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I'm reading a paper on the Schwarz D minimal surface, and I'm wondering whether the authors have made a mistake. They evaluate the integral $$ \int_0^z \frac{2t\;\mathrm{d}t}{\sqrt{t^8-14t^4+1}}, $$ with $z$ inside a domain $\Omega$ which contains no zeros of the octic polynomial, and say it equals $$ \frac{1}{4}F\left(\arcsin\left(\frac{4z^2}{z^4+1}\right),97-56\sqrt{3}\right), $$ with $F$ the incomplete elliptic integral of the first kind. However, Mathematica gives me $$ \frac{\sqrt{-z^4-4 \sqrt{3}+7} \sqrt{\left(4 \sqrt{3}-7\right) z^4+1}}{\sqrt{z^8-14 z^4+1}} F\left(\arcsin\left(\frac{z^2}{\sqrt{7-4 \sqrt{3}}}\right),97-56 \sqrt{3}\right). $$

The strange thing is that, after evaluating them numerically at a couple of points inside $\Omega$, the two expressions seem to follow one another quite closely (within about $10^{-3}$ for both the real and the imaginary part) without actually being equal. Is this due to Mathematica-related inaccuracies or are they really unequal. If the latter is the case, which of the two expressions (if any) is correct?

I'm not very knowledgeable about elliptic integrals, so any help would be warmly appreciated. Thanks for reading!

Parcly Taxel
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Kim Fierens
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  • It may help to increase the WorkingPrecision and MaxRecursion on NIntegrate to get an accurate result. – user111187 Feb 01 '15 at 12:25
  • I'm not currently using NIntegrate; I simply had Mathematica calculate the antiderivative symbolically. Edit: NIntegrate (with high settings for WorkingPrecision and MaxRecursion) gives a result somewhat closer to Mathematica's own antiderivative. But it still doesn't really decide the issue for me. – Kim Fierens Feb 01 '15 at 12:34
  • Interesting. Do you pass numerical or exact values of $z$ to the functions?. If numerical, try increasing their precision. I assume you have already taken into account that Mathematica's definition of the elliptic integral differs from the convention used in the literature. – user111187 Feb 01 '15 at 12:45
  • $x=t^2$ simplifies things a bit. – Lucian Feb 01 '15 at 14:49

3 Answers3

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I think the Mathematica answer is better. The original integrand $$ \frac{2t}{\sqrt{t^8-14t^2+1}}, $$ is imaginary for $t>0.267$, but the integral in the paper $$ \frac{1}{4}F\left(\arcsin\left(\frac{4z^2}{z^4+1}\right),97-56\sqrt{3}\right), $$ remains real until the arcsin reaches $1$, which is when $z=0.4696$. But the Mathematica result $$ \frac{\sqrt{-z^4-4 \sqrt{3}+7} \sqrt{\left(4 \sqrt{3}-7\right) z^4+1}}{\sqrt{z^8-14 z^4+1}} F\left(\arcsin\left(\frac{z^2}{\sqrt{7-4 \sqrt{3}}}\right),97-56 \sqrt{3}\right) $$ is real until $z=0.267$ again, because of its denominator.

added
the value $0.4696$ for the singularity of the solution of the paper is a solution of $$ \frac{4z^2}{z^4+1} = \sin 1 $$

GEdgar
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  • Great answser; thanks for pointing this out, GEdgar! By the way, do you have any idea why the two expressions agree rather closely on at least part of their common domain (other than mere coincidence)? – Kim Fierens Feb 01 '15 at 13:53
  • Was there a typo in the OP's integrand? Kindly check my comment/answer. – Tito Piezas III Nov 08 '16 at 15:04
  • I investigated the original paper (while trying to render the gyroid). It's supposed to have $-14t^4$ in the denominator, not $-14t^2$. Even with that, the provided solution is completely wrong. See my answer. – Parcly Taxel Oct 07 '20 at 07:58
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(Too long for a comment.) I think it is all because of a typo. The integrand supposedly is $$\frac{2t}{\sqrt{t^8-14t^2+1}}\tag1$$ but I think it should be $$\frac{2t}{\sqrt{t^8-14t^\color{blue}4+1}}\tag2$$ Note that the latter form is an octahedral invariant and is also expressible as, $$(t^2-i)^8+(2t)^8+(t^2+i)^8=2(t^8-14t^4+1)^2$$ with the imaginary unit $i$.

Tito Piezas III
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  • Interesting idea. Your (2) has its singularity at $0.5266$ which differs from the answer in the paper with singularity at $0.4696$. So even with this integrand, it still seems the paper is wrong. – GEdgar Nov 08 '16 at 18:02
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The paper referred to in the question is most likely Gandy et al. (1999), Exact computation of the triply periodic D ('Diamond') minimal surface, Chemical Physics Letters 314, 543–551. It is one of three related papers in which exact coordinates are derived for the Schwarz D/G/P surface family.

The integral (with my correction) appears in the Weierstrass–Enneper parametrisation of the surfaces. After scaling and taking the real part, it gives the $z$-coordinate of a point on a patch which then gets repeated to form the entire surface.

But I can confirm that $$\frac14F\left(\sin^{-1}\frac{4z^2}{z^4+1},97-56\sqrt3\right)$$ is dead wrong. The authors say they used Byrd and Friedman's Handbook of Elliptic Integrals for Engineers and Physicists – a reference that really helped me with finding the Willmore energy of an ellipsoid – but they must have applied the tables incorrectly, for a more careful calculation yields the correct answer as $$\int_0^z\frac{2t}{\sqrt{t^8-14t^4+1}}\,dt=\frac1{2+\sqrt3}F\left(\sin^{-1}\frac{z^2}{2-\sqrt3},97-56\sqrt3\right)$$ As it turns out, there are many, many typos in the three papers, including having $Q_6$ and $Q_7$ among $O_1$ and $O_2$ in a table in the gyroid paper and getting the W–E parametrisation equations themselves wrong in the Schwarz D paper.

Parcly Taxel
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