Prequel to “A Relentless Pursuit”

Fig. 1

Suppose we have two circular hoops of unit radius, centered on a common x-axis and a distance 2a apart. Suppose too, that a soap films extends between the two hoops, taking the form of a surface of revolution about the x-axis (see Fig. 2). Then if gravity is negligible the film takes up a state of stable, equilibrium in which its surface area is a minimum.

Fig. 2

Our problem is to find the function y(x), satisfying the boundary conditions

y(-a) = y(a) = 1,\quad\quad\quad(1)

which makes the surface area

A=2\pi\displaystyle\int\limits_{-a}^{a}y\sqrt{1+(y')^2}\;dx\quad\quad\quad(2)

a minimum.

Let

F(x,y, y') = 2\pi y \sqrt{1+(y')^2}.

We have

\frac{\partial F}{\partial y} = 2\pi \sqrt{1+(y')^2}

and

\frac{\partial F}{\partial y'} = 2 \pi y \cdot\frac{1}{2}\left(1+(y')^2\right)^{-\frac{1}{2}}\cdot 2y'=\frac{2 \pi y y'}{\sqrt{1+(y')^2}}.

The Euler-Lagrange equation

\frac{\partial F}{\partial y} - \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) = 0

becomes

\sqrt{1+(y')^2} - \frac{d}{dx}\left(\frac{y y'}{\sqrt{1+(y')^2}}\right) = 0.

Fig. 3

Using Omega CAS Explorer (see Fig. 3), it can be simplified to:

y \frac{d^2 y}{dx^2}- \left(\frac{dy}{dx}\right)^2=1.

This is the differential equation solved in “A Relentless Pursuit” whose solution is

y = C_1\cdot \cosh(\frac{x+C_2}{C_1}).

We must then find C_1 and C_2 subject to the boundary condition (1), i.e.,

C_1\cdot \cosh(\frac{a+C_2}{C_1}) = C_1\cdot\cosh(\frac{-a+C_2}{C_1})\implies \cosh(\frac{a+C_2}{C_1}) = \cosh(\frac{-a+C_2}{C_1}).

The fact that \cosh is an even function implies either

a+C_2 = -a+C_2\quad\quad\quad(3)

or

a+C_2 = -(-a+C_2).\quad\quad\quad(4)

While (3) is clearly false since it claims for all a >0, a = -a, (4) gives

C_2=0.

And so, the solution to boundary-value problem

\begin{cases} y \frac{d^2 y}{dx^2}- \left(\frac{dy}{dx}\right)^2=1,\\ y(-a)=y(a)=1\end{cases}\quad\quad\quad(5)

is

y = C_1\cdot \cosh(\frac{x}{C_1}).\quad\quad\quad(6)

To determine C_1, we deduce the following equation from the boundary conditions that y=1 at x=\pm a:

C_1\cdot \cosh(\frac{a}{C_1}) = 1.\quad\quad\quad(7)

This is a transcendental equation for C_1 that can not be solved explicitly. Nonetheless, we can examine it qualitatively.

Let

\mu = \frac{a}{C_1}

and express (7) as

\cosh(\mu) = \frac{\mu}{a}.\quad\quad\quad(8)

Fig. 4

A plot of (8)’s two sides in Fig. 4 shows that for sufficient small a, the curves z = \cosh(\mu) and z = \frac{\mu}{a} will intersect. However, as a increases, z=\frac{\mu}{a}, a line whose slope is \frac{1}{a} rotates clockwise towards \mu-axis. The curves will not intersect if a is too large. The critical case is when a=a^*, the curves touch at a single point, so that

\cosh(\mu) = \frac{\mu}{a^*}\quad\quad\quad(9)

and y=\frac{\mu}{a} is the tangent line of z=\cosh(\mu), i.e.,

\sinh(\mu) = \frac{1}{a^*}.\quad\quad\quad(10)

Dividing (9) by (10) yields

\coth(\mu) = \mu. \quad\quad\quad(11)

What the mathematical model (5) predicts then is, as we gradually move the rings apart, the soap film breaks when the distance between the two rings reaches 2a^*, and for a > a^*, there is no more soap film surface connects the two rings. This is confirmed by an experiment (see Fig. 1).

We compute the value of a^*, the maximum value of a that supports a minimum area soap film surface as follows.

Fig. 5

Solving (11) for \mu numerically (see Fig. 5), we obtain

\mu = 1.1997.

By (10), the corresponding value of

a^* = \frac{1}{\sinh(\mu)} = \frac{1}{\sinh(1.1997)} = 0.6627.

We also compute the surface area of the soap film from (2) and (6) (see Fig. 6). Namely,

A = 2\pi \displaystyle\int\limits_{-a}^{a} C_1 \cosh\left(\frac{x}{C_1}\right) \sqrt{1+\left(\frac{d}{dx}C_1\cosh\left(\frac{x}{C_1}\right)\right)^2}\;dx =  \pi C_1^2\left(\sinh\left(\frac{2a}{C_1}\right) + \frac{2a}{C_1}\right).

Fig. 6


Exercise-1 Given a=\frac{1}{2}, solve (7) numerically for C_1.

Exercise-2 Without using a CAS, find the surface area of the soap film from (2) and (6).

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