Past exam of the topics in applied mathematics course of the University of Oxford / 2007 / A / 4

At a homogeneous equilibrium, add the two reaction equations: $f+g=1-b=0$, so $b=1$. Then $f=0$ gives $a=3/4$. Thus $\boxed{(a_{\ast },b_{\ast })=(3/4,1)}$.

The reaction Jacobian isIts trace is $-1/2$ and determinant is $1$, so both eigenvalues have negative real parts. Explicitly they are $(-1\pm i\sqrt{15})/4$. Therefore the homogeneous reaction equilibrium is a stable focus in the absence of diffusion.

The cosine modes satisfy both zero-flux endpoint conditions and ${\partial }_x^2\cos (k_{n}x)=-k_n^2\cos (k_{n}x)$. Linearizing the reaction terms gives $f_a\delta a+f_b\delta b$ and $g_a\delta a+g_b\delta b$. Cancelling the common mode factor yieldsThe derivatives are exactly the entries of the matrix computed in (ii). The zero mode reproduces the stable homogeneous perturbation; positive integers give the possible spatial instabilities.

Put $q=k_n^2$. The mode matrix has trace $-1/2-(1+D)q<0$, so a growing eigenvalue occurs exactly when its determinant is negative:This upward quadratic has two positive roots only if $D<1/2$ and $D^2-5D+1/4>0$. Within $0<D<1$ this isFor such $D$, defineThe unstable modes satisfy $\boxed{\sqrt{q_{-}}<k_n<\sqrt{q_{+}}}$. These are strict inequalities: at either endpoint the leading eigenvalue is zero rather than positive.

If $D\ge 1/2$, both nonconstant coefficients of $\Delta (q)$ are nonnegative, so it is positive for $q\ge 0$. If $D_c<D<1/2$, its discriminant is negative, so it is again strictly positive. Therefore $\boxed{D>D_c⟹\text{no diffusion-driven instability}}$.

At $D=D_c$ the double root can only give a neutral zero eigenvalue if it coincides with an allowed spatial mode; it still gives no positive growth rate. This is the finite-wavenumber onset of a Turing instability, not an oscillatory crossing since the mode trace stays negative.

Small $L$ makes every nonzero $k_n=n\pi /L$ large, placing them above the finite instability band, where diffusion dominates reaction. The homogeneous zero mode is already stable.

For $0<D<D_c$, the first possible band entry as $L$ increases is $n=1$ crossing $k_1=\sqrt{q_{+}}$. Hence the critical length isAt equality the mode is neutral; it is unstable just above this value, while $\pi /\sqrt{q_{+}}<L<\pi /\sqrt{q_{-}}$. Thus strictly positive growth has infimum $L_c$, with no attained smallest length if “unstable” means $\mathrm{Re}\lambda >0$. The conventional threshold answer is $L_c$. Larger modes cannot enter earlier because their first-entry lengths are $n{L}_c$.