Tuesday, 21 March 2017

The slope of solutions to the Fisher equation

This blog post describes what I believe to be a common misconception about solutions to the Fisher equation, namely that the slope is inversely proportional to the wave speed. I realised this during supervision of a Bachelor thesis when the students couldn't find agreement between the analytical result and their numerical solutions.

The Fisher equation is partial differential equation of the form:
$$\frac{\partial u(x,t)}{\partial t}=D \nabla^2 u + r u(1-u)$$
where u(x,t) represents the density of cancer cells at time t and location x, and the parameters of the model are D, the diffusivity of cancer cells (i.e. how fast they migrate) and r, the rate of cell division.

The Fisher equations exhibits travelling wave solution, i.e. a fixed front profile that is being translated in space as time progresses. These solutions are typically characterised by their velocity c and slope s. It has been shown that the wave speed is given by

$$c=2\sqrt{Dr}.$$

A partial proof of this can be found in Mathematical Biology by James Murray. In the same book it is claimed that the slope

$$s=1/4c,$$

which implies that faster waves are less steep. This statement is accompanied by the below figure.

Substituting the expression for c one is lead to believe that

$$s=\frac{1}{8\sqrt{D r}}.$$

This expression is never mentioned in Murray's book, but I would claim that the presentation is quite misleading, because if one looks closer at the analysis that leads up to the statement s=1/4c, then one finds that the analysis is done on a non-dimensional version of the Fisher equation, which has wavespeed c = 2. This implies that the statement s=1/4c simply means s = 1/8.

If one instead carries out the exact same analysis on the dimensional Fisher equation one finds that

$$s=1/8\sqrt{r/D}.$$

I think this fact has been missed by many mathematical biologist and I hope this blog post can shed some light on this misunderstanding.

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