DIFFERENCE OPERATORS
N. C. APREUTESEI AND V. A. VOLPERTReceived 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−jmuj−m+···+a0juj+···+a
j
muj+m, j∈Z, (1.1)
and to obtain the solvability conditions for the equationLu=f, wherem≥0 is a given integer and f = {fj}∞j=−∞is an element of the Banach space
E=
u=uj∞j=−∞,uj∈R, supj ∈Z
uj<∞. (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.
The operatorL:E→Edefined in (1.1) can be regarded as (Lu)j=AjUj, where
Aj=a−jm,...,a0,j ...,amj, Uj=uj−m,...,uj,...,uj+m
(1.3)
are 2m+ 1-vectors,Ajis known, andUjis variable.
We suppose that there exist the limits of the coefficients of the operatorLasj→ ±∞
a±
l =jlim→±∞alj, l∈Z,−m≤l≤m, (1.4)
anda±±m=0.
Denote byL±:E→Ethe limiting operators
L±u
j=a±−muj−m+···+a±0uj+···+a±muj+m, j∈Z. (1.5)
Recall that a linear operatorL:E→Eisnormally solvableif its image ImLis closed. If Lis normally solvable with a finite-dimensional kernel and the codimension of its image is also finite, thenLis calledFredholm operator. Denoting byα(L) andβ(L) the dimen-sion of kerLand the codimension of ImL, respectively, we can definethe indexκ(L) of the operatorLasκ(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) andP−(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 operatorLis normally solvable with a finite-dimensional kernel. If moreover the polynomials have the same number of roots inside the unit circle, thenLis a Fredholm operator and its index is zero.
InSection 3we prove that under some conditions on the polynomialsP+andP− cor-responding to operatorLin (1.1), the bounded solutions of the equationLu=0 are ex-ponentially decreasing at +∞and−∞. The idea is to approximate the equationLu=0 at +∞with the problem on half-axis:
L+u
j=0, j≥1,
u1=a1,...,uk=ak,
(1.6)
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, forLin (1.1) and f = {fj}∞j=−∞∈Ebeing given.
Let L∗ be the formally adjoint of L, α(L∗)=dim(kerL∗), and vl= {vl
j}∞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 solutionsvl,l=1,...,α(L∗).
Section 5is devoted to a particular case of the operatorLrelated to discretization of a second-order differential equation on the real axis:
where{bj}∞j=−∞and{cj}∞j=−∞are given sequences of real numbers. If there exist the lim-itsb±=limj→±∞bj<0, andc±=limj→±∞cj≥ −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 moreoverc±∈[−2, 2], then the solvability conditions for the equationLu= f are applicable.
2. Limiting operators and normal solvability
Let Ebe the Banach space of all bounded real sequencesE= {u= {uj}∞j=−∞, uj∈R,
supj∈Z|uj|<∞}with the norm
u =sup
j∈Z
uj, (2.1)
and letL:E→Ebe the general linear difference operator (Lu)j=a−jmuj−m+···+a0juj+
···+amjuj+m, j∈Z, wherem≥0 is an integer anda−jm,...,a0,j ...,amj ∈Care given coef-ficients. Denote byL+:E→Ethe limiting operator
L+u
j=a+−muj−m+···+a+0uj+···+a+muj+m, j∈Z, (2.2)
where
a+
l =limj→∞alj, 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 formuj=exp(µj),j∈Z, and obtain
a+
−me−µm+···+a+−1e−µ+a+0+a+1eµ+···+a+meµm=0. (2.4)
One takesσ=eµand finds the polynomial associated toL+:
P+(σ)=a+
mσ2m+···+a+0σm+···+a+−m. (2.5)
Recall the following auxiliary result from [1].
Lemma2.1. The equationL+u=0has nonzero bounded solutions if and only if the corre-sponding algebraic polynomialP+has a rootσwith|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 coefficienta+
m=1. Consider the
polynomial with complex coefficients P(σ)=σn+a
1σn−1+···+an−1σ+an. (2.6)
such that
P0(σ)=P(σ), P1(σ)=σk−aσn−k−λ, (2.7)
and the polynomialPτ(σ)does not have roots with|σ| =1for any0≤τ≤1. Hereλ >1and
a <1are real numbers.
Proof. We represent the polynomialP(σ) in the form
P(σ)=σ−σ1
···σ−σn, (2.8)
where the rootsσ1,...,σk are inside the unit circle, and the other roots are outside it.
Consider the polynomial
Pτ(σ)=σ−σ1(τ)···σ−σn(τ) (2.9)
that depends on the parameterτthrough 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,...,σk with (σi)k=a,i=1,...,k (inside the unit circle) andn−k roots σk+1,...,σn such that (σi)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
is proved.
Using the associated polynomialsP+andP−ofL+andL−, we can study the normal solvability of the operatorL.
Theorem2.3. The operatorLis normally solvable with a finite-dimensional kernel if and only if the corresponding algebraic polynomialsP+andP−do not have rootsσwith|σ| =1. Proof
The necessity. Suppose that the polynomialsP+,P−do not have rootsσwith|σ| =1. We first show that the image ofLis closed. To do this, let{fn}be a sequence in ImLsuch
that fn→f and let{un}be a sequence with the propertyLun=fn.
Suppose in the beginning that{un}is bounded inE. We construct a convergent
sub-sequence. Sinceun =sup
j∈Z|unj| ≤c, then for every positive integerN, there exists a
subsequence{unk}of{un}and an elementu= {uj}N
j=−N∈Esuch that
sup −N≤j≤N
unk
j −uj−→0, (2.10)
that is,unk→uask→ ∞uniformly on each bounded interval of j. Using a diagonaliza-tion process, we extendujto allj∈Z.
It is clear that supj∈Z|uj| ≤c; that meansu∈E. Passing to the limit ask→ ∞in the
We show that the convergence in (2.10) is uniform with respect to allj∈Z. Supposing
Since the sequence{yk}is bounded inE, there exists a subsequence{ykl}which converges (sayykl→y0) uniformly with respect tojon bounded intervals. We may pass to the limit askl→ ∞in (2.11) and obtain via (2.3),
a+
−my0j−m+···+a+0y0j+···+a+my0j+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 convergenceunk
j −uj→0 is uniform
In order to prove that kerLhas a finite dimension, it suffices to show that every se-quenceunfromB∩kerL(whereBis the unit ball) has a convergent subsequence. The reasoning is similar to that of the first part, taking fn=0.
It is sufficient to prove that fn→0. Indeed, in this case, by hypothesis it follows that vn→0. But this is in contradiction with
vn=sup
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 boundednessun = u =1 lead to the convergence Now we are ready to establish the invertibility ofL+.
Proof. Lemma 2.2forP+implies the existence of a continuous deformationPτ(σ), 0≤ τ≤1, from the polynomialP0=P+toP1(σ)=(σk−a)(σ2m−k−λ) such thatPτ(σ) does not admit solutions with|σ| =1. Hereλ >1,a <1 are given. The operator which corre-sponds toP1isL+1 defined by
L+
1u
j=uj+k−auj−λuj+2k−2m+aλuj+k−2m. (2.19) Indeed, looking for the solution ofL+
1 in the formuj=eµj, we arrive at
eµk−a−λeµ(2k−2m)+aλeµ(k−2m)=0. (2.20)
We putσ=eµand get
σk−aσ2m−k−λ=0, (2.21)
soP1is the above polynomial. Takinga=1/λ, we obtain
L+
1u
j=(Mu)j−1λuj, (2.22)
where
(Mu)j=uj+k−λuj+2k−m+uj+k−2m (2.23)
is invertible for largeλ≥0 (see, e.g., [1, Lemma 4.9]). SinceL+1 is close toM (forλ≥0 large enough), one deduces thatL+1 is also invertible. Hence the index ofL+1 is zero.
Since the continuous deformationPτ 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, thereforeL+is invertible. The theorem is proved. Remark 2.5. An analogous result can be stated forL−.
As a consequence, we may study the Fredholm property ofLwith the aid of the poly-nomialsP+andP−.
Corollary2.6. If the limiting operatorsL+andL− for an operatorL are such that the corresponding polynomialsP+(σ)andP−(σ)do not have roots with|σ| =1and have the same number of roots inside the unit circle, thenL is a Fredholm operator with the zero index.
Proof. We construct a homotopy ofLin such a way thatL+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.
3. Exponential decay
We consider now the problem
L+u
j=0, j≥1, (3.1)
assuming that the corresponding polynomialP+(σ) does not have roots with|σ| =1 and haskroots with|σ|<1. One associates to (3.1) the boundary conditions
u1=a1,...,uk=ak, (3.2)
witha1,...,ak∈Rgiven.
Since there arek roots inside the unit circle, then there arek linearly independent solutions of the equationL+u=0 decaying asj→ ∞. Denote them byu1,...,uk. Consider their values forj=1,..., 2m:
u11,u12,...,u1 2m,
.. . uk
1,uk2,...,uk2m.
(3.3)
Each of these solutions is completely determined by the above values. Therefore the corre-spondingkvectors are linearly independent. Indeed, otherwise the solutions would have been linearly dependent. Therefore there existklinearly independent columns. Without loss of generality we can assume that these are the firstkcolumns. Hence the correspond-ingk×kmatrix is invertible.
Any bounded solution of (3.1) can be represented in the form
u=c1u1+···+ckuk. (3.4)
Substituting it in (3.2), we uniquely determine the coefficientsc1,...,ck. Therefore we
have proved the following result.
Proposition3.1. If the corresponding polynomialP+(σ)forL+ does not admit roots on the unit circle|σ| =1and haskroots with|σ|<1, then for each(a1,...,ak)∈Rk,
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.
Theorem3.2. Suppose that the polynomialsP+(σ)andP−(σ)corresponding toL+andL−, respectively, do not have roots with|σ| =1and have the same number of roots with|σ|<1. Then the bounded solutions of the equationLu=0are exponentially decreasing at±∞. Proof. Let ˜u= {u˜j}∞j=−∞ be a bounded solution of the equation Lu=0. Consider the problem
ForN sufficiently large this problem is uniquely solvable for anya1,...,ak since prob-lem (3.1)-(3.2) is uniquely solvable, and the operatorLis close to the operatorL+. If we putai=u˜N+i,i=1,...,k, then the solution of problem (3.5) coincides with ˜ufor j≥N.
Therefore it is sufficient to prove that the solution of problem (3.5) is exponentially de-creasing for anyai,i=1,...,k.
Consider the operatorSof multiplication by exp(µ1 +j2), that is,
(Su)j=eµ √
1+j2
uj, j=0,±1,.... (3.6)
Hereµ >0. LetLµ=SLS−1. Then
Lµuj=a−jmuj−meµ( √
1+j2−√1+(j−m)2)
+···+a0juj+···+a
j muj+meµ(
√
1+j2−√1+(j+m)2)
. (3.7)
Forµsufficiently small, the operatorLµis close to the operatorL. Therefore the problem
Lµvj=0, vN+1=b1,...,vN+k=bk (3.8)
is uniquely solvable for anyb1,...,bk.
If we putbi=exp(µ1 + (N+i)2)ai,i=1,...,k, then the solutionuof problem (3.5) can be expressed through the solutionvof problem (3.8):u=S−1v. Sincevis bounded, thenuis exponentially decreasing.
Thus we have proved that ˜uis exponentially decreasing as j→ ∞. Similarly it can be
proved forj→ −∞. The theorem is proved.
4. Solvability conditions
In this section, we establish solvability conditions for the equation
Lu=f . (4.1)
HereLis the operator in (1.1) and f = {fj}∞j=−∞is fixed inE.
For the operatorL, denoteα(L)=dim(kerL) andβ(L)=codim(ImL). If (u,v) is the inner product of two sequencesu= {uj}∞j=−∞,v= {vj}∞j=−∞in the sensel2, that is,
(u,v)= ∞
j=−∞
ujvj, (4.2)
then we may define the formally adjointL∗of the operatorLby the equality
(Lu,v)=u,L∗v. (4.3)
LetL+,L−andL+
∗,L−∗be the limiting operators associated withLandL∗, 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+
Corollary 2.6implies thatLandL∗are Fredholm operators with the zero index. Lemma4.1. Under hypothesis (H), it holds thatβ(L)≥α(L∗).
Proof. By the definition of the Fredholm operator it follows that (4.1) is solvable for a given f = {fj}∞j=−∞∈Eif and only if there exist linearly independent functionalsϕk∈ E∗,k=1,...,β(L) such that
ϕk(f)=0, k=1,...,β(L). (4.4)
On the other hand consider the functionalsψlgiven by
ψl(f)=
∞
j=−∞
fjvlj, l=1,...,αL∗, (4.5)
wherevl= {vl
j}∞j=−∞,l=1,...,α(L∗) are linearly independent solutions of the homoge-neous equationL∗v=0. We know fromTheorem 3.2that the valuesvljare exponentially
decreasing with respect toj. Therefore the functionalsψlare well defined.
Obviously,ψlis linear for eachl. If f(n)→f inE(in the norm supremum), then we may pass to the limit in (4.5) under the sum to find thatψl(f(n))→ψl(f), asn→ ∞, for
alll=1,...,α(L∗), that is,ψlare continuous. Thereforeψl∈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-tionalsψlthere exists at least one functional (sayψ1) which is linearly independent with
respect to allϕk,k=1,...,β(L). This means that there existsf ∈Esuch that (4.4) holds, but
ψ1(f)=
∞
j=−∞
fjv1j=0. (4.6)
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 sincev1is a solution of the equationL∗v=0, we deduce that (Lu,v1)=(u,L∗v1)=0. The contradiction we
arrive at, proves the lemma.
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 operatorL, we get
κ(L) +κL∗≤0. (4.7)
Since in our caseκ(L)=κ(L∗)=0, it follows that
Theorem4.3. Equation (4.1) is solvable if and only if
wherebj,cjare real sequences. It is a discretization of the second-order differential
equa-tion on the real axis. Here we find sufficient conditions on the coefficientsbj,cjin order
to establish the Fredholm property and the solvability conditions. Suppose that there exist the limits
Denote byL±andL±∗the operators with constant coefficients
They are the limiting operators associated withLandL∗, respectively. One defines the polynomialsP+,P−corresponding toL+,L−and the polynomialsP+
toL+
∗,L−∗. To this end, we search for the solution of the equationL+u=0 in the form uj=eµj. Denotingσ=eµ, one obtains the polynomialP+corresponding toL+,
P+(σ)=1 +c+σ2+b+−c+−2σ+ 1. (5.5)
Analogously, we have
P−(σ)=1 +c−σ2+b−−c−−2σ+ 1, P+
∗(σ)=(1−c+)σ2+
b++c+−2σ+ 1, P−
∗(σ)=
1−c−σ2+b−+c−−2σ+ 1.
(5.6)
We first show that, under some certain conditions on the coefficients, the polynomial P(σ)=(1 +c)σ2+ (b−c−2)σ+ 1 (5.7)
does not have solutionsσwith|σ| =1.
Lemma5.1. Ifb <0,c≥ −2, thenP(σ)does not admit roots on the unit circle and admits one and only one root inside the unit circle.
Proof. Ifc= −1, we getP(σ)=(b−1)σ+ 1, so in view of hypothesisb <0, one finds that the rootσ=1/(1−b) belongs to the unit circle, that is, to the interval (−1, 1).
Assume nowc= −1. Then we may divide the equationP(σ)=0 by 1 +cto obtain
σ2+pσ+q=0 withp=b−c−2 1 +c ,q=
1
1 +c. (5.8)
Since∆=p2−4q=(b−c)2−4b/(1 +c)2>0, (5.8) has two different real roots
σ1=
−p+p2−4q
2 , σ2=
−p−p2−4q
2 . (5.9)
Case 1 (c >−1). Fromb <0,c >−1, we deriveq >0 and p+q+ 1<0. Consequently p <−1. A simple computation leads to the conclusion thatσ1>1 andσ2∈(−1, 1). Case 2 (−2≤c <−1). In this case, q≤ −1 and p+q+ 1>0. This implies that p >0. Henceσ1∈(−1, 1) andσ2<−1.
The lemma is proved.
ApplyingTheorem 2.3,Theorem 2.4,Corollary 2.6, andTheorem 3.2, we deduce with the aid ofLemma 5.1for the polynomialsP+andP−, some properties of the operatorL given in (5.1).
Proposition 5.2. If b+<0 and c+≥ −2, then the limiting operator L+ does not have nonzero bounded solutions. In addition, it is invertible.
An analogous result can be deduced forL−.
Proposition5.4. Assume thatb+,b−<0andc+,c−≥ −2. Then every bounded solution of the equationLu=0is exponentially decreasing at±∞.
Similar results can be obtained for the adjoint operatorL∗ofL, defined in (5.3), under the hypotheses thatb+,b−<0 andc+,c−≤2.
Now we are ready to find the solvability conditions for the equationLu= f. Denote α(L∗)=dim(kerL∗).
Proposition5.5. Letb+,b−<0andc+,c−∈[−2, 2]. Then the equationLu=f is solvable for a given f ∈E, if and only if
∞
j=−∞
fjvlj=0, l=1,...,αL∗, (5.10)
where vl= {vl
j}∞j=−∞, l=1,...,α(L∗) are linearly independent solutions of the equation L∗v=0.
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N. C. Apreutesei: Department of Mathematics, “Gh. Asachi” Technical University of Iasi, 700506 Iasi, Romania
E-mail address:[email protected]
V. A. Volpert: Laboratoire de Math´ematiques Appliqu´ees, Universit´e Claude Bernard Lyon 1, 69622 Villeurbanne, France
