Why is finding a closed form of a geodesic so difficult? What is it about geodesics (does this term apply in $R^n$ or just $R^3$?) that makes finding a closed form of the curve so difficult?
My thoughts:
The minimal curve may not be unique which complicates solving the differential equation.
If the surface is not continuous, smooth or lacking a closed form then finding the geodesic that satisfies the conditions would be difficult but I don't really know why.
The way in which the minimal curve passes across the surface may be in such a way that it doesn't fit well into our set of functions.  I don't really know a better way to put this.
So what exactly makes this class of questions so difficult to find closed form solutions to?
Sorry if this is too general.
Thanks
 A: The term does apply in $\mathbb R^n$. Where it really applies is on manifolds. What I think you meant was a surface (which is a two dimensional manifold) embedded in $\mathbb R^3$, but of course we can have manifolds in any dimension, and they may or may not be embedded in $\mathbb R^n$. An $n$ dimensional manifold is a goemetric (well, really topological) structure for which "looks like $\mathbb R^n$" locally. Kind of like the earth is really round, but looks flat to us, because we are so close to it. 
For special types of manifolds (called Riemannian manifolds), there is a structure (called the metric) which allows us to construct geodesics. To denote points on a manifold, we use coordinates. A point $p$ in the manifold might be denoted $(x_1,...,x_n)$ where $x_i$ are numbers. It is important that we usually cannot use the same system of coordinates for the whole manifold; we have to make a change somewhere. The Riemannian structure gives rise to a family of functions called the Christoffel symbols. They are denoted $$\Gamma^k_{ij}(x_1,...,x_n),$$ where $i$, $j$, and $k$ are integers between $1$ and $n$ (thus, there are $n^3$ Christoffel symbols). A geodesic curve is denoted $\gamma(t)=(y_1(t),...,y_n(t))$, and has to satisfy the system of differential equations $$\frac{d^2y_k}{dt^2}+\sum_{i,j=1}^n\Gamma^k_{ij}(y_1(t),...,y_n(t))\frac{dy_i}{dt}\frac{dy_j}{dt}=0.$$ As you can see, this is a rather complicated nonlinear second order system of O.D.E.'s. Each equation has $n^2+1$ terms, all but one of which is nonlinear. This is horrible! Unless most of the Christoffel symbols are zero and $n$ is small, we don't stand a chance of solving this.
There are some situations where geodesics are not so hard to calculate. First of all, if the manifold is itself $\mathbb R^n$, then the Christoffel symbols are all zero, so the O.D.E. is linear $y''_i(t)=0$, a straight line. If the manifold can be written as the set of points $(x_1,...,x_n)$ in $\mathbb R^n$ such that a function $f(x_1,...,x_n)=0$, then you can review the answer I just gave here to see that the geodesic equation reduces to $\nabla f(x_1,...,x_n)=C \gamma''(t)$. This might be linear (depending on $f$). It can still be pretty tough to solve, but it's possible. Give it a shot!
Another complication is that finding a geodesics is inherently solving a boundary value problem. Once you have a general solution, you need to make sure it starts and ends at the right points. Most of the theory of differential equations, however, is built around solving initial value problems. There are big theorems regarding the existence and uniqueness of solutions to initial value problems which are simply not true for boundary value problems. So we don't know if a solution to the geodesic equation even exists, and if it does, we do not know if it is unique. At this point, we may decide to try to use a computer to find a numerical approximation. We're out of luck there too because it is much easier to solve initial value problems numerically than boundary value problems. It almost seems as though geodesics just don't want to be found!
