Chapter 4. -ry2 ] [ IXll ] -r/2. V ~ -Xl - X2 + r Xl X2 = -

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1 Chapter 4 () asymptotically stable (2) unstable (3) asympto~ica1ly stable (4) unstable (5) stable (6) unstable!. 4.2 Let f(x) = ax'p + g(x). Near the origin, the term ax'p is dominan~. Hence, sign(f(x)) = sign(ax'p). Consider the case when a < 0 and P is odd. With Vex) = ix2 as a Lyapuf~v function candidate, we have V = x[az'p +g(x)] $ azp+l + klxlp+2 Near the origin, the term axp+l is dominant. Hence, Vex) is negative definite and the origin is asymptotically stable. Consider now the case when a > 0 andp is odd. n. the neighbothood of the origin, sign(f(x» =. sign(x). Hence, a trajectory starting near z = 0 will be alwa,ys moving atay from x = O. This shows that the origin is unstable. When p is even, a similar behavior wiu take place n one side of the origin; namely, on the side x > 0 when a > 0 and x < 0 when a < O. Therefore, the ori~ is Let Vex) = (1/2)(xi + xu. i. n the set {lxll2 ~ r2}, we have Xll ~ r. Hence, T i. 2 2 V ~ -Xl - X2 + r Xl X2 = - [ Xll] X21 [ 1 -r/2 -ry2 ] [ Xll ] 11 X21 11 is negative ~efinite for r < 2. Thus, the origin is asymptotically stable.!to investigate global asymptotic stability, note that the solution of the second equation is X2{t) = exp(-t)f2 (O), which when substituted in the first equation yields i Xl = [-1 + exp(-t)x:ho)}xl i i This is a linear time-varying system whose solution does not have a finite e~cape time. After some finite time the coefficient of Xl on the right-hand side will be less than a negative n4mber. Hence, limt -+oo Xl (t) = O. Thus, the origin is globally asymptotically stable.! (2) Let Vex) = (1/2)(xi + x~) V = -(Xl + x2)(1 - Xl - X2) = -2V(1-2f)! n the region Vex) < 1/2, V is negative definite. Hence, the~.'gin is asymptotically stable. For V > 1/2, V is positive. Hence, trajectories starting in the region Vex) >1/2 cannot ~pproach the origin. n fact, they grew unbounded. Thus, the origin is not globally asymptoticcitlly stable. (3) Let Vex) =xt Px =Pllxi + 2P12XlX2 + P22X~, where P is a positive 4efinite symmetric matrix V = -2P12Xl + 2(Pu - P12 - P22)XX2-2CP22 - P12)X2 + ijigher order terms 59

2 60 CHAPTER 4. Near the origin, the quadratic term dominates the higher-order terms. Thus, V will be negative definite in the neighborhood of the origin if the quadratic term is negative definite. Choosing Pl2 = 1, P22 = 2, and Pu = 3 makes Vex) positive definite and ii(x) negative definite. Hence, the origin is asymptotically stable. t is not globally asymptotically stable since the origin is not the unique equilibrium point. The set {x~ = } is an equilibrium set. (4) Let V(x) = xi + (1/2)x~. Hence, the origin is globally asymptotically stable. 4.4 (a) Take V(w) = (1/2)(Jlwf + J2W~ + J3wi) as a Lyapunov function candidate. v = J1Wc:.h + J2W2W2 + J3W3W3 = (J2 - J3)WW2W3 + (Ja - J1 )WW2W3 + (J1 - J2)WW2W3 = 0 The origin is stable. t is not asymptotically stable since V is identically zero. The closed-loop state equation is J1Wl = (h - J3)W2W3 - kw J2W2 = (J3 - Jl )W3W - k2w2 J3W3 = (J1 - J2)WW2 - k3w3 Using the same function V(w) as in part (a), we obtain Thus, the origin is.globally asymptotically stable. 4.5 Let g(x) = VV; gi(x) = OV/OXi. Ogi a ov 0 2 V OXj = OXj {)xi = OXjOXi Similarly Hence ogj a ov {)2v OXi = OXi OXj ~ OXiOXj Alternatively, suppose Define Ogi ogj -;--=-0-, uxj UXi Vi,j=,...,n

3 61 Similarly, it can be shown av - =gi(x), 'Vi ax; To meet the symmetry requirement, take "y =(3. Take 8 = (3. 2 Vex) = -(3Xl + (a - 2(3)XX2 - (3(Xl + x2)h(x + X2) Taking a = 2(3 and (3 > 0 yields which is negative definite for all x E R2. Now g(x) = (3 [i ~] x ~ Px.::} V(x)';" 1% gt(y) dy = ~XTPx where P is positive definite. Thus, Vex) is a radially unbounded Lyapunov function and the origin is globally asymptotically stable. 4.7 (a) Let VV(x) =g(x). Then, V:= _gt(x)q4>(x). Choose g(x) =Px so that Vex) = (lj2)xtpx. We need to choose P = pt > 0 such that V = -xtpq4>(x) is negative definite. Choosing P =Q-l yields V = _xt4>{x) =- L n i=1 xi4>i(xi) V is negative definite in the neighborhood of the origin because y4>(y) > 0 for y ::f: O. Hence, the origin is asymptotically stable. The function Vex) is radially unbounded. The origin will be globally asymptotically stable if V is negative definite for all x. This will be the case if y4>i(y) > 0 for all y ::f: O. (c) The function 4>2 satisfies the condition y4>i(y) > 0 for all y ::f: O. The function 4>1 satisfies the condition only near y = 0 because 4>1 (y) vanishes at y = 1. Thus, we can only show asymptotic stability of the origin using the Lyapunov function Vex) = xtq-1x = x~ + 2X1X2 + 2x~.

4 Vex) = jt(x)pf(x) is positive semidefinite. To show that it is positive definite, we need to show that V == 0 :} x = O. Since P is positive definite, we need to show that f(x) == 0 if and only if x = 0; that is, the origin is the unique equilibrium point. Suppose, to the contrary, that there is p"# 0 such that f(p) = O. Then, 63 which is a contradiction. Hence, the origin is the unique equilibrium point. To show that V is radially unbounded, we note that for any x E Rn xtpf(x) _ 1 T T, 1 xll~ - 21xll~[x Pf(x) + j (x)px]::; -2' 't/ x E R n Suppose now that f(x)12 ::; cas xl Then But this is a contradiction since x T pf(x)12 < x T 121!F12c < 1!F12C -+ as xli xll~ - xll~ - xll2 2 xtpf(x) 1 xll~ ::; - 2' n 't/ x E R Thus, as xll , the magnitude of at least one component of f(x) must approach 00. This shows that Vex) as xll (c) We have shown that V(x) is positive definite and radially unbounded. Since f(x) = 0 :} x = 0, V(x) < 0, for all x ERn, x"# O. Thus, the origin is globally asymptotically stable Since V l (x) is not negative semidefinite, there exists a point Xo arbitrarily close to the origin such that V1(xo) > O. Let U = {x E Br Vi{x) > O}, where Br C D. Since V{x) is positive definite, we have V(x) > 0 for all x E U. Hence, the origin is unstable Since V1(x) is not negative semidefinite, there exists a point Xo arbitrarily close to the origin such that V (xo) > O. Let U = {x E Br Vi(x) > O}, where Br CD. Since W(x) ~ 0 for all xed, we have Hence, the origin is unstable. Vt(x) = W(x) +,\t}{x) > 0, 't/ x E U ~1) Apply Chetaev's theorem with Vex) = (1/2)(x~ - x~). The function V is positive at points arbitrarily close to the origin on the Xl-axis. Vex) = Xl (x~ + X~X2) - X2{-X2 + x~ + XX2 - X~) = (X~ + XX2)2 + x~{1 - X2 - Xl - x~) :For any 0 < c < 1, there is a domain around the origin where Hence, in this domain, we have 1 - X2 - Xl - x~ > c > 0

5 64 CHAPTER 4. 1 Xz Figure 4.1: Exercise 4.13 (2). The right-hand side of the preceding inequality is positive definite; hence all the conditions of Chetaev's theorem are satisfied and the equilibrium point at the origin is unstable. (2) The system Xl = -x~ + X2 =!l(x), X2 = x~ - x~ =hex} has two equilibrium points at (O,O) and (1,1). The set r ={0:5 Xl :5 1} n {X2 ~ xn n {X2 :5 xn is shown in Figure 4.1. On the boundary X2 = x, 12 = 0 and i > 0; hence, all trajectories on this boundary move into r. On the boundary X2 = x~, t = 0 and 12 > 0; hence, all trajectories on this boundary move into r. Thus, r is positively invariant. nside r, both h and h are positive. Thus all trajectories move toward the equilibriwn point (1,1). Since this happens for trajectories starting arbitrarily close to the origin, we conclude that the origin is unstable Therefore, 12:1 12:1 1 o yg(y) dy ~ 0 y dy = 2x~ V() T [1 1] x, ~ 2Xl +XlX2+ x2 = '2x 1 2 X The matrix of the quadratic form is positive definite. Hence, V(x) is positive definite for all x, and radially unbounded. v = Xlg(Xt}Xl + XX2 + X2Xl + 2X2X2 = g(xl)(x1x2 - x~ - XX2-2XX2-2x~) + x~ = -g(xd(x~ + 2XX2 + 2x~} + x~ = _g(xdxtqx + x~ where Q = [~ ;] is positive definite. Since g(xt} 2: 1 and xtqx 2: 0, we have V :5 -(x~ + 2XX2 + 2x~) + x~ = -(x~ + 2XX2 + x~) = -(Xl + X2)2 This shows that "i is negative semidefinite. We need to apply the invariance principle. V = 0 => 0:5 -(Xl +X2)2 => 02: (Xl +X2}2 => Xl +X2 =0

6 71 Hence, there is a unique equilibrium point at (1,1,1). 81 = [~3 ;3 ~:2] = [~ =! ~1] 8x :1:=(1,1,1) Xs - Xs :1:=(1,1,1) The eigenvalues are -1 and (-1 ± jv3)/2. Hence, the origin is asymptotically stable. The third state equation has equilibrium at X3 = O. Starting with the initial condition xs(o) = 0, we have xs(t) == O. Then, Xl = 1 and Xl(t) grows unbounded. Thus, the equilib'tium point is not globally asymptotieally stable (a) 0= Xl! Substitution of Xl = 0 in the second equation yields Hence, the origin is the unique equilibrium point. -x2(1 + x~) = 0 '* X2 = 0 Hence, the origin is asymptotically stable. (c) Let Vex) =XX2 Vex) = Xl%2 + %lx2 = (XX2-1)x1x~ + (XX2-1 + X~)X1X2 - XX2 V(X) = 4x~ > 0. :1:1:1:2=2 which implies that r is a positively invariant set. (d) The origin is not globally asymptotically stable since trajectories starting in r do not converge to the origin. ~ The equilibrium points are the roots of the equ8:tions X2 =3X '* Xl (4 - xn = 0 The equilibrium points are (0,0), (2,6), and (-2, -6). fj = [ 1-3x~,- 1 ] 8x 3-1 The equilibrium point (0,0) is unstable (saddle). fj / _ [1 1] '* ).2-4=0 '*..\=±2 8x :1:=(0,0) {){) = [-i 1!1]- '* ) ,,\ + 8=0'*..\ = x :1:=(2,6) The equilibrium point (2,6) is asymptotically stable (stable node). 811 [-11 1] 8x :1:=(-2,-6) = 3-1

7 72 CHAPTER 4. The equilibrium point (-2,-6) is asymptotically stable (stable node). (c) Let A = [-;1!1] and P be the solution of PA + ATp = -1. Using Matlab, P is found to be P = [~:~~~~ ~:!~~~]. The eigenvalues of Pare,xmoz{P) = and,xmin{p) = To estimate the region of attraction of (2, 6), shift the equilibrium point to the origin via the change of variables The state equation in the new coordinates is given by i l = -llxl + X2-6X~ - x~ i2 = 3Xl-X2 We use V = xt Px as a Lyapunov function candidate. The derivative V is given by V = -xtx - 2(pUXl +Pl2X2){6 + Xl)X~ ::; -lxll~ -12CPllXl + Pl2X2)X~ - 2Pl2Xfx2 ::; -lxll~ + 12VYtl + Yt211xll~ + p12l1xll~ ::; -( - 2.4r r2)lxll~, for xll2 ::; r Taking r = 0.4, we see that V(x) is negative in {lxll2 S r}. Choosing c <,xmin{p)r 2 = , ensures that {Vex) $ c} C xll2 S r because,xmin{p)xlli S V(x). Take c = Thus, the region of attraction is estimated by {xtpx::; 0.007}. The estimate of the region of attraction of (-2, -6) is done similarly and the constant c is chosen to be A less conservative estimate of the region of attraction can be obtained graphically by plotting the contour of Vex) = 0 in the :l:1-z2 plane and then choosing c and plotting the surface V (x) = c, with increasing c, until we obtain the largest c for which the surface V (x) = c is inside the region {Vex) < O}. The constant c is determined to be 0.1. The two estimates of the region of attraction are shown in Figure 4.2. (d) The phase portrait is shown in Figure 4.3 together with the estimates of the region of attraction obtained in part (c). The stable trajectories of the saddle form a separatrix that divides the plane into two halves, with the right half as the region of attraction of (2,6) and the left half the region of attraction of (-2, -6). Notice that the estimates of the regions of attraction are much smaller that the regions themselves (a) The equilibrium points are the roots of the equations There are three equilibrium points at (-f,-i), (0,0), and"'(l, 1). 8j = [ -(1(/4)sec2(1(Zl/2) 1 ] 8:1: 1 -(1(/4) sec1(1(:l:2/2) 811 _ [ -(1(/2) 1 ] 8:1: (-i,-i) - 1 -(1(/2)' Eigenvalues are - (1(/2) ± = [ -(1(/4) 1 ] Eigenvalues are - (1(/4) ± 1 8:1: (0,0) 1 -(1(/4)' 81 1 = [ -(1(/2) 1 ] Eigenvalues are - (1(/2) ± 1 8:1: (t,t) 1 -(1(/2)'

8 74 CHAPTER x"" 0 /" /! -0.5 '-,.,/ x, Figure 4.5: Exercise Phase portrait with esti mates of the regions of attraction. Figure 4.4: Exercise The dotted line is the contour of V(y) = 0 and the solid line is the contour of f(y) = (3) (4) ~e investigate stability of the origin using linearization. (1) - af = [-1 + 2Xl -1 2xs 0] A = - af = [-1 _ 1 ax _2Xl 0 l' ax ",=0 0 0 A has an eigenvalue at 1; hence, the origin is unstable. (2) Near the origin, sat(y) = y which implies that dec ( z = -2xs - sat y) = 2Xl + 5X2-4xs The eigenvalues of A are -1, -1, -2; hence, the origin is asymptotically stable. (3) of [-2+3Xf 0] Ofl [-2 0 -{) = 2Xl -1 0, A = -f) = -1 x -1 x :1:=0 0 0 The eigenvalues of A are -2, -1, -1; hence, the origin is asymptotically stable. (4) The eigenvalues of A are -1, -i ±j!v'3; hence, the origin is asymptotically stable.

9 77 t can be shown that the minimum eigenvalue of P is given by where 2 a=r+ R t can be easily seen that there are positive cqnstants Cl and C2 such that a 2-42:: C and AminCPCt» 2:: C2, for all t 2:: 0, which shows that Vet, x) is positive definite. Suppose t, 6, and R satisfy. 2 [. ( L C) L6 t] 2 Vet x) = R Xl - -2 ( 1 + -LR) x 2, C R2 2 RC R L R2 1 + R (~ _ C) + L6 _ t > C R2 2 RC R LR 1+ R2 2:: C2 for all t 2:: 0, for some positive constants C and C2' Then 2Cl 2 2C2 2 V(t,x) $ - k;x - k;x2 Hence, V(t, x) is negative definite. This shows that the origin is uniformly asymptotically stable. By linearity, it is exponentially stable. Alternatively, we can conclude that the origin is exponentially stable by noting that Vet, x) satisfies the conditions of Theorem ~ (a) Since get} 2:: a > a, Hence 1 + ag(t} - a 2 2:: 1 + aa - a 2 > 1 V(t,x} 2::!(asinx + X2) cos Xl which shows that V is positive definite in the region xl < 211'. Since get) $ (3, Vet, x) $!(asinxl + X2}2 + ( + a(3 + a 2 )11 - cosxd which shows that Vet, x) is decrescent. v = -[get) - acosxl]x~ - asin 2 Xl - a 2 (1- COSX)X2 sin Xl + ag(t)(l- cosxt} < -(a - a)x~ + a1'(l- cosxd - asin 2 Xl + O(lxl 3 ) = -(a - a)x~ - a(2-1')(1 - COS Xl) + a[2(1 - cosxl) - sin2x] + O(lx 3 ) The term [2(1 - cosxd - sin 2 Xl] is O(Xl3). Hence V $ -(a - a)x~ - a(2-1')(1- cosxd + O(l/x13) (c) The preceding inequality shows ~hat V is negative definite, since near the origin the negative definite quadratic term dominates the cubic qrder term. Thus, the origin is uniformly asymptotically stable.

10 Let Vex) ==!(x~ + x~). V = -xi + X1X2 + x1(xi + x~) sin t - X1X2 - x~ + x2(xi + x~) cost = -(xi + x~) + (xi + X~)(X1 sin t + X2 cos t) < -lxll~ + xll~j(sint)2 + (cost)2 == -lxll~ + xll~ < -(1 - r)lxll~, 'if xll2 :s; r, for any r < 1 Hence, by Theorem 4.10, the origin is exponenti~ stable. Since Vex) =!lxll~, can be estimated by the set {lxll2 :s; r} for any r < Use Vex) ==!(xi + x~) as a Lyapunov function candidate. V == xlh(t)x2 - g(t)x~] + x2[-h(t)x1 - g(t)x~] == -g(t)(xf + x~) :s; -k(xt + x~) the region of attraction Hence, the origin is uniformly asymptotically stable. Since all assumptions hold globally and V(x) is radially unbounded, the origin is globally uniformly asymptotically stable, which answers part (c). The conditions of Theorem 4.10 are not satisfied. Let us linearize the system [0 al h(t) ] A(t) = ax (t, 0) = -h(t) 0 Use V(x) ==!(xi + x~) as a Lyapunov function candidate for the linear system. V = x1h(t)x2 - x2h(t)xl == 0 This shows that solutions starting on the surface Vex) =c remain on that surface for all t. Hence, the origin of the linear system is not exponentially stable. This implies that the origin of the nonlinear system is not exponentially stable. (d) No Linearization at the origin yields the matrix all == [-1+3Xi+x~ -1+2X1X2] = [-1-1] ax.,= X1 X2-1 + xi + 3x~.,=0 1-1 whose eigenvalues are -1 ± j. Hence, the matrix is Hurwitz and the origin is exponentially stable. Consequently, it is asymptotically stable Let Vex) =!(bxi + ax~). V == -b</>(t)(x1 - ax2)2 - ac1/l(t)x~ :s; -b</>o(x - ax2)2 - ac1/lox~ ~f -W3(X) t can be verified that W3 (x) is positive definite for all x. Hence, by Theorem 4.9, the origin is globally uniformly asymptotically stable. Linearization at the origin yields the linear system The invertible change of variables results in the system %1 == 0, %2 = </>(t)zl - </>(t)(l + ab)z2 which has the solution Z1(t) == constant. This shows that the linear system is not exponentially stable. Therefore, the origin of the nonlinear system is not exponentially stable.

11 85 E'4.~t follows from Exercise 4.58, but an indepenent solution is given next. Let V(x) = tx4. V = _x 6 + x 3 e-t Starting from any time to, we have e-t ~ e-to for all t ~ to. Hence where 0 < f} < 1. t follows from Theorem 4.18 that there is a finite time tl ~ to such that x(t)1 ~ _to)1/3 ( T ' 'V t ~ tl Given 0 < < 1, take to = n(ljf} 3). Then there exists a finite time T such that x(t)1 ~ for all t ~ T. This shows that x(t) converges to zero as t tends to infinity Applying the non-global version of Theorem 4.18 shows that for any x(to) and any input u(t) such that the solution x(t) exists and satisfies x(t)1 ~,B(lx(to), t - to) + 'Y (;~fo lu(t)), 'V t ~ to Since the solution x(t) depends only on U(T) for to ~ T ~ t, the supremum on the right-hand side can be taken over [to, t], which yields (4.47) (a) Asymptotic stability can be shown by linearization which yields the Hurwitz matrix [of jox](o) = -1. Global asymptotic stability can be shown as follows. First we note that there is no finite escape time because /(x)1 ~ kllxll; see Exercise 3.6. We have X2(t) = e-tx2(o). Therefore, there exists a finite time T > 0 such that Consequently X1Xl ~ -(1-,B)x~, 'V t ~ T => Xl(t) -+ 0 as t The linear system X2 = -X2 +u is input-to-state stable. Hence, for any bounded u(t}, X2(t) is bounded. x,(t) = x, (0) +1.' X,(T) [(sin "~(T»)' -1] dr By Gronwall-Bellman inequality Let T be as defined in part (a). XXl ~ -(1 -,B)x~, 'V t ~ T => Xl (t)1 ~ Xl (T), V t ~ T => Xl (t)1 ~ Xl (O)le T, V t ~ 0