Solving the Lienard equation by differential transform method

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1 ISSN , England, U World Jornal of Modelling and Simlation Vol. 8 (2012) No. 2, pp Solving the Lienard eqation by differential transform method Mashallah Matinfar, Saber Rakhshan Bahar, Maryam Ghasemi Department of Mathematics, University of Mazandaran, Babolsar , Iran (Received Agst , Accepted May ) Abstract. In this paper, the differential transform method (DTM) is proposed for solving the Lienard eqation. The differential transform method is a nmerical method based on the Taylor series epansion which constrcts an analytical soltion in the form of a polynomial. Using this method, it is possible to find the eact soltion or an approimate soltion of the problem. eywords: differential transform method, lienard eqation 1 Introdction In this work, we consider the Lienard eqation + f() + g() = h(), (1) which is not only regarded as a generalization of the damped pendlm eqation or a damped spring-mass system (where f() is the damping force, g() is the restoring force, and h() is the eternal force), bt also sed as nonlinear models in many physically significant fields when taking different choices for f(), g() and h(). For eample, the choices f() = ɛ( 2 1), g() =, and h() = 0 lead Eq. (1) to the Van der Pol eqation served a a nonlinear model of electronic oscillation [2, 7]. Therefore, stdying Eq. (1) is of physical significance. In the general case, it is commonly believed that it is very difficlt to find its eact soltion by sal ways [3]. ong stdied the following special case of Eq. (1) [1, 4] : () + l() + m 3 () + n 5 () = 0, where l, m, n are real coefficients. Finding eplicit eact and nmerical soltions of nonlinear eqations efficiently is of major importance and has widespread applications in nmerical methods and applied mathematics. Implement the variational iteration method for Eq. (1) was known in the literatre [6]. The differential transform is an analytic method for Solving differential eqations. The concept of the differential transform was first introdced by Zho in [8]. Its main application therein is to solve both linear and non-linear initial vale problems in electric circit analysis. This method constrcts and analytical soltion in the form of a polynomial. It is different from the traditional higher order Taylor series method. The Taylor series method is comptationally epensive for large orders. The differential transform method is an alterative procedre for obtaining analytic Taylor series soltion of the differential eqations. By sing DTM, we get a series soltion, in practice a trncated series soltion. 2 Differential transform method Differential transformation of fnction () is defined as follows [5] Corresponding athor. Tel.: address: m.matinfar@mz.ac.ir Pblished by World Academic Press, World Academic Union

2 World Jornal of Modelling and Simlation, Vol. 8 (2012) No. 2, pp U(k) = 1 [ d k ] () k! d k. =0 (2) In Eq. (2), () is the original fnction and U(k) is the transformed fnction. Differential inverse transform of U(k) is defined as follows () = k U(k). (3) In fact, from Eqs. (2) and (3), we obtain () = k=0 k=0 k k! [ d k ] () d k. (4) =0 Eq. (4) implies that the concept of differential transformation is derived from the Taylor series epansion. From the definitions Eqs. (2) and (3), it is easy to obtain the following mathematical operations: (a) If () = v() ± w(), then U(k) = V (k) ± W (k). (b) If () = av(), a R, then U(k) = av (k). (c) If () = v()w(), then U(k) = k r=0 V (r)w (k r). (d) If () = n, then U(k) = δ(k n). (e) If () = 1 () 2 () n 1 () n (), then U(k) = k kn 1 k n 1 =0 k n 2 =0 k 3 k 2 =0 k2 k 1 =0 U 1(k 1 )U 2 (k 2 k 1 ) U n 1 (k n 1 k n 2 )U n (k k n 1 ). 3 Applications and reslts In this section we consider the Lienard eqation () + l() + m 3 () + n 5 () = 0, (5) with the initial conditions (0) = m, (0) = ll m m, (6) where m and l are arbitrary constants. Taking differential transform of Eq. (5), we can obtain 1 U(k + 2) = lu(k) m (k + 2)(k + 1) n k k 4 k 3 k 2 k 4 =0 k 3 =0 k 2 =0 k 1 =0 k k 2 k 2 =0 k 1 =0 And the transform of the initial conditions Eq. (6) are U(k 1 )U(k 2 k 1 )U(k k 2 ) U(k 1 )U(k 2 k 1 )U(k 3 k 2 )U(k 4 k 3 )U(k k 4 ). (7) U(0) = l, U(1) =. (8) m 2m Utilizing the recrrence relation Eq. (7) and the transformed initial conditions Eq. (8) and Matlab software, the approimate soltion of the Lienard Eq. (5) can be derived as WJMS for sbscription: info@wjms.org.k

3 144 M. Matinfar & S. Bahar & M. Ghasemi: Solving the Lienard eqation by differential transform method 1 (t) = 2 + D 1 [ l + m ] D 2 + D + n2 (2 + D) 2 t 2, () = ( m ) ( 1/2l 1/2m(/m)3/2 (2m) 21(/m) 1/2 1/2n(/m) 5/2 ) 2 + ( 5/12l 2 2 1/2 /m (1/2) + 5/3nl 3 2 1/2 /m 5/2 ) 3 + (221/30000l 1/12m(3/2(/m) 1/2 l 2 /m + 663/1250l/m) 1/12n(5l 2 /m(/m) 3/2 221/125l 2 /m 2 ) 4 + ( 1/20l( 5/12l 2 2 1/2 /m 1/2 + 5/3nl 3 2 1/2 /m 5/2 ) 1/20m ( 2( 5/4l 2 2 1/2 /m ( 1/2) + 5nl 3 2 1/2 /m 5/2 )l/m + 663/2500l2 1/2 /m 1/2 (/m) 1/2 1/4l 3 2 1/2 /m 3/2 ) 1/20n(4( 25/12l 2 2 1/2 /m 1/2 + 25/3nl 3 2 1/2 /m 5/2 )l 2 /m /250(/m) 3/2 l2 ( 1/2)/m 1/2 + 5l 4 2 1/2 /m 5/2 )) 5 + ( 313/300000l 1/30m ( 939/5000l/m /10000l2 1/2 /m 1/2 (/m) 1/ / (/m) 1/2 663/5000l 2 /m) 1/30n (313/500l 2 /m /1000(/m) 3/2 l2 1/2 /m 1/ / (/m) 3/ /250l 3 /m 2 + 5/4l 4 /m 2 (/m) 1/2 )) 6 +. Selecting the vales of m = 4, n = 3, l = 1, and sing the before eqation, we can write: () = Of corse, the eact soltion () (see [1]) is (1 + tanh( l)) () =. (9) m The nmerical reslts with sing differential transform method for Lienard eqation in comparison with the eact soltion of (m = 4, n = 3, l = 1 and = 0.1(0.1)1) are given in Tab. 1. Also in Fig. 1 we can see the comparison between The eact soltion and approimate soltion with sing DTM for the Lienard eqation for the vales of m = 4, n = 3 and l = 1. Table 1. The nmerical reslts for the first eample U Eact U DTM e e e e e e e e e e 002 Table 2. The nmerical reslts for the second eample U Eact U DTM e e e e e e e e e e 002 In the second eample, we will consider the Lienard Eq. (5) with the initial conditions (0) = 2 + D, (0) = 0, (10) where with arbitrary constants m and l WJMS for contribtion: sbmit@wjms.org.k

4 World Jornal of Modelling and Simlation, Vol. 8 (2012) No. 2, pp Eact soltion Approimate soltion Fig. 1. The eact soltion and approimate soltion 0.65 Eact soltiondata Approimate soltion Fig. 2. The eact soltion and approimate soltion = 4 3l 2 3m 2 16nl, D = 1 + 3m 3m 2 16nl. Transform of the initial conditions Eq. (10) are U(0) =, 2 + D U(1) = 0. (11) Utilizing the recrrence relation Eq. (7) and the transformed initial conditions Eq. (11) and compting with Matlab software, the approimate soltion of the Lienard Eq. (5) can be derived as () =2(3 1/2 (l 2 /(3m 2 16nl)) 1/2 /( /2 m/(3m 2 16nl) 1/2 )) 1/2 + ( l(3 1/2 (l 2 /(3m 2 16nl)) 1/2 /( /2 m/(3m 2 16nl) 1/2 )) 1/2 4m(3 1/2 (l 2 /(3m 2 16nl)) 1/2 /( /2 m/(3m 2 16nl) 1/2 )) 3/2 16n(3 1/2 (l 2 /(3m 2 16nl)) 1/2 /( /2 m/(3m 2 16nl) 1/2 )) 5/2 ) 2 + (1/3(112nl 2 96m3 1/2 ( l 2 /( 3m nl)) 1/2 nl + l(3m 2 16nl) 1/2 3 1/2 m + 3lm m 2 ( l 2 /( 3m nl)) 1/2 (3m 2 16nl) 1/2 + 18m 3 3 1/2 ( l 2 /( 3m nl)) 1/2 ) (3lm nl 2 + l(3m 2 16nl) 1/2 3 1/2 m + 6m 3 3 1/2 ( l 2 /( 3m nl)) 1/2 32m3 1/2 ( l 2 /( 3m nl)) 1/2 nl + 6m 2 ( l 2 /( 3m nl)) 1/2 (3m 2 16nl) 1/2 )3 1/4 (( l 2 / ( 3m nl)) 1/2 (3m 2 16nl) 1/2 /((3m 2 16nl) 1/ /2 m)) 1/2 /((3m 2 16nl) 1/ /2 m) 4 ) 4 + ( 32/453 1/4 (( l 2 /( 3m nl)) 1/2 (3m 2 16nl) 1/2 /((3m 2 16nl) 1/ /2 m)) 1/2 (3lm nl 2 + l(3m 2 16nl) 1/2 3 1/2 m + 6m 3 3 1/2 ( l 2 /( 3m nl)) 1/2 32m3 1/2 ( l 2 / WJMS for sbscription: info@wjms.org.k

5 146 M. Matinfar & S. Bahar & M. Ghasemi: Solving the Lienard eqation by differential transform method ( 3m 2 +16nl)) 1/2 nl+6m 2 ( l 2 /( 3m 2 +16nl)) 1/2 (3m 2 16nl) 1/2 )(284672n 4 l l 4 m3 1/2 ( l 2 /( 3m nl)) 1/2 n l 4 n 3 m l 4 n 3 (3m 2 16nl) 1/2 3 1/2 m l 3 n 3 m 3 3 1/2 ( l 2 /( 3m nl)) 1/ l 3 n 3 m 2 ( l 2 /( 3m nl)) 1/2 (3m 2 16nl) 1/ l 3 n 2 (3m 2 16nl) 1/2 3 1/2 m l 3 n 2 m l 2 n 2 ( l 2 /( 3m nl)) 1/2 m 5 3 1/ l 2 n 2 ( l 2 /( 3m nl)) 1/2 (3m 2 16nl) 1/2 m l 2 nm 5 (3m 2 16nl) 1/2 3 1/ l 2 nm ln( l 2 /( 3m 2 +16nl)) 1/2 m 7 3 1/ nl( l 2 /( 3m nl)) 1/2 (3m 2 16nl) 1/2 m lm l(3m 2 16nl) 1/2 m 7 3 1/ ( l 2 /( 3m 2 +16nl)) 1/2 (3m 2 16nl) 1/2 m ( l 2 /( 3m nl)) 1/2 m 9 3 1/2 )l/((3m 2 16nl) 1/ /2 m) 10 ) 6. Selecting the vales of m = 4, n = 3, l = 1, and sing the above eqation, we can write: () = 64359/ / / / The eact soltion () (see [1]) is () = sech 2 ( l) 2 + Dsech 2 ( l), where = 4 3l 2 3m 2 16nl, D = 1 + 3m 3m 2 16nl. The nmerical reslts with sing differential transform method for Lienard eqation in comparison with the eact soltion of (m = 4, n = 3, l = 1 and = 0.1(0.1)1) are given in Tab. 2. Also in Fig. 2 we can see the comparison between The eact soltion and approimate soltion with sing DTM for the Lienard eqation for the vales of m = 4, n = 3 and l = 1. 4 Conclsions In this paper, differential transform method was employed sccessflly for solving the Lienard eqation. The eact soltions are compared with the nmerical soltions in Tabs. 1 and 2. The reslts show that the differential transform method is a powerfl mathematical tool for finding the eact and approimate soltions of nonlinear eqations. In or work we se the Matlab to calclate the series obtained from the differential transform method. References [1] Z. Feng. On eplicit eact soltins for the Lienard eqation and its applications. Physics Letters A, 2002, 239: [2] J. Gckenheimer. Dynamics of the van der pol eqation. IEEE Transactions on Circits and Systems, 1980, 27: [3] J. Hale. Ordinary Differential Eqations, Wiley, New York, [4] D. ong. Eplicit eact soltins for the Lienard eqation and its applications. Physics Letters A, 1995, 196: [5] C. ang, H. He. Application of differential transformation to eigenvale problems. Applied Mathematics and Comptation, 1996, 79: [6] M. Matinfar, H. Hosseinzadeh, M. Ghanbari. A nmerical implementation of the variational iteration method for the Lienard eqation. World Jornal of Modelling and Simlation, 2008, 4(3): [7] Z. Zhang, T. Ding, H. W. Hang. Qalitative Theory of Differential Eqations. Science Press, Peking, In Chinese. [8] J. Zho. Differential Transformation and its Applications for Electrical Circits. Hzhong University Press, Whan, China, In Chinese. WJMS for contribtion: sbmit@wjms.org.k