1
$\begingroup$

I've seen this vector identity from the book[1] in page 89, $$ (\nabla p)\times\nu =0,\ \text{on}\ \partial\Omega,$$ where $\nu $ is the outer normal vector of $\partial \Omega$, $ p \in H_0^1(\Omega).$ I try to calculate directly $(\nabla p)\times \nu =(\nu_3\partial_2 p-\nu_2\partial_3 p,\nu_1\partial_3 p-\nu_3\partial_1 p,\nu_2\partial_1 p-\nu_1\partial_2 p)^T,$ but I don't know how to show it equals $0$.

[1] Monk, Peter, Finite Element Methods for Maxwell's Equations (Oxford, 2003; online edn, Oxford Academic, 1 Sept. 2007)

$\endgroup$
1
  • $\begingroup$ This depends on the context. For general $p$ this is certainly false. $\endgroup$
    – daw
    Commented Mar 5 at 15:47

1 Answer 1

Reset to default
0
$\begingroup$

Also, $(\nabla p)\times \nu=|\nabla p ||\nu|\sin(\theta)$ where $\theta$ is the angle between the two vectors. For the cross product to be zero the two vectors must be parallel. You must show that $\nabla p$ is parallel to $\nu$

$\endgroup$
3
  • 1
    $\begingroup$ This is a comment, not an answer. If you do not have sufficieint reputation to comment, then consider leaving a detailed answer that will increase your reputation. $\endgroup$
    – whpowell96
    Commented Mar 5 at 17:41
  • $\begingroup$ On $\partial \Omega, p=0$. The gradient $\nabla p$ is perpendicular to the level sets of $p $(the level sets are surfaces where $p$ is constant). The outer normal vector $ν$ is also perpendicular to the boundary. $\endgroup$
    – Du Xin
    Commented Mar 6 at 10:22
  • 1
    $\begingroup$ You must do more work to define level sets and $\nabla p$ on the boundary for functions in Sobolev spaces, which are not functions, but equivalence classes of integrable functions. $\endgroup$
    – whpowell96
    Commented Mar 6 at 16:26

Not the answer you're looking for? Browse other questions tagged .