Determining stability of equilibria of a nonlinear pendulum with torque ode system I am working with the system of ODE's or second order differential equation:
\begin{equation*}
  \theta'=v
\end{equation*}
\begin{equation*}
v'=-bv-\sin(\theta)+k
\end{equation*} 
with $b,k>0$ for physics reasons.
I determined that we have equilibria at \begin{equation*}
\begin{bmatrix}\theta\\v \end{bmatrix}=
\begin{bmatrix}\sin^{-1}(k)\\0 \end{bmatrix}
\end{equation*}
I want to determine the stability of these equilibrium points. Normally, I would linearize the system and use linear algebra tools to determine the stability of the linearized system and extrapolate. However, I am having trouble linearizing.
I could approximate the sine term via maclaurin expansion as just $\theta$ but this would be valid for $\theta$ near 0, not necessarily near $\sin^{-1}(k)$. Can you "ignore" trigonometric terms when linearizing the system? 
Should I instead take the jacobian of the system viewed as a function from $\mathbb{R}^2$ to itself? 
i.e. 
$D_{F}(\theta,v)=\begin{bmatrix} 0&1\\\cos\theta&-b \end{bmatrix}$
In that case, I am still unclear about how to treat the cosine term around a theta value which i have next to know information on. Should I taylor expand about $\sin^{-1}(k)$? 
 A: HINT :
in order to study the system for $\theta$ close to $\sin^{-1}(k)$ change of variable :
$$\theta(t)=\sin^{-1}(k)+\alpha(t)$$
$\sin(\theta)=\sin\left(\sin^{-1}(k)+\alpha \right)=k\cos(\alpha)+\sqrt{1-k^2}\sin(\alpha)$
$$\begin{cases}
\alpha'=v \\
v'=-bv-\left(k\cos(\alpha)+\sqrt{1-k^2}\sin(\alpha)\right)+k \\
\end{cases}$$
Now, with $\alpha$ close to $0$, you can approximate $\quad \sin(\alpha)\simeq \alpha\quad$ and $\quad\cos(\alpha)\simeq 1-\frac{1}{2}\alpha^2\simeq 1$
A: When $k>1$ obviously there is no equilibrium. When, $0<k<1$ there are infinite equilibrium points namely $(2m\pi+\theta_1,0)$, $(2m\pi+\theta_2,0)$, $m\in\mathbb{Z}$ with $\theta_1\in(0,\pi/2)$ and $\theta_2=\pi-\theta_1\in(\pi/2,\pi)$.  Define $\tilde{\theta}_{m,i}=\theta-(2m\pi+\theta_i)$ the error from the equilibrium $(2m\pi+\theta_i,0)$. Then,
$$\dot{v}=-bv-\sin\theta+\sin(\theta_i)\\ =-bv-2\cos\left(\frac{\tilde{\theta}_{m,i}}{2}+2m\pi+\theta_i\right)\sin\left(\frac{\tilde{\theta}_{m,i}}{2}\right)$$
The linearized error system around the equilibrium point $(2m\pi+\theta_i,0)$  is
$$\left[\matrix{\dot{\tilde{\theta}}_{m,i}\\ \dot{v}}\right]=\left[\matrix{0 & 1 \\ -\cos(\theta_i) & -b}  \right]\left[\matrix{{\tilde{\theta}}_{m,i}\\ {v}}\right]$$
The eigenvalues of the Jacobian of the nonlinear mapping are the roots of $$\lambda^2+b\lambda+\cos(\theta_i)=0$$
Whenever $\cos(\theta_i)<0$ then there is one unstable eigenvalue and the equilibrium is unstable. Thus, all $(2m\pi+\theta_2,0)$ are unstable equilibria. Points $(2m\pi+\theta_1,0)$ on the other hand are stable since the eigenvalues are 
$$\frac{-b\pm\sqrt{b^2-4\cos(\theta_1)}}{2}$$
in the case $b^2\geq 4\cos(\theta_1)$ which are both negative or
$$\frac{-b\pm j\sqrt{4\cos(\theta_1)-b^2}}{2}$$
in the case $b^2< 4\cos(\theta_1)$ which both have negative real part.
I am showing below the streamplot for the system with $b=1$ and $k=1/2$ which illustrates the above analysis
 
