PlanetPhysics/Equation of Catenary Via Calculus of Variations

Using the mechanical principle that the [[../CenterOfGravity/|centre of mass]] places itself as low as possible, determine the equation of the curve formed by a flexible homogeneous wire or a thin chain with length $l$ when supported at its ends in the points \,$P_1=(x_1,y_1)$\, and\, $P_2=(x_2,y_2)$.\\

We have an isoperimetric problem

$\begin{array}{c}\text{to minimise}\underset{P_1}{\overset{P_2}{\int }}yds\end{array}$

under the constraint

$\begin{array}{c}\underset{P_1}{\overset{P_2}{\int }}ds=l,\end{array}$

where both the path integrals are taken along some curve $c$.\, Using a Lagrange multiplier $\lambda$, the task changes to a free problem

$\begin{array}{c}\underset{P_1}{\overset{P_2}{\int }}(y-\lambda )ds=\underset{x_1}{\overset{x_2}{\int }}(y-\lambda )\sqrt{1+y{\text{'}}^2}|dx|=\text{min}!\end{array}$

(cf. example of calculus of variations).

The Euler--Lagrange differential equation, the necessary condition for (3) to give an extremal $c$, reduces to the Beltrami [[../Cod/|identity]] $(y-\lambda )\sqrt{1+y{\text{'}}^2}-y^{\prime }\cdot (y-\lambda )\cdot \frac{y^{\prime }}{\sqrt{1+y{\text{'}}^2}}\equiv \frac{y-\lambda }{\sqrt{1+y{\text{'}}^2}}=a,$ where $a$ is a constant of integration.\, After solving this equation for the derivative $y^{\prime }$ and [[../SeparationOfVariables/|separation of variables]], we get $\pm \frac{dy}{\sqrt{(y-\lambda {)}^2-a^2}}=\frac{dx}{a}$ which may become clearer by notating\, $y-\lambda :=u$;\, then by integrating $\pm \frac{du}{\sqrt{u^2-a^2}}=\frac{dx}{a}$ we choose the new constant of integration $b$ such that\, $x=b$\, when\, $u=a$: $\pm \underset{a}{\overset{u}{\int }}\frac{du}{\sqrt{u^2-a^2}}=\underset{b}{\overset{x}{\int }}\frac{dx}{a}$ We can write two equivalent results $\ln \frac{u+\sqrt{u^2-a^2}}{a}=+\frac{x-b}{a},\ln \frac{u-\sqrt{u^2-a^2}}{a}=-\frac{x-b}{a},$ i.e. $\frac{u+\sqrt{u^2-a^2}}{a}=e^{+\frac{x-b}{a}},\frac{u-\sqrt{u^2-a^2}}{a}=e^{-\frac{x-b}{a}}.$ Adding these allows to eliminate the [[../PiecewiseLinear/|square]] roots and to obtain $u=\frac{a}{2}(e^{\frac{x-b}{a}}+e^{-\frac{x-b}{a}}),$ or

$\begin{array}{c}y-\lambda =a\cosh \frac{x-b}{a}.\end{array}$

This is the sought form of the equation of the [[../Catenary/|chain curve]].\, The constants $\lambda ,a,b$ can then be determined for putting the curve to pass through the given points $P_1$ and $P_2$.