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Solution to the scalar wave equation in cylindrical coordinates

  1. Jun 23, 2013 #1
    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
    [tex]\left(\nabla^2+k^2\right)\,\phi(\vec{r})=0,[/tex]which in cylindrical coordinates is
    [tex]\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},[/tex]where the translation between cartesian and cylindrical coordinates is given by [itex]\rho=\sqrt{x^2+y^2}[/itex], [itex]\varphi=\arctan\left(y/x\right)[/itex], [itex]z=z[/itex].

    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
    [tex]\phi(\vec{r})=J_n\left(k_\rho \rho\right)\,e^{i\left(n \varphi+k_z z\right)},[/tex]where [itex]k^2=k_\rho^2+k_z^2[/itex], [itex]n\in\mathbb{Z}[/itex], and [itex]J_n[/itex] is the first-kind Bessel function of the [itex]n[/itex]-th order.

    I know very little about Bessel functions. I do know, however, that
    [tex]J_n(x)=\sum_{m=0}^{\infty}\frac{(-1)^m}{m!\,\Gamma(m+n+1)}\left(\frac{x}{2}\right)^{2m+n},[/tex]which, by writing [itex]\Gamma(m+n+1)[/itex] explicitly, becomes[tex]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}.[/tex]I also know that
    [tex]\frac{d}{dx}J_n(x)=\frac{1}{2}\left[J_{n-1}(x)-J_{n+1}(x)\right].[/tex]
    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:
    [tex]\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)}[/tex][itex]\Rightarrow[/itex][tex]\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)}[/tex][itex]\Rightarrow[/itex][tex]\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)}.[/tex]However, I don't know where to go from here.

    If I do
    [tex]\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)},[/tex]I get stuck as well.

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

    Thanks! :)
     
  2. jcsd
  3. Jun 23, 2013 #2
    The solutions to the differential equation

    [itex]x^2 \frac{d^2 y}{d x^2}+ x \frac{d y}{d x} + (x^2-n^2) y = 0 [/itex]

    are [itex] J_n (x) [/itex] and [itex] Y_n (x) [/itex].

    You can also rewrite the Bessel Differential Equation as

    [itex] \frac{d^2 y}{d x^2}+ \frac{1}{x} \frac{d y}{d x} + (1-\frac{n^2}{x^2}) y = 0 [/itex].

    Evaluate the derivatives of [itex] \phi [/itex] and [itex] z [/itex] first, then try and rewrite the resulting differential equation in [itex] r [/itex] in the above form.
     
  4. Jun 23, 2013 #3
    Hey! Thanks for your help.

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

    What am I doing wrong?

    Thanks again.
     
  5. Jun 23, 2013 #4
    You left out the [itex]k ^2 \phi[/itex] term in the equation
    [itex](\nabla ^2+k^2) \phi =0[/itex]
     
  6. Jun 24, 2013 #5
    But of course!

    So
    [tex]\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;[/tex]dividing by [itex]k+\rho^2\,\Phi\,Z[/itex],
    [tex]\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,[/tex]which is the Bessel differential equation.

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