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Math Help - Find the extremal to the functional and discuss whether they provide a max/min

  1. #1
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    Smile Find the extremal to the functional and discuss whether they provide a max/min

    I am having a hard time getting my head around Functionals and Calculus of Variations,

    My question is: Given a functional and using the Euler-Lagrange equation to find an extremal, how do we show that the extremal provides a min/max (if it does)

    The question I am working on is

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

    I found the extremal to be: y(x) = \frac{-1}{4}x^2 +\frac{5}{4}x which I am told is a minimum to the functional problem.

    However I am unsure on what is sufficient to show this, in the notes I have it is shown that:

    J(y+f) = J(y) + \int_{0}^{1}(f')^2dx \geq J(y) where f is continuously differentiable on the interval 0,1 with  y(0)=y(1)=0

    Thanks in advance!
    Last edited by dtwazere; October 6th 2012 at 01:03 PM.
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  2. #2
    GJA
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    Re: Find the extremal to the functional and discuss whether they provide a max/min

    Hi dtwazere.

    Calculus of Variations can seem a little daunting, however hopefully a few comments will make things seem a little less scary.

    1) You can think of the term "functional" in this case as something that sends functions to real numbers. The definition of a functional is actually a little more technical than that (it is a map from a vector space to its underlying scalar field - see Functional (mathematics) - Wikipedia, the free encyclopedia) - in your case the vector space is the space of continuously differentiable functions with y(0)=0 and y(1)=1 and the "field" is the real numbers. However, thinking of a functional as something whose domain is a certain set of functions (in your case continuously differentiable functions with y(0)=0 and y(1)=1) and whose range is the real numbers is good enough for now.

    2) You have actually done most of the hard work already by solving the Euler-Lagrange equation to find a possible minimizer. I will denote your possible minimizer by y_{min}; i.e. y_{min}(x)=-\frac{1}{4}x^{2}+\frac{5}{4}x. Note that my notation is a little dangerous because we haven't actually proved that y_{min} is actually a minimizer of J yet. The last thing we need to note is that the condition f(0)=0=f(1) was left out in the original post where you mentioned J(y+f). The condition f(0)=1=f(1) is essential; y_{min}+f represents a small pertubation of y_{min}, however we don't perturb the endpoints of y_{min}, which is where the condition f(0)=1=f(1) comes from.

    Now we compute

    J(y_{min}+f)=\int_{0}^{1}[(y_{min}+f)']^{2}-(y_{min}+f)dx

    expanding this we get

    J(y_{min}+f)=\int_{0}^{1}(y_{min}')^{2}-y_{min}dx+\int_{0}^{1}2y_{min}'f'dx-\int_{0}^{1}fdx+\int_{0}^{1}(f')^{2}dx

    Which becomes

    J(y_{min}+f)=J(y_{min})+\int_{0}^{1}2y_{min}'f'dx-\int_{0}^{1}fdx+\int_{0}^{1}(f')^{2}dx

    If we integrate by parts on the second term on the RHS we obtain
    2\int_{0}^{1}y_{min}'f'dx=2(y'f |_{0}^{1} -\int_{0}^{1}y_{min}''fdx)

    Now use f(0)=0=f(1) and y_{min}''=-\frac{1}{2} on the previous line to get

    2\int_{0}^{1}y_{min}'f'dx=\int_{0}^{1}fdx

    Using the last line in J(y_{min}+f) we have

    J(y_{min}+f)=J(y_{min})+\int_{0}^{1}fdx-\int_{0}^{1}fdx+\int_{0}^{1}(f')^{2}dx=J(y_{min})+  \int_{0}^{1}(f')^{2}dx\geq J(y_{min})

    This proves that y_{min} really is a minimizer of J for the class of functions that you're considering.

    Does this help? Let me know if anything is unclear.

    Good luck!
    Last edited by GJA; October 11th 2012 at 06:08 PM.
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  3. #3
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    Re: Find the extremal to the functional and discuss whether they provide a max/min

    This really helps, I think I understand most of what you put, thanks a lot!
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