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(Please note that I am a physics student, so please try to avoid rigorous mathematics in the explanation.)

I have the following proof for the Euler-Lagrange:

Consider the integral $$I=\int F(x,y,y').$$

Consider now a line $Y$ such that $Y=y+\epsilon h$ for some small $\epsilon$. Then $$\frac{dI}{d\epsilon}=\int_{x_1}^{x_2}\Big(\frac{\partial F}{\partial Y}\frac{\partial Y}{\partial \epsilon}+\frac{\partial F}{\partial Y'}\frac{\partial Y'}{\partial \epsilon}\Big)dx=\int_{x_1}^{x_2}\Big(\frac{\partial F}{\partial Y}h+\frac{\partial F}{\partial Y'}h'\Big).$$ Integrating by parts we get $$\int_{x_1}^{x_2}\frac{\partial F}{\partial Y'}h'dx=\Big[\frac{\partial F}{\partial Y'}h\Big]_{x_1}^{x_2}-\int_{x_1}^{x_2}h\frac{d}{dx}\Big(\frac{\partial F}{\partial Y'}\Big)dx.$$ Then $$\frac{dI}{d\epsilon}=\int_{x_1}^{x_2}\Big[\frac{\partial F}{\partial Y}h-h\frac{d}{dx}\Big(\frac{\partial F}{\partial Y'}\Big)\Big]dx=\int_{x_1}^{x_2}h\Big[\frac{\partial F}{\partial Y}-\frac{d}{dx}\Big(\frac{\partial F}{\partial Y'}\Big)\Big]dx.$$ $$\implies \frac{\partial F}{\partial y}-\frac{d}{dx}\Big(\frac{\partial F}{\partial y'}\Big)=0.$$ I understand everything up to the beginning of integration by parts. On the third line of equations (the first line after "Integrating by parts we get...") we must have $$\Big[\frac{\partial F}{\partial Y'}h\Big]_{x_1}^{x_2}=0$$ in order to get to the next line with the expression for $dI/d\epsilon$, but how is this so? I also don't understand how the last line implies the Euler-Lagrange.

I guess a helpful answer would be explaining everything from "Integration by parts" in detail.

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    $\begingroup$ Perhaps there are assumptions, where we choose $x_1=x_2$ so that it equals $0$. $\endgroup$ – Simply Beautiful Art May 15 '16 at 16:10
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As for the last line implying Euler-Lagrange, it follows since $h$ is an arbitrary perturbation. See also the lemma of duBois Reymond: https://en.wikipedia.org/wiki/Fundamental_lemma_of_calculus_of_variations

Also, I should add, we do want $dI/d\epsilon=0$ for that to happen, but that follows since we want to minimize the functional, so the usual calculus applies…

…and yes, Simple Art is right, we impose boundary conditions on $h$ (i.e. $h(x_1)=h(x_2)=0)$ to force $[h\frac{\partial F}{\partial Y'}]_{x_1}^{x_2}=0$. We can do that and still have the above hold.

For a full proof with a little more details, I refer to the Wikipedia page: https://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation

Look for "Derivation of one-dimensional Euler–Lagrange equation".

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The previous answer is correct. I going to simplify the answer. One consideration for $h$ is that $h(x_1)=h(x_2)=0$, then \begin{eqnarray} \left( h(x) \frac{\partial F(x)}{\partial Y'} \right|_{x_1}^{x_2} &=& \left(h(x_2) \frac{\partial F(x_2)}{\partial Y'} \right)-\left(h(x_1) \frac{\partial F(x_1)}{\partial Y'} \right)\\ &=& \left(0 \cdot \frac{\partial F(x_2)}{\partial Y'} \right)-\left(0\cdot \frac{\partial F(x_1)}{\partial Y'} \right)\\ &=& 0-0=0 \end{eqnarray}

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