Chains of Kinematic Points

Infinite Chains of Kinematic P oin ts ? Avraham F ein tuch a , Bruce F rancis b a Dep artment of Mathematics, Ben-Gurion University of the Ne gev, Isr ael b Dep artment of Ele ctric al and Computer Engine ering, University of T or onto, Canada Abstract In form ulating the stabilit y problem for an infinite chain of cars, state space is traditionally tak en to be the Hilb ert space ` 2 , wherein the displacemen ts of cars from their equilibria, or the velocities from their equilibria, are tak en to b e square summable. But this obliges the displacemen ts or velocity p erturbations of cars that are far down the chain to b e v anishingly small and leads to anomalous b eha viour. In this pap er an alternative formulation is prop osed wherein state space is the Banach space ` ∞ , allo wing the displacements or velocity perturbations of cars from their equilibria to b e merely b ounded. 1 In tro duction In studying the formation of a very large num b er of ve- hicles, one approac h is instead to mo del an infinite num- b er of v ehicles [1], [7], [15], [16]. The question then arises as to what mathematical framework to take so that the latter mo del correctly describ es the b ehaviour of the for- mer. The purp ose of this paper is to suggest that the Hilb ert space framework usually adopted is not alwa ys appropriate and to suggest an alternative. Consider the infinite chain of cars in Figure 1. The cars are mo delled as p oints on the real line R and are num- b ered by the integers. The p osition of car n is denoted b y q n ∈ R . W e take the simplest mo del of a car, a kine- matic p oint: ˙ q n = u n , n ∈ Z . With nearest-neigh b our interaction, the con trol v elo cit y w ould b e of the form u n = f ( q n +1 − q n , q n − 1 − q n ) , n ∈ Z , where t ypically the function f is linear and the same for all n . Thus ˙ q n = f ( q n +1 − q n , q n − 1 − q n ) . The cars are nominally spaced a unit distance apart. It is assumed that q n = n is an equilibrium of the system, ? This pap er was not presented at any IF A C meeting. The corresp onding author is Bruce F rancis. Email addr esses: abie@math.bgu.ac.il (Avraham F eintuc h), bruce.francis@utoronto.ca (Bruce F rancis). that is, f (1 , − 1) = 0 . Let p n denote the displacement of car n aw ay from its equilibrium p osition: p n = q n − n . Th us the nominal displacemen ts are p n = 0. With f linear it follows that p n satisfies the same equation as q n : ˙ p n = f ( p n +1 − p n , p n − 1 − p n ) , n ∈ Z , (1) The difference b etw een the tw o mo dels is that it is nat- ural to take b ounded initial conditions in the p -mo del. Th us the mo del is an infinite num b er of coupled differ- en tial equations. 0 1 − 1 2 ·· · ·· · Fig. 1. Infinite c hain of cars Let p ( t ) denote the infinite vector of displacements at time t , that is, the comp onen ts of p ( t ) are p n ( t ), n ∈ Z . In this pap er w e are in terested in the question of stability— do es p ( t ) con v erge as t → ∞ and if so in what sense? In the existing literature, e.g., [1], [7], the state space for p ( t ) is the Hilb ert space ` 2 of square-summable se- quences, the adv antage of this setting b eing that F ourier transforms can b e exploited. But this assumption re- quires that p n ( t ) → 0 as n go es to ±∞ , for every t . This seems to b e an unjustified assumption to make at the start of a stability theory , b efore anything has b een pro v ed: If we wan t to know ab out the b ehaviour of p ( t ) as t → ∞ there is no justification in limiting p (0) to sat- isfy p n (0) → 0 as n → ±∞ . Therefore we take the state space to be the Banach space ` ∞ of b ounded sequences. Then p n (0) can all b e of roughly equal magnitude, or they can be randomly distributed in an interv al, etc. The Preprin t submitted to Automatica 4 Marc h 2022 only requirement is that p n (0) lie in some interv al inde- p enden t of n . The goal of this pap er is to develop a sta- bilit y theory in this context, an ` ∞ theory , and to show that it is differen t from the ` 2 theory . W e illustrate with t w o examples . Example 1 Suppose each car heads to w ard the sum of the relative displacements to its tw o neighbours: ˙ q n = ( q n +1 − q n ) + ( q n − 1 − q n ) = q n +1 + q n − 1 − 2 q n . It follows that p n satisfies the same equation: ˙ p n = p n +1 + p n − 1 − 2 p n . Let p denote the infinite vector of displacements. Th us p = 0 is an equilibrium. If p (0) ∈ ` 2 , it turns out (pro ved in the pap er) that p ( t ) conv erges to zero, that is, the cars return to their original positions! But why should the infinite chain b eha v e in this wa y? After all, the cars are not fitted with global sensors to know where the origin is. This anomaly is caused entirely by taking p (0) in ` 2 . Example 2 Consider again an infinite c hain of cars, but where the control ob jective is to maintain a constan t distance b etw een the cars and a constant velocity for eac h car. Let the desired distance b etw een tw o consec- utiv e cars b e d and the desired velocity of each b e v d . Supp ose that for t < 0 the cars are spaced exactly dis- tance d apart and are all moving in the same direction at sp eed v d . That is an equilibrium situation. Let the v e- lo cities of the cars b e denoted v n ( t ), where n ranges o ver all integers. Now supp ose that at time t = 0 every car suddenly sp eeds up by 1%. This is a p erturbation aw ay from the equilibrium. A t t = 0 the velocity of every car is v d + 0 . 01 v d . The p erturbations ˆ v n (0) = 0 . 01 v d are not square-summable, that is, ∞ X n = −∞ ˆ v n (0) 2 = ∞ . Therefore the v ector of p erturbations ˆ v (0) is not in ` 2 but rather is in ` ∞ . That is the situation we are discussing in the pap er and that is not handled b y the ` 2 form ulation. The literature review is p ostp oned until the end of the pap er, where we hav e suitable notation. 2 Mathematical Preliminaries The signals that we deal with are denoted, for example, b y x ( t ), where t denotes time and x is a vector with an infinite n umber of components, x n , n ∈ Z . The meaning is that x n ( t ) is the state vector of car n . F or simplicity , the dimension of x n ( t ) is just 1. Thus for each t , x ( t ) is the state vector of the entire chain. 2.1 Sp ac es, Op er ators, and Sp e ctr a In this subsection, x , y , etc. will denote generic se- quences of real num b ers with comp onen ts x n , y n , etc. Let R ∞ denote the space of such sequences. The index n runs o v er the set of all in tegers. The space R ∞ is an infinite-dimensional vector space. It do es not ha v e a norm, though there is a natural top ology arising from comp onen twise conv ergence. The Hilb ert space ` 2 of square-summable se quences and the Banach space ` ∞ of b ounded sequences are b oth subspaces of R ∞ . The space ` 2 is based on the inner pro duct h x, y i = ∞ X −∞ x n y n . The induced norm is k x k 2 = X n x 2 n ! 1 / 2 . And ` ∞ is based on the norm k x k ∞ = sup n | x n | . Of course every square-summable sequence is b ounded and therefore ` 2 is a subset of ` ∞ . Let X b e a Banac h space. F or us it will be either ` 2 or ` ∞ . The space of b ounded linear op erators on X is denoted B ( X ), or just B . Let A ∈ B . A complex num b er λ is a regular p oint of A if ( λI − A ) − 1 ∈ B . The set of regular p oin ts is the resolven t set , and its complement, σ ( A ), the sp ectrum of A . An eigenv alue is, as for matrices, a complex num b er λ for whic h there exists a nonzero x in X such that Ax = λx . Eigenv alues, if they exist, certainly b elong to the sp ectrum, but λI − A may fail to hav e a b ounded in verse for other complex num b ers λ than eigenv alues. The sp ectrum is alwa ys nonempty , closed, and contained in the disk | z | ≤ r A , where the sp ectral radius r A is given by r A = lim sup n →∞ k A n k 1 /n . 2.2 The Bilater al Right Shift The bilateral righ t shift U is the linear transformation on R ∞ defined by y = U x, y n = x n − 1 . Its inv erse is the left shift: y = U − 1 x, y n = x n +1 . 2 When restricted either to ` 2 or to ` ∞ , U is a b ounded op erator, that is, U b elongs to b oth B ( ` 2 ) and B ( ` ∞ ). Its prop erties in these tw o spaces are somewhat different. Lemma 1 (Pr op erties of U ) (1) Consider U as an op er ator in B ( ` 2 ) . Its sp e ctrum e quals the unit cir cle, but U has no eigenvalues. The kernel (nul lsp ac e) of U − I e quals the zer o subsp ac e. (2) Consider U as an op er ator in B ( ` ∞ ) . Its sp e ctrum e quals the unit cir cle and every p oint in the sp e c- trum is an eigenvalue. The kernel of U − I is the 1- dimensional subsp ac e sp anne d by the ve ctor 1 , whose c omp onents ar e al l 1. Pro of (1) This result is standard, e.g., [9]. As an op- erator on ` 2 , U has no eigenv alues. T o see this, supp ose U x = λx , x ∈ ` 2 , x 6 = 0. Then x n − 1 = λx n , n ∈ Z . Without loss of generalit y starting with x 0 = 1, w e ha ve b y iterating backw ard in the index that x − m = λ m , m > 1. Since x ∈ ` 2 , so lim m →∞ λ m = 0 and so | λ | < 1. But by iterating forward in the index from x 0 = 1 we conclude that | λ | > 1. This inconsistency shows there is no λ . A nonzero vector in the kernel of U − I would b e an eigenv ector and 1 would b e an eigenv alue. But there are no eigenv alues. (2) Since k U n k = 1 for all n , the sp ectral radius is r U = lim n k U n k 1 /n = 1 . Th us σ ( U ) is contained in the closed unit disk. In fact, 1 is an eigenv alue, with eigenv ector all 1’s. Similarly , for every real θ , e j θ is an eigenv alue, with eigen v ector x = ( x n ) defined by x 0 = 1 , x n − 1 = e j θ x n . So the unit circle is contained in σ ( U ). The norm of U − n also equals 1 for ev ery n , and therefore the sp ectral radius of U − 1 equals 1 too. The sp ectra of U and U − 1 are recipro cals. This fact and the equalities r U = r U − 1 = 1 sho w that σ ( U ) has no points in | z | < 1. Th us σ ( U ) and σ ( U − 1 ) b oth equal the unit circle. Finally , the kernel of U − I is the eigenspace for the eigen v alue 1. It is easy to c hec k that all eigenv ectors are constant. This concludes the pro of. Let A ∈ B ( ` 2 ) ∩ B ( ` ∞ ). The sp ectrum of A ∈ B ( ` 2 ) and the sp ectrum of A ∈ B ( ` ∞ ) are not alwa ys equal (see Example 7) but, as w e saw, they are for A = U . And this extends to the case where A is a p olynomial in U, U − 1 , for example A = a 2 U 2 + a 1 U + a 0 I + a − 1 U − 1 + a − 2 U − 2 . The sp ectrum of this A equals { a 2 z 2 + a 1 z + a 0 + a − 1 z − 1 + a − 2 z − 2 : | z | = 1 } . 2.3 Differ ential Equations in Banach Sp ac e Here w e summarize the theory of Dalec ki ˘ ı and Kre ˘ ın [6]. The results are for a general Banac h space X . W e shall need the results for X = ` ∞ and ` 2 (a Hilb ert space is also a Banach space). The exp onential e A can b e defined by the series e A = I + A + 1 2! A 2 + · · · . The operator e A b elongs to B whenever A does. It follo ws that the function t 7→ e At is differentiable and satisfies d dt e At = A e At = e At A. It also satisfies e At   t =0 = I . Consequently for any x 0 ∈ X , x ( t ) = e At x 0 satisfies the equation ˙ x = Ax, x (0) = x 0 . In fact, it is the unique solution among differentiable functions. The sp ectral mapping theorem holds in this general con- text. Let A ∈ B . Let K A denote the class of functions φ ( z ) that are piecewise analytic on σ ( A ). This means that 1) the domain of definition of φ consists of a fi- nite num b er of ope n connected comp onents whose union con tains σ ( A ), each comp onent containing at least one p oin t of σ ( A ); and 2) The function φ is analytic in each comp onen t of its domain of definition. Then σ ( φ ( A )) = φ ( σ ( A )) , whic h says that the sp ectrum of the operator φ ( A ) equals the set of p oin ts φ ( z ) as z ranges ov er the sp ectrum of A . In particular, the sp ectrum of e A equals the set of p oin ts e z as z ranges ov er the sp ectrum of A . Theorem 4.1, page 26, of [6] provides the following key fact for studying stability . Theorem 1 L et A ∈ B . If σ ( A ) lies in the op en half- plane Re λ < a , then ther e exists a c onstant b such that k e At k ≤ b e at , t ≥ 0 . Conversely, if such b exists, then σ ( A ) lies in the close d half-plane Re λ ≤ a . 3 Th us, just as for matrices, if σ ( A ) lies in the op en half- plane Re λ < a and a is negative, then e At con v erges to 0 as t → ∞ , 2.4 Sp atial Invarianc e and the ` 2 -Induc e d Norm Recall that the F ourier transform of x ∈ ` 2 is the func- tion X (e j ω ) = X n x n e − j ω n . This function b elongs to the Hilb ert space, denoted by L 2 ( S 1 ), of square-integrable functions on the unit cir- cle, S 1 (the notation suggests the 1-dimensional unit sphere). The mapping F : ` 2 − → L 2 ( S 1 ) , F : x 7→ X is the F ourier op erator . It is an isomorphism of Hilb ert spaces. Consider an infinite chain mo delled b y the first-order equation ˙ x ( t ) = Ax ( t ) . The deriv ative is with resp ect to time t and A is an op- erator in B ( ` 2 ). The infinite c hain is said to be spatially in v ariant if A commutes with U , i.e., A is a T o eplitz op erator. Then F AF − 1 is the op erator on L 2 ( S 1 ) of m ultiplication b y a function G (e j ω ), the spatial transfer function. Lik ewise, F e At F − 1 is the op erator on L 2 ( S 1 ) of multiplication by e G (e j ω ) t . Then for each fixed t , the ` 2 -induced norm k e At k is given by k e At k = max ω    e G (e j ω ) t    = max ω e Re G (e j ω ) t = e max ω Re G (e j ω ) t . (2) 3 Serial Pursuit and Rendezvous In this section w e tak e the non-symmetric coupling where car n pursues car n − 1, for every n , according to ˙ q n = q n − 1 − q n . (3) This setup is not quite like the one in the introduction with symmetry , because q n (0) = n is not an equilibrium in (3). So w e shall not assume the cars are initially spread out to infinit y but rather are all within a bounded in ter- v al: | q n (0) | ≤ B for some B and all n . W e are interested in whether the cars rendezv ous , that is, all con verge to the same lo cation. The vector form of (3) is ˙ q = ( U − I ) q , (4) whose solution is q ( t ) = e ( U − I ) t q (0). This op erator U − I is represented by the infinite T o eplitz matrix                 . . . . . . . . . . . . . . . . . . − 1 0 0 0 0 . . . . . . 1 − 1 0 0 0 . . . . . . 0 1 − 1 0 0 . . . . . . 0 0 1 − 1 0 . . . . . . 0 0 0 1 − 1 . . . . . . . . . . . . . . . . . .                 . The vertical and horizontal lines in the matrix separate the index range n < 0 from the range n ≥ 0. The sp ec- trum of this operator is the circle of radius 1, centre − 1. Th us 0 is in the spectrum. Perhaps con trary to one’s in- tuition, − 1 is not in the sp ectrum. It’s enligh tening to compare the infinite c hain with a finite one, as we do in three examples. Example 3 Supp ose there are only finitely many cars, in fact, only three cars: n = 0 , 1 , 2. With only three cars w e need a b oundary condition for n = 0 because there is no n = − 1. One p ossibilit y is that car 0 is tethered and hence stationary: ˙ q = Aq , A =     0 0 0 1 − 1 0 0 1 − 1     . Cars 1 and 2 conv erge to the stationary car 0. Example 4 Contin uing with the same setup, tak e the b oundary condition to b e that car 0 can see the global origin and heads for it: ˙ q = Aq , A =     − 1 0 0 1 − 1 0 0 1 − 1     . All cars conv erge to the origin. Example 5 Contin uing still with the same setup, take the b oundary condition to b e that car 0 can see car 2 and heads for it: ˙ q = Aq , A =     − 1 0 1 1 − 1 0 0 1 − 1     . All cars con verge to the a verage of their starting p oin ts. 4 W e return now to the sub ject of the pap er—infinitely man y cars with no b oundary condition. W e shall see that if q (0) b elongs to ` 2 , then q ( t ) conv erges to 0, just as in Example 4. This b ehaviour, where cars without global sensing capabilit y rendezv ous at the origin of the global co ordinate system, is an anomaly caused by the Hilb ert space hypothesis. Theorem 2 With r efer enc e to (4), the ` 2 -induc e d norm of e ( U − I ) t satisfies k e ( U − I ) t k = 1 for al l t ≥ 0 . F or every q (0) ∈ ` 2 , the ` ∞ -norm of q ( t ) c onver ges to 0 as t tends to ∞ ; in addition, q ( t ) c onver ges to zer o we akly, that is, the ` 2 inner pr o duct h q ( t ) , y i c onver ges to zer o as t → ∞ for every y in ` 2 . The pro of is a minor mo dification of the pro of of Theo- rem 4 to follow, and hence is omitted. The case where q (0) instead b elongs to ` ∞ is significan tly more interesting. In this case all the p oin ts start merely in some ball cen tred at the origin, i.e., | q n (0) | ≤ k q (0) k ∞ . W e don’t ha v e a complete theory on this ` ∞ problem; what w e do hav e are six results presented in the subsec- tions to follow. 3.1 R endezvous with we akene d initial c onditions Our first result relaxes the assumption q (0) ∈ ` 2 to merely lim n →±∞ q n (0) = 0. W e will see that the cars again rendezvous at the origin. By linearity , this is the equiv alen t to sa ying that if lim n →±∞ q n (0) = c , then the cars rendezvous at the lo cation c . Lemma 2 Assume q (0) ∈ ` ∞ . If q n (0) tends to 0 as n tends to ±∞ , then q ( t ) c onver ges in ` ∞ to 0 as t → ∞ . Pro of F ollowing signal pro cessing notation, let δ denote the unit impulse in ` ∞ ; that is, δ n = 0 for all n except that δ 0 = 1. Consider the case where q (0) = δ . W e hav e k q ( t ) k ∞ = k e ( U − I ) t δ k ∞ = e − t k e U t δ k ∞ = e − t k δ + tU δ + ( t 2 / 2!) U 2 δ + · · · k ∞ = e − t sup k ≥ 0     t k k !     → 0 as t → ∞ . Next, if q (0) is a finite linear combination of co ordinate v ectors, that is, of { U k δ } , it follo ws by linearit y and the triangle inequality that k q ( t ) k ∞ → 0 as t → ∞ . The closed linear span of finite linear combinations of the co ordinate vectors in ` ∞ is the subspace c 0 of vectors f suc h that f n → 0 as n → ±∞ . Note that the semigroup { e U t } t ≥ 0 satisfies, for f ∈ ` ∞ , k e U t f k ∞ = k f + tU f + ( t 2 / 2!) U 2 f + · · · k ∞ ≤ k f k ∞ + t k U f k ∞ + ( t 2 / 2!) k U 2 f k ∞ + · · · = e t k f k ∞ . And so k e U t k ≤ e t . Thus k e ( U − I ) t k = e − t k e U t k ≤ 1. Since c 0 is inv ariant under U − I , and therefore under e ( U − I ) t , w e can apply the uniform b oundedness principle to obtain that for any q (0) ∈ c 0 , q ( t ) → 0 as t → ∞ . This completes the pro of. 3.2 Conver genc e of c ar n implies that of c ar n + 1 W e will see later that a car’s p osition do esn’t necessarily con v erge, as t → ∞ , under just the assumption that q (0) ∈ ` ∞ . Here w e sho w that if car n conv erges to some lo cation, then cars n, n + 1 , n + 2 , . . . rendezv ous. This is almost ob vious, because car n 0 + 1 is pursuing car n 0 , car n 0 + 2 is pursuing car n 0 + 1, and so on. Lemma 3 If the limit lim t →∞ q n ( t ) exists for n = n 0 , then it exists for every n > n 0 and al l the limits ar e e qual. Pro of A direct computation shows that the n th co or- dinate of q ( t ) is given by q n ( t ) = e − t ∞ X k =0 q n − k (0) t k k ! . (5) Alternativ ely q n ( t ) = h n ( t ) e t , h n ( t ) = ∞ X k =0 q n − k (0) t k k ! . Notice that ˙ h n +1 = h n . Therefore if lim t →∞ q n ( t ) = c , then 1 lim t →∞ q n +1 ( t ) = lim t →∞ h n +1 ( t ) e t = lim t →∞ ˙ h n +1 ( t ) e t b y l’Hˆ opital’s rule = lim t →∞ h n ( t ) e t = lim t →∞ q n ( t ) = c. 1 T o apply l’Hˆ opital’s rule in the equations to follow, we need h n +1 ( t ) → ∞ as t → ∞ . If this isn’t the case, sim- ply p erturb q n (0) to q n (0) + ε for a small p ositiv e ε , that is, translate all the p oin ts. Then h n +1 ( t ) is p erturbed to h n +1 ( t ) + ε e t . 5 This concludes the pro of. W e don’t know if the result extends to n < n 0 . 3.3 A sufficient c ondition for r endezvous F or every t > 0, the series (5) con v erges, that is, q n ( t ) is well-defined. How ever the limit lim t →∞ q n ( t ) may or ma y not exist, dep ending on q (0). Here w e give an ex- ample where rendezv ous o ccurs without the strong as- sumption on q (0) that is in Lemma 2. Example 6 T ake q (0) = ( . . . , q − 2 (0) , q − 1 (0) | q 0 (0) , q 1 (0) , . . . ) = ( . . . , 0 , 0 , 0 | 1 , 0 , − 1 , 0 , 1 , 0 , − 1 , . . . ) , for which q n ( t ) = Re e ( j − 1) t . Th us q n ( t ) conv erges to 0 as t → ∞ , i.e., the cars ren- dezv ous at the origin. Along a similar line, let a b e a real n um b er that is not a rational m ultiple of π . Kroneck er’s densit y theorem says that the sequence { e j na } n ∈ Z is dense on the unit circle, that is, every p oint on the circle is an accumulation p oin t for the sequence. F or q n (0) = e j na w e hav e lim t →∞ q n ( t ) = lim t →∞ e − t ∞ X k =0 e j ( n − k ) a t k k ! = lim t →∞ e − t e j na e ( e − j a ) t = lim t →∞ e (cos a − 1) t e j na e − j t sin a = 0 . So, the cars rendezvous at the origin even though they are initially densely disp ersed around the unit circle. 3.4 Conver genc e to the aver age starting p oint In the cyclic pursuit problem for a finite num b er of kine- matic cars, the cars rendezvous at the av erage of their starting p ositions—see Example 5. This turns out to b e true also in the infinite c hain serial pursuit problem pro- vided there is an appropriate av erage initial p osition. The av erage of the p oints { q m (0) , . . . , q m − N (0) } is a vg { q m − k (0) } N k =0 = 1 N + 1 N X k =0 q m − k (0) . Our assumption will b e that this a v erage con v erges at the rate 1 / √ N as N increases. That is, we will assume that there exists a num b er ¯ q such that a vg { q m − k (0) } N k =0 = ¯ q + o ( N − 1 / 2 ) . (6) This means that for ev ery constan t C there exists an in teger L such that if N > L then   a vg { q m − k (0) } N k =0 − ¯ q   ≤ C √ N . Lemma 4 Assume that for some m ther e exists a num- b er ¯ q such that (6) holds. Then lim t →∞ q n ( t ) = ¯ q for every n . The pro of uses a result of G. H. Hardy . Theorem 149 in [10] is as follows: Let { f n } ∞ n =0 b e a sequence of complex n um b ers such that there exists a constant ¯ f suc h that a vg { f 0 , . . . , f n } = ¯ f + o ( n − 1 / 2 ) . Then lim t →∞ e − t ∞ X k =0 f k t k k ! = ¯ f . Pro of of the Lemma Assume (6) holds. Then it holds for all other v alues of m . F or example a vg { q m +1 − k (0) } N k =0 = a vg { q m − k (0) } N k =0 + 1 N + 1 [ q m +1 (0) − q m +1 − N (0)] = ¯ q + o ( N − 1 / 2 ) + 1 N + 1 [ q m +1 (0) − q m +1 − N (0)] = ¯ q + o ( N − 1 / 2 ) . F rom Hardy’s theorem, then, for every m lim t →∞ e − t ∞ X k =0 q m − k (0) t k k ! = ¯ q . F rom (5), for every m lim t →∞ q m ( t ) = ¯ q . This concludes the pro of. 3.5 An example of non-c onver genc e It is more difficult to construct an example where lim t →∞ q n ( t ) doesn’t exist. W e turn to suc h an example no w. In this example, the points q n (0) are either 0 or 1. Then existence of the limit lim t →∞ q n ( t ) is related to some v ery interesting results of Diaconis and Stein on T aub e- rian theory [8], which w e now briefly describ e. Let A b e an infinite subset of non-negative integers, for example 6 the non-negativ e ev en in tegers, and consider the ques- tion of whether this limit exists: lim t →∞ e − t X k ∈ A t k k ! . (7) Existence of the limit is a prop ert y of the set A . No w let S n denote the num b er of heads that o ccur in n tosses of a coin and consider the question of whether this limit exists: lim n →∞ Pr( S n ∈ A ) . (8) Finally , let c ar d denote cardinality and consider the question of whether this limit exists for every ε > 0: lim n →∞ 1 ε √ n card { k : k ∈ A , n ≤ k < n + ε √ n } . (9) Remark ably , the three limits are intimately related: If either exists, then so do the other tw o and they are all equal. This is Theorem 1 in [8]. T o get an example where (7) fails, by taking n = m 2 and ε = 1, it suffices to get an example where 1 m card { k : k ∈ A , m 2 ≤ k < m 2 + m } do es not conv erge as m → ∞ . Defining γ ( m ) = card { k : k ∈ A , m 2 ≤ k < m 2 + m } , it suffices to choose A such that γ ( m ) = m − 1 for m ev en and γ ( m ) = 0 for m o dd. Returning to (5), take n = 0 and take the initial conditions q 0 (0) , q − 1 (0) , . . . as follows. F or m ≥ 0 ev en, set q − k (0) = 1 for m 2 ≤ k < m 2 + m , and for m ≥ 0 o dd, set q − k (0) = 0 for m 2 ≤ k < m 2 + m . F or other v alues of k , the v alue of q k (0) is irrelev an t and could b e set to 0. Then q 0 ( t ) fails to conv erge as t → ∞ . 3.6 Conver genc e of q ( t ) on a subsp ac e Our final result on this problem seems to b e particularly in teresting. Theorem 3 The ` ∞ -induc e d norm of ( U − I )e ( U − I ) t c onver ges to 0 as t → ∞ . Thus, for every q (0) ∈ ` ∞ , ˙ q ( t ) c onver ges to zer o in ` ∞ as t → ∞ , and, mor e over, if q (0) b elongs to ( U − I ) ` ∞ , the image sp ac e of U − I acting on ` ∞ , then q ( t ) c onver ges to zer o in ` ∞ as t → ∞ . Pro of Given q (0) ∈ ` ∞ , let r ( t ) = ( U − I )e ( U − I ) t q (0). T o simplify lay out, define ψ ( k , t ) = t k k ! − t k +1 ( k + 1)! . The n th comp onen t of r ( t ) is r n ( t ) = e − t " − q n (0) + ∞ X k =0 ψ ( k , t ) q n − ( k +1) (0) # . Th us k r ( t ) k ∞ = e − t sup n      − q n (0) + ∞ X k =0 ψ ( k , t ) q n − ( k +1) (0)      ≤ e − t " 1 + ∞ X k =0 | ψ ( k , t ) | # k q (0) k ∞ . It therefore suffices to show that lim t →∞ e − t ∞ X k =0 | ψ ( k , t ) | = 0 . It is elementary that for fixed t , the sequence { t k /k ! } has a maxim um at some k 0 ( t ), and that this in teger satisfies k 0 ( t ) ≤ t ≤ k 0 ( t ) + 1. Also, for k ≤ k 0 ( t ), the sequence is increasing and for k ≥ k 0 ( t ) the sequence is decreasing. Therefore, ∞ X k =0 | ψ ( k , t ) | = − k 0 ( t ) − 1 X k =0 ψ ( k , t ) + ∞ X k = k 0 ( t ) ψ ( k , t ) = 2 t k 0 ( t ) k 0 ( t )! − 1 . Th us e − t ∞ X k =0 | ψ ( k , t ) | ≤ 2e − t max k t k k ! , whic h approac hes zero as t → ∞ . This completes the pro of. Unfortunately , a simulation to illustrate the preceding result is not p ossible, b ecause one cannot simulate an infinite num b er of kinematic p oin ts. The result implies that for every q (0) in ` ∞ lim t →∞ sup n | q n − 1 ( t ) − q n ( t ) | = 0 . In tuitiv ely , the p oin ts { q n ( t ) } cluster around a point, but that p oint may itself not b e stationary . Theorem 3 raises the question of characterizing the im- age of the op erator U − I . A b ounded sequence y b e- longs to Im( U − I ) iff there exists a b ounded x such that y = ( I − U ) x , i.e., x = ( I + U + U 2 + · · · ) y . Th us Im( U − I ) equals the space of y such that y and ( I + U + U 2 + · · · ) y are b oth b ounded. 7 4 A Symmetric Chain W e turn now to kinematic cars where each is coupled to its tw o neighbours, not just one as in the preceding section. Thus the coupled equations ˙ p n = ( p n +1 − p n ) + ( p n − 1 − p n ) or in vector form ˙ p = Ap, A = U + U − 1 − 2 I . (10) If the state space is ` 2 , then of course U − 1 = U ∗ . The solution of (10) is p ( t ) = e At p (0). Theorem 4 (With r efer enc e to (10).) (1) The ` 2 -induc e d norm of e At satisfies k e At k = 1 for al l t ≥ 0 . F or every p (0) ∈ ` 2 , the ` ∞ -norm of p ( t ) c onver ges to 0 as t tends to ∞ ; in addition, p ( t ) c onver ges to zer o we akly, that is, the ` 2 inner pr o duct h p ( t ) , y i c onver ges to zer o as t → ∞ for every y ∈ ` 2 . (2) The ` ∞ -induc e d norm of e At satisfies k e At k = 1 for al l t ≥ 0 . Also k A e At k → 0 as t → ∞ . Thus, for every q (0) ∈ ` ∞ , ˙ q ( t ) c onver ges to zer o in ` ∞ as t → ∞ , and, mor e over, if q (0) b elongs to A` ∞ , the r ange sp ac e of A acting on ` ∞ , then q ( t ) c onver ges to zer o in ` ∞ as t → ∞ . Pro of (1) The spectrum of A is the real in terv al [ − 4 , 0]. The op erator F AF − 1 is multiplication by G (e j ω ) = e − j ω + e j ω − 2 = 2( cos ω − 1) . The maximum real part of G (e j ω ) equals 0. Thus from (2) k e At k = 1. T ake the spatial F ourier transform of the comp onen ts of p ( t ), holding t fixed: P (e j ω , t ) = X n p n ( t )e − j ω n . Then the function ω 7→ P (e j ω , t ), denoted for con v e- nience by P ( t ), b elongs to L 2 ( S 1 ). T aking F ourier trans- form of the differential equation gives ˙ P ( t ) = (e − j ω + e j ω − 2) P ( t ) . Solv e the equation: P ( t ) = exp[(e − j ω + e j ω − 2) t ] P (0) = e 2 t (cos ω − 1) P (0) . That is, P (e j ω , t ) = e 2 t (cos ω − 1) P (e j ω , 0) . (11) Recall that L 2 ( S 1 ) is a subspace of L 1 ( S 1 ). Using in turn the definition of the norm in L 1 ( S 1 ), equation (11), the Cauc hy-Sc hw arz inequality , and the mo dified Bessel function of the first kind I 0 . we hav e k P ( t ) k 1 = 1 2 π Z π − π | P (e j ω , t ) | dω = 1 2 π Z π − π e 2 t (cos ω − 1) | P (e j ω , 0) | dω ≤  1 2 π Z π − π e 4 t (cos ω − 1) dω  1 / 2 k P (0) k 2 =  1 2 π e − 4 t I 0 (4 t )  1 / 2 k P (0) k 2 . But e − 4 t I 0 (4 t ) con verges to 0 as t → ∞ (the easiest wa y to see this is to graph the function), and therefore so do es k P ( t ) k 1 . Since by definition p n ( t ) = 1 2 π Z π − π P (e j ω , t )e j ω n dω , so | p n ( t ) | ≤ 1 2 π Z π − π | P (e j ω , t ) | dω = k P ( t ) k 1 . Therefore k p ( t ) k ∞ ≤ k P ( t ) k 1 and therefore k p ( t ) k ∞ → 0 as t → ∞ . Let e n denote the n th basis v ector in ` 2 . W e pro ved abov e that p n ( t ) conv erges to zero as t → ∞ . Thus lim t →∞ h e At p (0) , e n i = 0 . Therefore lim t →∞ h e At p (0) , y i = 0 for every finite linear com bination y of { e n } . Also, k e At k ≤ 1. Therefore by the uniform b oundedness prin- ciple, lim t →∞ h e At p (0) , y i = 0 for every y in ` 2 . (2) Fix t ≥ 0. The op erator U is an isometry . F or every x in ` ∞ k e U t x k ∞ =     x + tU x + 1 2 t 2 U 2 x + · · ·     ∞ ≤ k x k ∞ + t k x k ∞ + 1 2 t 2 k x k 2 ∞ + · · · = e t k x k ∞ . And so k e U t k ≤ e t . Likewise for U − 1 . Thus k e At k ≤ e − 2 t k e U t kk e U − 1 t k ≤ e − 2 t e t e t = 1 . 8 T o conclude equality , apply e At to 1 . W e next show that the kernel of A = U + U − 1 − 2 I as an op erator on ` ∞ is the one-dimensional subspace spanned b y 1 . Let x b e a b ounded sequence in the kernel of A . Then x n − x n − 1 = x n +1 − x n . In particular x 0 − x − 1 = x 1 − x 0 x 1 − x 0 = x 2 − x 1 etc. and so x n +1 = x n + ( x 0 − x − 1 ) for all n > 0. If x 0 − x − 1 6 = 0, then x n gro ws without b ound as n → ∞ . Th us x 0 = x − 1 . Similarly x n = x n − 1 for all n . Recall that k e ( U − I ) t k ≤ 1 for all t ≥ 0. By symmetry the same holds for k e ( U − 1 − I ) t k . Thus k A e At k = k [( U − I ) + ( U − 1 − I )]e ( U − I ) t e ( U − 1 − I ) t k ≤ k e ( U − 1 − I ) t ( U − I )e ( U − I ) t k + k e ( U − I ) t ( U − 1 − I )e ( U − 1 − I ) t k ≤ k ( U − I )e ( U − I ) t k + k ( U − 1 − I )e U − 1 − I ) t k → 0 as t → ∞ . This completes the pro of. As our final contrast b etw een the ` 2 and ` ∞ cases, pre- sen ted next is a chain that is unstable in ` 2 but stable in ` ∞ . The chain is not spatially inv ariant. Example 7 Define the op erator B to rep eat ev ery com- p onen t. Thus B is defined by y = B x , y 2 n = y 2 n +1 = x n . As an operator on ` 2 , k B k = √ 2, r B = √ 2, and σ ( B ) = { λ : | λ | ≤ √ 2 } . As an operator on ` ∞ , k B k = 1, r B = 1, and σ ( B ) = { λ : | λ | ≤ 1 } . Let a b e a p ositiv e constant and consider the first-order system ˙ p n = u n u 2 n = − ap 2 n + p n u 2 n +1 = − ap 2 n +1 + p n . Then ˙ p = Ap, A = − aI + B . This chain is not a simple mass-spring-dashp ot system. The information flow structure is shown in Figure 2. F or example, for the comp onent ˙ p 5 = − ap 5 + p 2 0 1 2 3 4 5 6 7 8 -1 -2 etc. etc. Fig. 2. Information flo w in the example with spatial v ariation the graph shows an arrow from no de 2 to no de 5. The sp ectrum of A equals σ ( A ) = − a + σ ( B ) . Select a to lie in the in terv al 1 < a < √ 2. Then, as an op erator on ` 2 , σ ( A ) has a nonempty intersection with the closed righ t half-plane, and so the origin p = 0 is not asymptotically stable; while on the other hand as an op erator on ` ∞ , σ ( A ) is contained in the op en left half- plane, and so the origin p = 0 is asymptotically stable. That is to sa y , there exists an initial state p (0) in ` 2 suc h that k p ( t ) k ∞ con v erges to zero but k p ( t ) k 2 div erges to ∞ . 5 Literature review W e now briefly review the literature. Chains are 1- dimensional lattices; lattices o ccur in physics problems. The theory of the propagation of wa ves, and in particu- lar the application to determine the v elo cit y of sound, is due to Newton and was published in 1687. In Chapter I II of [3], Brillouin offers a mathematical treatmen t of w a ve propagation in a one-dimensional lattice of identi- cal particles. How ev er it is not mathematically rigorous. Reference [14] is t ypical of the physics literature. Kopell has a substan tial o euvre on chains of oscillators, for example [13]. In her work she do es study the situation when the length of the c hain go es to infinity . How ever the b oundary conditions are maintained. W e start with an infinite chain that therefore has no b oundary condi- tions. An early con tribution to optimal con trol of an infinite c hain of cars is that of Melzer and Kuo [15]. Their “in- finite ob ject problem” has the mo del ˙ x ( t ) = Ax ( t ) + B u ( t ) , where for eac h t , x ( t ) and u ( t ) belong to ` 2 and where A and B are spatially-inv ariant op erators on ` 2 . The pap er form ulates a linear-quadratic optimal control problem with a cost function in v olving the time-domain L 2 -norm of h x ( t ) , Qu ( t ) i + h u ( t ) , Ru ( t ) i , 9 where Q and R are spatially-inv ariant op erators on ` 2 . That is to say , the optimal control problem is form ulated in the space L 2 ( R , ` 2 ). The optimal control la w takes the form u = F x . The solution is derived via the F ourier transform. The work of Melzer and Kuo has b een gener- alized and extended, most notably by Bamieh et al. [1], D’Andrea and Dullerud [7], and Motee and Jadbabaie [16]. Curtain et al. [5] studied the LQR problem in the Hilb ert space context, addressing the question of trun- cating the infinite chain. Our pap er is related to that of Swaroop and Hedrick on string stabilit y [17]. The system in that pap er is a semi- infinite chain, that is, the cars are num b ered 0 , 1 , 2 , . . . and car 0 is therefore a b oundary , its dynamics b eing indep enden t of all others. This mo del is appropriate for a plato on (or conv oy) with a leader. By contrast, in the other references and in our pap er there is no b oundary car. On the other hand, reference [17] is the only refer- ence we found that prop oses ` ∞ for the state space. Other recent pap ers are [2], [4], [11], and [12]. 6 Conclusion An infinite chain of vehicles obviously do esn’t o ccur in realit y . It is intended to b e relev an t to the case of a finite but very large chain. W e hav e argued that the ` 2 - framew ork is not the righ t one, b ecause an infinite chain do es not behav e lik e a finite but large one. F or example, form ulating the rendezvous problem in ` 2 results in a rendezv ous at the origin, whereas a finite chain would not do so. The ` ∞ form ulation seems more appropriate in our opinion. How ever, the problems are harder. Designing con trollers that are ` ∞ -optimal is an op en problem. A cknow le dgements Thanks to Harry Dym for drawing our attention to [6], to Ronen Peretz for introducing us to and discussions on [8], to Joyce Poon for introducing us to the physics literature, and to Geir Dullerud and Bassam Bamieh for commen ts ab out the existing literature. References [1] B. Bamieh, F. Paganini, and M. Dahleh. 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