Skip to main content

differential equations - Derivative of function solved from NDSolve can't be plotted


I have used NDSolve to solve for a function (an angle of a triple pendulum), and now I wish to plot the derivative of that function (the angular velocity). Mathematica plots an empty plot. I had this issue before with a double pendulum system and ended up having to copy and paste cells into a new notebook one by one, and reevaluating everything. Then it suddenly started working (for the double pendulum). Doing that is not working in the case of my triple pendulum; the plot is still empty.


Needs["VariationalMethods`"]

ClearAll[Global]

variables = {Subscript[θ, 1][t], Subscript[θ, 2][t], Subscript[θ, 3][t]};
Subscript[r, 1] = 
  Subscript[l, 1] {Sin[Subscript[θ, 1][t]], -Cos[Subscript[θ, 1][t]]};
Subscript[r, 2] = Subscript[r, 1] + 
  Subscript[l, 2] {Sin[Subscript[θ, 2][t]], -Cos[Subscript[θ, 2][t]]};
Subscript[r, 3] = Subscript[r, 2] + 
  Subscript[l, 3] {Sin[Subscript[θ, 3][t]], -Cos[Subscript[θ, 3][t]]};


Subscript[l, 1] = 1;
Subscript[l, 2] = 1;
Subscript[l, 3] = 1;
Subscript[m, 1] = 1;
Subscript[m, 2] = 1;
Subscript[m, 3] = 1;
g = 9.81; tMax = 10;
initial = 
  {Subscript[θ, 1][0] == Pi/2, Subscript[θ, 2][0] == Pi, Subscript[θ, 3][0] == Pi/2,
   Subscript[θ, 1]'[0] == Pi/2, Subscript[θ, 2]'[0] == 0, Subscript[θ, 3]'[0] == 0};


lagrangian = (Subscript[m, 1]/2) D[Subscript[r, 1], t].D[Subscript[r, 
      1], t] + (Subscript[m, 2]/2) D[Subscript[r, 2], t].D[Subscript[
      r, 2], t] + (Subscript[m, 3]/2) D[Subscript[r, 3], t].D[
      Subscript[r, 3], t] - g {0, 1}.(Subscript[m, 1]*Subscript[r, 1] + 
       Subscript[m, 2]*Subscript[r, 2] + Subscript[m, 3]*Subscript[r, 3]);
eqs = EulerEquations[lagrangian, variables, t];

sol = First[NDSolve[Join[eqs, initial], variables, {t, 0, tMax}]];
Plot[Subscript[θ, 2]'[t] /. sol, {t, 0, 10}]


and the plot is blank. Like I said, I did the exact same code for the double pendulum, except the Lagrangian was slightly different (no third mass), and I was working with fewer variables. I had the same problem; there was no plot. After copying and pasting, it started working. But no amount of copying and pasting and renaming variables is working with the triple pendulum.


Notice Subscript[θ, 2][t] /. sol /. t -> 5 gives an answer of -5.75151. But notice Subscript[θ, 2]'[t] /. sol /. t -> 5 simply spits out Derivative[1][Subscript[θ, 2]][5] (the same thing). So the derivative has not been calculated at all, it seems.


Why is there no plot?


Note: the theta function plots fine. I'm wanting to plot the derivative of each theta function. That is not working.




Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more