Calculus of Variations/CHAPTER XVI

CHAPTER XVI: THE DETERMINATION OF THE CURVE OF GIVEN LENGTH AND GIVEN END-POINTS, WHOSE CENTER OF GRAVITY LIES LOWEST.

• 216 Statement of the problem.
• 217 The necessary conditions
• 218 The number of catenaries having a prescribed length that may be through two given points with respect to a fixed directrix.
• 219 The constants uniquely determined.

Article 216.
To solve the problem of this Chapter, let the $Y$-axis be taken vertically with the positive direction upward, and denote by $S$ the length of the whole curve. If the coordinates of the center of gravity are $x_0,y_0$, then $y_0$ is determined from the equation

$y_0=S\underset{t_0}{\overset{t_1}{\int }}y\sqrt{x{\text{'}}^2+y{\text{'}}^2}\text{d}t$, where $S=\underset{t_0}{\overset{t_1}{\int }}\sqrt{x{\text{'}}^2+y{\text{'}}^2}\text{d}t$

The problem is: So determine $x$ and $y$ as functions of $t$ that the first integral will be a minimum while the second integral retains a constant value. (See Art. 16).

The property that the center of gravity is to lie as low as possible must also be satisfied for every portion of the curve; for if this were not true, then we could replace a portion 1 2 of the curve by a portion of the same length but with a center of gravity that lies lower, with the result that the center of gravity of the whole curve could be shoved lower down, and consequently the original curve would not have the required minimal property.

We have here

$F^{(0)}=y\sqrt{x{\text{'}}^2+y{\text{'}}^2}{F}^{(1)}=\sqrt{x{\text{'}}^2+y{\text{'}}^2}$

$F=(y-\lambda )\sqrt{x{\text{'}}^2+y{\text{'}}^2}$

and therefore

$\frac{\partial F}{\partial x^{\prime }}=\frac{x^{\prime }(y-\lambda )}{\sqrt{x{\text{'}}^2+y{\text{'}}^2}}\frac{{\partial }^{2}F}{\partial x^{\prime }\partial y^{\prime }}=\frac{-x^{\prime }{y}^{\prime }(y-\lambda )}{(\sqrt{x{\text{'}}^2+y{\text{'}}^2}{)}^3}\frac{\partial F}{\partial y^{\prime }}=\frac{y^{\prime }(y-\lambda )}{\sqrt{x{\text{'}}^2+y{\text{'}}^2}}$

$F_1=\frac{y-\lambda }{(\sqrt{x{\text{'}}^2+y{\text{'}}^2}{)}^3}$

We exclude once for all the case where the two given points lie in the same vertical line, because then the integral for $S$ does not express for every case the absolute length of the curve ; for example, when a certain portion of the curve overlaps itself. Similarly we exclude the case where the given length $S$ is exactly equal to the length between the two points on a straight line ; for, in this case, the curve cannot be varied and at the same time retain the constant length.

Article 217.
Since $F_1$ must be positive, a minimum being required, it follows that $(y-\lambda )>0$. Since further, $\frac{\partial F}{\partial x^{\prime }}$ and $\frac{\partial F}{\partial y^{\prime }}$ vary in a continuous manner along the whole curve, and since these quantities differ from the direction-cosines only through the factor $y-\lambda$, which varies in a continuous manner, it follows that the curve changes everywhere its direction in a continuous manner.

The function $F$ is the same as the function $F$ which appeared in Art. 7, except that here we have $y-\lambda$ instead of $y$ in that problem. Since the differential equation here must be the same as in the problem just mentioned, we must have as the required curve

$x=\alpha \pm \beta ty=\lambda +\frac{\beta (e^t+e^{-t})}{2}$

the equation of a catenary.

Since $y-\lambda >0$, it follows that $\beta$ is a positive constant. For $S$ we have the value

$S=\underset{t_0}{\overset{t_1}{\int }}\sqrt{x{\text{'}}^2+y{\text{'}}^2}\text{d}t=\frac{\beta }{2}[e^{t_1}+e^{-t_1}-(e^{t_0}+e^{-t_0})]$

Article 218.
We have next to investigate whether and how often a catenary may be passed through two points and have the length $S$ that is, whether and in how many different ways it is possible to determine the constants $\alpha ,\beta ,\lambda$ in terms of $S$ and the coordinates of the given points. If we denote the coordinates of these points by $a_0,b_0,a_1,b_1$, then is

$a_0=\alpha \ne \beta t_0{a}_1=\alpha \ne \beta t_1$

$b_0=\lambda +\frac{\beta }{2}(e^{t_0}+e^{-t_0})b_1=\lambda +\frac{\beta }{2}(e^{t_1}+e^{-t_1})$

$S=\frac{\beta }{2}[e^{t_1}+e^{-t_1}-(e^{t_0}+e^{-t_0})]$

It follows that

$a_1-a_0=\pm \beta (t_1-t_0)b_1-b_0=\frac{\beta }{2}[e^{t_1}+e^{-t_1}-(e^{t_0}+e^{-t_0})]$

We have assumed that $t_1>t_0$, and consequently we have to take the upper or lower sign according as $a_1-a_0>0$ or $a_1-a_0<0$. It is clear that we may always take $a_1-a_0>0$, since we may interchange the point $a_1,b_1$ with the point $a_0,b_0$, and vice versa.

We shall accordingly take the upper sign. If we write

$\frac{t_1-t_0}{2}=\mu \frac{t_1+t_0}{2}=\nu$

then $\mu$ is a positive quantity and we have

$a_1-a_0=+2\mu \beta$

$b_1-b_0=\frac{\beta }{2}(e^{\mu }-e^{-\mu })(e^{\nu }-e^{-\nu })$

$S=\frac{\beta }{2}(e^{\mu }-e^{-\mu })(e^{\nu }+e^{\nu })$

$\frac{b_1-b_0}{S}=\frac{1-e^{-2\nu }}{1+e^{-2\nu }}=-\frac{1-e^{2\nu }}{1+e^{2\nu }}$

$\frac{\text{d}}{\text{d}\nu }\left(\frac{b_1-b_0}{S}\right)=\frac{4}{(e^{\nu }+e^{-\nu }{)}^2}$

Since this derivative is continuously positive, the expression $\frac{b_1-b_0}{S}$ varies in a continuous manner from $-1$ to $+1$, while $\nu$ increases from $-\mathrm{\infty }$ to $+\mathrm{\infty }$. Hence for every real value of $\nu$ there is one and only one real value of $\frac{b_1-b_0}{S}$ which is situated between $-1$ and $+1$, and vice versa to every value of $\frac{b_1-b_0}{S}$ situated between $-1$ and $+1$ there is one and only one real value of $\nu$. Since we excluded the case where $S$ was equal to the length along a straight line between the two given points, it follows that $S$ is always greater than $b_1-b_0$ and consequently $\frac{b_1-b_0}{S}$ is in reality a proper fraction. Hence $\nu$ is uniquely determined through $\frac{b_1-b_0}{S}$.

Article 219.
We have further

$\frac{S}{a_1-a_0}=\frac{e^{\mu }-e^{-\mu }}{2\mu }\frac{e^{\nu }+e^{-\nu }}{2}$

or

$\frac{2\mu }{e^{\mu }-e^{-\mu }}=\frac{a_1-a_0}{S\sqrt{1-{\left(\frac{b_1-b_0}{S}\right)}^2}}=\frac{a_1-a_0}{\sqrt{S^2-(b_1-b_0{)}^2}}$

The right-hand side is a given positive quantity which we may denote by $M$. It is seen that

$\frac{\text{d}}{\text{d}\mu }\left(\frac{2\mu }{e^{\mu }-e^{-\mu }}\right)=-2\frac{[(\mu -1)e^{\mu }+(\mu +1)e^{-\mu }]}{(e^{\mu }-e^{-\mu }{)}^2}$

By its definition $\mu$ is always greater than $0$. If $\mu$ is situated between 1 and $\mathrm{\infty }$, the right-hand side of the equation is always negative. Since further the differential quotient of the expression $(\mu -1)e^{\mu }+(\mu +1)e^{-\mu }$ is never less than $0$ while $\mu$ varies from $0$ to $1$, it is seen that this expression increases continuously when $\mu$ varies from $0$ to 1; hence the differential quotient of $\frac{2\mu }{e^{\mu }-e^{-\mu }}$ is continuously negative, and consequently

$\frac{\text{d}}{\text{d}\mu }\left(\frac{2\mu }{e^{\mu }-e^{-\mu }}\right)<0$ for $0<\mu <\mathrm{\infty }$

Consequently the expression $\frac{2\mu }{e^{\mu }-e^{-\mu }}$, or the quantity $M$, continuously decreases from 1 to 0 while $\mu$ takes the values from 0 to $\mathrm{\infty }$, and therefore to every value of $M$ lying between 0 and 1 there is one and only one value of $\mu$ situated between 0 and $\mathrm{\infty }$.

Since by hypothesis $M$ is always a positive proper fraction, it follows from the above that $\mu$ is uniquely determined through the given quantities. Through $\mu$ and $\nu$ and the other given quantities we may also determine uniquely $\alpha ,\beta ,\lambda$; and consequently if $S$ is taken sufficiently large, it is possible to lay one and only one catenary between the given points which satisfies the given conditions.

If, then, there exists a curve which is a solution of the problem, this curve is a catenary. We have not yet proved that in reality for this curve the first integral is a minimum. The sufficient criteria for this will be developed in the next Chapter.