Modified Energy Method for Transformed Fokker-Planck Equation (Tricky Integration by parts...)

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!

• also asked at mathoverflow.net/q/368619/11260 Mar 9 '21 at 20:48
• @CarloBeenakker I have not digested the answers there.... Thanks! Mar 9 '21 at 20:58

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.
• 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).... Aug 20 '20 at 21:40