Stability VS orbital stability in ODEs Let $\dot x=f(x)$ and ODE. 

We say that $x$ is a stable solution if $\forall \varepsilon >0$, $\exists \delta >0$ s.t for all solution $y(t)$ s.t. $\|x(t_0)-y(t_0)\|<\delta \implies \|x(t)-y(t)\|<\varepsilon .$

Let $\Phi_t$ denote the flow of the ODE, i.e. $x(t)=\Phi^{t,t_0}(x_0)$ for $x(t)$ is a solution.

We say that $x$ is orbitaly stable if $\forall \varepsilon >0$, $\exists \delta >0$ s.t. for all $\xi$ s.t. $\|x(t_0)-\Phi^{t_0,0}(\xi)\|<\delta \implies \text{dist}(\mathcal O^+(x_0,t_0),\Phi^{t,t_0}(\xi))<\varepsilon $ where $\mathcal O^+(x_0,t_0)=\{\Phi^{t,t_0}(x_0)\mid t\geq t_0\}$.

I see what is an stable solution, but I can't see what is an orbitaly stable solution. I don't really understand the example. Could someone explain ?
 A: Removing the orbit from the definition for orbitally stable gives

We say that x is orbitaly stable if $∀ε>0$, $∃δ>0$ s.t for any solution $y(t)$ with $∥x(t_0)−y(t_0)∥<δ$ it holds that for all $t>t_0$ there is some $s>t_0$ so that $∥x(s)−y(t)∥<ε$.

The distance in the second definition is to the curve, the set of points on the trajectory of $x$, not to a specific point on the path as in the first definition.

Take as example $\dot r = r(1-r^2)$, $\dot \theta=r$  in polar coordinates, that is, with $r=\sqrt{x_1^2+x_2^2}$
$$\dot x_1 = -rx_2+(1-r^2)x_1,\\ \dot x_2=~~~rx_1+(1-r^2)x_2,$$ with the closed loop solution $x(t)=(x_1(t),x_2(t))=(\cos t,\sin t)$. The general solution is given by
\begin{align}
%\frac{d}{dt}r(t)^{-2}&=2(1-r(t)^{-2})\implies (r(t)^{-2}-1)=e^{-2t}(r_0^{-2}-1)\\
r(t)&=\frac{r_0}{\sqrt{r_0^2+e^{-2t}(1-r_0^2)}}
=\frac{r_0e^t}{\sqrt{r_0^2(e^{2t}-1)+1}}
\end{align}
and 
$$
θ(t)
%=\int_0^t\frac{d(r_0e^t)}{\sqrt{(r_0^2e^t)^2-r_0^2+1}}
=t+\ln\left(r_0+\sqrt{e^{-2t}+r_0^2(1-e^{-2t})}\right)-\ln(r_0+1)
$$
While the orbit starting at $r(0)=r_0=x_{1,0}$, $θ(0)=0$ (with $x_{2,0}=0$) converges visually to the limit cycle, the angle evolution is initially very different from the unit speed on the limit cycle. Asymptotically $θ(t)\sim t+\ln(2r_0)-\ln(r_0+1)$, so that the angle difference stabilizes, but does not go to zero.
A: To understand the difference you need first carefully understand the difference between the integral curves, which are the graph of solutions in $\mathbb R\times \mathbb R^n$ and orbits, which are the images of the solutions in $\mathbb R^n$ parametrized by the time $t$.
For instance, take 
$$
\dot x=y,\\
\dot y=-x.
$$
Clearly your solutions will be 
$$
x(t)=C \cos t,\quad y(t)= C\sin t
$$
and you should sketch the corresponding graphs in $\mathbb R\times \mathbb R^2=\mathbb R^3$ and convince yourself that all the solutions are stable.
Now, the orbits are given by 
$$
x^2+y^2=C^2,
$$
which are circles, and it should be clear that each orbit starting close to another one stays close for any $t$, hence they are also orbitally stable.
Now take
$$
\dot x=y,\\
\dot y=-\sin x.
$$
The orbits will look almost the same (and since you are discussing orbital stability you should be able to sketch a phase portrait of this seen-in-any-texbook system), and therefore the corresponding solutions are orbitally stable. It is also well known (but not elementary to show) that each solution has its own period (depending on the initial conditions, this is a general property of nonlinear oscillators). This means that if you take two sufficiently close initial conditions the corresponding integral curves, given enough time, will part sufficiently far from each other and hence are not stable.
