Skip to main content

How to increase performance of this code for plotting a contour plot?


I have an equation which I need to triple integrate over a unit cube. The equation is


pot = NIntegrate[1/Sqrt[(x - h)^2 + (y - k)^2 + (z - l)^2], {h,-1,1}, {k,-1, 
1},{l,-1,1}];

As soon as I enter Shift+Enter it immediately processes the command. But now what I want is to plot its ContourPlot for different ${z}$ values (I chose $z=0.5$). So I give the command


ContourPlot[pot /. {z -> 0.5}, {x, -2, 2}, {y, -2, 2}]


But this piece of code takes just forever to process. I just keep on waiting and waiting but processing never ends (it takes really really long time). I am not sure that how is this such a computationally heavy task. For $z$ other than $0$ it takes longer time.


Is there something that I am doing wrong? I don't think this is a drawback of the device I am using. Is there a way to improve the performance of this code I am using?


P.S. It's been more than 10 minutes but the code for $z=0.5$ has not processed.


For your reference, I am attaching the contour plot for $z=0$.


enter image description here


This is the output for $z=0.5$ from the code above (it took about 10 minutes)


enter image description here



Answer



One helpfull rule to get fast integration is, to do analytical integration as much as you can.


int1 = Integrate[1/Sqrt[(x - h)^2 + (y - k)^2 + (z - l)^2], {l, -1, 1}, 

Assumptions -> -1 <= h <= 1 && -1 <= k <= 1 && x \[Element] Reals &&
y \[Element] Reals && z \[Element] Reals]

(* -Log[-z + Sqrt[h^2 + k^2 - 2 h x + x^2 - 2 k y + y^2 + z^2]] -
Log[z + Sqrt[h^2 + k^2 - 2 h x + x^2 - 2 k y + y^2 + z^2]] +
Log[1 - z + Sqrt[1 + h^2 + k^2 - 2 h x + x^2 - 2 k y + y^2 - 2 z + z^2]] +
Log[1 + z + Sqrt[1 + h^2 + k^2 - 2 h x + x^2 - 2 k y + y^2 + 2 z + z^2]] *)

I do the second integration with the rule based integrator (Rubi) by Albert Rich (see http://www.apmaths.uwo.ca/~arich/ ), because, in contrast to Mathematica, it gives an antiderivative without discontinuities.


rint2[x_, y_, z_, h_, k_] = Int[int1, k];


Take integration values at borders to get the definite integral.


rint2def[x_, y_, z_, h_] = 
rint2[x, y, z, h, 1] - rint2[x, y, z, h, -1] //
Simplify[#, Assumptions -> -1 <= h <= 1 && -1 <= k <= 1 &&
x \[Element] Reals && y \[Element] Reals && z \[Element] Reals] &

(* -h ArcTan[((-1 + y) (-1 + z))/((h - x) Sqrt[
2 + h^2 - 2 h x + x^2 - 2 y + y^2 - 2 z + z^2])] +
x ArcTan[((-1 + y) (-1 + z))/((h - x) Sqrt[

2 + h^2 - 2 h x + x^2 - 2 y + y^2 - 2 z + z^2])] +
h ArcTan[((1 + y) (-1 + z))/((h - x) Sqrt[
2 + h^2 - 2 h x + x^2 + 2 y + y^2 - 2 z + z^2])] -
x ArcTan[((1 + y) (-1 + z))/((h - x) Sqrt[
2 + h^2 - 2 h x + x^2 + 2 y + y^2 - 2 z + z^2])] -
h ArcTan[((1 - y) (1 + z))/((h - x) Sqrt[
2 + h^2 - 2 h x + x^2 - 2 y + y^2 + 2 z + z^2])] +
x ArcTan[((1 - y) (1 + z))/((h - x) Sqrt[
2 + h^2 - 2 h x + x^2 - 2 y + y^2 + 2 z + z^2])] -
h ArcTan[((1 + y) (1 + z))/((h - x) Sqrt[

2 + h^2 - 2 h x + x^2 + 2 y + y^2 + 2 z + z^2])] +
x ArcTan[((1 + y) (1 + z))/((h - x) Sqrt[
2 + h^2 - 2 h x + x^2 + 2 y + y^2 + 2 z + z^2])] - (-1 +
z) ArcTanh[(1 - y)/Sqrt[
2 + h^2 - 2 h x + x^2 - 2 y + y^2 - 2 z + z^2]] -
y ArcTanh[Sqrt[2 + h^2 - 2 h x + x^2 - 2 y + y^2 - 2 z + z^2]/(
1 - z)] -
ArcTanh[(-1 - y)/Sqrt[
2 + h^2 - 2 h x + x^2 + 2 y + y^2 - 2 z + z^2]] +
z ArcTanh[(-1 - y)/Sqrt[

2 + h^2 - 2 h x + x^2 + 2 y + y^2 - 2 z + z^2]] +
y ArcTanh[Sqrt[2 + h^2 - 2 h x + x^2 + 2 y + y^2 - 2 z + z^2]/(
1 - z)] +
ArcTanh[(1 - y)/Sqrt[
2 + h^2 - 2 h x + x^2 - 2 y + y^2 + 2 z + z^2]] +
z ArcTanh[(1 - y)/Sqrt[
2 + h^2 - 2 h x + x^2 - 2 y + y^2 + 2 z + z^2]] -
y ArcTanh[Sqrt[2 + h^2 - 2 h x + x^2 - 2 y + y^2 + 2 z + z^2]/(
1 + z)] -
ArcTanh[(-1 - y)/Sqrt[

2 + h^2 - 2 h x + x^2 + 2 y + y^2 + 2 z + z^2]] -
z ArcTanh[(-1 - y)/Sqrt[
2 + h^2 - 2 h x + x^2 + 2 y + y^2 + 2 z + z^2]] +
y ArcTanh[Sqrt[2 + h^2 - 2 h x + x^2 + 2 y + y^2 + 2 z + z^2]/(
1 + z)] - Log[-z + Sqrt[1 + h^2 - 2 h x + x^2 - 2 y + y^2 + z^2]] -
Log[-z + Sqrt[1 + h^2 - 2 h x + x^2 + 2 y + y^2 + z^2]] +
Log[1 - z + Sqrt[2 + h^2 - 2 h x + x^2 + 2 y + y^2 - 2 z + z^2]] +
Log[((1 - z + Sqrt[
2 + h^2 - 2 h x + x^2 - 2 y + y^2 - 2 z + z^2]) (1 + z + Sqrt[
2 + h^2 - 2 h x + x^2 - 2 y + y^2 + 2 z + z^2]) (1 + z + Sqrt[

2 + h^2 - 2 h x + x^2 + 2 y + y^2 + 2 z + z^2]))/((z + Sqrt[
1 + h^2 - 2 h x + x^2 - 2 y + y^2 + z^2]) (z + Sqrt[
1 + h^2 - 2 h x + x^2 + 2 y + y^2 + z^2]))] *)

The last integration has to be done numericaly.


rint3[x_, y_, z_] := NIntegrate[rint2def[x, y, z, h], {h, -1, 1}]

ContourPlot now finishes within 21 seconds.


ContourPlot[rint3[x, y, 1/2], {x, -2, 2}, {y, -2, 2}, 
ImageSize -> 400] // Timing


enter image description here


Comments

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...

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...

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-...