# Discrete matrix delayed exponential for two delays and its property

## Abstract

In recent papers, a discrete matrix delayed exponential for a single delay was defined and its main property connected with the solution of linear discrete systems with a single delay was proved. In the present paper, a generalization of the concept of discrete matrix delayed exponential is designed for two delays and its main property is proved as well.

## Introduction

Throughout the paper, we use the following notation. For integers s, t, $s≤t$, we define a set $Z s t :={s,s+1,…,t−1,t}$. Similarly, we define sets $Z − ∞ t :={…,t−1,t}$ and $Z s ∞ :={s,s+1,…}$. The function $⌊⋅⌋$ used below is the floor integer function. We employ the following property of the floor integer function:

$x−1<⌊x⌋≤x,$
(1)

where $x∈R$.

Define binomial coefficients as customary, i.e., for $n∈Z$ and $k∈Z$,

(2)

We recall that for a well-defined discrete function $f(k)$, the forward difference operator Δ is defined as $Δf(k)=f(k+1)−f(k)$. In the paper, we also adopt the customary notation $∑ i = i 1 i 2 g i =0$ if $i 2 < i 1$. In the case of double sums, we set

$∑ i = i 1 , j = j 1 i 2 , j 2 g i j =0$
(3)

if at least one of the inequalities $i 2 < i 1$, $j 2 < j 1$ holds.

In [1, 2], a discrete matrix delayed exponential for a single delay $m∈N$ was defined as follows.

Definition 1 For an $r×r$ constant matrix B, $k∈Z$, and fixed $m∈N$, we define the discrete matrix delayed exponential $e m B k$ as follows:

where Θ is an $r×r$ null matrix and I is an $r×r$ unit matrix.

Next, the main property (Theorem 1 below) of discrete matrix delayed exponential for a single delay $m∈N$ is proved in .

Theorem 1 Let B be a constant $r×r$ matrix. Then, for $k∈ Z − m ∞$,

$Δ e m B k =B e m B ( k − m ) .$
(4)

The paper is concerned with a generalization of the notion of discrete matrix delayed exponential for two delays and a proof of one of its properties, similar to the main property (4) of discrete matrix delayed exponential for a single delay.

## Discrete matrix delayed exponential for two delays and its main property

We define a discrete $r×r$ matrix function $e m n B C k$ called the discrete matrix delayed exponential for two delays $m,n∈N$, $m≠n$ and for two $r×r$ commuting constant matrices B, C as follows.

Definition 2 Let B, C be constant $r×r$ matrices with the property $BC=CB$ and let $m,n∈N$, $m≠n$ be fixed integers. We define a discrete $r×r$ matrix function $e m n B C k$ called the discrete matrix delayed exponential for two delays m, n and for two $r×r$ constant matrices B, C:

where

$p ( k ) := ⌊ k + m m + 1 ⌋ , q ( k ) := ⌊ k + n n + 1 ⌋ .$
(5)

The main property of $e m n B C k$ is given by the following theorem.

Theorem 2 Let B, C be constant $r×r$ matrices with the property $BC=CB$ and let $m,n∈N$, $m≠n$ be fixed integers. Then

$Δ e m n B C k =B e m n B C ( k − m ) +C e m n B C ( k − n )$
(6)

holds for $k≥0$.

Proof Let $k≥1$. From (1) and (5), we can see easily that, for an integer $k≥0$ satisfying

$( p ( k ) −1)(m+1)+1≤k≤ p ( k ) (m+1)∧( q ( k ) −1)(n+1)+1≤k≤ q ( k ) (n+1),$

the relation

$Δ e m n B C k =Δ [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j + 1 ) ]$

holds in accordance with Definition 2 of $e m n B C k$. Since $ΔI=Θ$, we have

$Δ e m n B C k =Δ [ ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j + 1 ) ] .$
(7)

Considering the increment by its definition, i.e.,

$Δ e m n B C k = e m n B C ( k + 1 ) − e m n B C k ,$
(8)

we conclude that it is reasonable to divide the proof into four parts with respect to the value of integer k. In case one, k is such that

$( p ( k ) −1)(m+1)+1≤k< p ( k ) (m+1)∧( q ( k ) −1)(n+1)+1≤k< q ( k ) (n+1),$

in case two

$k= p ( k ) (m+1)∧( q ( k ) −1)(n+1)+1≤k< q ( k ) (n+1),$

in case three

$( p ( k ) −1)(m+1)+1≤k< p ( k ) (m+1)∧k= q ( k ) (n+1)$

and in case four

$k= p ( k ) (m+1)∧k= q ( k ) (n+1).$

We see that the above cases cover all the possible relations between k, $p ( k )$ and $q ( k )$.

In the proof, we use the identities

$( n + 1 k ) = ( n k ) + ( n k − 1 ) ,$
(9)

where $n,k∈N$ and

$( i i ) = ( i − 1 i − 1 ) , ( j 0 ) = ( j − 1 0 ) , ( i + j i ) = ( i + j − 1 i − 1 ) + ( i + j − 1 i ) ,$
(10)

where $i,j∈N$, which are derived from (2) and (9).

### I. $( p ( k ) −1)(m+1)+1≤k< p ( k ) (m+1)∧( q ( k ) −1)(n+1)+1≤k< q ( k ) (n+1)$

From (1) and (5), we get

$p ( k − m ) = ⌊ k − m + m m + 1 ⌋ ≤ k m + 1 < p ( k ) , p ( k − m ) = ⌊ k − m + m m + 1 ⌋ > k m + 1 − 1 = k − m − 1 m + 1 > p ( k ) − 2 .$

Therefore, $p ( k − m ) = p ( k ) −1$ and, by Definition 2,

$e m n B C ( k − m ) =I+(B+C) ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k − m ) − 1 B i C j ( i + j i ) ( k − m − m i − n j i + j + 1 ) .$
(11)

Similarly, omitting details, we get (using (1), and (5)) $q ( k − n ) = q ( k ) −1$ and

$e m n B C ( k − n ) =I+(B+C) ∑ i = 0 , j = 0 p ( k − n ) − 1 , q ( k ) − 2 B i C j ( i + j i ) ( k − n − m i − n j i + j + 1 ) .$
(12)

Let $q ( k − m ) ≥1$. We show that

(13)

In accordance with (1),

$q ( k − m ) = ⌊ k − m + n n + 1 ⌋ > k − m + n n + 1 −1= k − m − 1 n + 1$

or

From the last inequality, we get

and (13) holds by (2). For that reason and since $q ( k − m ) ≤ q ( k )$, we can replace $q ( k − m )$ by $q ( k )$ in (11). Thus, we have

$e m n B C ( k − m ) =I+(B+C) ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) .$
(14)

It is easy to see that, due to (3), formula (14) can be used instead of (11) if $q ( k − m ) <1$ also.

Let $p ( k − n ) ≥1$. Similarly, we can show that

and, since $p ( k − n ) ≤ p ( k )$, we can replace $p ( k − n )$ by $p ( k )$ in (12). Thus, we have

$e m n B C ( k − n ) =I+(B+C) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 2 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) .$
(15)

It is easy to see that, due to (3), formula (15) can be used instead of (12) if $p ( k − n ) <1$, too.

Due to (1), we also conclude that

$p ( k + 1 ) = p ( k ) , q ( k + 1 ) = q ( k )$
(16)

because

$p ( k + 1 ) = ⌊ k + 1 + m m + 1 ⌋ ≤ k m + 1 +1< p ( k ) +1$

and

$p ( k + 1 ) = ⌊ k + 1 + m m + 1 ⌋ > k + 1 + m m + 1 −1= k m + 1 ≥ p ( k ) −1+ 1 m + 1 .$

The second formula can be proved similarly.

Now we are able to prove that

$Δ e m n B C k = B e m n B C ( k − m ) + C e m n B C ( k − n ) = B [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) ] + C [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 2 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) ] .$
(17)

With the aid of (7), (8), (9) and (16), we get

$Δ e m n B C k = e m n B C ( k + 1 ) − e m n B C k = I + ( B + C ) ∑ i = 0 , j = 0 p ( k + 1 ) − 1 , q ( k + 1 ) − 1 B i C j ( i + j i ) ( k + 1 − m i − n j i + j + 1 ) − I − ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j + 1 ) = I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k + 1 − m i − n j i + j + 1 ) − I − ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j + 1 ) = ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) [ ( k + 1 − m i − n j i + j + 1 ) − ( k − m i − n j i + j + 1 ) ] = ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j ) = ( B + C ) [ I + ∑ i = 1 p ( k ) − 1 B i C 0 ( i i ) ( k − m i i ) + ∑ j = 1 q ( k ) − 1 B 0 C j ( j 0 ) ( k − n j j ) + ∑ i = 1 , j = 1 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j ) ] .$

By (10), we have

$Δ e m n B C k = ( B + C ) [ I + ∑ i = 1 p ( k ) − 1 B i C 0 ( i − 1 i − 1 ) ( k − m i i ) + ∑ j = 1 q ( k ) − 1 B 0 C j ( j − 1 0 ) ( k − n j j ) + ∑ i = 1 , j = 1 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j − 1 i − 1 ) ( k − m i − n j i + j ) + ∑ i = 1 , j = 1 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j − 1 i ) ( k − m i − n j i + j ) ] = ( B + C ) [ I + ∑ i = 1 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j − 1 i − 1 ) ( k − m i − n j i + j ) + ∑ i = 0 , j = 1 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j − 1 i ) ( k − m i − n j i + j ) ] .$

Now in the first sum we replace the summation index i by $i+1$ and in the second sum we replace the summation index j by $j+1$. Then

$Δ e m n B C k = ( B + C ) [ I + ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k ) − 1 B i + 1 C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) + ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 2 B i C j + 1 ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) ] = B + B ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) + C + C ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 2 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) = B [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) ] + C [ I + C ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 2 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) ] = B e m n B C ( k − m ) + C e m n B C ( k − n ) .$

Due to (14) and (15), we conclude that formula (17) is valid.

### II. $k= p ( k ) (m+1)∧( q ( k ) −1)(n+1)+1≤k< q ( k ) (n+1)$

In this case,

$p ( k − m ) = ⌊ k − m + m m + 1 ⌋ = ⌊ k m + 1 ⌋ = p ( k ) , p ( k + 1 ) = ⌊ k + 1 + m m + 1 ⌋ ≤ k + 1 + m m + 1 = k m + 1 + 1 = p ( k ) + 1 , p ( k + 1 ) = ⌊ k + 1 + m m + 1 ⌋ > k + 1 + m m + 1 − 1 = k m + 1 = p ( k )$

and $p ( k + 1 ) = p ( k ) +1$. In addition to this (see relevant computations performed in case I), we have $q ( k − n ) = q ( k ) −1$ and $q ( k + 1 ) = q ( k )$.

Then

$e m n B C k = I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j + 1 ) , e m n B C ( k + 1 ) = I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) , q ( k ) − 1 B i C j ( i + j i ) ( k + 1 − m i − n j i + j + 1 )$

and

$e m n B C ( k − m ) =I+(B+C) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k − m ) − 1 B i C j ( i + j i ) ( k − m − m i − n j i + j + 1 ) ,$
(18)
$e m n B C ( k − n ) =I+(B+C) ∑ i = 0 , j = 0 p ( k − n ) − 1 , q ( k ) − 2 B i C j ( i + j i ) ( k − n − m i − n j i + j + 1 ) .$
(19)

Like with the computations performed in the previous part of the proof, we get

and

So, we can substitute $q ( k − m )$ by $q ( k )$ in (18) and $p ( k − n )$ by $p ( k )$ in (19).

Accordingly, we have

$e m n B C ( k − m ) =I+(B+C) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) ,$
(20)
$e m n B C ( k − n ) =I+(B+C) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 2 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) .$
(21)

It is easy to see that, due to (3), formula (20) can also be used instead of (18) if $q ( k − m ) <1$ and formula (21) can also be used instead of (19) if $p ( k − n ) <1$.

We have to prove

$Δ e m n B C k = B e m n B C ( k − m ) + C e m n B C ( k − n ) = B [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) ] + C [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 2 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) ] .$
(22)

Therefore,

$Δ e m n B C k = e m n B C ( k + 1 ) − e m n B C k = I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) , q ( k ) − 1 B i C j ( i + j i ) ( k + 1 − m i − n j i + j + 1 ) − I − ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j + 1 ) = ( B + C ) [ ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) [ ( k + 1 − m i − n j i + j + 1 ) − ( k − m i − n j i + j + 1 ) ] + ∑ j = 0 q ( k ) − 1 B p ( k ) C j ( p ( k ) + j p ( k ) ) ( k + 1 − m p ( k ) − n j p ( k ) + j + 1 ) ] .$

With the aid of the equation $k= p ( k ) (m+1)$, we get

and, by (9), we have

$Δ e m n B C k = ( B + C ) [ ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j ) + B p ( k ) ] = ( B + C ) [ I + ∑ i = 1 p ( k ) − 1 B i C 0 ( i i ) ( k − m i i ) + ∑ j = 1 q ( k ) − 1 B 0 C j ( j 0 ) ( k − n j j ) + ∑ i = 1 , j = 1 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j ) + B p ( k ) ] .$

By (10), we have

$Δ e m n B C k = ( B + C ) [ I + ∑ i = 1 p ( k ) − 1 B i C 0 ( i − 1 i − 1 ) ( k − m i i ) + ∑ j = 1 q ( k ) − 1 B 0 C j ( j − 1 0 ) ( k − n j j ) + ∑ i = 1 , j = 1 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j − 1 i − 1 ) ( k − m i − n j i + j ) + ∑ i = 1 , j = 1 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j − 1 i ) ( k − m i − n j i + j ) + B p ( k ) ] = ( B + C ) [ I + ∑ i = 1 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j − 1 i − 1 ) ( k − m i − n j i + j ) + ∑ i = 0 , j = 1 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j − 1 i ) ( k − m i − n j i + j ) + B p ( k ) ] .$

Now we replace in the first sum the summation index i by $i+1$ and in the second sum we replace the summation index j by $j+1$. Then

$Δ e m n B C k = ( B + C ) [ I + ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k ) − 1 B i + 1 C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) + ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 2 B i C j + 1 ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) + B p ( k ) ] = B + B ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) + B p ( k ) ( B + C ) + C + C ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 2 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) = B [ I + ( B + C ) ( ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) + B p ( k ) − 1 ) ] + C [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 2 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) ] .$

For $k= p ( k ) (m+1)$, we have

$B p ( k ) − 1 = ∑ j = 0 q ( k ) − 1 B p ( k ) − 1 C j ( p ( k ) − 1 + j p ( k ) − 1 ) ( k − m ( p ( k ) − 1 + 1 ) − n j p ( k ) − 1 + j + 1 ) ,$

where

Thus,

$Δ e m n B C k = B [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) ] + C [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 2 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) ] = B e m n B C ( k − m ) + C e m n B C ( k − n )$

and formula (22) is proved.

### III. $( p ( k ) −1)(m+1)+1≤k< p ( k ) (m+1)∧k= q ( k ) (n+1)$

In this case, we have (see relevant computations in cases I and II)

$p ( k − m ) = p ( k ) −1, p ( k + 1 ) = p ( k )$

and

$q ( k − n ) = q ( k ) , q ( k + 1 ) = q ( k ) +1.$

Then

$e m n B C k = I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j + 1 ) , e m n B C ( k + 1 ) = I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) B i C j ( i + j i ) ( k + 1 − m i − n j i + j + 1 )$

and

$e m n B C ( k − m ) =I+(B+C) ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k − m ) − 1 B i C j ( i + j i ) ( k − m − m i − n j i + j + 1 ) ,$
(23)
$e m n B C ( k − n ) =I+(B+C) ∑ i = 0 , j = 0 p ( k − n ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − n − m i − n j i + j + 1 ) .$
(24)

Like with the computations performed in case I, we can get

and

So, we can substitute $q ( k )$ for $q ( k − m )$ in (23) and $p ( k )$ for $p ( k − n )$ in (24).

Thus, we have

$e m n B C ( k − m ) =I+(B+C) ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) ,$
(25)
$e m n B C ( k − n ) =I+(B+C) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) .$
(26)

It is easy to see that, due to (3), formula (25) can also be used instead of (23) if $q ( k − m ) <1$ and formula (26) can also be used instead of (24) if $p ( k − n ) <1$.

Now we have to prove

$Δ e m n B C k = B e m n B C ( k − m ) + C e m n B C ( k − n ) = B [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) ] + C [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) ] .$
(27)

Considering the difference by its definition, we get

$Δ e m n B C k = e m n B C ( k + 1 ) − e m n B C k = I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) B i C j ( i + j i ) ( k + 1 − m i − n j i + j + 1 ) − I − ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j + 1 ) = ( B + C ) [ ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) [ ( k + 1 − m i − n j i + j + 1 ) − ( k − m i − n j i + j + 1 ) ] + ∑ i = 0 p ( k ) − 1 B i C q ( k ) ( i + q ( k ) i ) ( k + 1 − m i − n q ( k ) i + q ( k ) + 1 ) ] .$

With the aid of relation $k= q ( k ) (n+1)$, we get

and

$Δ e m n B C k = ( B + C ) [ ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j ) + C q ( k ) ] = ( B + C ) [ I + ∑ i = 1 p ( k ) − 1 B i C 0 ( i i ) ( k − m i i ) + ∑ j = 1 q ( k ) − 1 B 0 C j ( j 0 ) ( k − n j j ) + ∑ i = 1 , j = 1 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j ) + C q ( k ) ] .$

By (10), we have

$Δ e m n B C k = ( B + C ) [ I + ∑ i = 1 p ( k ) − 1 B i C 0 ( i − 1 i − 1 ) ( k − m i i ) + ∑ j = 1 q ( k ) − 1 B 0 C j ( j − 1 0 ) ( k − n j j ) + ∑ i = 1 , j = 1 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j − 1 i − 1 ) ( k − m i − n j i + j ) + ∑ i = 1 , j = 1 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j − 1 i ) ( k − m i − n j i + j ) + C q ( k ) ] = ( B + C ) [ I + ∑ i = 1 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j − 1 i − 1 ) ( k − m i − n j i + j ) + ∑ i = 0 , j = 1 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j − 1 i ) ( k − m i − n j i + j ) + C q ( k ) ] .$

Now we replace in the first sum the summation index i by $i+1$ and in the second sum we replace the summation index j by $j+1$. Then

$Δ e m n B C k = ( B + C ) [ I + ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k ) − 1 B i + 1 C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) + ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 2 B i C j + 1 ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) + C q ( k ) ] = B + B ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) + C + C ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 2 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) + C q ( k ) ( B + C ) = B [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) ] + C [ I + ( B + C ) ( ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 2 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) + C q ( k ) − 1 ) ] .$

For $k= q ( k ) (n+1)$, we have

$C q ( k ) − 1 = ∑ i = 0 p ( k ) − 1 B i C q ( k ) − 1 ( i + q ( k ) − 1 i ) ( k − m i − n ( q ( k ) − 1 + 1 ) i + q ( k ) − 1 + 1 ) ,$

where

Thus,

$Δ e m n B C k = B [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) ] + C [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) ] = B e m n B C ( k − m ) + C e m n B C ( k − n )$

and formula (27) is proved.

### IV. $k= p ( k ) (m+1)∧k= q ( k ) (n+1)$

In this case, we have (see similar combinations in cases II and III)

$p ( k − m ) = p ( k ) , p ( k + 1 ) = p ( k ) +1$

and

$q ( k − n ) = q ( k ) , q ( k + 1 ) = q ( k ) +1.$

Then

$e m n B C k = I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j + 1 ) , e m n B C ( k + 1 ) = I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) , q ( k ) B i C j ( i + j i ) ( k + 1 − m i − n j i + j + 1 )$

and

$e m n B C ( k − m ) =I+(B+C) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k − m ) − 1 B i C j ( i + j i ) ( k − m − m i − n j i + j + 1 ) ,$
(28)
$e m n B C ( k − n ) =I+(B+C) ∑ i = 0 , j = 0 p ( k − n ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − n − m i − n j i + j + 1 ) .$
(29)

As before,

and

So, we can substitute $q ( k )$ for $q ( k − m )$ in (28) and $p ( k )$ for $p ( k − n )$ in (29) and

$e m n B C ( k − m ) =I+(B+C) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) ,$
(30)
$e m n B C ( k − n ) =I+(B+C) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) .$
(31)

It is easy to see that, due to (3), formula (30) can also be used instead of (28) if $q ( k − m ) <1$ and formula (31) can also be used instead of (29) if $p ( k − n ) <1$.

Now it is possible to prove the formula

$Δ e m n B C k = B e m n B C ( k − m ) + C e m n B C ( k − n ) = B [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) ] + C [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) ] .$
(32)

By definition, we get

$Δ e m n B C k = e m n B C ( k + 1 ) − e m n B C k = I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) , q ( k ) B i C j ( i + j i ) ( k + 1 − m i − n j i + j + 1 ) − I − ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j + 1 ) = ( B + C ) [ ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) [ ( k + 1 − m i − n j i + j + 1 ) − ( k − m i − n j i + j + 1 ) ] + ∑ j = 0 q ( k ) B p ( k ) C j ( p ( k ) + j p ( k ) ) ( k + 1 − m p ( k ) − n j p ( k ) + j + 1 ) + ∑ i = 0 p ( k ) B i C q ( k ) ( i + q ( k ) i ) ( k + 1 − m i − n q ( k ) i + q ( k ) + 1 ) ] .$

With the aid of equations $k= p ( k ) (m+1)$, $k= q ( k ) (n+1)$, we get

and

$Δ e m n B C k = ( B + C ) [ ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j ) + B p ( k ) + C q ( k ) ] = ( B + C ) [ I + ∑ i = 1 p ( k ) − 1 B i C 0 ( i i ) ( k − m i i ) + ∑ j = 1 q ( k ) − 1 B 0 C j ( j 0 ) ( k − n j j ) + ∑ i = 1 , j = 1 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n j i + j ) + B p ( k ) + C q ( k ) ] .$

By (10), we have

$Δ e m n B C k = ( B + C ) [ I + ∑ i = 1 p ( k ) − 1 B i C 0 ( i − 1 i − 1 ) ( k − m i i ) + ∑ j = 1 q ( k ) − 1 B 0 C j ( j − 1 0 ) ( k − n j j ) + ∑ i = 1 , j = 1 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j − 1 i − 1 ) ( k − m i − n j i + j ) + ∑ i = 1 , j = 1 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j − 1 i ) ( k − m i − n j i + j ) + B p ( k ) + C q ( k ) ] = ( B + C ) [ I + ∑ i = 1 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j − 1 i − 1 ) ( k − m i − n j i + j ) + ∑ i = 0 , j = 1 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j − 1 i ) ( k − m i − n j i + j ) + B p ( k ) + C q ( k ) ] .$

We replace in the first sum the summation index i by $i+1$ and in the second sum we substitute the summation index j by $j+1$. Then

$Δ e m n B C k = ( B + C ) [ I + ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k ) − 1 B i + 1 C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) + ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 2 B i C j + 1 ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) + B p ( k ) + C q ( k ) ] = B + B ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) + B p ( k ) ( B + C ) + C + C ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 2 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) + C q ( k ) ( B + C ) = B [ I + ( B + C ) ( ∑ i = 0 , j = 0 p ( k ) − 2 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) + B p ( k ) − 1 ) ] + C [ I + ( B + C ) ( ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 2 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) + C q ( k ) − 1 ) ] .$

Because $k= p ( k ) (m+1)= q ( k ) (n+1)$, we can express $B p ( k ) − 1$ and $C q ( k ) − 1$ in the form

$B p ( k ) − 1 = ∑ j = 0 q ( k ) − 1 B p ( k ) − 1 C j ( p ( k ) − 1 + j p ( k ) − 1 ) ( k − m ( p ( k ) − 1 + 1 ) − n j p ( k ) − 1 + j + 1 ) , C q ( k ) − 1 = ∑ i = 0 p ( k ) − 1 B i C q ( k ) − 1 ( i + q ( k ) − 1 i ) ( k − m i − n ( q ( k ) − 1 + 1 ) i + q ( k ) − 1 + 1 ) ,$

where

Thus,

$Δ e m n B C k = B [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m ( i + 1 ) − n j i + j + 1 ) ] + C [ I + ( B + C ) ∑ i = 0 , j = 0 p ( k ) − 1 , q ( k ) − 1 B i C j ( i + j i ) ( k − m i − n ( j + 1 ) i + j + 1 ) ] = B e m n B C ( k − m ) + C e m n B C ( k − n ) .$

Therefore, formula (32) is valid.

We proved that formula (6) holds in each of the considered cases I, II, III and IV for $k≥1$. If $k=0$, the proof can be done directly because $p ( 0 ) = q ( 0 ) =0$, $p ( 1 ) = q ( 1 ) =1$,

$Δ e m n B C 0 = e m n B C 1 − e m n B C 0 = I + ( B + C ) ∑ i = 0 , j = 0 0 , 0 B i C j ( i + j i ) ( 1 − m i − n j i + j + 1 ) − I − ( B + C ) ∑ i = 0 , j = 0 − 1 , − 1 B i C j ( i + j i ) ( − m i − n j i + j + 1 ) = I + B + C − I = B + C$

and

$B e m n B C ( − m ) +C e m n B C ( − n ) =BI+CI=B+C.$

Formula (6) holds again. Theorem 2 is proved. □

## Open problems and concluding remarks

Formula (4) is valid for $k∈ Z − m ∞$. However, formula (6) holds for $k∈ Z 0 ∞$ only. Therefore, there is a difference between the definition domains of the formulas, and it is a challenge how to modify Definition 2 of discrete matrix delayed exponential for two delays in such a way that formula (6) will hold for $k∈ Z − max { m , n } ∞$. In  formula (4) is used to get a representation of the solution of the problems (both homogeneous and nonhomogeneous)

$Δ y ( k ) = B y ( k − m ) + f ( k ) , k ∈ Z 0 ∞ , y ( k ) = φ ( k ) , k ∈ Z − m 0 ,$

where $f: Z 0 ∞ → R r$, $y: Z − m ∞ → R r$ and $φ: Z − m 0 → R r$.

It is an open problem how to use formula (6) to get a representation of the solution of the homogeneous and nonhomogeneous problems

$Δ y ( k ) = B y ( k − m ) + C y ( k − n ) + f ( k ) , k ∈ Z 0 ∞ , y ( k ) = φ ( k ) , k ∈ Z − s 0 , s = max { m , n }$

if $BC=CB$.

Let us note that the first concept of matrix delayed exponential was given in  and the first concept of discrete matrix delayed exponential was given in . Further development of the delayed matrix exponentials method and its utilization to various problems can be found, e.g., in .

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## Acknowledgements

The first author was supported by Operational Programme Research and Development for Innovations, No. CZ.1.05/2.1.00/03.0097, as an activity of the regional Centre AdMaS.

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Diblík, J., Morávková, B. Discrete matrix delayed exponential for two delays and its property. Adv Differ Equ 2013, 139 (2013). https://doi.org/10.1186/1687-1847-2013-139 