Simplest Variational Problem with Subsidiary Conditions for Curve on Surface

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


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

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:


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

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:


 * $\displaystyle\Delta J\sqbrk{y;~\delta_1\map y x+\delta_2\map y x}=\paren{\frac{\delta F}{\delta y}\bigg\rvert_{x=x_1}+\epsilon_1}\Delta\sigma_1+\paren{\frac{\delta F}{\delta z}\bigg\rvert_{x=x_1}+\epsilon_2}\Delta\sigma_2$

where


 * $\displaystyle\Delta\sigma_1=\int_{a}^{b}\delta \map y x,~\Delta\sigma_2=\int_{a}^{b}\delta \map z x$

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 $g_y \rvert_{x=x_1}$ or $g_z \rvert_{x=x_1}$ is nonzero.

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

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


 * $\displaystyle\delta J=\paren{\frac{\delta F}{\delta y}\bigg\rvert_{x=x_1}-\paren{\frac{g_y}{g_z}\frac{\delta F}{\delta z} } \bigg\rvert_{x=x_1} }\Delta\sigma_1$

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

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

The latter equation can be rewritten as

$\displaystyle\frac{F_y-\frac \d {\d x}F_{y'} }{g_y}=\frac{F_z-\frac \d {\d x}F_{z'} }{g_z}$.

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