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I came across Villani's paper titled "Hypocoercive diffusion operators" and could not figure out a computation that is skipped in that paper. Specifically, consider the following transformed Fokker-Planck equation, where $h(t,x,v)$ is the unknown, $(x,v) \in \mathbb{R}^n \times \mathbb{R}^n$, $V(x)$ is some potential force: $$\partial_t h + v\cdot \nabla_x h - \nabla V(x)\cdot \nabla_v h = \Delta_v h - v\cdot \nabla_v h.$$ Notice that the Laplacian $\Delta_v$ is only a partial Laplacian in the sense that it only acts on the velocity variables $v$, and for the usual $L^2$ energy $\int h^2 d\mu$, where $d\mu = f_\infty(x,v) dxdv$ and $f_\infty(x,v) = \frac{\mathrm{e}^{-\left(V(x)+\frac{|v|^2}{2}\right)}}{Z}$ with $Z$ a normalization constant making $f_\infty$ a probability density in $(x,v) \in \mathbb{R}^n \times \mathbb{R}^n$, and we easily have $\frac{1}{2} \frac{d}{dt} \int h^2 d\mu = -\int |\nabla_v h|^2 d\mu$. Then the author says under suitable assumptions on $V$, we can find suitable constants $a,c, K>0$ so that $$\frac{d}{dt}\left(\int h^2 d\mu + a\int |\nabla_x h|^2 d\mu + c\int |\nabla_v h|^2 d\mu \right) \leq -K\left(\int |\nabla_v h|^2 d\mu + \int |\nabla_v\nabla_x h|^2 d\mu + \int |\nabla_v\nabla_v h|^2 d\mu\right). $$ However, I have no clue why the above inequality holds (and justifying it in 1D should be enough for me, i.e., in the case $(x,v) \in \mathbb{R}\times\mathbb{R}$). What I did is the following (in 1D setting): \begin{align*} \frac 12\frac{d}{dt}\left(a\int |\nabla_x h|^2 d\mu + c\int |\nabla_v h|^2 d\mu \right) &= -a\int |\partial_v\partial_x h|^2 d\mu - c\int |\partial_v\partial_v h|^2 d\mu - c\int |\partial_v h|^2 d\mu\\ &\quad \color{red}{+ a\int \partial_x h \partial_x\left(V'(x)\partial_v h\right) - v\partial_xh\partial_{xx}h~d\mu} \\ &\quad \color{red}{+c\int V'(x)\partial_vh\partial_{vv}h - \partial_vh\left(\partial_x h+v\partial_v\partial_xh\right)~d\mu} \end{align*} But I have no clue as to the treatment of the terms in red. Any help would be greatly appreciated!

Fei Cao
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1 Answers1

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A long comment: doesn't the paper give any clue about suitable conditions on $V$? In any case, in the 1D case I'd try repeated integration by parts, for example: $$ \begin{align} \int v\partial_xh\partial_{xx}h\,d\mu &= -\int\partial_xh\partial_{xx}h \partial_v(\frac{e^{-(V+v^2/2)}}{Z})\,dxdv \\ &= \int (\partial_{vx}h\partial_{xx}h + \partial_xh\partial_{vxx}h)\,d\mu \\ &= \int V'(x)\partial_xh\partial_{vx}h\,d\mu. \end{align} $$ You can cancel this against the first term in red, and the first line in red gets into $$ \int\partial_xh V''(x)\partial_v h\,d\mu. $$ We may group this term with $-\int\partial_xh\partial_vh\,d\mu$, which shows up in the second line. I'm not sure I got it right, but I'd play around integrating by parts and using Young inequality to adjust things.

user90189
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  • Thanks for your comment. The paper didn't give any clue about the suitable conditions on V (but I assume these conditions should be mild). I will try to follow your calculation though at this time... – Fei Cao Aug 20 '20 at 18:28
  • I agree with your computations so far, but I not sure how to handle the second line in red... – Fei Cao Aug 20 '20 at 21:14
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    Using a similar computation would give us $$\int V'(x)\partial_vh\partial_{vv}h - \partial_vh\left(\partial_x h+v\partial_v\partial_xh\right)~d\mu = -\int\partial_xh\partial_vh,d\mu,$$ so the term in red in combined reads $$\int \partial_xh\partial_vh\cdot(aV''(x)-c),d\mu.$$ I am guessing we need some Poincare inequality for $\mu$ (in the $x$ and $v$ variables).... – Fei Cao Aug 20 '20 at 21:40