Motion of a Rigid Body in a Special Lorentz Gas: Loss of Memory Effect

Kai Koike

doi:10.1007/s10955-018-2072-3

Journal of Statistical Physics

August 2018, Volume 172, Issue 3, pp 795–823 | Cite as

Motion of a Rigid Body in a Special Lorentz Gas: Loss of Memory Effect

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Kai Koike

Article

First Online: 01 June 2018

Received: 12 February 2018

Accepted: 25 May 2018

Abstract

Linear motion of a rigid body in a special kind of Lorentz gas is mathematically analyzed. The rigid body moves against gas drag according to Newton’s equation. The gas model is a special Lorentz gas consisting of gas molecules and background obstacles, which was introduced in Tsuji and Aoki (J Stat Phys 146:620–645, 2012). The specular boundary condition is imposed on the resulting kinetic equation. This study complements the numerical study by Tsuji and Aoki cited above—although the setting in this paper is slightly different from theirs, qualitatively the same asymptotic behavior is proved: The velocity V(t) of the rigid body decays exponentially if the obstacles undergo thermal motion; if the obstacles are motionless, then the velocity V(t) decays algebraically with a rate $t^{-\,5}$ independent of the spatial dimension. This demonstrates the idea that interaction of the molecules with the background obstacles destroys the memory effect due to recollision.

Keywords

Lorentz gas Moving boundary problem Memory effect

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Notes

Acknowledgements

I thank T. Tsuji for his comments on the references.

Appendix: Proof of Theorem 2

The basic strategy is the same as that of Theorem 1: (i) Define an appropriate function space $\mathcal {K}$. (ii) Prove decay estimates of $r_{W}^{\pm }(t)$ given $W\in \mathcal {K}$. (iii) Prove that $V_W$ again belongs to $\mathcal {K}$. (iv) Apply Schauder’s fixed point theorem to show that the map $\mathcal {K}\ni W\mapsto V_W \in \mathcal {K}$ has a fixed point $V\in \mathcal {K}$—this shows the existence part. (v) Prove that any solution (f, V) satisfies ineqs. (90) and that V is decreasing on the interval $[0,\infty )$.

Only steps (i) through (iii) are explained in this section; steps (iv) and (v) can be carried out similarly as in the proof of Theorem 1.

Step (i) is the following:

Definition 2

Let $\gamma >0$. A Lipschitz continuous function $W:[0,\infty )\rightarrow \mathbb {R}$ belongs to $\mathcal {K}=\mathcal {K}(\gamma )$ if $W(0)=\gamma $; W is decreasing on the interval $[0,\infty )$; and satisfies

$$\begin{aligned} \frac{1}{2}\gamma e^{-C_0 t}\le W(t)\le \gamma e^{-C_0 t} \end{aligned}$$

and $|dW(t)/dt|\le 1$.

Let $W\in \mathcal {K}$. Note that $r_{W}^{\pm }(t)\ge 0$ since $W(t)>0$. Moreover, $r_{W}^{+}(t)=0$ since W is decreasing on the interval $[0,\infty )$; therefore, only $r_{W}^{-}(t)$ needs analysis.

Step (ii) is to prove the following:

Proposition 6

Let $\kappa \ge 1$ and $\varepsilon \ge 2\kappa C_0$. If $W\in \mathcal {K}(\gamma )$, then

$$\begin{aligned} 0\le r_{W}^{-}(t)\le C\gamma ^3 \frac{e^{-C_0 t}}{(1+t)^{d+2}} \end{aligned}$$

(91)

for some positive constant C independent of $\gamma $.

Proof

Define $L_t$ by Eq. (46) and let

$$\begin{aligned} L_{t}^{*}=\{ (x,\xi )\in L_t \mid \tau _1 \le t/2 \}, \quad L_{t}^{**}=\{ (x,\xi )\in L_t \mid \tau _1>t/2 \}. \end{aligned}$$

Then $r_{W}^{-}(t)=r_{W,*}^{-}(t)+r_{W,**}^{-}(t)$, where

$$\begin{aligned} r_{W,*}^{-}(t)&=2\int _{L_{t}^{*}}(\xi _1-W(t))^2 (f_0-f_W)\, d\xi dS, \\ r_{W,**}^{-}(t)&=2\int _{L_{t}^{**}}(\xi _1-W(t))^2 (f_0-f_W)\, d\xi dS. \end{aligned}$$

Note first that $(x,\xi )\in L_t$ implies

$$\begin{aligned} 0<\xi _1 -W(t)<\xi _1 \le \frac{1}{t-\tau _1}\int _{\tau _1}^{t}\gamma e^{-C_0 s}\, ds=\gamma e^{-C_0 \tau _1}\frac{1-e^{-C_0(t-\tau _1)}}{C_0(t-\tau _1)}. \end{aligned}$$

Similar arguments as in the proof of Proposition 2 show

$$\begin{aligned} r_{W,*}^{-}(t)&\le C\gamma ^3 w_{\varepsilon ,\kappa ,d}(t)e^{-\frac{\varepsilon }{2\kappa }t}, \\ r_{W,**}^{-}(t)&\le C\gamma ^3 e^{-\frac{3}{2}C_0 t}. \end{aligned}$$

Since $\varepsilon /2\kappa \ge C_0$, these imply ineq. (91). $\square $

This is step (ii). Step (iii) is to prove the following:

Proposition 7

Let $\kappa \ge 1$ and $\varepsilon \ge 2\kappa C_0$. Then $W\in \mathcal {K}=\mathcal {K}(\gamma )$ implies $V_W \in \mathcal {K}$ for sufficiently small $\gamma $.

Proof

Since $r_{W}^{\pm }(t)\ge 0$, the upper bound

$$\begin{aligned} V_W(t)\le \gamma e^{-C_0 t} \end{aligned}$$

(92)

follows from Eq. (18).

To prove the lower bound

$$\begin{aligned} V_W(t)\ge \frac{1}{2}\gamma e^{-C_0 t}, \end{aligned}$$

(93)

I need first to show

$$\begin{aligned} e^{-\int _{0}^{t}K(W(s))\, ds}\ge \frac{3}{4}e^{-C_0 t} \end{aligned}$$

(94)

for sufficiently small $\gamma $: Note first that $D_{0}''(0)=0$ and $D_{0}'''(0)>0$ by Eq. (11); hence for $\gamma \le 1$,

$$\begin{aligned} 0<K(W(s))&=D_0(W(s))/W(s) \\&\le D_{0}'(W(s)) \\&\le C_0+C(W(s))^2 \\&\le C_0+C\gamma ^2 e^{-2C_0 s} \end{aligned}$$

for some positive constant C. This implies

$$\begin{aligned} e^{-\int _{0}^{t}K(W(s))\, ds}\ge e^{-\int _{0}^{t}(C_0+C\gamma ^2 e^{-2C_0 s})\, ds}=e^{-C\gamma ^2}e^{-C_0 t}. \end{aligned}$$

And taking $\gamma $ sufficiently small leads to ineq. (94).

Now ineq. (93) follows from eq. (18), Proposition 6 and ineq.(94) as follows:

$$\begin{aligned} V_W(t)&\ge \frac{3}{4}\gamma e^{-C_0 t}-C\gamma ^3 \int _{0}^{t}e^{-C_0(t-s)}\frac{e^{-C_0 s}}{(1+s)^{d+2}}\, ds \\&=\frac{3}{4}\gamma e^{-C_0 t}-C\gamma ^3 e^{-C_0 t}\int _{0}^{t} \frac{ds}{(1+s)^{d+2}} \\&\ge \frac{1}{2}\gamma e^{-C_0 t} \end{aligned}$$

if $\gamma $ is sufficiently small.

Next, I show that $V_W$ is decreasing on the interval $[0,\infty )$: Since $V_W>0$ (by ineq. (93)) and $r_{W}^{\pm }(t)\ge 0$,

$$\begin{aligned} \frac{dV_W(t)}{dt}=-K(W(t))V_W(t)-r_{W}^{+}(t)-r_{W}^{-}(t)<0 \end{aligned}$$

for $t\ge 0$.

Lastly, since $r_{W}^{+}(t)=0$, $|r_{W}^{-}(t)|\le C\gamma ^3$ (by Proposition 6) and $|V_W(t)|\le \gamma $ (by ineqs. (92) and (93)),

$$\begin{aligned} \left| \frac{dV_W(t)}{dt} \right| \le K(W(t))|V_W(t)|+|r_{W}^{+}(t)|+|r_{W}^{-}(t)|\le C_{\gamma }\gamma +C\gamma ^3 \le 1 \end{aligned}$$

for sufficiently small $\gamma $. $\square $

Steps (iv) and (v) can be carried out similarly to the proof of Theorem 1. This proves Theorem 2.

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Kai Koike
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1.School of Fundamental Science and Technology, Graduate School of Science and Technology Keio UniversityKohokuku, YokohamaJapan
2.RIKEN Center for Advanced Intelligence ProjectTokyoJapan

Motion of a Rigid Body in a Special Lorentz Gas: Loss of Memory Effect

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Acknowledgements

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