I am having some problems with the following question:
Show that the origin is not Lyapunov-stable for the differential equation $x'=x^3$
Now, if we solve for the flow $\phi_t(x)$, we obtain that
$$\phi_t(x) = \frac{1}{\sqrt{\frac{1}{x^2} - 2t}}$$
A point $x_0$ is Lyapunov-stable if for each $\epsilon > 0$ there is a $\delta > 0$ where $|\phi_t(x) - \phi_t(x_0)| <\epsilon$ if $|x - x_0| < \delta$
My problem is that $\phi_t(x)$ is not even defined at $x = 0$, so how can I go about disproving this? (Or is this fact enough?)
The origin is not stable for the following reason: Since $x'=x^3$, solutions starting at any $x_1\neq 0$ get farther and farther away from the origin ($x'(t)>0$ if $x_1>0$ or $x'(t)<0$ if $x_1< 0$). In other words, solutions starting arbitrarily close to the stationary solution $x=0$ do not stay close for all $t>0$. Given any $\varepsilon>0$, no matter what $\delta>0$ we choose, for any $x_1$ with $0<|x_1|<\delta$ we have $|x(t)|>\varepsilon$ for $t>t_0$.