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Show that the system

$\dot{x}=y-y^{3}$

$\dot{y}=-x-y^{2}$

has a non-linear center and plot the phase potrait.

My attempt:

The system is non-linear so we linearise it:

The approximation to the non-linear system is the Jacobian

$J=\begin{bmatrix} 0 & 1-3y^{2}\\ -1 &-2y \end{bmatrix}$

Note that the fixed points of this system are $\left ( 0,0 \right ),\left ( -1,1 \right ),\left ( -1,-1 \right ).$

$J \mid _{\left ( 0,0 \right )}$ is a neutrally stable center. This is a non-linear center as can be checked by time-reversal and spatial reversal symmetry.

$J \mid _{\left ( -1,1 \right )} and \space J \mid_{\left ( -1,-1 \right )}$ are both saddle points.

What is left is determining the direction of the trajectories and the stable and unstable manifold. The stable and unstable manifold is the vector span by the Eigenvector of the associated Eigenvalues. However, the Jacobian provides a linear approximation to the non-linear system. Could someone give me a leg up?

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  • $\begingroup$ Is the problem written correctly? $\endgroup$ – Moo Apr 24 '16 at 4:15
  • $\begingroup$ @Moo It has been edited. $\endgroup$ – Mathematicing Apr 24 '16 at 5:20
  • $\begingroup$ @Moo $\frac{dy}{dx}=\frac{-x-y^{2}}{y-y^{3}}$ $\par$ $\frac{dy}{dx}\mid _{\left ( -2,-2 \right )}=-3$ $\frac{dy}{dx}\mid _{\left ( -0.5,-0.5 \right )}=-1.5$ If the first derivative of a function is less than zero on some interval $\mathit{I}$ then the function is decreasing on that interval. $\endgroup$ – Mathematicing Apr 24 '16 at 6:26
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...and plot the phase po(r)trait:

enter image description here

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  • $\begingroup$ Is the second derivative necessary here? If a function is increasing, there are two slopes where it can increase. $\endgroup$ – Mathematicing Apr 24 '16 at 8:41
  • $\begingroup$ What? Sorry but this is the second completely mysterious comment you are posting... What are you talking about exactly? $\endgroup$ – Did Apr 24 '16 at 10:52

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