The Euler-Lagrange equation for solving for stationary point of functional $J$ with fixed endpoints $x_1$ and $x_2$ (i.e. $\delta$y($x_1$) = $\delta$y($x_2$) = 0) states:

$\hspace{5 cm}\Large\frac{\partial f}{\partial y(x)}-\frac{d}{dx}\left (\frac{\partial f}{\partial y'(x)} \right )\large=0, \hspace{1 cm}\large x_1 < x < x_2$

My question is, why is it a necessary condition for this expression to equal zero over the entire interval of $x_1 < x < x_2$? Could there not be a case where the functional $J$ increases for part of the interval ($x_1, x_2$) given $\delta y(x)$and decreases for the rest of the interval, such that $\delta J$ equals zero over the interval as a whole when integrated, and $J$ is therefore still stationary?


1 Answer 1


This is necessary because the action integral should vanish for arbitrary variations.


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