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Math Help - Finding the extremal for a functional

  1. #1
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    Finding the extremal for a functional

    Find the extremal for

     \int_0^T (\dot{x}^2 + 2x\dot{x} + 2x^2) dt\,when\,x(0)=1\,and\,T=2 (i.e\,x(2)\,free)

    Now I have done most of the question as follows

    \frac{\partial f}{\partial x} - \frac{d}{dt}*\frac{\partial f}{\partial \dot{x}} = 2\dot{x} + 4x - \frac{d}{dt}(2\dot{x}+2x) = 0

    which cancels down to

     -2\ddot{x}+4x = 0

    then when you solve the ODE you get

     x(t) = Ae^{sqrt(2)t} + Be^{-sqrt(2)t}

    then subbing in the first bound of  x(0)=1 we get  A+B=1

    I believe I have done this part correctly

    Now, I have got no clue how to do the other bound. Can somebody please show me how to do it
    Last edited by whiteboard; September 6th 2011 at 07:25 PM. Reason: fix code
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  2. #2
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    Re: Finding the extremal for a functional

    You now have the form of x(t). Why not sub. into the integral and integrate?
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