finding $a,b$ such that the equation $\frac{a}{x} = \cosh{\frac{b}{x}}$ only has one solution

Question

The equation can be rewritten (assuming I haven't made a mistake) as

$$\alpha y = \ln (y \pm \sqrt{y^2 - 1})$$

with $\alpha = \frac{b}{a}$ and $y=\frac{a}{x}$.

It seems that, for the +ve square root, the maximum $\alpha$ for which there is at least one solution for y is 0.6627434... (from numerical analysis). This appears to be very close to the Laplace Limit Constant.

Is there any way to more rigorously determine whether the Laplace Limit Constant is this 'critical' alpha value for which there is only one solution? For example, can the problem be reduced to any of the formulas given in the Wikipedia link for the Laplace Limit Constant?

Furthermore, is there any way to approach the problem for the -ve square root? Or perhaps a different way which doesn't involve logarithms at all?

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Side note on how I came across this problem:

If we wish to find the shape a soap bubble makes between two equal radii circles on the same axis, we need to minimise the total surface area

$$\int_b^b r(z) \sqrt{(\frac{dr}{dz})^2 + 1} \ dz$$

where $\pm b$ are the z coordinates of the circles, $r(z)$ is the distance from the z axis of the soap bubble and we constrain $r(\pm b) = a =$ the radius of the circles. Solving the Euler-Lagrange equation gives $r = c \cosh{\frac{z}{c}}$ , where c is a constant which satisfies $a = c \cosh{\frac{b}{c}}$. This final equation is the equation given in the title of this post.

For an example image & a write up of the problem see e.g. https://mathematicalgarden.wordpress.com/2014/09/06/soap-film-and-minimal-surface/

Answer 1

I think you got your substitution slightly wrong, if you set $\alpha = \frac{a}{b}$ and $y = \frac{b}{x}$ then the equation becomes

$$\alpha y = \cosh y$$

You can play around with the equations $y = \alpha x$ and $y = \cosh x$ in Desmos to get a feel for what happens as $\alpha$ varies, and what you will notice is that you get a single intersection point when the linear function is the tangent to the hyperbolic cosine curve.

So in other words, we are looking for the value(s) of $\alpha$ such that $\alpha x$ is tangent to $\cosh x$, or in other words the value that makes the tangent pass through the origin.

With some quick calculus, we find the equation of the tangent to $y = \cosh x$ at $(x_0, \cosh x_0)$ is $y = \sinh x_0 (x - x_0) + \cosh x_0$, which has a $y$-intercept at $(0, \cosh x_0 - x_0 \sinh x_0)$, and for that point to be the origin means we have $x_0 = \frac{\cosh x_0}{\sinh x_0} = \coth x_0$, which Wolfram Alpha estimates to have a solution around $\pm 1.19968$.

Now we're interested in $\alpha = \sinh^{-1} x_0$, so we can make that substitution and employ the identity $\coth^2 \alpha = \mathrm{csch}^2 \alpha + 1$ (and we'll ignore the negative solution because we know it's going to be symmetric in the end) to rearrange a bit to:

$$\begin{eqnarray} \sinh^{-1} \alpha & = & \coth(\sinh^{-1} \alpha) \\ & = & \sqrt{\mathrm{csch}^2 (\sinh^{-1} \alpha) + 1} \\ & = & \sqrt{\frac{1}{\alpha^2} + 1} \end{eqnarray}$$

If we then make the substitution $\beta = \frac{1}{\alpha}$, we get the equation

$$ \beta = \frac{1}{\sinh(\sqrt{1 + \beta^2})}$$

Which is tantalisingly close to the equation that defines the Laplace limit, and I wouldn't be surprised if you can do a bit of rearranging and/or further substitutions to get to the desired form.

Answer 2

Considering the function $$f(x)=\cosh \left(\frac{b}{x}\right)-\frac{a}{x}$$ Using, $x=\frac b y$ and $c=\frac{a}{b}$ we have $$g(y)=\cosh (y)-c y$$

To have a single root, it is equivalent to look for $y$ such that $\frac{\cosh (y)}{y}$ reaches its minimum value. This means that we look for the zero of the derivative; in other words, what is the zero $$y\,\tanh(y)=1\qquad \implies \qquad e^{-2y}=\frac{y-1}{y+1}$$ which has an explicit solution in terms of the generalized Lambert function (have a look at equation $(4)$).

Using the summation in terms of the generalized Laguerre polynomials,

$$y=1+2\sum_{n=1}^\infty \frac {e^{-2n}} n\, L_{n-1}^1(4 n)$$

the result is $$y=1.1996786402577338339163698486411419442614655655331\cdots$$ and then $$c=1.5088795615383199289098844881605785736942785890478\cdots$$ from which $$\frac 1 c=0.66274341934918158097474209710925290705623354911502\cdots$$ which is Laplace limit constant.