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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!

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    In the question [Treacherous Euler-Lagrange equation](http://math.stackexchange.com/questions/57407/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](http://math.stackexchange.com/questions/57994/resolved-differential-equation/58154#58154) 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.2011-08-18
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    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 $y^{(n)}(0)$.2011-08-18
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    @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 :)2011-08-18
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    @Alexander Thumm: Thanks, that is an interesting approach!2011-08-18
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    @scoobs: Yes, but it is also very ad hoc and computationally intensive.2011-08-18
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    As I told you there, the general solution requires those "complicated" beasties; but, if the initial conditions are "nice", the solutions *might* be elementary... if this is for school, and not for real applications, then I would presume the one who assigned you the question most likely did not intend you to invoke Jacobi's functions...2011-08-18
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    @J.M.: Thanks again! Your reference book was a nice read :)2011-08-18

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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!2011-08-18
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    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.2011-08-18
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    Hmm, on second thoughts, how would the constant solutions satisfy the limits at $\pm \infty$ which are not equal?2011-08-18
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    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
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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 $\pm\int\frac{dy}{\sqrt{C_1-2\cos y}}=x+C_2$.