How one can proof that
$\int_{\Omega}\nabla \mathbf{u} \cdot \mathbf{t} \,d\Omega = (\oint_{\partial\Omega}\mathbf{u} \otimes \mathbf{m} \,dL) \cdot \mathbf{t}$,
where $\mathbf{u}$ is a vector field, $\Omega$ a polygon (i.e., flat and has piecewise smooth boundary) in the 3d-Euclidean space with normal $\mathbf{n}$, $\mathbf{m}$ is outward-facing unit normal (to the boundary) field and the vector $\mathbf{t}$ is any constant vector such that $\mathbf{t} \cdot \mathbf{n}=0$?
A draft:
1) Suppose that $\Omega$ is a rectangle (the polygon case can easily be solved by subdividing it).
2) Apply a coordinate transformation such that:
- $\Omega$ becomes $\underline{\Omega}$, which is co-planar and aligned to the xy-plane;
- $\mathbf{t}$ becomes $\underline{\underline{\mathbf{t}}}$, which lies on $\underline{\Omega}$;
- $\mathbf{u}$ becomes $\underline{\underline{\mathbf{u}}}$.
3) Then,
$\int_{\Omega}\nabla \mathbf{u} \cdot \mathbf{t} \,d\Omega$ becomes $\int_{\underline{{\Omega}}} \nabla \underline{\underline{\mathbf{u}}} \cdot \underline{\underline{\mathbf{t}}} \,d\Omega \quad$ (I)
4) Write $\underline{\underline{\mathbf{u}}} = \underline{\underline{u}}_{t1} \mathbf{e}_{1} + \underline{\underline{u}}_{t2} \mathbf{e}_{2} + \underline{\underline{u}}_{n} \mathbf{e}_{3}$, where $\{\mathbf{e}_i\}$ is the canonical 3d-Euclidean base, $\underline{\underline{u}}_{t1} = \underline{\underline{\mathbf{u}}} \cdot \mathbf{e}_{1}$, $\underline{\underline{u}}_{t2} = \underline{\underline{\mathbf{u}}} \cdot \mathbf{e}_{2}$ and $\underline{\underline{u}}_{n} = \underline{\underline{\mathbf{u}}} \cdot \mathbf{e}_{3}$. Thus, and since $\underline{\underline{\mathbf{t}}}$ is constant,
$ \int_{\underline{{\Omega}}} \nabla \underline{\underline{\mathbf{u}}} \cdot \underline{\underline{\mathbf{t}}} \,d\Omega = \left[ \int_{\underline{{\Omega}}} \nabla \left( \underline{\underline{u}}_{t1} \mathbf{e}_{1} + \underline{\underline{u}}_{t2} \mathbf{e}_{2} + \underline{\underline{u}}_{n} \mathbf{e}_{3} \right) \,d\Omega \right] \cdot \underline{\underline{\mathbf{t}}} \quad $ (II).
5) As a corollary from 2d-divergence theorem, we have that
$\int_{\underline{\Omega}} \nabla \underline{u}(x,y) \,d\Omega = \oint_{\partial\underline{\Omega}} \underline{u}(x,y) \underline{\mathbf{m}} \,dL. \quad$ (III)
6) Focusing only on one piece of (II), we have that,
$ \left[ \int_{\underline{{\Omega}}} \nabla \left( \underline{\underline{u}}_{t1} \mathbf{e}_{1} \right) \,d\Omega \right] \cdot \underline{\underline{\mathbf{t}}} = \left[ \mathbf{e}_{1} \otimes \int_{\underline{{\Omega}}} \nabla \underline{\underline{u}}_{t1} \,d\Omega \right] \cdot \underline{\underline{\mathbf{t}}} = \left[ \int_{\underline{{\Omega}}} \nabla \underline{\underline{u}}_{t1} \cdot \underline{\underline{\mathbf{t}}} \,d\Omega \right] \mathbf{e}_{1} = \left[ \int_{\underline{{\Omega}}} \begin{bmatrix} \frac{\partial}{\partial x}\underline{\underline{u}}_{t1} \\ \frac{\partial}{\partial y}\underline{\underline{u}}_{t1} \\ \frac{\partial}{\partial z}\underline{\underline{u}}_{t1} \\ \end{bmatrix} \cdot \begin{bmatrix} \underline{\underline{t}}_x \\ \underline{\underline{t}}_y \\ 0 \\ \end{bmatrix} \,d\Omega \right] \mathbf{e}_{1} = \left[ \int_{\underline{{\Omega}}} \frac{\partial}{\partial x}\underline{\underline{u}}_{t1} \underline{\underline{t}}_x + \frac{\partial}{\partial y}\underline{\underline{u}}_{t1} \underline{\underline{t}}_y \,d\Omega \right] \mathbf{e}_{1} = \left[ \int_{\underline{{\Omega}}} \begin{bmatrix} \frac{\partial}{\partial x}\underline{\underline{u}}_{t1} \\ \frac{\partial}{\partial y}\underline{\underline{u}}_{t1} \\ \end{bmatrix} \cdot \begin{bmatrix} \underline{\underline{t}}_x \\ \underline{\underline{t}}_y \\ \end{bmatrix} \,d\Omega \right] \mathbf{e}_{1} . \quad$ (IV)
The last term of the equation above suggests us to define $ \underline{u}(x,y) = \underline{\underline{u}}(x,y,0) $, $ \underline{\mathbf{t}} = [\underline{\underline{t}}_x \; \underline{\underline{t}}_y ]^T $ and redefine $\mathbf{e}_{1}$ to be $[1 \; 0]^T$ in order to write the 2d version of this term. Now we can apply the corollary like this:
$ \left[ \int_{\underline{{\Omega}}} \nabla \underline{u}_{t1} \cdot \underline{\mathbf{t}} \,d\Omega \right] \mathbf{e}_{1} = \left[ \mathbf{e}_{1} \otimes \int_{\underline{{\Omega}}} \nabla \underline{u}_{t1} \,d\Omega \right] \cdot \underline{\mathbf{t}} \stackrel{\text{(III)}}{=} \left[ \mathbf{e}_{1} \otimes \oint_{\partial\underline{{\Omega}}} \underline{u}_{t1} \, \underline{\mathbf{m}} \,dL \right] \cdot \underline{\mathbf{t}} = \left[ \oint_{\partial\underline{{\Omega}}} (\underline{\mathbf{m}} \cdot \underline{\mathbf{t}}) \underline{u}_{t1} \mathbf{e}_{1}\, \,dL \right]. $
Of course the process is analogous for $\underline{u}_{t2}$. I will write the rest of the proof when someone says it is ok up until this point.
