For the Ricker population map $H(u)=ru{e}^{-u}$, fixed points satisfy $u=0$ or $r{e}^{-u}=1$. Thus zero exists for all $r>0$, while $u_{\ast }=\log r$ is positive exactly when $r>1$. Since $H^{\prime }(u)=r{e}^{-u}(1-u)$, the respective periodic-orbit multipliers are $r$ and $1-\log r$.

Consequently zero is linearly attracting for $0<r<1$ and repelling for $r>1$; the positive biological population is linearly attracting for $1<r<e^2$ and repelling for $r>e^2$. The periodic-orbit multiplier of the positive point is positive below $r=e$ and negative above it, distinguishing monotone from alternating local convergence. At $r=1$, the coincident zero state has unit periodic-orbit multiplier but $0<H(u)<u$ for every $u>0$, giving nonlinear attraction from the physical right side. At $r=e^2$ the periodic-orbit multiplier is $-1$ and linearization is inconclusive. Writing $u=2+\eta$, the second iterate has expansion $\eta \mapsto \eta -{\eta }^3/3+O({\eta }^4)$, so the critical point is weakly attracting; beyond it a period-doubling instability occurs.

Here $L_1=1$ and $L_2=1/2$, so a perturbation varying in both coordinates has $n,m\ge 1$ andWith the given unit coefficients the determinant becomes $D(s)=(1+s{)}^2-\beta s$. The trace remains negative, so instability occurs precisely when $D(s)<0$, orThe right-hand side is strictly increasing for $s>1$, since its derivative is $1-s^{-2}>0$. Its smallest allowed value is attained by the mode $(n,m)=(1,1)$, giving the onset thresholdAt this value the first mixed mode has a zero eigenvalue; it becomes unstable for $\beta >{\beta }_{\mathrm{crit}}$. Thus the critical value is the infimum of the strictly unstable range, rather than a strictly unstable point itself. Its spatial factor is $\cos (\pi x)\cos (2\pi y)$.

For the modal matrix $M(s)$, $s\ge 0$, its trace is always negative:Its determinant isThe proposed inequality is exactly $\beta \kappa \eta <\alpha \kappa +1$. It makes the coefficient of $s$ positive, while the leading and constant coefficients are already positive. Therefore $D(s)>0$ for every $s\ge 0$, independent of which discrete wave numbers the domain allows.

A real two-by-two matrix with negative trace and positive determinant has both eigenvalues with negative real part: if they are real they have positive product and negative sum, and if nonreal their common real part is half the trace. Every mode therefore decays, proving linear stability for all domain sizes under the stated sufficient condition.

For the cosine mode in the question, the Laplacian acts by multiplication by $-s$, whereThese modes obey the Neumann boundary conditions. In the chemotactic term, put $q=1+u$, $c=\eta +v$. Then $\nabla c=\nabla v$, and $q\nabla c=\nabla v+u\nabla v$; the second product is a quadratic polynomial in perturbation amplitude. Thus the linearized equations areSubstituting the modal time dependence givesSeparating this matrix into its transport part and the reaction Jacobian matrix gives the required form, with the same $J$ as above.