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About the factorization of rational matrix functions Juan C. S - - PowerPoint PPT Presentation
About the factorization of rational matrix functions Juan C. S - - PowerPoint PPT Presentation
About the factorization of rational matrix functions Juan C. S anchez Rodr guez jsanchez@ualg.pt Departamento de Matem atica Universidade do Algarve CEAF-Center of Functional Analysis and Applications QISG - 2012 - OLH AO About
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Section’s Contents
1 Introduction
Definition of factorization Riemann-Hilbert BVP Fundamental set of solution of Riemann-Hilbert problem
2 Factorization procedure 3 Examples 4 Historic remarks and references
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Definition of factorization If A is a m × m continuous matrix function on T, by (left) factorization of A in C(T) we mean the following representation: A(t) = A+(t)Λ(t)A−(t), t ∈ T, where Λ(t) = diag [tκ1, . . . , tκm], κi ∈ Z, κi ≥ κj, i < j, i, j = 1, . . . , m, and A±1
+ ∈ C m×m +
(T), A±1
− ∈ C m×m −
(T),
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Example of (left) factorization The matrix function A(t) = t + 1 1 t2 t
- ,
admits the factorization A(t) = A+(t)Λ(t)A−(t), with Λ(t) = diag
- t,
1
- and
A+(t) =
- −1
t + 1 −t + 1 t2
- ,
A−(t) = 1 1 1
- .
The integers κ1 = 1 and κ2 = 0, are the partial indices.
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Riemann-Hilbert BVP This problem consists in finding the m-vector function Φ(t) = ϕ+(t), t ∈ T+ ϕ−(t), t ∈ T− where ϕ± are analytic in T±, respectively, with the following boundary condition: ϕ+ = Aϕ−, ϕ−(∞) ∈ C, (1) and A is a given m × m matrix function.
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Fundamental set of solution Φ(1), . . . , Φ(m) are particular solutions of (1); −κi is the order at infinity of Φ(i), i = 1, . . . , m; Among the solutions of (1), −κ1 is the lowest order at infinity; Among the solutions of (1), which cannot be written in the form p1Φ(1) + · · · + psΦ(s), 1 ≤ s < m, −κs+1 is the lowest order at infinity; Let Φ−(t), t ∈ T−, be the matrix function whose columns are the vector functions Φ(i), i = 1, . . . , m. Then the factors of the factorization of A are A+ = AΦ−, Λ(t) = diag [tκ1, . . . , tκm], A− = (Φ−Λ)−1 .
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Section’s Contents
1 Introduction 2 Factorization procedure
Riemann-Hilbert problem and difference equations The Z-transform Solution of the difference equation Main Result
3 Examples 4 Historic remarks and references
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Riemann-Hilbert problem and difference equations It is very well known that the factorization of rational matrix-functions is, up to a multiplication by a polynomial, equivalent to the factorization of a polynomial matrix-function. Let A be the m × m matrix-function with polynomial entries aij = a(k)
ij tk + . . . + a(1) ij t + a(0) ij ,
then A admits the following representation A(t) = Aktk + ... + A1t + A0, where As is the square constant matrix As = {a(s)
ij }, s = 0, . . . , k.
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Riemann-Hilbert problem and difference equations If ϕ−(t) = ϕ0 +
∞
- n=1
ϕnt−n, the pair of vector functions ϕ± is a solution of the Riemann-Hilbert problem (1) if and only if enjoy the equality ϕ+ = Aϕ− = Aϕ0 +
k
- i=1
Ai
i
- j=1
ϕjtj−i+ +
∞
- n=1
(Akϕn+k + . . . + A1ϕn+1 + A0ϕn)t−n, and we can assert that the vector function ϕ+ is analytic in T+ if and only if the vector ϕn is a solution of the linear system of difference equations with constant coefficients Akϕn+k + . . . + A1ϕn+1 + A0ϕn = 0.
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The Z-transform By Z-transform of xn we mean the scalar function Z(xn) = x =
∞
- n=1
xnz−n. Some properties of the Z-transform Z(xn+k) = zkx − k
i=1 xizk−i
Z(δ(s)
n ) = 1 zs , where δ(s) n
= { 1, n = s 0, n = s is the Kronecker delta sequence, Z
- an−1
=
1 z−a, |z| > |a|,
Z
- nan−1
=
z (z−a)2 , |z| > |a|,
Z
- (n+s−2)!
(n−1)! an−1
= f (s)
a
(z), |z| > |a|, where f (s)
a
(z) = s!zs−1 (z − a)s , a = 0, s ∈ N.
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The Z-transform Applying the Z-transform to the linear system Akϕn+k + . . . + A1ϕn+1 + A0ϕn = 0. and using the Z-transform’s properties we can write the following identity Ak
- zkϕ −
k
- i=1
ϕizk−i
- + . . . + A1(zϕ − ϕ1) + A0ϕ = 0.
where ϕi = (ϕ1i, ..., ϕmi)⊤, i = 1, ..., k, are arbitrary constant vectors.
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The Z-transform Then ϕ, the Z-transform of ϕn, enjoy the following properties: ϕ = A−1(z) k
j=1
j
i=1 Ajϕizj−i;
ϕ is a column with rational entries; The poles of ϕ are zeros of det A; ϕ = l
i=1
di
j=1 Cijf (j) zi (z) + d i=1 Kiz−i, where
f (s)
a
(z) = s!zs−1 (z − a)s , a = 0, s ∈ N. ϕ(∞) = 0.
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Solution of the difference equation Applying the Z-transform inverse we get the following solution ϕn =
l
- i=1
di
- j=1
Cij (n + j − 2)! (n − 1)! zn−1
i
+
d
- i=1
Kiδ(i)
n
(2)
- f
Akϕn+k + . . . + A1ϕn+1 + A0ϕn = 0, where the vector-sequence ϕn must satisfy the initial conditions ϕn|n=i = ϕi, i = 1, . . . , k.
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Solution of the difference equation Theorem If zi is one of the zeros of det A(t) and |zi| > 1 the series
∞
- n=1
ϕnt−n with ϕn =
−di
- j=1
Cij (n + j − 2)! (n − 1)! zn−1
i
, converges in T−, if and only if, Cij = 0, j = 1, ..., di.
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Main Result Theorem The pair of functions ϕ− = ϕ0 +
∞
- n=1
ϕnt−n and ϕ+ = Aϕ− is the general solution of a Riemann-Hilbert problem, if and only if, the coefficients ϕn are given by (2), where the constants ϕsi, i = 1, ..., k; s = 1, ..., m must satisfy the additional conditions: ϕ(∞) = 0; ϕn|n=i = ϕi; Cij = 0, j = 1, ..., di. for each zero zi of det A, such that |zi| > 1.
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The factorization procedure Let A(t) = Aktk + ... + A1t + A0, to obtain the factorization of A, we can use the following procedure: Write Akϕn+k + . . . + A1ϕn+1 + A0ϕn = 0. Find the Z-transform of ϕn. Impose the condition ϕ(∞) = 0. Using the Z-transform inverse, obtain the solution ϕn of the linear system of difference equations. Consider the restrictions Cij = 0, j = 1, ..., di, for each zero zi of det A, such that |zi| > 1. Use the initial conditions ϕn|n=i = ϕi. Obtain the vector function ϕ−(t) = ϕ0 + ∞
n=1 ϕnt−n.
Find a fundamental set of solutions of the Riemann-Hilbert BVP. Construct the factors of the left factorization of A.
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Section’s Contents
1 Introduction 2 Factorization procedure 3 Examples
First example Second example
4 Historic remarks and references
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First example Let A be the 2 × 2 polynomial matrix-function A(t) = t + 1 1 t2 t
- = A2t2 + A1t + A0,
where A2 = 1
- ,
A1 = I2, A0 = 1 1
- .
In order to find the general solution of the Riemann-Hilbert BVP, we need to solve the respective system of difference equations A2ϕn+2 + A1ϕn+1 + A0ϕn = 0.
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First example Let ϕi =
- ai
bi ⊤ , i = 1, 2, be a constant vector. Applying the Z-transform: ϕ(z) =
- −(a2 + b1)z−1
a1 + a2 + b1 + (a2 + b1)z−1
- .
Taking into account ϕ(∞) = 0, we have the following equality a1 + a2 + b1 = 0, and consequently ϕ(z) = (a2 + b1)
- −1
1 ⊤ z−1. Using the inverse of the Z-transform we get ϕn = (a2 + b1)
- −1
1 ⊤ δ(1)
n .
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First example The sequence ϕn must satisfy the initial conditions ϕn|n=1 =
- a1
b1 ⊤ , ϕn|n=2 =
- a2
b2 ⊤ , then a1 = −(a2 + b1), a2 = 0, b1 = a2 + b1, b2 = 0, and making use of the obtained equality a1 + a2 + b1 = 0, we have ϕn = a1
- −1
1 ⊤ δ(1)
n .
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First example The general solution of the Riemann-Hilbert BVP is ϕ+ = Aϕ−, where ϕ− = ϕ0 +
∞
- n=1
ϕnt−n =
- a0
b0 ⊤ + a1t−1 −1 1 ⊤ . To construct the factorization of A we need to find a particular solution with the lowest order at infinity: Φ(1)(t) = ϕ(1)
− (t) = t−1
−1 1 ⊤ , t ∈ T−, with order −1 at infinity. So κ1 = 1. Next, we need to find another solution with the lowest order at
- infinity. This solution cannot be written in the form p1Φ1, where
p1 is a polynomial. One such solution, with the desired property is for instance Φ(2)(t) = ϕ(2)
− (t) =
- 1
⊤ , t ∈ T−, and obviously κ2 = 0.
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First example Then Λ(t) = diag
- t,
1
- ,
and Φ−(t) = −t−1 1 t−1
- .
Consequently A+(t) = A(t)Φ−(t) =
- −1
t + 1 −t + 1 t2
- ,
A−(t) = (Φ−(t)Λ(t))−1 = 1 1 1
- ,
are the factors of the left factorization of A(t) = t + 1 1 t2 t
- .
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Second example Next we will consider the factorization of the following polynomial matrix function A(t) = (t − 3)2 (t − 3)2 t3 t3 + t2(3t − 1)2
- .
(3) In this case, some of the calculations were done with the help of the software Wolfram Mathematica. The matrix function A can be written as follows A(t) = A4t4 + A3t3 + A2t2 + A1t + A0, where A0 = 9 9
- ,
A1 = −6 −6
- ,
A2 = 1 1 1
- ,
A3 = 1 −5
- ,
A4 = 9
- .
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Second example As we already know, we need to solve the system of difference equations: A4ϕn+4 + A3ϕn+3 + A2ϕn+2 + A1ϕn+1 + A0ϕn = 0, where ϕn =
- an
bn ⊤. The Z-transform ϕ of ϕn is given by ϕ = A−1(z)[A1ϕ1 + A2(ϕ1z + ϕ2) + A3(ϕ1z2 + ϕ2z + ϕ3)+ +A4(ϕ1z3 + ϕ2z2 + ϕ3z + ϕ4)], with the initial data ϕi =
- ai
bi ⊤ , i = 1, . . . , 4.
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Second example Then, using the software Wolfram Mathematica, we have that a and b, the Z-transforms of an and bn are, respectively a = M1 z + M2 z2 + M3 z − 1
3
+ M4z (z − 1
3)2 +
M5 z − 3 + M6 (z − 3)2 , b = N1 z + N2 z2 + N3 z − 1
3
+ N4z (z − 1
3)2 +
N5 z − 3 + N6 (z − 3)2 , with M5 = 1 256(283a1−5a2+283b1−5b2), M6 = −67 64(3a1−a2+3b1−b2), and N5 = 1 256(−27a1+5a2−27b1+5b2), N6 = − 3 64(3a1−a2+3b1−b2).
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Second example M5, M6, N5 and N6 must vanish, then b1 = −a1, b2 = −a2. The new expressions for the coefficients Ni, Mi, i = 1, . . . , 4 are M1 = a1 − 6a3 + 21b3 − 54b4, M2 = a2 − a3 + 5b3 − 9b4, M3 = 9(a3 − 3b3 + 9b4), M4 = −3(a3 − 2b3 + 9b4), Ni = −Mi, i = 1, . . . , 4. and a = M1 z + M2 z2 + M3 z − 1
3
+ M4z (z − 1
3)2 ,
b = N1 z + N2 z2 + N3 z − 1
3
+ N4z (z − 1
3)2 .
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Second example Using the inverse of Z-transform we can affirm that an = M1δ(1)
n
+ M2δ(2)
n
+ M3 3n−1 + M4 n 3n−1 , bn = −an. and b3 = −a3, b4 = −a4. So, the final expressions of the coefficients Ni, Mi, i = 1, . . . , 4 are M1 = a1 − 27a3 + 54a4, M2 = a2 − 6a3 + 9a4 M3 = 36a3 − 81a4, M4 = −9a3 + 27a4, and Ni = −Mi, i = 1, . . . , 4.
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Second example Finally, the general solution of the Riemann-Hilbert problem is ϕ+ = Aϕ−, ϕ− = ϕ0 +
∞
- n=1
ϕnt−n, where ϕn = an
- 1
−1 ⊤ , Then ϕ− =
- a0
b0 ⊤ + a(t)
- 1
−1 ⊤ with a(t) = M1 t + M2 t2 + M3 t − 1
3
+ M4 t (t − 1
3)2
and M1 = a1 − 27a3 + 54a4, M2 = a2 − 6a3 + 9a4, M3 = 36a3 − 81a4, M4 = −9a3 + 27a4.
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Second example To conclude the factorization of A we need to find the fundamental set of solutions. Among the solutions of the considered Riemann-Hilbert problem, the lowest order at infinity is −4, when ϕ0 = ϕ1 = ϕ2 = ϕ3 = 0, i.e., when a0 = b0 = a1 = a2 = a3 = 0 and a4 = 0. One solution with such order at infinity is, for instance Φ(1)(t) = ϕ(1)
− (t) =
1 t2(3t − 1)2
- 1
−1 ⊤ , t ∈ T−. Further, the solution Φ2 must have order equal to zero at infinity. Otherwise, Φ2 can be written in the form p1Φ2, where p1 is a
- polynomial. Among the solutions with the order equal to zero at
infinity we take Φ(2)(t) = ϕ(2)
− (t) =
- 1
⊤ , t ∈ T−. Now it is possible to assert that κ1 = 4 and κ2 = 0.
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Second example Then Λ = diag
- t4,
1
- ,
and taking into account that Φ−(t) =
- 1
t2(3t−1)2
1 −
1 t2(3t−1)2
- ,
we have that the external factors of the factorization of A are A+(t) = A(t)Φ−(t) =
- (t − 3)2
−1 t3
- A−(t) = (Φ−(t)Λ(t))−1 =
- − (3t−1)2
t2
1 1
- .
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Section’s Contents
1 Introduction 2 Factorization procedure 3 Examples 4 Historic remarks and references
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Historic remarks and references As far as we know, there are two methods of constructing the factorization of rational matrices: The splitting of zeros was proposed by Gakhov in the middle
- f XX century
The second method is based on the space theory for linear input-output system. Bart H., Gohberg I. and Kaashoek M. A. Minimal Factorization
- f Matrix and Operator Functions. Birkh¨