Super dumb question but I can't seem to figure out what I'm doing wrong. Let $E$ be some set and $E_r= \{ x \mid rx \in E\}$ for $r>0$. Then if $\phi \in C^1_c(\mathbb{R}^n;\mathbb{R}^n)$ $$\int_{E_r} \operatorname{div}(\phi(x))\,dx = \frac{1}{r^{n-1}}\int_E \operatorname{div}(\phi(y/r)) \, dy.$$ I applied the change of variables $x = y/r$ so then $dx = dy/r^n$ and the new domain of integration becomes $E$, but I don't know how to get this $n-1$ term in place of $n$.

  • $\begingroup$ What's the subscript $c$ in $C^1_c$? $\endgroup$ – Jackozee Hakkiuz Jun 5 at 21:20
  • $\begingroup$ compact support $\endgroup$ – asuuuka Jun 5 at 21:21
  • $\begingroup$ I edited my answer, I hope this helps! $\endgroup$ – Jackozee Hakkiuz Jun 5 at 22:10

$\renewcommand\div{\mathrm{div}}$ Remember that the change of variables says $$\int_{g(E)} f = \int_E (f\circ g) \cdot \;|\det g'|$$

Now let $g(y)=y/r$, so that $$E_r=\{x\mid rx\in E\}=\{g(y)\mid y\in E\}=g(E)$$ Then $$\begin{align*} \int_{g(E)} \div\phi &= \int_{E} ((\div\phi)\circ g)\cdot |\det g'| \end{align*}$$ and you can calculate that $$\begin{align*} \div(\phi\circ g) &= \frac{1}{r}(\div\phi)\circ g \\[1mm] |\det g'| &= \frac 1 {r^n} \end{align*}$$ Solving from the first formula, you get $(\div\phi)\circ g = r\cdot \div(\phi\circ g)$, so that $$\begin{align*} \int_{E_r} \div\phi &= \int_{E} r\cdot ((\div\phi)\circ g)\cdot \frac 1{r^n} \\ &= \frac{1}{r^{n-1}}\int_{E} (\div\phi)\circ g \end{align*}$$ as you wanted. Or, in the notation $\int_E f(x)\,dx=\int_E f$ $$\begin{align*} \int_{E_r} (\div\phi)(x)\,dx &= \frac{1}{r^{n-1}}\int_{E} (\div\phi)(g(y))\,dy. \end{align*}$$

I think a possible source of your confusion is the notation $\div(\phi(x))$. Let me explain.

What we do is that we first apply the divergence operator $\div$ to the function $\phi:\mathbb R^n\to\mathbb R^n$, which yields another function $\div\phi:\mathbb R^n\to\mathbb R$. Then you evaluate $\div\phi$ at $x\in E_r$, giving you a real number $(\div\phi)(x)$. I'm of the opinion that you should not denote $(\div\phi)(x)$ as $\div(\phi(x))$.

First: if you take notation seriously, then $\div(\phi(x))$ should be zero, since $\phi(x)$ is the value of $\phi$ at $x$, so that it is already a point in $\mathbb R^n$ and hence its derivative is zero.

Second, and this the worst: if you write $(\div\phi)(x)$ as $\div(\phi(x))$, now the notation $\div(\phi(y/r))$ is confusing, because some people may interpret it as $(\div(\phi\circ g))(y)$, where $g(y)=y/r$ (as before) and some other may interpret it as $(\div\phi)(g(y))$. My hypothesis is that this made you think that $\div(\phi\circ g)$ is the same as $(\div\phi)\circ g$, which is not true: as you have seen, they differ by a factor of $r$, which is precisely the missing $r$.


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