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

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?

• I am confused: for $F$ in $C^2(\Omega) \cap C^0(\bar{\Omega})$, and $\Omega$ bounded with smooth boundary, the integral expressions defining $U$ and $W$ make sense. Why isn't it sufficient to simply check that $- \mathrm{grad}(U) + \mathrm{curl}(W)$ in fact equals $F$? (This you can do using Gauss-Green carefully I think.) Oct 16, 2019 at 1:16
• I don’t see any assumption of compact support stated or needed in the Wikipedia article. Oct 16, 2019 at 5:30
• @Deane Yang My concern is due to fact that in that proof (from Wiki page) we start with an integral of $F$ with $\delta$, doesn't that assume that $F$ is a test field? Oct 16, 2019 at 6:27
• @Willie Wong I see, just to do everything in reverse... my concern was with the initial usage of the $\delta$ (as I've commented above), which I wanted to avoid. Oct 16, 2019 at 6:29
• No. Integrating against delta does not require the test function to be compactly supported since delta itself is compactly supported. But, on a closer look, I’m not sure about the switching of the integral and Laplacian. Is that valid? Oct 16, 2019 at 13:57

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.

• Great, exactly what I was looking for! Thanks! Oct 16, 2019 at 18:12