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Math Help - Calculus of variations

  1. #1
    Super Member Showcase_22's Avatar
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    Calculus of variations

    Write down the Euler-Lagrange equations for critical points of the functional I(y)=\int_{x_1}^{x_2}y^2+y'y'''-(y'')^2~dx
    I'm finding this odd since the only functions i've looked at are of the form g(x,y,y') not f(y',y'',y''').

    I only thought the Euler-Lagrange equations were defined the the first type of function, not the second.

    Would I do it like this:

    \frac{\partial f}{\partial y'''}=y'

    \frac{\partial f}{\partial y''}=2(y'')

    So the E-L equation is 2(y'')-\frac{d}{dy'}(y')=0

    Which becomes 2y''-1=0
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    I'm finding this odd since the only functions i've looked at are of the form g(x,y,y') not f(y',y'',y''').

    I only thought the Euler-Lagrange equations were defined the the first type of function, not the second.

    Would I do it like this:

    \frac{\partial f}{\partial y'''}=y'

    \frac{\partial f}{\partial y''}=2(y'')

    So the E-L equation is 2(y'')-\frac{d}{dy'}(y')=0

    Which becomes 2y''-1=0
    For higher order Lagrangians, the Euler-Lagrange equation is

     <br />
\frac{\partial L}{\partial y} - \frac{d}{d x}\left(\frac{\partial L}{\partial y'}\right) + \frac{d^2}{d x^2} \left(\frac{\partial L}{\partial y''} \right) - \cdots + (-1)^n \frac{d^n}{d x^n}\left(\frac{\partial L}{\partial y^{(n)}}\right) = 0.<br />
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