# 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 ${\displaystyle f_{1}(m)}$ and ${\displaystyle f_{2}(m)}$.
• 43 The discussion of the function ${\displaystyle f_{1}(m)}$.
• 44 The discussion of the function ${\displaystyle f_{2}(m)}$.
• 45 An approximate geometrical construction for the root of a transcendental equation.
• 46,47,48 Graphical representation of the functions ${\displaystyle f_{1}(m)}$ and ${\displaystyle f_{2}(m)}$.
• 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[1]

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

It follows at once that

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

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 ${\displaystyle x>x_{0}'}$, then is

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

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

${\displaystyle e^{(x-x_{0}')/m}

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

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

Article 35.
At every point of the curve we have

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

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

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

consequently

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

and integrating,

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

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

The geometrical locus of ${\displaystyle R}$ 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 ${\displaystyle QR}$, etc., of length ${\displaystyle m}$, 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 ${\displaystyle X}$-axis, the surface of rotation has a constant curvature,

Further, ${\displaystyle PN}$, the normal to the catenary,

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

or

${\displaystyle PN=PC}$ (see figure),

where ${\displaystyle PC}$ is the length of the radius of curvature.

Article 36.
The geometrical construction of the catenary. Take an ordinate equal to ${\displaystyle 2m}$. This determines the point ${\displaystyle P}$ (see figure). With ${\displaystyle P}$ as center and radius equal to ${\displaystyle m}$, describe a circle. This intersects ${\displaystyle PB}$ at a point ${\displaystyle A}$, say. On the circumference of this circle take a point ${\displaystyle A_{1}}$, very near ${\displaystyle A}$, and draw the line ${\displaystyle PA_{1}B_{1}}$, and on this line extended take ${\displaystyle P_{1}}$ such that ${\displaystyle P_{1}A_{1}=A_{1}B_{1}}$. With radius ${\displaystyle P_{1}A_{1}}$ draw another circle, and on this circle take a point ${\displaystyle A_{2}}$, very near the point ${\displaystyle A_{1}}$, and draw the line ${\displaystyle P_{1}A_{2}B_{2}}$. Take on this line extended the point ${\displaystyle P_{2}}$, so that ${\displaystyle P_{2}A_{2}=A_{2}B_{2}}$, etc. The locus of the points ${\displaystyle A}$ 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 ${\displaystyle {\bar {x}}}$,${\displaystyle {\bar {y}}}$ be a point through which passes a straight line, making with the ${\displaystyle X}$-axis an angle whose tangent is ${\displaystyle k}$. The conditions that a catenary pass through this point and have the given line as tangent are:

${\displaystyle {\bar {y}}={\frac {m}{2}}[e^{({\bar {x}}-x_{0}')/m}+e^{-({\bar {x}}-x_{0}')/m}]}$,
${\displaystyle k={\bar {y}}'={\frac {1}{2}}[e^{({\bar {x}}-x_{0}')/m}-e^{-({\bar {x}}-x_{0}')/m}]}$.

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

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

Hence,

${\displaystyle z^{2}-2kz-1=0}$;

therefore

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

and

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

We therefore have

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

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

${\displaystyle k=\tan(\alpha )}$,

we have

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

and

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

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

${\displaystyle {\frac {{\bar {x}}-x_{0}'}{m}}=\log(z)=\log {\frac {\sin(\alpha )+1}{\cos(\alpha )}}}$

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

Article 38.
In particular, consider the catenary that has the K-axis as the ${\displaystyle Y}$-axis of symmetry, and let the two points ${\displaystyle P_{0}}$ and ${\displaystyle P_{1}}$ be at equal heights on the curve so that their coordinates are, say ${\displaystyle (-a,b)}$ and ${\displaystyle (a,b)}$.

The equation of the catenary is now, since ${\displaystyle x_{0}'}$,

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

and consequently

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

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

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

Further

${\displaystyle 2\phi '(m)=e^{a/m}-e^{-a/m}-{\frac {a}{m}}(e^{a/m}-e^{-a/m})}$,

or

${\displaystyle \phi '(m)=1-{\frac {1}{2}}{\frac {a^{2}}{m^{2}}}-{\frac {3}{4!}}{\frac {a^{4}}{m^{4}}}-\cdots -{\frac {2n-1}{(2n)!}}{\frac {a^{2n}}{m^{2n}}}-\cdots }$

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

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

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

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

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

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

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

Article 39.
Returning to the catenary ${\displaystyle y={\frac {m}{2}}[e^{(x-x_{0}')/m}+e{-(x_{0}')/m}]}$, 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 ${\displaystyle y_{1}\geq y_{0}}$, ${\displaystyle x_{1}>x_{0}}$, we would only have to change the direction of the ${\displaystyle X}$-axis which we name positive and negative; or we might consider the case of ${\displaystyle P_{0}}$ and ${\displaystyle P_{1}'}$, where ${\displaystyle P_{1}'}$ is the image of ${\displaystyle P_{1}}$; that is, the point symmetrically situated to ${\displaystyle P_{1}}$ on the other side of the ${\displaystyle y_{0}}$-ordinate.

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

${\displaystyle y_{0}={\frac {m}{2}}[e^{(x_{0}-x_{0}')/m}+e^{-(x_{0}-x_{0}')/m}]}$,

and

${\displaystyle y_{0}^{2}-m^{2}={\frac {m^{2}}{4}}[e^{(x_{0}-x_{0}')/m}+e^{-(x_{0}-x_{0}')/m}]^{2}}$.

Therefore

${\displaystyle {\sqrt {y_{0}^{2}-m^{2}}}=\pm {\frac {m}{2}}[e^{(x_{0}-x_{0}')/m}+e^{-(x_{0}-x_{0}')/m}]{\text{;}}\qquad {\text{[I]}}}$

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

${\displaystyle {\frac {x_{0}-x_{0}'}{m}}=\pm \ln {\frac {y_{0}+{\sqrt {y_{0}^{2}}}}{m}}{\text{.}}\qquad {\text{[a]}}}$

Article 41.
Under the assumption that ${\displaystyle y_{1}\geq y_{0}}$, we must first show that such a figure as the one which follows cannot exist in the present discussion. We know that

${\displaystyle y_{1}={\frac {m}{2}}[e^{(x_{1}-x_{0}')/m}+e^{-(x_{1}-x_{0}')/m}]}$.

That ${\displaystyle x_{1}-x_{0}'}$ is necessarily positive is seen from the fact that the ordinate ${\displaystyle y_{0}'=m}$ corresponds to the value ${\displaystyle x_{0}'}$, and is a minimum. (See Art. 34.) Suppose that ${\displaystyle x_{0}'>x_{1}}$. By hypothesis ${\displaystyle y_{1}\geq y_{0}}$, and further ${\displaystyle m\leq y_{0}}$, and consequently ${\displaystyle m\leq y_{1}}$. The form of the curve is then that given in the figure; and we have within the interval ${\displaystyle x_{0}}$ to ${\displaystyle x_{0}'}$ a value of ${\displaystyle x}$, for which the ordinate ${\displaystyle y}$ is greater than it is at the end-points. ${\displaystyle y}$ must therefore have within this interval a maximum value. But we have shown (Art. 34) that there is no maximum value[2] of y;

hence,

${\displaystyle {\sqrt {y_{1}^{2}-m^{2}}}=+{\frac {m}{2}}[e^{(x_{1}-x_{0}')/m}-e^{-(x_{1}-x_{0}')/m}]}$,

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

${\displaystyle {\frac {x_{1}-x_{0}'}{m}}=+\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}-m^{2}}}}{m}}{\text{.}}\qquad {\text{[b]}}}$

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

${\displaystyle f_{1}(m)=\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}-m^{2}}}}{m}}-\ln {\frac {y_{0}+{\sqrt {y_{0}^{2}-m^{2}}}}{m}}-{\frac {x_{1}-x_{0}}{m}}}$,

and

${\displaystyle f_{1}(m)=\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}-m^{2}}}}{m}}+\ln {\frac {y_{0}+{\sqrt {y_{0}^{2}-m^{2}}}}{m}}-{\frac {x_{1}-x_{0}}{m}}}$,

two functions of a transcendental nature, which we have now to consider. We must see whether ${\displaystyle f_{1}(m)=0}$, ${\displaystyle f_{2}(m)=0}$ have roots with regard to ${\displaystyle m}$; that is whether it is possible to give to ${\displaystyle m}$ positive real values, so that the equations ${\displaystyle f_{1}(m)=0}$, ${\displaystyle f_{2}(m)=0}$ will be satisfied. If it is possible thus to determine ${\displaystyle m}$, we must then see whether the values ${\displaystyle x_{0}'}$ which may be derived from equations [a] and [b] are one-valued.

The first derivative ${\displaystyle f_{1}(m)}$ is

${\displaystyle f_{1}'(m)={\frac {1}{m}}\left[\right]{\text{.}}\qquad {\text{[c]}}}$

On the right-hand side of this expression ${\displaystyle {\frac {1}{m}}}$ is positive, also ${\displaystyle {\frac {x_{1}-x_{0}}{m}}}$ is positive, and

${\displaystyle {\frac {1}{\sqrt {1-{\frac {m^{2}}{y_{0}^{2}}}}}}-{\frac {1}{\sqrt {1-{\frac {m^{2}}{y_{1}^{2}}}}}}}$ is positive, if ${\displaystyle y_{1}>y_{0}}$.

Hence ${\displaystyle f_{1}'(m)}$ is positive in the interval ${\displaystyle 0\cdots y_{0}}$.

Further,

${\displaystyle f_{1}(0)=\ln(2y_{1})-\ln(m=0)-\ln(2y_{0})+\ln(m=0)-\left[{\frac {x_{1}-x_{0}}{m}}\right]_{m=0}=-\infty }$.

Article 43.
It is further seen that ${\displaystyle f_{1}(m)}$ continuously increases within the interval ${\displaystyle 0\cdots y_{0}}$, so that ${\displaystyle -\infty }$ is the least value that ${\displaystyle f_{1}(m)}$ can take.

Again

${\displaystyle f_{1}(y_{0})=\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}-y_{0}^{2}}}}{y_{0}}}-{\frac {x_{1}-x_{0}}{y_{0}}}{\text{.}}\qquad {\text{[II]}}}$

Then if

I ${\displaystyle f_{1}(y_{0})<0}$, ${\displaystyle f_{1}(m)}$ has no root;

II. ${\displaystyle f_{1}(y_{0})=0}$, ${\displaystyle f_{1}(m)}$ has one root, ${\displaystyle m=y_{0}}$;

III. ${\displaystyle f_{1}(y_{0})>0}$, ${\displaystyle f_{1}(m)}$ has a root, ${\displaystyle m.

When

${\displaystyle f_{1}(y_{0})<0}$, ${\displaystyle P_{1}}$ is outside of the catenary;
${\displaystyle f_{1}(y_{0})-0}$, ${\displaystyle P_{1}}$ is on the catenary;
${\displaystyle f_{1}(y_{0})>0}$, ${\displaystyle P_{1}}$ is within the catenary.

This may be shown as follows:

${\displaystyle y={\frac {y_{0}}{2}}\left[e^{(x-x_{0})/y_{0}}+e^{-(x-x_{0})/y_{0}}\right]}$;

since when ${\displaystyle y=m}$, ${\displaystyle x=x_{0}'}$; and, therefore, when ${\displaystyle y=y_{0}=m}$, ${\displaystyle x=x_{0}}$. We also have

${\displaystyle y_{2}-y_{0}^{2}={\frac {y_{0}^{2}}{4}}\left[e^{(x-x_{0})/y_{0}}-e^{-(x-x)_{0})/y_{0}}\right]^{2}}$.

Hence

${\displaystyle {\sqrt {y^{2}-y_{0}^{2}}}=\pm {\frac {y_{0}}{x}}\left[e^{(x-x_{0})/y_{0}}-e^{-(x-x_{0})/y_{0}}\right]}$;

where the positive sign is to be taken, when ${\displaystyle x>x_{0}}$, and the negative sign, when ${\displaystyle x.

We also have ${\displaystyle x-x_{0}=\ln {\frac {y+{\sqrt {y^{2}-y_{0}^{2}}}}{y_{0}}}}$. Comparing this equation with equation [II] above, and noticing the figure, it is seen that, when

${\displaystyle x_{1}-x_{0}=y_{0}\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}}}}{y_{0}}}}$, then ${\displaystyle P_{1}}$ is on the catenary,
${\displaystyle x_{1}-x_{0}>y_{0}\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}}}}{y_{0}}}}$, then ${\displaystyle P_{1}}$ is outside the catenary,
${\displaystyle x_{1}-x_{0}, then ${\displaystyle P_{1}}$ is within the catenary.

Hence, when ${\displaystyle f_{1}(y)>0}$, there is one and only one real root in the interval ${\displaystyle 0\cdots y_{0}}$, and we can draw through the points ${\displaystyle P_{1}}$ and ${\displaystyle P_{0}}$ a catenary, for which the abscissa of the lowest point is ${\displaystyle .

Article 44.
The discussion of ${\displaystyle f_{2}(m)}$. We saw (Art. 42) that

${\displaystyle f_{2}(m)=\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}-m^{2}}}}{m}}+\ln {\frac {y_{0}+{\sqrt {y_{0}^{2}-m^{2}}}}{m}}-{\frac {x_{1}-x_{0}}{m}}}$.

Therefore

${\displaystyle f_{2}'(m)=-{\frac {1}{m^{2}}}\left({\frac {y_{1}m}{\sqrt {y_{1}^{2}-m^{2}}}}+{\frac {y_{0}m}{\sqrt {y_{0}^{2}-m^{2}}}}-(x_{1}-x_{0})\right)}$.

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

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

${\displaystyle {\frac {y_{1}\mu }{\sqrt {y_{1}^{2}-\mu ^{2}}}}+{\frac {y_{0}\mu }{\sqrt {y_{0}^{2}-\mu ^{2}}}}-(x_{1}-x_{0})=0{\text{,}}\qquad {\text{[A]}}}$

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

Article 45.
An approximate geometrical construction for the root that lies between ${\displaystyle 0}$ and ${\displaystyle y_{0}}$. In the figure it is seen that the triangles ${\displaystyle P_{0}Q_{0}A_{0}}$ and ${\displaystyle P_{0}Q_{0}C_{0}}$ are similar, as are also the triangles ${\displaystyle P_{1}Q_{1}A_{1}}$ and ${\displaystyle P_{1}Q_{1}C_{1}}$; hence, if ${\displaystyle m}$ is the length of the line ${\displaystyle Q_{0}C_{0}=Q_{1}C_{1}}$, we have

${\displaystyle Q_{0}A_{0}={\frac {y_{0}m}{\sqrt {y_{0}^{2}-m^{2}}}}}$,

and

${\displaystyle Q_{1}A_{1}={\frac {y_{1}m}{\sqrt {y_{1}^{2}-m^{2}}}}}$.

By taking equal lengths ${\displaystyle Q_{0}C_{0}=Q_{1}C_{1}}$ on the two semi-circles and prolonging ${\displaystyle P_{0}C_{0}}$ and ${\displaystyle P_{1}C_{1}}$ until they intersect, we have as the locus of the intersections a certain curve. This curve must intersect the ${\displaystyle X}$-axis in a point ${\displaystyle S}$, say. Noting that

${\displaystyle Q_{0}S+Q_{1}S=Q_{0}Q_{1}=x_{1}-x_{0}}$,

it follows that

${\displaystyle {\frac {y_{0}\cdot Q_{0}B_{0}}{\sqrt {y_{0}^{2}+{\overline {Q_{0}B_{0}}}^{2}}}}+{\frac {y_{1}\cdot Q_{1}B_{1}}{\sqrt {y_{1}^{2}-{\overline {Q_{1}B_{1}}}^{2}}}}=x_{1}-x_{0}}$,

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

${\displaystyle Q_{0}B_{0}=Q_{1}B_{1}=\mu }$.

Article 46.
Graphical representation of the functions ${\displaystyle f_{1}(m)}$ and ${\displaystyle f_{2}(m)}$. The lengths ${\displaystyle m}$ are measured on the ${\displaystyle X}$-axis. Equation [c] gives ${\displaystyle f_{1}'(y_{0})=\infty }$; that is, the tangent to the curve ${\displaystyle y=f_{1}(x)}$ at the point ${\displaystyle y_{0}}$ is parallel to the axis of ${\displaystyle y}$. Further, ${\displaystyle f_{1}(0)=-\infty }$, so that the negative half of the axis of ${\displaystyle y}$ is asymptotic to the curve ${\displaystyle y=f_{1}(x)}$. The branch of the curve is here algebraic, since ${\displaystyle y=f_{1}(x)}$, for ${\displaystyle x=0}$, is algebraically infinite.

Article 47.
Consider next the cvrve ${\displaystyle y=f_{2}(m)}$. It is seen that ${\displaystyle f_{1}(y_{0})=f_{2}(y_{0})}$; and also ${\displaystyle f_{2}'(y_{0})=-\infty }$, so that the tangent at this point[3] is also parallel to the axis of the ${\displaystyle y}$. Further, the negative half of the axis of the ${\displaystyle y}$ is an asymptote to the curve; but the branch of the curve ${\displaystyle y=f_{2}(m)}$ is transcendental at the point ${\displaystyle m=0}$; because logarithms enter in the development of this function in the neighborhood of ${\displaystyle m=0}$, as may be seen as follows:

${\displaystyle f_{2}(m)=\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}-m^{2}}}}{m}}+\ln {\frac {y_{0}+{\sqrt {y_{0}^{2}-m^{2}}}}{m}}-{\frac {x_{1}-x_{0}}{m}}=-{\frac {x_{1}-x_{0}}{m}}-2\ln(m)+P(m)}$,

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

Article 48.
We saw that

${\displaystyle f_{2}'(m)=-{\frac {1}{m^{2}}}\left({\frac {y_{1}m}{\sqrt {y_{1}^{2}-m^{2}}}}+{\frac {y_{0}m}{\sqrt {y_{0}^{2}-m^{2}}}}-(x_{1}-x_{0})\right)}$.

For the value ${\displaystyle m=\mu }$ the expression within the brackets is zero, and when ${\displaystyle m=0}$, this expression becomes ${\displaystyle -(x_{1}-x_{0})}$, and is negative. As seen above in the interval ${\displaystyle m=0}$ to ${\displaystyle m=y_{0}}$, the expression

${\displaystyle {\frac {y_{1}m}{\sqrt {y_{1}^{2}m^{2}}}}+{\frac {y_{0}m}{\sqrt {y_{0}^{2}m^{2}}}}-(x_{1}-x_{0})}$

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

Furthermore, ${\displaystyle f_{2}'(m)}$ is positive between ${\displaystyle m=0}$ and ${\displaystyle m=\mu }$, and negative between ${\displaystyle m=\mu }$ and ${\displaystyle m=y_{0}}$.

Hence ${\displaystyle f_{2}(m)}$ increases between ${\displaystyle m=0}$ and ${\displaystyle m=\mu }$, and decreases between ${\displaystyle m=\mu }$ and ${\displaystyle m=y_{0}}$; and consequently ${\displaystyle f_{2}(\mu )}$ is a maximum.

Article 49.
We must consider the function ${\displaystyle f_{2}(m)}$ when ${\displaystyle m}$ is given different values and see how many catenaries may be laid between the points ${\displaystyle P_{0}}$ and ${\displaystyle P_{1}}$.

We have:

Case I. ${\displaystyle f_{2}(\mu )<0}$.

In this case ${\displaystyle f_{2}(m)}$ is nowhere zero, and there is no root of ${\displaystyle f_{2}(m)}$ which we can use. There is also no root of ${\displaystyle f_{1}(m)}$, since ${\displaystyle f_{2}(y_{0})<0}$ and ${\displaystyle f_{2}(y_{0})=f_{1}(y_{0})}$, so that ${\displaystyle f_{1}(y_{0})}$, and there is no root (see Art. 43).

Case II. ${\displaystyle f_{2}(\mu )=0}$.

All values of ${\displaystyle m}$ other than ${\displaystyle \mu }$ cause ${\displaystyle f_{2}(m)}$ to be negative, so that there is a root and only one root of the equation ${\displaystyle f_{2}(m)=0}$, and consequently only one catenary. In this case ${\displaystyle f_{1}(m)}$ can never be zero; since ${\displaystyle f_{2}(y_{0})<0}$, and ${\displaystyle f_{1}(y_{0})=f_{2}(y_{0})}$, so that ${\displaystyle f_{1}(y_{0})<0}$, with the result similar to that in Case I.

Case III. ${\displaystyle f_{2}(\mu )>0}$.

We have here two catenaries. One root ${\displaystyle f_{2}(m)=m}$ lies between ${\displaystyle 0}$ and ${\displaystyle \mu }$, and often another between ${\displaystyle \mu }$ and ${\displaystyle y_{0}}$, as is seen from what follows:

${\displaystyle f_{2}(+0)=-\infty }$ and ${\displaystyle f_{2}(\mu )>0}$.

Since ${\displaystyle f_{2}(m)}$ continuously increases in the interval ${\displaystyle +0\ldots \mu }$, it can take the value ${\displaystyle 0}$ only once within this interval.

In the interval ${\displaystyle \mu \ldots y_{0}}$, ${\displaystyle f_{2}(m)}$ continuously decreases, so that if ${\displaystyle f_{2}(y_{0})>0}$, there is no root of ${\displaystyle f_{2}(m)=0}$ within this interval; but if ${\displaystyle f_{2}(y_{0})\leq 0}$, 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 ${\displaystyle f_{1}(m)}$. When ${\displaystyle f_{2}(y_{0})}$, then is ${\displaystyle f_{1}(y_{0})<0}$, so that there is no root of ${\displaystyle f_{1}(m)=0}$. But when ${\displaystyle f_{2}(y_{0})=0}$, then ${\displaystyle f_{1}(y_{0})=0}$; and ${\displaystyle f_{1}(m)=0}$ has the root ${\displaystyle m=y_{0}}$, which was just considered.

Therefore:

A) When ${\displaystyle f_{2}(y_{0})<0}$, ${\displaystyle f_{2}(m)}$ has two roots; and when ${\displaystyle f_{2}(y_{0})=0}$, ${\displaystyle f_{2}(m)}$ has a root in addition to the root which belongs to ${\displaystyle f_{2}(y_{0})=f_{1}(y_{0})}$.

B) But when ${\displaystyle f_{2}(y_{0})>0}$, then there is only one root for ${\displaystyle f_{2}(m)=0}$, which lies between ${\displaystyle 0\ldots \mu }$; this root is denoted by ${\displaystyle m_{1}}$.

Article 50.
From the formula (Art. 42) for ${\displaystyle f_{1}(m)}$ and ${\displaystyle f_{2}(m)}$ we have:

${\displaystyle f_{2}(m)=f_{1}(m)+2\ln {\frac {y_{0}+{\sqrt {y_{0}^{2}-m^{2}}}}{m}}}$.

We consider the values of ${\displaystyle m}$ within the interval ${\displaystyle 0\ldots y_{0}}$; for ${\displaystyle m=0}$, ${\displaystyle {\frac {y_{0}+{\sqrt {y_{0}^{2}}}}{m}}=\infty }$; and for ${\displaystyle m=y_{0}}$, ${\displaystyle {\frac {y_{0}+{\sqrt {y_{0}^{2}-m^{2}}}}{m}}=1}$. Consequently within this interval ${\displaystyle \ln {\frac {y_{0}+{\sqrt {y_{0}^{2}-m^{2}}}}{m}}}$ is positive, and therefore also ${\displaystyle f_{2}(m)>f_{1}(m)}$; and since ${\displaystyle f_{2}(m_{1})=0}$, it follows that ${\displaystyle f_{1}(m_{1})<0}$.

On the other hand, ${\displaystyle f_{1}(y_{0})=f_{2}(y_{0})}$; and since ${\displaystyle f_{2}(y_{0})>0}$, we have ${\displaystyle f_{1}(y_{0})>0}$. Moreover, within the interval ${\displaystyle 0\ldots m_{1}}$, ${\displaystyle f_{1}(m)}$ continuously increases, and ${\displaystyle f_{1}(+0)<0}$, so that within the interval ${\displaystyle 0\ldots m_{1}}$, ${\displaystyle f_{1}(m)}$ has no root, and within the interval ${\displaystyle m_{1}\ldots y_{0}}$, one root.

Hence, under B), ${\displaystyle f_{2}(m)}$ has a root ${\displaystyle m_{1}}$, within the interval ${\displaystyle 0\ldots \mu }$, and only one root, and ${\displaystyle f_{1}(m)}$ has a root between ${\displaystyle m_{1}}$ and ${\displaystyle y_{0}}$, and only one, making a total under the heading B) of two catenaries.

We have the following summary:

${\displaystyle 1^{0}}$. ${\displaystyle f_{2}(\mu )<0}$ no catenary;
${\displaystyle 2^{0}}$. ${\displaystyle f_{2}(\mu )=0}$, one catenary;
${\displaystyle 3^{0}}$. ${\displaystyle f_{2}(\mu )>0}$, two catenaries.

Article 51.
On the consideration of the intersection of the tangents drawn to the catenary at the points ${\displaystyle P_{0}}$ and ${\displaystyle P_{1}}$.

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

Case II. ${\displaystyle f_{2}(\mu )=0}$.

Here the catenary enjoys the remarkable property that the tangents drawn at the points ${\displaystyle P_{0}}$ and ${\displaystyle P_{1}}$ intersect on the ${\displaystyle X}$-axis. In order to show this, we must return to the construction of the tangents at the points ${\displaystyle P_{0}}$ and ${\displaystyle P_{1}}$. It was seen (Art. 45) that points ${\displaystyle B_{0}}$ and ${\displaystyle B_{1}}$ were found on the semi-circumferences ${\displaystyle P_{0}B_{0}Q_{0}}$ and ${\displaystyle P_{1}B_{1}Q_{1}}$ such that ${\displaystyle Q_{0}B_{0}=Q_{1}B_{1}}$ (${\displaystyle m=\mu }$ in this case), and that then the lines ${\displaystyle P_{0}B_{0}}$ and ${\displaystyle P_{1}B_{!}}$ were the required tangents, which intersect on the ${\displaystyle X}$-axis.

Case III. ${\displaystyle f_{2}(\mu )>0}$.

A) ${\displaystyle f_{2}(y_{0})\leq 0}$.

Then, as already shown, ${\displaystyle f_{2}(m)=0}$ has two roots, one of which lies between ${\displaystyle 0}$ and ${\displaystyle \mu }$, and the other between ${\displaystyle \mu }$ and ${\displaystyle y_{0}}$. Let these roots be ${\displaystyle m_{1}}$ and ${\displaystyle m_{2}}$ respectively. For the root ${\displaystyle m_{1}}$, we have

${\displaystyle Q_{0}T_{0}={\frac {y_{0}m_{1}}{\sqrt {y_{0}^{2}-m_{1}^{2}}}}}$;
${\displaystyle Q_{1}T_{1}={\frac {y_{1}m_{1}}{\sqrt {y_{1}^{2}-m_{1}^{2}}}}}$.

We assert that here the intersection of the tangents at ${\displaystyle P_{0}}$ and ${\displaystyle P_{1}}$ lies on the other side of the ${\displaystyle X}$-axis from the curve.

In order to show this we need only prove that

${\displaystyle Q{0}T_{0}+Q_{1}T_{1}.

This is seen as follows:

${\displaystyle f_{2}'(m_{1})=-{\frac {1}{m^{2}}}\left({\frac {y_{1}m_{1}}{\sqrt {y_{1}^{2}-m_{1}^{2}}}}+{\frac {y_{0}m_{1}}{\sqrt {y_{0}^{2}-m_{1}^{2}}}}-(x_{1}-x_{0})\right)}$.

Now, since ${\displaystyle f_{2}'(m)}$ within the interval ${\displaystyle 0\ldots \mu }$ is positive, and since ${\displaystyle m_{1}}$ lies within this interval, it follows that ${\displaystyle f_{2}'(m_{1})}$ is positive. Therefore ${\displaystyle -m_{1}^{2}f_{2}'(m_{1})}$ is negative, and consequently ${\displaystyle q_{0}T_{0}+Q_{1}T_{1}-Q_{0}Q_{1}}$ is negative.

REMARK. In this consideration the whole interpretation depends upon the fact that the root lies in the interval ${\displaystyle 0\ldots \mu }$, and the same discussion is applicable to Case B), where ${\displaystyle f_{2}(y_{0})>0}$, and where the root lies between ${\displaystyle 0\ldots \mu }$.

Article 52.
On the consideration of the root ${\displaystyle m_{2}}$.

${\displaystyle 1^{\circ }}$. When ${\displaystyle f_{2}(y_{0})\leq 0}$.

The root lies within the interval ${\displaystyle \mu \ldots y_{0}}$ and here ${\displaystyle f_{2}'(m)}$ is negative within the interval; therefore ${\displaystyle -m^{2}f_{2}'(m)}$ is positive, and consequently

${\displaystyle {\frac {y_{1}m_{2}}{\sqrt {y_{1}^{2}m_{2}^{2}}}}+{\frac {y_{0}m_{2}}{\sqrt {y_{0}^{2}m_{2}^{2}}}}-(x_{1}-x_{0})>0}$;

therefore

${\displaystyle Q_{0}T_{0}+Q_{1}T_{1}>Q_{0}Q_{1}}$;

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

${\displaystyle 2^{\circ }}$. When ${\displaystyle f_{2}(y_{0})>0}$; then the root ${\displaystyle m_{2}}$ is a root of the equation ${\displaystyle f_{1}(m)=0}$, so we have here to consider the sign of

${\displaystyle {\frac {y_{1}m_{2}}{\sqrt {y_{1}^{2}-m_{2}^{2}}}}+{\frac {y_{0}m_{2}}{\sqrt {y_{0}^{2}-m_{2}^{2}}}}-(x_{1}-x_{0})}$

within the interval ${\displaystyle 0\ldots y_{0}}$.

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

${\displaystyle f_{1}'(m_{2})=-{\frac {1}{m^{2}}}\left({\frac {y_{1}m_{2}}{\sqrt {y_{1}^{2}-m_{2}^{2}}}}-{\frac {y_{0}m_{2}}{\sqrt {y_{0}^{2}-m_{2}^{2}}}}-(x_{1}-x_{0})\right)}$

is positive, it follows that

${\displaystyle {\frac {y_{1}m_{2}}{\sqrt {y_{1}^{2}-m_{2}^{2}}}}-{\frac {y_{0}m_{2}}{\sqrt {y_{0}^{2}-m_{2}^{2}}}}-(x_{1}-x_{0})}$

is negative. Hence

${\displaystyle {\frac {y_{1}m_{2}}{\sqrt {y_{1}^{2}-m_{2}^{2}}}}-{\frac {y_{0}m_{2}}{\sqrt {y_{0}^{2}-m_{2}^{2}}}}<(x_{1}-x_{0})}$.

Consequently

${\displaystyle {\frac {y_{0}m_{2}}{\sqrt {y_{0}^{2}-m_{2}^{2}}}}-{\frac {y_{1}m_{2}}{\sqrt {y_{1}^{2}-m_{2}^{2}}}}>(x_{1}-x_{0})}$.

Since ${\displaystyle {\frac {y_{1}m_{2}}{\sqrt {y_{1}^{2}-m_{2}^{2}}}}}$ is a positive quantity, it follows a fortiori that

${\displaystyle {\frac {y_{1}m_{2}}{\sqrt {y_{1}^{2}-m_{2}^{2}}}}+{\frac {y_{0}m_{2}}{\sqrt {y_{0}^{2}-m_{2}^{2}}}}>(x_{1}-x_{0})}$,

and the intersection lies on the same side of the ${\displaystyle X}$-axis as the curve.

Article 53.
We have seen that two catenaries having the same directrix cannot intersect in more than two points ${\displaystyle P_{0}}$ and ${\displaystyle P_{1}}$. Denote as above the smaller parameter of these two curves by ${\displaystyle m_{1}}$ and the larger by ${\displaystyle m_{2}}$. Then it is seen that ${\displaystyle C_{1}}$, the curve of smaller parameter, comes up from below and crosses ${\displaystyle C_{2}}$, the catenary of larger parameter, and, having crossed ${\displaystyle C_{2}}$, never finds its way out again. For, consider the tangent ${\displaystyle PT}$ to the curve ${\displaystyle C_{1}}$ as the point ${\displaystyle P}$ moves along this curve. This tangent must at first intersect ${\displaystyle C_{2}}$, but at the vertex it is parallel to the ${\displaystyle X}$-axis and evidently has no point in common with ${\displaystyle C_{2}}$. Hence, for some position between these two positions the tangent to ${\displaystyle C_{1}}$ must also be tangent to ${\displaystyle C_{2}}$ see that there are two tangents common to ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$, and we shall next show that they intersect on the directrix.

Article 54.
Draw the common tangent ${\displaystyle AT_{0}}$ and draw a tangent ${\displaystyle AT_{1}}$ to the curve ${\displaystyle C_{1}}$. Then between these lines we may lay an infinite number of catenaries that have the same directrix. One of these catenaries must be ${\displaystyle C_{2}}$, for it touches ${\displaystyle AT_{0}}$ and is the only catenary that can be drawn through the point of tangency made by ${\displaystyle AT_{0}}$ (Art. 37). Consequently ${\displaystyle AT_{1}}$ is the other common tangent to both curves.

We see also that the points ${\displaystyle P_{0}}$ and ${\displaystyle P_{1}}$ are beyond the points of contact of ${\displaystyle C_{1}}$, with the two common tangents, while for ${\displaystyle C_{2}}$ the points of contact of the tangents are beyond ${\displaystyle P_{0}}$ and ${\displaystyle P_{1}}$. It is also seen that, as the two curves ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$ tend to coincide, the common tangents to the distinct curve become tangents to the single curve at the points ${\displaystyle P_{0}}$ and ${\displaystyle P_{1}}$ (see Art. 51). If we call ${\displaystyle \mu }$ the value of ${\displaystyle m}$ corresponding to this latter curve we have ${\displaystyle m_{2}>\mu >m_{1}}$.

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

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

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

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

therefore

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

or

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

Therefore

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

and consequently

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

which are the coordinates of one point.

Article 56.
Lindelöf's Theorem (1860).

If we suppose the catenary to revolve around the ${\displaystyle X}$-axis, as also the lines ${\displaystyle P_{0}T}$ and ${\displaystyle P_{1}T}$, 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 ${\displaystyle P_{0}T}$ and ${\displaystyle P_{1}T}$ about the ${\displaystyle X}$-axis.

Suppose that with ${\displaystyle T}$ as center of similarity (Aehnlichkeits-punkt), the curve ${\displaystyle P_{0}P_{1}}$ is subjected to a strain so that ${\displaystyle P_{0}}$ goes into the point ${\displaystyle P_{0}'}$, and ${\displaystyle P_{1}}$ into the point ${\displaystyle P_{1}'}$, the distance ${\displaystyle P_{0}P_{0}'}$ being very small and equal, say, to ${\displaystyle P_{1}P_{1}'}$.

Then

${\displaystyle P_{)}T~:~P_{0}'T=1~:~1-\alpha }$.

To abbreviate, let

${\displaystyle M_{0}}$ denote the surface generated by ${\displaystyle P_{0}T}$;${\displaystyle M_{0}'}$ that generated by ${\displaystyle P_{0}'T}$; ${\displaystyle M_{1}}$ denote the surface generated by ${\displaystyle P_{1}T}$; ${\displaystyle M_{1}'}$ that generated by ${\displaystyle P_{1}'T}$; ${\displaystyle S}$ that by the catenary ${\displaystyle P_{0}P_{1}}$; ${\displaystyle S'}$ that by the catenary ${\displaystyle P_{0}'P_{1}'}$.

From the nature of the strain, the tangents ${\displaystyle P_{0}T}$ and ${\displaystyle P_{1}T}$ are tangents to the new curve at the points ${\displaystyle P_{0}'}$ and ${\displaystyle P_{1}'}$, so that we may consider ${\displaystyle P_{0}P_{0}'P_{1}'P_{1}}$ as a variation of the curve ${\displaystyle P_{0}P_{1}}$.

It is seen that

${\displaystyle S~:~S'=1~:~(1-\alpha )^{2}}$;
${\displaystyle M_{0}~:~M_{0}'=1~:~(1-\alpha )^{2}}$;
${\displaystyle M_{1}~:~M_{1}'=1~:~(1-\alpha )^{2}}$.

Now from the figure we have as the surface of rotation of ${\displaystyle P_{0}P_{0}'P_{1}P_{1}'}$

${\displaystyle (M_{0}-M_{0}')+S'+(M_{1}-M_{1}')+[(\alpha )^{2}]=S}$,

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

Therefore

${\displaystyle S-S'=(M_{0}-M_{0}')+(M_{1}-M_{1}')+[(\alpha )^{2}]}$.

Hence

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

and consequently

${\displaystyle 2\alpha S=2\alpha M_{0}+2\alpha M_{1}+[(\alpha ^{2})]}$,

or finally

${\displaystyle S=M_{0}'+M_{1}'}$,

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

In a similar manner

${\displaystyle S'=M_{0}'+M_{1}'}$;

so that

${\displaystyle S-S'=(M_{0}-M_{0}')+(M_{1}-M_{1}')}$;

or

${\displaystyle S=(M_{0}-M_{0}')+S'+(M_{1}-M_{1}')}$

is an expression which is absolutely correct.

Article 57.
Another proof.

We have seen that

${\displaystyle {\frac {y_{0}\mu }{\sqrt {y_{0}-\mu ^{2}}}}+{\frac {y_{1}\mu }{\sqrt {y_{1}-\mu ^{2}}}}-(x_{1}-x_{0})=0}$,

and (see Fig. in Art. 45)

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

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

${\displaystyle {\frac {y_{0}\cdot y_{0}^{2}\pi }{\sqrt {y_{0}-\mu ^{2}}}}}$ and ${\displaystyle {\frac {y_{1}\cdot y_{1}^{2}\pi }{\sqrt {y_{1}-\mu ^{2}}}}}$.

The surface generated by the catenary is

${\displaystyle \int _{x_{0}}^{x_{1}}2y\pi ~{\text{d}}s}$.

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

${\displaystyle \int _{x_{0}}^{x_{1}}2y\pi ~{\text{d}}s=\int _{x_{0}}^{x_{1}}{\frac {2y^{2}\pi ~{\text{d}}x}{m}}=2\pi \int _{x_{0}}^{x_{1}}{\frac {m^{2}}{4}}[e^{2(x-x_{0}')/m}+2+e^{-2(x-x_{0}')/m}]{\frac {{\text{d}}x}{m}}}$
${\displaystyle ={\frac {\pi m^{2}}{4}}\left[e^{2(x-x_{0}')/m}-e^{-2(x-x_{0}')/m}+{\frac {4x}{m}}\right]_{x_{0}}^{x_{1}}}$
${\displaystyle =\pi \left[{\frac {m}{2}}(e^{(x-x_{0}')/m}+e^{-(x-x_{0}')/m})\cdot {\frac {m}{2}}(e^{(x-x_{0}')/m}-e^{-(x-x_{0}')/m})+mx\right]_{x_{0}}^{x_{1}}\qquad {\text{[A]}}}$
${\displaystyle =\pi [\pm y{\sqrt {y^{2}-m^{2}}}+mx]_{x_{0}}^{x_{1}}}$
${\displaystyle =\pi [y_{1}{\sqrt {y_{1}-m^{2}}}+y_{0}{\sqrt {y_{0}-m^{2}}}]+m(x_{1}-x_{0}){\text{,}}\qquad {\text{[B]}}}$

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

But from [1]

${\displaystyle x_{1}-x_{0}={\frac {y_{1}\mu }{\sqrt {y_{1}^{2}-\mu ^{2}}}}+{\frac {y_{0}\mu }{\sqrt {y_{0}^{2}-\mu ^{2}}}}}$.

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

${\displaystyle \pi \left[y_{1}{\sqrt {y_{1}-\mu ^{2}}}+{\frac {y_{1}\mu ^{2}}{\sqrt {y_{1}^{2}-\mu ^{2}}}}+y_{0}{\sqrt {y_{0}^{2}-\mu ^{2}}}+{\frac {y_{0}\mu ^{2}}{\sqrt {y_{0}^{2}-\mu ^{2}}}}\right]=\pi \left[{\frac {y_{1}^{3}}{\sqrt {y_{1}^{2}-\mu ^{2}}}}+{\frac {y_{0}^{3}}{\sqrt {y_{0}^{2}-\mu ^{2}}}}\right]}$,

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

Article 58.
Let us consider[4] 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: ${\displaystyle 1^{\circ }}$ because, drawing tangents (in Art. 59 it is proved that there exists an infinite number) which intersect on the ${\displaystyle X}$-axis, it is seen that the rotation of ${\displaystyle P_{0}'P_{1}'}$ is the same as that of the two lines ${\displaystyle P_{0}'T}$ and ${\displaystyle P_{1}'T}$, as shown above, so that there are an infinite number of lines that may be drawn between ${\displaystyle P_{0}}$ and ${\displaystyle P_{1}}$ which give the same surface of revolution as the catenary between these points; ${\displaystyle 2^{\circ }}$ because between ${\displaystyle P_{0}}$ and ${\displaystyle P_{1}}$ lines may be drawn which, when caused to revolve about the ${\displaystyle X}$-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 ${\displaystyle P_{0}'P_{1}'}$ is the same as that generated by ${\displaystyle P_{0}'P_{0}''P_{1}''P_{1}'}$. But the straight lines ${\displaystyle P_{0}'P_{0}''}$ and ${\displaystyle P_{1}'P_{1}''}$ do not satisfy the differential equation of the problem, since they are not catenaries. Hence the first variation along these lines is ${\displaystyle \gtrless 0}$, so that between the points ${\displaystyle P_{0}'}$,${\displaystyle P_{0}''}$ and ${\displaystyle P_{1}'}$,${\displaystyle P_{1}''}$ curves may be drawn whose surface of rotation is smaller than that generated by the straight lines ${\displaystyle P_{0}'P_{0}''}$ and ${\displaystyle P_{1}'P_{1}''}$.

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 ${\displaystyle X}$-axis, affords also no minimal surface, since, as already seen, there are, by varying the quantity ${\displaystyle \alpha }$ (Art. 56), an infinite number of surfaces of revolution that have the same area.

Article 59.
In Case III we had two roots of ${\displaystyle m}$, which we called ${\displaystyle m_{1}}$ and ${\displaystyle m_{2}}$, where ${\displaystyle m_{2}>m_{1}}$. We consider first the catenary with parameter ${\displaystyle m_{1}}$. This parameter satisfies the inequality

${\displaystyle {\frac {y_{1}m_{1}}{\sqrt {y_{1}^{2}-m_{1}^{2}}}}+{\frac {y_{0}m_{1}}{\sqrt {y_{0}^{2}-m_{1}^{2}}}}

The equation of the tangent to the curve is

${\displaystyle {\frac {{\text{d}}y}{{\text{d}}x}}={\frac {y'-y}{x'-x}}}$,

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

${\displaystyle x'-x=-{\frac {y}{\frac {{\text{d}}y}{{\text{d}}x}}}}$, or ${\displaystyle x'=x-{\frac {y}{\frac {{\text{d}}y}{{\text{d}}x}}}}$;

i.e.,

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

Hence, when ${\displaystyle x=x_{0}'}$, ${\displaystyle x'=-\infty }$, and when ${\displaystyle x=+\infty }$, ${\displaystyle x'=+\infty }$.

On the other hand, ${\displaystyle {\frac {{\text{d}}x'}{{\text{d}}x}}}$ is always positive, so that ${\displaystyle x'}$ always increases when ${\displaystyle x}$ increases, and the tangent passes from ${\displaystyle -\infty }$ along the ${\displaystyle X}$-axis to ${\displaystyle +\infty }$, 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 ${\displaystyle P_{0}}$ and ${\displaystyle P_{1}}$ such that the tangents at any of these pairs of points intersect on the ${\displaystyle X}$-axis, and there can consequently be no minimum. Such pairs of points are known as conjugate points.

When ${\displaystyle m=m_{1}}$, the tangents intersect above the ${\displaystyle X}$-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 ${\displaystyle O}$ be midway between ${\displaystyle O'}$ and ${\displaystyle O''}$. The film is symmetric with respect to the ${\displaystyle OO''}$ and ${\displaystyle OL}$ axes and has the form of a surface of revolution about the ${\displaystyle OO''}$ axis, this surface being a catenoid. The line ${\displaystyle OL}$ is the axis of symmetry of the generating catenary. Construct the tangents ${\displaystyle OP''}$ and ${\displaystyle OP'}$ from the origin to the catenary. Only when ${\displaystyle P'}$ and ${\displaystyle P''}$ 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 ${\displaystyle \tan ^{-1}(3/2)}$ 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 ${\displaystyle OP'}$ and ${\displaystyle OP''}$ and have these lines as tangents. The arcs of these catenaries between their points of contact with ${\displaystyle OT'}$ and ${\displaystyle OT''}$ do not intersect one another. Through any point ${\displaystyle P_{0}}$ inside the angle ${\displaystyle T'OT''}$ will evidently pass one of these arcs, and the same arc (on account of the axis of symmetry ${\displaystyle OL}$ of the catenary) will contain the point ${\displaystyle P_{1}}$ symmetrical to ${\displaystyle P_{0}}$ on the other side of ${\displaystyle OL}$. The arc ${\displaystyle P_{0}P_{1}}$ 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 ${\displaystyle P_{0}}$ and ${\displaystyle P_{1}}$ 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 ${\displaystyle O}$. As long as ${\displaystyle P_{0}}$ and ${\displaystyle P_{1}}$ are situated within the angle ${\displaystyle T'OT''}$ (or what is the same thing, as long as ${\displaystyle P_{0}OP_{1}) then the tangents at ${\displaystyle P_{0}}$ and ${\displaystyle P_{1}}$ meet on the upper side of the ${\displaystyle X}$-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 ${\displaystyle P_{0}OP_{1}}$ becomes equal to or greater than ${\displaystyle T'OT''}$ 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.

1. Throughout this discussion the ${\displaystyle X}$-axis is taken as the directrix.
2. In other words, ${\displaystyle y_{1}}$ cannot be greater than ${\displaystyle y_{0}}$ and at the same time ${\displaystyle x_{0}'}$ greater than ${\displaystyle x_{1}}$.
3. The distance ${\displaystyle y_{0}}$ is, of course, measured on the ${\displaystyle X}$-axis.
4. See also Todhunter, Researches in the Calculus of Variations, p. 29.