stability and asymptotic stability: unstable but asymptotically convergent solution of nonlinear system Consider nonlinear systems of the form $X(t)'=F(X(t))$, where $F$ is smooth (assume $C^\infty$). Is it possible to construct such a system (preferably planar system) so that $X_0$ is an unstable equilibrium, but all nearby solution curves tend to $X_0$ as $t \to \infty$? If so, how? A conceptual construction is enough. What will the phase portrait look like?


Definition of stability:
Suppose $X^∗$ ∈ $\mathbb{R}^n$ is an equilibrium point for the differential equation
$$X′ = F(X).$$
Then $X^∗$ is a stable equilibrium if for every neighborhood $\mathcal{O}$ of $X^∗$ in $\mathbb{R}^n$ there is a neighborhood $\mathcal{O}_1$ of $X^∗$ in $\mathcal{O}$ such that every solution $X(t)$ with $X(0) = X_0$ in $\mathcal{O}_1$ is defined and remains in $\mathcal{O}$ for all $t > 0$.

and

Definition of asymptotic stability:
If $\mathcal{O}_1$ can be chosen so that, in addition to the properties for stability, we have $\lim_{t \to \infty} X(t) = X^∗$, then we say that $X^∗$ is asymptotically stable.

The exact problem:

In the definition of an asymptotically stable equilibrium point, we required that the equilibrium point also be stable. This requirement is not vacuous. Give an example of a phase portrait (a sketch is sufficient) that has an equilibrium point toward which all nearby solution curves (eventually) tend, but which is not stable.


The $C^1$ example I found with the aid of Mathematica (I guess $C^\infty$ is not possible):
StreamPlot[{x^2 + (-(x^2 + 1/16) (y - 1/2)^3) Boole[y >= 1/2], -y + (y - 1)^2 y Boole[y >= 1] - 500 (y - 2)^4 Boole[y >= 2] + (x + 1)^3 Boole[x <= -1] + (x - 1)^3 Boole[x >= 1]}, {x, -4, 4}, {y, -4, 4}]

The phase portrait is here (I don't have privilege to post image; you can always run Mathematica with the above code to get the phase portrait). The example is unprecedentedly ugly, but at least it works.
This is inspired by the system
$$x'=x^2,$$
$$y'=-y.$$
and in a small neighborhood of 0 (actually $[-1,1] \times [-\infty, 1/2]$), the system is exactly the above simple simple. All solutions nearby 0 goes back to 0, since nearby solutions in the left half plane naturally converge to 0, and nearby solutions in the right half plane first converge to the $x$-axis, then wind back to the left half plane.
