Skip to main content

equation solving - NDSolve with vectors


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]

Mathematica graphics


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.


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...

equation solving - Invert and fit implicitly defined curve

I need to fit an implicitly defined curve. I thought I could get some data out of Solve , and then using FindFit . Therefore, I would like to find the relation the parametric curve defined by $F(x,y)=0$: Solve[-(1/2) + 1/2 (0.41202 BesselK[0, 0.1 Sqrt[x^2 + y^2]] + (0.101483 x BesselK[1, 0.1 Sqrt[x^2 + y^2]])/Sqrt[x^2 + y^2]) == 0, y] But I can't get an output: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help. >> Edit: In particular, I would like to fit the data coming from the curve with the expression of another curve, and not with a function $f(x)$. In particular, since this clearly looks like a cardioid , I would like it to fit to something like it. What other strategies could I try?