Exercise using the derivation of the Euler-Lagrange equation

Question

Here is a exercice using the derivation of the Euler-Lagrange equation:

Here is the exercice:

For a given function $f(x,u,u')$ and constants $K_1, K_2$ minimize the functional using the Euler-Lagrange equation $$ F(u) = \int_{a}^{b}f(x,u(x),u'(x))dx - K_1 \cdot u(b) - \frac{1}{2}K_2u^2(b) $$

Here is the answer:

\begin{align*} F(u + \varepsilon\phi) = &\int_{a}^{b}f(x,u(x) + \varepsilon\phi(x),u'(x) + \varepsilon\phi'(x))\:dx - \\ &- K_1 \cdot(u(b) + \varepsilon\phi(b)) - \frac{1}{2}K_2 \cdot (u(b) + \varepsilon\phi(b))^2 \tag{1}\\ 0 = \frac{d}{d\varepsilon}F(u + \varepsilon\phi)\Big|_{\varepsilon = 0} = &\int_{a}^{b}f_{u}(x,u(x),u'(x))\phi(x) + f_{u'}(x,u(x),u'(x))\phi'(x)\:dx \\ &- K_1 \cdot\phi(b) - K_2 u(b) \cdot \phi(b) \tag{2}\\ = &\int_{a}^{b}f_{u}(x,u(x),u'(x))\phi(x) - \frac{d}{dx}(f_{u'}(x,u(x),u'(x)))\phi(x)\:dx \\ &-f_{u'}(a,u(a),u'(a))\phi(a) + f_{u'}(b,u(b),u'(b))\phi(b) - K_1\cdot \phi(b) - K_2u(b)\cdot \phi(b)\tag{3} \\ \end{align*}

Here are my questions regarding this proof:

1. The expression $0 = \frac{d}{d\varepsilon}F(u + \varepsilon\phi)\Big|_{\varepsilon = 0}$ can be read as "the derivativ of $F(u + \varepsilon\phi)$ with respect to $\varepsilon$ at the point $\varepsilon = 0$ correct?

2. How $- K_1 \cdot(u(b) + \varepsilon\phi(b))$ in (1) becomes $- K_1 \cdot\phi(b))$ in (2). Why doesn't it become $- K_1 \cdot u(b))$

3. What is beeing done between (2) and (3)? Is it integration by parts?

Answer 1

1. Yes.

2. Only the term linear in $\varepsilon$ contributes.

3. Yes.