Predator Prey System
with a stable periodic orbit
1st Session - Simple Analysis
Systems Dynamics Study Group
Ellis S. Nolley
11/7/2001
12/20/2001
Systems Dynamics Study Group
1
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)
12/20/2001
Systems Dynamics Study Group
2
Model
x = amount of prey,
y = amount of predator
g(x)
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
g(x)
k
g(x) is a growth function,
g(x), monotonic non-increasing, dg(x)/dx <=0, g(0)>0
p(x) is predation function
p(x), monotonic increasing, dp(x)/dx >0 , p(0)=0
12/20/2001
Systems Dynamics Study Group
3
x
Fixed Points
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
3 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 point
2) y* > 0, in order to have a system
3) If there is a k, g(k)=0, then x* <k,
So, we have a fixed point, (x*,y*), x*>0, y*>0
12/20/2001
Systems Dynamics Study Group
4
Fixed Points (Cont’d)
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
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*<k
dy/dx (k) = y[-s+cp(k)]/[-yp(k)]
= [-s+cp(k)]
numerator >0
-p(k)
denominator <0
dy/dx <0, slope is negative
Since, -s+cp(x) >0, then delta y>0 (numerator)
So, the vectors are coming in.
12/20/2001
Systems Dynamics Study Group
y
a
(x*,y*)
x
k
5
Analysis at Fixed Points
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
(0,0)
What happens at x=0 (y axis)?
dy/dt= y(-s) <0
At y=0, (x axis),
dx/dt=xg(x)>0
y
So, (0,0) is a saddle point.
(0,0)
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Systems Dynamics Study Group
x
6
Analysis at
Carrying Capacity
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
(k,0)
At (k,0), g(k) = 0
dx/dt = xg(x) – yp(x)
= xg(x)
g(x) is monotonic non-increasing.
For x<k, g(x) >0
For x>k, g(x)<0
y
From p. 13, at x=k,
[-s+cp(x)] > 0
So dy/dt = y[-s + cp(x)] > 0 for y>0, x=k
x
(k,0) is a saddle point
12/20/2001
Systems Dynamics Study Group
k
7
Prey Isocline
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
y
At the prey isocline, dx/dt = 0
y(0)=g(0)/p’(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
(x*,y*)
x
k
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
12/20/2001
Systems Dynamics Study Group
8
The Vector Space
Since 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=k
Vectors can only turn around at the critical pt.
At x=x* above y*, dx/dt<0
At x=x* below y*, dx/dt>0
Left 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
12/20/2001
Systems Dynamics Study Group
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
y
(k,x**)
(x*,y*)
k
x
9
Periodic Orbit
y
• Fixed points are unstable.
• All vectors enter the region
and move away from the boundary.
• Stable periodic orbit exists around
the unstable fixed point.
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
(k,x**)
(x*,y*)
x
k
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Systems Dynamics Study Group
10
Next Session
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
Simple Mathematics – 1st session
Rigorous Mathematics – 2nd session, 11/27
Simulation Results – 3rd session
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Systems Dynamics Study Group
11
Thank you!
12/20/2001
Systems Dynamics Study Group
12
Predator Prey System
with a stable periodic orbit
2nd Session - Rigorous Analysis
Systems Dynamics Study Group
Ellis S. Nolley
11/27/2001
12/20/2001
Systems Dynamics Study Group
13
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 & Eigenvalues
• Poincare-Bendixon Theorem
4 key slides: #23 – 26
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14
References
• McGehee & Armstrong, Journal of Differential Equations, 23, 30-52 (1977)
• Morris Hirsch & Stephen Smale, Differential Equations, Dynamical Systems
and Linear Algebra, 1974, Academic Press
Ch 3-5, Linear Systems, Eigenvalues & Exponentials of Operators
Ch 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, NorthHolland Publishing Company
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15
Where to find
these References
Vincent Hall,
206 Church Street,
Mpls, MN 55455
Mathematics Library,
Vincent Hall,
3rd Floor,
University of MN
http://onestop.umn.edu/Maps/VinH/VinH-map.html
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16
Model
x = amount of prey,
y = amount of predator
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
g(x)
g(x) is a growth function,
g(x), monotonic non-increasing, dg(x)/dx <=0, g(0)>0
p(x) is predation function
p(x), monotonic increasing, dp(x)/dx >0 , p(0)=0
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Systems Dynamics Study Group
g(x)
k
17
x
Jacobian & Eigenvalue Review
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
dx/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)/∂znk]
|
…
| (z0,1,…,z0,n)
[∂kfn(z1,..,zn)/∂z1k, … , ∂kfn(z1,..,zn)/∂znk]
f(k)(z0) is the kth derivative of f(z0), f(1)(z0) is the Jacobian
12/20/2001
Systems Dynamics Study Group
18
Eigenvalues
determine stability
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
dz/dt = f(z)
If origin, is a fixed point, 0=(01, … ,0n)
then F(0)=0, f(0)=0
Note, if z0 is a fixed point of f(z),
f*(z) = f(z+z0)-z0 has 0 as fixed point.
y
dz/dt=f(z), eigenvalues λ are solution of
(0,0)
det [f(1)(z)-λI]=0 evaluated at fixed point z0
where I is identity matrix.
12/20/2001
Systems Dynamics Study Group
x
19
Eigenvalues (Cont’d)
f(1)(z0) =
[g(x0)+x0g’(x0)-y0p’(x0),
[cy0p’(x0),
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
dz/dt = f(z)
p(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
z0 is unstable
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if max (Re(λk), k=1, … , n) < 0
if max (Re(λk), k=1, … , n) > 0
Systems Dynamics Study Group
20
Why Re λ
determines stability
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
dz/dt = f(z)
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.
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21
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
dz/dt = f(z)
Fixed Points
det [f(1)(z0)-λI] =
(g(x0)+x0g’(x0)-y0p’(x0)-λ) (-s+cp(x0)-λ)
– cy0p’(x0)p(x0) = 0
dx/dt = xg(x) – yp(x) = 0
dy/dt = y[-s + cp(x)] = 0
y
1. (0,0),
2. (k,0),
3. (x*,y*),
12/20/2001
p(0) = 0
g(k) = 0
0<x*<k,
(x*,y*)
y*>0
Systems Dynamics Study Group
k
22
x
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
dz/dt = f(z)
(0,0)
det [f(1)(z0)-λI] =
(g(x0)+x0g’(x0)-y0p’(x0)-λ) (-s+cp(x0)-λ)
– cy0p’(x0)p(x0) = 0
(g(x0)+x0g’(x0)-y0p’(x0)-λ) (-s+cp(x0)-λ) – cy0p’(x0)p(x0) = 0
x0=y0=p(x0)=0
(g(0)-λ)(-s-λ)=0
λ=g(0),-s;
y
λ1= g(0) > 0, corresponds to x axis
λ2= -s < 0 , corresponds to y axis
(0,0)
x
(0,0) is unstable
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Systems Dynamics Study Group
23
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
dz/dt = f(z)
(k,0)
det [f(1)(z0)-λI] =
(g(x0)+x0g’(x0)-y0p’(x0)-λ) (-s+cp(x0)-λ)
– cy0p’(x0)p(x0) = 0
(g(x0)+x0g’(x0)-y0p’(x0)-λ) (-s+cp(x0)-λ) – cy0p’(x0)p(x0) = 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
y
λ1= kg’(k) < 0, corresponds to x axis
λ2= -sp(k) > 0, corresponds to y axis
(k,0) is unstable
x
k
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dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
dz/dt = f(z)
(x*,y*)
det [f(1)(z0)-λI] =
(g(x0)+x0g’(x0)-y0p’(x0)-λ) (-s+cp(x0)-λ)
– cy0p’(x0)p(x0) = 0
(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
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Systems Dynamics Study Group
y
(k,x**)
B>0
(x*,y*)
x
k
25
Poincaré-Bendixon
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
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.
y
B>0
C1 – has a continuous first
derivative
(x*,y*)
x
k
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dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
Poincaré-Bendixon Rationale
F(z+,t1)=z+1=(x1,y1)
F(z+,t2)=z+2=(x2,y2)
lim z+k -> z, as k->∞
– =(x ,y )
)=z
1
1
1 1
F(z–,t2)=z–2=(x2,y2)
lim F(z–k) -> z-, as
z- <= z
y
Z+2
Z
Z–2
Z–1
Z+1
B>0
F(z–,t
k->∞
x
(perhaps more than one periodic orbit?)
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y
Summary
B>0
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)],
(x*,y*)
x
s>0, c>0
k
g(x) is a growth function,
g(x), monotonic non-increasing, g’(x) <=0, g(0)>0, g(k)=0
p(x) is predation function
p(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*)=0
B = g(x*)+ x*g’(x*)-yp’(x*) > 0
Then, the dynamical system has a stable periodic orbit.
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Thank you!
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Systems Dynamics Study Group
29
Predator-Prey System
with a stable periodic orbit
Systems Dynamics Study Group
3rd Session – Simulation Results
Ellis S. Nolley
12/20/2001
12/20/2001
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30
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
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Topics
• Overview
– Simple Analysis – 1st session, 11/7/2001
– Rigorous Analysis – 2nd session, 11/27/2001
– Simulation Results – 3rd session, 12/11/2001
•
•
•
•
Model Parameters
Simulation Results
Vensim Techniques
Bifurcation
• Extra: Mathematics of Parameter Selection
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32
y
Model
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)],
B>0
(x*,y*)
s>0, c>0
x
k
g(x) is a growth function,
g(x), monotonic non-increasing, g’(x) <=0, g(0)>0, g(k)=0
p(x) is predation function
p(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*)=0
B = g(x*)+ x*g’(x*)-yp’(x*) > 0
Then, the dynamial system has a stable periodic orbit.
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dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
Model Parameters
g(0)>0, g’(x)<0
p(0)=0, p’(x)>0, B>0
g(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,
p(0) = 0,
g’(x)= -0.15<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
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Inside Orbit
X & Y Log Inside: time 0 - 40
400
300
200
100
0
Time Series of first 3 Periods
0
100
200
300
X - Prey
400
Amount X & Y
Y - Predator
500
120
100
80
60
40
20
0
x
y
0
0.5
1
1.5
2
2.5
Time
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Outside Orbit
X & Y Log Outside: time 0 - 40
400
300
200
100
0
400
X - Prey
Time Series of first 2 Periods
500
400
300
200
100
0
0.96
0.88
0.8
0.72
0.64
0.56
0.48
0.08
0
x
y
0.4
300
0.32
200
0.24
100
0.16
0
Amount X & Y
Y - Predator
500
Time
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Inside & Outside Orbits
X & Y Log Inside: time 0 - 40
500
Y - Predator
X & Y Log Outside: time 0 - 40
400
Y - Predator
500
300
200
100
400
0
0
100
200
300
400
X - Prey
300
200
100
0
0
100
200
300
400
X - Prey
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Combined Inside & Outside Orbits
LogOutside:
Inside: time
XX&&YYLog
time00- -40
40
Y - Predator
500
500
400
400
300
300
200
200
100
100
00
00
100
200
300
300
400
400
X - Prey
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Vensim Model Layout
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Systems Dynamics Study Group
dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
39
g(x)
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40
p(x)
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dx/dt
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x
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dy/dt
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y
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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
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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
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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
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Example
Excel
Result
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dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
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Bifurcation
Systems Dynamics Study Group
g(0)>0, g’(x)<0
p(0)=0, p’(x)>0, B>0
50
a0=54
X & Y Log Outside: time 0 - 40
500
500
400
400
Y - Predator
Y - Predator
X & Y Log Outside: time 0 - 40
300
200
100
300
200
100
0
0
0
100
200
300
0
400
100
X & Y Log Outside: time 0 - 40
a0=40
Y - Predator
200
150
100
50
0
100
200
300
400
350
300
250
200
150
100
50
0
0
X - Prey
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400
300
X & Y Log Outside: time 0 - 40 a0=49.7
250
0
200
X - Prey
X - Prey
Y - Predator
a0=51
100
200
300
400
X - Prey
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Projection of
Stable Attractor onto X Axis
x
Actual
boundary shape
is not described
~ 147.4
a0
~ 22.1
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~ 49.7
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dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
g(0)>0, g’(x)<0
p(0)=0, p’(x)>0
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
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•
•
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Existence Proof
Simulation Results
Vensim techniques
Bifurcation
y
B>0
(x*,y*)
x
k
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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
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Predator-Prey models Competition
Typical Competition Beliefs
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•
•
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•
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?
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Thank you!
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Extra
Mathematics of Parameter Selection
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dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
Polynomials
g(0)>0, g’(x)<0
p(0)=0, p’(x)>0, B>0
Recall 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>0
B= 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
and its polynomial has deg >= 3
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bk<0 for some k:1< k <n,
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dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
Log (Ln)
g(0)>0, g’(x)<0
p(0)=0, p’(x)>0, B>0
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)
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dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
Log (Cont’d)
g(0)>0, g’(x)<0
p(0)=0, p’(x)>0, B>0
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 Parameters
1)
Select s,c,b so that s > cb>cb[x/(x+1)], since x<x+1
2)
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) >0
5)
Verify that B>0
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dx/dt = xg(x) – yp(x)
dy/dt = y[-s + cp(x)]
Log (Cont’d)
g(0)>0, g’(x)<0
p(0)=0, p’(x)>0, B>0
Model Parameter Selections
1) b=4, c=10, s > bc=40, Let s=200
2) 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 > 54
4) y* = x*g(x*)/p(x*) = 147.4(200-0.15*147.4)/(200/10) =
~ 235.0 > 0
5) 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
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Thank you!
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