= Solution
Use the usual bounded <Lipschitz domain> assumptions so that the <Sobolev trace> is defined. Multiplication by $v\in H^1(\Omega)$ and <Green's first identity> give
$$
\int_\Omega\nabla u\cdot\nabla v-\int_{\partial\Omega}\frac{\partial u}{\partial n}v=\int_\Omega fv.
$$
The <Robin boundary condition> substitutes $\partial_nu=-u$ in the boundary integral. The resulting <Weak formulation of the Robin Poisson problem> is therefore: find $u\in H^1(\Omega)$ such that
$$
\boxed{a(u,v):=\int_\Omega\nabla u\cdot\nabla v+\int_{\partial\Omega}uv=\int_\Omega fv\quad\text{for every }v\in H^1(\Omega).}
$$
The boundary values in the integral mean <Sobolev traces>; no classical normal <derivative> is required of the weak solution. For uniqueness, subtract two solutions and test by their difference $w$. This yields $\int_\Omega|\nabla w|^2+\int_{\partial\Omega}w^2=0$. Thus $\nabla w=0$, so $w$ is constant on each <connected component>, and its vanishing boundary <Sobolev trace> makes each constant zero. \b[The weak solution is unique.]
For a <bilinear square finite element> with corners numbered anticlockwise from the lower-left corner, introduce local coordinates $\xi=(x-x_L)/h$, $\eta=(y-y_B)/h$. The four <shape functions> are
$$
N_1=(1-\xi)(1-\eta),\quad N_2=\xi(1-\eta),\quad N_3=\xi\eta,\quad N_4=(1-\xi)\eta.
$$
The local <stiffness matrix> has entries $K_{ij}=\int_E\nabla N_j\cdot\nabla N_i$. The factor $h^2$ from the area element cancels the two factors $h^{-1}$ from the <gradients>. Symmetry leaves three types: diagonal, adjacent corners, and opposite corners. For representatives,
$$
\begin{aligned}
K_{11}&=\int_0^1\!\int_0^1[(1-\eta)^2+(1-\xi)^2],d\xi\,d\eta=2/3,\\
K_{12}&=\int_0^1\!\int_0^1[-(1-\eta)^2+\xi(1-\xi)],d\xi\,d\eta=-1/6,\\
K_{13}&=\int_0^1\!\int_0^1[-\eta(1-\eta)-\xi(1-\xi)],d\xi\,d\eta=-1/3.
\end{aligned}
$$
Consequently
$$
\boxed{K_E=\frac16\begin{pmatrix}4&-1&-2&-1\\-1&4&-1&-2\\-2&-1&4&-1\\-1&-2&-1&4\end{pmatrix}.}
$$
For an element whose left edge lies on $x=0$, the boundary term in the <weak formulation> adds $B_{ij}=\int_eN_jN_i\,ds$. Only $N_1=1-\eta$ and $N_4=\eta$ are nonzero on that edge. Integrating $h(1-\eta)^2$, $h\eta(1-\eta)$ and $h\eta^2$ gives the extra contribution
$$
\boxed{B_e=\frac h6\begin{pmatrix}2&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&2\end{pmatrix}.}
$$
Its entries are added at the corresponding global node numbers during <finite element assembly>. Additional boundary edges contribute their own analogous edge matrices. The <Robin boundary condition> is represented by these integrals, so boundary nodes remain unknowns.
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