Simplifying the Willmore energy of an ellipsoid

Question

Willmore energy measures how "non-spherical" a smooth surface $S$ is. It is defined by $$W(S)=\int_SH^2\,dA$$ where $H$ is the mean curvature.

For a torus of revolution with major and minor radii $a$ and $b$ respectively where $a>b$, if we let $p=b/a$ then its Willmore energy is easily shown to be $\frac{\pi^2}{p\sqrt{1-p^2}}$, which attains its minimum at $p=1/\sqrt2$. The (proved) Willmore conjecture states that the torus thus obtained has the minimum energy among all genus-$1$ surfaces of $2\pi^2$. I took Blender out for a ride and produced a render of this "perfect doughnut":

Now I want to calculate the Willmore energy of an ellipsoid. In this genus-$0$ case the extremal results are easy to get: $W(S)\ge4\pi$ and equality is attained iff $S$ is a sphere. But I still want numerical results for the fun of it. Using fundamental forms (and cross-checking with this), I found that for an ellipsoid $E$ with semi-axes $1,a,b$: $$W(E)=\frac{a^2b^2}4\int_0^{2\pi}\int_0^\pi\frac{(a^2+b^2+(1-(a\cos u)^2-(b\sin u)^2)\sin^2v)^2\sin v}{((ab\cos v)^2+((a\sin u)^2+(b\cos u)^2)\sin^2v)^{5/2}}\,dv\,du$$ which I cannot seem to simplify further.

Does the double integral above have a simpler or even closed form?

Or am I doing it wrong, whereby I would have an easier time using implicit equations as mooted here?

Edit: Using Zhou's ellipsoidal coordinates as suggested by Jean Marie in the comments I have got an expression using only single integrals. For an ellipsoid $E$ with semi-axes $a>b>c>0$ let $$R_\eta(k)=\int_{c^2}^{b^2}\frac{\eta^k}{\sqrt{(a^2-\eta)(b^2-\eta)(\eta-c^2)\eta}}\,d\eta$$ and $$R_\zeta(k)=\int_{b^2}^{a^2}\frac{\zeta^k}{\sqrt{(a^2-\zeta)(\zeta-b^2)(\zeta-c^2)\zeta}}\,d\zeta$$ Then $$W(E)=\frac{(abc)^2}2(R_\eta(-2)R_\zeta(1)+R_\eta(-1)R_\zeta(0)-R_\eta(0)R_\zeta(-1)-R_\eta(1)R_\zeta(-2))$$ Edit 2: The above expression can be simplified to $$W(E)=\frac{(abc)^2}2(R_\eta(-2)R_\zeta(1)-R_\eta(1)R_\zeta(-2))+\pi$$

Answer 1

For an ellipsoid $E$ with semi-axes $a\ge b\ge c>0$, define $$A=a^2,B=b^2,C=c^2,\varphi=\cos^{-1}\frac ca$$ $$g=\sqrt{(A-C)B},m=\frac{(B-C)A}{(A-C)B}$$ Then $$\color{red}{\begin{align}W(E)=\frac\pi{3ABC}&\Big(C(5AB+2AC+2BC)\\ &+2(AB+AC+BC)E(\varphi,m)g\\ &+BC(AB+AC+2BC-A^2)F(\varphi,m)/g\Big)\end{align}}$$ If $b=c$ (the spheroidal case) the above formula simplifies to $$W(E)=\frac{\pi A\varphi}g+\frac\pi3\left(7+\frac{2B}A\right)$$ while if $a=b=c$ then $E$ is a sphere and $W(E)=4\pi$. For the spheroidal case, whether $a>b$ does not matter – the complex parts cancel out.

All elliptic integrals in this answer use the same argument interpretation as their implementations in Mathematica and mpmath.

---

This is the result of a long and tedious simplification of the eight integrals in the expression resulting from the use of Zhou's ellipsoidal coordinates. In the equations below, $g=\frac2{\sqrt{(A-C)B}}$, and where $m$ appears in a formula for $R_\zeta(\cdot)$ its value is $1-m=\frac{(A-B)C}{(A-C)B}$ and not $\frac{(B-C)A}{(A-C)B}$. $$R_\eta(1)=Cg\Pi\left(\frac{B-C}B,m\right)$$ $$R_\zeta(1)=g\left((B-C)\Pi\left(\frac{A-B}{A-C},m\right)+CK(m)\right)$$ $$R_\eta(0)=gK(m)$$ $$R_\zeta(0)=gK(m)$$ $$R_\eta(-1)=\frac gA\left(K(m)+\frac{A-C}CE(m)\right)$$ $$R_\zeta(-1)=\frac gC\left(K(m)+\frac{C-A}AE(m)\right)$$ $$R_\eta(-2)=\frac g{3A^2BC^2}\begin{pmatrix}C(2BC+AB+AC-A^2)K(m)\\+2(A-C)(AB+AC+BC)E(m)\end{pmatrix}$$ $$R_\zeta(-2)=\frac g{3A^2BC^2}\begin{pmatrix}A(2AB+AC+BC-C^2)K(m)\\+2(C-A)(AB+AC+BC)E(m)\end{pmatrix}$$ The terms $R_\eta(-1)R_\zeta(0)-R_\eta(0)R_\zeta(-1)$ in the main expression simplify to just $\frac{2\pi}{ABC}$. The other two terms simplify to a more complicated expression, but still one not involving any integral signs.

---

Here is Python code to calculate the Willmore energy:

from mpmath import *
def W_ellipsoid(a, b, c):
    c, b, a = sorted([fabs(a), fabs(b), fabs(c)])
    if c == 0: return inf
    if a == c: return 4pi
    A, B, C = aa, bb, cc
    g = sqrt((A-C)B)
    phi = acos(c/a)
    m = ((B-C)A) / ((A-C)B)
    return pi/(3ABC) * ( C(5AB + 2AC + 2BC)
                          + 2(AB + AC + BC)ellipe(phi, m)g
                          + BC(AB + AC + 2BC - AA)*ellipf(phi, m)/g )
def W_spheroid(a, b):
    a, b = fabs(a), fabs(b)
    if a == 0 or b == 0: return inf
    if a == b: return 4pi
    A, B = aa, bb
    g = sqrt((A-B)B)
    phi = acos(b/a)
    return re(pi(Aphi/g + (7+2*B/A)/3))

---

Edit: I got around to symmetrising the elliptic integrals and obtained the following formula for the Willmore energy, where $A,B,C$ keep their meanings from above: $$W(E)=\frac\pi3\left(3+4\left(\frac1A+\frac1B+\frac1C\right)R_G(AB,CA,BC)-(A+B+C)R_F(AB,CA,BC)\right)$$