A decomposition of incompressible vector fields

Question

In Andrew, Majda- Vorticity and incompressible flow page 93, there is a theorem which is not proved:

Take a smooth incompressible (free divergence) vector field $v$ in $\mathbb{R}^2$. Call $w$ its vorticity (its curl), and suppose it is $L^1(\mathbb{R}^2)$. Then $v$ admits a radial energy decomposition, which means:

There exists a radially symmetric smooth scalar field $\overline{w}$ such that we can write $v = u + b$ where $u$ and $b$ satisfy:

$u$ is divergence free and $u \in L^2(\mathbb{R}^2)$. $b(x)= x^{\perp}|x|^{-2}\int_0^{|x|} s\overline{w}(s)ds $.

In the page before we prove that for a smooth incompressible vector field, using the biot savart law: $v(x) = (K_2 \ast w) (x) $ and making a taylor expansion of order 1 of the kernel $K_2$: for large $x$, $v(x)= x^{\perp}|x|^{-2}\int_{\mathbb{R}^2} w + O(|x|^{-2})$ and it is supposed to be a consequence of that, but I don't see why !!

Answer 1

Take $v$ a smooth incompressible vector field and call $w$ its vorticity.

If $w\in L^1(\mathbb{R}^2)$, the number \begin{align*} \alpha := \int_{\mathbb{R}^2} w \,dx, \end{align*} is well-defined. Looking at Example 2.1. on p.47 (so-called "radial eddies"), we now that for any function $\overline{w}:\mathbb{R}_+\rightarrow\mathbb{R}$, the vector field $b$ that you defined is divergence free, with curl $x\mapsto \overline{w}(|x|)$, so you just have to choose $\overline{\omega}$ properly to insure that \begin{align*} \int_{\mathbb{R}^2} \overline{w}(|x|)\,dx = 2\pi \int_0^\infty r \,\overline{w}(r)dr = \alpha, \end{align*} and you're done since $u:=v-b$ is then divergence-free vector field having a vorticity with $0$ mean on $\mathbb{R}^2$ : the Taylor expansion you mentionned allows to conclude that $u\in L^2(\mathbb{R}^2)$.