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!

  • $\begingroup$ 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

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 $0$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.