I follow the lecture of Calculus of variations, the topic is Euler-Lagrange's equations.
$$ \mathcal{L}: \ A\times\mathbb{R}\times\mathbb{R}^n\to \mathbb{R}, \ x\in A, \ u\in \mathbb{R}, \ \xi\in \mathbb{R}^n,\ \text{open} \ A\subset \mathbb{R}^n, \ n\geq 1 $$ If it is well defined, we let $F:\ \mathcal{C}^1\to\mathbb{R}$ $$ F(u)=\int_{A}\mathcal{L}(x,u(x),\nabla u(x))dx $$ We assume that $u\in \mathcal{C}^1(A)$ is a minimizer of $F$ w.r.t. compact variation $$ F(u)\leq F(u+\varphi) \; \forall \varphi\in \mathcal{C}_c^1(A) $$ $$ \text{fix} \; \varphi\in \mathcal{C}_c^1(A) \; \text{and let} \; f:\mathbb{R}\to\mathbb{R} \; \text{such that} \; f(\varepsilon)=F(u+\varepsilon\varphi) $$
I didn't get why we need to introduce $f(\varepsilon)$. Could you please explain? thanks.
I guess that the idea of introducing the new function $f$ is to have the $f'(0)=0$ construction which can be deduced since $f(0)=F(u)$ and $u$ is a minimizer, and therefore we have $f'(0)=F'(u)=0$