Modelo de Gray-Scott

Descrição do Modelo

O modelo de Gray-Scott descreve uma reação autocatalítica. Sejam duas substâncias químicas cujas concentrações em um dado ponto do espaço são dadas pelas variáveis $u$ e $v$, a reação pode ser representada como

$u+2v\to 3v$

Isso significa que uma molécula da substância $u$ é transformada em uma molécula da substância $v$ por meio da ação de outras duas moléculas da substância $v$, ou seja, $v$ é um catalisador de sua própria produção (daí o termo autocatálise). Além dessa reação, ambas substâncias se difundem pelo meio (por isso esse modelo pertence à classe mais geral de modelos reativos-difusivos) e, portanto, as concentrações $u$ e $v$ mudam com o tempo e diferem em cada ponto. Por simplicidade, assume-se que a reação reversa (i.e., $3v\to u+2v$) não ocorre. Há reposição de $u$ a uma taxa $F$ (taxa de alimentação, feed rate) e remoção de $v$ a uma taxa ligeiramente mais rápida do que a reposição de $u$.

O comportamento geral do sistema pode ser descrito pelas equações abaixo:

${\frac {\partial {u}}{\partial {t}}}=-uv^{2}+F(1-u)+D_{u}\nabla ^{2}u$

${\frac {\partial {v}}{\partial {t}}}=uv^{2}-(F+k)v+D_{v}\nabla ^{2}v$

The first equation tells how quickly the quantity u increases. There are three terms. The first term, Du∇2u, is the diffusion term. It specifies that u will increase in proportion to the Laplacian (a sort of multidimensional second derivative giving the amount of local variation in the gradient) of U. When the quantity of U is higher in neighboring areas, u will increase. ∇2u will be negative when the surrounding regions have lower concentrations of U, and in such cases the diffusion term is negative and u decreases. If we made an equation for u with only the first term, we would have ∂u/∂t = Du∇2u, which is a diffusion-only system equivalent to the heat equation.

The second term is -uv2. This is the reaction rate. The first reaction shown above requires one U and two V; such a reaction takes place at a rate proportional to the concentration of U times the square of the concentration of V6. Also, it converts U into V: the increase in v is equal to the decrease in u (as shown by the positive uv2 in the second equation). There is no constant on the reaction terms, but the relative strength of the other terms can be adjusted through the constants Du, Dv, F and k, and the choice of the arbitrary time unit implicit in ∂t.

The third term, F(1-u), is the replenishment term. Since the reaction uses up U and generates V, all of the chemical U will eventually get used up unless there is a way to replenish it. The replenishment term says that u will be increased at a rate proportional to the difference between its current level and 1. As a result, even if the other two terms had no effect, 1 would be the maximum value for u. The constant F is the feed rate and represents the rate of replenishment.

In the systems this equation is modeling, the area where the reaction occurs is physically adjacent to a large supply of U and separated by something that limits its flow, such as a semi-permeable membrane; replenishment takes place via diffusion across the membrane, at a rate proportional to the concentration difference Δ[U] across the membrane. The value 1 represents the concentration of U in this supply area, and F corresponds to the permeability of the membrane.

It is useful to note here that in biological systems, such as the skin of a developing embryo, this "supply" could be from the bloodstream, or from cells in an adjacent layer of tissue that continuously generate the needed chemical(s), with rates regulated by enzymes.

Comparing the two equations, the biggest difference is in the third term, where we have F(1-u) in one equation and (F+k)v in the other. The third term in the v equation is the diminishment term. Without the diminishment term, the concentration of V could increase without limit. In practice, V could be allowed to accumulate for a long time without interfering with further production of more V, but it naturally diffuses out of the system through the same (or a similar) process as that which introduces the new supply of U. The diminishment term is proportional to the concentration of V that is currently present, and also to the sum of two constants F and k. F, as above, represents the permeability of the membrane to U, and k represents the difference between this rate and that for V.

In the original stirred tank model, "F" stands for "feed" or "flow", and is the rate at which a pure-U supply is pumped into a tank. k is the rate at which the reaction V→P takes place; Karl Sims [13] calls it the "kill" rate9. The k times v being subtracted in the second equation is the result of the V→P reaction converting (and therefore "removing") some of the V chemical.

The tank does not fill or empty; there is an exit pipe that carries liquid out at the same rate as the inflow. This outflow is represented in the equations by the minus F times u in the first equation and the minus F times v in the second equation.

Since the tank does not fill or empty, during the time it takes for 1 unit of U to be pumped in, the outflow is a mixture of U, V, and P adding up to 1 unit. Since the tank is stirred continuously, the outflow represents the current ratio of the three chemicals in the tank. If the current concentration of U, V, and P in the tank are u, v, and p respectively, then u+v+p = 1 (where 1 represents 100% concentration). None of them can be negative (examining the equations makes this clear) and since they add up to 1, none of them can be greater than 1. Also, since there are three variables adding up to 1, there are two degrees of freedom: u and v can be anywhere in the triangular region bounded by u = v = 0 at one corner, (u,v) = (0,1) at a second corner, and (u,v) = (1,0) at the third.

There is nothing in the equations that states whether the system exists in a two-dimensional space (like a Petri dish) or in three dimensions, or even some other number of dimensions. In fact, any number of dimensions is possible, and the resulting behavior is fairly similar. The only significant difference is that in higher dimensions, there are more directions for diffusion to happen in and the first term of the equation becomes relatively stronger. It is for this reason that phenomena depending on diffusion for their action (such as gradient-sustained stable "spots") occur at higher k values in the 2-D system as compared to the 1-D system, and at yet higher values for the 3-D system (this relationship can be seen in Lidbeck's Java simulator1). The use of a uniform grid is not essential either, for example see the "amorphous layout" of the simulations by Abelson et al.2

The original Gray-Scott model, with only a single quantity for the concentration of U and V that is equal throughout the tank, can effectively be thought of as the zero-dimensional case. In the 2-D simulations shown here, the "zero-dimensional" behavior is seen in any oscillatory or dynamic phenomena that have a long wavelength in space. For example, at F=0.026, k=0.049, the U changes to V pretty quickly, but there is a sort of damped oscillation as the system approaches the steady-state equilibrium concentrations; and the same is seen in the original Gray-Scott system.

Análise de estabilidade

Nota: A análise em toda esta seção pressupõe sempre que os parâmetros e coeficientes de difusão são positivos.

Estado de Equilíbrio Trivial

O modelo de Gray-Scott depende dos parâmetros $F,k$ e dos coeficientes de difusão $D_{u},D_{v}$. É fácil mostrar que, ignorando os termos de difusão, o sistema possui estado de equilíbrio estável em $(u^{*},v^{*})=(1,0)$ para quaisquer valores dos parâmetros.

Demonstração. O sistema de equações do modelo, com ${\frac {\partial u}{\partial t}}=0$ e ${\frac {\partial v}{\partial t}}=0$, fazendo $D_{u}=D_{v}=0$, é dado por

$0=-uv^{2}+F(1-u)$

$0=uv^{2}-(F+k)v\quad (1)$

Logo, é trivial que o sistema acima é satisfeito quando $(u^{*},v^{*})=(1,0)$. Esse estado de equilíbrio é estável porque a matriz jacobiana $J|_{(u^{*},v^{*})=(1,0)}=\left({\begin{array}{cc}-F&0\\0&-F-k\end{array}}\right)$ possui traço negativo e determinante positivo.

Se agora incluímos os termos de difusão $D_{u}$ e $D_{v}$, deve-se levar em consideração a matriz $\left(J-D\omega ^{2}\right){\Bigg |}_{f=f_{eq}}$. Aqui, $J$ é a matriz jacobiana dos termos de reação, $D$ é a matriz diagonal dos termos de difusão e $\omega$ é o parâmetro que determina a frequência espacial das perturbações. A demonstração da validade desse método pode ser encontrada na referência. Aplicando ao modelo de Gray-Scott em $(u^{*},v^{*})=(1,0)$:

$\left(\left({\begin{array}{cc}-F-v^{2}&-2uv\\v^{2}&-F-k+2uv\end{array}}\right)-\left({\begin{array}{cc}D_{u}&0\\0&D_{v}\end{array}}\right)\omega ^{2}\right){\Bigg |}_{(u^{*},v^{*})=(1,0)}=\left({\begin{array}{cc}-F-D_{u}\omega ^{2}&0\\0&-F-k-D_{v}\omega ^{2}\end{array}}\right)$

Para que o estado de equilíbrio $(u^{*},v^{*})=(1,0)$ seja estável é necessário que o determinante da matriz acima seja positivo e o traço seja negativo. Obtém-se então

$(F+D_{u}\omega ^{2})(F+k+D_{v}\omega ^{2})>0$

$-2F-(D_{u}+D_{v})\omega ^{2}-k<0$

Ambas desigualdades são imediatamente satisfeitas para quaisquer valores de $F,k,D_{u}$, e $D_{v}$. Portanto, o estado de equilíbrio $(u^{*},v^{*})=(1,0)$ permanece estável no modelo de Gray-Scott mesmo após a inclusão dos coeficientes de difusão, sejam quais forem os valores desses coeficientes (lembrando que estamos nos restringindo a valores positivos dos parâmetros e coeficientes).

Esse é um resultado à primeira vista surpreendente. Em geral, o surgimento de padrões complexos e não homogêneos em sistemas reativos-difusivos está relacionado à desestabilização de um ou mais estados de equilíbrio homogêneo causada pela introdução dos coeficientes de difusão (conhecida como instabilidade de Turing).

Entretanto, no caso do modelo de Gray-Scott, o surgimento de padrões complexos e não homogêneos não decorre da instabilidade de Turing, uma vez que o surgimento de padrões não triviais nesse modelo ocorre mesmo quando apenas o estado de equilíbrio trivial $(u^{*},v^{*})=(1,0)$ está presente .

Estados de Equilíbrio Não Triviais

Há outros dois estados de equilíbrio que são soluções não triviais do sistema de equações (1). Desde que seja obedecida a condição $F\geq 4(F+k)^{2}$, esses estados são $(u_{+},v_{-})$ e $(u_{-},v_{+})$, com

$u_{\pm }={\frac {1}{2}}(1\pm {\sqrt {1-4\gamma ^{2}F}})$

$v_{\mp }={\frac {1}{2\gamma ^{2}}}(1\mp {\sqrt {1-4\gamma ^{2}F}})$

$\gamma ={\frac {F+k}{F}}$