Green's first identity

Question

Good morning/evening to everybody.

I'm interested in proving this proposition from the Green's first identity, which reads that, for any sufficiently differentiable vector field $\mathbf{\Gamma}$ and scalar field $\psi$ it holds:

$$ \int_U \nabla \cdot \mathbf{\Gamma} \, \psi \, dU = \int_{\partial U} (\mathbf{\Gamma} \cdot \mathbf{n}) \, \psi \, dS - \int_U \mathbf{\Gamma} \cdot \nabla \psi \, dU.$$

I've been told that, for $\mathbf{u}, \vec{\omega} \in \mathbb{R}^2$, it is true that (which I can prove quite easily using index notation):

$$ \int_U \Delta u_j \omega_j \, dU = \int_{\partial U} \frac{\partial u_j}{\partial n} \, \omega_j \, dS - \int_U \nabla u_j \cdot \nabla \omega_j \, dU,$$ which is equivalent to:

$$\int_U (\nabla \cdot \nabla \mathbf{u}) \cdot \, \vec{\omega} \, dU = \int_{\partial U} \frac{\partial {\mathbf{u}}}{\partial n} \cdot \vec{\omega}\, dS - \int_U \nabla \mathbf{u} : \nabla \vec{\omega} \, dU, \quad j =1,2 . $$

In every one of the equations above, $U$ is a closed region and $\partial U$ its border, being $\mathbf{n}$ its outer normal unit vector and $:$ stands for tensor contraction. I would like to prove that, for any given tensor $\mathbf{T}$, it holds:

$$\int_U (\nabla \cdot \mathbf{T}) \cdot \, \vec{\omega} \, dU = \int_{\partial U} (\mathbf{T} \cdot \mathbf{n}) \cdot \vec{\omega}\, dS - \int_U \mathbf{T} : \nabla \vec{\omega} \, dU , $$

but I'm not very proficient at dealing with index notation, yet. How could I proceed?

Any help will be much appreciated.

Cheers!

Edit and hint: Maybe divergence theorem is useful here? Since the Green's first identity is derived from it.

Answer 1

It appears that I misread the question the first time. In any case, this seems to follow from the generalisation of the divergence theorem for rank $n$ tensors:

$$\int_{U}{\frac{\partial T_{i_1i_2 \ldots i_q \ldots i_n}}{\partial x_{i_q}} dU}=\int_{\partial U}{T_{i_1i_2 \ldots i_q \ldots i_n}n_{i_q} dS}$$

applied to $T \cdot \omega$ instead of $T$.

\begin{align} \int_{\partial U}{T_{i_1i_2 \ldots i_q \ldots i_r \ldots i_n}n_{i_q} \omega_{i_r} dS}& =\int_{U}{\frac{\partial (T_{i_1i_2 \ldots i_q \ldots i_r \ldots i_n}\omega_{i_r})}{\partial x_{i_q}} dU} \\ &=\int_{U}{\frac{\partial T_{i_1i_2 \ldots i_q \ldots i_r \ldots i_n}}{\partial x_{i_q}}\omega_{i_r} dU}+\int_{U}{T_{i_1i_2 \ldots i_q \ldots i_r \ldots i_n}\frac{\partial \omega_{i_r}}{\partial x_{i_q}} dU} \end{align}

Hopefully I haven't messed up again.

Answer 2

Let me answer myself so I can see if I got this right.

Divergence theorem tells us that:

$$\int_S d \vec{s} \cdot \square = \int_V dV \, \nabla \cdot \square, $$

in the particular case of a given second order tensor, $\mathbf{T}$, and a vector field $\omega$, it follows:

$$\int_S d \vec{s} \cdot (\mathbf{T} \cdot \vec{\omega}) = \int_V dV \, \nabla \cdot (\mathbf{T} \cdot \vec{\omega}),$$

since, and now I will use index notation:

$$ \nabla \cdot (\mathbf{T} \cdot \vec{\omega}) = \partial_q (T_{pq} \omega_p) = T_{pq} \partial_q\omega_p + (\partial_q T_{pq}) \omega_p , $$

so I come up with the vector identity:

$$\nabla \cdot(\mathbf{T} \cdot \vec{\omega} ) = T:\nabla\vec{\omega} + (\nabla \cdot \mathbf{T}) \cdot\vec{\omega}, $$

so my original assumption holds.

Am I right?

Cheers and thanks!