Observer design for a general class of triangular systems

1st International Symposium on Mathematical Theory of Networks and Systems July 7-11, 014. Observer design for a general class of triangular systems Dimitris Boskos 1 John Tsinias Abstract The paper deals with the observer design problem for a wide class of triangular nonlinear systems. Our main results generalize those obtained in the recent author s works [] and [3]. Index Terms nonlinear triangular systems, observer design, switching dynamics. I. INTRODUCTION We derive sufficient conditions for the solvability of the observer design problem for time-varying single output systems of the form ẋ i = f i (t, x 1,..., x i+1 ), i = 1,..., n 1 ẋ n = f n (t, x 1,..., x n ), y = x 1, (x 1,..., x n ) R n (1a) (1b) where the functions f i ( ), i = 1,..., n, are continuous and (locally) Lipschitz. It is known from Theorem.1 of Chapter 3 in [5] that for every x 0 R n and for each fixed input, every single output control system which has a uniform canonical flag ([5, Chapter, Definition.1]) can be locally transformed in the above canonical form (1) in such a way that x i+1 f i (t, x 1,..., x i+1 ) 0, x = (x 1,..., x n ) U, t 0, i = 1,..., n 1 () for certain neighborhood U of x 0. We also mention the works [4] and [8] where algebraic type necessary and sufficient conditions are established for feedback equivalence between a single input system and a triangular system whose dynamics have p-normal form. In our recent work [3], the observer design problem is studied for a subclass of systems (1) whose dynamics have p-normal form. The result of present work constitutes a generalization of previous results in the literature dealing with the observer design problem for triangular systems (see for instance [1], [5], [6] and relative references therein) and particularly generalizes the main result of the recent author s work in [3]. We make the following assumption for the right hand side of system (1) which constitutes a (global) generalization of assumption (). *This work was not supported by any organization 1 John Tsinias is with Faculty of Mathematics, Department of Mathematical and Physical Sciences, National Technical University of Athens, Zografou Campus 15780, Athens, Greece jtsin@central.ntua.gr Dimitris Boskos with the Department of Mathematical and Physical Sciences, National Technical University of Athens, Zografou Campus 15780, Athens, Greece dmposkos@central.ntua.gr H1. For each (t; x 1,..., x i ) R 0 R i, i = 1,..., n 1, the function R z f i (t, x 1,..., x i, z) R is strictly monotone. The paper is organized as follows. We first provide notations and various concepts, including the concept of the switching observer that has been originally introduced in [], for general time-varying systems: ẋ = f(t, x), (t, x) R 0 R n y = h(t, x), y R n (3a) (3b) where y( ) plays the role of output. We then provide the precise statement of our main result (Theorem 1.1) concerning solvability of the observer design problem for (1). Section II contains some preliminary results concerning solvability of the observer design problem for the case (3) with linear output (Propositions.1 and.). In Section III we use the results of Section II, in order to prove our main result. Notations and definitions: We adopt the following notations. For a given vector x R n, x denotes its transpose and x its Euclidean norm. We use the notation A := max{ Ax : x R n ; x = 1} for the induced norm of a matrix A R m n. By N we denote the class of all increasing C 0 functions ϕ : R 0 R 0. For given R > 0, denote by B R the closed ball of radius R > 0, centered at 0 R n. Consider a pair of metric spaces X 1, X and a set-valued map X 1 x Q(x) X. We say that Q( ) satisfies the Compactness Property (CP), if for every sequence (x ν ) ν N X 1 and (q ν ) ν N X with x ν x X 1 and q ν Q(x ν ), there exist a subsequence (x νk ) k N and q Q(x) such that q νk q. We also invoke the well known fact, see [7], that the time-varying system (3a) is forward complete, if and only if there exists a function β NN such that the solution x( ) := x(, t 0, x 0 ) of (3a) initiated from x 0 at time t = t 0 satisfies: x(t) β(t, x 0 ), t t 0 0, x 0 R n (4) provided that the dynamics of (3a) are C 0 and Lipschitz on x R n. It turns out that, under these regularity assumptions plus forward completeness for (3a), for each t 0 0 and x 0 R n the corresponding output y(t) = h(t, x(t, t 0, x 0 )) of (3) is defined for all t t 0. For each t 0 0 and nonempty subset M of R n, we may consider the set O(t 0, M) of all outputs of (3), corresponding to initial state x 0 M and initial time t 0 0: O(t 0, M) := {y : [t 0, ) R n : y(t) = h(t, x(t, t 0, x 0 )); t t 0, x 0 M} (5) ISBN: 978-90-367-631-9 35

A(t, q, x, e, y) := (A C1 (t, q, x, e 1, y), A C (t, q, x, e 1, e, y),, A Cm (t, q, x, e 1,..., e m, y)) (10) For given = M R n, we say that the Observer Design Problem (ODP) is solvable for (3) with respect to M, if for every t 0 0 there exist a continuous map G := G t0 (t, z, w) : [t 0, ) R n R n R n and a nonempty set M R n such that for every z 0 M and output y O(t 0, M) the corresponding trajectory z( ) := z(, t 0, z 0 ; y); z(t 0 ) = z 0 of the observer ż = G(t, z, y) exists for all t t 0 and the error e( ) := x( ) z( ) between the trajectory x( ) := x(, t 0, x 0 ), x 0 M of (3a) and the trajectory z( ) := z(, t 0, z 0 ; y) of the observer satisfies: lim e(t) = 0 (6) t We say that the Switching Observer Design Problem (SODP) is solvable for (3) with respect to M, if for every t 0 0 there exist a strictly increasing sequence of times (t k ) k N with t 1 = t 0 and lim k t k =, a sequence of continuous mappings G k := G k,tk 1 (t, z, w) : [t k 1, t k+1 ] R n R n R n, k N and a nonempty set M R n such that the solution z k ( ) of the system ż k = G k (t, z k, y), t [t k 1, t k+1 ] (7) with initial z(t k 1 ) M and output y O(t 0, M), is defined for every t [t k 1, t k+1 ] and in such a way that, if we consider the piecewise continuous map Z : [t 0, ) R n defined as Z(t) := z k (t), t [t k, t k+1 ), k N, where for each k N the map z k ( ) denotes the solution of (7), then the error e( ) := x( ) Z( ) between the trajectory x( ) := x(, t 0, x 0 ), of (3a) and Z( ) satisfies (6). Our main result is the following theorem. Theorem 1.1: For the system (1), assume that Hypothesis H1 is satisfied and (1a) is forward complete, i.e. there exists a function β NN, such that the solution x( ) := x(, t 0, x 0 ) of (1a) satisfies the estimation (4). Then: (i) the SODP is solvable for (1) with respect to R n. (ii) if in addition we assume that it is a priori known, that the initial states of (1) belong to a given ball B R of radius R > 0 centered at zero 0 R n, then the ODP is solvable for (1) with respect to B R. Theorem 1.1 constitutes a generalization of Theorem 1.1 in [3] for systems (1), whose dynamics are C 1 and have the particular form f i (t, x 1,..., x i+1 ) := f i (t, x 1,..., x i ) + a i (t, x 1 )x mi i+1, for certain f i C 1 (R 0 R i ; R) and a i C 1 (R 0 R; R), i = 1,..., n 1, where the constants m i, i = 1,..., n 1 are odd integers and the functions a i (, ), i = 1,..., n 1 satisfy the condition a i (t, y) > 0, t R 0, y R. II. PRELIMINARY RESULTS The proof of our main result concerning the case (1), is based on some preliminary results concerning the case of systems (3) with linear output: ẋ = f(t, x) := F (t, x, H(t)x), (t, x) R 0 R n y = h(t, x) := H(t)x, y R n (8a) (8b) where H : R 0 R n n is C 0 and F : R 0 R n R n R n is C 0 and Lipschitz on (x, y) R n R n. We assume that system (8a) is forward complete, namely, the solution x( ) := x(, t 0, x 0 ) of (8a) satisfies (4) for certain β NN, hence, for every R > 0 and t 0 we can define: Y R (t) := {y R n : y = H(t)x(t, t 0, x 0 ), for certain t 0 [0, t] and x 0 B R } (9) where H( ) is given in (8b). Obviously, Y R (t) for all t 0 and, if (4) holds, the set-valued map [0, ) t Y R (t) R n satisfies the CP and further y(t) Y R (t), for every t t 0 0 and y O(t 0, B R ); (see notations). Also, given integers l, m, n, n N and a map A : R 0 R l R n R m R n R m m, we say that A(,,,, ) satisfies Property P1, if the following holds: P1. A(t, q, x, e, y) has the form (10) on top of the page, where each mapping A Ci : R 0 R l R n R i R n R m 1, i = 1,..., m is continuous on R 0 R l R n {(e 1,..., e i ) R i : e i 0} R n and bounded on every compact subset of R 0 R l R n R i R n. We make the following hypothesis: Hypothesis.1. There exist a function g C 1 (R 0 ; R) satisfying: 0 < g(t) < 1, t 0; (11a) ġ(t) g(t), t 0; lim g(t) = 0 t (11b) (11c) an integer l N, a map A : R 0 R l R n R n R n R n n satisfying P1 and constants L > 1, c 1, c > 0, R > 0 with c 1 1 (1) such that the following properties hold: A1. For every ξ 1 there exists a set-valued map [0, ) t Q R (t) := Q R,ξ (t) R l (13) with Q R (t) for any t 0, satisfying the CP and such that t 0, y Y R (t), x, z R n with x β(t, R) and x z ξ F (t, x, z, y) := F (t, x, y) F (t, z, y) = A(t, q, x, x z, y)(x z) for certain q Q R (t) (14) with Y R ( ) as given by (9). A. For every ξ 1 there exists a set-valued map Q R := Q R,ξ as in Hypothesis A1, in such a way that for every t 0 0, a time-varying symmetric matrix P R := 36

P R,ξ,t0 C 1 ([t 0, ); R n n ) and a function d R := d R,ξ,t0 C 0 ([t 0, ); R) can be found, satisfying: P R (t) I n n, t t 0 ; P R (t 0 ) L; (15a) t d R (t) > c 1, t t 0 +1; d R (s)ds > c, t t 1 t 0 ; t 1 (15b) e P R (t)a(t, q, x, e, y)e + 1 e P R (t)e d R (t)e P R (t)e, t t 0, q Q R (t), x R n, e ker H(t), y Y R (t) : x β(t, R), e ξ, e P R (t)e g(t) (15c) with Y R ( ) as given by (9). The following proposition provides a Luenberger type observer, for the case where it is a priori known, that the initial state of the system belongs to a given nonempty bounded subset of the state space. This result constitutes a slight generalization of Proposition.1 in [3]. Proposition.1: Consider the system (8) and assume that it is forward complete, namely, (4) holds for certain β NN and satisfies Hypothesis.1. Then, the following hold: For each t 0 t 0 0 and constant ξ satisfying ξ L exp{c }β( t 0, R), R := R + 1 (16) there exists a function ϕ R := ϕ R,ξ, t 0 C 1 ([ t 0, ); R >0 ), such that the solution z( ) of system ż =G t 0 (t, z, y) :=F (t, z, y) + ϕ R (t)p 1 R (t)h (t)(y H(t)z) (17a) with initial z( t 0 ) = 0 (17b) where P R ( ) := P R,ξ, t 0 ( ) is given in A, is defined for all t t 0 and the error e( ) := x( ) z( ) between the trajectory x( ) := x(, t 0, x 0 ) of (8a), initiated from x 0 B R at time t 0 0 and the trajectory z( ) := z(, t 0, 0; y) of (17) satisfies: e(t) < ξ, t t 0 ; e(t) max{ξ exp{ c 1 (t ( t 0 + 1))}, g(t)}, t t 0 + 1 (18a) (18b) It follows from (11c) and (18b), that for t 0 := t 0 the ODP is solvable for (8) with respect to B R ; particularly the error e( ) between the trajectory x( ) := x(, t 0, x 0 ), x 0 B R of (8a) and the trajectory z( ) := z(, t 0, z 0 ; y), z 0 = 0 of the observer ż = G t0 (t, z, y) satisfies (6). The following proposition constitutes a slight generalization of Proposition. in [3]. It establishes sufficient conditions for the existence of a switching observer exhibiting the state determination of (8), without any a priori information concerning the initial condition. We make the following hypothesis: Hypothesis.: There exist constants L > 1, c 1, c > 0 such that (1) holds, an integer l N, a function g C 1 (R 0 ; R) satisfying (11) and a map A : R 0 R l R n R n R n R n n satisfying P1, in such a way that for every R > 0 Hypothesis.1 is fulfilled, namely, both A1 and A hold. Proposition.: In addition to the hypothesis of forward completeness for (8a), assume that system (8) satisfies Hypothesis.. Then the SODP is solvable for (8) with respect to R n. The proofs of Propositions.1 and. are based on the same approaches with those used for proofs of Propositions.1 and. in [3], respectively, plus some elementary modifications. Moreover a preliminary technical result (Lemma.1 below) is required, which also constitutes a slight modification of Lemma.1 in [3]. In order to present the precise statement of this preliminary result we consider a pair (H, A) of mappings: H := H(t) : R 0 R n m ; (19a) A := A(t, q, x, e, y) : R 0 R l R n R m R n R m m (19b) for certain l, m, n, n N, where H( ) is continuous and A(,,,, ) satisfies Property P1. We make the following hypothesis: Hypothesis.3: Let g( ) C 0 (R 0 ; R) satisfying (11a) and assume that for certain constant R > 0 and for every ξ 1, there exist a function β R := β R,ξ N and setvalued mappings [0, ) t Y R (t) := Y R,ξ (t) R n and [0, ) t Q R (t) := Q R,ξ (t) R l with Y R (t) and Q R (t) for all t 0, satisfying the CP, in such a way that for every t 0 0, a time-varying symmetric matrix P R := P R,ξ,t0 C 1 ([t 0, ); R m m ) can be found, satisfying P R (t) I m m, t t 0 and a function d R := d R,ξ,t0 C 0 ([t 0, ); R), in such a way that e P R (t)a(t, q, x, e, y)e + 1 e P R (t)e d R (t)e P R (t)e, t t 0, q Q R (t), x R n, e ker H(t), y Y R (t) : x β R (t), e ξ, e P R (t)e g(t) (0) Lemma.1: Consider the pair (H, A) of the time-varying mappings in (19) and assume that Hypothesis.3 is fulfilled for certain R > 0. Then for every ξ 1, t 0 0 and d R := d R,ξ,t0 C 0 ([t 0, ); R) with d R (t) < d R (t), t t 0, there exists a function ϕ R := ϕ R,ξ,t0 C 1 ([t 0, ); R >0 ) such that e P R (t)a(t, q, x, e, y)e + 1 e P R (t)e ϕ R (t) H(t)e d R (t)e P R (t)e, t t 0, q Q R (t), x R n, e R m, y Y R (t) : x β R (t), e ξ, e P R (t)e g(t) (1) III. PROOF OF THEOREM 1.1 In this section we apply the results of Section II to prove our main result concerning the solvability of the SODP(ODP) for triangular systems (1). Proof of Theorem 1.1: The proof of both statements is based on the results of Propositions.1 and. and is based on a generalization of the methodology employed in the proof of the main result in [3]. Without loss of generality, we may assume that, instead of Assumption H1 it holds: 37

δ i, (t, y, x,..., x i+1, e ) 1 i n; n δ i,j (t, y, x,..., x i+1, e,..., e j ) 3 j i n; n 3 { δ i,i+1 (t, y, x,..., x i+1, e,..., e i+1 ) i n 1; n 3 := fi(t,y,x,x3,...,xi+1) fi(t,y,x e,x3,...,xi+1) e, for e 0 and i n; n 3 := fi(t,y,x) fi(t,y,x e) e, for e 0 and i = 1, ; n = or i = 1; n 3 := 0, for e = 0 and n, i, j in each case above (t; y; x,..., x i+1 ; e ) R 0 R R i R (3a) := f i(t,y,x e,...,x j 1 e j 1,x j,x j+1,...,x i+1 ) f i (t,y,x e,...,x j 1 e j 1,x j e j,x j+1,...,x i+1 ) e j, for e j 0 and 3 i n 1; 3 j i; n 4 or i = n; 3 j n 1; n 4 := fn(t,y,x e,...,xn 1 en 1,xn) fn(t,y,x e,...,xn 1 en 1,xn en) e n, for e n 0 and i = j = n; n 3 := 0, for e j = 0, and n, i, j in each case above (t; y; x,..., x i+1 ; e,..., e j ) R 0 R R i R j 1 := f i(t,y,x e,...,x i e i,x i+1 ) f i (t,y,x e,...,x i e i,x i+1 e i+1 ) e i+1, for e i+1 0 := 0, for e i+1 = 0 (t; y; x,..., x i+1 ; e,..., e i+1 ) R 0 R R i R i, (3b) (3c) H1. For each (t; x 1,..., x i ) R 0 R i, i = 1,..., n 1, the function R z f i (t, x 1,..., x i, z) R is strictly increasing. The proof of the first statement, is based on the establishment of Hypothesis. for system (1). Hence, we show that there exist an integer l N, a map A : R 0 R l R n R n R R n n satisfying P1, constants L > 1, c 1, c > 0 such that (1) holds and a function g( ) satisfying (11), in such a way that for each R > 0, both A1 and A hold for (1). Let R > 0, ξ 1 and define: F (t, x, y) :=(f 1 (t, y, x ), f (t, y, x, x 3 ),..., f n 1 (t, y, x,..., x n ), f n (t, y, x,..., x n )), (t, x, y) R 0 R n R () Also, for every pair of indices (i, j) with j n, j 1 i n we define the functions δ i,j ( ) as in (3a)-(3c) on top of the page (where we have used the notation x i+1 i=n := x n and R i i=n := R n 1, in (3a), (3b)). By exploiting the Lipschitz continuity assumption for the functions f i ( ), i = 1,..., n it follows that for every j n, j 1 i n, the following properties hold: S1. δ i,j ( ) is continuous on the set O i,j := R 0 R R i {(e,..., e j ) R j 1 : e j 0} (4) S. δ i,j ( ) is bounded on every compact subset of R 0 R R i R j 1. (where we have used the notation R i i=n := R n 1 in both S1 and S) Furthermore, from H1, (4), (3a) and (3c) it follows: S3. δ i,i+1 (t, y, x,..., x i+1, e,..., e i+1 ) > 0, (5) (t; y; x,..., x i+1 ; e,..., e i+1 ) O i,i+1, i = 1,..., n 1 From (3a)-(3c) we deduce that for i = 1,..., n it holds: f i (t, y, x,..., x i+1 ) f i (t, y, z,..., z i+1 ) δ i,j (t, y, x,..., x i+1, x z,..., x j z j )(x j z j ) i+1 = j= t R 0, y R, (x,..., x i+1 ), (z,..., z i+1 ) R i (6) (where we have used the notation i+1 j= i=n := n j=, x i+1 i=n := x n and z i+1 i=n := z n ). Also, by invoking Property S, the following property holds. S4. There exists a function σ R := σ R,ξ N C 1 ([0, ); R) satisfying: σ R (t) n i= j= i sup{ δ i,j (t, y, x,..., x i+1, e,..., e j ) : y β(t, R), (x,..., x i+1 ) β(t, R), (e,..., e j ) ξ}, t 0 (7) (where we have used the notation x i+1 i=n := x n ). Next, consider the set-valued map [0, ) t Q R (t) := Q R,ξ (t) R l, l := n(n+1) defined as Q R (t) := {q = (q 1,1 ; q,1, q, ;...; q n,1, q n,,..., q n,n ) R l : q σ R (t)} (8) that obviously satisfies the CP and consider the set valued mappings R 0 (0, ξ] (t, r) Q R,i (t, r) := Q R,ξ,i (t, r) R R i R i, i = 1,..., n 1 given as Q R,i (t, r) := {(y; x,..., x i+1 ; e,..., e i+1 ) R R i R i : y β(t, R), (x,..., x i+1 ) β(t, R), (e,..., e i+1 ) ξ, e i+1 r} (9) 38

Finally, for i = 1,..., n 1 define: D R,ξ,i (t, r) := D R,i (t, r) = min{δ i,i+1 (t, y, x,..., x i+1, e,..., e i+1 ) : (y; x,..., x i+1 ; e,..., e i+1 ) Q R,i (t, r)} (t, r) R 0 (0, ξ] (30) By exploiting (9), (30) and Properties S1, S3, it follows that the functions D R,i (, ), i = 1,..., n 1 are continuous and the following hold: 0 < D R,i (t, r) δ i,i+1 (t, y, x,..., x i+1, e,..., e i+1 ), (y; x,..., x i+1 ; e,..., e i+1 ) Q R,i (t, r), (t, r) R 0 (0, ξ] (31) D R,i (t, r 1 ) D R,i (t, r ), t R 0, r 1, r (0, ξ] with r 1 < r (3) Now, let Y R ( ) as given by (9) with H := (1, 0,..., 0) R 1 n (33) and notice that, due to (4), (9) and (33), it holds: y β(t, R), for every y Y R (t), t 0 (34) From (), (6), (8), (34) and Property S4, it follows that for every t 0, y Y R (t) and x, z R n with x β(t, R) and x z ξ we have: F (t, x, y) F (t, z, y) = A(t, q, x, x z, y)(x z), for some q Q R (t) with q i,1 = 0, i = 1,..., n; (35a) with A(,,,, ) as given in (35b) on top of the next page. Notice that this map has the form (10), and, due to S1 and S, satisfies Property P1. Hence, A1 is satisfied. In order to establish A, we prove that there exist constants L > 1, c 1, c > 0 such that (1) holds and a function g( ) satisfying (11), in such a way that for every R > 0, ξ 1 and t 0 0, a time-varying symmetric matrix P R := P R,ξ,t0 C 1 ([t 0, ); R n n ) and a function d R := d R,ξ,t0 C 0 ([t 0, ); R) can be found satisfying all conditions (15a), (15b), (15c) with H, A(,,,, ), Y R ( ) and Q R ( ), as given by (33), (35b), (9) and (8), respectively and with β(, ) as given in (4) for the case of system (1). We proceed by induction as follows. Pick L > 1, c 1 := 1, c := n and let g( ) be a C 1 function satisfying (11). Also, let R > 0 and for k =,..., n define: H k :=(1, 0,..., 0) R 1 k, e := (e n k+1 ; ê ) R R k 1 ê :=(e n k+,..., e n ) R k 1 (36a) and consider the map A k : R 0 R l R n R k R R k k with components: (A k (t, q, x, e, y)) i,j (36b) := q n k+i,n k+j, for j i := δ n k+i,n k+j (t, y, x,..., x n k+j, 0,..., 0, e n k+1,..., e n k+j ), for j = i + 1; k < n 1 := δ n k+i,n k+j (t, y, x,..., x n k+j, e,..., e n k+j ), for j = i + 1; k = n 1, n := 0, for j > i + 1 Claim 1 (Induction Hypothesis): Let k N with k n. Then for L, R and g( ) as above and for every ξ 1 and t 0 0, there exist a time-varying symmetric matrix P R,k := P R,ξ,t0,k C 1 ([t 0, ); R k k ) and a mapping d R,k := d R,ξ,t0,k C 0 ([t 0, ); R), in such a way that the following hold: P R,k (t) > I k k, t t 0 ; P R,k (t 0 ) L; (37a) t d R,k (t) > n k + 1, t t 0 + 1; d R,k (s)ds > k, t 1 t t 1, t 1, t [t 0, t 0 + 1] (37b) e P R,k (t)a k (t, q, x, e, y)e + 1 e P R,k (t)e d R,k (t)e P R,k (t)e, t t 0, q Q R (t), x R n, e ker H k, y Y R (t) : x β(t, R), e ξ, e P R,k (t)e g(t) (37c) with H k, A k (,,,, ), Y R ( ) and Q R ( ) as given in (36a), (36b), (9) and (8), respectively. By taking into account (36), it follows that the mappings H n and A n (,,,, ) coincide with H and A(,,,, ) as given by (33) and (35b), respectively, hence, A is a consequence of Claim 1 with H := H n and A(,,,, ) := A n (,,,, ) and with d R := d R,n and P R := P R,n as given in (37b), (37a). Indeed, relations (15a), (15c) follow directly from (37a), (37c) and both inequalities of (15b) are a consequence of (37b) with k = n, if we take into account that c 1 = 1 and c = n. Proof of Claim 1 for k := : For reasons of notational simplicity, we may assume that n > 3. In that case we may define: H := (1, 0), e := (e n 1, e n ) R (38a) A (t, q, x, e, y) := ( ) qn 1,n 1 δ n 1,n (t, y, x,..., x n, 0,..., 0, e n 1, e n ) q n,n 1 q n,n (38b) Also, consider the constants L, R and the function g( ) as above and let ξ 1 and t 0 0. We establish existence of a time-varying symmetric matrix P R, := P R,ξ,t0, C 1 ([t 0, ); R ) and a mapping d R, := d R,ξ,t0, C 0 ([t 0, ); R) in such a way that P R, (t) > I, t t 0 ; P R, (t 0 ) L; (39a) t d R, (t) > n 1, t t 0 + 1; d R, (s)ds >, t 1 t t 1, t 1, t [t 0, t 0 + 1]; (39b) e P R, (t)a (t, q, x, e, y)e + 1 e P R, (t)e d R, (t)e P R, (t)e, t t 0, q Q R (t), x R n, e = (e n 1, e n ) R, y Y R (t) : x β(t, R), e ker H, e ξ, e P R, (t)e g(t) (39c) 39

q 1,1 δ 1, (t, y, x, e ) 0 0. q,1 q, δ,3 (t, y, x, x 3, e, e 3 )... A(t, q, x, e, y) :=...... 0 q n 1,1 q n 1, q n 1,3 δ n 1,n (t, y, x,..., x n, e,..., e n ) q n,1 q n, q n,3 q n,n (35b) with H, A (,,,, ), Y R ( ) and Q R ( ) as given in (38a), (38b), (9) and (8), respectively. Define: ( ) pr,1 (t) p P R, (t) := R (t), t t p R (t) L 0 (40) for certain p R,1, p R C 1 ([t 0, ); R), to be determined in the sequel and notice, that due to (38a) and (40), we have: {e ker H : e ξ and e P R, (t)e g(t)} = {e = (0, e n ) : g(t)/l e n ξ} (41) Then, by taking into account (38), (40) and (41), the desired (39c) is written: p R (t)δ n 1,n (t, y, x,..., x n, 0,..., 0, e n ) + Lq n,n Ld R, (t), t t 0, q Q R (t), x R n, e = (0, e n ) R, y Y R (t) : x β(t, R), g(t)/l e n ξ (4) By invoking (8), (9), (34) and the equivalence between (39c) and (4), it follows that, in order to prove (39c), it suffices to determine p R,1, p R C 1 ([t 0, ); R) and d R, C 0 ([t 0, ); R) in such a way that (39a) and (39b) are fulfilled, and further: p R (t)δ n 1,n (t, y, x,..., x n, 0,..., 0, e n ) + Lσ R (t) Ld R, (t), t t 0, (y; x,..., x n ; 0,..., 0, e n ) Q R,n 1 (t, g(t)/l) (43) We also require, that the candidate function p R ( ) satisfies: p R (t) 0, t t 0 ; p R (t 0 ) = 0 (44) Then, by taking into account (31), (44) and (43), it suffices to prove: p R (t)d R,n 1 (t, g(t)/l) + Lσ R (t) Ld R, (t), t t 0 (45) for certain p R,1, p R C 1 ([t 0, ); R) and d R, C 0 ([t 0, ); R) satisfying (39a), (39b) and (44). Construction of p R and d R, : First, notice that the mapping t D R,n 1 (t, g(t)/l), t t 0 is continuous, and due to (31) we can find a function µ C 1 ([t 0, ); R), satisfying: Let 0 < µ (t) D R,n 1 (t, g(t)/l), for every t t 0 (46) M := max{σ R (t) : t [t 0, t 0 + 1 { } ]} 1 τ := min, 1 M (47a) (47b) and define θ := θ R,ξ,t0 C 1 ([t 0, ); R), p R C 1 ([t 0, ); R) and d R, C 0 ([t 0, ); R) as follows: := 0, t = t 0 θ(t) [0, 1], t [ ] t 0, t 0 + τ := 1, t [ t 0 + τ, ) (48) p R (t) := θ(t) L(n + σ R(t)), t t 0 (49) µ (t) := M, t [ t 0, t 0 + τ ] d R, (t) [ M, n], t [ t 0 + τ, t ] 0 + τ (50) := n, t [t 0 + τ, ) We show that (39b), (44) and (45) are fulfilled, with p R ( ) and d R, ( ) as given by (49) and (50), respectively. Indeed, (44) follows directly by recalling (46), (48) and (49). Both inequalities of (39b) are a direct consequence of (47b) and (50). We next show that (45) holds as well, with p R ( ) and d R, ( ) as above and consider two cases: Case 1: t [t 0, t 0 + τ ]. In that case, (45) follows directly from (44), (46), (47) and (50). Case : t [t 0 + τ, ). Then from (46), (48), (49) and (50) it follows that: L(n + σ R(t)) D R,n 1 (t, g(t)/l) + Lσ R (t) µ (t) Ln Ld R, (t) namely, (45) again holds for all t [t 0 + τ, ). We therefore conclude that (45) is fulfilled for all t t 0. Finally, the construction of p R,1 ( ) included in (40), is the same with that given in proof of Theorem 1.1 in [3] and is omitted. This completes the proof of Claim 1 for k =. Proof of Claim 1 (general step of induction procedure): Assume now that Claim 1 is fulfilled for certain integer k with k < n. We prove that Claim 1 also holds for k := k + 1. Consider the pair (H, A) as given in (19) with H(t) := H k, A(t, q, x, e, y) := A k (t, q, x, e, y), l = n(n+1), m := k, n := n and n := 1, where H k and A k are defined by (36a) and (36b), respectively. Notice, that the map A k as given by (36b) has the form (10) and due to S1 and S, satisfies Property P1. Hence, by the first inequality of (37a) and (37c), we conclude that Hypothesis.3 of the previous section holds, with R and g( ) as above, Y R ( ), Q R ( ) and β R ( ) := β(, R) as given in (9), (8) and (4), respectively, and with d R ( ) := d R,k ( ) and P R ( ) := P R,k ( ) as given in (37a), (37b). Finally, for every ξ 1 and t 0 0, consider the function d R,k := d R,ξ,t0,k defined as: d R,k (t) := d R,k (t) 1, t t 0 (51) 330

q n k,n k δ n k,n k+1 (t, y, x,..., x n k+1, 0,..., 0, e n k+1 ) 0 0 q n k+1,n k A k+1 (t, q, x, e, y) =. A k (t, q, x, ê, y) q n,n k (56) which satisfies dr,k (t) < d R,k (t) for all t t 0. It follows that all requirements of Lemma.1 are fulfilled and therefore, there exists a function ϕ R,k := ϕ R,ξ,t0,k C 1 ([t 0, ); R >0 ) such that e P R,k (t)a k (t, q, x, e, y)e + 1 e P R,k (t)e ϕ R,k (t) H k e d R,k (t)e P R,k (t)e, t t 0, q Q R (t), x R n, e R k, y Y R (t) : x β(t, R), e ξ, e P R,k (t)e g(t) (5) Furthermore, due to (37b) and (51), the map d R,k ( ) satisfies: t d R,k (t) > n k + 1, t t 0 + 1; dr,k (s)ds > (k + 1 ), t 1 t t 1, t 1, t [t 0, t 0 + 1] (53) In the sequel, we exploit (5) and (53), in order to establish that Claim 1 is fulfilled for k = k + 1. Specifically, for the same L, R and g( ) as above and for any ξ 1 and t 0 0, we show that there exist a time-varying symmetric matrix P R,k+1 C 1 ([t 0, ); R (k+1) (k+1) ) and a map d R,k+1 C 0 ([t 0, ); R), such that both (37a) and (37b) are fulfilled with k = k + 1 and further: e P R,k+1 (t)a k+1 (t, q, x, e, y)e + 1 e P R,k+1 (t)e d R,k+1 (t)e P R,k+1 (t)e, t t 0, q Q R (t), x R n where e ker H k+1, y Y R (t) : x β(t, R), e ξ, e P R,k+1 (t)e g(t) (54) H k+1 :=(1, 0,..., 0) R 1 (k+1), e := (e n k ; ê ) R R k, ê :=(e n k+1,..., e n ) R k (55a) the components of the map A k+1 : R 0 R l R n R k+1 R R (k+1) (k+1) are defined as: (A k+1 (t, q, x, e, y)) i,j (55b) := q n k 1+i,n k 1+j, for j i := δ n k 1+i,n k 1+j (t, y, x,..., x n k+i, 0,..., 0, e n k,..., e n k+i ), for j = i + 1; k < n := δ n k 1+i,n k 1+j (t, y, x,..., x n k+i, e,..., e n k+i ), for j = i + 1; k = n, n 1 := 0, for j > i + 1 p R,1 (t) p R (t) 0 0 p R (t) P R,k+1 (t) := 0 P R,k (t). 0 (55c) and where Y R ( ), Q R ( ) are given in (9) and (8), respectively. Again, for reasons of notational simplicity, we may assume that k < n 1. It then follows from (36b) and (55b) that for every e = (0, ê ) = (0, e n k+1,..., e n ) ker H k+1 := (1, 0,..., 0) R 1 (k+1), the map A k+1 (,,,, ) takes the form (56) on top of the page. Let ξ 1 and t 0 0. We determine functions p R,1, p R C 1 ([t 0, ); R) and d R,k+1 C 0 ([t 0, ); R) such that (37a) and (37b) are fulfilled with k = k+1, and further (54) holds, with H k+1, A k+1 (,,,, ) and P R,k+1 ( ) as given by (55). By taking into account (56), it follows that (54) is equivalent to: e n k+1p R (t)δ n k,n k+1 (t, y, x,..., x n k+1, 0,..., 0, e n k+1 ) + ê P R,k (t)a k (t, q, x, ê, y)ê + 1 ê P R,k (t)ê d R,k+1 ê P R,k (t)ê, t t 0, q Q R (t), x R n, e := (e n k ; ê ) R R k, y Y R (t) : x β(t, R), e ker H k+1, e ξ, e P R,k+1 (t)e g(t) (57) Notice that, according to (55a) and (55c), we have e P R,k+1 (t)e = ê P R,k (t)ê for every e = (0, ê ) = (0, e n k+1,..., e n ) ker H k+1, thus, by taking into account (36a) and (5), it suffices, instead of (57), to show that e n k+1(p R (t)δ n k,n k+1 (t, y, x,..., x n k+1, 0,..., 0, e n k+1 ) + ϕ R,k (t)) ( d R,k (t) d R,k+1 (t))ê P R,k (t)ê, t t 0, x R n, ê R k, y Y R (t) : x β(t, R), ê ξ, ê P R,k (t)ê g(t) (58) Establishment of (58) plus (37a) and (37b) for k = k + 1: We impose the following additional requirements for the candidate functions p R ( ) and d R,k+1 ( ): p R (t) 0, t t 0 ; p R (t 0 ) = 0; d R,k+1 (t) d R,k (t), t t 0 (59a) (59b) Then, by taking into account (59b) and the fact that the desired inequality in (58) should be valid for those ê R k for which ê ξ and ê P R,k (t)ê g(t), it follows that, in order to show (58) and that (37a), (37b) are valid with k = k + 1, it suffices to show that e n k+1(p R (t)δ n k,n k+1 (t, y, x,..., x n k+1, 0,..., 0, e n k+1 ) + ϕ R,k (t)) ( d R,k (t) d R,k+1 (t))g(t), t t 0, x R n, e n k+1 R, y Y R (t) : x β(t, R), e n k+1 ξ (60) for suitable functions p R,1, p R C 1 ([t 0, ); R) and d R,k+1 C 0 ([t 0, ); R), in such a way that (37a), (37b) hold with k = k + 1, and in addition p R ( ) and d R,k+1 ( ) 331

satisfy (59). We proceed to the explicit construction of these functions. Due to (9), (31), (34), (59a) and the fact that, due to requirement (59b), equation (60) holds trivially for e n k+1 = 0, it suffices, instead of (60), to show that r (p R (t)d R,n k (t, r) + ϕ R,k (t)) ( d R,k (t) d R,k+1 (t))g(t), t t 0, r (0, ξ] (61) Construction of the mappings p R and d R,k+1 : Let { M k+1 := max d R,k (t) + 1 } 4 + ξ ϕ R,k (t) : t [t 0, t 0 + 1 g(t) ] (6a) { 1 τ k+1 := min, 1 } 4M k+1 (6b) and consider a function θ := θ R,ξ,t0 C 1 ([t 0, ); R) defined as: := 0, t = t 0 θ(t) [0, 1], t [t 0, t 0 + τ k+1 ] (63) := 1, t [t 0 + τ k+1, ) By taking into account (6), it follows that: d R,k (t) 1 4 M k+1, t [t 0, t 0 + τ k+1 ] (64) hence, by exploiting (64), we can construct a function d R,k+1 C 0 ([t 0, ); R), satisfying: d R,k+1 (t) := M k+1, t [t 0, t 0 + τ k+1 ] [ M k+1, d R,k (t) 1 4 ], t [t 0 + τ k+1, t 0 + τ k+1 ] := d R,k (t) 1 4, t [t 0 + τ k+1, ) (65) Notice that (59b), follows from (64) and (65). Also, define: ζ(t) := 1 g(t) ϕ R,k (t) + 1, t t 0 (66) Due to (30) and (31), the map t D R,n k (t, ζ(t)), t t 0 is continuous and there exists a function µ k+1 C 1 ([t 0, ); R) satisfying: 0 < µ k+1 (t) D R,n k (t, ζ(t)), for every t t 0 (67) Finally, define p R C 1 ([t 0, ); R) as: p R (t) := θ(t)ϕ R,k(t), t t 0 (68) µ k+1 (t) which due to (63) and (67), satisfies (59a). Proof of (61): We consider two cases: Case 1: t [t 0, t 0 + τ k+1 ]. By taking into account (6a), it follows that M k+1 d R,k (t) + ϕ R,k(t)ξ g(t) for every t [t 0, t 0 + 1 ], which in conjunction with (6b) and (65) imply: d R,k (t) d R,k+1 (t) ϕ R,k(t)ξ [, t t 0, t 0 + τ ] k+1 g(t) (69) Hence, from (31), (59a) and (69) we deduce that r (p R (t)d R,n k (t, r) + ϕ R,k (t)) r ϕ R,k (t) ξ ϕ R,k (t) g(t) ( g(t) d R,k (t) d R,k+1 (t))g(t), r (0, ξ] which implies (61) for t [t 0, t 0 + τ k+1 ]. Case : t [t 0 + τ k+1, ). We consider two further subcases. Subcase 1: 0 < r ζ(t). Due to (65), it holds d R,k (t) d R,k+1 (t) 1 4 for every t [t 0 + τ k+1, ), hence, by exploiting (31), (59a) and (66) we have r (p R (t)d R,n k (t, r) + ϕ R,k (t)) ζ (t)ϕ R,k (t) ( d R,k (t) d R,k+1 (t))g(t) (70) Subcase : ζ(t) r ξ. By taking into account (59b), (63), (3), (66), (67) and (68), we deduce that r (p R (t)d R,n k (t, r) + ϕ R,k (t)) r ϕ R,k (t) ( D ) R,n k(t, ζ(t)) + 1 0 ( µ k+1 (t) d R,k (t) d R,k+1 (t))g(t) The latter, in conjunction with (70) asserts that (61) is fulfilled for every t [t 0 + τ k+1, ). Both cases above guarantee that (61) holds for all t [t 0, ) as required. Proof of (37a) and (37b) for k = k + 1: The proof of (37b) is the same with that given in proof of Theorem 1.1 in [3] and is omitted. Finally, the proof of (37a) is based on the construction of the map p R,1 ( ) as involved in (55c), and is also the same with that given in proof of Theorem 1.1 in [3]. We have shown that all requirements of Claim 1 hold, which, as was pointed out, establishes that for every R > 0 Hypothesis A is fulfilled. We therefore conclude, that for system (1) Hypothesis. is satisfied hence, by invoking the result of Proposition., it follows that the SODP is solvable for (1) with respect to R n. The establishment of the second statement of Theorem 1.1, follows directly from Proposition.1. Example: Consider the two dimensional system ẋ 1 = x 3 a 1 (t, x 1 ) + x a (t, x 1 ) + a 3 (t, x 1 ); ẋ = x b(t, x 1 ); y = x 1, where a i (t, x 1 ) > 0, i = 1,, 3, (t, x 1 ) R 0 R and assume that there exist C N and a constant L > 0 such that a i (t, x 1 ) C(t)( x 1 + L), i = 1,, 3; b(t, x 1 ) L, (t, x 1 ) R 0 R. It is easy to check that the system satisfies all conditions of Theorem 1.1, therefore the SODP is solvable with respect to R. REFERENCES [1] V. Andrieu, L. Praly, and A. Astolfi, High gain observers with updated gain and homogeneous correction terms, Automatica, vol. 45, pp. 4-48, 009. [] D. Boskos, and J. Tsinias, Sufficient conditions on the existence of switching observers for nonlinear time-varying systems, Eur. J. Control, vol. 19, pp. 87-103, 013. [3] D. Boskos, and J. Tsinias, Observer design for nonlinear triangular systems with unobservable linearisation, Int. J. Control, vol 86, pp. 71-739, 013. [4] D. Chen, and W. Lin, On p-normal Forms of Nonlinear Systems, IEEE Trans. Automat. Control, vol. 48, pp. 14-148, 003. [5] J. P. Gauthier, and I. Kupka, Deterministic Observation Theory and Applications. Cambridge: Cambridge University Press, 001. [6] H. Hammouri, B. Targui, and F. Armanet, High gain observer based on a triangular structure, Int. J. Robust Nonlinear Control vol. 1, pp. 497-518, 00. [7] I. Karafyllis, Non-uniform in time robust global asymptotic output stability, Systems Control Lett., vol. 54, pp. 181-193, 005. [8] W. Respondek, Transforming a Single-Input System to a p-normal Form via Feedback, in Proc. 4nd IEEE Conf. Decision Control, Maui, HI, 003, pp. 1574-1579. 33