# Solution to the scalar wave equation in cylindrical coordinates

1. Jun 23, 2013

### IridescentRain

Hello.

I don't know how to prove that a certain function is a solution to the scalar wave equation in cylindrical coordinates.

The scalar wave equation is
$$\left(\nabla^2+k^2\right)\,\phi(\vec{r})=0,$$which in cylindrical coordinates is
$$\frac{1}{\rho}\frac{\partial}{\partial\rho}\left(\rho\frac{\partial\phi}{\partial\rho}\right)+\frac{1}{\rho^2}\frac{\partial^2\phi}{\partial \varphi^2}+\frac{\partial^2\phi}{\partial z^2},$$where the translation between cartesian and cylindrical coordinates is given by $\rho=\sqrt{x^2+y^2}$, $\varphi=\arctan\left(y/x\right)$, $z=z$.

According to Scattering of electromagnetic waves: theories and applications by Tsang L, Kong J A and Ding K-H, a solution to this is the function
$$\phi(\vec{r})=J_n\left(k_\rho \rho\right)\,e^{i\left(n \varphi+k_z z\right)},$$where $k^2=k_\rho^2+k_z^2$, $n\in\mathbb{Z}$, and $J_n$ is the first-kind Bessel function of the $n$-th order.

I know very little about Bessel functions. I do know, however, that
$$J_n(x)=\sum_{m=0}^{\infty}\frac{(-1)^m}{m!\,\Gamma(m+n+1)}\left(\frac{x}{2}\right)^{2m+n},$$which, by writing $\Gamma(m+n+1)$ explicitly, becomes$$J_n(x)=\sum_{m=0}^{\infty}\frac{(-1)^m}{m!\,\int_0^{\infty} t^{m+n}\,e^{-t}\,dt}\left(\frac{x}{2}\right)^{2m+n}.$$I also know that
$$\frac{d}{dx}J_n(x)=\frac{1}{2}\left[J_{n-1}(x)-J_{n+1}(x)\right].$$
So I set out to prove that this is indeed a solution to the wave equation in cylindrical coordinates. However, I didn't get very far. Here's what I did:
$$\frac{\partial\phi}{\partial\rho}=\frac{k_\rho}{2}\left[J_{n-1}(k_\rho \rho)-J_{n+1}(k_\rho \rho)\right]\,e^{i\,(n \varphi+k_z z)}$$$\Rightarrow$$$\left(\nabla^2+k^2\right)\,\phi=\frac{1}{\rho}\frac{\partial}{\partial \rho}\left[\frac{k_\rho \rho}{2}J_{n-1}(k_\rho \rho)-\frac{k_\rho \rho}{2}J_{n+1}(k_\rho \rho)\right]\,e^{i\,(n \varphi+k_z z)}-\left(\frac{n^2}{\rho^2}+k_z^2\right)\,J_n(k_\rho \rho)\,e^{i\,(n \varphi+k_z z)}$$$\Rightarrow$$$\left(\nabla^2+k^2\right)\,\phi=\left[\frac{k_\rho^2}{4}J_{n-2}(k_\rho \rho)+\frac{k_\rho}{2\rho}J_{n-1}(k_\rho \rho)-\left(\frac{k_\rho^2}{2}+k_z^2+\frac{n^2}{\rho^2}\right)\,J_n(k_\rho \rho)-\frac{k_\rho}{2\rho}J_{n+1}(k_\rho \rho)-\frac{k_\rho^2}{4}J_{n+2}(k_\rho \rho)\right]\,e^{i\,(n \varphi+k_z z)}.$$However, I don't know where to go from here.

If I do
$$\frac{\partial\phi}{\partial\rho}=\sum_{m=0}^{\infty}\frac{(-1)^m}{m!\,\int_0^{\infty} t^{m+n}\,e^{-t}\,dt}\left(\frac{k_\rho}{2}\right)^{2m+n}\,(2m+n)\,\rho^{2m+n-1}\,e^{i\,(n \varphi+k_z z)},$$I get stuck as well.

How should I approach the problem of proving that the above function $\phi(\vec{r})$ is a solution to the wave equation in cylindrical coordinates?

Thanks! :)

2. Jun 23, 2013

### the_wolfman

The solutions to the differential equation

$x^2 \frac{d^2 y}{d x^2}+ x \frac{d y}{d x} + (x^2-n^2) y = 0$

are $J_n (x)$ and $Y_n (x)$.

You can also rewrite the Bessel Differential Equation as

$\frac{d^2 y}{d x^2}+ \frac{1}{x} \frac{d y}{d x} + (1-\frac{n^2}{x^2}) y = 0$.

Evaluate the derivatives of $\phi$ and $z$ first, then try and rewrite the resulting differential equation in $r$ in the above form.

3. Jun 23, 2013

### IridescentRain

Hey! Thanks for your help.

All right, I have
$$\phi(\vec{r})=R(\rho)\,\Phi(\varphi)\,Z(z),$$where
$$R(\rho)=J_n(k_\rho\rho),$$$$\Phi(\varphi)=e^{in\varphi},$$$$Z(z)=e^{ik_zz}.$$
Therefore,
$$\frac{1}{\rho^2}\frac{\partial^2\phi}{\partial\varphi^2}=-\frac{n^2}{\rho^2}\phi,$$$$\frac{\partial^2\phi}{\partial z^2}=-k_z^2\,\phi,$$$$\frac{1}{\rho}\frac{\partial}{\partial\rho}\left(\rho\frac{\partial\phi}{\partial\rho}\right)=\frac{1}{\rho}\frac{\partial (k_\rho\rho)}{\partial\rho}\frac{\partial}{\partial(k_\rho\rho)} \left(\rho \frac{\partial(k_\rho\rho)}{\partial\rho}\frac{\partial\phi}{\partial(k_\rho\rho)}\right)=k_\rho^2\frac{\partial^2\phi}{\partial(k_\rho\rho)^2}+ \frac{k_\rho}{\rho} \frac{\partial\phi}{\partial(k_\rho\rho)}.$$
Putting all three together, I get
$$\left[k_\rho^2\frac{d^2R}{d(k_\rho\rho)^2}+\frac{k_\rho}{\rho}\frac{dR}{d (k_\rho \rho)}\right]\,\Phi\,Z+R\frac{1}{\rho^2}\frac{d^2\Phi}{d\varphi^2}Z+R\,\Phi\frac{d^2Z}{dz^2}=\left[k_\rho^2\frac{d^2R}{d(k_\rho\rho)^2}+\frac{k_\rho}{\rho}\frac{dR}{d (k_\rho \rho)}\right]\,\Phi\,Z-\left[\frac{n^2}{\rho^2}+k_z^2\right]\,R\,\Phi\,Z=0.$$
Let $x:=k_\rho\rho$. Since $\Phi(\varphi)$ and $Z(z)$ are never zero and $k_\rho\neq0$, I may divide everything by $k_\rho^2\,\Phi\,Z$:
$$\frac{d^2R}{dx^2}+\frac{1}{x}\frac{dR}{dx}-\left[\frac{n^2}{x^2}+\frac{k_z^2}{k_\rho^2}\right]\,R=0.$$
Comparing this with the equation you provided, $R(x)=J_n(x)$ only if $k_z^2/k_\rho^2=-1$, which is a really strange condition to impose on $k$ (recall that $k^2=k_\rho^2+k_z^2$; if I impose $k_z^2/k_\rho^2=-1$, then $k^2=k_\rho^2-k_\rho^2=0$, which is surely very silly).

What am I doing wrong?

Thanks again.

4. Jun 23, 2013

### the_wolfman

You left out the $k ^2 \phi$ term in the equation
$(\nabla ^2+k^2) \phi =0$

5. Jun 24, 2013

### IridescentRain

But of course!

So
$$\left[k_\rho^2\frac{d^2R}{d(k_\rho\rho)^2}+\frac{k_\rho}{\rho}\frac{dR}{d (k_\rho \rho)}\right]\,\Phi\,Z-\left[\frac{n^2}{\rho^2}+k_z^2\right]\,R\,\Phi\,Z+\left(k_\rho^2+k_z^2\right)\,R\,\Phi\,Z=0;$$dividing by $k+\rho^2\,\Phi\,Z$,
$$\frac{d^2R}{dx^2}+\frac{1}{x}\frac{dR}{dx}-\left[\frac{n^2}{x^2}+\frac{k_z^2}{k_\rho^2}-\frac{k_\rho^2+k_z^2}{k_\rho^2}\right]\,R=\frac{d^2R}{dx^2}+\frac{1}{x}\frac{dR}{dx}+\left[1-\frac{n^2}{x^2}\right]\,R=0,$$which is the Bessel differential equation.

I completely forgot about that $k^2$ in the original wave equation. Thanks for pointing it out!