Solution to the scalar wave equation in cylindrical coordinates

[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, becomesJ_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! :)

 

[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.

 

[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.

 

[the_wolfman]

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

 

[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!