equation of catenary via calculus of variations

Using the mechanical principle that the centre of mass itself as low as possible, determine the equation of the curve formed by a $l$ 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

 $\displaystyle\mbox{to minimise}\quad\int_{P_{1}}^{P_{2}}\!y\,ds$ (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

 $\displaystyle\int_{P_{1}}^{P_{2}}\!(y\!-\!\lambda)\,ds\;=\;\int_{x_{1}}^{x_{2}% }(y\!-\!\lambda)\sqrt{1\!+\!y^{\prime 2}}\,|dx|\;=\;\mbox{min}!$ (3)

(cf. example of calculus of variations).

The Euler–Lagrange differential equation (http://planetmath.org/EulerLagrangeDifferentialEquation), the necessary condition for (3) to give an extremal $c$, reduces to the Beltrami identity

 $(y\!-\!\lambda)\sqrt{1\!+\!y^{\prime 2}}-y^{\prime}\!\cdot\!(y\!-\!\lambda)\!% \cdot\!\frac{y^{\prime}}{\sqrt{1\!+\!y^{\prime 2}}}\;\equiv\;\frac{y\!-\!% \lambda}{\sqrt{1\!+\!y^{\prime 2}}}\;=\;a,$

where $a$ is a constant of integration.  After solving this equation for the derivative $y^{\prime}$ and 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 equivalent (http://planetmath.org/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

 $\displaystyle y\!-\!\lambda\;=\;a\cosh\frac{x\!-\!b}{a}.$ (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 $P_{2}$.

References

• 1 E. Lindelöf: Differentiali- ja integralilasku ja sen sovellutukset IV. Johdatus variatiolaskuun.  Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1946).
Title equation of catenary via calculus of variations EquationOfCatenaryViaCalculusOfVariations 2013-03-22 19:12:07 2013-03-22 19:12:07 pahio (2872) pahio (2872) 10 pahio (2872) Derivation msc 49K05 msc 49K22 msc 47A60 Catenary CalculusOfVariations LeastSurfaceOfRevolution