Simple Variable End Point Problem/Endpoints on Curves

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Theorem

Let $y$, $F$, $\phi$ and $\psi$ 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 Euler and transversality equations:

$ \begin{cases} &F_y - \dfrac {\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}$

Proof

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

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

Since the curve $y = \map y x$ is an extremum of $\map J y$, 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.

$\blacksquare$


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