Necessary Condition for Integral Functional to have Extremum for given Function/Dependent on Nth Derivative of Function

Theorem
Let $F \paren {x, y, z_1, \ldots, z_n}$ be a function in differentiability class $C^2$ all its variables.

Let $y = y \paren x \in C^n \openint a b$ such that:


 * $y \paren a = A_0, y' \paren a = A_1, \ldots, y^{\paren {n - 1} } \paren a = A_{n - 1}$

and:


 * $y \paren b = B_0, y' \paren b = B_1, \ldots, y^{\paren {n - 1} } \paren b = B_{n - 1}$

Let $J \sqbrk y$ be a functional of the form


 * $ \displaystyle J \sqbrk y = \int_a^b F \paren {x, y, y', \ldots, y^{\paren n} } \rd x$

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


 * $F_y - \dfrac \d {\d x} F_{y'} + \dfrac {\d^2} {\d x^2} F_{y''} - \cdots + \paren {-1}^n \dfrac {\d^n} {\d x^n} F_{y^{\paren n} } = 0$

Proof
From Condition for Differentiable Functional to have Extremum we have


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

For the variation to exist it has to satisfy the requirement for a differentiable functional.

Note that the endpoints of $y \paren x$ are fixed.

$h \paren x$ is not allowed to change values of $y \paren x$ at those points.

Hence $h^{\paren i} \paren a = 0$ and $h^{\paren i} \paren b = 0$ for $i = \paren {1, \ldots, n}$.

We will start from the increment of a functional:



Using multivariate Taylor's theorem, one can expand $F \paren {x, \ldots, y^{\paren i} + h^{\paren i}, \ldots}$ with respect to $h^{\paren i}$:


 * $\displaystyle F \paren {x, \ldots, y^{\paren i} + h^{\paren i}, \ldots} = F \paren {x, \ldots, y^{\paren i} + h^{\paren i}, \ldots} \bigg \rvert_{h^{\paren i} = 0, i = \paren {0, \ldots, n} } + \sum_{i \mathop = 0}^n \frac {\partial F \paren {x, \ldots, y^{\paren i} + h^{\paren i}, \ldots} } {\partial y^{\paren i} } \bigg \rvert_{ h^{\paren i} = 0, i = \paren {0, \ldots, n} } h^{\paren i} + \mathcal O \paren {h^{\paren i} h^{\paren j}, i, j = \paren {0, \ldots, n} }$

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.


 * $\displaystyle \Delta J \sqbrk {y; h} = \int_a^b \sum_{i \mathop = 0}^n \paren {F \paren {x, \ldots, y^{\paren i}, \ldots}_{y^{\paren i} } h^{\paren i} + \mathcal O \paren {h^{\paren i} h^{\paren j}, i, j = \paren {0, \ldots, n} } } \rd x$

Terms in $\mathcal O \paren { h^{\paren i} h^{\paren j}, i, j = \paren {0, \ldots, n} }$ represent terms of order higher than 1 with respect to $h^{\paren i}$.

Now, suppose we expand $\displaystyle \int_a^b \mathcal O \paren {h^{\paren i} h^{\paren j}, i, j = \paren {0, \ldots, n} } \rd x$.

By definition, the integral not counting in $\mathcal O \paren {h^{\paren i} h^{\paren j}, i, j = \paren {0, \ldots, n} }$ is a variation of functional:


 * $\displaystyle \delta J \sqbrk {y; h} = \int_a^b \sum_{i \mathop = 0}^n F_{y^{\paren i} } h^{\paren i} \rd x$

Application of Generalized Integration by Parts, together with boundary values for $h^{\paren i}$ yields:


 * $\displaystyle \int_a^b h \sum_{i \mathop = 0}^n \paren {-1}^i \frac {\d^i} {\d x^i} F_{ y^{\paren i} } \rd x$

Since If Definite Integral of a(x)h(x) vanishes for any C^0 h(x) then C^0 a(x) vanishes, then for any $h \paren x$ the variation vanishes if:


 * $F_y - \dfrac \d {\d x} F_{y'} + \dfrac {\d^2} {\d x^2} F_{y''} - \cdots + \paren {-1}^n \dfrac {\d^n} {\d x^n} F_{ y^{\paren n} } = 0$