Differential equation describing diffusion to a partially covered sphere

Question

I'm trying to work through a paper by Zwanzig (1990, Diffusion controlled ligand binding to spheres partially covered by receptors: An effective medium treatment). There is one section that I can't follow the math, here it is:

The steady state diffusion equation:

$$D~ \bigtriangledown ^{2}C = 0$$

[Where $~D~$ is the diffusion coefficient and $~C~$ is concentration.]

"This is to be solved with appropriate boundary conditions on the surface of the sphere. To an observer far from the sphere, the surface appears to be uniform but neither perfectly reflecting nor perfectly absorbing. This suggests the use of a "radiation boundary condition,"

$$D~\frac{\partial C}{\partial r} = k~C$$

on

$$r = R$$

If $~k = 0~$, the surface is perfectly reflecting, and if $~k~$ goes to $~\infty~$, the surface is perfectly absorbing. Then the appropriate solution of the diffusion equation is

$$C = 1 - \frac{\alpha}{r}$$

and the boundary condition on R determines the coefficient

$$\alpha = \frac{k}{\frac{k}{R} + \frac{D}{R^{2}}}$$

I'm stuck on how to get that solution, any help is appreciated!

Answer 1

Let me understand your question(s).

If you start from the step

$\displaystyle C = 1 - \frac{\alpha}{r}$

and note that

$\displaystyle \left[ C \right]_{r=R} = 1 - \frac{\alpha}{R}$

and

$\displaystyle \left[ \frac{\partial C}{\partial r} \right]_{r=R} = \frac{\alpha}{R^2}$

Substituting in the original equation

$\displaystyle D \frac{\partial C}{\partial r} = kC$

we have (when $r = R$)

$\displaystyle D \frac{\alpha}{R^2} = k \left(1 - \frac{\alpha}{R} \right)$

$\implies \displaystyle \alpha = \frac{k}{\frac{k}{R} + \frac{D}{R^2}}$

What if the form of the solution was not given?

This is easier to solve. In this case, $C$ is a function of $r$ only and hence the partial derivatives will become ordinary derivatives.

$\displaystyle D \frac{dC}{dr} = kC$

or, $\displaystyle D \frac{dC}{C} = k ~dr$

or, $\displaystyle \ln|C| = \frac{k}{D} r + C_1^\prime$

or, $\displaystyle C = C_1 e^{\frac{kr}{D}}$

Here, $\displaystyle C_1^\prime$ is the constant of integration and $\displaystyle C_1 = e^{C_1^\prime}$

Does it help? If not , please feel free to ask. I'll explain further.

Answer 2

Use the polar coordinates. Nothing depends on angles here, so the equation becomes a second order ODE. Its general solution can be easily written out. Putting it in the boundary condition gives the answer.