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

Theorem
Let $F \left ( { x, y, z_1,~...~,z_n } \right ) $ be a function in $ C^2 $ all its variables.

Let $ y = y \left ( { x } \right ) \in C^n \left( { a \,. \,. \, b } \right ) $ such that


 * $ y \left ( { a } \right ) = A_0, y' \left ( { a } \right ) = A_1,~...,y^{ \left ( { n - 1 } \right ) } \left ( { a } \right ) = A_{n-1} $

and


 * $ y \left ( { b } \right ) = B_0, y' \left ( { b } \right ) = B_1,~...,y^{ \left ( { n - 1 } \right ) } \left ( { b } \right ) = B_{n-1} $

Let $ J \left [ { y } \right ] $ be a functional of the form


 * $ \displaystyle J \left [ { y } \right ] = \int_{ a }^{ b } F \left ( { x, y, y',~...~,y^{ \left ( { n } \right ) } } \right ) \mathrm d x $

Then a necessary condition for $ J \left [ { y } \right ] $ to have an extremum (strong or weak) for a given function $ y \left ( { x } \right ) $ is that $ y \left ( { x } \right ) $ satisfy Euler's equation

$ \displaystyle F_y - \frac{ \mathrm d }{ \mathrm d x } F_{ y' } + \frac{ \mathrm{ d^2 }{} }{ \mathrm{ d }{ x^2 } } F_{ y'' } - ... + \left ( { -1 } \right )^n \frac{ \mathrm{ d^n }{} }{ \mathrm{ d }{ x^n } } F_{ y^{ \left ( { n } \right ) } } = 0 $

Proof
From Condition for Differentiable Functional to have Extremum we have


 * $ \displaystyle \delta J \left [ { y; h } \right ] \bigg \rvert_{ y = \hat y } = 0 $

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

Note that the endpoints of $ y \left ( { x } \right ) $ are fixed. $ h \left ( { x } \right ) $ is not allowed to change values of $ y \left ( { x } \right ) $ at those points.

Hence $ h^{ \left ( { i } \right ) } \left ( { a } \right ) = 0 $ and $ h^{ \left ( { i } \right ) } \left ( { b } \right ) = 0 $ for $ i = \left ( { 1,~...,~n } \right ) $.

We will start from the increment of a functional:



Using multivariate Taylor's theorem, one can expand $ F \left ( { x,~...,y^{ \left ( { i } \right ) } + h^{ \left ( { i } \right ) },~... } \right ) $ with respect to $ h^{ \left ( { i } \right ) } $:


 * $ \displaystyle

F \left ( { x,~...,y^{ \left ( { i } \right ) } + h^{ \left ( { i } \right ) },~... } \right ) = F \left ( { x,~..., y^{ \left ( { i } \right ) } + h^{ \left ( { i } \right ) },~... } \right ) \bigg \rvert_{ h^{ \left ( { i } \right ) } = 0,~i = \left ( { 0,~...,~n } \right ) } + \sum_{ i = 0 }^n \frac{ \partial{ F \left ( { x,~...,y^{ \left ( { i } \right ) } + h^{ \left ( { i } \right ) },~... } \right ) } }{ \partial{ y^{ \left ( { i } \right ) } } } \bigg \rvert_{ h^{ \left ( { i } \right ) } = 0,~i = \left ( { 0,~...,~n } \right ) } h^{ \left ( { i } \right ) } + \mathcal{ O } \left ( { h^{ \left ( { i } \right ) } h^{ \left ( { j } \right ) },~i,j = \left ( { 0,~...,~n } \right ) } \right) $

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.


 * $\displaystyle\Delta J[y;h]=\int_{a}^{b}\sum_{i=0}^n\left[F\left(x,~...,~y^{(i)},~...\right)_{y^{(i)} } h^{(i)} + \mathcal{O}\left(h^{(i)}h^{(j)},~i,j=(0,~...,~n\right)\right]\mathrm{d}{x}$

Terms in $\mathcal{O}\left(h^{(i)}h^{(j)},~i,j=(0,~...,~n\right)$ represent terms of order higher than 1 with respect to $h^{(i)}$.

Now, suppose we expand $\int_{a}^{b}\mathcal{O}\left(h^{(i)}h^{(j)},~i,j=(0,~...,~n)\right)\mathrm{d}{x}$.

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


 * $\displaystyle \delta J[y;h]=\int_{a}^{b}\sum_{i=0}^n F_{y^{(i)} } h^{(i)}\mathrm{d}{x}$

Application of Generalized Integration by Parts, together with boundary values for $h^{(i)}$ yields


 * $\displaystyle \int_{a}^{b}h\sum_{i=0}^n (-1)^i \frac{\mathrm{d^i} }{ \mathrm{d}{x^i } }F_{y^{(i)} } \mathrm{d}{x}$

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 $h(x)$ the variation vanishes if


 * $\displaystyle F_y-\frac{\mathrm{d}{}}{\mathrm{d}{x}}F_{y'}+\frac{\mathrm{d^2}{}}{\mathrm{d}{x^2}}F_{y''}-...+(-1)^n\frac{\mathrm{d^n}{}}{\mathrm{d}{x^n} }F_{y^{(n)}}=0$