Skip to main content

differential equations - Debugging NDSolve to see numerical values at each time-step


I wish to debug my NDSolve function and this is my first time using the Mathematica debugger. I have read around and attempted various different ways of debugging but I cannot figure it out. I want to be able to go step by step through NDSolve and see values for x[t], y[t], t, etc.. My code is as follows:


(* Define the \[Theta] terms via piecewise functions *)
\[Theta]North \
:= Piecewise[{{ArcTan[x[t] - L/2 , y[t] - H],

x[t] > L/2 && y[t] > H}, {ArcTan[x[t] - L/2, H - y[t]],
x[t] > L/2 && y[t] < H}, {ArcTan[L/2 - x[t], y[t] - H],
x[t] < L/2 && y[t] > H}, {ArcTan[L/2 - x[t], H - y[t]],
x[t] < L/2 && y[t] < H}}]


(* Define the force terms in the x and y directions using piecewise \
functions *)
Fnx :=
Piecewise[{{Cn*Abs[H - y[t]]*Cos[\[Theta]North]*Sign[L/2 - x[t]],

x[t] != L/2 && y[t] != H}, {Cn*(L/2 - x[t]), y[t] == H}, {0,
x[t] == L/2}}]
Fny := Piecewise[{{Cn*(H - y[t])*Sin[\[Theta]North],
y[t] != H && x[t] != L/2}, {Cn*(H - y[t]), x[t] == L/2}, {0,
y[t] == H}}]

(* Define frictional terms *)
Ffx := -B*Sign[x'[t]]
Ffy := -B*Sign[y'[t]]


solution =
NDSolve[{x''[t] == (1/M)*(Fnx + Ffx), y''[t] == (1/M)*(Fny + Ffy),
x[0] == x0, x'[0] == vx0, y[0] == y0, y'[0] == vy0}, {x, y, Fnx,
Fny, \[Theta]North}, {t, 0, simTime}];

Is there a way to peer inside of NDSolve one step at a time? (I hope so that is kind of the point of a debugger).


Edit 1: Adding Parameter values:


(* Define the constants for simulation *)
(* Define the size of the \
box *)

L = 5;
H = 5;
(* Define Spring Constant *)
Cn = 0.3;
(* Define initial conditions *)
x0 = 0;
y0 = 0;
vx0 = 0;
vy0 = 0;
(* Define magnitude of sliding friction *)

B = 0.1;
(* Define mass of object *)
M = 1;
(* Define the simulation length *)
simTime = 50;

Answer



The steps are stored in the InterpolatingFunction results. Here's a way to view five steps at a time:


stepdata = MapThread[
Function[{tt, xx, yy, xp, yp},
Block[{x, y, t},

x[t] = xx; x'[t] = xp;
y[t] = yy; y'[t] = yp;
{First@tt, {x[t], y[t]}, {x'[t], y'[t]}, θNorth, {Fnx,
Fny}} /. solution]
],
{x["Grid"], x["ValuesOnGrid"], y["ValuesOnGrid"],
x'["ValuesOnGrid"], y'["ValuesOnGrid"]} /. solution
];

Manipulate[

TableForm[
Map[Pane[#, {100, 40}] &, stepdata[[n ;; n + 4]], {2}],
TableHeadings -> {None,
HoldForm /@
Unevaluated@{First@t, {x[t], y[t]}, {x'[t], y'[t]}, θNorth, {Fnx, Fny}}}],
{n, 1, Length@stepdata - 4}]

Mathematica graphics


Comments

Popular posts from this blog

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...

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...