= Solution
{c}
Let $T:H^1(U)\to L^2(\partial U)$ be the trace map from the <Sobolev trace theorem>. The <weak Robin problem for a uniformly elliptic operator> uses all of $H^1(U)$ as the test space, since a <Robin boundary condition> does not require a zero trace. Define
$$
\begin{aligned}
B_\lambda[u,v]={}&\int_U\left(\sum_{i,j}a^{ij}u_{x_i}v_{x_j}
+\sum_i b^iu_{x_i}v+(c+\lambda)uv\right)dx\\
&+\int_{\partial U}\beta\,Tu\,Tv\,dS.
\end{aligned}
$$
The <weak formulation> is
$$
\boxed{u\in H^1(U),\qquad B_\lambda[u,v]=\int_Ufv\,dx
\quad\text{for every }v\in H^1(U).}
$$
For a <classical solution>, <integration by parts> produces the boundary term $-\int_{\partial U}(\sum a^{ij}u_{x_i}\nu_j)v$. Substituting the <Robin boundary condition> replaces it by $\int_{\partial U}\beta uv$, giving the displayed identity.
Conversely, if $u\in C^2(\overline U)$ satisfies the identity, tests $v\in C_c^\infty(U)$ first show $(L+\lambda)u=f$ in the sense of <distributions>. Subtracting this interior equality after <integration by parts> gives
$$
\int_{\partial U}\left(\sum_{i,j}a^{ij}u_{x_i}\nu_j+\beta u\right)Tv\,dS=0.
$$
Every smooth boundary function is the trace of a smooth function on $\overline U$. The boundary residual is continuous, so this proves the boundary equation pointwise. Thus the formulations agree for $C^2$ solutions. If $f$ is continuous the interior equation holds pointwise; for merely $f\in L^2$, equality with that given representative of $f$ is understood <almost everywhere>.
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