In this example, the constants *k*_{1} and *k*_{0} which are arbitrarily selectable when initial conditions are zero are tried to be chosen so that both the second-order system *C* and its commutative subsystems *A* and *B* are all stable. Let *C* be in the form

$$\begin{array}{cc}\text{C:}& \stackrel{\mathrm{..}}{\ddot{y}}(t)+2(2-\mathrm{cos}t)\dot{y}(t)+{c}_{0}(t)y(t)=x(t).\end{array}$$(50)

Obviously, *c*_{2}(*t*) = 1 and *c*_{1}(*t*) = 4 −2*cost*. Hence, (10) yields that *c*_{0}(*t*) will be in the form

$${c}_{0}(t)=(2-\mathrm{cos}\phantom{\rule{thinmathspace}{0ex}}t{)}^{2}+\mathrm{sin}\phantom{\rule{thinmathspace}{0ex}}t-\frac{{k}_{0}^{2}}{4{k}_{1}}.$$(51)

Using (6), (7) and (8), (9), the subsystems *A* and *B* become

$$\begin{array}{cc}\text{A:}& \frac{1}{\sqrt{{k}_{1}}}{\dot{y}}_{1}(t)+\left(\frac{2-\mathrm{cos}t}{\sqrt{{k}_{1}}}-\frac{{k}_{0}}{2{k}_{1}}\right){y}_{1}(t)={x}_{1}\left(t\right),\end{array}$$(52)

$$\begin{array}{cc}\text{B:}& \sqrt{{k}_{1}}{\dot{y}}_{2}(t)+\left((2-\mathrm{cos}t)\sqrt{{k}_{1}}-\frac{{k}_{0}}{2}\right){y}_{2}(t)={x}_{2}(t).\end{array}$$(53)

The characteristic values of *A* and *B* are obviously

$${\text{D}}_{A}=\frac{{k}_{0}}{2\sqrt{{k}_{1}}}-2+\mathrm{cos}t,$$(54)

$${\text{D}}_{B}=\frac{{k}_{0}}{2\sqrt{{k}_{1}}}-2-\mathrm{cos}t.$$(55)

It can be shown very easily that both of these values remain negative for all *t* ≥ 0 if

$$-2\sqrt{{k}_{1}}<{k}_{0}<2\sqrt{{k}_{1}}.$$(56)

Hence, when (*k*_{0}, *k*_{1}) is chosen inside of the parabola ${k}_{1}=\frac{{k}_{0}^{2}}{4}\left({k}_{0}=\mp 2\sqrt{{k}_{1}}\right)$ as shown in Fig. 2, both *D*_{A} and *D*_{B} will always remain negative (this may not be sufficient for stability [16, 18]) and *A* and *B* most likely be stable subsystems, so is *C* = *AB* = *BA*. When the initial conditions are not zero, the additional equation (12) should also be satisfied for commutativity, the line *k*_{0} + *k*_{1} = 1 is also shown in Fig. 2. For two different choices of *k*_{0}, *k*_{1}, the outputs of the systems *C*, *AB* and *BA* are observed on the MATLAB Scope and plotted as shown in Fig. 3.

In Case I, *k*_{1} = 4, *k*_{0} = −3 (point P in Fig. 2), the subsystems *A*, *B* and their cascade connections *C* = *AB* = *BA* as found from (52), (53) and (51) become

$$\begin{array}{cc}\text{A:}& \frac{1}{2}{\dot{y}}_{1}(t)+(1.375-0.5\mathrm{cos}t){y}_{1}(t)={x}_{1}(t),\text{\hspace{0.17em}}{y}_{1}\left(0\right)=1,\end{array}$$(57)

$$\begin{array}{cc}\text{B:}& 2{\dot{y}}_{2}(t)+(2.5-2\mathrm{cos}t){y}_{2}(t)={x}_{2}(t),\text{\hspace{0.17em}}{y}_{2}\left(0\right)=1,\end{array}$$(58)

$$\begin{array}{cc}\text{C:}& \ddot{y}(t)+(4-2\mathrm{cos}t)\dot{y}(t)+(4\mathrm{cos}t-0.5\mathrm{cos}2t-\mathrm{sin}t-3.9375)\text{\hspace{0.17em}}y(t)=x(t);\\ & y(0)=1,\text{\hspace{0.17em}}\dot{y}(0)=1.\end{array}$$(59)

Fig. 2 Different regions of *k*_{1} − *k*_{0} plane for possibly stable choices of *C* and its stable subsystems *A*, *B*.

Fig. 3 Outputs of Example 2 for Cases I and II.

Note that *y*_{1}(0) = *y*_{2}(0) = *y*(0) and *ẏ*(0) for *C* is computed from (38); hence, all the conditions of the main theorem and its corollaries are satisfied. All of the systems *C*, *AB* and *BA* are excited by *x*(*t*) = −5 + 10*sint* superimposed by a square wave of amplitude 20, period 5 and pulse with % 5. It is observed in Fig. 3 that all the outputs are the same (*CI* = *ABI* = *BAI*) and verify the theory. It is pertinent to note that when the simulation time is extended far beyond *t* = 10, all the systems’ outputs remain bounded, that is they are all stable as expected.

In Case II, *k*_{1} = 4, *k*_{0} = 1 as shown in Fig. 2 by point *Q*. For this case, the subsystems *A*, *B* and the second-order system *C* become

$$\begin{array}{cc}\mathrm{A}:& \frac{1}{2}{\dot{y}}_{1}(t)+\frac{1-\phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}\phantom{\rule{thinmathspace}{0ex}}t}{2}{y}_{1}(t)={x}_{1}(t),\phantom{\rule{thickmathspace}{0ex}}{y}_{1}\left(0\right)=1,\end{array}$$(60)

$$\begin{array}{l}\mathrm{B}:2{\dot{y}}_{2}(t)+(6-2\mathrm{cos}t){y}_{2}(t)={x}_{2}(t),\phantom{\rule{thickmathspace}{0ex}}{y}_{2}\left(0\right)=0,\\ \mathrm{C}:\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\ddot{\mathit{y}}(t)+2(2-\text{cos}\phantom{\rule{thinmathspace}{0ex}}t)\dot{y}(t)+\left(3+{\text{cos}}^{2}\phantom{\rule{thinmathspace}{0ex}}t-\mathrm{sin}t-4\phantom{\rule{thinmathspace}{0ex}}\text{cos}\phantom{\rule{thinmathspace}{0ex}}t\right)y(t)=x(t)\end{array}$$(61)

$$y(0)=\dot{y}(0)=0,$$(62)

which are found as mentioned in the previous case. For the same input of Case I, all the outputs are observed to be equal as in the first case and are shown in Fig. 3 (*CII* 0 = *ABII* 0 = *BAII* 0). Note that *k*_{1} + *k*_{0} ≠ 1 for this case, which is not a necessary condition for commutativity in the case of zero initial conditions. We observe in this figure that due to the strong stability of the system, the initial condition response dies away quickly so that the total response of Case I approaches to that of the input response (response with zero initial conditions) of Case II.

Case III is exactly the same as Case II except nonzero initial conditions are assigned to *A*, *B* and *C*. In fact *A*, *B* and *C* are assigned to have the same initial conditions for their outputs, that is *y*_{1}(0) = *y*_{2}(0) = *y*(0) = 4 and hence, (11) of the main Theorem is satisfied; *ẏ*(0) = 2 as to satisfy the condition (38) of Corollary 2.3. Since the condition of (12) of this Corollary is not satisfied (*k*_{1} + *k*_{0} ≠ 1), we should not have equal outputs for *C*, *AB* and *BA*; see Fig. 4 (*CIII* ≠ *ABIII* ≠ *BAIII*). We note again that due to the strong stability of the systems *C*, *AB*, *BA*, the initial condition responses die away quickly and the outputs become equal as *t* → ∞. This is because of the decomposition of *C* into the commutative pair *AB* (or *BA*) is valid under zero initial conditions.

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