EquationOfCatenaryViaCalculusOfVariations 2010-04-18 15:07:27 2026-09-05 21:00:24 Derivation introduction to calculus of variations fixe the bad latex but needs a once over % this is the default PlanetPhysics preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Using the mechanical principle that the centre of mass \PMlinkescapetext{places} itself as low as possible, determine the equation of the curve formed by a \PMlinkescapetext{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} \text{to minimize} \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 path integrals are taken along some curve $c$. Using a Lagrange multiplier $\lambda$, the task changes to the free problem \begin{align} \int_{P_1}^{P_2} (y-\lambda)\,ds &= \int_{x_1}^{x_2} (y-\lambda)\sqrt{1+(y^{\prime})^2}\,|dx| \\ &= \text{minimum}. \end{align} The \PMlinkname{Euler--Lagrange differential equation}{EulerLagrangeDifferentialEquation}, the necessary condition for the preceding functional to give an extremal $c$, reduces to the Beltrami identity \[ (y-\lambda)\sqrt{1+(y^{\prime})^2} - y^{\prime}(y-\lambda) \frac{y^{\prime}}{\sqrt{1+(y^{\prime})^2}} \equiv \frac{y-\lambda}{\sqrt{1+(y^{\prime})^2}} = a, \] where $a$ is a constant of integration. After solving this equation for the derivative $y^{\prime}$ and separating variables, we get \[ \pm\frac{dy}{\sqrt{(y-\lambda)^2-a^2}} = \frac{dx}{a}. \] This may become clearer by writing $u := y-\lambda$; then \[ \pm\frac{du}{\sqrt{u^2-a^2}} = \frac{dx}{a}. \] Choose the new constant of integration $b$ such that $x=b$ when $u=a$. Then \[ \pm\int_a^u \frac{du}{\sqrt{u^2-a^2}} = \int_b^x \frac{dx}{a}. \] We can write two \PMlinkname{equivalent}{Equivalent3} results: \[ \ln\frac{u+\sqrt{u^2-a^2}}{a} = \frac{x-b}{a}, \qquad \ln\frac{u-\sqrt{u^2-a^2}}{a} = -\frac{x-b}{a}. \] Thus \[ \frac{u+\sqrt{u^2-a^2}}{a} = e^{(x-b)/a}, \qquad \frac{u-\sqrt{u^2-a^2}}{a} = e^{-(x-b)/a}. \] Adding these equations eliminates the square roots and gives \[ u = \frac{a}{2} \left(e^{(x-b)/a}+e^{-(x-b)/a}\right), \] or \begin{align} y-\lambda = a\cosh\frac{x-b}{a}. \end{align} This is the sought form of the equation of the chain curve. The constants $\lambda$, $a$, and $b$ can then be determined by requiring 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}