# Find the equilibrium points and their stability in the system

## Homework Statement

Find the equilibrium points and their stability in the system

xdot = xy - 2y - x + 2

ydot = xy + x

## Homework Equations

Jacobian matrix = g'(x)

## The Attempt at a Solution

first find the points that a critical point would satisfy

rewrite xdot = x(y-1) - 2(y-1) = 0
rewrite ydot = x(y+1) = 0

if x = 2 then y = -1
if y = 1 then x = 0

This yields two critical points:

(2,-1)
(0,1)

construct the jacobian matrix and determine eigenvalues for each critical point

g'(x) = [(y-1) (x-2); (y+1) x]

A_1 = g'((2,-1)) = (-2 0; 0 2)

A_2 = g'((0,1)) = (0 -2; 2 0)

det(A_1-lambda*I) = (lambda-2)(lambda+2)

yields the roots: lambda_1 = 2, lambda_2 = -2

Since the roots are real and unequal with the opposite sign, the critical point is a
saddle point and is unstable

det(A_2-lambda*I) = (lambda-2i)(lambda+2i)

yields the roots: lambda_1 = 2i, lambda_2 = -2i

Since the roots are pure imaginary, the solution is an ellipse that is marginally stable?

I don't know how to classify the stability of pure imaginary eigenvalues. Do I call this marginally stable? Is this still called a center point even though it is centered around (0,1) rather than the origin?

Also I am concerned about how much work I am showing and the completeness of the solution. The if x then y procedure seems especially informal..I want to provide the most complete solution possible to demonstrate I understand the concept.

Any advice or criticism is encouraged and appreciated! Thank you

#### Attachments

• phase portrait.png
68.5 KB · Views: 517

## Answers and Replies

Eigenvalues which are pure imaginary are "centers" no matter where they're located and are stable since they remain bounded however this may not always reflect the behavior of the non-linear system. That is, the non-linear system may in fact be a spiral source or sink which only does so slowly. Nice plot.