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 $\Delta \sigma_1, \Delta \sigma_2 \to 0$. {{Explain|support above statement with an actual calculation}} 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: {{begin-eqn}} {{eqn | l = 0       | r = \int_a^b \paren {\map g {x, y^*, z^*} - \map g {x, y, z} } \rd x       | c =  }} {{eqn | r = \int_a^b \paren {\overline {g_y} \delta y + \overline {g_z} \delta z} }} {{eqn | r = \paren {\bigvalueat {g_y} {x \mathop = x_1} + \epsilon'_1} \Delta \sigma_1 + \paren {\bigvalueat {g_z} {x \mathop = x_1} + \epsilon'_2} \Delta \sigma_2 }} {{end-eqn}} 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. {{Explain|how?}} 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: {{begin-eqn}} {{eqn | l = \Delta \sigma_2       | r = -\Delta \sigma_1 \frac {\bigvalueat {g_y} {x \mathop = x_1} + \epsilon'_1} {\bigvalueat {g_z} {x \mathop = x_1} + \epsilon'_2}       | c =  }} {{eqn | r = -\Delta \sigma_1 \frac {\bigvalueat {g_y} {x \mathop = x_1} + \epsilon'_1} {\bigvalueat {g_z} {x \mathop = x_1} \paren {1 + \frac {\epsilon'_2} {\bigvalueat {g_z} {x \mathop = x_1} } } } }} {{eqn | r = -\Delta \sigma_1 \frac {\bigvalueat {g_y} {x \mathop = x_1} + \epsilon'_1} {\bigvalueat {g_z} {x \mathop = x_1} } \sum_{n \mathop = 0}^\infty \paren {\frac {\epsilon_2'} {\bigvalueat {g_z} {x \mathop = x_1} } }^n        | c = Sum of Infinite Geometric Sequence, holds for $\size {\epsilon_2'} \to 0$ }} {{eqn | r = -\Delta \sigma_1 \frac {\bigvalueat {g_y} {x \mathop = x_1} } {\bigvalueat {g_z} {x \mathop = x_1} } - \Delta \sigma_1 \frac {\epsilon'_1} {\bigvalueat {g_z} {x \mathop = x_1} } - \Delta \sigma_1 \frac {\bigvalueat {g_y} {x \mathop = x_1} + \epsilon'_1} {\bigvalueat {g_z} {x \mathop = x_1} } \sum_{n \mathop = 1}^\infty \paren { {\frac {\epsilon_2'} {\bigvalueat {g_z} {x \mathop = x_1} } } }^n  }} {{eqn | ll= \leadsto       | l = \Delta \sigma_2       | r = -\paren {\frac {\bigvalueat {g_y} {x \mathop = x_1} } {\bigvalueat {g_z} {x \mathop = x_1} } + \epsilon'} \Delta \sigma_1       | c =  }} {{end-eqn}} where $\epsilon'\to 0$ as $\Delta \sigma_1 \to 0$. Substitute this back into the equation for $\Delta J \sqbrk {y, z}$  {{begin-eqn}} {{eqn | l = \Delta J       | r = \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {\valueat {\frac {\delta F} {\delta z} } {x \mathop = x_1} + \epsilon_2} \Delta \sigma_2       | c = }} {{eqn | r = \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {\valueat {\frac {\delta F} {\delta z} } {x \mathop = x_1} + \epsilon_2} \paren {-\paren {\frac {\bigvalueat {g_y} {x \mathop = x_1} } {\bigvalueat {g_z} {x \mathop = x_1} } + \epsilon'} \Delta \sigma_1} }} {{eqn | r = \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} - \valueat {\paren {\frac {g_y} {g_z} \frac {\delta F} {\delta z} } } {x \mathop = x_1} } \Delta \sigma_1 + \paren {\epsilon_1 - \epsilon_2 \frac {\bigvalueat {g_y} {x \mathop = x_1} } {\bigvalueat {g_z} {x \mathop = x_1} } - \epsilon' \valueat {\frac {\delta F} {\delta z} } {x \mathop = x_1} - \epsilon_2 \epsilon'} \Delta \sigma_1 }} {{eqn | r = \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} - \valueat {\paren {\frac {g_y} {g_z} \frac {\delta F} {\delta z} } } {x \mathop = x_1} } \Delta \sigma_1 + \epsilon \Delta \sigma_1 }} {{end-eqn}} 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. {{qed}}