The geodesic line on Poincare half plane I was calculating the geodesic lines on Poincare half plane but I found I somehow missed a parameter. It would be really helpful if someone could help me find out where my mistake is. 
My calculation is the following: 
Let $ds^2=\frac{a^2}{y^2}(dx^2+dy^2)$, then we could calculate the nonvanishing Christoffel symbols which are $\Gamma^x_{xy}=\Gamma^x_{yx}=-\frac{1}{y}, \Gamma^y_{xx}=\frac{1}{y}, \Gamma^y_{yy}=-\frac{1}{y}$. From these and geodesic equations, we have $$\ddot{x}-y^{-1}\dot{x}\dot{y}=0$$ $$\ddot{y}+y^{-1}\dot{x}^2=0$$ $$\ddot{y}-y^{-1}\dot{y}^2=0$$
From the last equation, it's straightforward that $y=Ce^{\omega\lambda}$, where $C$ and $\lambda$ are integral constants. Then substitute the derivative of $y$ into the first equation, we have, $$\ddot{x}-\omega\dot{x}=0$$ Therefore we have $x=De^{\omega\lambda}+x_0$ where $D, x_0$ are integral constants. However, by the second equation, we have, assuming $C$ is nonzero, $$C^2+D^2=0$$ And this leads to a weird result which is $$(x-x_0)^2+y^2=0$$
But the actual result should be $(x-x_0)^2+y^2=l^2$, where $l$ is another constant. 
 A: You say $C,\lambda$ are constants of integration, but that gives $\ddot{x}-\lambda\dot{x} = 0$ instead. Since your followup would be inconsistent, I will assume you meant $C,\omega$ are constants of integration.
You should not have three components to the geodesic equation, but rather two:
$$\ddot{y} + \Gamma^y_{xx}\dot{x}\dot{x} + \Gamma^y_{xy}\dot{x}\dot{y} + \Gamma^y_{yx}\dot{y}\dot{x} +\Gamma^y_{yy}\dot{y}\dot{y}  = 0\text{.}$$
You are also missing a factor of $2$ for your $\ddot{x}$ equation.
I will give you a further hint to say that since the $x$-coordinate is cyclic, $\dot{x} = Ey^2$ for some constant $E$. If you're not familiar with Killing vector fields, you can see this from the Euler-Lagrange equations on $L = \frac{1}{2}g(u,u)$, where $u^\mu = (\dot{x},\dot{y})$, which is also a nice way to find geodesics.
A: Although the following doesn't directly answer your question (which of course is a request for a review of the "GR" method of calculating geodesics and which Stan Liou did perfectly ), I can't resist writing down the following elegant little characterisation of geodesics in the hyperbolic plane. I believe thoughts along the following lines, although very simple and specialised, help grow an intuition for hyperbolic geometry and for some of the basic behaviours of negatively curved spacetime. Naturally it does not replace the "GR" method as it is much more specialised.
I'll simply sketch the proof as it is not quite as compact as I recalled when written out in full, but nonetheless it is a sequence of little jewels: readily grasped, simple and clear landmarks that one can keep in one's head when thinking about this kind of thing. If you need me to fill details in, I'll naturally add these to my answer.
The hyperbolic plane: 
$$\mathbb{H}^2 = \{z\in\mathbb{C}: {\rm Im}(z)>0\}\tag{1}$$ 
is kitted with the hyperbolic metric defined by:
$$\mathrm ds^2 = \frac{\mathrm dx^2 + \mathrm dy^2}{y^2}\tag{2}$$
which is the same as your metric, aside from a scaling constant. One wontedly studies $\mathbb{H}^2$ together with the Poincaré disk:
$$\mathbb{D}^2 = \{z\in\mathbb{C}: |z|<1\}\tag{3}$$ 
that is the isometric image of $\mathbb{H}^2$ under the bilinear transformation:
$$T:\mathbb{H}^2\to \mathbb{D}^2;\quad T(z) = \frac{1+i\,z}{z+i}\tag{4}$$
and you can readily show that the metric in $\mathbb{D}^2$ is defined by:
$$\mathrm ds^2 = \frac{4\,|\mathrm d\omega|^2}{(1-|\omega|^2)^2}\tag{5}$$
where $|\mathrm d\omega|$ is the everyday Euclidean metric in $\mathbb{D}^2$. 
Witness that the following are clearly isometries of $\mathbb{H}^2$:


*

*"Sideways translations", i.e. 
$$T_\lambda(z) = \lambda + z; \,\lambda \in \mathbb{R}\tag{6}$$

*"Dilations", i.e. 
$$D_\rho(z) = \rho\,z;\,\rho\in \mathbb{R},\,\rho>0\tag{7}$$


and also the transformation that corresponds to a rotation through any angle $\theta$ about the origin in the disk $\mathbb{D}^2$, i.e. $\omega\mapsto e^{i\theta} \omega$, which corresponds to the bilinear transformation:
$$R_\theta(z) = \frac{z \cos\left(\frac{\theta}{2}\right) + \sin\left(\frac{\theta}{2}\right)}{-z \sin\left(\frac{\theta}{2}\right) + \cos\left(\frac{\theta}{2}\right)}\tag{8}$$
because, from (5), it is clearly an isometry in $\mathbb{D}^2$ and $\mathbb{D}^2$ and $\mathbb{H}^2$ are isometrically equivalent.
So now we call on the following:
Theorem (Poincaré): The group $PSL(2, \mathbb{R})$ of billinear transformations of the form:
$$f:\mathbb{H}^2\to\mathbb{H}^2;\;f(z) = \frac{\alpha\,z+\beta}{\gamma\,z+\delta};\;\alpha,\,\beta,\,\gamma,\,\delta \in \mathbb{R};\;\alpha\delta-\beta\gamma=1\tag{9}$$
is precisely the group of isometries of $\mathbb{H}^2$; that is, every transformation of this kind is an isometry, and all isometries are of this kind.$\square$
Indeed, although not relevant here, $PSL(2, \mathbb{R})$ is also precisely the group of conformal transformations of $\mathbb{H}^2$, i.e. all maps of this kind are conformal, and any globally conformal bijection $\mathbb{H}^2\to \mathbb{H}^2$ is of this kind.
To prove the first part, one shows that all such bilinear maps can be decomposed into the following composition of known distance-preserving maps above:
$$f = T_{\frac{\alpha}{\gamma}} \circ D_{\frac{\alpha\delta-\beta\gamma}{\gamma}} \circ R_\pi \circ D_\gamma\tag{10}$$
To prove to converse, one shows that any distance preserving map is determined by the images of three non-collinear points $A, B, C\in\mathbb{H}^2$ because any other point $D\in \mathbb{H}^2$ is set by its distance from the reference points $A, B, C$ and their images.
Now one considers line segments $PQ$ between any pair of points $P$ and $Q$ on the imaginary axis in $\mathbb{H}^2$.

Clearly from (2) for any $C^0$ path $\Gamma$ between $P$ and $Q$, $ds^\prime \geq ds$ when we project the path onto the imaginary axis as shown. Therefore the unique geodesic linking two points on imaginary axis is simply a segment of the imagine axis between those two points.
It is then not hard to show that any two points $A,B\in\mathbb{H}^2$ are the images of two points $P$ and $Q$ on the imaginary axis under a mapping in the group $PSL(2,\mathbb{R})$ of isometries of $\mathbb{H}^2$ and this group member is uniquely defined by $A$ and $B$.
Therefore, by Poincaré's theorem, the image of the line segment $PQ$ along the imaginary axis defined by this unique map is the unique geodesic between $A$ and $B$. Since bilinear transformations map circles to circles (i.e. "circle" as defined in the wonted Euclidean space), the geodesic between $A$ and $B$ is the arc of a Euclidean circle. Indeed it is the arc between the two points of the unique Euclidean circle passing through $A$ and $B$ with its centre on the real axis.
Thus you can find the geodesics you need!
A: Just a note, the map $z\mapsto -\overline{z} $ is also an isometry of the upper half plane model, so the above statement about the group of isometries of $\mathbb{H}^2$ being $PSL_2(\mathbb{R})$ is not quite correct - these are precisely the orientation preserving isometries (they are indeed the only conformal/holomorphic ones). One needs to look at the bigger group $PSL^+_2(\mathbb{R})$ which also includes isometries of the form $\frac{a\overline{z}+b}{c\overline{z}+d}$ where $a,b,c,d\in\mathbb{R}$ and $ad-bc=-1$; clearly $PSL_2(\mathbb{R})$ is an index two subgroup of $Isom(\mathbb{H}^2)=PSL^+_2(\mathbb{R})$. 
