Quaternion-Valued Linear Impulsive Differential Equations

Abstract

In this paper, we transfer the approach of finding the explicit solutions to linear impulsive differential equations (LIDEs) to quaternion-valued case. We develop the similar idea to find the solutions to LIDEs and give the details to derive the representations of the solutions to quaternion-valued LIDEs in the sense of complex numbers and quaternion. In addition, we present that two formulas are equivalent to each other. Finally, some examples are given to illustrate the theory approach.

Appendices

Appendix 1. Proof of Theorem 3.3

Proof

According to Lemma 2.3, for $t\in [0,t_{1}],$ since $x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)$ are linear independent solutions of $\dot{x}(t)=Ax(t),$ then each solution x(t) in the interval $[0,t_{1}]$ has the form

$$\begin{aligned} x(t) =x_{1}(t)c_{1}+x_{2}(t)c_{2}+\ldots +x_{n}(t)c_{n} =(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) ,\qquad \end{aligned}$$

(34)

where $c_{1},c_{2},\ldots ,c_{n}$ are arbitrary quaternions.

From (34), we can obtain

$$\begin{aligned} x(t_{1}^{-}) =x(t_{1}) =(x_{1}(t_{1}),x_{2}(t_{1}),\ldots ,x_{n}(t_{1})) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) . \end{aligned}$$

(35)

For $t\in (t_{1},t_{2}],$ each solution x(t) in the interval $(t_{1},t_{2}]$ has the form

$$\begin{aligned} x(t)= & {} x_{1}(t)a_{21}+x_{2}(t)a_{22}+\ldots +x_{n}(t)a_{2n} =(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \nonumber \\&\left( \begin{array}{cccccc} a_{21} \\ a_{22} \\ \vdots \\ a_{2n} \\ \end{array} \right) , \end{aligned}$$

(36)

where $a_{21},a_{22},\ldots ,a_{2n}$ are arbitrary quaternions.

From (36), we can obtain

$$\begin{aligned} x(t_{1}^{+}) =(x_{1}(t_{1}^{+}),x_{2}(t_{1}^{+}),\ldots ,x_{n}(t_{1}^{+})) \left( \begin{array}{cccccc} a_{21} \\ a_{22} \\ \vdots \\ a_{2n} \\ \end{array} \right) . \end{aligned}$$

(37)

Due to $\Delta x(t_{1})=Bx(t_{1}^{-}),$ we get $x(t_{1}^{+})=(E+B)x(t_{1}^{-}).$ From (35) and (37),

$$\begin{aligned} (x_{1}(t_{1}^{+}),x_{2}(t_{1}^{+}),\ldots ,x_{n}(t_{1}^{+})) \left( \begin{array}{cccccc} a_{21} \\ a_{22} \\ \vdots \\ a_{2n} \\ \end{array} \right) =(E+B) (x_{1}(t_{1}),x_{2}(t_{1}),\ldots ,x_{n}(t_{1})) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) , \end{aligned}$$

we have

$$\begin{aligned} \left( \begin{array}{cccccc} a_{21} \\ a_{22} \\ \vdots \\ a_{2n} \\ \end{array} \right) =(E+B) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) . \end{aligned}$$

(38)

From (36) and (38), each solution x(t) in the interval $(t_{1},t_{2}]$ can be expressed as

$$\begin{aligned} x(t) =&\,(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(E+B) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \nonumber \\ =&\,(E+B)(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) . \end{aligned}$$

(39)

From (39), we can obtain

$$\begin{aligned} x(t_{2}^{-}) =x(t_{2}) =(E+B)(x_{1}(t_{2}),x_{2}(t_{2}),\ldots ,x_{n}(t_{2})) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) . \end{aligned}$$

(40)

For $t\in (t_{2},t_{3}],$ each solution x(t) in the interval $(t_{2},t_{3}]$ has the form

$$\begin{aligned} x(t) =x_{1}(t)a_{31}+x_{2}(t)a_{32}+\ldots +x_{n}(t)a_{3n} =(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left( \begin{array}{cccccc} a_{31} \\ a_{32} \\ \vdots \\ a_{3n} \\ \end{array} \right) , \end{aligned}$$

(41)

where $a_{31},a_{32},\ldots ,a_{3n}$ are arbitrary quaternions.

From (41), we can obtain

$$\begin{aligned} x(t_{2}^{+}) =(x_{1}(t_{2}^{+}),x_{2}(t_{2}^{+}),\ldots ,x_{n}(t_{2}^{+})) \left( \begin{array}{cccccc} a_{31} \\ a_{32} \\ \vdots \\ a_{3n} \\ \end{array} \right) . \end{aligned}$$

(42)

Due to $\Delta x(t_{2})=Bx(t_{2}^{-}),$ we get $x(t_{2}^{+})=(E+B)x(t_{2}^{-}).$ From (40) and (42),

$$\begin{aligned}&(x_{1}(t_{2}^{+}),x_{2}(t_{2}^{+}),\ldots ,x_{n}(t_{2}^{+})) \left( \begin{array}{cccccc} a_{31} \\ a_{32} \\ \vdots \\ a_{3n} \\ \end{array} \right) \\&\quad =(E+B) (E+B)(x_{1}(t_{2}),x_{2}(t_{2}),\ldots ,x_{n}(t_{2})) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \\&\quad =(E+B)^{2} (x_{1}(t_{2}),x_{2}(t_{2}),\ldots ,x_{n}(t_{2})) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) , \end{aligned}$$

we have

$$\begin{aligned} \left( \begin{array}{cccccc} a_{31} \\ a_{32} \\ \vdots \\ a_{3n} \\ \end{array} \right) = (E+B)^{2} \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) . \end{aligned}$$

(43)

From (41) and (43), each solution x(t) in the interval $(t_{2},t_{3}]$ can be expressed as

$$\begin{aligned} x(t) =&\,(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(E+B)^{2} \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \\ =&\,(E+B)^{2}(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) . \end{aligned}$$

Repeating this process, we obtain each solution x(t) in the interval $(t_{k-1},t_{k}]$ can be expressed as

$$\begin{aligned} x(t) =(E+B)^{k-1}(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) . \end{aligned}$$

(44)

From (44), we can obtain

$$\begin{aligned} x(t_{k}^{-})=x(t_{k}) =(E+B)^{k-1}(x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k})) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) . \end{aligned}$$

(45)

For $t\in (t_{k},t_{k+1}],$ each solution x(t) in the interval $(t_{k},t_{k+1}]$ has the form

$$\begin{aligned} x(t) =x_{1}(t)a_{k+1,1}+x_{2}(t)a_{k+1,2}+\ldots +x_{n}(t)a_{k+1,n} =(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left( \begin{array}{cccccc} a_{k+1,1} \\ a_{k+1,2} \\ \vdots \\ a_{k+1,n} \\ \end{array} \right) , \end{aligned}$$

(46)

where $a_{k+1,1},a_{k+1,2},\ldots ,a_{k+1,n}$ are arbitrary quaternions.

From (46), we can obtain

$$\begin{aligned} x(t_{k}^{+}) =(x_{1}(t_{k}^{+}),x_{2}(t_{k}^{+}),\ldots ,x_{n}(t_{k}^{+})) \left( \begin{array}{cccccc} a_{k+1,1} \\ a_{k+1,2} \\ \vdots \\ a_{k+1,n} \\ \end{array} \right) . \end{aligned}$$

(47)

Due to $\Delta x(t_{k})=Bx(t_{k}^{-}),$ we get $x(t_{k}^{+})=(E+B)x(t_{k}^{-}).$ From (45) and (47),

$$\begin{aligned}&(x_{1}(t_{k}^{+}),x_{2}(t_{k}^{+}),\ldots ,x_{n}(t_{k}^{+})) \left( \begin{array}{cccccc} a_{k+1,1} \\ a_{k+1,2} \\ \vdots \\ a_{k+1,n} \\ \end{array} \right) \\&\quad =(E+B)(E+B)^{k-1} (x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k})) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \\&\quad =(E+B)^{k} (x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k})) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) , \end{aligned}$$

we have

$$\begin{aligned} \left( \begin{array}{cccccc} a_{k+1,1} \\ a_{k+1,2} \\ \vdots \\ a_{k+1,n} \\ \end{array} \right) =(E+B)^{k} \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) . \end{aligned}$$

(48)

From (46) and (48), each solution x(t) in the interval $(t_{k},t_{k+1}]$ can be expressed as

$$\begin{aligned} x(t) =&\,(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(E+B)^{k} \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \nonumber \\ =&\,(E+B)^{k}(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) . \end{aligned}$$

(49)

Due to $x(t),x_{i}(t)\in \mathbb {H}^{n\times 1},(E+B)^{k}\in \mathbb {H}^{n\times n},c_{i}\in \mathbb {H},i=1,\dots ,n.$ For $x(t),(E+B)^{k}$ and each $x_{i}(t),c_{i},i=1,\dots ,n,$ they can be expressed as

$$\begin{aligned}&x(t)=u_{1}(t)+u_{2}(t)\varvec{j},\nonumber \\&(E+B)^{k}=(E+B)^{k}_{1}+(E+B)^{k}_{2}\varvec{j},\nonumber \\&x_{i}(t)=x_{i1}(t)+x_{i2}(t)\varvec{j},\nonumber \\&c_{i}=c_{i1}+c_{i2}\varvec{j}, \end{aligned}$$

(50)

respectively. Combining (49) and (50), we get

(51)

Equation (51) is equal to

$$\begin{aligned} \left\{ \begin{aligned} u_{1}(t) =&\,(E+B)_{1}^{k}(x_{11}(t)c_{11}-x_{12}(t)\overline{c_{12}}+\ldots +x_{n1}(t)c_{n1}-x_{n2}(t)\overline{c_{n2}})\\&+(E+B)_{2}^{k}(-\overline{x_{11}(t)}\overline{c_{12}} -\overline{x_{12}(t)}c_{11}-\ldots -\overline{x_{n1}(t)}\overline{c_{n2}} -\overline{x_{n2}(t)}c_{n1}),\\ u_{2}(t) =&\,(E+B)_{1}^{k}(x_{11}(t)c_{12}+x_{12}(t)\overline{c_{11}} +\ldots +x_{n1}(t)c_{n2}+x_{n2}(t)\overline{c_{n1}})\\&+(E+B)_{2}^{k}(\overline{x_{11}(t)}\overline{c_{11}} -\overline{x_{12}(t)}c_{12}+\ldots +\overline{x_{n1}(t)}\overline{c_{n1}} -\overline{x_{n2}(t)}c_{n2}).\\ \end{aligned} \right. \end{aligned}$$

(52)

The system (52) is equal to

$$\begin{aligned} \left\{ \begin{aligned} u_{1}(t) =&\,(E+B)_{1}^{k}(x_{11}(t)c_{11}-x_{12}(t)\overline{c_{12}}+\ldots +x_{n1}(t)c_{n1}-x_{n2}(t)\overline{c_{n2}})\\&+(E+B)_{2}^{k}(-\overline{x_{11}(t)}\overline{c_{12}} -\overline{x_{12}(t)}c_{11}-\ldots -\overline{x_{n1}(t)}\overline{c_{n2}} -\overline{x_{n2}(t)}c_{n1}),\\ -\overline{u_{2}(t)} =&\,\overline{(E+B)_{1}^{k}}(-\overline{x_{11}(t)}\overline{c_{12}}-\overline{x_{12}(t)}c_{11} -\ldots -\overline{x_{n1}(t)}\overline{c_{n2}}-\overline{x_{n2}(t)}c_{n1})\\&-\overline{(E+B)_{2}^{k}}(x_{11}(t)c_{11}-x_{12}(t)\overline{c_{12}}+\ldots +x_{n1}(t)c_{n1} -x_{n2}(t)\overline{c_{n2}}).\\ \end{aligned} \right. \end{aligned}$$

(53)

The system (53) is equal to

$$\begin{aligned} \left( \begin{array}{cccccc} u_{1}(t) \\ -\overline{u_{2}(t)} \\ \end{array} \right) =&\, \left( \begin{array}{cccccc} (E+B)_{1}^{k} &{} (E+B)_{2}^{k} \\ -\overline{(E+B)_{2}^{k}} &{} \overline{(E+B)_{1}^{k}} \\ \end{array} \right) \nonumber \\&\quad \left( \begin{array}{cccccc} x_{11}(t)c_{11}-x_{12}(t)\overline{c_{12}}+\ldots +x_{n1}(t)c_{n1} -x_{n2}(t)\overline{c_{n2}}\\ -\overline{x_{11}(t)}\overline{c_{12}}-\overline{x_{12}(t)}c_{11} -\ldots -\overline{x_{n1}(t)}\overline{c_{n2}}-\overline{x_{n2}(t)}c_{n1}\\ \end{array} \right) \nonumber \\ =&\, \left( \begin{array}{cccccc} (E+B)_{1}^{k} &{} (E+B)_{2}^{k} \\ -\overline{(E+B)_{2}^{k}} &{} \overline{(E+B)_{1}^{k}} \\ \end{array} \right) \nonumber \\&\quad \left[ \left( \begin{array}{cccccc} x_{11}(t) \\ -\overline{x_{12}(t)} \\ \end{array} \right) c_{11} + \left( \begin{array}{cccccc} x_{12}(t) \\ \overline{x_{11}(t)} \\ \end{array} \right) (-\overline{c_{12}}) +\ldots +\left( \begin{array}{cccccc} x_{n1}(t) \\ -\overline{x_{n2}(t)} \\ \end{array} \right) c_{n1} \right. \nonumber \\&\qquad \left. + \left( \begin{array}{cccccc} x_{n2}(t) \\ \overline{x_{n1}(t)} \\ \end{array} \right) (-\overline{c_{n2}}) \right] \nonumber \\ =&\, \left( \begin{array}{cccccc} (E+B)_{1}^{k} &{} (E+B)_{2}^{k} \\ -\overline{(E+B)_{2}^{k}} &{} \overline{(E+B)_{1}^{k}} \\ \end{array} \right) \nonumber \\&\qquad \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} x_{11}(t) \\ -\overline{x_{12}(t)} \\ \end{array} \right) &{} \left( \begin{array}{cccccc} x_{12}(t) \\ \overline{x_{11}(t)} \\ \end{array} \right) &{} \ldots &{} \left( \begin{array}{cccccc} x_{n1}(t) \\ -\overline{x_{n2}(t)} \\ \end{array} \right) &{} \left( \begin{array}{cccccc} x_{n2}(t) \\ \overline{x_{n1}(t)} \\ \end{array} \right) \end{array} \right) \nonumber \\&\qquad \left( \begin{array}{cccccc} c_{11} \\ \vdots \\ c_{n1} \\ -\overline{c_{12}}\\ \vdots \\ -\overline{c_{n2}}\\ \end{array} \right) . \end{aligned}$$

(54)

Due to

$$\begin{aligned}&\varphi (x(t))= \left( \begin{array}{cccccc} u_{1}(t) \\ -\overline{u_{2}(t)} \\ \end{array} \right) ,\,\,\, \zeta ((E+B)^{k})= \left( \begin{array}{cccccc} (E+B)_{1}^{k} &{} (E+B)_{2}^{k} \\ -\overline{(E+B)_{2}^{k}} &{} \overline{(E+B)_{1}^{k}} \\ \end{array} \right) ,\\&\varphi (x_{i}(t))= \left( \begin{array}{cccccc} x_{i1}(t) \\ -\overline{x_{i2}(t)} \\ \end{array} \right) ,\,\,\, \varphi (x_{i}(t))^{*}= \left( \begin{array}{cccccc} x_{i2}(t) \\ \overline{x_{i1}(t)} \\ \end{array} \right) ,\\&\zeta (x_{1}(t)\,\,\,\ldots \,\,\,x_{n}(t)) =(\varphi (x_{1}(t))\,\,\,\ldots \,\,\,\varphi (x_{n}(t))\,\,\,\varphi (x_{1}(t))^{*}\,\,\,\ldots \,\,\,\varphi (x_{n}(t))^{*}),\\&\varphi \left( \begin{array}{cccccc} c_{1} \\ \vdots \\ c_{n} \\ \end{array} \right) = \left( \begin{array}{cccccc} c_{11} \\ \vdots \\ c_{n1} \\ -\overline{c_{12}}\\ \vdots \\ -\overline{c_{n2}}\\ \end{array} \right) , \end{aligned}$$

Lemma 3.2 and invertibility of $\zeta (x_{1}(t),\ldots ,x_{n}(t))$, Eq. (54) can be expressed by

$$\begin{aligned} \varphi (x(t)) =\zeta ((E+B)^{k})\zeta (x_{1}(t)\,\,\,\ldots \,\,\,x_{n}(t)) \varphi \left( \begin{array}{cccccc} c_{1} \\ \vdots \\ c_{n} \\ \end{array} \right) , \end{aligned}$$

(55)

where $c_{1},\ldots ,c_{n}$ are arbitrary quaternions.

Taking the inverse isomorphism $\varphi ^{-1}$ on both side of Eq. (55), we get

$$\begin{aligned} x(t) =\varphi ^{-1}\left\{ \zeta ((E+B)^{k})\zeta (x_{1}(t)\,\,\,\ldots \,\,\,x_{n}(t)) \varphi \left( \begin{array}{cccccc} c_{1} \\ \vdots \\ c_{n} \\ \end{array} \right) \right\} . \end{aligned}$$

The proof is finished. $\square$

Appendix 2. Proof of Theorem 3.5

Proof

According to Lemma 2.4, for $t\in [0,t_{1}],$ since $e^{At}$ is the fundamental matrix of $\dot{x}(t)=Ax(t),$ then each solution x(t) in the interval $[0,t_{1}]$ has the form

$$\begin{aligned} x(t)=e^{At}c, \end{aligned}$$

(56)

where c is a constant quaternion vector.

From (56), we can obtain that

$$\begin{aligned} x(t_{1}^{-})=x(t_{1})=e^{At_{1}}c. \end{aligned}$$

(57)

For $t\in (t_{1},t_{2}],$ each solution x(t) in the interval $(t_{1},t_{2}]$ has the form

$$\begin{aligned} x(t)=e^{At}a_{2}, \end{aligned}$$

(58)

where $a_{2}$ is a constant quaternion vector.

From (58), we can obtain

$$\begin{aligned} x(t_{1}^{+})=e^{At_{1}^{+}}a_{2}. \end{aligned}$$

(59)

Due to $\Delta x(t_{1})=Bx(t_{1}^{-}),$ we get $x(t_{1}^{+})=(E+B)x(t_{1}^{-}).$ From (57) and (59),

$$\begin{aligned} e^{At_{1}^{+}}a_{2}=(E+B)e^{At_{1}}c, \end{aligned}$$

we have

$$\begin{aligned} a_{2}=(E+B)c. \end{aligned}$$

(60)

From (58) and (60), each solution x(t) in the interval $(t_{1},t_{2}]$ can be expressed as

$$\begin{aligned} x(t) =&\,e^{At}(E+B)c\nonumber \\ =&\,(E+B)e^{At}c. \end{aligned}$$

(61)

From (61), we can obtain that

$$\begin{aligned} x(t_{2}^{-})=x(t_{2})=(E+B)e^{At_{2}}c. \end{aligned}$$

(62)

For $t\in (t_{2},t_{3}],$ each solution x(t) in the interval $(t_{2},t_{3}]$ has the form

$$\begin{aligned} x(t)=e^{At}a_{3}, \end{aligned}$$

(63)

where $a_{3}$ is a constant quaternion vector.

From (63), we can obtain

$$\begin{aligned} x(t_{2}^{+})=e^{At_{2}^{+}}a_{3}. \end{aligned}$$

(64)

Due to $\Delta x(t_{2})=Bx(t_{2}^{-}),$ we get $x(t_{2}^{+})=(E+B)x(t_{2}^{-}).$ From (62) and (64),

$$\begin{aligned} e^{At_{2}^{+}}a_{3}=(E+B)(E+B)e^{At_{2}}c=(E+B)^{2}e^{At_{2}}c, \end{aligned}$$

we have

$$\begin{aligned} a_{3}=(E+B)^{2}c. \end{aligned}$$

(65)

From (63) and (65), each solution x(t) in the interval $(t_{2},t_{3}]$ can be expressed as

$$\begin{aligned} x(t) =&\,e^{At}(E+B)^{2}c\\ =&\,(E+B)^{2}e^{At}c. \end{aligned}$$

Repeating this process, we obtain each solution x(t) in the interval $(t_{k-1},t_{k}]$ can be expressed as

$$\begin{aligned} x(t)=(E+B)^{k-1}e^{At}c. \end{aligned}$$

(66)

From (66), we can obtain that

$$\begin{aligned} x(t_{k}^{-})=x(t_{k})=(E+B)^{k-1}e^{At_{k}}c. \end{aligned}$$

(67)

For $t\in (t_{k},t_{k+1}],$ each solution x(t) in the interval $(t_{k},t_{k+1}]$ has the form

$$\begin{aligned} x(t)=e^{At}a_{k+1}, \end{aligned}$$

(68)

where $a_{k+1}$ is a constant quaternion vector

From (68), we can obtain

$$\begin{aligned} x(t_{k}^{+})=e^{At_{k}^{+}}a_{k+1}. \end{aligned}$$

(69)

Due to $\Delta x(t_{k})=Bx(t_{k}^{-}),$ we get $x(t_{k}^{+})=(E+B)x(t_{k}^{-}).$ From (67) and (69),

$$\begin{aligned} e^{At_{k}^{+}}a_{k+1}=(E+B)(E+B)^{k-1}e^{At_{k}}c=(E+B)^{k}e^{At_{k}}c, \end{aligned}$$

we have

$$\begin{aligned} a_{k+1}=(E+B)^{k}c. \end{aligned}$$

(70)

From (68) and (70), each solution x(t) in the interval $(t_{k},t_{k+1}]$ can be expressed as

$$\begin{aligned} x(t) =&\,e^{At}(E+B)^{k}c\\ =&\,(E+B)^{k}e^{At}c. \end{aligned}$$

The proof is finished. $\square$

Appendix 3. Proof of Theorem 3.9

Proof

According to Lemma 2.3 and Lemma 2.5, for $t\in [0,t_{1}],$ since $x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)$ are linear independent solutions of $\dot{x}(t)=Ax(t),$ then each solution x(t) in the interval $[0,t_{1}]$ has the form

$$\begin{aligned} x(t) =&\,(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \nonumber \\&+(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \int _{0}^{t}(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds, \end{aligned}$$

(71)

where $c_{1},c_{2},\ldots ,c_{n}$ are arbitrary quaternions.

From (71), we can obtain

$$\begin{aligned} x(t_{1}^{-}) =x(t_{1}) =&\,(x_{1}(t_{1}),x_{2}(t_{1}),\ldots ,x_{n}(t_{1})) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \nonumber \\&+(x_{1}(t_{1}),x_{2}(t_{1}),\ldots ,x_{n}(t_{1})) \int _{0}^{t_{1}}(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \nonumber \\&\quad \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds. \end{aligned}$$

(72)

For $t\in (t_{1},t_{2}],$ each solution x(t) in the interval $(t_{1},t_{2}]$ has the form

$$\begin{aligned} x(t) =&\,(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left( \begin{array}{cccccc} a_{21} \\ a_{22} \\ \vdots \\ a_{2n} \\ \end{array} \right) \nonumber \\&+(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \int _{t_{1}}^{t}(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds, \end{aligned}$$

(73)

where $a_{21},a_{22},\ldots ,a_{2n}$ are arbitrary quaternions.

From (73), we can obtain

$$\begin{aligned} x(t_{1}^{+}) =(x_{1}(t_{1}^{+}),x_{2}(t_{1}^{+}),\ldots ,x_{n}(t_{1}^{+})) \left( \begin{array}{cccccc} a_{21} \\ a_{22} \\ \vdots \\ a_{2n} \\ \end{array} \right) . \end{aligned}$$

(74)

Due to $\Delta x(t_{1})=Bx(t_{1}^{-}),$ we get $x(t_{1}^{+})=(E+B)x(t_{1}^{-}).$ From (72) and (74),

$$\begin{aligned}&(x_{1}(t_{1}^{+}),x_{2}(t_{1}^{+}),\ldots ,x_{n}(t_{1}^{+})) \left( \begin{array}{cccccc} a_{21} \\ a_{22} \\ \vdots \\ a_{2n} \\ \end{array} \right) \\&\quad =(E+B) \left[ (x_{1}(t_{1}),x_{2}(t_{1}),\ldots ,x_{n}(t_{1})) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \right. \\&\qquad \left. +\,(x_{1}(t_{1}),x_{2}(t_{1}),\ldots ,x_{n}(t_{1})) \int _{0}^{t_{1}}(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds \right] , \end{aligned}$$

we have

$$\begin{aligned} \left( \begin{array}{cccccc} a_{21} \\ a_{22} \\ \vdots \\ a_{2n} \\ \end{array} \right) =&\,(E+B) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) +\int _{0}^{t_{1}}(E+B)(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds. \end{aligned}$$

(75)

From (73) and (75), each solution x(t) in the interval $(t_{1},t_{2}]$ can be expressed as

$$\begin{aligned} x(t) =&\,(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \nonumber \\&\quad \left[ (E+B)\left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) +\int _{0}^{t_{1}}(E+B)(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\right] \nonumber \\&+(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \int _{t_{1}}^{t}(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\ =&\,(E+B)(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \nonumber \\&+\int _{0}^{t_{1}}(E+B) (x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&+\int _{t_{1}}^{t} (x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds. \end{aligned}$$

(76)

From (76), we can obtain

$$\begin{aligned} x(t_{2}^{-})=x(t_{2}) =&\,(E+B)(x_{1}(t_{2}),x_{2}(t_{2}),\ldots ,x_{n}(t_{2})) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \nonumber \\&+\int _{0}^{t_{1}}(E+B) (x_{1}(t_{2}),x_{2}(t_{2}),\ldots ,x_{n}(t_{2}))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1}\nonumber \\&\quad \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&+\int _{t_{1}}^{t_{2}} (x_{1}(t_{2}),x_{2}(t_{2}),\ldots ,x_{n}(t_{2}))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1}\nonumber \\&\quad \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds. \end{aligned}$$

(77)

For $t\in (t_{2},t_{3}],$ each solution x(t) in the interval $(t_{2},t_{3}]$ has the form

$$\begin{aligned} x(t) =&\,(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left( \begin{array}{cccccc} a_{31} \\ a_{32} \\ \vdots \\ a_{3n} \\ \end{array} \right) \nonumber \\&+(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \int _{t_{2}}^{t}(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds, \end{aligned}$$

(78)

where $a_{31},a_{32},\ldots ,a_{3n}$ are arbitrary quaternions.

From (78), we can obtain

$$\begin{aligned} x(t_{2}^{+}) =(x_{1}(t_{2}^{+}),x_{2}(t_{2}^{+}),\ldots ,x_{n}(t_{2}^{+})) \left( \begin{array}{cccccc} a_{31} \\ a_{32} \\ \vdots \\ a_{3n} \\ \end{array} \right) . \end{aligned}$$

(79)

Due to $\Delta x(t_{2})=Bx(t_{2}^{-}),$ we get $x(t_{2}^{+})=(E+B)x(t_{2}^{-}).$ From (77) and (79),

$$\begin{aligned}&(x_{1}(t_{2}^{+}),x_{2}(t_{2}^{+}),\ldots ,x_{n}(t_{2}^{+})) \left( \begin{array}{cccccc} a_{31} \\ a_{32} \\ \vdots \\ a_{3n} \\ \end{array} \right) \\&\quad =(E+B)\left[ (E+B)(x_{1}(t_{2}),x_{2}(t_{2}),\ldots ,x_{n}(t_{2})) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \right. \\&\qquad +\int _{0}^{t_{1}}(E+B) (x_{1}(t_{2}),x_{2}(t_{2}),\ldots ,x_{n}(t_{2}))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\\&\qquad \left. +\int _{t_{1}}^{t_{2}} (x_{1}(t_{2}),x_{2}(t_{2}),\ldots ,x_{n}(t_{2}))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\right] \\&\quad =(E+B)^{2}(x_{1}(t_{2}),x_{2}(t_{2}),\ldots ,x_{n}(t_{2})) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \\&\qquad +\int _{0}^{t_{1}}(E+B)^{2} (x_{1}(t_{2}),x_{2}(t_{2}),\ldots ,x_{n}(t_{2}))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\\&\qquad +\int _{t_{1}}^{t_{2}}(E+B) (x_{1}(t_{2}),x_{2}(t_{2}),\ldots ,x_{n}(t_{2}))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds, \end{aligned}$$

we have

$$\begin{aligned} \left( \begin{array}{cccccc} a_{31} \\ a_{32} \\ \vdots \\ a_{3n} \\ \end{array} \right) =&\, (E+B)^{2} \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \nonumber \\&+\int _{0}^{t_{1}}(E+B)^{2}(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&+\int _{t_{1}}^{t_{2}}(E+B)(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds. \end{aligned}$$

(80)

From (78) and (80), each solution x(t) in the interval $(t_{2},t_{3}]$ can be expressed as

$$\begin{aligned} x(t) =&\,(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left[ (E+B)^{2} \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \right. \\&+\int _{0}^{t_{1}}(E+B)^{2}(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\\&\left. +\int _{t_{1}}^{t_{2}}(E+B)(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\right] \\&+(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \int _{t_{2}}^{t}(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\\ =&\,(E+B)^{2}(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \\&+\int _{0}^{t_{1}}(E+B)^{2}(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\\&+\int _{t_{1}}^{t_{2}}(E+B)(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\\&+\int _{t_{2}}^{t}(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds, \end{aligned}$$

Repeating this process, we obtain each solution x(t) in the interval $(t_{k-1},t_{k}]$ can be expressed as

$$\begin{aligned} x(t) =&\,(E+B)^{k-1}(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \nonumber \\&+\sum \limits _{i=1}^{k-2}\int _{t_{i-1}}^{t_{i}}(E+B)^{k-i} (x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1}\nonumber \\&\quad \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&+\int _{t_{k-2}}^{t_{k-1}}(E+B) (x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1}\nonumber \\&\quad \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&+\int _{t_{k-1}}^{t} (x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds. \end{aligned}$$

(81)

From (81), we can obtain

$$\begin{aligned}&x(t_{k}^{-})=x(t_{k}) =(E+B)^{k-1}(x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k})) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \nonumber \\&\quad +\sum \limits _{i=1}^{k-2}\int _{t_{i-1}}^{t_{i}}(E+B)^{k-i} (x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k}))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1}\nonumber \\&\qquad \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&\quad +\int _{t_{k-2}}^{t_{k-1}}(E+B) (x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k}))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1}\nonumber \\&\qquad \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&\quad +\int _{t_{k-1}}^{t_{k}} (x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k}))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds. \end{aligned}$$

(82)

For $t\in (t_{k},t_{k+1}],$ each solution x(t) in the interval $(t_{k},t_{k+1}]$ has the form

$$\begin{aligned} x(t) =&\,(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left( \begin{array}{cccccc} a_{k+1,1} \\ a_{k+1,2} \\ \vdots \\ a_{k+1,n} \\ \end{array} \right) \nonumber \\&+(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \int _{t_{k}}^{t}(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds, \end{aligned}$$

(83)

where $a_{k+1,1},a_{k+1,2},\ldots ,a_{k+1,n}$ are arbitrary quaternions.

From (83), we can obtain

$$\begin{aligned} x(t_{k}^{+}) =(x_{1}(t_{k}^{+}),x_{2}(t_{k}^{+}),\ldots ,x_{n}(t_{k}^{+})) \left( \begin{array}{cccccc} a_{k+1,1} \\ a_{k+1,2} \\ \vdots \\ a_{k+1,n} \\ \end{array} \right) . \end{aligned}$$

(84)

Due to $\Delta x(t_{k})=Bx(t_{k}^{-}),$ we get $x(t_{k}^{+})=(E+B)x(t_{k}^{-}).$ From (82) and (84),

$$\begin{aligned}&(x_{1}(t_{k}^{+}),x_{2}(t_{k}^{+}),\ldots ,x_{n}(t_{k}^{+})) \left( \begin{array}{cccccc} a_{k+1,1} \\ a_{k+1,2} \\ \vdots \\ a_{k+1,n} \\ \end{array} \right) \\&\quad =(E+B)\left[ (E+B)^{k-1}(x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k})) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \right. \\&\qquad +\sum \limits _{i=1}^{k-2}\int _{t_{i-1}}^{t_{i}}(E+B)^{k-i} (x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k}))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1}\nonumber \\&\qquad \quad \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\\&\qquad +\int _{t_{k-2}}^{t_{k-1}}(E+B) (x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k}))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1}\nonumber \\&\qquad \quad \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\\&\qquad \left. +\int _{t_{k-1}}^{t_{k}} (x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k}))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds \right] \\&\quad =(E+B)^{k}(x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k})) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \\&\qquad +\sum \limits _{i=1}^{k-2}\int _{t_{i-1}}^{t_{i}}(E+B)^{k-i+1} (x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k}))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1}\nonumber \\&\qquad \quad \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\\&\qquad +\int _{t_{k-2}}^{t_{k-1}}(E+B)^{2} (x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k}))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1}\nonumber \\&\qquad \quad \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\\&\qquad +\int _{t_{k-1}}^{t_{k}}(E+B) (x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k}))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1}\nonumber \\&\qquad \quad \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds, \end{aligned}$$

we have

$$\begin{aligned} \left( \begin{array}{cccccc} a_{k+1,1} \\ a_{k+1,2} \\ \vdots \\ a_{k+1,n} \\ \end{array} \right) =&\, (x_{1}(t_{k}^{+}),x_{2}(t_{k}^{+}),\ldots ,x_{n}(t_{k}^{+}))^{-1}\nonumber \\&\quad \left[ (E+B)^{k}(x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k})) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \right. \nonumber \\&+\sum \limits _{i=1}^{k-2}\int _{t_{i-1}}^{t_{i}}(E+B)^{k-i+1} (x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k}))\nonumber \\&\quad (x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&+\int _{t_{k-2}}^{t_{k-1}}(E+B)^{2} (x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k}))\nonumber \\&\quad (x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&+\int _{t_{k-1}}^{t_{k}}(E+B) (x_{1}(t_{k}),x_{2}(t_{k}),\ldots ,x_{n}(t_{k}))\nonumber \\&\quad \left. (x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds \right] \nonumber \\ =&\, (E+B)^{k} \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \nonumber \\&+\sum \limits _{i=1}^{k-2}\int _{t_{i-1}}^{t_{i}}(E+B)^{k-i+1} (x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&+\int _{t_{k-2}}^{t_{k-1}}(E+B)^{2} (x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&+\int _{t_{k-1}}^{t_{k}}(E+B) (x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds. \end{aligned}$$

(85)

From (83) and (85), each solution x(t) in the interval $(t_{k},t_{k+1}]$ can be expressed as

$$\begin{aligned} x(t) =&\,(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left[ (E+B)^{k} \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \right. \nonumber \\&+\sum \limits _{i=1}^{k-2}\int _{t_{i-1}}^{t_{i}}(E+B)^{k-i+1} (x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&+\int _{t_{k-2}}^{t_{k-1}}(E+B)^{2} (x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&\left. +\int _{t_{k-1}}^{t_{k}}(E+B) (x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\right] \nonumber \\&+(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \int _{t_{k}}^{t}(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\ =&\,(E+B)^{k}(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \nonumber \\&+\sum \limits _{i=1}^{k-2}\int _{t_{i-1}}^{t_{i}}(E+B)^{k-i+1} (x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1}\nonumber \\&\quad \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&+\int _{t_{k-2}}^{t_{k-1}}(E+B)^{2} (x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1}\nonumber \\&\quad \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&+\int _{t_{k-1}}^{t_{k}}(E+B) (x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1}\nonumber \\&\quad \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&+\int _{t_{k}}^{t} (x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\ =&\,(E+B)^{k}(x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)) \left( \begin{array}{cccccc} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \\ \end{array} \right) \nonumber \\&+\sum \limits _{i=1}^{k-1}\int _{t_{i-1}}^{t_{i}}(E+B)^{k-i+1} (x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1}\nonumber \\&\quad \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&+\int _{t_{k-1}}^{t_{k}}(E+B) (x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1}\nonumber \\&\quad \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&+\int _{t_{k}}^{t} (x_{1}(t),x_{2}(t),\ldots ,x_{n}(t))(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1} \left( \begin{array}{cccccc} f_{1}(s) \\ f_{2}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds. \end{aligned}$$

(86)

Supposing

$$\begin{aligned} (x_{1}(t),x_{2}(t),\ldots ,x_{n}(t)&= \left( \begin{array}{cccccc} y_{1}(t) \\ y_{2}(t) \\ \vdots \\ y_{n}(t) \\ \end{array} \right) ,\\ (x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))^{-1}&=(z_{1}(s),z_{2}(s),\ldots ,z_{n}(s)), \end{aligned}$$

where $x_{i}(t),x_{i}(s),z_{i}(s)\in \mathbb {H}^{n\times 1},y_{i}(t)\in \mathbb {H}^{1\times n},i=1,2,\ldots ,n.$

Due to $y_{i}(t)\in \mathbb {H}^{1\times n},z_{i}(s)\in \mathbb {H}^{n\times 1},f_{i}(s)\in \mathbb {H},i=1,\dots ,n.$ For each $y_{i}(t),z_{i}(s),f_{i}(s),i=1,\dots ,n,$ they can be expressed as

$$\begin{aligned}&y_{i}(t)=y_{i1}(t)+y_{i2}(t)\varvec{j},\nonumber \\&z_{i}(s)=z_{i1}(s)+z_{i2}(s)\varvec{j},\nonumber \\&f_{i}(s)=f_{i1}(s)+f_{i2}(s)\varvec{j}, \end{aligned}$$

(87)

respectively.

Combining (86) and (87), we get

(88)

Equation (88) is equal to

(89)

The system (89) is equal to

(90)

The system (90) is equal to

(91)

Due to

$$\begin{aligned}&\zeta \left( \begin{array}{cccccc} y_{1}(t) \\ \vdots \\ y_{n}(t) \\ \end{array} \right) = \left( \begin{array}{cccccc} y_{11}(t) &{} y_{12}(t) \\ \vdots &{} \vdots \\ y_{n1}(t) &{} y_{n2}(t) \\ -\overline{y_{12}(t)} &{} \overline{y_{11}(t)} \\ \vdots &{} \vdots \\ -\overline{y_{n2}(t)} &{} \overline{y_{n1}(t)} \\ \end{array} \right) ,\\&\varphi (z_{i}(s))= \left( \begin{array}{cccccc} z_{i1}(s) \\ -\overline{z_{i2}(s)} \\ \end{array} \right) ,\,\,\, \varphi (z_{i}(s))^{*}= \left( \begin{array}{cccccc} z_{i2}(s) \\ \overline{z_{i1}(s)} \\ \end{array} \right) ,\\&\zeta (z_{1}(s)\,\,\,\ldots \,\,\,z_{n}(s)) =(\varphi (z_{1}(s))\,\,\,\ldots \,\,\,\varphi (z_{n}(s))\,\,\,\varphi (z_{1}(s))^{*}\,\,\,\ldots \,\,\,\varphi (z_{n}(s))^{*}),\\&\varphi \left( \begin{array}{cccccc} f_{1}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) = \left( \begin{array}{cccccc} f_{11}(s) \\ \vdots \\ f_{n1}(s) \\ -\overline{f_{12}(s)}\\ \vdots \\ -\overline{f_{n2}(s)}\\ \end{array} \right) , \end{aligned}$$

Equation (91) can be expressed by

$$\begin{aligned} \varphi (x(t)) =&\,\zeta ((E+B)^{k})\zeta (x_{1}(t)\,\,\,\ldots \,\,\,x_{n}(t)) \varphi \left( \begin{array}{cccccc} c_{1} \\ \vdots \\ c_{n} \\ \end{array} \right) \nonumber \\&+\sum \limits _{i=1}^{k-1}\int _{t_{i-1}}^{t_{i}}\zeta ((E+B)^{k-i+1}) \zeta \left( \begin{array}{cccccc} y_{1}(t) \\ \vdots \\ y_{n}(t) \\ \end{array} \right) \zeta (z_{1}(s)\,\,\,\ldots \,\,\,z_{n}(s)) \varphi \left( \begin{array}{cccccc} f_{1}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&+\int _{t_{k-1}}^{t_{k}}\zeta ((E+B)^{k}) \zeta \left( \begin{array}{cccccc} y_{1}(t) \\ \vdots \\ y_{n}(t) \\ \end{array} \right) \zeta (z_{1}(s)\,\,\,\ldots \,\,\,z_{n}(s)) \varphi \left( \begin{array}{cccccc} f_{1}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\nonumber \\&+\int _{t_{k}}^{t} \zeta \left( \begin{array}{cccccc} y_{1}(t) \\ \vdots \\ y_{n}(t) \\ \end{array} \right) \zeta (z_{1}(s)\,\,\,\ldots \,\,\,z_{n}(s)) \varphi \left( \begin{array}{cccccc} f_{1}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds, \end{aligned}$$

(92)

where $c_{1},\ldots ,c_{2}$ are arbitrary quaternions.

Taking the inverse isomorphism $\varphi ^{-1}$ on both side of Eq. (92), we get

$$\begin{aligned} x(t) =&\,\varphi ^{-1} \left\{ \zeta ((E+B)^{k})\zeta (x_{1}(t)\,\,\,\ldots \,\,\,x_{n}(t)) \varphi \left( \begin{array}{cccccc} c_{1} \\ \vdots \\ c_{n} \\ \end{array} \right) \right. \\&+\sum \limits _{i=1}^{k-1}\int _{t_{i-1}}^{t_{i}}\zeta ((E+B)^{k-i+1}) \zeta \left( \begin{array}{cccccc} y_{1}(t) \\ \vdots \\ y_{n}(t) \\ \end{array} \right) \zeta (z_{1}(s)\,\,\,\ldots \,\,\,z_{n}(s)) \varphi \left( \begin{array}{cccccc} f_{1}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\\&+\int _{t_{k-1}}^{t_{k}}\zeta ((E+B)^{k}) \zeta \left( \begin{array}{cccccc} y_{1}(t) \\ \vdots \\ y_{n}(t) \\ \end{array} \right) \zeta (z_{1}(s)\,\,\,\ldots \,\,\,z_{n}(s)) \varphi \left( \begin{array}{cccccc} f_{1}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds\\&\left. +\int _{t_{k}}^{t} \zeta \left( \begin{array}{cccccc} y_{1}(t) \\ \vdots \\ y_{n}(t) \\ \end{array} \right) \zeta (z_{1}(s)\,\,\,\ldots \,\,\,z_{n}(s)) \varphi \left( \begin{array}{cccccc} f_{1}(s) \\ \vdots \\ f_{n}(s) \\ \end{array} \right) ds \right\} . \end{aligned}$$

The proof is finished. $\square$

Appendix 4. Proof of Theorem 3.11

Proof

According to Lemma 2.5, for $t\in [0,t_{1}],$ since $e^{At}$ is the fundamental matrix of $\dot{x}(t)=Ax(t),$ then each solution x(t) in the interval $[0,t_{1}]$ has the form

$$\begin{aligned} x(t)=e^{At}c+e^{At}\int _{0}^{t}e^{-As}f(s)ds, \end{aligned}$$

(93)

where c is a constant quaternion vector.

From (93), we can obtain

$$\begin{aligned} x(t_{1}^{-})=x(t_{1}) =e^{At_{1}}c+e^{At_{1}}\int _{0}^{t_{1}}e^{-As}f(s)ds. \end{aligned}$$

(94)

For $t\in (t_{1},t_{2}],$ each solution x(t) in the interval $(t_{1},t_{2}]$ has the form

$$\begin{aligned} x(t)=e^{At}a_{2}+e^{At}\int _{t_{1}}^{t}e^{-As}f(s)ds, \end{aligned}$$

(95)

where $a_{2}$ is a constant quaternion vector.

From (95), we can obtain

$$\begin{aligned} x(t_{1}^{+})=e^{At_{1}^{+}}a_{2}. \end{aligned}$$

(96)

Due to $\Delta x(t_{1})=Bx(t_{1}^{-}),$ we get $x(t_{1}^{+})=(E+B)x(t_{1}^{-}).$ From (94) and (96),

$$\begin{aligned} e^{At_{1}^{+}}a_{2}=(E+B)[e^{At_{1}}c+e^{At_{1}}\int _{0}^{t_{1}}e^{-As}f(s)ds], \end{aligned}$$

we have

$$\begin{aligned} a_{2}=(E+B)c+\int _{0}^{t_{1}}(E+B)e^{-As}f(s)ds. \end{aligned}$$

(97)

From (95) and (97), each solution x(t) in the interval $(t_{1},t_{2}]$ can be expressed as

$$\begin{aligned} x(t) =&\,e^{At}\left[ (E+B)c+\int _{0}^{t_{1}}(E+B)e^{-As}f(s)ds\right] +e^{At}\int _{t_{1}}^{t}e^{-As}f(s)ds\nonumber \\ =&\,(E+B)e^{At}c+\int _{0}^{t_{1}}(E+B)e^{A(t-s)}f(s)ds+\int _{t_{1}}^{t}e^{A(t-s)}f(s)ds. \end{aligned}$$

(98)

From (98), we can obtain

$$\begin{aligned} x(t_{2}^{-})=x(t_{2})=(E+B)e^{At_{2}}c+\int _{0}^{t_{1}}(E+B)e^{A(t_{2}-s)}f(s)ds +\int _{t_{1}}^{t_{2}}e^{A(t_{2}-s)}f(s)ds. \end{aligned}$$

(99)

For $t\in (t_{2},t_{3}],$ each solution x(t) in the interval $(t_{2},t_{3}]$ has the form

$$\begin{aligned} x(t)=e^{At}a_{3}+e^{At}\int _{t_{2}}^{t}e^{-As}f(s)ds, \end{aligned}$$

(100)

where $a_{3}$ is a constant quaternion vector.

From (100), we can obtain

$$\begin{aligned} x(t_{2}^{+})=e^{At_{2}^{+}}a_{3}. \end{aligned}$$

(101)

Due to $\Delta x(t_{2})=Bx(t_{2}^{-}),$ we get $x(t_{2}^{+})=(E+B)x(t_{2}^{-}).$ From (99) and (101),

$$\begin{aligned} e^{At_{2}^{+}}a_{3} =&\,(E+B)\left[ (E+B)e^{At_{2}}c+\int _{0}^{t_{1}}(E+B)e^{A(t_{2}-s)}f(s)ds\right. \\&\qquad \left. +\int _{t_{1}}^{t_{2}}e^{A(t_{2}-s)}f(s)ds\right] \\ =&\,(E+B)^{2}e^{At_{2}}c+\int _{0}^{t_{1}}(E+B)^{2}e^{A(t_{2}-s)}f(s)ds\\&\qquad +\int _{t_{1}}^{t_{2}}(E+B)e^{A(t_{2}-s)}f(s)ds], \end{aligned}$$

we have

$$\begin{aligned} a_{3}=(E+B)^{2}c+\int _{0}^{t_{1}}(E+B)^{2}e^{-As}f(s)ds +\int _{t_{1}}^{t_{2}}(E+B)e^{-As}f(s)ds. \end{aligned}$$

(102)

From (100) and (102), each solution x(t) in the interval $(t_{2},t_{3}]$ can be expressed as

$$\begin{aligned} x(t) =&\,e^{At}\left[ (E+B)^{2}c+\int _{0}^{t_{1}}(E+B)^{2}e^{-As}f(s)ds +\int _{t_{1}}^{t_{2}}(E+B)e^{-As}f(s)ds\right] \\&+e^{At}\int _{t_{2}}^{t}e^{-As}f(s)ds\\ =&\,(E+B)^{2}e^{At}c+\int _{0}^{t_{1}}(E+B)^{2}e^{A(t-s)}f(s)ds +\int _{t_{1}}^{t_{2}}(E+B)e^{A(t-s)}f(s)ds\\&+\int _{t_{2}}^{t}e^{A(t-s)}f(s)ds. \end{aligned}$$

Repeating this process, we obtain each solution x(t) in the interval $(t_{k-1},t_{k}]$ can be expressed as

$$\begin{aligned} x(t) =&\,(E+B)^{k-1}e^{At}c+\sum \limits _{i=1}^{k-2}\int _{t_{i-1}}^{t_{i}}(E+B)^{k-i}e^{A(t-s)}f(s)ds\nonumber \\&+\int _{t_{k-2}}^{t_{k-1}}(E+B)e^{A(t-s)}f(s)ds\nonumber \\&+\int _{t_{k-1}}^{t}e^{A(t-s)}f(s)ds. \end{aligned}$$

(103)

From (103), we can obtain

$$\begin{aligned} x(t_{k}^{-})=x(t_{k})=\,&(E+B)^{k-1}e^{At_{k}}c+\sum \limits _{i=1}^{k-2}\int _{t_{i-1}}^{t_{i}}(E+B)^{k-i}e^{A(t_{k}-s)}f(s)ds\nonumber \\&+\int _{t_{k-2}}^{t_{k-1}}(E+B)e^{A(t_{k}-s)}f(s)ds+\int _{t_{k-1}}^{t_{k}}e^{A(t_{k}-s)}f(s)ds. \end{aligned}$$

(104)

For $t\in (t_{k},t_{k+1}],$ each solution x(t) in the interval $(t_{k},t_{k+1}]$ has the form

$$\begin{aligned} x(t) =e^{At}a_{k+1}+e^{At}\int _{t_{k}}^{t}e^{-As}f(s)ds, \end{aligned}$$

(105)

where $a_{k+1}$ is a constant quaternion vector.

From (105), we can obtain

$$\begin{aligned} x(t_{k}^{+})=e^{At_{k}^{+}}a_{k+1}. \end{aligned}$$

(106)

Due to $\Delta x(t_{k})=Bx(t_{k}^{-}),$ we get $x(t_{k}^{+})=(E+B)x(t_{k}^{-}).$ From (104) and (106),

$$\begin{aligned} e^{At_{k}^{+}}a_{k+1} =&\,(E+B)\left[ (E+B)^{k-1}e^{At_{k}}c+\sum \limits _{i=1}^{k-2}\int _{t_{i-1}}^{t_{i}}(E+B)^{k-i}e^{A(t_{k}-s)}f(s)ds\right. \\&\left. +\int _{t_{k-2}}^{t_{k-1}}(E+B)e^{A(t_{k}-s)}f(s)ds+\int _{t_{k-1}}^{t_{k}}e^{A(t_{k}-s)}f(s)ds\right] \\ =&\,(E+B)^{k}e^{At_{k}}c+\sum \limits _{i=1}^{k-2}\int _{t_{i-1}}^{t_{i}}(E+B)^{k-i+1}e^{A(t_{k}-s)}f(s)ds\\&+\int _{t_{k-2}}^{t_{k-1}}(E+B)^{2}e^{A(t_{k}-s)}f(s)ds+\int _{t_{k-1}}^{t_{k}}(E+B)e^{A(t_{k}-s)}f(s)ds, \end{aligned}$$

we have

$$\begin{aligned} a_{k+1} =&\,(E+B)^{k}c+\sum \limits _{i=1}^{k-2}\int _{t_{i-1}}^{t_{i}}(E+B)^{k-i+1}e^{-As}f(s)ds\nonumber \\&+\int _{t_{k-2}}^{t_{k-1}}(E+B)^{2}e^{-As}f(s)ds+\int _{t_{k-1}}^{t_{k}}(E+B)e^{-As}f(s)ds. \end{aligned}$$

(107)

From (105) and (107), each solution x(t) in the interval $(t_{k},t_{k+1}]$ can be expressed as

$$\begin{aligned} x(t) =&\,e^{At}\left[ (E+B)^{k}c+\sum \limits _{i=1}^{k-2}\int _{t_{i-1}}^{t_{i}}(E+B)^{k-i+1}e^{-As}f(s)ds\right. \\&\left. +\int _{t_{k-2}}^{t_{k-1}}(E+B)^{2}e^{-As}f(s)ds+\int _{t_{k-1}}^{t_{k}}(E+B)e^{-As}f(s)ds\right] \\&+e^{At}\int _{t_{k}}^{t}e^{-As}f(s)ds\\ =&\,(E+B)^{k}e^{At}c+\sum \limits _{i=1}^{k-1}\int _{t_{i-1}}^{t_{i}}(E+B)^{k-i+1}e^{A(t-s)}f(s)ds\\&+\int _{t_{k-1}}^{t_{k}}(E+B)e^{A(t-s)}f(s)ds+\int _{t_{k}}^{t}e^{A(t-s)}f(s)ds. \end{aligned}$$

The proof is finished. $\square$

Appendix 5. (11) to (13)

Combining (11) and (12), we get

(108)

Equation (108) is equal to

$$\begin{aligned} \left\{ \begin{aligned} u_{1}(t)=&\, \left( \begin{array}{cccccc} 2 &{} 0 \\ 0 &{} 2 \\ \end{array} \right) \bigg [\left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) c_{11} - \left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) \overline{c_{12}} + \left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) c_{21} - \left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) \overline{c_{22}}\bigg ]\\&+ \left( \begin{array}{cccccc} 0 &{} 0 \\ 0 &{} \varvec{i} \\ \end{array} \right) \bigg [ - \overline{\left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) }\overline{c_{12}} - \overline{\left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) }c_{11} - \overline{\left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) }\overline{c_{22}} - \overline{\left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) }c_{21}\bigg ],\\ u_{2}(t)=&\, \left( \begin{array}{cccccc} 2 &{} 0 \\ 0 &{} 2 \\ \end{array} \right) \bigg [\left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) c_{12} + \left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) \overline{c_{11}} + \left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) c_{22} + \left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) \overline{c_{21}}\bigg ]\\&+ \left( \begin{array}{cccccc} 0 &{} 0 \\ 0 &{} \varvec{i} \\ \end{array} \right) \bigg [\overline{\left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) }\overline{c_{11}} - \overline{\left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) }c_{12} + \overline{\left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) }\overline{c_{21}} - \overline{\left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) }c_{22}\bigg ].\\ \end{aligned} \right. \end{aligned}$$

(109)

The system (109) is equal to

$$\begin{aligned} \left\{ \begin{aligned} u_{1}(t)=&\, \left( \begin{array}{cccccc} 2 &{} 0 \\ 0 &{} 2 \\ \end{array} \right) \bigg [\left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) c_{11} - \left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) \overline{c_{12}} + \left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) c_{21} - \left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) \overline{c_{22}}\bigg ]\\&+ \left( \begin{array}{cccccc} 0 &{} 0 \\ 0 &{} \varvec{i} \\ \end{array} \right) \bigg [ - \overline{\left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) }\overline{c_{12}} - \overline{\left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) }c_{11} - \overline{\left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) }\overline{c_{22}} - \overline{\left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) }c_{21}\bigg ],\\ -\overline{u_{2}(t)}=&\, \overline{\left( \begin{array}{cccccc} 2 &{} 0 \\ 0 &{} 2 \\ \end{array} \right) } \bigg [-\overline{\left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) }\overline{c_{12}} - \overline{\left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) }c_{11} - \overline{\left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) }\overline{c_{22}} - \overline{\left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) }c_{21}\bigg ]\\&- \overline{\left( \begin{array}{cccccc} 0 &{} 0 \\ 0 &{} \varvec{i} \\ \end{array} \right) } \bigg [\left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) c_{11} - \left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) \overline{c_{12}} + \left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) c_{21} - \left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) \overline{c_{22}}\bigg ].\\ \end{aligned} \right. \end{aligned}$$

(110)

The system (110) is equal to

$$\begin{aligned} \left( \begin{array}{cccccc} u_{1}(t) \\ -\overline{u_{2}(t)} \\ \end{array} \right) =&\, \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 2 &{} 0 \\ 0 &{} 2 \\ \end{array} \right) &{} \left( \begin{array}{cccccc} 0 &{} 0 \\ 0 &{} \varvec{i} \\ \end{array} \right) \\ -\overline{\left( \begin{array}{cccccc} 0 &{} 0 \\ 0 &{} \varvec{i} \\ \end{array} \right) } &{} \overline{\left( \begin{array}{cccccc} 2 &{} 0 \\ 0 &{} 2 \\ \end{array} \right) }\\ \end{array} \right) \nonumber \\&\qquad \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) c_{11} - \left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) \overline{c_{12}} + \left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) c_{21} - \left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) \overline{c_{22}}\\ - \overline{\left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) }\overline{c_{12}} - \overline{\left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) }c_{11} - \overline{\left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) }\overline{c_{22}} - \overline{\left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) }c_{21} \end{array} \right) \nonumber \\ =&\, \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 2 &{} 0 \\ 0 &{} 2 \\ \end{array} \right) &{} \left( \begin{array}{cccccc} 0 &{} 0 \\ 0 &{} \varvec{i} \\ \end{array} \right) \\ -\overline{\left( \begin{array}{cccccc} 0 &{} 0 \\ 0 &{} \varvec{i} \\ \end{array} \right) } &{} \overline{\left( \begin{array}{cccccc} 2 &{} 0 \\ 0 &{} 2 \\ \end{array} \right) }\\ \end{array} \right) \nonumber \\&\qquad \left[ \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) \\ - \overline{\left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) } \end{array} \right) c_{11} + \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) \\ - \overline{\left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) } \end{array} \right) c_{21} + \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) \\ \overline{\left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) } \end{array} \right) (-\overline{c_{12}}) \right. \nonumber \\&\quad \left. + \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) \\ \overline{\left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) } \end{array} \right) (-\overline{c_{22}}) \right] \nonumber \\ =&\, \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 2 &{} 0 \\ 0 &{} 2 \\ \end{array} \right) &{} \left( \begin{array}{cccccc} 0 &{} 0 \\ 0 &{} \varvec{i} \\ \end{array} \right) \\ -\overline{\left( \begin{array}{cccccc} 0 &{} 0 \\ 0 &{} \varvec{i} \\ \end{array} \right) } &{} \overline{\left( \begin{array}{cccccc} 2 &{} 0 \\ 0 &{} 2 \\ \end{array} \right) }\\ \end{array} \right) \nonumber \\&\qquad \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) \\ - \overline{\left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) } \end{array} \right) &{} \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) \\ - \overline{\left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) } \end{array} \right) &{} \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) \\ \overline{\left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) } \end{array} \right) &{} \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) \\ \overline{\left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) } \end{array} \right) \end{array} \right) \nonumber \\&\quad \left( \begin{array}{cccccc} c_{11} \\ c_{21} \\ -\overline{c_{12}}\\ -\overline{c_{22}}\\ \end{array} \right) . \end{aligned}$$

(111)

Due to

$$\begin{aligned}&\varphi ({x(t)})= \left( \begin{array}{cccccc} u_{1}(t)\\ -\overline{u_{2}(t)}\\ \end{array} \right) , \zeta \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 2 &{} 0 \\ 0 &{} 2+\varvec{k}\\ \end{array} \right) \end{array} \right) = \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 2 &{} 0 \\ 0 &{} 2 \\ \end{array} \right) &{} \left( \begin{array}{cccccc} 0 &{} 0 \\ 0 &{} \varvec{i} \\ \end{array} \right) \\ -\overline{\left( \begin{array}{cccccc} 0 &{} 0 \\ 0 &{} \varvec{i} \\ \end{array} \right) } &{} \overline{\left( \begin{array}{cccccc} 2 &{} 0 \\ 0 &{} 2 \\ \end{array} \right) }\\ \end{array} \right) ,\\&\varphi \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{(1+\varvec{j})t}\\ 0 \\ \end{array} \right) \end{array} \right) = \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) \\ - \overline{\left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) } \end{array} \right) , \varphi \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) \end{array} \right) = \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) \\ - \overline{\left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) } \end{array} \right) ,\\&\varphi \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{(1+\varvec{j})t}\\ 0 \\ \end{array} \right) \end{array} \right) ^{*} = \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) \\ \overline{\left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) } \end{array} \right) , \varphi \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) \end{array} \right) ^{*} = \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) \\ \overline{\left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) } \end{array} \right) ,\\&\zeta \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{(1+\varvec{j})t} &{} 0 \\ 0 &{} e^{t} \\ \end{array} \right) \end{array} \right) = \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) \\ - \overline{\left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) } \end{array} \right) &{} \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) \\ - \overline{\left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) } \end{array} \right) &{} \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{t}\sin t \\ 0 \\ \end{array} \right) \\ \overline{\left( \begin{array}{cccccc} e^{t}\cos t \\ 0 \\ \end{array} \right) } \end{array} \right) &{} \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) \\ \overline{\left( \begin{array}{cccccc} 0 \\ e^{t} \\ \end{array} \right) } \end{array} \right) \end{array} \right) ,\\&\varphi \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} c_{1}\\ c_{2}\\ \end{array} \right) \end{array} \right) = \left( \begin{array}{cccccc} c_{11}\\ c_{21}\\ -\overline{c_{12}}\\ -\overline{c_{22}}\\ \end{array} \right) . \end{aligned}$$

Equation (111) can be expressed by

$$\begin{aligned} \varphi (x(t)) =&\,\zeta \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 2 &{} 0 \\ 0 &{} 2+\varvec{k}\\ \end{array} \right) \end{array} \right) \zeta \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{(1+\varvec{j})t} &{} 0 \\ 0 &{} e^{t} \\ \end{array} \right) \end{array} \right) \varphi \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} c_{1}\\ c_{2}\\ \end{array} \right) \end{array} \right) . \end{aligned}$$

Appendix 6. $\Phi ^{-1}(t)$

Owing to

$$\begin{aligned} \Phi (t)= \left( \begin{array}{cccccc} e^{(1+\varvec{j})t} &{} 0 \\ 0 &{} e^{t} \\ \end{array} \right) , \end{aligned}$$

and it is obvious that

$$\begin{aligned} \Phi ^{+}(t)= \left( \begin{array}{cccccc} e^{(1-\varvec{j})t} &{} 0 \\ 0 &{} e^{t} \\ \end{array} \right) . \end{aligned}$$

According to Definition 2.1,

$$\begin{aligned} ddet\Phi (t)=&\,{\det }_{p}(\Phi ^{+}(t)\Phi (t))\\ =&\,{\det }_{p} \bigg [ \left( \begin{array}{cccccc} e^{(1-\varvec{j})t} &{} 0 \\ 0 &{} e^{t} \\ \end{array} \right) \left( \begin{array}{cccccc} e^{(1+\varvec{j})t} &{} 0 \\ 0 &{} e^{t} \\ \end{array} \right) \bigg ] ={\det }_{p} \left( \begin{array}{cccccc} e^{2t} &{} 0 \\ 0 &{} e^{2t} \\ \end{array} \right) =e^{4t}\ne 0, \end{aligned}$$

therefore, $\Phi (t)$ has inverse matrix $\Phi ^{-1}(t).$

If we represent $\Phi ^{-1}(t)=(b_{jk}),$ according to Lemma 2.2,

$$\begin{aligned}&w_{11}={\det }_{p} \bigg [ \left( \begin{array}{cccccc} 0 &{} e^{t} \\ 1 &{} 0 \\ \end{array} \right) \left( \begin{array}{cccccc} 0 &{} e^{(1+\varvec{j})t} \\ e^{t} &{} 0 \\ \end{array} \right) \bigg ] ={\det }_{p} \left( \begin{array}{cccccc} e^{2t} &{} 0 \\ 0 &{} e^{(1+\varvec{j})t} \\ \end{array} \right) =e^{(3+\varvec{j})t},\\&\overline{b_{11}}=\frac{w_{11}}{ddet\Phi (t)}=\frac{e^{(3+\varvec{j})t}}{e^{4t}}=e^{(-1+\varvec{j})t}, b_{11}=e^{(-1-\varvec{j})t},\\&w_{12}={\det }_{p} \bigg [ \left( \begin{array}{cccccc} e^{(1-\varvec{j})t} &{} 0 \\ 1 &{} 0 \\ \end{array} \right) \left( \begin{array}{cccccc} e^{(1+\varvec{j})t} &{} 0 \\ 0 &{} e^{t} \\ \end{array} \right) \bigg ] ={\det }_{p} \left( \begin{array}{cccccc} e^{2t} &{} 0 \\ e^{(1+\varvec{j})t} &{} 0 \\ \end{array} \right) =0,\\&\overline{b_{21}}=\frac{w_{12}}{ddet\Phi (t)}=0, b_{21}=0,\\&w_{21}={\det }_{p} \bigg [ \left( \begin{array}{cccccc} 0 &{} e^{t} \\ 0 &{} 1 \\ \end{array} \right) \left( \begin{array}{cccccc} 0 &{} e^{(1+\varvec{j})t} \\ e^{t} &{} 0 \\ \end{array} \right) \bigg ] ={\det }_{p} \left( \begin{array}{cccccc} e^{2t} &{} 0 \\ e^{t} &{} 0 \\ \end{array} \right) =0,\\&\overline{b_{12}}=\frac{w_{21}}{ddet\Phi (t)}=0, b_{12}=0,\\&w_{22}={\det }_{p} \bigg [ \left( \begin{array}{cccccc} e^{(1-\varvec{j})t} &{} 0 \\ 0 &{} 1 \\ \end{array} \right) \left( \begin{array}{cccccc} e^{(1+\varvec{j})t} &{} 0 \\ 0 &{} e^{t} \\ \end{array} \right) \bigg ] ={\det }_{p} \left( \begin{array}{cccccc} e^{2t} &{} 0 \\ 0 &{} e^{t} \\ \end{array} \right) =e^{3t},\\&\overline{b_{22}}=\frac{w_{22}}{ddet\Phi (t)}=\frac{e^{3t}}{e^{4t}}=e^{-t}, b_{22}=e^{-t}, \end{aligned}$$

and it is obvious that

$$\begin{aligned} \Phi ^{-1}(t)= \left( \begin{array}{cccccc} e^{(-1-\varvec{j})t} &{} 0 \\ 0 &{} e^{-t} \\ \end{array} \right) . \end{aligned}$$

Appendix 7. (26) to (28)

Combining (26) and (27), we get

(112)

Equation (112) is equal to

(113)

The system (113) is equal to

(114)

The system (114) is equal to

(115)

Due to

$$\begin{aligned}&\zeta \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{(1+\varvec{j})t} &{} 0 \\ 0 &{} e^{t} \\ \end{array} \right) \end{array} \right) = \left( \begin{array}{cccccc} &{}\left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{t}\cos t &{} 0 \\ \end{array} \right) &{} \left( \begin{array}{cccccc} e^{t}\sin t &{} 0 \\ \end{array} \right) \end{array} \right) \\ &{}\left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 0 &{} e^{t} \\ \end{array} \right) &{} \left( \begin{array}{cccccc} 0 &{} 0 \\ \end{array} \right) \\ \end{array} \right) \\ &{}\left( \begin{array}{cccccc} - \overline{\left( \begin{array}{cccccc} e^{t}\sin t &{} 0 \\ \end{array} \right) } &{} \overline{\left( \begin{array}{cccccc} e^{t}\cos t &{} 0 \\ \end{array} \right) } \end{array} \right) \\ &{}\left( \begin{array}{cccccc} - \overline{\left( \begin{array}{cccccc} 0 &{} 0 \\ \end{array} \right) } &{} \overline{\left( \begin{array}{cccccc} 0 &{} e^{t} \\ \end{array} \right) }\\ \end{array} \right) \end{array} \right) ,\\&\varphi \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{(-1-\varvec{j})s}\\ 0 \\ \end{array} \right) \end{array} \right) = \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{-s}\cos s \\ 0 \\ \end{array} \right) \\ - \overline{\left( \begin{array}{cccccc} -e^{-s}\sin s \\ 0 \\ \end{array} \right) } \end{array} \right) , \varphi \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 0 \\ e^{-s} \\ \end{array} \right) \end{array} \right) = \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 0 \\ e^{-s} \\ \end{array} \right) \\ - \overline{\left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) } \end{array} \right) ,\\&\varphi \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{(-1-\varvec{j})s}\\ 0 \\ \end{array} \right) \end{array} \right) ^{*} = \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} -e^{-s}\sin s \\ 0 \\ \end{array} \right) \\ \overline{\left( \begin{array}{cccccc} e^{-s}\cos s \\ 0 \\ \end{array} \right) } \end{array} \right) , \varphi \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 0 \\ e^{-s} \\ \end{array} \right) \end{array} \right) ^{*} = \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) \\ \overline{\left( \begin{array}{cccccc} 0 \\ e^{-s} \\ \end{array} \right) } \end{array} \right) ,\\&\zeta \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{(-1-\varvec{j})s} &{} 0 \\ 0 &{} e^{-s} \\ \end{array} \right) \end{array} \right) = \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{-s}\cos s \\ 0 \\ \end{array} \right) \\ - \overline{\left( \begin{array}{cccccc} -e^{-s}\sin s \\ 0 \\ \end{array} \right) } \end{array} \right) &{} \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 0 \\ e^{-s} \\ \end{array} \right) \\ - \overline{\left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) } \end{array} \right) &{} \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} -e^{-s}\sin s \\ 0 \\ \end{array} \right) \\ \overline{\left( \begin{array}{cccccc} e^{-s}\cos s \\ 0 \\ \end{array} \right) } \end{array} \right) \\ &{}\quad \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 0 \\ 0 \\ \end{array} \right) \\ \overline{\left( \begin{array}{cccccc} 0 \\ e^{-s} \\ \end{array} \right) } \end{array} \right) \end{array} \right) ,\\&\varphi \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} s\varvec{i} \\ (1+s)\varvec{j}\\ \end{array} \right) \end{array} \right) = \left( \begin{array}{cccccc} s\varvec{i} \\ 0\varvec{i} \\ -\overline{0} \\ -\overline{(1+s)} \\ \end{array} \right) . \end{aligned}$$

Equation (115) can be expressed by

$$\begin{aligned} \varphi (x(t)) =&\,\zeta \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 2 &{} 0 \\ 0 &{} 2+\varvec{k}\\ \end{array} \right) \end{array} \right) \zeta \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{(1+\varvec{j})t} &{} 0 \\ 0 &{} e^{t} \\ \end{array} \right) \end{array} \right) \varphi \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} c_{1}\\ c_{2}\\ \end{array} \right) \end{array} \right) \\&+\int _{0}^{t_{1}} \zeta \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} 2 &{} 0 \\ 0 &{} 2+\varvec{k}\\ \end{array} \right) \end{array} \right) \zeta \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{(1+\varvec{j})t} &{} 0 \\ 0 &{} e^{t} \\ \end{array} \right) \end{array} \right) \zeta \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{(-1-\varvec{j})s} &{} 0 \\ 0 &{} e^{-s} \\ \end{array} \right) \end{array} \right) \\&\quad \varphi \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} s\varvec{i} \\ (1+s)\varvec{j}\\ \end{array} \right) \end{array} \right) ds,\\&+\int _{t_{1}}^{t} \zeta \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{(1+\varvec{j})t} &{} 0 \\ 0 &{} e^{t} \\ \end{array} \right) \end{array} \right) \zeta \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} e^{(-1-\varvec{j})s} &{} 0 \\ 0 &{} e^{-s} \\ \end{array} \right) \end{array} \right) \varphi \left( \begin{array}{cccccc} \left( \begin{array}{cccccc} s\varvec{i} \\ (1+s)\varvec{j}\\ \end{array} \right) \end{array} \right) ds, \end{aligned}$$

where $c_{1},c_{2}$ are arbitrary quaternions.