I need to find the velocity potential $\Phi$ defined by $\vec{u}=\nabla\Phi$ in the domain $D=\{(x,y,z) : x^2+y^2\leq R^2, z\geq0\}$. We are considering sound propagation so $\Phi$ satisfies the wave equation $\partial_{tt}\Phi=c^2\Delta\Phi$.
I am told that I can separate variables into $\Phi(x,y,z,t)=R(r)e^{in\theta}\psi(z,t)$.
Here is what I have done so far
If I put this into the wave equation, after a few lines of algebra I end up with
$R''+\frac{1}{r}R'+R(\frac{\partial_{zz}\psi}{\psi}-\frac{n^2}{r^2}-\frac{1}{c^2}\frac{\partial_{tt}\psi}{\psi})=0 \implies \frac{1}{R}\left(R''+\frac{1}{r}R'\right)-\frac{n^2}{r^2}=-\left(\frac{\partial_{zz}\psi}{\psi}-\frac{1}{c^2}\frac{\partial_{tt}\psi}{\psi}\right)$
The left side is dependent on $r$, the right side is dependent on $z,t$, so I can set both sides to be equal to a constant $C$.
I can deal with the right side
$\frac{\partial_{zz}\psi}{\psi}-\frac{1}{c^2}\frac{\partial_{tt}\psi}{\psi}=C$
by separation of variables once again, say $\psi(z,t)=A(z)B(t)$. I can show
$\frac{A''(z)}{A(z)}-\frac{1}{c^2}\frac{B''(t)}{B(t)}=C \implies \frac{A''(z)}{A(z)}=\frac{1}{c^2}\frac{B''(t)}{B(t)}+C$.
Again I can play the same game, the left side is dependent on $z$, the right side on $t$, so both sides are equal to a constant, call it $k_z$.
Both are simple second order ODEs to solve,
$A(z)=C_1e^{-ik_zz}+C_2e^{ik_zz}$
$B(t)=C_3e^{-i\omega t}+C_4e^{i\omega t}$
Where $\omega, k_z$ and the $C_i$ are constants.
Here is where I get a bit lost
For this semi infinite pipe the boundary conditions would be (I think) $u_z=0$ at $z=0$, $u_r=0$ for $r=R$. I also think there might be a pressure condition at the open end at $\infty$, say $p=0$ at $z\rightarrow\infty$, but I am not sure.
The $u_z$ condition tells me that the $z$ constants have to be equal, that is $C_1=C_2=D$. I am not sure how to eliminate any of the other constants here though, since the initial conditions are confusing me.
I am also confused about the R solution
If I sub $\psi$ back into the equation $R''+\frac{1}{r}R'+R(\frac{\partial_{zz}\psi}{\psi}-\frac{n^2}{r^2}-\frac{1}{c^2}\frac{\partial_{tt}\psi}{\psi})=0$ which I derived at the start, I get
$R''+\frac{1}{r}R'+R\left(-k_z^2-\frac{n^2}{r^2}+\frac{\omega^2}{c^2}\right)=0$
which looks sort of like a Bessel ODE, but I am not sure what the correct transformation is to get it into the Bessel form.
I think from here I could get the general solution, but I am confused about (1) the initial/boundary conditions and (2) the Bessel ODE here.