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If I have an Euler-Lagrange equation: $(y')^2 = 2 (1-\cos(y))$ where $y$ is a function of $x$ subjected to boundary conditions $y(x) \to 0$ as $x \to -\infty$ and $y(x) \to 2\pi$ as $x \to +\infty$, how might I find all its solutions?

I can't seem to directly integrate the equation and sub in the conditions... Please help!

Thanks.

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up vote 5 down vote accepted

Complementary to solutions of joriki and Sivaram you could differentiate your equation once to get

$$2 y''(x) y'(x) = 2 y'(x) \sin y(x)$$

which implies $y''(x) = \sin y(x)$. After substitution $y(x)=\pi - \theta(x)$ this translates into $\theta''(x) = -\sin \theta(x)$ which is the pendulum equation. Your boundary condition require that $\lim\limits_{x\to-\infty} \theta(x) = \pi$ and $\lim\limits_{x\to+\infty} \theta(x) = -\pi$. Hence the solution is not periodic.

This trajectory is described by the Gudermannian function.

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@Sasha: Thanks! I have never met the pendulum equation before... So what are "all the solutions" to the problem? – scoobs Aug 15 '11 at 4:35
    
Generally @scoobs, you'll need elliptic integrals and elliptic functions to solve the problem. The Gudermannian pops up since it's a special case of the Jacobi amplitude function: $\mathrm{am}(u\mid 1)=\mathrm{gd}(u)$. – J. M. Aug 15 '11 at 4:39
    
In any event, this is a bit dated as a reference, but you would want to see the first chapter where the pendulum equation is dealt with. – J. M. Aug 15 '11 at 4:44
    
@J.M.: Thanks! Hmm... elliptic integrals/functions? They are alien to me too... Would these particular boundary conditions somehow simplify the general proceedure to allow a special simple(r) case? – scoobs Aug 15 '11 at 4:46
    
@scoobs: The Gudermannian is pretty elementary :) The thing is that the elliptic integrals and elliptic functions become elementary when a certain quantity called the parameter takes on either of the values 0 or 1. In any event, you really might want to try looking at the book, it should be readable even if you haven't dealt with these beasties previously. – J. M. Aug 15 '11 at 4:50

$1-\cos(y) = 2 \sin^2 \left( \frac{y}{2} \right)$. Hence, $y'^2 = 4 \sin^2 \left( \frac{y}{2} \right) \implies y' = \pm 2 \sin \left( \frac{y}{2} \right)$

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Use $1-\cos y=1-(\cos^2\frac y2-\sin^2\frac y2)=2\sin^2\frac y2$.

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Shouldn't that be $2\sin^2\frac{y}{2}$? – J. M. Aug 14 '11 at 14:16
    
@J.M.: Indeed. Corrected. – joriki Aug 14 '11 at 14:19

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