# Diffusion in different regions

1. Feb 25, 2015

### bobred

1. The problem statement, all variables and given/known data
A slab of radioactive material of thickness L lies in the x-y plane surrounded by a material that can be thought of as extending to +/- infinity from -L/2 and +L/2. The system is in steady. Find the general solution for the diffusion equations in each region(ok). Find the form of the general solution in Region 1 that is an even function and in 2 that is physically meaningful(ok). Find the general solution that satisifes the boundary conditions(?)

2. Relevant equations
Region 1
$$0=D_1 \frac{\partial^2 c}{\partial x^2}+H$$
for $\left| z \right|<L/2$
Region 2
$$0=D_2 \frac{\partial^2 c}{\partial x^2}-Rc$$
for $\left| z \right|>L/2$

3. The attempt at a solution
The general steady state solutions for each region are
$$c=-\frac{H}{2D_1}z^2+Az+B$$
and
$$c=Ce^{\lambda z}+De^{-\lambda z}$$
$\lambda=\pm \sqrt{\frac{R}{D_2}}$

In region 1 to be even A=0, for region 2 the concentration must tend to zero as z approaches +/- infinity, so

$$c=-\frac{H}{2D_1}z^2+B$$

$$c(z)=\begin{cases} De^{-\lambda z} & z>\tfrac{L}{2}\\ Ce^{\lambda z} & z<-\tfrac{L}{2} \end{cases}$$

The flux is the same on both sides and concentration continuous. I am having difficultly picturing these, what is happening at z=0 and z=+/-L/2.
Any pointers would be greatly appreciated.
James

2. Feb 27, 2015

### Astronuc

Staff Emeritus
At the outer boundaries, are they nonreentrant, i.e., once a particle leaves, it's gone, so there is no return current.

What is the boundary condition at the interface? Think about the current.

3. Feb 28, 2015

### Staff: Mentor

At z = L/2, $D_1\frac{dc}{dz} = -\frac{HL}{2}$. This is the other boundary condition that follows from integrating the differential equation once.

Also, at z > L/2, express c as : $c(z) = Ec^{-λ(z-\frac{L}{2})}$

The mass fluxes in the two regions must match at z = L/2.

The continuity of concentration at the boundary gives:
$$E=B-\frac{HL^2}{8D_1}$$

Physically, what's happening is that, by symmetry, there is no mass flow across the boundary z = 0, and all the mass generated between z = 0 and z = L/2 has to pass through the boundary at L/2. This is HL/2. This mass flow is then reacted away in the region beyond z = L/2.

Chet

Last edited: Feb 28, 2015
4. Mar 4, 2015

### bobred

Hi
I ended up down the lines you have mentioned. Needed to think about the flux at the boundary and the equations being continuous at the boundary.
Thanks
James