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An activator in an activator-inhibitor reaction–diffusion system is a species whose local feedback enhances activation. Near a homogeneous equilibrium, positive self-feedback is reflected in the corresponding diagonal entry of the reaction Jacobian matrix. Cross-coupling to an inhibitor and differing diffusion rates can generate a Turing instability.

A quadratic activator can have both basal production and activation inhibited by a second species. A normalized example has reaction rates $f(u,v)=1+r{u}^2/v-u$ and $g(u,v)=q(u^2-v)$, with $r,q>0$ and positive concentrations. At its spatially homogeneous equilibrium, $v=u^2$, so $u_{\ast }=1+r$ and $v_{\ast }=(1+r{)}^2$. Writing $a=(r-1)/(r+1)$, its reaction Jacobian matrix is $J=\left(\begin{array}{cc}a& -r/(1+r{)}^2\\ 2q(1+r)& -q\end{array}\right)$. Thus $\mathrm{tr}J=a-q$ and $\det J=q>0$, giving linear asymptotic stability when $q>a$ and instability when $q<a$. Basal production makes $a$ change sign at $r=1$; a positive activation feedback needed for a conventional Turing instability requires $r>1$.

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.

An inhibitor in an activator-inhibitor reaction–diffusion system suppresses the activator. Its diffusion and reaction feedback can stabilize a spatially uniform state while allowing some nonzero wavenumbers to grow through a Turing instability.

For a reaction-stable two-species equilibrium with both Jacobian matrix diagonal entries negative and positive diagonal diffusivities, each spatial mode has more negative trace after diffusion is added. Its determinant is the positive reaction determinant plus positive coefficients of $k^2$ and $k^4$. Hence every mode stays stable. Unequal diffusion alone is insufficient for a Turing instability; the reaction feedback must also permit a destabilizing weighted diagonal sum.

Assume $Q>1$, so $T(k)<0$ for every wavenumber. A Turing instability therefore occurs exactly when $D(k)<0$ for some $k$. Put $z=k^2\ge 0$. The upward quadratic $1+(1-P)z+P{z}^2$ has a negative minimum on this half-line precisely whenCombining these givesThe unstable band is $z_{-}<k^2<z_{+}$, whereThe degradation-limited diffusion distances are proportional to ${\ell }_u=\sqrt{D_u/{\beta }_u}$ and ${\ell }_v=\sqrt{D_v/{\beta }_v}$. Thus $P=({\ell }_v/{\ell }_u{)}^2$ and the criterion says the inhibitor must travel more than $(1+\sqrt{2})$ times as far as the activator before degradation. Short-range activation and longer-range inhibition permit a local peak to reinforce itself. On a finite domain the allowed wavenumbers must also intersect the unstable band; the displayed condition assumes continuously available modes.

A Fourier mode with $q=k^2\ge 0$ evolves under $J-q\mathrm{diag}({\kappa }_1,{\kappa }_2)$. Its trace remains negative for a homogeneously stable positive branch. Its determinant isIf $u_0{\kappa }_2\le s{\kappa }_1$, this is positive for every $q\ge 0$. Otherwise its minimum is attained at $q_{\ast }=(u_0{\kappa }_2-s{\kappa }_1)/(2{\kappa }_1{\kappa }_2)$. Stability requires $D(q_{\ast })>0$, orIn terms of $d=\sqrt{{\kappa }_2/{\kappa }_1}$, the boundary is the positive root of $u_0{d}^2-s=2d\sqrt{s(2u_0^2-u_0)}$:Within $1/2<u_0<s$, all modes decay for $d<d_c$; a finite band grows for $d>d_c$, giving a Turing instability. Equality is neutral at one nonzero wavenumber.

For $s=2$, the positive lower branch $0<u_0<1/2$ is unstable for every $d$. The upper branch is homogeneously stable for $1/2<u_0<2$: it is stable to all modes to the left of $d_c=2+\sqrt{4-2/u_0}$ and unstable to spatial modes to the right. For $u_0>2$ it is already homogeneously unstable. The zero branch is stable to every spatial mode when $r<0$, because its mode eigenvalues are $r-{\kappa }_1{k}^2$ and $-s-{\kappa }_2{k}^2$.

At onset the determinant has a double root, soSquaring the neutral-boundary equation and canceling the positive factors gives $u_0=s/[d(2\sqrt{2s}-d)]$. Substitution yieldsFor a positive branch with $u_0>1/2$, the actual neutral curve lies strictly between $\sqrt{2s}$ and $2\sqrt{2s}$. If $d<\sqrt{2s}$, then $d<d_c(u_0)$ for every such branch: a homogeneously stable state cannot have this Turing instability. If $d>2\sqrt{2s}$, then $d>d_c(u_0)$ for every finite positive $u_0$, so every homogeneously stable positive state is spatially unstable. The negative value yielded by the displayed formula outside its boundary domain is not a physical critical wavenumber and does not imply renewed stability. One must also keep $u_0<s$, which can restrict the attainable neutral curve further.

The species $u$ is the activator, since increasing $u$ increases its own production $u^2/v$; $v$ is the inhibitor, since increasing $v$ reduces that production. The unique positive uniform steady state solves $v=u^2$ and $u^2/v=bu$, giving $\boxed{u_{\ast }=1/b,v_{\ast }=1/b^2}$.

The reaction Jacobian matrix at this state isIts eigenvalues solve ${\lambda }^2+(1-b)\lambda +b=0$. Since $b>0$, both have negative real part precisely for $0<b<1$. This proves stability of the spatially uniform reaction kinetics.

A perturbation proportional to $e^{\lambda t+ikx}$ replaces $J$ by $J-k^2\mathrm{diag}(1,d)$. Put $q=k^2$. Its trace is $b-1-(1+d)q<0$ in the kinetically stable range, and its determinant isA growing spatial mode exists exactly when this determinant is negative for some $q>0$. Its minimum occurs at $q=(bd-1)/(2d)$, which is positive only if $bd>1$, and the minimum is negative precisely if $(bd-1{)}^2>4bd$. Together these conditions reduce to the Turing threshold of the quadratic activator-inhibitor model,The unstable band isAt the Turing instability threshold the band shrinks to its double root, yielding $\boxed{k_c^2=(1+\sqrt{2})/d}$. On a bounded spatial interval, an allowed wavenumber must additionally lie in the unstable band; the displayed parameter condition assumes the continuous spatial spectrum.

First find the positive spatially homogeneous equilibrium. Setting the reactions to zero gives $v=u^2$ and $u/v=b$, and hence the unique positive equilibrium isThe zero pair is not an admissible positive state and is singular in the $u^2/v$ term. For homogeneous deviations, the reaction Jacobian matrix at this equilibrium isThus $\mathrm{tr}J=b-1$, $\det J=b$, and its eigenvalues areFor $0<b<1$, both real parts are negative: a real pair has positive product and negative sum, while a complex pair has real part $(b-1)/2$. The reaction equilibrium is therefore asymptotically stable, with no growing oscillatory mode. This does not exclude damped oscillations. Indeed, for $3-2\sqrt{2}<b<1$ the eigenvalues are complex; for example, $b=1/2$ gives $-1/4\pm i\sqrt{7}/4$, a stable focus with oscillatory transients. Thus a literal exclusion of every oscillatory solution is false; the intended exclusion of sustained or growing oscillations is the meaningful statement.

One can strengthen this to exclude nonlinear periodic oscillations in the positive quadrant without relying only on the local linear stability analysis. For the reactions $f=u^2/v-bu$, $g=u^2-v$, use the Dulac function $H=u^{-2}$. ThenWere there a periodic orbit wholly in the positive quadrant, the flux of $H(f,g)$ through it would vanish, since this vector is tangent to the orbit; Green's theorem would equate that flux to a strictly negative interior integral. This contradiction proves there is no sustained positive periodic oscillation when $0<b<1$, the positive-quadrant Dulac multiplier for quadratic activation argument. It leaves the damped transients entirely consistent with the conclusion.

Now introduce a perturbation $(\delta u,\delta v)=\mathbf{a}{e}^{\sigma t+ikx}$ in the quadratic activator-inhibitor model. The Fourier mode of squared wavenumber $q=k^2$ evolves underIts matrix trace and determinant areA Turing instability must start with reaction stability, requiring $0<b<1$. Then $\tau (q)<0$ for every $q\ge 0$, so a mode is unstable exactly when $D(q)<0$: its two real eigenvalues have opposite signs. This also proves the diffusion-driven onset is stationary rather than oscillatory.

The upward-opening quadratic $D(q)$ has its minimum at $q_m=(db-1)/(2d)$. To have a negative value at a physically allowed positive squared wavenumber, one needsWriting $r=db$, the second inequality is $r^2-6r+1>0$, with roots $3\pm 2\sqrt{2}$. The first inequality discards the lower branch, givingThese are the Turing threshold of the quadratic activator-inhibitor model conditions. Equivalently, in the $(b,d)$ plane the instability region lies strictly above $d=(3+2\sqrt{2})/b$ and strictly to the left of $b=1$, with $b>0$. The boundary diverges as $b\to 0^{+}$ and approaches $3+2\sqrt{2}$ as $b\to 1^{-}$.

At equality, the determinant has a double root and one growth rate is zero, while the other remains negative. Substituting $db=3+2\sqrt{2}$ into its minimum location givesBeyond onset, the unstable band isThe stated parameter region assumes continuously available wavenumbers, as on an infinite domain. With specified finite-domain boundary conditions, an allowed nonzero spatial eigenmode must additionally lie in this band; homogeneous stability alone does not ensure that a finite box admits the critical wavelength.

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.

At a positive spatially homogeneous equilibrium point, $v=u^2$ and $2+u-v=0$. Hence $(u,v)=(2,4)$. The reaction Jacobian matrix there isThe two-dimensional Routh-Hurwitz stability criterion gives strict linear asymptotic stability for $a>2$, and instability for $0<a<2$.

For a spatial mode with $s=k^2$, the reaction-diffusion linearization is $J_s=J-\mathrm{diag}(s,ds)$. When $a>2$, its trace is negative for every $s\ge 0$, whileA growing mode exists precisely when this quadratic is negative at some $s>0$. Its minimum must occur at positive $s$, and must be strictly negative, givingWith $z=d/a$, the latter inequality becomes $4z^2-28z+1>0$. Selecting the root compatible with $2d>a$ yields the Turing instability thresholdThe unstable band isOn a bounded domain a boundary condition must permit a mode in this band. Equality at the threshold gives a neutral mode, not positive growth.

At $a=2$, the homogeneous linearization has purely imaginary eigenvalues, so it is not strictly linearly asymptotically stable. If “stable” is read as nonlinear stability, the boundary deserves attention. Put $x=u-2$, $y=(v-4-x)/\sqrt{3}$, and $x=r\cos \theta$, $y=r\sin \theta$. The exact equations give $\dot{r}=r^{2}F(\theta )$, $\dot{\theta }=\omega +rG(\theta )$, where $\omega =2\sqrt{3}$,Expanding $dr/d\theta$ through cubic order, ${\int }_0^{2\pi }F=0$ cancels the quadratic displacement. The correction from the quadratic change in $r$ also integrates to zero, because it is proportional to $(\int F{)}^2$. Since ${\int }_0^{2\pi }FG=\pi /(2\sqrt{3})$, the Poincaré return map isThis cubic return-map stability of a weak focus shows that $a=2$ is a weak attracting focus. The strict linear range used in the Turing instability calculation is $a>2$; the positive homogeneous equilibrium is also nonlinearly asymptotically stable at the boundary $a=2$. For that boundary, $\mathrm{tr}{J}_s=-(1+d)s<0$ for every nonzero spatial mode. The same strict Turing instability threshold and unstable band therefore also apply there, with the homogeneous mode understood through nonlinear stability.

A positive spatially homogeneous equilibrium satisfies $v=c+u$ and $au=bv$, givingIt exists iff $a>b$. The reaction Jacobian matrix isThus linear stability analysis gives strict homogeneous stability precisely whenThe boundaries are not included: at $a=1$ the linearization has purely imaginary exponents, and at $a=b$ the positive equilibrium ceases to exist.

For a mode $\cos (kx)$, put $q=k^2$. The perturbation matrix is $J-\mathrm{diag}(q,dq)$. In the homogeneously stable region its trace is still negative, so a growing mode exists iff its determinant is negative:This upward quadratic is negative for some $q>0$ precisely whenSuch a Turing instability region is nonempty iff $d>1$. The unstable band isFor a finite domain only the admissible boundary-condition wavenumbers in this band grow. The region shown below describes a continuous wavenumber spectrum.

At onset the minimum of $D$ is zero and $q_c=u_{\ast }(d-a)/(2d)$. On the threshold $b=4d{a}^2/(d+a{)}^2$, substitution gives $u_{\ast }=4dac/(d-a{)}^2$. HenceThis last expression is the threshold wavenumber; away from onset the fastest-growing mode need not coincide with the minimum of the determinant.

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

At a positive spatially homogeneous equilibrium,soThe Jacobian matrix of the reaction terms there isThe linear stability analysis of a two-dimensional system requires negative trace and positive determinant. Since $a>0$, the homogeneous equilibrium is stable exactly when

For a spatial Fourier mode proportional to $\cos (kx)$, put $q=k^2$. The linearized matrix isIts trace is still negative when $0<a<1$, whileA Turing instability occurs for some $q>0$ exactly when this quadratic has a negative minimum. This requiresWriting $d=\sqrt{D}$, the second inequality reduces to $d<a(\sqrt{2}-1)$. Hence, for continuously available wave numbers, the condition isOn a bounded spatial domain, one additionally requires an allowed wave number $k$ for which $\det J_{k^2}<0$.

If $d=1$, the condition would requireBut homogeneous stability requires $f_u+g_v<0$. These inequalities are incompatible, so equal diffusivities cannot produce a Turing instability.

Put $y=k^2\ge 0$. At the threshold of a Turing instability, the maximum oftouches zero. HenceThe second equation gives $\rho =D(1+y_c{)}^2$. Substitution in the first givesThereforeand

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.