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Simpler Euler-Lagrange equation

  1. Jul 13, 2009 #1
    If we have a functional
    [tex]
    J(y)=\int L(y,y',x)dx
    [/tex]

    then the first variation is
    [tex]
    \delta J=\int\left(\frac{\partial L}{\partial y}\eta(x)+\frac{\partial L}{\partial y'}\eta'(x)\right)dx,
    [/tex]
    where [itex]\eta(x)[/itex] is the variation of the stationary solution. Now, if [itex]L[/itex] is independent of [itex]y(x)[/itex], then [itex]\frac{\partial L}{\partial y}=0[/itex] and I have
    [tex]
    \delta J=\int\left(\frac{\partial L}{\partial y'}\eta'(x)\right)dx.
    [/tex]

    At this point, why can't I simply say that [itex]\eta'(x)[/itex] is arbitrary and hence the E-L equation is [itex]\frac{\partial L}{\partial y'}=0[/itex] ?
     
  2. jcsd
  3. Jul 14, 2009 #2
    eta-prime is not arbitrary, because eta is zero at both end points. If eta prime were arbitrary then the condition that eta is zero at one boundary would imply that eta at the other boundary is arbitrary and thus not necessarily equal to zero.
     
  4. Jul 14, 2009 #3
    You can also treat the fact that eta-prime is not arbitrary everywhere by introducing a contstraint: The integral of eta-prime over the interval has to be zero. You can then introduce a Lagrange multiplier to take that into acount and you then find that:


    [tex]\frac{\partial L}{\partial y'} = \lambda[/tex]

    which is the same as saying that:

    [tex]\frac{d}{dt}\frac{\partial L}{\partial y'} = 0[/tex]
     
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