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Helmholtz theorem states that given a smooth vector field $\mathbf{H}$, there are a scalar field $\phi$ and a vector field $\mathbf{G}$ such that

$\mathbf{H}=\nabla \phi +\nabla \times \mathbf{G}$


$\nabla \mathbf{\cdot G}=0$

Is this decomposition unique? That is, given $\mathbf{H}$, are the fields $\phi$, $\mathbf{G}$ satisfying the above equations unique?

Edit: Unique, up to an additive constant.


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Do you mean "unique up to an additive constant"? (A latex-suggestion, use \mathbf{G} $\mathbf{G}$ instead of "poor man's bold" \pmb{G} $\pmb{G}$ which looks ugly if you do something wrong like $\pmb{\cdot G}$) – Myself May 28 '11 at 17:29
@Myself: Yes, unique up to an additive constant. Edited my post to correct that. – becko May 28 '11 at 17:38
Look here – user24703 Feb 9 '12 at 18:58
@Alexandr What's the problem? The Helmholtz/Hodge decompositions only make sense when you have a Riemannian metric anyway, and in the Helmholtz type statement, like most statements in classical vector calculus, the Euclidean metric is implicitly used to identify vectors and covectors. – Willie Wong Feb 9 '12 at 19:19
cross-post: – becko May 30 '13 at 20:56
up vote 3 down vote accepted

The decomposition is not unique without further conditions. You can add linear terms to $\phi$ and $\mathbf G$ that yield constant contributions to $\mathbf H$ that cancel:

$$ \begin{eqnarray} \phi &\to& \phi + z\;, \\ \nabla\phi &\to& \nabla\phi + \mathbf e_z\;, \\ \mathbf G &\to& \mathbf G + \frac{1}{2}(y\mathbf e_x-x\mathbf e_y)\;, \\ \nabla\times\mathbf G &\to& \nabla\times\mathbf G - \mathbf e_z\;. \end{eqnarray} $$

However, I think that if you impose suitable conditions that the fields decay at infinity, the decomposition should be unique.

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Thanks. Yes, I realize now that boundary conditions are essential. After boundary conditions have been specified, the resolution is unique. – becko May 29 '11 at 0:07

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