- #1
bjohnson2001
- 15
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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