Editing Helmholtz decomposition (section)

== Three-dimensional space ==

Many physics textbooks restrict the Helmholtz decomposition to the three-dimensional space and limit its application to vector fields that decay sufficiently fast at infinity or to [[bump function]]s that are defined on a [[bounded domain]]. Then, a [[vector potential]] <math>A</math> can be defined, such that the rotation field is given by <math>\mathbf{R} = \nabla \times \mathbf{A}</math>, using the [[Curl (mathematics)|curl]] of a vector field.<ref name="petrascheck2015" />

Let <math>\mathbf{F}</math> be a vector field on a bounded domain <math>V\subseteq\mathbb{R}^3</math>, which is twice continuously differentiable inside <math>V</math>, and let <math>S</math> be the surface that encloses the domain <math>V</math> with outward surface normal <math> \mathbf{\hat{n}}' </math>. Then <math>\mathbf{F}</math> can be decomposed into a curl-free component and a divergence-free component as follows:<ref name="vermont" />

<math display="block">\mathbf{F}=-\nabla \Phi+\nabla\times\mathbf{A},</math>
where
<math display="block">
\begin{align}
\Phi(\mathbf{r}) & =\frac 1 {4\pi} \int_V \frac{\nabla'\cdot\mathbf{F} (\mathbf{r}')}{|\mathbf{r} -\mathbf{r}'|} \, \mathrm{d}V' -\frac 1 {4\pi} \oint_S \mathbf{\hat{n}}' \cdot \frac{\mathbf{F} (\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} \, \mathrm{d}S' \\[8pt]
\mathbf{A}(\mathbf{r}) & =\frac 1 {4\pi} \int_V \frac{\nabla' \times \mathbf{F}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} \, \mathrm{d}V' -\frac 1 {4\pi} \oint_S \mathbf{\hat{n}}'\times\frac{\mathbf{F} (\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} \, \mathrm{d}S'
\end{align}
</math>

and <math>\nabla'</math> is the [[nabla operator]] with respect to <math>\mathbf{r'}</math>, not <math> \mathbf{r} </math>.

If <math>V = \R^3</math> and is therefore unbounded, and <math>\mathbf{F}</math> vanishes faster than <math>1/r</math> as <math>r \to \infty</math>, then one has<ref name="griffiths1999"/>

<math display="block">\begin{align}
\Phi(\mathbf{r}) & =\frac{1}{4\pi}\int_{\R^3} \frac{\nabla' \cdot \mathbf{F} (\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} \, \mathrm{d}V' \\[8pt]
\mathbf{A} (\mathbf{r}) & =\frac{1}{4\pi}\int_{\R^3} \frac{\nabla'\times\mathbf{F} (\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} \, \mathrm{d}V'
\end{align}</math>
This holds in particular if <math>\mathbf F</math> is twice continuously differentiable in <math>\mathbb R^3</math> and of bounded support.

=== Derivation ===

{{math proof| proof =
Suppose we have a vector function <math>\mathbf{F}(\mathbf{r})</math> of which we know the curl, <math>\nabla\times\mathbf{F}</math>, and the divergence, <math>\nabla\cdot\mathbf{F}</math>, in the domain and the fields on the boundary. Writing the function using the [[delta function]] in the form
<math display="block">\delta^3(\mathbf{r}-\mathbf{r}')=-\frac 1 {4\pi} \nabla^2 \frac{1}{|\mathbf{r}-\mathbf{r}'|}\, ,</math>
where <math>\nabla^2</math> is the [[Laplacian]] operator, we have

<math display="block">\begin{align}
\mathbf{F}(\mathbf{r}) &= \int_V \mathbf{F}\left(\mathbf{r}'\right)\delta^3 (\mathbf{r}-\mathbf{r}') \mathrm{d}V' \\
&=\int_V\mathbf{F}(\mathbf{r}')\left(-\frac{1}{4\pi}\nabla^2\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}\right)\mathrm{d}V' \end{align}</math>

Now, changing the meaning of <math>\nabla^2</math> to the [[vector Laplacian]] operator (we have the right to do so because this laplacian is with respect to <math>\mathbf{r}</math> therefore it sees the vector field <math>\mathbf{F}(\mathbf{r'})</math> as a constant), we can move <math>\mathbf{F}(\mathbf{r'})</math> to the right of the<math>\nabla^2</math>operator.

<math display="block">\begin{align}\mathbf{F}(\mathbf{r})&=\int_V-\frac{1}{4\pi}\nabla^2\frac{\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V' \\
&=-\frac{1}{4\pi}\nabla^2 \int_V \frac{\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V' \\
&=-\frac{1}{4\pi}\left[\nabla\left(\nabla\cdot\int_V\frac{\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)-\nabla\times\left(\nabla\times\int_V\frac{\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)\right] \\
&= -\frac{1}{4\pi} \left[\nabla\left(\int_V\mathbf{F}(\mathbf{r}')\cdot\nabla\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)+\nabla\times\left(\int_V\mathbf{F}(\mathbf{r}')\times\nabla\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)\right] \\
&=-\frac{1}{4\pi}\left[-\nabla\left(\int_V\mathbf{F}(\mathbf{r}')\cdot\nabla'\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)-\nabla\times\left(\int_V\mathbf{F} (\mathbf{r}')\times\nabla'\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)\right]
\end{align}</math>

where we have used the vector Laplacian identity:
<math display="block">\nabla^{2}\mathbf{a}=\nabla (\nabla\cdot\mathbf{a})-\nabla\times (\nabla\times\mathbf{a}) \ ,</math>

differentiation/integration with respect to <math>\mathbf r'</math>by <math>\nabla'/\mathrm dV',</math> and in the last line, linearity of function arguments:
<math display="block"> \nabla\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}=-\nabla'\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}\ .</math>

Then using the vectorial identities

<math display="block">\begin{align}
\mathbf{a}\cdot\nabla\psi &=-\psi(\nabla\cdot\mathbf{a})+\nabla\cdot (\psi\mathbf{a}) \\
\mathbf{a}\times\nabla\psi &=\psi(\nabla\times\mathbf{a})-\nabla \times (\psi\mathbf{a})
\end{align}</math>

we get
<math display="block">\begin{align}
\mathbf{F}(\mathbf{r})=-\frac{1}{4\pi}\bigg[
&-\nabla\left(-\int_{V}\frac{\nabla'\cdot\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'+\int_{V}\nabla'\cdot\frac{\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)
\\&
-\nabla\times\left(\int_{V}\frac{\nabla'\times\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'
- \int_{V}\nabla'\times\frac{\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)\bigg].
\end{align}</math>

Thanks to the [[divergence theorem]] the equation can be rewritten as

<math display="block">\begin{align}
\mathbf{F} (\mathbf{r}) 
&=
-\frac{1}{4\pi}
\bigg[
    -\nabla\left(
        -\int_{V}
        \frac{
            \nabla'\cdot\mathbf{F}\left(\mathbf{r}'\right)
        }{
            \left|\mathbf{r}-\mathbf{r}'\right|
        } \mathrm{d}V'
        +
        \oint_{S}\mathbf{\hat{n}}'\cdot
        \frac{
            \mathbf{F}\left(\mathbf{r}'\right)
        }{
            \left|\mathbf{r}-\mathbf{r}'\right|
        }\mathrm{d}S'
    \right)
\\
&\qquad\qquad
-\nabla\times\left(\int_{V}\frac{\nabla'\times\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'
-\oint_{S}\mathbf{\hat{n}}'\times\frac{\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}S'\right)
\bigg]
\\
&= 
-\nabla\left[
  \frac{1}{4\pi}\int_{V}
  \frac{
    \nabla'\cdot\mathbf{F}\left(\mathbf{r}'\right)
  }{\left|
    \mathbf{r}-\mathbf{r}'
  \right|}
  \mathrm{d}V' 
  - 
  \frac{1}{4\pi} 
  \oint_{S}\mathbf{\hat{n}}' \cdot
  \frac{
    \mathbf{F}\left(\mathbf{r}'\right)
  }{
     \left|
       \mathbf{r}-\mathbf{r}'
     \right|
   }
 \mathrm{d}S'
\right]
\\
&\quad
+
\nabla\times
\left[
  \frac{1}{4\pi}\int_{V}
  \frac{
    \nabla '\times\mathbf{F}\left(\mathbf{r}'\right)
  }{
     \left|
       \mathbf{r}-\mathbf{r}'
     \right|
  }
  \mathrm{d}V' 
  -
  \frac{1}{4\pi}\oint_{S}
  \mathbf{\hat{n}}'
  \times
  \frac{
    \mathbf{F}\left(\mathbf{r}'\right)
  }{
    \left|
      \mathbf{r}-\mathbf{r}'
    \right|
   }
  \mathrm{d}S'
\right]
\end{align}</math>

with outward surface normal <math> \mathbf{\hat{n}}' </math>.

Defining

<math display="block">\Phi(\mathbf{r})\equiv\frac{1}{4\pi}\int_{V}\frac{\nabla'\cdot\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'-\frac{1}{4\pi}\oint_{S}\mathbf{\hat{n}}'\cdot\frac{\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}S'</math>
<math display="block">\mathbf{A}(\mathbf{r})\equiv\frac{1}{4\pi}\int_{V}\frac{\nabla'\times\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'-\frac{1}{4\pi}\oint_{S}\mathbf{\hat{n}}'\times\frac{\mathbf{F}\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}S'</math>

we finally obtain
<math display="block">\mathbf{F}=-\nabla\Phi+\nabla\times\mathbf{A}.</math>
}}

===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>.

=== Fields with prescribed divergence and curl ===
The term "Helmholtz theorem" can also refer to the following. Let {{math|'''C'''}} be a [[solenoidal vector field]] and ''d'' a scalar field on {{math|'''R'''<sup>3</sup>}} which are sufficiently smooth and which vanish faster than {{math|1/''r''<sup>2</sup>}} at infinity. Then there exists a vector field {{math|'''F'''}} such that

<math display="block">\nabla \cdot \mathbf{F} = d \quad \text{ and } \quad \nabla \times \mathbf{F} = \mathbf{C};</math>

if additionally the vector field {{math|'''F'''}} vanishes as {{math|''r'' → ∞}}, then {{math|'''F'''}} is unique.<ref name="griffiths1999"/>

In other words, a vector field can be constructed with both a specified divergence and a specified curl, and if it also vanishes at infinity, it is uniquely specified by its divergence and curl. This theorem is of great importance in [[electrostatics]], since [[Maxwell's equations]] for the electric and magnetic fields in the static case are of exactly this type.<ref name="griffiths1999"/> The proof is by a construction generalizing the one given above: we set

<math display="block">\mathbf{F} = \nabla(\mathcal{G} (d)) - \nabla \times (\mathcal{G}(\mathbf{C})),</math>

where <math>\mathcal{G}</math> represents the [[Newtonian potential]] operator. (When acting on a vector field, such as {{math|∇ × '''F'''}}, it is defined to act on each component.)

=== Weak formulation ===
The Helmholtz decomposition can be generalized by reducing the regularity assumptions (the need for the existence of strong derivatives). Suppose {{math|Ω}} is a bounded, simply-connected, [[Lipschitz domain]]. Every [[square-integrable]] vector field {{math|'''u''' ∈ (''L''<sup>2</sup>(Ω))<sup>3</sup>}} has an [[orthogonality|orthogonal]] decomposition:<ref name="amrouche1998" /><ref name="dautray1990" /><ref name="girault1986" />

<math display="block">\mathbf{u}=\nabla\varphi+\nabla \times \mathbf{A}</math>

where {{mvar|φ}} is in the [[Sobolev space]] {{math|''H''<sup>1</sup>(Ω)}} of square-integrable functions on {{math|Ω}} whose partial derivatives defined in the [[distribution (mathematics)|distribution]] sense are square integrable, and {{math|'''A''' ∈ ''H''(curl, Ω)}}, the Sobolev space of vector fields consisting of square integrable vector fields with square integrable curl.

For a slightly smoother vector field {{math|'''u''' ∈ ''H''(curl, Ω)}}, a similar decomposition holds:

<math display="block">\mathbf{u}=\nabla\varphi+\mathbf{v}</math>

where {{math|''φ'' ∈ ''H''<sup>1</sup>(Ω), '''v''' ∈ (''H''<sup>1</sup>(Ω))<sup>''d''</sup>}}.

=== Derivation from the Fourier transform ===
Note that in the theorem stated here, we have imposed the condition that if <math>\mathbf{F}</math> is not defined on a bounded domain, then <math>\mathbf{F}</math> shall decay faster than <math>1/r</math>. Thus, the [[Fourier transform]] of <math>\mathbf{F}</math>, denoted as <math>\mathbf{G}</math>, is guaranteed to exist. We apply the convention
<math display="block">\mathbf{F}(\mathbf{r}) = \iiint \mathbf{G}(\mathbf{k}) e^{i\mathbf{k} \cdot \mathbf{r}} dV_k </math>

The Fourier transform of a scalar field is a scalar field, and the Fourier transform of a vector field is a vector field of same dimension.

Now consider the following scalar and vector fields:
<math display="block">\begin{align}
G_\Phi(\mathbf{k}) &= i \frac{\mathbf{k} \cdot \mathbf{G}(\mathbf{k})}{\|\mathbf{k}\|^2} \\
\mathbf{G}_\mathbf{A}(\mathbf{k}) &= i \frac{\mathbf{k} \times \mathbf{G}(\mathbf{k})}{\|\mathbf{k}\|^2}  \\ [8pt]
\Phi(\mathbf{r}) &= \iiint G_\Phi(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{r}} dV_k \\
\mathbf{A}(\mathbf{r}) &= \iiint \mathbf{G}_\mathbf{A}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{r}} dV_k
\end{align} </math>

Hence

<math display="block">\begin{align}
\mathbf{G}(\mathbf{k}) &= - i \mathbf{k} G_\Phi(\mathbf{k}) + i \mathbf{k} \times \mathbf{G}_\mathbf{A}(\mathbf{k}) \\ [6pt]
\mathbf{F}(\mathbf{r}) &= -\iiint i \mathbf{k} G_\Phi(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{r}} dV_k + \iiint i \mathbf{k} \times \mathbf{G}_\mathbf{A}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{r}} dV_k \\
&= - \nabla \Phi(\mathbf{r}) +  \nabla \times \mathbf{A}(\mathbf{r})
\end{align}</math>

=== Longitudinal and transverse fields ===
A terminology often used in physics refers to the curl-free component of a vector field as the '''longitudinal component''' and the divergence-free component as the '''transverse component'''.<ref name="stewart2011"/> This terminology comes from the following construction: Compute the three-dimensional [[Fourier transform]] <math>\hat\mathbf{F}</math> of the vector field <math>\mathbf{F}</math>. Then decompose this field, at each point '''k''', into two components, one of which points longitudinally, i.e. parallel to '''k''', the other of which points in the transverse direction, i.e. perpendicular to '''k'''. So far, we have

<math display="block">\hat\mathbf{F} (\mathbf{k}) = \hat\mathbf{F}_t (\mathbf{k}) + \hat\mathbf{F}_l (\mathbf{k})</math>
<math display="block">\mathbf{k} \cdot \hat\mathbf{F}_t(\mathbf{k}) = 0.</math>
<math display="block">\mathbf{k} \times \hat\mathbf{F}_l(\mathbf{k}) = \mathbf{0}.</math>

Now we apply an inverse Fourier transform to each of these components. Using properties of Fourier transforms, we derive:

<math display="block">\mathbf{F}(\mathbf{r}) = \mathbf{F}_t(\mathbf{r})+\mathbf{F}_l(\mathbf{r})</math>
<math display="block">\nabla \cdot \mathbf{F}_t (\mathbf{r}) = 0</math>
<math display="block">\nabla \times \mathbf{F}_l (\mathbf{r}) = \mathbf{0}</math>

Since <math>\nabla\times(\nabla\Phi)=0</math> and <math>\nabla\cdot(\nabla\times\mathbf{A})=0</math>,

we can get

<math display="block">\mathbf{F}_t=\nabla\times\mathbf{A}=\frac{1}{4\pi}\nabla\times\int_V\frac{\nabla'\times\mathbf{F}}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'</math>
<math display="block">\mathbf{F}_l=-\nabla\Phi=-\frac{1}{4\pi}\nabla\int_V\frac{\nabla'\cdot\mathbf{F}}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'</math>

so this is indeed the Helmholtz decomposition.<ref name="littlejohn"/>