Dynamical system with infinite vector field

Question

I have a system of the form: $[\dot{x},\dot{\alpha}] = [\text{tan}(\alpha),1]$. When my system goes to $\alpha = \frac{\pi}{2}$ or to $\alpha = \frac{3\pi}{2}$ the first component of the vector field goes to infinity. This can be seen as an instability or I have to impose some initial conditions?

Answer 1

When you are given a vector field $$ \begin{pmatrix}\dot{x} \\ \dot{\alpha}\end{pmatrix} = \begin{pmatrix}\tan\alpha \\ 1\end{pmatrix} $$ you cannot say anything about the behaviour "at" $\alpha=n \frac{\pi}{2}$ for $n$ odd. This is because the vector field is not defined at those points. As a result we can only consider solutions to the initial value problem where $\alpha\neq n \frac{\pi}{2}.$

You can verify that solutions to the differential equation, assuming an initial condition $(x, \alpha)\in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\times\mathbb{R}$ exhibit finite-escape time. For example, suppose $x(0)=0$ and $\alpha(0) = 0.$ Then $$ \begin{pmatrix} x(t)\\ \alpha(t) \end{pmatrix} = \begin{pmatrix} \ln(\sec(t)) \\ t \end{pmatrix} $$ is only valid for $t\in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ The key here is that any solution $(x(t), \alpha(t))$, given an initial condition, will be constrained in time $t$ even though (in some sense) $\alpha$ has a global solution by itself.