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In order to determine the geodesics between two points, one must solve the geodesic differential equations, which are as following \begin{align} p &= u'(s)\\ q &= v'(s)\\ p' + \Gamma^0_{00}p^2 + 2\Gamma^0_{01}pq + \Gamma^0_{11}q^2 &= 0\\ q' + \Gamma^1_{00}p^2 + 2\Gamma^1_{01}pq + \Gamma^1_{11}q^2 &= 0 \end{align} where $\Gamma^i_{jk}$ is the Christoffel symbol of second kind, and the initial conditions are $u(s_0) = u_0,\, \, v(s_0) = v_0,\, \, u'(s_0) = p_0,\, \, v'(s_0) = q_0$.

In order to solve these equations, one can only provide one point $(u,v)$, and the direction $(u',v')$. How is it possible to determine the geodesics between two points ?

For a sphere, both points will lie on the great circle. So here it is sufficient to provide the initial point and direction. But is it possible to accomplish this by only providing both points, if information such as direction is not known a priori ?

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    $\begingroup$ You need to look up "boundary value problem for ODEs". Short summary: there is an established theory for solving second order ODEs by specifying the values at the left and right end points, instead of specifying initial position and velocity. On geodesically complete manifolds the existence of a solution is guaranteed. In general this is an optimization problem and is studied in calculus of variations. $\endgroup$
    – Willie Wong
    Commented Nov 2, 2015 at 20:12
  • $\begingroup$ @Willie-Wong Phrased in a different manner : how can I determine the initial directions u',v', if I know at prior, two points that exist upon the geodesic : (u0,v0) and (u1,v1). For example on a sphere I can determine the directions geometrically, since I know that the geodesic curve connects the two points along the great circle. I thought that the entire purpose of determining the geodesics, originally, was to establish the shortest distance between two points (underline the two). $\endgroup$
    – imranal
    Commented Nov 2, 2015 at 20:40
  • $\begingroup$ (1) You don't need to determine the initial directions to solve the geodesic equation, which it is soluble. (2) You can get the initial directions by first solving the geodesic equation and then taking derivative at initial time. (3) You seem to be very interested in "solving" the geodesic equation. What concept of a solution are you using? $\endgroup$
    – Willie Wong
    Commented Nov 2, 2015 at 20:54
  • $\begingroup$ I ask because if you want to solve "numerically" then you may want to look into geometric flows like the curve-shortening flow with fixed end points, or the harmonic map heat flow with fixed end points. $\endgroup$
    – Willie Wong
    Commented Nov 2, 2015 at 20:57
  • $\begingroup$ The boundary value problem seems superfluous, from a practical point of view (the shortest distance in Euclidean space is a straight line that connects the two points - why is it not possible to restate this for curved space?). What is the practical point if the problem can not be restated as a way of determining the geodesics that connect two points (i.e any method which can determine u' and v' initial values for two points - provided the two points (u0,v0) and (u1,v1) are valid - unlike the case Igor stated). $\endgroup$
    – imranal
    Commented Nov 2, 2015 at 21:09

3 Answers 3

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I assume that imranal asks how to find numerically a geodesic connecting two given points if the connection is given. One way to do it is to implement the solution of the ODE system he wrote in his question numerically (there are many effective ways for it) and then use this implimintation to find the correct initial velocity vector $(u', v')$ such that the geodesic starting from $(u, v)$ in the direction $(u', v')$ hits the second point. One can slighly improve this search but the principle remains the same.

Alternative method which actually needs that the connection you have comes from a Riemannian metric would be to use the curve shortening flow: start with any curve connecting these two points (say, a straight line) and then start to deform the curve such that the velocity vector of the deformation is orthogonal to the velocity vector of the curve and its length is the geodesic curvature times some fixed function which vanishes at the first and last points of the curve. This deformation should normally converge to a geodesic.

I expect that there should be effective implimentation of this idea as a numerical scheme but I am not an expert in these questions.

Of course geodesic does not always exist (as explained by Igor) and may be is not unique (as explained by Narasimham) or/and the second procedure may converge to a geodesic which is not a minimal one.

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  • $\begingroup$ I have in fact managed to implement a numerical solution for the system of ODE's : scicomp.stackexchange.com/questions/21103/… . I was thinking about geodesics between two points, since I want to improve the code, as to permit the user more options. I have made a brute force method where I try to find the second point by iterating a "gazillion" (u',v') values. I halt the execution after finding the first pair (u',v'). Here is the code : pastebin.com/WJ5XdSxS $\endgroup$
    – imranal
    Commented Nov 3, 2015 at 14:42
  • $\begingroup$ Is you connection the Levi-Civita connection of a Riemannian metric? Otherwise your way is essentially the only one $\endgroup$
    – Vladimir S Matveev
    Commented Nov 3, 2015 at 14:45
  • $\begingroup$ It is. I will take a look at curve shortening flow. $\endgroup$
    – imranal
    Commented Nov 3, 2015 at 15:27
  • $\begingroup$ If many geodesic are found by the brute force procedure, how does one determine the geodesic with the shortest path ? $\endgroup$
    – imranal
    Commented Nov 3, 2015 at 19:20
  • $\begingroup$ I possibly misunderstood your question since the trivial answer is to compute the length of all such geodesics, compare, and choose the shortest one, or in fact your programm should already do it since the length is simply the time $s$ one need to go along the geodesic starting at the first point to hit the second point divided by the length of the initial velocity vector. $\endgroup$
    – Vladimir S Matveev
    Commented Nov 4, 2015 at 9:15
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This is not always possible, consider for example the punctured (at $(0,0)$) plane with the usual flat metric, and the points $(-1, 0)$ and $(1, 0).$

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  • $\begingroup$ Since I can not determine the proper directions between these two points, how do I then determine the geodesics (i.e do I solve the above equations). $\endgroup$
    – imranal
    Commented Nov 2, 2015 at 20:43
  • $\begingroup$ You can't. In Igor's example, there does not exist a geodesic passing through those two points. $\endgroup$
    – Willie Wong
    Commented Nov 2, 2015 at 20:48
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If two points $P,Q$ which are not umbilical are given, then there are in general several geodesics but only one that is shortest among them all. Their $ u^{'}, v^{'}$ are different, but unique for any one of the geodesic choices.

If $P,u^{'}, v^{'}$ are given there is no guarantee that the line so defined passes through $Q.$

These can be physically verified between two points on a cylinder/cone using a taut thread for example.

EDIT1:

It can be compared to the dynamic geodesic trajectory situation when a gun is fired with a given velocity and inclination from start point $P$. The initial velocity and direction should be adjusted to make trajectory pass through another desired point $Q$ in the vertical plane, as it cannot meet a desired point with arbitrary setting of dynamic parameters like $ u^{'}, v^{'} $ in case of geometrical geodesics.

EDIT2:

The solution also reflects in numerical procedures adopted. We have both initial value and boundary value problems. The former is straight forward and for the latter position/ derivative values at $P$ and $Q$ are given as input and numerical iterative (again!) shoot-through technique is employed.

EDIT3:

Actually the answer to your question is determination of the common geodesic invariant through $P,Q$. Geodesic curvature $k_g$ vanishes for all geodesics, so condition for them to be on the same geodesic line is that they be on same filament of the tangent fibre bundle sharing the same invariant. The parameters of the two points go to determine it..

For a surface of revolution the condition leads to Clairaut's Law.. a particularly easy relation. Geodesic constant condition is derivable as Liouville's relation generally for the $k_g=0$ condition from Christoffel symbols.

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