# Variation of S with fixed end points

1. Jan 22, 2013

### Leb

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 $\delta S$ 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!

2. Jan 22, 2013

### HallsofIvy

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

3. Jan 22, 2013

### 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 ?

Last edited: Jan 22, 2013