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I was working through an energy calculation for the Fokker-Planck equation:

$$ \partial_t \rho + \nabla\cdot(\mathbf{v}\rho) = 0 $$where

$$ \mathbf{v} = -(\frac{1}{\rho}\nabla\rho + \nabla V). $$ At equilibrium, the density $\rho(\mathbf{x},\,t)$ no longer changes with time, $\rho(\mathbf{x},\,t) = \rho_{\inf}$.

We get that $$ \rho_{inf} (\mathbf{x},\,t) = Ce^{-V(\mathbf{x})} $$

Let $\phi : [0, inf) \rightarrow [0,inf)$ be a smooth function such that $\phi(1) = \phi’(1) = 0.$

$$ E(\rho) = \int_{\Omega}\phi(\frac{\rho}{\rho_{inf}})\rho_{inf} dx. $$

In the solution, I was given the following line: $$ \frac{dE}{dt} = \int_{\Omega} \phi’(\frac{\rho}{\rho_{inf}})\partial_t\rho dx = -\int_{\Omega}\nabla\phi’(\frac{\rho}{\rho_{inf}}) \cdot (\nabla \rho + \rho\nabla V) dx $$

I am not sure how they arrive at the last equality. When I asked my professor, he said it was integration by parts/Green’s identity.

Cheers.

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  • $\begingroup$ Is the difficulty in understanding the IBP step? $\endgroup$
    – Sean Roberson
    Commented Apr 17 at 7:20
  • $\begingroup$ Where is the "last inequality"? $\endgroup$
    – MathArt
    Commented Apr 17 at 7:36
  • $\begingroup$ Your function $\phi : [0, inf) \rightarrow [0,inf)$ is also dependent on $\mathbf{x}$? Note that $\nabla\cdot(\mathbf{v}\rho)=\nabla\rho\cdot\mathbf{v}+\rho\nabla\cdot\mathbf{v}$ and $\int_{\Omega} \phi’(\frac{\rho}{\rho_{inf}})\partial_t\rho d\mathbf{x}=-\phi'\rho\mathbf{v}|_0^1+\int_{\Omega}\rho\mathbf{v}\nabla\phi 'd\mathbf{x}=\int_{\Omega}\nabla(\frac{\rho}{\rho_{inf}}) \cdot (\nabla \rho + \rho\nabla V)d\mathbf{x}$ $\endgroup$
    – MathArt
    Commented Apr 17 at 8:25
  • $\begingroup$ Does $\nabla(\rho/\rho_\inf) $refers to $\nabla \phi'(\rho/\rho_\inf) $ ? Also there seems to be a sign issue ? $\endgroup$
    – Steph
    Commented Apr 17 at 9:16
  • $\begingroup$ @SeanRoberson yes $\endgroup$
    – Samarth Chirania
    Commented Apr 17 at 10:11

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