Pome

Proposition 11.23.8AE4D0608184F

Proposition 11.23.8.AE4D0608184F Let $\Phi \in C^2((0,+\infty ),\mathbb R)$, and choose a function $G$ with $G'(r)=\Phi '(r)/r$. Along (11.23.7),

313276789DFD 11.23.9
\begin{equation} \begin{split} \frac{\mathrm{d}}{\mathrm{d}t}\int _ X\Phi (\rho _ t)\nu ={}& \frac nM\int _ XG(\rho _ t)\mathrm{dd}^{\mathrm{c}}(\omega _{u_ t}^{n-1})\\ & -\frac nM\int _ X\frac{\Phi ''(\rho _ t)}{\rho _ t} \mathrm{d}\rho _ t\wedge \mathrm d^{\mathrm c}\rho _ t\wedge \omega _{u_ t}^{n-1}. \end{split} \end{equation}

In particular, for $p>1$,

0BC0A4E1D2A4 11.23.10
\begin{equation} \begin{split} \frac{\mathrm{d}}{\mathrm{d}t}\int _ X\rho _ t^ p\nu ={}& \frac{np}{(p-1)M}\int _ X\rho _ t^{p-1}\mathrm{dd}^{\mathrm{c}}(\omega _{u_ t}^{n-1})\\ & -\frac{np(p-1)}M\int _ X\rho _ t^{p-3} \mathrm{d}\rho _ t\wedge \mathrm d^{\mathrm c}\rho _ t\wedge \omega _{u_ t}^{n-1}. \end{split} \end{equation}

Proof. Using $\partial _ t\rho _ t=\rho _ t\Delta _ t\log \rho _ t$ and $\rho _ t\nu =M^{-1}\omega _{u_ t}^ n$, we find that the derivative of the entropy is

\[ \frac nM\int _ X\Phi '(\rho _ t)\mathrm{dd}^{\mathrm{c}}\log \rho _ t \wedge \omega _{u_ t}^{n-1}. \]

The chain rule and the definition of $G$ give

\[ \Phi '(r)\mathrm{dd}^{\mathrm{c}}\log r =\mathrm{dd}^{\mathrm{c}}G(r)-\frac{\Phi ''(r)}r\mathrm{d}r\wedge \mathrm d^{\mathrm c}r, \]

and lemma 11.1.2 applies to the first term. Adding a constant to $G$ does not change the result, since $\int _ X\mathrm{dd}^{\mathrm{c}}(\omega _{u_ t}^{n-1})=0$. Taking $\Phi (r)=r^ p$ and $G(r)=p r^{p-1}/(p-1)$ proves the second formula. $\square$

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