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The Catenary as an Isoperimetric Problem

April 9th, 2009 |

Author: Merrin

An isoperimetric problem is a calculus of variations problem with a definite

integral as a constraint. An isoperimetric problem is to minimize

$J=\int _{x_{1} }^{x_{2} }f(x,y,\dot{y})dx$

subject to the constraint that

$L=\int _{x_{1} }^{x_{2} }g(x,y,\dot{y})dx$

To solve this type of problem we apply the Euler Lagrange equations to

$h=f-\lambda g$

The constant lambda is called the Lagrange multiplier. A problem that fits this mould is the hanging chain problem. Two poles a distance 2X apart which are each of height H, support a chain of length L. Of course the length of the chains has to be bigger than 2x. We can write a constraint equation for the length of the chain.

$L=\int_{x_1}^{x_2}g(x,y,\dot{y})dx=\int _{-X}^{X}\sqrt{1+\dot{y}^{2} } dx$

The function we want to minimize is the center of gravity of the chain.

$J=\int_{x_1}^{x_2}f(x,y,\dot{y})dx=\int _{-X}^{X}mgyds =\int _{-X}^{X}mgy\sqrt{1+\dot{y}^{2} } dy$

The constant in front doesn’t change anything so we can just drop that.

To find the differential equation we need to solve, we could apply the Euler Lagrange equations to

$\cfrac{d}{dx} \cfrac{\partial }{\partial \dot{y}} (f-\lambda g)-\cfrac{\partial (f-\lambda g)}{\partial y} =0$

Since the functions depend on y we have hit a small obstacle. The

usual approach to reverse the integration variables also hits an obstacle.

The new functional integrand wouldn’t pass the vertical line test so it would

have to be split into pieces. Luckily we have the Beltrami identity to give

us the correct first integral anyway. Verify that the Beltrami identity gives the following differential equation.

$\cfrac{(y-\lambda )\dot{x}}{\sqrt{1+\dot{x}^{2} } } =C$

This is a separable differential equation.

$dx=\cfrac{Cdy}{\sqrt{(y-\lambda)^{2} -C^{2} } }$

Integrating once and using an integration constant of -B

$x-B=C\text{arccosh}\,\left(\cfrac{y-\lambda }{C} \right)\quad \text{or}\quad \, y=\lambda +C\cosh \left(\cfrac{x-B}{C} \right)$

Now our task is to find the constants lambda, B, C so we can plot the curve. We have a number of constraints such as the endpoints of the chain and the length of the chain. We can first plug in at the endpoints of the chains (-x,H) and (x,H).

${H=\lambda +C\cosh \left(\cfrac{X-B}{C} \right)} \\ {H=\lambda +C\cosh \left(\cfrac{-X-B}{C} \right)}$

These equations imply that B = 0. So we are down to

$H=\lambda +C\cosh \left(\cfrac{X}{C} \right)$

With the equation of the chain being

$y=\lambda +C\cosh (x/C)$

Now we will use the constraint on the length of the chain.

${L} = {\int _{-X}^{X}\sqrt{1+y'^{2} } dx =\int _{-X}^{X}\sqrt{1+\sinh ^{2} (x/C)} dx }$

Let the variable

$\gamma =\cfrac{1}{C}$

then we have

$L\gamma =2\sinh (X\gamma )$

This equation can be solved numerically, graphically, or with Newton’s method. The goal is to eliminate the last variable from the equation.

$L\gamma _{0} =2\sinh (X\gamma _{0} )$

We have for our Lagrange multiplier that

$\lambda =H-\cfrac{1}{\gamma _{0} } \cosh \left(X\gamma _{0} \right)$

So our full solution is then

$y=\lambda +C\cosh (x/C)=\lambda +\cfrac{1}{\gamma _{0} } \cosh(x\gamma _{0} )y(x)$

$y==H+\cfrac{1}{\gamma _{0} } \left[\cosh (x\gamma _{0} )-\cosh \left(\gamma _{0} \right)\right]$

The equation for the lowest point of the catenary is

$y(0)=H+\cfrac{1}{\gamma _{0} } \left[1-\cosh \left(X\gamma _{0} \right)</p> <p style=$ \right]” style=”vertical-align:-20%;” class=”tex” alt=”y(0)=H+\cfrac{1}{\gamma _{0} } \left[1-\cosh \left(X\gamma _{0} \right)