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Minimization with cos

  1. Mar 12, 2015 #1
    I'm trying to find a function for x in [0, L] that minimizes this:

    [tex] \int_0^{L} A \phi(x) \frac{ d \phi(x) }{dx} + B cos(\phi(x))\ d\mbox{x} [/tex]

    For real (given) positve numbers A and B.

    with
    [itex] \phi(0) = 0 [/itex]
    [itex] \phi(x) [/itex] is an increasing positve function.

    Can somebody point me in the right direction?
     
    Last edited: Mar 12, 2015
  2. jcsd
  3. Mar 12, 2015 #2

    jfizzix

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    Sounds like a calculus of variations problem.

    I'm no expert, but the principal thing to do would be to find the Euler-Lagrange equation for [itex]\phi(x)[/itex]. The solution (or solutions) to that equation will be the function which minimizes that integral.

    I don't exactly know how to do this, but if you look at the Wikipedia article on the calculus of variations, and go down to the section on the Euler-Lagrange equation, you should find some useful tips there.
     
  4. Mar 12, 2015 #3
    Yes I get the zero function then but that is not what I am looking for.
     
  5. Mar 12, 2015 #4

    jfizzix

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    the Euler-Lagrange equation only has zero as a solution?
     
  6. Mar 12, 2015 #5
    Yes I get the zero function then but that is not what I am looking for.

    [tex] \frac { \partial L } { \partial \phi } = A \frac{d \phi}{dx} -C sin ( \phi(x)) [/tex]

    [tex] \frac{d}{dx } \left( \frac { \partial L } { \partial \phi' } \right) = A \frac{d \phi}{dx} [/tex]

    [tex] \frac { \partial L } { \partial \phi } - \frac{d}{dx } \left( \frac { \partial L } { \partial \phi' } \right) = C sin ( \phi(x)) = 0 [/tex]
     
  7. Mar 12, 2015 #6

    jfizzix

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    well that sure is interesting...

    How do you know that the zero function isn't actually the solution you're looking for?
     
  8. Mar 12, 2015 #7

    jfizzix

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    Also, since you're integrand doesn't depend explicitly on [itex]x[/itex], you can try the simplified Euler Lagrange equation instead:
    [itex]L-\frac{\partial L}{\partial \phi}\frac{d \phi}{d x} = const[/itex].
    Here, that would translate to
    [itex]A \phi \phi' + B Cos(\phi) -(A \phi' - B Sin(\phi))\phi'= const[/itex]
    or rather
    [itex]A (\phi-\phi') \phi' + B \big(Cos(\phi) -\frac{d}{dx}( Cos(\phi)\big)= const[/itex]

    I don't know how to solve this off the top of my head, but this equation probably admits more than zero as a solution.
     
  9. Mar 12, 2015 #8
    Yes, you are right but I do not see how this is going to help me. This might be easier than an integral but this a non linear DE and I do not think an analytical solution exists.
     
  10. Mar 12, 2015 #9

    pasmith

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    Since [itex]\phi\phi' = \frac12 (\phi^2)'[/itex] you are trying to minimize [tex]
    \frac{A\phi(L)^2}{2} + \int_0^L B\cos(\phi(x))\,dx.[/tex] The Euler-Lagrange equation is then [tex]B\sin(\phi) = 0[/tex] so that indeed [itex]\phi(x) = 0[/itex].

    Do you have reason to believe that any other (differentiable) solutions exist? Also, although [itex]\phi(x) = 0[/itex] is not a strictly positive strictly increasing function, it is a positive increasing function.

    (For weak solutions, you can take [tex]
    \phi(x) = \begin{cases} 0 & 0 \leq x < x_0 \\
    (2n+1)\pi & x_0 \leq x \leq L \end{cases}[/tex] for some positive integer [itex]n[/itex] and if [itex]0 < 2L - \frac{A(2n+1)^2\pi^2}{2B} < L[/itex] then for any [itex]0 < x_0 < 2L - \frac{A(2n+1)^2\pi^2}{2B}[/itex] you have [tex]
    \frac{A\phi(L)^2}{2} + \int_0^L B \cos(\phi(x))\,dx = \frac{A(2n+1)^2\pi^2}{2} - B(L - x_0) < BL,[/tex] the last being the value attained when [itex]\phi[/itex] is zero.)
     
  11. Mar 13, 2015 #10
    I made a mistake with the model. I will create a new thread.
     
    Last edited: Mar 13, 2015
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