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Math Help - Stationary Point of a functional

  1. #1
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    Stationary Point of a functional

    Okay, I have worked this problem a few times and I end up with a solution that I don't think is correct. Any feedback would be awsome.

    If possible, find the stationary point of the functional

    \mathcal{J}(y)=\int_{0}^{1}(y^2+x^2y')dx, y(0)=0, y(1)=\alpha

    So I use the Euler-Lagrange Equation

    \mathcal{L}_y-\frac{d}{dx}\mathcal{L}_{y'}=0 with

    \mathcal{L}(y,y',x)=y^2+x^2y'

    so \frac{\partial}{\partial y}(y^2+x^2y')=2y

    and \frac{\partial}{\partial y'}(y^2+x^2y')=x^2

    and \frac{d}{dx}\bigg[\frac{\partial}{\partial y'}(y^2+x^2y') \bigg]=\frac{d}{dx}x^2=2x

    So now I end up with

    2y-2x=0 \implies y=x

    I don't think this is correct and it would only be consistant with the boundry values if \alpha=1

    Thanks
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  2. #2
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    That looks correct to me. Why do you think it is wrong? After all, the question says "If possible, find the stationary point", and your answer is that it is not possible, unless α=1.
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    Okay, I have worked this problem a few times and I end up with a solution that I don't think is correct. Any feedback would be awsome.



    If possible, find the stationary point of the functional

    \mathcal{J}(y)=\int_{0}^{1}(y^2+x^2y')dx, y(0)=0, y(1)=\alpha

    So I use the Euler-Lagrange Equation

    \mathcal{L}_y-\frac{d}{dx}\mathcal{L}_{y'}=0 with

    \mathcal{L}(y,y',x)=y^2+x^2y'

    so \frac{\partial}{\partial y}(y^2+x^2y')=2y

    and \frac{\partial}{\partial y'}(y^2+x^2y')=x^2

    and \frac{d}{dx}\bigg[\frac{\partial}{\partial y'}(y^2+x^2y') \bigg]=\frac{d}{dx}x^2=2x

    So now I end up with

    2y-2x=0 \implies y=x

    I don't think this is correct and it would only be consistant with the boundry values if \alpha=1


    Thanks
    Usually one gets differential equations (is that what you were expecting?). Check and make sure the Lagrangian is correct.
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  4. #4
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    I was expecting a differential equatin of some kind. Thank again for the feedback.
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