Fractional Hypocoercivity

Abstract

This paper is devoted to kinetic equations without confinement. We investigate the large time behaviour induced by collision operators with fat tailed local equilibria. Such operators have an anomalous diffusion limit. In the appropriate scaling, the macroscopic equation involves a fractional diffusion operator so that the optimal decay rate is determined by a fractional Nash type inequality. At kinetic level we develop an $\mathrm {L}^2$-hypocoercivity approach and establish a rate of decay compatible with the fractional diffusion limit.

Steady States and Force Field for the Fractional Laplacian with Drift

This appendix is devoted to the Case ${\mathsf {L}}={\mathsf {L}}_3$ of the collision operator ${\mathsf {L}}$, that is, to ${\mathsf {L}}_3 f=\Delta _v^{\sigma /2}f+\nabla _v\cdot \left( E\,f\right)$. Our goal here is to prove that the collision frequency $\nu (v)$ behaves like $|v|^{- \beta }$ with $\beta = \sigma -\gamma$ as $|v|\rightarrow +\infty$, as claimed in Sect. 1. By Definition (9) of the force field E, we know that

$$\begin{aligned} \nabla _v\cdot (E\,F)=-\Delta _v^{\sigma /2} F=-\,\nabla _v\cdot \left( \nabla _v(-\Delta _v)^\frac{\sigma -2}{2}\,F\right) , \end{aligned}$$

and this implies that, up to an additive constant,

$$\begin{aligned} E\,F=-\,\nabla _v(-\Delta _v)^\frac{\sigma -2}{2}\,F=-\,\nabla _v\left( \frac{C_{d,\sigma }}{|v|^{d+\sigma -2}}*\frac{c_\gamma }{\left\langle v\right\rangle ^{d+\gamma }}\right) \end{aligned}$$

where $c_\gamma$ and $C_{d,\sigma }$ are given respectively by (2) and (20).

Proposition 7

Assume that $\gamma >0$, $\sigma \in (0,2)$ and let $\beta =\sigma - \gamma$. There is a positive function $G\in \mathrm {L}^\infty ({{\mathbb {R}}}^d)$ with $1/G\in \mathrm {L}^\infty (B_0^c(1))$ such that E is given by

$$\begin{aligned} \forall \,v\in {{\mathbb {R}}}^d,\quad E(v)=G(v) \left\langle v\right\rangle ^{- \beta } v. \end{aligned}$$

Proof

Let $u(v)=-\nabla _v\,\big (\frac{1}{|v|^{d+\sigma -2}}*\frac{1}{\left\langle v\right\rangle ^{d+\gamma }}\big )(v)$ so that $E(v)=C_{d,\sigma } \left\langle v\right\rangle ^{d+\gamma } u(v)$. Since

$$\begin{aligned} u(v)=(d+\gamma )\left( \frac{1}{|v|^{d+\sigma -2}}*\frac{v}{\left\langle v\right\rangle ^{d+\gamma +2}}\right) \end{aligned}$$

where $\left\langle v\right\rangle ^{-(d+\gamma +2)} v\in C^\infty ({{\mathbb {R}}}^d)\cap \mathrm {L}^1(\mathrm dv)$, and $\sigma <2$, one has $u\in C^1_{\mathrm {loc}}({{\mathbb {R}}}^d)$ and $u(0)=0$ which proves the result in $B_1(0)$. We look for an estimate of $u(v)\cdot v$ from above and below on $B_0^c(1)$. Notice that u can also be written as

$$\begin{aligned} u(v)=\left( d+\sigma -2\right) \left( \frac{v}{|v|^{d+\sigma }}*\frac{1}{\left\langle v\right\rangle ^{d+\gamma }}\right) . \end{aligned}$$

(29)

Depending on the integrability at infinity of $v/|v|^{d+\sigma }$, that is, whether $\sigma \in (0,1)$ or not, we have to distinguish two cases.

$\bullet$ Case $\sigma \in (0,1)$. Using (29), we have the estimates

$$\begin{aligned} \left| \int _{|w|\ge \left\langle v\right\rangle /2}\frac{w}{|w|^{d+\sigma }}\frac{\mathrm dw}{\left\langle w-v\right\rangle ^{d+\gamma }}\right|&\le \frac{2^{d+\sigma -1}}{\left\langle v\right\rangle ^{d+\sigma -1}}\int _{{{\mathbb {R}}}^d}\frac{\mathrm dw}{\left\langle w\right\rangle ^{d+\gamma }},\\ \left| \int _{|w|<\left\langle v\right\rangle /2}\frac{w}{|w|^{d+\sigma }}\frac{\mathrm dw}{\left\langle w-v\right\rangle ^{d+\gamma }}\right|&\le \left( \int _{|w|<\left\langle v\right\rangle /2}\frac{\mathrm dw}{|w|^{d+\sigma -1}}\right) \frac{2^{d+\sigma -1}}{|v|^{d+\gamma }}\\&\le \frac{2^{d+\sigma -1}\,\omega _d}{(1-\sigma )\,|v|^{d+\gamma +\sigma -1}}, \end{aligned}$$

and obtain

$$\begin{aligned} \forall \,v\in {{\mathbb {R}}}^d,\quad |u(v)\cdot v|\le |u(v)|\,|v|\lesssim |v|^{-(d+\sigma -2)}. \end{aligned}$$

To get a bound from below on $u(v)\cdot v$, we cut the integral in two pieces and use the fact that $|v|>1$ and $|w-v|<1/2$ implies $w\cdot v>0$. First

$$\begin{aligned}&\int _{\begin{array}{c} |w-v|>1/2\\ |w+v|>1/2 \end{array}}\frac{w\cdot v}{|w|^{d+\sigma }}\frac{\mathrm dw}{\left\langle w-v\right\rangle ^{d+\gamma }}=\left( \int _{\begin{array}{c} |w-v|>1/2\\ w\cdot v>0 \end{array}}+\int _{\begin{array}{c} |w+v|>1/2\\ w\cdot v<0 \end{array}}\right) \frac{w\cdot v}{|w|^{d+\sigma }}\frac{\mathrm dw}{\left\langle w-v\right\rangle ^{d+\gamma }}\\&\quad =\int _{\begin{array}{c} |w-v|>1/2\\ w\cdot v>0 \end{array}}\left( \frac{1}{\left\langle w-v\right\rangle ^{d+\gamma }} -\frac{1}{\left\langle w+v\right\rangle ^{d+\gamma }}\right) \frac{w\cdot v}{|w|^{d+\sigma }}\,\mathrm dw, \end{aligned}$$

which is positive since $\left\langle w+v\right\rangle ^2-\left\langle w-v\right\rangle ^2=2\,w\cdot v\ge 0$. The remaining terms are dealt with as follows

$$\begin{aligned}&\int _{\begin{array}{c} |w-v|\le 1/2\\ \text{ or }\\ |w+v|\le 1/2 \end{array}}\frac{w\cdot v}{|w|^{d+\sigma }}\frac{\mathrm dw}{\left\langle w-v\right\rangle ^{d+\gamma }}\\&\quad =\int _{|w-v|<\frac{1}{2}}\left( \frac{1}{\left\langle w-v\right\rangle ^{d+\gamma }}-\frac{1}{\left\langle w+v\right\rangle ^{d+\gamma }}\right) \frac{w\cdot v}{|w|^{d+\sigma }}\,\mathrm dw\\&\quad \ge \left( (4/5)^{d+\gamma }-(2/5)^{d+\gamma }\right) \int _{|w-v|<\frac{1}{2}}\frac{w\cdot v}{|w|^{d+\sigma }}\,\mathrm dw, \end{aligned}$$

since $|w+v|\ge 2\,|v|-|w-v|\ge \frac{3}{2}$. Finally, if $|v|>1$ and $|w-v|<\frac{1}{2}$, we get

$$\begin{aligned} 2\,w\cdot v&=|v|^2+|w|^2-|w-v|^2\ge |v|^2-\frac{1}{2}\ge \frac{|v|^2}{2},\\ |w|&\le |v|+|w-v|\le 2\,|v|, \end{aligned}$$

so that

$$\begin{aligned} \int _{|w-v|<\frac{1}{2}}\frac{w\cdot v}{|w|^{d+\sigma }}\,\mathrm dw\ge \frac{|B_0(1/2)|}{2^{d+\sigma +2}}\frac{1}{|v|^{d+\sigma -2}}. \end{aligned}$$

This implies $u(v)\cdot v\ge C\,|v|^{-(d+\sigma -2)}$ for some $C>0$. Since u is radial, we proved that

$$\begin{aligned} u(v)=G(v)\,\frac{v}{|v|^{d+\sigma }} \end{aligned}$$

where $G\in \mathrm {L}^\infty ({{\mathbb {R}}}^d)$ and $G^{-1}\in \mathrm {L}^\infty (B_0^c(1))$ and the conclusion holds with $\beta =\sigma - \gamma$.

$\bullet$ Case $\sigma \in [1,2)$. The gradient of $v\mapsto |v|^{2-d-\sigma }$ is a distribution of order 1 that can be defined as a principal value. Indeed, in the sense of distributions, for any $\varphi \in {\mathcal {D}}({{\mathbb {R}}}^d)$, we have

$$\begin{aligned} \left\langle \nabla _v|v|^{2-d-\sigma },\varphi \right\rangle _{{\mathcal {D}}',{\mathcal {D}}}&=-\int _{{{\mathbb {R}}}^d}\frac{\nabla _v\varphi (v)}{|v|^{d+\sigma -2}}\,\mathrm dv=-\int _{{{\mathbb {R}}}^d}\frac{\nabla _v(\varphi (v)-\varphi (0))}{|v|^{d+\sigma -2}}\,\mathrm dv\\&=(d+\sigma -2)\int _{{{\mathbb {R}}}^d}\frac{v}{|v|^{d+\sigma }}\left( \varphi (v)-\varphi (0)\right) \mathrm dv\\&=: (d+\sigma -2) \left\langle {{\,\mathrm{pv}\,}}\!\left( \frac{v}{|v|^{d+\sigma }}\right) \!,\varphi \right\rangle _{{\mathcal {D}}',{\mathcal {D}}}. \end{aligned}$$

Identity (29) is replaced by

$$\begin{aligned} \frac{u(v)}{d+\sigma -2}={{\,\mathrm{pv}\,}}\!\left( \frac{v}{|v|^{d+\sigma }}\right) *\frac{1}{\left\langle v\right\rangle ^{d+\gamma }}=\int _{{{\mathbb {R}}}^d}\frac{w}{|w|^{d+\sigma }}\left( \frac{1}{\left\langle v-w\right\rangle ^{d+\gamma }}-\frac{1}{\left\langle v\right\rangle ^{d+\gamma }}\right) \mathrm dw, \end{aligned}$$

so that, after computations like the ones in the proof of Lemma 10,

$$\begin{aligned} \frac{|u(v)|}{d+\sigma -2}\le \int _{{{\mathbb {R}}}^d}\frac{1}{|w-v|^{d+\sigma -1}}\left| \frac{1}{\left\langle w\right\rangle ^{d+\gamma }}-\frac{1}{\left\langle v\right\rangle ^{d+\gamma }}\right| \mathrm dw\lesssim \frac{1}{\left\langle v\right\rangle ^{d+\sigma -2}}. \end{aligned}$$

Now estimate $u(v)\cdot v$ by

$$\begin{aligned}&\int _{|w|\ge \frac{1}{2}}\frac{w\cdot v}{|w|^{d+\sigma }}\left( \frac{1}{\left\langle v-w\right\rangle ^{d+\gamma }}-\frac{1}{\left\langle v\right\rangle ^{d+\gamma }}\right) \mathrm dw\\&\quad =\int _{|w|\ge \frac{1}{2}}\frac{w\cdot v}{|w|^{d+\sigma }}\frac{1}{\left\langle v-w\right\rangle ^{d+\gamma }}\,\mathrm dw\gtrsim \frac{1}{\left\langle v\right\rangle ^{d+\sigma -2}}. \end{aligned}$$

and

$$\begin{aligned}&\left| \int _{|w|<\frac{1}{2}}\frac{w\cdot v}{|w|^{d+\sigma }}\left( \frac{1}{\left\langle v-w\right\rangle ^{d+\gamma }}-\frac{1}{\left\langle v\right\rangle ^{d+\gamma }}\right) \mathrm dw \right| \\&\quad \le \sup _{w\in B_v(1/2)}\frac{(d+\gamma )\,|v|}{\left\langle w\right\rangle ^{d+\gamma +1}}\int _{|w|<\frac{1}{2}}\frac{\mathrm dw}{|w|^{d+\sigma -2}}\lesssim \frac{1}{\left\langle v\right\rangle ^{d+\gamma }}. \end{aligned}$$

The result follows from the fact that $d+\gamma>d>d+\sigma -2$. $\quad \square$