При решении уравнения для геометрии в сферических координатах возникает следующее уравнение для радиуса
где γ \gamma γ k k k
Сделаем подстановку R ( r ) = 1 r ⋅ R ^ ( r ) R(r) = \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \widehat{R}(r) R ( r ) = r 1 ⋅ R ( r )
d R ( r ) d r = d d r ( 1 r ⋅ R ^ ( r ) ) = − 1 2 ⋅ r ⋅ r ⋅ R ^ ( r ) + 1 r ⋅ d R ^ ( r ) d r , \frac{\displaystyle d R(r)}{\displaystyle dr} = \frac{\displaystyle d}{\displaystyle dr} \left( \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \widehat{R}(r) \right) = - \frac{\displaystyle 1}{\displaystyle 2 \cdot r \cdot \sqrt r} \cdot \widehat{R}(r) + \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr}, d r d R ( r ) = d r d ( r 1 ⋅ R ( r ) ) = − 2 ⋅ r ⋅ r 1 ⋅ R ( r ) + r 1 ⋅ d r d R ( r ) , d 2 R ( r ) d r 2 = − 1 2 ⋅ d d r ( 1 r ⋅ r ⋅ R ^ ( r ) ) + d d r ( 1 r ⋅ d R ^ ( r ) d r ) , \frac{\displaystyle d^2 R(r)}{\displaystyle dr^2} = - \frac{\displaystyle 1}{\displaystyle 2} \cdot \frac{\displaystyle d}{\displaystyle dr} \left( \frac{\displaystyle 1}{\displaystyle r \cdot \sqrt r} \cdot \widehat{R}(r) \right) + \frac{\displaystyle d}{\displaystyle dr} \left( \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr} \right), d r 2 d 2 R ( r ) = − 2 1 ⋅ d r d ( r ⋅ r 1 ⋅ R ( r ) ) + d r d ( r 1 ⋅ d r d R ( r ) ) , − 1 2 ⋅ d d r ( 1 r ⋅ r ⋅ R ^ ( r ) ) = 3 4 ⋅ 1 r 2 ⋅ r ⋅ R ^ ( r ) − 1 2 ⋅ 1 r ⋅ r ⋅ d R ^ ( r ) d r , - \frac{\displaystyle 1}{\displaystyle 2} \cdot \frac{\displaystyle d}{\displaystyle dr} \left( \frac{\displaystyle 1}{\displaystyle r \cdot \sqrt r} \cdot \widehat{R}(r) \right) = \frac{\displaystyle 3}{\displaystyle 4} \cdot \frac{\displaystyle 1}{\displaystyle r^2 \cdot \sqrt r} \cdot \widehat{R}(r) - \frac{\displaystyle 1}{\displaystyle 2} \cdot \frac{\displaystyle 1}{\displaystyle r \cdot \sqrt r} \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr}, − 2 1 ⋅ d r d ( r ⋅ r 1 ⋅ R ( r ) ) = 4 3 ⋅ r 2 ⋅ r 1 ⋅ R ( r ) − 2 1 ⋅ r ⋅ r 1 ⋅ d r d R ( r ) , d d r ( 1 r ⋅ d R ^ ( r ) d r ) = − 1 2 ⋅ r ⋅ r ⋅ d R ^ ( r ) d r + 1 r ⋅ d 2 R ^ ( r ) d r 2 , \frac{\displaystyle d}{\displaystyle dr} \left( \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr} \right) = - \frac{\displaystyle 1}{\displaystyle 2 \cdot r \cdot \sqrt r} \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr} + \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \frac{\displaystyle d^2 \widehat{R}(r)}{\displaystyle dr^2}, d r d ( r 1 ⋅ d r d R ( r ) ) = − 2 ⋅ r ⋅ r 1 ⋅ d r d R ( r ) + r 1 ⋅ d r 2 d 2 R ( r ) , d 2 R ( r ) d r 2 = 1 r ⋅ d 2 R ^ ( r ) d r 2 − 1 r ⋅ r ⋅ d R ^ ( r ) d r + 3 4 ⋅ 1 r 2 ⋅ r ⋅ R ^ ( r ) , \frac{\displaystyle d^2 R(r)}{\displaystyle dr^2} = \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \frac{\displaystyle d^2 \widehat{R}(r)}{\displaystyle dr^2} - \frac{\displaystyle 1}{\displaystyle r \cdot \sqrt r} \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr} + \frac{\displaystyle 3}{\displaystyle 4} \cdot \frac{\displaystyle 1}{\displaystyle r^2 \cdot \sqrt r} \cdot \widehat{R}(r), d r 2 d 2 R ( r ) = r 1 ⋅ d r 2 d 2 R ( r ) − r ⋅ r 1 ⋅ d r d R ( r ) + 4 3 ⋅ r 2 ⋅ r 1 ⋅ R ( r ) , r 2 ⋅ d 2 R ( r ) d r 2 = r ⋅ r ⋅ d 2 R ^ ( r ) d r 2 − r ⋅ d R ^ ( r ) d r + 3 4 ⋅ 1 r ⋅ R ^ ( r ) , r^2 \cdot \frac{\displaystyle d^2 R(r)}{\displaystyle dr^2} = r \cdot \sqrt r \cdot \frac{\displaystyle d^2 \widehat{R}(r)}{\displaystyle dr^2} - \sqrt r \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr} + \frac{\displaystyle 3}{\displaystyle 4} \cdot \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \widehat{R}(r), r 2 ⋅ d r 2 d 2 R ( r ) = r ⋅ r ⋅ d r 2 d 2 R ( r ) − r ⋅ d r d R ( r ) + 4 3 ⋅ r 1 ⋅ R ( r ) , r 2 ⋅ d 2 R ( r ) d r 2 + 2 ⋅ r ⋅ d R ( r ) d r = r ⋅ r ⋅ d 2 R ^ ( r ) d r 2 + r ⋅ d R ^ ( r ) d r − 1 4 ⋅ 1 r ⋅ R ^ ( r ) , r^2 \cdot \frac{\displaystyle d^2 R(r)}{\displaystyle dr^2} + 2 \cdot r \cdot \frac{\displaystyle d R(r)}{\displaystyle dr} = r \cdot \sqrt r \cdot \frac{\displaystyle d^2 \widehat{R}(r)}{\displaystyle dr^2} + \sqrt r \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr} - \frac{\displaystyle 1}{\displaystyle 4} \cdot \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \widehat{R}(r), r 2 ⋅ d r 2 d 2 R ( r ) + 2 ⋅ r ⋅ d r d R ( r ) = r ⋅ r ⋅ d r 2 d 2 R ( r ) + r ⋅ d r d R ( r ) − 4 1 ⋅ r 1 ⋅ R ( r ) , разделим на r ⋅ r r \cdot \sqrt r r ⋅ r G.1 r 2 ⋅ d 2 R ( r ) d r 2 + 2 ⋅ r ⋅ d R ( r ) d r + ( r 2 ⋅ γ 2 − k ⋅ ( k + 1 ) ) ⋅ R ( r ) = 0 , r^2 \cdot \frac{\displaystyle d^2 R(r)}{\displaystyle dr^2} + 2 \cdot r \cdot \frac{\displaystyle d R(r)}{\displaystyle dr} + \left( r^2 \cdot \gamma^2 - k \cdot (k + 1) \right) \cdot R(r) = 0, r 2 ⋅ d r 2 d 2 R ( r ) + 2 ⋅ r ⋅ d r d R ( r ) + ( r 2 ⋅ γ 2 − k ⋅ ( k + 1 ) ) ⋅ R ( r ) = 0 ,
ζ 1 ⋅ d 2 R ^ ( r ) d r 2 + ζ 2 ⋅ d R ^ ( r ) d r + ζ 3 ⋅ R ^ ( r ) = 0 , \zeta_1 \cdot \frac{\displaystyle d^2 \widehat{R}(r)}{\displaystyle dr^2} + \zeta_2 \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr} + \zeta_3 \cdot \widehat{R}(r) = 0, ζ 1 ⋅ d r 2 d 2 R ( r ) + ζ 2 ⋅ d r d R ( r ) + ζ 3 ⋅ R ( r ) = 0 , а теперь определим коэффициенты, предварительно преобразовав свободный член
γ 2 − k ⋅ ( k + 1 ) r 2 − 1 4 ⋅ r 2 = γ 2 − ( k + 1 / 2 ) 2 r 2 , \gamma^2 - \frac{\displaystyle k \cdot (k + 1)}{\displaystyle r^2} - \frac{\displaystyle 1}{\displaystyle 4 \cdot r^2} = \gamma^2 - \frac{\displaystyle (k + 1/2)^2}{\displaystyle r^2}, γ 2 − r 2 k ⋅ ( k + 1 ) − 4 ⋅ r 2 1 = γ 2 − r 2 ( k + 1/2 ) 2 , ζ 1 = 1 , ζ 2 = 1 r , ζ 3 = γ 2 − ( k + 1 / 2 ) 2 r 2 , \zeta_1 = 1, \quad \zeta_2 = \frac{\displaystyle 1}{\displaystyle r}, \quad \zeta_3 = \gamma^2 - \frac{\displaystyle (k + 1/2)^2}{\displaystyle r^2}, ζ 1 = 1 , ζ 2 = r 1 , ζ 3 = γ 2 − r 2 ( k + 1/2 ) 2 , таким образом, уравнение (G.1 r 2 ⋅ d 2 R ( r ) d r 2 + 2 ⋅ r ⋅ d R ( r ) d r + ( r 2 ⋅ γ 2 − k ⋅ ( k + 1 ) ) ⋅ R ( r ) = 0 , r^2 \cdot \frac{\displaystyle d^2 R(r)}{\displaystyle dr^2} + 2 \cdot r \cdot \frac{\displaystyle d R(r)}{\displaystyle dr} + \left( r^2 \cdot \gamma^2 - k \cdot (k + 1) \right) \cdot R(r) = 0, r 2 ⋅ d r 2 d 2 R ( r ) + 2 ⋅ r ⋅ d r d R ( r ) + ( r 2 ⋅ γ 2 − k ⋅ ( k + 1 ) ) ⋅ R ( r ) = 0 ,
d 2 R ^ ( r ) d r 2 + 1 r ⋅ d R ^ ( r ) d r + ( γ 2 − ( k + 1 / 2 ) 2 r 2 ) ⋅ R ^ ( r ) = 0. \frac{\displaystyle d^2 \widehat{R}(r)}{\displaystyle dr^2} + \frac{\displaystyle 1}{\displaystyle r} \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr} + \left( \gamma^2 - \frac{\displaystyle (k + 1/2)^2}{\displaystyle r^2} \right) \cdot \widehat{R}(r) = 0. d r 2 d 2 R ( r ) + r 1 ⋅ d r d R ( r ) + ( γ 2 − r 2 ( k + 1/2 ) 2 ) ⋅ R ( r ) = 0. Получили уравнение Бесселя с полуцелым индексом k + 1 / 2 k + 1/2 k + 1/2 R ^ ( r ) = J k + 1 / 2 ( γ ⋅ r ) \widehat{R}(r) = J_{k + 1/2}(\gamma \cdot r) R ( r ) = J k + 1/2 ( γ ⋅ r ) k ∈ ( 0.. ∞ ) k \in (0..\infty) k ∈ ( 0..∞ ) Y k + 1 / 2 ( γ ⋅ r ) Y_{k + 1/2}(\gamma \cdot r) Y k + 1/2 ( γ ⋅ r ) (B.2) , и отбрасываются из условия ограниченности. Таким образом, решение исходного уравнения (G.1 r 2 ⋅ d 2 R ( r ) d r 2 + 2 ⋅ r ⋅ d R ( r ) d r + ( r 2 ⋅ γ 2 − k ⋅ ( k + 1 ) ) ⋅ R ( r ) = 0 , r^2 \cdot \frac{\displaystyle d^2 R(r)}{\displaystyle dr^2} + 2 \cdot r \cdot \frac{\displaystyle d R(r)}{\displaystyle dr} + \left( r^2 \cdot \gamma^2 - k \cdot (k + 1) \right) \cdot R(r) = 0, r 2 ⋅ d r 2 d 2 R ( r ) + 2 ⋅ r ⋅ d r d R ( r ) + ( r 2 ⋅ γ 2 − k ⋅ ( k + 1 ) ) ⋅ R ( r ) = 0 ,
где J k + 1 / 2 ( γ ⋅ r ) J_{k + 1/2}(\gamma \cdot r) J k + 1/2 ( γ ⋅ r ) j k ( γ ⋅ r ) = π 2 ⋅ γ ⋅ r ⋅ J k + 1 / 2 ( γ ⋅ r ) j_k(\gamma \cdot r) = \sqrt{\frac{\displaystyle \pi}{\displaystyle 2 \cdot \gamma \cdot r}} \cdot J_{k + 1/2}(\gamma \cdot r) j k ( γ ⋅ r ) = 2 ⋅ γ ⋅ r π ⋅ J k + 1/2 ( γ ⋅ r )