19.3.6.2 Solution pnifd-b

Question:

Derive the Poisson integral formula in three dimensions as given in the previous subsection.

Answer:

In two dimensions the source distribution drops out completely. In three dimensions, both the source and dipole distributions stay.

In particular, you cannot get $\partial(u-u_{\rm {out}})/\partial{r}$ to be zero. You can however take $A$ so that it only involves the given $u$ on the boundary, not the unknown radial derivative of $u$.

Then you will need to combine

$$2 \frac{\partial G}{\partial n_{\vec\xi}} + G$$

and clean up that combination using similar procedures as in three dimensions.

If you cannot find the unit vector ${\hat\imath}_r$ in spherical coordinates, evaluate it as $\vec{r}/r$ with $\vec{r}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(x,y,z)$ expressed in spherical coordinates. Use trig to clean up the dot product ${\hat\imath}_r\cdot{\hat\imath}_\rho$ a bit.

Also, in an earlier homework you, hopefully, showed that in spherical coordinates

$$\vec n {\rm d}S = \frac{\nabla F}{F_r} r^2 \sin\theta{ \rm d}\theta\phi$$

where $F$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 describes the surface, in this case $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. The gradient in spherical coordinates is in table books.