From Condition for Differentiable Functional of N Functions to have Extremum: :$\displaystyle \bigvalueat {\delta J \sqbrk {\mathbf y; \mathbf h} } {\mathbf y \mathop = \hat{\mathbf y} } = 0$ For the variation to exist it has to satisfy the requirement for a differentiable functional. Note that the endpoints of $\map {\mathbf y} x$ are fixed. $\map {\mathbf h} x$ is not allowed to change values of $\map {\mathbf y} x$ at those points. Hence $\map {\mathbf h} a = 0$ and $\map {\mathbf h} b = 0$. We will start from the increment of a functional: {{begin-eqn}} {{eqn | l = \Delta J \sqbrk {\mathbf y; \mathbf h}       | r = J \sqbrk {\mathbf y + \mathbf h} - J \sqbrk {\mathbf y} }} {{eqn | r = \int_a^b \map F {x, \mathbf y + \mathbf h, \mathbf y' + \mathbf h'} \rd x - \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x }} {{eqn | r = \int_a^b \paren {\map F {x, \mathbf y + \mathbf h, \mathbf y' + \mathbf h'} - \map F {x, \mathbf y, \mathbf y'} } \rd x }} {{end-eqn}} Using multivariate Taylor's theorem, one can expand $\map F {x, \mathbf y + \mathbf h, \mathbf y' + \mathbf h'}$ with respect to functions $\map {\mathbf h} x$ and $\map {\mathbf h'} x$: {{begin-eqn}} {{eqn | l = \map F {x, \mathbf y + \mathbf h, \mathbf y' + \mathbf h'}       | r = \bigvalueat {\map F {x, \mathbf y + \mathbf h, \mathbf y' + \mathbf h'} } {\mathbf h \mathop = \mathbf 0, \mathbf h' \mathop = \mathbf 0}       | c =  }} {{eqn | o =        | ro= +       | r = \valueat {\sum_{i \mathop = 1}^n \frac {\partial {\map F {x, \mathbf y + \mathbf h, \mathbf y' + \mathbf h'} } } {\partial {y_i} } } {h_i \mathop = 0, h_i' \mathop = 0} h_i       | c =  }} {{eqn | o =        | ro= +       | r = \valueat {\sum_{i \mathop = 1}^n \frac {\partial {\map F {x, \mathbf y + \mathbf h, \mathbf y + \mathbf h} } } {\partial {y_i}'} } {h_i \mathop = 0, h_i' \mathop = 0} {h_i}'       | c =  }} {{eqn | o =        | ro= +       | r = \map \OO {h_i h_j, h_i h_j', h_i' h_j'}       | c = for $i, j \in \openint 1 n$ }} {{end-eqn}} We can substitute this back into the integral. Note that the first term in the expansion and the negative one in the integral will cancel out. Hence: :$\displaystyle \Delta J \sqbrk {\mathbf y; \mathbf h} = \int_a^b \sum_{i \mathop = 1}^n \paren {F_{y_i} h_i + F_{y_i'} h_i' + \map \OO {h_i h_j,h_i h_j',h_i' h_j'} } \rd x$ for $i, j \in \openint 1 n$ {{Explain|Finish single function case, then return here}} By definition, the integral not counting in $\map \OO {h_i h_j,h_i h_j',h_i' h_j'}$ for $i, j \in \openint 1 n$ is a variation of functional: {{explain|The link is to Definition:Differentiable Functional but the label says "variation of functional" which is not defined on that page}} :$\displaystyle \delta J \sqbrk {\mathbf y; \mathbf h} = \int_a^b \paren {F_{\mathbf y} \mathbf h + F_{\mathbf y'} \mathbf h'} \rd x$ The variation vanishes if for all functions $h_i$ every term containing $h_i$ vanishes independently. Therefore, we discover a set of Euler's Equations being satisfied simultaneously:    :$F_{\mathbf y} - \dfrac \d {\d x} F_{\mathbf y'} =  0$ {{qed}} {{improve|find a better way to deal with notation of dependence on multiple functions (probably through mapping to avoid explicit arguments); build related pages}}