Is it possible to find $x^2$ in terms of $a$ in $x^4 \cosh (\pi x)-\left(x^2-1\right)^4 \sinh (3 x+a)-x^2=0$?

Question

I have this equation: $$x^4 \cosh (\pi x)-\left(x^2-1\right)^4 \sinh (3 x+a)-x^2=0$$ where $0<a<\infty$ and $0<x<\infty$. Is it possible to apply some change of variable or something to find $x^2$ in terms of $a$?

Answer 1

There is no possible closed form solution but we can do a few things

Consider the implicit equation$$x^4 \cosh (\pi x)-\left(x^2-1\right)^4 \sinh (3 x+a)-x^2=0\tag 1$$ Fist, let $x^2=t$ to get $$t^2 \cosh \left(\pi\sqrt{t}\right)-(t-1)^4 \sinh \left(3 \sqrt{t}+a\right)-t=0\tag 2$$ From $(2)$, we have $$a=\sinh ^{-1}\left(\frac{t \left(t \cosh \left(\pi \sqrt{t}\right)-1\right)}{(t-1)^4}\right)-3 \sqrt{t}\tag 3$$ which gives the relation between $a$ and $x^2$.

The problem is that $a$ can be positive only if $0.440506 \leq t \leq 2.76039$ and, in this domain, $a$ varies between $0$ and $\infty$. Because of the vertical asymptote, for a given value of $a$, there are two solutions in $t$.

I give you below a table $$\left( \begin{array}{cc} t & a \\ 0.45 & 0.15708 \\ 0.50 & 0.94034 \\ 0.55 & 1.68159 \\ 0.60 & 2.41815 \\ 0.65 & 3.17995 \\ 0.70 & 3.99655 \\ 0.75 & 4.90465 \\ 0.80 & 5.95923 \\ 0.85 & 7.25830 \\ 0.90 & 9.01713 \\ 0.95 & 11.9171 \\ 1.00 & \infty \\ 1.05 & 12.1481 \\ 1.10 & 9.48116 \\ 1.15 & 7.95937 \\ 1.20 & 6.90374 \\ 1.25 & 6.10181 \\ 1.30 & 5.45912 \\ 1.35 & 4.92546 \\ 1.40 & 4.47092 \\ 1.45 & 4.07630 \\ 1.50 & 3.72855 \\ 1.55 & 3.41839 \\ 1.60 & 3.13899 \\ 1.65 & 2.88522 \\ 1.70 & 2.65309 \\ 1.75 & 2.43945 \\ 1.80 & 2.24179 \\ 1.85 & 2.05805 \\ 1.90 & 1.88655 \\ 1.95 & 1.72588 \\ 2.00 & 1.57485 \\ 2.05 & 1.43247 \\ 2.10 & 1.29786 \\ 2.15 & 1.17029 \\ 2.20 & 1.04911 \\ 2.25 & 0.93376 \\ 2.30 & 0.82375 \\ 2.35 & 0.71864 \\ 2.40 & 0.61805 \\ 2.45 & 0.52164 \\ 2.50 & 0.42909 \\ 2.55 & 0.34014 \\ 2.60 & 0.25453 \\ 2.65 & 0.17204 \\ 2.70 & 0.09247 \\ 2.75 & 0.01564 \end{array} \right)$$ So, for a given value of $a$, you can pick an estimate of $t$ and safely use Newton method.

Let us try for $a=5$; the iterates will be $$\left( \begin{array}{cc} n & t_n \\ 0 & 0.75000000 \\ 1 & 0.75480447 \\ 2 & 0.75488029 \\ 3 & 0.75488031 \end{array} \right)$$

$$\left( \begin{array}{cc} n & t_n \\ 0 & 1.35000000 \\ 1 & 1.34282664 \\ 2 & 1.34248914 \\ 3 & 1.34248842 \end{array} \right)$$