Lately, I've been trying to find the period of an angle included in the following differential equations, but only could with the basic model:
Basic or original: $$\mathrm{For}\ (\Phi (0), \Omega (0))=(\Phi_{o},0),\ \frac{d^2\Phi}{dt^2}= \frac{g}{\ell_{o}}\sin{\Phi}-\frac{g}{\ell_{o}}\zeta\ \mathrm{sgn\ \Phi}\ ;$$
Modified: $$\mathrm{For\ the\ same\ initial\ conditions},\ \frac{d^2\Phi}{dt^2}= \frac{g}{\ell_{o}}\frac{\sin{\Phi}}{f(\Phi)}-\frac{g}{\ell_{o}}\zeta \frac{\mathrm{sgn\ \Phi}}{f(\Phi)}\ -2\dot{\Phi}^2 \frac{f'(\Phi)}{f(\Phi)}.$$
Where $g$ is gravity, $\ell_{o}$ is the length of the inverted pendulum, $\zeta$ a group of other constants, $\operatorname{sgn}\left(\cdot\right)$ is the signum function, $\dot{\Phi}=\Omega=\frac{d\Phi}{dt}$, $f(\Phi)=\sqrt[3]{1-\eta\cos{\Phi}}$ ($\eta$ is another constant) and $f'(\Phi)=\frac{df(\Phi)}{d\Phi}$.
And so, the method I used to get the period was basically this:
Let $F(\Phi)= \frac{g}{\ell_{o}}\sin{\Phi}-\frac{g}{\ell_{o}}\zeta\ \mathrm{sgn\ \Phi}$ , then the diff. eq. reduces to $\frac{d^2\Phi}{dt^2}=F(\Phi).$ And now I just proceed. \begin{align} \int \frac{d^2\Phi}{dt^2}d\Phi &= \int F(\Phi)\ d\Phi\\ \frac{1}{2}\dot{\Phi}^2 &= \int F(\Phi)\ d\Phi\ +C\\ \dot{\Phi} &= \frac{d\Phi}{dt} = \sqrt{2\int F(\Phi)\ d\Phi +C}\\ \frac{T}{4}=\int_{t_{o}}^{t_{1}}dt &= \int_{0}^{\Phi_{o}}\frac{d\Phi}{\sqrt{2\int F(\Phi)\ d\Phi +C}}\\ T &=2\sqrt{2} \int_{0}^{\Phi_{o}}\frac{d\Phi}{\sqrt{\int F(\Phi)\ d\Phi +C}}. \end{align}
This worked for the basic model; but didn't for the modified one. The issue was the integral of $F(\Phi)$ since in the modified version it included all terms divided by $f(\Phi)$ and also the $\dot{\Phi}^2 \frac{f'(\Phi)}{f(\Phi)}$ one too. Can someone tell me any easier way to attain the period of this modified system? Or what approximation could I use to make it easier to deal with?
I just found a really really good approximation, but couldn't find its analitical expression in the form of an integral...
For anyone wondering, here it is:
Since from the last integration step we see that the period is expressed in terms of the initial angle, I thought about taking the average of $\ell$, $\frac{1}{\ell}$ and $\ell'(\Phi)$ and setting integration limits at $-\Phi_{o}$ and $\Phi_{o}$ (the two maximum angles at which the inverted pendulum swings). Only averages I could find were the ones for $\ell'(\Phi)$ and $\frac{1}{\ell}$. So, briefly:
$$\langle \ell'(\Phi)\rangle = \frac{\ell_{o}}{2\Phi_{o}} \int_{-\Phi_{o}}^{\Phi_{o}} f'(\Phi)\ d\Phi = \frac{\ell_{o}}{2\Phi_{o}}(f(\Phi_{o})-f(-\Phi_{o}))=0$$ $$\bigg\langle \frac{1}{\ell} \bigg\rangle = \frac{1}{2\ell_{o}\Phi_{o}} \int_{-\Phi_{o}}^{\Phi_{o}} \frac{1}{f(\Phi)}\ d\Phi = \frac{1}{\ell_{o}\Phi_{o}} \int_{0}^{\Phi_{o}} \frac{1}{f(\Phi)}\ d\Phi$$
Since for the integration of $\frac{1}{f(\Phi)}=\frac{1}{\sqrt[3]{1-\eta \cos{\Phi}}}$ the solution was undefined at $0$, doing the approximation $\cos{\Phi} ≈ 1-\frac{\Phi^2}{2}$, which is very precise for $|\Phi|<90º$, we could actually get a solution which is not undefined at $0$ and evaluated at which is $0$ too:
$$\Phi\sqrt[3]{\frac{1}{1-\eta}}\ \ {}_{2}F_{1} \bigg(\frac{1}{3},\frac{1}{2};\frac{3}{2};\frac{\eta}{2\eta-2} \Phi^2 \bigg) +C$$
Thus: $$\bigg\langle \frac{1}{\ell} \bigg\rangle=\frac{1}{\ell_{o}}\sqrt[3]{\frac{1}{1-\eta}}\ \ {}_{2}F_{1} \bigg(\frac{1}{3},\frac{1}{2};\frac{3}{2};\frac{\eta}{2\eta-2} \Phi_{o}^2 \bigg)$$
Now, using both averages, the modified differential equation reduces to this:
\begin{align} \frac{d^2\Phi}{dt^2} & =g \bigg\langle \frac{1}{\ell} \bigg\rangle\sin{\Phi}-g \zeta \bigg\langle \frac{1}{\ell} \bigg\rangle\ \mathrm{sgn\ \Phi}\ -2\dot{\Phi}^2 \langle \ell'(\Phi)\rangle \bigg\langle \frac{1}{\ell} \bigg\rangle\\ & =g \bigg\langle \frac{1}{\ell} \bigg\rangle\sin{\Phi}-g \zeta \bigg\langle \frac{1}{\ell} \bigg\rangle\ \mathrm{sgn\ \Phi} \end{align}
Finally, using the same method, mentioned in the original question, with which I found the approximated period of the basic differential equation, I solve the one I previously got. Coincidentally, the equation for the period is just the same for the basic model but the $\frac{1}{\ell}$ converts into $\big\langle \frac{1}{\ell} \big\rangle$:
Basic: $$T = 4\sqrt{\frac{\ell_{o}}{g}} \ln{\Bigg|\frac{\Phi_{o}-\zeta}{\sqrt{2\zeta\Phi_{o}-\Phi_{o}^2}-\zeta}\Bigg|}$$ Modified: $$T = 4\sqrt{\frac{1}{g \bigg\langle \frac{1}{\ell} \bigg\rangle}} \ln{\Bigg|\frac{\Phi_{o}-\zeta}{\sqrt{2\zeta\Phi_{o}-\Phi_{o}^2}-\zeta}\Bigg|}$$
Looks like the plot of the approximated and original period really agrees with these calculations.