*Helmholtz equation*

$$\nabla^2\psi+k^2\psi=0\ \ \ \ \ \mbox{(1)}$$

is extremely important in physics. Solving many physically important partial differential equations such as heat equation, wave equation (Klein-Gordon equation), Maxwell’s equations, and SchrÃ¶dinger equation, etc. often require solving Helmholtz equation (1).

In this notes, we discuss how to solve Helmholtz equation using separation of variables in rectangular, cylindrical, and spherical coordinate systems. The solutions we discuss here will be used when you solve boundary value problems associated with Helmholtz equation.

**Helmholtz Equation in Rectangular Coordinates**

Assume that $\psi(x,y,z)=X(x)Y(y)Z(z)$. Then the equation (1) becomes

$$YZ\frac{d^2X}{dx^2}+XZ\frac{d^2Y}{dy^2}+XY\frac{d^2Z}{dz^2}+k^2XYZ=0.\ \ \ \ \ \mbox{(2)}$$

Dividing (2) by $XYZ$, we obtain

$$\frac{1}{X}\frac{d^2X}{dx^2}+\frac{1}{Y}\frac{d^2Y}{dy^2}+\frac{1}{Z}\frac{d^2Z}{dz^2}+k^2=0.\ \ \ \ \mbox{(3)}$$

Let us write (3) as

$$\frac{1}{X}\frac{d^2X}{dx^2}=-\frac{1}{Y}\frac{d^2Y}{dy^2}-\frac{1}{Z}\frac{d^2Z}{dz^2}-k^2.\ \ \ \ \ \mbox{(4)}$$

Now we have a paradox. The LHS of (4) depends only on the $x$-variable while the RHS of (4) depends on $y$ and $z$-variables. One way to to avoid this paradox is to assume that the LHS and the RHS of (4) is a constant, say $-l^2$. If you are wondering why we choose a negative constant, the reason comes from physics. For a physical reason, we need an oscillating solution which can be obtained by choosing a negative separation constant. Often boundary conditions for Helmholtz equation lead to a trivial solution for a positive separation constant. Continuing a similar process, we separate Helmholtz equation into three ordinary differential equations:

\begin{align*}

\frac{1}{X}\frac{d^2 X}{dx^2}&=-l^2,\\

\frac{1}{Y}\frac{d^2Y}{dy^2}&=-m^2,\\

\frac{1}{Z}\frac{d^2Z}{dz^2}&=-n^2,

\end{align*}

where $k^2=l^2+m^2+n^2$.

Each mode is given by

$$\psi_{lmn}(x,y,z)=X_l(x)Y_m(y)Z_n(z)$$ and the most general solution is given by the linear combination of the modes

$$\psi(x,y,z)=\sum_{i,m,n}a_{lmn}\psi_{lmn}(x,y,z).$$

**Helmholtz Equation in Cylindrical Coordinates**

In cylindrical coordinate system $(\rho,\varphi,z)$, Helmholtz equation (1) is written as

$$\frac{1}{\rho}\frac{\partial}{\partial\rho}\left(\rho\frac{\partial\psi}{\partial\rho}\right)+\frac{1}{\rho^2}\frac{\partial^2\psi}{\partial\varphi^2}+\frac{\partial^2\psi}{\partial z^2}+k^2\psi=0.\ \ \ \ \ \mbox{(5)}$$

We assume that $\psi(\rho,\varphi,z)=P(\rho)\Phi(\varphi)Z(z)$. Then (5) can be written as

$$\frac{\Phi Z}{\rho}\frac{\partial}{\partial\rho}\left(\rho\frac{\partial\psi}{\partial\rho}\right)+\frac{PZ}{\rho^2}\frac{\partial^2\psi}{\partial\varphi^2}+P\Phi\frac{\partial^2\psi}{\partial z^2}+k^2=0.\ \ \ \ \ \mbox{(6)}$$

As we have done in rectangular coordinate system, by introducing the separation constants we can separate (6) into three ordinary differential equations

\begin{align*}

\frac{d^2Z}{dz^2}=l^2z,\\

\frac{d^2\Phi}{d\phi^2}=-m^2\Phi,\\

\rho\frac{d}{d\rho}\left(\rho\frac{dP}{d\rho}\right)+(n^2\rho^2-m^2)P=0,\ \ \ \ \ \mbox{(7)}

\end{align*}

where $n^2=k^2+l^2$. The last equation (7) is *Bessel’s differential equation*.

The general solution of Helmholtz equation in cylindrical coordinates is given by

$$\psi(\rho,\varphi,z)=\sum_{m,n}a_{mn}P_{mn}(\rho)\Phi_m(\varphi)Z_n(z).$$

**Helmholtz Equation in Spherical Coordinates**

In spherical coordinates $(r,\theta,\varphi)$, Helmholtz equation (1) is written as

$$\frac{1}{r^2\sin\theta}\left[\sin\theta\frac{\partial}{\partial r}\left(r^2\frac{\partial\psi}{\partial r}\right)+\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial\psi}{\partial\theta}\right)+\frac{1}{\sin\theta}\frac{\partial^2\psi}{\partial\varphi^2}\right]=-k^2\psi.\ \ \ \ \ \mbox{(8)}$$

Assume that $\psi(r,\theta,\varphi)=R(r)\Theta(\theta)\phi(\varphi)$. Then (8) can be written as

$$\frac{1}{Rr^2}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\frac{1}{\Theta r^2\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)+\frac{1}{\Phi r^2\sin^2\theta}\frac{d^2\Phi^2}{d\varphi^2}=-k^2.\ \ \ \ \ \mbox{(9)}$$

By introducing separation constants, (9) is separated into three ordinary differential equations

\begin{align*}

\frac{1}{\Phi}\frac{d^2\Phi}{d\varphi^2}=-m^2,\\

\frac{1}{\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)+\left(Q-\frac{m^2}{\sin^2\theta}\right)\Theta=0,\ \ \ \ \ \mbox{(10)}\\

\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\left(k^2-\frac{Q}{r^2}\right)R=0.\ \ \ \ \ \mbox{(11)}

\end{align*}

The second equation (10) is the *associated Legendre equation* with $Q=l(l+1)$. The third equation (11) is *spherical Bessel equation* with $k^2>0$.

The general solution of Helmholtz equation (8) is then given by

$$\psi(r,\theta,\varphi)=\sum_{Q,m}R_Q(r)\Theta_{Qm}(\theta)\Phi_m(\varphi).$$

The restriction that $k^2$ be a constant is unnecessary. For instance the separation process will still be possible for $k^2=f(r)$. If $k^2=f(r)$, (11) is the *associated Laguerre equation*. The associated Laguerre equation is appeared in the hydrogen atom problem in quantum mechanics.