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:

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

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:

\(\ds F_y - \dfrac \d {\d x} F_{y'}\) \(=\) \(\ds 0\)
\(\ds \bigvalueat {\paren {F + \paren {\phi' - y'} F_{y'} } } {x \mathop = x_0}\) \(=\) \(\ds 0\)
\(\ds \bigvalueat {\paren {F + \paren {\psi' - y'} F_{y'} } } {x \mathop = x_1} = 0\) \(=\) \(\ds 0\)


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Proof

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

$\ds \delta J \sqbrk{y; h} = \int_{x_0}^{x_1} \intlimits {\paren {F_y - \dfrac \d {\d x} F_{y'} } \map h x + F_{y'} \delta y} {x \mathop = x_0} {x \mathop = x_1} + \bigintlimits {\paren {F - y'F_{y'} } \delta x} {x \mathop = x_0} {x \mathop = x_1}$

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

$\ds \delta J \sqbrk {y; h} = \bigintlimits {F_{y'} \delta y} {x \mathop = x_0} {x \mathop = x_1} + \bigintlimits {\paren {F - y'F_{y'} } \delta x} {x \mathop = x_0} {x \mathop = x_1}$

According to Taylor's Theorem:

$\delta y_0 = \paren {\map {\phi'} {x_0} + \epsilon_0} \delta x_0$
$\delta y_1 = \paren {\map {\phi'} {x_1} + \epsilon_1} \delta x_1$

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


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

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

$y = \map y x$ 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.

$\blacksquare$


Sources

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