I'm having trouble understanding the derivation in my Calculus of Variations course material and I was hoping if someone could clarify this out. Here is my reference (as I have rewritten it, the original material can be found on the bottom of this post). I'm having trouble only in the final part (Integration by parts):
$$J(u)=\int_{\Omega}F(\textbf{x}, u, u')\;d\textbf{x},$$
where $\textbf{x}\in \mathbb{R}^n, u=u(\textbf{x}), u'=u'(x)=\frac{du}{d\textbf{x}}=\nabla u(\textbf{x}), \;\Omega$ is a bounded open set in $\mathbb{R}^n,$ $u: \Omega\rightarrow\mathbb{R}$ is a function of class $C^2$ and $F(x,u,u')$ is a real valued function of class $C^2$ with respect to all its arguments.
Now we define:
$$\Phi(\epsilon):=J(u+\epsilon\eta),$$
where $\eta\in C^{\infty}_0(\Omega, \mathbb{R})$ and $\epsilon$ is a small number. $\Phi(\epsilon)$ is of class $C^2$ on some interval $(-\epsilon_0, \epsilon_0)$. Now we want to calculate the derivative of $\Phi(\epsilon)$ at the point $\epsilon =0$ and equate it to $0$. We get:
$$\Phi'(0)=\int_\Omega\left(F_u(\textbf{x},u,u')\eta+F_{u'}(\textbf{x},u,u')\cdot \eta'\right)\;d\textbf{x}=$$$$\int_\Omega \left(F_u(\textbf{x}, u, u')\eta+\sum_{i=1}^n\frac{\partial F(\textbf{x},u,u')}{\partial u'_i}\eta'_i \right)\;d\textbf{x}=0,$$
where $$\eta'=\nabla\eta(\textbf{x}), \;\eta'_i=\frac{\partial \eta }{\partial x_i},\;u'_i=\frac{\partial u}{\partial x_i}.\;$$
Now my lecturer uses integration by parts and he gets:
$$\int_\Omega \left(F_u(\textbf{x}, u, u')\eta+\sum_{i=1}^n\frac{\partial F(\textbf{x},u,u')}{\partial u'_i}\eta'_i \right)\;d\textbf{x}=\;\;\;\;\;\;\;(1)$$
$$F_u(\textbf{x}, u, u')-\sum_{i=1}^n\frac{d}{dx_i}\left(\frac{\partial F(\textbf{x},u,u')}{\partial u'_i}\right)=0.\;\;\;\;\;\;\;\;\;\;(2)$$
Now what happened from $(1)$ to $(2)$? I'm a bit confused. Can someone explain the integration by parts part...
P.S. here is also the original lecture material (the notation is a bit different to what I used):
The second term of the integral is integrated by parts using Gauss's divergence theorem: \begin{align} \int\limits_\Omega \mbox{grad}_p F \cdot \mbox{grad}_x \eta \, dx &= \int\limits_\Omega \left(\mbox{div}_x(\eta \, \mbox{grad}_p F ) - \eta \, \mbox{grad}_x \cdot \mbox{grad}_p F\right) \, dx \\ &= \int\limits_{\partial\Omega} \underbrace{\eta \, \mbox{grad}_p F}_{\eta|_{\partial\Omega = 0}} \cdot dA - \int\limits_\Omega \eta \, \mbox{grad}_x \cdot \mbox{grad}_p F \, dx \\ &= -\int\limits_\Omega \eta \, \mbox{grad}_x \cdot \mbox{grad}_p F \, dx \end{align} So \begin{align} 0 = \delta J = \int\limits_\Omega \left( \eta \, \mbox{grad}_u F + \mbox{grad}_p F \cdot \mbox{grad}_x \eta \right) \, dx &= \int\limits_\Omega \left( \eta \, \mbox{grad}_u F - \eta \, \mbox{grad}_x \cdot \mbox{grad}_p F \right) \, dx \\ &= \int\limits_\Omega \eta \, \left( \mbox{grad}_u F - \mbox{grad}_x \cdot \mbox{grad}_p F \right) \, dx \end{align} which for arbitrary $\eta$ (within $\Omega$) implies the term in parentheses to vanish. (For a rigorous proof, this probably has to be shown as well).
The last step seems to be the result of looking at the dependencies of $F$ which are given as $F(x, u, Du)$ and applying the partial derivatives regarding $x$: $$ \sum_i \partial_{x_i} \partial_{p_i} F(x, u, Du) $$ that would yield the three sums of second order derivatives, one for each variable.
Note: your text uses total derivatives, so I might be sloppy/wrong. To be more precise I would need to better understand your $D$ operator and its connection to the $p$ variable.