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Help in deriving series expansion of Helmholtz Equation.

  1. Nov 20, 2013 #1
    Orthogonality of spherical bessel functions

    1. The problem statement, all variables and given/known data

    Proof of orthogonality of spherical bessel functions

    The book gave
    [tex]\int_{0}^{a}\int_{0}^{2\pi}\int_{0}^{\pi} j_{n} (\lambda_{n,j}r) j_{n'} (\lambda_{n',j'}r) Y_{n,m}(\theta,\phi)\overline{Y}_{n',m'}(\theta,\phi) \sin\theta \;dr d\theta d\phi=\frac{a^{3}}{2}j^2_{n+1}(\alpha_{n+\frac{1}{2},j})[/tex]
    For [itex](n=n')[/itex], [itex](j=j')[/itex] and [itex](m= m')[/itex]

    I got only
    [tex]\int_{0}^{a}\int_{0}^{2\pi}\int_{0}^{\pi} j_{n} (\lambda_{n,j}r) j_{n'} (\lambda_{n',j'}r) Y_{n,m}(\theta,\phi)\overline{Y}_{n',m'}(\theta,\phi) \sin\theta \;dr d\theta d\phi=\frac{a^{2}}{2}j^2_{n+1}(\alpha_{(n+\frac{1}{2},j)})[/tex]

    2. Relevant equations

    Helmholtz equation: [itex]\nabla^2 u(r,\theta,\phi)=-k u(r,\theta,\phi)[/itex] Where [itex]u_{n,m}(r,\theta,\phi)=R_{n}(r)Y_{n,m}(\theta,\phi)[/itex]

    Where [itex]Y_{n,m}(\theta,\phi)=\sqrt{\frac{(2n+1)(n-m)!}{4\pi(n+m)!}} P_{n}^{m}(\cos\theta)e^{jm\phi}[/itex] is the Spherical Harmonics.

    And [itex]R_{n}(r)=j_{n}(\lambda_{n,j} r)=\sqrt{\frac{\pi}{2\lambda_{n,j} r}} J_{n+1/2}(\lambda_{n,j} r)[/itex] (1) is the Spherical Bessel function.


    Orthogonal properties stated that

    For [itex]0\leq r \leq a [/itex] where [itex]R(0)[/itex] is finite and [itex]R(a)=0[/itex]:

    [itex]R(a)=0\Rightarrow\; \lambda{n,j}=\frac{\alpha_{n,j}}{a}[/itex]

    [itex]\int_{0}^{2\pi}\int_{0}^{\pi}Y_{n,m}(\theta,\phi)\overline{Y}_{n',m'}(\theta,\phi) \sin\theta \;d\theta d\phi=0[/itex] For [itex](n\neq n')[/itex] and [itex](m\neq m')[/itex]

    [itex]\int_{0}^{2\pi}\int_{0}^{\pi}Y_{n,m}(\theta,\phi)\overline{Y}_{n',m'}(\theta,\phi) \sin\theta \;d\theta d\phi=1 [/itex] For [itex](n=n')[/itex] and [itex](m= m')[/itex](2)

    And [itex]\int_{0}^{a} r J_n^2(\lambda_{n,j} r)dr=\frac {a^{2}}{2}J_{n+1}^2(\alpha_{n,j})[/itex] (3) where [itex]\alpha_{n,j}[/itex] is the j zero of the Bessel function.


    3. The attempt at a solution

    For [itex](n=n')[/itex], [itex](j=j')[/itex] and [itex](m= m')[/itex]

    [itex]\int_{0}^{a}\int_{0}^{2\pi}\int_{0}^{\pi} j_{n} (\lambda_{n,j}r) j_{n'} (\lambda_{n',j'}r) Y_{n,m}(\theta,\phi)\overline{Y}_{n',m'}(\theta,\phi) \sin\theta \;dr d\theta d\phi=\int_{0}^{a} j_{n}^{2} (\lambda_{n,j}r)dr\;\int_{0}^{2\pi}\int_{0}^{\pi}|Y_{n,m}(\theta,\phi)|^{2}\sin\theta \;d\theta d\phi=\int_{0}^{a} j_{n}^{2} (\lambda_{n,j}r)dr[/itex]

    As the two have different independent variables and from (2), [itex]\int_{0}^{2\pi}\int_{0}^{\pi}|Y_{n,m}(\theta,\phi)|^{2}\sin\theta \;d\theta d\phi=1[/itex]



    Using (1), (3)
    [tex]\int_{0}^{a}\int_{0}^{2\pi}\int_{0}^{\pi} j_{n} (\lambda_{(n,j)}r) j_{n'} (\lambda_{n',j'}r) Y_{n,m}(\theta,\phi)\overline{Y}_{n',m'}(\theta,\phi) \sin\theta \;dr d\theta d\phi = \int_{0}^{a} j_{n}^{2} (\lambda_{(n,j)}r)rdr= \frac{\pi}{2\lambda_{(n,j)}}\int_{0}^{a} J_{n+\frac{1}{2}}^{2}(\lambda_{(n,j)}r)rd r[/tex]

    [itex]R(a)=0\;\Rightarrow\;\lambda_{(n,j)}=\frac{\alpha_{(n+\frac{1}{2},j)}}{a}[/itex] as [itex]\alpha_{(n+\frac{1}{2},j)}[/itex] is the [itex]j^{th}[/itex] zero of [itex]J_{n+\frac{1}{2}}(\lambda_{(n,j)})[/itex]

    [tex] \frac{\pi}{2\lambda_{(n,j)}}\int_{0}^{a} J_{n+\frac{1}{2}}^{2}(\lambda_{(n,j)}r)rd r= \frac{\pi}{2\lambda_{(n,j)}} \frac{a^2}{2}J_{n+\frac{3}{2}}(\alpha_{(n+\frac{1}{2},j)})=\frac{a^{2}}{2}j^2_{n+1}(\alpha_{(n+\frac{1}{2},j)})[/tex]

    I am missing the [itex]a[/itex]. Thanks
     
    Last edited: Nov 21, 2013
  2. jcsd
  3. Nov 21, 2013 #2
    I already find proof that for [itex]0\leq r \leq a [/itex] where [itex]R(0)[/itex] is finite and [itex]R(a)=0[/itex]:

    [tex]\int_{0}^{a}\int_{0}^{2\pi}\int_{0}^{\pi} j_{n} (\lambda_{(n,j)}r) j_{n'} (\lambda_{n',j'}r) Y_{n,m}(\theta,\phi)\overline{Y}_{n',m'}(\theta,\phi) \sin\theta \;dr d\theta d\phi = \int_{0}^{a} j_{n}^{2} (\lambda_{(n,j)}r)rdr[/tex]

    So all I need to proof is
    [tex]\int_{0}^{a} j_{n}^{2} (\lambda_{(n,j)}r)rdr=\frac{a^{3}}{2}j^2_{n+1}(\alpha_{(n+\frac{1}{2},j)})[/tex]


    But as in the last post:
    [tex]\int_{0}^{a} j_{n}^{2} (\lambda_{(n,j)}r)rdr= \frac{\pi}{2\lambda_{(n,j)}}\int_{0}^{a} J_{n+\frac{1}{2}}^{2}(\lambda_{(n,j)}r)rd r[/tex]

    [itex]R(a)=0\;\Rightarrow\;\lambda_{(n,j)}=\frac{\alpha_{(n+\frac{1}{2},j)}}{a}[/itex] as [itex]\alpha_{(n+\frac{1}{2},j)}[/itex] is the [itex]j^{th}[/itex] zero of [itex]J_{n+\frac{1}{2}}(\lambda_{(n,j)})[/itex]

    [tex] \frac{\pi}{2\lambda_{(n,j)}}\int_{0}^{a} J_{n+\frac{1}{2}}^{2}(\lambda_{(n,j)}r)rd r= \frac{\pi}{2\lambda_{(n,j)}} \frac{a^2}{2}J_{n+\frac{3}{2}}(\alpha_{(n+\frac{1}{2},j)})=\frac{a^{2}}{2}j^2_{n+1}(\alpha_{(n+\frac{1}{2},j)})[/tex]

    I am missing the [itex]a[/itex]. This time, it's a lot simpler and more focus, please help me on this.

    Thanks
     
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