I am stuck with an equation I've been trying to solve for a while in different ways. Tried searching on the internet about the properties of arccosh (or cosh-1) and cosh but still couldn't find the way. I get the feeling the solution to this is going to be easy and stupid, but I feel I've been going in circles here.
I need to solve it by hand because I will need to solve this kind of equation in an exam.
It should be solvable, but in case it's not, I need to say that I've got this after a simplification, in which I don't think I made any mistakes but just in case here is the simplification I made: Here
Please note that for U1 and U2 all the variables are unknown but they remain the same for these two cases, except Z. That's why I could simplify it in this way.
In case someone is wondering, this is a problem about sea waves. I was given the speed of particles in a given place (at 2 different depths) and I'm using Airy's linear theory to solve it.
Thanks in advance!
Edit: $$\frac{0.34}{0.44} = \frac{cosh(\frac{90\pi}{x})}{cosh(\frac{110\pi}{x})}$$
Edit2: I suppose I could solve it using iterations, but I will need to do it several times during the exam and I'd rather know if there is a better way to solve it.
Rewriting the equation, you need to find the zero of $$f(x)=34 \cosh \left(\frac{110 \pi }{x}\right)-44 \cosh \left(\frac{90 \pi }{x}\right)$$ If you try to plot it, you will notice that the fnction varies extremely fast and it is difficult to locate where is more or less the root.
So, consider instead $$g(x)=\log \left(34 \cosh \left(\frac{110 \pi }{x}\right)\right)-\log \left(44 \cosh \left(\frac{90 \pi }{x}\right)\right)$$ which is better conditioned (looking more or less like an hyperbola) and, graphing, you will see that there is a root close to $x=200$.
At this point, using $x_0=200$, apply Newton method with $$g'(x)=\frac{90 \pi \tanh \left(\frac{90 \pi }{x}\right)}{x^2}-\frac{110 \pi \tanh \left(\frac{110 \pi }{x}\right)}{x^2}$$ You will then get the following iterates : $$\left( \begin{array}{cc} n & x_n \\ 0 & 200.0000000 \\ 1 & 216.3366146 \\ 2 & 217.6456022 \\ 3 & 217.6530075 \\ 4 & 217.6530077 \end{array} \right)$$ which is the solution for ten significant figures.
If you want a shortcut, expand $f(x)$ as a Taylor series for large values of $x$. This would give $$f(x)=-10+\frac{27500 \pi ^2}{x^2}+\frac{261387500 \pi ^4}{3 x^4}+O\left(\frac{1}{x^6}\right)$$ and, ignoring the higher order terms, this would give a quadratic in $\frac 1 {x^2}$ for which the solution is $$x^2=\frac{25}{3} \left(165+\sqrt{152691}\right) \pi ^2$$ which makes $x\approx 213.797$