Topological classification of linear control systems—An elementary analytic approach

J. Math. Anal. Appl. 402 (2013) 84–102

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Topological classification of linear control systems—An elementary analytic approach✩ Jing Li a , Zhixiong Zhang b,∗ a

School of Economic Mathematics, Southwestern University of Finance and Economics, Chengdu 610074, China

b

School of Mathematics, Sichuan University, Chengdu 610064, China

article

abstract

info

Article history: Received 24 August 2011 Available online 14 January 2013 Submitted by David Russell

In this paper, we present a unified treatment on the linear, differentiable and topological equivalences for time-invariant linear control systems governed by ordinary differential equations (ODEs). Especially, we give a new solution to the topological classification problem for linear control systems by means of an elementary analytic approach, which differs considerably from J.C. Willems’ algebraic method developed in (Willems, 1980) [17]. More precisely, based on P. Brunovsky’s results in (Brunovsky, 1970) [2], we determine all topological invariants for the linear control system. As an interesting by-product, we show that there exist only finitely many topological equivalence classes for any stabilizable system with given numbers of state and control variables, which might be useful for some engineering applications. © 2013 Elsevier Inc. All rights reserved.

Keywords: Linear control systems Classification Topological equivalence Differentiable equivalence Linear equivalence

1. Introduction and main results In this paper, we study the classification problem for the following time-invariant linear control system x˙ (t ) = Ax(t ) + Bu(t ),

t ≥ 0,

(1.1)

where x˙ = dx , x(t ) ∈ Rn (n ∈ N) is the state variable, u(t ) ∈ Rm (m ∈ N) is the control variable, and A and B are real matrices dt of dimensions n × n and n × m, respectively. We choose the admissible control set to be L1loc ([0, +∞); Rm ). Since (1.1) is uniquely determined by the pair of matrices A and B, we also denote it simply by (A, B). As in other branches of mathematics, classification of control systems is a fundamental problem in mathematical control theory. Indeed, elements in the same equivalent class have some similar properties (such as controllability, stabilizability and the number of the efficient controls of control systems in our case). Therefore, in order to know the properties of all elements in the same class, we only need to study a special element in this class (e.g., the system with the Brunovsky canonical form that will be introduced later). The classification problem for completely controllable linear systems was studied by P. Brunovsky [2]. In [2], P. Brunovsky introduced the concept of feedback equivalence (see also Definition 2.1 in Section 2) and showed that there are only finitely many feedback equivalence classes and each of them can be represented by a considerably simple canonical form. Later, B.C. Liang [8] extended this result to general linear control systems (which may be not completely controllable). There exist other further works on the classification of control systems by means of the linear equivalence transformation (cf. [1,9,12]).

✩ This work was partially supported by the PCSIRT of the Ministry of Education of China under grant IRT1273 and the NSF of China under grants 11231007 and 11001188. ∗ Corresponding author. E-mail address: [email protected] (Z. Zhang).

0022-247X/$ – see front matter © 2013 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2013.01.011

J. Li, Z. Zhang / J. Math. Anal. Appl. 402 (2013) 84–102

85

As for the nonlinear and smooth transformations, we refer readers to [5,16]. In this paper, we mainly consider the nonsmooth transformation, i.e., the equivalence transformation is only continuous. Let us consider the following two systems x˙ (t ) = A1 x(t ) + B1 u(t )

(1.2)

and y˙ (t ) = A2 y(t ) + B2 v(t ).

(1.3)

Here, x(t ), y(t ) ∈ R are state variables, u(t ), v(t ) ∈ R are control variables, and Ai , Bi (i = 1, 2) are real matrices of dimensions n × n and n × m, respectively. We introduce the following definition of linear, differentiable and topological equivalences. n

m

Definition 1.1. (1) Systems (1.2) and (1.3) are said to be topologically equivalent if there exists a (vector-valued) function F (x, u) := (H (x, u), G(x, u)), where H (x, u) ∈ C (Rn × Rm ; Rn ) and G(x, u) ∈ C (Rn × Rm ; Rm ), such that (i) F (·, ·) is a homeomorphism from Rn × Rm to itself (henceforth we denote the inverse function of F (x, u) by F −1 (y, v) := (Z (y, v), W (y, v))); (ii) Transformation (y(t ), v(t )) = (H (x(t ), u(t )), G(x(t ), u(t ))) brings (1.2) to (1.3), and transformation (x(t ), u(t )) = (Z (y(t ), v(t )), W (y(t ), v(t ))) brings (1.3) to (1.2). (2) Systems (1.2) and (1.3) are said to be differentiably (resp., linearly) equivalent if the above function F (x, u) is a C 1 diffeomorphism (resp., a linear isomorphism) from Rn × Rm to itself. Several remarks are in order. Remark 1.1. ‘‘Transformation (y(t ), v(t )) = (H (x(t ), u(t )), G(x(t ), u(t ))) brings (1.2) to (1.3)’’ means that: if x(t ) is the solution of (1.2) with initial datum x(0) = x0 and control u(t ) ∈ L1loc ([0, +∞); Rm ) being right-continuous at t = 0, then by transformation F (·, ·) introduced in Definition 1.1, y(t ) = H (x(t ), u(t )) is the solution of (1.3) with initial datum y(0) = H (x(0), u(0)) = H (x0 , u(0)) and control v(t ) = G(x(t ), u(t )), provided that v(t ) ∈ L1loc ([0, +∞); Rm ). Remark 1.2. It is easy to check that the linear, differentiable and topological equivalences are equivalence relations, by which we mean, as usual, they are symmetric, reflexive and transitive. In the sequel, we call (H (x, u), G(x, u)) the equivalence transformation from system (1.2) to system (1.3). Also, it is clear that the following relation holds: Linear equivalence ⇒ Differentiable equivalence ⇒ Topological equivalence. Remark 1.3. From Proposition 2.1 in the first subsection of Section 2, we know that actually H (x, u) ≡ H (x) and Z (y, v) ≡ H −1 (y) in Definition 1.1. For this reason, we shall analyze the equivalence classes of system (1.1) in the sense of Definition 1.1 with H (x, u) replaced by H (x). Remark 1.4. By Theorem 1.1 to appear later in this section, we conclude that the linear equivalence defined in Definition 1.1 is actually the same as the feedback equivalence introduced by P. Brunovsky in [2] in 1970. For the reader’s convenience, we restate Brunovsky’s definition as Definition 2.1 in the third subsection of Section 2. Hence, due to Theorem 1.1 and Remark 1.2, we can take the canonical form of system (1.1) in the sense of Brunovsky’s feedback equivalence as our starting point for further discussion. We know from [2,8] that any system (A, B) is feedback equivalent (i.e., linearly equivalent) to a system ( A,  B) in the form:

 C  A=



  D  B= ,

0 , M

0

0

where M is a (n − k) × (n − k) (k is the Kalman rank of (A, B)) Jordan matrix in the form M− M :=  0 0



0 M+ 0



0 0 , M0

(1.4)

for which the real parts of eigenvalues of M − , M + and M 0 are negative, positive and zero, respectively. Denote by n− , n+ and n0 , respectively, the numbers of matrix M’s eigenvalues with negative real parts, positive real parts and zero real parts. C and D are respectively k × k and k × m matrices in the forms

.

.. .

··· ··· .. .

0

0

···

J

p1

0 C :=   ..

0 Jp2

0 0 



, ..  . 

Jpr0

0 ep2

0 0

··· ···

.. .

0 0

0 0

.

··· ··· .. .

0

0

···

epr0

0

···

0

e

p1

0 D :=   ..

.. .

.. .



, ..  .

86

J. Li, Z. Zhang / J. Math. Anal. Appl. 402 (2013) 84–102

where Jν and eν are the following ν × ν and ν × 1 matrices 0 0



. Jν :=   .. 0

1 0

0 1

.. .

··· ··· .. .

0 0

0 0

··· ···

.. .

0

pi m i =1

respectively; { } definitions).

0 0



0

   ..  . eν :=   ,

..   . ,  1

0 1

0

= P (A, B) are the Brunovsky indices, or Kronecker indices, controllability indices (see Section 2 for their

Remark 1.5. In the sequel, we call the pair ( A,  B) as the Brunovsky canonical form of (A, B). Clearly, the Brunovsky canonical form ( A,  B) is composed of a completely controllable subsystem

ξ˙ (t ) = C ξ (t ) + Du(t ) and a completely uncontrollable subsystem, i.e. an ODE

η( ˙ t ) = M η(t ). Similarly, systems (Ai , Bi ) can be linearly equivalent to the following Brunovsky canonical forms:

( Ai ,  Bi ) =





Ci 0

0 , Mi

  Di 0

,

i = 1, 2,

(1.5)

where Ci , Di and Mi are ki × ki , ki × m and (n − ki ) × (n − ki ) matrices, respectively. Our main results in this article are stated as follows. Theorem 1.1. Systems (1.2) and (1.3) are linearly equivalent if and only if there are matrices O, Q and L of dimensions n × n, m × m and m × n, respectively, with O and Q being nonsingular, such that A2 = O−1 A1 O + O−1 B1 L,

B2 = O−1 B1 Q .

(1.6)

Theorem 1.2. Assume that the Brunovsky canonical forms of systems (1.2) and (1.3) are given by (1.5). Then systems (1.2) and (1.3) are linearly (resp., differentiably, topologically) equivalent if and only if completely controllable subsystems x˙ 1 (t ) = C1 x1 (t ) + D1 u(t ),

y˙ 1 (t ) = C2 y1 (t ) + D2 v(t )

and ODE subsystems x˙ 2 (t ) = M1 x2 (t ),

y˙ 2 (t ) = M2 y2 (t )

are linearly (resp., differentiably, topologically) equivalent respectively. Theorem 1.3. Assume that the Brunovsky canonical forms of systems (1.2) and (1.3) are given by (1.5). Then (i) systems (1.2) and (1.3) are linearly or differentiably equivalent if and only if R(A1 , B1 ) = R(A2 , B2 )

(or equivalently P (A1 , B1 ) = P (A2 , B2 )),

(1.7)

and matrices M1 and M2 are similar; (ii) systems (1.2) and (1.3) are topologically equivalent if and only if R(A1 , B1 ) = R(A2 , B2 )

(n1 , n1 , −

and matrices

+

n01

M10

(or equivalently P (A1 , B1 ) = P (A2 , B2 )),

) = (n2 , n2 ,

and

−

M20

+

n02

),

(1.8) (1.9)

are similar.

The definition of R(A, B) in Theorem 1.3 will be given later in Section 2. Theorem 1.3 indicates that for system (1.1), the differentiable classification is the same as its linear classification. In addition, when system (1.1) is completely controllable, there is no difference between the linear, differentiable and topological classifications. However, when system (1.1) is not completely controllable, it is the ODE subsystem that brings difference between the topological classification and the linear classification, as illustrated by the following example.

J. Li, Z. Zhang / J. Math. Anal. Appl. 402 (2013) 84–102

87

Example 1.1. Systems x˙ (t ) =



0 0



 

0 1 x(t ) + u( t ) a 0

and y˙ (t ) =



0 0



 

0 1 y(t ) + v(t ) −1 0

are topologically equivalent, but not linearly equivalent. Here the real number a < 0 and a ̸= −1. Since any ODE can be regarded as a control system without effective controls (i.e. B = 0 in (1.1)), the classification results in this paper also apply to ODEs. For autonomous linear ODEs, our results coincide with those in [7,13]. Also, Theorem 1.1 shows that the linear equivalence defined in this paper is actually the same as the feedback equivalence introduced in [2]. The novelty in our Definition 1.1 for the linear equivalence is that we discuss the equivalence by means of the property of transformations, and therefore, three types of classifications are defined in a unified way. As for the topological equivalence defined in this paper, we remark that it is a little different from the one introduced by J.C. Willems in [17]. In fact, the topological equivalence introduced in [17] is from a point of view of linear flows, and the control variable u is suppressed. By contrast, we consider the pair (x, u), and allow the explicit appearance of u in the equivalence transformation. Hence, J.C. Willems’ definition requires only the existence of a homeomorphism H (x), while in our Definition 1.1, the two control variables u and v are further related by v = G(x, u) for some continuous function G(·, ·). As indicated in the conclusion section of [17], our Definition 1.1 is ‘‘another suitable starting point for classifying linear systems’’. It turns out that our topological classification result coincides with that of J.C. Willems [17]. Nevertheless, unlike J.C. Willems’ algebraic method in [17], we develop in this work a new elementary analytic approach to solve the topological classification problem for linear control systems. It seems that our approach is more suitable for other problems without the usual algebraic structure. The key technical part of this paper is to determine the following two principle topological invariants for system (1.1): (i) the structure of the ODE subsystem, i.e. (n− , n+ , n0 ) and the Jordan matrix M 0 , and (ii) the structure of the completely controllable subsystem, i.e. R(A, B) or P (A, B). The determination of the second topological invariant is the most difficult part, which is equivalent to showing that completely controllable systems are topologically equivalent if and only if they are linearly equivalent. To do this, we need to show first that the following are also topological invariants for system (1.1): (iii) the Kalman rank (and hence the dimension of the controllable subspace); (iv) stabilizability; (v) the rank of the controller B, i.e., the numbers of effective controls. Although these topological invariants are subject to the principle topological invariants for system (1.1), they have remarkable and obvious control meaning. For example, since the stabilizability is a topological invariant for the linear control system governed by the ODE, it is easy to deduce that the number of topological equivalence classes for any stabilizable system is always finite, while the number of smooth equivalence classes for the same system is infinite unless it is completely controllable. This observation should be useful for some engineering application problems. Indeed, from the point of view of engineering, only stabilizable system works well (and therefore is reasonable) in practice. In this paper, we focus on linear control systems governed by ODEs, for which we obtain complete solutions to the linear, differentiable and topological classification problems. For time-variant linear control systems, nonlinear control systems and control systems governed by partial differential equations, or stochastic differential equations, or even stochastic partial differential equations, the classification problems, especially the topological classification problem, deserve much more efforts. We refer the reader to [3,4,10,15,19–22] for more detailed materials in this respect. On the other hand, it seems also quite interesting to study the classification problem for other control problems, such as the general input–output problems, disturbance decoupling problems, optimal control problems, and so on (see [11,14,18], etc. for some background materials). We believe that the analytic approach developed in this article can be adopted to solve some of these problems. The rest of this paper is organized as follows. Some preliminary knowledge is collected in Section 2. In Section 3, we analyze the differentiable equivalence for system (1.1). Section 4 is devoted to giving the proof of Theorem 1.2. Finally, in Section 5, we study the topological classification for linear control systems, and then give a proof of Theorem 1.3. 2. Some preliminaries In this section, we present some useful preliminary results. 2.1. Property of function H (x, u) in Definition 1.1 Proposition 2.1. If F (x, u) = (H (x, u), G(x, u)) is a topological equivalence transformation from system (1.2) to system (1.3), then H (x, u) is independent of u. Furthermore, H (x) := H (x, 0) is a homeomorphism from Rn to Rn .

88

J. Li, Z. Zhang / J. Math. Anal. Appl. 402 (2013) 84–102

Proof. The proof is divided into two steps. Step 1. First of all, we prove that H (x, u) is independent of u by the contradiction argument. Hypotheses that there exist an x0 ∈ Rn and a u0 ∈ Rm \ {0} such that H x0 , u0 ̸= H x0 , 0 .









Denote ε0 = H x0 , u0 − H x0 , 0 . It suffices to find an x1 ∈ Rn satisfying the following two estimates (the existence of x1 will be proved in Step 2)

 







  1 0   H x , u − H x0 , u0  < ε0 ,

(2.1)

  1 0   H x , u − H x0 , 0  < ε0 .

(2.2)

3

3

     2ε Indeed, by (2.1) and (2.2), it follows that H x0 , u0 − H x0 , 0  < 30 , which is a contradiction. Step 2. We now show that there is an x1 ∈ Rn such that (2.1) and (2.2) hold simultaneously.   Since H (x, u) is continuous in x, there exists a constant δ > 0 such that for any x − x0  < δ , it holds

  0   H x, u − H x0 , u0  < ε0 .

(2.3)

3

Construct a control as follows

 0  tu ,  u(t ) = t  0 u ,

t ∈ [0, t ], t ∈ ( t , +∞),

where  t > 0 will be given later. Clearly,

  |u(t )| ≤ u0  ,

∀t ∈ [0, +∞).

(2.4)

The solution of system (1.2) associated to x(0) = x0 and u(t ) is expressed as follows x(t ) = eA1 t x0 +

t



eA1 (t −τ ) B1 u(τ )dτ .

(2.5)

0

By (2.4), one can find a t1 > 0 (which is independent of  t) small enough such that

  x(t ) − x0  < δ,

∀t ∈ [0, t1 ].

(2.6)

By (2.3) and (2.6), we obtain

     H x(t ), u0 − H x0 , u0  < ε0 ,

∀t ∈ [0, t1 ].

3

(2.7)

On the other hand, from (2.6), we get

  |x(t )| < x0  + δ,

∀t ∈ [0, t1 ].

(2.8)

Since G(x, u) maps a bounded set in R × R to a bounded set in R , (2.4) and (2.8) imply that there exists a constant K > 0 satisfying n

|v(t )| = |G(x(t ), u(t ))| < K ,

m

m

∀t ∈ [0, t1 ].

(2.9)

The solution of system (1.3) corresponding to x(t ) and u(t ) is expressed as follows H (x(t ), u(t )) = y(t ) = eA2 t y0 +

t



eA2 (t −τ ) B2 v(τ )dτ .

(2.10)

0

Taking t = 0 in (2.10), we see that H x0 , u(0) = y0 . Recalling u(0) = 0, (2.10) can be rewritten as



H (x(t ), u(t )) = eA2 t H x0 , 0 +







t



eA2 (t −τ ) B2 v(τ )dτ .

0

The above formula and (2.9) imply that there exists a t2 ∈ (0, t1 ) (which is independent of  t) such that

   H (x(t ), u(t )) − H x0 , 0  < ε0 , 3

∀t ∈ [0, t2 ].

(2.11)

J. Li, Z. Zhang / J. Math. Anal. Appl. 402 (2013) 84–102

89

Setting  t = t2 /2, from the construction of u(t ) we have u(t2 ) = u0 . Taking t = t2 , x1 = x(t2 ) in (2.7) and (2.11), we conclude that inequalities (2.1) and (2.2) hold simultaneously. Step 3. Since H (x, u) is independent of u, we simply denote it as H (x). Then we show that H (x) is a homeomorphism from Rn to Rn . Recall that the inverse transformation of F (x, u) is F −1 (y, v) = (Z (y, v), W (y, v)). Similar to Step 1, we conclude that Z (y, v) ≡ Z (y),

∀y ∈ R n , v ∈ R m .

We claim that H (x) is invertible and H −1 = Z . In fact, since F (x, u) is a homeomorphism from Rn × Rm to itself, we obtain

(x, u) = F −1 (F (x, u)) = (Z (H (x)), W (H (x), G(x, u))),

∀x ∈ Rn , u ∈ Rm ,

which yields x = Z (H (x)),

∀x ∈ Rn .

Similarly, we have y = H (Z (y)),

∀y ∈ Rn .

Hence, H −1 = Z . Furthermore, noting H and Z are continuous functions, we conclude that H (x) is a homeomorphism from Rn to Rn .



Remark 2.1. Proposition 2.1 shows that the state variable and the control variable play different roles in the topological transformation. Indeed, in the present case, it is easy to see that if u changes a lot in a very short time duration, by (2.5) (resp., (2.10)), x (resp., y) might change little. This inspires us to guess that y does not depend on u. Remark 2.2. To prove Proposition 2.1, we only need the facts that G(x, u) maps a bounded set in Rn × Rm to a bounded set in Rm , H (x, u) is continuous with respect to x and Z (y, v) is continuous with respect to y. Remark 2.3. For any topological equivalence transformation (H (x), G(x, u)) from system (1.2) to system (1.3), without loss of generality, we can always assume H (0) = 0 in the following discussion. Otherwise, one may choose (H (x) − H (0), G(x, u) − G(0, 0)) to be the new topological equivalence transformation. 2.2. Classification of linear ordinary differential equations The following result concerning the classification of ODEs is well known in the literature; see e.g. [7,13]. Lemma 2.1. (1) ODEs (M1 , 0) and (M2 , 0) are linearly or differentiably equivalent if and only if matrices M1 and M2 are similar. + − + 0 0 0 0 (2) ODEs (M1 , 0) and (M2 , 0) are topologically equivalent if and only if (n− 1 , n1 , n1 ) = (n2 , n2 , n2 ), and matrices M1 and M2 are similar. Here, Mi (i = 1, 2) are matrices in the form of (1.4). 2.3. Brunovsky’s feedback equivalence and proof of Theorem 1.1 The following definition of feedback equivalence was first introduced by P. Brunovsky [2] for linear control systems. Definition 2.1. Systems (1.2) and (1.3) are called feedback equivalent if there exist matrices O, Q and L of dimensions n × n, m × m and m × n, respectively, with O and Q being nonsingular, such that A2 = O−1 A1 O + O−1 B1 L,

B2 = O−1 B1 Q .

Proof of Theorem 1.1. Sufficiency. It is easy to check that

  y

v

 =

H (x) O−1 = G(x, u) −Q −1 LO−1





0 Q −1

  x u

is the equivalence transformation from system (1.2) to system (1.3). Therefore these two systems are linearly equivalent. Necessity. Assume systems (1.2) and (1.3) are linearly equivalent and denote the equivalence transformation by

  y

where

v 

L1 L2

 = 0 L3

H (x) L = 1 G(x, u) L2







0 L3

 

x , u

is a nonsingular (n + m) × (n + m) constant matrix. Hence L1 and L3 are nonsingular. Set

1 O := L− 1 ,

1 Q := L− 3 ,

1 −1 L := −L− 3 L2 L1 ,

then Ai , Bi (i = 1, 2) satisfy relation (1.6).



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J. Li, Z. Zhang / J. Math. Anal. Appl. 402 (2013) 84–102

For the pair (A, B) (appeared in (1.1)), define the Kalman rank as follows k := rank B, AB, . . . , An−1 B .





Clearly, k ≤ n. It is well known that system (1.1) is completely controllable if and only if k = n (e.g. [6,14,18]). The definition of R(A, B) and P (A, B) can be found in the literature (e.g. [2]). For the sake of completeness, we present it here. First of all, put r0 := rankB, rj := rank B, AB, . . . , Aj B − rank B, AB, . . . , Aj−1 B ,







j = 1, . . . , n − 1.



Define R(A, B) := {rj }jn=−01 . It is easy to check that 0 ≤ rj ≤ m(j = 0, . . . , n − 1) and

n−1 j=0

rj = k. Second, denote by Lj the linear subspace of Rn

spanned by the column vectors of (B, AB, . . . , Aj B . Denote by Λj , the orthogonal complement of Lj−1 in Lj . Denote by πj (b), the orthogonal projection of a vector b ∈ Lj into Λj . Then, rj is the dimension of Λj , which is equal to rank(πj (Aj B)). Noting



that, for any j = 0, . . . , n − 1 and any vectors b, b1 , . . . , bm ∈ Rn , Aj b = j +1

A

b=

j−1 m

ν=1 kµν A

µ=1

µ+1

j−1 m µ=1

ν=1 kµν A

µ

bν (for kµν ∈ R) implies

bν , we see that

r0 ≥ r1 ≥ · · · ≥ rn−1 , and we can find k linear independent vectors from the column vectors of B, AB, . . . , An−1 B in the following way. First, choose r0 linear independent vectors from the column vectors of B to construct a set {bi ; i ∈ I0 }, where I0 ⊂ {1, 2, . . . , m}. Next, for j = 1, we can choose r1 linear independent vectors from the set {Abi ; i ∈ I0 } to construct a set {Abi ; i ∈ I1 } with I1 ⊂ I0 such that the vectors {π1 (Abi ); i ∈ I1 } span Λ1 . Similarly, forj = 2, . . . , n − 1, one can choose rj linear    independent vectors from the set Aj bi ; i ∈ Ij−1 to construct a set Aj bi ; i ∈ Ij with Ij ⊂ Ij−1 such that the vectors πj (Aj bi ); i ∈ Ij span Λj . Put





S := {bi ; i ∈ I0 } ∪ {Abi ; i ∈ I1 } ∪ · · · ∪ An−1 bi ; i ∈ In−1 .





From the construction of S , we see that if Aj0 bi0 ̸∈ S for some i0 ∈ {1, . . . , m} and j0 ∈ {0, 1, . . . , n − 1}, then Aj0 +1 bi0 ̸∈ S . Associate each column vector bi of B with a number pi defined by pi := min{j; Aj bi ̸∈ S , 0 ≤ j ≤ n − 1},

i = 1 , . . . , m.

Clearly, A bi ∈ S for 0 ≤ j ≤ pi − 1 but A bi ̸∈ S whenever pi > 0, while pi = 0 means that bi can be linearly expressed by the vectors in {bi ; i ∈ I0 }. By re-ordering suitably the columns of B (if necessary), one can achieve that j

pi

p1 ≥ p2 ≥ · · · ≥ pm . Define P (A, B) := {pi }m i =1 . The following two well-known results are also due to [2]. Lemma 2.2. For any pair (A, B), the finite sequences R(A, B) = {rj }nj=−01 and P (A, B) = {pi }m i=1 have the following properties: (1) 0 ≤ rj ≤ m, r0 ≥ r1 ≥ · · · ≥ rp1 −1 > 0, rj = 0 for j ≥ p1 ,

n−1 r = k; m j=0 j (2) 0 ≤ pi ≤ n, p1 ≥ p2 ≥ · · · ≥ pr0 > 0, pi = 0 for i > r0 , i=1 pi = k; (3) P (A1 , B1 ) = P (A2 , B2 ) if and only if R(A1 , B1 ) = R(A2 , B2 ). Lemma 2.3. Assume that systems (1.2) and (1.3) are completely controllable. Then they are feedback equivalent if and only if R(A1 , B1 ) = R(A2 , B2 ) (or equivalently P (A1 , B1 ) = P (A2 , B2 )). 3. Differentiable classification of linear control systems Proposition 3.1. Systems (1.2) and (1.3) are differentiably equivalent if and only if they are linearly equivalent. Proof. By Remark 1.2, it suffices to show the ‘‘only if’’ part. Denote by (H (x), G(x, u)) the differentiable equivalence transformation from system (1.2) to system (1.3). The solutions of systems (1.2) and (1.3) can be expressed respectively by x(t , x ) = e 0

A1 t 0

t



x + 0

eA1 (t −τ ) B1 u(τ )dτ

J. Li, Z. Zhang / J. Math. Anal. Appl. 402 (2013) 84–102

91

and y(t , y0 ) = H (x(t , x0 )) = eA2 t y0 +

t



eA2 (t −τ ) B2 G(x(τ , x0 ), u(τ ))dτ .

(3.1)

0

Taking t = 0 in (3.1), one sees that H x0 = y0 . Hence, (3.1) can be rewritten as

 

H (x(t , x0 )) = eA2 t H x0 +

 

t



eA2 (t −τ ) B2 G(x(τ , x0 ), u(τ ))dτ .

(3.2)

0

Setting u(t ) ≡ 0, differentiating (3.2) with respect to x0 , and then setting x0 = 0, we arrive at H ′ (0)eA1 t = eA2 t H ′ (0) +



t

eA2 (t −τ ) B2 Dx G(0, 0)eA1 τ dτ .

(3.3)

0

Furthermore, differentiating (3.3) with respect to t and setting t = 0, we get H ′ (0)A1 = A2 H ′ (0) + B2 Dx G(0, 0).

(3.4)

Similarly, setting x = 0 and u(t ) ≡ u , differentiate (3.2) with respect to t, and then setting t = 0, it follows that 0

0

H ′ (0)B1 u0 = A2 H (0) + B2 G 0, u0 .





Differentiating the above formula with respect to u0 and setting u0 = 0, we find H ′ (0)B1 = B2 Du G(0, 0).

(3.5)

Since F (x, u) is a diffeomorphism from R × R to itself, we see that F D(x,u) F (x, u) = I, which implies n

m

−1

(F (x, u)) = (x, u). Consequently, D(y,v) F

−1

(y, v)·

det D(y,v) F −1 (y, v) × det D(x,u) F (x, u) = 1. From this formula, we deduce that the matrix H ′ (0) D(x,u) F (0, 0) = Dx G(0, 0)





0 Du G(0, 0)

is nonsingular. Hence, both H ′ (0) and Du G(0, 0) are nonsingular. Therefore, (3.4) and (3.5) yield A1 = H ′ (0)−1 A2 H ′ (0) + H ′ (0)−1 B2 Dx G(0, 0),

(3.6)

B1 = H ′ (0)−1 B2 Du G(0, 0).

(3.7)

and

Combining (3.6) and (3.7), and noting Theorem 1.1, we arrive at the desired result.



4. Proof of Theorem 1.2 The main purpose of this section is to prove Theorem 1.2. As a byproduct, we obtain some topological invariants for two topological equivalence systems (A1 , B1 ) and (A2 , B2 ). We begin with the following proposition, which shows that the Kalman rank is an invariant under the topological equivalence transformation, and therefore the topological equivalence keeps the controllability of the control system. Denote the Kalman ranks of (A1 , B1 ) and (A2 , B2 ) (i.e. systems (1.2) and (1.3)) by k1 and k2 respectively. That is ki := rank Bi , Ai Bi , . . . , Ani −1 Bi ,





i = 1, 2.

Proposition 4.1. If systems (1.2) and (1.3) are topologically equivalent, then k1 = k2 . Proof. We use the contradiction argument. Without loss of generality, suppose that k1 > k2 . By Remark 1.4, we assume that systems (Ai , Bi )(i = 1, 2) are in the Brunovsky canonical forms (1.5). Denote by (H (x), G(x, u)) the topological equivalence transformation from (1.2) to (1.3). Let x(0) = 0 and fix any t3 > 0. Since system (1.2) is composed by a completely controllable system (C1 , D1 ) and an ODE (M1 , 0), it is easy to see that

   Θ1 := x (t3 ; 0, u(·)) ; u(·) ∈ C [0, +∞); Rm = Rk1 × {0} ⊂ Rn .

(4.1)

Similarly, for the initial datum y(0) = 0, one finds

   Θ2 := y(t3 ; 0, v(·)); v(·) ∈ C [0, +∞); Rm = Rk2 × {0} ⊂ Rn .

(4.2)

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J. Li, Z. Zhang / J. Math. Anal. Appl. 402 (2013) 84–102

Note that H : Θ1 → H (Θ1 ) is a homeomorphism. Hence (4.1) implies that H (Θ1 ) is a k1 -dimensional topological manifold. On the other hand, by Remark 2.3 and noting that transformation (y(t ), v(t )) = (H (x(t )), G(x(t ), u(t ))) brings system (1.2) to system (1.3), one has H (Θ1 ) ⊂ Θ2 . From (4.2), we conclude that the dimension of H (Θ1 ) is at most k2 . Noting however that k1 > k2 , we arrive at a contradiction.  Thanks to Proposition 4.1, when systems (1.2) and (1.3) are topologically equivalent, we can denote their common Kalman rank as follows k := rank Bi , Ai Bi , . . . , Ani −1 Bi ,

i = 1, 2.





Assume that the Brunovsky canonical forms of systems (1.2) and (1.3) are given by (1.5). Naturally, we introduce the following two systems determined by (1.5): x1 ( t ) D + 1 u(t ) 0 x2 ( t )

(4.3)

y1 (t ) D + 2 v(t ), y2 (t ) 0

(4.4)

C x˙ 1 (t ) = 1 x˙ 2 (t ) 0

0 M1



y˙ 1 (t ) C = 2 y˙ 2 (t ) 0

0 M2















and













where Ci , Di and Mi (i = 1, 2) are real matrices of dimensions k × k, k × m and (n − k) × (n − k), respectively. The following result says that the transformation function H (x) appeared in Definition 1.1 has another important property. Proposition 4.2. Assume that system (4.3) is topologically equivalent to system (4.4), and the topological equivalence transformation from (4.3) to (4.4) is given by

(y1 , y2 , v) = F (x1 , x2 , u) = (h1 (x1 , x2 ), h2 (x1 , x2 ), G(x1 , x2 , u)). Then, y2 = h2 (x1 , x2 ) is independent of x1 and h2 (x2 ) := h2 (0, x2 ) is a homeomorphism from Rn−k to Rn−k . Proof. The proof is divided into two steps. Step 1. We show that h2 (x1 , x2 ) is independent of x1 , that is h2 (x1 , x2 ) ≡ h2 (x2 ),

∀x1 ∈ Rk , ∀x2 ∈ Rn−k .

In fact, if this is not the case, we deduce that there exist vectors a0 ∈ Rk , b0 ∈ Rk and x2 ∈ Rn−k such that h2 a0 , x2 ̸= h2 b0 , x2 .









(4.5)

We use (x1 (t ), x2 (t )) to denote the solution of (4.3) with initial datum x01 , x2 and control u(t ). Denote by ( x1 (t ), x2 (t ))



 0

the solution of (4.3) with initial datum  x01 , x02 and control  u(t ). Similar notations are for system (4.4).





Fix some t4 > 0 and some x01 ∈ Rk . Taking  x01 = x01 and  x02 = x02 = e−M1 t4 x2 , and noting the special structure of (4.3) and (4.4), one finds

 x2 (t4 ) = eM1 t4 x02 = x2 = eM1 t4 x02 = x2 (t4 ).

(4.6)

Since system x˙ 1 (t ) = C1 x1 (t )+ D1 u(t ) is completely controllable, there exist two controls u(t ) and  u(t ) such that x1 (t4 ) = a0 and  x1 (t4 ) = b0 . Hence, combining (4.5) and (4.6), it follows

     y2 (t4 ) = h2 ( x1 (t4 ), x2 (t4 )) = h2 b0 , x2 ̸= h2 a0 , x2 = h2 (x1 (t4 ), x2 (t4 )) = y2 (t4 ).

(4.7)

However, by

 0 0    y02 = h2  x1 , x2 = h2 x01 , x02 = y02 , we obtain

 y2 (t4 ) = eM2 t4 y02 = eM2 t4 y02 = y2 (t4 ), which contradicts (4.7). Therefore, h2 (x1 , x2 ) is independent of x1 . Hence we simply denote h2 (x1 , x2 ) by h2 (x2 ). Step 2. We now show that h2 (x2 ) is a homeomorphism from Rn−k to Rn−k . First we prove that h2 is invertible. Indeed, we denote by (z1 (y1 , y2 ), z2 (y1 , y2 )) the inverse function of (h1 (x1 , x2 ), h2 (x2 )) (recall that, by Proposition 2.1, (h1 (x1 , x2 ), h2 (x2 )) is invertible). Similar to Step 1, one finds z2 (y1 , y2 ) ≡ z2 (y2 ),

∀y1 ∈ Rk , ∀y2 ∈ Rn−k .

J. Li, Z. Zhang / J. Math. Anal. Appl. 402 (2013) 84–102

93

Therefore,

(x1 , x2 ) = (z1 (h1 (x1 , x2 ), h2 (x2 )), z2 (h2 (x2 ))),

∀x1 ∈ Rk , ∀x2 ∈ Rn−k ,

which yields x2 = z2 (h2 (x2 )),

∀x2 ∈ Rn−k .

Similarly, we have y2 = h2 (z2 (y2 )),

∀y2 ∈ Rn−k .

1 Hence h− 2 = z2 . Noting that h2 and z2 are continuous functions, we conclude that h2 (x2 ) is a homeomorphism from Rn−k to Rn−k .



As we shall see later, Proposition 4.2 is the basis of the analysis in the sequel. By Proposition 4.2, we can get the following interesting result, which is Theorem 1.2 in the linear equivalence case. Besides, it seems that this result has not appeared explicitly in the literature. Hence, the study of topological classification may stimulate some new insights even for linear classification problem. Corollary 4.1. Systems (4.3) and (4.4) are linearly equivalent if and only if completely controllable subsystems x˙ 1 (t ) = C1 x1 (t ) + D1 u(t ),

y˙ 1 (t ) = C2 y1 (t ) + D2 v(t )

(4.8)

and ODE subsystems x˙ 2 (t ) = M1 x2 (t ),

y˙ 2 (t ) = M2 y2 (t )

(4.9)

are linearly equivalent respectively. Proof. It suffices  to show the ‘‘only if’’ part. Since systems (4.3) and (4.4) are linearly equivalent, there is a nonsingular n × n

matrix



O1 O3

O2 O4

such that

y1 (t ) O = 1 y2 (t ) O3





O2 O4

x1 (t ) . x2 (t )





Proposition 4.2 implies that O3 = 0 and therefore



C2 0

0 M2



 =  =  =

O1 0

O2 O4



O1 0

O2 O4



C1 0

C1 0

0 M1



0 M1



O1 0

1 O1 C1 O− 1 + O1 D1 L1 0

O1 0

O2 O4

 −1



−1 O1

=

 O2

−1

0

O4



1 −1 (with  O2 = −O− 1 O2 O4 ). It follows from



Theorem 1.1 that there exist an m × n matrix L1





L2 and an m × m nonsingular matrix Q satisfying O2 O4

1 O− 1 0

−1

 O2

1 O− 4

 +



O1 0



O + 1 0

O2 O4



O2 O4



D1  L1 0



L2





D1  L1 0

L2



1 O1 C1 O2 + O2 M1 O− 4 + O1 D1 L2 , −1 O4 M1 O4



and



D2 0





O = 1 0

O2 O4









D1 O1 D1 Q Q = . 0 0

Hence 1 C2 = O1 C1 O− 1 + O1 D1 L1 ,

D2 = O1 D1 Q ,

1 M2 = O4 M1 O− 4 .

Using Theorem 1.1 again, we see that the two systems in (4.8) are linearly equivalent. On the other hand, by Lemma 2.1, the two ODEs in (4.9) are linearly equivalent as well.  Proof of Theorem 1.2. Due to Corollary 4.1 and Proposition 3.1, we only need to show that: systems (4.3) and (4.4) are topologically equivalent if and only if completely controllable subsystems x˙ 1 (t ) = C1 x1 (t ) + D1 u(t )

(4.10)

y˙ 1 (t ) = C2 y1 (t ) + D2 v(t )

(4.11)

and

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J. Li, Z. Zhang / J. Math. Anal. Appl. 402 (2013) 84–102

and ODE subsystems x˙ 2 (t ) = M1 x2 (t )

(4.12)

y˙ 2 (t ) = M2 y2 (t )

(4.13)

and

are topologically equivalent respectively. Obviously, it suffices to prove the ‘‘only if’’ part. By Proposition 4.2, we suppose that the topological equivalence transformation from (4.3) to (4.4) is given by

(y1 , y2 , v) = F (x1 , x2 , u) = (h1 (x1 , x2 ), h2 (x2 ), G(x1 , x2 , u)). It is easy to see that the transformation y2 (t ) = h2 (x2 (t )) brings (4.12) to (4.13). Note that h2 (x2 ) is a homeomorphism from Rn−k to Rn−k . Consequently, y2 = h2 (x2 ) is an equivalence transformation from system (4.12) to system (4.13), and therefore these two ODEs are topologically equivalent. We now show that system (4.10) is topologically equivalent to system (4.11). Again from Proposition 4.2, we deduce that the inverse function of (h1 (x1 , x2 ), h2 (x2 ), G(x1 , x2 , u)) is in the form (z1 (y1 , y2 ), z2 (y2 ), W (y1 , y2 , v)). First, we claim that (h1 (x1 , 0), G(x1 , 0, u)) is a homeomorphism from Rk × Rm to Rk × Rm and its inverse function is (z1 (y1 , 0), W (y1 , 0, v)). In fact, Remark 2.3 implies h2 (0) = 0. Since h2 (x2 ) is a homeomorphism, x2 = 0 if and only if y2 = 0. Noting (h1 (x1 , x2 ), h2 (x2 ), G(x1 , x2 , u)) is a homeomorphism on Rn × Rm , we see that (h1 (x1 , 0), h2 (0), G(x1 , 0, u)) is a homeomorphism on Rk × {0} × Rm and its inverse function is (z1 (y1 , 0), z2 (0), W (y1 , 0, v)). Therefore, the desired assertion follows. Next, for any (x1 (t ), u(t )) satisfying system (4.10), it is easy to check that (x1 (t ), 0, u(t )) satisfies system (4.3). Since the transformation (h1 (x1 , x2 ), h2 (x2 ), G(x1 , x2 , u)) brings system (4.3) to system (4.4), we conclude that h1 (x1 (t ), 0), h2 (0), G(x1 (t ), 0, u(t )) satisfies (4.4). Hence (h1 (x1 (t ), 0), G(x1 (t ), 0, u(t ))) satisfies (4.11), which shows that the transformation (h1 (x1 , 0), G(x1 , 0, u)) brings system (4.10) to system (4.11). Similarly, the inverse transformation (z1 (y1 , 0), W (y1 , 0, v)) of (h1 (x1 , 0), G(x1 , 0, u)) brings (4.11) to (4.10). Hence, (y1 , v) = (h1 (x1 , 0), G(x1 , 0, u)) is a topological equivalence transformation from system (4.10) to system (4.11). Therefore these two control systems are topologically equivalent.  Remark 4.1. As Remark 1.5 indicates, in the sense of linear equivalence, system (1.1) can be reduced to a completely controllable system ξ˙ (t ) = C ξ (t ) + Du(t ) and an ODE η( ˙ t ) = M η(t ). Theorem 1.2 tells us further that in order to study the classification problem of system (1.1), one only need to investigate the classification problem of the completely controllable subsystem and that of the ODE subsystem respectively. 5. Topological classification of control systems and proof of Theorem 1.3 This section is devoted to giving a complete proof for Theorem 1.3. The main difficulty lies in the topological classification. More precisely, we need to show that P (Ai , Bi ) or R(Ai , Bi ) is a topological invariant. In order to achieve our final goal, we need to prepare some preliminary results. Lemma 5.1. Assume that (H (x), G(x, u)) is a topological equivalence transformation from system ( A1 ,  B1 ) to system ( A2 ,  B2 ), and

⊤ p1 −1 times p2 −1 times pr −1 times n−k times             x = x1 , 0, . . . , 0, xp1 +1 , 0, . . . , 0, . . . , xp1 +···+pr −1 +1 , 0, . . . , 0, 0, . . . , 0 , 

(5.1)

then the corresponding point y = H (x) has the following form

⊤ q1 −1 times q2 −1 times qs −1 times n−k times             y = y1 , 0, . . . , 0, yq1 +1 , 0, . . . , 0, . . . , yq1 +···+qs−1 +1 , 0, . . . , 0, 0, . . . , 0 . 

(5.2)

Here, x⊤ denotes the transpose of a vector (or matrix) x and ( Ai ,  Bi ) (i = 1, 2) are the Brunovsky canonical forms given by (1.5). Proof. By the special forms of matrices  A1 and  B1 in (1.5), and noting the special form of x in (5.1), the solution of system ( A1 ,  B1 ) associated to x(0) = x and u(t ) ≡ 0 reads x(t ) ≡ x,

∀ t ≥ 0.

Therefore, the corresponding solution of system ( A2 ,  B2 ): y(t ) = H (x(t )) ≡ H (x) and the control: v(t ) = G(x(t ), u(t )) ≡ 0 ⊤ G(x, 0) are two constants. We denote y = (y1 , . . . , yn )⊤ := H (x) = (h1 (x), . . . , hn (x))⊤ and v 0 = (v10 , . . . , vm ) := G(x, 0).

J. Li, Z. Zhang / J. Math. Anal. Appl. 402 (2013) 84–102

95

Substituting ξ (t ) ≡ (y1 , . . . , yq1 +···+qs )⊤ and v(t ) ≡ v 0 into system ξ˙ (t ) = C2 ξ (t ) + D2 v(t ), we deduce that

 ⊤ (0, . . . , 0)⊤ = y2 , . . . , yq1 , v10 , yq1 +2 , . . . , yq1 +q2 , v20 , . . . , yq1 +···+qs−1 +2 , . . . , yq1 +···+qs , vs0 , which gives

⊤ q1 −1 times q2 −1 times qs −1 times          = y1 , 0, . . . , 0, yq1 +1 , 0, . . . , 0, . . . , yq1 +···+qs−1 +1 , 0, . . . , 0 . 



y1 , . . . , yq1 +···+qs

⊤

(5.3)

Proposition 4.2 shows that (yq1 +···+qs +1 , . . . , yn )⊤ is independent of (x1 , . . . , xp1 +···+pr )⊤ . Hence



yq1 +···+qs +1 , . . . , yn

⊤

⊤  ⊤  = hq1 +···+qs +1 (x), . . . , hn (x) = hq1 +···+qs +1 (0), . . . , hn (0) .

Since Remark 2.3 implies H (0) = 0, the above formula yields



yq1 +···+qs +1 , . . . , yn

⊤

= (0, . . . , 0)⊤ .

(5.4)

Combining (5.3) and (5.4), we arrive at the desired equation (5.2). This completes the proof of Lemma 5.1.



Also, noting the special structure of matrices ( Ai ,  Bi ) (i = 1, 2) in (1.5), it is easy to establish the following result. Hence we omit the proof. Lemma 5.2. Assume that ( Ai ,  Bi ) (i = 1, 2) are the Brunovsky canonical forms given by (1.5), it holds the following. (1) The solution x(t ) = (x1 (t ), . . . , xn (t ))⊤ and the control u(t ) = (u1 (t ), . . . , um (t ))⊤ of system ( A1 ,  B1 ) satisfy: for i = 1, . . . , r, x˙ p1 +···+pi (t ) = ui (t ),

x˙ p1 +···+pi −j (t ) = xp1 +···+pi −j+1 (t ),

j = 1, . . . , pi − 1;

(2) The solution y(t ) = (y1 (t ), . . . , yn (t ))⊤ and the control v(t ) = (v1 (t ), . . . , vm (t ))⊤ of system ( A2 ,  B2 ) satisfy: for i = 1, . . . , s, y˙ q1 +···+qi (t ) = vi (t ),

y˙ q1 +···+qi −j (t ) = yq1 +···+qi −j+1 (t ),

j = 1, . . . , qi − 1.

The following proposition shows that the rank of the controller in system (1.1) is another invariant under the topological equivalence transformation. Denote r := rankB1 , s := rankB2 . Proposition 5.1. If systems (1.2) and (1.3) are topologically equivalent, then r = s. Proof. We use the contradiction argument and suppose that r > s. By Remark 1.4, we still assume that systems (Ai , Bi )(i = 1, 2) are in the Brunovsky canonical forms (1.5). Denote by (H (x), G(x, u)) the equivalence transformation from (1.2) to (1.3). Noting the special form of the canonical form (1.5), we see that for any fixed t5 > 0,

  Θ3 := x(t5 ; x(0), u(·)); u(t ) ≡ 0, x(0) = x, x is of form (5.1) and x1 , xp1 +1 , . . . , xp1 +···+pr −1 +1 ∈ R   = x; x is of form (5.1) and x1 , xp1 +1 , . . . , xp1 +···+pr −1 +1 ∈ R .

(5.5)

Thanks to Lemma 5.1, we have H (Θ3 ) ⊂ y; y is of form (5.2) and y1 , yq1 +1 , . . . , yq1 +···+qs−1 +1 ∈ R .





(5.6)

Since H is a homeomorphism from Θ3 to H (Θ3 ), (5.5) implies that H (Θ3 ) is an r-dimensional topological manifold. On the other hand, (5.6) indicates that H (Θ3 ) is at most an s-dimensional topological manifold, which contradicts the fact r > s.  Remark 5.1. Proposition 5.1 guarantees that the numbers of effective controls of two topological equivalence systems coincide. Therefore, an ODE (rank B = 0) cannot be topologically equivalent to any control system with effective control (rank B ̸= 0). As mentioned before, the main purpose of this section is to prove Theorem 1.3. First, we prove Theorem 1.3 with the hypothesis that the systems under consideration are completely controllable. Proposition 5.2. If systems (1.2) and (1.3) are both completely controllable, then they are topologically equivalent if and only if they are linearly equivalent.

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J. Li, Z. Zhang / J. Math. Anal. Appl. 402 (2013) 84–102

By Remark 1.2, it suffices to prove the ‘‘only if’’ part of Proposition 5.2. Since systems (1.2) and (1.3) are supposed to be completely controllable and topologically equivalent, we assume that Ai and Bi (i = 1, 2) are already in the Brunovsky canonical forms as follows (recall r = s in Proposition 5.1)

.

.. .

··· ··· .. .

0

0

···

Jpr

Jq1 0

0 Jq2

0 0 Jqr

0 Jp2

Jp1 0



A1 =   ..

0 0

0 ep2

.. .

0 0

.

··· ··· .. .

0

0

···

epr

0

···

0

eq1 0

0 eq2

0 0

··· ···

.. .

0 0

0 0

.

··· ··· .. .

0

0

···

e qr

ep1 0





..  , .

B1 =   ..

··· ···

0 0

.. .

.. .

0 0



..  , .

(5.7)

and

.

.. .

··· ··· .. .

0

0

···



A2 =   ..





..  , .

B2 =   ..

.. .

.. .

···

0



..  . .

(5.8)

0

Recalling Lemma 2.2 and noting k = n, r0 = r, we have p1 ≥ p2 ≥ · · · ≥ pr > 0,

q1 ≥ q2 ≥ · · · ≥ qr > 0,

r 

pi =

i=1

r 

qi = n.

i =1

Remark 5.2. By Lemma 2.3, in order to prove Proposition 5.2, one only needs to show that P (A1 , B1 ) = P (A2 , B2 ), i.e. pi = qi ,

i = 1, . . . , r .

(5.9)

The following lemma is a preliminary result for proving (5.9), and we use N+ to denote the set of all positive integers. Lemma 5.3. For any k ∈ N+ and l ∈ N+ , we denote the determinant D(k, l) by

        D(k, l) :=       

1

1

1

l+1 1

l 1

(l + 2)(l + 1) .. .

(l + 1)l .. .

···

k+l 1

···

(k + l + 1)(k + l) .. .

..

1

(2k + l)(2k + l − 1) · · · (k + l)

.

···

1

1

(k + l + 1)(k + l) · · · (l + 1)

(k + l)(k + l − 1) · · · l

        .       

Then D(k, l) ̸= 0 for any k, l ∈ N+ . Proof. For any l ∈ N+ , we prove D(k, l) ̸= 0, k = 1, 2, . . . by mathematical induction. When k = 1,

    D(1, l) =   

1

1

l+1 1

l 1

(l + 2)(l + 1)

(l + 1)l

    1 =  l(l + 1)2 (l + 2) ̸= 0,  

∀ l ∈ N+ .

Suppose that for some positive integer k ≥ 1, D(k, l) ̸= 0, ∀ l ∈ N+ . Then we continue to show that D(k + 1, l) ̸= 0, ∀ l ∈ N+ . In fact, for any l ∈ N+ ,         D(k + 1, l) =       

1 k+l+1 1

(k + l + 2)(k + l + 1) . . . 1

··· ··· ..

.

1

1

l+1 1

l 1

(l + 2)(l + 1) . . .

(l + 1)l . . .

1

1

               

··· (2k + l + 2) · · · (k + l + 1) (k + l + 2) · · · (l + 1) (k + l + 1) · · · l    1 ··· 1   1 1  · · ·  1 k+l+2 l+2  = . . .. (k + l + 1) · · · (l + 1)l  . . . . .    1 1  ··· (2k + l + 2) · · · (k + l + 2) (k + l + 2) · · · (l + 2)

1 1 l+1

. . .

1

(k + l + 1) · · · (l + 1)

             

J. Li, Z. Zhang / J. Math. Anal. Appl. 402 (2013) 84–102

         (l − 1)!  =  (k + l + 1)!         

1

0

1

1

k+l+2

(k + l + 2)(k + l + 1)

1

2

(k + l + 3)(k + l + 2) . . .

(k + l + 3)(k + l + 2)(k + l + 1) . . .

97

···

0

···

(l + 3)(l + 2)

···

(l + 4)(l + 3)(l + 2) . . .

..

.

k+1

1 2

k+1 ··· (2k + l + 2) · · · (k + l + 2) (2k + l + 2) · · · (k + l + 1) (k + l + 3) · · · (l + 2)     1 1 1   ···   (k + l + 2)(k + l + 1) (l + 3)(l + 2) (l + 2)(l + 1)       1 1 1   ··· (k + 1)!(l − 1)!  (l + 4)(l + 3)(l + 2) (l + 3)(l + 2)(l + 1)  =  (k + l + 3)(k + l + 2)(k + l + 1)  (k + l + 1)!  . . . ..   . . .   . . . .       1 1 1   ··· (2k + l + 2) · · · (k + l + 1) (k + l + 3) · · · (l + 2) (k + l + 2) · · · (l + 1)     1 1 1   · · ·   k+l+2 l+3 l+2       1 1 1   · · · (k + 1)! (l − 1)! l!  (k + l + 3)(k + l + 2)  (l + 4)(l + 3) (l + 3)(l + 2) =    (k + l + 1)!(k + l + 1)!  . . . .   . .. . .   . . .       1 1 1   ··· (2k + l + 2) · · · (k + l + 2) (k + l + 3) · · · (l + 3) (k + l + 2) · · · (l + 2) (k + 1)! l! (l − 1)! = D(k, l + 2) ̸= 0. (k + l + 1)!(k + l + 1)! 1

Therefore, we conclude that D(k, l) ̸= 0 for any k, l ∈ N+ .

      1   (l + 2)(l + 1)    2  (l + 3)(l + 2)(l + 1)   .  .  .    k+1  (k + l + 2) · · · (l + 1) 0



Before giving our next key lemma for proving Proposition 5.2, we introduce some new notations. Denote by ℓ and ν two arbitrary positive integers satisfying 1 ≤ ℓ ≤ r and 1 ≤ ν ≤ r. Put n1 := p1 + · · · + pℓ−1 ,

n2 := pℓ ,

n3 := pℓ+1 + · · · + pr ,

 n1 := q1 + · · · + qν−1 ,

 n2 := qν ,

 n3 := qν+1 + · · · + qr .

and Here, we remark that n1 = 0 when ℓ = 1, and n3 = 0 when ℓ = r. A similar remark applies to  n1 and  n3 . In what follows, we assume that (µ1 , . . . , µn1 ) vanishes automatically when n1 = 0, and (µn1 +1 , . . . , µn1 +n3 ) vanishes automatically when n3 = 0. Lemma 5.4. Assume that systems (A1 , B1 ) and (A2 , B2 ) are topologically equivalent, and the topological equivalence transformation is (H (x), G(x, u)). If pℓ < qν (i.e. n2 <  n2 ), then for any (µ1 , . . . , µn1 +n3 ) ∈ Rn1 +n3 and any (λ1 , . . . , λn2 ) ∈ Rn2 , it holds

 n2 −1 times    µ1 , . . . , µn1 , λ1 , 0, . . . , 0, µn1 +1 , . . . , µn1 +n3  . h n1 +2 (µ1 , . . . , µn1 , λ1 , . . . , λn2 , µn1 +1 , . . . , µn1 +n3 ) ≡ h n 1 +2 

Proof. The proof is divided into two steps. Step 1. We consider the special situation of ℓ = ν = 1. Now n1 =  n1 = 0. We need to show that for any (λ1 , . . . , λn2 ) ∈ Rn2 n3 and any (µ1 , . . . , µn3 ) ∈ R ,

 n2 −1 times    h2 (λ1 , . . . , λn2 , µ1 , . . . , µn3 ) ≡ h2 λ1 , 0, . . . , 0, µ1 , . . . , µn3  . 

When n2 = 1, the above formula holds automatically. Thus we assume that n2 ≥ 2. In what follows, we will prove a more general result: for k = 0, 1, . . . , n2 − 1, it holds

 k times    hs (λ1 , . . . , λn2 , µ1 , . . . , µn3 ) ≡ hs λ1 , . . . , λn2 −k , 0, . . . , 0, µ1 , . . . , µn3  , 

1 ≤ s ≤ n2 + 1 − k.

(5.10)

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J. Li, Z. Zhang / J. Math. Anal. Appl. 402 (2013) 84–102

In fact, (5.10) can be achieved by mathematical induction. When k = 0, (5.10) holds automatically for s = 1, 2, . . . , n2 +1. Suppose that (5.10) is valid for some integer k, 0 ≤ k ≤ n2 − 2. Then we continue to show (5.10) is valid for k + 1, that is,

 k+1 times    hs (λ1 , . . . , λn2 , µ1 , . . . , µn3 ) ≡ hs λ1 , . . . , λn2 −k−1 , 0, . . . , 0, µ1 , . . . , µn3  , 

s = 1, 2, . . . , n2 − k.

Making use of our hypothesis and denoting vectors a := (λ1 , . . . , λn2 −k−1 ) ∈ Rn2 −k−1 , b := (µ1 , . . . , µn3 ) ∈ Rn3 , we finally need to prove: for any a ∈ Rn2 −k−1 , any λn2 −k ∈ R and any b ∈ Rn3 ,

    k+1 times  k times           hs a, λn −k , 0, . . . , 0, b − hs a, 0, . . . , 0, b = 0, 2    

s = 1, 2, . . . , n2 − k.

(5.11)

Consider system (A1 , B1 ) in the time interval t ∈ [0, 1]. Let the initial datum

⊤ k+1 times    x(0) = a, 0, . . . , 0, b . 

(5.12)

We use ‘‘t0 ’’ to denote any real number lying in (0, 1), and we use u(t0 , t ) to denote the control in the following form:

⊤  ⊤ m−1 times m−1 times k+1     t k+1−i    u(t0 , t ) = u1 (t0 , t ), 0, . . . , 0 =  ci 2k+2−i , 0, . . . , 0 

i=1

t0

for t ∈ [0, t0 ].

(5.13)

As we will show later, here ci , i = 1, 2, . . . , k + 1 are another k + 1 real numbers which can be uniquely determined by λn2 −k . Using x(t0 , t ) to denote the state of system (A1 , B1 ) with initial datum (5.12) and control (5.13), we claim that: there exists a unique control in the form (5.13), such that the following requirements can be satisfied: xn2 −k (t0 , t0 ) = λn2 −k+1 ,

xn2 −k+1 (t0 , t0 ) = xn2 −k+2 (t0 , t0 ) = · · · = xn2 (t0 , t0 ) = 0.

(5.14)

Notice that the first n2 components of x(t0 , t ) satisfy

 d 

dt



x1 (t0 , t )





.. .

x1 (t0 , t )



   ..  = Jn2   + en2 u1 (t0 , t ) . xn2 (t0 , t ) xn2 (t0 , t ) 

with (x1 (t0 , 0), . . . , xn2 (t0 , 0)) =

k+1 times     a, 0, . . . , 0 . A straightforward computation shows that (5.14) is equivalent to the

following identity: k+1 





ci

 (k + 2 − i)t0k  i=1  k+1 xn2 (t0 , t0 )  ci xn2 −1 (t0 , t0 )     ( k + 3 − i )( k + 2 − i)t0k−1 .. i=1  =  . ..  .  xn2 −k (t0 , t0 )  k+1  ci 



i=1

(2k + 2 − i) · · · (k + 2 − i)

     0    0     =  ..  .  .   λn2 −k  

Note that t0 > 0, the above identity is further equivalent to the linear equations as follows:

          

1

1

k+1 1

k 1

(k + 2)(k + 1) .. .

(k + 1)k .. .

1

1

(2k + 1) · · · (k + 1)

(2k) · · · k

 ··· ··· ..

.

···

1 1 2

.. .

1

(k + 1)!

  c1   0     c2   0       ..   =  ..  .  . .   ck+1 λn2 −k 

J. Li, Z. Zhang / J. Math. Anal. Appl. 402 (2013) 84–102

99

By Lemma 5.3, we know that D(k, 1) ̸= 0 for any k ∈ N+ . This implies that the above linear equations have a unique solution for any k ≥ 0, and we still denote this real solution by (c1 , c2 , . . . , ck+1 )⊤ . Thus, with such a choice of control only depending on λn2 −k , but not depending on t0 , requirements (5.14) can be fulfilled. In addition, we can continue to prove that there exists some common real constant K > 0 only depending on a, λn2 −k and b, such that for t ∈ [0, t0 ],

∥ (x1 (t0 , t ), x2 (t0 , t ), . . . , xn2 −k (t0 , t )) ∥≤ K

(5.15)

and

∥ (xn2 +1 (t0 , t ), . . . , xn (t0 , t )) ∥≤ K .

(5.16)

⊤ m−1 times    As a matter of fact, recalling x(0) = (a, 0, b)⊤ and u(t0 , t ) = u1 (t0 , t ), 0, . . . , 0 , we know that for t ∈ [0, 1], the last 

n3 components of x(t0 , t ) (i.e. xn2 +1 (t0 , t ), . . . , xn (t0 , t )) satisfy xn2 +1 (t0 , t )



 d 

dt



.. .

xn (t0 , t )



Jp2

  .. = .

0

··· .. . ···

0

xn2 +1 (t0 , t )





..   . 

Jpr

.. .

xn (t0 , t )

 

with initial datum (xn2 +1 (t0 , 0), . . . , xn (t0 , 0)) = b. By the well-posedness of the above ODE system, we arrive at (5.16) with the constant K1 only depending on b, and we have limt0 →0 (xn2 +1 (t0 , t0 ), . . . , xn (t0 , t0 )) = b. Next, consider x1 (t0 , t ), . . . , xn2 −k (t0 , t ) in the time interval t ∈ [0, t0 ]. Since

 xn

  k+1    ci t 2k+2−i    (t , t ) =   2 −k 0  i=1 (2k + 2 − i) · · · (k + 2 − i)t02k+2−i  ≤

|c1 | |c2 | |ck+1 | + + ··· + , (2k + 1) · · · (k + 1) (2k) · · · k (k + 1)!

and recall that for t ∈ [0, 1] and j = n2 − k − 1, n2 − k − 2, . . . , 1, xj (t0 , t ) = λj +

t



xj+1 (t0 , τ )dτ , 0

we can conclude that (5.15) holds for some constant K2 eventually depending on a and λn2 −k . Besides, we have limt0 →0 (x1 (t0 , t0 ), . . . , xn2 −k (t0 , t0 )) = a. Denote K = max{K1 , K2 }, then (5.15) and (5.16) hold simultaneously for real constant K . By our hypothesis, one has

 k times    hs (λ1 , . . . , λn2 , µ1 , . . . , µn3 ) ≡ hs λ1 , . . . , λn2 −k , 0, . . . , 0, µ1 , . . . , µn3  , 

s = 1, 2, . . . , n2 + 1 − k.

This combining with (5.15), (5.16) and the continuity of hs yields that: there exists some positive constant K such that for t ∈ [0, t0 ],

|hs (x(t0 , t ))| ≤ K ,

s = 1, 2, . . . , n2 + 1 − k.

⊤ k+1 times    Recalling Lemma 5.2, x(0) = a, 0, . . . , 0, b and (5.14), the above inequality implies that for s = 1, 2, . . . , n2 − k, 

    k+1 times  k times           hs a, λn −k , 0, . . . , 0, b − hs a, 0, . . . , 0, b = lim |hs (x(t0 , t0 )) − hs (x(t0 , 0))| 2   t0 →0   = lim |ys (t0 , t0 ) − ys (t0 , 0)| t0 →0  t    0   = lim  ys+1 (t0 , t )dt  = lim  t0 →0 t0 → 0 0

≤ lim K · t0 = 0. t0 →0

This is (5.11), and we obtain the validity of (5.10).

t0 0

 

hs+1 (x(t0 , t ))dt 

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J. Li, Z. Zhang / J. Math. Anal. Appl. 402 (2013) 84–102

Lemma 5.4 with ℓ = ν = 1 is just a special case of (5.10) with k = n2 − 2 and s = 2. Step 2. Consider the general situation with 1 ≤ ℓ ≤ r and 1 ≤ ν ≤ r. There exist invertible matrices L1 and Q1 , such that through linear equivalence transformations  x = L1 x and  u = Q1 u, system (A1 , B1 ) changes into Jp1  0

0 Jp2

ep1 0 0  0

0 ep2

.. .

··· ··· .. .

0 0

0 0

··· ···

.. .

··· ··· .. .

0 0

0

0

···

Jpr

0

0

···

epr

0

···

0



    A1 ,  B1 :=   .. .

 

 ..   ,  .. . .

.. .

.. .



 ..   , .

where ( p1 , p2 , . . . , pr ) = (pℓ , p1 , . . . , pℓ−1 , pℓ+1 , . . . , pr ). Similarly, there exist invertible matrices L2 and Q2 , such that through linear transformations  y = L2 y and  v = Q2 v , system (A2 , B2 ) changes into Jq1  0

0 Jq2

eq1 0 0  0

0 eq2

.. .

··· ··· .. .

0 0

0 0

··· ···

.. .

··· ··· .. .

0 0

0

0

···

Jqr

0

0

···

eqr

0

···

0



    A2 ,  B2 :=   .. .

 

 ..   ,  .. . .

.. .

.. .



 ..   , .

where ( q1 , q2 , . . . , qr ) = (qν , q1 , . . . , qν−1 , qν+1 , . . . , qr ). Denote

  1   −1    −1 . y, v) =  F ( x, u) = ( H ( x),  G( x, u)) := L2 H L− ( 1 x , Q2 G L1 x, Q1 u

(5.17)

Then systems ( A1 ,  B1 ) and ( A2 ,  B2 ) are topologically equivalent with the topological equivalence transformation ( H ( x),  G( x, u)). Notice that  pi and  qi do not satisfy inequalities  p1 ≥  p2 ≥ · · · ≥  pr > 0 and  q1 ≥  q2 ≥ · · · ≥  qr > 0 anymore. However, inequalities p1 ≥ p2 ≥ · · · ≥ pr > 0 and q1 ≥ q2 ≥ · · · ≥ qr > 0 are not required in the proof of step 1. Thus, recalling  p1 = pℓ < qν =  q1 and applying the conclusion of step 1 to systems ( A1 ,  B1 ) and ( A2 ,  B2 ), we obtain  n2 −1 times      h2 ( x1 , . . . , xn2 , xn2 +1 , . . . , xn ) ≡  h2  x1 , 0, . . . , 0,  xn2 +1 , . . . , xn  . 1 From (5.17), one has H (x) = L− 2 H (L1 x). This together with the above formula yields

 n2 −1 times    x1 , . . . , xn1 , xn1 +1 , 0, . . . , 0, xn1 +n2 +1 , . . . , xn  . h n 1 +2 n1 +2 (x1 , . . . , xn1 , xn1 +1 , . . . , xn1 +n2 , xn1 +n2 +1 , . . . , xn ) ≡ h 

Therefore, Lemma 5.4 with 1 ≤ ℓ ≤ r and 1 ≤ ν ≤ r is valid.



We are now in a position to give the proof of Proposition 5.2. Proof of Proposition 5.2. By Remark 5.2, it suffices to prove (5.9), i.e. pi = qi , i = 1, . . . , r . Note that (5.9) is obviously true for r = 1. We assume that r ≥ 2 in what follows. We use the contradiction argument to prove (5.9). Hypothesize that (5.9) is false, then there exists an integer ℓ ∈ {2, . . . , r } such that ℓ = max{j; pj ̸= qj , j = 1, . . . , r } (ℓ ̸= 1 is due to the fact p1 + · · · + pr = q1 + · · · + qr = n). Without loss of generality, we assume that pℓ < qℓ . Noting that p1 ≥ p2 ≥ · · · ≥ pr > 0 and q1 ≥ q2 ≥ · · · ≥ qr > 0 by Lemma 2.2, we have pi < qj for any pair (i, j) satisfying ℓ ≤ i ≤ r , 1 ≤ j ≤ ℓ. With the aid of our key Lemma 5.4, we obtain that for ℓ ≤ i ≤ r , 1 ≤ j ≤ ℓ,

 pi −1 times    hq1 +···+qj−1 +2 (x1 , . . . , xn ) ≡ hq1 +···+qj−1 +2 x1 , . . . , xp1 +···+pi−1 , xp1 +···+pi−1 +1 , 0, . . . , 0, xp1 +···+pi +1 , . . . , xn  . 

The above formula implies that for 1 ≤ j ≤ ℓ, hq1 +···+qj−1 +2 (x1 , . . . , xn ) ≡ hq1 +···+qj−1 +2



 pℓ+1 −1 times pl −1 times pr −1 times            × x1 , . . . , xp1 +···+pℓ−1 , xp1 +···+pℓ−1 +1 , 0, . . . , 0, xp1 +···+pℓ +1 , 0, . . . , 0, . . . , xp1 +···+pr −1 +1 , 0, . . . , 0 .

(5.18)

J. Li, Z. Zhang / J. Math. Anal. Appl. 402 (2013) 84–102

101

Letting x(0) = 0 and choosing u(t ) in the special form of

 ℓ−1 times ⊤ m−r times       u(t ) = 0, . . . , 0, uℓ (t ), . . . , ur (t ), 0, . . . , 0 ∈ L1loc ([0, +∞); Rm ), we get the solution of system (A1 , B1 ) (given by (5.7)) in the following form



⊤

p1 +···+pℓ−1 times

    x(t ) = (x1 (t ), . . . , xn (t ))⊤ =  0, . . . , 0,

xp1 +···+pℓ−1 +1 (t ), . . . , xn (t ) .



(5.19)

In this situation, by use of (5.18), the solution of system (A2 , B2 ) (given by (5.8)) y(t ) = H (x(t )) = (h1 (x(t )), . . . , hn (x(t )))⊤ satisfies that for 1 ≤ j ≤ ℓ, yq1 +···+qj−1 +2 (t ) = hq1 +···+qj−1 +2 (x(t )) ≡ hq1 +···+qj−1 +2



 pℓ+1 −1 times pl −1 times pr −1 times          xp1 +···+pℓ−1 +1 (t ), 0, . . . , 0, xp1 +···+pℓ +1 (t ), 0, . . . , 0, . . . , xp1 +···+pr −1 +1 (t ), 0, . . . , 0 .

p1 +···+pℓ−1 times

    ×  0, . . . , 0,

By Lemma 5.1, the above formula implies yq1 +···+qj−1 +2 (t ) ≡ 0 for 1 ≤ j ≤ ℓ. Recalling y(0) = H (x(0)) = 0 and making use of Lemma 5.2, we obtain that for 1 ≤ j ≤ ℓ and t ∈ [0, +∞), yq1 +···+qj−1 +1 (t ) =

t



yq1 +···+qj−1 +2 (τ )dτ ≡ 0,

·

yq1 +···+qj−1 +3 (t ) = yq1 +···+qj−1 +2 (t ) ≡ 0, . . . ,

0

(qj −2)

yq1 +···+qj−1 +qj (t ) = yq1 +···+qj−1 +2 (t ) ≡ 0. This means that for 1 ≤ i ≤ q1 + · · · + qℓ , yi (t ) ≡ 0. Therefore, for t ∈ [0, +∞),

q1 +···+qℓ times ⊤    y(t ) =  0, . . . , 0, yq1 +···+qℓ +1 (t ), . . . , yn (t ) .

(5.20)

Since system (A1 , B1 ) given by (5.7) is completely controllable, by (5.19), for any fixed t6 > 0, we see that Θ4 := {x(t6 ;

 ℓ−1 times m−r times ⊤       0, u(·)); u(t ) = 0, . . . , 0, uℓ (t ), . . . , ur (t ), 0, . . . , 0 ∈ L1loc ([0, +∞); Rm )} is a (pℓ + · · · + pr )-dimensional topological

manifold contained in Rn . On the other hand, (5.20) implies that H (Θ4 ) is at most a (qℓ+1 +· · ·+ qr )-dimensional topological manifold contained in Rn . Note that qℓ+1 + · · · + qr = pℓ+1 + · · · + pr < pℓ + pℓ+1 + · · · + pr . This contradicts the fact that H (x) is a homeomorphism from Rn to Rn . The proof of Proposition 5.2 is complete.  Proof of Theorem 1.3. By Remark 4.1, Theorem 1.3 is a direct consequence of Lemmas 2.1, 2.3, Propositions 3.1 and 5.2.



Acknowledgment The authors would like to thank the anonymous referee for his/her careful reading and constructive suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]

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