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12/20/2001Systems Dynamics Study Group1 Predator Prey System with a stable periodic orbit 1 st Session - Simple Analysis Systems Dynamics Study Group Ellis S. Nolley 11/7/2001

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12/20/2001Systems Dynamics Study Group2 Topics Overview –Simple Analysis – 1 st session, 11/7/2001 –Rigorous Analysis – 2 nd session, 11/27/2001 –Simulation Results – 3 rd session, 12/11/2001 Mathematical Model Fixed Points Stable Periodic Orbit Reference: McGehee & Armstrong, Journal of Differential Equations, 23, 30-52 (1977)

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12/20/2001Systems Dynamics Study Group3 Model x = 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 p(x) is predation function p(x), monotonic increasing, dp(x)/dx >0, p(0)=0 g(x) x k

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12/20/2001Systems Dynamics Study Group4 Fixed Points 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*

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12/20/2001Systems Dynamics Study Group5 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*

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12/20/2001Systems Dynamics Study Group6 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) dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)]

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12/20/2001Systems Dynamics Study Group7 Analysis at Carrying Capacity (k,0) At (k,0), g(k) = 0 dx/dt = xg(x) – yp(x) = xg(x) g(x) is monotonic non-increasing. For x 0 For x>k, g(x)<0 From p. 13, at x=k, [-s+cp(x)] > 0 So dy/dt = y[-s + cp(x)] > 0 for y>0, x=k (k,0) is a saddle point k x y dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)]

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12/20/2001Systems Dynamics Study Group8 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 dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] k x 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

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12/20/2001Systems Dynamics Study Group9 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 dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] k x y (k,x**) (x*,y*)

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12/20/2001Systems Dynamics Study Group10 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. dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] k x y (k,x**) (x*,y*)

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12/20/2001Systems Dynamics Study Group11 Next Session Simple Mathematics – 1 st session Rigorous Mathematics – 2 nd session, 11/27 Simulation Results – 3 rd session dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)]

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12/20/2001Systems Dynamics Study Group12

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12/20/2001Systems Dynamics Study Group13 Predator Prey System with a stable periodic orbit 2 nd Session - Rigorous Analysis Systems Dynamics Study Group Ellis S. Nolley 11/27/2001

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12/20/2001Systems Dynamics Study Group14 Topics Overview –Simple Analysis – 1 st session, 11/7/2001 –Rigorous Analysis – 2 nd session, 11/27/2001 –Simulation Results – 3 rd session, 12/11/2001 Mathematical Model Fixed Points & Eigenvalues Poincare-Bendixon Theorem 4 key slides: #23 – 26

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12/20/2001Systems Dynamics Study Group15 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, North- Holland Publishing Company

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12/20/2001Systems Dynamics Study Group16 Where to find these References Mathematics Library, Vincent Hall, 3 rd Floor, University of MN Vincent Hall, 206 Church Street, Mpls, MN 55455 http://onestop.umn.edu/Maps/VinH/VinH-map.html

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12/20/2001Systems Dynamics Study Group17 Model x = 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 p(x) is predation function p(x), monotonic increasing, dp(x)/dx >0, p(0)=0 g(x) x k

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12/20/2001Systems Dynamics Study Group18 Jacobian & Eigenvalue Review dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] z’(t) = f(z) = [f 1 (z 1,…z n ), …, f n (z 1 …,z n )] Note above: z 1 =x, z 2 =y, f 1 (z)=xg(x)-yp(x), f 2 (z)=y[-s+cp(x)] dF(z,t)/dt = f(z); F (n) (z)=d n F(z)/dz n,n=0, … ∞; F (0) (z)=F(z) If z є B(z 0,ε) ={z|z-z 0 |<ε}, then the Taylor Series is: F(z) = k=0 ∞ Σ F (k) (z 0 )(z-z 0 ) k /k! = F(z 0 )+ k=0 ∞ Σ f (k) (z 0 )(z-z 0 ) k+1 /(k+1)! where f (k) (z 0 ) = [∂ k f 1 (z 1,..,z n )/∂z 1 k, …, ∂ k f 1 (z 1,..,z n )/∂z n k ] |… | (z 0,1,…,z 0,n ) [∂ k f n (z 1,..,z n )/∂z 1 k, …, ∂ k f n (z 1,..,z n )/∂z n k ] f (k) (z 0 ) is the kth derivative of f(z 0 ), f (1) (z 0 ) is the Jacobian dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)]

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12/20/2001Systems Dynamics Study Group19 Eigenvalues determine stability If origin, is a fixed point, 0=(0 1, …,0 n ) then F(0)=0, f(0)=0 Note, if z 0 is a fixed point of f(z), f*(z) = f(z+z 0 )-z 0 has 0 as fixed point. dz/dt=f(z), eigenvalues λ are solution of det [f (1) (z)-λI]=0 evaluated at fixed point z 0 where I is identity matrix. dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] dz/dt = f(z) x y (0,0)

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12/20/2001Systems Dynamics Study Group20 Eigenvalues (Cont’d) f (1) (z 0 ) =[g(x 0 )+x 0 g’(x 0 )-y 0 p’(x 0 ), p(x 0 )] [cy 0 p’(x 0 ), -s+cp(x 0 )] det [f (1) (z 0 )-λI] = 0 = det[g(x 0 )+x 0 g’(x 0 )-y 0 p’(x 0 )-λ, p(x 0 )] [cy 0 p’(x 0 ), -s+cp(x 0 )-λ] (g(x 0 )+x 0 g’(x 0 )-y 0 p’(x 0 )-λ) (-s+cp(x 0 )-λ) – cy 0 p’(x 0 )p(x 0 ) = 0 z 0 is stable if max (Re(λ k ), k=1, …, n) < 0 z 0 is unstable if max (Re(λ k ), k=1, …, n) > 0 dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] dz/dt = f(z)

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12/20/2001Systems Dynamics Study Group21 Why Re λ determines stability z’ = f(z); f(z 0 )=0, z 0 fixed point, λ k eigenvalues. Suppose λ j has Re λ j >0. Pick z close to z 0 f(z) = k=0 ∞ Σ f (k) (z 0 )(z-z 0 ) k /k! = f(z 0 ) + f (1) (z 0 )(z-z 0 ) + … Taylor Series ~ f (1) (z 0 ) (z-z 0 ) = (z-z 0 )Σc k λ k ; d f(z)/z ~ Σc k λ k dt ln f(z) ~ Σc k λ k t; f(z) ~ c*e Σλkt |f(z)| ~ |c*| |e λjt | |e Σλkt |; λ = Re λ + i Im λ ; |e i w |= |Cos(Im w) + i Sin(Im w)| = 1 lim |f(z)| ~ lim(|c*| |e Re(λj)t | |e Σλkt |) as t-> ∞ Then, |e Re(λj)t | -> large because Re λ j >0 So, z 0 is an unstable fixed point. If all Re λ j 0 as t-> ∞ So, z 0 is a stable fixed point. dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] dz/dt = f(z)

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12/20/2001Systems Dynamics Study Group22 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 0 dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] dz/dt = f(z) det [f (1) (z 0 )-λI] = (g(x 0 )+x 0 g’(x 0 )-y 0 p’(x 0 )-λ) (-s+cp(x 0 )-λ) – cy 0 p’(x 0 )p(x 0 ) = 0 k x y (x*,y*)

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12/20/2001Systems Dynamics Study Group23 (0,0) (g(x 0 )+x 0 g’(x 0 )-y 0 p’(x 0 )-λ) (-s+cp(x 0 )-λ) – cy 0 p’(x 0 )p(x 0 ) = 0 x 0 =y 0 =p(x 0 )=0 (g(0)-λ)(-s-λ)=0 λ=g(0),-s; λ 1 = g(0) > 0, corresponds to x axis λ 2 = -s < 0, corresponds to y axis dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] dz/dt = f(z) det [f (1) (z 0 )-λI] = (g(x 0 )+x 0 g’(x 0 )-y 0 p’(x 0 )-λ) (-s+cp(x 0 )-λ) – cy 0 p’(x 0 )p(x 0 ) = 0 x y (0,0) (0,0) is unstable

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12/20/2001Systems Dynamics Study Group24 (k,0) (g(x 0 )+x 0 g’(x 0 )-y 0 p’(x 0 )-λ) (-s+cp(x 0 )-λ) – cy 0 p’(x 0 )p(x 0 ) = 0 Note: x 0 =k, y 0 =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* 0, p(x*) < p(k) -s+cp(k) > 0 λ 1 = kg’(k) < 0, corresponds to x axis λ 2 = -sp(k) > 0, corresponds to y axis dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] dz/dt = f(z) det [f (1) (z 0 )-λI] = (g(x 0 )+x 0 g’(x 0 )-y 0 p’(x 0 )-λ) (-s+cp(x 0 )-λ) – cy 0 p’(x 0 )p(x 0 ) = 0 (k,0) is unstable k x y

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12/20/2001Systems Dynamics Study Group25 (x*,y*) (g(x 0 )+x 0 g’(x 0 )-y 0 p’(x 0 )-λ) (-s+cp(x 0 )-λ) – cy 0 p’(x 0 )p(x 0 ) = 0 Note: -s+cp(x 0 )=0; x 0 =x*, y 0 =y* (g(x*)+x*g’(x*)-y*p’(x*)-λ)(-λ) – cy*p’(x*)p(x*) = 0 λ 2 - [g(x*)+x*g’(x*)-y*p’(x*)]λ – cy*p’(x*)p(x*) = 0 BC > 0 λ = (B +/– sqrt(B 2 + 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 λ 1 = [B – sqrt(B 2 + 4C)]/2 < 0 λ 2 = [B + sqrt(B 2 + 4C)]/2 > 0 If B < 0, (x*,y*) is stable. λ 1 = [B – sqrt(B 2 + 4C)]/2 < 0 λ 2 = [B + sqrt(B 2 + 4C)]/2 < 0 dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] dz/dt = f(z) det [f (1) (z 0 )-λI] = (g(x 0 )+x 0 g’(x 0 )-y 0 p’(x 0 )-λ) (-s+cp(x 0 )-λ) – cy 0 p’(x 0 )p(x 0 ) = 0 (k,x**) k x y (x*,y*) B > 0

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12/20/2001Systems Dynamics Study Group26 Poincaré-Bendixon Theorem: A nonempty compact limit set of a C 1 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. C 1 – has a continuous first derivative dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] k x y (x*,y*) B > 0

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12/20/2001Systems Dynamics Study Group27 Poincaré-Bendixon Rationale F(z +,t 1 )=z + 1 =(x 1,y 1 ) F(z +,t 2 )=z + 2 =(x 2,y 2 ) lim z + k -> z, as k->∞ F(z –,t 1 )=z – 1 =(x 1,y 1 ) F(z –,t 2 )=z – 2 =(x 2,y 2 ) lim F(z – k ) -> z -, as k->∞ z - <= z (perhaps more than one periodic orbit?) dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] x y B > 0 Z+1Z+1 Z+2Z+2 Z–2Z–2 Z–1Z–1 Z

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12/20/2001Systems Dynamics Study Group28 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(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. k x y (x*,y*) B > 0

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12/20/2001Systems Dynamics Study Group29

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12/20/2001Systems Dynamics Study Group30 Predator-Prey System with a stable periodic orbit Systems Dynamics Study Group 3 rd Session – Simulation Results Ellis S. Nolley 12/20/2001

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Systems Dynamics Study Group31 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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12/20/2001Systems Dynamics Study Group32 Topics Overview –Simple Analysis – 1 st session, 11/7/2001 –Rigorous Analysis – 2 nd session, 11/27/2001 –Simulation Results – 3 rd session, 12/11/2001 Model Parameters Simulation Results Vensim Techniques Bifurcation Extra: Mathematics of Parameter Selection

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12/20/2001Systems Dynamics Study Group33 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(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. k x y (x*,y*) B > 0

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12/20/2001Systems Dynamics Study Group34 Model Parameters g(x) = a 0 +a 1 x, x*~147.4, a 0 =54, a 1 = -0.15 p(x) = b ln(x+1), b=4, s=200, c=10 g(0) = 54>0, g’(x)= -0.15<0 p(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 dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] g(0)>0, g’(x)<0 p(0)=0, p’(x)>0, B>0

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12/20/2001Systems Dynamics Study Group35 Inside Orbit

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12/20/2001Systems Dynamics Study Group36 Outside Orbit

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12/20/2001Systems Dynamics Study Group37 Inside & Outside Orbits

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12/20/2001Systems Dynamics Study Group38 Combined Inside & Outside Orbits

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12/20/2001Systems Dynamics Study Group39 Vensim Model Layout dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)]

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12/20/2001Systems Dynamics Study Group40 g(x)

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12/20/2001Systems Dynamics Study Group41 p(x)

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12/20/2001Systems Dynamics Study Group42 dx/dt

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12/20/2001Systems Dynamics Study Group43 x

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12/20/2001Systems Dynamics Study Group44 dy/dt

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12/20/2001Systems Dynamics Study Group45 y

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12/20/2001Systems Dynamics Study Group46 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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12/20/2001Systems Dynamics Study Group47 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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12/20/2001Systems Dynamics Study Group48 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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12/20/2001Systems Dynamics Study Group49 Example Excel Result

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12/20/2001Systems Dynamics Study Group50 Bifurcation dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] g(0)>0, g’(x)<0 p(0)=0, p’(x)>0, B>0

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12/20/2001Systems Dynamics Study Group51 a 0 =54 a 0 =49.7 a 0 =40 a 0 =51

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12/20/2001Systems Dynamics Study Group52 Projection of Stable Attractor onto X Axis x a0a0 ~ 147.4 ~ 49.7 Actual boundary shape is not described ~ 22.1

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12/20/2001Systems Dynamics Study Group53 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 dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] g(0)>0, g’(x)<0 p(0)=0, p’(x)>0 k x y (x*,y*) B > 0

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12/20/2001Systems Dynamics Study Group54 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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12/20/2001Systems Dynamics Study Group55 Predator-Prey models Competition Typical 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?

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12/20/2001Systems Dynamics Study Group57 Extra Mathematics of Parameter Selection

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12/20/2001Systems Dynamics Study Group58 Polynomials 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)= 0 Σ m a k x k, a 0 >0, a 1 0, g’(x) 0 p(x)= 1 Σ n b k x k, b 0 =0, b 1 >0, b n >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* + - - = 0 Σ m a k x k +x( 1 Σ m ka k x k-1 )-( 0 Σ m a k x k )( 1 Σ n kb k x k )/( 1 Σ n b k x k ) = 0 Σ m a k x k + 1 Σ m ka k x k -( 0 Σ m a k x k )( 1 Σ n kb k x k )/( 1 Σ n b k x k ) = a 0 [1-( 1 Σ n kb k x k )/( 1 Σ n b k x k )]+ 1 Σ m ka k x k -( 1 Σ m a k x k )( 1 Σ n kb k x k )/( 1 Σ n b k x k ) Note: if a j 0 & b k >0 for all k>0, then ( 1 Σ n kb k x k )/( 1 Σ n b k x k )>1, then B<0. Then, no stable periodic orbit exists. Therefore, if there is a stable periodic orbit, then a j >0 for some j:1< j

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12/20/2001Systems Dynamics Study Group59 Log (Ln) g(x)= 0 Σ m a k x k, a 0 >0, a 1 <0, a m <0, p(x)=b ln(x + 1), B= g(x)+xg’(x)-xg(x)p’(x)/p(x), for x=x* = 0 Σ m a k x k + 1 Σ m ka k x k -( 0 Σ m a k x k )(xb/(x+1))/(s/c), since p(x*)=s/c = a 0 (1- (cb/s)[x/(x+1)])+ 1 Σ m a k x k + 1 Σ m ka k x k –(cb/s)[x/(x+1)]( 1 Σ m a k x k ) = a 0 (1- (cb/s)[x/(x+1)])+ 1 Σ m a k (1+k-(cb/s)[x/(x+1)])x k x*= e [s/(bc)] - 1 y*= xg(x)/p(x) = ( 0 Σ m a k x k+1 )/(s/c) = ( 0 Σ m a k x k+1 )(c/s) Let g(x) = a 0 +a 1 x, a o >0, a 1 0, p(0)=0 Find x=x*, s/c = b ln(x +1), x*= e [s/(bc)] - 1 y*= (a 0 x+a 1 x 2 )(c/s) dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] g(0)>0, g’(x)<0 p(0)=0, p’(x)>0, B>0

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12/20/2001Systems Dynamics Study Group60 Log (Cont’d) B= g(x)+xg’(x)-xg(x)p’(x)/p(x), for x=x* = a 0 + a 1 x + a 1 x – x(a 0 + a 1 x)b/[x+1] /(b [ln (x+1)]) = a 0 (1-(cb/s)[x/(x+1)]) + 2a 1 x - [(a 0 + a 1 x)b[x/(x+1)]/(s/c)], since p(x*)=s/c = a 0 (1-(cb/s)[x/(x+1)]) + a 1 x(2-(cb/s)[x/(x+1)]) > 0, a 0 > - a 1 x[(2s-cb)[x/(x+1)]/s][s/(s-cb[x/(x+1)])] a 0 > - a 1 x[(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 s<(cb/2)[x/(x+1)] Selecting Model Parameters 1)Select s,c,b so that s > cb>cb[x/(x+1)], since x

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12/20/2001Systems Dynamics Study Group61 Log (Cont’d) 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) a 0 > - a 1 x(2s-cb[x/(x+1)])/(s-cb[x/(x+1]) = - a 1 *147.4(2*200 - 40[1])/(200 - 40[1]), since [x/(x+1)]~1 = - a 1 *147.4(360/160), = - a 1 (332.7) let a 1 = -0.15, a 0 > 49.7, Let a 0 > 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) = = (a 0 +a 1 x) + 2a 1 x – b[x/(x+1)] (a 0 +a 1 x) = [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 dx/dt = xg(x) – yp(x) dy/dt = y[-s + cp(x)] g(0)>0, g’(x)<0 p(0)=0, p’(x)>0, B>0

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