I'm stumped. I'm trying to write this using vectors, but the 2nd derivative isn't being expanded like I expected it to be. This is a system of equations for a projectile with quadratic drag and gravity (the linear drag is ignored for now). Negative Z is down, X and Y are the horizontal plane. If I write it as 9 equations, one for each coordinate, it works fine, but I'd rather use vectors since it is shorter and (to at least me) more obvious what is going on. Plus since I am new to Mathematica it would be good to learn more/better ways to use it.
gravity = 10;
withDrag[p0_, v0_, drag_] :=
NDSolve[{
p[0] == p0,
p'[0] == v0,
p''[t] == drag * Norm[p'[t]] * p'[t] + {0,0,-gravity}},
{p}, {t, 0, 5}]
withDrag[{0,0,0}, {0,10^4,10}, 0.001]
I get:
NDSolve::ndfdmc: Computed derivatives do not have dimensionality consistent with the initial conditions. >>
NDSolve[{
p[0] == {0, 0, 0},
p'[0] == {0, 10000, 10},
p''[t] == {
0.001 Norm[p'[t]] p'[t],
0.001 Norm[p'[t]] p'[t],
-10 + 0.001 Norm[p'[t]] p'[t]}},
{p}, {t,0,5}]
I formatted the output to make the error more obvious. Each of elements of the p'' vector has all three elements of p'[t]. Each one should really be p'[t][[dim]] (or something like that).
Any clues as to what I'm doing wrong?
Answer
Mathematica doesn't have vector variables (yet). That is to say, you can assign a list to a variable, but you cannot use a variable in a function like NDSolve and let Mathematica work out its dimensions or let the dimensions be undetermined.
If you change your function to this:
gravity = 10;
withDrag[p0_, v0_, drag_] :=
Module[{p},
p[t_] := {p1[t], p2[t], p3[t]};
p[t] /.
NDSolve[
Thread /@ {
p[0] == p0,
p'[0] == v0,
p''[t] == drag*Norm[p'[t]]*p'[t] + {0, 0, -gravity}} // Flatten,
p[t],
{t, 0, 5}
]// First
]
it works. What is does is defining your p as a vector (list) of functions. Thread takes care of distributing == over the vector components and Flatten makes a single list of equations from all this.
track[t_] = withDrag[{0, 0, 0}, {0, 10^2, 10}, 0.001];
ParametricPlot3D[track[t], {t, 0, 5}, BoxRatios -> 1]

Note that I reduced the starting value of v0[[2]] to 10^2 because 10^4 yields a 'stiff' system. Also note that I used BoxRatios -> 1 to prevent the box from becoming flat.
While under the hood this method still provides Mathematica with the 9 equations that you already tried manually, it has the advantage that it leaves your vector equations intact.
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