Q: Determine whether the solution of pendulum equation without friction is stable Let us assume we have the simplest DE of a pendulum, without friction:
$$\frac{d^2\phi}{d t^2}+\omega^2sin(\phi)=0$$
where $\phi$ is the angle of alteration. Boundary and initial conditions: $\phi(t=0)=\phi_0$ and $\frac{d\phi}{d t}=0$. For the sake of simplicity: $\omega=1$. I would like to find a stationary solution, so $$\frac{d}{d t}=0, \frac{d^2\phi}{d t^2}=0$$ Therefore, the next step is to find a solution of the following equation: $$sin(\phi)=0$$ which gives me two points: $$\phi=0, \phi=\pi/2$$ Now I want to examine the stability of these points.
If the angle between $\phi$ and the normal is zero, I suppose it is an absolutely stable state to an external perturbation, and if $\phi=\pi/2$ the pendulum is placed upside down and this particular state is absolutely unstable to an external perturbation.
I tried to find a way to represent this assumption in analytical mathematical form, but I could not get anything. There are many sources to explain what stability is, Lyapunov stability and how to find the stability of the ODE system, but this very example I failed to examine.
Could someone please help me to prove that $\phi=0$ is a stable state and $\phi=\pi/2$ is unstable to external perturbations..
Thank you in advance!
