08 finite elements basis functions

8/2/2019 08 Finite Elements Basis Functions

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1Finite element method basis functions

Finite Elements: Basis functions

1-D elements

coordinate transformation 1-D elements

linear basis functions quadratic basis functions cubic basis functions

2-D elements

coordinate transformation triangular elements

linear basis functions quadratic basis functions

rectangular elements linear basis functions quadratic basis functions

Scope: Understand the origin and shape of basis functions used in classical

finite element techniques.


2/20


1-D elements: coordinate transformation

We wish to approximate a function u(x) defined inan interval [a,b] by some set of basis functions

==

n

i

iicxu1

)(

where i is the number of grid points (the edges ofour elements) defined at locations xi. As the basis

functions look the same in all elements (apart fromsome constant) we make life easier by moving to alocal coordinate system

ii

i

xx

xx

=+1

so that the element is defined for x=[0,1].


3/20


1-D elements linear basis functions

There is not much choice for the shape of a(straight) 1-D element! Notably the length can varyacross the domain.

We require that our function u() be approximatedlocally by the linear function

21)( ccu +=

Our node points are defined at 1,2=0,1 and werequire that

212

11

212

11

uuc

uc

ccu

cu

+=

=

+=

=Auc =

= 11-

01

A


4/20


1-D elements linear basis functions

As we have expressed the coefficients ci as afunction of the function values at node points 1,2we can now express the approximate functionusing the node values

)()(

)1(

)()(

211

21

211

NNu

uu

uuuu

+=

+=++=

.. and N1,2(x) are the linearbasis functions for 1-Delements.


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1-D quadratic elements

Now we require that our function u(x) beapproximated locally by the quadratic function

2

321)( cccu ++=

Our node points are defined at 1,2,3=0,1/2,1 andwe require that

3213

3212

11

25.05.0

cccu

cccu

cu

++=

++=

=

Auc =

=

242

143

001

A


6/206Finite element method basis functions

1-D quadratic basis functions

... again we can now express our approximatedfunction as a sum over our basis functionsweighted by the values at three node points

... note that now we reusing three grid pointsper element ...

Can we approximate aconstant function?

=

=

++++=++=3

1

2

3

2

2

2

1

2

321

)(

)2()44()231()(

i

iiNu

uuucccu



1-D cubic basis functions

... using similar arguments the cubic basisfunctions can be derived as

32

4

32

3

32

2

32

1

3

4

2

321

)(

23)(

2)(

231)(

)(

+=

=

+=+=

+++=

N

N

N

N

ccccu

... note that here weneed derivative

information at theboundaries ...

How can we

approximate a constantfunction?




Let us now discuss the geometry and basisfunctions of 2-D elements, again we want toconsider the problems in a local coordinatesystem, first we look at triangles

P3

P2

P1

x

y

P3

P2P1

before after




Any triangle with corners Pi(xi,yi), i=1,2,3 can betransformed into a rectangular, equilateral trianglewith

P3

P2

P1

P1 (0,0)

P3 (0,1)

P2 (1,0)

)()(

)()(

13121

13121

yyyyyy

xxxxxx

++=

++=

using counterclockwise numbering. Note that if

=0, then these equations are equivalent to the 1-D tranformations. We seek to approximate afunction by the linear form

321),( cccu ++=

we proceed in the same way as in the 1-D case



2-D elements: coefficients

... and we obtain P3

P2

P1

P1 (0,0)

P3 (0,1)

P2 (1,0)

... and we obtain the coefficients as a function of the

values at the grid nodes by matrix inversion

313

212

11

)1,0(

)0,1(

)0,0(

ccuu

ccuu

cuu

+==

+==

==

Auc =

=

101

011

001

A containing the1-D case

=

11-

01A



triangles: linear basis functions

from matrix A we can calculate the linear basisfunctions for triangles

P3

P2

P1

P1 (0,0)

P3 (0,1)

P2 (1,0)

==

=

),(

),(

1),(

3

2

1

N

N

N



triangles: quadratic elements

Any function defined on a triangle can be approximated by the quadraticfunction 2

65

2

4321),( yxyxyxyxu +++++=and in the transformed system we obtain

2

65

2

4321),( ccccccu +++++=

as in the 1-D casewe need additionalpoints on theelement.

P3

P2

P1

P1 (0,0)

P3 (0,1)P2 (1,0)

P5 (1/2,1/2)

P4 (1/2,0)

P6 (0,1/2)

++

+

+ +

+

P5

P4

P6



triangles: quadratic elements

To determine the coefficients we calculate thefunction u at each grid point to obtain

6316

6543215

4214

6313

4212

11

6/12/1

4/14/14/12/12/1

4/12/1

cccu

ccccccu

cccu

cccu

cccu

cu

++=

+++++=

++=

++=

++=

=

P3

P2P1

P1 (0,0)

P3 (0,1)P2 (1,0)

P5 (1/2,1/2)P4 (1/2,0)

P6 (0,1/2)

++

+

+ +

+

P5

P4

P6

... and by matrix inversion we can calculate the coefficients as a function ofthe values at Pi

Auc =



triangles: basis functions

=

400202

444004004022

400103

004013

000001

A

P3

P2P1

P1 (0,0)

P3 (0,1)P2 (1,0)

P5 (1/2,1/2)P4 (1/2,0)

P6 (0,1/2)

++

+

+ +

+

P5

P4

P6

... to obtain the basis functions

Auc =

)1(4),(

4),(

)1(4),()12(),(

)12(),(

)221)(1(),(

2

5

4

3

2

1

=

=

==

=

=

N

N

NN

N

N

... and they look like ...



triangles: quadratic basis functions

P3

P2P1

P1 (0,0)

P3 (0,1)P2 (1,0)

P5 (1/2,1/P4 (1/2,0)

P6 (0,1/2)

++

+

+ +

+

P5

P4

P6The first three quadratic basis functions ...



triangles: quadratic basis functions

P3

P2

P1

P1 (0,0)

P3 (0,1)P2 (1,0)

P5 (1/2,1/P4 (1/2,0)

P6 (0,1/2)

++

+

+ +

+

P5

P4

P6.. and the rest ...



rectangles: transformation

Let us consider rectangular elements, and transform them into a localcoordinate system

P3

P2P1

x

y

P3

P2P1

beforeafter

P4

P4



rectangles: linear elements

With the linear Ansatz

4321

),( ccccu +++=

we obtain matrix A as

=

1111

1001

0011

0001

A

and the basis functions

)1(),(

),(

)1(),(

)1)(1(),(

4

3

2

1

=

=

==

N

N

N

N



rectangles: quadratic elements

With the quadratic Ansatz

2

8

2

7

2

65

2

4321),( ccccccccu +++++++=

we obtain an 8x8 matrix A ... and a basis function lookse.g. like

P3

P2P1

P4

+

+

+

+

P5

P6

P7

P8

)1)(1(4),(

)221)(1)(1(),(

5

1

==

N

N

N1 N2


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1-D and 2-D elements: summary

The basis functions for finite element problems can be obtained by:

Transforming the system in to a local (to the element) system Making a linear (quadratic, cubic) Ansatzfor a function defined across

the element.

Using the interpolation condition (which states that the particular basisfunctions should be one at the corresponding grid node) to obtain thecoefficients as a function of the function values at the grid nodes.

Using these coefficients to derive the nbasis functions for the nnode

points (or conditions).

08 finite elements basis functions

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