Asymptotics for repulsive aggregation(-diffusion) equation

Question

Consider the aggregation-diffusion equation $$ \frac{\partial \rho}{\partial t} = \nabla (\rho \nabla(W\star \rho)) + \nu \Delta \rho, $$ where $W:\mathbb{R}^d \to \mathbb{R}$ is a twice continuously differentiable interaction potential, $\nu \ge 0$, and where $\star$ denotes convolution, that is, $(W\star \rho) (x) = \int_{\mathbb{R}^d} W(x-y) \rho(y) \mathrm{d}y$. It seems that the case where $W$ is an attractive potential (that is $W(x)$ increases as $x\to \infty$) is well-studied, see for example the article I cite below. I suppose that makes sense as it is called aggregation equation.

However, I am interested in the case where $W$ is repulsive, on which I have found very little. In that case the solution should converge vaguely to zero, and I would be interested in asymptotic upper bounds on $\rho_t(0)$ in the case where $\rho$ is started from something nice and symmetric (say the standard normal density or something compactly supported), and where $W$ is a symmetric repulsive potential, like a standard normal density, a power law, anything really. I would even be interested in the case where $\nu = 0$, and specifically in $d = 2$.

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Carrillo, José A.; Craig, Katy; Yao, Yao, Aggregation-diffusion equations: dynamics, asymptotics, and singular limits, Bellomo, Nicola (ed.) et al., Active particles, Volume 2. Advances in theory, models, and applications. Cham: Birkhäuser. Model. Simul. Sci. Eng. Technol., 65-108 (2019). ZBL1451.76117..

Answer 1

(Too long to be a comment.) A hand-waving, physicist calculation of a specific example for $d=3$ goes as follows. It may be put in more rigours terms for higher but not for lower dimensions. Let's take

\begin{equation} W(\mathbf{x}-\mathbf{x}')=\frac{\alpha}{|\mathbf{x}-\mathbf{x}'|}, \end{equation}

with $\alpha>0$ for a repulsive force. The one can show that $\nabla^2 W(\mathbf{x}-\mathbf{x}')=-4\pi\alpha\delta(\mathbf{x}-\mathbf{x}')$. Since the potential is spherically symmetric let's assume that $\rho(\mathbf{x})=\rho(r)$ and just for notation omit the time dependence, then

\begin{align} \nabla\cdot\left(\rho(\mathbf{x})\nabla\left( W\star \rho\right)(\mathbf{x}) \right)&=\nabla\rho\cdot\nabla\left( W\star \rho\right)+\rho\nabla^2\left( W\star \rho\right)\\ &=\nabla\rho\cdot\int d^3x'\:\rho(r') \nabla W(\mathbf{x}-\mathbf{x}')+\rho \int d^3x'\:\rho(r')\nabla^2 W(\mathbf{x}-\mathbf{x}')\\ &=-\nabla\rho\cdot\int d^3x'\:\rho(r') \nabla' W(\mathbf{x}-\mathbf{x}')-4\pi\alpha\:\rho \int d^3x'\:\rho(r')\delta(\mathbf{x}-\mathbf{x}')\\ &=\alpha\int d^3x'\frac{\nabla\rho(r)\cdot\nabla'\rho(r')}{|\mathbf{x}-\mathbf{x}'|}-4\pi \alpha\:\rho^2(r). \end{align}

It's not hard to show that the remaining integral vanishes by explicit calculation. Then we have

\begin{align} \frac{\partial\rho(r;t)}{\partial t}=-4\pi\alpha\: \rho^2(r;t)+\frac{\nu}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial \rho(r;t)}{\partial r} \right). \end{align}

Note that we cannot put $\nu=0$ (i.e. vanishing diffusion) because then we lose the solution. Since $4\pi\:\alpha>0$ you can see that the solution indeed dies away very rapidly, much more that typically linear Fokker–Planck solutions, but the non-linearity makes it hard to see analytically how this happens.