I am looking at the Deal.ii finite element package docs, and was reading the section about vector valued problems. These are simply systems of PDEs that are solved together.
I am trying to derive the weak form of the mixed Laplace equation below, but I am missing one piece of the derivation. I was hoping that someone might be able to fill in the relevant vector calculus identity.
So the equations are as follows.
$$ u + \nabla p = 0 $$
$$ -\nabla \cdot u = f $$
To find the weak form, we first setup the problem in vector form.
$$ \left(\begin{array}{cc} \mathbf{1} & \nabla \\ -\nabla^T & 0 \end{array}\right)\left(\begin{array}{l} \mathbf{u} \\ p \end{array}\right)=\left(\begin{array}{l} \mathbf{0} \\ f \end{array}\right) $$
In this case, $u$ is a velocity field that is 2 dimensional, and $p$ is a scalar pressure field.
We can define a test function $V$ to be of the same dimension as the system $V=\left(\begin{array}{l} \mathbf{v} \\ q \end{array}\right)$
Now, when we multiply the test function by the strong from, we obtain.
$$ \int_{\Omega}\left(\begin{array}{c} \mathbf{v} \\ q \end{array}\right)^T\left(\begin{array}{cc} 1 & \nabla \\ -\nabla^T & 0 \end{array}\right)\left(\begin{array}{l} \mathbf{u} \\ p \end{array}\right) d x=\int_{\Omega}\left(\begin{array}{l} \mathbf{v} \\ q \end{array}\right)^T\left(\begin{array}{l} \mathbf{0} \\ f \end{array}\right) d x $$
After some simplification and an adjustment to bracket notation for the inner products, the author end up with.
$$ (\mathbf{v}, \mathbf{u})+(\mathbf{v}, \nabla p)-(q, \operatorname{div} \mathbf{u})=(q, f) $$
Now, here is the part that is tripping me up. The author indicates that he uses integration by parts on the second term to move the derivative from the pressure variable to the test function. The result is.
$$ (\mathbf{v}, \mathbf{u})-(\operatorname{div} \mathbf{v}, p)-(q, \operatorname{div} \mathbf{u})=(q, f)-(\mathbf{n} \cdot \mathbf{v}, p)_{\partial \Omega} . $$
So I am just trying to understand the derivation of that second term and how integration by parts works--where $u$ is a vector field and $v$ is also a vector.
Thanks for any assistance.
Integration by parts tells us $$ \int_\Omega \mathbf{v}\nabla p\,dx = \int_{\partial\Omega}\mathbf{v}\cdot\mathbf{n}p\, dS - \int_\Omega \nabla\cdot\mathbf{v} p \,dx, $$ which is exactly what the author did.