Periods of Jacobi elliptic functions – Part 1

Posted on by arkajad

This one is another purely technical post. The point is that I am not very happy with all the existing expositions about Jacobi elliptic functions. I want to organize it my way which, I am sure, is better than all other ways. Of course what is better and what not depends on the interests of the particular person and on the purpose of the study. I am not completely sure what my purpose is, but somehow I am navigating and, at some point, I will be able to shout: Tierra! Tierra! But not yet. So I continue.

First: a summary of the essential properties of the elliptic functions \sn,\cn,\dn.

In fact the properties below are sufficient to completely define these elliptic functions. This is how these functions are defined in an old but good (and freely avilable) monograph by Dixon, The elementary properties of the elliptic functions, with examples, 1894.

Definition of elliptic functions \sn,\cn,\dn:

(1)   \begin{equation*} \frac{d}{du}\,\sn\, u=\cn\, u\,\dn\, u,\end{equation*}

(2)   \begin{equation*} \cn^2 u+\sn^2 u=1,\end{equation*}

(3)   \begin{equation*} \dn^2\,u+k^2\,\sn^2\,u=1.\end{equation*}

(4)   \begin{equation*}\sn\, 0=0,\quad \cn\,0=\dn\,0=1.\end{equation*}

Quite often the argument m is omitted, it simplifies the notation. Sometimes we use m sometimes k. For instance in the formulas above \sn u should be written as \sn(u,m), but some people will write it as \sn(u,k), where k^2=m. After some little training there should be no confusion.

Previously, we have also found the addition formula for \sn(u+v). We could also prove the addition formulas for \cn(u+v) and \dn(u+v), but I will not do it. I am just copying, for instance Eqs (48)-(50) in Jacobi Elliptic Functions, on Wolfram’s site

Addition formulas for Jacobi elliptic functions:

(5)   \begin{eqnarray*} \sn (u+v)&=&\frac{\sn\,u\,\cn\,v\,\dn\,v+\sn\,v\,\cn\,u\,\dn\,u}{1-k^2\,\sn^2 u\,\sn^2 v},\\ \cn (u+v)&=&\frac{\cn\,u\,\cn\,v-\sn\,u\,\sn\,v\,\dn\,u\,\dn\,v}{1-k^2\,\sn^2 u\,\sn^2 v},\\ \dn (u+v)&=&\frac{\dn\,u\,\dn\,v-k^2\sn\,u\,\sn\,v\,\cn\,u\,\cn\,v}{1-k^2\,\sn^2 u\,\sn^2 v}. \end{eqnarray*}

We have also defined F(\phi,m)

(6)   \begin{equation*}F(\phi,m)=\int_0^\phi\frac{d\theta}{\sqrt{1-m\,\sin^2\theta}},\end{equation*}

and, for m\leq 1, we have defined the function \am (u,m) as the inverse function of F(\phi,m): If u=\int_0^\phi\frac{d\theta}{\sqrt{1-m\,\sin^2\theta}}, then \am (u,m)=\phi. Moreover

(7)   \begin{equation*}\sn (u,m)=\sin \am (u,m),\quad \cn (u,m)=\cos \am(u,m).\end{equation*}

The function F(\phi,m) is called “the incomplete elliptic integral of the first kind“. From the shape of the function \sin^2\theta

it is clear that

    \[F(2\pi,m)=2F(\pi,m)=4F(\pi/2,m).\]

The value F(\pi/2,m) is called the complete elliptic integral of the first kind, and denoted as K

(8)   \begin{equation*}K=K(m)=K(k)=F(\pi/2,m)=\int_0^{\pi/2}\frac{d\theta}{\sqrt{1-k^2\,\sin^2\theta}}.\end{equation*}

Thus we can write \am\, K=\pi/2, and therefore \sn\, K=\sin \pi/2=1,\, \cn\,K=\cos\pi/2=0. This way we can build the following table

where

    \[k'=\sqrt{1-k^2}.\]

Then from addition formulas we obtain:

The case of m>1

When m>1 the expression 1-m\sin^2\theta under the square root in the integrand of the integral (8) defining K(m) goes through the negative values.  Thus K(m), for m>1 is, in general, a complex number. During the discussion under the last post Period of a pendulum, the following conjecture has appeared:

Conjecture

Assuming k>1,\, m= k^2, the following identity holds:

(9)   \begin{equation*}K(m)=\frac{K(\frac{1}{m})-i\,K(1-\frac{1}{m})}{k}.\end{equation*}

I do not have a rigorous proof of that conjecture. What I have is a plot, using Mathematica, of the numerical difference between the left and the right hand side of Eq. (9). Here it is:

As you can see it looks like a solid zero! One can also verify that both, the real and the imaginary part of the right hand side, as a function of k satisfy the differential equation that Km) is supposed to satisfy – Eq. (19) in Complete Elliptic Integral of the First Kind:

But that is not yet a complete proof. Nevertheless I will assume that the above conjecture is true. To proof that the real part of K(m) is right is, in fact, easy. Here is how I do it: Let \phi_1 be the angle \phi_1<\pi/2, for which m\sin^2\phi_1=1. Then the real part of K(m) is given by

    \[Re(K(m))=\int_0^{\phi_1}\frac{d\theta}{\sqrt{1-k^2\sin^2\theta}}.\]

Introduce \tau defined by k\sin\theta=\sin\tau. Then \tau(\phi_1)=\pi/2 and k\cos\theta d\theta=\cos\tau d\tau, thus d\theta=\frac{\cos\tau d\tau}{k \cos\theta}. Moreover \sqrt{1-k^2\sin^2\theta}=\sqrt{1-\sin^2\tau}=\cos\tau. Since \cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{1}{m}\sin^2\tau}, we obtain

(10)   \begin{equation*}Re(K(m))=\frac{1}{k}\int_0^{\pi/2}\frac{d\tau}{\sqrt{1-\frac{1}{m}\sin^2\tau}}=K(1/m)/k.\end{equation*}

I suspect that the proof for the imaginary part should not be much more complicated, but I did not succeeded doing it yet. Therefore at this point I will make a break. Till tomorrow.

Update

The conjectured formula is well known. In the tables by I. S. Gradshteyn, I. M. Ryzhik we find

Here K'(k)=K(k').
There is a condition Im(k)>0, but there is no such restriction in the source that is quoted – MO 131. Some has added this condition later, for security reasons. We extend it to the case of Im(k)=0. Then, solving for K(k) we get our formula.

In fact I can almost get this formula by changes of integration variables, except that one has to be careful what to do with the square root of a negative number. A convention is needed. The sign of the imaginary part that I am getting depends on this convention.

Update 2, Sunday Jan. 29

I decided to share my derivation for the imaginary part. It is a continuation of the real part derivation at the end of the post. The imaginary part is given by the integral

    \[Im(K(m))i=\int_{\phi_1}^{\pi/2}\frac{d\theta}{\sqrt{1-k^2\sin^2\theta}}.\]

Here the expression under the square root is negative all the time. So we write

    \[\sqrt{1-k^2\sin^2\theta}=\sqrt{-1(k^2\sin^2\theta-1)}=i\sqrt{k^2\sin^2\theta-1}\]

and therefore (since 1/i=-i)

    \[Im(K(m))i=-i\int_{\phi_1}^{\pi/2}\frac{d\theta}{\sqrt{k^2\sin^2\theta-1}}=\frac{-i}{k}\int_{\phi_1}^{\pi/2}\frac{d\theta}{\sqrt{\sin^2\theta-\frac{1}{m}}}.\]

Now we change the integration variable \theta by setting \theta=\tau+\pi/2. Then \sin^2\theta=\cos^2\tau=1-\sin^2\tau and so

    \[Im(K(m))=\frac{-1}{k}\int_{-\tau_1}^{0}\frac{d\tau}{\sqrt{1-\frac{1}{m}-\sin^2\tau}}=\frac{-1}{k}\int_{0}^{\tau_1}\frac{d\tau}{\sqrt{1-\frac{1}{m}-\sin^2\tau}},\]

where \tau_1 is the positive value of \tau such that \sin\tau=k_1, where

    \[k_1=\sqrt{1-\frac{1}{m}}\]

Now we do the same trick that we did in the real case. We introduce \psi so that

    \[\sin\tau/k_1=\sin\psi\]

This is possible since the integration is within the range of \tau where \sin\tau/k_1\leq 1. Then \cos\tau d\tau=k_1\cos\psi d\psi, so

    \[d\tau=\frac{k_1\cos\psi d\psi}{\cos\tau},\]

and

    \[\sqrt{k_1^2-\sin^2\tau}=k_1\sqrt{1-\sin^2\psi}=k_1\cos\psi\,\]

    \[\cos\tau=\sqrt{1-\sin^2\tau}=\sqrt{1-k_1^2\sin^2\psi.\]

So, finally

(11)   \begin{equation*}Im(K(m))=\frac{-1}{k}\int_0^{\pi/2}\frac{d\psi}{\sqrt{1-k_1^2\sin^2\psi}}=\frac{-1}{k}K(1-1/m).\end{equation*}

Here and there I was skipping explanation and justifications, but these small cracks in the proof can be easily filled in.
The conjecture appears to have been finally proved. So, it is not a conjecture any more, it is a proven property! Uff!

9 thoughts on “Periods of Jacobi elliptic functions – Part 1”

  2. Do you know yet what John Baez had in mind with his gravity towards the sky comment? That paper I linked to in a comment earlier relates the real vs imaginary time to an S-duality which is like an electric field vs a magnetic field so I’m thinking is Baez referring to a gravitomagnetic field?

    After looking gravitoelectromagnetism up on Wikipedia, pointing straight up from the earth to the sky would only work for the gravitomagnetic field if you are standing on the North Pole.

    1. “Do you know yet what John Baez had in mind with his gravity towards the sky comment? ”

      I think it was purely mathematically formal. No deeper reasons. John Baez is distancing himself from any weirdness (me included) as possible.

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