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Sometimes our lecturer forgets to prepare his lecture beforehand and thus his notes at the blackboard seem to be more his stream of thoughts than study material. I have already seen a couple of correct and full formulations of the Divergence Theorem, there really are not many variations of that.

However for the homework exercises we are advised to strictly follow the lecturers notes and thus would like to know if the following transcript of the lecturer's note can be salvaged (especially, if it is equivalent to a "usual" definition of the Divergence Theorem") - (if this is not possible, I will try to contact either my lecturer or his assistant)

In lecture we defined the Divergence theorem as follows:

$\Omega, U \subset \mathbb{R}^n $

$ \{g<0\} = \Omega \cap U, \{g = 0\} = U \cap \partial \Omega , \{ g>0 \} = U \setminus \Omega $

$ \nu(x) = \frac{\nabla g(x)}{|\nabla g(x)|} $ with: $ g : U \to \mathbb{R} $

$ u : \Omega \to \mathbb{R}^n, C^1 function $

$ v: \to \mathbb{R}, C^1 function $

Then the Divergence theorem states: $ \int\limits_{\Omega} div(u*v) = \int\limits_{\partial\Omega} v(u*\nu) - \int\limits_{\Omega} div(u* \nabla v) $

Checking wikipedia (or any further calculus book) gives a slightly different formula of the Divergence Theorem. None of the version I have seen included this part: $ - \int\limits_{\Omega} div(u* \nabla v) $

Does this part suffices a special purpose?

Any constructive hint, comment or answer is appreciated.

Imago
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1 Answers1

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What you've written doesn't make sense. Notice for instance that $u$ is a vector-valued function and $v$ is scalar-valued, so $u\cdot\nabla v$ is a scalar, which you cannot take the (three-dimensional) divergence of.

The "usual" divergence theorem for vector fields $w$ is $$\int_{\Omega} \nabla \cdot w \,dV = \int_{\partial \Omega} w\cdot \hat{n}\,dA.$$ One guess for what is happening in the notes is that your professor is plugging in $w=uv$, where $u$ is a vector and $v$ is a scalar, and using the product rule: $$\nabla \cdot (vu) = v\nabla \cdot u +\nabla v \cdot u.$$

However this would give

$$\int_{\Omega} v\nabla \cdot u \,dV = \int_{\partial \Omega} vu\cdot \hat{n}\,dA - \int_{\Omega} u\cdot \nabla v \,dV$$ which has some terms in common with your notes, but I can't fully reconcile it with your formula.

user7530
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