What is the Euler-Lagrange equation for:
I got that it was u'' = sin(u) but I cannot solve this nicely and this is a question that I should be able to do.
Well, in general, if is the integrand of your functional , then the Euler-Lagrange equations are
the same as you. To integrate this, I would multiply through by and then integrate once. You might be able to integrate the resulting first-order DE at that point. Make sense?
I understand this, and get:
I have to find all solutions where the limit as x tends to infinity of u is 2pi and as x tends to minus infinity is 0. What can I do here? Clearly I can set the integral out for this equation however I don't think that this integral will have a nice expression.
Yeah, off-hand I'd say you're likely going to end up with an elliptic. (checking...) Yep, you've got yourself an elliptic integral there. Not very pretty.
In the original DE, can you make the small-angle approximation with u? That is, sin(u) is approximately u?
Do you have any initial conditions? Or are you dealing with asymptotic conditions?
The OP has given me a link to the original problem, which I will now type out in full.
I guess one question I have is this: when you are finding all the solutions which satisfy the Euler-Lagrange equation, are you already allowed to assume they are ? If so, I think you can simplify a bit. For a function, I think it's true thatWrite down the Euler-Lagrange equation for the functional
and find all solutions which satisfy and Show that if satisfies and
Deduce that a lower bound for amongst such functions is , and give a first order differential equation which must satisfy in order to realize this lower bound. Show that any solution of this first order equation solves the Euler-Lagrange equation you derived in the first part of the question. Give all the functions satisfying
So, supposing it is allowed to assume that you're restricted to finding functions as solutions, we might be able to get a handle on that first constant of integration.
The first integration yielded
Suppose we take the limit of this equation as On the LHS we'll get , and we'll get on the RHS. Therefore, Now, I think you can see that the integration becomes much more manageable. Can you continue from here?