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Gauss Quadrature-Numerical Methods in Engineering-Lecture 19 Slides-Civil Engineering and Geological Sciences, Slides of Numerical Methods in Engineering

University of Notre Dame du Lac (ND)Numerical Methods in Engineering

Concept: Let’s allow the placement of the integration points to vary such that we further increase the degree of the polynomial we can integrate exactly for a given number of integration points. Gauss Quadrature, Gauss Legendre Formulae, Hermite Interpolation Formulae, Cubic Hermite Interpolating Function, Two Point Gauss Legendre Integration, Newton Cotes, Integration Points

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CE 341/441 - Lecture 19 - Fall 2004
p. 19.1
LECTURE 19
GAUSS QUADRATURE
• In general for Newton-Cotes (equispaced interpolation points/ data points/ integration
points/ nodes).
84
• Note that for Newton-Cotes formulae only the weighting coefficients were unknown
and the were fixed
fx()xd
xS
xE
∫hw'
ofow'
1f1…w'
NfN
+++[]E+=
f0
f1f2
fN
h
= x0
xsx1x2= xN
xE
closed formula
wi
xi
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a
pf1b
pf1c
pf1d
pf1e
pf1f
pf20
pf21
pf22
pf23
pf24
pf25
pf26
pf27
pf28

Partial preview of the text

Download Gauss Quadrature-Numerical Methods in Engineering-Lecture 19 Slides-Civil Engineering and Geological Sciences and more Slides Numerical Methods in Engineering in PDF only on Docsity!

CE 341/441 - Lecture 19 - Fall 2004

p. 19.

LECTURE 19GAUSS QUADRATURE • In general for Newton-Cotes (equispaced interpolation points/ data points/ integration

points/ nodes).

84 • Note that for Newton-Cotes formulae only the weighting coefficients

were unknown

and the

were fixed

f

x (

x d

x

S x

E ∫^

h w

' o^

f

o

w

f

^1

w

' N

f

N

[

]

E

f

0

f

1

f

2

f

N

h

= x

0

x

s

x

1

x

2

= x

N

x

E

closed formula

w

i

x

i

CE 341/441 - Lecture 19 - Fall 2004

p. 19.

  • However the number of and placement of the integration points influences the accuracy

of the Newton-Cotes formulae:

even

→

degree interpolation function exactly integrates an

degree

polynomial

→

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 we

further 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

N

th

N

th

N

N

th

N

th

N

N

CE 341/441 - Lecture 19 - Fall 2004

p. 19.

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

(no

loss in generality since we can always transform coordinates).

  • Substituting in the form of

⇒

w

o

x

o

f

x (

Ax

B

A B

[

]

x

S

x

E

f

x (

x d

1

+ 1^ ∫ –

w

o

f

x

o

(

f

x (

Ax

B

x d

1

+ 1^ ∫ –

w

o

Ax

o

B

CE 341/441 - Lecture 19 - Fall 2004

p. 19.

⇒

  • 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-Cotes

we needed two points!

A

x

2 ----^2

Bx

  • 1

1

w

o

Ax

o

B

A

B

A x

o

w

o

B w

o

A

B

x

o

w

o

w

o

x

o

w

o

N

f^0

f(x)

x

0

-

+

CE 341/441 - Lecture 19 - Fall 2004

p. 19.

⇒

⇒

  • In order for this to be true for

any

third degree polynomial (i.e. all arbitrary coefficients,

), we must have:

Ax

3

Bx

2

Cx

D

[

]

x d

1

+ 1^ ∫ –

w

o

Ax

(^3) o

Bx

2 o

C x

o

D

[

]

w

1

Ax

31

Bx

21

C x

1

D

[

]

Ax

4

4

Bx

3 3

C x

2

2

Dx

  • 1

1

w

o

Ax

3 o

Bx

2 o

C x

o

D

w

1

Ax

(^31)

Bx

21

C x

1

D

A w

o

x

3 o

w

1

x

31

[

]

B w

o

x

2 o

w

1

x

21

C w

o

x

o

w

1

x

1

[

]

D w

o

w

1

[

]

A

B

C

D

w

o

x

3 o

w

1

x

31

w

o

x

2 o

w

1

x

21

w

o

x

o

w

1

x

1

w

o

w

1

CE 341/441 - Lecture 19 - Fall 2004

p. 19.

  • 4 nonlinear equations

→

4 unknowns

and

and

  • All polynomials of degree 3 or less will be

exactly

integrated with a Gauss-Legendre 2

point formula.

w

o

w

1

x

o

x

1

CE 341/441 - Lecture 19 - Fall 2004

p. 19.

,

N

N

x

i

i

N

w

i

Exact for

polynomials of

degree

CE 341/441 - Lecture 19 - Fall 2004

p. 19.

  • Notes

= the number of integration points

  • Integration points are symmetrical on• Formulae can be applied on any interval using a coordinate transformation•

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 point

locations to vary to match an integral exactly

⇒

more d.o.f.

⇒

allows you to

match a higher degree polynomial!

  • An alternative way of looking at Gauss-Legendre integration formulae is that we

use Hermite interpolation instead of Lagrange interpolation! (How can this besince Hermite interpolation involves derivatives

→

let’s examine this!)

N

[

]

N

N

N

N

th

N

th

N

CE 341/441 - Lecture 19 - Fall 2004

p. 19.

where

p

N

x (

x

x

o

x

x

1

x

x

N

l

iN

x (

p

N

x (

x

x

i

p

1 (

)

N

x

i

(

i

N

t

i

x (

x

x

i

l

iN

1 (

)

x

i

(

s

i

x (

x

x

i

CE 341/441 - Lecture 19 - Fall 2004

p. 19.

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

⇒

[

]

f

0

f

1

x

0

=

−

1

x

1

= +

f

0

(1) f

1

(1)

x

N

N

p

N

x (

p

1

x (

x

x

o

x

x

1

p

1 (

)

1

x (

x

x

o

x

x

1

CE 341/441 - Lecture 19 - Fall 2004

p. 19.

  • Let • Substitute in values for
  • Taking derivatives

i

l

11

x (

x

x

o

x

x

1

x

x

1

x

1

x

o

[

]

x

o

x

1

l

11

x (

x

l

1 (

)

o

1

x (

l

1 (

)

11

x (

CE 341/441 - Lecture 19 - Fall 2004

p. 19.

  • Establish

⇒

⇒

⇒

⇒

  • Establish

t

i

x (

t

o

x (

x

x

o

l

1 (

)

o

1

x

o

(

t

o

x (

x

t

o

x (

x

t

1

x (

x

x

1

l

1 (

)

11

x

1

t

1

x (

x

t

1

x (

x

s

i

x (

s

o

x (

x

s

1

x (

x

CE 341/441 - Lecture 19 - Fall 2004

p. 19.

  • Establish

⇒

⇒

⇒

⇒

  • In general

β

i

x (

β

o

x (

s

o

x (

l

o

1

x (

l

o

1

x (

β

o

x (

x

x

x

β

o

x (

x

x

2

x

3

β

1

x (

s

1

x (

l

11

x (

l

11

x (

β

1

x (

x

x

x

β

1

x (

x

x

2

x

3

g x

α

o

x (

f

o

α

1

x (

f

^1

β

o

x (

f

1 (

)

o

β

1

x (

f

1 (

)

1

CE 341/441 - Lecture 19 - Fall 2004

p. 19.

90 • These functions satisfy the constraints

β

1

α

(x)i

α

1

(x)

α

0

(x)

x

β

(x)i

β

0

(x)

β

1

(x)

x

α

i

(x)

α

1

(x)

α

0

(x)

(1)

(1)

(1)

x

β

i

(x) (1)

β

0

(1)

(1)

x

α

i^

x

j

(

δ

ij

α

1 (

)

i^

x

j

(

β

i

x

j

(

β

1 (

)

i^

x

j

(

δ

ij