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I would like to solve the integral:

$ f(a)=\iiint\limits_{\mathcal{R}(a)}\frac{\sqrt{12\left(\left(\frac{x-y}{x+y}\right)^2+\left(\frac{y-z}{y+z}\right)^2+\left(\frac{x-z}{x+z}\right)^2\right)}}{\exp(x+y+z)\sqrt{\left(\frac{4y(x-y)}{(x+y)^3}+\frac{4z(x-z)}{(x+z)^3}\right)^2+\left(\frac{-4x(x-y)}{(x+y)^3}+\frac{4z(y-z)}{(y+z)^3}\right)^2+\left(\frac{-4y(y-z)}{(y+z)^3}-\frac{4x(x-z)}{(x+z)^3}\right)^2}}\rm{d} x \rm{d} y \rm{d}z $

The region of integration $\mathcal{R}(a)$ is the surface:

$ \mathcal{R}(a)=\left\{(x,y,z)\in \mathbb{R}^{3}, x\geq 0, y\geq 0, z\geq 0,z\leq x,z\leq y, \sqrt{\frac{1}{3}\left(\left(\frac{x-y}{x+y}\right)^2+\left(\frac{y-z}{y+z}\right)^2+\left(\frac{x-z}{x+z}\right)^2\right)}=a\right\} $

With parameter $a\in[0,1]$.

All my efforts to solve this analytically have failed. I'm now struggling to get an accurate numerical solution.

I implemented it in Mathematica (v. 10.3) in this manner:

region[a_?NumericQ] =  ImplicitRegion[Sqrt[1/3 ((x - y)^2/(x + y)^2 + (x - z)^2/(x + z)^2 + (y - z)^2/(y + z)^2)] == a && x > 0 && y > 0 &&  z > 0 && z <= x && z <= y, {x, y, z}]

result = Table[{a,Re[NIntegrate[Sqrt[12 ((x - y)^2/(x + y)^2 + (x - z)^2/(x + z)^2 + (y - z)^2/(y + z)^2)]/(E^(x + y + z) Sqrt[(-((4 x (x - z))/(x + z)^3) - (4 y (y - z))/(y + z)^3)^2 + ((4 z (y - z))/(y + z)^3 - (4 x (x - y))/(x + y)^3)^2 + ((4 y (x - y))/(x + y)^3 + (4 z (x - z))/(x + z)^3)^2]), {x, y, z} [\Element] region[a]]]}, {a, 0, 1, .01}]

But I'm having all sort of convergence problems, 1/0 errors, accuracy, instabilities, etc you name it...

Do you know what I might be doing wrong? Do you know any work-around? Any advice is welcome.

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    $\begingroup$ I might suggest a strategy: start with an analogous two-dimensional problem and get it working there first. Once you have the bugs ironed out of that, generalize to 3D. $\endgroup$ – bill s Oct 29 '15 at 12:44
  • $\begingroup$ Welcome to Mathematica.SE! 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$ – user9660 Oct 29 '15 at 12:48
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I have rewritten your definition of the region as

region2[a_?NumericQ]:=ImplicitRegion[2 x^3 (y-z)^2 (y+z)+2 x y (y-z)^2 z (y+z)+x^4 (3 y^2+2 y z+3 z^2)+y^2 z^2 (3 y^2+2 y z+3 z^2)+x^2 (3 y^4-2 y^3 z-18 y^2 z^2-2 y z^3+3 z^4)==3 a^2 (x+y)^2 (x+z)^2 (y+z)^2&&x>0&&y>0&&z>0&&z<=x&&z<=y,{x,y,z}]

That makes at least ContourPlot3D for these regions considerably faster.

Then Mathematica 10.3 can do the integration. Of course, it is time consuming.

NIntegrate[ Sqrt[12 ((x-y)^2/(x+y)^2+(x-z)^2/(x+z)^2+(y-z)^2/(y+z)^2)]/(E^(x+y+z) Sqrt[(-((4 x (x-z))/(x+z)^3)-(4 y (y-z))/(y+z)^3)^2+((4 z (y-z))/(y+z)^3-(4 x (x-y))/(x+y)^3)^2+((4 y (x-y))/(x+y)^3+(4 z (x-z))/(x+z)^3)^2]), {x,y,z} \[Element] region2[0.3]] // Timing

(* {166.765, 0.321009 -1.75152*10^-8 I} *)
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  • $\begingroup$ Thanks for your interest! Your strategy really improved the calculation! The problem is that it does not work well for every value of $a$. You tried 0.3, and it works. But for very low values close to zero or higher ones, $a\leq 0.7$ for instance, then it fails. Anyway, I observed that NIntegrate fails to converge for a certain value of $a$, but it manages for a value of $a$ close to that. I will do the "artisan work" and try to get a smooth curve with this strategy. Something like a manual adaptive algorithm. Luckily enough, the solution $f(a)$ is quite smooth and nice. $\endgroup$ – Ram Oct 30 '15 at 8:45

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