= Solution
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:
$$
L(\gamma)=\int_0^1s(t)\,dt,\qquad E(\gamma)=\int_0^1s(t)^2\,dt.
$$
The <Cauchy-Schwarz inequality> here has the direct variance form
$$
E-L^2=\int_0^1(s(t)-L)^2\,dt\geq0.
$$
Equality 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 $L^2\leq2E$.
In the upper half-plane write $\gamma=u+iv$, $v>0$, and use $ds^2=(du^2+dv^2)/v^2$. Let the endpoints be $P=ip$, $Q=iq$, $p,q>0$. Then
$$
L=\int_0^1\frac{\sqrt{\dot u^2+\dot v^2}}v\,dt
\geq\int_0^1\frac{|\dot v|}v\,dt
\geq\left|\log\frac qp\right|.
$$
The 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 \b[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 $F=(\dot u^2+\dot v^2)/v^2$:
$$
\frac d{dt}\left(\frac{\dot u}{v^2}\right)=0,\qquad v\ddot v-\dot v^2+\dot u^2=0.
$$
Thus $\dot u=cv^2$ for a constant $c$. Since $u(1)-u(0)=0$, integration gives $c\int_0^1v^2dt=0$, hence $c=0$. The second equation then becomes $(\dot v/v)'=0$. Therefore
$$
\boxed{\gamma(t)=ip\exp\left[t\log(q/p)\right],\qquad \dot v/v=\log(q/p).}
$$
It has constant hyperbolic speed, is one of the absolute length minimizers, and also has minimum <energy of a curve> $E=[\log(q/p)]^2$.
Back to article page
