From Condition for Differentiable Functional to have Extremum we have :$\bigvalueat {\delta J \sqbrk {y h} } {y \mathop = \hat y} = 0$ For the variation to exist it has to satisfy the requirement for a differentiable functional. Note that the endpoints of $\map y x$ are fixed. $\map h x$ is not allowed to change values of $\map y x$ at those points. Hence the higher derivative $\map {h^{\paren i} } a = 0$ and $\map {h^{\paren i} } b = 0$ for $i \in \openint 1 n$. We will start from the increment of a functional: {{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, \ldots, y^{\paren i} + h^{\paren i}, \ldots} \rd x - \int_a^b \map F {x, \ldots, y^{\paren i}, \ldots} \rd x       | c = where $i \in \set {0, 1, \ldots, n}$ }} {{eqn | r = \int_a^b \paren {\map F {x, \ldots, y^{\paren i} + h^{\paren i}, \ldots} - \map F {x, \ldots, y^{\paren i}, \ldots} } \rd x       | c = bringing under the same integral }} {{end-eqn}} Using multivariate Taylor's theorem, one can expand $\map F {x, \ldots, y^{\paren i} + h^{\paren i}, \ldots}$ with respect to $h^{\paren i}$: :$\displaystyle \map F {x, \ldots, y^{\paren i} + h^{\paren i}, \ldots} = \valueat {\map F {x,\ldots,y^{\paren i} + h^{\paren i}, \ldots} } {h^{\paren i} = 0, i \in \openint 0 n} + \valueat {\sum_{i \mathop = 0}^n \frac {\partial \map F {x, \ldots, y^{\paren i} + h^{\paren i}, \ldots} } {\partial y^{\paren i} } } {h^{\paren i} = 0, i \in \openint 0 n} h^{\paren i} + \map \OO {h^{\paren i} h^{\paren j}, i, j \in \openint 0 n}$ 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: :$\displaystyle \Delta J \sqbrk {y; h} = \int_a^b \sum_{i \mathop = 0}^n \paren {\map F {x, \ldots, y^{\paren i}, \ldots}_{y^{\paren i} } h^{\paren i} + \map \OO {h^{\paren i} h^{\paren j}, i, j \in \openint 0 n} } \rd x$ {{explain|What is the meaning of the subscript $y^{\paren i}$ in the middle of the above?}} Terms in $\map \OO {h^{\paren i} h^{\paren j}, i, j \in \openint 0 n}$ represent terms of order higher than 1 with respect to $h^{\paren i}$. Now, suppose we expand $\displaystyle \int_a^b \map \OO {h^{\paren i} h^{\paren j}, i, j \in \openint 0 n} \rd x$. {{Explain|Proof that this goes to zero faster than $\size h$, as definition of variation requires}} By definition, the integral not counting in $\map \OO {h^{\paren i} h^{\paren j}, i, j \in \openint 0 n}$ 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}} :$\displaystyle \delta J \sqbrk {y ; h} = \int_a^b \sum_{i \mathop = 0}^n F_{y^{\paren i} } h^{\paren i} \rd x$ Application of Generalized Integration by Parts, together with boundary values for $h^{\paren i}$ yields: :$\displaystyle \int_a^b h \sum_{i \mathop = 0}^n \paren {-1}^i \frac {\d^i} {\d x^i} F_{y^{\paren i} }\rd x$ From If Definite Integral of a(x)h(x) vanishes for any C^0 h(x) then C^0 a(x) vanishes, then for any $\map h x$ the variation vanishes if: :$F_y - \dfrac \d {\d x} F_{y'} + \dfrac {\d^2} {\d x^2} F_{y''} - \cdots + \paren {-1}^n \dfrac {\d^n} {\d x^n} F_{ y^{\paren n} } = 0$ {{qed}} {{improve|minor refinements, existence of derivatives of F}}