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Jacobi Davidson method and its applications
Tsung-Ming Huang
Department of Mathematics National Taiwan Normal University, Taiwan
October 31, 2008
T.M. Huang (Taiwan Normal Univ.) JD method and applications October 31, 2008 1 / 71
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Outline
1 Some basic theorems
2 Jacobi-Davidson method for linear eigenvalue problems
3 Polynomial eigenvalue problems
4 Locking method for computing λ2, . . . , λ`
5 Non-equivalence deflation of quadratic eigenproblems
6 Non-equivalence deflation for cubic polynomial eigenproblems
7 Applications
8 Numerical experiments for linear eigenproblems
9 Numerical experiments for polynomial eigenproblems
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Some basic theorems
Eigenproblems
Standard eigenproblems: Ax = λx Generalized eigenproblems: Ax = λBx
Higher order poly. eigenproblems: (A0+ λA1+ .... + λnAn)x = 0 Eigenproblems of λ-matrices: F (λ)x = 0
What do we care ?
(i) In theory: eigenstructure, spectrum decomposition, canonical form, . . . , etc.
(ii) In computation: eigenvalues, eigenvectors, invariant subspaces, . . . , etc.
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Theorem (Fischer)
Let the Hermitian matrix A have eigenvalues λ1 ≥ λ2 ≥ · · · ≥ λn. Then
λi = min dim(W)=n−i +1
w ∈W,kw kmax2=1wHAw
and
λi = max dim(W)=i
min
w ∈W,kw k2=1wHAw
.
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Theorem
Let X be an eigenspace of A and let X be a basis for X . Then there is a unique matrix L such that
AX = XL.
The matrix L is given by
L = XIAX , where XI is a matrix satisfying XIX = I .
If (λ, x ) is an eigenpair of A with x ∈ X , then (λ, XIx ) is an eigenpair of L.
Conversely, if (λ, u) is an eigenpair of L, then (λ, Xu) is an eigenpair of A.
T.M. Huang (Taiwan Normal Univ.) JD method and applications October 31, 2008 5 / 71
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Proof:
Let
X = [x1· · · xk] and Y = AX = [y1· · · yk] .
Since yi ∈ X and X is a basis for X , there is a unique vector `i such that yi = X `i.
If we set L = [`1· · · `k], then AX = XL and L = XIXL = XIAX .
Now let (λ, x ) be an eigenpair of A with x ∈ X . Then there is a unique vector u such that x = Xu. However, u = XIx . Hence
λx = Ax = AXu = XLu ⇒ λu = λXIx = Lu.
Conversely, if Lu = λu, then
A(Xu) = (AX )u = (XL)u = X (Lu) = λ(Xu), so that (λ, Xu) is an eigenpair of A.
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Theorem (Optimal residuals) Let [X X⊥] be unitary. Let
R = AX − XL and SH = XHA − LXH. Then kRk and kS k are minimized when
L = XHAX , in which case
(a) kRk = kX⊥HAX k, (b) kSk = kXHAX⊥k, (c) XHR = 0.
T.M. Huang (Taiwan Normal Univ.) JD method and applications October 31, 2008 7 / 71
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Proof:
Set
XH X⊥H
A
X X⊥ =
ˆL H
G M
. Then
XH X⊥H
R =
ˆL H
G M
XH X⊥H
X −
XH X⊥H
XL =
L − Lˆ G
. It implies that
kRk =
XH X⊥H
R
=
ˆL − L G
, which is minimized when L = ˆL = XHAX and
min kRk = kG k = kX⊥HAX k.
The proof for S is similar. If L = XHAX , then
XHR = XHAX − XHXL = XHAX − L = 0.
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Definition
Let X be of full column rank and let XI be a left inverse of X . Then XIAX is a Rayleigh quotient of A.
Theorem
Let X be orthonormal, A be Hermitian and R = AX − XL.
If `1, . . . , `k are the eigenvalues of L, then there are eigenvalues λj1, . . . , λjk of A such that
|`i − λji| ≤ kRk2 and v u u t
k
X
i =1
(`i − λji)2 ≤√
2kRkF.
T.M. Huang (Taiwan Normal Univ.) JD method and applications October 31, 2008 9 / 71
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Jacobi’s orthogonal component correction (JOCC), 1846
Consider the eigenvalue problem A
1 z
≡
α cT b F
1 z
= λ
1 z
, (1)
where A is diagonal dominant and α is the largest diagonal element.
(1) is equivalent to
λ = α + cTz, (F − λI )z = −b.
Jacobi iteration : (with z1 = 0)
θk = α + cTzk,
(D − θkI )zk+1 = (D − F )zk − b (2) where D = diag(F ).
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Davidson’s method (1975)
Algorithm (Davidson’s method) Given unit vector v , set V = [v ] Iterate until convergence
Compute desired eigenpair (θ, s) ofVTAV. Computeu = Vs andr = Au − θu.
If (k r k2< ε), stop.
Solve (DA− θI )t = r.
Orthog. t ⊥ V → v , V = [V , v ] end
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Let uk = (1, zkT)T. Then
rk = (A − θkI )uk =
α − θk+ cTzk (F − θkI )zk + b
Substituting the residual vector rk into linear systems (DA− θkI )tk = −rk, where DA =
α 0
0 D
, we get
(D − θkI )yk = −(F − θkI )zk − b
= (D − F )zk − (D − θkI )zk− b From (2) and above equality, we see that
(D − θkI )(zk+ yk) = (D − F )zk − b = (D − θkI )zk+1
This implies that zk+1 = zk+ yk as one step of JOCC starting with zk.
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Polynomial eigenvalue problems
Polynomial eigenvalue problems:
A(λ)x ≡ λτAτ + · · · + +λ2A2+ λA1+ A0 x = 0. (3) Enlarged linear eigenvalue problem: (e.g. cubic polynomial)
0 I 0
0 0 I
A0 A1 A2
x λx λ2x
= λ
I 0 0
0 I 0
0 0 −A3
x λx λ2x
. Disadvantages:
I The order of the larger matrices are tripled
I The condition number of eigenvalues and eigenvectors may increase
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Consider the quadratic eigenvalue problem1 Q(λ)x ≡ λ2M + λC + K x = 0 with
M = 1
2In, K = diag1≤j ≤n j2π2(j2π2+ τ − κv2)/2 , C = −CT = (cij) with cij =
( 4ij
j2−i2v if i + j is odd 0 otherwise.
Take v = 10, κ = 0.8 and τ = 77.9.
Enlarged linear eigenvalue problem:
K 0 0 I
x λx
= λ
−C −M
I 0
x λx
1F. Tisseur and K. Meerbergen, SIAM Rev. Vol. 43, pp. 235-286, 2001
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−1 0 1
x 10−9
−4
−3
−2
−1 0 1 2 3
4x 104 n = 60
polyeig exact sol
Figure: The spectrum of Q(λ)
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Polynomial Jacobi-Davidson method
(θk, uk): approx. eigenpair of A(λ), θk ≈ λ, with
uk = Vksk, VkTA(λ)Vksk = 0 and k sk k2= 1.
Let
rk = A(θk)uk. Then
ukTrk = uTkA(θk)uk = skTVkTA(θk)Vksk = 0 ⇒ rk ⊥ uk Find the correction t such that
A(λ)(uk+ t) = 0.
That is
A(λ)t = −A(λ)uk = −rk + (A(θk) − A(λ))uk.
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Let
pk = A0(θk)uk ≡
τ
X
i =1
i θi −1k Ai
! uk.
A(λ) = A − λI :
pk = −uk,
(A(θk) − A(λ))uk = (λ − θk)uk = (θk − λk)pk A(λ) = A − λB:
pk = −Buk,
(A(θk) − A(λ))uk = (λ − θk)Buk = (θk− λ)pk A(λ) =Pτ
i =0λiAi with τ ≥ 2:
(A(θk) − A(λ))uk =
(θk− λ)A0(θk) −1
2(θk − λ)2A00(ξk)
uk
= (θk − λ)pk −1
2(θk − λ)2A00(ξk)uk
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Hence
A(λ)t = −rk+ (θk − λ)pk− 1
2(θk− λ)2A00(ξk)uk
Since rk ⊥ uk, we have
I −pkukT ukTpk
A(λ)t = −rk −1
2(θk− λ)2
I − pkukT ukTpk
A00(ξk)uk. Correction equation:
I − pkuTk ukTpk
A(θk)(I − ukukT)t = −rk and t ⊥ uk,
or
I − pkukT ukTpk
(A − θkB)
I − ukpTk pkTuk
t = −rk and t ⊥B uk,
with symmetric positive definite matrix B.
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Solving correction vector t
Correction equation:
I − pkuTk ukTpk
A(θk)(I − ukukT)t = −rk. (4) Method I:
Use preconditioning iterative approximations, e.g., GMRES, to solve (4).
Use a preconditioner Mp≡
I − pkuTk ukTpk
M
I − ukuTk
≈
I − pkukT ukTpk
A(θk)
I − ukukT
, where M is an approximation of A(θk).
In each of the iterative steps, it needs to solve the linear system
Mpt = y , t ⊥ uk (5)
for a given y .
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Since t ⊥ uk, Eq. (5) can be rewritten as
I − pkuTk ukTpk
Mt = y ⇒ Mt = ukTMt
ukTpk pk+ y ≡ ηkpk + y . Hence
t = M−1y + ηkM−1pk, where
ηk = − ukTM−1y ukTM−1pk.
SSOR preconditioner: Let A(θk) = L + D + U. Then M = (D + ωL)D−1(D + ωU).
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Method II: Since t ⊥ uk, Eq. (4) can be rewritten as A(θk)t = uTkA(θk)t
ukTpk pk − rk ≡ εpk − rk. (6) Let t1 and t2 be approximated solutions of the following linear
systems:
A(θk)t = −rk and A(θk)t = pk, respectively. Then the approximated solution ˜t for (6) is
˜t = t1+ εt2 for ε = −ukTt1 ukTt2
. The approximated solution ˜t for (6) is
˜t = −M−1rk+ εM−1pk for ε = uTkM−1rk ukTM−1pk, where M is an approximation of A(θk).
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Method III:
Eq. (6) implies that
t = εA(θk)−1pk − A(θk)−1rk = εA(θk)−1pk − uk. Let t1 be approximated solution of the following linear system:
A(θk)t = pk. Then the approximated solution ˜t for (6) is
˜t = εt1− uk for ε =
ukTt1−1
.
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Algorithm (Jacobi-Davidson Algorithm for solving A(λ)x = 0) Choose an n-by-m orthonormal matrix V0
Do k = 0, 1, 2, · · ·
Compute all the eigenpairs of VkTA(λ)Vk = 0.
Select the desired (target) eigenpair (θk, sk) with kskk2= 1.
Computeuk = Vksk,rk = A(θk)uk andpk = A0(θk)uk. If (krkk2 < ε), λ = θk, x = uk, Stop
Solve (approximately) a tk ⊥ uk from (I − pku
T k
uTkpk)A(θk)(I − ukuTk)t = −rk. Orthogonalize tk ⊥ Vk → vk+1, Vk+1 = [Vk, vk+1]
T.M. Huang (Taiwan Normal Univ.) JD method and applications October 31, 2008 23 / 71
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Restarting
A double precision real variable needs to 8 bytes to save in Memory.
125 double precision real variables ≈ 1 KB 125, 000 double precision real variables ≈ 1 MB
Keep the locked Schur vectors as well as the Schur vectors of interest in the subspace and throw away those we are not interested.
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(v) Solve correct equation (approximately) to obtain a t ⊥ uk by the method determined below.
If ( krkk2> 0.1 and k ≤ 9 ) then
Use {BiCGSTAB, No precond., 7, 10−3} else
Use {GMRES, SSOR, 30, 10−3} End if
Figure: The heuristic strategy for computing the first target eigenvalue.
T.M. Huang (Taiwan Normal Univ.) JD method and applications October 31, 2008 25 / 71
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Locking for Ax = λx
Vk with Vk∗V = Ik are convergentSchur vectors, i.e., AVk = VkTk
for someupper triangular Tk. SetV = [Vk, Vq]with V∗V = Ik+q in k + 1-th iteration of Jacobi-Davidson Algorithm. Then
V∗AV =
Vk∗AVk Vk∗AVq Vq∗AVk Vq∗AVq
=
Tk Vk∗AVq Vq∗VkTk Vq∗AVq
=
Tk Vk∗AVq 0 Vq∗AVq
.
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Locking Polynomial Jacobi–Davidson Method (0) Given A(λ) =Pτ
i =0λiAi and the number of desired eigenvalues σ.
(1) Initialize V = [Vini] as an orthonormal matrix and Vx = [].
(2) For j = 1, 2, ..., σ
(2.1) Iterate until convergence
(i) Compute thej th desired eigenpairs (θ, s) of VTA(θ)V . (ii) Compute u = Vs, p = A0(θ)u, and r = A(θ)u.
(iii) If (||r ||2 < ε), Set λj = θ, xj = u, Goto Step (2.2).
(iv) Solve (approximately) a t ⊥ u from (I − puuTTp)A(θ)(I − uuT)t = −r . (v) Orthogonalize t ⊥ V → v , V = [V , v ] (2.2) Orthogonalizexj ⊥ Vx → xj; Vx = [Vx, xj]
(2.3) Choose an orthonormal matrixVini ⊥ Vx; Set V = [Vx, Vini] Ref: G. L. G. Sleijpen, G. L. Booten, D. R. Fokkema and H. A. van der Vorst, BIT, 36:595-633, 1996
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(v) Solve correct equation (approximately) to obtain a t ⊥ uk by the method determined below.
If ( k rk k2> 0.1 and k < 10 ) then
Use {BiCGSTAB, No precond., 7, 10−3} else if ( k rk k2≥ 0.1 and k > 14 ) then
Use {GMRES, SSOR, 30, 10−3}
else if ( k rk k2< 0.1 and k rk−1 k2/ k rk k2< 4) then Set j = min(30, j + 2) and use {GMRES, SSOR, j, 10−3} else
Use {GMRES, SSOR, j, 10−3} End if
Figure: The heuristic strategy for computing eigenvalues other than the first convergent eigenvalue.
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Non-equivalence deflation of quadratic eigenproblems
Let λ1 be a real eigenvalue of Q(λ) and x1, z1 be the associated right and left eigenvectors, respectively, with z1TKx1 = 1. Let
θ1 = (z1TMx1)−1. We introduce a deflated quadratic eigenproblem
Q(λ)x ≡e h
λ2M + λ ee C + eKi x = 0, where
Me = M − θ1Mx1z1TM, Ce = C + θ1
λ1
(Mx1z1TK + Kx1z1TM), Ke = K − θ1
λ21Kx1z1TK .
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Complex deflation
Let λ1 = α1+ i β1 be a complex eigenvalue of Q(λ) and x1 = x1R+ ix1I, z1 = z1R+ iz1I be the associated right and left eigenvectors, respectively, such that
Z1TKX1 = I2, where X1 = [x1R, x1I] and Z1 = [z1R, z1I]. Let
Θ1 = (Z1TMX1)−1.
Then we introduce a deflated quadratic eigenproblem with Me = M − MX1Θ1Z1TM,
Ce = C + MX1Θ1Λ−T1 Z1TK + KX1Λ−11 ΘT1Z1TM, Ke = K − KX1Λ−11 Θ1Λ−T1 Z1TK
in which Λ1 =
α1 β1
−β1 α1
.
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Theorem
(i) Let λ1 be a simple real eigenvalue of Q(λ). Then the spectrum of Q(λ) is given bye
(σ(Q(λ)){λ1}) ∪ {∞}
provided that λ21 6= θ1.
(ii) Let λ1 be a simple complex eigenvalue of Q(λ). Then the spectrum of eQ(λ) is given by
σ(Q(λ)){λ1, ¯λ1} ∪ {∞, ∞}
provided that Λ1ΛT1 6= Θ1.
Furthermore, in both cases (i) and (ii), if λ2 6= λ1 and (λ2, x2) is an eigenpair of Q(λ) then the pair (λ2, x2) is also an eigenpair of eQ(λ).
Ref: T.-M. Hwang, W.-W. Lin and V. Mehrmann, SIAM J. Sci. Comput.
T.M. Huang (Taiwan Normal Univ.) JD method and applications October 31, 2008 31 / 71
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Suppose that M, C , K are symmetric. Given an eigenmatrix pair
(Λ1, X1) ∈ Rr ×r × Rn×r of Q(λ), where Λ1 is nonsingular and X1 satisfies X1TKX1 = Ir, Θ1:= (X1TMX1)−1.
We define ˜Q(λ) := λ2M + λ ˜˜ C + ˜K , where M˜ := M − MX1Θ1X1TM,
C˜ := C + MX1Θ1Λ−T1 X1TK + KX1Λ−11 Θ1X1TM, K˜ := K − KX1Λ−11 Θ1Λ−T1 X1TK .
Theorem
Suppose that Θ1− Λ1ΛT1 is nonsingular. Then the eigenvalues of the real symmetric quadratic pencil ˜Q(λ) are the same as those of Q(λ) except that the eigenvalues of Λ1, which are closed under complex conjugation, are replaced by r infinities.
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Proof:
Since (Λ1, X1) is an eigenmatrix pair of Q(λ), i.e., MX1Λ21+ CX1Λ1+ KX1 = 0, we have
Q(λ)˜ = Q(λ) + [MX1(λIr + Λ1) + CX1] Θ1Λ−T1 (X1TK − λΛT1X1TM)
= Q(λ) + Q(λ)X1(λIr − Λ1)−1Θ1Λ−T1 (X1TK − λΛT1X1TM).
By using the identity
det(In+ RS ) = det(Im+ SR), where R, ST ∈ Rn×m, we have
det[ ˜Q(λ)]
= det[Q(λ)]det[I + X1(λIr − Λ1)−1Θ1Λ−T1 (X1TK − λΛT1X1TM)]
= det[Q(λ)]det[Ir + (λIr − Λ1)−1Θ1Λ−T1 (Ir − λΛT1Θ−11 )]
= det[Q(λ)]
det(λIr − Λ1)det(Θ1Λ−T1 − Λ1).
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Since (Θ1− Λ1ΛT1) ∈ Rr ×r is nonsingular, we have det(Θ1Λ−T1 − Λ1) 6= 0.
Therefore, ˜Q(λ) has the same eigenvalues as Q(λ) except that r eigenvalues of Λ1 are replaced by r infinities.
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Non-equivalence deflation for cubic polynomial eigenproblems
Let (Λ, Vu) ∈ Rr ×r × Rn×r be an eigenmatrix pair of A(λ) with VuTVu= Ir and 0 /∈ σ(Λ), i.e.,
A3VuΛ3+ A2VuΛ2+ A1VuΛ + A0Vu= 0. (7) Define a new deflated cubic eigenvalue problem by
A(λ)u = (λe 3A˜3+ λ2A˜2+ λ˜A1+ ˜A0)u = 0, (8) where
A˜0 = A0,
A˜1 = A1− (A1VuVuT+ A2VuΛVuT + A3VuΛ2VuT), A˜2 = A2− (A2VuVuT+ A3VuΛVuT),
A˜3 = A3− A3VuVuT.
(9)
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Lemma
Let A(λ) and eA(λ) be cubic pencils given by (3) and (8), respectively.
Then it holds
A(λ) = A(λ)e
In− λVu(λIr − Λ)−1VuT
. (10)
Theorem
Let (Λ, Vu) be an eigenmatrix pair of A(λ) with VuTVu= Ir. Then (i) (σ(A(λ))σ(Λ)) ∪ {∞} = σ(eA(λ)).
(ii) Let (µ, z) be an eigenpair of A(λ) with k z k2= 1 and µ /∈ σ(Λ).
Define
˜
z = (In− µVuΛ−1VuT)z ≡ T (µ)z. (11) Then (µ, ˜z) is an eigenpair of eA(λ).
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Proof of Lemma:
Using (9) and (7), and the fundamental matrix calculation, we have A(λ)e = A(λ) − λ
λ2A3VFVFT + λA2VFVFT + λA3VFΛVFT+ A1VFVFT +A2VFΛVFT + A3VFΛ2VFT
= A(λ) − λ
A3VF(λIr − Λ)3(λIr − Λ)−1VFT +3A3VFΛ(λIr − Λ)2(λIr − Λ)−1VFT +3A3VFΛ2(λIr − Λ)(λIr − Λ)−1VFT +A2VF(λIr − Λ)2(λIr − Λ)−1VFT +2A2VFΛ(λIr − Λ)(λIr − Λ)−1VFT
+A1VF(λIr − Λ)(λIr − Λ)−1VFT
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A(λ)e = A(λ) − λA3VF(λ3Ir − Λ3) + A2VF(λ2Ir − Λ2) +A1VF(λIr − Λ) + A0VF − A0VF] (λIr − Λ)−1VFT
o
= A(λ) − λh
A(λ)VF(λIr − Λ)−1VFTi
= A(λ)h
In− λVF(λIr − Λ)−1VFTi .
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Proof of Theorem
: (i) Using the identity
det(In+ RS ) = det(Im+ SR) and Lemma 8, we have
det( eA(λ)) = det(A(λ)) det
In− λVF(λIr − Λ)−1VFT
= det(A(λ)) det In− λ(λIr − Λ)−1
= det(A(λ)) det(λIr − Λ)−1det(−Λ).
Since 0 /∈ σ(Λ), det(−Λ) 6= 0. Thus, eA(λ) and A(λ) have the same finite spectrum except the eigenvalues in σ(Λ). Furthermore, dividing Eq. (8) by λ3 and using the fact that
Ae3VF = (A3− A3VFVFT)VF = 0,
we see that (diagr{∞, · · · , ∞}, VF) is an eigenmatrix pair of eA(λ) corresponding to infinite eigenvalues.
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(ii) Since µ /∈ σ(Λ), the matrix T (µ) = (I − µVFΛ−1VFT) in (11) is invertible with the inverse
T (µ)−1= In− µVF(µIr − Λ)−1VFT. (12) From Lemma 8, we have
A(µ)˜e z = A(µ) h
In− µVF(µIr − Λ)−1VFT i h
In− µVFΛ−1VFT i
z = 0.
This completes the proof.
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Applications
Quantum well (2 dim.)
Quantum wire (1 dim.)
Quantum dot (0 dim.)
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Nanometer
1nm = 10−9m
Nano-scale ≈ 1–100 nm A semiconductor QD ≈ 10 nm QD : hair ≈ 1 : 10000
Why consider quantum effects?
Small devices imply significant quantum effect
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Quantum dot
Cross-sections of hetero-structure InAs/GaAs QDs by Transmission Electron Microscope [Schoenfeld, 00]
150 Å
300 Å 150 Å
300 Å 150 Å
300 Å 150 Å
300 Å
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Molecular beam epitaxy
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Quantum dot
Cross-sections of hetero-structure InAs/GaAs QDs by Transmission Electron Microscope [Schoenfeld, 00]
150 Å
300 Å 150 Å
300 Å 150 Å
300 Å 150 Å
300 Å
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Energy levels (eigenvalues)
Confined Discrete Energy Levels Confined Discrete
Energy Levels
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Numerical experiments for linear eigenproblems
TheSchr¨odinger equationfor Semiconductor:
−∇ · (α∇u) + Vu = λu,
where α =
( α−≡ 2m~2
1 inside, α+≡ 2m~2
2 outside, V =
V−= V1 inside, V+= V2 outside
~: Plank constant m`: parabolic effective mass V`: confinement potential λ: total energy
Interface condition:
α−∂u
∂n ∂D−
= α+∂u
∂n ∂D+
Dirichlet boundary conditions
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Symmetric eigenvalue problem
Ax = λx where A is a symmetric matrix.
Reference:
Tsung-Min Hwang, Wen-Wei Lin, Wei-Cheng Wang and Weichung Wang, Numerical simulation of three dimensional pyramid quantum dot, Journal of Computational Physics, Vol 196, pp. 208-232, 2004.
Figure: PRB, 54, 8743, (1996)
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Three dimensional pyramid quantum dot
H
W W
Pyramid Dot Cuboid Matrix i
2 1. 5 1 0. 5 0 0.5 1 1.5 2
1 0. 5 0 0.5 1 1.5 2
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 4 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 4 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 4 1 1 2 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 4 1 1 1 2 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 4 1 1 1 1 2 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 4 1 1 1 1 1 2 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 4 1 1 1 1 1 1 2 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 4 1 1 1 1 1 1 1 5 0 0 0 0 0 0 0
0 0 0 0 0 0 0 4 1 1 1 1 1 1 3 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 4 1 1 1 1 1 3 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 4 1 1 1 1 3 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 4 1 1 1 3 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 4 1 1 3 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 4 1 3 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 4 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
T.M. Huang (Taiwan Normal Univ.) JD method and applications October 31, 2008 49 / 71
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Finite volume discretized scheme Symmetric eigenvalue problems:
Ax = λx
Second order convergent rate Correction vector with SSOR preconditioner:
t = −MA−1r + εMA−1p
with
ε = uTMA−1r uTMA−1p,
whereMA= (D + ωL)D−1(D + ωU).
H
W W
Pyramid Dot Cuboid Matrix i
Figure: Structure schema of a pyramid quantum dot.
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The width of the QD (matrix) base is 12.4 nm (24.8 nm); the height of the QD (matrix) is 6.2 nm (18.6 nm).
InAs QD: α1 = 0.024me and V1 = 0.0 GaAs matrix: α2= 0.067me and V2= 0.70.
Stopping criteria: residual < 10−10 Convergent rate
(L,M,N) Mtx. dim. λ1 Rate λ2 Rate
(16, 16, 12) 2,475 0.4226 - 0.6527 - (32, 32, 24) 22,103 0.4001 - 0.6423 - (64, 64, 48) 186,543 0.3934 1.744 0.6391 1.708 (128,128, 96) 1,532,255 0.3916 1.905 0.6383 1.866 (256,256,192) 12,419,775 0.3911 1.954 0.6380 1.912 Table: Convergent rate =log2 (λ(4h)− λ(2h))/(λ(2h)− λ(h))
T.M. Huang (Taiwan Normal Univ.) JD method and applications October 31, 2008 51 / 71
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dim(A) = 1, 532, 255on PC with P4 1.8 GHz CPUand1 GB of main memory.
Value Ite. no. CPU time (sec.)
λ1 0.3916 72 278.0
λ2 0.6383 72 284.3
λ3 0.6383 125 521.4
dim(A) = 32, 401, 863on HP workstation with a 1.0 GHz Intel Itanium II CPUand 12 GBsof main memory.
Value Ite. no. CPU time (sec.)
λ1 0.3910 138 5,852
λ2 0.6380 133 5,354
λ3 0.6380 220 8,511
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Unsymmetric eigenvalue problem
Ax = λx where A is a unsymmetric matrix.
Reference:
Tsung-Min Hwang, Wei-Hua Wang and Weichung Wang, Efficient numerical schemes for electronic states in coupled quantum dots, accepted for publication in Journal of Nanoscience and
Nanotechnology.
JAP, 90-12, (2001)-
Figure: JAP, 90-12, (2001)
T.M. Huang (Taiwan Normal Univ.) JD method and applications October 31, 2008 53 / 71
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Vertically aligned quantum dot array
(a) Nonuniform mes h (b) Uniform mesh
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Reduce three-dimensional systems to two-dimensional systems by Fourier transformation.
Finite volume discretized scheme Unsymmetric eigenvalue problems:
Ax = λx
Second order convergent rate
Method II with SSOR preconditioner:
MA= (D + ωL)D−1(D + ωU).
Figure: Structure schema of a cylindrical vertically aligned quantum dot array.
T.M. Huang (Taiwan Normal Univ.) JD method and applications October 31, 2008 55 / 71
師大
Reduce three-dimensional systems to two-dimensional systems by Fourier transformation.
Finite volume discretized scheme Unsymmetric eigenvalue problems:
Ax = λx
Second order convergent rate
Method II with SSOR preconditioner:
MA= (D + ωL)D−1(D + ωU). Figure: Structure schema of a cylindrical vertically aligned quantum dot array.
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Reduce three-dimensional systems to two-dimensional systems by Fourier transformation.
Finite volume discretized scheme Unsymmetric eigenvalue problems:
Ax = λx
Second order convergent rate
Method II with SSOR preconditioner:
MA= (D + ωL)D−1(D + ωU).
Figure: Structure schema of a cylindrical vertically aligned quantum dot array.
T.M. Huang (Taiwan Normal Univ.) JD method and applications October 31, 2008 55 / 71
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gap = 6nm and dim(A) = 12, 288, 000on HP workstation with a 1.0 GHz Intel Itanium II CPU and12 GBsof main memory.
r2 = 3.198 r2 = 3.223
Value Ite. no. Time (sec.) Value Ite. no. Time (sec.)
λ1 0.1587 83 20,287 0.1587 84 23,840
λ2 0.35553 109 26,030 0.3526 91 31,612
λ3 0.3558 53 13,831 0.3555 61 25,706
gap = 3nm and dim(A) = 11, 059, 200on HP workstation with a 1.0 GHz Intel Itanium II CPU and12 GBsof main memory.
r2 = 3.198 r2 = 3.223
Value Ite. no. Time (sec.) Value Ite. no. Time (sec.)
λ1 0.1586 85 20,072 0.1586 83 18,455
λ2 0.3540 257 56,811 0.3515 130 29,506
λ3 0.3564 53 12,257 0.3558 54 13,082
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Generalized eigenvalue problem
Ax = λBx
where A is symmetric positive definiteand B is a positive diagonalmatrix.
Reference:
Tsung-Min Hwang, Wei-Cheng Wang and Weichung Wang, Numerical schemes for three dimensional irregular shape quantum dots over curvilinear coordinate systems, accepted for publication in Journal of Computational Physics.
T.M. Huang (Taiwan Normal Univ.) JD method and applications October 31, 2008 57 / 71
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Three dimensional arbitrary shape quantum dots
Quantum dot Matrix
ZA ZB
ZC
1 M
Rmtx
0 H
N1
N2
N
H HC
HRmtx
R adial coordinate
Axial coordinate
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Curvilinear coordinate system
Reduce three-dimensional systems to two-dimensional systems by Fourier transformation.
Jump condition capturing scheme (nine points)
Generalized eigenvalue problems:
Ax = λBx , A issymmetric positive definite and B is apositive diagonal matrix.
Second order convergent rate
Method I with SSOR preconditioner:
MA= (D + ωL)D−1(D + ωU).
Quantum dot Matrix
Figure: Structure schema of the quantum dot model.
T.M. Huang (Taiwan Normal Univ.) JD method and applications October 31, 2008 59 / 71