上海理工大学学报  2019, Vol. 41 Issue (5): 422-428   PDF    
四阶方程的有理Legendre函数全对角化谱方法
李珊, 栗巧玲     
上海理工大学 理学院,上海 200093
摘要: 针对四阶椭圆型方程,提出了在半直线域上全对角化的有理Legendre谱方法。构造了Sobolev正交的Legendre有理基函数,并导出了相应的全对角化的离散代数方程组。与此同时,微分方程的真解和数值解都表示为Fourier级数形式及其截断形式。数值结果表明了该方法的高效性并保持谱精度。
关键词: 四阶椭圆型方程     有理Legendre谱方法     Sobolev正交     半直线    
A Fully Diagonalized Legendre Rational Spectral Method for Solving Fourth Order Equations
LI Shan, LI Qiaoling     
College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China
Abstract: A fully diagonalized Legendre rational spectral method for solving fourth order elliptic equations on the half line was proposed. Some Sobolev orthogonal Legendre rational basis functions were constructed which leaded to the diagonalization of discrete systems. Accordingly, both the exact solutions and the approximate solutions could be represented as infinite and truncated Fourier series. Numerical results demonstrate the effectiveness and the spectral accuracy
Key words: fourth order elliptic equation     Legendre rational spectral method     Sobolev orthogonal function     half line    
1 问题提出

谱方法是一种古老的数值方法,广泛应用于数学物理和流体力学等理论研究中[1-2]。谱方法最吸引人的地方是它的“谱精度”,它的收敛性只与所逼近问题的光滑性有关,所求问题的解越光滑,收敛率越高。但是对于奇异问题以及无界区域问题,谱方法有时会出现不稳定的现象,可能会造成精度的损失。由于科学和工程中的微分方程常常设定在无界区域上,如何准确、高效地解决这些问题是一个非常重要的课题。近几十年,谱方法在无界区域上得到了迅速的发展,关于这一研究课题的大量文献也已经出现[3-4],意味着谱方法的研究取得了进展和突破。

针对无界区域上的问题,通常采用的第一种方法是人工边界法;第二种方法是使用正交多项式相关的谱逼近,例如Laguerre多项式和Hermite多项式;第三类方法是变量变换法,即将无界区域问题变换成有界区域上的奇异问题,再利用Jacobi谱方法等方法进行数值求解;第四种方法是有理谱方法[4-7]。前3种方法已相对成熟,但由于其各自的局限性,使得第四种方法在近年快速发展起来。

众所周知,Fourier函数是Laplace算子[8]的特征函数,它具有很好的正交性,这个特性使得Fourier谱方法在求解周期边值问题时形成了对角化的代数系统。同时,对于非周期边值问题通常采用Legendre谱方法。然而,与高度稀疏(例如三对角线,五对角线)和条件数良好的正交多项式函数的代数系统相比,仍然希望得到一组完全对角化的Fourier型的基函数,由此建立Fourier型的Sobolev正交多项式的基函数。本文的主要目的是实现半直线域上的四阶微分方程的有理Legendre谱方法的全对角化,构造Fourier型的Sobolev正交多项式基函数[9-10],将原问题的真解展开成Fourier级数,同时将数值解表示成级数的截断形式,通过数值算例可以看出该算法的可行性以及高精度性。

2 预备知识 2.1 Legendre多项式

首先回顾一下Legendre多项式。令 $I = ( - 1,1)$ ${L_k}(y)$ 是指数为 $k$ 次的Legendre多项式,且是奇异的Sturm-Liouville问题的特征函数:

${\partial _y}((1 - {y^2}){\partial _y}{L_k}(y)) + k(k + 1){L_k}(y) = 0,\;\;k \geqslant 0$ (1)
$\int_I {{L_k}(y){L_l}(y){\rm{d}}y = \frac{2}{{2k + 1}}{\delta _{k,l}}} $ (2)
$\int_I {{\partial _y}} {L_k}(y){\partial _y}{L_l}(y)(1 - {y^2}){\rm{d}}y = \frac{{2k(k + 1)}}{{2k + 1}}{\delta _{k,l}}$ (3)

其中, ${\delta _{k,l}}$ 是Kronecker符号。Legendre多项式满足如下递推式:

$ \begin{split}&{L_0}(y) = 1,\;{L_1}(y) = y\\ &(k + 1){L_{k + 1}}(y) = (2k + 1)y{L_k}(y) - k{L_{k - 1}}(y)\end{split}$ (4)
$(2k + 1){L_k}(y) = {\partial _y}{L_{k + 1}}(y) - {\partial _y}{L_{k - 1}}(y)$ (5)
$\begin{split}&\quad(2k + 1)(1 - {y^2}){\partial _y}{L_k}(y) =\\ & \qquad k(k + 1)({L_{k - 1}}(y) - {L_{k + 1}}(y))\end{split}$ (6)

这里 $k \geqslant 1$ ,且 ${L_k}( \pm 1) = {( \pm 1)^k}$ $ {\partial _y}{L_k}( \pm 1) = {( \pm 1)^{k - 1}}{\text{·}}$ $\dfrac{k}{2}(k + 1)$

2.2 Legendre有理函数

其次回顾一下修正的Legendre有理函数。令 $\varLambda = (0,\infty )$ $(u,v)$ ${L^2}(\varLambda )$ 空间的内积。

定义1[8]  记 ${R_k}(x)$ $k$ 次Legendre有理函数,满足

$ {R_k}(x) = \frac{{\sqrt 2 }}{{x + 1}}{L_k}\left(\frac{{x - 1}}{{x + 1}}\right),\;\;x \in \varLambda = (0,\infty ),\;\;k \geqslant 0 $

为了方便起见, $k < 0$ 时, ${R_k}(x) \equiv 0$

Legendre有理函数满足如下递推式:

$ \begin{split}&{R_0}(x) = \frac{{\sqrt 2 }}{{x + 1}},\;\;{R_1}(x) =\dfrac{{\sqrt 2 (x - 1)}}{{{{(x + 1)}^2}}},\;\;(k + 1){R_{k + 1}}(x) =\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\\ & \quad (2k + 1)\frac{{x - 1}}{{x + 1}}{R_k}(x) - k{R_{k - 1}}(x),\;\;k \geqslant 1 \end{split} $ (7)
$ \begin{split}&2(2k + 1){R_k}(x) = (x + 1)({R_{k + 1}}(x) - {R_{k - 1}}(x)) +\\ & \quad {(x + 1)^2}({\partial _x}{R_{k + 1}}(x) - {\partial _x}{R_{k - 1}}(x)),\;\;k \geqslant 1 \end{split} $ (8)

且具有正交性:

$\int_\varLambda {{R_k}} (x){R_l}(x){\rm{d}}x = \frac{2}{{2k + 1}}{\delta _{k,l}}$ (9)

因此,Legendre有理函数展开式为

$v(x) = \sum\limits_{k = 0}^\infty {{{\hat v}_k}} {R_k}(x),\;\;{\hat v_k} = \left(k + \frac{1}{2}\right)\int_\varLambda {v(x){R_k}} (x){\rm{d}}x$

接下来,对 $N \in {N^ + }$

${{\cal{R}} _N}(\varLambda ) = {\rm{span}}\{ {R_0}(x),{R_1}(x), \cdots ,{R_N}(x)\} $

为了方便起见,可以用下列符号表示:

$ \begin{split}&{\varphi _k}(x) = {R_k}(x) + {R_{k + 1}}(x),{\rm{ }}{\psi _k}(x) = (k + 2)({R_k}(x) + \\ &\quad {R_{k + 1}}(x)) + (k + 1)({R_{k + 1}}(x) + {R_{k + 2}}(x)) \end{split} $

引理1  对 $k \geqslant 0$ ,有如下等式成立

$ \begin{split}&{\partial _x}{R_k}(x) = \dfrac{{7{k^2} - 2}}{{4k + 2}}{R_{k - 1}}(x) - \dfrac{{4{k^2} + 4k + 1}}{{4k + 2}}{R_k}(x) +\\ &\;\;\;\; \dfrac{{{k^2} + 2k + 1}}{{4k + 2}}{R_{k + 1}}(x) \!+\! \sum\limits_{j = 0}^{k \!-\! 2} {{{( - 1)}^{j \!+\! k \!+\! 1}}} (2j \!-\! 3){R_j}(x) \end{split} \!\!\!\!$ (10)
$\begin{split}&{\partial _x}{\varphi _k}(x) = - \frac{{{k^2}}}{{2(2k + 1)}}{R_{k - 1}}(x) +\frac{{3{k^2} + 6k + 2}}{{2(2k + 3)}}{R_k}(x) -\!\!\!\!\!\!\!\!\!\!\!\!\!\\ &\quad \frac{{3{k^2} + 6k + 2}}{{2(2k + 1)}}{R_{k + 1}}(x) + \frac{{{{(k + 2)}^2}}}{{2(2k + 3)}}{R_{k + 2}}(x)\end{split} $ (11)

证明  首先用数学归纳法来验证式(10),由定义1可知

${\partial _x}{R_0}(x) = - \frac{1}{2}{R_0}(x) + \frac{1}{2}{R_1}(x)$
${\partial _x}{R_1}(x) = \frac{5}{6}{R_0}(x) - \frac{3}{2}{R_1}(x) + \frac{2}{3}{R_2}(x)$
${\partial _x}{R_2}(x) = - {R_0}(x) + \frac{{13}}{5}{R_1}(x) - \frac{5}{2}{R_2}(x) + \frac{9}{{10}}{R_3}(x)$

$k = 0,1,2$ 时,式 (10)成立。假设 $k \leqslant n$ 时,式(10)成立,只需验证 $k = n + 1$ 时式(10)成立即可。由式(7)可知

$ \begin{split}&(k + 1){\partial _x}{R_{k + 1}}(x) = - k{\partial _x}{R_{k - 1}}(x) + \frac{{2(2k + 1)}}{{{{(x + 1)}^2}}}{R_k}(x) +\\ &\quad (2k + 1){\partial _x}{R_k}(x) - \frac{{2(2k + 1)}}{{x + 1}}{\partial _x}{R_k}(x) \end{split} $

$\begin{split} &\dfrac{1}{{x + 1}}{R_k}(x) = - \dfrac{k}{{2(2k + 1)}}{R_{k - 1}}(x) + \dfrac{1}{2}{R_k}(x) -\\ &\quad \dfrac{{k + 1}}{{2(2k + 1)}}{R_{k + 1}}(x)\end{split}$

$ \begin{split}&\!\!\!\!\dfrac{1}{{{{(x + 1)}^2}}}{R_k}(x) = \dfrac{{k(k - 1)}}{{2(2k + 1)(2k - 1)}}{R_{k - 2}}(x) - \\ &\dfrac{k}{{2(2k + 1)}}{R_{k - 1}}(x) + \dfrac{{3{k^2} + 3k + 2}}{{2(2k + 3)(2k - 1)}}{R_k}(x)- \\ &\dfrac{{k + 1}}{{2(2k + 1)}}{R_{k + 1}}(x) + \dfrac{{(k + 1)(k + 2)}}{{4(2k + 3)(2k + 1)}}{R_{k + 2}}(x) \end{split}$

因此,由以上计算和归纳假设直接可计算得到 $k = n + 1$ 的结果,由式(10)可以很容易推导出式(11)的结果,即证。

引理2  对 $k \geqslant 0$ ,有

$\begin{split}&{\partial _x}{\psi _k}(x) = - \dfrac{{{k^2}(k + 2)}}{{4k + 2}}{R_{k - 1}}(x) - \dfrac{{3(2{k^3} + 9{k^2} + 11k + 3)}}{{2(2k + 2)(2k + 5)}}\cdot\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\\ &\quad{R_{k + 1}}(x)+\dfrac{{{k^2} + 3k + 1}}{2}{R_k}(x) - \dfrac{{{k^2} + 3k + 1}}{2}{R_{k + 2}}(x) + \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\\ &\quad\dfrac{{{{(k + 3)}^2}(k + 1)}}{{2(2k + 5)}}{R_{k + 3}}(x)\end{split}$ (12)
$ \begin{split}&\partial _x^2{\psi _k}(x) = \dfrac{{({k^3} - 3k + 2){k^2}}}{{4(4{k^2} - 1)}}{R_{k - 2}}(x) - \dfrac{{(3{k^2} + 6k - 1){k^2}}}{{4(2k + 1)}}\cdot\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\\ &\quad {R_{k - 1}}(x)+ \dfrac{{15{k^5} + 75{k^4} + 97{k^3} + {k^2} - 20k - 8}}{{4(2k - 1)(2k + 5)}}{R_k}(x)-\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\\ &\quad \dfrac{{10{k^5} + 75{k^4} + 204{k^3} + 243{k^2} + 124k + 24}}{{2(2k + 1)(2k + 5)}}{R_{k + 1}}(x)+\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\\ &\quad \dfrac{{15{k^5} + 150{k^4} + 547{k^3} + 872{k^2} + 568k + 128}}{{4(2k + 1)(2k + 7)}}{R_{k + 2}}(x)-\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\\ &\quad \dfrac{{{{(k + 3)}^2}(3{k^2} + 12k + 8)}}{{4\left( {2k + 5} \right)}}{R_{k + 3}}(x) + \\ &\quad \dfrac{{{{(k + 3)}^2}(k + 1){{(k + 4)}^2}}}{{4\left( {2k + 5} \right)(2k + 7)}}{R_{k + 4}}(x) \end{split}$ (13)

证明  由式(10)知,当 $k \geqslant 2$ 时,有

$\begin{split} &\!\!\!\!{\partial _x}{\psi _k}(x) = (k + 2){\partial _x}{\varphi _k}(x) + (k + 1){\partial _x}{\varphi _{k + 1}}(x)=\\ & (k + 2)\left( - \frac{{{k^2}}}{{2(2k + 1)}}{R_{k - 1}}(x) + \frac{{3{k^2} + 6k + 2}}{{2(2k + 3)}}{R_k}(x)\right. -\\ &\left. \frac{{3{k^2} + 6k + 2}}{{2(2k + 1)}}{R_{k + 1}}(x)+ \frac{{{{(k + 2)}^2}}}{{2(2k + 3)}}{R_{k + 2}}(x)\right) + \\ \end{split}$
$\begin{split} &(k + 1)\left( - \frac{{{{(k + 1)}^2}}}{{2(2k + 3)}}{R_k}(x)+\! \frac{{3{{(k \!+\! 1)}^2}\!+ 6(k + 1) + 2}}{{2(2k + 5)}}{R_{k + 1}}(x)-\right. \\ &\left.\frac{{3{{(k + 1)}^2} + 6(k + 1) + 2}}{{2(2k + 3)}}{R_{k + 2}}(x) + \frac{{{{(k + 3)}^2}}}{{2(2k + 5)}}{R_{k + 3}}(x)\right)= \\ & - \frac{{{k^2}(k + 2)}}{{4k + 2}}{R_{k - 1}}(x) + \frac{{{k^2} + 3k + 1}}{2}{R_k}(x) - \\ &\frac{{3(2{k^3} + 9{k^2} + 11k + 3)}}{{2(2k + 1)(2k + 5)}}{R_{k + 1}}(x)- \frac{{{k^2} + 3k + 1}}{2}{R_{k + 2}}(x) +\\ &\frac{{{{(k + 3)}^2}(k + 1)}}{{2(2k + 5)}}{R_{k + 3}}(x) \end{split}$

根据式(10)和式(11),当 $k \geqslant 3,$

$\begin{split} &\partial _x^2{\psi _k}(x) = (k + 2)\partial _x^2{\varphi _k}(x) + (k + 1)\partial _x^2{\varphi _{k + 1}}(x)=\\ & (k + 2)\left( - \dfrac{{{k^2}}}{{2(2k + 1)}}{\partial _x}{R_{k - 1}}(x) + \dfrac{{3{k^2} + 6k + 2}}{{2(2k + 3)}}{\partial _x}{R_k}(x) -\right.\\ & \left.\dfrac{{3{k^2} + 6k + 2}}{{2(2k + 1)}}{\partial _x}{R_{k + 1}}(x)+ \dfrac{{{{(k + 2)}^2}}}{{2(2k + 3)}}{\partial _x}{R_{k + 2}}(x)\right)+\\ & (k + 1)\left( - \dfrac{{{{(k + 1)}^2}}}{{2(2k + 3)}}{\partial _x}{R_k}(x) + \dfrac{{{{(k + 3)}^2}}}{{2(2k + 5)}}{\partial _x}{R_{k + 3}}(x)\right.+\\ & \dfrac{{3{{(k \!+\! 1)}^2} \!+\! 6(k \!+\! 1) \!+\! 2}}{{2(2k\! +\! 5)}}{\partial _x}{R_{k \!+\! 1}}(x) \!-\! \dfrac{{3{{(k \!+\! 1)}^2} \!+\! 6(k \!+\! 1) \!+\! 2}}{{2(2k \!+\! 3)}}\cdot\\ & {\partial _x}{R_{k + 2}}(x){\Bigg{)}}= \dfrac{{({k^3} - 3k + 2){k^2}}}{{4(4{k^2} - 1)}}{R_{k - 2}}(x) -\dfrac{{(3{k^2} + 6k + 1){k^2}}}{{4(2k + 1)}}\cdot\\ &{R_{k - 1}}(x)+\dfrac{{15{k^5} + 75{k^4} + 97{k^3} + {k^2} - 20k - 8}}{{4(2k - 1)(2k + 5)}}{R_k}(x)-\\ & \dfrac{{10{k^5} + 75{k^4} + 204{k^3} + 243{k^2} + 124k + 24}}{{2(2k + 5)(2k + 1)}}{R_{k + 1}}(x)+\\ & \dfrac{{15{k^5} + 150{k^4} + 547{k^3} + 872{k^2}+568k + 128}}{{4(2k + 1)(2k + 7)}}{R_{k + 2}}(x)-\\ & \dfrac{{{{(k \!+\! 3)}^2}(3{k^2} \!+\! 12k \!+\! 8)}}{{4(2k \!+\! 5)}}{R_{k \!+\! 3}}(x) \!+\! \dfrac{{{{(k \!+\! 3)}^2}(k \!+\! 1){{(k \!+\! 4)}^2}}}{{4(2k \!+\! 5)(2k \!+\! 7)}}{R_{k \!+\! 4}}(x)\end{split} $

所以对于 $k \geqslant 3$ 可直接得到结果。因此,对于 $0 \leqslant $ $k \leqslant 3,$ 由式(10)和式(11)很容易验证符合式(13),即证。

3 四阶Dirichlet边值问题

本节主要研究四阶Dirichlet边值问题的全对角化的有理Legendre谱方法,构造相应的Fourier型Sobolev正交基函数,使得方程的真解和数值解应展开成Fourier级数的形式。考虑如下的四阶Dirichlet边值问题:

$\left\{\!\!\!\begin{array}{l} {u^{(4)}}(x) - \alpha u''(x) + \beta u(x) = f(x),\alpha ,\beta \geqslant 0,x \in \varLambda \\ u(0) = u'(0) = 0,\mathop {\lim }\limits_{x \to + \infty } u(x) = \mathop {\lim }\limits_{x \to + \infty } u'(x) = 0 \end{array} \right.\!\!\!\!\!$ (14)

方程(14)的弱形式是找到 $u \in H_0^2(\varLambda )$ ,使得

$\begin{split}&{D_{\alpha ,\beta }}(u,v): = (\partial _x^2u,\partial _x^2v) + \alpha ({\partial _x}u,{\partial _x}v) +\\ &\quad\beta (u,v) = (f,v),\;\;\;v \in H_0^2(\varLambda )\end{split}$ (15)

显然,若 $f \in ({H^2}(\varLambda ))'$ ,由Lax-Milgram引理可知,式(15)就存在唯一解。

Legendre有理谱方法是寻找到 ${u_N} \in {{\cal{R}} _N}(\varLambda )$ ,使得

${D_{\alpha ,\beta }}({u_N},\varphi ) = (f,\varphi ),\;\;\;\varphi \in {{\cal{R}} _N}(\varLambda )$ (16)

为了对式(16)提出一种完全对角化逼近格式,需要在 ${{\cal{R}} _N}(\varLambda )$ 中构造一组新的基函数 ${\left\{ {{P_k}(x)} \right\}_{0 \leqslant k \leqslant N}}$ ,这组基函数与Sobolev内积 ${D_{\alpha ,\beta }}( \cdot ,\cdot )$ 是相互正交的。

引理3  令 ${P_k}(x) \in {{\cal{R}} _{k + 2}}(\varLambda )$ 是Sobolev空间下的正交Legendre有理函数,则 ${P_k}(x) - \dfrac{{{\psi _k}(x)}}{{{{(k + 1)}^2}}} \in {{\cal{R}} _{k + 1}}(\varLambda )$ 并且有

${D_{\alpha ,\beta }}({P_k},{P_l}) = {\sigma _k}{\delta _{k,l}},\;\;\;k,l \geqslant 0$ (17)

那么,可以得到如下递推关系:

$\begin{split}&{P_k}(x) = \frac{{{\psi _k}(x)}}{{{{(k + 1)}^2}}} - {a_k}{P_{k - 1}}(x) - {b_k}{P_{k - 2}}(x) -{c_k}{P_{k - 3}}(x) -\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\\ & \quad{d_k}{P_{k - 4}}(x) - {e_k}{P_{k - 5}}(x) - {f_k}{P_{k - 6}}(x),k \geqslant {\rm{6}}\end{split}$ (18)

其中, ${P_k}(x) \equiv 0(k < 0) $ $ {\sigma _k} = 0(k < 0) $ ${a_k} = 0(k < 1),$ $ {b_k} = 0(k < 2)$ $ {c_k} = 0(k < 3)$ ${d_k} = 0(k < 4) $ ${e_k} = 0(k < 5)$ ${f_k} = 0(k < 6)$ ,记:

$\begin{split} &{\tilde a_{k - i}} = {a_{k - i}}{a_{k - i - 1}} - {b_{k - i}},\;\;{\tilde b_{k - i}} = {a_{k - i}}{b_{k - i - 1}} - {c_{k - i}} ,\;\;{\tilde c_{k - i}} = {a_{k - i}}{c_{k - i - 1}} - {d_{k - i}} ,\;\;{\tilde d_{k - i}} = {a_{k - i}}{d_{k - i - 1}} - {e_{k - i}},\\ &{\hat a_{k - i}} = {a_{k - i - 2}}{\tilde a_{k - i}} - {\tilde b_{k - i}},\;\;{\hat b_{k - i}} = {\tilde a_{k - i}}{\tilde a_{k - i - 2}} - {a_{k - i - 3}}{\tilde b_{k - i}} + {\tilde c_{k - i}},\;\;{\hat c_{k - i}} = {\tilde a_{k - i - 3}}{\tilde b_{k - i}} - {a_{k - i - 4}}{\tilde c_{k - i}} + {\tilde d_{k - i}},\;\;{\bar a_{k - 1}} = {\tilde a_{k - 1}}{\hat a_{k - 3}} - {\hat c_{k - 1}},\\ & {Q_k} = 2k + 1,\;\;{M_k} = {Q_{k + 2}}{Q_k}{Q_{k - 1}}{Q_{k - 2}}{Q_{k - 3}},{\tilde M_k} = {M_k}{Q_{k + 3}},\;\;{\hat M_k} = {M_k}{Q_{k - 4}},\;\;{N_k} = {k^2},{R_k} = k\\ &{S_1} = 3(154{k^{11}} + 2\;541{k^{10}} + 16\;737{k^9} + 54\;432{k^8} + 79\;212{k^7} - 13\;671{k^6})\\ &{S_2} = - 3(216\;427{k^5} + 289\;902{k^4} + 128\;220{k^3} - 31\;752{k^2} - 44\;064k - 10\;368),\;\;S = ({S_1} + {S_2}){Q_{k + 1}}\\ & \tilde A = \frac{{3(33{k^8} + 264{k^7} + 598{k^6} - 108{k^5} - 1\;739{k^4} - 1\;308{k^3} + 532{k^2} + 576k - 192){Q_k}{Q_{k - 3}}}}{{2{{\tilde M}_k}{N_{k + 1}}{N_k}}}\\ & \tilde B = \frac{{3(165{k^{10}} + 825{k^9} - 670{k^8} - 7\;630{k^7} - 4\;931{k^6} + 15\;377{k^5} + 19\;564{k^4} + 2\;948{k^3} - 17\;392{k^2} - 14\;784k - 3\;840){Q_{k - 1}}}}{{8{{\tilde M}_k}{N_{k + 1}}{N_{k - 1}}}} \\ &\tilde C = \frac{{{k^3}(55{k^6} - 570{k^4} + 1\;263k + 980){Q_{k - 2}}}}{{4{M_k}{N_{k + 1}}{N_{k - 2}}}}\\ &\tilde D = \frac{{{k^2}(33{k^8} - 165{k^7} - 158{k^6} + 1\;622{k^5} - 1\;135{k^4} - 2\;965{k^3} + 5\;420{k^2} - 3\;676k + 1\;024){Q_{k - 3}}}}{{4{{\hat M}_k}{N_{k + 1}}{N_{k - 3}}}}\\ &\tilde E = \frac{{(3{k^2} - 6k - 25){R_{k - 1}}{Q_{k + 2}}{N_k}{N_{k - 1}}{N_{k - 2}}}}{{4{M_k}{N_{k + 1}}{N_{k - 4}}}},\;\;\tilde F = \frac{{({k^3} - 3k + 2){R_{k - 5}}{Q_{k + 2}}{N_k}{N_{k - 2}}{N_{k - 3}}}}{{8{{\hat M}_k}{N_{k + 1}}{N_{k - 5}}}},\;\; \tilde G = \frac{{{N_k}{N_{k - 1}}{R_{k + 2}}{R_{k - 3}}}}{{2{Q_k}{Q_{k - 1}}{Q_{k - 2}}{N_{k + 1}}{N_{k - 3}}}}\\ &\tilde H = \frac{{{k^3}({k^2} - 5)}}{{{Q_k}{Q_{k - 1}}{N_{k + 1}}{N_{k + 2}}}},\;\;\tilde I = \frac{{(2{k^6} + 6{k^5} - 11{k^4} - 32{k^3} + 17{k^2} + 34k + 12){Q_{k - 1}}{Q_{k - 3}}}}{{{M_k}{N_{k + 1}}{N_{k - 1}}}},\;\;\tilde J = \frac{{2{R_{k + 2}}{R_{k - 1}}}}{{{Q_k}{N_{k + 1}}{N_{k - 1}}}}\\ &\tilde K = \frac{{{k^5} + 5{k^4} + 9{k^3} + 7{k^2} - 2k + 4}}{{{Q_{k + 2}}{Q_{k - 1}}{N_{k + 1}}{N_k}}},\;\;\tilde L = \frac{{4{R_{k + 1}}}}{{{N_{k + 1}}{N_k}}},\;\;\tilde S = \frac{S}{{2{Q_{k + 2}}{{\tilde M}_{k + 1}}N_{k + 1}^2}}\\ &\tilde T = \frac{{(10{k^7} + 105{k^6} + 417{k^5} + 765{k^4} + 589{k^3} + 18{k^2} - 176k - 48){Q_{k - 2}}{Q_{k - 3}}}}{{{{\tilde M}_k}N_{k + 1}^2}},\;\;\tilde W = \frac{{12(2{k^3} + 9{k^2} + 13k + 6)}}{{{Q_{k + 2}}{Q_k}N_{k + 1}^2}} \end{split}$

a. ${\sigma _k} = - {({a_k})^2}{\sigma _{k - 1}} - {({b_k})^2}{\sigma _{k - 2}} - {({c_k})^2}{\sigma _{k - 3}} - {({d_k})^2}{\sigma _{k - 4}} - {({e_k})^2}{\sigma _{k - 5}} - {({f_k})^2}{\sigma _{k - 6}} + \tilde S + \alpha \tilde T + \beta \tilde W,\;\;k \geqslant 0$

b. ${a_k} = - \dfrac{1}{{{\sigma _{k - 1}}}}(\tilde A + \tilde B{a_{k - 1}} + \tilde C{\tilde a_{k - 1}} + \tilde D{\hat a_{k - 1}} \!+\! \tilde E{\hat b_{k - 1}} \!+\! \tilde F{\bar a_{k - 1}} \!-\! \alpha \tilde G{\hat a_{k - 1}} \!-\! \alpha \tilde H{\tilde a_{k - 1}} \!-\! \alpha \tilde I{a_{k - 1}} + \beta \tilde J{a_{k - 1}} + \alpha \tilde K - \beta \tilde L),\;\;k \geqslant 1$

c. ${b_k} = \dfrac{1}{{{\sigma _{k - 2}}}}(\tilde B + \tilde C{a_{k - 2}} + \tilde D{\tilde a_{k - 2}} + \tilde E{\hat a_{k - 2}} + \tilde F{\hat b_{k - 2}} - \alpha \tilde G{\tilde a_{k - 2}} - \alpha \tilde H{a_{k - 2}} - \alpha \tilde I + \beta \tilde J),\;\;k \geqslant 2$

d. ${c_k} = - \dfrac{1}{{{\sigma _{k - 3}}}}(\tilde C + \tilde D{a_{k - 3}} + \tilde E{\tilde a_{k - 3}} + \tilde F{\hat a_{k - 3}} - \alpha \tilde G{a_{k - 3}} - \alpha \tilde H),\;\;k \geqslant 3$

e. ${d_k} = \dfrac{1}{{{\sigma _{k - 4}}}}(\tilde D + \tilde E{a_{k - 4}} + \tilde F{\tilde a_{k - 4}} - \alpha \tilde G),\;\;k \geqslant 4$

f. ${e_k} = - \dfrac{1}{{{\sigma _{k - 5}}}}(\tilde E + \tilde F{a_{k - 5}}),\;\;k \geqslant 5$

g. ${f_k} = \dfrac{{\tilde F}}{{{\sigma _{k - 6}}}},\;\;k \geqslant 6$

证明  首先用数学归纳法验证式(14),由正交性有

$ \begin{split} & {P_k}(x) = \dfrac{{{\psi _k}(x)}}{{{{(k + 1)}^2}}} - \sum\limits_{m = 0}^{k - 1} {\dfrac{{{D_{\alpha ,\beta }}\left(\dfrac{{{\psi _k}(x)}}{{{{(k + 1)}^2}}},{P_m}(x)\right)}}{{{\sigma _m}}}} {P_m}(x),\\ & \quad k \geqslant 1 \\[-11pt] \end{split} $ (19)

根据式(12),(13)和(15),可以得到

$ \begin{split} &{D_{\alpha ,\beta }}\left(\frac{{{\psi _1}(x)}}{4},{P_0}(x)\right) = \left(\partial _x^2\frac{{{\psi _1}(x)}}{4},\partial _x^2{P_0}(x)\right) + \\ &\quad \alpha \left({\partial _x}\frac{{{\psi _1}(x)}}{4},{\partial _x}{P_0}(x)\right) + \beta \left(\frac{{{\psi _1}(x)}}{4},{P_0}(x)\right)=\\ &\quad - 16 - \frac{{4\alpha }}{7} + 2\beta \end{split}$

因此 ${a_1} = \dfrac{{ - 16 + 2\beta }}{{{\sigma _0}}} - \dfrac{{4\alpha }}{{7{\sigma _0}}}$ ,故 ${P_1}(x) = \dfrac{{{\psi _1}(x)}}{4} - $ $ {a_1}{P_0}(x)$ ,用同样的方法,可以得到

$\begin{split} &{D_{\alpha , \beta }}\left(\frac{{{\psi _2}(x)}}{9},{P_0}(x)\right) = \left(\partial _x^2\frac{{{\psi _2}(x)}}{9},\partial _x^2{P_0}(x)\right) +\\ & \quad \alpha \left({\partial _x}\frac{{{\psi _2}(x)}}{9},{\partial _x}{P_0}(x)\right) + \beta \left(\frac{{{\psi _2}(x)}}{9},{P_0}(x)\right)= \\ & \quad \frac{{6\;112 - 44\alpha + 88\beta }}{{495}}\end{split} $
$\begin{split}&{D_{\alpha ,\beta }}\left(\frac{{{\psi _2}(x)}}{9},{P_1}(x)\right) = - \frac{{1\;744}}{{99}} - \frac{{6\;112{a_1}}}{{495}} +\\ &\quad \alpha \left(\frac{{4{a_1}}}{{45}} - \frac{{17}}{{81}}\right) + \beta \left(\frac{1}{3} - \frac{{8{a_1}}}{{45}}\right)\end{split}$

所以,这里 $ {a_2} = - \dfrac{{1\;744}}{{99{\sigma _1}}} - \dfrac{{6\;112{a_1}}}{{495{\sigma _1}}} +\alpha \left(\dfrac{{4{a_1}}}{{45{\sigma _1}}} -\right. $ $ \left.\dfrac{{17}}{{81{\sigma _1}}}\right) + \beta \left(\dfrac{1}{{3{\sigma _1}}} - \dfrac{{8{a_1}}}{{45{\sigma _1}}}\right)$ ${b_2} = \dfrac{{6\;112 - 44\alpha + 88\beta }}{{495{\sigma _0}}}$ ,故 ${P_2}(x) = \dfrac{{{\psi _2}(x)}}{9} - {a_2}{P_1}(x) - {b_2}{P_0}(x)$ ,相似地,当 $n = 3,4, $ $5,6$ 时,式(18)仍成立,其中

$\begin{split} &{a_3} = - \frac{{43\;072}}{{2\;145{\sigma _2}}} - \frac{{12\;696{a_2}}}{{1\;001{\sigma _2}}} - \frac{{378{{\tilde a}_2}}}{{55{\sigma _2}}} - \alpha \left(\frac{{59}}{{495{\sigma _2}}} - \frac{{17{a_2}}}{{176{\sigma _2}}} - \frac{{27{{\tilde a}_2}}}{{140{\sigma _2}}}\right) + \beta \left(\frac{1}{{9{\sigma _2}}} - \frac{{5{a_2}}}{{112{\sigma _2}}}\right)\\ &{b_3} = \frac{{12\;696}}{{1\;001{\sigma _1}}} + \frac{{378{a_1}}}{{55{\sigma _1}}} - \alpha \left(\frac{{17}}{{176{\sigma _1}}} + \frac{{27{a_1}}}{{140{\sigma _1}}}\right) + \frac{{5\beta }}{{112{\sigma _1}}},{c_3} = - \frac{{378}}{{55{\sigma _0}}} + \frac{{27\alpha }}{{140{\sigma _0}}}\\ & {a_4} = - \frac{{37\;092}}{{1\;625{\sigma _3}}} - \frac{{204\;416{a_3}}}{{14\;625{\sigma _3}}} - \frac{{2\;128{{\tilde a}_3}}}{{325{\sigma _3}}} + ({{\tilde b}_3} - {a_1}{{\tilde a}_3})\frac{{29\;248}}{{11\;375{\sigma _3}}}+ \alpha \left( - \frac{{149}}{{1\;820{\sigma _3}}} + \frac{{9\;892{a_3}}}{{131\;625{\sigma _3}}} + \frac{{176{{\tilde a}_3}}}{{1\;575{\sigma _3}}} -\right. \\ &\quad\quad\left.({{\tilde b}_3} - {a_1}{{\tilde a}_3})\frac{{48}}{{875{\sigma _3}}}\right) + \beta \left(\frac{1}{{20{\sigma _3}}} - \frac{{4{a_3}}}{{225{\sigma _3}}}\right)\\ &{b_4} = \frac{{204\;416}}{{14\;625{\sigma _2}}} + \frac{{2\;128{a_2}}}{{325{\sigma _2}}} + \frac{{29\;248{{\tilde a}_2}}}{{11\;375{\sigma _2}}} - \alpha \left(\frac{{9\;892}}{{131\;625{\sigma _2}}} + \frac{{176{a_2}}}{{1\;575{\sigma _2}}} - \frac{{48{{\tilde a}_2}}}{{875{\sigma _2}}}\right) + \frac{{4\beta }}{{225{\sigma _2}}}\\ &{c_4} = - \frac{{2\;128}}{{325{\sigma _1}}} - \frac{{29\;248{a_1}}}{{11\;375{\sigma _1}}} + \alpha \left(\frac{{176}}{{1\;575{\sigma _1}}} + \frac{{48{a_1}}}{{875{\sigma _1}}}\right),\;\;{d_4} = \frac{{29\;248}}{{11\;375{\sigma _0}}} - \frac{{48\alpha }}{{875{\sigma _0}}}\\ &{a_5} = - \frac{{3\;440\;624}}{{133\;875{\sigma _4}}} - \frac{{611\;441{a_4}}}{{39\;270{\sigma _4}}} - \frac{{6\;200{{\tilde a}_4}}}{{891{\sigma _4}}} + ({\tilde b_4} - {a_2}{\tilde a_4})\frac{{1\;580{{\tilde a}_3}}}{{693{\sigma _4}}} - ({\tilde a_2}{\tilde a_4} - {a_1}{\tilde b_4} + {c_4})\frac{{400}}{{693{\sigma _4}}}+ \beta \left(\frac{2}{{75{\sigma _4}}} - \frac{{7{a_4}}}{{792{\sigma _4}}}\right) -\\ &\quad\quad\alpha \left(\frac{{628}}{{10\;125{\sigma _4}}} - \frac{{43{a_4}}}{{720{\sigma _4}}} - \frac{{625{{\tilde a}_4}}}{{8\;019{\sigma _4}}} + ({\tilde b_4} - {a_2}{\tilde a_4})\frac{{25}}{{891{\sigma _4}}}\right)\\ &{b_5} = \frac{{611\;441}}{{39\;270{\sigma _3}}} + \frac{{6\;200{a_3}}}{{891{\sigma _3}}} + \frac{{1\;580{{\tilde a}_3}}}{{693{\sigma _3}}} - ({\tilde b_3} - {a_1}{\tilde a_3})\frac{{400}}{{693{\sigma _3}}} - \alpha \left(\frac{{43}}{{720{\sigma _3}}} + \frac{{625{a_3}}}{{8\;019{\sigma _3}}} + \frac{{25{{\tilde a}_3}}}{{891{\sigma _3}}}\right) + \frac{{7\beta }}{{792{\sigma _3}}}\\ &{c_5} = - \frac{{6\;200}}{{891{\sigma _2}}} - \frac{{1\;580{a_2}}}{{693{\sigma _2}}} - \frac{{400{{\tilde a}_2}}}{{693{\sigma _2}}} + \alpha \left(\frac{{625}}{{8\;019{\sigma _2}}} + \frac{{25{a_2}}}{{891{\sigma _2}}}\right),\;\;{d_5} = \frac{{1\;580}}{{693{\sigma _1}}} + \frac{{400{a_1}}}{{693{\sigma _1}}} - \frac{{25\alpha }}{{891{\sigma _1}}},\;\;{e_5} = - \frac{{400}}{{693{\sigma _0}}}\\ &{a_6} = - \frac{{6\;411\;544}}{{223\;839{\sigma _5}}} - \frac{{622\;689\;488{a_5}}}{{36\;006\;425{\sigma _5}}} - \frac{{372\;006{{\tilde a}_5}}}{{49\;049{\sigma _5}}} + ({{\tilde b}_5} - {a_3}{{\tilde a}_5})\frac{{2\;508\;320}}{{1\;072\;071{\sigma _5}}} - ({{\tilde a}_3}{{\tilde a}_5} - {a_2}{{\tilde b}_5} + {{\tilde c}_5})\frac{{23\;500}}{{49\;049{\sigma _5}}} +({a_1}{{\tilde a}_3}{{\tilde a}_5} + \\ &\quad\quad {{\tilde a}_5}{{\tilde b}_3} \!+\!{{\tilde a}_2}{{\tilde b}_5} \!+\! {a_1}{{\tilde c}_5} \!-\! {{\tilde d}_5})\frac{{2\;880}}{{49\;049{\sigma _5}}} \!-\! \alpha \left(\!\frac{{587}}{{11\;781{\sigma _5}}} \!-\! \frac{{13\;292{a_5}}}{{270\;725{\sigma _5}}} \!-\! \frac{{837{{\tilde a}_5}}}{{14\;014{\sigma _5}}} \!+\! ({{\tilde b}_5} \!-\! {a_3}{{\tilde a}_5})\frac{{400}}{{21\;021{\sigma _5}}}\!\right) \!+\! \beta \left(\frac{1}{{63{\sigma _5}}} \!-\! \frac{{16}}{{3\;185{\sigma _5}}}\right) \\ &{b_6} = \frac{{622\;689\;488}}{{36\;006\;425{\sigma _4}}} + \frac{{372\;006{a_4}}}{{49\;049{\sigma _4}}} + \frac{{2\;508\;320{{\tilde a}_4}}}{{1\;072\;071{\sigma _4}}} - ({{\tilde b}_4} - {a_2}{{\tilde a}_4})\frac{{23\;500}}{{49\;049{\sigma _4}}} + ({{\tilde a}_2}{{\tilde a}_4} - {a_1}{{\tilde b}_4} + {{\tilde c}_4})\frac{{2\;880}}{{49\;049{\sigma _4}}} +\\ &\quad\quad\alpha \left(\frac{{13\;292}}{{270\;725{\sigma _4}}} - \frac{{837{a_4}}}{{14\;014{\sigma _4}}} - \frac{{400{{\tilde a}_4}}}{{21\;021{\sigma _4}}}\right) + \frac{{16\beta }}{{3\;185{\sigma _4}}} \\ &{c_6} = - \frac{{372\;006}}{{49\;049{\sigma _3}}} - \frac{{2\;508\;320{a_3}}}{{1\;072\;071{\sigma _3}}} - \frac{{23\;500{{\tilde a}_3}}}{{49\;049{\sigma _3}}} + ({\tilde b_3} - {a_1}{\tilde a_3})\frac{{2\;880}}{{49\;049{\sigma _3}}} + \alpha \left(\frac{{837}}{{14\;014{\sigma _3}}} + \frac{{400{a_3}}}{{21\;021{\sigma _3}}}\right)\\ &{d_6} = \frac{{2\;508\;320}}{{1\;072\;071{\sigma _2}}} + \frac{{23\;500{a_2}}}{{49\;049{\sigma _2}}} + \frac{{2\;880{{\tilde a}_2}}}{{49\;049{\sigma _2}}} - \frac{{400\alpha }}{{21\;021{\sigma _2}}},\;\;{e_6} = - \frac{{23\;500}}{{49\;049{\sigma _1}}} - \frac{{2\;880{a_1}}}{{49\;049{\sigma _1}}},\;\;{f_6} = \frac{{2\;880}}{{49\;049{\sigma _0}}} \end{split}$

进一步,可以假设,当 $1 \leqslant m \leqslant k - 1$ $k > 7$ 时,

$\begin{split}&\!\!\!\!{P_l}(x) = \frac{{{\psi _l}(x)}}{{{{(l + 1)}^2}}} - {a_l}{P_{l - 1}}(x) - {b_l}{P_{l - 2}}(x)- \\ &{c_l}{P_{l - 3}}(x) -{d_l}{P_{l - 4}}(x) - {e_l}{P_{l - 5}}(x) - {f_l}{P_{l - 6}}(x)\end{split}$

接下来,只需要证明,当 $k \geqslant 7$

$ \begin{split}&{P_k}(x) = \frac{{{\psi _k}(x)}}{{{{(k + 1)}^2}}} - {a_k}{P_{k - 1}}(x) - {b_k}{P_{k - 2}}(x)- {c_k}{P_{k - 3}}(x) -\\ &\quad {d_k}{P_{k - 4}}(x) - {e_k}{P_{k - 5}}(x) - {f_k}{P_{k - 6}}(x)\end{split} $

事实上,由式(12),(13)和(15),当 $k > m \geqslant 0,$ $\;\;k \geqslant 7$ ,则

$\begin{split} &{D_{\alpha ,\beta }}\left(\frac{{{\psi _k}}}{{{{(1 + k)}^2}}},{P_m}\right)= \left(\partial _x^2\frac{{{\psi _k}}}{{{{(1 + k)}^2}}},\partial _x^2{P_m}\right) + \alpha \left({\partial _x}\frac{{{\psi _k}}}{{{{(1 + k)}^2}}},{\partial _x}{P_m}\right) + \beta \left(\frac{{{\psi _k}}}{{{{(1 + k)}^2}}},{P_m}\right)=\\ &\quad \left(\partial _x^2\frac{{{\psi _k}}}{{{{(1 + k)}^2}}},\partial _x^2\frac{{{\psi _m}}}{{{{(1 + m)}^2}}} - {a_m}\partial _x^2{P_{m - 1}} - {b_m}\partial _x^2{P_{m - 2}} - {c_m}\partial _x^2{P_{m - 3}} - {d_m}\partial _x^2{P_{m - 4}} - {e_m}\partial _x^2{P_{m - 5}} - {f_m}\partial _x^2{P_{m - 6}}\right)+\\ &\quad { \alpha \left({\partial _x}\frac{{{\psi _k}}}{{{{(1 + k)}^2}}},{\partial _x}\frac{{{\psi _m}}}{{{{(1 + m)}^2}}} - {a_m}{\partial _x}{P_{m - 1}} - {b_m}{\partial _x}{P_{m - 2}} - {c_m}{\partial _x}{P_{m - 3}} - {d_m}{\partial _x}{P_{m - 4}} - {e_m}{\partial _x}{P_{m - 5}} - {f_m}{\partial _x}{P_{m - 6}}\right)}+\\ &\quad \beta \left({\frac{{{\psi _k}}}{{{{(1 + k)}^2}}},\frac{{{\psi _m}}}{{{{(1 + m)}^2}}} - {a_m}{P_{m - 1}} - {b_m}{P_{m - 2}} - {c_m}{P_{m - 3}}}- {d_m}{P_{m - 4}} - {e_m}{P_{m - 5}} - {f_m}{P_{m - 6}}\right)=\\ &\quad \left(\partial _x^2\frac{{{\psi _k}}}{{{{(1 + k)}^2}}},\partial _x^2\frac{{{\psi _m}}}{{{{(1 + m)}^2}}} -\right. {a_m}\partial _x^2\frac{{{\psi _{m - 1}}}}{{{m^2}}} + {\tilde a_m}\partial _x^2\frac{{{\psi _{m - 2}}}}{{{{(m - 1)}^2}}}+ ({\tilde b_m} - {a_{m - 2}}{\tilde a_m})\partial _x^2\frac{{{\psi _{m - 3}}}}{{{{(m - 2)}^2}}}+ ({\tilde a_m}{\tilde a_{m - 2}} - {a_{m - 3}}{\tilde b_m} + {\tilde c_m})\partial _x^2\frac{{{\psi _{m - 4}}}}{{{{(m - 3)}^2}}}+\\ &\quad { ( - {a_{m - 4}}{{\tilde a}_m}{{\tilde a}_{m - 2}} + {{\tilde a}_m}{{\tilde b}_{m - 2}} + {{\tilde a}_{m - 3}}{{\tilde b}_m} - {a_{m - 4}}{{\tilde c}_m} + {{\tilde d}_m}){\partial _x}^2\frac{{{\psi _{m - 5}}}}{{{{(m - 4)}^2}}}}+ \beta \left(\frac{{{\psi _k}}}{{{{(k + 1)}^2}}},\frac{{{\psi _m}}}{{{{(m + 1)}^2}}} - {a_m}\frac{{{\psi _{m - 1}}}}{{{m^2}}}\right)+\\ &\quad \alpha \left({{\partial _x}\frac{{{\psi _k}}}{{{{(1 + k)}^2}}},{\partial _x}\frac{{{\psi _m}}}{{{{(1 + m)}^2}}} - {a_m}{\partial _x}\frac{{{\psi _{m - 1}}}}{{{m^2}}} + {{\tilde a}_m}{\partial _x}\frac{{{\psi _{m - 2}}}}{{{{(m - 1)}^2}}}}+ ({\tilde b_m} - {a_{m - 2}}{\tilde a_m}){\partial _x}\frac{{{\psi _{m - 3}}}}{{{{(m - 2)}^2}}}\right)\end{split} $

$m = k - 1,\;\;k - 2,\;\;k - 3,\;\;k - 4,\;\;k - 5,\;\;k - 6$ ,可以得到结论 ${\rm{b}} \sim {\rm{g}}$

接下来计算常数 ${\sigma _k}$ ,由式(12),(13),(15)和(17),当 $k \geqslant 7$ 时,有

$\begin{split} &{D_{\alpha ,\beta }}\left(\frac{{{\psi _k}}}{{{{(k + 1)}^2}}},\frac{{{\psi _k}}}{{{{(k + 1)}^2}}}\right)= \left(\partial _x^2\frac{{{\psi _k}}}{{{{(k + 1)}^2}}},\partial _x^2\frac{{{\psi _k}}}{{{{(k + 1)}^2}}}\right) + \\ &\quad \alpha \left({\partial _x}\frac{{{\psi _k}}}{{{{(k + 1)}^2}}},{\partial _x}\frac{{{\psi _k}}}{{{{(k + 1)}^2}}}\right) + \beta \left(\frac{{{\psi _k}}}{{{{(k + 1)}^2}}},\frac{{{\psi _k}}}{{{{(k + 1)}^2}}}\right)=\\ &\quad \frac{{3(154{k^{11}} + 2\;541{k^{10}} + 16\;737{k^9} + 54\;432{k^8})}}{{2(2k + 9)(2k + 7)(2k + 5)(2k - 3)(4{k^2} - 1)}}+\\ &\quad \frac{{3( - 128\;220{k^3} + 31\;752{k^2} + 44\;064k + 10\;368)}}{{2(2k + 9)(2k + 7)(2k + 5)(2k - 3)(4{k^2}- 1)}}+\\ &\quad \alpha \frac{{10{k^7} \!+\! 105{k^6} \!+\! 417{k^5} \!+\! 765{k^4} \!+\! 589{k^3} \!+\! 18{k^2} \!-\! 176k \!-\! 48}}{{(2k + 7)(2k + 5)(4{k^2} - 1)}}+ \\ &\quad \beta \frac{{12(2{k^3} + 9{k^2} + 13k + 6)}}{{(2k + 5)(2k + 1)}}=\tilde S + \alpha \tilde T + \beta \tilde W\end{split} $

另一方面

$\begin{split} &{D_{\alpha ,\beta }}\left(\frac{{{\psi _k}}}{{{{(k + 1)}^2}}},\frac{{{\psi _k}}}{{{{(k + 1)}^2}}}\right)={D_{\alpha ,\beta }}({P_k} + {a_k}{P_{k - 1}} + {b_k}{P_{k - 2}} + \\ &\quad {c_k}{P_{k - 3}} + {d_k}{P_{k - 4}} + {e_k}{P_{k - 5}} + {f_k}{P_{k - 6}},\;\;{P_k} + {a_k}{P_{k - 1}} +\\ &\quad {b_k}{P_{k - 2}} + {c_k}{P_{k - 3}} + {d_k}{P_{k - 4}} + {e_k}{P_{k - 5}} + {f_k}{P_{k - 6}})= \\ &\quad {\sigma _k} + {({a_k})^2}{\sigma _{k - 1}} + {({b_k})^2}{\sigma _{k - 2}} + {({c_k})^2}{\sigma _{k - 3}}+ {({d_k})^2}{\sigma _{k - 4}} +\\ &\quad {({e_k})^2}{\sigma _{k - 5}} + {({f_k})^2}{\sigma _{k - 6}}\end{split} $

由此可以得到结论a,即证。

那么由式(15)和式(16)以及 $\{ {{{P}}_k}\} _{k = 0}^\infty $ 的正交性可以推出定理1。

定理1  设 $u\left( x \right)$ ${u_N}\left( x \right)$ 分别是方程(15)和有理谱格式(16)的真解和数值解,那么 $u\left( x \right)$ ${u_N}\left( x \right)$ 分别可以展开成基函数 $\{ {{{P}}_k}\} _{k = 0}^\infty $ 的级数形式和级数截断形式,即

$\begin{gathered} u\left( x \right) = \sum\limits_{k = 0}^\infty {{{\hat u}_k}{P_k}(x)} ,\;\;\;\;\;{u_N}(x) = \sum\limits_{k = 0}^N {{{\hat u}_k}{P_k}(x)} \\ {{\hat u}_k} = \frac{1}{{{\eta _k}}}{D_{\alpha ,\beta }}\left( {u,{P_k}} \right) = \frac{1}{{{\eta _k}}}(f,{P_k}),\;\;\;k \geqslant 0 \\ \end{gathered} $
4 数值实验

在这一节中针对对角化Legendre有理谱方法在半直线上求解椭圆型方程的有效性和精确度进行了检验。并考察了四阶方程(14)在 $\mu = 1,\;\;\lambda = 1$ 时测试函数具有不同衰减速度时的收敛性。

针对震荡且无穷远处指数衰减的测试函数 $u(x) = {{\rm{e}}^{ - {x^2}}}\sin (2{x^2})$ 图1绘制了谱格式(16)取自然对数后的离散 ${L^2}$ 误差和 ${H^1}$ 误差与 $N$ 之间的关系。从图中可以观察到谱格式(16)具有几何收敛速度,即保持了谱方法的指数收敛优点。

针对震荡且无穷远处代数衰减的测试函数 $u(x) = \dfrac{{{x^2}}}{{{{(1 + {x^2})}^2}}}$ 图2绘制了谱格式(16)取自然对数后的离散 ${L^2}$ 误差和 ${H^1}$ 误差与 $N$ 之间的关系。从图中可以观察到谱格式(16)具有几何收敛速度,即保持了谱方法的指数收敛优点。


图 1 指数衰减时谱格式(16)的误差 Fig. 1 Errors of scheme (16) with exponential decay function

图 2 代数衰减时谱格式(16)的误差 Fig. 2 Errors of scheme (16) with algebraic decay function
参考文献
[1]
BERNARDI C, MADAY Y. Spectral methods, in handbook of numerical analysis[M]//CIARLET P G, LIONS J L. Techniques of Scientific Computing. Amsterdam: Elsevier, 1997: 209-486.
[2]
SHEN J, TANG T, WANG L L. Spectral methods: algorithms, analysis and applications[M]. Berlin: Springer, 2011.
[3]
CHRISTOV C I. A complete orthonormal system of functions in L2 (-∞, ∞) space [J]. SIAM Journal on Applied Mathematics, 1982, 42(6): 1337-1344. DOI:10.1137/0142093
[4]
LIU F J, WANG Z Q, LI H Y. A fully diagonalized spectral method using generalized Laguerre functions on the half line[J]. Advances in Computational Mathematics, 2017, 43(6): 1227-1259. DOI:10.1007/s10444-017-9522-3
[5]
GUO B Y. Error estimation of Hermite spectral method for nonlinear partial differential equations[J]. Mathematics of Computation, 1999, 68(227): 1067-1078. DOI:10.1090/S0025-5718-99-01059-5
[6]
GUO B Y, SHEN J. Laguerre-Galerkin method for nonlinear partial differential equations on a semi-infinite interval[J]. Numerische Mathematik, 2000, 86(4): 635-654. DOI:10.1007/PL00005413
[7]
BOYD J P. Spectral methods using rational basis functions on an infinite interval[J]. Journal of Computational Physics, 1987, 69(1): 112-142. DOI:10.1016/0021-9991(87)90158-6
[8]
GUO B Y, SHEN J. On spectral approximations using modified Legendre rational functions: application to the Korteweg-de Vries equation on the half line[J]. Indiana University Mathematics Journal, 2001, 50(1): 181-204. DOI:10.1512/iumj.2001.50.2090
[9]
GUO B Y, SHEN J, WANG Z Q. A rational approximation and its applications to differential equations on the half line[J]. Journal of Scientific Computing, 2000, 15(2): 117-147. DOI:10.1023/A:1007698525506
[10]
LIU F J, LI H Y, WANG Z Q. Spectral methods using generalized Laguerre functions for second and fourth order problems[J]. Numerical Algorithms, 2017, 75(4): 1005-1040. DOI:10.1007/s11075-016-0228-2