MATH 250 Homework 4: Due May 4, 2017

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1 Due May 4, 17 Answer the following questions to the best of your ability. Solutions should be typed. Any plots or graphs should be included with the question (please include the questions in your solutions). For class we looked at the populations of two species, a prey denoted by y 1 and a predator denoted by y 2, can be modeled by the autonomous, nonlinear ODE ( ) ( ) y y = 1 y1 (α y 2 = 1 β 1 y 2 ) = f(y). y 2 ( α 2 + β 2 y 1 ) This model is known as the Lotka-Voltera model. The parameters α 1 and α 2 are birth and death rates in isolation for prey and predators respectively. The parameters β 1 and β 2 determine the effects of the interactions of the two species. Write a program that uses the classical fourth order Runge-Kutta method to solve the Lotka-Voltera system. Integrate from t = 0 to t = 25. For class we will use the parameters α 1 = 0.5, β 1 = 0.1, α 2 = 1.0, β 2 = 0.02, and the initial populations y 1 (0) = 100, and y 2 (0) = 10. We plotted each of the two populations as a function of time, and on a separate graph plot the trajectory of the point (y 1 (t), y 2 (t)) in the plane as a function of time and created a phase portrait. For this assignment try other initial populations. Observe the results using the same type of graphs used in class. Can you find nonzero initial populations that allow for either of the populations to become extinct? Can you find a non-zero initial population that never changes. For this assignment we can consider the following piece of python code adapted to work in sage so that it may be contained fully inside a L A TEX document. from math import import numpy as np #import m a t p l o t l i b. pyplot as p l t from m a t p l o t l i b. backends. backend agg import FigureCanvasAgg as FigureCanv from m a t p l o t l i b. pyplot import f i g u r e, show # Define the f o r c i n g f u n c t i o n s : alpha1 = 0. 5 alpha2 = 1. 0 beta1 = 0. 1 beta2 = 0.02 def f 1 ( y1, y2 ) : return ( alpha1 beta1 y2 ) y1 def f 2 ( y1, y2 ) : return ( alpha2 + beta2 y1 ) y2 # Define a 4 th order ODE s o l v e r. def ODE4f1(h, y1, y2 ) : 1

2 K1 = f 1 ( y1, y2 ) K2 = f 1 ( y1+(k1/ 2. 0 ), y2 ) K3 = f 1 ( y1+(k2/ 2. 0 ), y2 ) K4 = f 1 ( y1+k3, y2 ) y1new = y1 + ( h / 6. 0 ) (K K K3 + K4) return y1new def ODE4f2(h, y1, y2 ) : K1 = f 2 ( y1, y2 ) K2 = f 2 ( y1, y2+(k1/ 2. 0 ) ) K3 = f 2 ( y1, y2+(k2/ 2. 0 ) ) K4 = f 2 ( y1, y2+k3) y2new = y2 + ( h / 6. 0 ) (K K K3 + K4) return y2new # Lets s l o w l y advance the s o l u t i o n in time. y1 = 100 # people y2 = 10 # zombies t = 0. 0 # h = 0.01 # D i s c r e t e time step. m = 25 n = 100 m y1pts = [ ] y2pts = [ ] t p t s = [ ] # Make a loop to populate our v e c t o r s over time. f o r i in range (n 1): t = t + h y1 = ODE4f1(h, y1, y2 ) y2 = ODE4f2(h, y1, y2 ) f i g = f i g u r e ( ) ax = f i g. add subplot (111) A = l i s t p l o t ( z i p ( tpts, y2pts ) ) + l i s t p l o t ( z i p ( tpts, y1pts ), c o l o r = red ) B = l i s t p l o t ( z i p ( y1pts, y2pts ) ) The following figure represents the provided code run with the parameters listed in the caption. Note that time is depicted on the horizontal axis, and the vertical axis denotes the size of each of the populations over time. 2

3 Figure 1: A figure denoting the interaction of y 1 initially at 100 in red and y 2 initially at 10 in blue with α 1 = 0.5, α 2 = 1.0, β 1 = 0.1, and β 2 = Figure 2: A figure denoting phase diagram of y 1 initially at 100 in red and y 2 initially at 10 in blue with α 1 = 0.5, α 2 = 1.0, β 1 = 0.1, and β 2 =

4 Looking at the relationship denoted between the populations we see that if α 1 β 1 y 2 = 0 and α 2 + β 2 y 1 = 0 then the population will not be changing. Further simplification yields: y 2 = α 1 β 1 and y 1 = α 2 β 2 and using the values specified in the initial conditions above: y 2 = = 5 and y 1 = = 50 should produce a non changing population in the model Figure 3: A figure denoting the interaction of y 1 initially at 50 in red and y 2 initially at 5 in blue with α 1 = 0.5, α 2 = 1.0, β 1 = 0.1, and β 2 = A phase diagram with these initial conditions would be useless as it would graph as a single point. Repeat your work using the Leslie-Gower model: y 1 = y 1 (α 1 β 1 y 2 ) and y 2 = y 2 (α 2 (β 2 y 2 )/y 1 ) use the same parameters given above with the exception of setting β 2 = 10. How does the solution differ between the two models? The following code represents an implementation of the Leslie-Gower model: The following code listing will solve the Leslie-Gower model with the altered initial parameters. 4

5 \ begin { s a g e s i l e n t } # Define the f o r c i n g f u n c t i o n s : alpha1 = 0. 5 alpha2 = 1. 0 beta1 = 0. 1 beta2 = 10 def f1l ( y1, y2 ) : return ( alpha1 beta1 y2 ) y1 def f2l ( y1, y2 ) : return ( alpha2 ( ( beta2 y2 )/ y1 )) y2 # Define a 4 th order ODE s o l v e r. def ODE4f1L(h, y1, y2 ) : K1 = f1l ( y1, y2 ) K2 = f1l ( y1+(k1/ 2. 0 ), y2 ) K3 = f1l ( y1+(k2/ 2. 0 ), y2 ) K4 = f1l ( y1+k3, y2 ) y1new = y1 + ( h / 6. 0 ) (K K K3 + K4) return y1new def ODE4f2L(h, y1, y2 ) : K1 = f2l ( y1, y2 ) K2 = f2l ( y1, y2+(k1/ 2. 0 ) ) K3 = f2l ( y1, y2+(k2/ 2. 0 ) ) K4 = f2l ( y1, y2+k3) y2new = y2 + ( h / 6. 0 ) (K K K3 + K4) return y2new y1 = 100 # people y2 = 10 # zombies t = 0. 0 # h = 0.01 # D i s c r e t e time step. m = 25 n = 100 m y1pts = [ ] y2pts = [ ] t p t s = [ ] # Make a loop to populate our v e c t o r s over time. f o r i in range (n 1): t = t + h y1 = ODE4f1L(h, y1, y2 ) y2 = ODE4f2L(h, y1, y2 ) 5

6 f i g = f i g u r e ( ) ax = f i g. add subplot (111) E = l i s t p l o t ( z i p ( tpts, y2pts ) ) + l i s t p l o t ( z i p ( tpts, y1pts ), c o l o r = red ) F = l i s t p l o t ( z i p ( y1pts, y2pts ) ) \end{ s a g e s i l e n t } The following figure represents the provided code run with the parameters listed in the caption. Note that time is depicted on the horizontal axis, and the vertical axis denotes the size of each of the populations over time Figure 4: A figure denoting the interaction of y 1 initially at 100 in red and y 2 initially at 10 in blue with α 1 = 0.5, α 2 = 1.0, β 1 = 0.1, and β 2 = 0.02 for the Leslie-Gower model. Note that using this model the two populations seem to seek out a point of balance. A balance point that changes depending on the initial conditions. If we look at the original model Leslie-Gower model will start in a point of equilibrium if the prey, y 1 = 0, and the predator y 2 = 10 initially. y 1 = y 1 (1 0.1y 2 ) = y 2 = 10 y 2 = y 2 (0.5 10y 2 /y 1 ) = y 1 = 0 6

7 Figure 5: A figure denoting phase diagram of y 1 initially at 100 in red and y 2 initially at 10 in blue with α 1 = 0.5, α 2 = 1.0, β 1 = 0.1, and β 2 = Thus, the big difference between the two models is how they view the populations interaction. The Lotka-Volterra model sees the two populations in constant competition, going through constant cycles of near extinction, followed by growth. The Leslie-Gower model has the two populations seeking a point of equilibrium, at which neither population changes. 7

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