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[parent] equation of catenary via calculus of variations (Derivation)

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

to minimise$\displaystyle \quad \int_{P_1}^{P_2}\!y\,ds$ (1)
under the constraint
$\displaystyle \int_{P_1}^{P_2}\!ds \;=\; l,$ (2)
where both the path integrals are taken along some curve $c$.  Using a Lagrange multiplier $\lambda$, the task changes to a free problem
$\displaystyle \int_{P_1}^{P_2}\!(y\!-\!\lambda)\,ds \;=\; \int_{x_1}^{x_2}(y\!-\!\lambda)\sqrt{1\!+\!y'^2}\,\vert dx\vert \;=\;$   min$\displaystyle !$ (3)
(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

$\displaystyle (y\!-\!\lambda)\sqrt{1\!+\!y'^2}-y'\!\cdot\!(y\!-\!\lambda)\!\cdo... ...{\sqrt{1\!+\!y'^2}} \;\equiv\; \frac{y\!-\!\lambda}{\sqrt{1\!+\!y'^2}} \;=\; a,$
where $a$ is a constant of integration.  After solving this equation for the derivative $y'$ and separation of variables, we get

$\displaystyle \pm\frac{dy}{\sqrt{(y\!-\!\lambda)^2\!-\!a^2}} \;=\; \frac{dx}{a}$
which may become clearer by notating  $y\!-\!\lambda := u$;  then by integrating

$\displaystyle \pm\frac{du}{\sqrt{u^2\!-\!a^2}} \;=\; \frac{dx}{a}$
we choose the new constant of integration $b$ such that  $x = b$  when  $u = a$:

$\displaystyle \pm\int_a^u\frac{du}{\sqrt{u^2\!-\!a^2}} \;=\; \int_b^x\frac{dx}{a}$
We can write two equivalent results

$\displaystyle \ln\frac{u\!+\!\sqrt{u^2\!-\!a^2}}{a} \;=\; +\frac{x\!-\!b}{a}, \qquad \ln\frac{u\!-\!\sqrt{u^2\!-\!a^2}}{a} \;=\; -\frac{x\!-\!b}{a},$
i.e.

$\displaystyle \frac{u\!+\!\sqrt{u^2\!-\!a^2}}{a} \;=\; e^{+\frac{x-b}{a}}, \qquad \frac{u\!-\!\sqrt{u^2\!-\!a^2}}{a} \;=\; e^{-\frac{x-b}{a}}.$
Adding these allows to eliminate the square roots and to obtain

$\displaystyle u \;=\; \frac{a}{2}\!\left(e^{\frac{x-b}{a}}+e^{-\frac{x-b}{a}}\right),$
or
$\displaystyle y\!-\!\lambda \;=\; a\cosh\frac{x\!-\!b}{a}.$ (4)
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 $P_2$.

Bibliography

1
E. LINDELÖF: Differentiali- ja integralilasku ja sen sovellutukset IV. Johdatus variatiolaskuun.  Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1946).



"equation of catenary via calculus of variations" is owned by pahio.

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Cross-references: chain curve, square, separation of variables, identity, centre of mass

This is version 1 of equation of catenary via calculus of variations, born on 2010-04-18.
Object id is 852, canonical name is EquationOfCatenaryViaCalculusOfVariations.
Accessed 508 times total.

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Physics Classification: 02.30.Xx (Calculus of variations)
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