Lagrangian for nonlinear small oscillations

My original Lagrangian is this, but I want to obtain nonlinear terms considering small oscillations : $$L = ma^2[\dot \theta^2(1+ 2\sin^2\theta) + \Omega^2\sin^2\theta + 2\Omega_0\cos\theta] .$$ Now, equilibrium point of potential energy $$U$$ is $$\cos\theta_0 = \frac{\Omega_0^2}{\Omega^2}$$. Now if $$\Omega_0 = \Omega$$ and, $$x = \theta - \theta_0$$ where $$x$$ is angular displacement. Then Taylor expansion around equilibrium point $$\theta_0$$ for potential energy is: $$U = ma^2(-2 +\frac{x^4}{4})$$ and Kinetic enrgy is: $$T=ma^2\dot x^2(1+ sin^2x)=ma^2\dot x^2(1 + 2x^2)$$ and finally Lagrangian is:$$L= T - U =ma^2\dot x^2(1 + 2x^2) - ma^2(-2 +\frac{x^4}{4})$$ is it procces correct, and if so, can I solve with successive approximation?

The potential energy divided by $$m\,a^2$$ is:

$$U={\Omega}^{2} \left( \sin \left( \vartheta \right) \right) ^{2}+2\, \Omega_{{0}}\cos \left( \vartheta \right)$$

the Taylor series for a small $$\vartheta$$ at $$\vartheta_0$$ and $$\vartheta^n=0~,n=3,4,...~$$is:

$$U_T=U(\vartheta_0)+\frac{\partial U}{\partial \vartheta} \bigg|_{\vartheta_0}\left(\vartheta-\vartheta_0\right)+\frac{1}{2}\,\frac{\partial^2 U}{\partial \vartheta^2}\bigg|_{\vartheta_0}\left(\vartheta-\vartheta_0\right)^2$$

edit

notice that for $$~\vartheta=\vartheta_0~~$$ you don't obtain static equilibrium

Why:

the equation of motion :

$$\frac{d}{dt}\left(\frac{\partial T_T}{\partial \dot \vartheta}\right)+\underbrace{\left(\frac{\partial U_T}{\partial \vartheta}\right)}_{-F}=0$$

thus F is:

$$F=f_1(\vartheta_0)+f_2(\vartheta_0)\,(\vartheta-\vartheta_0)$$

with $$f_1=U'\bigg|_{\vartheta_0}~,f_2=U''\bigg|_{\vartheta_0}$$

now for $$~\vartheta=\vartheta_0~,F=f_1(\vartheta_0)$$

thus to obtain static equilibrium at$$~\vartheta=\vartheta_0~$$ you have to add to the equation of motion the static force $$f_1(\vartheta_0)$$

• Presumably the fixed point is at $\vartheta=\vartheta_0$ so you'd need to keep the quadratic term in the series for $U$, so the EOM is linear and non-trivial. Commented Sep 25, 2020 at 15:48
• yes i will correct it!!
– Eli
Commented Sep 25, 2020 at 15:49