Introduction

The Gray-Scott system is a reaction-diffusion system. This means that it models a process that consists of a reaction and diffusion. In the case of the Gray-Scott model the reaction is a chemical reaction between two substances \(U\) and \(V\), both of which diffuse over time. In the reaction \(U\) gets used up, while \(V\) is produced. The densities of the substances \(u\) and \(v\) are what is represented in the simulation.

The system is characterized by two parameters: \(F\) is the rate at which \(U\) is replenished and \(k\) controls the rate at which \(V\) is removed from the system. Varying these parameters leads to a wide range of interesting patterns, some of which look quite familiar.

Content

Overview

The Gray-Scott system models the chemical reaction \begin{equation} U+2V \rightarrow 3V\ . \\ \end{equation} This reaction consumes \(U\) and produces \(V\), so the amount of both substances needs to be controlled to maintain the reaction. This is done by adding \(U\) at the feed rate \(F\) and removing \(V\) at the "kill rate" \(k\). The removal of \(V\) can also be desribed by another chemical reaction, \begin{equation} V\rightarrow P\ ,\\ \end{equation} where \(P\) is an inert product, meaning it doesn't react and therefore does not contribute to our observations [1]. In this case the Parameter \(k\) controlls the rate of the second reaction.

Both substances diffuse over time at the diffusion rates \(D_u\) and \(D_v\). In this simulation \(D_u\) is tiwce as high as \(D_v\). This is simply to ensure that patterns can actually form.

You can control the feed rate \(F\) and the kill rate \(k\). Most possible combinations of the two will not lead to any patterns, which is why we have given you a number of presets that lead to interesting results. Simply choose one of these and observe what happens. Of course you can also experiment yourself, using the sliders for \(F\) and \(k\).

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Simulation

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Description

Theory

The Gray-Scott system is defined by two equations that describe the behaviour of two reacting substances: \begin{align} \dot{u}&=-uv^2+F(1-u)+D_u\Delta u\\ \dot{v}&=uv^2-(F+k)v+D_v\Delta v\\ \end{align} The variables in these equations \(u\) and \(v\) are the concentrations of the two reacting substances \(U\) and \(V\). On the left-hand side of each equation is the time derivative of one of these concentrations, describing the rate at which it changes. The right-hand sides of the equations both contain three seperate terms. The first desribes the reaction between the two substances. Since one \(U\) and two \(V\) react the corresponding term includes \(u\) to the power of one and \(v\) to the power of two: \(uv^2\). As \(U\) gets consumed by the reaction, the term has a negative sign in the first equation, and it has a positive sign in the second equation, as \(V\) is produced in the reaction.

The second term of the first equation describes the rate at which \(U\) is replenished externally. This is necessary, as \(U\) would otherwise simply be used up. The feed rate is given by the parameter \(F\). \(F\) is multiplied by \((1-u)\) to ensure that \(u\) is replenished at a rate dependent on the current concentration, which never exeeds \(1\). \(V\) doesn't need to be replenished, since it is produced in the reaction. Instead, it needs to be removed in order maintain the reaction. The rate of removal, the "kill rate", is controlled by the parameter \(k\). \(k\) is added to \(F\) so \(V\) is removed faster than \(U\) is added, and then mutiplied by \(v\), since the removal of \(V\) is also supposed to be dependant on it's concentration.

The last term in both equations describes the diffusion of \(U\) and \(V\) respectively.

Phase diagram of the Gray-Scott system.[1]

As you can see in the simulation above, there is a variety of patterns that can form for different values for \(F\) and \(k\). However, most possible combinations of values for those parameters lead to rather boring results: For combinations of \(F\) and \(k\) on one side of a certain threshold the screen is completely white, indicating that it is completely filled with substance \(U\). There is no substance \(V\) left to react with, and so the system is in equilibrium. The system will always return to this state for these values of \(F\) and \(k\).

For combinations of \(F\) and \(k\) on the other side of this threshold the screen completely fills with black and stays that way. Virtually all of the added substance \(U\) reacts immediately, so the area is almost completely filled with substance \(V\). In this case the system is in equilibrium, too, and will always return to this state.

The interesting region, the region we're looking at in the simulation, is close to this threshold. Here the system changes from one stable state to the other: The system potentially has three fixed points. Which ones of these actually exists depends on the parameters \(F\) and \(k\): The trivial fixed point of the system is at \(u=1\) and \(v=0\), and it always exists, since the reaction cannot take place without \(v\). The two non-trivial fixed points exist beyond a bifurcation line at \(F=4(F+k)^2\)(the solid line in the figure). One of these is always a saddle, meaning it is stable in one direction and unstable in the other. The other one changes it's stability at the dotted line. [2] These changes in number and stability of fixed points are what creates the patterns.

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Example Pictures

Some snapshots of the patterns after they filled the whole simulation-frame.

Flower-pattern: \(F\)=0.055 ; \(k\)=0.062

Mazes-pattern: \(F\)=0.029 ; \(k\)=0.057

Mitosis-pattern: \(F\)=0.028 ; \(k\)=0.062

Solitons-pattern: \(F\)=0.03 ; \(k\)=0.06