Editing Helmholtz decomposition (section)

== Uniqueness of the solution ==

In general, the Helmholtz decomposition is not uniquely defined.
A [[harmonic function]] <math>H(\mathbf{r})</math> is a function that satisfies <math>\Delta H(\mathbf{r}) = 0</math>.
By adding <math>H(\mathbf{r})</math> to the scalar potential <math>\Phi(\mathbf{r})</math>, a different Helmholtz decomposition can be obtained:

<math display="block">\begin{align}
\mathbf{G}'(\mathbf{r}) &= \nabla (\Phi(\mathbf{r}) + H(\mathbf{r})) = \mathbf{G}(\mathbf{r}) + \nabla H(\mathbf{r}),\\
\mathbf{R}'(\mathbf{r}) &= \mathbf{R}(\mathbf{r}) - \nabla H(\mathbf{r}).
\end{align}</math>

For vector fields <math>\mathbf{F}</math>, decaying at infinity, it is a plausible choice that scalar and rotation potentials also decay at infinity. 
Because <math>H(\mathbf{r}) = 0</math> is the only harmonic function with this property, which follows from [[Liouville's theorem (complex analysis)|Liouville's theorem]], this guarantees the uniqueness of the gradient and rotation fields.<ref name="axler1992" />

This uniqueness does not apply to the potentials: In the three-dimensional case, the scalar and vector potential jointly have four components, whereas the vector field has only three. The vector field is invariant to gauge transformations and the choice of appropriate potentials known as [[gauge fixing]] is the subject of [[gauge theory]]. Important examples from physics are the [[Lorenz gauge condition]] and the [[Coulomb gauge]]. An alternative is to use the [[poloidal–toroidal decomposition]].