The math is sound. The method is not.
We show that the order- \(n\) Hankel transform in \(r\) diagonalizes the Bessel-type radial operator \(\mathscr{L}_n\), mapping the PDE \(\left(\mathscr{L}_n+\partial_z^2\right) \varphi_n=0\) to the transformed equation \(\left(\partial_z^2-k^2\right) \hat{\varphi}_n=0\). The key ingredients are the Sturm-Liouville form and self-adjointness of \(\mathscr{L}_n\), together with the Bessel differential equation.
Claim
Consider the scalar field \(\varphi_n(r, z)\) and the radial Bessel operator \[ \mathscr{L}_n:=\partial_r^2+\frac{1}{r} \partial_r-\frac{n^2}{r^2} . \] The \(n\)-th order Hankel transform (in the radial variable \(r\) ) of the equation
\[ \left(\partial_r^2+\frac{1}{r} \partial_r-\frac{n^2}{r^2}\right) \varphi_n+\partial_z^2 \varphi_n=0 \]
yields the transformed equation
\[ -k^2 \hat{\varphi}_n(k, z)+\partial_z^2 \hat{\varphi}_n(k, z)=0 . \]
Here the order- \(n\) Hankel transform is
\[ \mathscr{H}_n[f](k):=\int_0^{\infty} f(r) J_n(k r) r d r, \quad k>0 \]
and we write \(\hat{\varphi}_n(k, z):=\mathscr{H}_n\left[\varphi_n(\cdot, z)\right](k)\).
Radial Bessel Operator and Sturm-Liouville Form
We first rewrite \(\mathscr{L}_n\) in Sturm-Liouville form: \[ \mathscr{L}_n f=\frac{1}{r} \frac{d}{d r}\left(r \frac{d f}{d r}\right)-\frac{n^2}{r^2} f . \] This representation shows that \(\mathscr{L}_n\) is formally self-adjoint with respect to the weighted inner product
\[ \langle u, v\rangle:=\int_0^{\infty} u(r) v(r) r d r \]
[[kelvin-wedge-self-adjoint-operators]]
Self-Adjointness and Integration by Parts
We compute the Hankel transform of \(\mathscr{L}_n f\) : \[ \begin{aligned} \mathscr{H}_n\left[\mathscr{L}_n f\right](k) & =\int_0^{\infty} r J_n(k r)\left(\mathscr{L}_n f\right) d r \\ & =\int_0^{\infty} J_n(k r) \frac{d}{d r}\left(r \frac{d f}{d r}\right) d r-\int_0^{\infty} \frac{n^2}{r} J_n(k r) f(r) d r \end{aligned} \]
By self-adjointness of \(\mathscr{L}_n\) with respect to the weight \(r\), we may transfer \(\mathscr{L}_n\) from \(f\) to \(J_n(k r)\) :
\[ \int_0^{\infty} r J_n(k r)\left(\mathscr{L}_n f\right) d r=\int_0^{\infty}\left(\mathscr{L}_n\left[J_n(k r)\right]\right) f(r) r d r \]
provided that the boundary terms at \(r=0\) and \(r \rightarrow \infty\) vanish. This is the crucial self-adjointness property.
Notes
Detailed integration by parts:
For completeness, we verify \[ \int_0^{\infty} r J_n(k r)\left(\mathscr{L}_n f\right) d r=\int_0^{\infty} \mathscr{L}_n\left[J_n(k r)\right] f(r) r d r \]
Starting from the Sturm-Liouville form,
\[ \begin{aligned} \int_0^{\infty} r J_n(k r)\left(\mathscr{L}_n f\right) d r & =\int_0^{\infty} r J_n(k r)\left\{\frac{1}{r} \frac{d}{d r}\left(r \frac{d f}{d r}\right)-\frac{n^2}{r^2} f\right\} d r \\ & =\int_0^{\infty}\left[J_n(k r) \frac{d}{d r}\left(r \frac{d f}{d r}\right)-\frac{n^2}{r} J_n(k r) f\right] d r \\ & =\left[J_n(k r) r f^{\prime}(r)\right]_0^{\infty}-\int_0^{\infty} r f^{\prime}(r) \frac{d}{d r} J_n(k r) d r-\int_0^{\infty} \frac{n^2}{r} J_n(k r) f d r \end{aligned} \]
Assuming the boundary term vanishes, we apply integration by parts again:
\[ -\int_0^{\infty} r f^{\prime}(r) J_n^{\prime}(k r) d r=-\left[f(r) r J_n^{\prime}(k r)\right]_0^{\infty}+\int_0^{\infty} f(r) \frac{d}{d r}\left(r J_n^{\prime}(k r)\right) d r \] so that
\[ \begin{aligned} \int_0^{\infty} r J_n(k r)\left(\mathscr{L}_n f\right) d r & =\int_0^{\infty} f(r)\left\{\frac{d}{d r}\left(r J_n^{\prime}(k r)\right)-\frac{n^2}{r} J_n(k r)\right\} d r \\ & =\int_0^{\infty} f(r)\left\{\frac{1}{r} \frac{d}{d r}\left(r \frac{d}{d r} J_n(k r)\right)-\frac{n^2}{r^2} J_n(k r)\right\} r d r \\ & =\int_0^{\infty} f(r) \mathscr{L}_n\left[J_n(k r)\right] r d r \end{aligned} \]
Action of \(\mathscr{L}_n\) on Bessel Functions
The Bessel function \(J_n(k r)\) satisfies the Bessel differential equation \[ J_n^{\prime \prime}(k r)+\frac{1}{k r} J_n^{\prime}(k r)+\left(1-\frac{n^2}{k^2 r^2}\right) J_n(k r)=0 . \]
Applying \(\mathscr{L}_n\) to \(J_n(k r)\), we obtain
\[ \begin{aligned} \mathscr{L}_n\left[J_n(k r)\right] & =\frac{d^2}{d r^2} J_n(k r)+\frac{1}{r} \frac{d}{d r} J_n(k r)-\frac{n^2}{r^2} J_n(k r) \\ & =k^2\left[J_n^{\prime \prime}(k r)+\frac{1}{k r} J_n^{\prime}(k r)-\frac{n^2}{k^2 r^2} J_n(k r)\right] \end{aligned} \]
Using the Bessel equation in the form
\[ J_n^{\prime \prime}(k r)+\frac{1}{k r} J_n^{\prime}(k r)=-\left(1-\frac{n^2}{k^2 r^2}\right) J_n(k r), \]
we deduce
\[ \begin{aligned} \mathscr{L}_n\left[J_n(k r)\right] & =k^2\left[-\left(1-\frac{n^2}{k^2 r^2}\right) J_n(k r)-\frac{n^2}{k^2 r^2} J_n(k r)\right] \\ & =-k^2 J_n(k r) \end{aligned} \] Substituting this into the self-adjointness identity gives
\[ \begin{aligned} \mathscr{H}_n\left[\mathscr{L}_n f\right](k) & =\int_0^{\infty} \mathscr{L}_n\left[J_n(k r)\right] f(r) r d r \\ & =\int_0^{\infty}\left(-k^2 J_n(k r)\right) f(r) r d r \\ & =-k^2 \int_0^{\infty} f(r) J_n(k r) r d r \\ & =-k^2 \hat{f}_n(k) \end{aligned} \]
Thus the Hankel transform diagonalizes the radial Bessel operator:
\[ \mathscr{H}_n\left[\mathscr{L}_n f\right](k)=-k^2 \hat{f}_n(k) . \]
Conclusion for the PDE
Applying \(\mathscr{H}_n\) in \(r\) to \[ \mathscr{L}_n \varphi_n(r, z)+\partial_z^2 \varphi_n(r, z)=0 \]
and using linearity in \(z\), we obtain
\[ -k^2 \hat{\varphi}_n(k, z)+\partial_z^2 \hat{\varphi}_n(k, z)=0 \]
In summary: 1. The boundary conditions at \(r=0\) and \(r \rightarrow \infty\) are essential for the self-adjointness of \(\mathscr{L}_n\) and for discarding boundary terms. 2. The order- \(n\) Hankel transform diagonalizes the Bessel operator \(\mathscr{L}_n\). 3. The eigenvalue associated with radial wavenumber \(k\) is \(-k^2\), leading to the transformed PDE \(\left(\partial_z^2-\right. \left.k^2\right) \hat{\varphi}_n=0\).