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We have

$$ \begin{cases} \dot{x} = - y - y^3 \\ \dot{y} = x \end{cases} $$

where $x,y \in \mathbb{R}$. Show that the critical point for the linear system is a $\mathbf{center}$. Prove that the type of the critical point is the same for the $\mathbf{nonlinear}$ system.

TRY:

Notice $( \dot{x}, \dot{y} ) = (0,0)$ iff $x = 0 $, $y + y^3 =0 \iff y(1+y^2) = 0$. Thus, the only critical point is $(0,0)$. Lets linearize the system. The jacobian is

$$ J(x,y) = \left( \begin{matrix} 0 & -1 - 3y^2 \\ 1 & 0 \end{matrix} \right) $$

We have

$$ J(0,0) = \left( \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right) $$

And the eigenvalues of this matrix are $\lambda = \pm i $ showing that indeed the critical point is a $\mathbf{center}$. But, Im stuck on showing that the same is true for the nonlinear system. How do we show the existence of closed orbits near the critical point?

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  • $\begingroup$ Can you briefly define "a center"? [otherwise, your work looks good.] $\endgroup$ – Michael Dec 15 '16 at 23:42
  • $\begingroup$ Your reasoning for using the linearized system is not valid for eigenvalues with zero real part. Lyapunov's linearization theorem states that if all eigenvalues have positive real parts -> unstable equilibrium for the nonlinear system, negative real parts -> asymptotically stable equilibrium for the nonlinear system and real part equal to zero -> no conclusion possible. For the last case you can come up with a Lyapunov function or you can determine stability by the center manifold theorem. $\endgroup$ – MrYouMath Feb 19 at 21:10
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You can use the following Lyapunov function candidate

$$V(x,y)=1/2x^2+1/2y^2+1/4y^4,$$

which is positive definite and $V(x=0,y=0)=0$.

The derivative is given by

$$\dot{V}=x\dot{x}+y\dot{y}+y^3\dot{y}=x[-y-y^3]+yx+y^3[x]\equiv0.$$

This condition implies that the origin is a center.

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The differential equation is the same as y"+(y+y^3)=0, which describes free vibration of an oscillator with a nonlinear spring. The system is conserve in energy. So its orbit is a closed curve.

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