mathematical biology - background, key issues · 2014-04-15 · overview of week ahead single...

Overview of week aheadSingle Population Models

Coupled population modelsSummary

Mathematical Biology - Background, Key Issues

Stuart Townley

University of Exeter, UK

March 17, 2014

Stuart Townley Math Biol - Basics 1/ 54

Lecture 1: Mathematical biology for one and two dimensionalmodels

Lecture 2: Population Projection Models - A Lecture by“Crowd Sourcing”Built into this topic is a group research project - Presentationson Friday

Lecture 3: A Feedback Control Approach to NonlinearPopulation Dynamics

Lecture 4: Diffusion Driven Instability and links to SwitchedSystems

Lecture 5: Further and future topics in “Systems Theory forMathematical Biology”

IntroductionContinuous time modelsDiscrete Population Models

Birth, growth, logistic growth

Hysteresis effect

Harvesting

Cob-web models

Let N(t) denote the total population of a species at time t.Then

dt= “Births− deaths + migration”.

Malthus (1798) proposed a simplified model for the right handside with

birth and death proportional to N

dt= αN − β N, for constants α (fecundity), β (mortality).

Separating variables

αN − βN= dt

solving for N by integrating left and right hand sides yields:

N(t) = N0e(α−β)t, for initial N0.

ThenN(t) →∞ if α > β

→ 0 if α < β .

We will return to this simplified view of birth and deathprocesses in a context of multiple age- or stage-structuredpopulations involving also growth between stages (See MBLecture 2).

αN − βN= dt

→ 0 if α < β .

αN − βN= dt

→ 0 if α < β .

αN − βN= dt

→ 0 if α < β .

In this single stage/species model we question thissimplification and introduce self-limited growth, Verhulst(1836):

dt= rN

(1− N

), r,K > 0 (1)

This is called logistic growth; r is called the linear birth rateand K is the carrying capacity.

Separation of variables in (1) and integration yields:

N(t) = N0Kert

K+N0(ert−1)→ K as t→∞

dt= rN

(1− N

), r,K > 0 (1)

N(t) = N0Kert

dt= rN

(1− N

), r,K > 0 (1)

N(t) = N0Kert

The steady states of (1) are N = 0 and N = K. ClearlyN = 0 is unstable, while N = K is exponentially stable.

0 5 10 150

Label the curves accordingto their respective initialconditions

Single species models with generalised population dynamics

A natural extension of the logistic growth model (1) is

dt= f(N) (2)

with nonlinear birth-death-migration dynamics f(N).

Equilibrium populations N∗ are determined from f(N∗) = 0.

We can determine whether N∗ is stable by considering smallperturbations around N∗ so that N(t) = N∗ + n(t) with n(t)small.

dt= f(N) (2)

For N∗ an equilibrium, f(N∗) = 0. Therefore theperturbation n(t) satisfies

n ≈ f ′(N∗)n

so thatn(t) ≈ n0e[f

′(N∗)t]

and, analogously with the linear birth-death model,

n(t)→∞ if f ′(N∗) > 0 and n(t)→ 0 if f ′(N∗) < 0 .

N∗ is unstable if f ′(N∗) > 0 and

exponentially stable if f ′(N∗) < 0.

n ≈ f ′(N∗)n

′(N∗)t]

n ≈ f ′(N∗)n

′(N∗)t]

Define the time-response of the equilibrium population N∗ tobe the interval of time in which the perturbed population n ofthe unstable, resp. stable, linearised dynamics, increases, resp.decreases, by a factor e.

time-response =1

|f ′(N∗)|.

Typically, f(N) = 0 has several solutions, thus the populationdynamics exhibit multiple equilibria. A sketch of the graph off(N) readily determines their stability type simply byinspecting the sign of f ′(N∗).

time-response =1

|f ′(N∗)|.

time-response =1

|f ′(N∗)|.

Modelling an Outbreak of Spruce Budworm

A major problem in Canadian Forests is caused by the SpruceBudworm devouring the foliage of fir trees.

Let N(t) be the number of budworms. A plausible model forN(t) is:

dt= rBN

(1− N

)− p(N) (3)

Here rB is the birth rate,KB is the carrying capacity,p(N) is a nonlinear function modelling predation by birds.Typical characteristics of p(N) are depicted in the graphbelow:

One possibility for p(N) is p(N) = BN2

A2+N2 .

dt= rBN

(1− N

)− p(N) (3)

A2+N2 .

dt= rBN

(1− N

)− p(N) (3)

A2+N2 .

0 2 4 6 8 100

saturation level

To start our analysis we first non-dimensionalise, so as toreduce the number of crucial parameters. In fact with

Aand τ =

we end up with

dτ=ArBB

(1− A

)− u2

1 + u2.

This suggests a rescaling of parameters

A, r =

so yielding a two-parameter model:

dτ= ru

(1− u

)− u2

1 + u2:= f(u; r; q) . (4)

Aand τ =

we end up with

dτ=ArBB

(1− A

)− u2

1 + u2.

A, r =

dτ= ru

(1− u

)− u2

1 + u2:= f(u; r; q) . (4)

Aand τ =

we end up with

dτ=ArBB

(1− A

)− u2

1 + u2.

A, r =

dτ= ru

(1− u

)− u2

1 + u2:= f(u; r; q) . (4)

The equilibria of (4) are given by f(u; r; q) = 0, i.e.

(1− u

1 + u2.

Clearly u = 0 is always a solution.The non-zero equilibria are given by:

(1− u

1 + u2. (5)

How many non-zero equilibria will depend on r and q. This isbest analysed graphically by plotting both the left and righthand sides of (5) on the same graph. This is shown below.

(1− u

1 + u2.

(1− u

1 + u2. (5)

(1− u

1 + u2.

(1− u

1 + u2. (5)

0 1 2 3 4 5 6 7 8 9 100

One non−zero Eqm, r=a

Two non−zero Eqm, r=b, d

Three non−zero Eqm, r=c

r=br=c

We see from this sketch that if q is large enough, then

r < d there is one small stable non-zero equilibrium

r = d there are two non-zero equilibria

d < r < b there are three non-zero equilibria

r = b there are two non-zero equilibria

r > b there is one large stable non-zero equilibrium

As we change the parameter, we get differing numbers ofequilibria and in this example this induces tipping points andhysteresis effect.

We see from this sketch that if q is large enough, then

r < d there is one small stable non-zero equilibrium

r = d there are two non-zero equilibria

d < r < b there are three non-zero equilibria

r = b there are two non-zero equilibria

r > b there is one large stable non-zero equilibrium

As we change the parameter, we get differing numbers ofequilibria and in this example this induces tipping points andhysteresis effect.

non−

d < r < b, three equilibriar < d,one eqm

STABLE

UNSTABLE

STABLE

r > bone eqm

r=d r=b

Harvesting a single natural population

Harvested populations are everywhere - bees, cod (and otherfish species), whale, rabbits, etc.

Harvesting is desirable - as a food resource - and necessary -in order to control species numbers.

Typically we look to maximise sustainable yield whilstminimising harvesting effort.

To illustrate harvesting, consider a model system which,without harvesting, obeys a logistics model with birth rate rand carrying capacity K.

Add to this a harvesting term, hN , proportional to population:

dt= rN

(1− N

)− hN ≡ f(N) . (6)

The steady states are

N = 0, Nh = K

(1− h

)> 0 if r > h .

The steady state Nh gives a yield hNh = hK(1− h

Maximum steady state yield is achieved at h = r/2 (bystandard calculus) giving

maximum yield Ymax = Y (h)|h=r/2 = rK4

at Nh|Ymax = 12K

To illustrate harvesting, consider a model system which,without harvesting, obeys a logistics model with birth rate rand carrying capacity K.Add to this a harvesting term, hN , proportional to population:

dt= rN

(1− N

)− hN ≡ f(N) . (6)

N = 0, Nh = K

(1− h

)> 0 if r > h .

at Nh|Ymax = 12K

dt= rN

(1− N

)− hN ≡ f(N) . (6)

N = 0, Nh = K

(1− h

)> 0 if r > h .

at Nh|Ymax = 12K

dt= rN

(1− N

)− hN ≡ f(N) . (6)

N = 0, Nh = K

(1− h

)> 0 if r > h .

at Nh|Ymax = 12K

Does a dynamical analysis give us anything different?

Linearising (6) about the non-zero equilibrium Nh gives:

dt= (f ′(Nh))n = (h− r)n .

Thus the non-zero equilibrium is linearly stable if h < r andNh is attracting. For n ≈ 0 (i.e. we are close to Nh),

n(t) ≈ exp [(h− r)t] .

This gives a recovery time as a function of h

TR(h) =1

r − hwith relative recovery time

TR(0)=

r − h.

In the case of maximal yield, i.e. h = r/2, this gives

TR(h = r/2)

TR(0)= 2 .

dt= (f ′(Nh))n = (h− r)n .

n(t) ≈ exp [(h− r)t] .

TR(h) =1

TR(0)=

r − h.

TR(h = r/2)

TR(0)= 2 .

dt= (f ′(Nh))n = (h− r)n .

n(t) ≈ exp [(h− r)t] .

TR(h) =1

TR(0)=

r − h.

TR(h = r/2)

TR(0)= 2 .

dt= (f ′(Nh))n = (h− r)n .

n(t) ≈ exp [(h− r)t] .

TR(h) =1

TR(0)=

r − h.

TR(h = r/2)

TR(0)= 2 .

dt= (f ′(Nh))n = (h− r)n .

n(t) ≈ exp [(h− r)t] .

TR(h) =1

TR(0)=

r − h.

TR(h = r/2)

TR(0)= 2 .

In this problem we are interested in yield Y = hN . At thenon-zero harvesting equilibrium N = Nh this means a yield

Y = hK(1− h

Solving for harvesting h in terms of yield Y gives:

h2 − rh+rY

K=⇒ h =

√1− 4Y

]But Ymax = rK

4 . So

√1− Y

]and relative recovery time:

TR(Y )

TR(0)=

1±√

1− YYmax

Y = hK(1− h

h2 − rh+rY

K=⇒ h =

√1− 4Y

But Ymax = rK4 . So

√1− Y

TR(Y )

TR(0)=

1±√

1− YYmax

Y = hK(1− h

h2 − rh+rY

K=⇒ h =

√1− 4Y

]But Ymax = rK

4 . So

√1− Y

TR(Y )

TR(0)=

1±√

1− YYmax

0 0.5 1 1.50

Y/Ymax

Starting with a small harvesting effort h ≈ 0, then Nh ≈ K,yield hNh is small, but the relative recovery time TR(Y )

TR(0) isclose to one and shortest

Then increasing h we follow the L+ branch in the figureabove. As h increases, Nh approaches K/2, the point A onthe graph, corresponding to maximum yield Y = Ymax, andrelative recovery time 2.

Increasing h beyond point A means we are now on the L−branch and the yield Y begins to decrease and recovery timeincreases.

So an optimal harvesting strategy is to choose h below r/2 sothat Nh is close to, but above, K/2. Getting too close toK/2 and with additional harvesting/model uncertainty risksswitching the dynamics onto the L− branch on which yieldthen diminishes.

Many populations exhibit little overlap between successivegenerations (or more generally stage classes , see MBL2s).

In this case it is natural to model populations in discrete timesteps. Typically, the time step is a year, or a few days, or evenmilli-seconds.

Ignoring age- and/or stage-structured effects we arrive atmodels of the form:

Nt+1 = f(Nt) (8)

Different models for f will yield radically different populationdynamics.

Nt+1 = f(Nt) (8)

In general, crowding and self-regulation effects will lead to fhaving a maximum at some critical population density Nm

with f decreasing for n > Nm.

f Comments/Usefulness

rN(1− N

)Max. at N = K

2 , f(N) < 0 if N > K

rN exp[r(1− N

)]f(N) > 0, f decreases N large

In general, crowding and self-regulation effects will lead to fhaving a maximum at some critical population density Nm

with f decreasing for n > Nm.

f Comments/Usefulness

rN(1− N

)Max. at N = K

2 , f(N) < 0 if N > K

rN exp[r(1− N

)]f(N) > 0, f decreases N large

Cob-Webbing

The steady-states, N∗, of (8) satisfy f(N∗) = N∗

0 1 2 3 4 5 6 7 8 9 100

f(N) = N

We can determine the solution Nt, t ≥ 0 using a graphicalCob-Webbing technique:

0 0.5 1 1.5 2 2.5 30

y=f(N)

N0 N1 N2

It is apparent that the slope of the graph N 7→ f(N) plays acrucial role in determining the qualitative dynamics of themodel.

The dynamics are distinguished by whether

f ′(N∗) ∈ (0, 1) (N∗ is stable),

f ′(N∗) ∈ (−1, 0) (N∗ is stable),

f ′(N∗) = −1 (N∗ is oscillatory), or

f ′(N∗) < −1 (N∗ is unstable).

This can be visualised using cob-web diagrams.

f ′(N∗) ∈ (0, 1) (N∗ is stable),

f ′(N∗) ∈ (−1, 0) (N∗ is stable),

f ′(N∗) ∈ (0, 1) (N∗ is stable),

f ′(N∗) ∈ (−1, 0) (N∗ is stable),

f ′(N∗) is a crucial quantity. We call it λ, the eigenvalue ofthe system

Nt+1 = f(Nt).

Ifλ ∈ (−1, 1) N∗ is an attracting eqilibrium,

|λ| > 1 N∗ is an unstable, repelling,

λ = −1 N∗ is a bifurcation value

λ = +1 N∗ is a bifurcation value.

Some extra reading: find out about bifurcations - saddle,pitchfork, ...

Nt+1 = f(Nt).

Chaos in the logistics map.

Setting ut = Nt/K, then the discrete logistics map:

Nt+1 = rNt

(1− Nt

)becomes:

ut+1 = rut(1− ut) .

With f(u) = ru(1− u), f ′(u) = r(1− 2u).

We have steady-states u∗ given by

u∗ = ru∗(1− u∗), i.e. u∗ = 0, u∗ =r − 1

At u∗ = 0, λ = f ′(0) = r. At u∗ = r−1r , with necessarily

r > 1, λ = f ′( r−1r ) = 2− r < 1.

Nt+1 = rNt

(1− Nt

)becomes:

ut+1 = rut(1− ut) .

With f(u) = ru(1− u), f ′(u) = r(1− 2u).

u∗ = ru∗(1− u∗), i.e. u∗ = 0, u∗ =r − 1

r > 1, λ = f ′( r−1r ) = 2− r < 1.

Nt+1 = rNt

(1− Nt

)becomes:

ut+1 = rut(1− ut) .

With f(u) = ru(1− u), f ′(u) = r(1− 2u).

u∗ = ru∗(1− u∗), i.e. u∗ = 0, u∗ =r − 1

r > 1, λ = f ′( r−1r ) = 2− r < 1.

Nt+1 = rNt

(1− Nt

)becomes:

ut+1 = rut(1− ut) .

With f(u) = ru(1− u), f ′(u) = r(1− 2u).

u∗ = ru∗(1− u∗), i.e. u∗ = 0, u∗ =r − 1

r > 1, λ = f ′( r−1r ) = 2− r < 1.

We can analyse this model using r as a bifurcation parameter,starting with small r ≈ 0 and gradually increasing it.

0 < r < 1: u∗ = 0 is a unique equilibrium with λ = r < 1. Sou∗ in this case is STABLE.

r = 1. First bifurcation. u∗ loses its stability and we gain anew non-zero equilbrium u∗ = r−1

1 < r < 3. u∗ = 0 is unstable. u∗ = r−1r , with λ = 2− r, is

stable.

r = 3. Here there is a pitchfork bifurcation. u∗ = r−1r loses

stability. What happens next?

stable.

Let’s seek period 2 solutions - i.e. solutions with

ut+2 = ut, but with ut+1 not necessarily equal to ut

So we need to look for

ut+2 = f(ut+1) = f(f(ut))

= f(rut[1− ut])

= r(rut[1− ut])(1− rut[1− ut])

This iteration has equilibria ut+2 = ut = u∗ given by

u∗ = r2u∗[1− u∗](1− ru∗[1− u∗]) (9)

ut+2 = f(ut+1) = f(f(ut))

= f(rut[1− ut])

= r(rut[1− ut])(1− rut[1− ut])

u∗ = r2u∗[1− u∗](1− ru∗[1− u∗]) (9)

ut+2 = f(ut+1) = f(f(ut))

= f(rut[1− ut])

= r(rut[1− ut])(1− rut[1− ut])

u∗ = r2u∗[1− u∗](1− ru∗[1− u∗]) (9)

The quartic equation

u∗ = r2u∗[1− u∗](1− ru∗[1− u∗])contains the equilibrium solutions u∗ = 0, u∗ = r−1

In fact (9) factorises to give

u∗[ru∗ − (r − 1)][r2(u∗)2 − r(r + 1)u∗ + (r + 1)] = 0

In addition to the equilibria u∗ = 0, and if r > 1, u∗ = r−1r ,

this yields period 2 solutions

u∗ =(r + 1)±

√(r + 1)(r − 3)

2r, valid for r > 3 .

As r passes through r = 3, we develop TWO period 2solutions. Are these stable or not? We need to linearise theperiod two dynamics:

ut+2 = f(f(ut)) = g(ut), with g(u) = r2u[1−u](1−ru[1−u]) .

r .In fact (9) factorises to give

u∗[ru∗ − (r − 1)][r2(u∗)2 − r(r + 1)u∗ + (r + 1)] = 0

u∗ =(r + 1)±

√(r + 1)(r − 3)

u∗[ru∗ − (r − 1)][r2(u∗)2 − r(r + 1)u∗ + (r + 1)] = 0

u∗ =(r + 1)±

√(r + 1)(r − 3)

u∗[ru∗ − (r − 1)][r2(u∗)2 − r(r + 1)u∗ + (r + 1)] = 0

u∗ =(r + 1)±

√(r + 1)(r − 3)

The eigenvalues for these period 2 solutions are given by

λ = g′(u)|u=u∗

One can show (quite easily) that λ ∈ (−1,+1), for3 < r < 1 +

√6 = 3.4495.

For r > 3.4495, we seek stable period 4 solutions, which as rincreases further still break down into period 8, 16, ...solutions.

This period doubling process continues as r increases toproduce stable solutions of higher and higher periods. There isa critical value r ≈ 3.89 beyond which all period 2n solutions,for all n, are unstable. This is the chaotic regime.

λ = g′(u)|u=u∗

√6 = 3.4495.

λ = g′(u)|u=u∗

√6 = 3.4495.

λ = g′(u)|u=u∗

√6 = 3.4495.

IntroductionPredator–prey modelsCompetition and Cooperation

1 Pedator–prey models

2 Competition and mutualism

Revision: Think phase plane analysis

Many species interact. For two species these interactions takeone of three distinct forms:

If the growth rate of one of the species is decreased by the(presence of the) other we call this a PREDATOR-PREYsystem.

If the growth rates of each species are decreased by eachother we call this a COMPETITIVE system.

If the growth rates of each species are enhanced by each otherwe call this a MUTUALISTIC system.

The Lotka-Volterra predator–prey system

In 1926 Lotka/Volterra proposed a simple model to explainthe oscillatory levels of certain fish catches in the Adriatic.If N(t) and P (t) are the populations of the prey and thepredator then

dt= N(a− bP ) dP

dt= P (cN − d) (10)

a, b, c and d are constant parameters and

In the absence of predators, dNdt ∝ N

Predation reduces the prey population growth rate by anamount proportional to both N and P .

The presence of prey increases the population growth rate ofpredators by an amount proportional to N and P .

In the absence of prey, dPdt ∝ P .

dt= N(a− bP ) dP

dt= P (cN − d) (10)

dt= N(a− bP ) dP

dt= P (cN − d) (10)

dt= N(a− bP ) dP

dt= P (cN − d) (10)

dt= N(a− bP ) dP

dt= P (cN − d) (10)

We non-dimensionalize to reduce parameters by writing

u(τ) =c

dN(t) ν(τ) =

aP (t) τ = at α =

This gives

dτ= u(1− v) dv

dτ= αv(u− 1) (11)

The non-zero equilibrium is u = v = 1

In a phase plane dvdu = αv(u−1)

u(1−v) . Separating variable implies

(1− v)v

dv = α(u− 1)

αu+ v − ln(uαv) = H (H is a constant)

For a given H > 1+α these are closed trajectories, see figure.

u(τ) =c

dN(t) ν(τ) =

aP (t) τ = at α =

This gives

dτ= u(1− v) dv

dτ= αv(u− 1) (11)

(1− v)v

dv = α(u− 1)

u(τ) =c

dN(t) ν(τ) =

aP (t) τ = at α =

This gives

dτ= u(1− v) dv

dτ= αv(u− 1) (11)

(1− v)v

dv = α(u− 1)

u(τ) =c

dN(t) ν(τ) =

aP (t) τ = at α =

This gives

dτ= u(1− v) dv

dτ= αv(u− 1) (11)

(1− v)v

dv = α(u− 1)

0 0.5 1 1.5 2 2.5 30

Competition Models

Here two or more species compete for the same limited foodsource or in some way inhibit each others growth.

For example:

dN1dt = r1N1

[1− N1

K1− b12N2

]dN2dt = r2N2

[1− N2

K2− b21N1

] (12)

Here both species grow logistically in the absence of the otherthe positive parameters b12 and b21 indicate the inhibitioneffect

Competition Models

For example:

dN1dt = r1N1

[1− N1

K1− b12N2

]dN2dt = r2N2

[1− N2

K2− b21N1

] (12)

Competition Models

For example:

dN1dt = r1N1

[1− N1

K1− b12N2

]dN2dt = r2N2

[1− N2

K2− b21N1

] (12)

We nondimensionalise by writing

ui =Ni

Ki, τ = r1t, ρ =

α12 =b12K2

K1α21 =

du1dτ = u1[1− u1 − α12u2] = f(u1, u2)

du2dτ = ρu2[1− u2 − α21u1] = g(u1, u2)

The steady states are (exercise)

(u∗1, u∗2) = (0, 0), (1, 0), (0, 1),

(1− α12

1− α12α21,

1− α21

1− α12α21

)The null clines (lines on which f = 0 or g = 0) can bearranged in 4 ways depending on the relative sizes of α12 andα21

ui =Ni

Ki, τ = r1t, ρ =

α12 =b12K2

K1α21 =

du1dτ = u1[1− u1 − α12u2] = f(u1, u2)

du2dτ = ρu2[1− u2 − α21u1] = g(u1, u2)

(u∗1, u∗2) = (0, 0), (1, 0), (0, 1),

(1− α12

1− α12α21,

1− α21

1− α12α21

ui =Ni

Ki, τ = r1t, ρ =

α12 =b12K2

K1α21 =

du1dτ = u1[1− u1 − α12u2] = f(u1, u2)

du2dτ = ρu2[1− u2 − α21u1] = g(u1, u2)

(u∗1, u∗2) = (0, 0), (1, 0), (0, 1),

(1− α12

1− α12α21,

1− α21

1− α12α21

s - stable equilibrium, u - unstable equilibrium.(a) Top Left α12, α21 > 1 (b) Top Right α12, α21 < 1 (c) BottomLeft α12 < 1, α21 > 1 (d) Bottom Right α12 > 1, α21 < 1

In cases c) and d) we have no co-existent steady states, (i.e.when an equilibrium (u∗1, u

∗2) has u∗1 and u∗2 both non-zero)

If dudt = f(u) then to look at the stability of the various states

we need to look at the community matrix

(∂f∂u1

∂f∂u2

∂g∂u1

∂g∂u2

(1− 2u1 − α12u2 −α12u1−ρα21u2 ρ(1− 2u2 − α21u1)

In cases c) and d) we have no co-existent steady states, (i.e.when an equilibrium (u∗1, u

∗2) has u∗1 and u∗2 both non-zero)

If dudt = f(u) then to look at the stability of the various states

we need to look at the community matrix

(∂f∂u1

∂f∂u2

∂g∂u1

∂g∂u2

(1− 2u1 − α12u2 −α12u1−ρα21u2 ρ(1− 2u2 − α21u1)

(0, 0). Now A =

(1 00 ρ

)=⇒ λ = 1, ρ. (0, 0) is an

UNSTABLE NODE

(1, 0). Now A =

(−1 −α12

0 ρ(1− α21)

)so λ = −1, ρ(1− α21)

(1, 0) is an STABLE NODE if α21 > 1(1, 0) is an UNSTABLE SADDLE if α21 < 1

(0, 1). Now A =

(1− α12 0−ρα −ρ

)=⇒ λ = −ρ, (1− α12)

(0, 1) is a STABLE NODE if α12 > 1(0, 1) is an UNSTABLE SADDLE if α12 < 1

(0, 0). Now A =

(1 00 ρ

)=⇒ λ = 1, ρ. (0, 0) is an

UNSTABLE NODE

(1, 0). Now A =

(−1 −α12

0 ρ(1− α21)

)so λ = −1, ρ(1− α21)

(0, 1). Now A =

(1− α12 0−ρα −ρ

)=⇒ λ = −ρ, (1− α12)

(0, 0). Now A =

(1 00 ρ

)=⇒ λ = 1, ρ. (0, 0) is an

UNSTABLE NODE

(1, 0). Now A =

(−1 −α12

0 ρ(1− α21)

)so λ = −1, ρ(1− α21)

(0, 1). Now A =

(1− α12 0−ρα −ρ

)=⇒ λ = −ρ, (1− α12)

For the equilibrium:(

1−α121−α12α21

, 1−α211−α12α21

). Needs

α12, α21 < 1 or α21, α12 > 1

(−u∗1 −α12u

−ρα21u∗2 −ρu∗2

Trace(A) = −u∗1 − ρu∗2 = λ1 + λ2 < 0Det(A) = ρu∗1u

∗2 − ρα12α21u)1

∗u∗2= ρu∗1u

∗2(1− α12α21)

= λ1λ2

if α12α21 < 1 Re(λ) < 0 =⇒ STABLEif α12α21 > 1 λ ∈ R with λ1 < 0 < λ2 =⇒ UNSTABLESADDLE

1−α121−α12α21

, 1−α211−α12α21

). Needs

α12, α21 < 1 or α21, α12 > 1

(−u∗1 −α12u

−ρα21u∗2 −ρu∗2

∗2 − ρα12α21u)1

∗u∗2= ρu∗1u

∗2(1− α12α21)

= λ1λ2

1−α121−α12α21

, 1−α211−α12α21

). Needs

α12, α21 < 1 or α21, α12 > 1

(−u∗1 −α12u

−ρα21u∗2 −ρu∗2

∗2 − ρα12α21u)1

∗u∗2= ρu∗1u

∗2(1− α12α21)

= λ1λ2

1−α121−α12α21

, 1−α211−α12α21

). Needs

α12, α21 < 1 or α21, α12 > 1

(−u∗1 −α12u

−ρα21u∗2 −ρu∗2

∗2 − ρα12α21u)1

∗u∗2= ρu∗1u

∗2(1− α12α21)

= λ1λ2

The fact that the stable steady state is either (0, 1) or (1, 0)illustrates the principle of competitive exclusion

If two species compete according to (12), then one of thembecomes extinct.

The fact that the stable steady state is either (0, 1) or (1, 0)illustrates the principle of competitive exclusion

If two species compete according to (12), then one of thembecomes extinct.

Mutualism

There are many real-life situations where the interaction oftwo or more species is to the advantage of all.

A typical model is

dN1dt = r1N1

(1− N1

K1+ b12

)dN2dt = r2N2

(1− N2

K2+ b21

)The +N1N2 in both RHS models that both species increasein the presence of the other.

Let ui =NiKi

τ = r1t ρ = r2r1

α12 =b12K2K1

α21 =b21K1K2

thendu1dτ = u1(1− u1 + α12u2)du2dτ = ρu2(1− u2 + α21u1)

Mutualism

A typical model is

dN1dt = r1N1

(1− N1

K1+ b12

)dN2dt = r2N2

(1− N2

K2+ b21

Let ui =NiKi

τ = r1t ρ = r2r1

α12 =b12K2K1

α21 =b21K1K2

Mutualism

A typical model is

dN1dt = r1N1

(1− N1

K1+ b12

)dN2dt = r2N2

(1− N2

K2+ b21

Let ui =NiKi

τ = r1t ρ = r2r1

α12 =b12K2K1

α21 =b21K1K2

(0, 0), (1, 0), (0, 1) and

(1 + α12

1− α12α21,

1 + α21

1− α12α21

)(if α12α21 < 1)

Community matrix

[1− 2u1 + α12u2 α12u1

ρα21u2 ρ(1− 2u2 + α21u1)

(0, 0), (1, 0), (0, 1) and

(1 + α12

1− α12α21,

1 + α21

1− α12α21

)(if α12α21 < 1)

Community matrix

[1− 2u1 + α12u2 α12u1

ρα21u2 ρ(1− 2u2 + α21u1)

(0, 0) A =

(1 00 ρ

)λ = 1, ρ. (0, 0) is an UNSTABLE

(1, 0) A =

(−1 α12

0 ρ(1 + α21)

)λ = −1, ρ(1 + α21)

(1, 0) is an UNSTABLE SADDLE

(0, 1) A =

(1 + α12 0ρα21 −ρ

)λ = −ρ, 1 + α12 (0, 1) is an

UNSTABLE SADDLE

(u∗1, u∗2) =

(1+α12

1−α12α21, 1+α211−α12α21

)=⇒ A =

(−u∗1 α12u

ρα12u∗2 −ρu∗2

)traceA < 0, det(A) = ρ(1− α12α21)u

∗1u∗2 > 0

so when it exists it is STABLE

(0, 0) A =

(1 00 ρ

(1, 0) A =

(−1 α12

0 ρ(1 + α21)

)λ = −1, ρ(1 + α21)

(0, 1) A =

(1 + α12 0ρα21 −ρ

)λ = −ρ, 1 + α12 (0, 1) is an

UNSTABLE SADDLE

(u∗1, u∗2) =

(1+α12

1−α12α21, 1+α211−α12α21

)=⇒ A =

(−u∗1 α12u

ρα12u∗2 −ρu∗2

∗1u∗2 > 0

(0, 0) A =

(1 00 ρ

(1, 0) A =

(−1 α12

0 ρ(1 + α21)

)λ = −1, ρ(1 + α21)

(0, 1) A =

(1 + α12 0ρα21 −ρ

)λ = −ρ, 1 + α12 (0, 1) is an

UNSTABLE SADDLE

(u∗1, u∗2) =

(1+α12

1−α12α21, 1+α211−α12α21

)=⇒ A =

(−u∗1 α12u

ρα12u∗2 −ρu∗2

∗1u∗2 > 0

(0, 0) A =

(1 00 ρ

(1, 0) A =

(−1 α12

0 ρ(1 + α21)

)λ = −1, ρ(1 + α21)

(0, 1) A =

(1 + α12 0ρα21 −ρ

)λ = −ρ, 1 + α12 (0, 1) is an

UNSTABLE SADDLE

(u∗1, u∗2) =

(1+α12

1−α12α21, 1+α211−α12α21

)=⇒ A =

(−u∗1 α12u

ρα12u∗2 −ρu∗2

∗1u∗2 > 0

u - unstable equilibrium s - stable equilibrium

For the mutualistic model

du1dτ = u1(1− u1 + α12u2)

du2dτ = ρu2(1− u2 + α21u1)

if α12α21 > 1, all equilibrium points unstable and unboundedgrowth in each species.

if α12α21 < 1 new steady state is stable and system evolves toa co-existent state.

du1dτ = u1(1− u1 + α12u2)

du2dτ = ρu2(1− u2 + α21u1)

du1dτ = u1(1− u1 + α12u2)

du2dτ = ρu2(1− u2 + α21u1)

single (scalar) and two species models

continuous and discrete time

nonlinearity, i.e density dependence

equilibria and local stability

harvesting, i.e. control

Overview of week

X Lecture 1: Mathematical biology for one and two dimensionalmodels

Lecture 2: Population Projection Models - A Lecture by“Crowd Sourcing” Built into this topic is a group researchproject - Presentations on Friday

mathematical biology - background, key issues · 2014-04-15 · overview of week ahead single...

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