Let $J \sqbrk y$ be a functional, for which $y = \map y x$ is an extremal with the boundary conditions $\map y a = A$ and $\map y b = B$. Choose two points, $x_1$ and $x_2$ from the interval $\closedint a b$. Let $\delta_1 \map y x$ and $\delta_2 \map y x$ be functions, different from zero only in the neighbourhood of $x_1$ and $x_2$ respectively. Then we can exploit the definition of variational derivative in the following way: :$\Delta J \sqbrk {y; \, \delta_1 \map y x + \delta_2 \map y x} = \paren {\valueat {\dfrac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {\valueat {\dfrac {\delta F} {\delta y} } {x \mathop = x_2} + \epsilon_2} \Delta \sigma_2$ where: :$\ds \Delta \sigma_1 = \int_a^b \delta_1 \map y x$ :$\ds \Delta \sigma_2 = \int_a^b \delta_2 \map y x$ 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 $y^* = \map y x + \delta_1 \map y x + \delta_2 \map y x$ satisfies the condition $K \sqbrk {y^*} = K \sqbrk y$. This way we limit arbitrary varied curves to those who still satisfy the given functional constraint. Similarly, write down $\Delta K \sqbrk y$ as: :$\ds \Delta K \sqbrk y = K \sqbrk{y^*} - K \sqbrk y = \valueat {\paren {\dfrac {\delta G} {\delta y} } {x \mathop = x_1} + \epsilon_1'} \Delta \sigma_1 + \valueat {\paren {\dfrac {\delta G} {\delta y} } {x \mathop = x_2} + \epsilon_2'} \Delta\sigma_2$ where $\epsilon_1', \epsilon_2' \to 0$ as $\Delta \sigma_1, \Delta \sigma_2 \to 0$. Suppose $x_2$ is such that: :$\valueat {\dfrac {\delta G} {\delta y} } {x \mathop = x_2} \ne 0$ Such a point exists, because by assumption $\map y x$ is not an extremal of $K \sqbrk y$. Since $\Delta K = 0$, the previous equation can be rewritten as :$\Delta \sigma_2 = -\paren {\dfrac {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} } {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} } + \epsilon'} \Delta \sigma_1$ where $\epsilon' \to 0$ as $\Delta \sigma_1 \to 0$. {{Explain|how?}} Set : :$\lambda = -\dfrac {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_2} } {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} }$ Substitute this into the formula for $\Delta J$: {{begin-eqn}} {{eqn | l = \Delta J \sqbrk {y; \delta_1 \map y x + \delta_2 \map y x}       | r = \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_2} + \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 {\lambda \frac {\delta G} {\delta y} } {x \mathop = x_2} + \epsilon_2} \paren {-\paren {\frac {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} } {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} } + \epsilon'} \Delta \sigma_1} }} {{eqn | r = \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \epsilon_1} \Delta \sigma_1 + \paren {\lambda \valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} + \lambda \valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} \epsilon' - \epsilon_2 \frac {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_1} } {\valueat {\frac {\delta G} {\delta y} } {x \mathop = x_2} } - \epsilon_2 \epsilon'} \Delta \sigma_1 }} {{eqn | r = \paren {\valueat {\frac {\delta F} {\delta y} } {x \mathop = x_1} + \lambda \valueat {\frac {\delta G} {\delta y} } {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: :$\delta J = \paren {\valueat {\dfrac {\delta F} {\delta y} } {x \mathop = x_1} + \lambda \valueat {\dfrac {\delta G} {\delta y} } {x \mathop = x_1} } \Delta \sigma$ A necessary condition for $\delta J$ to vanish for any $\Delta \sigma$ and arbitrary $x_1$ is: :$\dfrac {\delta F} {\delta y} + \lambda \dfrac {\delta G} {\delta y} = F_y - \dfrac \d {\d x} F_{y'} + \lambda \paren {G_y - \dfrac \d {\d x} G_{y'} } = 0$ {{qed}}