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Supplement. Discrete Operators Supplement. Discrete Operators Infinite systems of algebraic equations can arise in some applications or as a re- sult of discretization of differential equations in unbounded domains. Though the results obtained for elliptic problems in unbounded domains cannot be directly applied to infinite-dimensional discrete operators, the methods developed in this book can be adapted for them. In particular, we will define limiting problems in order to formulate conditions of normal solvability. We will discuss solvability con- ditions and some other properties of discrete operators. One of the results concerns the generalization of the Perron-Frobenius theorem about the principal eigenvalue of the matrices with non-negative off-diagonal elements to infinite matrices. We will also see that conditions of normal solvability are related to stability of fi- nite difference approximations of differential equations. The representation below follows the works [28]–[30]. Some related questions are discussed in [443]. 1 One-parameter equations 1.1 Limiting operators and normal solvability Let E be the Banach space of all bounded real sequences ∞ E = u = {uj}j=−∞ ,uj ∈ R, sup |uj| < ∞ (1.1) j∈Z with the norm u =sup|uj| , j∈Z and L : E → E be the linear difference operator j j j (Lu)j = a−muj−m + ···+ a0uj + ···+ amuj+m,j∈ Z, (1.2) j j j where m ≥ 0 is an integer and a−m,...,a0,...,am ∈ R are given coefficients. In some cases, it will be also convenient for us to consider complex coefficients. This operator acts on sequences of numbers depending on one integer parameter j.In this sense, we call such operators and the corresponding equations one-parameter V. Volpert, Elliptic Partial Differential Equations: Volume 1: Fredholm Theory of Elliptic 527 Problems in Unbounded Domains, Monographs in Mathematics 101, DOI 10.1007/978-3-0346-0537-3, © Springer Basel AG 2011 528 Supplement. Discrete operators operators and equations. They can arise as a result of discretization of differential equations on the real axis. Denote by L+ : E → E the limiting operator + + ··· + ··· + ∈ Z L u j = a−muj−m + + a0 uj + + amuj+m,j , (1.3) where j a+ = lim a ,l∈ Z, −m ≤ l ≤ m. (1.4) l j→∞ l We are going to define the associated polynomial for the operator L+.To do this, we are looking for the solution of the equation L+u = 0 under the form uj =exp(µj), j ∈ Z and obtain + −µm + −µ + + µ + µm a−me + ···+ a−1e + a0 + a1 e + ···+ ame =0. Let σ = eµ and + + 2m + m + P (σ)=amσ + ···+ a0 σ + ···+ a−m. (1.5) We present without proof the following auxiliary result (see [28]). Lemma 1.1. The equation L+u =0has nonzero bounded solutions if and only if the corresponding algebraic polynomial P + has a root σ with |σ| =1. We will find conditions in terms of P + for the limiting operator L+ to be invertible. We begin with an auxiliary result concerning continuous deformations + + of the polynomial P . Without loss of generality, we can assume that am =1. Consider the polynomial with complex coefficients n n−1 P (σ)=σ + a1σ + ···+ an−1σ + an. (1.6) Lemma 1.2. Suppose that a polynomial P (σ) does not have roots with |σ| =1and it has k roots with |σ| < 1, 0 ≤ k ≤ n. Then there exists a continuous deformation Pτ (σ)0≤ τ ≤ 1, such that k n−k P0(σ)=P (σ),P1(σ)=(σ − a)(σ − λ), and the polynomial Pτ (σ) does not have roots with |σ| =1for any 0 ≤ τ ≤ 1. Here λ>1 and a<1 are real numbers. Proof. Let us represent the polynomial P (σ)intheform P (σ)=(σ − σ1) ...(σ − σn), where the roots σ1,...,σk are inside the unit circle, and the other roots are outside it. Consider the polynomial Pτ (σ)=(σ − σ1(τ)) ...(σ − σn(τ)) 1. One-parameter equations 529 that depends on the parameter τ 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 k polynomial, for τ = 1 it has the roots σ1,...,σk with (σi) = a, i =1,...,k(inside n−k the unit circle) and n−k roots σk+1,...,σn such that (σi) = 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 polynomials P + and P − of L+ and L−, we can study normal solvability of operator L. Theorem 1.3. 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. Suppose that the polynomials P +,P− do not have roots σ with |σ| =1. Let {f n} be a sequence in the image Im L of the operator L such that f n → f and let {un} be such that Lun = f n. Suppose in the beginning that {un} is bounded in E. We construct a conver- n n gent subsequence. Since ||u || =supj∈Z |uj |≤c, then for every positive integer n n N, there exists a subsequence {u k } of {u } and u = {uj}∈E such that nk sup uj − uj → 0, (1.7) −N≤j≤N that is unk → u as k →∞uniformly on each bounded interval of j.Usinga diagonalization process, we extend uj to all j ∈ Z. It is clear that supj∈Z |uj|≤c, that means u ∈ E. Passing to the limit as k →∞in the equation Lunk = f nk ,we get Lu = f,sothatf ∈ Im L. We show that the convergence in (1.7) is uniform with respect to all j ∈ Z. nk k →∞ − j ≥ If, by contradiction, there exists j such that ujk u k ε>0, then the k nk k nk − j+j | − j |≥ sequence yj = uj+jk u k verifies the inequality y0 = ujk u k ε and the equation j+j j+j k k ··· k k ··· j+jk k nk − ∈ Z a−m yj−m + + a0 yj + + am yj+m = fj+jk fj+jk ,j . (1.8) ( ) ( ) Since the sequence yk is bounded in E, there exists a subsequence ykl which converges to some y0 ∈ E uniformly with respect to j on bounded intervals. We pass to the limit as kl →∞in (1.8) and obtain + 0 + 0 + 0 a−myj−m + ···+ a0 yj + ···+ amyj+m =0,j∈ Z. Thus, the limiting equation L+u = 0 has a nonzero bounded solution and Lemma nk 1.1 leads to a contradiction. Therefore the convergence uj − uj → 0 is uniform with respect to all j ∈ Z.SinceLu = f,thenImL is closed. 530 Supplement. Discrete operators We note that in order to prove that ker L has a finite dimension, it suffices to show that every sequence un from B ∩ ker L (where B is the unit ball) has a convergent subsequence. We prove this using the same reasoning with f n =0. We analyze now the case where {un} is unbounded in E.Thenwewrite un = xn + yn,with{xn}∈Ker L and {yn} in the supplement of Ker L.Then Lyn = f n.If{yn} is bounded in E, then it follows as above that Im L is closed. If not, then we repeat the above reasoning for zn = yn/||yn|| and gn = f n/||yn||. n 0 Passing to the limit on a subsequence nk (such that z k → z ) in the equality Lznk = gnk and using the convergence gnk → 0, one obtains that z0 belongs to the kernel of the operator L and to its supplement. This contradiction finishes the proof of the closeness of the image. Assume now that Im L is closed and dim Ker L is finite. Suppose, by con- tradiction, that one of the polynomials, for certainty P +, has a root on the unit ∞ + circle. Then there exists a solution u = {uj}j=−∞ of the equation L u =0,where iξj ∈ R ∈ Z uj = e , ξ , j . ( ) ( ) { }∞ N N ∞ N N ∞ Let α = αj j=−∞, β = βj j=−∞, γ = γj j=−∞ be a partition of N N unity (αj + βj + γj =1)givenby # # # 1,j≤ 0 N 1, 1 ≤ j ≤ N N 1,j≥ N +1 αj = ,βj = ,γj = . 0,j≥ 1 0,j≤ 0,j≥ N +1 0,j≤ N Consider a sequence εn → 0asn →∞.Forafixedεn, put n i(ξ+εn)j n n n n uj = e ,vj =(1− αj) uj − uj ,fj = Lvj ,j∈ Z. n It is clear that uj → uj as n →∞uniformly on every bounded interval of integers j. It is sufficient to prove that f n → 0. Indeed, in this case, since the image of the operator is closed and the kernel is finite dimensional, then vn → 0. But this is in contradiction with vn =supei(ξ+εn)j − eiξj ≥ m>0, j>0 for some m. n n In order to show that f → 0asn →∞,werepresentfj in the form - . n N N N N n fj = αj + βj + γj L β + γ (u − u) - . - j . N N n N N N n = αj L β + γ (u − u) + βj L β + γ (u − u) - . j j N N n N + N n + γj L β (u − u) + γj L − L [γ (u − u)] j j N + N n − + γj L [γ (u u)] j . (1.9) 1. One-parameter equations 531 A simple computation shows that the first three terms in the right-hand side of the last equality tend to zero as n →∞uniformly with respect to all integer j.
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