# 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 $x_1,x_2$...

## 1 Answer

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.

The correct way to solve this is to first find the equilibrium solution of the non-linear differential equation and then use linearisation in order to find the stability of the system at those equilibrium solutions.

• 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