e = fx x hw f +++ w f w f e · ce 30125 - lecture 16 p. 16.3 • assume that for gauss quadrature...
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CE 30125 - Lecture 16
p. 16.1
LECTURE 16
GAUSS QUADRATURE
• In general for Newton-Cotes (equispaced interpolation points/ data points/ integrationpoints/ nodes).
84
• Note that for Newton-Cotes formulae only the weighting coefficients were unknownand the were fixed
f x xd
xS
xE
h w'o fo w'1 f1 w'N fN+ + + E+=
f0
f1f2
fN
h
= x0xs x1 x2 = xNxE
closed formula
wi
xi
CE 30125 - Lecture 16
p. 16.2
• However the number of and placement of the integration points influences the accuracyof the Newton-Cotes formulae:
• even degree interpolation function exactly integrates an degree poly-nomial This is due to the placement of one of the data points.
• odd degree interpolation function exactly integrates an degree polyno-mial.
• Concept: Let’s allow the placement of the integration points to vary such that wefurther increase the degree of the polynomial we can integrate exactly for a givennumber of integration points.
• In fact we can integrate an degree polynomial exactly with only integra-tion points
N Nth N 1+ th
N Nth Nth
2N 1+ N 1+
CE 30125 - Lecture 16
p. 16.3
• Assume that for Gauss Quadrature the form of the integration rule is
85
• In deriving (not applying) these integration formulae
• Location of the integration points, are unknown
• Integration formulae weights, are unknown
• unknowns we will be able to exactly integrate any degree polyno-mial!
f x xdxS
xE
wo fo w1 f1 wN fN+ + + E+=
f0
f1
f2
fN
x0xs x1 x2 xN xE
f3
x3
xi i O N=
wi i O N=
2 N 1+ 2N 1+
CE 30125 - Lecture 16
p. 16.4
Derivation of Gauss Quadrature by Integrating Exact Polynomials and Matching
Derive 1 point Gauss-Quadrature
• 2 unknowns , which will exactly integrate any linear function
• Let the general polynomial be
where the coefficients can equal any value
• Also consider the integration interval to be such that and (noloss in generality since we can always transform coordinates).
• Substituting in the form of
wo xo
f x Ax B+=
A B
1 + 1– xS 1–= xE + 1=
f x xd
1–
+ 1
wo f xo =
f x
Ax B+ xd
1–
+ 1
wo Axo B+ =
CE 30125 - Lecture 16
p. 16.5
• In order for this to be true for any 1st degree polynomial (i.e. any and ).
• Therefore , for 1 point Gauss Quadrature. 86
• We can integrate exactly with only 1 point for a linear function while for Newton-Coteswe needed two points!
Ax2
2----- Bx+
+ 1
1–wo Axo B+ =
A 0 B 2 + A xowo B wo +=
A B
0 xowo=
2 wo =
xo 0= wo 2= N 1=
f0f(x)
x0-1 +1
CE 30125 - Lecture 16
p. 16.6
Derive a 2 point Gauss Quadrature Formula 87
• The general form of the integration formula is
• , , , are all unknowns
• 4 unknowns we can fit a 3rd degree polynomial exactly
• Substituting in for into the general form of the integration rule
x0 x1-1 +1
I wo fo w1 f1+=
wo xo w1 x1
f x Ax3 Bx2 Cx D+ + +=
f x
f x xd
1–
+ 1
wo f xo w1 f x1 +=
CE 30125 - Lecture 16
p. 16.7
• In order for this to be true for any third degree polynomial (i.e. all arbitrary coefficients,, , , ), we must have:
Ax3 Bx2 Cx D+ + + xd
1–
+ 1
wo Axo3
Bx2o Cxo D+ + + w1 Ax3
1 Bx21 Cx1 D+ + + +=
Ax4
4---------
Bx3
3---------
Cx2
2--------- Dx+ + +
+ 1
1–wo Ax3
o Bx2o Cxo D+ + + w1 Ax3
1 Bx21 Cx1 D+ + + +=
A wox3o w1x3
1+ B wox2o w1x2
123---–+ C woxo w1x1+ D wo w1 2–+ + + + 0=
A B C D
wox3o w1x3
1+ 0=
wox2o w1x2
123---–+ 0=
woxo w1x1+ 0=
wo w1 2–+ 0=
CE 30125 - Lecture 16
p. 16.8
• 4 nonlinear equations 4 unknowns
and
and
• All polynomials of degree 3 or less will be exactly integrated with a Gauss-Legendre 2point formula.
wo 1= w1 1=
xo13---–= x1 + 1
3---=
CE 30125 - Lecture 16
p. 16.9
Gauss Legendre Formulae
,
0 1 0 2 1
1 2,
1, 1 3
2 3 -0.774597, 0, +0.774597
0.5555, 0.8889, 0.5555
5
I f x xd
1–
+1
wi fi
i 0=
N
E+= =
N N 1+xi
i 0 N=wi
Exact forpolynomials of
degree
13---– +
13---
N N 1+ 2N 1+
CE 30125 - Lecture 16
p. 16.10
,
3 4 -0.86113631-0.33998104 0.33998104 0.86113631
0.347854850.652145150.652145150.34785485
7
4 5 -0.90617985-0.53846931 0.00000000 0.53846931 0.90617985
0.236926890.478628670.568888890.478628670.23692689
9
5 6 -0.93246951-0.66120939-0.23861919 0.23861919 0.66120939 0.93246951
0.171324490.360761570.467913930.467913930.360761570.17132449
11
N N 1+xi
i 0 N=wi
Exact forpolynomials of
degree
CE 30125 - Lecture 16
p. 16.11
• Notes
• = the number of integration points
• Integration points are symmetrical on
• Formulae can be applied on any interval using a coordinatetransformation
• integration points will integrate polynomials of up to degree exactly.
• Recall that Newton Cotes integration points only integrates an degree polynomial exactly depending on being odd or even.
• For Gauss-Legendre integration, we allowed both weights and integration pointlocations to vary to match an integral exactly more d.o.f. allows you tomatch a higher degree polynomial!
• An alternative way of looking at Gauss-Legendre integration formulae is that weuse Hermite interpolation instead of Lagrange interpolation! (How can this besince Hermite interpolation involves derivatives let’s examine this!)
N 1+
1 +1–
N 1+ 2N 1+
N 1+
Nth/N 1th+ N
CE 30125 - Lecture 16
p. 16.12
Derivation of Gauss Quadrature by Integrating Hermite Interpolating Functions
Hermite interpolation formulae
• Hermite interpolation which matches the function and the first derivative at inter-polation points is expressed as:
88
• It can be shown that in general for non-equispaced points
N 1+
g x i x fi
i 0=
N
i x fi1
i 0=
N
+=
f0 f1 f2 fNf4
x0 x1 x2 x3
f0(1)
f1 f2(1)
f3 f3(1)
x4 x5
f4(1)
f5 f5(1)
fN(1)
xN
0 1 2 3 4 5 N
(1)
i x ti x liN x liN x = i 0 N=
i x si x liN x liN x = i 0 N=
CE 30125 - Lecture 16
p. 16.13
where
pN x x xo– x x1– x xN–
liN x p
Nx
x xi– p 1 N
xi --------------------------------------- i 0 N=
ti x 1 x xi– 2 liN1
xi –
si x x xi–
CE 30125 - Lecture 16
p. 16.14
Example of defining a cubic Hermite interpolating function
• Derive Hermite interpolating functions for 2 interpolation points located at and for the interval .
89
points
• Establish
1– + 11 + 1–
f0 f1
x0 = 1 x1 = +1
f0 (1)f1
(1)
x
N 1+ 2= N 1=
pN x
p1
x x xo– x x1– =
p 1 1
x x xo– x x1– +=
CE 30125 - Lecture 16
p. 16.15
• Establish
• Let
• Substitute in and
liN x
li1 x p1 x
x xi– p11
xi -----------------------------------------=
li1 x x xo– x x1–
x xi– xi xo– xi x1– + ---------------------------------------------------------------------=
i 0=
lo1 x x xo– x x1–
x xo– 0 xo x1– + ------------------------------------------------------=
lo1 x x x1–
xo x1–----------------=
xo 1–= x1 + 1 =
lo1 x 12--- 1 x– =
CE 30125 - Lecture 16
p. 16.16
• Let
• Substitute in values for ,
• Taking derivatives
i 1=
l11 x x xo– x x1–
x x1– x1 xo– 0+ ------------------------------------------------------=
xo x1
l11 x 12--- 1 x+ =
l1
o1x
12---–=
l1
11x +
12----=
CE 30125 - Lecture 16
p. 16.17
• Establish
• Establish
ti x
to x 1 x xo– 2l1
o1 xo –=
to x 1 x 1+ 2 – 12---–
=
to x 2 x+=
t1 x 1 x x1– 2 l1
11 x1 –=
t1 x 1 x 1– 212---
–=
t1 x 2 x–=
si x
so x x 1+=
s1 x x 1–=
CE 30125 - Lecture 16
p. 16.18
• Establish
i x
o x to x lo1 x lo1 x =
o x 2 x+ 12--- 1 x– 1
2--- 1 x– =
o x 14--- 2 3x– x3+ =
1 x t1 x l11 x l11 x =
1 x 2 x– 12--- 1 x+ 1
2--- 1 x+ =
1 x 14--- 2 3x x3–+ =
CE 30125 - Lecture 16
p. 16.19
• Establish
• In general
i x
o x so x lo1 x lo1 x =
o x x 1+ 12--- 1 x– 1
2--- 1 x– =
o x 14--- 1 x– x2– x3+ =
1 x s1 x l11 x l11 x =
1 x x 1– 12--- 1 x+ 1
2--- 1 x+ =
1 x 14--- 1– x– x2 x3+ + =
g x o x fo 1 x f1 o x f 1 o
1 x f 1 1
+ + +=
CE 30125 - Lecture 16
p. 16.20
90
• These functions satisfy the constraints
1
i(x)
1(x)0(x)
x
i(x)
0(x)
1(x)
x
i (x)
1 (x)
0 (x)
(1)
(1)
(1)
x
i (x)(1)
0 (1) (1)
x
i xj ij= 1 i
xj 0=
i xj 0= 1 i xj ij=
CE 30125 - Lecture 16
p. 16.21
Gauss-Legendre Quadrature by integrating Hermite interpolating polynomials
• Notes
• Use without loss of generality we can always transform the interval.
• Approximation for is exact for degree polynomials
• We can derive all Gauss-Legendre quadrature formulae by approximating with an degree Hermite interpolating function using specially selected integration/
interpolation points.
where
I f x xd
1–
+ 1
wi fi E+
i 0=
N
= =
1 +1–
I 2N 1+
f x 2N 1th+ N
I g x x E+d
1–
+ 1
=
g x i x fi
i 0=
N
i x fi1
i 0=
N
+=
CE 30125 - Lecture 16
p. 16.22
• Thus
where
and
• Furthermore we can show that
I i x fi
i 0=
N
i x fi1
i 0=
N
+
1–
+ 1
dx= E+
I Ai fi
i 0=
N
Bi fi1
i 0=
N
E+ +=
Ai i x xd
1–
+ 1
Bi i x xd
1–
+ 1
EpN 1+
2x
2N 2+ !----------------------- f
2N 2+ xo + H.O.T. xd
1–
+ 1
=
CE 30125 - Lecture 16
p. 16.23
• Note that we are assuming Taylor series expansions about and using higher orderterms in the expansion.
• Therefore for any polynomial of degree or less!
• The problem that we encounter is that the integration formula as it now stands ingeneral requires us to know both functional and first derivative values at the nodes!
• Let us select such that
xo
E 0= 2N 1+
xo x1 x2 xN
Bi 0= i 0 N=
i x xd
1–
+ 1
0= i 0 N=
si x liN x liN x xd
1–
+ 1
0= i 0 N=
x xi– pN x
x xi– pN1
xi ------------------------------------ liN x xd
1–
+ 1
0= i 0 N=
CE 30125 - Lecture 16
p. 16.24
polynomial of degree
polynomial of degree
• Therefore we require to be orthogonal on to all polynomials of degree or less any multiple of Legendre-Polynomials will satisfy this.
• Let
where
= the Legendre polynomial of degree
is required to normalize the leading coefficient of
1
pN1
xi ------------------ pN x liN x
1–
+ 1
dx 0= i 0 N=
pN x N 1+
liN x N
pN x 1 +1– N
pN x 2N 1+ N 1+ ! 2
2 N 1+ !-----------------------------------------PN 1+ x =
pN x x xo– x x1– x x2– x xN– =
PN 1+ N 1+
2N 1+ N 1+ ! 2
2 N 1+ !----------------------------------------- PN 1+ x
CE 30125 - Lecture 16
p. 16.25
• What have we done by defining in this way we have selected the integration/interpolation/data points to be the roots of .
• In general
.
.
.
pN x xo x1 xN PN 1+ x
Pn x 12nn!----------
dn x2 1– n
dxn---------------------------=
Po x 1=
P1 x x=
P2 x 12--- 3x2 1– =
P3 x 12--- 5x3 3x– =
CE 30125 - Lecture 16
p. 16.26
• So far we have established
• Selecting to be proportional to the Legendre Polynomial of degree this satisfies the orthogonality condition which will lead to:
As a result terms will not appear in the Gauss-Legendre integration formula.
• If we select to be the Legendre Polynomial of degree the roots ofthat polynomial will represent the interpolating/integration/data points since
has been set equal to
• Now we must find the weights of the integration formula. Note that will represent theweights!
pN x N 1+
i x xd
1–
+ 1
0=
f 1 i
pN x N 1+
pN x x xo– x x1– x xN– = CPN 1+ x
Ai
Ai i x xd
1–
+ 1
CE 30125 - Lecture 16
p. 16.27
where
and where are the roots of the Legendre polynomial of degree or
Ai ti x liN x liN x 1–
+ 1
= dx
ti x 1 x xi– 2 liN1
xi –=
liN x pN x
x xi– pN1
xi ------------------------------------=
pN x x xo– x xN– =
xo xN N 1+
pN x 2N 1+ N 1+ ! 2
2 N 1+ !-----------------------------------------PN 1+ x =
CE 30125 - Lecture 16
p. 16.28
Two point Gauss-Legendre integration
Develop a 2 point Gauss-Legendre integration formula for . Let
• Thus
1 +1–
g x i x fi
j 0=
1
i x fi1
j 0=
1
+=
g x o x fo 1 x f1 o x fo1 1 x f1
1 + + +=
I g x x E+d
1–
+ 1
=
I o x fo xd
1–
+1
1 x f1 xd
1–
+1
o x fo1
xd
1–
+1
1 x f11
xd
1–
+1
+ + +=
I fo o x xd
1–
+1
f1 1 x xd
1–
+1
fo1 o x xd
1–
+1
f11 1 x xd
1–
+1
+ + +=
CE 30125 - Lecture 16
p. 16.29
Step 1 - Establish interpolating points
• Interpolation points will be the roots of the Legendre Polynomial of order 2.
P2 x 12--- 3x2 1– =
12--- 3x2 1– 0=
3x2 1=
x2 13---=
x0 113---=
x0 1 0.57735=
CE 30125 - Lecture 16
p. 16.30
• Checking these roots
P2 x 1222!----------
d2 x2 1– 2
dx2---------------------------=
P2 x 18---
d2
dx2-------- x4 2x2– 1+ =
P2 x 18--- 12x2 4– =
P2 x 12--- 3x2 1– =
p1 x 22 2! 2
2 2 !------------------ P2 x =
p1 x 4 44 3 2 ------------------
12--- 3x2 1– =
p1 x x2 13---–=
CE 30125 - Lecture 16
p. 16.31
• From formula which defines using the integration points
Step 2 - Establish the coefficients of the derivative terms in the integration formula
• Let’s demonstrate that with the roots we will satisfy
and
• First develop and by developing
p1 x
p1 x x13---+
x13---–
x2 13---–= =
xo 1 0.57735=
o x 1–
+ 1
dx 0= 1 x xd
1–
+ 1
0=
o x 1 x p1 x , p11
x , lo1 x , l11 x , so x and s1 x
p1 x x xo– x x1– =
p11
x x xo– x x1– +=
CE 30125 - Lecture 16
p. 16.32
lj1 x p1 x
x xj– p 1 1
xj ---------------------------------------= j 0 1=
lj1 x x xo– x x1–
x xj– xj xo– xj x1– + ---------------------------------------------------------------------=
lo1 x x xo– x x1–
x xo– xo xo– xo x1–+ --------------------------------------------------------------=
lo1 x x x1–
xo x1–----------------=
l11 x x xo– x x1–
x x1– x1 xo– x1 x1– + ------------------------------------------------------------------------=
l11 x x xo–
x1 xo–----------------=
CE 30125 - Lecture 16
p. 16.33
• Now we can establish
• Noting that ,
so x x xo–=
s1 x x x1–=
o x
o x so x lo1 x lo1 x =
o x x xo– x x1–
xo x1–---------------- x x1–
xo x1–---------------- =
xo13---–= x1
13---=
o x x13---+
=x
13---–
x13---–
13---–
13---–
2
-------------------------------------------
o x 34--- x3 1
3---x2 1
3---x–
13--- 3 2/
+–=
CE 30125 - Lecture 16
p. 16.34
• Similarly for
• Substituting ,
1 x
1 x s1 x l11 x l11 x =
1 x x x1– x xo– x1 xo–
---------------------x xo– x1 xo–
---------------------=
xo13---–= x1
13---=
1 x x
13---–
x13---+
x13---+
13---
13---+
2
------------------------------------------------------------------=
1 x 34--- x3 1
3---x2 1
3---x–
13--- 3 2/
–+=
CE 30125 - Lecture 16
p. 16.35
• Now we can develop
o x xd1–
+1
o x xd1–
+1
34--- x3 1
3---x2–
13---x–
13--- 3 2/
+ xd1–
+1
=
o x xd1–
+1
34---
x4
4-----
13--- 3 2/
x3–16---x2–
13--- 3 2/
x++1
1–=
o x xd1–
+1
34---
14---
13--- 3 2/
–16---–
13--- 3 2/
+ 1
4---
13--- 3 2/ 1
6---–
13--- 3 2/
–+ –=
o x xd1–
+1
0=
CE 30125 - Lecture 16
p. 16.36
• Develop
1 x xd1–
+1
1 x xd1–
+1
34--- x3 1
3---x2 1
3---x–
13--- 3 2/
–+ xd1–
+1
=
1 x xd1–
+1
34---
x4
4-----
13--- 3 2/
x3 16---x2–
13--- 3 2/
x–++1
1–=
1 x xd1–
+1
34---
14---
13--- 3 2/ 1
6---–
13--- 3 2/
–+ 1
4---
13--- 3 2/
–16---–
13--- 3 2/
+ –=
1 x xd1–
+1
0=
CE 30125 - Lecture 16
p. 16.37
• Now our integration formula reduces to:
where
and
Step 3 - Develop ,
• Establish
I fo o x x f1 1 x xd1–
+1
+d1–
+1
=
I Aofo A1f1+=
Ao o x xd1–
+1
A1 1 x xd1–
+1
Ao A1
o x
o x to x lo1 x lo1 x =
o x 1 x xo– 2lo11
xo – lo1 x lo1 x =
CE 30125 - Lecture 16
p. 16.38
o x 1 x xo– 2xo x1–----------------–
x x1–
xo x1–---------------- x x1–
xo x1–---------------- =
o x 1 x13---+
2
13---–
13---–
-------------------------–
x
13---–
x13---–
13---–
13---–
13---–
13---–
--------------------------------------------------------------=
o x 34--- 1 x
13---+
2
213---–
--------------
–
x2 213---x–
13---+
=
o x 3 34
----------2
3------- x+
x2 213---x–
13---+
=
o x 34--- 3 x3 x– 2
13--- 3 2/
+
=
CE 30125 - Lecture 16
p. 16.39
• Develop
o x xd1–
+1
o x xd1–
+1
34--- 3 x3 x– 2
13--- 3 2/
+
1–
+1
dx=
o x xd1–
+1
34--- 3
x4
4-----
x2
2-----– 2
13--- 3 2/
+ x+1
1–=
o x xd1–
+1
34--- 3=
14---
12---– 2
13--- 3 2/
+ 1
4---
12---– 2
13--- 3 2/
– –
o x xd1–
+1
34--- 3 4
13---
13---
=
o x xd1–
+1
1=
Ao 1=
CE 30125 - Lecture 16
p. 16.40
• Similarly we can show that
• Thus we have established the two point Gauss Quadrature rule
where and are the integration points and
• We note that this integration rule was established by defining a Hermite cubic interpo-lating function and defining the integration points , such that
and
• Therefore the functional derivative values drop out of the Gauss Legendre integra-tion formula!
A1 1 x xd1–
+1
1= =
I f x xd1–
+1
wo fo w1 f1+= =
xo13---–= x1 +
13---= wo w1 1= =
xo x1
o x xd1–
+1
0= 1 x xd1–
+1
0=