To first order about the fixed point,
$$
\boxed{u_t=D_nu_{xx}-\chi n_*v_{xx}-\delta u},
\qquad
\boxed{v_t=D_cv_{xx}+\alpha u-\beta v}.
$$
For a mode proportional to $e^{\sigma t+ikx}$, the linear matrix is
$$
M(k)=\begin{pmatrix}
-\delta-D_nk^2&\chi n_*k^2\\
\alpha&-\beta-D_ck^2
\end{pmatrix}.
$$
Its trace is always negative, so instability occurs exactly when its determinant is negative. Writing $q=k^2$,
$$
\det M=D_nD_cq^2+
\left(\delta D_c+\beta D_n-\frac{\alpha\chi\gamma}{\delta}\right)q
+\delta\beta.
$$
At onset this quadratic touches zero, requiring
$$
\boxed{\chi_c=\frac\delta{\alpha\gamma}
\left(\sqrt{\delta D_c}+\sqrt{\beta D_n}\right)^2}.
$$
The repeated positive root is
$$
q_c=\sqrt{\frac{\delta\beta}{D_nD_c}},
$$
and therefore
$$
\boxed{k_c=\left(\frac{\delta\beta}{D_nD_c}\right)^{1/4}}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 355 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-355/2/b/solution