Proof of Equivalence of Eigenvalues for 2D and 3D Helmholtz Equations

Consider the Helmholtz equation for the electric field:

$$\begin{array}{c}({\nabla }^2+k^2)E=0\end{array}$$

The can be written as the following eigenvalue equation for the Laplacian operator

$$\begin{array}{c}{\nabla }^{2}E=-k^{2}E\end{array}$$

Let $E^{(3)}(r,\theta ,z)=R(r)\Theta (\theta )Z(z)$ where $E^{(3)}$ is any one of the transverse components of the 3D electric field $\text{E}$ (i.e. either $E_r$ or $E_{\theta }$). Prove (or disprove) that the eigenvalues $\mu$ of $E^{(2)}=\mathcal{R}(r)\mathcal{Z}(z)$ where $E^{(2)}$ satisfies $({\nabla }^2+k^2)E^{(2)}=0$ are identical to that of $E^{(3)}$ for the fundamental mode of $\Theta$, assuming the following conditions:

1. Sommerfeld radiation condition holds (corresponding to an open boundary where waves can freely pass through at $r=C$, where $R$ is an arbitrary positive constant, and can radiate out to infinity):

$$\begin{array}{c}\underset{r\to \infty }{\lim }r(\frac{\partial }{\partial r}-ik)E^{(3)}(r,\theta ,z)=r(\frac{\partial }{\partial r}-ik)E^{(3)}(r,\theta ,z){|}_{r=C}=0\end{array}$$

2. $E^{(3)}$ and $E^{(2)}$ are at least $C^2$ continuous functions at every point in space (that is, for $z\in [0,\infty ]$, $r\in [0,\infty ]$, and $\theta \in [0,2\pi ]$). In particular, $E^{(3)}$ and $E^{(2)}$ must stay bounded at every point in space.
3. $Z(z)=0$ at $z=0$ and $z=L$

Do not assume in general any boundary conditions on $\Theta (\theta )$, $Z(z)$, or $R(r)$ other than the above given conditions. The idea is that since $E^{(3)}$ and $E^{(2)}$ are product of independent functions $R(r)$, $\Theta (\theta )$, $Z(z)$ (in the case of the former) and $\mathcal{R}(r)$, $\mathcal{Z}(z)$ (in the case of the latter), then as long as the electric field is not quantized along $r$ (as stated by the first condition) and we are considering only the lowest (fundamental) mode of $\Theta$, then the eigenvalues are completely determined by the boundary conditions on $z$. If this can be proven, and it can also be proven $Z(z)=\mathcal{Z}(z)$, then the eigenvalues are thus identical. If this is too hard to show, we can consider the limiting case in which $\Theta (\theta )$ is azimuthally symmetric. The proof of this is essential since it allows us to solve the more limited case of the 2D Helmholtz equation, which is both computationally and analytically much simpler, and which has a known analytical solution for the possible values of the eigenvalues $k^2$:

$$\begin{array}{c}{k}^2=\frac{n^2{\pi }^2}{L^2},n=1,2,3,\dots \end{array}$$

First, substituting for $E=E^{(3)}$,

$$\begin{array}{c}({\nabla }^2+k^2)E^{(3)}=0\end{array}$$ $$\begin{array}{c}{\nabla }^2{E}^{(3)}+k^2{E}^{(3)}=0\end{array}$$

Because $E^{(3)}$ is dependent on cylindrical coordinates,

$$\begin{array}{c}\frac{1}{r}(r{E}_r^{(3)}{)}_r+\frac{1}{r^2}{E}_{\theta \theta }^{(3)}+E_{zz}^{(3)}+k^2{E}^{(3)}=0\end{array}$$

Substituting $E^{(3)}=R(r)\Theta (\theta )Z(z)$,

$$\begin{array}{r}\frac{\Theta Z}{r}(r{R}^{\prime }{)}^{\prime }+\frac{RZ}{r^2}{\Theta }^{\prime \prime }+R\Theta Z^{\prime \prime }+k^{2}R\Theta Z=0\end{array}$$

Multiply throughout by $\frac{r^2}{R\Theta Z}$, so

$$\begin{array}{rl}& \frac{r}{R}(r{R}^{\prime }{)}^{\prime }+\frac{{\Theta }^{\prime \prime }}{\Theta }+\frac{r^2{Z}^{\prime \prime }}{Z}+k^2{r}^2=0\\ & \frac{{\Theta }^{\prime \prime }}{\Theta }=-\frac{r}{R}(r{R}^{\prime }{)}^{\prime }-\frac{r^2{Z}^{\prime \prime }}{Z}-k^2{r}^2=-\lambda \end{array}$$

Joining the middle and right side of the equation,

$$\begin{array}{rl}& -\frac{r}{R}(r{R}^{\prime }{)}^{\prime }-\frac{r^2{Z}^{\prime \prime }}{Z}-k^2{r}^2=-\lambda \\ & \frac{r}{R}(r{R}^{\prime }{)}^{\prime }+\frac{r^2{Z}^{\prime \prime }}{Z}+k^2{r}^2=\lambda \\ & \frac{1}{rR}(r{R}^{\prime }{)}^{\prime }+\frac{Z^{\prime \prime }}{Z}+k^2=\frac{\lambda }{r^2}\\ & \frac{Z^{\prime \prime }}{Z}=-\frac{1}{rR}(r{R}^{\prime }{)}^{\prime }-k^2+\frac{\lambda }{r^2}=-\mu \end{array}$$

Joining the left and right side of the equation,

$$\begin{array}{rl}& \frac{Z^{\prime \prime }}{Z}=-\mu \\ & Z^{\prime \prime }=-\mu Z\\ & Z^{\prime \prime }+\mu Z=0\end{array}$$

Suppose $Z=e^{zt}$. Then,

$$\begin{array}{rl}& t^2{e}^{zt}+\mu e^{zt}=0\\ & t^2+\mu =0\\ & t^2=-\mu \\ & t=\pm i\sqrt{\mu }\end{array}$$

Therefore, substituting the expression above,

$$\begin{array}{c}Z(z)=A{e}^{iz\sqrt{\mu }}+B{e}^{-iz\sqrt{\mu }}\end{array}$$

Using Euler’s formula $e^{i\theta }=\cos (\theta )+i\sin (\theta )$,

$$\begin{array}{rl}Z(z)& =A(\cos (\sqrt{\mu }z)+i\sin (\sqrt{\mu }z))+B(\cos (-\sqrt{\mu }z)+i\sin (-\sqrt{\mu }z))\\ & =A\cos (\sqrt{\mu }z)+Ai\sin (\sqrt{\mu }z)+B\cos (-\sqrt{\mu }z)+Bi\sin (-\sqrt{\mu }z)\\ & =A\cos (\sqrt{\mu }z)+Ai\sin (\sqrt{\mu }z)+B\cos (\sqrt{\mu }z)-Bi\sin (\sqrt{\mu }z)\\ & =(A+B)\cos (\sqrt{\mu }z)+(A-B)i\sin (\sqrt{\mu }z)\\ & =A\cos (\sqrt{\mu }z)+B\sin (\sqrt{\mu }z)\end{array}$$

Using the given boundary conditions,

$$\begin{array}{c}Z(0)=Z(L)=0\end{array}$$ $$\begin{array}{r}Z(0)=B=0\end{array}$$ $$\begin{array}{rl}& Z(L)=A\sin (\sqrt{\mu }L)=0\\ & \sqrt{\mu }L=\pi n\\ & \mu =(\frac{n\pi }{L}{)}^2\end{array}$$

where $n=1,2,\dots$. On the other hand, substituting for $E=E^{(2)}$,

$$\begin{array}{c}({\nabla }^2+k^2)E^{(2)}=0\end{array}$$ $$\begin{array}{c}{\nabla }^2{E}^{(2)}+k^2{E}^{(2)}=0\end{array}$$

Because $E^{(2)}$ is dependent on cartesian coordinates,

$$\begin{array}{c}{E}_{rr}^{(2)}+E_{zz}^{(2)}+k^2{E}^{(2)}=0\end{array}$$

Substituting $E^{(2)}=R(r)Z(z)$,

$$\begin{array}{rl}& R^{\prime \prime }Z+R{Z}^{\prime \prime }+k^{2}RZ=0\\ & \frac{R^{\prime \prime }}{R}+\frac{Z^{\prime \prime }}{Z}+k^2=0\\ & -\frac{R^{\prime \prime }}{R}-k^2=\frac{Z^{\prime \prime }}{Z}=-\mu \end{array}$$

Joining the middle and right side,

$$\begin{array}{rl}& \frac{Z^{\prime \prime }}{Z}=-\mu \\ & Z^{\prime \prime }=-\mu Z\\ & Z^{\prime \prime }+\mu Z=0\end{array}$$

Suppose $Z(z)=e^{zt}$. Then,

$$\begin{array}{rl}& t^2{e}^{zt}+\mu e^{zt}=0\\ & t^2+\mu =0\\ & t^2=-\mu \\ & t=\pm i\sqrt{\mu }\end{array}$$

Therefore,

$$\begin{array}{c}Z(z)=A\sin (\sqrt{\mu }z)+B\cos (\sqrt{\mu }z)\end{array}$$

Using the boundary conditions provided,

$$\begin{array}{c}Z(0)=Z(L)=0\end{array}$$ $$\begin{array}{c}Z(0)=B=0\end{array}$$ $$\begin{array}{c}Z(L)=A\sin (\sqrt{\mu }L)=0\end{array}$$

Because $A\ne 0$ and $\sin (n\pi )=0$ when $n=1,2,\dots$,

$$\begin{array}{rl}& \sqrt{\mu }L=n\pi \\ & \sqrt{\mu }=\frac{n\pi }{L}\\ & \mu =(\frac{n\pi }{L}{)}^2\end{array}$$

Therefore, the eigenvalues $\mu$ of $E^{(2)}$ are identical to that of $E^{(3)}$.