I am currently working on some differential geometry and I came across the following question.
Let $H = \{(x,y)\in \mathbb{R}^2|y>0\}$, this is the hyperbolic plane (the upper half-plane model). The first fundamental form is
$$
\frac{dx^2+dy^2}{y^2}.
$$
Write down the geodesic equation of the hyperbolic plane and show that geodesics in $H$ are either straight lines with constant $x$ or a half circle with center on the $x$-axis.
I have found that the geodesic equations are
$$
\frac{d}{dt}\bigg[ \frac{\overset{.}{x}}{y^2} \bigg] =0
$$
$$
\frac{d}{dt}\bigg[ \frac{\overset{.}{y}}{y^2} \bigg] = \frac{-1}{y^3}(\overset{.}{x}^2+\overset{.}{y}^2).
$$
From this its easy to see that $\overset{.}{x} = Cy^2$ for some constant $C$, and so if $C=0$ we find the straight-line solutions. However I am struggling to show that there are also circular solutions. Using the second geodesic equation I am also able to show that
$$
\overset{..}{y} = \frac{\overset{.}{y}^2}{y}-\frac{\overset{.}{x}^2}{y}.
$$
However I have not been able to make any more progress after this.
Best Answer
Let the geodesic has unit speed, so $$ \frac{\dot{x}^2+\dot{y}^2}{y^2}=1 $$ i.e., $$ \dot{y}^2=y^2-\dot{x}^2. $$ You know $\dot{x}=Cy^2$, so this gives $$ \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2=\frac{\dot y^2}{\dot x^2}=\frac{y^2}{\dot x^2}-1=\frac{R^2}{y^2}-1=\frac{R^2-y^2}{y^2} $$ where $R^2=C^{-2}$ (remember we are assuming $C\neq 0$). Solving the last equation gives the circle centered at $x$-axis of radius $R$, and of course only half of it lies on the upper half space.