Mathematical methods in biology Master of Mathematics, second year

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1 Mathematical methos in biology Master of Mathematics, secon year Benoît Perthame Sorbonne Université, UMR 7598, Laboratoire J.-L. Lions, BC187 4, place Jussieu, F Paris ceex 5 October 26, 2018

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3 Contents 1 Lotka-Volterra equations Movement an growth Nonnegativity principle Operator monotonicity for competition or cooperative systems Challenges Reaction kinetics an entropy Reaction rate equations The law of mass action Elementary examples Small numbers of molecules Slow-fast ynamics an enzymatic reactions Slow-fast ynamics (monostable) Slow-fast ynamics (general) Enzymatic reactions Belousov-Zhabotinskii reaction Some material for parabolic equations Bounary conitions The spectral ecomposition of Laplace operators (Dirichlet) The spectral ecomposition of Laplace operators (Neumann) The Poincaré an Poincaré-Wirtinger inequalities Rectangles: explicit solutions Brownian motion an the heat equation Relaxation, perturbation an entropy methos Asymptotic stability by perturbation methos (Dirichlet) Asymptotic stability by perturbation methos (Neuman) Entropy an relaxation Entropy: chemostat an SI system of epiemiology The Lotka-Volterra prey-preator system with iffusion (Problem)

4 CONTENTS Problem Blow-up an extinction of solutions Semilinear equations; the metho of the eigenfunction Semilinear equations; the energy metho Keller-Segel system; the metho of moments Non-extinction Linear instability, Turing instability an pattern formation Turing instability in linear reaction-iffusion systems Spots on the boy an stripes on the tail The simplest nonlinear example: the non-local Fisher/KPP equation Phase transition: what is NOT Turing instability Gallery of parabolic systems giving Turing patterns A cell polarity system The CIMA reaction The iffusive Fisher/KPP system The Brusselator The Gray-Scott system (2) Schnakenberg system The FitzHugh-Nagumo system with iffusion The Gierer-Meinhart system Moels from ecology Competing species an Turing instability Prey-preator system Prey-preator system with Turing instability (problem) Keller-Segel with growth Transport an avection equations Transport equation; metho of caracteristics Avection an volumes transformation Avection (Dirac masses) Examples an relate equations Long time behaviour Nonlinear avection The renewal equation The Fokker-Planck equation Other stochastic aspects

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6 Chapter 1 Lotka-Volterra equations As first examples, we present two general classes of equations that are use in biology: Lotka-Volterra systems an chemical reactions. These are reaction-iffusion equations, or in a mathematical classification, semilinear equations. Our goal is to explain what mathematical properties follow from the set-up of the moel: nonnegativity properties, monotonicity an entropy inequalities. These are very general examples, for more material an more biologically oriente textbooks, the reaer can also consult for instance [33, 32, 38, 22, 9, 1]. 1.1 Movement an growth Class I are use in the area of population biology, an ecological interactions, an are characterize by birth an eath. For 1 i I, we enote by n i (t, x) the population ensities of I interacting species at the location x R ( = 2 for example). We assume that these species move ranomly accoring to Brownian motion with a bias accoring to a velocity U i (t, x) R an that they have growth rates R i (t, x) (meaning birth minus eath). Then, the system escribing the ynamic of these population ensities is t n i D i n i }{{} ranom motion + iv(u i n i ) }{{} = n i R i, }{{} oriente rift growth an eath t 0, x R, i = 1,..., I. (1.1) In the simplest cases, the iffusion coefficients D i > 0 are constants epening on the species (they represent active motion of iniviuals) an the bulk velocity U i vanishes. Here we insist on the birth an eath rates R i ; they epen strongly on the interactions between species an we express this fact as a nonlinearity R i (t, x) = R i ( n1 (t, x), n 2 (t, x),..., n I (t, x) ). (1.2) A stanar family of such nonlinearities results in quaratic interactions an is written R i (n 1, n 2,..., n I ) = r i + I c ij n j, i=1

7 2 Relaxation an the energy metho with r i the intrinsic growth rate of the species i (it can be positive or negative) an c ij the interaction effect of species j on species i. The coefficients c ij are usually neither symmetric nor nonnegative. One can istinguish c ij < 0, c ji > 0: species i is a prey for j, species j is a preator for i. When several species eat no other foo, these are calle trophic chains/interactions, c ij > 0, c ji > 0: a mutualistic interaction (both species help the other an benefit from it) to be istinguishe from symbiosis (association of two species whose survival epens on each other), c ij < 0, c ji < 0: a irect competition (both species compete for example for the same foo), c ii < 0 is usually assume to represent intra-specific competition. The quaratic aspect is relate to the necessary binary encounters for the interaction to occur. Better moels inclue saturation effects: for instance the effect of too numerous preators is to ecrease their iniviual efficiency. This leas ecologists rather to use (with c ii = c ii > 0) R i (n 1, n 2,..., n I ) = r i c ii n i + I j=1, j i c ij n j 1 + n j, the terminology Holing II is use in the case where j refers to a prey, i to a preator. Ecological networks are escribe by such systems: iversity refers to the number of species I itself, connectance refers to the proportion of interacting species, i.e. such that c ij 0. Connectance is low for trophic chain an higher for a foo web (species use several foos). The original prey-preator system of Lotka an Volterra has two species, I = 2 (i = 1 the prey, i = 2 the preator). The prey (small fishe) can fee on abunant zooplankton an thus r 1 > 0, while preators (sharks) will ie out without small fishe to eat (r 2 < 0). The sharks eat small fishs in proportion to the number of sharks (c 12 < 0), while the shark population grows proportionally to the small fishe they can consume c 21 > 0). Therefore, we fin the rule r 1 > 0, r 2 < 0, c 11 = c 22 = 0, c 12 < 0, c 21 > Nonnegativity principle In all generality, solutions of the Lotka-Volterra system (1.1) satisfy very few qualitative properties; there are neither conservation laws (because they contain birth an eath), nor entropy properties, a concept which oes not seem to be relevant for ecological systems. As we shall see the quaratic aspect may lea to blow-up (solutions exist only for a finite time, see Chapter 6). Let us only mention here, that the moel is consistent with the property that population ensity be nonnegative. Lemma 1.1 (Nonnegativity principle) Assume that the initial ata n 0 i are nonnegative functions of L 2 (R ), that U i 0 an that there is a locally boune function Γ(t) such that R i (t, x) Γ(t). Then, the weak solutions in C ( R + ; L 2 (R ) ) to the Lotka-Volterra system (1.1) satisfy n i (t, x) 0. The efinition an usual properties of weak solutions are given in Chapter 4, but we o not require

8 1.3. OPERATOR MONOTONICITY FOR COMPETITION OR COOPERATIVE SYSTEMS 3 that theoretical backgroun to show the formal manipulations leaing to this result. Proof. (Formal) Here, we use the metho of Stampacchia. Set p i = n i, then we have t p i D i p i = p i R i. Multiply by (p i ) + := max(0, p i ) an integrate by parts. We fin 1 ( pi (t, x) ) 2 2 t + x + D i (p i ) + 2 ( = pi (t, x) ) 2 + R i (1.3) R R R an thus Therefore, we have 1 2 t R ( pi (t, x) ) 2 + x Γ(t) R ( pi (t, x) ) 2 +. ( pi (t, x) ) 2 x t + e2 0 Γ(s)s ( p 0 i (x) ) 2 x. + R R But, the assumption n 0 i 0 implies that R ( p 0 i (x) ) 2 + x = 0, an thus, for all times we have R ( pi (t) ) 2 + x = 0, which means n i (t, x) 0. (Rigorous) The ifficulty here stems from the linear efinition of weak solutions of (1.1), which oes not allow for nonlinear manipulations such as multiplication by (p i ) + because this is not a smmoth function. Section?? explains how to resolve this ifficulty. Exercise. Perform the same formal proof with the rift terms U i. Which assumption is neee on U i? Exercise. In the formal context of the Lemma 1.1, show that n i (t, x)x = R i (t, x)n i (t, x)x, t R R n i (t, x) 2 x (n 0 i (x) 2 x e 2 R R t 0 Γ(s)s. 1.3 Operator monotonicity for competition or cooperative systems The positivity property is general but a rather weak property. Comparison principle an monotonicity principle are much stronger properties but require either competition or cooperative systems. For simplicity we consier only two cooperative species, I = 2 in equation (1.1) (1.2), an we make the cooperative assumption n 2 R 1 (n 1, n 2 ) 0, n 1 R 2 (n 1, n 2 ) 0. (1.4)

9 4 Relaxation an the energy metho Lemma 1.2 (Monotonicity principle for cooperative systems) Consier a cooperative system, that is, (1.4) an two solutions (n 1, n 2 ), (m 1, m 2 ) such that for some constants 0 n i Γ 1, 0 m i Γ 1 an R i Γ 2, n i R j Γ 3. Then, the Lotka-Volterra system (1.1) (1.2) is orere for all times n 0 i m 0 i, i = 1, 2 = n i (t) m i (t), t 0, i = 1, 2. Proof. Subtracting the two equations for n 1 an m 1, we fin successively, by aing an subtracting convenient terms, t (n 1 m 1 ) D 1 (n 1 m 1 ) = (n 1 m 1 )R 1 (n 1, n 2 ) + m 1 ( R1 (n 1, n 2 ) R 1 (m 1, m 2 ) ), t (n 1 m 1 ) 2 + D 1 (n 1 m 1 ) + (n 1 m 1 ) = (n 1 m 1 ) 2 2 +R 1 (n 1, n 2 ) ( +m 1 (n 1 m 1 ) + R1 (n 1, n 2 ) R 1 (m 1, n 2 ) ) The first an secon lines are controlle by bouns on R +m 1 (n 1 m 1 ) + ( R1 (m 1, n 2 ) R 1 (m 1, m 2 ) ). (n 1 m 1 ) D 1 (n 1 m 1 ) + t R 2 R 2 (Γ 2 + Γ 3 ) (n 1 m 1 ) 2 + R ( + m 1 (n 1 m 1 ) + R1 (m 1, n 2 ) R 1 (m 1, m 2 ) ), R We now use 0 m 1 γ 1, n 2 R 1 (n 1, n 2 ) 0 an a Lipschitz constant of R 1 (n 1, n 2 ), to conclue (n 1 m 1 ) 2 + (Γ 2 + Γ 3 ) (n 1 m 1 ) Γ 1 Γ 3 (n 1 m 1 ) + (n 2 m 2 ) +. t R 2 R R The same argument on n 2 m 2 leas to the similar inequality (n 2 m 2 ) 2 + (Γ 2 + Γ 3 ) (n 2 m 2 ) Γ 1 Γ 3 (n 1 m 1 ) + (n 2 m 2 ) +, t R 2 R R an aing these two inequalities, for u(t) = [(n R 1 m 1 ) (n 2 m 2 ) 2 +] an 2ab a 2 + b 2, we obtain for some Γ the inequality u(t) Γu(t). t Since u(0) = 0 from our initial orer assumption, we conclue that u(t) = 0 for all t 0. That is the conclusion of the lemma. Exercise. Show that the result of Lemma 1.2 hols true for the system t n i D i n i = R i ( n1 (t, x), n 2 (t, x),..., n I (t, x) ).

10 1.4. CHALLENGES 5 Exercise. For a cooperative system, assume that we are given an initial ata such that (n 0 1, n 0 2) is a smooth subsolution of the steay state equation. Write the equation for ni(t) t an prove that n i (t) t 0. Exercise. Write the proof that orer is preserve for a general k k cooperative system. Exercise. In the case of 2 by 2 competitive systems, that means, n 2 R 1 (n 1, n 2 ) 0, n 1 R 2 (n 1, n 2 ) 0. (1.5) Show that the orer 0 n 1 m 1, 0 m 2 n 2 is preserve. There is no such general statement for larger systems. 1.4 Challenges Variability. There is always a large variability in living populations. For that reason, solutions of moels with fixe parameters, for movement or growth, usually fit poorly with experiments or observations. However, the moels are useful for explaining qualitatively these observations but not for preicting iniviual behavior, a major problem when ealing with meical applications for example. A usual point of view is that these parameters are istribute; one can use a statistical representation (see the software Monolix at in pharmacology. One can also use so-calle structure population moels, see [32, 34, 38]. When moeling Darwinian evolution (see aaptive evolution in Section??), the parameters are part of the moel solution an mutations are seen as changes in the moel coefficients selecte by aaptation to the environment. Small numbers. In physics an chemistry, is the normal orer of magnitue for a number of molecules. Populations that are stuie with Lotka-Volterra equations are much smaller an 10 6 is alreay a large number. This means that exponentially ecaying tails are very quickly meaningless because they represent a number of iniviuals, less than one. The interpretation of such tails shoul be questione carefully an several types of corrections can be inclue; emographic stochasticity is use in Monte-Carlo simulations, which correspon to the survival threshol in PDEs 1. 1 Gauuchon, M. an Perthame, B. Survival threshols an mortality rates in aaptive ynamic: conciliating eterministic an stochastic simulations. Mathematical Meicine an Biology 27 (2010), no. 3,

11 6 Relaxation an the energy metho

12 Chapter 2 Reaction kinetics an entropy When large numbers of molecules are involve in a chemical reaction, the kinetics are well escribe by reaction rate equations. These are nonlinear equations escribing the population ensities (concentrations) of the reacting molecules an the specific form of these nonlinearities is usually prescribe by the law of mass action. This section eals with this aspect of reaction kinetics. The erivation of these equations is postpone to Section?? where we introuce the chemical master equation, which better escribes a small number of reacting molecules, an important topic in cell biology. 2.1 Reaction rate equations The general form of the equations leas to a particular structure for the right han sie of semi-linear parabolic equations. They are written t n i D i n i }{{} molecular iffusion + n i L i = G }{{} i, t 0, x R, i = 1, 2,..., I. (2.1) reaction The quantities n i 0 are molecular concentrations, the loss terms L i 0 epen on all the molecules n j (with which the molecule i can react) an the gain terms G i 0, which also epens on the n j s, enote the rates of prouction of n i from the other reacting molecules. The molecular iffusion rate of these molecules is D i an can be compute accoring to the Einstein rule from the molecular size. For a single reaction, the nonlinearities L i an G i take the form n i L i = reactions p a pi k p I j=1 n apj j, G i = reactions p b pi k p I j=1 n apj j. (2.2) The powers a pj, b pj N represent the number of molecules j necessary for the reaction p; this is the law of mass action. Nonnegativity property. We have factore the term n i in front of L i to ensure that this term vanishes at n i = 0. The loss term L i is not singular here because the prouct contains a pi 1 when 7

13 8 CHAPTER 2. REACTION KINETICS AND ENTROPY k pi > 0 (the reactant i really reacts, otherwise the corresponing L i vanishes). For that reason, as for the Lotka-Volterra systems, the nonnegativity property hols true. Lemma 2.1 A weak solution of (2.1) with G i 0 an nonnegative initial ata satisfies n i (t, x) 0, i = 1, 2,..., I. Proof. Aapt the proof of Lemma 1.1. Even though we o not know that G i 0, this lemma is useful because one may argue as follows. Solutions are first built using the positive part of G i, then the lemma tells us that the n i s are positive. From formula (2.2), the positivity of the n j s implies that G i is positive an thus that we have built the solution to the correct problem. Conservation of atoms. The secon main property of these moels comes from the conservation of atoms, which asserts that some quantities shoul be constant in time. For each atom k = 1,..., K, one efines the number N ki of atoms k in the molecule i. Then all reactions shoul preserve the number of atoms an, for all k {1,..., K}, the coefficients α, β, k an k shoul be such that This implies that I N ki [n i L i G i ] = 0, (n i ) R I. (2.3) i=1 t R t I N ki n i = i=1 I N ki D i n i, (2.4) i=1 I N ki n i (t, x)x = 0, 1 k K, i=1 an thus, a priori, the weighte L 1 boun hols R i=1 I N ki n i (t, x)x = R i=1 I N ki n 0 i (x)x, 1 k K. (2.5) Except when all the iffusion rates D i s are equal (then its main part is the heat equation), it is not easy to extract conclusions from equation (2.4) except the conservation law (2.5). Very few general tools, such as M. Pierre s uality estimate, are available, see the survey 1. There is a thir property, the entropy issipation that we shall iscuss later. 2.2 The law of mass action Irreversible reaction. To begin with, consier molecular species S i unergoing the single irreversible reaction a 1 S a I S I k + b1 S b I S I, 1 Pierre, M. Global existence in reaction-iffusion systems with control of mass: a survey. Milan J. Math. 78 (2) (2010),

14 2.2. THE LAW OF MASS ACTION 9 where a i an b i N, (a i ) i=1,...,i (b i ) i=1,...,i, enote the number of molecules involve as reactants an proucts. Then, using the law of mass action, the equation for the population number ensities n i of species i is written t n i D i n i + a i k + I j=1 n aj j }{{} loss of molecules by reaction = b i k + I j=1 n aj j }{{} gain of reaction prouct, i = 1, 2,..., I. From the conservation of atoms, it is impossible that for all i we have b i a i (resp. a i b i ). Therefore, at least one reactant (a i < b i ) shoul isappear an one prouct (b i > a i ) be prouce. Reversible reaction. More interesting is when several reactions occur. Then they are represente by a sum of such terms over each chemical reaction. For example, still with a i b i N, the reversible reaction k + a 1 S a I S I b 1 S b I S I, k leas to reaction rate equations t n i D i n i loss of forwar reacting molecules {}}{ I + a i k + j=1 n aj j + = b i k + I A more convenient form is t n i D i n i = (b i a i ) k + j=1 n aj j }{{} gain of forwar reaction prouct I j=1 n aj j k loss of backwar reacting molecules { }} { I b i k j=1 n bj j + a i k I I j=1 j=1 n bj j. }{{} gain of backwar reaction prouct n bj j, i = 1, 2,..., I. (2.6) With this form, one can check the funamental entropy property for reversible reactions. Define S(t, x) := I [ n i ln(ni ) + σ i 1 ], i=1 with I σ i (a i b i ) = ln k + ln k. (2.7) There are ifferent possible choices for the constant σ i, which can be more or less convenient epening on the case. In all cases, one can check the Lemma 2.2 (Entropy property for reversible reactions) The entropy issipation equality hols t [ D(t, x) := ln(k+ S(t, x)x = I j=1 n aj j I i=1 ) ln(k i=1 D i ni 2 n i x D(t, x) 0, I j=1 n bj j )] [ k + I j=1 n aj j k I j=1 ] n bj j x 0.

15 10 CHAPTER 2. REACTION KINETICS AND ENTROPY The entropy issipation property is very important because it ictates the long term behavior of the reaction. Mathematically it is also useful to prove a priori estimates for the quantity ni 2 n i but also for integrability properties of powers of n i, which are necessary to efine solutions of the reaction-iffusion equation. This has been use recently in a series of papers by L. Desvillettes an K. Fellner 2 for estimates an relaxation properties, an by T. Gouon an A. Vasseur 3 for regularity properties. 2.3 Elementary examples Isomerization. The reversible isomerization reaction is the simplest chemical reaction. The atoms within the molecule are not change but only their spatial arrangement. The reaction is represente as S 1 k + k S 2 an correspons, in (2.6), to a 1 = 1, b 1 = 0, a 2 = 0, b 2 = 1, which meas to t n 1 D 1 n 1 + k + n 1 = k n 2, t n 2 D 2 n 2 + k n 2 = k + n 1. (2.8) The conserve quantity is simply [n 1 (t, x) + n 2 (t, x)]x = 0, t [n 1 (t, x) + n 2 (t, x)]x = [n 0 1(x) + n 0 2(x)]x. The formula (2.7) for the entropy (with a 1 = b 2 = 1, b 1 = a 2 = 0) gives an one can check the entropy issipation relation S(t, x) = n 1 [ ln(k+ n 1 ) 1 ] + n 2 [ ln(k n 2 ) 1 ], (2.9) t S(t, x)x = R [ n 1 2 n D1 R n 1 + D 2 ] 2 2 n 2 x R [ ln(k+ n 2 ) ln(k n 1 ) ][ k n 2 k + n 1 ] x 0. Dioxygen issociation. The stanar egraation reaction of ioxygen to monoxygen is usually associate with hyperbolic moels for flui flows rather than iffusion. This is because it is a very energetic reaction occurring at very high temperature (for atmosphere re-entry vehicles for example) an with reaction rates epening critically on this temperature, see [17]. But for our purpose here, we forget these limitations an consier the issociation rate k + > 0 of n 1 = [O 2 ] in n 2 = [O], an conversely its recombination with rate k > 0 O 2 k + O + O. k 2 L. Desvillettes an K. Fellner, Entropy methos for reaction-iffusion equations: Slowly growing a-priori bouns. Rev. Mat. Iberoamericana, 24 (2) (2008), T. Gouon an A. Vasseur, regularity analysis for systems of reaction-iffusion equations, Annales e l ENS, 43 (1) (2010),

16 2.3. ELEMENTARY EXAMPLES 11 This leas to a 1 = 1, b 1 = 0, a 2 = 0, b 2 = 2, an thus we obtain t n 1 D 1 n 1 + k + n 1 = k (n 2 ) 2, t n 2 D 2 n 2 + 2k (n 2 ) 2 = 2k + n 1, (2.10) with initial ata n 0 1 0, n Accoring to the law of mass action, the term (n 2 ) 2 arises because the encounter of two atoms of monoxygen is require for the reaction. We erive the conservation law (number of atoms is constant) by a combination of the equations t [2n 1 + n 2 ] [2D 1 n 1 + D 2 n 2 ] = 0, which implies that for all t 0 [ 2n1 (t, x) + n 2 (t, x) ] [ x = M := 2n 0 1 (x) + n 0 2(x) ] x. (2.11) R R For the simple case of the reaction (2.10), the formula (2.7) for the entropy (with a 1 = 1, b 2 = 2, b 1 = a 2 = 0) gives S(t, x) = n 1 [ ln(k+ n 1 ) 1 ] + n 2 [ ln(k 1/2 n 2 ) 1 ]. (2.12) One can reaily check that Lemma 2.3 (Entropy inequality) t S(t, x)x = R [ n 1 2 n D1 R n 1 + D 2 ] 2 2 n 2 x R [ ln(k (n 2 ) 2 ) ln(k + n 1 ) ][ k (n 2 ) 2 k + n 1 ] x 0. Exercise. Deuce from (2.11) an n 1 0, n 2 0 the a priori boun T 2k (n 2 ) 2 xt M(1 + 2k + T ). 0 R Hint. In (2.10), integrate the equation for n 1. Hemoglobin oxiation. reaction Hemoglobin Hb can bin to ioxygen O 2 to form HbO 2 accoring to the k + Hb + O 2 HbO 2. k This reaction is important in brain imaging because the magnetic properties of Hb an HbO 2 are ifferent; the consumption of oxygen by neurons generates esoxyhemoglobin that can be etecte by MRI (magnetic resonance imaging) an thus, inirectly, inicates the location of neural activity. The resulting system of PDEs, for n 1 = [Hb], n 2 = [O 2 ] an n 3 = [HbO 2 ], is t n 1 D 1 n 1 + k + n 1 n 2 = k n 3, t n 2 D 2 n 2 + k + n 1 n 2 = k n 3, t n 3 D 3 n 3 + k n 3 = k + n 1 n 2.

17 12 CHAPTER 2. REACTION KINETICS AND ENTROPY From this system, we can erive two conservation laws for the total number of molecules Hb an O 2, t [n 1 + n 3 ] [D 1 n 1 + D 3 n 3 ] = 0, t [n 2 + n 3 ] [D 2 n 2 + D 3 n 3 ] = 0. these imply an L 1 control M := [n 1 (t) + n 2 (t) + n 3 (t) + n 4 (t)] = [n n n n 0 4]. As in the case of ioxygene issociation, integrating the thir equation, one conclues the quaratic estimate k + n 1 n 2 x t M(1 + k T ). 0 R From (2.7) (with a 1 = 1, b 2 = 2, b 1 = a 2 = 0), this also comes with an entropy S(t, x) = n 1 [ln(k 1/2 + n 1 1] + n 2 [ln(k 1/2 + n 2 1] + n 3 [ln(k n 3 1]. Mathematical references an stuies on this system can be foun in B. Anreianov an H. Labani 4. Exercise. Another simple an generic example is the reversible reaction A + B k+ C + D t n 1 D 1 n 1 + k + n 1 n 2 = k n 3 n 4, t n 2 D 2 n 2 + k + n 1 n 2 = k n 3 n 4, t n 3 D 3 n 3 + k n 3 n 4 = k + n 1 n 2, t n 4 D 4 n 4 + k n 3 n 4 = k + n 1 n 2. with the n i 0 an the single atom conservation law [ n1 (t, x) + n 2 (t, x) + n 3 (t, x) + n 4 (t, x) ] [ x = M := n 0 1 (x) + n 0 2(x) + n 0 3(x) + n 0 4(x) ] x. R R Choose k i so that S = reaction (2.13). k (2.13) 4 n i ln(k i n i ) is a convex entropy an write the entropy inequality for the chemical i=1 2.4 Small numbers of molecules It may happen that in biological systems, e.g. intracellular bio-molecular interactions, the number of molecules is not large so that the reaction rate equations are not applicable. The alternative is to use the so-calle chemical master equations that escribe the ynamic of iniviual molecules. These moels are close to the jump processes escribe in Chapter?? an with a brief presentation in Section??. They allow for a number of asymptotic limits (from iscrete to continuous, from continuous jumps to rift-iffusion). For these reasons, there is a large literature on this subject, on both the theoretical an numerical aspects 5,6,7. 4 B. Anreianov an H. Labani, Preconitioning operators an L attractor for a class of reaction-iffusion systems. Comm. Pure Appl. Anal. Vol. 11, No. 6 (2012) A. De Masi, E. Presutti, Mathematical methos for hyroynamic limits. Springer-Verlag, Berlin, D. T. Gillespie, Exact stochastic simulation of couple chemical reactions. J. Phys. Chem. 81(25) (1977) 7 J.-L. Doob, Markoff chains- Denumerable case. Trans. Amer. Math. Soc. 58(3), (1945)

18 Chapter 3 Slow-fast ynamics an enzymatic reactions In many situations, one can encounter very ifferent time scales for ifferent reactions, an more generally phenomena. When a time scale is very short compare to the other, one wishes to consier that the fast phenomenon is at equilibrium. This chapter gives circumstances where this is legitimate. A typical example is the case of enzymatic reactions, where a low level of enzymes regulates a reaction. Other examples arise in the spiking ynamic of neurons. Throughout this chapter we ignore iffusion an work with orinary ifferential equations. parameter ε > 0, we stuy the system of two equations with a fast an a slow time scale Formally, the limiting system shoul be written as ε uε(t) t = f ( u ε (t), v ε (t) ), u ε (t = 0) = u 0, v ε(t) t = g ( u ε (t), v ε (t) ), v ε (t = 0) = v 0. 0 = f ( ū(t), v(t) ), v(t) t = g ( ū(t), v(t) ), v(t = 0) = v 0. For a (3.1) (3.2) There are however several objections to such a general aymptotic result. Firstly, to pass to the limit in nonlinear expressions as f ( u ε (t), v ε (t) ), one nees compactness an for u ε (t) the time erivative is not uner control. Seconly, the loss of an initial conition for ū(t) shoul be explaine. Thirly, the relation 0 = f ( ū(t), v(t) ) oes not necessarily efine ū(t) as a function of v(t) so as to insert it in the ifferential equation for v(t). This last observation leas us to istinguish two cases uf(u, v) < 0, then we can hope to fin a smooth mapping F (v) such that f(f (v), v) = 0. We call it the monostable case. Otherwise, some organizing principle shoul be foun to ecie which branch of solution of f(u, v) = 0 is selecte an we present some general results in section 3.2.

19 14 Slow-fast ynamics an enzymatic reactions The sign of matrix is uf(u, v) also appears when one stuies the stability of a steay state. The linearize 1 ε u f(u, v) 1 ε u (gu, v) v f(u, v) v g(u, v) = 1 ε Again stability is relate to the property uf(u, v) < 0. uf(u, v) 3.1 Slow-fast ynamics (monostable) v f(u, v) O(1). We begin with the simplest situation where a fast scale can be consiere at equilibrium, meaning that we can invert the relation 0 = f ( u, v ) u = F (v), with F a smooth function from R to R. Then, we en up with the single equation v(t) t = g ( F ( v(t)), v(t) ), v(t = 0) = v 0. In particular the initial ata for ū is lost, a phenomena calle initial layer. We are going to assume f, g are Lipschitzian, (3.3) f (u, v) α < 0 for some constant α > 0. (3.4) u The secon assumption implies that the equation on u ε is monostable, as one can see it in the Exercise. With the assumption (3.4), for the ifferential equation with v fixe u(t) t 1. prove that u(t) t f(u 0, v) e αt, 2. show that there is a limit u(t) U as t. Hint. Write the equation on z(t) = u(t) t. = f ( u(t), v ), u(t = 0) = u 0, Theorem 3.1 We assume (3.3) (3.4) an that for some T > 0 there is a constant M such that u ε (t) + v ε (t) M. Then, (i) v ε (t) is Lipschitzian, uniformly in ε, for t [0, T ], (ii) u ε (t) is Lipschitzian, uniformly in ε, for t [τ, ), for all τ ]0, T ], (iii) there are ū(t), v(t) C([0, T ]) such that, as ε 0, u ε (t) ū(t) uniformly in [τ, T ] an v ε (t) v(t), uniformly in [0, T ] an (3.2) hols. Proof. The statement (i) is obvious since g(u ε (t), v ε (t)) is boune. For statement (ii), we efine z ε := u ε(t). t Differentiating the equation on u ε, we compute using the chain rule z ε (t) t f = z ε u (u ε, v ε ) + f v (u ε, v ε ) v ε(t), t

20 3.1. SLOW-FAST DYNAMICS (MONOSTABLE) 15 1 z ε (t) 2 2 t Thanks to assumption (3.3) an (3.4), we fin an thus This proves (ii). z ε (t) 2 t = zε 2 f u (u f ε, v ε ) + z ε v (u ε, v ε ) g(u ε, v ε ). 2αz 2 ε + αz 2 ε + 1 α [f v (u ε, v ε ) g(u ε, v ε )] 2 αz 2 ε + K z ε (t) 2 z 2 0e αt/ε + K α = f(u0, v 0 ) 2 ε 2 e αt/ε + K α. For (iii), we use the Ascoli-Arzela theorem an conclue that for a subsequence, we have uniform convergence. Then we may pass to the limit in the integral form. The full family converges because of uniqueness of solutions of (3.2). Exercise. if the intital ata is well-prepare, that means f(u 0, v 0 ) = 0, prove that u ε (t) is uniformly lipschitzian on [0, ). Exercise. Consier the system, with Lipschitz continous f an g, ε uε(t) t = f ( u ε (t), v ε (t) ), u ε (t = 0) = u 0 (0, 1), v ε(t) t = g ( u ε (t), v ε (t) ), v ε (t = 0) = v 0, assuming that f(0, v) = f(1, v) = 0, f(u, v) 0 for u (0, 1). 1. Show that 0 < u ε (t) < 1 for all times. 2. Show that uε(t) 0 t t 1 an conclue that, after extracting a subsequence, u ε (t) ū(t) as ε 0, almost everywhere an in all L p (0, ), 1 p <. 3. Show that ū = 1 an v(t) t = g ( 1, v(t) ), v(t = 0) = v 0. We may also prove uniform bouns on u ε (t) an v ε (t). Theorem 3.2 With the assumptions (3.3) (3.4), for all T > 0, there is a constant M(T ) such that v ε (t) M(T ), 0 t T, Proof. We can write, ε u ε(t) t = f ( 0, v ε (t) ) uε(t), + 0 f ( σ, v ε (t) ) u ε u ε(t) 2 2 t u ε (t)f ( 0, v ε (t) ) αu ε (t) 2 σ, α 2 u ε(t) 2 + C[1 + v ε (t) 2 ] αu ε (t) 2 because f(, ) is Lipschitz continuous. In the same way, we fin ε v ε(t) 2 2 t C v ε (t) [1 + u ε (t) + v ε (t) ] α 2 u ε(t) 2 + C[1 + v ε (t) 2 ]

21 16 Slow-fast ynamics an enzymatic reactions Aing up these two inequalities rises 1 2 t [v ε(t) 2 + ε u ε (t) 2 ] C[1 + v ε (t) 2 ] C[1 + v ε (t) 2 + ε u ε (t) 2 ] an the Gronwall inequality gives us that v ε (t) 2 + ε u ε (t) 2 is locally boune. To boun u ε (t) 2 itself, we may now write the inequation for u ε (t) 2 as ε u ε (t) 2 α 2 t 2 u ε(t) 2 + C[1 + M(T )] which proves the result because max ( (u 0 ) 2. 2C[1+M(T ) ) α is a solution with a larger initial ata. 3.2 Slow-fast ynamics (general) The assumption (3.4) is very strong an one can exten the result to more general systems but with weaker conclusions. We keep the system ε uε(t) t = f ( u ε (t), v ε (t) ), u ε (t = 0) = u 0, v ε(t) t = g ( u ε (t), v ε (t) ), v ε (t = 0) = v 0, (3.5) f, g are Lipschitz continuous an C 1. (3.6) Theorem 3.3 With the assumption (3.6) an assuming that u ε (t), v ε (t) are boune on [0, T ], we have T 0 f(u ε (t), v ε (t)) 2 t C(T, K)ε. If, for all v, f(u, v) oes not vanish on an interval in u, then after extaction of a subsequence an the relations hols Proof. 1. We efine an write Since we conclue that T 0 u ε(n) ū(t) a.e. v ε(n) v(t) uniformly on [0, T ] T 0 v(t) t = g ( ū(t), v(t) ), f ( ū(t), v(t) ) = 0. F (u, v) = u T f(u ε (t), v ε (t)) 2 t = ε 0 F (u(t), v(t)) t f(u ε (t), v ε (t)) 2 t 0 f(σ, v)σ = f(u(t), v(t)) u(t) t f(u ε (t), v ε (t)) u ε(t) t. t + T [ F (uε (t), v ε (t)) = ε 0 t F (u, v) v v(t), t F (u ] ε, v ε ) g(u ε, v ε ) t v = ε [ F (u ε (T ), v ε (T )) F (u 0, v 0 ) + T K ].

22 3.2. SLOW-FAST DYNAMICS (GENERAL) 17 Because u ε (t), v ε (t) are locally uniformly boune, we conclue. 2. We efine u G(u, v) = f(σ, v) 2 σ. We have t G(u ε(t), v ε (t)) = f(u ε (t), v ε (t)) 2 u ε(t) + 2g(u ε (t), v ε (t)) t = f(u ε(t), v ε (t)) 2 f(u ε (t), v ε (t)) + 2g(u ε (t), v ε (t)) ε 0 uε(t) 0 uε(t) 0 f(σ, v ε (t)) v f(σ, v ε (t))σ f(σ, v ε (t)) v f(σ, v ε (t))σ an thus G(u ε (t), v ε (t)) is locally BV. As a consequence for a subsequence, there is Ḡ(t) such that G(u ε(n) (t), v ε(n) (t)) Ḡ(t) a.e. as n. But because v ε is uniformly Lipschitzian, by the Ascoli theorem, we also have v ε(n) (t) v(t) an thus G(u ε(n) (t), v(t)) Ḡ(t) a.e. as n. 3. Because G(u, v) is strictly increasing in u from our assumption, the above convergence implies that u ε(n) (t) converges a.e. to some ū(t) L (0, T ). The en follows immeiately. Exercise. (Morris-Lecar system) Consier the system { ε gε(t) t = G(v ε (t)) g ε (t), = g ( v ε (t) ) + g ε (t)(v v ε (t)). v ε(t) t Ientify the limiting system an show convergence. Exercise. (Bistable case) Consier the case when, for a smooth function a : R (0, 1) we have f(0, v) = f(1, v) = f(a(v), v) = 0, < 0 for 0 < u < a(v), f(u, v) = > 0 for a(v) < u < Prove that T 0 f(u ε(t), v ε (t)) t C(T )ε. 2. Show that u ε is uniformly with boune total variation on [0, T ]. 3. Show that, for a subsequence, as k, v εk (t) v (uniformly), an u εk (t) ū(t) (in L p (0, T ), with 1 p < ), with ū, v a solution of (3.2) (almost everywhere). [Hint.] Use the explicit formula for ϕ(u, v) := u sgn(f(σ, v)σ 0 Exercise. (The FitzHugh-Nagumo system). This system, use in the neuroscience, is written ε uε(t) t = F ( u ε (t) ) v ε (t), u ε (t = 0) = u 0, v ε(t) t = g ( u ε (t), v ε (t) ), v ε (t = 0) = v 0, an F (u) = u(u a)(u + a) is increasing (an thus unstable, see section 3.1) in the interval ( a 3, a 3 ). 1. Show that T 0 f(u ε(t), v ε (t)) t C(T )ε. (3.7)

23 18 Slow-fast ynamics an enzymatic reactions 2. Show that, for a subsequence, as k, v εk (t) v (uniformly), an u εk (t) ū(t) (in L p (0, T ), with 1 p < ), with ū, v a solution (almost everywhere) of F ( ū(t) ) = v, v(t) t = g ( ū(t), v(t) ), v ε (t = 0) = v 0, (3.8) Figure 3.1: Solution of the FitzHugh-Nagumo system (3.8) with F (u) = u(3 u)(3 + u) an g(u, v) = u.5. Upper left u component, upper right v component. Lower re curve the nullcline v = F (u) an black curve the solution u(t), v(t) in the phase plane (u, v). 3.3 Enzymatic reactions More representative of biology are the enzymatic reactions, which are usually associate with the names of Michaelis an Menten 1. A substrate S can be transforme into a prouct P, without molecular change as in the isomerization reaction, but this reaction occurs only if an enzyme E is present. The process consists first of the formation of a complex SE, that can be itself issociate into P + E. The formation 1 Michaelis, L. an Menten, M. I., Die Kinetic er Invertinwirkung. Biochem. Z., 49, (1913)

24 3.3. ENZYMATIC REACTIONS 19 of the complex is a reversible reaction, but the conversion to P + E is an irreversible reaction, leaing to the representation S + E k+ k C = SE kp P + E. Still using the law of mass action, this leas to the equations (we o not consier the space variable an molecular iffusion, this is not the point here) t n S = k n C k + n S n E, t n E = (k + k p )n C k + n S n E, t n C = k + n S n E (k + k p )n C, t n P = k p n C. This reaction comes with the initial ata n 0 S > 0, n0 E > 0 an n0 C = n0 P the substrate is entirely converte into the prouct = 0. One can easily verify that lim t n P (t) = n 0 S, lim n S (t) = 0, t lim n E (t) = n 0 t E, lim n C (t) = 0. t A conclusion that is incorrect if n 0 E = 0. This is a funamental observation in enzymatic reactions that a very small amount of enzyme is sufficient to convert the substrate into the prouct. Two conservation laws hol true in these equations an this helps us to unerstan the above limit, n E (t) + n C (t) = n 0 E + n0 C = n0 E, (3.9) n S (t) + n C (t) + n P (t) = n 0 S + n0 C + n0 P = n0 S. Because the properties of the ynamic o not epen on n P, the system is equivalent to a 2 2 system t n S = k n C k + n S (n 0 E n C), (3.10) t n C = k + n S (n 0 E n C) (k + k p )n C. Briggs an Halane(1925) arrive to the conclusion that this system can be further simplifie with the quasi-static approximation on n C. That means that n C (t) can be compute algebraically from n S (t), thus leaing to a single ODE for n S (t) t n S = k n C k + n S (n 0 E n C), (3.11) 0 = k + n S (n 0 E n C) (k + k p )n C, still with the initial ata n S (t = 0) = n 0 S (but no initial conition on n C). We can of course write this in a more conense way. Since n C = k+n Sn 0 E k +k ++k +n S, an after aing the two equations, we fin the typical enzymatic reaction kinetic equation t n S = n 0 E k p k + n S k + k p + k + n S (3.12) In other wors, the law of mass action oes not apply in this approximation an one fins the so-calle Michaelis-Menten law for the irreversible reactions, a Hill function rather than a polynomial. Although not obvious at first glance, this result makes rigorous sense an we have the

25 20 Slow-fast ynamics an enzymatic reactions is sufficiently small. The solu- Proposition 3.4 (Valiity of Michaelis-Menten law) Assume n 0 E tions of (3.10) an of (3.12) satisfy for some constant C(n 0 S ) inepenant of n0 E. sup n S (t) n S (t) C(n 0 S) n 0 E t 0 We recall that n C (t = 0) = u ε C (t = 0) = 0 an n S(t = 0) = u ε S (t = 0) > 0. Proof. We reuce the system to slow-fast ynamic (see [13] for an extene presentation of the subject). To simplify the notations, we set ε := n 0 E an we efine the slow time an new unknowns as ε := n 0 E, s = εt, u ε S(s) = n S (t) 0, u ε C(s) = n C(t) n 0 E 0. Then the two systems are respectively written as s uε S = k u ε C k +u ε S (1 uε C ), ε s uε C = k +u ε S (1 uε C ) (k + k p )u ε C, s u S = k u C k + u S (1 u C ), 0 = k + u S (1 u C ) (k + k p )u C. (3.13) With these notations the result in Proposition 3.4 is equivalent to sup u ε S(s) u S (s) Cε. s 0 The situation is the favorable case of Section 3.1 with the aition that the convergence is uniform globally an not only locally. The etails of the proof are left to the reaer because it is a general conclusion from the theory of slow-fast ynamic. We shall make use of the two consequences of (3.13) s [uε S + εu ε C] = k p u ε C, We have from (3.9) an u ε C (t = 0) = 0, s u S = k p u C. (i) 0 u ε C 1, 0 u C 1, (ii) 0 u ε S n0 S, 0 u S n 0 S, (iii) 0 u ε C um < 1, 0 u C u M < 1 with u M := k +n 0 S k +n 0 S +k +kp inepenent of ε, (iv) s uε C (s) K for some K(n0 S ), (v) From step (iv), introuce the boune quantity r ε (s) := s uε C (s) + k +u ε C (s)2, an R ε (s) = u ε S(s) + εu ε C(s) u S (s). Compute that u ε C(s) = εrε (s) k + u ε C Rε + k + u ε S k + u S + k + k p. u ε C u C = εrε + k + (1 u ε C )Rε εk + u ε C k + u S + k + k p.

26 3.4. BELOUSOV-ZHABOTINSKII REACTION 21 Conclue that k + (1 u ε s Rε C (s) + k ) p R ε r ε + k + u ε C (s) = εk p. k + u S + k + k p k + u S + k + k p (vi) Using steps (iii) an (v), conclue that R ε (s) Cε for a constant C. Hints. (ii) u ε S + uε C ecreases. (iii) Write s uε C k +n 0 S (1 uε C ) (k + k p )u ε C. (iv) Write the equation for z(s) = s uε C (s) as ε s z + λ(s)z(s) = µ(s) with λ > k + k p an µ boune. (vi) Write 1 2 s Rε (s) 2 + AR ε (s) 2 εb R ε (s), with A > 0, B > 0 two constants. 3.4 Belousov-Zhabotinskii reaction Figure 3.2: Perioic regime of the system of ODEs (3.14) that mimics the Belousov-Zhabotinskii reaction. The parameters are k A = k B = 1 an k C = 2 (Left), k C = 20 (Right). The solution n B is the top (re) curve, n A is in blue an n C in green. In abscissae is time. For all complex chemical reactions, the etaile escription of all elementary reactions is not realistic. Then, as for the enzymatic reaction, one simplifies the system by assuming that some reactions are much faster than others, or that some components are in higher concentrations than others. These manipulations may violate mass conservation an entropy inequality may be lost; this is a conition to obtain pattern formation as in the example of the CIMA reaction in Section The famous Belousov-Zhabotinskii reaction is known as the first historical example to prouce perioic patterns. This was iscovere in 1951 by Belousov, it remaine unpublishe because no respectable chemist at that time coul accept this iea. Belousov an Zhabotinskii receive the Lenin Prize in 1980, a ecae after Belousov s eath an the iscovery in the USA of other perioic reactions, simpler to reprouce.

27 22 Slow-fast ynamics an enzymatic reactions We borrow a simple example from A. Turner 2, this avois entering the etails of a real chemical reaction (refer to [32] for a complete treatment). The simple example consists of three reactants enote as A, B, C, an three irreversible reactions (because of this feature, entropy issipation an relaxation o not hol). A + B k A 2A, B + C k B 2B, C + A k C 2C. Therefore, following the general formalism in (2.6), the system is written as t n A = n A (k A n B k C n C ), t n B = n B (k B n C k A n A ), t n C = n C (k C n A k B n B ). (3.14) Notice that n A (t) > 0, n B (t) > 0, n C (t) > 0, n A (t) + n B (t) + n C (t) = n 0 A + n 0 B + n 0 C. Numerical simulations are presente in Figure 3.2. For k c = 20, which is very large compare to k A an k B one can estimate that the equations imply that k C n A n C = O(1) than thus n A n C 0. This is oberve on the simulation on the right. 2 A. Turner, A Simple Moel of the Belousov-Zhabotinsky Reaction from First Principles, 2009,

28 Chapter 4 Some material for parabolic equations 4.1 Bounary conitions When working in a omain Ω (a connecte open set with sufficiently smooth bounary), we encounter two elementary types of bounary conitions. The reaction-iffusion systems (1.1) or (2.1) are complete either by the Dirichlet or the Neumann bounary conitions. Dirichlet bounary conition on Ω. n i = 0 on Ω. This bounary conition means that iniviuals or molecules, when they reach the bounary are efinitively remove from the population Ω, which therefore iminishes. This interpretation stems from the Brownian motion unerlying the iffusion equation (see below). But we can see that inee, if we consier the conservative chemical reactions (2.1) with (2.3), then t Ω I n i (t, x)x = i=1 I D i i=1 Ω ν n i, with ν the outwar normal to the bounary. But with n i (t, x) 0 in Ω an n i (t, x) = 0 in Ω, we have ν n i 0, therefore the total mass iminishes, for all t 0, t Ω I n i (t, x)x 0, i=1 Neumann bounary conition on Ω. Ω i=1 I n i (t, x)x ν n i = 0 on Ω, Ω i=1 I n 0 i (x)x. still with ν the outwar normal to the bounary. This means that iniviuals or molecules are reflecte when they hit the bounary. In the computation above for (2.1) with (2.3), the normal erivative vanishes

29 24 Some material for parabolic equations an we fin irectly mass conservation t Ω I n i (t, x)x = 0, i=1 I I n i (t, x)x = n 0 i (x)x. Ω i=1 Ω i=1 Mixe or Robin bounary conition on Ω. αn i + ν n i = 0 on Ω, α 0. When reaching the bounary, there is either reflexion or removal with a propability relate to α (α = 0 no removal, α = no reflexion). The mass balance reas t Ω I n i (t, x)x = i=1 I D i i=1 Ω ν n i = α I D i i=1 Ω n i < 0. Full space. There is a large ifference between the case of the full space R an the case of a boune omain. This can be seen by the results of spectral analysis in Section 4.2, which o not hol in the same form in R. The countable spectral basis w k (x) is replace by e ix.ξ, an leas to the Fourier transform. 4.2 The spectral ecomposition of Laplace operators (Dirichlet) The spectral ecomposition of the Laplace operator is going to play an important role in subsequent chapters, therefore we introuce some material now. We begin with the Dirichlet bounary conition u = f in Ω, (4.1) u = 0 on Ω. Theorem 4.1 (Dirichlet) Consier a boune connecte open set Ω, then there is a spectral basis (λ k, w k ) k 1 for (4.1), that is, (i) λ k is a non-ecreasing sequence with 0 < λ 1 < λ 2 λ 3... λ k... an λ k, k (ii) (λ k, w k ) are eigenelements, i.e., for all k 1 we have w k = λ k w k in Ω, w k = 0 on Ω, (iii) (w k ) k 1 is an orthonormal basis of L 2 (Ω), (iv) we have w 1 (x) > 0 in Ω an the first eigenvalue λ 1 is simple, an for k 2, the eigenfunction w k changes sign an can be multiple. Remark 1. The hypothesis that Ω is connecte is simply use to guarantee that the first eigenvalue is simple an the corresponing eigenfunction is positive in Ω. Otherwise, we have several aitional nonnegative eigenfunctions with a first eigenfunction in one component an 0 in the others. Remark 2. The sequence w k is also orthogonal in H 1 0 (Ω) (for Dirichlet bounary conitions) an H 1 (Ω)

30 4.2. THE SPECTRAL DECOMPOSITION OF LAPLACE OPERATORS (DIRICHLET) 25 for Neumann bounary conitions (see below). Inee, if w k is orthogonal to w j in L 2 (Ω), then from the Laplace equation for w k an the Stokes formula w j. w k = λ k w j w k = 0. Ω Therefore, orthogonality in L 2 implies orthogonality in H0 1 or H 1. Remark 3. Note that for k 2, the eigenfunction w k changes sign because Ω w 1w k = 0 an w 1 has a sign. Remark 4. Notice the formula w i (x)w i (x ) = δ(x x ). Inee, this means that for all test function in ϕ C( Ω) L 2 (Ω), i=1 i=1 w i (x) w i (x )ϕ(x )x = ϕ(x )δ(x x )x = ϕ(x), Ω Ω which is the ecomposition of ϕ on the orthonormal basis (w i ) i 1 Ω Proof of Theorem 4.1. We only sketch the proof. For aitional matters see [15] Ch. 5, [6] p. 96. The result is base on two ingreients. (i) The spectral ecomposition of self-ajoint compact linear mappings on Hilbert spaces is a general theory that extens the case of symmetric matrices. (ii) The simplicity of the first eigenvalue with a positive eigenfunction is also a consequence of the Krein-Rutman theorem (infinite imension version of the Perron-Froebenius theorem). First step. 1st eigenelements. In the Hilbert space H = L 2 (Ω), we consier the linear subspace V = H0 1 (Ω). Then we efine the minimum on V λ 1 = min u 2 x. (4.2) Ω u 2 =1 This minimum is achieve because a minimizing sequence (u n ) will converge strongly in H an weakly in V to w 1 V with Ω w 1 2 = 1, by the Rellich compactness theorem (see [10, 6]). Therefore, Ω w 1 2 x lim inf n Ω u n 2 x = λ 1. This implies equality an that λ 1 > 0. The variational principle associate to this minimization problem says that Ω w 1 = λ 1 w 1, which implies that w 1 is smooth in Ω (by elliptic regularity) Secon step. Positivity. Because in V, u 2 = u 2 a.e., the construction above tells us that w 1 is also a first eigenfunction an we may assume that w 1 is nonnegative. By the strong maximum principle applie to the Laplace equation, we obtain that w 1 is positive insie Ω (because it is connecte). This also proves that all the eigenfunctions associate with λ 1 have a sign in Ω because, in a connecte open set, w 1 cannot satisfy the three properties (i) be smooth, (ii) change sign an (iii) w 1 be positive also.

31 26 Some material for parabolic equations Thir step. Simplicity. Finally, we can euce the simplicity of this eigenfunction because if there were two inepenent eigenfunction, a linear combination woul allow us to buil one which changes sign (by orthogonality to w 1 for example) an this is impossible by the above positivity argument. Fourth step. Other eigenelements. We may iterate the construction. Denote E k the finite imensional subspace generate by the k-th first eigenspaces. We work on the close subspace Ek of H, an we may efine λ k+1 = min u 2 x. u Ek V, Ω u 2 =1 Ω This minimum is achieve by the same reason as before. The variational form gives that the minimizers are solutions of the k + 1-th eigenproblem. They can form a multiimensional space but always finite imensional; otherwise we woul have an infinite imensional subspace of L 2 (Ω), whose unit ball is compact thanks to the Rellich compactness theorem, since Ω u 2 x λ k+1 in this ball Also λ k as k because one can easily buil (with oscillations or sharp graients) functions satisfying Ω u n 2 = 1 an Ω u n 2 x n. 4.3 The spectral ecomposition of Laplace operators (Neumann) The same metho applies to the Neumann bounary conition u = f in Ω, We state without proof the (4.3) u ν = 0 on Ω. Theorem 4.2 (Neumann) Consier a C 1 boune connecte open set Ω, then there is a spectral basis (λ k, w k ) k 1 for (4.3), i.e., (i) λ k is a non-ecreasing sequence with 0 = λ 1 < λ 2 λ 3... λ k... an λ k, k (ii) (λ k, w k ) are eigenelements, i.e., for all k 1 we have w k = λ k w k in Ω, w k ν = 0 on Ω, (iii) (w k ) k 1 is an orthonormal basis of L 2 (Ω) (iv) w 1 (x) = 1 > 0, an for k 2, the eigenfunction w Ω 1/2 k changes sign. 4.4 The Poincaré an Poincaré-Wirtinger inequalities As an outcome of the proof in Section 4.2, we recover the Poincaré inequality for ω a boune omain u 2 u 2 x, u H0 1 (Ω), (4.4) λ D 1 Ω Ω

32 4.5. RECTANGLES: EXPLICIT SOLUTIONS 27 where λ D 1 enotes the first eigenvalue of the Laplace-Dirichlet equation (4.1). This inequality is just another statement equivalent to (4.2). The Poincaré-Wirtinger inequality states that λ N 2 u u 2 u 2 x, u H 1 (Ω), (4.5) Ω Ω where λ N 2 is the first positive eigenvalue for the Laplace-Neumann equation (4.3) an u = 1 u(x)x, Ω Ω is the L 2 projection on the first eigenfunction with Neumann bounary conitions (a constant). Therefore this statement is nothing but the characterization of the secon eigenvalue given in the proof in Section Rectangles: explicit solutions In one imension, one can compute explicitly the spectral basis because the solutions of u = λu are all known. On Ω = (0, 1) we have w k = a k sin(kπx), λ k = (kπ) 2, (Dirichlet) w k = b k cos ( (k 1)πx ), λ k = ( (k 1)π ) 2, (Neuman) an a k an b k are normalization constants that ensure 1 0 w k 2 = 1. In two imensions, on a rectangle (0, L 1 ) (0, L 2 ) we see that the family is better escribe by two inices k 1 an l 1 an we have w kl = a kl sin(kπ x 1 L 1 ) sin(lπ x 2 L 2 ), λ kl = ( ( k L 1 ) 2 + ( l L 2 ) 2) π 2, (Dirichlet) w kl = b kl cos ( (k 1)π x L 1 ) cos ( (l 1)π x L 2 ), λkl = ( ( k 1 L 1 ) 2 + ( l 1 L 2 ) 2) π 2, (Neuman) These examples inicate that The first positive eigenvalue is of the orer of 1/ max(l 1, L 2 ) 2 = 1 iam(ω) 2. For a large omain (even with a large aspect ratio) we can expect that the first eigenvalues are close to zero an that the eigenvalues are close to each other. Except the first one, eigenvalues can be multiple (take L 1 /L 2 N). Large eigenvalues are associate with highly oscillating eigenfunctions.

33 28 Some material for parabolic equations Figure 4.1: Two sample paths of one imensional Brownian motion accoring to the approximation (4.6). Abscissae t k, orinates X k. Figure 4.2: Two sample paths of two imensional Brownian motion simulate accoring to (4.6). 4.6 Brownian motion an the heat equation The relationship between Brownian motion an the heat equation explains, in a simple framework, why a ranom walk of iniviuals leas to the terms D n for the population ensity n(t, x). It is rather ifficult to construct rigorously Brownian motion. However, it is easy to give an intuitive approximation, which is sufficient to buil the probability law of Brownian trajectories. To o so, we follow the Euler numerical metho for an ODE with a time step t that uses iscrete times t k = k t. In a probability space with a measure enote P (ω), the initial position X 0 R being given with a probability law n 0 (x), we efine iteratively a iscrete trajectory X k (ω) R as follows. We set X k+1 (ω) = X k (ω) + t Y k (ω) (4.6) where Y k (ω) enotes a -imensional ranom variable inepenent of X k (one speaks of inepenent

34 4.6. BROWNIAN MOTION AND THE HEAT EQUATION 29 increments) an with a normal law N(y). We recall that inepenent means Ef(X k, Y k ) := f ( X k (ω), Y k (ω) ) P (ω) = f(x, y)n k (x) N(y)y (4.7) R R with N(y) = 1 (2π) /2 e y 2 /2, the normal law, n k (x) the law of the process X k, efine as P ( Φ(X k (ω)) ) = Φ(x)n k (x). Two simulations are presente. Figure 4.1 epicts, in the one imensional case (t k, X k ), two iterates X k (for two ifferent ω). Figure 4.2 shows two iterates X k in two imensions. Our purpose is to compute the law n k (x) at the limit t 0. To o so, we first use a C 3 function u, with u, Du, D 2 u an D 3 u boune. We use the Taylor expansion of u to compute an then u(x k+1 ) = u(x k ) + tdu(x k ).Y k + t 2 D2 u(x k ).(Y k, Y k ) + O ( t Y k ) 3/2, Eu(X k+1 ) = Eu(X k ) + t 2 ED2 u(x k ).(Y k, Y k ) + O ( t ) 3/2. (4.8) Inee, because N( ) is raially symmetric, the formula (4.7) use with f(x, y) = Du(X k ).Y k yiels EDu(X k ).Y k = EDu(X k ). yn(y)y = 0. R Because y R i y j N(y)y = δ ij, a further use of the formulas (4.8), (4.7) an the efinition of the probability law mentione earlier, give u(x)n k+1 (x) = u(x)n k (x) + t u(x)n k (x) + O ( t ) 3/2. R R 2 R After iviing by t an integration by parts of the term u, we may rewrite this as R u [ n k+1 n k t 1 2 nk ] = O ( t ) 1/2. This hols true for any smooth function u an this means that, in the weak sense, n k+1 n k 1 t 2 nk = O ( t ) 1/2. At the limit in the sense of istributions, as t 0, we obtain a probability law with a ensity n(t, x) that satisfies t n(t, x) 1 2 n(t, x) = 0, n(0, x) = n 0 (x). In particular, even though n 0 is a probability measure, it follows from the regularizing effects of the heat equation that n(t, x) is a (smooth) function an not only a measure as soon as t > 0.

35 30 Some material for parabolic equations The equation for n(t, x) (the heat equation here) is generically calle the Kolmogorov equation for the limiting process (Brownian motion). Exercise. Prove that the ensity of the probability law n k satisfies the integral equation n k+1 (x) n k (x) = [n k (x + t y) n k (x) ] N(y)y. (4.9) Derive the heat equation for the limit n(t, x), as t 0, if it exists. Show that this erivation uses only two y-moments of N an not the full normal law. See also the scattering equation in Chapter??.

36 Chapter 5 Relaxation, perturbation an entropy methos Solutions of nonlinear parabolic equations an systems presente in Chapter 1 can exhibit various an sometimes complex behavior, a phenomena usually calle pattern formation. In which circumstances can such complex behavior happen? A first answer is given in this chapter by inicating some conitions for relaxation to trivial steay states; then nothing interesting can happen! We present relaxation results by perturbation methos (small nonlinearity) an entropy methos. Inee, before we can unerstan how patterns occur in parabolic systems, a necessary step is to unerstan why patterns shoul not appear in principle! Solutions of parabolic equations unergo regularization effects that lea them to constant or simple states. Several asymptotic results can be state in this irection an we present some of them in this chapter. Sections 5.1 an 5.2 have been very much influence by the book [21]. 5.1 Asymptotic stability by perturbation methos (Dirichlet) The simplest possible long term behavior for a semilinear parabolic equation is simply the relaxation towars a stable steay state (that we choose to be 0 here). This is possible when the following two features are combine the nonlinear part is a small perturbation of a main (linear ifferential) operator, this main linear operator has a positive ominant eigenvalue. Of course this simple relaxation behavior is rather boring an appears as the opposite of pattern formation when Turing instability occurs, an this will be escribe later, see Chapter 7. To illustrate this, we consier, in a boune omain Ω, the semi-linear heat equation with a Dirichlet

37 32 Relaxation an the energy metho bounary conition t i(t, x) D i u i (t, x) = F i (t, x; u 1,..., u I ), 1 i I, x Ω, u i (t, x) = 0, x Ω, u i (t = 0, x) = u 0 i (x) L2 (Ω). We assume that F (t, x; 0) = 0, so that u 0 is a steay state solution. Is it a stable an attractive state? Is the number of iniviuals leaving the omain through Ω compensate by the birth term? (5.1) We shall use here a technical result. The Laplace operator (with Dirichlet bounary conition) has a first eigenvalue λ 1 > 0, associate with a positive eigenfunction, w 1 (x), which is unique except multiplication by a constant, w 1 = λ 1 w 1, w 1 H 1 0 (Ω). (5.2) This eigenvalue is characterize as being the best constant in Poincaré inequality (see Section 4.2 or the book [10]) λ 1 v(x) 2 v 2, v H0 1 (Ω), with equality only for v = µw 1, µ R. Ω Ω Theorem 5.1 (Asymptotic stability) Assume min i D i = D > 0 an that there is a (small) constant L > 0 such that u R I, t 0, x Ω, F (t, x; u) L u, or more generally, F (t, x; u) u L u 2, (5.3) then, u i (t, x) vanishes exponentially as t, namely, u(t, x) 2 e 2δt Ω δ := Dλ 1 L > 0, (5.4) Proof. We multiply the parabolic equation (5.1) by u i an integrate by parts 1 u i (t) 2 + D i u i (t) 2 ( ) = u i (t)f i t, x; u(t), 2 t an using the characterization (5.2) of λ 1, we conclue 1 I I u i (t) 2 + Dλ 1 2 t Ω Ω i=1 The result follows from the Gronwall lemma. Ω Ω i=1 Ω u 0 (x) 2. (5.5) Ω u i (t) 2 L Ω I u i (t) 2. Exercise. Consier a smooth boune omain Ω R, a real number λ > 0 an two smooth an Lipschitz continuous functions R(u, v), Q(u, v), such that R(0, 0) = Q(0, 0) = 0. Let ( u(t, x), v(t, x) ) be solutions of the system tu u = R(u(t, x), v(t, x)), t 0, x Ω, u(t, x) = 0 sur Ω, tv + λv = Q(u(t, x), v(t, x)). i=1

38 5.2. ASYMPTOTIC STABILITY BY PERTURBATION METHODS (NEUMAN) Recall the Poincaré inequality for u. 2. Assume R(u, v) L( u + v ) an Q(u, v) L( u + v ), give a size conition for L such that for all initial ata, the solution (u, v) converges exponentially to (0, 0) for t. Solution. A simple answer is min(λ 1, µ) 2L =: δ > 0. A more elaborate conition is base on the positive real number such that λ 1 µ = a a 1 an is λ 1 L a 1 = µ L a =: δ > Asymptotic stability by perturbation methos (Neuman) The next simplest possible long term behavior for a parabolic equation is relaxation towars an homogeneous (i.e., inepenent of x) solution, which is not constant in time. This is possible when two features are combine the nonlinear part is a small perturbation of a main (ifferential) operator, this main operator has a non-empty kernel (0 is the first eigenvalue). Consier again, in a boune omain Ω with outwar unit normal ν, the semi-linear parabolic equation with a Neumann bounary conition t u i(t, x) D i u i (t, x) = F i (t; u 1,..., u I ), 1 i I, x Ω, ν(x) u i(t, x) = 0, x Ω, (5.6) u i (t = 0, x) = u 0 i (x) L2 (Ω). The Laplace operator (with a Neumann bounary conition) has λ 1 = 0 as a first eigenvalue, associate with the constants w 1 (x) = 1/ Ω as eigenfunction. We shall use its secon eigenvalue λ 2, characterize by the Poincaré-Wirtinger inequality (see [10] an Section 4.2) λ 2 v(x) v 2 v 2, v H 1 (Ω), (5.7) with the average of v(x) efine as Ω Ω v = 1 v. Ω Ω Note that this is also the L 2 projection on the eigenspace spanne by w 1. Theorem 5.2 (Relaxation to a homogeneous solution) Assume min i D i = D > 0 an ( F (u) F (v) ) (u v) L u v 2, u, v R I, (5.8) δ := Dλ 2 L > 0, (5.9) then, u i (t, x) tens to become homogeneous with an exponential rate, namely, u(t, x) u(t) 2 e 2δt u 0 (x) u 0 2. (5.10) Ω Ω

39 34 Relaxation an the energy metho Proof. Integrating in x equation (5.6), we fin therefore, Thus, using assumption (5.8), we fin 1 2 t t u i = F i (t; u), t [u i u i ] D i [u i u i ] = F i (t; u) F i (t; u). Ω u u 2 + D Ω ( u u ) 2 = Ω ( F (t; u) F (t; u) ) (u u ) = F (t; u) (u u ) Ω = Ω ( F (t; u) F (t; u ) ) (u u ) L Ω u u 2. Therefore, with notation (5.9) u u 2 2δ u u 2. t Ω Ω The result (5.10) follows irectly. Exercise. Explain why we cannot allow a epenency on x in the nonlinearity F i (t; u). Exercise. Let v H 2 (Ω) satisfy v ν = 0 on Ω (Neumannconition). 1. Prove, using the Poincaré-Wirtinger ineqality (5.7), that λ 2 2. In the context of Theorem5.2, assume that inequality, prove that Ω Ω v 2 Ω v 2. (5.11) I D j F i (t; u)ξ i ξ j L ξ 2, ξ R I. Using the above i,j=1 u i (t, x) 2 e 2δt i Ω u 0 i (x) 2. (5.12) 3. Deuce a variant of Theorem 5.2. Hints. 1. Integrate by parts the expression Ω v Use the equation for u i t x l. i 5.3 Entropy an relaxation We have seen in Lemma 2.3 that reaction kinetic equations as in (2.10) are enowe with an entropy (2.12). This originates from the microscopic N-particle stochastic systems from which reaction kinetics are erive (see Section??).

40 5.3. ENTROPY AND RELAXATION 35 This entropy inequality is also very useful because it can be use to show relaxation to the steay state, inepenently of the size of the constants k 1, k 2. To o that we consier a Neumann bounary conition in a boune omain Ω t n 1 D 1 n 1 + k 1 n 1 = k 2 (n 2 ) 2, t, x Ω, t n 2 D 2 n 2 + 2k 2 (n 2 ) 2 = 2k 1 n 1, (5.13) ν n 1 = ν n 2 = 0 on Ω. Theorem 5.3 The solutions of (5.13), with n 0 i 0, n0 i L1 (Ω), n 0 i ln(n 0 i ) L1 (Ω), satisfy that n i (t, x) N i, as t with N i the constants efine uniquely by 2N 1 + N 2 = [2n 0 1(x) + n 0 2(x)]x, k 2 (N 2 ) 2 = k 1 N 1. Ω Proof. Then S(t, x) = n 1 [ ln(k1 n 1 ) 1 ] + n 2 [ ln(k 1/2 2 n 2 ) 1 ] satisfies, following Lemma 2.3, t S(t, x)x Ω = Ω Ω [ n 1 2 n D1 n 1 + D 2 ] 2 2 n 2 x [ ln(k2 n 2 2) ln(k 1 n 1 ) ][ k 2 (n 2 ) 2 k 1 n 1 ] x. An, because S is boune from below, it also gives a boun on the entropy issipation 0 Ω [ n 1 2 n D1 n 1 + D 2 ] 2 2 n 2 xt + [ ln(k2 n 2 0 Ω 2) ln(k 1 n 1 ) ][ ] k 2 (n 2 ) 2 k 1 n 1 xt C(n 0 1, n 0 2). (5.14) This is again a better integrability estimate in x than the L 1 log estimate (erive from mass conservation) because of the quaratic term (n 2 ) 2. From a qualitative point of view, this says that the chemical reaction shoul lea the system to a spatially homogeneous equilibrium state. Inee, formally at least, the integral (5.14) can be boune only if n 1 = n 2 0 as t, k 2 (n 2 ) 2 k 1 n 1, n 1 = n 2 0 as t. The first conclusion says that the ynamic becomes homogeneous in x (but may epen on t). The secon conclusion, combine with the mass conservation relation (2.11) shows that there is a unique possible asymptotic homogeneous state because the constant state satisfies k 2 (N 2 ) 2 = k 1 N 1 = k 1 ( M Ω N 2), which has a unique positive root. Exercise. Consier (2.10) with a Neumann bounary conition an set S(t, x) = 1 k 1 Σ 1 ( k1 n 1 (t, x) ) + 1 2k 1/2 2 Σ 2 ( k 1/2 2 n 2 (t, x) ).

41 36 Relaxation an the energy metho 1. We assume that Σ 1, Σ 2 satisfy Σ 1(u) = Σ 2(u 1/2 ). Show that Σ 1 is convex if, an only if, Σ 2 is convex. 2. Uner the conitions in question 1, show that the equation (2.10) issipates entropy. 3. Aapt Stampacchia s metho to prove L bouns on n 1, n 2 (an which is the appropriate quantity for the maximum principle). What are the intuitive L p bouns. Exercise. Consier (2.10) with Dirichlet bounary conitions an n 0 i Using the inequality nj ν 0, which hols because n j 0, show that M(t) ecreases where [ M(t) = 2n1 (t, x) + n 2 (t, x) ] x. Ω 2. Consier the entropies of the previous exercise with the aitional conition Σ i (0) = 0. Show that the equation (2.10) issipates entropy. (iii) Show that solutions of (2.10) with Dirichlet bounary conitions ten to 0 as t. 5.4 Entropy: chemostat an SI system of epiemiology Examples of entropy also appear in biological moels. We treat here an example that arises as a first moeling stage in two applications: 1. ecology an the moel of the chemostat (the u represents a nutrient, v a population that consumes the nutrient), 2. epiemiology with the celebrate Suceptible- Infecte moel (SI system). In both cases the moel is t u = B µ uu ruv, t v = ruv µ vv, (5.15) with B > 0, r > 0, µ u > 0 an µ v > 0 parameters. In the chemostat B represents the renewal of nutrients u, µ the removal or egraation of nutrients an r the consumption rate by the population an µ v is the mortality an removal rate of the population. In epiemiology, B represents the newborn, µ u the mortality rate, r the encounter rate between susceptible an infecte iniviuals (these encounters are responsible of new infecte), µ v is the mortality of the infecte (an recovery rate in the SIR moel). There is always a trivial (healthy) steay state an a positive steay state ū 0 = B/µ u, v 0 = 0, ū = µ v /r, v = rb µ uµ v rµ v if rb > µ u µ v. (5.16) Depening on the sign of v we may have two ifferent Lyapunov functionals (entropies) Lemma 5.4 A ssume rb > µ u µ v an efine S(u, v) = ū ln(u) v ln(v) + u + v,

42 5.5. THE LOTKA-VOLTERRA PREY-PREDATOR SYSTEM WITH DIFFUSION (PROBLEM) 37 we have t S = 1 u ( ūb u vr + µ u ) 2. (5.17) Lemma 5.5 We assume rb µ u µ v an efine S(u, v) = ū 0 ln(u) + u + v, we have t S = v (µ v µ u rb) 1 µ u µ u u (B µ uu) 2. (5.18) We leave the proofs of these lemmas to the reaer an go irectly to the conclusion Proposition 5.6 Solutions of the system (5.15) behave as follows If Br > µ u µ v, then the entropy S is convex an all solutions with v 0 > 0 converge as t to the positive steay state. If Br µ u µ v, then solutions become extinct (they converge to the trivial steay state as t ). The proof is stanar an left to the reaer. The steps are (i) u(t) is boune, (ii) S(t) ecreases, in the case v < 0, the limit can be an this means that v(t) vanishes an the result (ii) follows, otherwise S stays boune an converges to a finite value, (iii) we conclue thanks to the right han sie of (5.17). 5.5 The Lotka-Volterra prey-preator system with iffusion (Problem) In the case of the Lotka-Volterra prey-preator system we can show relaxation towars a homogeneous solution. The coefficients of the moel nee not be small as is require in Theorem 5.2. This is because the moel comes with a physically relevant quantity (such as entropy), which gives a global control. Exercise. Consier the prey-preator Lotka-Volterra system without iffusion t n 1 = n 1 [ r 1 an 2 ], t n 2 = n 2 [ r 2 + bn 1 ], where r 1, r 2, a an b are positive constants an the initial ata n 0 i are positive. 1. Show that there are local solutions an that they remain positive. 2. Show that the entropy (Lyapunov functional) E(t) = r 1 ln n 2 + an 2 r 2 ln n 1 + bn 1,

43 38 Relaxation an the energy metho is constant. Show that E is boune from below an that E as n 1 + n 2. Conclue that solutions are global. 3. What is the unique steay state solution? 4. Show, using question 2, that the solutions are perioic (trajectories are close). Solution 1. By local Lipschitz regularity of the right han sie, the Cauchy-Lipschitz theorem asserts that there is a local solution, i.e., efine on a maximal interval [0, T ]. We can write an, same thing for n 2. n 1 (t) = n 0 1e t 0 [r1 an2(s)]s > 0, 2. Set ϕ i = ln n i an write t ϕ 1 = r 1 an 2 = r 1 ae ϕ2, t ϕ 2 = r 2 bn 1 = r 2 + be ϕ1. This is a Hamiltonian system an the Hamiltonian is constant along trajectories H(t) = r 1 ϕ 2 (t) + ae ϕ2(t) r 2 ϕ 1 (t) + be ϕ1(t) ] = H(0). This proves that the ϕ i (t) remain boune from above an below, an thus that solutions are global. 3. n 1 = r 2 /b, n 2 = r 1 /a. 4. In each quarant efine by the origin ( n 1, n 2 ), we can write n 2 as a function of n 1 (or vice versa) an conclue from that. Exercise. Let Ω be a smooth boune omain. Consier smooth positive solutions of the Lotka-Volterra equation with iffusion an a Neumann bounary conition t n 1 1 n 1 = n 1 [ r 1 an 2 ], t n 2 2 n 2 = n 2 [ r 2 + bn 1 ], n i ν = 0 on Ω, i = 1, 2, where 1, 2, r 1, r 2, a an b are positive constants an the initial ata n 0 i are positive. 1. Consier the quantity m(t) = Ω [bn 1(t, x) + an 2 (t, x)]x. Show that m(t) m(0)e rt an fin the value r. 2. Show that the convex entropy E(t) = [ r 1 ln n 2 + an 2 r 2 ln n 1 + bn 1 ]x, a) is boune from below, b) is ecreasing. Conclue that m(t) is boune. 3. What finite integral o we obtain from the entropy issipation? 3. Assume that the quantities ln n i (t, x) converge, as t, a. What are their limits? b. What can you conclue about the behavior of n i (t, x) as t? Ω

44 5.6. PROBLEM Problem The goal of this problem is to show that for rank-1 nonlinearities, the long term behaviour is etermine by simple states in x without a size conition. Let Ω a smooth boune omain. Consier a smooth positive solution of the Lotka-Volterra system with iffusion an Neumann bounary conition t n i D i n i + a i (x)n i = n i r i (t), t 0, x Ω, i = 1, 2,..., I, n i ν = 0 on Ω, i = 1, 2,..., I, n i (t = 0, x) = n 0 i (x), with the nonlinearity efine, for some given positive an smooth functions ( ψ i (x) ) i=1,...,i, by ( r i (t) = R i ψ 1 (x)n 1 (t, x)x,..., ψ I (x)n I (t, x)x ). Ω 1. We also efine the first eigenfunctions N i (x) > 0 for the eigenvalue λ i, efine by D i N i + a i (x)n i = λ i N i, x Ω, N i ν = 0 on Ω, Ω N i(x) 2 x = 1, i = 1, 2,..., I. Explain why the pair (N i, λ i ) exists an give the corresponing Poincaré-Wirtinger inequality. 2. We consier ñ i the solution of tñi D i ñ i + a i (x)ñ i = λ i ñ i, t 0, x Ω, i = 1, 2,..., I, ñ i ν = 0 on Ω, i = 1, 2,..., I, ñ i (t = 0, x) = n 0 i (x). Prove that Ω ñi(t, x)n i (x)x = Ω ñ0 i (x)n i(x)x. Fin the constant α i such that, as t, ñ i (t, x) α i N i (x) L2 (Ω) ñ 0 i (x) α i N i (x) L2 (Ω)e µit an ientify µ i. 3. We write the solution of (5.6) as n i (t, x) = ρ i (t)ñ i (t, x). Ientify the evolution equation giving t ρ i(t) in terms of the ρ j an ñ j. 4. Assume that for some M > 0 we have R i ( Y1,...Y I ) < λi whenever Y i > M, i = 1,..., I. Show that the ρ i (t) are boune an that for some µ > 0 we have n i (t, x) ρ i (t)α i N i (x) L 2 (Ω) C 0 e µt. 5. In imension 1, assuming R(+ ) < λ, R( ) > λ an R < 0. Show that the long term ynamic is given by that of the equation ϱ(t) = ϱ(t) R ( ϱ(t)α ψ(x)n(x)x ), an that n(t, x) converges generically to a steay state ρn(x) as t. Ω [Hint.] ρ i (t) = ρ i (t)(r i λ i )

45 40 Relaxation an the energy metho

46 Chapter 6 Blow-up an extinction of solutions We know from Chapter 5 that for small nonlinearities, the solutions of parabolic systems relax to an elementary state. When the nonlinearity is too large, it is possible that solutions of nonlinear parabolic equations o not exist for all times or simply vanish, two scenarios that are the first signs of visible nonlinear effects. The mechanisms can be seen in simple orinary ifferential equations. Blow-up means that the solution becomes pointwise larger an larger an eventually becomes infinite in finite time. To see this, consier the equation ż(t) = z(t) q, z(0) > 0, q > 1. Its solution z(t) > 0 is given by z(t) q 1 = z(0) q 1 / ( 1 (q 1)tz(0) q 1) an thus it tens to infinity at t = T 1 := (q 1)z(0). For q = 2, it is a typical illustration of the q 1 alternative arising in the Cauchy-Lipschitz Theorem; solutions can only ten to infinity in finite time (case z(0) > 0), or they are globally efine (case z(0) < 0). The mechanism of extinction is illustrate by the equation with the opposite sign ż(t) = z(t) q, z(0) > 0, q > 1. Its solution z(t) q 1 = z(0) q 1 / ( 1 + (q 1)tz(0) q 1) (6.1) vanishes as t. The purpose of this Chapter is to stuy in which respect these phenomena can occur or not when iffusion is ae. We begin with the semilinear parabolic equation with Dirichlet bounary conitions t u u = u2. We present several methos for proving blow-up in finite time. The first two methos are for semi-linear parabolic equations, the thir metho is illustrate by the Keller-Segel system for chemotaxis; this is 41

47 42 Blow-up an extinction of solutions more interesting because it blows-up in all L p norms, for all p > 1, but the L 1 norm is conserve (this represents the total number of cells in a system where cell multiplication is ignore). The last section is evote to a counter-intuitive result with respect to extinction. 6.1 Semilinear equations; the metho of the eigenfunction To stuy the case of nonlinear parabolic equations, we consier the moel t u u = u2, t 0, x Ω, u(x) = 0, x Ω, u(t = 0, x) = u 0 (x) 0. (6.2) Here we treat the case when Ω is a boune omain. To efine a istributional solution is not completely obvious because the right han sie u 2 shoul be well efine, that is why we require that u belongs to L 2 ; then it remains to solve the heat equation with a right han sie in L 1. That is why we call solution of (6.2), a function satisfying for some T > 0, u L 2( (0, T ) Ω ), u C ( [0, T ); L 1 (Ω) ). (6.3) The question then is to know if effects of iffusion are able to overcome the effects of quaratic nonlinearity an (6.2) coul have global solutions. The answer is given by the next theorems Theorem 6.1 Assume that Ω is a boune omain, u 0 0, u 0 is sufficiently large (in a weighte L 1 space introuce below), then there is a time T for which the solution of (6.2) satisfies u L2 ((0,T ) R ). T T Of course, this result means that u(t) also blows up in all L p norms, 2 p because we work in a boune omain. It is also possible to prove that the blow-up time is the same for all these norms [35]. Proof of Theorem 6.1. We are going to arrive at a contraiction in the hypothesis that a function u L 2( (0, T ) Ω ) can be a solution of (6.2) when T excees an explicit value compute later. First we, notice that u(t, x) 0 because u 0 0. Because we are working in a boune omain, the smallest eigenvalue of the operator exists an is associate with a positive eigenfunction (see Section 4.2) w 1 = λ 1 w 1, w 1 > 0 in Ω, w 1 (x) = 0, x Ω, Ω (w 1) 2 = 1. (6.4)

48 6.1. SEMILINEAR EQUATIONS; THE METHOD OF THE EIGENFUNCTION 43 For a solution, we can multiply equation (6.2) by w 1 an integrate by parts. We arrive at t Ω u(t, x) w 1(x) x = Ω u(t, x) w 1(x) x + Ω u(t, x)2 w 1 (x) x = Ω u(t, x) w 1(x) x + Ω u(t, x)2 w 1 (x) x = λ 1 Ω u(t, x) w 1(x) x + Ω u(t, x)2 w 1 (x) x λ 1 Ω u(t, x) w 1(x) x + ( Ω u(t, x) w 1(x) x ) 2 ( Ω w 1(x) x ) 1 (after using the Cauchy-Schwarz inequality). We set z(t) = e λ1t Ω u(t, x) w 1(x) an, with a = ( Ω w 1(x) x ) 1, the above inequality reas t z(t) ae λ1t z(t) 2, an we obtain: 1 t z(t) ae λ1t, Assume now the size conition 1 z(t) 1 e λ1t a1. z0 λ 1 z 0 > λ 1 a. (6.5) The above inequality contraicts that z(t) > 0 for e λ1t 1 λ1 az 0. Therefore, the computation, an thus the assumption that u L 2 ((0, T ) R ), fails before that finite time. The size conition is necessary. There are various ways to unerstan this. For 5 it follows from a general elliptic equation Theorem 6.2 There is a steay state solution ū > 0 in Ω to u = u p, x Ω, u(x) = 0, x Ω, when p satisfies 1 < p < We refer to [10] for a proof of this theorem an relate results (as non-existence for p > +2 2 ). One can see more irectly that the size conition is neee. λ We choose µ = min 1 Ω w 1(x) an set w = µw 1. This is a supersolution of (6.2) because t w w = λ 1 w w 2. One conclues that, when u 0 w, solutions of (6.2) satisfy u(t) w for all times t, as long as the solution exists (an this is sufficient to prove that the solution is global). Therefore, we have the

49 44 Blow-up an extinction of solutions Lemma 6.3 Uner the smallness conition u 0 min Ω λ 1 w 1( ) w 1, there is a global solution of (6.2) an u(t) w, t 0. Proof. We subtract a solution u to w an fin t [u w] [u w] u2 w 2 = [u w][u + w], with u w(t = 0) 0. From the comparison principle, we conclue that u w(t) 0 for all times where the solution exists. Therefore, by continuation methos, there is a global solution. Exercise. Prove blow-up in finite time for the case of a general nonlinearity tu u = f(u), t 0, x Ω, u(x) = 0, x Ω, u(t = 0, x) = u 0 (x) > 0 large enough, (6.6) an f(u) cu α with α > 1. Exercise. (Neumann bounary conitions) A solution in (0, T ) of the equation t u u = u2, t 0, x Ω, ν u(x) = 0, x Ω, u(t = 0, x) = u 0 (x) 0, Ω u0 > 0, (6.7) is a istributional solution satisfying u L 2 ((0, T ) Ω). Prove that there is no such solution after some time T an that u(t ) L 1 (R ) (no size conition is require here). T T [Hint] Because Ω u(t ) = Ω u0 + T 0 Ω u2 (t, x)x t, both u L 2 ((0, T ) Ω) an u(t ) L 1 (R ) blow-up simultaneously. Conclue using the Cauchy-Schwarz inequality. Exercise. For = 1 an Ω =] 1, 1[, construct a unique, even an non-zero solution of v = v 2, v(±1) = 0. Hint. Reuce it to 1 2 (v ) 2 = 1 3 v3 + c 0 an fin a positive real number c 1 such that v = c u Semilinear equations; the energy metho Still consiering semilinear parabolic equations, we present another metho leaing to a ifferent size conition. This uses the intrinsic properties of the equation better an oes not use the sign conition.

50 6.2. SEMILINEAR EQUATIONS; THE ENERGY METHOD 45 We consier a more general case with p > 1 t u u = u p 1 u, t 0, x Ω, u(x) = 0, x Ω, u(t = 0, x) = u 0 (x) 0. (6.8) Before we state our result, it is useful to recall the energy principle unerlying this equation. We efine the energy E(u) = 1 u 2 1 u p+1. 2 Ω p + 1 Ω Note that this has a mechanical interpretation: the term 1 2 Ω u 2 is the kinetic energy an the term 1 p+1 Ω u p+1 is the potential energy. One can easily see that this energy ecreases with time (in fact this comes from the structure of a graient flow for (6.8)). Inee, using the chain rule an integration by parts, we have t E( u(t) ) = [ Ω u. u t ] u p 1 u u t = [ u Ω t u + u p 1 u ] = [ Ω u + u p 1 u ] 2 0. Consequently, we have E(t) E 0 := E(u 0 ). (6.9) This explains the central role of the energy in Theorem 6.4 For u 0 H 1 0 (Ω) satisfying E(u 0 ) 0, there are no global solution of (6.8) with boune energy. Proof. We efine α = p+1 > 0. We combine the L2 -estimate with the energy ecay an obtain 1 4 t Ω u2 x = 1 2 Ω [ u 2 u p+1 ]x = E(u) + α Ω u p+1 E(u 0 ) + α Ω (1 p)/2 ( Ω u2) (p+1)/2. The last inequality uses Jensen s inequality for the probability measure x/ Ω ( u 2 x ) (p+1)/2 Ω Ω u Ω p+1 x This proves blow-up for E(u 0 ) 0 because the positive function z(t) := Ω u2 satisfies z(t) t Ω. βz(t) (p+1)/2, β := α Ω (1 p)/2 > 0, an thus it blows-up in finite time, as shown in the introuction.

51 46 Blow-up an extinction of solutions Exercise. Fin functions u 0 H 1 0 (Ω) that satisfy the conition E(u 0 ) 0. Hint. Use the fact that the two terms in the efinition of the energy have ifferent scales in u an consier expressions as rv (x). Exercise. Prove blow-up for ( Ω (u0 ) 2) (p+1)/2 > 2 β E(u0 ). The topic of blow-up is very rich an many moalities of blow-up, ifferent blow-up rates, blow-up for ifferent norms an regularizing effects that prevent blow-up are also possible. See [35]. 6.3 Keller-Segel system; the metho of moments We come back to the Keller-Segel moel use to escribe chemotaxis as mentione in Section??. We recall that it consists of a system which escribes the evolution of the population ensity of cells (bacteria, amoebia,...) n(t, x), t 0, x R an the concentration c(t, x) of the attracting molecules release by the cells themselves, t n n + iv(n χ c) = 0, t 0, x R, c + τ c = n, (6.10) n(t = 0) = n 0 L L 1 +(R ). The first equation expresses the ranom (Brownian) iffusion of the cells with a bias irecte by the chemoattractant concentration with a sensitivity χ. The case χ = 0 means the cells o not react. The chemoattractant c is irectly release by the cell, iffuses on the substrate an is egrae with a coefficient τ that scales in a such a way that τ 1/2 represents the activation length. The notation L 1 + means nonnegative integrable functions, an the parabolic equation for n gives nonnegative solutions (as expecte for the cell ensity, see Chapter??) n(t, x) 0, c(t, x) 0. (6.11) Another property we use is the conservation of the total number of cells m 0 := n 0 (x) x = R n(t, x) x. R (6.12) In particular, solutions cannot blow-up in L 1. But we have the Theorem 6.5 In R 2, take τ = 0 an assume x 2 n 0 (x)x <. R 2 (i) (Blow-up) When the initial mass satisfies m 0 := n 0 (x)x > m crit := 8π/χ, (6.13) R 2 then any solution of (6.10) becomes a singular measure in finite time. (ii) When the initial ata satisfies n 0 (x) log(n 0 (x)) x < an R 2 m 0 := n 0 (x)x < m crit := 8π/χ, (6.14) R 2

52 6.4. NON-EXTINCTION 47 there are weak solutions of (6.10) satisfying the a priori estimates R 2 n[ ln(n(t)) + x 2 ] x C(t), n(t) L p (R 2 ) C(p, t, n 0 ) for n 0 L p (R 2 ) <, 1 < p <. Here we only explain the argument for blow-up. We refer to [34, 7] for the complete proof of Theorem 6.5. Proof. We follow Nagai s argument base on the metho of moments, assuming sufficient ecay in x at infinity. This is base on the formula x y c(t, x) = λ 2 x y 2 n(t, y)y, λ 2 = 1 2π. Then, we consier the secon x moment We have, from (6.10), R 2 R2 x 2 m 2 (t) := n(t, x)x. 2 t m 2(t) = x 2 R 2 2 [ n iv(n χ c)]x = R 2 [2n + χnx c]x = 2m 0 χλ 2 = 2m 0 χ λ 2 2 n(t, x)n(t, y) x (x y) R 2 R 2 x y 2 R 2 R 2 n(t, x)n(t, y) (x y) (x y) x y 2 (this last equality simply follows by a symmetry argument, interchanging x an y in the integral). This yiels finally, t m 2(t) = 2m 0 (1 χ 8π m0 ). Therefore, if we have m 0 > 8π/χ, we arrive at the conclusion that m 2 (t) shoul become negative in finite time, which is impossible since n is nonnegative. Consequently, the solution cannot be smooth until that time. 6.4 Non-extinction We consier now a nonlinearity with the negative sign, still with p > 1, t u u = up, t 0, x R, u(t = 0, x) = u 0 (x) > 0. (6.15) We recall that the solution remains positive u(t, x) > 0.

53 48 Blow-up an extinction of solutions As pointe out in the introuction, in the absence of iffusion, the solution (6.1) vanishes in infinite time. Does iffusion change this effect? To analyze this issue, we efine the total mass M(t) = u(t, x)x R which clearly ecreases t M(t) = u p (t, x)x < 0. (6.16) R Following [24], we are going to prove non-extinction when iffusion is present Theorem 6.6 Assume p > an u0 L 1 L (R ), then the solution of (6.17) satisfies lim M(t) > 0. t There is a nice biological interpretation in [24] relate to the phenomena of so-calle broacast-spawning. This is an external fertilization strategy use by various benthic invertebrates (sea urchins, anemones, corals, jellyfish) whereby males an females release at the same time, short live sperm an egg gametes into the surrouning flow. The gametes are positively buoyant, an rise to the surface of the ocean. The fertilize gametes form larva an are negatively buoyant an sink to the bottom of the ocean floor to start a new colony. For the coral spawning problem, fiel measurements of the fertilization rates are often as high as 90%. To arrive at such high rates, rapi ispersion on the ocean surface is require for successful encounters an, as shown above, iffusion is clearly not sufficient, aitional chemotactic attraction is certainly involve. The moel at han supposes that sperm an eggs have the same ensity (but this assumption can be release in a system of two parabolic equations, arriving at the same conclusion, see below). Chemotaxis is omitte. The parameter p = 2 represents binary interactions leaing to fertilize eggs that are withrawn from the balance equation. The theorem shows that, in the absence of chemotaxis, not all the eggs are fertilize. In [24] it is prove that chemoattraction increases this rate but aitional flow mixing oes not. Proof. Our assumption for p is equivalent to δ = (p 1)/2 1 > 0. First step. A lower boun. For all t > τ > 0, we can estimate, thanks to (??) with initial time τ, R u p (t, x)x C(, p) M(τ) p (t τ) 1 δ. Therefore, we can write for T τ + a, an a > 0, T T M(T ) = M(τ + a) u p (t, x)xt M(τ + a) C(, p)m(τ) p (t τ) 1 δ t, τ+a R τ+a an thus, M(T ) M(τ + a) C(, p)m(τ) p a δ.

54 6.4. NON-EXTINCTION 49 Secon step. A upper boun. Let ũ be the solution to the linear equation tũ ũ = 0, t 0, x R, ũ(t = 0, x) = u 0 (x) > 0. (6.17) We have u ũ an thus, thanks to (??), Therefore, we fin u(t) Lp (R ) C(, p)m(0) p t 1 δ. M(τ + a) M(τ) Because M( ) is non-increasing, we finally obtain τ+a τ R u p (t, x)xt. M(τ + a) M(τ) a u 0 p 1 L (R ) M(τ). Thir step. Conclusion. Combining these two step, an choosing a with the rule a u 0 p 1 = 1/2, we L (R ) obtain M(T ) 1 2 M(τ) C(, p)m(τ)p a δ. If we ha M(T ) 0 as T, we woul conclue that C(, p)m(τ) p 1 a δ 1 2, which contraicts that M(τ) can vanish for large values of τ. This proves the result. Exercise. From the above argument erive an explicit boun for the minimum mass (epening on M(0), p, ). Exercise. Assume u 0 L 1 L (R ) an consier the equation without iffusion Show that R u(t, x)x 0 as t. t u = up, t 0, x R. Exercise. Consier the system for eggs an other gametes (sperm) t e e e = (eg) p/2, t 0, x R, t g g g = (eg) p/2, e(t = 0, x) = e 0 (x) > 0. g(t = 0, x) = g 0 (x) > 0 Show that both lim t e(t, x)x an lim R t g(t, x)x are positive. R See [24] again.

55 50 Blow-up an extinction of solutions

56 Chapter 7 Linear instability, Turing instability an pattern formation In his seminal paper 1 A. Turing suggests that a system of chemical substances reacting together an iffusing through a tissue, is aequate to account for the main phenomena of morphogenesis. He introuces several concepts associate with the chemical basis of morphogenesis (an the name morphogen itself), spatial chemical patterns, an what is now calle Diffusion Driven Instability. The concept of Turing instability has become stanar an the aim of this chapter is to escribe what it is (an what it is not!). The first numerical simulations of a system exhibiting Turing patterns was publishe in 1972 involving the celebrate system of Gierer an Meinhart [16] (see also Section 7.5.8). It was only 20 years later that the first experimental evience for a chemical reaction exhibiting spatial patterns explaine by these principles was obtaine. This reaction is name the CIMA reaction after the name of the reactants use by P. De Kepper et al 2,3. See also Section In 1995, S. Kono an R. Asai 4 foun an explanation of the patterns arising uring the evelopment of animals, proposing that the moel shou be set in a growing omain, thus opening up a larger class of possible patterns. Spots, which are a usual steay state for Turing systems in a fixe omain, can more easily give way to bans in a growing omain. Meanwhile, several nonlinear parabolic systems exhibiting Turing Patterns have been stuie. Some have been aime at moeling particular examples of morphogenesis such as the moels evelope in [29, 30]. Some have been erive as the simplest possible moels exhibiting Turing instability conitions. Nowaays, the biological interest in morphogenesis has evolve towars molecular cascaes, pathways an genetic networks. Some biologists oubt that, within cells or tissues, iffusion provies an aequate escription of molecular spreaing. However, Turing s mechanism remains both the simplest explanation 1 A. M. Turing, The chemical basis of morphogenesis, Phil. Trans. Roy. Soc. Lonon, B237, 37 7 (1952). 2 P. De Kepper, V. Castets, E. Dulos an J. Boissonae., Turing-type chemical patterns in the chlorite-ioie-malonic aci reaction, Physica D 49 (1991), V. Castets, E. Dulos, J. Boissonae an P. De Kepper, Experimental evience of a sustaine staning Turing-type nonequilibrium chemical pattern. Phys. Rev. Letters 64(24) (1990). 4 S. Kono an R. Asai, A reaction-iffusion wave on the skin of the marine anglefish Pomocanthus, Nature (1995) 51

57 52 Linear instability, Turing instability an pattern formation for pattern formation, an one of the most counter-intuitive results in the fiel of Partial Differential Equations. This chapter presents Turing s theory an several examples of iffusion riven instabilities. We begin with the historical example of reaction-iffusion systems where the linear theory shows its exceptional originality. Then we present nonlinear examples. The simplest of the latter is the non-local Fisher/KPP equation, while some more stanar parabolic systems are also presente. 7.1 Turing instability in linear reaction-iffusion systems An amazingly counter-intuitive observation is the instability mechanism propose by A. Turing [39]. Consier a linear 2 2 O.D.E. system u t = au + bv, with real constant coefficients a, b, c an. We assume that Consequently, we have (7.1) v t = cu + v, T := a + < 0, D := a bc > 0. (7.2) (u, v) = (0, 0) is a stable attractive point for the system (7.1). (7.3) In other wors, the matrix A = a c b has two eigenvalues with negative real parts (or a single negative eigenvalue an a Joran form). Inee, its characteristic polynomial is (a X)( X) bc = X 2 XT + D, an the two complex roots are X ± = 1 2 [T ± T 2 4D]. Now consier a boune omain Ω of R an a iffusion to the system (7.1), u t σ u u = au + bv, x Ω, v t σ v v = cu + v, (7.4) with either a Neumann or a Dirichlet bounary conition. In both cases, the state (u, v) = (0, 0) is still a steay solution, of (7.4). In principle, aing iffusion to the ifferential system (7.1) shoul increase stability. But surprisingly we have

58 7.1. TURING INSTABILITY IN LINEAR REACTION-DIFFUSION SYSTEMS 53 Theorem 7.1 (Turing instability theorem) Consier the system (7.4) where we fix the omain Ω an the matrix A an σ v > 0. We assume (7.2) with a > 0, < 0. Then, for σ u sufficiently small, the steay state (u, v) = (0, 0) is linearly unstable. Moreover, only a finite number of eigenmoes are unstable. The usual interpretation of this result is as follows. Because a > 0 an < 0, the quantity u is calle an activator an v an inhibitor. On the other han, fixing a unit of time, the σ s are scale as the square of length. The result can be extene as the Turing instability alternative. Turing instability short range activator, long range inhibitor. Traveling waves long range activator, short range inhibitor. There is no general proof of this statement, which can only be applie to nonlinear systems. But it is a general observation that can be verifie case by case. Also, the statement shoul be written with the notion of stable traveling waves, because traveling waves can also connect an unstable state to a Turing unstable state 5. However, they are unstable in the sense that the ynamic creates perioic patterns. Proof of Theorem 7.1. We consier the Laplace operator, with a Dirichlet or a Neumann conition accoring to those consiere for the system (7.4). It has an orthonormal basis of eigenfunctions (w k ) k 1 associate with positive eigenvalues λ k, We recall that we know that λ k k w k = λ k w k.. We use this basis to ecompose u(t) an v(t), i.e., u(t) = k=1 α k (t) w k, v(t) = k=1 β k (t) w k. We can project the system (7.4) on these eigenfunctions an arrive to α k (t) t β k (t) t + σ u λ k α k (t) = aα k (t) + bβ k (t), + σ v λ k β k (t) = cα k (t) + β k (t). (7.5) Now, we look for solutions with exponential growth in time, i.e., α k (t) = e λt α k, β k (t) = e λt β k with λ > 0 (in fact a complex number with Re(λ) > 0 is sufficient, but this oes not change the conitions we fin below). The system is again reuce to λ α k + σ u λ k α k = aα k + bβ k, (7.6) λ β k + σ v λ k β k = cα k + β k. 5 G. Nain, B. Perthame an M. Tang, Can a traveling wave connect two unstable states?the case of the nonlocal Fisher equation. C. R. Aca. Sci. Paris, Ser. I 349 (2011)

59 54 Linear instability, Turing instability an pattern formation This is a 2 2 linear system for α k, β k an it has a nonzero solution if, an only if, its eterminant vanishes λ + σ u λ k a b 0 = et c λ + σ v λ k Hence, there is a solution with exponential growth for those eigenvalues λ k for which there is a root λ > 0 to (λ + σ u λ k a)(λ + σ v λ k ) bc = 0. (7.7) This conition can be further reuce to the ispersion relation λ 2 + λ [ (σ u + σ v )λ k T ] + σ u σ v (λ k ) 2 λ k ( σ u + a σ v ) + D = 0. Because the first orer coefficient of this polynomial is positive, it can have a positive root if, an only if, the zeroth orer term is negative an we arrive to the final conition σ u σ v (λ k ) 2 λ k ( σ u + a σ v ) + D < 0, (λ k ) 2 λ k ( σ v + a σ u ) + D σ u σ v < 0. (7.8) D Because λ k > 0 an σ uσ v > 0, this polynomial can take negative values only for σ v + a σ u > 0 an sufficiently large, with sufficiently small. It is ifficult to give an accurate general characterization D σ uσ v in terms of σ u an σ v when (a, b, c, ) are fixe because we o not know, in general, the istribution of the eigenvalues. To go further, we set θ = σ u σ v, an we write explicitly the roots of the above polynomial an we require that λ k [Λ, Λ + ], Λ ± = 1 2σ v θ [ ] (θ ) 2 θ + a ± + a 4Dθ. (7.9) We can restrict ourselves to the regime θ small, then the Taylor expansion gives, [ ] Λ ± = θ + a 1 ± 1 4Dθ 2σ v θ (θ + a) 2, an thus Λ Λ ± a [ 1 ± [ 1 2Dθ ] ] 2σ v θ (θ + a) 2, D = O(1), Λ + a a σ v σ v θ 1. In the regime σ u small, σ v of the orer of 1, the interval [Λ, Λ + ] becomes very large, hence we know that some eigenvalues λ k will belong to this interval.

60 7.2. SPOTS ON THE BODY AND STRIPES ON THE TAIL 55 Note however that, because lim k λ k = +, there are only a finite number of unstable moes λ k. In principle one will observe the moe w k 0 corresponing to the largest possible λ in (7.7) among the λ k s that satisfy the conition (7.9). This oes not correspon necessarily to the largest value of λ k that satisfy the inequality (7.8). Exercise. Compute the first two terms in the expansion of λ in λ k. Solution. λ σ u λ k + T. Exercise. Check how the conition (7.8) is generalize if we only impose the more general instability criteria that (7.7) hols with λ C an Re(λ) > Spots on the boy an stripes on the tail Figure 7.1: Examples of animals with spots an stripes. Left: Discus ( Right: My cat. Several papers 6,7 (see also [32]) give a striking explanation of how Turing instability provies us with a possible explanation of why so many animals have spots on the boy an stripes on the tail, see Figure 7.1. In short, in a long an narrow omain (a tail), typical eigenfunctions are bans, an with a more spherically shape omain (a mathematical square boy), the eigenfunctions are spots or chessboars. A much better an more etaile iscussion with precise biological cases emonstrate, an also the original papers where this iea is first expresse, can be foun in [32] Vol. II Chapter 3. To explain this, we consier the Neumann bounary conition an use our computations of eigenvalues from Section 4.5. In one imension, in a omain [0, L], the eigenvalues an eigenfunctions are ( ) 2 ( ) πk πkx λ k =, w k (x) = cos, k N. L L 6 Oster, G.F.; Shubin, N.; Murray, J. D. an Alberch P. Evolution an morphogenetic rules: the shape of the vertebrate limb in ontogeny an phylogeny Evolution 45, (1988) 7 Maini, P. K. How the mouse got its stripes. PNAS 100(17): (2003). DOI: /pnas

61 56 Linear instability, Turing instability an pattern formation In a rectangle [0, L 1 ] [0, L 2 ], we obtain the eigenelements, for k, l N ( ) 2 ( ) 2 ( ) πk πl πkx λ kl = +, w kl(x, y) = cos cos L 1 L 2 Consier a narrow stripe, say L 2 0 an L 1 1. The conition (7.9), namely λ kl [Λ, Λ + ], will impose l = 0 otherwise λ kl will be very large an cannot fit the interval [Λ, Λ + ]. The corresponing eigenfunctions are bans parallel to the y axis. L 1 ( πly L 2 ). When L 2 L 1, the istribution of sums of square integers generically etermines that the istribution of the λ kl [Λ, Λ + ] will be for l k. To conclue this section, we point out that growing omains uring evelopment also very strongly influence the topic of pattern formation. Again, we refer to [32] for a etaile analysis of the Turing patterns, an their interpretation in evelopment biology. 7.3 The simplest nonlinear example: the non-local Fisher/KPP equation As a simple nonlinear example to explain what is Turing instability, we consier the non-local Fisher/KPP equation t u ν 2 u = r u(1 K u), t 0, x R, (7.10) x2 still with ν > 0, r > 0, given parameters. For the convolution kernel K, we choose a smooth probability ensity function K( ) 0, K(x)x = 1, K L (R) (at least). R Compare to the Fisher/KPP equation, this takes into account that competition for resources can be of long range (the size of the support of K) an not only local. This iea has been propose in ecology as an improvement of the Fisher equation that takes into account long range competition for resources (N. F. Britton 8,9 ). In semi-ari regions the roots of trees, in competition for water, can cover, up to ten times the external size of the tree itself (while in temperate regions the ratio is roughly one to one). This leas to the so-calle tiger bush lanscape [26]. The same equation has also been propose as a simple moel of aaptive evolution. Then x represents a phenotypical trait, see S. Génieys et al 10. The Laplace term represents mutations an the right han sie, growth an competition. The convolution kernel is use to express that competition is higher between iniviuals, whose traits are close to each other (see also Section??). 8 Britton, N. F. Spatial structures an perioic traveling waves in an integro-ifferential reaction- iffusion population moel. SIAM J. Appl. Math. 50(6), (1990) 9 Gourley, S. A. Travelling front solutions of a non-local Fisher equation. J. Math. Biol. 41, (2000) 10 Génieys, S., Volpert, V. an Auger, P., Pattern an waves for a moel in population ynamic with non-local consumption of resources. Math. Moel. Nat. Phenom. 1 (2006), no. 1, 65 82

62 7.3. THE SIMPLEST NONLINEAR EXAMPLE: THE NON-LOCAL FISHER/KPP EQUATION 57 The convolution term has a rastic effect for solutions; it can inuce solutions that exhibit a behavior quite ifferent from those of the Fisher/KPP equation. The reason is mainly that the maximum principle is lost with the non-local term. Again, we notice that the steay state u 0 is unstable, that u 1 is also unstable because it inuces a strong ecay. In one imension, for a general reaction function f(u) the conitions rea f(0) = 0, f (0) > 0 an f(u) < 0 for u large. Consequently there is a point u 0 satisfying (generically) f(u 0 ) = 0, f (u 0 ) < 0, i.e. a stable steay state shoul be foun between the unstable ones. This is the case of the nonlinearities arising in the Fisher/KPP equation that we have alreay encountere. In the infinite imensional framework at han, we shall see that uner certain circumstances, the steay state u 1 can be unstable in the sense of Definition 7.2 The steay state u 1 is calle linearly unstable if there are perturbations such that the linearize system has exponential growth in time. Then, the following conitions are satisfie Definition 7.3 A steay state u 0 is sai to form Turing patterns if (i) there is no blow-up, no extinction (it is between two unstable states as above), (ii) it is linearly unstable, (iii) the corresponing growth moes are boune (no high frequency oscillations). Obviously when Turing instability occurs, solutions shoul exhibit strange behavior because they remain boune away from the two extreme steay states, they cannot converge to the steay state u 0 or oscillate rapily. In other wors, they shoul exhibit Turing patterns. See Figure 7.2 for a numerical solution of (7.10). In practice, to check linear instability we use a spectral basis. In compact omains the concept can be hanle using eigenfunctions of the Laplace operator as we i in Section 7.1. In the full line, we may use the generalize eigenfunctions, these are the Fourier moes. We efine the Fourier transform as û(ξ) = u(x)e ix ξ x. Theorem 7.4 Assume the conition R ξ 0 such that K(ξ 0 ) < 0, (7.11) then, for ν/r sufficiently small (epening on ξ 0 an K(ξ 0 )), the non-local Fisher/KPP equation (7.10) is nonlinearly Turing unstable. A practical consequence of this theorem is that solutions shoul create Turing patterns as mentione earlier. This can easily be observe in numerical simulations, see Figure 7.2. The non-local Fisher equation also gives an example of the alreay mentione Turing instability alternative. Turing instability K( ) is long range. Traveling waves K( ) is short range.

63 58 Linear instability, Turing instability an pattern formation Figure 7.2: Two steay state solutions of the non-local Fisher/KPP equation (7.10) in 2 imensions with ifferent iffusion coefficients. Inee the non-local term K u is the inhibitor (negative) term. The iffusion represents the activator (with a coefficient normalize to 1). The limit of very short range is the case of K = δ, a Dirac mass, an we recover the Fisher/KPP equation. More on this is prove in [4]. Proof. (i) The state u 0 an u are inee formally both unstable (to prove this rigorously is not so easy for u.) (ii) The linearize equation aroun u 1 is obtaine by setting u = 1 + ũ an keeping the first orer terms, we obtain tũ ν 2 ũ = r K ũ. x2 An we look for solutions of the form ũ(t, x) = e λt v(x) with λ > 0. This means that we shoul fin eigenfunctions associate with the positive eigenvalue λ, that means solution v(x) of λv ν 2 v = r K v. x2 We look for a possible Fourier moe v(x) = e ix ξ1 that we insert in the previous equation. Then we obtain the conition λ + νξ 2 1 = r K(ξ 1 ), for some λ > 0. (7.12) An it is inee possible to obtain such a λ an a ξ 1 = ξ 0 uner the conitions of the Theorem. (iii) The possible unstable moes ξ 0 are obviously boune because K is boune as the Fourier transform of a probability ensity ( K 1). Note however that the moe ξ 1 we observe in practice is that with the highest growth rate λ. 7.4 Phase transition: what is NOT Turing instability What happens if the thir conition in Definition 7.3 oes not hol? The system remains boune away from zero an infinity by conition (i) an it is unstable by conition (ii). But it might blow-up by high frequency oscillations. As an example of such an unstable system, which is not Turing unstable, we consier the phase

64 7.5. GALLERY OF PARABOLIC SYSTEMS GIVING TURING PATTERNS 59 transition moel also use in Section??, u t A(u) = 0, x Ω, ν u = 0 on Ω, (7.13) with A(u) = u (3 u) 2. (7.14) Because A (u) = 3(3 u)(1 u) < 0, the equation (7.13) is a backwar-parabolic equation in the interval u (1, 3). We expect that linear instability occurs in this interval. We take ū = 2 an set u = 2 + ũ, Inserting this in the above equation we fin the linearize equation for ũ(t, x) ũ t A (2) ũ = 0, x Ω, ν ũ = 0 on Ω. We set γ = A (2) > 0 an we look for solutions ũ(t, x) = e λt w(x), which are unstable, i.e., λ > 0. These are given by λw + γ w = 0, an thus they stem from the Neumann eigenvalue problem in Theorem 4.1. We have λ = λ i γ, w = w i. We can see that all the eigenvalues of the Laplace operator generate possible unstable moes an thus they can be of very high frequency. An we expect to see the moe corresponing to the largest λ, i.e., to the largest λ i which of course oes not exist because λ i. In the space variable these correspon i to highly oscillatory eigenfunctions w i that we can observe numerically. Figure 7.3 gives numerical solutions of (7.13) (7.14) corresponing to two ifferent gris; high frequency solutions are obtaine that epen on the gri. This efect explains why we require boune unstable moes in the efinition of Turing instability. It also explains why in Section?? we have introuce a relaxation system. More on the subject of phase transitions can be foun in Chapter?? on the Stefan problem. 7.5 Gallery of parabolic systems giving Turing patterns Many examples of nonlinear parabolic systems exhibiting Turing instabilities (an patterns) have been wiely stuie. This section gives several examples but it is in no way complete; the theories are far too complicate to be presente here an their use in biology an other applications are far too numerous.

65 60 Linear instability, Turing instability an pattern formation Figure 7.3: Two numerical solutions of the phase transition system (7.13) (7.14) for (Right) 80 gri points an (Left) 150 gri points. The oscillation frequencies epen on the gri an these are not Turing patterns A cell polarity system We take the following system from Morita an Ogawa 11 t u σ 1 u = f(u) + v, t 0, x Ω, t v σ 2 v = f(u) v, u(t,x) ν = v(t,x) ν = 0 on Ω. We take a function f C 2 (R + ; R + ) that, for a given value u c > 0, satisfies (7.15) f(0) = 0, f (u) > 0 for u [0, u c [, 1 < f (u) < 0 for u > u c, f(+ ) = f 0. This system is mass conservative since we have u(t, x) > 0, v(t, x) > 0, [u(t, x) + v(t, x)]x = 0. t Ω These properties give us the non extinction/non blow-up (in L 1 ) conitions. We analyze Turing instability. The ifferential system is stable. The corresponing ifferential system is tu = f(u) + V, t 0, tv = f(u) V, U(0) > 0, V (0) > 0. It satisfies U(t) + V (t) = U(0) + V (0) =: M 0 > 0 an thus can be also written as t U = f(u) + M 0 U(t) := G(U(t)). 11 Morita Y. an Ogawa T. Stability an bifurcation of nonconstant solutions of a reaction-iffusion system with conservation of mass, Nonlinearity 23 (2010)

66 7.5. GALLERY OF PARABOLIC SYSTEMS GIVING TURING PATTERNS 61 Therefore, it preserves the positive cone U(t) > 0, V (t) > 0, inee, at the first point t 0 where u(t 0 ) = 0 we woul have t U(t 0) = M 0 > 0, which is a contraiction (same argument to hanle the function V ). Since G (u) = f (u) 1 < 0, G(0) > 0 an G(+ ) =, there is a single steay state (ū, v) characterize by G(ū) = 0 that is equivalent to ū + v = M 0, v = f(ū). (7.16) Because G(U) > 0 for U ū, G(U) < 0 for U ū, it is very clear that (monotonically) U(t) t ū, V (t) t v. Turing instability. Consier a steay state (7.16). We compute the ifferential matrix for the right han sie ( ) f (ū) 1 f (ū) 1 To fit the assumptions of Theorem 7.1, we have to check T := f (ū) 1 < 0, D := 0 (a egenerate case, corresponing to mass conservation). The only possibility is that u is the activator, which imposes f (ū) < 0 ū > u c. Then, we fin that unstable moes e λt (αw k, βw k ) exist if there is a positive root to the polynomial λ 2 + λ[λ k (σ 1 + σ 2 ) + f (ū) + 1] + (σ 1 λ k + f (ū))(σ 2 λ k + 1) f (ū) = 0. As in Section 7.1, this is equivalent to (σ 1 λ k + f (ū))(σ 2 λ k + 1) f (ū) = σ 1 σ 2 λ 2 k + σ 1 λ k + σ 2 f (ū)λ k < 0, λ k + 1 σ 2 f (ū) σ 1. This is clearly satisfie for some eigenvalues λ k when σ 1 is sufficiently small; this is the same result as in Theorem The CIMA reaction As mentione earlier, the first experimental evience of Turing instability was obtaine with the CIMA (chlorite-ioie-malonic aci) chemical reaction. It was moele by I. Lengyel an I. R. Epstein 12 who propose the system (with c = 1) u t σ u u = a u 4uv 1 + u 2, v t σ v v = bc u cuv (7.17) 1 + u Moeling of Turing structure in the chlorite-ioie-malonic aci-starch reaction system, Science 251 (1991)

67 62 Linear instability, Turing instability an pattern formation Figure 7.4: Labyrinth an spot patterns in the CIMA reaction (7.17). The solutions have been compute with the software FreeFEM++ [14, 19] by S. Kaber. Here u (the activator) enotes the ioie (I ) concentration an v (the inhibitor) the chlorite (ClO 2 ) concentration. Existence of steay states was analyze 13,14. Here we consier this system with a > 0, b > 0, c > 0. There is a single homogeneous steay state ū = a 4b + 1, v = b(1 + ū2 ). The simple invariant region for solutions (that means the bouns are satisfie for all times if initially true) are given by the maximum principle 0 u a, b v b(1 + a 2 ) := v M. Also solutions cannot become extinct because we can use the upper boun on v to go further an fin u u min, u min (1 + 4v M 1 + u 2 ) = a, min (in fact we can go even further an fin a more restrictive invariant region, iterating the argument). These a priori bouns are consequences of the maximum principle an we skip the erivation. Therefore, accoring to our theory of Turing patterns, solutions cannot become extinct or blow-up an it remains to stuy the linearize operator aroun the steay state (ū, v). Lemma 7.5 The CIMA reaction system (7.17) is Turing unstable if 4b > 1, c > 2 16b 2 1, 4b + 1 4b 1 < ū < c + c2 4(16b 2 1), 2(4b 1) an u is the activator, v the inhibitor. Consequently, for this range of parameters, σ v of the orer of 1, an σ u sufficiently small, there will be Turing patterns. 13 W. Ni an M. Tang. Turing patterns in the Lengyel-Epstein system for the CIMA reaction. Trans. Amer. Math. Soc 357 (2005) F. Yi, J. Wei an J. Shi. Diifusion-riven instability an bifurcation in the Lengyel-Epstein reaction-iffusion system. Nonl. Anal. Real Worl Appl. 9 (2008)