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
$\mathbf{H}=\nabla \phi +\nabla \times \mathbf{G}$
and
$\nabla \mathbf{\cdot G}=0$
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.
Thanks
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.