Rewriting PDE as "push-forward"

Question

Suppose that we have the following PDE $$\partial_t \mu_t = \nabla\cdot \left(\nabla \mu_t - (b*\mu_t)\mu_t\right), \tag{1}$$ with $\mu_0$ being a (smooth) probability measure/density on $\mathbb{R}^d$ and with the notation $$b*\mu(x) := \int_{\mathbb{R}^d} b(x,z)\,\mu(z)\,\mathrm{d}z.$$ For $N \in \mathbb{N}_+$ denote $N$ times tensorized product of $\mu$ by $\mu^{\otimes N}$. It is stated (without proof) in page 5 of this paper that $\mu^{\otimes N}_t$ can be seen as the pushed forward of $\mu^{\otimes N}_0$ by the application $\zeta^{\otimes N}_{t,0}$ defined as the solution to the following ODE $$\zeta_{t,0}(x) = x + \int_0^t \left(b*\mu_s - \nabla \ln \mu_s\right)\,\mathrm{d}s. \tag{2} $$ However, I have no clue as to deriving $(2)$ from $(1)$. I understand the definition of push forward of measures (see for instance here) but I really need someone to help me out on this particular problem. Thank you very much for your kind help!

Answer 1

Let's focus on $N=1$ only, the case $N>1$ is just a tensorization of the argument below.

The whole argument is actually unrelated to the specific (aggregation-diffusion) PDE or gradient flow: As soon as you have a curve of probability measures $\mu_t$ satisfying the continuity equation $$ \partial_t\mu_t+\nabla\cdot(v_t\mu_t)=0 $$ for some (smooth enough) vector-field $v=v_t(x)$, then in fact you have the representation $$ \mu_t=\Phi_t\#\mu_0, $$ where $\Phi_t(x)$ is the flow-map associated to the ODE $\begin{cases}\dot X_t=v_t(X_t)\\ X_0=x\end{cases}$. This is easy to check formally, and for a precise reference I can suggest reading chapter 8 in [1] (e.g. lemma 8.1.6 or prop. 8.1.8).

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Now, for your specific context, it is just a matter of rewriting the PDE into a continuity equation. Factoring out and explointing the classical identity $\Delta\mu=\nabla\cdot\nabla\mu=\nabla\cdot(\frac{\nabla\mu}{\mu}\mu)=\nabla\cdot((\nabla\log \mu) \mu)$ you see that $\mu$ solves the continuity equation $$ \partial_t\mu_t=\nabla\cdot(\nabla\mu_t-(b\ast\mu_t)\mu_t) \quad\Leftrightarrow \quad \partial_t\mu_t+\nabla\Big((\underbrace{b\ast\mu_t-\nabla\log\mu_t}_{v_t})\mu_t\Big)=0. $$ Then the flow-map $\Phi_t$ can be simply represented from the implicit Duhamel formula as $$ \begin{cases}\dot X_t=v_t(X_t)\\ X_0=x\end{cases} \quad\Leftrightarrow\quad X_t=x+\int_0^t v_s(X_s)ds. $$ Plugging in the explicit formula for the velocity field $v_t=b\ast\mu_t-\nabla\log\mu_t$ gives exactly your expression (2).

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[1] Ambrosio, L., Gigli, N., & Savaré, G. (2008). Gradient flows: in metric spaces and in the space of probability measures. Springer Science & Business Media.