p. 1/5 Introduction to Nonlinear Control Lecture # 3 Time-Varying and Perturbed Systems
p. 2/5 Time-varying Systems ẋ = f(t, x) f(t, x) is piecewise continuous in t and locally Lipschitz in x for all t 0 and all x D. The origin is an equilibrium point at t = 0 if f(t, 0) = 0, t 0 While the solution of the autonomous system ẋ = f(x), x(t 0 ) = x 0 depends only on (t t 0 ), the solution of may depend on both t and t 0 ẋ = f(t, x), x(t 0 ) = x 0
p. 3/5 Comparison Functions A scalar continuous function α(r), defined for r [0, a) is said to belong to class K if it is strictly increasing and α(0) = 0. It is said to belong to class K if it defined for all r 0 and α(r) as r A scalar continuous function β(r, s), defined for r [0, a) and s [0, ) is said to belong to class KL if, for each fixed s, the mapping β(r, s) belongs to class K with respect to r and, for each fixed r, the mapping β(r, s) is decreasing with respect to s and β(r, s) 0 as s
p. 4/5 Example α(r) = tan 1 (r) is strictly increasing since α (r) = 1/(1 + r 2 ) > 0. It belongs to class K, but not to class K since lim r α(r) = π/2 < α(r) = r c, for any positive real number c, is strictly increasing since α (r) = cr c 1 > 0. Moreover, lim r α(r) = ; thus, it belongs to class K α(r) = min{r, r 2 } is continuous, strictly increasing, and lim r α(r) =. Hence, it belongs to class K
p. 5/5 β(r, s) = r/(ksr + 1), for any positive real number k, is strictly increasing in r since β r = 1 (ksr + 1) 2 > 0 and strictly decreasing in s since β s = kr2 (ksr + 1) 2 < 0 Moreover, β(r, s) 0 as s. Therefore, it belongs to class KL β(r, s) = r c e s, for any positive real number c, belongs to class KL
p. 6/5 Definition: The equilibrium point x = 0 of ẋ = f(t, x) is uniformly stable if there exist a class K function α and a positive constant c, independent of t 0, such that x(t) α( x(t 0 ) ), t t 0 0, x(t 0 ) < c uniformly asymptotically stable if there exist a class KL function β and a positive constant c, independent of t 0, such that x(t) β( x(t 0 ), t t 0 ), t t 0 0, x(t 0 ) < c globally uniformly asymptotically stable if the foregoing inequality is satisfied for any initial state x(t 0 )
p. 7/5 exponentially stable if there exist positive constants c, k, and λ such that x(t) k x(t 0 ) e λ(t t 0), x(t 0 ) < c globally exponentially stable if the foregoing inequality is satisfied for any initial state x(t 0 )
p. 8/5 Theorem: Let the origin x = 0 be an equilibrium point for ẋ = f(t, x) and D R n be a domain containing x = 0. Suppose f(t, x) is piecewise continuous in t and locally Lipschitz in x for all t 0 and x D. Let V (t, x) be a continuously differentiable function such that (1) W 1 (x) V (t, x) W 2 (x) (2) V t + V x f(t, x) 0 for all t 0 and x D, where W 1 (x) and W 2 (x) are continuous positive definite functions on D. Then, the origin is uniformly stable
p. 9/5 Theorem: Suppose the assumptions of the previous theorem are satisfied with V t + V x f(t, x) W 3(x) for all t 0 and x D, where W 3 (x) is a continuous positive definite function on D. Then, the origin is uniformly asymptotically stable. Moreover, if r and c are chosen such that B r = { x r} D and c < min x =r W 1 (x), then every trajectory starting in {x B r W 2 (x) c} satisfies x(t) β( x(t 0 ), t t 0 ), t t 0 0 for some class KL function β. Finally, if D = R n and W 1 (x) is radially unbounded, then the origin is globally uniformly asymptotically stable
p. 10/5 Theorem: Suppose the assumptions of the previous theorem are satisfied with k 1 x a V (t, x) k 2 x a V t + V x f(t, x) k 3 x a for all t 0 and x D, where k 1, k 2, k 3, and a are positive constants. Then, the origin is exponentially stable. If the assumptions hold globally, the origin will be globally exponentially stable.
p. 11/5 Example: ẋ = [1 + g(t)]x 3, g(t) 0, t 0 V (x) = 1 2 x2 V (t, x) = [1 + g(t)]x 4 x 4, x R, t 0 The origin is globally uniformly asymptotically stable Example: ẋ 1 = x 1 g(t)x 2 ẋ 2 = x 1 x 2 0 g(t) k and ġ(t) g(t), t 0
p. 12/5 V (t, x) = x 2 1 + [1 + g(t)]x2 2 x 2 1 + x2 2 V (t, x) x2 1 + (1 + k)x2 2, x R2 V (t, x) = 2x 2 1 + 2x 1x 2 [2 + 2g(t) ġ(t)]x 2 2 2 + 2g(t) ġ(t) 2 + 2g(t) g(t) 2 V (t, x) 2x 2 1 + 2x 1x 2 2x 2 2 = xt [ 2 1 1 2 The origin is globally exponentially stable ] x
p. 13/5 Perturbed Systems Nominal System: ẋ = f(x), f(0) = 0 Perturbed System: ẋ = f(x) + g(t, x), g(t, 0) = 0 Case 1: The origin of the nominal system is exponentially stable c 1 x 2 V (x) c 2 x 2, V x c 4 x V x f(x) c 3 x 2
p. 14/5 Use V (x) as a Lyapunov function candidate for the perturbed system Assume that V (t, x) = V x f(x) + V x g(t, x) g(t, x) γ x, γ 0 V (t, x) c 3 x 2 + V x g(t, x) c 3 x 2 + c 4 γ x 2
p. 15/5 γ < c 3 c 4 V (t, x) (c 3 γc 4 ) x 2 The origin is an exponentially stable equilibrium point of the perturbed system
p. 16/5 Example ẋ 1 = x 2 ẋ 2 = 4x 1 2x 2 + βx 3 2, β 0 A = [ ẋ = Ax + g(x) ] 0 1, g(x) = 4 2 [ 0 βx 3 2 ] The eigenvalues of A are 1 ± j 3 PA + A T P = I P = 3 2 1 8 1 8 5 16
p. 17/5 V (x) = x T Px, V x Ax = xt x c 3 = 1, c 4 = 2 P = 2λ max (P) = 2 1.513 = 3.026 g(x) = β x 2 3 g(x) satisfies the bound g(x) γ x over compact sets of x. Consider the compact set Ω c = {V (x) c} = {x T Px c}, c > 0 k 2 = max x T P x c x 2 x 3 2 k2 2 x 2
p. 18/5 k 2 = max x T P x c [0 1]x = max y T y c = c [0 1]P 1/2 = 1.8194 c c [0 1]P 1/2 y g(x) β c (1.8194) 2 x, x Ω c g(x) γ x, x Ω c, γ = β c (1.8194) 2 γ < c 3 c 4 β < 1 3.026 (1.8194) 2 c 0.1 c β < 0.1/c V (x) (1 10βc) x 2 Hence, the origin is exponentially stable and Ω c is an estimate of the region of attraction
p. 19/5 Alternative Bound on β V (x) = x 2 + 2x T Pg(x) x 2 + 1 8 βx3 2 ([2 5]x) x 2 + 29 8 βx2 2 x 2 Over Ω c, x 2 2 (1.8194)2 c ( ) V (x) 1 29 8 β(1.8194)2 c ( = 1 βc ) x 2 0.448 x 2 If β < 0.448/c, the origin will be exponentially stable and Ω c will be an estimate of the region of attraction
Remark: The inequality β < 0.448/c shows a tradeoff between the estimate of the region of attraction and the estimate of the upper bound on β p. 20/5
p. 21/5 Application to Linearization ẋ = f(x) = [A + G(x)]x A = f x (x), G(x) 0 as x 0 x=0 Theorem: The origin of ẋ = f(x) is exponentially stable if and only if A is Hurwitz
p. 22/5 Proof Of sufficiency: Suppose A is Hurwitz. Choose Q = Q T > 0 and solve the Lyapunov equation PA + A T P = Q for P Use V (x) = x T Px as a Lyapunov function candidate for ẋ = f(x) V (x) = x T Pf(x) + f T (x)px = x T P[A + G(x)]x + x T [A T + G T (x)]px = x T (PA + A T P)x + 2x T PG(x)x = x T Qx + 2x T PG(x)x
p. 23/5 V (x) x T Qx + 2 P G(x) x 2 For any γ > 0, there exists r > 0 such that G(x) < γ, x < r x T Qx λ min (Q) x 2 x T Qx λ min (Q) x 2 Choose V (x) < [λ min (Q) 2γ P ] x 2, x < r γ < λ min(q) 2 P The origin of ẋ = f(x) is exponentially stable
p. 24/5 Proof of Necessity: Suppose the origin of ẋ = f(x) is exponentially stable. View the system ẋ = Ax = f(x) G(x)x as a perturbation of ẋ = f(x). Recall that G(x) < γ, x < r Because the origin of ẋ = f(x) is exponentially stable, let V (x) be the function provided by the converse Lyapunov theorem over a domain { x < r 0 }. Use V (x) as a Lyapunov function candidate for ẋ = Ax.
p. 25/5 In the domain { x < min{r 0, r}}, we have V x V V Ax = f(x) x x G(x)x c 3 x 2 + c 4 γ x 2 = (c 3 c 4 γ) x 2 Take γ < c 3 /c 4, and set λ = (c 3 c 4 L) > 0 V x Ax λ x 2, x < min{r 0, r} The origin of ẋ = Ax is exponentially stable
p. 26/5 Case 2: The origin of the nominal system is asymptotically stable V (t, x) = V x f(x)+ V x g(t, x) W 3(x)+ V g(t, x) x Under what condition will the following inequality hold? V g(t, x) x < W 3(x) Special Case: Quadratic-Type Lyapunov function V x f(x) c 3φ 2 (x), V x c 4φ(x)
p. 27/5 V (t, x) c 3 φ 2 (x) + c 4 φ(x) g(t, x) If g(t, x) γφ(x), with γ < c 3 c 4 V (t, x) (c 3 c 4 γ)φ 2 (x)
p. 28/5 Example ẋ = x 3 + g(t, x) V (x) = x 4 is a quadratic-type Lyapunov function for the nominal system ẋ = x 3 V x ( x3 ) = 4x 6, V x = 4 x 3 φ(x) = x 3, c 3 = 4, c 4 = 4 Suppose g(t, x) γ x 3, x, with γ < 1 V (t, x) 4(1 γ)φ 2 (x) Hence, the origin is a globally uniformly asymptotically stable
p. 29/5 Remark: A nominal system with asymptotically, but not exponentially, stable origin is not robust to smooth perturbations with arbitrarily small linear growth bounds Example ẋ = x 3 + γx The origin is unstable for any γ > 0
p. 30/5 Ultimate Boundedness Definition: The solutions of ẋ = f(t, x) are uniformly bounded if c > 0 and for every 0 < a < c, β = β(a) > 0 such that x(t 0 ) a x(t) β, t t 0 0 uniformly ultimately bounded with ultimate bound b if b and c and for every 0 < a < c, T = T(a, b) 0 such that x(t 0 ) a x(t) b, t t 0 + T Globally if a can be arbitrarily large Drop uniformly if ẋ = f(x)
p. 31/5 Lyapunov Analysis: Let V (x) be a cont. diff. positive definite function and suppose that the sets Ω c = {V (x) c}, Ω ε = {V (x) ε}, Λ = {ε V (x) c} are compact for some c > ε > 0 Λ Ω ε Ω c
p. 32/5 Suppose V (t, x) = V x f(t, x) W 3(x), x Λ, t 0 W 3 (x) is continuous and positive definite Ω c and Ω ε are positively invariant k = min x Λ W 3(x) > 0 V (t, x) k, x Λ, t t 0 0 V (x(t)) V (x(t 0 )) k(t t 0 ) c k(t t 0 ) x(t) enters the set Ω ε within the interval [t 0, t 0 + (c ε)/k]
p. 33/5 Suppose V (t, x) W 3 (x), µ x r, t 0 Choose c and ε such that Λ {µ x r} B µ Ω ε Ω c B r
p. 34/5 Let α 1 and α 2 be class K functions such that α 1 ( x ) V (x) α 2 ( x ) V (x) c α 1 ( x ) c x α 1 1 (c) What is the ultimate bound? c = α 1 (r) Ω c B r x µ V (x) α 2 (µ) ε = α 2 (µ) B µ Ω ε V (x) ε α 1 ( x ) ε x α 1 1 (ε) = α 1 1 (α 2(µ))
p. 35/5 Theorem (special case of Thm 4.18): Suppose V α 1 ( x ) V (x) α 2 ( x ) x f(t, x) W 3(x), x µ > 0 t 0 and x r, where α 1, α 2 K, W 3 (x) is continuous & positive definite, and µ < α 1 2 (α 1(r)). Then, for every initial state x(t 0 ) { x α 1 2 (α 1(r))}, there is T 0 (dependent on x(t 0 ) and µ) such that x(t) α 1 1 (α 2(µ)), t t 0 + T If the assumptions hold globally and α 1 K, then the conclusion holds for any initial state x(t 0 )
p. 36/5 Remarks: The ultimate bound is independent of the initial state The ultimate bound is a class K function of µ; hence, the smaller the value of µ, the smaller the ultimate bound. As µ 0, the ultimate bound approaches zero
p. 37/5 Example ẋ 1 = x 2, ẋ 2 = (1 + x 2 1 )x 1 x 2 + M cos ωt, M 0 With M = 0, ẋ 2 = (1 + x 2 1 )x 1 x 2 = h(x 1 ) x 2 V (x) = x T 1 2 1 2 1 2 1 V (x) = x T x + 2 3 2 1 2 1 2 1 x1 0 x + 1 2 x4 1 (y + y 3 ) dy (Example 4.5) def = x T Px + 1 2 x4 1
p. 38/5 λ min (P) x 2 V (x) λ max (P) x 2 + 1 2 x 4 α 1 (r) = λ min (P)r 2, α 2 (r) = λ max (P)r 2 + 1 2 r4 V = x 2 1 x4 1 x2 2 + (x 1 + 2x 2 )M cos ωt x 2 x 4 1 + M 5 x = (1 θ) x 2 x 4 1 θ x 2 + M 5 x (0 < θ < 1) (1 θ) x 2 x 4 1, x M 5/θ def = µ The solutions are GUUB by b = α 1 1 (α λ max (P)µ 2 + µ 4 /2 2(µ)) = λ min (P)
p. 39/5 Perturbed Systems: Nonvanishing Perturbation Nominal System: Perturbed System: ẋ = f(x), f(0) = 0 ẋ = f(x) + g(t, x), g(t, 0) 0 Case 1: The origin of ẋ = f(x) is exponentially stable c 1 x 2 V (x) c 2 x 2 V x f(x) c 3 x 2, x B r = { x r} V x c 4 x
p. 40/5 Use V (x) to investigate ultimate boundedness of the perturbed system Assume V (t, x) = V x g(t, x) δ, V (t, x) c 3 x 2 + f(x) + V x V x g(t, x) t 0, x B r g(t, x) c 3 x 2 + c 4 δ x = (1 θ)c 3 x 2 θc 3 x 2 + c 4 δ x 0 < θ < 1 (1 θ)c 3 x 2, x δc 4 /(θc 3 ) def = µ
p. 41/5 Apply Theorem 4.18 x(t 0 ) α 1 2 (α 1(r)) x(t 0 ) r c1 µ < α 1 2 (α 1(r)) δc 4 c1 < r δ < c 3 c1 θr θc 3 c 2 c 4 c 2 b = α 1 c2 1 (α 2(µ)) b = µ b = δc 4 c2 c 1 θc 3 c 1 For all x(t 0 ) r c 1 /c 2, the solutions of the perturbed system are ultimately bounded by b c 2
p. 42/5 Example ẋ 1 = x 2, ẋ 2 = 4x 1 2x 2 + βx 3 2 + d(t) β 0, d(t) δ, t 0 V (x) = x T Px = x T 3 2 1 8 1 8 5 16 x V (t, x) = x 2 + 2βx 2 ( 1 2 8 x 1x 2 + 5 ) 16 x2 2 + 2d(t) ( 1 8 x 1 + 5 16 x ) 2 29 29δ x 2 + 8 βk2 2 x 2 + x 8
p. 43/5 k 2 = max x T P x c x 2 = 1.8194 c Suppose β 8(1 ζ)/( 29k2 2 ) (0 < ζ < 1) V (t, x) ζ x 2 + 29δ 8 x (1 θ)ζ x 2, x 29δ 8ζθ def = µ (0 < θ < 1) If µ 2 λ max (P) < c, then all solutions of the perturbed system, starting in Ω c, are uniformly ultimately bounded by 29δ λ max (P) b = 8ζθ λ min (P)
p. 44/5 Case 2: The origin of ẋ = f(x) is asymptotically stable α 1 ( x ) V (x) α 2 ( x ) V x f(x) α 3( x ), V x k x B r = { x r}, α i K, i = 1, 2, 3 V (t, x) α 3 ( x ) + V x g(t, x) α 3 ( x ) + δk (1 θ)α 3 ( x ) θα 3 ( x ) + δk (1 θ)α 3 ( x ), x α 1 3 0 < θ < 1 ( δk θ ) def = µ
p. 45/5 Apply Theorem 4.18 µ < α 1 2 (α 1(r)) α 1 3 ( ) δk θ < α 1 2 (α 1(r)) δ < θα 3(α 1 2 (α 1(r))) k Example V (x) = x 4 Compare with δ < c 3 c1 θr c 4 c 2 ẋ = x 1 + x 2 V [ x ] x 1 + x 2 = 4x4 1 + x 2 α 1 ( x ) = α 2 ( x ) = x 4 ; α 3 ( x ) = 4 x 4 1 + x 2; k = 4r3
p. 46/5 The origin is globally asymptotically stable θα 3 (α 1 2 (α 1(r))) k = θα 3(r) k rθ 1 + r 2 0 as r ẋ = x 1 + x 2 + δ, δ > 0 δ > 1 2 lim t x(t) = = rθ 1 + r 2
p. 47/5 Input-to-State Stability (ISS) Definition: The system ẋ = f(x, u) is input-to-state stable if there exist β KL and γ K such that for any initial state x(t 0 ) and any bounded input u(t) ( ) x(t) β( x(t 0 ), t t 0 ) + γ ISS of ẋ = f(x, u) implies BIBS stability sup t 0 τ t u(τ) x(t) is ultimately bounded by a class K function of sup t t0 u(t) lim t u(t) = 0 lim t x(t) = 0 The origin of ẋ = f(x, 0) is GAS
p. 48/5 Theorem (Special case of Thm 4.19): Let V (x) be a continuously differentiable function such that V α 1 ( x ) V (x) α 2 ( x ) x f(x, u) W 3(x), x ρ( u ) > 0 x R n, u R m, where α 1, α 2 K, ρ K, and W 3 (x) is a continuous positive definite function. Then, the system ẋ = f(x, u) is ISS with γ = α 1 1 α 2 ρ Proof: Let µ = ρ(sup τ t0 u(τ) ); then V x f(x, u) W 3(x), x µ
p. 49/5 Choose ε and c such that V x f(x, u) W 3(x), x Λ = {ε V (x) c} Suppose x(t 0 ) Λ and x(t) reaches Ω ε at t = t 0 + T. For t 0 t t 0 + T, V satisfies the conditions for the uniform asymptotic stability. Therefore, the trajectory behaves as if the origin was uniformly asymptotically stable and satisfies x(t) β( x(t 0 ), t t 0 ), for some β KL For t t 0 + T, x(t) α 1 1 (α 2(µ))
p. 50/5 x(t) β( x(t 0 ), t t 0 ) + α 1 1 (α 2(µ)), t t 0 x(t) β( x(t 0 ), t t 0 ) + γ ( sup u(τ) τ t 0 ), t t 0 Since x(t) depends only on u(τ) for t 0 τ t, the supremum on the right-hand side can be taken over [t 0, t]
p. 51/5 Example ẋ = x 3 + u The origin of ẋ = x 3 is globally asymptotically stable V = 1 2 x2 V = x 4 + xu = (1 θ)x 4 θx 4 + xu (1 θ)x 4, x ( u θ ) 1/3 0 < θ < 1 The system is ISS with γ(r) = (r/θ) 1/3
p. 52/5 Example ẋ = x 2x 3 + (1 + x 2 )u 2 The origin of ẋ = x 2x 3 is globally exponentially stable V = 1 2 x2 V = x 2 2x 4 + x(1 + x 2 )u 2 = x 4 x 2 (1 + x 2 ) + x(1 + x 2 )u 2 x 4, x u 2 The system is ISS with γ(r) = r 2
p. 53/5 Example ẋ 1 = x 1 + x 2 2, ẋ 2 = x 2 + u Investigate GAS of ẋ 1 = x 1 + x 2 2, ẋ 2 = x 2 V (x) = 1 2 x2 1 + 1 4 x4 2 V = x 2 1 + x 1x 2 2 x4 2 = (x 1 1 2 x2 2 )2 ( 1 1 4 ) x 4 2 Now u 0, V = 1 2 (x 1 x 2 2 )2 1 2 (x2 1 + x4 2 ) + x3 2 u 1 2 (x2 1 + x4 2 ) + x 2 3 u V 1 2 (1 θ)(x2 1 + x4 2 ) 1 2 θ(x2 1 + x4 2 ) + x 2 3 u (0 < θ < 1)
p. 54/5 1 2 θ(x2 1 + x4 2 ) + x 2 3 u 0 if x 2 2 u or x 2 2 u and x 1 θ θ if x 2 u ( ) 2 u 2 1 + θ θ ρ(r) = 2r ( ) 2r 2 1 + θ θ ( 2 u θ ) 2 V 1 2 (1 θ)(x2 1 + x4 2 ), x ρ( u ) The system is ISS