advection-diffusion equation with origin-directed drift

Question

Is there a solution (and if yes, what is the solution) to the 3D diffusion equation with a drift velocity directed to the origin $\vec{v}(\vec{x}) = a|\vec{x}|^{-1}\vec{x}$, with $a>0$? Thus a solution to $$ \partial_t c = \nabla^2c +\nabla \cdot \Big(\vec{v}c\Big) =\nabla^2c +a\nabla \cdot \Big(\frac{\vec{x}}{|\vec{x}|}c\Big) $$ with initial condition $$ c(\vec{x},0) = g(\vec{x}-\vec{x}_0) $$ where $g$ is a function, that has arbitrary small support around $\vec{x}=0$, and might be thought of as an approximation to the delta distribution $g\to\delta$.

To simplify further, the spatial domain of interest is a 3D sphere $S_r(0)\subset\mathbb{R}^3$ with $\vec{x}_0\in S_r(0)$ of small radius $r$ at the origin and a the temporal domain is also small $T\ll1$ in some sense: $$ c:~~S_r(0)\times[0,T]\ni(\vec{x},t) \mapsto c(\vec{x},t)\in \mathbb{R}_+ $$

Edit: Corrected the sign of the drift.

Answer 1

This is not an answer but rather some non-formal consideration of how the solution will look like.

One can derive many properties from the equation above, giving clear hints to the equations behaviour. To see those more clearly resolve the divergence term: $$ \partial_t c = \nabla^2c +\frac{2ac}{|\vec{x}|} + a\frac{\vec{x}}{|\vec{x}|}\cdot \nabla c $$

• The overall L1 norm is conserved (since it's a diffusion equation)
• The Laplace term decreases the function $c$, wherever it looks gauss-like. Think of 1D and the expansion of the 1D-Laplacian $\partial_x^2c(x) = \lim_{\epsilon\to0} \epsilon^{-2}(c(x+\epsilon) -2 c(x)+c(x-\epsilon))$. Thus $\partial_x^2c$ will be negative when $c$ is concave, positive when convex. This is the known smoothing behaviour.
• The second (source) term constantly increases $c$ at the origin leading to a cusp there (seen by the stationary solution).
• The last term increases or decreases $c$ depending on the angle of the gradient (vector of steepest ascent) and the vector pointing to the origin. $$ \vec{x}\cdot \nabla c \propto \cos(\angle\vec{x},\nabla c) $$

  • When the gradient points away from the origin, aligning with $\frac{\vec{x}}{|\vec{x}|}$ the last term gives positive contribution and increases $c$

  • When the gradients points to the origin, opposite to $\frac{\vec{x}}{|\vec{x}|}$, $c$ decreases

  • This makes any initial gaussian-like concentration package $c(\vec{x},0)$ not centered around the origin move to the origin by lowering its outer flank (gradient pointing to origin) and increasing its inner flank (gradient pointing away from origin).

• Finally a stationary normalized solution (but not the general solution) is easy to obtain in spherical coordinates, reading $$ c(\vec{x}) = \frac{a^3}{8\pi} \exp(-a|\vec{x}|) $$
• Any (smooth) initial state concentrated at some $\vec{x}_0$ would travel to the origin in the time evolution (more precise its center of mass would travel there), forming a cusp as in the stationary solution. The cusp would most likely appear immediately after $t=0$, since it would be very strange if it formed after a finite time?!

What I am missing is the exact form. But I am sure from the considerations above that there should be a nice one.

Edit: Taylor-expanding the solution for a small time step and assuming the initial condition is a gaussian $$ g_{\vec{\mu}\sigma}(\vec{x}) \propto \exp\Big(-\frac{|\vec{x}-\vec{\mu}|^2}{2\sigma^2}\Big) $$ yields a solution $$ c(\vec{x},t) \propto \exp\Bigg(-\frac{|\vec{x}-\vec{\mu}|^2+2at|\vec{x}|}{2\sigma^2 + 4 t}\Bigg) + \frac{2a}{|\vec{x}|}g_{\vec{\mu}\sigma}(\vec{x}) $$