Nonlinear Dynamics: Nullclines and phase plane of a nonlinear system

[amk0713]

Homework Statement

Find the fixed points and classify them using linear analysis. Then sketch the nullclines, the vector field, and a plausible phase portrait.

dx/dt = x(x-y), dy/dt = y(2x-y)

Homework Equations

The Attempt at a Solution

f₁(x,y) = x(x-y)

x-nullcline: x(x-y) = 0 \Rightarrow x = 0

f₂(x,y) = y(2x-y)

y-nullcline: y(2x-y) = 0 \Rightarrow y = 0

Fixed point: (0,0)

J(x,y) =

(2x-y -x )
(2y 2x-2y)

Therefore,

J(0,0) =

(0 0)
(0 0)

Thus,

0 = |J(0,0) - λI| = (-λ)(-λ) = λ² = 0 \Rightarrow λ_1,2 = 0
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Now this is where I get stuck; I have no idea where to go from here when λ₁ = λ₂ = 0.

I would really appreciate at least a little nudge in the right direction. Thank you.

 

[jackmell]

There are some cases where linearization fails to capture the dynamics of a non-linear system near it's fixed points and I think one case is when the eigenvalues are zero. Need to check out a good reference like "Differential Equations" by Blanchard Hall and Devaney. Also, whenever working on these types of problems, in my opinion, it is essential to have a CAS like Mathematica to help you with it unless you like to suffer. Now unfortunately you can't use StreamPlot with Wolfram Alpha but if you could find a machine at school running Mathematica, you could very easily answer three of the questions with the code:

Show[{StreamPlot[{x^2 - x y, 2 x y - y^2}, {x, -1, 1}, {y, -1, 1}],
Plot[{x, 2 x}, {x, -1, 1}, PlotStyle -> Thickness[0.008]]}]

I don't know how to classify the fixed-point at the origin but if I had to say something, I would say it's some type of sink. Maybe someone else can help us further.