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Suppose I have three differential equations systems, each one of them has 4 equations. I find the 4 solutions of each one, let's call them x,y,z,w. Now, I want to take $x_1$,$x_2$,$x_3$ and put them in an array so I can plot it by letting t vary. This array would be: $$ \alpha(t) = \{x_1(t),x_2(t),x_3(t)\} $$ Then I also want to take $y_i$,$z_i$ and $w_i$ to to the same. If you are curious $\alpha$ would be my curve and then the other is the Frenet tried.

However I am struggling to to this, for alpha I would do this in Mathematica:

alpha[t_] := {x[t] /. Part[solution1[[1]], 1], 
              y[t] /. Part[solution2[[1]], 1], 
              z[t] /. Part[solution3[[1]], 1]}

However It doesn't seem to work, because I then try to plot Alpha like this:

ParametricPlot3D[Evaluate[alpha[t]], {t, 0, 6}]

But I get an empty cube.

Solution is given from the NDSolve. The code Part[solution1[[1]],1] is mean to take the array of solutions, and then the first one because that would be x.

I also tried to do: x[t] /. solution[[1]] etc etc, that seemed too work just fine for $\alpha$, but then failed for the other things because it keeps saying that the solution[2], solution[[3]] and solution[[4]] do not exist (even though I have those).

Do you see what did I do wrong?

EDIT: Added picture of form of solution1 enter image description here Where v[t],k1[t], k2[t] are given functions of t.

And this is what solution1 looks like: enter image description here

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  • $\begingroup$ does ParametricPlot3D[Evaluate[Through[alpha[t]]], {t, 0, 6}] work? $\endgroup$
    – kglr
    Dec 19, 2018 at 22:23
  • $\begingroup$ What are the forms of the solutions, solution1,...? To figure out what's with someone else's code often requires the complete code. $\endgroup$
    – Michael E2
    Dec 20, 2018 at 0:39
  • $\begingroup$ @kglr No it doesn't. $\endgroup$
    – qcc101
    Dec 20, 2018 at 6:36
  • $\begingroup$ @MichaelE2 Edited the question with more info, let me know if you need more. $\endgroup$
    – qcc101
    Dec 20, 2018 at 6:37
  • $\begingroup$ What happens if you get rid of Part and use x[t] /. solution1[[1]], etc.? Alternatively, change the y component to y[t] /. Part[solution2[[1]], 2] or equivalently, y[t] /. solution2[[1, 2]]; similarly for z. $\endgroup$
    – Michael E2
    Dec 20, 2018 at 14:07

1 Answer 1

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Always share complete information along with copyable Mathematica syntax.

I suspect that the empty plot is because of extra curls around the entries in alpha[t]. If we go to the documentation of ParametricPlot3D, the syntax is ParametricPlot3D[{fx, fy, fz},{u, a, b}]. I'm guessing that your alpha[t] structure is {{fx},{fy},{fz}}, which is causing the problem.

Anyways, here is an example to follow,

v[t_] = Sin[t];

k1[t_] = Cos[t];

k2[t_] = 5;

a = 0; b = 10;

Eq1 = x'[t] - v[t]*y[t] == 0;

Eq2 = y'[t] - v[t]*k1[t]*z[t] == 0;

Eq3 = z'[t] + v[t]*k1[t]*y[t] + v[t]*k2[t]*w[t] == 0;

Eq4 = w'[t] - v[t]*k2[t]*z[t] == 0;

sol1 = NDSolve[{Eq1, Eq2, Eq3, Eq4, x[0] == 0, y[0] == 1, z[0] == 0, 
    w[0] == 0}, {x, y, z, w}, {t, a, b}];

Eq41 = w'[t] - 1/2*v[t]*k2[t]*z[t] == 0;

sol2 = NDSolve[{Eq1, Eq2, Eq3, Eq41, x[0] == 0, y[0] == 1, z[0] == 0, 
    w[0] == 0}, {x, y, z, w}, {t, a, b}];

Eq21 = 2*y'[t] - v[t]*k1[t]*z[t] == 0;

sol3 = NDSolve[{Eq1, Eq21, Eq3, Eq41, x[0] == 0, y[0] == 1, z[0] == 0,
     w[0] == 0}, {x, y, z, w}, {t, a, b}];

alpha[t_] = {x[t] /. sol1, y[t] /. sol2, z[t] /. sol3};

ParametricPlot3D[Evaluate[Flatten[alpha[t]]], {t, a, b}]
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