How to go about solving this first-order nonlinear differential equation?

  • #1

Main Question or Discussion Point

I saw this post at stackexchange:
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.
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.

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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Answers and Replies

  • #2
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[itex]\frac{\alpha}{2\pi}=\frac{\eta \quad \dot{\eta}^2}{\sqrt{1+\dot{\eta}^2}}-\sqrt{1+\dot{\eta}^2}[/itex]
That should be [tex]
\frac{\alpha}{2\pi}=\frac{\eta \dot{\eta}^2}{\sqrt{1+\dot{\eta}^2}}-\eta\sqrt{1+\dot{\eta}^2} = \eta [\frac{\quad \dot{\eta}^2}{\sqrt{1+\dot{\eta}^2}}-\sqrt{1+\dot{\eta}^2}]
[/tex]

Then you square the equation: [tex]
(\frac{\alpha}{2\pi})^2= \eta^2 [\frac{\quad \dot{\eta}^4}{1+\dot{\eta}^2} -2 \dot{\eta}^2 + 1 + \dot{\eta}^2] = \eta^2 [\frac{\quad \dot{\eta}^4}{1+\dot{\eta}^2} -\dot{\eta}^2 + 1]

\\ (\frac{\alpha}{2\pi})^2= \eta^2 \frac{\quad \dot{\eta}^4 -\dot{\eta}^2(1+\dot{\eta}^2) + (1+\dot{\eta}^2)} {1+\dot{\eta}^2}
= \eta^2 \frac{\quad \dot{\eta}^4 -\dot{\eta}^2 - \dot{\eta}^4 + 1+\dot{\eta}^2} {1+\dot{\eta}^2}
= \eta^2 \frac{1} {1+\dot{\eta}^2}[/tex]

I assume you can continue from here.
 
  • #3
Thanks a lot! That made the problem a lot simpler.

I can't believe I didn't carry that [itex]\eta[/itex] over.
 
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