The Non-Eigenvalue Form of Liouville’s Formula and α-Matrix Exponential Solutions for Combined Matrix Dynamic Equations on Time Scales

Abstract

In this paper, the non-eigenvalue forms of Liouville’s formulas for delta, nabla and $\alpha$-diamond matrix dynamic equations on time scales are given and proved. Meanwhile, a diamond matrix exponential function (or $\alpha$-matrix exponential function) is introduced and some classes of homogenous linear diamond-$\alpha$ dynamic equations which possess the $\alpha$-matrix exponential solutions is studied. The difference and relation of non-eigenvalue forms of Liouville’s formulas among these representative types of dynamic equations is investigated. Moreover, we establish some sufficient conditions to guarantee transformational relation of Liouville’s formulas and exponential solutions among these types of matrix dynamic equations. In addition, we provide several examples on various time scales to illustrate the effectiveness of our result.

1. Introduction

As an effective and powerful tool, time scale calculus is initiated to unify continuous and discrete analysis and is extensively applied to study dynamic equations [1,2,3,4,5]. Liouville’s formula and Liouville’s problems are very important topic in ordinary differential equations [6,7,8,9,10]. In Theorem 5.28 from [9], Liouville’s formula of dynamic equations on time scales is given by Bohner and Peterson. The theorem is given as follows:

Theorem 1.

Let $A\in \mathcal{R}$ be a $2\times 2$-matrix-valued function and assume that X is a solution of $X^{\Delta }=A\left(t\right)X$. Then X satisfies Liouville’s formula

$$detX\left(t\right)=e_{trA+\mu detA}(t,t_0)detX\left(t_0\right),fort\in \mathbb{T}.$$

(1)

Please note that Liouville’s formula given by the form (1) is very convenient to use due to the main reason that $trA+\mu detA$ is independent of the eigenpolynomial and eigenvalue of A. The Liouville’s formula for $n\times n$-matrix dynamic equations on time scales were studied in [11,12]. In [11,12], the authors provided the nice form of Liouville’s formula by considering the eigenpolynomial and eigenvalue of the coefficient matrix of the dynamic equations. However, if A is a $n\times n$-matrix-valued function for n sufficiently large, the calculation of eigenpolynomial and eigenvalue of A becomes a complicated task and cannot be always achieved, so it will be a better way to provide a matrix form of Liouville’s formula for the case of $A:\mathbb{T}\to {\mathbb{R}}^{n\times n}$ to avoid the calculation of eigenpolynomial and eigenvalue of A, similar to (1). Unfortunately, for $n\ge 3$, the matrix X that is a solution of dynamic equations $X^{\Delta }=A\left(t\right)X$ does not satisfy (1), i.e., the nice form (1) of Liouville’s formula will not hold for $n\ge 3$.

On the other hand, the combined dynamic derivatives on time scales was proposed in [13], which can unify $\Delta$-dynamic derivative ($\alpha =1$) and ∇-dynamic derivative ($\alpha =0$). Moreover, this combined type of derivative includes the hybrid dynamic derivatives between $\Delta$ and ∇ cases and was used to study various complex dynamic equations and inequalities on time scales [14,15,16,17,18].

However, since there is no Liouville’s formula of diamond-$\alpha$ matrix dynamic equations, in this paper, we make the following contributions:

(i) 

the combined matrix exponential function is introduced and studied;

(ii) 

Liouville’s formula of diamond-$\alpha$ matrix dynamic equations is obtained without considering the eigenpolynomial and eigenvalue;

(iii) 

some classes of diamond-$\alpha$ matrix dynamic equations which have $\alpha$-matrix exponential solutions are investigated;

(iv) 

the obtained results are completely new even for $\Delta$ and ∇-matrix dynamic equations and several examples on various time scales are provided.

In particular, for the $2\times 2$ coefficient matrix and $\alpha =1$, Liouville’s formula happens to be (1).

2. Liouville’S Formula for $\mathbf{\Delta }$-Dynamic Equations

In this section, we will derive the non-eigenvalue form of Liouville’s formula for $\Delta$-matrix dynamic equations which will be used to study the combined dynamic equations on time scales.

Definition 1

([9]). Define the forward jump operator $\sigma :\mathbb{T}\to \mathbb{T}$ by $\sigma \left(t\right)=inf\{s\in \mathbb{T}:s>t\}$; the backward jump operator $\rho :\mathbb{T}\to \mathbb{T}$ by $\rho \left(t\right)=sup\{s\in \mathbb{T}:s<t\}$; the graininess function $\mu :\mathbb{T}\to [0,+\infty )$ by $\mu \left(t\right)=\sigma \left(t\right)-t$ ( denoted by μ); the $\nu \left(t\right)$ by $\nu \left(t\right)=t-\rho \left(t\right)$ ( denoted by ν).

Definition 2

([9]). Definite $e_A(t,t_0)$ by the unique matrix solution of the initial value problem:

$$X^{\Delta }\left(t\right)=A\left(t\right)X\left(t\right),X_0=I.$$

(2)

Lemma 1.

Let A be an upper triangular $n\times n$-matrix-valued function, then A is regressive iff each diagonal element of A is regressive.

Proof. 

Let

$$A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \dots & a_{1n}\\ 0& a_{22}& \dots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & a_{nn}\end{array}\right].$$

Hence

$$\begin{array}{ccc}\hfill det[I+\mu A]& =& \left|\begin{array}{cccc}1+\mu a_{11}& \mu a_{12}& \dots & \mu a_{1n}\\ 0& 1+\mu a_{22}& \dots & \mu a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & 1+\mu a_{nn}\end{array}\right|=\prod _{i=1}^n\left(1+\mu a_{ii}\right).\hfill \end{array}$$

Therefore A is regressive iff each diagonal element of A is regressive. □

Remark 1.

Let $A:\mathbb{T}\to {\mathbb{R}}^{n\times n}$ be an upper triangular matrix, we can obtain

$$detA=\prod _{i=1}^n{a}_{ii},trA=\sum _{i=1}^n{a}_{ii},det\left(I+\mu A\right)=\prod _{i=1}^n\left(1+\mu a_{ii}\right).$$

Hence for $n>2$,

$$1+\mu \left(trA+\mu detA\right)=1+\mu (\sum _{i=1}^n{a}_{ii}+\mu \prod _{i=1}^n{a}_{ii})\ne \prod _{i=1}^n\left(1+\mu a_{ii}\right),$$

which implies that $I+\mu A$ is invertible does not be equivalent to $trA+\mu detA$ is regressive, i.e., Liouville’s Formula (1) is not suitable for $n\ge 3$.

It is easy to check the following $\Delta$-derivative formula of determinant function by using the determinant algorithm and Theorem 1.20 from [9].

Lemma 2.

Let $A:\mathbb{T}\to {\mathbb{R}}^{n\times n}$ be the following function matrix:

$$A\left(t\right)=\left|\begin{array}{ccccccc}{a}_{11}& a_{12}& \dots & a_{1i}& a_{1(i+1)}& \dots & a_{1n}\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{(i-1)1}& a_{(i-1)2}& \dots & a_{(i-1)i}& a_{(i-1)(i+1)}& \dots & a_{(i-1)n}\\ a_{i1}& a_{i2}& \dots & a_{ii}& a_{i(i+1)}& \dots & a_{in}\\ a_{(i+1)1}& a_{(i+1)2}& \dots & a_{(i+1)i}& a_{(i+1)(i+1)}& \dots & a_{(i+1)n}\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}& a_{n2}& \dots & a_{ni}& a_{n(i+1)}& \dots & a_{nn}\end{array}\right|,$$

then

$$A^{\Delta }\left(t\right)=\sum _{i=1}^n\left|\begin{array}{ccccccc}{a}_{11}^{\sigma }& a_{12}^{\sigma }& \dots & a_{1i}^{\sigma }& a_{1(i+1)}^{\sigma }& \dots & a_{1n}^{\sigma }\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{(i-1)1}^{\sigma }& a_{(i-1)2}^{\sigma }& \dots & a_{(i-1)i}^{\sigma }& a_{(i-1)(i+1)}^{\sigma }& \dots & a_{(i-1)n}^{\sigma }\\ a_{i1}^{\Delta }& a_{i2}^{\Delta }& \dots & a_{ii}^{\Delta }& a_{i(i+1)}^{\Delta }& \dots & a_{in}^{\Delta }\\ a_{(i+1)1}& a_{(i+1)2}& \dots & a_{(i+1)i}& a_{(i+1)(i+1)}& \dots & a_{(i+1)n}\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}& a_{n2}& \dots & a_{ni}& a_{n(i+1)}& \dots & a_{nn}\end{array}\right|.$$

(3)

Remark 2.

In Lemma 2, notice that $a_{ij}$ denotes the element that is located in the ith row and the jth column, $i.e.$, $1\le i,j\le n$. Hence $A^{\Delta }\left(t\right)$ equals to the following:

$$\begin{array}{cc}\hfill A^{\Delta }\left(t\right)=& \left|\begin{array}{ccccc}{a}_{11}^{\Delta }\left(t\right)& a_{12}^{\Delta }\left(t\right)& a_{13}^{\Delta }\left(t\right)& \dots & a_{1n}^{\Delta }\left(t\right)\\ a_{21}\left(t\right)& a_{22}\left(t\right)& a_{23}\left(t\right)& \dots & a_{2n}\left(t\right)\\ a_{31}\left(t\right)& a_{32}\left(t\right)& a_{33}\left(t\right)& \dots & a_{3n}\left(t\right)\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{i1}\left(t\right)& a_{i2}\left(t\right)& a_{i3}\left(t\right)& \dots & a_{in}\left(t\right)\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}\left(t\right)& a_{n2}\left(t\right)& a_{n3}\left(t\right)& \dots & a_{nn}\left(t\right)\end{array}\right|+\left|\begin{array}{ccccc}{a}_{11}^{\sigma }\left(t\right)& a_{12}^{\sigma }\left(t\right)& a_{13}^{\sigma }\left(t\right)& \dots & a_{1n}^{\sigma }\left(t\right)\\ a_{21}^{\Delta }\left(t\right)& a_{22}^{\Delta }\left(t\right)& a_{23}^{\Delta }\left(t\right)& \dots & a_{2n}^{\Delta }\left(t\right)\\ a_{31}\left(t\right)& a_{32}\left(t\right)& a_{33}\left(t\right)& \dots & a_{3n}\left(t\right)\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{i1}\left(t\right)& a_{i2}\left(t\right)& a_{i3}\left(t\right)& \dots & a_{in}\left(t\right)\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}\left(t\right)& a_{n2}\left(t\right)& a_{n3}\left(t\right)& \dots & a_{nn}\left(t\right)\end{array}\right|+\dots \hfill \\ & +\left|\begin{array}{ccccccc}{a}_{11}^{\sigma }\left(t\right)& a_{12}^{\sigma }\left(t\right)& \dots & a_{1i}^{\sigma }\left(t\right)& a_{1(i+1)}^{\sigma }\left(t\right)& \dots & a_{1n}^{\sigma }\left(t\right)\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{(i-1)1}^{\sigma }\left(t\right)& a_{(i-1)2}^{\sigma }\left(t\right)& \dots & a_{(i-1)i}^{\sigma }\left(t\right)& a_{(i-1)(i+1)}^{\sigma }\left(t\right)& \dots & a_{(i-1)n}^{\sigma }\left(t\right)\\ a_{i1}^{\Delta }\left(t\right)& a_{i2}^{\Delta }\left(t\right)& \dots & a_{ii}^{\Delta }\left(t\right)& a_{i(i+1)}^{\Delta }\left(t\right)& \dots & a_{in}^{\Delta }\left(t\right)\\ a_{(i+1)1}\left(t\right)& a_{(i+1)2}\left(t\right)& \dots & a_{(i+1)i}\left(t\right)& a_{(i+1)(i+1)}\left(t\right)& \dots & a_{(i+1)n}\left(t\right)\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}\left(t\right)& a_{n2}\left(t\right)& \dots & a_{ni}\left(t\right)& a_{n(i+1)}\left(t\right)& \dots & a_{nn}\left(t\right)\end{array}\right|+\dots \hfill \\ & +\left|\begin{array}{ccccccc}{a}_{11}^{\sigma }\left(t\right)& a_{12}^{\sigma }\left(t\right)& \dots & a_{1i}^{\sigma }\left(t\right)& a_{1(i+1)}^{\sigma }\left(t\right)& \dots & a_{1n}^{\sigma }\left(t\right)\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{i1}^{\sigma }\left(t\right)& a_{i2}^{\sigma }\left(t\right)& \dots & a_{ii}^{\sigma }\left(t\right)& a_{i(i+1)}^{\sigma }\left(t\right)& \dots & a_{in}^{\sigma }\left(t\right)\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{(n-1)1}^{\sigma }\left(t\right)& a_{(n-1)2}^{\sigma }\left(t\right)& \dots & a_{(n-1)i}^{\sigma }\left(t\right)& a_{(n-1)(i+1)}^{\sigma }\left(t\right)& \dots & a_{(n-1)n}^{\sigma }\left(t\right)\\ a_{n1}^{\Delta }\left(t\right)& a_{n2}^{\Delta }\left(t\right)& \dots & a_{ni}^{\Delta }\left(t\right)& a_{n(i+1)}^{\Delta }\left(t\right)& \dots & a_{nn}^{\Delta }\left(t\right)\end{array}\right|,\hfill \end{array}$$

where $3\le i\le n-1$. For convenience, we denote the sum by (3). Similarly, the determinant symbol is with the same meaning for all theorems, lemmas and examples.

Theorem 2

(Liouville’s formula). Let $A\in \mathcal{R}$ be an upper triangular $n\times n$-matrix-valued function and assume that X is a solution of $X^{\Delta }=A\left(t\right)X$. Then X satisfies Liouville’s formula

$$detX\left(t\right)=e_{{\displaystyle \sum _{i=1}^n}{\displaystyle \prod _{j=1}^{i-1}}\left(1+\mu a_{jj}\right)a_{ii}}(t,t_0)det{X}_0,fort\in \mathbb{T}.$$

Proof. 

For the matrix A, by Lemma 1, we can obtain A is regressive iff each diagonal element of A is regressive. Let

$$A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \dots & a_{1n}\\ 0& a_{22}& \dots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & a_{nn}\end{array}\right],X=\left[\begin{array}{cccc}{x}_{11}& x_{12}& \dots & x_{1n}\\ 0& x_{22}& \dots & x_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & x_{nn}\end{array}\right]$$

$$\begin{array}{ccc}\hfill {\left(detX\left(t\right)\right)}^{\Delta }& =& {\left|\begin{array}{ccccc}{x}_{11}& x_{12}& x_{13}& \dots & x_{1n}\\ 0& x_{22}& x_{23}& \dots & x_{2n}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \dots & x_{nn}\end{array}\right|}^{\Delta }\hfill \\ & =& \left|\begin{array}{ccccc}{x}_{11}^{\Delta }& x_{12}^{\Delta }& x_{13}^{\Delta }& \dots & x_{1n}^{\Delta }\\ 0& x_{22}& x_{23}& \dots & x_{2n}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \dots & x_{nn}\end{array}\right|+\sum _{i=2}^{n-1}\left|\begin{array}{ccccccc}{x}_{11}^{\sigma }& x_{12}^{\sigma }& \dots & x_{1i}^{\sigma }& x_{1(i+1)}^{\sigma }& \dots & x_{1n}^{\sigma }\\ 0& x_{22}^{\sigma }& \dots & x_{2i}^{\sigma }& x_{2(i+1)}^{\sigma }& \dots & x_{2n}^{\sigma }\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ \dots & \dots & \dots & x_{ii}^{\Delta }& x_{i(i+1)}^{\Delta }& \dots & x_{in}^{\Delta }\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 0& 0& 0& 0& 0& \dots & x_{nn}\end{array}\right|\hfill \\ & =& \sum _{i=1}^n\left|\begin{array}{ccccccc}{x}_{11}^{\sigma }& x_{12}^{\sigma }& \dots & x_{1i}^{\sigma }& x_{1(i+1)}^{\sigma }& \dots & x_{1n}^{\sigma }\\ 0& x_{22}^{\sigma }& \dots & x_{2i}^{\sigma }& x_{2(i+1)}^{\sigma }& \dots & x_{2n}^{\sigma }\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ \dots & \dots & \dots & x_{ii}^{\Delta }& x_{i(i+1)}^{\Delta }& \dots & x_{in}^{\Delta }\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 0& 0& 0& 0& 0& \dots & x_{nn}\end{array}\right|\hfill \\ & =& \sum _{i=1}^n\prod _{j=1}^{i-1}{x}_{jj}^{\sigma }{x}_{ii}^{\Delta }\prod _{k=i+1}^n{x}_{kk}=\left[\sum _{i=1}^n\prod _{j=1}^{i-1}\left(1+\mu a_{jj}\right)a_{ii}\right]detX.\hfill \end{array}$$

This completes the proof. □

Remark 3.

Let A be a $2\times 2$-matrix-valued function. Then A is regressive iff the scalar-valued function $trA+\mu detA$ is regressive (where $trA$ denotes the trace of the matrix A, $i.e.$, the sum of the diagonal elements of A). In fact, let

$$A=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right],$$

then we can obtain

$$\begin{array}{ccc}\hfill det(I+\mu A)& =& det\left[\begin{array}{cc}1+\mu a_{11}& \mu a_{12}\\ \mu a_{21}& 1+\mu a_{22}\end{array}\right]\hfill \\ & =& (1+\mu a_{11})(1+\mu a_{22})-\mu a_{12}\mu a_{21}\hfill \\ & =& 1+\mu (a_{11}+a_{22})+{\mu }^2(a_{11}{a}_{22}-a_{12}{a}_{21})\hfill \\ & =& 1+\mu (trA+\mu detA).\hfill \end{array}$$

Hence A is regressive iff the scalar-valued function $trA+\mu detA$ is regressive.

Remark 4.

Assume that for $n=k$, we can obtain similar characterizations of Remark 3. For $n=k+1$, let

$$\begin{array}{ccc}\hfill A_{k+1}& =& \left[\begin{array}{cc}{A}_k& 0\\ 0& a_0\end{array}\right],\hfill \end{array}$$

where $A_{k+1}$ is $(k+1)\times (k+1)$-matrix-valued function, $A_k$ is $k\times k$-matrix-valued function. Then

$$\begin{array}{ccc}\hfill det(I+\mu A_{k+1})& =& det\left[\begin{array}{cc}{I}_k+\mu A_k& 0\\ 0& 1+\mu a_0\end{array}\right],\hfill \\ & =& (1+\mu a_0)det(I+\mu A_k),\hfill \\ \hfill 1+\mu (tr{A}_{k+1}+\mu det{A}_{k+1})& =& 1+\mu [tr{A}_k+a_0+\mu a_{0}det{A}_k]\hfill \\ & =& 1+\mu tr{A}_k+\mu a_0+\mu a_0\mu det{A}_k.\hfill \end{array}$$

Hence there exists $A_{k+1}$ such that $A_{k+1}$ is regressive and so is $tr{A}_{k+1}+\mu det{A}_{k+1}$.

Theorem 3

(Liouville’s formula). Let $A\in \mathcal{R}$ be $n\times n$-matrix-valued matrix function and assume that X is a solution of $X^{\Delta }=A\left(t\right)X$. Then X satisfies Liouville’s formula

$$detX\left(t\right)=e_{{\displaystyle \sum _{i=1}^n}det{A}_i}(t,t_0)det{X}_0,fort\in \mathbb{T},$$

where

$$A_i=\left[\begin{array}{cccccccc}1+\mu a_{11}& \mu a_{12}& \dots & \mu a_{1(i-1)}& \mu a_{1i}& 0& \dots & 0\\ \mu a_{21}& 1+\mu a_{22}& \dots & \mu a_{2(i-1)}& \mu a_{2i}& 0& \dots & 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ \mu a_{(i-1)1}& \mu a_{(i-1)2}& \dots & 1+\mu a_{(i-1)(i-1)}& \mu a_{(i-1)i}& 0& 0& 0\\ a_{i1}& a_{i2}& \dots & a_{i(i-1)}& a_{ii}& 0& 0& 0\\ 0& 0& \dots & 0& 0& 1& 0& 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 0& 0& \dots & 0& 0& 0& \dots & 1\end{array}\right].$$

Proof. 

For $n=2$, by Remark 3, A is regressive implies $trA+\mu detA$ is regressive. Let

$$A\left(t\right)=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right],X\left(t\right)=\left[\begin{array}{cc}{x}_{11}& x_{12}\\ x_{21}& x_{22}\end{array}\right].$$

Then

$$\begin{array}{ccc}\hfill {\left(detX\left(t\right)\right)}^{\Delta }& =& \left|\begin{array}{cc}{x}_{11}^{\Delta }& x_{12}^{\Delta }\\ x_{21}& x_{22}\end{array}\right|+\left|\begin{array}{cc}{x}_{11}^{\sigma }& x_{12}^{\sigma }\\ x_{21}^{\Delta }& x_{22}^{\Delta }\end{array}\right|\hfill \\ & =& \left|\begin{array}{cc}{a}_{11}{x}_{11}+a_{12}{x}_{21}& a_{11}{x}_{12}+a_{12}{x}_{22}\\ x_{21}& x_{22}\end{array}\right|+\left|\begin{array}{cc}{x}_{11}+\mu x_{11}^{\Delta }& x_{12}+\mu x_{12}^{\Delta }\\ x_{21}^{\Delta }& x_{22}^{\Delta }\end{array}\right|\hfill \\ & =& a_{11}detX\left(t\right)+\left|\begin{array}{cc}{x}_{11}+(a_{11}{x}_{11}+a_{12}{x}_{21})\mu & x_{12}+(a_{11}{x}_{12}+a_{12}{x}_{22})\mu \\ a_{21}{x}_{11}+a_{22}{x}_{21}& a_{21}{x}_{12}+a_{22}{x}_{22}\end{array}\right|\hfill \\ & =& a_{11}detX\left(t\right)+\left|\begin{array}{cc}1+\mu a_{11}& a_{12}\mu \\ a_{21}& a_{22}\end{array}\right|detX\hfill \\ & =& \left(trA+\mu detA\right)detX\hfill \\ & =& \left\{\left|\begin{array}{cc}{a}_{11}& 0\\ 0& 1\end{array}\right|+\left|\begin{array}{cc}1+\mu a_{11}& a_{12}\mu \\ a_{21}& a_{22}\end{array}\right|\right\}detX.\hfill \end{array}$$

For $n\ge 3$, let

$$A\left(t\right)=\left[\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \dots & a_{1n}\\ a_{21}& a_{22}& a_{23}& \dots & a_{2n}\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}& a_{n2}& a_{n3}& \dots & a_{nn}\end{array}\right],X\left(t\right)=\left[\begin{array}{ccccc}{x}_{11}& x_{12}& x_{13}& \dots & x_{1n}\\ x_{21}& x_{22}& x_{23}& \dots & x_{2n}\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ x_{n1}& x_{n2}& x_{n3}& \dots & x_{nn}\end{array}\right].$$

Hence by Lemma 2, we have

$$\begin{array}{cc}& {\left(X\left(t\right)\right)}^{\Delta }={\left|\begin{array}{ccccc}{x}_{11}& x_{12}& x_{13}& \dots & x_{1n}\\ x_{21}& x_{22}& x_{23}& \dots & x_{2n}\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ x_{n1}& x_{n2}& x_{n3}& \dots & x_{nn}\end{array}\right|}^{\Delta }=\sum _{i=1}^n\left|\begin{array}{ccccc}{x}_{11}^{\sigma }& x_{12}^{\sigma }& x_{13}^{\sigma }& \dots & x_{1n}^{\sigma }\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ x_{(i-1)1}^{\sigma }& x_{(i-1)2}^{\sigma }& x_{(i-1)3}^{\sigma }& \dots & x_{(i-1)n}^{\sigma }\\ x_{i1}^{\Delta }& x_{i2}^{\Delta }& x_{i3}^{\Delta }& \dots & x_{in}^{\Delta }\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ x_{n1}& x_{n2}& x_{n3}& \dots & x_{nn}\end{array}\right|\hfill \\ =& \sum _{i=1}^n\left|\begin{array}{ccccc}{x}_{11}+\mu x_{11}^{\Delta }& x_{12}+\mu x_{12}^{\Delta }& x_{13}+\mu x_{13}^{\Delta }& \dots & x_{1n}+\mu x_{1n}^{\Delta }\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ x_{(i-1)1}+\mu x_{(i-1)1}^{\Delta }& x_{(i-1)2}+\mu x_{(i-1)2}^{\Delta }& x_{(i-1)3}+\mu x_{(i-1)3}^{\Delta }& \dots & x_{(i-1)n}+\mu x_{(i-1)n}^{\Delta }\\ x_{i1}^{\Delta }& x_{i2}^{\Delta }& x_{i3}^{\Delta }& \dots & x_{in}^{\Delta }\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ x_{n1}& x_{n2}& x_{n3}& \dots & x_{nn}\end{array}\right|\hfill \\ =& {\displaystyle \sum _{i=1}^n}\left|\begin{array}{cccc}{x}_{11}+\mu {\displaystyle \sum _{j=1}^n}{a}_{1j}{x}_{j1}& x_{12}+\mu {\displaystyle \sum _{j=1}^n}{a}_{1j}{x}_{j2}& \dots & x_{1n}+\mu {\displaystyle \sum _{j=1}^n}{a}_{1j}{x}_{jn}\\ ⋮& ⋮& ⋮& ⋮\\ x_{(i-1)1}+\mu {\displaystyle \sum _{j=1}^n}{a}_{(i-1)j}{x}_{j1}& x_{(i-1)2}+\mu {\displaystyle \sum _{j=1}^n}{a}_{(i-1)j}{x}_{j2}& \dots & x_{(i-1)n}+\mu {\displaystyle \sum _{j=1}^n}{a}_{(i-1)j}{x}_{jn}\\ {\displaystyle \sum _{j=1}^n}{a}_{ij}{x}_{j1}& {\displaystyle \sum _{j=1}^n}{a}_{ij}{x}_{j2}& \dots & {\displaystyle \sum _{j=1}^n}{a}_{ij}{x}_{jn}\\ ⋮& ⋮& ⋮& ⋮\\ x_{n1}& x_{n2}& \dots & x_{nn}\end{array}\right|\hfill \\ =& {\displaystyle \sum _{i=1}^n}\left|\begin{array}{cccc}{x}_{11}+\mu {\displaystyle \sum _{j=1}^i}{a}_{1j}{x}_{j1}& x_{12}+\mu {\displaystyle \sum _{j=1}^i}{a}_{1j}{x}_{j2}& \dots & x_{1n}+\mu {\displaystyle \sum _{j=1}^i}{a}_{1j}{x}_{jn}\\ ⋮& ⋮& ⋮& ⋮\\ x_{(i-1)1}+\mu {\displaystyle \sum _{j=1}^i}{a}_{(i-1)j}{x}_{j1}& x_{(i-1)2}+\mu {\displaystyle \sum _{j=1}^i}{a}_{(i-1)j}{x}_{j2}& \dots & x_{(i-1)n}+\mu {\displaystyle \sum _{j=1}^i}{a}_{(i-1)j}{x}_{jn}\\ {\displaystyle \sum _{j=1}^i}{a}_{ij}{x}_{j1}& {\displaystyle \sum _{j=1}^i}{a}_{ij}{x}_{j2}& \dots & {\displaystyle \sum _{j=1}^i}{a}_{ij}{x}_{jn}\\ ⋮& ⋮& ⋮& ⋮\\ x_{n1}& x_{n2}& \dots & x_{nn}\end{array}\right|\hfill \\ =& \sum _{i=1}^{n}det\left(A_{i}X\right)=\sum _{i=1}^{n}det{A}_{i}detX.\hfill \end{array}$$

Then

$$X\left(t\right)=\left(e_{{\displaystyle \sum _{i=1}^n}det{A}_i}(t,t_0)\right)det{X}_0.$$

This completes the proof. □

Theorem 4.

Let A be a $n\times n$-matrix-valued function. Then A is regressive iff the scalar-valued function ${\displaystyle \sum _{i=1}^n}det{A}_i$ is regressive, where $A_i$ is defined by Theorem 3.

Proof. 

By Theorem 3, for $n=2$ we can obtain

$$det(I+\mu A)=1+\mu \left\{\left|\begin{array}{cc}{a}_{11}& 0\\ 0& 1\end{array}\right|+\left|\begin{array}{cc}1+\mu a_{11}& a_{12}\mu \\ a_{21}& a_{22}\end{array}\right|\right\}.$$

Next, let

$$\begin{array}{ccc}\hfill A\left(t\right)& =& \left[\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \dots & a_{1n}\\ a_{21}& a_{22}& a_{23}& \dots & a_{2n}\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}& a_{n2}& a_{n3}& \dots & a_{nn}\end{array}\right]=\left[\begin{array}{ccccc}& & & & a_{1n}\\ & & A_{\left(0\right)}& & a_{2n}\\ & & & & ⋮\\ a_{n1}& a_{n2}& a_{n3}& \dots & a_{nn}\end{array}\right],\hfill \end{array}$$

where $A_{\left(0\right)}$ is a $(n-1)\times (n-1)$-valued-matrix function. Assume that $(n-1)\times (n-1)$-matrix-value function is regressive iff the scalar-valued function ${\displaystyle \sum _{i=1}^{n-1}}det{A}_i$ is regressive. That is

$$det(I+\mu A_{\left(0\right)})=1+\mu \sum _{i=1}^{n-1}det{A}_i.$$

Hence

$$\begin{array}{cc}& 1+\mu \left(\sum _{i=1}^{n-1}det{A}_i+det{A}_n\right)=1+\mu \sum _{i=1}^{n-1}det{A}_i+\mu det{A}_n=det\left(I+\mu A_{\left(0\right)}\right)+\mu det{A}_n\hfill \\ =& det\left(I+\mu A_{\left(0\right)}\right)+\mu \left|\begin{array}{ccccc}& & & & \mu a_{1n}\\ & & I+\mu A_{\left(0\right)}& & \mu a_{2n}\\ & & & & \dots \\ a_{n1}& a_{n2}& a_{n3}& \dots & a_{nn}\end{array}\right|\hfill \\ =& \left|\begin{array}{ccccc}& & & & \mu a_{1n}\\ & & I+\mu A_{\left(0\right)}& & \mu a_{2n}\\ & & & & \dots \\ 0& 0& 0& \dots & 1\end{array}\right|+\left|\begin{array}{ccccc}& & & & \mu a_{1n}\\ & & I+\mu A_{\left(0\right)}& & \mu a_{2n}\\ & & & & \dots \\ \mu a_{n1}& \mu a_{n2}& \mu a_{n3}& \dots & \mu a_{nn}\end{array}\right|\hfill \\ =& \left|\begin{array}{ccccc}& & & & \mu a_{1n}\\ & & I+\mu A_{\left(0\right)}& & \mu a_{2n}\\ & & & & \dots \\ \mu a_{n1}& \mu a_{n2}& \mu a_{n3}& \dots & 1+\mu a_{nn}\end{array}\right|=det(I+\mu A).\hfill \end{array}$$

Therefore A is regressive iff the scalar-valued function ${\displaystyle \sum _{i=1}^n}det{A}_i$ is regressive. This completes the proof. □

Example 1.

For (2), let $\mathbb{T}=\mathbb{R}$, Liouville’s formula can be given as:

$$detX=det{X}_0{e}^{{\int }_{t_0}^{t}trA\left(\tau \right)\mathrm{d}ø},trA\left(\tau \right)=\sum _{i=1}^n{a}_{ii}\left(\tau \right).$$

In fact, let $\mathbb{T}=\mathbb{R}$, then $\mu \left(t\right)=0$, $X^{\sigma }\left(t\right)=X\left(t\right)$ for any $t\in \mathbb{T}$, hence $X^{\Delta }\left(t\right)=X^{\prime }$, therefore

$$\begin{array}{ccc}\hfill {\left(detX\right)}^{\prime }& =& {\left|\begin{array}{ccccc}{x}_{11}& x_{12}& x_{13}& \dots & x_{1n}\\ x_{21}& x_{22}& x_{23}& \dots & x_{2n}\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ x_{n1}& x_{n2}& x_{n3}& \dots & x_{nn}\end{array}\right|}^{\prime }=\sum _{i=1}^n\left|\begin{array}{ccccc}{x}_{11}& x_{12}& x_{13}& \dots & x_{1n}\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ x_{(i-1)1}& x_{(i-1)2}& x_{(i-1)3}& \dots & x_{(i-1)n}\\ x_{i1}^{\prime }& x_{i2}^{\prime }& x_{i3}^{\prime }& \dots & x_{in}^{\prime }\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ x_{n1}& x_{n2}& x_{n3}& \dots & x_{nn}\end{array}\right|\hfill \\ & =& {\displaystyle \sum _{i=1}^n}\left|\begin{array}{cccc}{x}_{11}& x_{12}& \dots & x_{1n}\\ ⋮& ⋮& ⋮& ⋮\\ x_{(i-1)1}& x_{(i-1)2}& \dots & x_{(i-1)n}\\ {\displaystyle \sum _{j=1}^n}{a}_{ij}{x}_{j1}& {\displaystyle \sum _{j=1}^n}{a}_{ij}{x}_{j2}& \dots & {\displaystyle \sum _{j=1}^n}{a}_{ij}{x}_{jn}\\ ⋮& ⋮& ⋮& ⋮\\ x_{n1}& x_{n2}& \dots & x_{nn}\end{array}\right|\hfill \\ & =& {\displaystyle \sum _{i=1}^n}\left|\begin{array}{cccc}{x}_{11}& x_{12}& \dots & x_{1n}\\ ⋮& ⋮& ⋮& ⋮\\ x_{(i-1)1}& x_{(i-1)2}& \dots & x_{(i-1)n}\\ a_{ii}{x}_{i1}& a_{ii}{x}_{i2}& \dots & a_{ii}{x}_{in}\\ ⋮& ⋮& ⋮& ⋮\\ x_{n1}& x_{n2}& \dots & x_{nn}\end{array}\right|={\displaystyle \sum _{i=1}^n}{a}_{ii}\left(detX\right).\hfill \end{array}$$

Hence

$$detX=det{X}_0{e}^{{\int }_{t_0}^{t}trA\left(\tau \right)\mathrm{d}ø}.$$

Example 2.

For (2), let $\mathbb{T}=\mathbb{Z}h$, $h>0$, the solution of (2) can be given as:

$$X\left(t\right)=\prod _{k=\frac{t-t_0}{h}-1}^0\left[I+hA(t_0+kh)\right]X_0.$$

Furthermore, by Theorem 3,

$$\begin{array}{ccc}\hfill detX\left(t\right)& =& \prod _{k=\frac{t-t_0}{h}-1}^{0}det\left[I+hA(t_0+kh)\right]det{X}_0=e_{{\displaystyle \sum _{i=1}^n}det{A}_i}(t,t_0)det{X}_0,fort\in \mathbb{T},\hfill \end{array}$$

where

$$A_i=\left[\begin{array}{cccccccc}1+h{a}_{11}& h{a}_{12}& \dots & h{a}_{1(i-1)}& h{a}_{1i}& 0& \dots & 0\\ h{a}_{21}& 1+h{a}_{22}& \dots & h{a}_{2(i-1)}& h{a}_{2i}& 0& \dots & 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ h{a}_{(i-1)1}& h{a}_{(i-1)2}& \dots & 1+h{a}_{(i-1)(i-1)}& h{a}_{(i-1)i}& 0& 0& 0\\ a_{i1}& a_{i2}& \dots & a_{i(i-1)}& a_{ii}& 0& 0& 0\\ 0& 0& \dots & 0& 0& 1& 0& 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 0& 0& \dots & 0& 0& 0& \dots & 1\end{array}\right].$$

In fact, let $\mathbb{T}=\mathbb{Z}h$, $h>0$, by $X^{\sigma }\left(t\right)-X\left(t\right)=\mu \left(t\right)A\left(t\right)X\left(t\right)$ and $\mu \left(t\right)=h$, we have $X(t+h)-X\left(t\right)=hA\left(t\right)X\left(t\right)$. Hence $X(x+h)=\left(I+hA\left(t\right)\right)X\left(t\right)$. Therefore

$$X\left(t\right)=\prod _{k=\frac{t-t_0}{h}-1}^0\left[I+hA(t_0+kh)\right]X_0.$$

Example 3.

For (2), let $\mathbb{T}=\overline{q^{\mathbb{Z}}}$, $q>1$, the solution of (2) can be given as:

$$X\left(t\right)=\prod _{k=\frac{lnt-ln{t}_0}{lnq}-1}^0\left[I+(q-1)q^k{t}_{0}A\left(q^k{t}_0\right)\right]X_0.$$

Furthermore, by Theorem 3,

$$\begin{array}{ccc}\hfill detX\left(t\right)& =& \prod _{k=\frac{lnt-ln{t}_0}{lnq}-1}^{0}det\left[I+(q-1)q^k{t}_{0}A\left(q^k{t}_0\right)\right]det{X}_0\hfill \\ & =& e_{{\displaystyle \sum _{i=1}^n}det{A}_i}(t,t_0)det{X}_0,fort\in \mathbb{T},\hfill \end{array}$$

where

$$A_i=\left[\begin{array}{cccccccc}1+(q-1)t{a}_{11}& (q-1)t{a}_{12}& \dots & (q-1)t{a}_{1(i-1)}& (q-1)t{a}_{1i}& 0& \dots & 0\\ (q-1)t{a}_{21}& 1+(q-1)t{a}_{22}& \dots & (q-1)t{a}_{2(i-1)}& (q-1)t{a}_{2i}& 0& \dots & 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ (q-1)t{a}_{(i-1)1}& (q-1)t{a}_{(i-1)2}& \dots & 1+(q-1)t{a}_{(i-1)(i-1)}& (q-1)t{a}_{(i-1)i}& 0& 0& 0\\ a_{i1}& a_{i2}& \dots & a_{i(i-1)}& a_{ii}& 0& 0& 0\\ 0& 0& \dots & 0& 0& 1& 0& 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 0& 0& \dots & 0& 0& 0& \dots & 1\end{array}\right].$$

In fact, let $\mathbb{T}=\overline{q^{\mathbb{Z}}}$, $q>1$, by $X^{\sigma }\left(t\right)-X\left(t\right)=\mu \left(t\right)A\left(t\right)X\left(t\right)$ and $\mu \left(t\right)=(q-1)t$, we have $X\left(qt\right)-X\left(t\right)=(q-1)tA\left(t\right)X\left(t\right)$. Hence $X\left(qx\right)=\left(I+(q-1)tA\left(t\right)\right)X\left(t\right)$. Therefore

$$X\left(t\right)=\prod _{k=\frac{lnt-ln{t}_0}{lnq}-1}^0\left[I+(q-1)q^k{t}_{0}A\left(q^k{t}_0\right)\right]X_0.$$

3. Liouville’S Formula for ∇-Dynamic Equations and Some Lemmas

In the following, we will obtain Liouville’s formula for ∇-dynamic equations and some lemmas which will be used to discuss $\alpha$-diamond dynamic equations are established.

Lemma 3.

By Definition 2, the matrix function $e_A(t,t_0)$ is nabla differentiable at t with

$$e_A^{\nabla }(t,t_0)=A\left(\rho \left(t\right)\right){\left(I+\nu \left(t\right)A\left(\rho \left(t\right)\right)\right)}^{-1}{e}_A(t,t_0)$$

Proof. 

For $e_A^{\Delta }(t,t_0)=\frac{e_A(\sigma \left(t\right),t_0)-e_A(t,t_0)}{\mu \left(t\right)}=A\left(t\right)e_A(t,t_0)$, we can obtain

$$e_A^{\nabla }(t,t_0)=\frac{e_A(t,t_0)-e_A(\rho \left(t\right),t_0)}{\nu \left(t\right)}=A\left(\rho \left(t\right)\right)e_A(\rho \left(t\right),t_0).$$

On the other hand, by $f\left(t\right)-f\left(\rho \left(t\right)\right)=f^{\nabla }\left(t\right)\nu \left(t\right)$, we can obtain

$$e_A(t,t_0)-e_A(\rho \left(t\right),t_0)=\nu \left(t\right)A\left(\rho \left(t\right)\right)e_A(\rho \left(t\right),t_0),$$

that is $e_A(t,t_0)=\left(I+\nu \left(t\right)A\left(\rho \left(t\right)\right)\right)e_A(\rho \left(t\right),t_0)$. Hence

$$e_A^{\nabla }(t,t_0)=A\left(\rho \left(t\right)\right){\left(I+\nu \left(t\right)A\left(\rho \left(t\right)\right)\right)}^{-1}{e}_A(t,t_0).$$

The proof is complete. □

Next, we consider the dynamic equations by nabla derivative.

Lemma 4.

Let A be an upper triangular $n\times n$-matrix-valued function, then A is regressive iff each diagonal element of A is regressive.

Proof. 

Let

$$A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \dots & a_{1n}\\ 0& a_{22}& \dots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & a_{nn}\end{array}\right].$$

Hence

$$\begin{array}{ccc}\hfill det[I-\nu A]& =& \left|\begin{array}{cccc}1-\nu a_{11}& -\nu a_{12}& \dots & -\nu a_{1n}\\ 0& 1-\nu a_{22}& \dots & -\nu a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & 1-\nu a_{nn}\end{array}\right|=\prod _{i=1}^n\left(1-\nu a_{ii}\right).\hfill \end{array}$$

Therefore A is regressive iff each diagonal element of A is regressive. □

Remark 5.

Let $A:\mathbb{T}\to {\mathbb{R}}^{n\times n}$ be an upper triangular matrix, we can obtain

$$detA=\prod _{i=1}^n{a}_{ii},trA=\sum _{i=1}^n{a}_{ii},det\left(I-\nu A\right)=\prod _{i=1}^n\left(1-\nu a_{ii}\right).$$

Hence for $n>2$,

$$1-\nu \left(trA-\nu detA\right)=1-\nu (\sum _{i=1}^n{a}_{ii}-\nu \prod _{i=1}^n{a}_{ii})\ne \prod _{i=1}^n\left(1-\nu a_{ii}\right),$$

which implies that $I+\nu A$ is invertible does not be equivalent to $trA+\nu detA$ is regressive.

It is easy to check the following ∇-derivative formula of determinant function.

Lemma 5.

Let $A:\mathbb{T}\to {\mathbb{R}}^{n\times n}$ be the following function matrix:

$$A\left(t\right)=\left|\begin{array}{ccccccc}{a}_{11}& a_{12}& \dots & a_{1i}& a_{1(i+1)}& \dots & a_{1n}\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{(i-1)1}& a_{(i-1)2}& \dots & a_{(i-1)i}& a_{(i-1)(i+1)}& \dots & a_{(i-1)n}\\ a_{i1}& a_{i2}& \dots & a_{ii}& a_{i(i+1)}& \dots & a_{in}\\ a_{(i+1)1}& a_{(i+1)2}& \dots & a_{(i+1)i}& a_{(i+1)(i+1)}& \dots & a_{(i+1)n}\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}& a_{n2}& \dots & a_{ni}& a_{n(i+1)}& \dots & a_{nn}\end{array}\right|,$$

then

$$A^{\nabla }\left(t\right)=\sum _{i=1}^n\left|\begin{array}{ccccccc}{a}_{11}^{\rho }& a_{12}^{\rho }& \dots & a_{1i}^{\rho }& a_{1(i+1)}^{\rho }& \dots & a_{1n}^{\rho }\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{(i-1)1}^{\rho }& a_{(i-1)2}^{\rho }& \dots & a_{(i-1)i}^{\rho }& a_{(i-1)(i+1)}^{\rho }& \dots & a_{(i-1)n}^{\rho }\\ a_{i1}^{\nabla }& a_{i2}^{\nabla }& \dots & a_{ii}^{\nabla }& a_{i(i+1)}^{\nabla }& \dots & a_{in}^{\nabla }\\ a_{(i+1)1}& a_{(i+1)2}& \dots & a_{(i+1)i}& a_{(i+1)(i+1)}& \dots & a_{(i+1)n}\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}& a_{n2}& \dots & a_{ni}& a_{n(i+1)}& \dots & a_{nn}\end{array}\right|.$$

Theorem 5

(Liouville’s formula). Let $A\in \mathcal{R}$ be an upper triangular $n\times n$-matrix-valued function and assume that X is a solution of $X^{\nabla }=A\left(t\right)X$. Then X satisfies Liouville’s formula

$$detX\left(t\right)=e_{{\displaystyle \sum _{i=1}^n}{\displaystyle \prod _{j=1}^{i-1}}\left(1-\nu a_{jj}\right)a_{ii}}(t,t_0)det{X}_0,fort\in \mathbb{T}.$$

Proof. 

For the matrix A, by Lemma 4, we can obtain A is regressive iff each diagonal element of A is regressive. Let

$$A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \dots & a_{1n}\\ 0& a_{22}& \dots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & a_{nn}\end{array}\right]X=\left[\begin{array}{cccc}{x}_{11}& x_{12}& \dots & x_{1n}\\ 0& x_{22}& \dots & x_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & x_{nn}\end{array}\right]$$

$$\begin{array}{ccc}\hfill {\left(detX\left(t\right)\right)}^{\nabla }& =& {\left|\begin{array}{ccccc}{x}_{11}& x_{12}& x_{13}& \dots & x_{1n}\\ 0& x_{22}& x_{23}& \dots & x_{2n}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \dots & x_{nn}\end{array}\right|}^{\nabla }\hfill \\ & =& \left|\begin{array}{ccccc}{x}_{11}^{\nabla }& x_{12}^{\nabla }& x_{13}^{\nabla }& \dots & x_{1n}^{\nabla }\\ 0& x_{22}& x_{23}& \dots & x_{2n}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \dots & x_{nn}\end{array}\right|+\sum _{i=2}^{n-1}\left|\begin{array}{ccccccc}{x}_{11}^{\rho }& x_{12}^{\rho }& \dots & x_{1i}^{\rho }& x_{1(i+1)}^{\rho }& \dots & x_{1n}^{\rho }\\ 0& x_{22}^{\rho }& \dots & x_{2i}^{\rho }& x_{2(i+1)}^{\rho }& \dots & x_{2n}^{\rho }\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ \dots & \dots & \dots & x_{ii}^{\nabla }& x_{i(i+1)}^{\nabla }& \dots & x_{in}^{\nabla }\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 0& 0& 0& 0& 0& \dots & x_{nn}\end{array}\right|\hfill \\ & =& \sum _{i=1}^n\left|\begin{array}{ccccccc}{x}_{11}^{\rho }& x_{12}^{\rho }& \dots & x_{1i}^{\rho }& x_{1(i+1)}^{\rho }& \dots & x_{1n}^{\rho }\\ 0& x_{22}^{\rho }& \dots & x_{2i}^{\rho }& x_{2(i+1)}^{\rho }& \dots & x_{2n}^{\rho }\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ \dots & \dots & \dots & x_{ii}^{\nabla }& x_{i(i+1)}^{\nabla }& \dots & x_{in}^{\nabla }\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 0& 0& 0& 0& 0& \dots & x_{nn}\end{array}\right|\hfill \\ & =& \sum _{i=1}^n\prod _{j=1}^{i-1}{x}_{jj}^{\rho }{x}_{ii}^{\nabla }\prod _{k=i+1}^n{x}_{kk}=\left[\sum _{i=1}^n\prod _{j=1}^{i-1}\left(1-\nu a_{jj}\right)a_{ii}\right]detX.\hfill \end{array}$$

This completes the proof. □

Remark 6.

Let A be a $2\times 2$-matrix-valued function. Then A is regressive iff the scalar-valued function $trA-\nu detA$ is regressive (where $trA$ denotes the trace of the matrix A, $i.e.$, the sum of diagonal elements of A). In fact, let

$$A=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right],$$

then we can obtain

$$\begin{array}{ccc}\hfill det(I-\nu A)& =& det\left[\begin{array}{cc}1-\nu a_{11}& -\nu a_{12}\\ -\nu a_{21}& 1-\nu a_{22}\end{array}\right]\hfill \\ & =& (1-\nu a_{11})(1-\nu a_{22})-\nu a_{12}\nu a_{21}\hfill \\ & =& 1-\nu (a_{11}+a_{22})+{\nu }^2(a_{11}{a}_{22}-a_{12}{a}_{21})\hfill \\ & =& 1-\nu (trA-\nu detA).\hfill \end{array}$$

Hence A is regressive iff the scalar-valued function $trA-\nu detA$ is regressive.

Remark 7.

Assume that for $n=k$, we can obtain similar characterizations of Remark 6. For $n=k+1$, let

$$\begin{array}{ccc}\hfill A_{k+1}& =& \left[\begin{array}{cc}{A}_k& 0\\ 0& a_0\end{array}\right],\hfill \end{array}$$

where $A_{k+1}$ is $(k+1)\times (k+1)$-matrix-valued function, $A_k$ is $k\times k$-matrix-valued function. Then

$$\begin{array}{ccc}\hfill det(I-\nu A_{k+1})& =& det\left[\begin{array}{cc}{I}_k-\nu A_k& 0\\ 0& 1-\nu a_0\end{array}\right]\hfill \\ & =& (1-\nu a_0)det(I-\nu A_k),\hfill \\ \hfill 1-\nu (tr{A}_{k+1}-\nu det{A}_{k+1})& =& 1-\nu [tr{A}_k+a_0-\nu a_{0}det{A}_k]\hfill \\ & =& 1-\nu tr{A}_k-\nu a_0+\nu a_0\nu det{A}_k.\hfill \end{array}$$

Hence there exists $A_{k+1}$ such that $A_{k+1}$ is regressive and so is $tr{A}_{k+1}-\nu det{A}_{k+1}$.

Theorem 6

(Liouville’s formula). Let $A\in \mathcal{R}$ be $n\times n$-matrix-valued matrix function and assume that X is a solution of $X^{\nabla }=A\left(t\right)X$. Then X satisfies Liouville’s formula

$$detX\left(t\right)=e_{{\displaystyle \sum _{i=1}^n}det\widetilde{A_i}}(t,t_0)det{X}_0,fort\in \mathbb{T},$$

where

$$\widetilde{A_i}=\left[\begin{array}{cccccccc}1-\nu a_{11}& -\nu a_{12}& \dots & -\nu a_{1(i-1)}& -\nu a_{1i}& 0& \dots & 0\\ -\nu a_{21}& 1-\nu a_{22}& \dots & -\nu a_{2(i-1)}& -\nu a_{2i}& 0& \dots & 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ -\nu a_{(i-1)1}& -\nu a_{(i-1)2}& \dots & 1-\nu a_{(i-1)(i-1)}& -\nu a_{(i-1)i}& 0& 0& 0\\ a_{i1}& a_{i2}& \dots & a_{i(i-1)}& a_{ii}& 0& 0& 0\\ 0& 0& \dots & 0& 0& 1& 0& 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 0& 0& \dots & 0& 0& 0& \dots & 1\end{array}\right].$$

Proof. 

For $n=2$, by Remark 4, A is regressive implies $trA-\nu detA$ is regressive. Let

$$A\left(t\right)=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right],X\left(t\right)=\left[\begin{array}{cc}{x}_{11}& x_{12}\\ x_{21}& x_{22}\end{array}\right].$$

Then

$$\begin{array}{ccc}\hfill {\left(detX\left(t\right)\right)}^{\nabla }& =& \left|\begin{array}{cc}{x}_{11}^{\nabla }& x_{12}^{\nabla }\\ x_{21}& x_{22}\end{array}\right|+\left|\begin{array}{cc}{x}_{11}^{\rho }& x_{12}^{\rho }\\ x_{21}^{\nabla }& x_{22}^{\nabla }\end{array}\right|\hfill \\ & =& \left|\begin{array}{cc}{a}_{11}{x}_{11}+a_{12}{x}_{21}& a_{11}{x}_{12}+a_{12}{x}_{22}\\ x_{21}& x_{22}\end{array}\right|+\left|\begin{array}{cc}{x}_{11}+\mu x_{11}^{\nabla }& x_{12}+\mu x_{12}^{\nabla }\\ x_{21}^{\nabla }& x_{22}^{\nabla }\end{array}\right|\hfill \\ & =& a_{11}detX\left(t\right)+\left|\begin{array}{cc}{x}_{11}+(a_{11}{x}_{11}+a_{12}{x}_{21})(-\nu )& x_{12}+(a_{11}{x}_{12}+a_{12}{x}_{22})(-\nu )\\ a_{21}{x}_{11}+a_{22}{x}_{21}& a_{21}{x}_{12}+a_{22}{x}_{22}\end{array}\right|\hfill \\ & =& a_{11}detX\left(t\right)+\left|\begin{array}{cc}1-\nu a_{11}& a_{12}(-\nu )\\ a_{21}& a_{22}\end{array}\right|detX\hfill \\ & =& \left(trA-\nu detA\right)detX\hfill \\ & =& \left\{\left|\begin{array}{cc}{a}_{11}& 0\\ 0& 1\end{array}\right|+\left|\begin{array}{cc}1-\nu a_{11}& a_{12}(-\nu )\\ a_{21}& a_{22}\end{array}\right|\right\}detX.\hfill \end{array}$$

For $n\ge 3$, let

$$A\left(t\right)=\left[\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \dots & a_{1n}\\ a_{21}& a_{22}& a_{23}& \dots & a_{2n}\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}& a_{n2}& a_{n3}& \dots & a_{nn}\end{array}\right],X\left(t\right)=\left[\begin{array}{ccccc}{x}_{11}& x_{12}& x_{13}& \dots & x_{1n}\\ x_{21}& x_{22}& x_{23}& \dots & x_{2n}\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ x_{n1}& x_{n2}& x_{n3}& \dots & x_{nn}\end{array}\right].$$

Hence by Lemma 5, we have

$$\begin{array}{cc}& {\left(X\left(t\right)\right)}^{\nabla }={\left|\begin{array}{ccccc}{x}_{11}& x_{12}& x_{13}& \dots & x_{1n}\\ x_{21}& x_{22}& x_{23}& \dots & x_{2n}\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ x_{n1}& x_{n2}& x_{n3}& \dots & x_{nn}\end{array}\right|}^{\nabla }={\displaystyle \sum _{i=1}^n}\left|\begin{array}{ccccc}{x}_{11}^{\rho }& x_{12}^{\rho }& x_{13}^{\rho }& \dots & x_{1n}^{\rho }\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ x_{(i-1)1}^{\rho }& x_{(i-1)2}^{\rho }& x_{(i-1)3}^{\rho }& \dots & x_{(i-1)n}^{\rho }\\ x_{i1}^{\nabla }& x_{i2}^{\nabla }& x_{i3}^{\nabla }& \dots & x_{in}^{\nabla }\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ x_{n1}& x_{n2}& x_{n3}& \dots & x_{nn}\end{array}\right|\hfill \\ =& {\displaystyle \sum _{i=1}^n}\left|\begin{array}{ccccc}{x}_{11}-\nu x_{11}^{\nabla }& x_{12}-\nu x_{12}^{\nabla }& x_{13}-\nu x_{13}^{\nabla }& \dots & x_{1n}-\nu x_{1n}^{\nabla }\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ x_{(i-1)1}-\nu x_{(i-1)1}^{\nabla }& x_{(i-1)2}-\nu x_{(i-1)2}^{\nabla }& x_{(i-1)3}-\nu x_{(i-1)3}^{\nabla }& \dots & x_{(i-1)n}-\nu x_{(i-1)n}^{\nabla }\\ x_{i1}^{\nabla }& x_{i2}^{\nabla }& x_{i3}^{\nabla }& \dots & x_{in}^{\nabla }\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ x_{n1}& x_{n2}& x_{n3}& \dots & x_{nn}\end{array}\right|\hfill \\ =& {\displaystyle \sum _{i=1}^n}\left|\begin{array}{cccc}{x}_{11}-\nu {\displaystyle \sum _{j=1}^n}{a}_{1j}{x}_{j1}& x_{12}-\nu {\displaystyle \sum _{j=1}^n}{a}_{1j}{x}_{j2}& \dots & x_{1n}-\nu {\displaystyle \sum _{j=1}^n}{a}_{1j}{x}_{jn}\\ ⋮& ⋮& ⋮& ⋮\\ x_{(i-1)1}-\nu {\displaystyle \sum _{j=1}^n}{a}_{(i-1)j}{x}_{j1}& x_{(i-1)2}-\nu {\displaystyle \sum _{j=1}^n}{a}_{(i-1)j}{x}_{j2}& \dots & x_{(i-1)n}-\nu {\displaystyle \sum _{j=1}^n}{a}_{(i-1)j}{x}_{jn}\\ {\displaystyle \sum _{j=1}^n}{a}_{ij}{x}_{j1}& {\displaystyle \sum _{j=1}^n}{a}_{ij}{x}_{j2}& \dots & {\displaystyle \sum _{j=1}^n}{a}_{ij}{x}_{jn}\\ ⋮& ⋮& ⋮& ⋮\\ x_{n1}& x_{n2}& \dots & x_{nn}\end{array}\right|\hfill \\ =& {\displaystyle \sum _{i=1}^n}\left|\begin{array}{cccc}{x}_{11}-\nu {\displaystyle \sum _{j=1}^i}{a}_{1j}{x}_{j1}& x_{12}-\nu {\displaystyle \sum _{j=1}^i}{a}_{1j}{x}_{j2}& \dots & x_{1n}-\nu {\displaystyle \sum _{j=1}^i}{a}_{1j}{x}_{jn}\\ ⋮& ⋮& ⋮& ⋮\\ x_{(i-1)1}-\nu {\displaystyle \sum _{j=1}^i}{a}_{(i-1)j}{x}_{j1}& x_{(i-1)2}-\nu {\displaystyle \sum _{j=1}^i}{a}_{(i-1)j}{x}_{j2}& \dots & x_{(i-1)n}-\nu {\displaystyle \sum _{j=1}^i}{a}_{(i-1)j}{x}_{jn}\\ {\displaystyle \sum _{j=1}^i}{a}_{ij}{x}_{j1}& {\displaystyle \sum _{j=1}^i}{a}_{ij}{x}_{j2}& \dots & {\displaystyle \sum _{j=1}^i}{a}_{ij}{x}_{jn}\\ ⋮& ⋮& ⋮& ⋮\\ x_{n1}& x_{n2}& \dots & x_{nn}\end{array}\right|\hfill \\ =& {\displaystyle \sum _{i=1}^n}det\left(\widetilde{A_i}X\right)={\displaystyle \sum _{i=1}^n}det\widetilde{A_i}detX.\hfill \end{array}$$

Then

$$X\left(t\right)=\left(e_{{\displaystyle \sum _{i=1}^n}det\widetilde{A_i}}(t,t_0)\right)det{X}_0.$$

This completes the proof. □

Theorem 7.

Let A be a $n\times n$-matrix-valued function. Then A is regressive iff the scalar-valued function ${\displaystyle \sum _{i=1}^n}det\widetilde{A_i}$ is regressive, where $\widetilde{A_i}$ is defined by Theorem 6.

Proof. 

By Theorem 6, for $n=2$ we can obtain

$$det(I-\nu A)=1-\nu \left\{\left|\begin{array}{cc}{a}_{11}& 0\\ 0& 1\end{array}\right|+\left|\begin{array}{cc}1-\nu a_{11}& a_{12}(-\nu )\\ a_{21}& a_{22}\end{array}\right|\right\}.$$

Next, let

$$\begin{array}{ccc}\hfill A\left(t\right)& =& \left[\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \dots & a_{1n}\\ a_{21}& a_{22}& a_{23}& \dots & a_{2n}\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ a_{n1}& a_{n2}& a_{n3}& \dots & a_{nn}\end{array}\right]=\left[\begin{array}{ccccc}& & & & a_{1n}\\ & & A_{\left(0\right)}& & a_{2n}\\ & & & & ⋮\\ a_{n1}& a_{n2}& a_{n3}& \dots & a_{nn}\end{array}\right],\hfill \end{array}$$

where $A_{\left(0\right)}$ is a $(n-1)\times (n-1)$-valued-matrix function. Assume that $(n-1)\times (n-1)$-matrix-value function is regressive iff the scalar-valued function ${\displaystyle \sum _{i=1}^{n-1}}det\widetilde{A_i}$ is regressive. That is

$$det(I-\nu A_{\left(0\right)})=1-\nu \sum _{i=1}^{n-1}det\widetilde{A_i}.$$

Hence

$$\begin{array}{cc}& 1-\nu \left(\sum _{i=1}^{n-1}det{A}_i+det{A}_n\right)=1-\nu \sum _{i=1}^{n-1}det{A}_i-\nu det{A}_n=det\left(I-\nu A_{\left(0\right)}\right)-\nu det{A}_n\hfill \\ =& det\left(I-\nu A_{\left(0\right)}\right)-\nu \left|\begin{array}{ccccc}& & & & -\nu a_{1n}\\ & & I-\nu A_{\left(0\right)}& & -\nu a_{2n}\\ & & & & \dots \\ a_{n1}& a_{n2}& a_{n3}& \dots & a_{nn}\end{array}\right|\hfill \\ =& \left|\begin{array}{ccccc}& & & & -\nu a_{1n}\\ & & I-\nu A_{\left(0\right)}& & -\nu a_{2n}\\ & & & & \dots \\ 0& 0& 0& \dots & 1\end{array}\right|+\left|\begin{array}{ccccc}& & & & -\nu a_{1n}\\ & & I-\nu A_{\left(0\right)}& & -\nu a_{2n}\\ & & & & \dots \\ -\nu a_{n1}& -\nu a_{n2}& -\nu a_{n3}& \dots & -\nu a_{nn}\end{array}\right|\hfill \\ =& \left|\begin{array}{ccccc}& & & & -\nu a_{1n}\\ & & I-\nu A_{\left(0\right)}& & \mu a_{2n}\\ & & & & \dots \\ \mu a_{n1}& -\nu a_{n2}& -\nu a_{n3}& \dots & 1-\nu a_{nn}\end{array}\right|=det(I-\nu A).\hfill \end{array}$$

Therefore A is regressive iff the scalar-valued function ${\displaystyle \sum _{i=1}^n}det\widetilde{A_i}$ is regressive. This completes the proof. □

Definition 3

([19]). Define ${\widehat{e}}_A(t,t_0)$ by the unique matrix solution of the initial value problem:

$$X^{\nabla }\left(t\right)=A\left(t\right)X\left(t\right),X_0=I.$$

(4)

Lemma 6.

By Definition 3, the matrix function ${\widehat{e}}_A(t,t_0)$ is delta differentiable at t with

$${\widehat{e}}_A^{\Delta }(t,t_0)=A\left(\sigma \left(t\right)\right){\left(I-\mu \left(t\right)A\left(\sigma \left(t\right)\right)\right)}^{-1}{\widehat{e}}_A(t,t_0).$$

Proof. 

For ${\widehat{e}}_A^{\nabla }(t,t_0)=\frac{{\widehat{e}}_A(t,t_0)-{\widehat{e}}_A(\rho \left(t\right),t_0)}{\nu \left(t\right)}=A\left(t\right)e_A(t,t_0)$, we can obtain

$${\widehat{e}}_A^{\Delta }(t,t_0)=\frac{{\widehat{e}}_A(\sigma \left(t\right),t_0)-{\widehat{e}}_A(t,t_0)}{\mu \left(t\right)}=A\left(\sigma \left(t\right)\right){\widehat{e}}_A(\sigma \left(t\right),t_0).$$

On the other hand, by $f\left(\sigma \left(t\right)\right)-f\left(t\right)=f^{\Delta }\left(t\right)\mu \left(t\right)$, we can obtain

$${\widehat{e}}_A(\sigma \left(t\right),t_0)-{\widehat{e}}_A(t,t_0)=\mu \left(t\right)A\left(\sigma \left(t\right)\right){\widehat{e}}_A(\sigma \left(t\right),t_0),$$

that is ${\widehat{e}}_A(t,t_0)=\left(I-\mu \left(t\right)A\left(\sigma \left(t\right)\right)\right){\widehat{e}}_A(\sigma \left(t\right),t_0)$. Hence

$${\widehat{e}}_A^{\Delta }(t,t_0)=A\left(\sigma \left(t\right)\right){\left(I-\mu \left(t\right)A\left(\sigma \left(t\right)\right)\right)}^{-1}{\widehat{e}}_A(t,t_0).$$

The proof is complete. □

Example 4.

For (4), let $\mathbb{T}=\mathbb{R}$, Liouville’s formula can be given as:

$$detX=det{X}_0{e}^{{\int }_{t_0}^{t}trA\left(\tau \right)\mathrm{d}ø},trA\left(\tau \right)=\sum _{i=1}^n{a}_{ii}\left(\tau \right).$$

In fact, let $\mathbb{T}=\mathbb{R}$, we can obtain $\nu \left(t\right)=0$ for any $t\in \mathbb{T}$, hence $X^{\nabla }\left(t\right)=X^{\prime }$. The result is obvious.

Example 5.

For (4), let $\mathbb{T}=\mathbb{Z}h$, $h>0$, the solution of (4) can be given as:

$$X\left(t\right)=\prod _{k=\frac{t-t_0}{h}}^1{\left[I-hA(t_0+kh)\right]}^{-1}{X}_0.$$

Furthermore, by Theorem 6,

$$\begin{array}{ccc}\hfill detX\left(t\right)& =& \prod _{k=\frac{t-t_0}{h}}^{1}det{\left[I-hA(t_0+kh)\right]}^{-1}det{X}_0\hfill \\ & =& {\tilde{e}}_{{\displaystyle \sum _{i=1}^n}det{\tilde{A}}_i}(t,t_0)det{X}_0,fort\in \mathbb{T},\hfill \end{array}$$

where

$${\overline{A}}_i=\left[\begin{array}{cccccccc}1-h{a}_{11}& -h{a}_{12}& \dots & -h{a}_{1(i-1)}& -h{a}_{1i}& 0& \dots & 0\\ -h{a}_{21}& 1-h{a}_{22}& \dots & -h{a}_{2(i-1)}& -h{a}_{2i}& 0& \dots & 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ -h{a}_{(i-1)1}& -h{a}_{(i-1)2}& \dots & 1-h{a}_{(i-1)(i-1)}& -h{a}_{(i-1)i}& 0& 0& 0\\ a_{i1}& a_{i2}& \dots & a_{i(i-1)}& a_{ii}& 0& 0& 0\\ 0& 0& \dots & 0& 0& 1& 0& 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 0& 0& \dots & 0& 0& 0& \dots & 1\end{array}\right].$$

In fact, let $\mathbb{T}=\mathbb{Z}h$, $h>0$, by $X\left(t\right)-X^{\rho }\left(t\right)=\nu \left(t\right)X^{\nabla }\left(t\right)=\nu \left(t\right)A\left(t\right)X\left(t\right)$ and $\nu \left(t\right)=h$, we have $X(t+h)-X\left(t\right)=hA(t+h)X(t+h)$. Hence $X(x+h)={\left(I-hA(t+h)\right)}^{-1}X\left(t\right)$. Therefore

$$X\left(t\right)=\prod _{k=\frac{t-t_0}{h}}^1{\left[I-hA(t_0+kh)\right]}^{-1}{X}_0.$$

Example 6.

For (4), let $\mathbb{T}=\overline{q^{\mathbb{Z}}}$, $q>1$, the solution of (4) can be given as:

$$X\left(t\right)=\prod _{k=\frac{lnt-ln{t}_0}{lnq}}^1{\left[I-(q-1)q^k{t}_{0}A\left(q^k{t}_0\right)\right]}^{-1}{X}_0.$$

Furthermore, by Theorem 6,

$$\begin{array}{ccc}\hfill detX\left(t\right)& =& \prod _{k=\frac{lnt-ln{t}_0}{lnq}}^{1}det{\left[I-(q-1)q^k{t}_{0}A\left(q^k{t}_0\right)\right]}^{-1}det{X}_0\hfill \\ & =& e_{{\displaystyle \sum _{i=1}^n}det{\tilde{A}}_i}(t,t_0)det{X}_0,fort\in \mathbb{T},\hfill \end{array}$$

where

$$A_i=\left[\begin{array}{cccccccc}1-(1-\frac{1}{q})t{a}_{11}& -(1-\frac{1}{q})t{a}_{12}& \dots & -(1-\frac{1}{q})t{a}_{1(i-1)}& -(1-\frac{1}{q})t{a}_{1i}& 0& \dots & 0\\ -(1-\frac{1}{q})t{a}_{21}& 1-(1-\frac{1}{q})t{a}_{22}& \dots & -(1-\frac{1}{q})t{a}_{2(i-1)}& -(1-\frac{1}{q})t{a}_{2i}& 0& \dots & 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ -(1-\frac{1}{q})t{a}_{(i-1)1}& -(1-\frac{1}{q})t{a}_{(i-1)2}& \dots & 1-(1-\frac{1}{q})t{a}_{(i-1)(i-1)}& -(1-\frac{1}{q})t{a}_{(i-1)i}& 0& 0& 0\\ a_{i1}& a_{i2}& \dots & a_{i(i-1)}& a_{ii}& 0& 0& 0\\ 0& 0& \dots & 0& 0& 1& 0& 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ 0& 0& \dots & 0& 0& 0& \dots & 1\end{array}\right].$$

In fact, let $\mathbb{T}=\overline{q^{\mathbb{Z}}}$, $q>1$, by $X\left(t\right)-X^{\rho }\left(t\right)=\nu \left(t\right)A\left(t\right)X\left(t\right)$ and $\nu \left(t\right)=(1-\frac{1}{q})t$, we have $X\left(qt\right)-X\left(t\right)=(q-1)qtA\left(qt\right)X\left(qt\right)$. Hence $X\left(qt\right)={\left(I-(q-1)qtA\left(qt\right)\right)}^{-1}X\left(t\right)$. Therefore

$$X\left(t\right)=\prod _{k=\frac{lnt-ln{t}_0}{lnq}}^1{\left[I-(q-1)q^k{t}_{0}A\left(q^k{t}_0\right)\right]}^{-1}{X}_0.$$

4. Liouville’S Formula of Diamond-$\mathit{\alpha }$ Dynamic Equations

In the sequel, we will introduce the $\alpha$-matrix exponential function and obtain Liouville’s formula of diamond-$\alpha$ dynamic equations. Several examples are provided on various time scales.

Definition 4

([13]). Let $\mathbb{T}$ be a time scale and $f\left(t\right)$ be differentiable on $\mathbb{T}$ in the Δ and ∇ sense. For $t\in \mathbb{T}$ we define the diamond-α dynamic derivative $f^{{♢}_{\alpha }}\left(t\right)$ by

$$f^{{♢}_{\alpha }}\left(t\right)=\alpha f^{\Delta }\left(t\right)+(1-\alpha )f^{\nabla }\left(t\right),0\le \alpha \le 1.$$

Thus f is diamond-α differentiable if and only if f is Δ and ∇ differentiable.

Definition 5

([14]). Let $f(·):\mathbb{T}\to \mathbb{R}$, we define the generalized diamond exponential function by

$${\ddot{e}}_f(t,t_0)=\alpha e_f(t,t_0)+(1-\alpha ){\widehat{e}}_f(t,t_0),0\le \alpha \le 1.$$

Theorem 8.

The generalized diamond exponential function is the solution of the diamond α-dynamic equation as follows:

$$x^{{♢}_{\alpha }}=\left(\alpha f\left(t\right)+(1-\alpha )\frac{f^{\rho }\left(t\right)}{1+\nu \left(t\right)f^{\rho }\left(t\right)}\right)\alpha e_f(t,t_0)+\left(\alpha \frac{f^{\sigma }\left(t\right)}{1-\mu f^{\sigma }\left(t\right)}+(1-\alpha )f\left(t\right)\right)(1-\alpha ){\widehat{e}}_f(t,t_0),$$

(5)

where $0\le \alpha \le 1$ with the initial condition $x\left(t_0\right)=1$. When $\alpha =1$, the dynamic equation is Δ-dynamic equation, and when $\alpha =0$, the dynamic equation is ∇-dynamic equation.

Proof. 

By Definitions 4 and 5, we can obtain

$$\begin{array}{ccc}\hfill {\left({\ddot{e}}_f(t,t_0)\right)}^{{♢}_{\alpha }}\left(t\right)& =& \alpha \left(\alpha e_f^{\Delta }(t,t_0)+(1-\alpha ){\widehat{e}}_f^{\Delta }(t,t_0)\right)+(1-\alpha )\left(\alpha e_f^{\nabla }(t,t_0)+(1-\alpha ){\widehat{e}}_f^{\nabla }(t,t_0)\right)\hfill \\ & =& \alpha \left(\alpha f\left(t\right)e_f(t,t_0)+(1-\alpha )\frac{f^{\rho }\left(t\right)}{1+\nu \left(t\right)f^{\rho }\left(t\right)}{\widehat{e}}_f(t,t_0)\right)\hfill \\ & & +(1-\alpha )\left(\alpha \frac{f^{\sigma }\left(t\right)}{1-\mu \left(t\right)f^{\sigma }\left(t\right)}{e}_f(t,t_0)+(1-\alpha )f\left(t\right){\widehat{e}}_f(t,t_0)\right)\hfill \\ & =& \left(\alpha f\left(t\right)+(1-\alpha )\frac{f^{\rho }\left(t\right)}{1+\nu \left(t\right)f^{\rho }\left(t\right)}\right)\alpha e_f(t,t_0)\hfill \\ & & +\left(\alpha \frac{f^{\sigma }\left(t\right)}{1-\mu f^{\sigma }\left(t\right)}+(1-\alpha )f\left(t\right)\right)(1-\alpha ){\widehat{e}}_f(t,t_0).\hfill \end{array}$$

Hence the generalized diamond exponential function ${\ddot{e}}_f(t,t_0)$ is the solution of (5). Further when $\alpha =1$, we obtain the dynamic Equation (5) as follows:

$$x^{{♢}_1}\left(t\right)=x^{\Delta }\left(t\right)=f\left(t\right)e_f(t,t_0),$$

(6)

for the solution of (6) is $x=e_f(t,t_0)$, hence (6) can be denoted by $x^{\Delta }\left(t\right)=f\left(t\right)x\left(t\right)$. Similarly, we can also obtain the case for $\alpha =0$. The proof is complete. □

Theorem 9.

Let $f(·):\mathbb{T}\to \mathbb{T}$, then $f\left(t\right)$ is a solution of the following equation

$$f\left(t\right)=e^{{\int }_{t_0}^t\frac{log\left(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right)\right)}{\mu \left(\tau \right)}\Delta \tau },$$

(7)

iff (5) can be reduced to

$$x^{{♢}_{\alpha }}\left(t\right)=f\left(t\right)x\left(t\right),0\le \alpha \le 1.$$

(8)

Proof. 

For $f\left(t\right)=e^{{\int }_{t_0}^t\frac{1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right)}{\mu \left(\tau \right)}\Delta \tau }$, we can obtain

$$\begin{array}{ccc}\hfill f^{\Delta }\left(t\right)& =& \frac{e^{{\int }_{t_0}^{\sigma \left(t\right)}\frac{log\left(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right)\right)}{\mu \left(\tau \right)}\Delta \tau }-e^{{\int }_{t_0}^t\frac{log\left(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right)\right)}{\mu \left(\tau \right)}\Delta \tau }}{\mu \left(t\right)}\hfill \\ & =& \frac{e^{{\int }_t^{\sigma \left(t\right)}\frac{log\left(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right)\right)}{\mu \left(\tau \right)}\Delta \tau }-1}{\mu \left(t\right)}{e}^{{\int }_{t_0}^t\frac{log\left(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right)\right)}{\mu \left(\tau \right)}\Delta \tau }\hfill \\ & =& \frac{1-\mu \left(t\right)f^{\sigma }\left(t\right)-1}{\mu \left(t\right)}{e}^{{\int }_{t_0}^t\frac{log\left(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right)\right)}{\mu \left(\tau \right)}\Delta \tau }=-f^{\sigma }\left(t\right)f\left(t\right),\hfill \end{array}$$

(9)

$$\begin{array}{ccc}\hfill f^{\nabla }\left(t\right)& =& \frac{e^{{\int }_{t_0}^t\frac{log\left(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right)\right)}{\mu \left(\tau \right)}\Delta \tau }-e^{{\int }_{t_0}^{\rho \left(t\right)}\frac{log\left(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right)\right)}{\mu \left(\tau \right)}\Delta \tau }}{\nu \left(t\right)}\hfill \\ & =& \frac{e^{{\int }_{\rho \left(t\right)}^t\frac{log\left(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right)\right)}{\mu \left(\tau \right)}\Delta \tau }-1}{\nu \left(t\right)}{e}^{{\int }_{t_0}^{\rho \left(t\right)}\frac{log\left(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right)\right)}{\mu \left(\tau \right)}\Delta \tau }\hfill \\ & =& \frac{1-\nu \left(t\right)f\left(t\right)-1}{\nu \left(t\right)}{e}^{{\int }_{t_0}^{\rho \left(t\right)}\frac{log\left(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right)\right)}{\mu \left(\tau \right)}\Delta \tau }=-f\left(t\right)f^{\rho }\left(t\right),\hfill \end{array}$$

(10)

hence we have the solution of (9) and (10) with the initial value $f\left(t_0\right)=1$ is equivalent to the solution of (7).

On the other hand, by (5), we can obtain

$$\begin{array}{ccc}\hfill \alpha f\left(t\right)+(1-\alpha )\frac{f^{\rho }\left(t\right)}{1+\nu \left(t\right)f^{\rho }\left(t\right)}& =& f\left(t\right)+(1-\alpha )\frac{f^{\rho }\left(t\right)-f\left(t\right)-\nu \left(t\right)f^{\rho }\left(t\right)f\left(t\right)}{1+\nu \left(t\right)f^{\rho }\left(t\right)}\hfill \\ & =& f\left(t\right)+(1-\alpha )\frac{-\nu \left(t\right)f^{\nabla }\left(t\right)-\nu \left(t\right)f^{\rho }\left(t\right)f\left(t\right)}{1+\nu \left(t\right)f^{\rho }\left(t\right)}=f\left(t\right)\hfill \\ \hfill \alpha \frac{f^{\sigma }\left(t\right)}{1-\mu f^{\sigma }\left(t\right)}+(1-\alpha )f\left(t\right)& =& \alpha \frac{f^{\sigma }\left(t\right)-f\left(t\right)+\mu \left(t\right)f\left(t\right)f^{\sigma }\left(t\right)}{1-\mu f^{\sigma }\left(t\right)}+f\left(t\right)\hfill \\ & =& \alpha \frac{\mu \left(t\right)f^{\Delta }\left(t\right)+\mu \left(t\right)f\left(t\right)f^{\sigma }\left(t\right)}{1-\mu f^{\sigma }\left(t\right)}+f\left(t\right)=f\left(t\right).\hfill \end{array}$$

Therefore the dynamic Equation (5) can be reduced to

$$x^{{♢}_{\alpha }}=f\left(t\right)x\left(t\right).$$

Conversely, if $x^{{♢}_{\alpha }}=f\left(t\right)x\left(t\right)$ with $x\left(t_0\right)=1$, then

$$\alpha f\left(t\right)+(1-\alpha )\frac{f^{\rho }\left(t\right)}{1+\nu \left(t\right)f^{\rho }\left(t\right)}=\alpha \frac{f^{\sigma }\left(t\right)}{1-\mu f^{\sigma }\left(t\right)}+(1-\alpha )f\left(t\right),$$

that is

$$-\nu \left(t\right)(1-\alpha )\frac{f^{\nabla }\left(t\right)+f^{\rho }\left(t\right)f\left(t\right)}{1+\nu \left(t\right)f^{\rho }\left(t\right)}=\mu \left(t\right)\alpha \frac{f^{\Delta }\left(t\right)+f\left(t\right)f^{\sigma }\left(t\right)}{1-\mu f^{\sigma }\left(t\right)},$$

for any $0\le \alpha \le 1$, thus $f^{\nabla }\left(t\right)+f^{\rho }\left(t\right)f\left(t\right)=f^{\Delta }\left(t\right)+f\left(t\right)f^{\sigma }\left(t\right)=0$, so we have

$$f^{\nabla }\left(t\right)=-f^{\rho }\left(t\right)f\left(t\right),f^{\Delta }\left(t\right)=-f\left(t\right)f^{\sigma }\left(t\right)$$

i.e., $f\left(t\right)$ with $f\left(t_0\right)=1$ is a solution of (7). The proof is complete. □

Example 7.

For (7), let $\mathbb{T}=\mathbb{Z}h$, $h>0$, by Theorem 9, then the solution of (7) can be given by

$$f\left(t\right)=\left\{\begin{array}{c}\frac{{\displaystyle \prod _{k=0}^{\frac{t-t_0}{h}-2}}\left(1-h{f}^{\sigma }(t_0+kh)\right)}{1+h{\displaystyle \prod _{k=0}^{\frac{t-t_0}{h}-2}}\left(1-h{f}^{\sigma }(t_0+kh)\right)},t>t_0,\hfill \\ {\displaystyle \prod _{k=0}^{\frac{t-t_0}{h}+1}}{\left(1-hf(t_0+kh)\right)}^{-1},t<t_0.\hfill \end{array}\right.$$

In fact, by $f\left(t\right)=e^{{\int }_{t_0}^t\frac{log(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right))}{\mu \left(\tau \right)}\Delta \tau }$, we can obtain

$$\begin{array}{ccc}\hfill f\left(t\right)& =& e^{{\int }_{t_0}^t\frac{log(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right))}{\mu \left(\tau \right)}\Delta \tau }=e^{{\displaystyle \sum _{k=0}^{\frac{t-t_0}{h}-1}}{\int }_{t_0+kh}^{t_0+(k+1)h}\frac{log(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right))}{\mu \left(\tau \right)}\Delta \tau }\hfill \\ & =& {\displaystyle \prod _{k=0}^{\frac{t-t_0}{h}-1}}{e}^{{\int }_{t_0+kh}^{\sigma (t_0+kh)}\frac{log(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right))}{\mu \left(\tau \right)}\Delta \tau }={\displaystyle \prod _{k=0}^{\frac{t-t_0}{h}-1}}\left(1-h{f}^{\sigma }(t_0+kh)\right)\hfill \\ & =& \left(1-hf\left(t\right)\right){\displaystyle \prod _{k=0}^{\frac{t-t_0}{h}-2}}\left(1-h{f}^{\sigma }(t_0+kh)\right),\hfill \end{array}$$

that is

$$f\left(t\right)=\frac{{\displaystyle \prod _{k=0}^{\frac{t-t_0}{h}-2}}\left(1-h{f}^{\sigma }(t_0+kh)\right)}{1+h{\displaystyle \prod _{k=0}^{\frac{t-t_0}{h}-2}}\left(1-h{f}^{\sigma }(t_0+kh)\right)}.$$

For $t\in \mathbb{T}$, $t<t_0$, we can obtain

$$\begin{array}{ccc}\hfill f\left(t\right)& =& e^{{\int }_{t_0}^t\frac{log(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right))}{\mu \left(\tau \right)}\Delta \tau }=e^{-{\displaystyle \sum _{k=0}^{\frac{t-t_0}{h}+1}}{\int }_{t_0+(k-1)h}^{t_0+kh}\frac{log(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right))}{\mu \left(\tau \right)}\Delta \tau }\hfill \\ & =& {\displaystyle \prod _{k=0}^{\frac{t-t_0}{h}+1}}{e}^{-{\int }_{t_0+(k-1)h}^{\sigma (t_0+(k-1)h)}\frac{log(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right))}{\mu \left(\tau \right)}\Delta \tau }={\displaystyle \prod _{k=0}^{\frac{t-t_0}{h}+1}}{\left(1-h{f}^{\sigma }(t_0+(k-1)h)\right)}^{-1}\hfill \\ & =& {\displaystyle \prod _{k=0}^{\frac{t-t_0}{h}+1}}{\left(1-hf(t_0+kh)\right)}^{-1},\hfill \end{array}$$

for any $0\le k\le \frac{t-t_0}{h}+1$, $t_0+kh>t$. Therefore

$$f\left(t\right)=\prod _{k=0}^{\frac{t-t_0}{h}+1}{\left(1-hf(t_0+kh)\right)}^{-1}.$$

Example 8.

For (7), let $\mathbb{T}=\overline{q^{\mathbb{Z}}}$, $q>1$, by Theorem 9, then the solution of (7) can be given by

$$f\left(t\right)=\left\{\begin{array}{c}\frac{{\displaystyle \prod _{k=0}^{\frac{lnt-ln{t}_0}{lnq}-2}}\left(1-(q-1)t_0{q}^k{f}^{\sigma }\left(t_0{q}^k\right)\right)}{1+(1-\frac{1}{q})t{\displaystyle \prod _{k=0}^{\frac{lnt-ln{t}_0}{lnq}-2}}\left(1-(q-1)t_0{q}^k{f}^{\sigma }\left(t_0{q}^k\right)\right)},t>t_0,\hfill \\ {\displaystyle \prod _{k=0}^{\frac{lnt-ln{t}_0}{lnq}+1}}{\left(1-(q-1)t_0{q}^{k-1}f\left(t_0{q}^k\right)\right)}^{-1},t<t_0.\hfill \end{array}\right.$$

In fact, by $f\left(t\right)=e^{{\int }_{t_0}^t\frac{log(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right))}{\mu \left(\tau \right)}\Delta \tau }$, we can obtain

$$\begin{array}{ccc}\hfill f\left(t\right)& =& e^{{\int }_{t_0}^t\frac{log(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right))}{\mu \left(\tau \right)}\Delta \tau }=e^{{\displaystyle \sum _{k=0}^{\frac{t-t_0}{h}-1}}{\int }_{t_0{q}^k}^{t_0{q}^{k+1}}\frac{log(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right))}{\mu \left(\tau \right)}\Delta \tau }\hfill \\ & =& {\displaystyle \prod _{k=0}^{\frac{lnt-ln{t}_0}{lnq}-1}}{e}^{{\int }_{t_0{q}^k}^{\sigma \left(t_0{q}^k\right)}\frac{log(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right))}{\mu \left(\tau \right)}\Delta \tau }={\displaystyle \prod _{k=0}^{\frac{lnt-ln{t}_0}{lnq}-1}}\left(1-(q-1)t_0{q}^k{f}^{\sigma }\left(t_0{q}^k\right)\right)\hfill \\ & =& \left(1-\left(1-\frac{1}{q}\right)tf\left(t\right)\right){\displaystyle \prod _{k=0}^{\frac{lnt-ln{t}_0}{lnq}-2}}\left(1-(q-1)t_0{q}^k{f}^{\sigma }\left(t_0{q}^k\right)\right),\hfill \end{array}$$

that is

$$f\left(t\right)=\frac{{\displaystyle \prod _{k=0}^{\frac{lnt-ln{t}_0}{lnq}-2}}\left(1-(q-1)t_0{q}^k{f}^{\sigma }\left(t_0{q}^k\right)\right)}{1+(1-\frac{1}{q})t{\displaystyle \prod _{k=0}^{\frac{lnt-ln{t}_0}{lnq}-2}}\left(1-(q-1)t_0{q}^k{f}^{\sigma }\left(t_0{q}^k\right)\right)}.$$

For $t\in \mathbb{T}$, $t<t_0$, we can obtain

$$\begin{array}{ccc}\hfill f\left(t\right)& =& e^{{\int }_{t_0}^t\frac{log(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right))}{\mu \left(\tau \right)}\Delta \tau }=e^{-{\displaystyle \sum _{k=0}^{\frac{lnt-ln{t}_0}{lnq}+1}}{\int }_{t_0{q}^{k-1}}^{t_0{q}^k}\frac{log(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right))}{\mu \left(\tau \right)}\Delta \tau }\hfill \\ & =& {\displaystyle \prod _{k=0}^{\frac{lnt-ln{t}_0}{lnq}+1}}{e}^{-{\int }_{t_0{q}^{k-1}}^{\sigma \left(t_0{q}^{k-1}\right)}\frac{log(1-\mu \left(\tau \right)f^{\sigma }\left(\tau \right))}{\mu \left(\tau \right)}\Delta \tau }={\displaystyle \prod _{k=0}^{\frac{lnt-ln{t}_0}{lnq}+1}}{\left(1-(q-1)t_0{q}^{k-1}{f}^{\sigma }\left(t_0{q}^{k-1}\right)\right)}^{-1}\hfill \\ & =& {\displaystyle \prod _{k=0}^{\frac{lnt-ln{t}_0}{lnq}+1}}{\left(1-(q-1)t_0{q}^{k-1}f\left(t_0{q}^k\right)\right)}^{-1},\hfill \end{array}$$

for any $0\le k\le \frac{lnt-ln{t}_0}{lnq}+1$, $t_0{q}^k>t$. Therefore

$$f\left(t\right)=\prod _{k=0}^{\frac{lnt-ln{t}_0}{lnq}+1}{\left(1-(q-1)t_0{q}^{k-1}f\left(t_0{q}^k\right)\right)}^{-1}.$$

Next we consider the diamond $\alpha$-dynamic equations on time scale as follows:

$$\begin{array}{ccc}\hfill X^{{♢}_{\alpha }}& =& \left(\alpha A\left(t\right)+(1-\alpha )A^{\rho }\left(t\right){\left(I+\nu \left(t\right)A^{\rho }\left(t\right)\right)}^{-1}\right)\alpha e_A(t,t_0)\hfill \\ & & +\left(\alpha A^{\sigma }\left(t\right){\left(1-\mu \left(t\right)A^{\sigma }\left(t\right)\right)}^{-1}+(1-\alpha )A\left(t\right)\right)(1-\alpha ){\widehat{e}}_A(t,t_0),0\le \alpha \le 1,\hfill \end{array}$$

(11)

where $A:\mathbb{T}\to {\mathbb{R}}^{n\times n}$, $e_A(t,t_0)$ is solution of (2), ${\widehat{e}}_A(t,t_0)$ is solution of (3) with the initial value

$$X\left(t_0\right)=X_0\in {\mathbb{R}}^{n\times n}.$$

Definition 6.

Let $\mathbb{T}$ be a time scale and $A(·):\mathbb{T}\to {\mathbb{R}}^{n\times n}$ be Δ and ∇-differentiable on $\mathbb{T}$. For $t\in \mathbb{T}$ we define the diamond-α dynamic derivative $A^{{♢}_{\alpha }}\left(t\right)$ as

$$A^{{♢}_{\alpha }}\left(t\right)=\alpha A^{\Delta }\left(t\right)+(1-\alpha )A^{\nabla }\left(t\right),0\le \alpha \le 1.$$

Obviously, $A\left(t\right)$ is diamond-α differentiable if and only if $A\left(t\right)$ is Δ and ∇ differentiable.

Definition 7.

Let $A(·):\mathbb{T}\to {\mathbb{R}}^{n\times n}$, we define the generalized diamond matrix exponential function (i.e., α-matrix exponential function) by

$${\ddot{e}}_A(t,t_0)=\alpha e_A(t,t_0)+(1-\alpha ){\widehat{e}}_A(t,t_0),0\le \alpha \le 1.$$

Remark 8.

For the diamond α-dynamic Equation (11), if $\alpha =1$, then (11) turns into (2); if $\alpha =0$, then (11) turns into (3).

Theorem 10.

The generalized α-matrix exponential function ${\ddot{e}}_A(t,t_0)$ is the solution of (11).

Proof. 

Since

$$\begin{array}{ccc}\hfill {\left({\ddot{e}}_A(t,t_0)\right)}^{{♢}_{\alpha }}& =& \alpha \left(\alpha e_A^{\Delta }(t,t_0)+(1-\alpha ){\widehat{e}}_A^{\Delta }(t,t_0)\right)\hfill \\ & & +(1-\alpha )\left(\alpha e_A^{\nabla }(t,t_0)+(1-\alpha ){\widehat{e}}_A^{\nabla }(t,t_0)\right)\hfill \end{array}$$

by Lemma 3 and Lemma 6, we can obtain

$$\begin{array}{ccc}\hfill {\left({\ddot{e}}_A(t,t_0)\right)}^{{♢}_{\alpha }}& =& \alpha \left[\alpha A\left(t\right)e_A(t,t_0)+(1-\alpha )A\left(\sigma \left(t\right)\right){\left(I-\mu \left(t\right)A\left(\sigma \left(t\right)\right)\right)}^{-1}{\widehat{e}}_A(t,t_0)\right]\hfill \\ & & +(1-\alpha )\left[A\left(\rho \left(t\right)\right){\left(I+\nu \left(t\right)A\left(\rho \left(t\right)\right)\right)}^{-1}{e}_A(t,t_0)+(1-\alpha )A{\widehat{e}}_A(t,t_0)\right]\hfill \\ & =& \left[\alpha A\left(t\right)+(1-\alpha )A\left(\rho \left(t\right)\right){\left(I+\nu \left(t\right)A\left(\rho \left(t\right)\right)\right)}^{-1}\right]\alpha e_A(t,t_0)\hfill \\ & & +\left[\alpha A\left(\sigma \left(t\right)\right){\left(I-\mu \left(t\right)A\left(\sigma \left(t\right)\right)\right)}^{-1}+(1-\alpha )A\left(t\right)\right](1-\alpha ){\widehat{e}}_A(t,t_0).\hfill \end{array}$$

The proof is complete. □

Theorem 11.

Let $A(·):\mathbb{T}\to {\mathbb{R}}^{n\times n}$, then the diamond α-dynamic Equation (11) can be written as:

$$X^{{}^{{♢}_{\alpha }}}=A\left(t\right)X\left(t\right),0\le \alpha \le 1,$$

iff $A\left(t\right)$ satisfies

$$A\left(t\right)=e_{-A^{\sigma }}(t,t_0).$$

(12)

Proof. 

If $A\left(t\right)$ is solution of the equation $A\left(t\right)=e_{-A^{\sigma }}(t,t_0)$, then

$$\begin{array}{cc}& \alpha A\left(t\right)+(1-\alpha )A\left(\rho \left(t\right)\right){\left(I+\nu \left(t\right)A\left(\rho \left(t\right)\right)\right)}^{-1}\hfill \\ =& A\left(t\right)+(1-\alpha )\left(A\left(\rho \left(t\right)\right)-A\left(t\right)-\nu \left(t\right)A\left(t\right)A\left(\rho \left(t\right)\right)\right){\left(I+\nu \left(t\right)A\left(\rho \left(t\right)\right)\right)}^{-1}\hfill \\ =& A\left(t\right)-\nu \left(t\right)(1-\alpha )\left(A^{\nabla }\left(t\right)+A\left(t\right)A\left(\rho \left(t\right)\right)\right){\left(I+\nu \left(t\right)A\left(\rho \left(t\right)\right)\right)}^{-1},\hfill \\ & \alpha A\left(\sigma \left(t\right)\right){\left(I-\mu \left(t\right)A\left(\sigma \left(t\right)\right)\right)}^{-1}+(1-\alpha )A\left(t\right)\hfill \\ =& \alpha {(-\mu \left(t\right))}^{-1}(-\mu \left(t\right))A^{\sigma }\left(t\right){\left(I-\mu \left(t\right)A^{\sigma }\left(t\right)\right)}^{-1}+(1-\alpha )A\left(t\right)\hfill \\ =& \alpha {(-\mu \left(t\right))}^{-1}\left(-I+I-\mu \left(t\right)A^{\sigma }\left(t\right)\right){\left(I-\mu \left(t\right)A^{\sigma }\left(t\right)\right)}^{-1}+(1-\alpha )A\left(t\right)\hfill \\ =& \alpha {(-\mu \left(t\right))}^{-1}\left[-I{\left(I-\mu \left(t\right)A^{\sigma }\left(t\right)\right)}^{-1}+\left(I-\mu \left(t\right)A^{\sigma }\left(t\right)\right){\left(I-\mu \left(t\right)A^{\sigma }\left(t\right)\right)}^{-1}\right]\hfill \\ & +(1-\alpha )A\left(t\right)\hfill \\ =& \alpha {(-\mu \left(t\right))}^{-1}\left[{\left(I-\mu \left(t\right)A^{\sigma }\left(t\right)\right)}^{-1}(-I)+{\left(I-\mu \left(t\right)A^{\sigma }\left(t\right)\right)}^{-1}\left(I-\mu \left(t\right)A^{\sigma }\left(t\right)\right)\right]\hfill \\ & +(1-\alpha )A\left(t\right)\hfill \\ =& \alpha {(-\mu \left(t\right))}^{-1}{\left(I-\mu \left(t\right)A^{\sigma }\left(t\right)\right)}^{-1}(-\mu \left(t\right))A^{\sigma }\left(t\right)+(1-\alpha )A\left(t\right)\hfill \\ =& \alpha {\left(I-\mu \left(t\right)A^{\sigma }\left(t\right)\right)}^{-1}\left(A^{\sigma }\left(t\right)-A\left(t\right)+\mu \left(t\right)A^{\sigma }\left(t\right)A\left(t\right)\right)+A\left(t\right)\hfill \\ =& \alpha \mu \left(t\right){\left(I-\mu \left(t\right)A^{\sigma }\left(t\right)\right)}^{-1}\left(A^{\Delta }\left(t\right)+A^{\sigma }\left(t\right)A\left(t\right)\right)+A\left(t\right).\hfill \end{array}$$

On the other hand, for $A\left(t\right)=e_{-A^{\sigma }}(t,t_0)$, we can obtain

$$\begin{array}{ccc}\hfill A^{\Delta }\left(t\right)& =& -A^{\sigma }\left(t\right)e_{-A^{\sigma }}(t,t_0)=-A^{\sigma }\left(t\right)A\left(t\right),\hfill \\ \hfill A^{\nabla }\left(t\right)& =& -A^{\sigma }\left(\rho \left(t\right)\right)e_{-A^{\sigma }}(\rho \left(t\right),t_0)=-A\left(t\right)A^{\rho }\left(t\right),\hfill \end{array}$$

hence the dynamic Equation (11) can turn into:

$$X^{{}^{{♢}_{\alpha }}}=A\left(t\right)X\left(t\right),0\le \alpha \le 1.$$

If the diamond $\alpha$-dynamic Equation (11) can be given as:

$$X^{{}^{{♢}_{\alpha }}}=A\left(t\right)X\left(t\right),0\le \alpha \le 1,$$

then

$$\alpha A\left(t\right)+(1-\alpha )A^{\rho }\left(t\right){\left(I+\nu \left(t\right)A^{\rho }\left(t\right)\right)}^{-1}=\alpha A^{\sigma }\left(t\right){\left(I-\mu \left(t\right)A^{\sigma }\left(t\right)\right)}^{-1}+(1-\alpha )A\left(t\right),$$

hence

$$(1-\alpha )\nu \left(t\right)\left(A^{\nabla }\left(t\right)+A\left(t\right)A^{\rho }\left(t\right)\right){\left(I+\nu \left(t\right)A^{\rho }\left(t\right)\right)}^{-1}=\alpha \mu \left(t\right)\left(A^{\Delta }+A^{\sigma }\left(t\right)A\left(t\right)\right){\left(I-\mu \left(t\right)A^{\sigma }\left(t\right)\right)}^{-1},$$

for any $0\le \alpha \le 1$. Therefore

$$A^{\nabla }\left(t\right)+A\left(t\right)A^{\rho }\left(t\right)=A^{\Delta }+A^{\sigma }\left(t\right)A\left(t\right)=0,$$

thus $A\left(t\right)=e_{-A^{\sigma }}(t,t_0)$. The proof is complete. □

Example 9.

For (12), let $\mathbb{T}=\mathbb{Z}h$, $h>0$, by Theorem 11, then the solution of (12) can be given by

$$A\left(t\right)=\left\{\begin{array}{c}{\displaystyle \prod _{k=\frac{t-t_0}{h}-2}^0}\left(I-h{A}^{\sigma }(t_0+kh)\right)\left[I+h{\displaystyle \prod _{k=\frac{t-t_0}{h}-2}^0}\left(I-h{A}^{\sigma }(t_0+kh)\right)\right],t>t_0,\hfill \\ {\displaystyle \prod _{k=\frac{t-t_0}{h}+1}^0}{\left(I-hA(t_0+kh)\right)}^{-1},t<t_0.\hfill \end{array}\right.$$

In fact, by $A\left(t\right)=e_{-A^{\sigma }}(t,t_0)$, we can obtain

$$\begin{array}{ccc}\hfill A\left(t\right)& =& \prod _{k=\frac{t-t_0}{h}-1}^0\left(I-h{A}^{\sigma }(t_0+kh)\right)=\left(I-hA\left(t\right)\right)\prod _{k=\frac{t-t_0}{h}-2}^0\left(I-h{A}^{\sigma }(t_0+kh)\right),\hfill \end{array}$$

that is

$$A\left(t\right)=\prod _{k=\frac{t-t_0}{h}-2}^0\left(I-h{A}^{\sigma }(t_0+kh)\right)\left[I+h\prod _{k=\frac{t-t_0}{h}-2}^0\left(I-h{A}^{\sigma }(t_0+kh)\right)\right].$$

For $t\in \mathbb{T}$, $t<t_0$, we can obtain

$$\begin{array}{ccc}\hfill A\left(t\right)& =& \prod _{k=\frac{t-t_0}{h}+1}^0{\left(I-h{A}^{\sigma }(t_0+(k-1)h)\right)}^{-1}=\prod _{k=\frac{t-t_0}{h}+1}^0{\left(1-hA(t_0+kh)\right)}^{-1},\hfill \end{array}$$

for any $0\le k\le \frac{t-t_0}{h}+1$, $t_0+kh>t$. Therefore

$$A\left(t\right)=\prod _{k=\frac{t-t_0}{h}+1}^0{\left(I-hA(t_0+kh)\right)}^{-1}.$$

Example 10.

For (12), let $\mathbb{T}=\overline{q^{\mathbb{Z}}}$, $q>1$, by Theorem 11, then the solution of (12) can be given by

$$A\left(t\right)=\left\{\begin{array}{c}{\displaystyle \prod _{k=\frac{lnt-ln{t}_0}{lnq}-2}^0}\left(I-(q-1)t_0{q}^k{A}^{\sigma }\left(t_0{q}^k\right)\right)\hfill \\ \times {\left[I+(1-\frac{1}{q})t{\displaystyle \prod _{k=\frac{lnt-ln{t}_0}{lnq}-2}^0}\left(I-(q-1)t_0{q}^k{A}^{\sigma }\left(t_0{q}^k\right)\right)\right]}^{-1},t>t_0,\hfill \\ {\displaystyle \prod _{k=\frac{lnt-ln{t}_0}{lnq}+1}^0}{\left(I-(q-1)t_0{q}^{k-1}A\left(t_0{q}^k\right)\right)}^{-1},t<t_0.\hfill \end{array}\right.$$

In fact, by $A\left(t\right)=e_{-A^{\sigma }}(t,t_0)$, we can obtain

$$\begin{array}{ccc}\hfill A\left(t\right)& =& \prod _{k=\frac{lnt-ln{t}_0}{lnq}-1}^0\left(I-(q-1)t_0{q}^k{A}^{\sigma }\left(t_0{q}^k\right)\right)\hfill \\ & =& \left(I-(1-\frac{1}{q})tA\left(t\right)\right)\prod _{k=\frac{lnt-ln{t}_0}{lnq}-2}^0\left(I-(q-1)t_0{q}^k{A}^{\sigma }\left(t_0{q}^k\right)\right)\hfill \end{array}$$

that is

$$A\left(t\right)=\prod _{k=\frac{lnt-ln{t}_0}{lnq}-2}^0\left(I-(q-1)t_0{q}^k{A}^{\sigma }\left(t_0{q}^k\right)\right){\left[I+(1-\frac{1}{q})t\prod _{k=\frac{lnt-ln{t}_0}{lnq}-2}^0\left(I-(q-1)t_0{q}^k{A}^{\sigma }\left(t_0{q}^k\right)\right)\right]}^{-1}.$$

For $t\in \mathbb{T}$, $t<t_0$, we can obtain

$$\begin{array}{ccc}\hfill A\left(t\right)& =& \prod _{k=\frac{lnt-ln{t}_0}{lnq}+1}^0{\left(I-(q-1)t_0{q}^{k-1}{A}^{\sigma }\left(t_0{q}^{k-1}\right)\right)}^{-1}\hfill \\ & =& \prod _{k=\frac{lnt-ln{t}_0}{lnq}+1}^0{\left(I-(q-1)t_0{q}^{k-1}A\left(t_0{q}^k\right)\right)}^{-1},\hfill \end{array}$$

for any $0\le k\le \frac{lnt-ln{t}_0}{lnq}+1$, $t_0{q}^k>t$. Therefore

$$A\left(t\right)=\prod _{k=\frac{lnt-ln{t}_0}{lnq}+1}^0{\left(I-(q-1)t_0{q}^{k-1}A\left(t_0{q}^k\right)\right)}^{-1}.$$

Theorem 12

(Liouville’s formula). Let $A(·):\mathbb{T}\to {\mathbb{R}}^{n\times n}$, X is a solution of (11), then X satisfies Liouville’s formula

$$detX\left(t\right)=det\left(\alpha X_1+(1-a)X_2\right)det{X}_0,fort\in \mathbb{T},$$

where $X_1$ is solution of $X^{\Delta }=A\left(t\right)X\left(t\right)$, $X_2$ is solution of $X^{\nabla }=A\left(t\right)X\left(t\right)$.

Proof. 

By Theorem 10, we can obtain $\alpha X_1+(1-a)X_2$ is a solution of (11), hence

$$detX\left(t\right)=det\left(\alpha X_1+(1-a)X_2\right)det{X}_0,fort\in \mathbb{T}.$$

The proof is complete. □

Example 11.

For (11), let $\mathbb{T}=\mathbb{R}$, any $t\in \mathbb{T}$ are dense points, according to Theorem 12, we have (11) can be written as

$$X^{{}^{{♢}_{\alpha }}}=X^{\prime }=A\left(t\right)X\left(t\right),0\le \alpha \le 1.$$

Furthermore, Liouville’s formula can be given by

$$detX=det{X}_0{e}^{{\int }_{t_0}^{t}trA\left(\tau \right)\mathrm{d}ø},trA\left(\tau \right)=\sum _{i=1}^n{a}_{ii}\left(\tau \right).$$

In fact, $\mu \left(t\right)=\nu \left(t\right)=0$ for any $t\in \mathbb{T}$, hence $X^{\Delta }\left(t\right)=X^{\nabla }\left(t\right)=X^{\prime }$, therefore $X^{{}^{{♢}_{\alpha }}}=X^{\prime }$ for any $0\le \alpha \le 1$. By Examples 1 and 4 we can obtain desired results.

Example 12.

For (11), let $\mathbb{T}=\mathbb{Z}h$, $h>0$, by Theorem 12, the solution of (11) can be given as:

$$X\left(t\right)=\left\{\alpha \prod _{k=\frac{t-t_0}{h}-1}^0\left[I+hA(t_0+kh)\right]+(1-\alpha )\prod _{k=\frac{t-t_0}{h}}^1{\left[I-hA(t_0+kh)\right]}^{-1}\right\}{X}_0,0\le \alpha \le 1.$$

In fact, by Theorem 10, we can obtain $X\left(t\right)=\alpha e_A(t,t_0)+(1-\alpha ){\widehat{e}}_A(t,t_0)$. On the other hand, by Examples 2 and 5, we can obtain

$$e_A(t,t_0)=\prod _{k=\frac{t-t_0}{h}-1}^0[I+hA(t_0+kh),{\widehat{e}}_A(t,t_0)=\prod _{k=\frac{t-t_0}{h}}^1{\left[I-hA(t_0+kh)\right]}^{-1}.$$

Example 13.

For (11), let $\mathbb{T}=q^{\mathbb{Z}}$, $q>1$, by Theorem 12, the solution of (11) can be given as:

$$X\left(t\right)=\left\{\alpha \prod _{k=\frac{lnt-ln{t}_0}{lnq}-1}^0\left[I+(q-1)q^k{t}_{0}A\left(q^k{t}_0\right)\right]+(1-\alpha )\prod _{k=\frac{lnt-ln{t}_0}{lnq}}^1{\left[I-(q-1)q^k{t}_{0}A\left(q^k{t}_0\right)\right]}^{-1}\right\}{X}_0,$$

where $0\le \alpha \le 1$. In fact, by Theorem 10, we can obtain $X\left(t\right)=\alpha e_A(t,t_0)+(1-\alpha ){\widehat{e}}_A(t,t_0)$. On the other hand, by Examples 3 and 6, we can obtain

$$e_A(t,t_0)=\prod _{k=\frac{lnt-ln{t}_0}{lnq}-1}^0\left[I+(q-1)q^k{t}_{0}A\left(q^k{t}_0\right)\right],{\widehat{e}}_A(t,t_0)=\prod _{k=\frac{lnt-ln{t}_0}{lnq}}^1{\left[I-(q-1)q^k{t}_{0}A\left(q^k{t}_0\right)\right]}^{-1}.$$