12/20/2001Systems Dynamics Study Group1 Predator Prey System with a stable periodic orbit 1 st Session - Simple Analysis Systems Dynamics Study Group Ellis

12/20/2001 Systems Dynamics Study Group 1

Predator Prey System with a stable periodic orbit

1st Session - Simple Analysis

Systems Dynamics Study Group

Ellis S. Nolley11/7/2001


Topics• Overview

– Simple Analysis – 1st session, 11/7/2001– Rigorous Analysis – 2nd session, 11/27/2001– Simulation Results – 3rd session, 12/11/2001

• Mathematical Model • Fixed Points• Stable Periodic Orbit

Reference: McGehee & Armstrong, Journal of Differential Equations, 23, 30-52 (1977)


Modelx = amount of prey, y = amount of predator

dx/dt = xg(x) – yp(x)dy/dt = y[-s + cp(x)]

g(x) is a growth function, g(x), monotonic non-increasing, dg(x)/dx <=0, g(0)>0

p(x) is predation functionp(x), monotonic increasing, dp(x)/dx >0 , p(0)=0

g(x)

x

g(x)

k


Fixed Points3 Fixed points: (x*,y*), (0,0), (k,0)

(x*,y*)dy/dt = 0, dx/dt = 0 for (x*,y*)At dy/dt=0, y>0, then p(x*) = s/c, y*=x*g(x*)/p(x*)

Assume Lim p(x) = a, as x-> inf+1) x* > s/c, otherwise there is no fixed point2) y* > 0, in order to have a system3) If there is a k, g(k)=0, then x* <k,

So, we have a fixed point, (x*,y*), x*>0, y*>0



Fixed Points (Cont’d)

Let’s look at the slope on x=k

dy/dx = (dy/dt)/(dx/dt)At x=k, g(k)=0 Recall: -s+cp(x*)=0, p’(x)>0, x*<kdy/dx (k) = y[-s+cp(k)]/[-yp(k)]

= [-s+cp(k)] numerator >0 -p(k) denominator <0

dy/dx <0, slope is negativeSince, -s+cp(x) >0, then delta y>0 (numerator)So, the vectors are coming in.

kx

y

a(x*,y*)



Analysis at Fixed Points

(0,0)

What happens at x=0 (y axis)?

dy/dt= y(-s) <0

At y=0, (x axis),

dx/dt=xg(x)>0

So, (0,0) is a saddle point. x

y

(0,0)



Analysis at Carrying Capacity

(k,0)At (k,0), g(k) = 0dx/dt = xg(x) – yp(x)

= xg(x)g(x) is monotonic non-increasing.For x<k, g(x) >0For x>k, g(x)<0

From p. 13, at x=k, [-s+cp(x)] > 0So dy/dt = y[-s + cp(x)] > 0 for y>0, x=k

(k,0) is a saddle point kx

y



Prey Isocline

At the prey isocline, dx/dt = 0

y= xg(x)/p(x)

and goes through (k,0) and (x*,y*).

To find y(0): by L’Hospital’s Rule,

y(0) = [x g’(0) + g(0)]/p’(0) = g(0)/p’(0) > 0


kx

y

(x*,y*)

y(0)=g(0)/p’(0)

Recall: L’Hospitals Rule: if f(x) & g(x) both go to either 0 or infinity as x->a,Then lim f(x)/g(x)] = lim [df(x)/dx]/[dg(x)/dx], as x-> a


The Vector SpaceSince delta x<0 on y axis

then delta x<0 near y axis.Since delta x>0 near x=k,

then vector is up near x=kVectors can only turn around at the critical pt.

At x=x* above y*, dx/dt<0At x=x* below y*, dx/dt>0Left of x*, dy/dt<0 because p is an increasing function & crosses zero at p(x*)Right of x*, dy/dt > 0

(x*,y*) is unstable if tangent is positive.Pick a line tangent to dy/dx at (k,x**)

All vectors cross it inward


kx

y

(k,x**)

(x*,y*)


Periodic Orbit

• Fixed points are unstable.• All vectors enter the region

and move away from the boundary.• Stable periodic orbit exists around

the unstable fixed point.


kx

y

(k,x**)

(x*,y*)


Next Session

Simple Mathematics – 1st session

Rigorous Mathematics – 2nd session, 11/27

Simulation Results – 3rd session



Thank you!


Predator Prey System with a stable periodic orbit

2nd Session - Rigorous Analysis


Ellis S. Nolley

11/27/2001


Topics• Overview


• Mathematical Model • Fixed Points & Eigenvalues• Poincare-Bendixon Theorem

4 key slides: #23 – 26


References

• McGehee & Armstrong, Journal of Differential Equations, 23, 30-52 (1977)• Morris Hirsch & Stephen Smale, Differential Equations, Dynamical Systems

and Linear Algebra, 1974, Academic PressCh 3-5, Linear Systems, Eigenvalues & Exponentials of OperatorsCh 9-12, Stability, Differential Equations on Electrical Systems, Poincare-Bendixon

Theorem, Ecology• Michael Spivak, Calculus on Manifolds, 1965, W.A Benjamin• Raghavan Narasimhan, Analysis on Real & Complex Manifolds, 1968, North-

Holland Publishing Company


Where to find these References

Mathematics Library, Vincent Hall, 3rd Floor, University of MN

Vincent Hall, 206 Church Street, Mpls, MN 55455

http://onestop.umn.edu/Maps/VinH/VinH-map.html


Modelx = amount of prey, y = amount of predator


g(x) is a growth function, g(x), monotonic non-increasing, dg(x)/dx <=0, g(0)>0

p(x) is predation functionp(x), monotonic increasing, dp(x)/dx >0 , p(0)=0

g(x)

x

g(x)

k


Jacobian & Eigenvalue Reviewdx/dt = xg(x) – yp(x)dy/dt = y[-s + cp(x)]

z’(t) = f(z) = [f1(z1,…zn), …, fn(z1…,zn)]

Note above: z1=x, z2=y, f1(z)=xg(x)-yp(x), f2(z)=y[-s+cp(x)]

dF(z,t)/dt = f(z); F(n)(z)=dnF(z)/dzn,n=0, … ∞; F(0)(z)=F(z)

If z є B(z0,ε) ={z|z-z0|<ε}, then the Taylor Series is:

F(z) = k=0∞Σ F(k)(z0)(z-z0)k/k! = F(z0)+ k=0

∞Σ f(k)(z0)(z-z0)k+1/(k+1)!

where f(k)(z0) = [∂kf1(z1,..,zn)/∂z1k, … , ∂kf1(z1,..,zn)/∂zn

k] | … | (z0,1,…,z0,n)

[∂kfn(z1,..,zn)/∂z1k, … , ∂kfn(z1,..,zn)/∂zn

k]

f(k)(z0) is the kth derivative of f(z0), f(1)(z0) is the Jacobian



Eigenvalues determine stability

If origin, is a fixed point, 0=(01, … ,0n)

then F(0)=0, f(0)=0Note, if z0 is a fixed point of f(z),

f*(z) = f(z+z0)-z0 has 0 as fixed point.

dz/dt=f(z), eigenvalues λ are solution of

det [f(1)(z)-λI]=0 evaluated at fixed point z0

where I is identity matrix.


dz/dt = f(z)

x

y

(0,0)

http://www.watchofthelord.com/articles/stability-art-hdr-pg.jpg


Eigenvalues (Cont’d)f(1)(z0) = [g(x0)+x0g’(x0)-y0p’(x0), p(x0)]

[cy0p’(x0), -s+cp(x0)]

det [f(1)(z0)-λI] = 0

= det [g(x0)+x0g’(x0)-y0p’(x0)-λ, p(x0)]

[cy0p’(x0), -s+cp(x0)-λ]

(g(x0)+x0g’(x0)-y0p’(x0)-λ) (-s+cp(x0)-λ) – cy0p’(x0)p(x0) = 0

z0 is stable if max (Re(λk), k=1, … , n) < 0

z0 is unstable if max (Re(λk), k=1, … , n) > 0


dz/dt = f(z)


Why Re λ determines stability

z’ = f(z); f(z0)=0, z0 fixed point, λk eigenvalues.

Suppose λj has Re λj >0. Pick z close to z0

f(z) = k=0∞Σ f(k)(z0)(z-z0)k/k! = f(z0) + f(1)(z0)(z-z0) + … Taylor Series

~ f(1)(z0) (z-z0) = (z-z0)Σckλk ; d f(z)/z ~ Σckλk dt

ln f(z) ~ Σckλkt; f(z) ~ c*eΣλkt

|f(z)| ~ |c*| |eλjt| |eΣλkt|; λ = Re λ + i Im λ ; |ei w|= |Cos(Im w) + i Sin(Im w)| = 1

lim |f(z)| ~ lim(|c*| |eRe(λj)t | |eΣλkt|) as t-> ∞

Then, |eRe(λj)t | -> large because Re λj >0 So, z0 is an unstable fixed point.

If all Re λj <0, then lim |f(z)| ->0 as t-> ∞ So, z0 is a stable fixed point.


dz/dt = f(z)


Fixed Points

dx/dt = xg(x) – yp(x) = 0

dy/dt = y[-s + cp(x)] = 0

1. (0,0), p(0) = 0

2. (k,0), g(k) = 0

3. (x*,y*), 0<x*<k, y*>0


dz/dt = f(z)

det [f(1)(z0)-λI] =(g(x0)+x0g’(x0)-y0p’(x0)-λ) (-s+cp(x0)-λ)

– cy0p’(x0)p(x0) = 0

kx

y

(x*,y*)


(0,0)


x0=y0=p(x0)=0

(g(0)-λ)(-s-λ)=0

λ=g(0),-s;

λ1= g(0) > 0, corresponds to x axis

λ2= -s < 0 , corresponds to y axis


dz/dt = f(z)


– cy0p’(x0)p(x0) = 0

x

y

(0,0)

(0,0) is unstable


(k,0)


Note: x0=k, y0=g(k)=0

(kg’(k)-λ)(-s+cp(k)-λ)=0

λ=kg’(k), -s+cp(k); recall g’(x) < 0,

Note: -s+cp(x*)=0, x*<k, p’(x) > 0, p(x*) < p(k)

-s+cp(k) > 0

λ1= kg’(k) < 0, corresponds to x axis

λ2= -sp(k) > 0, corresponds to y axis


dz/dt = f(z)


– cy0p’(x0)p(x0) = 0

(k,0) is unstablek

x

y


(x*,y*)(g(x0)+x0g’(x0)-y0p’(x0)-λ) (-s+cp(x0)-λ) – cy0p’(x0)p(x0) = 0

Note: -s+cp(x0)=0; x0=x*, y0=y*(g(x*)+x*g’(x*)-y*p’(x*)-λ)(-λ) – cy*p’(x*)p(x*) = 0 λ2 - [g(x*)+x*g’(x*)-y*p’(x*)]λ – cy*p’(x*)p(x*) = 0

B C > 0 λ = (B +/– sqrt(B2 + 4C))/2

Note: slope of prey isocline, (dy/dt) at (x*,y*) = d(dx/dt)dx = g(x)+xg’(x)-yp’(x) = B

If B > 0, (x*,y*) is unstable λ1 = [B – sqrt(B2 + 4C)]/2 < 0 λ2 = [B + sqrt(B2 + 4C)]/2 > 0

If B < 0, (x*,y*) is stable. λ1 = [B – sqrt(B2 + 4C)]/2 < 0 λ2 = [B + sqrt(B2 + 4C)]/2 < 0


dz/dt = f(z)


– cy0p’(x0)p(x0) = 0

(k,x**)

kx

y

(x*,y*)

B > 0


Poincaré-Bendixon

Theorem: A nonempty compact limit set of a C1 planar dynamical system, which contains no equilibrium point, is a closed orbit.

compact limit set – The limit of a closed bounded set when mapped through time. Since it is the limit set, it is stable.

C1 – has a continuous first derivative


kx

y

(x*,y*)

B > 0


Poincaré-Bendixon Rationale F(z+,t1)=z+

1=(x1,y1)

F(z+,t2)=z+2=(x2,y2)

lim z+k -> z, as k->∞

F(z–,t1)=z–1=(x1,y1)

F(z–,t2)=z–2=(x2,y2)

lim F(z–k) -> z-, as k->∞

z- <= z (perhaps more than one periodic orbit?)


x

y

B > 0

Z+1

Z+2

Z–2

Z–1

Z


Summary

dx/dt = xg(x) – yp(x)dy/dt = y[-s + cp(x)], s>0, c>0

g(x) is a growth function, g(x), monotonic non-increasing, g’(x) <=0, g(0)>0, g(k)=0

p(x) is predation functionp(x), monotonic increasing, p’(x) >0 , p(0)=0

(x*,y*) fixed point, x*>0, y*,>0 => x*g(x*)-y*p(x*)=0; -s+cp(x*)=0B = g(x*)+ x*g’(x*)-yp’(x*) > 0

Then, the dynamical system has a stable periodic orbit.

kx

y

(x*,y*)

B > 0


Thank you!


Predator-Prey System with a stable periodic orbit


3rd Session – Simulation Results

Ellis S. Nolley

12/20/2001


References

• McGehee & Armstrong, Journal of Differential Equations, 23, 30-52 (1977)

• Vensim ® PLE software (free for educational use) www.vensim.com/download.html

• Vensim Tutorial by Craig Kirkwood, Arizona State University

www.public.asu.edu/~kirkwood/sysdyn/SDRes.htm

• Vensim User Guide

www.vensim.com/ffiles/venple.pdf


Topics• Overview


• Model Parameters• Simulation Results• Vensim Techniques• Bifurcation

• Extra: Mathematics of Parameter Selection


Model

dx/dt = xg(x) – yp(x)dy/dt = y[-s + cp(x)], s>0, c>0

g(x) is a growth function, g(x), monotonic non-increasing, g’(x) <=0, g(0)>0, g(k)=0

p(x) is predation functionp(x), monotonic increasing, p’(x) >0 , p(0)=0

(x*,y*) fixed point, x*>0, y*,>0 => x*g(x*)-y*p(x*)=0; -s+cp(x*)=0B = g(x*)+ x*g’(x*)-yp’(x*) > 0

Then, the dynamial system has a stable periodic orbit.

kx

y

(x*,y*)

B > 0


Model Parametersg(x) = a0+a1x, x*~147.4, a0=54, a1= -0.15 p(x) = b ln(x+1), b=4, s=200, c=10 g(0) = 54>0, g’(x)= -0.15<0p(0) = 0, p’(x) = b/(x+1)>0

B= g(x)+xg’(x)-xg(x)p’(x)/p(x), for x=x*, p(x*)=s/c ~ 54+2(-0.15)147.4+4(1)[54-0.15(147.4)]/(200/10)

since x/(x+1) ~ 1~ 54 - 44.2 - 6.4 = 3.4 > 0


g(0)>0, g’(x)<0p(0)=0, p’(x)>0, B>0


Inside Orbit

Time Series of first 3 Periods

020406080100120

0 0.5 1 1.5 2 2.5

TimeA

mo

un

t X

& Y

x

y

X & Y Log Inside: time 0 - 40

0

100

200

300

400

500

0 100 200 300 400

X - Prey

Y -

Pre

da

tor


Outside OrbitX & Y Log Outside: time 0 - 40

0

100

200

300

400

500

0 100 200 300 400

X - Prey

Y -

Pre

da

tor

Time Series of first 2 Periods

0100200300400500

0

0.08

0.16

0.24

0.32 0.4

0.48

0.56

0.64

0.72 0.8

0.88

0.96

Time

Am

ou

nt

X &

Y

x

y


Inside & Outside Orbits

X & Y Log Outside: time 0 - 40

0

100

200

300

400

500

0 100 200 300 400

X - Prey

Y -

Pre

da

tor

X & Y Log Inside: time 0 - 40

0

100

200

300

400

500

0 100 200 300 400

X - Prey

Y -

Pre

da

tor


Combined Inside & Outside Orbits


0

100

200

300

400

500

0 100 200 300 400

X - Prey

Y -

Pre

da

tor

0

100

200

300

400

500

0 100 200 300 400


Vensim Model Layout dx/dt = xg(x) – yp(x)dy/dt = y[-s + cp(x)]


g(x)


p(x)


dx/dt


x


dy/dt


y


Other Vensim Techniques

• Select Runge Kutta Integration (RK4).• Select initial points (x,y)=(1,1) for an outside orbit

and (x,y)=(125,200) for an inside orbit.• Select 0.005 for a step size in Model/Settings• Select a custom graph/table to export to Excel

– Control Panel, Graphs, New, Name title, select variables x & y, click on scale between them

– Click on As Table, click on “running down”– Click on Ok, close


Run and Export Text File

• Click on Run Simulation

• Click on Control Panel

• Click on graph name, click on Display

• Click on File, then Save As


Import Text File into Excel• Run Excel• Click Open, select txt type, select file, click Open, Finish.• Click on Chart Wizard, XY (scatter), click on Data

Source icon (to right of data range), click and drag over x & y data, click on Data Source icon, complete the chart.

• Create a time series chart using t,x,y data the same way as above, dragging over several periods of data.

• Then, alter step size and initial points in Vensim• Create other charts for parameter changes by edit/copy

sheet, run new simulation & copy/paste simulation data onto new sheet’s data region


ExampleExcelResult


Bifurcationdx/dt = xg(x) – yp(x)dy/dt = y[-s + cp(x)]

g(0)>0, g’(x)<0p(0)=0, p’(x)>0, B>0



0

100

200

300

400

500

0 100 200 300 400

X - Prey

Y -

Pre

da

tor

a0=54


050100150200250300350400

0 100 200 300 400

X - Prey

Y -

Pre

da

tor

a0=49.7X & Y Log Outside: time 0 - 40

0

50

100

150

200

250

0 100 200 300

X - Prey

Y -

Pre

da

tor

a0=40


0

100

200

300

400

500

0 100 200 300 400

X - Prey

Y -

Pre

da

tor

a0=51


Projection of Stable Attractor onto X Axis

x

a0

~ 147.4

~ 49.7

Actual boundary shape is not described

~ 22.1


Summary• Generalized Lotka-Volterra Predator-Prey Model

• Internal Fixed Point (x*,y*)– Stable when B<0

– Unstable, surrounded by Stable Periodic Orbit when B>0

• Existence Proof

• Simulation Results

• Vensim techniques

• Bifurcation


g(0)>0, g’(x)<0p(0)=0, p’(x)>0

kx

y

(x*,y*)

B > 0


Another Reference

• C. Neuhauser, “Mathematical Challenges in Spatial Ecology,” Notices of the American Mathematical Society, 48, 1304-1314 (Dec 2001)

http://www.ams.org/notices/200111/fea-neuhauser.pdf

University of Minnesota, EEB dept of CBS


Predator-Prey models CompetitionTypical Competition Beliefs

• Survival of the fittest• Competition develops excellence• Diversity increases stability• Complexity decreases stability• One competitor per niche• Good designs stabilize desirable behavior and

destabilize undesirable behavior

What are likely outcomes of well defined systems?

What systems produce specific outcomes?


Thank you!


Extra

Mathematics of Parameter Selection


PolynomialsRecall that any continuous function within a closed bounded region can be uniformly

approximated by polynomials. (Stone-Weierstrauss)

Let g(x) ε R(m), p(x) ε R(n) real polynomials of degree m & n

g(x)=0Σmakxk, a0>0, a1<0, am<0, since g(0)>0, g’(x)<0 for x>0

p(x)= 1Σnbkxk, b0=0, b1>0, bn>0, since p(0)=0, p’(x)>0 for x>0B= g(x)+xg’(x)-xg(x)p’(x)/p(x), for x=x* + - -

= 0Σmakxk+x(1Σmkakxk-1)-(0Σmakxk)(1Σnkbkxk)/(1Σnbkxk)

= 0Σmakxk+1Σmkakxk -(0Σmakxk)(1Σnkbkxk)/(1Σnbkxk)

= a0[1-(1Σnkbkxk)/(1Σnbkxk)]+ 1Σmkakxk -(1Σmakxk)(1Σnkbkxk)/(1Σnbkxk)

Note: if aj<0 for all j>0 & bk>0 for all k>0, then (1Σnkbkxk)/(1Σnbkxk)>1, then B<0. Then, no stable periodic orbit exists.

Therefore, if there is a stable periodic orbit,

then aj>0 for some j:1< j <m or bk<0 for some k:1< k <n,

and its polynomial has deg >= 3


g(0)>0, g’(x)<0p(0)=0, p’(x)>0, B>0

http://gamba.math.ubc.ca/coursedoc/math152/lab3/poly1.gif


Log (Ln)

g(x)= 0Σmakxk , a0>0, a1<0, am<0, p(x)=b ln(x + 1),

B= g(x)+xg’(x)-xg(x)p’(x)/p(x), for x=x*

= 0Σmakxk+1Σmkakxk -(0Σmakxk)(xb/(x+1))/(s/c), since p(x*)=s/c

= a0(1- (cb/s)[x/(x+1)])+ 1Σmakxk+1Σmkakxk –(cb/s)[x/(x+1)](1Σmakxk)

= a0(1- (cb/s)[x/(x+1)])+ 1Σmak(1+k-(cb/s)[x/(x+1)])xk

x*= e[s/(bc)] - 1

y*= xg(x)/p(x) = (0Σmakxk+1 )/(s/c) = (0Σmakxk+1 )(c/s)

Let g(x) = a0+a1x, ao>0, a1<0, p(x) = b ln(x +1), p,c>0, p(0)=0

Find x=x*, s/c = b ln(x +1), x*= e[s/(bc)] - 1

y*= (a0x+a1x2 )(c/s)


g(0)>0, g’(x)<0p(0)=0, p’(x)>0, B>0

http://www.dai.ed.ac.uk/HIPR2/pixlogb.gif


Log (Cont’d)B= g(x)+xg’(x)-xg(x)p’(x)/p(x), for x=x*

= a0 + a1x + a1x – x(a0 + a1x)b/[x+1] /(b [ln (x+1)])

= a0 (1-(cb/s)[x/(x+1)]) + 2a1x - [(a0 + a1x)b[x/(x+1)]/(s/c)], since p(x*)=s/c

= a0 (1-(cb/s)[x/(x+1)]) + a1x(2-(cb/s)[x/(x+1)]) > 0,

a0 > - a1x[(2s-cb)[x/(x+1)]/s][s/(s-cb[x/(x+1)])]

a0 > - a1x[(2s-cb[x/(x+1)])/(s-cb[x/(x+1)])] 2s>cb[x/(x+1)] and s>cb[x/(x+1)] => s>cb[x/(x+1)] or 2s<cb[x/(x+1)] and s<cb[x/(x+1)] => s<(cb/2)[x/(x+1)] Selecting Model Parameters1) Select s,c,b so that s > cb>cb[x/(x+1)], since x<x+12) Find x* = e[s/(bc)] - 1

3) Select a0, a1 so that a0 > - a1x[(2s-cb[x/(x+1)])/(s-cb[x/(x+1)])]

4) Verify y* = (a0x+a1x2 )(c/s) >05) Verify that B>0


g(0)>0, g’(x)<0p(0)=0, p’(x)>0, B>0


Log (Cont’d)

Model Parameter Selections1) b=4, c=10, s > bc=40, Let s=2002) x* = e[200/(4*10)] –1 = e 5 –1 ~ 148.4 – 1 = 147.4

3) a0 > - a1x(2s-cb[x/(x+1)])/(s-cb[x/(x+1])

= - a1*147.4(2*200 - 40[1])/(200 - 40[1]) , since [x/(x+1)]~1

= - a1*147.4(360/160),

= - a1(332.7)

let a1= -0.15, a0 > 49.7 , Let a0 > 544) y* = x*g(x*)/p(x*) = 147.4(200-0.15*147.4)/(200/10) =

~ 235.0 > 05) B = g(x) + xg’(x) – xg(x)p’(x)/p(x) =

= (a0+a1x) + 2a1x – b[x/(x+1)] (a0+a1x) = [54 – 0.15(147.4)] + 4(-0.15)[54 – 0.15(147.4)] , since [x/(x+1)]~1 = 54 – 44.2 – 6.4 = 3.4 > 0


g(0)>0, g’(x)<0p(0)=0, p’(x)>0, B>0

http://yak.iuma.com/IUMA/ftp/volume98/LOGARITHM/images/sm-LOGARITHM.gif


Thank you!

Documents

12/20/2001Systems Dynamics Study Group1 Predator Prey System with a stable periodic orbit 1 st Session - Simple Analysis Systems Dynamics Study Group Ellis