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Math Help - Extremal - Is my solution correct?

  1. #1
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    Extremal - Is my solution correct?

    Hi,
    This is my first post here, and I was wondering if someone could see something wrong with my answer to the following question.

    Find the extremal for \displaystyle \ \int_0^T (\dot{x}^2 + 2x \dot{x}) \ dt when x(0) = 1 and T = 2 i.e. x(2) free)

    My Solution (sorry, it goes for a while )

    Let \displaystyle \ f(t, x, \dot{x}) = \dot{x}^2 + 2x \dot{x} . Then the Euler-Lagrange equation is \displaystyle \ \frac{\partial f}{\partial x} - \frac{d}{dx} \Big( \frac{\partial f}{\partial \dot{x}} \Big) = 0

    which implies \displaystyle \ 2 \dot{x} - \frac{d}{dx} \Big( 2 \dot{x} + 2x \Big) = 0

    \displaystyle \ \therefore \ 2 \dot{x} - \ddot{x} + 2 \dot{x} = 0

    \displaystyle \ \therefore \ 2 \ddot{x} - 4 \dot{x} = 0 which has general solution x = A + Be^{2t}.

    Now using the endpoint conditions x(0) = 1 = A + B \ \Rightarrow \ B = 1 - A

    Transversality Condition \Rightarrow

    \displaystyle \frac{\partial f}{\partial \dot{x}} (T) = 0 \ \Rightarrow \frac{\partial f}{\partial \dot{x}} (2) = 0

    \displaystyle \therefore \ 2 \dot{x} (2) + 2x(2) = 0

    Recall that x = A + Be^{2t} so \dot{x} = At + \frac{1}{2} Be^{2t}

    \displaystyle \ 2A + Be^2 + 2A + 2Be^2 = 0

    \displaystyle 4A + 3Be^2 = 0

    Sub in \displaystyle B = 1 - A

    \displaystyle 4A + 3(1-A)e^2 = 0

    \displaystyle 4A + 3e^2 - 3Ae^2 = 0

    \displaystyle A(4 - 3e^2) = -3e^2

    \displaystyle \therefore \ A = \frac{-3e^2}{4 - 3e^2}

    Now B = 1 - A so \displaystyle B = \frac{-4}{-4 + 3e^2}

    Therefore the extremal is

    \displaystyle x = -3\,{\frac {{{\rm e}^{2}}}{4-3\,{{\rm e}^{2}}}}-4\,{\frac {{{\rm e}^{2<br />
\,t}}}{-4+3\,{{\rm e}^{2}}}}<br />

    \displaystyle = {\frac {3\,{{\rm e}^{2}}-4\,{{\rm e}^{2\,t}}}{-4+3\,{{\rm e}^{2}}}}

    Have I done this correctly?
    Thank-you
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  2. #2
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    Quote Originally Posted by OliviaB View Post
    Let \displaystyle \ f(t, x, \dot{x}) = \dot{x}^2 + 2x \dot{x} . Then the Euler-Lagrange equation is \displaystyle \ \frac{\partial f}{\partial x} - \frac{d}{dx} \Big( \frac{\partial f}{\partial \dot{x}} \Big) = 0 Should be d/dt, not d/dx.

    which implies \displaystyle \ 2 \dot{x} - \frac{d}{dx} \Big( 2 \dot{x} + 2x \Big) = 0

    \displaystyle \ \therefore \ 2 \dot{x} - \ddot{x} + 2 \dot{x} = 0 Middle coefficient is –2, not –1 (corrected in following line).

    \displaystyle \ \therefore \ 2 \ddot{x} - 4 \dot{x} = 0 which has general solution x = A + Be^{2t}.

    Now using the endpoint conditions x(0) = 1 = A + B \ \Rightarrow \ B = 1 - A

    Transversality Condition \Rightarrow

    \displaystyle \frac{\partial f}{\partial \dot{x}} (T) = 0 \ \Rightarrow \frac{\partial f}{\partial \dot{x}} (2) = 0

    \displaystyle \therefore \ 2 \dot{x} (2) + 2x(2) = 0

    Recall that x = A + Be^{2t} so \dot{x} = At + \frac{1}{2} Be^{2t} You should be differentiating x, not integrating it !
    From that point on, the calculation will need amending. But the method looks correct.
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  3. #3
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    Quote Originally Posted by OliviaB View Post
    Hi,

    . Then the Euler-Lagrange equation is \displaystyle \ \frac{\partial f}{\partial x} - \frac{d}{dt} \Big( \frac{\partial f}{\partial \dot{x}} \Big) = 0 Should be d/dt, not d/dx.
    Thanks - typo

    Recall that x = A + Be^{2t} so \dot{x} = At + \frac{1}{2} Be^{2t} -You should be differentiating x, not integrating it !
    Ahh, thank-you.

    So here is what I have now:


    Recall that x = A + Be^{2t} so \dot{x} = 2B e^{2t}

    \displaystyle 4Be^4 + 2A + 2Be^4 = 0

    \displaystyle \therefore \ 2A + 6Be^4 = 0

    Sub in \displaystyle B = 1

    \displaystyle 4A + 6(1-A)e^4 = 0

    \displaystyle 4A + 6e^4 - 6Ae^4 = 0

    \displaystyle A(4 - 6e^4) = -6e^4

    \displaystyle \therefore \ A = \frac{-6e^4}{4 - 6e^4}

    Now B = 1 - A so \displaystyle \  \frac{-2}{-2 + 3e^4}

    So the extremal is

    \displaystyle x =  -6\,{\frac {{{\rm e}^{4}}}{4-6\,{{\rm e}^{4}}}}+2\,{\frac {{{\rm e}^{2<br />
\,t}}}{-2+3\,{{\rm e}^{4}}}}<br />

    or \displaystyle x = {\frac {3\,{{\rm e}^{4}}+2\,{{\rm e}^{2\,t}}}{-2+3\,{{\rm e}^{4}}}}
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  4. #4
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    Quote Originally Posted by OliviaB View Post
    \displaystyle \therefore \ 2A + 6Be^4 = 0

    Sub in \displaystyle B = 1 A (corrected in next line)

    \displaystyle 4A + 6(1-A)e^4 = 0 Should be 2A, not 4A. Better still, divide through by 2 and use the equation A + 3(1A)e^4 = 0.
    I think the moral here is that you should always check your work carefully before submitting it.
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  5. #5
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    Sorry, this equation gives
    \displaystyle \ 2 \dot{x} - \frac{d}{dt} \Big( 2 \dot{x} + 2x \Big) = 0

    2x'-2x''-2x'=0

    x''=0
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  6. #6
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    Quote Originally Posted by Opalg View Post
    I think the moral here is that you should always check your work carefully before submitting it.
    Quote Originally Posted by zzzoak View Post
    Sorry, this equation gives
    \displaystyle \ 2 \dot{x} - \frac{d}{dt} \Big( 2 \dot{x} + 2x \Big) = 0

    2x'-2x''-2x'=0

    x''=0
    That makes two if us who don't check things carefully enough.
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  7. #7
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    Quote Originally Posted by Opalg View Post
    I think the moral here is that you should always check your work carefully before submitting it.
    Yes, I agree.

    Now with \ddot{x} = 0

    we have the general solution x = At + B

    Applying endpoint conditions

    x(0) = 1 = B = 1

    After applying the transversality condition and seeing that  x = At + B so \dot{x} = A we have

    2A + 4A + 2B = 0

    6A + 2B = 0

    Sub in B = 1 and we get A = \frac{-1}{3}

    so the extremal is x = \frac{-t}{3} + 1
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