# Help in deriving series expansion of Helmholtz Equation.

1. Nov 20, 2013

### yungman

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

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

2. Relevant 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.

3. 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

Last edited: Nov 21, 2013
2. Nov 21, 2013

### 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