Consider the system,

\begin{align} \dot{x}&=x^2 \\ \dot y&=-y \end{align}

I am trying to show that this system has infinitely many local center manifolds.

Here is what I have done so far: Clearly the system has a rest point at the origin. I linearized the system at the origin and got that $\lambda=0$ and $\lambda=-1$ to be the eigenvalues of the linearized operator. Turns out that the corresponding eigenvectors are $(1,0)^T$ and $(0,1)^T$ respectively. So solutions of the linearized system can be written as follows.

\begin{align} x&=c_1\\ y&=c_2e^{-t} \end{align}

When $c_1=0$, $y\to0$ as $t\to \infty$. Thus, by definition of the stable manifold the $y$ axis must be a stable manifold. Also the unstable manifold is the trivial set $(0,0)$.

Questions: How do I go about showing that there are infinitely many center manifolds?. Does solving the system explicitly have anything to do with answering this question?. In particular any orbit of the system not on the stable manifold $(0,y)$ satisfies an equation of the form $\displaystyle y=c_2e^{\frac{1}{x}-c_1}$. I feel like I am not seeing something simple. Can somebody give an explanation?

Context: In the ODE book I am using (C. Chicone) this problem is given as an example to show that the center manifold need not be unique. (After proving the invariant manifold theorem).


The center subspace is the line $y=0$. The stable subspace is the line $x=0$. For every point $(x_0,y_0)$, the solution passing through $(x_0,y_0)$ is such that, for every $t$ in the interval of definition around $0$, $$x(t)=\frac{x_0}{1-x_0t},\qquad y(t)=y_0\mathrm e^{-t}.$$ Right half-plane: If $y_0\ne0$, $x_0\gt0$, then $x(t)\to+\infty$ and $y(t)\to y_0\exp(-1/x_0)\ne0$ when $t\to1/x_0$ the upper limit of its interval of definition.

Thus, no point $(x_0,y_0)$ such that $y_0\ne0$ and $x_0\gt0$ belongs to a center manifold.

Left half-plane: If $y_0\ne0$, $x_0\lt0$, then $x(t)\to0$ and $y(t)\to0$ when $t\to+\infty$. The solution stays on the curve with equation $$y=y_0\exp(1/x-1/x_0),\qquad x\lt0,$$ hence $y\ll x$ when $x\to0$, $x\lt0$, that is, the curve is tangent at $(0,0)$ to the line $y=0$.

Thus, every point $(x_0,y_0)$ such that $y_0\ne0$ and $x_0\lt0$ belongs to a center manifold with associated center subspace $(y=0,x\lt0)$. On the diagram below, these are the trajectories in the left half-plane pointing to $(0,0)$, the axis $y=0$ excepted.

$\qquad\qquad\qquad\qquad$ Phase portrait around $(0,0)$

  • $\begingroup$ Makes sense.Thanks for the explanation. Do you mind telling me what you used to draw the phase portrait diagram?. $\endgroup$ – minibuffer Aug 21 '14 at 21:13
  • $\begingroup$ Exported the result of streamplot x^2,-y from W|A. $\endgroup$ – Did Aug 21 '14 at 21:18

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