Mtg 45: Fri, 3 Dec10

Difference Eq.(1)p.44-5 :

${\displaystyle \left[(1-\mu ^{2})P_{n}'\right]'+n\left[\mu \ P_{n}'+P_{n}\right]-nP_{n-1}'=0\ }$

${\displaystyle \displaystyle {\begin{aligned}(RR1)\Rightarrow \ \left[(1-\mu ^{2})P_{n}'\right]'+n(n+1)P_{n}=0\end{aligned}}}$

(1)

Where: ${\displaystyle \left[(1-\mu ^{2})P_{n}'\right]\rightarrow \ (1-\mu ^{2})P_{n}''-2\mu \ P_{n}'\ }$

and ${\displaystyle P_{n-1}'=\mu \ P_{n}'-nP_{n}\ }$

Legendre differential equation Eq.(1)p.5-4 and Eq.(1)p.24-1

Gauss-Legendre (GL) quadrature

Important application, e.g. Finite Element

The word quadrature comes from quadrilateral (note use of quad in both words).



Greeks: measure areas, where Area equals sum of quadrilaterals

The word cubature comes from the word cube:

Volume is equal to the sum of cubes

${\displaystyle \displaystyle {\begin{aligned}I(f):=\int _{x=-1}^{x=+1}f(x)\,dx\end{aligned}}}$

(1)

${\displaystyle \displaystyle {\begin{aligned}I_{n}(f):=\sum _{j=1}^{n}w_{j}f(x_{j})\end{aligned}}}$

(2)

with ${\displaystyle \left\{x_{j}\right\}\ }$ being roots of ${\displaystyle P_{n}(x)\Rightarrow \ P_{n}(x_{j})=0\ }$ for ${\displaystyle j=1,...,n\ }$ Where the sub n is defined as degree "n"

${\displaystyle -1<x_{1}<x_{2}<...<x_{n}<+1\ }$

THEOREM :

${\displaystyle \displaystyle {\begin{aligned}I(f)=I_{n}(f)+E_{n}(f)\end{aligned}}}$

(3)

${\displaystyle \displaystyle {\begin{aligned}w_{j}={\frac {-2}{(n+1)P_{n}'(x_{j})P_{n+1}(x_{j})}}\end{aligned}}}$

(4)

${\displaystyle \displaystyle {\begin{aligned}E_{n}(f)={\frac {2^{2n+1}(n!)^{4}}{(2n+1)\left[(2n)!\right]^{2}}}{\frac {f^{(2n)}(y)}{(2n)!}},y\in \left[-1,1\right]\end{aligned}}}$

(5)

END THEOREM

Example: ${\displaystyle n=2\ }$ (2 integration points)

Eq.(4)p.36-2 : ${\displaystyle P_{2}(x)={\frac {1}{2}}(3x^{2}-1)\ }$

${\displaystyle \Rightarrow \ x_{1,2}=\pm {\frac {1}{\sqrt {3}}}\ }$



Eq.(5)p.36-2 : ${\displaystyle P_{2}'(x)=3x,P_{3}(x)={\frac {1}{2}}(5x^{3}-3x)\ }$

${\displaystyle w_{1}={\frac {-2}{(2+1)^{(3)}(-{\frac {1}{\sqrt {3}}}){\frac {1}{2}}\left[5(-{\frac {1}{\sqrt {3}}})^{3}-3(-{\frac {1}{\sqrt {3}}})\right]}}=1\ }$

HW8.2 ${\displaystyle w_{2}=1\ }$ . More generally, verify the table for Gauss Legendre quadrature in wikipedia. Analyze expression for ${\displaystyle \left\{x_{j}\right\}\ }$ and ${\displaystyle \left\{w_{j}\right\}\ }$ , ${\displaystyle j=1,...,n\ }$ and ${\displaystyle n=1,...,6\ }$ ater verifying the expression for ${\displaystyle P_{n}(x)\ }$ ,with ${\displaystyle n=1,...,6\ }$. See Fall 2009 HW. pg.35-2 and HW7.9 P41-2 . Evaluate ${\displaystyle \left\{x_{j}\right\}\ }$ and ${\displaystyle \left\{w_{j}\right\}\ }$ and compare results. END HW8.2

NIST Handbook

Gaussian quadrature

Number of points, n	Points, xi	Weights, wi
1	0	2
2	${\displaystyle \pm 1/{\sqrt {3}}}$	1
3	0	8⁄9
	${\displaystyle \pm {\sqrt {3/5}}}$	5⁄9
4	${\displaystyle \pm {\sqrt {{\Big (}3-2{\sqrt {6/5}}{\Big )}/7}}}$	${\displaystyle {\tfrac {18+{\sqrt {30}}}{36}}}$
	${\displaystyle \pm {\sqrt {{\Big (}3+2{\sqrt {6/5}}{\Big )}/7}}}$	${\displaystyle {\tfrac {18-{\sqrt {30}}}{36}}}$
5	0	128⁄225
	${\displaystyle \pm {\tfrac {1}{3}}{\sqrt {5-2{\sqrt {10/7}}}}}$	${\displaystyle {\tfrac {322+13{\sqrt {70}}}{900}}}$
	${\displaystyle \pm {\tfrac {1}{3}}{\sqrt {5+2{\sqrt {10/7}}}}}$	${\displaystyle {\tfrac {322-13{\sqrt {70}}}{900}}}$

HW8.4 Back to HW7.4 P38-5. Complete the first three non-zero coefficients using Gauss Legendre quadrature up to within 5% accuracy. END HW8.4

see Fall 2009 P34-2

Question: How does Gauss Legendre quadrature compare to other quadratures, e.g. trapezoidal rule?

Answer: Look at ${\displaystyle E_{n}(f)\ }$ Eq.(5) P45-2.

Consider ${\displaystyle f\in \mathrm {P} _{2n-1}\ }$ (set of polynomials of degree less than or equal ${\displaystyle 2n-1\ }$ )

${\displaystyle \Rightarrow \ f^{(2n)}(x)=0\Rightarrow \ E_{n}(f)=0\ }$ i.e. can integrate exactly any polynomial of degree les than or equal to ${\displaystyle 2n-1\ }$ using only n integration points (almost half)

Example: ${\displaystyle 2n-1=3\Rightarrow \ n=2\Rightarrow \ 2n=4\ }$

${\displaystyle f\in \mathrm {P} _{3}\Rightarrow \ f(x)=\sum _{j=0}^{3}c_{j}x^{j}\Rightarrow \ f^{(4)}(x)=0\ }$

Trapezoidal Rule (more details in EGM6341)

${\displaystyle I(f)=\int _{a}^{b}f(x)\,dx\ }$

${\displaystyle h={\frac {b-a}{n}}\ }$

Where n=number of trapezoidal panels



${\displaystyle E_{n}(f)=-{\frac {(b-a)h^{2}}{12}}f^{(2)}(y),y\in \left[a,b\right]\ }$

Trapezoidal rule can only integrate exactly a straight line. Also ${\displaystyle E_{n}(f)\rightarrow \ 0\ }$ as ${\displaystyle h\rightarrow \ 0\ }$ (${\displaystyle \Leftrightarrow n\rightarrow \ \infty \ }$ ), even for a simple polynomial of degree 2 (parabola)