Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.243j (Fall 2003): DYNAMICS OF NONLINEAR SYSTEMS by A. Megretski Lecture 12: Local Controllability 1 In this lecture, nonlinear ODE models with an input are considered. Partial answers to the general controllability question (which states can be reached in given time from a given state by selecting appropriate time-dependent control action) are presented. More precisely, we consider systems described by x˙(t) = a(x(t),u(t)), (12.1) where a : R n ￡ R m is a given continuously difierentiable function, and u = u(t)?? R n is an m-dimensional time-varying input to be chosen to steer the solution x = x(t) in a desired direction. Let U be an open subset of R n , ?x 0 ? R n . The reachable set for a given T > 0 the (U-locally) reachable set R U (?x 0 ,T) is deflned as the set of all x(T) where x : [0,T] , u : [0,T] is a bounded solution of (12.1) such that x(0) = ?x 0 ?? R n ?? R m and x(t) ? U for all t ? [0,T]. Our task is to flnd conditions under which R U (?x 0 ,T) is guaranteed to contain a neig- borhood of some point in R n , or, alternatively, conditions which guarante that R U (?x 0 ,T) has an empty interior. In particular, when ?x 0 is a controlled equilibrium of (12.1), i.e. x 0 , ? u 0 ? R m , complete local controllability of (12.1) at ?a(? u 0 ) = 0 for some ? x 0 means that for every ? > 0 and T > 0 there exists – > 0 such that R U (? x 0 ) for every x,T) B – (? ? x 0 ), where U = B ? (?x ? B – (? x 0 ) and x) = {? xB r (? x 1 ? R n : x 1 ? ?|? |? r} denotes the ball of radius r centered at ?x. 1 Version of October 31, 2003 2 12.1 Systems with controllable linearizations A relatively straightforward case of local controllability analysis is deflned by systems with controllable linearizations. 12.1.1 Controllability of linearized system Let x 0 : [0,T] , u 0 : [0,T] be a bounded solution of (12.1). The standard ?? R n ?? R m linearization of (12.1) around the solution (x 0 (￠),u 0 (￠)) describes the dependency of small state increments – x (t) = x(t) ? x 0 (t) + o(– x (t)) on small input increments – u (t) = u(t) ? – u (t): – ˙ x (t) = A(t)– x (t) + B(t)– u (t), (12.2) where fl fl da fl da fl A(t) = fl , B(t) = fl (12.3) fl dx fl du x=x 0 (t),u=u 0 (t) x=x 0 (t),u=u 0 (t) are bounded measureable matrix-valued functions of time. Let us call system (12.2) controllable on time interval [0,T] if for every – ? 0 – ? T ? R n x , x there exists a bounded measureable function – u : [0,T] such that the solution of ?? R m ? (12.2) with – x (0) = – ? 0 satisfles – x (T) = – T . The following simple criterion of controllability x x is well known from the linear system theory. Theorem 12.1 System (12.2) is controllable on interval [0,T] if and only if the matrix ? T W c = M(t) ?1 B(t)B(t) ? (M(t) ? ) ?1 dt 0 is positive deflnite, where M = M(t) is the evolution matrix of (12.2), deflned by ˙ M(t) = A(t)M(t), M(0) = I. Matrix W c is frequently called the Grammian, or Gram matrix of (12.2) over [0,T]. It is easy to see why Theorem 12.1 is true: the variable change – x (t) = M(t)z(t) reduces (12.2) to z˙(t) = M(t) ?1 B(t)– u (t). Moreover, since ? T z(T) = M(t) ?1 B(t)– u (t)dt 0 is a linear integral dependence, function – u can be chosen to belong to any subclass which is dense in L 1 (0,T). For example, – u (t) can be selected from the class of polynomials, class of piecewise constant functions, etc. Note that controllability over an interval ￠ implies controllability over every interval ￠ + containing ￠, but in general does not imply controllability over all intervals ￠ ? contained in ￠. Also, system (12.2) in which A(t) = A 0 and B(t) = B 0 are constant is equivalent to controllability of the pair (A,B). ? 3 12.1.2 Consequences of linearized controllability Controllability of linearization implies local controllability. The converse is not true: a nonlinear system with an uncontrollable linearization can easily be controllable. Theorem 12.2 Let a : R n ￡ R m be continuously difierentiable. Let x 0 : [0,T] R n , u 0 : ?? R n ?? [0,T] be a bounded solution of (12.1). Assume that system (12.2), deflned by (12.3), is controllable over [0,T]. Then for every ? > 0 there exists – > 0 such that for ?? R m x 0 , ?all ? x T satisfying x 0 ? x 0 (0) < –, x T ? x 0 (T) < – |? | |? | there exist functions x : [0,T] , u : [0,T] satisfying the ODE in (12.2) and ?? R n ?? R m conditions x 0 , x(T) = ?x(0) = ? x T , x(t) ? x 0 (t) < ?, u(t) ? u 0 (t) < ? ? t ? [0,T].| | | In other words, if linearization around a trajectory (x 0 ,u 0 ) is controllable then from every point in a su–ciently small neigborhood of x 0 (0) the solution of (12.1) can be steered to every point in a su–ciently small neigborhood of x 0 (T) by applying a small perturbation u = u(t) of the nominal control u 0 (t). In particular, this applies when x 0 , u 0 (t) · ?x 0 (t) · ? u 0 is a conditional equilibrium, in which case A,B are constant, and hence controllability of (12.2) is easy to verify. When system (12.2) is not controllable, system (12.1) could still be: for example, the second order ODE model 3 x˙ 1 = x 2 , x˙ 2 = u has an uncontrollable linearization around the equilibrium solution x 1 · 0, x 2 · 0, but is nevertheless locally controllable. The proof of Theorem 12.2 is based on the implicit mapping theorem. Let e 1 ,...,? n be the standard basis in R n . Let – u = – k be the controls which cause the solution of u (12.2) wit – x (0) = 0 to reach – x (T) = e k . For ? > 0 let x ? R n : xB ? = {? ? < ?}.| | The function S : B ? ￡ B ? , which maps w = [w 1 ; w 2 ; ...; w n ] ? B ? and v ? B ? to?? R n S(w,v) = x(T), where x = x(t) is the solution of (12.1) with x(0) = x 0 (0) + v and n u(t) = u 0 + w k – k u (t), k=1 is well deflned and continuously difierentiable when ? > 0 is su–ciently small. The derivative of S with respect to w at w = v = 0 is identity. Hence, by the implicit mapping theorem, equation S(w,v) = x has a solution w … 0 whenever v and ?? x ? x 0 (T) are| | | | small enough. ? 4 12.2 Controllability of driftless models In this section we consider ODE models in which the right side is linear with respect to the control variable, i.e. when (12.1) has the special form m x˙(t) = g(x(t))u(t) = g k (x(t))u(t), x(0) = ?x 0 , (12.4) k=1 where g k : are given C ? (i.e. having continuous derivatives of arbitrary order) X 0 ?? R n functions deflned on an open subset X 0 of R n , and u(t) = [u 1 (t); ...; u m (t)] is the vector control input. Note that linearization (12.2) of (12.4) around every equilibrium solution x 0 (t) · x 0 = const, u 0 (t) = 0 yields A = 0 and B = g(?? x 0 ), which means that the linearization is never controllable unless m = n. Nevertheless, it turns out that, for a “generic” function g, system (12.4) is expected to be completely controllable, as long as m > 1. 12.2.1 Local controllability and Lie brackets Let us say that system (12.4) is locally controllable at a point ? if for every ? > 0,x 0 ? X 0 T > 0, and ? x? ?x ? X 0 such that ? x 0 | < ? there exists a bounded measureable function u : [0,T] deflning a solution of (12.4) with x(0) = ? x and | x 0 such that x(T) = ??? R m x 0 < ? ? t ? [0,T].x(t) ? ?| | The local controlability conditions to be presented in this section are based on the notion of a Lie bracket. Let us write h 3 = [h 1 ,h 2 ] (which reads as “h 3 is the Lie bracket of h 1 and h 2 ”) when h k : are continuous functions deflned on an open subset X 0 ?? R n X 0 of R n , functions h 1 ,h 2 are continuously difierentiable on X 0 , and h 3 (? x)h 2 (? x)h 1 (?x) = h ˙ 1 (? x) ? h ˙ 2 (? x) x ? X 0 , where h ˙ k (? xfor all ? x) denotes the Jacobian of h k at ?. The reasoning behind the deflnition, as well as a more detailed study of the properties of Lie brackets, will be postponed until the proof of the controllability results of this subsection. Let us call a set of functions h k : , (k = 1,...,q) complete at a point ?x ? X 0 X 0 ?? R n if either the vectors h i (? ?? R n x) with i = 1,...,m span the whole R n or there exist functions h k : X 0 , (k = q + 1,...,N), such that for every k > q we have h k = [h i ,h s ] for some i,s < k, and the vectors h i (?x) with i = 1,...,N span the whole R n . Theorem 12.3 If C ? functions g k : X 0 ?? R n form a complete set at ? ? X 0 thenx 0 system (12.4) is locally controllable at ?x 0 . ? ? ? ? 5 Theorem 12.3 provides a su–cient criterion of local controllability in terms of the span of all vector flelds which can be generated by applying repeatedly the Lie bracket operation to g k . This condition is not necessary, as can be seen from the following example: the second order system x˙ 1 = u 1 , x˙ 2 = `(x 1 )u 2 , where function ` : R ?? R is inflnitely many times continuously difierentiable and such that `(0) = 0, `(y) > 0 for y = 0, ` (k) (0) = 0 ? k, ∈ is locally controllable at every point ?x 0 ? R n despite the fact that the corresponding set of vector flelds 1 0 g 1 (x) = , g 2 (x) = 0 `(x 1 ) is not complete at ?x = 0. On the other hand, the example of the system x˙ = xu, which is not locally controlable at ?x = 0, but is deflned by a (single element) set of vector ?flelds which is complete at every point except x = 0, shows that there is little room for relaxing the su–cient conditions of Theorem 12.3. 12.2.2 Proof of Theorem 12.3 Let S denote the set of all continuous functions s : ? s ?? X 0 , where ? s is an open subset of R ￡X 0 containing {0}￡X 0 (? s is allowed to depend on s). Let S k ? S be the elements of S deflned by x) = x(?) : x˙(t) = g k (x(t)), x(0) = ?S k (?, ? x. Let S g be subset of S which consists of all functions which can be obtained by recursion x,?) = S fi(k) (s k (? x,?) = ?s k+1 (? x,?),` k (?)), 0 (? x, where fi(k) ? {1,2,...,m} and ` k : R ?? R are continuous functions such that ` k (0) = 0. One can view elements of S g as admissible state transitions in system (12.2) with piecewise constant control depending on parameter ? in such a way that ? = 0 corresponds to the identity transition. Note that for every s ? S g there exists an “inverse” s ? ? S g such that s(s ? (? x x,?) ? ? s ? ,x,?),?) = ? ? (? deflned by applying inverses S fi(k) (￠,?` k (?)) of the basic transformations S fi(k) (￠,` k (?)) in the reverse order. ? 6 Let us call a C ? function h : X 0 ?? R n implementable in control system (12.4) if for every integer k > 0 there exists a function which is k times continuously s ? S g difierentiable in the region ? ? 0 and in the region ? ? 0, such that s(? x + ?h(?x,?) = ? x) + o(?) (12.5) as ? ? 0,? 0 for all ? x) of an implementable x ? X 0 . One can say that the value h(?? function h(￠) at a given point ? describes a direction in which solutions of (12.4) x ? X 0 can be steered from ?.x We will prove Theorem 12.3 by showing that Lie bracket of two implementable vec- tor flelds is also an implementable vector fleld. After this is done, an implicit function argument similar to one used in the proof of Theorem 12.2 shows local controllability of (12.4). Now we need to prove two intermediate statements concerning the set of implementable vector flelds. Remember that for the difierential ow (t, ? (?x,t) deflned by a smooth x) ?? S h vector fleld h we have x,t 1 ,t 2 ) = S h (?S h (S h (? x,t 1 + t 2 ), which, in particular, implies that t 2 S h (? x + th(? x)h(?x,t) = ? x) + h ˙ (? x) + O(t 3 ) 2 as t ? 0. This is not necessarily true for a general transition s from the deflnition of an implementable vector fleld h. However, the next Lemma shows that s can always be chosen to match the flrst k Taylor coe–cients of S h . Lemma 12.1 If h is implementable then for every integer k > 0 there exists a k times continuously difierentiable function s ? S g such that x,?) = S h (?s(? x,?) + O(? k ). (12.6) Proof By assumption, (12.6) holds for k = 2 and ? ? 0, where s is N times continuously difierentiable in the region ? ? 0, and N can be chosen arbitrarily large. Assume that for ? 0? s(? x,?) + ? k w(?x,?) = S h (? x) + O(? k+1 , (which is implied by (12.6)), where 2 ? k < N and w is continuously difierentiable. Then s (? x,??) ? ? k w(?x,?) = S h (? x) + O(? k+1 , and hence for every a,b > 0 the function x,?) = s(s ? (s(?s a,b (? x,a?),b?),a?) ? 7 satisfles s a,b (? x,(2 ?a?b)?) + (2a k ?b k )? k w(?x,?) = S h (? x) + O(? k+1 . Since k ? 2, one can choose a,b in such way that 2a?b = 1, 2a k = b k , which yields (12.6) with k increased by 1. After (12.6) is established for ? ? 0, s can be deflned for negative arguments by x,??) = s ? (?s(? x,?), ? 0,? which makes it k ?1 times continuously difierentiable. Next lemma is a key result explaining the importance of Lie brackets in controllability analysis. Lemma 12.2 If vector flelds h 1 ,h 2 are implementable then so is their Lie bracket h = [h 2 ,h 1 ]. Proof By Lemma 12.1, there exist 2 ?k+ 2 times continuously difierentiable (for ? = 0) ∈ functions s 1 ,s 2 ?S g such that s i (? x + ?h i (? x)h i (?x,?) = ? x) + ? 2 h ˙ i (? x) + o(? 2 ). Hence (check this!), s 3 ?S g deflned by s 3 (? x,?),?),??),??),x,?) = s 2 (s 1 (s 2 (s 1 (? satisfles s 3 (? x + ? 2 h(?x,?) = ? x) + o(? 2 ). Now for i = 3,4,...,2 ?k + 2 let s i+1 (? x,?/ ≈ 2),??/ ≈ 2).x,?) = s i (s i (? By induction, i s i+2 (? x + ? 2i fl iq (?x,?) = ? x) + o(? 2i ), q=1 i.e. the transformation from s i to s i+1 removes the smallest odd power of ? in the Taylor expansion for s i . Hence x,?) = s 2k+2 (?s(? x, ≈ ?), ? 0? deflnes a k times continuously difierentiable function for su–ciently small ? ? 0, and s(? x + ?h(?x,?) = ? x) + o(?) for ? ? 0, ? ? 0. 8 12.2.3 Frobenius Theorem Let g k : R n , k = 1,...,m, be k times (k ? 1) continuously difierentiable functions. ?? R n We will say that g k deflne a regular C k distribution D({g k }) at a point ?x 0 ? R n if vectors g k (?x 0 ) are linearly independent. Let X 0 be an open subset of R n . The distribution D({g k }) is called involutive on X 0 if the value g ij (?x) of every Lie bracket g ij = [g i ,g j ] x) for every ?belongs to the linear span of g k (? x ? X 0 . Finally, distribution D({g k }) is called completely C k integrable over X 0 if there exists a set a set of k times continuously difierentiable functions h k : X 0 ?? R, k = 1,...,n? m, such that the gradients ≡h k (?x) are linearly independent for all ?x ? X 0 , and ≡h i (? x) = 0 ? ?x)g j (? x ? X 0 . The following classical result gives a partial answer to the question of what happens to controllability when the Lie brackets of vector flelds g k do not span R n . Theorem 12.4 Let D({g k }) deflne a C r distribution (r 1) which is regular at ?x 0 ? R n .? Then the following conditions are equivalent: (a) there exists an open set X 0 containing ?x 0 such that D({g k }) is completely C r inte- grable over X 0 ; (b) there exists an open set X 0 containing ?x 0 such that D({g k }) is involutive on X 0 . Essentially, the Frobenius theorem states that in the neigborhood of a point where the dimension of the vector flelds generated by Lie brackets of a given driftless control system is maximal but still less than n, there exist non-constant functions of the state vector which remain constant along all solutions of the system equations. The condition of regularity in Theorem 12.4 is essential. For example, when ? ?? ? ? x 2 n = 2, m = 1, ?x 0 = 0 ? R 2 , g x 1 = , x 2 ?x 1 the distribution deflned by g is smooth and involutive (because [g,g] = 0 for every vector fleld g), but not regular at x 0 . Consequently, the conclusion of Theorem 12.4 does not ? hold at ?x 0 = 0, but is nevertheless valid in a neigborhood of all other points. The “locality” of complete integrability is also essential for the theorem. For example, the vector fleld ?? ?? ? ? 2 2 x 1 x 1 + (1 ? x 2 2 ? x 2 ) 2 g ?? ?? = ? ? x 2 x 3 x 3 ?x 2 deflnes a smooth regular involutive distribution on the whole R 3 . However, the distri- bution is not completely integrable over R 3 , while it is still completely integrable in a neigborhood of every point. ? ? ? 9 12.2.4 Proof of Theorem 12.4 The implication (a)≤(b) follows straightforwardly from the reachability properties of Lie brackets. Let us prove the implication (b)≤(a). Let S ? denote the difierential ow map associated with g k , i.e. S k ? (?x) = x(?), where k x = x(t) is the solution of x˙(t) = g k (x(t)), x(0) = ?x. Let ￠(? x),. .. , g m (?x) denote the span of g 1 (? x). The following stetement, which relies on m k (?both regularity and involutivity of the family {g k } k=1 , states that the Jacobian D t x) of S t at x maps ￠(? x)). This a generalization of the (obvious) fact that, for ? x) onto ￠(S k t (? k a single vector fleld g : R n , moving the initial condition x(0) by g(x(0))– of a ?? R n solution x = x(t) of dx/dt = g(x) results in x(t) shifted by g(x(t))– + o(–). Lemma 12.3 Under the assumptions of Theorem 12.4, k (? x) = ￠(S k t (?D t x)￠(? x)). Proof According to the rules for difierentiation with respect to initial conditions, for a flxed ? k (?x, D k (t) = D t x) satisfles the ODE d D k (t) = g˙ k (x(t))D k (t), D k (0) = I, dt where x(t) = S k t (? x) denotes the Jacobian of g at ?x), and ˙g k (? x. Hence ? D k (t) = D k (t)g(?x), where g(?x) = [g 1 (?x) g 2 (?x) . .. g m (?x)], satisfles d ? D k (t) = ˙g k (x(t)) ? D k (t), ? D k (0) = g(?x). (12.7) dt Note that the (12.7) is an ODE with a unique solution. Hence, it is su–cient to show that (12.7) has a solution of the form m ? D k (t) = g(x(t))–(t) = g i (x(t))– k (t), (12.8) i=1 where – = –(t) is a continuously difierentiable m-by-m matrix valued function of time, and – i (t) is the i-th row of –(t). Indeed, substituting into (12.7) yields [g˙ i (x(t))g k (x(t))– k (t) + g i (x(t))– ˙ k (t)] = g˙ k (x(t)) g i (x(t))– k (t) i i and –(0) = I. Equivalently, g(x(t))– ˙ (t) = A(t)–(t), 10 where A(t) is the n-by-m matrix with columns g ki (x(t)), g ki = [g k ,g i ]. By involutivity and regularity, A(t) = g(x(t))a(t) for some continuous m-by-m matrix valued function a = a(t). Thus, the equation for –(t) becomes – ˙ (t) = a(t)–(t), –(0) = I, ? hence existence of –(t) such that D k (t) = g(x(t))–(t) is guaranteed. x 0 ),...,g n (?Let g m+1 ,...,g n be C ? smooth functions g i : R n ?? R n such that vectors g 1 (? x 0 ) form a basis in R n . (For example, the functions g i with i > m can be chosen constant). Consider the map F(z) = S z 1 (S z 2 (...(S z n (?x 0 )) ...)), 1 2 n deflned and k times continuously difierentiable for z = [z 1 ,...,z n ] in a neigborhood of zero in R n . F is a k times difierentiable map deflned in a neigborhood of z = 0, ? ?x = x 0 . Since the Jacobian F ˙ (0) of F at zero, given by x 0 ) g 2 (? x 0 )]F ˙ (0) = [g 1 (? x 0 ) ... g n (? is not singular, by the implicit mapping theorem there exists a k times continuously difierentiable function z = H(x) = [h n (x); h n?1 (x); ...; h 1 (x)] deflned in a neigborhood of ?x 0 , such that F(H(x)) · x. Let us show that functions h i = h i (x) satisfy the requirements of Theorem 12.4. Indeed, difierentiating the identity F(H(x)) · x yields F ˙ (H(x))H ˙ (x) = I, (12.9) where H ˙ (x) = [≡h n (x); ≡h n?1 (x); ...; ≡h 1 (x)] is the Jacobian of H at x, and F ˙ (z) = [f 1 (z) f 2 (z) ... f n (x)] is the Jacobian of F at z. Hence vectors f i (z) form a basis, as well as the co-vectors ≡h i (x)h. By Lemma 12.3, vectors f i (z) with i ? m belong to ￠(F(z)) = ￠(x) (and hence, by linear independence, form a basis in ￠(x)). On the other hand, (12.9) implies r (x)f i (H(x)) = 0 for r ? n? m and i ? m. Hence ≡h r (x) for r ? n? m are linearly ≡h independent and orthogonal to all g i (x) for i ? m.