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Asked: 2026-05-15T11:10:24+00:00

I’m trying to use Mathematica’s NDSolve[] to compute a geodesic along a sphere using

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I’m trying to use Mathematica’s NDSolve[] to compute a geodesic along a sphere using the coupled ODE:

x" - (x" . x) x = 0

The problem is that I can only enter initial conditions for x(0) and x'(0) and the solver is happy with the solution where x” = 0. The problem is that my geodesic on the sphere has the initial condition that x”(0) = -x(0), which I have no idea how to tell mathematica. If I add this as a condition, it says I’m adding True to the list of conditions.

Here is my code:

s1 = NDSolve[{x1''[t] - (x1[t] * x1''[t] + x2[t] * x2''[t] + x3[t]*x3''[t]) * x1[t] == 0, x2''[t] -  (x1[t] * x1''[t] + x2[t] * x2''[t] + x3[t]*x3''[t]) * x2[t] == 0, x3''[t] - (x1[t] * x1''[t] + x2[t] * x2''[t] + x3[t]*x3''[t]) * x3[t] == 0, x1[0] == 1, x2[0] == 0, x3[0] == 0, x1'[0] == 0, x2'[0] == 0, x3'[0] == 1} , { x1, x2, x3}, {t, -1, 1}][[1]]

I would like to modify this so that the initial acceleration is not zero but -x(0).

Thanks

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1 Answer

  1. Editorial Team

    I fixed this problem through a mathematical rearrangement rather than addressing my original issue:

    Let V(t) be a vector field along x(t).

    x . V = 0 implies d/dt (x . V) = (x’ . V) + (x . V’) = 0

    So the equation D/dt V = V’ – (x . V’) x = V’ + (x’ . V) x holds
    This means the geodesic equation becomes: x” + (x’ . x’) x = 0 and so it can be solved using the initial conditions I originally had.

    Thanks a lot Janus for going through and pointing out the various problems I was having including horrible code layout, I learnt a lot through your re-writing as well.

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