How do I derive the following identity? I'm following a math proof on the Kuramoto model and the following identity is stated:
$$ \partial_t \int_\psi^{\psi+\delta\psi} p(\psi'|\omega)d\psi'=p(\psi|\omega)\dot{\psi}(\psi)-p(\psi+\delta\psi|\omega)\dot{\psi}(\psi+\delta\psi) $$
where $p(\psi|\omega)$ is some unknown probability distribution with normalization $\int_{-\pi}^{\pi}p(\psi|\omega)d\psi=1$ and
$$\dot{\psi}(\psi) = (\omega-\Omega) - KR\sin\psi$$
with $K,R,\omega,\Omega$ constants, is the equation of motion for $\psi$.
The two terms in the RHS represent the fluxes of probability into and out of the control interval $[\psi, \psi+\delta\psi]$, respectively.
I don't know how to derive this seemingly obvious statement (since it is taken for granted in the proof I'm following). What I have done is move the partial time derivative inside the integral, apply the chain rule and integrate by parts as the following:
$$ \int_\psi^{\psi+\delta\psi}\partial_t [p(\psi'|\omega)]d\psi' = \int_\psi^{\psi+\delta\psi} \dot{\psi}(\psi') \partial_{\psi'} [p(\psi'|\omega)]d\psi' $$
assigning $u=\dot{\psi}(\psi')$ and $v'=\partial_{\psi}[p(\psi'|\omega)]$ and using
$$\int_a^buv'dx = uv|_a^b - \int_a^b u'vdx$$
I get:
$$ \partial_t \int_\psi^{\psi+\delta\psi} p(\psi'|\omega)d\psi'=p(\psi'|\omega)\dot{\psi}(\psi')|_{\psi}^{\psi+\delta\psi} - \int_{\psi}^{\psi+\delta\psi}(-KR\cos\psi')p(\psi'|\omega)d\psi' $$
If the second term on the RHS is zero then I get the original identity, but I don't see any reason why this integral should be zero, fundamentally or particularly to the Kuramoto model.
Thanks.
Moving the time derivative inside the integral is actually a mistake, and also I think there is a sign error in the original proof. So I think the correct answer goes like this:
$$ \partial_t \int_\psi^{\psi+\delta\psi} p(\psi'|\omega)d\psi'=\partial_t \left[ P(\psi+\delta\psi|\omega)-P(\psi|\omega) \right] $$
where $P$ is the anti-derivative of $p$. Using the chain rule we get $$ \partial_t [P(\psi)] = \frac{\partial P(\psi)}{\partial \psi}\frac{\partial \psi}{\partial t} $$ and thus:
$$ \partial_t \left[ P(\psi+\delta\psi|\omega)-P(\psi|\omega) \right]=\frac{\partial P(\psi+\delta\psi|\omega)}{\partial(\psi+\delta\psi)}\frac{\partial(\psi+\delta\psi)}{\partial t} - \frac{\partial P(\psi|\omega)}{\partial\psi}\frac{\partial \psi}{\partial t}\\ =p(\psi+\delta\psi|\omega)\dot{\psi}(\psi+\delta\psi) - p(\psi|\omega)\dot{\psi}(\psi) $$
which is the sought expression but with opposite sign.