Help in deriving series expansion of Helmholtz Equation.

[yungman]

Orthogonality of spherical bessel functions

Homework Statement

Proof of orthogonality of spherical bessel functions

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)})

Homework Equations

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+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.

The Attempt at a Solution

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 the a. Thanks

 

[yungman]

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