The problem in calculus of variations is to find a curve $y[]:[x_1,x_2]\to \mathbb{R}$, that minimizes, maximizes, or renders stationary the functional
\[I\left[y\left[\cdot\right]\right]=\int_{x_1}^{x_2}f\left[x,y[x],y'[x]\right]\, dx.\]where $f:[x_1,x_2]\times \mathbb{R} \times \mathbb{R}\to \mathbb{R}$, i.e., $f$ is a function of three arguments. In class we considered a simpler functional of the form
\[I\left[y\left[\cdot\right]\right]=\int_{x_1}^{x_2}f\left[y[x],y'[x]\right]\, dx,\]i.e., the integrand does not explicitly depend of ``$x$’’.
Additionally, the function $y[\cdot]$ is required to satisfy the ``boundary conditions’’
\[\begin{align} y[x_1]&=y_1\\ y\left[x_2 \right]&=y_2 \end{align}\]Thus, in general we want a function $y\left[\cdot \right]$ to belong to the set of functions that map real numbers between and including $x_1$ and $x_2$ to real numbers; are well-behaved (in terms of continuity and smoothness) so that the integral $I\left[\cdot \right]$ is well-defined; and at $x_1$ and $x_2$ take values $y_1$ and $y_2$, respectively. This set is often denoted as $\mathcal{V}$, and referred to as the set of admissible functions. It is mathematically denoted as
\[\mathcal{V}:=\{y\left[\cdot \right]\in H^{1}(\mathcal{D}; \mathbb{R})\ :\ y\left[x_1 \right]=y_1,\ y\left[x_2 \right]=y_2 \},\]where $\mathcal{D}=(x_1,x_2)$ (open interval) is called the domain of $y[\cdot]$, and $H^{1}(\mathcal{D}; \mathbb{R})$ is a so called ``Sobolev space of functions’’ which map points in $\mathcal{D}$ to points in $\mathbb{R}$. A space of functions that is closely related to $\mathcal{V}$ is $T\mathcal{V}$, called the space of admissible variations, and defined as
\[\begin{equation} T\mathcal{V}:= \{y\left[\cdot \right]\in H^{1}(\mathcal{D}; \mathbb{R})\ :\ y\left[x_1 \right]=0,\ y\left[x_2 \right]=0 \}. \end{equation}\]Having defined $\mathcal{V}$, we can more rigorously define $I\left[\cdot \right]$ as $I:\mathcal{V} \to \mathbb{R}$, where
\[\begin{equation} I\left[y\left[\cdot\right]\right]=\int_{x_1}^{x_2}f\left[x,y[x],y'[x]\right]\, dx. \end{equation}\]We consider functions $y_{\epsilon}\left[\cdot \right]$ that lie around the candidate function $y\left[\cdot \right]$. We can decompose $y_{\epsilon}\left[\cdot \right]$ as
\[\begin{equation} y_{\epsilon}[\cdot]=y[\cdot]+\epsilon \delta y[\cdot], \end{equation}\]where $\delta y\in T\mathcal{V}$, and $\epsilon$ is a real number. The value of $\epsilon$ controls how close $y_{\epsilon}\left[\cdot \right]$ is to $y\left[ \cdot \right]$. As $\epsilon \to 0$ the function $y_{\epsilon}\left[\cdot \right]\to y\left[ \cdot\right]$.
As motivated in class the condition for $I[]$ to be stationary at $y[\cdot]$ is that
\[\begin{equation} \left\langle I[y],\delta y\right\rangle=0, \label{eq:StationarityCondition} \end{equation}\]for all $\delta y \in T\mathcal{V}$, where
\[\begin{equation} \left\langle I[y],\delta y\right\rangle:= \left. \frac{d}{d\epsilon} I[y_{\epsilon}(\cdot)]\right|_{\epsilon=0}, \end{equation}\]and
\[\begin{equation} I[y_{\epsilon}(\cdot)]=\int_{x_1}^{x_2}f(x,y_{\epsilon}(x),y'_{\epsilon}(x))\, dx \end{equation}\]Moving the derivative to be under the integral sign we get that
\[\left\langle I[y],\delta y\right\rangle=\begin{equation} \int_{x_1}^{x_2} \left[\partial_{y} f(x, y(x),y'(x))\delta y(x)+ \partial_{y'} f(x, y(x),y'(x))\delta y'(x)\right]\, dx \end{equation}\]Integrating the second term by parts we get that
\[\begin{equation} \left\langle I[y],\delta y\right\rangle=\int_{x_1}^{x_2}\partial_{y}f(y(x),y'(x))\delta y(x)\,dx+\\ \left. \partial_{y'} f(y(x),y'(x))\delta y(x)\right|_{x_1}^{x_2} \\-\int_{x_1}^{x_2}\frac{d}{dx} \partial_{y'} f(y(x),y'(x))\delta y(x)\, dx \end{equation}\]Noting that, since $\delta y\in T\mathcal{V}$, $\delta y[x_1]=\delta y[x_2]=0$, the second term in the above equation vanishes leads us to
\[\begin{equation} \left\langle I[y],\delta y\right\rangle=\int_{x_1}^{x_2}\partial_{y}f(y(x),y'(x))\delta y(x)\,dx-\int_{x_1}^{x_2}\frac{d}{dx} \partial_{y'} f(y(x),y'(x))\delta y(x)\, dx \end{equation}\]Simplifying the above equation we get
\[\begin{equation} \left\langle I[y],\delta y\right\rangle=\int_{x_1}^{x_2} \left[ \partial_{y}f(y(x),y'(x))-\frac{d}{dx} \partial_{y'} f(y(x),y'(x))\right]\delta y(x)\, dx \end{equation}\]The above equation in conjunction with \eqref{eq:StationarityCondition} implies that
\[\begin{equation} \partial_{y}f(y(x),y'(x))-\frac{d}{dx} \partial_{y'} f(y(x),y'(x))=0, \end{equation}\]for all $x\in \mathcal{D}$.