Re: computing geodesics, geodesic distance algorithms

In article <4394eb11@xxxxxxxxxxxxx>,
Thomas Mautsch <mautsch@xxxxxxxxxxxx> wrote:

>I see! - You suggest something like this:
>
> a:= dsolve(
> {D(D(u))(s) + (2+cos(u(s)))*sin(u(s))*(D(v)(s))^2 = 0,
> D(D(v))(s) - 2*sin(u(s))/(2 + cos(u(s)))*D(u)(s)*D(v)(s) = 0,
> u(0) = 0, v(0) = 0, u(1) =Pi/2,v(1) = Pi},
> numeric, output=listprocedure):
> uu := eval(u(s),a):
> vv := eval(v(s),a):
> plots[display](
> plot3d([(2 + cos(u))*cos(v), (2 + cos(u))*sin(v),sin(u)],
> u=0..2*Pi,v=0..2*Pi),
> plot3d([(2 + cos(uu(t)))*cos(vv(t)), (2 + cos(uu(t)))*sin(vv(t)),
> sin(uu(t))],t=0..1,s=0..1,thickness=4),
> scaling=constrained);
> duu:= eval(diff(u(s),s),a):
> dvv:= eval(diff(v(s),s),a):
> print("curve length:",
> seq(sqrt(duu(t)^2+((2+cos(uu(t)))*dvv(t))^2),t=[0,0.5,1]));

Something like that, but it's better to use odeplot rather than plot3d
to plot the curve. Use plot3d for surfaces.

>Is there a method to control/test
>whether the solution that Maple comes up with (if it comes up with one)
>is really an approximate geodesic?

>I mean: The equations are non-linear boundary value problems,
>and there could be things like "resonances" or such happening
>for certain boundary values.

The 'abserr' option gives an (absolute) error tolerance for the
solution. Of course, this is estimated, but the help page says
"in all but exceptional cases the true solution should be within
the error tolerance value of the continuous approximate solution
obtained by the method". Of course, since the solution may be
nonunique, that should say "a true solution", not "the true
solution".

The 'approxsoln' option may also be useful, to specify an initial
approximate discrete solution to be used as a starting point. This
should be a good way to determine which geodesic you get when
there are several with the same starting and ending points.

See the help pages ?dsolve,numeric,BVP and
?dsolve,numeric_bvp,advanced.

>Is it really appropriate to do such calculations
>on surfaces with points of negative Gauss curvature
>(like the torus, in contrast to, e.g., ellipsoids),
>where geodesics kind of "spread out"?

I think it's negative curvature that makes the geodesics
"spread out" more, while positive curvature will tend to
focus them. So on a sphere you get a continuum of geodesics
from a point to its antipode, while on a complete simply-connected
manifold of negative curvature you get a unique geodesic
joining two given points.

>Besides, I cannot imagine why mathematicians
>might be interested in obtaining geodesics between two given points
>that do not minimize distance...

Why not?

I don't know, I'm not a differential geometer. But I do know that
e.g. in optics, the path of a ray of light through a medium with
variable refractive index n(x,y,z) can be any geodesic for the metric
g_{ij} = n^2 delta_{ij}, not just one that minimizes distance.

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada


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