Predator-Prey Systems Homework: Equilibrium, Phase-Portraits, and Bifurcations, Assignments of Spanish Language

Download Predator-Prey Systems Homework: Equilibrium, Phase-Portraits, and Bifurcations and more Assignments Spanish Language in PDF only on Docsity! Solutions to Homework 7 By H̊akan Nordgren Problem 11.2: Sketch the phase-portrait for the following predator-prey system: ẋ = x(1 − x) − xy ẏ = y ( 1 − y x ) , defined for x > 0 and y ≥ 0. Solution: The x-nullcline is the curve Cx = {(x, y) ∈ R 2 : y = 1 − x}. The y-nullcline is the curve C1y = {(x, y) ∈ R2 : y = 0} and the curve C2y = {(x, y) ∈ R 2 : y = x}. Along the x-nullcline Cx, we have ẏ = (1 − x)(1 − 1−x x ) = (x−1)(2x−1) x . Thus ẏ < 0 for 0 < x < 12 ; and ẏ > 0 for 1 2 < x < 1, along Cx. Along the y-nullcline C1y , we have ẋ = x(1 − x). Thus ẋ > 0 for x with 0 < x < 1 and ẋ < 0 for x > 1, along C 1 y . Along the y-nullcline C2y , we have ẋ = x(1 − x) − x 2 = x(1 − 2x). Thus ẋ > 0 for x with 0 < x < 12 and ẋ < 0 for x > 12 , along C 2 y . The sketch is on a separate pdf. Problem 11.4: Consider the predator-prey equations given by ẋ = x(1 − x) − xy x + b ẏ = y(1 − y), where b > 0 is a parameter. 1. Find all equilibrium points and classify them. 2. Sketch the phase-portrait for different values of b. 3. Describe any bifurcations which occur. Solution: 1. We begin by finding equilibrium points. We have ẏ = 0 for all (x, y) which satisfy 0 = y(1 − y). Thus the y-nullclines are the curves C0y = {(x, y) ∈ R 2 : y = 0} and C1y = {(x, y) ∈ R 2 : y = 1}. We have ẋ = 0 for all (x, y) which satisfy the equation 0 = x(1 − x) − xy x + b , so we have equilibrium points for all (x, 0) which satisfy 0 = x(1 − x), and for all (x, 1) which satisfy 0 = x(−x2 + (1 − b)x + (b − 1)) x + b . That is, (0, 0), (1, 0), and (0, 1) are equilibrium points. And so are points (x, 1) where x = (1 − b) ± √ (b − 1)(3 + b) 2 . 1 For b with b < 1 there is no such equilibrium point. For b = 1 there is one, but it is one which we have already counted. For b > 1, there are two. To find the type of these equilbrium points we find the linearised system. Define F (x, y) = ( x(1 − x) − xy x+b y(1 − y) ) . Then D(F )(x, y) = ( 1 − 2x − by(x+b)2 − x x−b 0 1 − 2y ) . Thus D(F ) will have eigenvalues 1 − 2x − by(x+b)2 and 1 − 2y, which means that (0, 0) is a source; (1, 0) is a saddle; (0, 1) is a sink for b < 1, it is not hyperbolic for b = 1 and for b > 1 it is a saddle. 2. The sketches are on a separate pdf. 3. There is bifurcation when b = 1. Problem 11.5: Consider the competing-species equations ẋ = xM(x, y) ẏ = yN(x, y), where M(x, y) = a − x − ay and N(x, y) = b − bx − y, for a, b > 0. 1. Show that we have ∂M ∂y < 0 and ∂N ∂x < 0; 2. that there is K > 0 such that if x ≥ K or y ≥ K, then we have M(x, y) < 0 and N(x, y) < 0; and 3. that there is c, d > 0 such that M(x, 0) > 0 for x < c, and M(x, 0) < 0 for x > c; and N(0, y) > 0 for y < d, and N(0, y) < 0 for y > d. Draw the phase-portrait of this system. Explain why these equations make it possible, but unlikely, that both species should survive. Solution: We have ∂M ∂y = −a < 0 and ∂N ∂x = −b < 0; also if K = max{a, b, 1}, then for x ≥ K or y ≥ K, then we have M(x, y) < 0 and N(x, y) < 0. Also, M(x, 0) = a − x so for x < a we have M(x, 0) < 0, and for x > a we have M(x, 0) > 0. Similarly, for N . The diagrams for this system will be similar to diagrams drawn in homework one, so I will refer to these. It is possible to pick a and b such that the two lines intersect at (x, y) with x, y > 0. Problem 11.7: Consider the equations ẋ = xM(x, y) ẏ = yN(x, y), where we have the following conditions on M and N : 1. ∂M ∂y > 0 and ∂N ∂x > 0 for x, y ≥ 0. 2. There is A, B > 0 such that if x > A, then M(x, y) < 0, and if y > B, then we N(x, y) < 0. 3. M(0, 0) > 0 and N(0, 0) > 0. Let M−1(0) ∩ N−1(0) be finite, and let all the intersections be transverse. Prove the following: 1. Let the point (x, y) satisfy x < A and y < B. Show that φt(x, y) tends to an equilibrium point. 2