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

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).



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