Matrix square root and interpolation spaces
Mario Arioli and Daniel Loghin
m.arioli@rl.ac.uk
STFC-Rutherford Appleton Laboratory, and University of Birmingham
Householder, Berlin, 2008 – p.1/23
Outline
Norms and duality in finite dimensional Hilbert spaces
Discrete Interpolation Norms
The continuous case and finite-element approximation
An example: the biharmonic operator
Summary and open problems
Householder, Berlin, 2008 – p.2/23
TRUTH may be misleading
In a finite dimensional space all norms are equivalent i.e.
Householder, Berlin, 2008 – p.3/23
TRUTH may be misleading
In a finite dimensional space all norms are equivalent i.e.
c(N )||v||1 ≤ ||v||2 ≤ C(N )||v||1
Householder, Berlin, 2008 – p.3/23
TRUTH may be misleading
In a finite dimensional space all norms are equivalent i.e.
c(N )||v||1 ≤ ||v||2 ≤ C(N )||v||1
Identify the norms for which we have
c||v||1 ≤ ||v||2 ≤ C||v||1
Householder, Berlin, 2008 – p.3/23
Finite dimensional Hilbert spaces and IRN
(·, ·) : H p
× H → IR scalar product and
kukH = (u, u)
∀u ∈ H norm.
Householder, Berlin, 2008 – p.4/23
Finite dimensional Hilbert spaces and IRN
(·, ·) : H p
× H → IR scalar product and
kukH = (u, u)
∀u ∈ H norm.
∃{ψi }i=1,...,N a basis for H
PN
∀u ∈ H
u = i=1 ui ψi
ui ∈ IR
i = 1, . . . , N
Householder, Berlin, 2008 – p.4/23
Finite dimensional Hilbert spaces and IRN
(·, ·) : H p
× H → IR scalar product and
kukH = (u, u)
∀u ∈ H norm.
∃{ψi }i=1,...,N a basis for H
PN
∀u ∈ H
u = i=1 ui ψi
ui ∈ IR
i = 1, . . . , N
Representation of scalar product in IRN .
PN
PN
Let u = i=1 ui ψi and v = i=1 vi ψi .
Then
N X
N
X
(u, v) =
ui vj (ψi , ψj ) = vT Hu
i=1 j=1
where Hij = Hji = (ψi , ψj ) and u, v ∈ IRN .
Moreover, uT Hu > 0
iff
u 6= 0 and, thus H SPD.
Householder, Berlin, 2008 – p.4/23
Dual space H′
f ∈ H′ : H → IR (functional);
f (αu + βv) = αf (u) + βf (v)
∀u, v ∈ H
H′ is the space of the linear functionals on H
kf kH′
f (u)
= sup
u6=0 kukH
Householder, Berlin, 2008 – p.5/23
Dual space H′
f ∈ H′ : H → IR (functional);
f (αu + βv) = αf (u) + βf (v)
∀u, v ∈ H
H′ is the space of the linear functionals on H
f (u)
kf kH′ = sup
u6=0 kukH
PN
If H finite dimensional and u = i=1 ui ψi , then
PN
f (u) = i=1 ui f (ψi ) = f T u
Householder, Berlin, 2008 – p.5/23
Dual space H′
f ∈ H′ : H → IR (functional);
f (αu + βv) = αf (u) + βf (v)
∀u, v ∈ H
H′ is the space of the linear functionals on H
f (u)
kf kH′ = sup
u6=0 kukH
PN
If H finite dimensional and u = i=1 ui ψi , then
PN
f (u) = i=1 ui f (ψi ) = f T u
Dual vector
Let u ∈ H, u 6= 0, then ∃fu ∈ H′ such that
fu (u) = kukH
(Hahn-Banach).
Householder, Berlin, 2008 – p.5/23
Dual space H′
Let H be a Hilbert finite dimensional space and H the real N × N
matrix identifying the scalar product.
Householder, Berlin, 2008 – p.6/23
Dual space H′
Let H be a Hilbert finite dimensional space and H the real N × N
matrix identifying the scalar product.
fu (u) = f T u = (uT Hu)1/2
The dual vector of u has the following representation:
Householder, Berlin, 2008 – p.6/23
Dual space H′
Let H be a Hilbert finite dimensional space and H the real N × N
matrix identifying the scalar product.
fu (u) = f T u = (uT Hu)1/2
The dual vector of u has the following representation:
Hu
f=
kukH
and
kfu k2H′ = uT Hu = f T H−1 f
Householder, Berlin, 2008 – p.6/23
Linear operator
A : H → V H and V finite dimensional Hilbert spaces.
Householder, Berlin, 2008 – p.7/23
Linear operator
A : H → V H and V finite dimensional Hilbert spaces.
kAkH,V = max
u6=0
kAukV
= kV1/2 AH−1/2 k2
kukH
Householder, Berlin, 2008 – p.7/23
Linear operator
A : H → V H and V finite dimensional Hilbert spaces.
kAkH,V = max
u6=0
kAukV
= kV1/2 AH−1/2 k2
kukH
The result follows from the generalized eigenvalue problem in IRN
AT VAu = λHu
Householder, Berlin, 2008 – p.7/23
Linear operator
A : H → V H and V finite dimensional Hilbert spaces.
kAkH,V = max
u6=0
kAukV
= kV1/2 AH−1/2 k2
kukH
The result follows from the generalized eigenvalue problem in IRN
AT VAu = λHu
κH (M ) = kM kH,H −1 kM −1 kH −1 ,H .
Householder, Berlin, 2008 – p.7/23
Linear operator
A : H → V H and V finite dimensional Hilbert spaces.
kAkH,V = max
u6=0
kAukV
= kV1/2 AH−1/2 k2
kukH
The result follows from the generalized eigenvalue problem in IRN
AT VAu = λHu
κH (M ) = kM kH,H −1 kM −1 kH −1 ,H .
The interesting case is κH (M ) independent of N
Householder, Berlin, 2008 – p.7/23
Interpolation spaces
H = (IRN , (u, v)H = uT Hv)
M = (IRN , (u, v)M = uT M v)
(u, v)H = (u, Sv)M = (Su, v)M
S = M −1 H
S self-adjoint in the good scalar product!
n
o
Sx = µx ⇔ Hx = µM x ⇒ µ = δ 2 > 0
∃W s.t.
M = W T W,
H = W T ∆2 W,
∆
diagonal
Householder, Berlin, 2008 – p.8/23
Interpolation spaces
H = (IRN , (u, v)H = uT Hv)
M = (IRN , (u, v)M = uT M v)
(u, v)H = (u, Sv)M = (Su, v)M
S = M −1 H
S self-adjoint in the good scalar product!
n
o
Sx = µx ⇔ Hx = µM x ⇒ µ = δ 2 > 0
∃W s.t.
M = W T W,
H = W T ∆2 W,
∆
diagonal ∆ ≥ 0
Householder, Berlin, 2008 – p.8/23
Interpolation spaces
Λ = W −1 ∆W
Λ1/2 = W −1 ∆1/2 W
S = M −1 H = W −1 W −T W T ∆2 W
= W −1 ∆W W −1 ∆W
= Λ2
M Λ = W T W W −1 ∆W −T W T W = ΛT M ⇒ (u, Λv)M = (Λu, v)M
(Λ1/2 u, Λ1/2 u)M = (u, Λu)M
Householder, Berlin, 2008 – p.9/23
Interpolation spaces
h
1/2 o
i
n
H, M = u ∈ IRN ; (u, u)M + (u, S 1−ϑ u)M
h
ϑ
i
H, M
1/2
n
= u ∈ IRN ; (u, u)M + (u, Λu)M
1/2 o
Householder, Berlin, 2008 – p.10/23
Interpolation spaces
h
1/2 o
i
n
H, M = u ∈ IRN ; (u, u)M + (u, S 1−ϑ u)M
h
ϑ
i
H, M
1/2
n
= u ∈ IRN ; (u, u)M + (u, Λu)M
1/2 o
||v||2ϑ = ||v||2Hϑ = v T M + M S 1−ϑ v
Hϑ = M I + S 1−ϑ = W T I + ∆2(1−ϑ) W
Householder, Berlin, 2008 – p.10/23
Interpolation spaces (duality)
M′ and H′ dual spaces of M and H
h
i′
h
i
H, M = M′ , H′
ϑ
1−ϑ
S ′ = M H −1 = W T ∆−2 W −T
Householder, Berlin, 2008 – p.11/23
Interpolation spaces (duality)
M′ and H′ dual spaces of M and H
h
i′
h
i
H, M = M′ , H′
ϑ
1−ϑ
S ′ = M H −1 = W T ∆−2 W −T
−1
′
= H1−ϑ
= W −1 ∆−2ϑ W −T
H1−ϑ
Householder, Berlin, 2008 – p.11/23
Interpolation spaces (∞ dimension case)
X, Y two Hilbert spaces with X ⊂ Y , X dense and continuously
embedded in Y . h·, ·iX , h·, ·iY and k · kX , k · kY the respective norms.
(Riesz representation theory) ∃J : X → Y positive and self-adjoint
with respect to h·, ·iY such that hu, viX = hu, J viY .
E = J 1/2 : X → Y ,
2
2 1/2
.
X = D(E) with kukX ∼ kukE := kukY + kEukY
1/2
2
1−θ
2
kukθ := kukY + kE
ukY
.
1−θ
The interpolation space of index θ [X, Y ]θ :=
D(E
), 0 ≤ θ ≤ 1,
with the inner-product hu, viθ = hu, viY + u, E 1−θ v Y is a Hilbert
space (Lions Magenes 1968).
[X, Y ]0 = X and [X, Y ]1 = Y . If 0 < θ1 < θ2 < 1 then
X ⊂ [X, Y ]θ1 ⊂ [X, Y ]θ2 ⊂ Y.
∀θ ∈ (0, 1) π ∈ L(X; X ) ∩ L(Y ; Y) =⇒ π ∈ L([X, Y ]θ ; [X , Y]θ ).
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Interpolation spaces (∞ dimension case)
Ω ⊂ Rn open bounded with smooth boundary Γ and let α denote a
multi-index of order m where m is a positive integer
m
α
2
H (Ω) = u : D u ∈ L (Ω), |α| ≤ m
(H 0 (Ω) = L2 (Ω))
H s (Ω) := [H m (Ω), H 0 (Ω)]1−s/m
H0s (Ω) completion of C0∞ (Ω) in H m (Ω), where s > 0. For 0 ≤ s2 < s1 ,
(
(1−θ)s1 +θs2
(Ω)
if (1 − θ)s1 + θs2 6= k + 1/2 (k
[H0s1 (Ω), H0s2 (Ω)]θ = H0
k+1/2
[H0s1 (Ω), H0s2 (Ω)]θ = H00
k+1/2
(Ω) ⊂ H0
if (1 − θ)s1 + θs2 = k + 1/2 (k
H −s (Ω) = (H0s (Ω))′ s > 0
If (1 − θ)s1 + θs2 = 1/2
′
1/2
[H −s1 (Ω), H −s2 (Ω)]θ = H00 (Ω) .
Householder, Berlin, 2008 – p.13/23
Finite-element example
1/2
H00 (Ω) = [H01 (Ω), L2 (Ω)]1/2 .
Let Xh ⊂ H01 (Ω), Yh ⊂ L2 (Ω). Let {φi }1≤i≤n ∈ Xh be a spanning set for
Yh and let Lk ∈ Rn×n denote the Grammian matrices corresponding to the
h·, ·iH0k (Ω) -inner product (H 0 (Ω) = L2 (Ω)):
(Lk )ij = hφi , φj iH k (Ω) .
0
H = L1 , M = L0 and
H1/2,h
Moreover, we have
1/2
= L0 I + (L−1
L
)
1
0
H1/2,h ∼ H1/2 = L0 L−1
0 L1
1/2
Householder, Berlin, 2008 – p.14/23
Finite-element example
′
1/2
H00 (Ω)
= [H
−(1)
0
(Ω), H (Ω)]1/2 =
′
1
0
[H0 (Ω), H0 (Ω)]1/2 .
Let Xh , Yh be defined as above. Let Yh′ ⊂ Xh′ = span {ψi }1≤i≤n where ψi
are basis functions dual to φi , i.e.,hψi , φj iH −1 (Ω)×H01 (Ω) = δij .
If Yh′ ∋ z =
n
X
zi ψi =
i=1
δij = hψi , φj iH −1 (Ω)×H01 (Ω) =
n
X
wi φi , φl =
i=1
n
X
n
X
Kli ψi
i=1
Kil−1 hφl , φj iH 1 (Ω) =
0
i=1
n
X
Kil−1 (L1 )lj
i=1
′ = kwk
,
kzk
and the
so that z = L0 w and kzkHY′ = kwkL−1
H
L0 L−1
X
1 L0
0
′
matrix representation of HY′ , HX
are respectively L0 and L0 L−1
1 L0 .
′
−1/2
H1/2
= L0 (L−1
L
)
= H−1/2 .
1
0
Householder, Berlin, 2008 – p.15/23
Evaluation of Hθ z
Generalised Lanczos
HX Vk = HY Vk Tk + βk+1 Hy vk+1 eTk , VkT HY Vk = Ik
(Tk tridiagonal).
Householder, Berlin, 2008 – p.16/23
Evaluation of Hθ z
Generalised Lanczos
HX Vk = HY Vk Tk + βk+1 Hy vk+1 eTk , VkT HY Vk = Ik
(Tk tridiagonal).
v0 = z
Hθ z ≈ HY Vk Tk1−θ e1 kzkHY and
Hθ,h z ≈ HY Vk (Ik + Tk1−θ )e1 kzkHY .
Householder, Berlin, 2008 – p.16/23
Evaluation of Hθ z
Generalised Lanczos
HX Vk = HY Vk Tk + βk+1 Hy vk+1 eTk , VkT HY Vk = Ik
(Tk tridiagonal).
v0 = z
Hθ z ≈ HY Vk Tk1−θ e1 kzkHY and
Hθ,h z ≈ HY Vk (Ik + Tk1−θ )e1 kzkHY .
v0 = HY−1 z
Hθ−1 z ≈ Vk Tkθ−1 e1 kzkHY−1 and
−1
Hθ,h
z ≈ Vk (Ik + Tk1−θ )−1 e1 kzkHY−1 .
Householder, Berlin, 2008 – p.16/23
Evaluation of Hθ z
Generalised Lanczos
HX Vk = HY Vk Tk + βk+1 Hy vk+1 eTk , VkT HY Vk = Ik
(Tk tridiagonal).
v0 = z
Hθ z ≈ HY Vk Tk1−θ e1 kzkHY and
Hθ,h z ≈ HY Vk (Ik + Tk1−θ )e1 kzkHY .
v0 = HY−1 z
Hθ−1 z ≈ Vk Tkθ−1 e1 kzkHY−1 and
−1
Hθ,h
z ≈ Vk (Ik + Tk1−θ )−1 e1 kzkHY−1 .
Alternative: N. Hale, and N. J. Higham and L. N. Trefethen, SIAM J. Numer. Anal.
Householder, Berlin, 2008 – p.16/23
An example: biharmonic operator
Consider the biharmonic problem in a polygonal convex open domain
Ω ⊂ R2 with boundary Γ = ∪K
i=1 Γi .
(
∆2 u = f
in Ω,
u = ∂u/∂n = 0
on Γ.



−∆u = f
v + ∆u = 0


u = ∂u/∂n = 0
in Ω,
in Ω,
on Γ.
Householder, Berlin, 2008 – p.17/23
An example: biharmonic operator
(Pironneau-Glowinski)
(i)
(
Sλ = ∂u0 /∂ν
(ii)
(iii)
−∆v0 = f
v0 = 0
(
−∆v1 = 0
v1 = λ
in Ω,
on Γ,
(
−∆u0 = v0
u0 = 0
in Ω,
on Γ,
(
−∆u1 = v1
u1 = 0
in Ω,
on Γ,
on Γ,
in Ω,
on Γ,
Householder, Berlin, 2008 – p.18/23
An example: biharmonic operator
S is a boundary operator which is defined on H −1/2 (Γ) and which induces
a bilinear form
s(·, ·) : H −1/2 (Γ) × H −1/2 (Γ)
s(·, ·) is symmetric, positive-definite and H −1/2 (Γ)−elliptic, i.e.,
c1 kλk2H −1/2 (Γ) ≤ s(λ, λ) ≤ c2 kλk2H −1/2 (Γ) .
Householder, Berlin, 2008 – p.19/23
An example: biharmonic operator
L
ZT
Z
−MBB
!
x
vB
!
=
!
g
0
The Schur complement associated with L in the matrix is
S = −MBB − Z T L−1 Z.
Let Xh ⊂ H 1 (Γ) denote the space spanned by the restriction of the basis
functions of VIh to the boundary Γ. If λh ∈ Xh has a vector of coefficients
λ , then
λ.
s(λh , λh ) = λ T Sλ
Householder, Berlin, 2008 – p.20/23
An example: biharmonic operator
A discrete H −1/2 -norm on Xh can be defined as a sum of norms
corresponding to each open segment of the polygonal boundary Γ:
!1/2
K
X
kλh kH −1/2 (Γ) :=
kλh k2H −1/2 (Γi )
.
i=1
1/2
In particular, H −1/2 (Γi ) = (H00 (Γi ))′ .
λk2H{−1/2,h}
kλh k2H −1/2 (Γ) = kλ
for λh ∈ Xh where
H{−1/2,h} =
K
M
λk2H{−1/2}
or = kλ
′{i}
H1/2,h
i=1
K
M
{i}
H{−1/2} =
H−1/2 ,
i=1
{i}
H−1/2
−1/2
L
)
= L0,i (L−1
1,i
0,i
Householder, Berlin, 2008 – p.21/23
An example: biharmonic operator
P =
L
Z
PS
!
Householder, Berlin, 2008 – p.22/23
An example: biharmonic operator
P =
L
Z
PS
!
n
m
I
H{−1/2,h}
H{−1/2}
b {−1/2}
H
84,610
640
26
10
12
12
337,154
1,280
30
9
11
11
1,346,050
2,560
36
9
11
11
FGMRES iterations for model problem .
Householder, Berlin, 2008 – p.22/23
Open problems
Domain decomposition: preliminary results show total independence
from mesh size
Householder, Berlin, 2008 – p.23/23
Open problems
Domain decomposition: preliminary results show total independence
from mesh size
Strange domains: 1D simplex (wirebasket) and QUANTUM GRAPH
Householder, Berlin, 2008 – p.23/23
Open problems
Domain decomposition: preliminary results show total independence
from mesh size
Strange domains: 1D simplex (wirebasket) and QUANTUM GRAPH
3D PDEs
Householder, Berlin, 2008 – p.23/23
Open problems
Domain decomposition: preliminary results show total independence
from mesh size
Strange domains: 1D simplex (wirebasket) and QUANTUM GRAPH
3D PDEs
Utilization in Image Denoising
Householder, Berlin, 2008 – p.23/23
Open problems
Domain decomposition: preliminary results show total independence
from mesh size
Strange domains: 1D simplex (wirebasket) and QUANTUM GRAPH
3D PDEs
Utilization in Image Denoising
Hs s > 1
Householder, Berlin, 2008 – p.23/23
Open problems
Domain decomposition: preliminary results show total independence
from mesh size
Strange domains: 1D simplex (wirebasket) and QUANTUM GRAPH
3D PDEs
Utilization in Image Denoising
Hs s > 1
Generalization to Banach spaces (K-method Peetre)
Householder, Berlin, 2008 – p.23/23
Open problems
Domain decomposition: preliminary results show total independence
from mesh size
Strange domains: 1D simplex (wirebasket) and QUANTUM GRAPH
3D PDEs
Utilization in Image Denoising
Hs s > 1
Generalization to Banach spaces (K-method Peetre)
For which class of matrices the Schur complement is spectrally
equivalent to an algebraic interpolation space matrix
Householder, Berlin, 2008 – p.23/23