Flatness and Defect  Theory and Examples

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FLATNESS AND DEFECT OF NONLINEAR SYSTEMS:INTRODUCTORY THEORY AND EXAMPLES Michel FliessJean LvinePhilippe MartinPierre RouchonCAS internal report A284, January 1994.We introduce at systems, which are equivalent to linear ones via a special type of feedbackcalled endogenous. Their physical properties are subsumed by a linearizing output and they might beregarded as providing another nonlinear extension of Kalmans controllability. The distance to atnessis measured by a nonnegative integer, the defect. We utilize differential algebra which suits well tothe fact that, in accordance with Willems standpoint, atness and defect are best dened withoutdistinguishing between input, state, output and other variables. Many realistic classes of examplesare at. We treat two popular ones: the crane and the car with n trailers, the motion planning ofwhich is obtained via elementary properties of planar curves. The three nonat examples, the simple,double and variable length pendulums, are borrowed fromnonlinear physics. Ahigh frequency controlstrategy is proposed such that the averaged systems become at.This work was partially supported by the G.R. Automatique of the CNRS and by the D.R.E.D. of the Minist` ere delEducation Nationale.1 1 IntroductionWe present here ve casestudies: the control of a crane, of the simple, double and variable lengthpendulums and the motion planning of the car with ntrailers. They are all treated within the frameworkof dynamic feedback linearization which, contrary to the static one, has only been investigated by fewauthors (Charlet et al. 1989, Charlet et al. 1991, Shadwick 1990). Our point of view will be probablybest explained by the following calculations where all vector elds and functions are realanalytic.Consider x = f (x, u) (x Rn, u Rm), (1)where f (0, 0) = 0 and rank fu(0, 0) = m. The dynamic feedback linearizability of (1) means,according to (Charlet et al. 1989), the existence of1. a regular dynamic compensator_ z = a(x, z, v)u = b(x, z, v) (z Rq, v Rm) (2)where a(0, 0, 0) = 0, b(0, 0, 0) = 0. The regularity assumption implies the invertibility1ofsystem (2) with input v and output u.2. a diffeomorphism = (x, z) ( Rn+q) (3)such that (1) and (2), whose (n +q)dimensional dynamics is given by_ x = f (x, b(x, z, v)) z = a(x, z, v),becomes, according to (3), a constant linear controllable system = F + Gv.Up to a static state feedback and a linear invertible change of coordinates, this linear system maybe written in Brunovsky canonical form (see, e.g., (Kailath 1980)),___y(1)1 = v1...y(m)m = vmwhere 1, . . ., m are the controllability indices and (y1, . . . , y(11)1 , . . . , ym, . . . , y(m1)m ) is another basis of the vector space spanned by the components of . Set Y = (y1, . . . , y(11)1 , . . . , ym, . . . , y(m1)m );1See (Li and Feng 1987) for a denition of this concept via the structure algorithm. See (Di Benedetto et al. 1989,Delaleau and Fliess 1992) for a connection with the differential algebraic approach.2 thus Y = T where T is an invertible (n + q) (n + q) matrix. Otherwise stated, Y = T(x, z).The invertibility of yields_ xz_= 1(T1Y). (4)Thus from (2) u = b_
1(T1Y), v_. From vi = y(i)i , i = 1, . . . , m, u and x can be expressedas realanalytic functions of the components of y = (y1, . . . , ym) and of a nite number of theirderivatives:_ x = A(y, y, . . . , y())u = B(y, y, . . . , y()). (5)The dynamic feedback (2) is said to be endogenous if, and only if, the converse holds, i.e., if, andonly if, any component of y can be expressed as a realanalytic function of x, u and a nite number ofits derivatives:y = C(x, u, u, . . . , u( )). (6)Note that, according to (4), this amounts to expressing z as a function of (x, u, u, . . . , u()) forsome . In other words, the dynamic extension does not contain exogenous variables, which areindependent of the original system variables and their derivatives. This justies the word endogenous. Note that quasistatic feedbacks, introduced in the context of dynamic inputoutput decoupling (Delaleau and Fliess 1992), share the same property.A dynamics (1) which is linearizable via such an endogenous feedback is said to be (differentially) at; y, which might be regarded as a ctitious output, is called a linearizing or at output. The terminology at is due to the fact that y plays a somehow analogous role to the at coordinates in the differential geometric approach to the Frobenius theorem (see, e.g., (Isidori 1989,Nijmeijer and van der Schaft 1990)). A considerable amount of realistic models are indeed at. Wetreat here twocasestudies, namelythe crane (DAndr eaNovel and L evine 1990, Marttinen et al. 1990)and the car with n trailers (Murray and Sastry 1993, Rouchon et al. 1993a). Notice that the use of alinearizing output was already known in the context of static state feedback (see (Claude 1986) and(Isidori 1989, page 156)).One major property of differential atness is that, due to formulas (5) and (6), the state and inputvariables can be directly expressed, without integrating any differential equation, in terms of the atoutput and a nite number of its derivatives. This general idea can be traced back to works by D. Hilbert(Hilbert 1912) and E. Cartan (Cartan 1915) on underdetermined systems of differential equations,where the number of equations is strictly less than the number of unknowns. Let us emphasize on thefact that this property may be extremely usefull when dealing with trajectories: from y trajectories,x and u trajectories are immediately deduced. We shall detail in the sequel various applications ofthis property from motion planning to stabilization of reference trajectories. The originality of ourapproach partly relies on the fact that the same formalism applies to study systems around equilibriumpoints as well as around arbitrary trajectories.As demonstrated by the crane, atness is best dened by not distinguishing between input, state,output and other variables. The equations moreover might be implicit. This standpoint, which matcheswell with Willems approach (Willems 1991), is here taken into account by utilizing differential3 algebra which has already helped clarifying several questions in control theory (see, e.g., (Diop 1991,Diop 1992, Fliess 1989, Fliess 1990a, Fliess and Glad 1993)).Flatness might be seen as another nonlinear extension of Kalmans controllability. Such anassertion is surprising when having in mind the vast literature on this subject (see (Isidori 1989,Nijmeijer and van der Schaft 1990) and the references therein). Remember, however, Willems trajectory characterization (Willems 1991) of linear controllability which can be interpreted as the freenessof the module associated to a linear system (Fliess 1992). A linearizing output now is the nonlinearanalogue of a basis of this free module.We know from (Charlet et al. 1989) that any singleinput dynamics which is linearizable by adynamic feedback is also linearizable by a static one. This implies the existence of nonat systemswhich verify the strong accessibility property (Sussmann and Jurdjevic 1972). We introduce a nonnegative integer, the defect, which measures the distance from atness.These new concepts and mathematical tools are providing the common formalism and the underlying structure of ve physically motivated case studies. The rst two ones, i.e., the control of a craneand the motion planning of a car with ntrailers, which are quite concrete, resort from at systems.The three others, i.e., the simple and double Kapitsa pendulums and the variablelength pendulumexhibit a non zero defect.The characterization of the linearizing output in the crane is obvious when utilizing a nonclassicrepresentation, i.e., a mixture of differential and nondifferential equations, where there are no distinction between the system variables. It permits a straightforward tracking of a reference trajectoryvia an openloop control. We do not only take advantage of the equivalence to a linear system but alsoof the decentralized structure created by assuming that the engines are powerful with respect to themasses of the trolley and the load.The motion planning of the car with ntrailer is perhaps the most popular example of path planningof nonholonomic systems (Laumond 1991, Murray and Sastry 1993, Monaco and NormandCyrot 1992,Rouchon et al. 1993a, Tilbury et al. 1993, Martin and Rouchon 1993, Rouchon et al. 1993b). It is aat systemwhere the linearizing output is the middle of the axle of the last trailer. Once the linearizingoutput is determined, the path planning problem becomes particularly easy: the reference trajectoryas well as the corresponding openloop control can be expressed in terms of the linearizing output anda nite number of its derivatives. Let us stress that no differential equations need to be integrated toobtain the openloop control. The relative motions of the various components of the system are thenobtained thanks to elementary geometric properties of plane curves. The resulting calculations, whichare presented in the twotrailer case, are very fast and have been implemented on a standard personalmicrocomputer under MATLAB.The control of the three nonat systems is based on high frequency control and approximations by averaged and at systems (for other approaches, see, e.g., (Baillieul 1993, Bentsman 1987,Meerkov 1980)). We exploit here anidea due tothe Russianphysicist Kapitsa (Bogaevski and Povzner 1991,Landau and Lifshitz 1982) for stabilizing these three systems in the neighborhood of quite arbitrary positions and trajectories, and in particular positions which are not equilibriumpoints. This idea is closelyrelated to a curiosity of classical mechanics that a double inverted pendulum (Stephenson 1908), andeven the N linked pendulums which are inverted and balanced on top of one another (Acheson 1993),4 can be stabilized in the same way. Closedloop stabilization around reference averaged trajectories becomes straightforward by utilizing the endogenous feedback equivalence to linear controllablesystems.The paper is organized as follows. After some differential algebraic preliminaries, we dene equivalence by endogenous feedback, atness and defect. Their implications for uncontrolled dynamicsand linear systems are examined. We discuss the link between atness and controllability. In orderto verify that some systems are not linearizable by dynamic feedback, we demonstrate a necessarycondition of atness, which is of geometric nature. The last two sections are devoted respectively tothe at and nonat examples.First drafts of various parts of this article have beenpresentedin(Fliess et al. 1991, Fliess et al. 1992b,Fliess et al. 1992a, Fliess et al. 1993b, Fliess et al. 1993c).2 The algebraic frameworkWe consider variables related by algebraic differential equations. This viewpoint, which possessa nice formalisation via differential algebra, is strongly related to Willems behavioral approach(Willems 1991), where trajectories play a key role. We start with a brief review of differential elds(see also (Fliess 1990a, Fliess and Glad 1993)) and we refer to the books of Ritt (Ritt 1950) andKolchin (Kolchin 1973) and Seidenbergs paper (Seidenberg 1952) for details. Basics on the customary (nondifferential) eld theory may be found in (Fliess 1990a, Fliess and Glad 1993) as wellas in the textbook by Jacobson (Jacobson 1985) and Winter (Winter 1974) (see also (Fliess 1990a,Fliess and Glad 1993)); they will not be repeated here.2.1 Basics on differential eldsAn (ordinary) differential ring R is a commutative ring equipped with a single derivation ddt = suchthata R, a = dadt Ra, b R,ddt(a +b) = a + bddt(ab) = ab +ab.A constant c R is an element such that c = 0. A ring of constants only contains constant elements.An (ordinary) differential eld is an (ordinary) differential ring which is a eld.A differential eld extension L/K is given by two differential elds, K and L, such that K Land such that the restriction to K of the derivation of L coincides with the derivation of K.An element L is said to be differentially Kalgebraic if, and only if, it satises an algebraicdifferential equation over K, i.e., if there exists a polynomial K[x0, x1, . . . , x], = 0, such that(, , . . . , ()) = 0. The extension L/K is said to be differentially algebraic if, and only if, anyelement of L is differentially Kalgebraic.5 An element L is said to be differentially Ktranscendental if, and only if, it is not differentiallyKalgebraic. The extension L/K is said to be differentially transcendental if, and only if, there existsat least one element of L that is differentially Ktranscendental.A set {i  i I } of elements in L is said to be differentially Kalgebraically independent if,and only if, the set of derivatives of any order, {()i  i I, = 0, 1, 2, . . .}, is Kalgebraicallyindependent. Such an independent set which is maximal with respect to inclusion is called a differentialtranscendence basis of L/K. Two such bases have the same cardinality, i.e., the same number ofelements, which is called the differential transcendence degree of L/K: it is denoted by diff tr d0L/K.Notice that L/K is differentially algebraic if, and only if, diff tr d0L/K = 0.Theorem 1 For a nitely generated differential extension L/K, the next two properties are equivalent:(i) L/K is differentially algebraic;(ii) the (nondifferential) transcendence degree of L/K is nite, i.e., tr d0L/K < .More details and some examples may be found in (Fliess and Glad 1993).2.2 Systems 2Let k be a given differential ground eld. A system is a nitely generated differential extension D/k 3.Such a denition corresponds to a nite number of quantities which are related by a nite number ofalgebraic differential equations over k 4. We do not distinguish in this setting between input, state,output and other types of variables. This eldtheoretic language therefore ts Willems standpoint(Willems 1991) on systems. The differential order of the systemD/k is the differential transcendencedegree of the extension D/k.Example Set k = R; D/k is the differential eld generated by the four unknowns x1, x2, x3, x4related by the two algebraic differential equations: x1 + x3 x4 = 0, x2 +(x1 + x3x4)x4 = 0. (7)Clearly, diff tr d0D/k = 2: it is equal to the number of unknowns minus the number of equations.Denote by k < u > the differential eld generated by k and by a nite set u = (u1, . . . , um) ofdifferential kindeterminates: u1, . . ., um are differentially kalgebraically independent, i.e.,2See also (Fliess 1990a, Fliess and Glad 1993).3Two systems D/k and D/k are, of course, identied if, and only if, there exists a differential kisomorphism betweenthem (a differential kisomorphism commutes with d/dt and preserves every element of k).4It is a standard fact in classic commutative algebra and algebraic geometry (c.f. (Hartshorne 1977)) that one needsprime ideals for interpreting concrete equations in the language of eld theory. In our differential setting, we of courseneed differential prime ideals (see (Kolchin 1973) and also (Fliess and Glad 1993) for an elementary exposition). Theverication of the prime character of the differential ideals corresponding to all our examples is done in appendix A.6 diff tr d0k < u > /k = m. A dynamics with (independent) input u is a nitely generated differentiallyalgebraic extension D/k < u >. Note that the number m of independent input channels is equal tothe differential order of the corresponding system D/k. An output y = (y1, . . . , yp) is a nite set ofdifferential quantities in D.According to theorem 1, there exists a nite transcendence basis x = (x1, . . . , xn) ofD/k < u >. Consequently, any component of x = ( x1, . . . , xn) and of y is k < u >algebraicallydependent on x, which plays the role of a (generalized) state. This yields:___A1( x1, x, u, u, . . . , u(1)) = 0...An( xn, x, u, u, . . . , u(n)) = 0B1(y1, x, u, u, . . . , u(1)) = 0...Bp(yp, x, u, u, . . . , u(p)) = 0(8)where the Ais and Bjs are polynomial over k. The integer n is the dimension of the dynamicsD/k < u >. We refer to (Fliess and Hasler 1990, Fliess et al. 1993a) for a discussion of suchgeneralized statevariable representations (8) and their relevance to practice.Example (continued) Set u1 = x3 and u2 = x4. The extension D/R < u > is differentiallyalgebraic and yields the representation___ x1 = u1u2 x2 = (x1 + u1x4)x4 x4 = u2.(9)The dimension of the dynamics is 3 and (x1, x2, x4) is a generalized state. It would be 5 if we setu1 = x3 and u2 = x4, and the corresponding representation becomes causal in the classical sense.Remark 1 Take the dynamics D/k < u > and a nitely generated algebraic extension D/D. Thetwo dynamics D/k < u > and D/k < u >, which are of course equivalent, have the same dimensionand can be given the same state variable representation (11). In the sequel, a system D/k < u > willbe dened up to a nitely generated algebraic extension of D.2.3 Modules and linear systems 5Differential elds are to general for linear systems which are specied by linear differential equations.They are thus replaced by the following appropriate modules.5See also (Fliess 1990b).7 Let k be again a given differential ground eld. Denote by k_ddt_ the ring of linear differentialoperators of the type
niteaddt (a k).This ring is commutative if, and only if, k is a eld of constants. Nevertheless, in the generalnoncommutative case, k_ddt_still is a principal ideal ring and the most important properties of leftk_ddt_modules mimic those of modules over commutative principal ideal rings (see (Cohn 1985)).Let M be a left k_ddt_ module. An element m M is said to be torsion if, and only if, there exists k_ddt_, = 0, such that m = 0. The set of all torsion elements of M is a submodule T, whichis called the torsion submodule of M. The module M is said to be torsion if, and only if, M = T. Thefollowing result can regarded as the linear counterpart of theorem 1.Proposition 1 For a nitely generated left k_ddt_module M, the next two properties are equivalent:(i) M is torsion;(ii) the dimension of M as a kvector space is nite.A nitely generated module M is free if, and only if, its torsion submodule T is trivial, i.e., T = {0}6.Any nitely generated module M can be written M = T where T is the torsion submodule of Mand is a free module. The rank of M, denoted by rk M, is the cardinality of any basis of . Thus,M is torsion if, and only if, rk M = 0.Alinear systemis, by denition, a nitely generated left k_ddt_module . We are thus dealing witha nite number of variables which are related by a nite number of linear homogeneous differentialequations and our setting appears to be strongly related to Willems approach (Willems 1991). Thedifferential order of is the rank of .A linear dynamics with input u = (u1, . . . , um) is a linear system which contains u suchthat the quotient module /[u] is torsion, where [u] denotes the left k_ddt_module spanned by thecomponents of u. The input is assumed to be independent, i.e., the module [u] is free. This impliesthat the differential order of is equal to m. Aclassical Kalman state variable representation is alwayspossible:ddt___x1...xn___= A___x1...xn___+ B___u1...um___(10)where the dimension n of the state x = (x1, . . . , xn), which is called the dimension of the dynamics, isequal to the dimension of the torsion module /[u] as a kvector space.6This is not the usual denition of free modules, but a characterization which holds for nitely generated modules overprincipal ideal rings, where any torsionfree module is free (see (Cohn 1985)).8 the matrices A and B, of appropriate sizes, have their entries in k.An output y = (y1, . . . , yp) is a set of elements in . It leads to the following output map:___y1...yp___= C___x1...xn___+
niteDddt___u1...um___.The controllability of (10) can be expressed in a moduletheoretical language which is independentof any denomination of variables. Controllability is equivalent to the freeness of the module . Thisjust is an algebraic counterpart (Fliess 1992) of Willems trajectory characterization (Willems 1991).When the systemis uncontrollable, the torsion submodule corresponds to the Kalman uncontrollabilitysubspace.Remark 2 The relationship with the general differential eld setting is obtained by producing a formalmultiplication. The symmetric tensor product (Jacobson 1985) of a linear system, where is viewedas a kvector space, is an integral differential ring. Its quotient eld D, which is a differential eld,corresponds to the nonlinear eld theoretic description of linear systems.2.4 Differentials and tangent linear systemsDifferential calculus, which plays such a role in analysis and in differential geometry, admits a niceanalogue in commutative algebra (Kolchin 1973, Winter 1974), which has been extended to differentialalgebra by Johnson (Johnson 1969).To a nitely generated differential extension L/K, associate a mapping dL/K : L L/K, called(K ahler) differential 7and where L/K is a nitely generated left L_ddt_module, such thata L dL/K_dadt_= ddt_dL/Ka_a, b L dL/K(a +b) = dL/Ka +dL/KbdL/K(ab) = bdL/Ka +adL/Kbc K dL/Kc = 0.Elements of K behave like constants with respect to dL/K. Properties of the extension L/K can betranslated into the linear moduletheoretic framework of L/K: A set = (1, . . . , m) is a differential transcendence basis of L/K if, and only if, dL/K =(dL/K1, . . . , dL/Km) is a maximal set of L_ddt_linearly independent elements in L/K. Thus,diff tr d0L/K = rk L/K.7For any a L, dL/Ka should be intuitively understood, like in analysis and differential geometry, as a small variationof a.9 The extension L/K is differentially algebraic if, and only if, the module L/K is torsion. A setx = (x1, . . . , xn) is a transcendence basis of L/K if, and only if, dL/Kx = (dL/Kx1, . . . , dL/Kxn)is a basis of L/K as Lvector space. The extension L/K is algebraic if, and only if, L/K is trivial, i.e., L/K = {0}.The tangent (or variational) linear system associated to the system D/k is the left D_ddt_module
D/k. To a dynamics D/k < u > is associated the tangent (or variational) dynamics D/k with thetangent (or variational) input dL/Ku = (dL/Ku1, . . . , dL/Kum). The tangent (or variational) outputassociated to y = (y1, . . . , yp) is dL/Ky = (dL/Ky1, . . . , dL/Kyp).3 Equivalence, atness and defect3.1 Equivalence of systems and endogenous feedbackTwo systems D/k and D/k are said to be equivalent or equivalent by endogenous feedback if, andonly if, any element of D (resp. D) is algebraic over D (resp. D)8. Two dynamics, D/k < u > andD/k < u >, are said to be equivalent if, and only if, the corresponding systems, D/k and D/k, areso.Proposition 2 Two equivalent systems (resp. dynamics) possess the same differential order, i.e., thesame number of independent input channels.Proof Denote by K the differential eld generated by D and D: K/D and K/ D are algebraicextensions. Therefore,diff tr d0D/k = diff tr d0K/k = diff tr d0 D/k.Consider two equivalent dynamics, D/k < u >and D/k < u >. Let n (resp. n) be the dimensionof D/k < u > (resp. D/k < u >). In general, n = n. WriteAi( xi, x, u, u, . . . , u(i)) = 0, i = 1, . . . , n (11)andAi( xi, x, u, u, . . . , u( i)) = 0, i = 1, . . . , n (12)the generalized state variable representations of D/k < u > and D/k < u >, respectively. Thealgebraicity of any element of D (resp. D) over D (resp. D) yields the following relationships8According to footnote 3, this denition of equivalence can also be read as follows: two systems D/k and D/k areequivalent if, and only if, there exist two differential extensions D/D and D/D which are algebraic (in the usual sense),and a differential kautomorphism between D/k and D/k.10 between (11) and (12):___i(ui, x, u, u, . . . , u(i)) = 0 i = 1, . . . , m(x, x, u, u, . . . , u()) = 0 = 1, . . . , n i( ui, x, u, u, . . . , u( i)) = 0 i = 1, . . . , m ( x, x, u, u, . . . , u( )) = 0 = 1, . . . , n(13)where the is, s, is and s are polynomials over k.The two dynamic feedbacks corresponding to (13) are called endogenous as they do not necessitatethe introduction of any variable that is transcendental over D and D (see also (Martin 1992)). If weknow x (resp. x), we can calculate u (resp. u) from u (resp. u) without integrating any differentialequation. The relationship with general dynamic feedbacks is given in appendix B.Remark 3 The tangent linear systems (see subsection 2.4) of two equivalent systems are stronglyrelated and, in fact, are almost identical. Take two equivalent systems D1/k and D2/k and denote byDthe smallest algebraic extension of D1 and D2. It is straightforward to check that the three left D_ddt_modules D/k, DD1
D1/k and DD2
D2/k are isomorphic (see (Hartshorne 1977, Jacobson 1985)).3.2 Flatness and defectLike in the nondifferential case, a differential extension L/K is said to be purely differentially transcendental if, and only if, there exists a differential transcendence basis = {i  i I } of L/K suchthat L = K < >. A system D/k is called purely differentially transcendental if, and only if, theextension D/k is so.A system D/k is called (differentially) at if, and only if, it is equivalent to a purely differentiallytranscendental system L/k. A differential transcendence basis y = (y1, . . . , ym) of L/k such thatL = k < y > is called a linearizing or at output of the system D/k.Example (continued) Let us prove that y = (y1, y2) withy1 = x2 + (x1 + x3x4)22x(3)3, y2 = x3.is a linearizing output for (7). Set = x1+ x3x4. Differentiating y1 = x2+2/2y(3)2 , we have, using(7), 2= 2 y1(y(3)2 )2y(4)2. Thus x2 = y1 22y(3)2is an algebraic function of (y1, y1, y(3)2 , y(4)2 ). Sincex4 = x2and x1 = y2x4, x4 and x1 are algebraic functions of (y1, y1, y1, y2, y(3)2 , y(4)2 , y(5)2 ).Remark there exist many other linearizing outputs such as y = ( y1, y2) = (2y1y(3)2 , y2), the inversetransformation being y = ( y1/2 y(3)2 , y2).11 Take an arbitrary system D/k of differential order m. Among all the possible choices of setsz = (z1, . . . , zm) of m differential kindeterminates which are algebraic over D, take one such thattr d0D < z > /k < z > is minimum, say . This integer is called the defect of the systemD/k. Thenext result is obvious.Proposition 3 A system D/k is at if, and only if, its defect is zero.Example The defect of the system generated by x1 and x2 satisfying x1 = x1 + ( x2)3is one. Itsgeneral solution cannot be expressed without the integration of, at least, one differential equation.3.3 Basic examples3.3.1 Uncontrolled dynamical systemsAn uncontrolled dynamical system is, in our eldtheoretic language (Fliess 1990a), a nitely generated differentially algebraic extension D/k: diff tr d0D/k = 0 implies the nonexistence of anydifferential kindeterminate algebraic over D. Thus, the defect of D/k is equal to tr d0D/k, i.e., tothe dimension of the dynamical system D/k, which corresponds to the state variable representationAi( xi, x) = 0, where x = (x1, . . . , xn) is a transcendence basis of D/k. Flatness means that D/k isalgebraic in the (nondifferential) sense: the dynamics D/k is then said to be trivial.3.3.2 Linear systemsThe defect of is, by denition, the defect of its associated differential eld extension D/k (seeremark 2).Theorem 2 The defect of a linear system is equal to the dimension of its torsion submodule, i.e.,to the dimension of its Kalman uncontrollable subspace. A linear system is at if, and only if, it iscontrollable.Proof Take the decomposition = T , of section 2.3, where T is the torsion submodule and a free module. A basis b = (b1, . . . , bm) of plays the role of a linearizing output when isfree: the system then is at. When T = {0}, the differential eld extension T /k generated by T isdifferentially algebraic and its (nondifferential) transcendence degree is equal to the dimension of Tas kvector space. The conclusion follows at once.Remark 4 The above arguments can be made more concrete by considering a linear dynamics overR. If it is controllable, we may write it, up to a static feedback, in its Brunovsky canonical form:y(i)= ui, (i = 1, . . . , m)12 where the is are the controllability indices and y = (y1, . . . , ym) is a linearizing output. In theuncontrollable case, the defect d is the dimension of the uncontrollable subspace:ddt___1...d___= M___1...d___where M is a d d matrix over R.3.4 A necessary condition for atnessConsider the system D/k where D = k < w > is generated by a nite set w = (w1, . . . , wq). Thewis are related by a nite set, (w, w, . . . , w()) = 0, of algebraic differential equations. Dene thealgebraic variety S corresponding to (0, . . . , ) = 0 in the ( +1)qdimensional afne space withcoordinates j= ( j1, . . . , jq), j = 0, 1, . . . , .Theorem 3 If the system D/k is at, the afne algebraic variety S contains at each regular point astraight line parallel to the axes.Proof The components of w, w, . . ., w(1)are algebraically dependent on the components of alinearizing output y = (y1, . . . , ym) and a nite number of their derivatives. Let be the highest orderof these derivatives. The components of w()depend linearly on the components of y(+1), which playthe role of independent parameters for the coordinates 1, . . ., q.The above condition is not sufcient. Consider the systemD/Rgenerated by (x1, x2, x3) satisfying x1 = ( x2)2+ ( x3)3. This system does not satisfy the necessary condition: it is not at. The samesystem D can be dened via the quantities (x1, x2, x3, x4) related by x1 = (x4)2+( x3)3and x4 = x2.Those new equations now satisfy our necessary criterion.3.5 Flatness and controllabilitySussmann and Jurdjevic (Sussmann and Jurdjevic 1972) have introduced in the differential geometricsetting the concept of strong accessibility for dynamics of the form x = f (x, u). Sontag (Sontag 1988)showed that strong accessibility implies the existence of controls such that the linearized systemarounda trajectory passing through a point a of the statespace is controllable. Coron (Coron 1994) and Sontag(Sontag 1992) demonstrated that, for any a, those controls are generic.The above considerations with those of section 2.3 and 2.4 lead in our context to the followingdenition of controllability, which is independent of any distinction between variables: a systemD/kis said to be controllable (or strongly accessible) if, and only if, its tangent linear systemis controllable,i.e., if, and only if, the module D/k is free.Remark 3 shows that this denition is invariant under our equivalence via endogenous feedback.Proposition 4 A at system is controllable13 Proof It sufces to prove it for a purely differentially transcendental extensions k < y > /k, wherey = (y1, . . . , ym). The module k/k, which is spanned by dk/ky1, . . ., dk/kym, is necessarilyfree.The converse is false as demonstrated by numerous examples of strongly accessible singleinputdynamics x = f (x, u) which are not linearizable by static feedback and therefore neither by dynamicones (Charlet et al. 1989).Flatness which is equivalent to the possibility of expressing any element of the system as a function of the linearizing output and a nite number of its derivatives, may be viewed as the nonlinearextension of linear controllability, if the latter is characterized by free modules. Whereas the strongaccessibility property only is an innitesimal generalization of linear controllability, atness shouldbe viewed as a more global and, perhaps, as a more tractable one. This will be enhanced in section5 where controllable systems of nonzero defect are treated using highfrequency control that enablesto approximate them by at systems for which the control design is straightforward.4 Examples and control of at systemsThe verication of the prime character of the differential ideals corresponding to all our examplesis done in appendix A. This means that the equations dening all our examples can be rigorouslyinterpreted in the language of differential eld theory.4.1 The 2D craneDRxzXZmgOFigure 1: The two dimensional crane.14 Consider the crane displayed on gure 1 which is a classical object of control study (see, e.g.,(DAndr eaNovel and L evine 1990), (Marttinen et al. 1990)). The dynamics can be divided into twoparts. The rst part corresponds to the motor drives and industrial controllers for trolley travels androlling up and down the rope. The second part is relative to the trolley load, the behavior of which isvery similar to the pendulum one. We concentrate here on the pendulum dynamics by assuming that the traversing and hoisting are control variables, the trolley load remains in a xed vertical plane OXZ, the rope dynamics are negligible.A dynamic model of the load can be derived by Lagrangian formalism. It can also be obtained,in a very simple way, by writing down all the differential (Newton law) and algebraic (geometricconstraints) equations describing the pendulum behavior:___m x = T sin m z = T cos +mgx = R sin + Dz = R cos (14)where (x, z) (the coordinates of the load m), T (the tension of the rope) and (the angle between therope and the vertical axis OZ) are the unknown variables; D (the trolley position) and R (the rope length) are the input variables.From (14), it is clear that sin , T, D and R are algebraic functions of (x, z) and their derivatives:sin = x DR , T = mR(g z)z , ( z g)(x D) = xz, (x D)2+ z2= R2that is___D = x xz z gR2= z2+_ xz z g_2.(15)Thus, system (14) is at with (x, z) as linearizing output.Remark 5 Assume that the modeling equations (14) are completed with the following traversing andhoisting dynamics:___M D = F D + T sin J2R = C R T (16)15 where the new variables F and C are, respectively the external force applied to the trolley and thehoisting torque. The other quantities (M, J, , , ) are constant physical parameters. Then (14,16)is also at with the same linearizing output (x, z). This explains without any additional computationwhy the systemconsidered in (DAndr eaNovel and L evine 1990) is linearizable via dynamic feedback.Let us now address the following question which is one of the basic control problems for a crane:how can one carry a load m from the steadystate R = R1 > 0 and D = D1 at time t1, to thesteadystate R = R2 > 0 and D = D2 at time t2 > t1 ?It is clear that any motion of the load induces oscillations that must be canceled at the end of theload transport. We propose here a very simple answer to this question when the crane can be describedby (14). This answer just consists in using (15).Consider a smooth curve [t1, t2] t ((t ), (t )) R]0, +[ such that for i = 1, 2, ((ti), (ti)) = (Di, Ri), and drdtr (, )(ti) = 0 with r = 1, 2, 3, 4. for all t [t1, t2], (t ) < g.Then the solution of (14) starting at time t1 from the steadystate D1 and R1, and with the controltrajectory dened, for t [t1, t2], by___D(t ) = (t ) (t ) (t ) (t ) gR(t ) =_2(t ) +_ (t ) (t ) (t ) g_2 (17)and, for t > t2, by (D(t ), R(t )) = (D2, R2), leads to a load trajectory t (x(t ), z(t )) such that(x(t ), z(t )) = ((t ), (t )) for t [t1, t2] and (x(t ), z(t )) = (D2, R2) for t t2. Notice that, since forall t [t1, t2], z(t ) < g, the rope tension T = mR(g z)z remains always positive and the descriptionof the system by (14) remains reasonable.This results from the following facts. The generalized state variable description of the system isthe following (Fliess et al. 1991, Fliess et al. 1993a):R = 2 R D cos g sin .Since and are smooth, D and R are at least twice continuously differentiable. Thus, the classicalexistence and uniqueness theoremensures that the above ordinary differential equation admits a uniquesmooth solution that is nothing but (t ) = arctan((t ) D(t ))/ (t )).The approximation of the crane dynamics by (14) implies that the motor drives and industrial lowlevel controllers (trolley travels and rolling up and down the rope) produce fast and stable dynamics (seeremark 5). Thus, if these dynamics are stable and fast enough, classical results of singular perturbationtheory of ordinary differential equation (see, e.g., (Tikhonov et al. 1980)), imply that the control (17)leads to a nal conguration close to the steadystate dened by D2 and R2.16 In the simulations displayed here below, we have veried that the addition of reasonable fast andstable regulator dynamics modies only slightly the nal position (R2, D2). Classical proportionalintegral controller for D and R are added to (14). The typical regulator time constants are equal toone tenth of the period of small oscillations ( 110 2_R/g 0.3 s) (see (Fliess et al. 1991)).10987650 10 20x (m)z (m)load trajectory051015200 5 10 15time (s)(m)trolley position56789100 5 10 15time (s)(m)rope length0.40.200.20.40 5 10 15time (s)(rd)vertical deviation angleFigure 2: Simulation of the control dened by (17) without (solid lines) and with (dot lines) ideallowlevel controllers for D and R.For the simulations presented in gure 2, the transport of the load m may be considered as a ratherfast one: the horizontal motion of D is of 10 m in 3.5 s; the vertical motion of R is up to 5 m in 3.5 s.Compared with the lowlevel regulator time constants (0.1 and 0.3 s), such motions are not negligible.This explains the transient mismatch between the ideal and nonideal cases. Nevertheless, the nalcontrol performances are not seriously altered: the residual oscillations of the load after 7 s admit lessthan 3 cm of horizontal amplitude. Such small residual oscillations can be canceled via a simple PID17 regulator with the vertical deviation as input and the setpoint of D as output.The simulations illustrate the importance of the linearizing output (x, z). When the regulationsof R and D are suitably designed, it is possible to use the control given in (17) for fast transports ofthe load m from one point to another. The simplicity and the independence of (17) with respect to thesystem parameters (except g) constitute its main practical interests.Remark 6 Similar calculations can be performed when a second horizontal direction OX2, orthogonal to OX1 = OX, is considered. Denoting then by (x1, x2, z) the cartesian coordinates of the load, Rthe rope length and (D1, D2) the trolley horizontal position, the system is described by___( z g)(x1 D1) = x1z( z g)(x2 D2) = x2z(x1 D1)2+(x2 D2)2+ z2= R2.This system is clearly at with the cartesian coordinates of the load, (x1, x2, z), as at output.Remark 7 In (DAndr eaNovel et al. 1992b), the control of a body of mass m around a rotationaxle of constant direction is investigated. This system is at as a consequence of the followingconsiderations. According to an old result due to Huygens (see, e.g. (Whittaker 1937, p. 131132)),the equations describing the motion are equivalent to those of a pendulum of the same mass m and oflength l = Jmd where d = 0 is the vertical distance between the mass center G and the axle , J isthe inertial moment around . Denoting by u and v, respectively, the vertical and horizontal positionsof , the equations of motion are the following (compare to (15)):___ uu x = v gv z(u x)2+(v z)2= l2where (x, z) are the horizontal and vertical coordinates of the Huygens oscillation center. Clearly(x, z) is a linearizing output.Remark 8 The examples corresponding to the crane, Huygens oscillation center (see remark 7) andthe car with ntrailers here below, illustrate the fact that linearizing outputs admit most often a clearphysical interpretation.4.2 The car with ntrailers4.2.1 Modeling equationsSteeringa car withn trailers is nowthe object of active researches (Laumond 1991, Murray and Sastry 1993,Monaco and NormandCyrot 1992, Rouchon et al. 1993a, Tilbury et al. 1993). The atness of a basicmodel9of this system combined with the use of Frnet formula lead to a complete and simple solution9More realistic models where trailer i is not directly hitched to the center of the axle of trailer i 1 are considered in(Martin and Rouchon 1993, Rouchon et al. 1993b).18 xnynPn dnnPn1n1P1d11P0d0 Q0Figure 3: The kinematic car with n trailers.of the motion planning problem without obstacles. Notice that most of nonholonomic mobile robotsare at (DAndr eaNovel et al. 1992a, Campion et al. 1992).The hitch of trailer i is attached to the center of the rear axle of trailer i 1. The wheels are alignedwith the body of the trailer. The two control inputs are the driving velocity (of the rear wheels of thecar) and the steering velocity (of the front wheels of the car). The constraints are based on allowingthe wheels to roll and spin without slipping. For the steering front wheels of the car, the derivation issimplied by assuming them as a single wheel at the midpoint of the axle. The resulting dynamicsare described by the following equations (the notations are those of (Murray and Sastry 1993) andsummarized on gure 3):___ x0 = u1 cos 0 y0 = u1 sin 0 = u20 = u1d0tan i = u1di_i 1
j =1cos(j 1 j)_ sin(i 1 i) for i = 1, . . . , n(18)where (x0, y0, , 0, . . . , n) R2] /2, +/2[(S1)n+1is the state, (u1, u2) is the control andd0, d1, . . ., dn are positive parameters (lengths). As displayed on gure 3, we denote by Pi, the mediumpoint of the wheel axle of trailer i , for i = 1, . . . , n. The medium point of the rear (resp. front) wheelaxle of the car is denoted by P0 (resp. Q).19 4.2.2 Cartesian coordinates of Pn as at outputDenote by (xi, yi) the cartesian coordinates of Pi, i = 0, 1, . . . , n:xi = x0 i
j =1dj cos jyi = y0 i
j =1dj sin j.A direct computation shows that tan i = yi xi. Since, for i = 0, . . . , n 1, xi = xi +1 + di +1 cos i +1and yi = yi +1+di +1 sin i +1, the variables n, xn1, yn1, n1, . . ., 1, x0, y0 and 0 are functions of xnand yn and their derivatives up to the order n +1. But u1 = x0/ cos 0, tan = d00/u1 and u2 = .Thus, the entire state and the control are functions of xn and yn and their derivatives up to order n +3.This proves that the car with n trailers described by (18) is a at system: the linearizing outputcorresponds to the cartesian coordinates of the point Pn, the medium point of the wheel axle of thelast trailer.Flatness implies that for generic values of the state, the strong accessibility rank associated to thecontrol system(18) is maximumand equal to its statespace dimension: the systemis thus controllable.The singularity which might occur when dividing by xi = 0 in tan i = yi/ xi, can be avoided bythe following developments.4.2.3 Motion planing using atnessIn (Rouchon et al. 1993a, Rouchon et al. 1993b), the following result was sketched.Proposition 5 Consider (18) and two different statespace congurations: p = ( x0, y0, , 0, . . . , n)and p = (x0, y0, , 0, . . . , n). Assume that the angles i 1 i, i = 1, . . . , n, , i 1 i,i = 1, . . . , n, and belong to ] /2, /2[. Then, there exists a smooth openloop control [0, T] t (u1(t ), u2(t )) steering the systemfrom p at time 0 to p at time T > 0, such that the angles i 1i,i = 1, . . . , n, and (i = 1, . . . , n) always remain in ] /2, /2[ and such that (u1(t ), u2(t )) = 0for t = 0, T.The conditions i 1i ] /2, /2[ (i = 1, . . . , n) and ] /2, /2[ are meant for avoidingsome undesirable geometric congurations: trailer i should not be in front of trailer i 1.The detailed proof is given in the appendix and relies basically on the fact that the system is at. Itis constructive and gives explicitly (u1(t ), u2(t )). The involved computations are greatly simplied bya simple geometric interpretation of the rolling without slipping conditions and the use of the Frnetformula. Here, we just recall this geometric construction and give the explicit formula for parking acar with two trailers. The Frnet formula are recalled in the appendix.Denote by Ci the curve followed by Pi, i = 0, . . . , n. As displayed on gure 4, the point Pi 1belongs to the tangent to Ci at Pi and at the xed distance di from Pi:Pi 1 = Pi +dii20 Ci 1CiPiiPi 1i 1i 1Pi 1iPiFigure 4: The geometric interpretation of the rolling without slipping conditions.with i the unitary tangent vector to Ci. Differentiating this relation with respect to si, the arc lengthof Ci, leads toddsiPi 1 = i +diiiwhere i is the unitary vector orthogonal to i and i is the curvature of Ci. Since ddsiPi 1 gives thetangent direction to Ci 1, we havetan(i 1 i) = dii.4.2.4 Parking simulations of the 2trailer systemWe now restrict to the particular case n = 2. We show how the previous analysis can be employed tosolve the parking problem. The simulations of gures 5 and 6 have been written in MATLAB. Theycan be obtained upon request from the fourth author via electronic mail ([email protected]).The car and its trailers are initially in A with angles 2 = 1 = 0 = /6, = 0. The objectiveis to steer the system to C with nal angles (2, 1, 0, ) = 0. We consider the two smooth curvesCAB and CCB of the gure 5, dened by their natural parameterizations [0, LAB] s PAB(s) and[0, LCB] s PCB(s), respectively (PAB(0) = A, PCB(0) = C, LAB is the length of CAB and LCB thelength of CCB). Their curvatures are denoted by AB(s) and CB(s). These curves shall be followed byP2. The initial and nal systemconguration in A and C impose AB(0) = ddsAB(0) = d2ds2AB(0) = 0and CB(0) = ddsCB(0) = d2ds2CB(0) = 0. We impose additionally that AB and CB are tangent at B21 CABCCBABCbeginendoooFigure 5: parking the car with two trailers from A to B via C.andAB(LAB) = ddsAB(LAB) = d2ds2AB(LAB) = CB(LCB) = ddsCB(LCB) = d2ds2CB(LCB) = 0.It is straightforward to nd curves satisfying such conditions. For the simulation of gure 6, we takepolynomial curves of degree 9.Proposition 5 implies that, if P2 follows CAB and CCB as displayed on gure 5, then the initial andnal states will be as desired. Take a smooth function [0, T] t s(t ) [0, LAB] such that s(0) = 0,s(T) = LABand s(0) = s(T) = 0. This leads to smooth control trajectories [0, T] t u1(t ) 0and [0, T] t u2(t ) steering the system from A at time t = 0 to B at time t = T. Similarly,[T, 2T] t s(t ) [0, LCB] such that s(T) = LCB, s(2T) = 0 and s(T) = s(2T) = 0 leads tocontrol trajectories [T, 2T] t u1(t ) 0 and [T, 2T] t u2(t ) steering the system from Bto C. This gives the motions displayed on gure 6 with forwards motions from A to B, backwardsmotions from B to C and a stop in B.Let us detail the calculation of the control trajectories for the motion from A to B. Similarcalculations can be done for the motion from B to C. The curve CAB corresponds to the curve Ci ofgure 4 with i = 2. Assume that CAB is given via the regular parameterization, y = f (x) ((x, y) arethe cartesian coordinates and f is a polynomial of degree 9). Denote by si the arc length of curve Ci,i = 0, 1, 2. Then ds2 =_1 +(d f /dx)2dx and the curvature of C2 is given by2 = d2f /dx2(1 +(d f /dx)2)3/2.22 ABCbeginendoooo o o o o o o ooooooooo o o o o oo o o o ooooooooooooo o o o oFigure 6: the successive motions of the car with two trailers.We have1 = 1_1 +d2222_2 + d21 +d2222d2ds2_and ds1 =_1 +d2222 ds2. Similarly,0 = 1_1 +d2121_1 + d11 +d2121d1ds1_and ds0 =_1 +d2121 ds1. Thus u1 is given explicitly byu1 = ds0dt =_1 +d2121_1 +d2222_1 +(d f /dx)2 x(t )where [0, T] t x(t ) is anyincreasingsmoothtime function. (x(0), f (x(0))) (resp. (x(T), f (x(T))))are the coordinates of A (resp. B) and x(0) = x(T) = 0. Since tan() = d00, we getu2 = ddt = d01 +d2020d0ds0u1.Here, we are not actually concerned with obstacles. The fact that the internal conguration dependsonly on the curvature results from the general following property: a plane curve is entirely dened (upto rotation and translation) by its curvature. For the ntrailer case, the angles nn1, . . ., 10 and describing the relative conguration of the system are only functions of n and its rst nderivativeswith respect to sn.23 Consequently, limitations due to obstacles can be expressed up to a translation (dened by Pn)and a rotation (dened by the tangent direction d Pndsn) via n and its rst nderivatives with respect tosn. Such considerations can be of some help in nding a curve avoiding collisions. More details onobstacle avoidance can be found in (Laumond et al. 1993) where a car without trailer is considered.The multisteering trailer systems considered in (Bushnell et al. 1993), (Tilbury et al. 1993),(Tilbury and Chelouah 1993) are also at: the at output is then obtained by adding to the Cartesiancoordinates of the last trailer, the angles of the trailers that are directly steered. This generalization isquite natural in view of the geometric construction of gure 4.5 Highfrequency control of nonat systemsWe address here a method for controlling nonat systems via their approximations by averagedand at ones. More precisely, we develop on three examples an idea due to the Russian physicistKapitsa (Bogaevski and Povzner 1991, Landau and Lifshitz 1982, Sagdeev et al. 1988). He considersthe motion of a particle in a highly oscillating eld and proposes a method for deriving the equationsof the averaged motion and potential. He shows that the inverted position of a single pendulum isstabilized when the suspension point oscillates rapidly. Notice that some related calculations maybe found in (Baillieul 1993). For the use of highfrequency control in different contexts see also(Bentsman 1987, Meerkov 1980, Sussmann and Liu 1991).(Acheson 1993, Stephenson 1908)5.1 The Kapitsa pendulumz lgmFigure 7: The Kapitsa pendulum: the suspension point oscillates rapidly on a vertical axis.24 The notation are summarized on gure 7. We assume that the vertical velocity z = u of the suspensionpoint is the control. The equations of motion are:___ = p + ul sin p =_gl u2l2 cos _sin ul p cos z = u(19)where p is proportional to the generalized impulsion; g and l are physical constants. This system is not at since it admits only one control variable and is not linearizable via static feedback(Charlet et al. 1989). However it is strongly accessible.We stateu = u1 +u2 cos(t /)where u1 and u2 are auxiliary control and 0 < l/g. It is then natural to consider the followingaveraged control system:___ = p + u1l sin p =_gl (u1)2l2 cos (u2)22l2 cos _sin u1l p cos z = u1.(20)It admits two control variables, u1 and u2, whereas the original system (19) admits only one, u.Moreover (20) is at with (, z) as linearizing output.The endogenous dynamic feedback___ = v1u1 = u2 =_ 2lcos (g +v1) 2l2cos sin v2(21)transforms (20) into_ z = v1 = v2. (22)Set___v1 = _11+ 12_ 112(z zsp)v2 = _11+ 12__p + l sin _ 112( sp)(23)where the parameters 1, 2 > 0 and sp] /2, /2[/{0}. Then, the closedloop averaged system(20,21,23) admits an hyperbolic equilibrium point characterized by (zsp, sp) that is asymptoticallystable.25 Consider now (19) and the highfrequency control u = u1 +u2 sin(t /) with 0 < l/g and(u1, u2) given by (21,23) where , p and z are replaced by , p and z. Then, the corresponding averagedsystemis nothing but (22) with v1 and v2 given by (23). Since the averaged systemadmits a hyperbolicasymptotically stable equilibrium, the perturbed system admits an hyperbolic asymptotically stablelimit cycle around (, p, z) = (sp, 0, zsp) (Guckenheimer and Holmes 1983, theorem 4.1.1, page168): such control maintains (z, ) near (zsp, sp). Moreover this control method is robust in thefollowing sense: the existence and the stability of the limit cycle is not destroyed by small static errorsin the parameters l and g and in the measurements of , p, z and u.As illustrated by the simulations of gure 8, the generalization to trajectory tracking for and zis straightforward. These simulations give also a rough estimate of the errors that can be tolerated.The system parameter values are l = 0.10 m and g = 9.81 ms2. The design control parameters are = 0.025/2 s and 1 = 2 = 0.10 s. For the two upper graphics of gure 8, no error is introduced:control is computed with l = 0.10 m and g = 9.81 ms2. For the two lower graphics of gure 8,parameter errors are introduced: control is computed with with l = 0.11 m and g = 9.00 ms2.26 0.40.60.811.20 1 2time (s)(rd)no parameter error00.510 1 2time (s)v (m)no parameter error0.40.60.811.20 1 2time (s)(rd)parameter error00.510 1 2time (s)v (m)parameter errorzzFigure 8: Robustness test of the highfrequency control for the inverted pendulum.27 5.2 The variablelength pendulumgravityuqOFigure 9: pendulum with variablelength.Let us consider the variablelength pendulum of (Bressan and Rampazzo 1993). The notations aresummarized on gure 9. We assume as in (Bressan and Rampazzo 1993) that the velocity u = v isthe control. The equations of motion are:___ q = p p = cos u +qv2 u = v(24)where mass and gravity are normalized to 1.This system is not at since it admits only one control variable and is not linearizable via staticfeedback (Charlet et al. 1989). It is, however, strongly accessible.As for the Kapitsa pendulum, we setv = v1 +v2 cos(t /)where v1 and v2 are auxiliary controls, 0 < 1 .We consider the averaged control system:___q = pp = cos u +q(v1)2+q(v2)2/2u = v1.(25)This system is obviously linearizable via static feedback with (q, u) as linearizing output.The static feedback___v1 = w1v2 =_2_w2 +cos uq (w1)2_ (26)28 transforms (25) into_ u = w1q = w2. (27)Set___w1 = u usp1w2 = _11+ 12_ p 112(q qsp)(28)with 1, 2 > 0, usp] /2, /2[, qsp> 0 . The closedloop averaged system (25,26,28) admits anhyperbolic equilibrium point (usp, qsp), which is asymptotically stable.Similarly to the Kapitsa pendulum, the control law is as follows: v = v1+v2 sin(t /), 0 < 1;(v1, v2) is given by (26,28) where q, p and u are replaced by q, p and u. This control strategy leads toa small and attractive limit cycle. As illustrated by the simulations of gure 10, the size of these limitcycle is an increasing function of and tends to 0 as tends to 0+. The design control parameters are1 = 0.5, 2 = 0.4.29 11.21.41.61.820 2 4 6timeq= 0.020.500.511.50 2 4 6timeu= 0.020.511.520 2 4 6timeq= 0.040.500.511.50 2 4 6timeu= 0.04 Figure 10: highfrequency control for the variablelength pendulum.30 5.3 The inverted double pendulum12guvxzObeam 1beam 2Figure 11: The inverted double pendulum: the horizontal velocity u and vertical velocity v of thesuspension point are the two control variables.The double inverted pendulum of gure 11 moves in a vertical plane. Assume that u (resp. v) thehorizontal (resp. vertical) velocity of the suspension point (x, z) is a control variable. The equationsof motion are (implicit form):___p1 = I1 1 + I 2 cos(1 2) +n1 x cos 1 n1 z sin 1p2 = I 1 cos(1 2) + I2 2 +n2 x cos 2 n2 z sin 2 p1 = n1g sin 1 n1 1 x sin 1 n1 1 z cos 1 p2 = n2g sin 2 n2 2 x sin 2 n2 2 z cos 2 x = u z = v(29)where p1 and p2 are the generalized impulsions associated to the generalized coordinates 1and 2,respectively. The quantities g, I , I1, I2, n1 and n2 are constant physical parameters:I1 =_m13 +m2_ (l1)2, I2 = m23 (l2)2, I = m22 l1l2, n1 =_m12 +m2_ l1, n2 = m22 l2,where m1 and m2 (resp. l1 and l2) are the masses (resp. lengths) of beams 1 and 2 which are assumedto be homogeneous.Proposition 6 System (29) with the two control variables u and v, is not at.31 Proof The proof is just an application of the necessary atness condition of theorem 3. Since u = xand v = z, (29) is at if, and only if, the reduced system,___p1 = I1 1 + I 2 cos(1 2) +n1 x cos 1 n1 z sin 1p2 = I 1 cos(1 2) + I2 2 +n2 x cos 2 n2 z sin 2 p1 = n1g sin 1 n1 1 x sin 1 n1 1 z cos 1 p2 = n2g sin 2 n2 2 x sin 2 n2 2 z cos 2(30)is at. Denote symbolically by F(, ) = 0 the equations (30) where = (1, 2, x, z, p1, p2).Consider (, ) such that F(, ) = 0. We are looking for a vector a = (a1, a2, ax, az, ap1, ap2) suchthat, for all R, F(, +a) = 0. The second order conditions, d2d2=0F(, +a) = 0, leadtoa1(ax sin 1 +az cos 1) = 0, a2(ax sin 2 +az cos 2) = 0.Two rst order conditions, dd=0F(, +a) = 0, are___ax cos 1 +az sin 1 = I1n1a1 + In1cos(1 2)a2ax cos 2 +az sin 2 = In2cos(1 2)a1 + I2n2a2Simple computations show that, if In1= I2n2and I1n1= In2(these conditions are always satised forhomogeneous identical beams), then (a1, a2, ax, az) = 0. The two remaining rst order conditionsimply that (ap1, ap2) = 0. Thus a = 0 and the inverted double pendulum is not at.The same control method as the one explained in details for the Kapitsa pendulum (19) can bealso used for the double pendulum. The only difference relies on the calculations that are here moretedious. We just sketch some simulations (Fliess et al. 1993b).To approximate the nonat system (29) by a at one, we set u = u1 + u2 cos(t /) and v =v1+v2 cos(t /) where 0 < min__ I1n1g,_ I2n2g_and u1, u2, v1, v2 are new control variables. Thisleads to a at averaged system with (1, 2, x, z) as the linearizing output. The endogenous dynamicfeedback that linearized the averaged system provides then (u1, u2, v1, v2). For the simulations ofgure 12, the angles 1 and 2 follow approximately prescribed trajectories whereas, simultaneously,the suspension point (x, z) is maintained approximately constant.32 0.50.450.40.350.30.250.20.150.10 5 10time (s)(rd)vertical deviation of beam 10.20.30.40.50.60.70.80.910 5 10time (s)(rd)vertical deviation of beam 212Figure 12: Simulation of the inverted double pendulum via highfrequency control.33 6 ConclusionOur ve examples, as well as other ones in preparation in various domains of engineering, indicate thatatness and defect ought to be considered as physical and/or geometric properties. This explains whyat systems are so often encountered in spite of the nongenericity of dynamic feedback linearizabilityin some customary mathematical topologies (Tcho n 1994, Rouchon 1994).We hope to have convinced the reader that atness and defect bring a new theoretical and practicalinsight in control. We briey list some important open problems: Ritts work (Ritt 1950) shows that differential algebra provides powerful algorithmic means (see(Diop 1991, Diop 1992) for a survey and connections with control). Can atness and defect bedetermined by this kind of procedures ? great progress have recently been made in nonlinear timevarying feedback stabilization (see,e.g., (Coron 1992, Coron 1994)). Most of the examples which were considered happen to beat (see, e.g., (Coron and DAndr eaNovel 1992)). The utilization of this property is related tothe understanding of the notion of singularity (see, e.g., (Martin 1993) for a rst step in thisdirection and the references therein). the two averaged systems associated to highfrequency control are at. Can this result begeneralized to a large class of devices ? differential algebra is not the only possible language for investigating atness and defect. The extensionof the differential algebraic formalismtosmoothandanalytic functions (Jakubczyk 1992)and the differential geometric approach (Martin 1992, Fliess et al. 1993d, Fliess et al. 1993e,Pomet 1993) should also be examined in this context.ReferencesAcheson, D. 1993. A pendulum theorem. Proc. R. Soc. Lond. A 443, 239245.Baillieul, J. 1993. Stable average motions of mechanical systems subject to periodic forcing. preprint.Bentsman, J. 1987. Vibrational control of a class of nonlinear multiplicative vibrations. IEEE Trans.Automat. Control 32, 711716.Bogaevski, V. and A. Povzner 1991. 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It is then immediate that the differential idealcorresponding to the tutorial example (7) is prime: set x = (x1, x2) and u = (x3, x4). Let us now listour ve casestudies.39 Kapitsa pendulum (19) Let us replace by = tan(/2). Then, using = 1 +22 , cos = 1 21 +2, sin = 21 +2,the equations (19) become explicit and rational___ = 1 +22_p + 2ul(1 +2)_ p =_gl u2(1 2)l2(1 +2)_ 21 +2 pu(1 2)l(1 +2) z = u.The associateddifferential ideal is thus prime andleads toa nitelygenerateddifferential eldextensionover R.Variablelength pendulum (24) Similar computations with = tan(u/2) prove that the associateddifferential ideal is prime.Double pendulum (29) Similar computations with 1 = tan(1/2) and 2 = tan(2/2) prove thatthe associated differential ideal is prime.Car with ntrailers (18) Similar computations with = tan(/2) and i = tan(i/2) prove thatthe associated differential ideal is prime.Crane (17) Analogous calculations on the generalized state variable equation R = 2 R D cos g sin given in (Fliess et al. 1991, Fliess et al. 1993a) lead to a prime differential ideal.Another more direct way for obtaining the differential eld corresponding to the crane is thefollowing. Take (17) and consider the differential eld R < x, z > generated by the two differentialindeterminates x and z. The variable D belongs to R < x, z > and the variable R belongs to anobvious algebraic extension D of R < x, z >, which denes the system.B Dynamic feedbacks versus endogenous feedbacksA dynamic feedback between two systems D/k and D/k consists in a nitely differential extensionE/k such that D E and D E. Assume moreover that the extension E/ D is differentially algebraic.According to theorem 1, the (nondifferential) transcendence degree of E/ D is nite, say . Choose atranscendence basis z = (z1, . . . , z) of E/ D. It yields like (8):A( z, z) = 0 = 1, . . . , B(, z) = 040 where is any element of E and the As and B are polynomials over D.The above formulas are the counterpart in the eld theoretic language of the usual ones for dening general dynamic feedbacks (see, e.g., (Isidori 1989, Nijmeijer and van der Schaft 1990)). Thedynamic feedback is said to be regular if, and only if, E/D and E/ D are both differentially algebraic.The following generalization of proposition 2 is immediate: the systems D/k and D/k possess thesame differential order, i.e., the same number of independent input channels.The situation of endogenous feedbacks is recovered when E/D and E/ D are both algebraic, i.e., = 0.C Proof of proposition 51/P(s)Figure 13: Frnet frame (, ) and curvature of a smooth planar curve.The Frnet formula Let us recall some terminology and relations relative to planar smooth curvesthat are displayed on gure 13 (see, e.g., (Dubrovin et al. 1984)). A curve parameterization R s P(s) R2is called regular if, and only if, for all s, d Pds = 0. A curve is called smooth if, and onlyif, it admits a regular parameterization. A parameterization is called natural if, and only if, for all s,____d Pds____= 1 where denotes the Euclidian norm. For smooth curves with a natural parameterizations P(s), its signed curvature is dened by dds = , where = d Pds is the unitary tangent vectorand is the oriented normal vector ((, ) is a direct orthonormal frame of the oriented Euclidian planeR2). Notice that dds = . Every smooth curve admits a natural parameterization: every regularparameterization t P(t ) leads to a natural parameterization s P(s) via the differential relationds =____d Pdt____dt .Lemma Consider a trajectory of (18) such that the curve Cn followed by Pn is smooth with thenatural parameterization [0, Ln] sn Pn(sn): sn = 0 (resp. sn = Ln) corresponds to the startingpoint (resp. end point); Ln is the length of Cn. Assume also that for sn = 0, i 1 i (i = 1, . . . , n)and belong to ] /2, /2[. Then,41 (i) for all sn [0, Ln], i 1 i (i = 1, . . . , n) and belong to ] /2, /2[.(ii) the curves Ci and C followed by Pi and Q are smooth (i = 0, 1, . . . , n).(iii) tan(i 1 i) = dii (i = 1, . . . , n) and tan = d00, where i and 0 are the curvatures of Ciand C0, respectively;(iv) the curvature i can be expressed as a smooth function of n and of its rst n i derivatives withrespect to sn; moreover the mapping (which is independent of sn)________ndndsn...dnndsnn_____________nn1...0_____is a global diffeomorphism from Rn+1to Rn+1.Proof of the lemma As displayed on gure 4, the point Pi 1 belongs to the tangent to Ci at Pi and atthe xed distance di from Pi. By assumption n = d Pndsnadmits the good orientation: Pn1 = Pn+dnn(we do not have Pn1 = Pndnn). Thus Cn1 is given by the parameterization sn Pn+dnn whichis regular since___d Pn1dsn___ =_1 +d2n2n. A natural parameterization sn1 Pn1 is given bydsn1 =_1 +d2n2n dsn. (31)The unitary tangent vector, n1, is given by_1 +d2n2n n1 = n +dnn n,where n is the oriented normal to Cn. The angle n1 n is the angle between n and n1. Thustan(n1 n) = dnn. Since n is always nite and n1 n belongs ] /2, /2[ for sn = 0,n1n cannot escapes from ] /2, /2[ for any sn [0, Ln]. The oriented normal to Cn1, n1,is given by_1 +d2n2n n1 = dnn n +n,and the signed curvature n1 of Cn1 is, after some calculations,n1 = 1_1 +d2n2n_n + dn1 +d2n2ndndsn_. (32)Since n1 n remains in ] /2, /2[, the unitary tangent vector n1 has the good direction,i.e., Pn2 = Pn1+dn1n1. The analysis can be continued for Pn2, . . ., P0 and Q. This proves (i),(ii) and (iii).42 Assertion (iv) comes from the following formula derived from (32) and (31) (i = 1, . . . , n):i 1 = 1_1 +d2i 2i_i + di1 +d2i 2ididsi_ (33)where si 1 is the natural parameterization of Ci 1 dened bydsi 1 =_1 +d2i 2i dsi. (34)Consequently, i is an algebraic function of n and its rst ni derivatives with respect to sn. Moreover,the dependence with respect to dnindsninis linear via the termdi +1(1 +d2i +12i +1)3/2di +2(1 +d2i +22i +2)3/2 . . .dn(1 +d2n2n)3/2dnindsninThe map of assertion (iv) has a triangular structure with a diagonal dependence that is linear and alwaysinvertible: it is a global diffeomorphism.Proof of proposition 5 Denote by ( xn, yn) and (xn, yn) the cartesian coordinates of Pn and Pn,the initial and nal positions of Pn. There always exists a smooth planar curve Cn with a naturalparameterization sn Pn(sn) satisfying the following constraints: Pn(0) = Pn and Pn(Ln) = Pn for some Ln > 0. the direction of tangent at Pn (resp. Pn) is given by the angle n (resp. n); the rst n derivatives of the signed curvature n at points Pn and Pn have prescribed values.According to (iii) and (iv) of the above lemma, the initial and nal values of the angles (i = 1, . . . , n)i 1i and dene entirely the initial and nal rst n derivatives of n. It sufces now to choose asmooth function [0, T] t sn(t ) [0, Ln] such that sn(0) = 0, sn(T) = Ln and sn(0) = sn(Ln) =0, to obtain the desired control trajectory via the relations (the notations are those of the above lemma):___ s0 = u1 =_ n
i =1_1 +d2i 2i_ snu2 =_ n
i =1_1 +d2i 2i_ d01 +d2020d0ds0 sn.43