# General Explicit Solution of Planar Weakly Delayed Linear Discrete Systems and Pasting Its Solutions

Authors: Josef Diblík; Hana Halfarová
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

Planar linear discrete systems with constant coefficients and delaysx(k+1)=Ax(k)+∑l=1n‍Blxl(k-ml) are considered where k∈ℤ0∞:={0,1,…,∞}, m1,m2,…,mn are constant integer delays, 0<m1<m2<⋯<mn, A,B1,…,Bn are constant 2×2 matrices, and x:ℤ-mn∞→ℝ2. It is assumed that the considered system is weakly delayed. The characteristic equations of such systems are identical with those for the same systems but without delayed terms. In this case, after several steps, the space of solutions with a given starting dimension 2(mn+1) is pasted into a space with a dimension less than the starting one. In a sense, this situation is analogous to one known in the theory of linear differential systems with constant coefficients and special delays when the initially infinite dimensional space of solutions on the initial interval turns (after several steps) into a finite dimensional set of solutions. For every possible case, explicit general solutions are constructed and, finally, results on the dimensionality of the space of solutions are obtained.


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## Body

## 1. Introduction

### 1.1. Preliminary Notions and Properties

We use the following notation: for integerss, q, s≤q, we define ℤsq:={s,s+1,…,q}, where s=-∞ or q=∞ is admitted, too. Throughout this paper, using notation ℤsq, we always assume s≤q. In the paper, we deal with the discrete planar system
(1)x(k+1)=Ax(k)+∑l=1n‍Blxl(k-ml),
where m1,m2,…,mn are constant integer delays, 0<m1<m2<⋯<mn, k∈ℤ0∞, A,B1,...,Bn are constant 2 × 2 matrices, A=(aij), Bl=(bijl), i,j=1,2, l=1,2,…,n, and x:ℤ-mn∞→ℝ2. Throughout the paper, we assume that
(2)Bl≠Θ,
where l=1,2,…,n and Θ is 2 × 2 zero matrix. Together with (1), we consider an initial (Cauchy) problem
(3)x(k)=φ(k),
where k=-mn,-mn+1,…,0 with φ:ℤ-mn0→ℝ2. Theexistence anduniqueness of the solution of the initial problem (1), (3) on ℤ-mn∞ are obvious. We recall that thesolutionx:ℤ-mn∞→ℝ2 of (1), (3) is defined as aninfinite sequence(4){x(-mn)=φ(-mn),x(-mn+1)=φ(-mn+1),…,x(0)=φ(0),x(1),x(2),…,x(k),…x(-mn)=φ(-mn),}
such that, for any k∈ℤ0∞, equality (1) holds.The space of all initial data (3) with φ:ℤ-mn0→ℝ2 is obviously 2(mn+1)-dimensional. Below, we describe the fact that, among system (1), there are such systems that their space of solutions, being initially 2(mn+1)-dimensional, on a reduced interval turns into a space having a dimension less than 2(mn+1). The problem under consideration (pasting property of solutions) is exactly formulated in Section 1.4.


### 1.2. Weakly Delayed Systems

We consider system (1) and look for a solution having the form x(k)=ξλk, where k∈ℤ-mn∞, λ=constant with λ≠0, and ξ=(ξ1,ξ2)T is a nonzero constant vector. The usual procedure leads to a characteristic equation
(5)D∶=det⁡(A+∑l=1n‍λ-mlBl-λI)=0,
where I is the unit 2 × 2 matrix. Together with (1), we consider a system with the terms containing delays omitted:
(6)x(k+1)=Ax(k)
and its characteristic equation
(7)det⁡(A-λI)=0.Definition 1.
System (1) is called a weakly delayed system if characteristic equations (5), (7) corresponding to systems (1) and (6) are equal, that is, if, for every λ∈ℂ∖{0},
(8)D=det⁡(A+∑l=1n‍λ-mlBl-λI)=det⁡(A-λI).
We consider a linear transformation
(9)x(k)=𝒮y(k),
with a nonsingular 2 × 2 matrix 𝒮, then the discrete system for y is
(10)y(k+1)=A𝒮y(k)+∑l=1n‍B𝒮ly(k-ml),
with A𝒮=𝒮-1A𝒮, B𝒮l=𝒮-1Bl𝒮, where l=1,2,…,n. We show that a system’s property of being one weakly delayed is preserved by every nonsingular linear transformation.Lemma 2.
If system (1) is a weakly delayed system, then its arbitrary linear nonsingular transformation (9) again leads to a weakly delayed system (10).Proof.
It is easy to show that(11)det⁡(A𝒮+∑l=1n‍λ-mlB𝒮l-λI)=det⁡⁡(A𝒮-λI)
holds since
(12)det⁡(A𝒮+∑l=1n‍λ-mlB𝒮l-λI)=det⁡(A𝒮+λ-m1B𝒮1+λ-m2B𝒮2+⋯+λ-mnB𝒮n-λIλ-m1B𝒮1)=det⁡(𝒮-1A𝒮+λ-m1𝒮-1B1𝒮+λ-m2𝒮-1B2𝒮+⋯+λ-mn𝒮-1Bn𝒮-λ𝒮-1I𝒮λ-m1𝒮-1B1𝒮)=det⁡[𝒮-1(A+λ-m1B1+λ-m2B2+⋯+λ-mnBn-λIλ-m1B1)𝒮λ-m1B1]=det⁡𝒮-1det⁡(A+λ-m1B1+λ-m2B2+⋯+λ-mnBn-λIλ-m2B2)det⁡𝒮=det⁡(A+λ-m1B1+λ-m2B2+⋯+λ-mnBn-λIλ-m2B2)=det⁡(A+∑l=1n‍λ-mlBl-λI),det⁡(A𝒮-λI)=det⁡(𝒮-1A𝒮-λ𝒮-1I𝒮)=det⁡[𝒮-1(A-λI)𝒮]=det⁡𝒮-1det⁡(A-λI)det⁡𝒮=det⁡(A-λI),det⁡(A+∑l=1n‍λ-mlBl-λI)=det⁡(A-λI);
that is, equality (8) is assumed.


### 1.3. Necessary and Sufficient Conditions Determining Weakly Delayed Systems

In the next theorem, we give conditions, in terms of determinants, indicating whether a system is weakly delayed.Theorem 3.
System (1) is a weakly delayed system if and only if the following 3n+n(n-1)/2 conditions hold simultaneously:
(13)b11l+b22l=0,(14)|b11lb12lb21lb22l|=0,(15)|a11a12b21lb22l|+|b11lb12la21a22|=0,(16)|b11lb12lb21vb22v|+|b11vb12vb21lb22l|=0,
where l,v=1,2,…,n and v>l.Proof.
We start with computing determinantD defined by (5). We get
(17)D=|D1D2D3D4|,
where
(18)D1=a11+b111λ-m1+b112λ-m2+⋯+b11nλ-mn-λ,D2=a12+b121λ-m1+b122λ-m2+⋯+b12nλ-mn,D3=a21+b211λ-m1+b212λ-m2+⋯+b21nλ-mn,D4=a22+b221λ-m1+b222λ-m2+⋯+b22nλ-mn-λ.
Expanding the determinant on the right-hand side along summands of the first column, we get(19)D=|a11a12+b121λ-m1+b122λ-m2+⋯+b12nλ-mna21a22+b221λ-m1+b222λ-m2+⋯+b22nλ-mn-λ|+λ-m1|b111a12+b121λ-m1+b122λ-m2+⋯+b12nλ-mnb211a22+b221λ-m1+b222λ-m2+⋯+b22nλ-mn-λ|+λ-m2|b112a12+b121λ-m1+b122λ-m2+⋯+b12nλ-mnb212a22+b221λ-m1+b222λ-m2+⋯+b22nλ-mn-λ|+⋯+λ-mn|b11na12+b121λ-m1+b122λ-m2+⋯+b12nλ-mnb21na22+b221λ-m1+b222λ-m2+⋯+b22nλ-mn-λ|+λ|-1a12+b121λ-m1+b122λ-m2+⋯+b12nλ-mn0a22+b221λ-m1+b222λ-m2+⋯+b22nλ-mn-λ|.
Expanding each of the above determinants along summands of the second column, we have(20)D=|a11a12a21a22|+λ-m1|a11b121a21b221|+λ-m2|a11b122a21b222|+⋯+λ-mn|a11b12na21b22n|+λ|a110a21-1|+λ-m1[|b111a12b211a22|+λ-m1|b111b121b211b221|+λ-m2|b111b122b211b222|+⋯+λ-mn|b111b12nb211b22n|+λ|b1110b211-1|]+λ-m2[|b112a12b212a22|+λ-m1|b112b121b212b221|+λ-m2|b112b122b212b222|+⋯+λ-mn|b112b12nb212b22n|+λ|b1120b212-1|]+⋯+λ-mn[|b11na12b21na22|+λ-m1|b11nb121b21nb221|+λ-m2|b11nb122b21nb222|+⋯+λ-mn|b11nb12nb21nb22n|+λ|b11n0b21n-1|]+λ[|-1a120a22|+λ-m1|-1b1210b221|+λ-m2|-1b1220b222|+⋯+λ-mn|-1b12n0b22n|+λ|-100-1|].
After simplification, we get(21)D=|a11-λa12a21a22-λ|-λ-m1+1(b111+b221)-λ-m2+1(b112+b222)+⋯-λ-mn+1(b11n+b22n)+λ-m1[|a11a12b211b221|+|b111b121a21a22|]+λ-m2[|a11a12b212b222|+|b112b122a21a22|]+⋯+λ-mn[|a11a12b21nb22n|+|b11nb12na21a22|]+λ-m1-m2[|b111b121b212b222|+|b112b122b211b221|]+⋯+λ-m1-mn[|b111b121b21nb22n|+|b11nb12nb211b221|]+λ-m2-m3[|b112b122b213b223|+|b113b123b212b222|]+⋯+λ-m2-mn[|b112b122b21nb22n|+|b11nb12nb212b222|]+⋯+λ-mn-1-mn[|b11n-1b12n-1b21nb22n|+|b11nb12nb21n-1b22n-1|]+λ-2m1|b111b121b211b221|+λ-2m2|b112b122b212b222|+⋯+λ-2mn|b11nb12nb21nb22n|.
Now we see that for (8) to hold; that is,
(22)D=det⁡(A+∑l=1n‍λ-mlBl-λI)=det⁡(A-λI)=|a11-λa12a21a22-λ|,
conditions (13)–(16) are both necessary and sufficient.Lemma 4.
Conditions (13)–(16) are equivalent to
(23)tr⁡Bl=det⁡Bl=0,(24)det⁡(A+Bl)=det⁡A,(25)det⁡(Bl+Bv)=0,
where l,v=1,2,…,n and v>l.Proof.
(I) We show that assumptions (13)–(16) imply (23)–(25). It is obvious that condition (23) is equivalent to (13), (14). Now we consider
(26)det⁡(A+Bl)=|a11+b11la12+b12la21+b21la22+b22l|.
Expanding the determinant on the right-hand side along summands of the first column and then expanding each of the determinants along summands of the second column, we have
(27)det⁡(A+Bl)=|a11a12+b12la21a22+b22l|+|b11la12+b12lb21la22+b22l|=|a11a12a21a22|+|a11b12la21b22l|+|b11la12b21la22|+|b11lb12lb21lb22l|=[dueto(15)and(16)]=det⁡A.
Now we consider(28)det⁡(Bl+Bv)=|b11l+b11vb12l+b12vb21l+b21vb22l+b22v|.
Expanding the determinant on the right-hand side along summands of the first column and then expanding each of the determinants along summands of the second column, we have
(29)det⁡(Bl+Bv)=|b11lb12l+b12vb21lb22l+b22v|+|b11vb12l+b12vb21vb22l+b22v|=|b11lb12lb21lb22l|+|b11lb12vb21lb22v|+|b11vb12lb21vb22l|+|b11vb12vb21vb22v|=[dueto(15),(17)]=0.
(II) Now we prove that assumptions (23)–(25) imply (13) and (16). Due to equivalence of (13) and (14) with (23), it remains to be shown that (23)–(25) imply (15) and (16).
If (24) holds, then, from computations in (27), we see that
(30)|a11b12la21b22l|+|b11la12b21la22|+|b11lb12lb21lb22l|=0,
and because of (23) we get (15).
Finally, we show that (23) and (25) imply (16). From (29) (using (23)) we get
(31)det⁡(Bl+Bv)=|b11lb12vb21lb22v|+|b11vb12lb21vb22l|=0,
that is, (16) holds.


### 1.4. Problem under Consideration

The aim of this paper is to give explicit formulas for solutions of weakly delayed systems and to show that, after several steps, the dimension of the space of all solutions, being initially equal to the dimension2(mn+1) of the space of initial data (3) generated by discrete functions φ, is reduced to a dimension less than the initial one on an interval of the form ℤs∞ with an s>0. In other words, we will show that the 2(mn+1)-dimensional space of all solutions of (1) is pasted to a less-dimensional space of solutions on ℤs∞. This problem is solved directly by explicitly computing the corresponding solutions of the Cauchy problems with each of the cases arising being considered. The underlying idea for such investigation is simple. If (1) is a weakly delayed system, then the corresponding characteristic equation has only two eigenvalues instead of 2(mn+1) eigenvalues in the case of systems with nonweak delays. This explains why the dimension of the space of solutions becomes less than the initial one. The final results (Theorems 10–13) provide the dimension of the space of solutions. Our results generalize the results in [1, 2], where system (1) with n=1 and n=2 was analyzed.


### 1.5. Auxiliary Formula

For the reader’s convenience, we recall one explicit formula (see, e.g., [3]) for the solutions of linear scalar discrete nondelayed equations used in this paper. We consider initial-value problem for the first order linear discrete nonhomogeneous equation
(32)w(k+1)=aw(k)+g(k),w(k0)=w0,k∈ℤk0∞,
with a∈ℂ and g:ℤk0∞→ℂ. Then, it is easy to verify that unique solution of this problem is
(33)w(k)=ak-k0w0+∑r=k0k-1‍ak-1-rg(r),k∈ℤk0+1∞.
Throughout the paper, we adopt the customary notation for the sum: ∑i=ℓ+sℓ‍ℱ(i)=0, where ℓ is an integer, s is a positive integer, and “ℱ” denotes the function considered independently of whether it is defined for indicated arguments or not.Note that the formula (33) is used many times in recent literature to analyze asymptotic properties of solutions of various classes of difference equations, including nonlinear equations. We refer, for example, to [4–8] and to relevant references therein.


## 1.1. Preliminary Notions and Properties

We use the following notation: for integerss, q, s≤q, we define ℤsq:={s,s+1,…,q}, where s=-∞ or q=∞ is admitted, too. Throughout this paper, using notation ℤsq, we always assume s≤q. In the paper, we deal with the discrete planar system
(1)x(k+1)=Ax(k)+∑l=1n‍Blxl(k-ml),
where m1,m2,…,mn are constant integer delays, 0<m1<m2<⋯<mn, k∈ℤ0∞, A,B1,...,Bn are constant 2 × 2 matrices, A=(aij), Bl=(bijl), i,j=1,2, l=1,2,…,n, and x:ℤ-mn∞→ℝ2. Throughout the paper, we assume that
(2)Bl≠Θ,
where l=1,2,…,n and Θ is 2 × 2 zero matrix. Together with (1), we consider an initial (Cauchy) problem
(3)x(k)=φ(k),
where k=-mn,-mn+1,…,0 with φ:ℤ-mn0→ℝ2. Theexistence anduniqueness of the solution of the initial problem (1), (3) on ℤ-mn∞ are obvious. We recall that thesolutionx:ℤ-mn∞→ℝ2 of (1), (3) is defined as aninfinite sequence(4){x(-mn)=φ(-mn),x(-mn+1)=φ(-mn+1),…,x(0)=φ(0),x(1),x(2),…,x(k),…x(-mn)=φ(-mn),}
such that, for any k∈ℤ0∞, equality (1) holds.The space of all initial data (3) with φ:ℤ-mn0→ℝ2 is obviously 2(mn+1)-dimensional. Below, we describe the fact that, among system (1), there are such systems that their space of solutions, being initially 2(mn+1)-dimensional, on a reduced interval turns into a space having a dimension less than 2(mn+1). The problem under consideration (pasting property of solutions) is exactly formulated in Section 1.4.


## 1.2. Weakly Delayed Systems

We consider system (1) and look for a solution having the form x(k)=ξλk, where k∈ℤ-mn∞, λ=constant with λ≠0, and ξ=(ξ1,ξ2)T is a nonzero constant vector. The usual procedure leads to a characteristic equation
(5)D∶=det⁡(A+∑l=1n‍λ-mlBl-λI)=0,
where I is the unit 2 × 2 matrix. Together with (1), we consider a system with the terms containing delays omitted:
(6)x(k+1)=Ax(k)
and its characteristic equation
(7)det⁡(A-λI)=0.Definition 1.
System (1) is called a weakly delayed system if characteristic equations (5), (7) corresponding to systems (1) and (6) are equal, that is, if, for every λ∈ℂ∖{0},
(8)D=det⁡(A+∑l=1n‍λ-mlBl-λI)=det⁡(A-λI).
We consider a linear transformation
(9)x(k)=𝒮y(k),
with a nonsingular 2 × 2 matrix 𝒮, then the discrete system for y is
(10)y(k+1)=A𝒮y(k)+∑l=1n‍B𝒮ly(k-ml),
with A𝒮=𝒮-1A𝒮, B𝒮l=𝒮-1Bl𝒮, where l=1,2,…,n. We show that a system’s property of being one weakly delayed is preserved by every nonsingular linear transformation.Lemma 2.
If system (1) is a weakly delayed system, then its arbitrary linear nonsingular transformation (9) again leads to a weakly delayed system (10).Proof.
It is easy to show that(11)det⁡(A𝒮+∑l=1n‍λ-mlB𝒮l-λI)=det⁡⁡(A𝒮-λI)
holds since
(12)det⁡(A𝒮+∑l=1n‍λ-mlB𝒮l-λI)=det⁡(A𝒮+λ-m1B𝒮1+λ-m2B𝒮2+⋯+λ-mnB𝒮n-λIλ-m1B𝒮1)=det⁡(𝒮-1A𝒮+λ-m1𝒮-1B1𝒮+λ-m2𝒮-1B2𝒮+⋯+λ-mn𝒮-1Bn𝒮-λ𝒮-1I𝒮λ-m1𝒮-1B1𝒮)=det⁡[𝒮-1(A+λ-m1B1+λ-m2B2+⋯+λ-mnBn-λIλ-m1B1)𝒮λ-m1B1]=det⁡𝒮-1det⁡(A+λ-m1B1+λ-m2B2+⋯+λ-mnBn-λIλ-m2B2)det⁡𝒮=det⁡(A+λ-m1B1+λ-m2B2+⋯+λ-mnBn-λIλ-m2B2)=det⁡(A+∑l=1n‍λ-mlBl-λI),det⁡(A𝒮-λI)=det⁡(𝒮-1A𝒮-λ𝒮-1I𝒮)=det⁡[𝒮-1(A-λI)𝒮]=det⁡𝒮-1det⁡(A-λI)det⁡𝒮=det⁡(A-λI),det⁡(A+∑l=1n‍λ-mlBl-λI)=det⁡(A-λI);
that is, equality (8) is assumed.


## 1.3. Necessary and Sufficient Conditions Determining Weakly Delayed Systems

In the next theorem, we give conditions, in terms of determinants, indicating whether a system is weakly delayed.Theorem 3.
System (1) is a weakly delayed system if and only if the following 3n+n(n-1)/2 conditions hold simultaneously:
(13)b11l+b22l=0,(14)|b11lb12lb21lb22l|=0,(15)|a11a12b21lb22l|+|b11lb12la21a22|=0,(16)|b11lb12lb21vb22v|+|b11vb12vb21lb22l|=0,
where l,v=1,2,…,n and v>l.Proof.
We start with computing determinantD defined by (5). We get
(17)D=|D1D2D3D4|,
where
(18)D1=a11+b111λ-m1+b112λ-m2+⋯+b11nλ-mn-λ,D2=a12+b121λ-m1+b122λ-m2+⋯+b12nλ-mn,D3=a21+b211λ-m1+b212λ-m2+⋯+b21nλ-mn,D4=a22+b221λ-m1+b222λ-m2+⋯+b22nλ-mn-λ.
Expanding the determinant on the right-hand side along summands of the first column, we get(19)D=|a11a12+b121λ-m1+b122λ-m2+⋯+b12nλ-mna21a22+b221λ-m1+b222λ-m2+⋯+b22nλ-mn-λ|+λ-m1|b111a12+b121λ-m1+b122λ-m2+⋯+b12nλ-mnb211a22+b221λ-m1+b222λ-m2+⋯+b22nλ-mn-λ|+λ-m2|b112a12+b121λ-m1+b122λ-m2+⋯+b12nλ-mnb212a22+b221λ-m1+b222λ-m2+⋯+b22nλ-mn-λ|+⋯+λ-mn|b11na12+b121λ-m1+b122λ-m2+⋯+b12nλ-mnb21na22+b221λ-m1+b222λ-m2+⋯+b22nλ-mn-λ|+λ|-1a12+b121λ-m1+b122λ-m2+⋯+b12nλ-mn0a22+b221λ-m1+b222λ-m2+⋯+b22nλ-mn-λ|.
Expanding each of the above determinants along summands of the second column, we have(20)D=|a11a12a21a22|+λ-m1|a11b121a21b221|+λ-m2|a11b122a21b222|+⋯+λ-mn|a11b12na21b22n|+λ|a110a21-1|+λ-m1[|b111a12b211a22|+λ-m1|b111b121b211b221|+λ-m2|b111b122b211b222|+⋯+λ-mn|b111b12nb211b22n|+λ|b1110b211-1|]+λ-m2[|b112a12b212a22|+λ-m1|b112b121b212b221|+λ-m2|b112b122b212b222|+⋯+λ-mn|b112b12nb212b22n|+λ|b1120b212-1|]+⋯+λ-mn[|b11na12b21na22|+λ-m1|b11nb121b21nb221|+λ-m2|b11nb122b21nb222|+⋯+λ-mn|b11nb12nb21nb22n|+λ|b11n0b21n-1|]+λ[|-1a120a22|+λ-m1|-1b1210b221|+λ-m2|-1b1220b222|+⋯+λ-mn|-1b12n0b22n|+λ|-100-1|].
After simplification, we get(21)D=|a11-λa12a21a22-λ|-λ-m1+1(b111+b221)-λ-m2+1(b112+b222)+⋯-λ-mn+1(b11n+b22n)+λ-m1[|a11a12b211b221|+|b111b121a21a22|]+λ-m2[|a11a12b212b222|+|b112b122a21a22|]+⋯+λ-mn[|a11a12b21nb22n|+|b11nb12na21a22|]+λ-m1-m2[|b111b121b212b222|+|b112b122b211b221|]+⋯+λ-m1-mn[|b111b121b21nb22n|+|b11nb12nb211b221|]+λ-m2-m3[|b112b122b213b223|+|b113b123b212b222|]+⋯+λ-m2-mn[|b112b122b21nb22n|+|b11nb12nb212b222|]+⋯+λ-mn-1-mn[|b11n-1b12n-1b21nb22n|+|b11nb12nb21n-1b22n-1|]+λ-2m1|b111b121b211b221|+λ-2m2|b112b122b212b222|+⋯+λ-2mn|b11nb12nb21nb22n|.
Now we see that for (8) to hold; that is,
(22)D=det⁡(A+∑l=1n‍λ-mlBl-λI)=det⁡(A-λI)=|a11-λa12a21a22-λ|,
conditions (13)–(16) are both necessary and sufficient.Lemma 4.
Conditions (13)–(16) are equivalent to
(23)tr⁡Bl=det⁡Bl=0,(24)det⁡(A+Bl)=det⁡A,(25)det⁡(Bl+Bv)=0,
where l,v=1,2,…,n and v>l.Proof.
(I) We show that assumptions (13)–(16) imply (23)–(25). It is obvious that condition (23) is equivalent to (13), (14). Now we consider
(26)det⁡(A+Bl)=|a11+b11la12+b12la21+b21la22+b22l|.
Expanding the determinant on the right-hand side along summands of the first column and then expanding each of the determinants along summands of the second column, we have
(27)det⁡(A+Bl)=|a11a12+b12la21a22+b22l|+|b11la12+b12lb21la22+b22l|=|a11a12a21a22|+|a11b12la21b22l|+|b11la12b21la22|+|b11lb12lb21lb22l|=[dueto(15)and(16)]=det⁡A.
Now we consider(28)det⁡(Bl+Bv)=|b11l+b11vb12l+b12vb21l+b21vb22l+b22v|.
Expanding the determinant on the right-hand side along summands of the first column and then expanding each of the determinants along summands of the second column, we have
(29)det⁡(Bl+Bv)=|b11lb12l+b12vb21lb22l+b22v|+|b11vb12l+b12vb21vb22l+b22v|=|b11lb12lb21lb22l|+|b11lb12vb21lb22v|+|b11vb12lb21vb22l|+|b11vb12vb21vb22v|=[dueto(15),(17)]=0.
(II) Now we prove that assumptions (23)–(25) imply (13) and (16). Due to equivalence of (13) and (14) with (23), it remains to be shown that (23)–(25) imply (15) and (16).
If (24) holds, then, from computations in (27), we see that
(30)|a11b12la21b22l|+|b11la12b21la22|+|b11lb12lb21lb22l|=0,
and because of (23) we get (15).
Finally, we show that (23) and (25) imply (16). From (29) (using (23)) we get
(31)det⁡(Bl+Bv)=|b11lb12vb21lb22v|+|b11vb12lb21vb22l|=0,
that is, (16) holds.


## 1.4. Problem under Consideration

The aim of this paper is to give explicit formulas for solutions of weakly delayed systems and to show that, after several steps, the dimension of the space of all solutions, being initially equal to the dimension2(mn+1) of the space of initial data (3) generated by discrete functions φ, is reduced to a dimension less than the initial one on an interval of the form ℤs∞ with an s>0. In other words, we will show that the 2(mn+1)-dimensional space of all solutions of (1) is pasted to a less-dimensional space of solutions on ℤs∞. This problem is solved directly by explicitly computing the corresponding solutions of the Cauchy problems with each of the cases arising being considered. The underlying idea for such investigation is simple. If (1) is a weakly delayed system, then the corresponding characteristic equation has only two eigenvalues instead of 2(mn+1) eigenvalues in the case of systems with nonweak delays. This explains why the dimension of the space of solutions becomes less than the initial one. The final results (Theorems 10–13) provide the dimension of the space of solutions. Our results generalize the results in [1, 2], where system (1) with n=1 and n=2 was analyzed.


## 1.5. Auxiliary Formula

For the reader’s convenience, we recall one explicit formula (see, e.g., [3]) for the solutions of linear scalar discrete nondelayed equations used in this paper. We consider initial-value problem for the first order linear discrete nonhomogeneous equation
(32)w(k+1)=aw(k)+g(k),w(k0)=w0,k∈ℤk0∞,
with a∈ℂ and g:ℤk0∞→ℂ. Then, it is easy to verify that unique solution of this problem is
(33)w(k)=ak-k0w0+∑r=k0k-1‍ak-1-rg(r),k∈ℤk0+1∞.
Throughout the paper, we adopt the customary notation for the sum: ∑i=ℓ+sℓ‍ℱ(i)=0, where ℓ is an integer, s is a positive integer, and “ℱ” denotes the function considered independently of whether it is defined for indicated arguments or not.Note that the formula (33) is used many times in recent literature to analyze asymptotic properties of solutions of various classes of difference equations, including nonlinear equations. We refer, for example, to [4–8] and to relevant references therein.


## 2. General Solution of Weakly Delayed System

If (8) holds, then (5) and (7) have only two (and the same) roots simultaneously. In order to prove the properties of the family of solutions of (1) formulated in the introduction, we will discuss each combination of roots, that is, the cases of two real and distinct roots, a pair of complex conjugate roots, and, finally, a double real root.Although computations in Sections1.2 and 1.3 were performed under assumption that λ≠0, results of this part remain valid also if one or both roots of characteristic equation (7) are zero.


### 2.1. Jordan Forms of the MatrixA and Corresponding Solutions of Problem (1) and (3)

It is known that, for every matrixA, there exists a nonsingular matrix S transforming it to the corresponding Jordan matrix form Λ. This means that
(34)Λ=S-1AS,
where Λ has the following four possible forms (denoted below as Λ1,Λ2,Λ3,Λ4), depending on the roots of the characteristic equation (7), that is, on the roots of
(35)λ2-(a11+a22)λ+(a11a22-a12a21)=0.
If (35) has two real distinct roots λ1, λ2, then
(36)Λ1=(λ100λ2),
if the roots are complex conjugate, that is, λ1,2=p±iq with q≠0, then
(37)Λ2=(pq-qp)
and, finally, in the case of one double real root λ1,2=λ, we have either
(38)Λ3=(λ00λ)
or
(39)Λ4=(λ10λ).
The transformation y(k)=S-1x(k) transforms (1) into a system
(40)y(k+1)=Λy(k)+∑l=1n‍B*ly(k-ml),k∈ℤ0∞
with B*l=S-1BlS, B*l=(bij*l), l=1,…,n, and i,j=1,2. Together with (40), we consider an initial problem
(41)y(k)=φ*(k),k∈ℤ-mn0 with φ*:ℤ-mn0→ℝ2 where φ*(k)=S-1φ(k) is the initial function corresponding to the initial function φ in (3).Next, we consider all four possible cases (36)–(39) separately.We define(42)Φ1(k):=(0,φ1*(k))T,Φ2(k):=(φ2*(k),0)T,223232323223312232321111122223232k∈ℤ-mn0.
Assuming that (1) is a weakly delayed system, by Lemma 2, the system (40) is weakly delayed system again.


#### 2.1.1. Case (36) of Two Real Distinct Roots

In this case, we haveΛ=Λ1 and Λ1k=diag⁡(λ1k,λ2k). The necessary and sufficient conditions (13)–(16) for (40) turn into
(43)b11*l+b22*l=0,(44)|b11*lb12*lb21*lb22*l|=b11*lb22*l-b12*lb21*l=0,(45)|λ10b21*lb22*l|+|b11*lb12*l0λ2|=λ1b22*l+λ2b11*l=0,(46)|b11*lb12*lb21*vb22*v|+|b11*vb12*vb21*lb22*l|=0.
Since λ1≠λ2, (43) and (45) yield b11*l=b22*l=0, then, from (44), we get b12*lb21*l=0, so that either b21*l=0 or b12*l=0. In view of assumptions Bl≠Θ, l=1,2,…,n, we conclude that only the following cases I, II are possible:(I)
b11*l=b22*l=b21*l=0, b12*l≠0, l=1,2,…,n,(II)
b11*l=b22*l=b12*l=0, b21*l≠0, l=1,2,…,n.In Theorem5 both cases I, II are analyzed.Theorem 5.
Let (1) be a weakly delayed system and (35) has two real distinct roots λ1, λ2. If case (I) holds, then the solution of the initial problem (1), (3) is x(k)=Sy(k), k∈ℤ-mn∞, where y(k) has the form
(47)y(k)={φ*(k)ifk∈ℤ-mn0,Λ1kφ*(0)+∑r=0k-1‍λ1k-1-r[∑l=1n‍b12*lΦ2(r-ml)]ifk∈ℤ1m1+1,⋮Λ1kφ*(0)+∑r=0k-1‍λ1k-1-r[∑l=s+1n‍b12*lΦ2(r-ml)]+∑l=1s‍b12*l[∑r=ml+1k-1‍∑r=0ml‍λ1k-1-rΦ2(r-ml)+Φ2(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml]ifk∈ℤms+2ms+1+1,s=1,2,…,n-1,⋮Λ1kφ*(0)+∑l=1n‍b12*l[∑r=ml+1k-1‍∑r=0ml‍λ1k-1-rΦ2(r-ml)+Φ2(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml]ifk∈ℤmn+2∞.
If case (II) is true, then the solution of initial problem (1), (3) is x(k)=Sy(k), k∈ℤ-mn∞, where y(k) has the form
(48)y(k)={φ*(k)ifk∈ℤ-mn0,Λ1kφ*(0)+∑r=0k-1‍λ2k-1-r[∑l=1n‍b21*lΦ1(r-ml)]ifk∈ℤ1m1+1,⋮Λ1kφ*(0)+∑r=0k-1‍λ2k-1-r[∑l=s+1n‍b21*lΦ1(r-ml)]+∑l=1s‍b21*l[∑r=ml+1k-1∑r=0ml‍λ2k-1-rΦ1(r-ml)+Φ1(0)∑r=ml+1k-1‍λ1r-mlλ2k-1-r]0000000000000000000000ifk∈ℤms+2ms+1+1,00000000000000000000s=1,2,…,n-1,⋮Λ1kφ*(0)+∑l=1n‍b21*l[∑r=ml+1k-1λ1r-mlλ2k-1-r‍∑r=0ml‍λ2k-1-rΦ1(r-ml)+Φ1(0)∑r=ml+1k-1λ1r-mlλ2k-1-r‍]0000000000000000000000000ifk∈ℤmn+2∞.Proof.
If case (I) is true, then the transformed system (40) takes the form
(49)y1(k+1)=λ1y1(k)+∑l=1n‍b12*ly2(k-ml),(50)y2(k+1)=λ2y2(k),k∈ℤ0∞,
and if case (II) holds, then (40) takes the form
(51)y1(k+1)=λ1y1(k),(52)y2(k+1)=λ2y2(k)+∑l=1n‍b21*ly1(k-ml),k∈ℤ0∞.
We investigate only the initial problem (49), (50), (41) since the initial problem (51), (52), (41) can be examined in a similar way.
From (50), (41), we get
(53)y2(k)={φ2*(k)ifk∈ℤ-mn0,λ2kφ2*(0)ifk∈ℤ1∞,
then (49) becomes
(54)y1(k+1)={λ1y1(k)+∑l=1n‍b12*lφ2*(k-ml)ifk∈ℤ0m1,λ1y1(k)+b12*1λ2k-m1φ2*(0)+∑l=2n‍b12*lφ2*(k-ml)ifk∈ℤm1+1m2,λ1y1(k)+∑l=12‍b12*lλ2k-mlφ2*(0)+∑l=3n‍b12*lφ2*(k-ml)ifk∈ℤm2+1m3,⋮λ1y1(k)+∑l=1s‍b12*lλ2k-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(k-ml)ifk∈ℤms+1ms+1,s=3,4,…,n-1,⋮λ1y1(k)+∑l=1n‍b12*lλ2k-mlφ2*(0)ifk∈ℤmn+1∞.
First, we solve this equation for k∈ℤ0m1. This means that we consider the problem
(55)y1(k+1)=λ1y1(k)+∑l=1n‍b12*lφ2*(k-ml),k∈ℤ0m1,y1(0)=φ1*(0).
With the aid of formula (33), we get
(56)y1(k)=λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=1n‍b12*lφ2*(r-ml)],1111111111111111111111111k∈ℤ1m1+1.
Now we solve (54) for k∈ℤm1+1m2 with initial data deduced from (56); that is, we consider the problem
(57)y1(k+1)=λ1y1(k)+b12*1λ2k-m1φ2*(0)+∑l=2n‍b12*lφ2*(k-ml),k∈ℤm1+1m2,y1(m1+1)=λ1m1+1φ1*(0)+∑r=0m1‍λ1m1-r[∑l=1n‍b12*lφ2*(r-ml)].
Applying formula (33) we get (for k∈ℤm1+2m2+1)
(58)y1(k)=λ1k-(m1+1)y1(m1+1)+∑r=m1+1k-1‍λ1k-1-r[∑l=2n‍b12*lφ2*(r-ml)b12*1λ2r-m1φ2*(0)+∑r=m1+1k-1‍λ1k-1-r1212+∑l=2n‍b12*lφ2*(r-ml)]=λ1k-m1-1[[∑l=1n‍b12*lφ2*(r-ml)]λ1m1+1φ1*(0)+∑r=0m1‍λ1m1-r[∑l=1n‍b12*lφ2*(r-ml)]]+∑r=m1+1k-1‍λ1k-1-r[+∑l=2n‍b12*lφ2*(r-ml)b12*1λ2r-m1φ2*(0)+∑r=m1+1k-1‍λ1k-1-r++∑l=2n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0m1‍λ1k-1-r[∑l=1n‍b12*lφ2*(r-ml)]+∑r=m1+1k-1‍λ1k-1-r[+∑l=2n‍b12*lφ2*(r-ml)b12*1λ2r-m1φ2*(0)+∑r=m1+1k-1‍λ1k-1-r+++∑l=2n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1k-1‍∑r=0m1‍λ1k-1-rφ2*(r-m1)+b12*1+b12*1+φ2*(0)∑r=m1+1k-1‍λ1k-1-rλ2r-m1].
Now we solve (54) for k∈ℤm2+1m3 with initial data deduced from (58); that is, we consider the problem
(59)y1(k+1)=λ1y1(k)+∑l=12‍b12*lλ2k-mlφ2*(0)+∑l=3n‍b12*lφ2*(k-ml),k∈ℤm2+1m3,y1(m2+1)=λ1m2+1φ1*(0)+∑r=0m2‍λ1m2-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1m2‍∑r=m1+1m2‍λ1m2-rλ2r-m1∑r=0m1‍λ1m2-rφ2*(r-m1)+φ2*+φ2*+φ2*(0)∑r=m1+1m2‍λ1m2-rλ2r-m1].
Applying formula (33) yields (for k∈ℤm2+2m3+1)
(60)y1(k)=λ1k-(m2+1)y1(m2+1)+∑r=m2+1k-1‍λ1k-1-r[∑l=12‍b12*lλ2r-mlφ2*(0)+∑l=3n‍b12*lφ2*(r-ml)]=λ1k-m2-1[∑r=m1+1m2‍λ1m2+1φ1*(0)+∑r=0m2‍λ1m2-r[∑l=2n‍b12*lφ2*(r-ml)]111111111+b12*1[∑r=m1+1m2‍∑r=0m1‍λ1m2-rφ2*(r-m1)111111111111+φ2*(0)∑r=m1+1m2‍λ1m2-rλ2r-m1]∑r=m1+1m2‍]+∑r=m2+1k-1‍λ1k-1-r[∑l=12‍b12*lλ2r-mlφ2*(0)+∑l=3n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0m2‍λ1k-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1m2‍∑r=0m1‍λ1k-1-rφ2*(r-m1)1111111+φ2*(0)∑r=m1+1m2‍λ1k-1-rλ2r-m1]+∑r=m2+1k-1‍λ1k-1-r[∑l=12‍b12*lλ2r-mlφ2*(0)11111111111111+∑l=3n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=3n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1k-1‍∑r=0m1‍λ1k-1-rφ2*(r-m1)11111111+φ2*(0)∑r=m1+1k-1‍λ1k-1-rλ2r-m1]+b12*2[∑r=m2+1k-1‍∑r=m2+1k-1λ1k-1-rλ2r-m2∑r=0m2‍λ1k-1-rφ2*(r-m2)111111111+φ2*(0)∑r=m2+1k-1λ1k-1-rλ2r-m2].
From (56), (58), and (60) we deduce that expected form of the solution of the initial problem for k∈ℤms-1+1ms with initial data derived from the solution of previous equation for k∈ℤms-2+1ms-1 is
(61)y1(k)=λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=ml+1k-1‍λ1k-1-rλ2r-ml∑r=0ml‍λ1k-1-rφ2*(r-ml)+∑l=1s-1‍b12*l+∑l=1s-1‍+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml]+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml+ifk∈ℤms-1+2ms+1.
We solve (54) for k∈ℤms+1ms+1 with initial data deduced from (61); that is, we consider the problem
(62)y1(k+1)=λ1y1(k)+∑l=1s‍b12*lλ2k-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(k-ml),k∈ℤms+1ms+1,y1(ms+1)=λ1ms+1φ1*(0)+∑r=0ms‍λ1ms-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=ml+1ms‍λ1ms-rλ2r-ml∑r=0ml‍λ1ms-rφ2*(r-ml)+∑l=1s-1‍b12*l+∑l=1s-1‍+φ2*(0)∑r=ml+1ms‍λ1ms-rλ2r-ml].
Applying formula (33) yields (for k∈ℤms+2ms+1+1)
(63)y1(k)=λ1k-(ms+1)y1(ms+1)+∑r=ms+1k-1‍λ1k-1-r[∑l=1s‍b12*lλ2r-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(r-ml)]=λ1k-(ms+1)[[∑r=ml+1ms‍λ1ms-rλ2r-ml]λ1ms+1φ1*(0)+∑r=0ms‍λ1ms-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=ml+1ms‍λ1ms-rλ2r-ml∑r=0ml‍λ1ms-rφ2*(r-ml)111111111144444444+φ2*(0)∑r=ml+1ms‍λ1ms-rλ2r-ml]∑r=0ms‍λ1ms-r[∑l=sn‍b12*lφ2*(r-ml)]]+∑r=ms+1k-1‍λ1k-1-r[∑l=1s‍b12*lλ2r-mlφ2*(0)111111111144444+∑l=s+1n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0ms‍λ1k-1-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=ml+1ms‍λ1k-1-rλ2r-ml∑r=ml+1ms‍∑r=0ml‍λ1k-1-rφ2*(r-ml)11111111111+φ2*(0)∑r=ml+1ms‍λ1k-1-rλ2r-ml]+∑r=ms+1k-1‍λ1k-1-r[∑l=1s‍b12*lλ2r-mlφ2*(0)111111111144444+∑l=s+1n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0ms‍λ1k-1-r[∑l=s+1n‍b12*sφ2*(r-ms)11111111114444444444+∑l=s+1n‍b12*lφ2*(r-ml)]+b12*1[∑r=ml+1ms‍λ1k-1-rλ2r-ml∑r=0m1‍λ1k-1-rφ2*(r-m1)11111111114+φ2*(0)∑r=m1+1ms‍λ1k-1-rλ2r-m1]+b12*2[∑r=m2+1ms‍∑r=0m2‍λ1k-1-rφ2*(r-m2)111111114+φ2*(0)∑r=m2+1ms‍λ1k-1-rλ2r-m2]+⋯+b12*s-1[∑r=ms-1+1ms‍∑r=0ms-1‍λ1k-1-rφ2*(r-ms-1)11111111114+φ2*(0)∑r=ms-1+1ms‍λ1k-1-rλ2r-ms-1]+b12*1[φ2*(0)∑r=ms+1k-1‍λ1k-1-rλ2r-m1]+b12*2[φ2*(0)∑r=ms+1k-1‍λ1k-1-rλ2r-m2]+…+b12*s[φ2*(0)∑r=ms+1k-1‍λ1k-1-rλ2r-ms]+∑r=ms+1k-1‍λ1k-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+∑l=1s‍b12*l[∑r=ml+1k-1‍∑r=0ml‍λ1k-1-rφ2*(r-ml)11111111114+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml].
In the end we solve (54) for k∈ℤmn+1∞ with initial data deduced from (63); that is, we consider the problem
(64)y1(k+1)=λ1y1(k)+∑l=1n‍b12*lλ2k-mlφ2*(0),k∈ℤmn+1∞,y1(mn+1)=λ1mn+1φ1*(0)+∑r=0mn‍λ1mn-rb12*nφ2*(r-mn)+∑l=1n-1‍b12*l[∑r=ml+1mn‍λ1mn-rλ2r-ml∑r=0ml‍λ1mn-rφ2*(r-ml)+∑l=1n-1‍b12*l+222+φ2*(0)∑r=ml+1mn‍λ1mn-rλ2r-ml].
Applying formula (33) yields (for k∈ℤmn+2∞)
(65)y1(k)=λ1k-(mn+1)y1(mn+1)+∑r=mn+1k-1‍λ1k-1-r[∑l=1n‍b12*lλ2r-mlφ2*(0)]=λ1k-(mn+1)[[∑r=ml+1mn‍λ1mn-rλ2r-ml]λ1mn+1φ1*(0)+∑r=0mn‍λ1mn-rb12*nφ2*(r-mn)λ1k-(mn+1)λ1+∑l=1n-1‍b12*l[∑r=ml+1mn‍λ1mn-rλ2r-ml∑r=0ml‍λ1mn-rφ2*(r-ml)λ1k-(mn+1)1λ1k-(mn+1)λ1+φ2*(0)∑r=ml+1mn‍λ1mn-rλ2r-ml]]+∑r=mn+1k-1‍λ1k-1-r[∑l=1n‍b12*lλ2r-mlφ2*(0)]=λ1kφ1*(0)+∑r=0mn‍λ1k-1-rb12*nφ2*(r-mn)+∑l=1n-1‍b12*l[∑r=ml+1mn‍λ1k-1-rλ2r-ml∑r=0ml‍λ1k-1-rφ2*(r-ml)+∑l=1n-1‍b12*l+∑l=1n-1‍b12*l+φ2*(0)∑r=ml+1mn‍λ1k-1-rλ2r-ml]+∑r=mn+1k-1‍λ1k-1-r[∑l=1n‍b12*lλ2r-mlφ2*(0)]=λ1kφ1*(0)+∑r=0mn‍λ1k-1-rb12*nφ2*(r-mn)+b12*1[∑r=m1+1ms‍λ1k-1-rλ2r-m1∑r=0m1‍λ1k-1-rφ2*(r-m1)+b12*1+b12*1+φ2*(0)∑r=m1+1ms‍λ1k-1-rλ2r-m1]+b12*2[∑r=m2+1ms‍λ1k-1-rλ2r-m2∑r=0m2‍λ1k-1-rφ2*(r-m2)+b12*2+b12*2+φ2*(0)∑r=m2+1ms‍λ1k-1-rλ2r-m2]+⋯+b12*n-1[∑r=mn-1+1ms‍λ1k-1-rλ2r-mn-1∑r=0mn-1‍λ1k-1-rφ2*(r-mn-1)+b12*n-1+b12*n-1+φ2*(0)∑r=mn-1+1ms‍λ1k-1-rλ2r-mn-1]+b12*1[φ2*(0)∑r=ms+1k-1‍λ1k-1-rλ2r-m1]+b12*2[φ2*(0)∑r=ms+1k-1‍λ1k-1-rλ2r-m2]+⋯+b12*n[φ2*(0)∑r=mn+1k-1‍λ1k-1-rλ2r-mn]=λ1kφ1*(0)+∑l=1n‍b12*l[∑r=ml+1k-1‍λ1k-1-rλ2r-ml∑r=0ml‍λ1k-1-rφ2*(r-ml)∑l=1n‍b12*l∑l=1n‍b12*l∑l=1n‍b12*l∑l=1n‍+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml].
Summing up all particular cases (56)–(65) we have
(66)y1(k)={φ1*(k)ifk∈ℤ-mn0,λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=1n‍b12*lφ2*(r-ml)]00ifk∈ℤ1m1+1,λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1k-1‍∑r=0m1‍λ1k-1-rφ2*(r-m1)+φ2*(0)∑r=m1+1k-1‍λ1k-1-rλ2r-m1]0ifk∈ℤm1+2m2+1,λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=3n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1k-1‍∑r=0m1‍λ1k-1-rφ2*(r-m1)+φ2*(0)∑r=m1+1k-1‍λ1k-1-rλ2r-m1]+b12*2[∑r=m2+1k-1‍∑r=0m2‍λ1k-1-rφ2*(r-m2)+φ2*(0)∑r=m2+1k-1‍λ1k-1-rλ2r-m2]00000ifk∈ℤm2+2m3+1,⋮λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+∑l=1s‍b12*l[∑r=ml+1k-1‍∑r=0ml‍λ1k-1-rφ2*(r-ml)+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml]000ifk∈ℤms+2ms+1+1,⋮λ1kφ1*(0)+∑l=1n‍b12*l[∑r=ml+1k-1‍∑r=0ml‍λ1k-1-rφ2*(r-ml)+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml]00000ifk∈ℤmn+2∞.
Now, taking into account (42), formula (47) is a consequence of (53) and (66). Formula (48) can be proved in a similar way.
Finally, we note that both formulas (47), (48) remain valid for b12*l=b21*l=0. In this case, the transformed system (1) reduces to a system without delays. This possibility is excluded by condition (2).


#### 2.1.2. Case (37) of Two Complex Conjugate Roots

The necessary and sufficient conditions (13)–(16) take the forms (43), (44), (46), and
(67)|pqb21*lb22*l|+|b11*lb12*l-qp|=p(b11*l+b22*l)+q(b12*l-b21*l)=0,
where l,v=1,2,…,n and v>l.The system of conditions (43), (44), and (67) gives b12*l=b21*l, (b11*l)2=-(b12*l)2 and admits only one possibility; namely,
(68)b11*l=b22*l=b12*l=b21*l=0.Consequently,B*l=Θ, Bl=Θ.The initial problem (1), (3) reduces to a problem without delay
(69)x(k+1)=Ax(k),x(k)=φ(k),k∈ℤ-mn0
and, obviously,
(70)x(k)={φ(k)ifk∈ℤ-mn0,Akφ(0)ifk∈ℤ1∞.
From this discussion, the next theorem follows.Theorem 6.
There exists no weakly delayed system (1) if Λ has the form (37).Finally, we note that the assumption (2) alone excludes this case.


#### 2.1.3. Case (38) of Double Real Root

In this case we haveΛ=Λ3 and Λ3k=diag⁡(λk,λk). For (40), the necessary and sufficient conditions (13)–(16) are reduced to (43), (44), (46), and
(71)|λ0b21*lb22*l|+|b11*lb12*l0λ|=λ(b11*l+b22*l)=0,
where l=1,2,…,n.From (43), (44), and (71), we get b12*lb21*l=-(b11*l)2. From the condition (46) we get
(72)b11*lb22*v-b12*lb21*v+b22*lb11*v-b21*lb12*v=0,
where l,v=1,2,…,n and v>l. Multiplying (72) by b12*lb12*v, we have
(73)b11*lb22*vb12*lb12*v-(b12*l)2b21*vb12*v+b22*lb11*vb12*lb12*v-b21*lb12*l(b12*v)2=0.
Substituting b12*lb21*l=-(b11*l)2, b12*vb21*v=-(b11*v)2 into (73) and using (43) we obtain
(74)-b11*lb11*vb12*lb12*v+(b12*l)2(b11*v)2-b11*lb11*vb12*lb12*v+(b11*l)2(b12*v)2=0.
The equation (74) can be written as
(75)(b12*lb11*v-b12*vb11*l)2=0,b12*lb11*v=b12*vb11*l.Now we will analyse the two possible cases:b12*lb21*l=0 and b12*lb21*l≠0.For the caseb12*lb21*l=0, we have from (43), (44) that b11*l=b22*l=0 and b12*l=0 or b21*l=0. For b12*l=0 and b21*l≠0, condition (46) gives b12*v=0, where l,v=1,2,…,n and v>l. Then, from (43), (44) for l=v, we get b11*v=b22*v=0 and b21*v≠0.Forb21*l=0 and b12*l≠0, condition (46) gives b21*v=0, where l,v=1,2,…,n and v>l, then, from (43), (44) for l=v, we get b11*v=b22*v=0 and b12*v≠0.Now we discuss the caseb12*lb21*l≠0. From conditions (43), (44), we have b12*lb21*l=-(b11*l)2 and b11*lb22*l≠0. This yields b11*l≠0, b22*l≠0 and, from (75), we have b11*v≠0, b12*v≠0. By conditions (43), (44) for v=l, we get b22*v≠0, b21*v≠0.From the assumptionsBl≠Θ, we conclude that only the following cases ((I), (II), (III)) are possible:(I)
b11*l=b22*l=b21*l=0, b12*l≠0,(II)
b11*l=b22*l=b12*l=0, b21*l≠0,(III)
b12*lb21*l≠0,wherel=1,2,…,n.


#### 2.1.4. Caseb12*lb21*l=0

Theorem 7.
Let (1) be a weakly delayed system, (35) has a twofold root λ1,2=λ, b12*lb21*l=0 and the matrix Λ has the form (38). Then the solution of the initial problem (1), (3) is x(k)=Sy(k), k∈ℤ-mn∞, where in case b21*l=0, y(k) has the form
(76)y(k)={φ*(k)ifk∈ℤ-mn0,Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍b12*lΦ2(r-ml)]ifk∈ℤ1m1+1,⋮Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b12*lΦ2(r-ml)]+∑l=1s‍b12*l[∑r=0ml‍λk-1-rΦ2(r-ml)+(k-1-ml)λk-1-mlΦ2(0)∑r=0ml‍]ifk∈ℤms+2ms+1+1,s=1,2,…,n-1,⋮Λ3kφ*(0)+∑l=1n‍b12*l[∑r=0ml‍λk-1-rΦ2(r-ml)+(k-1-ml)λk-1-mlΦ2(0)∑r=0ml‍]ifk∈ℤmn+2∞.
If b12*l=0 is true then the solution of initial problem (1), (3) is x(k)=Sy(k), k∈ℤ-mn∞, where y(k) has the form
(77)y(k)={φ*(k)ifk∈ℤ-mn0,Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍b21*lΦ1(r-ml)]ifk∈ℤ1m1+1,⋮Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b21*lΦ1(r-ml)]+∑l=1s‍b21*l[∑r=0ml‍λk-1-rΦ1(r-ml)+(k-1-ml)λk-1-mlΦ1(0)∑r=0ml‍]ifk∈ℤms+2ms+1+1,s=1,2,…,n-1,⋮Λ3kφ*(0)+∑l=1n‍b21*l[∑r=0ml‍∑r=0ml‍λk-1-rΦ1(r-ml)+(k-1-ml)λk-1-mlΦ1(0)∑r=0ml‍]ifk∈ℤmn+2∞.Proof.
Case (I) means thatb12*l≠0. Then (40) turns into the system
(78)y1(k+1)=λy1(k)+∑l=1n‍b12*ly2(k-ml),k∈ℤ0∞y2(k+1)=λy2(k),
and, if b21*l≠0, (40) turns into the system
(79)y1(k+1)=λy1(k),y2(k+1)=λy2(k)+∑l=1n‍b21*ly1(k-ml),k∈ℤ0∞.
System (78) can be solved in much the same way as the systems (49), (50) if we put λ1=λ2=λ, and the discussion of the system (79) goes along the same lines as that of the systems (51), (52) with λ1=λ2=λ. Formulas (76) and (77) are consequences of (47), (48).


#### 2.1.5. Caseb12*lb21*l≠0

Fork∈ℤ-mn0, we define
(80)Φl*(k):=((b11*l)2b12*l[φ1*(k)+b12*1b11*1φ2*(k)]b11*l[φ1*(k)+b12*1b11*1φ2*(k)],-(b11*l)2b12*l[φ1*(k)+b12*1b11*1φ2*(k)])T.Theorem 8.
Let system (1) be a weakly delayed system, (35) admits two repeated roots λ1,2=λ, b12*lb21*l≠0 and the matrix Λ3 has the form (38). Then the solution of the initial problem (1), (3) is given by x(k)=Sy(k), k∈ℤ-mn∞, where y(k) has the form
(81)y(k)={φ*(k)ifk∈ℤ-mn0,Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍Φl*(r-ml)]ifk∈ℤ1m1+1,⋮Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍Φl*(r-ml)]+∑l=1s‍[∑r=0ml‍λk-1-rΦl*(r-ml)+(k-1-ml)λk-1-mlΦl*(0)∑r=0ml‍]ifk∈ℤms+2ms+1+1,s=1,2,…,n-1,⋮Λ3kφ*(0)+∑l=1n‍[∑r=0ml‍λk-1-rΦl*(r-ml)+(k-1-ml)λk-1-mlΦl*(0)∑r=0ml‍]ifk∈ℤmn+2∞.Proof.
In this case, all the entries ofB*l are nonzero and, from (43), (44), and (71), we get
(82)B*=(b11*lb12*l-(b11*l)2b12*l-b11*l),
where l=1,2,…,n, then, the system (40) reduces to
(83)y1(k+1)=λy1(k)+∑l=1n‍[b11*ly1(k-ml)+b12*ly2(k-ml)],(84)y2(k+1)=λy2(k)-∑l=1n‍[(b11*l)2b12*ly1(k-ml)+b11*ly2(k-ml)],
where k∈ℤ0∞. It is easy to see (multiplying (84) by b12*1/b11*1 and summing both equations) that
(85)y1(k+1)+b12*1b11*1y2(k+1)=λ[y1(k)+b12*1b11*1y2(k)],1111111111111k∈ℤ0∞.
Equation (85) is a homogeneous equation with respect to the unknown expression
(86)y1(k)+(b12*1b11*1)y2(k),
then, using (33), we obtain
(87)y1(k)+b12*1b11*1y2(k)={φ1*(k)+b12*1b11*1φ2*(k)ifk∈ℤ-mn0,λk[φ1*(0)+b12*1b11*1φ2*(0)]ifk∈ℤ1∞.
With the aid of (87), we rewrite the systems (83), (84) as follows:
(88)y1(k+1)={λy1(k)+∑l=1n‍b11*l[b12*1b11*1φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤ0m1,λy1(k)+b11*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=2n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤm1+1m2,λy1(k)+∑l=12‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=3n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤm2+1m3,⋮λy1(k)+∑l=1s‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=s+1n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤms+1ms+1,s=3,4,…,n-1,⋮λy1(k)+∑l=1n‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]ifk∈ℤmn+1∞,y2(k+1)={λy2(k)-∑l=1n‍(b11*l)2b12*l[b12*1b11*1φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤ0m1,λy2(k)-(b11*1)2b12*1λk-m1×[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(k-m1)+b12*1b11*1φ2*(k-m1)]ifk∈ℤm1+1m2,λy2(k)-∑l=12‍(b11*l)2b12*l[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=3n‍(b11*l)2b12*l[b12*1b11*1φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤm2+1m3,⋮λy2(k)-∑l=1s‍(b11*l)2b12*l[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=s+1n‍(b11*l)2b12*l[b12*1b11*1φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤms+1ms+1,s=3,4,…,n-1,⋮λy2(k)-∑l=1n‍(b11*l)2b12*lλk-ml×[φ1*(0)+b12*1b11*1φ2*(0)]ifk∈ℤmn+1∞.
First, we solve this system for k∈ℤ0m1 and consider the problems
(89)y1(k+1)=λy1(k)+∑l=1n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]11111111111111111111111000ifk∈ℤ0m1,y1(0)=φ1*(0),y2(k+1)=λy2(k)-∑l=1n‍(b11*l)2b12*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]111111111111111111111111111100ifk∈ℤ0m1,y2(0)=φ2*(0).
With the aid of formula (33), we get
(90)y1(k)=λkφ1*(0)+∑r=0k-1‍λk-1-r×(∑l=1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]),11111111111111111111111111111k∈ℤ1m1+1,(91)y2(k)=λkφ2*(0)-∑r=0k-1‍λk-1-r×(∑l=1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]),1111111111111111111111111111111k∈ℤ1m1+1.
Now we solve system (88) for k∈ℤm1+1m2; that is, we consider the problem (with initial data deduced from (90), (91))
(92)y1(k+1)=λy1(k)+b11*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=2n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]1111111111111111111111111ifk∈ℤm1+1m2,y1(m1+1)=λm1+1φ1*(0)+∑r=0m1‍λm1-r×(∑l=1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]),y2(k+1)=λy2(k)-(b11*1)2b12*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=2n‍(b11*l)2b12*l[φ1*(k-m1)+b12*1b11*1φ2*(k-m1)]1111111111111111111111111111ifk∈ℤm1+1m2,y2(m1+1)=λm1+1φ2*(0)-∑r=0m1‍λm1-r×(∑l=1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]).
Formula (33) yields (for k∈ℤm1+2m2+1)
(93)y1(k)=λk-(m1+1)y1(m1+1)+∑r=m1+1k-1‍λk-1-r×(∑l=2n‍∑l=2n‍b11*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=2n‍b11*l[φ1*(r-ml)b12*1b11*11111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍[φ1*(0)+b12*1b11*1φ2*(0)])=λk-m1-1[([+b12*1b11*1φ2*(r-ml)]∑l=1n‍b11*l)λm1+1φ1*(0)+∑r=0m1‍λm1-rλk-m1-1λk-m1×(∑l=1n‍∑l=1n‍b11*l[b12*1b11*1φ1*(r-ml)11111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1n‍)]+∑r=m1+1k-1‍λk-1-r(b11*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑r=m1+1k-1‍λk-1∑r=m1+1k-1‍+∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111+b12*1b11*1φ2*(r-ml)])=λkφ1*(0)+∑r=0m1‍λk-1-r(∑l=1n‍b11*l[b12*1b11*1φ1*(r-ml)1111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1n‍)+∑r=m1+1k-1‍λk-1-r(∑l=2n‍b11*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑r=m1+1k-1‍+∑r=m1+1k-1‍+∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111+b12*1b11*1φ2*(r-ml)][φ1*(0)+b12*1b11*1φ2*(0)]∑l=2n‍)=λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍)+∑r=0m1‍λk-1-rb11*1[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)b11*1λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]=λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍)+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)λk-1-m1[b12*1b11*1φ1*(0)11111111111111111111111111111+b12*1b11*1φ2*(0)]∑r=0m1‍),(94)y2(k)=λk-(m1+1)y2(m1+1)-∑r=m1+1k-1‍λk-1-r×((b11*1)2b12*1λr-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)11111111111111111+b12*1b11*1φ2*(r-ml)][φ1*(0)+b12*1b11*1φ2*(0)](b11*1)2b12*1)=λk-m1-1[(∑l=1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])λm1+1φ2*(0)-∑r=0m1‍λm1-r=λk-112=×(∑l=1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)1111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1n‍(b11*l)2b12*l)(∑l=1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])]-∑r=m1+1k-1‍λk-1-r((b11*1)2b12*1λr-m1[φ1*(0)+b12*1b11*1φ2*(0)]-∑r=m1+1k-1‍λk-1-r-12+∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111111+b12*1b11*1φ2*(r-ml)](b11*1)2b12*1λr-m1[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ2*(0)-∑r=0m1‍λk-1-r(∑l=1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1n‍∑l=1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml))-∑r=m1+1k-1‍λk-1-r((b11*1)2b12*1λr-m1[φ1*(0)+b12*1b11*1φ2*(0)]-∑r=m1+1k-1‍λk-1-r-12+∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111+b12*1b11*1φ2*(r-ml)](b11*1)2b12*1λr-m1[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍(b11*l)2b12*l)-∑r=0m1‍λk-1-r(b11*1)2b12*1[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]-(k-1-m1)λk-1-m1(b11*1)2b12*1[φ1*(0)+b12*1b11*1φ2*(0)]=λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111111+b12*1b11*1b12*lφ2*(r-ml)]∑l=2n‍(b11*l)2b12*l)-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]-(b11*1)2b12*1-+(k-1-m1)λk-1-m1×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍).
Now we solve (88) for k∈ℤm2+1m3; that is, we consider the problem (with initial data deduced from (93), (94))
(95)y1(k+1)=λy1(k)+∑l=12‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=3n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]+∑l=3n‍b11*l+∑l=3n‍b11*l+1∑l=3n‍b11*l11∑l=3n‍ifk∈ℤm2+1m3,y1(m2+1)=λm2+1φ1*(0)+∑r=0m2‍λm2-r×(∑l=2n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+b11*1(∑r=0m1‍λm2-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+b11*1+b11*1+(m2-m1)λm2-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍),y2(k+1)=λy2(k)-∑l=12‍(b11*l)2b12*l[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=3n‍(b11*l)2b12*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]-∑l=3n‍(b11*l)2b12*l-∑l=3n‍(b11*l)2b12*l111111-∑l=3n‍ifk∈ℤm2+1m3,y2(m2+1)=λm2+1φ2*(0)-∑r=0m2‍λm2-r×(∑l=2n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1b12*lφ2*(r-ml)])-(b11*1)2b12*1(∑r=0m1‍λm2-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]-(b11*1)2b12*1-+(m2-m1)λm2-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍).
Applying formula (33) yields (for k∈ℤm2+2m3+1)
(96)y1(k)=λk-(m2+1)y1(m2+1)+∑r=m2+1k-1‍λk-1-r×(∑l=12‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]2222+∑l=3n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])=λk-m2-1[λm2+1φ1*(0)+∑r=0m2‍λm2-r222222222×(∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml))222222222+b11*1(∑r=0m1‍λm2-r[b12*1b11*1φ1*(r-m1)11111111111111111111111+b12*1b11*1φ2*(r-m1)]22222222222222+(m2-m1)λm2-m111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])∑r=0m1‍]+∑r=m2+1k-1‍λk-1-r(∑l=12‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]222222222222222+∑l=3n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=12‍)=λkφ1*(0)+∑r=0m2‍λk-1-r(∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)11111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍)+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]2222222+(m2-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍λk-1-r)+∑r=m2+1k-1‍λk-1-r(∑l=12‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]1111111111111111+∑l=3n‍b11*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=12‍)=λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=3n‍b11*l[b12*1b11*1φ1*(r-ml)11111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=3n‍)+∑r=0m2‍λk-1-rb11*2[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+b11*1+b11*1+(m2-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)])+(k-1-m2)(b11*1λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)](k-1-m2)(k-1)+b11*2λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)]b11*1λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=3n‍b11*l[b12*1b11*1φ1*(r-ml)11111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=3n‍)+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+b11*1+b11*1+(k-1-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m2‍)+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+b11*2+1+(k-1-m2)λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m2‍),(97)y2(k)=λk-(m2+1)y2(m2+1)-∑r=m2+1k-1‍λk-1-r×(∑l=12‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=3n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])=λk-m2-1[λm2+1φ2*(0)-∑r=0m2‍λm2-r×(∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)11111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml))-(b11*l)2b12*1(b12*1b11*1∑r=0m1‍λm2-r[b12*1b11*1φ1*(r-m1)111111111111111111111111111+b12*1b11*1φ2*(r-m1)]1111111111111111111+(m2-m1)λm2-m11111111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍)]-∑r=m2+1k-1‍λk-1-r(∑l=12‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]1111111111111111+∑l=3n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=12(b11*l)2b12*l‍)=λkφ2*(0)-∑r=0m2‍λk-1-r(∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍(b11*l)2b12*l)-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]-111b12*1-111b12*1-+(m2-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍)-∑r=m2+1k-1‍λk-1-r(∑l=12‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]-∑r=m2+1k-1‍λk-1-r1111-+∑l=3n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=12‍(b11*l)2b12*l)=λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=3n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=3n(b11*l)2b12*l‍)-∑r=0m2‍λk-1-r(b11*2)2b12*2[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]-(b11*1)2b12*1(∑r=0m1‍λk-1-r[b12*1b11*1φ1*(r-m1)11111111111111111111+b12*1b11*1φ2*(r-m1)]11111111111+(m2-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍)-(k-1-m2)×((b11*1)2b12*1λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]11111+(b11*2)2b12*2λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)](b11*1)2b12*1)=λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=3n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)11111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=3n‍(b11*l)2b12*l)-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]111111111111+(k-1-m1)λk-1-m1111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]1111111111111+(k-1-m2)λk-1-m21111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m2‍).
From (93)–(97) we deduce that expected form of the solution of the initial problem for k∈ℤms-1+1ms with initial data derived from the solution of previous equation for k∈ℤms-2+1ms-1 is
(98)y1(k)=λkφ1*(0)+∑r=0k-1‍λk-1-r×(∑l=sn‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+∑l=1s-1‍[b11*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]111111111111+(k-1-ml)λk-1-ml111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)]+(k-1-ml)λk-1-ml+11ifk∈ℤms-1+2ms+1,y1(k)=λkφ2*(0)-∑r=0k-1‍λk-1-r×(∑l=sn‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-∑l=1s-1[(b11*l)2b12*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]‍220000000001100+(k-1-ml)λk-1-ml111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)(b11*l)2b12*l]23232113213213213213213112ifk∈ℤms-1+2ms+1.
We solve (88) for k∈ℤms+1ms+1 with initial data deduced from (98); that is, we consider the problem
(99)y1(k+1)=λy1(k)+∑l=1s‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=s+1n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]∑l=s+1n‍∑l=s+1n‍b11*l∑l=s+1n‍b11*l∑l=s+1n‍b11*l111111ifk∈ℤms+1ms+1,y1(ms+1)=λms+1φ1*(0)+∑r=0ms‍λms-r(∑l=sn‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+∑l=1s-1‍b11*l(∑r=0ml‍λms-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]11111111111+(ms-ml)λms-ml11111111111×[φ1*(0)+b12*1b11*1φ2*(0)]),y2(k+1)=λy2(k)-∑l=1s‍(b11*l)2b12*l[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=s+1n‍(b11*l)2b12*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]∑l=s+1n‍(b11*l)2b12*l∑l=s+1n‍(b11*l)2b12*l∑l=s+1n‍(b11*l)2b12*l11ifk∈ℤms+1ms+1,y2(ms+1)=λms+1φ2*(0)-∑r=0ms‍λms-r(∑l=sn‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-∑l=1s-1‍(b11*l)2b12*l(∑r=0ml‍λms-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]-∑l=1s-1‍(b11*l)2b12*l-11+(ms-ml)λms-ml[φ1*(0)+b12*1b11*1φ2*(0)]).
Applying formula (33) yields (for k∈ℤms+2ms+1+1)
(100)y1(k)=λk-(ms+1)y1(ms+1)+∑r=ms+1k-1‍λk-1-r×(∑l=1s‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=s+1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])=λk-ms-1[λms+1φ1*(0)+∑r=0ms‍λms-r=λk-ms-1×(∑l=sn‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])=λk-ms-1=+∑l=1s-1‍b11*l(∑r=0ml‍λms-r[b12*1b11*1φ1*(r-ml)11111111111111111111111111+b12*1b11*1φ2*(r-ml)]111111111111111111+(ms-ml)λms-ml111111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)]+∑r=ms+1k-1‍λk-1-r(∑l=1s‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑r=ms+1k-1‍λk-1-r+11+∑l=s+1n‍b11*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1s‍)=λkφ1*(0)+∑r=0ms‍λk-1-r×(∑l=sn‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+∑l=1s-1‍b11*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]1111111111+(ms-ml)λk-1-ml1111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+∑r=ms+1k-1‍λk-1-r(∑l=1s‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑r=ms+1k-1‍λk-1-r+11+∑l=s+1n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111+b12*1b11*1φ2*(r-ml)b12*1b11*1]∑l=1s‍)=λkφ1*(0)+∑r=0k-1‍λk-1-r×(∑l=s+1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+b11*s∑r=0ms‍λk-1-r[φ1*(r-ms)+b12*1b11*1φ2*(r-ms)]+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+b11*1+b11*1+(ms-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+b11*2+b11*2+(ms-m2)λk-1-m2111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m2‍)+⋯+b11*s-1(∑r=0ms-1‍λk-1-r[b12*1b11*1φ1*(r-ms-1)1111111111111111111+b12*1b11*1φ2*(r-ms-1)]111111111+(ms-ms-1)λk-1-ms-1111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ms-1‍)+(k-1-ms)×(λk-1-m1b11*1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-m2b11*2[φ1*(0)+b12*1b11*1φ2*(0)]+⋯+λk-1-ms-1b11*s-1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-msb11*s[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ1*(0)+∑r=0k-1‍λk-1-r×(∑l=s+1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]111111111+(k-1-m1)λk-1-m1111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]111111111+(k-1-m2)λk-1-m2111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯+b11*s-1(∑r=0ms-1‍λk-1-r[b12*1b11*1φ1*(r-ms-1)1111111111111111111+b12*1b11*1φ2*(r-ms-1)]111111111+(k-1-ms-1)λk-1-ms-1111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+b11*s(∑r=0ms‍λk-1-r[φ1*(r-ms)+b12*1b11*1φ2*(r-ms)]+b11*s+b11*s+(k-1-ms)λk-1-ms[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ1*(0)+∑r=0k-1‍λk-1-r×(∑l=s+1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+∑l=1s‍b11*l(∑r=0ml‍λk-1-r[b12*1b11*1φ1*(r-ml)gggggggggggggggg+b12*1b11*1φ2*(r-ml)]1111111111+(k-1-ml)λk-1-ml1111111111×[φ1*(0)+b12*1b11*1φ2*(0)]),(101)y2(k)=λk-(ms+1)y2(ms+1)-∑r=ms+1k-1‍λk-1-r×(∑l=1s‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=s+1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1s‍(b11*l)2b12*l)=λk-ms-1[λms+1φ2*(0)-∑r=0ms‍λms-r=λk-ms-1=×(∑l=sn‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)11111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=sn‍(b11*l)2b12*l)1111111111-∑l=1s-1‍(b11*l)2b12*l(∑r=0ml‍λms-r[b12*1b11*1φ1*(r-ml)11111111111111111111111111111+b12*1b11*1φ2*(r-ml)]111111111111111111111+(ms-ml)λms-ml111111111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)]-∑r=ms+1k-1‍λk-1-r(∑l=1s‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]-∑r=ms+1k-1‍λk-∑r=ms+1k-1‍11+∑l=s+1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1s‍(b11*l)2b12*l)=λkφ2*(0)-∑r=0ms‍λk-1-r×(∑l=sn‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-∑l=1s-1‍(b11*l)2b12*l(∑r=0ml‍λk-1-r[b12*1b11*1φ1*(r-ml)1111111111111111111111+b12*1b11*1φ2*(r-ml)]-∑l=1s-1‍-∑l=1s-1‍-11+(ms-ml)λk-1-ml[φ1*(0)+b12*1b11*1φ2*(0)])-∑r=ms+1k-1‍λk-1-r(∑l=1s‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]-∑r=ms+1k-1‍λk-1-r-11+∑l=s+1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1s‍(b11*l)2b12*l)=λkφ2*(0)-∑r=0k-1‍λk-1-r×(∑l=s+1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-(b11*s)2b12*s∑r=0ms‍λk-1-r[φ1*(r-ms)+b12*1b11*1φ2*(r-ms)]-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]λkφ2*λkφ2*λk+(ms-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]λk-1-m2λk-11+(ms-m2)λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)])+⋯-(b11*s-1)2b12*s-1(∑r=0ms-1‍λk-1-r[φ1*(r-ms-1)+b12*1b11*1φ2*(r-ms-1)]111111111111+(ms-ms-1)λk-1-ms-1[φ1*(0)+b12*1b11*1φ2*(0)])-(k-1-ms)×(λk-1-m1(b11*1)2b12*1[φ1*(0)+b12*1b11*1φ2*(0)]11111+λk-1-m2(b11*2)2b12*2[φ1*(0)+b12*1b11*1φ2*(0)]+⋯11111+λk-1-ms-1(b11*s-1)2b12*s-1[φ1*(0)+b12*1b11*1φ2*(0)]11111+λk-1-ms(b11*s)2b12*s[φ1*(0)+b12*1b11*1φ2*(0)]λk-1-m1(b11*1)2b12*1[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ2*(0)-∑r=0k-1‍λk-1-r×(∑l=s+1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]11111111111+(k-1-m1)λk-1-m111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]11111111111+(k-1-m2)λk-1-m211111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯-(b11*s-1)2b12*s-1(∑r=0ms-1‍λk-1-r[b12*1b11*1φ1*(r-ms-1)1111111111111111111111+b12*1b11*1φ2*(r-ms-1)]11111111111111+(k-1-ms-1)λk-1-ms-111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ms-1‍)∑l=s+1n‍∑l=s+1n‍-(b11*s)2b12*s(∑r=0ms‍λk-1-r[b12*1b11*1φ1*(r-ms)1111111111111111111111111+b12*1b11*1φ2*(r-ms)]∑l=s+1n‍111111111∑l=s+1n‍+(k-1-ms)λk-1-ms∑l=s+1n‍111111111∑l=s+1n‍×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ms‍)=λkφ2*(0)-∑r=0k-1‍λk-1-r×(∑l=s+1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-∑l=1s‍(b11*l)2b12*l(∑r=0ml‍λk-1-r[b12*1b11*1φ1*(r-ml)11111111111111111111111+b12*1b11*1φ2*(r-ml)]1111111111111111+(k-1-ml)λk-1-ml1111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]).
In the end, we solve (88) for k∈ℤmn+1∞ with initial data deduced from (100) and (101); that is, we consider the problem
(102)y1(k+1)=λy1(k)+∑l=1n‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]=λy1(k)+∑l=1n‍b11*lλk-mlλy1(k)ifk∈ℤmn+1∞,y1(mn+1)=λmn+1φ1*(0)+∑r=0mn‍λmn-rb11*n×[φ1*(r-mn)+b12*1b11*1φ2*(r-mn)]+∑l=1n-1‍b11*l(∑r=0ml‍λmn-r[b12*1b11*1φ1*(r-ml)1111111111111111111+b12*1b11*1φ2*(r-ml)]+(mn-ml)λmn-ml×[φ1*(0)+b12*1b11*1φ2*(0)]),y2(k+1)=λy2(k)-∑l=1n‍(b11*l)2b12*lλk-ml×[φ1*(0)+b12*1b11*1φ2*(0)]ifk∈ℤmn+1∞,y2(mn+1)=λmn+1φ2*(0)-∑r=0mn‍λmn-r(b11*n)2b12*n×[φ1*(r-mn)+b12*1b11*1φ2*(r-mn)]-∑l=1n-1‍(b11*l)2b12*l(∑r=0ml‍λmn-r[b12*1b11*1φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]+(mn-ml)λmn-ml×[φ1*(0)+b12*1b11*1φ2*(0)]).
Applying formula (33) yields (for k∈ℤmn+2∞)
(103)y1(k)=λk-(mn+1)y1(mn+1)+∑r=mn+1k-1‍λk-1-r×(∑l=1n‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)])=λk-mn-1[[φ1*(0)+b12*1b11*1φ2*(0)]λmn+1φ1*(0)=λk-mn-111+∑r=0mn‍λmn-rb11*n[b12*1b11*1φ1*(r-mn)11111111111111111111111+b12*1b11*1φ2*(r-mn)]=λk-mn-111+∑l=1n-1‍b11*l(∑r=0ml‍λmn-r[b12*1b11*1φ1*(r-ml)11111111111111111111111111+b12*1b11*1φ2*(r-ml)]111111111111111111+(mn-ml)λmn-ml111111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)]+∑r=mn+1k-1‍λk-1-r(∑l=1n‍b11*lλr-ml1111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑l=1n‍)=λkφ1*(0)+∑r=0mn‍λk-1-rb11*n[b12*1b11*1φ1*(r-mn)11111111111111111111111+b12*1b11*1φ2*(r-mn)]+∑l=1n-1‍b11*l(∑r=0ml‍λk-1-r[b12*1b11*1φ1*(r-ml)1111111111111111111+b12*1b11*1φ2*(r-ml)]+∑l=1n-1‍b11*l+∑l=1n-1‍+(mn-ml)λk-1-ml+∑l=1n-1‍b11*l+∑l=1n-1‍×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)+∑r=mn+1k-1‍λk-1-r(∑l=1n‍b11*lλr-ml1111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)=λkφ1*(0)+∑r=0mn‍λk-1-rb11*n[b12*1b11*1φ1*(r-mn)1111111111111111111111+b12*1b11*1φ2*(r-mn)]+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+b11*1+b11*1+(mn-m1)λk-1-m1+b11*1+b11*1×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+b11*2b11*2+(mn-m2)λk-1-m2+b11*2b11*2×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯+b11*n-1(∑r=0mn-1‍λk-1-r[b12*1b11*1φ1*(r-mn-1)1111111111111111111+b12*1b11*1φ2*(r-mn-1)]+b11*n-1+b11*n+(mn-mn-1)λk-1-mn-1+b11*n-1+b11*n×[φ1*(0)+b12*1b11*1φ2*(0)])+(k-1-mn)×(λk-1-m1b11*1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-m2b11*2[φ1*(0)+b12*1b11*1φ2*(0)]+⋯+λk-1-mn-1b11*n-1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-mnb11*n[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ1*(0)+b11*1(∑r=0m1‍λk-1-r[b12*1b11*1φ1*(r-m1)1111111111111111111111111+b12*1b11*1φ2*(r-m1)]111111111111111+(k-1-m1)λk-1-m1111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]11111111+(k-1-m2)λk-1-m211111111×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯+b11*n-1(∑r=0mn-1‍λk-1-r[b12*1b11*1φ1*(r-mn-1)1111111111111111111+b12*1b11*1φ2*(r-mn-1)]1111111111+(k-1-mn-1)λk-1-mn-11111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0mn-1‍)+b11*n(∑r=0mn‍λk-1-r[φ1*(r-mn)+b12*1b11*1φ2*(r-mn)]+b11*n+b11*n+(k-1-mn)λk-1-mn111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0mn‍)=λkφ1*(0)+∑l=1n‍b11*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]1111111111+(k-1-ml)λk-1-ml1111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍),(104)y2(k)=λk-(mn+1)y2(mn+1)-∑r=mn+1k-1‍λk-1-r×(∑l=1n‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)])=λk-mn-1×[λmn+1φ2*(0)-∑r=0mn‍λmn-r(b11*n)2b12*n×[φ1*(r-mn)+b12*1b11*1φ2*(r-mn)]-∑l=1n-1‍(b11*l)2b12*l×(∑r=0ml‍λmn-r[b12*1b11*1φ1*(r-ml)11111111111111111+b12*1b11*1φ2*(r-ml)]+(mn-ml)λmn-ml×[φ1*(0)+b12*1b11*1φ2*(0)])∑r=0mn‍λmn-r(b11*n)2b12*n]-∑r=mn+1k-1‍λk-1-r(∑l=1n‍(b11*l)2b12*lλr-ml11111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑l=1n‍(b11*l)2b12*l)=λkφ2*(0)-∑r=0mn‍λk-1-r(b11*n)2b12*n×[φ1*(r-mn)+b12*1b11*1φ2*(r-mn)]-∑l=1n-1‍(b11*l)2b12*l(∑r=0ml‍λk-1-r[b12*1b11*1φ1*(r-ml)1111111111111111111111+b12*1b11*1φ2*(r-ml)]11111111111111+(mn-ml)λk-1-ml11111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)-∑r=mn+1k-1‍λk-1-r(∑l=1n‍(b11*l)2b12*lλr-ml1111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑l=1n‍(b11*l)2b12*l)=λkφ2*(0)-∑r=0mn‍λk-1-r(b11*n)2b12*n[b12*1b11*1φ1*(r-mn)1111111111111111111111111+b12*1b11*1φ2*(r-mn)]-(b11*1)2b12*1(∑r=0m1‍λk-1-r[b12*1b11*1φ1*(r-m1)11111111111111111111+b12*1b11*1φ2*(r-m1)]11111111111+(mn-m1)λk-1-m111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[b12*1b11*1φ1*(r-m2)11111111111111111111+b12*1b11*1φ2*(r-m2)]111111111111+(mn-m2)λk-1-m2111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯-(b11*n-1)2b12*n-1(∑r=0mn-1‍λk-1-r[b12*1b11*1φ1*(r-mn-1)1111111111111111111111+b12*1b11*1φ2*(r-mn-1)]111111111111+(mn-mn-1)λk-1-mn-1111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(k-1-mn)×(λk-1-m1(b11*1)2b12*1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-m2(b11*2)2b12*2[φ1*(0)+b12*1b11*1φ2*(0)]+⋯+λk-1-mn-1(b11*n-1)2b12*n-1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-mn(b11*n)2b12*n[φ1*(0)+b12*1b11*1φ2*(0)][φ1*(0)+b12*1b11*1φ2*(0)](b11*1)2b12*1)=λkφ2*(0)-(b11*1)2b12*1×(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]11111+(k-1-m1)λk-1-m111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[b12*1b11*1φ1*(r-m2)11111111111111111111+b12*1b11*1φ2*(r-m2)]11111111111+(k-1-m2)λk-1-m211111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯-(b11*n-1)2b12*n-1(∑r=0mn-1‍λk-1-r[b12*1b11*1φ1*(r-mn-1)1111111111111111111111+b12*1b11*1φ2*(r-mn-1)]1111111111111+(k-1-mn-1)λk-1-mn-11111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*n)2b12*n(∑r=0mn‍λk-1-r[b12*1b11*1φ1*(r-mn)11111111111111111111+b12*1b11*1φ2*(r-mn)]11111111111+(k-1-mn)λk-1-mn11111111111×[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ2*(0)-∑l=1n‍(b11*l)2b12*l×(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]11111+(k-1-ml)λk-1-ml11111×[φ1*(0)+b12*1b11*1φ2*(0)]).
Summing up all particular cases (90), (93), (96), (100), and (103) we have(105)y1(k)={φ1*(k)ifk∈ℤ-mn0,λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])ifk∈ℤ1m1+1,λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=2n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤm1+2m2+1,λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=3n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+(k-1-m2)λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤm2+2m3+1,⋮λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=s+1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+∑l=1s‍b11*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]+(k-1-ml)λk-1-ml[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤms+2ms+1+1,⋮λkφ1*(0)+∑l=1n‍b11*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]+(k-1-ml)λk-1-ml[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤmn+2∞,and from cases (91), (94), (97), (101), and (104) we conclude that(106)y2(k)={φ2*(k)ifk∈ℤ-mn0,λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])ifk∈ℤ1m1+1,λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=2n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1b12*lφ2*(r-ml)])-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤm1+2m2+1,λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=3n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+(k-1-m2)λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤm2+2m3+1,⋮λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=s+1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-∑l=1s‍(b11*l)2b12*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]+(k-1-ml)λk-1-ml[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤms+2ms+1+1,⋮λkφ2*(0)-∑l=1n‍(b11*l)2b12*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]+(k-1-ml)λk-1-ml[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤmn+2∞.Formula (81) is now a direct consequence of (105), (106), and (80).


#### 2.1.6. Case (39) of a Double Real Root

If the matrixΛ has the form (39), the necessary and sufficient conditions (13)–(16), for (40), are reduced to (43), (44), (46), and
(107)|λ1b21*lb22*l|+|b11*lb12*l0λ|=λ(b11*l+b22*l)-b21*l=0.Then (43), (44), and (107) give b11*l=b22*l=b21*l=0.Theorem 9.
Let (1) be a weakly delayed system, (35) has a double root λ1,2=λ and the matrix Λ has the form (39). Then b11*l=b22*l=b21*l=0 and the solution of the initial problem (1), (3) is x(k)=𝒮y(k), y(k)=(y1(k),y2(k))T, and
(108)y1(k)={φ1*(k)ifk∈ℤ-mn0,λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍b12*lφ2*(r-ml)]ifk∈ℤ1m1+1,⋮λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+∑l=1s‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍]ifk∈ℤms+2ms+1+1,s=1,2,…,n-1,⋮λkφ1*(0)+kλk-1φ2*(0)+∑l=1n‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍]ifk∈ℤmn+2∞,(109)y2(k)={φ2*(k)ifk∈ℤ-mn0,λkφ2*(0)ifk∈ℤ1∞.Proof.
The system (40) can be written as
(110)y1(k+1)=λy1(k)+y2(k)+∑l=1n‍b12*ly2(k-ml),(111)y2(k+1)=λy2(k),k∈ℤ0∞.
Solving (111), we get
(112)y2(k)={φ2*(k)ifk∈ℤ-mn0,λkφ2*(0)ifk∈ℤ1∞,
then (110) turns into
(113)y1(k+1)={λy1(k)+λkφ2*(0)+∑l=1n‍b12*lφ2*(k-ml)ifk∈ℤ0m1,λy1(k)+λkφ2*(0)+b12*1λk-m1φ2*(0)+∑l=2n‍b12*lφ2*(k-ml)ifk∈ℤm1+1m2,λy1(k)+λkφ2*(0)+∑l=12‍b12*lλk-mlφ2*(0)+∑l=3n‍b12*lφ2*(k-ml)ifk∈ℤm2+1m3,⋮λy1(k)+λkφ2*(0)+∑l=1s‍b12*lλk-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(k-ml)ifk∈ℤms+1ms+1,s=3,4,…,n-1,⋮λy1(k)+λkφ2*(0)+∑l=1n‍b12*lλk-mlφ2*(0)ifk∈ℤmn+1∞.
Equation (113) can be solved in a way similar to that of (54) in the proof of Theorem 5 using (33).
First we solve (113) for k∈ℤ0m1. This means that we consider the problem
(114)y1(k+1)=λy1(k)+λkφ2*(0)+∑l=1n‍b12*lφ2*(k-ml)ifk∈ℤ0m1,y1(0)=φ1*(0).
With the aid of formula (33), we get
(115)y1(k)=λkφ1*(0)+∑r=0k-1‍λk-1-r[λrφ2*(0)+∑l=1n‍b12*lφ2*(r-ml)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍b12*lφ2*(r-ml)],k∈ℤ1m1+1.
Now we solve (113) for k∈ℤm1+1m2 with initial data deduced from (115); that is, we consider the problem
(116)y1(k+1)=λy1(k)+λkφ2*(0)+b12*1λk-m1φ2*(0)+∑l=2n‍b12*lφ2*(k-ml),k∈ℤm1+1m2,y1(m1+1)=λm1+1φ1*(0)+(m1+1)λm1φ2*(0)+∑r=0m1‍λm1-r[∑l=1n‍b12*lφ2*(r-ml)].
Applying formula (33) we get (for k∈ℤm1+2m2+1)
(117)y1(k)=λk-(m1+1)y1(m1+1)+∑r=m1+1k-1‍λk-1-r[∑l=2n‍λrφ2*(0)+b12*1λr-m1φ2*(0)+∑l=2n‍b12*lφ2*(r-ml)]=λk-m1-1[+∑r=0m1‍λm1-r[∑l=1n‍b12*lφ2*(r-ml)]]λm1+1φ1*(0)+(m1+1)λm1φ2*(0)=λk-m1-1111+∑r=0m1‍λm1-r[∑l=1n‍b12*lφ2*(r-ml)]]+∑r=m1+1k-1‍λk-1-r[+∑l=2n‍b12*lφ2*(r-ml)λrφ2*(0)+b12*1λr-m1φ2*(0)+∑r=m1+1k-1‍λk-1-r11111+∑l=2n‍b12*lφ2*(r-ml)]=λkφ1*(0)+(m1+1)λk-1φ2*(0)+∑r=0m1‍λk-1-r[∑l=1n‍b12*lφ2*(r-ml)]+(k-m1-1)λk-1φ2*(0)+∑r=m1+1k-1‍λk-1-r[b12*1λr-m1φ2*(0)+∑l=2n‍b12*lφ2*(r-ml)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+b12*1+b12*1+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍λk-1-rφ2*(r-m1)].
Now we solve (113) for k∈ℤm2+1m3 with initial data deduced from (117); that is, we consider the problem
(118)y1(k+1)=λy1(k)+λkφ2*(0)+∑l=12‍b12*lλk-mlφ2*(0)+∑l=3n‍b12*lφ2*(k-ml)k∈ℤm2+1m3,y1(m2+1)=λm2+1φ1*(0)+(m2+1)λm2φ2*(0)+∑r=0m2‍λm2-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λm2-rφ2*(r-m1)+b12*1+ll+(m2-m1)λm2-m1φ2*(0)∑r=0m1‍λm2-rφ2*(r-m1)].
Applying formula (33) yields (for k∈ℤm2+2m3+1)
(119)y1(k)=λk-(m2+1)y1(m2+1)+∑r=m2+1k-1‍λk-1-r[λrφ2*(0)+∑l=12‍b12*lλr-mlφ2*(0)+∑l=3n‍b12*lφ2*(r-ml)]=λk-m2-1[+∑r=0m2‍λm2-r[∑l=2n‍b12*lφ2*(r-ml)]λm2+1φ1*(0)+(m2+1)λm2φ2*(0)=λk-m2-1=+∑r=0m2‍λm2-r[∑l=2n‍b12*lφ2*(r-ml)]=λk-m2-1=+b12*1[∑r=0m1‍λm2-rφ2*(r-m1)=λk-m2-1=+(m2-m1)λm2-m1φ2*(0)∑r=0m1‍λm2-rφ2*(r-m1)]]+∑r=m2+1k-1‍λk-1-r[λrφ2*(0)+∑l=12‍b12*lλr-mlφ2*(0)+∑r=m2+1k-1‍λk-1-r+1+∑l=3n‍b12*lφ2*(r-ml)]=λkφ1*(0)+(m2+1)λk-1φ2*(0)+∑r=0m2‍λk-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+b12*1++(m2-m1)λk-1-m1φ2*(0)∑r=0m1‍λk-1-rφ2*(r-m1)]+(k-1-m2)λk-1φ2*(0)+∑r=m2+1k-1‍λk-1-r[∑l=12‍b12*lλr-mlφ2*(0)+∑l=3n‍b12*lφ2*(r-ml)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=3n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+b12*1+b12*1+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍λk-1-rφ2*(r-m1)]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)+b12*2+b12*2+(k-1-m2)λk-1-m2φ2*(0)∑r=0m2‍λk-1-rφ2*(r-m2)].
From (115), (117), and (119) we deduce that expected form of the solution of the initial problem for k∈ℤms-1+1ms with initial data derived from the solution of previous equation for k∈ℤms-2+1ms-1 is
(120)y1(k)=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+∑l=1s-1‍b12*l+∑l=1s-1‍+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍λk-1-rφ2*(r-ml)]+∑l=1s-1‍b12*l+∑l=1s-1‍b12*l+∑l=1s-1‍b12*lifk∈ℤms-1+2ms+1.
We solve (113) for k∈ℤms+1ms+1 with initial data deduced from (120); that is, we consider the problem
(121)y1(k+1)=λy1(k)+λkφ2*(0)+∑l=1s‍b12*lλk-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(k-ml),k∈ℤms+1ms+1,y1(ms+1)=λms+1φ1*(0)+(ms+1)λmsφ2*(0)+∑r=0ms‍λms-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=0ml‍λms-rφ2*(r-ml)+∑l=1s-1‍b12*l+∑l=1s-1‍+(ms-ml)λms-mlφ2*(0)∑r=0ml‍λms-rφ2*(r-ml)].
Applying formula (33) yields (for k∈ℤms+2ms+1+1)
(122)y1(k)=λk-(ms+1)y1(ms+1)+∑r=ms+1k-1‍λk-1-r[λrφ2*(0)+∑l=1s‍b12*lλr-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(r-ml)]=λk-ms-1×[∑r=0ml‍λms+1φ1*(0)+(ms+1)λmsφ2*(0)+∑r=0ms‍λms-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=0ml‍λms-rφ2*(r-ml)111111111111+(ms-ml)λms-mlφ2*(0)∑r=0ml‍]]+∑r=ms+1k-1‍λk-1-r[λrφ2*(0)+∑l=1s‍b12*lλr-mlφ2*(0)111111111111111+∑l=s+1n‍b12*lφ2*(r-ml)]=λkφ1*(0)+(ms+1)λk-1φ2*(0)+∑r=0ms‍λk-1-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)11111111111+(ms-ml)λk-1-mlφ2*(0)∑r=0ml‍]+(k-1-ms)λk-1φ2*(0)+∑r=ms+1k-1‍λk-1-r[∑l=1s‍b12*lλr-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(r-ml)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r×[∑l=s+1n‍b12*lφ2*(r-ml)]+∑r=0ms‍λk-1-rb12*sφ2*(r-ms)+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)11111111111+(ms-m1)λk-1-m1φ2*(0)∑r=0m1‍]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)11111111111+(ms-m2)λk-1-m2φ2*(0)∑r=0m2‍]+⋯+b12*s-1[∑r=0ms-1‍λk-1-rφ2*(r-ms-1)111111111111+(ms-ms-1)λk-1-ms-1φ2*(0)∑r=0ms-1‍]+(k-1-ms)[λk-1-m1b12*1φ2*(0)1111111111111111+λk-1-m2b12*2φ2*(0)11111111111111+⋯+λk-1-ms-1b12*s-1φ2*(0)11111111111111+λk-1-m2b12*sφ2*(0)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)11111111111+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)11111111111+(k-1-m2)λk-1-m2φ2*(0)∑r=0m2‍]+⋯+b12*s-1[∑r=0ms-1‍λk-1-rφ2*(r-ms-1)11111111111+(k-1-ms-1)λk-1-ms-1φ2*(0)∑r=0ms-1‍]+b12*s[∑r=0ms‍λk-1-rφ2*(r-ms)111111111+(k-1-ms)λk-1-msφ2*(0)∑r=0ms‍]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+∑l=1s‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)111111111111+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍].
In the end, we solve (113) for k∈ℤmn+1∞ with initial data deduced from (122); that is, we consider the problem
(123)y1(k+1)=λy1(k)+λkφ2*(0)+∑l=1n‍b12*lλk-mlφ2*(0)ifk∈ℤmn+1∞,y1(mn+1)=λmn+1φ1*(0)+(mn+1)λmnφ2*(0)+∑r=0mn‍λmn-rb12*nφ2*(r-mn)+∑l=1n-1‍b12*l[∑r=0ml‍λmn-rφ2*(r-ml)+∑l=1n-1‍b12*l+∑l=1n-1‍+(mn-ml)λmn-mlφ2*(0)∑r=0ml‍].
Applying formula (33) yields (for k∈ℤmn+2∞)
(124)y1(k)=λk-(mn+1)y1(mn+1)+∑r=mn+1k-1‍λk-1-r[λrφ2*(0)+∑l=1n‍b12*lλr-mlφ2*(0)]=λk-mn-1[∑r=0ml‍λmn+1φ1*(0)+(mn+1)λmnφ2*(0)=1111λk-mn+∑r=0mn‍λmn-rb12*nφ2*(r-mn)=1111λk-mn+∑l=1n-1‍b12*l[∑r=0ml‍λmn-rφ2*(r-ml)=λk-mn=1λk-mn=λk-+(mn-ml)λmn-mlφ2*(0)∑r=0ml‍]]+∑r=mn+1k-1‍λk-1-r[λrφ2*(0)+∑l=1n‍b12*lλr-mlφ2*(0)]=λkφ1*(0)+(mn+1)λk-1φ2*(0)+∑r=0mn‍λk-1-rb12*nφ2*(r-mn)+∑l=1n-1‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+(mn-ml)λk-1-mlφ2*(0)∑r=0ml‍]+(k-1-mn)λk-1φ2*(0)+∑r=mn+1k-1‍λk-1-r[∑l=1n‍b12*lλr-mlφ2*(0)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0mn‍λk-1-rb12*nφ2*(r-mn)+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+b12*1+b12*1+(mn-m1)λk-1-m1φ2*(0)∑r=0m1‍]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)+b12*2+b12*2+(mn-m2)λk-1-m2φ2*(0)∑r=0m2‍]+⋯+b12*n-1[∑r=0mn-1‍λk-1-rφ2*(r-mn-1)+b12*n-1+b12*n-1+(mn-mn-1)λk-1-mn-1φ2*(0)∑r=0mn-1‍]+(k-1-mn)×[λk-1-m1b12*1φ2*(0)+λk-1-m2b12*2φ2*(0)+⋯+λk-1-mn-1b12*n-1φ2*(0)+λk-1-mnb12*nφ2*(0)]=λkφ1*(0)+kλk-1φ2*(0)+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+b12*1+b12*1+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)+b12*1+b12*1+(k-1-m2)λk-1-m2φ2*(0)∑r=0m2‍]+⋯+b12*n-1[∑r=0mn-1‍λk-1-rφ2*(r-mn-1)+b12*1+b12*1+(k-1-mn-1)λk-1-mn-1φ2*(0)∑r=0mn-1‍]+b12*n[∑r=0mn‍λk-1-rφ2*(r-mn)+b12*1+b12*1+(k-1-mn)λk-1-mnφ2*(0)∑r=0mn‍]=λkφ1*(0)+kλk-1φ2*(0)+∑l=1n‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+b12*1+b12*1+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍].
Summing up all particular cases (115)–(124), we get
(125)y1(k)={φ1*(k)ifk∈ℤ-mn0,λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍b12*lφ2*(r-ml)]ifk∈ℤ1m1+1,λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍]ifk∈ℤm1+2m2+1,λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=3n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)+(k-1-m2)λk-1-m2φ2*(0)∑r=0m2‍]ifk∈ℤm2+2m3+1,⋮λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+∑l=1s‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍]ifk∈ℤms+2ms+1+1,⋮λkφ1*(0)+kλk-1φ2*(0)+∑l=1n‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍]ifk∈ℤmn+2∞.
Formulas (108) and (109) are consequences of (125), (112).


## 2.1. Jordan Forms of the MatrixA and Corresponding Solutions of Problem (1) and (3)

It is known that, for every matrixA, there exists a nonsingular matrix S transforming it to the corresponding Jordan matrix form Λ. This means that
(34)Λ=S-1AS,
where Λ has the following four possible forms (denoted below as Λ1,Λ2,Λ3,Λ4), depending on the roots of the characteristic equation (7), that is, on the roots of
(35)λ2-(a11+a22)λ+(a11a22-a12a21)=0.
If (35) has two real distinct roots λ1, λ2, then
(36)Λ1=(λ100λ2),
if the roots are complex conjugate, that is, λ1,2=p±iq with q≠0, then
(37)Λ2=(pq-qp)
and, finally, in the case of one double real root λ1,2=λ, we have either
(38)Λ3=(λ00λ)
or
(39)Λ4=(λ10λ).
The transformation y(k)=S-1x(k) transforms (1) into a system
(40)y(k+1)=Λy(k)+∑l=1n‍B*ly(k-ml),k∈ℤ0∞
with B*l=S-1BlS, B*l=(bij*l), l=1,…,n, and i,j=1,2. Together with (40), we consider an initial problem
(41)y(k)=φ*(k),k∈ℤ-mn0 with φ*:ℤ-mn0→ℝ2 where φ*(k)=S-1φ(k) is the initial function corresponding to the initial function φ in (3).Next, we consider all four possible cases (36)–(39) separately.We define(42)Φ1(k):=(0,φ1*(k))T,Φ2(k):=(φ2*(k),0)T,223232323223312232321111122223232k∈ℤ-mn0.
Assuming that (1) is a weakly delayed system, by Lemma 2, the system (40) is weakly delayed system again.


### 2.1.1. Case (36) of Two Real Distinct Roots

In this case, we haveΛ=Λ1 and Λ1k=diag⁡(λ1k,λ2k). The necessary and sufficient conditions (13)–(16) for (40) turn into
(43)b11*l+b22*l=0,(44)|b11*lb12*lb21*lb22*l|=b11*lb22*l-b12*lb21*l=0,(45)|λ10b21*lb22*l|+|b11*lb12*l0λ2|=λ1b22*l+λ2b11*l=0,(46)|b11*lb12*lb21*vb22*v|+|b11*vb12*vb21*lb22*l|=0.
Since λ1≠λ2, (43) and (45) yield b11*l=b22*l=0, then, from (44), we get b12*lb21*l=0, so that either b21*l=0 or b12*l=0. In view of assumptions Bl≠Θ, l=1,2,…,n, we conclude that only the following cases I, II are possible:(I)
b11*l=b22*l=b21*l=0, b12*l≠0, l=1,2,…,n,(II)
b11*l=b22*l=b12*l=0, b21*l≠0, l=1,2,…,n.In Theorem5 both cases I, II are analyzed.Theorem 5.
Let (1) be a weakly delayed system and (35) has two real distinct roots λ1, λ2. If case (I) holds, then the solution of the initial problem (1), (3) is x(k)=Sy(k), k∈ℤ-mn∞, where y(k) has the form
(47)y(k)={φ*(k)ifk∈ℤ-mn0,Λ1kφ*(0)+∑r=0k-1‍λ1k-1-r[∑l=1n‍b12*lΦ2(r-ml)]ifk∈ℤ1m1+1,⋮Λ1kφ*(0)+∑r=0k-1‍λ1k-1-r[∑l=s+1n‍b12*lΦ2(r-ml)]+∑l=1s‍b12*l[∑r=ml+1k-1‍∑r=0ml‍λ1k-1-rΦ2(r-ml)+Φ2(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml]ifk∈ℤms+2ms+1+1,s=1,2,…,n-1,⋮Λ1kφ*(0)+∑l=1n‍b12*l[∑r=ml+1k-1‍∑r=0ml‍λ1k-1-rΦ2(r-ml)+Φ2(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml]ifk∈ℤmn+2∞.
If case (II) is true, then the solution of initial problem (1), (3) is x(k)=Sy(k), k∈ℤ-mn∞, where y(k) has the form
(48)y(k)={φ*(k)ifk∈ℤ-mn0,Λ1kφ*(0)+∑r=0k-1‍λ2k-1-r[∑l=1n‍b21*lΦ1(r-ml)]ifk∈ℤ1m1+1,⋮Λ1kφ*(0)+∑r=0k-1‍λ2k-1-r[∑l=s+1n‍b21*lΦ1(r-ml)]+∑l=1s‍b21*l[∑r=ml+1k-1∑r=0ml‍λ2k-1-rΦ1(r-ml)+Φ1(0)∑r=ml+1k-1‍λ1r-mlλ2k-1-r]0000000000000000000000ifk∈ℤms+2ms+1+1,00000000000000000000s=1,2,…,n-1,⋮Λ1kφ*(0)+∑l=1n‍b21*l[∑r=ml+1k-1λ1r-mlλ2k-1-r‍∑r=0ml‍λ2k-1-rΦ1(r-ml)+Φ1(0)∑r=ml+1k-1λ1r-mlλ2k-1-r‍]0000000000000000000000000ifk∈ℤmn+2∞.Proof.
If case (I) is true, then the transformed system (40) takes the form
(49)y1(k+1)=λ1y1(k)+∑l=1n‍b12*ly2(k-ml),(50)y2(k+1)=λ2y2(k),k∈ℤ0∞,
and if case (II) holds, then (40) takes the form
(51)y1(k+1)=λ1y1(k),(52)y2(k+1)=λ2y2(k)+∑l=1n‍b21*ly1(k-ml),k∈ℤ0∞.
We investigate only the initial problem (49), (50), (41) since the initial problem (51), (52), (41) can be examined in a similar way.
From (50), (41), we get
(53)y2(k)={φ2*(k)ifk∈ℤ-mn0,λ2kφ2*(0)ifk∈ℤ1∞,
then (49) becomes
(54)y1(k+1)={λ1y1(k)+∑l=1n‍b12*lφ2*(k-ml)ifk∈ℤ0m1,λ1y1(k)+b12*1λ2k-m1φ2*(0)+∑l=2n‍b12*lφ2*(k-ml)ifk∈ℤm1+1m2,λ1y1(k)+∑l=12‍b12*lλ2k-mlφ2*(0)+∑l=3n‍b12*lφ2*(k-ml)ifk∈ℤm2+1m3,⋮λ1y1(k)+∑l=1s‍b12*lλ2k-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(k-ml)ifk∈ℤms+1ms+1,s=3,4,…,n-1,⋮λ1y1(k)+∑l=1n‍b12*lλ2k-mlφ2*(0)ifk∈ℤmn+1∞.
First, we solve this equation for k∈ℤ0m1. This means that we consider the problem
(55)y1(k+1)=λ1y1(k)+∑l=1n‍b12*lφ2*(k-ml),k∈ℤ0m1,y1(0)=φ1*(0).
With the aid of formula (33), we get
(56)y1(k)=λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=1n‍b12*lφ2*(r-ml)],1111111111111111111111111k∈ℤ1m1+1.
Now we solve (54) for k∈ℤm1+1m2 with initial data deduced from (56); that is, we consider the problem
(57)y1(k+1)=λ1y1(k)+b12*1λ2k-m1φ2*(0)+∑l=2n‍b12*lφ2*(k-ml),k∈ℤm1+1m2,y1(m1+1)=λ1m1+1φ1*(0)+∑r=0m1‍λ1m1-r[∑l=1n‍b12*lφ2*(r-ml)].
Applying formula (33) we get (for k∈ℤm1+2m2+1)
(58)y1(k)=λ1k-(m1+1)y1(m1+1)+∑r=m1+1k-1‍λ1k-1-r[∑l=2n‍b12*lφ2*(r-ml)b12*1λ2r-m1φ2*(0)+∑r=m1+1k-1‍λ1k-1-r1212+∑l=2n‍b12*lφ2*(r-ml)]=λ1k-m1-1[[∑l=1n‍b12*lφ2*(r-ml)]λ1m1+1φ1*(0)+∑r=0m1‍λ1m1-r[∑l=1n‍b12*lφ2*(r-ml)]]+∑r=m1+1k-1‍λ1k-1-r[+∑l=2n‍b12*lφ2*(r-ml)b12*1λ2r-m1φ2*(0)+∑r=m1+1k-1‍λ1k-1-r++∑l=2n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0m1‍λ1k-1-r[∑l=1n‍b12*lφ2*(r-ml)]+∑r=m1+1k-1‍λ1k-1-r[+∑l=2n‍b12*lφ2*(r-ml)b12*1λ2r-m1φ2*(0)+∑r=m1+1k-1‍λ1k-1-r+++∑l=2n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1k-1‍∑r=0m1‍λ1k-1-rφ2*(r-m1)+b12*1+b12*1+φ2*(0)∑r=m1+1k-1‍λ1k-1-rλ2r-m1].
Now we solve (54) for k∈ℤm2+1m3 with initial data deduced from (58); that is, we consider the problem
(59)y1(k+1)=λ1y1(k)+∑l=12‍b12*lλ2k-mlφ2*(0)+∑l=3n‍b12*lφ2*(k-ml),k∈ℤm2+1m3,y1(m2+1)=λ1m2+1φ1*(0)+∑r=0m2‍λ1m2-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1m2‍∑r=m1+1m2‍λ1m2-rλ2r-m1∑r=0m1‍λ1m2-rφ2*(r-m1)+φ2*+φ2*+φ2*(0)∑r=m1+1m2‍λ1m2-rλ2r-m1].
Applying formula (33) yields (for k∈ℤm2+2m3+1)
(60)y1(k)=λ1k-(m2+1)y1(m2+1)+∑r=m2+1k-1‍λ1k-1-r[∑l=12‍b12*lλ2r-mlφ2*(0)+∑l=3n‍b12*lφ2*(r-ml)]=λ1k-m2-1[∑r=m1+1m2‍λ1m2+1φ1*(0)+∑r=0m2‍λ1m2-r[∑l=2n‍b12*lφ2*(r-ml)]111111111+b12*1[∑r=m1+1m2‍∑r=0m1‍λ1m2-rφ2*(r-m1)111111111111+φ2*(0)∑r=m1+1m2‍λ1m2-rλ2r-m1]∑r=m1+1m2‍]+∑r=m2+1k-1‍λ1k-1-r[∑l=12‍b12*lλ2r-mlφ2*(0)+∑l=3n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0m2‍λ1k-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1m2‍∑r=0m1‍λ1k-1-rφ2*(r-m1)1111111+φ2*(0)∑r=m1+1m2‍λ1k-1-rλ2r-m1]+∑r=m2+1k-1‍λ1k-1-r[∑l=12‍b12*lλ2r-mlφ2*(0)11111111111111+∑l=3n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=3n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1k-1‍∑r=0m1‍λ1k-1-rφ2*(r-m1)11111111+φ2*(0)∑r=m1+1k-1‍λ1k-1-rλ2r-m1]+b12*2[∑r=m2+1k-1‍∑r=m2+1k-1λ1k-1-rλ2r-m2∑r=0m2‍λ1k-1-rφ2*(r-m2)111111111+φ2*(0)∑r=m2+1k-1λ1k-1-rλ2r-m2].
From (56), (58), and (60) we deduce that expected form of the solution of the initial problem for k∈ℤms-1+1ms with initial data derived from the solution of previous equation for k∈ℤms-2+1ms-1 is
(61)y1(k)=λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=ml+1k-1‍λ1k-1-rλ2r-ml∑r=0ml‍λ1k-1-rφ2*(r-ml)+∑l=1s-1‍b12*l+∑l=1s-1‍+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml]+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml+ifk∈ℤms-1+2ms+1.
We solve (54) for k∈ℤms+1ms+1 with initial data deduced from (61); that is, we consider the problem
(62)y1(k+1)=λ1y1(k)+∑l=1s‍b12*lλ2k-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(k-ml),k∈ℤms+1ms+1,y1(ms+1)=λ1ms+1φ1*(0)+∑r=0ms‍λ1ms-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=ml+1ms‍λ1ms-rλ2r-ml∑r=0ml‍λ1ms-rφ2*(r-ml)+∑l=1s-1‍b12*l+∑l=1s-1‍+φ2*(0)∑r=ml+1ms‍λ1ms-rλ2r-ml].
Applying formula (33) yields (for k∈ℤms+2ms+1+1)
(63)y1(k)=λ1k-(ms+1)y1(ms+1)+∑r=ms+1k-1‍λ1k-1-r[∑l=1s‍b12*lλ2r-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(r-ml)]=λ1k-(ms+1)[[∑r=ml+1ms‍λ1ms-rλ2r-ml]λ1ms+1φ1*(0)+∑r=0ms‍λ1ms-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=ml+1ms‍λ1ms-rλ2r-ml∑r=0ml‍λ1ms-rφ2*(r-ml)111111111144444444+φ2*(0)∑r=ml+1ms‍λ1ms-rλ2r-ml]∑r=0ms‍λ1ms-r[∑l=sn‍b12*lφ2*(r-ml)]]+∑r=ms+1k-1‍λ1k-1-r[∑l=1s‍b12*lλ2r-mlφ2*(0)111111111144444+∑l=s+1n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0ms‍λ1k-1-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=ml+1ms‍λ1k-1-rλ2r-ml∑r=ml+1ms‍∑r=0ml‍λ1k-1-rφ2*(r-ml)11111111111+φ2*(0)∑r=ml+1ms‍λ1k-1-rλ2r-ml]+∑r=ms+1k-1‍λ1k-1-r[∑l=1s‍b12*lλ2r-mlφ2*(0)111111111144444+∑l=s+1n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0ms‍λ1k-1-r[∑l=s+1n‍b12*sφ2*(r-ms)11111111114444444444+∑l=s+1n‍b12*lφ2*(r-ml)]+b12*1[∑r=ml+1ms‍λ1k-1-rλ2r-ml∑r=0m1‍λ1k-1-rφ2*(r-m1)11111111114+φ2*(0)∑r=m1+1ms‍λ1k-1-rλ2r-m1]+b12*2[∑r=m2+1ms‍∑r=0m2‍λ1k-1-rφ2*(r-m2)111111114+φ2*(0)∑r=m2+1ms‍λ1k-1-rλ2r-m2]+⋯+b12*s-1[∑r=ms-1+1ms‍∑r=0ms-1‍λ1k-1-rφ2*(r-ms-1)11111111114+φ2*(0)∑r=ms-1+1ms‍λ1k-1-rλ2r-ms-1]+b12*1[φ2*(0)∑r=ms+1k-1‍λ1k-1-rλ2r-m1]+b12*2[φ2*(0)∑r=ms+1k-1‍λ1k-1-rλ2r-m2]+…+b12*s[φ2*(0)∑r=ms+1k-1‍λ1k-1-rλ2r-ms]+∑r=ms+1k-1‍λ1k-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+∑l=1s‍b12*l[∑r=ml+1k-1‍∑r=0ml‍λ1k-1-rφ2*(r-ml)11111111114+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml].
In the end we solve (54) for k∈ℤmn+1∞ with initial data deduced from (63); that is, we consider the problem
(64)y1(k+1)=λ1y1(k)+∑l=1n‍b12*lλ2k-mlφ2*(0),k∈ℤmn+1∞,y1(mn+1)=λ1mn+1φ1*(0)+∑r=0mn‍λ1mn-rb12*nφ2*(r-mn)+∑l=1n-1‍b12*l[∑r=ml+1mn‍λ1mn-rλ2r-ml∑r=0ml‍λ1mn-rφ2*(r-ml)+∑l=1n-1‍b12*l+222+φ2*(0)∑r=ml+1mn‍λ1mn-rλ2r-ml].
Applying formula (33) yields (for k∈ℤmn+2∞)
(65)y1(k)=λ1k-(mn+1)y1(mn+1)+∑r=mn+1k-1‍λ1k-1-r[∑l=1n‍b12*lλ2r-mlφ2*(0)]=λ1k-(mn+1)[[∑r=ml+1mn‍λ1mn-rλ2r-ml]λ1mn+1φ1*(0)+∑r=0mn‍λ1mn-rb12*nφ2*(r-mn)λ1k-(mn+1)λ1+∑l=1n-1‍b12*l[∑r=ml+1mn‍λ1mn-rλ2r-ml∑r=0ml‍λ1mn-rφ2*(r-ml)λ1k-(mn+1)1λ1k-(mn+1)λ1+φ2*(0)∑r=ml+1mn‍λ1mn-rλ2r-ml]]+∑r=mn+1k-1‍λ1k-1-r[∑l=1n‍b12*lλ2r-mlφ2*(0)]=λ1kφ1*(0)+∑r=0mn‍λ1k-1-rb12*nφ2*(r-mn)+∑l=1n-1‍b12*l[∑r=ml+1mn‍λ1k-1-rλ2r-ml∑r=0ml‍λ1k-1-rφ2*(r-ml)+∑l=1n-1‍b12*l+∑l=1n-1‍b12*l+φ2*(0)∑r=ml+1mn‍λ1k-1-rλ2r-ml]+∑r=mn+1k-1‍λ1k-1-r[∑l=1n‍b12*lλ2r-mlφ2*(0)]=λ1kφ1*(0)+∑r=0mn‍λ1k-1-rb12*nφ2*(r-mn)+b12*1[∑r=m1+1ms‍λ1k-1-rλ2r-m1∑r=0m1‍λ1k-1-rφ2*(r-m1)+b12*1+b12*1+φ2*(0)∑r=m1+1ms‍λ1k-1-rλ2r-m1]+b12*2[∑r=m2+1ms‍λ1k-1-rλ2r-m2∑r=0m2‍λ1k-1-rφ2*(r-m2)+b12*2+b12*2+φ2*(0)∑r=m2+1ms‍λ1k-1-rλ2r-m2]+⋯+b12*n-1[∑r=mn-1+1ms‍λ1k-1-rλ2r-mn-1∑r=0mn-1‍λ1k-1-rφ2*(r-mn-1)+b12*n-1+b12*n-1+φ2*(0)∑r=mn-1+1ms‍λ1k-1-rλ2r-mn-1]+b12*1[φ2*(0)∑r=ms+1k-1‍λ1k-1-rλ2r-m1]+b12*2[φ2*(0)∑r=ms+1k-1‍λ1k-1-rλ2r-m2]+⋯+b12*n[φ2*(0)∑r=mn+1k-1‍λ1k-1-rλ2r-mn]=λ1kφ1*(0)+∑l=1n‍b12*l[∑r=ml+1k-1‍λ1k-1-rλ2r-ml∑r=0ml‍λ1k-1-rφ2*(r-ml)∑l=1n‍b12*l∑l=1n‍b12*l∑l=1n‍b12*l∑l=1n‍+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml].
Summing up all particular cases (56)–(65) we have
(66)y1(k)={φ1*(k)ifk∈ℤ-mn0,λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=1n‍b12*lφ2*(r-ml)]00ifk∈ℤ1m1+1,λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1k-1‍∑r=0m1‍λ1k-1-rφ2*(r-m1)+φ2*(0)∑r=m1+1k-1‍λ1k-1-rλ2r-m1]0ifk∈ℤm1+2m2+1,λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=3n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1k-1‍∑r=0m1‍λ1k-1-rφ2*(r-m1)+φ2*(0)∑r=m1+1k-1‍λ1k-1-rλ2r-m1]+b12*2[∑r=m2+1k-1‍∑r=0m2‍λ1k-1-rφ2*(r-m2)+φ2*(0)∑r=m2+1k-1‍λ1k-1-rλ2r-m2]00000ifk∈ℤm2+2m3+1,⋮λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+∑l=1s‍b12*l[∑r=ml+1k-1‍∑r=0ml‍λ1k-1-rφ2*(r-ml)+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml]000ifk∈ℤms+2ms+1+1,⋮λ1kφ1*(0)+∑l=1n‍b12*l[∑r=ml+1k-1‍∑r=0ml‍λ1k-1-rφ2*(r-ml)+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml]00000ifk∈ℤmn+2∞.
Now, taking into account (42), formula (47) is a consequence of (53) and (66). Formula (48) can be proved in a similar way.
Finally, we note that both formulas (47), (48) remain valid for b12*l=b21*l=0. In this case, the transformed system (1) reduces to a system without delays. This possibility is excluded by condition (2).


### 2.1.2. Case (37) of Two Complex Conjugate Roots

The necessary and sufficient conditions (13)–(16) take the forms (43), (44), (46), and
(67)|pqb21*lb22*l|+|b11*lb12*l-qp|=p(b11*l+b22*l)+q(b12*l-b21*l)=0,
where l,v=1,2,…,n and v>l.The system of conditions (43), (44), and (67) gives b12*l=b21*l, (b11*l)2=-(b12*l)2 and admits only one possibility; namely,
(68)b11*l=b22*l=b12*l=b21*l=0.Consequently,B*l=Θ, Bl=Θ.The initial problem (1), (3) reduces to a problem without delay
(69)x(k+1)=Ax(k),x(k)=φ(k),k∈ℤ-mn0
and, obviously,
(70)x(k)={φ(k)ifk∈ℤ-mn0,Akφ(0)ifk∈ℤ1∞.
From this discussion, the next theorem follows.Theorem 6.
There exists no weakly delayed system (1) if Λ has the form (37).Finally, we note that the assumption (2) alone excludes this case.


### 2.1.3. Case (38) of Double Real Root

In this case we haveΛ=Λ3 and Λ3k=diag⁡(λk,λk). For (40), the necessary and sufficient conditions (13)–(16) are reduced to (43), (44), (46), and
(71)|λ0b21*lb22*l|+|b11*lb12*l0λ|=λ(b11*l+b22*l)=0,
where l=1,2,…,n.From (43), (44), and (71), we get b12*lb21*l=-(b11*l)2. From the condition (46) we get
(72)b11*lb22*v-b12*lb21*v+b22*lb11*v-b21*lb12*v=0,
where l,v=1,2,…,n and v>l. Multiplying (72) by b12*lb12*v, we have
(73)b11*lb22*vb12*lb12*v-(b12*l)2b21*vb12*v+b22*lb11*vb12*lb12*v-b21*lb12*l(b12*v)2=0.
Substituting b12*lb21*l=-(b11*l)2, b12*vb21*v=-(b11*v)2 into (73) and using (43) we obtain
(74)-b11*lb11*vb12*lb12*v+(b12*l)2(b11*v)2-b11*lb11*vb12*lb12*v+(b11*l)2(b12*v)2=0.
The equation (74) can be written as
(75)(b12*lb11*v-b12*vb11*l)2=0,b12*lb11*v=b12*vb11*l.Now we will analyse the two possible cases:b12*lb21*l=0 and b12*lb21*l≠0.For the caseb12*lb21*l=0, we have from (43), (44) that b11*l=b22*l=0 and b12*l=0 or b21*l=0. For b12*l=0 and b21*l≠0, condition (46) gives b12*v=0, where l,v=1,2,…,n and v>l. Then, from (43), (44) for l=v, we get b11*v=b22*v=0 and b21*v≠0.Forb21*l=0 and b12*l≠0, condition (46) gives b21*v=0, where l,v=1,2,…,n and v>l, then, from (43), (44) for l=v, we get b11*v=b22*v=0 and b12*v≠0.Now we discuss the caseb12*lb21*l≠0. From conditions (43), (44), we have b12*lb21*l=-(b11*l)2 and b11*lb22*l≠0. This yields b11*l≠0, b22*l≠0 and, from (75), we have b11*v≠0, b12*v≠0. By conditions (43), (44) for v=l, we get b22*v≠0, b21*v≠0.From the assumptionsBl≠Θ, we conclude that only the following cases ((I), (II), (III)) are possible:(I)
b11*l=b22*l=b21*l=0, b12*l≠0,(II)
b11*l=b22*l=b12*l=0, b21*l≠0,(III)
b12*lb21*l≠0,wherel=1,2,…,n.


### 2.1.4. Caseb12*lb21*l=0

Theorem 7.
Let (1) be a weakly delayed system, (35) has a twofold root λ1,2=λ, b12*lb21*l=0 and the matrix Λ has the form (38). Then the solution of the initial problem (1), (3) is x(k)=Sy(k), k∈ℤ-mn∞, where in case b21*l=0, y(k) has the form
(76)y(k)={φ*(k)ifk∈ℤ-mn0,Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍b12*lΦ2(r-ml)]ifk∈ℤ1m1+1,⋮Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b12*lΦ2(r-ml)]+∑l=1s‍b12*l[∑r=0ml‍λk-1-rΦ2(r-ml)+(k-1-ml)λk-1-mlΦ2(0)∑r=0ml‍]ifk∈ℤms+2ms+1+1,s=1,2,…,n-1,⋮Λ3kφ*(0)+∑l=1n‍b12*l[∑r=0ml‍λk-1-rΦ2(r-ml)+(k-1-ml)λk-1-mlΦ2(0)∑r=0ml‍]ifk∈ℤmn+2∞.
If b12*l=0 is true then the solution of initial problem (1), (3) is x(k)=Sy(k), k∈ℤ-mn∞, where y(k) has the form
(77)y(k)={φ*(k)ifk∈ℤ-mn0,Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍b21*lΦ1(r-ml)]ifk∈ℤ1m1+1,⋮Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b21*lΦ1(r-ml)]+∑l=1s‍b21*l[∑r=0ml‍λk-1-rΦ1(r-ml)+(k-1-ml)λk-1-mlΦ1(0)∑r=0ml‍]ifk∈ℤms+2ms+1+1,s=1,2,…,n-1,⋮Λ3kφ*(0)+∑l=1n‍b21*l[∑r=0ml‍∑r=0ml‍λk-1-rΦ1(r-ml)+(k-1-ml)λk-1-mlΦ1(0)∑r=0ml‍]ifk∈ℤmn+2∞.Proof.
Case (I) means thatb12*l≠0. Then (40) turns into the system
(78)y1(k+1)=λy1(k)+∑l=1n‍b12*ly2(k-ml),k∈ℤ0∞y2(k+1)=λy2(k),
and, if b21*l≠0, (40) turns into the system
(79)y1(k+1)=λy1(k),y2(k+1)=λy2(k)+∑l=1n‍b21*ly1(k-ml),k∈ℤ0∞.
System (78) can be solved in much the same way as the systems (49), (50) if we put λ1=λ2=λ, and the discussion of the system (79) goes along the same lines as that of the systems (51), (52) with λ1=λ2=λ. Formulas (76) and (77) are consequences of (47), (48).


### 2.1.5. Caseb12*lb21*l≠0

Fork∈ℤ-mn0, we define
(80)Φl*(k):=((b11*l)2b12*l[φ1*(k)+b12*1b11*1φ2*(k)]b11*l[φ1*(k)+b12*1b11*1φ2*(k)],-(b11*l)2b12*l[φ1*(k)+b12*1b11*1φ2*(k)])T.Theorem 8.
Let system (1) be a weakly delayed system, (35) admits two repeated roots λ1,2=λ, b12*lb21*l≠0 and the matrix Λ3 has the form (38). Then the solution of the initial problem (1), (3) is given by x(k)=Sy(k), k∈ℤ-mn∞, where y(k) has the form
(81)y(k)={φ*(k)ifk∈ℤ-mn0,Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍Φl*(r-ml)]ifk∈ℤ1m1+1,⋮Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍Φl*(r-ml)]+∑l=1s‍[∑r=0ml‍λk-1-rΦl*(r-ml)+(k-1-ml)λk-1-mlΦl*(0)∑r=0ml‍]ifk∈ℤms+2ms+1+1,s=1,2,…,n-1,⋮Λ3kφ*(0)+∑l=1n‍[∑r=0ml‍λk-1-rΦl*(r-ml)+(k-1-ml)λk-1-mlΦl*(0)∑r=0ml‍]ifk∈ℤmn+2∞.Proof.
In this case, all the entries ofB*l are nonzero and, from (43), (44), and (71), we get
(82)B*=(b11*lb12*l-(b11*l)2b12*l-b11*l),
where l=1,2,…,n, then, the system (40) reduces to
(83)y1(k+1)=λy1(k)+∑l=1n‍[b11*ly1(k-ml)+b12*ly2(k-ml)],(84)y2(k+1)=λy2(k)-∑l=1n‍[(b11*l)2b12*ly1(k-ml)+b11*ly2(k-ml)],
where k∈ℤ0∞. It is easy to see (multiplying (84) by b12*1/b11*1 and summing both equations) that
(85)y1(k+1)+b12*1b11*1y2(k+1)=λ[y1(k)+b12*1b11*1y2(k)],1111111111111k∈ℤ0∞.
Equation (85) is a homogeneous equation with respect to the unknown expression
(86)y1(k)+(b12*1b11*1)y2(k),
then, using (33), we obtain
(87)y1(k)+b12*1b11*1y2(k)={φ1*(k)+b12*1b11*1φ2*(k)ifk∈ℤ-mn0,λk[φ1*(0)+b12*1b11*1φ2*(0)]ifk∈ℤ1∞.
With the aid of (87), we rewrite the systems (83), (84) as follows:
(88)y1(k+1)={λy1(k)+∑l=1n‍b11*l[b12*1b11*1φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤ0m1,λy1(k)+b11*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=2n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤm1+1m2,λy1(k)+∑l=12‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=3n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤm2+1m3,⋮λy1(k)+∑l=1s‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=s+1n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤms+1ms+1,s=3,4,…,n-1,⋮λy1(k)+∑l=1n‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]ifk∈ℤmn+1∞,y2(k+1)={λy2(k)-∑l=1n‍(b11*l)2b12*l[b12*1b11*1φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤ0m1,λy2(k)-(b11*1)2b12*1λk-m1×[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(k-m1)+b12*1b11*1φ2*(k-m1)]ifk∈ℤm1+1m2,λy2(k)-∑l=12‍(b11*l)2b12*l[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=3n‍(b11*l)2b12*l[b12*1b11*1φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤm2+1m3,⋮λy2(k)-∑l=1s‍(b11*l)2b12*l[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=s+1n‍(b11*l)2b12*l[b12*1b11*1φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤms+1ms+1,s=3,4,…,n-1,⋮λy2(k)-∑l=1n‍(b11*l)2b12*lλk-ml×[φ1*(0)+b12*1b11*1φ2*(0)]ifk∈ℤmn+1∞.
First, we solve this system for k∈ℤ0m1 and consider the problems
(89)y1(k+1)=λy1(k)+∑l=1n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]11111111111111111111111000ifk∈ℤ0m1,y1(0)=φ1*(0),y2(k+1)=λy2(k)-∑l=1n‍(b11*l)2b12*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]111111111111111111111111111100ifk∈ℤ0m1,y2(0)=φ2*(0).
With the aid of formula (33), we get
(90)y1(k)=λkφ1*(0)+∑r=0k-1‍λk-1-r×(∑l=1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]),11111111111111111111111111111k∈ℤ1m1+1,(91)y2(k)=λkφ2*(0)-∑r=0k-1‍λk-1-r×(∑l=1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]),1111111111111111111111111111111k∈ℤ1m1+1.
Now we solve system (88) for k∈ℤm1+1m2; that is, we consider the problem (with initial data deduced from (90), (91))
(92)y1(k+1)=λy1(k)+b11*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=2n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]1111111111111111111111111ifk∈ℤm1+1m2,y1(m1+1)=λm1+1φ1*(0)+∑r=0m1‍λm1-r×(∑l=1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]),y2(k+1)=λy2(k)-(b11*1)2b12*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=2n‍(b11*l)2b12*l[φ1*(k-m1)+b12*1b11*1φ2*(k-m1)]1111111111111111111111111111ifk∈ℤm1+1m2,y2(m1+1)=λm1+1φ2*(0)-∑r=0m1‍λm1-r×(∑l=1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]).
Formula (33) yields (for k∈ℤm1+2m2+1)
(93)y1(k)=λk-(m1+1)y1(m1+1)+∑r=m1+1k-1‍λk-1-r×(∑l=2n‍∑l=2n‍b11*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=2n‍b11*l[φ1*(r-ml)b12*1b11*11111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍[φ1*(0)+b12*1b11*1φ2*(0)])=λk-m1-1[([+b12*1b11*1φ2*(r-ml)]∑l=1n‍b11*l)λm1+1φ1*(0)+∑r=0m1‍λm1-rλk-m1-1λk-m1×(∑l=1n‍∑l=1n‍b11*l[b12*1b11*1φ1*(r-ml)11111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1n‍)]+∑r=m1+1k-1‍λk-1-r(b11*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑r=m1+1k-1‍λk-1∑r=m1+1k-1‍+∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111+b12*1b11*1φ2*(r-ml)])=λkφ1*(0)+∑r=0m1‍λk-1-r(∑l=1n‍b11*l[b12*1b11*1φ1*(r-ml)1111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1n‍)+∑r=m1+1k-1‍λk-1-r(∑l=2n‍b11*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑r=m1+1k-1‍+∑r=m1+1k-1‍+∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111+b12*1b11*1φ2*(r-ml)][φ1*(0)+b12*1b11*1φ2*(0)]∑l=2n‍)=λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍)+∑r=0m1‍λk-1-rb11*1[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)b11*1λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]=λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍)+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)λk-1-m1[b12*1b11*1φ1*(0)11111111111111111111111111111+b12*1b11*1φ2*(0)]∑r=0m1‍),(94)y2(k)=λk-(m1+1)y2(m1+1)-∑r=m1+1k-1‍λk-1-r×((b11*1)2b12*1λr-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)11111111111111111+b12*1b11*1φ2*(r-ml)][φ1*(0)+b12*1b11*1φ2*(0)](b11*1)2b12*1)=λk-m1-1[(∑l=1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])λm1+1φ2*(0)-∑r=0m1‍λm1-r=λk-112=×(∑l=1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)1111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1n‍(b11*l)2b12*l)(∑l=1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])]-∑r=m1+1k-1‍λk-1-r((b11*1)2b12*1λr-m1[φ1*(0)+b12*1b11*1φ2*(0)]-∑r=m1+1k-1‍λk-1-r-12+∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111111+b12*1b11*1φ2*(r-ml)](b11*1)2b12*1λr-m1[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ2*(0)-∑r=0m1‍λk-1-r(∑l=1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1n‍∑l=1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml))-∑r=m1+1k-1‍λk-1-r((b11*1)2b12*1λr-m1[φ1*(0)+b12*1b11*1φ2*(0)]-∑r=m1+1k-1‍λk-1-r-12+∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111+b12*1b11*1φ2*(r-ml)](b11*1)2b12*1λr-m1[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍(b11*l)2b12*l)-∑r=0m1‍λk-1-r(b11*1)2b12*1[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]-(k-1-m1)λk-1-m1(b11*1)2b12*1[φ1*(0)+b12*1b11*1φ2*(0)]=λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111111+b12*1b11*1b12*lφ2*(r-ml)]∑l=2n‍(b11*l)2b12*l)-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]-(b11*1)2b12*1-+(k-1-m1)λk-1-m1×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍).
Now we solve (88) for k∈ℤm2+1m3; that is, we consider the problem (with initial data deduced from (93), (94))
(95)y1(k+1)=λy1(k)+∑l=12‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=3n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]+∑l=3n‍b11*l+∑l=3n‍b11*l+1∑l=3n‍b11*l11∑l=3n‍ifk∈ℤm2+1m3,y1(m2+1)=λm2+1φ1*(0)+∑r=0m2‍λm2-r×(∑l=2n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+b11*1(∑r=0m1‍λm2-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+b11*1+b11*1+(m2-m1)λm2-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍),y2(k+1)=λy2(k)-∑l=12‍(b11*l)2b12*l[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=3n‍(b11*l)2b12*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]-∑l=3n‍(b11*l)2b12*l-∑l=3n‍(b11*l)2b12*l111111-∑l=3n‍ifk∈ℤm2+1m3,y2(m2+1)=λm2+1φ2*(0)-∑r=0m2‍λm2-r×(∑l=2n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1b12*lφ2*(r-ml)])-(b11*1)2b12*1(∑r=0m1‍λm2-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]-(b11*1)2b12*1-+(m2-m1)λm2-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍).
Applying formula (33) yields (for k∈ℤm2+2m3+1)
(96)y1(k)=λk-(m2+1)y1(m2+1)+∑r=m2+1k-1‍λk-1-r×(∑l=12‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]2222+∑l=3n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])=λk-m2-1[λm2+1φ1*(0)+∑r=0m2‍λm2-r222222222×(∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml))222222222+b11*1(∑r=0m1‍λm2-r[b12*1b11*1φ1*(r-m1)11111111111111111111111+b12*1b11*1φ2*(r-m1)]22222222222222+(m2-m1)λm2-m111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])∑r=0m1‍]+∑r=m2+1k-1‍λk-1-r(∑l=12‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]222222222222222+∑l=3n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=12‍)=λkφ1*(0)+∑r=0m2‍λk-1-r(∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)11111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍)+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]2222222+(m2-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍λk-1-r)+∑r=m2+1k-1‍λk-1-r(∑l=12‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]1111111111111111+∑l=3n‍b11*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=12‍)=λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=3n‍b11*l[b12*1b11*1φ1*(r-ml)11111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=3n‍)+∑r=0m2‍λk-1-rb11*2[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+b11*1+b11*1+(m2-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)])+(k-1-m2)(b11*1λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)](k-1-m2)(k-1)+b11*2λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)]b11*1λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=3n‍b11*l[b12*1b11*1φ1*(r-ml)11111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=3n‍)+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+b11*1+b11*1+(k-1-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m2‍)+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+b11*2+1+(k-1-m2)λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m2‍),(97)y2(k)=λk-(m2+1)y2(m2+1)-∑r=m2+1k-1‍λk-1-r×(∑l=12‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=3n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])=λk-m2-1[λm2+1φ2*(0)-∑r=0m2‍λm2-r×(∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)11111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml))-(b11*l)2b12*1(b12*1b11*1∑r=0m1‍λm2-r[b12*1b11*1φ1*(r-m1)111111111111111111111111111+b12*1b11*1φ2*(r-m1)]1111111111111111111+(m2-m1)λm2-m11111111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍)]-∑r=m2+1k-1‍λk-1-r(∑l=12‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]1111111111111111+∑l=3n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=12(b11*l)2b12*l‍)=λkφ2*(0)-∑r=0m2‍λk-1-r(∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍(b11*l)2b12*l)-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]-111b12*1-111b12*1-+(m2-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍)-∑r=m2+1k-1‍λk-1-r(∑l=12‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]-∑r=m2+1k-1‍λk-1-r1111-+∑l=3n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=12‍(b11*l)2b12*l)=λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=3n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=3n(b11*l)2b12*l‍)-∑r=0m2‍λk-1-r(b11*2)2b12*2[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]-(b11*1)2b12*1(∑r=0m1‍λk-1-r[b12*1b11*1φ1*(r-m1)11111111111111111111+b12*1b11*1φ2*(r-m1)]11111111111+(m2-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍)-(k-1-m2)×((b11*1)2b12*1λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]11111+(b11*2)2b12*2λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)](b11*1)2b12*1)=λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=3n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)11111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=3n‍(b11*l)2b12*l)-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]111111111111+(k-1-m1)λk-1-m1111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]1111111111111+(k-1-m2)λk-1-m21111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m2‍).
From (93)–(97) we deduce that expected form of the solution of the initial problem for k∈ℤms-1+1ms with initial data derived from the solution of previous equation for k∈ℤms-2+1ms-1 is
(98)y1(k)=λkφ1*(0)+∑r=0k-1‍λk-1-r×(∑l=sn‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+∑l=1s-1‍[b11*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]111111111111+(k-1-ml)λk-1-ml111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)]+(k-1-ml)λk-1-ml+11ifk∈ℤms-1+2ms+1,y1(k)=λkφ2*(0)-∑r=0k-1‍λk-1-r×(∑l=sn‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-∑l=1s-1[(b11*l)2b12*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]‍220000000001100+(k-1-ml)λk-1-ml111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)(b11*l)2b12*l]23232113213213213213213112ifk∈ℤms-1+2ms+1.
We solve (88) for k∈ℤms+1ms+1 with initial data deduced from (98); that is, we consider the problem
(99)y1(k+1)=λy1(k)+∑l=1s‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=s+1n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]∑l=s+1n‍∑l=s+1n‍b11*l∑l=s+1n‍b11*l∑l=s+1n‍b11*l111111ifk∈ℤms+1ms+1,y1(ms+1)=λms+1φ1*(0)+∑r=0ms‍λms-r(∑l=sn‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+∑l=1s-1‍b11*l(∑r=0ml‍λms-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]11111111111+(ms-ml)λms-ml11111111111×[φ1*(0)+b12*1b11*1φ2*(0)]),y2(k+1)=λy2(k)-∑l=1s‍(b11*l)2b12*l[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=s+1n‍(b11*l)2b12*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]∑l=s+1n‍(b11*l)2b12*l∑l=s+1n‍(b11*l)2b12*l∑l=s+1n‍(b11*l)2b12*l11ifk∈ℤms+1ms+1,y2(ms+1)=λms+1φ2*(0)-∑r=0ms‍λms-r(∑l=sn‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-∑l=1s-1‍(b11*l)2b12*l(∑r=0ml‍λms-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]-∑l=1s-1‍(b11*l)2b12*l-11+(ms-ml)λms-ml[φ1*(0)+b12*1b11*1φ2*(0)]).
Applying formula (33) yields (for k∈ℤms+2ms+1+1)
(100)y1(k)=λk-(ms+1)y1(ms+1)+∑r=ms+1k-1‍λk-1-r×(∑l=1s‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=s+1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])=λk-ms-1[λms+1φ1*(0)+∑r=0ms‍λms-r=λk-ms-1×(∑l=sn‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])=λk-ms-1=+∑l=1s-1‍b11*l(∑r=0ml‍λms-r[b12*1b11*1φ1*(r-ml)11111111111111111111111111+b12*1b11*1φ2*(r-ml)]111111111111111111+(ms-ml)λms-ml111111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)]+∑r=ms+1k-1‍λk-1-r(∑l=1s‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑r=ms+1k-1‍λk-1-r+11+∑l=s+1n‍b11*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1s‍)=λkφ1*(0)+∑r=0ms‍λk-1-r×(∑l=sn‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+∑l=1s-1‍b11*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]1111111111+(ms-ml)λk-1-ml1111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+∑r=ms+1k-1‍λk-1-r(∑l=1s‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑r=ms+1k-1‍λk-1-r+11+∑l=s+1n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111+b12*1b11*1φ2*(r-ml)b12*1b11*1]∑l=1s‍)=λkφ1*(0)+∑r=0k-1‍λk-1-r×(∑l=s+1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+b11*s∑r=0ms‍λk-1-r[φ1*(r-ms)+b12*1b11*1φ2*(r-ms)]+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+b11*1+b11*1+(ms-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+b11*2+b11*2+(ms-m2)λk-1-m2111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m2‍)+⋯+b11*s-1(∑r=0ms-1‍λk-1-r[b12*1b11*1φ1*(r-ms-1)1111111111111111111+b12*1b11*1φ2*(r-ms-1)]111111111+(ms-ms-1)λk-1-ms-1111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ms-1‍)+(k-1-ms)×(λk-1-m1b11*1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-m2b11*2[φ1*(0)+b12*1b11*1φ2*(0)]+⋯+λk-1-ms-1b11*s-1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-msb11*s[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ1*(0)+∑r=0k-1‍λk-1-r×(∑l=s+1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]111111111+(k-1-m1)λk-1-m1111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]111111111+(k-1-m2)λk-1-m2111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯+b11*s-1(∑r=0ms-1‍λk-1-r[b12*1b11*1φ1*(r-ms-1)1111111111111111111+b12*1b11*1φ2*(r-ms-1)]111111111+(k-1-ms-1)λk-1-ms-1111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+b11*s(∑r=0ms‍λk-1-r[φ1*(r-ms)+b12*1b11*1φ2*(r-ms)]+b11*s+b11*s+(k-1-ms)λk-1-ms[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ1*(0)+∑r=0k-1‍λk-1-r×(∑l=s+1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+∑l=1s‍b11*l(∑r=0ml‍λk-1-r[b12*1b11*1φ1*(r-ml)gggggggggggggggg+b12*1b11*1φ2*(r-ml)]1111111111+(k-1-ml)λk-1-ml1111111111×[φ1*(0)+b12*1b11*1φ2*(0)]),(101)y2(k)=λk-(ms+1)y2(ms+1)-∑r=ms+1k-1‍λk-1-r×(∑l=1s‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=s+1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1s‍(b11*l)2b12*l)=λk-ms-1[λms+1φ2*(0)-∑r=0ms‍λms-r=λk-ms-1=×(∑l=sn‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)11111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=sn‍(b11*l)2b12*l)1111111111-∑l=1s-1‍(b11*l)2b12*l(∑r=0ml‍λms-r[b12*1b11*1φ1*(r-ml)11111111111111111111111111111+b12*1b11*1φ2*(r-ml)]111111111111111111111+(ms-ml)λms-ml111111111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)]-∑r=ms+1k-1‍λk-1-r(∑l=1s‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]-∑r=ms+1k-1‍λk-∑r=ms+1k-1‍11+∑l=s+1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1s‍(b11*l)2b12*l)=λkφ2*(0)-∑r=0ms‍λk-1-r×(∑l=sn‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-∑l=1s-1‍(b11*l)2b12*l(∑r=0ml‍λk-1-r[b12*1b11*1φ1*(r-ml)1111111111111111111111+b12*1b11*1φ2*(r-ml)]-∑l=1s-1‍-∑l=1s-1‍-11+(ms-ml)λk-1-ml[φ1*(0)+b12*1b11*1φ2*(0)])-∑r=ms+1k-1‍λk-1-r(∑l=1s‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]-∑r=ms+1k-1‍λk-1-r-11+∑l=s+1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1s‍(b11*l)2b12*l)=λkφ2*(0)-∑r=0k-1‍λk-1-r×(∑l=s+1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-(b11*s)2b12*s∑r=0ms‍λk-1-r[φ1*(r-ms)+b12*1b11*1φ2*(r-ms)]-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]λkφ2*λkφ2*λk+(ms-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]λk-1-m2λk-11+(ms-m2)λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)])+⋯-(b11*s-1)2b12*s-1(∑r=0ms-1‍λk-1-r[φ1*(r-ms-1)+b12*1b11*1φ2*(r-ms-1)]111111111111+(ms-ms-1)λk-1-ms-1[φ1*(0)+b12*1b11*1φ2*(0)])-(k-1-ms)×(λk-1-m1(b11*1)2b12*1[φ1*(0)+b12*1b11*1φ2*(0)]11111+λk-1-m2(b11*2)2b12*2[φ1*(0)+b12*1b11*1φ2*(0)]+⋯11111+λk-1-ms-1(b11*s-1)2b12*s-1[φ1*(0)+b12*1b11*1φ2*(0)]11111+λk-1-ms(b11*s)2b12*s[φ1*(0)+b12*1b11*1φ2*(0)]λk-1-m1(b11*1)2b12*1[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ2*(0)-∑r=0k-1‍λk-1-r×(∑l=s+1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]11111111111+(k-1-m1)λk-1-m111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]11111111111+(k-1-m2)λk-1-m211111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯-(b11*s-1)2b12*s-1(∑r=0ms-1‍λk-1-r[b12*1b11*1φ1*(r-ms-1)1111111111111111111111+b12*1b11*1φ2*(r-ms-1)]11111111111111+(k-1-ms-1)λk-1-ms-111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ms-1‍)∑l=s+1n‍∑l=s+1n‍-(b11*s)2b12*s(∑r=0ms‍λk-1-r[b12*1b11*1φ1*(r-ms)1111111111111111111111111+b12*1b11*1φ2*(r-ms)]∑l=s+1n‍111111111∑l=s+1n‍+(k-1-ms)λk-1-ms∑l=s+1n‍111111111∑l=s+1n‍×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ms‍)=λkφ2*(0)-∑r=0k-1‍λk-1-r×(∑l=s+1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-∑l=1s‍(b11*l)2b12*l(∑r=0ml‍λk-1-r[b12*1b11*1φ1*(r-ml)11111111111111111111111+b12*1b11*1φ2*(r-ml)]1111111111111111+(k-1-ml)λk-1-ml1111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]).
In the end, we solve (88) for k∈ℤmn+1∞ with initial data deduced from (100) and (101); that is, we consider the problem
(102)y1(k+1)=λy1(k)+∑l=1n‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]=λy1(k)+∑l=1n‍b11*lλk-mlλy1(k)ifk∈ℤmn+1∞,y1(mn+1)=λmn+1φ1*(0)+∑r=0mn‍λmn-rb11*n×[φ1*(r-mn)+b12*1b11*1φ2*(r-mn)]+∑l=1n-1‍b11*l(∑r=0ml‍λmn-r[b12*1b11*1φ1*(r-ml)1111111111111111111+b12*1b11*1φ2*(r-ml)]+(mn-ml)λmn-ml×[φ1*(0)+b12*1b11*1φ2*(0)]),y2(k+1)=λy2(k)-∑l=1n‍(b11*l)2b12*lλk-ml×[φ1*(0)+b12*1b11*1φ2*(0)]ifk∈ℤmn+1∞,y2(mn+1)=λmn+1φ2*(0)-∑r=0mn‍λmn-r(b11*n)2b12*n×[φ1*(r-mn)+b12*1b11*1φ2*(r-mn)]-∑l=1n-1‍(b11*l)2b12*l(∑r=0ml‍λmn-r[b12*1b11*1φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]+(mn-ml)λmn-ml×[φ1*(0)+b12*1b11*1φ2*(0)]).
Applying formula (33) yields (for k∈ℤmn+2∞)
(103)y1(k)=λk-(mn+1)y1(mn+1)+∑r=mn+1k-1‍λk-1-r×(∑l=1n‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)])=λk-mn-1[[φ1*(0)+b12*1b11*1φ2*(0)]λmn+1φ1*(0)=λk-mn-111+∑r=0mn‍λmn-rb11*n[b12*1b11*1φ1*(r-mn)11111111111111111111111+b12*1b11*1φ2*(r-mn)]=λk-mn-111+∑l=1n-1‍b11*l(∑r=0ml‍λmn-r[b12*1b11*1φ1*(r-ml)11111111111111111111111111+b12*1b11*1φ2*(r-ml)]111111111111111111+(mn-ml)λmn-ml111111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)]+∑r=mn+1k-1‍λk-1-r(∑l=1n‍b11*lλr-ml1111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑l=1n‍)=λkφ1*(0)+∑r=0mn‍λk-1-rb11*n[b12*1b11*1φ1*(r-mn)11111111111111111111111+b12*1b11*1φ2*(r-mn)]+∑l=1n-1‍b11*l(∑r=0ml‍λk-1-r[b12*1b11*1φ1*(r-ml)1111111111111111111+b12*1b11*1φ2*(r-ml)]+∑l=1n-1‍b11*l+∑l=1n-1‍+(mn-ml)λk-1-ml+∑l=1n-1‍b11*l+∑l=1n-1‍×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)+∑r=mn+1k-1‍λk-1-r(∑l=1n‍b11*lλr-ml1111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)=λkφ1*(0)+∑r=0mn‍λk-1-rb11*n[b12*1b11*1φ1*(r-mn)1111111111111111111111+b12*1b11*1φ2*(r-mn)]+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+b11*1+b11*1+(mn-m1)λk-1-m1+b11*1+b11*1×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+b11*2b11*2+(mn-m2)λk-1-m2+b11*2b11*2×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯+b11*n-1(∑r=0mn-1‍λk-1-r[b12*1b11*1φ1*(r-mn-1)1111111111111111111+b12*1b11*1φ2*(r-mn-1)]+b11*n-1+b11*n+(mn-mn-1)λk-1-mn-1+b11*n-1+b11*n×[φ1*(0)+b12*1b11*1φ2*(0)])+(k-1-mn)×(λk-1-m1b11*1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-m2b11*2[φ1*(0)+b12*1b11*1φ2*(0)]+⋯+λk-1-mn-1b11*n-1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-mnb11*n[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ1*(0)+b11*1(∑r=0m1‍λk-1-r[b12*1b11*1φ1*(r-m1)1111111111111111111111111+b12*1b11*1φ2*(r-m1)]111111111111111+(k-1-m1)λk-1-m1111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]11111111+(k-1-m2)λk-1-m211111111×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯+b11*n-1(∑r=0mn-1‍λk-1-r[b12*1b11*1φ1*(r-mn-1)1111111111111111111+b12*1b11*1φ2*(r-mn-1)]1111111111+(k-1-mn-1)λk-1-mn-11111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0mn-1‍)+b11*n(∑r=0mn‍λk-1-r[φ1*(r-mn)+b12*1b11*1φ2*(r-mn)]+b11*n+b11*n+(k-1-mn)λk-1-mn111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0mn‍)=λkφ1*(0)+∑l=1n‍b11*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]1111111111+(k-1-ml)λk-1-ml1111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍),(104)y2(k)=λk-(mn+1)y2(mn+1)-∑r=mn+1k-1‍λk-1-r×(∑l=1n‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)])=λk-mn-1×[λmn+1φ2*(0)-∑r=0mn‍λmn-r(b11*n)2b12*n×[φ1*(r-mn)+b12*1b11*1φ2*(r-mn)]-∑l=1n-1‍(b11*l)2b12*l×(∑r=0ml‍λmn-r[b12*1b11*1φ1*(r-ml)11111111111111111+b12*1b11*1φ2*(r-ml)]+(mn-ml)λmn-ml×[φ1*(0)+b12*1b11*1φ2*(0)])∑r=0mn‍λmn-r(b11*n)2b12*n]-∑r=mn+1k-1‍λk-1-r(∑l=1n‍(b11*l)2b12*lλr-ml11111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑l=1n‍(b11*l)2b12*l)=λkφ2*(0)-∑r=0mn‍λk-1-r(b11*n)2b12*n×[φ1*(r-mn)+b12*1b11*1φ2*(r-mn)]-∑l=1n-1‍(b11*l)2b12*l(∑r=0ml‍λk-1-r[b12*1b11*1φ1*(r-ml)1111111111111111111111+b12*1b11*1φ2*(r-ml)]11111111111111+(mn-ml)λk-1-ml11111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)-∑r=mn+1k-1‍λk-1-r(∑l=1n‍(b11*l)2b12*lλr-ml1111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑l=1n‍(b11*l)2b12*l)=λkφ2*(0)-∑r=0mn‍λk-1-r(b11*n)2b12*n[b12*1b11*1φ1*(r-mn)1111111111111111111111111+b12*1b11*1φ2*(r-mn)]-(b11*1)2b12*1(∑r=0m1‍λk-1-r[b12*1b11*1φ1*(r-m1)11111111111111111111+b12*1b11*1φ2*(r-m1)]11111111111+(mn-m1)λk-1-m111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[b12*1b11*1φ1*(r-m2)11111111111111111111+b12*1b11*1φ2*(r-m2)]111111111111+(mn-m2)λk-1-m2111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯-(b11*n-1)2b12*n-1(∑r=0mn-1‍λk-1-r[b12*1b11*1φ1*(r-mn-1)1111111111111111111111+b12*1b11*1φ2*(r-mn-1)]111111111111+(mn-mn-1)λk-1-mn-1111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(k-1-mn)×(λk-1-m1(b11*1)2b12*1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-m2(b11*2)2b12*2[φ1*(0)+b12*1b11*1φ2*(0)]+⋯+λk-1-mn-1(b11*n-1)2b12*n-1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-mn(b11*n)2b12*n[φ1*(0)+b12*1b11*1φ2*(0)][φ1*(0)+b12*1b11*1φ2*(0)](b11*1)2b12*1)=λkφ2*(0)-(b11*1)2b12*1×(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]11111+(k-1-m1)λk-1-m111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[b12*1b11*1φ1*(r-m2)11111111111111111111+b12*1b11*1φ2*(r-m2)]11111111111+(k-1-m2)λk-1-m211111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯-(b11*n-1)2b12*n-1(∑r=0mn-1‍λk-1-r[b12*1b11*1φ1*(r-mn-1)1111111111111111111111+b12*1b11*1φ2*(r-mn-1)]1111111111111+(k-1-mn-1)λk-1-mn-11111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*n)2b12*n(∑r=0mn‍λk-1-r[b12*1b11*1φ1*(r-mn)11111111111111111111+b12*1b11*1φ2*(r-mn)]11111111111+(k-1-mn)λk-1-mn11111111111×[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ2*(0)-∑l=1n‍(b11*l)2b12*l×(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]11111+(k-1-ml)λk-1-ml11111×[φ1*(0)+b12*1b11*1φ2*(0)]).
Summing up all particular cases (90), (93), (96), (100), and (103) we have(105)y1(k)={φ1*(k)ifk∈ℤ-mn0,λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])ifk∈ℤ1m1+1,λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=2n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤm1+2m2+1,λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=3n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+(k-1-m2)λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤm2+2m3+1,⋮λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=s+1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+∑l=1s‍b11*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]+(k-1-ml)λk-1-ml[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤms+2ms+1+1,⋮λkφ1*(0)+∑l=1n‍b11*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]+(k-1-ml)λk-1-ml[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤmn+2∞,and from cases (91), (94), (97), (101), and (104) we conclude that(106)y2(k)={φ2*(k)ifk∈ℤ-mn0,λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])ifk∈ℤ1m1+1,λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=2n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1b12*lφ2*(r-ml)])-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤm1+2m2+1,λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=3n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+(k-1-m2)λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤm2+2m3+1,⋮λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=s+1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-∑l=1s‍(b11*l)2b12*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]+(k-1-ml)λk-1-ml[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤms+2ms+1+1,⋮λkφ2*(0)-∑l=1n‍(b11*l)2b12*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]+(k-1-ml)λk-1-ml[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤmn+2∞.Formula (81) is now a direct consequence of (105), (106), and (80).


### 2.1.6. Case (39) of a Double Real Root

If the matrixΛ has the form (39), the necessary and sufficient conditions (13)–(16), for (40), are reduced to (43), (44), (46), and
(107)|λ1b21*lb22*l|+|b11*lb12*l0λ|=λ(b11*l+b22*l)-b21*l=0.Then (43), (44), and (107) give b11*l=b22*l=b21*l=0.Theorem 9.
Let (1) be a weakly delayed system, (35) has a double root λ1,2=λ and the matrix Λ has the form (39). Then b11*l=b22*l=b21*l=0 and the solution of the initial problem (1), (3) is x(k)=𝒮y(k), y(k)=(y1(k),y2(k))T, and
(108)y1(k)={φ1*(k)ifk∈ℤ-mn0,λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍b12*lφ2*(r-ml)]ifk∈ℤ1m1+1,⋮λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+∑l=1s‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍]ifk∈ℤms+2ms+1+1,s=1,2,…,n-1,⋮λkφ1*(0)+kλk-1φ2*(0)+∑l=1n‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍]ifk∈ℤmn+2∞,(109)y2(k)={φ2*(k)ifk∈ℤ-mn0,λkφ2*(0)ifk∈ℤ1∞.Proof.
The system (40) can be written as
(110)y1(k+1)=λy1(k)+y2(k)+∑l=1n‍b12*ly2(k-ml),(111)y2(k+1)=λy2(k),k∈ℤ0∞.
Solving (111), we get
(112)y2(k)={φ2*(k)ifk∈ℤ-mn0,λkφ2*(0)ifk∈ℤ1∞,
then (110) turns into
(113)y1(k+1)={λy1(k)+λkφ2*(0)+∑l=1n‍b12*lφ2*(k-ml)ifk∈ℤ0m1,λy1(k)+λkφ2*(0)+b12*1λk-m1φ2*(0)+∑l=2n‍b12*lφ2*(k-ml)ifk∈ℤm1+1m2,λy1(k)+λkφ2*(0)+∑l=12‍b12*lλk-mlφ2*(0)+∑l=3n‍b12*lφ2*(k-ml)ifk∈ℤm2+1m3,⋮λy1(k)+λkφ2*(0)+∑l=1s‍b12*lλk-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(k-ml)ifk∈ℤms+1ms+1,s=3,4,…,n-1,⋮λy1(k)+λkφ2*(0)+∑l=1n‍b12*lλk-mlφ2*(0)ifk∈ℤmn+1∞.
Equation (113) can be solved in a way similar to that of (54) in the proof of Theorem 5 using (33).
First we solve (113) for k∈ℤ0m1. This means that we consider the problem
(114)y1(k+1)=λy1(k)+λkφ2*(0)+∑l=1n‍b12*lφ2*(k-ml)ifk∈ℤ0m1,y1(0)=φ1*(0).
With the aid of formula (33), we get
(115)y1(k)=λkφ1*(0)+∑r=0k-1‍λk-1-r[λrφ2*(0)+∑l=1n‍b12*lφ2*(r-ml)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍b12*lφ2*(r-ml)],k∈ℤ1m1+1.
Now we solve (113) for k∈ℤm1+1m2 with initial data deduced from (115); that is, we consider the problem
(116)y1(k+1)=λy1(k)+λkφ2*(0)+b12*1λk-m1φ2*(0)+∑l=2n‍b12*lφ2*(k-ml),k∈ℤm1+1m2,y1(m1+1)=λm1+1φ1*(0)+(m1+1)λm1φ2*(0)+∑r=0m1‍λm1-r[∑l=1n‍b12*lφ2*(r-ml)].
Applying formula (33) we get (for k∈ℤm1+2m2+1)
(117)y1(k)=λk-(m1+1)y1(m1+1)+∑r=m1+1k-1‍λk-1-r[∑l=2n‍λrφ2*(0)+b12*1λr-m1φ2*(0)+∑l=2n‍b12*lφ2*(r-ml)]=λk-m1-1[+∑r=0m1‍λm1-r[∑l=1n‍b12*lφ2*(r-ml)]]λm1+1φ1*(0)+(m1+1)λm1φ2*(0)=λk-m1-1111+∑r=0m1‍λm1-r[∑l=1n‍b12*lφ2*(r-ml)]]+∑r=m1+1k-1‍λk-1-r[+∑l=2n‍b12*lφ2*(r-ml)λrφ2*(0)+b12*1λr-m1φ2*(0)+∑r=m1+1k-1‍λk-1-r11111+∑l=2n‍b12*lφ2*(r-ml)]=λkφ1*(0)+(m1+1)λk-1φ2*(0)+∑r=0m1‍λk-1-r[∑l=1n‍b12*lφ2*(r-ml)]+(k-m1-1)λk-1φ2*(0)+∑r=m1+1k-1‍λk-1-r[b12*1λr-m1φ2*(0)+∑l=2n‍b12*lφ2*(r-ml)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+b12*1+b12*1+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍λk-1-rφ2*(r-m1)].
Now we solve (113) for k∈ℤm2+1m3 with initial data deduced from (117); that is, we consider the problem
(118)y1(k+1)=λy1(k)+λkφ2*(0)+∑l=12‍b12*lλk-mlφ2*(0)+∑l=3n‍b12*lφ2*(k-ml)k∈ℤm2+1m3,y1(m2+1)=λm2+1φ1*(0)+(m2+1)λm2φ2*(0)+∑r=0m2‍λm2-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λm2-rφ2*(r-m1)+b12*1+ll+(m2-m1)λm2-m1φ2*(0)∑r=0m1‍λm2-rφ2*(r-m1)].
Applying formula (33) yields (for k∈ℤm2+2m3+1)
(119)y1(k)=λk-(m2+1)y1(m2+1)+∑r=m2+1k-1‍λk-1-r[λrφ2*(0)+∑l=12‍b12*lλr-mlφ2*(0)+∑l=3n‍b12*lφ2*(r-ml)]=λk-m2-1[+∑r=0m2‍λm2-r[∑l=2n‍b12*lφ2*(r-ml)]λm2+1φ1*(0)+(m2+1)λm2φ2*(0)=λk-m2-1=+∑r=0m2‍λm2-r[∑l=2n‍b12*lφ2*(r-ml)]=λk-m2-1=+b12*1[∑r=0m1‍λm2-rφ2*(r-m1)=λk-m2-1=+(m2-m1)λm2-m1φ2*(0)∑r=0m1‍λm2-rφ2*(r-m1)]]+∑r=m2+1k-1‍λk-1-r[λrφ2*(0)+∑l=12‍b12*lλr-mlφ2*(0)+∑r=m2+1k-1‍λk-1-r+1+∑l=3n‍b12*lφ2*(r-ml)]=λkφ1*(0)+(m2+1)λk-1φ2*(0)+∑r=0m2‍λk-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+b12*1++(m2-m1)λk-1-m1φ2*(0)∑r=0m1‍λk-1-rφ2*(r-m1)]+(k-1-m2)λk-1φ2*(0)+∑r=m2+1k-1‍λk-1-r[∑l=12‍b12*lλr-mlφ2*(0)+∑l=3n‍b12*lφ2*(r-ml)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=3n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+b12*1+b12*1+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍λk-1-rφ2*(r-m1)]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)+b12*2+b12*2+(k-1-m2)λk-1-m2φ2*(0)∑r=0m2‍λk-1-rφ2*(r-m2)].
From (115), (117), and (119) we deduce that expected form of the solution of the initial problem for k∈ℤms-1+1ms with initial data derived from the solution of previous equation for k∈ℤms-2+1ms-1 is
(120)y1(k)=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+∑l=1s-1‍b12*l+∑l=1s-1‍+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍λk-1-rφ2*(r-ml)]+∑l=1s-1‍b12*l+∑l=1s-1‍b12*l+∑l=1s-1‍b12*lifk∈ℤms-1+2ms+1.
We solve (113) for k∈ℤms+1ms+1 with initial data deduced from (120); that is, we consider the problem
(121)y1(k+1)=λy1(k)+λkφ2*(0)+∑l=1s‍b12*lλk-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(k-ml),k∈ℤms+1ms+1,y1(ms+1)=λms+1φ1*(0)+(ms+1)λmsφ2*(0)+∑r=0ms‍λms-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=0ml‍λms-rφ2*(r-ml)+∑l=1s-1‍b12*l+∑l=1s-1‍+(ms-ml)λms-mlφ2*(0)∑r=0ml‍λms-rφ2*(r-ml)].
Applying formula (33) yields (for k∈ℤms+2ms+1+1)
(122)y1(k)=λk-(ms+1)y1(ms+1)+∑r=ms+1k-1‍λk-1-r[λrφ2*(0)+∑l=1s‍b12*lλr-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(r-ml)]=λk-ms-1×[∑r=0ml‍λms+1φ1*(0)+(ms+1)λmsφ2*(0)+∑r=0ms‍λms-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=0ml‍λms-rφ2*(r-ml)111111111111+(ms-ml)λms-mlφ2*(0)∑r=0ml‍]]+∑r=ms+1k-1‍λk-1-r[λrφ2*(0)+∑l=1s‍b12*lλr-mlφ2*(0)111111111111111+∑l=s+1n‍b12*lφ2*(r-ml)]=λkφ1*(0)+(ms+1)λk-1φ2*(0)+∑r=0ms‍λk-1-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)11111111111+(ms-ml)λk-1-mlφ2*(0)∑r=0ml‍]+(k-1-ms)λk-1φ2*(0)+∑r=ms+1k-1‍λk-1-r[∑l=1s‍b12*lλr-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(r-ml)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r×[∑l=s+1n‍b12*lφ2*(r-ml)]+∑r=0ms‍λk-1-rb12*sφ2*(r-ms)+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)11111111111+(ms-m1)λk-1-m1φ2*(0)∑r=0m1‍]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)11111111111+(ms-m2)λk-1-m2φ2*(0)∑r=0m2‍]+⋯+b12*s-1[∑r=0ms-1‍λk-1-rφ2*(r-ms-1)111111111111+(ms-ms-1)λk-1-ms-1φ2*(0)∑r=0ms-1‍]+(k-1-ms)[λk-1-m1b12*1φ2*(0)1111111111111111+λk-1-m2b12*2φ2*(0)11111111111111+⋯+λk-1-ms-1b12*s-1φ2*(0)11111111111111+λk-1-m2b12*sφ2*(0)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)11111111111+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)11111111111+(k-1-m2)λk-1-m2φ2*(0)∑r=0m2‍]+⋯+b12*s-1[∑r=0ms-1‍λk-1-rφ2*(r-ms-1)11111111111+(k-1-ms-1)λk-1-ms-1φ2*(0)∑r=0ms-1‍]+b12*s[∑r=0ms‍λk-1-rφ2*(r-ms)111111111+(k-1-ms)λk-1-msφ2*(0)∑r=0ms‍]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+∑l=1s‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)111111111111+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍].
In the end, we solve (113) for k∈ℤmn+1∞ with initial data deduced from (122); that is, we consider the problem
(123)y1(k+1)=λy1(k)+λkφ2*(0)+∑l=1n‍b12*lλk-mlφ2*(0)ifk∈ℤmn+1∞,y1(mn+1)=λmn+1φ1*(0)+(mn+1)λmnφ2*(0)+∑r=0mn‍λmn-rb12*nφ2*(r-mn)+∑l=1n-1‍b12*l[∑r=0ml‍λmn-rφ2*(r-ml)+∑l=1n-1‍b12*l+∑l=1n-1‍+(mn-ml)λmn-mlφ2*(0)∑r=0ml‍].
Applying formula (33) yields (for k∈ℤmn+2∞)
(124)y1(k)=λk-(mn+1)y1(mn+1)+∑r=mn+1k-1‍λk-1-r[λrφ2*(0)+∑l=1n‍b12*lλr-mlφ2*(0)]=λk-mn-1[∑r=0ml‍λmn+1φ1*(0)+(mn+1)λmnφ2*(0)=1111λk-mn+∑r=0mn‍λmn-rb12*nφ2*(r-mn)=1111λk-mn+∑l=1n-1‍b12*l[∑r=0ml‍λmn-rφ2*(r-ml)=λk-mn=1λk-mn=λk-+(mn-ml)λmn-mlφ2*(0)∑r=0ml‍]]+∑r=mn+1k-1‍λk-1-r[λrφ2*(0)+∑l=1n‍b12*lλr-mlφ2*(0)]=λkφ1*(0)+(mn+1)λk-1φ2*(0)+∑r=0mn‍λk-1-rb12*nφ2*(r-mn)+∑l=1n-1‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+(mn-ml)λk-1-mlφ2*(0)∑r=0ml‍]+(k-1-mn)λk-1φ2*(0)+∑r=mn+1k-1‍λk-1-r[∑l=1n‍b12*lλr-mlφ2*(0)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0mn‍λk-1-rb12*nφ2*(r-mn)+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+b12*1+b12*1+(mn-m1)λk-1-m1φ2*(0)∑r=0m1‍]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)+b12*2+b12*2+(mn-m2)λk-1-m2φ2*(0)∑r=0m2‍]+⋯+b12*n-1[∑r=0mn-1‍λk-1-rφ2*(r-mn-1)+b12*n-1+b12*n-1+(mn-mn-1)λk-1-mn-1φ2*(0)∑r=0mn-1‍]+(k-1-mn)×[λk-1-m1b12*1φ2*(0)+λk-1-m2b12*2φ2*(0)+⋯+λk-1-mn-1b12*n-1φ2*(0)+λk-1-mnb12*nφ2*(0)]=λkφ1*(0)+kλk-1φ2*(0)+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+b12*1+b12*1+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)+b12*1+b12*1+(k-1-m2)λk-1-m2φ2*(0)∑r=0m2‍]+⋯+b12*n-1[∑r=0mn-1‍λk-1-rφ2*(r-mn-1)+b12*1+b12*1+(k-1-mn-1)λk-1-mn-1φ2*(0)∑r=0mn-1‍]+b12*n[∑r=0mn‍λk-1-rφ2*(r-mn)+b12*1+b12*1+(k-1-mn)λk-1-mnφ2*(0)∑r=0mn‍]=λkφ1*(0)+kλk-1φ2*(0)+∑l=1n‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+b12*1+b12*1+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍].
Summing up all particular cases (115)–(124), we get
(125)y1(k)={φ1*(k)ifk∈ℤ-mn0,λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍b12*lφ2*(r-ml)]ifk∈ℤ1m1+1,λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍]ifk∈ℤm1+2m2+1,λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=3n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)+(k-1-m2)λk-1-m2φ2*(0)∑r=0m2‍]ifk∈ℤm2+2m3+1,⋮λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+∑l=1s‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍]ifk∈ℤms+2ms+1+1,⋮λkφ1*(0)+kλk-1φ2*(0)+∑l=1n‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍]ifk∈ℤmn+2∞.
Formulas (108) and (109) are consequences of (125), (112).


## 2.1.1. Case (36) of Two Real Distinct Roots

In this case, we haveΛ=Λ1 and Λ1k=diag⁡(λ1k,λ2k). The necessary and sufficient conditions (13)–(16) for (40) turn into
(43)b11*l+b22*l=0,(44)|b11*lb12*lb21*lb22*l|=b11*lb22*l-b12*lb21*l=0,(45)|λ10b21*lb22*l|+|b11*lb12*l0λ2|=λ1b22*l+λ2b11*l=0,(46)|b11*lb12*lb21*vb22*v|+|b11*vb12*vb21*lb22*l|=0.
Since λ1≠λ2, (43) and (45) yield b11*l=b22*l=0, then, from (44), we get b12*lb21*l=0, so that either b21*l=0 or b12*l=0. In view of assumptions Bl≠Θ, l=1,2,…,n, we conclude that only the following cases I, II are possible:(I)
b11*l=b22*l=b21*l=0, b12*l≠0, l=1,2,…,n,(II)
b11*l=b22*l=b12*l=0, b21*l≠0, l=1,2,…,n.In Theorem5 both cases I, II are analyzed.Theorem 5.
Let (1) be a weakly delayed system and (35) has two real distinct roots λ1, λ2. If case (I) holds, then the solution of the initial problem (1), (3) is x(k)=Sy(k), k∈ℤ-mn∞, where y(k) has the form
(47)y(k)={φ*(k)ifk∈ℤ-mn0,Λ1kφ*(0)+∑r=0k-1‍λ1k-1-r[∑l=1n‍b12*lΦ2(r-ml)]ifk∈ℤ1m1+1,⋮Λ1kφ*(0)+∑r=0k-1‍λ1k-1-r[∑l=s+1n‍b12*lΦ2(r-ml)]+∑l=1s‍b12*l[∑r=ml+1k-1‍∑r=0ml‍λ1k-1-rΦ2(r-ml)+Φ2(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml]ifk∈ℤms+2ms+1+1,s=1,2,…,n-1,⋮Λ1kφ*(0)+∑l=1n‍b12*l[∑r=ml+1k-1‍∑r=0ml‍λ1k-1-rΦ2(r-ml)+Φ2(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml]ifk∈ℤmn+2∞.
If case (II) is true, then the solution of initial problem (1), (3) is x(k)=Sy(k), k∈ℤ-mn∞, where y(k) has the form
(48)y(k)={φ*(k)ifk∈ℤ-mn0,Λ1kφ*(0)+∑r=0k-1‍λ2k-1-r[∑l=1n‍b21*lΦ1(r-ml)]ifk∈ℤ1m1+1,⋮Λ1kφ*(0)+∑r=0k-1‍λ2k-1-r[∑l=s+1n‍b21*lΦ1(r-ml)]+∑l=1s‍b21*l[∑r=ml+1k-1∑r=0ml‍λ2k-1-rΦ1(r-ml)+Φ1(0)∑r=ml+1k-1‍λ1r-mlλ2k-1-r]0000000000000000000000ifk∈ℤms+2ms+1+1,00000000000000000000s=1,2,…,n-1,⋮Λ1kφ*(0)+∑l=1n‍b21*l[∑r=ml+1k-1λ1r-mlλ2k-1-r‍∑r=0ml‍λ2k-1-rΦ1(r-ml)+Φ1(0)∑r=ml+1k-1λ1r-mlλ2k-1-r‍]0000000000000000000000000ifk∈ℤmn+2∞.Proof.
If case (I) is true, then the transformed system (40) takes the form
(49)y1(k+1)=λ1y1(k)+∑l=1n‍b12*ly2(k-ml),(50)y2(k+1)=λ2y2(k),k∈ℤ0∞,
and if case (II) holds, then (40) takes the form
(51)y1(k+1)=λ1y1(k),(52)y2(k+1)=λ2y2(k)+∑l=1n‍b21*ly1(k-ml),k∈ℤ0∞.
We investigate only the initial problem (49), (50), (41) since the initial problem (51), (52), (41) can be examined in a similar way.
From (50), (41), we get
(53)y2(k)={φ2*(k)ifk∈ℤ-mn0,λ2kφ2*(0)ifk∈ℤ1∞,
then (49) becomes
(54)y1(k+1)={λ1y1(k)+∑l=1n‍b12*lφ2*(k-ml)ifk∈ℤ0m1,λ1y1(k)+b12*1λ2k-m1φ2*(0)+∑l=2n‍b12*lφ2*(k-ml)ifk∈ℤm1+1m2,λ1y1(k)+∑l=12‍b12*lλ2k-mlφ2*(0)+∑l=3n‍b12*lφ2*(k-ml)ifk∈ℤm2+1m3,⋮λ1y1(k)+∑l=1s‍b12*lλ2k-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(k-ml)ifk∈ℤms+1ms+1,s=3,4,…,n-1,⋮λ1y1(k)+∑l=1n‍b12*lλ2k-mlφ2*(0)ifk∈ℤmn+1∞.
First, we solve this equation for k∈ℤ0m1. This means that we consider the problem
(55)y1(k+1)=λ1y1(k)+∑l=1n‍b12*lφ2*(k-ml),k∈ℤ0m1,y1(0)=φ1*(0).
With the aid of formula (33), we get
(56)y1(k)=λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=1n‍b12*lφ2*(r-ml)],1111111111111111111111111k∈ℤ1m1+1.
Now we solve (54) for k∈ℤm1+1m2 with initial data deduced from (56); that is, we consider the problem
(57)y1(k+1)=λ1y1(k)+b12*1λ2k-m1φ2*(0)+∑l=2n‍b12*lφ2*(k-ml),k∈ℤm1+1m2,y1(m1+1)=λ1m1+1φ1*(0)+∑r=0m1‍λ1m1-r[∑l=1n‍b12*lφ2*(r-ml)].
Applying formula (33) we get (for k∈ℤm1+2m2+1)
(58)y1(k)=λ1k-(m1+1)y1(m1+1)+∑r=m1+1k-1‍λ1k-1-r[∑l=2n‍b12*lφ2*(r-ml)b12*1λ2r-m1φ2*(0)+∑r=m1+1k-1‍λ1k-1-r1212+∑l=2n‍b12*lφ2*(r-ml)]=λ1k-m1-1[[∑l=1n‍b12*lφ2*(r-ml)]λ1m1+1φ1*(0)+∑r=0m1‍λ1m1-r[∑l=1n‍b12*lφ2*(r-ml)]]+∑r=m1+1k-1‍λ1k-1-r[+∑l=2n‍b12*lφ2*(r-ml)b12*1λ2r-m1φ2*(0)+∑r=m1+1k-1‍λ1k-1-r++∑l=2n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0m1‍λ1k-1-r[∑l=1n‍b12*lφ2*(r-ml)]+∑r=m1+1k-1‍λ1k-1-r[+∑l=2n‍b12*lφ2*(r-ml)b12*1λ2r-m1φ2*(0)+∑r=m1+1k-1‍λ1k-1-r+++∑l=2n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1k-1‍∑r=0m1‍λ1k-1-rφ2*(r-m1)+b12*1+b12*1+φ2*(0)∑r=m1+1k-1‍λ1k-1-rλ2r-m1].
Now we solve (54) for k∈ℤm2+1m3 with initial data deduced from (58); that is, we consider the problem
(59)y1(k+1)=λ1y1(k)+∑l=12‍b12*lλ2k-mlφ2*(0)+∑l=3n‍b12*lφ2*(k-ml),k∈ℤm2+1m3,y1(m2+1)=λ1m2+1φ1*(0)+∑r=0m2‍λ1m2-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1m2‍∑r=m1+1m2‍λ1m2-rλ2r-m1∑r=0m1‍λ1m2-rφ2*(r-m1)+φ2*+φ2*+φ2*(0)∑r=m1+1m2‍λ1m2-rλ2r-m1].
Applying formula (33) yields (for k∈ℤm2+2m3+1)
(60)y1(k)=λ1k-(m2+1)y1(m2+1)+∑r=m2+1k-1‍λ1k-1-r[∑l=12‍b12*lλ2r-mlφ2*(0)+∑l=3n‍b12*lφ2*(r-ml)]=λ1k-m2-1[∑r=m1+1m2‍λ1m2+1φ1*(0)+∑r=0m2‍λ1m2-r[∑l=2n‍b12*lφ2*(r-ml)]111111111+b12*1[∑r=m1+1m2‍∑r=0m1‍λ1m2-rφ2*(r-m1)111111111111+φ2*(0)∑r=m1+1m2‍λ1m2-rλ2r-m1]∑r=m1+1m2‍]+∑r=m2+1k-1‍λ1k-1-r[∑l=12‍b12*lλ2r-mlφ2*(0)+∑l=3n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0m2‍λ1k-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1m2‍∑r=0m1‍λ1k-1-rφ2*(r-m1)1111111+φ2*(0)∑r=m1+1m2‍λ1k-1-rλ2r-m1]+∑r=m2+1k-1‍λ1k-1-r[∑l=12‍b12*lλ2r-mlφ2*(0)11111111111111+∑l=3n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=3n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1k-1‍∑r=0m1‍λ1k-1-rφ2*(r-m1)11111111+φ2*(0)∑r=m1+1k-1‍λ1k-1-rλ2r-m1]+b12*2[∑r=m2+1k-1‍∑r=m2+1k-1λ1k-1-rλ2r-m2∑r=0m2‍λ1k-1-rφ2*(r-m2)111111111+φ2*(0)∑r=m2+1k-1λ1k-1-rλ2r-m2].
From (56), (58), and (60) we deduce that expected form of the solution of the initial problem for k∈ℤms-1+1ms with initial data derived from the solution of previous equation for k∈ℤms-2+1ms-1 is
(61)y1(k)=λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=ml+1k-1‍λ1k-1-rλ2r-ml∑r=0ml‍λ1k-1-rφ2*(r-ml)+∑l=1s-1‍b12*l+∑l=1s-1‍+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml]+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml+ifk∈ℤms-1+2ms+1.
We solve (54) for k∈ℤms+1ms+1 with initial data deduced from (61); that is, we consider the problem
(62)y1(k+1)=λ1y1(k)+∑l=1s‍b12*lλ2k-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(k-ml),k∈ℤms+1ms+1,y1(ms+1)=λ1ms+1φ1*(0)+∑r=0ms‍λ1ms-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=ml+1ms‍λ1ms-rλ2r-ml∑r=0ml‍λ1ms-rφ2*(r-ml)+∑l=1s-1‍b12*l+∑l=1s-1‍+φ2*(0)∑r=ml+1ms‍λ1ms-rλ2r-ml].
Applying formula (33) yields (for k∈ℤms+2ms+1+1)
(63)y1(k)=λ1k-(ms+1)y1(ms+1)+∑r=ms+1k-1‍λ1k-1-r[∑l=1s‍b12*lλ2r-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(r-ml)]=λ1k-(ms+1)[[∑r=ml+1ms‍λ1ms-rλ2r-ml]λ1ms+1φ1*(0)+∑r=0ms‍λ1ms-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=ml+1ms‍λ1ms-rλ2r-ml∑r=0ml‍λ1ms-rφ2*(r-ml)111111111144444444+φ2*(0)∑r=ml+1ms‍λ1ms-rλ2r-ml]∑r=0ms‍λ1ms-r[∑l=sn‍b12*lφ2*(r-ml)]]+∑r=ms+1k-1‍λ1k-1-r[∑l=1s‍b12*lλ2r-mlφ2*(0)111111111144444+∑l=s+1n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0ms‍λ1k-1-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=ml+1ms‍λ1k-1-rλ2r-ml∑r=ml+1ms‍∑r=0ml‍λ1k-1-rφ2*(r-ml)11111111111+φ2*(0)∑r=ml+1ms‍λ1k-1-rλ2r-ml]+∑r=ms+1k-1‍λ1k-1-r[∑l=1s‍b12*lλ2r-mlφ2*(0)111111111144444+∑l=s+1n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0ms‍λ1k-1-r[∑l=s+1n‍b12*sφ2*(r-ms)11111111114444444444+∑l=s+1n‍b12*lφ2*(r-ml)]+b12*1[∑r=ml+1ms‍λ1k-1-rλ2r-ml∑r=0m1‍λ1k-1-rφ2*(r-m1)11111111114+φ2*(0)∑r=m1+1ms‍λ1k-1-rλ2r-m1]+b12*2[∑r=m2+1ms‍∑r=0m2‍λ1k-1-rφ2*(r-m2)111111114+φ2*(0)∑r=m2+1ms‍λ1k-1-rλ2r-m2]+⋯+b12*s-1[∑r=ms-1+1ms‍∑r=0ms-1‍λ1k-1-rφ2*(r-ms-1)11111111114+φ2*(0)∑r=ms-1+1ms‍λ1k-1-rλ2r-ms-1]+b12*1[φ2*(0)∑r=ms+1k-1‍λ1k-1-rλ2r-m1]+b12*2[φ2*(0)∑r=ms+1k-1‍λ1k-1-rλ2r-m2]+…+b12*s[φ2*(0)∑r=ms+1k-1‍λ1k-1-rλ2r-ms]+∑r=ms+1k-1‍λ1k-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]=λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+∑l=1s‍b12*l[∑r=ml+1k-1‍∑r=0ml‍λ1k-1-rφ2*(r-ml)11111111114+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml].
In the end we solve (54) for k∈ℤmn+1∞ with initial data deduced from (63); that is, we consider the problem
(64)y1(k+1)=λ1y1(k)+∑l=1n‍b12*lλ2k-mlφ2*(0),k∈ℤmn+1∞,y1(mn+1)=λ1mn+1φ1*(0)+∑r=0mn‍λ1mn-rb12*nφ2*(r-mn)+∑l=1n-1‍b12*l[∑r=ml+1mn‍λ1mn-rλ2r-ml∑r=0ml‍λ1mn-rφ2*(r-ml)+∑l=1n-1‍b12*l+222+φ2*(0)∑r=ml+1mn‍λ1mn-rλ2r-ml].
Applying formula (33) yields (for k∈ℤmn+2∞)
(65)y1(k)=λ1k-(mn+1)y1(mn+1)+∑r=mn+1k-1‍λ1k-1-r[∑l=1n‍b12*lλ2r-mlφ2*(0)]=λ1k-(mn+1)[[∑r=ml+1mn‍λ1mn-rλ2r-ml]λ1mn+1φ1*(0)+∑r=0mn‍λ1mn-rb12*nφ2*(r-mn)λ1k-(mn+1)λ1+∑l=1n-1‍b12*l[∑r=ml+1mn‍λ1mn-rλ2r-ml∑r=0ml‍λ1mn-rφ2*(r-ml)λ1k-(mn+1)1λ1k-(mn+1)λ1+φ2*(0)∑r=ml+1mn‍λ1mn-rλ2r-ml]]+∑r=mn+1k-1‍λ1k-1-r[∑l=1n‍b12*lλ2r-mlφ2*(0)]=λ1kφ1*(0)+∑r=0mn‍λ1k-1-rb12*nφ2*(r-mn)+∑l=1n-1‍b12*l[∑r=ml+1mn‍λ1k-1-rλ2r-ml∑r=0ml‍λ1k-1-rφ2*(r-ml)+∑l=1n-1‍b12*l+∑l=1n-1‍b12*l+φ2*(0)∑r=ml+1mn‍λ1k-1-rλ2r-ml]+∑r=mn+1k-1‍λ1k-1-r[∑l=1n‍b12*lλ2r-mlφ2*(0)]=λ1kφ1*(0)+∑r=0mn‍λ1k-1-rb12*nφ2*(r-mn)+b12*1[∑r=m1+1ms‍λ1k-1-rλ2r-m1∑r=0m1‍λ1k-1-rφ2*(r-m1)+b12*1+b12*1+φ2*(0)∑r=m1+1ms‍λ1k-1-rλ2r-m1]+b12*2[∑r=m2+1ms‍λ1k-1-rλ2r-m2∑r=0m2‍λ1k-1-rφ2*(r-m2)+b12*2+b12*2+φ2*(0)∑r=m2+1ms‍λ1k-1-rλ2r-m2]+⋯+b12*n-1[∑r=mn-1+1ms‍λ1k-1-rλ2r-mn-1∑r=0mn-1‍λ1k-1-rφ2*(r-mn-1)+b12*n-1+b12*n-1+φ2*(0)∑r=mn-1+1ms‍λ1k-1-rλ2r-mn-1]+b12*1[φ2*(0)∑r=ms+1k-1‍λ1k-1-rλ2r-m1]+b12*2[φ2*(0)∑r=ms+1k-1‍λ1k-1-rλ2r-m2]+⋯+b12*n[φ2*(0)∑r=mn+1k-1‍λ1k-1-rλ2r-mn]=λ1kφ1*(0)+∑l=1n‍b12*l[∑r=ml+1k-1‍λ1k-1-rλ2r-ml∑r=0ml‍λ1k-1-rφ2*(r-ml)∑l=1n‍b12*l∑l=1n‍b12*l∑l=1n‍b12*l∑l=1n‍+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml].
Summing up all particular cases (56)–(65) we have
(66)y1(k)={φ1*(k)ifk∈ℤ-mn0,λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=1n‍b12*lφ2*(r-ml)]00ifk∈ℤ1m1+1,λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1k-1‍∑r=0m1‍λ1k-1-rφ2*(r-m1)+φ2*(0)∑r=m1+1k-1‍λ1k-1-rλ2r-m1]0ifk∈ℤm1+2m2+1,λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=3n‍b12*lφ2*(r-ml)]+b12*1[∑r=m1+1k-1‍∑r=0m1‍λ1k-1-rφ2*(r-m1)+φ2*(0)∑r=m1+1k-1‍λ1k-1-rλ2r-m1]+b12*2[∑r=m2+1k-1‍∑r=0m2‍λ1k-1-rφ2*(r-m2)+φ2*(0)∑r=m2+1k-1‍λ1k-1-rλ2r-m2]00000ifk∈ℤm2+2m3+1,⋮λ1kφ1*(0)+∑r=0k-1‍λ1k-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+∑l=1s‍b12*l[∑r=ml+1k-1‍∑r=0ml‍λ1k-1-rφ2*(r-ml)+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml]000ifk∈ℤms+2ms+1+1,⋮λ1kφ1*(0)+∑l=1n‍b12*l[∑r=ml+1k-1‍∑r=0ml‍λ1k-1-rφ2*(r-ml)+φ2*(0)∑r=ml+1k-1‍λ1k-1-rλ2r-ml]00000ifk∈ℤmn+2∞.
Now, taking into account (42), formula (47) is a consequence of (53) and (66). Formula (48) can be proved in a similar way.
Finally, we note that both formulas (47), (48) remain valid for b12*l=b21*l=0. In this case, the transformed system (1) reduces to a system without delays. This possibility is excluded by condition (2).


## 2.1.2. Case (37) of Two Complex Conjugate Roots

The necessary and sufficient conditions (13)–(16) take the forms (43), (44), (46), and
(67)|pqb21*lb22*l|+|b11*lb12*l-qp|=p(b11*l+b22*l)+q(b12*l-b21*l)=0,
where l,v=1,2,…,n and v>l.The system of conditions (43), (44), and (67) gives b12*l=b21*l, (b11*l)2=-(b12*l)2 and admits only one possibility; namely,
(68)b11*l=b22*l=b12*l=b21*l=0.Consequently,B*l=Θ, Bl=Θ.The initial problem (1), (3) reduces to a problem without delay
(69)x(k+1)=Ax(k),x(k)=φ(k),k∈ℤ-mn0
and, obviously,
(70)x(k)={φ(k)ifk∈ℤ-mn0,Akφ(0)ifk∈ℤ1∞.
From this discussion, the next theorem follows.Theorem 6.
There exists no weakly delayed system (1) if Λ has the form (37).Finally, we note that the assumption (2) alone excludes this case.


## 2.1.3. Case (38) of Double Real Root

In this case we haveΛ=Λ3 and Λ3k=diag⁡(λk,λk). For (40), the necessary and sufficient conditions (13)–(16) are reduced to (43), (44), (46), and
(71)|λ0b21*lb22*l|+|b11*lb12*l0λ|=λ(b11*l+b22*l)=0,
where l=1,2,…,n.From (43), (44), and (71), we get b12*lb21*l=-(b11*l)2. From the condition (46) we get
(72)b11*lb22*v-b12*lb21*v+b22*lb11*v-b21*lb12*v=0,
where l,v=1,2,…,n and v>l. Multiplying (72) by b12*lb12*v, we have
(73)b11*lb22*vb12*lb12*v-(b12*l)2b21*vb12*v+b22*lb11*vb12*lb12*v-b21*lb12*l(b12*v)2=0.
Substituting b12*lb21*l=-(b11*l)2, b12*vb21*v=-(b11*v)2 into (73) and using (43) we obtain
(74)-b11*lb11*vb12*lb12*v+(b12*l)2(b11*v)2-b11*lb11*vb12*lb12*v+(b11*l)2(b12*v)2=0.
The equation (74) can be written as
(75)(b12*lb11*v-b12*vb11*l)2=0,b12*lb11*v=b12*vb11*l.Now we will analyse the two possible cases:b12*lb21*l=0 and b12*lb21*l≠0.For the caseb12*lb21*l=0, we have from (43), (44) that b11*l=b22*l=0 and b12*l=0 or b21*l=0. For b12*l=0 and b21*l≠0, condition (46) gives b12*v=0, where l,v=1,2,…,n and v>l. Then, from (43), (44) for l=v, we get b11*v=b22*v=0 and b21*v≠0.Forb21*l=0 and b12*l≠0, condition (46) gives b21*v=0, where l,v=1,2,…,n and v>l, then, from (43), (44) for l=v, we get b11*v=b22*v=0 and b12*v≠0.Now we discuss the caseb12*lb21*l≠0. From conditions (43), (44), we have b12*lb21*l=-(b11*l)2 and b11*lb22*l≠0. This yields b11*l≠0, b22*l≠0 and, from (75), we have b11*v≠0, b12*v≠0. By conditions (43), (44) for v=l, we get b22*v≠0, b21*v≠0.From the assumptionsBl≠Θ, we conclude that only the following cases ((I), (II), (III)) are possible:(I)
b11*l=b22*l=b21*l=0, b12*l≠0,(II)
b11*l=b22*l=b12*l=0, b21*l≠0,(III)
b12*lb21*l≠0,wherel=1,2,…,n.


## 2.1.4. Caseb12*lb21*l=0

Theorem 7.
Let (1) be a weakly delayed system, (35) has a twofold root λ1,2=λ, b12*lb21*l=0 and the matrix Λ has the form (38). Then the solution of the initial problem (1), (3) is x(k)=Sy(k), k∈ℤ-mn∞, where in case b21*l=0, y(k) has the form
(76)y(k)={φ*(k)ifk∈ℤ-mn0,Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍b12*lΦ2(r-ml)]ifk∈ℤ1m1+1,⋮Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b12*lΦ2(r-ml)]+∑l=1s‍b12*l[∑r=0ml‍λk-1-rΦ2(r-ml)+(k-1-ml)λk-1-mlΦ2(0)∑r=0ml‍]ifk∈ℤms+2ms+1+1,s=1,2,…,n-1,⋮Λ3kφ*(0)+∑l=1n‍b12*l[∑r=0ml‍λk-1-rΦ2(r-ml)+(k-1-ml)λk-1-mlΦ2(0)∑r=0ml‍]ifk∈ℤmn+2∞.
If b12*l=0 is true then the solution of initial problem (1), (3) is x(k)=Sy(k), k∈ℤ-mn∞, where y(k) has the form
(77)y(k)={φ*(k)ifk∈ℤ-mn0,Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍b21*lΦ1(r-ml)]ifk∈ℤ1m1+1,⋮Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b21*lΦ1(r-ml)]+∑l=1s‍b21*l[∑r=0ml‍λk-1-rΦ1(r-ml)+(k-1-ml)λk-1-mlΦ1(0)∑r=0ml‍]ifk∈ℤms+2ms+1+1,s=1,2,…,n-1,⋮Λ3kφ*(0)+∑l=1n‍b21*l[∑r=0ml‍∑r=0ml‍λk-1-rΦ1(r-ml)+(k-1-ml)λk-1-mlΦ1(0)∑r=0ml‍]ifk∈ℤmn+2∞.Proof.
Case (I) means thatb12*l≠0. Then (40) turns into the system
(78)y1(k+1)=λy1(k)+∑l=1n‍b12*ly2(k-ml),k∈ℤ0∞y2(k+1)=λy2(k),
and, if b21*l≠0, (40) turns into the system
(79)y1(k+1)=λy1(k),y2(k+1)=λy2(k)+∑l=1n‍b21*ly1(k-ml),k∈ℤ0∞.
System (78) can be solved in much the same way as the systems (49), (50) if we put λ1=λ2=λ, and the discussion of the system (79) goes along the same lines as that of the systems (51), (52) with λ1=λ2=λ. Formulas (76) and (77) are consequences of (47), (48).


## 2.1.5. Caseb12*lb21*l≠0

Fork∈ℤ-mn0, we define
(80)Φl*(k):=((b11*l)2b12*l[φ1*(k)+b12*1b11*1φ2*(k)]b11*l[φ1*(k)+b12*1b11*1φ2*(k)],-(b11*l)2b12*l[φ1*(k)+b12*1b11*1φ2*(k)])T.Theorem 8.
Let system (1) be a weakly delayed system, (35) admits two repeated roots λ1,2=λ, b12*lb21*l≠0 and the matrix Λ3 has the form (38). Then the solution of the initial problem (1), (3) is given by x(k)=Sy(k), k∈ℤ-mn∞, where y(k) has the form
(81)y(k)={φ*(k)ifk∈ℤ-mn0,Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍Φl*(r-ml)]ifk∈ℤ1m1+1,⋮Λ3kφ*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍Φl*(r-ml)]+∑l=1s‍[∑r=0ml‍λk-1-rΦl*(r-ml)+(k-1-ml)λk-1-mlΦl*(0)∑r=0ml‍]ifk∈ℤms+2ms+1+1,s=1,2,…,n-1,⋮Λ3kφ*(0)+∑l=1n‍[∑r=0ml‍λk-1-rΦl*(r-ml)+(k-1-ml)λk-1-mlΦl*(0)∑r=0ml‍]ifk∈ℤmn+2∞.Proof.
In this case, all the entries ofB*l are nonzero and, from (43), (44), and (71), we get
(82)B*=(b11*lb12*l-(b11*l)2b12*l-b11*l),
where l=1,2,…,n, then, the system (40) reduces to
(83)y1(k+1)=λy1(k)+∑l=1n‍[b11*ly1(k-ml)+b12*ly2(k-ml)],(84)y2(k+1)=λy2(k)-∑l=1n‍[(b11*l)2b12*ly1(k-ml)+b11*ly2(k-ml)],
where k∈ℤ0∞. It is easy to see (multiplying (84) by b12*1/b11*1 and summing both equations) that
(85)y1(k+1)+b12*1b11*1y2(k+1)=λ[y1(k)+b12*1b11*1y2(k)],1111111111111k∈ℤ0∞.
Equation (85) is a homogeneous equation with respect to the unknown expression
(86)y1(k)+(b12*1b11*1)y2(k),
then, using (33), we obtain
(87)y1(k)+b12*1b11*1y2(k)={φ1*(k)+b12*1b11*1φ2*(k)ifk∈ℤ-mn0,λk[φ1*(0)+b12*1b11*1φ2*(0)]ifk∈ℤ1∞.
With the aid of (87), we rewrite the systems (83), (84) as follows:
(88)y1(k+1)={λy1(k)+∑l=1n‍b11*l[b12*1b11*1φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤ0m1,λy1(k)+b11*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=2n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤm1+1m2,λy1(k)+∑l=12‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=3n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤm2+1m3,⋮λy1(k)+∑l=1s‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=s+1n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤms+1ms+1,s=3,4,…,n-1,⋮λy1(k)+∑l=1n‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]ifk∈ℤmn+1∞,y2(k+1)={λy2(k)-∑l=1n‍(b11*l)2b12*l[b12*1b11*1φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤ0m1,λy2(k)-(b11*1)2b12*1λk-m1×[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(k-m1)+b12*1b11*1φ2*(k-m1)]ifk∈ℤm1+1m2,λy2(k)-∑l=12‍(b11*l)2b12*l[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=3n‍(b11*l)2b12*l[b12*1b11*1φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤm2+1m3,⋮λy2(k)-∑l=1s‍(b11*l)2b12*l[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=s+1n‍(b11*l)2b12*l[b12*1b11*1φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]ifk∈ℤms+1ms+1,s=3,4,…,n-1,⋮λy2(k)-∑l=1n‍(b11*l)2b12*lλk-ml×[φ1*(0)+b12*1b11*1φ2*(0)]ifk∈ℤmn+1∞.
First, we solve this system for k∈ℤ0m1 and consider the problems
(89)y1(k+1)=λy1(k)+∑l=1n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]11111111111111111111111000ifk∈ℤ0m1,y1(0)=φ1*(0),y2(k+1)=λy2(k)-∑l=1n‍(b11*l)2b12*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]111111111111111111111111111100ifk∈ℤ0m1,y2(0)=φ2*(0).
With the aid of formula (33), we get
(90)y1(k)=λkφ1*(0)+∑r=0k-1‍λk-1-r×(∑l=1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]),11111111111111111111111111111k∈ℤ1m1+1,(91)y2(k)=λkφ2*(0)-∑r=0k-1‍λk-1-r×(∑l=1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]),1111111111111111111111111111111k∈ℤ1m1+1.
Now we solve system (88) for k∈ℤm1+1m2; that is, we consider the problem (with initial data deduced from (90), (91))
(92)y1(k+1)=λy1(k)+b11*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=2n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]1111111111111111111111111ifk∈ℤm1+1m2,y1(m1+1)=λm1+1φ1*(0)+∑r=0m1‍λm1-r×(∑l=1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]),y2(k+1)=λy2(k)-(b11*1)2b12*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=2n‍(b11*l)2b12*l[φ1*(k-m1)+b12*1b11*1φ2*(k-m1)]1111111111111111111111111111ifk∈ℤm1+1m2,y2(m1+1)=λm1+1φ2*(0)-∑r=0m1‍λm1-r×(∑l=1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]).
Formula (33) yields (for k∈ℤm1+2m2+1)
(93)y1(k)=λk-(m1+1)y1(m1+1)+∑r=m1+1k-1‍λk-1-r×(∑l=2n‍∑l=2n‍b11*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=2n‍b11*l[φ1*(r-ml)b12*1b11*11111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍[φ1*(0)+b12*1b11*1φ2*(0)])=λk-m1-1[([+b12*1b11*1φ2*(r-ml)]∑l=1n‍b11*l)λm1+1φ1*(0)+∑r=0m1‍λm1-rλk-m1-1λk-m1×(∑l=1n‍∑l=1n‍b11*l[b12*1b11*1φ1*(r-ml)11111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1n‍)]+∑r=m1+1k-1‍λk-1-r(b11*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑r=m1+1k-1‍λk-1∑r=m1+1k-1‍+∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111+b12*1b11*1φ2*(r-ml)])=λkφ1*(0)+∑r=0m1‍λk-1-r(∑l=1n‍b11*l[b12*1b11*1φ1*(r-ml)1111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1n‍)+∑r=m1+1k-1‍λk-1-r(∑l=2n‍b11*1λk-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑r=m1+1k-1‍+∑r=m1+1k-1‍+∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111+b12*1b11*1φ2*(r-ml)][φ1*(0)+b12*1b11*1φ2*(0)]∑l=2n‍)=λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍)+∑r=0m1‍λk-1-rb11*1[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)b11*1λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]=λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍)+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)λk-1-m1[b12*1b11*1φ1*(0)11111111111111111111111111111+b12*1b11*1φ2*(0)]∑r=0m1‍),(94)y2(k)=λk-(m1+1)y2(m1+1)-∑r=m1+1k-1‍λk-1-r×((b11*1)2b12*1λr-m1[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)11111111111111111+b12*1b11*1φ2*(r-ml)][φ1*(0)+b12*1b11*1φ2*(0)](b11*1)2b12*1)=λk-m1-1[(∑l=1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])λm1+1φ2*(0)-∑r=0m1‍λm1-r=λk-112=×(∑l=1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)1111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1n‍(b11*l)2b12*l)(∑l=1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])]-∑r=m1+1k-1‍λk-1-r((b11*1)2b12*1λr-m1[φ1*(0)+b12*1b11*1φ2*(0)]-∑r=m1+1k-1‍λk-1-r-12+∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111111+b12*1b11*1φ2*(r-ml)](b11*1)2b12*1λr-m1[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ2*(0)-∑r=0m1‍λk-1-r(∑l=1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1n‍∑l=1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml))-∑r=m1+1k-1‍λk-1-r((b11*1)2b12*1λr-m1[φ1*(0)+b12*1b11*1φ2*(0)]-∑r=m1+1k-1‍λk-1-r-12+∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111+b12*1b11*1φ2*(r-ml)](b11*1)2b12*1λr-m1[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍(b11*l)2b12*l)-∑r=0m1‍λk-1-r(b11*1)2b12*1[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]-(k-1-m1)λk-1-m1(b11*1)2b12*1[φ1*(0)+b12*1b11*1φ2*(0)]=λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111111+b12*1b11*1b12*lφ2*(r-ml)]∑l=2n‍(b11*l)2b12*l)-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]-(b11*1)2b12*1-+(k-1-m1)λk-1-m1×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍).
Now we solve (88) for k∈ℤm2+1m3; that is, we consider the problem (with initial data deduced from (93), (94))
(95)y1(k+1)=λy1(k)+∑l=12‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=3n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]+∑l=3n‍b11*l+∑l=3n‍b11*l+1∑l=3n‍b11*l11∑l=3n‍ifk∈ℤm2+1m3,y1(m2+1)=λm2+1φ1*(0)+∑r=0m2‍λm2-r×(∑l=2n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+b11*1(∑r=0m1‍λm2-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+b11*1+b11*1+(m2-m1)λm2-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍),y2(k+1)=λy2(k)-∑l=12‍(b11*l)2b12*l[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=3n‍(b11*l)2b12*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]-∑l=3n‍(b11*l)2b12*l-∑l=3n‍(b11*l)2b12*l111111-∑l=3n‍ifk∈ℤm2+1m3,y2(m2+1)=λm2+1φ2*(0)-∑r=0m2‍λm2-r×(∑l=2n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1b12*lφ2*(r-ml)])-(b11*1)2b12*1(∑r=0m1‍λm2-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]-(b11*1)2b12*1-+(m2-m1)λm2-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍).
Applying formula (33) yields (for k∈ℤm2+2m3+1)
(96)y1(k)=λk-(m2+1)y1(m2+1)+∑r=m2+1k-1‍λk-1-r×(∑l=12‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]2222+∑l=3n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])=λk-m2-1[λm2+1φ1*(0)+∑r=0m2‍λm2-r222222222×(∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml))222222222+b11*1(∑r=0m1‍λm2-r[b12*1b11*1φ1*(r-m1)11111111111111111111111+b12*1b11*1φ2*(r-m1)]22222222222222+(m2-m1)λm2-m111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])∑r=0m1‍]+∑r=m2+1k-1‍λk-1-r(∑l=12‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]222222222222222+∑l=3n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=12‍)=λkφ1*(0)+∑r=0m2‍λk-1-r(∑l=2n‍b11*l[b12*1b11*1φ1*(r-ml)11111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍)+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]2222222+(m2-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍λk-1-r)+∑r=m2+1k-1‍λk-1-r(∑l=12‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]1111111111111111+∑l=3n‍b11*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=12‍)=λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=3n‍b11*l[b12*1b11*1φ1*(r-ml)11111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=3n‍)+∑r=0m2‍λk-1-rb11*2[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+b11*1+b11*1+(m2-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)])+(k-1-m2)(b11*1λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)](k-1-m2)(k-1)+b11*2λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)]b11*1λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=3n‍b11*l[b12*1b11*1φ1*(r-ml)11111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=3n‍)+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+b11*1+b11*1+(k-1-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m2‍)+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+b11*2+1+(k-1-m2)λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m2‍),(97)y2(k)=λk-(m2+1)y2(m2+1)-∑r=m2+1k-1‍λk-1-r×(∑l=12‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=3n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])=λk-m2-1[λm2+1φ2*(0)-∑r=0m2‍λm2-r×(∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)11111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml))-(b11*l)2b12*1(b12*1b11*1∑r=0m1‍λm2-r[b12*1b11*1φ1*(r-m1)111111111111111111111111111+b12*1b11*1φ2*(r-m1)]1111111111111111111+(m2-m1)λm2-m11111111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍)]-∑r=m2+1k-1‍λk-1-r(∑l=12‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]1111111111111111+∑l=3n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=12(b11*l)2b12*l‍)=λkφ2*(0)-∑r=0m2‍λk-1-r(∑l=2n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=2n‍(b11*l)2b12*l)-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]-111b12*1-111b12*1-+(m2-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍)-∑r=m2+1k-1‍λk-1-r(∑l=12‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]-∑r=m2+1k-1‍λk-1-r1111-+∑l=3n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=12‍(b11*l)2b12*l)=λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=3n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=3n(b11*l)2b12*l‍)-∑r=0m2‍λk-1-r(b11*2)2b12*2[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]-(b11*1)2b12*1(∑r=0m1‍λk-1-r[b12*1b11*1φ1*(r-m1)11111111111111111111+b12*1b11*1φ2*(r-m1)]11111111111+(m2-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m1‍)-(k-1-m2)×((b11*1)2b12*1λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]11111+(b11*2)2b12*2λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)](b11*1)2b12*1)=λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=3n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)11111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=3n‍(b11*l)2b12*l)-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]111111111111+(k-1-m1)λk-1-m1111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]1111111111111+(k-1-m2)λk-1-m21111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m2‍).
From (93)–(97) we deduce that expected form of the solution of the initial problem for k∈ℤms-1+1ms with initial data derived from the solution of previous equation for k∈ℤms-2+1ms-1 is
(98)y1(k)=λkφ1*(0)+∑r=0k-1‍λk-1-r×(∑l=sn‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+∑l=1s-1‍[b11*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]111111111111+(k-1-ml)λk-1-ml111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)]+(k-1-ml)λk-1-ml+11ifk∈ℤms-1+2ms+1,y1(k)=λkφ2*(0)-∑r=0k-1‍λk-1-r×(∑l=sn‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-∑l=1s-1[(b11*l)2b12*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]‍220000000001100+(k-1-ml)λk-1-ml111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)(b11*l)2b12*l]23232113213213213213213112ifk∈ℤms-1+2ms+1.
We solve (88) for k∈ℤms+1ms+1 with initial data deduced from (98); that is, we consider the problem
(99)y1(k+1)=λy1(k)+∑l=1s‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=s+1n‍b11*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]∑l=s+1n‍∑l=s+1n‍b11*l∑l=s+1n‍b11*l∑l=s+1n‍b11*l111111ifk∈ℤms+1ms+1,y1(ms+1)=λms+1φ1*(0)+∑r=0ms‍λms-r(∑l=sn‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+∑l=1s-1‍b11*l(∑r=0ml‍λms-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]11111111111+(ms-ml)λms-ml11111111111×[φ1*(0)+b12*1b11*1φ2*(0)]),y2(k+1)=λy2(k)-∑l=1s‍(b11*l)2b12*l[φ1*(0)+b12*1b11*1φ2*(0)]-∑l=s+1n‍(b11*l)2b12*l[φ1*(k-ml)+b12*1b11*1φ2*(k-ml)]∑l=s+1n‍(b11*l)2b12*l∑l=s+1n‍(b11*l)2b12*l∑l=s+1n‍(b11*l)2b12*l11ifk∈ℤms+1ms+1,y2(ms+1)=λms+1φ2*(0)-∑r=0ms‍λms-r(∑l=sn‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-∑l=1s-1‍(b11*l)2b12*l(∑r=0ml‍λms-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]-∑l=1s-1‍(b11*l)2b12*l-11+(ms-ml)λms-ml[φ1*(0)+b12*1b11*1φ2*(0)]).
Applying formula (33) yields (for k∈ℤms+2ms+1+1)
(100)y1(k)=λk-(ms+1)y1(ms+1)+∑r=ms+1k-1‍λk-1-r×(∑l=1s‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=s+1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])=λk-ms-1[λms+1φ1*(0)+∑r=0ms‍λms-r=λk-ms-1×(∑l=sn‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])=λk-ms-1=+∑l=1s-1‍b11*l(∑r=0ml‍λms-r[b12*1b11*1φ1*(r-ml)11111111111111111111111111+b12*1b11*1φ2*(r-ml)]111111111111111111+(ms-ml)λms-ml111111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)]+∑r=ms+1k-1‍λk-1-r(∑l=1s‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑r=ms+1k-1‍λk-1-r+11+∑l=s+1n‍b11*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1s‍)=λkφ1*(0)+∑r=0ms‍λk-1-r×(∑l=sn‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+∑l=1s-1‍b11*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]1111111111+(ms-ml)λk-1-ml1111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+∑r=ms+1k-1‍λk-1-r(∑l=1s‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑r=ms+1k-1‍λk-1-r+11+∑l=s+1n‍b11*l[b12*1b11*1φ1*(r-ml)111111111111111111111111+b12*1b11*1φ2*(r-ml)b12*1b11*1]∑l=1s‍)=λkφ1*(0)+∑r=0k-1‍λk-1-r×(∑l=s+1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+b11*s∑r=0ms‍λk-1-r[φ1*(r-ms)+b12*1b11*1φ2*(r-ms)]+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+b11*1+b11*1+(ms-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+b11*2+b11*2+(ms-m2)λk-1-m2111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0m2‍)+⋯+b11*s-1(∑r=0ms-1‍λk-1-r[b12*1b11*1φ1*(r-ms-1)1111111111111111111+b12*1b11*1φ2*(r-ms-1)]111111111+(ms-ms-1)λk-1-ms-1111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ms-1‍)+(k-1-ms)×(λk-1-m1b11*1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-m2b11*2[φ1*(0)+b12*1b11*1φ2*(0)]+⋯+λk-1-ms-1b11*s-1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-msb11*s[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ1*(0)+∑r=0k-1‍λk-1-r×(∑l=s+1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]111111111+(k-1-m1)λk-1-m1111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]111111111+(k-1-m2)λk-1-m2111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯+b11*s-1(∑r=0ms-1‍λk-1-r[b12*1b11*1φ1*(r-ms-1)1111111111111111111+b12*1b11*1φ2*(r-ms-1)]111111111+(k-1-ms-1)λk-1-ms-1111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+b11*s(∑r=0ms‍λk-1-r[φ1*(r-ms)+b12*1b11*1φ2*(r-ms)]+b11*s+b11*s+(k-1-ms)λk-1-ms[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ1*(0)+∑r=0k-1‍λk-1-r×(∑l=s+1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+∑l=1s‍b11*l(∑r=0ml‍λk-1-r[b12*1b11*1φ1*(r-ml)gggggggggggggggg+b12*1b11*1φ2*(r-ml)]1111111111+(k-1-ml)λk-1-ml1111111111×[φ1*(0)+b12*1b11*1φ2*(0)]),(101)y2(k)=λk-(ms+1)y2(ms+1)-∑r=ms+1k-1‍λk-1-r×(∑l=1s‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]+∑l=s+1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1s‍(b11*l)2b12*l)=λk-ms-1[λms+1φ2*(0)-∑r=0ms‍λms-r=λk-ms-1=×(∑l=sn‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)11111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=sn‍(b11*l)2b12*l)1111111111-∑l=1s-1‍(b11*l)2b12*l(∑r=0ml‍λms-r[b12*1b11*1φ1*(r-ml)11111111111111111111111111111+b12*1b11*1φ2*(r-ml)]111111111111111111111+(ms-ml)λms-ml111111111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)]-∑r=ms+1k-1‍λk-1-r(∑l=1s‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]-∑r=ms+1k-1‍λk-∑r=ms+1k-1‍11+∑l=s+1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)1111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1s‍(b11*l)2b12*l)=λkφ2*(0)-∑r=0ms‍λk-1-r×(∑l=sn‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-∑l=1s-1‍(b11*l)2b12*l(∑r=0ml‍λk-1-r[b12*1b11*1φ1*(r-ml)1111111111111111111111+b12*1b11*1φ2*(r-ml)]-∑l=1s-1‍-∑l=1s-1‍-11+(ms-ml)λk-1-ml[φ1*(0)+b12*1b11*1φ2*(0)])-∑r=ms+1k-1‍λk-1-r(∑l=1s‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)]-∑r=ms+1k-1‍λk-1-r-11+∑l=s+1n‍(b11*l)2b12*l[b12*1b11*1φ1*(r-ml)111111111111111111111111111+b12*1b11*1φ2*(r-ml)]∑l=1s‍(b11*l)2b12*l)=λkφ2*(0)-∑r=0k-1‍λk-1-r×(∑l=s+1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-(b11*s)2b12*s∑r=0ms‍λk-1-r[φ1*(r-ms)+b12*1b11*1φ2*(r-ms)]-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]λkφ2*λkφ2*λk+(ms-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]λk-1-m2λk-11+(ms-m2)λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)])+⋯-(b11*s-1)2b12*s-1(∑r=0ms-1‍λk-1-r[φ1*(r-ms-1)+b12*1b11*1φ2*(r-ms-1)]111111111111+(ms-ms-1)λk-1-ms-1[φ1*(0)+b12*1b11*1φ2*(0)])-(k-1-ms)×(λk-1-m1(b11*1)2b12*1[φ1*(0)+b12*1b11*1φ2*(0)]11111+λk-1-m2(b11*2)2b12*2[φ1*(0)+b12*1b11*1φ2*(0)]+⋯11111+λk-1-ms-1(b11*s-1)2b12*s-1[φ1*(0)+b12*1b11*1φ2*(0)]11111+λk-1-ms(b11*s)2b12*s[φ1*(0)+b12*1b11*1φ2*(0)]λk-1-m1(b11*1)2b12*1[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ2*(0)-∑r=0k-1‍λk-1-r×(∑l=s+1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]11111111111+(k-1-m1)λk-1-m111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]11111111111+(k-1-m2)λk-1-m211111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯-(b11*s-1)2b12*s-1(∑r=0ms-1‍λk-1-r[b12*1b11*1φ1*(r-ms-1)1111111111111111111111+b12*1b11*1φ2*(r-ms-1)]11111111111111+(k-1-ms-1)λk-1-ms-111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ms-1‍)∑l=s+1n‍∑l=s+1n‍-(b11*s)2b12*s(∑r=0ms‍λk-1-r[b12*1b11*1φ1*(r-ms)1111111111111111111111111+b12*1b11*1φ2*(r-ms)]∑l=s+1n‍111111111∑l=s+1n‍+(k-1-ms)λk-1-ms∑l=s+1n‍111111111∑l=s+1n‍×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ms‍)=λkφ2*(0)-∑r=0k-1‍λk-1-r×(∑l=s+1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-∑l=1s‍(b11*l)2b12*l(∑r=0ml‍λk-1-r[b12*1b11*1φ1*(r-ml)11111111111111111111111+b12*1b11*1φ2*(r-ml)]1111111111111111+(k-1-ml)λk-1-ml1111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]).
In the end, we solve (88) for k∈ℤmn+1∞ with initial data deduced from (100) and (101); that is, we consider the problem
(102)y1(k+1)=λy1(k)+∑l=1n‍b11*lλk-ml[φ1*(0)+b12*1b11*1φ2*(0)]=λy1(k)+∑l=1n‍b11*lλk-mlλy1(k)ifk∈ℤmn+1∞,y1(mn+1)=λmn+1φ1*(0)+∑r=0mn‍λmn-rb11*n×[φ1*(r-mn)+b12*1b11*1φ2*(r-mn)]+∑l=1n-1‍b11*l(∑r=0ml‍λmn-r[b12*1b11*1φ1*(r-ml)1111111111111111111+b12*1b11*1φ2*(r-ml)]+(mn-ml)λmn-ml×[φ1*(0)+b12*1b11*1φ2*(0)]),y2(k+1)=λy2(k)-∑l=1n‍(b11*l)2b12*lλk-ml×[φ1*(0)+b12*1b11*1φ2*(0)]ifk∈ℤmn+1∞,y2(mn+1)=λmn+1φ2*(0)-∑r=0mn‍λmn-r(b11*n)2b12*n×[φ1*(r-mn)+b12*1b11*1φ2*(r-mn)]-∑l=1n-1‍(b11*l)2b12*l(∑r=0ml‍λmn-r[b12*1b11*1φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]+(mn-ml)λmn-ml×[φ1*(0)+b12*1b11*1φ2*(0)]).
Applying formula (33) yields (for k∈ℤmn+2∞)
(103)y1(k)=λk-(mn+1)y1(mn+1)+∑r=mn+1k-1‍λk-1-r×(∑l=1n‍b11*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)])=λk-mn-1[[φ1*(0)+b12*1b11*1φ2*(0)]λmn+1φ1*(0)=λk-mn-111+∑r=0mn‍λmn-rb11*n[b12*1b11*1φ1*(r-mn)11111111111111111111111+b12*1b11*1φ2*(r-mn)]=λk-mn-111+∑l=1n-1‍b11*l(∑r=0ml‍λmn-r[b12*1b11*1φ1*(r-ml)11111111111111111111111111+b12*1b11*1φ2*(r-ml)]111111111111111111+(mn-ml)λmn-ml111111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)]+∑r=mn+1k-1‍λk-1-r(∑l=1n‍b11*lλr-ml1111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑l=1n‍)=λkφ1*(0)+∑r=0mn‍λk-1-rb11*n[b12*1b11*1φ1*(r-mn)11111111111111111111111+b12*1b11*1φ2*(r-mn)]+∑l=1n-1‍b11*l(∑r=0ml‍λk-1-r[b12*1b11*1φ1*(r-ml)1111111111111111111+b12*1b11*1φ2*(r-ml)]+∑l=1n-1‍b11*l+∑l=1n-1‍+(mn-ml)λk-1-ml+∑l=1n-1‍b11*l+∑l=1n-1‍×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)+∑r=mn+1k-1‍λk-1-r(∑l=1n‍b11*lλr-ml1111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)=λkφ1*(0)+∑r=0mn‍λk-1-rb11*n[b12*1b11*1φ1*(r-mn)1111111111111111111111+b12*1b11*1φ2*(r-mn)]+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+b11*1+b11*1+(mn-m1)λk-1-m1+b11*1+b11*1×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+b11*2b11*2+(mn-m2)λk-1-m2+b11*2b11*2×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯+b11*n-1(∑r=0mn-1‍λk-1-r[b12*1b11*1φ1*(r-mn-1)1111111111111111111+b12*1b11*1φ2*(r-mn-1)]+b11*n-1+b11*n+(mn-mn-1)λk-1-mn-1+b11*n-1+b11*n×[φ1*(0)+b12*1b11*1φ2*(0)])+(k-1-mn)×(λk-1-m1b11*1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-m2b11*2[φ1*(0)+b12*1b11*1φ2*(0)]+⋯+λk-1-mn-1b11*n-1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-mnb11*n[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ1*(0)+b11*1(∑r=0m1‍λk-1-r[b12*1b11*1φ1*(r-m1)1111111111111111111111111+b12*1b11*1φ2*(r-m1)]111111111111111+(k-1-m1)λk-1-m1111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]11111111+(k-1-m2)λk-1-m211111111×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯+b11*n-1(∑r=0mn-1‍λk-1-r[b12*1b11*1φ1*(r-mn-1)1111111111111111111+b12*1b11*1φ2*(r-mn-1)]1111111111+(k-1-mn-1)λk-1-mn-11111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0mn-1‍)+b11*n(∑r=0mn‍λk-1-r[φ1*(r-mn)+b12*1b11*1φ2*(r-mn)]+b11*n+b11*n+(k-1-mn)λk-1-mn111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0mn‍)=λkφ1*(0)+∑l=1n‍b11*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]1111111111+(k-1-ml)λk-1-ml1111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍),(104)y2(k)=λk-(mn+1)y2(mn+1)-∑r=mn+1k-1‍λk-1-r×(∑l=1n‍(b11*l)2b12*lλr-ml[φ1*(0)+b12*1b11*1φ2*(0)])=λk-mn-1×[λmn+1φ2*(0)-∑r=0mn‍λmn-r(b11*n)2b12*n×[φ1*(r-mn)+b12*1b11*1φ2*(r-mn)]-∑l=1n-1‍(b11*l)2b12*l×(∑r=0ml‍λmn-r[b12*1b11*1φ1*(r-ml)11111111111111111+b12*1b11*1φ2*(r-ml)]+(mn-ml)λmn-ml×[φ1*(0)+b12*1b11*1φ2*(0)])∑r=0mn‍λmn-r(b11*n)2b12*n]-∑r=mn+1k-1‍λk-1-r(∑l=1n‍(b11*l)2b12*lλr-ml11111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑l=1n‍(b11*l)2b12*l)=λkφ2*(0)-∑r=0mn‍λk-1-r(b11*n)2b12*n×[φ1*(r-mn)+b12*1b11*1φ2*(r-mn)]-∑l=1n-1‍(b11*l)2b12*l(∑r=0ml‍λk-1-r[b12*1b11*1φ1*(r-ml)1111111111111111111111+b12*1b11*1φ2*(r-ml)]11111111111111+(mn-ml)λk-1-ml11111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑r=0ml‍)-∑r=mn+1k-1‍λk-1-r(∑l=1n‍(b11*l)2b12*lλr-ml1111111111111111×[φ1*(0)+b12*1b11*1φ2*(0)]∑l=1n‍(b11*l)2b12*l)=λkφ2*(0)-∑r=0mn‍λk-1-r(b11*n)2b12*n[b12*1b11*1φ1*(r-mn)1111111111111111111111111+b12*1b11*1φ2*(r-mn)]-(b11*1)2b12*1(∑r=0m1‍λk-1-r[b12*1b11*1φ1*(r-m1)11111111111111111111+b12*1b11*1φ2*(r-m1)]11111111111+(mn-m1)λk-1-m111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[b12*1b11*1φ1*(r-m2)11111111111111111111+b12*1b11*1φ2*(r-m2)]111111111111+(mn-m2)λk-1-m2111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯-(b11*n-1)2b12*n-1(∑r=0mn-1‍λk-1-r[b12*1b11*1φ1*(r-mn-1)1111111111111111111111+b12*1b11*1φ2*(r-mn-1)]111111111111+(mn-mn-1)λk-1-mn-1111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(k-1-mn)×(λk-1-m1(b11*1)2b12*1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-m2(b11*2)2b12*2[φ1*(0)+b12*1b11*1φ2*(0)]+⋯+λk-1-mn-1(b11*n-1)2b12*n-1[φ1*(0)+b12*1b11*1φ2*(0)]+λk-1-mn(b11*n)2b12*n[φ1*(0)+b12*1b11*1φ2*(0)][φ1*(0)+b12*1b11*1φ2*(0)](b11*1)2b12*1)=λkφ2*(0)-(b11*1)2b12*1×(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]11111+(k-1-m1)λk-1-m111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[b12*1b11*1φ1*(r-m2)11111111111111111111+b12*1b11*1φ2*(r-m2)]11111111111+(k-1-m2)λk-1-m211111111111×[φ1*(0)+b12*1b11*1φ2*(0)])+⋯-(b11*n-1)2b12*n-1(∑r=0mn-1‍λk-1-r[b12*1b11*1φ1*(r-mn-1)1111111111111111111111+b12*1b11*1φ2*(r-mn-1)]1111111111111+(k-1-mn-1)λk-1-mn-11111111111111×[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*n)2b12*n(∑r=0mn‍λk-1-r[b12*1b11*1φ1*(r-mn)11111111111111111111+b12*1b11*1φ2*(r-mn)]11111111111+(k-1-mn)λk-1-mn11111111111×[φ1*(0)+b12*1b11*1φ2*(0)])=λkφ2*(0)-∑l=1n‍(b11*l)2b12*l×(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]11111+(k-1-ml)λk-1-ml11111×[φ1*(0)+b12*1b11*1φ2*(0)]).
Summing up all particular cases (90), (93), (96), (100), and (103) we have(105)y1(k)={φ1*(k)ifk∈ℤ-mn0,λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])ifk∈ℤ1m1+1,λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=2n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤm1+2m2+1,λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=3n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+b11*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])+b11*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+(k-1-m2)λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤm2+2m3+1,⋮λkφ1*(0)+∑r=0k-1‍λk-1-r(∑l=s+1n‍b11*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])+∑l=1s‍b11*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]+(k-1-ml)λk-1-ml[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤms+2ms+1+1,⋮λkφ1*(0)+∑l=1n‍b11*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]+(k-1-ml)λk-1-ml[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤmn+2∞,and from cases (91), (94), (97), (101), and (104) we conclude that(106)y2(k)={φ2*(k)ifk∈ℤ-mn0,λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])ifk∈ℤ1m1+1,λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=2n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1b12*lφ2*(r-ml)])-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤm1+2m2+1,λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=3n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-(b11*1)2b12*1(∑r=0m1‍λk-1-r[φ1*(r-m1)+b12*1b11*1φ2*(r-m1)]+(k-1-m1)λk-1-m1[φ1*(0)+b12*1b11*1φ2*(0)])-(b11*2)2b12*2(∑r=0m2‍λk-1-r[φ1*(r-m2)+b12*1b11*1φ2*(r-m2)]+(k-1-m2)λk-1-m2[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤm2+2m3+1,⋮λkφ2*(0)-∑r=0k-1‍λk-1-r(∑l=s+1n‍(b11*l)2b12*l[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)])-∑l=1s‍(b11*l)2b12*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]+(k-1-ml)λk-1-ml[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤms+2ms+1+1,⋮λkφ2*(0)-∑l=1n‍(b11*l)2b12*l(∑r=0ml‍λk-1-r[φ1*(r-ml)+b12*1b11*1φ2*(r-ml)]+(k-1-ml)λk-1-ml[φ1*(0)+b12*1b11*1φ2*(0)])ifk∈ℤmn+2∞.Formula (81) is now a direct consequence of (105), (106), and (80).


## 2.1.6. Case (39) of a Double Real Root

If the matrixΛ has the form (39), the necessary and sufficient conditions (13)–(16), for (40), are reduced to (43), (44), (46), and
(107)|λ1b21*lb22*l|+|b11*lb12*l0λ|=λ(b11*l+b22*l)-b21*l=0.Then (43), (44), and (107) give b11*l=b22*l=b21*l=0.Theorem 9.
Let (1) be a weakly delayed system, (35) has a double root λ1,2=λ and the matrix Λ has the form (39). Then b11*l=b22*l=b21*l=0 and the solution of the initial problem (1), (3) is x(k)=𝒮y(k), y(k)=(y1(k),y2(k))T, and
(108)y1(k)={φ1*(k)ifk∈ℤ-mn0,λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍b12*lφ2*(r-ml)]ifk∈ℤ1m1+1,⋮λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+∑l=1s‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍]ifk∈ℤms+2ms+1+1,s=1,2,…,n-1,⋮λkφ1*(0)+kλk-1φ2*(0)+∑l=1n‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍]ifk∈ℤmn+2∞,(109)y2(k)={φ2*(k)ifk∈ℤ-mn0,λkφ2*(0)ifk∈ℤ1∞.Proof.
The system (40) can be written as
(110)y1(k+1)=λy1(k)+y2(k)+∑l=1n‍b12*ly2(k-ml),(111)y2(k+1)=λy2(k),k∈ℤ0∞.
Solving (111), we get
(112)y2(k)={φ2*(k)ifk∈ℤ-mn0,λkφ2*(0)ifk∈ℤ1∞,
then (110) turns into
(113)y1(k+1)={λy1(k)+λkφ2*(0)+∑l=1n‍b12*lφ2*(k-ml)ifk∈ℤ0m1,λy1(k)+λkφ2*(0)+b12*1λk-m1φ2*(0)+∑l=2n‍b12*lφ2*(k-ml)ifk∈ℤm1+1m2,λy1(k)+λkφ2*(0)+∑l=12‍b12*lλk-mlφ2*(0)+∑l=3n‍b12*lφ2*(k-ml)ifk∈ℤm2+1m3,⋮λy1(k)+λkφ2*(0)+∑l=1s‍b12*lλk-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(k-ml)ifk∈ℤms+1ms+1,s=3,4,…,n-1,⋮λy1(k)+λkφ2*(0)+∑l=1n‍b12*lλk-mlφ2*(0)ifk∈ℤmn+1∞.
Equation (113) can be solved in a way similar to that of (54) in the proof of Theorem 5 using (33).
First we solve (113) for k∈ℤ0m1. This means that we consider the problem
(114)y1(k+1)=λy1(k)+λkφ2*(0)+∑l=1n‍b12*lφ2*(k-ml)ifk∈ℤ0m1,y1(0)=φ1*(0).
With the aid of formula (33), we get
(115)y1(k)=λkφ1*(0)+∑r=0k-1‍λk-1-r[λrφ2*(0)+∑l=1n‍b12*lφ2*(r-ml)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍b12*lφ2*(r-ml)],k∈ℤ1m1+1.
Now we solve (113) for k∈ℤm1+1m2 with initial data deduced from (115); that is, we consider the problem
(116)y1(k+1)=λy1(k)+λkφ2*(0)+b12*1λk-m1φ2*(0)+∑l=2n‍b12*lφ2*(k-ml),k∈ℤm1+1m2,y1(m1+1)=λm1+1φ1*(0)+(m1+1)λm1φ2*(0)+∑r=0m1‍λm1-r[∑l=1n‍b12*lφ2*(r-ml)].
Applying formula (33) we get (for k∈ℤm1+2m2+1)
(117)y1(k)=λk-(m1+1)y1(m1+1)+∑r=m1+1k-1‍λk-1-r[∑l=2n‍λrφ2*(0)+b12*1λr-m1φ2*(0)+∑l=2n‍b12*lφ2*(r-ml)]=λk-m1-1[+∑r=0m1‍λm1-r[∑l=1n‍b12*lφ2*(r-ml)]]λm1+1φ1*(0)+(m1+1)λm1φ2*(0)=λk-m1-1111+∑r=0m1‍λm1-r[∑l=1n‍b12*lφ2*(r-ml)]]+∑r=m1+1k-1‍λk-1-r[+∑l=2n‍b12*lφ2*(r-ml)λrφ2*(0)+b12*1λr-m1φ2*(0)+∑r=m1+1k-1‍λk-1-r11111+∑l=2n‍b12*lφ2*(r-ml)]=λkφ1*(0)+(m1+1)λk-1φ2*(0)+∑r=0m1‍λk-1-r[∑l=1n‍b12*lφ2*(r-ml)]+(k-m1-1)λk-1φ2*(0)+∑r=m1+1k-1‍λk-1-r[b12*1λr-m1φ2*(0)+∑l=2n‍b12*lφ2*(r-ml)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+b12*1+b12*1+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍λk-1-rφ2*(r-m1)].
Now we solve (113) for k∈ℤm2+1m3 with initial data deduced from (117); that is, we consider the problem
(118)y1(k+1)=λy1(k)+λkφ2*(0)+∑l=12‍b12*lλk-mlφ2*(0)+∑l=3n‍b12*lφ2*(k-ml)k∈ℤm2+1m3,y1(m2+1)=λm2+1φ1*(0)+(m2+1)λm2φ2*(0)+∑r=0m2‍λm2-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λm2-rφ2*(r-m1)+b12*1+ll+(m2-m1)λm2-m1φ2*(0)∑r=0m1‍λm2-rφ2*(r-m1)].
Applying formula (33) yields (for k∈ℤm2+2m3+1)
(119)y1(k)=λk-(m2+1)y1(m2+1)+∑r=m2+1k-1‍λk-1-r[λrφ2*(0)+∑l=12‍b12*lλr-mlφ2*(0)+∑l=3n‍b12*lφ2*(r-ml)]=λk-m2-1[+∑r=0m2‍λm2-r[∑l=2n‍b12*lφ2*(r-ml)]λm2+1φ1*(0)+(m2+1)λm2φ2*(0)=λk-m2-1=+∑r=0m2‍λm2-r[∑l=2n‍b12*lφ2*(r-ml)]=λk-m2-1=+b12*1[∑r=0m1‍λm2-rφ2*(r-m1)=λk-m2-1=+(m2-m1)λm2-m1φ2*(0)∑r=0m1‍λm2-rφ2*(r-m1)]]+∑r=m2+1k-1‍λk-1-r[λrφ2*(0)+∑l=12‍b12*lλr-mlφ2*(0)+∑r=m2+1k-1‍λk-1-r+1+∑l=3n‍b12*lφ2*(r-ml)]=λkφ1*(0)+(m2+1)λk-1φ2*(0)+∑r=0m2‍λk-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+b12*1++(m2-m1)λk-1-m1φ2*(0)∑r=0m1‍λk-1-rφ2*(r-m1)]+(k-1-m2)λk-1φ2*(0)+∑r=m2+1k-1‍λk-1-r[∑l=12‍b12*lλr-mlφ2*(0)+∑l=3n‍b12*lφ2*(r-ml)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=3n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+b12*1+b12*1+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍λk-1-rφ2*(r-m1)]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)+b12*2+b12*2+(k-1-m2)λk-1-m2φ2*(0)∑r=0m2‍λk-1-rφ2*(r-m2)].
From (115), (117), and (119) we deduce that expected form of the solution of the initial problem for k∈ℤms-1+1ms with initial data derived from the solution of previous equation for k∈ℤms-2+1ms-1 is
(120)y1(k)=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+∑l=1s-1‍b12*l+∑l=1s-1‍+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍λk-1-rφ2*(r-ml)]+∑l=1s-1‍b12*l+∑l=1s-1‍b12*l+∑l=1s-1‍b12*lifk∈ℤms-1+2ms+1.
We solve (113) for k∈ℤms+1ms+1 with initial data deduced from (120); that is, we consider the problem
(121)y1(k+1)=λy1(k)+λkφ2*(0)+∑l=1s‍b12*lλk-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(k-ml),k∈ℤms+1ms+1,y1(ms+1)=λms+1φ1*(0)+(ms+1)λmsφ2*(0)+∑r=0ms‍λms-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=0ml‍λms-rφ2*(r-ml)+∑l=1s-1‍b12*l+∑l=1s-1‍+(ms-ml)λms-mlφ2*(0)∑r=0ml‍λms-rφ2*(r-ml)].
Applying formula (33) yields (for k∈ℤms+2ms+1+1)
(122)y1(k)=λk-(ms+1)y1(ms+1)+∑r=ms+1k-1‍λk-1-r[λrφ2*(0)+∑l=1s‍b12*lλr-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(r-ml)]=λk-ms-1×[∑r=0ml‍λms+1φ1*(0)+(ms+1)λmsφ2*(0)+∑r=0ms‍λms-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=0ml‍λms-rφ2*(r-ml)111111111111+(ms-ml)λms-mlφ2*(0)∑r=0ml‍]]+∑r=ms+1k-1‍λk-1-r[λrφ2*(0)+∑l=1s‍b12*lλr-mlφ2*(0)111111111111111+∑l=s+1n‍b12*lφ2*(r-ml)]=λkφ1*(0)+(ms+1)λk-1φ2*(0)+∑r=0ms‍λk-1-r[∑l=sn‍b12*lφ2*(r-ml)]+∑l=1s-1‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)11111111111+(ms-ml)λk-1-mlφ2*(0)∑r=0ml‍]+(k-1-ms)λk-1φ2*(0)+∑r=ms+1k-1‍λk-1-r[∑l=1s‍b12*lλr-mlφ2*(0)+∑l=s+1n‍b12*lφ2*(r-ml)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r×[∑l=s+1n‍b12*lφ2*(r-ml)]+∑r=0ms‍λk-1-rb12*sφ2*(r-ms)+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)11111111111+(ms-m1)λk-1-m1φ2*(0)∑r=0m1‍]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)11111111111+(ms-m2)λk-1-m2φ2*(0)∑r=0m2‍]+⋯+b12*s-1[∑r=0ms-1‍λk-1-rφ2*(r-ms-1)111111111111+(ms-ms-1)λk-1-ms-1φ2*(0)∑r=0ms-1‍]+(k-1-ms)[λk-1-m1b12*1φ2*(0)1111111111111111+λk-1-m2b12*2φ2*(0)11111111111111+⋯+λk-1-ms-1b12*s-1φ2*(0)11111111111111+λk-1-m2b12*sφ2*(0)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)11111111111+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)11111111111+(k-1-m2)λk-1-m2φ2*(0)∑r=0m2‍]+⋯+b12*s-1[∑r=0ms-1‍λk-1-rφ2*(r-ms-1)11111111111+(k-1-ms-1)λk-1-ms-1φ2*(0)∑r=0ms-1‍]+b12*s[∑r=0ms‍λk-1-rφ2*(r-ms)111111111+(k-1-ms)λk-1-msφ2*(0)∑r=0ms‍]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+∑l=1s‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)111111111111+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍].
In the end, we solve (113) for k∈ℤmn+1∞ with initial data deduced from (122); that is, we consider the problem
(123)y1(k+1)=λy1(k)+λkφ2*(0)+∑l=1n‍b12*lλk-mlφ2*(0)ifk∈ℤmn+1∞,y1(mn+1)=λmn+1φ1*(0)+(mn+1)λmnφ2*(0)+∑r=0mn‍λmn-rb12*nφ2*(r-mn)+∑l=1n-1‍b12*l[∑r=0ml‍λmn-rφ2*(r-ml)+∑l=1n-1‍b12*l+∑l=1n-1‍+(mn-ml)λmn-mlφ2*(0)∑r=0ml‍].
Applying formula (33) yields (for k∈ℤmn+2∞)
(124)y1(k)=λk-(mn+1)y1(mn+1)+∑r=mn+1k-1‍λk-1-r[λrφ2*(0)+∑l=1n‍b12*lλr-mlφ2*(0)]=λk-mn-1[∑r=0ml‍λmn+1φ1*(0)+(mn+1)λmnφ2*(0)=1111λk-mn+∑r=0mn‍λmn-rb12*nφ2*(r-mn)=1111λk-mn+∑l=1n-1‍b12*l[∑r=0ml‍λmn-rφ2*(r-ml)=λk-mn=1λk-mn=λk-+(mn-ml)λmn-mlφ2*(0)∑r=0ml‍]]+∑r=mn+1k-1‍λk-1-r[λrφ2*(0)+∑l=1n‍b12*lλr-mlφ2*(0)]=λkφ1*(0)+(mn+1)λk-1φ2*(0)+∑r=0mn‍λk-1-rb12*nφ2*(r-mn)+∑l=1n-1‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+(mn-ml)λk-1-mlφ2*(0)∑r=0ml‍]+(k-1-mn)λk-1φ2*(0)+∑r=mn+1k-1‍λk-1-r[∑l=1n‍b12*lλr-mlφ2*(0)]=λkφ1*(0)+kλk-1φ2*(0)+∑r=0mn‍λk-1-rb12*nφ2*(r-mn)+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+b12*1+b12*1+(mn-m1)λk-1-m1φ2*(0)∑r=0m1‍]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)+b12*2+b12*2+(mn-m2)λk-1-m2φ2*(0)∑r=0m2‍]+⋯+b12*n-1[∑r=0mn-1‍λk-1-rφ2*(r-mn-1)+b12*n-1+b12*n-1+(mn-mn-1)λk-1-mn-1φ2*(0)∑r=0mn-1‍]+(k-1-mn)×[λk-1-m1b12*1φ2*(0)+λk-1-m2b12*2φ2*(0)+⋯+λk-1-mn-1b12*n-1φ2*(0)+λk-1-mnb12*nφ2*(0)]=λkφ1*(0)+kλk-1φ2*(0)+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+b12*1+b12*1+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)+b12*1+b12*1+(k-1-m2)λk-1-m2φ2*(0)∑r=0m2‍]+⋯+b12*n-1[∑r=0mn-1‍λk-1-rφ2*(r-mn-1)+b12*1+b12*1+(k-1-mn-1)λk-1-mn-1φ2*(0)∑r=0mn-1‍]+b12*n[∑r=0mn‍λk-1-rφ2*(r-mn)+b12*1+b12*1+(k-1-mn)λk-1-mnφ2*(0)∑r=0mn‍]=λkφ1*(0)+kλk-1φ2*(0)+∑l=1n‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+b12*1+b12*1+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍].
Summing up all particular cases (115)–(124), we get
(125)y1(k)={φ1*(k)ifk∈ℤ-mn0,λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=1n‍b12*lφ2*(r-ml)]ifk∈ℤ1m1+1,λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=2n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍]ifk∈ℤm1+2m2+1,λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=3n‍b12*lφ2*(r-ml)]+b12*1[∑r=0m1‍λk-1-rφ2*(r-m1)+(k-1-m1)λk-1-m1φ2*(0)∑r=0m1‍]+b12*2[∑r=0m2‍λk-1-rφ2*(r-m2)+(k-1-m2)λk-1-m2φ2*(0)∑r=0m2‍]ifk∈ℤm2+2m3+1,⋮λkφ1*(0)+kλk-1φ2*(0)+∑r=0k-1‍λk-1-r[∑l=s+1n‍b12*lφ2*(r-ml)]+∑l=1s‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍]ifk∈ℤms+2ms+1+1,⋮λkφ1*(0)+kλk-1φ2*(0)+∑l=1n‍b12*l[∑r=0ml‍λk-1-rφ2*(r-ml)+(k-1-ml)λk-1-mlφ2*(0)∑r=0ml‍]ifk∈ℤmn+2∞.
Formulas (108) and (109) are consequences of (125), (112).


## 3. Dimension of the Set of Solutions

Since all the possible cases of the planar system (1) with weak delay have been analysed, we are ready to formulate results concerning the dimension of the space of solutions of (1) assuming that initial condition (3) is variable. Although case b11*l=b22*l=b12*l=b21*l=0 does not lead to a weakly delayed system and is excluded by (2), for completeness of analysis we incorporate such possibility in our analysis as well (such a case can be considered as a degenerated weakly delayed system). Before formulation we remark that if an assumption in the following theorem is assumed to be valid for a fixed index l∈{1,2,…,n}, it is easy to see that it must be valid for all indices l=1,2,…,n.Theorem 10.
Let (1) be a weakly delayed system and let (35) having both roots different from zero and l∈{1,2,…,n} be fixed. Then the space of solutions, being initially 2(mn+1)-dimensional, becomes on ℤmn+2∞ only (1)
(mn+2)-dimensional if (35) has
(a)
two real distinct roots and(b12*l)2+(b21*l)2>0,(b)
a double real root,b12*lb21*l=0, and (b12*l)2+(b21*l)2>0.(c)
a double real root andb12*lb21*l≠0,(2)
2-dimensional if (35) has
(a)
two real distinct roots andb12*l=b21*l=0,(b)
a pair of complex conjugate roots,(c)
a double real root andb12*l=b21*l=0.Proof.
We will carefully go through all the theorems considered (Theorems5–9) adding the case of a pair of complex conjugate roots and our conclusion will hold at least on ℤmn+2∞ (some of the statements hold on a larger interval).
(a) Analysing the statement of Theorem5 (case (36) of two real distinct roots), we obtain the following subcases.(a1)
Ifb11*l=b22*l=b21*l=0, b12*l≠0, then the dimension of the space of solutions on ℤmn+2∞ equals mn+2 since the last formula in (47) uses only mn+2 arbitrary parameters:
(126)φ1*(0),φ2*(-mn),φ2*(-mn+1),…,φ2*(0).(a2)
Ifb11*l=b22*l=b12*l=0, b21*l≠0, then the dimension of the space of solutions on ℤmn+2∞ equals mn+2 since the last formula in (48) uses only mn+2 arbitrary parameters:
(127)φ1*(-mn),φ1*(-mn+1),…,φ1*(0),φ2*(0).(a3)
Ifb12*l=b21*l=0, then b11*l=b22*l=0 and Theorem 5 is not applicable. The dimension of the space of solutions on ℤmn+2∞ equals 2 since the solution is determined only by 2 arbitrary parameters
(128)φ1*(0),φ2*(0).This means that all the cases considered are covered by conclusions (1)(a) and (2)(a) of Theorem10.
(b) In case (37) of two complex conjugate roots, we have b11*l=b22*l=b12*l=b21*l=0 (i.e., we deal not with a weakly delayed system, as noted previosly) and the formula (70) uses only 2 arbitrary parameters
(129)φ1*(0),φ2*(0)
for every k∈ℤ1∞. This is covered by case (2)(b) of Theorem 10.
(c) Analysing the statement of Theorems7 and 8 (case (38) of a double real root), we obtain the following subcases.(c1)
Ifb21*l=0, b12*l≠0, then the dimension of the space of solutions on ℤmn+2∞ equals mn+2 since the last formula in (76) uses only mn+2 arbitrary parameters:
(130)φ1*(0),φ2*(-mn),φ2*(-mn+1),…,φ2*(0).(c2)
Ifb12*l=0, b21*l≠0, then the dimension of the space of solutions on ℤmn+2∞ equals mn+2 since the last formula in (77) uses only mn+2 arbitrary parameters:
(131)φ1*(-mn),φ1*(-mn+1),…,φ1*(0),φ2*(0).(c3)
Ifb12*l=b21*l=0 (degenerated weakly delayed system), then the dimension of the space of solutions on ℤmn+2∞ equals 2and solutions are determined only by 2 arbitrary parameters:
(132)φ1*(0),φ2*(0).(c4)
Ifb12*lb21*l≠0, then the dimension of the space of solutions on ℤmn+2∞ equals mn+2 since the last formula in (81) uses only mn+2 arbitrary parameters:
(133)C(-mn),C(-mn+1),…,C(0),φ1*(0),where(134)C(k):=[φ1*(k)+b12*1b11*1φ2*(k)],k∈ℤ-mn0.The parameterφ2*(0) cannot be seen as independent since it depends on the independent parameters φ1*(0) and C(0).
All the cases considered are covered by conclusions (1)(b), (1)(c), and (2)(c) of Theorem10.
(d) Analysing the statement of Theorem9 (case (39) of a double real root), we obtain the following subcases:(d1)
Ifb11*l=b22*l=b21*l=0, b12*l≠0, then the dimension of the space of solutions on ℤmn+2∞ equals mn+2 since the last formula in (108) uses only mn+2 arbitrary parameters:
(135)φ1*(0),φ2*(-mn),φ2*(-mn+1),…,φ2*(0)and the last formula in (109) provides no new information.(d2)
Ifb11*l=b22*l=b21*l=b12*l=0 (degenerated weakly delayed system), then the dimension of the space of solutions on ℤmn+2∞ equals 2 since solutions are determined only by 2 arbitrary parameters
(136)φ1*(0),φ2*(0).Both cases are covered by conclusions (1)(b) and (2)(c) of Theorem10.
Since there are no cases other than cases (a)–(d), the proof is finished.Theorem10 can be formulated simply as follows.Theorem 11.
Let (1) be a weakly delayed system and let (35) have both roots different from zero, then the space of solutions, being initially 2(mn+1)-dimensional, is on ℤmn+2∞ only(1)
(mn+2)-dimensional if (b12*l)2+(b21*l)2>0,(2)
2-dimensional if b12*l=b21*l=0.We omit the proofs of the following two theorems since, again, they are much the same as those of Theorems5–9.Theorem 12.
Let (1) be a weakly delayed system and let (35) have a simple root λ=0, then the space of solutions, being initially 2(mn+1)-dimensional, is either (mn+1)-dimensional or 1-dimensional on ℤmn+2∞.Theorem 13.
Let (1) be a weakly delayed system and let (35) have a double root λ=0, then the space of solutions, being initially 2(mn+1)-dimensional, turns into a 0-dimensional space on ℤmn+2∞, namely, into the zero solution.


## 4. Concluding Remarks

To our best knowledge, weakly delayed systems were firstly defined in [9] for systems of linear delayed differential systems with constant coefficients and in [1] for planar linear discrete systems with a single delay (in these papers such systems are called systems with a weak delay). The weakly delayed systems analyzed in this paper can be simplified and their solutions can be found in explicit analytical forms (results obtained generalize those in [1, 2]). Consequently, analytical forms of solutions can be used directly to solve several problems for weakly delayed systems, for example, problems of asymptotical behavior of their solutions, boundary-value problems, or some problems of control theory (using different methods, such problems have recently been investigated e.g., in [10–18]). For an alternative approach to differential-difference equations using the variational iteration method and new analytical and asymptotic methods see, for example, [19–21].In the case of discrete systems of two equations investigated in this paper, to obtain the corresponding eigenvalues, it is sufficient to solve only a second-order polynomial equation rather than a polynomial equation of order2mn.



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