# How to numerical solve equation with 3x-integrals

Posted 4 months ago
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 Good day everyone. I have a big problem with Mathematica. I cannot calculate equation with 3x-integrals. Equation will be needed solve numerically. Please, help me, I don't know how to solve this.I need to find Delta: Delta = mu + 2*U*(sigma - rho) where rho = (1/2)* NIntegrate[(eps + Delta)/Ek - 1, {kx, -Pi, Pi}, {ky, -Pi, Pi}, {kz, -Pi, Pi}] sigma = -(1/2)* NIntegrate[Delta/Ek, {kx, -Pi, Pi}, {ky, -Pi, Pi}, {kz, -Pi, Pi}] where m = 0.02 J = 50 muB = 0.67 Dst = 7.1 U = 313 g = 2.06 mu = muB*g*Hext - Dst eps = J*(3 - Cos[kx] - Cos[ky] - Cos[kz]) Ek = eps^(1/2)*(eps + Delta)^(1/2) Here Hext should change from 6 to 12 with the step 0.5Please, help me. Answer
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Posted 4 months ago
 I get the impression that there is no solution of your problem.And there are problems: may be your integrand has a singularity in the region of integration. I did not check this. I tried the following m = 0.02 J = 50 muB = 0.67 Dst = 7.1 U = 313 g = 2.06 mu = muB*g*Hext - Dst eps = J*(3 - Cos[kx] - Cos[ky] - Cos[kz]) Ek[Delta_] := eps^(1/2)*(eps + Delta)^(1/2) (Note the error messages in what follows pointing to an unfriendly integrand. And it could be that the results are simply wrong) trho = Table[ (1/2)*NIntegrate[(eps + Delta)/Ek[Delta] - 1, {kx, -Pi, Pi}, {ky, -Pi, Pi}, {kz, -Pi, Pi}], {Delta, 0, 2, .2}] tsigma = Table[ -(1/2)*NIntegrate[Delta/Ek[Delta], {kx, -Pi, Pi}, {ky, -Pi, Pi}, {kz, -Pi, Pi}], {Delta, 0, 2, .2}] And then mu + 2*U*(tsigma - trho)[] /. Hext -> 5.14418 Here you can see that Delta = 0 is a solution if Hext = 5.14418 Answer
Posted 4 months ago
 Thank you for your answer. It will be helpful for me. I try to simplify the integral, previously. Thank you. Answer
Posted 4 months ago
 Ok. Your integrand has indeed a singularity at the origin. See below. Are you sure that your model / equations are correct? m = 0.02 J = 50 muB = 0.67 Dst = 7.1 U = 313 g = 2.06 mu = muB*g*Hext - Dst eps[kx_, ky_, kz_] := J*(3 - Cos[kx] - Cos[ky] - Cos[kz]) Ek[Delta_, kx_, ky_, kz_] := eps[kx, ky, kz]^(1/2)*(eps[kx, ky, kz] + Delta)^(1/2) eps[0, 0, 0] Ek[del, 0, 0, 0] So the expression containing Ek will tend to infinitiy whern Ek is evaulated at the origin fac = .2; Plot3D[((eps[kx, ky, 0] + 1)/Ek[2, kx, ky, 0] - 1) Boole[ kx^2 + ky^2 + 0 > .0004], {kx, -Pi fac, Pi fac}, {ky, -Pi fac, Pi fac}, PlotRange -> All, PlotPoints -> 20] Plot[((eps[kx, 0, 0] + 1)/Ek[2, kx, 0, 0] - 1), {kx, -Pi fac, Pi fac}, PlotRange -> All, PlotStyle -> {Thick, Red}, AxesOrigin -> {0, 0}] Answer
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Posted 4 months ago
 This I think re-inforces what Hans showed. sol = Series[eq1, {kx, 0, 3}, {ky, 0, 3}, {kz, 0, 3}] // Normal // ExpandAll; For either eq1 or eq2 is a "typical" series with factors kz Delta in the denominator but sporadically powered so (i believe) neither can be made to eliminate the other in all terms of the series.Limit[sol, kz -> 0] ~= kx^(n-1) * ky^(n-1) * Sqrt[Delta] * InfinityNIntegrate will approach inf when kz or Delta -> 0. All assuming real values (Nintegrate does not handle complex values, and complex analysis of the problem would be much different). Answer