Several explicit constructions of the Helmholtz decomposition

4 minute read

Here we present two approaches to construct Helmholtz decompositions.

Let Ω be open and simply connected, and bounded region in R3. Then the smooth vector field E:ΩR3 can be expressed by f+×A for some f(scalar field), A(vector field).

Solving a Neumann boundary value problemPermalink

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 ΔΦ=δ(x), by Green’s second identity and convolution formula: for xΩ

Ω(u(y)ΔΦ(yx)Φ(yx)Δu(y))dy=Ω(u(y)Φn(yx)Φ(yx)un(y))dS(y),

we have

u(x)=ΩΦ(yx)Δu(y)dy+Ω(Φ(yx)un(y)u(y)Φn(yx))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:

{ΔΦ=δ(x)in Ω,Φn=1|Ω|on Ω,

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 Ω. I myself prefer to set something like:

{ΔΦ=δ(x)1|Ω|in Ω,Φn=0on Ω,

so that we have:

u(x)=1|Ω|Ωu(y)dyΩΦ(yx)Δu(y)dy+ΩΦ(yx)un(y)dS(y).

This means the solution is also unique up to a constant, that constant is the average of u in Ω, 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:

ΩΔu=Ωu=ΩundS=ΩEndS.

Consider u is smooth and Ωu=0, then

u(x)=ΩΦ(yx)(yE(y))dy+ΩΦ(yx)(E(y)n)dS(y).

On the whole spacePermalink

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 Ω is convex with respect to a point x0R3:

E=ϕ+A,

where

A=(xx0)×10t×E(x0+t(xx0))dt,

and

ϕ=(xx0)10tE(x0+t(xx0)).

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 functionPermalink

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

{ΔF=Ein Ω,F=0on Ω,

Then for the Dirichlet Green function:

{ΔΦ=δ(x)in Ω,Φ=0on Ω.

Use the Green representation formula component-wise, we have

Fi(x)=ΩΦ(yx)ΔFi(y)dyΩFi(y)Φn(yx)dS(y),

and this is

Fi(x)=ΩΦ(yx)Ei(y)dy.

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

E(x)=ΔF(x)=××FF=x×(x×ΩΦ(yx)E(y)dy)x(xΩΦ(yx)E(y)dy),

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

E(x)=x×(Ωx×(Φ(yx)E(y))dy)x(Ωx(Φ(yx)E(y))dy)=x×(ΩxΦ(yx)×E(y)dy)x(ΩxΦ(yx)E(y)dy)=x×(ΩyΦ(yx)×E(y)dy)+x(ΩyΦ(yx)E(y)dy)=x×(Ω(Φ(yx)y×E(y)y×(Φ(yx)E(y)))dy)+x(Ω(Φ(yx)yE(y)+y(Φ(yx)E(y)))dy)=x×(ΩΦ(yx)y×E(y)dyΩΦ(yx)n×E(y)dS(y))+x(ΩΦ(yx)yE(y)dy+ΩΦ(yx)E(y)ndS(y)).

So that for any xΩ

E=ϕ+×A,

where

ϕ(x)=ΩΦ(yx)yE(y)dy+ΩΦ(yx)E(y)ndS(y),A(x)=ΩΦ(yx)y×E(y)dyΩΦ(yx)n×E(y)dS(y).
  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 

Comments