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Calculus of Variations

  1. Jan 21, 2009 #1
    Just did this in class today and was doing a problem to see if I understood it and I'm not sure I did. Thanks for any help


    Solve the Euler equation to make the following integral:

    (integral from x1->x2)

    ∫ [(y')² + y²] dx

    2. Relevant equations

    Euler-Lagrange equation

    ∂F/∂y - d/dx (∂F/∂y') = 0

    3. The attempt at a solution

    Clearly F = (y')² + y²

    In class we had been rearranging the integral so that ∂F/∂y = 0, which made the problem much simpler. I was unable to do this here as there was no apparent way to introduce ds to swap y' for x'. This was my concern as, although I didn't think all equations would be this simple, I thought most would reduce.

    So instead I proceeded as follows:

    ∂F/∂y - d/dx (∂F/∂y') = 2y - d/dx (2y') = 2y - 2y'' = 0 => y'' = y

    And this is easy enough to solve. But I'm concerned I've made a mistake getting there.

    Have I made a mistake up to here or is there a better way to proceed?

  2. jcsd
  3. Jan 21, 2009 #2


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    AFAIK this is the best you can do. You can't eliminate the dF/dy term in general.
  4. Jan 21, 2009 #3
    Cheers. I'm a bit rusty on my calculus, but is there anything wrong with the step:

    d/dx(2y') = 2y''

    My lecturer made note to be careful of the d/dx part, but I don't think that applies here, just want to check.
  5. Jan 21, 2009 #4


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    That's just fine.

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