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

For fixed endpoint variations, the Euler-Lagrange equations areIndeed the first variation is $\int {\sum }_i(L_{q_i}\delta q_i+L_{{\dot{q}}_i}\delta {\dot{q}}_i)dt$. Integration by parts removes the derivatives from the variations; their zero endpoint values remove the boundary term. The fundamental lemma then makes every remaining coefficient vanish.

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.

For the surface of revolution $r(\eta ,\zeta )=(\eta ,f(\eta )\cos \zeta ,f(\eta )\sin \zeta )$, the coordinate tangent vectors have squared lengths $1+f^{\prime 2}$ and $f^2$, and their dot product is zero. Hence its first fundamental form isUse the energy Lagrangian $L=A{\dot{\eta }}^2+f^2{\dot{\zeta }}^2$. Its Euler-Lagrange equations areFor a meridian $\zeta ={\zeta }_0$, choose arc length as parameter, so $\dot{\eta }=1/\sqrt{A}$. Differentiating gives $2A\ddot{\eta }+A^{\prime }{\dot{\eta }}^2=0$, while the second equation is automatic. Part (b) makes this path an extremal of length. Thus every meridian is a geodesic.

For a nonconstant parallel $\eta ={\eta }_0$, choose constant nonzero $\dot{\zeta }$. The second equation holds, and the first reduces to $f({\eta }_0)f^{\prime }({\eta }_0){\dot{\zeta }}^2=0$. Since $f>0$, this is equivalent toConversely that condition makes both equations hold at constant speed, hence supplies a length extremal. Geometrically, a parallel is geodesic exactly at a stationary radius of the surface.