# Reaction diffusion problem concentric spheres

## Main Question or Discussion Point

Hello,

I am trying to (numerically) solve the following reaction-diffusion equation for the probability density of the a pair, $$\rho (\vec{r}_1,\vec{r}_2)$$:

$$\dot{\rho} (\vec{r}_1,\vec{r}_2,t) = D_1 \nabla^2_1 \rho (\vec{r}_1,\vec{r}_2,t) + D_2 \nabla^2_2 \rho (\vec{r}_1,\vec{r}_2,t) - k \left( \left\| \vec{r}_1 - \vec{r}_2 \right\| \right)$$,

where the subscripts refer to the first and second particle, respectively. In 2D and polar coordinates, $$r_i$$ and $$\theta_i$$:

$$\nabla^2_i = \frac{1}{r_i} \frac{\partial}{\partial r_i} r_i \frac{\partial}{\partial r_i} + \frac{1}{r_i^2} \frac{\partial}{\partial \theta_i}$$.

The domain is confined by two concentric spheres: $$0 \leq \left\| \vec{r}_1 \right\| \leq R$$ and $$\left\| \vec{r}_2 \right\| \geq R$$. The initial condition are spherically symmetric, i.e. only depends on the $$r_i$$s. The reaction term is a function of the distance of the two particles, i.e. in 2D $$k( \left\| \vec{r}_1 - \vec{r}_2 \right\| ) = k( \sqrt{r_1^2 + r_2^2 - 2 r_1 r_2 \cos ( \theta_1-\theta_2)} )$$. I hoped to get rid of at least 1 coordinate by a variable transformation and separation of variables. However, so far I just could not come up with a separable problem. Do I really have to retain all 4 variables? Any suggestions of how to reduce this problem to something manageable are highly welcome. Eventually I will be interested in 3D and 4D as well.

Thank you,
Daniel