Laplace eq. in cylindrical coordinates and boundary conditions

[giulianinimat]

TL;DR

I have a 3D problems having a cylindrical symmetry. My domain is an hollow cylinder where i want to know if it is possible to find an analytical solution of the homogeneous Poisson equation, and if it is possible which is the potential in the whole domain. Due to the symmetry with respect to the angle, the domain is just a section of the cylinder and i have 4 different boundary conditions for the 4 segments of the boundary

 

[BvU]

Hello @giulianinimat ,

!​

Three questions

1. Is this homework ?
2. Did you read the guidelines ? (for homework we need you to post your own attempt at solution)

And finally:

3. What is your question ?​

 

[pasmith]

The answet is "Yes", in the sense that a solution will exist in terms of a Fourier-Bessel series. However in practice solving the problem numerically is probably more efficient than trying to solve the Sturm-Liouville problem for the radial basis functions or calculating the coefficients.

The boundary condition on $\Gamma_2$ is not a self-adjoint condition, but you can solve that by taking $\phi = \psi + \frac{K_2}{K_1}$ so that $\psi$ satisfies Laplace's equation together with $\psi = V_1 - \frac{K_2}{K_1}$ on $\Gamma_1$, $\psi = V_2 - \frac{K_2}{K_1}$ on $\Gamma_3$, $\partial \psi/\partial r = 0$ on $\Gamma_4$ and $\partial \psi /\partial r + K_1 \psi = 0$ on $\Gamma_2$.

Since $r = 0$ is not in the domain, this is one of those rare instances where we will need to use the Bessel function of the second kind $Y_0$ as well as $J_0$, and the radial dependence must be $$\rho_n(r) = \cos \alpha_n J_0(k_nr) + \sin \alpha_nY_0(k_nr)$$ where the eigenvalues $k_n$, $n = 0, 1, \cdots,$ satisfy $$\left| \begin{array}{cc} k_nJ_0'(k_nR_1) & k_nY_0'(k_nR_1) \\ k_nJ_0'(k_nR_2) + K_1 J_0(k_nR_2) & k_nY_0'(k_nR_2) + K_1Y_0(k_nR_2)<br /> \end{array} \right| = 0$$ and $$\tan \alpha_n = -\frac{J_0'(k_nR_1)}{Y_0'(k_nR_1)}.$$ The $\rho_n$ are orthogonal with respect to the inner product
$$\langle f, g \rangle = \int_{R_1}^{R_2} f(r) g(r) r\,dr.$$