligned}$$

where

$$\begin{aligned} I_1= & {} -\int _{{\mathbb R}}\int _I\left( 1 + |v|^k\right) \partial _xu\,\partial _vf\,\partial _xf\,{\textrm{d}}x\,{\textrm{d}}v,\\ I_2= & {} \int _{{\mathbb R}}\int _I\left( 1 + |v|^k\right) \left| \partial _xf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v \end{aligned}$$

and

$$\begin{aligned} I_3 = -\frac{1}{2}\int _{{\mathbb R}}\int _I\left( 1 + |v|^k\right) \left( u - v\right) \partial _v\left| \partial _xf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v. \end{aligned}$$

After using the Young’s inequality in $I_1$, we obtain

$$\begin{aligned} I_1 \le \Vert u_x\Vert _{L^\infty (I)}\int _{{\mathbb R}}\int _I\left( 1 + |v|^k\right) \left( \left| \partial _xf\right| ^2 + \left| \partial _vf\right| ^2\right) \,{\textrm{d}}x\,{\textrm{d}}v. \end{aligned}$$

An integration by parts yields

$$\begin{aligned} I_3 = -\frac{1}{2}\int _{{\mathbb R}}\int _I\left( 1 + |v|^k\right) \left| \partial _xf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v + I_4 \end{aligned}$$

with

$$\begin{aligned} I_4 = \frac{k}{2}\int _{{\mathbb R}}\int _I\,\vert v\vert ^{k-2}v\left( u - v\right) \left| \partial _xf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v. \end{aligned}$$

A use of the Young’s inequality shows

$$\begin{aligned} \begin{aligned} I_4&\le \frac{k}{2}\Vert u\Vert _{L^\infty (I)}\int _{{\mathbb R}}\int _I\vert v\vert ^{k-1}\left| \partial _xf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v + \frac{k}{2}\int _{{\mathbb R}}\int _I\,\vert v\vert ^k\left| \partial _xf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v \\&\le \frac{k}{2}\Vert u\Vert _{L^\infty (I)}\int _{{\mathbb R}}\int _I\left( \frac{k-1}{k}\vert v\vert ^{k} + \frac{1}{k}\right) \left| \partial _xf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v + \frac{k}{2}\int _{{\mathbb R}}\int _I\,\vert v\vert ^k\left| \partial _xf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v \\&\le C\left( 1 + \Vert u\Vert _{L^\infty (I)}\right) \int _{{\mathbb R}}\int _I\left( 1 + \vert v\vert ^k\right) \left| \partial _xf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v. \end{aligned} \end{aligned}$$

Next, consider the equation for $\partial _vf$

$$\begin{aligned} \partial _t\partial _vf + v\partial _x\partial _vf + \partial _xf - \partial _vf + \partial _v\left( \left( u - v\right) \partial _vf\right) = 0. \end{aligned}$$

Multiplying the above equation by $\left( 1 + \vert v\vert ^k\right) \partial _vf$ and integrating with respect to x, v yields

$$\begin{aligned} \frac{1}{2}\int _{{\mathbb R}}\int _I\left( 1 + \vert v\vert ^k\right) \left| \partial _vf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v = I_5 + I_6 + I_7 \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} I_5&= -\int _{{\mathbb R}}\int _I\left( 1 + \vert v\vert ^k\right) \partial _xf\,\partial _vf\,{\textrm{d}}x\,{\textrm{d}}v\\ I_6&= \int _{{\mathbb R}}\int _I\left( 1 + \vert v\vert ^k\right) \left| \partial _vf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v\\ I_7&= -\int _{{\mathbb R}}\int _I\partial _v\left( \left( u - v\right) \partial _vf\right) \left( 1 + \vert v\vert ^k\right) \partial _vf\,{\textrm{d}}x\,{\textrm{d}}v. \end{aligned} \end{aligned}$$

After using the Young’s inequality in $I_5$, we obtain

$$\begin{aligned} I_5 \le \int _{{\mathbb R}}\int _I\left( 1 + |v|^k\right) \left( \left| \partial _xf\right| ^2 + \left| \partial _vf\right| ^2\right) \,{\textrm{d}}x\,{\textrm{d}}v. \end{aligned}$$

An integration by parts yields

$$\begin{aligned} I_7 = \int _{{\mathbb R}}\int _I \left( 1 + \vert v\vert ^k\right) \left| \partial _vf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v + I_8 \end{aligned}$$

with

$$\begin{aligned} I_8 = -\frac{1}{2}\int _{{\mathbb R}}\int _I\left( 1 + \vert v\vert ^k\right) \left( u - v\right) \partial _v\left| \partial _vf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v. \end{aligned}$$

An integration by parts shows

$$\begin{aligned} I_8 = -\frac{1}{2}\int _{{\mathbb R}}\int _I\left( 1 + \vert v\vert ^k\right) \left| \partial _vf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v - \frac{k}{2}\int _{{\mathbb R}}\int _I\,\vert v\vert ^k\left| \partial _vf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v + I_9 \end{aligned}$$

with

$$\begin{aligned} I_9 = \frac{k}{2}\int _{{\mathbb R}}\int _I\,\vert v\vert ^{k-2}vu\,\vert \partial _vf\vert ^2\,{\textrm{d}}x\,{\textrm{d}}v. \end{aligned}$$

A use of the Young’s inequality leads to

$$\begin{aligned} \begin{aligned} I_9&\le \Vert u\Vert _{L^\infty (I)}\frac{k}{2}\int _{{\mathbb R}}\int _I\,\vert v\vert ^{k-1}\left| \partial _vf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v\\&\le \Vert u\Vert _{L^\infty (I)}\frac{k}{2}\int _{{\mathbb R}}\int _I\left( \frac{k-1}{k}\vert v\vert ^{k} + \frac{1}{k}\right) \left| \partial _vf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v \\&\le C\int _{{\mathbb R}}\int _I\left( 1 + \vert v\vert ^k\right) \left| \partial _vf\right| ^2\,{\textrm{d}}x\,{\textrm{d}}v. \end{aligned} \end{aligned}$$

Altogether, we obtain

$$\begin{aligned} \begin{aligned}&\frac{{\textrm{d}}}{{\textrm{d}}t}\left( \int _{{\mathbb R}}\int _I\left( 1 + \vert v\vert ^k\right) \left( \left| \partial _xf\right| ^2 + \left| \partial _vf\right| ^2\right) \,{\textrm{d}}x\,{\textrm{d}}v\right) \le C\left( 1 + \Vert u\Vert _{W^{1,\infty }(I)}\right) \\&\quad \int _{{\mathbb R}}\int _I\left( 1 + \vert v\vert ^k\right) \left( \left| \partial _x f\right| ^2 + \left| \partial _vf\right| ^2\right) \,{\textrm{d}}x\,{\textrm{d}}v. \end{aligned} \end{aligned}$$

Let $F(t) = \int _{{\mathbb R}}\int _I\left( 1 + \vert v\vert ^k\right) \left( \left| \partial _xf\right| ^2 + \left| \partial _vf\right| ^2\right) \,{\textrm{d}}x\,{\textrm{d}}v$, then

$$\begin{aligned} \begin{aligned} F'(t) \le C\left( 1 + \Vert u(t,x)\Vert _{W^{1,\infty }(I)}\right) F(t) \end{aligned} \end{aligned}$$

this implies

$$\begin{aligned} F(t) \le CF(0)e^{\Vert u\Vert _{L^1(0,T;W^{1,\infty }(I))}}. \end{aligned}$$

This completes our proof. $\square$

Theorem A.4

(Existence and Uniqueness of strong solution) Let the initial data $\left( f_0,u_0\right)$ be such that

$$\begin{aligned}{} & {} f_0 \in L^1(I\times {\mathbb R})\cap L^\infty (I\times {\mathbb R})\cap H^1(I\times {\mathbb R}), \quad f_0 \ge 0,\\{} & {} \quad \int _{{\mathbb R}}\int _I\,\vert v\vert ^p\left\{ f_0 + \left| \partial _xf_0\right| ^2 + \left| \partial _vf_0\right| ^2\right\} \,{\textrm{d}}x\,{\textrm{d}}v \le C, \end{aligned}$$

for $0 \le p \le 4$ and $u_0 \in H^1(I)$. Then, there exists a unique global-in-time strong solution $\left( f,u\right)$ to the Vlasov-viscous Burgers’ system (1.1)–(1.2).

Proof

Let $0< T < \infty$ and set $X = L^\infty (0,T;H^1(I))$, with norm

$$\begin{aligned} \Vert u\Vert ^2_{X} = \sup _{t \in [0,T]}\left( \int _I\left( u^2 + |\partial _x u|^2\right) \,{\textrm{d}}x\right) . \end{aligned}$$

We now consider the map

$$\begin{aligned} \begin{aligned}&\mathcal {T} : X \rightarrow X\\&\quad u^* \longmapsto u = \mathcal {T}(u^*) \end{aligned} \end{aligned}$$

(A.4)

defined by the following scheme.

• Solve the Vlasov equation

  $$\begin{aligned} \partial _t f + v\,\partial _x f + \partial _v \left( \left( u^* - v\right) f\right) = 0, \end{aligned}$$

  (A.5)

  with initial data $f_0$.

• Solve the viscous Burgers’ equation

  $$\begin{aligned} \begin{aligned} \left( u_t,\phi \right) + \left( uu_x, \phi \right) + \epsilon \left( u_x,\phi _x\right) = \left( \rho V - \rho u, \phi \right) , \quad \forall \,\, \phi \in H^1(I) \end{aligned} \end{aligned}$$

  (A.6)

  with initial data $u_0$. Here $\rho$ and $\rho V$ are the local density and the momentum associated with the solution f of (A.5).

To begin with, we show that the above map $\mathcal {T}$ is well-defined. For a given $u^* \in X$ and a given initial datum $f_0$, the Vlasov equation (A.5) is uniquely solvable (see Lemma A.5 below for details). Having solved (A.5) for $f(u^*)$, one gather that the corresponding local density $\rho \in L^\infty$ (see Lemma 2.5) and the corresponding momentum $\rho V \in L^2$ (see Lemma 2.3). Hence, classical theory for the viscous Burgers’ problem [15] yields a unique solution $u \in X$ for the problem (A.6). Thus, the map $\mathcal {T}: X \rightarrow X$ that takes $u^*$ to $\mathcal {T}(u^*) = u$ is well-defined. Our next step in the proof is to show that $\mathcal {T}$ is a contraction map and that has been demonstrated in Lemma A.6 below. Therefore, an application of the Banach fixed-point theorem ensures the existence of a unique solution (f, u) in a short time interval $(0,T^0)$. As the solution (f, u) stays bounded at $t = T^0$, thanks to a priori estimates, we can employ continuation argument to extend the interval of existence upto (0, T]. As T is arbitrary, we get global-in-time well-posedness of our system. $\square$

Next we deal with Lemmata A.5 and A.6 which played a crucial role in the above proof.

Lemma A.5

Let $u^* \in X$ and let $f_0 \in L^1(I\times {\mathbb R})\cap L^\infty (I\times {\mathbb R})$. Then, there exists a unique solution $f \in L^\infty (0,T;L^1(I\times {\mathbb R})\cap L^\infty (I\times {\mathbb R}))$ to (A.5).

Proof

Note that (A.5) can be rewritten as

$$\begin{aligned} \partial _t f + b \cdot \nabla _{x,v}f - f =0, \end{aligned}$$

where $b = (v, u^*(t,x) - v)$, which lies in

$$\begin{aligned} L^1(0,T;H^1(I \times (-K,K))), \quad 0< K < \infty . \end{aligned}$$

Note that div$_{x,v}b = -1 \in L^\infty ((0,T) \times \Omega )$. Furthermore, $|b|/(1 + |v|)$ is bounded. This setting appeals to the general results in [8]. In particular, we can apply [8, Corollaries II-1 and II-2, p.518] to arrive at the existence of the unique solution. $\square$

Lemma A.6

The map defined by (A.5) and (A.6) is a contraction map.

Proof

Let for given $u_i^*$ there exist a unique solution $f_i$ for equation (A.5). Define $\bar{u} = u_1 - u_2, \bar{u}^* = u_1^* - u_2^*$ and $\bar{f} = f_1 - f_2$, then from (A.5) and (A.6) we find that

$$\begin{aligned} \bar{f}_t + v\bar{f}_x + \partial _v\left( \bar{u}^*f_1 + u_2^*\bar{f} - v\bar{f}\right) = 0, \quad \text{ with } \quad \bar{f}(0,x,v) = 0, \end{aligned}$$

(A.7)

and

$$\begin{aligned} \left( \bar{u}_t, \phi \right) _I - \epsilon \,\left( \bar{u}_{xx}, \phi \right) _I = \left( \int _{{\mathbb R}}\left( v\bar{f} + u_2\bar{f} - \bar{u}f_1\right) \,{\textrm{d}}v, \phi \right) _I - \left( \left( \bar{u}\partial _xu_1 + u_2\bar{u}_x\right) , \phi \right) _I, \end{aligned}$$

(A.8)

with $\quad \bar{u}(0,x) = 0.$

Choose $\phi = \bar{u} - \bar{u}_{xx}$ in (A.8) to obtain

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\partial _t\Vert \bar{u}\Vert ^2_{L^2(I)} + \frac{1}{2}\partial _t\Vert \bar{u}_x\Vert ^2_{L^2(I)} + \epsilon \Vert \bar{u}_x\Vert ^2_{L^2(I)} + \epsilon \,\Vert \bar{u}_{xx}\Vert ^2_{L^2(I)}\\&\quad = \left( \int _{{\mathbb R}}\left( v\bar{f} + u_2\bar{f} - \bar{u}f_1\right) \,{\textrm{d}}v, \bar{u} - \bar{u}_{xx}\right) _I\\&\quad - \left( \left( \bar{u}\partial _xu_1 + u_2\bar{u}_x\right) , \bar{u} - \bar{u}_{xx}\right) _I \end{aligned} \end{aligned}$$

a use of the Hölder inequality with the Young’s inequality, a kickback argument and integration in time yields

$$\begin{aligned} \begin{aligned}&\Vert \bar{u}\Vert ^2_{X} + \epsilon \Vert \bar{u}_x\Vert ^2_{L^2([0,T]\times I)} + \epsilon \,\Vert \bar{u}_{xx}\Vert _{L^2([0,T]\times I)}^2 \\&\quad \le 4\epsilon ^{-1}\,\left\| \int _{{\mathbb R}}\left( v\bar{f} + u_2\bar{f} - \bar{u}f_1\right) \,{\textrm{d}}v\right\| ^2_{L^2([0,T]\times I)}\\&\quad + 4\epsilon ^{-1}\,\Vert \bar{u}\partial _xu_1 + u_2\bar{u}_x\Vert ^2_{L^2([0,T] \times I)} + \epsilon T\Vert \bar{u}\Vert ^2_X. \end{aligned} \end{aligned}$$

(A.9)

Now, the Hölder inequality followed by Sobolev imbedding shows

$$\begin{aligned} \begin{aligned}&\left\| \int _{{\mathbb R}}\left( v\bar{f} + u_2\bar{f} - \bar{u}f_1\right) \,{\textrm{d}}v\right\| _{L^2([0,T]\times I)} \le \left\| \int _{{\mathbb R}}v\bar{f}\,{\textrm{d}}v\right\| _{L^2([0,T]\times I)} \\&\quad + T\Vert \bar{u}\Vert _{X}\Vert m_0f_1\Vert _{L^\infty (0,T;L^2(I))} + \Vert u_2\Vert _{X }\Vert m_0\bar{f}\Vert _{L^2([0,T] \times I)}, \end{aligned} \end{aligned}$$

(A.10)

and

$$\begin{aligned} \begin{aligned} \Vert \bar{u}\partial _xu_1 + u_2\bar{u}_x\Vert _{L^2([0,T]\times I)}&\le \Vert \bar{u}\Vert _{L^\infty ([0,T]\times I)}\Vert \partial _xu_1\Vert _{L^2([0,T]\times I)} \\&\quad + \Vert u_2\Vert _{L^2([0,T];L^\infty (I))}\Vert \bar{u}\Vert _{L^\infty ([0,T];H^1(I))} \\&\le T^\frac{1}{2}\left( \Vert u_1\Vert _{X} + \Vert u_2\Vert _X\right) \Vert \bar{u}\Vert _{X}. \end{aligned} \end{aligned}$$

(A.11)

For a sufficiently small $T > 0$

$$\begin{aligned} \begin{aligned} 4\epsilon ^{-1}\,T\left( \Vert m_0f_1\Vert ^2_{L^\infty (0,T;L^2(I))} + \Vert u_1\Vert ^2_{X} + \Vert u_2\Vert ^2_X + \frac{\epsilon ^2}{4}\,\right) \le \frac{1}{2}. \end{aligned} \end{aligned}$$

(A.12)

Using Eqs. (A.9)–(A.12), we obtain

$$\begin{aligned} \Vert \bar{u}\Vert _{X}^2 \le C\left\| \int _{{\mathbb R}}v\bar{f}\,{\textrm{d}}v\right\| ^2_{L^2([0,T]\times I)} + C\Vert u_2\Vert ^2_{X}\Vert m_0\bar{f}\,{\textrm{d}}v\Vert ^2_{L^2([0,T] \times I)}. \end{aligned}$$

(A.13)

Now, a similar calculation as in proof of Lemma 2.3 shows

$$\begin{aligned} \left\| \int _{{\mathbb R}}v\bar{f}\,{\textrm{d}}v\right\| _{L^2([0,T] \times I)} \le C\int _0^T\int _{{\mathbb R}}\int _I|v|^3|\bar{f}|\,{\textrm{d}}x\,{\textrm{d}}v\,{\textrm{d}}t, \end{aligned}$$

(A.14)

and

$$\begin{aligned} \left\| \int _{{\mathbb R}}\bar{f}\,{\textrm{d}}v\right\| _{L^2([0,T]\times I)} \le C\int _0^T\int _{{\mathbb R}}\int _I|v||\bar{f}|\,{\textrm{d}}x\,{\textrm{d}}v\,{\textrm{d}}t. \end{aligned}$$

(A.15)

Multiplying Eq. (A.7) by $|v|^k\frac{\bar{f}}{\sqrt{\bar{f}^2 + \delta }}$; $k \ge 1$, we obtain

$$\begin{aligned} \begin{aligned}&|v|^k\partial _t\left( \sqrt{\bar{f}^2 + \delta }\right) + |v|^k\,v\,\partial _x\left( \sqrt{\bar{f}^2 + \delta }\right) + \bar{u}^*|v|^k\frac{\bar{f}}{\sqrt{\bar{f}^2 + \delta }}\,\partial _vf_1 \\&\qquad + u_2^*|v|^k\,\partial _v\left( \sqrt{\bar{f}^2 + \delta }\right) - |v|^k\,\frac{\bar{f}}{\sqrt{\bar{f}^2 + \delta }}\,\partial _v\left( v\bar{f}\,\right) = 0. \end{aligned} \end{aligned}$$

Then, an integrate with respect to x, v yields

$$\begin{aligned} \begin{aligned}&\partial _t\int _{\mathbb R}\int _I |v|^k\left( \sqrt{\bar{f}^2 + \delta }\right) \,{\textrm{d}}x\,{\textrm{d}}v - k\int _{\mathbb R}\int _I\bar{u}^*\,|v|^{k-2}v\,\frac{\bar{f}}{\sqrt{\bar{f}^2 + \delta }}\,f_1\,{\textrm{d}}x\,{\textrm{d}}v \\&\qquad - \int _{\mathbb R}\int _I|v|^k\,\bar{u}^*\,f_1\,\partial _v\left( \frac{\bar{f}}{\sqrt{\bar{f}^2 + \delta }}\right) \,{\textrm{d}}x\,{\textrm{d}}v - k\int _{\mathbb R}\int _I |v|^{k-2}v\,u^*_2\,\left( \sqrt{\bar{f}^2 + \delta }\right) \,{\textrm{d}}x\,{\textrm{d}}v \\&\qquad + k\int _{\mathbb R}\int _I|v|^k\,\frac{\bar{f}^2}{\sqrt{\bar{f}^2 + \delta }} \,{\textrm{d}}x\,{\textrm{d}}v + \int _{\mathbb R}\int _I |v|^k\partial _v\left( \frac{\bar{f}}{\sqrt{\bar{f}^2 + \delta }}\right) v\bar{f}\,{\textrm{d}}x\,{\textrm{d}}v = 0. \end{aligned} \end{aligned}$$

A use of Sobolev inequality with integration in time shows

$$\begin{aligned} \begin{aligned}&\sup _{t \in [0,T]}\int _{{\mathbb R}}\int _I|v|^k\left( \sqrt{\bar{f}^2 + \delta }\right) \,{\textrm{d}}x\,{\textrm{d}}v + k\int _0^T\int _{{\mathbb R}}\int _I|v|^k\frac{\bar{f}^2}{\sqrt{\bar{f}^2 + \delta }}\,{\textrm{d}}x\,{\textrm{d}}v\,{\textrm{d}}t \\&\quad \le \int _0^Tk\Vert \bar{u}^*\Vert _{H^1(I)}\Vert m_{k-1}f_1\Vert _{L^1(I)} \,{\textrm{d}}t + |T_k^1| + |T^2_k| \\&\qquad + \int _{0}^T k\Vert u_2^*\Vert _{H^1(I)}\left\| \int _{{\mathbb R}}|v|^{k-1}\left( \sqrt{\bar{f}^2 + \delta }\right) \,{\textrm{d}}v\right\| _{L^1(I)}\,{\textrm{d}}t \\&\quad \le T\Vert \bar{u}^*\Vert _{X}\Vert m_{k-1}f_1\Vert _{L^\infty (0,T;L^1(I))} + |T_k^1| + |T^2_k| \\&\qquad + T\Vert u_2^*\Vert _{X}\left\| \int _{{\mathbb R}}|v|^{k-1}\left( \sqrt{\bar{f}^2 + \delta }\right) \,{\textrm{d}}v\right\| _{L^\infty (0,T;L^1(I))}. \end{aligned} \end{aligned}$$

(A.16)

Here,

$$\begin{aligned} T^1_k= & {} \int _0^T\int _{\mathbb R}\int _I|v|^k\,\bar{u}^*\,f_1\,\partial _v\left( \frac{\bar{f}}{\sqrt{\bar{f}^2 + \delta }}\right) \,{\textrm{d}}x\,{\textrm{d}}v\,{\textrm{d}}t\\= & {} \int _0^T\int _{\mathbb R}\int _I|v|^k\,\bar{u}^*\,f_1\,\frac{\delta \,\bar{f}_v}{\left( \bar{f}^2+\delta \right) ^\frac{3}{2}}{\textrm{d}}x\,{\textrm{d}}v\,{\textrm{d}}t \end{aligned}$$

and

$$\begin{aligned} T^2_k = \int _0^T\int _{\mathbb R}\int _I |v|^k\partial _v\left( \frac{\bar{f}}{\sqrt{\bar{f}^2 + \delta }}\right) v\bar{f}\,{\textrm{d}}x\,{\textrm{d}}v\,{\textrm{d}}t = \int _0^T\int _{\mathbb R}\int _I |v|^k v\bar{f}\frac{\delta \,\bar{f}_v}{\left( \bar{f}^2+\delta \right) ^\frac{3}{2}}\,{\textrm{d}}x\,{\textrm{d}}v\,{\textrm{d}}t. \end{aligned}$$

As $\bar{f} \in L^1(0,T;L^\infty (I\times {\mathbb R}))$ and as fourth order velocity moments of $\left| \partial _vf\right| ^2$ and $\bar{f}$ are bounded (see Lemma A.3), $\left| T_k^1\right| \rightarrow 0$ and $\left| T_k^2\right| \rightarrow 0$ as $\delta \rightarrow 0$ for $k = 1,2,3$. Next, we multiply Eq. (A.7) by $\frac{\bar{f}}{\sqrt{\bar{f}^2+\delta }}$ and integrate with respect to x, v and t variables to obtain

$$\begin{aligned} \begin{aligned}&\int _{\mathbb R}\int _I\sqrt{\bar{f}^2+\delta }\,{\textrm{d}}x\,{\textrm{d}}v - \int _0^T\int _{\mathbb R}\int _I\frac{\delta \,\bar{f}_v}{\left( \bar{f}^2+\delta \right) ^\frac{3}{2}}\left( \bar{u}^*f_1 + u_2^*\bar{f} - v\bar{f}\right) \,{\textrm{d}}x\,{\textrm{d}}v\\&\quad = \int _{\mathbb R}\int _I\sqrt{\bar{f}^2(0,x,v)+\delta }\,{\textrm{d}}x\,{\textrm{d}}v, \end{aligned} \end{aligned}$$

with $\bar{f}(0,x,v) = 0$. Arguing as we did with the $T_k^1$ and $T_k^2$ terms, in the $\delta \rightarrow 0$ limit in above equation, we arrive at

$$\begin{aligned} \int _{\mathbb R}\int _I|\bar{f}|\,{\textrm{d}}x\,{\textrm{d}}v = 0. \end{aligned}$$

(A.17)

A use of Eq. (A.16) for $k = 1$ in Eq. (A.15) and $\delta \rightarrow 0$ shows

$$\begin{aligned} \left\| \int _{{\mathbb R}}\bar{f}\,{\textrm{d}}v\right\| _{L^2([0,T]\times I)} \le T\Vert \bar{u}^*\Vert _{X}\Vert f_0\Vert _{L^1(\Omega )}. \end{aligned}$$

(A.18)

Using Eq. (A.14) and by induction from Eq. (A.16), we deduce

$$\begin{aligned} \begin{aligned} \left\| \int _{{\mathbb R}}v\bar{f}\,{\textrm{d}}v\right\| _{L^2([0,T]\times I)}&\lesssim T\Vert \bar{u}^*\Vert _{X}\left( \Vert m_2f_1\Vert _{L^\infty (0,T;L^1(I))} + \Vert u_2^*\Vert _{X}\Vert m_1f_1\Vert _{L^\infty (0,T;L^1(I))}\right. \\&\quad \left. + \Vert u_2^*\Vert ^2_{X}\Vert m_0f_1\Vert _{L^\infty (0,T;L^1(I))}\right) . \end{aligned} \end{aligned}$$

(A.19)

Using (A.18)–(A.19) for a sufficiently small $T > 0$ from (A.13), we obtain

$$\begin{aligned} \Vert \bar{u}\Vert _{X} \le \alpha _0\Vert \bar{u}^*\Vert _{X}, \quad \alpha _0 < 1. \end{aligned}$$

This shows that $\mathcal {T}$ is a contraction map. $\square$