Reaction–diffusion system

A reaction-diffusion system combines local reaction kinetics with spatial diffusion,

An autocatalytic species promotes its own growth while consuming a substrate that promotes that growth. Its reaction Jacobian has the displayed signs after choosing the activator first. This differs from the activator-inhibitor sign pattern $\left(\begin{array}{cc}+& -\\ +& -\end{array}\right)$, where the activator stimulates its inhibitor. Appropriate magnitudes, diffusivities and admissible spatial modes are needed for a Turing instability; signs alone are insufficient.

This nondimensional two-variable reaction–diffusion system combines a fast cubic local response with slow negative feedback. With $D>0$, $0<\alpha <1$, $\beta >0$ and $\epsilon \ll 1$, its spatially uniform rest state at $I=0$ is linearly stable: the Jacobian has trace $-\alpha -\epsilon \beta$ and determinant $\epsilon (1+\alpha \beta )$. For suitable parameters a finite perturbation can nevertheless cause a large excursion followed by recovery. Frozen recovery gives a bistable cubic reaction-diffusion equation; the full coupled travelling-wave boundary-value problem determines whether front pairs form a pulse. This is a generic excitable-medium formulation of the FitzHugh-Nagumo model, without assigning any biological variable or application.

For reaction Jacobian $J=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ with $\mathrm{tr}J<0$ and $\det J>0$, a Fourier mode evolves under $J-q^2\mathrm{diag}(D_u,D_v)$. Its determinant is $D_u{D}_v{q}^4-(D_{v}a+D_{u}d)q^2+\det J$. A finite unstable band requires $D_{v}a+D_{u}d>0$ and $(D_{v}a+D_{u}d{)}^2>4D_u{D}_v\det J$, with an allowed domain mode in that band.

If both diffusion coefficients equal $D$, each spatial mode has matrix $J-D{q}^{2}I$. Its eigenvalues are those of $J$ shifted by the negative real number $D{q}^2$. Thus a reaction-stable homogeneous state remains stable under this equal-diffusion linear mechanism.

For $u_t=u_{xx}-a(t)u$, the displayed substitution converts the equation exactly into the heat equation $v_t=v_{xx}$. The scalar prefactor preserves homogeneous Dirichlet boundary conditions. Consequently each spatial eigenfunction $X_n$ with $X_n^{\prime \prime }=-{\lambda }_n{X}_n$ acquires temporal factor $\exp [-{\lambda }_{n}t-{\int }_0^{t}a(s)ds]$. The same algebra works with an evolution variable named $y$; calling that variable spatial does not alter the separation of variables. For $a(t)=1+\sin t$, the prefactor normalized to one at zero is $e^{-t+\cos t-1}$.

For nonnegative nonzero integrable initial data $f$ on $(0,L)$, the first coefficient in the normalized Fourier sine series is strictly positive, because the first sine eigenfunction is positive throughout the interval. For ${\rho }_t=D{\rho }_{xx}+\gamma \rho$ with zero Dirichlet boundary conditions, the first decay rate is ${\lambda }_1=D{\pi }^2/L^2-\gamma$. Every higher rate exceeds it by $D{\pi }^2(n^2-1)/L^2$, so smoothing for positive time and the spectral gap imply $e^{{\lambda }_{1}t}\rho (x,t)\to a_1\sqrt{2/L}\sin (\pi x/L)$ uniformly. Thus all interior points share the growth/decay threshold $\gamma =D{\pi }^2/L^2$; the endpoints remain zero. At equality the limiting profile is nonzero and stationary.

For $n_t=\mu \Delta n+\gamma n$ in a rectangular box with zero Dirichlet boundary conditions, separation of variables gives spatial eigenfunctions $\sin (p\pi x/L_x)\sin (q\pi y/L_y)\sin (r\pi z/L_z)$ and temporal rates $\gamma -\mu {\pi }^2(p^2/L_x^2+q^2/L_y^2+r^2/L_z^2)$. The largest rate is the displayed one. Every nonzero nonnegative initial density has positive projection onto this first eigenfunction, so its sign determines eventual exponential growth or decay; equality leaves a stationary first mode. At fixed volume, write $a_j=L_j^{-2}$: their product is fixed, and the arithmetic-geometric mean inequality minimizes their sum when all three are equal. Thus the cube maximizes the growth rate among rectangular boxes of fixed volume; making a box sufficiently thin suppresses growth.

The scalar equation $u_t=D{u}_{xx}+u(1-u)(u-r)$ with $D>0$ and $0<r<1$ has two stable homogeneous phases $u=0,1$ and an unstable intervening equilibrium $u=r$. Indeed the reaction derivatives at these three equilibria are $-r$, $-(1-r)$ and $r(1-r)$, respectively. A travelling wave can connect the two stable phases. At the equal-depth value $r=1/2$, its decreasing heteroclinic orbit is stationary. Away from this value the phases have unequal potential depths and the front propagates.

For the decreasing front connecting $1$ to $0$, the balanced stationary travelling-wave reduction of a reaction-diffusion system has first integral $D{U}_x^2/2-U^2(1-U{)}^2/4=0$. Choosing the negative derivative gives $U_x=-U(1-U)/\sqrt{2D}$ and hence $U=[1+e^{(x-q)/\sqrt{2D}}{]}^{-1}$. This logistic profile remains an exact travelling wave when the reaction parameter differs from $1/2$: using $z=(x-ct-q_0)/\sqrt{D}$, its derivative is $U_z=-U(1-U)/\sqrt{2}$, and the wave equation is satisfied for $c=\sqrt{D/2}(1-2r)$. Thus the speed changes with the potential bias while the shape, apart from translation, does not.

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 $r>1$.

At $q=a=(r-1)/(r+1)>0$, the reaction eigenvalues are $\pm i\sqrt{a}$. This linear marginal case is a weakly attracting focus, not a nonlinear center. To check it directly, put $u=(1+r)(1+x)$, $v=(1+r{)}^2(1+y)$, $h=(1+a)/2$, $\beta =(1-a)/2$, $\omega =\sqrt{a}$, and $X=x$, $Y=(hy-ax)/\omega$. The equations become $\dot{X}=-\omega Y+F_2+F_3+O(4)$, $\dot{Y}=\omega X+G_2+G_3+O(4)$, withOn the unit circle write $A=\cos \theta F_2+\sin \theta G_2$, $B=\cos \theta F_3+\sin \theta G_3$, $C=\cos \theta G_2-\sin \theta F_2$. In polar coordinates, $d\varrho /d\theta =(A/\omega ){\varrho }^2+(B/\omega -AC/{\omega }^2){\varrho }^3+O({\varrho }^4)$. The quadratic angular average vanishes. Integrating the quadratic correction over a whole cycle contributes its final squared integral, also zero, leaving the cubic average. Using the even trigonometric moments gives $⟨B⟩=a/4$ and $⟨AC⟩/\omega =(a+1)/8$. The Poincaré map is therefore $\varrho \mapsto \varrho +[2\pi /\omega ](a-1){\varrho }^3/8+O({\varrho }^4)$. Since $0<a<1$, it contracts sufficiently small positive radii and proves nonlinear asymptotic stability at the threshold.

Add diffusivities $1,p$ to the basally forced quadratic activator-inhibitor model. A Fourier mode with $s=k^2$ has matrix $J-s\mathrm{diag}(1,p)$ and determinant $D(s)=p{s}^2+(q-pa)s+q$. When $q>a$, its trace is negative for every $s\ge 0$. A positive-growth mode therefore occurs precisely when $D(s)<0$ for some positive $s$. The parabola has a positive minimizer only if $pa>q$, and its minimum is negative exactly when $(pa-q{)}^2>4pq$. Thus the ordinary Turing instability requires $a>0$, $q>a$ and $p>q/(\sqrt{1+a}-1{)}^2$. At onset, $k_c^2=(pa-q)/(2p)=\sqrt{q/p}$ and the critical wavelength is $2\pi (p/q{)}^{1/4}$. A bounded spatial domain further requires an allowed mode inside the unstable band.

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.

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.

For positive concentrations and positive parameters, this reaction–diffusion system has homogeneous equilibrium $(u_{\ast },v_{\ast })=(1/b,1/b^2)$ and reaction Jacobian matrix $\left(\begin{array}{cc}b& -b^2\\ 2/b& -1\end{array}\right)$. Its trace is $b-1$ and determinant is $b$, making the equilibrium reaction-stable exactly when $0<b<1$. Stable complex eigenvalues permit damped oscillations, which must be distinguished from sustained periodic oscillations.

A mode with squared wavenumber $q$ in the quadratic activator-inhibitor model has negative trace $b-1-(1+d)q$ when $0<b<1$. Its determinant is $D(q)=d{q}^2+(1-db)q+b$. A positive minimum location and a negative minimum require $db>1$ and $(db-1{)}^2>4db$, equivalent to $db>3+2\sqrt{2}$. At onset, the double determinant root is $q_c=(1+\sqrt{2})/d$. In a bounded domain an admissible nonzero spatial mode must additionally lie in the open negative-determinant band.

For the diffusionless quadratic activator-inhibitor model, choose the Dulac function $u^{-2}$. Then $u^{-2}f=1/v-b/u$ and $u^{-2}g=1-v/u^2$, giving divergence $(b-1)/u^2$. For $0<b<1$ it is strictly negative throughout the simply connected positive quadrant. Green's theorem rules out a periodic orbit: its boundary flux would be zero because the rescaled vector field is tangent to the orbit, while its interior divergence integral would be negative.

A spatially homogeneous equilibrium of a reaction–diffusion system is a constant state $u_{\ast }$ satisfying $f(u_{\ast })=0$. Since its spatial derivatives vanish, its stability to homogeneous perturbations is determined by the Jacobian matrix $Df(u_{\ast })$; diffusion can additionally destabilize nonconstant modes.

A morphogen reaction-diffusion equation models a spatial concentration by diffusion and local production or decay. Linearization at a homogeneous equilibrium $C=0$ replaces $f(C)$ by $f^{\prime }(0)C$.

It is a transport-and-reaction model for a morphogen, rather than the identity of the signaling molecule concept.

The eigenfunctions satisfying $X(0)=0$ and $X^{\prime }(L)=0$ are

For $C_t=D{C}_{xx}+aC$ with $a>0$, $C(0,t)=0$, and $C_x(L,t)=0$, every mode decays exactly when

If a fast inhibitor satisfies $0=v_{xx}+u-v$, then its spatial Fourier transform obeysSubstitution produces a nonlocal scalar equation for the activator.

For activator diffusion $D$, local linear decay $r$, and coupling $\rho (u-v)$, eliminating the fast inhibitor givesIts pattern-forming threshold is

A Turing pattern is a spatially structured state produced by interactions between local reactions and diffusion. A Turing instability explains linear pattern onset when a homogeneous equilibrium is stable without diffusion but unstable to a nonzero spatial mode.

A Turing, or diffusion-driven, instability occurs when a spatially homogeneous equilibrium is stable to homogeneous perturbations but unstable to a nonzero spatial Fourier mode after unequal diffusion is introduced.

If a one-dimensional reaction-diffusion model has a strict unstable squared-wavenumber band $(q_{-},q_{+})$, its nonzero Neumann modes on $[0,L]$ have $q_n=(n\pi /L{)}^2$. Increasing $L$ from zero first reaches the band with $n=1$ at the displayed threshold. Equality is neutral; the first mode is strictly unstable for $\pi /\sqrt{q_{+}}<L<\pi /\sqrt{q_{-}}$. Thus positive growth has an infimum in domain length rather than an attained smallest length. The zero mode must be checked separately through reaction stability.

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.

Suppose a two-species reaction-diffusion instability has an unstable squared-wave-number band $(q_{-},q_{+})$. On a rectangle with Neumann boundary conditions, $q=(n_x\pi /L_x{)}^2+(n_y\pi /L_y{)}^2$, with both indices nonnegative. The displayed thin-width condition excludes every $n_y\ge 1$ mode but can leave $n_y=0$ modes in the band for sufficiently large $L_x$. Omitting zero indices would incorrectly remove the longitudinal pattern. Linear onset selects allowed directions; it does not alone establish the final nonlinear pattern.

For $u_t=D{u}_{xx}-u(u-r)(u-1)-\rho (v-u)$ and $0=v_{xx}-v+u$, a Fourier mode perturbation about zero has $v=u/(1+k^2)$ and growth rate $\sigma (k)=-r-D{k}^2+\rho k^2/(1+k^2)$. If $\rho >D$, its maximum is $-r+(\sqrt{\rho }-\sqrt{D}{)}^2$ at $k^2=\sqrt{\rho /D}-1$; if $\rho \le D$, its maximum is $-r$ at zero. Replacing $(u,v,r)$ by $(1-u,1-v,1-r)$ gives the corresponding criterion at the other homogeneous equilibrium.

Let a stable two-species reaction equilibrium have Jacobianand diffusion matrix $\mathrm{diag}(D_u,D_v)$. A mode with $q=k^2$ has determinantSince its trace decreases with $q$, diffusion-driven instability occurs exactly when

When $D_u=D_v=D$, every spatial Fourier mode replaces the reaction Jacobian $J$ by $J-D{k}^{2}I$. Each eigenvalue is shifted left by $D{k}^2$, so diffusion cannot destabilize a stable homogeneous equilibrium.

For reaction Jacobianthe critical diffusivity ratio isIt approaches one as the stable reaction system approaches marginality.

The Brusselator is a two-species autocatalytic reaction model. In the parametrizationits positive homogeneous equilibrium is $(u_{\ast },v_{\ast })=(\alpha ,\beta /\alpha )$.

For $u^{\prime }=1-(b+1)u+a{u}^{2}v$, $v^{\prime }=bu-a{u}^{2}v$, the equilibrium $(1,b/a)$ has a Hopf bifurcation at $b=1+a$. Coordinates $x=u-1$, $y=-\sqrt{a}[(u-1)+(v-b/a)]$ give the threshold equations $x^{\prime }=-\sqrt{a}y+(1-a)x^2-2\sqrt{a}xy-a{x}^3-\sqrt{a}{x}^{2}y$, $y^{\prime }=\sqrt{a}x$. In the usual planar radial normalization the cubic coefficient is $-(a+2)/8<0$, proving a nondegenerate supercritical bifurcation. The leading small-amplitude angular frequency is $\sqrt{a}$.

For the spatially homogeneous Brusselator with parameters $r,s>0$, set $\alpha =r/(1+s)$. The quadrilateralis a trapping region. The four inward-pointing tests use $\dot{x}\ge 0$ on $x=\alpha$, $\dot{y}\ge 0$ on $y=0$, $\dot{y}\le 0$ on $y=s/\alpha$, and $\dot{x}+\dot{y}=r-x\le 0$ on the sloping edge.

The spatially homogeneous Brusselator has its unique positive equilibrium at $(r,s/r)$. Its Jacobian there has determinant $r^2$ and trace $s-1-r^2$. If $s-1>r^2$, the equilibrium is a repeller inside the Brusselator trapping region; the Poincaré-Bendixson theorem then supplies a periodic orbit.

When $u$ and $v$ have diffusivities $D$ and $1$, respectively, and the homogeneous Brusselator equilibrium is stable, a Turing instability occurs precisely whenAt onset,

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.

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.