NDSolve with functions of vectors

[John Lee]

Hi,
I'm trying to integrate a function with NDSolve (ExplicitRungeKutta method), and it doesn't seem to recognize vector variables (I keep getting an incorrect dimension error). In addition, I would like to integrate over an absolute distance and I don't know how to do that in Mathematica. For simplicity's sake, I will post a random example of what I'm talking about:
U[x] = {2*x[[2]],x[[1]]/3,x[[3]]} is the derivative
X0 = {1,1,1} is the initial condition
so if a'[s] = U[x[s]], I'd like to solve for a[s] where s is the total distance. And s is to be integrated from 0 to 6 with a[0]=X0.
Ultimately, I have a derivative that depends on a 3-component position vector. I want to integrate that derivative over a total distance traveled from a starting point.

Any advice would be helpful. Thanks,
John

[Jens-Peer Kuska]

Hi,
U[x_List] := {2*x[[2]], x[[1]]/3, x[[3]]}

X0 = {1, 1, 1};

and

sol = NDSolve[{a'[s] == U[a[s]], a[0] == X0}, a[s], {s, 0, 1}]

will do it. If yoy wish to use the path length you have to transform
a[s]== a[Sqrt[x[s]^2+y[s]^2+z[s]^2]].

Regards
Jens

[John Lee]

I'd like to clarify or make this question simpler:

M0=NDSolve[{a'[s]==U[a[s]],a[0]==X},a,{s,0,6},Method->"ExplicitRungeKutta"]

Where X = {1,0,0} and U = {f(x,y,z),g(x,y,z),h(x,y,z)}

That does not work and I can't figure out why. The error I get is "Part::partw: Part 2 of a[s] does not exist." Same with "Part 3".

-John

[Steve Luttrell]

This does what you want:

solution =
NDSolve[{x'[s] == {2*x[s][[2]], x[s][[1]]/3, x[s][[3]]},
x[0] == {1, 1, 1}}, x[s], {s, 0, 1}]

gives

{{x[s]->InterpolatingFunction[{{0.,1.}},<>][s]}}

Then

ParametricPlot3D[Expand[x[s]/.solution[[1]]],{s,0,1}]

gives a 3D plot of the trajectory for 0<=s<=1.

--
Steve Luttrell
West Malvern, UK

"John Lee" <johnl...@hotmail.com> wrote in message
news:fgca49$9ho$1...@smc.vnet.net...

[Jerry]

Hi, I tried your code and got these errors. What did I miss?
Thanks.

Part::partw: Part 2 of x[s] does not exist. >>

Part::partw: Part 3 of x[s] does not exist. >>

[DrMajorBob]

If I interpret your question properly...

Clear[x, y, z, a, u, arc, s, t]
a[t_] = {x[t], y[t], z[t]};
u[{x_, y_, z_}] = {2 y, x/3, z};
ds[{x_, y_, z_}] = Sqrt[#.#] &@u[{x, y, z}];
eqns = Thread[
Flatten@{a'[t] - u[a[t]], a[0] - {1, 1, 1}, arc'[t] - ds[a[t]],
arc[0]} == 0];
{x, y, z, arc} = {x, y, z, arc} /.
Last@NDSolve[eqns, {x, y, z, arc}, {t, 0, 6}];

The "arc" function is s in terms of t. To get x, y, and z in terms of s,
we invert the "arc" function:

Clear[si]
si = si /. First@
NDSolve[{si[arc[0]] == 0, si'[u] == 1/arc'[si@u]},
si, {u, arc@0, arc@6}];

Plot[si@arc[t], {t, 0, 6}]

Plot[arc@si[t], {t, arc@0, arc@6}]

Finally, here's a plot of x, y, and z in terms of arc length:

Plot[a[si[s]], {s, arc@0, arc@6}]

or

ParametricPlot3D[a[s], {s, arc@0, arc@6}]

Bobby

On Fri, 02 Nov 2007 03:28:04 -0500, John Lee <johnl...@hotmail.com>
wrote:

--

DrMaj...@bigfoot.com

[Steve Luttrell]

Now I have had a look at my Messages window I see that I am getting the same
warning messages as you. But the final result appears to be the correct
solution to the differential equation.

I have $Version = "6.0 for Microsoft Windows (32-bit) (April 20, 2007)".

--
Steve Luttrell
West Malvern, UK

"Jerry" <JLK...@yahoo.com> wrote in message
news:fghb2n$i9j$1...@smc.vnet.net...

[John Lee]

Thank you so much! This worked. For the longest time, I did not notice U[x_List]. That is what fixed my problem. Thank you everyone else for helping me.