# Simplest Variational Problem with Subsidiary Conditions

## Theorem

Let $J \sqbrk y$ and $K \sqbrk y$ be functionals, such that

$\displaystyle J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$
$\displaystyle K \sqbrk y = \int_a^b \map G {x, y, y'} \rd x = l$

where $l$ is a constant.

Let $y = \map y x$ be an extremum of $F \sqbrk y$, and satisfy boundary conditions:

$\map y a = A$
$\map y b = B$

Then, if $y = \map y x$ is not an extremal of $K \sqbrk y$, there exists a constant $\lambda$ such that $y = \map y x$ is an extremal of the functional:

$\displaystyle \int_a^b \paren {F + \lambda G} \rd x$

or, in other words, $y = \map y x$ satisfies:

$F_y - \dfrac {\d} {\d x} F_{y'} + \lambda \paren {G_y - \dfrac {\d} {\d x} G_{y'} } = 0$

## Proof

Let $J \sqbrk y$ be a functional, for which $y = \map y x$ is an extremal with the boundary conditions $\map y a = A$ and $\map y b = B$.

Choose two points, $x_1$ and $x_2$ from the interval $\closedint a b$.

Let $\delta_1 \map y x$ and $\delta_2 \map y x$ be functions, different from zero only in the neighbourhood of $x_1$ and $x_2$ respectively.

Then we can exploit the definition of variational derivative in the following way:

$\Delta J \sqbrk {y; \, \delta_1 \map y x + \delta_2 \map y x} = \paren {\valueat {\dfrac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {\valueat {\dfrac {\delta F} {\delta y} } {x \mathop = x_2} + \epsilon_2} \Delta \sigma_2$

where:

$\displaystyle \Delta \sigma_1 = \int_a^b \delta_1 \map y x$
$\displaystyle \Delta \sigma_2 = \int_a^b \delta_2 \map y x$

and $\epsilon_1, \epsilon_2 \to 0$ as $\Delta \sigma_1, \Delta \sigma_2 \to 0$.

We now require that the varied curve $y^* = \map y x + \delta_1 \map y x + \delta_2 \map y x$ satisfies the condition $K \sqbrk {y^*} = K \sqbrk y$.

This way we limit arbitrary varied curves to those who still satisfy the given functional constraint.

Similarly, write down $\Delta K \sqbrk y$ as:

$\displaystyle \Delta K \sqbrk y = K \sqbrk{y^*} - K \sqbrk y = \valueat {\paren {\dfrac {\delta G} {\delta y} } {x \mathop = x_1} + \epsilon_1'} \Delta \sigma_1 + \valueat {\paren {\dfrac {\delta G} {\delta y} } {x \mathop = x_2} + \epsilon_2'} \Delta\sigma_2$

where $\epsilon_1', \epsilon_2' \to 0$ as $\Delta \sigma_1, \Delta \sigma_2 \to 0$.

Suppose $x_2$ is such that:

$\valueat {\dfrac {\delta G} {\delta y} } {x \mathop = x_2} \ne 0$

Such a point exists, because by assumption $\map y x$ is not an extremal of $K \sqbrk y$.

Since $\Delta K = 0$, the previous equation can be rewritten as

$\Delta \sigma_2 = -\paren {\dfrac {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} } {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} } + \epsilon'} \Delta \sigma_1$

where $\epsilon' \to 0$ as $\Delta \sigma_1 \to 0$.

Set :

$\lambda = -\dfrac {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_2} } {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} }$

Substitute this into the formula for $\Delta J$:

 $\displaystyle \Delta J \sqbrk {y; \delta_1 \map y x + \delta_2 \map y x}$ $=$ $\displaystyle \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_2} + \epsilon_2} \Delta \sigma_2$ $\displaystyle$ $=$ $\displaystyle \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {-\valueat {\lambda \frac {\delta G} {\delta y} } {x \mathop = x_2} + \epsilon_2} \paren {-\paren {\frac {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} } {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} } + \epsilon'} \Delta \sigma_1}$ $\displaystyle$ $=$ $\displaystyle \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {\lambda \valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} + \lambda \valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} \epsilon' - \epsilon_2 \frac {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} } {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} } - \epsilon_2 \epsilon'} \Delta \sigma_1$ $\displaystyle$ $=$ $\displaystyle \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \lambda \valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} } \Delta \sigma_1 + \epsilon \Delta \sigma_1$

where $\epsilon \to 0$ as $\Delta \sigma_1 \to 0$.

Then the variation of the functional $J \sqbrk y$ at the point $x_1$ is:

$\delta J = \paren {\valueat {\dfrac {\delta F} {\delta y} } {x \mathop = x_1} + \lambda \valueat {\dfrac {\delta G} {\delta y} } {x \mathop = x_1} } \Delta \sigma$

A necessary condition for $\delta J$ to vanish for any $\Delta \sigma$ and arbitrary $x_1$ is:

$\dfrac {\delta F} {\delta y} + \lambda \dfrac {\delta G} {\delta y} = F_y - \dfrac \d {\d x} F_{y'} + \lambda \paren {G_y - \dfrac \d {\d x} G_{y'} } = 0$

$\blacksquare$