= Solution
The nonzero homogeneous equilibrium is $\boxed{n_*=1,\ a_*=\alpha/\beta}$. Without transport its <Jacobian matrix> is
$$
\begin{pmatrix}-1&0\\\alpha&-\beta\end{pmatrix},
$$
with negative <eigenvalues> $-1,-\beta$.
Zero chemical flux gives $a_x=0$ at the endpoints; zero bacterial flux then gives $n_x=0$. Linear perturbations therefore have <Neumann boundary conditions> and modes $\cos(k_mx)$, $k_m=m\pi$. For each mode, the amplitude matrix is
$$
J_m=\begin{pmatrix}-1-D_nk_m^2&\chi k_m^2\\\alpha&-\beta-D_ak_m^2\end{pmatrix}.
$$
Its trace is always negative. Its <determinant> is
$$
\boxed{\Delta(k)=D_nD_ak^4+(D_a+\beta D_n-\alpha\chi)k^2+\beta.}
$$
A mode grows exactly when $\Delta(k_m)<0$. In particular, any nonzero mode is destabilized by sufficiently large $\chi$. For continuously variable $k$, a nonempty unstable band exists precisely when
$$
\boxed{\alpha\chi>(\sqrt{D_a}+\sqrt{\beta D_n})^2.}
$$
Writing $C=\alpha\chi-D_a-\beta D_n$, its squared-wavenumber endpoints are $[C\pm\sqrt{C^2-4\beta D_nD_a}]/(2D_nD_a)$. On the finite interval, at least one allowed $k_m^2$ must lie inside this band. The uniform $m=0$ mode stays stable. This is the <chemotactic pattern-instability criterion>.
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