Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!


share|cite|improve this question
up vote 6 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.

share|cite|improve this answer
@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

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

share|cite|improve this answer
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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.