Show: $\int_{\mathbb{R}^n}\mbox{div }a(x)\, d^nx=0$

Question

For a function $f\colon\mathbb{R}^n\to\mathbb{R}$ we write $f=o(r^{\alpha})$ if to any $\varepsilon>0$ there exists a $R>0$ so that $$ \lvert f(x)\rvert\leq\varepsilon\lVert x\rVert^{\alpha}\mbox{ for }\lVert x\rVert\geq R. $$ Let $a\colon\mathbb{R}^n\to\mathbb{R}^n$ be a continiously differentiable vector field with $$ a_k=o(r^{1-n}),~~~~~~~~~~\frac{\partial a_k}{\partial x_k}=o(r^{-n}),~~~~~~~~~~k=1,...,n. $$ Show that $$ \int_{\mathbb{R}^n}\mbox{div }a(x)\ d^nx=0. $$

I thought about how to show that and came to the following:

Set $$ \varepsilon:=\max\left\{\varepsilon_i:\left\lvert\frac{\partial a_i}{\partial x_i}(x)\right\rvert\leq\varepsilon_i\lVert x\rVert^{-n}\mbox{ for }\lVert x\rVert\geq R_i(\varepsilon_i), i=1,...,n\right\} $$ and $$ R(\varepsilon):=\min\left\{R_i(\varepsilon_i):\left\lvert\frac{\partial a_i}{\partial x_i}(x)\right\rvert\leq\varepsilon_i\lVert x\rVert^{-n}\mbox{ for }\lVert x\rVert\geq R_i(\varepsilon_i), i=1,...,n\right\}. $$

Then, because of $$ \lvert\mbox{div }a(x)\rvert=\left\lvert\sum\limits_{i=1}^n\frac{\partial a_i}{\partial x_i}(x)\right\rvert\leq\sum\limits_{i=1}^n\left\lvert\frac{\partial a_i}{\partial x_i}(x)\right\rvert\leq\sum\limits_{i=1}^n\epsilon\lVert x\rVert^{-n}=n\epsilon\lVert x\rVert^{-n}\leq\frac{n\epsilon}{R(\epsilon)^n} $$

I get

$$ \left\lvert\int_{\mathbb{R}^n}\mbox{div }a(x)\, d^nx\right\rvert\leq\int_{\mathbb{R}^n}\frac{n\epsilon}{R(\epsilon)^n}\, d^nx=\frac{n\epsilon}{R(\epsilon)^n}\int_{\mathbb{R}^n}1\, d^nx\to 0\mbox{ for }\varepsilon\to 0, $$

which means

$$ \int_{\mathbb{R}^n}\mbox{div }a(x)\, d^nx=0\mbox{ for }\varepsilon\to 0. $$

I do not know if this makes any sense.

Maybe one additionaly needs the integral theorem of Gauss, but if yes I do not know how...

Answer 1

I don't follow your argument - an upper bound on $|\operatorname{div} a|$ cannot possibly prove this unless $\operatorname{div}a=0$. The divergence theorem is the way to go - write

$$ \int_{\mathbb{R}^n} \operatorname{div} a= \lim_{r \to \infty} \int_{B_r} \operatorname{div} a = \lim_{r \to \infty} \int_{\partial B_r} a \cdot \nu $$

for $B_r$ the ball of radius $r$ centred at the origin and $\nu$ the outwards unit normal to this ball. You can now easily estimate the integrand to get $a \cdot \nu = o(r^{1-n})$. From here the fact that the measure of $\partial B_r$ is proportional to $r^{n-1}$ will give you the convergence to zero.