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I am trying to solve Equation number (1.2) numerically in MATHEMATICA. This equation is solved in the papers https://arxiv.org/pdf/2205.05193.pdf, https://arxiv.org/pdf/2202.13264.pdf, and https://arxiv.org/pdf/2005.05718.pdf. This equation is quite similar to I am trying to solve nonlinear Schrödinger equation with dipolar interaction which @AlexTrounev solved. I like to find the figures posted below and energy especially kinetic and dipolar energy. This paper might be a little help https://arxiv.org/pdf/1506.03283.pdf as Equation number (1.3) goes to infinity at the center.

enter image description here enter image description here

Here I paste my failed attempt to modify the code from @AlexTrounev

Needs["NumericalDifferentialEquationAnalysis`"];

np = 4; xg = GaussianQuadratureWeights[np, -L, L]; xpoints = 
 xg[[All, 1]];
xweights = xg[[All, 2]];
yg = GaussianQuadratureWeights[np, -L, L]; ypoints = yg[[All, 1]];
yweights = yg[[All, 2]];
zg = GaussianQuadratureWeights[np, -L, L]; zpoints = zg[[All, 1]];
zweights = zg[[All, 2]];

sol[0][x_, y_, z_, t_] := 
  E^(- (x^2 + y^2 + z^2)/2);
\[Omega]x = 45; \[Omega]y = 45; \[Omega]z = 133; gmean\[Omega] = (\
\[Omega]x \[Omega]y \[Omega]z)^(
 1/3); \[Gamma] = \[Omega]x/gmean\[Omega]; \[Lambda] = \
\[Omega]y/gmean\[Omega]; \[Nu] = \[Omega]z/gmean\[Omega];
a0 = 5.29*10^-5; a = 85 a0; add = 131 a0; \[Epsilon]dd = add/a;
\[Gamma]QF = 
 128/3 Sqrt[\[Pi] a^5] (1 + 1.5 \[Epsilon]dd^2); angle = ((
  3 Cos[\[Phi]]^2 - 1)/2); \[Phi] = 0;
Nat = 10^6; L = 10; dt = 1/100; nt = 100; T = 
 Table[i dt, {i, 0, 1001}];


nkx[0] = Table[
   Sum[xweights[[i]] Exp[
      I xpoints[[j]] xpoints[[i]]] Abs[
       sol[0][xpoints[[i]], 0, 0, 0]]^2, {i, Length[xg]}], {j, 
    Length[xg]}];
nky[0] = Table[
   Sum[yweights[[i]] Exp[
      I ypoints[[j]] ypoints[[i]]] Abs[
       sol[0][0, ypoints[[i]], 0, 0]]^2, {i, Length[yg]}], {j, 
    Length[yg]}];
nkz[0] = Table[
   Sum[zweights[[i]] Exp[
      I zpoints[[j]] zpoints[[i]]] Abs[
       sol[0][0, 0, zpoints[[i]], 0]]^2, {i, Length[zg]}], {j, 
    Length[zg]}];
finalnk[0] = nkx[0] nky[0] nkz[0];

intn[0] = 
  1/(6 \[Pi]^2)
    angle Table[{xpoints[[j]], ypoints[[k]], zpoints[[m]], 
     Re[Sum[xweights[[j]] yweights[[k]] zweights[[
         m]] Exp[-I xpoints[[j]] xpoints[[i]]]  Exp[-I ypoints[[
           k]] ypoints[[i]]]  Exp[-I zpoints[[m]] zpoints[[i]]] ((
          3 zpoints[[m]]^2)/(
          xpoints[[j]]^2 + ypoints[[k]]^2 + zpoints[[m]]^2) - 
          1) finalnk[0][[i]], {i, Length[xg]}]]}, {j, Length[xg]}, {k,
      Length[yg]}, {m, Length[zg]}];


Vdd[0] = Interpolation[
  Join[{{{-L, -L, -L, intn[0][[1, 1, 1, 4]]}}}, 
   intn[0], {{{L, L, L, intn[0][[np, np, np, 4]]}}}]]


Do[sol[s] = 
  NDSolveValue[{-I D[\[Psi][x, y, z, t], t] - 
       1/2 Laplacian[\[Psi][x, y, z, t], {x, y, z}] + 
       1/2 (\[Gamma]^2 x^2 + \[Nu]^2 y^2 + \[Lambda]^2 z^2) \[Psi][x, 
         y, z, t] + 
       4 \[Pi] a Nat Abs[\[Psi][x, y, z, t]]^2 \[Psi][x, y, z, t] + 
       3 add Nat Vdd[s - 1][x, y, z] \[Psi][x, y, z, 
         t] + \[Gamma]QF Nat^(3/2)
         Abs[\[Psi][x, y, z, t]]^3 \[Psi][x, y, z, t] == 0,
     \[Psi][x, y, z, T[[s]]] == 
      sol[s - 1][x, y, z, T[[s]]], \[Psi][L, y, z, t] == 
      0, \[Psi][-L, y, z, t] == 0,
     \[Psi][x, L, z, t] == 0, \[Psi][x, -L, z, t] == 
      0, \[Psi][x, y, L, t] == 0, \[Psi][x, y, -L, t] == 
      0}, \[Psi], {t, T[[s]], T[[s + 1]]}, {x, -L, L}, {y, -L, 
     L}, {z, -L, L}, 
    Method -> {"MethodOfLines", 
      "SpatialDiscretization" -> {"TensorProductGrid", 
        "MinPoints" -> 40, "MaxPoints" -> np, 
        "DifferenceOrder" -> "Pseudospectral"}}] // Quiet;
 
 
 
 nkx[s] = 
  Table[Sum[
    xweights[[i]] Exp[
      I xpoints[[j]] xpoints[[i]]] Abs[
       sol[s][xpoints[[i]], 0, 0, T[[s + 1]]]]^2, {i, 
     Length[xg]}], {j, Length[xg]}];
 nky[s] = 
  Table[Sum[
    yweights[[i]] Exp[
      I ypoints[[j]] ypoints[[i]]] Abs[
       sol[s][0, ypoints[[i]], 0, T[[s + 1]]]]^2, {i, 
     Length[yg]}], {j, Length[yg]}];
 nkz[s] = 
  Table[Sum[
    zweights[[i]] Exp[
      I zpoints[[j]] zpoints[[i]]] Abs[
       sol[s][0, 0, zpoints[[i]], T[[s + 1]]]]^2, {i, 
     Length[zg]}], {j, Length[zg]}];
 finalnk[s] = nkx[s] nky[s] nkz[s];
 
 
 intn[s] = 
  1/(6 \[Pi]^2)
    angle Table[{xpoints[[j]], ypoints[[k]], zpoints[[m]], 
     Re[Sum[xweights[[j]] yweights[[k]] zweights[[
         m]] Exp[-I xpoints[[j]] xpoints[[i]]]  Exp[-I ypoints[[
           k]] ypoints[[i]]]  Exp[-I zpoints[[m]] zpoints[[i]]] ((
          3 zpoints[[m]]^2)/(
          xpoints[[j]]^2 + ypoints[[k]]^2 + zpoints[[m]]^2) - 
          1) finalnk[s][[i]], {i, Length[xg]}]]}, {j, Length[xg]}, {k,
      Length[yg]}, {m, Length[zg]}];
 
 Vdd[s] = 
  Interpolation[
   Join[{{{-L, -L, -L, intn[s][[1, 1, 1, 4]]}}}, 
    intn[s], {{{L, L, L, intn[s][[np, np, np, 4]]}}}]];, {s, 1, 15}]

DensityPlot[
 Abs[sol[10][x, y, 0, T[[10 + 1]]]]^2, {x, -L, L}, {y, -L, L}, 
 ColorFunction -> "BlueGreenYellow", PlotPoints -> 200, Frame -> True,
  FrameTicks -> Automatic, LabelStyle -> {24, Bold, Large, Black}, 
 FrameLabel -> {{Style["y", FontFamily -> "Times New Roman", 
     FontSlant -> "Italic", FontWeight -> Bold, FontSize -> 30], 
    None}, {Style["x", FontFamily -> "Times New Roman", 
     FontSlant -> "Italic", FontWeight -> Bold, FontSize -> 30], 
    None}}]
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  • $\begingroup$ The first error: you declare: intn[0] = 1/(6 [Pi]^2) but later you use: intn[0][[1, 1, 1, 4]] $\endgroup$ Commented Sep 20, 2022 at 10:20
  • $\begingroup$ From what paper did you get picture? $\endgroup$ Commented Sep 21, 2022 at 9:05
  • $\begingroup$ @AlexTrounev I posted the pictures from arxiv.org/pdf/2205.05193.pdf. Though this work arxiv.org/pdf/2202.13264.pdf also deals with the same equation. The formulation to avoid the singularity at the center is given in arxiv.org/pdf/1506.03283.pdf. Singularity arrives for large np in intn[0] in the code and I could not figure out the interpolation in vdd[0]. $\endgroup$ Commented Sep 24, 2022 at 16:31
  • $\begingroup$ @ArghaDebnath It is not clear how they compute 3D states. As I remember we used Mathematica FEM in our report researchgate.net/publication/… $\endgroup$ Commented Sep 25, 2022 at 8:03
  • $\begingroup$ @AlexTrounev I will study the paper. I have one question regarding your previous solution mathematica.stackexchange.com/questions/209951/…. Why did you put f[m]'[t] instead of f[m][t]' as it is a time derivative d/dt in your "Now we can make a system of equations" section. $\endgroup$ Commented Sep 25, 2022 at 8:17

1 Answer 1

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3D case can be solved with a linear Mathematica FEM in imaginary time. But there are no stationary states for the chosen set of parameters. We need about 1 min to compute 1 step. First step looks like picture in the paper.

Needs["NDSolve`FEM`"];
Needs["NumericalDifferentialEquationAnalysis`"];


Lx = 8; Ly = 8; Lz = 8; np = 10; xg = 
 GaussianQuadratureWeights[np, -Lx, Lx]; xpoints = xg[[All, 1]];
xweights = xg[[All, 2]];
yg = GaussianQuadratureWeights[np, -Ly, Ly]; ypoints = yg[[All, 1]];
yweights = yg[[All, 2]];
zg = GaussianQuadratureWeights[np, -Lz, Lz]; zpoints = zg[[All, 1]];
zweights = zg[[All, 2]]; mesh = 
 ToElementMesh[Cuboid[{-Lx, -Ly, -Lz}, {Lx, Ly, Lz}], 
  MaxCellMeasure -> 0.25]


sol[0][x_, y_, z_] := E^(-(x^2 + y^2 + z^2)/4);
\[Omega]x = 45; \[Omega]y = 45; \[Omega]z = 133; gmean\[Omega] = (\
\[Omega]x \[Omega]y \[Omega]z)^(1/3); \[Gamma] = \[Omega]x/
  gmean\[Omega]; \[Lambda] = \[Omega]z/
  gmean\[Omega]; \[Nu] = \[Omega]y/gmean\[Omega];
a0 = 5.29*10^-5; a = 85 a0; add = 131 a0; \[Epsilon]dd = add/a;
\[Gamma]QF = 
 128/3 Sqrt[\[Pi] a^5] (1 + 
    1.5 \[Epsilon]dd^2); angle = ((3 Cos[\[Phi]]^2 - 1)/2); \[Phi] = 0;
Nat = 10^6; nt = 10; dt = 1/10; c0 = angle 4 Pi/3/(2 Pi)^3;


nk[0] = Table[
   Sum[xweights[[i1]] yweights[[i2]] zweights[[i3]] Exp[
      I xpoints[[j1]] xpoints[[i1]] + I ypoints[[j2]] ypoints[[i2]] + 
       I zpoints[[j3]] zpoints[[i3]]] Abs[
       sol[0][xpoints[[i1]], ypoints[[i2]], zpoints[[i3]]]]^2, {i1, 
     Length[xg]}, {i2, Length[yg]}, {i3, Length[zg]}], {j1, 
    Length[xg]}, {j2, Length[yg]}, {j3, Length[zg]}];
int = Flatten[
   Table[{{xpoints[[j]], ypoints[[k]], zpoints[[m]]}, 
     c0 Re[Sum[
        xweights[[i1]] yweights[[i2]] zweights[[
          i3]] Exp[-I xpoints[[j]] xpoints[[i1]]] Exp[-I ypoints[[
            k]] ypoints[[i2]]] Exp[-I zpoints[[m]] zpoints[[
            i3]]] ((3 zpoints[[i3]]^2)/(xpoints[[i1]]^2 + 
              ypoints[[i2]]^2 + zpoints[[i3]]^2) - 1) nk[0][[i1, i2, 
          i3]], {i1, Length[xg]}, {i2, Length[yg]}, {i3, 
         Length[zg]}]]}, {j, Length[xg]}, {k, Length[yg]}, {m, 
     Length[zg]}], 2];
intn[0] = 
 Join[Flatten[Table[{{-Lx, y, z}, 0}, {y, ypoints}, {z, zpoints}], 1],
   Flatten[Table[{{x, -Ly, z}, 0}, {x, xpoints}, {z, zpoints}], 1], 
  Flatten[Table[{{x, y, -Lz}, 0}, {x, xpoints}, {y, ypoints}], 1], 
  int, Flatten[Table[{{Lx, y, z}, 0}, {y, ypoints}, {z, zpoints}], 1],
   Flatten[Table[{{x, Ly, z}, 0}, {x, xpoints}, {z, zpoints}], 1], 
  Flatten[Table[{{x, y, Lz}, 0}, {x, xpoints}, {y, ypoints}], 1]]; 
Vdd[0] = Interpolation[intn[0], InterpolationOrder -> 1];



Do[sol[s] = 
    NDSolveValue[{ (\[Psi][x, y, z] - sol[s - 1][x, y, z])/dt - 
         1/2 Laplacian[\[Psi][x, y, z], {x, y, z}] + 
         1/2 (\[Gamma]^2 x^2 + \[Nu]^2 y^2 + \[Lambda]^2 z^2) \[Psi][
           x, y, z] + 
         4 \[Pi] a Nat Abs[sol[s - 1][x, y, z]]^2 \[Psi][x, y, z] + 
         3 add Nat Vdd[s - 1][x, y, z] \[Psi][x, y, 
           z] + \[Gamma]QF Nat^(3/2) Abs[
            sol[s - 1][x, y, z]]^3 \[Psi][x, y, z] == 0, 
       DirichletCondition[\[Psi][x, y, z] == sol[0][x, y, z], 
        True]}, \[Psi], Element[{x, y, z}, mesh]] // Quiet;
   nk[s] = 
    Table[Sum[
      xweights[[i1]] yweights[[i2]] zweights[[i3]] Exp[
        I xpoints[[j1]] xpoints[[i1]] + 
         I ypoints[[j2]] ypoints[[i2]] + 
         I zpoints[[j3]] zpoints[[i3]]] Abs[
         sol[s][xpoints[[i1]], ypoints[[i2]], zpoints[[i3]]]]^2, {i1, 
       Length[xg]}, {i2, Length[yg]}, {i3, Length[zg]}], {j1, 
      Length[xg]}, {j2, Length[yg]}, {j3, Length[zg]}];
   int = 
    Flatten[Table[{{xpoints[[j]], ypoints[[k]], zpoints[[m]]}, 
       c0 Re[Sum[
          xweights[[i1]] yweights[[i2]] zweights[[
            i3]] Exp[-I xpoints[[j]] xpoints[[i1]]] Exp[-I ypoints[[
              k]] ypoints[[i2]]] Exp[-I zpoints[[m]] zpoints[[
              i3]]] ((3 zpoints[[i3]]^2)/(xpoints[[i1]]^2 + 
                ypoints[[i2]]^2 + zpoints[[i3]]^2) - 1) nk[s][[i1, i2,
             i3]], {i1, Length[xg]}, {i2, Length[yg]}, {i3, 
           Length[zg]}]]}, {j, Length[xg]}, {k, Length[yg]}, {m, 
       Length[zg]}], 2];
   intn[s] = 
    Join[Flatten[Table[{{-Lx, y, z}, 0}, {y, ypoints}, {z, zpoints}], 
      1], Flatten[Table[{{x, -Ly, z}, 0}, {x, xpoints}, {z, zpoints}],
       1], Flatten[
      Table[{{x, y, -Lz}, 0}, {x, xpoints}, {y, ypoints}], 1], int, 
     Flatten[Table[{{Lx, y, z}, 0}, {y, ypoints}, {z, zpoints}], 1], 
     Flatten[Table[{{x, Ly, z}, 0}, {x, xpoints}, {z, zpoints}], 1], 
     Flatten[Table[{{x, y, Lz}, 0}, {x, xpoints}, {y, ypoints}], 1]]; 
   Vdd[s] = Interpolation[intn[s], InterpolationOrder -> 1];, {s, 1, nt}] // AbsoluteTiming

Visualization in cross-section $z=0$

Table[DensityPlot[Abs[sol[i][x, y, 0]], {x, -Lx, Lx}, {y, -Ly, Ly}, 
  ColorFunction -> Hue, PlotLegends -> Automatic, PlotLabel -> i, 
  PlotRange -> All, PlotPoints -> 50], {i, 0, nt}] 

Figure 1

Visualization in cross-section $y=0$

Table[DensityPlot[Abs[sol[i][x, 0, z]], {x, -Lx, Lx}, {z, -Lz, Lz}, 
  ColorFunction -> Hue, PlotLegends -> Automatic, PlotLabel -> i, 
  PlotRange -> All, PlotPoints -> 50], {i, 0, nt}]

Figure 2

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  • $\begingroup$ Thanks for your solution @AlexTrounev. I like how you used join in int[0] and discretized the time derivative. Indeed the first step looks like the figures in the paper but after some iterations, the segregated parts are again rejoined with two maxima. Any idea why? Which iteration step will be safe to say it is the final solution? If you can help me further with the energy terms: Ekin and Edip and the phase, it will be very helpful. $\endgroup$ Commented Sep 28, 2022 at 7:20
  • $\begingroup$ @ArghaDebnath In the paper they did not discuss the numerical method used to compute results. We can try to reproduce what they did, but it takes time. $\endgroup$ Commented Oct 3, 2022 at 5:23
  • $\begingroup$ It's alright @AlexTrounev. Thanks again for what you have done. $\endgroup$ Commented Oct 3, 2022 at 5:42

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