Variation of S with fixed end points

[Leb]

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

Homework Statement

Let S be a functional.

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

For functional S
S[\gamma]=\int_{x_{0}}^{x_{1}}L(x,y(x),y'(x))dx
the variation is defined by
\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}}

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

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

\frac{\delta S}{\delta \gamma}= \frac{\partial{L}}{\partial{y}}-\frac{d}{dx}\frac{\partial{L}}{\partial{y'}}

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!

 

[HallsofIvy]

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

Homework Statement

Let S be a functional.

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

For functional S
S[\gamma]=\int_{x_{0}}^{x_{1}}L(x,y(x),y'(x))dx
the variation is defined by
\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}}

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

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

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

\frac{\delta S}{\delta \gamma}= \frac{\partial{L}}{\partial{y}}-\frac{d}{dx}\frac{\partial{L}}{\partial{y'}}

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!

 

[Leb]

Thanks for your input.I get how the last term vanishes. But how do you differentiate the first term ? How does \frac{\delta S}{\delta \gamma} appear ?