Then, the following theorem holds.

*For the decomposition of system* (3) *as the cascade connected first-order commutative pair of subsystems* (4) *and* (5), *that is C* = *AB* = *BA, it is necessary and sufficient that*

$${a}_{1}\left(t\right)=\sqrt{\frac{{c}_{2}\left(t\right)}{{k}_{1}},}$$(6)

$${a}_{0}\left(t\right)=\frac{2{c}_{1}\left(t\right)-{\dot{c}}_{2}\left(t\right)}{4\sqrt{{k}_{1}{c}_{2}\left(t\right)}}-\frac{{k}_{0}}{2{k}_{1}},$$(7)

$${b}_{1}\left(t\right)=\sqrt{{k}_{1}{c}_{2}\left(t\right),}$$(8)

$${b}_{0}\left(t\right)=\frac{2{c}_{1}\left(t\right)-{\dot{c}}_{2}\left(t\right)}{4}\sqrt{\frac{{k}_{1}}{{c}_{2}\left(t\right)}}+\frac{{k}_{0}}{2},$$(9)

*where k*_{0} *and k*_{1} *are arbitrary constants in R satisfying*

$$\frac{3{\left({\dot{c}}_{2}\left(t\right)\right)}^{2}}{16{c}_{2}\left(t\right)}-\frac{{\ddot{c}}_{2}\left(t\right)}{4}+\frac{{c}_{1}\left(t\right)-2{\dot{c}}_{2}\left(t\right)}{4{c}_{2}\left(t\right)}{c}_{1}\left(t\right)+\frac{{\dot{c}}_{1}\left(t\right)}{2}-\frac{{k}_{0}^{2}}{4{k}_{1}}={c}_{0}\left(t\right).$$(10)

*Further if the equivalence is to be valid under non-zero initial conditions y*(0), *ẏ*(0). *the constraint*

$${y}_{1}\left(0\right)={y}_{2}\left(0\right)=y\left(0\right)$$(11)

*between the initial conditions, and the relation*

$${k}_{1}+{k}_{0}=1$$(12)

*between the arbitrary constants should be included in the necessary and sufficient conditions in* (6), (7), (8), (9), (10).

Consider the connection *AB*; that is the input *x* of *AB* is that of *A*, the output of *A* is equal to the input of *B*, the output of *B* is that of *AB*; which can be expressed as

$$x={x}_{1},$$(13)

$${y}_{1}={x}_{2},$$(14)

$${y}_{2}=y.$$(15)

Differentiating (5), we obtain

$${\dot{b}}_{1}{\dot{y}}_{2}+{b}_{1}{\ddot{y}}_{2}+{\dot{b}}_{0}{y}_{2}+{b}_{0}{\dot{y}}_{2}={\dot{x}}_{2}.$$(16)

From (14), and then from (4)

$${\dot{x}}_{2}={\dot{y}}_{1}=\frac{{x}_{1}-{a}_{0}{y}_{1}}{{a}_{1}}.$$(17)

Inserting the value *x*_{1} = *x* from (13), and the value

$${y}_{1}={x}_{2}={b}_{1}{\dot{y}}_{2}+{b}_{0}{y}_{2}$$(18)

from (14), and (5) all in to (17), we obtain

$${\dot{x}}_{2}=\frac{x-{a}_{0}\left({b}_{1}{\dot{y}}_{2}+{b}_{0}{y}_{2}\right)}{{a}_{1}}.$$(19)

Finally, substituting this expression in (16) and letting *y*_{2} = *y* from (15), (16) is transformed to

$${b}_{1}{a}_{1}\ddot{y}+\left[{\dot{b}}_{1}{a}_{1}+{b}_{1}{a}_{0}+{b}_{0}{a}_{1}\right]\dot{y}+\left[{\dot{b}}_{0}{a}_{1}+{b}_{0}{a}_{0}\right]y=x$$(20)

after rearrangements. This equation describes the interconnection *AB* with the initial conditions

$$y\left(0\right)={y}_{2}\left(0\right),$$(21)

$$\dot{y}\left(0\right)={\dot{y}}_{2}\left(0\right)=\frac{{x}_{2}\left(0\right)-{b}_{0}\left(0\right){y}_{2}\left(0\right)}{{b}_{1}\left(0\right)}=\frac{{y}_{1}\left(0\right)-{b}_{0}\left(0\right){y}_{2}\left(0\right)}{{b}_{1}\left(0\right)},$$(22)

where the first equation follows from (15); the subsequent equalities in the second equation follow from (15), (5), and (14).

When the interconnection *BA* is considered, a similar procedure can be followed to derive

$${a}_{1}{b}_{1}\ddot{y}+\left[{\dot{a}}_{1}{b}_{1}+{a}_{1}{b}_{0}+{a}_{0}{b}_{1}\right]\dot{y}+\left[{\dot{a}}_{0}{b}_{1}+{b}_{0}{a}_{0}\right]y=x,$$(23)

$$y\left(0\right)={y}_{1}\left(0\right),$$(24)

$$\dot{y}\left(0\right)=\frac{{y}_{2}\left(0\right)-{a}_{0}\left(0\right){y}_{1}\left(0\right)}{{a}_{1}\left(0\right)}.$$(25)

These equations can also be obtained straightforwardly from (20), (21) and (22), respectively, by interchanging *a*_{i} ↔ *b*_{i}, (*i* = 0, 1) and *y*_{1} ↔ *y*_{2}.

The first commutativity condition for subsystems *A* and *B* is reduced from the general theorem (Koksal 1) of [12] as

$$\begin{array}{c}\left[\begin{array}{c}{b}_{1}\\ {b}_{0}\end{array}\right]\end{array}=\left[\begin{array}{cc}{a}_{1}& 0\\ {a}_{0}& 1\end{array}\right]\left[\begin{array}{c}{k}_{1}\\ {k}_{0}\end{array}\right],$$(26)

where *k*_{1} and *k*_{0} are arbitrary constants. And the second commutativity condition is reduced from Eq. 3.2a of [12] as

$$\left[\begin{array}{c}{y}_{1}\left(0\right)\\ {y}_{2}\left(0\right)-{a}_{0}\left(0\right){y}_{1}\left(0\right)\\ {a}_{1}\left(0\right)\end{array}\right]=\left[\begin{array}{c}{y}_{2}\left(0\right)\\ {y}_{1}\left(0\right)-{b}_{0}\left(0\right){y}_{2}\left(0\right)\\ {b}_{1}\left(0\right)\end{array}\right].$$(27)

For the case of first-order subsystems considered in the present proof, (27) results from the comparison of (21), (22), and (24), (25). Equation (26) is obvious by equating the coefficients of (20) and (23). In fact the equality of the coefficients of the first derivatives of *ẏ* yields

$${\dot{a}}_{1}{b}_{1}={b}_{1}{a}_{1}\to {b}_{1}={k}_{1}{a}_{1},$$(28)

which is the first line equation in (26). The second line equation can be obtained by equating the coefficients of y in (20) and (23), which yields together with (28)

$${\dot{a}}_{0}{b}_{1}={\dot{b}}_{0}{a}_{1}\to {b}_{0}={k}_{1}{a}_{0}+{k}_{0}.$$(29)

Before proceeding further, we focus on the second line equation in (27). For nonzero initial conditions, this equation together with the first line equation and (26) yields

$$\frac{1-{a}_{0}\left(0\right)}{{a}_{1}\left(0\right)}{y}_{1}\left(0\right)=\frac{1-{b}_{0}\left(0\right)}{{b}_{1}\left(0\right)}{y}_{1}\left(0\right)=\frac{1-{a}_{0}\left(0\right){k}_{1}-{k}_{0}}{{a}_{1}\left(0\right){k}_{1}}{y}_{1}\left(0\right),$$(30)

which results in

$${k}_{1}+{k}_{0}=1$$(31)

for the nonzero initial conditions *y*_{1} (0) = *y*_{2}(0).

The intermediate step of the proof is to eliminate the coefficients *b*_{1} (*t*), *b*_{0}(*t*) and the initial condition *y*_{2}(0) from the description (20), (21), (22) of the interconnection *AB*. In fact, using (26) and (27) in (20), (21), (22), we obtain

$${a}_{1}^{2}{k}_{1}\ddot{y}+\left[{\dot{a}}_{1}{a}_{1}{k}_{1}+2{a}_{1}{a}_{0}{k}_{1}+{a}_{1}{k}_{0}\right]\dot{y}+\left[{a}_{1}{\dot{a}}_{0}{k}_{1}+{a}_{0}^{2}{k}_{1}+{a}_{0}{k}_{0}\right]y=x,$$(32)

$$y\left(0\right)={y}_{1}\left(0\right),$$(33)

$$\dot{y}\left(0\right)=\frac{1-{k}_{0}-{a}_{0}\left(0\right){k}_{1}}{{a}_{1}\left(0\right){k}_{1}}{y}_{1}\left(0\right)=\frac{1-{a}_{0}\left(0\right)}{{a}_{1}\left(0\right)}{y}_{1}\left(0\right)=\frac{1-{a}_{0}\left(0\right)}{{a}_{1}\left(0\right)}y\left(0\right).$$(34)

Since it is well known that (3) and (32) will have identical input-output pairs (*x*(*t*), *y*(*t*)) for all inputs *x*(*t*) and the initial conditions *y*(0) and *ẏ*(0) if and only if they have the same continuous time-varying coefficients, for their equivalence it is required that

$${a}_{1}^{2}{k}_{1}={c}_{2},$$(35)

$${\dot{a}}_{1}{a}_{1}{k}_{1}+2{a}_{1}{a}_{0}{k}_{1}+{a}_{1}{k}_{0}={c}_{1},$$(36)

$${a}_{1}{\dot{a}}_{0}{k}_{1}+{a}_{0}^{2}{k}_{1}+{a}_{0}{k}_{0}={c}_{0}.$$(37)

In fact (35) yields (6). Moreover, (36) yields (7) since ${\dot{a}}_{1}=\frac{{\dot{c}}_{2}}{2{a}_{1}{k}_{1}}$ from (35). Inserting (6) and (7) in (26) yields (8) and (9) respectively. Finally, substituting (6) and (7) in (37) yields (10).

Note that (11) and (12) have already been proved as seen in (21), (24) and (31), respectively. Hence the proof of the theorem is completed.

Theorem 2.1 and its proof reveal some important results which are expressed by the following corollaries. ◻

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