# Differential equation with some constraints

I posted this to stackexchange, and after some hours got a comment that was so pessimistic about finding some neat orderly solution, that I'm posting it here too. (In case anyone cares, this is related to this question, which I posted here earlier.)

I'd like $\alpha,\beta,\gamma$ as functions of $t$, satisfying the following conditions: \begin{align} \alpha+\beta+\gamma & = 0 \\ \sin^2\alpha + \sin^2\beta + \sin^2\gamma & = c^2 \\ \left| \frac{d}{dt}(\sin\alpha,\sin\beta,\sin\gamma)\right| & = 1 \end{align} I'm thinking of $c^2$ as small. At the very least that means $<2$, and intuitively it means $\ll 2$. Some geometry shows that there is a qualitative change in the nature of the solutions when $c^2$ goes from $<2$ to $>2$.

Later edit: The question above asks for a parametrization by arc-length. Here's an ugly parametrization by something quite remote from arc length: $$\beta = \frac{\arccos\left(\frac{1 + \sin^2\alpha - c^2}{\cos\alpha}\right)-\alpha}{2}$$ And then $\gamma = \pi - \alpha - \beta$. In order to get the whole curve, you'd need a multiple-valued arccosine and then you'd pick the right value for the particular point on the curve. One thing that fails to be obvious to me just from the way the function above is written, giving $\beta$ as a function of $\alpha$, is that that function is its own inverse.

So here's a less demanding question that the one above: Is there some nice pleasant way of parametrizing the curve that, if not by arc-length, at least treats $\alpha$, $\beta$, and $\gamma$ equally, so that it's perfectly self-evident from the way it's written that the whole expression is symmetric in $\alpha,\beta,\gamma$?

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BTW, the first two constraints imply that $0 \le c^2 \le 9/4$. –  Michael Hardy Aug 28 '11 at 19:29

A solution is in effect an arc-length parametrization of a space curve. Let $\vec v =(x,y,z) = (\sin \alpha, \sin \beta, \sin \gamma)$. The first equation is then the somewhat complicated algebraic surface, call it $S_1$: $$S_1: 2(y^2z^2+z^2x^2+x^2y^2) - (x^4+y^4+z^4) = 4(xyz)^2.$$ More precisely, it's a quarter of $S_1$, because $S_1$ also contains the loci where one of $\alpha,\beta,\gamma$ is the sum of the other two. You may recognize the left-hand side from Hero(n)'s formula: it factors as $(x+y+z)(-x+y+z)(x-y+z)(x+y-z)$. The second equation then intersects this $S_1$ with the sphere $\Sigma_c: \|\vec v\| = c$. The final equation says that $\vec v$ depends on $t$ and its derivative has norm $1$. This makes $\vec v(t)$ the arc-length parametrization of the curve.

The following might give you a handle on what happens for small $c$. Scaling by $c$ yields the intersection of the unit sphere $\Sigma_1$ with the varying surface $$S_c: (x+y+z)(-x+y+z)(x-y+z)(x+y-z) = 4c^2(xyz)^2.$$ Now $S_0 \cap \Sigma_1$ is the union of the four great circles $\{ x \pm y \pm z = 0 \} \cap \Sigma_1$, each of which has an easy arc-length parametrization. The one that corresponds to $\alpha + \beta + \gamma = 0$ is $x+y+z=0$. To get at your problem with small $c$, you might start from these parametrizations of $S_0 \cap \Sigma_1$ and consider the curves $S_c \cap \Sigma_1$ as deformations of that great circle, then at the end speed up the resulting arc-length parametrizations by a factor $1/c$ to undo the scaling.

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The hour is late and I will look at this tomorrow. But even before digesting everything above, I've up-voted it because the identity following "$S_1$" looks just like what I wrote in another stackexchange posting (except for a factor of 2.....): math.stackexchange.com/questions/59508/trigonometric-identity And that one arose from a product of three sines, whereas this one arose from a sum of products of two sines, and there ought to be certain connections. So are you suggesting that there is a nice neat solution after all? Or only that there is one in a limiting case? –  Michael Hardy Aug 28 '11 at 3:35
The arc-length parametrization is the inverse function of an arc-length integral, which involves a square root and can only rarely have an elementary formula (already for an ellipse we famously get elliptic integral). I see that Robert Bryant already worked out what happens here, and verified that there's almost never an elementary formula; so it won't get any more nice or neat than inverting the integral of the square root of a rational function. Did you have a reason to expect or hope for a particularly nice form here? –  Noam D. Elkies Aug 28 '11 at 17:18
@Noam: Well, now I've started looking at this answer while awake. First I wondered how you got $S_1$ from the first constraint. Then I saw how it can be done, if not how you did it, which I suspect is different. To be continued....... –  Michael Hardy Aug 29 '11 at 17:55

Building on Noam's suggestion, you could try using the inherent symmetry of the problem: Set $\sigma_1 = x^2 + y^2 + z^2$, $\sigma_2 = x^2y^2+y^2z^2+z^2x^2$, and $\sigma_3 = x^2y^2z^2$. Then your conditions become $\sigma_1 = c^2$ and $\sigma_2 = \sigma_3 + \tfrac14c^4$. Meanwhile, you have $$dt^2 = dx^2 + dy^2 + dz^2 = \frac{d(x^2)^2}{4x^2}+\frac{d(y^2)^2}{4y^2}+\frac{d(z^2)^2}{4z^2},$$ and this latter expression, being symmetric in $x^2,y^2,z^2$, can be expressed as a differential expression in $\sigma_1, \sigma_2, \sigma_3$. I won't write out the details, but a short computation (using Maple) shows that, taking advantage of the relations $\sigma_1 = c^2$ and $\sigma_2 = \sigma_3 + \tfrac14c^4$, this leads to the relation $$dt^2 = \frac{\bigl(c^6(c^2{-}2)-4(36{-}52c^2{+}21c^4{-}2c^6)\sigma_3+16(c^2{-}1){\sigma_3}^2\bigr)}{16\sigma_3\bigl(c^6(2{-}c^2)-4(27{-}18c^2{+}2c^4)\sigma_3-16{\sigma_3}^2\bigr)}\bigl(d\sigma_3\bigr)^2.$$ Now, for example, you can see why $c^2=2$ is special. The integral that gives $t$ will simplify dramatically in this case; in fact, it becomes an elementary integral. For general values of $c$, though, this is a hyperelliptic integral, and you won't find any simple relation between $t$ and $\sigma_3$, so, a fortiori, none between $t$ and $x$, $y$, and $z$. There are various special values of $c$ for which the roots and poles of the rational expression cancel, such as $c=0$, $c = \pm\sqrt{2}$, and $c = \pm\frac32$, and, for these, you'd expect the integral to simplify considerably, but, otherwise, you don't expect any nice relation.

Added remark: By the way, you can get from this more directly to the relation between $t$ and $x$, $y$, and $z$, since, for example, letting $u$ represent any one of $x^2$, $y^2$, or $z^2$, one has the relation $u^3 - c^2 u^2 + (\sigma_3 + \tfrac14 c^4)u - \sigma_3 = 0$, which can clearly be solved for $\sigma_3$ as a rational function of $u$. Substituting this into the above relation gives a differential equation directly relating $t$ and, say, $u = x^2$. It's not a particularly nice relation, though. Ultimately, this gives a relation of the form $x = F(t,c)$ where $F$ is some function periodic of period $3\tau(c)$ in the first variable for some $\tau(c)>0$. Then one finds that $y = F(t + \tau(c),c)$ and $z = F(t-\tau(c),c)$. This is, of course, a very symmetric expression, though it's not explicit.

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That $c^2 = 0$ and $c^2 = (3/2)^2$ are the two opposite extreme values of the sum of squares of the three sines follows from the first constraint. That $c^2 = 2$ gives an exceptionally well-behaved situation is actually seen just be thinking about secondary-school-level trigonometry. It's striking how a little trigonometry problem leads straight into moderately exotic (by comparison to this sort of trigonometry) functions, but I suppose the same can be said of other things that we've all seen. I'm going to print out the two answers posted so far and think about them before saying much more. –  Michael Hardy Aug 28 '11 at 19:38

OK, the "less demanding" question does seem more tractable; a few possible answers follow, though none is clearly the most "nice pleasant way of parametrizing" your curve. One direction leads to the trigonometric solution of a cubic equation with all roots real; the other leads to an elliptic curve with 6-torsion, and even to an extremal elliptic K3 surface! Which if any of these is best for you is a matter of taste and of what you're trying to do with these curves.

Let $(X,Y,Z) = (\sin^2 \alpha, \phantom.\sin^2 \beta, \phantom.\sin^2 \gamma)$. Then $(X,Y,Z)$ are coordinates of an algebraic curve $$E_c : X+Y+Z = c^2, \phantom{=} X^2+Y^2+Z^2 - 2(YZ+ZX+XY) + 4XYZ = 0.$$ So far we've preserved the $S_3$ symmetry, and can recover the original variables via $\alpha = \arcsin X^{1/2} = \frac12 \arccos(1-2X)$ and likewise for $\beta,\gamma$. But this begs the question of what $E_c$ looks like, and leaves us with multivalued arcsines or arccosines. The latter problem seems inherent in another symmetry of the equation: we can translate $\alpha,\beta,\gamma$ by $a\pi,b\pi,c\pi$ for any integers $a,b,c$ with $a+b+c=0$. But we can try to do more with $E_c$.

One direction is to express everything in terms of elementary symmetric functions $\sigma_1,\sigma_2,\sigma_3$ of $X,Y,Z$, as R.Bryant did: the first equation says $\sigma_1=c^2$, and the second says $\sigma_1^2 = 4 (\sigma_3 - \sigma_2)$; so $(\sigma_1,\sigma_2,\sigma_3)$ are parametrized in terms of $\sigma_3$, and then $X,Y,Z$ are the three roots of $$0 = u^3 - \sigma_1 u^2 + \sigma_2 u - \sigma_3 = u^3 - c^2 u^2 + (\sigma_3 + \tfrac14 c^4)u - \sigma_3.$$ This is still manifestly symmetric but rather implicit. We we can solve the cubic; since it has three real roots the solution will involve trisecting some auxiliary angle $\theta$, itself given as the arccosine of some explicit but complicated algebraic function of $c$ and $\sigma_3$. The roots will then be given in terms of $c$, $\sigma_3$, and the cosines of $\theta/3$, $(\theta+2\pi)/3$, and $(\theta+4\pi)/3$, and the action of $S_3$ will correspond to replacing $\theta$ by the equivalent $\pm (\theta+2\pi n)$ for some $n \bmod 3$. This will be far from nice and pleasant (compare with the formulas for constructing a regular 13-gon using an angle trisector, as in p.192 of Gleason's Monthly article), but it will have the advantage of leaving the symmetry close to the surface.

Another direction is to consider $E_c$ on its own terms. It is an elliptic curve, so rational functions on it like $x,y,z$ can be parametrized by elliptic functions like $\wp$ and $\wp'$. Moreover $E_c$ inherits the $S_3$ action so the resulting formulas must retain this symmetry; and the periodicity of $\wp,\wp'$ may even cancel out the ambiguity in the arcsine or arccosine. That's great if you love elliptic curves, not so great if you regard $\wp$ as yet another obscure transcendental function... At least these elliptic curves are rather nice: the cyclic permutations of $X,Y,Z$ are translations by 3-torsion points of $E_c$, and there's also a 2-torsion point because switching two of the variables, say $Y \leftrightarrow Z$, has a rational fixed point where the third variable vanishes (this corresponds to taking $\alpha = 0$ and $\beta+\gamma=0$ in the original equation). So $E_c$ actually has 6-torsion. If I did this right, an equivalent equation for $E_c$ is $$y^2 = x^3 + ((c^2-3) x - (c^2-2)^2)^2,$$ which exhibits the 3-torsion points where $x=0$, and has 2-torsion at $(x,y) = -((c^2-2)^2,0)$. As it happens $E_c$ is not far from the universal elliptic curve with a 6-torsion point, which is given by $y^2 = x^3 + ((h-3)x - (h-2)^2)^2$. What's more, our substitution $h=c^2$ produces an elliptic K3 surface whose fiber $E_c$ becomes singular at the familiar points $c = 0, \phantom.\pm \sqrt2, \phantom.\pm \frac32$, and also $c=\infty$ — and the multiplicities at $c=0,\phantom.\pm\sqrt2,\phantom.\infty$ are large enough that this elliptic K3 surface is "extremal" (finite Mordell-Weil group, maximal Picard number)! Such surfaces have attracted considerable attention over the years, starting with the Miranda-Persson list of semistable extremal surfaces (Math. Z. 201 (1989), 339–361), which includes ours with multiplicity vector $[1,1,4,6,6,6]$. This makes your family of curves very nice in that context, even if it doesn't do much to answer your motivating question...

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I tried to include this link to the Miranda-Persson paper, which however does not seem to work in the above answer: springerlink.com/content/u30268141h636x04/fulltext.pdf –  Noam D. Elkies Aug 29 '11 at 5:48
Very nice, Noam! I especially like the relation with the K3 surface. I knew that the curve on $x^2$, $y^2$, $z^2$ was an elliptic curve with some symmetries, but I didn't figure out the properties of the ($8$-fold) branched cover that represents the original $xyz$-curve or the branched cover of that on which $dt$ is actually a meromorphic differential. –  Robert Bryant Aug 29 '11 at 11:54
Thanks, Robert! Yes, the original curve, to say nothing of the $dt$ double cover, looks much more complicated, but M.Hardy seems willing to extract square roots of a function of one variable for free in his setting $-$ in any case he'll have to take some inverse trig function to recover his original variables $\alpha,\beta,\gamma$. –  Noam D. Elkies Aug 29 '11 at 12:57
It's more a case of my not having decided what I should be willing to do........ –  Michael Hardy Aug 29 '11 at 18:07
.....but I am willing to use $\wp$ and things like Jacobi's elliptic functions. If we momentarily adopt Euler's willingness to say that if $\alpha$ is an infinitely small positive number then $\sin\alpha=\alpha$, then when $c$ is an infinitely small positive number, the curve can be parametrized using the ordinary sine and cosine functions. So if $c$ is merely very small, then one should get periodic functions that are approximately those. –  Michael Hardy Aug 29 '11 at 18:21