# Error Estimate of Eigenvalues of Perturbed Higher-Order Discrete Vector Boundary Value Problems

Authors: Haiyan Lv; Yuming Shi; Guojing Ren
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

This paper is concerned with the eigenvalues of perturbed higher-order discrete vector boundary value problems. A suitable admissible function space is first introduced, a new variational formula of eigenvalues is then established under certain nonsingularity conditions, and error estimates of eigenvalues of problems with small perturbation are finally given by using the variational formula. As a direct consequence, continuous dependence of eigenvalues on boundary value problems is obtained under the nonsingularity conditions. In addition, two special perturbed cases are discussed.


---

## Body

## 1. Introduction

Consider the following2n-order vector difference equation:
(1)∑i=0n‍Δi[ri(t)Δiy(t-i)]=λw(t)y(t),t∈[n,N+n],
with the boundary condition
(2)R(-u(0,y)u(N+1,y))+S(v(0,y)v(N+1,y))=0,
whereΔis the forward difference operator; that is, Δy(t)=y(t+1)-y(t); y(t) is a d-dimensional column vector-valued function on interval [0,N+2n] of integer, n≥1 and N≥2n-1; ri(t) (t∈[n,N+n+i],0≤i≤n) and w(t) (t∈[n,N+n]) are d×d Hermitian matrices, w(t)>0 (t∈[n,N+n]);
(3)rn(t)isnonsingularon⁡[n,2n-1]∪[N+n+1,N+2n];R and S are 2nd×2nd matrices satisfying the following self-adjoint condition [1, Lemma 2.1]:
(4)rank⁡(R,S)=2nd,RS*=SR*;uT(t,y)=(u1T(t,y),…,unT(t,y)), vT(t,y)=(v1T(t,y),…,vnT(t,y)) are nd-dimensional vectors;
(5)ui(t,y)=Δi-1y(t+n-i),vi(t,y)=(-1)i-1∑k=in‍Δk-i[rk(t+n)Δky(t+n-k)];yT denotes the transpose of y; R* denotes the complex conjugate transpose of R; and λ∈ℂ is the spectral parameter.Higher-order discrete linear problems also have been investigated by some scholars besides second-order discrete Sturm-Liouville problems and discrete linear Hamiltonian systems (cf. [2–14] and their references). Zhou [15] and Grzegorczyk and Werbowski [7] studied a higher-order linear difference equation in which the leading coefficient is equal to 1 and established some criteria for the oscillation of solutions. Shi and Chen [1] investigated higher-order discrete linear boundary value problems (1)-(2) and obtained some spectral results, including Rayleigh’s principle, the minimax theorem, the dual orthogonality, and the number of eigenvalues. These results establish the theoretical foundation for our further research. Ren and Shi [16] discussed the defect index of singular symmetric linear difference equations of order 2n with real coefficients and one singular endpoint and showed that the defect index d satisfies the inequalities n≤d≤2n and that all values of d in this range are realized. However, because of the characteristics of higher-order difference equations, compared with the research of second-order difference equations and discrete Hamiltonian systems, it is more difficult to study higher-order difference equations. Thus, there are few references in higher-order difference equations. For more information about higher-order discrete linear problems, the reader is referred to [6, 12, 14].Recently, we have studied second-order discrete Sturm-Liouville problems and obtained error estimates of eigenvalues of perturbed problem under some hypotheses in [17]. Motivated by the ideas and methods used in [17], we extend the results to 2n-order discrete vector boundary value problems (1)-(2) by means of the results obtained in [1]. Although the method is similar, the problems we investigate in this paper are more complex, since they are not only of higher order but also of higher dimension.Ifrn(t) is nonsingular on [n,N+n], then the 2n-order vector difference equation (1) can be converted into the discrete linear Hamiltonian system studied in [18]:
(6)Δx(t)=A(t)x(t+1)+B(t)u(t),Δu(t)=[C(t)-λ𝒲(t)]x(t+1)-A*(t)u(t),1111111111111111111111111111111t∈[0,N],
where
(7)A(t)=(0Id(n-1)00),B(t)=diag⁡{0,…,0,(-1)n-1rn-1(t+n)},C(t)=diag⁡{-r0(t+n),r1(t+n),…,(-1)nrn-1(t+n)},𝒲(t)=diag⁡{w(t+n),0,…,0}.However, hypothesis (3) does not require the leading coefficient rn(t) to be always nonsingular in [n,N+n]. So, the coefficient B(t) and the weight functions 𝒲(t) of the corresponding discrete linear Hamiltonian system do not satisfy assumption (2.1) and the positive definiteness of the weight function in [18]. Hence the Hamiltonian system considered in [18] does not include the equation we discuss in this paper.In the present paper, we study error estimate of eigenvalues of (1)-(2) under small perturbation. By employing a variational property—the minimax theorem established in [1]—an error estimate of eigenvalues of all perturbed problems sufficiently close to problem (1)-(2) is given under certain nonsingularity conditions. The continuous dependence of eigenvalues on problems is consequently obtained from the error estimate under the nonsingularity conditions. The continuous dependence of eigenvalues on problems may not hold in general. It is under certain nonsingularity conditions that we get the related result. In addition, the minimax theorem [1, Theorem 3.5] was established in an admissible function space, which is dependent on boundary condition (2). Hence, it is difficult to apply to the case that some perturbation occurs in boundary condition (2). So we will first establish a minimax theorem in an admissible function space with a new weight function that includes the data of (1) and boundary condition (2) by [1, Theorem 3.5]. Then, employing the new minimax theorem, we study the error estimate of eigenvalues of perturbed problem. Another difficulty results from the complicated calculations since the problem is not only of higher order but also of higher dimension and needs to estimate the norms of inverses of some perturbed matrices.The setup of this paper is as follows. In the next section, we recall some useful existing results, introduce a new suitable admissible function space, and establish a new minimax theorem in it. In Section3, we give the main results that provide error estimates of eigenvalues of perturbed problems of (1)-(2) under certain nonsingularity conditions. Finally, We discuss two special perturbed problems in Section 4.


## 2. Preliminaries

In this section, we first introduce some notations and results for convenience in the following discussion, then give a suitable admissible function space, and establish a new variational property of eigenvalues for (1)-(2) in this space.Consider the following linear space:(8)L[0,N+2n]:={y={y(t)}t=0N+2n⊂ℂd}.Obviously,dim⁡L[0,N+2n]=(N+2n+1)d. Let ℒ denote the following difference operator:
(9)(ℒy)(t):=w-1(t)∑i=0n‍Δi[ri(t)Δiy(t-i)],t∈[n,N+n].For convenience, fory∈L[0,N+2n], we write y∈ℬ if y satisfies boundary condition (2). Denote
(10)L^[0,N+2n]:={y∈L[0,N+2n]:y∈ℬ}.Lemma 1 (see, [1, Lemma 2.2]).
Assume that (3) and (4) hold. Then y∈ℬ if and only if there exists a unique vector ξ∈ℂ2nd such that
(11)(-u(0,y)u(N+1,y))=-S*ξ,(v(0,y)v(N+1,y))=R*ξ.
Let(12)YT(t,k):=(yT(t+k-1),…,yT(t)),111111t∈[0,N+n+1],k∈[1,n],(13)Y(t):=Y(t,n).In particular,(14)YT(0)=(yT(n-1),…,yT(0)),YT(n)=(yT(2n-1),…,yT(n)),YT(N+1)=(yT(N+n),…,yT(N+1)),YT(N+n+1)=(yT(N+2n),…,yT(N+n+1)).Expressu and v in terms of Y:
(15)u(0,y)=LY(0),v(0,y)=AY(n)+BY(0),u(N+1,y)=LY(N+1),v(N+1,y)=A1Y(N+n+1)+B1Y(N+1),
where L=(lij), A=(aij), B=(bij), and A1 and B1 are nd×nd matrices; A1 and B1 are matrices about ri(t+N+1), which are the shifts of variable t of ri(t) in A and B to the right with N+1 units, respectively. More precisely, for 1≤i, j≤n,
(16)lij={0d,i<j,(-1)j-1Ci-1j-1Id,i≥j,(17)aij={(-1)i-1∑k=0j-i∑h=0k(-1)kCn-j+kh×Ci+n-j+kk-hri+n-j+k(2n-j+k-h),i≤j,0d,i>j,(18)bij={-∑k=0n-j∑h=0k(-1)kCj-i+kk-hCj+khrj+k(n+k-h),i≤j,(-1)i-j+1∑k=0n-i∑h=0k(-1)kCkhCi+kk-h+j×ri+k(n+k-h),i>j.Obviously,L is nonsingular and L-1=L. In upper triangular matrix A, aii=(-1)i-1rn(2n-i). So if (3) holds, then A and A1 are nonsingular. By Proposition  2.1 in [1], L*B, L*B1, BL, and B1L are Hermitian matrices.Denote(19)R=(R1,R2),S=(S1,S2),
where Rj and Sj (j=1,2) are 2nd×nd matrices. Substitute (15) into (2) to get
(20)Ωdiag⁡{L,-A1}(Y(0)Y(N+n+1))=(S1A,R2L+S2B1)(Y(n)Y(N+1)),
where
(21)Ω:=(R1-S1BL-1,S2)=(R1-S1BL,S2).Next, we always assume that(22)Ωisnonsingular.Let(23)〈x,y〉:=∑t=nN+n‍y*(t)w(t)x(t),x,y∈L[0,N+2n].When (22) holds, L^[0,N+2n] is an (N+1)d-dimensional Hilbert space with the inner product 〈·,·〉 by Theorem  2.3 in [1]. In this case, L^[0,N+2n] is the same as the admissible function space Lω2[0,N+2n] in [1].A series of spectral results including the variational properties of eigenvalues for problem (1)-(2) have been established by Shi and Chen in [1]. We state some of these results for future use.The following lemma is Theorem  3.1 in [1] in the special case that (22) holds.Lemma 2.
Assume that (3), (4), and (22) hold. Then problem (1)-(2) has exactly (N+1)d real eigenvalues (multiplicity included) arranged as
(24)λ1≤λ2≤⋯≤λ(N+1)d.The Rayleigh quotient for the difference operatorℒ on L^[0,N+2n] with 〈·,·〉 is defined by
(25)R(y):=〈ℒy,y〉〈y,y〉,
where y∈L^[0,N+2n] and y′={y(t)}t=nN+n≠0.The following lemma is the minimax theorem—Theorem  3.5 in [1] in the special case that (22) holds.Lemma 3.
Assume that (3), (4), and (22) hold. Then, for each k, 1≤k≤(N+1)d,
(26)λk=max⁡{f(z(1),…,z(k-1)):z(j)∈L^111111111×[0,N+2n],1≤j≤k-1z(1)}
with f(z(1),…,z(k-1))=min⁡{R(y):y∈L^[0,N+2n],y⊥z(j),1≤j≤k-1,y′≠0}, where y⊥z(j) denotes 〈y,z(j)〉=0.Since the perturbation of (1) and (2) affects the space L^[0,N+2n], we need a new suitable admissible function space and a variational formula to apply (26).For anyy∈L^[0,N+2n], by Lemma 1 and (15) there exists a unique vector ξ∈ℂ2nd such that
(27)Y(0)=LS1*ξ,Y(n)=A-1(R1*-BLS1*)ξ,Y(N+1)=-LS2*ξ,Y(N+n+1)=A1-1(R2*+B1LS2*)ξ;
that is,
(28)(Y(0)Y(n)Y(N+1)Y(N+n+1))=(00L0A-10-A-1BL0000-L0A1-10A1-1B1L)(R,S)*ξ.From above we know that{Y(0),Y(n),Y(N+1),Y(N+n+1)}, and then {y(0),…,y(2n-1),y(N+1),…,y(N+2n)} can be uniquely determined by ξ. Hence, we introduce the following new admissible function space:
(29)X:={z={ξ,y(2n),…,y(N)}:ξ∈ℂ2nd,111111y(t)∈ℂd,2n≤t≤N}.Sincew(t)>0 (t∈[n,N+n]), it follows from (3) and (22) that
(30)W:=Ωdiag⁡{A-1*W1A-1,L*W2L}Ω*>0,
where
(31)W1=diag⁡{w(2n-1),…,w(n)},W2=diag⁡{w(N+n),…,w(N+1)}.Thus, we can define an inner product onX by
(32)〈z1,z2〉1:=η*Wξ+∑t=2nN‍y*(t)w(t)x(t),
where z1={ξ,x(2n),…,x(N)}, z2={η,y(2n),…,y(N)}∈X. Denote its induced norm by
(33)∥z1∥1:=(〈z1,z1〉1)1/2.Obviously,(X,〈·,·〉1) is also an (N+1)d-dimensional Hilbert space. Note that the elements of the space X are independent of (1) and boundary condition (2), which are partly put in the new weight function {W}∪{w(t)}t=2nN.In order to establish a connection betweenX and L^[0,N+2n], we define a linear map
(34)T1:X⟶L^[0,N+2n],
by T1(z)=y={y(t)}t=0N+2n∈L^[0,N+2n] with {Y(0),Y(n),Y(N+1),Y(N+n+1)} determined by (28) for z={ξ,y(2n),…,y(N)}∈X. Evidently, T1 is an invertible linear map. Moreover, for any z1={ξ,x(2n),…,x(N)}, z2={η,y(2n),…,y(N)}∈X, set T1(z1)=x and T1(z2)=y. Then, from (23), (27), and (30), we have
(35)〈T1(z1),T1(z2)〉=∑t=nN+n‍y*(t)w(t)x(t)〈T1(z1),T1(z2)〉=Y*(n)W1X(n)+Y*(N+1)〈T1(z1),T1(z2)〉=×W2X(N+1)+∑t=2nN‍y*(t)w(t)x(t)〈T1(z1),T1(z2)〉=(Y(n)Y(N+1))*(W100W2)(X(n)X(N+1))〈T1(z1),T1(z2)〉=+∑t=2nN‍y*(t)w(t)x(t)〈T1(z1),T1(z2)〉=η*Wξ+∑t=2nN‍y*(t)w(t)x(t)〈T1(z1),T1(z2)〉=〈z1,z2〉1;
that is, T1 is a product-preserving map.Next, we introduce the Rayleigh quotient corresponding toℒ on X with 〈·,·〉1 as follows:
(36)ℛ(z):=P(z)〈z,z〉1,z={ξ,y(2n),…,y(N)}∈X,z≠0,
where P(z)=〈ℒ(T1(z)),T1(z)〉 and T1(z)=y. By a direct calculation we have from (9) and (23) that
(37)P(z)=〈ℒ(T1(z)),T1(z)〉=〈ℒy,y〉P(z)=∑t=nN+n‍y*(t){∑i=0n‍Δi[ri(t)Δiy(t-i)]}P(z)=∑t=nN+n‍y*(t){∑i=0n‍Di(t)y(t+i)P(z)=111111111+∑i=1n‍Di(t-i)y(t-i)}P(z)=Y*(n)L1Y(n)+Y*(N+1)L2Y(N+1)P(z)=+Y*(n)L3Y(0)+Y*(N+1)L4Y(N+n+1)P(z)=+2Re{Y*(N-n+1)L5Y(N+1)}P(z)=+2Re{Y*(n)L6Y(2n)}P(z)=+∑t=2nN‍y*(t)D0(t)y(t)P(z)=+2Re{∑i=1n‍∑t=2nN-i‍y*(t)Di(t)y(t+i)},
where
(38)Di(t)=∑k=0n-i‍∑h=0k‍(-1)kCi+kk-hCi+khri+k(t+i+k-h),111111111111111n-i≤t≤N+n,0≤i≤n.For any1≤i, j≤n, and  1≤p≤6, Lp=(lijp) are nd×nd matrices, lijp are d×d matrices, and
(39)lij1={Dj-i(2n-j),i≤j,Di-j(2n-i),i>j,lij2={Dj-i(N+n-j+1),i≤j,Di-j(N+n-i+1),i>j,lij3={0,i<j,Dn-i+j(n-j),i≥j,lij4={Dn+i-j(N+n-i+1),i≤j,0,i>j,lij5={Dn+i-j(N-i+1),i≤j,0,i>j,lij6={Dn+i-j(2n-i),i≤j,0,i>j.Further, it follows from (27) that
(40)P(z)=ξ*(-S2L*L4A1-1(R2*+B1LS2*))ΩMΩ*+(R1-S1BL)A-1*L3LS1*P(z)=111-S2L*L4A1-1(R2*+B1LS2*))ξP(z)=-2Re{Y*(N-n+1)L5LS2*ξ}P(z)=+2Re{ξ*(R1-S1BL)A-1*L6Y(2n)}P(z)=+∑t=2nN‍y*(t)D0(t)y(t)P(z)=+2Re{∑i=1n‍∑t=2nN-i‍y*(t)Di(t)y(t+i)},
where
(41)M=diag⁡{A-1*L1A-1,L*L2L}.On the basis of the above discussion, we obtain the following variational formula of eigenvalues for (1)-(2) on X by Lemma 3, which plays an important role in the next section.Theorem 4.
Assume that (3), (4), and (22) hold. Then, for each k, 1≤k≤(N+1)d,
(42)λk=max⁡{g(z(1),…,z(k-1)):z(j)∈X,1≤j≤k-1}
with g(z(1),…,z(k-1))=min⁡{ℛ(z):z∈X,z⊥1z(j),1≤j≤k-1,z≠0}, where z⊥1z(j) denotes 〈z,z(j)〉1=0.At the end of this section, we quote two lemmas about matrices and their perturbation. For convenience, we introduce the following notation for an invertible matrixA=(aij)∈ℂd×d:
(43)h(A):=|det⁡A|2dd!(∥A∥+1)d-1,
where the norm of matrix A is defined by
(44)∥A∥:=(∑i,j=1d‍|aij|2)1/2.With the aid of [9, Corollary 7.8.2] we have immediately the following results.Lemma 5.
For any matrixA=(aij)∈ℂd×d, |det⁡A|≤∥A∥d.Lemma 6 ([17, Lemma 2.5]).
LetA∈ℂd×d be invertible. If a matrix A~∈ℂd×d satisfies
(45)∥A~-A∥≤min⁡{h(A),1},
then A~ is invertible, and
(46)∥A~-1∥≤2d(∥A∥+1)d-1|det⁡A|.


## 3. Main Results

In this section, we discuss eigenvalues of perturbed problems sufficiently close to problem (1)-(2) and give error estimates of them.For convenience, introduce the following notations and several constant matrices:(47)w=max⁡n≤t≤N+n∥w(t)∥,w0=min⁡n≤t≤N+n|det⁡w(t)|,r=∥R∥,s=∥S∥,l=∥L∥,r^=max⁡{∥ri(t)∥:t∈[n,N+n+i],0≤i≤n},r0=min⁡{|det⁡rn(t)|:t∈[n,2n-1]11111111∪[N+n+1,N+2n]|det⁡rn(t)|},di=∑k=0n-i‍∑h=0k‍Ci+kk-hCi+kh,0≤i≤n,d^=d0+2∑i=1n‍di.For any0≤i, j≤n,
(48)e1=(∑i≤j‍|dj-i|2+∑i>j‍|di-j|2)1/2,e2=(∑i≤j‍|dn+i-j|2)1/2,aij0={∑k=0j-i∑h=0kCn-j+khCi+n-j+kk-h,i≤j,0,i>j,bij0={∑k=0n-j∑h=0kCj-i+kk-hCj+kh,i≤j,∑k=0n-i∑h=0kCkhCi+kk-h+j,i>j,a=∥(aij0)n×n∥,b=∥(bij0)n×n∥,s^=r+s+sbr^l,(49)β=min⁡{min⁡σ(w(t)):t∈[n,N+n]},γ=min⁡σ(W),
where min⁡σ(w(t)) denotes the minimum value of all eigenvalues of w(t) and W is the same as in (30). It is evident that β>0 and γ>0.Based on the above discussion, we know(50)a<b,∥A∥≤ar^,∥A1∥≤ar^,∥B∥≤br^,∥B1∥≤br^,∥L1∥≤e1r^,∥L2∥≤e1r^,∥Lp∥≤e2r^(3≤p≤6),∥Di(t)∥≤dir^(0≤i≤n).Now, we consider the following perturbed problem of (1)-(2):
(1)′∑i=0n‍Δi[r~i(t)Δiy(t-i)]=λw~(t)y(t),t∈[n,N+n],(2)′R~(-u(0,y)u(N+1,y))+S~(v(0,y)v(N+1,y))=0,
where r~i(t) (t∈[n,N+n+i],0≤i≤n) and w~(t) (t∈[n,N+n]) are d×d Hermitian matrices, w~(t)>0 (t∈[n,N+n]), and R~ and S~ are 2nd×2nd matrices and satisfy
(51)R~S~*=S~R~*.In the following, we will prove that if the perturbation is sufficiently small in norm, then(52)r~n(t)isnonsingularon⁡[n,2n-1]∪[N+n+1,N+2n],(53)rank⁡(R~,S~)=2nd,Ω~=(R~1-S~1B~L,S~2)isinvertible,
where B~ has the same form of B with ri(t) in (18) replaced by r~i(t). The matrices B~1, A~, A~1, D~i(t) (0≤i≤n), L~p (1≤p≤6) are the perturbations of the matrices B, A, A1, Di(t) (0≤i≤n), Lp (1≤p≤6), respectively.Proposition 7.
Let(54)ε1:=min⁡{22h(D),h(Ω)(r^+s+1)bl+2,1(r^+s+1)bl+2,1111111111r0n2nd(nd)!(ar^+1)nd-1a22},
where D is a 2nd×2nd nonsingular submatrix of (R,S). For any 0<ε≤ε1, if
(55)∥R~-R∥≤ε,∥S~-S∥≤ε,(56)∥r~i(t)-ri(t)∥≤ε,t∈[n,N+n+i],0≤i≤n,
then(i)
(52) holds,A~andA~1are nonsingular, and(57)∥A~-1∥≤m,∥A~1-1∥≤m,where(58)m=2nd(ar^+1)nd-1r0n;(ii)
(53) holds, and(59)∥Ω~-1∥≤4nd(s^+1)2nd-1|det⁡Ω|-1.Proof.
(i) We only prove thatA~ is invertible. The invertibility of A~1 can be similarly proved. Since
(60)ε≤r0n2nd(nd)!(ar^+1)nd-1aε≤|det⁡A|2nd(nd)!(∥A∥+1)nd-1a=h(A)a,
we have
(61)∥A~-A∥≤aε≤min⁡{h(A),1}.
Thus,A~ is invertible by Lemma 6, and
(62)∥A~-1∥≤2nd(∥A∥+1)nd-1|det⁡A|≤m.
In addition, since(63)det⁡A~=(-1)(n(n-1)d)/2det⁡r~n(n)⋯det⁡r~n(2n-1),
then r~n(t) is nonsingular on [n,2n-1], which, together with the invertibility of A~1, yields that (52) holds. So (i) is proved.
(ii) LetD~ be a 2nd×2nd submatrix of (R~,S~) and let its position be the same as that of D in (R,S). Since
(64)∥D~-D∥2≤∥R~-R∥2+∥S~-S∥2≤2ε2,
that is,
(65)∥D~-D∥≤2ε≤2ε1≤min⁡{h(D),1},
D
~ is invertible by Lemma 6 and, consequently, rank⁡(R~,S~)=2nd.
In addition,(66)∥S~1B~L-S1BL∥≤(∥S~1∥∥B~-B∥+∥S~1-S1∥∥B∥)∥L∥≤((s+1)bε+br^ε)l=(r^+s+1)blε.
Then we have(67)∥Ω~-Ω∥2=∥R~1-S~1B~L-R1+S1BL∥2+∥S~2-S2∥2≤(∥R~1-R1∥+∥S~1B~L-S1BL∥)2+∥S~2-S2∥2≤(ε+(r^+s+1)blε)2+ε2≤((r^+s+1)bl+2)2ε2.
Thus,(68)∥Ω~-Ω∥≤((r^+s+1)bl+2)ε∥Ω~-Ω∥≤((r^+s+1)bl+2)ε1∥Ω~-Ω∥≤min⁡{h(Ω),1}.
Hence,Ω~ is invertible and ∥Ω~-1∥≤4nd(∥Ω∥+1)2nd-1|det⁡Ω|-1 by Lemma 6. Further,
(69)∥Ω∥2=∥R1-S1BL∥2+∥S2∥2∥Ω∥2≤(r+sbr^l)2+s2∥Ω∥2≤(r+s+sbr^l)2=s^2,
which yields that (59) holds. The proof is complete.Under the assumptions of Proposition7, A~ and Ω~ are invertible. So, we can define the following inner product on X corresponding to problem (1)′-(2)′:
(70)〈z1,z2〉2:=η*W~ξ+∑t=2nN‍y*(t)w~(t)x(t)
for any z1={ξ,x(2n),…,x(N)}, z2={η,y(2n),…,y(N)}∈X, where
(71)W~=Ω~diag⁡{(A~-1)*W~1A~-1,L*W~2L}Ω~*>0,W~1=diag⁡{w~(2n-1),…,w~(n)},W~2=diag⁡{w~(N+n),…,w~(N+1)}.The corresponding induced norm is denoted by(72)∥z∥2:=(〈z,z〉2)1/2,z={ξ,y(2n),…,y(N)}∈X.Similarly,(X,〈·,·〉2) is also an (N+1)d-dimensional Hilbert space.Under the hypotheses of Proposition7, if further (51) holds, then, by Lemma 2, the perturbed problem (1)′-(2)′ has also (N+1)d real eigenvalues (multiplicity included) arranged as
(73)λ~1≤λ~2≤⋯≤λ~(N+1)d.Notice that the multiplicity ofλ~k, the kth eigenvalue of (1)′-(2)′, may be different from that of the kth eigenvalue λk of (1)-(2) in general.Similarly, The Rayleigh quotient corresponding to the difference operator for(1)′-(2)′(74)(ℒ~y)(t)=w~-1(t)∑i=0n‍Δi[r~i(t)Δiy(t-i)],111111111111111111111111t∈[n,N+n],
on X with 〈·,·〉2 can be defined by
(75)ℛ~(z):=P~(z)〈z,z〉2,z={ξ,y(2n),…,y(N)}∈Xwithz≠0,
where
(76)P~(z)=ξ*(Ω~M~Ω~*+(R~1-S~1B~L)(A~-1)*L~3LS~1*1111111111-S~2L*L~4A~1-1(R~2*+B~1LS~2*)(A~-1)*)ξP~(z)=-2Re{Y*(N-n+1)L~5LS~2*ξ}P~(z)=+2Re{ξ*(R~1-S~1B~L)(A~-1)*L~6Y(2n)}P~(z)=+∑t=2nN‍y*(t)D~0(t)y(t)P~(z)=+2Re{∑i=1n‍∑t=2nN-i‍y*(t)D~i(t)y(t+i)},
in which
(77)M~=diag⁡{(A~-1)*L~1A~-1,L*L~2L}.According to the above discussion, if (51), (55), and (56) hold, then we can get the following variational formula of eigenvalues for (1)′-(2)′ on X in a similar way to Theorem 4: for each k, 1≤k≤(N+1)d,
(78)λ~k=max⁡{g~(z(1),…,z(k-1)):z(j)∈X,1≤j≤k-1}
with g~(z(1),…,z(k-1))=min⁡{ℛ~(z):z∈X,z⊥2z(j),1≤j≤k-1,z≠0}, where z⊥2z(j) denotes 〈z,z(j)〉2=0.In order to obtain an error estimate of eigenvalues for the perturbed problem by applying the above variational formula of eigenvalues, we will discuss the relationship between⊥2 and ⊥1 and then give another form of variational formula of eigenvalues for (1)′-(2)′ on X. Now we introduce the following linear transformation:
(79)T2:X⟶X,
where, for any z={ξ,y(2n),…,y(N)}∈X,(80)T2(z)={W~-1Wξ,w~-1(2n)w(2n)y(2n),11111111…,w~-1(N)w(N)y(N)}.Obviously,T2 is invertible, and
(81)〈z1,z2〉2=〈T2-1(z1),z2〉1,∀z1,z2∈X.So, for anyz(1),…,z(k-1)∈X, we get
(82)g~(z(1),…,z(k-1))=min⁡{ℛ~(z):z∈X,z⊥2z(j),11111111111111111111111≤j≤k-1,z≠0}g~(z(1),…,z(k-1))=min⁡{ℛ~(z):z∈X,T2-1(z)⊥1z(j),11111111111111111111111≤j≤k-1,z≠0}g~(z(1),…,z(k-1))=min⁡{ℛ~(T2(z)):z∈X,z⊥1z(j),11111111111111111111111≤j≤k-1,z≠0}.Thus, the variational formula of eigenvalues for(1)′-(2)′ on X can be restated as follows: if (51), (55), and (56) hold, then, for each k, 1≤k≤(N+1)d,
(83)λ~k=max⁡{g~(z(1),…,z(k-1)):z(j)∈X,1≤j≤k-1}
with g~(z(1),…,z(k-1))=min⁡{ℛ~(T2(z)):z∈X,z⊥1z(j),1≤j≤k-1,z≠0}.Before giving the main results, we prepare some estimates.Lemma 8.
For any0<ε≤ε1, if (3) and (56) hold, then
(84)∥M~-M∥≤(m2e1(2mar^+1)+l2e1)ε,(85)∥M∥≤e1r^(m4+l4)1/2,(86)∥M~∥≤e1(r^+1)(m4+l4)1/2,
where ε1, m, M, and M~ are the same as in (54), (58), (41), and (77), respectively.Proof.
A
a
=
(
A
i
j
)
n
d
×
n
d denotes the adjoint matrix of A. Then, by Lemma 5, we get
(87)|Aij|≤∥A∥nd-1,
which yields
(88)∥Aa∥=(∑i,j=1nd|Aij|2‍)1/2≤nd∥A∥nd-1≤nd(ar^)nd-1.
So,(89)∥A-1∥=∥Aa∥|det⁡A|≤nd(ar^)nd-1r0n≤m.
Similarly, one gets(90)∥A1-1∥≤m.
Hence, we have from (57) and (89) that
(91)∥A~-1-A-1∥=∥A-1AA~-1-A-1A~A~-1∥∥A~-1-A-1∥≤∥A-1∥∥A-A~∥∥A~-1∥∥A~-1-A-1∥≤m2aε.
Similarly, we obtain(92)∥A~1-1-A1-1∥≤m2aε.
It follows from (41) and (77) that
(93)∥M~-M∥2=∥(A~-1)*L~1A~-1-A-1*L1A-1∥2∥M~-M∥2=+∥L*L~2L-L*L2L∥2.
From (91) one can get
(94)∥(A~-1)*L~1A~-1-A-1*L1A-1∥≤∥A~-1∥∥L~1A~-1-L1A-1∥+∥A~-1-A-1∥∥L1∥∥A-1∥≤∥A~-1∥(∥L~1-L1∥∥A~-1∥11111111111+∥L1∥∥A~-1-A-1∥)+∥A~-1-A-1∥∥L1∥∥A-1∥≤m(me1ε+m2ae1r^ε)+m3ae1r^ε≤m2e1(2mar^+1)ε.
In addition,(95)∥L*L~2L-L*L2L∥≤∥L∥2∥L~2-L2∥≤l2e1ε.
Therefore, we have(96)∥M~-M∥≤(m2e1(2mar^+1)+l2e1)ε.
It is easy to get from (41) that
(97)∥M∥=[∥A-1*L1A-1∥2+∥L*L2L∥2]1/2∥M∥≤[∥A-1∥4∥L1∥2+∥L∥4∥L2∥2]1/2∥M∥≤(m4e12r^2+l4e12r^2)1/2∥M∥=e1r^(m4+l4)1/2.
Inequality (86) can be obtained by a similar argument. The proof is complete.Now, we study|P(z)| for any z∈X.Proposition 9.
For anyz∈X, if (3) holds, then
(98)|P(z)|≤(G1γ-1+G2β-1)∥z∥12,
where β and γ are defined as in (49),
(99)G1:=s^2e1r^(m4+l4)1/2+me2r^(r+sbr^l)(2sl+1)+e2r^ls,(100)G2:=(r+sbr^l)me2r^+e2r^ls+d^r^.Proof.
For any givenz={ξ,y(2n),…,y(N)}∈X, it follows from (40) that
(101)|P(z)|≤∥ξ∥2(∥A1-1∥∥Ω∥2∥M∥+∥R1-S1BL∥|P(z)|≤111111×∥A-1∥∥L3∥∥L∥∥S1∥|P(z)|≤111111+∥S2∥∥L∥∥L4∥|P(z)|≤111111×∥A1-1∥∥R2*+B1LS2*∥)|P(z)|≤1+2∥ξ∥∥Y(N-n+1)∥|P(z)|≤1×∥L5∥∥L∥∥S2∥+2∥ξ∥∥Y(2n)∥|P(z)|≤1×∥R1-S1BL∥∥A-1∥∥L6∥|P(z)|≤1+∑t=2nN‍∥y(t)∥2∥D0(t)∥+2∑i=1n‍∑t=2nN-i‍|P(z)|≤1×∥y(t)∥∥y(t+i)∥∥Di(t)∥,
where ∥x∥ is the Euclidean norm of x∈ℂd; that is,
(102)∥x∥=(∑i=1d‍|xi|2)1/2.
Further, from (50), (69), (85), (89), and (90) we have
(103)|P(z)|≤∥ξ∥2(s^2e1r^(m4+l4)1/2|P(z)|≤111111+2slme2r^(r+sbr^l)(s^2e1r^(m4+l4)1/2)|P(z)|≤+(∥ξ∥2+∥Y(N-n+1)∥2)e2r^ls|P(z)|≤+(∥ξ∥2+∥Y(2n)∥)(r+sbr^l)me2r^+d^r^∑t=2nN‍∥y(t)∥2|P(z)|≤∥ξ∥2(s^2e1r^(m4+l4)1/2|P(z)|≤11111+me2r^(r+sbr^l)(2sl+1)+e2r^ls(m4+l4)1/2)|P(z)|≤+(∥Y(2n)∥2+∥Y(N-n+1)∥2)|P(z)|≤×((r+sbr^l)me2r^+e2r^ls)+d^r^∑t=2nN‍∥y(t)∥2|P(z)|≤G1∥ξ∥2+G2∑t=2nN‍∥y(t)∥2|P(z)|≤G1γ-1ξ*Wξ+G2β-1∑t=2nN‍y*(t)w(t)y(t)|P(z)|≤(G1γ-1+G2β-1)∥z∥12.
This completes the proof.Next, we study the difference between∥T2(z)∥22 and ∥z∥12 for any z∈X.Proposition 10.
Let(104)ε2:=min⁡{ε1,w02dd!(w+1)d-1},
whereε1is defined as in (54). For any 0<ε≤ε2, if (55) and (56) hold and
(105)∥w~(t)-w(t)∥≤ε,t∈[n,N+n],
then
(106)|∥T2(z)∥22-∥z∥12|≤(G3γ-1+G4β-1)∥z∥12ε,∀z∈X,
where
(107)G3:=([(m4+l4)(a4(r^+1)4+l4)]1/232g1n3d3(s^+1)4nd-2(w+1)d-1ws^2G3:=11×[(m4+l4)(a4(r^+1)4+l4)]1/2)G3:=×(|det⁡Ω|2w0)-1,(108)g1:=n(s^+1)2(m2(2mwa+1)+l2)g1:=+nw(2s^+1)((r^+s+1)bl+2)g1:=×(m4+l4)1/2,(109)G4:=2d(w+1)d-1ww0-1.Proof.
It follows from (80) that, for any given z={ξ,y(2n),…,y(N)}∈X,
(110)∥T2(z)∥22=ξ*WW~-1Wξ∥T2(z)∥22=+∑t=2nN‍y*(t)w(t)w~-1(t)w(t)y(t),
which, together with (33), yields that
(111)|∥T2(z)∥22-∥z∥12|=|∑t=2nN‍ξ*W(W~-1W-I2nd)ξ11111+∑t=2nN‍y*(t)w(t)(w~-1(t)w(t)-Id)y(t)|≤∥ξ∥2∥W∥∥W~-1∥∥W-W~∥+∑t=2nN‍∥y(t)∥2∥w(t)∥111111111×∥w~-1(t)∥∥w(t)-w~(t)∥.
Since(112)ε≤w02dd!(w+1)d-1≤h(w(t)),t∈[n,N+n],
it follows from Lemma 6 that
(113)∥w~-1(t)∥≤2d(∥w(t)∥+1)d-1|det⁡w(t)|∥w~-1(t)∥≤2d(w+1)d-1w0-1,t∈[n,N+n].
Thus,(114)∥W~1-1∥=∥diag⁡{w~-1(2n-1),…,w~-1(n)}∥∥W~1-1∥≤2nd(w+1)d-1w0-1.
Similarly, we have(115)∥W~2-1∥≤2nd(w+1)d-1w0-1.
In addition, from (59), (71), and (113) we get
(116)∥W~-1∥=∥Ω~*-1diag⁡{A~W~1-1A~*,LW~2-1L*}Ω~-1∥∥W~-1∥≤∥Ω~-1∥2[(∥A~∥2∥W~1-1∥)21111111111111111+(∥L∥2∥W~2-1∥)2[(∥A~∥2∥W~1-1∥)2]1/2∥W~-1∥≤((a4(r^+1)4+l4)1/232n5/2d3(s^+1)4nd-2(w+1)d-111111111111×(a4(r^+1)4+l4)1/2)∥W~-1∥=×(|det⁡Ω|2w0)-1.
Now, we are in a position to estimate∥W~-W∥. Let
(117)K~=diag⁡{(A~-1)*W~1A~-1,L*W~2L},K=diag⁡{A-1*W1A-1,L*W2L}.
Then, from (89) we obtain
(118)∥K∥=[∥A-1*W1A-1∥2+∥L*W2L∥2]1/2∥K∥≤nw(m4+l4)1/2.
With a similar argument to that for (93), we get
(119)∥K~-K∥≤n(m2(2mwa+1)+l2)ε.
From (67) one has
(120)∥Ω~-Ω∥≤((r^+s+1)bl+2)ε∥Ω~-Ω∥≤((r^+s+1)bl+2)ε1≤1,
which, together with (69), implies that
(121)∥Ω~∥≤∥Ω∥+1≤s^+1.
Hence, it follows from (30), (71), and (121) that
(122)∥W~-W∥=∥Ω~K~Ω~*-ΩKΩ*∥∥W~-W∥≤∥Ω~∥∥K~Ω~*-KΩ*∥+∥Ω~-Ω∥∥KΩ*∥∥W~-W∥≤∥Ω~∥(∥K~-K∥∥Ω~*∥+∥K∥∥Ω~*-Ω*∥)∥W~-W∥=+∥Ω~-Ω∥∥K∥∥Ω*∥∥W~-W∥≤(s^+1)2∥K~-K∥+∥K∥(2s^+1)∥W~-W∥=×((r^+s+1)bl+2)ε∥W~-W∥≤g1ε,
whereg1is the same as in (108). It can be easily concluded from (69) that
(123)∥W∥=∥ΩKΩ*∥≤∥Ω∥2∥K∥∥W∥≤nws^2(m4+l4)1/2.
Therefore, from (113), (116), (122), and (123) we have
(124)|∥T2(z)∥22-∥z∥12|≤(([(m4+l4)(a4(r^+1)4+l4)]1/232g1n3d3(s^+1)4nd-2(w+1)d-1ws^21111111×[(m4+l4)(a4(r^+1)4+l4)]1/2)×(|det⁡Ω|2w0)-1(([(m4+l4)(a4(r^+1)4+l4)]1/232g1n3d3(s^+1)4nd-2(w+1)d-1ws^2)ε∥ξ∥2+2d(w+1)d-1ww0-1ε∑t=2nN‍∥y(t)∥2=(G3∥ξ∥2+G4∑t=2nN‍∥y(t)∥2)ε≤(G3γ-1+G4β-1)∥z∥12ε.
Consequently, (106) holds and the proof is complete.The following result is about the estimate of difference betweenP~(T2(z))andP(z).Proposition 11.
For any0<ε≤ε2, in whichε2is defined as in (104), if (55), (56), and (105) hold, then
(125)|P~(T2(z))-P(z)|≤(G5γ-1+G6β-1)∥z∥12ε,∀z∈X,
where
(126)G5:=g2+G3(G3+g1)g1nws^2(m4+l4)1/2G5≔×(r^+1)(e1(s^+1)2(m4+l4)1/2G5≔111111111+2mle2(s+1)(r+sbr^l+1)(m4+l4)1/2)G5≔+(s^+1)2(m2e1(2mar^+1)+l2e1)G5≔+(2s^+1)((r^+s+1)bl+2)G5≔×e1r^(m4+l4)1/2+2mle2G5≔×((r^+s+1)(r+2sbr^l+1)11111111+sar^m(r+sbr^l+1)+sr^((r^+s+1)(r+2sbr^l+1))G5≔+(r^+s+1)le2+[(r^+1)((r^+s+1)bl+1)G5≔111111111111111+(r+sbr^l)(mar^+1)((r^+s+1)bl+1)]me2,(127)G6:=G4(G4+1)d^(r^+1)w-1+d^G6:=+g2+(r^+s+1)le2G6:=+[(r^+1)((r^+s+1)bl+1)G6:=11+(r+sbr^l)(mar^+1)((r^+s+1)bl+1)]me2,(128)g2:={(((a4(r^+1)4+l4)1/264g1n3d4(s^+1)4nd-2(w+1)2d-2g2:=111×w(a4(r^+1)4+l4)1/2)g2:=11×(|det⁡Ω|2w02)-1(((a4(r^+1)4+l4)1/264g1n3d4(s^+1)4nd-2(w+1)2d-2)g2:=1+2nd(w+1)d-1w0-1(((a4(r^+1)4+l4)1/264g1n3d4(s^+1)4nd-2(w+1)2d-2}g2:=×(r^+1)((s+1)l+(r+sbr^l+1)m)e2,
andG3,g1, andG4are the same as in (107), (108), and (109), respectively.Proof.
It follows from (76) and (80) that, for any z={ξ,y(2n),…,y(N)}∈X,
(129)P~(T2(z))=ξ*WW~-1(Ω~M~Ω~*+(R~1-S~1B~L)(A~-1)*L~3LS~1*11111111111111111111-S~2L*L~4A~1-1(R~2*+B~1LS~2*)(A~-1)*)W~-1WξP~(T2(z))=-2Re{Y*(N-n+1)W3L~5LS~2*W~-1Wξ}P~(T2(z))=+2Re{ξ*WW~-1(R~1-S~1B~L)111111111111111111×(A~-1)*L~6W4Y(2n)}P~(T2(z))=+∑t=2nN‍y*(t)w(t)w~-1(t)D~0(t)w~-1(t)w(t)y(t)P~(T2(z))=+2Re{∑i=1n‍∑t=2nN-i‍y*(t)w(t)w~-1(t)D~i(t)w~-1P~(T2(z))=1111111×(t+i)w(t+i)y(t+i)∑i=1n‍},
where
(130)W3=diag⁡{w(N)w~-1(N),…,w(N-n+1)w~-111111111111×(N-n+1){w(N)w~-1(N),…,w(N-n+1)w~-1},W4=diag⁡{w~-1(3n-1)w(3n-1),…,w~-11111111111×(2n)w(2n){w~-1(3n-1)w(3n-1),…,w~-1}.
So we get from (40) and (129) that
(131)|P~(T2(z))-P(z)|≤∥ξ∥2Δ1+2∥ξ∥∥Y(N-n+1)∥Δ2+2∥ξ∥∥Y(2n)∥Δ3+∑t=2nN‍∥y(t)∥2Δ4(t)+2∑i=1n‍∑t=2nN-i‍∥y(t)∥∥y(t+i)∥Δ5(t)≤∥ξ∥2(Δ1+Δ2+Δ3)+∑t=2nN‍∥y(t)∥2(Δ2+Δ3+Δ4(t))+2∑i=1n‍∑t=2nN-i‍∥y(t)∥∥y(t+i)∥Δ5(t),
where
(132)Δ1=∥WW~-1(Ω~M~Ω~*+(R~1-S~1B~L)(A~-1)*L~3LS~1*Δ1=111111111-S~2L*L~4A~1-1(R~2*+B~1LS~2*)(A~-1)*)W~-1WΔ1=1-(-S2L*L4A1-1(R2*+B1LS2*))ΩMΩ*+(R1-S1BL)A-1*L3LS1*Δ1=11111-S2L*L4A1-1(R2*+B1LS2*))∥WW~-1(Ω~M~Ω~*+(R~1-S~1B~L)(A~-1)*L~3LS~1*∥,Δ2=∥W3L~5LS~2*W~-1W-L5LS2*∥,Δ3=∥WW~-1(R~1-S~1B~L)(A~-1)*L~6W4111111-(R1-S1BL)A-1*L6∥WW~-1(R~1-S~1B~L)(A~-1)*L~6W4∥,Δ4(t)=∥w(t)w~-1(t)D~0(t)w~-1(t)w(t)-D0(t)∥,Δ5(t)=∥w(t)w~-1(t)D~i(t)w~-1(t+i)w(t+i)-Di(t)∥.
In the following we discussΔj, 1≤j≤5, term by term. It follows from the first relation in (132) that
(133)Δ1≤∥WW~-1Ω~M~Ω~*W~-1W-ΩMΩ*∥Δ1≤+∥WW~-1(R~1-S~1B~L)(A~-1)*L~3LS~1*W~-1W11111111-(R1-S1BL)A-1*L3LS1*(A~-1)*∥Δ1≤+∥WW~-1S~2L*L~4A~1-1(R~2*+B~1LS~2*)W~-1W11111111-S2L*L4A1-1(R2*+B1LS2*)∥WW~-1S~2L*L~4A~1-1(R~2*+B~1LS~2*)W~-1W∥,
in which the right-hand side is a sum of three terms. Now, we calculate the first term.
(134)∥WW~-1Ω~M~Ω~*W~-1W-ΩMΩ*∥≤∥W-W~∥∥W~-1Ω~M~Ω~*W~-1W∥+∥Ω~M~Ω~*W~-1∥∥W-W~∥+∥Ω~M~Ω~*-ΩMΩ*∥≤∥W~-1∥∥W-W~∥×(∥W~-1W∥+1)∥Ω~M~Ω~*∥+∥Ω~M~Ω~*-ΩMΩ*∥.
From (67), (69), and (121) we have
(135)∥Ω~M~Ω~*-ΩMΩ*∥≤∥Ω~∥∥M~Ω~*-MΩ*∥∥Ω~M~Ω~*-ΩMΩ*∥=+∥Ω~-Ω∥∥MΩ*∥∥Ω~M~Ω~*-ΩMΩ*∥≤∥Ω~∥(∥M~-M∥∥Ω~*∥∥Ω~M~Ω~*-ΩMΩ*∥1111111+∥M∥∥Ω~*-Ω*∥)∥Ω~M~Ω~*-ΩMΩ*∥=+∥Ω~-Ω∥∥M∥∥Ω*∥∥Ω~M~Ω~*-ΩMΩ*∥≤(s^+1)2∥M~-M∥+∥M∥∥Ω~M~Ω~*-ΩMΩ*∥=×(2s^+1)((r^+s+1)bl+2)ε,
which, together with (84) and (85) in Lemma 8, implies that
(136)∥Ω~M~Ω~*-ΩMΩ*∥≤{(m4+l4)1/2(s^+1)2(m2e1(2mar^+1)+l2e1)11111+(2s^+1)((r^+s+1)bl+2)11111×e1r^(m4+l4)1/2}ε.
In addition, from (86) and (121) we get
(137)∥Ω~M~Ω~*∥≤∥Ω~∥2∥M~∥∥Ω~M~Ω~*∥≤(s^+1)2e1(r^+1)(m4+l4)1/2.
Hence, it follows from (134)–(137) that
(138)∥WW~-1Ω~M~Ω~*W~-1W-ΩMΩ*∥≤∥W~-1∥∥W-W~∥(∥W~-1W∥+1)×(s^+1)2e1(r^+1)(m4+l4)1/2+{(m4+l4)1/2(s^+1)2(m2e1(2mar^+1)+l2e1)1111111+(2s^+1)((r^+s+1)bl+2)1111111×e1r^(m4+l4)1/2}ε.
Next, we study the second term in the right-hand side of (133):
(139)∥WW~-1(R~1-S~1B~L)(A~-1)*L~3LS~1*W~-1W1-(R1-S1BL)A-1*L3LS1*(A~-1)*∥≤∥W-W~∥∥W~-1(R~1-S~1B~L)(A~-1)*11111111111111×L~3LS~1*W~-1W(A~-1)*∥+∥(R~1-S~1B~L)(A~-1)*L~3LS~1*W~-1∥×∥W-W~∥+∥(R~1-S~1B~L)(A~-1)*L~3LS~1*11111111111111111-(R1-S1BL)A-1*L3LS1*(A~-1)*∥≤∥W~-1∥∥W-W~∥(∥W~-1W∥+1)×∥(R~1-S~1B~L)(A~-1)*L~3LS~1*∥+∥(R~1-S~1B~L)(A~-1)*L~3LS~1*1111111-(R1-S1BL)A-1*L3LS1*(A~-1)*∥.
Sinceε≤ε1≤1/((r^+s+1)bl+1), from (66) we have
(140)∥R~1-S~1B~L-R1+S1BL∥≤((r^+s+1)bl+1)ε≤1.
So,(141)∥R~1-S~1B~L∥≤r+sbr^l+1,
which, together with (57), (89), and (91), yields that
(142)∥(R~1-S~1B~L)(A~-1)*L~3LS~1*-(R1-S1BL)A-1*L3LS1*∥≤∥R~1-S~1B~L∥∥(A~-1)*L~3LS~1*-A-1*L3LS1*∥+∥R~1-S~1B~L-R1+S1BL∥∥A-1*L3LS1*∥≤∥R~1-S~1B~L∥(∥A~-1∥∥L~3LS~1*-L3LS1*∥1111111111111111+∥A~-1-A-1∥∥L3LS1*∥)+∥R~1-S~1B~L-R1+S1BL∥∥A-1*L3LS1*∥≤∥R~1-S~1B~L∥[∥A~-1∥(∥L~3L∥∥S~1*-S1*∥11111111111111111111111+∥L~3-L3∥∥LS1*∥∥S~1*-S1*∥)1111111111111111+∥A~-1-A-1∥∥L3LS1*∥]+∥R~1-S~1B~L-R1+S1BL∥∥A-1*L3LS1*∥≤mle2((r^+s+1)(r+2sbr^l+1)111111111+sar^m(r+sbr^l+1)+sr^((r^+s+1)(r+2sbr^l+1))ε.
Hence, it follows from (139)–(142) that
(143)∥WW~-1(R~1-S~1B~L)(A~-1)*L~3LS~1*W~-1W1-(R1-S1BL)A-1*L3LS1*(A~-1)*∥≤∥W~-1∥∥W-W~∥(∥W~-1W∥+1)×mle2(r^+1)(s+1)(r+sbr^l+1)+mle2((r^+s+1)(r+2sbr^l+1)1111111111+sar^m(r+sbr^l+1)+sr^((r^+s+1)(r+2sbr^l+1))ε.
With a similar argument, one can obtain an estimate of the third term in the right-hand side of (133), which is the same as (143). Then, from (116), (122), (123), (133), (138), and (143), one can get
(144)Δ1≤{×[(m4+l4)(a4(r^+1)4+l4)]1/2)((×(w+1)d-1(a4(r^+1)4+l4)1/2)32g1n5/2d3(s^+1)4nd-211111111×(w+1)d-1(a4(r^+1)4+l4)1/2)1111111×(|det⁡Ω|2w0)-1((×(w+1)d-1(a4(r^+1)4+l4)1/2)32g1n5/2d3(s^+1)4nd-2)111111×((×[(m4+l4)(a4(r^+1)4+l4)]1/2)32n3d3(s^+1)4nd-2(w+1)d-1ws^211111111111×[(m4+l4)(a4(r^+1)4+l4)]1/2)1111111111×(|det⁡Ω|2w0)-1+1((×[(m4+l4)(a4(r^+1)4+l4)]1/2)32n3d3(s^+1)4nd-2(w+1)d-1ws^2)111111×(r^+1)(e1(s^+1)2(m4+l4)1/21111111111111111+2mle2(s+1)(r+sbr^l+1)(m4+l4)1/2)111111+(s^+1)2(m2e1(2mar^+1)+l2e1)111111+(2s^+1)((r^+s+1)bl+2)111111×e1r^(m4+l4)1/2111111+2mle2((r^+s+1)(r+2sbr^l+1)11111111111111+sar^m(r+sbr^l+1)+sr^((r^+s+1)(r+2sbr^l+1)){×[(m4+l4)(a4(r^+1)4+l4)]1/2)(32g1n5/2d3(s^+1)4nd-2}ε.
Next, we consider the second relation in (132). It is evident that
(145)Δ2≤∥W3L~5LS~2*W~-1∥∥W-W~∥Δ2≤+∥W3-Ind∥∥L~5LS~2*∥Δ2≤+∥L~5LS~2*-L5LS2*∥.
From (50) we have
(146)∥L~5LS~2*-L5LS2*∥≤∥L~5-L5∥∥LS~2*∥∥L~5LS~2*-L5LS2*∥=+∥L5L∥∥S~2-S2∥∥L~5LS~2*-L5LS2*∥≤(r^+s+1)le2ε.
It follows from the expression ofW3 that
(147)∥W3∥≤2nd(w+1)d-1ww0-1,∥W3-Ind∥≤2nd(w+1)d-1w0-1ε.
Additionally,(148)∥L~5LS~2*∥≤(r^+1)(s+1)le2.
Further, from (116) and (122) one has
(149)Δ2≤{{((a4(r^+1)4+l4)1/264g1n3d4(s^+1)4nd-2((a4(r^+1)4+l4)1/264g1n3d4(s^+1)4nd-211111111×(w+1)2d-2w(a4(r^+1)4+l4)1/2)111111×(|det⁡Ω|2w02)-1111111+2nd(w+1)d-1w0-1{((a4(r^+1)4+l4)1/264g1n3d4(s^+1)4nd-2}11111×(r^+1)(s+1)le2ε+(r^+s+1)le2ε.
From the third relation in (132) we get
(150)Δ3≤∥W-W~∥∥W~-1(R~1-S~1B~L)(A~-1)*L~6W4∥Δ3≤+∥(R~1-S~1B~L)(A~-1)*L~6∥∥W4-Ind∥Δ3≤+∥(R~1-S~1B~L)(A~-1)*L~6-(R1-S1BL)A-1*L6∥.
From (57), (66), (89), and (91) we obtain
(151)∥(R~1-S~1B~L)(A~-1)*L~6-(R1-S1BL)A-1*L6∥≤∥R~1-S~1B~L-R1+S1BL∥∥(A~-1)*L~6∥+∥R1-S1BL∥(∥(A~-1)*∥∥L~6-L6∥11111111111111111+∥(A~-1)*-A-1*∥∥L6∥)≤[(r^+1)((r^+s+1)bl+1)11111+(r+sbr^l)(mar^+1)((r^+s+1)bl+1)]me2ε.
According to the expression ofW4, we know that it has the same estimate as W3 in (147). Thus, we have
(152)Δ3≤{((a4(r^+1)4+l4)1/264g1n3d4(s^+1)4nd-21111111×(w+1)2d-2w(a4(r^+1)4+l4)1/2)111111×(|det⁡Ω|2w02)-1111111+2nd(w+1)d-1w0-1{((a4(r^+1)4+l4)1/264g1n3d4(s^+1)4nd-2}Δ3≤×(r^+1)(r+sbr^l+1)me2εΔ3≤+[(r^+1)((r^+s+1)bl+1)1111111+(r+sbr^l)(mar^+1)((r^+s+1)bl+1)]me2ε.
It follows from (113) and (132) that, for any t∈[2n,N],
(153)Δ4(t)≤∥w(t)-w~(t)∥Δ4(t)≤×∥w~-1(t)D~0(t)w~-1(t)w(t)∥Δ4(t)≤+∥D~0(t)w~-1(t)∥∥w(t)-w~(t)∥Δ4(t)≤+∥D~0(t)-D0(t)∥Δ4(t)≤d0(r^+1)∥w~-1(t)∥Δ4(t)≤×(∥w~-1(t)∥w+1)ε+d0εΔ4(t)≤(G4(G4+1)d0(r^+1)w-1+d0)ε.
Similarly, it can be concluded that(154)Δ5(t)≤(G4(G4+1)di(r^+1)w-1+di)ε.
So, by the assumptions and the Hölder inequality, we have(155)∑i=1n‍∑t=2nN-i‍∥y(t)∥∥y(t+i)∥Δ5(t)≤∑i=1n‍(G4(G4+1)di(r^+1)w-1+di)ε×∑t=2nN‍∥y(t)∥2.
Therefore, from (144) and (149)–(155) we obtain
(156)|P~(T2(z))-P(z)|≤(G5∥ξ∥2+G6∑t=2nN‍∥y(t)∥2)ε,
which, together with (49), implies that (125) holds. The proof is complete.Now we give the main result of the present paper—an error estimate of eigenvalues of the perturbed problem(1)′-(2)′.Theorem 12.
Assume that (3), (4), (22), and (51) hold. Let
(157)ε0:=min⁡{ε2,βγ2(G3β+G4γ)},
where ε2, β, γ, G3, and G4 are the same as in (104), (49), (107), and (109), respectively. For any 0<ε≤ε0, if (55), (56), and (105) hold, then thekth eigenvalue λ~k of (1)′-(2)′ and the kth eigenvalue λk of (1)-(2) (in the increasing order as in (73) and (24), resp.) satisfy
(158)|λ~k-λk|≤2Γε,1≤k≤(N+1)d,
where
(159)Γ=G5γ-1+G6β-1+(G1γ-1+G2β-1)(G3γ-1+G4β-1)
andG1, G2, G5, and G6 are the same as in (99), (100), (126), and (127), respectively.Proof.
By Propositions9–11, we have that, for any z∈X with z≠0,
(160)|ℛ~(T2(z))-ℛ(z)|=|P~(T2(z))∥T2(z)∥22-P(z)∥z∥12||ℛ~(T2(z))-ℛ(z)|≤(|P(z)||∥T2(z)∥22-∥z∥12|∥z∥12|P~(T2(z))-P(z)|111111111111111111111+|P(z)||∥T2(z)∥22-∥z∥12|∥z∥12)|ℛ~(T2(z))-ℛ(z)|=×1∥T2(z)∥22≤Γ∥z∥12∥T2(z)∥22ε.
Since(161)ε≤ε0≤βγ2(G3β+G4γ),
we have from (106) that
(162)|∥T2(z)∥22∥z∥12-1|=|∥T2(z)∥22-∥z∥12|∥z∥12|∥T2(z)∥22∥z∥12-1|≤(G3γ-1+G4β-1)ε≤12,
which implies that ∥T2(z)∥22/∥z∥12≥1/2; that is, ∥z∥12/∥T2(z)∥22≤2. Hence, it follows from (160) that
(163)|ℛ~(T2(z))-ℛ(z)|≤2Γε.
Therefore, for eachk, 1≤k≤(N+1)d and for any z(1),…,z(k-1)∈X, we get from Theorem 4 and (82) that
(164)|g~(z(1),…,z(k-1))-g(z(1),…,z(k-1))|=|min⁡{ℛ~(T2(z)):z∈X,z⊥1z(j),1≤j≤k-1,1111111111z≠0(T2(z))}111111-min⁡{ℛ(z):z∈X,z⊥1z(j),1≤j≤k-1,1111111111111z≠0z(j)}(T2(z))}|≤2Γε,
which, together with (83), yields that (158) holds. The proof is complete.The following result is a direct consequence of Theorem12.Corollary 13.
Assume that all the assumptions in Theorem12 hold. Then each eigenvalue of problem (1)-(2) is continuously dependent on the coefficients and weight function of (1) and the coefficients of the boundary condition (2).Remark 14.
The nonsingularity assumption (22) for Ω can be illustrated by giving examples. Since 2n-order discrete vector boundary value problems include second-order discrete boundary value problems and the necessity of the nonsingularity assumption forΩhas been clarified through an example in [17]. Here we will not discuss it.


## 4. Two Special Cases

In this section, we consider two special perturbed problems. The error estimates will be simpler for these two special cases.Case 1.
The perturbed problem consists of (1)-(2)′; that is, only the coefficients of boundary condition (2) are perturbed, and the coefficients and weight function of (1) are invariant. Since the method of proof is similar to that of Theorem 12, only the related result is given.Theorem 15.
Assume that (3), (4), (22), and (51) hold. Let
(165)ε^0:=min⁡{22h(D),h(Ω)br^l+2,1br^l+2,γ2G^3},
where D is a 2nd×2nd nonsingular submatrix of (R,S),
(166)G^3=(((m4+l4)(a4r^4+l4))1/216g^1n3d3(s^+1)4nd-21111111×wds^2((m4+l4)(a4r^4+l4))1/2)G^3=×(|det⁡Ω|2w0)-1,g^1=nw(2s^+1)(br^l+2)(m4+l4)1/2.
For any0<ε≤ε^0, if (55) holds, then thekth eigenvalue λ^k of (1)-(2)′ and thekth eigenvalueλkof (1)-(2) satisfy
(167)|λ^k-λk|≤2(G^5γ-1+G^6β-1+(G1γ-1+G2β-1)G^3γ-1)ε,
where 1≤k≤(N+1)d, β, γ, G1, and G2 are the same as in (49), (99), and (100), respectively,
(168)G^5=g^2+G^3(G^3+g^1)g^1nws^2(m4+l4)1/2G^5=×r^(e1(s^+1)2(m4+l4)1/2G^5=1111+2mle2(s+1)(r+sbr^l+1)(m4+l4)1/2)G^5=+(2s^+1)(br^l+2)e1r^(m4+l4)1/2G^5=+2mle2r^(2sbr^l+r+s+1)G^5=+e2r^(l+(br^l+1)m),G^6=g^2+e2r^(l+(br^l+1)m),g^2=((a4r^4+l4)1/216g^1n5/2d3(s^+1)4nd-2wd-1e2r^111111×((s+1)l+(r+sbr^l+1)m)(a4r^4+l4)1/2)g^2=×(|det⁡Ω|2w0)-1.Case 2.
The perturbed problem consists of(1)′-(2); that is, only the coefficients and weight function of (1) are perturbed, and the coefficients of boundary condition (2) are invariant.Since boundary condition contains the coefficientsri(t+n-1) and ri(t+N+n) (1≤i≤n and 1≤t≤i) of equation, the coefficients are invariant in this case; then A, A1, B, B1, L3, and L4 are invariant.In addition, since in this case the admissible function spaceL^[0,N+2n] of perturbed problem is the same as that for the original problem, it can be directly applied instead of the space X. However, since the weight function is perturbed, the inner product on L^[0,N+2n] for the perturbed problem changes with it. Define an inner product on L^[0,N+2n] for the perturbed problem by
(169)〈x,y〉0:=∑t=nN+n‍y*(t)w~(t)x(t),x,y∈L^[0,N+2n],
and the following induced norm
(170)∥y∥0:=(〈y,y〉0)1/2,y∈L^[0,N+2n].Obviously,L^[0,N+2n] is still an (N+1)d-dimensional Hilbert space with the inner product 〈·,·〉0 by [1, Theorem 2.3].For convenience, we now introduce the Rayleigh quotient corresponding to the difference operatorℒ~ on L^[0,N+2n] with 〈·,·〉0 as follows:
(171)R-(x):=〈ℒ~y,y〉0〈y,y〉0,y∈L^[0,N+2n]withy′={y(t)}t=nN+n≠0,
where ℒ~ is the same as in (74).By Lemma2, problem (1)′-(2) has also (N+1)d real eigenvalues (multiplicity included) arranged as
(172)λ-1≤λ-2≤⋯≤λ-(N+1)d.The variational property (26) of eigenvaluesλ-kfor perturbed problem (1)′-(2) on L^[0,N+2n] still holds, where λk, f, R(y), ⊥ and 〈·,·〉 are replaced by λ-k, g-, R-(y), ⊥0, and 〈·,·〉0, respectively.In a similar way to the discussion in Section3, we first discuss the relation between ⊥0 and ⊥ and then give another form of variational formula of eigenvalues for problem (1)′-(2) on L^[0,N+2n]. Now we introduce the following linear transformation:
(173)T3:L^[0,N+2n]⟶L^[0,N+2n];
for any y={y(t)}t=0N+2n∈L^[0,N+2n], we have
(174)T3(y)(t)=w~-1(t)w(t)y(t),t∈[n,N+n],YT3(y)(0)=LS1*Ω*-1diag⁡{A,-L}(W~1-1W1Y(n)W~2-1W2Y(N+1)),YT3(y)(N+n+1)=A1-1(R2*+B1LS2*)Ω*-1diag⁡{A,-L}YT3(y)(N+n+1)=×(W~1-1W1Y(n)W~2-1W2Y(N+1)),
where W1, W2, W~1, and W~2 are the same as in (31) and (71), respectively, and YT3(y)(t) has the same definition as Y(t) in (13) only with y(t) replaced by T3(y)(t).Evidently,T3 is invertible and
(175)〈y1,y2〉0=〈T3-1(y1),y2〉,∀y1,y2∈L^[0,N+2n].
Hence, for any z(1),…,z(k-1)∈L^[0,N+2n], we get
(176)g-(z(1),…,z(k-1))=min⁡{R-(y):y∈L^[0,N+2n],y⊥0z(j),1111111111≤j≤k-1,y′≠0z(j)}=min⁡{R-(y):y∈L^[0,N+2n],T3-1(y)⊥z(j),1111111111≤j≤k-1,y′≠0z(j)}=min⁡{R-(T3(y)):y∈L^[0,N+2n],y⊥z(j),1111111111≤j≤k-1,y′≠0z(j)}.Therefore, the variational property (26) of eigenvalues λ-k for problem (1)′-(2) on L^[0,N+2n] still holds, where λk, f, and R(y) are replaced by λ-k, g-, and R-(T3(y)), respectively.Now, we give an error estimate of eigenvalues of the perturbed problem(1)′-(2).Theorem 16.
Assume that (3), (4), and (22) hold. Let
(177)ε*=min⁡{βw04d(w+1)d-1w,w02dd!(w+1)d-1,1},
where β is the same as in (49). For any 0<ε≤ε*, if (56) and (105) hold, then the kth eigenvalue λ-k of (1)′-(2) and the kth eigenvalue λk of (1)-(2) (in the increasing order as in (172) and (24), resp.) satisfy
(178)|λ-k-λk|≤2(M1+M2)ε,1≤k≤(N+1)d,
where
(179)M1:=[2nG4(G4n+1)1111111×e2r^c(ls+m(r+sbr^l))(a2r^2+l2)1/21111111+G4(G4+1)d^(r^+1)+d^w]β-1w-1,M2:=[e2r^c(ls+m(r+sbr^l))(a2r^2+l2)1/2+d^r^]G4β-2,c:=2nds^2nd-1|det⁡Ω|-1,
and G4 is the same as in (109).Proof.
It follows from (25) and (171) that, for any y∈L^[0,N+2n] with y′≠0,
(180)|R-(T3(y))-R(y)|=|〈ℒ~(T3(y)),T3(y)〉0〈T3(y),T3(y)〉0-〈ℒy,y〉〈y,y〉|≤(|〈ℒy,y〉||∥T3(y)∥02-∥y∥*2|∥y∥*2〈ℒ~(T3(y)),T3(y)〉0-〈ℒy,y〉111111+|〈ℒy,y〉||∥T3(y)∥02-∥y∥*2|∥y∥*2)1∥T3(y)∥02,
where ∥y∥*=(〈y,y〉)1/2, and
(181)〈ℒy,y〉=∑t=nN+n‍y*(t){∑i=0n‍Δi[ri(t)Δiy(t-i)]}〈ℒy,y〉=∑t=nN+n‍y*(t){∑i=0n‍Di(t)y(t+i)1111111111111111111+∑i=1n‍Di(t-i)y(t-i)}〈ℒy,y〉=Y*(n)L3Y(0)+Y*(N+1)〈ℒy,y〉=×L4Y(N+n+1)+∑t=nN+n‍y*(t)D0(t)y(t)〈ℒy,y〉=+2Re{∑i=1n‍∑t=nN+n-i‍y*(t)Di(t)y(t+i)}.
It follows from (27) that
(182)(Y(n)Y(N+1))=diag⁡{A-1,-L}Ω*ξ;
that is,
(183)ξ=Ω*-1diag⁡{A,-L}(Y(n)Y(N+1)).
So,(184)Y(0)=LS1*Ω*-1diag⁡{A,-L}(Y(n)Y(N+1)),Y(N+n+1)=A1-1(R2*+B1LS2*)Ω*-1Y(N+n+1)=×diag⁡{A,-L}(Y(n)Y(N+1)).
Hence,(185)|Y*(n)L3Y(0)|=|Y*(n)L3LS1*Ω*-1diag⁡{A,-L}(Y(n)Y(N+1))|≤∥L3LS1*Ω*-1diag⁡{A,-L}∥∥Y(n)∥×∥Y(n)∥2+∥Y(N+1)∥2≤e2r^ls(a2r^2+l2)1/2∥Ω-1∥×2∥Y(n)∥2+∥Y(N+1)∥22≤e2r^ls(a2r^2+l2)1/2∥Ω-1∥∑t=nN+n‍∥y(t)∥2.
Similarly,(186)|Y*(N+1)L4Y(N+n+1)|≤me2r^(r+sbr^l)(a2r^2+l2)1/2×∥Ω-1∥∑t=nN+n‍∥y(t)∥2.
From (69) and (88), we have
(187)∥Ω-1∥=∥Ωa∥|det⁡Ω|≤2nd∥Ω∥2nd-1|det⁡Ω|≤2nds^2nd-1|det⁡Ω|=c.
Thus, it follows from (185)–(187) that
(188)|〈ℒy,y〉|≤e2r^c(ls+m(r+sbr^l))(a2r^2+l2)1/2|〈ℒy,y〉|=1×∑t=nN+n‍∥y(t)∥2+d^r^∑t=nN+n‍∥y(t)∥2|〈ℒy,y〉|≤[e2r^c(ls+m(r+sbr^l))(a2r^2+l2)1/2|〈ℒy,y〉|111(a2r^2+l2)1/2+d^r^]β-1∥y∥*2.
In addition, from (174), we get
(189)|∥T3(y)∥02-∥y∥*2|=|∑t=nN+n‍(w~-1(t)w(t)y(t))*w~(t)11111111×(w~-1(t)w(t)y(t))-∑t=nN+n‍y*(t)w(t)y(t)|=|∑t=nN+n‍y*(t)(w(t)w~-1(t)w(t)-w(t))y(t)|≤∑t=nN+n‍∥y(t)∥2∥w(t)∥∥w~-1(t)∥∥w(t)-w~(t)∥,
which, together with (113), yields that
(190)|∥T3(y)∥02-∥y∥*2|≤2d(w+1)d-1ww0-1ε∑t=nN+n‍∥y(t)∥2=G4ε∑t=nN+n‍∥y(t)∥2≤G4β-1ε∥y∥*2.
By the assumptionε≤βw0/(4d(w+1)d-1w), one can easily obtain
(191)∥y∥*2≤2∥T3(y)∥02.
With a similar argument to that used in the proof of Proposition11, from (174) and (184) one can get that
(192)|〈ℒ~(T3(y)),T3(y)〉0-〈ℒy,y〉|=|∑t=nN+n-i‍(Y*(n)W1W~1-1L3LS1*+Y*(N+1)111111×W2W~2-1L4A1-1(R2*+B1LS2*))11111×Ω*-1diag⁡{A,-L}(W~1-1W1Y(n)W~2-1W2Y(N+1))11111-Y*(n)L3Y(0)-Y*(N+1)L4Y(N+n+1)11111+∑t=nN+n‍y*(t)(w(t)w~-1(t)D~0(t)w~-1(t)w(t)11111111111111111-D0(t)(w(t)w~-1(t)D~0(t)w~-1(t)w(t))y(t)11111+2Re{∑i=1n‍∑t=nN+n-i‍y*(t)(w(t)w~-1(t)D~i(t)w~-11111111111111111111111111×(t+i)w(t+i)1111111111111111111111111-Di(t)(w(t)w~-1(t)D~i(t)w~-1)y(t+i)∑t=nN+n-i‍}|≤M1∥y∥*2ε,
which, together with (180) and (188)–(191), implies that
(193)|R-(T3(y))-R(y)|≤2(M1+M2)ε.
By Theorem4, we have
(194)|λ-k-λk|≤2(M1+M2)ε,1≤k≤(N+1)d.
This completes the proof.Remark 17.
Sinceε*in Theorem 16 is greater than ε0 in Theorem 12, the perturbed amplitude in Theorem 16 is even bigger.Remark 18.
The error estimate of eigenvalues of the special perturbed problem(1)′-(2) can be deduced from the proof of Theorem 12. Here, we give the proof instead of using the method of the space transformation T1 from L^[0,N+2n] into X. The proof here is simpler and more direct.Remark 19.
The estimate obtained in Theorem16 does not involve γ of (49), so we do not need to calculate the eigenvalues of matrix W when Theorem 16 is applied.



---

*Source: 437506-2014-02-17.xml*