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Calculus of variations with circular boundary conditions

  1. Dec 6, 2015 #1
    The Euler-Lagrange equations give a necessary condition for the action be extremal given some lagrangian which depends on some function to be varied over. The basic form assumes fixed endpoints for the function to be varied over, but we can extend to cases in which one or both endpoints are free to vary by introducing additional transversality conditions. I would like to know what the transversality condition(s) would be in the case that the values at the two boundaries are equal, but otherwise free to vary. The reason for this is that I want to consider functions which map the circle onto the real numbers, and use the calculus of variations approach to maximise an action based on a lagrangian of this function.

    Thanks!
     
  2. jcsd
  3. Dec 6, 2015 #2

    pasmith

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    If you have [tex]
    I[y] = \int_0^a F(y,y')\,dx[/tex] then [tex]
    I[y + \eta] - I[y] = \int_0^a \eta \left(\frac{\partial F}{\partial y} - \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right)\right)\,dx + \left[ \eta \frac{\partial F}{\partial y'} \right]_0^a.[/tex] Now if [itex]y[/itex], [itex]y'[/itex] and the perturbation [itex]\eta[/itex] are all periodic with period [itex]a[/itex] then so is [itex]\eta \frac{\partial F}{\partial y'}[/itex], and the boundary term vanishes.
     
  4. Dec 6, 2015 #3
    Thanks, that was really helpful. However, my lagrangian F depends also explicitly on x. I guess if the dependence of F on x is also periodic with period a then the result should still hold.

    Edit: it seems to me that the assumption of periodicity is suspect here. Unless we impose this as a constraint when performing the optimisation then there is no reason it should hold. Imposing the periodicity constraint is basically what I was trying to achieve by including these boundary conditions.
     
  5. Dec 7, 2015 #4

    pasmith

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    Requiring that [itex]f : [0,a] \to \mathbb{R}[/itex] satisfies [itex]f(0) = f(a)[/itex] is equivalent to requiring that [itex]g : \mathbb{R} \to \mathbb{R}[/itex] is periodic with period [itex]a[/itex] - and it is natural that if [itex]y[/itex] is periodic then only periodic perturbations need to be considered.
     
  6. Dec 7, 2015 #5
    Ok, buy [itex]y[/itex] is a function that we obtain by solving the relevant Euler-Lagrange equations, and we have not included any periodicity constraint in there, so we have no reason to believe that the function [itex]y[/itex] that we obtain should be periodic.
     
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