Post on 06-Apr-2018
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.
8/2/2019 08 Finite Elements Basis Functions
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2Finite element method basis functions
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].
8/2/2019 08 Finite Elements Basis Functions
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3Finite element method basis functions
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
8/2/2019 08 Finite Elements Basis Functions
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4Finite element method basis functions
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.
8/2/2019 08 Finite Elements Basis Functions
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5Finite element method basis functions
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
8/2/2019 08 Finite Elements Basis Functions
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
8/2/2019 08 Finite Elements Basis Functions
7/207Finite element method basis functions
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?
8/2/2019 08 Finite Elements Basis Functions
8/208Finite element method basis functions
2-D elements: coordinate transformation
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
8/2/2019 08 Finite Elements Basis Functions
9/209Finite element method basis functions
2-D elements: coordinate transformation
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
8/2/2019 08 Finite Elements Basis Functions
10/2010Finite element method basis functions
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
8/2/2019 08 Finite Elements Basis Functions
11/2011Finite element method basis functions
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
8/2/2019 08 Finite Elements Basis Functions
12/2012Finite element method basis functions
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
8/2/2019 08 Finite Elements Basis Functions
13/2013Finite element method basis functions
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 =
8/2/2019 08 Finite Elements Basis Functions
14/2014Finite element method basis functions
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 ...
8/2/2019 08 Finite Elements Basis Functions
15/2015Finite element method basis functions
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 ...
8/2/2019 08 Finite Elements Basis Functions
16/2016Finite element method 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 ...
8/2/2019 08 Finite Elements Basis Functions
17/2017Finite element method basis functions
rectangles: transformation
Let us consider rectangular elements, and transform them into a localcoordinate system
P3
P2P1
x
y
P3
P2P1
beforeafter
P4
P4
8/2/2019 08 Finite Elements Basis Functions
18/2018Finite element method basis functions
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
8/2/2019 08 Finite Elements Basis Functions
19/2019Finite element method basis functions
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).