# Solution of diffusion equation with spherical sink

I hope this question is not too basic, but I have no experience with partial differential equations and would like to ask for some hints on how to solve the following problems:

The visual idea is to describe the diffusion of some dilute chemical around a spherical sink or a sink at some point.

It would be nice to obtain a time evolution when starting with a uniform density (this is only possible in problems 1) and 5)), but I would already be satisfied with a "nice" steady-state solution.

1) The simplest case would be a diffusion equation in one dimension with a sink at $$x=0$$, i.e. $$\frac{\partial{\rho}(x)} {\partial{t}} = - \nabla \cdot J(x), \qquad x \in \mathbb{R}$$ with $$J(x) = -D \nabla \rho(x)$$ for $$x \neq 0$$ and $$J(x) = -D \nabla \rho(x) -k\rho(x)$$ for $$x = 0$$ with $$k$$ some large depletion constant.

Moreover, I would like to require that $$\rho(x) \rightarrow c$$ for $$|x| \rightarrow \infty$$ for some constant $$c >0$$.

The initial conditions are not so important, let's say $$\rho(x;t=0) = c_0$$ for all $$x \in \mathbb{R}$$. Actually, I am not as much interested in the time evolution as in finding a nice steady-state solution, which does not diverge for $$x \rightarrow \infty$$. Thus, some more appropriate initial conditions could be chosen.

2) Alternatively, One could just require the boundary condition $$\rho(x=0;t) = 0$$ for all $$t$$ in the above situation. Then the initial condition needs to be adjusted accordingly.

3) The problem in 3D is the same: to solve the diffusion equation with $$x \in \mathbb{R}^3$$ and boundary conditions $$\rho(x) = 0$$ for $$x = 0$$ and $$\rho(x) \rightarrow c$$ for $$|x| \rightarrow \infty$$ with some $$c >0$$. I guess with spherically symmetric initial conditions, this case is completely equivalent to 2).

4) Now I would like to have a real sphere in $$\mathbb{R}^3$$ of radius $$r >0$$, i.e. the boundary conditions $$\rho(x) = 0$$ for $$|x| < r$$ and $$\rho(x) \rightarrow c$$ for $$|x| \rightarrow \infty$$ with some $$c >0$$.

5) Maybe 4) is simpler when formulated with a finite depletion rate $$k$$ as in 1) instead of the boundary condition around $$0$$. Physically, this is very similar anyway.

Any literature suggestions are also highly appreciated.

• Look at this. – Deep Oct 24 '18 at 6:17
• In eq (1), the right-hand side would be $D \nabla^2 \rho + k \nabla \cdot \rho$ at $x = 0$. Naïvely, this looks like you're adding a scalar to a vector. Can you clarify what you mean by editing the question? – Michael Seifert Oct 24 '18 at 12:53
• Also, see this old answer of mine for a cautionary tale on setting a "boundary at $r = 0$" in cylindrical coordinates. The same argument would apply to a spherical coordinate system. – Michael Seifert Oct 24 '18 at 12:57
• @MichaelSeifert You are right, I should have added the depletion term to $\partial{\rho}/\partial{t}$, i.e. $\partial{\rho}/\partial{t} = D\nabla^2 \rho - k \rho$ at $x=0$ – Dimitri Oct 24 '18 at 13:21

For steady state diffusion in an unbounded region toward a spherical sink of radius $$r_0$$, we have $$4\pi r^2 D\frac{dc}{dr}=Q$$where Q is the diffusion rate of the species toward the sink (a constant). The solution to this equation is $$c=c_{\infty}-\frac{Q}{4\pi rD}$$where $$c_{\infty}$$ is the concentration far from the sphere. Applying the boundary condition $$c=c_0$$ at $$r=r_0$$ for the surface of the sphere, we have: $$Q=4\pi r_0D(c_{\infty}-c_0)$$Therefore, the concentration profile is $$\frac{(c_{\infty}-c)}{(c_{\infty}-c_0)}=\frac{r_0}{r}$$