EquationOfCatenaryViaCalculusOfVariations 2010-04-18 15:07:27 2010-04-18 15:07:27 Derivation introduction to calculus of variations The likely problem is this part from my rewrite: y'(y-\lambda)\frac{y'}{\sqrt{1+y'^2}} make4ht/TeX4ht can be touchier about prime syntax followed by another superscript-ish token. Use braced primes everywhere: % 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} \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 \PMlinkname{Euler--Lagrange differential equation}{EulerLagrangeDifferentialEquation}, 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 \PMlinkname{equivalent}{Equivalent3} results $$\ln\frac{u\!+\!\sqrt{u^2\!-\!a^2}}{a} \;=\; +\frac{x\!-\!b}{a}, \qquad \ln\frac{u\!-\!\sqrt{u^2\!-\!a^2}}{a} \;=\; -\frac{x\!-\!b}{a},$$ 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{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 $\lambda,\,a,\,b$ 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}