# Simple Variable End Point Problem

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## Theorem

Let $y$ and $F$ be mappings.

Suppose endpoints of $y$ lie on two given vertical lines $x=a$ and $x=b$.

Suppose $J$ is a functional of the form

 $(1):\quad$ $\displaystyle J\sqbrk{y}$ $=$ $\displaystyle \int_a^b\map F {x,y,y'} \rd x$

and has an extremum for a certain function $\hat y$.

Then $y$ satisfies the system of equations

$\begin{cases} & F_y-\frac\d{\d x}F_{y'}=0\\ & F_{y'}\big\rvert_{x=a}=0\\ & F_{y'}\big\rvert_{x=b}=0 \end{cases}$

## Proof

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

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

We will start from the increment of a functional:

 $\displaystyle \Delta J\sqbrk{y;h}$ $=$ $\displaystyle J\sqbrk{y+h}-J\sqbrk y$ definition $\displaystyle$ $=$ $\displaystyle \int_a^b\map F{x,y+h,y'+h'}\rd x-\int_a^b\map F{x,y,y'}\rd x$ $(1)$ $\displaystyle$ $=$ $\displaystyle \int_a^b\sqbrk{\map F{x,y+h,y'+h'}-\map F{x,y,y'} }\rd x$

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

$\displaystyle F\paren{x,y+h,y'+h'}=F\paren{x,y+h,y'+h'}\bigg\rvert_{h=0,~h'=0}+\frac{\partial F\paren{x,y+h,y'+h'} }{\partial y} \bigg \rvert_{h=0,~h'=0}h+\frac{\partial{F\paren{x,y+h,y'+h'} } }{\partial y'}\bigg\rvert_{h=0,~h'=0}h'+\map {\mathcal{O}} {h^2,hh',h'^2}$

Substitute this back into the integral.

Note that the first term in the expansion and the negative one in the integral will cancel out:

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

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

Now, suppose we expand $\int_a^b\map {\mathcal{O}} {h^2,hh',h'^2}\rd x$.

Every term in this expansion will be of the form

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

where $m,~n\in\N$ and $m+n\ge 2$

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

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

Now, integrate by parts and note that $\map h x$ does not necessarily vanish at the endpoints:

 $\displaystyle \delta J \sqbrk{y;h}$ $=$ $\displaystyle \int_a^b\sqbrk{F_y-\frac{\d}{\d x}F_{y'} }\map h x\rd x+F_{y'}\map h x\big\rvert_{x=a}^{x=b}$ $\displaystyle$ $=$ $\displaystyle \int_a^b\sqbrk{F_y-\frac{\d}{\d x}F_{y'} }\map h x\rd x+F_{y'}\big\rvert_{x=b}\map h b-F_{y'}\big\rvert_{x=a}\map h a$

Then, for arbitrary $\map h x$, $J$ has an extremum if

$\begin{cases} & F_y-\frac{\d}{\d x}F_{y'}=0\\ & F_{y'}\big\rvert_{x=a}=0\\ & F_{y'}\big\rvert_{x=b}=0 \end{cases}$

$\blacksquare$