Introduction

Throughout the paper, we use the following notation. For integers s, t, st, we define a set Z s t :={s,s+1,,t1,t}. Similarly, we define sets Z t :={,t1,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:

x1<xx,
(1)

where xR.

Define binomial coefficients as customary, i.e., for nZ and kZ,

( n k ) :={ n ! k ! ( n k ) ! if  n k 0 , 0 otherwise .
(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 mN was defined as follows.

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

e m B k :={ Θ if  k Z m 1 , I + j = 1 B j ( k m ( j 1 ) j ) if  = 0 , 1 , 2 , , k Z ( 1 ) ( m + 1 ) + 1 ( m + 1 ) ,

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 mN is proved in [1].

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,nN, mn 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,nN, mn 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:

e m n B C k :={ Θ if  k Z max { m , n } 1 , I if  k Z max { m , n } 0 , 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 ) if  k Z 1 ,

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,nN, mn 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 k0.

Proof Let k1. From (1) and (5), we can see easily that, for an integer k0 satisfying

( p ( k ) 1)(m+1)+1k p ( k ) (m+1)( q ( k ) 1)(n+1)+1k 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)+1k< p ( k ) (m+1)( q ( k ) 1)(n+1)+1k< q ( k ) (n+1),

in case two

k= p ( k ) (m+1)( q ( k ) 1)(n+1)+1k< q ( k ) (n+1),

in case three

( p ( k ) 1)(m+1)+1k< 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,kN 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,jN, which are derived from (2) and (9).

I. ( p ( k ) 1)(m+1)+1k< p ( k ) (m+1)( q ( k ) 1)(n+1)+1k< 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

( k m m i n j i + j + 1 ) =0if i0,j q ( k m ) .
(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

km<(n+1) q ( k m ) +1(m+1)i+(n+1)j+1if i0,j q ( k m ) .

From the last inequality, we get

kmminj<i+j+1if i0,j q ( k m )

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

( k n m i n j i + j + 1 ) =0if i p ( k n ) ,j0

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)+1k< 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

( k m m i n j i + j + 1 ) =0if i0,j q ( k m )

and

( k n m i n j i + j + 1 ) =0if i p ( k n ) ,j0.

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

( k + 1 m p ( k ) n j p ( k ) + j + 1 ) = ( p ( k ) + 1 n j p ( k ) + 1 + j ) =0if j>0

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

( k m ( p ( k ) 1 + 1 ) n j p ( k ) 1 + j + 1 ) = ( k m p ( k ) n j p ( k ) + j ) =0if j>0.

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)+1k< 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

( k m m i n j i + j + 1 ) =0if i0,j q ( k m )

and

( k n m i n j i + j + 1 ) =0if i p ( k n ) ,j0.

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

( k + 1 m i n q ( k ) i + q ( k ) + 1 ) = ( q ( k ) + 1 m i q ( k ) + 1 + i ) =0if i>0

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

( k m i n ( q ( k ) 1 + 1 ) i + q ( k ) 1 + 1 ) = ( k m i n q ( k ) i + q ( k ) ) =0if i>0.

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,

( k m m i n j i + j + 1 ) =0if i0,j q ( k m )

and

( k n m i n j i + j + 1 ) =0if i p ( k n ) ,j0.

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

( k + 1 m p ( k ) n j p ( k ) + j + 1 ) = ( p ( k ) + 1 n j p ( k ) + 1 + j ) = 0 if  j > 0 , ( k + 1 m i n q ( k ) i + q ( k ) + 1 ) = ( q ( k ) + 1 m i q ( k ) + 1 + i ) = 0 if  i > 0

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

( k m ( p ( k ) 1 + 1 ) n j p ( k ) 1 + j + 1 ) = ( k m p ( k ) n j p ( k ) + j ) = 0 if  j > 0 , ( k m i n ( q ( k ) 1 + 1 ) i + q ( k ) 1 + 1 ) = ( k m i n q ( k ) i + q ( k ) ) = 0 if  i > 0 .

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 k1. 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 [1] 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 [3] and the first concept of discrete matrix delayed exponential was given in [1]. Further development of the delayed matrix exponentials method and its utilization to various problems can be found, e.g., in [416].