Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 3 / 13B / Solution

For a small Fourier mode $u=U{e}^{ikx+\sigma t}$, $v=V{e}^{ikx+\sigma t}$, the inhibitor equation gives $V=U/(1+k^2)$. The linearization of the cubic reaction at zero is $-rU$. Thus the fast-inhibitor cubic activator growth rate isIt is an even function of $k$, starts at $\sigma (0)=-r<0$, and tends to $-\infty$ as $|k|\to \infty$. In terms of $q=k^2$, its derivative is $-D+\rho /(1+q{)}^2$. If $\rho >D$, the maxima occur at $k=\pm \sqrt{\sqrt{\rho /D}-1}$, withIf $\rho \le D$, the maximum is $-r$ at $k=0$ and all modes decay. For $\rho >D$, the zero state is unstable when $(\sqrt{\rho }-\sqrt{D}{)}^2>r$, and the stability boundary iswhere this curve lies in the allowed parameter region.

Putting $\tilde{u}=1-u$, $\tilde{v}=1-v$ and $\tilde{r}=1-r$ negates both original equations and restores precisely the same form, because $\tilde{u}(\tilde{u}-\tilde{r})(\tilde{u}-1)=-u(u-r)(u-1)$. Therefore the state $u=v=1$ has the corresponding boundary $\boxed{\rho =(\sqrt{1-r}+\sqrt{D}{)}^2}$.