Finite element approximation of eigenproblems
Daniele Boffi and Lucia Gastaldi
http://www-dimat.unipv.it/boffi/
http://dm.ing.unibs.it/gastaldi/
Outline
References
Preliminary examples
General abstract analysis
D. Boffi
Finite element approximation of eigenvalue problems
Acta Numerica, 19 (2010) 1-120.
D. Boffi, F. Gardini, and L. Gastaldi
Some remarks on eigenvalue approximation by finite elements
In Frontiers in Numerical Analysis - Durham 2010, Springer
Lecture Notes in Computational Science and Engineering, 85
(2012), pp. 1-77.
1D Laplace eigenvalue problem
Problem
Find λ and u 6= 0 such that
− u 00 = λu in Ω =]0, π[
u(0) = u(π) = 0
Known facts
I λ = 1, 4, 9, 16, 25, . . . that is λ(k) = k 2 for k ∈ N;
I u (k) (x) = sin(kx) for k ∈ N.
Variational formulation
Find λ ∈ R and u ∈ H01 (Ω) with u 6= 0 such that
Z π
Z π
0
0
u (x)v (x) dx = λ
u(x)v (x) dx ∀v ∈ H01 (Ω)
0
0
Finite element discretization
Vh ⊆ V finite dimensional
For example, subdivide Ω into n sub interval of length h and
consider
Vh = {v ∈ H01 (Ω) : v piecewise linear polynomials}
Discrete problem
Find λh ∈ R and uh ∈ Vh with uh 6= 0 such that
Z π
Z π
uh0 (x)v 0 (x) dx = λh
uh (x)v (x) dx
0
0
Generalized algebraic eigenvalue problem
Au = λh Mu
∀v ∈ Vh
1D Laplace: P1
1
4
9
16
25
n=8
n = 16
n = 32
1.0129160450588
4.2095474481529
10.0802909335883
19.4536672593288
33.2628304890884
1.0032168743567
4.0516641802355
9.2631305555446
16.8381897926118
27.0649225609802
1.0008034482562
4.0128674974272
9.0652448637285
16.2066567209423
25.5059230069702
1D Laplace: P1
1
4
9
16
25
1
4
9
16
25
n=8
n = 16
n = 32
1.0129160450588
4.2095474481529
10.0802909335883
19.4536672593288
33.2628304890884
1.0032168743567
4.0516641802355
9.2631305555446
16.8381897926118
27.0649225609802
1.0008034482562
4.0128674974272
9.0652448637285
16.2066567209423
25.5059230069702
n = 64
n = 128
n = 256
1.0002008137390
4.0032137930241
9.0162763381719
16.0514699897078
25.1257489536113
1.0000502004122
4.0008032549556
9.0040668861371
16.0128551720960
25.0313903532369
1.0000125499161
4.0002008016414
9.0010165838380
16.0032130198251
25.0078446408520
Error analysis
We can show that:
(k)
uh (ih) = sin(πih)
(k)
λh =
for i = 0, . . . , n;
6 1 − cos(kh)
.
h2 2 + cos(kh)
with the following asymptotic behavior
(k)
λh = k 2 +
That is:
k4 2
h + O(k 6 h4 )
12
(k)
as h → 0.
λ(k) ≤ λh ≤ λ(k) + C (k)h2 .
P2
1
4
9
16
25
36
49
64
81
100
#
n=8
n = 16
1.0000
1.0000
4.0020
4.0001
9.0225
9.0015
16.1204 16.0082
25.4327 25.0307
37.1989 36.0899
51.6607 49.2217
64.8456 64.4814
95.7798 81.9488
124.9301 101.7308
15
31
(4.0)
(4.0)
(3.9)
(3.9)
(3.8)
(3.7)
(3.6)
(0.8)
(4.0)
(3.8)
Computed (rate)
n = 32
n = 64
1.0000 (4.0)
1.0000
4.0000 (4.0)
4.0000
9.0001 (4.0)
9.0000
16.0005 (4.0) 16.0000
25.0020 (3.9) 25.0001
36.0059 (3.9) 36.0004
49.0148 (3.9) 49.0009
64.0328 (3.9) 64.0021
81.0659 (3.8) 81.0042
100.1229 (3.8) 100.0080
63
127
n = 128
(4.0)
1.0000
(4.0)
4.0000
(4.0)
9.0000
(4.0) 16.0000
(4.0) 25.0000
(4.0) 36.0000
(4.0) 49.0001
(4.0) 64.0001
(4.0) 81.0003
(3.9) 100.0005
255
(4.0)
(4.0)
(4.0)
(4.0)
(4.0)
(4.0)
(4.0)
(4.0)
(4.0)
(4.0)
2D Laplace eigenvalue problem
Ω ⊆ R2
Problem
Find λ and u 6= 0 such that
− ∆u = λu in Ω
u = 0 on ∂Ω
Variational formulation
Find λ ∈ R and u ∈ H01 (Ω) with u 6= 0 such that
Z
Z
∇u∇v dx = λ uv dx ∀v ∈ H01 (Ω)
Ω
Ω
Finite element discretization
Vh ⊆ V finite dimensional
Vh = {v ∈ H01 (Ω) : v piecewise linear polynomials}
Discrete problem
Find λh ∈ R and uh ∈ Vh with uh 6= 0 such that
Z
Z
∇uh ∇v dx = λh
uh v dx ∀v ∈ Vh
Ω
Ω
Example
Let Ω be the square ]0, π[×]0, π[ then it is known that for m, n ∈ N
λm,n = m2 + n2
um,n (x, y ) = sin(mx) sin(ny )
UNSTRUCTURED
UNIFORM
CRISS−CROSS
Unstructured mesh
2
5
5
8
10
10
13
13
17
17
#
N=4
2.2468
6.5866
6.6230
10.2738
12.7165
14.3630
19.7789
24.2262
34.0569
9
N=8
2.0463 (2.4)
5.2732 (2.5)
5.2859 (2.5)
8.7064 (1.7)
11.0903 (1.3)
11.1308 (1.9)
14.8941 (1.8)
14.9689 (2.5)
20.1284 (2.4)
20.2113
56
Computed (rate)
N = 16
N
2.0106 (2.1)
2.0025
5.0638 (2.1)
5.0154
5.0643 (2.2)
5.0156
8.1686 (2.1)
8.0402
10.2550 (2.1)
10.0610
10.2595 (2.1)
10.0622
13.4370 (2.1)
13.1046
13.4435 (2.2)
13.1053
17.7468 (2.1)
17.1771
17.7528 (2.1)
17.1798
257
1106
= 32
(2.1)
(2.0)
(2.0)
(2.1)
(2.1)
(2.1)
(2.1)
(2.1)
(2.1)
(2.1)
N
2.0006
5.0038
5.0038
8.0099
10.0152
10.0153
13.0258
13.0258
17.0440
17.0443
4573
= 64
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
Uniform mesh
2
5
5
8
10
10
13
13
17
17
#
N=4
2.3168
6.3387
7.2502
12.2145
15.5629
16.7643
20.8965
26.0989
32.4184
9
N=8
2.0776 (2.0)
5.3325 (2.0)
5.5325 (2.1)
9.1826 (1.8)
11.5492 (1.8)
11.6879 (2.0)
15.2270 (1.8)
17.0125 (1.7)
21.3374 (1.8)
21.5751
49
Computed (rate)
N = 16
N
2.0193 (2.0)
2.0048
5.0829 (2.0)
5.0207
5.1302 (2.0)
5.0324
8.3054 (2.0)
8.0769
10.3814 (2.0)
10.0949
10.3900 (2.1)
10.0955
13.5716 (2.0)
13.1443
13.9825 (2.0)
13.2432
18.0416 (2.1)
17.2562
18.0705 (2.1)
17.2626
225
961
= 32
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
N
2.0012
5.0052
5.0081
8.0193
10.0237
10.0237
13.0362
13.0606
17.0638
17.0653
3969
= 64
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
Crisscross mesh
2
5
5
8
10
10
13
13
17
17
18
20
20
25
25
#
N=4
2.0880
5.6811
5.6811
9.4962
12.9691
12.9691
17.1879
17.1879
25.1471
38.9073
38.9073
38.9073
38.9073
38.9073
38.9073
25
N=8
2.0216 (2.0)
5.1651 (2.0)
5.1651 (2.0)
8.3521 (2.1)
10.7578 (2.0)
10.7578 (2.0)
14.0237 (2.0)
14.0237 (2.0)
19.3348 (1.8)
19.3348 (3.2)
19.8363 (3.5)
22.7243 (2.8)
22.7243 (2.8)
28.7526 (1.9)
28.7526 (1.9)
113
Computed (rate)
N = 16
N
2.0054 (2.0)
2.0013
5.0408 (2.0)
5.0102
5.0408 (2.0)
5.0102
8.0863 (2.0)
8.0215
10.1865 (2.0)
10.0464
10.1865 (2.0)
10.0464
13.2489 (2.0)
13.0617
13.2489 (2.0)
13.0617
17.5733 (2.0)
17.1423
17.5733 (2.0)
17.1423
18.4405 (2.1)
18.1089
20.6603 (2.0)
20.1634
20.6603 (2.0)
20.1634
25.8940 (2.1)
25.2201
25.8940 (2.1)
25.2201
481
1985
= 32
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
N
2.0003
5.0025
5.0025
8.0054
10.0116
10.0116
13.0154
13.0154
17.0355
17.0355
18.0271
20.0407
20.0407
25.0548
25.0548
8065
= 64
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
(2.0)
Multiple eigenfunctions
3
3
3
2.5
2.5
2.5
2
2
2
1.5
1.5
1.5
1
1
1
0.5
0.5
0.5
0
0
0
0.5
1
1.5
2
2.5
3
0
0
0.5
1
1.5
2
2.5
3
3
3
3
2.5
2.5
2.5
2
2
2
1.5
1.5
1.5
1
1
1
0.5
0.5
0.5
0
0
0
0.5
1
1.5
2
2.5
3
0
0.5
1
1.5
2
2.5
3
0
0.5
1
1.5
2
2.5
3
0
0
0.5
1
1.5
2
2.5
3
Multiple eigenfunctions (cont’ed)
Exact solutions
Multiple eigenfunctions (cont’ed)
Uniform mesh
3
3
2.5
2.5
2
2
1.5
1.5
1
1
0.5
0.5
0
0
0
0.5
1
1.5
2
2.5
3
0
0.5
1
1.5
2
2.5
3
0
0.5
1
1.5
2
2.5
3
Uniform mesh (reversed)
3
3
2.5
2.5
2
2
1.5
1.5
1
1
0.5
0.5
0
0
0
0.5
1
1.5
2
2.5
3
Laplace eigenproblem with Neumann boundary conditions
Ω L-shaped domain with vertices (0, 0), (1, 0), (1, 1), (1, 1), (1, 1),
and (0, 1)
− ∆u = λu
∂u
=0
∂n
in Ω
on Ω
Remark First eigenvalue λ1 = 0 with eigenfunction u1 = constant
L-shaped domain
P1 elements (Neumann boundary conditions)
0
1.48
3.53
9.87
9.87
11.39
#
N=4
N=8
-0.0000
0.0000
1.6786
1.5311 (1.9)
3.8050
3.5904 (2.3)
12.2108 10.2773 (2.5)
12.5089 10.3264 (2.5)
13.9526 12.0175 (2.0)
20
65
Computed (rate)
N = 16
N
-0.0000
-0.0000
1.4946 (1.5)
1.4827
3.5472 (2.1)
3.5373
9.9692 (2.0)
9.8935
9.9823 (2.0)
9.8979
11.5303 (2.2) 11.4233
245
922
= 32
N
-0.0000
(1.4)
1.4783
(2.0)
3.5348
(2.1)
9.8755
(2.0)
9.8767
(2.1) 11.3976
3626
= 64
(1.4)
(2.0)
(2.0)
(2.0)
(2.1)
Variationally posed eigenproblem
(Laplace operator)
V (= H01 (Ω)), H (= L2 (Ω)) Hilbert spaces with V ⊂ H with dense
and continuous embedding.
Eigenvalue problem
Find λ ∈ R such that for some u ∈ V with u 6≡ 0 it holds
a(u, v ) = λb(u, v )
∀v ∈ V
Assumptions
R
a(u, v ) : V × V → R
(= Ω ∇u · ∇v dx)
bilinear, continuous, symmetric, coercive
a(v , v ) ≥ αkv kV
∀v ∈ V ,
for some α > 0;
b(u, v ) : H × H → R
(= (u, v ))
bilinear, continuous, symmetric, equivalent to a scalar product in H
Resolvent operator
Resolvent operator T : H → H
For all f ∈ H, Tf ∈ V ⊂ H is such that
a(Tf , v ) = b(f , v )
∀v ∈ V .
Proposition
(λ, u) is an eigenpair of the variational problem if and only if
Tu = µu with µ = 1/λ.
Assumption T : H → H is a compact operator. T (H) ⊂ V
V = H01 (Ω) ⊂ H = L2 (Ω) is a compact embedding then T is
compact.
Properties of the spectrum
I
I
The spectrum is a countable set.
The eigenvalues (numbered such that all have multiplicity
one) are positive and form a diverging sequence
0 < λ(1) ≤ λ(2) ≤ · · · ≤ λ(i) ≤ · · ·
I
The eigenfunctions are orthogonal both respect to a and b:
a(u (m) , u (n) ) = b(u (m) , u (n) ) = 0
I
for m 6= n;
The eigenfunctions are normalized:
b(u (n) , u (n) ) = 1 a(u (n) , u (n) ) = λ(n) .
Characterization of the eigenvalues
a(v , v )
b(v , v )
= min R(v ),
Rayleigh quotient: R(v ) =
λ(1)
∀v ∈ H01 (Ω).
v ∈H 1 (Ω)
0
v 6=0
λ(i) =
min
(
v∈
⊥
Li−1
E
j=1 j
v 6=0
)
R(v ) = min max R(v )
E ∈V (i) v ∈E
where V (i) ⊂ H01 (Ω) with dim(V (i) ) = i
Finite element discretization
Discrete eigenproblem
Find λh ∈ R such that for some uh ∈ Vh with uh 6≡ 0 it holds
∀v ∈ Vh .
a(uh , v ) = λh b(uh , v )
Discrete (compact) resolvent operator Th : H → H
For all f ∈ H, Th f ∈ Vh ⊂ H is such that
a(Th f , v ) = b(f , v )
∀v ∈ Vh .
Properties of the discrete spectrum
(1)
(2)
(i)
I
λh ≤ λh ≤ · · · ≤ λh ≤ · · · ≤ λN(h),h
I
a(uh , uh ) = b(uh , uh ) = 0 for m 6= n;
I
Normalization: b(uh , uh ) = 1 and a(uh , uh ) = λh ;
I
Vh = ⊕ Ei,h with Ei,h = span(uh ).
(m)
(n)
N(h)
i=1
(m)
(n)
(n)
(n)
(n)
(i)
(n)
(n)
Characterization of the discrete eigenvalues
Rayleigh quotient: Rh (v ) =
ah (v , v )
b(v , v )
∀v ∈ Vh .
(1)
λh = min Rh (v ),
v ∈Vh
v 6=0
(i)
λh =
min
(
v∈
(i)
⊥
Li−1
E
j=1 j,h
v 6=0
Rh (v ) = min max Rh (v )
)
(i)
where Vh ⊂ Vh with dim(Vh ) = i
(i)
E ∈Vh
v ∈E
Relation between discrete and continuous eigenvalues
A first easy consequence
(1)
λ(1) = min R(v ),
λh = min R(v )
v ∈V
v ∈Vh
(1)
Vh ⊂ V implies λh ≥ λ(1) .
From the min-max characterization of eigenvalues we have also
that
(i)
λ(i) ≤ λh .
Convergence of eigenvalues and eigenfunctions
Some notation
m : N → N such that
λm(1) < λm(2) < · · · < λm(N) < . . .
m(1) is the multiplicity of the first eigenvalue
m(2) − m(1) is the multiplicity of the second eigenvalue
......
δ̂(E , F ) = max(δ(E , F ), δ(F , E )), where E , F subspaces of H
δ(E , F ) =
sup
inf ||u − v ||H
u∈E , ||u||H =1 v ∈F
Definition of convergence
∀ε > 0, ∀N ∈ N, ∃h0 > 0 such that ∀h ≤ h0
max
i=1,...,m(N)
δ̂
|λi − λi,h | ≤ ε
m(N)
m(N)
i=1
i=1
⊕ Ei , ⊕ Ei,h
≤ε
Convergence of eigenvalues and eigenfunctions
Uniform convergence
||T − Th ||L(H,H) → 0
Theorem
If T is selfadjoint and compact
Uniform convergence ⇐⇒ Eigenmodes convergence
Galerkin approximation of compact operators
<Bramble–Osborn ’73>
<Osborn ’75>
<Kolata ’78>
Elliptic projection Ph : V → Vh projection w.r.t. bilinear form a,
that is:
for all v ∈ V , Ph v ∈ Vh such that
a(v − Ph v , w ) = 0 ∀w ∈ Vh .
Proposition
Th = Ph T
T − Th = (I − Ph )T
Theorem
If T is compact from H to V and Ph converges strongly to the
identity operator from V to H, then the uniform convergence
lim kT − Th kL(H,H) = 0
h→0
holds true.
Error estimate for Laplace eigenproblem
Theorem
For h < h0
(i)
λ(i) ≤ λh ≤ λ(i) + Ch2
(i)
ku (i) − uh kL2 (Ω) ≤ Ch2
(i)
(i)
(i)
ku (i) − uh kH 1 (Ω) ≤ C λ(i) ku (i) − uh kL2 (Ω) − (λ(i) − λh )
Order of convergence
<Kolata ’78>
<Babuška–Osborn ’91>
Assume that T is autoadjoint, that is T t = T .
If
||T − Th ||L(V ,V ) = ε(h)
then
|λi − λi,h | ≤ C ε(h)2
δ̂(Ei , Ei,h ) ≤ C ε(h).