Simplest Variational Problem with Subsidiary Conditions for Curve on Surface

Theorem
Let $J \left[{y, z}\right]$ be a functional of the form:


 * $\displaystyle J \left[{y}\right] = \int_a^b F \left({x, y, z, y', z'}\right) \mathrm d x$

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


 * $g \left({x, y, z}\right) = 0$

which satisfy boundary conditions:


 * $y \left({a}\right) = A_1, y \left({b}\right) = B_1$
 * $z \left({a}\right) = A_2, z \left({b}\right) = B_2$

Let $J \left[{y, z}\right]$ have an extremum for the curve $y = y \left({x}\right), z = z \left({x}\right)$.

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

Then there exists a function $\lambda \left({x}\right)$ such that the curve $y = y \left({x}\right), z = z \left({x}\right)$ is an extremal of the functional:


 * $\displaystyle \int_a^b \left({F + \lambda \left({x}\right) g}\right) \mathrm d x$

In other words, $y = y \left({x}\right)$ satisfies the differential equations:

Proof
Let $J[y]$ be a functional, for which the curve $y=y(x),~z=z(x)$ is an extremal with the boundary conditions $y(a)=A,~y(b)=B$ as well as $g\left(x,~y,~z\right)=0$.

Choose an arbitrary point $x_1$ from the interval $\left[{{a}\,.\,.\,{b}}\right]$.

Let $\delta y(x)$ and $\delta z(x)$ 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:


 * $\displaystyle\Delta J\left[y;~\delta_1y(x)+\delta_2y(x)\right]=\left(\frac{\delta F}{\delta{y}}\bigg\rvert_{x=x_1}+\epsilon_1\right)\Delta\sigma_1+\left(\frac{\delta F}{\delta{z}}\bigg\rvert_{x=x_1}+\epsilon_2\right)\Delta\sigma_2$

where


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

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

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

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 $\Delta J[y,~z]$

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

Then the variation of the functional $J[y]$ at the point $x_1$ is


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

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

$\displaystyle \frac{ \delta F }{ \delta y }- \frac{ g_y }{ g_z } \frac{ \delta F }{ \delta z }=F_y-\frac{\mathrm{d} }{\mathrm{d}{x} }F_{y'}-\frac{g_y}{g_z}\left( F_z- \frac{\mathrm{d} }{\mathrm{d}{x} }F_{z'}\right)=0$

The latter equation can be rewritten as

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

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