Euler-Lagrange equations

What are the stationary paths of the functional

S [y] = \int_{a}^{b} F (x, y, y^{'}) d x, y (a) = A, y (b) = B, y^{'} = \frac{d y}{d x}

With $\tilde{y} = y + ϵ g$ for arbitrary smooth function $g$ with $g (a) = g (b) = 0$ and real $ϵ$ , we have

\begin{aligned} {\frac{d}{d ϵ} S [\tilde{y}] |}_{ϵ = 0} & = {\frac{d}{d ϵ} \int_{a}^{b} F (x, \tilde{y}, {\tilde{y}}^{'}) d x |}_{ϵ = 0} \\ = \int_{a}^{b} {\frac{d}{d ϵ} F (x, \tilde{y}, {\tilde{y}}^{'}) |}_{ϵ = 0} d x \\ = \int_{a}^{b} {\frac{d}{d ϵ} F (x, \tilde{y}, {\tilde{y}}^{'}) |}_{ϵ = 0} d x \\ = \int_{a}^{b} {(\frac{\partial F}{\partial \tilde{y}} \frac{d \tilde{y}}{d ϵ} + \frac{\partial F}{\partial {\tilde{y}}^{'}} \frac{d {\tilde{y}}^{'}}{d ϵ}) |}_{ϵ = 0} d x \\ = \int_{a}^{b} (\frac{\partial F}{\partial y} g + \frac{\partial F}{\partial y^{'}} g^{'}) d x \\ = \int_{a}^{b} (\frac{\partial F}{\partial y} - \frac{d}{d x} (\frac{\partial F}{\partial y^{'}})) g d x + {[g (x) \frac{\partial F}{\partial y^{'}}]}_{a}^{b}^{0} \end{aligned}

Setting this to zero (for stationary $y$ ), we finally get the condition that the coefficient of $g$ must be uniformly zero, i.e.,

\frac{\partial F}{\partial y} - \frac{d}{d x} (\frac{\partial F}{\partial y^{'}}) = 0

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