Download
# Chapter Lyapunov Stability I Autonomous Systems Denitions Consider the autonomous system Denition is an equilibrium point of if We want to know whether or not the trajecories near an equilib PDF document - DocSlides

trish-goza | 2014-12-15 | General

### Presentations text content in Chapter Lyapunov Stability I Autonomous Systems Denitions Consider the autonomous system Denition is an equilibrium point of if We want to know whether or not the trajecories near an equilib

Show

Page 1

Chapter 3: Lyapunov Stability I: Autonomous Systems 1 Deﬁnitions Consider the autonomous system (1) Deﬁnition 1 is an equilibrium point of (1) if ) = 0 We want to know whether or not the trajecories near an equilib rium point are “well behaved”. Deﬁnition 2 is said to be stable if for each > (0) < ⇒k < otherwise, the equilibrium point is said to be unstable. IMPORTANT : This notion applies to the equilibrium, not “the system.” A dynamical system can have several equilibrium points.

Page 2

&% '$ Figure 1: Stable equilibrium point. &% '$ Figure 2: Asymptotically stable equilibrium point. Deﬁnition 3 of the system (1) is said to be convergent if there exists 0 : (0) < lim ) = Equivalently, is convergent if for any given such that (0) < ⇒k < T. Deﬁnition 4 is said to be asymptotically stable if it is both stable and co n- vergent.

Page 3

Deﬁnition 5 is said to be (locally) exponentially stable if there exist t wo real constants α,λ> such that k (0) λt t> 0 (2) whenever (0) < . It is said to be globally exponentially stable if (2) holds for any Note : Clearly, exponential stability implies asymptotic stabi lity. The con- verse is, however, not true. Remarks : We can always assume that = 0. Given any other equilibrium point we can make a change of variables and deﬁne a new system w ith an equilibrium point at = 0. Deﬁne: = ) = Thus, the equilibrium point of the new systems ) is = 0, since (0) = (0 + ) = ) = 0 Thus, is stable for the system ) if and only if = 0 is stable for the system ).

Page 4

2 Positive Deﬁnite Functions Deﬁnition 6 is positive semi deﬁnite in if (i) and (0) = 0 (ii) xinD −{ is positive deﬁnite in if (ii’) inD −{ is negative deﬁnite (semi deﬁnite)in if is positive deﬁnite (semi deﬁnite).

Page 5

Example 1 : (Quadratic form) ) : Qx, Q , Q Since , its eigenvalues ,i = 1 , are all real. Thus positive deﬁnite = 1 , n positive semideﬁnite = 1 , n negative deﬁnite = 1 , n negative semideﬁnite = 1 , n Notation : Given a dynamical system and a function we will denote ) = dV dt ∂V ∂x dx dt ∂V ∂x ∂V ∂x ∂V ∂x Example 2 : Let ax bx + cos and deﬁne . Thus, we have ) = ∂V ∂x ) = [2 ax bx + cos = 2 ax + 2 bx + 2 cos Notice that the ) depends on the system’s equation ) and thus it will be diﬀerent for diﬀerent systems.

Page 6

Note: Qx, where if sym asym sym asym Then, sym asym where, asym a b i.e. skew symmetric and asym = 0 sym Q is always symmetric Theorem Rayleigh-Ritz theorem min }k Qx max }k

Page 7

3 Stability Theorems Theorem 1 : (Lyapunov Stability Theorem) Let = 0 be an equilibrium point of , and let be a continuously diﬀerentiable function such that (i) (0) = 0 (ii) in D −{ (iii) inD −{ thus = 0 is stable. Theorem 2 : (Asymptotic Stability Theorem) Under the conditions of Th eorem 1, if is such that (i) (0) = 0 (ii) in D −{ (iii) inD −{ thus = 0 is asymptotically stable.

Page 8

Proof of theorem 1 : Choose r> 0 and deﬁne k } is contained in , so deﬁned is a closed and bounded (“compact”) set (a sphere). We now construct a Lyapunov surface inside and show that all trajec- tories starting near = 0 remain inside the surface. Let = min ) (thus α> 0) Now choose (0 , ) and denote Thus, by construction, . Assume now that (0) (0)) Trajectories starting in at = 0 stays inside for all 0. By the continuity of ), δ > 0: < < It then follows that (0) < t> and then (0) < ⇒k which means that the equilibrium = 0 is stable. Proof of theorem 2 : Similar, only that in D implies that is “shrinking” until eventually it becomes the single point = 0.

Page 9

mg Figure 3: Pendulum without friction. 4 Examples Example 3 : (Pendulum Without Friction) Using Newton’s second law of motion we have, ma mg sin l where is the length of the pendulum, and is the angular acceleration. Thus ml mg sin = 0 or sin = 0 choosing state variables we have sin To study the stability of the equilibrium at the origin, we ne ed to propose a Lyapunov function candidate . This is diﬃcult! In this case we “try the total energy, which is a positive function. We have kinetic plus potential energy ωl mgh

Page 10

where (1 cos ) = (1 cos Thus ml mgl (1 cos We now deﬁne ) = . We see that because of the periodicity of cos( we have that ) = 0 whenever = ( ,x = (2 kπ, 0) = 1 Thus, is not positive deﬁnite. However, restricting the domain of to the interval π, ; i.e., we take , with = (( π, ,R We have that R> is indeed positive deﬁnite. Also ) = ∂V ∂x ∂V ∂x ,f )] = [ mgl sin ,ml ][ sin mglx sin mglx sin = 0 Thus ) = 0 and the origin is stable by Theorem 1. 10

Page 11

Example 4 : (Pendulum with Friction) We now modify the previous exampl by adding the friction force kl ml mg sin kl deﬁning the same state variables as in example 3 we have sin Again = 0 is an equilibrium point. The energy is the same as in Example 3. Thus ) = ml mgl (1 cos in D −{ ) = ∂V ∂x ∂V ∂x ,f )] = [ mgl sin ,ml ][ sin kl Thus is negative semi-deﬁnite . It is not negative deﬁnite since ) = 0 for = 0 , regardless of the value of (thus ) = 0 along the axis). We conclude that the origin is stable by Theorem 1, but cannot conclude asymptotic stability. The result is disappointing. We know that a pendulum with fri ction cin- verges to = 0. This example emphasizes the fact that all of the Lyapunov theorems provide suﬃcient but not necessary conditions for stability. 11

Page 12

Example 5 : Consider the following system: ) + To study the equilibrium point at the origin, we deﬁne ) = 1 2( We have ) = = [ ,x ][ ) + )] ) + = ( )( Thus, and , provided that < , and it follows that the origin is an asymptotically stable equilibrium poi nt. 12

Page 13

5 Asymptotic Stability in the Large Deﬁnition 7 : The equilibrium state is globally asymptotically stable (or A.S. in the large), if it is stable and every motion converges to th e equilibrium as Question Can we infer that if the conditions of Theorem 2 hold in the who le space , then the asymptotic stability of the equilibrium is global Answer : No! and the next example illustrates this. 13

Page 14

> > Figure 4: The curves ) = Example 6 : Consider the following positive deﬁnite function: ) = 1 + The region is closed for values of β < . However, if β > , the surface is open. Figure 4 shows that an initial state can dive rge from the equilibrium state at the origin while moving towards lower e nergy curves. Deﬁnition 8 : Let be a continuously diﬀerentiable function. Then is said to be radially unbounded if as k Theorem 3 : (Global Asymptotic Stability) Under the conditions of The orem 2, if is radially unbounded then = 0 is globally asymptotically stable. 14

Page 15

6 Positive Deﬁnite Functions Revisited Deﬁnition 9 : A continuous function : [0 ,a is said to be in the class if (i) (0) = 0 (ii) It is strictly increasing. is said to be in the class if in addition and as In the sequel, represents the ball k Lemma 4 is positive deﬁnite if and only if there exists class functions and such that D. Moreover, if and is radially unbounded then and can be chosen in the class Example 7 : Let ) = Px , where is a constant positive deﬁnite sym- metric matrix. Denote min and max the minimum and maximum eigenvalues of , respectively. We have: min Px max min max Thus, , : [0 , and are deﬁned by ) = min ) = max 15

Page 16

Lemma 5 = 0 is stable if and only if there exists a class function and a constant such that (0) < ⇒k k (0) (3) A stronger class of functions is needed in the deﬁnition of as ymptotic stability. Deﬁnition 10 : A continuous function : [0 ,a is said to be in the class KL if (i) For ﬁxed s, r,s is in the class with respect to (ii) For ﬁxed r, r,s is decreasing with respect to (iii) r,s as Lemma 6 : The equilibrium = 0 of the system (1) is asymptotically stable if and only if there exists a class KL function and a constant such that (0) < ⇒k k (0) ,t (4) 16

Page 17

6.1 Exponential Stability Theorem 7 : Suppose that all the conditions of Theorem 2 are satisﬁed, a nd in addition assume that there exist positive constants , K , K and such that Then the origin is exponentially stable. Moreover, if the co nditions hold glob- ally, the = 0 is globally exponentially stable. Proof : According to the assumptions of Theorem 7, the function ) sat- isﬁes Lemma 4 with ) and ), satisfying somewhat strong conditions. Indeed, by assumption i.e. /K ⇒k k /p /K /p or k≤k /p /ρK 17

Page 18

7 The Invariance Principle Asymptotic stability is always more desirable that stabili ty. Lyapunov func- tions often fail to identify asymptotic stability. We now st udy an improve- ment over the Lyapunov theorems studied earlier. Deﬁnition 11 : A set is said to be an invariant set with respect to the dy- namical system if: (0) Example 8 : Any equilibrium point is an invariant set, since if at = 0 we have (0) = , then ) = Example 9 : For autonomous systems, any trajectory is an invariant set Example 10 : A limit cycle is an invariant set (a special case of Example 9 ). Example 11 : If is continuously diﬀerentiable (not necessarily positive deﬁnite) and satisﬁes along the solutions of , then the set deﬁned by is an invariant set. 18

Page 19

Theorem 8 : The equilibrium point = 0 of the autonomous system (1) is asymptotically stable if there exists a function satisfying (i) positive deﬁnite , where we assume that (ii) is negative semi deﬁnite in a bounded region (iii) does not vanish identically along any trajectory in , other than the null solution = 0 Example 12 : Consider again the pendulum with friction of Example 4: (5) sin (6) Again π, R, ) = kl (7) which is negative semi deﬁnite since ) = 0 for all = [ 0] (so = 0 stable but cannot conclude AS). We now apply Theorem 8. Condi tions (i) and (ii) of Theorem 8 are satisﬁed in the region with π < x < , and a < x < a , for any . We now check condition (iii), that is, we check whether can vanish identically along the trajectories trapped in , other than the null solution. By (7) we have ) = 0 0 = kl = 0 thus x = 0 = 0 and by (6), we obtain 0 = sin andthusx = 0 sin = 0 restricting to π, we have that the last condition is satisﬁed if and only if = 0 . Thus, ) = 0 does not vanish identically along any trajectory other than = 0 . Thus = 0 is asymptotically stable by Theorem 8. 19

Page 20

Theorem 9 : The null solution = 0 of the autonomous system (1) is asymp- totically stable in the large if the assumptions of theorem 8 hold in the entire state space and is radially unbounded. Example 13 αx where is a positive scalar. To study the stability of = 0 we deﬁne ) = αx (Radially unbounded). Thus diﬀerentiating ) = ∂V ∂x [1 + ( and and since ) = 0 for = ( 0) . Assume now that = 0 = 0 = 0 , x = 0 = 0 = 0 αx = 0 and considering the fact that = 0 , the last equation implies that = 0 It follows that does not vanish identically along any solution other than = [0 0] . Thus, = 0 is globally asymptotically stable. Theorem 10 : (LaSalle’s theorem) Let be a continuously diﬀeren- tiable function and assume that (i) is a compact set, invariant with respect to the solutions of ( 1). (ii) in (iii) M,and = 0 ; that is, is the set of all points of such that = 0 (iv) : is the largest invariant set in Then every solution starting in approaches as 20

Page 21

8 Region of Attraction Example 14 : Consider the system deﬁned by = 3 We are interested in the stability of = 0 . Consider ) = 12 + 6 + 6 = 3( + 2 + 9 + 3 (8) ) = 30 + 6 (9) According to Theorem 2, if and V < in −{ , then = 0 is “locally” asymptotically stable. Studying and we conclude that deﬁning by (10) we have that and V < −{ Question : Can we conclude that any solution starting in converges to the origin? No! Plotting the trajectories as shown in we see that, for exampl e, the trajectory initiating at the point = 0 ,x = 4 is quickly divergent from = 0 even though the point (0 4) The problem is this: is not an invariant set and there are no guarantees that trajectories starting in will remain within . Thus, once a trajec- tory crosses the border there are no guarantees that will be negative. We now study how to estimate the region of attraction. Deﬁnition 12 : Let x,t be the trajectories of the systems (1) with initial condition at = 0 . The region of attraction to the equilibrium point denoted , is deﬁned by x,t , as t →∞} We nos estimate this region based on LaSalle’s Theorem. 21

Page 22

Theorem 11 : Let be a continuous diﬀerentiable function and assume that is an equilibrium point and (i) is a compact set containing , invariant with respect to the solutions of (1). (ii) is such that V < M. = 0 if x Under these conditions we have that Example 15 kx kx βx for k>x βx for k> (0) >V βx for k> (0) βx for k>x (0) βV for k>x (0) (0) exp( βt for K >x (0) exp( βt |≤| (0) exp( βt for K > (0) 22

Page 23

9 Analysis of Linear Time-Invariant Systems Consider the autonomous linear time-invariant system give n by Ax, A (11) and let ) be deﬁned as follows ) = Px (12) where is (i) symmetric and (ii) positive deﬁnite (iii) P is a consta nt. Thus ) is positive deﬁnite. Also = Px PA or Qx (13) PA Q. (14) Here the matrix is symmetric, since PA AP ) = If is positive deﬁnite, then ) is negative deﬁnite and the origin is (glob- ally) asymptotically stable. To analyze the positive deﬁni teness of the pair of matrices ( P,Q ) we need two steps: (i) Choose an arbitrary symmetric, positive deﬁnite matrix (ii) Find that satisﬁes equation (14) and verify that it is positive de ﬁnite. Equation (14) appears very frequently in the literature and is called Algebraic Lyapunov equation The procedure described above for the stability analysis ba sed on the pair P,Q ) depends on the existence of a unique solution of the Lyapuno v equation for a given matrix . The following theorem guarantees the existence of such a solution. 23

Page 24

Theorem 12 : The eigenvalues of a matrix satisfy if and only if for any given symmetric positive deﬁnite matrix there exists a unique positive deﬁnite symmetric matrix satisfying the Lyapunov equation (14) Proof : Assume ﬁrst that given Q > P > 0 satisfying (14). Thus Px > 0 and Qx < 0 and asymptotic stability follows from Theorem 2. For the converse assume that 0 and given , deﬁne as follows: Qe At dt which is symmetric. We claim that it is also positive deﬁnite . To see this, assume the opposite: i.e. that =0 such that Px = 0. But then Px = 0 Qe At xdt = 0 Qydt = 0 withy At At = 0 = 0 since At is nonsingular . This contradicts the assumption. Thus P > 0. We now show that satisﬁes the Lyapunov equation PA Qe At Adt Qe At dt dt Qe At dt Qe At To complete the proof, there remains to show that this is unique. Suppose that there is another solution . Then ) = 0 At = 0 dt At = 0 which implies that At is constant . This can be the case if and only if = 0, or equivalently, 24

Page 25

10 Instability Theorem 13 : (Chetaev) Consider the autonomous dynamical systems (1) a nd assume that = 0 is an equilibrium point. Let have the following properties: (i) (0) = 0 (ii) , arbitrarily close to = 0 , such that (iii) V > , where the set is deﬁned as follows: k ,andV Under these conditions, = 0 is unstable. Example 16 : Consider again the system of Example 3.20 (textbook) The origin of this system is an unstable equilibrium point. W e now verify this result using Chetaev’s result. Let ) = 1 2( . Thus we have that (0) = 0 , and moreover =0 , ı.e., is positive deﬁnite. Also = ( ,x = ( )( Deﬁning the set by k , << we have that U,x =0 , and V > U, x =0 . Thus the origin is unstable, by Chetaev’s result. 25

De64257nition 2 is said to be stable if for each 57359 0 8658k otherwise the equilibrium point is said to be unstable IMPORTANT This notion applies to the equilibrium not the system A dynamical system can have several equilibrium points brPage 2b ID: 24217

- Views :
**309**

**Direct Link:**- Link:https://www.docslides.com/trish-goza/chapter-lyapunov-stability-i
**Embed code:**

Download this pdf

DownloadNote - The PPT/PDF document "Chapter Lyapunov Stability I Autonomous..." is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.

Page 1

Chapter 3: Lyapunov Stability I: Autonomous Systems 1 Deﬁnitions Consider the autonomous system (1) Deﬁnition 1 is an equilibrium point of (1) if ) = 0 We want to know whether or not the trajecories near an equilib rium point are “well behaved”. Deﬁnition 2 is said to be stable if for each > (0) < ⇒k < otherwise, the equilibrium point is said to be unstable. IMPORTANT : This notion applies to the equilibrium, not “the system.” A dynamical system can have several equilibrium points.

Page 2

&% '$ Figure 1: Stable equilibrium point. &% '$ Figure 2: Asymptotically stable equilibrium point. Deﬁnition 3 of the system (1) is said to be convergent if there exists 0 : (0) < lim ) = Equivalently, is convergent if for any given such that (0) < ⇒k < T. Deﬁnition 4 is said to be asymptotically stable if it is both stable and co n- vergent.

Page 3

Deﬁnition 5 is said to be (locally) exponentially stable if there exist t wo real constants α,λ> such that k (0) λt t> 0 (2) whenever (0) < . It is said to be globally exponentially stable if (2) holds for any Note : Clearly, exponential stability implies asymptotic stabi lity. The con- verse is, however, not true. Remarks : We can always assume that = 0. Given any other equilibrium point we can make a change of variables and deﬁne a new system w ith an equilibrium point at = 0. Deﬁne: = ) = Thus, the equilibrium point of the new systems ) is = 0, since (0) = (0 + ) = ) = 0 Thus, is stable for the system ) if and only if = 0 is stable for the system ).

Page 4

2 Positive Deﬁnite Functions Deﬁnition 6 is positive semi deﬁnite in if (i) and (0) = 0 (ii) xinD −{ is positive deﬁnite in if (ii’) inD −{ is negative deﬁnite (semi deﬁnite)in if is positive deﬁnite (semi deﬁnite).

Page 5

Example 1 : (Quadratic form) ) : Qx, Q , Q Since , its eigenvalues ,i = 1 , are all real. Thus positive deﬁnite = 1 , n positive semideﬁnite = 1 , n negative deﬁnite = 1 , n negative semideﬁnite = 1 , n Notation : Given a dynamical system and a function we will denote ) = dV dt ∂V ∂x dx dt ∂V ∂x ∂V ∂x ∂V ∂x Example 2 : Let ax bx + cos and deﬁne . Thus, we have ) = ∂V ∂x ) = [2 ax bx + cos = 2 ax + 2 bx + 2 cos Notice that the ) depends on the system’s equation ) and thus it will be diﬀerent for diﬀerent systems.

Page 6

Note: Qx, where if sym asym sym asym Then, sym asym where, asym a b i.e. skew symmetric and asym = 0 sym Q is always symmetric Theorem Rayleigh-Ritz theorem min }k Qx max }k

Page 7

3 Stability Theorems Theorem 1 : (Lyapunov Stability Theorem) Let = 0 be an equilibrium point of , and let be a continuously diﬀerentiable function such that (i) (0) = 0 (ii) in D −{ (iii) inD −{ thus = 0 is stable. Theorem 2 : (Asymptotic Stability Theorem) Under the conditions of Th eorem 1, if is such that (i) (0) = 0 (ii) in D −{ (iii) inD −{ thus = 0 is asymptotically stable.

Page 8

Proof of theorem 1 : Choose r> 0 and deﬁne k } is contained in , so deﬁned is a closed and bounded (“compact”) set (a sphere). We now construct a Lyapunov surface inside and show that all trajec- tories starting near = 0 remain inside the surface. Let = min ) (thus α> 0) Now choose (0 , ) and denote Thus, by construction, . Assume now that (0) (0)) Trajectories starting in at = 0 stays inside for all 0. By the continuity of ), δ > 0: < < It then follows that (0) < t> and then (0) < ⇒k which means that the equilibrium = 0 is stable. Proof of theorem 2 : Similar, only that in D implies that is “shrinking” until eventually it becomes the single point = 0.

Page 9

mg Figure 3: Pendulum without friction. 4 Examples Example 3 : (Pendulum Without Friction) Using Newton’s second law of motion we have, ma mg sin l where is the length of the pendulum, and is the angular acceleration. Thus ml mg sin = 0 or sin = 0 choosing state variables we have sin To study the stability of the equilibrium at the origin, we ne ed to propose a Lyapunov function candidate . This is diﬃcult! In this case we “try the total energy, which is a positive function. We have kinetic plus potential energy ωl mgh

Page 10

where (1 cos ) = (1 cos Thus ml mgl (1 cos We now deﬁne ) = . We see that because of the periodicity of cos( we have that ) = 0 whenever = ( ,x = (2 kπ, 0) = 1 Thus, is not positive deﬁnite. However, restricting the domain of to the interval π, ; i.e., we take , with = (( π, ,R We have that R> is indeed positive deﬁnite. Also ) = ∂V ∂x ∂V ∂x ,f )] = [ mgl sin ,ml ][ sin mglx sin mglx sin = 0 Thus ) = 0 and the origin is stable by Theorem 1. 10

Page 11

Example 4 : (Pendulum with Friction) We now modify the previous exampl by adding the friction force kl ml mg sin kl deﬁning the same state variables as in example 3 we have sin Again = 0 is an equilibrium point. The energy is the same as in Example 3. Thus ) = ml mgl (1 cos in D −{ ) = ∂V ∂x ∂V ∂x ,f )] = [ mgl sin ,ml ][ sin kl Thus is negative semi-deﬁnite . It is not negative deﬁnite since ) = 0 for = 0 , regardless of the value of (thus ) = 0 along the axis). We conclude that the origin is stable by Theorem 1, but cannot conclude asymptotic stability. The result is disappointing. We know that a pendulum with fri ction cin- verges to = 0. This example emphasizes the fact that all of the Lyapunov theorems provide suﬃcient but not necessary conditions for stability. 11

Page 12

Example 5 : Consider the following system: ) + To study the equilibrium point at the origin, we deﬁne ) = 1 2( We have ) = = [ ,x ][ ) + )] ) + = ( )( Thus, and , provided that < , and it follows that the origin is an asymptotically stable equilibrium poi nt. 12

Page 13

5 Asymptotic Stability in the Large Deﬁnition 7 : The equilibrium state is globally asymptotically stable (or A.S. in the large), if it is stable and every motion converges to th e equilibrium as Question Can we infer that if the conditions of Theorem 2 hold in the who le space , then the asymptotic stability of the equilibrium is global Answer : No! and the next example illustrates this. 13

Page 14

> > Figure 4: The curves ) = Example 6 : Consider the following positive deﬁnite function: ) = 1 + The region is closed for values of β < . However, if β > , the surface is open. Figure 4 shows that an initial state can dive rge from the equilibrium state at the origin while moving towards lower e nergy curves. Deﬁnition 8 : Let be a continuously diﬀerentiable function. Then is said to be radially unbounded if as k Theorem 3 : (Global Asymptotic Stability) Under the conditions of The orem 2, if is radially unbounded then = 0 is globally asymptotically stable. 14

Page 15

6 Positive Deﬁnite Functions Revisited Deﬁnition 9 : A continuous function : [0 ,a is said to be in the class if (i) (0) = 0 (ii) It is strictly increasing. is said to be in the class if in addition and as In the sequel, represents the ball k Lemma 4 is positive deﬁnite if and only if there exists class functions and such that D. Moreover, if and is radially unbounded then and can be chosen in the class Example 7 : Let ) = Px , where is a constant positive deﬁnite sym- metric matrix. Denote min and max the minimum and maximum eigenvalues of , respectively. We have: min Px max min max Thus, , : [0 , and are deﬁned by ) = min ) = max 15

Page 16

Lemma 5 = 0 is stable if and only if there exists a class function and a constant such that (0) < ⇒k k (0) (3) A stronger class of functions is needed in the deﬁnition of as ymptotic stability. Deﬁnition 10 : A continuous function : [0 ,a is said to be in the class KL if (i) For ﬁxed s, r,s is in the class with respect to (ii) For ﬁxed r, r,s is decreasing with respect to (iii) r,s as Lemma 6 : The equilibrium = 0 of the system (1) is asymptotically stable if and only if there exists a class KL function and a constant such that (0) < ⇒k k (0) ,t (4) 16

Page 17

6.1 Exponential Stability Theorem 7 : Suppose that all the conditions of Theorem 2 are satisﬁed, a nd in addition assume that there exist positive constants , K , K and such that Then the origin is exponentially stable. Moreover, if the co nditions hold glob- ally, the = 0 is globally exponentially stable. Proof : According to the assumptions of Theorem 7, the function ) sat- isﬁes Lemma 4 with ) and ), satisfying somewhat strong conditions. Indeed, by assumption i.e. /K ⇒k k /p /K /p or k≤k /p /ρK 17

Page 18

7 The Invariance Principle Asymptotic stability is always more desirable that stabili ty. Lyapunov func- tions often fail to identify asymptotic stability. We now st udy an improve- ment over the Lyapunov theorems studied earlier. Deﬁnition 11 : A set is said to be an invariant set with respect to the dy- namical system if: (0) Example 8 : Any equilibrium point is an invariant set, since if at = 0 we have (0) = , then ) = Example 9 : For autonomous systems, any trajectory is an invariant set Example 10 : A limit cycle is an invariant set (a special case of Example 9 ). Example 11 : If is continuously diﬀerentiable (not necessarily positive deﬁnite) and satisﬁes along the solutions of , then the set deﬁned by is an invariant set. 18

Page 19

Theorem 8 : The equilibrium point = 0 of the autonomous system (1) is asymptotically stable if there exists a function satisfying (i) positive deﬁnite , where we assume that (ii) is negative semi deﬁnite in a bounded region (iii) does not vanish identically along any trajectory in , other than the null solution = 0 Example 12 : Consider again the pendulum with friction of Example 4: (5) sin (6) Again π, R, ) = kl (7) which is negative semi deﬁnite since ) = 0 for all = [ 0] (so = 0 stable but cannot conclude AS). We now apply Theorem 8. Condi tions (i) and (ii) of Theorem 8 are satisﬁed in the region with π < x < , and a < x < a , for any . We now check condition (iii), that is, we check whether can vanish identically along the trajectories trapped in , other than the null solution. By (7) we have ) = 0 0 = kl = 0 thus x = 0 = 0 and by (6), we obtain 0 = sin andthusx = 0 sin = 0 restricting to π, we have that the last condition is satisﬁed if and only if = 0 . Thus, ) = 0 does not vanish identically along any trajectory other than = 0 . Thus = 0 is asymptotically stable by Theorem 8. 19

Page 20

Theorem 9 : The null solution = 0 of the autonomous system (1) is asymp- totically stable in the large if the assumptions of theorem 8 hold in the entire state space and is radially unbounded. Example 13 αx where is a positive scalar. To study the stability of = 0 we deﬁne ) = αx (Radially unbounded). Thus diﬀerentiating ) = ∂V ∂x [1 + ( and and since ) = 0 for = ( 0) . Assume now that = 0 = 0 = 0 , x = 0 = 0 = 0 αx = 0 and considering the fact that = 0 , the last equation implies that = 0 It follows that does not vanish identically along any solution other than = [0 0] . Thus, = 0 is globally asymptotically stable. Theorem 10 : (LaSalle’s theorem) Let be a continuously diﬀeren- tiable function and assume that (i) is a compact set, invariant with respect to the solutions of ( 1). (ii) in (iii) M,and = 0 ; that is, is the set of all points of such that = 0 (iv) : is the largest invariant set in Then every solution starting in approaches as 20

Page 21

8 Region of Attraction Example 14 : Consider the system deﬁned by = 3 We are interested in the stability of = 0 . Consider ) = 12 + 6 + 6 = 3( + 2 + 9 + 3 (8) ) = 30 + 6 (9) According to Theorem 2, if and V < in −{ , then = 0 is “locally” asymptotically stable. Studying and we conclude that deﬁning by (10) we have that and V < −{ Question : Can we conclude that any solution starting in converges to the origin? No! Plotting the trajectories as shown in we see that, for exampl e, the trajectory initiating at the point = 0 ,x = 4 is quickly divergent from = 0 even though the point (0 4) The problem is this: is not an invariant set and there are no guarantees that trajectories starting in will remain within . Thus, once a trajec- tory crosses the border there are no guarantees that will be negative. We now study how to estimate the region of attraction. Deﬁnition 12 : Let x,t be the trajectories of the systems (1) with initial condition at = 0 . The region of attraction to the equilibrium point denoted , is deﬁned by x,t , as t →∞} We nos estimate this region based on LaSalle’s Theorem. 21

Page 22

Theorem 11 : Let be a continuous diﬀerentiable function and assume that is an equilibrium point and (i) is a compact set containing , invariant with respect to the solutions of (1). (ii) is such that V < M. = 0 if x Under these conditions we have that Example 15 kx kx βx for k>x βx for k> (0) >V βx for k> (0) βx for k>x (0) βV for k>x (0) (0) exp( βt for K >x (0) exp( βt |≤| (0) exp( βt for K > (0) 22

Page 23

9 Analysis of Linear Time-Invariant Systems Consider the autonomous linear time-invariant system give n by Ax, A (11) and let ) be deﬁned as follows ) = Px (12) where is (i) symmetric and (ii) positive deﬁnite (iii) P is a consta nt. Thus ) is positive deﬁnite. Also = Px PA or Qx (13) PA Q. (14) Here the matrix is symmetric, since PA AP ) = If is positive deﬁnite, then ) is negative deﬁnite and the origin is (glob- ally) asymptotically stable. To analyze the positive deﬁni teness of the pair of matrices ( P,Q ) we need two steps: (i) Choose an arbitrary symmetric, positive deﬁnite matrix (ii) Find that satisﬁes equation (14) and verify that it is positive de ﬁnite. Equation (14) appears very frequently in the literature and is called Algebraic Lyapunov equation The procedure described above for the stability analysis ba sed on the pair P,Q ) depends on the existence of a unique solution of the Lyapuno v equation for a given matrix . The following theorem guarantees the existence of such a solution. 23

Page 24

Theorem 12 : The eigenvalues of a matrix satisfy if and only if for any given symmetric positive deﬁnite matrix there exists a unique positive deﬁnite symmetric matrix satisfying the Lyapunov equation (14) Proof : Assume ﬁrst that given Q > P > 0 satisfying (14). Thus Px > 0 and Qx < 0 and asymptotic stability follows from Theorem 2. For the converse assume that 0 and given , deﬁne as follows: Qe At dt which is symmetric. We claim that it is also positive deﬁnite . To see this, assume the opposite: i.e. that =0 such that Px = 0. But then Px = 0 Qe At xdt = 0 Qydt = 0 withy At At = 0 = 0 since At is nonsingular . This contradicts the assumption. Thus P > 0. We now show that satisﬁes the Lyapunov equation PA Qe At Adt Qe At dt dt Qe At dt Qe At To complete the proof, there remains to show that this is unique. Suppose that there is another solution . Then ) = 0 At = 0 dt At = 0 which implies that At is constant . This can be the case if and only if = 0, or equivalently, 24

Page 25

10 Instability Theorem 13 : (Chetaev) Consider the autonomous dynamical systems (1) a nd assume that = 0 is an equilibrium point. Let have the following properties: (i) (0) = 0 (ii) , arbitrarily close to = 0 , such that (iii) V > , where the set is deﬁned as follows: k ,andV Under these conditions, = 0 is unstable. Example 16 : Consider again the system of Example 3.20 (textbook) The origin of this system is an unstable equilibrium point. W e now verify this result using Chetaev’s result. Let ) = 1 2( . Thus we have that (0) = 0 , and moreover =0 , ı.e., is positive deﬁnite. Also = ( ,x = ( )( Deﬁning the set by k , << we have that U,x =0 , and V > U, x =0 . Thus the origin is unstable, by Chetaev’s result. 25

Today's Top Docs

Related Slides