Solution of Additional Exercises for Chapter 4

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1 1 1. (1) Try V (x) = 1 (x 1 + x ). Solution of Additional Exercises for Chapter 4 V (x) = x 1 ( x 1 + x ) x = x 1 x + x 1 x In the neighborhood of the origin, the term (x 1 + x ) dominates. Hence, the origin is asymptotically stable. Moreover x (t) = e t x x 1 (t) = e t x 1 + t e (t s) e s ds x = e t x 1 + e t e t x For all x, x(t) as t, which implies that the origin is globally asymptotically stable. () Try V (x) = ax 1 + bx, a >, b >. V (x) = ax 1 (x 1 x )(x 1+x 1)+bx (x 1 +x )(x 1+x 1) = ax 1 (x 1 x )+bx (x 1 +x )(x 1+x 1) Let a = b. V (x) = a(x 1 + x )1 (x 1 + x ) For x 1 + x < 1, V (x) is negative definite. Hence, the origin is asymptotically stable. It is not globally asymptotically stable since there other equilibrium points on the unit circle. (3) Try V (x) = 1 (ax 1 + bx ), a >, b >. V (x) = ax 1 ( x 1 + x 1x ) + bx ( x + x 1 ) = ax 1 + bx 1 x bx + ax 3 1x = x T Qx + ax 3 1x a.5b where Q =. The matrix Q is positive definite when ab b.5b b /4 >. Choose b = a = 1. Near the origin, the quadratic term x T Qx dominates the fourth-order term x 3 1x. Thus, V (x) is negative definite and the origin is asymptotically stable. It is not globally asymptotically stable since there are other equilibrium points at (1, 1) and ( 1, 1).. Consider V (x) = x 1 + x as a Lyapunov function candidate. (a) V (x) = x 1(k x 1 x ) + x 1 x (x 1 + x + k ) x 1 x (x 1 + x + k ) + x (k x 1 x ) = (x 1 + x )(k x 1 x ) k = V (x) = (x 1 + x ) The origin is globally asymptotically stable. (b) k V (x) >, for < x 1 + x < k By Chetaev s theorem, the origin is unstable. αx1 + βx 3. Try g(x) =. To meet the symmetry requirement, take γ = β. γx 1 + δx V (x) = (αx 1 + βx )x (βx 1 + δx )(x 1 + x ) + sin(x 1 + x ) Take δ = β. V (x) = βx 1 + (α β)x 1 x β(x 1 + x ) sin(x 1 + x ) Taking α = β and β > yields V (x) = βx 1 β(x 1 + x ) sin(x 1 + x )

2 which is negative definite in the set { x 1 + x < π}. Now 1 g(x) = β x def = P x V (x) = 1 1 where P is positive definite. 4. For x 1 + x 1, we have ẋ = x x g T (y) dy = 1 xt P x This matrix is Hurwitz. Hence, the origin is asymptotically stable. It is not globally asymptotically stable because there is another equilibrium point at (1, ). 5. Take V (x) = 1 x x4. V (x) = x 5 1ẋ 1 + x 3 ẋ = x 1 + x x 5 1x + x 3 x 1 Near the origin x 5 1x + x 3 x ( 1 1 x 1 + x ) Hence V (x) ( 1 1 ) ( x 1 + x ) which shows that V (x) is positive definite. Application of Chetaev s theorem shows that the origin is unstable.. Since zg(z) > for z, we have g(z) > for z > and g(z) < for z <. Since g(z) is continuous, g() =. At the origin (z =, y = ), a i y i =, h(, )y i =, and g() =. Hence, the origin is an equilibrium point. Consider the Lyapunov function candidate 7. V (z, y) = α z g(s) ds + V (z, y) is positive definite and radially unbounded. β i yi, α >, β i > V = αg(z)ż + β i y i ẏ i = ( αa i + β i b i )y i g(z) β i h(z, y)yi Choose β i = αa i /b i, to obtain Thus V is negative semidefinite. V = β i h(z, y)yi, (z, y) V = y i (t), i ẏ i (t), i g(z(t)) z(t) By LaSalle s theorem (Corollary 4.), the origin is globally asymptotically stable. = x = x 1 x sat(x x 3) x 1 = = x 3 sat(x x 3) x 3 sat( x 3) = x 3 =

3 3 Hence, the origin is the unique equilibrium point. consider V (x) = x T x = x 1 + x + x 3 V (x) = x 1 x x 1 x x sat(x x 3) + x 3sat(x x 3) = (x x 3)sat(x x 3) V is negative semidefinite. V = x (t) x 3(t) ẋ 3 (t) Hence, both x (t) and x 3 (t) must be constants. This implies that ẋ (t). From the second state equation we conclude that x 1 (t). Then, the first state equation implies that x (t). Consequently, x 3 (t). By LaSalle s theorem (Corollary 4.), the origin is globally asymptotically stable. 8. V (x) = 1 4 x x is positive definite and radially unbounded. V (x) = x 3 1 kh(x)x 1 + x + x h(x)x x 3 1 = kx 4 1h(x) x h(x) (1) k >, h(x) > x D. In this case V (x) is negative definite. Hence, the origin is asymptotically stable. () k >, h(x) > x R. In this case V (x) <, x. Hence, the origin is globally asymptotically stable. (3) k >, h(x) < x D. In this case V (x) is positive definite. Hence, by Chetaev s theorem, the origin is unstable. (4) k >, h(x) = x D. In this case V (x) =. Hence, the origin is stable. It is not asymptotically stable because trajectories starting on the level surface V (x) = c remain on the surface for all future time. (5) k =, h(x) > x D. In this case V (x) = x h(x), x D. Moreover V (x) = x (t) ẋ (t) x 1 (t) Hence, by LaSalle s theorem (Corollary 4.1), the origin is asymptotically stable. () k =, h(x) > x R. The same as part (5), except that all the conditions hold globally. Hence, the origin is globally asymptotically stable. 9. (a) = x 1 + g(x 3 ), = g(x 3 ), = ax 1 + bx cx 3 From the properties of g( ) we know that g(x 3 ) = has an isolated solution x 3 =. Substituting x 3 = in the foregoing equations we obtain x 1 = x =. Hence, the origin is an isolated equilibrium point. (b) V (x) = a x 1 + b x3 x + g(y) dy V (x) = ax 1 x 1 + g(x 3 ) bx g(x 3 ) + g(x 3 ) ax 1 + bx cg(x 3 ) = ax 1 cg (x 3 ) V (x) = x 1 (t) and x 3 (t) ẋ 3 (t) From the third state equation we see that x (t). Hence, by LaSalle s theorem (Corollary 4.1), the origin is asymptotically stable. (c) To conclude that the origin is globally asymptotically stable, we need to know that V (x) is radially unbounded. But this is not guaranteed since yg(y) >, y = x g(y) dy as x.

4 4 Consider, for example, g(y) = (1 e y )e y sgn(y). For x >, we have 1. x (1 e y )e y dy = 1 e x 1 (1 e x ) 1 as x Thus we cannot conclude that the origin is globally asymptotically stable. V (x) = a(1 cos x 1 ) + kx 1 + x + px 3 kx 1 + x + px 3 Hence, V is positive definite and radially unbounded. V (x) = ( dx cx x 3 px 3 + px x 3 ) Taking p = c, we obtain V = dx px 3, x R 3 V x (t) & x 3 (t) a sin x 1 (t) + kx 1 (t) Since k > a, a sin x 1 (t) + kx 1 (t) x 1 (t) Using LaSalle s theorem (Corollary 4.), we conclude that the origin is globally asymptotically stable = x x 3 = x 1 x x 1 = x = x 3x 3 x 3 = 1 3 x Substitution of x 1 and x 3 in the first equation yields x + x 3 = x = 3 ± 15 Thus there is only one equilibrium point in the region x i ; namely, x 1 = , x =, x 3 = 1 1 f x = (1+x 3 ) 1 = 1 3 x 3 = The eigenvalues are ± j.449 and Hence, the equilibrium point is asymptotically stable. 1. (1) = x 1 + x x 1 = x = (x 1 + x ) sin x 1 3x Thus x 1 ( sin x 1 3) = x 1 = x =

5 5 Hence, the origin is the unique equilibrium point. f x = 1 1 sin x 1 + (x 1 + x ) cos x 1 sin x 1 3 ; A = f x = x= A is Hurwitz; hence, the origin is asymptotically stable. V (x) = 1 (x 1 + x ). To show global asymptotic stability, let V (x) = x 1 + x 1 x (1 + sin x 1 ) (3 sin x 1 )x x 1 + x 1 x x T x1 1 1 x1 = <, x x 1 x Hence, the origin is globally asymptotically stable. () = x x = ax 1 bx x = b a x 1 x 1 (x 1 + a/b) = x 1 = Hence, the origin is the unique equilibrium point. f 3x x = 1 1 ; A = f 1 a b x = x= a b A is Hurwitz; hence, the origin is asymptotically stable. V (x) = 1 (x 1 + αx ), α >. To show global asymptotic stability, let V (x) = x x 1 x (1 aα) bαx Taking α = 1/a, we obtain V (x) = x 4 1 b a x <, x Hence, the origin is globally asymptotically stable. 13. Take V (x) = 1 (x 1 + x ). V = x 1 x α(t)x + x α(t)x 1 x 3 = x 4 1 x 4 Hence, the origin is globally uniformly asymptotically stable. Linearization at the origin yields α(t) A(t) = α(t) The origin of the linear system is not exponentially stable since, with V (x) = 1 (x 1 + x ), we have V = which implies that V (x(t)) is constant along the solution. Thus, x(t) does not converge to the origin as t tends to infinity. From Theorem 4.15, we conclude that the origin of the nonlinear system is not exponentially stable. 14. View this system as a perturbation of the autonomous system ẋ 1 = x, ẋ = x 1 x The matrix A of the autonomous system is Hurwitz. Find a Lyapunov function V (x) = x T P x for the 3 1 autonomous system by solving the Lyapunov equation P A + A T P = I, to obtain P = 1. 1

6 Now use V (x) = x T P x = 3 x 1 + x 1 x + x as a Lyapunov function candidate for the perturbed system with b. V = x 1 x b cos t x 1 x b cos t x x 1 x + b x 1 x + b x T x1 1 b / x1 = x b / 1 b x The right hand side is negative definite if (1 b b /4) >, which is the case if b <.47. Taking b =.47, we conclude that the origin is exponentially stable for all b < b. 15. Try V (x) = 1 (x 1 + x ). V = x 1 x g(t)x 1(x 1 + x ) x 1 x g(t)x (x 1 + x ) = g(t)(x 1 + x ) 4kV (x) Hence, V (t, x) is negative definite and the origin is uniformly asymptotically stable. From the inequality V 4kV, we cannot conclude exponential stability. Let us try linearization. 3g(t)x 1 g(t)x 1 g(t)x 1 x A(t) = f x (t, x) = x= 1 g(t)x 1 x 3g(t)x g(t)x 1 x= = 1 1 A is a constant matrix that is not Hurwitz. Hence, by Theorem 4.15, we conclude that the origin is not exponentially stable. 1. Let A 1 = f x () be the linearization of (1). To find the linearization of (), set g(x) = h(x)f(x). Then Hence g i = h(x) f i + h f i (x) g i () = h() f i () + h ()f i () = h() f i () Hence A = g x () = h()a 1 Since h() >, A 1 is Hurwitz if and only if A is Hurwitz. By Theorem 4.15 (1) is exp. stable A 1 is Hurwitz A is Hurwitz () is exp. stable 17. It is not input-to-state stable because with u = the origin is not globally asymptotically stable (notice that the unforced system has equilibrium points on the unit circle).