You should be able to find the solution using separation of variables. Be more specific if you need help.

This is Laplace’s Equation in Polar Coordinates.

Usually we use Separation of variables.

Using w=ln z you can map the given domain onto the rectangle [ln a, ln b] x [0, \pi/2].

Then try to see http://www.robots.ox.ac.uk/~jmb/lectures/pdelecture5.pdf

On R=[0,a]x [o,b] with the boundary condition u=f(x) on I= [ib, a +ib] and 0 on the rest of the boundary of R, the solution has the form

u= f(x)= sum_1^infty    K_n sin n pi x /a sinh n pi y /a                           and

u= f(x)= sum_1^infty K_n sin n pi x /a sinh n pi b /a for x in I .

Then

and as usual we use orthogonality formulae/HLT (Fourier series) to find the K_n.

go to the Miodrag's link and look at the page 25 and below

@Miodrag Mateljević:

Thanks for your response. :)

What if I am not interested in mapping! I want to solve the problem by direct use of separation of variables!

This special case which has some special boundary conditions is not found in usual text books!

I have problems for finding radial eigenfunctions! Can you provide me that what are the radial eigenfunctions?

@Michael Medvinsky:

Thank you for the attention.

I know that the usual process is using the separation of variables method. However, I have problem in finding the radial eigenfunctions in this BVP. Can you tell me that what are these functions?

I guess I got your problem now.

1) The Sturm-Liouville problem in this case is radial which is an Euler Ordinary Differential Equation - you can look for examples at http://tutorial.math.lamar.edu/Classes/DE/EulerEquations.aspx

A very close example to what you need can be found at http://www.math.ubc.ca/~peirce/M257_316_2012_Lecture_30.pdf

2) If you know to solve inhomogeneous problem you can also use another way to solve it(which is not necessary simpler, but may be more familiar to you). Define Z(r,\theta) = f(r)(\theta pi/2 - 1)

and solve inhomogeneous Laplace BVP for

u = \phi(r,\theta) +Z(r,\theta), i.e. you solve L(u) =  L(Z), where L is the Laplace operator. This is since L(u) = L(\phi) + L(Z) = L(Z) and L(\phi)=0

the BC for the new problem you derive in a similar way, i.e. from the BC given for \phi, e.g.vvu(a,\theta) = \phi(a,theta) +Z(0,\theta) = Z(0,\theta)vvetc.

once you obtain the solution for u you get \phi(r,\theta) = u(r,\theta) - Z(r,\theta), 

@ Michael Medvinsky :

Many thanks, Although so simple, it was exactly what I was looking for. :)

At this moment I can say the following. Set u= R \varphi. We are also going to have to multiply by to completely separate variables. So, doing all that, moving each term to one side of the equal sign and introduction a separation constant gives, u= R \varphi, we get

r/R (rR'_r)'_r= -1/\varphi \varphi''= \lambda .

The eigenvalues and eigenfunctions for this problem are, \lambda_n=n^2, \varphi_n(\theta)= sin n \theta, \varphi_n(\theta)= cos n \theta;  and

R(r)= a_1 +a_2 ln r, n=0; R(r)=c_1r^n + c_2 r^{-n}, n=1,2,3,...

u(r,\theta)= a_1 +a_2 ln r +\sum_{n=0}^infty A_n r^n cos (n \theta) + \sum_{n=1}^infty B_n r^n sin ( n \theta).  

@Miodrag Mateljević: 

Thank you professor Mateljević. :)

 I found my answer regarding Michael Medvinsky. However, I think that what you obtained here is not appropriate for this problem since we need a radial eigenfunction rather than an azimuthal one. So Sin(n*thetha) and Cos(n*thetha) are useless here!

Dear  Hosein Rahnama   & Michael Medvinsky,

It is interesting to consider    the problem from theoretical point of view.

It seems that we need to assume something about the function $f$ if we wish to get rigorous mathematical solution. I will try to explain it.

Laplacian in Polar Coordinates has the form

$\triangle u= u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2} u_{\theta\theta}$.

We set $u= R \Theta$ and consider $\triangle u= 0$.

We are also going to have to multiply by to completely separate variables.

So, doing all that, moving each term to one side of the equal sign and introduction a separation constant gives, $u= R (r) \Theta(\theta)$, we get

$$\frac{r^2 R'' + r R'}{R} = - \frac{\Theta'' }{\Theta } = - \lambda^2\,.$$

Hence (1) $\Theta'' - \lambda^2 \Theta=0$ and (2) $r^2 R'' + r R'\lambda^2 R=0$.

The solution of (1) is $\Theta= c_1 \cosh\theta + c_2 \sinh \theta$; denote this as $\Theta^\lambda $.

The solution of (2) is $R=d_1 r^{i \lambda} + d_2 r^{- i \lambda}= A \cos(\lambda \ln r) + B \sin(\lambda \ln r)$; denote this as $R^\lambda$.

Is $R^\lambda\Theta^\lambda $ a harmonic function on $C^*=C\setminus \{0\}$?

Example 1. Let $D=\{(r,\theta): A

@ Miodrag Mateljević:

Thank you for further explanation professor. I was wondering that what happens if we change the domain of solution,i.e., to solve the following boundary value problem (see the attached pic). In this case I am unable to determine the radial eigenfunctions! To be specific, I can not specify them by just considering R(b)=0. Also, the radial eigenfunctions seem strange at r=0 which may cause problems for satisfying the unhomogenuous boundary condition (the third one) at r=0.

try the second way I've suggested it is still valid here...i.e. define that Z do all the required transformations and solve the resulting inhomogeneous problem.

@ Michael Medvinsky:

Good idea, I will try to do the details and let you know. Do you have any comments about solving the problem without using change of variables and auxiliary functions ? 

nope, but fixed a typo, the problem will be inhomogeneous...

@Michael Medvinsky:

Thanks for comprehensive guides. :)

That's interesting.

When you try to solve the problem by direct use of separation of variables method, it leads to an infinite series of products of hyperbolic functions of thetha and radial eigenfunctions which are solutions of the Euler equation. However, the radial eigenfunctions are not well behaved at the origin and you can not ensure that the non-homogeneous boundary condition will be satisfied at r=0.

On the other side, when we make use of auxiilary function to transform the problem to a Poisson equation with homogeneous boundary conditions, everything is just OK and you can be sure that all boundary conditions are satisfied in both the original and auxiliary problems, specially at r=0. The solution of the original problem in this method is the sum of auxiliary function and infinite series of products of bessel functions of "r" and trigonometric functions of thetha.

I guess for the direct use of SOV you should use the continuity argument to eliminate the "bad" eigenfunctions, i.e. the coefficient of every radial eigenfunction with a (non removable) singularity at the origin must be zero.

In that case they do not form a complete set over the interval [0,a] and everything is ruined! Also, all of the eigenfunctions sin(alpha*ln(r)) and cos(alpha*ln(r)) are not defined at zero and even the behavior near zero is very oscillatory that their limit does not exist!

On the other hand, finding the auxiliary functions for complicated boundary conditions is not always as easy as this simple problem!

I think that's why we finally end up with numerical methods like FDM, FEM, BEM, etc! :D

Dear followers,

The discussion here raises question when we can use method of separate variables and what is theoretical basis for it.

Example 2. Let $D=\{(r,\theta): 0