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](https://www.aimsciences.org/article/doi/10.3934/dcds.2020243) 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](https://math.stackexchange.com/questions/1970164/push-forward-measure-equivalent)) but I really need someone to help me out on this particular problem. Thank you very much for your kind help!



Rewriting PDE as "push-forward"
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.


