I'm working with a problem from Electromagnetics where I'm supposed to calculate the potential distribution in two different intervals. However, I'm very unfamiliar with the boundary conditions and I couldn't proceed solving the problem. The configuration of the problem is as follows:
Suppose we have an infinitely long and conducting cylinder of radius $b$. The potentials of on the conducting cylinder is given by:
$$ V(b,\phi) = 0$$
for $\phi \in [\pi/2, \pi) \cup (3\pi/2,2\pi)$ and
$$ V(b,\phi) = V_0 $$
for $\phi \in [0, \pi/2)$ and
$$ V(b,\phi) = -V_0 $$
for $\phi \in (\pi, 3\pi/2]$.
We want to find the potential distribution of this system, both inside and outside of the cylinder. In order do to this, we can use Laplace's equation, since we have no free charges in the system. Also, cylindrical coordinates suits us well, so let's calculate the Laplacian in this set of coordinates instead.
$$\nabla^2 V = 0 \Leftrightarrow \frac{1}{r}\frac{\partial}{\partial r}(r \frac{\partial V}{\partial r}) + \frac{1}{r^2} \frac{\partial^2V}{\partial\phi^2} = 0$$
I decided to exclude the $z$-term in the laplacian since the potential is independent of this variable either way since we have translation symmetry along the $z$-axis. We notice that we get a PDE, so let's try solving this using separation of variables:
$$V(r,\phi) = R(r) \Phi(\phi)$$
Substituing this into the equation above yields us:
$$(R'(r)+rR''(r))\Phi(\phi) = \frac{\partial}{\partial r}(r R'(r)\Phi(\phi)) = - \frac{1}{r} (R(r)\Phi''(\phi))$$
Dividing through by $\Phi(\phi) R(r)$ and multiplying by $r$ on both sides gives us:
$$\frac{rR'(r)+r^2R''(r)}{R(r)} = - \frac{\Phi''(\phi)}{\Phi(\phi)} := \lambda$$
Since the LHS depends solely on $r$ wheras the RHS depends solely on $\phi$, this must be equal to a constant $\lambda$. We first solve for $R(r)$. Rearranging for $R(r)$ in the equation above gives us Euler's differential equation:
$$ r^2 R''(r) + rR'(r) - \lambda R(r) = 0$$
Assigning the solution $R(r) = C r^n$ gives us:
$$ C r^n (n(n-1)+n-\lambda) = 0$$
For which we get $\lambda = n^2$. Now $R(r) = Cr^{n}$.
I have some doubts here too since $R(r) = Cr^n + Dr^{-n}$ is also a solution, but I can't really see how to use my initial conditions to show that $D = 0$ with the argument below.
Either way, we also have some condition on $R$, namely $R(b) = b$, so $C = b^{1-n}$. We now go on to solve for $\Phi(\phi)$:
$$\Phi''(\phi) + \lambda \Phi(\phi) = \Phi''(\phi) + n^2\Phi(\phi) = 0$$
where for $n \geq 1 $ we can let $$\Phi(\phi) = a_n \cos(n\phi) + b_n \sin(n\phi)$$
The total solution can be written as:
$$ V(r,\phi) = b\sum_{n\geq 1} (a_n \cos(n\phi) + b_n \sin(n\phi) ) \left(\frac{r}{b}\right)^n$$
Now, I don't really know how to apply my boundary conditions since they're not something I'm familiar with from before. I'd like to get help here so that I can find my coefficients $a_n$ and $b_n$. Also, I might have missed some details before that you freely can point out so I can correct for them.
Thank you in advance.
$D=0$ is required for regularity at the axis of the cylinder (otherwise the solution blows up at the axis). This is sort of like a boundary condition, even though in the original coordinates there is no boundary there.
After that, if you have done your other calculations correctly, you are just left to find the Fourier series of the given $V(b,\phi)$ by doing the usual integrals, then you can read off $a_n,b_n$ in your notation from there.