Editing Mellin transform (section)

==Problems with Laplacian in cylindrical coordinate system==

In the Laplacian in cylindrical coordinates in a generic dimension (orthogonal coordinates with one angle and one radius, and the remaining lengths) there is always a term:
<math display="block">\frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial f}{\partial r} \right) = f_{rr} + \frac{f_r}{r}</math>

For example, in 2-D polar coordinates the Laplacian is:
<math display="block">\nabla^2 f = \frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial f}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 f}{\partial \theta^2}</math>
and in 3-D cylindrical coordinates the Laplacian is,
<math display="block"> \nabla^2 f = \frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial f}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 f}{\partial \varphi^2} + \frac{\partial^2 f}{\partial z^2}.</math>

This term can be treated with the Mellin transform,<ref>Bhimsen, Shivamoggi, Chapter 6: The Mellin Transform, par. 4.3: Distribution of a Potential in a Wedge, pp. 267–8</ref> since:
<math display="block">\mathcal M \left(r^2 f_{rr} + r f_r,  r \to s \right) = s^2 \mathcal M \left(f,  r \to s \right) = s^2 F</math>

For example, the 2-D [[Laplace equation]] in polar coordinates is the PDE in two variables:
<math display="block"> r^2 f_{rr} +  r f_r + f_{\theta \theta} = 0</math>
and by multiplication:
<math display="block">\frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial f}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 f}{\partial \theta^2} = 0</math>
with a Mellin transform on radius becomes the simple [[harmonic oscillator]]:
<math display="block"> F_{\theta \theta} + s^2 F = 0</math>
with general solution:
<math display="block"> F (s, \theta) = C_1(s) \cos (s\theta) + C_2(s) \sin (s \theta)</math>

Now let's impose for example some simple wedge [[boundary conditions]] to the original Laplace equation:
<math display="block"> f(r,-\theta_0) = a(r), \quad  f(r,\theta_0) = b(r) </math>
these are particularly simple for Mellin transform, becoming:
<math display="block"> F(s,-\theta_0) = A(s), \quad  F(s,\theta_0) = B(s) </math>

These conditions imposed to the solution particularize it to:
<math display="block"> F (s, \theta) = A(s) \frac {\sin(s (\theta_0 - \theta))}{\sin (2 \theta_0 s)}+ B(s) \frac {\sin(s (\theta_0 + \theta))}{\sin (2 \theta_0 s)}</math>

Now by the convolution theorem for Mellin transform, the solution in the Mellin domain can be inverted:
<math display="block"> f(r, \theta) = \frac{r^m \cos (m \theta)}{2 \theta_0} \int_0^\infty \left ( \frac{a(x)}{x^{2m} + 2r^m x^m \sin(m \theta) + r^{2m}} + \frac{b(x)}{x^{2m} - 2r^m x^m \sin(m \theta) + r^{2m}} \right ) x^{m-1} \, dx </math>
where the following inverse transform relation was employed:
<math display="block">\mathcal M^{-1} \left( \frac {\sin (s \varphi)}{\sin (2 \theta_0 s)}; s \to r \right) = \frac 1 {2 \theta_0} \frac{r^m \sin (m \varphi)}{1+2r^m \cos(m \varphi) + r^{2m}}</math>
where <math>m= \frac \pi {2 \theta_0}</math>.