Assuming I have a second-order ordinary differential equation
$$[\alpha – {\beta^2}\cos^2\theta] {\theta}'' + [{\beta^2}\cos\theta\sin\theta] {\theta '}^2 + \gamma \sin\theta = 0$$
Where $\beta^2>>\alpha>0$, and $\gamma<0$ are constants. How can I analytically determine if its solution is periodic and calculate its period?
I tried to calculate the numerical solution ($\beta^2 = 10;\alpha = 0.1;\gamma = -16;$). When $\theta(0)=\pi/3$ and $\theta'(0)=0$ are given, the numerical solution is periodic.
Is there any analytical method to prove it?
Thank you in advance for your comments and help!
Update
This example is indeed too complicated, so I plan to assume that ($\beta^2 = 10;\alpha = 0.1;\gamma = -16;$) and $\theta(0)\in (-\frac{\pi}{2},\frac{\pi}{2})$ or $\theta(0)\in (-\frac{\pi}{4},\frac{\pi}{4})$ to study it.
My current rough idea is to draw the vector field diagram (make an intuitive judgment first), and then use Poincare Bendixson theorem and Dulac's theorem/Criterion.
Firstly, convert it into a first-order equation. Let
$$x(t) = \theta(t), y(t) = \theta(t)'$$
so we have
$$x'(t)= y(t)$$
$$ y(t)' = – (\gamma \sin(x(t)) + \beta \cos(x(t)) \sin(x(t)) y(t)^2) / (\alpha – \beta \cos(x(t))^2)$$
Best Answer
You can multiply with $2θ'$ and integrate once to get a first integral, that is, an expression that is constant along each solution curve, $$ (α-β^2\cos^2θ)θ'^2+4γ\sin^2(θ/2)=C $$ As $γ$ is negative, the negated version gives a sum-of-squares with positive coefficients in some neighborhood around the origin $(θ,θ')=(0,0)$. The level sets are closed curves around there. As the solutions follow the level sets, this results in periodic solutions for all initial values in that open set. The frequency depends on the level curve close to the origin it will be around $\omega ^2=\frac{γ}{β^2-α}$.