Simple Variable End Point Problem/Endpoints on Curves

From ProofWiki
Jump to: navigation, search


Let $y$ and $F$ be smooth real functions.

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

$\displaystyle J\sqbrk y=\int_{x_0}^{x_1}\map F {x,y,y'}\rd x$

Let $P_0$, $P_1$ be the endpoints of the curve $y$.

Suppose $P_0$, $P_1$ lie on curves $y=\map {\phi} x$, $y=\map {\psi} x$.

Then the extremum of $J\sqbrk y$ is a curve which satisfies the following system of equations:

$ \begin{cases} &F_y-\frac {\d} {\d x} F_{y'}=0\\ & \sqbrk{F+\paren{\phi'-y'} F_{y'} }\big\rvert_{x=x_0}=0\\ & \sqbrk{F+\paren{\psi'-y'}F_{y'} }\big\rvert_{x=x_1}=0 \end{cases}$

The last two equations are known as the transversality conditions.


By general variation of integral functional with $n=1$:

$\displaystyle\delta J\sqbrk{y;h}=\int_{x_0}^{x_1} \paren {F_y-\frac{\d}{\d x}F_{y'} }\map h x+ F_{y'}\delta y\bigg \rvert_{x=x_0}^{x=x_1}+\paren {F- y'F_{y'} }\delta x\bigg\rvert_{x=x_0}^{x=x_1}$

Since the curve $y=\map y x$ is an extremum of $J\sqbrk y$, the integral term vanishes:

$\displaystyle\delta J\sqbrk{y;h}=F_{y'}\delta y\bigg\rvert_{x=x_0}^{x=x_1}+\paren {F- y'F_{y'} }\delta x\bigg\rvert_{x=x_0}^{x=x_1}$

According to Taylor's theorem:

$\displaystyle\delta y_0=\sqbrk {\map {\phi'} {x_0}+\epsilon_0}\delta x_0,\quad\delta y_1=\sqbrk{\map {\phi'} {x_1} +\epsilon_1}\delta x_1$

where $\epsilon_0\to 0$ as $\delta x_0\to 0$ and $\epsilon_1\to 0$ as $\delta x_1\to 0$.

Substitution of $\delta y_0$ and $\delta y_1$ into $\delta J$ leads to

$\displaystyle \delta J\sqbrk {y;h}=\paren { F_{y'}\psi'+F-y'F_{y'} }\big\rvert_{x=x_1}\delta x_1-\paren {F_{y'}\psi'+F-y'F_{y'} } \big\rvert_{x=x_0}\delta x_0$

$y=\map y x$ is an extremal of $J\sqbrk y$, thus $ \delta J=0$.

$\delta x_0$ and $\delta x_1$ are independent increments.

Hence both remaining terms in $\delta J$ vanish independently.