Necessary Condition for Integral Functional to have Extremum for given function/Dependent on N Functions

Theorem
Let $\mathbf y$ be an $n$-dimensional real vector.

Let $J \sqbrk {\mathbf y}$ be a functional of the form:

$\displaystyle J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$

Let:


 * $\mathbf y \in C^1 \closedint a b$

where $C^1 \closedint a b$ denotes that $\mathbf y$ is continuously differentiable in $\closedint a b$

Let $\mathbf y$ satisfy boundary conditions:


 * $\map {\mathbf y} a = \mathbf A$


 * $\map {\mathbf y} b = \mathbf B$

where $\mathbf A$, $\mathbf B$ are real vectors.

Then a necessary condition for $J \sqbrk {\mathbf y}$ to have an extremum (strong or weak) for a given $\mathbf y$ is that they satisfy Euler's equations:

$F_{\mathbf y} - \dfrac \d {\d x} F_{\mathbf y'} = 0$

Proof
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:

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 $\map {\mathbf h'} x$:

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 $i, j \in \openint 1 n$

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:


 * $\displaystyle \delta J \sqbrk {\mathbf y; \mathbf h} = \int_a^b \paren {F_{\mathbf y} \mathbf h + F_{\mathbf y'} \mathbf h'} \rd x$

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:
 * $F_{\mathbf y} - \dfrac \d {\d x} F_{\mathbf y'} = 0$