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Introduction to the Modeling and Analysis of Complex Systems

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128 CHAPTER 7. CONTINUOUS-TIME MODELS II: ANALYSISExercise 7.15 Consider <strong>the</strong> logistic growth model (r > 0, K > 0):dx(1dt = rx − x )K(7.70)Conduct a linear stability analysis <strong>to</strong> determine whe<strong>the</strong>r this model is stable or notat each <strong>of</strong> its equilibrium points x eq = 0, K.Exercise 7.16 Consider <strong>the</strong> following differential equations that describe <strong>the</strong> interactionbetween two species called commensalism (species x benefits from <strong>the</strong>presence <strong>of</strong> species y but doesn’t influence y):dxdt = −x + rxy − x2 (7.71)dy= y(1 − y)dt(7.72)x ≥ 0, y ≥ 0, r > 1 (7.73)1. Find all <strong>the</strong> equilibrium points.2. Calculate <strong>the</strong> Jacobian matrix at <strong>the</strong> equilibrium point where x > 0 <strong>and</strong> y > 0.3. Calculate <strong>the</strong> eigenvalues <strong>of</strong> <strong>the</strong> matrix obtained above.4. Based on <strong>the</strong> result, classify <strong>the</strong> equilibrium point in<strong>to</strong> one <strong>of</strong> <strong>the</strong> following:Stable point, unstable point, saddle point, stable spiral focus, unstable spiralfocus, or neutral center.Exercise 7.17Consider <strong>the</strong> differential equations <strong>of</strong> <strong>the</strong> SIR model:dS= −aSIdt(7.74)dI= aSI − bIdt(7.75)dR= bIdt(7.76)As you see in <strong>the</strong> equations above, R doesn’t influence <strong>the</strong> behaviors <strong>of</strong> S <strong>and</strong> I,so you can safely ignore <strong>the</strong> third equation <strong>to</strong> make <strong>the</strong> model two-dimensional.Do <strong>the</strong> following:

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