There are two different Stokes' theorems. One is used typically in analysis and can be stated as follows: for each $u\in C^1(\bar{U})$ ,with $U\subset \Bbb{R}^n$ with boundary $C^1$,and $U$ bounded we have
$$\int_{U} \operatorname{div} \mathbf{u}\, d x=\int_{\partial U} \mathbf{u} \cdot \nu \,d S$$
where $\nu$ takes the outer normal direction of $U$.
Another Stokes' theorem typically appears in differential geometry and says
let $M$ be an oriented smooth $n$-manifold with boundary, and let $\omega$ be a compactly supported smooth $(n-1)$-form on $M$. Then
$$
\int_{M} d \omega=\int_{\partial M} \omega
$$
Here are some questions about them:
-
when we consider $U\subset \Bbb{R}^n$ as manifold,there is no notion of manifold boundary if $U$ is open, why we can consider the "boundary" here?
-
the second theorem needs compact supported condition, whereas the first one does not need. What's the difference between them?
-
Can we remove the bounded condition for $U$ in theorem 1?
Best Answer
To apply Stokes, you cannot just have an $\omega$ that's defined only over $U$, instead $\omega$ has to be extended to $\partial U$ (at least $C^1$-ly), so the boundary is implicitly there. In practice, when we integrate an explicitly expressed function, it's usually defined almost everywhere. For example, integerate div $\textbf{u}$ on the open unit ball where $\textbf{u} = (x, y, z)$. We can instead integrate $\textbf{u}$ on the unit sphere. Here we are implicitly using the fact that the vector field $\textbf{u}$ can be extended to the sphere by the same formula.
Compactly supported is not the best condition, but some condition to make the integrals finite is necessary. The first version can be false if no such condition is posed. In calculus, we usually only apply Stokes when $\overline{U}$ is bounded, therefore compact. If we remove the boundedness, we have to then assume something like $\omega$ has compact support like in the second version, which is a weaker condition than $\overline{U}$ is bounded.