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.

  • $\begingroup$ Multiply through by $h$, integrate, and use integration by parts. $\endgroup$ Mar 7, 2021 at 1:36
  • $\begingroup$ @Aruralreader Can you elaborate more? For instance, multiply "who" through $h$? $\endgroup$
    – Fei Cao
    Mar 7, 2021 at 1:42

1 Answer 1


There is a worked-out proof in page 10 and following of Hérau's lecture notes. 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 + Young inequality $$2|c\langle\partial_v f,\partial_x f\rangle|\leq 2c \|\partial_v f \|\, \|\partial_x f \|\leq c^2 \|\partial_v f \|^2+ \|\partial_x f \|^2.$$ 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.

$^\ast$ 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}

  • $\begingroup$ Thanks! But my problem is to derive inequality (11) in details. (If you read my problem statement carefully...) $\endgroup$
    – Fei Cao
    Mar 7, 2021 at 17:13
  • $\begingroup$ Thanks! I will digest them and examine in details of your calculation, I will finish this within today $\endgroup$
    – Fei Cao
    Mar 9, 2021 at 17:53
  • 1
    $\begingroup$ The only difference introducing the $V$ term is that $dI / dt$ includes additionally the term $+2a \int \partial_x h \partial_v h V'' ~d\mu$; if you have good control on the Hessian of $V$ then this term can be treated in the same way as the $-2c \langle \partial_x h, \partial_v h\rangle$ term. $\endgroup$ Mar 9, 2021 at 19:42
  • $\begingroup$ @CarloBeenakker Actually I have several questions. (1) I do not see why $\langle v\partial_x f,f\rangle=0$ (2) the meaning of $[\partial_v v,\partial_x]=1$ is not clear (3) for $ n = 1$ and $V = 0$ the PDE reads as $\partial_t h+v\partial_x h + v\partial_v h - \Delta_v h=0$, not $\partial_t h+v\partial_x h+(-\partial_v+v)h=0.$ $\endgroup$
    – Fei Cao
    Mar 9, 2021 at 21:06
  • $\begingroup$ (1) $\langle v\partial_xf,f\rangle=-\langle f,v\partial_x f\rangle=\langle f,v\partial_x f\rangle\Rightarrow \langle v\partial_x f,f\rangle=0$; (2) $[\partial_v v,\partial_x]\equiv \partial_v v\partial_x-\partial_x\partial_v v =\partial_x$; (the "1" was a typo); (3) that was also a typo; there may be more typo's, it's quite a lengthy calculation. $\endgroup$ Mar 9, 2021 at 21:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.