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Homework Help: Variation of S with fixed end points

  1. Jan 22, 2013 #1


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    Not really a homework question, just the notes are confusing me.

    1. The problem statement, all variables and given/known data
    Let S be a functional.

    (Given without proof)
    If S is differentiable its derivative [itex]\delta S[/itex] is uniquely defined as
    [itex]\delta S = \int_{x_{0}}^{x_{1}}\frac{\delta S}{\delta \gamma} \delta \gamma dx[/itex] where [itex]\frac{\delta S}{\delta \gamma}[/itex]

    For functional S
    the variation is defined by
    [itex]\delta S=\int_{x_{0}}^{x_{1}}\left( \frac{\partial{L}}{\partial{y}}-\frac{d}{dx}\frac{\partial{L}}{\partial{y'}}\right)\delta \gamma dx + \left.\frac{\partial{L}}{\partial{y'}}\delta \gamma \right|_{x_{0}}^{x_{1}}[/itex]

    Define [itex]\delta \gamma = \epsilon h(x)[/itex] where [itex]\epsilon =const.<< 1[/itex] and h(x) is an arbitrary (perturbation) function.

    Now, the notes given by the lecturer say, that if end points are fixed i.e. [itex]h(x_{0})=h(x_{1})=0[/itex] the variation simplifies to

    [itex]\frac{\delta S}{\delta \gamma}= \frac{\partial{L}}{\partial{y}}-\frac{d}{dx}\frac{\partial{L}}{\partial{y'}}[/itex]

    I do not understand, how do we get this (I get how the last term in the variation vanishes). Is this the functional derivative now or just delta S divided by delta gamma ?

  2. jcsd
  3. Jan 22, 2013 #2


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    Science Advisor

    If [itex]h(x_0)= h(x_1)= 0[/itex], that last term is 0 and the result follows by differentiating both sides.

  4. Jan 22, 2013 #3


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    Thanks for your input.I get how the last term vanishes. But how do you differentiate the first term ? How does [itex]\frac{\delta S}{\delta \gamma}[/itex] appear ?
    Last edited: Jan 22, 2013
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