Proof of Helmholtz-Hodge decomposition, poor man's version

Question

Helmholtz (-Hodge) decomposition commonly used in physics includes decomposition of a (sufficiently smooth) vector field $F = -\mathrm{grad}(U) + \mathrm{curl}(W)$ on bounded simply connected domain $\Omega \subseteq \mathbb{R}^3$ (with smooth boundary), with scalar $U$ and vector field $W$ which are explicitly given by integrals on Wiki page https://en.wikipedia.org/wiki/Helmholtz_decomposition

The proof presented on the same Wiki page (an early instance of which appeared in Aris: "Vectors, Tensors, and the Basic Equations of Fluid Mechanics"; see also Griffiths: "Introduction to Electrodynamics", Appendix C) silently assumes, among other things, that vector field $F$ is test (smooth and compactly supported), whereas one would like to have decomposition for vector fields which are merely $C^2(\Omega) \cap C^0(\overline{\Omega})$.

Is there some standard reference where the decomposition, with explicit expressions for $U$ and $W$, is proven in this modest setting?

Answer 1

For avoiding $\delta$, you should be able to just do as follows:

Let $G(x)$ denote the Newton potential $\frac{1}{4\pi |x|}$, and $G_y(x) = G(x-y)$

Let $r'(x)$, for $x \in \Omega$, denote $\frac12 d(x, \Omega^c)$.

Given $y\in \Omega$, consider the integral for $\lambda\in (0,1)$

$$ \tilde{F}(\lambda,y) = - \int_{\partial B(y,\lambda r'(y))} \partial_\nu G_y(z) F(z) ~\mathrm{d}\sigma(z) $$ Here $\partial_\nu$ is the derivative in the direction of the outward unit normal, and $\sigma$ is the induced surface measure.

As $F$ is $C^2$, by Taylor's theorem we know that $F(z) = F(y) + O(|z - y|)$. Therefore we have $$ \lim_{\lambda \to 0} \tilde{F}(\lambda,y) = F(y)$$ uniformly.

Applying divergence theorem to the integral defining $\tilde{F}$ you get

$$ \tilde{F}(\lambda,y) = \int_{\Omega \setminus B(y,\lambda r'(y))} (\nabla G_y(z) \cdot \nabla) F(z) ~\mathrm{d}z - \int_{\partial\Omega} \partial_n G_y(z) F(z) ~\mathrm{d}\sigma(z). $$

The volume integral you can use the vector triple product formula to write

$$ (\nabla G_y \cdot \nabla)F = (\nabla\cdot F) \nabla G_y + \nabla \times (F \times \nabla G_y ) $$

Since $\nabla G_y$ blows up no faster than $1 / |z-y|^2$, which is integrable in $\mathbb{R}^3$, we can take the limit $\lambda \to 0$ of the integral. This gives

$$ F(y) = \int_{\Omega} (\nabla \cdot F) \nabla G_y + \nabla \times (F \times \nabla G_y) ~\mathrm{d}z - \int_{\partial\Omega} \partial_n G_y F ~\mathrm{d}\sigma $$

after which you can continue following exactly the derivation on Wikipedia.