# Several explicit constructions of the Helmholtz decomposition

Here we present two approaches to construct Helmholtz decompositions.

Let $\Omega$ be open and simply connected, and bounded region in $\mathbb R^3$. Then the smooth vector field $E : \Omega \to \mathbb R^3$ can be expressed by $\nabla f + \nabla \times A$ for some $f(\text{scalar field})$, $A(\text{vector field})$.

## Solving a Neumann boundary value problem

The existence and uniqueness of the pure Neumann boundary value problem for smooth data can be proved using the Green representation formula, explicitly.

First notice that for $-\Delta \Phi = \delta(x)$, by Green’s second identity and convolution formula: for $x\in \Omega$

$\int_{\Omega} \Big(u(y) \Delta \Phi(y-x) - \Phi(y-x)\Delta u(y)\Big) dy = \int_{\partial \Omega}\left(u(y)\frac{\partial \Phi}{\partial n}(y-x) - \Phi(y-x)\frac{\partial u}{\partial n}(y) \right) dS(y),$

we have

$u(x) = -\int_{\Omega} \Phi(y-x)\Delta u(y) \, dy + \int_{\partial \Omega}\left(\color{red}{\Phi(y-x)}\frac{\partial u}{\partial n}(y) - u(y)\color{cyan}{\frac{\partial \Phi}{\partial n}(y-x)} \right) dS(y).$

For Dirichlet problem, a correction function is constructed so that red term vanishes.

For Neumann problems, Green function is constructed so that $u(x)$ on each point is unique up to a constant. Some people prefer to set:

\left\{\begin{aligned} -\Delta \Phi &= \delta(x) \quad \text{in }\Omega, \\ \frac{\partial \Phi}{\partial n} &= \frac{1}{|\partial \Omega|} \quad \text{on } \partial\Omega, \end{aligned}\right.

where absolute value just means the Lebesgue measure of codimension $1$, i.e., surface area, so that the constant difference is the average of $u$ on $\partial \Omega$. I myself prefer to set something like:

\left\{\begin{aligned} -\Delta \Phi &= \delta(x) - \frac{1}{|\Omega|}\quad \text{in }\Omega, \\ \frac{\partial \Phi}{\partial n} &=0 \quad \text{on } \partial\Omega, \end{aligned}\right.\tag{\star}

so that we have:

$u(x) = \frac{1}{|\Omega|}\int_{\Omega}u(y)dy -\int_{\Omega} \Phi(y-x)\Delta u(y) \, dy + \int_{\partial \Omega} \Phi(y-x)\frac{\partial u}{\partial n}(y)dS(y).$

This means the solution is also unique up to a constant, that constant is the average of $u$ in $\Omega$, in other words, once we fix this average, $u$ is pinned down to just one function. The existence of the Green function is another story, which relies on functional analysis1.

The uniqueness of $u$ up to a constant can be either proved by the maximum principle or an energy method, but for Helmholtz decomposition does not address the uniqueness, we don’t have to address this issue here.

Now for our equation, we know that the compatibility condition is satisfied because:

$\int_{\Omega}\Delta u = \int_{\Omega}\nabla \cdot u = \int_{\partial \Omega} \frac{\partial u}{\partial n} dS = \int_{\partial \Omega} E\cdot n\,dS.$

Consider $u$ is smooth and $\int_{\Omega} u = 0$, then

$u(x) = -\int_{\Omega} \Phi(y-x) (\nabla_y \cdot E(y)) \, dy + \int_{\partial \Omega} \Phi(y-x) (E(y)\cdot n) dS(y).$

## On the whole space

The Neumann boundary approach is good for less regular vector field (say the one has component in some Sobolev space), for smooth vector field, using vector potentials would be more preferable (at least for me).

The first one is using a pointwisely held formula if $\Omega$ is convex with respect to a point $x_0 \in \mathbb{R}^3$:

$E = \nabla \phi + A,$

where

$A = -(x - x_0)\times \int^1_0 t \nabla \times E\big(x_0 + t(x- x_0)\big)\,dt ,$

and

$\phi = (x - x_0)\cdot \int^1_0 tE\big(x_0 + t(x- x_0)\big).$

Though this does not bear the exact same form, but the formula gives us more physical intuition. This is using the Poincare lemma2.

## Using Green’s function

The second one is using the Dirichlet problem’s Green function. Now we want to solve a vector Poisson equation:

\left\{\begin{aligned} -\Delta F &= E\quad \text{in }\Omega, \\ F &=0 \quad \text{on } \partial\Omega, \end{aligned}\right.

Then for the Dirichlet Green function:

\left\{\begin{aligned} -\Delta \Phi &= \delta(x) \quad \text{in }\Omega, \\ \Phi &=0 \quad \text{on } \partial\Omega. \end{aligned}\right.

Use the Green representation formula component-wise, we have

$F_i(x) = -\int_{\Omega} \Phi(y-x)\Delta F_i(y) \, dy - \int_{\partial \Omega} F_i(y) \frac{\partial \Phi}{\partial n}(y-x) dS(y),$

and this is

$F_i(x) = \int_{\Omega} \Phi(y-x)E_i(y) \, dy.$

Now use the vector Laplacian’s identity for smooth vector fields

$E(x) = -\Delta F(x) = \nabla \times \nabla \times F - \nabla \nabla \cdot F \\ =\nabla_x \times \left(\nabla_x \times \int_{\Omega} \Phi(y-x)E (y) \, dy\right) - \nabla _x\left(\nabla_x \cdot \int_{\Omega} \Phi(y-x)E(y) \, dy\right),$

where the subscript $x$ means taking gradient/div/curl w.r.t. $x$. Further moving the derivative into the integral:

$E(x) = \nabla_x \times \left(\int_{\Omega} \nabla_x \times \Big(\Phi(y-x)E (y)\Big) \, dy\right) - \nabla_x\left(\int_{\Omega} \nabla_x \cdot \Big(\Phi(y-x)E(y)\Big) \, dy\right) \\ = \nabla_x \times \left(\int_{\Omega} \nabla_x \Phi(y-x)\times E (y) \, dy\right) - \nabla_x\left(\int_{\Omega} \nabla_x \Phi(y-x)\cdot E(y) \, dy\right) \\ = - \nabla_x \times \left(\int_{\Omega} \nabla_y \Phi(y-x)\times E (y) \, dy\right) + \nabla_x\left(\int_{\Omega} \nabla_y \Phi(y-x)\cdot E(y) \, dy\right) \\ = \nabla_x \times \left(\int_{\Omega} \Big( \Phi(y-x) \nabla_y \times E (y) - \nabla_y \times (\Phi(y-x)E(y))\Big) \, dy\right) \\ + \nabla_x\left(\int_{\Omega} \Big( -\Phi(y-x) \nabla_y \cdot E (y) + \nabla_y \cdot (\Phi(y-x)E(y))\Big) \, dy\right) \\ = \nabla_x \times \left(\int_{\Omega} \Phi(y-x) \nabla_y \times E (y)\,dy - \int_{\partial \Omega} \Phi(y-x)n\times E(y)\,dS(y) \right) \\ + \nabla_x\left(-\int_{\Omega} \Phi(y-x) \nabla_y \cdot E (y) \, dy +\int_{\partial \Omega}\Phi(y-x)E(y)\cdot n\,dS(y) \right).$

So that for any $x \in \Omega$

$E = \nabla \phi + \nabla \times A,$

where

\begin{aligned} \phi(x) = -\int_{\Omega} \Phi(y-x) \nabla_y \cdot E (y) \, dy +\int_{\partial \Omega}\Phi(y-x)E(y)\cdot n\,dS(y), \\ A(x) = \int_{\Omega} \Phi(y-x) \nabla_y \times E (y)\,dy - \int_{\partial \Omega} \Phi(y-x)n\times E(y)\,dS(y). \end{aligned}
1. http://www.ams.org/journals/proc/1952-003-04/S0002-9939-1952-0051380-0/S0002-9939-1952-0051380-0.pdf

2. http://www.math.cornell.edu/~bterrell/pl.pdf

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