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Math Help - Calculus of Variations problem

  1. #1
    Super Member Showcase_22's Avatar
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    Calculus of Variations problem

    Find solutions of the Euler-Lagrange equations for critical points of the following functionals with the given endpoint conditions.

    I(y)=\int_1^2 x(y')^2~dx where y(1)=1 and y(2)=2
    So I decided to use the Euler Lagrange equations:

    \frac{\partial f}{\partial y}=0 and \frac{\partial f}{\partial y'}=2(y')x

    Inputting this into the Euler-Lagrange equations gives:

    -\frac{d}{dx}(2y'x)=0 \ \Rightarrow -2y'=0 \Rightarrow y'=0


    So integrating both sides w.r.t x gives y=A where A is a positive constant.

    From here I was expecting to use the conditions on y to work out what A is. Unfortunately this doesn't work! y can't equal a constant and yet change from 1 to 2 as described in the conditions.

    What's going wrong?
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  2. #2
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    No , we should take total differential not partial ...

    It is

     \frac{d}{dx}( 2xy' ) = 0

    Actually , we obtain 2xy' = c immediately but in other situations , we should be careful that what we are doing is taking total differential .

     \frac{d}{dx}( 2xy' ) = 2 \frac{d}{dx}( xy' )

     = 2 [\frac{\partial }{\partial x}(xy') \frac{dx}{dx} + \frac{\partial }{\partial y'}(xy') \frac{dy'}{dx}]

     = 2 ( y' + x y'' )
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    So I decided to use the Euler Lagrange equations:

    \frac{\partial f}{\partial y}=0 and \frac{\partial f}{\partial y'}=2(y')x

    Inputting this into the Euler-Lagrange equations gives:

    -\frac{d}{dx}(2y'x)=0 \ \Rightarrow -2y'=0 \Rightarrow y'=0


    So integrating both sides w.r.t x gives y=A where A is a positive constant.

    From here I was expecting to use the conditions on y to work out what A is. Unfortunately this doesn't work! y can't equal a constant and yet change from 1 to 2 as described in the conditions.

    What's going wrong?
    Just because a functional has an extremal does not mean that it is actually a minimum or maximum, just like how the first derivative equals zero does not imply you have a maximum there. In fact, the E-L equations can be thought very informally as an infinite dimensional analogue of a gradient.
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