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I am curious if a method exists for solving for $k$ or $h$ in this implicit equation:

$$\frac{k^2}{h} \mathrm{sech}^2(k) \sqrt{1 + \left(\frac{k}{h} \tanh(k)\right)^2} = \ln\left( \frac{k}{h} \tanh(k) + \sqrt{1 + \left(\frac{k}{h} \tanh(k)\right)^2}\right)$$

Both $k=0$ and $h=\pm k\, \mathrm{sech}(k)$ are solutions. I can easily check that these work, but I have been unable to find a way to manipulate the implicit equation to get these solutions. I have tried writing the hyperbolic functions in terms of their equivalent exponential functions (like $\tanh(x)=\frac{e^{2x}-1}{e^{2x}+1}$, etc.), raising $e$ to the power of both sides (to remove the log), and (unsuccessfully) factoring the part under the root. Even Wolfram Alpha says, "Standard computation time exceeded..."!

Again, my question is about what techniques could be used to find these explicit solutions (especially $h=k\, \mathrm{sech}(k)$) if I did not already know them.

Background

I am currently taking Calculus 2. A couple weeks ago, for fun, I decided to try to calculate the shape of a soap film between two parallel rings and the maximum distance between the rings before the bubble would pop. Since the soap tries to minimize surface area, it forms the shape of a catenoid (the surface of revolution of hyperbolic cosine).

I let the radius of the rings equal $1$, $h$ equal half the distance between the rings, and $r$ equal the radius halfway between the rings (the minimum of $cosh$). I found that letting $k = \mathrm{sech}^{-1}(r)$ simplified things. Using integration, I found the surface area of $y=\mathrm{sech}(k)\cosh\left(\frac{k}{h}x\right)$ revolved around the X-axis between $-h$ and $h$. The bubble should then converge to the nearest local minimum of this area. Taking the derivative of this with respect to $k$ and setting it equal to zero gave me the above equation.

After trying to solve the equation for a while, I found this site that says the maximum separation between the rings was the maximum height of $\frac{x}{\cosh(x)}$. I tried this expression in my equation, and, to my amazement, it worked.

David Brock
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  • Not certain, but what I would try is substituting the hyperbolic functions for their equivalents and see what you come up with. Like, $\tanh(x) = \frac{e^{2x} - 1}{e^{2x} +1}$ and $\mathrm{sech}(x) = \frac{2e^x}{e^{2x} + 1}$. – johnnyb Feb 10 '21 at 03:56
  • Thanks for the suggestion! I had actually already tried that, but without much success. I just edited the question to make this clearer. – David Brock Feb 10 '21 at 04:20
  • I think you're stuck with something that has no clean solution. Try setting $z = \tanh k$, consequently $\text{sech}^2k =1 - z^2$, and you end up with an equation of the form $f(z)/h = 1$. – A rural reader Feb 10 '21 at 04:30
  • If you let $ u = \frac{k}{h} \tanh(k) $, then the RHS simplifies to $ \ln \left(u+\sqrt{1+u^2}\right) = \sinh^{-1} (u) $. Not sure if that helps though. – Amaan M Feb 10 '21 at 04:31
  • What about embedding trig substitutions as well? If you take Amaan's solution and say $u = \mathrm{sec}(q)$, then $\sqrt{1 + \mathrm{sec}^2(q)} = \sqrt{\tan^2(q)} = \tan(q)$. Again, not sure if it helps. – johnnyb Feb 10 '21 at 04:40

1 Answers1

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I will prove that these are the only solutions: $k=0$ and $h(k)=\pm k\operatorname{sech}k$.

It is easy to see that $k=0$ is a solution so herein we let $k\ne0$.

Let $h(k)$ be a function such that $$\frac{k^2}{h(k)\cosh^2k}\sqrt{1+\left(\frac{k\tanh k}{h(k)}\right)^2}=\operatorname{arsinh}\frac{k\tanh k}{h(k)}.$$ This is equivalent to your equation as your RHS is the expanded version of $\operatorname{arsinh}$.

Let $g(k)=(k\tanh k)/h(k).$ Then we have $$\operatorname{arsinh}g(k)=\frac{k^2g(k)\sqrt{1+g(k)^2}}{k\tanh k\cosh^2k}=\frac{2kg(k)\sqrt{1+g(k)^2}}{\sinh2k}.$$

Let $g(k)=\sinh f(k)$, which is permitted as $\sinh$ is bijective over the reals. Then $$f(k)=\frac{2k\sinh f(k)\sqrt{1+\sinh^2f(k)}}{\sinh2k}=\frac{k\sinh2f(k)}{\sinh2k}\iff\frac{\sinh2f(k)}{f(k)}=\frac{\sinh2k}k.$$ The function $\sinh(2k)/k$ is symmetric about the $y$-axis and strictly increasing for positive $k$. Therefore, the equation only holds when $f(k)=\pm k$, in which case $$h(k)=\frac{k\tanh k}{\sinh f(k)}=\pm k\operatorname{sech}k.$$

TheSimpliFire
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