Mathematician Alan Turing proposed a possible explanation for shapes and structures found in nature. Using reaction-diffusion equations an example of emergent order in nature was seen. The patterns created are now called Turing patterns. By adding a diffusive term to a model of a stable chemical reaction, he could cause it to become unstable. This is called Turing instability and it causes a transition in which patterns evolve into a new, spatially varying steady-state solution.
An example of Turing instability is the Brusselator model. It consists of two coupled partial differential equations, each representing one species of a two species chemical reaction, given by
with conditions
Using the given code we were able to recreate Example 8.15. We ran the code until \(time = 1000\) for \(K = 7\) and this took well over an hour to complete. So, we only ran the other K values until \(time = 100\).
It is easily seen that with some basic algebra that the right sides of both equations will be zero.
Now we're asked the question: "For parameter settings \(Dp = 1,Dq = 8,C = 4.5\) of the Brusselator, for what values of K is the equilibrium solution \(p \equiv C,q \equiv K/C\) stable?"
We were given that a Turing instability is encountered when:
Therefore, we will have stability when:
Plugging in the appropriate values gives:
Solve the Brusselator equations for \(Dp = 1,Dq = 8,C = 4.5 \text{ and } \text{(a) } K = 4 \text{ (b) } K = 5 \text{ (c) } K = 6 \text{ (d) } K = 6.5.\) Using homogeneous Neumann boundary conditions and initial conditions \(p(x, y,0) = 1 + \cos(\pi x) \cos(\pi y), q(x, y,0) = 2 + \cos(2\pi x) \cos(2\pi y),\) estimate the least value T for which \(\lvert p(x,y,t) - C \rvert < 0.01 \text{ for all } t > T.\)
Changing the inital conditions in our code and inserting a while loop until \(\lvert p(x,y,t) - C \rvert < 0.01 \) gives: 8.4.5 Code
| K | T |
|---|---|
| 4 | 10 |
| 5 | 16 |
| 6 | 29 |
| 6.5 | 63 |
These videos of \(K = 4, 5, 6, 6.5\) are shown below on \([0,40]\times [0,40]\) with \(N = M = 40\)
Here, we went back to the original information given in the introduction of the problem, and tried to fill in the gaps between the first two figures in Example 8.15, i.e. between \(K=7\) and \(K=8\). Below are the animations on the grid \([0,20]\times [0,20]\), with \(M = N = 40\), and \(\Delta t = 1\). The code used to generate these figures is given here.
| \(K = 7.2,\; 0 \leq t \leq 100\) | \(K = 7.4,\; 0 \leq t \leq 100\) | \(K = 7.6,\; 0 \leq t \leq 100\) | \(K = 7.8,\; 0 \leq t \leq 100\) |
Because the time it took to generate these videos was very long even only going to \(t = 100\), we opted to only generate a photo for the time \(t=2000\) for these values of \(K\). Additionally, we narrowed the grid to \([0,10]\times [0,10]\), with \(N = 20, M=20\), to reduce the waiting time and still keep our grid step sizes at 0.5.
| \(K = 7.2,\; t = 2000\) | \(K = 7.4,\; t = 2000\) | \(K = 7.6,\; t = 2000\) | \(K = 7.8,\; t = 2000\) |
Lastly, we made a video showing the Turing patterns at \(T = 100\) with \(K\) values varying between 6.8 and 9.8 with steps of 0.1. You can see that the higher \(K\) gets, i.e. the farther from the instability, the more the spots morph into stripes. The code used to generate this video is given here.