Consider an ODE $x'=f(x)$, where $f:U\to\mathbb{R}^n$ is $C^1$ and $U$ is some open subset of $R^n$.
Supose $x^*\in U$ is an unstable singularity of the ODE, that is:
$f(x^*)=0$ and there is a neighborhood $W$ of $x^*$ such that there are solutions with initial condition arbitrarily close to $x^*$ that leave $W$.
How can I prove that there is at least one non-trivial solution $x$ such that $\displaystyle \lim_{t\to-\infty}x(t)=x^*$?
I'm not sure this existence is necessary, but every example of unstable singularity that I know of satisfies this property.
My approach so far was the following:
$~$
Let $\Phi(t,x)$ be the flow of the ODE and WLOG we can consider $W=B(x^*,r)$ for some radius $r>0$.
By continuity, the hypothesis of there being initial conditions $x_n\to x^*$ that leave $W$ implies that for each $n$ there is a $t_n>0$ where $\Phi(t_n,x_n)=c_n\in S$ where $S$ is the border of $W$.
Since $S$ is compact, by using a subsequence if necessary, we can consider $c_n\to c\in S$.
We can also rewrite $\Phi(t_n,x_n)=c_n$ as $\Phi(-t_n,c_n)=x_n$.
So, in short, I have the following relevant properties:
$\bullet~c_n\to c$ ;
$\bullet~\Phi(-t_n,c_n)=x_n$ ;
$\bullet~x_n\to x^*$ ;
$\bullet~\Phi$ is differentiable (with respect to both $t$ and $x$) .
The last step would be to show that $x(t):=\Phi(t,c)$ is the solution we are looking for, but I couldn't manage to do it.
Turns out it is not only false, but also, unlike I said, there is a linear counterexample in $\mathbb{R}^2$:
$x'= \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}x $ .
All the solutions are of the form: $x(t)= \begin{bmatrix} x_0+ty_0\\ y_0 \end{bmatrix}~$ , with $~x_0,y_0\in\mathbb{R}$ .
Clearly, the limit in question only converges if $y_0=0$, in that case $x(t)\equiv \begin{bmatrix} x_0\\ y_0 \end{bmatrix}$ , but we were looking for non-trivial solutions, so there are none.