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

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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:_{$^\ast$ 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}_{\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}}

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}

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}}

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edited Mar 10, 2021 at 20:52

Carlo Beenakker

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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 $$\partial_t h+v\partial_x h + v\partial_v h - \partial_v\partial_v h=0.$$$$\partial_t h+Lh=0,\;\;L=v\partial_x + (-\partial_v+v)\partial_v .$$ This implies theThe 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$^{\ast}$ $$\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle=- \|\partial_x h \|^2+2\langle(-\partial_v+v)\partial_v h,\partial_v\partial_x h\rangle-\langle\partial_x h,\partial_v h\rangle,$$$$\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle=- \|\partial_x h \|^2-2\langle\partial_v\partial_x h,\partial_v\partial_v h\rangle-\langle\partial_x h,\partial_v h\rangle,$$ hence $$\frac{d}{dt}J\equiv\frac{d}{dt}\biggl( \|h \|^2+a \|\partial_x h \|^2+b\langle\partial_x h,\partial_v h\rangle+c \|\partial_v h \|^2\biggr)=$$ $$\qquad=-2 \|\partial_v h \|^2-2a \|\partial_v\partial_x h \|^2-b \|\partial_x h \|^2-2c \|(-\partial_v +v)\partial_v h \|^2+2b\langle(-\partial_v+v)\partial_v h,\partial_v\partial_x h\rangle-(b+2c)\langle\partial_x h,\partial_v h\rangle.$$$$\qquad=-2(c+1) \|\partial_v h \|^2-2a \|\partial_v\partial_x h \|^2-b \|\partial_x h \|^2-2c\|\partial_v\partial_v h\|^2-2b\langle\partial_v\partial_x h,\partial_v\partial_v h\rangle-(b+2c)\langle\partial_x h,\partial_v h\rangle.$$ The first four terms on the right-hand-side have a fixed sign, the last two terms arecan be bounded by the Cauchy-Schwartz + Young inequality, $$\frac{d}{dt}J\leq -(2a-1) \|\partial_v\partial_x h \|^2-(b-1/2) \|\partial_x h \|^2-(2c-b^2) \|(-\partial_v +v)\partial_v h \|^2-(1-c-b/2) \|\partial_v h \|^2.$$.

For $b=1/2$ we thus find $$\frac{d}{dt}J\leq-(2a-1) \|\partial_v\partial_x h \|^2-(2c-1/4) \|\partial_v^2 h \|^2-(1/2+c) \|\partial_v h \|^2$$ $$\qquad \leq-K\biggl( \|\partial_v\partial_x h \|^2+ \|\partial_v^2 h \|^2+ \|\partial_v h \|^2\biggr)$$ for some $K$ if $a>1/2$,$^\ast$ The result for $c>1/8$$(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}

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 $$\partial_t h+v\partial_x h + v\partial_v h - \partial_v\partial_v h=0.$$ This implies the derivatives $$-\frac{1}{2}\frac{d}{dt} \|h \|^2= \|\partial_v h \|^2,$$ $$-\frac{1}{2}\frac{d}{dt} \|\partial_x h \|^2= \|\partial_v\partial_x h \|^2,$$ $$-\frac{1}{2}\frac{d}{dt} \|\partial_v h \|^2=\langle\partial_x h,\partial_v h\rangle+ \|(-\partial_v +v)\partial_v h \|^2=\langle\partial_x h,\partial_v h\rangle+ \|\partial_v^2 h \|^2+ \|\partial_v h \|^2.$$

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 $$\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle=- \|\partial_x h \|^2+2\langle(-\partial_v+v)\partial_v h,\partial_v\partial_x h\rangle-\langle\partial_x h,\partial_v h\rangle,$$ hence $$\frac{d}{dt}J\equiv\frac{d}{dt}\biggl( \|h \|^2+a \|\partial_x h \|^2+b\langle\partial_x h,\partial_v h\rangle+c \|\partial_v h \|^2\biggr)=$$ $$\qquad=-2 \|\partial_v h \|^2-2a \|\partial_v\partial_x h \|^2-b \|\partial_x h \|^2-2c \|(-\partial_v +v)\partial_v h \|^2+2b\langle(-\partial_v+v)\partial_v h,\partial_v\partial_x h\rangle-(b+2c)\langle\partial_x h,\partial_v h\rangle.$$ The first four terms on the right-hand-side have a fixed sign, the last two terms are bounded by the Cauchy-Schwartz + Young inequality, $$\frac{d}{dt}J\leq -(2a-1) \|\partial_v\partial_x h \|^2-(b-1/2) \|\partial_x h \|^2-(2c-b^2) \|(-\partial_v +v)\partial_v h \|^2-(1-c-b/2) \|\partial_v h \|^2.$$

For $b=1/2$ we thus find $$\frac{d}{dt}J\leq-(2a-1) \|\partial_v\partial_x h \|^2-(2c-1/4) \|\partial_v^2 h \|^2-(1/2+c) \|\partial_v h \|^2$$ $$\qquad \leq-K\biggl( \|\partial_v\partial_x h \|^2+ \|\partial_v^2 h \|^2+ \|\partial_v h \|^2\biggr)$$ for some $K$ if $a>1/2$, $c>1/8$.

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}

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edited Mar 10, 2021 at 7:14

Carlo Beenakker

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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$$ \|\partial_x h \|^2=\int (\partial h/\partial x)^2\,d\mu$ and $\langle\partial_x h,\partial_v h\rangle=\int (\partial h/\partial x)(\partial h/\partial v)\,d\mu$. We will make use of the identities $$\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.$$$$2|c\langle\partial_v f,\partial_x f\rangle|\leq 2c \|\partial_v f \|\, \|\partial_x f \|\leq c^2 \|\partial_v f \|^2+ \|\partial_x f \|^2.$$ The Fokker-Planck equation for $n=1$, $V=0$ reads $$\partial_t h+v\partial_x h + v\partial_v h - \partial_v\partial_v h=0.$$ This implies the derivatives $$-\frac{1}{2}\frac{d}{dt}||h||^2=||\partial_v h||^2,$$$$-\frac{1}{2}\frac{d}{dt} \|h \|^2= \|\partial_v h \|^2,$$ $$-\frac{1}{2}\frac{d}{dt}||\partial_x h||^2=||\partial_v\partial_x h||^2,$$$$-\frac{1}{2}\frac{d}{dt} \|\partial_x h \|^2= \|\partial_v\partial_x h \|^2,$$ $$-\frac{1}{2}\frac{d}{dt}||\partial_v h||^2=\langle\partial_x h,\partial_v h\rangle+||(-\partial_v +v)\partial_v h||^2=\langle\partial_x h,\partial_v h\rangle+||\partial_v^2 h||^2+||\partial_v h||^2.$$$$-\frac{1}{2}\frac{d}{dt} \|\partial_v h \|^2=\langle\partial_x h,\partial_v h\rangle+ \|(-\partial_v +v)\partial_v h \|^2=\langle\partial_x h,\partial_v h\rangle+ \|\partial_v^2 h \|^2+ \|\partial_v h \|^2.$$

So the derivative $dI/dt$ in the OP reduces to $$\frac{dI}{dt}=-2a||\partial_v\partial_x h||^2-2c\biggl(\langle\partial_x h,\partial_v h\rangle+||\partial_v^2 h||^2+||\partial_v h||^2\biggr)$$$$\frac{dI}{dt}=-2a \|\partial_v\partial_x h \|^2-2c\biggl(\langle\partial_x h,\partial_v h\rangle+ \|\partial_v^2 h \|^2+ \|\partial_v h \|^2\biggr)$$ $$\qquad\leq -2a||\partial_v\partial_x h||^2-2c||\partial_v^2 h||^2-(2c-c^2)||\partial_v h||^2+||\partial_x h||^2.$$$$\qquad\leq -2a \|\partial_v\partial_x h \|^2-2c \|\partial_v^2 h \|^2-(2c-c^2) \|\partial_v h \|^2+ \|\partial_x h \|^2.$$ It remains to bound $||\partial_x h||^2$$ \|\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$$-b \|\partial_x h \|^2$ on the right-hand-side to dominate. Let me work that out, using the derivative $$\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle=-||\partial_x h||^2+2\langle(-\partial_v+v)\partial_v h,\partial_v\partial_x h\rangle-\langle\partial_x h,\partial_v h\rangle,$$$$\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle=- \|\partial_x h \|^2+2\langle(-\partial_v+v)\partial_v h,\partial_v\partial_x h\rangle-\langle\partial_x h,\partial_v h\rangle,$$ hence $$\frac{d}{dt}J\equiv\frac{d}{dt}\biggl(||h||^2+a||\partial_x h||^2+b\langle\partial_x h,\partial_v h\rangle+c||\partial_v h||^2\biggr)=$$$$\frac{d}{dt}J\equiv\frac{d}{dt}\biggl( \|h \|^2+a \|\partial_x h \|^2+b\langle\partial_x h,\partial_v h\rangle+c \|\partial_v h \|^2\biggr)=$$ $$\qquad=-2||\partial_v h||^2-2a||\partial_v\partial_x h||^2-b||\partial_x h||^2-2c||(-\partial_v +v)\partial_v h||^2+2b\langle(-\partial_v+v)\partial_v h,\partial_v\partial_x h\rangle-(b+2c)\langle\partial_x h,\partial_v h\rangle.$$$$\qquad=-2 \|\partial_v h \|^2-2a \|\partial_v\partial_x h \|^2-b \|\partial_x h \|^2-2c \|(-\partial_v +v)\partial_v h \|^2+2b\langle(-\partial_v+v)\partial_v h,\partial_v\partial_x h\rangle-(b+2c)\langle\partial_x h,\partial_v h\rangle.$$ The first four terms on the right-hand-side have a fixed sign, the last two terms are bounded by the Cauchy-Schwartz + Young inequality, $$\frac{d}{dt}J\leq -(2a-1)||\partial_v\partial_x h||^2-(b-1/2)||\partial_x h||^2-(2c-b^2)||(-\partial_v +v)\partial_v h||^2-(1-c-b/2)||\partial_v h||^2.$$$$\frac{d}{dt}J\leq -(2a-1) \|\partial_v\partial_x h \|^2-(b-1/2) \|\partial_x h \|^2-(2c-b^2) \|(-\partial_v +v)\partial_v h \|^2-(1-c-b/2) \|\partial_v h \|^2.$$

For $b=1/2$ we thus find $$\frac{d}{dt}J\leq-(2a-1)||\partial_v\partial_x h||^2-(2c-1/4)||\partial_v^2 h||^2-(1/2+c)||\partial_v h||^2$$$$\frac{d}{dt}J\leq-(2a-1) \|\partial_v\partial_x h \|^2-(2c-1/4) \|\partial_v^2 h \|^2-(1/2+c) \|\partial_v h \|^2$$ $$\qquad \leq-K\biggl(||\partial_v\partial_x h||^2+||\partial_v^2 h||^2+||\partial_v h||^2\biggr)$$$$\qquad \leq-K\biggl( \|\partial_v\partial_x h \|^2+ \|\partial_v^2 h \|^2+ \|\partial_v h \|^2\biggr)$$ for some $K$ if $a>1/2$, $c>1/8$.

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 $$\partial_t h+v\partial_x h + v\partial_v h - \partial_v\partial_v h=0.$$ This implies the derivatives $$-\frac{1}{2}\frac{d}{dt}||h||^2=||\partial_v h||^2,$$ $$-\frac{1}{2}\frac{d}{dt}||\partial_x h||^2=||\partial_v\partial_x h||^2,$$ $$-\frac{1}{2}\frac{d}{dt}||\partial_v h||^2=\langle\partial_x h,\partial_v h\rangle+||(-\partial_v +v)\partial_v h||^2=\langle\partial_x h,\partial_v h\rangle+||\partial_v^2 h||^2+||\partial_v h||^2.$$

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 $$\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle=-||\partial_x h||^2+2\langle(-\partial_v+v)\partial_v h,\partial_v\partial_x h\rangle-\langle\partial_x h,\partial_v h\rangle,$$ hence $$\frac{d}{dt}J\equiv\frac{d}{dt}\biggl(||h||^2+a||\partial_x h||^2+b\langle\partial_x h,\partial_v h\rangle+c||\partial_v h||^2\biggr)=$$ $$\qquad=-2||\partial_v h||^2-2a||\partial_v\partial_x h||^2-b||\partial_x h||^2-2c||(-\partial_v +v)\partial_v h||^2+2b\langle(-\partial_v+v)\partial_v h,\partial_v\partial_x h\rangle-(b+2c)\langle\partial_x h,\partial_v h\rangle.$$ The first four terms on the right-hand-side have a fixed sign, the last two terms are bounded by the Cauchy-Schwartz + Young inequality, $$\frac{d}{dt}J\leq -(2a-1)||\partial_v\partial_x h||^2-(b-1/2)||\partial_x h||^2-(2c-b^2)||(-\partial_v +v)\partial_v h||^2-(1-c-b/2)||\partial_v h||^2.$$

For $b=1/2$ we thus find $$\frac{d}{dt}J\leq-(2a-1)||\partial_v\partial_x h||^2-(2c-1/4)||\partial_v^2 h||^2-(1/2+c)||\partial_v h||^2$$ $$\qquad \leq-K\biggl(||\partial_v\partial_x h||^2+||\partial_v^2 h||^2+||\partial_v h||^2\biggr)$$ for some $K$ if $a>1/2$, $c>1/8$.

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 $$\partial_t h+v\partial_x h + v\partial_v h - \partial_v\partial_v h=0.$$ This implies the derivatives $$-\frac{1}{2}\frac{d}{dt} \|h \|^2= \|\partial_v h \|^2,$$ $$-\frac{1}{2}\frac{d}{dt} \|\partial_x h \|^2= \|\partial_v\partial_x h \|^2,$$ $$-\frac{1}{2}\frac{d}{dt} \|\partial_v h \|^2=\langle\partial_x h,\partial_v h\rangle+ \|(-\partial_v +v)\partial_v h \|^2=\langle\partial_x h,\partial_v h\rangle+ \|\partial_v^2 h \|^2+ \|\partial_v h \|^2.$$

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 $$\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle=- \|\partial_x h \|^2+2\langle(-\partial_v+v)\partial_v h,\partial_v\partial_x h\rangle-\langle\partial_x h,\partial_v h\rangle,$$ hence $$\frac{d}{dt}J\equiv\frac{d}{dt}\biggl( \|h \|^2+a \|\partial_x h \|^2+b\langle\partial_x h,\partial_v h\rangle+c \|\partial_v h \|^2\biggr)=$$ $$\qquad=-2 \|\partial_v h \|^2-2a \|\partial_v\partial_x h \|^2-b \|\partial_x h \|^2-2c \|(-\partial_v +v)\partial_v h \|^2+2b\langle(-\partial_v+v)\partial_v h,\partial_v\partial_x h\rangle-(b+2c)\langle\partial_x h,\partial_v h\rangle.$$ The first four terms on the right-hand-side have a fixed sign, the last two terms are bounded by the Cauchy-Schwartz + Young inequality, $$\frac{d}{dt}J\leq -(2a-1) \|\partial_v\partial_x h \|^2-(b-1/2) \|\partial_x h \|^2-(2c-b^2) \|(-\partial_v +v)\partial_v h \|^2-(1-c-b/2) \|\partial_v h \|^2.$$

For $b=1/2$ we thus find $$\frac{d}{dt}J\leq-(2a-1) \|\partial_v\partial_x h \|^2-(2c-1/4) \|\partial_v^2 h \|^2-(1/2+c) \|\partial_v h \|^2$$ $$\qquad \leq-K\biggl( \|\partial_v\partial_x h \|^2+ \|\partial_v^2 h \|^2+ \|\partial_v h \|^2\biggr)$$ for some $K$ if $a>1/2$, $c>1/8$.