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For the Brusselator, the stationary spatial threshold is $b_T=(1+a\sqrt{D_u/D_v}{)}^2$ and the homogeneous oscillatory threshold is $b_H=1+a^2$. A genuine Turing instability appears first only if $b_T<b_H$, which rearranges to the displayed diffusivity-ratio bound. Equality gives simultaneous stationary spatial and homogeneous oscillatory neutral modes. Otherwise homogeneous loss of stability occurs first.

On a specified finite domain, only its boundary-compatible wavenumbers can grow. Solve the Brusselator mode determinant for its neutral value of $b$ and minimize over those nonzero modes. Minimizing over all continuous $k$ instead gives the usual threshold and $k_c^2=a/\sqrt{D_u{D}_v}$. The finite-mode minimum must still precede the homogeneous threshold to describe a Turing instability.

By the law of mass action, the net changes in the intermediate concentrations areThe autocatalytic reaction increases $X$ by one and decreases $Y$ by one; its reaction-rate convention is $k_3{X}^{2}Y$. With the specified common concentration scale and time scale, chooseSubstitution gives the nondimensional Brusselator system. Adding its two steady equations gives $u=1$, and then the second gives $v=b/a$. Its Jacobian matrix there isThe equilibrium is asymptotically stable for $b<1+a$ and unstable for $b>1+a$. At $b_c=1+a$ the eigenvalues are $\pm i\sqrt{a}$, crossing the imaginary axis with nonzero speed $\frac{d}{db}\mathrm{Re}\lambda =1/2$.

The nonlinear Hopf bifurcation is nondegenerate. At the threshold, put $x=u-1$ and $y=-\sqrt{a}[(u-1)+(v-b/a)]$. The translated equations areThe Hopf coefficient of the Brusselator in the planar radial normalization is $-(a+2)/8<0$: the cubic contribution is $-6a/16$ and the quadratic contribution is $[-2\sqrt{a}2(1-a)]/(16\sqrt{a})$. Consequently the bifurcation is supercritical, producing small stable oscillations on the unstable-equilibrium side. Their limiting period is

Adding the two homogeneous reaction rates gives $a-u$, so the unique positive Brusselator equilibrium isThe reaction Jacobian matrix at that state isA real two-dimensional linear system has strict asymptotic stability exactly when $\tau <0$ and $\Delta >0$: the eigenvalues solve ${\lambda }^2-\tau \lambda +\Delta =0$. Real eigenvalues then have positive product and negative sum; a complex conjugate pair has real part $\tau /2$. Conversely, stable eigenvalues require those signs.

Since $a>0$, the Brusselator is homogeneously stable precisely forAt $b_H=1+a^2$, the linear eigenvalues are $\pm ia$, so strict decay is lost through a homogeneous Hopf bifurcation threshold. For larger $b$ the equilibrium is unstable. The source's matrix label $A$ is here written $J$ to distinguish it from the filament modulus in the preceding question.

For linear stability analysis of a general stable two-species reaction–diffusion system, take Fourier modes proportional to $e^{\lambda t+ikx}$. Their matrix is $J_k=J-k^2\mathrm{diag}(D_u,D_v)$. Put $s=k^2$, $\tau =\mathrm{tr}J$, $\Delta =\det J$ and $B=D_v{a}_{11}+D_u{a}_{22}$. ThenThe full system is stable precisely when ${\tau }_k<0$ and ${\Delta }_k>0$ for every allowed mode. If the homogeneous state is stable, the trace already decreases with $s$; only the determinant can change sign. If $B\le 0$, its minimum for $s\ge 0$ is at zero. If $B>0$, it is at $s_{\ast }=B/(2D_u{D}_v)$. Thus the two-species Turing criterion for instability on a continuum of wavenumbers isStrict stability has $B<2\sqrt{D_u{D}_v\Delta }$ together with the homogeneous conditions. Equality at positive $B$ marks a stationary neutral mode. At that first onset,Beyond threshold, the growing band lies between the positive roots $s_{\pm }=[B\pm \sqrt{B^2-4D_u{D}_v\Delta }]/(2D_u{D}_v)$, as long as the homogeneous trace stays negative.

For the Brusselator, $B=D_v(b-1)-D_u{a}^2$. Write $r=D_u/D_v$. The stationary determinant threshold gives the Turing threshold of the Brusselator in dimensional diffusivities:For this to be a genuine diffusion-driven loss of stability, it must occur before the homogeneous Hopf bifurcation threshold. The Brusselator Turing-before-Hopf diffusivity condition isIndeed $a^{2}r+2a\sqrt{r}<a^2$ is equivalent to $\sqrt{r}<(\sqrt{1+a^2}-1)/a$. With this strict inequality, $b_T<b<b_H$ contains a spatially unstable but homogeneously stable interval. If the ratio condition fails, the first linear instability as $b$ increases is homogeneous at $b_H$, rather than a Turing instability; equality gives simultaneous neutral homogeneous and finite-wavenumber modes. Equal diffusivities cannot produce a Turing instability.

These continuum formulas assume the onset wavenumber is available. On a finite domain with a specified discrete mode set, the discrete-mode Turing threshold for the Brusselator isprovided this minimum is below $b_H$. The question supplies no boundary geometry, so the continuous-mode onset above is the usual answer; this last expression states how a finite-size restriction changes it.

For a spatially homogeneous equilibrium of the Brusselator, the reaction terms obeyPositivity gives $uv=\beta$, and substitution into the first equation gives

The reaction Jacobian there issoThe linear stability analysis therefore makes the equilibrium asymptotically stable. It is a stable node whenand a stable focus when the reverse strict inequality holds. Since the upper-right entry of $J$ is positive and the lower-left entry is negative, focus trajectories rotate clockwise. Thus the local phase portrait consists of trajectories approaching the fixed point, either directly as a node or while spiralling clockwise as a focus.