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I'm self studying a little bit of physics at the moment and for that I needed the derivation of the Euler Lagrange Equation. I understand everything but for a little step in the proof, maybe someone can help me. That's were I am: $$ \frac{dJ(\varepsilon=0 )}{d\varepsilon } = \int_{a}^{b}\eta(x)\frac{\partial F}{\partial y}+\eta'(x)\frac{\partial F}{\partial y'}dx = 0 $$ Then the second term is integrated by parts: $$ \frac{dJ(\varepsilon =0)}{d\varepsilon } = \int_{a}^{b}\eta(x)\frac{\partial F}{\partial y}dx + \left [ \frac{\partial F}{\partial y'}\eta(x) \right ]_a^b-\int_{a}^{b}\frac{d}{dx}(\frac{\partial F}{\partial y'})\eta(x) = 0 $$ And the equation is simplified to: $$ \frac{dJ(\varepsilon =0)}{d\varepsilon } = \int_{a}^{b}\frac{\partial F}{\partial y}-\frac{d}{dx}(\frac{\partial F}{\partial y'})\eta(x)dx = 0 $$ What I don't understand is why you just can omit the $$ \left [ \frac{\partial F}{\partial y'}\eta(x) \right ]_a^b $$ Why does that equal zero, but the integral following it which also goes from a to b isn't left out? I think it's pretty obvious, but I'm just to stupid to see it. I'd appreciate if someone could help me!

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  • $\begingroup$ Here $\eta(b),\eta(a)=0$ from the beginning. For the extemal curve $\zeta(t)$, a small variation of $\zeta(t)$ is given by $\tilde{\zeta}(t)=\zeta(t)+\epsilon \eta(t)$, and we want the end points of the small variation be fixed. $\endgroup$ – cjackal Jul 21 '16 at 12:51
  • $\begingroup$ I think there is an error in your second equation. The last term should have the $\eta(x)$ outside the pararentheses. (There is also a missing parenthesis in the third equation.) $\endgroup$ – smcc Jul 21 '16 at 12:53
  • $\begingroup$ @cjackal Perfect, now I understand it! Thanks for your explanation, I think I get the derivation now! $\endgroup$ – Jannik Pitt Jul 21 '16 at 12:56
  • $\begingroup$ @smcc Yes you're right, I'll fix it $\endgroup$ – Jannik Pitt Jul 21 '16 at 12:56
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In the Euler-Lagrange equation, the function $\eta$ has by hypothesis the following properties:

  • $\eta$ is continuously differentiable (for the derivation to be rigorous)
  • $\eta$ satisfies the boundary conditions $\eta(a) = \eta(b) = 0$.

In addition, $F$ should have continuous partial derivatives.

This is why $\left [ \frac{\partial F}{\partial y'}\color{red}{\eta(x)} \right ]_a^b$ simplifies to $0$.

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  • $\begingroup$ Perfect, I get it now! Thanks, I really overlooked that! $\endgroup$ – Jannik Pitt Jul 21 '16 at 12:58

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