REVIEWREVIEW
A A Six stableSix states stable states Initial conditionInitial condition I I II II III III
Uniform, Uniform,stationary stationary Uniform, Uniform,oscillating oscillating StationaryStationary waves with waves with extremelyextremely short wavelength short wavelength
IV IV V V VI VI Both morphogensBoth morphogens diffuse anddiffuse react and react with eachwith other each other
OscillatoryOscillatory cases cases OscillatoryOscillatory cases cases StationaryStationary waves with waves with with extremelywith extremely short short with finitewith finite finite wavelengthfinite wavelength wavelengthwavelength wavelengthwavelength (Turing pattern(Turing) pattern)
Case V Case V Case VI (CaseTuring VI pattern (Turing) pattern) B B on February 28, 2013 on February 28, 2013
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Fig. 2. SchematicFig. 2. Schematic drawing showing drawing the showing mathematical the mathematical analysis of analysis the RD of theReproduction RD Reproduction of biological of biological patterns cr patternseated by cr modifiedeated by RD modified mechanisms. RD mechanisms. With With system andsystem the patterns and the generated patterns generated by the simulation. by the simulation. (A)Sixstablestates (A)Sixstablestatesmodification,modification, the RD mechanism the RD mechanism can generate can more generate complex more patterns complex such patterns as such as toward whichtoward the which two-factorLendert the two-factor RD system Gelens RD can system converge. can converge. (B)Two-dimensional (B)Two-dimensionalthose seenthose in the seen real in organism. the real organism. Simulation Simulation images are images courtesy are of courtesy H. Meinhardt of H. Meinhardt patterns generatedpatterns generated by the Turing by the model. Turing These model. patterns These were patterns made were by made an by[sea an shell[sea pattern shell (5 pattern)] and A. (5 R.)] and Sandersen A. R. Sandersen [fish pattern [fish (13 pattern)]. Photos (13)]. of Photos actual of actual identical equationidentical with equation slightly with differe slightlynt parameter different parameter values. These values. simulations These simulationsseashells areseashells from are Bishougai-HP from Bishougai-HP (http://shell.kwansei.ac.jp/~shell/). (http://shell.kwansei.ac.jp/~shell/). Images of Images of were calculatedwere calculated by theKU software by Leuven the provided software asprovided- supportingVrije as supporting Universiteit online material. online (material.C) popperBrussel (C) fishpopper are courtesy fish are of courtesy Massimo of Boyer Massimo (www.edge-of-reef.com). Boyer (www.edge-of-reef.com). www.gelenslab.org 1618 1618 24 SEPTEMBER24 SEPTEMBER 2010 VOL 2010 329 VOLSCIENCE 329 SCIENCEwww.sciencemag.orgwww.sciencemag.org
Nonlinear dynamics course - VUB What is pattern formation?
Systems that have the ability to self-organize into spatially structured states from initially unstructured (spatially homogeneous) states. 1PatternFormation
The last two examples of spatially distributed oscillators serve as a good intro to one of the most important aspects in complex systems: pattern formation. By pattern formation we mean that certain systemsExamples have of the pattern ability formation: to self-organize sand intopatterns spatially structured states from intitially unstructured or spatially homogenous states. This behavior takes place all over the place, in physics in chemistry in biology in social science etc. Here we will discuss the basic ingredient that are necessary for these processes to occur. One of the patterns that we see in any desert or any beach are ripples in the sand: - driven by wind/water (energy put in)
- positive feedback
- negative feedback
One can understand the basic ingredients for pattern formation in this system. Laminar flows of wind move accross the sand and move individual sand grains with it. Slight pertubations in the surface have a higherhttps://www.youtube.com/watch?v=uY2QdZLLRP8 likelihood of collecting these grains increasing the magnitude of the pertubation and thus increasing the likelihood of catching more sand. This is a positive feedback effect which is necessary in almost all pattern forming systems. This example also shows that one needs to put energy into pattern forming systems, in this case wind. Right behind the pertubation a dip forms, as the probability of grains collecting there becomes lower. Of course the increasing hills cannot increase forever and will be eroded at the top. This is the negative feedback that is also required in pattern formation. The same thing happens when river beds are formed. There’s a continuous precipi- tation that falls onto the land and the water streams it forms follows gravity downhill. Small streams carve a path and the following water will preferentially follow those paths, too digging deeper into the soil. Of course this can’t go on foverever. The process can produce valleys that collapse and new valleys will form:
1 Let’s discuss a few examples that all show these type of ingredient.
1.1 Examples of pattern forming in nature
1.1.1 Physics The most known example of spontaneous pattern formation in physics is probably Benard convection. This happens when a liquid is heated from below at such a high rate that the heat cannot dissipate through the system fast enough. In this case convetion rolls or cells emerge that transport the heat to the cooler regions of the liquid where heat is given off. TheExamples cooled liquid of pattern then getsformation: pushed convection to the bottom cells to be reheated:
- driven by temperature difference This experiment can be easily done by heating oil on the stove and adding metal dust to the oil. Again we have to add energy to the system and once certain parts of the liquid start moving upwardshttps://www.youtube.com/watch?v=gSTNxS96fRg this process will be accelerated and other regions have to move downward to keep the liquid in place. Images of the suns surface show granules which are exactly these kind of convetion pockets.
1.1.2 Chemistry There’s an abundance of chemical reactions that, if they occur in a not-well-stirred sce- nario, have the potential of generating patterns. One such reaction, the most famous one, is the Belousov–Zhabotinsky reaction. This reaction uses essentially five reactions,
in which particular chemicals in it start oscillating in their concentration. If the reac- tion takes place in a petri dish, spiral and target wave form. This is because in its essence this reaction is an activator inhibitor system in which an autocatalytic reaction increases the abundance of a chemical which in turn increases the abundance of a chemical that turns this reaction off. Here’s a snapshot of a pattern that the BZ reaction can generate:
2 Examples of pattern formation: chemical reactions
(the Belousov-Zhabotinsky reaction) This looks very similar to the patterns we generated with the spatially distributed pulse-coupled oscillators. Andhttps://www.youtube.com/watch?v=3JAqrRnKFHo in fact the mechanisms in both systems are very similar. We will come back to this later.
1.1.3 Biology Biology is full of pattern forming systems, so we are going to look at numerous examples.
1.1.3.1 cAMP signaling in Dictyostelium discoideum Dictyostelium discoideum is an interesting organism. It’s a single cell amoeba that, when nutrient run low excretes a signalling molecule called cAMP which other individuals in the neighborhood respond to in two ways. 1.) they move up the gradient towards the source, getting closer to the orignator of the signal and, 2.) the start excreting cAMP, too. This is positive feedback. Eventually this leads to the aggregation of thousands of these organisms into a multicellualr organism, a slug like creature that crawls away and develops into a differentiated mold that forms a fruiting body and a stem, spores that get carried away by wind. If one measures the concentration of cAMP one sees spiral waves, just like in the BZ reaction and the pulse coupled oscillators:
1.1.3.2 Orientation Maps in the visual cortex of primates An interesting pattern is seen on the visual cortex of primates and other higher mammals that have binocular vision. In the visual cortex signals that come in from the receptors
3 This looks very similar to the patterns we generated with the spatially distributed pulse-coupled oscillators. And in fact the mechanisms in both systems are very similar. We will come back to this later.
1.1.3 Biology Biology is full of pattern forming systems, so we are going to look at numerous examples.
1.1.3.1 cAMP signaling in Dictyostelium discoideum Dictyostelium discoideum is an interesting organism. It’s a single cell amoeba that, when nutrient run low excretes a signalling molecule called cAMP which other individuals in the neighborhood respond to in two ways. 1.) they move up the gradient towards the source, getting closer to the orignator of the signal and, 2.) the start excreting cAMP, too. This is positive feedback. Eventually this leads to the aggregation of thousands of these organisms into a multicellualr organism, a slug like creature that crawls away and develops into a differentiated mold that forms a fruiting body and a stem, spores that get carried away by wind. If one measures the concentration of cAMP one sees spiral Examples of pattern formation: slime mold (Dictyostelium) waves, just like in the BZ reaction and the pulse coupled oscillators:
https://www.youtube.com/watch?v=uqi_WTllG7A 1.1.3.2 Orientation Maps in thehttps://www.youtube.com/watch?v=bkVhLJLG7ug visual cortex of primates An interesting pattern is seen on the visual cortex of primates and other higher mammals that have binocular vision. In the visual cortex signals that come in from the receptors
3 in the eyes are processed. It turns out that the visual cortex has a bunch of cells that respond best to stimuli of a particular orientation, for instance contrast contours that are horizontal or vertical etc. If one records response strengths from neurons in the visual cortex one can color code the locations in the visual cortex according to the prefered stimulus orientation and this is what one gets:
We see a pattern that is reminiscent of the pinwheel pattern we saw in the phase coupled oscillator lattice. We see singularties, the pinwheels where all orientation meet. The cool part about thes visual maps is that animals aren’t born with them. These maps develop after birth and in response to visual stimuli.
1.1.3.3 Patterns on sea shell Here’s a picture of natural patterns of different snail shells. These shells grow slowly over time by continuous deposits of material and the distribution of types of material Examples of pattern formation: shells and interactions between regions in the organism that make the hard material generate different patterns. This is an example of a growth process that yields complex patterns. Something we will come back to.
This is an interesting example because not only the patterns on the shels are interesting but the patterns of the shells themselves which appear to be following some very basic formation rules. In fact, a whole book was written by one of the pioneers of pattern formation in developmental biology: Hans Meinhardt. He developed a model for pattern formation which we will discuss in detail.
4 1.1.3.4 Patterns in animal fur Examples of pattern formation: animal coating And of course, when we think of patterns in organism we have to mention animal fur:
This picture shows different types of animal fur. Some patterns are spotty, some consist of stripes of different wavelength. It turns out that will very few dynamic ingredient one can devise a model that is capable of generating all these patterns at different parameter values.
1.1.3.5 Biological Morphogenesis The above pattern in animals are a special case of something a lot more fundamental. The problem of how multicelluar organisms with a great diversity of function and cell morphology can develop from a single, homogeneous fertilized egg. First, all cells in a multicellular organism have the same genome despite the variety in shape and function. This can be the case because these cells differ in what genes are expressed, i.e. active or repressed, i.e. inactive. On the way towards an adult organsism, cells differentiate by a sequential switching off an on of genes. This mechanism is responsible for spontaneouly introducing differences in embryos. For example in the drosophila melanogaster (fruitfly) embryo, different genes are expressed in different regions. The combination of expression levels introduces different cell fates for the cells in specific locations which eventually governs the development of a full blown fly:
5 Pattern formation: outline
Systems that have the ability to self-organize into spatially structured states from initially unstructured (spatially homogeneous) states.
1. Formation periodic patterns - Turing patterns
2. Formation of domains and moving fronts Turing patterns
- Alan Turing (1952): The chemical basis of morphogenesis
- Claim: reaction and diffusion of chemicals in an initially uniform state could explain morphogenesis (biological patterns that arise during growth)
- Insights:
1. At least two interacting chemicals are needed for pattern formation
2. Diffusion can have a destabilizing influence
3. A diffusion-induced instability can lead to growth of a structure at a particular wavelength: mechanism for producing biological patterns!
4. Patterns form if diffusion coefficients of two reagents differ sufficiently phylum), or zebra stripes. A fourth insight (which was not clearly stated until after Turing’s paper) is that pattern formation in a chemical system will not occur unless the diffusion coefficients of at least two reagents differ substantially.phylum), or zebra The difficulty stripes. A offourth satisfying insight (which this condition was not clearly for chemicals stated until in after solution Turing’s partially paper) is explains that pattern formation in a chemical system will not occur unless the diffusion coefficients of at least two reagents why nearly 40phylum), years or passed zebra stripes. after Turing’s A fourth insight paper (which before was experiments not clearly stated were until able after to Turing’s demonstrate paper) is that the truth of his ideas. differpattern substantially. formation in The a chemical difficulty system of will satisfying not occur this unless condition the diffusion for coefficients chemicals of in at solution least two partially reagents explains whydiffer nearly substantially. 40 years passed The difficulty after Turing’s of satisfying paper this before condition experiments for chemicals were in able solution to demonstrate partially explains the truth of his ideas.why nearly 40 years passed after Turing’s paper before experiments were able to demonstrate the truth of his Reaction-Diffusionideas. Equations Reaction-Diffusion Equations We will studyReaction-Diffusion the Turing model Equations2 for two reacting and diffusion chemicals of the form: Reaction-diffusion equations: the Turing model 2 We will study the Turing model2 for two reacting and diffusion chemicals of the form: We will study the Turing model for two reacting and diffusion chemicals2 of the form: ∂t u1 f1 (u1, u2) D1∂x u1, (1a) = + 2 2 ∂t∂ut1u1 f1 (fu11(,uu12), u2)D1∂ Du112,∂x u1, (1a) (1a) ∂ u =f =(u , u )+ +Dx ∂ u , (1b) t 2 2 1 2 2 x 22 ∂t∂u=2u f2 (fu1(,uu2), u+)D2∂ Du2,∂ u , (1b) (1b) t =2 = 2 1 +2 +x 2 x 2 or in vector form or inor vector in vector form form 2 2 ∂t u∂t u ff((uu)) DD∂x∂ux,u2, (2) (2) =∂t=u f(++u) D∂x u, (2) where we have introduced a diagonal 2 2 diffusion= matrix+ D defined by where wewhere have we introduced have introduced a diagonal a diagonal 2 2 2×diffusion2 diffusion matrix matrix D Ddefineddefined by by × × D 0 D 1 . (3) = D0 DD1 02 0 D D ! 1 " . . (3) (3) = 0 D2 Eq. (1) describes the evolution of two concentration= !0 fieldsD2u (t, x") on the real line < x < .The Homogeneous! steady statei "solutions? −∞ ∞ nonlinear functions f (u , u ) are the reaction rates of the two chemicals while the D are the corresponding Eq. (1) describes thei evolution1 2 of twoLinear concentration stability (in time fields and space!)?ui (t, x) on the reali line < x < .The Eq. (1) describesdiffusion the coefficients. evolution The of simplest two concentration possible model is obtained fields u byi (t assuming, x) on that the there real is line no prior−∞ spatiotem-< x <∞ .The nonlinear functions fi (u1, u2) are the reaction rates of the two chemicals while the Di are the corresponding poral structure in the system so that the functions f and the diffusion coefficients D do not depend−∞ explicitly ∞ nonlineardiffusion functions coefficients.fi (u1, u2) Theare simplest the reaction possible rates modeli of the is obtained two chemicals by assuming whilei that the thereDi isare no the prior corresponding spatiotem- on time t or on position x.Forsimplicity,wefurtherassumethatthediffusioncoefficientsareconstantsand diffusionporal coefficients. structure in The the simplest system so possible that the functions model isf obtainedand the diffusion by assuming coefficients thatD theredo not is no depend prior explicitly spatiotem- so do not depend on the field values ui .Theseassumptionsareallquitereasonableformanyexperimentali i poral structureon timesituations. int theor on system position sox that.Forsimplicity,wefurtherassumethatthediffusioncoefficientsareconstantsand the functions fi and the diffusion coefficients Di do not depend explicitly so do not depend on the field values u .Theseassumptionsareallquitereasonableformanyexperimental on time t or onEquations position (1) cannotx.Forsimplicity,wefurtherassumethatthediffusioncoefficientsareconstantsand accurately describei a sustained nonequilibrium chemical system since they incorporate so do notsituations. dependno way on to feed the reactants field values into andui remove.Theseassumptionsareallquitereasonableformanyexperimental products from the system. The neglect of a transverse confined situations.Equationscoordinate (1) along cannot which accurately such feed describe might occur a sustained is a major nonequilibrium simplification. An chemical accurate system treatment since of the they feed incorporate nodirection way to feed introduces reactants complicated into and spatial remove structure. products Actually from many the early system. experiments The neglect on pattern of a formation transverse in confined Equations (1)chemical cannot reactions accurately could describebe rather well a sustained approximated nonequilibrium by ignoring the confined chemical coordinate. system Typically since they these incorporate coordinate along which such feed might occur is a major simplification. An accurate treatment of the feed no way to feedexperiments reactants were into done and using remove a thin layer products of chemicals from in the a petri system. dish, or chemicals The neglect soaked of in a filter transverse paper. confined directionThe variation introduces of chemical complicated concentration spatial across structure. the layer Actually or thickness many of early filter paper experiments is perhaps on small pattern in these formation in coordinatechemical alongexperiments which reactions (typically such could feed the be conditions might rather well occur are not approximated is well a controlled, major by simplification. so ignoring this is just the an confined assumption). An accurate coordinate. A reduction treatment Typically to of the these feed directionexperiments introducesequations describing complicated were done just using the spatial spatial a thin variation structure. layer of in thechemicals Actually plane (ie. in many Eqs. a petri (1) early but dish, with experiments or∂2 chemicals2 ∂ on2 soaked pattern∂2)would in filter formation paper. in x →∇ = x + y chemicalThe reactionsbe variation a reasonable could of chemical approximation. be rather concentration well However, approximated in across these theexperiments, layer by ignoring or since thickness there the is of confined no filter feed paper of coordinate.refreshed is perhaps chemicals small Typically in these these experimentsexperimentsto were sustain done the (typically reactions, using the a any thin conditions pattern layer formation of are chemicals not or well dynamics controlled, in is a a petritransient so dish, this,andeventuallythesystemwould is or just chemicals an assumption). soaked A in reduction filter paper. to equationsapproach describing a uniform chemical just the spatial equilibrium. variation This difficulty in the plane may be(ie. hidden Eqs. in (1) the but simple with reduced∂2 equations2 ∂2 (1)∂2)would The variationby of approximating chemical concentration some dynamical chemical across concentrations the layer or as thickness constants in of the filter reaction paper termx →∇ isf( perhapsu).= x + smally in these experimentsbe a (typically reasonable the approximation. conditions However, are not well in these controlled, experiments, so this since is there just is an no assumption). feed of refreshed A reduction chemicals to to sustain the reactions, any pattern formation or dynamics is a transient,andeventuallythesystemwould equations describing just the spatial variation in the plane (ie. Eqs. (1) but with ∂2 2 ∂2 ∂2)would approachLinear a Stability uniform chemical Analysis equilibrium. This difficulty may be hidden in the simplex →∇ reduced= equationsx + y (1) be a reasonableby approximating approximation. some dynamical However, chemical in these concentrations experiments, as since constants there in is the no reaction feed of term refreshedf(u). chemicals to sustain theWe reactions, now perform any the pattern linear stability formation analysis or of dynamics uniform solutions is a transient of the two-chemical,andeventuallythesystemwould reaction-diffusion model Eq. (1). Turing’s surprising and important discovery was that there are conditions under which approach a uniform chemical equilibrium. This difficulty may be hidden in the simple reduced equations (1) Linear2 Stability Analysis In his paper, Turing examined two kinds of models, spatially coupled odes that modelled discrete biological cells and coupled( ) by approximatingpdes of the some form that dynamical we analyze here, chemical that treated the concentrations tissue as a continuous as medium. constants in the reaction term f u . We now perform the linear stability analysis of uniform solutions of the two-chemical reaction-diffusion model Eq. (1). Turing’s surprising and important discovery was that there are conditions under which Linear Stability Analysis 2 2In his paper, Turing examined two kinds of models, spatially coupled odes that modelled discrete biological cells and coupled We now performpdes of the form the that linear we analyze stability here, analysisthat treated the of tissue uniform as a continuous solutions medium. of the two-chemical reaction-diffusion model Eq. (1). Turing’s surprising and important discovery was that there are conditions under which 2 2In his paper, Turing examined two kinds of models, spatially coupled odes that modelled discrete biological cells and coupled pdes of the form that we analyze here, that treated the tissue as a continuous medium.
2 the spatially uniform state is stable in the absence of diffusion3 but can become unstable to nonuniform the spatially uniform state is stable in the absence ofperturbations diffusion3 preciselybut can because become ofunstable diffusion.to Further, nonuniform for many conditions the instability first occurs at a perturbations precisely because of diffusion. Further,finite for wave many length conditions and so a the cellular instability pattern startsfirst occurs to appear. at a finite wave length and so a cellular pattern starts to appear.For simplicity, we will discuss the one-dimensional case. If we assume that the evolution equations have the spatially uniform state is stable in the absence of diffusion3 but can become unstable to nonuniform For simplicity, we will discuss the one-dimensionalrotational case. If wesymmetry assume in that higher the dimensions, evolution the equations two- and have three-dimensional cases are identical mutatis mu- perturbations precisely because of diffusion. Further, for many conditions the instabilitytandis first:theone-dimensionalLaplacian occurs at a ∂2 becomes a higher-dimensional Laplacian 2,thecombinationqx rotational symmetry in higher dimensions, the two- and three-dimensional cases are identicalx mutatis mu- ∇ finite wave length and so a cellular pattern starts to appear. becomes everywhere a dot product q x with a wave vector q,andthewavenumbersquareq2 becomes the tandis:theone-dimensionalLaplacian∂2 becomes a higher-dimensional Laplacian 2·,thecombinationqx For simplicity, we will discuss the one-dimensional case. If we assumex that the evolutionquantity equationsq q. have ∇ becomes everywhere a dot product q x with a wave vector q,andthewavenumbersquare· q2 becomes the rotational symmetry in higher dimensions, the two- and three-dimensional· cases are identicalThe followingmutatis discussion mu- is not mathematically difficult but has many details. You will likely best appreciate tandis:theone-dimensionalLaplacianquantity q ∂q2 .becomes a higher-dimensional Laplacian 2,thecombinationqx · x ∇ the discussion if you take your time and derive the results for yourself in parallel with the text. The goal of becomes everywhere a dot product q x with a wave vector q,andthewavenumbersquareq2 becomes the The following· discussion is not mathematically difficultthe but discussion has many is to details. derive, and You then will to likely understand best appreciate physically, conditions that are sufficient for the real parts quantity q q. · the discussion if you take your time and derive the resultsof all growth for yourself rates to in be parallel negative. with When the these text. conditions The goal of are first violated and instability occurs, it is then The following discussion isthe not discussion mathematically is to difficult derive, but and has then many to details. understand You will physically,important likely best conditions toappreciate think about that the are values sufficient of the wave for the numbers real parts corresponding to the fastest growing modes. the discussion if you take your time and derive the results for yourself in parallel with the text. The goal of of all growth rates to be negative. When these conditionsHomogeneousWe begin are by first assuming steady-state violated that and we solutions instability have somehow of occurs, the found Turing it a is stationary thenmodel uniform base solution u (u , u ).This the discussion is to derive, and then to understand physically, conditions that are sufficient for the real parts b = 1b 2b important to think about the values of the wave numberssatisfies corresponding the Turing model to the with fastest all partial growing derivatives modes. set to zero, leading to f(u ) 0 or of all growth rates to be negative. When these conditions are first violated and instability occurs, it is then b = important to think about the values of the wave numbers corresponding to the fastest growing modes. We begin by assuming that we have somehow found a stationary uniform base solution ub (uf1b(,uu2,bu).This) 0, (4a) = 1 1b 2b = We begin by assuming thatsatisfies we have somehowthe Turing found model a stationary with all uniform partial base derivatives solution u setb to(u zero,1b, u2b leading).This to f(ub) 0 or = f2(u1b, u2b) 0. (4b) satisfies the Turing model with all partial derivatives set to zero, leading to f(u ) 0 or = = b = f1(u1b, Theseu2b) are0, two nonlinear equations in two unknowns. Finding(4a) a uniform solution can be hard since there is no f1(u1b, u2b) 0, = (4a) = systematic way to find even a single solution of a set of nonlinear equations. Numerical methods such as the f (u , u ) 0. f2(u1b, u2b) 0. (4b) (4b) 2 1b 2b = Newton= method can find accurate approximations to solutions of nonlinear equations but only if a good guess These are two nonlinear equationsThese are in two two unknowns. nonlinear Finding equations a uniform in two solution unknowns. can beHowfor Finding hard to a since solutionsolve a there uniformsystem is is already noof solutiontwo known.nonlinear can For equations? be two hard nonlinear since there equations is no like Eq. (4), a graphical way to find solutions that is sometimes useful is to plot the nullclines of each equation. An equation of the form f (u , u ) 0 systematic way to find evensystematic a single solution way to of find a set even of nonlinear a single equations. solution Numerical of a set of methods nonlinear such equations. as the Numerical methods such as the 1 1 2 = - Analytically:defines an implicit often not relation possibleu g (u ) between the two variables u and u called the nullcline of that Newton method can find accurateNewton approximations method can find to solutions accurate of nonlinear approximations equations to but solutions only if a good of nonlinear guess equations2 = but1 only1 if a good guess 1 2 for a solution is already known. For two nonlinear equations like Eq. (4), a graphical- Numerically:equation. way to find If solutions thee.g. Newton-Rhapson functions fi (u1, umethod2) are sufficiently simple, their nullclines gi (u1) can sometimes be found for a solution is already known. For two nonlinear- equations Graphically: like plot Eq. nullclines (4), a graphical way to find solutions that is sometimes useful is to plot the nullclines of each equation. An equation of theexplicitly form f1(u and1, u2 then) 0 plotted on a single plot with axes labelled by u1 and u2.Anyintersectionofthetwo that is sometimes useful is to plot the nullclines of eachnullclines equation. is then= An a solution equation of the of the nonlinear form equations.f1(u1, u2) 0 defines an implicit relation u2 g1(u1) between the two variables u1 and u2 called the nullcline of that = defines= an implicit relation u g (u ) between the two variables u and u called the nullcline of that equation. If the functions fi (u1, u2) are sufficiently simple,2 their nullclines1 1 gi (u1) can sometimes be found1 2 = By linearizing about the base state ub,youcanshowthatanarbitraryinfinitesimalperturbationδu(t, x) explicitly and then plottedequation. on a single If plot the withfunctions axes labelledfi (u1, u by2)uare1 and sufficientlyu2.Anyintersectionofthetwo simple, their nullclines gi (u1) can sometimes be found = (δu1(t, x), δu2(t, x)) of the base state will evolve in time according to the following linear constant-coefficient nullclines is then a solution of the nonlinear equations. explicitly and then plotted on a single plot with axesevolution labelled equations: by u1 and u2.Anyintersectionofthetwo nullclines is then a solution of the nonlinear equations. By linearizing about the base state ub,youcanshowthatanarbitraryinfinitesimalperturbationδu(t, x) = 2 (δu1(t, x), δu2(t, x)) of the base state will evolve in time according to the following linear constant-coefficient ∂t δu1 a11δu1 a12δu2 D1∂x δu1, (5a) By linearizing about the base state ub,youcanshowthatanarbitraryinfinitesimalperturbation= δu+(t, x) + evolution equations: ∂ δu a δu a δu= D ∂2δu . (5b) (δu1(t, x), δu2(t, x)) of the base state will evolve in time according to the following lineart constant-coefficient2 = 21 1 + 22 2 + 2 x 2 2 evolution∂t δ equations:u1 a11δu1 a12δu2 D1∂ δu1, (5a) = + + x The constant coefficients aij come from the 2 2JacobianmatrixA ∂f/∂u evaluated at the constant base 2 × = ∂t δu2 a21δu1 a22δu2 D2∂ δu2. solution u , (5b) = + + x b 2 ∂t δu1 a11δu1 a12δu2 D1∂x δu1, ∂ fi(5a) = + + aij . (6) The constant coefficients aij come from the 2 2JacobianmatrixA ∂f/∂u evaluated at the constant base2 ∂ δ δ δ ∂ δ . = ∂u j × t =u2 a21 u1 a22 u2 D2 x u2 (5b)!ub solution ub, = + + ! ∂ f The mathematical structure of Eqs. (5) can be clarified by! writing them in vector form: i ! The constant coefficientsaij aij come. from the 2 2JacobianmatrixA (6)∂f/∂u evaluated at the constant base = ∂u j !ub × = 2 solution ub, ! ∂t δu Aδu D∂ δu, (7) The mathematical structure of Eqs. (5) can be clarified by! writing them in vector form: = + x ! ∂ fi aij where D.is the 2 2diffusionmatrixpreviouslyintroducedinEq.(3).BecauseEq.(7)islinearwithconstant(6) 2 = ∂u j ∂t δu Aδu D∂x δu, !ub ×(7) = + coefficients! and because the boundaries are periodic or at infinity, we can use translational symmetry to seek The mathematical structure of Eqs. (5) can be clarified by! writing them in vector form: where D is the 2 2diffusionmatrixpreviouslyintroducedinEq.(3).BecauseEq.(7)islinearwithconstant3 ! × Diffusion can be suppressed in several ways. Mathematically, we simply set the diffusion coefficients to zero. Experimentally, coefficients and because the boundaries are periodic or at infinity, we can use translationalwe can symmetry stir2 the chemicals to seek to eliminate spatial nonuniformity. ∂t δu Aδu D∂x δu, (7) 3Diffusion can be suppressed in several ways. Mathematically, we simply set the diffusion coefficients= to+ zero. Experimentally, we can stir the chemicals to eliminatewhere spatialD is nonuniformity. the 2 2diffusionmatrixpreviouslyintroducedinEq.(3).BecauseEq.(7)islinearwithconstant × 3 coefficients and because the boundaries are periodic or at infinity, we can use translational symmetry to seek 3 3Diffusion can be suppressed in several ways. Mathematically, we simply set the diffusion coefficients to zero. Experimentally, we can stir the chemicals to eliminate spatial nonuniformity.
3 the spatially uniform state is stable in the absence of diffusion3 but can become unstable to nonuniform perturbations precisely because of diffusion. Further, for many conditions the instability first occurs at a finite wave length and so a cellular pattern starts to appear. For simplicity, we will discuss the one-dimensional case. If we assume that the evolution equations have rotational symmetry in higher dimensions, the two- and three-dimensional cases are identical mutatis mu- tandis:theone-dimensionalLaplacian∂2 becomes a higher-dimensional Laplacian 2,thecombinationqx x ∇ becomes everywhere a dot product q x with a wave vector q,andthewavenumbersquareq2 becomes the the spatially uniform state is ·stable in the absence of diffusion3 but can become unstable to nonuniform quantity q q. perturbations· precisely because of diffusion. Further, for many conditions the instability first occurs at a The followingfinite wave discussion length is and not so mathematically a cellular pattern difficult starts to but appear. has many details. You will likely best appreciate the discussion if you take your time and derive the results for yourself in parallel3 with the text. The goal of For simplicity,the spatially we will uniform discuss state the is stable one-dimensionalin the absence case. of If diffusion we assumebut can that become the evolutionunstable equationsto nonuniform have the discussion is to derive, and then to understand physically, conditions that are sufficient for the real parts rotationalperturbations symmetry precisely in higher because dimensions, of diffusion. the two- Further, and three-dimensional for many conditions cases the are instability identical firstmutatis occurs mu- at a of all growth ratesfinite to wave be negative. length and When so a cellular these2 conditionspattern starts are to appear. first violated and instability occurs,2 it is then tandis:theone-dimensionalLaplacian∂x becomes a higher-dimensional Laplacian ,thecombinationqx important to think about the values of the wave numbers corresponding to the fastest growing∇ modes.2 becomesFor everywhere simplicity, a we dot will product discussq thex with one-dimensional a wave vector case.q,andthewavenumbersquare If we assume that the evolutionq equationsbecomes the have · quantityrotationalq q. symmetry in higher dimensions, the two- and three-dimensional cases are( identical, ) mutatis mu- We begin by assuming· that we have somehow found a stationary uniform base solution ub u1b u2b .This tandis:theone-dimensionalLaplacian∂2 becomes a higher-dimensional Laplacian= 2,thecombinationqx satisfiesThe the Turing following model discussion with all is partial not mathematically derivativesx set difficult to zero, but leading has many to f details.(ub) You0 or will∇ likely best appreciate becomes everywhere a dot product q x with a wave vector q,andthewavenumbersquare= q2 becomes the the discussion if you take your time and derive· the results for yourself in parallel with the text. The goal of quantity q q. f (u , u ) 0, (4a) the discussion is to· derive,the spatially and then uniform to understand1 1b state2b is physically,=stable in conditions the absence that of are diffusion sufficient3 but for can the realbecome partsunstable to nonuniform the spatially uniform state is stableofin all the growthThe absence following rates to ofperturbations discussion be diffusion negative. is not3 When preciselybut mathematicallyf2( canu these1b, because becomeu2 conditionsb) difficult0 of.unstable diffusion. are but first hasto violatedmany Further, nonuniform details. and for instability You many will conditions likely occurs, best(4b) it appreciate the is then instability first occurs at a the discussion if you take your time and derive= the results for yourself in parallel with the text. The goal of perturbations precisely because ofimportant diffusion. to think Further, aboutfinite for the wave many values length conditions of the and wave so a numbers the cellular instability corresponding pattern startsfirst occurs to to appear. the fastest at a growing modes. These are two nonlinearthe discussion equations is to derive, in two and unknowns. then to understand Finding a physically, uniform solution conditions can that be are hard sufficient since there for the is real no parts finite wave length and so a cellularWe pattern begin starts by assuming to appear. that we have somehow found a stationary uniform base solution ub (u1b, u2b).This systematic wayof to all find growth even rates aFor single simplicity, to be solution negative. we of When awill set discuss of these nonlinear conditions the one-dimensional equations. are first violatedNumerical case. and methods If instability we assume= such occurs, as that the it the is then evolution equations have the spatially uniform state is stable in the absence of diffusion3 satisfiesbut can the become Turingunstable model withto nonuniform all partial derivatives set to zero, leading to f(u ) 0 or Newton methodimportant can find toaccurate thinkrotational about approximations the symmetry values of in to the solutions higher wave numbers dimensions, of nonlinear corresponding the equations two- to and theb but three-dimensional fastest= only if growing a good modes. guess cases are identical mutatis mu- perturbations precisely becauseFor simplicity, of diffusion. we Further, will discuss for many the conditions one-dimensional the instability case. first If occurs we assume at a that the evolution2 equations have 2 for a solution isWe already begin by known. assumingtandis For:theone-dimensionalLaplacian that two we nonlinear have somehow equations found like a stationary Eq.∂x (4),becomes uniform a graphical a higher-dimensionalbase solution way tou find solutions(u Laplacian, u ).This ,thecombinationqx rotational symmetry in higher dimensions, the two- and three-dimensionalf1( casesu1b, u2 areb) identical0, mutatis mu- b 1b 2b (4a)∇ finite wave length and so a cellular pattern starts to appear. becomes everywhere a dot product q x with a wave vector q,andthewavenumbersquare= q2 becomes the that is sometimes2satisfies useful the is Turing to plot model the nullclines with all partialof each derivatives equation. set=2 to An zero, equation leading of to thef(ub form) 0for1(u1, u2) 0 tandis:theone-dimensionalLaplacian∂ becomes a higher-dimensional Laplacianf2(u1b, u2b) ·,thecombination0. qx = = (4b) For simplicity, we will discuss the one-dimensional case. If we assumex that the evolutionquantity equations(q )q. have defines an implicit relation u2 g1 u·1 between the two variables∇= u1 and2 u2 called the nullcline of that rotational symmetry in higherbecomes dimensions, everywhere the two- a dot and product three-dimensionalq x with cases a wave are identical vector= qmutatis,andthewavenumbersquare mu- f1(u1b, u2b) 0,q becomes the (4a) equation.These If· the are functions two nonlinearfi The(u1, equations followingu2) are sufficiently in discussion two unknowns. simple, is not Finding mathematically their nullclines a= uniform difficultg solutioni (u1) can but can sometimes has be many hard since details. be found there You is no will likely best appreciate tandis:theone-dimensionalLaplacianquantity q ∂q2 .becomes a higher-dimensional Laplacian 2,thecombinationqx f (u , u ) 0. (4b) · x explicitlysystematic and then way plotted to∇ find onthe even a discussion single a single plot if solution with you takeaxes of a your labelled set2 of time1b nonlinear by2 andb u=1 deriveand equations.u the2.Anyintersectionofthetwo results Numerical for yourself methods in such parallel as the with the text. The goal of becomes everywhere a dot product q x with a wave vector q,andthewavenumbersquareq2 becomes the The following· discussion isnullclines not mathematicallyNewton is then method a solution difficult can offindthe the but discussion accurate nonlinear has many approximations is equations. to details. derive, and You to solutions then will to likely understand of nonlinear best appreciate physically, equations but conditions only if a thatgood are guess sufficient for the real parts quantity q q. These are two nonlinear equations in two unknowns. Finding a uniform solution can be hard since there is no for a solution is already known. For two nonlinear equations like Eq. (4), a graphical way to find solutions · the discussion if you take yourBy linearizing time and aboutsystematic derive the the base way resultsof to state all find growth foru evenb,youcanshowthatanarbitraryinfinitesimalperturbation yourself a rates single to solution in be parallel negative. of a set with ofWhen nonlinear the these text. equations. conditions The goal Numerical of are first methods violatedδu(t, x such) and as instability the occurs, it is then The following discussion is not mathematically difficult but has manythat details. is sometimesNewton You method will usefulimportant likely can is find best to plot accurate toappreciate think the nullclines approximations about theof values each to solutions equation. of the wave of An nonlinear numbers equation equations corresponding of the but form onlyf1( if tou a1 the, goodu=2) fastest guess0 growing modes. the discussion is to derive,( andδu1( thent, x), toδu understand2(t, x)) of the physically, base state will conditions evolve in time that according are sufficient to the for following the real linear parts constant-coefficient = the discussion if you take your time and derive the results for yourselfdefines infor parallel an a implicit solution with relation is the already text.u2 Theknown. goalg1(u For1 of) twobetween nonlinear the two equations variables likeu Eq.1 and (4),u a2 graphicalcalled the way nullcline to find solutions of that of all growth rates to be negative.evolution When equations: these conditionsLinearWe begin stability? are by= first assuming violated that and we instability have somehow occurs, found it a is stationary then uniform base solution ub (u1b, u2b).This the discussion is to derive, and then to understand physically, conditionsequation.that that If is are the sometimes sufficient functions useful forfi ( theu is1, real tou2 plot) partsare the sufficientlynullclines simple,of each their equation. nullclines An equationgi (u1) can of the sometimes form f1(u be1, u found2) 0 = important to think about the values of the wave numberssatisfies corresponding the Turing model to the with fastest all partial growing derivatives modes. set to zero, leading to f(ub) = 0 or of all growth rates to be negative. When these conditions are firstexplicitly violateddefines and and instability an then implicit plotted occurs, relation on a singleitu is2 then plotg1(u1 with) between axes labelled the two2 variables by u1 andu1uand2.Anyintersectionofthetwou2 called the nullcline of that ∂t δu1 a=11δu1 a12δu2 D1∂x δu1, (5a) = important to think about the values of the wave numbers correspondingnullclines toequation. the is fastest then If a the solution growing functions of modes. thef =( nonlinearu , u ) are+ equations. sufficiently+ simple, their nullclines g (u ) can sometimes be found We begin by assuming that we have somehow found a stationaryi uniform1 2 base solution u 2 (u , u ).Thisi 1 ∂t δu2 a21δu1 a22δu2 D2∂b δu2. f11b(u1b2,bu2b) 0, (5b) (4a) explicitly and then plotted on= a single+ plot with+ axes labelledx = by u1 and u2.Anyintersectionofthetwo= We begin by assuming thatsatisfies we have somehowthe Turing found model a stationary with all uniform partialBy linearizing base derivatives solution aboutu set theb tobase(u zero,1b state, u2b leading)u.Thisb,youcanshowthatanarbitraryinfinitesimalperturbation to f(ub) 0 or δu(t, x) nullclines is then= a solution of the nonlinear equations.= f2(u1b, u2b) 0. = (4b) satisfies the Turing model with all partial derivatives set to zero, leading(δu1(t, tox),f(δuub2)(t, x0))orof the base state will evolve in time according∂ to/ the∂ following= linear constant-coefficient The constant coefficients=aij come from the 2 2JacobianmatrixA f u evaluated at the constant base evolutionBy equations: linearizing about the base state×ub,youcanshowthatanarbitraryinfinitesimalperturbation= δu(t, x) solution ub, f1(u1b, Theseu2b) are0, two nonlinear equations in two unknowns. Finding(4a) a uniform solution can= be hard since there is no f1(u1b, u2b) 0, (δu (t, x), δu (t, x))=of the base state(4a) will evolve in time according to the following linear constant-coefficient = 1 2systematic way to find even∂ fi a single solution of a set of nonlinear equations. Numerical methods such as the f2(u1b, u2b) 0. aij . 2 (4b) (6) f2(u1b, u2b) 0. evolution equations: ∂t δ(4b)u1 a11δu1 a12δu2 D1∂x δu1, (5a) Newton= method can= find= accurate∂u j +u approximations+ to solutions of nonlinear equations but only if a good guess = ! b 2 for a solution∂ ist δ alreadyu2 a21 known.δu1 ! a22 Forδu2 twoD nonlinear2∂x δu2 . equations like Eq. (4), a graphical(5b) way to find solutions These are two nonlinear equationsThe mathematical in two unknowns. structure of Finding Eqs. (5) a can uniform be clarified∂=t δ solutionu1 a by11!+δ canwritingu1 bea12 hard themδ+u2 since inD vector1∂x δ thereu1, form: is no (5a) These are two nonlinear equations in two unknowns. Finding a uniform solution can be hard since there is no = ! + + that is sometimes useful is to plot the nullclines2 of each equation. An equation of the form f1(u1, u2) 0 systematic way to find evensystematic a single solution way to of find a set even of nonlinear a single equations. solution Numerical of a set of methods nonlinear such equations. as the∂t δu2 Numericala21δu1 a22δ methodsu2 D2∂x such∂δu/2∂. as the (5b) The constant coefficients aij come from the 2 =2Jacobianmatrix2+ +A f u evaluated at the constant base = Newton method can find accurate approximations to solutions of nonlinear equations butdefines only if an a good implicit∂ guesst δu relationAδ×u uD2 ∂ δug1,(u1) between= the two variables u1 and(7)u2 called the nullcline of that Newton method can find accurate approximationssolution ub, to solutions of nonlinear= equations+ =x but only if a good guess for a solution is already known. For two nonlinear equations like Eq. (4),The a graphical constant coefficientsequation. way to finda Ifij solutions thecome functions from the 2fi (u2Jacobianmatrix1∂, uf 2) are sufficientlyA ∂ simple,f/∂u evaluated their nullclines at the constantgi (u1 base) can sometimes be found for a solution is already known. For two nonlinear equations like Eq. (4), a graphical× i way. to find= solutions where D is the 2solution2diffusionmatrixpreviouslyintroducedinEq.(3).BecauseEq.(7)islinearwithconstantub, explicitly and then plottedaij on a single plot with axes labelled by u1 and u2.Anyintersectionofthetwo(6) that is sometimes useful is to plot the nullclines of each equation. An equation of the form f1(u1, u2) 0 = ∂u j that is sometimes useful is to plot the nullclines× of each equation.= An equation of the!∂ubfi form f1(u1, u2) 0 defines an implicit relation u2 g1(u1) between the twocoefficients variables andu1 and becauseu2 called the boundariesnullclines the nullcline is are then ofwith periodic that a Jacobian: solution or at of infinity,aij the nonlinear! we can. equations. use translational symmetry to seek (6) = = !∂u = defines an implicit relation u2 gThe1(u mathematical1) between structure the two of variables Eqs. (5) canu1 beand clarifiedu2 called by writingj theub nullcline them in vector of that form: equation. If the functions fi (u1, u2) are sufficiently simple,3 their nullclines gi (u1) can sometimes be found ! ! δ ( , ) ( Diffusion,= ) can be suppressed inBy several linearizing ways. Mathematically, about the base we( simply state) u setb,youcanshowthatanarbitraryinfinitesimalperturbation the! diffusion coefficients to zero. Experimentally, u t x explicitly and then plottedequation. on a single If plot the withfunctions axes labelledfwei u can1 stiru by2 theuare1 chemicalsand sufficientlyTheu2 mathematical.Anyintersectionofthetwo to eliminate simple, spatial structure their nonuniformity. of nullclines Eqs. (5) cang bei u clarified1 can by sometimes2! writing them be in found vector form: = (δu1(t, x), δu2(t, x))∂oft δ theu baseAδu stateD will∂x!δu evolve, in time according to the following(7) linear constant-coefficient nullclines is then a solutionexplicitly of the nonlinear and then equations. plotted on a single plot with axes labelled by u1 and u=2.Anyintersectionofthetwo+ evolution equations: ∂ δu Aδu D∂2δu, (7) nullclines is then a solution of the nonlinear equations. t x By linearizing about the base state ub,youcanshowthatanarbitraryinfinitesimalperturbationwhere D is the 2 2diffusionmatrixpreviouslyintroducedinEq.(3).BecauseEq.(7)islinearwithconstantδu(t, x) = + × = 2 (δu1(t, x), δu2(t, x)) of the base state will evolve in time according tocoefficients the followingwhere andD is linear because the 2 constant-coefficient2diffusionmatrixpreviouslyintroducedinEq.(3).BecauseEq.(7)islinearwithconstant the boundaries are periodic3 or at∂ infinity,t δu1 wea11δ canu1 usea translational12δu2 D1∂ symmetryx δu1, to seek (5a) By linearizing about the base state ub,youcanshowthatanarbitraryinfinitesimalperturbation× = δu+(t, x) + evolution equations: coefficients and because the boundaries are periodic or at infinity, we can use translational2 symmetry to seek 3Diffusion can be suppressed in several ways. Mathematically, we∂ simplyt δu2 seta the21δ diffusionu1 a22 coefficientsδu=2 D to2∂ zero.δu2 Experimentally,. (5b) (δu1(t, x), δu2(t, x)) of the base state will evolve in time according to the following linear constant-coefficient= + + x we2 can stir the3Diffusion chemicals can to be eliminate suppressed spatial in several nonuniformity. ways. Mathematically, we simply set the diffusion coefficients to zero. Experimentally, evolution∂t δ equations:u1 a11δu1 a12δu2 D1∂ δu1, (5a) = + + x we can stir the chemicalsThe constant to eliminate coefficients spatial nonuniformity.aij come from the 2 2JacobianmatrixA ∂f/∂u evaluated at the constant base 2 × = ∂t δu2 a21δu1 a22δu2 D2∂ δu2. solution u , (5b) = + + ∂xδ δ δ b ∂2δ , t u1 a11 u1 a12 u2 D1 x u1 3 ∂ fi(5a) = + + aij . (6) The constant coefficients aij come from the 2 2JacobianmatrixA ∂f/∂u evaluated at the constant base2 3 ∂ δ δ δ ∂ δ . = ∂u j × t =u2 a21 u1 a22 u2 D2 x u2 (5b)!ub solution ub, = + + ! ∂ f The mathematical structure of Eqs. (5) can be clarified by! writing them in vector form: i ! The constant coefficientsaij aij come. from the 2 2JacobianmatrixA (6)∂f/∂u evaluated at the constant base = ∂u j !ub × = 2 solution ub, ! ∂t δu Aδu D∂ δu, (7) The mathematical structure of Eqs. (5) can be clarified by! writing them in vector form: = + x ! ∂ fi aij where D.is the 2 2diffusionmatrixpreviouslyintroducedinEq.(3).BecauseEq.(7)islinearwithconstant(6) 2 = ∂u j ∂t δu Aδu D∂x δu, !ub ×(7) = + coefficients! and because the boundaries are periodic or at infinity, we can use translational symmetry to seek The mathematical structure of Eqs. (5) can be clarified by! writing them in vector form: where D is the 2 2diffusionmatrixpreviouslyintroducedinEq.(3).BecauseEq.(7)islinearwithconstant3 ! × Diffusion can be suppressed in several ways. Mathematically, we simply set the diffusion coefficients to zero. Experimentally, coefficients and because the boundaries are periodic or at infinity, we can use translationalwe can symmetry stir2 the chemicals to seek to eliminate spatial nonuniformity. ∂t δu Aδu D∂x δu, (7) 3Diffusion can be suppressed in several ways. Mathematically, we simply set the diffusion coefficients= to+ zero. Experimentally, we can stir the chemicals to eliminatewhere spatialD is nonuniformity. the 2 2diffusionmatrixpreviouslyintroducedinEq.(3).BecauseEq.(7)islinearwithconstant × 3 coefficients and because the boundaries are periodic or at infinity, we can use translational symmetry to seek 3 3Diffusion can be suppressed in several ways. Mathematically, we simply set the diffusion coefficients to zero. Experimentally, we can stir the chemicals to eliminate spatial nonuniformity.
3 the spatially uniform state is stable in the absence of diffusion3 but can become unstable to nonuniform perturbations precisely because of diffusion. Further, for many conditions the instability first occurs at a finite wave length and so a cellular pattern starts to appear. For simplicity, we will discuss the one-dimensional case. If we assume that the evolution equations have rotational symmetry in higher dimensions, the two- and three-dimensional cases are identical mutatis mu- tandis:theone-dimensionalLaplacian∂2 becomes a higher-dimensional Laplacian 2,thecombinationqx x ∇ becomes everywhere a dot product q x with a wave vector q,andthewavenumbersquareq2 becomes the · quantity q q. · The following discussion is not mathematically difficult but has many details. You will likely best appreciate the discussion if you take your time and derive the results for yourself in parallel with the text. The goal of the discussion is to derive, and then to understand physically, conditions that are sufficient for the real parts of all growth rates to be negative. When these conditions are first violated and instability occurs, it is then important to think about the values of the wave numbers corresponding to the fastest growing modes. We begin by assuming that we have somehow found a stationary uniform base solution u (u , u ).This b = 1b 2b satisfies the Turing model with all partial derivatives set to zero, leading to f(u ) 0 or b = f (u , u ) 0, (4a) 1 1b 2b = f (u , u ) 0. (4b) 2 1b 2b = These are two nonlinear equations in two unknowns. Finding a uniform solution can be hard since there is no systematic way to find even a single solution of a set of nonlinear equations. Numerical methods such as the Newton method can find accurate approximations to solutions of nonlinear equations but only if a good guess for a solution is already known. For two nonlinear equations like Eq. (4), a graphical way to find solutions that is sometimes useful is to plot the nullclines of each equation. An equation of the form f (u , u ) 0 1 1 2 = defines an implicit relation u g (u ) between the two variables u and u called the nullcline of that 2 = 1 1 1 2 equation. If the functions fi (u1, u2) are sufficiently simple, their nullclines gi (u1) can sometimes be found explicitly and then plotted on a single plot with axes labelled by u1 and u2.Anyintersectionofthetwo nullclines is then a solution of the nonlinear equations. By linearizing about the base state u ,youcanshowthatanarbitraryinfinitesimalperturbationδu(t, x) b = (δu1(t, x), δu2(t, x)) of the base state will evolve in time according to the following linear constant-coefficient evolution equations: aparticularsolutionδu(t, x) that is a constant vector δuq times an exponential in time times an exponential in space: ∂ δu a δu a δu D ∂2δu , (5a) t 1 = 11 1 + 12 2 + 1 x 1 2 δ ∂ δu a δu aσqδtu iqx D ∂ δuu.1q σq t iqx (5b) t 2 =δu21 δ1u+q e22 e2 + 2 x 2 e e , (8) = = δu2q ! " The constant coefficients aij come from the 2 2JacobianmatrixA ∂f/∂u evaluated at the constant base with growth rate σq and waveδ number( , ) × q. Note that both components=δ of the perturbation vector δu have the solution ub, aparticularsolution u t x that is a constant vector uq times an exponential in time times an exponential same dependencein space: on time and space. Only∂ withf this assumption can the spatial and temporal dependencies aparticularsolutionδu(t, x) that is a constant vectora δuiq times. an exponential in time times an exponential(6) be eliminated completely from the linearizedij evolution equationsδu and a simple solution found. in space: = ∂u j u σq t iqx 1q σq t iqx Linear δstability?u δu! qb e e e e , (8) = !δu = δu2q TheIf mathematical we substitute structure Eq. (8) of Eqs.δ intou (5) Eq.δu cane (7), beσq t clarifiede divideiqx by out! writing the1q exponentials, themeσq t eiqx in! vector, " form: and collect some terms,(8) we obtain the q !δ followingaparticularsolutionwith eigenvalue growth rate problemδu(σtq,=andx) that wave is number a constant= q. Note vectoru2q thatδu bothq times components an exponential of the perturbation in time times vector an exponentialδu have the ∂ δ δ !aparticularsolution∂2δ ", δu(t, x) that is a constant vector δuq times an exponential in time times an exponential in space:same dependence on time andt u space.A uOnlyD withx u this assumption can the spatial and temporal(7) dependencies with growth rate σq and wave number q. Note that= bothA+in componentsq space:δuq σq δ ofuq the, perturbation vector δu have the (9) be eliminated completely from the linearized evolution= δu equations and a simple solution found. same dependence on time and space. Only with this assumptionσq t iqx can the1 spatialq σ andq t iqx temporal dependencies δ where D is the 2 2diffusionmatrixpreviouslyintroducedinEq.(3).BecauseEq.(7)islinearwithconstantδu δuq e e e e , σq t iqx u1q (8)σq t iqx where the the× 2 2realmatrixAq is defined by δu δu δuq e e e e , (8) be eliminatedcoefficients completely andIf because we substitute× from the boundaries the Eq. linearized (8) are into periodic evolution Eq.= (7), or at divide equations infinity, out= we and the can exponentials, a usesimple2q translational solution and symmetry found. collect= some to seek terms,= weδ obtainu2q the ! " ! " following eigenvalue problem 2 3 with growth rate σq and wave number q. Note thata bothD componentsq σ a of the perturbation vector δu have the δ If we substituteDiffusion can Eq. be suppressed (8) into in Eq. several (7), ways. divide Mathematically,= outperturbation the2 exponentials, wewith with simply growth 11growth set the and raterate diffusion1 collect q andand coefficients wave wave some12 number tonumber terms, zero.. q. Experimentally,q we. Note obtain that the both components of the perturbation vector u have the we can stir the chemicals to eliminate spatialA nonuniformity.q A Dq Aq −δuq σq δuq, 2 (10)(9) followingsame eigenvaluedependence problem on time and= space.− Only= withsame thisdependencea assumption21 = ona22 time can andD the2q space. spatialOnly and withtemporal this assumption dependencies can the spatial and temporal dependencies ! − " be eliminatedwhere the completely the 2 2realmatrix from theA linearizedδAuq is definedσbe evolutionδ eliminatedu , by equations completely and from a simple the linearized solution evolution found.(9) equations and a simple solution found. × q q q q Eq. (9) tells us that the growth rate σq and= constant vector δuq form an eigenvalue-eigenvector pair of the where theIf the we 2 substitute2realmatrix Eq.A (8)is into defined Eq. (7), by divide3 If out we the substitute exponentials, Eq.2 (8) into and Eq. collect (7), divide some out terms, the we exponentials, obtain the and collect some terms, we obtain the matrix A×q,andthatthereisonesuch2q 2eigenvalueproblemforeachwavenumber2 a11 D1q a12 q.Theeigenvalue following eigenvalue problem Aq ×A Dfollowingq eigenvalue− problem 2 . (10) problem for a given q has generally two= linearly− = independenta21 eigenvectorsa22 D2q that we will denote by δu 2 ! − " A δu σ δu , iq (9) 2 a11 DA1qqδuq σaq12δuq, q q q q (9) for i 1, 2. If the correspondingAq A Dq eigenvalues− are σiq,theparticularsolutionwithwavenumber= 2 . (10)= q will have = Eq. (9) tells us= that− the growth=For rateeach σwaveaq21and number constanta q22 the vectoreigenvalueD2q δu qproblemform an gives eigenvalue-eigenvector solution of the form: pair of the thewhere form: the the 2 2realmatrixA is! definedwhere by the the− 2 2realmatrix" Aq is defined by matrix Aq,andthatthereisonesuch2q 2eigenvalueproblemforeachwavenumber× q.Theeigenvalue × σ1q t σ2q t iqx Eq. (9) tells us that the growth rate σ and constantc1q δu1 vectorq e× δu cform2q δu2 anq e eigenvalue-eigenvectore , pair of the 2 (11) problem for a givenqq has generally two linearlyq independent eigenvectors that we willa denoteD q by δuiqa + 2 2 11 1 12 . matrix Aq,andthatthereisonesuch2 2eigenvalueproblemforeachwavenumber2 a11 D1q a12Aq qA.TheeigenvalueDq − 2 (10) for i 1, 2. If the correspondingA A eigenvaluesDq are σiq,theparticularsolutionwithwavenumber= −. = a21 q 4willa have(10)22 D2q where the coefficients= ciq are complex×q # constants, that− depend on$ the initial2 perturbation! at t 0 .This− " problem for a giventhe form:q has generally two= linearly− independent= eigenvectorsa21 a22 thatD we2q will denote by δuiq = solution decays if Re(σiq)<0fori 1, 2.! An arbitrary perturbation− δ"u(t, x) is a superposition of σ1q t σ2q t iqx σ δ for i 1, 2. If the corresponding eigenvalues are σ=iq,theparticularsolutionwithwavenumberc1qEq.δu1 (9)q e tells usc2q thatδu2 theq e growthe , rate q andq constantwill have vector uq form(11) an eigenvalue-eigenvector pair of the = expressionsEq. (9) tells like us Eq. that (11) the over growth all wave rate σ numbersand constantq.Theuniformsolution vector+ δu form anu eigenvalue-eigenvectorb is stable if both eigenvalues pair ofσ theiq the form: q matrix Aq,andthatthereisonesuch2q 2eigenvalueproblemforeachwavenumberq.Theeigenvalue where the coefficients c are complex constants, that depend on the initial perturbation× at t 04.This havematrix negativeAq,andthatthereisonesuch2 real parts for all waveiq numbersσ1q t # 2eigenvalueproblemforeachwavenumberproblemq,i.e.,ifmaxσ2q t foriqx ai givenmaxqqRehas($σiq generally)<0. two linearlyq.Theeigenvalue independent eigenvectors that we will denote by δu c1q δu1q e ×c2q δu2q e e , (11) = iq solution decays if Re(σiq)<0fori 1, 2. An arbitrary perturbation δu(t, x) is a superposition of problem for a given q has generally+ two linearlyfor i 1 independent, 2. If the corresponding eigenvectors eigenvalues that we are willσiq denote,theparticularsolutionwithwavenumber by δuiq q will have The characteristic polynomial for the eigenvalue= problem Eqs. (9,10) can be written 4 where the coefficientsexpressionsciq are like complex Eq.# (11) constants, over all wave that dependnumbers= onq$.Theuniformsolution the initial perturbationub atist stable0 if.This both eigenvalues σiq for i 1, 2. If the corresponding eigenvaluesthe are form:σiq,theparticularsolutionwithwavenumber= q will have solution decays= ifhave Re( negativeσ )< real0for partsi for1 all, 2. wave An numbers arbitraryq,i.e.,ifmax perturbationi maxδu(qt,Rex)(σisiq)< a superposition0. σ1q t of σ2q t iqx the form: iq 2 c1q δu1q e c2q δu2q e e , (11) 0= det Aq σq I σq (trAq )σq det Aq (12) expressions like Eq. (11) over all wave numbers q.Theuniformsolutionσ t u σ ist stableiqx if both eigenvalues +σ The characteristic polynomial= forc theδ−u eigenvaluee 1q = c problem−δu e Eqs.b 2q (9,10)+e , can be written iq (11) 4 1q 1qwhere the2 coefficientsq 2q ciq are complex# constants, that depend on$ the initial perturbation at t 0 .This have negative real parts for all wave numbers q,i.e.,ifmax# $i max+ q Re(σiq)<0. where tr(Aq ) denotes the trace (sum of diagonal elements) of the matrix Aq and det(Aq ) the determinant. = where the coefficients c are complex constants,solution that decays depend2 if Re on(σiq the)< initial0for perturbationi 1, 2. An at t arbitrary04.This perturbation δu(t, x) is a superposition of iq 0# det Aq σq I σq (trA$q )σq det Aq = (12) The characteristicThe eigenvalues polynomial are then for theσ1q eigenvalueand σ2q given= problem byexpressions Eqs.− (9,10)= like can Eq.− be (11) written over+ all wave numbers q.Theuniformsolution= ub is stable if both eigenvalues σiq solution decays if Re(σiq)<0fori 1, 2. An arbitrary perturbation δu(t, x) is a superposition of ( ) = #have negative$ real parts for all wave numbers q(,i.e.,ifmax) i maxq Re(σiq)<0. expressionswhere like tr A Eq.q denotes (11) over the all trace wave (sum numbers1 of2 diagonalq1.Theuniformsolution elements) of the matrixub isAq stableand det if bothAq the eigenvalues determinant.σiq 0 det Aq σq I σq (trAq )σq det2 Aq (12) The eigenvalues= are then σ−1qσqand σ=2qtrgivenAq− by (trA+q ) 4detAq (13) have negative real parts for all wave= numbers2 Theq±,i.e.,ifmax characteristic2 i −max polynomialq Re(σiq for)< the0. eigenvalue problem Eqs. (9,10) can be written # $ % where tr(Aq ) denotes the trace (sum of diagonal elements)1 of the1 matrix Aq and det(Aq ) the determinant. 2 The characteristic polynomial for the eigenvalue problem Eqs. (9,10)2 can0 bedet writtenAq σq I σ (trAq )σq det Aq (12) The eigenvalues are then σ and σ given by σq trAq (trAq ) 4detAq q (13) The regions of stability1q (both2q Re σq negative)= and2 instability± 2 (at least− one Re=σq positive)− in the= tr A−q -detAq + 2% # $ plane are shown in Fig. 1. From0 thisdet figureAq orwhere theσq I expression tr(Aσqq) denotes(tr Eq.Aq (13)) theσq trace a simpledet (sumAq criterion of diagonal can elements) be derived of the that(12) matrix Aq and det(Aq ) the determinant. 1 = 1 − 2 = − + determinesThe when regions the of real stabilityσ partsq (both oftrA bothq Re σ eigenvaluesq negative)(trTheAq ) eigenvalues and are4det instability negative:Aq are then (atthe leastσ trace1q and oneσ Re of2q theσgivenq positive) matrix by must(13) in the be tr A negativeq -detAq = 2 ± #2 $− andwhere the determinant trplane(Aq ) aredenotes shown of the the in Fig. matrix trace 1. (sum From must of this be diagonal positive figure or.Forthematrix elements) the expression of the Eq. matrixA (13)Eq. aA simple (10),q and these criterion det(A criteriaq ) canthe be for determinant. derived stability that % q 1 1 The eigenvaluesdetermines are when then theσ realand partsσ ofgiven both by eigenvalues are negative: the trace of the matrix must be negative2 The regionstake of the stability explicit (both form: Re σ negative)1q and2q instability (at least one Re σ positive) inσq the tr AtrA-detq A (trAq ) 4detAq (13) and the determinantq of the matrix must be positive.Forthematrixq A Eq. (10),= 2 theseq ± criteria2 q for stability− q % plane are shown in Fig. 1. From this figure or the expression1 Eq.1 (13) a simple2 criterion can be derived that take the explicit form:trAq a11 a22 (D1 D2)q2 < 0, (14a) σq trAq (trAq ) 4detAq σ (13) σ determines when the real parts of both eigenvalues= =+ are2 The negative:−± regions2 +the of trace stability− of (both the matrix Re q negative) must be negative and instability (at least one Re q positive) in the tr Aq -detAq 2 2 2 and the determinant of the matrix mustdet beAq positivetr(Aa11.ForthematrixaDplane1q a)( area%22 shown(DADq2 inEq.qD Fig.)) (10),q 1.a<12 From thesea021, > this criteria0. figure for or stability the expression Eq.(14b)(14a) (13) a simple criterion can be derived that = q =− 11 + 22 − − 1 + 2 − take the explicit form: determines2 when the real2 parts of both eigenvalues are negative: the trace of the matrix must be negative The regions of stability (both Re σdetq negative)Aq (a11 andD instability1q )(a22 (atD least2q ) onea12 Rea21σ>q positive)0. in the tr Aq -det(14b)Aq If both conditions hold for all wave numbers= qand,thestationaryuniformbasestate− the determinant− of the− matrix mustub beis positive linearly.Forthematrix stable. Aq Eq. (10), these criteria for stability plane are shown in Fig. 1. From this figure or the expression2 Eq. (13) a simple criterion can be derived that trAq a11 a22 (D1 takeD2) theq explicit< 0, form: (14a) Ifdetermines Eq. (2)If instead both when conditions involved the real= holdN partsinteracting for+ of all both wave− eigenvalueschemicals, numbers+ q,thestationaryuniformbasestate we are would negative: needthe to trace solve a ofN theu matrixNb iseigenvalue linearly must stable. be problem negative 2 2 × Eq. (9) for each wavedet numberAq (aq11.ForDN1q )(3,a22 analyticalD2q ) criteriaa12a21 that> 0 all. the eigenvalues of a(14b)N N matrix2 A and theIf determinant Eq. (2) instead of the involved= matrix−N mustinteracting be positive− chemicals,.Forthematrix− we would needAq totrEq.A solveq (10),a a11N thesea22N criteriaeigenvalue(D1 forD2 stability) problemq
4 aparticularsolutionδu(t, x) that is a constant vector δuq times an exponential in time times an exponential in space: δ aparticularsolutionδu(t, x) that is a constantσq t iqx vector δuu1q timesσq t aniqx exponential in time times an exponential δu δuq e e q e e , (8) = = δu2q in space: ! " δ with growth rate σq and wave number q. Note thatσq t bothiqx componentsu1q ofσ theq t iqx perturbation vector δu have the δu δuq e e e e , (8) = = δu2q same dependence on time and space. Only with this assumption! can" the spatial and temporal dependencies be eliminated completely from the linearized evolution equations and a simple solution found. with growth rate σq and wave number q. Note that both components of the perturbation vector δu have the If wesame substitutedependence Eq. (8) on into time Eq. and (7), space. divideOnly out with the thisexponentials, assumption and can collect the spatial some and terms, temporal we obtain dependencies the followingbe eliminated eigenvalue completely problem from the linearized evolution equations and a simple solution found. A δu σ δu , (9) If we substitute Eq. (8) into Eq. (7), divideq outq = theq exponentials,q and collect some terms, we obtain the wherefollowing the the 2 eigenvalue2realmatrix problemA is defined by × q Aq δuq σq δuq, (9) = 2 2 a11 D1q a12 where the the 2 2realmatrixAq A AqDisq defined by − 2 . (10) × = − = a21 a22 D2q ! − " 2 σ 2 a11 δD1q a12 Eq. (9) tells us that the growth rateAq q Aand constantDq vector− uq form an eigenvalue-eigenvector2 . pair of the (10) matrix A ,andthatthereisonesuch2= −2eigenvalueproblemforeachwavenumber= a21 a22 D2q q.Theeigenvalue q × ! − " problem for a given q has generally two linearly independent eigenvectors that we will denote by δuiq Eq. (9) tells us that the growth rate σq and constant vector δuq form an eigenvalue-eigenvector pair of the for i 1, 2. If the corresponding eigenvalues are σiq,theparticularsolutionwithwavenumberq will have matrix= Aq,andthatthereisonesuch2 2eigenvalueproblemforeachwavenumberq.Theeigenvalue the form: × problem for a given q has generally two linearly independent eigenvectors that we will denote by δuiq σ1q t σ2q t iqx c1q δu1q e c2q δu2q e e , (11) for i 1, 2. If the corresponding eigenvalues+ are σiq,theparticularsolutionwithwavenumberq will have = 4 wherethe the form: coefficients ciq are complex# constants, that depend on$ the initial perturbation at t 0 .This σ1q t σ2q t iqx = solution decays if Re(σ )<0fori c1q1,δ2.u1q e An arbitraryc2q δu2 perturbationq e e ,δu(t, x) is a superposition of (11) iq = + expressions like Eq. (11) over all wave numbers q.Theuniformsolutionub is stable if both eigenvalues σ4iq where the coefficients ciq are complex# constants, that depend on$ the initial perturbation at t 0 .This (σ )< = havesolution negative decays real parts if forRe( allσ wave)< numbers0fori q,i.e.,ifmax1, 2. An arbitraryi maxq Re perturbationiq 0. δu(t, x) is a superposition of iq = The characteristicexpressions like polynomial Eq. (11) over for the all eigenvalue wave numbers problemq.Theuniformsolution Eqs. (9,10) can be writtenub is stable if both eigenvalues σiq have negative real parts for all wave numbers q,i.e.,ifmaxi maxq Re(σiq)<0. Linear stability? Solving2 the eigenvalue problem. 0 det Aq σq I σ (trAq )σq det Aq (12) The characteristic polynomial= for the eigenvalue− = problemq − Eqs.+ (9,10) can be written ( ) # $ ( ) where tr Aq denotes the trace (sum of diagonal elements)2 of the matrix Aq and det Aq the determinant. 0 det Aq σq I σ (trAq )σq det Aq (12) The eigenvalues are then σ1q and σ2q=given by − = q − + # $ where tr(Aq ) denotes the trace (sum1 of diagonal1 elements) of the matrix Aq and det(Aq ) the determinant. σ trA (trA )2 4detA (13) The eigenvalues are then σ1q andq =σ2q2givenq ± by2 q − q % 1 1 2 The regions of stability (both Re σq negative)σ andtrA instability(tr (atA least)detA one4det ReAσq positive) in the tr Aq -detAq (13) q = 2 q ± 2 q −q q plane are shown in Fig. 1. From this figure or the expressionstable % Eq. (13) a simple criterion can be derived that Re σˇ˛˝¸< 0 Re σˇ˛˝¸> 0 determines when the real parts of both eigenvalues are negative:Im σˇ˛˝¸≠ 0 the traceIm σˇ˛˝¸ of≠ 0 the matrix must be negative The regions of stability (both Re σq negative) and instability (at least one Re σq positive) in the tr Aq -detAq and the determinant of the matrix must be positive.ForthematrixoscillatoryA Eq. (10), these criteria for stability plane are shown in Fig. 1. From this figure or the expression Eq.q (13) a simple criterion can be derived that Im σ1,2 = 0 take the explicit form: stationary Re σ > 0 determines when the real parts of both eigenvalues are negative: the trace of the1,2 matrix must be negative trA and the determinant of thetrA matrixa musta be positive(D .ForthematrixD )q2 < 0, Aq Eq. (10), theseq criteria for(14a) stability q = 11 + 22 − 1 + 2 take the explicit form: 2 2 Im σ1,2 = 0 det Aq (a11 D1q )(a22 D2q )Re σ1 a> 012 Rea σ212 < 0> 0. (14b) = − − −2 trAq a11 a22 (D1 D2)q < 0, (14a) If both conditions hold for all wave numbers= q+,thestationaryuniformbasestate− + ub is linearly stable. 2 2 det Aq (a11 D1q )(a22 D2q ) a12a21 > 0. (14b) If Eq. (2) instead involved N interacting= chemicals,− we would− need to− solve a N N eigenvalue problem Figure 1: Stability regions of the Turing system in the trAq -detAq plane. The plot shows regions of different characteristics of the two eigenvalues σ and σ calculated from Eq. (12).× The parabola det A 1 trA Eq. (9)If both for each conditions wave number hold forq.For all waveN numbers3, analyticalq,thestationaryuniformbasestate criteria1 that2 all the eigenvalues ofu aisN linearlyqN= matrix4 stable.q Aq divides≥ the plane into two halves: above this curve the two eigenvalues are complexb (and× are complex conjugates of one another), below this curve both eigenvalues are real. The shaded region is the stable 4 Arealsolutioncanbeobtainedasusualbyaddingthecomplexconjugatesolution.If Eq. (2) instead involved N interactingregion, Re σ1,2 chemicals,< 0. Over the unshaded we would portion there need is at least to onesolve eigenvalue a N with positiveN eigenvalue real part. A problem stationary instability occurs passing through the negative tr Aq axis to negative values× of det Aq ,whereasan Eq. (9) for each wave number q.ForN 3, analytical criteria that all the eigenvalues of a N N matrix Aq oscillator instability≥ occurs passing through the positive det Aq axis to positive values of tr Aa . × 4 4Arealsolutioncanbeobtainedasusualbyaddingthecomplexconjugatesolution. have negative real parts become cumbersome to derive and to work with5 and so the case N 2thatTuring = discussed in his paper hits the mathematical sweet spot of being manageable and leading to interesting results. For experiments with N > 3reactingchemicals,itisusuallyeasiesttostudythecorrespondingmodelby numerical methods. With modern4 computers and modern numerical algorithms, it is straightforward to find all the eigenvalues of a N N matrix quickly for N as large as 10,000. As you can imagine, it would be × exceedingly difficult to map out the reaction rates for so many interacting chemicals. Progress in studying the linear stability of chemical systems with large N is therefore limited by scientific knowledge, not by the ability to calculate eigenvalues. We now discuss the physical meaning and implications of the mathematical criterion Eq. (14) in the context of pattern formation6.Therearemanyabstractsymbolsandequationshereandsomecarefulthinkingis needed to see how to extract some physical insights. A first step is to identify the precise scientific question of interest, not just the technical mathematical question of when some base state ub becomes linearly unstable. Turing’s insight was that diffusion of chemicals may somehow cause a pattern-forming instability. If so, then the starting point scientifically is to imagine that somehow the diffusion has been turned off (mathematically by setting the diffusion coefficients or the wave number q to zero, experimentally by stirring the solutions at high speed) and then we slowly turn on the diffusion to see if instability ensues. If we adopt this as our strategy, then we need to assume that the reacting chemicals form a stable stationary state in the absence of
5The book by Murray [4] has an appendix that discusses some necessary and sufficient analytical criteria that all the eigenvalues of a real N N matrix have negative real parts. The Routh-Hurwitz criterion states that a certain sequence of determinants from × size 1 to N all have to be positive. Determinants are difficult to work with symbolically since they involve a sum of N!productsof matrix elements. 6Our discussion here follows that of a paper by Segel and Jackson [6] that clarified Turing’s original analysis.
5 aparticularsolutionδu(t, x) that is a constant vector δuq times an exponential in time times an exponential in space:
σq t iqx δu1q σq t iqx δu δuq e e e e , (8) = = δu2q ! " with growth rate σq and wave number q. Note that both components of the perturbation vector δu have the same dependence on time and space. Only with this assumption can the spatial and temporal dependencies be eliminated completely from the linearized evolution equations and a simple solution found. If we substitute Eq. (8) into Eq. (7), divide out the exponentials, and collect some terms, we obtain the following eigenvalue problem A δu σ δu , (9) q q = q q where the the 2 2realmatrixA is defined by × q 2 2 a11 D1q a12 Aq A Dq − 2 . (10) = − = a21 a22 D2q ! − "
Eq. (9) tells us that the growth rate σq and constant vector δuq form an eigenvalue-eigenvector pair of the matrix A ,andthatthereisonesuch2 2eigenvalueproblemforeachwavenumberq.Theeigenvalue q × problem for a given q has generally two linearly independent eigenvectors that we will denote by δuiq for i 1, 2. If the corresponding eigenvalues are σ ,theparticularsolutionwithwavenumberq will have = iq the form: c δu eσ1q t c δu eσ2q t eiqx, (11) 1q 1q + 2q 2q 4 where the coefficients ciq are complex# constants, that depend on$ the initial perturbation at t 0 .This = solution decays if Re(σ )<0fori 1, 2. An arbitrary perturbation δu(t, x) is a superposition of iq = expressions like Eq. (11) over all wave numbers q.Theuniformsolutionub is stable if both eigenvalues σiq have negative real parts for all wave numbers q,i.e.,ifmaxi maxq Re(σiq)<0. The characteristic polynomial for the eigenvalue problem Eqs. (9,10) can be written
0 det A σ I σ 2 (trA )σ det A (12) = q − q = q − q q + q # $ where tr(Aq ) denotes the trace (sum of diagonal elements) of the matrix Aq and det(Aq ) the determinant. The eigenvalues are then σ1q and σ2q given by 1 1 σ trA (trA )2 4detA (13) q = 2 q ± 2 q − q %
The regions of stability (both Re σq negative) and instability (at least one Re σq positive) in the tr Aq -detAq plane are shown in Fig. 1. From this figure or the expression Eq. (13) a simple criterion can be derived that determines when the real parts of both eigenvalues are negative: the trace of the matrix must be negative Linear stability? Stability conditions. and the determinant of the matrix must be positive.ForthematrixAq Eq. (10), these criteria for stability take the explicit form:
trA a a (D D )q2 < 0, (14a) q = 11 + 22 − 1 + 2 det A (a D q2)(a D q2) a a > 0. (14b) q = 11 − 1 22 − 2 − 12 21
If both conditions hold for all wave numbers q,thestationaryuniformbasestateub is linearly stable.
If Eq. (2) instead involved N interacting chemicals,detA we would need to solve a N N eigenvalue problem q × Eq. (9) for each wave number q.ForN 3, analytical criteria that all the eigenvalues of a N N matrix Aq stable Re < 0 Re > 0 ≥ σˇ˛˝¸ σˇ˛˝¸ × Im σˇ˛˝¸≠ 0 Im σˇ˛˝¸≠ 0 4Arealsolutioncanbeobtainedasusualbyaddingthecomplexconjugatesolution. oscillatory
Im σ1,2 = 0 stationary Re σ1,2 > 0 4 trAq
Im σ1,2 = 0 Re σ1 > 0 Re σ2 < 0
Figure 1: Stability regions of the Turing system in the trAq -detAq plane. The plot shows regions of different characteristics of the two eigenvalues σ and σ calculated from Eq. (12). The parabola det A 1 trA 1 2 q = 4 q divides the plane into two halves: above this curve the two eigenvalues are complex (and are complex conjugates of one another), below this curve both eigenvalues are real. The shaded region is the stable region, Re σ1,2 < 0. Over the unshaded portion there is at least one eigenvalue with positive real part. A stationary instability occurs passing through the negative tr Aq axis to negative values of det Aq ,whereasan oscillator instability occurs passing through the positive det Aq axis to positive values of tr Aa .
have negative real parts become cumbersome to derive and to work with5 and so the case N 2thatTuring = discussed in his paper hits the mathematical sweet spot of being manageable and leading to interesting results. For experiments with N > 3reactingchemicals,itisusuallyeasiesttostudythecorrespondingmodelby numerical methods. With modern computers and modern numerical algorithms, it is straightforward to find all the eigenvalues of a N N matrix quickly for N as large as 10,000. As you can imagine, it would be × exceedingly difficult to map out the reaction rates for so many interacting chemicals. Progress in studying the linear stability of chemical systems with large N is therefore limited by scientific knowledge, not by the ability to calculate eigenvalues. We now discuss the physical meaning and implications of the mathematical criterion Eq. (14) in the context of pattern formation6.Therearemanyabstractsymbolsandequationshereandsomecarefulthinkingis needed to see how to extract some physical insights. A first step is to identify the precise scientific question of interest, not just the technical mathematical question of when some base state ub becomes linearly unstable. Turing’s insight was that diffusion of chemicals may somehow cause a pattern-forming instability. If so, then the starting point scientifically is to imagine that somehow the diffusion has been turned off (mathematically by setting the diffusion coefficients or the wave number q to zero, experimentally by stirring the solutions at high speed) and then we slowly turn on the diffusion to see if instability ensues. If we adopt this as our strategy, then we need to assume that the reacting chemicals form a stable stationary state in the absence of
5The book by Murray [4] has an appendix that discusses some necessary and sufficient analytical criteria that all the eigenvalues of a real N N matrix have negative real parts. The Routh-Hurwitz criterion states that a certain sequence of determinants from × size 1 to N all have to be positive. Determinants are difficult to work with symbolically since they involve a sum of N!productsof matrix elements. 6Our discussion here follows that of a paper by Segel and Jackson [6] that clarified Turing’s original analysis.
5 diffusion.SettingthediffusionconstantsDi to zero in Eq. (14), we obtain the following criterion for linear stabilityQ. When of the uniform do we state have in thea stable absence homogeneous of diffusion: solution that is
destabilizeddiffusion by.Settingthediffusionconstants diffusion? Di to zero in Eq. (14), we obtain the following criterion for linear a a < 0, (15a) stability of the uniform state in the absence11 + 22 of diffusion: a a a a > 0. (15b) 11 22 − 12 21 a11 a22 < 0, (15a) If thisHomogeneous criterion is satisfied, solutions a well-mixed (D = 0) stable two-chemical when: solution will+ remain stable and uniform. Comparing a11a22 a12a21 > 0. 2 (15b) Eq. (15a) with Eq. (14a) and remembering that diffusion constants− Di and the quantity q are non-negative, we conclude that If this criterion is satisfied, a well-mixed two-chemical2 solution will remain stable and uniform. Comparing If so: trAq a11 a22 (D1 D2)q < a11 a22 < 0, (16)2 Eq. (15a) with Eq. (14a)= and+ remembering− + that diffusion+ constants Di and the quantity q are non-negative, so the tracewe of the conclude matrix A thatq is always negative. We conclude that the only way for diffusion to destabilize the 2 detA uniform state is for the second criterion Eq.trA (14b)q a to11 becomeq a22 reversed(D1 soD2 that)q the< a determinant11 a22 < of0, Aq becomes (16) stable = + − + + negative. Re σˇ˛˝¸< 0 Re σˇ˛˝¸> 0 Im σ 0 Im σ 0 so the trace of the matrix Aq is alwaysˇ˛˝¸ negative.≠ ˇ˛˝¸≠ We conclude that the only way for diffusion to destabilize the The next step is therefore to figure out when the determinantoscillatory det Aq changes sign from positive to negative. uniform state is for the second criterion Eq. (14b) toIm become σ1,2 = 0 reversed so that the determinant of Aq becomes stationary Re σ > 0 2 No oscillatory instabilities 1,2 Eq. (14b) tellsnegative. us that the determinant det Aq is a parabola in the quantity q that opens upwards, being trA positive for q2 0byEq.(15b)andpositiveforlargeq2.Aconditionforlinearinstabilityinthepresenceq = Im σ = 0 The next step is therefore to figure? out when1,2 the determinant det Aq changes sign from positive to negative. of diffusion is then obtained by asking when the minimumRe σ1 > 0 Re σ2 < 0 value of this parabola first becomes negative. 2 2 Setting theEq. derivative (14b) of tells det usAq thatwith the respect determinant to q to det zero,Aq weis learn a parabola that the in minimum the quantity occursq atthat the opens wave upwards, being number q positivegiven by: for q2 0byEq.(15b)andpositiveforlargeq2.Aconditionforlinearinstabilityinthepresence m = Figure 1: Stability regions of the TuringD systema in the trAq -detAD q aplane. The plot shows regions of different of diffusion is thencharacteristics obtained of the two by eigenvalues2 askingσ and1 σ 22calculated when from2 the Eq.11 (12). minimum The parabola det A value1 trA of this parabola first becomes negative. 1 2 . q = 4 q divides the plane into twoqm halves: above this curve+ the two eigenvalues2 are complex (and are complex (17) Setting the derivativeconjugates of of one det another),Aq belowwith= this curve respect both2D eigenvalues1 D2 to areq real. Theto shaded zero, region is we the stable learn that the minimum occurs at the wave region, Re σ1,2 < 0. Over the unshaded portion there is at least one eigenvalue with positive real part. A The correspondingnumber valueqm given of det by:stationaryA at instability this occurs minimum passing through the is negative tr Aq axis to negative values of det Aq ,whereasan oscillatorq instability occurs passing through the positive det Aq axis to positive values of tr Aa . 2 D1a22 D2a11 q + 2 . (17) have negative real parts become cumbersome to derive andm to work with5 and so the case N 2thatTuring (=D1a22 2DD12Da=112 ) detdiscussedAq in his papera hits11 thea mathematical22 a sweet12a spot21 of being manageable and+ leading to interesting results.. (18) For experimentsm = with N > 3reactingchemicals,itisusuallyeasiesttostudythecorrespondingmodelby− − 4D D The correspondingnumerical value methods. of With det modernAq computersat this and modern minimum numerical algorithms, is it1 is straightforward2 to find all the eigenvalues of a N N matrix quickly for N as large as 10,000. As you can imagine, it would be × exceedingly difficult to map out the reaction rates for so many interacting chemicals. Progress in studying This expression is negative whenthe linear the stability inequality of chemical systems with large N is therefore limited by scientific knowledge, not by the 2 ability to calculate eigenvalues. (D1a22 D2a11) We now discuss the physicaldet meaningAq and implicationsa11 ofa the22 mathematicala12 criteriona21 Eq. (14) in the context + . (18) of pattern formation6.Therearemanyabstractsymbolsandequationshereandsomecarefulthinkingism Dneeded1a to22 see how toD extract2a some11 physical> 2 insights.=D A1 firstD step2(a is to−11 identifya22 the precisea12 scientific−a21 question), of4D1 D2 (19) interest, not just+ the technical mathematical question of when some base state −ub becomes linearly unstable. Turing’s insight was that diffusion of chemicals may somehow cause a pattern-forming instability. If so, then This expression is negativethe starting point scientifically when is to the imagine! inequality that somehow the diffusion has been turned off (mathematically is satisfied. The term inside theby square setting the diffusion root coefficients is positive or the wave number becauseq to zero, experimentally of Eq. by (15b). stirring the solutions As a corollary, Eq. (19) implies at high speed) and then we slowly turn on the diffusion to see if instability ensues. If we adopt this as our that strategy, then we need to assume that the reacting chemicals form a stable stationary state in the absence of 5The book by Murray [4] has anD appendix1a22 that discusses someD2 necessarya11 and> sufficient2 analyticalD criteria1 D that2( alla the11 eigenvaluesa22 a12a21), (19) of a real N N matrix have negativeD real1a parts.22 The Routh-HurwitzD2a11 criterion>states0 that, a certain sequence of determinants from (20) × + − size 1 to N all have to be positive. Determinants are+ difficult to work with symbolically since they involve a sum of N!productsof matrix elements. 6Our discussion here follows that of a paper by Segel and Jackson [6]2 that clarified Turing’s! original analysis. which can alsois satisfied. be deduced The directly term inside from the Eq. square (17) since rootq ism positiveis a non-negative because of real Eq. number. (15b). From As a corollary, Eqs. (15a) Eq. (19) implies and (20), wethat see that one of the quantities a11 and a22 must5 be positive and the other negative. For concreteness, let us choose a11 > 0anda22 < 0inthesubsequentdiscussion.ThenEq.(15b)furtherimpliesthattheD1a22 D2a11 > 0, (20) quantities a12 and a21 must also have opposite signs. + 2 which can also be deduced directly from Eq. (17) since qm is a non-negative real number. From Eqs. (15a) Eq. (19) is a necessary and sufficient condition for linear instability of a uniform state that is stable in the and (20), we see that one of the quantities a11 and a22 must be positive and the other negative. For concreteness, absence of diffusion Eq. (15). As some experimental knob is turned, the matrix elements aij will change their let us choose a11 > 0anda22 < 0inthesubsequentdiscussion.ThenEq.(15b)furtherimpliesthatthe values smoothly through their dependence on the experimental parameter. (Again, diffusion constants D quantities a and a must also have opposite signs. i can be considered constant12 for21 many experiments and so usually do not play the role of an easily varied bifurcationEq. parameter.) (19) is a At necessary some parameter and sufficient value, the condition inequality for Eq. linear (19) instability may become of true a uniform and the state uniform that is stable in the state will becomeabsence unstable of diffusion to perturbations Eq. (15). As growing some experimentalwith a wave number knob is close turned, to the the value matrixqm elementsin Eq. (17).aij will change their The conditionvalues Eq. (19) smoothly can be through expressed their alternatively dependence in terms on the of experimental two diffusion parameter. lengths (Again, diffusion constants Di can be considered constant for many experiments and so usually do not play the role of an easily varied bifurcation parameter.) At some parameter value, the inequality Eq. (19) may become true and the uniform D1 D2 state will become unstablel1 to perturbationsand growingl2 with a, wave number close to the value(21)qm in Eq. (17). = "a11 = " a22 − The condition Eq. (19) can be expressed alternatively in terms of two diffusion lengths
6 D1 D2 l1 and l2 , (21) = "a11 = " a22 −
6 diffusion.SettingthediffusionconstantsDi to zero in Eq. (14), we obtain the following criterion for linear stability of the uniform state in the absence of diffusion:
a a < 0, (15a) 11 + 22 a a a a > 0. (15b) 11 22 − 12 21 If this criterion is satisfied, a well-mixed two-chemical solution will remain stable and uniform. Comparing 2 Eq. (15a) with Eq. (14a) and remembering that diffusion constants Di and the quantity q are non-negative, we conclude that 2 diffusion.SettingthediffusionconstantstrAD to zeroa in Eq.a (14),( weD obtainD the)q following< a criteriona < for0 linear, (16) diffusion.SettingthediffusionconstantsDqi to zero11 in Eq.22 (14), we1 obtain2 the following11 criterion22 for linear stability of the uniform state in the absencei of= diffusion+: − + + stability of the uniform state in the absence of diffusion: so the trace of the matrix Aq is always negative. We conclude that the only way for diffusion to destabilize the diffusion.SettingthediffusionconstantsDi to zero in Eq. (14),a we obtaina < the0 following, criterion for linear (15a) uniform state is for the second criterion11 Eq.22 (14b)< , to become reversed so that the determinant of Aq becomes stability of the uniform state in the absence of diffusion: a11 + a22 0 (15a) a11a22 a+12a21 > 0. (15b) negative. a11a22 − a12a21 > 0. (15b) a a < 0,− (15a) If this criterion is satisfied, a well-mixed11 two-chemical22 solution will remain stable and uniform. Comparing The next step is therefore to figure+ out when the determinant det Aq changes sign from positive to negative. If this criterion is satisfied, a well-mixeda11a22 a12 two-chemicala21 > 0. solution will remain stable and(15b) uniform.2 Comparing Eq. (15a) with Eq. (14a) and remembering− that diffusion constants Di and the quantity q 2are non-negative,2 Eq.Eq. (15a) (14b) with Eq. tells (14a) us and that remembering the determinant that diffusion det constantsA is aD parabolai and the quantity in theq quantityare non-negative,q that opens upwards, being we conclude that q If this criterion is satisfied, a well-mixed2 two-chemical solution will remain stable and uniform.2 Comparing wepositive conclude that for q 0byEq.(15b)andpositiveforlarge2 q .Aconditionforlinearinstabilityinthepresence trAq a11 a22 (D1 D2)q 2 < a11 a222 < 0, (16) Eq. (15a) with Eq. (14a) and remembering= trA that= diffusiona + a constants− (D +Di Dand)q the< quantitya + aq are< 0 non-negative,, (16) of diffusion is then obtainedq = 11 + by22 asking− 1 + when2 the minimum11 + 22 value of this parabola first becomes negative. we concludeso that the trace of the matrix Aq is always negative. We conclude that the only way for diffusion to destabilize the so the trace of the matrix Aq is always negative.2 We conclude that the2 only way for diffusion to destabilize the uniformSetting state the istr forA derivativeq thea second11 a22 criterion of det(D1 A Eq.qD2 (14b)with)q < to respecta11 becomea22 < to reversed0q, to so zero, that the we determinant learn(16) that of A thebecomes minimum occurs at the wave uniform state is for the= second+ criterion− + Eq. (14b) to become+ reversed so that the determinant of Aq becomes negative.number q given by: q so the tracenegative. of the matrix Aqmis always negative. We conclude that the only way for diffusion to destabilize the uniform state is for the second criterion Eq. (14b) to become reversed so that theD determinant1a22 D of2aA11becomes The next step is therefore to figure out when the determinantq2 det Aq changes+ signq from. positive to negative. (17) negative. The next step is therefore toQ. figure When out do when we thehave determinant a stablem homogeneous det Aq changes solution sign2 from that positive is to negative. Eq. (14b) tells us that the determinant det Aq is a parabola= in the quantity2D1 D2q 2 that opens upwards, being Eq. (14b) tells2 us that the determinantdestabilized det byA diffusion?q is a parabola 2 in the quantity q that opens upwards, being The next steppositive is therefore for q2 to figure0byEq.(15b)andpositiveforlarge out when the determinant det Aq changesq 2.Aconditionforlinearinstabilityinthepresence sign from positive to negative. positiveThe for correspondingq = 0byEq.(15b)andpositiveforlarge value of det Aq at thisq minimum.Aconditionforlinearinstabilityinthepresence2 is Eq. (14b) tellsof diffusion us that the is= determinant then obtained det byAq askingis a parabola when the in the minimum quantity valueq that of opens this parabola upwards, first being becomes negative. positive forofq2 diffusion0byEq.(15b)andpositiveforlarge is then obtained by asking whenq2.Aconditionforlinearinstabilityinthepresence the2 minimum value of this parabola first becomes negative. Setting the derivative of det Aq with respect to q 2 to zero, we learn that the minimum occurs at the wave Setting= the derivative of det Aq with respect to q to zero, we learn that the minimum occurs at2 the wave of diffusionnumber is thenq obtainedm given by by: asking whenDestabilized the minimum by diffusion value when of thisdeterminant parabola becomes first becomes negative.(D1a negative.22 D2a11) 2 det A a a 2 a a + . (18) Setting thenumber derivativeqm ofgiven det A by:with respectThis todeterminantq to zero, is weaq mparabolaD learn1a2211 that in Dq22 the2 witha11 minimum minimum:12 21 occurs at the wave q q2 D=a + D a−. − 4D1 D2 (17) number q given by: m2 1 22 2 11 m qm = 2D+1 D2 . (17) 2 D1a22 D=2a11 2D1 D2 TheThis corresponding expression value is of negative detqmAq at when this+ minimum the inequality. is (17) The corresponding value of det A=q at this2D minimum1 D2 is 2 The corresponding value of det Aq at this minimum is (D a D a ) 1 22 2 11 2 det Aqm a11Da122a22 a12aD212a11(D>1a222 + DD12aD112)(.a11a22 a12a21), (18) (19) det A = a a − a+ a − 4D+1 D2 . − (18) qm 11 22(D a 12 21D a )2 = −1 22 −2 11 4D1 D2 det Aqm a11a22 a12a21 + . ! (18) Thisis expression satisfied. is The negative term= whenDeterminant inside− the inequality the becomes− square negative4D root1 D2 isfor: positive because of Eq. (15b). As a corollary, Eq. (19) implies This expression is negative when the inequality This expressionthat is negative when the inequality D1a22 D2a11 > 2 D1 D2(a11a22 a12a21), (19) D1a22 + D2a11 > 2 D1 DD2(a1a1122a22 − Da122aa2111),> 0, (19) (20) D1a22 D2a11 > 2+D1 D2(a11a22! a12a21), +− (19) is satisfied. The term inside+ the square root is positive− because of Eq. (15b). As a corollary, Eq. (19) implies is satisfied.which canThe term also inside be deduced the square directly root is positive from! because Eq. (17) of Eq. since (15b).q2 Asis a a corollary, non-negative Eq. (19) real implies number. From Eqs. (15a) is satisfied.that The term inside the square root is positive! because of Eq. (15b). As a corollary,m Eq. (19) implies that that and (20), we see that one of the quantitiesD1a22 Da211a11and> 0a,22 must be positive and the other(20) negative. For concreteness, D1a22 + D2a11 > 0, (20) whichlet can us choose also be deduceda11 > directly0andD1a22 froma22D2 Eq.a<11 > (17)0inthesubsequentdiscussion.ThenEq.(15b)furtherimpliesthatthe0,+ since q2 is a non-negative real number.(20) From Eqs. (15a) + m2 whichand (20), can we also see be that deduced one of the directly quantities froma Eq.and2 (17)a sincemust beqm positiveis a non-negative and the other real negative. number. For From concreteness, Eqs. (15a) which can alsoquantities be deduced directlya12 and froma21 Eq.must (17) since also11q havem is a22 non-negative opposite signs. real number. From Eqs. (15a) and (20), we see that one of the quantities a11 and a22 must be positive and the other negative. For concreteness, and (20), welet see us that choose one ofa the11 quantities> 0andaa1122and< a0inthesubsequentdiscussion.ThenEq.(15b)furtherimpliesthatthe22 must be positive and the other negative. For concreteness, let us choose a > 0anda < 0inthesubsequentdiscussion.ThenEq.(15b)furtherimpliesthatthe let us choosequantitiesaEq.11 > (19)0anda12 and is11a22 aa<21 necessary0inthesubsequentdiscussion.ThenEq.(15b)furtherimpliesthatthemust also22 have and opposite sufficient signs. condition for linear instability of a uniform state that is stable in the quantities aquantitiesand a amust12 and alsoa21 havemust opposite also have signs. opposite signs. Eq.12 absence (19)21 is a necessary of diffusion and sufficient Eq. (15). condition As some for experimental linear instability knob of a uniform is turned, state the that matrix is stable elements in the aij will change their Eq. (19) isEq. aabsence necessaryvalues (19) of is diffusion a and smoothly necessary sufficient Eq. and (15). through condition sufficient As some for their condition linear experimental dependence instability for linear knob of a on is instability uniform turned, the experimental statethe of matrix a that uniform is elements stable state parameter. ina the thatij will is change stable (Again, in their the diffusion constants Di absence of diffusion Eq. (15). As some experimental knob is turned, the matrix elements a will change their absence of diffusionvaluescan smoothly beEq. (15). considered through As some their experimental constant dependence for knob many on is turned, the experimental experiments the matrix elements parameter. anda soij will (Again, usually change diffusion their doij not constants play theDi role of an easily varied values smoothlyvalues through smoothly their through dependence their on dependence the experimental on the parameter. experimental (Again, parameter. diffusion (Again, constants diffusionD constants D canbifurcation be considered parameter.) constant for many At some experiments parameter and so value, usually the do inequalitynot play the role Eq. of (19)i an easily may variedbecomei true and the uniform can be consideredcanbifurcation be considered constant parameter.) for constant many At experiments some for many parameter and experiments so value, usually the and do inequality not so usually play Eq. the do role (19) not of may play an become easily the role varied true of and an easily the uniform varied bifurcation parameter.)state will At some become parameter unstable value, the to perturbations inequality Eq. (19) growing may become with true a and wave the uniform number close to the value q in Eq. (17). bifurcationstate will become parameter.) unstable At some to perturbations parameter growing value, the with inequality a wave Eq. number (19) close may become to the value trueq andm in the Eq. uniform (17). m state will becomestate will unstable become to perturbations unstable to perturbations growing with growinga wave number with a close wave to number the value closeqm in to Eq. the (17). value q in Eq. (17). TheThe condition condition Eq. (19) Eq. can (19) be expressed can be alternatively expressed in alternatively terms of two diffusion in terms lengths of two diffusionm lengths The conditionThe Eq. condition (19) can Eq. be expressed (19) can be alternatively expressed in alternatively terms of two indiffusion terms of lengths two diffusion lengths
D1 D2 D1 l1 andD2D1l2 , D2 (21) D1 D2 l1 and= "a11l2l1 , = "anda22 , l2 (21) , (21) = "a11 l1 = "anda22 l2 − (21) = "a11 =−"a11 = " a22 = " a22 − −
6 6 6 6 diffusion.SettingthediffusionconstantsDi to zero in Eq. (14), we obtain the following criterion for linear stability of the uniform state in the absence of diffusion:
a11 a22 < 0, (15a) diffusion.SettingthediffusionconstantsDi to zero in Eq. (14), we obtain the following criterion+ for linear stability of the uniform state in the absence of diffusion: a a a a > 0. (15b) 11 22 − 12 21 a a < 0, (15a) If this criterion is11 satisfied,+ 22 a well-mixed two-chemical solution will remain stable and uniform. Comparing a11a22 a12a21 > 0. (15b) 2 Eq. (15a) with Eq.− (14a) and remembering that diffusion constants Di and the quantity q are non-negative, If this criterion is satisfied, a well-mixedwe conclude two-chemical that solution will remain stable and uniform. Comparing 2 2 Eq. (15a) with Eq. (14a) and remembering that diffusion constantstrADq i anda11 the quantitya22 q(Dare1 non-negative,D2)q < a11 a22 < 0, (16) we conclude that = + − + + 2 trsoA thea tracea of the(D matrixD )qAq
7 diffusion.SettingthediffusionconstantsDi to zero in Eq. (14), we obtain the following criterion for linear stability of the uniform state in the absence of diffusion:
a a < 0, (15a) 11 + 22 a a a a > 0. (15b) 11 22 − 12 21 If this criterion is satisfied, a well-mixed two-chemical solution will remain stable and uniform. Comparing 2 Eq. (15a) with Eq. (14a) and remembering that diffusion constants Di and the quantity q are non-negative, we conclude that trA a a (D D )q2 < a a < 0, (16) q = 11 + 22 − 1 + 2 11 + 22 so the trace of the matrix Aq is always negative. We conclude that the only way for diffusion to destabilize the uniform state is for the second criterion Eq. (14b) to become reversed so that the determinant of Aq becomes negative.
The next step is therefore to figure out when the determinant det Aq changes sign from positive to negative. 2 Eq. (14b) tells us that the determinant det Aq is a parabola in the quantity q that opens upwards, being positive for q2 0byEq.(15b)andpositiveforlargeq2.Aconditionforlinearinstabilityinthepresence = of diffusion is then obtained by asking when the minimum value of this parabola first becomes negative. 2 Setting the derivative of det Aq with respect to q to zero, we learn that the minimum occurs at the wave number qm given by: 2 D1a22 D2a11 qm + . (17) = 2D1 D2
The corresponding value of det Aq at this minimum is
2 (D1a22 D2a11) det Aqm a11a22 a12a21 + . (18) = − − 4D1 D2 This expression is negative when the inequality
D a D a > 2 D D (a a a a ), (19) 1 22 + 2 11 1 2 11 22 − 12 21 is satisfied. The term inside the square root is positive! because of Eq. (15b). As a corollary, Eq. (19) implies that D a D a > 0, (20) 1 22 + 2 11 2 which can also be deduced directly from Eq. (17) since qm is a non-negative real number. From Eqs. (15a) and (20), we see that one of the quantities a11 and a22 must be positive and the other negative. For concreteness, let us choose a11 > 0anda22 < 0inthesubsequentdiscussion.ThenEq.(15b)furtherimpliesthatthe quantities a12 and a21 must also have opposite signs. Eq. (19) is a necessary and sufficient condition for linear instability of a uniform state that is stable in the absence of diffusion Eq. (15). As some experimental knob is turned, the matrix elements aij will change their values smoothly through their dependence on the experimental parameter. (Again, diffusion constants Di can be considered constant for many experiments and so usually do not play the role of an easily varied bifurcation parameter.) At someQ. parameter When value, do thewe inequality have a Eq. stable (19) may homogeneous become true and thesolution uniform that is state will become unstable to perturbations growing with a wave number close to the value q in Eq. (17). destabilized by diffusion? m The condition Eq. (19) can be expressed alternatively in terms of two diffusion lengths
D1 D2 in the form l1 and l2 , (21) = "a11 = " a22 1 1 1 − a11a22 a12a21 q2 > − . (22) m = 2 l2 − l2 D D in the form ! 1 2 " # 1 2 This implies that the length l2 must1 1 be sufficiently1 largera a thana thea length l1.Nowourassumptionthata11 > 0 2 6 > 11 22 12 21 . implies that chemical 1q enhancesm its2 own2 instability and− so could be called an activator.Similarly,since(22) = 2 l − l D1 D2 a < 0, chemical 2 inhibits its own! 1 growth2 " and# could be called an inhibitor.Thenecessaryconditionl > l This implies22 that the length l must be sufficiently larger than the length l .Nowourassumptionthata > 0 2 1 for a Turing instability2 is then sometimes referred to as “local activation1 with long range inhibition.”11 implies that chemical 1 enhances its own instability and so could be called an activator.Similarly,since The condition l > l So,whenexpressedintheequivalentform diffusion length l2 > l1 or equivalently D /D >( a /a ) partly explains why a22 < 0, chemical 2 inhibits2 its1 own growth and could be called an inhibitor2.Thenecessarycondition1 − 22 11 l2 > l1 for a Turingexperimentalists instability is then had sometimessuch a hard referred time finding to as a“local laboratory activation example with of long a Turing range inhibition.” instability. The diffusion coefficient D2 of the inhibitorAnd chemical has to exceed 1 is an the activator diffusion (a11 coefficient > 0), whileD 1chemicalof the inhibitor 2 is an byinhibitor a factor (a(22 a<22 0))/ a11 The condition l > l ,whenexpressedintheequivalentformD /D >( a /a ) partly explains why− which2 can exceed1 10 for some realistic models of reaction-diffusion2 1 experiments.− 22 11 Since the diffusion coeffi- experimentalists had such a hard time finding a laboratory example of a Turing9 2 instability. The diffusion cients of most small ionsHence: in water have the same value of about 10− m /sec, some ingenuity is required to coefficient D of the inhibitor has to exceed the diffusion coefficient D of the inhibitor by a factor ( a )/a create2 a Turing instability. Experimentalists found (by accident!)1 that one way to achieve− a large22 11 disparity which canin exceed diffusion 10 for coefficients some realistic wascondition to models introduce for Turing of reaction-diffusion a third instability molecule requires (such experiments. local as starch activation Since in the and the CDIMA long diffusion range reaction) coeffi-inhibition that was 9 2 cients of mostfixed small to an immobile ions in water matrix have in the the same solution value (the of walls about of 10 the− porousm /sec, gel). some The ingenuity effective is diffusion required coefficient to create a Turingfor a chemical instability. that Experimentalists reversibly binds to found this immobile (by accident!) molecule that is one substantially way to achieve smaller a than large that disparity for chemicals in diffusionthat coefficients do not bind. was to introduce a third molecule (such as starch in the CDIMA reaction) that was fixed to an immobile matrix in the solution (the walls of the porous gel). The effective diffusion coefficient The criteria Eq. (15) and Eq. (19) for the instability of a uniform state are rather abstract and so we now for a chemical that reversibly binds to this immobile molecule is substantially smaller than that for chemicals apply these criteria to a simple two-variable mathematical model known as the Brusselator7 to illustrate the that do notideas. bind. The Brusselator is a reaction-diffusion model that describes the evolution of two chemical concentra- The criteriations Eq.u1 (15)(t, x) andand Eq.u2(t (19), x): for the instability of a uniform state are rather abstract and so we now apply these criteria to a simple two-variable mathematical model known as the Brusselator7 to illustrate the ∂ u a (b 1)u u2u D ∂2u , (23a) ideas. The Brusselator is a reaction-diffusiont 1 model= − that+ describes1 + the1 2 evolution+ 1 x 1 of two chemical concentra- 2 2 ∂t u2 bu1 u u2 D2∂ u2. (23b) tions u1(t, x) and u2(t, x): = − 1 + x The parameters a, b, D1,andD2 are positive constants.2 Although2 invented with the goal of understanding ∂t u1 a (b 1)u1 u u2 D1∂ u1, (23a) the Belousov-Zhabotinsky reaction= − and+ although+ successful1 + inx producing uniform oscillations and travelling ∂ u bu u2u D ∂2u . (23b) waves, this model was intendedt 2 = not1 to− describe1 2 + a2 specificx 2 chemical experiment but to show how an invented plausible sequence of chemical reactions could reproduce qualitative but difficult to understand features of The parameters a, b, D1,andD2 are positive constants. Although invented with the goal of understanding the Belousov-Zhabotinskyactual experiments. reaction The context and although in which successful this model inwas producing invented suggests uniform assigning oscillations the and following travelling parameter values waves, this model was intended not to describe a specific chemical experiment but to show how an invented a 1.5, D 2.8, D 22.4, (24) plausible sequence of chemical reactions could= reproduce1 qualitative= but2 difficult= to understand features of actual experiments.and varying The the context parameter in whichb as the this bifurcation model was parameter. invented suggests We now assigning show how the to predict following analytically parameter for what values value of b auniformbasestatebecomeslinearlyunstable. Astationaryuniformbasestatea 1.5,u (Du1 , u2.8),can beD found2 22 by.4, looking for solutions of Eq. (23)(24) with all = b = 1b= 2b = and varyingpartial the parameter derivativesb setas to the zero. bifurcation It is straightforward parameter. We to nowshow show that there how to is only predict one analytically uniform state for given what by value of b auniformbasestatebecomeslinearlyunstable. b u1b a, u2b . (25) Astationaryuniformbasestateu (u , u ) can be= found by looking= a for solutions of Eq. (23) with all b = 1b 2b partial derivativesPlease keep set into zero.mind thatIt is itstraightforward is rarely this easy to show to find that the there stationary is only uniform one uniform state state of some given set by of nonlinear evolution equations! b , . By linearizing the Brusselator modelu1b arounda this baseu2b state, you can show that the Jacobian matrix A(25) ∂f/∂u = = a = is given by: Please keep in mind that it is rarely this easy to find the stationary uniform2 state of some set of nonlinear a11 a12 b 1 a evolution equations! A − 2 . (26) = a21 a22 = b a ! " ! − − " By linearizing the Brusselator model around this base state, you can show that the Jacobian matrix A ∂f/∂u 7Two of the more widely studied models of reaction-diffusion dynamics are named after the geographical= location where the is given by:model was invented. Thus the Brusselator is named after Brussels, Belgium, and the Oregonator is named after the state of Oregon 2 in the United States. a11 a12 b 1 a A − 2 . (26) = a21 a22 = b a ! " ! − − " 7Two of the more widely studied models of reaction-diffusion dynamics7 are named after the geographical location where the model was invented. Thus the Brusselator is named after Brussels, Belgium, and the Oregonator is named after the state of Oregon in the United States.
7 in thein form the form 2 1 1 1 a11a22 a12a21 q2 1 1 1 > a11−a22 a.12a21 (22) q m = 2 l2 − l2 > D D − . (22) m ! 1 2 2 "2 # 1 2 = 2 l1 − l2 D1 D2 This implies that the length l2 must be sufficiently! larger" than# the length l1.Nowourassumptionthata11 > 0 This impliesimplies that chemical the length 1 enhancesl2 must its be own sufficiently instability larger and so than could the be length called anl1.Nowourassumptionthatactivator.Similarly,since a11 > 0 impliesa22 that< 0, chemical chemical 2 1 inhibits enhances its own its growth own and instability could be and called so an couldinhibitor be.Thenecessarycondition called an activator.Similarly,sincel2 > l1 for a Turing instability is then sometimes referred to as “local activation with long range inhibition.” a22 < 0, chemical 2 inhibits its own growth and could be called an inhibitor.Thenecessaryconditionl2 > l1 for a TuringThe condition instabilityl2 > isl1,whenexpressedintheequivalentform then sometimes referred to as “localD2 activation/D1 >( a with22/a11 long) partly range explains inhibition.” why − experimentalists had such a hard time finding a laboratory example of a Turing instability. The diffusion The conditioncoefficientlD2 >of thel1,whenexpressedintheequivalentform inhibitor has to exceed the diffusion coefficient D Dof2/ theD inhibitor1 >( bya22 a/ factora11)(partlya )/a explains why 2 1 − − 22 11 experimentalistswhich can exceed had 10 such for some a hard realistic time models finding of a reaction-diffusion laboratory example experiments. of a Turing Since the instability. diffusion coeffi- The diffusion 9 2 coefficientcientsD of mostof the small inhibitor ions in has water to have exceed the same the diffusion value of about coefficient 10− m /Dsec,of some the inhibitor ingenuity is by required a factor to( a )/a 2 1 − 22 11 whichcreate can exceed a Turing 10 instability. for some Experimentalists realistic models found of (by reaction-diffusion accident!) that one experiments. way to achieve Since a large the disparity diffusion coeffi- in diffusion coefficients was to introduce a third molecule (such as starch9 in the2/ CDIMA reaction) that was cientsfixed of most to an small immobile ions matrix in water in the have solution the (the same walls value of the of porous about gel). 10− Them effectivesec, some diffusion ingenuity coefficient is required to createfor a Turing a chemical instability. that reversibly Experimentalists binds to this immobile found molecule (by accident!) is substantially that smaller one way than to that achieve for chemicals a large disparity in diffusionthat do coefficients not bind. was to introduce a third molecule (such as starch in the CDIMA reaction) that was fixed toThe an criteria immobile Eq. (15) matrix and Eq. in the (19) solution for the instability (the walls of a of uniform the porous state are gel). rather The abstract effective and so diffusion we now coefficient for a chemicalapply these that criteria reversibly to a simple binds two-variable to this immobile mathematical molecule model known is substantially as the Brusselator smaller7 to than illustrate that the for chemicals that doideas. not bind. The Brusselator is a reaction-diffusion model that describes the evolution of two chemical concentra- tions u1(t, x) and u2(t, x): The criteria Eq. (15) and Eq. (19) for the instability of a uniform state are rather abstract and so we now 2 2 ∂t u1 a (b 1)u1 u u2 D1∂ u1, 7(23a) apply these criteria to a simple two-variableTuring= patterns− mathematical+ in+ a 1more model+ concretex known as example: the Brusselator to illustrate the ∂ u bu u2u D ∂2u . (23b) ideas. The Brusselator is a reaction-diffusionthet 2Brusselator= 1 − model1 2 + (Toy that2 modelx describes2 for the the Belousov-Zhabotinsky evolution of two reaction) chemical concentra- tions uThe1(t, parametersx) and u2a(,tb,,xD)1:,andD2 are positive constants. Although invented with the goal of understanding the Belousov-Zhabotinsky reaction and although successful in producing uniform oscillations and travelling 2 2 waves, this model was intended not∂t u to1 describea ( ab specific1)u1 chemicalu u2 experimentD1∂ u but1, to show how an invented (23a) = − + + 1 + x plausible sequence of chemical reactions∂ u bu could reproduceu2u qualitativeD ∂2u . but difficult to understand features of (23b) actual experiments. The context int which2 = this1 model− 1 was2 + invented2 x suggests2 assigning the following parameter values The parameters a, b, D1,andD2 are positive constants. Although invented with the goal of understanding a 1.5, D 2.8, D 22.4, b is bifurcation parameter(24) the Belousov-Zhabotinsky reaction= and although1 = successful in2 = producing uniform oscillations and travelling waves,and this varying model the was parameter intendedb as the not bifurcation to describe parameter. a specific We chemicalnow show how experiment to predict analytically but to show for how what an invented value of b auniformbasestatebecomeslinearlyunstable. plausible sequence of chemical reactionsExercise to could explore reproduce Turing patterns. qualitative but difficult to understand features of Astationaryuniformbasestateu (u , u ) can be found by looking for solutions of Eq. (23) with all actual experiments. The context inb which= 1b this2b model was invented suggests assigning the following parameter partial derivatives set to zero. It is straightforward to show that there is only one uniform state given by values 1. Calculate homogeneous steady state solutions and their stability for D = 0 b a 1.5,u aD, 1 u2.8, . D2 22.4, (25) 2 (24) 2. = Calculate1b = Turing= instability2b = a and plot= growth rates i.f.o. wave number q . What is the critical wave number? and varyingPlease keep the parameter in mind thatb itas is the rarely bifurcation this easy to parameter. find the stationary We now uniform show state how of to some predict set of analytically nonlinear for what value ofevolutionb auniformbasestatebecomeslinearlyunstable. equations! 3. Explore dynamics numerically. By linearizing the Brusselator model around( , this base) state, you can show that the Jacobian matrix A ∂f/∂u Astationaryuniformbasestateub u1b u2b can be found by looking for solutions of= Eq. (23) with all is given by: = partial derivatives set to zero. It is straightforward to show that there2 is only one uniform state given by a11 a12 b 1 a A − 2 . (26) = a21 a22 = b ba ! u1b "a, ! −u2b − . " (25) 7Two of the more widely studied models of reaction-diffusion= dynamics are= nameda after the geographical location where the model was invented. Thus the Brusselator is named after Brussels, Belgium, and the Oregonator is named after the state of Oregon Pleasein keep the United in mind States. that it is rarely this easy to find the stationary uniform state of some set of nonlinear evolution equations! By linearizing the Brusselator model around this base7 state, you can show that the Jacobian matrix A ∂f/∂u = is given by: 2 a11 a12 b 1 a A − 2 . (26) = a21 a22 = b a ! " ! − − " 7Two of the more widely studied models of reaction-diffusion dynamics are named after the geographical location where the model was invented. Thus the Brusselator is named after Brussels, Belgium, and the Oregonator is named after the state of Oregon in the United States.
7 in the form 1 1 1 a a a a q2 > 11 22 − 12 21 . (22) m = 2 l2 − l2 D D ! 1 2 " # 1 2 This implies that the length l2 must be sufficiently larger than the length l1.Nowourassumptionthata11 > 0 implies that chemical 1 enhances its own instability and so could be called an activator.Similarly,since in the forma < 0, chemical 2 inhibits its own growth and could be called an inhibitor.Thenecessaryconditionl > l 22 1 1 1 a a a a 2 1 for a Turing instability isq then2 sometimes referred> 11 to as22 − “local12 21 activation. with long range inhibition.”(22) m = 2 l2 − l2 D D ! 1 2 " # 1 2 The condition l2 > l1,whenexpressedintheequivalentformD2/D1 >( a22/a11) partly explains why This implies that the length l2 must be sufficiently larger than the length l1.Nowourassumptionthat− a11 > 0 impliesexperimentalists that chemical 1 enhances had such its a hard own instability time finding and a so laboratory could be called example an ofactivator a Turing.Similarly,since instability. The diffusion coefficient D of the inhibitor has to exceed the diffusion coefficient D of the inhibitor by a factor ( a )/a a22 < 0, chemical 2 inhibits2 its own growth and could be called an inhibitor.Thenecessarycondition1 l2 > l1 − 22 11 for a Turingwhich instability can exceed is then 10 for sometimes some realistic referred models to as “local of reaction-diffusion activation with long experiments. range inhibition.” Since the diffusion coeffi- cients of most small ions in water have the same value of about 10 9 m2/sec, some ingenuity is required to The condition l > l ,whenexpressedintheequivalentformD /D >(−a /a ) partly explains why create a2 Turing1 instability. Experimentalists found (by accident!)2 1 − that22 one11 way to achieve a large disparity experimentalists had such a hard time finding a laboratory example of a Turing instability. The diffusion in thein form the form in diffusion coefficients was to introduce a third molecule (such as starch in the CDIMA reaction) that was coefficient 2D of1 the1 inhibitor1 hasa11 toa22 exceeda12a21 the diffusion coefficient D of the inhibitor by a factor ( a )/a q2 2 1 1 1 > a11−a22 a.12a21 1 (22) 22 11 fixedq m = to2 anl immobile2 − l2 matrix> D inD the− solution. (the walls of the porous gel).(22) The effective− diffusion coefficient which canm exceed= 2! 101l2 for−2 somel"2 # realistic1 models2D D of reaction-diffusion experiments. Since the diffusion coeffi- for a chemical1 that2 reversibly binds1 to2 this immobile molecule9 2 is substantially smaller than that for chemicals This implies that the lengthcientsl2 ofmust most be sufficiently small! ions larger in" water than# the have length thel1 same.Nowourassumptionthat value of about 10 a11m>/0sec, some ingenuity is required to − > This impliesimplies that chemical the lengthcreate 1 enhancesl2 must athat Turing its be do own sufficiently not instability. instability bind. larger and Experimentalists so than could the be length called found anl1.Nowourassumptionthatactivator (by accident!).Similarly,since that one waya11 to achieve0 a large disparity < > impliesa22 that0, chemical chemical 2 1 inhibits enhancesin diffusion its own its growth coefficients own and instability could was be to and called introduce so an couldinhibitor a third be.Thenecessarycondition called molecule an activator (such as.Similarly,since starchl2 l1 in the CDIMA reaction) that was for a Turing instability is thenThe sometimes criteria referred Eq. (15)to as “local and Eq. activation (19) with for the long instability range inhibition.” of a uniform state are rather abstract and so we now a22 < 0, chemical 2 inhibits its own growth and could be called an inhibitor.Thenecessaryconditionl2 > l1 fixed toapply an immobile these criteria matrix to in a the simple solution two-variable (the walls mathematical of the porous model gel). The known effective as the diffusionBrusselator coefficient7 to illustrate the for a TuringThe condition instabilityl2 > isl1,whenexpressedintheequivalentform then sometimes referred to as “localD2 activation/D1 >( a with22/a11 long) partly range explains inhibition.” why for a chemical that reversibly binds to this immobile− molecule is substantially smaller than that for chemicals experimentalists had such a hardideas. time The finding Brusselator a laboratory is example a reaction-diffusion of a Turing instability. model that The diffusion describes the evolution of two chemical concentra- that do not bind. The conditioncoefficientlD2 >of thel1,whenexpressedintheequivalentform inhibitor hastions to exceedu (t, thex) diffusionand u (t coefficient, x): D Dof2/ theD inhibitor1 >( bya22 a/ factora11)(partlya )/a explains why 2 1 2 1 − − 22 11 experimentalistswhich can exceed had 10 such forThe some a hard criteria realistic time Eq. models finding (15) of and a reaction-diffusion laboratory Eq. (19) for example the experiments. instability of a Turing Since of a the uniform instability. diffusion state coeffi- The are diffusion rather abstract and so we now 9 2 2 2 coefficientcientsD of mostof the small inhibitor ions in has water to have exceed the same the diffusion value of about coefficient 10− m∂t u/Dsec,1 of somea the inhibitor ingenuity(b 1)u is by1 required a factoru1u2 to( Da1∂)/x ua1, 7 (23a) 2 apply these criteria to a simple two-variable mathematical1= − + model known+ as+ the− 22Brusselator11 to illustrate the whichcreate can exceed a Turing 10 instability. for some Experimentalists realistic models found of (by reaction-diffusion accident!) that one experiments. way to achieve2 Since a large the disparity diffusion2 coeffi- ideas. The Brusselator is a reaction-diffusion∂t u2 modelbu1 thatu describes1u2 D the2∂x evolutionu2. of two chemical concentra- (23b) in diffusion coefficients was to introduce a third molecule (such as starch9 in= the2/ CDIMA− reaction)+ that was cientsfixed of most to an small immobile ions matrixtions in wateru in1( thet, havex solution) and theu (the2 same(t, wallsx) value: of the of porous about gel). 10− Them effectivesec, some diffusion ingenuity coefficient is required to The parameters a, b, D1,andD2 are positive constants. Although invented with the goal of understanding createfor a Turing a chemical instability. that reversibly Experimentalists binds to this immobile found molecule (by accident!) is substantially that smaller one way than to that2 achieve for chemicals a2 large disparity the Belousov-Zhabotinsky∂t u reaction1 a and(b although1)u1 u successful1u2 D1∂ inx u producing1, uniform oscillations(23a) and travelling in diffusionthat do coefficients not bind. was to introduce a third molecule (such= as− starch+ in the+ CDIMA+ reaction) that was 2 2 fixed to an immobile matrix inwaves, the solution this model (the walls was of intended∂ thet u2 porous notbu1 to gel). describeu u The2 effectiveD a2∂ specificu2. diffusion chemical coefficient experiment but to show(23b) how an invented The criteria Eq. (15) and Eq. (19) for the instability of a uniform= state are− rather1 + abstractx and so we now for a chemical that reversibly bindsplausible to this sequence immobile of molecule chemical is substantiallyreactions could smaller reproduce7 than that qualitative for chemicals but difficult to understand features of apply these criteria to aThe simple parameters two-variablea, b mathematical, D ,andD modelare known positive as the constants.Brusselator Althoughto illustrate invented the with the goal of understanding that doideas. not bind. The Brusselator is a reaction-diffusionactual experiments. model1 that The describes2 context the in evolution which this of two model chemical was concentra- invented suggests assigning the following parameter the Belousov-Zhabotinsky reaction and although successful in producing uniform oscillations and travelling tions u1(t, x) and u2(t, x): values The criteria Eq. (15) andwaves, Eq. (19) this for model the was instability intended of not a uniform to describe state a specific are rather chemical abstract experiment and so we but now to show how an invented 2 2 ∂t u1 a (b 1)u1 u u2 D1∂au1, 1.5, D1 2.8, 7(23a)D2 22.4, (24) apply these criteria to a simpleplausible two-variableTuring sequence= patterns− of mathematical chemical+ in+ a reactions1more model+ concretex known could= reproduce as example: the Brusselator qualitative= butto illustrate difficult= theto understand features of ∂ u bu u2u D ∂2u . (23b) ideas. The Brusselator isactual a reaction-diffusion experiments.andthet varying 2Brusselator= 1 the− The model1 parameter context2 + (Toy that2 model inx describes which2b as for the thisthe bifurcation the Belousov-Zhabotinsky model evolution was parameter. invented of two reaction) chemical suggests We now assigning concentra- show how the following to predict parameter analytically for what tions uThe1(t, parametersx) and u2a(,tb,,xDvalues)1:,andvalueD2 are of positiveb auniformbasestatebecomeslinearlyunstable. constants. Although invented with the goal of understanding the Belousov-Zhabotinsky reaction and although successful ina producing1.5, uniformD1 oscillations2.8, andD2 travelling22.4, (24) Astationaryuniformbasestate=2 u 2(u =, u ) can be found= by looking for solutions of Eq. (23) with all waves, this model was intended not∂t u to1 describea ( ab specific1)u1 chemicalu u2 experimentbD1∂ u but11,b to2 showb how an invented (23a) and varying the= parameter− + b as the+ bifurcation1 + = parameter.x We now show how to predict analytically for what plausible sequence of chemicalpartial reactions derivatives could reproduce2 set to qualitative zero.2 It is but straightforward difficult to understand to show features that of there is only one uniform state given by value of b∂tauniformbasestatebecomeslinearlyunstable.u2 bu1 u1u2 D2∂x u2. (23b) actual experiments. The context in which= this model− was+ invented suggests assigning the following parameterb values The parameters a, b, D1Astationaryuniformbasestate,andD2 are positive constants.u Althoughb (u1b, u invented2b) canu1 beb with founda the, bygoalu looking2 ofb understanding for. solutions of Eq. (23) with all (25) a 1.5, D1 2.8, D=2 22.4, b is bifurcation= parameter=(24)a the Belousov-Zhabotinskypartial reaction derivatives= and although set to zero.= successful It is straightforward in= producing uniform to show oscillationsthat there is andonly travelling one uniform state given by waves,and this varying model the was parameter intendedb asPlease the not bifurcation to keep describe in parameter. mind a specific that We it chemicalnow is show rarely how experiment this to predict easy to analytically but find to show the for stationary how what an invented uniform state of some set of nonlinear value of b auniformbasestatebecomeslinearlyunstable. b plausible sequence of chemicalevolution reactionsHomogeneous equations! could reproducesolutions: qualitativeu1b buta, difficultu2b to understand. features of (25) Astationaryuniformbasestateu (u , u ) can be found by looking for= solutions of Eq.= (23)a with all actual experiments. The contextBy inb linearizingwhich= 1b this2b model the Brusselator was invented model suggests around assigning this base the state, following you can parameter show that the Jacobian matrix A ∂f/∂u partial derivatives set toPlease zero. Itkeep is straightforward in mind that to it show is rarely that there this is easy only to one find uniform the stationary state given by uniform state of some set of nonlinear = values is given by: evolution equations! b a 1.5,u aD, 1 u2.8, . D2 22.4a, a b (25)1 a2 (24) =Jacobian:1b = = 2b = a = 11 12 . The off-diagonalBy elements linearizing have the Brusselator opposite model signs around as requiredA this base for state, a Turing you can show instability,− that the Jacobian2 but the matrix diagonalA ∂ elementsf/∂u (26) and varying the parameter b as the bifurcation parameter. We now show= howa21 toa predict22 = analyticallyb fora what = Please keep in mind thatis given it is rarely by: this easy to find the stationary uniform! state of some" set of! nonlinear− − " valuehave of b oppositeauniformbasestatebecomeslinearlyunstable. signs only if evolution equations! 7Two of the more widely studied modelsa a of reaction-diffusionb 1 dynamicsa2 are named after the geographical location where the Signs of the terms: A 11 12 . (26) By linearizing the Brusselator modelmodel around was( invented., this base) Thus state, the you Brusselator can showb is that> named the1. Jacobian after Brussels, matrix− A Belgium,∂2f/∂ andu the Oregonator is named after the state(27) of Oregon Astationaryuniformbasestateub u1b u2b can be found= bya21 lookinga22 for= solutionsb of= a Eq. (23) with all is given by: in the= United States. ! " ! − − " partial derivatives set to zero.7 It is straightforward to show that there2 is only one uniform state given by When this inequalityTwo holds, of the more wea11 widely concludea12 studied that modelsb 1 chemical ofa reaction-diffusion 1 is an dynamics activator are named (a11 after> the0) geographical and that location chemical where 2 the is an A − 2 . (26) model was invented.= a Thus21 a the22 Brusselator= b is namedba after Brussels, Belgium, and the Oregonator is named after the state of Oregon inhibitor (a22 < 0). in the United States.! u1b "a, ! −u2b − . " 7 (25) 7Two of the more widely studied models of reaction-diffusion= dynamics are= nameda after the geographical location where the PleaseThemodel keep uniform was in invented. mind state Thus that the it is Brusselator is stable rarely is innamed this the easy after absence Brussels, to find Belgium, the of stationary diffusion and the Oregonator uniform whenis named Eq. state after (15) of the some state holds, of Oregon set of i.e. nonlinear in the United States. 7 evolution equations! trA < 0 b < 1 a2 3.25, (28a) By linearizing the Brusselator model around this base7 state, you!⇒ can show that the+ Jacobian= matrix A ∂f/∂u 2 = is given by: det A > 0 a > 0. (28b) !⇒ 2 a11 a12 b 1 a A − 2 . (26) = a21 a22 = b a Only the first condition leads! to a constraint," ! that− the− parameter" b must be smaller than 3.25. Using the matrix 2 7elements Eq. (26) and the fact that a a a a a ,theSegel-Jacksoncriterionforlinearinstability Two of the more widely studied models of reaction-diffusion11 22 dynamics− 12 are21 named= after the geographical location where the modelEq. was (19) invented. can Thus be the manipulated Brusselator is named into after the Brussels, form Belgium, and the Oregonator is named after the state of Oregon in the United States. 2 7 D1 b 1 a . (29) ≥ ! + " D2 #
The critical value bc is determined by equality and the parameter values Eq. (24) imply
b 2.34. (30) c ≈
The corresponding wave number qc at instability is given by Eq. (17)
D1a22 D2a11 qc + 0.435, (31) = " 2D1 D2 ≈ which corresponds to a wave length of 2π/q 14.5. c ≈ The onset of instability can be understood visually by plotting the maximum growth rate curve maxi Reσiq as afunctionofthewavenumberq for values of the parameter b below, equal to, and above the critical value bc (see Fig. 2). The maximum growth rate can be calculated explicitly as the maximum of the real part of the two eigenvalues σiq associated with each wave number q (recall the discussion associated with Eq. (11)). We summarize the results in Fig. 2. The growth rate σq is actually complex for small q (the dotted line in the figure indicates the imaginary part of the eigenvalue with the largest real part) but becomes real for larger q. In particular, the imaginary part is zero near the peak corresponding to the fastest growing mode. For a given parameter b,notehowthecurvemaxi Re(σiq) has a kink —the slope changes discontinuously—because the eigenvalues switch from complex to real at this point. Since the Brusselator does not describe an actual experiment, you may wonder whether it is possible to test independently the above predictions of the critical parameter bc and critical wave number qc.Itis straightforward to write a computer code that integrates the evolution equations Eq. (23) in a large periodic interval. The parameter values could then be set to those of Eq. (24) and the initial conditions of the fields u1 and u2 set to be the uniform values Eq. (25) plus some random noise of tiny amplitude. For b < bc,the small-amplitude noise should decay exponentially and the fields will converge towards their uniform values. For b just larger than bc,theuniformstateshouldbeunstableandacellularstructurewithwavenumberclose to Eq. (31) should appear. What happens in the long term as the exponential growth starts to saturate is not predicted by the linear stability analysis but would be revealed by the numerical integration. The unstable uniform state could evolve onto a stationary, periodic, quasiperiodic, or chaotic attractor.
8 The off-diagonal elements have opposite signs as required for a Turing instability, but the diagonal elements have opposite signs only if b > 1. (27)
The off-diagonal elements haveWhen opposite this inequality signs as holds, required we conclude for a Turing that chemical instability, 1 is but an the activator diagonal (a11 elements> 0) and that chemical 2 is an have opposite signs only ifinhibitor (a22 < 0). b > 1. (27) The uniform state is stable in the absence of diffusion when Eq. (15) holds, i.e.
When this inequality holds, we conclude that chemical 1 is an activator (a11 > 0) and that chemical 2 is an trA < 0 b < 1 a2 3.25, (28a) inhibitor (a22 8 Pattern formation: outline Systems that have the ability to self-organize into spatially structured states from initially unstructured (spatially homogeneous) states. 1. Formation periodic patterns - Turing patterns 2. Formation of domains and moving fronts Domains and moving fronts v? Domain 1 Front Domain 2 - Each domain can consist of any stable solution: homogeneous or spatially structured (e.g. Turing pattern) - Example of possible dynamics: “coarsening dynamics” , domains merge until only one domain is left https://www.youtube.com/watch?v=6qzQIsg5eTE then heads toward the stable steady state (Figure 2D, red curve). where the system has a higher membrane potential (u(t = 0) 1). In The result is that some initial conditions will result in pulses of Figure 3D, the oscillatory case, we also assumed that the frequency depolarization (Figure 2C, red curve), whereas others will not (Figure of the oscillations is higher in the middle of the tube than in the 2C, blue curve). The separatrix dividing the trajectories that pulse other regions. We then assumed that either there was no diffusive from those that do not is similar in shape to that which divided the coupling (top, D = 0) or there was diffusive coupling (bottom, D = 1) basins of attraction in the bistable case (Figure 2B). and examined how the systems evolved with time. We represent the value of u at each point in space and time by a heat map color OSCILLATIONS IN THE FHN MODEL scale. When b is decreased below ^1.2, so that the v-nullcline intersects If there is no reaction and no diffusion, then the middle region the middle limb of the u-nullcline rather that the lower limb, there is remains red and the outer regions remain blue indefinitely (Figure only one steady state, and it is unstable (Figure 2F). The result is that 3A, top). Adding diffusion makes the high-potential region both the system is now oscillatory rather than excitable, and this type of spread out and decrease in amplitude with time (Figure 3A, bot- oscillation in which the trajectory switches back and forth between tom), as would be expected intuitively. The high-u region has a sig- the two limbs of the u-nullcline is termed a relaxation oscillation. nificant effect only on the regions close to it and only for a limited From some initial conditions, the trajectories aim upward (Figure 2F, period of time. pink region), and from others, they aim downward (blue region), but they always approach the same stable limit cycle (Figure 2F). This BISTABLE, EXCITABLE, AND OSCILLATORY yields an unending succession of alternations between high and low TRIGGER WAVES membrane potentials (Figure 2E), which are very similar to repeated The situation is very different when we assume that the reactions of trains of the pulses that were generated when the model was in its u and v are described by the FHN model. If the system is bistable excitable regime (Figure 2C, red curve). (b 2), so that in the absence of diffusion the two regions settle into Thus the FHN model can exhibit three types of behavior: bista- the two different stable steady states (Figure 3B, top), then adding bility, excitability, and oscillations. When combined with a spatial diffusion of the appropriate strength allows the high-u state to prop- coupling mechanism like diffusion, each of these responses can be agate up and down the tube at constant speed (Figure 3B, bottom). propagated as a trigger wave. Eventually the entire tube flips from the low-u state to the high-u state. If the system is excitable (Figure 3C), diffusion allows a pulse REACTION-DIFFUSION DYNAMICS IN THE FHN MODEL of high u to spread up and down the tube at constant speed The ordinary differential equations of the FHN model describe ei- (Figure 3C, bottom). If the tube is oscillatory, with the phase ad- ther the behavior of a well-stirred, spatially homogeneous system vanced and frequency higher in the center of the tube relative the or a spatially inhomogeneous system with no coupling between rest of the tube, the fast oscillations spread up and down from the points in space. To generate trigger waves, we need the system to center at constant speed until they meet up with the delayed oscil- be both inhomogeneous and spatially coupled. Action potentials, lations (Figure 3D, bottom). With each successive oscillation, the calcium waves, and mitotic waves all arise out of spatial inhomoge- trigger wave of fast oscillations spreads further up and down the neities in either the initial conditions of the system or the parame- tube (Figure 3D, bottom). In each of these cases, a dynamical phe- ters of the system. Because they all occur within a shared cyto- nomenon—bistable switching, excitable spikes, or oscillations— plasm, we will assume here that diffusion provides the spatial propagates through space via trigger waves. coupling. To get an idea of why the combination of FHN reactions and In the absence of any local reaction process, a locally elevated diffusion allows for the generation of trigger waves, first ignore concentration of molecules will spread through a cell and eventually diffusion and recall that there is a separatrix or threshold built approach a homogeneous spatial distribution. Fick’s second law de- into the reactions (Figure 2). For a given initial value of v, if you scribes how the concentrations change in time through diffusion. begin with a value of u above the threshold, the trajectory will For one spatial dimension (e.g., for diffusion in a thin tube like an have one sort of fate, and if you begin below it, another. In the axon) this is bistable case (Figure 2, A and B), the threshold separates the trajectories that approach the high-u steady state from those that tux(,t)(tux2 ,)t D (5) approach the low-u steady state; in the excitable case (Figure 2, tt t2 x C and D), it separates the trajectories that include an upward where D is the diffusion coefficient, typically ^300 µm2/s for a small pulse of u from those that do not; and in the oscillatory case 2 (Figure 2, E and F), it separates the oscillations that initially head Domainsmolecule and like moving IP3 and ^fronts:10 µm /s example for a rapidly diffusing cytosolic pro- tein or for Ca2+, which spends most of its time bound to proteins toward the upper limb of the u-nullcline from those that head The Fitzhugh-Nagumo(Allbritton et al., equation 1992). We (toy can model add for diffusion neuronal terms activity) to the FHN toward the lower limb. model to yield a system of partial differential equations: Diffusion provides a mechanism for crossing the threshold. This is illustrated in Figure 4 for the case of the bistable FHN system. Dif- tu t2u D uu3 v fusion mixes nearby values such that a point in space within the tt 2 (6) tx D = 1; a =0.1; ✏ =0.01 low-u region will have its value of u initially increase as the high-u tv t2v region mixes with it and then eventually fall back down (Figure 4, D E()ubva tt tx 2 (7) A–C). The higher the diffusion coefficient, the faster is the initial in- crease, but the fall back down is faster as well (Figure 4, A–C). Diffu- These can be solved numerically for some choice of parameters sion can therefore allow the value of u to increase above the thresh- Exerciseand initialto explore conditions. front dynamics.For each of the simulations shown in Figure 3, old (or, in phase space, to cross the separatrix) for some period of we assumed that we have a long, one-dimensional tube (like an time. If the time is sufficient, the FHN reactions can convert that re- 1. Calculateaxon) with homogeneous the system in steady a low- stateu state solutions (u(t = 0) and ^ –0.6, their v (stabilityt = 0) ^ –0.3)for D = 0;gion of space into an even higher level of u (Figure 4, C and D). The Use graphical method using nullclines. Influence parameter b ? everywhere except for a small region in the middle of the tube, entire process is repeated in the next region of space and then the 2. Verify (local, D=0) time evolution dynamics numerically and plot the trajectories3490 | inL. phase Gelens space, et al. together with the nullclines. Molecular Biology of the Cell 3. Add diffusion (D=1) and evaluate how one or more domains (use non- homogeneous initial conditions) evolve in time. The FitzHugh-Nagumo equation: local dynamics a backward S, which is critical to the model’s behavior (see later A 1.5 B 1.5 discussion), and not because they correspond exactly to the situa- tion in nerves or other biological oscillators. The second variable, v, is related to the potassium gradient. The = 2) u 0 u 0 b ( resting situation, where the extracellular concentration of K+ is lower Bistable than the intracellular concentration, corresponds to a negative value -1.5 -1.5 of v. The processes that regulate v are all assumed to occur on a 1.5 1.5 slow time scale. The first term represents the activation of the volt- C D age-gated potassium channel by the membrane potential u, which increases the value of v. The second term, –Ebv, represents the u 0 u 0 + = 1.5) pumping of K out of the neuron. The third term on the right-hand b ( Excitable side, Ea, roughly corresponds to a potassium leak current. The FHN model is less directly related to calcium signaling and -1.5 -1.5 Cdk1 activation, but one can still draw analogies: for calcium signal- E 1.5 F 1.5 ing, u can represent the free cytosolic calcium, and v can represent the pumps and leaks that regulate calcium concentration more u 0 u 0 slowly. For mitosis, u can represent cyclin B-Cdk1 activity and v can = 1.0) b ( represent the more slowly varying cyclin B concentration. Note, Oscillatory however, that the enduring significance of the FHN model for biolo- -1.5 -1.5 gists does not come from a precise correspondence between the 0 200 400 -1 01 individual terms of the two equations and any particular biological Time v process. Instead, it lies in the fact that the model is a simple way of FIGURE 2: Different types of dynamics from the FHN model. generating bistability, excitability, and relaxation oscillations, three (A, C, E) Time course; (B, D, F) phase plots. (A, B) Bistability. For b 2, systems-level behaviors known to arise in a numberGelens of biological et al., MBoC the 25, system 3486 is (2014)bistable, with two stable steady states (B, filled circles) contexts. and one saddle point (B, open circle). For the value of v(t = 0) assumed here (v(t = 0) –0.3), trajectories beginning above a BISTABILITY IN THE FHN MODEL threshold value of u (A, dashed line) go to the high-u stable steady Here we assume a 0.1 and E 0.01 and change the behavior by state, whereas those beginning below the threshold go to the low-u varying the parameter b. Note that all parameters and variables are stable steady state. In the phase plane, a separatrix (dashed curve) dimensionless throughout. When b is relatively large (b 1.8), the divides the starting points that approach the high-u stable steady state (pink area) from those that go to the low-u steady state. system is bistable. Depending on the initial conditions, the system (C, D) Excitability. For b 1.5, there is a single stable steady state plus will settle down into one of two alternative stable steady states, one a saddle point and an unstable steady state. Trajectories beginning with a negative membrane potential (Figure 2A) and one with a above the threshold (C) or the separatrix (D) yield a pulse of u and positive membrane potential. For the initial value of v assumed here circle the unstable steady state before settling down to a low (v 0.3), all trajectories that start with u 0.3 (the threshold shown steady-state value of u. Those beginning below the threshold do not by the dashed line in Figure 2A) will approach the positive-potential yield a pulse of high u. (E, F) Oscillations. For b 1.0, the single steady state (Figure 2A, red curve), and all trajectories that start with steady state is unstable. From all initial conditions (except starting u < 0.3 will end up at the negative-potential steady state (Figure right on the unstable steady state), the trajectories approach the 2A, blue curve). Thus a small perturbation that pushes the system same stable limit cycle, although from above the threshold, they go across the threshold will be amplified into a large difference in the first to the upper limb of the u-nullcline, and below the threshold, they go first to the lower limb. A, C, and E are time courses; B, D, and system’s ultimate fate. F are phase plots. In each case, a 0.1, E 0.01, v(t = 0) 0.3, and The origin of the bistability can be understood by examining a u(t = 0) 0.25 (red trajectories) or 0.35 (blue trajectories). phase plot of the two-variable system, where the values of u and v (each of which is a function of time) are plotted in the uv-plane (Figure 2B). Stable steady states are attained when both du/dt and steady states (filled circles), and one is an unstable saddle point dv/dt are equal to zero: (open circle). This means that the system is bistable. Some initial conditions produce trajectories that approach the uu 3 v 0 (3) negative-potential steady state (Figure 2, A and B, blue region) and others that approach the positive-potential steady state (Figure 2, A E()ub va 0 (4) and B, pink region). A separatrix (dashed curve) divides the basins of Equation 3 defines a curve in the uv plane, the u-nullcline, and it attraction for the two stable steady states. can be thought of as the steady-state response of u to various con- stant levels of v (Figure 2B, gray curve). The positive feedback and EXCITABILITY IN THE FHN MODEL cubic negative feedback terms give the nullcline a characteristic As b is decreased, the slope of the v-nullcline decreases. This moves backward-S shape that is critical for the behavior of the model. one of the stable steady states from the upper limb of the u-nullcline Equation 4 is the v-nullcline, a straight line that describes the steady- to the middle limb, and it changes from being stable to unstable. state response of v to u (Figure 2B, black line). Wherever the two This leaves a single remaining stable steady state—the low-mem- nullclines intersect, both time derivatives (Eqs. 1 and 2) are equal to brane-potential steady state (Figure 2D). The system is now monos- zero and the system is in steady state. For the particular value of b table but excitable. For some initial conditions, u and v will proceed chosen in Figure 2, A and B (b 2), the two nullclines intersect at more or less directly to the remaining stable steady state (Figure 2D, three points (Figure 2B). Two of the intersection points can be shown blue curve), whereas for others, the trajectory first shoots to the through linear stability analysis (Strogatz, 1994) to represent stable u-nullcline, circles around the unstable steady state, and only Volume 25 November 5, 2014 Spatial trigger waves | 3489 The FitzHugh-Nagumo equation: front dynamics no diffusion diffusion D Fronts ( “trigger waves”) move with speed ~ , where ⌧ is reaction time scale r ⌧ Gelens et al., MBoC 25, 3486 (2014) • Positive feedback allows to generate robust trigger waves • Positive feedback is likely to be a general motif used in biology for spatial cell signaling over larger distances Membrane potential nucleus axon terminals Excitation axon hillock axon + Action + Position Na fast slow K potential cell body (soma) dendrites Time e.g. action potentials in neurons • Positive feedback allows to generate robust trigger waves • Positive feedback is likely to be a general motif used in biology for spatial cell signaling over larger distances sperm sperm entry point PLC Calcium concentration 2+ fast Cytosolic Ca IP3 slow Ca2+ pump Time IP R high calcium concentration 3 e.g. calcium waves in the cell • Positive feedback allows to generate robust trigger waves • Positive feedback is likely to be a general motif used in biology for spatial cell signaling over larger distances Nuclear fluorescence centrosome cyclin Cdc25 Cdk1-cyclin Position fast slow APC high Cdk1-cyclin concentration Wee1 Time Nuclear envelope breakdown e.g. Cdk1 trigger waves in a Xenopus embryo Spatial organization of mitotic entry and exit can be studied in cell-free extracts Experiment (mitosis ~ cell division) Interphase Mitosis Interphase Time J. Chang and J.E. Ferrell, Nature (2013) Mitotic entry and exit are spatially coordinated via trigger waves Experiment S phase M phase Position Time