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?

  • $\begingroup$ 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.) $\endgroup$ – Willie Wong Oct 16 '19 at 1:16
  • $\begingroup$ I don’t see any assumption of compact support stated or needed in the Wikipedia article. $\endgroup$ – Deane Yang Oct 16 '19 at 5:30
  • $\begingroup$ @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? $\endgroup$ – Ivica Smolić Oct 16 '19 at 6:27
  • $\begingroup$ @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. $\endgroup$ – Ivica Smolić Oct 16 '19 at 6:29
  • $\begingroup$ 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? $\endgroup$ – Deane Yang Oct 16 '19 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.

  • $\begingroup$ Great, exactly what I was looking for! Thanks! $\endgroup$ – Ivica Smolić Oct 16 '19 at 18:12

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