Example of an ODE with an asymptotically stable equilibrium which is unstable in the corresponding linearization I am looking for an example of an ODE $x'=f(x)$ with an asymptotically stable equilibrium (WLOG $x=0$) such that $0$ is an unstable equilibrium of the linearized equation $x'=Ax$ where $A=Df(0)$. The strategy is to choose $f$ so that zero is an eigenvalue of $A$ whose algebraic multiplicity exceeds its geometric multiplicity (leading to linearly growing solutions), but whose nonlinear parts promote stability.
The example I came up with is $$f:\mathbb R^2\to\mathbb R^2, \quad f(x_1,x_2)=(x_2-x_1^3,-x_2^3),\quad A=Df(0)=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}.$$ In the above example, solutions to the linearized problem $x'=Ax$ have the form $x(t)=(c_2 t+c_1,c_2)$ for constants $c_1$ and $c_2$, so $0$ is an unstable equilibrium. The nonlinear problem, however, has the following phase diagram:

This phase diagram seems to strongly suggest that the origin is asymptotically stable in the nonlinear system, but I cannot prove it with, e.g., a Lyapunov function. How can I show the origin is asymptotically stable in this or a similar example?
 A: I think a direct argument just based on the direction of the vector field should work.
Let me write $(x,y)$ instead of $(x_1,x_2)$, for simplicity. Then the ODEs are $\dot x = y-x^3$ and $\dot y = -y^3$. So it's clear that all trajectories have the property that $y(t) \to 0$ monotonically as $t \to \infty$.
Now consider to begin with the region $x>0$ and $0<y<x^3$, i.e., the region between the positive $x$-axis (which is a $y$-nullcline) and the right half of the curve $y=x^3$ (which is an $x$-nullcline). In that region, $\dot x$ and $\dot y$ are both negative, so a trajectory $(x(t),y(t))$ starting in that region cannot exit it (since it can't cross those nullclines; the vector field isn't pointing out of the region anywhere on those curves). So $x(t)$ is bounded from below (by zero) and decreasing, and must therefore have a limit $x^* \ge 0$ as $t \to \infty$. Thus, $(x(t),y(t)) \to (x^*,0)$, and the only possibility is then that $x^*=0$, since the points $(x^*,0)$ with $x^*>0$ aren't equilibria.
Next, if you start in the upper half plane, but to the left of the curve $y=x^3$, you have $\dot x>0$ and $\dot y<0$, so you will go downwards to the right. Then either $(x(t),y(t)) \to (0,0)$ directly, monotonically in both $x$ and $y$ (which I don't think actually can happen, but that doesn't matter), or you will eventually cross the curve $y=x^3$ and enter into the region covered in the first case, which will force you to tend to the origin from there.
Similarly if you start anywhere in the lower half plane, because of the rotational symmetry of the system. And if you start on the $x$-axis, it's also clear that you tend to the origin.
So all trajectories tend to the origin.
And the origin is a stable equilibrium to begin with; this wasn't immediately clear to me, but I convinced myself as follows: given a neighbourhood $U$, one can take a rectangle $V = (-a,a) \times (-a^3,a^3)$ with $a>0$ small enough so that $V \subseteq U$, and then it can be seen by similar arguments as above that trajectories starting in $V$ will stay in $V$, and hence in $U$.
