# Derivation of the Euler Lagrange Equation

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!

• 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. – cjackal Jul 21 '16 at 12:51
• 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.) – smcc Jul 21 '16 at 12:53
• @cjackal Perfect, now I understand it! Thanks for your explanation, I think I get the derivation now! – Jannik Pitt Jul 21 '16 at 12:56
• @smcc Yes you're right, I'll fix it – 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 $0$.