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## Main Question or Discussion Point

I saw this post at stackexchange:

Here is the simplest form I have managed to get it in:

[itex]\frac{\alpha}{2\pi}=\frac{\eta \quad \dot{\eta}^2}{\sqrt{1+\dot{\eta}^2}}-\sqrt{1+\dot{\eta}^2}[/itex]

Where [itex]\alpha[/itex] is a constant. For the simplification, [itex]\eta = (y-y_0)[/itex] & [itex]λ \equiv -2\pi y_0[/itex] is used from the quote above.

I don't know where to even begin to solve this. Any help would be good.

Thanks.

-edit

Some details that I thought I should add.

Since [itex]h=f+\lambda g≠h(x)[/itex], I used the special case of the E-L equation:

[itex]\dot{y}\frac{\partial h}{\partial \dot{y}}-h=const.[/itex]

But after the substituions to trim it up the special E-L turns into this:

[itex]\dot{\eta}\frac{\partial h}{\partial \dot{\eta}}-h=const.=\alpha[/itex]

Where [itex]h[/itex] now is: [itex]h=2\pi\eta\sqrt{1+\dot{\eta}^2}[/itex]

I ran across this post when trying to solve a homework problem. But I have no idea how he got that solution for that. When I use the Euler-Lagrange, I get this diff eq below.The fact that curve length is fixed amounts to a functional constraint. So we need to use both the Euler-Lagrange equation and Lagrange multipliers. The functional to minimise is [itex]\int^a_{-a}(2\pi y-\lambda)\sqrt{1+y'^2}dx[/itex], where λ is a Lagrange multiplier. For this functional, the first integral of the Euler-Lagrange equation gives a different ODE from what I got above, which has the solution [itex]y(x)=k(\cosh\frac{x}{k}-\cosh\frac{a}{k})[/itex] for some k depending on l, as required.

Here is the simplest form I have managed to get it in:

[itex]\frac{\alpha}{2\pi}=\frac{\eta \quad \dot{\eta}^2}{\sqrt{1+\dot{\eta}^2}}-\sqrt{1+\dot{\eta}^2}[/itex]

Where [itex]\alpha[/itex] is a constant. For the simplification, [itex]\eta = (y-y_0)[/itex] & [itex]λ \equiv -2\pi y_0[/itex] is used from the quote above.

I don't know where to even begin to solve this. Any help would be good.

Thanks.

-edit

Some details that I thought I should add.

Since [itex]h=f+\lambda g≠h(x)[/itex], I used the special case of the E-L equation:

[itex]\dot{y}\frac{\partial h}{\partial \dot{y}}-h=const.[/itex]

But after the substituions to trim it up the special E-L turns into this:

[itex]\dot{\eta}\frac{\partial h}{\partial \dot{\eta}}-h=const.=\alpha[/itex]

Where [itex]h[/itex] now is: [itex]h=2\pi\eta\sqrt{1+\dot{\eta}^2}[/itex]

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