Wave-equation on a horn Consider a conical horn, generated by the curve $y=Kx$ ($0<x<L$) rotating about the x-axis. The intersection surface of the horn is $a(x)=\pi K^2x^2$.
The related wave-equation is
\begin{equation}
u_{tt}+c^2\mathcal{A}u=0
\end{equation}
and where the identity:
\begin{equation}
\frac{1}{x^2}\frac{\partial}{\partial x}\bigg(x^2\frac{\partial u}{\partial x}\bigg)=\frac{1}{x}\frac{\partial}{\partial x^2}(xu)
\end{equation}
The horn is finite, hence $u$ is bounded and $u_x(L,t)=0$.
The function $u$ is actually $u=\frac{p-p_0}{p_0}$ where  $p$ is the pressure from the sound waves and $p_0$ is the equilibrium pressure.
How do I find a scalar product  with respect to which eigenfunctions to A with the given B.C. which are pairwise orthogonal?
My attempt:
Since the conditions are clearly Neumann conditions, I have the ansatz: $u(x,t)=u(t)\cos\frac{n\pi}{L}x$. So a scalar product would be
$\sum_{n=1}^\infty u_n(t)\cos\frac{n\pi}{L}x$
But is this really so simple?
Then I should find the eigenfunctions and eigenvalues for the operator $\mathcal{A}$ under the given conditions.
My attempt
Here i simply solved the stationary problem:
\begin{equation}
\mathcal{A}u=0\\
-\frac{1}{\pi K^2 x^2}\frac{\partial^2}{\partial x^2}(\pi K^2 xu)=0\\
\frac{\partial}{\partial x}(\pi K^2 xu)=C\\
(\pi K^2 xu)=Cx+D\\
u(x)=\frac{Cx+D}{\pi K^2 x}, \ \ \ \ \ \  I.C.: u(L)=0: \\
u(x)=-\frac{\frac{D}{L}x+D}{\pi K^2 x} 
\end{equation}
But this is giving only a partially revealed eigenfunction for the operator $\mathcal{A}u$.
Then , how do I find the eigenfrequencies, when I clearly cannot solve this wave equation without getting the nonsolvable PDE from inserting the ansatz in the original wave equation:
\begin{equation}
u_{tt}+\bigg[\frac{c^2}{x}-\frac{c^2n}{K^2xL}\cot\frac{n\pi}{L}x\bigg]u(t)=0
\end{equation}
Any hints appreciated!
Thanks
 A: I think a complete solution will help to show the general methods.
Let us study the wave equation on the surface of the cone:
$$\tag{1}
\nabla^2u(r,\phi,t)=\partial_{tt}u(r,\phi,t) \qquad, \quad0\leq r\leq R,\quad 0\leq\phi\leq 2\pi,\quad t\geq 0 
$$
Where $r$ and $\phi$ are the spherical radial and azimuthal co-ordinates, and $\theta=\alpha$ a constant (the internal half-angle of the cone, your constant $L=R \cos(\alpha)$). I've set $c=1$ for my convenience. My cone is aligned with the $z$ not $x$ axis. I'll choose the initial and boundary conditions to be
$$\tag{2}
u(r,\phi,0)=f(r) \qquad,\quad \partial_t u(r,\phi,0)=0,\quad \partial_ru(R,\phi,t)=0
$$
That is: an initial axisymmetric displacement $f$, no initial velocity, and a Neumann condition at the circular opening of the cone. At $r=0$ we must have $\partial_r u=0$ to ensure that $u$ is differentiable. Restricting the Laplacian in (1) to the cone surface yields
$$\tag{3}
\nabla^2u=\frac{1}{r^2}\partial_r(r^2\partial_r u)+\frac{1}{\sin(\alpha)r^2}\partial_{\phi \phi}u 
$$
For axisymmetric solutions the second term on the RHS of (3) vanishes, and separation of varaiables $u(r,t)=S(r)T(t)$ yields
$$\tag{4}
\frac{1}{r^2 S}\partial_r(r^2\partial_r S)=\frac{T''}{T}:=-k^2
$$
The eigenvalue problem for $S(r)$ is
$$\tag{5}
\partial_r(r^2\partial_r S)=-r^2 k^2 S
$$
This is a Sturm-Liouville problem with eigenvalues $k^2$, and weight function $w(r)=r^2$. Equation (5) may be solved directly using the transformation $v(r)=rS(r)$. The solutions are
$$\tag{6}
S(r)=\frac{1}{r}\left[A\sin(kr)+B\cos(kr) \right]
$$
Using $S'(0)=0$ sets $B=0$ and $S'(R)=0$ sets the allowed values of $k$ to be solutions to the transcendental equation
$$\tag{7}
kR=\tan(kR)
$$
Call the nonzero roots of (7) $k_n$ with $n=1,2,3,\dots$. Call the eigenfunctions of (5) $S_n(r)$. They will be orthogonal with respect to weight $w(r)$ and the following inner product
$$\tag{8}
\int\limits_0^R dr \ w(r)S_n(r)S_m(r)=\delta_{nm}
$$
The eigenvalues are really $k_n^2$ so both $k_n$ and $-k_n$ are to be identified with the same eigenfunction. The normalized eigenfunctions are
$$\tag{9}
S_n(r)=\frac{1}{\sin(k_nR)}\sqrt{\frac{2}{R }}\frac{\sin(k_n r)}{r} \qquad,\quad n=1,2,3,\dots \\
S_0(r)=\sqrt{\frac{3}{R^3}}=\text{constant}
$$
Note we deal with the $k=0$ eigenfunction separately. The initial condition $f(r)$ may be expanded in the $S_n$ basis by setting $f=c_0 S_0+\sum\limits_{n=1}^\infty c_n S_n$ and applying (8)
$$\tag{10}
\int\limits_0^R dr \ r^2 S_n(r) f(r)=c_n \qquad,\quad n=1,2,3,\dots \\
c_0=S_0\int\limits_0^R dr \ r^2 f(r)
$$
The $T(t)$ eigenfunctions may be found from (4). The ones with $T'(0)=0$ are
$$\tag{11}
T_n(t)=\cos(k_n t)
$$
The $k=0$ eigenfunction has no time dependance either: it is a constant. Using (9), (10), (11) and the definition of $k_n$ completes the solution:
$$\tag{12}
u(r,t)=c_0 S_0+\sum\limits_{n=1}^\infty c_n S_n(r) \cos(k_n t)
$$
If you want to study non-asixymmetric problems, ie. $f=f(r,\phi)$, then you simply go back to (3) and keep the term involving $\partial_\phi$. You will then have two separation constants in (4)- but the overall method remains the same. The $\phi$ part of the eigenfunctions will be $e^{im\phi}$ for integer $m$.
