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Find the equilibrium points and their stability in the system

  • #1

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
 

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Answers and Replies

  • #2
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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.
 

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