Deriving a type of Stokes theorem for vectors on R^2 using differential forms

139 Views Asked by At

I would like to derive a relationship expressing an integral over a region $\Omega \subset \mathbb{R}^2$ as an integral over $\partial \Omega$.

To be more specific, define the div operator in two dimensions on $p=(p_1,p_2)$ as $$\nabla \cdot p=\partial_xp_1+\partial_yp_2$$I would like to express

$$\iint_\Omega \nabla \cdot p \quad dxdy \tag{1}$$

where p is a function of $x$ and $y$, as an integral of $p$ over the boundary of $\Omega$. I would also like to do this using differential forms. Here is my attempt;

First we start by expressing the integrand of $(1)$ as a two form;

$$\iint_\Omega \nabla \cdot p \quad dx \wedge dy =\iint_\Omega \star d \star p \quad dx \wedge dy \tag{2}$$ By expressing the divergence operator in terms of the hodge dual and the exterior derivative.

I'd like to now use stokes theorem $$\int_\Omega d \omega =\int_{\partial\Omega}\omega $$ But i am unable to do so because of the left-most Hodge dual in $(2)$.

Does anyone have any suggestions / can show me how to proceed?

1

There are 1 best solutions below

2
On BEST ANSWER

You're one step away. Get rid of the last (outer) star and the $dx\wedge dy$. After all, the last star turned the $2$-form into a $0$-form. Leave it just as an exact $2$-form and then use Stokes's Theorem.