Answer to Question #163080 in Differential Equations for Ajay

Question

A model corresponding to the cooperative interaction between two species x and y

is given by

dx/dt=(4-2x+y)x

dy/dt=(4+x-2y)y

Find all the equilibrium points of the system and discuss the stability of the system at

these points.

Answer 1

Equilibrium points occur when both dx/dt = 0 and dy/dt = 0.

$$(4-2x+y)x=0$$

$$(4+x-2y)y=0$$

$x=0: 4+0-2y=0=>y=2$

Point $(0, 2)$

$y=0: 4-2x+0=0=>x=2$

Point $(2, 0)$

$$\begin{matrix} 4-2x+y=0 \\ 4+x-2y=0 \end{matrix}$$

$x=4, y=4$

Point $(4,4)$

$$J=\begin{bmatrix} 4-4x+y & x \\ y & 4+x-4y \end{bmatrix}$$

Point $(0, 2)$

$$J=\begin{bmatrix} 6 & 0 \\ 2 & -4 \end{bmatrix}$$

Find the eigenvalues

$$\begin{vmatrix} 6-\lambda & 0 \\ 2 & -4-\lambda \end{vmatrix}$$

$$( 6-\lambda)( -4-\lambda)=0$$

$$\lambda_1=6>0, \lambda_2=-4<0$$

The equilibrium point $(0, 2)$ is unstable.

Point $(2, 0)$

$$J=\begin{bmatrix} -4 & 2 \\ 0 & 6 \end{bmatrix}$$

Find the eigenvalues

$$\begin{vmatrix} -4-\lambda & 2 \\ 0 & 6-\lambda \end{vmatrix}$$

$$( -4-\lambda)(6-\lambda)=0$$

$$\lambda_1=6>0, \lambda_2=-4<0$$

The equilibrium point $(2, 0)$ is unstable.

Point $(4, 4)$

$$J=\begin{bmatrix} -8 & 4 \\ 4 & -8 \end{bmatrix}$$

Find the eigenvalues

$$\begin{vmatrix} -8-\lambda & 4 \\ 4 & -8-\lambda \end{vmatrix}$$

$$( -8-\lambda)(-8-\lambda)-16=0$$

$$\lambda_1=-12<0, \lambda_2=-4<0$$

The equilibrium point $(4, 4)$ is stable.