Simplest Variational Problem with Subsidiary Conditions for Curve on Surface

Theorem
Let $J[y,~z]$ be a functional of the form


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

Suppose admissible curves $y,~z$ lie on the surface


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

and satisfy boundary conditions


 * $y(a)=A_1,~y(b)=B_1,~z(a)=A_2,~z(b)=B_2$

Suppose, $J[y,~z]$ has an extremum for the curve $y=y(z),~z=z(x)$.

Then, if $g_y$ and $g_z$ are not simultaneously vanishing at any point of the surface $g=0$,

then there exists a function $\lambda(x)$,

such that the curve $y=y(x),~z=z(x)$ is an extremal of the functional


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

or, in other words, $y=y(x)$ satisfies the differential equations

$\begin{split} &\displaystyle F_y+\lambda g_y-\frac{\mathrm{d} }{\mathrm{d}{x} }F_{y'}=0\\ &\displaystyle F_z+\lambda g_z-\frac{\mathrm{d} }{\mathrm{d}{x} }F_{z'}=0\\ \end{split}$

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 $g(x,~y^*,~z^*)=0$.

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


 * $\displaystyle\Delta\sigma_2=-\left(\frac{g_y\rvert_{x=x_1} }{ g_z\rvert_{x=x_1} }+\epsilon'\right)\Delta\sigma_1$

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

Substitute this back into the equation for $\Delta J[y,~z]$


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

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

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


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

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

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

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