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State feedback controller design for the synchronization of Boolean networks with time delays
Accepted Manuscript State feedback controller design for the synchronization of Boolean networks with time delays Fangfei Li, Jianning Li, Lijuan Shen
PII: DOI: Reference:
S0378-4371(17)30774-4 http://dx.doi.org/10.1016/j.physa.2017.08.041 PHYSA 18487
To appear in:
Physica A
Received date : 16 February 2017 Revised date : 1 June 2017 Please cite this article as: F. Li, J. Li, L. Shen, State feedback controller design for the synchronization of Boolean networks with time delays, Physica A (2017), http://dx.doi.org/10.1016/j.physa.2017.08.041 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
State feedback controller design for the synchronization of Boolean networks with time delays Fangfei Li1,2∗, Jianning Li1∗, Lijuan Shen3∗ 1 School of Automation, Hangzhou Dianzi University, Hangzhou, 310018, P.R.China 2 Department of Mathematics, East China University of Science and Technology, Shanghai 200237, PR China 3 School of Mathematical Sciences, Tongji University, Shanghai 200092, China
Abstract State feedback control design to make the response Boolean network synchronize with the drive Boolean network is far from being solved in the literature. Motivated by this, this paper studies the feedback control design for the complete synchronization of two coupled Boolean networks with time delays. A necessary condition for the existence of a state feedback controller is derived first. Then the feedback control design procedure for the complete synchronization of two coupled Boolean networks is provided based on the necessary condition. Finally, an example is given to illustrate the proposed design procedure. Keywords: Boolean network, time delay, semi-tensor product, feedback control design. 1. Introduction Recently, the study of logical systems has attracted attention, because it can provide a simple and proper model to describe genetic regulatory networks, neural networks, etc, [1–4]. Boolean networks (BNs) as a kind of two valued logical system has been studied widely [5–7]. In a Boolean network, the state of each Boolean variable is 1 or 0, which means the state is on or off. Every Boolean variable updates its state according to a logical relationship, ∗
Corresponding author. Email address: [email protected] (Fangfei Li1,2 )
Preprint submitted to Physica A
August 28, 2017
given in the form of a Boolean function. Up to now, many significant results about the topological structure of Boolean network, including the attractors and transient time have been obtained, for example, see [8–11]. With the development of the semi-tensor product (STP) of matrices, an algebraic form of a Boolean network can be obtained. Then, some interesting progresses have been achieved, including the controllability and observability of Boolean networks, disturbance decoupling controller design for Boolean control networks, stability and stabilization of Boolean networks with impulsive effects etc, see e.g. [12–28]. In the past decades, synchronization of dynamic systems has been widely investigated, and still draw much research attention [29–32]. Synchronization of two Boolean networks is meaningful too. For example, the results on the synchronization of two coupled Boolean networks can be applied to the synchronization of two lasers [33]. The study of synchronized BNs can provide useful information on the coevolution of several biological species whose genetic dynamics influence each other [34]. Based on the semi-tensor product of matrices, necessary and sufficient conditions of the complete synchronization for two coupled Boolean networks are obtained [35]. Later, the results are generalized to the complete synchronization of Boolean networks with time delays [36]. Though some results on the synchronization of Boolean networks have been obtained in the literature, we can note that these results are control-free. To design controller to make the response BN synchronize with the drive BN is significant, which is one of the motivation of this paper. As we know, time delay phenomenon is so common in the real world, and is very important in the analysis and control of dynamic systems [37, 38]. In addition, time delay happens frequently and is unavoidable in biological and physiological systems [39]. For example, time delays of genetic regulatory networks are caused by slow processes of transcription, translation and so on. Since we usually use BNs to model genetic regulatory networks, the analysis and control of BNs with time delays are meaningful. Although there exist some results on the control problems of BNs with time delays, however to design controller for the synchronization of BNs with time delays is still lacking in the literature, to the best of our knowledge. Motivated by this, in the paper, we study the state feedback control design for the synchronization of two coupled BNs with time delays. As to the feedback control design for the synchronization analysis of two coupled Boolean networks with time delays, one main difficulty lies in the controller design. In this paper, time delay phenomenon is taken into ac2
count. Different from [36], where the time delays for each node are the same, in this paper, we consider the case that the time delays for each node are different. Thus, compared with [36] and [40], the system structure in this paper is more complex. In addition, to calculate the set consisting of all the states that can be reachable to the desirable state is much more complicated due to time delay. Moreover, a much more difficult situation arises since the response dynamic appear with different model structures for the reason that it is affected by the drive system. To overcome these difficulties, in this paper, a necessary condition for the existence of a state feedback controller to achieve the complete synchronization is first established. Then, we see the response system as a switched BN, and based on the necessary condition, the feedback law is constructed. The rest of this paper is organized as follows. In section 2, we provide some preliminary results. Section 3 gives a necessary condition to achieve synchronization. Then based on the necessary condition, feedback control design is obtained for the complete synchronization of two coupled Boolean networks. An example is presented in section 4. Finally, section 5 gives a brief conclusion. 2. Preliminaries In this section, some necessary preliminary results are given. Definition 2.1. ([41]). For M ∈ Rm×n and N ∈ Rp×q , their semi-tensor product (STP), denoted by M n N , is defined as follows: M n N := (M ⊗ Is/n )(N ⊗ Is/p ), where s is the least common multiple of n and p, and ⊗ is the Kronecker product. Remark 2.1. The matrix product we use in this paper is the semi-tensor product (STP). It can be verified that when n = p, the semi-tensor product of matrices M and N becomes the conventional matrix product. Hence, STP is a generalization of conventional matrix product. All the fundamental properties of conventional matrix product remain true. Based on this, we can omit the symbol “ n ” if no confusion raises. There are also some basic properties of STP, for detail, see ([42]). 3
Next, some notations are provided. (1) D := {1, 0}. (2) ∆k := {δki |1 ≤ i ≤ k}, where δki is the ith column of the identity matrix Ik . For compactness, ∆2 = ∆. (3) Denote the ith column of matrix A by Coli (A). (4) Denote by Mm×n the set of all m × n matrices. A matrix A ∈ Mm×n is called a logical matrix, if the columns of A are elements of ∆m . Denote the set of logical matrices by L; the set of n × s logical matrices is defined by Ln×s . (6) Assume that there is a matrix M = [δni1 , δni2 , · · · , δnis ], for notational compactness, denote by M := δn [i1 , i2 , · · · , is ]. By identifying True = 1 ∼ δ21 , False = 0 ∼ δ22 , the logical variable A(t) takes value from these two vectors, i.e. A(t) ∈ ∆ = {δ21 , δ22 }, where p ∼ q denotes the logical equivalence of p and q. The following lemma is fundamental for the matrix expression of the logical function. Lemma 2.1. ([41]) Any logical function L(A1 , · · · An ) with logical arguments A1 , · · · , An ∈ ∆ can be expressed in a multi-linear form as L(A1 , . . . , An ) = ML A1 A2 · · · An , where ML ∈ L2×2n is unique, called the structure matrix of L. 3. Main results Two Boolean networks coupled in the drive-response configuration are considered in this paper, where the response Boolean network is with time delays. The dynamic is described as follows. xi (t + 1) = fi (x1 (t), · · · , xn (t)), (1) yi (t + 1) = gi (u1 (t), · · · , um (t), x1 (t), · · · , xn (t), y1 (t), · · · , yn (t), y1 (t − 1), · · · , yn (t − 1), · · · , y1 (t − τ ), · · · , yn (t − τ )), (2) i = 1, 2, · · · , n, where xi , yi ∈ D, i = 1, 2, · · · , n, are the nodes of the drive BN (1) and the response BN (2) respectively, fi , gi are Boolean functions, u1 , u2 , · · · , 4
um ∈ D are the control inputs. The objective of this paper is to find a feedback law of the form uj (t) = kj (x1 (t), · · · , xn (t), y1 (t), · · · , yn (t), y1 (t − 1), · · · , yn (t − 1), · · · , yn (t − τ ), · · · , yn (t − τ )), j = 1, 2, · · · , m, (3) where k1 ,· · · , km are Boolean functions such that under this feedback control the response BN (2) can be globally synchronized with the drive BN (1). By using lemma 2.1, we denote x(t) = nni=1 xi (t), y(t) = nni=1 yi (t), u(t) = nm j=1 uj (t) and can express (1), (2) and (3) as xi (t + 1) = Fi x(t), yi (t + 1) = Gi u(t)x(t)y(t)y(t − 1) · · · y(t − τ ), uj (t) = Kj x(t)y(t)y(t − 1) · · · y(t − τ ),
(4) (5) (6)
where Fi , Gi and Kj are the structure matrix of fi , gi and kj respectively, i = 1, 2, · · · , n, j = 1, 2, · · · , m. Moreover, (4), (5) and (6) can further be converted into the following discrete-time system x(t + 1) = F x(t), y(t + 1) = Gu(t)x(t)y(t)y(t − 1) · · · y(t − τ ), u(t) = Kx(t)y(t)y(t − 1) · · · y(t − τ ),
(7) (8) (9)
where equations (7), (8) and (9) are called the algebraic representations of the BN (1), (2) and the feedback law (3) respectively, and matrix K is called the state feedback matrix. Let z(t) = nτi=0 zi (t), where z0 (t) = y(t), z1 (t) = y(t − 1), · · · ,zτ (t) = y(t − τ ), (8) can be rewritten as y(t + 1) = Gu(t)x(t)z(t). z(t + 1) = nτi=0 zi (t + 1) = y(t + 1)y(t) · · · y(t − τ + 1) =Gu(t)x(t)z(t)y(t) · · · y(t − τ + 1) =Gu(t)x(t)y(t)y(t − 1) · · · y(t − τ )y(t) · · · y(t − τ + 1) =Gu(t)x(t)W[2nτ ,2n(τ +1) ] (y(t) · · · y(t − τ + 1))2 y(t − τ ) =Gu(t)x(t)W[2nτ ,2n(τ +1) ] Φnτ z(t)
=G(I2m+n ⊗ W[2nτ ,2n(τ +1) ] Φnτ )u(t)x(t)z(t) ¯ ,Gu(t)x(t)z(t). 5
To see the definition of W[m,n] and Φnτ , it refers to [42]. Note that ¯ ¯ [2n ,2m ] x(t)u(t)y(t) z(t + 1) =Gu(t)x(t)z(t) = GW ,Hx(t)u(t)z(t), then (8) can be rewritten as z(t + 1) = Hx(t)u(t)z(t).
(10)
We split H into 2n equal dimension blocks as H = [H1 , H2 , · · · , H2n ]. Note that, if x(t) = δ2i n , z(t + 1) = Hx(t)u(t)z(t) = Hδ2i n u(t)z(t) = Hi u(t)z(t). Hence, (10) can be interpreted as a switched Boolean system switching between 2n possible subsystems. Assume that the switching signal is σ(t) : N → Ω = {1, 2, · · · , 2n }. Then under the switching signal σ = i, (that is x(t) = δ2i n ), the drive system (10) is z(t + 1) = Hi u(t)z(t). In the sequel, we denote El (δ2rn1 δ2rn2 · · · δ2rns ) under the switching signal sequence {σ(1), σ(2), · · · , σ(l − 1)} by the set consisting of all the states that can be steered to δ2rn1 n δ2rn2 n · · · n δ2rns in l steps under the switching signal sequence {σ(1), σ(2), · · · , σ(l − 1)}. Definition 3.1. An ordered sequence of distinct vectors (δ2i1n , δ2i2n , · · · , δ2ikn ) is a limit cycle C of the BN (7) if x(0) = δ2iln for some l ∈ [1, k] ensures that the corresponding state trajectory x(t) is periodic of period k and, for every i x(t) is periodic of period k and, for every t ∈ Z+ , x(t) = δ2jn , where j ∈ [1, k] and j ≡ (t + l) mod k. A limit cycle of unitary length is an equilibrium point of the BN. il
In the sequel, assume that Ci = (δ2i1n , δ2i2n , · · · , δ2ni ) is the ith cycle of the BN (7), and the length is li . il
Theorem 3.1. Let Ci = (δ2i1n , δ2i2n , · · · , δ2ni ) be the ith cycle of the drive BN (1) (equivalently (7)). If the response BN (2) (equivalently (10)) can be 6
completely synchronized with the drive BN (1) (equivalently (7)), then the state of response BN (10) satisfies il −τ +2 il −τ +1 il ) under the switching signal i1 ; ∈ E1 (δ2i2n δ2i1n · · · δ2ni (i) δ2i1n δ2ni · · · δ2ni ili −τ +3 ili −τ +2 i 3 i2 i2 i1 il i ) under the switching signal i2 ; ∈ E1 (δ2n δ2n · · · δ2n δ2n δ2n δ2n · · · δ2n .. . il
il
il
il
il
il
il
) under the switching signal ili ; δ2ni δ2ni · · · δ2ni ∈ E1 (δ2i1n δ2ni · · · δ2ni il il −τ +1 ) = ∆2n(τ +1) (ii) there exists an integer N , such that EN (δ2i1n δ2ni · · · δ2ni under the switching signal sequences {· · · i1 , · · · , ili }. −1
−τ
−τ +1
il
) under the ∈ E1 (δ2i2n δ2i1n · · · δ2ni Proof. We prove that δ2i1n δ2ni · · · δ2ni switching signal i1 first. Note that the drive system (7) will reach the cycle Ci , then it will reach the state x(t) = δ2i1n , that is the switching signal is σ = i1 . If the complete synchronization is achieved, we can find an integer T , such that x(T ) = δ2i1n , y(T ) = δ2i1n . Moreover, x(T + 1) = y(T + 1) = δ2i2n , x(T − 1) = il il −1 il −τ +1 y(T −1) = δ2ni , x(T −2) = y(T −2) = δ2ni , · · · , x(T −τ ) = y(T −τ ) = δ2ni . ili −τ +1 i1 il i Then z(T ) = y(T )y(T − 1) · · · y(T − τ ) = δ2n δ2n · · · δ2n , z(T + 1) = ili −τ +2 il −τ +1 i 2 i1 i1 ili y(T + 1) · · · y(T − τ + 1) = δ2n δ2n · · · δ2n . That is δ2n δ2n · · · δ2ni ∈ ili −τ +2 i2 i1 E1 (δ2n δ2n · · · δ2n ) under the switching signal i1 . Similarly, we can prove that condition (i) is satisfied. Now, we will show that condition (ii) is satisfied. If the complete synchronization of (7) and (10) is achieved, we can find an integer T , such that il −1 il x(T ) = y(T ) = δ2i1n , x(T − 1) = y(T − 1) = δ2ni , x(T − 2) = y(T − 2) = δ2ni , il −τ +1 · · · , x(T − τ ) = y(T − τ ) = δ2ni , and all the states of the response system il il −τ +1 (10) can reach z(T ) = y(T )y(T − 1) · · · y(T − τ ) = δ2i1n δ2ni · · · δ2ni . That is condition (ii) is satisfied. This completes the proof. We give the following algorithm to construct matrix K to achieve the complete synchronization. Split K into 2n square blocks as K = [K1 , K2 , · · · , K2n ]. Algorithm 3.1. Suppose that conditions (i) and (ii) in Theorem 3.1 hold. 1. Find the [(pi1 −1)2n(τ +1) +(i1 −1)2nτ +(ili −1)2n(τ −1) +· · ·+(ili −τ +1)]th column of Hi1 such that Col(pi1 −1)2n(τ +1) +(i1 −1)2nτ +(il −1)2n(τ −1) +···+(il −τ +1)] (Hi1 ) il
−τ +1
i
−τ +2
i
= δ2i2n δ2i1n · · · δ2ni . Let the [(i1 − 1)2nτ + (ili − 1)2n(τ −1) + · · · + (ili − τ + 1)]th pi1 column of Ki1 be δ2m . Find the [(pi2 −1)2n(τ +1) +(i2 −1)2nτ +(i1 −1)2n(τ −1) + · · · + (ili − τ + 2)]th column of Hi2 such that −τ +2
7
Col[(pi2 −1)2n(τ +1) +(i2 −1)2nτ +(i1 −1)2n(τ −1) +···+(il
i
il −τ +3 . δ2i3n δ2i2n · · · δ2ni
−τ +2)] (Hi2 )
Let the [(i2 − 1)2nτ + (i1 − 1)2n(τ −1) + · · · + (ili − τ + 2)]th = p i2 column of Ki2 be δ2m .· · · Find the [(pili − 1)2n(τ +1) + (ili − 1)2nτ + (ili −1 − 1)2n(τ −1) + · · · + (ili − τ )]th column of Hili such that Col[(pi −1)2n(τ +1) +(il −1)2nτ +(il −1 −1)2n(τ −1) +···+(il −τ )] (Hili ) li
i
i
il δ2i1n δ2ni
i
il −τ +1 · · · δ2ni .
Let the [(ili − 1)2 + (ili −1 − 1)2n(τ −1) + · · · + (ili − τ )]th = p il column of Kili be δ2mi . 2. Find all the [(pili − 1)2n(τ +1) + qili ]th column of Hili such that Col[(pi qil
li
nτ
−1)2n(τ +1) +qil ] (Hili ) i
il
il
= δ2i1n δ2ni · · · δ2ni
−τ +1
p il
il
il
. Let E1 (δ2i1n δ2ni · · · δ2ni
−τ +1
) =
{δ2n(τi +1) } and the qili th column of Kili be δ2mi . 3. Find all the [(pili −1 − 1)2n(τ +1) + qili −1 ]th column of Hili −1 such that
Col[(pi {δ
qil
li −1
i −1 2n(τ +1)
−1)2n(τ +1) +qil
i −1
] (Hili −1 )
il
il
∈ E1 (δ2i1n δ2ni · · · δ2ni pil
−τ +1
i −1 2m
il
il
). Let E2 (δ2i1n δ2ni · · · δ2ni
−τ +1
} and the qili −1 th column of Kili −1 be δ . ··· 4. Find all the [(pi1 − 1)2n(τ +1) + qi1 ]th column of Hi1 such that il il −τ +1 il il −τ +1 ). Let Eli (δ2i1n δ2ni · · · δ2ni )= Col[(pi1 −1)2n(τ +1) +qi1 ] (Hi1 ) ∈ Eli −1 (δ2i1n δ2ni · · · δ2ni qi1 p i1 {δ2n(τ +1) } and the qi1 th column of Ki1 be δ2m . 5. Doing it repeatedly until there exists an integer N such that il −τ +1 il ) = ∆2n(τ +1) . EN (δ2i1n δ2ni · · · δ2ni Theorem 3.2. Suppose that conditions (i) and (ii) in Theorem 3.1 hold. Then under the feedback law with the state feedback matrix K obtained from Algorithm 3.1, the response BN (2) (equivalently (10)) can be completely synchronized with the drive BN (1) (equivalently (7)). Proof. Note that the drive system will eventually enter the cycle Ci , that is the switching signal sequence will eventually be σ = {i1 , · · · , ili }. Assume that condition (ii) in theorem 3.1 is satisfied, then there exists il il −τ +1 N such that EN (δ2i1n δ2ni · · · δ2ni ) = ∆2n(τ +1) under the switching signal sequences {· · · , i1 , · · · , ili }. il il −τ +1 q1 (I) Assume that N = li . That is Eli (δ2i1n δ2ni · · · δ2ni ) = {δ2in(τ +1) } = ∆2n(τ +1) under the switching signal {i1 , · · · , ili }.
8
)=
q
1 Then, for any initial state z(0) = δ2in(τ +1) , it can be verified that
z(1) = Hx(0)u(0)z(0) = Hδ2i1n u(0)z(0) = Hi1 Kx(0)z(0)z(0) q
q
i1 1 = Hi1 Kδ2i1n δ2in(τ +1) δ2n(τ +1)
q
q
i1 1 = Hi1 Ki1 δ2in(τ +1) δ2n(τ +1)
q
1 = Hi1 Colqi1 (Ki1 )δ2in(τ +1)
p
q
i1 1 = Hi1 δ2m δ2in(τ +1)
il
il
= Col[(pi1 −1)2n(τ +1) +qi1 ] (Hi1 ) ∈ Eli −1 (δ2i1n δ2ni · · · δ2ni il
il
Note that Eli −1 (δ2i1n δ2ni · · · δ2ni
−τ +1
−τ +1
).
q
q
i2 2 ) = {δ2in(τ +1) }. Hence, z(1) = δ2n(τ +1) . Doing
il
il
it repeatedly, it can be verified that z(N −1) = z(li −1) ∈ E1 (δ2i1n δ2ni · · · δ2ni qil That is z(N − 1) = z(li − 1) = δ2n(τi +1) . It leads to
−τ +1
).
z(N ) =Hx(N − 1)u(N − 1)z(N − 1) il
=Hδ2ni u(N − 1)z(N − 1) =Hili u(N − 1)z(N − 1)
=Hili Kx(N − 1)z(N − 1)z(N − 1) qil
il
qil
=Hili Kδ2ni δ2n(τi +1) δ2n(τi +1) qil
qil
=Hili Kili δ2n(τi +1) δ2n(τi +1) qil
=Hili Colqil (Kili )δ2n(τi +1) =Hili δ
p il
i
i 2m
=Col[(pi
li
qil
δ2n(τi +1)
−1)2n(τ +1) +qil ] (Hili ) i
il
il
= δ2i1n δ2ni · · · δ2ni
−τ +1
.
Hence, the response system can globally reachable at the state z(N ) = z(li ) = il il −τ +1 δ2i1n δ2ni · · · δ2ni . Then y(li ) = δ2i1n under the switching signal {i1 , · · · , ili }. That is the drive and response system will reach δ2i1n at the same time. From the step 1 of algorithm 3.1, we can see that the response system (10) will stay at cycle Ci forever. The reason is as follows. 9
Note that z(N + 1) =Hx(N )u(N )z(N ) =Hδ2i1n Kx(N )z(N )z(N ) =Hδ2i1n Kx(N )y(N )y(N − 1) · · · y(N − τ )y(N )y(N − 1) · · · y(N − τ ) il
il
=Hi1 Kδ2i1n δ2i1n δ2ni · · · δ2ni il
il
−τ +1
il
il
δ2i1n δ2ni · · · δ2ni il
il
−τ +1
δ2i1n δ2ni · · · δ2ni =Hi1 Ki1 δ2i1n δ2ni · · · δ2ni =Hi1 Col[(i1 −1)2nτ +(il −1)2n(τ −1) +···+il −τ +1 ] (Ki1 ) −τ +1
i
i
il il −τ +1 δ2i1n δ2ni · · · δ2ni il −τ +1 p i 1 i1 il i =Hi1 δ2m δ2n δ2n · · · δ2ni
=Col[(pi1 −1)2n(τ +1) +(i1 −1)2nτ +(il
i
il
=δ2i2n δ2i1n · · · δ2ni
−τ +2
−τ +1
−1)2n(τ −1) +···+ili −τ +1 ] (Hi1 )
,
that is y(N + 1) = δ2i2n = x(N + 1). ···
z(N + li ) =Hx(N + li − 1)u(N + li − 1)z(N + li − 1) =Hx(N + li − 1)Kx(N + li − 1)y(N + li − 1)y(N + li − 2) · · · y(N + li − τ − 1) y(N + li − 1)y(N + li − 2) · · · y(N + li − τ − 1) il
=Hδ2ni Kx(N + li − 1)y(N + li − 1)y(N + li − 2) · · · y(N + li − τ − 1) y(N + li − 1)y(N + li − 2) · · · y(N + li − τ − 1) il
il
il
=Hili Kδ2ni δ2ni δ2ni il
il
=Hili Kili δ2ni δ2ni =Hili Col[(il pil
i
il
−1
il
li
il
il
−τ
il
· · · δ2ni δ2ni δ2ni il
−τ
il
il
· · · δ2ni δ2ni δ2ni
−1
−1
il
· · · δ2ni il
· · · δ2ni
−τ
−τ
ili ili −1 −1)2nτ +(ili −1 −1)2n(τ −1) +···+ili −τ ] (Kili )δ2n δ2n il
=Hili δ2mi δ2ni δ2ni =Col[(pi
−1
−1
il
· · · δ2ni
il
· · · δ2ni
−τ
−τ
−1)2n(τ +1) +(ili −1)2nτ +···+ili −τ ] (Hili ) il
=δ2i1n δ2ni · · · δ2ni
−τ +1
.
That is y(N + li ) = δ2i1n = x(N + li ). Then the response system (10) will stay at Ci forever. 10
(II) Assume that N < li . The proof is similar to (I). (III) Assume that N > li . il −τ +1 ) under the switching signal For Ek (δ2i1n δ2i2n · · · δ2ni il −τ +1 ) under the switching signal {i(li −k+1) , · · · , i(li −1) , ili }, Ek+mli (δ2i1n δ2i2n · · · δ2ni {i(li −k+1) , · · · , i(li −1) , ili · · · , i1 , · · · , ili , · · · , i1 , · · · , ili }, it can be proved that il −τ +1 il −τ +1 ), where k = 1, 2, · · · , li , m ∈ ) ⊆ Ek+mli (δ2i1n δ2i2n · · · δ2ni Ek (δ2i1n δ2i2n · · · δ2ni N. The reason is as follows. Without loss of generality, we prove that il −τ +1 il −τ +1 ) under the switching signal ili ) ⊆ E1+li (δ2i1n δ2i2n · · · δ2ni E1 (δ2i1n δ2i2n · · · δ2ni and switching signal sequence {ili , i1 , · · · , ili } respectively. Note that if condiil −τ +1 il il −τ il il −1 ) tion (i) in theorem 3.1 is satisfied, then δ2ni δ2ni · · · δ2ni ∈ E1 (δ2i1n δ2ni · · · δ2ni ili ili −1 ili −τ ili −τ +1 i1 il i under the switching signal ili , that is δ2n δ2n · · · δ2n can reach δ2n δ2n · · · δ2n il −τ +1 il ∈ in one step under the switching signal ili . Moreover, δ2i1n δ2ni · · · δ2ni ili −τ +2 ili −τ +1 i2 i1 i1 il i E1 (δ2n δ2n · · · δ2n ) under the switching signal i1 , that is δ2n δ2n · · · δ2n il −τ +2 i2 i1 can reach δ2n δ2n · · · δ2ni in one step under the switching signal i1 . Doing it il il −1 il −τ il il −τ +1 repeatedly, we can see that δ2ni δ2ni · · · δ2ni can reach δ2i1n δ2ni · · · δ2ni in 1+ li steps under the switching signal {ili , i1 , · · · , ili }. Hence, for any initial state il −τ +1 il −τ +1 il il ) ) = ∆2n(τ +1) , we have z(1) ∈ EN −1 (δ2i1n δ2ni · · · δ2ni z(0) ∈ EN (δ2i1n δ2ni · · · δ2ni ili −τ +1 i1 il i or z(1) ∈ EN −1−˜ali (δ2n δ2n · · · δ2n ). Doing it repeatedly, there exists T such ili −τ +1 i 1 il i )), where ˜b ≤ li . Then similar to the proof that z(T ) ∈ E˜b (δ2n δ2n · · · δ2n in (I) and (II), we can draw the conclusion. 4. Example Example 4.1. Assume that the drive-response system is x1 (t + 1) = x1 (t) ∨ (¬x2 (t)), x2 (t + 1) = ¬x2 (t), y1 (t + 1) = x1 (t) ∨ y2 (t − 1), y2 (t + 1) = u(t) ∧ y1 (t), u(t) = k(x1 (t), x2 (t), y1 (t), y2 (t), y1 (t − 1), y2 (t − 1)).
(11) (12) (13)
By letting x(t) = n2i=1 xi (t), y(t) = n2i=1 yi (t), z(t) = y(t)y(t − 1), the algebraic form of the drive-response system is x(t + 1) = F x(t), z(t + 1) = Hx(t)u(t)z(t), 11
where F = δ4 [2, 1, 4, 1] and H = δ4 [1, 1, 1, 1, 2, 2, 2, 2, 7, 7, 7, 7, 8, 8, 8, 8, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 1, 1, 1, 1, 2, 2, 2, 2, 7, 7, 7, 7, 8, 8, 8, 8, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 1, 9, 1, 9, 2, 10, 2, 10, 7, 15, 7, 15, 8, 8, 8, 8, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 1, 9, 1, 9, 2, 10, 2, 10, 7, 15, 7, 15, 8, 16, 8, 16, 5, 13, 5, 13, 6, 14, 6, 14, 7, 15, 7, 15, 8, 16, 8, 16]. We can see that the cycle of (11) is (δ41 , δ42 ). Split H into 2n = 4 equal blocks as H = [H1 , H2 , H3 , H4 ]. It can be verified that, for x(t) = y(t) = δ41 , y(t − 1) = δ42 , when u(t) = δ22 , 5 z(t + 1) = Hx(t)u(t)z(t) = Hδ41 δ22 δ41 δ42 = δ16 = δ42 δ41 .
That is δ41 δ42 ∈ E1 (δ42 δ41 ) under the switching signal σ = 1. In the same way, we can verify that δ42 δ41 ∈ E1 (δ41 δ42 ) under the switching signal σ = 2. 5 6 7 8 we can see that E1 (δ41 δ42 ) = {δ16 , δ16 , δ16 , δ16 } under the switching signal 1 2 σ = 2. E2 (δ4 δ4 ) = ∆16 under the switching signal σ = {1, 2}. Hence, conditions (i) and (ii) of Theorem 3.1 are satisfied. Now, we use Algorithm 3.1 to obtain the feedback matrix K. For the cycle Ci = (δ41 , δ42 ). 1. Find the [(2 − 1)16 + (1 − 1)4 + 2] = 18th column of H1 such that 5 Col18 (H1 ) = δ42 δ41 = δ16 . Let the (1-1)4+2=2nd column of K1 be δ22 . Find the 2 [(1−1)16+(2−1)4+1] = 5th column of H2 such that Col5 (H2 ) = δ41 δ42 = δ16 . 1 Let the (2-1)4+1=5th column of K2 be δ2 . 2. Find the [(1 − 1)16 + 5] = 5th, [(1 − 1)16 + 6] = 6th, [(1 − 1)16 + 7] = 7th, [(1 − 1)16 + 8] = 8th column of H2 such that Col5 (H2 ) = Col6 (H2 ) = 2 5 6 7 8 Col7 (H2 ) = Col8 (H2 ) = δ41 δ42 = δ16 . Let E1 (δ41 δ42 ) = {δ16 , δ16 , δ16 , δ16 } and the 1 5th, 6th, 7th and 8th column of K2 be δ2 . 3. Find the [(2 − 1)16 + 1] = 17th, [(2 − 1)16 + 2] = 18th, [(2 − 1)16 + 3] = 19th, [(2 − 1)16 + 4] = 20th column of H1 such that Col17 (H1 ) = Col18 (H1 ) = 5 Col19 (H1 ) = Col20 (H1 ) = δ16 ∈ E1 (δ41 δ42 ). Let the 1st, 2nd, 3th and 4th column of K1 be δ22 . Find the [(2 − 1)16 + 5] = 21th, [(2 − 1)16 + 6] = 22th, [(2 − 1)16 + 7] = 23th, [(2 − 1)16 + 8] = 24th column of H1 such that 6 Col21 (H1 ) = Col22 (H1 ) = Col23 (H1 ) = Col24 (H1 ) = δ16 ∈ E1 (δ41 δ42 ). Let the 2 5th, 6th, 7th and 8th column of K1 be δ2 . Find the [(1 − 1)16 + 9] = 9th, [(1 − 1)16 + 10] = 10th, [(1 − 1)16 + 11] = 11th, [(1 − 1)16 + 12] = 12th 12
column of H1 such that Col9 (H1 ) = Col10 (H1 ) = Col11 (H1 ) = Col12 (H1 ) = 7 δ16 ∈ E1 (δ41 δ42 ). Let the 9th, 10th, 11th and 12th column of K1 be δ21 . Find the [(1 − 1)16 + 13] = 13th, [(1 − 1)16 + 14] = 14th, [(1 − 1)16 + 15] = 15th, [(1 − 1)16 + 16] = 16th column of H1 such that Col13 (H1 ) = Col14 (H1 ) = 8 ∈ E1 (δ41 δ42 ). Let the 13th, 14th, 15th and 16th Col15 (H1 ) = Col16 (H1 ) = δ16 column of K1 be δ21 . In summary, the feedback matrix K is K = δ2 [2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, ∗, ∗, ∗, ∗, 1, 1, 1, 1, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗ ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗] Example 4.2. Consider the D. melanogaster segmentation polarity gene network introduced in Example 4.2 of [43]. If the gene wg4 (0) = 1, wg1 (0) = 0, then wg4 (t) = 1, for t ≥ 1. That is ¬wg4 (t) ≡ 0, for t ≥ 0. Hence wg1 (t) = 0, for t ≥ 0. Then, by letting wg2 = x1 , wg3 = x2 , P T C1 = y1 , P T C2 = y2 , the reduced model is as follows x1 (t + 1) = x1 (t) ∧ ¬x2 (t), x2 (t + 1) = x2 (t), y1 (t + 1) = y1 (t) ∧ ¬x2 (t), y2 (t + 1) = ¬x2 (t).
(14) (15)
In this paper, we consider the reduced model with time delay and controls as follows x1 (t + 1) = x1 (t) ∧ ¬x2 (t), x2 (t + 1) = x2 (t), y1 (t + 1) = u1 (t) ↔ (y1 (t − 1) ∧ ¬x2 (t)), y2 (t + 1) = u2 (t) ↔ (¬x2 (t)).
(16) (17)
By letting x(t) = x1 (t) n x2 (t), it can be verified that x(t + 1) = F x(t), where F = δ4 [3, 2, 3, 4]. By letting y(t) = y1 (t) n y2 (t), z(t) = y(t) n y(t − 1) it leads to z(t + 1) = Hx(t)u(t)z(t),
13
where H = δ16 [13, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 1, 1, 9, 9, 2, 2, 10, 10, 3, 3, 11, 11, 4, 4, 12, 12, 5, 5, 13, 13, 6, 6, 14, 14, 7, 7, 15, 15, 8, 8, 16, 16, 9, 9, 13, 13, 10, 10, 14, 14, 11, 11, 15, 15, 12, 12, 16, 16, 13, 13, 5, 5, 14, 14, 6, 6, 15, 15, 7, 7, 16, 16, 8, 8, 13, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 1, 1, 9, 9, 2, 2, 10, 10, 3, 3, 11, 11, 4, 4, 12, 12, 5, 5, 13, 13, 6, 6, 14, 14, 7, 7, 15, 15, 8, 8, 16, 16, 9, 9, 1, 1, 10, 10, 2, 2, 11, 11, 3, 3, 12, 12, 4, 4, 13, 13, 5, 5, 14, 14, 6, 6, 15, 15, 7, 7, 16, 16, 8, 8]. We can see that the cycle of (14) is (δ42 ), (δ43 ), (δ44 ). It can be verified that conditions (i) and (ii) in Theorem 3.1 are satisfied. Now, we will use algorithm 3.1 to obtain the feedback matrix K. 1. For the cycle C1 = (δ42 ). Find the [(2−1)16+(2−1)4+2] = 22th of H2 6 = δ42 δ42 . Let the (2 − 1)4 + 2 = 6th column of K2 be such that Col22 (H2 ) = δ16 δ42 . Find all the [(2−1)16+5] = 21th, [(2−1)16+6] = 22th column of H2 such 6 5 6 } and the , δ16 . Let E1 (δ42 δ42 ) = {δ16 that Col21 (H2 ) = Col22 (H2 ) = δ42 δ42 = δ16 5th, 6th column of K2 be δ42 . Find [(2−1)16+1] = 17th, [(2−1)16+2] = 18th, [(4 − 1)16 + 3] = 51th, [(4 − 1)16 + 4] = 52th column of H2 such that 5 Col17 (H2 ) = Col18 (H2 ) = Col51 (H2 ) = Col52 (H2 ) = δ16 ∈ E1 (δ42 δ42 ). Let 6 5 2 3 4 1 } and the 1st, 2nd, 3th and 4th column of , δ16 , δ16 , δ16 , δ16 , δ16 E2 (δ42 δ42 ) = {δ16 2 4 4 2 K2 be δ4 , δ4 , δ4 , δ4 respectively. Similarly, let the 7th, 8th, 9th, 10th, 11th, 12th, 13th, 14th, 15th, 16th column of K2 be δ42 , δ42 , δ41 , δ41 , δ44 , δ44 , δ41 , δ41 , δ44 , δ44 respectively. 2. For the cycle C2 = (δ43 ). Let the 1st to the 16th column of K3 be δ42 respectively. 3. For the cycle C3 = (δ44 ). Let the 1st, 2nd, 3th, 4th, · · · , 16th column of K4 be δ43 , δ43 , δ41 , δ41 , δ43 , δ43 , δ41 , δ41 , δ44 , δ44 , δ42 , δ42 , δ44 , δ44 , δ42 , δ42 respectively. From the above, we can conclude that K is K = δ4 [∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, 2, 2, 4, 4, 2, 2, 2, 2, 1, 1, 4, 4, 1, 1, 4, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 1, 1, 3, 3, 1, 1, 4, 4, 2, 2, 4, 4, 2, 2]. 14
Without loss of generality, we assume that K = δ4 [2, 2, 4, 4, 2, 2, 2, 2, 1, 1, 4, 4, 2, 2, 4, 4, 2, 2, 4, 4, 2, 2, 2, 2, 1, 1, 4, 4, 1, 1, 4, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 1, 1, 3, 3, 1, 1, 4, 4, 2, 2, 4, 4, 2, 2]. By using [12], we have u1 (t) =[x1 (t) ∧ ((y1 (t) ∧ (y1 (t − 1) ∨ ¬y2 (t)) ∨ (¬y1 (t) ∧ y1 (t − 1)))] ∨ [¬x1 (t) ∧ (x2 (t) ∨ ¬y1 (t − 1)], u2 (t) =[x1 (t) ∧ (¬y1 (t) ∧ y1 (t − 1)] ∧ [¬x1 (t) ∧ (¬x2 (t) ∧ y1 (t)]. It thus follows that the complete synchronization of the drive-response systems can be achieved. For example, assume that the initial states of the drive system and response system be x(0) = δ41 and y(0) = δ43 , y(−1) = δ44 respectively, then the trajectories are as follows: (x1 , x2 ) : (1, 1) → (0, 1) → (0, 1) → · · · (y1 , y2 ) : (0, 1) → (1, 1) → (0, 1) → · · · . 5. Conclusion This paper has investigated the feedback control design approach for the complete synchronization of two coupled Boolean networks with time delay. A necessary condition for the existence of a state feedback controller to achieve the complete synchronization has been established. Then based on the necessary condition, a constructive procedure for control design has been provided. An example has been worked out to illustrate the effectiveness of this paper. In this paper, the time delay τ should be known when designing controller. In practical situations, the time delay τ should be identified from microarray data. However, if the time delay is unknown, our method can not be applied to. Thus, we shall leave this problem for future study. Acknowledgement This work is supported by National Natural Science Foundation of China under Grant 61403139, Open Foundation of first level Zhejiang key in key discipline of Control Science and Engineering and the Fundamental Research Funds for the Central Universities. 15
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Highlights:
1.
A necessary condition for the existence of a state feedback controller for the complete synchronization of two coupled Boolean networks with time delays is derived.
2.
Feedback control design procedure for the complete synchronization of two coupled Boolean networks is provided based on the necessary condition.
3.
An example is given to illustrate the proposed design procedure.
PII: DOI: Reference:
S0378-4371(17)30774-4 http://dx.doi.org/10.1016/j.physa.2017.08.041 PHYSA 18487
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Physica A
Received date : 16 February 2017 Revised date : 1 June 2017 Please cite this article as: F. Li, J. Li, L. Shen, State feedback controller design for the synchronization of Boolean networks with time delays, Physica A (2017), http://dx.doi.org/10.1016/j.physa.2017.08.041 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
State feedback controller design for the synchronization of Boolean networks with time delays Fangfei Li1,2∗, Jianning Li1∗, Lijuan Shen3∗ 1 School of Automation, Hangzhou Dianzi University, Hangzhou, 310018, P.R.China 2 Department of Mathematics, East China University of Science and Technology, Shanghai 200237, PR China 3 School of Mathematical Sciences, Tongji University, Shanghai 200092, China
Abstract State feedback control design to make the response Boolean network synchronize with the drive Boolean network is far from being solved in the literature. Motivated by this, this paper studies the feedback control design for the complete synchronization of two coupled Boolean networks with time delays. A necessary condition for the existence of a state feedback controller is derived first. Then the feedback control design procedure for the complete synchronization of two coupled Boolean networks is provided based on the necessary condition. Finally, an example is given to illustrate the proposed design procedure. Keywords: Boolean network, time delay, semi-tensor product, feedback control design. 1. Introduction Recently, the study of logical systems has attracted attention, because it can provide a simple and proper model to describe genetic regulatory networks, neural networks, etc, [1–4]. Boolean networks (BNs) as a kind of two valued logical system has been studied widely [5–7]. In a Boolean network, the state of each Boolean variable is 1 or 0, which means the state is on or off. Every Boolean variable updates its state according to a logical relationship, ∗
Corresponding author. Email address: [email protected] (Fangfei Li1,2 )
Preprint submitted to Physica A
August 28, 2017
given in the form of a Boolean function. Up to now, many significant results about the topological structure of Boolean network, including the attractors and transient time have been obtained, for example, see [8–11]. With the development of the semi-tensor product (STP) of matrices, an algebraic form of a Boolean network can be obtained. Then, some interesting progresses have been achieved, including the controllability and observability of Boolean networks, disturbance decoupling controller design for Boolean control networks, stability and stabilization of Boolean networks with impulsive effects etc, see e.g. [12–28]. In the past decades, synchronization of dynamic systems has been widely investigated, and still draw much research attention [29–32]. Synchronization of two Boolean networks is meaningful too. For example, the results on the synchronization of two coupled Boolean networks can be applied to the synchronization of two lasers [33]. The study of synchronized BNs can provide useful information on the coevolution of several biological species whose genetic dynamics influence each other [34]. Based on the semi-tensor product of matrices, necessary and sufficient conditions of the complete synchronization for two coupled Boolean networks are obtained [35]. Later, the results are generalized to the complete synchronization of Boolean networks with time delays [36]. Though some results on the synchronization of Boolean networks have been obtained in the literature, we can note that these results are control-free. To design controller to make the response BN synchronize with the drive BN is significant, which is one of the motivation of this paper. As we know, time delay phenomenon is so common in the real world, and is very important in the analysis and control of dynamic systems [37, 38]. In addition, time delay happens frequently and is unavoidable in biological and physiological systems [39]. For example, time delays of genetic regulatory networks are caused by slow processes of transcription, translation and so on. Since we usually use BNs to model genetic regulatory networks, the analysis and control of BNs with time delays are meaningful. Although there exist some results on the control problems of BNs with time delays, however to design controller for the synchronization of BNs with time delays is still lacking in the literature, to the best of our knowledge. Motivated by this, in the paper, we study the state feedback control design for the synchronization of two coupled BNs with time delays. As to the feedback control design for the synchronization analysis of two coupled Boolean networks with time delays, one main difficulty lies in the controller design. In this paper, time delay phenomenon is taken into ac2
count. Different from [36], where the time delays for each node are the same, in this paper, we consider the case that the time delays for each node are different. Thus, compared with [36] and [40], the system structure in this paper is more complex. In addition, to calculate the set consisting of all the states that can be reachable to the desirable state is much more complicated due to time delay. Moreover, a much more difficult situation arises since the response dynamic appear with different model structures for the reason that it is affected by the drive system. To overcome these difficulties, in this paper, a necessary condition for the existence of a state feedback controller to achieve the complete synchronization is first established. Then, we see the response system as a switched BN, and based on the necessary condition, the feedback law is constructed. The rest of this paper is organized as follows. In section 2, we provide some preliminary results. Section 3 gives a necessary condition to achieve synchronization. Then based on the necessary condition, feedback control design is obtained for the complete synchronization of two coupled Boolean networks. An example is presented in section 4. Finally, section 5 gives a brief conclusion. 2. Preliminaries In this section, some necessary preliminary results are given. Definition 2.1. ([41]). For M ∈ Rm×n and N ∈ Rp×q , their semi-tensor product (STP), denoted by M n N , is defined as follows: M n N := (M ⊗ Is/n )(N ⊗ Is/p ), where s is the least common multiple of n and p, and ⊗ is the Kronecker product. Remark 2.1. The matrix product we use in this paper is the semi-tensor product (STP). It can be verified that when n = p, the semi-tensor product of matrices M and N becomes the conventional matrix product. Hence, STP is a generalization of conventional matrix product. All the fundamental properties of conventional matrix product remain true. Based on this, we can omit the symbol “ n ” if no confusion raises. There are also some basic properties of STP, for detail, see ([42]). 3
Next, some notations are provided. (1) D := {1, 0}. (2) ∆k := {δki |1 ≤ i ≤ k}, where δki is the ith column of the identity matrix Ik . For compactness, ∆2 = ∆. (3) Denote the ith column of matrix A by Coli (A). (4) Denote by Mm×n the set of all m × n matrices. A matrix A ∈ Mm×n is called a logical matrix, if the columns of A are elements of ∆m . Denote the set of logical matrices by L; the set of n × s logical matrices is defined by Ln×s . (6) Assume that there is a matrix M = [δni1 , δni2 , · · · , δnis ], for notational compactness, denote by M := δn [i1 , i2 , · · · , is ]. By identifying True = 1 ∼ δ21 , False = 0 ∼ δ22 , the logical variable A(t) takes value from these two vectors, i.e. A(t) ∈ ∆ = {δ21 , δ22 }, where p ∼ q denotes the logical equivalence of p and q. The following lemma is fundamental for the matrix expression of the logical function. Lemma 2.1. ([41]) Any logical function L(A1 , · · · An ) with logical arguments A1 , · · · , An ∈ ∆ can be expressed in a multi-linear form as L(A1 , . . . , An ) = ML A1 A2 · · · An , where ML ∈ L2×2n is unique, called the structure matrix of L. 3. Main results Two Boolean networks coupled in the drive-response configuration are considered in this paper, where the response Boolean network is with time delays. The dynamic is described as follows. xi (t + 1) = fi (x1 (t), · · · , xn (t)), (1) yi (t + 1) = gi (u1 (t), · · · , um (t), x1 (t), · · · , xn (t), y1 (t), · · · , yn (t), y1 (t − 1), · · · , yn (t − 1), · · · , y1 (t − τ ), · · · , yn (t − τ )), (2) i = 1, 2, · · · , n, where xi , yi ∈ D, i = 1, 2, · · · , n, are the nodes of the drive BN (1) and the response BN (2) respectively, fi , gi are Boolean functions, u1 , u2 , · · · , 4
um ∈ D are the control inputs. The objective of this paper is to find a feedback law of the form uj (t) = kj (x1 (t), · · · , xn (t), y1 (t), · · · , yn (t), y1 (t − 1), · · · , yn (t − 1), · · · , yn (t − τ ), · · · , yn (t − τ )), j = 1, 2, · · · , m, (3) where k1 ,· · · , km are Boolean functions such that under this feedback control the response BN (2) can be globally synchronized with the drive BN (1). By using lemma 2.1, we denote x(t) = nni=1 xi (t), y(t) = nni=1 yi (t), u(t) = nm j=1 uj (t) and can express (1), (2) and (3) as xi (t + 1) = Fi x(t), yi (t + 1) = Gi u(t)x(t)y(t)y(t − 1) · · · y(t − τ ), uj (t) = Kj x(t)y(t)y(t − 1) · · · y(t − τ ),
(4) (5) (6)
where Fi , Gi and Kj are the structure matrix of fi , gi and kj respectively, i = 1, 2, · · · , n, j = 1, 2, · · · , m. Moreover, (4), (5) and (6) can further be converted into the following discrete-time system x(t + 1) = F x(t), y(t + 1) = Gu(t)x(t)y(t)y(t − 1) · · · y(t − τ ), u(t) = Kx(t)y(t)y(t − 1) · · · y(t − τ ),
(7) (8) (9)
where equations (7), (8) and (9) are called the algebraic representations of the BN (1), (2) and the feedback law (3) respectively, and matrix K is called the state feedback matrix. Let z(t) = nτi=0 zi (t), where z0 (t) = y(t), z1 (t) = y(t − 1), · · · ,zτ (t) = y(t − τ ), (8) can be rewritten as y(t + 1) = Gu(t)x(t)z(t). z(t + 1) = nτi=0 zi (t + 1) = y(t + 1)y(t) · · · y(t − τ + 1) =Gu(t)x(t)z(t)y(t) · · · y(t − τ + 1) =Gu(t)x(t)y(t)y(t − 1) · · · y(t − τ )y(t) · · · y(t − τ + 1) =Gu(t)x(t)W[2nτ ,2n(τ +1) ] (y(t) · · · y(t − τ + 1))2 y(t − τ ) =Gu(t)x(t)W[2nτ ,2n(τ +1) ] Φnτ z(t)
=G(I2m+n ⊗ W[2nτ ,2n(τ +1) ] Φnτ )u(t)x(t)z(t) ¯ ,Gu(t)x(t)z(t). 5
To see the definition of W[m,n] and Φnτ , it refers to [42]. Note that ¯ ¯ [2n ,2m ] x(t)u(t)y(t) z(t + 1) =Gu(t)x(t)z(t) = GW ,Hx(t)u(t)z(t), then (8) can be rewritten as z(t + 1) = Hx(t)u(t)z(t).
(10)
We split H into 2n equal dimension blocks as H = [H1 , H2 , · · · , H2n ]. Note that, if x(t) = δ2i n , z(t + 1) = Hx(t)u(t)z(t) = Hδ2i n u(t)z(t) = Hi u(t)z(t). Hence, (10) can be interpreted as a switched Boolean system switching between 2n possible subsystems. Assume that the switching signal is σ(t) : N → Ω = {1, 2, · · · , 2n }. Then under the switching signal σ = i, (that is x(t) = δ2i n ), the drive system (10) is z(t + 1) = Hi u(t)z(t). In the sequel, we denote El (δ2rn1 δ2rn2 · · · δ2rns ) under the switching signal sequence {σ(1), σ(2), · · · , σ(l − 1)} by the set consisting of all the states that can be steered to δ2rn1 n δ2rn2 n · · · n δ2rns in l steps under the switching signal sequence {σ(1), σ(2), · · · , σ(l − 1)}. Definition 3.1. An ordered sequence of distinct vectors (δ2i1n , δ2i2n , · · · , δ2ikn ) is a limit cycle C of the BN (7) if x(0) = δ2iln for some l ∈ [1, k] ensures that the corresponding state trajectory x(t) is periodic of period k and, for every i x(t) is periodic of period k and, for every t ∈ Z+ , x(t) = δ2jn , where j ∈ [1, k] and j ≡ (t + l) mod k. A limit cycle of unitary length is an equilibrium point of the BN. il
In the sequel, assume that Ci = (δ2i1n , δ2i2n , · · · , δ2ni ) is the ith cycle of the BN (7), and the length is li . il
Theorem 3.1. Let Ci = (δ2i1n , δ2i2n , · · · , δ2ni ) be the ith cycle of the drive BN (1) (equivalently (7)). If the response BN (2) (equivalently (10)) can be 6
completely synchronized with the drive BN (1) (equivalently (7)), then the state of response BN (10) satisfies il −τ +2 il −τ +1 il ) under the switching signal i1 ; ∈ E1 (δ2i2n δ2i1n · · · δ2ni (i) δ2i1n δ2ni · · · δ2ni ili −τ +3 ili −τ +2 i 3 i2 i2 i1 il i ) under the switching signal i2 ; ∈ E1 (δ2n δ2n · · · δ2n δ2n δ2n δ2n · · · δ2n .. . il
il
il
il
il
il
il
) under the switching signal ili ; δ2ni δ2ni · · · δ2ni ∈ E1 (δ2i1n δ2ni · · · δ2ni il il −τ +1 ) = ∆2n(τ +1) (ii) there exists an integer N , such that EN (δ2i1n δ2ni · · · δ2ni under the switching signal sequences {· · · i1 , · · · , ili }. −1
−τ
−τ +1
il
) under the ∈ E1 (δ2i2n δ2i1n · · · δ2ni Proof. We prove that δ2i1n δ2ni · · · δ2ni switching signal i1 first. Note that the drive system (7) will reach the cycle Ci , then it will reach the state x(t) = δ2i1n , that is the switching signal is σ = i1 . If the complete synchronization is achieved, we can find an integer T , such that x(T ) = δ2i1n , y(T ) = δ2i1n . Moreover, x(T + 1) = y(T + 1) = δ2i2n , x(T − 1) = il il −1 il −τ +1 y(T −1) = δ2ni , x(T −2) = y(T −2) = δ2ni , · · · , x(T −τ ) = y(T −τ ) = δ2ni . ili −τ +1 i1 il i Then z(T ) = y(T )y(T − 1) · · · y(T − τ ) = δ2n δ2n · · · δ2n , z(T + 1) = ili −τ +2 il −τ +1 i 2 i1 i1 ili y(T + 1) · · · y(T − τ + 1) = δ2n δ2n · · · δ2n . That is δ2n δ2n · · · δ2ni ∈ ili −τ +2 i2 i1 E1 (δ2n δ2n · · · δ2n ) under the switching signal i1 . Similarly, we can prove that condition (i) is satisfied. Now, we will show that condition (ii) is satisfied. If the complete synchronization of (7) and (10) is achieved, we can find an integer T , such that il −1 il x(T ) = y(T ) = δ2i1n , x(T − 1) = y(T − 1) = δ2ni , x(T − 2) = y(T − 2) = δ2ni , il −τ +1 · · · , x(T − τ ) = y(T − τ ) = δ2ni , and all the states of the response system il il −τ +1 (10) can reach z(T ) = y(T )y(T − 1) · · · y(T − τ ) = δ2i1n δ2ni · · · δ2ni . That is condition (ii) is satisfied. This completes the proof. We give the following algorithm to construct matrix K to achieve the complete synchronization. Split K into 2n square blocks as K = [K1 , K2 , · · · , K2n ]. Algorithm 3.1. Suppose that conditions (i) and (ii) in Theorem 3.1 hold. 1. Find the [(pi1 −1)2n(τ +1) +(i1 −1)2nτ +(ili −1)2n(τ −1) +· · ·+(ili −τ +1)]th column of Hi1 such that Col(pi1 −1)2n(τ +1) +(i1 −1)2nτ +(il −1)2n(τ −1) +···+(il −τ +1)] (Hi1 ) il
−τ +1
i
−τ +2
i
= δ2i2n δ2i1n · · · δ2ni . Let the [(i1 − 1)2nτ + (ili − 1)2n(τ −1) + · · · + (ili − τ + 1)]th pi1 column of Ki1 be δ2m . Find the [(pi2 −1)2n(τ +1) +(i2 −1)2nτ +(i1 −1)2n(τ −1) + · · · + (ili − τ + 2)]th column of Hi2 such that −τ +2
7
Col[(pi2 −1)2n(τ +1) +(i2 −1)2nτ +(i1 −1)2n(τ −1) +···+(il
i
il −τ +3 . δ2i3n δ2i2n · · · δ2ni
−τ +2)] (Hi2 )
Let the [(i2 − 1)2nτ + (i1 − 1)2n(τ −1) + · · · + (ili − τ + 2)]th = p i2 column of Ki2 be δ2m .· · · Find the [(pili − 1)2n(τ +1) + (ili − 1)2nτ + (ili −1 − 1)2n(τ −1) + · · · + (ili − τ )]th column of Hili such that Col[(pi −1)2n(τ +1) +(il −1)2nτ +(il −1 −1)2n(τ −1) +···+(il −τ )] (Hili ) li
i
i
il δ2i1n δ2ni
i
il −τ +1 · · · δ2ni .
Let the [(ili − 1)2 + (ili −1 − 1)2n(τ −1) + · · · + (ili − τ )]th = p il column of Kili be δ2mi . 2. Find all the [(pili − 1)2n(τ +1) + qili ]th column of Hili such that Col[(pi qil
li
nτ
−1)2n(τ +1) +qil ] (Hili ) i
il
il
= δ2i1n δ2ni · · · δ2ni
−τ +1
p il
il
il
. Let E1 (δ2i1n δ2ni · · · δ2ni
−τ +1
) =
{δ2n(τi +1) } and the qili th column of Kili be δ2mi . 3. Find all the [(pili −1 − 1)2n(τ +1) + qili −1 ]th column of Hili −1 such that
Col[(pi {δ
qil
li −1
i −1 2n(τ +1)
−1)2n(τ +1) +qil
i −1
] (Hili −1 )
il
il
∈ E1 (δ2i1n δ2ni · · · δ2ni pil
−τ +1
i −1 2m
il
il
). Let E2 (δ2i1n δ2ni · · · δ2ni
−τ +1
} and the qili −1 th column of Kili −1 be δ . ··· 4. Find all the [(pi1 − 1)2n(τ +1) + qi1 ]th column of Hi1 such that il il −τ +1 il il −τ +1 ). Let Eli (δ2i1n δ2ni · · · δ2ni )= Col[(pi1 −1)2n(τ +1) +qi1 ] (Hi1 ) ∈ Eli −1 (δ2i1n δ2ni · · · δ2ni qi1 p i1 {δ2n(τ +1) } and the qi1 th column of Ki1 be δ2m . 5. Doing it repeatedly until there exists an integer N such that il −τ +1 il ) = ∆2n(τ +1) . EN (δ2i1n δ2ni · · · δ2ni Theorem 3.2. Suppose that conditions (i) and (ii) in Theorem 3.1 hold. Then under the feedback law with the state feedback matrix K obtained from Algorithm 3.1, the response BN (2) (equivalently (10)) can be completely synchronized with the drive BN (1) (equivalently (7)). Proof. Note that the drive system will eventually enter the cycle Ci , that is the switching signal sequence will eventually be σ = {i1 , · · · , ili }. Assume that condition (ii) in theorem 3.1 is satisfied, then there exists il il −τ +1 N such that EN (δ2i1n δ2ni · · · δ2ni ) = ∆2n(τ +1) under the switching signal sequences {· · · , i1 , · · · , ili }. il il −τ +1 q1 (I) Assume that N = li . That is Eli (δ2i1n δ2ni · · · δ2ni ) = {δ2in(τ +1) } = ∆2n(τ +1) under the switching signal {i1 , · · · , ili }.
8
)=
q
1 Then, for any initial state z(0) = δ2in(τ +1) , it can be verified that
z(1) = Hx(0)u(0)z(0) = Hδ2i1n u(0)z(0) = Hi1 Kx(0)z(0)z(0) q
q
i1 1 = Hi1 Kδ2i1n δ2in(τ +1) δ2n(τ +1)
q
q
i1 1 = Hi1 Ki1 δ2in(τ +1) δ2n(τ +1)
q
1 = Hi1 Colqi1 (Ki1 )δ2in(τ +1)
p
q
i1 1 = Hi1 δ2m δ2in(τ +1)
il
il
= Col[(pi1 −1)2n(τ +1) +qi1 ] (Hi1 ) ∈ Eli −1 (δ2i1n δ2ni · · · δ2ni il
il
Note that Eli −1 (δ2i1n δ2ni · · · δ2ni
−τ +1
−τ +1
).
q
q
i2 2 ) = {δ2in(τ +1) }. Hence, z(1) = δ2n(τ +1) . Doing
il
il
it repeatedly, it can be verified that z(N −1) = z(li −1) ∈ E1 (δ2i1n δ2ni · · · δ2ni qil That is z(N − 1) = z(li − 1) = δ2n(τi +1) . It leads to
−τ +1
).
z(N ) =Hx(N − 1)u(N − 1)z(N − 1) il
=Hδ2ni u(N − 1)z(N − 1) =Hili u(N − 1)z(N − 1)
=Hili Kx(N − 1)z(N − 1)z(N − 1) qil
il
qil
=Hili Kδ2ni δ2n(τi +1) δ2n(τi +1) qil
qil
=Hili Kili δ2n(τi +1) δ2n(τi +1) qil
=Hili Colqil (Kili )δ2n(τi +1) =Hili δ
p il
i
i 2m
=Col[(pi
li
qil
δ2n(τi +1)
−1)2n(τ +1) +qil ] (Hili ) i
il
il
= δ2i1n δ2ni · · · δ2ni
−τ +1
.
Hence, the response system can globally reachable at the state z(N ) = z(li ) = il il −τ +1 δ2i1n δ2ni · · · δ2ni . Then y(li ) = δ2i1n under the switching signal {i1 , · · · , ili }. That is the drive and response system will reach δ2i1n at the same time. From the step 1 of algorithm 3.1, we can see that the response system (10) will stay at cycle Ci forever. The reason is as follows. 9
Note that z(N + 1) =Hx(N )u(N )z(N ) =Hδ2i1n Kx(N )z(N )z(N ) =Hδ2i1n Kx(N )y(N )y(N − 1) · · · y(N − τ )y(N )y(N − 1) · · · y(N − τ ) il
il
=Hi1 Kδ2i1n δ2i1n δ2ni · · · δ2ni il
il
−τ +1
il
il
δ2i1n δ2ni · · · δ2ni il
il
−τ +1
δ2i1n δ2ni · · · δ2ni =Hi1 Ki1 δ2i1n δ2ni · · · δ2ni =Hi1 Col[(i1 −1)2nτ +(il −1)2n(τ −1) +···+il −τ +1 ] (Ki1 ) −τ +1
i
i
il il −τ +1 δ2i1n δ2ni · · · δ2ni il −τ +1 p i 1 i1 il i =Hi1 δ2m δ2n δ2n · · · δ2ni
=Col[(pi1 −1)2n(τ +1) +(i1 −1)2nτ +(il
i
il
=δ2i2n δ2i1n · · · δ2ni
−τ +2
−τ +1
−1)2n(τ −1) +···+ili −τ +1 ] (Hi1 )
,
that is y(N + 1) = δ2i2n = x(N + 1). ···
z(N + li ) =Hx(N + li − 1)u(N + li − 1)z(N + li − 1) =Hx(N + li − 1)Kx(N + li − 1)y(N + li − 1)y(N + li − 2) · · · y(N + li − τ − 1) y(N + li − 1)y(N + li − 2) · · · y(N + li − τ − 1) il
=Hδ2ni Kx(N + li − 1)y(N + li − 1)y(N + li − 2) · · · y(N + li − τ − 1) y(N + li − 1)y(N + li − 2) · · · y(N + li − τ − 1) il
il
il
=Hili Kδ2ni δ2ni δ2ni il
il
=Hili Kili δ2ni δ2ni =Hili Col[(il pil
i
il
−1
il
li
il
il
−τ
il
· · · δ2ni δ2ni δ2ni il
−τ
il
il
· · · δ2ni δ2ni δ2ni
−1
−1
il
· · · δ2ni il
· · · δ2ni
−τ
−τ
ili ili −1 −1)2nτ +(ili −1 −1)2n(τ −1) +···+ili −τ ] (Kili )δ2n δ2n il
=Hili δ2mi δ2ni δ2ni =Col[(pi
−1
−1
il
· · · δ2ni
il
· · · δ2ni
−τ
−τ
−1)2n(τ +1) +(ili −1)2nτ +···+ili −τ ] (Hili ) il
=δ2i1n δ2ni · · · δ2ni
−τ +1
.
That is y(N + li ) = δ2i1n = x(N + li ). Then the response system (10) will stay at Ci forever. 10
(II) Assume that N < li . The proof is similar to (I). (III) Assume that N > li . il −τ +1 ) under the switching signal For Ek (δ2i1n δ2i2n · · · δ2ni il −τ +1 ) under the switching signal {i(li −k+1) , · · · , i(li −1) , ili }, Ek+mli (δ2i1n δ2i2n · · · δ2ni {i(li −k+1) , · · · , i(li −1) , ili · · · , i1 , · · · , ili , · · · , i1 , · · · , ili }, it can be proved that il −τ +1 il −τ +1 ), where k = 1, 2, · · · , li , m ∈ ) ⊆ Ek+mli (δ2i1n δ2i2n · · · δ2ni Ek (δ2i1n δ2i2n · · · δ2ni N. The reason is as follows. Without loss of generality, we prove that il −τ +1 il −τ +1 ) under the switching signal ili ) ⊆ E1+li (δ2i1n δ2i2n · · · δ2ni E1 (δ2i1n δ2i2n · · · δ2ni and switching signal sequence {ili , i1 , · · · , ili } respectively. Note that if condiil −τ +1 il il −τ il il −1 ) tion (i) in theorem 3.1 is satisfied, then δ2ni δ2ni · · · δ2ni ∈ E1 (δ2i1n δ2ni · · · δ2ni ili ili −1 ili −τ ili −τ +1 i1 il i under the switching signal ili , that is δ2n δ2n · · · δ2n can reach δ2n δ2n · · · δ2n il −τ +1 il ∈ in one step under the switching signal ili . Moreover, δ2i1n δ2ni · · · δ2ni ili −τ +2 ili −τ +1 i2 i1 i1 il i E1 (δ2n δ2n · · · δ2n ) under the switching signal i1 , that is δ2n δ2n · · · δ2n il −τ +2 i2 i1 can reach δ2n δ2n · · · δ2ni in one step under the switching signal i1 . Doing it il il −1 il −τ il il −τ +1 repeatedly, we can see that δ2ni δ2ni · · · δ2ni can reach δ2i1n δ2ni · · · δ2ni in 1+ li steps under the switching signal {ili , i1 , · · · , ili }. Hence, for any initial state il −τ +1 il −τ +1 il il ) ) = ∆2n(τ +1) , we have z(1) ∈ EN −1 (δ2i1n δ2ni · · · δ2ni z(0) ∈ EN (δ2i1n δ2ni · · · δ2ni ili −τ +1 i1 il i or z(1) ∈ EN −1−˜ali (δ2n δ2n · · · δ2n ). Doing it repeatedly, there exists T such ili −τ +1 i 1 il i )), where ˜b ≤ li . Then similar to the proof that z(T ) ∈ E˜b (δ2n δ2n · · · δ2n in (I) and (II), we can draw the conclusion. 4. Example Example 4.1. Assume that the drive-response system is x1 (t + 1) = x1 (t) ∨ (¬x2 (t)), x2 (t + 1) = ¬x2 (t), y1 (t + 1) = x1 (t) ∨ y2 (t − 1), y2 (t + 1) = u(t) ∧ y1 (t), u(t) = k(x1 (t), x2 (t), y1 (t), y2 (t), y1 (t − 1), y2 (t − 1)).
(11) (12) (13)
By letting x(t) = n2i=1 xi (t), y(t) = n2i=1 yi (t), z(t) = y(t)y(t − 1), the algebraic form of the drive-response system is x(t + 1) = F x(t), z(t + 1) = Hx(t)u(t)z(t), 11
where F = δ4 [2, 1, 4, 1] and H = δ4 [1, 1, 1, 1, 2, 2, 2, 2, 7, 7, 7, 7, 8, 8, 8, 8, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 1, 1, 1, 1, 2, 2, 2, 2, 7, 7, 7, 7, 8, 8, 8, 8, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 1, 9, 1, 9, 2, 10, 2, 10, 7, 15, 7, 15, 8, 8, 8, 8, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 1, 9, 1, 9, 2, 10, 2, 10, 7, 15, 7, 15, 8, 16, 8, 16, 5, 13, 5, 13, 6, 14, 6, 14, 7, 15, 7, 15, 8, 16, 8, 16]. We can see that the cycle of (11) is (δ41 , δ42 ). Split H into 2n = 4 equal blocks as H = [H1 , H2 , H3 , H4 ]. It can be verified that, for x(t) = y(t) = δ41 , y(t − 1) = δ42 , when u(t) = δ22 , 5 z(t + 1) = Hx(t)u(t)z(t) = Hδ41 δ22 δ41 δ42 = δ16 = δ42 δ41 .
That is δ41 δ42 ∈ E1 (δ42 δ41 ) under the switching signal σ = 1. In the same way, we can verify that δ42 δ41 ∈ E1 (δ41 δ42 ) under the switching signal σ = 2. 5 6 7 8 we can see that E1 (δ41 δ42 ) = {δ16 , δ16 , δ16 , δ16 } under the switching signal 1 2 σ = 2. E2 (δ4 δ4 ) = ∆16 under the switching signal σ = {1, 2}. Hence, conditions (i) and (ii) of Theorem 3.1 are satisfied. Now, we use Algorithm 3.1 to obtain the feedback matrix K. For the cycle Ci = (δ41 , δ42 ). 1. Find the [(2 − 1)16 + (1 − 1)4 + 2] = 18th column of H1 such that 5 Col18 (H1 ) = δ42 δ41 = δ16 . Let the (1-1)4+2=2nd column of K1 be δ22 . Find the 2 [(1−1)16+(2−1)4+1] = 5th column of H2 such that Col5 (H2 ) = δ41 δ42 = δ16 . 1 Let the (2-1)4+1=5th column of K2 be δ2 . 2. Find the [(1 − 1)16 + 5] = 5th, [(1 − 1)16 + 6] = 6th, [(1 − 1)16 + 7] = 7th, [(1 − 1)16 + 8] = 8th column of H2 such that Col5 (H2 ) = Col6 (H2 ) = 2 5 6 7 8 Col7 (H2 ) = Col8 (H2 ) = δ41 δ42 = δ16 . Let E1 (δ41 δ42 ) = {δ16 , δ16 , δ16 , δ16 } and the 1 5th, 6th, 7th and 8th column of K2 be δ2 . 3. Find the [(2 − 1)16 + 1] = 17th, [(2 − 1)16 + 2] = 18th, [(2 − 1)16 + 3] = 19th, [(2 − 1)16 + 4] = 20th column of H1 such that Col17 (H1 ) = Col18 (H1 ) = 5 Col19 (H1 ) = Col20 (H1 ) = δ16 ∈ E1 (δ41 δ42 ). Let the 1st, 2nd, 3th and 4th column of K1 be δ22 . Find the [(2 − 1)16 + 5] = 21th, [(2 − 1)16 + 6] = 22th, [(2 − 1)16 + 7] = 23th, [(2 − 1)16 + 8] = 24th column of H1 such that 6 Col21 (H1 ) = Col22 (H1 ) = Col23 (H1 ) = Col24 (H1 ) = δ16 ∈ E1 (δ41 δ42 ). Let the 2 5th, 6th, 7th and 8th column of K1 be δ2 . Find the [(1 − 1)16 + 9] = 9th, [(1 − 1)16 + 10] = 10th, [(1 − 1)16 + 11] = 11th, [(1 − 1)16 + 12] = 12th 12
column of H1 such that Col9 (H1 ) = Col10 (H1 ) = Col11 (H1 ) = Col12 (H1 ) = 7 δ16 ∈ E1 (δ41 δ42 ). Let the 9th, 10th, 11th and 12th column of K1 be δ21 . Find the [(1 − 1)16 + 13] = 13th, [(1 − 1)16 + 14] = 14th, [(1 − 1)16 + 15] = 15th, [(1 − 1)16 + 16] = 16th column of H1 such that Col13 (H1 ) = Col14 (H1 ) = 8 ∈ E1 (δ41 δ42 ). Let the 13th, 14th, 15th and 16th Col15 (H1 ) = Col16 (H1 ) = δ16 column of K1 be δ21 . In summary, the feedback matrix K is K = δ2 [2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, ∗, ∗, ∗, ∗, 1, 1, 1, 1, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗ ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗] Example 4.2. Consider the D. melanogaster segmentation polarity gene network introduced in Example 4.2 of [43]. If the gene wg4 (0) = 1, wg1 (0) = 0, then wg4 (t) = 1, for t ≥ 1. That is ¬wg4 (t) ≡ 0, for t ≥ 0. Hence wg1 (t) = 0, for t ≥ 0. Then, by letting wg2 = x1 , wg3 = x2 , P T C1 = y1 , P T C2 = y2 , the reduced model is as follows x1 (t + 1) = x1 (t) ∧ ¬x2 (t), x2 (t + 1) = x2 (t), y1 (t + 1) = y1 (t) ∧ ¬x2 (t), y2 (t + 1) = ¬x2 (t).
(14) (15)
In this paper, we consider the reduced model with time delay and controls as follows x1 (t + 1) = x1 (t) ∧ ¬x2 (t), x2 (t + 1) = x2 (t), y1 (t + 1) = u1 (t) ↔ (y1 (t − 1) ∧ ¬x2 (t)), y2 (t + 1) = u2 (t) ↔ (¬x2 (t)).
(16) (17)
By letting x(t) = x1 (t) n x2 (t), it can be verified that x(t + 1) = F x(t), where F = δ4 [3, 2, 3, 4]. By letting y(t) = y1 (t) n y2 (t), z(t) = y(t) n y(t − 1) it leads to z(t + 1) = Hx(t)u(t)z(t),
13
where H = δ16 [13, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 1, 1, 9, 9, 2, 2, 10, 10, 3, 3, 11, 11, 4, 4, 12, 12, 5, 5, 13, 13, 6, 6, 14, 14, 7, 7, 15, 15, 8, 8, 16, 16, 9, 9, 13, 13, 10, 10, 14, 14, 11, 11, 15, 15, 12, 12, 16, 16, 13, 13, 5, 5, 14, 14, 6, 6, 15, 15, 7, 7, 16, 16, 8, 8, 13, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 1, 1, 9, 9, 2, 2, 10, 10, 3, 3, 11, 11, 4, 4, 12, 12, 5, 5, 13, 13, 6, 6, 14, 14, 7, 7, 15, 15, 8, 8, 16, 16, 9, 9, 1, 1, 10, 10, 2, 2, 11, 11, 3, 3, 12, 12, 4, 4, 13, 13, 5, 5, 14, 14, 6, 6, 15, 15, 7, 7, 16, 16, 8, 8]. We can see that the cycle of (14) is (δ42 ), (δ43 ), (δ44 ). It can be verified that conditions (i) and (ii) in Theorem 3.1 are satisfied. Now, we will use algorithm 3.1 to obtain the feedback matrix K. 1. For the cycle C1 = (δ42 ). Find the [(2−1)16+(2−1)4+2] = 22th of H2 6 = δ42 δ42 . Let the (2 − 1)4 + 2 = 6th column of K2 be such that Col22 (H2 ) = δ16 δ42 . Find all the [(2−1)16+5] = 21th, [(2−1)16+6] = 22th column of H2 such 6 5 6 } and the , δ16 . Let E1 (δ42 δ42 ) = {δ16 that Col21 (H2 ) = Col22 (H2 ) = δ42 δ42 = δ16 5th, 6th column of K2 be δ42 . Find [(2−1)16+1] = 17th, [(2−1)16+2] = 18th, [(4 − 1)16 + 3] = 51th, [(4 − 1)16 + 4] = 52th column of H2 such that 5 Col17 (H2 ) = Col18 (H2 ) = Col51 (H2 ) = Col52 (H2 ) = δ16 ∈ E1 (δ42 δ42 ). Let 6 5 2 3 4 1 } and the 1st, 2nd, 3th and 4th column of , δ16 , δ16 , δ16 , δ16 , δ16 E2 (δ42 δ42 ) = {δ16 2 4 4 2 K2 be δ4 , δ4 , δ4 , δ4 respectively. Similarly, let the 7th, 8th, 9th, 10th, 11th, 12th, 13th, 14th, 15th, 16th column of K2 be δ42 , δ42 , δ41 , δ41 , δ44 , δ44 , δ41 , δ41 , δ44 , δ44 respectively. 2. For the cycle C2 = (δ43 ). Let the 1st to the 16th column of K3 be δ42 respectively. 3. For the cycle C3 = (δ44 ). Let the 1st, 2nd, 3th, 4th, · · · , 16th column of K4 be δ43 , δ43 , δ41 , δ41 , δ43 , δ43 , δ41 , δ41 , δ44 , δ44 , δ42 , δ42 , δ44 , δ44 , δ42 , δ42 respectively. From the above, we can conclude that K is K = δ4 [∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, ∗, 2, 2, 4, 4, 2, 2, 2, 2, 1, 1, 4, 4, 1, 1, 4, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 1, 1, 3, 3, 1, 1, 4, 4, 2, 2, 4, 4, 2, 2]. 14
Without loss of generality, we assume that K = δ4 [2, 2, 4, 4, 2, 2, 2, 2, 1, 1, 4, 4, 2, 2, 4, 4, 2, 2, 4, 4, 2, 2, 2, 2, 1, 1, 4, 4, 1, 1, 4, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 1, 1, 3, 3, 1, 1, 4, 4, 2, 2, 4, 4, 2, 2]. By using [12], we have u1 (t) =[x1 (t) ∧ ((y1 (t) ∧ (y1 (t − 1) ∨ ¬y2 (t)) ∨ (¬y1 (t) ∧ y1 (t − 1)))] ∨ [¬x1 (t) ∧ (x2 (t) ∨ ¬y1 (t − 1)], u2 (t) =[x1 (t) ∧ (¬y1 (t) ∧ y1 (t − 1)] ∧ [¬x1 (t) ∧ (¬x2 (t) ∧ y1 (t)]. It thus follows that the complete synchronization of the drive-response systems can be achieved. For example, assume that the initial states of the drive system and response system be x(0) = δ41 and y(0) = δ43 , y(−1) = δ44 respectively, then the trajectories are as follows: (x1 , x2 ) : (1, 1) → (0, 1) → (0, 1) → · · · (y1 , y2 ) : (0, 1) → (1, 1) → (0, 1) → · · · . 5. Conclusion This paper has investigated the feedback control design approach for the complete synchronization of two coupled Boolean networks with time delay. A necessary condition for the existence of a state feedback controller to achieve the complete synchronization has been established. Then based on the necessary condition, a constructive procedure for control design has been provided. An example has been worked out to illustrate the effectiveness of this paper. In this paper, the time delay τ should be known when designing controller. In practical situations, the time delay τ should be identified from microarray data. However, if the time delay is unknown, our method can not be applied to. Thus, we shall leave this problem for future study. Acknowledgement This work is supported by National Natural Science Foundation of China under Grant 61403139, Open Foundation of first level Zhejiang key in key discipline of Control Science and Engineering and the Fundamental Research Funds for the Central Universities. 15
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Highlights:
1.
A necessary condition for the existence of a state feedback controller for the complete synchronization of two coupled Boolean networks with time delays is derived.
2.
Feedback control design procedure for the complete synchronization of two coupled Boolean networks is provided based on the necessary condition.
3.
An example is given to illustrate the proposed design procedure.