Minimization of the sums of complicated parametric integrals

Question

I have to find the values of the parametersω and λ that minimize the function Int[ω, λ] defined as the sum of five quite complicated integrals of a trial function f[r, ω, λ], namely

a = -74.04252664070837;
b = 208.01432471151327;
d = -65.08706834153939;
A = 1.56692098226;
X0 = 3.9468;
X1 = 2.4323;
X2 = 3.6348;
f[r_, ω_, λ_] = r/Sqrt[r^2 + 2] (1 + 4 E^(-λ r) Sin[ω r])
Df[r_, ω_, λ_] = D[f[r, ω, λ], r];
j[x_, r_] = 2 A^2 r x;
g0[x_, r_] =  
  a + b/2 + 3 d/4 + A^2 (b + d) (x^2 + r^2) + d A^4 ((x^2 + r^2)^2 + 4 x^2 r^2);
g1[x_, r_] = j[x, r] (b + 2 d) + 4 d A^4 r x (r^2 + x^2);

Int1[ω_, λ_] = NIntegrate[(Df[r, ω, λ]^2 + f[r, ω, λ]^2/r^2) r, {r, 0, 100}];
Int2[ω_, λ_] = 
  Pi^1.5/A 
    NIntegrate[
      ((1 - f[r, ω, λ]^2) (1 - f[x, ω, λ]^2) r x E^(-A^2 (r^2 + x^2)) 
        (g0[x, r] BesselI[0, j[x, r]] - g1[x, r] BesselI[1, j[x, r]])), 
      {x, 0, Infinity}, {r, 0, Infinity}, 
      Method -> "LevinRule"] 
Int3[ω_, λ_] = 
  -1/2 X0 
    NIntegrate[(1 - f[r, ω, λ]^2)^2 r, {r, 0, Infinity}, Method -> "LevinRule"];
Int4[ω_, λ_] = 
  -1/3 X1 
     NIntegrate[((1 - f[r, ω, λ]^2)^2 (2 + f[r, ω, λ]^2)) r, {r, 0, Infinity}, 
       Method -> "LevinRule"];
Int5[ω_, λ_] = 
  1/4 X2 
    NIntegrate[
      (1 -f[r, ω, λ]^2)^2 (f[r, ω, λ]^4 + 2 f[r, ω, λ]^2 + 3) r, {r, 0, Infinity}, 
      Method -> "LevinRule"] 
Int[ω_, λ_] = Int1[ω, λ] + Int2[ω, λ] + Int3[ω, λ] + Int4[ω, λ] + Int5[ω, λ];

I can't use any minimization algorithm since Mathematica doesn't seem to be able to minimize $Int[\omega,\lambda]$. At least I would like to plot or evaluate (also a table) $Int[\omega,\lambda]$ in a reasonable region for the parameters $0<\omega<2$ and $0<\lambda<2$ but Mathematica returns the values of very slowly $Int[\omega,\lambda]$ and for certain values in this region crashes and my pc with it.

Also, the output shows many errors. Since I'm quite new to Mathematica, I actually don't know how to solve the problem nor the best method of integration for Int2 and Int5 which seem to be the most difficult integrals to evaluate. I used also "LevinRule" because the output of Mathematica seemed to suggest that option.

Answer 1

As always, you should carefully verify that any result is correct.

On a fairly old slow desktop machine with only 12gb with a fresh clean evaluation of this:

a = -74.04252664070837;
b = 208.01432471151327;
d = -65.08706834153939;
A = 1.56692098226;
X0 = 3.9468;
X1 = 2.4323;
X2 = 3.6348;
Clear[r, ω, λ, f, Df, j, g0, g1];
f[r_, ω_, λ_] = r/Sqrt[r^2 + 2] (1 + 4 E^(-λ r) Sin[ω r]);
Df[r_, ω_, λ_] = D[f[r, ω, λ], r];
j[x_, r_] = 2 A^2 r x;
g0[x_, r_] = a + b/2 + 3 d/4 + A^2 (b + d) (x^2 + r^2) + d A^4 ((x^2 + r^2)^2 + 4 x^2 r^2);
g1[x_, r_] = j[x, r] (b + 2 d) + 4 d A^4 r x (r^2 + x^2);
Clear[Int1, Int2, Int3, Int4, Int5];
exp1 = Simplify[(Df[r, ω, λ]^2 + f[r, ω, λ]^2/r^2) r];
exp2 = Simplify[((1 - f[r, ω, λ]^2) (1 -  f[x, ω, λ]^2)* r x E^(-A^2 (r^2 + x^2))*
  (g0[x, r] BesselI[0, j[x, r]] - g1[x, r]*BesselI[1, j[x, r]]))];
exp3 = Simplify[(1 - f[r, ω, λ]^2)^2 r];
exp4 = Simplify[((1 - f[r, ω, λ]^2)^2 (2 + f[r, ω, λ]^2)) r];
exp5 = Simplify[(1 - f[r, ω, λ]^2)^2 (f[r, ω, λ]^4 + 2 f[r, ω, λ]^2 + 3) r];
bound=10;
Int1[ω_, λ_] := NIntegrate[exp1, {r, 0, bound}, Method -> "MonteCarlo"];
Int2[ω_, λ_] := Pi^1.5/A*NIntegrate[exp2, {x,0,bound}, {r,0,bound}, Method->"MonteCarlo"];
Int3[ω_, λ_] := -1/2 X0*NIntegrate[exp3, {r, 0, bound}, Method -> "MonteCarlo"];
Int4[ω_, λ_] := -1/3 X1*NIntegrate[exp4, {r, 0, bound}, Method -> "MonteCarlo"];
Int5[ω_, λ_] := 1/4 X2*NIntegrate[exp5, {r, 0, bound}, Method -> "MonteCarlo"];
Int[ω_, λ_] := Int1[ω, λ] + Int2[ω, λ] + Int3[ω, λ] + Int4[ω, λ] + Int5[ω, λ];

Look for a minimum near ω==0.5, λ==1.2

SortBy[Flatten[Table[{ω, λ, Int[ω, λ]}, {ω, 1/4, 3/4, 1/16}, {λ, 1, 3/2, 1/16}], 1], Last]
ListPlot3D[%]

and there does not appear to be any clear minimum near your estimate.

Can you do the work to determine what upper bound is sufficient for your integrands to be very close to zero beyond that?