ROCHESTER ME 406 - Study Notes - Predator-Prey Example Part 1 - Equilibrium

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In[322]:= sysidMathematica 4.1, DynPac 10.65, 2ê16ê2002ME 406Predator-Prey ExamplePart 1 - Equilibriumü IntroductionThis is a simple model of an ecological system with three components: a plant, a small mammal called the Murat which eats the plant, and a carnivorous predator called the Vekton which eats the mammal. We are using the model to test our belief that we can control this ecosystem by controlling the plant population. Our goal is to maintain both species at healthy numbers. We denote the population of the plant eater by M and the population of the carnivore by V. The basic equations governing this system were set-up in class. They are dMÅÅÅÅÅÅÅÅÅÅÅÅdt= rM J1 -MÅÅÅÅÅÅÅÅAN -bMVÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅM + H,dVÅÅÅÅÅÅÅÅÅÅÅdt=bMVÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅM + H- cV .The parameters were explained in class, so we just review them very briefly here. The parameter A is the maximum sustainable population of the plant-eaters in the absence of the carnivores, and the model incorporates this via a logistic law. The parameter A is the one we control, by controlling the plant population, and this is our only management tool for this system. In the limit M << A, the constraint represented by A is unimportant, and the net birth rate of of the plant eaters is r, and they will grow like ertuntil M becomes comparable with A. The second, negative, term in the M-equation models the effects of the predators. The loss is proportional to the product of the populations, because this product is proportional to the encounter rate. The denominator is a saturation effect, accounting for the fact that if the population of plant eaters is large, predators won't be as hungry and there will be fewer kills per encounter. In the predator equation, the first term models the birth rate, and accounts for the fact that it will be higher if food is more plentiful. If food is very plentiful (M much larger than H), this birth term saturates with a rate b*V, so b is the natural birth rate with ample food. It is an artificial simplification of the model that the saturation parameter has the same value H in both equations. ü Defining the System for MathematicaWe define the system for DynPac.In[323]:= setstate@8M, V<D;In[324]:= setparm@8A<D;In[325]:= slopevec = 8r * M * H1 - M ê AL - Hb*M * VLêHM + HL, Hb * M * VLêHM + HL - c * V<;predpt1.nb 1With so many parameters, the exploration of the system could take a very long time. To keep things simple, we choose values for all of the parameters except A (our management variable), and then study the system as we vary A. This is why we have included only A in the parameter vector. The units for the population quantities V, M, H, and A are in millions of individuals. The units of the time constants r, beta, b and c are in inverse years. We will assume that A can be varied from 0 up to a maximum of 15. We specify now all of the parameters that will be fixed throughout this study.In[326]:= r = 12 H** yr-1**L;In[327]:= b=20 H** yr-1**L;In[328]:= b = 4H** yr-1**L;In[329]:= c = 8 ê 5 H** yr-1**L;In[330]:= H = 4 H** millions **L;In our evaluations below of the effects of changing A, we adopt the following criterion for a healthy ecosystem: if the population of each species remains above 1 million, we count this as a successful preservation strategy. ü Locating the Equilibrium StatesWe first find the equilibrium states in terms of the parameter A. In[331]:= eqstates = findpolyeqOut[331]= 980, 0<, 8A, 0<, 98ÅÅÅÅ3,4 H-8 + 3ALÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ3A==We see that there are three equilibrium states. The first is {0,0}, in which no animals of either species are present. The second is {A,0}, in which the Vektons are absent, and the Murats are living free of fear at their maximum sustainable population A. In the third state, both species are present. We give each of these states a name.In[332]:= nulleq = eqstates@@1DDOut[332]= 80, 0<In[333]:= murateq = eqstates@@2DDOut[333]= 8A, 0<In[334]:= coeq = eqstates@@3DDOut[334]= 98ÅÅÅÅ3,4 H-8 + 3ALÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ3A=From these formulas, we can deduce the existence of some interesting transitions. For example, when A = 8/3, coeq reduces to murateq. In other words, at A = 8/3, those two equilibria coincide. For A < 8/3, coeq has a negative value of V. Such a state exists mathematically, but can never be reached in the model as long as we start in the first quadrant (because the V and M axes are orbits and cannot be crossed). Thus for 0 < A < 8/3, there are two equilibrium states in the relevant first quadrant -- nulleq and murateq. At A = 8/3, coeq moves up from the fourth quadrant and coalesces with murateq. For any A > 8/3, all three equilibria exist in the first quadrant. In particular, there is a state (coeq) with both species present. Whether these states are relevant depends on their stability. We examine that next. predpt1.nb 2ü Stability of EquilibriumWe begin by calculating the derivative matrix at the three equilibria.In[335]:= dmatnull = dermat ê. Thread@statevec -> nulleqDOut[335]= 9812, 0<, 90, -8ÅÅÅÅ5==In[336]:= dmatmurat = dermat ê. Thread@statevec -> murateqDOut[336]= 99-12, -20 AÅÅÅÅÅÅÅÅÅÅÅÅ4 + A=, 90, -8ÅÅÅÅ5+4AÅÅÅÅÅÅÅÅÅÅÅÅ4 + A==In[337]:= dmatco = dermat ê. Thread@statevec -> coeqDOut[337]= 9912 J1 -8ÅÅÅÅÅÅÅÅ3AN-32ÅÅÅÅÅÅÅA-12 H-8 + 3ALÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ5A, -8=, 912 H-8 + 3ALÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ25 A,0==We first see if we can draw any general stability conclusions for all values of A.In[338]:= Eigensystem@dmatnullDOut[338]= 99-8ÅÅÅÅ5,12=, 880, 1<, 81, 0<<=Here is our first result: the {0,0} equilibrium is always a saddle point and hence always unstable. If any Murats are introduced, they will increase in number. If both Vektons and Murats are introduced, they will increase. If only Vektons are introduced, they will die out (no food), and thus the Vekton axis is the stable manifold for this saddle.Now we try the Murat equilibrium.In[339]:= Eigensystem@dmatmuratDOut[339]= 99-12, -8ÅÅÅÅ5+4AÅÅÅÅÅÅÅÅÅÅÅÅ4 + A=, 981, 0<, 9-25 AÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 H26 + 9AL,1===We see from the eigenvalues that this is a strictly stable node for A < 8/3, and is a saddle


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