Theorem: Simple Variable End Point Problem

Let $y$ and $F$ be mappings.
{{explain|Define their domain and codomain}}
Suppose the endpoints of $y$ lie on two given vertical lines $x = a$ and $x = b$.
Suppose $J$ is a functional of the form
:$(1): \quad J \sqbrk y = \ds \int_a^b \map F {x, y, y'} \rd x$
and has an extremum for a certain function $\hat y$.
Then $y$ satisfies the system of equations
:$\begin {cases}
F_y - \dfrac \d {\d x} F_{y'} = 0 \\
\bigvalueat {F_{y'} } {x \mathop = a} = 0 \\
\bigvalueat {F_{y'} } {x \mathop = b} = 0
\end {cases}$

Proof:
From Condition for Differentiable Functional to have Extremum we have
:$\bigvalueat {\delta J \sqbrk {y; h} } {y \mathop = \hat y} = 0$
The variation exists if $J$ is a differentiable functional.
We will start from the increment of a functional:
{{explain|make the above link point to a page dedicated to the appropriate definition}}
{{begin-eqn}}
{{eqn | l = \Delta J \sqbrk {y; h}
      | r = J \sqbrk {y + h} - J \sqbrk y
      | c = definition
}}
{{eqn | r = \int_a^b \map F {x, y + h, y' + h'} \rd x - \int_a^b \map F {x, y, y'} \rd x
      | c = $(1)$
}}
{{eqn | r = \int_a^b \paren {\map F {x, y + h, y' + h'} - \map F {x, y, y'} } \rd x
}}
{{end-eqn}}
{{explain|"definition" in the above -- point to what it is a definition of}}
Using multivariate Taylor's theorem, one can expand $\map F {x, y + h, y' + h'}$ with respect to $h$ and $h'$:
:$\map F {x, y + h, y' + h'} = \bigvalueat {\map F {x, y + h, y' + h'} } {h \mathop = 0, \, h' \mathop = 0} + \valueat {\dfrac {\partial \map F {x, y + h, y' + h'} } {\partial y} } {h \mathop = 0, \, h' \mathop = 0} h + \valueat {\dfrac {\partial {\map F {x, y + h, y' + h'} } } {\partial y'} } {h \mathop = 0, \, h' \mathop = 0} h' + \map \OO {h^2, h h', h'^2}$
Substitute this back into the integral. 
Note that the first term in the expansion and the negative one in the integral will cancel out:
:$\ds \Delta J \paren {y; h} = \int_a^b \paren {\map F {x, y, y'}_y h + \map F {x, y, y'}_{y'} h' + \map \OO {h^2, h h', h'^2} } \rd x$
{{explain|What do the subscripted $y$ and $y'$ mean in the above?}}
Terms in $\map \OO {h^2, h'^2}$ represent terms of order higher than 1 with respect to $h$ and $h'$.
Now we expand:
:$\ds \int_a^b\map \OO {h^2, h h', h'^2} \rd x$
Every term in this expansion will be of the form:
:$\ds \int_a^b \map A {m, n} \frac {\partial^{m + n} \map F {x, y, y'} } {\partial y^m \partial {y'}^n} h^m h'^n \rd x$
where $m, n \in \N$ and $m + n \ge 2$
{{Explain|How to convert powers of $h'$ into $h$? Integration by parts is the only obvious candidate, but how precisely? Also check, if this is necessary}}
By definition, the integral not counting in $\map \OO {h^2, h h', h'^2}$ is a variation of functional.
:$\ds \delta J \sqbrk {y; h} = \int_a^b \paren {F_y h + F_{y'} h'} \rd x$
Now, integrate by parts and note that $\map h x$ does not necessarily vanish at the endpoints:
{{begin-eqn}}
{{eqn | l = \delta J \sqbrk {y; h}
      | r = \int_a^b \paren {F_y - \frac \d {\d x} F_{y'} } \map h x \rd x + \bigintlimits {F_{y'} \map h x} {x \mathop = a} {x \mathop = b}
}}
{{eqn | r = \int_a^b \paren {F_y - \frac \d {\d x} F_{y'} } \map h x \rd x + \bigvalueat {F_{y'} } {x \mathop = b} \map h b - \bigvalueat {F_{y'} } {x \mathop = a} \map h a
}}
{{end-eqn}}
Then, for arbitrary $\map h x$, $J$ has an extremum if:
:$ \begin {cases}
F_y - \dfrac \d {\d x} F_{y'} = 0 \\
\bigvalueat {F_{y'} } {x \mathop = a} = 0\\
\bigvalueat {F_{y'} } {x \mathop = b} = 0
\end {cases}$
{{qed}}