Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 14A (Mathematical Biology)

For a positive homogeneous equilibrium,sorequiring $\alpha >\beta$. The reaction Jacobian there isIts determinant is ${\beta }^2/(\alpha -\beta )>0$, while its trace is $\beta (1-\beta )/(\alpha -\beta )$. Stability therefore requiresThis is the region above the line $\alpha =\beta$ and to the right of $\beta =1$ in the positive quadrant.

For $q=k^2$, the mode matrix is $J-q\mathrm{diag}(D_1,D_2)$. Its trace remains negative. Its determinant isIt is negative for some $q>0$ exactly whenTogether with homogeneous stability, the spatial-instability region isFor fixed $\beta$, this is the region below the last curve, above $\alpha =\beta$, and left of $\lambda =1/\beta$.

If $D_1=D_2=D$, the linearized matrix for wavenumber $k$ isIts eigenvalues are those of the homogeneous Jacobian shifted left by $D{k}^2$, so a stable homogeneous equilibrium remains stable for every spatial mode. Thus equal diffusivities cannot produce a Turing instability. In part (b), equal diffusivities mean $\lambda =1$; because homogeneous stability requires $\beta >1$, the necessary condition $1-\lambda \beta >0$ fails, consistently ruling out instability.