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Math Help - Calculus of variations - showing extremal is minimum

  1. #1
    Newbie Naranja's Avatar
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    Smile Calculus of variations - showing extremal is minimum

    Hi,

    J[y] = \int_0^1{e^y(y')^2 dx
    y(0) = 1, y(1) = 2ln(2).

    I need to find the extremal and show whether it provides a maximum or minimum.

    In this case I only have problems with the second part. I have the extremal as y(x) = 2ln(x(2- \sqrt{e})+ \sqrt{e}).

    To find whether its a maximum or minimum the process I follow is; consider v(x) such that v(0)=0 and v(1)=0, and find the sign of:

    J[y+v] - J[y] = \int_0^1{e^{y+v}((y+v)')^2 dx - \int_0^1{e^y(y')^2 dx.

    I have that this should come out as positive in my notes and thus y provides a minimum - but I don't know how to get there. Squaring out the J[y+v] integral bracket gives me terms I don't know how to integrate, and integration by parts has proved unable to alleviate this problem. I fear I may be going wrong with my approach. Does all of what I've written make sense? Can someone offer any pointers?

    Thanks!
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    Re: Calculus of variations - showing extremal is minimum

    Hey Naranja.

    Subtituting u = y + v gives e^u * u' for J[v+y]. You can then integrate by parts to get e^u * u which is solve-able. The limits will be y(0) + v(0) to y(1) + v(1) for this transformation (since we are using u(x) = y(x)+ v(x)).
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    Newbie Naranja's Avatar
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    Re: Calculus of variations - showing extremal is minimum

    Thanks for your reply,

    I should make clear though that I am not actually looking to do any integration here. y and v are functions of x - I can't integrate these with respect to x. I am just looking to manipulate the entire expression to show that it is positive. I.e. by cancelling as many terms as possible.
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    Re: Calculus of variations - showing extremal is minimum

    You can evaluate the integral if its in terms of u since you are in u-space rather than x-space (as is done by doing a substitution). [Remember that the point of the substitution is to go from x to u(x) just like we usually go from say x to u in a normal substitution).
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    Newbie Naranja's Avatar
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    Re: Calculus of variations - showing extremal is minimum

    Hi,

    As I said in the previous post - I am not looking to do any integration. That may be an option but it is not the intended route for this problem, and not the method I am required to learn. This is purely manipulation of the integral to show it must be positive.
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    Re: Calculus of variations - showing extremal is minimum

    OK then no worries: hope you get what you were looking for!
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