Calculus of Variations/CHAPTER III

CHAPTER III: PROPERTIES OF THE CATENARY.

• 33 Preliminary remarks.
• 34 The general equation of the catenary.
• 35 Geometrical construction of its tangent.
• 36 Geometrical construction of the catenary.
• 37 The catenary uniquely determined when a point on it and the direction of the tangent at this point are given.
• 38 Limits within which the catenary must lie.
• 39,40,41 The number of catenaries that may be drawn through two fixed points.
• 42 The functions $f_1(m)$ and $f_2(m)$.
• 43 The discussion of the function $f_1(m)$.
• 44 The discussion of the function $f_2(m)$.
• 45 An approximate geometrical construction for the root of a transcendental equation.
• 46,47,48 Graphical representation of the functions $f_1(m)$ and $f_2(m)$.
• 49,50 The different cases that arise and the corresponding number of catenaries.
• 50,51,52 The position of the intersection of the tangents through the two fixed points for each case.
• 53,54 The common tangents to two catenaries.
• 55 Catenaries having the same parameter which intersect in only one point.
• 56 Lindelöf's Theorem.
• 57 A second proof of the same theorem.
• 58,59 Discussion of the several cases for the possibility of a minimal surface of rotation.
• 60,61 Application to soap bubbles.

Article 33.
Owing to certain theorems that have been discovered by Lindelöf and other writers, some of the very characteristics of a minimal surface of rotation, which are sought in the Calculus of Variations, may be obtained for the case of the revolution of the catenary without the use of that theory. We shall give these results here, as they offer a handy method of comparison when we come to the results that have been derived through the methods of the Calculus of Variations.

In presenting the subject-matter of this Chapter, the lectures given by Prof. Schwarz at Berlin are followed rather closely. The results are derived by Todhunter in a somewhat different form in his Researches in the Calculus of Variations, p. 54; see also the prize essay of Goldschmidt, Monthly Notices of the Royal Astronomical Society, Vol. 12, p. 84; Jellett, Calculus of Variations, 1850, p. 145; Moigno et Lindelöf, Calcul des Variations, 1861 p. 204; etc.

Article 34.
Take the equation of the catenary which was given in the preceding Chapter, Art. 30 in the form

$y=\frac{1}{2}m[e^{(x-x_{0^{\prime }})/m}+e^{-(x-x_{0^{\prime }})/m}]$.

It follows at once that

$m\frac{\text{d}y}{\text{d}x}=\pm \sqrt{y^2-m^2}=\frac{1}{2}m[e^{(x-x_{0^{\prime }})/m}+e^{-(x-x_{0^{\prime }})/m}]$.

On the right-hand side of the equation stands a one-valued function, but on the left-hand side, a two-valued function. It is therefore necessary to define the left-hand side so that it will be a one-valued function corresponding to the right-hand side.

If we make $x>x_{0^{\prime }}$, then is

$e^{(x-x_{0^{\prime }})/m}>e^{-(x-x_{0^{\prime }})/m}$

and consequently $\sqrt{y^2-m^2}$ is positive. But when $x<x_{0^{\prime }}$, it is seen that

$e^{(x-x_{0^{\prime }})/m}<e^{-(x-x_{0^{\prime }})/m}$

and then $\sqrt{y^2-m^2}$ is negative. It therefore follows that there is only one root of $\frac{\text{d}y}{\text{d}x}=0$, and this is for the value $x=x_{0^{\prime }}$. The corresponding value of $y$ is $m$.

This value $m$ is the smallest value that $y$ can have; for $\frac{\text{d}y}{\text{d}x}=0$ is the condition for a maximum or a minimum value and since $\frac{{\text{d}}^{2}y}{\text{d}{x}^2}$ is positive for $x=x_{0^{\prime }}$, it follows that $m$ is a minimum value of $y$. Further, since $\sqrt{y^2-m^2}$ is continuously positive or continuously negative, there is no maximum value of $y$. The tangent to the curve at the point $x=x_0$, $y=m$ is parallel to the $X$-axis, since at this point $\frac{\text{d}y}{\text{d}x}=0$.

Article 35.
At every point of the curve we have

$\frac{\text{d}y}{\text{d}x}=\tan (\tau )=\frac{\sqrt{y^2-m^2}}{m}$.

Hence, to construct a tangent at any point of the catenary, for example at $P$, drop the perpendicular $PQ$, and describe the semi-circle on $PQ$ as diameter. Then, with radius equal to $m$, draw a circle from $Q$ as center, which cuts the semi-circle at $R$; join $R$ and $P$. The line $RP$ is the required tangent.

Again $\text{d}{s}^2=\text{d}{x}^2+\text{d}{y}^2=(1+\frac{y^2-m^2}{m^2})\text{d}{x}^2=\frac{y^2\text{d}{x}^2}{m^2}$;

consequently

$\text{d}s=\frac{y\text{d}x}{m}=\frac{1}{2}[e^{(x-x_{0^{\prime }})/m}+e^{-(x-x_{0^{\prime }})/m}]\text{d}x$;

and integrating,

$s-s_{0^{\prime }}=\frac{1}{2}[e^{(x-x_{0^{\prime }})/m}+e^{-(x-x_{0^{\prime }})/m}]=\sqrt{y^2-m^2}$

where $s_{0^{\prime }}$ denotes that the arc is measured from the lowest point of the catenary.

The geometrical locus of $R$ is a curve which cuts all the tangents to the catenary at right angles, and is therefore the orthogonal trajectory of this system of tangents. This trajectory has the remarkable property that the perpendiculars $QR$, etc., of length $m$, which are employed in the construction of the tangents to the catenary, are themselves tangent to the trajectory.

This trajectory possesses also the remarkable property that, if we rotate it around the $X$-axis, the surface of rotation has a constant curvature,

Further, $PN$, the normal to the catenary,

$=y\sec (\tau )=\frac{y^2}{m}$, and

$\rho =\frac{{[1+{\left(\frac{\text{d}y}{\text{d}x}\right)}^2]}^{3/2}}{\frac{{\text{d}}^{2}y}{\text{d}{x}^2}}=\frac{{\left(\frac{\text{d}x}{\text{d}x}\right)}^3}{\frac{{\text{d}}^{2}y}{\text{d}{x}^2}}=\frac{{\left(\frac{y}{m}\right)}^3}{\frac{y}{m^2}}=\frac{y^2}{m}$,

or

$PN=PC$ (see figure),

where $PC$ is the length of the radius of curvature.

Article 36.
The geometrical construction of the catenary. Take an ordinate equal to $2m$. This determines the point $P$ (see figure). With $P$ as center and radius equal to $m$, describe a circle. This intersects $PB$ at a point $A$, say. On the circumference of this circle take a point $A_1$, very near $A$, and draw the line $P{A}_1{B}_1$, and on this line extended take $P_1$ such that $P_1{A}_1=A_1{B}_1$. With radius $P_1{A}_1$ draw another circle, and on this circle take a point $A_2$, very near the point $A_1$, and draw the line $P_1{A}_2{B}_2$. Take on this line extended the point $P_2$, so that $P_2{A}_2=A_2{B}_2$, etc. The locus of the points $A$ is the required catenary.

The accompanying figure shows approximately the relative positions of the catenary, its evolute and the trajectory.

Article 37.
It appears trom the previous article that a catenary is completely determined when we know any point on it and the tangent at this point. This may be proved analytically as follows:

Let $\overline{x}$,$\overline{y}$ be a point through which passes a straight line, making with the $X$-axis an angle whose tangent is $k$. The conditions that a catenary pass through this point and have the given line as tangent are:

$\overline{y}=\frac{m}{2}[e^{(\overline{x}-x_{0^{\prime }})/m}+e^{-(\overline{x}-x_{0^{\prime }})/m}]$,

$k={\overline{y}}^{\prime }=\frac{1}{2}[e^{(\overline{x}-x_{0^{\prime }})/m}-e^{-(\overline{x}-x_{0^{\prime }})/m}]$.

For brevity write $e^{(\overline{x}-x_{0^{\prime }})/m}=z$, so that the above conditions become

$\overline{y}=\frac{m}{2}(z+z^{-1}),k=\frac{1}{2}(z-z^{-1})$.

Hence,

$z^2-2kz-1=0$;

therefore

$z=k\pm \sqrt{1+k^2}$

and

$z^{-1}=-k\pm \sqrt{1+k^2}$.

We therefore have

$\overline{y}=\pm m\sqrt{1+k^2}$.

Since $\overline{y}$ and $m$ are both positive, it follows that we may take only the upper sign. Consequently, if we write

$k=\tan (\alpha )$,

we have

$z=\tan (\alpha )+\sqrt{1+{\tan }^2(\alpha )}=\frac{\sin (\alpha )+1}{\cos (\alpha }$,

$-z^{-1}=\tan (\alpha )-\sqrt{1+{\tan }^2(\alpha )}=\frac{\sin (\alpha )-1}{\cos (\alpha }$,

and

$m=\frac{\overline{y}}{\sqrt{1+{\tan }^2(\alpha )}}=\overline{y}\cos (\alpha )$.

Further, since $\log (z)$ has one and only one real value for a definite value of $z$, the constant $x_{0^{\prime }}$ is determined uniquely from

$\frac{\overline{x}-x_{0^{\prime }}}{m}=\log (z)=\log \frac{\sin (\alpha )+1}{\cos (\alpha )}$

and the quantities $x_{0^{\prime }}$ and $m$ determine uniquely a catenary which has the given line as tangent at the point $\overline{x}$,$\overline{y}$.

Article 38.
In particular, consider the catenary that has the K-axis as the $Y$-axis of symmetry, and let the two points $P_0$ and $P_1$ be at equal heights on the curve so that their coordinates are, say $(-a,b)$ and $(a,b)$.

The equation of the catenary is now, since $x_{0^{\prime }}$,

$y=\frac{m}{2}(e^{x/m}+e^{-x/m})$;

and consequently

$b=\frac{m}{2}(e^{a/m}+e^{-a/m})=\varphi (m)\text{,}\text{[1]}$

say, where we regard $\alpha$ as constant and $m$ variable.

We wish to determine whether this last equation gives a real value or real values for $m$. We see that $\varphi (m)$ is infinite when $m=0$ and also when $m=\mathrm{\infty }$.

Further

$2{\varphi }^{\prime }(m)=e^{a/m}-e^{-a/m}-\frac{a}{m}(e^{a/m}-e^{-a/m})$,

or

${\varphi }^{\prime }(m)=1-\frac{1}{2}\frac{a^2}{m^2}-\frac{3}{4!}\frac{a^4}{m^4}-\cdots -\frac{2n-1}{(2n)!}\frac{a^{2n}}{m^{2n}}-\cdots$

so that ${\varphi }^{\prime }(m)$ is negative infinity when $m$ is zero; is unity when $m$ is infinite, and changes sign once and only once as $m$ passes from zero to infinity. The least value that $\varphi (m)$ can have is for the value of $m$ that satisfies ${\varphi }^{\prime }(m)=0$.

If, then, the given value of $b$ is greater than the least value of $\varphi (m)$, there are two values of $m$ which satisfy [1]; if the given value of $b$ be equal to the least value of $\varphi (m)$, there is only one value of $m$; and if the given value of $b$ is less than the least value of $\varphi (m)$, there is no possible value of $m$.

Moigno and Lindelöf have shown that the value of $\frac{a}{m}$ which satisfies

$e^{a/m}+e^{-a/m}-\frac{a}{m}(e^{a/m}-e^{-a/m})=0$

is approximately $\frac{a}{m}=1.19968....$; and then from [1] it follows that $\frac{b}{m}=1.81017...$; and therefore $\frac{b}{a}=1.50888...=\tan (56^{\circ }28^{\prime })$ approximately (see Todhunter, loc. cit.. Art. 60). Thus there are two catenaries satisfying the prescribed conditions, or one or none according as $\frac{b}{a}$is greater than, equal to, or less than 1.50888...

If we write $k=\frac{b}{a}=\tan (56^{\circ }28^{\prime })$, it is seen that $y=kx$ and $y=-kx$ are the two tangents to the catenary that may be drawn through the origin.

As the ratio $\frac{b}{a}$ is independent of $m$ it also follows that all the catenaries of the form $y=\frac{m}{2}(e^{x/m}+e^{-x/m})$, which may be derived by varying $m$, have the same two tangent lines through the origin, the points of contact being $x=\pm 1.19968....m$ and $y=1.181017....m$.

Article 39.
Returning to the catenary $y=\frac{m}{2}[e^{(x-x_{0^{\prime }})/m}+e-(x_{0^{\prime }})/m]$, we shall see that also here there are three cases which come under investigation according as:

  I. Two catenaries may be drawn through the fixed points;

 II. One catenary may be drawn through these points;

III. No catenary may be drawn through the two points.

We may assume that $y_1\ge y_0$, $x_1>x_0$, we would only have to change the direction of the $X$-axis which we name positive and negative; or we might consider the case of $P_0$ and $P_{1^{\prime }}$, where $P_{1^{\prime }}$ is the image of $P_1$; that is, the point symmetrically situated to $P_1$ on the other side of the $y_0$-ordinate.

Article 40.
From the equation of the catenary it follows that

$y_0=\frac{m}{2}[e^{(x_0-x_{0^{\prime }})/m}+e^{-(x_0-x_{0^{\prime }})/m}]$,

and

$y_0^2-m^2=\frac{m^2}{4}[e^{(x_0-x_{0^{\prime }})/m}+e^{-(x_0-x_{0^{\prime }})/m}{]}^2$.

Therefore

$\sqrt{y_0^2-m^2}=\pm \frac{m}{2}[e^{(x_0-x_{0^{\prime }})/m}+e^{-(x_0-x_{0^{\prime }})/m}]\text{;}\text{[I]}$

and from this relation it is seen that $\sqrt{y_0^2-m^2}$ has a positive or negative sign according as $x_0-x_{0^{\prime }}\gtrless 0$. Hence, also,

$\frac{x_0-x_{0^{\prime }}}{m}=\pm \ln \frac{y_0+\sqrt{y_0^2}}{m}\text{.}\text{[a]}$

Article 41.
Under the assumption that $y_1\ge y_0$, we must first show that such a figure as the one which follows cannot exist in the present discussion. We know that

$y_1=\frac{m}{2}[e^{(x_1-x_{0^{\prime }})/m}+e^{-(x_1-x_{0^{\prime }})/m}]$.

That $x_1-x_{0^{\prime }}$ is necessarily positive is seen from the fact that the ordinate $y_{0^{\prime }}=m$ corresponds to the value $x_{0^{\prime }}$, and is a minimum. (See Art. 34.) Suppose that $x_{0^{\prime }}>x_1$. By hypothesis $y_1\ge y_0$, and further $m\le y_0$, and consequently $m\le y_1$. The form of the curve is then that given in the figure; and we have within the interval $x_0$ to $x_{0^{\prime }}$ a value of $x$, for which the ordinate $y$ is greater than it is at the end-points. $y$ must therefore have within this interval a maximum value. But we have shown (Art. 34) that there is no maximum value of y;

hence,

$\sqrt{y_1^2-m^2}=+\frac{m}{2}[e^{(x_1-x_{0^{\prime }})/m}-e^{-(x_1-x_{0^{\prime }})/m}]$,

and there cannot be the minus sign as in equation [I]; hence,

$\frac{x_1-x_{0^{\prime }}}{m}=+\ln \frac{y_1+\sqrt{y_1^2-m^2}}{m}\text{.}\text{[b]}$

Article 42.
Eliminate $x_{0^{\prime }}$ from [a] and [b] and noting that in [a] there is the $\pm$ sign, we have two different functions of $m$, which may be written:

$f_1(m)=\ln \frac{y_1+\sqrt{y_1^2-m^2}}{m}-\ln \frac{y_0+\sqrt{y_0^2-m^2}}{m}-\frac{x_1-x_0}{m}$,

and

$f_1(m)=\ln \frac{y_1+\sqrt{y_1^2-m^2}}{m}+\ln \frac{y_0+\sqrt{y_0^2-m^2}}{m}-\frac{x_1-x_0}{m}$,

two functions of a transcendental nature, which we have now to consider. We must see whether $f_1(m)=0$, $f_2(m)=0$ have roots with regard to $m$; that is whether it is possible to give to $m$ positive real values, so that the equations $f_1(m)=0$, $f_2(m)=0$ will be satisfied. If it is possible thus to determine $m$, we must then see whether the values $x_{0^{\prime }}$ which may be derived from equations [a] and [b] are one-valued.

The first derivative $f_1(m)$ is

$f_{1^{\prime }}(m)=\frac{1}{m}\left[\right]\text{.}\text{[c]}$

On the right-hand side of this expression $\frac{1}{m}$ is positive, also $\frac{x_1-x_0}{m}$ is positive, and

$\frac{1}{\sqrt{1-\frac{m^2}{y_0^2}}}-\frac{1}{\sqrt{1-\frac{m^2}{y_1^2}}}$ is positive, if $y_1>y_0$.

Hence $f_{1^{\prime }}(m)$ is positive in the interval $0\cdots y_0$.

Further,

$f_1(0)=\ln (2y_1)-\ln (m=0)-\ln (2y_0)+\ln (m=0)-{\left[\frac{x_1-x_0}{m}\right]}_{m=0}=-\mathrm{\infty }$.

Article 43.
It is further seen that $f_1(m)$ continuously increases within the interval $0\cdots y_0$, so that $-\mathrm{\infty }$ is the least value that $f_1(m)$ can take.

Again

$f_1(y_0)=\ln \frac{y_1+\sqrt{y_1^2-y_0^2}}{y_0}-\frac{x_1-x_0}{y_0}\text{.}\text{[II]}$

Then if

  I $f_1(y_0)<0$, $f_1(m)$ has no root;

 II. $f_1(y_0)=0$, $f_1(m)$ has one root, $m=y_0$;

III. $f_1(y_0)>0$, $f_1(m)$ has a root, $m<y_0$.

When

$f_1(y_0)<0$, $P_1$ is outside of the catenary;

$f_1(y_0)-0$, $P_1$ is on the catenary;

$f_1(y_0)>0$, $P_1$ is within the catenary.

This may be shown as follows:

$y=\frac{y_0}{2}[e^{(x-x_0)/y_0}+e^{-(x-x_0)/y_0}]$;

since when $y=m$, $x=x_{0^{\prime }}$; and, therefore, when $y=y_0=m$, $x=x_0$. We also have

$y_2-y_0^2=\frac{y_0^2}{4}{[e^{(x-x_0)/y_0}-e^{-(x-x{)}_0)/y_0}]}^2$.

Hence

$\sqrt{y^2-y_0^2}=\pm \frac{y_0}{x}[e^{(x-x_0)/y_0}-e^{-(x-x_0)/y_0}]$;

where the positive sign is to be taken, when $x>x_0$, and the negative sign, when $x<x_0$.

We also have $x-x_0=\ln \frac{y+\sqrt{y^2-y_0^2}}{y_0}$. Comparing this equation with equation [II] above, and noticing the figure, it is seen that, when

$x_1-x_0=y_0\ln \frac{y_1+\sqrt{y_1^2}}{y_0}$, then $P_1$ is on the catenary,

$x_1-x_0>y_0\ln \frac{y_1+\sqrt{y_1^2}}{y_0}$, then $P_1$ is outside the catenary,

$x_1-x_0<y_0\ln \frac{y_1+\sqrt{y_1^2}}{y_0}$, then $P_1$ is within the catenary.

Hence, when $f_1(y)>0$, there is one and only one real root in the interval $0\cdots y_0$, and we can draw through the points $P_1$ and $P_0$ a catenary, for which the abscissa of the lowest point is $<x_0$.

Article 44.
The discussion of $f_2(m)$. We saw (Art. 42) that

$f_2(m)=\ln \frac{y_1+\sqrt{y_1^2-m^2}}{m}+\ln \frac{y_0+\sqrt{y_0^2-m^2}}{m}-\frac{x_1-x_0}{m}$.

Therefore

$f_{2^{\prime }}(m)=-\frac{1}{m^2}(\frac{y_{1}m}{\sqrt{y_1^2-m^2}}+\frac{y_{0}m}{\sqrt{y_0^2-m^2}}-(x_1-x_0))$.

When $m$ changes from $0$ to $y_0$, the quantity $\sqrt{\frac{y_0^2}{m^2-1}}$ continuously decreases, and consequently $\frac{y_0}{\frac{y_0^2}{m^2-1}}$ becomes greater and greater. Hence if the expression $-m^2{f}_{2^{\prime }}(m)$ takes the value $0$, it takes it only once in the interval from $0$ to $y_0$. That this expression does take the value $0$ within this interval is seen from the fact that, for $m=0$, $-m^2{f}_{2^{\prime }}(m)=-(x_1-x_0)$, where $x_1-x_0>0$, so that $-m^2{f}_{2^{\prime }}(y_0)$ has a negative value; but, for $m=y_0$, $-m^2{f}_{2^{\prime }}(y_0)=+\mathrm{\infty }$, so that the expression must take the value zero between these two values of $m$.

Let $\mu$ be this value of $m$ which satisfies the equation, so that

$\frac{y_1\mu }{\sqrt{y_1^2-{\mu }^2}}+\frac{y_0\mu }{\sqrt{y_0^2-{\mu }^2}}-(x_1-x_0)=0\text{,}\text{[A]}$

which is an algebraical equation of the eight degree in $\mu$, or an algebraical equation of the fourth degree in ${\mu }^2$.

Article 45.
An approximate geometrical construction for the root that lies between $0$ and $y_0$. In the figure it is seen that the triangles $P_0{Q}_0{A}_0$ and $P_0{Q}_0{C}_0$ are similar, as are also the triangles $P_1{Q}_1{A}_1$ and $P_1{Q}_1{C}_1$; hence, if $m$ is the length of the line $Q_0{C}_0=Q_1{C}_1$, we have

$Q_0{A}_0=\frac{y_{0}m}{\sqrt{y_0^2-m^2}}$,

and

$Q_1{A}_1=\frac{y_{1}m}{\sqrt{y_1^2-m^2}}$.

By taking equal lengths $Q_0{C}_0=Q_1{C}_1$ on the two semi-circles and prolonging $P_0{C}_0$ and $P_1{C}_1$ until they intersect, we have as the locus of the intersections a certain curve. This curve must intersect the $X$-axis in a point $S$, say. Noting that

$Q_{0}S+Q_{1}S=Q_0{Q}_1=x_1-x_0$,

it follows that

$\frac{y_0\cdot Q_0{B}_0}{\sqrt{y_0^2+\stackrel{2}{\stackrel{‾}{Q_0{B}_0}}}}+\frac{y_1\cdot Q_1{B}_1}{\sqrt{y_1^2-\stackrel{2}{\stackrel{‾}{Q_1{B}_1}}}}=x_1-x_0$,

which, compared with the equation [A] above, shows that

$Q_0{B}_0=Q_1{B}_1=\mu$.

Article 46.
Graphical representation of the functions $f_1(m)$ and $f_2(m)$. The lengths $m$ are measured on the $X$-axis. Equation [c] gives $f_{1^{\prime }}(y_0)=\mathrm{\infty }$; that is, the tangent to the curve $y=f_1(x)$ at the point $y_0$ is parallel to the axis of $y$. Further, $f_1(0)=-\mathrm{\infty }$, so that the negative half of the axis of $y$ is asymptotic to the curve $y=f_1(x)$. The branch of the curve is here algebraic, since $y=f_1(x)$, for $x=0$, is algebraically infinite.

Article 47.
Consider next the cvrve $y=f_2(m)$. It is seen that $f_1(y_0)=f_2(y_0)$; and also $f_{2^{\prime }}(y_0)=-\mathrm{\infty }$, so that the tangent at this point is also parallel to the axis of the $y$. Further, the negative half of the axis of the $y$ is an asymptote to the curve; but the branch of the curve $y=f_2(m)$ is transcendental at the point $m=0$; because logarithms enter in the development of this function in the neighborhood of $m=0$, as may be seen as follows:

$f_2(m)=\ln \frac{y_1+\sqrt{y_1^2-m^2}}{m}+\ln \frac{y_0+\sqrt{y_0^2-m^2}}{m}-\frac{x_1-x_0}{m}=-\frac{x_1-x_0}{m}-2\ln (m)+P(m)$,

where $P(m)$ denotes a power series in positive and integral ascending powers of $m$ hence; the function behaves in the neighborhood of $m=0$ as a logarithm.

Article 48.
We saw that

$f_{2^{\prime }}(m)=-\frac{1}{m^2}(\frac{y_{1}m}{\sqrt{y_1^2-m^2}}+\frac{y_{0}m}{\sqrt{y_0^2-m^2}}-(x_1-x_0))$.

For the value $m=\mu$ the expression within the brackets is zero, and when $m=0$, this expression becomes $-(x_1-x_0)$, and is negative. As seen above in the interval $m=0$ to $m=y_0$, the expression

$\frac{y_{1}m}{\sqrt{y_1^2{m}^2}}+\frac{y_{0}m}{\sqrt{y_0^2{m}^2}}-(x_1-x_0)$

becomes greater and greater, so that between the value $m=0$ and $m=\mu$, it is negative.

Furthermore, $f_{2^{\prime }}(m)$ is positive between $m=0$ and $m=\mu$, and negative between $m=\mu$ and $m=y_0$.

Hence $f_2(m)$ increases between $m=0$ and $m=\mu$, and decreases between $m=\mu$ and $m=y_0$; and consequently $f_2(\mu )$ is a maximum.

Article 49.
We must consider the function $f_2(m)$ when $m$ is given different values and see how many catenaries may be laid between the points $P_0$ and $P_1$.

We have:

Case I. $f_2(\mu )<0$.

In this case $f_2(m)$ is nowhere zero, and there is no root of $f_2(m)$ which we can use. There is also no root of $f_1(m)$, since $f_2(y_0)<0$ and $f_2(y_0)=f_1(y_0)$, so that $f_1(y_0)$, and there is no root (see Art. 43).

Case II. $f_2(\mu )=0$.

All values of $m$ other than $\mu$ cause $f_2(m)$ to be negative, so that there is a root and only one root of the equation $f_2(m)=0$, and consequently only one catenary. In this case $f_1(m)$ can never be zero; since $f_2(y_0)<0$, and $f_1(y_0)=f_2(y_0)$, so that $f_1(y_0)<0$, with the result similar to that in Case I.

Case III. $f_2(\mu )>0$.

We have here two catenaries. One root $f_2(m)=m$ lies between $0$ and $\mu$, and often another between $\mu$ and $y_0$, as is seen from what follows:

$f_2(+0)=-\mathrm{\infty }$ and $f_2(\mu )>0$.

Since $f_2(m)$ continuously increases in the interval $+0\dots \mu$, it can take the value $0$ only once within this interval.

In the interval $\mu \dots y_0$, $f_2(m)$ continuously decreases, so that if $f_2(y_0)>0$, there is no root of $f_2(m)=0$ within this interval; but if $f_2(y_0)\le 0$, then there is one and only one root within this interval, and in the latter case there are two catenaries.

We must next consider the roots of $f_1(m)$. When $f_2(y_0)$, then is $f_1(y_0)<0$, so that there is no root of $f_1(m)=0$. But when $f_2(y_0)=0$, then $f_1(y_0)=0$; and $f_1(m)=0$ has the root $m=y_0$, which was just considered.

Therefore:

A) When $f_2(y_0)<0$, $f_2(m)$ has two roots; and when $f_2(y_0)=0$, $f_2(m)$ has a root in addition to the root which belongs to $f_2(y_0)=f_1(y_0)$.

B) But when $f_2(y_0)>0$, then there is only one root for $f_2(m)=0$, which lies between $0\dots \mu$; this root is denoted by $m_1$.

Article 50.
From the formula (Art. 42) for $f_1(m)$ and $f_2(m)$ we have:

$f_2(m)=f_1(m)+2\ln \frac{y_0+\sqrt{y_0^2-m^2}}{m}$.

We consider the values of $m$ within the interval $0\dots y_0$; for $m=0$, $\frac{y_0+\sqrt{y_0^2}}{m}=\mathrm{\infty }$; and for $m=y_0$, $\frac{y_0+\sqrt{y_0^2-m^2}}{m}=1$. Consequently within this interval $\ln \frac{y_0+\sqrt{y_0^2-m^2}}{m}$ is positive, and therefore also $f_2(m)>f_1(m)$; and since $f_2(m_1)=0$, it follows that $f_1(m_1)<0$.

On the other hand, $f_1(y_0)=f_2(y_0)$; and since $f_2(y_0)>0$, we have $f_1(y_0)>0$. Moreover, within the interval $0\dots m_1$, $f_1(m)$ continuously increases, and $f_1(+0)<0$, so that within the interval $0\dots m_1$, $f_1(m)$ has no root, and within the interval $m_1\dots y_0$, one root.

Hence, under B), $f_2(m)$ has a root $m_1$, within the interval $0\dots \mu$, and only one root, and $f_1(m)$ has a root between $m_1$ and $y_0$, and only one, making a total under the heading B) of two catenaries.

We have the following summary:

$1^0$. $f_2(\mu )<0$ no catenary;

$2^0$. $f_2(\mu )=0$, one catenary;

$3^0$. $f_2(\mu )>0$, two catenaries.

Article 51.
On the consideration of the intersection of the tangents drawn to the catenary at the points $P_0$ and $P_1$.

Case I. As shown above, ft there is no catenary, so that the consideration of the tangents is without interest.

Case II. $f_2(\mu )=0$.

Here the catenary enjoys the remarkable property that the tangents drawn at the points $P_0$ and $P_1$ intersect on the $X$-axis. In order to show this, we must return to the construction of the tangents at the points $P_0$ and $P_1$. It was seen (Art. 45) that points $B_0$ and $B_1$ were found on the semi-circumferences $P_0{B}_0{Q}_0$ and $P_1{B}_1{Q}_1$ such that $Q_0{B}_0=Q_1{B}_1$ ($m=\mu$ in this case), and that then the lines $P_0{B}_0$ and $P_1{B}_{!}$ were the required tangents, which intersect on the $X$-axis.

Case III. $f_2(\mu )>0$.

A) $f_2(y_0)\le 0$.

Then, as already shown, $f_2(m)=0$ has two roots, one of which lies between $0$ and $\mu$, and the other between $\mu$ and $y_0$. Let these roots be $m_1$ and $m_2$ respectively. For the root $m_1$, we have

$Q_0{T}_0=\frac{y_0{m}_1}{\sqrt{y_0^2-m_1^2}}$;

$Q_1{T}_1=\frac{y_1{m}_1}{\sqrt{y_1^2-m_1^2}}$.

We assert that here the intersection of the tangents at $P_0$ and $P_1$ lies on the other side of the $X$-axis from the curve.

In order to show this we need only prove that

$Q0T_0+Q_1{T}_1<Q_0{Q}_1$.

This is seen as follows:

$f_{2^{\prime }}(m_1)=-\frac{1}{m^2}(\frac{y_1{m}_1}{\sqrt{y_1^2-m_1^2}}+\frac{y_0{m}_1}{\sqrt{y_0^2-m_1^2}}-(x_1-x_0))$.

Now, since $f_{2^{\prime }}(m)$ within the interval $0\dots \mu$ is positive, and since $m_1$ lies within this interval, it follows that $f_{2^{\prime }}(m_1)$ is positive. Therefore $-m_1^2{f}_{2^{\prime }}(m_1)$ is negative, and consequently $q_0{T}_0+Q_1{T}_1-Q_0{Q}_1$ is negative.

REMARK. In this consideration the whole interpretation depends upon the fact that the root lies in the interval $0\dots \mu$, and the same discussion is applicable to Case B), where $f_2(y_0)>0$, and where the root lies between $0\dots \mu$.

Article 52.
On the consideration of the root $m_2$.

$1^{\circ }$. When $f_2(y_0)\le 0$.

The root lies within the interval $\mu \dots y_0$ and here $f_{2^{\prime }}(m)$ is negative within the interval; therefore $-m^2{f}_{2^{\prime }}(m)$ is positive, and consequently

$\frac{y_1{m}_2}{\sqrt{y_1^2{m}_2^2}}+\frac{y_0{m}_2}{\sqrt{y_0^2{m}_2^2}}-(x_1-x_0)>0$;

therefore

$Q_0{T}_0+Q_1{T}_1>Q_0{Q}_1$;

so that $T$ is on the same side of the $X$-axis as the curve.

$2^{\circ }$. When $f_2(y_0)>0$; then the root $m_2$ is a root of the equation $f_1(m)=0$, so we have here to consider the sign of

$\frac{y_1{m}_2}{\sqrt{y_1^2-m_2^2}}+\frac{y_0{m}_2}{\sqrt{y_0^2-m_2^2}}-(x_1-x_0)$

within the interval $0\dots y_0$.

We have proved that within this interval $f_{1^{\prime }}(m)$ is positive, and since

$f_{1^{\prime }}(m_2)=-\frac{1}{m^2}(\frac{y_1{m}_2}{\sqrt{y_1^2-m_2^2}}-\frac{y_0{m}_2}{\sqrt{y_0^2-m_2^2}}-(x_1-x_0))$

is positive, it follows that

$\frac{y_1{m}_2}{\sqrt{y_1^2-m_2^2}}-\frac{y_0{m}_2}{\sqrt{y_0^2-m_2^2}}-(x_1-x_0)$

is negative. Hence

$\frac{y_1{m}_2}{\sqrt{y_1^2-m_2^2}}-\frac{y_0{m}_2}{\sqrt{y_0^2-m_2^2}}<(x_1-x_0)$.

Consequently

$\frac{y_0{m}_2}{\sqrt{y_0^2-m_2^2}}-\frac{y_1{m}_2}{\sqrt{y_1^2-m_2^2}}>(x_1-x_0)$.

Since $\frac{y_1{m}_2}{\sqrt{y_1^2-m_2^2}}$ is a positive quantity, it follows a fortiori that

$\frac{y_1{m}_2}{\sqrt{y_1^2-m_2^2}}+\frac{y_0{m}_2}{\sqrt{y_0^2-m_2^2}}>(x_1-x_0)$,

and the intersection lies on the same side of the $X$-axis as the curve.

Article 53.
We have seen that two catenaries having the same directrix cannot intersect in more than two points $P_0$ and $P_1$. Denote as above the smaller parameter of these two curves by $m_1$ and the larger by $m_2$. Then it is seen that $C_1$, the curve of smaller parameter, comes up from below and crosses $C_2$, the catenary of larger parameter, and, having crossed $C_2$, never finds its way out again. For, consider the tangent $PT$ to the curve $C_1$ as the point $P$ moves along this curve. This tangent must at first intersect $C_2$, but at the vertex it is parallel to the $X$-axis and evidently has no point in common with $C_2$. Hence, for some position between these two positions the tangent to $C_1$ must also be tangent to $C_2$ see that there are two tangents common to $C_1$ and $C_2$, and we shall next show that they intersect on the directrix.

Article 54.
Draw the common tangent $A{T}_0$ and draw a tangent $A{T}_1$ to the curve $C_1$. Then between these lines we may lay an infinite number of catenaries that have the same directrix. One of these catenaries must be $C_2$, for it touches $A{T}_0$ and is the only catenary that can be drawn through the point of tangency made by $A{T}_0$ (Art. 37). Consequently $A{T}_1$ is the other common tangent to both curves.

We see also that the points $P_0$ and $P_1$ are beyond the points of contact of $C_1$, with the two common tangents, while for $C_2$ the points of contact of the tangents are beyond $P_0$ and $P_1$. It is also seen that, as the two curves $C_1$ and $C_2$ tend to coincide, the common tangents to the distinct curve become tangents to the single curve at the points $P_0$ and $P_1$ (see Art. 51). If we call $\mu$ the value of $m$ corresponding to this latter curve we have $m_2>\mu >m_1$.

Article 55.
Suppose we have two catenaries which are not coincident and which have the same parameter $m$. Denote their equations by

$y=\frac{m}{2}[e^{(x-x_{0^{\prime }})/m}+e^{-(x-x_{0^{\prime }})/m}]$,

$y=\frac{m}{2}[e^{(x-x_{0^{″}})/m}+e^{-(x-x_{0^{″}})/m}]$.

These catenaries intersect in only one point. For we have at once

$e^{(x-x_{0^{\prime }})/m}+e^{-(x-x_{0^{\prime }})/m}=e^{(x-x_{0^{″}})/m}+e^{-(x-x_{0^{″}})/m}$,

therefore

$e^{x/m}[e^{-x_{0^{\prime }}/m}-e^{-x_{0^{″}}/m}]=e^{-x/m}[e^{x_{0^{″}}/m}-e^{x_{0^{\prime }}/m}]$,

or

$e^{2x/m}=\frac{e^{x_{0^{″}}/m}-e^{x_{0^{\prime }}/m}}{e^{-x_{0^{\prime }}/m}-e^{-x_{0^{″}}/m}}=\frac{1}{e^{-(x_{0^{\prime }}+x_{0^{″}})/m}}\left(\frac{e^{x_{0^{″}}/m}-e^{x_{0^{\prime }}/m}}{-e^{x_{0^{\prime }}/m}+e^{x_{0^{″}}/m}}\right)$.

Therefore

$e^{2x/m}=e^{(x_{0^{\prime }}+x_{0^{″}})/m}$,

and consequently

$x=\frac{x_{0^{\prime }}+x_{0^{″}}}{2}y=\frac{m}{2}[e^{(x_{0^{″}}-x_{0^{\prime }})/(2m)}+e^{-(x_{0^{″}}-x_{0^{\prime }})/(2m))}]$,

which are the coordinates of one point.

Article 56.
Lindelöf's Theorem (1860).

If we suppose the catenary to revolve around the $X$-axis, as also the lines $P_{0}T$ and $P_{1}T$, then the surface area generated by the revolution of the catenary is equal to the sum of the surface areas generated by the revolution of the two lines $P_{0}T$ and $P_{1}T$ about the $X$-axis.

Suppose that with $T$ as center of similarity (Aehnlichkeits-punkt), the curve $P_0{P}_1$ is subjected to a strain so that $P_0$ goes into the point $P_{0^{\prime }}$, and $P_1$ into the point $P_{1^{\prime }}$, the distance $P_0{P}_{0^{\prime }}$ being very small and equal, say, to $P_1{P}_{1^{\prime }}$.

Then

$P_{)}T:P_{0^{\prime }}T=1:1-\alpha$.

To abbreviate, let

$M_0$ denote the surface generated by $P_{0}T$;$M_{0^{\prime }}$ that generated by $P_{0^{\prime }}T$; $M_1$ denote the surface generated by $P_{1}T$; $M_{1^{\prime }}$ that generated by $P_{1^{\prime }}T$; $S$ that by the catenary $P_0{P}_1$; $S^{\prime }$ that by the catenary $P_{0^{\prime }}{P}_{1^{\prime }}$.

From the nature of the strain, the tangents $P_{0}T$ and $P_{1}T$ are tangents to the new curve at the points $P_{0^{\prime }}$ and $P_{1^{\prime }}$, so that we may consider $P_0{P}_{0^{\prime }}{P}_{1^{\prime }}{P}_1$ as a variation of the curve $P_0{P}_1$.

It is seen that

$S:S^{\prime }=1:(1-\alpha {)}^2$;

$M_0:M_{0^{\prime }}=1:(1-\alpha {)}^2$;

$M_1:M_{1^{\prime }}=1:(1-\alpha {)}^2$.

Now from the figure we have as the surface of rotation of $P_0{P}_{0^{\prime }}{P}_1{P}_{1^{\prime }}$

$(M_0-M_{0^{\prime }})+S^{\prime }+(M_1-M_{1^{\prime }})+[(\alpha {)}^2]=S$,

where $[(\alpha {)}^2]$ denotes a variation of the second order.

Therefore

$S-S^{\prime }=(M_0-M_{0^{\prime }})+(M_1-M_{1^{\prime }})+[(\alpha {)}^2]$.

Hence

$S[1-(1-\alpha {)}^2]=M_0[1-(1-\alpha {)}^2]+M_1[1-(1-\alpha {)}^2]+[({\alpha }^2)]$,

and consequently

$2\alpha S=2\alpha M_0+2\alpha M_1+[({\alpha }^2)]$,

or finally

$S=M_{0^{\prime }}+M_{1^{\prime }}$,

a result which is correct to a differential of the first order.

In a similar manner

$S^{\prime }=M_{0^{\prime }}+M_{1^{\prime }}$;

so that

$S-S^{\prime }=(M_0-M_{0^{\prime }})+(M_1-M_{1^{\prime }})$;

or

$S=(M_0-M_{0^{\prime }})+S^{\prime }+(M_1-M_{1^{\prime }})$

is an expression which is absolutely correct.

Article 57.
Another proof.

We have seen that

$\frac{y_0\mu }{\sqrt{y_0-{\mu }^2}}+\frac{y_1\mu }{\sqrt{y_1-{\mu }^2}}-(x_1-x_0)=0$,

and (see Fig. in Art. 45)

$P_{0}S=\frac{y_0^2}{\sqrt{y_0-{\mu }^2}}$; $P_{1}S=\frac{y_1^2}{\sqrt{y_1-{\mu }^2}}$

The surfaces of the two cones are, therefore, equal to

$\frac{y_0\cdot y_0^2\pi }{\sqrt{y_0-{\mu }^2}}$ and $\frac{y_1\cdot y_1^2\pi }{\sqrt{y_1-{\mu }^2}}$.

The surface generated by the catenary is

$\underset{x_0}{\overset{x_1}{\int }}2y\pi \text{d}s$.

In the catenary $\text{d}s=\frac{y}{m}\text{d}x$ (see Art. 35), so that

$\underset{x_0}{\overset{x_1}{\int }}2y\pi \text{d}s=\underset{x_0}{\overset{x_1}{\int }}\frac{2y^2\pi \text{d}x}{m}=2\pi \underset{x_0}{\overset{x_1}{\int }}\frac{m^2}{4}[e^{2(x-x_{0^{\prime }})/m}+2+e^{-2(x-x_{0^{\prime }})/m}]\frac{\text{d}x}{m}$

$=\frac{\pi m^2}{4}{[e^{2(x-x_{0^{\prime }})/m}-e^{-2(x-x_{0^{\prime }})/m}+\frac{4x}{m}]}_{x_0}^{x_1}$

$=\pi {[\frac{m}{2}(e^{(x-x_{0^{\prime }})/m}+e^{-(x-x_{0^{\prime }})/m})\cdot \frac{m}{2}(e^{(x-x_{0^{\prime }})/m}-e^{-(x-x_{0^{\prime }})/m})+mx]}_{x_0}^{x_1}\text{[A]}$

$=\pi [\pm y\sqrt{y^2-m^2}+mx{]}_{x_0}^{x_1}$

$=\pi [y_1\sqrt{y_1-m^2}+y_0\sqrt{y_0-m^2}]+m(x_1-x_0)\text{,}\text{[B]}$

where we have taken the $+$ sign with $y_0\sqrt{y_0^2-m^2}$ because $x_0-x_{0^{\prime }}$ is negative, hence $e^{(x-x_{0^{\prime }})/m}-e^{-(x-x_{0^{\prime }})/m}$ in [A] is negative.

But from [1]

$x_1-x_0=\frac{y_1\mu }{\sqrt{y_1^2-{\mu }^2}}+\frac{y_0\mu }{\sqrt{y_0^2-{\mu }^2}}$.

Substituting in [B], we have, after making $m=\mu$, for the area generated by the revolution of the catenary

$\pi [y_1\sqrt{y_1-{\mu }^2}+\frac{y_1{\mu }^2}{\sqrt{y_1^2-{\mu }^2}}+y_0\sqrt{y_0^2-{\mu }^2}+\frac{y_0{\mu }^2}{\sqrt{y_0^2-{\mu }^2}}]=\pi [\frac{y_1^3}{\sqrt{y_1^2-{\mu }^2}}+\frac{y_0^3}{\sqrt{y_0^2-{\mu }^2}}]$,

which, as shown above, is the sum of the surface areas of the two cones.

Article 58.
Let us consider again the following figure, in which the strain is represented. In order to have a minimum surface of revolution, the curve which we rotate must satisfy the differential equation of the problem. If, then, we had a minimum, this would be brought about by the rotation of the catenary; for the catenary is the curve which satisfies the differential equation. But in our figure this curve can produce no minimal surface of revolution for two reasons: $1^{\circ }$ because, drawing tangents (in Art. 59 it is proved that there exists an infinite number) which intersect on the $X$-axis, it is seen that the rotation of $P_{0^{\prime }}{P}_{1^{\prime }}$ is the same as that of the two lines $P_{0^{\prime }}T$ and $P_{1^{\prime }}T$, as shown above, so that there are an infinite number of lines that may be drawn between $P_0$ and $P_1$ which give the same surface of revolution as the catenary between these points; $2^{\circ }$ because between $P_0$ and $P_1$ lines may be drawn which, when caused to revolve about the $X$-axis, would produce a smaller surface area than that produced by the revolution of the catenary. For the surface area generated by the revolution of $P_{0^{\prime }}{P}_{1^{\prime }}$ is the same as that generated by $P_{0^{\prime }}{P}_{0^{″}}{P}_{1^{″}}{P}_{1^{\prime }}$. But the straight lines $P_{0^{\prime }}{P}_{0^{″}}$ and $P_{1^{\prime }}{P}_{1^{″}}$ do not satisfy the differential equation of the problem, since they are not catenaries. Hence the first variation along these lines is $\gtrless 0$, so that between the points $P_{0^{\prime }}$,$P_{0^{″}}$ and $P_{1^{\prime }}$,$P_{1^{″}}$ curves may be drawn whose surface of rotation is smaller than that generated by the straight lines $P_{0^{\prime }}{P}_{0^{″}}$ and $P_{1^{\prime }}{P}_{1^{″}}$.

The Case II, given above and known as the transition case, i.e., where the point of intersection of the tangents pass from one side to the other side of the $X$-axis, affords also no minimal surface, since, as already seen, there are, by varying the quantity $\alpha$ (Art. 56), an infinite number of surfaces of revolution that have the same area.

Article 59.
In Case III we had two roots of $m$, which we called $m_1$ and $m_2$, where $m_2>m_1$. We consider first the catenary with parameter $m_1$. This parameter satisfies the inequality

$\frac{y_1{m}_1}{\sqrt{y_1^2-m_1^2}}+\frac{y_0{m}_1}{\sqrt{y_0^2-m_1^2}}<x_1-x_0\text{.}\text{[A]}$

The equation of the tangent to the curve is

$\frac{\text{d}y}{\text{d}x}=\frac{y^{\prime }-y}{x^{\prime }-x}$,

where $x^{\prime }$ and $y^{\prime }$ are the running coordinates. The intersection of this line with the $X$-axis is

$x^{\prime }-x=-\frac{y}{\frac{\text{d}y}{\text{d}x}}$, or $x^{\prime }=x-\frac{y}{\frac{\text{d}y}{\text{d}x}}$;

i.e.,

$x^{\prime }=x-m\left[\frac{e^{(x-x_{0^{\prime }})/m}+e^{-(x-x_{0^{\prime }})/m}}{e^{(x-x_{0^{\prime }})/m}-e^{-(x-x_{0^{\prime }})/m}}\right]$.

Hence, when $x=x_{0^{\prime }}$, $x^{\prime }=-\mathrm{\infty }$, and when $x=+\mathrm{\infty }$, $x^{\prime }=+\mathrm{\infty }$.

On the other hand, $\frac{\text{d}{x}^{\prime }}{\text{d}x}$ is always positive, so that $x^{\prime }$ always increases when $x$ increases, and the tangent passes from $-\mathrm{\infty }$ along the $X$-axis to $+\mathrm{\infty }$, and never passes twice through the same point. It is thus seen that there are an infinite number of pairs of points on the catenary between the points $P_0$ and $P_1$ such that the tangents at any of these pairs of points intersect on the $X$-axis, and there can consequently be no minimum. Such pairs of points are known as conjugate points.

When $m=m_1$, the tangents intersect above the $X$-axis, and there is in reality a minimum, as will be seen later.

Article 60.
Application. Suppose we have two rings of equal size attached to the same axis which passes perpendicularly through their centers. If the rims of these rings are connected by a free film of liquid (soap solution), what form does the film take?

By a law in physics the film has a tendency to make its area as small as possible. Hence, only as a minimal surface will the film be in a state of equilibrium. Let $O$ be midway between $O^{\prime }$ and $O^{″}$. The film is symmetric with respect to the $O{O}^{″}$ and $OL$ axes and has the form of a surface of revolution about the $O{O}^{″}$ axis, this surface being a catenoid. The line $OL$ is the axis of symmetry of the generating catenary. Construct the tangents $O{P}^{″}$ and $O{P}^{\prime }$ from the origin to the catenary. Only when $P^{\prime }$ and $P^{″}$ are situated beyond the rims of the circles will the generating arc of the catenary be free from conjugate points, and only then will we have a minimal surface and a position of stable equilibrium of the film.

Article 61.
We saw (Art. 38) that all catenaries having the same axis of symmetry and the same directrix may be laid between two lines inclined approximately at an angle ${\tan }^{-1}(3/2)$ to the directrix and which pass through the intersection of the directrix and the axis of symmetry. All catenaries under consideration then are ensconced within the lines $O{P}^{\prime }$ and $O{P}^{″}$ and have these lines as tangents. The arcs of these catenaries between their points of contact with $O{T}^{\prime }$ and $O{T}^{″}$ do not intersect one another. Through any point $P_0$ inside the angle $T^{\prime }O{T}^{″}$ will evidently pass one of these arcs, and the same arc (on account of the axis of symmetry $OL$ of the catenary) will contain the point $P_1$ symmetrical to $P_0$ on the other side of $OL$. The arc $P_0{P}_1$ contains no conjugate point (Chap. IX, Art. 128), and therefore generates a minimal surface of revolution. Further, this is the only arc of a catenary through the points $P_0$ and $P_1$ which generates a minimal surface.

Suppose that we started out with our two rings in contact and shoved them along the axis at the same rate and in opposite directions from the point $O$. As long as $P_0$ and $P_1$ are situated within the angle $T^{\prime }O{T}^{″}$ (or what is the same thing, as long as $P_{0}O{P}_1<T^{\prime }O{T}^{″}$) then the tangents at $P_0$ and $P_1$ meet on the upper side of the $X$-axis and there exists an arc of a catenary which gives a minimal surface of revolution and the film has a tendency to take a definite position and hold itself there. But as soon as the angle $P_{0}O{P}_1$ becomes equal to or greater than $T^{\prime }O{T}^{″}$ this tendency ceases and the equilibrium of the film becomes unstable. As a matter of fact (see Art. 101), the only minimum which now exists is that given by the surface of the two rings, the film having broken and gone into this form.