Show that $(1,0)$ is not Liapunov stable I have a couple of questions regarding an example.
Given the system 
$$\begin{cases}
x' = x-y-x(x^2+y^2) + \frac{xy}{\sqrt{x^2+y^2}} \\[2ex] 
y' = x+y-y(x^2+y^2) - \frac{x^2}{\sqrt{x^2+y^2}}, \\[2ex] 
\end{cases}
$$
we want to show that $(1,0)$ is not stable.
So first, transforming this into polar coordinates, we have $r' = r(1-r^2)$ and $\theta' = 2\sin(\theta/2)^2$. We see that since $r' > 0$ when $0 < r < 1$ and $r' < 0$ when $r > 1$, we must have $r(t) \rightarrow 1$ as $t \rightarrow \infty$. So far so good, but the first claim that I don't understand is that this is true as long as $r(0) \neq 0$. Similarly, we have $\theta(t) \rightarrow 2\pi$. So every solution converges to $(1,0)$ except solutions starting at the origin. This too confuses me and I'm guessing that it is related to the previously mentioned claim. (I understand why the solutions converge to $(1,0)$, but not why solutions starting at the origin are expections.)
I believe I will be able to follow the rest of the example once I've figured these two issues out. 
 A: I suppose it boils down to the solution of your system of ODEs.
Solving $r' = r(1-r^2)$, we have
$$ r(t) = \pm \frac{e^t}{\sqrt{c_1 + e^{2t}}}.$$
If $r(0)=0$, we have a problem, because there we have 
$$0 = \frac{1}{\sqrt{c_1+1}}$$
$$ \implies c_1 = 0,$$
and the solution to the ODE is $f(t) = 0$, constantly. This certainly does not tend to $(1,0)$.
A: The functions on the right-hand sides of your ODEs are undefined at the point $(x,y)=(0,0)$, because of the division by $\sqrt{x^2+y^2}$. But they tend to zero as $(x,y)\to (0,0)$, so if you interpret the system as
$$
x'=F(x,y)
,\qquad
y'=G(x,y)
$$
where
$$
F(x,y) = 
\begin{cases}
x-y - x(x^2+y^2) - \frac{xy}{\sqrt{x^2+y^2}},
& (x,y) \neq (0,0),
\\
0,
& (x,y)=(0,0)
,
\end{cases}
$$
and similarly for $G(x,y)$, then you have a system with continuous right-hand sides at least, and actually Lipschitz continuous (as can be checked) so that you have existence and uniqueness of solutions. And for this system, $(0,0)$ is an equilibrium, since $F(0,0)=G(0,0)=0$. So the (unique) solution starting at $(0,0)$ must stay at $(0,0)$ forever, and consequently can never reach $(1,0)$.
