При решении уравнения для геометрии в сферических координатах возникает следующее уравнение для радиуса
где γ \gamma — собственное значение, k k — целое неотрицательное число.
Сделаем подстановку R ( r ) R(r) = 1 r ⋅ R ^ ( r ) = \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \widehat{R}(r) и проведём преобразования
d R ( r ) d r \displaystyle \frac{\displaystyle d R(r)}{\displaystyle dr} = d d r ( 1 r ⋅ R ^ ( r ) ) \displaystyle = \frac{\displaystyle d}{\displaystyle dr} \left( \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \widehat{R}(r) \right) = − 1 2 ⋅ r ⋅ r ⋅ R ^ ( r ) \displaystyle = - \frac{\displaystyle 1}{\displaystyle 2 \cdot r \cdot \sqrt r} \cdot \widehat{R}(r) + 1 r ⋅ d R ^ ( r ) d r , \displaystyle + \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr}, d 2 R ( r ) d r 2 \displaystyle \frac{\displaystyle d^2 R(r)}{\displaystyle dr^2} = − 1 2 ⋅ d d r ( 1 r ⋅ r ⋅ R ^ ( r ) ) \displaystyle = - \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) + d d r ( 1 r ⋅ d R ^ ( r ) d r ) , \displaystyle + \frac{\displaystyle d}{\displaystyle dr} \left( \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr} \right), − 1 2 ⋅ d d r ( 1 r ⋅ r ⋅ R ^ ( r ) ) \displaystyle - \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) = 3 4 ⋅ 1 r 2 ⋅ r ⋅ R ^ ( r ) \displaystyle = \frac{\displaystyle 3}{\displaystyle 4} \cdot \frac{\displaystyle 1}{\displaystyle r^2 \cdot \sqrt r} \cdot \widehat{R}(r) − 1 2 ⋅ 1 r ⋅ r ⋅ d R ^ ( r ) d r , \displaystyle - \frac{\displaystyle 1}{\displaystyle 2} \cdot \frac{\displaystyle 1}{\displaystyle r \cdot \sqrt r} \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr}, d d r ( 1 r ⋅ d R ^ ( r ) d r ) \displaystyle \frac{\displaystyle d}{\displaystyle dr} \left( \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr} \right) = − 1 2 ⋅ r ⋅ r ⋅ d R ^ ( r ) d r \displaystyle = - \frac{\displaystyle 1}{\displaystyle 2 \cdot r \cdot \sqrt r} \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr} + 1 r ⋅ d 2 R ^ ( r ) d r 2 , \displaystyle + \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \frac{\displaystyle d^2 \widehat{R}(r)}{\displaystyle dr^2}, d 2 R ( r ) d r 2 \displaystyle \frac{\displaystyle d^2 R(r)}{\displaystyle dr^2} = 1 r ⋅ d 2 R ^ ( r ) d r 2 \displaystyle = \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \frac{\displaystyle d^2 \widehat{R}(r)}{\displaystyle dr^2} − 1 r ⋅ r ⋅ d R ^ ( r ) d r \displaystyle - \frac{\displaystyle 1}{\displaystyle r \cdot \sqrt r} \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr} + 3 4 ⋅ 1 r 2 ⋅ r ⋅ R ^ ( r ) , \displaystyle + \frac{\displaystyle 3}{\displaystyle 4} \cdot \frac{\displaystyle 1}{\displaystyle r^2 \cdot \sqrt r} \cdot \widehat{R}(r), r 2 ⋅ d 2 R ( r ) d r 2 \displaystyle r^2 \cdot \frac{\displaystyle d^2 R(r)}{\displaystyle dr^2} = r ⋅ r ⋅ d 2 R ^ ( r ) d r 2 \displaystyle = r \cdot \sqrt r \cdot \frac{\displaystyle d^2 \widehat{R}(r)}{\displaystyle dr^2} − r ⋅ d R ^ ( r ) d r \displaystyle - \sqrt r \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr} + 3 4 ⋅ 1 r ⋅ R ^ ( r ) , \displaystyle + \frac{\displaystyle 3}{\displaystyle 4} \cdot \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \widehat{R}(r), r 2 ⋅ d 2 R ( r ) d r 2 \displaystyle r^2 \cdot \frac{\displaystyle d^2 R(r)}{\displaystyle dr^2} + 2 ⋅ r ⋅ d R ( r ) d r \displaystyle + 2 \cdot r \cdot \frac{\displaystyle d R(r)}{\displaystyle dr} = r ⋅ r ⋅ d 2 R ^ ( r ) d r 2 \displaystyle = r \cdot \sqrt r \cdot \frac{\displaystyle d^2 \widehat{R}(r)}{\displaystyle dr^2} + r ⋅ d R ^ ( r ) d r \displaystyle + \sqrt r \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr} − 1 4 ⋅ 1 r ⋅ R ^ ( r ) , \displaystyle - \frac{\displaystyle 1}{\displaystyle 4} \cdot \frac{\displaystyle 1}{\displaystyle \sqrt r} \cdot \widehat{R}(r), разделим на r ⋅ r r \cdot \sqrt r и запишем уравнение (G.1 r 2 ⋅ d 2 R ( r ) d r 2 \displaystyle r^2 \cdot \frac{\displaystyle d^2 R(r)}{\displaystyle dr^2} + 2 ⋅ r ⋅ d R ( r ) d r \displaystyle + 2 \cdot r \cdot \frac{\displaystyle d R(r)}{\displaystyle dr} + ( r 2 ⋅ γ 2 − k ⋅ ( k + 1 ) ) ⋅ R ( r ) \displaystyle + \left( r^2 \cdot \gamma^2 - k \cdot (k + 1) \right) \cdot R(r) = 0 , \displaystyle = 0, ) через коэффициенты
ζ 1 ⋅ d 2 R ^ ( r ) d r 2 \displaystyle \zeta_1 \cdot \frac{\displaystyle d^2 \widehat{R}(r)}{\displaystyle dr^2} + ζ 2 ⋅ d R ^ ( r ) d r \displaystyle + \zeta_2 \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr} + ζ 3 ⋅ R ^ ( r ) \displaystyle + \zeta_3 \cdot \widehat{R}(r) = 0 , \displaystyle = 0, а теперь определим коэффициенты, предварительно преобразовав свободный член
γ 2 \displaystyle \gamma^2 − k ⋅ ( k + 1 ) r 2 \displaystyle - \frac{\displaystyle k \cdot (k + 1)}{\displaystyle r^2} − 1 4 ⋅ r 2 \displaystyle - \frac{\displaystyle 1}{\displaystyle 4 \cdot r^2} = γ 2 \displaystyle = \gamma^2 − ( k + 1 / 2 ) 2 r 2 , \displaystyle - \frac{\displaystyle (k + 1/2)^2}{\displaystyle r^2}, ζ 1 \displaystyle \zeta_1 = 1 , ζ 2 \displaystyle = 1, \quad \zeta_2 = 1 r , ζ 3 \displaystyle = \frac{\displaystyle 1}{\displaystyle r}, \quad \zeta_3 = γ 2 \displaystyle = \gamma^2 − ( k + 1 / 2 ) 2 r 2 , \displaystyle - \frac{\displaystyle (k + 1/2)^2}{\displaystyle r^2}, таким образом, уравнение (G.1 r 2 ⋅ d 2 R ( r ) d r 2 \displaystyle r^2 \cdot \frac{\displaystyle d^2 R(r)}{\displaystyle dr^2} + 2 ⋅ r ⋅ d R ( r ) d r \displaystyle + 2 \cdot r \cdot \frac{\displaystyle d R(r)}{\displaystyle dr} + ( r 2 ⋅ γ 2 − k ⋅ ( k + 1 ) ) ⋅ R ( r ) \displaystyle + \left( r^2 \cdot \gamma^2 - k \cdot (k + 1) \right) \cdot R(r) = 0 , \displaystyle = 0, ) принимает вид
d 2 R ^ ( r ) d r 2 \displaystyle \frac{\displaystyle d^2 \widehat{R}(r)}{\displaystyle dr^2} + 1 r ⋅ d R ^ ( r ) d r \displaystyle + \frac{\displaystyle 1}{\displaystyle r} \cdot \frac{\displaystyle d \widehat{R}(r)}{\displaystyle dr} + ( γ 2 − ( k + 1 / 2 ) 2 r 2 ) ⋅ R ^ ( r ) \displaystyle + \left( \gamma^2 - \frac{\displaystyle (k + 1/2)^2}{\displaystyle r^2} \right) \cdot \widehat{R}(r) = 0. \displaystyle = 0. Получили уравнение Бесселя с полуцелым индексом k k + 1 / 2 + 1/2 . Его ограниченные в нуле решения — функции Бесселя первого рода R ^ ( r ) \widehat{R}(r) = J k + 1 / 2 ( γ ⋅ r ) = J_{k + 1/2}(\gamma \cdot r) , k k ∈ ( 0.. ∞ ) \in (0..\infty) ; решения второго рода Y k + 1 / 2 ( γ ⋅ r ) Y_{k + 1/2}(\gamma \cdot r) бесконечны в нуле, что видно на рисунке (B.2) , и отбрасываются из условия ограниченности. Таким образом, решение исходного уравнения (G.1 r 2 ⋅ d 2 R ( r ) d r 2 \displaystyle r^2 \cdot \frac{\displaystyle d^2 R(r)}{\displaystyle dr^2} + 2 ⋅ r ⋅ d R ( r ) d r \displaystyle + 2 \cdot r \cdot \frac{\displaystyle d R(r)}{\displaystyle dr} + ( r 2 ⋅ γ 2 − k ⋅ ( k + 1 ) ) ⋅ R ( r ) \displaystyle + \left( r^2 \cdot \gamma^2 - k \cdot (k + 1) \right) \cdot R(r) = 0 , \displaystyle = 0, ) имеет вид
где J k + 1 / 2 ( γ ⋅ r ) J_{k + 1/2}(\gamma \cdot r) — функция Бесселя первого рода с полуцелым индексом. С точностью до постоянного множителя это сферическая функция Бесселя: j k ( γ ⋅ r ) j_k(\gamma \cdot r) = π 2 ⋅ γ ⋅ r ⋅ J k + 1 / 2 ( γ ⋅ r ) = \sqrt{\frac{\displaystyle \pi}{\displaystyle 2 \cdot \gamma \cdot r}} \cdot J_{k + 1/2}(\gamma \cdot r) .
F. Норма полиномов Лежандра H. Норма сферической функции Бесселя (Нейман)