Integr Equat Oper Th Vol. 18 (1994)

0378-620X/94/030277-2651.50 +0.20/0 (c) 1994 Birkhiuser Verlag, Basel

A METHOD OF EXPLICIT FACTORIZATION OF MATRIX FUNCTIONS AND APPLICATIONS

I. Feldman, I. Gohberg, N. Krupnik

A method of explicit factorization of matrix functions of second order is proposed. The method consists of reduction of this problem to two scalar barrier problems and a finite system of linear equations. Applications to various classes of singular integral equations and equations with Toeplitz and Hankel matrices are given.

Introduction.

In this paper we describe a method of explicit factorization of matrix functions of second order. This method can be applied to a wide variety of matrices of the form

1

b(z) 1

G(z)= c(z) d(z)J' where b, c, d are measurable essential bounded functions on the unit circle (real line) and

b(z)

admits a meromorphic extension into the interior of the unit disk (upper half-

plane). The problem of factorization of matrices of this type appears in various applications, for example, in soliton theory [TF]. The m e t h o d consists of reduction of this problem to factorization of two scalar functions and to solving of a finite system of linear equations.

The idea of such a

method was proposed by one of the authors [F] and recently also was suggested in [AKM]. The detailed description of this method is presented in Chapter 1, which contains also illustrative examples. Here, for simplicity, we restrict ourselves to functions b, c, d from Wiener algebras. The second chapter contains examples of applications to various classes of singular integral equations and equations with Toeplitz and Hankel matrices.

278

Feldman, Gohberg and Krupnik CHAPTER I

w

Barrier problem. Let W (= W(T)) be the Wiener algebra of all functions on the unit circle T of the

form a(z)=

~

(Elan]
anz n

and W+ (W_) the subalgebra of W consisting of those functions that a , = 0 for n < 0 (for n > 0). Denote by A the following matrix function

(1.1)

1

A(z)=

b(z)]

c(z)

d(z)J

where b, c, d E W. We suppose that the function b can be represented in the form (1.2)

b(z) = p(z) q(z)

k

where p E W+, q(z) = 1-I(z - c~j) mj (]~jl < 1) and the numbers c~j (j = 1 , 2 , . . . , k ) j=l

are distinct. In this section we consider the barrier problem (1.3)

F- = AF +

where A is a matrix of the form (1.1). Recall some notations and results from [CG], Ch.II, w Let D+ = {z E C, ]z I < 1}, D_ = C \ D + , f + , g + E W+, f - - q 1 E W_, g - -q2 E W_ for some polynomials ql, q2 and F + = If+, g+]T. A piecewise holomorphic vector function F ( z ) = F + ( z ) (z E D+) is called a solution of (1.3) if the equality (1.3) holds in T. If Fl(Z), F2(z) . . . . . F,.(z) are piecewise holomorphic vector functions which solve the barrier problem (1.3) and Pl,P2,... P,- are polynomials, then r

F(z) =

p (z)Fj(z)

j=l

also solves the barrier problem (1.3). The following method allows to find every solution of (1.3):

Feldman, Gohberg and Krupnik

279

T h e o r e m 1.1. Let the matrix function (1.1) satisfy condition (1.2) and det A ( z ) # 0 ( Izl = 1). The pair F +, F - of vectors F • = [f • g a:]T is a solu tion of problem (1.3) if and only if the following three conditions hold: 1) the pair of ftmctions f - and h + = q f + + pg+ is a solution of problem (1.4)

q f - = h +,

2) the pair of functions g• are a soluton of problem (1.5)

g - = (d - bc)g + + c f - ,

3) The function u = h + - pg+ vanishes in zeros of the polynomial q(z) : (1.6)

u ( r ) ( a / ) = 0 (r = 0 , 1 , . . . m j

- 1; j = 1 , 2 , . . . , k ) .

Proof. Let vectors F :t: = [f+,g+] be a solution of problem (1.3). The problem (1.3)

can be written in a equivalent form (1.7)

q f - = q f + + p9 + g - - c f - = (d - bc)g +.

Evidently, the function q f + + pg+ belongs to W+. Denote this function by h +. From the first equality (1.7) we obtain (1.4): q f - = h +. After solving this barrier problem we obtain from the second equality (1.7) the non-homogeneous barrier problem (1.5) for functions g+ : g - = (d - bc)g + + c f - . From the equality q f + + pg+ = h + it follows that the function u(z) = h + ( z ) - p(z)g+(z) must vanish in zeros of the polynomial q(z) and we come to the equalities (1.6). By virtue of these equalities the function (1.8)

f + ( z ) = h+(z) - P(z)g+(z) q(z)

belongs to W+. Conversely, let the functions f - and h + be a solution of problem (1.4), the functions g+ be a solution of problem (1.5) and the equalities (1.6) be true. Passing all the steps in reverse order we get that the pair of vectors F + = If+, g+] is a solution of the barrier problem (1.3). Here the function f + is defined by equality (1.8).

[]

The function u(z) = h + ( z ) - p(z)g+(z) depends on several arbitrary constants. The equalities (1.6) represent a linear system with respect to these constants. So the

280

Feldman, Gohberg and Krupnik

barrier problem (1.3) can be reduced to the scalar problems (1.4), (1.5) and the linear system (1.6). More explicitly, suppose that we look for solutions F(z) of the barrier problem (1.3) possessing at most a pole of order e at infinity. The functions

e+m h+(z) = E cJzJ' ,=o

f-(z)-

h+(z ) q(z) '

where cj (j = O, 1,...,~ + m) are arbitrary constants, give the general solution of problem (1.4) in the corresponding class. Hence problem (1.5) has the form

c(z) l+m g- = (6 -- be)g+ + ~ E c, zi. j=O

(1.9)

Assume for simplicity that i n d ( d - bc) = 0 and let the equality d - bc = A _ A + be a factorization of the function d - bc. Then (1.9) implies the equality

(1.1o)

t+rn t+rn A:19- - ~ cs~;(z) = A+g+ + Z c,~+(z) j=0

where

j=0

A=I(z)c(z)z j ~+(z) = P A-l(z)c(z)zj ~;(z) =Q q(z) ' q(z)

and the projections P, Q are defined by

P ~-~ ajzJ=~-~ajz j, j=-oc

Q=I-P.

j=O

Since the function in the left side of (1.1) has at most a pole of order g at infinity, both sides of (1.10) coincide with a polynomial t

r(z) = E Cm+t+l+JZJ" j=0

Now it follows from (1.10) that ~+m

t+m

j=0

j=0

So the function u(z) = h+(z) - p(z)g+(z) depends on m + 2g + 2 arbitrary constants

cj (j = 0, 1 , . . . , m + 2g + 1) and equalities (1.6) represent a homogeneous system of m linear equations with respect to these constants.

Feldman, Gohberg and Krupnik

w

281

Connection with factorization. Recall the well-known connection between barrier problems and factorization of

matrix functions [CG, Ch.II, w

We restrict ourselves to the case of 2 x 2 matrices.

Let A = A_DA+ be a factorization of a matrix A(C W 2x2) and let F1,F2 be piecewise holomorphic vector functions associated with this factorization:

[Fl(z),F2(z)]=

( A+l(z) A_(z)D(z)

zED+ zED_.

Then every solution of the barrier problem

F- = AF +

(2.1)

has the form pl(Z)Fl(z) -~-p2(z)F2(z) for some polynomials pl(z) and p2(z). Conversely, if det A(z) 7~ 0 ([z[ = 1) and all solutions of the barrier problem (2.1) are known, then the factorization of A can be obtained in the following way. We need the following definition. Let r

be a C2-valued holomorphic function

which possesses a series expansion

r

= ~kz k + ,Pk+lZ T M + . . . (~ot ~ C 2, ~ok # O)

in a neighborhood of infinity. Then r has order k at oo and we will write ord r = k. Let the solutions Fl(z) and F2(z) of the barrier problem (2.1) satisfy the following conditions a) det[Fl(0, F2(0)] # 0 ,

kl >_ k2 (kj =ordFj)

b) (kl, k2) is a minimal element in a lexicographic ordering of the set (ord r for all pairs of solutions r

r

ord r

of (2.1), which satisfy condition a).

Then the matrix A admits a factorization A = A_DA+ with (2.2)

A_ = [z-k'F~-,z-k2F2],

A+ 1 = [F~,F~],

D = diag(z kl , zk2). Therefore, in order to obtain a factorization of matrix A it is enough to find all solutions F of problem (2.1) such that ord F _ kl. The difficulty lies in the fact that usually kl is not known apriori. If we have an estimation kl _< s then we can restrict ourselves to solutions F of the barrier problem (2.1) such that o r d F < s. The following theorem gives such an estimation for the matrix (1.1).

282

Feldman, Gohberg and Krupnik

Theorem 2.1. Let A(z) be a matrix of the form (1.1) and det A(z) # 0 (Izl = 1). Then (2.3)

tr 1 < max(0, m + 5),

where g l , ~2 (1r --~ g2) ~

[;he partial indices of A(z),

5 = ind det A(z) and m is the

degree o f the polynomial q( z ).

Proof. Denote by F~- = (u-, v - ) T a solution of problem (1.3), a p p e a r i n g in relations (2.2). We s t a r t with the case u - ( z ) ~ O. Since u - is a solution of p r o b l e m (1.4) (see section 1), u - ( z ) = r ( z ) q - l ( z ) where r(z) is a polynomial. Hence o r d u - >_ - m and ~2 = o r d F 2 - > o r d u - > - m . Therefore, ~1 = 6 - ~2 < 5 + m . Let now u - ( z ) =_O. Then the m a t r i x A _ D has the form

d_(z)D(z) =

z~,ah(~) 0 ] z~,~;l(Z) z~'a;~(z)

It follows from A = A _ D A + t h a t z~'ah(z)a+~(z) = 1. Since ( a S ) -~ E W _ we have gl < 0, i.e., the e s t i m a t e (2.3) is true.

[]

In the sequel we need the following simple statement.

Remark 2.1. Let A(z) E W 2•

A ( z ) = d e t A ( z ) r 0 (Izl = 1), i n d A = 0, and let

the equality A ( z ) = A _ ( z ) A + ( z ) be a factorization of A ( z ) . If F [ , F 2- are solutions of the barrier p r o b l e m F - = A F +, o r d F ~ = 0 (j = 1,2) a n d d e t [ F ~ - , F ~ ] = c A ( z ) (c = const r 0) then A a d m i t s the canonical factorization A -= A_A+ with factors A+ = A--1A.

A _ = [F1- , F2-],

w

Factorization relative to the real line. Denote by W ( R ) ( W + ( R ) , w _ ( n ) ) the algebra of all functions r

(-~

< ~ < ~)

of the form

/5

(X)

(

/0

f

OO

where c is a constant and ~ is an a r b i t r a r y function from L1. In this section we consider a barrier problem with a m a t r i x function

(3.1)

(a,b, e c W(R)) J

)

Feldman, Gohbergand Krupnik

283

and the factorization of this matrix. Here the function c can be represented in the form c(A) = p(~)

(3.2)

q()O

where p E W+(R) and (Imc U > 0 , 7 j > O ; a j # a r f o r j # r ) .

q(A) = H \)t + i71 ] j=l

The matrix A(A) is written here in the form (3.1) instead of (1.1) because such a form appears in some examples. Let f+,g+ G W+(R),

f-(A)-

rl

~

i)

E

W_(R),

g - ( A ) - r2 ~

E

W_(R) for some polynomials ra,r2 and F + = [f+,g+]T. A piecewise holomorphic vector function S F+(A) F-(A)

F(A)

Im)~ > 0 Im), < 0

is a solution of the barrier problem F - ( A ) = A(A)F+()O

(3.3)

if the equality (3.3) holds on R. In order to solve this problem and to factorize the matrix A(A) we use the same method as in sections 1 and 2. Note only that conditions a) from section 2 must be replaced by a') det[Fl(i), F2(i)] ~ O,

kl > k2 (k i = o r d Fj) where the order of F / i s taken with respect to the point A = - i , and the equalities (2.2) must be replaced by A_=

~

F;-,k,A+i]

D=diag

]

]

F;,

'\A+i]

A+I=[F+,F+],

]"

Now let us formulate the analogy of theorem 1.1. T h e o r e m 3.1. Let the matrix function 0.1) satisfy condition (3.2) and det A(A) # 0 (-oo <_ A <_ oo). The pair of vectors F + = [f• is a solution of the barrier

problem (3.3) if and only if 1) the pair of functions 9- and h + = p f + + qg+ is a solution of the problem (3.4)

qg- = h +,

284

Feldman, Gohberg and Krupnik 2) the pair of functions f • is a solution of the problem

(3.5)

f - = (a - bc)f + -4- bg-,

a) the function u = h + - p f + vanishes in zeros of the function q( A ) : u(')(aj)=O

(r=0,1,...,mj-1;

j =l,2,...,k).

Note that theorem 2.1 and remark 2.1 remain true also in the case of the real line. In the analogy of the theorem, m denotes the total number of zeros of the function q(A) in the upper half-plane. In conclusion of this section, consider a matrix function of the form

=

(3.7)

[1 r17 ]

<

< oo)

r2 e - i x A

where rj(A) are rational functions without poles on the real line, rj(cr = 0 and x is a real parameter. This matrix has the form (3.1) since for x > 0 (x < 0) the function rl(A)e i ~ (r2(A)e -i~A) admits the representation (3.2). Indeed, let, for example, x > O , rl(.X) = k

Pl(A)

( I m a j > 0 , Im/3j < O) andpl(A) is apolynomial. k ~ _ aj

Then r](A)e I*~ = p(A)/q(A) where q(A) = 1-1 "A~-'7 '

Pl(A)ei*~

p(A) = (A + i) k fi(A - / 3 j )

w+(a). Here we use the first from the following two simple relations: elba

if I m a > 0 a n d b > 0 t h e n

A + a EW+(R); 1

ifaECandb>Othen

-

e ib(~+~) A+a C W+(R),

which follows from the equalities eiCa eiAtdt _ ieibc~eib.______~

A+a

b eica eiAtdt = i 1 -- e ib('~+c*)

'

Matrices of the form (3.7) arise from soliton theory [TF].

A+a

E

Feldman, Gohberg and Krupnik w

285

Examples of factorization relative to the circle. 1. We start with the factorization of the matrix

I A(z) =

1

(1-z

z +1 (1 - z ) e x p l " - - z

-1, z+l] )exp~L--~_1

.

1 - - z - - z -1

The function ( 1 - z -1) exp Z +__1 admits representation (1.2) and p(z) = ( z - l ) e x p z + 1 z-1 z-l' q(z) = z. Since det A(z) = - 1 it follows from theorem 2.1 that ~ < 1. Let us find all the solution of the barrier problem z+l + f - = f + + (1 - z - l ) exp ~--~_l g

(4.1)

g - = (1 - z ) e x p z1 -+ zl f + + ( 1 -

z - z-1)g +

possessing at most a pole of first order at ~ . We use the method suggested in theorem 1.1. Problem (1.4) has a form z f - = h +. Its general solution in the required class is: f - = coz -1 + cl + c2z, h + = co + cl z + c2z 2, where Co, cl, c2 are arbitrary constants. Problem (1.5) has a form l+z 9 - = - g + + (1 - z ) ( c o z -~ + e, + e~z) exp 1--~" Its general solution is l+z

9 - = (1 - z ) ( c o z -1 + cl + e ~ z ) e x p ~ -

z + c ~ e - l z ~ + rl(z),

g+ = - c 2 e - l z 2 - rl(z) where rl(z) is an arbitrary polynomial of the first degree. In this case system (1.6) consists of one equation: h+(0) - p(0)g+(0) = 0 and this implies rl(0) = coe. Thus,

rl(z) = coe + caz, where c3 is an arbitrary constant. T h e general form of vector F - , which solves problem (4.1) is:

F-(z) = = [c~ 3

j=o

+ c l 4-c2z'(1-- z ) ( c ~

+ c l + C 2 z ) e x p l + Zz + c 2 e - l z 2 4 - c 3 z 4 - c ~

286

Feldman, Gohberg and Krupnik

where (4.2)

[ r

z--l'

[

l_Zexpl+ z ]T Z 1~----'~ + e ' r

r m. z , z ( 1 - z ) e x P l _

l+zz

+ e--lz2

]r

[

l+z]T l'(1--z)exPl--zJ

r

= [0, Z] T.

Now we have to select two linear independent vector F [ and F 2- in the subspace generated by vectors (4.2) such that the pair (kl, k2) (kj = ord F/--) is a minimal element. It is clear that ordr = ord(Cl+e-lCa) = 0 and det[r r162 = c - e -1 ( # 0). Hence we can take F 1 = r F2" = r + e - l e a . So we obtain the canonical factorization A = A _ A + with factors A_ = [F1, F2-], A+ = A-_IA, i.e., z -1 1-z l+z e+ exp ~ z 1-z

A_ =

1

(1 - z)exp 1 + z + e_lz 1-z

e-1(1 - z ) e x p Z + l z 1

_e-lz

e

1

1] ~

A+ - e2 _ 1

e Z - 1 exp z + 1 z - 1 z z-1 +

Now we see that it was sufficient to look for solution of problem (4.1) of order * = 0. 2. Let us factorize the matrix 1

z-2 lx/q--Z~-z ]

A(z) =

vlI-z

z-2(z 2 - z + l ) J

Here, detA(z) = 1, p(z) = x / 1 - z, q(z) = z 2, and by theorem 2.1 ~1 -< 2. So the possible cases are: ~l = 2, ~1 = 1, ~1 = 0, and we have to solve the barrier problem in one of the corresponding classes. Let us try to solve it in the simplest class of functions F(z) with o r d F = 0. Problem (1.4) has the form z2f - = h + and its solution is: f-

= c2 + c'L1 + c o Z

Z2 '

h + = co + c l z +

Problem (1.5) has the form _

g =g++

( Cl lv/~--~-z c 2 + - - +

z

Co) 7

c2z 2.

Feldman, Gohberg and Krupnik

287

and its solution is: g - = c3 +

g+ = where the function r

1( -

1)

Cl

-

Z

-

~Co

"~-

co - -

Z2

1 c2z + clz + c2z 2) + ~C1 -- e2 + C3 + 2

r

1

is defined by the equality x/1 - z = 1 - ~z - z2r

System

(1.6) has the form 7 1 (u(0) =) g c 0 - ~cl + c2 - c 3 = 0 9

( U ' ( 0 ) = ) gC 1 -- C2 +

12c3 = 0.

7 5 7 7 This yields c3 = ~c0 + ~Cl, c2 = ~c0 + ~Cl. The general form of the vector F - ( z ) which solves the corresponding barrier problem is: f-(z)

= Co

+ z-2' -47_ l z _ 1 "~ Z - - 2

+ el

+ z_l, "4 "~- Z - - 1

= clF~-(z) + coFs 63 Here det[Fl(Z), F2-(z)] = ~ , and by remark 2.1 the matrix A(z) admits a canonical factorization A = A _ A + , where !+z_

7 2 -8 + z -

A_(z) = "4 + z - 1

w

47

]

1 -1 + z -2 ~z

A n e x a m p l e o f f a c t o r i z a t l o n r e l a t i v e t o t h e real llne.

Consider the following matrix of form (3.7) 1 (5.1)

A(A)=

i~/1 - ~2 ] A + is e2i~A

i~/1-r A - i-----~e-2i*~

(0
We factorize it by means of theorem 3.1. We start with the more difficult case x < 0. The function i~/1__e 2 e_2i~ admits a presentation (3.2) where A - ie

-

iv~

+

~:2e-2izA,

A - ie q(A) = A + i s "

288

Feldman, Gohberg and Krupnik

A2+l A~ + e and i n d d e t A(A) = 0. By the analogy of t h e o r e m 2.1 the

Here det A(A) problem F-

= A F + have to be solved in the class of functions F with ord F = 1 or

ord F = 0. We t r y again to restrict ourselves with ord F = 0. P r o b l e m (3.4) has the form A -- ie --gA+ie

and its solution is:

= h +,

clA + c2 g

A-is

h+

= c 1 A 71- C2

'

A+ie

'

where cl, c2 are a r b i t r a r y constants. Furthermore, p r o b l e m (3.5) has a form

A2+l f-

ivff- ~2(cla + c2) ~i~,

+

= " "2~ e

+

A2 + e 2

and its solution is A-i f - = (c3 + r a - ( A ) ) ~ - - - ~- ,

f+

= (c3 +

X+ic

m+(A))i~ ~

where

(~--+~(~,A L]~7-~ e2~X(1_A -i--~e2iZ(x+ie))]1 ' 1 -

re-(A) =

[e 2ira

e

+ c~)

+

/ T - cclA + cz ~

m+()~)

VT+ e

A+ie

e

,

and ca is an a r b i t r a r y constant. System (3.6) consists of one equation h + ( i e ) - p ( i ~ ) f + ( i e ) 0 and we o b t a i n iEcl + c2 1 - ~2

c3 =

2i~

~ 1/'f'~Te2~,:

---7--- where d~= V

(< 1).

The vector F - ( A ) = [ f - ( A ) , g-(A)] T can be written in the form F - ( A ) = C l P I ( A ) + c z F ~ ( X ) where

FC(~)= [(1-~; [\

2~

)~-i +,~w

A] T

~--i-e' , x - i t

'

),,-, 11 ~

L~--U+~

~ ~-i-2

j "

=

Feldman, Gohbergand Krupnik

289

1-6 2 A-i It is easy to see that ordF~- = o r d F 2- = 0 and det[F1, F2- ] = 2is6 X - i s " By remark 2.1, the matrix A admits a canonical factorization A = A_A+ where

A_(X) =

\ ~ + X W

)

1--6

~-V~

X - is

o]

The ease x > 0 is simple:

[ 1 A_(X)=

"~ )~--i ]

)~-is

i~/1-e 2

i C ' l ' - ~2 21~X]

1

X-i

X - i-------'--e-2i=x -e

~--~

0

ST(TT~ X+i X+ie

1. | J

T h e factorizaton of matrix (5.1) (and matrices of more general form) was obtained by another m e t h o d in [AKM]. In [AKM] the problem of factorization was also reduced to two scalar barrier problems and a linear system. Note that in [AKM] a condition A_(c~) = I was taken which corresponds (for x < 0) to the choice of parameters: 6 1+62 cl = 0, c2 = 2is 1---~-~ and ca = 1, cz = - i e l - - L ~ . In our factorization the choice is: c1=1,

c2=0andcl

=0, cz-1. CHAPTER II

w Integral equation with Hilbert kernel. Consider in the space Lp(O, 2~r] (1 < p < co) the integral equation A~ = r (6.1)

A~(x) = a(x)v(x ) + ~

cot

~(s)ds

where

(a, b 9 Lo~[0,27rl).

It is well-known (see, for example [GK2, pp. 141-145]) that the operator A is similar to the operator A0 = B - K , acting in Lp(T), where A0 is a characteristic singular integral operator (6.2)

bo(t) [ f(r)dt Aof(t) = ao(t)f(t) + 7ri Jr -~----( '

K is the one-dimensional operator

bo(t)f~ f(T)ldT l,

K f(t) = ~

and for any c C L~[0, 27r] we set co(exp(ix)) := c(x) (x 9 [0, 2~r]).

290

Feldman, Gohbergand Krupnik b'Yom here it follows that the operator A is Fredholm in Lp[0, 2~r] if and only if

(6.3)

a2(x) - b2(z) ~ O,

and the function y(t) = (ao(t) - bo(t))-X(ao(t) + bo(t)) admits a factorization

y(t) = y_(t)t~y+(t)

(6.4)

in the space Lp(T). If these conditions hold, then (6.5)

indA = -~.

In order to find the numbers dimker A, dimcoker A and explicit solutions of the equation A~ = r we assign to operator A a characteristic singular integral operator .4 = cP + dQ with matrix coefficients

tt--[11 tbo(t)

-t-1 J

1

_t-1

-~tbo(t)

ao(t)-lbo(t)]

E d(t) =

ao(t) + bo(t)

and a matrix function 1

G(t) =

(6.7)

-t-1

bo(t)t

1

ao(t)

a0( -L(t) Note that this matrix is of the form (1.1). T h e o r e m 6.1. Let the operator A, defined by (6.1) be a Fredholm operator. Then:

The matrix G admits a factorization in the space Lp(T) and dimkerA = -

(6.8)

E

t~j,

nj <0

where ~1, n2 are the partial indices of the matirx G.

Equation A~ = ~ is solvable in Lp[0, 2~r] iff the equation (6.9) sol able in L (T) •

.~ ( u 0 ) ( 0 ) v0 -- r

(~b0(exp(ix)) = r

Feldman, Gohberg and Krupnik

291

Let (uo, VO)T be the genera/ solution to equation (6.9), then v( x ) is the generM solution to equation (6.1). Proof. The operator fi, can be represented in the form

(6.10)

lr

- T -1

lboTS

ao + ~boS

"4 =

] (Tf(t) = tf(t))

1

[' :]I' 0][, l boTS

0

Let ~0 be a solution to equation A~0 = r

Ao

0

I.

then the vector (t~0, ~P0) T is a solution to

equation (6.9). Conversely, let (u0, vo) T be a solution to equation (6.9). It is readily checked that Aovo = r It remains to prove equality (6.8). The left and right factors in the right-hand side of (6.10) are invertible operators; hence dimkerA0 = dimkerA. It is well-known that dimker f~ = -

~ kj<0

kj,

where kl, k2 are the partial indices of the matrix

1

[ o0(t)t-lb0(t)]

(d-lc)(t) - ao(t)- bo(t) [ tbo(t)

ao(t)

J"

This matrix can be represented in the form

1 and hence ~1

=

a(t)

kj.

It follows from this theorem that the numbers dimkerA, dimcoker A and explicit solutions to the equation A~ = r can be found as soon as the explicit factorization of matrix (6.7) is obtained. The matrix G can be rewritten in the form

(6.11)

ao + bo a(t)

=

t

y(t~ + 1

u -

ao----Fo

'

and in the next part of this section we factorize the matrix (6.11) by the method suggested in this paper. Note that the matrix can be also factorized by reduction to triangular matrix. We suppose that the function y(t) admits a generalized factorization

292

Feldman, Gohberg and Krupnik

in Lp(T). By t h e o r e m 6.1 the m a t r i x G also admits a generalized factorization in Lp(T). For the definition of genrealized factorization see [S],[CG]. We mentioned t h a t the matrix G(z) is of the form (1.1) and can be factorized by the m e t h o d suggested in C h a p t e r I. We restrict ourselves here to the cases i n d y = 0 and ind V = - 1 . Let ind y = 0 and y = Y-Y+. The solution (of order zero) to equation (1.4) is given by h+(z) = Co + c l z , f - ( z )

= cl + CoZ-1. Problem (1.5) has the form g

_

y-1 = yg+ + - - - ~ ( C l Z + co),

a n d after some simple c o m p u t a t i o n s can be reduced to (6.12)

2 g - y -1 = y+(2g + --~ co + ClZ) - (c0 + ClZ)(y -1 - ~]-1(oo))

Co + ClZ

v_(m)

It follows from here t h a t

g-- = ~[C2 -- (Co -~- C l Z ) ( y =

2g+

1 --

v--l(oo))]

+c,z = v;' (c2+ c~ +c'z ) v_(m) '

where cz is an a r b i t r a r y constant. Here equation (1.6) means t h a t co + g+(0) -- 0 and hence c~ =

Vector F - = ( f - , g - ) T

co(l + y_(e<))y+(O) y_(m)

can be represented by F - = c 0 F o + clF~-, where

[F1,Fol =

i

z_, ]

2(y_ - y_(or

Y- - Y-(oo) - 7Y-

'

a n d ff = 1 + y _ ( o c ) y + ( 0 ) . Set G _ = [F~-,Fo], then det G _ ( z ) = - T y _ ( z ) / 2 y _ ( o c ) .

If

7 # 0 then (see r e m a r k 2.1) G ( z ) admits the canonical faetorization G = G _ G + , where G _ is defined above. If 7 = 0 then the subspace generated by the vectors F o and F lis one dimensional, a n d therefore tr ~r

=

> 0. By theorem 2.1 ~1 < 1, hence ~x = 1 and

-1.

Let now y ( z ) = y - ( z ) z - l y + ( z ) .

Denote by gl, a2 (g2 _< ~1) the p a r t i a l indices of

G. By t h e o r e m 2.1, ~1 _< 0 and since gl +/'1:2

:

--1 we have: gl = 0 and n2 = - 1 .

Feldman, Gohberg and Krupnik

293

Hence we look for solutions of order zero to problem (1.3). In this case (as well as in the previous one) the pair of functions h+(z) = co + clz, f - ( z ) = Cl + coz -1 solves "

l

problem (1.5). We write problem (1.5) in the form g - = yg+ + ~-~-..~-h+ or 2 g - y - X z -y+(2g + + h +) - zh+y -1. We obtain the functions ~ - = 2 g - y -1 and ~+ = y+(2g + + h +) as a solution of the barrier problem

(6.13)

Z~p- = ~+ -- (CoZ + ClZ2)y -1. oo

Set y_-i(z) = 1 + ~ ~ . The solution of problem (6.13) can be rewritten in the form k=l

~-(z)=

rl(z) --(coz+clz2)y=l(z) +Cl Z2 +(Cl~ 1 --Fc0)z Z

~+(z)=r,(z)+c,z2+(cx*x+co)z where rl(z ) is an arbitrary polynomial of first degree. It follows from here that

g

1 [r 1 --~Clz2 --~ (Cle, + Co)zly- - (cxz 2 + CoZ) = 2~ Y- = 2z

g + __ ~ + y + X __ h + _ [r 1 _{_ ClZ2 + (Cl~ 1 ..~ c 0 ) z ] y + l _ (c 0 + C l Z )

2

2

Here equality (1.6) means that g+(0) + h+(0) = 0 and hence rl(0 ) = --c0y+(0 ). Now we have f + ( z ) = g+(z) + h+(z) and vectors F Z

(f-,g-)T,

F + = ( f + , g+)T, which

solve the barrier problem F - = G F + : C0 C1 - } - - Z

F-(z)=

[

(Z+ellY_(Z)-Z

Cl

2

+ co (z - y+(0llv_(z) - z + c~v_(z) 2z 2

where c2 is a coefficient of the polynomial rl(z ) = --c0y+(O) + c2z;

F+(z) =

c2y+l(z el ~-[(Z -']'-~1)~1+1(Z) + 1] + ~z[(Z -- y+(O))y+l(z) + 1] + - 2 2 [ ( Z 2 "+ ~ I Z ) ~ + I ( z ) -- Z] + 2 [ ( Z -- y + ( 0 ) ) y + l ( z )

-- 11 -'~

c~zy+l(z) 2

In the case nl :/: ~2 the factors of factorization are defined not only up to invertible constant matrix. We propose one of possible factorizations taking arbitrary constant c2 = Clgl and two collections of constants

294

Feldman, Gohhergand Krupnik

1) cl = 1, Co = 0 for vectors F I- and F +, 2) cl = 0, co = 1 for vectors F 0- and F +. So we have obtained the following relations:

[ 1

]z

z(y_(z)- 1) (z - y+(O))y_(z)- z 2 2z zy+l(z) + 1 (z - y+(O))y+l(z) + 1 [F+,F+ ] = 2 2z = z2y+l(z) - z (z y+(O))y_~l(z) 1

o]

[FI'F~ =

-

-

-

2

Z--1

9

G+l(z).

-

2

Finally,

a(z) = G_(z) where

1 a+(z)

-

diag(1, z - 1 ) G + ( z ) ,

"y+(z)+y+(O)--z

~+(0)

z~

-

z-l(y+(z)--y+(O)+l]

y+(z)z

-~+(z)

-

z

Note that equation (6.1) in the case ~ = 0 was investigated in [GK2, pp. 141-145] without factorization of the matrix function. For arbitrary ~ this equation has not been investigated yet. w

Singular integral equations with transformed argument. In this section we consider a class of equations A~ = %5in

(7.1)

A~(t)

=

al(t)~(t ) Jr- a2(t) fV ~(t)dt vr--T

~

+

bl(t)~(-t)

Lp(T),

where

bz(t) [ ~(t)dt

+ ~

Jv

T + t

and aj, bj e Loo('r) (j = 1, 2). It is well-known (see [GK1]) that A is a Fredholm operator in Lp(T) iff (7.2)

essinf

]cl(t)c~(-t) - dl(t)dl(-t)l ~ O,

(7.2)

essi~f

{c2(t)c;(-t) - d2(t)d2(-t)l # O,

tET

where c = a + b, d = a - b, and the matrix

G(t)

r c~(t)

d~(t) l-' r c'(t)

Ld2(-t)

c2(-t)J

[ d,(-t)

dl(Ol c,(-OJ

admits factorization in the space Lp('~). In the sequel we need the following notations:

P--I(I+S),

O=I-P.

1 f~ ~(t)dt sv(~) = -=~ 7--~'

Feldman, Gohberg and Krupnik

295

T h e o r e m 7.1. Let the matr/x G admit a canonical factorization. Then the operator (7.1) is invertible in Lp(V) and A -1 = I-II(G+Ip + G_Q)G-1H-1II2.

(7.5) Here

H(t) = [ c2(t) d2(t) ] [ dz(-t) ~(-t),

and for any ~, r e Lp(T)

( n ~ ) ( t ) = (~(t), ~(-t)) r,

n,(~, r = ~.

Proof. We assign to the operator A a characteristic singular integral operator A = F P + H Q with matrix coefficients F(t)

Set B = a l I + a 2 S ,

[ el(t)

dl(t) ]

H(t) = [ c2(t)

d2(t) 1

Ld , ( - t )

c,(-t)J'

Ld~(-t)

~(-t)J"

C = blI + b2S, where

I~

s~(t) = ~Z

9~(t)dt

-;--2-7

and W ~ ( t ) = ~ ( - t ) . It is readily checked that A = B + CW, (7.6)

~i=~

w

-w

0

B-

W

-W

and 1

I

I

Since the matrix G = H - 1 F admits a canonical factorization in the space Lp(T) the operator A is invertible in Lp(T) • Lp(T). From the equality (7.6) it follows that the operator A (= B + C W ) is invertible and A -1 is given by (7.5). [] Example

7.1. Consider the equation

(7.8)

~(t) + ( 1 - - ~ T ) 9 ~

j~

~- + t - r

(t e v).

The operator A in the left-hand side of this equation is of the form (7.1) with cl(t) = c2(t) - x2t/ 2+- 1't filled.

dl(t) = -d2(t) =

x/2 + t 2(1 - 2t-------~"The conditions (7.2) and (7.3) are ful-

296

Fcldman, Gohberg and Krupnik

Consider matrix (7.4). Here

5 a(t) = g

4x,/'2+t(1+2t) 5~/2-t(1-2t) 1

1 42V~Z'~-t(1-2t)

52W-4--/(1+2t) This matrix is of the form (1.1) with

p(t) = 4 2x/~-+ t(1 + 2t) , 52x/~-2~_t

q(t) = 1 - 2 t and it can

be factorized by the method described in section 2. It turns out that the factorization is canonical:

G = G_G+, where 1 +2t 1 - 2t

a_(t) :

1 i -

2t

v~

4

1

4

,/Tg(1+2t)

8,/ig

and

a+(t) =

5

X / ~ + t ) ( 3 1 -2t) - 3 2 X / ~ - t (l+2t) 5(V/g(g-gt) 32v~

) 4(31 -2t)v~+t-4Ox/i-~-t ) 5(1 - 2 t ) , / 2 - t

- t) - 3 0 ~ / g ~ + t) v / ~ + t)

8(1 + 2t)~/Tg(5 - t) - 3 2,/g-;-/ (1 - 2t) 2,/5-:7-~

It follows from theorem 7.1 that equation (7.8) has a unique solution for any function r E Lv('ll'), and this solution coincides with the first coordinate of the vector (fl) f2 where

H-l(t)=

[

=(G+Ip + G _ Q ) G - 1 H - I ( ~ )

a~(t)

b:(t) ]-1

b~(-t) a~(-t)

=

4

l+2t

l+2t

1 - 2t

1 - 2t

w

Inversion of Toeplitz plus Hankel operators. Let {a/} and {bj} (j E Z) be the sequences of Fourier coefficients of the functions a and b (E Lo~('r). Denote by N the operator defined on g2 by the sum N = T(a) + H(b) of Toeplitz and Hankel matrices (8.1)

T(a) = [aj_k]j,~k=o, H(b)

= [aj+k+l]jTk= O.

Feldman, Gohberg and Krupnik

T h e o r e m 8.1").

297

Let a, b E Lot(T) and essinf la(t)l # 0.

(8.2)

tE'~"

Suppose that the matrix

admits a canonical factorization G(t) = G_(t)G+(t) in L~(T). Then the operator N = T(a) + H(b) is invertible in ~2 and (8.4)

N -1 = T(g+l)(T(y11) + H(y21)) + T(g+ )(T(y12) + H(y~2)),

where GA_1 : : [gjk]j,k~-l,

+,

G + M - ' = [yjk]j k=, and M(t) = [ a(t) [b(1/t) ,,

~ 9

Proof. Consider the operator F = P a P + PbVP + Q in the space L2(T) where P and QaretheprojectionsdefinedintheprevioussectionandV~(t)=l~2(1).Notethat the operator F in standard basis has the form

=

[, 0

0 ] T(a) +H(b) "

It is convenient to represent the operator F in the form F = RL where

R = (aP + Q) + bQY and L = I - Q(a + bV)P. Note that L is invertible and L -1 = I + Q(a -4- bV)P. We assign to R the operator

(8.5)

[aP -4- Q /~:= [ VbQV

bQ ] = M ( P + G-1Q) = MG+I(G+P + G-1Q). V(aP+Q)V

This operator is invertible and

k -1 = (G+IP + G - Q ) G + M - 1. 9) The connection between the properties of the operator T(a)TH(b) and the factorization matrix (8.3) was also investigated in [LMT].

298

Feldman, Gohbergand Krupnik It is readily checked that

where R1 = aP + Q - bQV. It follows from here that R is invertible and

R -1 = III(G+IP + G_Q)G+M-1IIa, where n 3 r = (r Vr r ~ d

rll(v, r

= ~.

Let ~+ be an arbitrary function from the Hardy space I-/2 and (PaP + PbVP)~+ = r

then F~0+ = r

i.e., RL~o+ = r

and

LV+ -- I I I ( G + I p + G_Q)G+M-1IIar Note that PLqo+ = ~+, hence

~+ = II1G+I PG+M-IH3~+

= g+P(Yn + y12V)r

Py22 Vr + + g21P(y21 + Y22V)r

Thus, in the space/-/2 we have

(PaP + PbYP) -1 = Pg+P(Yu + YI2V)P + Pg+lP(y21 + Y22V)P. This involves the equality (8.4).

[]

Note that the conditions of theorem 8.1 are not necessary for the operator T(a) +

H(b) to be invertible in g2. Indeed, the operator N {~m}m=0 ~ = (2~0, ~1, ~2,. .. ) is invertible in ~2 and can be represented as T(a) + H(b) with a(t) ==-1 and b(t) = t. Here

a(t)=

[0:] t 1

=

_t-1

011 0]

01]

and the factorization is not canonical.

Example 8.1. Consider in the space g2 operator N = T(a) + H(b) with iz a ( z ) - z2_ 3z + l,

1-z

z+l

b(z)- z2_32+lexp~.z_1

Fcldman, Gohberg and Krupnik

299

Here t h e m a t r i x (8.3) is

G,z, Eo ll 1 zl'ex':'l z,1 9 1 (1 - z ) e x p 1 - z 3 - z - z -1 a n d it c a n b e factorized following the scheme in section 2:

1

[

-e

x

--

z e

1

(1

.

]

Z -1

z) e. x p l + Z.

.

z

+ ~ e x p z z-1 1-z z+l 1+ exp e z-1

1 -.Z e x p .l + Z

1

z

e

1-z

z

where a = e(e + 1) -1 ( c o m p a r e with the factorization in e x a m p l e 1, w

We o b t a i n e d a

c a n o n i c a l f a c t o r i z a t i o n , hence the o p e r a t o r N is invertible. In order to give t h e explicit f o r m u l a for N - 1 we n e e d t h e explicit form of the following matrices:

(8.6) G + 1 = - ~ - 1

e

z----~

1 z

z

Z exZ,1] z-1

=

g

g+

J

(8.7) G + M -1 =

e ( z 2 - 3z + 1)

1- z

iz

iz

z 2 - 3z + 1 ie

=

Yll Y21

+

1-z i

z+ 1 e(1 - z) 2 + - exPl_z zi

z+l (l-z) + ~ exp '1 -- z ie

2

1 z

e(1 - z) z + 1 + - exp z

1-z 1+ ~ e

z-1

z+l exp - z -- 1

3

Y12[ . Y22 J

Finally, N -1 = T ( g + l ) ( T ( y 1 1 ) + H ( y 2 1 ) ) + T ( g + l ) ( T ( y 1 2 ) + H(y22)), where the f u n c t i o n g+k a n d Yjk are d e t e r m i n e d in (8.6) a n d (8.7). Now we are going to describe a wide class of f u n c t i o n s a, b for which t h e m a t r i x (8.3) is of t h e form (1.1) a n d can be factorized by the m e t h o d p r o p o s e d in C h a p t e r I. Let T be a n i n n e r f u n c t i o n [H, Ch.5]. T h e f u n c t i o n ~ c a n b e r e p r e s e n t e d in t h e form

: ( z ) = z , ,,=, 1-[

~- ] f~~. - z

,-

exp

-

_

d#(t)

,

300

Feldman, Gohberg and Krupnik

where d# is a positive singular measure. If a , E ( - 1 , I) and dp(z -1) = d#(z), then

~(Z--1)~(Z) ~ 1 (z E V).

(8.8) Let 7 E C \ Z a n d r

"r ( = e x p ( i T 0 ) , 0 _ < 0 < l ) . T h e n

(8.9)

= const

r162

(z E T).

Consider the operator N = T(a) + H(b) in the space e2 where a(z) = ri(z)cpl(z)r

b(z) = r2(z)q~2(z); r~ and r2 are some rational functions without poles on T and ~pl(z), ~?2(z) are some inner functions which possess property (8.8). In this case the function r a ( z ) : = a(z)fi(z) - b(z)b(z)

(a(z):= a(z-'))

is rational and

a(z) = g(~

~(z)

Suppose that r3(z) r 0 (Izl = 1), then G(z) can be represented in the form (1.1) (up to scalar factor): (8.10)

G ( z ) = r3(z) [ c(z)

P(Zd)~:r

(pEH~

Example 8.2. Consider a function a(z) = z ~ (7 E C \ Z) and b(z) = ~o(z)/z, where c2 E H a and ~(z)~(z) - 1 (Izl = 1). Up to scalar factor, matrix 8.10 has the form

~(z) O~

M(z) =

z z~(z)

(,~ r o).

1

We supposed that det M ( z ) r 0 (Izl = 1), i.e., c~ # - 1 . If ~(0) = 0 then z - i ~ ( z ) E H ~176 and the equality

M(z)=

z@(z)

o~+1

a -1

is a canonical factorization of M. Let now ~?(0) r 0. Following the scheme proposed in section 2 we solve the barrier problem z f - = h + and obtain its solution h+(z) = co+ciz, f - ( z ) = c0z -1 + cl. The second problem

_ _ a + l g + + _i(r + ~ z ) ~

Feldman, Gohberg and Krupnik

301

has the solution

1

g - ( z ) = ~(c0 + c l z ) ( ~ ( z ) - ~(0)) + c2 ~c2 g+(z) - a + l

~(0) (c0 + c,~). a+l

Using the equality h+(O) 4- ~p(O)g+(O) = 0 we obtain

c2 =

c0(~2(0) - a - 1) ~o(0)

Now the vector F - = ( f - , g - ) T can be written in the form F - = caF~ + coF~- where

[F~,F~] =

[

i

o~-Iz(~(z)

- -

~(0))

Oz-I(~(Z)

- -

~(0)) 4- o~--l~--l(0)(~p2(0)

- -

Ol

- -

1)

1

Denote this matrix by M_(z). If ~2(0) # a + l then det M_(z) = a-l~-l(0)(~v2(0)-a 1) # 0 and by theorem 2.1 M(z) admits canonical factorization M(z) = M_(z)M+(z). If %v2(0) = a 4- 1 then the vectors F 1- and F 2- are linear dependent and tr > 0. By theorem 2.1 Xl < 1, hence Xl = 1, tr = -1. The problem of solving the integral equation v(~) +

/5

kl(~ - ~)~(~)d~ +

/?

k~(t + ~)v(~)d~ =

r

(kl,k2 9 k2(t)=Ofort
1

~:2(-A)

]

-kz(A) (I 4- k1(A))(1+ k~(-A)) - k2(A)k2(-A) where k(i) denotes the Fourier transform of the function k (see, for example, [F]). If R(A) = (1 +~h(X))(14-/q(-A))- k 2 ( i ) k 2 ( - i ) i s a rational function and R(A) # 0 ( - o o < ), < oo), the matrix R-I(A)G(A) has the form (3.1) and can be factorized following the scheme in section 3.

302

Feldman, Gohberg and Krupnik

REFERENCES

[AKM] Aktosun, T., Klaus, M., Van der Mee, C., [CG] [F]

Ezphcit Wiener-Hopffactorization for certain nonrational mairiz functions, IEOT 15 (1992), 879-900. Clancey, K., Gohberg, I., Factorization of Malriz Functions and Singular Integral Operators, 0713, Birkhs Verlag, Basel, 1981. Feldman, I., On effective solution of some integral equations on a line and a half-line, Izv. AN

MSSR 10 (1961), 16-26 (Russian).

[GKI] Gohberg, I., Krupnik, N.,

On certain one-dimensional singular integral operators with a shift,

Izv. AN arm. SSR 8, N1 (1973), 3-12 (Russian). [GK2] Gohberg, I., Krupnik, N., One-dimensional Linear Singular Integral Equations, vol. II, General Theory and Applications, O T 54, Birkhhuser, Basel, 1992. [U] Hoffman, K., Banach Spaces of Analytic Functions, Prentice-Hall, Englewood Cliffs, 1962. [LMT] Lebre, A.B., Meister, E., Teixeira, F.S., Some results on the invertibility of Wiener-HopfHankel operators, Zeitschrift fiir Analysis und ihre Anwendungen 11, 1 (1992), 57-76. Simonenko, I., Some general questions on the theory of the Riemann boundary problem, [zv. IS] AN SSSR, Ser. Math. 32 (1968), 1138-1146 (Russian). [TF] Tahtajan, L.A., Faddeev, L.D., Hamilton Approach in Soliton Theory, Nauka, Moscow, 1986 (Russian). I. F e l d m a n D e p a r t m e n t of M a t h e m a t i c s and Computer Science Bar-Ilan University 52900 R a m a t - G a n Israel I. Gohberg School of M a t h e m a t i c a l Sciences R a y m o n d and Beverly Sackler Faculty of Exact Sciences Tel-Aviv University 69978 T e l - A r t y Israel N. K r u p n i k D e p a r t m e n t of M a t h e m a t i c s and C o m p u t e r Science B a r - I l a n University 52900 R a m a t - G a n Israel

MSC 47A79 Submitted:

November 17, 1993