Solution for a functional equation

Question

I'm searching for a solution to the following functional equation:

$$f(u)f(u+\lambda)=\prod_{i=1}^L\rho(u-u_i)\rho(u_i-u)+\prod_{i=1}^L\rho(u+\lambda-u_i)\rho(u_i-u-\lambda)$$

where $f$ is the unknown function, $\rho(u)=\sin(u-\lambda)/\sin(\lambda)$, $\lambda=\pi/4$ and there are $L$ arbitrary real parameters $\{u_i\}_{i=1}^L$ where $L$ is an even number.

This equation comes from a $TQ$-equation for a solvable lattice model. If necessary, I can give more information but the equation itself is independent from its physical origin. I hope someone can help.

Answer 1

This is not a complete solution, but just presenting a few ideas.

Let's work on $\rho$ first: \begin{align*} \rho(u)&=\frac{\sin(u-\lambda)}{\sin(\lambda)}\\ &=\frac{\sin(u)\cos(\lambda)-\cos(u)\sin(\lambda)}{\sin(\lambda)}\\ &=\sin(u)-\cos(u), \end{align*} since $\lambda=\pi/4$ and $\sin(\pi/4)=\cos(\pi/4).$

Next, we look at what happens when we have $\rho(u)\cdot\rho(-u).$ We have \begin{align*} \rho(u)\cdot\rho(-u)&=[\sin(u)-\cos(u)][-\sin(u)-\cos(u)] \\ &=-\sin^2(u)-\sin(u)\cos(u)+\sin(u)\cos(u)+\cos^2(u)\\ &=\cos^2(u)-\sin^2(u)\\ &=\cos(2u). \end{align*}

From this, we gather that \begin{align*}f(u)\,f(u+\lambda)&=\prod_{i=1}^L\rho(u-u_i)\,\rho(u_i-u)+\prod_{i=1}^L\rho(u+\lambda-u_i)\,\rho(u_i-u-\lambda)\\ &=\prod_{i=1}^L\cos(2(u_i-u))+\prod_{i=1}^L\cos(2(u_i-u-\lambda)) \\ &=\prod_{i=1}^L\cos(2(u_i-u))+\prod_{i=1}^L\cos(2(u_i-u)-\pi/2),\;\text{or} \\ f(u)\,f(u+\pi/4)&=\prod_{i=1}^L\cos(2(u_i-u))+\prod_{i=1}^L\sin(2(u_i-u)). \end{align*} From this, we can at least see that if $u=u_i$ for any $1\le i\le L,$ then the second product drops out and you get $$f(u_i)\,f(u_i+\pi/4)=\prod_{\begin{array}{c}j=1\\ j\not=i\end{array}}^L\cos(2(u_j-u)).$$ Conversely, if $u_i-u=(2k+1)\pi/4$ for some $k\in\mathbb{Z},$ then the first product series drops out and you only get the second one.

Finally, it's always worthwhile plugging in $u=0$ to arrive at $$f(0)\,f(\pi/4)=\prod_{i=1}^L\cos(2u_i)+\prod_{i=1}^L\sin(2u_i).$$

The RHS of $$f(u)\,f(u+\pi/4)=\prod_{i=1}^L\cos(2(u_i-u))+\prod_{i=1}^L\sin(2(u_i-u))$$ is $\pi$ periodic, and hence the LHS is as well.