Simplest Variational Problem with Subsidiary Conditions
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Theorem
Let $J \sqbrk y$ and $K \sqbrk y$ be (real) functionals, such that
- $\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$
- $\ds K \sqbrk y = \int_a^b \map G {x, y, y'} \rd x = l$
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:
- $\ds \int_a^b \paren {F + \lambda G} \rd x$
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:
- $\ds \Delta \sigma_1 = \int_a^b \delta_1 \map y x$
- $\ds \Delta \sigma_2 = \int_a^b \delta_2 \map y x$
and $\epsilon_1, \epsilon_2 \to 0$ as $\Delta \sigma_1, \Delta \sigma_2 \to 0$.
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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:
- $\ds \Delta K \sqbrk y = K \sqbrk{y^*} - K \sqbrk y = \valueat {\paren {\dfrac {\delta G} {\delta y} } {x \mathop = x_1} + \epsilon_1'} \Delta \sigma_1 + \valueat {\paren {\dfrac {\delta G} {\delta y} } {x \mathop = x_2} + \epsilon_2'} \Delta\sigma_2$
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 $\Delta \sigma_1 \to 0$.
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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 $\Delta J$:
\(\ds \Delta J \sqbrk {y; \delta_1 \map y x + \delta_2 \map y x}\) | \(=\) | \(\ds \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_2} + \epsilon_2} \Delta \sigma_2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {-\valueat {\lambda \frac {\delta G} {\delta y} } {x \mathop = x_2} + \epsilon_2} \paren {-\paren {\frac {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} } {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} } + \epsilon'} \Delta \sigma_1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {\lambda \valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} + \lambda \valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} \epsilon' - \epsilon_2 \frac {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} } {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} } - \epsilon_2 \epsilon'} \Delta \sigma_1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \lambda \valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} } \Delta \sigma_1 + \epsilon \Delta \sigma_1\) |
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:
- $\dfrac {\delta F} {\delta y} + \lambda \dfrac {\delta G} {\delta y} = F_y - \dfrac \d {\d x} F_{y'} + \lambda \paren {G_y - \dfrac \d {\d x} G_{y'} } = 0$
$\blacksquare$
Sources
- 1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 2.12$: Variational Problems with Subsidiary Conditions