= Solution
The Euler--Lagrange equation is
$$
2ma^2(1-\cos\phi)\ddot\phi+ma^2\sin\phi\,\dot\phi^2-mga\sin\phi=0.
$$
Putting $u=\cos(\phi/2)$ and simplifying gives
$$
\ddot u+\omega^2u=0,\qquad \omega^2=\frac{g}{4a}.
$$
Since $1-\cos\phi=2(1-u^2)$, $1+\cos\phi=2u^2$, and $\dot u^2=(1-u^2)\dot\phi^2/4$, the transformed functional is
$$
\widehat S[u]=\int_0^T(8ma^2\dot u^2-2mga u^2)dt.
$$
Its Euler--Lagrange equation is $\ddot u+(g/4a)u=0$, exactly the preceding equation.
For endpoint-vanishing $\eta$, the <second variation> is
$$
\delta^2\widehat S(\eta)=16ma^2\int_0^T(\dot\eta^2-\omega^2\eta^2)dt.
$$
Writing $\eta=\sum_{n\geq1}c_ne_n$ and using orthonormality gives
$$
\delta^2\widehat S=16ma^2\sum_{n\geq1}\left[\left(\frac{n\pi}{T}\right)^2-\omega^2\right]c_n^2.
$$
It is positive definite when $T<\pi/\omega$ and has a negative $e_1$ direction when $T>\pi/\omega$. Therefore
$$
\boxed{t_0=\frac\pi\omega=2\pi\sqrt{\frac ag}.}
$$
Back to article page
