# Modified energy method for transformed Fokker-Planck equation (tricky integration by parts…)

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.

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

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}

• Thanks! But my problem is to derive inequality (11) in details. (If you read my problem statement carefully...) Mar 7 at 17:13
• Thanks! I will digest them and examine in details of your calculation, I will finish this within today Mar 9 at 17:53
• 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. Mar 9 at 19:42
• @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 $\partial_t h+v\partial_x h+(-\partial_v+v)h=0.$ Mar 9 at 21:06
• (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. Mar 9 at 21:26