solve the equation $\tanh(\mu \pi)=\frac{2\mu}{a}$ I am trying to find what condition it must meet for the equation to have a real solution for $\mu$.
I understand that the function $\tanh(x)$ is bounded in the interval $ (- 1,1) $ if I make use of this I find the following:
$-1<\tanh(\mu \pi)<1$ then $-1< \frac{2\mu}{a}< 1$
where do i get that
$a>2\mu$
But I was reading the procedure for this and it tells me the following:
A simple sketch shows that this equation can only have a real solution for $\mu$ if
the slope of $f(\mu) = \tanh(\mu \pi)$ at $\mu = 0$ is greater than the slope of $g(\mu)=\frac{2µ}{a}$ at
the same place. The former slope is π and the latter $\frac{2}{a}$. Thus the condition for
negative eigenvalues of the original equation is $a > \frac{2}{\pi}$.
But the latter I cannot fully understand. Could you explain me?
 A: First of all, we always have the trivial solution $\mu = 0$. $\DeclareMathOperator{sech}{sech}$
But what about other solutions? We confine ourselves to considering solutions $\mu > 0$ (since if $\mu$ is a solution, so too is $-\mu$).
We note that for $\mu > 0$, $\tanh$ is a concave-down function. This can be easily verified by taking its second derivative. $\frac{d}{dx} \tanh(x) = \sech^2(x)$ and $\frac{d}{dx} \sech^2(x) = 2 \sech(x) (-\sech(x) \tanh(x)) = -2 \sech^2(x) \tanh(x) < 0$ when $x > 0$.
Therefore, the function $f(\mu) = \tanh(\pi \mu) - \frac{2 \mu}{a}$ is a concave down function.
Now suppose we have $f(\mu) = 0$ for some $\mu > 0$. Then by the Rolle's theorem, we can find some $x \in (0, \mu)$ such that $f'(\mu) = 0$. But $f'$ is a decreasing function, so this means $f'(0) < 0$. This means that $\pi - \frac{2}{a} > 0$; that is, $\pi > \frac{2}{a}$; that is, $a > \frac{2}{\pi}$.
Now on the other hand, if we have $a > \frac{2}{\pi}$, then we see that $f'(0) > 0$. So for sufficiently small $x > 0$, we have $f(x) > 0$. But we also see that for very large $x$, we have $f(x) < 0$. So by the intermediate value theorem, there must exist some solution $\mu$ of the equation $f(\mu) = 0$.
So there is a non-zero solution of $\tanh(\pi \mu) = \frac{2 \mu}{a}$ if, and only if, $\frac{2}{\pi} < a$.
A: Consider that you look for the zero of funtion
$$f(\mu)=\tanh(\mu \pi)-\frac{2\mu}{a}$$ Let $x=\mu \pi$ and $k=\frac{2a}\pi$ and, instead, consider the function
$$g(x)=\frac{\tanh (x)}{x}-k$$ The maximum value of $\frac{\tanh (x)}{x}$ is $1$; so in order to have a solution, you must have $0 < k \leq 1$ that is to say $0 < a \leq \frac \pi 2$. Outside this range, there will not exist any real solution.
Now, if you want to solve for $x$ $g(x)=0$, there is no analytical solution and you will need some numerical method (such as Newton which is the simplest) and, as usual, a reasonable starting guess.
It is possible to do it building around $x=0$ the $[4,4]$ Padé approximant
$$\frac{\tanh (x)}{x}\sim \frac{x^4+105 x^2+945}{15 \left(x^4+28 x^2+63\right)}$$ which means that the estimate will be otained at the price of a quadratic equation in $t=x^2$
$$(15 k-1) t^2+105(4k-1) t-945 (1-k)=0$$ giving as estimate
$$x_0=\sqrt{\frac{3 \left( \sqrt{35(380 k^2-88 k+23)}-140 k+35\right)}{2 (15 k-1)}}$$ which would work only for $k >\frac 1 {15}$ corresponding to $x>15$.
A few values for comparison
$$\left(
\begin{array}{ccc}
k & x_0 & \text{solution} \\
 0.1 & 11.7601 &  10.0000 \\
 0.2 & 5.03871 &  4.99955 \\
 0.3 & 3.32735 &  3.32471 \\
 0.4 & 2.46434 &  2.46406 \\
 0.5 & 1.91504 &  1.91501 \\
 0.6 & 1.51223 &  1.51222 \\
 0.7 & 1.18376 &  1.18376 \\
 0.8 & 0.88802 &  0.88802 \\
 0.9 & 0.58381 &  0.58381
\end{array}
\right)$$
