# Pendulum system, stability on stationary points, check my answer

The formula that describes the swing of a pendulum is $\ddot{\theta}(t)+\gamma \dot{\theta}(t)+\omega ^2 \sin (\theta (t))=0$ where $\gamma = \frac{c}{mL}$ and $\omega = \frac{g}{L}$.

We want to represent this formula as a system of linear equations, find the stationary points, and find if the system is stable, asymptotically stable, or unstable at those points.

Firstly, let's assume that $\theta \approx 0$, so we can say that $\sin (\theta (t)) \approx \theta (t)$ and now our equation is $\ddot{\theta}(t)+\gamma \dot{\theta}(t)+\omega ^2 \theta (t)=0$.

I defined $x_1=\theta$, $x_2 = \dot{x_1}=\dot{\theta}$.

Then we have the following equations: $\dot{x_1}=x_2$, and $\dot{x_2} + \gamma x_2 +\omega ^2 x_1 =0$, or in other words, $\dot{x_2} = -\omega ^2 x_1 -\gamma x_2$.

So the linear system is:

$\begin{pmatrix} \dot{x_1} \\ \dot{x_2}\end{pmatrix}=\begin{pmatrix}0 & 1 \\ -\omega ^2 & -\gamma \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$

At the stationary points we have $\dot{x_1} = \dot{x_2}=0$, but $\dot{x_1}=x_2$, so we can infer that in any stationary point, we will have $x_2=0$.

Furthermore, we need $\dot{x_2}=-\omega^2 x_1-\gamma x_2 = -\omega ^2 x_1 =0$. Since $\omega$ is known and is not zero, we must have $x_1=0$. so the only stationary point of this system is $(x_1,x_2)=(0,0)$.

Is this correct? And how can we check the stability? I know it has something to do with the eigenvalues, but the eigenvalues of the system are independent of $x_1,x_2$...

The first assumption you applied assumes that $\theta\approx 0$, which allows you to linearize the differential equation and turns it into a linear autonomous and homogeneous differential equation, however such differential equation will always have an equilibrium solution at zero.
• If I understood you correctly, then at our original equation we plug in $\ddot{\theta} = \dot{\theta}=0$ since we are interested in stationary points, and we are left with $\omega ^2 \sin (\theta (t))=0$ which implies $\theta (t) =\pi k$, or in other words, we have a stationary point at $t_k=\pi k$, $k \in \mathbb Z$. And now I linearize the system and check the stability there? – Rick Joker Sep 30 '15 at 19:05
• @RickJoker Yes ($\theta=k\pi,\ k\in\mathbb{Z}$, I am not sure what you mean by $t_k$, it should not mean time) but notice that $\sin(\theta(t))$ is periodic with period $2\pi$, so you only have to find the stability of $k=0$ and $k=\pi$, because the first stability will be the same as $k=2n\pi, \ n\in\mathbb{Z}$ and the second stability will be the same as $k=\pi+2n\pi, \ n\in\mathbb{Z}$. – Kwin van der Veen Sep 30 '15 at 19:12
• @RickJoker For stationary solution you are correct that $\theta(t)=k\pi,\ k\in\mathbb{Z}$ are solutions, but a stationary solution means that all derivatives of $\theta(t)$ with respect to time are zero $\forall t$ and thus should be equal to a constant, namely $k\pi,\ k\in\mathbb{Z}$. – Kwin van der Veen Sep 30 '15 at 19:21