Asymptotical stability for $x'=-x^3$ and $x'=x^3$ The task is to check that the stationary point $x_0=0$  is asymptotically stable for the equation $x'=-x^3$,but is not asymptotically stable for the equation $x'=x^3$, despite the fact that that linearised equation in both cases is $x'=0$. 
I know that $x_0$ is asymptotically stable if:


*

*$x_0$ is stable

*$ \exists \delta >0$ that  $ ||x-x_0|| < \delta \Rightarrow \lim_{t\to\infty}||\varphi (t,x) - x_0|| =0$
I've tried to find solution of both equations and then use this definition but it led me nowhere. How should I proceed with this type of tasks? 
 A: You can use the Lyapunov function $V(x)=x^2$. For the system
$$\tag{1}
\dot x= -x^3
$$
the derivative along the trajectories is
$$
\dot V\Big|_{(1)}= 2x\dot x= -2x^4.
$$
It is negative definite, so the the equilibrium point $x=0$ is asymptotically stable.
For the system
$$\tag{2}
\dot x= x^3
$$
the derivative along the trajectories is
$$
\dot V\Big|_{(1)}= 2x\dot x= 2x^4.
$$
It is positive definite, so the the equilibrium point $x=0$ is unstable.
Alternatively, you can notice the fact that for the system (1) $x(t)$ decreases (because $\dot x<0$) when $x(t)>0$ and increases (because $\dot x>0$) when $x(t)<0$. Hence, the solutions always move toward zero, that is, the norm $\|x(t)-x_0\|=|x(t)|$ decreases. This implies stability of the origin. In order to prove asymptotic stability, consider the general solution to (1)
$$
x(t)=\frac1{\sqrt{C+2t}}.
$$
It tends to zero when $t\to+\infty$, thus, the origin is asymptotically stable.
As for the system (2), $x(t)$ increases when $x(t)>0$ and decreases when $x(t)<0$. Hence, the direction is away from zero, the norm increases, hence, the origin is unstable. 
A: You can still use stability analysis for problems like this. Suppose we have $\dot{x} = f(x)$, with $f(x_0) = f^{(1)}(x_0) = ... = f^{(n-1)}(x_0) = 0$ but $f^{(n)}(x_0) \neq 0$. To examine the stability of $x_0$ let $x = x_0 + \varepsilon(t)$ where $\varepsilon$ is some small perturbation from the fixed point at $x = x_0$. Expanding $f(x) = f(x_0+\varepsilon)$ into a Taylor series about $\varepsilon = 0$ gives 
$$ f(\varepsilon) = f(x_0) + f^{(1)}(x_0)\varepsilon + \frac{1}{2}f^{(2)}(x_0)\varepsilon^2+...+\frac{1}{n!}f^{(n)}(x_0)\varepsilon^n + O(\varepsilon^{n+1})$$
In other words $$\dot{\varepsilon} \approx \frac{1}{n!}f^{(n)}(x_0)\varepsilon^n$$ 
This gives you a differential equation which you can solve and then look at the behaviour of the perturbation as $t \rightarrow \infty$. Do be careful though because sometimes the behaviour can depend on the initial sign of $\varepsilon$, in which case you have a semistable fixed point at $x = x_0$. 
Linear stability analysis is just a special case of this. 
In the problem that you mention, you would have $f(0) = f^{(1)}(0) = f^{(2)}(0) = 0$ and $f^{(3)}(0) = -6$.
