Any boundary conditions missing from this problem? Recently I was solving some boundary value problems in Electrostatics. I stumbled upon a problem with an infinitely long cylinder (axis along the $z$-direction and radius $a$) with a plate inside it (centered at $z=0$). The plate is perpendicular to the axis of the cylinder and has the same radius as the cylinder. The plate is maintained at a constant potential $V_0$. And the surface of the cylinder is maintained at a potential $V(\varphi, z$). It is asked to find $\Phi(\rho,\varphi,z)$ inside the cylinder.
Since it is an infinitely long cylinder I've used eigenvalues of the form $e^{ikz}$ and $e^{-ikz}$ and put the boundary conditions accordingly. But I'm missing a Boundary condition. Also, I'm considering one of the modified Bessel functions $I_\nu (x)$ as the region of consideration is bounded to inside the cylinder. Can someone help me with this?
Edit :  $\Phi(\rho,\varphi,z)$ is the electrostatic potential. $\Phi(a,\varphi,z) = V(\varphi,z) $ and $V(\rho,\varphi,0) = V_0$ are the two Boundary conditions. Since $V(\varphi,z) $ is a general function, I think I'm missing one boundary condition
 A: I'm using my own symbols, for readability and writability but it'll be clear what's what, I hope.

Steady state wave equation in cylindrical coordinates:
$$\nabla^2u(r,\varphi,z)=0$$
$$\frac1r \partial_r (ru_r)+\frac{1}{r^2}u_{\varphi 
\varphi}+u_{zz}=0$$
$$\frac1r u_r+u_{rr}+\frac{1}{r^2}u_{\varphi \varphi}+u_{zz}=0$$
The domain for $R$, $[0,+a]$, for $\Phi$, $[0,2\pi]$, for $Z$, $[-\infty,+\infty]$.
Boundary conditions as stated:
$$u(a,\varphi,z)=V(\varphi,z)$$
and:
$$u(r,\varphi,0) =V_0$$
Ansatz:
$$u(r,\varphi,z)=R(r)\Phi(\varphi)Z(z)$$
$$\frac1r ZR'+ZR''+\frac{1}{r^2}RZ\Phi''+RZ''=0$$
$$\frac{R'}{rR}+\frac{R''}{R}+\frac{\Phi''}{r^2\Phi}+\frac{Z''}{Z}=0$$
$$\frac{R'}{rR}+\frac{R''}{R}+\frac{\Phi''}{r^2\Phi}=\frac{Z''}{Z}=-k^2$$
$$Z''+k^2Z=0$$

$$Z''+k^2Z=0$$
has the solution:
$$Z(z)=A\sin kx +B\cos kx$$
The boundary condition $u(r,\varphi,0) =V_0$ is first transformed to $u(r,\varphi,0) =0$, so that the BC is rendered homogeneous:
$$R(r)\Phi(\varphi)Z(0)=0 \Rightarrow Z(0)=0$$
This implies that $B=0$ and:
$$Z(z)=A\sin kx$$
This does not imply any 'sense', that is both $z\geq 0$ and $z\leq 0$ are covered.

$$\frac{R'}{rR}+\frac{R''}{R}+\frac{\Phi''}{r^2\Phi}=-k^2$$
$$\frac{rR'}{R}+\frac{r^2R''}{R}+\frac{\Phi''}{\Phi}=-k^2r^2$$
$$\frac{rR'}{R}+\frac{r^2R''}{R}+k^2r^2=-\frac{\Phi''}{\Phi}=-m^2$$

$$\Phi''-m^2\Phi=0$$
has the solution:
$$\Phi(\varphi)=Ce^{m\varphi}+De^{-m\varphi}$$
However, **no boundary conditions** to resolve $C$ and $D$ have been provided.

$$\frac{rR'}{R}+\frac{r^2R''}{R}+k^2r^2=-m^2$$
$$r^2R''+rR'+(k^2r^2+m^2)R=0$$
Solution to this DE:
$$R(r)=c_1 J_{im}(kx)+c_2Y_{im}$$
Where $J$ is the Bessel function of the first kind and $Y$ is the Bessel function of the second kind.
And here's a 'hidden' BC you thought you didn't have. Note that for $r\to 0$ the Bessel function of the second kind $Y\to -\infty$, therefore, $c_2=0$ and:
$$R(r)=c_1 J_{im}(kx)$$

A second boundary condition would stem from:
$$u(a,\varphi,z)=V(\varphi,z)-V_0$$
$$R(a)\Phi(\varphi)Z(z)=V(\varphi,z)-V_0$$
But I don't see the use of the latter equation, unless (implausibly) $\Phi(\varphi)=1$ and $Z(z)=1$.

In general, a second order PDE in three independent variables, after separation of variables, yields three second order ODEs, each ODE requiring two boundary conditions (BCs). The total number of BCs required is thus $6$.
A: Correct me if I'm missing something @HeyDosa, but for the Laplace equation, $\nabla^2 \phi =0$, $\phi$ has the uniqueness property that if it is specified for the boundary of the region(volume) where you want to find it, then it is uniquely determined. Your region of interest is (in cylindrical coordinates), $S = [0,a]\times [0,2\pi]\times [-\infty,0] \cup  [0,a]\times [0,2\pi]\times [0,\infty]$.
Its boundary is simply given by
$$\partial S =  \{(a,\phi,z) \cup (\rho,\phi,0) | z \in \mathbb{R}, \phi \in [0,2\pi], \rho \in [0,a]\}$$ (pardon my sloppy notation, but I hope you get the idea.)
The boundary conditions $\phi(a,\phi,z)=V(\phi,z)$ takes care of the first half of $\partial S$ and $\phi(\rho,\phi,0) =V_0$ takes care of the second half. With these two, the whole boundary is taken care of and therefore the solution must be unique (so mathematically no other degree of freedom is left to be fixed by an additional BC).
