Necessary Condition for Integral Functional to have Extremum/Two Variables

Theorem
Let $S$ be a set of real mappings such that:


 * $S = \set {\map z {x, y}: \paren {z: S_1 \subseteq \R^2 \to S_2 \subseteq \R}, \paren {\map z {x, y} \in C^1 \closedint a b}, \paren {\map z \Gamma = 0} }$

Let $J \sqbrk z: S \to S_3 \subseteq \R$ be a functional of the form:


 * $\displaystyle \int \int_D \map F {x, y, z, z_x, x_y} \rd x \rd y$

Then a necessary condition for $J \sqbrk y$ to have an extremum (strong or weak) for a given function $\map y x$ is that $\map y x$ satisfy Euler's equation:


 * $F_z - \dfrac \partial {\partial x} F_{z_x} - \dfrac \partial {\partial y} F_{z_y} = 0$

Proof
From Condition for Differentiable Functional to have Extremum we have


 * $\delta J \sqbrk {y; h} \bigg \rvert_{y = \hat y} = 0$

The variation exists if $J$ is a differentiable functional.

The endpoints of $\map y x$ are fixed.

Hence:


 * $\map h a = 0$


 * $\map h b = 0$.

From the definition of increment of a functional:



Using multivariate Taylor's theorem, expand $\map F {x, y, z + h, z_x + h_x, z_y + h_y}$ $h$, $h_x$ and $h_y$.

Denote the ordered tuples $\tuple {h, h_x, h_y}$ as $\mathbf h$ and $\tuple {z, z_x, z_y}$ as $\mathbf z$ respectively:

where the last term includes all terms of the order not lesser than 2 the elements of $\mathbf h$.

Substitute this back into the integral:


 * $\displaystyle \Delta J \sqbrk {y; h} = \int_a^b \paren {\map F {x, y, y'}_y h + \map F {x, y, y'}_{y'} h' + \mathcal O \paren {h^2, h h', h'^2} } \rd x$

Terms in $\mathcal O \paren {h^2, h'^2}$ represent terms of order higher than 1 with respect to $h$ and $h'$.

Suppose we expand $\displaystyle \int_a^b \mathcal O \paren {h^2, h h', h'^2} \rd x$.

Every term in this expansion will be of the form:


 * $\displaystyle \int_a^b \map A {m, n} \frac {\partial^{m + n} \map F {x, y, y'} } {\partial y^m \partial y'^n} h^m h'^n \rd x$

where $m, n \in \N: m + n \ge 2$

By definition, the integral not counting in $\mathcal O \paren {h^2, h h', h'^2}$ is a variation of functional:


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

Use lemma.

Then for any $\map h x$ variation vanishes if:


 * $F_z - \dfrac \partial {\partial x} F_{z_x} - \dfrac \partial {\partial y} F_{z_y} = 0$