一类具有RiemannLiouville 分数阶导数的线性.doc

精品论文一类具有Riemann-Liouville 分数阶导数的线性时不变微分系统的完全能控性杨玲, 周先锋, 蒋威安徽大学数学科学学院，合肥230039 摘要：本文研究一类具有Riemann-Liouville分数阶导数的线性时不变微分系统的完全能控性。 首先得到了关于古典意义上状态方程初值问题的解，然后建立的关于系统能控性的判别准则是 充分必要条件，并提供两个例子说明结果。关键词：完全能控性, Riemann-Liouville 分数阶导数, 线性的, 微分系统中图分类号： O175.9Complete controllability of a fractional linear time-invariant differential system with Riemann-Liouville derivativeYANG Ling, ZHOU Xian-Feng, JIANG WeiSchool of Mathematical Sciences, Anhui University , Hefei 230039Abstract: This paper is concerned with the complete controllability of a fractional linear time-invariant differential system with Riemann-Liouville derivative. The solution of the state equation with classical initial value is first derived. Then two criteria on controllability for the system, which are sufficient and necessary, are established. Two examples illustrate the results.Key words:Complete controllability; Riemann-Liouvilles fractional derivative, linear, differential system0 IntroductionIn past few decades, theories of fractional differential equations have gained considerable developments, see the monographs1, 2, 3, papers4, 5, 6, 7, 8, 9 etc and the reference therein. The concept of controllability plays an important role in the analysis and design of control systems. Along with the development of theories of fractional differential equations, much more attention are paid to the controllability of fractional differential systems 10, 11, 12, 13,14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24. The main method used is fixed point theorem or to construct a suitable control input function. The fractional derivatives, which have been involved in the fractional differential systems, are Caputos derivative. However, to our best knowledge, little attention has been paid to the controllability of fractional differential system with Riemann-Liouville derivative.基金项目： Natural Science Foundation of China (No.11071001)，the Doctoral Fund of Ministry of Education of China (No.20093401110001 ), Ma jor Program of Educational Commission of Anhui Province of China (No. KJ2010ZD02), Program of Natural Science Research in Anhui Colleges and Universities( KJ2011A020) and the Scientific Research Starting Fund for Dr of Anhui University (No.023033190001)作者简介： Yang Ling, （1989-），female，postgraduate student，ma jor research direction：Functional differential equations. Correspondence author：Zhou Xian-Feng, (1968-), male, associate professor, major research direction: Functional differential equations.Motivated by the fact and the paper 23, 18, this paper is concerned with the complete controllability of the following systemRLDt > 0,(0.1)y(t) =Ex(t) + F u(t)(0.2)0,t x(t) =Ax(t) + Gu(t),0,tin n-dimensional Euclidean space, where RL Ddenotes Riemann-Liouvilles derivative of order with the lower limit 0, 0 < < 1, A are n order known constant matrix, the constant matrices G, E and F satisfy G Rn×p , E Rm×n , F Rm×p , state variable x(t) Rn , control input u(t) Rp .t0 ,tt0 ,tThroughout this paper, Df (t) or RL Df (t) denotes Riemann-Liouvilles derivative oft0 ,torder with the lower limit t0 for function f , I f (t) denotes integral of order with lowerlimit t0 for function f , f¯(s) = Lf (t); s denotes the Laplace transform of function f (t), and”|M |” denotes the norm of the matrix ”M ”, the matrix M denotes the transpose of the matrixM .1 PreliminariesIn this section, we first recall some definitions and lemmas. For more detail, please refer to 1, 2.Definition 1.1. (1) The fractional integral of order with the lower limit 0 for function fis defined as1Z tI 10,t f (t) = ()(t s)0f (s)ds,t > 0, > 0(1.3)provided the right-hand side is pointwise defined on 0, +), where is the gamma function.Definition 1.2. (1)Riemann-Liouvilles derivative of order with the lower limit 0 for a function f : 0, ) R can be written asRL D1dnZ t(t s)n1 f (s)ds = (I n f (t)(n) , n 1 < n. (1.4)00,t f (t) = (n ) dtn0,tParticularly, when 0 < < 1, it holdsRL D1d Z t(t s) f (s)ds =d (I 1 f (t),t > 0. (1.5)00,t f (t) = (1 ) dtdt 0,tDefinition 1.3. (1) The two-parameter Mittag-Leffler function is defined as zkE, (z) = X , > 0, > 0. (1.6)(k + )k=0The Laplace transform of Mittag-Leffler function is, (±atL htk+1 E(k)Z ); sik!s1, (±at=est tk+1 E(k)0)dt =, Re(s) > a | |(s a)k+1. (1.7)where Re(s) denotes the real parts of s.In addition, the Laplace transform of t1 isLt1 ; s = ()s , Re(s) > 0. (1.8)Definition 1.4. 1 Suppose that 0 < < 1. If the function f 0 (t) is continuous on a, t, thenGru¨nwald-Letnikov -order derivative with the lower limit a for f (t) is defined asRL DGLa,t f (t) =f (a)(t a)(1 )1+(1 )Z t0(t ) f ( )d (1.9)aRemark 1.1. If the function f 0 (t) is continuous on a, t, thena,tRL D f (t) =GLa,tD f (t)2 Main resultsDefinition 2.1. (complete controllability)The system (0.1)-(0.2) is said to be completely con- trollable on the interval J = 0, T if, for every Z Rn and ti R+ , there exists an admissible control input u(t) such that the state variable x(t) of the system (0.1)-(0.2) satisfies x(ti ) = Z .Lemma 1. Suppose that x(t) is a continuous derivable solution of the free state equation (0.1). Namely, u(t) = 0 in state equation (0.1) . Then x(t) 0.Proof. Since u(t) = 0, the state equation (0.1) can be reduced to1d Z t(t s) x(s)ds = Ax(t).(2.10)(1 ) dt 0By the fact that x(t) is continuous derivable and using the formula (1.9), we haveZ tThat is(t s) x (s)ds + t x(0) = (1 )Ax(t). (2.11)0t x (t) + t x(0) = (1 )Ax(t), (2.12)here denotes convolution. An application of the Laplace transform on both side of equality(2.12) together with the formula (1.8) and Convolution Theorem yieldsLt ; sLx (t); s + Lt x(0); s = (1 )ALx(t); s.(2.13) Reducing Eq.(2.13) yields(s I A)x¯(s) = 0, (2.14)here x¯(s) is the Laplace transform of x(t). Thus, x¯(s) = 0. An application of the inverseLaplace transform yields the desired result.Remark 2.1. Under the assumption that the control input u(t) = 0, there doesnt exist any nonzero continuous derivable solution for the state equation (0.1). However, there exists con- trol inputs u(t) such that the state equation (0.1) possesses a nonzero solution x(t), which is.continuous derivable on 0, . For example, if u(t) = (t, t, · · · , t) Rp , then 1 x(t) = XAk 1 1+(k+1)k=0tG(k + 1) + 2) 1(2.15)is a nonzero continuous derivable solution of the state equation (0.1).In fact, inserting Eq.(2.15) into the left of Eq.(0.1) yields1d Z t Ak 1 1 RL D(t s) X G . s1+(k+1)00,t x(t) = (1 ) dtk=0(k + 1) + 2) ds.11d Ak 1 1 Z t=X G . (t s)s1+(k+1)ds . (2.16)(1 ) dt k=0By computation, we haveZ t(k + 1) + 2) .01(t s) s1+(k+1) ds = tk+2 B(1 , (k + 1) + 2), (2.17)0here B denotes Beta function. Inserting (2.17) into (2.16) yields P 1 k 1 RL D1 d A k+2 0,t =(1) dt k=0(k+1)+2) G t.1B(1 , (k + 1) + 2)P Ak 1 1 k+1=k=0(k+2) G t . 1P Ak 1 1 k+1 1 1 (2.18)=k=1(k+2) G .1 t 1 + G .1 t 1 P Ak 1 1+(k+1) 1 = Ak=0(k+1)+2) G t.1+ G .t1= Ax(t) + Gu(t).This show that x(t) denoted by (2.15) is a nonzero solution of the state equation (0.1). Obvi- ously, x(t) is continuous and derivable.Lemma 2. There exist control inputs u(t) such that the solution x(t) of the state equation (0.1)can be written asx(t) =Z t(t s)1 E, (A(t s) )Gu(s)ds.(2.19)0Proof. The state equation (0.1) can be written as1d Z t(t s) x(s)ds = Ax(t) + Gu(t).(2.20)(1 ) dt 0By the Gru¨nwald-Letnikov derivative formula, we haveZ tThat is(t s) x (s)ds + t x(0) = (1 )(Ax(t) + Gu(t).(2.21)0t x (t) + t x(0) = (1 )(Ax(t) + Gu(t),(2.22)here ”*” denotes convolution. Applying the Laplace transform on both side of Eq.(2.22) yieldsx¯(s) = (s I A)1 LGu(t); s.(2.23) Applying the inverse Laplace transform on both side of (2.23) together with the formula (1.7) yields Eq.(2.19).Remark 2.2. There exists a control input u(t) such that the continuous and derivable solutionx(t) of the state equation (0.1) have nothing to do with its initial state x(0).Theorem 1. The system (0.1)-(0.2) is completely controllable on 0, tf if and only if Gramian matrixWc 0, tf =Z tf0(tf s)1 E, (A(tf s) )GGE, (A(tf s) )ds (2.24)is nonsingular for some tf (0, ). Here * donotes the matrix transpose.cProof. Sufficiency. Suppose that Wc 0, tf is nonsingular, then W 1 0, tf is well-defined. Foran arbitrary Z0 Rn , Choose a control input u(t) ascu(t) = GE, (A(tf t) )W 1 0, tf Z0 . (2.25) By the formula (2.19), it is easy to check that x(tf ) = Z0 . Thus the system (0.1)-(0.2) iscompletely controllable on 0, tf .Necessity. Suppose that the system (0.1)-(0.2) is completely controllable. Now we provethat Wc 0, tf is nonsingular. In fact, if Wc 0, tf is singular, then there exists a nonzero vectorz0 such thatz0 Wc 0, tf z0 = 0. (2.26)That isZ tfz10Then it follows0 (tf s)E, (A(tf s)z)GG E, (A (tf s)z0 ds = 0. (2.27)0 E, (A(tf s)G = 0(2.28)on s 0, tf . Since the system (0.1)-(0.2) is completely controllable, so there exists a control input u1 (t) such thatThat isx(tf ) =Z tf0(tf s)1 E, (A(tf s) )Gu1 (s)ds = z0 . (2.29)z0 Z tf0(tf s)1 E, (A(tf s) )Gu1 (s)ds = 0. (2.30)0Multiplying z on both side of Eq.(2.30), we getz0 z0 Z tf0(tf s)10zE, (A(tf s)G(u1 (s) u2 (s)ds = 0. (2.31)0By the equality (2.28), we have zz0 = 0. Thus z0 = 0. This is contradiction. The proof iscomplete.Theorem 2. The system (0.1)-(0.2) is completely controllable on 0, tf if and only ifrank(G|AG| · · · |An1 G) = n. (2.32)Proof. By Cayley-Hamilton theorem we haven1t1 E, (At ) = X ck (t)Ak , (2.33)k=0where ck (t) are functions in t. Combining the formula (2.19) with the formula (2.33) yieldsf0 fx(t ) = R tf (t s)n11E, (A(tf s)Gu(s)ds= P Ak G R tf c (t s)u(s)dsk=00 k f d0 (2.34),= (G|AG| · · · |An1 G) d1 · · ·dn1tfwhere dk = R0 ck (tf s)u(s)ds, k = 0, 1, 2, · · · , n 1. Obviously, for an arbitrary x(tf ) , the sufficient and necessary condition to have a control input u(t) satisfying Eq.(2.34) is that rank(G|AG| · · · |An1 G) = n.3 Illustrate examples2Example 3.1. Consider the time-invariant system (0.1)-(0.2). Choose = 1 , A = 01 00, 2 G =1. Now we apply Theorem 1 to prove that the system (0.1)-(0.2) is controllable. First,Z 1 111Wc 0, 1 =(1 s) 2 E 1 1 (A(1 s) 2 )G)GE 1 1 (A(1 s) 2 )ds. (3.35)0By computation,2 , 22 , 211 Ak (1 s)k 211 01 12 , 2E 1 1 (A(1 s) 2 ) = X(k 1 + 1 )= ( 1 ) I +(1)00(1 s) 2 , (3.36)here I = 10 01.k=0 2 2 2GG = 2 211 42 =21, (3.37)11 (A)k (1 s)k 211 00 12 , 2E 1 1 (A(1 s) 2 ) = X(k 1 + 1 )= ( 1 ) I +(1)10(1 s) 2 . (3.38)k=0 2 2 2Inserting Eq.(3.36), Eq.(3.37) and Eq.(3.38) into Eq.(3.35) yieldsWc 0, 1 = 842 + 0.5 + 341 + 0.5 . (3.39)412+ 0.5 It is obvious that Wc 0, 1 is nonsingular. Thus by Theorem 1, the system is completely con- trollable.Example 3.2. Consider the time-invariant system (0.1)-(0.2). Choose 142 2 A = 061 , = 1 , G = 0 . By computation, we have31711 246 rank(G|AG|A2 G) = rank 017 = 3. (3.40)1112By Theorem 2, the system is completely controllable.参考文献（References）1 I.Podlubny. Fractional Differential EquationsM. Academic Press, New York, 1993.2 A.A.Kilbas, Hari M.Srivastava, Juan J.Trujillo. Theory and Applications of FractionalDifferential EquationsM. Elsevier Science B. V., Amsterdan, 2006.3 V.Lakshmikantham, A.S.Vatsala. Basic theory of fractional differential equationsJ. Non- linear Analysis, 2008, 69:2677-2682.4 R.P.Agarwal, Y.Zhou, Y.He. Existence of fractional neutral functional differential equa- tionsJ. Computers and Mathematics with Applications, 2007, 59:1095-1100.5 Y. Zhou, F.Jiao, J.Li. Existence and uniqueness for fractional neutral differential equations with infinite delayJ. Nonlinear Analysis, 2009, 71:3249-3256.6 Y.Zhou, F.Jiao, J.Li. Existence and uniqueness for p-type fractional neutral differential equationsJ. 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