Is there an exact solution to $x \sinh\Big(\frac{1}{x}\Big) = a$?

Question

Is there an exact formula for solutions to the equation $x \sinh\Big(\frac{1}{x}\Big) = a$ where $a,x \in \mathbb{R}^+$? And if not, why?

I tried to rearrange to apply Lambert W somewhere to no avail.

If it does exist, does it also exist for $x \sinh\Big(\frac{k}{x}\Big) = a$ where $k\in \mathbb{R}^+$ and is a known constant.

Wolfram only gives approximant solutions and no explanation, which is unfortunate. https://www.wolframalpha.com/input?i2d=true&i=x+sinh%5C%2840%29Divide%5B1%2Cx%5D%5C%2841%29%3D2

Answer 1

$$x\sinh\left(\frac{k}{x}\right)=a\tag{1}$$ $$x\left(\frac{1}{2}e^{\frac{k}{x}}-\frac{1}{2}e^{-\frac{k}{x}}\right)=a$$ $x\to\frac{k}{t}$: $$\frac{1}{2}k(e^t)^2-ate^t-\frac{1}{2}k=0\tag{2}$$

We see, equation (2) is a polynomial equation of more than one algebraically independent monomials ($t,e^t$) and with no univariate factor. We therefore don't know how to rearrange the equation for $t$ by applying only finite numbers of elementary functions (operations) we can read from the equation.

We see, for algebraic $a$ and $k$, equation (2) is an irreducible algebraic equation of $t$ and $e^t$ simultaneously. According to the theorems in [Lin 1983] and [Chow 1999], such kind of equations cannot have solutions except $0$ that are elementary numbers or explicit elementary numbers respectively. For algebraic $a$ and $k$, $x$ cannot be an elementary number except $0$ therefore. $0$ isn't a solution of equation (1).
Because $x$ is not an elementary number, we can prove that equation (1) cannot have partial inverses over non-discrete domains that are elementary functions.

We see, because equation (2) is a polynomial equation of $t$ and different powers of $e^t$ greater than $1$, the equation is not in a form for applying Lambert W or Generalized Lambert W.

$$x\sinh\left(\frac{k}{x}\right)=a\tag{1}$$ $$x=\frac{ki}{t}$$ $$ki\frac{\sinh(-it)}{t}=a$$ $$t-\frac{k}{a}\sin(t)=0$$

This is Kepler's equation. See e.g. How to solve Kepler's equation $M=E-\varepsilon \sin E$ for $E$?

Answer 2

Let $$ f(x)=x\sinh(\frac kx)=x(e^{\frac kx}-(e^{-\frac kx})/2=\sum_{n=1}^\infty\frac{k^{2n-1}}{(2n-1)!x^{2n-2}}. $$ Clearly $f(x)$ is continuous and decreasing in $(0,\infty)$. Note $$ \lim_{x\to0^+}=\infty, \lim_{x\to\infty}f(x)=k $$ and hence if $a>k$, the equation $f(x)=a$ has a unique solution; and $a\le k$, the equation has no solution.

Answer 3

One thing you could do it to let $x=\frac 1y$ and face the problem of finding the zero of $$f(y)=\sinh(y)-ay$$ If $a$ is "large" , a first estimate of the solution is given by $$\frac 12 e^y=a y \quad \implies \quad y_0=-W_{-1}\left(-\frac{1}{2 a}\right)$$

For $a=2$, this would give $y_0=2.15329$ that is to say $x_0=0.464405$.

By Darboux theorem, $y_0$ is an under estimate of the solution if $a>3.151$.

Now, compute $y_1^{(n)}$, the first iterate of a Newton like method of order $n$. The formula is totally explicit.

For $a=2$, the results. $$\left( \begin{array}{ccc} n & y_1^{(n)} & \text{method} \\ 2 & 2.177565142 & \text{Newton} \\ 3 & 2.177317986 & \text{Halley} \\ 4 & 2.177318988 & \text{Householder} \\ 5 & 2.177318985 & \text{no name} \\ \end{array} \right)$$