Finite element method – basis functions 1 Finite Elements: Basis functions 1-D elements ...
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1 Finite 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.
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Transcript of Finite element method – basis functions 1 Finite Elements: Basis functions 1-D elements ...
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.
2Finite element method – basis functions
1-D elements: coordinate transformation
We wish to approximate a function u(x) defined in an interval [a,b] by some set of basis functions
n
iiicxu
1
)(
where i is the number of grid points (the edges of our elements) defined at locations xi. As the basis functions look the same in all elements (apart from some constant) we make life easier by moving to a local coordinate system
ii
i
xx
xx
1
so that the element is defined for x=[0,1].
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 vary across the domain.
We require that our function u() be approximated locally by the linear function
21)( ccu Our node points are defined at 1,2=0,1 and we require that
212
11
212
11
uuc
uc
ccu
cu
Auc
11-
01A
4Finite element method – basis functions
1-D elements – linear basis functions
As we have expressed the coefficients ci as a function of the function values at node points 1,2 we can now express the approximate function using the node values
)()(
)1(
)()(
211
21
211
NNu
uu
uuuu
.. and N1,2(x) are the linear basis functions for 1-D elements.
5Finite element method – basis functions
1-D quadratic elements
Now we require that our function u(x) be approximated locally by the quadratic function
2321)( cccu
Our node points are defined at 1,2,3=0,1/2,1 and we require that
3213
3212
11
25.05.0
cccu
cccu
cu
Auc
242
143
001
A
6Finite element method – basis functions
1-D quadratic basis functions
... again we can now express our approximated function as a sum over our basis functions weighted by the values at three node points
... note that now we re using three grid points per element ...
Can we approximate a constant function?
3
1
23
22
21
2321
)(
)2()44()231()(
iiiNu
uuucccu
7Finite element method – basis functions
1-D cubic basis functions
... using similar arguments the cubic basis functions can be derived as
324
323
322
321
34
2321
)(
23)(
2)(
231)(
)(
N
N
N
N
ccccu
... note that here we need derivative information at the boundaries ...
How can we approximate a constant function?
8Finite element method – basis functions
2-D elements: coordinate transformation
Let us now discuss the geometry and basis functions of 2-D elements, again we want to consider the problems in a local coordinate system, first we look at triangles
P3
P2
P1
x
y
P3
P2P1
before after
9Finite element method – basis functions
2-D elements: coordinate transformation
Any triangle with corners Pi(xi,yi), i=1,2,3 can be transformed into a rectangular, equilateral triangle with
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 a function by the linear form
321),( cccu
we proceed in the same way as in the 1-D case
10Finite 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 the 1-D case
11-
01A
11Finite element method – basis functions
triangles: linear basis functions
from matrix A we can calculate the linear basis functions for triangles
P3
P2
P1
P1 (0,0)
P3 (0,1)
P2 (1,0)
),(
),(
1),(
3
2
1
N
N
N
12Finite element method – basis functions
triangles: quadratic elements
Any function defined on a triangle can be approximated by the quadratic function 2
652
4321),( yxyxyxyxu and in the transformed system we obtain
265
24321),( ccccccu
as in the 1-D case we need additional points on the element.
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
13Finite element method – basis functions
triangles: quadratic elements
To determine the coefficients we calculate the function 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
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
... and by matrix inversion we can calculate the coefficients as a function of the values at Pi
Auc
14Finite element method – basis functions
triangles: basis functions
400202
444004
004022
400103
004013
000001
A
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
... to obtain the basis functions
Auc
)1(4),(
4),(
)1(4),(
)12(),(
)12(),(
)221)(1(),(
2
5
4
3
2
1
N
N
N
N
N
N
... and they look like ...
15Finite element method – basis functions
triangles: quadratic basis functions
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
P6The first three quadratic basis functions ...
16Finite element method – basis functions
triangles: quadratic basis functions
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 the rest ...
17Finite element method – basis functions
rectangles: transformation
Let us consider rectangular elements, and transform them into a local coordinate system
P3
P2P1
x
y
P3
P2P1
before after
P4
P4
18Finite 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
19Finite element method – basis functions
rectangles: quadratic elements
With the quadratic Ansatz
28
27
265
24321),( ccccccccu
we obtain an 8x8 matrix A ... and a basis function looks e.g. like
P3
P2P1
P4
+
+
+
+
P5
P6
P7
P8
)1)(1(4),(
)221)(1)(1(),(
5
1
N
N
N1 N2
20Finite element method – basis functions
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) Ansatz for a function defined across
the element. Using the interpolation condition (which states that the particular basis
functions should be one at the corresponding grid node) to obtain the coefficients as a function of the function values at the grid nodes.
Using these coefficients to derive the n basis functions for the n node points (or conditions).
