Critical point analysis for x ′ = y , y ′ = x 2 − y − ϵ

Question

Critical point analysis for       x    ′    =  y  ,      y    ′    =      x    2    −  y  −  ϵ

Tasinazzokbc · Accepted Answer

There is a theorem which says so long as the critical (or equilibrium) point is not a center then the nature of this critical point in the system There is a theorem which says so long as the critical (or equilibrium) point is not a center then the nature of this critical point in the system x′,y′, is the same as that point in the linearization i.e,, is the same as that point in the linearization i.e,

{\tilde{x}}^{'} = p_{x} (x_{0}, y_{0}) \cdot x + p_{y} (x_{0}, y_{0}) \cdot y

{\tilde{y}}^{'} = q_{x} (x_{0}, y_{0}) \cdot x + q_{y} (x_{0}, y_{0}) \cdot y

where

x^{'} = p (x, y)

and

y^{'} = q (x, y)

. Now you have a linear system. Find the associated matrix to this system, compute it's eigenvalues and use the Painleve Analysis.

Critical point analysis for x′=y,y′=x2−y−epsilon

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