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I am trying to solve the following problem.

Determine a center manifold for the rest point at the origin of the system

\begin{align} \dot x &=-xy \\ \dot y&= -y+x^2-2y^2 \end{align} a) Deternine a center manifold for the rest point at the origin of the system and a differential equation for the dynamics on this center manifold.

b) Show that every solution of the system is attracted to the center manifold.

c) Determine the stability type of the rest point at the origin.

Progress so far: By the Invariant manifold theorem (IMT) we look for an invariant manifold as a graph of a function of the form,

$$ y=g(x)=a_0+a_1x+\alpha x^2+ \beta x^3+\gamma x^4+\delta x^5+\theta x^6+ \mathcal O(x^7) \tag{A}$$ By IMT we get that the constant term and the $x$ term to be zero. By invariance we need to have that $\dot y =g'(x)\dot x$. After some tedious computations and comparing coefficients of $x$ of both sides we get,

$$g(x)=x^2-4x^4+16x^6-64x^8+\cdots$$ Thus we can conclude that (using properties of geometric series)

$$ g(x)=\frac{x^2}{1+2x^2}$$ for $x$ near zero. This is the required center manifold. A differential equation for the dynamics of the center manifold is $\dot x=-xg(x)$ with $g(x)$ as above.

Note I am able to answer part c) because $\dot x =-x^3+\mathcal O(x^5)$ and since the coefficient of $x^3$ is negative it follows that the origin is an asymptotically stable rest point for the center manifold and hence for the original system.

Question:I am a little confused on what part b) is asking me?. How do I go about answering part b). I am pretty sure it has something to do with the reduced system in the center manifold

Can someone give an explanation?. Thanks.

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  • $\begingroup$ I guess that if you start the motion with $x(t=0)=x_0$ and $y(t=0)=y_0 \not =g(x_0)$, then $y(t)$ will eventually reach $g(x(t))$. $\endgroup$ – mike Aug 22 '14 at 23:30
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Recall that the center manifold of a dynamical system at an equilibrium point is made of the orbits tangent to the center subspace at this point, where the center subspace is spanned by the eigenvectors of the linearized system at this point, corresponding to eigenvalues with real part zero.

Here one considers the equilibrium point $(0,0)$, the linearized system at this point is $$\dot x=0,\qquad \dot y=-y,$$ hence the two eigenvalues at $(0,0)$ are $0$ with eigenvector $(1,0)$, which yields the center subspace $\{(x,y)\mid y=0\}$, and $-1$ with eigenvector $(0,1)$, which yields the stable subspace $\{(x,y)\mid x=0\}$.

Regarding the center manifold $C$ at $(0,0)$, after some tedious computations and comparing coefficients of $x$ of both sides... one gets a rather different result, namely, in your notations, $$g(x)=x^2.$$ Thus, $C=\{(x,y)\mid y=x^2\}$ and the dynamics on $C$ is $$\dot x=-xg(x)=-x^3.$$ To solve question (b), change coordinates by considering $$y=g(x)\cdot z=x^2\cdot z,$$ which only excludes, as is natural, the stable manifold $S=\{(x,y)\mid x=0\}$. Then the $(x,z)$ differential system is $$\dot x=-x^3\cdot z,\qquad \dot z=-z+1,$$ in particular, $$z(t)\to1.$$ Thus, every solution of the system is attracted to the center manifold $C=\{(x,y)\mid y=g(x)\}$ in the sense that, for every initial condition $(x(0),y(0))$ not in the stable manifold $S=\{(x,y)\mid g(x)=0\}$, $$\lim_{t\to+\infty}\frac{y(t)}{g(x(t))}=1.$$

$\hspace{4cm}$enter image description here

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You know from the eigenvalues of the Jacobian that one eigenvalue is $0$ and the other is $-1$. That is, you have a one dimensional centre manifold and a one dimensional stable manifold. Since both manifolds intersect transversally, every nearby solution follow such invariant manifolds. Therefore, locally, every solution away from a centre manifold follows the stable manifold, which implies that the orbits approach the centre manifold. See the figure below.

enter image description here

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  • $\begingroup$ Thanks for your answer. is this image from a book you are using?. Would you mind telling me what it is?. It would be helpful If I could see some more examples. $\endgroup$ – minibuffer Aug 24 '14 at 1:30
  • $\begingroup$ I made it on Inkscape together with an extension called text ext, which allows you to "write Latex". $\endgroup$ – PepeToro Aug 24 '14 at 11:16
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    $\begingroup$ **tex text is the extension. $\endgroup$ – PepeToro Aug 26 '14 at 10:31

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