Representation of Solutions to a Two-Sided Matrix Delay Differential Equations

Abstract

In this paper, we investigate the representation of solution for the following linear matrix delayed differential equation $\dot{Y}\left(t\right)=A_{1}Y\left(t\right)+Y\left(t\right)A_2+BY(t-\tau )+Y(t-\tau )C+F\left(t\right),t\in [0,\infty )$ where t is an independent variable, $Y\left(t\right)$ is an $n\times n$ unknown variable matrix, $\tau >0$ is a delay, $A_1,A_2,B$, and C are given $n\times n$ constant matrices, and $F\left(t\right)$ is a given $n\times n$ variable matrix. Without requiring any commutativity condition between the coefficient matrices and $\mathrm{\Phi }$ and F, we establish a formula for the initial problem $Y\left(t\right)=\mathrm{\Phi }\left(t\right)$, $t\in [-\tau ,0]$, where $\mathrm{\Phi }\left(t\right)$ is an $n\times n$ variable matrix. The proof uses the vectorization technique and the method of steps. Our result settles Open Problems 1 and 2 posed by Diblík.

1. Introduction

Delay differential equations (DDEs) provide a powerful mathematical framework for modeling phenomena with historical dependence or time lags, finding broad application in control theory, neural networks, and ecological dynamics. For current research on this topic, consult [1,2,3,4,5,6].

In the theory of DDEs, the method of steps is a fundamental technique for solving initial value problems (for instance, see [7,8,9]). Considerable recent effort has been devoted to deriving explicit solution formulas for linear delay equations. Notably, Khusainov and Shuklin [10] introduced the new concept of delayed exponential matrix functions, which enabled the derivation of exact solution expressions for linear autonomous time-delay systems. In 2008, Khusainov et al. [11] further extended this approach by establishing delayed matrix sine and cosine functions, providing a solution to the Cauchy problem for second-order oscillatory systems with pure delay. Li and Wang [12] obtained a representation of a solution to the initial value problem for a linear fractional delay differential equation with Riemann–Liouville derivative via a delayed Mittag-Leffler type matrix function. Mahmudov [13] developed solution representations for linear nonhomogeneous fractional delay differential equations by employing delayed perturbation of Mittag-Leffler type matrix function. In the discrete case, Diblík derived the representations of solutions for linear discrete equations with first-order differences [14,15] and with second-order differences [16,17]. For further references on this topic, see [18,19,20,21] and the literature cited therein.

Research on the application of matrix delay differential equations not only advances the development of stability analysis, control theory, and numerical computation methods, but also provides a key tool for understanding complex systems such as ecosystem balance, gene regulation, and market evolution. This underscores its role as a bridge connecting mathematical theory with the real world. For the latest advancements in areas such as the finite-time stability of differential equations, the asymptotic stability of nonlinear multidelay differential equations, and the relative controllability of pure time-delay systems, one may refer to the existing literature [22,23,24,25,26] and the references therein.

Recently, Diblík [27] obtained the representation of solutions to a linear matrix first-order delay differential equation

$$\dot{X}\left(t\right)=BX(t-\tau )+X(t-\tau )C+F\left(t\right),t\in [0,\infty ),$$

(1)

with an initial condition

$$X\left(t\right)=\mathrm{\Psi }\left(t\right),t\in [-\tau ,0],$$

(2)

where t is the independent variable, $\tau$ is a positive constant delay, X is an $n\times n$ unknown real dependent variable matrix, and B, C are given $n\times n$ constant real matrices, F is a continuous $n\times n$ real matrix on $[0,\infty )$, and $\mathrm{\Psi }$ is a continuously differentiable $n\times n$ real matrix, defined on $[-\tau ,0]$.

In the concluding section, Diblík [27] raised two open questions as follows:

• Open Problem 1. Is it possible to do without the commutativity assumptions needed in Theorems 3–5?
• Open Problem 2. Consider a matrix equation, more general than Equation (1),

$$\dot{X}\left(t\right)=A_{1}X\left(t\right)+X\left(t\right)A_2+\sum _{i=1}^l{B}_{i}X(t-{\tau }_i)+X(t-{\tau }_i)C_i+F\left(t\right),t\in [0,\infty ),$$

(3)

where l is a positive integer, with constant matrices $A_1,A_2,B_i,C_i$. Is it possible to find formulas for representing the solutions of initial problems for the cases all matrices being pairwise permutable or even, in the general case, not assuming their permutability?

This paper addresses and resolves these two Open Problems. For the sake of brevity and clarity, we focus exclusively on the case where $l=1$. The general case for arbitrary l can be analyzed but is more complex; hence, it is omitted here due to the similar line of reasoning. Accordingly, we consider the following linear matrix first-order delay differential equation:

$$\dot{Y}\left(t\right)=A_{1}Y\left(t\right)+Y\left(t\right)A_2+BY(t-\tau )+Y(t-\tau )C+F\left(t\right),t\in [0,\infty )$$

(4)

with an initial condition

$$Y\left(t\right)=\mathrm{\Phi }\left(t\right),t\in [-\tau ,0],$$

(5)

where t is the independent variable, $\tau$ is a positive constant delay, X is an $n\times n$ unknown real dependent variable matrix, $A_1,A_2,B$, and C are given $n\times n$ constant real matrices, F is a continuous $n\times n$ real matrix on $[0,\infty )$, and $\mathrm{\Phi }$ is a continuous $n\times n$ real matrix on $[-\tau ,0]$. Equations (4) and (5) have a unique solution $Y=Y\left(t\right)$ in the sense that $Y\left(t\right)$ is continuous on $[-\tau ,\infty )$ and continuously differentiable on $[0,\infty )$ (as customary, we set $\dot{Y}\left(0\right)=\dot{Y}(0+)$). It is worth pointing out that a main tool in our study is matrices and their operations, particularly matrix multiplication. There exist a variety of fast algorithms for matrix multiplication, and a clear, accessible, and comprehensive overview is provided in the survey article by Respondek in [28].

In this paper, we employ the vectorization operator to derive an explicit solution formula for the linear matrix delay differential Equation (4) with initial condition (5). This solves Open Problem 2. Furthermore, our derivation does not impose commutativity conditions between the coefficient matrices and $\mathrm{\Phi }$ and F, thereby also solving Open Problem 1.

The remainder of this paper is organized as follows. In Section 2, we introduce the necessary definitions and preliminaries, and derive the solution to Equations (4) and (5). This solution constitutes the main result of the paper. Section 3 provides an illustrative example to demonstrate the theoretical results. The paper concludes in Section 4.

2. Main Result

In this section, using the vectorization method and the method of steps, we study the explicit expression for the solution of problem (4) with a general initial matrix. We begin with the key concepts and formulas of the vectorization method.

Let $A=[a_1,a_2,\cdots ,a_n]\in {\mathbb{R}}^{m\times n}$, where $a_i\in {\mathbb{R}}^m$ is the ith column of the matrix A, $i=1,2,\cdots ,n$, and the vec operator of A is defined to be $\mathrm{vec}\left(A\right)={[a_1^T,a_2^T,\cdots ,a_n^T]}^T\in {\mathbb{R}}^{mn\times 1}$. For $A=\left(a_{ij}\right)\in {\mathbb{R}}^{m\times n}$, $B=\left(b_{ij}\right)\in {\mathbb{R}}^{p\times q}$, the symbol $A\otimes B=\left(a_{ij}B\right)\in {\mathbb{R}}^{mp\times nq}$ stands for the Kronecker product of A and B.

For an $n\times n$ matrix A, the matrix exponential is defined by the power series:

$$e^A=exp\left(A\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{A^k}{k!}}=E+A+{\displaystyle \frac{A^2}{2!}}+{\displaystyle \frac{A^3}{3!}}+\cdots$$

Regarding the linear equations and matrix-vector operators over the real number field, there are the following well-known Kronecker formulations

Lemma 1

([29,30]). Let $A\in {\mathbb{R}}^{m\times n}$, $B\in {\mathbb{R}}^{n\times s}$, and $C\in {\mathbb{R}}^{s\times t}$. Then

$$\mathrm{vec}\left(ABC\right)=(C^T\otimes A)\mathrm{vec}\left(B\right).$$

For convenience, let $y\left(t\right)=vec\left(Y\right(t\left)\right)$, and

$$M=E\otimes A_1+A_2^T\otimes E,N=E\otimes B+C^T\otimes E,$$

(6)

where E is $n\times n$ unit matrix.

By utilizing the operator vec, and the method of steps, a solution to Equations (4) and (5) is formulated as the following theorem.

Theorem 1.

Let $A_1,A_2,B$, C be $n\times n$ constant real matrices. Assume that the matrixs $\mathrm{\Phi }\left(t\right)$ and $F\left(t\right)$ are continuous on $[-\tau ,0]$ and $[0,\infty )$, respectively. Then, $Y\left(t\right)={vec}^{-1}\left(y\left(t\right)\right)$ solves Equations (4) and (5), where

$$y\left(t\right):=\left\{\begin{array}{cc}vec\left(\mathrm{\Phi }\right(t\left)\right),\hfill & \mathrm{if}t\in [-\tau ,0],\hfill \\ {\displaystyle e^{Mt}vec\left(\mathrm{\Phi }\left(0\right)\right)+{\int }_0^t{e}^{M(t-s)}[Nvec\left(\mathrm{\Phi }(s-\tau )\right)+vec\left(F\left(s\right)\right)]ds,}\hfill & \mathrm{if}t\in (0,\tau ],\hfill \\ {\displaystyle e^{M(t-\tau )}y\left(\tau \right)+{\int }_{\tau }^t{e}^{M(t-s)}[Ny(s-\tau )+vec\left(F\left(s\right)\right)]ds,}\hfill & \mathrm{if}t\in (\tau ,2\tau ],\hfill \\ ⋮\hfill \\ {\displaystyle e^{M(t-k\tau )}y\left(k\tau \right)+{\int }_{k\tau }^t{e}^{M(t-s)}[Ny(s-\tau )+vec\left(F\left(s\right)\right)]ds,}\hfill & \mathrm{if}t\in (k\tau ,(k+1\left)\tau \right],\hfill \\ ⋮\hfill \end{array}\right.$$

(7)

Proof. 

Since the Equation (4) involves a delay $\tau$, we solve it step-by-step on intervals of length $\tau$. For convenience, the proof is divided into three steps.

Step 1. For $t\in (0,\tau ]$. In this interval, $t-\tau \in (-\tau ,0]$, so $Y(t-\tau )=\mathrm{\Phi }(t-\tau )$. The Equation (4) becomes

$$\dot{Y}\left(t\right)=A_{1}Y\left(t\right)+Y\left(t\right)A_2+B\mathrm{\Phi }(t-\tau )+\mathrm{\Phi }(t-\tau )C+F\left(t\right).$$

(8)

This is a linear matrix differential equation with nonhomogeneous term. Apply the vectorization operation to Equation (8), we have

$${\displaystyle \frac{d}{dt}}\mathrm{vec}\left(Y\left(t\right)\right)=M\mathrm{vec}\left(Y\left(t\right)\right)+N\mathrm{vec}\left(\mathrm{\Phi }(t-\tau )\right)+\mathrm{vec}\left(F\left(t\right)\right),t\ge 0,$$

that is

$${\displaystyle \frac{d}{dt}}y\left(t\right)=My\left(t\right)+N\mathrm{vec}\left(\mathrm{\Phi }(t-\tau )\right)+\mathrm{vec}\left(F\left(t\right)\right),t\ge 0,$$

(9)

where M and N are as in (6). Clearly, $y\left(0\right)=\mathrm{vec}\left(\mathrm{\Phi }\right(0\left)\right)$. Then for the non-homogeneous Equation (9), using the method of variation of parameters yields

$$y\left(t\right)=e^{Mt}\mathrm{vec}\left(\mathrm{\Phi }\left(0\right)\right)+{\int }_0^t{e}^{M(t-s)}[N\mathrm{vec}\left(\mathrm{\Phi }(s-\tau )\right)+\mathrm{vec}\left(F\left(s\right)\right)]ds,t\in (0,\tau ].$$

(10)

Step 2. For $t\in (\tau ,2\tau ]$. Apply the vectorization operation to Equation (4), we deduce that

$${\displaystyle \frac{d}{dt}}\mathrm{vec}\left(Y\left(t\right)\right)=M\mathrm{vec}\left(Y\left(t\right)\right)+N\mathrm{vec}\left(Y(t-\tau )\right)+\mathrm{vec}\left(F\left(t\right)\right),$$

(11)

where M and N are as in (6). Since $t-\tau \in (0,\tau ]$, $y(t-\tau )$ is known from the previous step (11), that is

$$y(t-\tau )=e^{M(t-\tau )}\mathrm{vec}\left(\mathrm{\Phi }\left(0\right)\right)+{\int }_0^{t-\tau }{e}^{M(t-\tau -s)}[N\mathrm{vec}\left(\mathrm{\Phi }(s-\tau )\right)+\mathrm{vec}\left(F\left(s\right)\right)]ds,t\in (\tau ,2\tau ],$$

(12)

which implies that $\mathrm{vec}\left(Y\right(t-\tau \left)\right)$ independent of Y for $t\in (\tau ,2\tau ]$. Thus, from (11), we obtain by using the method of variation of parameters that

$$y\left(t\right)=e^{M(t-\tau )}y\left(\tau \right)+{\int }_{\tau }^t{e}^{M(t-s)}[Ny(s-\tau )+\mathrm{vec}\left(F\left(s\right)\right)]ds,t\in (\tau ,2\tau ].$$

(13)

From (12), we can further explicitly express y on $[\tau ,2\tau )$ as follows:

$$\begin{array}{cc}y\left(t\right)=\hfill & {\displaystyle e^{Mt}\mathrm{vec}\left(\mathrm{\Phi }\left(0\right)\right)+{\int }_0^{\tau }{e}^{M(t-s)}[N\mathrm{vec}\left(\mathrm{\Phi }(s-\tau )\right)+\mathrm{vec}\left(F\left(s\right)\right)]ds}\hfill \\ \hfill & {\displaystyle +{\int }_{\tau }^t{e}^{M(t-s)}[N{e}^{M(s-\tau )}\mathrm{vec}\left(\mathrm{\Phi }\left(0\right)\right)+\mathrm{vec}\left(F\left(s\right)\right)]ds}\hfill \\ \hfill & {\displaystyle +{\int }_{\tau }^t{e}^{M(t-s)}{\int }_0^{s-\tau }N{e}^{M(s-\tau -\xi )}[N\mathrm{vec}\left(\mathrm{\Phi }(\xi -\tau )\right)+\mathrm{vec}\left(F\left(\xi \right)\right)]d\xi ds,t\in (\tau ,2\tau ].}\hfill \end{array}$$

(14)

Step 3. General Pattern. This process can be continuous recursively. For $t\in (k\tau ,(k+1\left)\tau \right]$ ($k\ge 2$), the solution of Equation (4) is

$$y\left(t\right)=e^{M(t-k\tau )}y\left(k\tau \right)+{\int }_{k\tau }^t{e}^{M(t-s)}[Ny(s-\tau )+\mathrm{vec}\left(F\left(s\right)\right)]ds.$$

(15)

Here, $y\left(k\tau \right)$ is known from the previous interval, and for $s\in [k\tau ,t]$, $s-\tau \in \left(\right(k-1)\tau ,k\tau ]$, so $y(s-\tau )$ is also known from earlier steps. The proof is complete. □

Remark 1.

Using Lemma 1, we get

$$Nvec\left(E\right)=(E\otimes B)vec\left(E\right)+(C^T\otimes E)vec\left(E\right)=vec(B+C),$$

(16)

and

$$\begin{array}{cc}\hfill & {\displaystyle Nvec(B+C)=Nvec\left(B\right)+Nvec\left(C\right)}\hfill \\ \hfill & {\displaystyle =(E\otimes B)vec\left(B\right)+(C^T\otimes E)vec\left(B\right)+(E\otimes B)vec\left(C\right)+(C^T\otimes E)vec\left(C\right)}\hfill \\ \hfill & {\displaystyle =vec(B^2+2BC+C^2).}\hfill \end{array}$$

(17)

Let $\mathrm{\Phi }\left(t\right)=E$ and $F\left(t\right)=\mathrm{\Theta }$. If $A_1=A_2=\mathrm{\Theta }$ ($\mathrm{\Theta }={\mathrm{\Theta }}_{n\times n}$ is the $n\times n$ zero matrix), then $M={\mathrm{\Theta }}_{n^2\times n^2}$ (${\mathrm{\Theta }}_{n^2\times n^2}$ is the $n^2\times n^2$ zero matrix). Thus, (10) and (14) are reduce to

$$y\left(t\right)=vec\left(E\right)+{\int }_0^{t}Nvec\left(E\right)ds,t\in (0,\tau ],$$

(18)

and

$$\begin{array}{cc}\hfill & {\displaystyle y\left(t\right)=vec\left(E\right)+{\int }_0^{\tau }vec(B+C)ds+{\int }_{\tau }^{t}Nvec\left(E\right)ds}\hfill \\ \hfill & {\displaystyle +{\int }_{\tau }^t{\int }_0^{s-\tau }Nvec(B+C)d\xi ds,t\in (\tau ,2\tau ].}\hfill \end{array}$$

(19)

From (16) and (18), we have

$$\begin{array}{cc}\hfill & {\displaystyle vec\left(Y\left(t\right)\right)=vec\left(E\right)+{\int }_0^{t}vec(B+C)ds=vec\left(E\right)+tvec(B+C)}\hfill \\ \hfill & {\displaystyle =vec(E+t(B+C\left)\right),t\in (0,\tau ],}\hfill \end{array}$$

which implies that

$$Y\left(t\right)=E+t(B+C),t\in (0,\tau ].$$

For $t\in (\tau ,2\tau ]$, from (16), (17) and (19), we obtain

$$\begin{array}{cc}\hfill & {\displaystyle vec\left(Y\right(t\left)\right)=vec\left(E\right)+\tau vec(B+C)+(t-\tau )vec(B+C)}\hfill \\ \hfill & {\displaystyle +\frac{{(t-\tau )}^2}{2}vec(B^2+2BC+C^2)}\hfill \\ \hfill & {\displaystyle =vec\left(E+\frac{t}{1!}(B+C)+\frac{{(t-\tau )}^2}{2!}(B^2+2BC+C^2)\right),}\hfill \end{array}$$

that is

$$Y\left(t\right)=E+{\displaystyle \frac{t}{1!}}(B+C)+{\displaystyle \frac{{(t-\tau )}^2}{2!}}(B^2+2BC+C^2),t\in (\tau ,2\tau ].$$

Using induction, for $t\in \left(\right(k-1)\tau ,k\tau ]$, we can demonstrate that

$$\begin{array}{cc}\hfill & {\displaystyle Y\left(t\right)=E+\frac{t}{1!}(B+C)+\frac{{(t-\tau )}^2}{2!}(B^2+2BC+C^2)}\hfill \\ \hfill & {\displaystyle +\cdots +\frac{{(t-(k-1)\tau )}^k}{k!}\sum _{s=0}^k\left(\begin{array}{c}k\\ s\end{array}\right)B^{k-s}{C}^s.}\hfill \end{array}$$

Thus, if $\mathrm{\Phi }\left(t\right)=E$, $F\left(t\right)=\mathrm{\Theta }$ and $A_1=A_2=\mathrm{\Theta }$, then (7) reduces to Equation (4) in [27].

Remark 2.

Our main result (Theorem 1) establishes a solution formula for the initial value problems (4) and (5). This formula is applicable to arbitrary matrices without any permutability assumptions, thus resolving the two open problems posed by Diblík [27].

3. An Example

In this section, a numerical example is presented to verify the theoretical result.

Example 1.

Let $n=2$, $\tau =1$ and let (4) and (5) be reduced to

$$\begin{array}{cc}\hfill & {\displaystyle \dot{Y}\left(t\right)=A_{1}Y\left(t\right)+Y\left(t\right)A_2+BY(t-1)+Y(t-1)C+F\left(t\right),t\in [0,\infty ),}\hfill \\ \hfill & {\displaystyle Y\left(t\right)=\mathrm{\Phi }\left(t\right),t\in [-1,0],}\hfill \end{array}$$

(20)

i.e., the matrices in (20) are specified as follows:

$$A_1=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),A_2=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),B=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),C=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right),$$

$$F\left(t\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),\mathrm{\Phi }\left(t\right)=\left(\begin{array}{cc}t& 0\\ 0& t\end{array}\right).$$

Let us compute the matrix solution $Y\left(t\right)$ of (20) on interval $(0,3]$. Since

$$M=E\otimes A_1+A_2^T\otimes E=\left(\begin{array}{cccc}0& 1& 1& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 1\\ 0& 0& 0& 0\end{array}\right),$$

and

$$N=E\otimes B+C^T\otimes E=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 2& 0\\ 0& 0& 0& 1\end{array}\right),$$

from (7), we can derive

$$vec\left(Y\left(t\right)\right)=\left(\begin{array}{c}{\displaystyle \frac{1}{12}}{t}^4+{\displaystyle \frac{1}{2}}{t}^2\\ {\displaystyle \frac{1}{6}}{t}^3\\ {\displaystyle \frac{1}{6}}{t}^3\\ {\displaystyle \frac{1}{2}}{t}^2\end{array}\right),t\in (0,1],$$

$$vec\left(Y\left(t\right)\right)=\left(\begin{array}{c}{\displaystyle \frac{1}{20}}{t}^5-{\displaystyle \frac{1}{4}}{t}^4+t^3-{\displaystyle \frac{3}{2}}{t}^2+{\displaystyle \frac{25}{12}}t-{\displaystyle \frac{4}{5}}\\ {\displaystyle \frac{1}{24}}{t}^4-{\displaystyle \frac{1}{6}}{t}^3+{\displaystyle \frac{3}{4}}{t}^2-{\displaystyle \frac{2}{3}}t+{\displaystyle \frac{5}{24}}\\ {\displaystyle \frac{1}{8}}{t}^4-{\displaystyle \frac{1}{2}}{t}^3+{\displaystyle \frac{5}{4}}{t}^2-t+{\displaystyle \frac{7}{24}}\\ {\displaystyle \frac{1}{6}}{t}^3-{\displaystyle \frac{1}{2}}{t}^2+{\displaystyle \frac{3}{2}}t-{\displaystyle \frac{2}{3}}\end{array}\right),t\in (1,2],$$

$$vec\left(Y\left(t\right)\right)=\left(\begin{array}{c}{\displaystyle \frac{7}{360}}{t}^6-{\displaystyle \frac{7}{30}}{t}^5+{\displaystyle \frac{35}{24}}{t}^4-{\displaystyle \frac{83}{18}}{t}^3+{\displaystyle \frac{55}{6}}{t}^2-{\displaystyle \frac{539}{60}}t+{\displaystyle \frac{182}{45}}\\ {\displaystyle \frac{1}{120}}{t}^5-{\displaystyle \frac{1}{12}}{t}^4+{\displaystyle \frac{1}{2}}{t}^3-{\displaystyle \frac{11}{12}}{t}^2+{\displaystyle \frac{4}{3}}t-{\displaystyle \frac{29}{40}}\\ {\displaystyle \frac{7}{120}}{t}^5-{\displaystyle \frac{7}{12}}{t}^4+{\displaystyle \frac{17}{6}}{t}^3-{\displaystyle \frac{77}{12}}{t}^2+{\displaystyle \frac{23}{3}}t-{\displaystyle \frac{143}{40}}\\ {\displaystyle \frac{1}{24}}{t}^4-{\displaystyle \frac{1}{3}}{t}^3+{\displaystyle \frac{3}{2}}{t}^2-{\displaystyle \frac{11}{6}}t+{\displaystyle \frac{4}{3}}\end{array}\right),t\in (2,3].$$

Thus, we get

$$Y\left(t\right)=\left(\begin{array}{cc}{\displaystyle \frac{1}{12}}{t}^4+{\displaystyle \frac{1}{2}}{t}^2& {\displaystyle \frac{1}{6}}{t}^3\\ {\displaystyle \frac{1}{6}}{t}^3& {\displaystyle \frac{1}{2}}{t}^2\end{array}\right),t\in (0,1],$$

$$Y\left(t\right)=\left(\begin{array}{cc}{\displaystyle \frac{1}{20}}{t}^5-{\displaystyle \frac{1}{4}}{t}^4+t^3-{\displaystyle \frac{3}{2}}{t}^2+{\displaystyle \frac{25}{12}}t-{\displaystyle \frac{4}{5}}& {\displaystyle \frac{1}{8}}{t}^4-{\displaystyle \frac{1}{2}}{t}^3+{\displaystyle \frac{5}{4}}{t}^2-t+{\displaystyle \frac{7}{24}}\\ {\displaystyle \frac{1}{24}}{t}^4-{\displaystyle \frac{1}{6}}{t}^3+{\displaystyle \frac{3}{4}}{t}^2-{\displaystyle \frac{2}{3}}t+{\displaystyle \frac{5}{24}}& {\displaystyle \frac{1}{6}}{t}^3-{\displaystyle \frac{1}{2}}{t}^2+{\displaystyle \frac{3}{2}}t-{\displaystyle \frac{2}{3}}\end{array}\right),t\in (1,2],$$

$$Y\left(t\right)=\left(\begin{array}{cc}{\displaystyle \frac{7}{360}}{t}^6-{\displaystyle \frac{7}{30}}{t}^5+{\displaystyle \frac{35}{24}}{t}^4-{\displaystyle \frac{83}{18}}{t}^3+{\displaystyle \frac{55}{6}}{t}^2-{\displaystyle \frac{539}{60}}t+{\displaystyle \frac{182}{45}}& {\displaystyle \frac{7}{120}}{t}^5-{\displaystyle \frac{7}{12}}{t}^4+{\displaystyle \frac{17}{6}}{t}^3-{\displaystyle \frac{77}{12}}{t}^2+{\displaystyle \frac{23}{3}}t-{\displaystyle \frac{143}{40}}\\ {\displaystyle \frac{1}{120}}{t}^5-{\displaystyle \frac{1}{12}}{t}^4+{\displaystyle \frac{1}{2}}{t}^3-{\displaystyle \frac{11}{12}}{t}^2+{\displaystyle \frac{4}{3}}t-{\displaystyle \frac{29}{40}}& {\displaystyle \frac{1}{24}}{t}^4-{\displaystyle \frac{1}{3}}{t}^3+{\displaystyle \frac{3}{2}}{t}^2-{\displaystyle \frac{11}{6}}t+{\displaystyle \frac{4}{3}}\end{array}\right),t\in (2,3].$$

It is easy to verify that the solution obtained above satisfies Equation (20).

4. Conclusions

The main result of this work is presented in Theorem 1, which provides a solution to Equations (4) and (5) using Formula (7). The derivation does not require matrix commutativity. This paper overcomes the limitations of existing techniques through a vectorization method. It transforms the two-sided matrix delay differential Equation (4) into a one-sided matrix delay differential equation-essentially the standard form and subsequently applies the method of steps, thereby solving the two open problems posed by Diblk [27]. The effectiveness of the proposed criterion is demonstrated through a numerical example.

Future work will focus on deriving solution representations for discrete matrix equations and linear delayed fractional matrix differential equations.