Past exam of the options II course of the University of Oxford / 2013 / G2 / b

The autonomous Lagrangian has the Beltrami identityHere $L$ is homogeneous of degree two in the velocities, so $\dot{u}{L}_{\dot{u}}+\dot{v}{L}_{\dot{v}}=2L$. Thus $L$ itself is constant, say $C\ge 0$.

For $C>0$, put $\ell =\sqrt{L}$. Its Euler-Lagrange expression along the curve isThe first term vanishes by stationarity for $L$, and the second by constancy of $L$. Thus the same path is stationary for its length functional $\int \sqrt{L}dt$.

On a regular segment described as a graph $v=v(u)$, reparametrize by $u$. Since $\dot{v}=v_u\dot{u}$,Choose the orientation with increasing $u$; reversing it does not change stationarity. Invariance of length under regular reparametrization therefore givesThis is the principle of length extremals from a constant quadratic energy. The stated nonnegativity also permits the degenerate case $C=0$. Then the length is zero, its absolute minimum since the integrand is nonnegative, so it is an extremal in the minimizing sense. The differentiated square-root Euler-Lagrange equation above is only used where $L>0$; it would not be legitimate to divide by $\sqrt{L}$ on a null curve. For a genuine surface metric and a regular nonconstant path, $C>0$ automatically.