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Help deriving the expansion of Helmholtz equation in spherical coordinates.

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Alan0354

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proof of orthogonality of spherical bessel fn in spherical coordinates.

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

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



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

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

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


Orthogonal properties stated that

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

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

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

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

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


Here is my work:
For \[(n=n')\], \[(j=j')\] and \[(m= m')\]

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

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

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

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

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

I am missing an \[a\]. Please help

Thanks
 
Last edited:

I already find proof that for \[0\leq \; r\leq \;a \] where \[R(0)\] is finite and \[R(a)=0\]:

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



So all I need to proof is

\[\int_{0}^{a} j_{n}^{2} (\lambda_{(n,j)}r)rdr=\frac{a^{3}}{2}j^2_{n+1}(\alpha_{(n+\frac{1}{2},j)})\]




But as in the last post:

\[\int_{0}^{a} j_{n}^{2} (\lambda_{(n,j)}r)rdr= \frac{\pi}{2\lambda_{(n,j)}}\int_{0}^{a} J_{n+\frac{1}{2}}^{2}(\lambda_{(n,j)}r)rd r\]

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

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

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

Thanks
 

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