The geodesic equation is a 2nd order differential equation given as $$\frac{\mathrm{d}^2 x^\alpha}{\mathrm{d} \lambda^2 }+\Gamma^\alpha_{\beta\gamma}\frac{\mathrm{d} x^\beta}{\mathrm{d} \lambda }\frac{\mathrm{d} x^\gamma}{\mathrm{d} \lambda }=0$$

The intuition says that we would need two initial conditions to uniquely solve for a solution of the equation. For example, the initial position or velocity or the other case where we have the initial and the final position. In the latter case, consider the following counter example. Imagine finding the geodesics on the surface of a 2-sphere where the initial and the final points are the north and the south pole. In that case we can get a family of curves(all the lines of longitudes) and hence we haven't solved for a unique curve. So do we need another piece of information to solve for a unique curve or am I missing something here?


As it has been suggested in the comments, knowing that we have a geodesic starting from north pole with surely pass thorough the south pole, we have not supplied two but only one piece of information. So we can't solve the ode uniquely. But how is one supposed to know that such a problem exists with the points in the given conditions? Also, is there a mathematical condition or theorem that enables us to see whether an ode can be solved uniquely with the given boundary conditions?

  • $\begingroup$ Any two boundary conditions are not guaranteed to be sufficient to uniquely determine a geodesic (or solve a second order differential equation in general). In the example you gave, each geodesic through the north pole is already guaranteed to go through the south pole, so the specification of two poles gives no more information than just one pole. $\endgroup$ – Puk Jul 16 at 9:50
  • $\begingroup$ @Puk In some cases though the boundary conditions are enough right? For example just choosing any two points on the sphere which are not the poles, gives us a unique line of longitude passing through them. So why is the case of poles mathematically different? $\endgroup$ – Naman Agarwal Jul 16 at 9:55
  • $\begingroup$ Sure, in some (most?) cases specifying two different points on the geodesic is enough to determine the geodesic. Unfortunately I can't give you a deeper answer than what I said in my previous comment (all geodesics through one pole are guaranteed to go through the other). $\endgroup$ – Puk Jul 16 at 10:02
  • $\begingroup$ @Puk Thanks for the answer. Your comment does give a very intuitive picture of why the boundary conditions in this case are not sufficient. But I am looking for a more mathematical treatment because in case of a sphere things are easy to imagine. But in case of more messed up geometries, how do we know which points are special and which aren't? $\endgroup$ – Naman Agarwal Jul 16 at 10:09
  • $\begingroup$ These are called conjugate points, but geodesics can be nonunique even if the points they connect are not conjugate points. For example, there are two geodesics connecting Chicago and Paris. One maximizes the distance, the other minimizes it. $\endgroup$ – Ben Crowell Jul 17 at 12:41

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