0
$\begingroup$

It has been a while since I posted the question Treacherous Euler-Lagrange equation. I have read the suggested text. But I was told that I shouldn't need Jacobi amplitude function and other beasties of the sort to solve the problem due to the non-arbitrary limits/boundary conditions. So I would really appreciate some help in finding the way! Here is a restatement of the problem:

$y^{\prime\prime}(x) = \sin y(x)$ subjected to boundary conditions $\lim_{x\to-\infty} y(x) =0$ $\lim_{x\to+\infty} y(x) = 2\pi$

Thanks in advance!

${}$

  • 0
    @J.M.: Thanks again! Your reference book was a nice read :)2011-08-18

2 Answers 2

5

Oh, well. $ y(x) = \pi + 2 \, \mbox{gd}(x - x_0), $ where gd refers to the Gudermannian function, and $x_0$ is the value of $x$ where $y=\pi.$ So $ y(x) = \pi + 2 \, \arctan \sinh(x - x_0)$ for real values of $x.$

Note that, in your original equation (y')^2 = 2 ( 1 - \cos y) there are constant solutions $y = 0, \; y = \pm 2 \pi,$ or any even multiple of $\pi.$

  • 0
    No, the constant solutions do not fit the limit conditions. The expressions with the $\arctan$ do so. You should graph several of those with different choices of $x_0.$ Also check very, very carefully that $y''(x) = \sin y(x)$ is satisfied, you will learn some things.2011-08-18
2

A possible approach is the following: since the equation doesn't contain $x$, substitute p(y)=y', then y''=p'\cdot y'=p'p. Now we have $\frac{dp}{dy}p=\sin y$, hence $pdp=\sin y\, dy$. By integrating both sides you get (y')^2=p^2=C_1-2\cos y. By isolating y' and integrating, you have $\pm\int\frac{dy}{\sqrt{C_1-2\cos y}}=x+C_2$.