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 $C^2$ w.r.t. all its variables.

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


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

and


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

Let $J[y]$ be a functional of the form


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

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

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

Proof
From Condition for Differentiable Functional to have Extremum we have


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

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

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

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

We will start from the increment of a functional:



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


 * $\displaystyle

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


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