# Still stuck on treacherous ODE

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 $$\lim_{x\to-\infty} y(x) =0$$ $$\lim_{x\to+\infty} y(x) = 2\pi$$

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In the question Treacherous Euler-Lagrange equation (which is decidedly mild as far as EL equations go), joriki and Sivaram show that the differential equation is equivalent to that of this recent question up to sign, namely $$y'=\pm2\sin(y/2)$$ which I already answered in the linked question. Are you the same account as "distressed" or are you in the same class? I'm this close to voting to close as a duplicate. –  anon Aug 18 '11 at 7:00
The boundary conditions tell you, that $y''(-\infty) = y''(\infty) = 0$. By MVT there is some $x_0 \in \mathbb R$, s.t. $y''(x_0) = 0$, so $y(x_0) = \pi k$ (since the equation does not contain $x$, we can assume $x_0 = 0$). You could now taylorexpand $y$ at $0$, using Faà di Bruno you should be able to prove some explicit formula for $y^{(n)}(0)$. –  Alexander Thumm Aug 18 '11 at 7:54
@anon: I don't think I know "distressed". But I won't be surprised that people from different schools get set similar questions...! Afterall, textbooks and various edu material is used all over the world! Haha, I wouldn't have posted the same question twice esp since "distressed" has already labeled his\her question "resolved"! :) Anyway, thanks for the link to "distressed"'s post. That is very helpful :) –  scoobs Aug 18 '11 at 7:55
@Alexander Thumm: Thanks, that is an interesting approach! –  scoobs Aug 18 '11 at 8:10
@scoobs: Yes, but it is also very ad hoc and computationally intensive. –  Alexander Thumm Aug 18 '11 at 8:14

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 $\pi.$

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Thanks, Will. I haven't thought of the constant solns! –  scoobs Aug 18 '11 at 8:13
That's what tells you to look for horizontal translates, meaning some constant $x_0$ as above. Also the fact that nothing depends explicitly on $x$ itself. So these are all solutions with your limit requirements. –  Will Jagy Aug 18 '11 at 8:19
Hmm, on second thoughts, how would the constant solutions satisfy the limits at $\pm \infty$ which are not equal? –  scoobs Aug 18 '11 at 9:28
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. –  Will Jagy Aug 18 '11 at 19:20

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 $\pm\int\frac{dy}{\sqrt{C_1-2\cos y}}=x+C_2$.

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Thanks, Dennis. –  scoobs Aug 18 '11 at 8:13