Heidelberg - Schwarz - NonlinearDynamics

8/13/2019 Heidelberg - Schwarz - NonlinearDynamics

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Ruprecht-Karls University of Heidelberg

Faculty of Physics and Astronomy

Non-linear Dynamics

Prof. Ulrich Schwarz

Winter term 2013/2014

Last update: January 29, 2014


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Contents

1 The central equation 1

2 Flow on a line 3

3 Bifurcations in 1d 83.1 Saddle-node bifurcation. . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.2 Transcritical bifurcation . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.3 Pitchfork bifurcation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.3.1 Supercritical pitchfork bifurcation . . . . . . . . . . . . . . . . . . 143.3.2 Subcritical pitchfork bifurcation . . . . . . . . . . . . . . . . . . . 15

3.4 Influence of high order terms. . . . . . . . . . . . . . . . . . . . . . . . . 163.5 Summary of 1d bifurcations . . . . . . . . . . . . . . . . . . . . . . . . . 17

4 Flow on a circle 19

5 Flow in linear 2d systems 255.1 General remarks. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255.2 Phase plane flow for linear systems . . . . . . . . . . . . . . . . . . . . . 27

6 Flow in non-linear 2d systems 30

7 Oscillations in 2d 39

8 Bifurcations in 2d 49

9 Excitable systems 55

10 Pattern formation in reaction-diffusion systems 64

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1 The central equation

1 The central equation

We consider dynamical systems of dimension d which are described by ODEs. Thisimplies that we use continuous time. (One alternative would be different equations with

discrete time.) Calling x the state vector of the system we consider the equationdx

dt = f(x)

with a vector-valued function fwhich can be non-linear. In case of a linear function fthe equation simplifies to

x= A xwith a matrix A and the system shows exponential behavior.

Examples

1. Harmonic oscillator x + 20x= 0

This looks like an one-dimensional system. But the following trick eliminatesthe second derivative and shows the linear but two-dimensional character of theharmonic oscillator:

Choosex1 = x and x2=v = xwith the velocity v. Then, the equation written in

the general form is

x=

x1x2

=

0 120 0

x.

2. Overdamped particle x+ k x= 0is the viscosity of the surrounding medium.Solving for x shows that the equation is al-ready in the general form:

x= k

x.The system is one-dimensional and linear.Because of this, no oscillations occur.

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2 Flow on a line

2 Flow on a line

In this chapter, we are looking at one-dimensional systems. Therefore, the central equa-tion becomes x= f(x) with an arbitrary function f.

The first example we want to discuss is non-linear: x = sin(x). The separation ofvariables leads to

dx

sin(x)= csc(x)dx= dt

which can be integrated with the result

t= ln

csc(x0) + cot(x0)

csc(x) + cot(x)

.

Even in this simple non-linear example, the behavior of the system is not easy to under-

stand from this solution. But graphical analysisshows the most important properties.

Plotting a phase portrait (left figure), stable and unstable fixed points can be deter-mined. In 1d, the systems dynamics corresponds to flow on the line. The correspondingtrajectories are shown in the right figure.

For a stable fixed point a little change in x drives the system back, whereas for anunstable fixed point it causes a flow away from the fixed point.

Choosing different starting points x the time-dependence of the acceleration computesas follows: for starting points|x |

2 the acceleration directly decreases. But if

x = xwith 2


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2 Flow on a line

The graphical analysis can be performed for the earlier examples as well:

Overdamped particle

x= xx = 0 , stable fixed point

Electrical curcuit

x = 0

The applied method works for any graph.

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2 Flow on a line

In an one-dimensional system, there are three possibilities in total the system can behave:

1. staying at a fixed point

2. flowing to a stable fixed point

3. flowing to infinity

There is also a mathematical method to analyze fixed points. It is called linear stabilityanalysis. Firstly, one determines the fixed points by solving x= f(x) = 0 forx. Takex

to be a fixed point. Then, the deviation from this fixed point is given by = x x.

The derivative can be written in dependence of the sum x + .

= x= f(x) =f(x + )

The first order Taylor expansion = f(x

)=0 + f

(x

) +O(2

) leads to a first order ODE

= f(x)

which can be integrated to a time-dependent deviation

(t) =0 exp(f(x) t)

with the starting deviation0 at t = 0. Introducing therelaxation timet0 = | 1f(x) | thisyields

(t) =0 exp (sign(f

(x

)) t/t0).Thus,f(x) hints towards the characteristics of a fixed point x.

Conclusion

If f(x)< 0: stable fixed point, exponential decayf(x)> 0: unstable fixed point, blow-upf(x) = 0: further investigations are needed

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2 Flow on a line

Examples

In both cases x= x3 the character of the fixed point is not clear from f.

For

x= x2 the system has a half-stablepoint at x= 0. If x is constant, the result is a lineof fixed points.

Uniqueness theorem

Iff(x) and f(x) are continuous on an open interval around x0, then a solution existsand is unique.

This leads to the impossibility of oscillations in an one-dimensional system. Instead,everything is overdamped. Consider a potential V. In 1d, we can write

x= f(x) = dV(x)dx

.

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3 Bifurcations in 1d

But why does x = r+ x2 describe all saddle-node bifurcations? We assume x to be afunction ofxand the parameter r.

Expanding x= f(x, r) around x= x and r= rc leads to

x f(x, rc) + fx

|(x,rc) (x x) + fr

|(x,rc) (r rc) +12

2f

x2|(x,rc) (x x)2.

Considering f(x, rc) = 0 and fx

|(x,r)= 0 the general form computes as

x= a(r rc) + b(x x)2 .

Example: Stability of adhesion cluster under constant force

We consider an adhesion cluster with bonds (receptor-ligand pairs) that can be eitheropen or closed. IfN(t) represents the time-dependent number of closed bonds and Nt

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the total number of bonds, the number of open bonds is given as Nt N. Two rateskoff, konare used to describe the systems dynamics. In contrast tokon,koffdepends on theacting force F. It can be written as koff= k0 exp( FF0 N) = k0 exp(

fN

). Furthermore

using = konk0

and the dimensionless force f= FF0

, the time-dependent number of bondsis given by

dN

dt =kon (Nt N) koff N

=kon (Nt N) k0 exp(fN) N.

N=dNd

= N exp(fN

) + (Nt N)

in dimensionless time = k0

t. The graphical analysis shows the different cases for

f < fc and f > fc. The fixed points are given by N exp(fN) = (Nt N).

Below, the bifurcation point fc is calculated using two equations.

Nc exp(fcNc ) = (Nt Nc) (3.1)

Nc exp(fcNc

)

1 fc

Nc

= Nc (3.2)

= fcNc

exp(fcNc

) (3.3)

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Equation (1) represents the fixed points. Dividing (2) by (1) results in1 fc

Nc

= Nc

Nt Nc fc= NtNcNt Nc

fcNc

= fcNt

+ 1

Using this,can be written as

= fcNt

exp(fcNt

+ 1) fcNt

exp(fcNt

) =

e fc = Nt plog

e

. In the last step, the function plog is used, defined by x exp(x) =Q x= plog(Q).

3.2 Transcritical bifurcation

The general form of a transcritical bifurcation x = r x x2

leads to the fixed pointx = 0 which exists for an arbitrary bifurcation parameter r and also represents thebifurcation pointrc = 0. The second fixed point x =r is stable for r >0 and unstablefor r 0. Theloss of photons is determined by the rate constant k >0.

n(t) = gain loss=G n N k n

Introducing the maximal possible number of excited atoms N0, the number of excitedatoms can be written as N(t) =N0 n(t). Hence, the rate n computes as

n(t) = (GN0 k) n G n2.

Thus, the fixed points aren1 = 0 andn

2 = GN0k

G . Since n describes a particle number,

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3.3.2 Subcritical pitchfork bifurcation

Now, f(x, r) = x= rx+x3. The fixed points are similar: x1 = 0 and x

2/3 =r if

r


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So, the two types of pitchfork bifurcation differ in the second and third fixed point whichare symmetrical in both cases. By now, the system is unstable. But it can be stabilizedby using high order terms.

3.4 Influence of high order terms

In order to stabilize pitchfork bifurcations, a fifth order term is used. For instance,consider the following equation:

x= rx+ x3 x5.

The fixed points are

x1= 0

x2/3= 1 + 1 + 4r

2 , exists for 1 + 4r >0

x4/5=

1 1 + 4r2

, exists for 1< 4r


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As well, a hysteresis effect is possible in this configuration. Starting at r > rc , x = 0and increasingr up tor >0 leads to an unstable condition. A small perturbation resultsin a transition to a stable branch at same r. Decreasing r again, the system remains onthe stable branch. This shows that the system does not come back to the original fixedpoint.

3.5 Summary of 1d bifurcations

The general form is f(x, r) = xand behaves as shown in the following table:

normal form f(x) fx(x, rc) f

r(x, rc) f

xx(x, rc) f

xr(x, rc) f

xxx(x, rc)

saddle-node x= r + x2 0 0 = 0 = 0bifurcation

transcritical x= rx

x2 0 0 0

= 0

= 0

bifurcation

pitchfork x= rx x3 0 0 0 0 0 = 0bifurcation

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Example: Overdamped bead on a rotating hoop

The hoop is rotating around the axis with an angular velocity . The acting forcesare the gravitational force FG, the centrifugal force FC and a friction force FR whichdescribes the system in a fluid. They are projected on the -plane.

FG=m g mg sin()FC=m 2 m2 cos()FR= b

Using = r sin(), the total acting force yields

F =m r= b mg sin() + mr2 sin()cos().

In order to receive a first order equation, the term m r shall be neglected. The time= t

Twith the timescale Tis introduced. In a second step, the equation is reformulated

dimensionless by dividing by the gravitational force FG.

mT2

2 d2d2

= bT

d

d mg sin() + mr2 sin()cos()

gT2

2d2

d2 = b

T mg

d

d sin() + r

2

g sin()cos()

How to define T so that 2 :=

gT2

2is negligible?

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4 Flow on a circle

Choosing the prefactor of the friction force bTmg

to be of order 1, T is defined to be

T = bmg

. With this, is negligible if = gT2

= m2gg2

1, so if the inertia is muchsmaller than the friction.

Set = 0 from now on and introduce := 2

g. The equation computes as

d

d= sin() ( cos() 1) .

The applied procedure has reduced the number of parameters from five to two: and. But setting = 0 transforms the equation to dimension one. So, only one initialcondition can be considered. Thus, the behavior of the system at the very beginningis neglected. After that, the system behaves as if it was of the order of 1.

The fixed points result from sin() = 0 and cos() 1

= 0. Therefore, the number offixed points depends on :

For|| >1 there are only the fixed points due to sin() = 0. For the bifurcation point|| = 1, one additional fixed point exists for each period of and for||


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4 Flow on a circle

with period T = 2

but without amplitude,(t) = t + 0.

Ths simplest non-trivial case is = a sin(). It has various applications in dif-ferent branches of science: e.g. Josephson junctions, electronics, biological oscillations,mechanics, etc.

ais a bifurcation parameter (a >0). Depending on its relation to , the following phaseportraits (lhs) and corresponding flow diagrams (rhs) exist.

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4 Flow on a circle

What is the oscillation period T?

T =

dt= 2

0

dt

dd=

20

1ddt

d= 2

0

1

a sin()d= 22 a2

Obviously, the oscillation period depends on a.

a= 0 T =2

a T = 2( a)(+ a)

= 22

a

The scaling is generic for a saddle-node bifurcation.

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4 Flow on a circle

A Taylor expansion around the critical value = 2

by introducing =

= a sin + 2

= a cos() a +1

2a 2

results in the normal form of saddle-node bifurcations:

x:=

a

2

1/2

r:=

a

2

a

1/2x= r + x2.

Most of the time will be spent close to .

T =

dt=

dt

dxdx=

2

a

1/2 dr

1

r+ x2 =

2

a

1/2 r

=

2

1/2 a

This is the same result as above. Therefore, extending the integration boundaries toinfinity is indeed not a problem.

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4 Flow on a circle

Examples

1. Driven overdamped pendulum b+ mgL sin() =

Dividing bymgLleads to the dimensionlessequation

b

mgL =0

+ sin() =

mgL=:.

The bifurcation parameter is the quotient of the applied torque to the maxi-mum gravitational torque. Having also introduced the dimensionless time0 =

t

the system is described by the equation

= d

d = sin().

At = 1 the pendulum stops its motion. For < 1 the external torque is tooweak to drive the pendulum around.

2. Firefly synchronization

Identify= 0 with the emission of the flash.

Without external stimulus, each firefly has = . Now, consider a periodic stimuluswith phase which satisfies

= . (4.1)

The basic model to simulate the fireflys re-action to the stimulus is given by

= + A sin( ) (4.2)

with the resetting or coupling strength A.

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4 Flow on a circle

The last equation can be expressed using the phase difference = . Sub-stracting (4.2) from (4.1) leads to = A sin(). Defining furthermore:=

A and =A t the final equation reads

=d

d

=

sin()

.

This leads to the following phase space diagrams:

For = 0, there is a perfect synchrony at = 0. If > 1 (or


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5 Flow in linear 2d systems


5.1 General remarks

In 2d, the varity of dynamical behavior is much larger than in 1d.

As a first step, we look at linear systems in two dimensions. Then, a complete classifi-cation is possible and starts from

x= A x=

a bc d

x1x2

.

In general, if x1(t) and x2(t) are solutions of the equation, so is c1 x1+ c2 x2.In addition,x= 0 is always a solution.

Graphical analysis can be done by drawing and analyzing the phase plane (x1, x2).

Examples

1. Harmonic oscillator: mx+ kx = 0

Defining the frequency 0 =

km

and choosing x1 = x, x2 = x = v as done in

section1the matrix isA=

0 120 0

.

Why are the trajectories ellipses?

x

v =

v

20 x 20 xdx= v dv 20 x2 v2 = const

This corresponds to energy conservation:

12

kx2 + 12

mv2 = const.

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2. A linear 2d system without oscillation:

xy

=

a 00 1

xy

For the given matrix A=

a 00 1

, the two equations are uncoupled. They can

be separately solved using an exponential ansatz.

x= a x x(t) =x0 exp(a t)y= y y(t) =y0 exp(t)

The phase portraits differ depending on a.

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3. Equal eigenvalues

The eigenvalues are equal if the discriminant is zero 2 4 = 0. Then, theeigenvectors exhibit the same velocity. They can either be different or the same.In both cases, stable and unstable phase portraits exist. As an example, the stableones are plotted.

1=2, v1=v2, degenerate node

4. At least one eigenvalue is zero

In this case, the phase portrait is a line or plane of fixed points.

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6 Flow in non-linear 2d systems

Summary in one scheme

Saddles, nodes and spirals are the major types of fixed points.


Non-linear systems show a much larger variety of flow behavior.

x=

f1(x)f2(x)

Example

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Recipe for phase space analysis:

1. Identify nullclines: lines with x= 0 or y = 0

2. Identify fixed points: intersections of nullclines

3. Linear stability analysis around fixed points

Example x= x+ exp(y), y= y

The phase portrait shows four different regions varying in the sign of xand y. At (-1,0),there is a saddle. Having drawn the nullclines it is easy to compute the remaining flowbehavior.

Linear stability analysis:

At a fixed point (x, y), we look at small deviations u = x x and v = y y usingCartesian coordinates. The derivatives are approximated in a Taylor expansion up tofirst order.

u= x = f1(x + u, y + v)

=f1(x, y) +

f1x

u+ f1y

v+ O(u2, v2, . . . )

v=y = f2(x + u, y + v)

=f2(x, y) +

f2x

u+ f2y

v+ O(u2, v2, . . . )

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a> 0 a< 0 a= 0

A center occurs only for a = 0. But linear stability analysis predicted this for all valuesofa. Instead, the typical case is a spiral.

The following biological model shows the power of phase space analysis.

Example: Phase space analysis of a biophysical model

Consider the Lotka-Volterra model for two competing species x, y. The variablesx andy name the population size of rabbits and sheep, respectively. The rabbit populationexhibits a faster logistic growth than the sheep population. As the sheep compete forgrass with the rabbits, the growth rate of the rabbits x decreases if more sheep existwhile the sheep suffer only little under more rabbits.

x= x(3 x 2y)y = y(2 y x)

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The Jacobian computes as A =

3 2x 2y 2x

y 2 2y x

. The character of the four

fixed points is different.

fixed point (0, 0) (0, 2) (3, 0) (1, 1)

A

3 00 2

1 02 2

3 60 1

1 21 1

5 -3 -4 -2

6 2 4 -1

1 3 -1 -2

2 1

2 2 -2 -2 (

2 1)

classification unstable node stable node stable node saddle

Furthermore, we have a brief look at two different types of special situations: con-servative (e.g. earth orbiting around the sun) and reversible systems (systems withtime-reversal).

1. Conservative systems

In a conservative system, the acting force Fcan be derived for a potential V. For

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example, in 1d we have:

m x= F(x) = dVdx

.

Multiplying with the velocity xleads to

m2 ddt(x2) = dVdt

ddt

(1

2mx2 + V

=E

) = 0.

There exists a quantityEwhich is constant along trajectories (but not in an openset inx). This corresponds to energy conservation.

Theorem: In a conservative system, an attractive fixed point cannot exist.

Proof: In such a case, there would be a bassin of attraction and thus Ecould notbe constant in a nontrivial way.

Example: Mexican hat V(x) = 12

x2 + 14

x4

The second derivative is x = x x3. Three fixed points exist: (0,0), (1,0) and(-1,0). Applying a simple trick x = y and y = x x3 the phase portrait can bedrawn.

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Example: Hamiltonian system H(q, p)

It is q= Hp

and p= Hq

. From this, energy conservation simply follows:

H=pH p+ qH q= 0.

We know from the theorem that there are no attractive fixed points in the system.Instead, a typical fixed point is a center and thus often oscillations occur in thesystem.

2. Reversible systems

Time reversal symmetry is more general than energy conservation. Reversible,non-conservative systems occur e.g. fluid flow, laser, superconductors, etc.

Mechanical systems are invariant under t t. Consider in 1d,m x = F. Weintroduce the velocity

v= x v= 1m

F(x).

Both (x(t), v(t)) and (x(t), v(t)) aresolutions of the system. In general,there is a twin for each trajectory. Notethe similarity to centers, which have

trajectories that have merged at theends.

Examples:

a) x= y y3 y= x y2

The system is invariant under t tand y y. There are three fixedpoints: two saddles and a center.There is mirror symmetry around thex-axis in regard to the flow lines (butnot the flow distribution).

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b) x= 2 cos(x) cos(y) y = 2cos(y) cos(x)

The system is invariant under t t,x xand y y.Four fixed points exist (x, y) =

2 , 2: two saddles, one stable nodeand one unstable node. Since the stablenode is an attractive fixed point, this isnot a conservative system.Now, there is mirror symmetry aroundthe bisector.

Numerical integration of ODEs

Several numerical integration methods for ODEs exist, differing in their accuracy. Theyare based on a Taylor expansion up to a certain order. In the following, we have a closerlook at three different methods.

1. Euler method:

The time t is discretized. Starting at tn, the next step is computed multiplyingthe velocity at tn with the time step t. This corresponds to a first order Taylorexpansion

xn+1=xn+ f(xn)t + O(t2).

Since this accuracy is not very good, higher order methods are often applied.

2. Runge-Kutta methods:

The Runge-Kutta methods combine several Euler-style steps. A second orderaccuracy is achieved by using the mid-point velocity of the integration step.

k1 = f(xn)t, k2=f

xn+

k12

t

xn+1=xn+ k2+ O(t3

)

We see that two function evaluations are needed. In an analogous manner, we

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maintain fourth order accuracy:

k1=f(xn)t, k2=f

xn+

k12

t,

k3=fxn+k2

2 t, k4=fxn+k3

2 t,

xn+1=xn+16

(k1+ 2k2+ 2k3+ k4) + O(t5)

Obviously, this requires four function evaluations. It is the a very good choice ifthe number of evaluations is not essential. To realize this, one standard choice isMatlab ODE 45 (see also on the web page).

3. Stormer-Verlet methods: (leaping frog)

Stormer-Verlet methods are especially suited for Hamiltonian systems, e.g. molec-ular dynamics. The simplest version is:

x= f(x) f(xn) = xn+1+ xn1 2xnt2

xn+1= 2xn xn1+ f(xn)t2

We now rewrite this as 2d system. We define the velocity v = xand discretize the

function f(x) = v. For each integration step, the position is updated for a freetime step and the velocity half a time step. The resulting equations are:

vn+1/2 =v1+t

2 f(xn)

xn+1=xn+ vn+1/2t

vn+1=vn+1/2+t

2 f(xn+1)

This procedure is called leaping frogbecause we have staggered jumps.

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7 Oscillations in 2d

Since

1 + is always real for 0, the only restriction for the trapping regioncomes from rmin: 0


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The nullclines are y = xa+x2

and

y = ba+x2

. So, there is a fixed point

(x, y) = (b, ba+b2

). In order to showthat a trapping region exists, we con-sider the region bounded by the greenlines. We know for the left part x = 0and 0 y b

a. So we get 0 x b

and 0 y b. The flow goes in-side. For the right part we calculatex (y) = x+ y = b x < 0 sincex > b. This yieldsy > x. Therefore,the flow is more negative than1 andgoes inside.

Now, we have found a trapping region. The Poincare-Bendixson theorem demands

that no fixed points exist. Hence, we make a hole around the fixed point and showthat no trajectory goes into the hole. This is equivialent to a repulsive (unstable)fixed point. The Jacobian is

A=

1 + 2xy a + x2

2xy (a + x2)

.

We need a positive trace and determinant of the matrix A.

=a + b2 >0

= b

4

+ (2a 1)b2

+ (a + a

2

)a + b2!

0

Thus, the boundary between stable and unstable fixed points = 0 is given byb2 = 1

2(1 2a) 1 8a. This can be represented via a state-diagram.

The arrow indicates an increasing b. Ifa is small enough, the system performsoscillations in a certain interval ofb. Wecall this a re-entrance process.

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Lienard systems / van der Pol oscillator

The following structure often occurs in mechanics and electronics:

xinertia + f(x) x damping + g(x) restoring force = 0.It is called Lienard-system. Note, that this is the equation of the harmonic oscillator forf= 0 and g= x.

We consider the most famous example of Lienard systems, the van der Pol oscillator:

f=(x2 1)g= x.

Forx

1 we have negative damping. The system can be driven by putting energy intothe system. For large x, the damping is positive. Energy is dissipated.

Example: Tetrode circuit (electronics)

The system is described as follows:

LI+ V + F(I) = 0

LI+ V + F(I)I= 0 LC I+ I+ CF(I) I= 0.

This is a van der Pol oscillator with f(I) =C F(I) and g= 1.

Lienard systems are very widespread: e.g.

1. neural activity, action potential

2. biological oscillators (ear, circadian rhythms)

3. stick-slip oscillations in sliding friction

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Lienard theorem:A Lienard system has a stable limit cycle around the origin at the phase plane if

1. g(x) = g(x), g(x)> 0 for x >02. f(

x) =f(x), F(x) =

x

0

f(x)dx has to have a zero at a >0

and F(x)< 0 for 0< x < a, F(x)> 0 for x > a, F() = .

Obviously, the first condition is fullfilled for the van der Pol oscillator. Consider thesecond condition:

F(x) =

x3

3 x

=

x

3

x2 3

a=

3.

As for the harmonic oscillator the deflection behaves sine-shaped, the deflection of thevan der Pol oscillator follows a sawtooth. The phase portrait shows a deformed circle.

Now, we analyze the van der Pol oscillator in two limits: 1 and 1.

1. 1: Lienard phase plane analysis

The dynamic equation is

x= x+ x(x2 1) = ddt

(x+ F(x)) =: d

dtw(x).

x= w

F(x)

In the last step, we are using dF(x)dt

=f(x) x. Define

y :=w

y= x

, x= (y F(x)).

Hence, the nullclines are x= 0 and y=F(x).

We assume the initial condition y F(x) O(1). Then, the velocity is very fastin x-direction but very slow in y-direction:

x O()y O(1/)

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The first part shows a fast movement tothe right which stops at the x-nullcline.In the second part,y

F(x)

O(1/2)

can be arbitrarily small. This leads tox O(1/) and y O(1/). Thus, wehave a slow movement along the null-cline. The third and fourth part are likethe first and second, respectively, onlywith changing signs in the velocities.

What is the oscillation period T? We neglect the fast paths. Hence, we have to

calculateT= 2(T(x= 2 x= 1)) = 2

t1t2

dt.

We knowdy

dt =F(x)

dx

dt = (x2 1) dx

dt = x

and from this

dt= dx(x2 1)

x .

Therefore, the oscillation period is T O():

T= 2 1

2dx

(x2 1)x

= 2

x2

2 ln(x)

21

=(3 2 ln(2))

T O().

2. 1:

This case is a small perturbation to the harmonic oscillator

x+ x+

h(x,x) = 0.

The systems dynamics depend on h(x,x).

a) Forh= (x2 1)x, the system is a van der Pol oscillator.

b) For h = 2x, we have a weakly damped harmonic oscillator. The system islinear.

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c) For h = x3, the system corresponds to an unharmonic spring with a springconstant k = 1 + x2 that increases by extending the spring. This is calledstrain stiffing. The system is a Duffing oscillator.

For simplicity, we consider the second case h= 2x.

For the initial conditions

x(0) = 0, x= 1,

the exact solution holds

x(t) = (1

2)1/2

exp(

t)

sin((1

2)1/2

t).

If we expand for 1, we find

x(t) = (1 t) sin(t) + O(2).

But, this is only valid for t < 1

and blows up for large times. Thus, the smallepsilon limit appears to be problematic.

We now try to solve the problem using regular perturbation theory. Plugging

x(t) =x0(t) + x1(t) + . . .

in the dynamic equation yields

d2

dt2(x0+ x1+ . . . ) + 2

d

dt(x0+ x1+ . . . ) + (x0+ x1+ . . . ) = 0.

Compare the parameters for different orders of.

O(1) : x0+ x0= 0

x0= sin(t)O() : x1+ 2 x0+ x1 = 0

x1+ x1 = 2 cos(t) x1(t) = t sin(t)

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Now, we seex= x0+ x1+ O(

2) (1 t) sin(t).Again, we end up with a term that is linear in t. The solution to this problemcomes from singular perturbation theory. We separate time scale into a fast time=t and a slow time T=t. This procedure is called two timing.

We use the new ansatz:

x(t) =x0(, T) + x1(, T) + O(2)

and therefore calculate the derivatives

x= xt + Txt T

= x+ T x = x0+ ( x1+ T x0) + O(

2)

x= x0+ ( x1+ 2T x0) + O(2)

and plug them into the dynamic equation

0 = x0+ ( x1+ 2T x0) + 2 x0+ (x0+ x1) + O(2).

O(1) : x0+ x0 = 0

O() : x1+ x1 = 2(T x0+ x0)O

(1) : x0=A(T) sin() + B(T)cos()O() : x1+ x1 = 2 (A(T) + A(T)) cos() + 2 (B(T) + B(T)) sin()

In order to end up with a well-behaved solution, demand the prefactors (A +A)and (B + B) to be zero.

A= A0 exp(T), B=B0 exp(T)

For the initial conditions x(0) = 0, x(0) = 1 B0 = 0, A0 = 1, the generalsolution is sine-shaped with an envelope decaying in time

x0= exp(T) sin() = exp(t) sin().

This is identical to the exact solution in order O(2).

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Application to van der Pol oscillator

The dynamic equationx+ (x2 1)x+ x= 0

holds [ x0+ ( x1+ 2T x0)] + (x20 1) x0+ (x0+ x1) = 0.

O(1) : x0+ x0= 0

O() : x1+ x1= 2T x0 (x20 1) x0

In polar coordinates:

x0=r(T) cos(+ (T)) x1+ x1= 2[r

sin(+ ) + r

cos(+ )] + r sin(+ )[r2 cos2(+ ) 1] [2r r+ 1

4r3] sin(+ ) + 2r cos(+ ) + 1

4r3 sin(3(+ )).

In the last step, we used the relation sin() cos2() = 14

[sin() + sin(3)]. In order toavoid a resonance catastrophe, demand the prefactors [2rr + 1

4r3] and 2r to be zero.

r =18

r(4 r2) and = 0

The result is a logistic growth in r .There is a fixed point forr = 2 and = const = 0.Note, that we get a limit cycle irrespec-tive of the value of. Therefore, is asingular perturbation.

Averaging method:

We now discuss a more general method to solve these kinds of problems. Consider again

x+ x+ h(x,x) = 0, which represents a large class of non-linear oscillators.

x0+ x0= 0

x1+ x1= 2T x0 h x0=r(T) cos(+ (T))

x1+ x1 = 2[r sin(+ ) + r cos(+ )] h

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Like in 1d, in 2d existence and stability of fixed points depend on the parameters of thesystem. In contrast to 1d, however, now also oscillations can be switched on and off. As

an example, look at the substrate-depletion-oscillator.

There are three types of bifurcations in 2d:

1. 1d-like bifurcations (4 types)

2. Hopf bifurcation (local switch on/off of oscillations)

3. global bifurcations

In the following, we have a closer look at them.

1. 1d-like bifurcations

All four types of 1d bifurcations exist in 2d (cf. center manifold theorem).

Example: Griffith model for genetic switch

Consider a gene that codes for a certain protein. The activity of the gene shall

be induced by the protein and its copies which are translated from the messengerRNA. The system is described as

x= y ax

y= x2

1 + x2 by,

where, xandy are the concentrations of the protein and the mRNA, respectively.

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The following figure shows a protein acting as transcription factor.

The nullclines are y = a x and y = x2b(1+x2)

. We see that the system depends on

the parameters a, b. For increasing a, the two upper fixed points approach eachother until they fall together when the nullclines intersect tangentially. For evenlarger a, only the fixed point in the origin remains.

The two upper fixed points are given by

x =ab(1 + x2

)

x =1

1 4a2b22ab

.

For 2ab= 1, the fixed points collide. The critical values are

ac= 1

2b xc = 1.

Ifa < ac, the system is bistable and acts like a genetic switch: depending on the

initial conditions, the gene is on or off.

We note that in this case, the 2d analysis gives essentially the same results likea 1d analysis along the x-nullcline. In fact, this is an example of a saddle-nodebifurcation.

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The prototype of a saddle-node bifurcationin 2d is

x= x2

y= y.The phase portrait varies with .

One fixed point is a saddle, x = (, 0), the other one a stable node, x =(+

, 0). The behavior of the saddle depends on the eigenvaluesof the Jacobian

A=

2x 0

0 1

withx = (, 0). The bifurcation is given for = 0. This iscalled a zero eigenvalue-bifurcation. It exists for all bifurcations from type 1).

Recall that the eigenvalues can be calculated using

1/2 = 2 4

2 .

Either both eigenvalues are real (shown on the left), which corresponds to (2 4) > 0, or they are complex (shown on the right), (2 4) < 0 1/2 =2 i.

The saddle-node bifurcation is of the first type. We now turn to the second type.

If the complex eigenvalues 1/2 =

2 i cross the yaxis from left to right,oscillations are switched on. This is called Hopf bifurcation.

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2. Hopf bifurcation

a) Supercritical Hopf bifurcation

r= r r3= + br2

The systems dynamics strongly depends on .

In order to analyze the behavior of the eigenvalues during the bifurcation,rewrite the system in Cartesian coordinates:

x= r cos()

y = r sin()

x= r cos() r sin()=x y + O(xr2, yr2)

y = x+ y

A=

1/2= i

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Thus, if we increase , the eigenvalues cross the imaginary axis from left toright as expected.

The structure is similar to the supercritical pitchfork bifurcation. But now,we have a supercritical Hopf bifurcation.

b) Subcritical Hopf bifurcation

r= r+ r3 r5= + br2

The term (r5) causes certain trajectories to drive away from the origin. For < 0, both exist a stable limit cycle and an attractive fixed point in theorigin. But for increasing, the attractive area around the origin decreases

and finally disappears for= 0. That is when the subcritical Hopf bifurcationoccurs. The origin is now unstable and there is an abrupt transition to large-amplitude oscillations.

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Note that a Hopf bifurcation theorem exists, demanding rigorous conditions on1/2 for a Hopf bifurcation to occur.

3. Global bifurcations

Apart from Hopf bifurcations, global bifurcations are another way in which limitcycles are created or destroyed. They are a combination of 1) and 2). Cycleinteractions with other fixed points exist. As well, there is a global change in flowstructure.

a) Saddle-node bifurcation of circles

This is a prototypical example of global bifurcations.

r= yr+ r3 r5= + br

2

c = 14

b) Infinite period bifurcation

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9 Excitable systems

A human neuron has a diameter ofr 50 m for the cell body and an axonthat is up to 1 m long. The axon trans-ports the signals calledaction potentialsorspikes. We think about them as trav-eling wavesV(x, t).

It is approximately cylindrically shapedcontaining mainly potassium (K+) andorganic anions (A) and being sur-rounded predominantly by sodium(Na+), calcium (Ca2+) and chlorine(Cl).

Typical values of the concentrations of the two dominant ions potassium and sodiumare given in the following table:

inside outside V[mM] [mM] [mV]

K+ 155 4 -98

Na+ 12 154 67

The axon membrane is typically 5 nm thick and made of dielectric material (= 2) whichis an insulator. To allow ion flow, there are many channels in the surface. Channelscan dynamically open and close. The resting potential is approximately V = 60 mV.Here is an equivalent electrical circuit:

A biological membrane acts as a capaci-tor for which the equation CA =

CA

= 0d

holds. Typically, CA

is around Fcm2

. The

timescale is = CAgA

= Cg 2 ms with

the conductivity per area gA= 5 1

m2.

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9 Excitable systems

Assuming equilibrium, the concentration c is proportional to a Boltzmann distributionc exp(eV/kBT). The ratio of the concentrations inside and outside of the axontherefore depends on the difference of the potential ci

co exp(e(Vi Vo)/kBT). Solving

for the difference of the potential, this yields the Nernst potential:

V = kBTe

ln cico .

This equation has been used to calculate V in the previous table (kBTe

= 25 mV). SinceV1= V2, the system is not in equilibrium. Therefore, we need dynamical models tounderstand.

The exact form of an action potential can be measured with a space clamp. Takinga giant axon of a squid and leading a silver wire into the axon, the action potentialV(t) =Vi(t) Vo(t) can be determined independent from spatial components.

Lets have a short look at the historical background.

1937/38 Hodgkin worked at Woods Hole and invented with others the space clamp.1947 He performed experiments at Plymouth together with the student Huxley.1952 They published five papers in the Journal of Physiology.1963 They got the Nobel prize for their work.1991 Neher and Sakmann got the Nobel prize for finding the first evidence

for ion channels using the patch clamp method.

Hodgkin-Huxley model (1952)

Starting with the electrical circuit,Hodgkin and Huxley invented a modelfor action potentials. They assumed alinear relation between voltage and cur-rent

INa=gNa(V VNa).

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9 Excitable systems

Faradays law states Q = C V I=C V.

V = 1C

gNa(V VNa) + gK(V VK) + gL(V VL)

leakage current

The leakage current contains all contributions except those from potassium and sodium.

This equations describes a linear systemwith a stable fixed point. If a little per-turbation occurs the system will relax-ing back. But this behavior does not fitto the experiments.

Hence, to get an action potential we consider a conductivity that depends on voltage,g = g(V). The basic idea is a two-state process for opening and closing of the gate.Such a process can be described in the following way using the opening and closingprobabilities n and n, respectively.

n= n(1 n) n n.

A closed gate corresponds ton= 0, an open one ton= 1. Therefore, the time-dependentprocesses are

opening: n(t) = 1 exp(t)closing: n(t) = exp(t).

A voltage clamp experiment showed agood agreement for the closing process.But the fit result of the opening pro-cess corresponds to power four. In gen-eral, higher powers and three gates areneeded to fit the data.

We use a four-dimensional model (V,n,m,h) defined by the additional equation

n= n(1 n) nnm= m(1 m) mmh= h(1 h) hh.

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9 Excitable systems

A more general NLD-analysis shows that the system can oscillate.

Up to now, we did not consider the effects of space.

Describing the signal as traveling wave, Ohms law holds in lateral direction

V(x+ x) V(x) = I(x) R.

Looking at a node of the circuit, current conservation demands

I(x x) I(x) = gA V(x).

Taking x 0, this yields

V

(x) =

r2 I(x)I(x) = gA 2r V(x)

V(x) = 12

V

with the radius r of the axon (r 0.25 mm for a squid), the conductivity of themedium and the resistance per length

r2 = 0.3 m. In the last step, the decay length

=

r2gA

= 9 mm has been introduced.

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9 Excitable systems

Injecting a voltage V0 at left, the space dependence is

V(x) =V0 exp(x

).

Because of this, the signal would just decay if the conductivities were independent from

the voltage.

We next combine the longitudinal equation with the full transverse Hodkin-Huxley cur-rent for the current conservation

I = [gNa(V VNaN ) + gK(V VKN) + gL(V VLN)] CU

t.

2V V = gNa(V VNaN ) + gK(V VKN) + gL(V VLN)

g .

The result is a non-linear PDE, the cable equation with time-dependent conductivity.

Considering one type of channel (e.g. Na) and injecting a voltage V0 at the left, a frontpropagates to the right. To get a wavepropagation, one has to add a counteractingprocess (e.g. opening of K channels). Hodgkin and Huxley showed the waves in 1952.To understand this better, we consider the bistable cable equation

V =V + f(V)

withf(V) = V(V )(V 1). We look for solutions of the form V(x, t) =U(x + c t)which describe waves propagating from right to left with velocityc. Definingy :=x + ct,

we can convert the system into an ODE.

yU c= 2yU+ f(U)

Now, we do a phase plane analysis and therefore define

W =yU yW =c W f(U), f(U) = U(U )(U 1).

A traveling front solution must connect the fixed points (U = 0, W = 0) and (U =1, W = 0) in the (U, W)-plane as we vary y from to +. There is a unique cwhich results in such a trajectory. It can be calculated analytically. For this, we guess

that the connection between the two resting states is given by

W(U) =dU

dy = BU(U 1).

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9 Excitable systems

So, we calculate

0 = yU c+ 2yU+ f(U)= W c+ yW+ f(U)=

cBU(U

1) + B2(2U

1)

U(U

1)

U(U

)(U

1)

=cB + B2(2U 1) (U )

and find

B= 1

2, c=

12

(1 2).

The speed is a decreasing function of and the direction of propagation changes at= 1

2. In this case, there is no propagation. The profile of the traveling front is found

by integrating the assumption

W(U) =

dU

dy = BU(U 1).

U(y) =12

1 + tanh

1

2

2y

.

A propagating wave is a trajectory which comes back to the original state. This occurs inthe Hodgkin-Huxley model as well as in the Fitzhugh-Nagumo model (which is discussedbelow) with diffusive coupling

V =V + f(V, W)

W =g(V, W).

Until now, we studied 1d wave propagation. The cable equation can also be extended to2d or 3d. We then obtain propagation fronts (planar or circular), but also spirals. In thecontext of heart and muscle biology, spirals are a sign of a pathophysiological situation.

Fitzhugh-Nagumo model

In depth analysis of the Hodgkin-Huxley model, Fitzhugh and Nagumo tried to analyzethe phase plane for a cut through phase space. On the first attempt, they used twovariablesV and mand kept n and h constant. The two equations then read

eV = gKn40(V VK) gNam3h0(V VNa) gL(V VL)m= (V)m (1 m) (V)m m

with the parameters g denoting constants.

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9 Excitable systems

The phase plane dependency ofmand V is shown in the figure below.

This analysis explains the threshold but not the relaxation. Because of this, on thesecond attempt Fitzhugh and Nagumo introduced one fast variablex=(V, m) responsiblefor the excitation and a slow variable y=(h, n) responsible for the relaxation. z= z(t)denotes the stimulus andcis an inverse time constant. This yields the Fitzhugh-Nagumomodel which is a simplification of the Hodgkin-Huxley model:

x= c (y+ x x3

3 + z(t))

y=x a + by

c .

For a= b = z= 0, this is a van der Pol oscillator which leads to relaxation oscillations.The phase portraits are shown in comparison: left the van der Pol oscillator and in themiddle and right the Fitzhugh-Naguma model with different parameters. The middlefigure shows a large excursion ending in a stable fixed point (the action potential) andthe right one that oscillations are possible for certain parameters.

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10 Pattern formation in reaction-diffusion systems


One of the most fascinating subjects in mathematical biology is called morphogenesis:a part of embryology leading to various patterns and forms. The pioneering paper was

published in 1952 by Alan Turing. He wondered under which condition a reaction-diffusion system produces a heterogeneous spatial pattern. To answer this question, heconsidered a two-dimensional system of the type:

A= F(A, B) + DAA

B=G(A, B) + DBB.

The simplest case was studied by J. Schnackenberg with the functions F and G

F=k1 k2A + k3A2BG= k4 k3A2B.

Thus,Ais autocatalytic and inhibitsB, whileBactivatesA. We first non-dimensionalizethe system:

u= A

k3k2

2, v= B

k3k2

2,

t=DAt

L2 , x= x

L, d=DBDA ,

a=k1k2

k3k2

1/2, b=

k4k2

k3k2

1/2,

=L2k2

DA

Note that the variables uand v are positive since they are concentrations of reactants.By introducing the variables above, the system is described as follows

u= (a

u+ u2v) =:f(u,v)

+ u

v=(b u2v) =:f(u,v)

+ dv

with the ratio of the diffusion d and the relative strength of the reaction versus thediffusion termswhich scales as L2. We start with a linear stability analysis of thisgeneral form with a zero-flux boundary condition and a given initial configurations for

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uand v . We first consider the case without diffusion D = 0 and ask for a homogeneousstate which is stable; we then ask under which conditions diffusion leads to an instability,the Turing instability.

Stabilizing diffusion. 1D reaction-diffusion system

For single reactantu(x, t) there is no Turing instability:

ut = Duxx+ f(u), x [0, l] + BC

An uniform (x- independent) ODE

u= f(u)

has either: stable m = f(u) < 0, or unstable m > 0 equilibrium at stationary point

u :f(u) = 0.

The diffusion will maintain a stable one, and it can stabilize the unstable one.To check it we take the linearized problem about u, for v(x, t) =u(x, t) u : vt = Dvxx+ mv, x [0, l]

m = f(u)

We use the Neumann BC (no flux):

xv0,l

= 0

and eigenfunction expansion.Separation of variables: look for the solution in the form:

v(x, t) =(t) (x)

Leading to: = (D + m )

=D + m

=:

Hence, the solution stability depends on the sign of:

(t) exp(t)

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The second equation:D + (m )= 0, (0, l) = 0

or

+ 2= 0, 2 =m

D

The eigenfunctions are:

k(x) = cos kx= coskx

l , k= 0, 1,...

Using the expression for we get:

= m D

kx

l

2

So:

stable case (m


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with wavenumber k= np

and wavelength = 2k

= 2pn

(n integer).

W(r, t) = k

ckexp(t) Wk(r)

Wk =AWk

Dk2 Wk

We now have to solve this eigenvalue problem. A Turing instability occurs if Re (k)> 0.Our side constraint is that the eigenvalue problem for D = 0 (only reactions) is assumedto be stable, that is Re (k= 0)< 0.

0 =2 + [k2(1 + d) tr A] + [dk4 (dfu+ gv)k2 + 2det A].

We first note that the coefficient of is always positive because k2(1 +d) > 0 and trA 0, the constant part has to be negative. Since the first and last terms arepositive, this implies

dfu+ gv >0 d = 1. (10.3)This is the main result by Turing: An instability can occur if one component diffusesfaster than the other. (10.3) is only a necessary, but not a sufficient condition. Werequire that the constant term as a function ofk2 has a negative minimum.

(dfu+ gv)2

4d >det A= fugv fvgu (10.4)

The critical wavenumber can be calculated to be

kc= det Adc 1/2

with the critical diffusion constant from

d2cf2u+ 2(2fvgu fugv)dc+ g2v = 0.

For d > dc, we have a band of instable wavenumbers. The relation= (k2) is called

the dispersion relation. The maximum singles out the fastest growing mode. This onedominates the solution

W(r, t) =

kckexp((k

2)t)

for larget. Note however, that in this case also non-linear effects will become importantand thus will determine the final pattern.

In summary, we have found four conditions (10.1) - (10.4) for the Turing instability tooccur. We now analyze the Schnackenberg-model in one spatial dimension. We alreadynoted thata + b >0 andb >0 for the steady state to make sense. From the phase planewe see that f > 0 for large u and f < 0 for small u. Hence,fu >0 around the steady

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state. Thus, b > a.

All in all, there are four relations:fu, fv >0 and gu, gv 1 in this case (the activationB diffuses faster in this model). In general, the conditions (10.1)-(10.4) define a domainin (a,b,d)space, the Turing space, in which the instabilities occurs. The structureof the matrix A tells us how this will happen: as u or v increases, u increases and vdecreases. So, the two species will grow out of phase.

The domain size p has to be large enough for a wavenumber k = np to be within the

range of the unstable wavenumbers ( L2):

L(a,b,d)

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