Find the stable, unstable, and center manifolds of the following system
\begin{align}
\dot{x}& = -x,\\
\dot{y}& = -2y + 2z^2(1-z^2), \\
\dot{z} &= -z^3
\end{align}
at the origin. Prove that origin is the local attractor.
Solving the above system I got: $$x = x_0 \ e^{-t}, \ y(t) = \frac{1}{2}\bigg[\frac{1}{t+z_0} + \bigg(\frac{2y_0 z_0 -1}{z_0}\bigg)e^{-2t} \bigg], \ z(t) = \sqrt{\frac{1}{2(t+z_0)}}$$
Stable manifold: $$W^{s}(0,0,0) := \{(x,y,z):x(t), y(t) \text{ and } z(t) \to 0 \text{ as } t \to +\infty\} =\{(x,0,z)\}$$
Unstable manifold: $$W^{u}(0,0,0) := \{(x,y,z):x(t), y(t) \text{ and } z(t) \to 0 \text{ as } t \to -\infty\} = \{(x,y,z):
x=0, yz=\frac{1}{2} \} $$
I do not whether my calculations are right or not. I do not know how to prove the second part.
Best Answer
First note that the third equation has the solution $$ z(t)^{-2}-z_0^{-2}=2t\implies z(t)=\frac{z_0}{\sqrt{1+2z_0^2t}} $$ where now indeed $z(0)=z_0$.
Next, you can be a little creative in solving the second equation by inserting the third to reduce the 4th degree term, $$ \dot y+2y=2z^2-2z^4=2z^2+2z\dot z=\frac{d}{dt}(z^2)+2z^2. $$ This means that "luckily" the terms turn out to be that the same differential operator $(D+2)$ is applied to both sides. The solution then is $$ y(t)-z(t)^2=(y_0-z_0^2)e^{-2t},\\ y(t)=\frac{z_0^2}{1+2z_0^2t}+(y_0-z_0^2)e^{-2t}. $$ I do not see how there can be any unstable manifold. From points outside the set with $x_0=0$ and $y_0=z_0^2$ the solutions will converge exponentially fast towards points within that set, and then slowly along that set towards the origin. In my mind this was the characteristic of a center manifold?