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In a book I study I found this exercise:

Let $U \subset \mathbb{R}^n$ open and $w: U \rightarrow \mathbb{R}^n$ vector field. Let $p \in U$.

$V_n$ is the volume of the n-dimensional unit sphere and $S_r(p) \subset \mathbb{R}^n$ the $n-1$-dimensional sphere around $p$ with radius $r > 0$ and $\nu$ the outer vector field of normals.

Now I want to show using Gauss's theorem that:

$(div \ w)(p) = \lim \limits_{r\rightarrow 0} \frac{1}{V_n r^n}\int_{S_r(p)} \langle w,\nu \rangle dS$

I proofed a similar problem in three dimensions. But now I don't even know how to start showing this relation.

Can I use the closed form for an n-sphere?:

$V_n=\begin{cases} \frac{\pi^k}{k!} & \text{for} \ \ \ n=2k\\ \frac{2k!(4\pi)^k}{(2k+1)!} & \text{for} \ \ \ n=2k+1 \end{cases}$

I hope you can help!

Gilfoyle
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  • why don't you show your proof in 3 dimensions? Here is a rather abstract version of your question, though it does not give a proof. – cactus314 Jun 08 '15 at 21:42

1 Answers1

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Instead of the sphere $S_\epsilon(p) = \{p: ||p' - p|| < \epsilon \}$ we are going to use cube: $ p + \tfrac{\epsilon}{2} [-1,1]^n$ with volume $\epsilon^n$? This will simplify calculations. I am claiming:

$$ (\nabla \cdot w)(p) = \lim \limits_{\epsilon \rightarrow 0} \frac{1}{V_n \epsilon^n}\int_{S_\epsilon(p)} \langle w,\nu \rangle dS = \lim_{\epsilon \to 0} \frac{1}{\epsilon^n} \int_{p + \tfrac{\epsilon}{2} [-1,1]^n} (w \cdot \nu) \,dS $$

  • The normal is always pointing outwards from the cube, so we get contributions from opposite faces where $w \cdot v$ is both positive and negative.

  • The volume of each face has volume $|dS| = \epsilon^{n-1}$. And we need to sum over each face. Integrating over the surfaces (and remembering the normal vector points outwards from the cube):

$$ \int_{p + \tfrac{\epsilon}{2} [-1,1]^n} (w \cdot \nu) \,dS \approx \sum_{k=1}^n \bigg[w_k(p + \epsilon_k) - w_k(p - \epsilon_k)\bigg]\epsilon^{n-1} $$

At this point we recover the divergence formula using infinitesimal analysis:

\begin{eqnarray} \frac{1}{\epsilon^n} \sum_{k=1}^n \bigg[w_k(p + \epsilon_k) - w_k(p - \epsilon_k)\bigg]\epsilon^{n-1} &\approx& \sum_{k=1}^n \frac{1}{\epsilon}\bigg[w_k(p + \epsilon_k) - w_k(p - \epsilon_k)\bigg] \\\\ &=&\nabla \cdot w\; + \;O(\epsilon)\end{eqnarray}

See the nice physics question: What is divergence?

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cactus314
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  • Sorry, I can not see a relation to my question in your answer. Could you please be more precise? I tried to understand your answer for the last twenty minutes, but did not get the point. Thank you! – Gilfoyle Jun 08 '15 at 21:09
  • Is there a reason why you use the $\epsilon$ instead of the $r$? Is this part of the simplification? – Gilfoyle Jun 08 '15 at 21:40
  • @Samuel I am sorry it's that $r \to 0$ so I always think of it as $\epsilon$. They are the same – cactus314 Jun 08 '15 at 21:42
  • I added the formula for the volume of an n-sphere. Is there an easy way to use this formula to show the relation? – Gilfoyle Jun 08 '15 at 22:38
  • @Samuel If you use a sphere instead of a cube, it changes how small $r$ has to be in order to approximate the divergence, but the limit is the same. We could rescale $r \mapsto \frac{1}{ \sqrt[n]{V_n}} , r $ – cactus314 Jun 08 '15 at 22:45
  • I worked today on a solution for my problem using your answer. I have still some questions. I hope you can explain some steps more precisely. 1) Why do you use $p+\frac{\epsilon}{2}[-1,1]^n$ and where does it come from? Do you mean an n-dim-cube around point $p$? 2) In your first equation I do not understand why the $V_n$ disappears. Did you use $V_n = \epsilon^n$? 3) Why can you say that $|dS| = \epsilon^{n-1}$? 4) Why do you use $f(p + \epsilon_k)$ and where does it come from? I tried a lot to understand your way but did not understand it perfectly. I hope you can help! – Gilfoyle Jun 09 '15 at 21:08
  • @Samuel 1) just a fancy way to say "cube of side length $\epsilon$ around $p$ 2) the cube has $V_n = \epsilon^n$ 3) the face has area $dS = \epsilon^{n-1}$. 4). $f(p+\epsilon_k)$ is slightly wrong notation. Since it's $n$ dimensions we can move $\epsilon$ units in any direction. I just mean $\pm \epsilon$ units in the $k$ direction. – cactus314 Jun 09 '15 at 21:12
  • Did you use Gauss's theorem (Divergence theorem) $\int_V div(w)dV = \int_S \langle w,\nu \rangle dS$? I can not see it. – Gilfoyle Jun 10 '15 at 06:37
  • @Samuel I fixed an error that I use $f$ instead of $\vec{w}$. Divergence can be written as a dot product $\mathrm{div} (w) = \nabla \cdot w$. – cactus314 Jun 10 '15 at 18:31