Solving $R\space \sinh\frac{D}{R}=k$ for $R$

Question

Does a solution exist for $R$ in this equation? I can't seem to solve it either analytically or numerically.

$$R\space \sinh\frac{D}{R}=k$$

Answer 1

$$R\space \sinh\frac{D}{R}=k$$ $$\text{Let}\qquad x=\frac{D}{R}\qquad;\qquad R=\frac{D}{x}\qquad;\qquad C=\frac{k}{D}$$ $$\boxed{\sinh(x)=C\:x}$$ This equation is well known. For example see a discussion about the existence of root(s) in : Solving $\sinh x = kx$

The root cannot be expressed with a finite number of standard functions.

On a formal view-point (of no interest in practice) an analytical form of solution is : $$R=\frac{D}{f^{-1}(k/D)}$$ where $f^{-1}$ is the inverse function of $f(x)=\frac{\sinh(x)}{x}$ . But this function $f^{-1}$ is not standard.

Of course, numerical solving is possible thanks to iterative methods for numerical solving of non-linear equations (For example Newton-Raphson method https://mathworld.wolfram.com/NewtonsMethod.html).

Answer 2

Starting from @JJacquelin's answer, you are looking for the zero of function $$f(x)=\frac{\sinh(x)}x-C$$ The algebraic expression varies very fast which is never very good for numerical methods.

Consider instead looking for the zero of function (assuming $C>0$) $$g(x)=\log \left(\frac{\sinh (x)}{x}\right)-\log(C)$$ which is much smoother and then better conditioned for the solver.

For sure, we need a starting point. A simple estimate could be $$x_0=2.35 \big[\log(C)\big]^{3/4}$$ For testing, let me try for $C=123456789$. The iterates of Newton method will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 21.07424431 \\ 1 & 22.43727831 \\ 2 & 22.43517918 \\ 3 & 22.43517917 \end{array} \right)$$