I am provided with $F(x,y(x),y'(x))$ which is a function of $x, y(x), y'(x)$ where $y(x)$ is function of $x$.
My goal is to find the stationary function of the quantity described below: $$ I[y] = \int_{x_1}^{x_2}F(x,y,y')\,dx + F_1(y_1,y_1') + F_2(y_2,y_2') $$ where $y_1 = y(x_1) \;,\; y_1' = y'(x_1) $ and $y_2 = y(x_2) \; , \; y_2' = y'(x_2)$. And $F_1$ is function of $y_1$ and $y_1'$ and similarly for $F_2$.
I am not provided with any boundary conditions and I want to solve for the stationary function of $I[f]$. Will there be any additional conditions imposed on the stationary function other than Euler-Lagrange equation?
Here is what I have concluded, please let me know whether the approach is correct or not:
Let $y(x)$ be the stationary function and $\eta(x)$ is any arbitrary function, such that $\bar{y}(x)$ any general satisfying the functional.
$\bar{y}(x) = y(x) + \varepsilon*\eta(x)$
Along the similar lines as derivation of Euler-Lagrange Equation :
\begin{equation*}
I(\varepsilon) = \int_{x_1}^{x_2} F(x,\bar{y},\bar{y}')\,dx + F_1(\bar{y}_1,\bar{y}_1') + F_2(\bar{y}_2,\bar{y}_2')
\end{equation*}
\begin{equation*}
\frac{dI}{d\varepsilon} = \int_{x_1}^{x_2} \frac{\partial}{\partial{\varepsilon}}F(x,\bar{y},\bar{y}')\,dx + \frac{d}{d\varepsilon}F_1(\bar{y}_1,\bar{y}_1') + \frac{d}{d\varepsilon}F_2(\bar{y}_2,\bar{y}_2')
\end{equation*}
For the first term that is the integral term, following steps same as Euler-Lagrange derivation and
$ \frac{d}{d\varepsilon}F_1(\bar{y}_1,\bar{y}_1') = \frac{\partial{F_1}}{\partial{\bar{y}_1}}\frac{\partial{\bar{y}_1}}{\partial{\varepsilon}} + \frac{\partial{F_1}}{\partial{\bar{y}_1'}}\frac{\partial{\bar{y}_1'}}{\partial{\varepsilon}}$
$\frac{\partial{\bar{y}_1}}{\partial{\varepsilon}} = \eta(x_1)$ and $\frac{\partial{\bar{y}_1'}}{\partial{\varepsilon}} = \eta'(x_1)$
$ \frac{d}{d\varepsilon}F_1(\bar{y}_1,\bar{y}_1') = \frac{\partial{F_1}}{\partial{\bar{y}_1}}\eta(x_1) + \frac{\partial{F_1}}{\partial{\bar{y}_1'}}\eta'(x_1)$ and $ \frac{d}{d\varepsilon}F_2(\bar{y}_2,\bar{y}_2') = \frac{\partial{F_2}}{\partial{\bar{y}_2}}\eta(x_2) + \frac{\partial{F_2}}{\partial{\bar{y}_2'}}\eta'(x_2)$
\begin{equation*}
\begin{split}
\frac{dI}{d\varepsilon} = \int_{x_1}^{x_2}\left[\frac{\partial{F}}{\partial{\bar{y}}} - \frac{d}{dx}\frac{\partial{F}}{\partial{\bar{y}'}}\right]\eta(x)\,dx +
\frac{\partial{F}}{\partial{\bar{y}'}}\bigg|_{x=x_2}\cdot\eta(x_2) - \frac{\partial{F}}{\partial{\bar{y}'}}\bigg|_{x=x_1}\cdot\eta(x_1) \\
+\frac{\partial{F_1}}{\partial{\bar{y}_1}}\eta(x_1) + \frac{\partial{F_1}}{\partial{\bar{y}_1'}}\eta'(x_1) + \frac{\partial{F_2}}{\partial{\bar{y}_2}}\eta(x_2) + \frac{\partial{F_2}}{\partial{\bar{y}_2'}}\eta'(x_2)
\end{split}
\end{equation*}
For $\varepsilon = 0 \;$, $\bar{y}(x) = y(x)$, that is stationary function so $\frac{dI}{d\varepsilon}\bigg|_{\varepsilon = 0} = 0$:
\begin{equation}
\begin{split}
\frac{dI}{d\varepsilon}\bigg|_{\varepsilon = 0} = \int_{x_1}^{x_2}\left[\frac{\partial{F}}{\partial{y}} - \frac{d}{dx}\frac{\partial{F}}{\partial{y'}}\right]\eta(x)\,dx +
\left[\frac{\partial{F}}{\partial{y'}}\bigg|_{x=x_2} + \frac{\partial{F_2}}{\partial{y_2}}\right]\cdot\eta(x_2) \\
+ \left[-\frac{\partial{F}}{\partial{y'}}\bigg|_{x=x_1}
+\frac{\partial{F_1}}{\partial{y_1}}\right]\cdot\eta(x_1) + \frac{\partial{F_1}}{\partial{y_1'}}\eta'(x_1) + \frac{\partial{F_2}}{\partial{y_2'}}\eta'(x_2) = 0
\end{split}
\end{equation}
Since the above equation holds for any arbitrary $\eta(x)$ therefore:
\begin{align*}
\frac{\partial{F}}{\partial{y}} - \frac{d}{dx}\frac{\partial{F}}{\partial{y'}} &= 0\\
\frac{\partial{F}}{\partial{y'}}\bigg|_{x=x_2} + \frac{\partial{F_2}}{\partial{y_2}} &= 0\\
-\frac{\partial{F}}{\partial{y'}}\bigg|_{x=x_1}
+\frac{\partial{F_1}}{\partial{y_1}} &= 0\\
\frac{\partial{F_1}}{\partial{y_1'}} &= 0\\
\frac{\partial{F_2}}{\partial{y_2'}} &= 0
\end{align*}