Are all Helmholtz decompositions related?

Question

Suppose $V\subset \mathbb{R}^3$ be non-empty and at least twice differentiable (Smooth) and let $S$ be the surface that encloses $V$ (for example a sphere). Let $\textbf{F}\in \mathbb{R}^3$ be a smooth vector field for all space. Let $\textbf{n}$ denote the normal to the surface $S$. It is well known using Helmhotz decompostion that we can decompose $\textbf{F}$ into two vector fields in $V$; $$\textbf{F} = \nabla \sigma + \nabla \times \Gamma$$ where $\nabla \sigma$ is called the irrotatioal part and $\nabla \times \Gamma$ the selonodial part of $\textbf{F}$. This decomposositon is known not to be unique. I want to know if the class of all solutions are related. For example take $\sigma_1$, $\sigma_2$, $\Gamma_1$, and $\Gamma_2$ so that $$\textbf{F} = \nabla \sigma_1 + \nabla \times \Gamma_1 = \nabla \sigma_2 + \nabla \times \Gamma_2$$ and then $$ \nabla (\sigma_1- \sigma_2) + \nabla \times (\Gamma_1- \Gamma_2) = \textbf{0}.$$ This should happen only if both these vector fields vanish. Or $$ \nabla \times (\Gamma_1- \Gamma_2) = \textbf{0}$$ $$ \nabla (\sigma_1- \sigma_2) = \textbf{0}.$$ Is there anything we can say about the solutions (i,e are they linear independent or multiples of one another).

Answer 1

Q: How are two Helmholtz decompositions related?

A: The scalar fields differ by a harmonic function.

Starting from a first decomposition $\sigma_1,\Gamma_1$, you can construct a second one by adding to $\sigma_1$ a harmonic function $h$, $$\sigma_2=\sigma_1+h,\;\;\text{with}\;\;\nabla^2 h=0.$$ Then the matching $\Gamma_2$ is determined by the equation $$\nabla\times(\Gamma_1-\Gamma_2)=\nabla h.$$