Get the equation catenary using variational method

1. Oct 14, 2009

kof9595995

I tried to solve the equation of catenary by variational method the other day. The integral we want to minimize is the potential energy:
$$U = \int_{{x_2}}^{{x_1}} {\rho gy\sqrt {1 + y{'^2}} } dx$$
Then I got stuck at the constraint problem, and in this book,page55:http://books.google.com.sg/books?id...epage&q=catenary, variational method&f=false"
It used the fact that the total length of the rope is constant:
$$l = \int_{{x_2}}^{{x_1}} {\sqrt {1 + y{'^2}} } dx$$
Then it just made use of lagrange multipliers,change the integral to
$$U + \lambda l = \int_{{x_1}}^{{x_2}} {(\rho gy\sqrt {1 + y{'^2}} } + \lambda \sqrt {1 + y{'^2}} )dx$$
and minimized it.
I got confused here, when I learnt lagrange multipliers in Lagrangian mechanics, if we want to minimize the action integral
$$I = \int_{{t_1}}^{{t_2}} L dt$$
with a constraint g(x,y...)=constant， we change the integral to
$$I = \int_{{t_1}}^{{t_2}} {(L} + \lambda g)dt$$,
e.g.,the constraint's equation goes inside the integral you want to minimize, so why not in this catenary problem change the integral into
$${U^*} = \int_{{x_1}}^{{x_2}} {(\rho gy\sqrt {1 + y{'^2}} } + \lambda l)dx = \int_{{x_1}}^{{x_2}} {(\rho gy\sqrt {1 + y{'^2}} } + \lambda \int_{{x_1}}^{{x_2}} {\sqrt {1 + y{'^2}} dx} )dx$$
And if I do it in this way, I can't see how to solve it.
(edit: typo in the title: equation of catenary )

Last edited by a moderator: Apr 24, 2017
2. Oct 14, 2009

lstellyl

It looks like you got confused when adding the lagrange multiplier...

your third line in your post is correct. you must integrate with respect to dx of the lagrangian = L+lambda*g where the x in your constraint is the same x that you are integrating over.

in essence, you simply apply the euler-lagrange equations to the full lagrangian (with the constraint). You will arrive with a second order (or in this case, first order with a constant if you apply the second form of the euler-lagrange equations) differential equation with a solution that has two constants of integration, and one lagrange multiplier.

to solve for these three constants you must use the original constraint equation (g=constant) as well as your two initial conditions

3. Oct 14, 2009

kof9595995

Em, I still don't get it. I tried to use varitional method by anology of what I learnt in classical mechanics, If you want to minimize the integral with a constraint g=constant, then g must go into the integrand, but in this case isn't $$g= \int_{{x_2}}^{{x_1}} {\sqrt {1 + y{single-quote^2}} } dx$$ ? So why don't we put this integral as part of the integrand of U?
(Just in case of confusion, note g here is the equation of constrant not the acceleration constant)

4. Oct 14, 2009

kof9595995

Seems that the latex output is messed up, the whole ysingle-quote thing is just y'.

5. Oct 15, 2009

lstellyl

so let's think about it this way:

if we were to do this:

$${U^*} = \int_{{x_1}}^{{x_2}} {(\rho gy\sqrt {1 + y{\single-quote^2}} } + \lambda l)dx = \int_{{x_1}}^{{x_2}} {(\rho gy\sqrt {1 + y{single-quote^2}} } + \lambda \int_{{x_1}}^{{x_2}} {\sqrt {1 + y{single-quote^2}} dx} )dx$$

then, all we are really doing is adding a constant to our integral, which does nothing for us.

the way i think about it is... we are minimizing the action (integral of L), but we have a constraint... what we want to do is add something to the integrand which we know will integrate to 0, and then multiply it by a constant (your lagrange multiplier, lambda) so that when we minimize the functional, we are still minimizing the action, just requiring that our constraint is 0 when we do it.

as a result, since this is what is called a global constraint (as opposed to a local constraint), we simply add the two together (since the limits of integration are the same), and then subtract by the constant that we require g to be (this ends up vanishing when we minimize it, since it is just a constant)

does that help at all? I'm sorry if it doesn't I will write up something a little more rigorous with latex and all if i finish my work anytime soon...

edit: yeah, wow.... something is wrong with the latex output

Last edited: Oct 15, 2009
6. Oct 15, 2009

kof9595995

I can see your point here, but isn't what you said still true if I put it in the intergrand?
I've no idea what a global constraint is, but I'll google it.
I believe we are making progress, and don't be sorry man, you are helping people.

7. Sep 28, 2010

dnquark

I don't understand why Lagrange multipliers are necessary. Why not take OP's first line and plug into Euler-Lagrange equations? You will end up getting a nonlinear 2nd order ODE that I don't know how to solve off the top of my head, but you can easily confirm that catenary is a solution.