Simple Variable End Point Problem/Endpoints on Curves

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

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


 * $\displaystyle J[y] = \int_{x_0}^{x_1}F \left({x, y, y'} \right) \mathrm d x$

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

Suppose $P_0$, $P_1$ lie on curves $y=\phi \left({x} \right)$, $y= \psi \left({x} \right)$.

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

$ \begin{cases} &F_y- \frac { \mathrm { d } } { \mathrm { d } { x } } F_{ y' }=0\\ & \left[ F+ \left( \phi' - y' \right) F_{y'} \right] \big \rvert_{x=x_0}=0\\ & \left[ F+ \left( \psi' - y' \right) F_{y'} \right] \big \rvert_{x=x_1}=0 \end{cases}$

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


 * $ \displaystyle \delta J[y;h]= \int_{x_0}^{x_1} \left({F_y-\frac{ \mathrm{d} }{ \mathrm{d}{x} }F_{y'} } \right)h \left({x } \right)+ F_{y'} \delta y \bigg \rvert_{x=x_0}^{x=x_1}+\left({F- y'F_{y'} } \right) \delta x\bigg\rvert_{x=x_0}^{x=x_1}$

Since the curve $y=y \left({ x} \right)$ is an extremum of $J[y]$, the integral term vanishes:


 * $ \displaystyle \delta J[y;h]= F_{y'} \delta y \bigg \rvert_{x=x_0}^{x=x_1}+\left({F- y'F_{y'} } \right) \delta x\bigg\rvert_{x=x_0}^{x=x_1}$

According to Taylor's theorem:


 * $\displaystyle \delta y_0= \left[{ \phi' \left({ x_0} \right) +\epsilon_0} \right] \delta x_0, \quad \delta y_1= \left[{ \phi' \left({ x_1} \right)+\epsilon_1} \right] \delta x_1$

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


 * $\displaystyle \delta J[{ y; h}]= \left({ F_{ y'} \psi' + F - y' F_{ y'} } \right) \big \rvert_{ x= x_1} \delta x_1 - \left({ F_{ y'} \psi' + F - y' F_{ y'} } \right) \big \rvert_{x=x_0} \delta x_0$

$y=y\left({ x} \right)$ is an extremal of $J[{ y}]$, thus $ \delta J =0$.

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

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