= Solution
The <Lax-Milgram theorem> says that if $H$ is a real <Hilbert space>, $B:H\times H\to\mathbb R$ is a bounded bilinear form satisfying
$$
B(v,v)\geq\alpha\lVert v\rVert_H^2
$$
for some $\alpha>0$, and $F\in H^*$, then there is a unique $u\in H$ such that $B(u,v)=F(v)$ for every $v\in H$.
For this problem take $H=H_0^1(U)$ and
$$
B(u,v)=\int_U\big(Du\cdot Dv+(D_nu)v+uv\big),
\qquad
F(v)=\int_Ufv.
$$
The <Cauchy-Schwarz inequality> and the continuous embedding $H_0^1(U)\hookrightarrow L^2(U)$ make both maps bounded. For smooth zero-boundary functions, integration by parts gives
$$
\int_U(D_nu)u=\frac12\int_U D_n(u^2)=0;
$$
density extends this identity to $H_0^1(U)$. Consequently
$$
B(u,u)=\lVert Du\rVert_2^2+\lVert u\rVert_2^2=\lVert u\rVert_{H^1}^2,
$$
so $B$ is coercive. Lax–Milgram supplies the unique weak solution.
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