Set up:
I am considering a plane autonomous system where there exists two ODEs, $\frac{dx}{dt}=X(x,y),\frac{dy}{dt}=Y(x,y)$. We then usually draw trajectories on the phase plane to indicate the solutions on the plane.
Question:
I often see trajectories sometimes cross at critical points, however I was wondering, if a trajectory does contain a critical point, wouldn't this trajectory just stay at that critical point rather than passing through that critical point since at that point $X=Y=0$ and so just becomes stationary?
I am not too sure if I made much sense with my question but any help is appreciated!
The simplest example is $${dx\over dt}=x,\quad {dy\over dt}=-y\ .$$ When ${\bf r}(0)=(x_0,0)$, $\>x_0\ne0$, then $${\bf r}(t)=(x_0e^t,0)\qquad(-\infty<t<\infty)\ ,$$ hence the half $x$-axis containing $x_0$ is a solution curve. Similarly, when ${\bf r}(0)=(0,y_0)$, $\>y_0\ne0$, then $${\bf r}(t)=(0,y_0e^{-t})\qquad(-\infty<t<\infty)\ ,$$ hence the half $y$-axis containing $y_0$ is a solution curve. But note that this curve reaches $(0,0)$ only in the limit $t\to\infty$. This means that the traveling point needs infinite time to reach the origin. And of course the initial condition ${\bf r}(0)=(0,0)$ leads to ${\bf r}(t)=(0,0)$ for all $t$.
Therefore no trajectory "traverses" the stationary point $(0,0)$.