I'm currently working on a problem related to heat transfer, and thus I encountered robin boundary condtions for the first time while working with the finite element method. Unlike neumann boundary conditions, which in the weak formulation $a(u,v) = l(v)$ are a part of the linear $l(v)$-term, I eventually figured out that the robin boundary condtitions end up in the bilinear term $a(u,v)$. So my question is this, how do we determine if the integral over the boundary belongs to the bilinear form, or the linear form?
For reference's sake, the weak formulation for a poisson problem with neumann boundaries look like this: $$ \int_{\Omega} \nabla u \cdot \nabla v d\Omega = \int_{\Omega} vf d\Omega + \int_{\partial \Omega} v \nabla u \cdot \vec{n} dS $$
While the robin case would look something like this;
$$ \int_{\Omega} \nabla u \cdot \nabla v d\Omega + \int_{\partial \Omega} v(u-T)dS = \int_{\Omega}vfd\Omega $$
with $T$ being the ambient temperature. The only difference I can spot is that the terms on the boudary related to $u$ is not differentiated for robin boundary conditions. So how does this work?
Thanks.