- #1

Leb

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## Homework Statement

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

[itex]S[\gamma]=\int_{x_{0}}^{x_{1}}L(x,y(x),y'(x))dx[/itex]

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 ?

Thanks!