Let $\phi:\mathbb{R}^3\to\mathbb{R}$ be a harmonic function that vanishes at infinity, fast enough for its gradient to have a Fourier transform.

Let $\vec{v} = \nabla\phi$. Then $\vec{v}$ has zero curl and zero divergence. By performing the Helmholtz decomposition using Fourier transform, as depicted here, we can write $$ \vec{v}=\vec{v}_{irrotational} + \vec{v}_{incompressible} $$ where $$ \vec{v}_{irrotational} = \frac{1}{(2\pi)^{3/2}}\iiint \vec{G}_\varphi e^{i\vec{k}\cdot\vec{r}}d^3\vec{k} $$

$$ \vec{v}_{incompressible} = \frac{1}{(2\pi)^{3/2}}\iiint \vec{G}_Ae^{i\vec{k}\cdot\vec{r}}d^3\vec{k} $$ with $$ \vec{G}_\varphi = \frac{\vec{k}\cdot\vec{G}}{\|{k}\|^2}\vec{k} \\ \vec{G}_A= -\vec{k}\times\left(\frac{\vec{k}\times\vec{G}}{\|{k}\|^2}\right) $$ where $\vec{G}$ is the Fourier transform of $\vec{v}$. In general, these two fields are not zero.

I am having trouble interpreting the result, in particular the uniqueness, as the field is divergence- and curl-free to begin with. To illustrate my confusion, I can always write $$ \vec{v} = \left(\vec{v}_{irrotational}+\frac{1}{2}\vec{v}_{incompressible}\right) + \frac{1}{2}\vec{v}_{incompressible} \equiv \widetilde{\vec{v}}_{irrotational}+\widetilde{\vec{v}}_{incompressible} $$ since if $\vec{v}_{incompressible}$ is divergence- and curl-free, so is any constant multiple. This is a different decomposition.


1 Answer 1


Unless you put other constraints on your Helmholtz decomposition, it is not unique in general. Take any vector field which is both divergence and curl free. You can add and subtract this vector field in any way you like in the the decomposition and still come up with a Helmholtz decomposition.

The simplest example I can come up with is replacing $v_{\mathrm{irrot.}}$ by $v_{\mathrm{irrot.}}+c$ and $v_{\mathrm{incomp.}}$ by $v_{\mathrm{incomp.}}-c$, where $c$ is any constant vector field. This is still a Helmholtz composition and adds up to your original vector field $v$. In your case, you just add and subtract a divergence- and curlfree field; i.e. you do $\pm \frac{1}{2} v_{\mathrm{incomp.}}$.

The formulas you are given provide you with a(!) Helmholtz decomposition, not the(!) Helmholtz decomposition.

In order to achieve uniqueness, some integrability constraints or boundary conditions usually help.

  • $\begingroup$ But there is a boundary condition - vanishing at inifnity. Isn't that enough to determine uniqueness? $\endgroup$
    – Joshhh
    Jan 16, 2023 at 10:19
  • $\begingroup$ I just noted something else: If $\phi$ is harmonic, so is $\nabla \phi=v$ component wise. You can show that any harmonic function on the whole $\mathbb{R}^3$ must be $0$, if it is either square integrable or if it vanishes as infinity (in the sense $v \to 0$ as $|x| \to 0$). $\endgroup$
    – F. Conrad
    Jan 16, 2023 at 11:25
  • $\begingroup$ I will leave this answer up for now, but if you want a precise answer on wether or not this boundary condition of admitting a Fourier transform is enough, you would have to deal with estimates for harmonic functions not being in any $L^p$ space. I wrote this answer since I had the feeling that you thought that the Helmholtz decomposition was unique without any further requirements. $\endgroup$
    – F. Conrad
    Jan 16, 2023 at 11:30
  • $\begingroup$ So basically there is no field $v=\nabla\phi$ defined on the whole domain that vanishes at infinity - is this what you are saying? $\endgroup$
    – Joshhh
    Jan 16, 2023 at 12:37
  • $\begingroup$ That is a delicated matter - having a Fourier transform is not equal to vanishing at infinity. However, for a function being harmonic AND having a Fourier transform might be equal to being the 0 function, but I would have to think about that. A separate question might be good for this. $\endgroup$
    – F. Conrad
    Jan 16, 2023 at 12:43

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