Mathematical Biology - Reaction Diffusion Models and Turing … · 2014. 4. 15. · Reaction Di...

Reaction Diffusion EquationsMatrix Pencils and Common Lyapunov Functions

CLFs and Turing InstabilityApplications

Mathematical Biology - Reaction Diffusion Modelsand Turing Instability

Stuart Townley

University of Exeter, UK

March 20, 2014

Stuart Townley Math Biol - Reaction Diffusion 1/ 33

DiffusionReaction Diffusion and Turing Mechanisms

In a collection of spatially distributed particles, be it bacteria,animals, chemicals etc., each particle often moves around in arandom way.The particles tend to spread out as a result of this irregularindividual motion

When this microscopic movement results in some macroscopicmotion of the group we call this a diffusion process.

In order to understand this diffusion via equations let usrestrict attention to one spatial dimension (for simplicity)

Consider a particle moving along the x-axis in a randomfashion - let us suppose that in time 4t the particle moveseither +4x or −4x.Denote by p(x, t) the probability that a particle released atx = 0, t = 0 reaches position x at time t.

At time t−4t the particle must have been at x+4x orx−4x so that

p(x, t) =1

2p(x−4x, t−4t) +

2p(x+4x, t−4t)

Expanding p(x, t) in a Taylor Series we get

p(x, t) = 12

[p(x, t)−4x ∂p∂x −4t

∂p∂t + 1

2(4x)2 ∂2p∂x2

+(4x)(4t) ∂2p

∂x∂t + 12(4t)2 ∂

2p∂t2

+ . . .

[p(x, t) +4x ∂p∂x −4t

∂p∂t + 1

2(4x)2 ∂2p∂x2

−(4x)(4t) ∂2p

∂x∂t + 12(4t)2 ∂

2p∂t2

+ . . .

]Simplifying this we obtain

0 = −4t∂p∂t

2(4x)2

∂x2+

2(4t)2∂

∂t2+ . . .

p(x, t) =1

2p(x−4x, t−4t) +

2p(x+4x, t−4t)

p(x, t) = 12

[p(x, t)−4x ∂p∂x −4t

∂p∂t + 1

2(4x)2 ∂2p∂x2

+(4x)(4t) ∂2p

∂x∂t + 12(4t)2 ∂

2p∂t2

+ . . .

[p(x, t) +4x ∂p∂x −4t

∂p∂t + 1

2(4x)2 ∂2p∂x2

−(4x)(4t) ∂2p

∂x∂t + 12(4t)2 ∂

2p∂t2

+ . . .

Simplifying this we obtain

0 = −4t∂p∂t

2(4x)2

∂x2+

2(4t)2∂

∂t2+ . . .

p(x, t) =1

2p(x−4x, t−4t) +

2p(x+4x, t−4t)

p(x, t) = 12

[p(x, t)−4x ∂p∂x −4t

∂p∂t + 1

2(4x)2 ∂2p∂x2

+(4x)(4t) ∂2p

∂x∂t + 12(4t)2 ∂

2p∂t2

+ . . .

[p(x, t) +4x ∂p∂x −4t

∂p∂t + 1

2(4x)2 ∂2p∂x2

−(4x)(4t) ∂2p

∂x∂t + 12(4t)2 ∂

2p∂t2

+ . . .

]Simplifying this we obtain

0 = −4t∂p∂t

2(4x)2

∂x2+

2(4t)2∂

∂t2+ . . .

So∂p

((4x)2

)∂2p

2(4t)∂

∂t2+ . . .

Assume 4x→ 0, 4t→ 0 such that

lim4t,4x→0

((4x)2

)= D (the diffusion coefficient)

∂t= D

the one-dimensional diffusion equation.

Extensions to higher-dimensions in space and more dependentvariables are natural extensions.

So∂p

((4x)2

)∂2p

2(4t)∂

∂t2+ . . .

lim4t,4x→0

((4x)2

∂t= D

So∂p

((4x)2

)∂2p

2(4t)∂

∂t2+ . . .

lim4t,4x→0

((4x)2

∂t= D

So∂p

((4x)2

)∂2p

2(4t)∂

∂t2+ . . .

lim4t,4x→0

((4x)2

∂t= D

Turing (1952) proposed that under certain circumstancesdiffusion may act as a destabilising mechanism.

More precisely in the absence of diffusion, linearly stableuniform steady states may persist but addition of diffusionmay drive instabilities.

Let us derive conditions under which this may happen.

Most reaction-diffusion systems of biological interest can benon dimensionalised and scaled to take the general form

ut = γf(u, v) +∇2u vt = γg(u, v) + d∇2v (1)

[The scale factor γ can obviously be incorporated into thereaction kinetics f and g but we keep it separate for now.]We typically have two spatial variables x and y and may havemore than two dependent variables u and v.

ut =∂u

∂t,∇2u =

∂x2+∂2u

∂y2, etc.

Suppose we have a homogeneous steady state (u0, v0) so that

f(u0, v0) = g(u0, v0) = 0

Let us linearise small perturbations from the steady state anddo so in the absence of diffusion.

Setu = u0 + u, v = v0 + v with |u|, |v| << 1

Thenut = γ[fuu+ fvv]vt = γ[guu+ gvv]

where all partial derivatives are evaluated at (u0, v0)

ut =∂u

∂t,∇2u =

∂x2+∂2u

∂y2, etc.

f(u0, v0) = g(u0, v0) = 0

ut =∂u

∂t,∇2u =

∂x2+∂2u

∂y2, etc.

f(u0, v0) = g(u0, v0) = 0

)then wt = γ

(fu fvgu gv

We look for solutions w ∝ eλt, where λ is an eigenvalue of thelinearised matrix. Then∣∣∣∣ γfu − λ γfv

γgu γgv − λ

∣∣∣∣ = 0 (2)

Recall that we are interested in cases where diffusion leads toinstability so we want the diffusion-free problem to be stable.

For the roots of (2) to have negative real parts we require

fu + gv < 0 (I)fugv − fvgu > 0 (II)

)then wt = γ

(fu fvgu gv

γgu γgv − λ

∣∣∣∣ = 0 (2)

)then wt = γ

(fu fvgu gv

γgu γgv − λ

∣∣∣∣ = 0 (2)

)then wt = γ

(fu fvgu gv

γgu γgv − λ

∣∣∣∣ = 0 (2)

Now we reconsider the linearised equations with diffusion andlook for instability

If u = u0 + u, v = v0 + v then on linearisation we have

ut = γ[fuu+ fvv] +∇2u

vt = γ[guu+ gvv] + d∇2v

So, if w =

), then

wt = γ

(fu fvgu gv

)w +D∇2w D =

(1 00 d

This is a PDE. So to make progress we need to specifyrelevant boundary conditions.

Frequently in biological applications zero flux conditions areappropriate (if in a particular application, some other boundarycondition is more natural, then the analysis is easily modified).

So, if w =

), then

wt = γ

(fu fvgu gv

)w +D∇2w D =

(1 00 d

So, if w =

), then

wt = γ

(fu fvgu gv

)w +D∇2w D =

(1 00 d

So, if w =

), then

wt = γ

(fu fvgu gv

)w +D∇2w D =

(1 00 d

So how do we solve the linearised PDE (3)?

We proceed by seeking separation of variable solutions usingeigenmodes of the Laplacian.

So let w be the time-independent solution of the eigenvalueproblem:

∇2w+k2w = 0 with boundary conditions. Here w =

)Here k is an eigenvalue and the corresponding eigenfunctionsgive us a basis in which to expand a solution.

Here k is an eigenvalue and the corresponding eigenfunctionsgive us a basis in which to expand a solution.

If we denote by wk the eigenfunctions of the eigenvalueproblem, we seek a solution of the original problem (3) of theform

w(x, y, t) = Σkckeλktwk(x, y) (4)

Here the constants, ck, are determined by a “Fourier”expansion of the initial conditions expressed in terms of thewk. We need not worry about this aspect!

We wish to know whether any of the λk have positive realpart (so that diffusion has caused instability).

Substituting (4) into (3) and comparing eigenfunctions gives

Σkckλkeλtwk = γA

(Σkcke

λktwk

)+D∇2

(Σkcke

λktwk

)where A =

(fu fvgu gv

), the community matrix of the

reaction part.

(Σkcke

λktwk

)+D∇2

(Σkcke

λktwk

)where A =

(fu fvgu gv

reaction part.

(Σkcke

λktwk

)+D∇2

(Σkcke

λktwk

)where A =

(fu fvgu gv

reaction part.

(Σkcke

λktwk

)+D∇2

(Σkcke

λktwk

)where A =

(fu fvgu gv

reaction part.

This can be written:

Σkckλkeλktwk = γA

(Σkcke

λktwk

(Σkcke

λktk2wk

)(Since ∇2wk + k2wk = 0 )

But wk are linearly independent.Hence λkwk = γAwk − k2Dwk or

(λkI − γA+ k2D)wk = 0.

Non-trivial solutions λ = λk exist if

det(λI − γA+ k2D) = 0, i.e.∣∣∣∣ λ− γfu + k2 −γfv−γgu λ− γgv + dk2

∣∣∣∣ = 0

(Σkcke

λktwk

(Σkcke

λktk2wk

∣∣∣∣ = 0

(Σkcke

λktwk

(Σkcke

λktk2wk

∣∣∣∣ = 0

So for non-trivial solutions we require:

λ2 + λ[k2(1 + d)− γ(fu + gv)] + h(k2) = 0 (5)

with h(k2) = dk4 − γ(dfu + gv)k2 + γ2(fugv − fvgu)

Remember that we originally wanted diffusion to drive thesystem unstable; that is we want problem (5) to havesolutions with real(λ) > 0 for some k2.

This can happen if either the coefficient of λ is negative or ifh(k2) < 0 for some k.The former is impossible as we have already requiredfu + gv < 0 (condition I)

Now since fugv − fvgu > 0 (condition II)h(k2) can only possibly be negative if

dfu + gv > 0 (III) ( =⇒ d 6= 1 by I)

λ2 + λ[k2(1 + d)− γ(fu + gv)] + h(k2) = 0 (5)

dfu + gv > 0 (III) ( =⇒ d 6= 1 by I)

λ2 + λ[k2(1 + d)− γ(fu + gv)] + h(k2) = 0 (5)

dfu + gv > 0 (III) ( =⇒ d 6= 1 by I)

λ2 + λ[k2(1 + d)− γ(fu + gv)] + h(k2) = 0 (5)

dfu + gv > 0 (III) ( =⇒ d 6= 1 by I)

Now observe that

h(k2) = d[k4 − γ

d (dfu + gv)k2 + γ2

d (fugv − fvgu)]

= d[(k2 − γ

2d(dfu + gv))2

d (fugv − fvgu)

− γ2

4d2(dfu + gv)

Therefore the minimum value of h occurs atk2 = γ

2d(dfu + gv). This is the wavenumber of the first modeto become unstable.

Consequently hmin = γ2(fugv + fvgu)− γ2

4d (dfu + gv)2

and so for instability we need hmin < 0, i.e.

(dfu + gv)2 > 4d(fugv − fvgu) (IV )

((I) and (III) =⇒ d 6= 1)

Now observe that

h(k2) = d[k4 − γ

d (fugv − fvgu)]

= d[(k2 − γ

2d(dfu + gv))2

d (fugv − fvgu)

− γ2

4d2(dfu + gv)

4d (dfu + gv)2

((I) and (III) =⇒ d 6= 1)

Now observe that

h(k2) = d[k4 − γ

d (fugv − fvgu)]

= d[(k2 − γ

2d(dfu + gv))2

d (fugv − fvgu)

− γ2

4d2(dfu + gv)

4d (dfu + gv)2

((I) and (III) =⇒ d 6= 1)

Thinking of the diffusion constant d as a parameter, we seethat there is a critical value of d = dc which leads toinstability.

By (IV) dc satisfies

(dcfu + gv)2 = 4dc(fugv − fvgu)

and the wavenumber of the first unstable mode satisfies

k2crit =γ

2dc(dcfu + gv) =

2dc2√dc√fugv − fvgu

[fugv − fvgu

Thinking of the diffusion constant d as a parameter, we seethat there is a critical value of d = dc which leads toinstability.

By (IV) dc satisfies

(dcfu + gv)2 = 4dc(fugv − fvgu)

and the wavenumber of the first unstable mode satisfies

k2crit =γ

2dc(dcfu + gv) =

2dc2√dc√fugv − fvgu

[fugv − fvgu

Consider the diffusion-driven activator-inhibitor system

ut =(u2

v − bu)

vt = (u2 − v) + dvxx

For homogeneous steady solutions u = u0, v = v0 with

= bu0, u20 = v0 =⇒ u0 =1

Reaction kinetics terms are

f = u2

v − bu g = u2 − vso

fu = 2uv − b fv = −u2

gu = 2u gv = −1

At the steady state

fu = b fv = −b2 gv =2

bgv = −1

Consider the diffusion-driven activator-inhibitor system

ut =(u2

v − bu)

vt = (u2 − v) + dvxx

For homogeneous steady solutions u = u0, v = v0 with

= bu0, u20 = v0 =⇒ u0 =1

Reaction kinetics terms are

f = u2

v − bu g = u2 − vso

fu = 2uv − b fv = −u2

gu = 2u gv = −1

At the steady state

fu = b fv = −b2 gv =2

bgv = −1

For stability in the absence of diffusion

fu + gv < 0 =⇒ b− 1 < 0 =⇒ b < 1fugv − fugv > 0 =⇒ −b+ 2b > 0 =⇒ b > 0

}therefore 0 < b < 1

Diffusion-driven instability necessitates

dfu + gv > 0 =⇒ db− 1 > 0 =⇒ db > 1

and (dfu + gv)2 > 4d(fugv − fvgu)

=⇒ (db− 1)2 > 4db

if X = db =⇒ (X − 1)2 > 4X =⇒ X2 − 6X + 1 > 0

dfu + gv > 0 =⇒ db− 1 > 0 =⇒ db > 1

=⇒ (db− 1)2 > 4db

if X = db =⇒ (X − 1)2 > 4X =⇒ X2 − 6X + 1 > 0

dfu + gv > 0 =⇒ db− 1 > 0 =⇒ db > 1

=⇒ (db− 1)2 > 4db

if X = db =⇒ (X − 1)2 > 4X =⇒ X2 − 6X + 1 > 0

So (X − 3)2 > 8so X − 3 > 2

or X − 3 < −2√

i.e. db > 3 + 2√

2 or db < 3− 2√

2. But d b > 1.

Hence for diffusion to be unstable we require 0 < b < 1 anddb > 3 + 2

We also know that the critical wavenumber kc satisfies

[fugv − fvgu

But at this point db = 3 + 2√

2 so bd = 3+2

d2and so

[3 + 2

1 +√

So (X − 3)2 > 8so X − 3 > 2

or X − 3 < −2√

i.e. db > 3 + 2√

2 or db < 3− 2√

2. But d b > 1.

Hence for diffusion to be unstable we require 0 < b < 1 anddb > 3 + 2

We also know that the critical wavenumber kc satisfies

[fugv − fvgu

But at this point db = 3 + 2√

2 so bd = 3+2

d2and so

[3 + 2

1 +√

Common Lyapunov functionsStability vs. existence of a CLF

We will use techniques based around common Lyapunovfunctions to test for DDI in reaction diffusion systems.

A matrix A is stable if all eigenvalues have negative real part.

We write P > 0 (P ≥ 0) to indicate a symmetric positive(positive semi-) definite matrix P ; P < 0 (P ≤ 0) means−P > 0 (−P ≥ 0).

Suppose that Ai, i = 1, . . . , n, is a family of n× n stablematrices. Consider the switching system

x = Aix, i ∈ {1, 2, ...n}. (6)

If there exists P > 0 so that

ATi P + PAi ≤ 0, for all i ∈ {1, 2, ...n}, (7)

then the quadratic function V (x) = xTPx is a CommonLyapunov function (CLF) for the switching system (6).

x = Aix, i ∈ {1, 2, ...n}. (6)

Shorten and Narendra derived necessary and sufficientconditions for the existence of a CLF for a finite number ofstable second order systems.

A pair of 2× 2 stable matrices A1 and A2 have a commonLyapunov function with strict inequality in (7) if, and only if,the matrices A1A2

−1 and A1A2 have no negative realeigenvalues.

We will apply the notion of CLF to the stable matrices A and−D where A is the linearised reaction matrix and D is thediffusion matrix in a reaction diffusion system.

Even when one matrix is diagonal, the existence of CLFs issubtle as illustrated by the following example.

Consider the following stable matrices supposed to arise fromlinearising a reaction system around a steady state:

(−1 0.20 −1

), A2 =

(−2 60 −3

), A3 =

(2 −18 −3

)with common diffusion matrix

(1 00 5

The identity is a CLF for A1 and −D but not for A2 and −DA3 and −D have no CLF (since −A3D

−1 has eigenvalues−1,−0.4).

(3 11 5

)is a CLF for A2 and D.

(−1 0.20 −1

), A2 =

(−2 60 −3

), A3 =

(2 −18 −3

(1 00 5

The identity is a CLF for A1 and −D but not for A2 and −D

A3 and −D have no CLF (since −A3D−1 has eigenvalues

−1,−0.4).

(3 11 5

(−1 0.20 −1

), A2 =

(−2 60 −3

), A3 =

(2 −18 −3

(1 00 5

(3 11 5

(−1 0.20 −1

), A2 =

(−2 60 −3

), A3 =

(2 −18 −3

(1 00 5

(3 11 5

DDI, Reactivity and CLFs

The systemdx

dt= Ax

is reactive i.e. exhibits initial growth, when A+AT has apositive eigenvalue, equivalently when I is not a Lypaunovmatrix for A.

When I is a Lyapunov function for A then I is a Lyapunovmatrix for A− k2D and so A−K2D is stable for all k andany diffusion matrix D.

Therefore DDI implies A is reactive (see Neubert, Caswell andMurray) and reactivity is necessary for DDI.

The systemdx

dt= Ax

The systemdx

dt= Ax

We are interested in whether the matrix A− k2D has aneigenvalue with positive real part for some (wave number) k.

When I is a Lypaunov matrix for A (and hence a CLF for Aand −D) then DDI is not possible.

If A and −D have a CLF, then similarly DDI is not possiblesince{ATP + PA < 0 and(−D)P + P (−D) < 0

=⇒ (A−k2D)TP+P (A−k2D)P < 0

for all k.

Whilst simple, this result is very powerful in applications.Why? Testing whether pairs of matrices share a CLF isdetermined using SEMI-DEFINITE PROGRAMMING e.g.using cvx in Matlab - even in high dimensions!

=⇒ (A−k2D)TP+P (A−k2D)P < 0

for all k.

=⇒ (A−k2D)TP+P (A−k2D)P < 0

for all k.

=⇒ (A−k2D)TP+P (A−k2D)P < 0

for all k.

To load cvx: Check out Steve Boyd at Stanford cvx_setup To run cvx: Key in A and D. Then run: cvx_begin sdp variable P(n,n) symmetric A'*P+P*A<zeros(n); -‐D'*P-‐P*D<zeros(n); P>=eye(n); cvx_end

Matlab Code

Gierer-Meinhardt ModelHost-parasite-hyperparasite interactionSchnakenberg diffusion-chemotaxis modelSummary

A simple inhibitor-activator PDE model describing the regenerativegrowth of Hydra:

∂a∂t = ρρ0 + c1ρ

h − µa+ d1∇2xa

∂h∂t = c2ρ

′a2 − νh+ d2∇2

a(x, t) and h(x, t) are the activator and inhibitor concentrations atposition x and time t.The parameters

Set 1: c1 = 0.005, c2 = 0.035, d1 = 0.03, d2 = 0.45,ρ′

= 0.075 and ρ = ρ0 = 3.2.

Set 2: c1 = 0.05, c2 = 0.025, d1 = 0.03, d2 = 0.45,ρ′

= 0.00075 and ρ = ρ0 = 3.2

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

I is a CLF

DDI region

Instability region

there exists a CLF ( not I )

Figure: For Set 1: Parameters (µ, ν) where there is DDI, I is a CLF andthere is a CLF.

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

1.2x 10 4

Instability region

I is a CLF

There exists a CLF ( not I )

Figure: For Set2: Parameters (µ, ν) where I is a CLF and there is a CLF.Here there is no DDI

Here we consider a three dependent variable model (and so thewell known parameter test for DDI does not apply). It describes aspatial model for the interaction of a host (u), a parasite (v) and ahyper-parasite (w).

∂t= τ [u(1− u

K)− uv] + d1∇2u

∂t= τµ[

1 + γ− vw

1 + γv] + d2∇2v

∂t= τd[v − w] + d3∇2w.

The parametersd1 = 0.02, d2 = 0.2, d3 = 1, µ = 15, K = 10 and τ = 1.

0 0.5 1 1.50

There exists a CLFInstability region No CLF

0 5 1040

(0.8,0.8)

0 5 1040

(0.8,2)

In our set up, indeed in the standard Turing instability framework,D is supposed to be diagonal. However, this is not an essentialingredient of our approach and the CLF-based approach stillapplies. This is useful for applications where other “movement”mechanisms are invoked.For example, a common movement mechanism is based on“chemotaxis” which mathematically brings in cross-derivatives.A simple model in which this arises is in the Schnakenbergdiffusion-chemotaxis model:

∂t= γ(a− u+ u2v) +∇2u− α∇.(u∇v),

∂t= γ(b− u2v) + d∇2v

Linearisation around the equilibrium (a+ b, b/(a+ b)2) leads to thereaction-diffusion-chemotaxis matrix pencil:

A− ||ω||2Dα,

( b−aa+b (a+ b)2

−2ba+b −(a+ b)2

), Dα =

(1 −α(a+ b)0 d

bd=100

I is a CLF

NO CLF

There exists a CLF ( not I )

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Figure: The parameter used here are d = 100,γ = 1 and α = 20.

0.2 0.4 0.6 0.8

d=1000

0.2 0.4 0.6 0.8

Figure: To the right of the red curve no pattern formation is possible asthe diffusion coefficient d is increased: d = 10, 25, 100 and 1000.

Using ideas from switched systems we present a strongnecessary condition for DDI or better put a strong sufficientcondition for there not to be DDI

If there exists a common Lyapunov function for A and thediffusion/chemotaxis coefficient matrix D, thendiffusion/chemotaxis driven pattern formation is not possible.

Testing for the existence of CLFs is a semi-definiteprogramming problem which can be solved numerically usingcvx in Matlab.

It provides a more systematic search for parameter domains onwhich pattern formation is possible compared to ad hocmethods based on computing the dispersion relation forspecific parameters

It is a simple, but powerful, generalisation of the “reactivity isnecessary” results of Neubert et al.

Mathematical Biology - Reaction Diffusion Models and Turing … · 2014. 4. 15. · Reaction Di...

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