A Candle in the Dark

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How the leopard may have really got her spots

with 3 comments

A mechanism of pattern formation in many animals, (including the zebra, leopard and the giraffe), was first suggested by Alan Turing in 1952.

To understand his mechanism, we’ll first look at a system where two reactive chemical species (morphogens) are present and they do not move about in space(ie, they don’t diffuse). If the concentration of the species are denoted by u and v, the rate at which the concentrations change will simply be the rate at which they react. That is,

\frac{\partial u}{\partial t} = f(u,v)

\frac{\partial v}{\partial t} = g(u,v)

where, f(u,v) and g(u,v) describe the rate law kinetics which the species obey. If we want to find the steady state concentrations of the species, we simply set the partial derivatives with time as 0, and solve for u and v.

What happens if f(u,v) and g(u,v) are non-linear equations with multiple solutions? The stability of each solution can be determined by a linear stability analysis, and the final stable solution will depend on the initial condition. Indeed, because of the homogeneity in the initial constraint, that the initial concentrations of the species are uniform everywhere and the they are spatially fixed, the final concentrations at steady state are again uniform in space.

Linearizing the above system of equations (ie, set u=u_{ss}+\tilde{u} and v=v_{ss}+\tilde{v}), one obtains

\left(\begin{array}{c} \frac{d \tilde{u}}{dt} \\ \frac{d \tilde{v}}{dt} \end{array}\right)=\left(\begin{array}{cc} f_u & f_v \\ g_u & g_v \end{array}\right) \left(\begin{array}{c} \tilde{u} \\ \tilde{v}\end{array}\right)

For the steady states to be stable, the condition is that the real parts of the eigenvalues of the jacobian are negative. For a two component system, this is true when the trace is negative and the determinant positive.

Now, what happens if we remove the spatial constraint and allow the chemical species to diffuse? The equations which describe the concentration of the species will be modified to account for diffusion. We’ll now have

\frac{\partial u}{\partial t} = D_u \nabla^2 u + f(u,v)

\frac{\partial v}{\partial t} = D_v \nabla^2 v + g(u,v)

where, D_u and D_v are the diffusivity of the chemical species. One of the observations which Turing made, was that under certain conditions, steady states which were initially spatially uniform could now show spatial variation due to diffusion. The natural question would be, under what conditions would we get these spatial variations? To answer that, we need to delve a bit into linear stability analysis.

However, before doing that, I’ll have to define what the boundary conditions and the geometry of the system is for the second set of equations. A very simple system would be a one-dimensional strip of space, where diffusion and reaction occurs. To make this system ‘self-contained’, we set the flux of the chemical species at the boundary to be 0.

Therefore, the system of equations are

\frac{\partial u}{\partial t} = D_u \frac{\partial^2 u}{\partial x^2} + f(u,v)

\frac{\partial v}{\partial t} = D_v \frac{\partial^2 v}{\partial x^2} + g(u,v)

Since we want to find out when the homogeneous solutions are unstable, we need to look at the linearized system in terms of the deviation variables from the steady state solutions, \tilde{u} = u - u_{ss} and \tilde{v} = v - v_{ss}. With these variables, the linearized equations are

\frac{\partial \tilde{u}}{\partial t} = D_u \frac{\partial^2 \tilde{u}}{\partial x^2} + f_u \tilde{u} + f_v\tilde{v}

\frac{\partial \tilde{v}}{\partial t} = D_v \frac{\partial^2 \tilde{v}}{\partial x^2} + g_u \tilde{u} + g_v\tilde{v}

The solutions to this equation can be shown to be of the form.

\tilde{u} = u^* e^{\sigma t} \sin{\alpha x}

\tilde{v} = v^* e^{\sigma t} \sin{\alpha x}

(Why? Substitute and check!)

Substituting to the equation, we get

u^* (\sigma +D_u\alpha^2 - f_u)  -f_v v^* = 0

v^* (\sigma +D_v\alpha^2 - g_v)  -g_u u^* = 0

Here’s the important argument. For a spatially varying pattern, we require non-zero solutions to u^* and v^*. The condition for that is

\left|\begin{array}{cc} \sigma +D_u\alpha^2 - f_u & -f_v \\ -g_u & \sigma +D_v\alpha^2 - g_v  \end{array}\right| = 0

This can be re-written as

det( \sigma I - A + \alpha^2 D) = 0


A= \left(\begin{array}{cc} f_u & f_v \\ g_u & g_v \end{array}\right)

D = \left(\begin{array}{cc} D_u & 0 \\ 0 & D_v \end{array}\right)

Now, the value of \sigma is the eigenvalue of the matrix R= (A - \alpha^2 D), and for the homogeneous solution to be unstable, one requires that the real part of the eigenvalues of R are positive for some \alpha^2

This condition can be simplified to,

\sigma^2 - Tr(R) \sigma + det(R) =0

For the eigenvalue to be positive, we require that det(R)<0 and Tr(R)>0 .

But, since Tr(R) = Tr(A - \alpha^2 D) = Tr(A) - \alpha^2 Tr(D) and Tr(A) is less than 0 because the homogeneous steady state is assumed to be stable, Tr(R) can never be positive. So, the criteria for instability of the homogeneous solution is that det(R)<0. One can also relate this to the ratio of the diffusivites of the two species u and v.

This means that under this condition, the homogeneous steady state solution is unstable, and the concentrations of u and v will increase till the non-linear terms we have neglected plays a role and causes spatial patterns.

I will edit and add pretty pictures displaying the simulation of this phenomenon once I figure out how to use the pdetool in MATLAB to solve non-linear parabolic partial differential equations in 2 variables.

The Chemical Basis of Morphogenisis, A.M Turing, Philosophical Transactions of the Royal Society of London. Series B, Biological Sciences, Vol. 237, No. 641. (Aug. 14, 1952),pp 37-72

Two-stage Turing model for generating pigment patterns on the leopard and the jaguar, Phys. Rev. E 74, 011914 (2006)

Mathematical Biology, JD Murray, 3rd edn, Springer (2002). chapter 2.4 &2.5


Written by parseval

April 14, 2008 at 4:28 pm

3 Responses

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  1. Spiffy! I’d been meaning to write about this stuff. Thanks, also, for the pointer to that Phys. Rev. E paper.

    Blake Stacey

    April 19, 2008 at 7:35 pm

  2. *De-Lurks to say:

    Oh, Brava! Eigenvalues and morphogens and kittens, oh my! Thank you for writing this and linking to the papers involved. You know, this inspires a LOL-cat macro or two…. would you mind terribly if I indulged?

    Maxwell's Demoness

    May 2, 2008 at 11:55 pm

  3. Feel free to indulge yourself.


    May 2, 2008 at 11:59 pm

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