# How to obtain the period of this nonlinear differential equation?

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?

• Welcome. I suggest a clearer title so that other users can easily decide at a glance on the home page whether or not they can answer / are interested in the question Jun 27, 2022 at 22:11
• Is it intentional that the second equation is not symmetric wrt. the rest position? Compare with an oscillator with quadratic (air-like) friction where the equation is $m\ddot x+c|\dot x|\dot x+kx=0$. Mar 17 at 13:09
• It is symmetric because, although $\dot\Phi^2>0$, the $f'/f$ term is positive for $\Phi>0$ and negative for $\Phi<0$ due to the sine in $f'$. You can check it by differentiation of $f(\Phi)=\sqrt[3]{1-\eta\cos\Phi}$. Also, I didn't mention that $\eta\in(0,1)$. Mar 17 at 14:04
• For further information, this term you mention is accounting for the centrifugal force which changes direction constantly depending on $\Phi$ as for $\Phi>0$ the skier would turn to the left and viceversa for $\Phi<0$. Mar 17 at 14:08

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.

After revisiting this post I forgot to add that I found and exact solution

\begin{align} \frac{\mathrm d^2\Phi}{\mathrm dt^2}&=\frac{g}{\ell_0}\frac{\sin\Phi}{f(\Phi)}-\frac{g}{\ell_0}\zeta\frac{\text{sgn}\,\Phi}{f(\Phi)}-2\dot\Phi^2\frac{f'(\Phi)}{f(\Phi)}\\ \ddot\Phi+2\dot\Phi^2\frac{f'(\Phi)}{f(\Phi)}&=\frac{g}{\ell_0f(\Phi)}(\sin\Phi-\zeta\text{sgn}\,\Phi)\\ f(\Phi)^4\ddot\Phi+2\dot\Phi^2f(\Phi)^3f'(\Phi)&=\frac{g}{\ell_0}f(\Phi)^3(\sin\Phi-\zeta\text{sgn}\,\Phi)\\ \frac{1}{2}\left(2f(\Phi)^4\frac{\dot\Phi\ddot\Phi}{\dot \Phi}+4\dot\Phi^2f(\Phi)^3f'(\Phi)\right) &= \frac{1}{2}\frac{\mathrm d}{\mathrm d\Phi}(f^4\dot\Phi^2)\\ &=\frac{g}{\ell_0}f(\Phi)^3(\sin\Phi-\zeta\text{sgn}\,\Phi)\\ \implies \int_{\Phi_{\text{0}}}^\Phi\frac{1}{2}\frac{\mathrm d}{\mathrm d\Phi}(f^4\dot\Phi^2)\,\mathrm d\Phi &= \frac{1}{2}f(\Phi)^4\dot\Phi^2 \\ &=\int_{\Phi_{\text{0}}}^\Phi\frac{g}{\ell_0}f(\Phi)^3(\sin\Phi-\zeta\text{sgn}\,\Phi)\, \mathrm d\Phi\\ \implies \dot\Phi &= \frac{\mathrm d\Phi}{\mathrm dt} \\ &= \sqrt{\frac{2g}{\ell_0f(\Phi)^4}\int_{\Phi_{\text{0}}}^\Phi(1-\eta\cos\Phi)(\sin\Phi-\zeta\text{sgn}\,\Phi)\, \mathrm d\Phi}\\ \int_0^{\frac{\mathcal T}{4}}\mathrm dt&=\int_0^{\Phi_{\text{0}}}\frac{\mathrm d\Phi}{\sqrt{\frac{2g}{\ell_0f(\Phi)^4}\displaystyle\int_{\Phi_{\text{0}}}^\Phi(1-\eta\cos\Phi)(\sin\Phi-\zeta\text{sgn}\,\Phi)\, \mathrm d\Phi}}\\ &\hspace{-0.6cm}\boxed{\mathcal T=2\sqrt 2\sqrt{\frac{\ell_0}{g}}\int_0^{\Phi_{\text{0}}}\frac{|1-\eta\cos\Phi|^{2/3}\mathrm d\Phi}{\sqrt{\displaystyle\int_{\Phi_{\text{0}}}^\Phi(1-\eta\cos\Phi)(\sin\Phi-\zeta\text{sgn}\,\Phi)\, \mathrm d\Phi}}} \end{align}