Open Access

On a singular system of fractional nabla difference equations with boundary conditions

Boundary Value Problems20132013:148

DOI: 10.1186/1687-2770-2013-148

Received: 20 February 2013

Accepted: 18 May 2013

Published: 19 June 2013

Abstract

In this article, we study a boundary value problem of a class of linear singular systems of fractional nabla difference equations whose coefficients are constant matrices. By taking into consideration the cases that the matrices are square with the leading coefficient matrix singular, square with an identically zero matrix pencil and non-square, we provide necessary and sufficient conditions for the existence and uniqueness of solutions. More analytically, we study the conditions under which the boundary value problem has a unique solution, infinite solutions and no solutions. Furthermore, we provide a formula for the case of the unique solution. Finally, numerical examples are given to justify our theory.

Keywords

boundary conditions singular systems fractional calculus nabla operator difference equations linear discrete time system

1 Introduction

Difference equations of fractional order have recently proven to be valuable tools in the modeling of many phenomena in various fields of science and engineering. Indeed, we can find numerous applications in viscoelasticity, electrochemistry, control, porous media, electromagnetism, and so forth [17]. There has been a significant development in the study of fractional differential/difference equations and inclusions in recent years; see the monographs of Baleanu et al. [1], Kaczorek [4], Klamka et al. [8], Malinowska et al. [5], Podlubny [7], and the survey by Agarwal et al. [9]. For some recent contributions on fractional differential/difference equations, see [1, 4, 5, 827] and the references therein. In this article we provide an introductory study for a boundary value problem of a class of singular fractional nabla discrete time systems. If we define N α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq1_HTML.gif by N α = { α , α + 1 , α + 2 , } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq2_HTML.gif, α integer, and n such that 0 < n < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq3_HTML.gif or 1 < n < 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq4_HTML.gif, then the nabla fractional operator in the case of Riemann-Liouville fractional difference of n th order for any Y k : N a R m × 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq5_HTML.gif is defined by, see [5, 1012, 23],
α n Y k = 1 Γ ( n ) j = α k ( k j + 1 ) n 1 ¯ Y j , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equa_HTML.gif
where the raising power function is defined by
k α ¯ = Γ ( k + α ) Γ ( k ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equb_HTML.gif
The following problem will then be considered. The singular fractional discrete time systems of the form
F 0 n Y k = G Y k , k = 1 , 2 , , N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ1_HTML.gif
(1)
with known boundary conditions
A 1 Y 0 = B 1 , A 2 Y N = B 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ2_HTML.gif
(2)
where F , G M ( r × m ; F ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq6_HTML.gif (i.e., the algebra of matrices with elements in the field ) with Y k , V k M ( m × 1 ; F ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq7_HTML.gif, A 1 M ( r 1 × m ; F ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq8_HTML.gif, A 2 M ( r 2 × m ; F ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq9_HTML.gif, B 1 M ( r 1 × 1 ; F ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq10_HTML.gif and B 2 M ( r 2 × 1 ; F ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq11_HTML.gif. For the sake of simplicity, we set M m = M ( m × m ; F ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq12_HTML.gif and M r m = M ( r × m ; F ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq13_HTML.gif. The matrices F and G can be non-square (when r m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq14_HTML.gif) or square ( r = m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq15_HTML.gif) and F singular ( det F = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq16_HTML.gif). The main purpose will be to provide necessary and sufficient conditions for the existence and uniqueness of solutions of the above boundary value problem, i.e., to study the conditions under which the system has unique, infinite and no solutions and to provide a formula for the case of the unique solution (if it exists). Many authors use matrix pencil theory to study linear discrete time systems with constant matrices; see, for instance, [2843]. A matrix pencil is a family of matrices s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif, parametrized by a complex number s, see [39, 41, 44, 45]. When G is square and F = I m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq18_HTML.gif, where I m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq19_HTML.gif is the identity matrix, the zeros of the function det ( s F G ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq20_HTML.gif are the eigenvalues of G. Consequently, the problem of finding the nontrivial solutions of the equation
s F X = G X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equc_HTML.gif
is called the generalized eigenvalue problem. Although the generalized eigenvalue problem looks like a simple generalization of the usual eigenvalue problem, it exhibits some important differences. In the first place, it is possible for F, G to be non-square matrices. Moreover, even with F, G square it is possible (in the case F is singular) for det ( s F G ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq20_HTML.gif to be identically zero, independent of s. Finally, even if we assume F, G square matrices with a non-zero pencil, it is possible (when F is singular) for the problem to have infinite eigenvalues. To see this, write the generalized eigenvalue problem in the reciprocal form
F X = s 1 G X . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equd_HTML.gif

If F is singular with a null vector X, then F X = 0 m , 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq21_HTML.gif, so that X is an eigenvector of the reciprocal problem corresponding to eigenvalue s 1 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq22_HTML.gif; i.e., s = https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq23_HTML.gif.

Definition 1.1 Given F , G M r m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq24_HTML.gif and an arbitrary s F https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq25_HTML.gif, the matrix pencil s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif is called:
  1. 1.

    Regular when r = m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq15_HTML.gif and det ( s F G ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq26_HTML.gif.

     
  2. 2.

    Singular when r m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq14_HTML.gif or r = m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq15_HTML.gif and det ( s F G ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq27_HTML.gif.

     

The paper is organized as follows. In Section 2, we study the existence of solutions of the system (1) when its pencil is regular. In Section 3 we study the case of the system (1) with a singular pencil, and Section 3 contains numerical examples.

2 Regular case

In this section, we consider the case of the system (1) with a regular pencil. The class of s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif is characterized by a uniquely defined element, known as complex Weierstrass canonical form, s F w Q w https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq28_HTML.gif, see [39, 41, 44, 45], specified by the complete set of invariants of s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif. This is the set of elementary divisors (e.d.) obtained by factorizing the invariant polynomials into powers of homogeneous polynomials irreducible over the field . In the case where s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif is regular, we have e.d. of the following type:

  • e.d. of the type ( s a j ) p j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq29_HTML.gifare called finite elementary divisors (f.e.d.), where a j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq30_HTML.gif is a finite eigenvalue of algebraic multiplicity p j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq31_HTML.gif;

  • e.d. of the type s ˆ q = 1 s q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq32_HTML.gif are called infinite elementary divisors (i.e.d.), where q is the algebraic multiplicity of the infinite eigenvalues.

We assume that i = 1 ν p j = p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq33_HTML.gif and p + q = m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq34_HTML.gif.

Definition 2.1 Let B 1 , B 2 , , B l https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq35_HTML.gif be elements of M m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq36_HTML.gif. The direct sum of them denoted by B 1 B 2 B l https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq37_HTML.gif is the blockdiag [ B 1 B 2 B l ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq38_HTML.gif.

From the regularity of s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif, there exist nonsingular matrices P , Q M m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq39_HTML.gif such that
P F Q = F w = I p H q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ3_HTML.gif
(3)
and
P G Q = G w = J p I q . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ4_HTML.gif
(4)
The complex Weierstrass form s F w Q w https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq40_HTML.gif of the regular pencil s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif is defined by
s F w Q w : = s I p J p s H q I q , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Eque_HTML.gif
where the first normal Jordan-type element is uniquely defined by the set of the finite eigenvalues of s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif and has the form
s I p J p : = s I p 1 J p 1 ( a 1 ) s I p ν J p ν ( a ν ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equf_HTML.gif
The second uniquely defined block s H q I q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq41_HTML.gif corresponds to the infinite eigenvalues of s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif and has the form
s H q I q : = s H q 1 I q 1 s H q σ I q σ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equg_HTML.gif
The matrix H q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq42_HTML.gif is a nilpotent element of M q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq43_HTML.gif with index q = max { q j : j = 1 , 2 , , σ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq44_HTML.gif, where
H q q = 0 q , q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equh_HTML.gif
and I p j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq45_HTML.gif, J p j ( a j ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq46_HTML.gif, H q j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq47_HTML.gif are defined as
I p j = [ 1 0 0 0 0 1 0 0 0 0 0 1 ] M p j , J p j ( a j ) = [ a j 1 0 0 0 a j 0 0 0 0 a j 1 0 0 0 a j ] M p j , H q j = [ 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 ] M q j . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equi_HTML.gif

For algorithms about the computations of the Jordan matrices, see [39, 41, 44, 45].

Definition 2.2 If for the system (1) with boundary conditions (2) there exists at least one solution, the boundary value problem (1)-(2) is said to be consistent.

For the regular matrix pencil of the system (1), there exist nonsingular matrices P , Q M m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq48_HTML.gif as applied in (3), (4). Let
Q = [ Q p Q q ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ5_HTML.gif
(5)

where Q p M m p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq49_HTML.gif is a matrix with columns p linear independent (generalized) eigenvectors of the p finite eigenvalues of s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif, and Q q M m q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq50_HTML.gif is a matrix with columns q linear independent (generalized) eigenvectors of the q infinite eigenvalues of s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif.

Lemma 2.1 Consider the system (1) with a regular pencil. Then the system (1) is divided into two subsystems:
0 n Z k p = J p Z k p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equj_HTML.gif
and
H q 0 n Z k q = Z k q . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equk_HTML.gif
Proof Consider the transformation
Y k = Q Z k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ6_HTML.gif
(6)
and by substituting (6) into (1), we obtain
F 0 n Q Z k = G Q Z k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equl_HTML.gif
or, equivalently,
F Q 0 n Z k = G Q Z k . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equm_HTML.gif
Whereby multiplying by P, we arrive at
F w 0 n Z k = G w Z k . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equn_HTML.gif
Moreover, let
Z k = [ Z k p Z k q ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equo_HTML.gif
where Z k p M p 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq51_HTML.gif, Z k q M q 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq52_HTML.gif, and by using (3) and (4), we obtain
[ I p 0 p , q 0 q , p H q ] α n [ Z k p Z k q ] = [ J p 0 p , q 0 q , p I q ] [ Z k p Z k q ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equp_HTML.gif
From the above expressions, we arrive easily at the subsystems
0 n Z k p = J p Z k p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ7_HTML.gif
(7)
and
H q 0 n Z k q = Z k q . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ8_HTML.gif
(8)

The proof is completed. □

Definition 2.3 With F n , n ( J p ( k + n ) n ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq53_HTML.gif we denote the discrete Mittag-Leffler function with two parameters defined by
F n , n ( J p ( k + n ) n ¯ ) = i = 0 J p i ( k + n ) i n ¯ Γ ( ( i + 1 ) n ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ9_HTML.gif
(9)

See [1012, 23, 46].

Proposition 2.1 The subsystem (7) has the solution
Z k p = ( k + 1 ) n 1 ¯ F n , n ( J p ( k + n ) n ¯ ) ( I p J p ) Z 0 p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ10_HTML.gif
(10)
if and only if
J p < 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ11_HTML.gif
(11)

where https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq54_HTML.gif is an induced matrix norm and F n , n ( J p ( k + n ) n ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq53_HTML.gif is the discrete Mittag-Leffler function with two parameters as defined by Definition  2.3.

Proof From [1012, 23, 46] the solution of (7) can be calculated and given by the formula
Z k p = ( k + 1 ) n 1 ¯ F n , n ( J p ( k + n ) n ¯ ) ( I p J p ) Z 0 p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equq_HTML.gif
or, equivalently, by
Z k p = ( k + 1 ) n 1 ¯ ( i = 0 J p i ( k + n ) i n ¯ Γ ( ( i + 1 ) n ) ) ( I p J p ) Z 0 p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equr_HTML.gif
The existence and uniqueness of the above solution depends on the convergence of the matrix power series
i = 0 J p i ( k + n ) i n ¯ Γ ( ( i + 1 ) n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equs_HTML.gif
or, equivalently, if and only if
lim i J p ( i + 1 ) ( k + n ) ( i + 1 ) n ¯ Γ ( ( i + 2 ) n ) J p i ( k + n ) i n ¯ Γ ( ( i + 1 ) n ) < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equt_HTML.gif
or, equivalently,
J p < lim i ( k + n ) i n ¯ Γ ( ( i + 1 ) n ) ( k + n ) ( i + 1 ) n ¯ Γ ( ( i + 2 ) n ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equu_HTML.gif
By using the property
Γ ( z + 1 ) = z Γ ( z ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equv_HTML.gif
we get
J p < lim i ( ( k 1 ) + ( i + 1 ) n ) ( 1 + ( i + 1 ) n ) ( ( i + 1 ) n ) ( ( k 1 ) + ( i + 2 ) n ) ( 1 + ( i + 2 ) n ) ( ( i + 2 ) n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equw_HTML.gif
or, equivalently,
J p < 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equx_HTML.gif

The proof is completed. □

Proposition 2.2 The subsystem (8) has the unique solution
Z k q = 0 q , 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ12_HTML.gif
(12)
Proof Let q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq55_HTML.gif be the index of the nilpotent matrix H q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq42_HTML.gif, i.e., H q q = 0 q , q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq56_HTML.gif. Then if we obtain the following equations:
H q 0 n Z k q = Z k q , H q 2 0 2 n Z k q = H q 0 n Z k q , H q 3 0 3 n Z k q = H q 2 0 2 n Z k q , H q 4 0 4 n Z k q = H q 3 0 3 n Z k q , H q q 1 0 ( q 1 ) n Z k q = H q q 2 0 ( q 2 ) n Z k q , H q q 0 q n Z k q = H q q 1 0 ( q 1 ) n Z k q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equy_HTML.gif

by taking the sum of the above equations and using the fact that H q q = 0 q , q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq57_HTML.gif, we arrive easily at the solution (12). The proof is completed. □

Theorem 2.1 Consider the system (1) with a regular pencil and boundary conditions of type (2). Then the boundary value problem (1)-(2) is consistent if and only if:
  1. 1.
    The pencil s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif has p distinct eigenvalues and all lie within the open disk
    | s | < 1 ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equz_HTML.gif
     
  2. 2.
    [ B 1 B 2 ] colspan [ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ13_HTML.gif
    (13)
     
Furthermore, when the boundary value problem (1)-(2) is consistent, it has a unique solution if and only if:
  1. 1.
    p r 1 + r 2 ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ14_HTML.gif
    (14)
     
  2. 2.
    rank [ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] = p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ15_HTML.gif
    (15)
     
In this case the unique solution is then given by
Y k = Q p ( k + 1 ) n 1 ¯ F n , n ( J p ( k + n ) n ¯ ) ( I p J p ) C , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ16_HTML.gif
(16)
where C is the unique solution of the algebraic system
[ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] C = [ B 1 B 2 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ17_HTML.gif
(17)
Proof By applying the transformation (6) into the system (1), we get the systems (7), (8) with solutions (10), (12) respectively. Note that from Proposition 2.1 the solution (10) exists if and only if
J p < 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equaa_HTML.gif
where J p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq58_HTML.gif is the Jordan matrix related to the p finite eigenvalues of the pencil s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif, which is equivalent to the fact that the finite eigenvalues of the pencil must be distinct and all lie within the unit disk | s | < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq59_HTML.gif. Based on these results, the solution of (1) can be written as
Y k = Q Z k = [ Q p Q q ] [ Z k p Z k q ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equab_HTML.gif
or, equivalently,
Y k = Q p Z k p + Q q Z k q , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equac_HTML.gif
or, equivalently, by using (10), (12)
Y k = Q p ( k + 1 ) n 1 ¯ F n , n ( J p ( k + n ) n ¯ ) ( I p J p ) Z 0 p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equad_HTML.gif
The initial value Z 0 p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq60_HTML.gif of the subsystem (7) is not known and can be replaced by a constant vector C M p 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq61_HTML.gif
Y k = Q p ( k + 1 ) n 1 ¯ F n , n ( J p ( k + n ) n ¯ ) ( I p J p ) C . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equae_HTML.gif
The above solution exists if and only if
A 1 Y 0 = B 1 , A 2 Y N = B 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equaf_HTML.gif
or, equivalently,
A 1 Q p C = B 1 , A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) C = B 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equag_HTML.gif
or, equivalently,
[ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] C = [ B 1 B 2 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equah_HTML.gif
For the above algebraic system, there exists at least one solution if and only if
[ B 1 B 2 ] colspan [ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equai_HTML.gif
The algebraic system (17) contains r 1 + r 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq62_HTML.gif equations and p unknowns. Hence the solution is unique if and only if
p r 1 + r 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equaj_HTML.gif
and
rank [ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] = p , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equak_HTML.gif
where C is then the unique solution of (17). This can be proved as follows. If we assume that the algebraic system has two solutions C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq63_HTML.gif and C 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq64_HTML.gif, then
[ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] C 1 = [ B 1 B 2 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equal_HTML.gif
and
[ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] C 2 = [ B 1 B 2 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equam_HTML.gif
or, equivalently,
[ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] ( C 1 C 2 ) = 0 p , 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equan_HTML.gif
But the matrix [ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq65_HTML.gif is left invertible since it is assumed to have p linear independent columns and r 1 + r 2 p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq66_HTML.gif and hence
C 1 = C 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equao_HTML.gif

The unique solution is then given from (16). The proof is completed. □

3 Singular case

In this section, we consider the case of the system (1) with a singular pencil. The class of s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif in this case is characterized by a uniquely defined element, s F K Q K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq67_HTML.gif, known as the complex Kronecker canonical form, see [39, 41, 44, 45], specified by the complete set of invariants of the singular pencil s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif. This is the set of the elementary divisors (e.d.) and the minimal indices (m.i.). Unlike the case of the regular pencils, where the pencil is characterized only from the e.d., the characterization of a singular matrix pencil apart from the set of the determinantal divisors requires the definition of additional sets of invariants, the minimal indices. The distinguishing feature of a singular pencil s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif is that either r m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq14_HTML.gif or r = m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq15_HTML.gif and det ( s F G ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq27_HTML.gif. Let N r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq68_HTML.gif, N l https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq69_HTML.gif be the right and the left null space of a matrix respectively. Then the equations
( s F G ) U ( s ) = 0 r , 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ18_HTML.gif
(18)
V T ( s ) ( s F G ) = 0 1 , m , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ19_HTML.gif
(19)
where ( ) T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq70_HTML.gif is the transpose tensor, have solutions in U ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq71_HTML.gif, V ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq72_HTML.gif, which are vectors in the rational vector spaces N r ( s F G ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq73_HTML.gif and N l ( s F G ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq74_HTML.gif respectively. The binary vectors U ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq71_HTML.gif and V T ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq75_HTML.gif express dependence relationships among the columns or rows of s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif respectively. Note that U ( s ) M m 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq76_HTML.gif and V ( s ) M r 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq77_HTML.gif are polynomial vectors. Let d = dim N r ( s F G ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq78_HTML.gif and t = dim N l ( s F G ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq79_HTML.gif. It is known, see [39, 41, 44, 45], that N r ( s F G ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq73_HTML.gif and N l ( s F G ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq74_HTML.gif as rational vector spaces are spanned by minimal polynomial bases of minimal degrees
ϵ 1 = ϵ 2 = = ϵ g = 0 < ϵ g + 1 ϵ d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ20_HTML.gif
(20)
and
ζ 1 = ζ 2 = = ζ h = 0 < ζ h + 1 ζ t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ21_HTML.gif
(21)

respectively. The set of minimal indices ϵ 1 , ϵ 2 , , ϵ d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq80_HTML.gif and ζ 1 , ζ 2 , , ζ t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq81_HTML.gif are known as column minimal indices (c.m.i.) and row minimal indices (r.m.i.) of s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif respectively. To sum up, in the case of a singular pencil, we have invariants of the following type:

  • finite elementary divisors of the type ( s a j ) p j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq29_HTML.gif;

  • infinite elementary divisors of the type s ˆ q = 1 s q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq32_HTML.gif;

  • column minimal indices of the type ϵ 1 = ϵ 2 = = ϵ g = 0 < ϵ g + 1 ϵ d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq82_HTML.gif;

  • row minimal indices of the type ζ 1 = ζ 2 = = ζ h = 0 < ζ h + 1 ζ t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq83_HTML.gif.

The Kronecker canonical form, see [39, 41, 44, 45], is defined by
s F K G K : = s I p J p s H q I q s F ϵ G ϵ s F ζ G ζ 0 h , g , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ22_HTML.gif
(22)
where s I p J p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq84_HTML.gif, s H q I q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq41_HTML.gif are defined as in Section 2. The matrices F ϵ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq85_HTML.gif, G ϵ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq86_HTML.gif, F ζ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq87_HTML.gif and G ζ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq88_HTML.gif are defined by
F ϵ = blockdiag { L ϵ g + 1 , L ϵ g + 2 , , L ϵ d } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ23_HTML.gif
(23)
where L ϵ = [ I ϵ 0 ϵ , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq89_HTML.gif for ϵ = ϵ g + 1 , , ϵ d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq90_HTML.gif
G ϵ = blockdiag { L ¯ ϵ g + 1 , L ¯ ϵ g + 2 , , L ¯ ϵ d } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ24_HTML.gif
(24)
where L ¯ ϵ = [ 0 ϵ , 1 I ϵ ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq91_HTML.gif for ϵ = ϵ g + 1 , , ϵ d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq90_HTML.gif. The matrices F ζ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq87_HTML.gif, G ζ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq88_HTML.gif are defined as
F ζ = blockdiag { L ζ h + 1 , L ζ h + 2 , , L ζ t } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ25_HTML.gif
(25)
where L ζ = [ I ζ 0 1 , ζ ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq92_HTML.gif for ζ = ζ h + 1 , , ζ t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq93_HTML.gif
G ζ = blockdiag { L ¯ ζ h + 1 , L ¯ ζ h + 2 , , L ¯ ζ t } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ26_HTML.gif
(26)

where L ¯ ζ = [ 0 1 , ζ I ζ ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq94_HTML.gif for ζ = ζ h + 1 , , ζ t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq93_HTML.gif.

For algorithms about the computations of these matrices, see [39, 41, 44, 45].

Following the above given analysis, there exist nonsingular matrices P, Q with P M r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq95_HTML.gif, Q M m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq96_HTML.gif such that
P F Q = F K , P G Q = G K . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ27_HTML.gif
(27)
Let
Q = [ Q p Q q Q ϵ Q ζ Q g ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ28_HTML.gif
(28)

where Q p M m p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq49_HTML.gif, Q q M m q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq50_HTML.gif, Q ϵ M m ( d g ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq97_HTML.gif, Q ζ M m ( t h ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq98_HTML.gif and Q g M m g https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq99_HTML.gif.

Lemma 3.1 The system (1) is divided into five subsystems:
0 n Z k p = J p Z k p , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ29_HTML.gif
(29)
the subsystem
H q 0 n Z k q = Z k q , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ30_HTML.gif
(30)
the subsystem
F ϵ 0 n Z k ϵ = G ϵ Z k ϵ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ31_HTML.gif
(31)
the subsystem
F ζ 0 n Z k ζ = G ζ Z k ζ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ32_HTML.gif
(32)
and the subsystem
0 h , g 0 n Z k + 1 g = 0 h , g Z k g . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ33_HTML.gif
(33)
Proof Consider the transformation
Y k = Q Z k . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ34_HTML.gif
(34)
Substituting the previous expression into (1), we obtain
F Q 0 n Z k = G Q Z k . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equap_HTML.gif
Whereby multiplying by P and using (27), we arrive at
F K 0 n Z k = G K Z k . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ35_HTML.gif
(35)
Moreover, let
Z k = [ Z k p Z k q Z k ϵ Z k ζ Z k g ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equaq_HTML.gif

where Z p k M p 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq100_HTML.gif, Z q k M q 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq101_HTML.gif, Z ϵ k M ( d g ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq102_HTML.gif, Z ζ k M ( t h ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq103_HTML.gif and Z g k M g 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq104_HTML.gif. Taking into account the above expressions, we arrive easily at the subsystems (29), (30), (31), (32), and (33). The proof is completed. □

Solving the system (1) is equivalent to solving subsystems (29), (30), (31), (32) and (33). The solutions of the systems (29), (30) are given by (10) and (12) respectively; see Propositions 2.1 and 2.2.

Proposition 3.1 The subsystem (31) has infinite solutions and can be taken arbitrarily
Z k ϵ = C k , 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ36_HTML.gif
(36)
Proof If we set
Z k ϵ = [ Z k ϵ g + 1 Z k ϵ g + 2 Z k ϵ d ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equar_HTML.gif
by using (23), (24), the system (31) can be written as
blockdiag { L ϵ g + 1 , , L ϵ d } [ 0 n Z k ϵ g + 1 0 n Z k ϵ g + 2 0 n Z k ϵ d ] = blockdiag { L ¯ ϵ g + 1 , , L ¯ ϵ d } [ Z k ϵ g + 1 Z k ϵ g + 2 Z k ϵ d ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ37_HTML.gif
(37)
Then, for the non-zero blocks, a typical equation from (37) can be written as
L ϵ i 0 n Z k ϵ i = L ¯ ϵ i Z k ϵ i , i = g + 1 , g + 2 , , d https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ38_HTML.gif
(38)
or, equivalently,
[ I ϵ i 0 ϵ i , 1 ] 0 n Z k ϵ i = [ 0 ϵ i , 1 I ϵ i ] Z k ϵ i , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equas_HTML.gif
or, equivalently,
[ 1 0 0 0 0 1 0 0 0 0 1 0 ] [ 0 n z k ϵ i , 1 0 n z k ϵ i , 2 0 n z k ϵ i , ϵ i 0 n z k ϵ i , ϵ i + 1 ] = [ 0 1 0 0 0 0 0 0 0 0 0 1 ] [ z k ϵ i , 1 z k ϵ i , 2 z k ϵ i , ϵ i z k ϵ i , ϵ i + 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equat_HTML.gif
or, equivalently,
0 n z k ϵ i , 1 = z k ϵ i , 2 + v k ϵ i , 1 , 0 n z k ϵ i , 2 = z k ϵ i , 3 + v k ϵ i , 2 , 0 n z k ϵ i , ϵ i = 0 n z k ϵ i , ϵ i + 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ39_HTML.gif
(39)
The system (39) is a regular-type system of difference equations with ϵ i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq105_HTML.gif equations and ϵ i + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq106_HTML.gif unknowns. It is clear from the above analysis that in every one of the d g https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq107_HTML.gif subsystems one of the coordinates of the solution has to be arbitrary by assigned total. The solution of the system can be assigned arbitrarily
Z k ϵ = C k , 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equau_HTML.gif

The proof is completed. □

Proposition 3.2 The solution of the system (32) is unique and is the zero solution
Z k ζ = 0 t h , 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ40_HTML.gif
(40)
Proof From (25), (26) the subsystem (32) can be written as
blockdiag { L ζ h + 1 , , L ζ t } [ 0 n Z k ζ h + 1 0 n Z k ζ h + 2 0 n Z k ζ t ] = blockdiag { L ¯ ζ h + 1 , , L ¯ ζ t } [ Z k ζ h + 1 Z k ζ h + 2 Z k ζ t ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ41_HTML.gif
(41)
Then for the non-zero blocks, a typical equation from (41) can be written as
L ζ j 0 n Z k ζ j = L ¯ ζ j Z k ζ j , j = h + 1 , h + 2 , , t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ42_HTML.gif
(42)
or, equivalently,
[ I ζ j 0 1 , ζ j ] 0 n Z k ζ j = [ 0 1 , ζ j I ζ j ] Z k ζ j , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equav_HTML.gif
or, equivalently,
[ 1 0 0 0 1 0 0 0 1 0 0 0 ] [ 0 n z k ζ j , 1 0 n z k ζ j , 2 0 n z k ζ j , ζ j ] = [ 0 0 0 1 0 0 0 0 0 0 0 1 ] [ z k ζ j , 1 z k ζ j , 2 z k ζ j , ζ j ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equaw_HTML.gif
or, equivalently,
0 n z k ζ j , 1 = 0 , 0 n z k ζ j , 2 = z k ζ j , 1 , 0 n z k ζ j , ζ j 1 = z k ζ j , ζ j 2 , 0 n z k ζ j , ζ j = z k ζ j , ζ j 1 , 0 = z k ζ j , ζ j . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ43_HTML.gif
(43)
We have a system of ζ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq108_HTML.gif+1 difference equations and ζ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq108_HTML.gif unknowns. Starting from the last equation, we get the solutions
z k ζ j , ζ j = 0 , z k ζ j , ζ j 1 = 0 , z k ζ j , ζ j 2 = 0 , z k ζ j , 1 = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equax_HTML.gif

which means that the solution of the system (32) is unique and is the zero solution. The proof is completed. □

Proposition 3.3 The subsystem (33) has an infinite number of solutions that can be taken arbitrarily
Z k g = C k , 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ44_HTML.gif
(44)
Proof It is easy to observe that the subsystem
0 h , g 0 n Z k + 1 g = 0 h , g Z k g https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equay_HTML.gif

does not provide any non-zero equations. Hence all its solutions can be taken arbitrarily. The proof is completed. □

We can now state the following theorem.

Theorem 3.1 Consider the system (1) with a singular pencil and known boundary conditions of type (2). Then the boundary value problem (1)-(2) is consistent if and only if:
  1. 1.
    J p < 1 ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equaz_HTML.gif
     
  2. 2.
    the column minimal indices are zero, i.e.,
    dim N r ( s F G ) = 0 ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ45_HTML.gif
    (45)
     
  3. 3.
    [ B 1 B 2 ] colspan [ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ46_HTML.gif
    (46)
     
Furthermore, when the boundary value problem (1)-(2) is consistent, it has a unique solution if and only if
  1. 1.
    p r 1 + r 2 ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ47_HTML.gif
    (47)
     
  2. 2.
    rank [ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] = p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ48_HTML.gif
    (48)
     
In this case the unique solution is given by the formula
Y k = Q p ( k + 1 ) n 1 ¯ F n , n ( J p ( k + n ) n ¯ ) ( I p J p ) C , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ49_HTML.gif
(49)
where C is the unique solution of the algebraic system
[ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] C = [ B 1 B 2 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ50_HTML.gif
(50)

In any other case the system has infinite solutions.

Proof First we consider that the system has non-zero column minimal indices and non-zero row minimal indices. By using the transformation (34), the solutions of the subsystems (29), (30), (31), (32) and (33) are given by (10), (12), (36), (40) and (44) respectively. Note that from Proposition 2.1 the solution (10) exists if and only if
J p < 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equba_HTML.gif
Furthermore, if
Z k = [ Z k p Z k q Z k ϵ Z k ζ Z k g ] = [ ( k + 1 ) n 1 ¯ F n , n ( J p ( k + n ) n ¯ ) ( I p J p ) Z 0 p 0 q , 1 C k , 1 0 t h , 1 C k , 2 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbb_HTML.gif
Since Z 0 p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq109_HTML.gif is unknown, it can be replaced with the unknown vector C. Then
Y k = Q Z k = [ Q p Q q Q ϵ Q ζ Q g ] [ ( k + 1 ) n 1 ¯ F n , n ( J p ( k + n ) n ¯ ) ( I p J p ) C 0 q , 1 C k , 1 0 t h , 1 C k , 2 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbc_HTML.gif
or, equivalently,
Y k = Q p ( k + 1 ) n 1 ¯ F n , n ( J p ( k + n ) n ¯ ) ( I p J p ) C + Q ϵ C k , 1 + Q g C k , 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbd_HTML.gif
Since C k , 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq110_HTML.gif and C k , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq111_HTML.gif can be taken arbitrarily, it is clear that the general singular discrete time system for every suitable defined boundary condition has an infinite number of solutions. It is clear that the existence of the column minimal indices is the reason that the systems (31) and consequently (33) exist. These systems as shown in Propositions 3.1 and 3.3 have always infinite solutions. Thus a necessary condition for the system to have a unique solution is not to have any column minimal indices which are equal to
dim N r ( s F G ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Eqube_HTML.gif
In this case the Kronecker canonical form of the pencil s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif has the following form:
s F K G K : = s I p J p s H q I q s F ζ G ζ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ51_HTML.gif
(51)
Then the system (1) is divided into three subsystems (29), (30), (32) with solutions (10), (12), (40) respectively. Thus
Y k = Q Z k = [ Q p Q q Q ζ ] [ ( k + 1 ) n 1 ¯ F n , n ( J p ( k + n ) n ¯ ) ( I p J p ) C 0 q , 1 0 t h , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbf_HTML.gif
or, equivalently,
Y k = Q p ( k + 1 ) n 1 ¯ F n , n ( J p ( k + n ) n ¯ ) ( I p J p ) C . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbg_HTML.gif
The solution exists if and only if
A 1 Y 0 = B 1 , A 2 Y N = B 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbh_HTML.gif
or, equivalently,
A 1 Q p C = B 1 , A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) C = B 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbi_HTML.gif
or, equivalently,
[ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] C = [ B 1 B 2 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbj_HTML.gif
For the above algebraic system, there exists at least one solution if and only if
[ B 1 B 2 ] colspan [ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbk_HTML.gif
The algebraic system (50) contains r 1 + r 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq112_HTML.gif equations and p unknowns. Hence the solution is unique if and only if
p r 1 + r 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbl_HTML.gif
and
rank [ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] = p , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbm_HTML.gif
where C is then the unique solution of (50). The uniqueness of C can be proved as follows. If we assume that the algebraic system has two solutions C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq63_HTML.gif and C 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq64_HTML.gif, then
[ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] C 1 = [ B 1 B 2 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbn_HTML.gif
and
[ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] C 2 = [ B 1 B 2 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbo_HTML.gif
or, equivalently,
[ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] ( C 1 C 2 ) = 0 p , 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbp_HTML.gif
But the matrix [ A 1 Q p A 2 Q p ( N + 1 ) n 1 ¯ F n , n ( J p ( N + n ) n ¯ ) ( I p J p ) ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq113_HTML.gif is left invertible since it is assumed to have p linear independent columns and r 1 + r 2 p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq66_HTML.gif and hence
C 1 = C 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbq_HTML.gif

The unique solution is then given from (49). The proof is completed. □

4 Numerical examples

Example 1

Assume the system (1) for k = 0 , 1 , 2 , 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq114_HTML.gif and n = 3 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq115_HTML.gif. Let
F = [ 0 1 0 0 0 0 0 0 0 1 1 1 0 1 1 1 1 0 2 1 0 0 0 0 0 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ52_HTML.gif
(52)
and
G = [ 0 0 1 1 1 0 0 0 0 1 4 1 2 1 2 0 1 2 1 2 1 2 1 2 1 1 2 3 4 0 0 0 1 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ53_HTML.gif
(53)
Then det ( s F G ) = ( s 1 2 ) s ( s 1 4 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq116_HTML.gif and the pencil is regular. We assume the boundary conditions (2) with
A 1 = [ 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 ] , A 2 = [ 1 0 0 0 1 ] , B 1 = [ 0 0 36 36 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbr_HTML.gif
and
B 2 = 24 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbs_HTML.gif
The three finite eigenvalues ( p = 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq117_HTML.gif) of the pencil are 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq118_HTML.gif, 0, 1 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq119_HTML.gif, and the Jordan matrix J p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq58_HTML.gif has the form
J p = [ 1 2 0 0 0 0 0 0 0 1 4 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbt_HTML.gif
It is easy to observe that
J p < 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbu_HTML.gif
By calculating the eigenvectors of the finite eigenvalues, we get the matrix Q p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq120_HTML.gif
Q p = [ 1 1 0 0 1 0 0 0 0 0 0 1 0 0 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbv_HTML.gif
Moreover,
A 1 Q p = [ 1 1 0 0 1 0 0 0 1 0 0 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ54_HTML.gif
(54)
From (9) we get the Mittag-Leffler function
F 3 2 , 3 2 ( J p ( 3 + 3 2 ) 3 2 ¯ ) = i = 0 J p i ( 3 + 3 2 ) i 3 2 ¯ Γ ( ( i + 1 ) 3 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbw_HTML.gif
or, equivalently,
F 3 2 , 3 2 ( J p ( 3 + 3 2 ) 3 2 ¯ ) = i = 0 J p i Γ ( 3 + 3 2 ( i + 1 ) ) Γ ( 3 + 3 2 ) Γ ( ( i + 1 ) 3 2 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbx_HTML.gif
or, equivalently,
F 3 2 , 3 2 ( J p ( 3 + 3 2 ) 3 2 ¯ ) = i = 0 J p i ( 2 + 3 2 ( i + 1 ) ) ( 1 + 3 2 ( i + 1 ) ) ( 3 2 ( i + 1 ) ) Γ ( 3 2 ( i + 1 ) ) Γ ( 3 + 3 2 ) Γ ( ( i + 1 ) 3 2 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equby_HTML.gif
or, equivalently,
F 3 2 , 3 2 ( J p ( 3 + 3 2 ) 3 2 ¯ ) = 1 Γ ( 3 + 3 2 ) i = 0 J p i ( 2 + 3 2 ( i + 1 ) ) ( 1 + 3 2 ( i + 1 ) ) ( 3 2 ( i + 1 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equbz_HTML.gif
and since J p < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq121_HTML.gif, by using the sum i = 0 x = ( 1 x ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq122_HTML.gif for x < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq123_HTML.gif, we calculate the sum of the matrix power series i = 0 J p i ( 2 + 3 2 ( i + 1 ) ) ( 1 + 3 2 ( i + 1 ) ) ( 3 2 ( i + 1 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq124_HTML.gif, and we get
F 3 2 , 3 2 ( J p ( 3 + 3 2 ) 3 2 ¯ ) = 1 Γ ( 3 + 3 2 ) ( 1 9 J p 2 + 20 9 J p + 35 9 I p ) ( I p J p ) 4 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equca_HTML.gif
And since
A 2 Q p = [ 1 1 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equcb_HTML.gif
by using the above expression
A 2 Q p ( 4 ) 1 2 ¯ F 3 2 , 3 2 ( J p ( 3 + 3 2 ) 3 2 ¯ ) ( I p J p ) = 1 36 [ 1 1 1 ] [ 358 0 0 0 35 0 0 0 24 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equcc_HTML.gif
or, equivalently,
A 2 Q p ( 4 ) 1 2 ¯ F 3 2 , 3 2 ( J p ( 3 + 3 2 ) 3 2 ¯ ) ( I p J p ) = 1 36 [ 358 35 24 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ55_HTML.gif
(55)
it is easy to observe that the conditions (13), (14) and (15) are satisfied. Thus from Theorem 2.1 the unique solution of the boundary value problem (1)-(2) is given by
Y k = Q p ( k + 1 ) 3 2 1 ¯ F 3 2 , 3 2 ( J p ( k + 3 2 ) 3 2 ¯ ) ( I p J p ) C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equcd_HTML.gif
or, equivalently, by
Y k = 1 Γ ( k + 1 ) i = 0 Γ ( k + 3 2 ( i + 1 ) ) Γ ( 3 2 ( i + 1 ) ) [ 1 2 i + 1 0 0 0 0 0 0 0 0 0 0 3 4 i + 1 0 0 3 4 i + 1 ] C , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equce_HTML.gif
where C is the unique solution of the algebraic system
[ A 1 Q p A 2 Q p ( 4 ) 1 2 ¯ F 3 2 , 3 2 ( J p ( 3 + 3 2 ) 3 2 ¯ ) ( I p J p ) ] C = [ B 1 B 2 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equcf_HTML.gif
or, equivalently,
C = [ 0 0 36 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equcg_HTML.gif
and thus the unique solution of the boundary value problem is
Y k = 27 Γ ( k + 1 ) i = 0 Γ ( k + 3 2 ( i + 1 ) ) Γ ( 3 2 ( i + 1 ) ) [ 0 0 0 1 4 i 1 4 i ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ56_HTML.gif
(56)

Example 2

We assume the system (1) as in Example 1 but with different boundary conditions. Let
A 1 = [ 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 ] , A 2 = [ 1 0 0 0 1 ] , B 1 = [ 0 0 0 36 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equch_HTML.gif
and
B 2 = 24 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equci_HTML.gif
It is easy to observe that
[ B 1 B 2 ] colspan [ A 1 Q p A 2 Q p ( 4 ) 1 2 ¯ F 3 2 , 3 2 ( J p ( 3 + 3 2 ) 3 2 ¯ ) ( I p J p ) ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equcj_HTML.gif
since
[ 0 0 0 36 24 ] colspan [ 36 36 0 0 36 0 0 0 36 0 0 36 358 35 24 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equck_HTML.gif

and thus from Theorem 2.1, and since (13) does not hold, the boundary value problem is not consistent.

Example 3

Consider the system (1) and let
F = [ 1 1 1 1 1 0 1 1 0 1 1 1 1 1 1 0 1 1 0 1 1 0 1 0 0 0 0 1 1 1 ] , G = [ 1 2 2 1 2 0 2 2 0 2 1 2 2 2 3 0 2 3 1 3 0 0 0 0 0 1 0 1 0 0 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equcl_HTML.gif
Since the matrices F, G are non-square, the matrix pencil s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif is singular and has invariants such as the finite elementary divisors s 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq125_HTML.gif, s 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq126_HTML.gif, an infinite elementary divisor of degree 1 and the row minimal indices ζ 1 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq127_HTML.gif, ζ 2 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq128_HTML.gif. Since the Jordan matrix has the form
J p = [ 2 0 0 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equcm_HTML.gif
with
J p > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equcn_HTML.gif

for every induced matrix norm, from Theorem 3.1 the boundary value problem (1)-(2) is non-consistent.

Example 4

Consider the system (1) for k = 0 , 1 , 2 , 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq114_HTML.gif and n = 3 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq115_HTML.gif. Let
F = [ 1 0 0 1 0 0 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ57_HTML.gif
(57)
and
G = [ 3 4 0 0 0 0 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ58_HTML.gif
(58)
Since the matrices F, G are non-square, the matrix pencil s F G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq17_HTML.gif is singular and has invariants such as a finite elementary divisor s 3 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq129_HTML.gif and the row minimal indices ζ 1 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq127_HTML.gif, ζ 2 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq128_HTML.gif. We assume the boundary conditions (2) with
A 1 = [ 1 1 ] , A 2 = [ 0 1 0 2 ] , B 1 = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equco_HTML.gif
and
B 2 = [ 0 0 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equcp_HTML.gif
The Jordan matrix is J p = 3 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq130_HTML.gif with J p < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq121_HTML.gif for every induced matrix norm. By calculating the matrix Q p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_IEq120_HTML.gif, we get
Q p = [ 1 0 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equcq_HTML.gif
Moreover,
A 1 Q p = 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ59_HTML.gif
(59)
and since
A 2 Q p = [ 0 0 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equcr_HTML.gif
we get
A 2 Q p ( 4 ) 1 2 ¯ F 3 2 , 3 2 ( J p ( 3 + 3 2 ) 3 2 ¯ ) ( I p J p ) = [ 0 0 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ60_HTML.gif
(60)
By using (59), (60), it is easy to observe that the conditions (45), (46), (47) and (48) are satisfied and thus from Theorem 3.1 the unique solution of the boundary value problem (1)-(2) is given by
Y k = Q p ( k + 1 ) 1 2 ¯ F 3 2 , 3 2 ( J p ( k + 3 2 ) 3 2 ¯ ) ( I p J p ) C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equcs_HTML.gif
or, equivalently, by
Y k = 1 Γ ( k + 1 ) i = 0 Γ ( k + 3 2 ( i + 1 ) ) Γ ( 3 2 ( i + 1 ) ) [ 3 i 4 i + 1 0 ] C , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equct_HTML.gif
where C is the unique solution of the algebraic system
[ A 1 Q p A 2 Q p ( 4 ) 1 2 ¯ F 3 2 , 3 2 ( J p ( 3 + 3 2 ) 3 2 ¯ ) ( I p J p ) ] C = [ B 1 B 2 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equcu_HTML.gif
or, equivalently,
C = 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equcv_HTML.gif
and thus the unique solution of the boundary value problem is
Y k = 2 Γ ( k + 1 ) i = 0 Γ ( k + 3 2 ( i + 1 ) ) Γ ( 3 2 ( i + 1 ) ) [ 3 i 4 i + 1 0 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-148/MediaObjects/13661_2013_Article_399_Equ61_HTML.gif
(61)

5 Conclusions

In this article, we study the boundary value problem of a class of a singular system of fractional nabla difference equations whose coefficients are constant matrices. By taking into consideration the cases that the matrices are square with the leading coefficient singular, square with an identically zero matrix pencil and non-square, we study the conditions under which the boundary value problem has unique, infinite and no solutions. Furthermore, we provide a formula for the case of the unique solution. As a further extension of this article, one can study the stability, the behavior under perturbation and possible applications in economics and engineering of singular matrix difference/differential equations of fractional order. For all this, there is already some research in progress.

Declarations

Acknowledgements

We would like to express our sincere gratitude to Professor GI Kalogeropoulos for his helpful and fruitful discussions that clearly improved this article. Moreover, we are very grateful to the anonymous referees for their valuable suggestions that improved the article.

Authors’ Affiliations

(1)
School of Mathematics and Maxwell Institute, The University of Edinburgh
(2)
Department of Mathematics and Computer Sciences, Cankaya University
(3)
Institute of Space Sciences
(4)
Department of Chemical and Materials Engineering, Faculty of Engineering, King Abdulaziz University

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© Dassios and Baleanu; licensee Springer. 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.