Whittaker's solution and separable variables

[matt_crouch]

So It is well known that the 2D solution to the Laplace equation can be obtained by changing to complex coordinates $u=x+iy$ and $v=x-iy$. This can be extended to n dimensions as long as the complex coordinates chosen also solve the Laplace equation. For example in 3D

$w=i\sqrt{2}z+xcos\vartheta +iysin(\vartheta)$

along with the complex conjugate of w. This allows the 3D laplace equation to be solved in the same way as the 2D case. I.e the solution of
$\nabla^{2}f=0$ can be found through

$f=g(w)+h(\tilde{w})$

A general solution to the Laplace equation is given by

$f=\int f(w,\vartheta)d\vartheta$

this is called Whittaker's solution and from this we can obtain

$f=\frac{2\pi x}{(x^{2}+y^{2}+z^{2})^{3/2}}$

which is a solution to the Laplace equation...Ok so my question is I am unsure how to show that they are equivalent?

Whittaker's solution must be able to be expressed as

$f=\int f(w,\vartheta)d\vartheta=g(w)+h(\tilde{w})=\frac{2\pi x}{(x^{2}+y^{2}+z^{2})^{3/2}}$

I am sure there is a simple explanation? probably involving a taylor expansion or something? but I am not sure how to get there...I believe these all have some context and relation to twistor theory and the like but before i delve cohmology etc I am hoping for an easier solution =]
Matt

 

[Strum]

As far as I could figure out by reading some old old document Whittaker shewed that solutions to the 3d laplace equation can be put on the form $\int_{0}^{2\pi} f(x + iy\cos(v) + iz\sin(v)) dv$. By pure guessing I found that

$\int_{0}^{2\pi} f(x,y,z,v) dv = \frac{2\pi x }{\left(x^2 + y^2 + z^2\right)^{3/2}}$ with $f(x,y,z,v) = \left(x + iz\cos(v) + iy\sin(v)\right)^{-2}$. Now you just have to find g(w) and h(w), right?

 

[matt_crouch]

As far as I could figure out by reading some old old document Whittaker shewed that solutions to the 3d laplace equation can be put on the form $\int_{0}^{2\pi} f(x + iy\cos(v) + iz\sin(v)) dv$. By pure guessing I found that

$\int_{0}^{2\pi} f(x,y,z,v) dv = \frac{2\pi x }{\left(x^2 + y^2 + z^2\right)^{3/2}}$ with $f(x,y,z,v) = \left(x + iz\cos(v) + iy\sin(v)\right)^{-2}$. Now you just have to find g(w) and h(w), right?

Pretty much, but where the solution to the Laplace equation is the addition of some function $g(w)$ and $h(\tilde{w})$ where $\tilde{w}$ is the complex conjugate of $w$...
The more I'm reading about it, it seems as if I need to look at twistor theory, global and local solutions etc...I wonder if take a laurent expansion would help, but I haven't really been able to get very far

 

[matt_crouch]

As far as I could figure out by reading some old old document Whittaker shewed that solutions to the 3d laplace equation can be put on the form $\int_{0}^{2\pi} f(x + iy\cos(v) + iz\sin(v)) dv$. By pure guessing I found that

$\int_{0}^{2\pi} f(x,y,z,v) dv = \frac{2\pi x }{\left(x^2 + y^2 + z^2\right)^{3/2}}$ with $f(x,y,z,v) = \left(x + iz\cos(v) + iy\sin(v)\right)^{-2}$. Now you just have to find g(w) and h(w), right?

Would you show me how you solved the integral? I appear to be having some trouble

 

[Strum]

Maple syntax.

f:=(x,y,z,v) -> (x + I*z*cos(v) + I*y*sin(v))^-2;
int(f(x,y,z,v),v=0..2*Pi) assuming(x,real,y,real,z,real,v,real)