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Modified Bessel Functionは以下の微分方程式を満たす：

$$K''_{\nu}(z) + \frac{1}{z} K'_{\nu}(z) - \left( 1+\frac{\nu^2}{z^2} \right) = 0 .$$ (47)

1 証明

Eq.(44),(45)より、

$$-\left( \frac{2\nu}{z} K_{\nu}(z) +2K_{'}(z){\nu} \right) =2K_{\n... ...grightarrow \quad -\left( \dI{z}+\frac{\nu}{z} \right) K_{\nu}(z) =K_{\nu-1}(z)$$   : 下降演算子$$$$

$$\frac{2\nu}{z} K_{\nu}(z) -2K_{'}(z){\nu} =2K_{\nu+1}(z) \quad \Longrightarrow \quad -\left( \dI{z}-\frac{\nu}{z} \right) K_{\nu}(z) =K_{\nu+1}(z)$$   : 上昇演算子$$$$

$$\therefore \quad \left( \dI{z}-\frac{... ...ft( \dI{z} + \frac{\nu}{z} \right) K_{\nu}(z) =K_{\nu}(z) \, \longrightarrow \,$$   Eq.(47)$$$$

著者: 茅根裕司 chinone_at_astr.tohoku.ac.jp