6-2

Bessel Function: $J_{{n}}(z)$ が次の微分方程式を満たすことを示せ。

$\displaystyle \dII{z}y_{n}(z) +\frac{1}{z}\dI{z}y_{n}(z) +\left(1-\frac{n^2}{z^2}\right)y_{n}(z) =0$

(42)

$\displaystyle \dII{z}J_{{n}}(z)$

$\displaystyle = \dI{z}\left(J_{{n-1}}(z)-J_{{n+1}}(z)\right) = \frac{1}{4} \lef... ...}(z) \right) =\frac{1}{4} \left( J_{{n-2}}(z)-2J_{{n}}(z) +J_{{n+2}}(z) \right)$

(43)

$\displaystyle \frac{1}{z}\dI{z}J_{{n}}(z)$	$\displaystyle = \frac{1}{2z} \left(J_{{n-1}}(z)-J_{{n+1}}(z)\right) = \frac{1}{... ...- \frac{1}{2z} \frac{z}{2\left(n+1\right)} \left(J_{{n}}(z)+J_{{n+2}}(z)\right)$
	$\displaystyle \frac{1}{4\left(n^2-1\right)} \left[ \left(n+1\right)J_{{n-2}}(z) +2J_{{n}}(z) -\left(n-1\right)J_{{n+2}}(z) \right]$	(44)

$\displaystyle J_{{n}}(z)$	$\displaystyle = \frac{z}{2n} \left(J_{{n-1}}(z)+J_{{n+1}}(z)\right) = \frac{z}{... ...) +\frac{z}{2n}\frac{z}{2\left(n+1\right)} \left(J_{{n}}(z)+J_{{n+2}}(z)\right)$
	$\displaystyle =\frac{z^2}{4n\left(n^2-1\right)} \left[ \left(n+1\right)J_{{n-2}... ...left(n-1\right)J_{{n+2}}(z) \right] + \frac{z^2}{2\left(n^2-1\right)}J_{{n}}(z)$
	$\displaystyle \quad \Longrightarrow \quad \left(1-\frac{z^2}{2\left(n^2-1\right... ...ht)} \left[ \left(n+1\right)J_{{n-2}}(z) + \left(n-1\right)J_{{n+2}}(z) \right]$
	$\displaystyle \quad \Longrightarrow \quad J_{{n}}(z) =\frac{z^2}{2n\left\{2\lef... ...}} \left\{ \left(n+1\right)J_{{n-2}}(z) + \left(n-1\right)J_{{n+2}}(z) \right\}$	(45)

	$\displaystyle \left[ \frac{1}{4} +\frac{1}{4\left(n-1\right)} \right]J_{{n-2}}(... ...frac{1}{2\left(n^2-1\right)} +\left(1-\frac{n^2}{z^2}\right) \right] J_{{n}}(z)$
	$\displaystyle = \left[ \frac{1}{4} +\frac{1}{4\left(n-1\right)} -\frac{n}{4\lef... ...\frac{1}{4\left(n+1\right)} -\frac{n}{4\left(n+1\right)} \right]J_{{n+2}}(z) =0$

$\displaystyle 2\dI{z}J_{{n}}(z) = J_{{n-1}}(z)-J_{{n+1}}(z) =\frac{2n}{z}J_{{n}}(z) -J_{{n+1}}(z)-J_{{n+1}}(z) =\frac{2n}{z}J_{{n}}(z) -2J_{{n}}(z)$

$\displaystyle \Longrightarrow \quad \left( -\dI{z} + \frac{n}{z} \right)J_{{n}}(z) =J_{{n+1}}(z)$ ：上昇演算子 $\displaystyle$

$\displaystyle 2\dI{z}J_{{n}}(z) = J_{{n-1}}(z)-J_{{n+1}}(z) =J_{{n-1}}(z) -\frac{2n}{z}J_{{n}}(z)+J_{{n-1}}(z) =-\frac{2n}{z}J_{{n}}(z)+2J_{{n+1}}(z)$

$\displaystyle \Longrightarrow \quad \left( \dI{z} + \frac{n}{z} \right)J_{{n}}(z) =J_{{n-1}}(z)$ ：下降演算子 $\displaystyle$

$% latex2html id marker 3336 $\displaystyle \therefore\quad \left(\dI{z}+\frac{n+... ...t) \left(-\dI{z}+\frac{n}{z}\right)J_{{n}}(z) =J_{{n}}(z) \, \longrightarrow \,$$ Eq.(42) $\displaystyle$

6-2

6-2解答