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. Next, condition (1.4) and the boundedness of the norms ||un|| and u lead to the convergence | N − + N n − |≤| N − + | · N n − → γj L L [γ (u u)] j γ L L 0 γ (u u) 0, as N →∞,where|·|0 is the norm of the operator. For a given N, one estimates iξj the last term in the right-hand side of (1.9). Since uj = e , j ∈ Z is a solution of the equation L+u =0, then + n − + n + n − iεnj + (L (u u))j =(L u )j =(L u )j e L u j = ei(ξ+εn)j[a+ e−iξm e−iεnm − 1 + ···+ a+ e−iξ(e−iεn − 1) −m −1 + iξ iεn + iξm iεnm + a1 e e − 1 + ···+ ame (e − 1)], so that
+ n i(ξ+εn)j + −iξm ib−m + −iξ ib−1 (L (u − u))j = iεne [a−m (−m) e e + ···+(−1)a−1e e
+ iξ ib1 + iξm ibm + a1 e .e + ···+ amme e ],j∈ Z, where bj, j =0, ±1,...,±m are some numbers. Thus, the last term in (2.9) goes to zero as n →∞and, therefore, f n → 0. This completes the proof.
We are now ready to establish the invertibility of L+. Theorem 1.4. If the operator L+ is such that the corresponding polynomial P +(σ) does not have roots with |σ| =1, then it is invertible.
Proof. Lemma 1.2 applied for P +(σ) implies the existence of a continuous defor- + mation Pτ (σ), 0 ≤ τ ≤ 1, from the polynomial P0 = P to k 2m−k P1 (σ)= σ − a σ − λ such that Pτ (σ) does not admit solutions with |σ| =1.Hereλ>1, a<1are + given. The operator which corresponds to P1 is L1 defined by the equality + − − L1 u j = uj+k auj λuj+2k−2m + aλuj+k−2m.
+ µj Indeed, looking for the solution of the equation L1 u =0intheformuj = e , we obtain eµk − a − λeµ(2k−2m) + aλeµ(k−2m) =0. We put σ = eµ and get σk − a σ2m−k − λ =0, 532 Supplement. Discrete operators so P1 is the above polynomial. Taking a =1/λ,wehave + − L1 u j =(Mu)j (1/λ) uj, where (Mu)j = uj+k − λuj+2k−m + uj+k−2m. The operator M is invertible for large λ ≥ 0. Indeed, 1 M = −λ(T − S), λ where (Tu)j = uj+2k−m,(Su)j = uj+k + uj+k−2m.SinceT is invertible, then T − S/λ is also invertible for λ large enough. + Hence L1 is also invertible for sufficiently large λ and its index is zero. Since the polynomial Pτ does not have solutions σ with |σ| = 1, for any 0 ≤ τ ≤ 1, then + the corresponding continuous deformation of the operator Lτ does not admit nonzero bounded solutions (see Lemma 1.1). By Theorem 1.3, one obtains that + Lτ is normally solvable with a finite-dimensional kernel. From the general theory of Fredholm operators, we know that the index does not change in the process of + + + such deformation. Since the index of L1 is κ L1 = 0, we deduce that κ (L )=0. This, together with the fact that the kernel of the operator L+ is empty, implies that it is invertible. The theorem is proved. A similar result can be stated for L−. As a consequence, we can study the Fredholm property of L with the help of the polynomials P + and P −. Corollary 1.5. 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+ and L− are reduced to the operator in Theorem 1.4. This is a homotopy in the class of the normally solvable operators with finite-dimensional kernels. Since the operators L+ and L− coincide, we finally reduce L to an operator with constant coefficients. According to Theorem 1.4, it is invertible. Therefore L is a Fredholm operator and has the index 0, as claimed. We note that if the polynomials P ±(σ) do not have roots with |σ| =1,then solutions of the equation Lu = 0 decay exponentially at infinity. This can be proved employing the properties of the holomorphic operator-functions similar to the proof in the case of elliptic operators (cf. Section 4, Chapter 5). 1. One-parameter equations 533
1.2 Solvability conditions In this section, we establish solvability conditions for the equation
Lu = f. (1.10)
Let α (L)=dimKerL and β (L)=codimImL,(u, v) the inner product in l2, ∞ (u, v)= ujvj. j=−∞
We define the formally adjoint L∗ of the operator L by the equality
(Lu, v)=(u, L∗v).
± ∗ ∗ Let L and L± be the limiting operators associated with L and L , respectively. We suppose that the following condition is satisfied. (H) The polynomials P +,P− corresponding to L+ and L− do not have roots with |σ| = 1 and have the same number of roots with |σ| < 1. Similarly for ∗ ∗ ∗ ∗ the polynomials P+ and P− corresponding to L+ and L−. Corollary 1.5 implies that L and L∗ are Fredholm operators with the index zero.
Lemma 1.6. If Condition H is satisfied, then β (L) ≥ α (L∗). Proof. By the definition of Fredholm operators, equation (1.10) is solvable if and only if ϕk (f)=0,k=1,...,β(L) (1.11) ∗ for some linearly independent functionals ϕk ∈ E , k =1,...,β(L). On the other hand, consider the functionals ψl given by ∞ l ∗ ψl (f)= fjvj,l=1,...,α(L ) , (1.12) j=−∞ where vl, l =1,...,α(L∗) are linearly independent solutions of the homogeneous ∗ l equation L v =0.Sincevj are exponentially decreasing with respect to j,then the functionals ψl are well defined. In order to prove that β (L) ≥ α (L∗), suppose that it is not true. Then among the functionals ψl there exists at least one functional (say ψ1) which is linearly independent with respect to all ϕk, k =1,...,β(L). This means that (∃) f ∈ E such that (1.11) holds, but ∞ 1 ψ1 (f)= fjvj =0 . (1.13) j=−∞ 534 Supplement. Discrete operators
1 From(1.11)it follows that equation (1.10) is solvable. We multiply it by v and find Lu, v1 = f,v1 . By (1.13) observe that the right-hand side is different 1 ∗ from zero. But since v is a solution of the equation L v = 0, we deduce that Lu, v1 = u, L∗v1 = 0. This contradiction proves the lemma.
Since L is formally adjoint to L∗, then similarly to the lemma we obtain β (L∗) ≥ α (L). Therefore, if κ (L)=α (L) − β (L) is the index of the operator L, then κ (L)+κ (L∗) ≤ 0. (1.14) Since in our case κ (L)=κ (L∗)=0, then it follows that
β (L)=α (L∗) ,β(L∗)=α (L) . (1.15)
Theorem 1.7. Equation (1.10) is solvable if and only if ∞ l ∗ fjvj =0,l=1,...,α(L ) , (1.16) j=−∞
l l ∞ ∗ where v = {vj}j=−∞, l =1,...,α(L ) are linearly independent solutions of the equation L∗v =0.
Proof. Equation (1.10) is solvable if and only if (1.11) holds for some functionals ∗ ∗ ϕk ∈ E , k =1,...,β(L). Consider the subspaces Φ and Ψ of E generated ∗ by the functionals ϕk, k =1,...,β(L)andbyψl from (1.12), l =1,...,α(L ), respectively. By (1.15) we deduce that their dimensions coincide. We show that actually Φ = Ψ. We first verify that Ψ ⊆ Φ. Indeed, if it is not the case, then there exists ψ ∈ Ψ, ψ/∈ Φ. Then ∃f ∈ E such that (1.11) holds, but at least one ψl (f) = 0, so we get the same contradiction as in the proof of Lemma 1.6. Therefore, Ψ ⊆ Φ and since they have the same dimensions, we get that Ψ = Φ. The theorem is proved.
1.3 Spectrum of difference and differential operators Consider the difference operator
(Lu)j = aj(uj+1 − 2uj + uj−1)+bj (uj+1 − uj)+cjuj, where aj, bj, cj arerealnumbers.Itcanbeconsidered as a discretization of the second-order differential equation on the real axis:
Mu = a(x)u + b(x)u + c(x)u.
We will discuss how the essential spectrum of the difference and of the differential operators are related to each other. Let us assume that the sequences aj,bj,cj 2. First-order systems 535 converge to a, b, c, respectively, as i →∞, and consider the infinite system of equations a(uj+1 − 2uj + uj−1)+b(uj+1 − uj)+cuj = λuj. iξj We substitute uj = e and obtain
λ(ξ)=(2a + b)cosξ + ib sin ξ − 2a − b + c.
Here ξ is a real parameter, λ(ξ) is the essential spectrum. If it crosses the origin, the operator L does not satisfy the Fredholm property. Let a = η2α, b = ηβ,
λ(ξ,η)=(2ηα + β)η cos ξ + iηβ sin ξ − 2η2α − ηβ + c.
Here η is a large parameter. This scaling corresponds to a finite difference ap- proximation of the first and second derivatives. If we consider λ as a function of ξ for a fixed η, then, as before, we obtain the essential spectrum of the operator L. We will now consider λ as a function of η and will show that it converges to the essential spectrum of the operator M as η →∞.Putλ = µ + iν.Equatingreal and imaginary parts in the last equality, we can express µ through ν and λ and exclude ξ: / (c − µ)2 2(c − µ) ν = bη − + . (2aη2 + bη)2 2aη2 + bη In the limit of large η,weobtain , c − µ ν = b . a
Therefore, µ = c − aν2/b2,
ν2 λ = µ + iν = −a + iν + c. b2 We finally put η = ν/b and obtain
λ = −aη2 + ibη + c.
If the coefficients of the operator M converge to a, b,andc at infinity, then the last formula gives the essential spectrum of the operator M.
2 First-order systems
Consider the linear algebraic system of equations
U(j) − U(j − 1) = A(j)U(j), (2.1) 536 Supplement. Discrete operators where A(j)aren × n matrices, U(j)aren-vectors, j ∈ Z.Wecallsuchasystem, for which only two consecutive values of the parameter j are present, first-order systems. Denote by Φ(j) the fundamental matrix of this system, that is the matrix whose columns are linearly independent solutions of (2.1). Suppose that there are n linearly independent solutions. Let Ψ(j)=(Φ−1(j))T , where the superscript T denotes the transposed matrix. Therefore, Ψ(j)isthe fundamental matrix of the system V (j) − V (j − 1) = −A(j)T V (j − 1), (2.2) which is adjoint to system (2.1). We note that V in the right-hand side of (2.2) is taken at j − 1, while U in (2.1) is taken at j. Consider next the nonhomogeneous equation W (j) − W (j − 1) = A(j)W (j)+f(j − 1). (2.3) Its solution can be given by the formula j−1 W (j)=Φ(j) Ψ(i)T f(i). (2.4) i=0 We will also use another form of the solution: ∞ W (j)=−Φ(j) Ψ(i)T f(i). (2.5) i=j
2.1 Solvability conditions
Assume that the elements ϕhk (j)(h, k =1,...,n) of the fundamental matrix Φ (j) (j ∈ Z) of the homogeneous system behave exponentially at infinity: ϕhk (j) ∼ ± λ±·j ± ± ± ahke k as j →±∞,whereλk =0 ,λk are different for different k and ahk are such that the limit matrix of Φ (j) is invertible for all j. Therefore, ± λ±j ± λ±j ± λ±j a11e 1 a12e 2 ... a1ne n ± λ±j ± λ±j ± λ±j a21e 1 a22e 2 ... a2ne n Φ(j)=(ϕhk(j)) ∼ (2.6) h,k=1,n ...... ± λ±j ± λ±j ± λ±j an1e 1 an2e 2 ... anne n as j →±∞.Then ± −λ±j ± −λ±j ± −λ±j b11e 1 b12e 2 ... b1ne n ± −λ±j ± −λ±j ± −λ±j −1 T b e 1 b e 2 ... b e n Ψ(j)=(Φ (j)) ∼ 21 22 2n , (2.7) ...... ± −λ±j ± −λ±j ± −λ±j bn1e 1 bn2e 2 ... bnne n 2. First-order systems 537 as j →±∞.Iffm (j)(m =1,...,n)aretheelementsoff (j), it follows that T Ψ(i) f (i) behaves like ± −λ±i s1 (i) e 1 ± −λ±i s2 (i) e 2 ± ± ± , where s (i)=b f1 (i)+···+ b fn (i) ,p=1,...,n. ... p 1p np ± −λ±i sn (i) e n
Let k, h, l, q be integers, 0 ≤ k, h, l, q ≤ n, such that k + h + l + q = n and:
+ − (i) λp < 0, λp > 0, (∀) p =1,...,k; + − (ii) λp < 0, λp < 0, (∀) p = k +1,...,k+ h; + − (iii) λp > 0, λp > 0, (∀) p = k + h +1,...,k+ h + l; + − (iv) λp > 0, λp < 0, (∀) p = k + h + l +1,...,k+ h + l + q(= n).
If this is not the case, we can rearrange the order of the numbers λp in such a way that (i)–(iv) hold. If one or several of the numbers k, h, l, q is zero, then we omit the corresponding line.
Denote by A1 (j) ,...,An (j)thecolumnsofΦ(j)andbyB1 (j) ,...,Bn (j) the columns of Ψ (j). Then A1 (j) ,...,Ak (j) are bounded at +∞ and −∞, Ak+1 (j) ,...,Ak+h (j) are bounded at +∞ and grow at −∞, Ak+h+1 (j), ..., Ak+h+l (j)growat+∞ and decay at −∞, while Ak+h+l+1 (j) ,...,An (j)grow at both +∞ and −∞.SincekerL is the subspace generated by the bounded (at both +∞ and −∞) columns of Φ (j), we find that dim ker L = k.
As a consequence of the behavior of A1 (j) ,...,An (j), we conclude that B1 (j) ,...,Bk (j) are exponentially growing at +∞ and −∞, Bk+1 (j), ..., Bk+h (j) are unbounded at +∞ and bounded at −∞ (decaying to 0), Bk+h+1 (j), ..., Bk+h+l (j) are bounded at +∞ and unbounded at −∞,andBk+h+l+1 (j), ..., Bn (j) are bounded at both +∞ and −∞. We put