I came across Villani's paper titled *"Hypocoercive diffusion operators"* and couldn't 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). Set
$$I(t):=\left(a\int |\nabla\_x h|^2 d\mu + c\int |\nabla\_v h|^2 d\mu \right).$$
Then
\begin{align\*}
\frac 12\frac{dI}{dt} &= -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!


Edit: I have also asked this question on Math Stack Exchange (the link is <https://math.stackexchange.com/questions/3782421/modified-energy-method-for-transformed-fokker-planck-equation-tricky-integratio>) and no satisfying answer is given as well.



Modified energy method for transformed Fokker-Planck equation (tricky integration by parts…)
There is a worked-out proof in page 10 and following of [Hérau's lecture notes.](https://hal.archives-ouvertes.fr/hal-01616979/document) The detailed steps are for $n=1$, $V=0$, but I assume once that is understood, the more general case would follow smoothly.


As a short-hand notation we write $ \|\partial\_x h \|^2=\int (\partial h/\partial x)^2\,d\mu$ and $\langle\partial\_x h,\partial\_v h\rangle=\int (\partial h/\partial x)(\partial h/\partial v)\,d\mu$. We will make use of the identities
$$\langle f,\partial\_v g\rangle=\langle(-\partial\_v+v)f,g\rangle,$$
$$\langle v\partial\_x f,f\rangle=0$$
$$[\partial\_v ,v\partial\_x]=\partial\_x,$$
and the [Cauchy-Schwartz](https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality) + [Young inequality](https://en.wikipedia.org/wiki/Young%27s_inequality_for_products#Elementary_case)
$$2|c\langle\partial\_v f,\partial\_x f\rangle|\leq 2c \|\partial\_v f \|\, \|\partial\_x f \|\leq c^2 \|\partial\_v f \|^2+ \|\partial\_x f \|^2.$$
The Fokker-Planck equation for $n=1$, $V=0$ reads
$$\partial\_t h+Lh=0,\;\;L=v\partial\_x + (-\partial\_v+v)\partial\_v .$$
The adjoint of $L$ is
$$L^\ast=-v\partial\_x + (-\partial\_v+v)\partial\_v.$$


The resulting derivatives are
$$-\frac{1}{2}\frac{d}{dt} \|h \|^2= \|\partial\_v h \|^2,$$
$$-\frac{1}{2}\frac{d}{dt} \|\partial\_x h \|^2= \|\partial\_v\partial\_x h \|^2,$$
$$-\frac{1}{2}\frac{d}{dt} \|\partial\_v h \|^2=\langle\partial\_x h,\partial\_v h\rangle+ \|(-\partial\_v +v)\partial\_v h \|^2=\langle\partial\_x h,\partial\_v h\rangle+ \|\partial\_v^2 h \|^2+ \|\partial\_v h \|^2.$$


So the derivative $dI/dt$ in the OP reduces to
$$\frac{dI}{dt}=-2a \|\partial\_v\partial\_x h \|^2-2c\biggl(\langle\partial\_x h,\partial\_v h\rangle+ \|\partial\_v^2 h \|^2+ \|\partial\_v h \|^2\biggr)$$
$$\qquad\leq -2a \|\partial\_v\partial\_x h \|^2-2c \|\partial\_v^2 h \|^2-(2c-c^2) \|\partial\_v h \|^2+ \|\partial\_x h \|^2.$$
It remains to bound $ \|\partial\_x h \|^2$.


In the lecture notes they do this by adding the mixed term $b\langle\partial\_x h,\partial\_v h\rangle$ to the left-hand-side of the inequality, which then gives a term $-b \|\partial\_x h \|^2$ on the right-hand-side to dominate. Let me work that out, using the derivative$^{\ast}$
$$\frac{d}{dt}\langle\partial\_x h,\partial\_v h\rangle=- \|\partial\_x h \|^2-2\langle\partial\_v\partial\_x h,\partial\_v\partial\_v h\rangle-\langle\partial\_x h,\partial\_v h\rangle,$$
hence
$$\frac{d}{dt}J\equiv\frac{d}{dt}\biggl( \|h \|^2+a \|\partial\_x h \|^2+b\langle\partial\_x h,\partial\_v h\rangle+c \|\partial\_v h \|^2\biggr)=$$
$$\qquad=-2(c+1) \|\partial\_v h \|^2-2a \|\partial\_v\partial\_x h \|^2-b \|\partial\_x h \|^2-2c\|\partial\_v\partial\_v h\|^2-2b\langle\partial\_v\partial\_x h,\partial\_v\partial\_v h\rangle-(b+2c)\langle\partial\_x h,\partial\_v h\rangle.$$
The first four terms on the right-hand-side have a fixed sign, the last two terms can be bounded by the Cauchy-Schwartz + Young inequality.




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$^\ast$ The result for $(d/dt)\langle\partial\_x h,\partial\_v h\rangle$ given on page 10 of the cited lecture notes is mistaken. Here is a derivation:
\begin{align}
\frac{d}{dt}\langle\partial\_x h,\partial\_v h\rangle&=-\langle\partial\_x Lh,\partial\_v h\rangle-\langle\partial\_x h,\partial\_v Lh\rangle\\
&=-\langle\partial\_x h,(L^\ast\partial\_v +\partial\_v L)h\rangle,\\
L^\ast\partial\_v +\partial\_v L&=\bigl(-v\partial\_x+(-\partial\_v+v)\partial\_v\bigr)\partial\_v+\partial\_v\bigl(v\partial\_x+(-\partial\_v+v)\partial\_v\bigr)\\
&=\partial\_x+\partial\_v+2(-\partial\_v+v)\partial\_v\partial\_v,\\
\Rightarrow\frac{d}{dt}\langle\partial\_x h,\partial\_v h\rangle&=-\|\partial\_x h\|^2-\langle\partial\_x h,\partial\_v h\rangle-2\langle\partial\_x h,(-\partial\_v+v)\partial\_v\partial\_v h\rangle\\
&=-\|\partial\_x h\|^2-\langle\partial\_x h,\partial\_v h\rangle-2\langle\partial\_v\partial\_x h,\partial\_v\partial\_v h\rangle.
\end{align}


