Editing Helmholtz decomposition (section)

===Solution space===

If <math>(\Phi_1, {\mathbf A_1})</math> is a Helmholtz decomposition of <math>\mathbf F</math>, then 
<math>(\Phi_2, {\mathbf A_2})</math> is another decomposition if, and only if,
:<math>\Phi_1-\Phi_2 = \lambda \quad </math> and <math>\quad \mathbf{A}_1 - \mathbf{A}_2 = {\mathbf A}_\lambda + \nabla \varphi,</math>
:where
:* <math> \lambda</math> is a [[harmonic function|harmonic scalar field]],
:* <math> {\mathbf A}_\lambda </math> is a vector field which fulfills <math>\nabla\times {\mathbf A}_\lambda = \nabla \lambda,</math>
:* <math> \varphi </math> is a scalar field.

Proof: 
Set <math>\lambda = \Phi_2 -  \Phi_1</math> and <math>{\mathbf B = A_2 - A_1}</math>. According to the definition
of the Helmholtz decomposition, the condition is equivalent to 
:<math> -\nabla \lambda + \nabla \times \mathbf B = 0 </math>. 

Taking the divergence of each member of this equation yields 
<math>\nabla^2 \lambda = 0</math>, hence <math>\lambda</math> is harmonic.

Conversely, given any harmonic function <math>\lambda</math>,
<math>\nabla \lambda </math> is solenoidal since
:<math>\nabla\cdot (\nabla \lambda) = \nabla^2 \lambda = 0.</math>

Thus, according to the above section, there exists a vector field <math>{\mathbf A}_\lambda</math> such that
<math>\nabla \lambda = \nabla\times {\mathbf A}_\lambda</math>.

If <math>{\mathbf A'}_\lambda</math> is another such vector field,	
then <math>\mathbf C = {\mathbf A}_\lambda -  {\mathbf A'}_\lambda</math> 
fulfills <math>\nabla \times {\mathbf C} = 0</math>, hence <math>C = \nabla \varphi</math>
for some scalar field <math>\varphi</math>.