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Use the homogeneous equilibrium point condition $k({\rho }_0){\rho }_0=a_{0}f({\rho }_0)$. Evaluate $D_1,D_2$ at $(a_0,{\rho }_0)$ and abbreviateThe quantity $c$ is the chemical relaxation rate with the cell density held fixed. Derivatives of $D_1,D_2$ do not enter this linear stability analysis: they multiply spatial derivatives of the homogeneous background or products of perturbations. For a Fourier mode with time dependence $e^{st}$, put $z=|\mathbf{q}{|}^2$ and writeConsequentlyAt $z=0$, the eigenvalues are $0$ and $-c$. The neutral cell-density mode reflects mass conservation, not decay of every homogeneous perturbation.

Write $a=D_2>0$, $b=D_{\rho }>0$, $C=D_1{f}_0$, and interpret production physically as $f_0\ge 0$. If $c\ge 0$, the trace-determinant stability criterion shows that an unstable nonzero wavenumber exists precisely when $C>ac$. The unstable band isIf $c<0$, the chemical field is already unstable at zero wavenumber; arbitrarily small positive wavenumbers are unstable too. Since $C\ge 0$, this also implies $C>ac$. Thus, on the infinite plane, the undivided Keller--Segel aggregation threshold isFor $\kappa >0$, dividing by $D_2\kappa$ gives the requested formAt equality there is no strictly growing mode when $c\ge 0$; nonzero spatial modes decay. On a finite domain, a permitted nonzero wavenumber must actually lie in the unstable band. If one allows a signed production function, the displayed matrix and trace-determinant stability criterion remain valid, but the simplification using $C\ge 0$ must be revisited.

The printed hypotheses do not ensure $\kappa >0$. Positivity of the degradation rate $k$ alone is insufficient. For a concrete counterexample, take $k(\rho )=e^{-2\rho }$, $f(\rho )=e^{-2}$, $a_0={\rho }_0=1$, and $D_1=D_2=D_{\rho }=1$. The homogeneous equilibrium point condition holds, all rates and transport coefficients are positive, but $\kappa =c=-e^{-2}$. The printed left-hand side equals $-1$, although $\det M=z(z-2e^{-2})<0$ for $0<z<2e^{-2}$. These spatial modes grow. If $\kappa =0$, the printed expression is undefined. The undivided criterion and the matrix above resolve both cases.

To find the fastest-growing Keller--Segel mode, use the larger eigenvalueFor $C>0$ the square root is real. Put $p=a+b$ and $d=b-a$. Differentiation givesA positive maximizing wavenumber exists in either of two growing cases: $c\ge 0$, $C>ac$, or $c<0$, $C>-bc$. Equivalently, $C>\max (ac,-bc)$. In this range $C+dc>0$, andThus the stationary point is the unique maximum. The derivative condition yieldsSolving it, with the root that satisfies the unsquared derivative equation, givesFor $c=0$, the unequal-diffusivity expression simplifies to $z_{\ast }=C/[\sqrt{ab}(\sqrt{a}+\sqrt{b}{)}^2]$, also agreeing with the equal-diffusivity limit. The formula maximizes the full two-field growth rate; it makes no instantaneous-chemical approximation. Every direction of $\mathbf{q}$ with this length is equivalent by rotational symmetry.

If $c<0$ but $0\le C\le -bc$, the maximum instead occurs at $z=0$: the fastest mode is homogeneous, with $s_{\ast }=-c$ and infinite wavelength. For $C>0$, its initial derivative is $-b+C/(-c)\le 0$. If $C+dc\ge 0$ the derivative thereafter decreases, while if $C+dc<0$ it increases towards the still-negative large-$z$ limit; either way no positive-$z$ maximum is missed. At $C=0$, the two eigenvalues are simply $-az$ and $-c-bz$, giving the same conclusion. In a finite box, maximize $s_{+}$ over the allowed Fourier modes; excluding the homogeneous mode can change the selected length. In a stable parameter range there is no fastest-growing mode.

The first ratio compares the positive feedback loop “more cells produce more attractant, which draws in more cells” with spreading by cell diffusion and removal of chemical perturbations. Its numerator $D_1{f}_0$ measures chemotaxis together with attractant production; its denominator $D_2\kappa$ measures dispersal together with incremental degradation. This competition between directed chemotaxis and cell diffusion produces spatial aggregation. The second ratio compares the concentration dependence of chemical production, $a_0{f}_0^{\prime }$, with incremental degradation $\kappa$. Positive $f_0^{\prime }$ amplifies chemical fluctuations, whereas negative $f_0^{\prime }$ suppresses them. It changes the chemical relaxation available to the chemotaxis feedback loop and can itself destabilize the homogeneous chemical field. Chemical diffusion $D_{\rho }$ suppresses short scales and sets the selected wavelength, but does not change the infinite-plane long-wave threshold. These interpretations as ratios of stabilizing and destabilizing processes assume $\kappa >0$.

The $u^2/v$ term makes $u$ locally autocatalytic, while increasing $v$ suppresses this production. Thus $u$ is the activator and $v$ is the inhibitor.

At a positive spatially homogeneous equilibrium,Hence $u=2bv$ and $v=u^2$, givingThe reaction Jacobian matrix there isBecause $b>0$, the trace-determinant stability criterion shows that the spatially uniform reaction kinetics are stable exactly when

For a perturbation proportional to $e^{\lambda t+ikx}$, put $q=k^2$. Its growth rates are the eigenvalues ofThe trace is $2b-1-(d+1)q<0$ in the reaction-stable regime, while the determinant isA Turing instability occurs exactly when this upward-opening quadratic is negative for some $q>0$. By the two-species diffusion-driven instability criterion, this requiresTogether these reduce toThus the unstable region in the $(b,d)$-plane lies below the straight line $d=(6-4\sqrt{2})b$ and to the left of $b=1/2$.

Inside this region, the unstable squared-wavenumber band isAt the bifurcation boundary the two roots coincide, soSubstituting $d=(6-4\sqrt{2})b$ yields

Eliminate $R=N-S-I$ to obtain the two-dimensional autonomous differential equationIts Jacobian matrix at the endemic equilibrium isSince $I^{\ast }>0$,The trace-determinant stability criterion shows that both eigenvalues have negative real part. The Endemic equilibrium of the SIR model with waning immunity is therefore locally asymptotically stable.

At an endemic equilibrium, $I^{\ast }>0$. The equation $\dot{I}=0$ then givesSince also $S^{\ast }>0$, the equation $\dot{S}=0$ givesThis infective population is positive precisely whenwhich is exactly the assumed instability condition for the disease-free equilibrium. Thus the endemic equilibrium of the susceptible-infective model with exponential birth exists.

At this equilibrium the relations above reduce the Jacobian matrix toHenceBy the trace-determinant stability criterion, both eigenvalues have negative real part. The linearization stability theorem therefore proves that the endemic equilibrium is locally asymptotically stable.