Numerical integration in more dimensions – part 2 · 16/10/2002 Seminar: ... Gauss approach in...

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Numerical integration in more dimensions – part 2 Remo Minero

Transcript of Numerical integration in more dimensions – part 2 · 16/10/2002 Seminar: ... Gauss approach in...

Numerical integration in more dimensions – part 2

Remo Minero

16/10/2002 Seminar: Numerical Integration in more dimensions 2

Outline

The role of a mapping function inmultidimensional integration

Gauss approach in more dimensions and quadrature rules

Critical analysis of acceptability of a given quadrature rule

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Problem definition

! We have and we want to compute I:ℜ→ℜ⊂Ω nf :

! We want to implement some numerical method, which ought to be (as usual) accurate and cheap (e.g. small number of operations)

∫Ω= xdfI

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Covering

! Step 1: covering of the domain with replicas of a basic geometry

Ω

( ) ( )∑∫∫ ≅Ω

lel

dfdf xxxx

Error 1: covering error

el

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Mapping function

! Step 2: introduction of a mapping function F" Example 1: squares

( ) ( ) ( ) ( )[ ] ( )

ll

l0

l0

l2

l0

l1l

Ay

x

y

xF

bx

vvvvv

+=

+

−−=

ˆˆ

ˆ|

ˆ

ˆ

x0

el

x10

y1

eFl

( )l2v ( )l

3vy

el( )l0v

( )l1v

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Mapping function

x0

y

( )l1v

el

" example 2: triangles

( ) ( ) ( ) ( )[ ] ( )

ll

l0

l0

l2

l0

l1l

Ay

x

y

xF

bx

vvvvv

+=

+

−−=

ˆˆ

ˆ|

ˆ

ˆ

Fl

x10

y1

e

( )l2v

el( )l0v

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Properties of the mapping function F

! Fl is affine: ( ) lll AF bxx += ˆˆlinear

constant

∑ ∑ =i i

iii 1x αα ,

F maps affine combinations

to affine combinations

Fl

Fl

e

e el

eltriangles are mapped to triangles, rectangles to parallelograms, etc.

…that is,

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The role of the mapping function

( ) ( )∑∫∫ ≅Ω

lel

dfdf xxxx

( ) ( )( ) ( )

( ) ( )( )∫∫ ∫

⋅=

=∂=

ell

ee

ll

dFfA

dFFfdfl

ˆ

ˆ

ˆˆdet

ˆdetˆ

xx

xxxx

! Step 1:

! Step 2:

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Quadrature

! Step 3: integration over the basic geometry

( )( ) ( ) ( )∑∫∫ ≅=i

iie

e l gwdgdFf xxxxx ˆˆˆˆˆˆ

ˆ

Error 2: quadrature error

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Integration over a surface

! Suppose we have a function defined over a surface

! Thanks to the properties of the mapping function, we can use the same approach:

( ) ( ) ( ) ( )[ ] ( )ll

l0

l0

l2

l0

l1l A

y

x

y

xF bxvvvvv +=+

−−=

ˆ

ˆ

ˆ|

ˆ

ˆ

…simply v(l) given in a suitable reference system…

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How can one reduce errors?

Covering error Quadrature error! More accurate formulas! Smaller volumes (where

necessary, depending on f)Ω

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Open problem

Given a basic geometry, find the least amount of points and weights such that

is exact for all monomials of degree d and lower

" Let’s look at some examples…

( ) ( )xxx ˆˆˆˆ

∑∫ ≅i

ie

gwdg

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d=1 in the square

3 equations # 1 point, 1 weight

Mathematical problem - Physical interpretation

1

x y

1

1

0

y

x0.5

0.5

==

==

==

e11

e11

e1

yw21dydxy

xw21dydxx

w1dydx1

ˆ

ˆ

ˆ

ˆ

ˆ 1w1 =ˆ

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d=1 in the triangle

3 equations # 1 point, 1 weight

Mathematical problem - Physical interpretation

1

x y

1

1

0

y

x

21w1 =ˆ

31

31

==

==

==

e11

e11

e1

yw61dydxy

xw61dydxx

w21dydx1

ˆ

ˆ

ˆ

ˆ

ˆ

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d=2 in the square

6 equations # 2 points, 2 weights

Mathematical problem - Physical interpretation

1

x y

x2 y2xy

x

y

1

1

0

+==

+==

+==

+==

+==

+==

e222111

e

222

211

2

e

222

211

2

e2211

e2211

e21

yxwyxw41dydxxy

ywyw31dydxy

xwxw31dydxx

ywyw21dydxy

xwxw21dydxx

ww1dydx1

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆˆˆˆ

ˆˆ

ˆˆ

ˆˆ

ˆˆ

...no Mathematica solution…...no Mathematica solution…

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d=3 (e.g. in the triangle)

10 equations # how many points?

3 points # too many equations?4 points # not enough

1

x y

x2 y2xy

x3 y3x2y xy2

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Biquadratic polynomials for the square

! Let’s choose two monomials p(x) and q(y) and let them be of degree d at most.

! If we choose d=3

! 16 equations # 4 points, 4 weightsx3

x3y

x3y2

x3y3x2y3xy3y3

x2x1

x2yxyy

x2y2xy2y2

...no Mathematica solution (in a reasonable time)…...no Mathematica solution (in a reasonable time)…

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Cross product Gauss

Gauss 1D in [0,1] Gauss 2D in [0,1]2

0 1

0

1

0 1

$ only for domains like [a,b]n

( ) ( ) ( ) ( )∑ ∫ ∑∑∫∫ ∫∑ ===i i j

jijiiii

ii xxgwwdyyxgwdyyxgwdxdyyxg ˆ,ˆ,ˆ,ˆ,

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Higher degree formulas

! Many of them in the literature

" Example 1: degree 6 in the triangle with 12 points

" Example 2:degree 20 in the triangle with 79 points!

D.A.Dunuvant, HIGH DEGREE EFFICIENT SYMMETRICAL GAUSSIAN QUADRATURE RULES FOR THE TRIANGLE, International Journal for Numerical Methods in Engineering, vol. 21, 1129-1148 (1985)

A.H. Stroud & D. Secrest, GAUSSIAN QUADRATURE FORMULA, Prentice-Hall, 1966

x10

y

1

1

25

4

3

67 8

9

10 1112

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Summary

Do the number of the unknows

correspond to the number of the

equations?

Does the non-linear system have

a solution?

Is the found solution

acceptable?

WE HAVE AN ACCEPTABLE QUADRATURE

METHOD!

Condition 1: Are all the points inside the element? Condition 2: Are all the weights wi positive?

ix

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Condition 1: ∀ iexi ˆˆ ∈

4x

1x

2x3x

F

∂Ω( )4F x

( )3F x( )1F x

( )2F x

Ω

What is the value of if does not belong to Ω?( )( )4Ff x ( )4F x

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Condition 2: wi ≥ 0, ∀ i

! To always have non-negative integrals for non-negative functions

0>∫ dxfb

a

( ) 07

1<∑

=i

ii xfw

Exact solution:

Approximation: if w2< 0

b

y

x0 a

w4 w6

f(x)

w5w3w2w1 w7

! Finite Elements Methods (FEM)

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Why weights always ≥ 0 in FEM

! Stiffness matrix A is positive definite

! Positive definite property is needed for the iterative solvers of Krylov type (all fast iterative solvers)

! Negative weights might cause positive definite property to be lost

0022>>+∇=== ∫∑∑ uifuuuau jij

i ji

T …Auu

∑=i

iiuu ϕ ( ) xda ijijij ∫ +∇∇= ϕϕϕϕ[ ]n1 uu ,,…=u

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Do we really get negative weights?

! Newton-Cotes approach in [-1,1]:

! Negative weights also in many formula using with Gauss approach

299376427368

299376260550

299376272400

29937648525

299376106300

2993761606711

141754540

1417510496

14175928

141755888

141759899

34

313211

AAAAAAn 654321

−−

−−

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Covering

" Example 2: triangles # more flexible

Ω