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Math Help - [SOLVED] Calculas of Variations problem

  1. #1
    Junior Member
    Joined
    Apr 2009
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    [SOLVED] Calculas of Variations problem

    Hey all

    I do Calculas of Variations, and have an exam in around nine hours, so would appreciate help before then. After is good too thhough.

    I got a solution, but it has a bit I don't understand, was hoping you guys could help.

    My problem is:
    Find extremal of J(y)= \int_0^1(y(x))dx,subject to  \int_0^1\sqrt(1+y'(x)^2)dx = L.
    y(0) = y(1) =0

    Solution that I have gotten is:
    F'= F + \lambda*G
    F= y(x), G =\sqrt(1+(y')^2).

    \frac{dF*}{dy}-\frac{d}{dx}(\frac{dF*}{dy'})=0
    1-\frac{d}{dx}(\lambda*y'/\sqrt(1+(y')^2)
    Integrate:
    \rightarrow \lambda*y'/\sqrt(1+(y')^2)=x+C C arbitary constant
    so
    (y')^2/(1+(y')^2)=(x+C)^2/\lambda^2
    y'^2 = ((x+c)^2/\lambda^2)/(1-(x+C)^2/\lambda^2);
    y'=(x+C/\lambda)/(\sqrt(1+(x+C)^2)/\lambda^2).
    u=1-(x+C)^2/\lambda^2

    Put u in, integrate, change the result around a small bit and you get
    (x+C)^2/\lambda^2+(y+D)^2/\lambda^2=1 D arbitary constant
    inset bound conditions, you get C = -1/2
    D=\sqrt(\lambda^2-1/4) =\sqrt(\lambda^2-C^2)

    So, (x-1/2)^2+(y+\sqrt(\lambda^2-1/4)^2=\lambda^2
    Diffretiate with respect to x;
    (x+1/2)+(y+\sqrt(\lambda^2-1/4)*y'=0<br />
=> y'=-(x-1/2)/(y+\sqrt(\lambda^2-1/4).

    Then (And herin lies the problem), I get
    1+y'^2=1+(x-1/2)^2/((y+\sqrt(\lambda^2-1/4))^2) = \lambda^2/(\lambda^2-(x-1/2)^2.

    How does he change 1+(x-1/2)^2/((y+\sqrt(\lambda^2-1/4))^2) into \lambda^2/(\lambda^2-(x-1/2)^2)?

    Thanks in advance.
    Last edited by Diemo; April 28th 2009 at 05:03 PM. Reason: bad notation
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  2. #2
    Junior Member
    Joined
    Apr 2009
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    Got answer, substitute from two lines up. Thanks all.
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