Question

1. The populations of two competing species x(t) and y(t) are governed by the non-linear system of differential equations dx dt 10x – x2 – 2xy, dy dt 5Y – 3y2 + xy. (a) Determine all of the critical points for the population model. (b) Determine the linearised system for each critical point in part (a) and discuss whether it can be used to approximate the behaviour of the non-linear system. (c) For the critical point at the origin: (i) Using eigenvalues and eigenvectors, determine the general solution of the linearised system. (ii) Sketch by hand, the phase portrait for the linearised system. Include all working of any special cases and the general case. (iii) What is the type and stability of the critical point? (d) Using PPLANE, sketch the global phase portrait of the non-linear system in a region that includes all of the critical points for the population model. (e) Based on the global phase portrait, discuss the survival prospects of each species in the long term.

Answer 1

The given model da lon- x² - 22y = f cory) 2 t dy ny = g(u,y) points at 54 - 34 for critical put and dy dt dt and lox-2²- sy- 3y + xy = 0 and y (5-34 tu) = 0 y=0 5-3y + 4zo if x=0 oor 2=0 or loud x-x-2xy = 0 & Cro-n-2ylo and -2y=0 y = Then y = 5 then =10 10-4-20₂0 and 5-34+420 solving we get a=4, y = 3 hence (0,0) 10,5) (10,0) ona (4,3) are four critical points Now by

will make b for we first this Jacobian af are 110-24-2y sy J= fra les ag y Я 5-64tr at (0,o) [ neah Coo) Linearized system don at dy at at Co, 5/3) (J) (2013 513 Linearized system near (o, 5/3) dal [/ n It dy 5 3

du [-lo -20 dt 15 0 dy [at -B hence corresponding at (lo, o [J] Linearized system near (10,0) at (43) -4 dy from kinear system the critical valid in close vicinity of the points in the nonlinear system, systery 3 -4 Linearized system near (4, 3) -8 dt 10 - 3 - 4 میله are Lab results con

the to be used approximate non linear Behavior of systen. (C) at origin Jacobian [J]- sro 5 O put we for eigen values JhIj = O 1 0 5 h lo-h) (5-4)=0 & h=105 eigenvector for h,=10 o :IN - 542 = 0 4₂ = 0 and n = 49 hat ni=1 then eigenvector v = (o

12=5 eigenvector for o 5 OJK) 1: 5), o - 2,0 2 = 2₂ Let n2 = 1 then eigenvector 2 Z/P lot =ce node hence and general solution = 4 en tu + C eh z tva +(!] + 4 st (o i) since cool is unstable move away from co, o). Also eigenvector are x and axis eigenectors. x = system arrows (i], hence can be thought Now Rewriting y'= sy linearized - lox dylat del at abc 5y 102 220 dae = 2 dny = lnx + enc 2

- longa - Inca y²-ca, which is parabola. Hence Phase portrait is as follows Rrete > > > tet hane plotted 2 such that =ce (For different c, we different parabolas y away from (0,0) arrows move have found in part ci (1) since both 10 and 5, that eigenvalues hence cool unntable tive one 9 node. Phase portrait including including all critical points is follows-

10.75 10 † 1 1 1 1 1 1 1 1 0.26 8.5 7.75 1 1 1 1 1 1 6.25 † 5.5 1 1 1 1 11 11 4.76 11 1 1 3.26 7 2.5 7 1 7 7 1.75 1 1 1 7 1 1.75 2.5 3.25 4.76 8.25 7.75 8.5 0.25 10 10.75 X

sol in the phase le You can easily portrait that all other critical points except (4,3) except (4,3) are unstable this means to say that for positive inital conolitions trajectory towards (4,3) critical point in which both species survive. This that in day teron behavior survival of each species take place every will ultermately more the means long succesfully in the habitat