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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

Letbe the reaction Jacobian at the homogeneous equilibrium. Stability without diffusion givesFor a spatial Fourier mode with $-{\nabla }^2$ eigenvalue $k^2$, the linearized matrix isIts trace is even more negative than $\mathrm{tr}M$, so instability occurs exactly when its determinant becomes negative. Put $z=D_v{k}^2$ and $d=D_u/D_v$. ThenThis upward-opening quadratic is negative for some $z>0$ exactly when its minimum occurs at positive $z$ and lies below zero:The second strict inequality implies the first when written with a positive square root, so the two-species diffusion-driven instability criterion becomes

Write the reaction terms asA nonzero homogeneous equilibrium satisfies $v_{\ast }=\beta u_{\ast }$ and $1+u_{\ast }-\gamma v_{\ast }=0$, henceFor physically positive populations it exists exactly when

The reaction Jacobian matrix at this equilibrium isIts trace and determinant areThe equilibrium is therefore stable to spatially uniform perturbations when

For a spatial Fourier mode of wavenumber $k$, put $q=k^2$. The linearized reaction-diffusion system has matrixIts trace is smaller than $\mathrm{tr}J$, whileThe two-species diffusion-driven instability criterion says that this upward-opening quadratic becomes negative for some $q>0$ precisely whenCombining all conditions, a Turing instability may occur in the region

At onset the discriminant vanishes, and the double root isThe threshold relation givesso the critical wavenumber isIf $d=1$, uniform stability requires $\beta >1$ whereas diffusion-driven instability requires $d>\beta$. These inequalities are incompatible, so the Turing region vanishes.

If $D=1$, both species have the same diffusion coefficient. For a perturbation proportional to $e^{\sigma t+ikx}$, diffusion replaces $J$ by $J-k^{2}I$, shifting both reaction eigenvalues to the left by $k^2$. Since the homogeneous equilibrium is already stable, every spatial mode remains stable.

For general $D$, put $q=k^2$. The mode matrix isIts trace iswhile its determinant isWith negative trace, instability occurs exactly when $\Delta (q)<0$ for some $q>0$. The two-species diffusion-driven instability criterion says that this upward-opening quadratic has a negative minimum precisely whenEquivalently, the condition isThis is the Turing instability region.