Skip to main content

performance tuning - Is there any way to speed up this code that's Maximizing a function got from numerical integration?


I have this code below, which is calculating the Binding energy of an electron in a Quantum Well Wire with a hydrogenic impurity in it. Well you don't have to care much about what kind of calculation it does, because it is returning the right number, my only problem is, that it's taking about 20 minutes for it to return a single value for the Eb function (you can try Eb[0.7, 1, 0.01]). I'm wondering, if there's a way to make this code run faster. As you can see I have written everything almost the same way as one would write on paper. I've searched and tried many different approaches to make it faster, but nothing has helped so far.


e = 4.803*10^-10;
m = 0.067*9.109*10^-28;
h = 1.054*10^-27;
c = 2.997*10^10;
e0 = 13.18;


O1[Om_] = 10^13*Om;

oH[h0_] = (e*10000*h0)/(m*c);

Oc[h0_, Om_] = Sqrt[oH[h0]^2 + 4*O1[Om]^2];

aH[h0_, Om_] = Sqrt[h/(m*Oc[h0, Om])];

r0[rho_, phi_, z_, rhoi_] = 
  Sqrt[rho^2 + rhoi^2 - 2*rho*rhoi*Cos[phi] + z^2];


Psi[rho_, h0_, Om_] = E^(-(1/2)*(rho^2*aH[1, Om]^2)/aH[h0, Om]^2);

MGamma[rho_, phi_, z_, rhoi_, lambda_, Om_] = 
  E^(-lambda*r0[rho, phi, z, rhoi]);

CPhi[rho_, phi_, z_, rhoi_, lambda_, h0_, Om_] = 
  Psi[rho, h0, Om]*MGamma[rho, phi, z, rhoi, lambda, Om];

intCPhiCPhi[rhoi_, lambda_, h0_, Om_] := 

  NIntegrate[
   Abs[CPhi[rho, phi, z, rhoi, lambda, h0, Om]]^2*rho, {rho, 0, 
    Infinity}, {phi, 0, 2*\[Pi]}, {z, -Infinity, +Infinity}];

leftover[rho_, phi_, z_, rhoi_, lambda_, h0_, Om_] = 
  CPhi[rho, phi, z, rhoi, lambda, h0, Om]*
    h^2/(2*m*(aH[1, Om])^2)*(Psi[rho, h0, Om]/rho*
       D[MGamma[rho, phi, z, rhoi, lambda, Om], rho] + 
      2*D[MGamma[rho, phi, z, rhoi, lambda, Om], rho]*
       D[Psi[rho, h0, Om], rho] + 

      Psi[rho, h0, Om]*
       D[MGamma[rho, phi, z, rhoi, lambda, Om], {rho, 2}] + 
  Psi[rho, h0, Om]/rho^2*
   D[MGamma[rho, phi, z, rhoi, lambda, Om], +{phi, 2}] + 
  Psi[rho, h0, Om]*
   D[MGamma[rho, phi, z, rhoi, lambda, Om], {z, 2}]) + (e^2*
  Abs[CPhi[rho, phi, z, rhoi, lambda, h0, Om]]^2)/(e0*
  r0[rho, phi, z, rhoi]*aH[1, Om]);

IntLeftover[rhoi_?NumericQ, lambda_?NumericQ, h0_?NumericQ, 

  Om_?NumericQ] := (2*e0^2*h^2)/(e^4*
     m)*((1/intCPhiCPhi[rhoi, lambda, h0, Om] )*
    NIntegrate[
     leftover[rho, phi, z, rhoi, lambda, h0, Om]*rho, {rho, 0, 
      Infinity}, {phi, 0, 2*\[Pi]}, {z, -Infinity, +Infinity}])

Eb[rhoi_, h0_, Om_] := 
  FindMaximum[IntLeftover[rhoi, lambda, h0, Om], lambda];

Answer



One way to speed up multidimensional integrals is to reduce the PrecisionGoal. That will be acceptable in some cases but not in others.



Modified OP's code:


Clear[e, m, h, c, e0]  (* I moved the parameter initialization to later (unimportant *)

O1[Om_] = 10^13*Om;
oH[h0_] = (e*10000*h0)/(m*c);
Oc[h0_, Om_] = Sqrt[oH[h0]^2 + 4*O1[Om]^2];
aH[h0_, Om_] = Sqrt[h/(m*Oc[h0, Om])];
r0[rho_, phi_, z_, rhoi_] = Sqrt[rho^2 + rhoi^2 - 2*rho*rhoi*Cos[phi] + z^2];
Psi[rho_, h0_, Om_] = E^(-(1/2)*(rho^2*aH[1, Om]^2)/aH[h0, Om]^2);
MGamma[rho_, phi_, z_, rhoi_, lambda_, Om_] = E^(-lambda*r0[rho, phi, z, rhoi]);

CPhi[rho_, phi_, z_, rhoi_, lambda_, h0_, Om_] = 
  Psi[rho, h0, Om]*MGamma[rho, phi, z, rhoi, lambda, Om];

intCPhiCPhi[rhoi_, lambda_, h0_, Om_] := 
  NIntegrate[Abs[CPhi[rho, phi, z, rhoi, lambda, h0, Om]]^2*rho, (* NB: Abs[..]^2=(..)^2 *)
   {phi, 0, 2*π}, {rho, 0, Infinity}, {z, -Infinity, +Infinity},
   PrecisionGoal -> 3];

leftover[rho_, phi_, z_, rhoi_, lambda_, h0_, Om_] = 
  CPhi[rho, phi, z, rhoi, lambda, h0, Om]*

    h^2/(2*m*(aH[1, Om])^2)*(Psi[rho, h0, Om]/rho*
       D[MGamma[rho, phi, z, rhoi, lambda, Om], rho] + 
      2*D[MGamma[rho, phi, z, rhoi, lambda, Om], rho]*D[Psi[rho, h0, Om], rho] + 
      Psi[rho, h0, Om]*D[MGamma[rho, phi, z, rhoi, lambda, Om], {rho, 2}] + 
      Psi[rho, h0, Om]/rho^2*D[MGamma[rho, phi, z, rhoi, lambda, Om], +{phi, 2}] + 
      Psi[rho, h0, Om]*D[MGamma[rho, phi, z, rhoi, lambda, Om], {z, 2}]) +
      (e^2*Abs[CPhi[rho, phi, z, rhoi, lambda, h0, Om]]^2) / 
       (e0*r0[rho, phi, z, rhoi]*aH[1, Om]);

IntLeftover[rhoi_?NumericQ, lambda_?NumericQ, h0_?NumericQ, 

  Om_?NumericQ] := (2*e0^2*h^2)/(e^4*
     m)*((1/intCPhiCPhi[rhoi, lambda, h0, Om])*
    NIntegrate[leftover[rho, phi, z, rhoi, lambda, h0, Om]*rho,
     {phi, 0, 2*π}, {rho, 0, Infinity}, {z, -Infinity, +Infinity},
      PrecisionGoal -> 3]);

Eb[rhoi_, h0_, Om_] := FindMaximum[IntLeftover[rhoi, lambda, h0, Om], lambda];

OP's example:


PrintTemporary@Dynamic@Clock@{0, Infinity};  (* running timer (helps preserve my sanity *)

Block[{  (* parameter initialization *)
  e = 4.803*10^-10,
  m = 0.067*9.109*10^-28,
  h = 1.054*10^-27,
  c = 2.997*10^10,
  e0 = 13.18},
 Eb[0.7, 1, 0.01] // AbsoluteTiming
 ]



FindMaximum::lstol: The line search decreased the step size to within the tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient increase in the function. You may need more than MachinePrecision digits of working precision to meet these tolerances.



(*  {11.0647, {1.16175, {lambda -> 2.28525}}}  *)

Playing with it, I sometimes did not get a FindMaximum::lstol warning, but I always got 2.28525 or 2.28526 for an answer. I suspect that the noise from the numerical integration is the source.


Comments

Post a Comment

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]
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

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.
Post a Comment
Read more