# Challenging problems concerning Jacobian elliptic functions with complex modulus

I study some qualitative properties of Jacobian elliptic functions. Consider, for example, function $sn(u,k)$. In most applications, modulus $k\in(0,1)$ and then everything is very clear, since $sn(u,k)$ is a real-valued function of $u\in\mathbb{R}$. For example, it holds $$|sn(u,k)|\leq1, \quad \forall u\in\mathbb{R},$$ as can be easily deduced, for example, from the identity: $sn^{2}(u,k)+cn^{2}(u,k)=1$ (since cn(u,k) is real-valued as well).

However, if $k$ is allowed to be complex, everything is much harder although it seems that similar properties still hold. In the complex case, one has to restrict the range for the argument on the ray $K\mathbb{R}$, where $K=K(k)$ is the complete elliptic integral of the first kind (a quantity very closely related to Jacobian elliptic functions in general).

For example, the following conjecture seems to be true (by numerical evidence), however, I was not able to find any proof.

Conjecture: Let $|k|\leq1$, $k\neq\pm1$, then it holds $$|sn(Ku,k)|\leq1, \quad \forall u\in\mathbb{R}.$$

Is anybody able to prove (or disprove) it?

Remarks:

1. Due to periodicity properties of $sn(\cdot,k)$, it would be sufficient to verify the conjecture for $u\in(0,1)$.

2. It is known that $sn(K,k)=1$, hence the inequality can not be improved.

3. I have asked for the proof of even stronger conjecture in: An extreme of Jacobi elliptic function on an interval. However, this seems to be far to be answered. This post is a significantly weakened version.

• can we prove it for $|k|=1$? Oct 12, 2015 at 5:46
• to Sergei: Currently, I do not have any proof even for the case $|k|=1$. With this restriction, the answer to the problem would be of interest to me, too.
– Twi
Oct 12, 2015 at 8:41
• the problem estimate is true for real $0<k<1$ and for imaginary $ik$ with $0<k<1$. If it will be proved also for $|k|=1$ it will be true for all $k$ due to the max principle for the corner, as it will be true on its three parts of boundary. Oct 12, 2015 at 9:15
• to Sergei: Good idea, nevertheless I am not sure if Maximum modulus principle applies even if $|sn(Ku,k)|\leq1$ for all $|k|=1, k\neq\pm1$. Function $k\mapsto sn(K(k)u,k)$ has a branch cut in $(-\infty,-1]\cup[1,\infty)$ and hence it is not analytic in points $k=\pm1$. Maybe one would have to known something like $\limsup_{|k|\leq1,k\to\pm1}|sn(Ku,k)|\leq1$.
– Twi
Oct 13, 2015 at 13:35
• to Sergei: I see now that the two points $k\neq\pm1$ would not cause any trouble: math.stackexchange.com/questions/234484/…
– Twi
Oct 13, 2015 at 20:58