# change of variables confusion

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 $$n$$.

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

## 1 Answer

$$\renewcommand\div{\mathrm{div}}$$ 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 $$r$$.