柱函数是 Bessel 方程的一类解的表达，三类Bessel 函数是柱函数。
满足如下两个递推关系的函数 Z ν ( z ) {\displaystyle Z_\nu(z)} Z ν − 1 + Z ν + 1 = 2 ν z − 1 Z ν , Z ν − 1 − Z ν + 1 = 2 Z ν ′ . {\displaystyle \begin{align} Z_{\nu-1} + Z_{\nu+1} &= 2\nu z^{-1} Z_{\nu}, \\ Z_{\nu-1} - Z_{\nu+1} & = 2Z'_\nu. \end{align}} 称为柱函数。第一类 Bessel 函数 J ν ( z ) {\displaystyle J_\nu(z)} 以及第二类 Bessel 函数 Y n ( z ) {\displaystyle Y_n(z)} 都是柱函数。可以证明柱函数必然是 Bessel 方程的解。
线性组合 a ν J ν ( z ) + b ν J ν ( z ) {\displaystyle a_\nu J_\nu(z) + b_\nu J_\nu(z)} 是柱函数当且仅当 a ν = a ν + 1 , b ν = b ν + 1 . {\displaystyle a_\nu = a_{\nu+1}, b_\nu = b_{\nu+1}.}
第三类 Bessel 函数也是柱函数。
∫ z z μ + 1 Z ν ( z ) d z = ( ν 2 − μ 2 ) ∫ z z μ − 1 Z ν ( z ) d z + [ z μ + 1 Z ν + 1 ( z ) − ( ν − μ ) z μ Z ν ( z ) ] . {\displaystyle \int^z z^{\mu+1} Z_\nu(z) \mathrm{d} z=\left(\nu^2-\mu^2\right) \int^z z^{\mu-1} Z_\nu(z) \mathrm{d} z+\left[z^{\mu+1} Z_{\nu+1}(z)-(\nu-\mu) z^\mu Z_\nu(z)\right] . } 假设 Z μ ( z ) , Z ¯ ν ( z ) {\displaystyle Z_\mu(z), \bar{Z}_\nu(z)} 分别是 μ {\displaystyle \mu} 阶和 ν {\displaystyle \nu} 阶柱函数, 那么 ∫ z { ( k 2 − l 2 ) z − μ 2 − ν 2 z } Z μ ( k z ) Z ¯ ν ( l z ) d z = z { k Z μ + 1 ( k z ) Z ¯ ν ( l z ) − l Z μ ( k z ) Z ¯ ν + 1 ( l z ) } − ( μ − ν ) Z μ ( k z ) Z ¯ ν ( l z ) . {\displaystyle \begin{aligned} &\int^z\left\{\left(k^2-l^2\right) z-\frac{\mu^2-\nu^2}{z}\right\} Z_\mu(k z) \bar{Z}_\nu(l z) \mathrm{d} z \\ &=z\left\{k Z_{\mu+1}(k z) \bar{Z}_\nu(l z)-l Z_\mu(k z) \bar{Z}_{\nu+1}(l z)\right\}-(\mu-\nu) Z_\mu(k z) \bar{Z}_\nu(l z) . \end{aligned} } 进一步可以证明 ∫ z z Z μ ( k z ) Z ¯ μ ( k z ) d z = − z 2 k { k z Z μ + 1 ( k z ) Z ¯ μ ′ ( k z ) − k z Z μ ( k z ) Z ¯ μ + 1 ( k z ) − Z μ ( k z ) Z ¯ μ + 1 ( k z ) } = z 2 4 { 2 Z μ ( k z ) Z ¯ μ ( k z ) − Z μ − 1 ( k z ˙ ) Z ¯ μ + 1 ( k z ) − Z μ + 1 ( k z ) Z ¯ μ − 1 ( z ) } . ∫ z Z μ ( k z ) Z ¯ μ ( k z ) d z z = k z 2 μ { Z μ + 1 ( k z ) ∂ ∂ μ Z ¯ μ ( k z ) − Z μ ( k z ) ∂ ∂ μ Z ¯ μ + 1 ( k z ) } + Z μ ( k z ) Z ¯ μ ( k z ) 2 μ . {\displaystyle \begin{align} \int^z z Z_\mu(k z) \bar{Z}_\mu(k z) \mathrm{d} z &= -\frac{z}{2 k}\left\{k z Z_{\mu+1}(k z) \bar{Z}_\mu^{\prime}(k z)-k z Z_\mu(k z) \bar{Z}_{\mu+1}(k z)-Z_\mu(k z) \bar{Z}_{\mu+1}(k z)\right\} \\ &= \frac{z^2}{4}\left\{2 Z_\mu(k z) \bar{Z}_\mu(k z)-Z_{\mu-1}(\dot{k z}) \bar{Z}_{\mu+1}(k z)-Z_{\mu+1}(k z) \bar{Z}_{\mu-1}(z)\right\}. \\ \int^z Z_\mu(k z) \bar{Z}_\mu(k z) \frac{\mathrm{d} z}{z} & = \frac{k z}{2 \mu}\left\{Z_{\mu+1}(k z) \frac{\partial}{\partial \mu} \bar{Z}_\mu(k z)-Z_\mu(k z) \frac{\partial}{\partial \mu} \bar{Z}_{\mu+1}(k z)\right\}+\frac{Z_\mu(k z) \bar{Z}_\mu(k z)}{2 \mu}.\end{align}} 注意微商 d { z ρ Z μ ( z ) Z ¯ ν ( z ) } d z {\displaystyle \dfrac{\mathrm{d}\left\{ z^\rho Z_\mu(z) \bar{Z}_\nu(z) \right\}}{\mathrm{d} z}} 和 d { z ρ Z μ + 1 ( z ) Z ¯ μ + 1 ( z ) } d z {\displaystyle \dfrac{\mathrm{d}\left\{z^\rho Z_{\mu+1}(z) \bar{Z}_{\mu+1}(z)\right\}}{\mathrm{d} z}} , 可以得到 ( ρ + μ + ν ) ∫ z z ρ − 1 Z μ ( z ) Z ¯ ν ( z ) d z + ( ρ − μ − ν − 2 ) ∫ z z ρ − 1 Z μ + 1 ( z ) Z ¯ ν + 1 ( z ) d z = z ρ { Z μ ( z ) Z ¯ ν ( z ) + Z μ + 1 ( z ) Z ¯ ν + 1 ( z ) } . {\displaystyle \begin{aligned} & \quad ~ (\rho+\mu+\nu) \int^z z^{\rho-1} Z_\mu(z) \bar{Z}_\nu(z) \mathrm{d} z+(\rho-\mu-\nu-2) \int^z z^{\rho-1} Z_{\mu+1}(z) \bar{Z}_{\nu+1}(z) \mathrm{d} z \\ & = z^\rho\left\{Z_\mu(z) \bar{Z}_\nu(z)+Z_{\mu+1}(z) \bar{Z}_{\nu+1}(z)\right\} . \end{aligned} } 于是进一步可得到 ∫ z z z − μ ν − 1 Z μ + 1 ( z ) Z ¯ ν + 1 ( z ) d z = − z − μ − ν 2 ( μ + ν + 1 ) { Z μ ( z ) Z ¯ ν ( z ) + Z μ + 1 ( z ) Z ¯ ν + 1 ( z ) } , ∫ z z μ + ν + 1 Z μ ( z ) Z ¯ ν ( z ) d z = z μ + ν + 2 2 ( μ + ν + 1 ) { Z μ ( z ) Z ¯ ν ( z ) + Z μ + 1 ( z ) Z ¯ ν + 1 ( z ) } , ∫ z Z n ( z ) Z ¯ n ( z ) d z z = − 1 2 n { Z 0 ( z ) Z ¯ 0 ( z ) + 2 ∑ m = 1 n − 1 Z m ( z ) Z ¯ m ( z ) + Z n ( z ) Z ¯ n ( z ) } , n ∈ N + . {\displaystyle \begin{aligned} &\int z^z z^{-\mu \nu-1} Z_{\mu+1}(z) \bar{Z}_{\nu+1}(z) \mathrm{d} z \\ &\qquad = -\frac{z^{-\mu-\nu}}{2(\mu+\nu+1)}\left\{Z_\mu(z) \bar{Z}_\nu(z)+Z_{\mu+1}(z) \bar{Z}_{\nu+1}(z)\right\}, \\ &\int^z z^{\mu+\nu+1} Z_\mu(z) \bar{Z}_\nu(z) \mathrm{d} z \\ &\qquad = \frac{z^{\mu+\nu+2}}{2(\mu+\nu+1)}\left\{Z_\mu(z) \bar{Z}_\nu(z)+Z_{\mu+1}(z) \bar{Z}_{\nu+1}(z)\right\}, \\ & \int^z Z_n(z) \bar{Z}_n(z) \frac{\mathrm{d} z}{z} \\ &\qquad= -\frac{1}{2 n}\left\{Z_0(z) \bar{Z}_0(z)+2 \sum_{m=1}^{n-1} Z_m(z) \bar{Z}_m(z)+Z_n(z) \bar{Z}_n(z)\right\}, n \in \N^+. \end{aligned}}
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称为柱函数。第一类 Bessel 函数
以及第二类 Bessel 函数
都是柱函数。可以证明柱函数必然是 Bessel 方程的解。
是柱函数当且仅当
![{\displaystyle \int ^{z}z^{\mu +1}Z_{\nu }(z)\mathrm {d} z=\left(\nu ^{2}-\mu ^{2}\right)\int ^{z}z^{\mu -1}Z_{\nu }(z)\mathrm {d} z+\left[z^{\mu +1}Z_{\nu +1}(z)-(\nu -\mu )z^{\mu }Z_{\nu }(z)\right].}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/2ec207f1a933bcedf8eec85f6bfa5be8c7529bb1)
假设
分别是 
阶和 
阶柱函数, 那么
进一步可以证明
注意微商
和
, 可以得到
于是进一步可得到
