Butterfly Wing Patterns: Saddle-node Bifurcation Problem

DK WC

10/14/2019

Butterfly wing patterns is one of most typical biological pattern formation. Explaining the development of these patterns of gene G switching from inactive to active status by a biochemical signal substance S can refer to the paper of Lewis et al. (1977). Gene may normally be inactive but can be switched on to produce a pigment or other gene product when the concentration of S exceeds a certain threshold. Let \(g(t)\) denote the concentration of the gene product, and assume that the concentration \(s_{0}\) of S is fixed. The model is \[\frac{dg}{dt} = k_{1}s_{0} -k_{2}g + \frac{k_{3}g^{2}}{k^{2}_{4} +g^{2}}\], which can be transformed to \[\frac{dx}{d\tau} = s-rx + \frac{x^{2}}{1+x^{2}}\], where \[\begin{align} x &=\frac{g}{k_{4}} \\ s &=\frac{k_{1}s_{0}}{k_{3}} \\ r &=\frac{k_{2}k_{4}}{k_{3}} \\ \tau &=\frac{k_{3}}{k_{4}} t \end{align}\]

There are three real fixed points: one is at zero, the other two real fixed points exit when \(1-4r^{2} \geq 0\) or \(r\leq r_{c}=\frac{1}{2}\) for all positive constants of k.

\[\begin{equation} x = \left\{ \begin{array}{cc} 0 :& (un)stable/semistable \\ \\ \frac{1+ \sqrt{1-4r^{2}}}{2r} :& unstable \\ \frac{1- \sqrt{1-4r^{2}}}{2r} :& stable \end{array} \right. \end{equation}\]

Interpretation: if increasing s from 0, and then returning back s = 0, what happen to the gene concentration or x :

  • for \(r> 1/2\), the semistable fixed point \(x=0\) increases very slowly with the increment of s; and decreases very slowly with the decrement of s when back to s= 0. Therefore, when s increases from 0 and then decreases back to s= 0, the gene concentration still remains zero and inactivated or still at \(x=0\).
  • for \(0\leq r< 1/2\), the stable fixed point of \(x=\frac{1- \sqrt{1-4r^{2}}}{2r}\) shifts to left(decreasing) with increment of s ; then shifts back to right(increasing) with decrement of s. Therefore, when s increases from 0 and then decreases back to s= 0, the gene concentration start decreasing from s = 0 and then increasing when back to s=0.
  • for \(r = 1/2\) the system will jump between the stable fixed point of \(x=0\) and the semi-stable fixed point at \(x=\frac{1}{2r} = 1\) and back to stable fixed point of \(x=0\) finally. The change rate of jump process depends on the change rate of s. Therefore, when s increases from 0 and then decreases back to s= 0, the gene concentration start to jump from 0(inactive) to 1(active) and back to 0(inactive). The switching rate of gene concentration depends on the change rate of s.

The saddle-node bifurcation curve in (r, s) space is as follow:

\[ \begin{equation} \left\{ \begin{array}{rl} s &= \frac{x^{2}(1-x^{2})}{(1+x^{2})^{2}} \\ \\ r &= \frac{2x}{(1+x^{2})^{2}} \end{array} \right. \end{equation}\]