In the Euler-Lagrange equation we consider the functional $$ I[y,y',x] = \int_a^b F(y,y',x) \operatorname{dx}$$ where $y = y(x)$ is a function of $x$.
It textbooks the stationary point of the functional $I$ is given such that $$ \frac{\operatorname{d}}{\operatorname{d\alpha}}I[y+\alpha \eta]\ \bigg |_{\alpha=0}=0 \ ,$$ where $\eta = \eta(x)$ may by chosen such that it coincides with the endpoints of $y$; $y(a)$ and $y(b)$.
What do we mean when we require that $I$ is stationary with respect to a function $y$?
Best Answer
In finding the function $y$ which makes $I$ stationary, we go about looking for which $y$ the first-order change in $I$ with respect to the function $y$ vanishes.
Hence a direct analogy can be made to stationary points of functions, as opposed to functionals.
The parameter $\alpha$ that we used is just a means by which we can look into changes in $I$ with respect to the function $y$ itself. This as opposed to changes in $y = y(x)$ with respect to the variable $x$, which isn't the topic at hand. Furthermore, if we where to Taylor expand $I$ with respect to $y$, we would go looking for changes of order $\operatorname{O}(\alpha^2)$.