Talk:PlanetPhysics/Equation of Catenary Via Calculus of Variations

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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: equation of catenary via calculus of variations %%% Primary Category Code: 02.30.Xx %%% Filename: EquationOfCatenaryViaCalculusOfVariations.tex %%% Version: 1 %%% Owner: pahio %%% Author(s): pahio %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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Using the mechanical principle that the \htmladdnormallink{centre of mass}{http://planetphysics.us/encyclopedia/CenterOfGravity.html} 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{align} \mbox{to minimise} \quad \int_{P_1}^{P_2}\!y\,ds \end{align} under the constraint \begin{align} \int_{P_1}^{P_2}\!ds \;=\; l, \end{align} where both the path integrals are taken along some curve $c$.\, Using a Lagrange multiplier $\lambda$, the task changes to a free problem \begin{align} \int_{P_1}^{P_2}\!(y\!-\!\lambda)\,ds \;=\; \int_{x_1}^{x_2}(y\!-\!\lambda)\sqrt{1\!+\!y'^2}\,|dx| \;=\; \mbox{min}! \end{align} (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 \htmladdnormallink{identity}{http://planetphysics.us/encyclopedia/Cod.html} $$(y\!-\!\lambda)\sqrt{1\!+\!y'^2}-y'\!\cdot\!(y\!-\!\lambda)\!\cdot\!\frac{y'}{\sqrt{1\!+\!y'^2}} \;\equiv\; \frac{y\!-\!\lambda}{\sqrt{1\!+\!y'^2}} \;=\; a,$$ where $a$ is a constant of integration.\, After solving this equation for the derivative $y'$ and \htmladdnormallink{separation of variables}{http://planetphysics.us/encyclopedia/SeparationOfVariables.html}, 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 \htmladdnormallink{square}{http://planetphysics.us/encyclopedia/PiecewiseLinear.html} roots and to obtain $$u \;=\; \frac{a}{2}\!\left(e^{\frac{x-b}{a}}+e^{-\frac{x-b}{a}}\right),$$ or \begin{align} y\!-\!\lambda \;=\; a\cosh\frac{x\!-\!b}{a}. \end{align} This is the sought form of the equation of the \htmladdnormallink{chain curve}{http://planetphysics.us/encyclopedia/Catenary.html}.\, The constants $\lambda,\,a,\,b$ can then be determined for putting the curve to pass through the given points $P_1$ and $P_2$.

\begin{thebibliography}{8} \bibitem{lindelof}{\sc E. Lindel\"of}: {\em Differentiali- ja integralilasku ja sen sovellutukset IV. Johdatus variatiolaskuun}.\, Mercatorin Kirjapaino Osakeyhti\"o, Helsinki (1946). \end{thebibliography}

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