SLIDE 1
1
Isogeometric Analysis for high-order two-point singularly perturbed - - PowerPoint PPT Presentation
Isogeometric Analysis for high-order two-point singularly perturbed - - PowerPoint PPT Presentation
Isogeometric Analysis for high-order two-point singularly perturbed problems of reaction-diffusion type Christos Xenophontos Department of Mathematics and Statistics University of Cyprus Outline The model problem Regularity assumptions
SLIDE 2
SLIDE 3
2
The Model Problem
Following [Sun and Stynes, 1995], let ν 1 be an integer and consider following problem: find u C2ν(I), Ι = (0, 1) such hat
( )
( 1) (2 ) 1 ( 1) 2( 1) 1 2 ( ) ( )
( 1) ( 1) in (0) (1) 0 , 0,1,..., 1
j j
L u u a u L u f I u u j
− − − −
− + − + = = = = −
SLIDE 4
2
The Model Problem
Following [Sun and Stynes, 1995], let ν 1 be an integer and consider following problem: find u C2ν(I), Ι = (0, 1) such hat where ε (0, 1] is a perturbation parameter and
( )
( ) 1 ( 1) ( ) 2( ) 1 2( ) 2
, 1 ( 1) , 1
k k k k k k k
L u a u a u
− − − + − − + − =
= − +
( )
( 1) (2 ) 1 ( 1) 2( 1) 1 2 ( ) ( )
( 1) ( 1) in (0) (1) 0 , 0,1,..., 1
j j
L u u a u L u f I u u j
− − − −
− + − + = = = = −
SLIDE 5
3
Examples: ν = 1
2
( ) ( ) ( ) , (0,1), (0) (1) u x u x f x x u u − + = = =
SLIDE 6
3
Examples: ν = 1 ν = 2
2
( ) ( ) ( ) , (0,1), (0) (1) u x u x f x x u u − + = = =
2 (4)( )
( ) ( ) ( ) , (0,1), (0) (1) (0) (1) u x u x u x f x x u u u u − + = = = = =
SLIDE 7
3
Examples: ν = 1 ν = 2 ν = 3
2
( ) ( ) ( ) , (0,1), (0) (1) u x u x f x x u u − + = = =
2 (4)( )
( ) ( ) ( ) , (0,1), (0) (1) (0) (1) u x u x u x f x x u u u u − + = = = = =
2 (6) (4)
( ) ( ) ( ) ( ) ( ) , (0,1), (0) (1) (0) (1) (0) (1) u x u x u x u x f x x u u u u u u − + − + = = = = = = =
etc.
SLIDE 8
4
2( 1)( )
[0,1] a x x
−
We assume that and
2( ) ( ) 1
1 ( ) ( ) [0,1], 2,..., 2
k k k
a x a x x k
− − + −
− =
1
0 , 2,...,
k j j
k
− =
=
for some constants αν – k , k = 1, …, ν, (αν – 1 = α), such that
SLIDE 9
4
The above conditions ensure coercivity of the associated bilinear form (as well as the absence of turning points)
[Sun and Stynes, 1995].
2( 1)( )
[0,1] a x x
−
1
0 , 2,...,
k j j
k
− =
=
We assume that and
2( ) ( ) 1
1 ( ) ( ) [0,1], 2,..., 2
k k k
a x a x x k
− − + −
− =
for some constants αν – k , k = 1, …, ν, (αν – 1 = α), such that
SLIDE 10
5
( ) ( ) ( ) ( )
! , !
i
n n n n i a f L I L I
a C n f C n n
We further assume that the data is analytic, i.e. the functions ai , i = 0, 1, …, 2(ν – 1), and the function f satisfy, for some positive constants C, γf , γai , i = 0, 1, …, 2(ν – 1) independent of ε,
SLIDE 11
6
Assumption 1: The BVP under study has a classical solution which can be decomposed as
( )
2
u C I
S BL R
u u u u
= + +
SLIDE 12
6
Assumption 1: The BVP under study has a classical solution which can be decomposed as
( )
2
u C I
S BL R
u u u u
= + +
smooth part (as smooth as the data allows)
SLIDE 13
6
Assumption 1: The BVP under study has a classical solution which can be decomposed as
( )
2
u C I
S BL R
u u u u
= + +
smooth part (as smooth as the data allows) boundary layers (have support only in a region near the boundary)
SLIDE 14
6
Assumption 1: The BVP under study has a classical solution which can be decomposed as
( )
2
u C I
S BL R
u u u u
= + +
smooth part (as smooth as the data allows) remainder (exponentially small in ε) boundary layers (have support only in a region near the boundary)
SLIDE 15
6
Assumption 1: The BVP under study has a classical solution which can be decomposed as
( )
2
u C I
S BL R
u u u u
= + +
smooth part (as smooth as the data allows)
and for all x [0, 1], n , there holds (under the analyticity of the data assumption)
remainder (exponentially small in ε) boundary layers (have support only in a region near the boundary)
SLIDE 16
7
, , ,
S BL
for some constants , independent of ε.
( )
1 2
( ) ( ) ( ) 1 dist( , )/ ( ) / ( ) ( ) ( )
!, ( ) ,
n n S S L I n n x I BL BL R R R H I L I L I
u C n u x C e u u u Ce
−
− − − −
+ +
SLIDE 17
7
( ) 1 ( )
max ,
n n n n L I
u CK n
− −
Moreover, there exist C, K > 0, such that for all n = 1, 2, 3, …
, , ,
S BL
for some constants , independent of ε.
( )
1 2
( ) ( ) ( ) 1 dist( , )/ ( ) / ( ) ( ) ( )
!, ( ) ,
n n S S L I n n x I BL BL R R R H I L I L I
u C n u x C e u u u Ce
−
− − − −
+ +
SLIDE 18
7
( ) 1 ( )
max ,
n n n n L I
u CK n
− −
Differentiability through asymptotic expansions Classical Differentiability
, , ,
S BL
for some constants , independent of ε. Moreover, there exist C, K > 0, such that for all n = 1, 2, 3, …
( )
1 2
( ) ( ) ( ) 1 dist( , )/ ( ) / ( ) ( ) ( )
!, ( ) ,
n n S S L I n n x I BL BL R R R H I L I L I
u C n u x C e u u u Ce
−
− − − −
+ +
SLIDE 19
8
Remark: Assumption 1 has been established for ν = 1,
[Melenk, 1997], and for ν = 2, [Constantinou, 2019].
SLIDE 20
9
IGA Galerkin Formulation
The variational formulation is given by: find such that
( , ) , ( )
I
u v f v v H I
= B
0 ( )
u H I
where, with the usual L2(I) inner product,
,
I
( ) ( ) ( 1) ( 1) 2( 1) 1 2
( , ) , , ( , )
I I
v w v w a v w v w
− − −
= + + B B
SLIDE 21
9
IGA Galerkin Formulation
The variational formulation is given by: find such that
( , ) , ( )
I
u v f v v H I
= B
0 ( )
u H I
where, with the usual L2(I) inner product,
,
I
( ) ( ) ( 1) ( 1) 2( 1) 1 2
( , ) , , ( , )
I I
v w v w a v w v w
− − −
= + + B B
1 ( 1) ( ) ( ) 2( ) 1 2( ) 2
, 1 ( , ) , , 1
I
k k k k k k
v w a v a v w
− + − − − + − =
= + B
SLIDE 22
10
At the discrete level, we seek
( ) s. t.
N N
u S H I
and there holds
,
N N E E
u u C u v v S C
+
− −
( , ) , ( )
N N I
u v f v v S H I
= B
SLIDE 23
10
At the discrete level, we seek
( ) s. t.
N N
u S H I
where the energy norm is defined as and there holds
,
N N E E
u u C u v v S C
+
− −
( , ) , ( )
N N I
u v f v v S H I
= B
1 2
2 2 2 ( ) ( ) ( ) 2 E H I L I
w w w
−
= +
SLIDE 24
11
In the FEM, the space SN consists of piecewise polynomials defined on some subdivision (mesh) of the domain Ω.
SLIDE 25
11
, 1 , ( ) N N p i i p
S S span B
=
=
In the FEM, the space SN consists of piecewise polynomials defined on some subdivision (mesh) of the domain Ω. In IGA, the space SN is defined using B-splines:
SLIDE 26
11
, 1 , ( ) N N p i i p
S S span B
=
=
In the FEM, the space SN consists of piecewise polynomials defined on some subdivision (mesh) of the domain Ω. In IGA, the space SN is defined using B-splines: The functions Bi,p are B-splines, defined as follows [Cotrell,
Hughes, Basilevs, 2009]:
SLIDE 27
12
Let
1 2 1
, ,...,
N p
+ +
=
be a knot vector, where is the i th knot, i = 1,…, N+p+1, p is the polynomial order and N is the total number of basis functions.
i
SLIDE 28
12
Let
1 2 1
, ,...,
N p
+ +
=
i
The numbers in Ξ are non-decreasing and may be repeated. If the first and last knot values appear p+1 times, the knot vector is called open. be a knot vector, where is the i th knot, i = 1,…, N+p+1, p is the polynomial order and N is the total number of basis functions.
SLIDE 29
13
With a knot vector Ξ at hand, the B-spline basis functions are defined recursively, starting with piecewise constants (p = 0):
1 ,0
1 , ( ) ,
i i i
B
- therwise
+
=
SLIDE 30
13
With a knot vector Ξ at hand, the B-spline basis functions are defined recursively, starting with piecewise constants (p = 0):
1 ,0
1 , ( ) ,
i i i
B
- therwise
+
=
For p = 1, 2, … they are defined by the Cox–de Boor formula:
1 , , 1 1, 1 1 1
( ) ( ) ( )
i p i i p i p i p i p i i p i
B B B
+ + − + − + + + +
− − = + − −
(The convention “ */0 = 0” is used.)
SLIDE 31
14
We also mention the formula for the kth derivative of a B-spline:
, , ,
! ( ) ( ) ( )!
k k i p k j i j p k k j
d p B B d p k
+ − =
= −
with
1,0 0,0 ,0 1 1, 1, 1 , 1 1, 1 , 1
1, , , 1,..., 1
k k i p k i k j k j k j i p j k i j k k k k i p i k
j k
− + − + − − − + + − + + − − + + +
= = − − = = − − − = −
SLIDE 32
15
e.g.
Uniform knot vector Ξ = [0,0.1,0.2,…,0.9,1]
SLIDE 33
15
e.g.
Uniform knot vector Ξ = [0,0.1,0.2,…,0.9,1]
SLIDE 34
16
If we assume we have ξ1,…,ξm distinct knots, each having multiplicity ri , then
1 2
1 1 2 2 times times times
,..., , ,..., ,..., ,...,
m
m m r r r
=
and there holds (r1 = rm = p + 1)
1
1.
m i i
r N p
=
= + +
SLIDE 35
16
If we assume we have ξ1,…,ξm distinct knots, each having multiplicity ri , then
1 2
1 1 2 2 times times times
,..., , ,..., ,..., ,...,
m
m m r r r
=
and there holds (r1 = rm = p + 1)
1
1.
m i i
r N p
=
= + +
We note that the B-spline has p–ri continuous derivatives at ξi , hence we define ki = p–ri + 1 as a measure of the regularity at ξi (k1 = km = 0).
SLIDE 36
17
B-splines form a partition of unity and they span the space
- f piecewise polynomials of degree p on the subdivision
{ξ1,…, ξm}.
SLIDE 37
17
B-splines form a partition of unity and they span the space
- f piecewise polynomials of degree p on the subdivision
{ξ1,…, ξm}. Each basis function is positive and has support in [ξi, ξi+p+1].
SLIDE 38
17
B-splines form a partition of unity and they span the space
- f piecewise polynomials of degree p on the subdivision
{ξ1,…, ξm}. Each basis function is positive and has support in [ξi, ξi+p+1]. So, we will use as our discrete space
, , 1
( )
N p N p i p i
S S S span B
=
=
k
where k = [k1, …,km].
SLIDE 39
18
For singularly perturbed problems, the hp version of the FEM
- n the Spectral Boundary Layer Mesh, performs extremely
well for 2nd and 4th order problems [X. et al 1998 – present].
SLIDE 40
18
κpε 1– κpε 1
For singularly perturbed problems, the hp version of the FEM
- n the Spectral Boundary Layer Mesh, performs extremely
well for 2nd and 4th order problems [X. et al 1998 – present].
SBLM for layers of width O(ε) near the endpoints,
.
+
SLIDE 41
19
1 times 1 times 1 times 1 times
0,...,0, 1,...,1 if 1/ 2 0,...,0, ,1 , 1,...,1 if 1/ 2
p p p p
p p p p
+ + + +
= −
So we analogously define the Spectral Boundary Layer knot vector
SLIDE 42
20
SLIDE 43
21
SLIDE 44
22
Error Analysis
( )
2 2
2( ) 2 2 ( ) ( ) ( ) 2 1, ( ) ( )
( )! ! 2 ( )!(2 )!
i i
s q j j j q s i q q L I L I
h q s j u u u q s q j
+ − + −
− − + −
Theorem: [Buffa, Sangalli, Schwab, 2014] Given the subdivision {0 = ξ1,…,ξm = 1} of the reference element I = (0, 1), let Ii = (ξi, ξi+1), hi = ξi+1– ξi, i = 1,…,m – 1, and assume u(q) Hs(I). Then, there exists a quasi-interpolation
- perator such that for i = 1,…,m – 1 ,
j = 0,…, q, and 0 s q,
2 2 1 – 1,
( ) :
q q s q q i
H S I
−
→
SLIDE 45
23
Recall that for our problem,
S BL R
u u u u
= + +
SLIDE 46
Recall that for our problem,
S BL R
u u u u
= + +
We approximate this (analytic) part by a quasi- interpolant on I = (0, 1) (given by the previous theorem)
23
SLIDE 47
Recall that for our problem,
S BL R
u u u u
= + +
We approximate this (analytic) part by a quasi- interpolant on I = (0, 1) (given by the previous theorem) Since the boundary layers have support only in a region near the boundary, the quasi-interpolant will be constructed only in the layer region. The boundary layers will not be approximated
- utside the layer region.
23
SLIDE 48
Recall that for our problem,
S BL R
u u u u
= + +
We approximate this (analytic) part by a quasi- interpolant on I = (0, 1) (given by the previous theorem) The remainder is exponentially small (in ε) and will not be approximated.
23
Since the boundary layers have support only in a region near the boundary, the quasi-interpolant will be constructed only in the layer region. The boundary layers will not be approximated
- utside the layer region.
SLIDE 49
Since the smooth part uS of the solution satisfies, Approximation of the smooth part: by the previous theorem we get, for j = 0, …, ν ( q)
n
24
( ) ( )
! , , ,
n n S S S L I
u Cn C
+
SLIDE 50
Since the smooth part uS of the solution satisfies, Approximation of the smooth part:
( ) ( )
! , , ,
n n S S S L I
u Cn C
+
by the previous theorem we get, for j = 0, …, ν ( q)
n
( )
2 2
2 2 ( ) ( ) ( ) 2 1, ( ) ( )
( )! ! ( )!(2 )!
j j q s S q q S S L I L I
q s j u u u q s q j
+ −
− − + −
24
( )
2 2( )
( )! ! ( )! ( )! !
q s S
q s C q s q s q
+
− + +
SLIDE 51
Since the smooth part uS of the solution satisfies, Approximation of the smooth part: by the previous theorem we get, for j = 0, …, ν ( q)
n
Choosing s = τq , with τ (0, 1) arbitrary and using Lemma 1
- f [Buffa, Sangalli, Schwab, 2014], we get
( )
2 2
2 2 ( ) ( ) ( ) 2 1, ( ) ( )
( )! ! ( )!(2 )!
j j q s S q q S S L I L I
q s j u u u q s q j
+ −
− − + −
24
( )
2( ) 2
! ( )! ( )! ( ) ! !
q s S
q s q C q s q s
+
− + +
( ) ( )
! , , ,
n n S S S L I
u Cn C
+
SLIDE 52
25
( )
2
2 ( ) ( ) 2( ) 2 1, ( ) 1 1
1 (1 ) ! 4 (1 ) !
j j q q S q q S S L q I
u u C q
+ − + −
+ − −
4 4
,
q q q S S q q
e C C Ce q e q
− + −
as q → .
SLIDE 53
25
In particular, in the energy norm, we have
( )
2 1
2 ( ) 2 2 ( ) 2 1, 2 1, ( ) 2 2 1, ( )
,
S q q S S q q S E L I S q q S H I q
u u u u u u Ce
−
− − − − +
− − + + −
as q → .
( )
2
2 ( ) ( ) 2( ) 2 1, ( ) 1 1
1 (1 ) ! 4 (1 ) !
j j q q S q q S S L q I
u u C q
+ − + −
+ − −
4 4
,
q q q S S q q
e C C Ce q e q
− + −
SLIDE 54
26
Since the boundary layer part
- f the solution satisfies
Approximation of the boundary layer: it will only be approximated in the layer region
( )
( ) 1 ( , )/
( ) , ,
n n n dist x I BL BL
u x C e n
− − − +
BL
u (0, ) (1 ,1) I = −
SLIDE 55
26
Since the boundary layer part
- f the solution satisfies
Approximation of the boundary layer: it will only be approximated in the layer region
( ) ( ) ( )
2 2
( ) ( 2 ( ) ( ) 2 1, 2( ) 2 ( ) )
( )! ! 2 ( )!(2 )!
j j BL q q BL L s q j q s BL L I I
u u q s j u q s q j
− + − +
− − + −
BL
u (0, ) (1 ,1) I = −
By the previous theorem, we get, for j = 0, …, ν ( q)
( )
( ) 1 ( , )/
( ) , ,
n n n dist x I BL BL
u x C e n
− − − +
SLIDE 56
27
( ) ( )
2
2 ( ) ( ) 2 1, 2( ) 2( ) ( 2( 1 ) 1 1 2 2 2( ) )
( )! ! 2 ( )! ! ( )! ! ( )! !
j j BL q q BL L s q j q s q s BL j q B I s L
u u q s C q s q q s C q s q
− + − + − − − + − + − +
− − + − +
SLIDE 57
27
( ) ( )
2
2 ( ) ( ) 2 1, 2( ) 2( ) ( 2( 1 ) 1 1 2 2 2( ) )
( )! ! 2 ( )! ( )! ( ! ! ) ! !
j j BL q q BL L s q j q s q s BL j BL I q s
u u q s C q s q C q q s q s
− + − + − − − + − + − +
− − + − +
Choosing s = τ΄q , τ΄ (0, 1) arbitrary, and using Stirling’s formula, we get
( ) ( )
2
2 ( ) ( ) 2 1, 1 2 2 1 2 2 1 ( ) (1 1 2 )
( ! 1 ) (1 )
j j BL q q BL L j q q B q I q L
u u C q e q
− − + − + − + − +
− + −
SLIDE 58
28
( ) ( )
2
2 ( ) ( ) 2 1, 1 2 2 1 2 2 1 ( ) (1 1 2 )
( ! 1 ) (1 )
j j BL q q BL L j q q B q I q L
u u C q e q
− − + − + − + − +
− + −
SLIDE 59
28
( ) ( )
2
2 ( ) ( ) 2 1, 1 2 2 1 2 2 1 ( ) (1 1 2 )
( ! 1 ) (1 )
j j BL q q BL L j q q B q I q L
u u C q e q
− − + − + − + − +
− + −
4 4 1 1 2 2 1
(1 ) (1 )
q j q BL q
e C q q
− − + − − +
− +
SLIDE 60
28
1 2 2
,
j q
C e
− + − − +
4 4 1 1 2 2 1
(1 ) (1 )
q j q BL q
e C q q
− − + − − +
− +
( ) ( )
2
2 ( ) ( ) 2 1, 1 2 2 1 2 2 1 ( ) (1 1 2 )
( ! 1 ) (1 )
j j BL q q BL L j q q B q I q L
u u C q e q
− − + − + − + − +
− + −
SLIDE 61
29
Outside the layer region, there holds
( )
2
2 2 ( ) ( ) 2 2( \ 1 ) 1 ( ) \
( )
I I I I j j j j q BL BL BL L
u u x dx C e
− − + −
=
2 1 2 2 j j q BL
C e
− +
− −
SLIDE 62
29
Outside the layer region, there holds so
( ) ( )
2
2 ( ) ( ) 1 2 2 2 1, ( ) j j j q BL q q BL L I
u u C e
− + − − −
−
( )
2
2 2 ( ) ( ) 2 2( \ 1 ) 1 ( ) \
( )
I I I I j j j j q BL BL BL L
u u x dx C e
− − + −
=
2 1 2 2 j j q BL
C e
− +
− −
SLIDE 63
29
Outside the layer region, there holds so
( ) ( )
2
2 ( ) ( ) 1 2 2 2 1, ( ) j j j q BL q q BL L I
u u C e
− + − − −
−
In particular, in the energy norm, we have
( )
2
2 2 ( ) ( ) 2 2( \ 1 ) 1 ( ) \
( )
I I I I j j j j q BL BL BL L
u u x dx C e
− − + −
=
2 1 2 2 j j q BL
C e
− +
− −
SLIDE 64
30
1/2 2 1, , q q q BL q q BL E I
u u C e e Ce
− − − −
− +
SLIDE 65
30
We assumed The remainder:
/ , ( ) R R E I L I
u u Ce
−
+
1/2 2 1, , q q q BL q q BL E I
u u C e e Ce
− − − −
− +
SLIDE 66
30
We assumed The remainder:
/ , ( ) R R E I L I
u u Ce
−
+
1/2 2 1, , q q q BL q q BL E I
u u C e e Ce
− − − −
− +
Putting it all together:
2 1, 2 1, S q q S BL q q BL R E E E N E
u u u u u u u
− −
− + − + −
SLIDE 67
30
We assumed The remainder:
/ , ( ) R R E I L I
u u Ce
−
+
1/2 2 1, , q q q BL q q BL E I
u u C e e Ce
− − − −
− +
Putting it all together:
2 1, 2 1, 1/2
1 ,
N S q q S BL q q BL R E E E p E p p
u u u u u C e e u e u C
− − − − + −
− + − + + − +
SLIDE 68
31
Numerical Results
We measure the
, ,
100
EXACT N E I EXACT E I
u u Error u − =
and plot it against the DOF in a semi-log scale.
SLIDE 69
31
Numerical Results
We consider the problem Example 1: ν = 1
2
1 in (0,1) (0) (1) u u I u u − + = = = =
We measure the
, ,
100
EXACT N E I EXACT E I
u u Error u − =
and plot it against the DOF in a semi-log scale.
SLIDE 70
32
SLIDE 71
33
We next consider the problem Example 2: ν = 2
2 (4) ( ) ( )
1 in (0,1) (0) (1) 0 , 0,1
j j
u u u I u u j − + = = = = =
SLIDE 72
34
SLIDE 73
35
Finally, we consider the problem Example 3: ν = 3
2 (6) (4) ( ) ( )
1 in (0,1) (0) (1) 0 , 0,1,2
j j
u u u u I u u j − + − + = = = = =
SLIDE 74
35
Finally, we consider the problem Example 3: ν = 3
2 (6) (4) ( ) ( )
1 in (0,1) (0) (1) 0 , 0,1,2
j j
u u u u I u u j − + − + = = = = =
… well … still working on it …
SLIDE 75
36
Closing Remarks
- Why are the error curves in the numerical experiments
not on top of each other, and instead, it appears that the error decreases as ε → 0?
SLIDE 76
36
Closing Remarks
Because the energy norm is too ‘weak’ and ‘does not see the layers’, meaning that
- Why are the error curves in the numerical experiments
not on top of each other, and instead, it appears that the error decreases as ε → 0?
SLIDE 77
36
Closing Remarks
- Why are the error curves in the numerical experiments
not on top of each other, and instead, it appears that the error decreases as ε → 0?
(1)
S E
u O =
1/2
( )
BL E
u O =
while Because the energy norm is too ‘weak’ and ‘does not see the layers’, meaning that so as ε → 0,
0.
BL E
u →
SLIDE 78
37
The norm
1 2
2 2 2 ( ) ( ) ( )
||| |||
H I L I
w w w
−
= +
is ‘balanced’, meaning
||| ||| || || (1)
S BL
u u O
SLIDE 79
37
The norm
1 2
2 2 2 ( ) ( ) ( )
||| |||
H I L I
w w w
−
= +
is ‘balanced’, meaning
||| ||| || || (1)
S BL
u u O
The analysis of the hp Galerkin FEM in this norm, for second order singularly perturbed problems in 1- and 2-D, appears in [MELENK & X., NUMER. ALG. 2017] where it was shown that
||| ||| ,
p N
u u Ce
− +