PlanetPhysics/Equation of Catenary Via Calculus of Variations

Using the mechanical principle that the 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 $$\begin{matrix} \mbox{to minimise} \quad \int_{P_1}^{P_2}\!y\,ds \end{matrix}$$ under the constraint $$\begin{matrix} \int_{P_1}^{P_2}\!ds \;=\; l, \end{matrix}$$ where both the path integrals are taken along some curve $$c$$.\, Using a Lagrange multiplier $$\lambda$$, the task changes to a free problem $$\begin{matrix} \int_{P_1}^{P_2}\!(y\!-\!\lambda)\,ds \;=\; \int_{x_1}^{x_2}(y\!-\!\lambda)\sqrt{1\!+\!y'^2}\,|dx| \;=\; \mbox{min}! \end{matrix}$$ (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 identity $$(y\!-\!\lambda)\sqrt{1\!+\!y'^2}-y'\!\cdot\!(y\!-\!\lambda)\!\cdot\!\frac{y'}{\sqrt{1\!+\!y'^2}} \;\equiv\; \frac{y\!-\!\lambda}{\sqrt{1\!+\!y'^2}} \;=\; a,$$ where $$a$$ is a constant of integration.\, After solving this equation for the derivative $$y'$$ and 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\, $$u = a$$: $$\pm\int_a^u\frac{du}{\sqrt{u^2\!-\!a^2}} \;=\; \int_b^x\frac{dx}{a}$$ We can write two equivalent results $$\ln\frac{u\!+\!\sqrt{u^2\!-\!a^2}}{a} \;=\; +\frac{x\!-\!b}{a}, \qquad \ln\frac{u\!-\!\sqrt{u^2\!-\!a^2}}{a} \;=\; -\frac{x\!-\!b}{a},$$ i.e. $$\frac{u\!+\!\sqrt{u^2\!-\!a^2}}{a} \;=\; e^{+\frac{x-b}{a}}, \qquad \frac{u\!-\!\sqrt{u^2\!-\!a^2}}{a} \;=\; e^{-\frac{x-b}{a}}.$$ Adding these allows to eliminate the square roots and to obtain $$u \;=\; \frac{a}{2}\!\left(e^{\frac{x-b}{a}}+e^{-\frac{x-b}{a}}\right),$$ or $$\begin{matrix} y\!-\!\lambda \;=\; a\cosh\frac{x\!-\!b}{a}. \end{matrix}$$ This is the sought form of the equation of the chain curve.\, The constants $$\lambda,\,a,\,b$$ can then be determined for putting the curve to pass through the given points $$P_1$$ and $$P_2$$.