Global Stability of a Fourth-Order Hybrid Affine System

Abstract

The hybrid affine system considered in the paper comes to existence when applying an impulse control of special form to the chain of four integrators. The goal of the control is to stabilize the system at the origin such that it approaches the equilibrium state along a given (desired) trajectory. The target trajectory is defined implicitly as that of the second-order integrator stabilized by means of a feedback in the form of nested saturators. The purpose of the study is to determine the range of the feedback coefficients for which this system is globally stable. The problem is shown to reduce to a simpler task of establishing stability of a second-order switched linear system with state-dependent switching law. It is proved that the latter system is stable for arbitrary switchings.

APPENDIX

Proof of Lemma 1. The proof does not depend on particular values of the control resource (${{k}_{4}}$) and the maximum velocity (${{k}_{2}}$). Therefore, without loss of generality, we set them equal to ones to simplify the calculations. Closing system (1.2) by feedback (1.4), we obtain the piecewise linear system given by equations

$${{\dot {x}}_{1}} = {{x}_{2}},\quad {{\dot {x}}_{2}} = - {{k}_{3}}{{k}_{1}}{{x}_{1}} - {{k}_{3}}{{x}_{2}}$$

(A.1)

in ${{D}_{1}}$ and equations

$${{\dot {x}}_{1}} = {{x}_{2}},\quad {{\dot {x}}_{2}} = - {{k}_{3}}({{x}_{2}} + {\text{sign}}{\kern 1pt} ({{x}_{1}})),$$

(A.2)

$${{\dot {x}}_{1}} = {{x}_{2}},\quad {{\dot {x}}_{2}} = - {\kern 1pt} {\text{sign}}{\kern 1pt} ({{x}_{2}} + {\text{sat}}{\kern 1pt} ({{k}_{1}}{{x}_{1}}))$$

(A.3)

in ${{D}_{2}}$ and ${{D}_{3}}$, respectively. The assumption of the lemma implies that the origin is a stable node of equation (A.1). Let the roots of the characteristic equation be $- {{\lambda }_{1}}$ and $- {{\lambda }_{2}}$, where $0 < {{\lambda }_{1}} \leqslant {{\lambda }_{2}}$.

The proof of the lemma relies on the following well-known fact from the theory of second-order dynamical systems [16]. The set of $\omega$-limiting points (i.e., limiting points of positive half-trajectories) of bounded trajectories of a system may consist of only equilibrium points, trajectories connecting equilibria, and closed trajectories. Since the system under consideration has only one equilibrium state, the stable node at the origin, it is required to prove that any trajectory, firstly, belongs to a bounded set and, secondly, cannot be a closed curve.

From equation (A.2) and Fig. 1, it is easily seen that any trajectory beginning in D₂ enters D₁ in a finite time. In D₃, integral curves of system (A.3) are parabolas ${{x}_{1}}(t) = - {\kern 1pt} {\text{sign}}{\kern 1pt} ({{x}_{2}} + {\text{sat}}{\kern 1pt} ({{k}_{1}}{{x}_{1}}))x_{2}^{2}(t){\text{/}}2 + C$, which bring the system to either D₁ or to D₂. In other words, any trajectory, in a finite time, necessarily occurs in D₁ (but not necessarily stays there forever). Hence, with regard to boundedness of the set D₁, it follows that no trajectory can go to infinity. Moreover, it also follows that no closed trajectory can entirely belong to ${{D}_{1}}$ or ${{D}_{2}}$, or to their union $D = {{D}_{1}} \cup {{D}_{2}}$. Thus, any closed trajectory, if exists, intersects twice set D, which separates ${{D}_{3}}$ into two disjoint sets $D_{3}^{ - }$ and $D_{3}^{ + }$ lying above and below D, in which control takes the limit values –1 and +1, respectively. For brevity, we refer to the trajectory segments belonging to the sets $D = {{D}_{1}} \cup {{D}_{2}}$ and ${{D}_{3}}$ as unsaturated and saturated segments, respectively. Hence, it follows that any closed trajectory must consist of four, two saturated and two unsaturated, segments, with the direction of motion along such a trajectory being clockwise. Only three variants of locating these segments on the phase plane are possible:

1. The unsaturated segments lie completely in the 2nd and 4th quadrants.

2. An unsaturated segment begins in the 1st (3d) quadrant in the set ${{D}_{1}}$ and ends in the 2nd (4th) quadrant.

3. An unsaturated segment begins in the 1st (3d) quadrant in the set ${{D}_{2}}$ and ends in the 2nd (4th) quadrant.

Consider these three cases separately.

1. Consider the function

$$V = {\text{|}}{{x}_{1}}{\text{|}} + \frac{1}{2}x_{2}^{2},$$

which is positive in the entire phase plane except for the origin, where it vanishes. The time derivative of V by virtue of system (1.2) is given by

$$\dot {V} = {{x}_{2}}({\text{sign}}({{x}_{1}}) + {{U}_{1}}(x)),$$

(A.4)

where ${\text{|}}{{U}_{1}}(x){\text{|}} \leqslant 1$. Let us find out what the sign of $\dot {V}$ is on each of the above segments. In the 1st quadrant, ${{U}_{1}}(x) = - 1$; therefore, $\dot {V} = 0$. The part of the curve lying in the 2nd quadrant generally consists of a part of the saturated segment with ${{U}_{1}}(x) = - 1$; an unsaturated segment; and a part of the saturated segment with ${{U}_{1}}(x) = 1$. On the first of them, $\dot {V} = 0$; on the second, $\dot {V} = {{x}_{2}}(1\; - \;{\text{|}}{{U}_{1}}(x){\text{|}}) < 0$, since ${{x}_{2}} < 0$; and, on the third, $\dot {V} = 2{{x}_{2}} < 0$. Similarly, we find that, on the segments lying in the 3d and 4th quadrants, the derivative is either negative or equal to zero. Then, it follows that, after passing all four segments, we arrive at the same point with a different value of V, which is impossible. Hence, no closed trajectory can exist.

2. Consider, for definiteness, the 1st quadrant. Let us prove that, if a trajectory enters the set ${{D}_{1}}$ in the 1st quadrant, then it does not leave this set anymore and, thus, cannot be closed. Having entered ${{D}_{1}}$ in the 1st quadrant, the trajectory will intersect the ${{x}_{1}}$-axis on the interval $(0,{{x}_{{10}}})$, where ${{x}_{{10}}} = 1{\text{/}}{{k}_{1}}{{k}_{3}}$ is the abscissa of the point where the upper boundary of the set ${{D}_{1}}$ intersects the ${{x}_{1}}$-axis (see Fig. 1). It can leave set ${{D}_{1}}$ only if it intersects the asymptote ${{x}_{2}} = - {{\lambda }_{1}}{{x}_{1}}$ of equation (1.6), which is impossible. Indeed, the asymptote intersects the lower boundary of set ${{D}_{1}}$ at the point with the coordinates $(1{\text{/}}\lambda _{1}^{2}, - 1{\text{/}}{{\lambda }_{1}})$, whose abscissa is greater than ${{x}_{{10}}}$. Hence, it follows that the entire segment of the asymptote belonging to the vertical strip $0 < {{x}_{1}} < {{x}_{{10}}}$ completely lies in the set ${{D}_{1}}$, so that the trajectory cannot leave ${{D}_{1}}$ and asymptotically approaches the origin.

3. In the third case, some part of the set ${{D}_{2}}$ belongs to the 1st and 3d quadrants. This is possible only for small values of ${{k}_{3}}$, when the distance from the straight line ${{x}_{2}} = - 1$ (${{x}_{2}} = 1$) to the upper (lower) boundary of ${{D}_{2}}$, which is equal to $1{\text{/}}{{k}_{3}}$, is greater than 1, i.e., for ${{k}_{3}} < 1$. Consider, for definiteness, the 1st quadrant. Having entered ${{D}_{2}}$ in the 1st quadrant, the trajectory will intersect the ${{x}_{1}}$-axis, and, then, moving in the strip $- 1 < {{x}_{2}} < 0$, will intersect in a finite time the boundary ${{x}_{1}} = 1{\text{/}}{{k}_{1}}$ between the sets ${{D}_{2}}$ and ${{D}_{1}}$. Note that the asymptote and the lower boundary of ${{D}_{1}}$ intersect this line at the points with the ordinates $x_{2}^{a} = - {{\lambda }_{1}}{\text{/}}{{k}_{1}} - 1 - {{\lambda }_{1}}{\text{/}}{{\lambda }_{2}}$, ${\text{|}}x_{2}^{a}{\text{|}} \leqslant 2$, and $x_{2}^{b} = - 1 - 1{\text{/}}{{k}_{3}}$, respectively. Since ${{k}_{3}} < 1$, we have ${\text{|}}x_{2}^{b}{\text{|}} > 2$ and, hence, ${\text{|}}x_{2}^{a}{\text{|}} < {\text{|}}x_{2}^{b}{\text{|}}$. Then, it follows that the entire segment of the asymptote belonging to the vertical strip $0 < {{x}_{1}} < 1{\text{/}}{{k}_{1}}$, completely lies in the set ${{D}_{1}}$, such that the trajectory cannot intersect it and, thus, leave${{D}_{1}}$, asymptotically approaching the origin, like in the previous case. The lemma is proved.

Proof of Lemma 2. Necessity. In order that system (2.3) be stable for any switchings, it is required that all three matrices ${{C}_{1}}$, ${{C}_{2}}$, and ${{C}_{3}}$ are Hurwitz. For $\lambda > 0$, matrix ${{C}_{3}}$ is Hurwitz for any $\xi > 0$; ${{C}_{2}}$, for $\xi > 1$ and $\xi \ne 2$, and ${{C}_{1}}$, for $\xi > 3$.

Sufficiency. In order that the zero solution of the $\delta$-subsystem (2.3) be globally stable, it is sufficient that there exists a common quadratic Lyapunov function $V = {{\delta }^{{\text{T}}}}P\delta$ for all three linear systems (2.3), i.e., there exists a positive definite matrix $\tilde {P}$ that simultaneously satisfy the three linear matrix inequalities (LMI) [7, 17]:

$${{Q}_{i}} \equiv - (\tilde {P}{{C}_{i}} + C_{i}^{{\text{T}}}\tilde {P}) \succ 0,\quad i = 1,2,3.$$

(A.5)

From the fact that C_i, $i = \overline {1,3}$, are Hurwitz matrices, it follows that each separate LMI in (A.5) has a solution. Let us prove that they have a common solution for any for $\xi > 3$.

We will seek matrix $\tilde {P}$ in the form $\tilde {P} = rP$, where $r > 0$ and [18]

$$P = \left[ {\begin{array}{*{20}{c}} {\lambda \xi }&{{{q}_{1}}{\text{/}}2} \\ {{{q}_{1}}{\text{/}}2}&{{{q}_{2}}{\text{/}}\lambda \xi } \end{array}} \right],$$

(A.6)

${{q}_{1}},{{q}_{2}} > 0$ and ${{q}_{2}} > q_{1}^{2}{\text{/}}4$. The last inequality ensures positive definiteness of P; it holds for all points in the plane $({{q}_{1}},{{q}_{2}})$ that lie above the parabola ${{q}_{2}} = q_{1}^{2}{\text{/}}4$ (Fig. 5). The set of solutions of each inequality in (A.5) in the space of variables $r,\;{{q}_{1}},\;{{q}_{2}}$ is a cone, with its projection on the plane r = 1 being the ellipse

$${{\Omega }_{i}} = \{ ({{q}_{1}},{{q}_{2}}):P{{C}_{i}} + C_{i}^{{\text{T}}}P \prec 0\} ,\quad i = \overline {1,3} ,$$

(A.7)

where P is defined by (A.6). A common Lyapunov function for all three systems (2.3) exists if the intersection of the three ellipses is not empty, i.e., there exists a point $(q_{1}^{*},q_{2}^{*})$ belonging to all three ellipses (A.7). Let us show that the center of ellipse ${{\Omega }_{1}}$ satisfies this condition for any $\xi > 3$.

Substituting (A.6) into the first LMI in (A.5), we obtain

$${{Q}_{1}} = \left[ {\begin{array}{*{20}{c}} {{{\lambda }^{2}}[({{\xi }^{2}} + 3){{q}_{1}} - 4\xi ]}&{\lambda [(\xi - 1){{q}_{1}} + \frac{{{{\xi }^{2}} + 3}}{\xi }{{q}_{2}} - \xi ]} \\ * &{4{{q}_{2}} - {{q}_{1}}} \end{array}} \right].$$

(A.8)

Matrix Q₁ is positive definite if and only if

$${{q}_{1}} > \frac{{4\xi }}{{{{\xi }^{2}} + 3}},$$

(A.9)

and its determinant is positive. The second condition, after simple but cumbersome calculations, reduces to the inequality

$${{q}^{{\text{T}}}}{{W}_{1}}q + 2{{q}^{{\text{T}}}}{{b}_{1}} + {{d}_{1}} < 0,$$

(A.10)

where $q = ({{q}_{1}},{{q}_{2}}{{)}^{{\text{T}}}}$,

$${{W}_{1}} = \left[ {\begin{array}{*{20}{c}} {2{{\xi }^{2}} - 2\xi + 4}&{ - \frac{1}{\xi }(\xi + 1)({{\xi }^{2}} + 3)} \\ { - \frac{1}{\xi }(\xi + 1)({{\xi }^{2}} + 3)}&{{{{({{\xi }^{2}} + 3)}}^{2}}{\text{/}}{{\xi }^{2}}} \end{array}} \right],$$

$${{b}_{1}} = - \left[ {\begin{array}{*{20}{c}} {\xi (\xi + 1)} \\ {{{\xi }^{2}} - 8\xi + 3} \end{array}} \right],\quad {{d}_{1}} = - {{\xi }^{2}}.$$

Comparing (A.10) with the canonical ellipse equation

$${{(q - {{O}_{1}})}^{{\text{T}}}}{{W}_{1}}(q - {{O}_{1}}) < {{R}_{1}},$$

where O₁ is the vector of coordinates of the ellipse center, we obtain

$${{O}_{1}} \equiv \left[ {\begin{array}{*{20}{c}} {q_{1}^{*}} \\ {q_{2}^{*}} \end{array}} \right] = - W_{1}^{{ - 1}}{{b}_{1}} = \left[ {\begin{array}{*{20}{c}} {\frac{{2\xi (\xi + 1)}}{{{{\xi }^{2}} + 3}}} \\ {\frac{{(3{{\xi }^{2}} - 4\xi + 5){{\xi }^{2}}}}{{{{{({{\xi }^{2}} + 3)}}^{2}}}}} \end{array}} \right].$$

(A.11)

The coordinate $q_{1}^{*}(\xi )$ obviously satisfies condition (A.9) for any $\xi > 3$.

Let us prove first that ${{O}_{1}}(\xi ) \in {{\Omega }_{3}}$ for $\xi > 3$. It is easy to verify (see, e.g., the proof in [18]) that ellipse ${{\Omega }_{3}}$ does not depend on the parameter $\lambda \xi$ and is given by the formula [18]

$${{\Omega }_{3}} = \{ ({{q}_{1}},{{q}_{2}}):({{q}_{2}} - {{q}_{1}} - {{1)}^{2}} + {{({{q}_{1}} - 2)}^{2}} < 4\} .$$

(A.12)

Figure 5 shows the boundary of the ellipse ${{\Omega }_{3}}$ and parabola ${{q}_{2}} = q_{1}^{2}{\text{/}}4$, which separates the domain where matrix P is positive definite P, as well as the curve described by the point ${{O}_{1}}(\xi )$ when ξ varies from 3 to infinity. As can be seen, the curve lies completely inside the ellipse, and, hence, ${{O}_{1}}(\xi ) \in {{\Omega }_{3}}$ for any $\xi > 3$. Note that the center ${{O}_{1}}(\xi )$ tends to the center ${{O}_{3}}{{ = [2,3]}^{{\text{T}}}}$ as $\xi \to \infty$.

Now, we prove that ${{O}_{1}}(\xi ) \in {{\Omega }_{2}}$ for any $\xi > 3$. Substituting (A.6) into the second LMI in (A.5), we get

$${{Q}_{2}} = \left[ {\begin{array}{*{20}{c}} {{{\lambda }^{2}}[({{\xi }^{2}} + 4){{q}_{1}} - 4\xi ]}&{\lambda \left[ {(\xi - 1){{q}_{1}} + \frac{{{{\xi }^{2}} + 4}}{\xi }{{q}_{2}} - \xi } \right]} \\ * &{4{{q}_{2}} - {{q}_{1}}} \end{array}} \right].$$

(A.13)

Condition ${{O}_{1}}(\xi ) \in {{\Omega }_{2}}(\xi )$ is satisfied if matrix $Q_{2}^{*}(\xi )$ resulting from the substitution of $q_{1}^{*}(\xi )$ and $q_{2}^{*}(\xi )$ into (A.13) for ${{q}_{1}}$ and ${{q}_{2}}$, respectively, is positive definite for any $\xi > 3$. From (A.9), it follows that the first diagonal element of $Q_{2}^{*}(\xi )$ is positive. Let us verify whether the determinant of matrix $Q_{2}^{*}(\xi )$ is positive. We have

$$\begin{gathered} \det {{Q}_{2}}(\xi ,{{q}_{1}},{{q}_{2}}) = {{\lambda }^{2}}\left[ { - ({{\xi }^{2}} + 4)q_{1}^{2} - {{{(\xi - 1)}}^{2}}q_{1}^{2} - \frac{{{{{({{\xi }^{2}} + 4)}}^{2}}}}{{{{\xi }^{2}}}}q_{2}^{2}} \right. \\ \, - \left. {\frac{{2(\xi - 1)({{\xi }^{2}} + 4)}}{{{{\xi }^{2}}}}{{q}_{1}}{{q}_{2}} + 4({{\xi }^{2}} + 4){{q}_{1}}{{q}_{2}} + 4\xi {{q}_{1}} + 2\xi (\xi - 1){{q}_{1}} - 16\xi {{q}_{2}} + 2({{\xi }^{2}} + 4){{q}_{2}} - {{\xi }^{2}}} \right]. \\ \end{gathered} $$

Substituting $q_{1}^{*}(\xi )$ and $q_{2}^{*}(\xi )$ for ${{q}_{1}}$ and ${{q}_{2}}$ into the last formula, we get

$$\det Q_{2}^{*}(\xi ) = {{\lambda }^{2}}R(\xi ){\text{/}}{{({{\xi }^{2}} + 3)}^{4}},$$

where $R(\xi )$ is the polynomial

$$R(\xi ) = 4{{\xi }^{8}} - 24{{\xi }^{7}} + 68{{\xi }^{6}} - 144{{\xi }^{5}} + 235{{\xi }^{4}} - 256{{\xi }^{3}} + 206{{\xi }^{2}} - 152\xi + 47,$$

(A.14)

and the proof reduces to proving strict positiveness of this polynomial for $\xi > 3$. It can be checked directly that $R(\xi )$ is transformed to the form

$$R(\xi ) = 4{{\xi }^{6}}{{(\xi - 3)}^{2}} + 16{{\xi }^{4}}(2\xi - 3)(\xi - 3) + {{\xi }^{2}}(91\xi - 74)(\xi - 2) + (58\xi + 22)(\xi - 3) + 113.$$

(A.15)

Since each term on the right-hand side of (A.15) is positive for $\xi > 3$, $R(\xi )$ is also positive. Lemma 2 is proved.