How to find equilibrium points of this differential equation I'm trying to find the equilibrium points for the vector field of the following differential equation:
$$\frac{d^2\Psi}{dt^2}= \frac{g}{\xi\ell}\sin{\xi\Psi}-\frac{V^2}{\xi \ell R}\frac{\cos{\xi \Psi}}{\cos{\Psi}}\mathrm{sgn}\Psi,$$ for $\ 0≤\xi≤1.$
I tried doing this:
At equilibrium points, conditions are: $\frac{d^2\Psi}{dt^2}=\frac{d\Psi}{dt}=0;\ \Psi=\Psi_{eq}$. Thus, the differential equation reduces to this:
$$0=\sin{\xi\Psi_{eq}}-\frac{V^2}{g R}\frac{\cos{\xi \Psi_{eq}}}{\cos{\Psi_{eq}}}\mathrm{sgn}\Psi_{eq},$$ where $0≤\frac{V^2}{g R}≤2\ \mathrm{or}\ 3.$
This most likely isn't solvable analitically, so I thought about doing an average or approximating. What do you suggest?
 A: Firstly, we can eliminate $\operatorname{sgn}(\Psi)$ by by noting that if $\Psi$ is a solution, so is $-\Psi$. We are studying the transcendental equation
$$\tag{1}
\sin(\xi \Psi)=\beta\frac{\cos(\xi\Psi)}{\cos(\Psi)}
$$
I've grouped the other constants into $\beta$. An analytical solution is not likely, but we can estimate the solutions for different extreme values of $\xi$.
As $\xi\to 0$, $\Phi \propto1/\xi\to\infty$, which suggests defining $f=\xi \Psi$ so that $f$ is of order unity. Then (1) becomes
$$\tag{2}
\tan(f)\cos(f/\xi)=\beta
$$
The solutions of (2) are not-quite $2\pi$ periodic.  I start by plotting the LHS and RHS of (2) as functions of $f$.

As $\xi\to0$, the solutions are contained in the interval
$$
\tan^{-1}(\beta)<f<\pi-\tan^{-1}(\beta)
$$
Every oscillation of $\cos(f/\xi)$ yields two solutions, and the interval length is $\pi-2\tan^{-1}(\beta)$, so there are approximately $N=\left(1-\frac{2}{\pi}\tan^{-1}(\beta)\right)/\xi$ solutions for every interval of length $\pi$ in $f$. The first root is approximately in the interval
$$
\tan^{-1}(\beta)\leq f\leq \tan^{-1}(\beta) +2\pi \xi
$$
As $\xi \to 1$ there are always solutions near $\Psi=\pi(n+1/2)$, as well as several other solutions that may exist depending on the size of $\beta$. Multiply (1) by $\cos(\Phi)$, and let and $\xi=1-\varepsilon$ to get
$$\tag{3}
\sin((1-\varepsilon)\Psi)\cos(\Psi)=\beta \cos((1-\varepsilon)\Psi)
$$
We can develop a perturbation series to the roots: let $\Psi=\sum\limits_n \varepsilon^n x_n$. Substitute into (3) and match coefficients at each order of $\varepsilon$. At $\varepsilon^0$ I find
$$
\cos(x_0)\left[\beta-\sin(x_0) \right]=0\\
\therefore \cos(x_0)=0 \qquad \text{or} \qquad \sin(x_0)=\beta
$$
The $\cos(x_0)=0$ possibility furnishes all the roots near $\pi(n+1/2)$, and the $\sin(x_0)=\beta$ finds the others (the ones that exist conditionally on $\beta$). At $\varepsilon^1$ I find
$$
x_1=x_0 \frac{\cos^2(x_0)+\beta\sin(x_0)}{\cos^2(x_0)+(\beta-\sin(x_0))\sin(x_0)}
$$
It is straightforward but tedious to continue to higher orders. Note that the approximation is not uniform in $\beta$. Here is a plot of the LHS and RHS of (3). The vertical red lines are first three approximate roots at $\Psi= x_0+\varepsilon x_1$.

