Get the equation catenary using variational method

[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 learned 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 )

 

[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

 

[kof9595995]

Em, I still don't get it. I tried to use varitional method by anology of what I learned 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)

 

[kof9595995]

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

 

[lstellyl]

so let's think about it this way:

if we were to do this:

<br /> {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<br />

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

 

[kof9595995]

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.

I can see your point here, but isn't what you said still true if I put it in the intergrand?

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)

I've no idea what a global constraint is, but I'll google it.

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...

I believe we are making progress, and don't be sorry man, you are helping people.

 

[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.