 Solution

The Euler--Lagrange equation isPutting $u=\cos (\varphi /2)$ and simplifying gives

Since $1-\cos \varphi =2(1-u^2)$, $1+\cos \varphi =2u^2$, and ${\dot{u}}^2=(1-u^2){\dot{\varphi }}^2/4$, the transformed functional isIts Euler--Lagrange equation is $\ddot{u}+(g/4a)u=0$, exactly the preceding equation.

For endpoint-vanishing $\eta$, the second variation isWriting $\eta ={\sum }_{n\ge 1}{c}_n{e}_n$ and using orthonormality givesIt is positive definite when $T<\pi /\omega$ and has a negative $e_1$ direction when $T>\pi /\omega$. Therefore