# Solving the diffusion equation with an absorbing boundary

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There is a one-dimensional diffusion process in which particles start running at $t = 0$ and from $x_o > 0$.

When particles reach x = 0 they are removed from the system, thus the total concentration is not conserved anymore.

I have to solve the diffusion equation, which is the following partial differential equation:

$$\frac{\partial P (R, t)}{\partial t} = D\triangledown^2P(R,t)$$

Where $P(R, t)$ is the probability that the particles arrive at R at time t.

We have the initial conditions:

$$c(x,0) = N\delta(x - x_o)$$

$$c(0, t) = 0$$

I have been doing some research in how to do so and I came across with a method which is based on a particular Gaussian function:

$$G (R, t) = (\frac{1}{4\pi Dt})^{\frac{d}{2}} e^{\frac{(R-R_0)^2}{4Dt}}$$

Where d is the dimensionality of the system.

But the issue here is that we are working with an 'absorbing boundary' that makes the condition $c(0, t) = 0$ useless because we work from $x_o > 0$.

Then how could I solve the probability (i.e this differential equation)? We have been suggested the method of images, but not sure how it works with differential equations.

NOTE: I know this is more a mathematical issue (solving a differential EQ.) but thought that as it is a mechanics problem it could be useful asking here.

retagged Oct 27, 2018

1 vote

In this situation the Method of Images would apply as follows :

At $t=0$ you have a dense concentration of particles at a particular point $+x_0$. Over time they diffuse outwards with a Gaussian probability density profile. However, this expanding distribution is disrupted by the absorbing boundary at $x=0$. The result is that the initially symmetrical Gaussian distribution becomes increasingly asymmetrical as it spreads out, as in the diagram below.

In order to take account of the absorbing boundary, the Method of Images places an equal dense concentration of anti-particles at the image point $-x_0$. These anti-particles have the same properties as the ordinary particles released at $+x_0$, so they diffuse outwards with the same speed hence the same Gaussian distribution. However, when they come into contact with ordinary particles they annihilate, just like electrons and positrons.

The two distributions are exactly anti-symmetrical about the boundary. There are always equal numbers of each type of particle at the boundary, so the sum at the boundary is always zero.

So if you have a solution $G (x,t)$ centred on $+x_0$ without the absorbing boundary, then to account for the absorbing boundary you should subtract the same solution $G (x,t)$ centred on $-x_0$.

As with the electrostatic and optical method of images, this trick only works on the object side of the boundary. It cannot tell you the distribution of ordinary particles on the image side, where there could be other sources and other absorbing boundaries. In the absence of such information, you must assume that you result applies only for $x \ge 0$ : for all $x \lt 0$ you must enforce the condition $G(x,t)=0$.

answered Oct 1, 2018 by (28,448 points)
edited Oct 8, 2018