Let $U$ be an open region in $\mathbb{R}^3$ and $G \subset U$ be another open set whose closure is compact in $U$.
Then, for any smooth divergence-free $u : U \to \mathbb{R}^3$, I ran into the statement that \begin{equation} u(x)=\frac{1}{4\pi} \int_{G} \Bigl[\nabla_y \frac{1}{\lvert x-y\rvert} \Bigr] \times \omega(y) dy+A(x) \text{ for } x \in G \end{equation} where $\omega=\nabla \times u$ and $A : G \to \mathbb{R}^3$ satisfies $\Delta A=0$ componentwise.
There is no detailed proof and just a brief mention of Hodge's Theorem. But, I cannot see how to make use of the theorem. Also, I have difficulty with proving the formula directly with vector calculus..
Could anyone please help me?
Edit : I forgot the divergence-free condition for $u$, so I add it to the main body.
If you integrate your expression by parts, your integrand becomes $-\frac{\nabla_y\times\omega(y)}{|x-y|}$, and the numerator is just the vector Laplacian of $u$ since
$$\nabla\times\nabla\times u = \nabla(\nabla\cdot u) - \Delta u = -\Delta u$$
due to the divergence free condition on $u$. So your statement is asserting that
$$u = K*\Delta u + A + \text{boundary term}$$
Where $K(x)=1/(4\pi |x|)$ is the Green’s function for the Laplacian. This is also known as Green’s representation formula, which you can find a proof of in Gilbarg and Trudinger’s Elliptic PDEs book, page 17-18. Clearly $A$ plus the boundary term (arising from integrating by parts) is harmonic, you might need to use the explicit representation given in Gilbarg-Trudinger to show that $A$ is. Morally speaking, convolution with $K$ inverts the Laplacian (at least on $\mathbb{R}^n$) and the presence of $A$ is to compensate for the fact there’s a boundary.