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Math Help - Find the extremal of this functional with given b.c.s

  1. #1
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    Find the extremal of this functional with given b.c.s

    I've used Euler-Lagrange and can't seem to get the right answer, please help if you can.
    Determine the extremal for the functional
    \int_{0}^{1}(xy+y^2-2y^2y') dx
    with y(0)=0, y(1)=2

    Using Euler-Lagrange I get,
    \frac{\partial f}{\partial y}+\frac{\mathrm{d}}{\mathrm{d} x}(\frac{\partial f}{\partial y'})=x+2y-4yy'+4yy'=x+2y=0
    but that isn't consistent with the given b.c.s

    Please help!
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  2. #2
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    Re: Find the extremal of this functional with given b.c.s

    Quote Originally Posted by featherbox View Post
    I've used Euler-Lagrange and can't seem to get the right answer, please help if you can.
    Determine the extremal for the functional
    \int_{0}^{1}(xy+y^2-2y^2y') dx
    with y(0)=0, y(1)=2

    Using Euler-Lagrange I get,
    \frac{\partial f}{\partial y}+\frac{\mathrm{d}}{\mathrm{d} x}(\frac{\partial f}{\partial y'})
    Actually, this should be

    \frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'},

    but you seem to have computed the correct expression here:

    =x+2y-4yy'+4yy'=x+2y=0.
    On the face of it, it doesn't surprise me that the term with y' drops out. After all,

    \int_{0}^{1}(xy+y^{2}-2y^{2}y')\,dx=\int_{0}^{1}(xy+y^{2})\,dx-2\int_{0}^{1}y^{2}y'\,dx

    =\int_{0}^{1}(xy+y^{2})\,dx-2\int_{0}^{2}y^{2}\,dy=\int_{0}^{1}(xy+y^{2})\,dx-\frac{16}{3}.

    So the problem reduces down to finding the extremal of

    \int_{0}^{1}(xy+y^{2})\,dx, subject to the boundary conditions. Since there is now no y' term, the Euler-Lagrange equation simplifies down to setting

    \frac{\partial L}{\partial y}=0, which implies

    x+2y=0, or y=-x/2, as before. And, as you've noted, this function does not satisfy the boundary conditions.

    Question: what is the domain of functions over which you're searching for a solution? Continuous? Differentiable? (I would assume probably differentiable, since you have a y' in the integrand; however, you might be interpreting that derivative in a weak sense, or in some other similarly exotic fashion.)

    If you require a differentiable function as your solution, then I would say your problem has no solution.
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