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Math Help - Calculus of Variations (Eulers Equation)

  1. #1
    Newbie
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    Mar 2011
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    Calculus of Variations (Eulers Equation)

    Hello, was wondering if you could help me with a question im stuck on...


    The functional I is defined by

    I[y] = integral from 1 to 0 of (1/2 y''^2) dx + (y(1))^2 + y'(1),

    where y(0) = y'(0) = 0 and y belongs to C^4[0,1].
    Show from first principles that a extremum y0(x) of I satisfies the euler equation y0'''' = 0
    together with the transversality conditions y0''(1) = -1, y0'''(1) = 2y0(1).
    show that y0= -1/10x^2(2+x).


    Thanks : ) x
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  2. #2
    Senior Member
    Joined
    Mar 2010
    Posts
    280
    May be this helps:

    <br />
\displaystyle<br />
I(y)= \frac{1}{2} \int_0^1 y'' \ ^2 dx+ (y(1))^2 + y'(1)<br />

    <br />
\displaystyle<br />
\delta I=I(y+\delta y)-I(y)= \frac{1}{2} \int_0^1 (y+\delta y)'' \ ^2 dx-\frac{1}{2} \int_0^1 (y)'' \ ^2 dx=\int_0^1 y'' \ \delta y'' \ dx \ .<br />

    Where

    <br />
\delta y(0)=\delta y(1)=0 \ .<br />

    I don't know if this is true:

    <br />
\delta y'(0)=\delta y'(1)=0 \ .<br />

    Then we may integrate two times by parts.
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