Deriving Two Inverse Functions

In “Instrumental Flying“, we defined function y=\sinh^{-1}(x) as

\{(x, y) | x = \frac{e^y-e^{-y}}{2}\}.\quad\quad\quad(1)

From x = \frac{e^y-e^{-y}}{2}, we obtain

(e^y)^2-2x\cdot e^y-1=0.

It means either e^y = x-\sqrt{x^2+1} or e^y = x+\sqrt{x^2+1}.

But e^y = x-\sqrt{x^2+1} suggests e^y < 0 (see Exercise-1), contradicts the fact that \forall t \in R, e^t > 0 (see “Two Peas in a Pod, Part 2“).

Therefore,

e^y = x+\sqrt{x^2+1} \implies y = \log(x + \sqrt{x^2+1}).

i.e.,

\sinh^{-1}(x) = \log(x + \sqrt{x^2+1}), \;\;x \in (-\infty, +\infty).

We also defined a non-negative valued function y = \cosh^{-1}(x):

\{(x, y) | x = \frac{e^y + e^{-y}}{2}, y \ge 0\}.\quad\quad\quad(2)

Simplifying x=\frac{e^y+e^{-y}}{2} yields

(e^y)^2-2x\cdot e^y+1=0.

It follows that either e^y = x-\sqrt{x^2-1} or e^y=x+\sqrt{x^2-1}.

For both expressions’ right-hand side to be valid requires that x \ge 1. However, when x = 2,

e^y = x-\sqrt{x^2-1} = 2 - \sqrt{3} < 1

suggests that y < 0 (see Exercise-2,3), contradicts (2).

Hence,

e^y = x+\sqrt{x^2-1} \implies y = \log(x+\sqrt{x^2-1}).

i.e.,

\cosh^{-1}(x) = \log(x+\sqrt{x^2-1}), \;\;x \in [1, +\infty).

See also “A Relentless Pursuit“.


Exercise-1 Show that \forall x \in R, x-\sqrt{x^2+1} < 0.

Exercise-2 Without calculator or CAS, show that 2-\sqrt{3} < 1.

Exercise-3 Prove \forall t \ge 0. e^t \ge 1 (hint: see “Two Peas in a Pod, Part 2“)

Beltrami’s Identity

The Beltrami’s identity

F - y' \frac{\partial F}{\partial y'} = C

where C is a constant, is a reduced form of Euler-Lagrange equation for the special case when F does not dependent explicitly on x.

For F = F(y, y'),

\frac{dF}{dx} = \frac{\partial F}{\partial y} y' + \frac{\partial F}{\partial y'} y''.\quad\quad\quad(1)

From Euler-Lagrange equation

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

we have

\frac{\partial F}{\partial y} = \frac{d}{dx}(\frac{\partial F}{\partial y'}).\quad\quad\quad(2)

Substituting (2) into (1) gives

\frac{dF}{dx} = \frac{d}{dx}(\frac{\partial F}{\partial y'}) y' + \frac{\partial F}{\partial y'} y''.\quad\quad\quad(3)

Consequently, when (3) is expressed as

\frac{dF}{dx} = \frac{d}{dx}(\frac{\partial F}{\partial y'}) y' + \frac{\partial F}{\partial y'} \frac{dy'}{dx},

it becomes clear that

\frac{dF}{dx} = \frac{d}{dx}(\frac{\partial F}{\partial y'}y').

Therefore,

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

i.e.,

F -  y' \frac{\partial F}{\partial y'} = C.

In “A Relentless Pursuit”, we derived differential equation

\frac{dy}{dx} = \frac{\sqrt{y^2-C_1^2}}{\pm C_1}\quad\quad\quad(4)

from Euler-Lagrange equation without taking advantage of the fact that F=y\sqrt{1+(y')^2} has no explicit dependency on x. The derivation was mostly done by a CAS.

Let’s derive (4) from Beltrami’s Identity. This time, we will not use CAS.

For F=y\sqrt{1+(y')^2}, the Beltrami’s Identity is

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

It simplifies to

\frac{y}{\sqrt{1+(y')^2}} = C.

Further simplification yields

C^2(y')^2 = y^2-C^2.

For C \ne 0,

(y')^2 = \frac{y^2-C^2}{C^2}.

Therefore,

\frac{dy}{dx} = \pm \sqrt{\frac{y^2-C^2}{C^2}}=\frac{\sqrt{y^2-C^2}}{\pm C}.

Prequel to “A Relentless Pursuit”

Fig. 1

Illustrated in Fig. 2 are two circular hoops of unit radius, centered on a common x-axis and a distance 2a apart. There is also a soap films extends between the two hoops, taking the form of a surface of revolution about the x-axis. 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).