Find $u \in M=C^1[0,1]$ such that $J(u) = \min_{y \in M} J(y)$, where
$$J(y) = \frac{1}{2}\int_0^1 (y'^2+y^2) \mathrm{d}x + \frac{1}{2}[y^2(0)+y^2(1)] - 2y(0).$$
Here is my attempt: let $u \in M$ be such that $J(u) = \min_{y \in M} J(y)$, then for any $\varepsilon > 0$ and $v \in M$,
$$\begin{aligned} j(\varepsilon) = J(u+\varepsilon v) &= \frac{1}{2} \int_0^1 [(u'+\varepsilon v')^2+(u+\varepsilon v)^2] \mathrm{d}x + \frac{1}{2} [u(0)+\varepsilon v(0)]^2 \\ &\quad + \frac{1}{2}[u(1)+\varepsilon v(1)]^2 -2[u(0)+\varepsilon v(0)] \end{aligned}$$
reaches minimum at $\varepsilon = 0$. Thus $j'(0) = 0$, that is
$$j'(0) = \int_0^1 (u'v'+uv) \mathrm{d}x + u(0)v(0) + u(1)v(1) - 2v(0) = 0$$
for any $v \in M=C^1[0,1]$.
I'm wondering if there is a way to continue my solution (maybe it is actually wrong), or a simpler way to solve the problem. Thanks for any help!
Hint: if the question is to minimize the functional by whatever method, i.e. the Euler-Lagrange equation is not the must, then one can use $$ \int_0^1 2yy'\,dx=y^2(1)-y^2(0) $$ to complete the squares in $y$ as $$ J(y)=\frac12\int_0^1|y+y'|^2\,dx+|y(0)-1|^2-1. $$ Now the minimization is obvious.