Answer to Question #286893 in Differential Equations for Mary

Question

Given x and y as species whose interaction is governed by

x

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

y

0 = y (( 50 + x x y),

(i) Identify the type of interaction represented by this system.

(ii) Determine the equilibrium points of this model and state the possible outcomes of

this interaction.

(iii) Linearise the system around each equilibrium point and hence discuss the nature of

the stability of each equilibrium point.

(iv) Sketch the phase portrait of the above system.

Answer 1

Solution:

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)