Reaction-Diffusion Systems
We present here a brief theory of reaction diffusion waves.
Law of Mass Action
The law of mass action states that equation rates are proportional to the concentration of reacting species and the ratio in which they combined. It is discussed in detail in Billingham and King 2000. We will present here a few simple examples.
Example 1: Simple Decay
Suppose we have of chemical [math]\displaystyle{ P }[/math] which decays to [math]\displaystyle{ A }[/math], i.e.
[math]\displaystyle{ P \to A }[/math]
with rate [math]\displaystyle{ k[P] }[/math] where [math]\displaystyle{ [P] }[/math] denotes concentration. Then if we set [math]\displaystyle{ p=[P] }[/math] and [math]\displaystyle{ a = [A] }[/math] we obtain the equations
[math]\displaystyle{ \frac{dp}{dt} = -kp\,\,\,\textrm{and}\,\,\, \frac{da}{dt} = kp }[/math]
which has solution
[math]\displaystyle{ p = p_0 e^{-kt}\,\,\,\textrm{and}\,\,\, a = a_0 + p_0(1-e^{-kt}) }[/math]
Example 2: Quadratic Autocatalysis
This example will be important when we consider reaction diffusion problems. We consider the reaction
[math]\displaystyle{ A + B \to 2B }[/math]
with rate proportional to [math]\displaystyle{ k[A][B] }[/math]. If we define [math]\displaystyle{ a = [A] }[/math] and [math]\displaystyle{ b = [B] }[/math] we obtain the following equations
[math]\displaystyle{ \frac{da}{dt} = -kab\,\,\,\textrm{and}\,\,\, \frac{db}{dt} = kab }[/math]
We can solve these equations by observing that
[math]\displaystyle{ \frac{d(a+b)}{dt} = 0 }[/math]
so that [math]\displaystyle{ a = b = a_0 + b_0 }[/math]. We can then eliminate [math]\displaystyle{ a }[/math] to obtain
[math]\displaystyle{ \frac{db}{dt} = k(a_0 + b_0 - b)b }[/math]
[math]\displaystyle{ \lt math\gt Insert formula here }[/math]</math>
which is separable with solution
[math]\displaystyle{ b = \frac{b_0(a_0 + b_0)e^{k(a_0 + b_0)t}}{a_0 + b_0e^{k(a_0 + b_0)t}} }[/math]
and
[math]\displaystyle{ a = \frac{a_0(a_0 + b_0)}{a_0 + b_0e^{k(a_0 + b_0)t}} }[/math]
Note that [math]\displaystyle{ a\to 0 }[/math] and [math]\displaystyle{ b\to a_0 + b_0 }[/math] as [math]\displaystyle{ t\to \infty. }[/math]
Diffusion
The equation for spatially homogeneous diffusion of a chemical with concentration [math]\displaystyle{ c }[/math] is
[math]\displaystyle{ \partial_t c = D\nabla^2 c }[/math]
which is the heat equation. We will consider this in only one spatial dimension. Consider it on the boundary [math]\displaystyle{ -\infty \lt x \lt \infty }[/math]. In this case we can solve by the Fourier transform and obtain
[math]\displaystyle{ \partial_t \hat{c} = -D k^2 \hat{c} }[/math]
where [math]\displaystyle{ \hat{c} }[/math] is the Fourier transform of [math]\displaystyle{ c }[/math]. This has solution
[math]\displaystyle{ \hat{c} = \hat{c}_0 e^{-D k^2 t} }[/math]
We can find the inverse transform using convolution and obtain
[math]\displaystyle{ c(x,t) = \frac{1}{\sqrt{4\pi D t}} \int_{-\infty}^{\infty} c_0(x) e^{(x-s)^2/4Dt}ds }[/math]
Solution of the dispersion equation using FFT
We can solve the dispersion equation using the discrete Fourier transform and its closely related numerical implementation the FFT. We consider the concentration on the finite domain [math]\displaystyle{ -L \leq x \leq L }[/math] and use a Fourier series expansion
[math]\displaystyle{ c(x,t) = \sum_{-\infty}^{infty} \hat{c}_n(t) e^{\mathrm{i} k_n x} }[/math]
where [math]\displaystyle{ k_n = \pi n /L }[/math]. If we substitute this into the diffusion equation we obtain
[math]\displaystyle{ c(x,t) = \sum_{-\infty}^{infty} \hat{c}_n(0)e^{-k_n^2 D t} e^{\mathrm{i} k_n x} }[/math]
Note that this is not the same solution as we obtained on the infinite domain because of the boundary conditions on the finite domain.
The key to the numerical solution of this equation is the use of the
