前言
在《n次Legendre(勒让德)多项式在区间(-1, 1)上根的分布及证明》这篇文章中，我们阐述了Legendre多项式在 [ − 1 , 1 ] [-1,1] [−1,1]上的根分布情况并给出了证明。本文将证明Legendre多项式在 [ − 1 , 1 ] [-1,1] [−1,1]上的正交性质。
正交多项式的定义
设 f n ( x ) , n ∈ N f_n(x),n\in \mathbb N fn(x),n∈N是定义在 [ a , b ] [a,b] [a,b]上的一列函数，若对于任意的自然数 m , n m,n m,n， f m ( x ) f n ( x ) f_m(x)f_n(x) fm(x)fn(x)在 [ a , b ] [a,b] [a,b]上可积，且满足：
∫ a b f m ( x ) f n ( x ) d x = { 0 , m ≠ n ∫ a b f n 2 ( x ) d x > 0 , m = n \int_{a}^{b}f_m(x)f_n(x) \mathrm{d}x=\begin{cases}0, &m\neq n \\\displaystyle \int_{a}^{b} f^2_n(x)\mathrm{d}x>0, &m=n\end{cases} ∫abfm(x)fn(x)dx=⎩ ⎨ ⎧0,∫abfn2(x)dx>0,m=nm=n
则称 { f n ( x ) } \{f_n(x)\} {fn(x)}是 [ a , b ] [a,b] [a,b]上的正交函数列。当 { f n ( x ) } \{f_n(x)\} {fn(x)}是 n n n次多项式时，则称 { f n ( x ) } \{f_n(x)\} {fn(x)}是 [ a , b ] [a,b] [a,b]上的正交多项式列。
n阶Legendre多项式在 [ − 1 , 1 ] [-1,1] [−1,1]上的正交性证明
n次Legendre多项式的定义如下：
p n ( x ) = 1 2 n n ! d n d x n ( x 2 − 1 ) n , n ∈ N p_{n}(x)=\frac{1}{2^n n!}\frac{\mathrm d^n}{\mathrm{d} x^n}(x^2-1)^n, n\in \mathbb{N} pn(x)=2nn!1dxndn(x2−1)n,n∈N
不妨设 n ≥ m n \geq m n≥m。首先构造如下函数
I m n = m ! n ! 2 m 2 n ∫ − 1 1 p m ( x ) p n ( x ) d x = ∫ − 1 1 d m d x m ( x 2 − 1 ) m ⋅ d n d x n ( x 2 − 1 ) n d x \begin{equation} I_{mn}=m!n!2^m2^n\int_{-1}^{1}p_{m}(x)p_{n}(x) \mathrm{d}x =\int_{-1}^{1}\frac{\mathrm d^m}{\mathrm{d} x^m}(x^2-1)^m \cdot \frac{\mathrm d^n}{\mathrm{d} x^n}(x^2-1)^n \mathrm{d}x \end{equation} Imn=m!n!2m2n∫−11pm(x)pn(x)dx=∫−11dxmdm(x2−1)m⋅dxndn(x2−1)ndx
用分部积分法对 ( 1 ) (1) (1)式进行积分，可以得到
I m n = ∫ − 1 1 d m d x m ( x 2 − 1 ) m d ( d n − 1 d x n − 1 ( x 2 − 1 ) n ) = d m d x m ( x 2 − 1 ) m ⋅ d n − 1 d x n − 1 ( x 2 − 1 ) n ∣ − 1 1 − ∫ − 1 1 d n − 1 d x n − 1 ( x 2 − 1 ) n ⋅ d m + 1 d x m + 1 ( x 2 − 1 ) m d x \begin{equation} \begin{align} I_{mn} &=\int_{-1}^{1}\frac{\mathrm d^m}{\mathrm{d} x^m}(x^2-1)^m \mathrm{d}(\frac{\mathrm d^{n-1}}{\mathrm{d} x^{n-1}}(x^2-1)^n) \nonumber \\ &=\left.\frac{\mathrm d^m}{\mathrm{d} x^m}(x^2-1)^m \cdot \frac{\mathrm d^{n-1}}{\mathrm{d} x^{n-1}}(x^2-1)^n \right |_{-1}^{1} \nonumber -\int_{-1}^{1}\frac{\mathrm d^{n-1}}{\mathrm{d} x^{n-1}}(x^2-1)^n \cdot \frac{\mathrm d^{m+1}}{\mathrm{d} x^{m+1}}(x^2-1)^m\mathrm{d}x \nonumber \\ \end{align} \end{equation} Imn=∫−11dxmdm(x2−1)md(dxn−1dn−1(x2−1)n)=dxmdm(x2−1)m⋅dxn−1dn−1(x2−1)n −11−∫−11dxn−1dn−1(x2−1)n⋅dxm+1dm+1(x2−1)mdx
这里引用《n次Legendre(勒让德)多项式在区间(-1, 1)上根的分布及证明》这篇文章里的结论：
当 k < n k<n k<n时， f k ( x ) = [ ( x 2 − 1 ) n ] ( k ) f_{k}(x)=[(x^2-1)^n]^{(k)} fk(x)=[(x2−1)n](k)的每一项都包含因式 x − 1 x-1 x−1与 x + 1 x+1 x+1
因此 d m d x m ( x 2 − 1 ) m ⋅ d n − 1 d x n − 1 ( x 2 − 1 ) n ∣ − 1 1 = 0 \displaystyle \left.\frac{\mathrm d^m}{\mathrm{d} x^m}(x^2-1)^m \cdot \frac{\mathrm d^{n-1}}{\mathrm{d} x^{n-1}}(x^2-1)^n \right |_{-1}^{1}=0 dxmdm(x2−1)m⋅dxn−1dn−1(x2−1)n −11=0。于是 ( 2 ) (2) (2)式可以写成：
I m n = − ∫ − 1 1 d n − 1 d x n − 1 ( x 2 − 1 ) n ⋅ d m + 1 d x m + 1 ( x 2 − 1 ) m d x \begin{equation} I_{mn}=-\int_{-1}^{1}\frac{\mathrm d^{n-1}}{\mathrm{d} x^{n-1}}(x^2-1)^n \cdot \frac{\mathrm d^{m+1}}{\mathrm{d} x^{m+1}}(x^2-1)^m\mathrm{d}x \end{equation} Imn=−∫−11dxn−1dn−1(x2−1)n⋅dxm+1dm+1(x2−1)mdx
继续用分部积分法对 ( 3 ) (3) (3)式重复上述过程，执行 n n n次后，得到
I m n = ( − 1 ) n ∫ − 1 1 d m + n d x m + n ( x 2 − 1 ) m ⋅ ( x 2 − 1 ) n d x \begin{equation} I_{mn}=(-1)^n\int_{-1}^{1} \frac{\mathrm d^{m+n}}{\mathrm{d} x^{m+n}}(x^2-1)^m \cdot (x^2-1)^n \mathrm{d}x \end{equation} Imn=(−1)n∫−11dxm+ndm+n(x2−1)m⋅(x2−1)ndx
下面分情况讨论。
- 若 n > m n>m n>m， d m + n d x m + n ( x 2 − 1 ) m = 0 \displaystyle \frac{\mathrm d^{m+n}}{\mathrm{d} x^{m+n}}(x^2-1)^m =0 dxm+ndm+n(x2−1)m=0，即 I m n = 0 I_{mn}=0 Imn=0，因此有
∫ − 1 1 p m ( x ) p n ( x ) d x = 0 \begin{equation} \int_{-1}^{1}p_{m}(x)p_{n}(x) \mathrm{d}x =0 \end{equation} ∫−11pm(x)pn(x)dx=0
- 若 n = m n=m n=m，根据高阶导数的Leibniz公式可以得到：
d m + n d x m + n ( x 2 − 1 ) m = ∑ i = 0 2 n C 2 n i [ ( x + 1 ) n ] ( i ) [ ( x − 1 ) n ] ( 2 n − i ) = C 2 n n [ ( x + 1 ) n ] ( n ) [ ( x − 1 ) n ] ( n ) = ( 2 n ) ! \begin{equation} \displaystyle \frac{\mathrm d^{m+n}}{\mathrm{d} x^{m+n}}(x^2-1)^m =\displaystyle \sum_{i=0}^{2n} C_{2n}^{i}[(x+1)^n]^{(i)}[(x-1)^n]^{(2n-i)}=C_{2n}^{n}[(x+1)^n]^{(n)}[(x-1)^n]^{(n)}=(2n)! \end{equation} dxm+ndm+n(x2−1)m=i=0∑2nC2ni[(x+1)n](i)[(x−1)n](2n−i)=C2nn[(x+1)n](n)[(x−1)n](n)=(2n)!
将 ( 6 ) (6) (6)式代入 ( 4 ) (4) (4)式，不断使用分部积分法后可以得到
I n n = ( 2 n ) ! ( − 1 ) n ∫ − 1 1 ( x − 1 ) n ( x + 1 ) n d x = ( 2 n ) ! ∫ − 1 1 ( 1 − x ) n d ( ( 1 + x ) n + 1 n + 1 ) = ( 2 n ) ! n + 1 ( 1 − x ) n ( 1 + x ) n + 1 ∣ − 1 1 + ( 2 n ) ! n n + 1 ∫ − 1 1 ( 1 − x ) n − 1 ( 1 + x ) n + 1 d x = ( 2 n ) ! n n + 1 ∫ − 1 1 ( 1 − x ) n − 1 ( 1 + x ) n + 1 d x = ( 2 n ) ! n ( n − 1 ) ( n + 1 ) ( n + 2 ) ∫ − 1 1 ( 1 − x ) n − 2 ( 1 + x ) n + 2 d x = . . . = ( n ! ) 2 ∫ − 1 1 ( 1 + x ) 2 n d x = ( n ! ) 2 2 2 n + 1 2 n + 1 \begin{equation} \begin{align} I_{nn} &= (2n)!(-1)^n\int_{-1}^{1} (x-1)^n (x+1)^n \mathrm{d}x \nonumber \\ &=(2n)!\int_{-1}^{1}(1-x)^n \mathrm{d}\left(\dfrac{(1+x)^{n+1}} {n+1}\right)\nonumber \\ &=\left.\dfrac{(2n)!}{n+1}(1-x)^n(1+x)^{n+1}\right|_{-1}^{1}+\dfrac{(2n)!n}{n+1}\int_{-1}^{1}(1-x)^{n-1}(1+x)^{n+1}\mathrm{d}x \nonumber \\ &=\dfrac{(2n)!n}{n+1}\int_{-1}^{1}(1-x)^{n-1}(1+x)^{n+1}\mathrm{d}x \nonumber \\ &=\dfrac{(2n)!n(n-1)}{(n+1)(n+2)}\int_{-1}^{1}(1-x)^{n-2}(1+x)^{n+2}\mathrm{d}x \nonumber \\ &=... \nonumber \\ &=(n!)^2\int_{-1}^{1}(1+x)^{2n}\mathrm{d}x =\dfrac{(n!)^2 2^{2n+1}}{2n+1}\nonumber \\ \end{align} \end{equation} Inn=(2n)!(−1)n∫−11(x−1)n(x+1)ndx=(2n)!∫−11(1−x)nd(n+1(1+x)n+1)=n+1(2n)!(1−x)n(1+x)n+1 −11+n+1(2n)!n∫−11(1−x)n−1(1+x)n+1dx=n+1(2n)!n∫−11(1−x)n−1(1+x)n+1dx=(n+1)(n+2)(2n)!n(n−1)∫−11(1−x)n−2(1+x)n+2dx=...=(n!)2∫−11(1+x)2ndx=2n+1(n!)222n+1
将 ( 7 ) (7) (7)式代入 ( 1 ) (1) (1)式，可得
∫ − 1 1 p m ( x ) p n ( x ) d x = I n n ( n ! ) 2 2 n = 2 2 n + 1 > 0 \begin{equation} \int_{-1}^{1}p_{m}(x)p_{n}(x) \mathrm{d}x =\dfrac{I_{nn}}{(n!)2^{2n}}=\dfrac{2}{2n+1}>0 \end{equation} ∫−11pm(x)pn(x)dx=(n!)22nInn=2n+12>0
结合 ( 5 ) , ( 8 ) (5),(8) (5),(8)式，我们得到了如下结论
∫ − 1 1 p m ( x ) p n ( x ) d x = { 0 , m ≠ n 2 2 n + 1 > 0 , m = n \int_{-1}^{1}p_{m}(x)p_{n}(x) \mathrm{d}x=\begin{cases}0, &m\neq n \\\displaystyle\dfrac{2}{2n+1}>0, &m=n\end{cases} ∫−11pm(x)pn(x)dx=⎩ ⎨ ⎧0,2n+12>0,m=nm=n
根据定义，我们得到 n n n次Legendre多项式列 { p n ( x ) } \{p_n(x)\} {pn(x)}是 [ − 1 , 1 ] [-1,1] [−1,1]上的正交多项式列。证毕。