\begin{align} \dot x&=\frac{x^2t}{z},&x(t_0)&=x^0=1,~~t_0=0,\\ ε\dot z&=-(z+xt)(z-2)(z-4),&z(t_0)&=z^0, \end{align} I am studying the book "Singular Perturbation Methods in Control Analysis and Design" and I am solving one of the examples (above) in which is required to check the following Assumptions:
where the system (3.6) is
The specific Example is
I don't understand where the boundaries on $z^0$, i.e. $z^0<2$ and $z^0>2$ are coming from. I am sure they are related to Assumption 3.1, but I don't see how to get the domain of attraction of $\hat{z}(\tau)=0$ of the system (3.6).
Thank you very much for help, and I am sorry if I put too many images, but it would have taken ages to rewrite everything down.
The general idea is that the second equation provides a very rapid motion towards one of the equilibria of $g$ in a timescale of $\epsilon^{1-\alpha}$ for any small $1>α>0$. Even if the equilibria of $g$ change over time, this change is glacial compared to this rapid movement of $z$.
So to get a qualitative picture of the flow of the system, you only need to consider the equilibria of the second equation for (almost) constant $t$ and $x$. Here you get $-xt, 2, 4$ and they keep that order for $t\ge 0$, as the plane $\{x=0\}$ is filled with solutions, so for non-zero $x_0$ the $x$ component can not change its sign.
The right side of the second equation has signs $+,-,+,-$ for the intervals outside and between the roots, so that $-xt$ and $4$ are stable, $2$ is unstable. If $z_0\ne 2$, the solution will thus move rapidly away from that plane $\{z=2\}$, towards $4$ for $z_0>2$ and towards $-xt$ for $z_0<2$.
In the first case $z=4$ the equilibrium solution leads to $\dot x=\frac{x^2t}4$, which gives $x^{-1}-1=-\frac{t^2}8\implies x(t)=\frac{8}{8-t^2}$.
In the case $z=-xt$ we get $\dot x=-x$, $x(t)=e^{-t}$