Для вычисления нормы решения уравнения для геометрии необходимо вычислить норму присоединённого полинома Лежандра, которая определяется следующим образом
где P k ( m ) ( z ) = ( 1 − z 2 ) m 2 ⋅ d m d z m P k ( z ) P_k^{(m)}(z) = (1 - z^2)^{\frac{\displaystyle m}{\displaystyle 2}} \cdot \frac{\displaystyle d^m}{\displaystyle dz^m} P_k(z) P k ( m ) ( z ) = ( 1 − z 2 ) 2 m ⋅ d z m d m P k ( z ) (F.1) ∫ − 1 1 [ P k ( m ) ( z ) ] 2 d z , \int_{-1}^1 \left[ P_k^{(m)}(z) \right]^2 \,dz, ∫ − 1 1 [ P k ( m ) ( z ) ] 2 d z ,
∫ − 1 1 [ P k ( m ) ( z ) ] 2 d z = ∫ − 1 1 ( 1 − z 2 ) m [ d m d z m P k ( z ) ] 2 d z . \int_{-1}^1 \left[ P_k^{(m)}(z) \right]^2 \,dz = \int_{-1}^1 (1 - z^2)^m \left[ \frac{d^m}{dz^m} P_k(z) \right]^2 \,dz. ∫ − 1 1 [ P k ( m ) ( z ) ] 2 d z = ∫ − 1 1 ( 1 − z 2 ) m [ d z m d m P k ( z ) ] 2 d z . Подставим формулу Родрига (E.2 Θ k ( z ) = P k ( z ) = 1 2 k ⋅ k ! ⋅ d k d z k ( z 2 − 1 ) k , \Theta_k(z) = P_k(z) = \frac{\displaystyle 1}{\displaystyle 2^k \cdot k!} \cdot \frac{\displaystyle d^k}{\displaystyle dz^k} \left(z^2 - 1\right)^k, Θ k ( z ) = P k ( z ) = 2 k ⋅ k ! 1 ⋅ d z k d k ( z 2 − 1 ) k , z z z
Понятно, что m ≤ k m \leq k m ≤ k k + m k + m k + m 2 ⋅ k 2 \cdot k 2 ⋅ k m = 0 m = 0 m = 0
Возьмём его по частям:
∫ − 1 1 [ d k d z k ( z 2 − 1 ) k ] 2 d z = ∣ u = d k d z k ( z 2 − 1 ) k , d u = d k + 1 d z k + 1 ( z 2 − 1 ) k d z d v = d k d z k ( z 2 − 1 ) k d z , v = d k − 1 d z k − 1 ( z 2 − 1 ) k ∣ \begin{aligned}\int_{-1}^1 \left[ \frac{d^k}{dz^k} \left(z^2 - 1 \right)^k \right]^2 \,dz =
\left |
\begin{array}{ll}
u = \frac{d^k}{dz^k} \left(z^2 - 1 \right)^k, &du = \frac{d^{k+1}}{dz^{k+1}} \left(z^2 - 1 \right)^k \,dz\\
dv = \frac{d^k}{dz^k} \left(z^2 - 1 \right)^k \,dz, &v = \frac{d^{k-1}}{dz^{k-1}} \left(z^2 - 1 \right)^k
\end{array}
\right |\end{aligned} ∫ − 1 1 [ d z k d k ( z 2 − 1 ) k ] 2 d z = u = d z k d k ( z 2 − 1 ) k , d v = d z k d k ( z 2 − 1 ) k d z , d u = d z k + 1 d k + 1 ( z 2 − 1 ) k d z v = d z k − 1 d k − 1 ( z 2 − 1 ) k Внеинтегральные члены равны нулю: производные d i d z i ( z 2 − 1 ) k \frac{\displaystyle d^i}{\displaystyle dz^i} \left(z^2 - 1 \right)^k d z i d i ( z 2 − 1 ) k i < k i < k i < k k k k
1 2 2 k ⋅ ( k ! ) 2 ⋅ ∫ − 1 1 [ d k d z k ( z 2 − 1 ) k ] 2 d z = ( − 1 ) k 2 2 k ⋅ ( k ! ) 2 ⋅ ∫ − 1 1 ( z 2 − 1 ) k ⋅ d 2 k d z 2 k ( z 2 − 1 ) k d z , \frac{\displaystyle 1}{\displaystyle 2^{2k} \cdot (k!)^2} \cdot \int_{-1}^1 \left[ \frac{d^k}{dz^k} \left(z^2 - 1 \right)^k \right]^2 \,dz = \frac{\displaystyle (-1)^k}{\displaystyle 2^{2k} \cdot (k!)^2} \cdot \int_{-1}^1 \left(z^2 - 1 \right)^k \cdot \frac{d^{2k}}{dz^{2k}} \left(z^2 - 1 \right)^k \,dz, 2 2 k ⋅ ( k ! ) 2 1 ⋅ ∫ − 1 1 [ d z k d k ( z 2 − 1 ) k ] 2 d z = 2 2 k ⋅ ( k ! ) 2 ( − 1 ) k ⋅ ∫ − 1 1 ( z 2 − 1 ) k ⋅ d z 2 k d 2 k ( z 2 − 1 ) k d z , теперь из формулы бинома Ньютона становится очевидно, что d 2 k d z 2 k ( z 2 − 1 ) k = ( 2 k ) ! \frac{d^{2k}}{dz^{2k}} \left(z^2 - 1 \right)^k = (2k)! d z 2 k d 2 k ( z 2 − 1 ) k = ( 2 k )! 2 ⋅ k 2 \cdot k 2 ⋅ k ( 2 k ) ! (2k)! ( 2 k )!
Осталось вычислить интеграл
∫ − 1 1 ( z 2 − 1 ) k d z = ∣ u = ( z 2 − 1 ) k , d u = 2 ⋅ k ⋅ z ⋅ ( z 2 − 1 ) k − 1 d z d v = d z , v = z ∣ \begin{aligned}\int_{-1}^1 \left(z^2 - 1 \right)^k \,dz =
\left |
\begin{array}{ll}
u = \left(z^2 - 1 \right)^k, &du = 2 \cdot k \cdot z \cdot (z^2 - 1)^{k-1} \,dz\\
dv = dz, &v = z
\end{array}
\right |\end{aligned} ∫ − 1 1 ( z 2 − 1 ) k d z = u = ( z 2 − 1 ) k , d v = d z , d u = 2 ⋅ k ⋅ z ⋅ ( z 2 − 1 ) k − 1 d z v = z очевидно, что внеинтегральный член равен нулю, поэтому
∫ − 1 1 ( z 2 − 1 ) k d z = − 2 ⋅ k ⋅ ∫ − 1 1 z 2 ⋅ ( z 2 − 1 ) k − 1 d z , \int_{-1}^1 \left(z^2 - 1 \right)^k \,dz = -2 \cdot k \cdot \int_{-1}^1 z^2 \cdot (z^2 - 1)^{k-1} \,dz, ∫ − 1 1 ( z 2 − 1 ) k d z = − 2 ⋅ k ⋅ ∫ − 1 1 z 2 ⋅ ( z 2 − 1 ) k − 1 d z , заменим z 2 z^2 z 2 z 2 − 1 + 1 z^2 - 1 + 1 z 2 − 1 + 1
∫ − 1 1 ( z 2 − 1 ) k d z = − 2 ⋅ k 2 ⋅ k + 1 ⋅ ∫ − 1 1 ( z 2 − 1 ) k − 1 d z , \int_{-1}^1 \left(z^2 - 1 \right)^k \,dz = - \frac{\displaystyle 2 \cdot k}{\displaystyle 2 \cdot k + 1} \cdot \int_{-1}^1 \left(z^2 - 1 \right)^{k-1} \,dz, ∫ − 1 1 ( z 2 − 1 ) k d z = − 2 ⋅ k + 1 2 ⋅ k ⋅ ∫ − 1 1 ( z 2 − 1 ) k − 1 d z , применяя эту рекуррентную формулу ещё k − 1 k - 1 k − 1
∫ − 1 1 ( z 2 − 1 ) k d z = ( − 1 ) k ⋅ 2 ⋅ k 2 ⋅ k + 1 ⋅ 2 ⋅ ( k − 1 ) 2 ⋅ ( k − 1 ) + 1 ⋅ 2 ⋅ ( k − 2 ) 2 ⋅ ( k − 2 ) + 1 ⋅ ⋯ ⋅ 2 ⋅ 1 2 ⋅ 1 + 1 ⋅ 2 , \int_{-1}^1 \left(z^2 - 1 \right)^k \,dz = (-1)^k \cdot \frac{\displaystyle 2 \cdot k}{\displaystyle 2 \cdot k + 1} \cdot \frac{\displaystyle 2 \cdot (k - 1)}{\displaystyle 2 \cdot (k - 1) + 1} \cdot \frac{\displaystyle 2 \cdot (k - 2)}{\displaystyle 2 \cdot (k - 2) + 1} \cdot \cdots \cdot \frac{\displaystyle 2 \cdot 1}{\displaystyle 2 \cdot 1 + 1} \cdot 2, ∫ − 1 1 ( z 2 − 1 ) k d z = ( − 1 ) k ⋅ 2 ⋅ k + 1 2 ⋅ k ⋅ 2 ⋅ ( k − 1 ) + 1 2 ⋅ ( k − 1 ) ⋅ 2 ⋅ ( k − 2 ) + 1 2 ⋅ ( k − 2 ) ⋅ ⋯ ⋅ 2 ⋅ 1 + 1 2 ⋅ 1 ⋅ 2 , ∫ − 1 1 ( z 2 − 1 ) k d z = ( − 1 ) k ⋅ 2 ⋅ 2 k ⋅ k ! ⋅ 2 k ⋅ k ! ( 2 ⋅ k + 1 ) ! . \int_{-1}^1 \left(z^2 - 1 \right)^k \,dz = (-1)^k \cdot 2 \cdot \frac{\displaystyle 2^k \cdot k! \cdot 2^k \cdot k!}{\displaystyle (2 \cdot k + 1)!}. ∫ − 1 1 ( z 2 − 1 ) k d z = ( − 1 ) k ⋅ 2 ⋅ ( 2 ⋅ k + 1 )! 2 k ⋅ k ! ⋅ 2 k ⋅ k ! . В итоге множители ( − 1 ) k (-1)^k ( − 1 ) k
Запишем общеизвестную рекуррентную формулу для присоединённых полиномов Лежандра, которая потребуется для следующего интегрирования по частям
для небольшого упрощения обозначим ϵ = ( k + m ) ⋅ ( k − m + 1 ) \epsilon = \left( k + m \right) \cdot \left( k - m + 1 \right) ϵ = ( k + m ) ⋅ ( k − m + 1 )
Вернёмся к интегралу (F.2 ∫ − 1 1 ( 1 − z 2 ) m [ d m d z m P k ( z ) ] 2 d z = 1 2 2 k ⋅ ( k ! ) 2 ⋅ ∫ − 1 1 ( 1 − z 2 ) m ⋅ [ d k + m d z k + m ( z 2 − 1 ) k ] 2 d z . \int_{-1}^1 (1 - z^2)^m \left[ \frac{d^m}{dz^m} P_k(z) \right]^2 \,dz = \frac{\displaystyle 1}{\displaystyle 2^{2k} \cdot (k!)^2} \cdot \int_{-1}^1 (1 - z^2)^m \cdot \left[ \frac{d^{k+m}}{dz^{k+m}} \left(z^2 - 1 \right)^k \right]^2 \,dz. ∫ − 1 1 ( 1 − z 2 ) m [ d z m d m P k ( z ) ] 2 d z = 2 2 k ⋅ ( k ! ) 2 1 ⋅ ∫ − 1 1 ( 1 − z 2 ) m ⋅ [ d z k + m d k + m ( z 2 − 1 ) k ] 2 d z .
∫ − 1 1 ( 1 − z 2 ) m ⋅ [ d m d z m P k ( z ) ] 2 d z = ∣ u = ( 1 − z 2 ) m ⋅ d m d z m P k ( z ) d u = ( 1 − z 2 ) ( m / 2 ) − 1 ⋅ [ − 2 ⋅ m ⋅ z ⋅ P k ( m ) ( z ) + 1 − z 2 ⋅ P k ( m + 1 ) ( z ) ] d z = ( 1 − z 2 ) m − 1 2 ⋅ [ − P k ( m + 1 ) ( z ) − ϵ ⋅ P k ( m − 1 ) ( z ) + P k ( m + 1 ) ( z ) ] d z = − ( 1 − z 2 ) m − 1 2 ⋅ ϵ ⋅ P k ( m − 1 ) ( z ) d z d v = d m d z m P k ( z ) d z , v = d m − 1 d z m − 1 P k ( z ) = ( 1 − z 2 ) 1 − m 2 ⋅ P k ( m − 1 ) ( z ) ∣ \begin{aligned}
&\int_{-1}^1 \left( 1 - z^2 \right)^m \cdot \left[ \frac{\displaystyle d^m}{\displaystyle dz^m} P_k(z) \right]^2 \,dz =\\
&\left |
\begin{aligned}
&u = \left( 1 - z^2 \right)^m \cdot \frac{\displaystyle d^m}{\displaystyle dz^m} P_k(z)\\
&du = \left( 1 - z^2 \right)^{(m/2)-1} \cdot \left[ -2 \cdot m \cdot z \cdot P_k^{(m)}(z) + \sqrt{1 - z^2} \cdot P_k^{(m+1)}(z) \right] \,dz=\\
&\left( 1 - z^2 \right)^{\frac{\displaystyle m-1}{\displaystyle 2}} \cdot \left[ - P_k^{(m+1)}(z) - \epsilon \cdot P_k^{(m-1)}(z) + P_k^{(m+1)}(z) \right] \,dz=\\
&- \left( 1 - z^2 \right)^{\frac{\displaystyle m-1}{\displaystyle 2}} \cdot \epsilon \cdot P_k^{(m-1)}(z) \,dz\\
&dv = \frac{\displaystyle d^m}{\displaystyle dz^m} P_k(z) \,dz,\\
&v = \frac{\displaystyle d^{m-1}}{\displaystyle dz^{m-1}} P_k(z) = \left( 1 - z^2 \right)^{\frac{\displaystyle 1-m}{\displaystyle 2}} \cdot P_k^{(m-1)}(z)
\end{aligned}
\right |
\end{aligned} ∫ − 1 1 ( 1 − z 2 ) m ⋅ [ d z m d m P k ( z ) ] 2 d z = u = ( 1 − z 2 ) m ⋅ d z m d m P k ( z ) d u = ( 1 − z 2 ) ( m /2 ) − 1 ⋅ [ − 2 ⋅ m ⋅ z ⋅ P k ( m ) ( z ) + 1 − z 2 ⋅ P k ( m + 1 ) ( z ) ] d z = ( 1 − z 2 ) 2 m − 1 ⋅ [ − P k ( m + 1 ) ( z ) − ϵ ⋅ P k ( m − 1 ) ( z ) + P k ( m + 1 ) ( z ) ] d z = − ( 1 − z 2 ) 2 m − 1 ⋅ ϵ ⋅ P k ( m − 1 ) ( z ) d z d v = d z m d m P k ( z ) d z , v = d z m − 1 d m − 1 P k ( z ) = ( 1 − z 2 ) 2 1 − m ⋅ P k ( m − 1 ) ( z ) внеинтегральный член содержит множитель 1 − z 2 \sqrt{1 - z^2} 1 − z 2
∫ − 1 1 P k ( m ) ( z ) ⋅ P k ( m ) ( z ) d z = ( k + m ) ⋅ ( k − m + 1 ) ⋅ ∫ − 1 1 P k ( m − 1 ) ( z ) ⋅ P k ( m − 1 ) ( z ) d z , \int_{-1}^1 P_k^{(m)}(z) \cdot P_k^{(m)}(z) \,dz = (k+m) \cdot (k-m+1) \cdot \int_{-1}^1 P_k^{(m-1)}(z) \cdot P_k^{(m-1)}(z) \,dz, ∫ − 1 1 P k ( m ) ( z ) ⋅ P k ( m ) ( z ) d z = ( k + m ) ⋅ ( k − m + 1 ) ⋅ ∫ − 1 1 P k ( m − 1 ) ( z ) ⋅ P k ( m − 1 ) ( z ) d z , последовательно применяя интегрирование по частям ещё m − 1 m - 1 m − 1 F.4 1 2 2 k ⋅ ( k ! ) 2 ⋅ ∫ − 1 1 [ d k d z k ( z 2 − 1 ) k ] 2 d z = 2 2 ⋅ k + 1 . \frac{\displaystyle 1}{\displaystyle 2^{2k} \cdot (k!)^2} \cdot \int_{-1}^1 \left[ \frac{d^k}{dz^k} \left(z^2 - 1 \right)^k \right]^2 \,dz = \frac{\displaystyle 2}{\displaystyle 2 \cdot k + 1}. 2 2 k ⋅ ( k ! ) 2 1 ⋅ ∫ − 1 1 [ d z k d k ( z 2 − 1 ) k ] 2 d z = 2 ⋅ k + 1 2 .
∫ − 1 1 P k ( m ) ( z ) ⋅ P k ( m ) ( z ) d z = 2 2 ⋅ k + 1 ⋅ ( k + m ) ⋅ ( k − m + 1 ) ⋅ ( k + m − 1 ) ⋅ ( k − m + 2 ) ⋯ , \int_{-1}^1 P_k^{(m)}(z) \cdot P_k^{(m)}(z) \,dz = \frac{\displaystyle 2}{\displaystyle 2 \cdot k + 1} \cdot (k+m) \cdot (k-m+1) \cdot (k+m-1) \cdot (k-m+2) \cdots, ∫ − 1 1 P k ( m ) ( z ) ⋅ P k ( m ) ( z ) d z = 2 ⋅ k + 1 2 ⋅ ( k + m ) ⋅ ( k − m + 1 ) ⋅ ( k + m − 1 ) ⋅ ( k − m + 2 ) ⋯ , множители с плюсом убывают от k + m k + m k + m k + 1 k + 1 k + 1 k − m + 1 k - m + 1 k − m + 1 k k k m ≤ k m \leq k m ≤ k
( k + m ) ⋅ ( k − m + 1 ) ⋅ ( k + m − 1 ) ⋅ ( k − m + 2 ) ⋯ = ( k + m ) ! k ! ⋅ k ! ( k − m ) ! = ( k + m ) ! ( k − m ) ! . (k+m) \cdot (k-m+1) \cdot (k+m-1) \cdot (k-m+2) \cdots = \frac{\displaystyle (k+m)!}{\displaystyle k!} \cdot \frac{\displaystyle k!}{\displaystyle (k-m)!} = \frac{\displaystyle (k+m)!}{\displaystyle (k-m)!}. ( k + m ) ⋅ ( k − m + 1 ) ⋅ ( k + m − 1 ) ⋅ ( k − m + 2 ) ⋯ = k ! ( k + m )! ⋅ ( k − m )! k ! = ( k − m )! ( k + m )! . Итоговая норма присоединённого полинома Лежандра имеет вид