# First Integral of Motion for Upper Half Plane Metric

The metric is $ds^2=\frac{dx^2+dy^2}{y^2}$. I have used the Euler-Lagrange equations to find the geodesics, and my equations are $\dot{x}=Ay^2$, $\ddot{y}+\frac{\dot{x}^2-\dot{y}^2}{y}=0$. I cannot seem to find the first integral for the second equation. I read online somewhere it is $\dot{y}=y\sqrt{1-Ay^2}$, but I can't seem to derive it. The only trick I currently know for doing these type of things is to multiple by $\dot{y}$ and then integrate, but that doesn't work here. Can anyone offer some guidance?

I tried it a slightly different way, but it doesn't seem to work for some reason: Instead of parametrizing, x=x(t), y=y(t) and minimizing, I just minimized $\frac{1+y'(x)^2}{y^2}$. Using the Euler-Lagrange equations, I get $y''y-y'^2+1=0$, and $y(x)=sinh(x)$ is a solution to this...but the geodesics are suppose to be half circles, and this doesn't give me a half circle..I am quite confused.

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The Lagrangian for the free motion in the Lobachevsky plane is $$L(t,x,y,\dot{x},\dot{y})=\frac{\dot x^2+\dot y^2}{y^2}.$$ It is autonomous, i.e. $\frac{\partial L}{\partial t}=0,$ and homogeneous of second degree in the velocities, therefore the energy is a first integral $$E\equiv\frac{\dot x^2+\dot y^2}{y^2}=c_1\tag{1}$$ Again it is invariant under $x\text{-translations}$, i.e. $\frac{\partial L}{\partial x}=0,$ therefore the associated momentum $\frac{\partial L}{\partial \dot x}$ is a first integral $$p_x\equiv\frac{\dot x}{y^2}=c_2\tag{2}.$$ In order to obtain the trajectories of motion we distinguish two cases $c_2=0$ and $c_2\neq 0.$
1. $c_2=0.$ From $(2)$ we get $\dot x=0,$ so the trajectories are implicitly defined by $x=x_0,$ where $x_0$ is an arbitrary constant of integration.
2. $c_2\neq 0.$ From (1) and (2) we get the new first integral $$\frac{E}{p_x^2}\equiv\frac{\dot x^2+\dot y^2}{\dot x^2}y^2=R^2\tag{3}$$
Considering $x$ as independent variable and denoting $\ '=\frac{\text{d}}{\text{d}x},$ now $(3)$ becomes $$(1+y'^2)y^2=R^2.$$ Integrating this latter by separation of variables, we get $(x-x_0)^2+y^2=R^2$ as the implicit equation for the trajectories, where $x_0$ is an arbitrary constant of integration.
A small correction to step 2: set $R = \sqrt{2c_1}/c_2$ and in the right-hand side of the ODE put $R^2$ instead of $R$. – Hans Lundmark Jun 3 '12 at 7:44