Suppression of Plasma Echoes and Landau Damping in Sobolev Spaces by Weak Collisions in a Vlasov-Fokker-Planck Equation

Abstract

In this paper, we study Landau damping in the weakly collisional limit of a Vlasov-Fokker-Planck equation with nonlinear collisions in the phase-space $(x,v) \in {{\mathbb {T}}}_x^n \times {\mathbb {R}}^n_v$. The goal is four-fold: (A) to understand how collisions suppress plasma echoes and enable Landau damping in agreement with linearized theory in Sobolev spaces, (B) to understand how phase mixing accelerates collisional relaxation, (C) to understand better how the plasma returns to global equilibrium during Landau damping, and (D) to rule out that collision-driven nonlinear instabilities dominate. We give an estimate for the scaling law between Knudsen number and the maximal size of the perturbation necessary for linear theory to be accurate in Sobolev regularity. We conjecture this scaling to be sharp (up to logarithmic corrections) due to potential nonlinear echoes in the collisionless model.

A Properties of the Dissipation Multiplier

The properties of the multiplier M are summarized in the next lemma. Properties (a) and (b) are essentially trivial, whereas properties (c) and (d) are not.

Lemma A.1

(Properties of M).    (a) There is a universal constant $c_m$ (in particular, uniform in $t,k,\eta ,$ and $\nu$) such that $c_m < M \le 1$.

(b) For $k \ne 0$ there holds,

$$\begin{aligned} \nu ^{1/3} \lesssim \partial _t M(t,k,\eta ) + \nu e^{2\nu t} \left| \eta - k\frac{1-e^{-\nu t}}{\nu }\right| ^2. \end{aligned}$$

(A.1)

(c) There holds (uniformly in $k,\ell ,\eta ,\nu$ and t)

$$\begin{aligned} \left| 1 - \frac{M(t,\ell ,\xi }{M(t,k,\eta )}\right|&\lesssim \left\langle k-\ell ,\eta -\xi \right\rangle ^3\frac{\left\langle t \right\rangle ^2}{\nu ^{1/3} \max \left( \left\langle \eta \right\rangle ,\left\langle k \right\rangle \right) }. \end{aligned}$$

(A.2)

(d) There holds for $\alpha \in {\mathbb {N}}^n$ such that $\left| \alpha \right| \ge 1$,

$$\begin{aligned} \left| D_\eta ^\alpha M\right|&\lesssim e^{(\left| \alpha \right| -1)\nu t} \nu ^{\left| \alpha \right| /3}. \end{aligned}$$

(A.3)

Remark 16

That (A.2) holds even when $\ell$ or k is zero is crucial to the proof.

Proof

Property (a) follows by integration and property (b) follows essentially by definition. Turn next to (A.2).

The case $k\ne 0$ and $\ell \ne 0$ :

First, observe that by Part (a),

$$\begin{aligned}&\left| 1 - \frac{M(t,\ell ,\xi )}{M(t,k,\eta )}\right| \nonumber \\&\quad \lesssim \nu ^{1/3} \int _0^t\left| \frac{1}{\left( 1 + \nu ^{2/3} e^{2\nu \tau }\left| \eta - k\frac{1-e^{-\nu \tau }}{\nu }\right| ^2 \right) } - \frac{1}{\left( 1 + \nu ^{2/3} e^{2\nu \tau }\left| \xi - \ell \frac{1-e^{-\nu \tau }}{\nu }\right| ^2 \right) } \right| {\, \mathrm d}\tau \nonumber \\&\quad = \nu ^{1/3}\int _0^t\left| \frac{\nu ^{2/3} e^{2\nu \tau }\left| \xi - \ell \frac{1-e^{-\nu \tau }}{\nu }\right| ^2 - \nu ^{2/3} e^{2\nu \tau }\left| \eta - k\frac{1-e^{-\nu \tau }}{\nu }\right| ^2}{\left( 1 + \nu ^{2/3} e^{2\nu \tau }\left| \eta -k \frac{1-e^{-\nu \tau }}{\nu }\right| ^2 \right) \left( 1 + \nu ^{2/3} e^{2\nu \tau }\left| \xi - \ell \frac{1-e^{-\nu \tau }}{\nu }\right| ^2 \right) } \right| {\, \mathrm d}\tau \nonumber \\&\quad \lesssim \nu ^{1/3} \int _0^t\frac{\left( \nu ^{1/3} e^{\nu \tau }\left| \eta - k\frac{1-e^{-\nu \tau }}{\nu }\right| + \nu ^{1/3} e^{\nu \tau }\left| \xi - \ell \frac{1-e^{-\nu \tau }}{\nu }\right| \right) }{\left( \left| k\right| ^2 + \nu ^{2/3} e^{2\nu \tau }\left| \eta - \frac{1-e^{-\nu \tau }}{\nu }\right| ^2 \right) \left( \left| \ell \right| ^2 + \nu ^{2/3} e^{2\nu \tau }\left| \xi - \frac{1-e^{-\nu \tau }}{\nu }\right| ^2 \right) } \nonumber \\&\qquad \times \left( \nu ^{1/3} \frac{e^{\nu \tau }-1}{\nu }\left| k-\ell \right| + \nu ^{1/3}e^{\nu \tau }\left| \eta -\xi \right| \right) {\, \mathrm d}\tau \nonumber \\&\quad \lesssim \int _0^t\frac{\left\langle k-\ell , \eta -\xi \right\rangle \nu ^{2/3}\left( e^{\nu \tau } + \left\langle \frac{e^{\nu \tau }-1}{\nu } \right\rangle \right) }{\left( 1 + \nu ^{2/3} e^{2\nu \tau }\left| \eta - k\frac{1-e^{-\nu \tau }}{\nu }\right| ^2 \right) ^{1/2}\left( 1 + \nu ^{2/3} e^{2\nu \tau }\left| \xi - \ell \frac{1-e^{-\nu \tau }}{\nu }\right| ^2 \right) } {\, \mathrm d}\tau \nonumber \\&\qquad + \int _0^t \frac{\left\langle k-\ell , \eta -\xi \right\rangle \nu ^{2/3} \left( e^{\nu \tau } + \left\langle \frac{e^{\nu \tau }-1}{\nu } \right\rangle \right) }{\left( 1 + \nu ^{2/3} e^{2\nu \tau }\left| \eta - k\frac{1-e^{-\nu \tau }}{\nu }\right| ^2 \right) \left( 1 + \nu ^{2/3} e^{2\nu \tau }\left| \xi - \ell \frac{1-e^{-\nu \tau }}{\nu }\right| ^2 \right) ^{1/2}} {\, \mathrm d}\tau . \end{aligned}$$

(A.4)

To extract a gain in k and $\ell$, simply make the change of variables $s = \left| \ell \right| \nu ^{1/3}\frac{e^{\nu \tau }-1}{\nu }$ in the former integral and in the latter use $s = \left| k\right| \nu ^{1/3}\frac{e^{\nu \tau }-1}{\nu }$, and we have

$$\begin{aligned} \left| 1 - \frac{M(t,\ell ,\xi )}{M(t,k,\eta )}\right|&\lesssim \frac{\left\langle k-\ell ,\eta -\xi \right\rangle ^2 \left\langle t \right\rangle }{\max (\left\langle k \right\rangle ,\left\langle \ell \right\rangle )}, \end{aligned}$$

which is sufficient. To extract a gain in $\eta$ and $\xi$ we multiply and divide (A.4) by $\left\langle k(1-e^{-\nu \tau })\nu ^{-1} \right\rangle$ or $\left\langle \ell (1-e^{-\nu \tau })\nu ^{-1} \right\rangle$:

$$\begin{aligned} \left| 1 - \frac{M(t,\ell ,\xi )}{M(t,k,\eta )}\right|&\lesssim \int _0^t\frac{\left\langle k-\ell , \eta -\xi \right\rangle \nu ^{2/3} \left( e^{\nu \tau } + \left\langle \frac{e^{\nu \tau }-1}{\nu } \right\rangle \right) \left\langle k\frac{1-e^{-\nu \tau }}{\nu } \right\rangle }{\nu ^{1/3} e^{\nu \tau } \left\langle \eta \right\rangle \left( 1 + \nu ^{2/3} e^{2\nu \tau }\left| \xi - \ell \frac{1-e^{-\nu \tau }}{\nu }\right| ^2 \right) } {\, \mathrm d}\tau \\&\quad + \int _0^t \frac{\left\langle k-\ell , \eta -\xi \right\rangle \nu ^{2/3} \left( e^{\nu \tau } + \left\langle \frac{e^{\nu \tau }-1}{\nu } \right\rangle \right) \left\langle \ell \frac{1-e^{-\nu \tau }}{\nu } \right\rangle }{\left( 1 + \nu ^{2/3} e^{2\nu \tau }\left| \eta - k\frac{1-e^{-\nu \tau }}{\nu }\right| ^2 \right) \nu ^{1/3} e^{\nu \tau } \left\langle \xi \right\rangle } {\, \mathrm d}\tau \end{aligned}$$

Estimate $\frac{1-e^{-\nu \tau }}{\nu } \le t$ on the support of the integrand and extract powers of k and $\ell$,

$$\begin{aligned} \left| 1 - \frac{M(t,\ell ,\xi )}{M(t,k,\eta )}\right|&\lesssim \frac{\left\langle k-\ell , \eta -\xi \right\rangle \left\langle t \right\rangle ^2}{\left\langle \eta \right\rangle } \int _0^t\frac{\nu ^{1/3}\left\langle k \right\rangle }{\left( 1 + \nu ^{2/3} e^{2\nu \tau }\left| \xi - \ell \frac{1-e^{-\nu \tau }}{\nu }\right| ^2 \right) } {\, \mathrm d}\tau \\&\quad + \frac{\left\langle k-\ell , \eta -\xi \right\rangle \left\langle t \right\rangle ^2}{\left\langle \xi \right\rangle } \int _0^t\frac{\nu ^{1/3}\left\langle \ell \right\rangle }{\left( 1 + \nu ^{2/3} e^{2\nu \tau }\left| \eta - k\frac{1-e^{-\nu \tau }}{\nu }\right| ^2 \right) } {\, \mathrm d}\tau . \end{aligned}$$

In this first integral, make the change of variables $s = \left| \ell \right| \nu ^{1/3}\frac{e^{\nu \tau }-1}{\nu }$ and in the latter use $s = \left| k\right| \nu ^{1/3}\frac{e^{\nu \tau }-1}{\nu }$, and therefore,

$$\begin{aligned} \left| 1 - \frac{M(t,\ell ,\xi )}{M(t,k,\eta )}\right|&\lesssim \frac{\left\langle k-\ell , \eta -\xi \right\rangle ^3 \left\langle t \right\rangle ^2}{\max (\left\langle \xi \right\rangle ,\left\langle \eta \right\rangle )} \end{aligned}$$

which is more than sufficient.

The case $k = 0$ or $\ell = 0$ :

By definition of M, (A.2) is trivial if $k = \ell = 0$. The two remaining cases are essentially equivalent, hence without loss of generality, assume that $k=0$ and $\ell \ne 0$. By part(a) and $\left| e^x - 1\right| \le x e^x$,

$$\begin{aligned} \left| 1 - M(t,\ell ,\xi )\right| \lesssim \int _0^t \frac{\nu ^{1/3}}{1 + \nu ^{2/3} e^{2\nu \tau }\left| \xi -\ell \frac{1-e^{-\nu \tau }}{\nu }\right| ^2} {\, \mathrm d}\tau . \end{aligned}$$

(A.5)

First, if $\left| \xi \right| > 2\left| \ell t\right|$, then by $(1-e^{-\nu \tau })/\nu \le \tau \le t$,

$$\begin{aligned} \left| 1 - M(t,\ell ,\xi )\right| \lesssim \int _0^t \frac{\nu ^{1/3}}{1 + \nu ^{2/3} e^{2\nu \tau }\left| \xi \right| ^2} {\, \mathrm d}\tau \lesssim \frac{\left\langle t \right\rangle }{\nu ^{1/3}\left\langle \xi \right\rangle ^2} \lesssim \frac{\left\langle t \right\rangle ^2}{\nu ^{1/3}\left\langle \xi \right\rangle \left\langle \ell \right\rangle }, \end{aligned}$$

which suffices. If $\left| \xi \right| \le 2\left| \ell t\right|$, then apply $s = \left| \ell \right| \nu ^{1/3} \frac{e^{\nu \tau }-1}{\nu }$ in (A.5) to deduce

$$\begin{aligned} \left| 1 - M(t,\ell ,\xi )\right| \lesssim \frac{1}{\left| \ell \right| } \lesssim \frac{\left\langle t \right\rangle }{\max (\left| \ell \right| ,\left\langle \xi \right\rangle )}. \end{aligned}$$

This covers all cases and completes the proof of (A.2).

Consider next the differentiation of M with respect to $\eta$. Compute a single derivative first:

$$\begin{aligned} \left| \nabla _\eta M(t,k,\eta )\right|&= \left| \int _0^t\frac{2 \nu e^{2\nu \tau }\left( \eta - k \frac{1-e^{-\nu \tau }}{\nu }\right) }{\left( 1 + \nu ^{2/3} e^{2\nu \tau }\left| \eta - k \frac{1-e^{-\nu \tau }}{\nu }\right| ^2\right) ^2} {\, \mathrm d}\tau M(t,k,\eta )\right| \\&\lesssim \int _0^t\frac{\nu ^{2/3} e^{\nu \tau } }{\left( 1 + \nu ^{2/3} e^{2\nu \tau }\left| \eta - k\frac{1-e^{-\nu \tau }}{\nu }\right| ^2\right) ^{3/2}} {\, \mathrm d}\tau . \end{aligned}$$

Via the change of variables $s = \left| k\right| \nu ^{1/3}e^{\nu \tau } \frac{1-e^{-\nu \tau }}{\nu }$, there holds,

$$\begin{aligned} \left| \nabla _\eta M\right|&\lesssim \nu ^{1/3}. \end{aligned}$$

Iterating the above argument gives for all $\left| \alpha \right| \ge 1$,

$$\begin{aligned} \left| D_\eta ^\alpha M\right|&\lesssim _{\left| \alpha \right| } e^{(\left| \alpha \right| -1)\nu t} \nu ^{\left| \alpha \right| /3}. \end{aligned}$$

$\square$