Simplest Variational Problem with Subsidiary Conditions for Curve on Surface

Theorem
Let $J \sqbrk {y, z}$ be a (real) functional of the form:


 * $\ds J \sqbrk y = \int_a^b \map F {x, y, z, y', z'} \rd x$

Let there exist admissible curves $y, z$ lying on the surface:


 * $\map g {x, y, z} = 0$

which satisfy boundary conditions:


 * $\map y a = A_1, \map y b = B_1$
 * $\map z a = A_2, \map z b = B_2$

Let $J \sqbrk {y, z}$ have an extremum for the curve $y = \map y x, z = \map z x$.

Let $g_y$ and $g_z$ not simultaneously vanish at any point of the surface $g = 0$.

Then there exists a function $\map \lambda x$ such that the curve $y = \map y x, z = \map z x$ is an extremal of the functional:


 * $\ds \int_a^b \paren {F + \map \lambda x g} \rd x$

In other words, $y = \map y x$ satisfies the differential equations:

Proof
Let $J \sqbrk y$ be a functional, for which the curve $y = \map y x, z = \map z x$ is an extremal with the boundary conditions $\map y a = A, \map y b = B$ as well as $\map g {x, y, z} = 0$.

Choose an arbitrary point $x_1$ from the interval $\closedint a b$.

Let $\delta \map y x$ and $\delta \map z x$ be functions, different from zero only in the neighbourhood of $x_1$.

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


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

where:


 * $\ds \Delta \sigma_1 = \int_a^b \delta \map y x$
 * $\ds \Delta \sigma_2 = \int_a^b \delta \map z x$

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

We now require that the varied curve $y^* = \map y x + \map {\delta_y} x$, $z^* = \map y x + \map {\delta_z} x$ satisfies the condition $\map g {x, y^*, z^*} = 0$.

This condition limits arbitrary varied curves only to those which still satisfy the original constraint on the surface.

By using constraints on $g$, we can follow the following chain of equalities:

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

and overbar indicates that corresponding derivatives are evaluated along certain intermediate curves.

By hypothesis, either $\bigvalueat {g_y} {x \mathop = x_1}$ or $\bigvalueat {g_z} {x \mathop = x_1}$ is nonzero.

Suppose $\bigvalueat {g_z} {x \mathop = x_1} \ne 0$.

Then the previous result can be rewritten as:

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

Substitute this back into the equation for $\Delta J \sqbrk {y, z}$

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:


 * $\delta J = \paren {\valueat {\dfrac {\delta F} {\delta y} } {x \mathop = x_1} - \valueat {\paren {\dfrac {g_y} {g_z} \frac {\delta F} {\delta z} } } {x \mathop = x_1} } \Delta \sigma_1$

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


 * $\dfrac {\delta F} {\delta y} - \dfrac {g_y} {g_z} \dfrac {\delta F} {\delta z} = F_y - \dfrac \d {\d x} F_{y'} - \dfrac {g_y} {g_z} \paren {F_z - \dfrac \d {\d x} F_{z'} } = 0$

The latter equation can be rewritten as


 * $\dfrac {F_y - \dfrac \d {\d x} F_{y'} } {g_y} = \dfrac {F_z - \dfrac \d {\d x} F_{z'} } {g_z}$

If we denote this ratio by $-\map \lambda x$, then this ratio can be rewritten as two equations presented in the theorem.