How to find roots of this equation?

Question

I have this equation and want to find its roots.

$\left(a^2+1\right) \cosh (a (c -b))- \cosh (c a)=0 $.

Any comment is welcome.

Answer 1

One has \begin{equation} 1 \le f(a) = \frac{(a^2-1)^2}{(a^2+1)^2}\cosh(2\pi a) + \frac{4 a^2}{(a^2+1)^2}< \cosh(2\pi a) \end{equation} because this expression is a convex combination of $\cosh(2\pi a)$ and $1$. Then the equation implies \begin{equation} b = \pi \pm \frac{1}{2a}\cosh^{-1}(f(a))\in (0, 2\pi) \end{equation} Hence for every $a$ there are two solutions $b$, except in the case where $f(a)=1$ which happens only when $a=1$. In this case there is only one solution $b = \pi$.

Plotting $b$ as a function of $a$ (with the $+$ sign) indicates that $b$ decreases from $2\pi$ to $\pi$ when $a$ varies from $0$ to $1$ and increases from $\pi$ to $2\pi$ when $a$ varies from $1$ to $+\infty$, so there should be two solutions $a$ for each $b\in (\pi, 2\pi)$, one of them in $(0, 1)$ and the other one in $(1, +\infty)$. The condition $a<10$ is irrelevant and should be removed.