Global Existence of Classical Solutions to the Inelastic Vlasov–Poisson–Boltzmann System

Abstract

We study the global existence and uniqueness of classical solutions to the inelastic Vlasov–Poisson–Boltzmann system for a soft potential in the near vacuum regime. For the global existence of classical solutions, we assume reasonable conditions on the restitution coefficient, which represents the character of inelastic collisions. We use the smallness of initial data and an algebraically decaying weight function in the spatial variable to control the self-consistence force and collision operator.

Appendix: Proof of the Second Half of Lemma 4.2

In the appendix, we estimate the term $N_1(f,g)$ in Lemma 4.2.

Recall that

$$\begin{aligned} N_1(f,g) := -\int \limits _0^t\int \limits _{{\mathbb {R}}^3\times {\mathbb {S}}^2}\frac{q(u,\omega )}{ ^{\prime } e^{\gamma } \cdot { ^{\prime } J} }f(X(s),V(s)-U_{\parallel },s)g(X(s),V(s)-U_{\perp },s)dud\omega ds. \end{aligned}$$

To estimate $N_1(f,g)$, we divide it into its small time part and large time part. That is,

$$\begin{aligned} |N_1(f,g)(t,x,v)|&\le \bigg |\int \limits _0^1\int \limits _{{\mathbb {R}}^3\times {\mathbb {S}}^2}[\cdots ]dud\omega ds\bigg |+\bigg |\int \limits _1^t\int \limits _{{\mathbb {R}}^3\times {\mathbb {S}}^2}[\cdots ]dud\omega ds\bigg |\\&\equiv N_{11}(f,g)+N_{12}(f,g). \end{aligned}$$

(Estimate of $N_{11}(f,g)$): Similar to the estimate for $N_{21}(f,g)$ in Lemma 4.1, we have

$$\begin{aligned} |N_{11}(f,g)| \le C\sup _s||f(s)||\sup _s ||g(s)||. \end{aligned}$$

(Estimate of $N_{12}(f,g)$): In this case, we use assumptions $({\mathcal A}1)$–$({\mathcal A}3)$ and the properties of $U_{\parallel }$, $U_{\perp }$, and $^{\prime } e$. Similar to the estimate given for the loss term, we have

$$\begin{aligned} |N_{12}(f,g)|&\le \int \limits _1^t ||f(s)||\cdot ||g(s)||\int \limits _{{\mathbb {R}}^3\times {\mathbb {S}}^2} \frac{q(u,\omega )}{ ^{\prime } e^{\gamma } \cdot { ^{\prime } J} }(1+|X_{\parallel }(0)|^2)^{-k}(1+|X_{\perp }(0)|^2)^{-k} \\&\quad \times e^{-\alpha |V_{\parallel }(0)|^2-\alpha |V_{\perp }(0)|^2} dud\omega ds\\&\le \sup _s||f(s)||\sup _s||g(s)|| \int \limits _1^t \int \limits _{{\mathbb {R}}^3\times {\mathbb {S}}^2} \frac{q(u,\omega )}{ ^{\prime } e^{\gamma } \cdot { ^{\prime } J}}(1+|X_{\parallel }(0)|^2)^{-k}(1+|X_{\perp }(0)|^2)^{-k} \\&\quad \times e^{-\alpha |V_{\parallel }(0)|^2-\alpha |V_{\perp }(0)|^2} dud\omega ds. \end{aligned}$$

It follows from Lemma 2.7 that

$$\begin{aligned} |N_{12}(f,g)|&\le \sup _s||f(s)||\sup _s||g(s)|| \int \limits _1^t \int \limits _{{\mathbb {R}}^3\times {\mathbb {S}}^2}\frac{q(u,\omega )}{ ^{\prime } e^{\gamma } \cdot { ^{\prime } J}}(1+|X_{\parallel }(0)|^2)^{-k}(1+|X_{\perp }(0)|^2)^{-k} \\&\quad \times e^{-\alpha (|V(0)|^2+|V(s)-u|^2+\frac{1-e^2}{2}|{ ^{\prime } u}\cdot \omega |^2)} dud\omega ds\\&\le \sup _s||f(s)||\sup _s||g(s)|| e^{-\alpha |V(0)|^2} \int \limits _1^t \int \limits _{{\mathbb {R}}^3\times {\mathbb {S}}^2} \frac{q(u,\omega )}{ ^{\prime } e^{\gamma }\cdot { ^{\prime } J}}e^{-\alpha \frac{1-e^2}{2}|{ ^{\prime } u}\cdot \omega |^2} \\&\quad \times (1+|X_{\parallel }(0)|^2)^{-k}(1+|X_{\perp }(0)|^2)^{-k} e^{-\alpha |V(s)-u|^2} du d\omega ds . \end{aligned}$$

Since the collision kernel $q$ satisfies (2.3) and $e$ satisfies $({\mathcal A}3)$ in Sect. 2.1, we obtain the following estimate for the collision potential term in the above integrand:

$$\begin{aligned}&\Big |\frac{q(u,\omega )}{ ^{\prime } e^{\gamma }\cdot { ^{\prime } J}}e^{-\alpha \frac{1-e^2}{2}|{ ^{\prime } u}\cdot \omega |^2}\Big | \\&\le C(1+|u|^{\gamma })\Big |\frac{1}{ ^{\prime } e^{\gamma } \cdot { ^{\prime } J}}e^{-\alpha \frac{1-e^2}{2}|{ ^{\prime } u}\cdot \omega |^2}\Big |\\&\le C\Big (||\psi _{\alpha /2,\gamma }||_{L^{\infty }}+\frac{|u|^{\gamma }}{ ^{\prime } e^{\gamma } \cdot { ^{\prime } J}}\Big )\\&\equiv q_1+q_2(u). \end{aligned}$$

We use the change of variable $\bar{u}=s u$ to calculate

$$\begin{aligned}&\int \limits _1^t \int \limits _{{\mathbb {R}}^3\times {\mathbb {S}}^2}\big [q_1+q_2(u)\big ] (1+|X_{\parallel }(0)|^2)^{-k}(1+|X_{\perp }(0)|^2)^{-k} e^{-\alpha |V(s)-u|^2} du d\omega ds \\&\quad =\int \limits _1^t \int \limits _{{\mathbb {R}}^3\times {\mathbb {S}}^2} \big [q_1+q_2(\bar{u}/s)\big ] (1+|X_{\parallel }(0)|^2)^{-k}(1+|X_{\perp }(0)|^2)^{-k} e^{-\alpha |V(s)-\bar{u}/s|^2} \frac{d\bar{u}}{s^3} d\omega ds\\&\quad \le \int \limits _1^t \int \limits _{{\mathbb {R}}^3\times {\mathbb {S}}^2} \big [q_1+q_2(\bar{u}/s)\big ] (1+|X_{\parallel }(0)|^2)^{-k}(1+|X_{\perp }(0)|^2)^{-k} \frac{d\bar{u}}{s^3} d\omega ds \\&\quad \equiv \Lambda (t). \end{aligned}$$

We claim that there exists a constant $C>0$ such that

$$\begin{aligned} \Lambda (t) \le C(1+|X(0)|^2)^{-k},\quad \text{ for } k>3/2\text{. } \end{aligned}$$

It follows from Lemma 2.6 that

$$\begin{aligned} \text{ either } \quad |X_{\parallel }(0)|^2 \ge \frac{|X(0)|^2}{4} \quad \text{ or } \quad |X_{\perp }(0)|^2 \ge \frac{|X(0)|^2}{4}. \end{aligned}$$

Case A $(|X_{\perp }(0)|^2\ge \frac{|X(0)|^2}{4})$: We decompose $\bar{u}$ as

$$\begin{aligned} \bar{u}=(\bar{u}\cdot \omega )\omega + (\bar{u}-(\bar{u}\cdot \omega )\omega ) \equiv \bar{u}_{\parallel }+\bar{u}_{\perp }. \end{aligned}$$

(7.1)

In (7.1), we consider $\bar{u}_{\parallel }$ as a three-dimensional variable with $|\bar{u}\cdot \omega |=|\bar{u}_{\parallel }|$ and $\bar{u}_{\parallel }=|\bar{u}_{\parallel }|\omega$. Since ${\mathbb {R}}^3\times {\mathbb {S}}^2$ is a five-dimensional surface, we use the following change of variable:

$$\begin{aligned} d\bar{u}d\omega =d\bar{u}_{\perp } d(\bar{u}\cdot \omega ) d\omega =\frac{1}{|\bar{u}_{\parallel }|^2}d\bar{u}_{\perp } d\bar{u}_{\parallel }. \end{aligned}$$

For simplicity, we define the following notation:

$$\begin{aligned} \bar{U}_{\parallel }=\frac{U_{\parallel }}{s}=\frac{1+ ^{\prime } e}{2 ^{\prime } e}\bar{u}_{\parallel },\quad \bar{U}_{\perp }=\frac{U_{\perp }}{s}=\frac{1+ ^{\prime } e}{2 ^{\prime } e}\bar{u}_{\perp } -\frac{1- ^{\prime } e}{2 ^{\prime } e}\bar{u}=\bar{u}_{\perp } -\frac{1- ^{\prime } e}{2 ^{\prime } e}\bar{u}_{\parallel }. \end{aligned}$$

By definition, we have

$$\begin{aligned}e=e(u\cdot \omega ), \quad ^{\prime } e=e({ ^{\prime } u}\cdot \omega ), \quad u\cdot \omega =-({ ^{\prime } u}\cdot \omega ) e({ ^{\prime } u}\cdot \omega ),\end{aligned}$$

which implies

$$\begin{aligned} u\cdot \omega =-\theta ({ ^{\prime } u}\cdot \omega ) \quad \text{ with } \quad \theta (z)=z e(z). \end{aligned}$$

Thus, we have the following relation:

$$\begin{aligned} ^{\prime } e=e({ ^{\prime } u}\cdot \omega )=\frac{u\cdot \omega }{\theta ^{-1}(u\cdot \omega )}=\frac{\bar{u}\cdot \omega /s}{\theta ^{-1}(\bar{u}\cdot \omega /s)}. \end{aligned}$$

This relation implies

$$\begin{aligned} \theta ^{-1}(|u_{\parallel }|)=\theta ^{-1}(u\cdot \omega )=\frac{u\cdot \omega }{ ^{\prime } e}=\frac{|u_{\parallel }|}{ ^{\prime } e}. \end{aligned}$$

Since $\bar{U}_{\parallel }$ depends on $u_{\parallel }$ , we can calculate the Jacobian $J_{U}$ using the inverse function theorem as follow:

$$\begin{aligned} J_{U}&= det \frac{\partial ( \bar{U}_{\perp },| \bar{U}_{\parallel }|,\omega )}{\partial (\bar{u}_{\perp }, |\bar{u}_{\parallel }|,\omega )}=\frac{d |\bar{U}_{\parallel }|}{d |\bar{u}_{\parallel }|}=\frac{d}{d |\bar{u}_{\parallel }|} (\frac{1+ ^{\prime } e}{2 ^{\prime } e}|\bar{u}_{\parallel }|)=\frac{1}{2}\frac{d}{d |\bar{u}_{\parallel }|} (\frac{|\bar{u}_{\parallel }|}{ ^{\prime } e})+\frac{1}{2}\nonumber \\&= \frac{1}{2}\frac{d}{d |u_{\parallel }|} (\frac{|u_{\parallel }|}{ ^{\prime } e})+\frac{1}{2}= \frac{1}{2\theta _z(|{ ^{\prime } u}\cdot \omega |)}+\frac{1}{2}. \end{aligned}$$

(7.2)

By assumption $({\mathcal A}2)$ of $e=e(z)$ and the above calculation, we arrive at the following lower bound for $J_U$:

$$\begin{aligned} \frac{1}{2}<det \frac{\partial ( \bar{U}_{\perp },| \bar{U}_{\parallel }|,\omega )}{\partial (\bar{u}_{\perp }, |\bar{u}_{\parallel }|,\omega )}=J_U. \end{aligned}$$

(7.3)

From $J_U$, we have the following relation:

$$\begin{aligned} d\bar{u}d\omega =d\bar{u}_{\perp } d(\bar{u}\cdot \omega ) d\omega =\frac{1}{J_{U}}d\bar{U}_{\perp } d(\bar{U}\cdot \omega ) d\omega =\frac{1}{J_{U}|\bar{U}_{\parallel }|^2}d\bar{U}_{\perp } d\bar{U}_{\parallel }. \end{aligned}$$

Using repeated integrals over $\bar{U}_{\perp }$ and $\bar{U}_{\parallel }$, we estimate $\Lambda (t)$. Specifically,

$$\begin{aligned} \Lambda (t)&= \int \limits _1^t \int \limits _{{\mathbb {R}}^3\times {\mathbb {S}}^2} \big [q_1+q_2(\bar{u}/s)\big ] (1+|X_{\parallel }(0)|^2)^{-k}(1+|X_{\perp }(0)|^2)^{-k} \frac{d\bar{u}}{s^3} d\omega ds\\&= \int \limits _1^t \int \limits _{{\mathbb {R}}^3\times {\mathbb {R}}^2} (1+|X_{\parallel }(0)|^2)^{-k}(1+|X_{\perp }(0)|^2)^{-k} \frac{\big [q_1+q_2(\bar{u}/s)\big ]}{s^3 J_{U}}d\bar{U}_{\perp } \frac{d\bar{U}_{\parallel }}{|\bar{U}_{\parallel }|^2} ds. \end{aligned}$$

For $\gamma \le 0$, it follows from (7.3) and Lemma 2.5 that

$$\begin{aligned} \frac{\big [q_{1}+q_{2}(\bar{u}/s)\big ]}{{s^{3}} {J_{U}}}&= C\Big (||\psi ||_{L^{\infty }}+\frac{|u|^{\gamma }}{{}^{\prime } e^{\gamma } \cdot {{}^{\prime }J}}\Big )\frac{1}{{s^{3}} {J_{U}}} = \frac{C||\psi ||_{L^{\infty }}}{{s^{3}} {J_{U}}} +C\frac{|u|^{\gamma }}{{}^{\prime } e^{\gamma } \cdot {{}^{\prime } J}}\frac{1}{{s^{3}}{J_{U}}}\\&\le \frac{C||\psi ||_{L^{\infty }}}{s^{3}} +C\Big (\frac{{}^{\prime }e|{\bar{U}_{\perp }}|}{s}\Big )^{\gamma }\frac{1}{s^{3}{}^{\prime }e^{\gamma } \cdot {{}^{\prime }J}{J_{U}}}. \end{aligned}$$

It follows from (7.2) and the definition of $\theta (z)$ in Lemma 2.1(ii) that

$$\begin{aligned} \frac{\big [q_1+q_2(\bar{u}/s)\big ]}{s^3 J_{U}}&\le \frac{C||\psi ||_{L^{\infty }}}{s^3} +C\Big (\frac{{}^{\prime } e|\bar{U}_{\perp }|}{s}\Big )^{\gamma }\frac{1}{s^3 {}^{\prime } e^{\gamma } \cdot { {}^{\prime } J} }\frac{2\theta _z(|{ {}^{\prime } u}\cdot \omega |)}{1+\theta _z(|{ {}^{\prime } u}\cdot \omega |)}\\&= \frac{C||\psi ||_{L^{\infty }}}{s^3 } +C\frac{2|\bar{U}_{\perp }|^{\gamma }}{s^{3+\gamma } [1+\theta _z(|{ {}^{\prime } u}\cdot \omega |)] } . \end{aligned}$$

Finally, assumption $(\mathcal A 2)$ and $s>1$ lead to

$$\begin{aligned} \frac{\big [q_1+q_2(\bar{u}/s)\big ]}{{s^{3}} {J_{U}}}&\le \frac{C||\psi ||_{L^{\infty }}}{s^3 } +C\frac{|\bar{U}_{\perp }|^{\gamma }}{s^{3+\gamma } } \le \frac{C}{s^{3+\gamma }}(1 +|\bar{U}_{\perp }|^{\gamma } ). \end{aligned}$$

(7.4)

From the upper bound of the collision potential in (7.4) and the Jacobian, we have

$$\begin{aligned} \Lambda (t)\le \int \limits _1^t \int \limits _{{\mathbb {R}}^3\times {\mathbb {S}}^2} \frac{C}{s^{3+\gamma }}(1 +|\bar{U}_{\perp }|^{\gamma } ) (1+|X_{\parallel }(0)|^2)^{-k}(1+|X_{\perp }(0)|^2)^{-k} d\bar{U}_{\perp } \frac{d\bar{U}_{\parallel }}{|\bar{U}_{\parallel }|^2} ds. \end{aligned}$$

Thus, it suffices to prove that the following integral is uniformly bounded with respect to a large time region $t>1$:

$$\begin{aligned} \int \limits _{{\mathbb {R}}^3\times {\mathbb {R}}^2} (1 +|\bar{U}_{\perp }|^{\gamma } ) (1+|X_{\parallel }(0)|^2)^{-k}(1+|X_{\perp }(0)|^2)^{-k} d\bar{U}_{\perp } \frac{d\bar{U}_{\parallel }}{|\bar{U}_{\parallel }|^2}, \end{aligned}$$

where $X_{\perp }(0)$ and $X_{\parallel }(0)$ depend only on $\bar{U}_{\perp }$ and $\bar{U}_{\parallel }$, respectively.

It follows from Lemma 2.3 that

$$\begin{aligned} X_{\perp }(0)-X(0)=X(0;s,X(s),V(s)-U_{\perp })-X(0)=[I+ {\mathcal O}(\varepsilon _0)]\bar{U}_{\perp }. \end{aligned}$$

Thus, we again use a change of variable, in this case $\bar{U}_{\perp }= X_{\perp }(0)$, to calculate the two-dimensional integral:

$$\begin{aligned} \int \limits _{{\mathbb {R}}^2} (1\!+\!|\bar{U}_{\perp }|^{\gamma })(1\!+\!|X_{\perp }(0)|^2)^{-k}d\bar{U}_{\perp } \!\le \! C \int \limits _{{\mathbb {R}}^2} (1\!+\!|X(0)-X_{\perp }(0)|^{\gamma })(1\!+\!|X_{\perp }(0)|^2)^{-k}dX_{\perp }. \end{aligned}$$

Because of the assumption in this case, i.e., $|X_{\perp }(0)|^2\ge |X(0)|^2/4$, the above integral is bounded by

$$\begin{aligned}&C \int \limits _{{\mathbb {R}}^2\cap \{|X_{\perp }(0)|\ge |X(0)|/4\}} (1+|X(0)-X_{\perp }(0)|^{\gamma })(1+|X_{\perp }(0)|^2)^{-k}dX_{\perp }\\& \le C \bigg \{\int \limits _{|X_{\perp }(0)-X(0)|\le (|X(0)|+1)/2}+\int \limits _{|X_{\perp }(0)-X(0)|>(|X(0)|+1)/2}\bigg \} ~[\cdots ]~ dX_{\perp } \\& \equiv I_1+I_2. \end{aligned}$$

The assumption that $|X_{\perp }(0)|^2\ge |X(0)|^2/4$ leads to $(1+|X_{\perp }(0)|^{-k})\le C(1+|X(0)|^2)^{-k}$, which, in turn, implies

$$\begin{aligned} |I_1|\le C\frac{(|X(0)|+1)^2+(|X(0)|+1)^{\gamma +2}}{(1+|X(0)|^2)^{k}} \end{aligned}$$

and

$$\begin{aligned} |I_2|\le C(1+(|X(0)|+1)^{\gamma })(1+|X(0)|^2)^{1-k}. \end{aligned}$$

Thus, we arrive at the following estimate for the two-dimensional integral:

$$\begin{aligned} \int \limits _{{\mathbb {R}}^2} (1+|\bar{u}|^{\gamma })(1+|X_{\perp }(0)|^2)^{-k}d\bar{U}_{\perp }\le C(1+|X(0)|^2)^{1-k}. \end{aligned}$$

For the three-dimensional integral with respect to $\bar{U}_{\parallel }$, we use the same method as above. Then it follows from Lemma 2.3 that we have

$$\begin{aligned} X_{\parallel }(0)-X(0)=X(0;s,X(s),V(s)-U_{\parallel })-X(0)=[I+O(\epsilon _0)]\bar{U}_{\parallel }. \end{aligned}$$

It follows from Lemma 2.9 that

$$\begin{aligned} \int \limits _{{\mathbb {R}}^3} (1+|X_{\parallel }(0)|^2)^{-k}\frac{1}{|\bar{U}_{\parallel }|^2}d\bar{U}_{\parallel }&\le C\int \limits _{{\mathbb {R}}^3} (1+|X_{\parallel }(0)|^2)^{-k}\frac{1}{|X(0)-X_{\parallel }(0)|^2}dX_{\parallel }(0)\\&\le C(1+|X(0)|)^{-2}. \end{aligned}$$

This completes Case A.

Case B $(|X_{\parallel }(0)|^2\ge |X(0)|^2/4)$: In this case, we use the following decomposition of $\bar{u}$:

$$\begin{aligned} \bar{u}&= (\bar{u}\cdot \omega )\omega + (\bar{u}-(\bar{u}\cdot \omega )\omega )=\bar{u}-|\bar{u}-(\bar{u}\cdot \omega )\omega |\eta + |\bar{u}-(\bar{u}\cdot \omega )\omega |\eta \\&= (u-|\bar{u}_{\perp }|\eta )+|\bar{u}_{\perp }|\eta \equiv \bar{u}_{\parallel }+\bar{u}_{\perp }. \end{aligned}$$

In other words, we consider $\bar{u}_{\perp }$ as a three-dimensional variable, which makes $\bar{u}_{\parallel }$ a two-dimensional variable. The independent variables of $\bar{u}_{\parallel }$ are $|\bar{u}_{\parallel }|$ and $\zeta$, where $\zeta$ is the angle between $\bar{u}$ and the plate generated by vector $\eta$. Note that $u_{\perp }$ and $u_{\parallel }$ are orthogonal. Therefore, the Jacobian is calculated as before as follows:

$$\begin{aligned} det \frac{\partial ( \bar{U}_{\perp },| \bar{U}_{\parallel }|,\zeta )}{\partial (\bar{u}_{\perp }, |\bar{u}_{\parallel }|,\zeta )}=\frac{d |\bar{U}_{\parallel }|}{d |\bar{u}_{\parallel }|}=\frac{d}{d |\bar{u}_{\parallel }|} (\frac{1+ ^{\prime } e}{2 ^{\prime } e}|\bar{u}_{\parallel }|)&= \frac{1}{2}\frac{d}{d |\bar{u}_{\parallel }|} (\frac{|\bar{u}_{\parallel }|}{ ^{\prime } e})+\frac{1}{2}. \end{aligned}$$

Thus, we obtain the same result in this case using a method similar to the one used in the previous case.