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Math Help - Solving an Euler Equation (Calculus of variations)...

  1. #1
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    Solving an Euler Equation (Calculus of variations)...

    I need to solve the Euler equation to make the following integral stationary:
    \int ( y'^2+\sqrt{y}) dx
    so I happily assume  F = y'^2+\sqrt{y} and find the partial derivative F wrt y' and F wrt y, put both into the Euler equation:
    \frac{d}{dx}(2y') -\frac{1}{2\sqrt{y}} = 0

    Route 1: simple rearrangement yield:
    y''- \frac{1}{4}y^{-\frac{1}{2}} = 0

    Route 2: intergate both side by x
    \int \sqrt{y}  dy' = \frac{x}{4}

    Then I'm stuck in both.... there is probably a very simple way of getting the solution out, but my brain just can't turn the concer~ or I've already made a mistake early in my calculations?

    Thanks for all the help!
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  2. #2
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    Re: Solving an Euler Equation (Calculus of variations)...

    Quote Originally Posted by QDsolarX View Post
    I need to solve the Euler equation to make the following integral stationary:
    \int ( y'^2+\sqrt{y}) dx
    so I happily assume  F = y'^2+\sqrt{y} and find the partial derivative F wrt y' and F wrt y, put both into the Euler equation:
    \frac{d}{dx}(2y') -\frac{1}{2\sqrt{y}} = 0

    Route 1: simple rearrangement yield:
    y''- \frac{1}{4}y^{-\frac{1}{2}} = 0

    Route 2: intergate both side by x
    \int \sqrt{y}  dy' = \frac{x}{4}
    This does not strike me as valid. You were good up to the point where you wrote down your second-order DE. I would go this route: whenever you have a DE of the form

    y''=f(y),

    you can multiply through by y' and integrate at least once immediately. In your case, you get

    y''y'=\frac{1}{4}\,y^{-1/2}y'

    \frac{(y')^{2}}{2}=\frac{\sqrt{y}}{2}+C_{1}.

    Now you have a first-order DE. Can you continue from here?

    Then I'm stuck in both.... there is probably a very simple way of getting the solution out, but my brain just can't turn the concer~ or I've already made a mistake early in my calculations?

    Thanks for all the help!
    Last edited by Ackbeet; June 28th 2011 at 07:40 AM.
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