Answer to Question #282697 in Differential Equations for Shanikwa

Question

Given x and y are species whose interaction is governed by x¹=x(-20-x+2y) and y¹=y(-50+x-y) ..... identify the type of interaction 2) determine equilibrium point of model and state possible outcomes of this interaction 3)linearise the system around each equilibrium point and discuss the nature of stability of each equilibrium point....4)sketch phase portrait of the above system

Answer 1

1)

x' and y' does not depend explicitly on time, it is autonomous system

also, this is nonlinear system

2)

$x(-20-x+2y)=0$

$y(-50+x-y)=0$

equilibrium points:

$(0,0),(0,-50),(-20,0),(120,70)$

Jacobian matrix:

$J=\begin{pmatrix} \partial f/ \partial x& \partial f/ \partial y \\ \partial g/ \partial x & \partial g/ \partial y \end{pmatrix}=\begin{pmatrix} -20-2x+2y & 2x \\ y& -50+x-2y \end{pmatrix}$

where

$f(x,y)=x(-20-x+2y)$

$g(x,y)=y(-50+x-y)$

for equilibrium points:

$J(0,0)=\begin{pmatrix} -20 & 0 \\ 0 & -50 \end{pmatrix}$

$\begin{pmatrix} x' \\ y' \end{pmatrix}=\begin{pmatrix} -20 & 0 \\ 0 & -50 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} -20x \\ -50y \end{pmatrix}$

$J(0,-50)=\begin{pmatrix} -120 & 0 \\ -50 & 50 \end{pmatrix}$

$\begin{pmatrix} x' \\ y' \end{pmatrix}=\begin{pmatrix} -120 & 0 \\ -50 & 50 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} -120x \\ -50x+50y \end{pmatrix}$

$J(-20,0)=\begin{pmatrix} 20 & -40 \\ 0 & -70 \end{pmatrix}$

$\begin{pmatrix} x' \\ y' \end{pmatrix}=\begin{pmatrix} 20 & -40 \\ 0 & -70 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} 20x-40y \\ -70y \end{pmatrix}$

$J(120,70)=\begin{pmatrix} -120 & 240 \\ 70 & -70 \end{pmatrix}$

$\begin{pmatrix} x' \\ y' \end{pmatrix}=\begin{pmatrix} -120 & 240 \\ 70 & -70 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} -120x+240y \\ 70x-70y \end{pmatrix}$

3)

find eigenvalues:

for $J(0,0)$ :

$(-20-r)(-50-r)=0$

$r_1=-20,r_2=-50$

eigenvalues are real, r₂ < r₂ < 0

so, (0,0) is an asymptotically stable node

for $J(0,-50)$ :

$(-120-r)(50-r)=0$

$r_1=-120,r_2=50$

eigenvalues are real, r₁ < 0 < r₂

so, (0,-50) is a saddle point

for $J(-20,0)$ :

$(20-r)(-70-r)=0$

$r_1=20,r_2=-70$

eigenvalues are real, r₂ < 0 < r₁

so, (0,-50) is a saddle point

for $J(120,70)$ :

$(-120-r)(-70-r)-240\cdot70=0$

$r^2+190r-8400=0$

$r=\frac{-190\pm \sqrt{190^2+4\cdot8400}}{2}$

$r_1=37,r_2=-227$

eigenvalues are real, r₂ < 0 < r₁

so, (120,70) is a saddle point

4)