Let $$S(\theta)=\int_0^{2\pi}\sqrt{\dot \theta^2(t)+\cos^2(\theta(t))}dt$$
a) How does the Euler-Lagrange equation for S look?
b) Show that the solutions of the Euler-Lagrange equation are given by $\theta (t)=\arctan(a\sin(t)+b\cos(t))$. Use the substitution $u=\tan(\theta)$.
What I did:
a) I've got: $$-\frac{\sin\theta\cos\theta}{\sqrt{\dot\theta^2+\cos^2\theta}}-\frac{d}{dt}\frac{\dot\theta}{\sqrt{\dot\theta^2+\cos^2\theta}}=0$$ Is this correct?
b) I've got for the Beltrami equation $\frac{\dot\theta^2}{\sqrt{\dot\theta^2+\cos^2\theta}}-\sqrt{\dot\theta^2+\cos^2\theta}=c$. So from this I get $-\cos^2\theta=c\sqrt{\dot\theta^2+\cos^2\theta}$. Now with $u=\tan\theta$ I get $1/c^2-1=u^2+\dot u^2$ with the chain rule. I think I did something wrong in the substitution. Is it ment $u(t)=\tan(\theta(t))$? Thats what I tried.
Does someone know a link where I could read through some examples similar to this porblem? Maybe some with a boundary condition too?
The derivation of Euler-Lagrange equation and Beltrami equation are correct. I also calculated the same expression
$$\frac{1}{c^2} -1 = u^2 + \dot u^2.$$
Rearranging gives
$$\frac{\dot u}{\sqrt{A^2 - u^2}} = 1$$
for some constant $A$. Integrating and using the substitution $u = A \sin\alpha $ gives
$$ \sin \left(\frac{u}{A}\right) = t+ B\Rightarrow u = A \sin (t+B) = a\sin t + b \cos t$$
for some $a, b$. Putting $\theta = \arctan u$ gives the correct answer.
Note that the boundary constraint does not affect the above calculations: the Euler-Lagrange equation and Beltrami equation are satisfied everywhere in the interval, irrespective on which boundary condition to use.
But if you are given some boundary constraints, you can go further and solves $a$ and $b$.