Past exam of the mathematics course of the University of Cambridge / 2005 / ib / Paper 2 / 12A (Geometry)

For a Riemannian metric $g$, write $s(t)=\sqrt{g_{\gamma (t)}(\dot{\gamma }(t),\dot{\gamma }(t))}$. Use the energy of a curve normalization compatible with the requested inequality:The Cauchy-Schwarz inequality here has the direct variance formEquality holds exactly when $s$ is constant almost everywhere, hence everywhere for a smooth curve. With the alternative energy of a curve convention containing a factor $1/2$, the same assertion reads $L^2\le 2E$.

In the upper half-plane write $\gamma =u+iv$, $v>0$, and use $d{s}^2=(d{u}^2+d{v}^2)/v^2$. Let the endpoints be $P=ip$, $Q=iq$, $p,q>0$. ThenThe first equality requires $\dot{u}=0$ throughout. Endpoint values force $u=0$. The second equality requires the derivative of $\log v$ to have one sign, so $v$ is monotone. Conversely every smooth monotone positive $v$ with these endpoints attains this bound. Consequently the absolute length minimizers are precisely the vertical curves with monotone height, including a constant curve when $P=Q$, and the distance is $|\log (q/p)|$.

For curves stationary for the energy of a curve, fixed-endpoint variations give the Euler–Lagrange equations for $F=({\dot{u}}^2+{\dot{v}}^2)/v^2$:Thus $\dot{u}=c{v}^2$ for a constant $c$. Since $u(1)-u(0)=0$, integration gives $c{\int }_0^1{v}^{2}dt=0$, hence $c=0$. The second equation then becomes $(\dot{v}/v{)}^{\prime }=0$. ThereforeIt has constant hyperbolic speed, is one of the absolute length minimizers, and also has minimum energy of a curve $E=[\log (q/p){]}^2$.