Vector calculus question regarding helmholtz theorem

[alejol]

How to get U(r) and W(r), why are them defined out of nowhere, is that the standard structure of that kind of functions ( what kind of functions are them??) ? T.Y

 

[Meir Achuz]

Use the div and curl equations to derive Poisson's equation for two functions (called U and W) with C and D as the source functions.

 

[vanhees71]

You can express any vector field as the sum of a potential field (curl free) and a solenoidal field (source free):

\vec{F}(\vec{x})=-\vec{\nabla} U(\vec{x})+\vec{\nabla} \times \vec{W}.

Taking the divergence of this equation, you find

\vec{\nabla} \cdot \vec{F}(\vec{x})=-\Delta U(\vec{x})=D(\vec{x}).

Assuming that all the fields vanish sufficiently quickly at infinity and that they are sufficiently well behaved everywhere, this equation (the Poisson-Laplace equation) is solved with help of the Green's function of the Laplace operator:

U(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{D(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.

The Green's function you find by solving the equation

\Delta_{\vec{x}} G(\vec{x}-\vec{x}')=-\delta^{(3)}(\vec{x}-\vec{x}').

This is done most easily by introducing spherical coordinates for \vec{r}=\vec{x}-\vec{x}' \neq {0}.

This yields together with the condition that the Green's function should vanish at infinity,

G(\vec{r})=\frac{C}{|\vec{r}|}.

The constant C is given by integrating the delta distribution over a ball of radius, R:

-1=-\int_{K_R} \mathrm{d}^3 \vec{r} \delta^{(3)}(\vec{r}) \int_{K_R} \mathrm{d}^3 \vec{r} \Delta G(\vec{r})=\int_{\partial K_R} \mathrm{d} \vec{A} \cdot \vec{\nabla} G(\vec{r})=-4 \pi C.

This means C=1/(4 \pi).

Now taking the curl of the above Helmholtz decomposition of the vector field yields

\vec{\nabla} \times \vec{F}=\vec{\nabla} \times (\vec{\nabla} \times \vec{W})=\vec{\nabla} (\vec{\nabla} \cdot \vec{W})-\Delta \vec{W}=\vec{C}.

Now \vec{W} is only determined up to the gradient of a scalar field, and thus we can demand one auxilliary condition (this is a special case of gauge invariance). Here, the condition

\vec{\nabla} \cdot \vec{W}=0

is most costumary since then we have

\Delta \vec{W}=-\vec{C},

and we can again use the Green's function of the Laplace operator to get

\vec{W}(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\vec{C}(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.