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´ ıguez jsanchez@ualg.pt Departamento de Matem´ atica Universidade do Algarve CEAF-Center of Functional Analysis and Applications QISG - 2012 - OLH˜ AO

About the factorization of rational matrix functions 1 Introduction Definition of factorization Riemann-Hilbert BVP Fundamental set of solution of Riemann-Hilbert problem 2 Factorization procedure Riemann-Hilbert problem and difference equations The Z-transform Solution of the difference equation Main Result 3 Examples First example Second example 4 Historic remarks and references 1

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 2

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 [ t κ 1 , . . . , t κ m ] , κ i ∈ Z , κ i ≥ κ j , i , j = 1 , . . . , m , i < j , and A ± 1 + ∈ C m × m A ± 1 − ∈ C m × m ( T ) , ( T ) , + − 3

Example of (left) factorization The matrix function � t + 1 � 1 A ( t ) = , t 2 t admits the factorization A ( t ) = A + ( t )Λ( t ) A − ( t ) , � � with Λ( t ) = diag t , 1 and � 0 � − 1 t + 1 � 1 � A + ( t ) = A − ( t ) = , . t 2 − t + 1 1 1 The integers κ 1 = 1 and κ 2 = 0, are the partial indices. 4

Riemann-Hilbert BVP This problem consists in finding the m -vector function � ϕ + ( t ) , t ∈ T + Φ( t ) = ϕ − ( t ) , t ∈ T − where ϕ ± are analytic in T ± , respectively, with the following boundary condition: ϕ + = A ϕ − , ϕ − ( ∞ ) ∈ C , (1) and A is a given m × m matrix function. 5

Fundamental set of solution Φ (1) , . . . , Φ ( m ) are particular solutions of (1); − κ i is the order at infinity of Φ ( i ) , i = 1 , . . . , m ; Among the solutions of (1), − κ 1 is the lowest order at infinity; Among the solutions of (1), which cannot be written in the form p 1 Φ (1) + · · · + p s Φ ( s ) , 1 ≤ s < m , − κ s +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 − = (Φ − Λ) − 1 . A + = A Φ − , Λ( t ) = diag [ t κ 1 , . . . , t κ m ] , 6

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 7

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 ij t k + . . . + a (1) a ij = a ( k ) ij t + a (0) ij , then A admits the following representation A ( t ) = A k t k + ... + A 1 t + A 0 , where A s is the square constant matrix A s = { a ( s ) ij } , s = 0 , . . . , k . 8

Riemann-Hilbert problem and difference equations If ∞ � ϕ n t − n , ϕ − ( t ) = ϕ 0 + n =1 the pair of vector functions ϕ ± is a solution of the Riemann-Hilbert problem (1) if and only if enjoy the equality k i � � ϕ j t j − i + ϕ + = A ϕ − = A ϕ 0 + A i i =1 j =1 ∞ � ( A k ϕ n + k + . . . + A 1 ϕ n +1 + A 0 ϕ n ) t − n , + n =1 and we can assert that the vector function ϕ + is analytic in T + if and only if the vector ϕ n is a solution of the linear system of difference equations with constant coefficients A k ϕ n + k + . . . + A 1 ϕ n +1 + A 0 ϕ n = 0 . 9

The Z-transform By Z-transform of x n we mean the scalar function ∞ � x n z − n . Z ( x n ) = x = n =1 Some properties of the Z-transform Z ( x n + k ) = z k x − � k i =1 x i z k − i = { 1 , n = s Z ( δ ( s ) z s , where δ ( s ) n ) = 1 is the Kronecker delta n 0 , n � = s sequence, a n − 1 � 1 � Z = z − a , | z | > | a | , � na n − 1 � z = ( z − a ) 2 , | z | > | a | , Z � ( n − 1)! a n − 1 � ( n + s − 2)! = f ( s ) Z ( z ) , | z | > | a | , where a ( z ) = s ! z s − 1 f ( s ) a � = 0 , s ∈ N . ( z − a ) s , a 10

The Z-transform Applying the Z-transform to the linear system A k ϕ n + k + . . . + A 1 ϕ n +1 + A 0 ϕ n = 0 . and using the Z-transform’s properties we can write the following identity � k � � z k ϕ − ϕ i z k − i A k + . . . + A 1 ( z ϕ − ϕ 1 ) + A 0 ϕ = 0 . i =1 where ϕ i = ( ϕ 1 i , ..., ϕ mi ) ⊤ , i = 1 , ..., k , are arbitrary constant vectors. 11

The Z-transform Then ϕ , the Z-transform of ϕ n , enjoy the following properties: � j ϕ = A − 1 ( z ) � k i =1 A j ϕ i z j − i ; j =1 ϕ is a column with rational entries; The poles of ϕ are zeros of det A ; j =1 C ij f ( j ) ϕ = � l � di z i ( z ) + � d i =1 K i z − i , where i =1 ( z ) = s ! z s − 1 f ( s ) a � = 0 , s ∈ N . ( z − a ) s , a ϕ ( ∞ ) = 0 . 12

Solution of the difference equation Applying the Z-transform inverse we get the following solution l di d ( n + j − 2)! K i δ ( i ) � � � z n − 1 ϕ n = C ij + (2) n i ( n − 1)! i =1 j =1 i =1 of A k ϕ n + k + . . . + A 1 ϕ n +1 + A 0 ϕ n = 0 , where the vector-sequence ϕ n must satisfy the initial conditions ϕ n | n = i = ϕ i , i = 1 , . . . , k . 13

Solution of the difference equation Theorem If z i is one of the zeros of det A ( t ) and | z i | > 1 the series ∞ � ϕ n t − n n =1 with − d i ( n + j − 2)! � z n − 1 ϕ n = C ij , i ( n − 1)! j =1 converges in T − , if and only if, C ij = 0 , j = 1 , ..., d i . 14

Main Result Theorem The pair of functions ∞ � ϕ n t − n ϕ − = ϕ 0 + n =1 and ϕ + = A ϕ − is the general solution of a Riemann-Hilbert problem, if and only if, the coefficients ϕ n are given by (2) , where the constants ϕ si , i = 1 , ..., k ; s = 1 , ..., m must satisfy the additional conditions: ϕ ( ∞ ) = 0 ; ϕ n | n = i = ϕ i ; C ij = 0 , j = 1 , ..., d i . for each zero z i of det A, such that | z i | > 1 . 15

The factorization procedure Let A ( t ) = A k t k + ... + A 1 t + A 0 , to obtain the factorization of A , we can use the following procedure: Write A k ϕ n + k + . . . + A 1 ϕ n +1 + A 0 ϕ n = 0. Find the Z-transform of ϕ n . Impose the condition ϕ ( ∞ ) = 0. Using the Z-transform inverse, obtain the solution ϕ n of the linear system of difference equations. Consider the restrictions C ij = 0 , j = 1 , ..., d i , for each zero z i of det A , such that | z i | > 1. Use the initial conditions ϕ n | n = i = ϕ i . Obtain the vector function ϕ − ( t ) = ϕ 0 + � ∞ n =1 ϕ n t − n . Find a fundamental set of solutions of the Riemann-Hilbert BVP. Construct the factors of the left factorization of A . 16

Section’s Contents 1 Introduction 2 Factorization procedure 3 Examples First example Second example 4 Historic remarks and references 17

First example Let A be the 2 × 2 polynomial matrix-function � t + 1 1 � = A 2 t 2 + A 1 t + A 0 , A ( t ) = t 2 t where � 0 � 1 � � 0 1 A 2 = , A 1 = I 2 , A 0 = . 1 0 0 0 In order to find the general solution of the Riemann-Hilbert BVP, we need to solve the respective system of difference equations A 2 ϕ n +2 + A 1 ϕ n +1 + A 0 ϕ n = 0 . 18

First example � ⊤ , i = 1 , 2, be a constant vector. Applying the � Let ϕ i = a i b i Z-transform: − ( a 2 + b 1 ) z − 1 � � ϕ ( z ) = . a 1 + a 2 + b 1 + ( a 2 + b 1 ) z − 1 Taking into account ϕ ( ∞ ) = 0, we have the following equality a 1 + a 2 + b 1 = 0 , and consequently � ⊤ z − 1 . � ϕ ( z ) = ( a 2 + b 1 ) − 1 1 Using the inverse of the Z-transform we get � ⊤ δ (1) � ϕ n = ( a 2 + b 1 ) − 1 1 n . 19

First example The sequence ϕ n must satisfy the initial conditions � ⊤ , � ⊤ , � � ϕ n | n =1 = a 1 b 1 ϕ n | n =2 = a 2 b 2 then a 1 = − ( a 2 + b 1 ) , a 2 = 0 , b 1 = a 2 + b 1 , b 2 = 0 , and making use of the obtained equality a 1 + a 2 + b 1 = 0, we have � ⊤ δ (1) � − 1 1 ϕ n = a 1 n . 20

First example The general solution of the Riemann-Hilbert BVP is ϕ + = A ϕ − , where ∞ � ⊤ + a 1 t − 1 � � ⊤ . ϕ n t − n = � � ϕ − = ϕ 0 + − 1 1 a 0 b 0 n =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 κ 1 = 1. Next, we need to find another solution with the lowest order at infinity. This solution cannot be written in the form p 1 Φ 1 , where p 1 is a polynomial. One such solution, with the desired property is for instance � ⊤ , Φ (2) ( t ) = ϕ (2) � − ( t ) = 1 0 t ∈ T − , and obviously κ 2 = 0. 21

First example Then � � Λ( t ) = diag t , 1 , and � − t − 1 � 1 Φ − ( t ) = . t − 1 0 Consequently � � − 1 t + 1 A + ( t ) = A ( t )Φ − ( t ) = , t 2 − t + 1 � 0 � 1 A − ( t ) = (Φ − ( t )Λ( t )) − 1 = , 1 1 are the factors of the left factorization of � t + 1 � 1 A ( t ) = . t 2 t 22