**4. Example**

Let X = U = -2 where -2 is the is the space of all sequences *x* = (*<sup>x</sup>*1, *x*2, ... , *xn*, ...) with the norm *x* = ∞∑*i*=1 <sup>|</sup>*xi*|<sup>2</sup>1/2. The space -2 is a separable Hilbert space ([51], § 56). Consider a linear control system:

$$
\dot{\mathbf{x}}(t) = Ax(t) + Bu(t), \quad t \in \mathbb{R}, \quad \mathbf{x} \in \mathfrak{X}, \quad u \in \mathfrak{U}, \tag{55}
$$

where

$$A \colon (\mathbf{x}\_1, \mathbf{x}\_2, \mathbf{x}\_3, \mathbf{x}\_4, \dots) \mapsto (-\mathbf{x}\_2, \mathbf{x}\_1, -\mathbf{x}\_4, \mathbf{x}\_3, \dots),\tag{56}$$

$$B: \ (\mathbf{x}\_1, \mathbf{x}\_2, \mathbf{x}\_3, \mathbf{x}\_4, \dots) \mapsto (\mathbf{x}\_1, 0, \mathbf{x}\_2, 0, \mathbf{x}\_3, 0, \dots). \tag{57}$$

Considering elements of -2 as column-vectors with an infinite number of coordinates, one can identify the operators *A* and *B* with the following matrices with an infinite number of rows and columns: 

$$A = \begin{bmatrix} 0 & -1 & 0 & 0 & \dots \\ 1 & 0 & 0 & 0 & \dots \\ 0 & 0 & 0 & -1 & \dots \\ 0 & 0 & 1 & 0 & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix}, \qquad B = \begin{bmatrix} 1 & 0 & 0 & \dots \\ 0 & 0 & 0 & \dots \\ 0 & 1 & 0 & \dots \\ 0 & 0 & 0 & \dots \\ \vdots & \vdots & \vdots & \ddots \end{bmatrix}. \tag{58}$$

Consider the matrices *F* = 0 −1 1 0 , *G* = 10. One can write the matrices (58) in the following

block-diagonal form:

It

state

$$A = \text{diag}\left\{ F, F, \dots, F, \dots \right\}, \quad B = \text{diag}\left\{ G, G, \dots, G, \dots \right\}. \tag{59}$$

We will use the following denotations for the matrices of the form (59):

$$A = \text{diag}\, [F]\_{\infty}, \quad B = \text{diag}\, [G]\_{\infty}.$$

Set *G*1 := *FG*. Then *G*1 = 01. Hence, *AB* = diag [*<sup>G</sup>*1]<sup>∞</sup>. Therefore, span {*B*U, *AB*U} = X. follows that system (55), (56), (57) is exactly controllable on [0, *ϑ*] for any *ϑ* > 0. Let us take *ϑ* = *π*. Let us show that the upper Bohl exponent of system (55), (56), (57) is arbitrarily assignable by linear feedback(18).

 Consider the system:

$$
\dot{y}(t) = Fy(t), \quad t \in \mathbb{R}, \quad y \in \mathbb{R}^2. \tag{60}
$$

The evolution operator <sup>Γ</sup>(*<sup>t</sup>*,*<sup>s</sup>*) of system (60) has the form <sup>Γ</sup>(*<sup>t</sup>*,*<sup>s</sup>*) = Γ(*t* − *s*) where Γ(*t*) = exp(*Ft*). Calculating the matrix exponent, we obtain that:

$$
\Gamma(t) = \begin{bmatrix}
\cos t & -\sin t \\
\sin t & \cos t
\end{bmatrix}.
$$

.

Let us construct the evolution operator <sup>Φ</sup>(*<sup>t</sup>*,*<sup>s</sup>*) of the free system:

$$
\dot{\mathfrak{x}}(t) = A\mathfrak{x}(t), \quad t \in \mathbb{R}.
$$

We obtain <sup>Φ</sup>(*<sup>t</sup>*,*<sup>s</sup>*) = Φ(*t* − *s*) where Φ(*t*) = diag [Γ(*t*)]<sup>∞</sup>. Hence, <sup>Φ</sup>(*τ* + *s*, *τ*) = diag [Γ(*s*)]<sup>∞</sup>. For any *y* = col(*y*1, *y*2) ∈ R<sup>2</sup> we have:

$$\left\|\Gamma(s)y\right\|^2 = \left(y\_1\cos s - y\_2\sin s\right)^2 + \left(y\_1\sin s + y\_2\cos s\right)^2 = y\_1^2 + y\_2^2 = \left\|y\right\|^2.$$

From this, it follows that <sup>Φ</sup>(*τ* + *s*, *τ*)*x* <sup>2</sup> = *x* <sup>2</sup> for all *τ*,*s* ∈ R and *x* ∈ X. Hence, <sup>Φ</sup>(*τ* + *s*, *τ*) = 1. So, ln <sup>Φ</sup>(*τ* + *s*, *τ*) = 0. Thus, κ(*A*) = 0.

Let an arbitrary *μ* ∈ R be given. Set *λ* := *μ* − κ(*A*) = *μ*. Let us construct *<sup>U</sup>*(·), by Theorem 2, that ensures equality (50), and, hence, by Corollary 1, kinematic similarity of systems (51) and (7), and, thus, the equality κ(*<sup>A</sup>* + *BU*) = *λ* = *μ*. Set *σ* := 2*ϑ* = 2*π*. We have:

$$\Phi(\sigma) = \Phi(2\pi) = \text{diag}\left[\Gamma(2\pi)\right]\_{\infty} = \text{diag}\left[I\_2\right]\_{\infty}$$

where *I*2 = 1 0 0 1, i.e., <sup>Φ</sup>(*σ*) = *I* ∈ L(<sup>X</sup>, <sup>X</sup>). Next,

$$
\Gamma(-s)G G^\* \Gamma^\*(-s) = \begin{bmatrix}
\cos^2 s & -\cos s \sin s \\
\end{bmatrix}.
$$

.

Set Σ(*t*) := *t* 0 <sup>Γ</sup>(−*<sup>s</sup>*)*GG*<sup>∗</sup>Γ<sup>∗</sup>(−*<sup>s</sup>*) *ds*. Then,

$$
\Sigma(t) = \begin{bmatrix}
\frac{t}{2} + \frac{\sin 2t}{4} & \frac{\cos 2t - 1}{4} \\
\frac{\cos 2t - 1}{4} & \frac{t}{2} - \frac{\sin 2t}{4}
\end{bmatrix} / 2
$$

and, hence, <sup>Σ</sup>(*σ*) = <sup>Σ</sup>(<sup>2</sup>*π*) = *<sup>π</sup>I*2. We have:

$$\begin{split} Q\_0(\sigma) &= \int\_0^{\sigma} \Phi(-s) B B^\* \Phi^\*(-s) \, ds = \int\_0^{\sigma} \text{diag} \left[ \Gamma(-s) \right]\_{\infty} \text{diag} \left[ G G^\* \right]\_{\infty} \text{diag} \left[ \Gamma^\*(s) \right]\_{\infty} \, ds \\ &= \int\_0^{\sigma} \text{diag} \left[ \Gamma(-s) G G^\* \Gamma^\*(s) \right]\_{\infty} \, ds = \text{diag} \left[ \Sigma(\sigma) \right]\_{\infty} = \pi I \in \mathcal{L}(\mathfrak{X}, \mathfrak{X}). \end{split}$$

So, *Q*−<sup>1</sup> 0 (*σ*) = 1*π I* ∈ L(<sup>X</sup>, <sup>X</sup>), hence, by (38), *H* = *e*2*πλ* − 1 *π I* ∈ L(<sup>X</sup>, <sup>X</sup>). Denote, *α* := *e*2*πλ* − 1 2*π* . (61)

By using (39), we obtain,

$$\mathcal{U}\_1(t) = \text{diag}\left[V\_1(t)\right]\_{\infty} \quad \text{where} \quad V\_1(t) = \begin{bmatrix} 2\pi\cos t \ -2\pi\sin t \end{bmatrix}.\tag{62}$$

Next, by (41), we have *R*(*t*) = diag [*K*(*t*)]<sup>∞</sup>, where *K*(*t*) = *I*2 + <sup>2</sup>*α*<sup>Σ</sup>(*t*). Set *S*(*t*) := <sup>Γ</sup>(*t*)*K*(*t*). Multiplying Γ(*t*) by *<sup>K</sup>*(*t*), we obtain that:

$$S(t) = \begin{bmatrix} (1+at)\cos t + a\sin t & -(1+at)\sin t \\ (1+at)\sin t & (1+at)\cos t - a\sin t \end{bmatrix} \tag{63}$$

By (40), we have,

$$\mathcal{Y}(t) = \Phi(t)\mathcal{R}(t) = \text{diag}\left[\Gamma(t)\right]\_{\mathbb{R}} \text{diag}\left[\mathcal{K}(t)\right]\_{\mathbb{R}} = \text{diag}\left[\Gamma(t)\mathcal{K}(t)\right]\_{\mathbb{R}} = \text{diag}\left[\mathcal{S}(t)\right]\_{\mathbb{R}}.\tag{64}$$

Finding Δ(*t*) := det *S*(*t*) from (63), we obtain that:

$$
\Delta(t) = (1+\alpha t)^2 - \alpha^2 \sin^2 t.
$$

It is easy to check that, for all *t* ∈ [0, *σ*]: Δ(*t*) > 0, if *α* > 0; Δ(*t*) < 0, if *α* < 0; and Δ(*t*) = 0, if *α* = 0. Hence, for all *t* ∈ [0, *σ*]: if *α* > 0, then Δ(*t*) ≥ Δ(0) = 1; if *α* < 0, then Δ(*t*) ≥ <sup>Δ</sup>(<sup>2</sup>*π*) = *e*4*πλ* > 0; if *α* = 0, then Δ(*t*) = 1. Thus, Δ(*t*) is separated from zero.

From (63), we obtain that

$$S^{-1}(t) = \frac{1}{\Delta(t)} \begin{bmatrix} (1+at)\cos t - a\sin t & (1+at)\sin t\\ -(1+at)\sin t & (1+at)\cos t + a\sin t \end{bmatrix} . \tag{65}$$

By (64), we have:

$$\mathcal{Y}^{-1}(t) = \text{diag}\left[\mathcal{S}^{-1}(t)\right]\_{\infty}.\tag{66}$$

Constructing *<sup>U</sup>*2(*t*) according to (42), by using (62), (66), and (65), we obtain:

$$lL\_2(t) = \text{diag}\left[V\_2(t)\right]\_{\approx \prime}, \quad V\_2(t) = \frac{1}{\Delta(t)}\left[2a(1+at) - 2a^2\sin t \cos t\_\prime - 2a^2\sin^2 t\right].\tag{67}$$

From (67) we obtain:

$$F + GV\_2(t) = \begin{bmatrix} 2a(1+at) - 2a^2 \sin t \cos t & -(1+at)^2 - a^2 \sin^2 t \\ \Delta(t) & \Delta(t) \\ 1 & 0 \end{bmatrix}.$$

One can check that the matrix (63) satisfies the following matrix differential equation:

$$\mathcal{S}(t) = (F + GV\_2(t))\mathcal{S}(t), \quad t \in [0, \sigma]. \tag{68}$$

Next, by (67), we have:

$$A + B\mathcal{U}\_2(t) = \text{diag}\left[F + G V\_2(t)\right]\_{\infty}.\tag{69}$$

Due to (68) and (69), the function (64) satisfies the system:

$$\dot{Y}(t) = (A + B\mathcal{U}\_2(t))Y(t), \quad t \in [0, \sigma],$$

and *Y*(0) = *I*. Hence, (45) and (46) holds. Constructing *U*(*t*) according to (47), we obtain:

$$M(t) = \text{diag}\left[V(t)\right]\_{\approx \prime}, \quad V(t) = \frac{1}{\Delta(t - 2\pi k)} \left[2a(1 + a(t - 2\pi k)) - 2a^2 \sin t \cos t, -2a^2 \sin^2 t\right], \tag{70}$$

$$t \in [2\pi k, 2\pi(k+1)), \quad k \in \mathbb{Z}.$$

By Theorem 2, the gain operator function (70) with *α* defined by (61) ensures equality (50), kinematic similarity of systems:

$$\dot{\mathbf{x}}(t) = (A + B\mathcal{U}(t))\mathbf{x}(t), \quad t \in \mathbb{R}, \tag{71}$$

and

$$\dot{\mathbf{x}}(t) = (A + \lambda I)\mathbf{x}(t), \quad t \in \mathbb{R}, \tag{72}$$

on R, and the equality κ(*<sup>A</sup>* + *BU*) = *λ* = *μ*.

For numerical simulation, let us construct the projection of systems (71) and (72) into the space R<sup>2</sup> = {(*<sup>x</sup>*1, *<sup>x</sup>*2), *x*1, *x*2 ∈ <sup>R</sup>}. We obtain the systems

$$\dot{y}(t) = (F + GV(t))y(t), \quad t \in \mathbb{R}, \quad y \in \mathbb{R}^2,\tag{73}$$

and

$$
\dot{y}(t) = (F + \lambda E)y(t), \quad t \in \mathbb{R}, \quad y \in \mathbb{R}^2. \tag{74}
$$

Here *E* is the identity (2 × 2)-matrix. Systems (73) and (74) are kinematically similar, hence, since κ(*F*) = 0, we have κ(*<sup>F</sup>* + *GV*) = κ(*<sup>F</sup>* + *λE*) = *λ*. Let us take, for example, *λ* = −1/4. The equality κ(*<sup>F</sup>* + *λE*) = −1/4 means that system (74) (and (73)) is uniformly exponentially stable with the decay rate 1/4.

Let <sup>Ξ</sup>(*<sup>t</sup>*,*<sup>s</sup>*) denote the evolution matrix of system (73) and <sup>Ω</sup>(*<sup>t</sup>*,*<sup>s</sup>*) denote the evolution matrix of system (74). Let,

$$\Sigma(t,0) = : \begin{bmatrix} \mathfrak{J}\_{11}(t) & \mathfrak{J}\_{12}(t) \\ \mathfrak{J}\_{21}(t) & \mathfrak{J}\_{22}(t) \end{bmatrix}, \qquad \Omega(t,0) = : \begin{bmatrix} \omega\_{11}(t) & \omega\_{12}(t) \\ \omega\_{21}(t) & \omega\_{22}(t) \end{bmatrix}.$$

We have <sup>Ξ</sup>(*<sup>t</sup>*, <sup>0</sup>)*<sup>t</sup>*=<sup>0</sup> = <sup>Ω</sup>(*<sup>t</sup>*, <sup>0</sup>)*<sup>t</sup>*=<sup>0</sup> = *E*. Let us construct the graphs of the functions *ξij*(*t*), *<sup>ω</sup>ij*(*t*) (see, Figures 1–4).

**Figure 1.** Graphs of the functions *A* = *<sup>ω</sup>*11(*t*), *B* = *<sup>ω</sup>*21(*t*).

**Figure 2.** Graphs of the functions *A* = *<sup>ω</sup>*12(*t*), *B* = *<sup>ω</sup>*22(*t*).

**Figure 3.** Graphs of the functions *A* = *ξ*11(*t*), *B* = *ξ*21(*t*).

**Figure 4.** Graphs of the functions *A* = *ξ*12(*t*), *B* = *ξ*22(*t*).

One can see from system (73) and (74) (and from the graphs) that:

$$\begin{aligned} \omega\_{11}(2\pi) &= \mathfrak{J}\_{11}(2\pi) = e^{-\pi/2} \approx 0.2079, &\omega\_{12}(2\pi) = \mathfrak{J}\_{12}(2\pi) = 0, \\ \omega\_{21}(2\pi) &= \mathfrak{J}\_{21}(2\pi) = 0, &\omega\_{22}(2\pi) = \mathfrak{J}\_{11}(2\pi) = e^{-\pi/2} \approx 0.2079. \end{aligned} \tag{75}$$

One can see also that the functions *<sup>ω</sup>ij* are smooth. Since the matrix of system (73) is piecewise continuous, the matrix function <sup>Ξ</sup>(*<sup>t</sup>*, 0) is piecewise smooth and its derivative can be discontinuous at the points *t* = 2*πk*. Calculating the one-sided limits from (73), we obtain that:

$$\begin{aligned} \left. \dot{\xi}\_{11}(t) \right|\_{t=2\pi-0} &= \left. (e^{-\pi/2} - 1)/\pi \approx -0.2521, \quad \left. \dot{\xi}\_{11}(t) \right|\_{t=2\pi+0} = (e^{-\pi/2} - 1)e^{-\pi/2}/\pi \approx -0.0524, \\\ \left. \dot{\xi}\_{12}(t) \right|\_{t=2\pi-0} &= \dot{\xi}\_{12}(t) \Big|\_{t=2\pi+0} = -e^{-\pi/2}, \quad \left. \dot{\xi}\_{21}(t) \right|\_{t=2\pi-0} = \dot{\xi}\_{21}(t) \Big|\_{t=2\pi+0} = e^{-\pi/2}, \\\ & \left. \dot{\xi}\_{22}(t) \right|\_{t=2\pi-0} = \dot{\xi}\_{22}(t) \Big|\_{t=2\pi+0} = 0, \end{aligned}$$

i.e., only the function *ξ*11(*t*) has the discontinuous derivative at the point *t* = 2*π*. This is confirmed by the graphs.

It follows from (75) that:

$$\Xi(2\pi,0) = \Omega(2\pi,0) = \begin{bmatrix} \varepsilon^{-\pi/2} & 0\\ 0 & \varepsilon^{-\pi/2} \end{bmatrix} \approx \begin{bmatrix} 0.2079 & 0\\ 0 & 0.2079 \end{bmatrix}.$$

By periodicity, we have:

$$\Xi(4\pi,0) = \Omega(4\pi,0) = \begin{bmatrix} e^{-\pi} & 0\\ 0 & e^{-\pi} \end{bmatrix} \approx \begin{bmatrix} 0.0432 & 0\\ 0 & 0.0432 \end{bmatrix} \uparrow$$

and so on, <sup>Ξ</sup>(<sup>2</sup>*πk*, 0) = <sup>Ω</sup>(<sup>2</sup>*πk*, 0) = *<sup>e</sup>*<sup>−</sup>*<sup>π</sup>k*/2*E*, *k* ∈ Z. The graphs confirm asymptotic equivalence of the behavior of solutions of systems (73) and (74).

**Remark 2.** *The advantage of the developed method is that it allows us to establish the exact asymptotics (i.e., exact equality* κ(*<sup>A</sup>* + *BU*) = *μ) for the closed-loop system, in contrast to, e.g., [35], from which one can only obtain the inequality* Λ(*A* + *BU*) ≤ κ *for the upper Lyapunov exponent* Λ *of the closed-loop system. The problem of exact assignment of the upper Bohl exponent for a system in infinite-dimensional space in the presented formulation has not been previously investigated. Moreover, the developed method allows us to assign exact values for other asymptotic invariants of the closed-loop system (central exponents, exponential exponents etc.). A disadvantage is that the analytical expressions for the controller (and for solutions of the closed-loop system) can be complicated, in contrast to the stabilization problem [35]. This method can be applied to any system with the property of exact controllability. The choice of matrices in the example in a rather simple form was made for illustrative purposes because in this case the analytical expressions for the controller and for the solutions of the closed-loop system is not very complicated.*
