Define $\mathcal F(u) = \int_U \frac{1}{2} |\nabla u|^2 - \int_U fu - \int_{\partial U} gu : C^\infty(\bar U) \to \mathbb R$, where $f \in C^\infty(\bar U), g \in C^\infty(\partial U)$. Suppose $u_0 \in C^\infty(\bar U)$ satisfies $\mathcal F(u_0) = \min_{u\in C^\infty(\bar U) \\ \int_U u^2=1}\mathcal F(u)$, prove
$$ -\Delta u_0 = f \text{ in } U \text{ and } \frac{\partial u_0}{\partial \vec{n}}=g \text{ on } \partial U $$
Here is my solution, but the result seems not the same as the first equation above. I am not sure what goes wrong.
Suppose $v \in C^\infty(\bar U), t \in \mathbb R $ and $u + tv \neq 0$. Since $u_0$ minimizes $\mathcal F$, we have
$$ 0 = \left .\frac{d}{dt}\right |_{t=0} \mathcal F \left ( \frac{u_0 + tv}{\lVert u_0 + tv \lVert_{L^2}} \right ) = \int_U \nabla u_0\cdot \nabla v - \mathcal F(u_0) \int_U u_0v - \int_U fv - \int_{\partial U} gv$$
Note $\int_U \nabla u_0 \cdot \nabla v = \int_{\partial U} v\frac{\partial u_0}{\partial \vec{n}} - \int_U v\Delta u_0$.
Now, choose $v$ such that $v|_{\partial U} = 0$, then we have
$$0 = -\int_Uv \Delta u_0-\mathcal F(u_0) \int_U u_0v - \int_U fv$$
Then we have $\Delta u_0 + \mathcal F(u_0) u_0 + f = 0$ in $U$.
Next, for $v \in C^\infty(\bar U)$, plug $\Delta u_0 = -\mathcal F(u_0) u_0 - f$ into the two equations above, we have $\int_{\partial U}\left ( \frac{\partial u_0}{\partial \vec{n}} - g\right )v=0$, then we have $\frac{\partial u_0}{\partial \vec{n}}=g$ on $\partial U$.
I checked the computation twice and couldn't find the mistake.
Update: Note that $U$ is a bounded, connected open set in $\mathbb R^n$ with $C^\infty$ boundary $\partial U$.