SOLVABILITY CONDITIONS FOR SOME DIFFERENCE OPERATORS N. C. APREUTESEI AND V. A. VOLPERT Received 24 docx

Under the assumption that the co-efficients of the operator have limits at infinity, limiting operators and associated polyno-mials are introduced.. Under some specific conditions on the p

DIFFERENCE OPERATORS

N C APREUTESEI AND V A VOLPERT

Received 24 June 2004

Infinite-dimensional difference operators are studied Under the assumption that the co-efficients of the operator have limits at infinity, limiting operators and associated polyno-mials are introduced Under some specific conditions on the polynopolyno-mials, the operator is Fredholm and has the zero index Solvability conditions are obtained and the exponential behavior of solutions of the homogeneous equation at infinity is proved

1 Introduction

Infinite-dimensional difference operators may not satisfy the Fredholm property, and the Fredholm-type solvability conditions are not necessarily applicable to them In other words, we do not know how to solve linear algebraic systems with infinite matrices Var-ious properties of linear and nonlinear infinite discrete systems are studied in [1,2,3,4,

5,6,7,8]

The goal of this paper is to establish the normal solvability for the difference operators

of the form

(Lu) j = a − j u j − +···+a0j u j+···+a m j u j+m, j ∈ Z, (1.1)

and to obtain the solvability conditions for the equationLu = f , where m ≥0 is a given integer and f = { f j } ∞

j =−∞is an element of the Banach space

E =

u =u j∞

j =−∞,u j ∈ R, sup

j ∈Z

u j< ∞. (1.2)

The right-hand side in (1.1) does not necessarily contain an odd number of summands

We use this form of the operator to simplify the presentation We will use here the ap-proaches developed for elliptic problems in unbounded domains [9,10] and adapt them for infinite-dimensional difference operators

Copyright©2005 Hindawi Publishing Corporation

Advances in Di fference Equations 2005:1 (2005) 1–13

DOI: 10.1155/ADE.2005.1

The operatorL : E → E defined in (1.1) can be regarded as (Lu) j = A j U j, where

A j =a − j , ,a0,j ,a m j

, U j =u j − , ,u j, ,u j+m

(1.3) are 2m + 1-vectors, A jis known, andU jis variable

We suppose that there exist the limits of the coefficients of the operator L as j→ ±∞

a ±

l = lim

j →±∞ a l j, l ∈ Z, − m ≤ l ≤ m, (1.4) anda ± =0

Denote byL ±:E → E the limiting operators



L ± uj = a ± u j − +···+a ±

0u j+···+a ± u j+m, j ∈ Z (1.5) Recall that a linear operatorL : E → E is normally solvable if its image ImL is closed If

L is normally solvable with a finite-dimensional kernel and the codimension of its image

is also finite, thenL is called Fredholm operator Denoting by α(L) and β(L) the

dimen-sion of kerL and the codimension of ImL, respectively, we can define the index κ(L) of

the operatorL as κ(L) = α(L) − β(L) It is known that the index does not change under

deformation in the class of Fredholm operators

InSection 2of this paper we introduce polynomialsP+(σ) and P −(σ) associated with

the limiting operatorsL+andL − We show that, ifP+andP −do not have roots on the unit circle, then the limiting operators are invertible and the operatorL is normally solvable

with a finite-dimensional kernel If moreover the polynomials have the same number of roots inside the unit circle, thenL is a Fredholm operator and its index is zero.

InSection 3we prove that under some conditions on the polynomialsP+andP − cor-responding to operatorL in (1.1), the bounded solutions of the equationLu =0 are ex-ponentially decreasing at +∞and−∞ The idea is to approximate the equation Lu =0 at +∞with the problem on half-axis:



L+uj =0, j ≥1,

and similarly at−∞ We first prove that (1.6) has a unique solution and that this solution

is exponentially decaying Then we deduce that the solution of the equationLu =0 is also exponentially decaying

Section 4deals with the solvability conditions for the equationLu = f , for L in (1.1) and f = { f j } ∞

j =−∞ ∈ E being given.

Let L ∗ be the formally adjoint of L, α(L ∗)=dim(kerL ∗), and v l = { v l } ∞

j =−∞, l =

1, ,α(L ∗) some linearly independent solutions of the equationL ∗ v =0 One states a result which is analogous to the continuous case: equationLu = f is solvable if and only

if f is orthogonal on all solutions v l,l =1, ,α(L ∗)

Section 5is devoted to a particular case of the operatorL related to discretization of a

second-order differential equation on the real axis:

(Lu) j = u j+1 −2u j+u j −1+c j

u j+1 − u j

+b j u j, j ∈ Z, (1.7)

where{ b j } ∞

j =−∞and{ c j } ∞

j =−∞are given sequences of real numbers If there exist the lim-itsb ± =limj →±∞ b j < 0, and c ± =limj →±∞ c j ≥ −2, then there are no roots of the

poly-nomials on the unit circle, and there is exactly one root inside it Therefore the bounded solution of the equationLv =0 is exponential decaying at±∞ If moreover c ± ∈[−2, 2], then the solvability conditions for the equationLu = f are applicable.

2 Limiting operators and normal solvability

Let E be the Banach space of all bounded real sequences E = { u = { u j } ∞

j =−∞, u j ∈ R,

supj ∈Z | u j | < ∞}with the norm

 u  =sup

j ∈Z

and letL : E → E be the general linear difference operator (Lu) j = a − j u j − +···+a0j u j+

···+a m j u j+m, j ∈ Z, where m ≥0 is an integer anda − j , ,a0,j ,a m j ∈ Care given coef-ficients Denote byL+:E → E the limiting operator



L+uj = a+

− u j − +···+a+

0u j+···+a+

m u j+m, j ∈ Z, (2.2) where

a+

l =lim

j →∞ a l j, l ∈ Z, − m ≤ l ≤ m. (2.3)

We are going to define the associated polynomial for the operatorL+ To do this, we are looking for the solution of the equationL+u =0 under the formu j =exp(µj), j ∈ Z,

and obtain

a+

− e − µm+···+a+

−1e − µ+a+

0+a+

1e µ+···+a+

m e µm =0. (2.4) One takesσ = e µand finds the polynomial associated toL+:

P+(σ) = a+

m σ2m+···+a+

0σ m+···+a+

Recall the following auxiliary result from [1]

Lemma 2.1 The equation L+u = 0 has nonzero bounded solutions if and only if the corre-sponding algebraic polynomial P+has a root σ with | σ | = 1.

We will find conditions in terms ofP+for the limiting operatorL+to be invertible One begins with an auxiliary result concerning continuous deformations of the polynomial

P+ Without loss of generality, we may assume that the coefficient a+

m =1 Consider the polynomial with complex coefficients

P(σ) = σ n+a1σ n −1+···+a n −1σ + a n (2.6)

Lemma 2.2 Suppose that a polynomial P(σ) does not have roots with | σ | = 1 and it has k roots with | σ | < 1, 0 ≤ k ≤ n Then there exists a continuous deformation P τ(σ), 0 ≤ τ ≤ 1,

such that

P0(σ) = P(σ), P1(σ) =σ k − aσ n − k − λ, (2.7)

and the polynomial P τ(σ) does not have roots with | σ | = 1 for any 0 ≤ τ ≤ 1 Here λ > 1 and

a < 1 are real numbers.

Proof We represent the polynomial P(σ) in the form

P(σ) =σ − σ1



···σ − σ n

where the rootsσ1, ,σ k are inside the unit circle, and the other roots are outside it Consider the polynomial

P τ(σ) =σ − σ1(τ)···σ − σ n(τ) (2.9)

that depends on the parameterτ through its roots This means that we change the roots

and find the coefficients of the polynomial through them We change the roots in such

a way that for τ =0 they coincide with the roots of the original polynomial; for τ =

1 it has the rootsσ1, ,σ k with (σ i)k = a, i =1, ,k (inside the unit circle) and n − k

roots σ k+1, ,σ n such that (σ i)n − k = λ, i = k + 1, ,n (outside of the unit circle) This

deformation can be done in such a way that there are no roots with| σ | =1 The lemma

Using the associated polynomialsP+andP −ofL+andL −, we can study the normal solvability of the operatorL.

Theorem 2.3 The operator L is normally solvable with a finite-dimensional kernel if and only if the corresponding algebraic polynomials P+and P − do not have roots σ with | σ | = 1 Proof

The necessity Suppose that the polynomials P+,P −do not have rootsσ with | σ | =1 We first show that the image ofL is closed To do this, let { f n }be a sequence in ImL such

that f n → f and let { u n }be a sequence with the propertyLu n = f n

Suppose in the beginning that{ u n }is bounded inE We construct a convergent

sub-sequence Since u n  =supj ∈Z | u n

j | ≤ c, then for every positive integer N, there exists a

subsequence{ u n k }of{ u n }and an elementu = { u j } N j =− N ∈ E such that

sup

− ≤ j ≤

u n k

that is,u n k → u as k → ∞uniformly on each bounded interval of j Using a

diagonaliza-tion process, we extendu jto allj ∈ Z.

It is clear that supj ∈Z | u j | ≤ c; that means u ∈ E Passing to the limit as k → ∞in the linear equationLu n k = f n k, we getLu = f , so f ∈ImL.

We show that the convergence in (2.10) is uniform with respect to allj ∈ Z Supposing

by contradiction that there exists j k → ∞such that| u n k

j k − u j k | ≥ ε > 0, observe that the

sequence y k = { y k

j } ∞

j =−∞, y k

j = u n k

j+j k − u j+j k verifies the inequality| y k0| = |u n k

j k − u j k | ≥ ε

and the equation

a j+j k

− y k

j − +···+a j+j k

0 y k

j+···+a j+j k

m y k j+m = f n k

j+j k − f j+j k, j ∈ Z (2.11)

Since the sequence{ y k }is bounded inE, there exists a subsequence { y k l }which converges (sayy k l → y0) uniformly with respect toj on bounded intervals We may pass to the limit

ask l → ∞in (2.11) and obtain via (2.3),

a+

− y0

j − +···+a+

0y0

j+···+a+

m y0

j+m =0, j ∈ Z (2.12) Thus, the limiting equationL+u =0 has a nonzero bounded solution y0= { y0

j } ∞

j =−∞

Lemma 2.1leads to a contradiction Therefore the convergenceu n k

j − u j →0 is uniform with respect to all j ∈ Zand, sinceLu = f , it follows that ImL is closed.

We analyze now the case when{ u n }is unbounded inE Then we write u n = x n+y n

with{ x n } ∈kerL and { y n }is in the supplement of kerL Then Ly n = f n

If{ y n }is bounded inE, it follows as above that ImL is closed If not, then we repeat

the above reasoning forz n = y n /  y n andg n = f n /  y n  Passing to the limit on a

sub-sequence n k (such thatz n k → z0) in the equalityLz n k = g n k and using the convergence

g n k →0, one obtains the contradiction thatz0∈kerL Therefore ImL is closed.

In order to prove that kerL has a finite dimension, it suffices to show that every

se-quenceu nfromB ∩kerL (where B is the unit ball) has a convergent subsequence The

reasoning is similar to that of the first part, taking f n =0

The sufficiency Assume that ImL is closed and dim(kerL) is finite By contradiction, one

supposes that eitherP+orP −(sayP+) has a root on the unit circle Then the correspond-ing solution ofL+u =0 has the formu = { u j } ∞

j =−∞, whereu j = e iξ j,ξ ∈ R, j ∈ Z.

Letα = { α j } ∞

j =−∞,β N = { β N

j } ∞

j =−∞,γ N = { γ N

j } ∞

j =−∞be a partition of unity (α j+β N

j +

γ N

j =1) given by

α j =







1, j ≤0,

0, j ≥1,

β N

j =







1, 1≤ j ≤ N,

0, j ≤0, j ≥ N + 1,

γ N

j =







1, j ≥ N + 1,

0, j ≤ N.

(2.13)

For a fixedε n →0 (asn → ∞), let u n = { u n

j } ∞

j =−∞,v n = { v n

j } ∞

j =−∞, f n = { f n

j } ∞

j =−∞be the sequences defined byu n

j = e i(ξ+ε n)j,v n

j =(1− α j)(u n

j − u j), and f n

j = Lv n

j,j ∈ Z It is clear

thatu n

j → u j(n → ∞) uniformly on every bounded interval of integers j.

It is sufficient to prove that fn →0 Indeed, in this case, by hypothesis it follows that

v n →0 But this is in contradiction with

v n =sup

j>0

e i(ξ+ε n)j − e iξ j  ≥ m > 0 (2.14)

for somem.

In order to show that f n →0 asn → ∞, observe that f n

j can be written under the form

f n

j =α j+β N

j +γ N

j

Lβ N+γ N

u n − u j

= α jLβ N+γ Nu n − u j+β N

jLβ N+γ Nu n − u j

+γ N

j 

Lβ N

u n − u j+γ N

j

L − L+ 

γ N

u n − u j

+γ N

j 

L+

γ N

u n − u j

(2.15)

A simple computation implies that the first three terms tend to zero asn → ∞, uniformly

with respect to all integersj.

Next, condition (2.3) and the boundedness u n  =  u  =1 lead to the convergence

γ N

j

L − L+ 

γ N

u n − u j  ≤  γ N

L − L+ 

0· γ N(u n − u) −→0, (2.16)

asN → ∞, where | · |0is the norm of the operator For a givenN, one estimates the last

term of (2.15) Sinceu j = e iξ j,j ∈ Zis a solution of the equationL+u =0, then



L+ 

u n − uj =L+u n

j =L+u n

j − e iε n j

L+uj

= e i(ξ+ε n)j

a+

− e − iξm

e − iε n m −1

+···+a+

−1e − iξ

e − iε n −1

+a+

1e iξ

e iε n −1

+···+a+

m e iξm

e iε n m −1 ,

(2.17)

so



L+ 

u n − uj = iε n e i(ξ+ε n)j

a+

− (−m)e − iξm e ib − m+··· − a+

−1e − iξ e ib −1

+a+

1e iξ e ib1+···+a+

m me iξm e ib m , j ∈ Z, (2.18)

whereb − , ,b −1,b1, ,b mare intermediate points Thus the last term in (2.15) goes to zero asn → ∞and therefore f n →0 This completes the proof

Now we are ready to establish the invertibility ofL+

Theorem 2.4 If the operator L+is such that the corresponding polynomial does not have roots with | σ | = 1, then it is invertible.

Proof. Lemma 2.2forP+implies the existence of a continuous deformationP τ(σ), 0 ≤

τ ≤1, from the polynomialP0= P+toP1(σ) =(σ k − a)(σ2m k − λ) such that P τ(σ) does

not admit solutions with| σ | =1 Hereλ > 1, a < 1 are given The operator which

corre-sponds toP1isL+

1 defined by



L+

1uj = u j+k − au j − λu j+2k −2m+aλu j+k −2m (2.19) Indeed, looking for the solution ofL+

1 in the formu j = e µj, we arrive at

e µk − a − λe µ(2k −2m)+aλe µ(k −2m) =0. (2.20)

We putσ = e µand get



σ k − aσ2m k − λ=0, (2.21)

soP1is the above polynomial

Takinga =1/λ, we obtain



L+

1uj =(Mu) j −1λ u j, (2.22)

where

(Mu) j = u j+k − λu j+2k − +u j+k −2m (2.23)

is invertible for largeλ ≥0 (see, e.g., [1, Lemma 4.9]) SinceL+

1 is close toM (for λ ≥0 large enough), one deduces thatL+

1 is also invertible Hence the index ofL+

1 is zero Since the continuous deformationP τ does not have solutionsσ with | σ | =1, we find that the corresponding continuous deformation of the operator L+

τ does not admit nonzero bounded solutions (seeLemma 2.1) ByTheorem 2.3one obtains thatL+

τis nor-mally solvable with a finite-dimensional kernel From the general theory of Fredholm operators, we know that the index of such homotopies does not change Since the index

ofL+

1 isκ(L+

1)=0, we deduce thatκ(L+)=0 This, together with the fact that kerL+=Φ, implies ImL+= E, therefore L+is invertible The theorem is proved

Remark 2.5 An analogous result can be stated for L −

As a consequence, we may study the Fredholm property ofL with the aid of the

poly-nomialsP+andP −

Corollary 2.6 If the limiting operators L+and L − for an operator L are such that the corresponding polynomials P+(σ) and P −(σ) do not have roots with | σ | = 1 and have the same number of roots inside the unit circle, then L is a Fredholm operator with the zero index.

Proof We construct a homotopy of L in such a way that L+andL −are reduced indepen-dently to the operator inTheorem 2.4 Then, this homotopy is in the class of the normally solvable operators with finite-dimensional kernels

Since at +∞ and −∞the operators L+ and L − coincide, we finally reduceL to an

operator with constant coefficients According toTheorem 2.4, it is invertible Therefore,

3 Exponential decay

We consider now the problem



assuming that the corresponding polynomialP+(σ) does not have roots with | σ | =1 and hask roots with | σ | < 1 One associates to (3.1) the boundary conditions

witha1, ,a k ∈ Rgiven

Since there arek roots inside the unit circle, then there are k linearly independent

solutions of the equationL+u =0 decaying asj → ∞ Denote them by u1, ,u k Consider their values forj =1, ,2m:

u1,u1, ,u1

m,

u k

1,u k

2, ,u k

2m

(3.3)

Each of these solutions is completely determined by the above values Therefore the corre-spondingk vectors are linearly independent Indeed, otherwise the solutions would have

been linearly dependent Therefore there existk linearly independent columns Without

loss of generality we can assume that these are the firstk columns Hence the

correspond-ingk × k matrix is invertible.

Any bounded solution of (3.1) can be represented in the form

Substituting it in (3.2), we uniquely determine the coefficients c1, ,c k Therefore we have proved the following result

Proposition 3.1 If the corresponding polynomial P+(σ) for L+ does not admit roots on the unit circle | σ | = 1 and has k roots with | σ | < 1, then for each (a1, ,a k)∈ R k , prob-lem ( 3.1 )-( 3.2 ) has a unique bounded solution In addition, this solution is exponentially decreasing.

This result holds also forL − Thus, for the solution ofLu =0, we may conclude the following

Theorem 3.2 Suppose that the polynomials P+(σ) and P −(σ) corresponding to L+and L − , respectively, do not have roots with | σ | = 1 and have the same number of roots with | σ | < 1 Then the bounded solutions of the equation Lu = 0 are exponentially decreasing at ±∞ Proof Let ˜ u = { u˜j } ∞

j =−∞ be a bounded solution of the equation Lu =0 Consider the problem

(Lu) j =0, u N+1 = a1, ,u N+k = a k (3.5)

ForN sufficiently large this problem is uniquely solvable for any a1, ,a k since prob-lem (3.1)-(3.2) is uniquely solvable, and the operatorL is close to the operator L+ If we puta i = u˜N+i,i =1, ,k, then the solution of problem (3.5) coincides with ˜u for j ≥ N.

Therefore it is sufficient to prove that the solution of problem (3.5) is exponentially de-creasing for anya i,i =1, ,k.

Consider the operatorS of multiplication by exp(µ1 +j2), that is,

(Su) j = e µ √

1+j2

Hereµ > 0 Let L µ = SLS −1 Then



L µ uj = a − j u j − e µ( √

1+j2− √

1+(j − m)2 )+···+a0j u j+···+a m j u j+m e µ( √

1+j2− √

1+(j+m)2 ).

(3.7) Forµ sufficiently small, the operator L µis close to the operatorL Therefore the problem



L µ vj =0, v N+1 = b1, ,v N+k = b k (3.8)

is uniquely solvable for anyb1, ,b k

If we putb i =exp(µ1 + (N + i)2)a i,i =1, ,k, then the solution u of problem (3.5) can be expressed through the solutionv of problem (3.8):u = S −1v Since v is bounded,

thenu is exponentially decreasing.

Thus we have proved that ˜u is exponentially decreasing as j → ∞ Similarly it can be

4 Solvability conditions

In this section, we establish solvability conditions for the equation

HereL is the operator in (1.1) and f = { f j } ∞

j =−∞is fixed inE.

For the operatorL, denote α(L) =dim(kerL) and β(L) =codim(ImL) If (u,v) is the

inner product of two sequencesu = { u j } ∞

j =−∞,v = { v j } ∞

j =−∞in the sensel2, that is,

(u,v) = ∞

j =−∞

then we may define the formally adjointL ∗of the operatorL by the equality

LetL+,L −andL+

∗,L −be the limiting operators associated withL and L ∗, respectively We work under the following hypothesis:

(H) the polynomialsP+,P −corresponding toL+andL −do not have roots with| σ | =

1 and have the same number of roots with| σ | < 1 Similarly for the polynomials

P+

∗andP −corresponding toL+

∗andL −

Corollary 2.6implies thatL and L ∗are Fredholm operators with the zero index.

Lemma 4.1 Under hypothesis (H), it holds that β(L) ≥ α(L ∗ ).

Proof By the definition of the Fredholm operator it follows that (4.1) is solvable for a given f = { f j } ∞

j =−∞ ∈ E if and only if there exist linearly independent functionals ϕ k ∈

E ∗,k =1, ,β(L) such that

On the other hand consider the functionalsψ lgiven by

ψ l(f ) =

∞



j =−∞

f j v l, l =1, ,αL ∗

wherev l = { v l } ∞

j =−∞,l =1, ,α(L ∗) are linearly independent solutions of the homoge-neous equationL ∗ v =0 We know fromTheorem 3.2that the valuesv lare exponentially decreasing with respect toj Therefore the functionals ψ lare well defined

Obviously,ψ lis linear for eachl If f(n) → f in E (in the norm supremum), then we

may pass to the limit in (4.5) under the sum to find thatψ l(f(n))→ ψ l(f ), as n → ∞, for

alll =1, ,α(L ∗), that is,ψ lare continuous Thereforeψ l ∈ E ∗,l =1, ,α(L ∗), where

E ∗denotes the dual space ofE.

In order to prove thatβ(L) ≥ α(L ∗), suppose that it is not true Then among the func-tionalsψ lthere exists at least one functional (sayψ1) which is linearly independent with respect to allϕ k,k =1, ,β(L) This means that there exists f ∈ E such that (4.4) holds, but

ψ1(f ) =

∞



j =−∞

f j v1

From (4.4) it follows that (4.1) is solvable We multiply it byv1and find (Lu,v1)=(f ,v1)

By (4.6) observe that the right-hand side is different from zero But since v1is a solution

of the equationL ∗ v =0, we deduce that (Lu,v1)=(u,L ∗ v1)=0 The contradiction we

Remark 4.2 Analogously we find β(L ∗)≥ α(L) Therefore, if one denotes by κ(L) = α(L) − β(L) the index of the operator L, we get

κ(L) + κL ∗

Since in our caseκ(L) = κ(L ∗)=0, it follows that

β(L) = αL ∗

, βL ∗

4 Solvability conditions< /b>

In this section, we establish solvability conditions for the equation

HereL is the operator in (1.1) and f = {... (3.5) coincides with ˜u for j ≥ N.< /i>

Therefore it is sufficient to prove that the solution of problem (3.5) is exponentially de-creasing for anya i,i...

poly-nomialsP+and< i>P −

Corollary 2.6 If the limiting operators L+and L − for an operator L are such that the