Science Journal of Applied Mathematics and Statistics 2019; 7(4): 51-55 http://www.sciencepublishinggroup.com/j/sjams doi: 10.11648/j.sjams.20190704.11 ISSN: 2376-9491 (Print); ISSN: 2376-9513 (Online)
A New Analytical Approach for Solving Van der Pol Oscillator
Md. Abul Kashem Mondal, Md. Helal Uddin Molla *, Md. Shamsul Alam
Department of Mathematics, Rajshahi University of Engineering and Technology (RUET), Rajshahi, Bangladesh
Email address: [email protected] (Md. A. K. Mondal), [email protected] (Md. H. U. Molla), [email protected] (Md. S. Alam) *Corresponding author To cite this article: Md. Abul Kashem Mondal, Md. Helal Uddin Molla, Md. Shamsul Alam. A New Analytical Approach for Solving Van der Pol Oscillator. Science Journal of Applied Mathematics and Statistics. Vol. 7, No. 4, 2019, pp. 51-55. doi: 10.11648/j.sjams.20190704.11
Received : July 20, 2019; Accepted : September 16, 2019; Published : October 9, 2019
Abstract: The Van der Pol oscillator is a nonlinear damping and non-conservative oscillator. Energy is generated at low amplitude and dissipated at high amplitude. This nonlinear oscillator was first introduced by Dutch electrical engineer and physicist B. Van der Pol and it was originally used to investigate vacuum tubes. Nowadays, it is used in both physical and biological sciences. It is also used in sociology and even in economics. It has a limit cycle and in earlier it was determined by the classical perturbation methods when the nonlinear term is small. Then the harmonic balance method was used to determine the limit cycle for stronger nonlinear case. Moreover, many researchers have been analyzed this oscillator by various numerical approaches. In this article, a new analytical approach based on harmonic balance method is presented to determine the limit cycle as well as approximate solutions of this nonlinear oscillator. The frequency as well as the limit cycle obtained by new approach has been compared with those obtained by other existing methods. The present method gives better result than other existing results and also close to the corresponding numerical result (considered to the exact result). Moreover, the present method is simpler than the existing harmonic balance method. Keywords: Van der Pol Oscillator, Limit Cycle, Harmonic Balance Method, Frequency
perturbation method which is originally developed for 1. Introduction handling weak nonlinear problems in which small Van der Pol oscillation has become the interest of many parameters exist [1-3]. The parameters are analytically researchers because of its various applications in human expanded into power series of the parameter. The coefficients activities, sciences, technologies and industrial applications. of the series are found as solutions of a set of linear algebraic Thus, in present time, nonlinear processes are one of the problem. However, in both science and engineering, there biggest challenges and not easy to control because the exist many nonlinear problems without small parameters. nonlinear characteristic of the system abruptly changes due Even if there exists such a parameter, the analytical solution to some small changes of valid parameters including time. determined by the perturbation method has a small validity. The general theory and solution technique for linear Thus, many approximation methods were developed for differential equations in both science and engineering as well solving strongly nonlinear oscillators, including modified as in other disciplines have extensively developed. Yet very Lindstedt-Poincare method [4-5], harmonic balance method often many oscillatory systems are governed by nonlinear [6-10], residue harmonic balance method [11], global residue differential equation. Although it is possible in many cases to harmonic balance method [12], iterative homotopy harmonic replace the nonlinear differential equation by a balance method [13], amplitude-frequency formulation [14] corresponding linear differential equation which and homotopy analysis method [15]. Moreover energy approximates the original equation, such linearization is not balance method is used technique for solving strongly always feasible or possible. In such cases, the actual nonlinear oscillators [16-21]. A detailed review on some differential equations must be directly dealt with. Many recently developed nonlinear analytical methods can be researchers have been determined the approximate solutions found in [22-28]. Also in a recent work, Rahman et al. has of nonlinear problems by using various methods such as used harmonic balance method to determine an approximate 52 Md. Abul Kashem Mondal et al. : A New Analytical Approach for Solving Van der Pol Oscillator
solution of Van der Pol oscillator [10]. But the method is not The obtained results have been compared with those a simple one. However, the approximate solutions obtained obtained by Rahman et al. [10]. by these methods are not significantly improved. On the other hand, any particular method is not suitable for all 2. Formulation and Solution Method problems i.e., the special method is appropriate for the special nonlinear problems. Also, the solution procedure of Let us consider a nonlinear differential equation some methods is laborious and difficult as well as results are &&+=ε & == & not so closed to exact results. Also, these methods involve xx fxx(,), x (0) ax , (0) 0 , (1) tedious derivations and computations, and they are difficult where, over dot denotes the derivative with respect to time t, to implement. In an attempt to improve the accuracy of the f( x , x& ) is a nonlinear function and ε is constant. Consider a existing analytical methods, a new analytical approach based on harmonic balance method has been presented to obtain periodic solution as in the form: the approximate solution of Van der Pol nonlinear oscillator.
tx )( = 2 cosϕ + δ cosϕ + p(sin 3ϕ − 3sin ϕ) + q(sin 5ϕ − 5sin ϕ ϕ)) (2) + u(cos 3ϕ − cos+ v(cos 5ϕ − cosϕ) + w(cos 7ϕ − cosϕ) + K) where 0<δ << 1,,,,,p q u v w are constants, φ= ω t and ω=+ δ + δ 2 + L ω= 2 π / T is a frequency of nonlinear oscillation, here T is a 1 c1 d 1 =δ + δ 2 + L period. Solution Eq.(2) readily satisfies the initial conditions u c2 d 2 =& = x(0) a , x (0) 0 . =δ + δ 2 + L v c3 d 3 Substituting Eq. (2) in Eq. (1) and expanding f( x , x& ) in a (5) w= cδ + d δ 2 + L Fourier series, it turns to an algebraic identity 4 4 =δ + δ 2 + L p c5 d 5 (−+++− 2uvwδω )(2 − 1)cos φ +− u (1 9 ω 2 )cos3 φ + L =δ + δ 2 + L q c6 d 6 . ++(3p 5 q )(ω2 − 1)sin φ +− p (1 9 ωφ 2 )sin 3 + L (3) =ε ω δφ + ω δφ + L [Fpq1 (,, ,,, uvw ,)cos Fpq 3 (,, ,,, uvw ,)cos3 Now Eqs. (4) can be solved for +ω δφ + ω δφ + LL δ Gpq1(,,,,,,)sin uvw Gpq 3 (,,,,,,)sin3 uvw ]. ccccccdd123456123456,,,,,,,,,,, d d d dand . Substituting ω the values of ,,,uv w , pq , and δ from Eqs. (5) into Eq. (1) Equating the coefficients of equal harmonics of Eq. (3), we obtain the algebraic equations are we obtain the solutions.
−+++−δ ω2 −= ε −ω2 = ε ( 2uvw )( 1) F 1 , u(1 9 ) F 3 , 3. Example
−ω2 = ε L To illustrate the accuracy of the proposed method one v(1 25 ) F 5 , example has been presented in this section. Consider the Van +ω2 − = ε −ω2 = ε der Pol nonlinear conservative oscillator as in the following (3p 5)( q 1) G 1 , p(1 9 ) G 3 , form q(1− 25ω2 ) = ε G ,L. (4) 5 &&xx+=−ε (1 xx2 ), & x (0) = ax ,(0) & = 0. (6) Since δ is much smaller than 1 so all the constants Consider a periodic solution as in the form: ω,,,u vw , pq ,, of Eqs.(4) can be expanded inpowers of δ as
tx )( = 2 cosϕ + δ cosϕ + p(sin 3ϕ − 3sin ϕ) + q(sin 5ϕ − 5sin ϕ ϕ)) (7) + u(cos 3ϕ − cos+ v(cos 5ϕ − cosϕ) + w(cos 7ϕ − cosϕ) + K)
Substituting Eq. (7) into Eq. (6) and expanding in a Fourier series and equating the coefficients of cosφ , cos3φ , cos5φ , cos7 φ and sinφ , sin 3 φ , sin 5 φ respectively, we obtained the following equations as
(84−++w 4δ εω (4 pp −− 4232 168 pq − 246 pq 23 −−+ 135 q 14( pqu 2) 2 −+ 3(2 pqv 5) 2 ++−−8pw 16 qw 8 pw2 13 qw 2 −−+ 8 pδ 20 q δ 4 pw δ +−− 8 qw δδ 2 p2 5)8 q δω 2 − 2 (8) +4wω2 − 4 δω 2 −++−+ 2(2 uqvw (6 9 11(2 δ )) εω + pw (6 −+ 5(2 δ )) εω − 2 ω 2 ) +2((23vp ++ wδ ) εω +++ q (10 w 5) δ εω + 2( ω2 −1))) / 4 = 0. Science Journal of Applied Mathematics and Statistics 2019; 7(4): 51-55 53
εω(18−qu232 εω + (84 p + 270 pqp +−++−−+− 3(887 q 22 12 vv 3 10 vww 3 2 4 δ +−+6vδδ2 ) 2(25 qqvww 222 − 3(7 +− 3(2 ++++ δδ ) 2(2 ) 2 3(4 vw −+ 3(2 δ )))))/4 (9) ++u(46(26 pvw −++ 3δ ) εω + 6(1015 q −++ vw 5) δ εω − 36 ω 2 )) = 0.
5(pqv+ 3)2εω − 5(12 p 32 + 15 pqp −+++−+++ (435 quww 22 8 4 3 2 4δ 2 w δ δ 2 −++−2(6uw 2 3))δ qquuw (512 + 19 2 − 2(4 +++−++ 7(2 δ )) 2(23 w 2 4 δδ 2 (10) ++3(2wδεω )))) +−−++ 2(25(47 vquw 3 2) δεω −−+− 5(23 pu δεωω ) 502 ) = 0.
7qw232εω + 7(3 p − 10 pqq +− (425 qu 22 −−+++−+ 6 24 vuvv 22 11 2 4δ 12 v δ δ 2 ) −pquvv(392 ++−+−++ 5 2 (83 4)2(24δ uv δεω ))) + 2(27( wpuv − − 2) εω (11) −7(26q −+ v δεω ) − 98 ω 2 ) = 0.
−q2εω(5462 −−−++ uvw 27 17 27) δ puw2 (14 +−+ 6 5(2 δ ) εω +++ (2 uvw3 3 3 8 −10ww23 +−+ 3 812δδδδδδ ww − 5 22232 −+ 6 3 w −+ vw (7 −++ 5(2)) δ uv 2 (4 +−+++2w 3(2δ )) vw (8722 ++− 12 δδ 3 8(2 w ++ δ ))2(23 uvwv ++ 22 2 − 3(2 (12) −+++−wδδδ)42 2(2 w + δεω ))) − 20(1 qp −+ ω 2 )2 ((22q− + 28 u + 6 v +−9w 11δεω ) −− 6(1 ω 2 )) = 0.
−3pu2 (28 −+−+ 3(62 vw 5 3))δ εω −+−+−+ (88 uvvwvww323 6 2 12 12 + 6 2 −−−6vwwq2 3 3(7097 2 −−−++−− uvw 10 24 35)12δ δ 3 v2 δ 12 wvw δ + 6 δ (13) +3w22232δδδδ +− 6 3 w ++ 3(2 uvw +−++ 3(2))3(3 δ uvw 22 +++ δδ (4) −++2(2vwδ ))) εω + 2(23(3450 pq + −+−+ uvw 3 19 17) δ εω − 1 8ω2 )= 0.
5(25q2 u−+ 51 v 25) wεω +−+−−+ 5(4 vv 823 3 v 4 w 12 vwvww − 5 222 +− 4 6 vw −+wu3 2(25 ++− vδ )5(2434 p 2 −+−+−+ uvw δδδδδ )8 vv 42 − 4 wvw + 6 (14) +2wvwuvvw2δδδ − 2 2 −− 2 (5 2 −−++−+ 2(64 3)(2 δ w δεω )))2 + 2(2 q −5(10p −+−+ 25 uvw 25 24 5)δεω − 50 ω 2 ) = 0.
δ 2 Since is much smaller than 1 so all the constants 2274δ 750917710871 δ ω,u ,, v w , p , q of Eqs.(8)-(14) can be expanded in powers u = + + L 527 7903611882 of δ as in the following form: δ δ 2 ω=+ δ + δ 2 + L 5035 2676227662505 1 c1 d 1 v =−− + L (16) 4743 426795041628 =δ + δ 2 + L u c2 d 2 =δ + δ 2 + L 455δ 4091164395829 δ 2 v c3 d 3 =−− + L (15) w =δ + δ 2 + L 4743 426795041628 w c4 d 4 p= cδ + d δ 2 + L 7016δ 2082541407358 δ 2 5 5 p =−− + L =δ + δ 2 + L 1581 35566253469 q c6 d 6 . 1040δ 1119969952960 δ 2 Substituting (15) into (8)-(12) and (14) and equating the q =−− + L. coefficients of δ we obtain a system of linear equations of 1581 35566253469 L L ccc123,,,,, ddd 123 ,, . Solving these equations and L L Substituting (16) into (13) we obtain another series. substituting the values of ccc123,,,,, ddd 123 ,, into (9) we obtain for ε = 1 :
1754δ 2958662347157 δ 2 ω =−−1 + L 1581 71132506938 54 Md. Abul Kashem Mondal et al. : A New Analytical Approach for Solving Van der Pol Oscillator
18146δ 44198163782443 δ2 2295665100755905642 δ 3 −++2 + +=L 0. (17) 527 47421671292 168690740203467
From this series (17) we obtain the value of δ and use the for Van der Pol oscillator together with numerical solution smallest δ in Eq.(16). Finally, substituting the value of all for ε = 1 has been provided in figure 2. The corresponding constants in Eq. (7) we obtain the required solution. numerical results have been calculated by Runge-Kutta fourth order formula. From the figures we observe that the 4. Results and Discussion present solutions are nicely agreement with the corresponding numerical solutions and gives almost similar To test the accuracy of an approximate solution, some results to those obtained by Rahman et al [10]. Though the authors compared analytical solutions to those obtained by present solution is almost similar to that of Rahman et al [10] the numerical techniques [9, 27]. Here, we have been but present method is much easier than that of Rahman et al presented a new analytical approach based on harmonic [10]. Moreover in the article of Rahman et al. [10] two balance method to determine the approximate solution of simultaneous algebraic-transcendental equations were solved Van der Pol nonlinear equation. The solutions have been which is laborious process. But in our present method a compared with numerical solution for ε = 1 which are cubical equation has been solved (Eq. (17)). presented in figure 1. Also the solution of Rahman et al [10]
2.5 2 1.5 1 X 0.5 0 t -0.5 0 2 4 6 8 10 12 -1
-1.5 Numerical results -2 Present results -2.5
Figure 1. Comparison of present results with numerical resultsfor ε = 1 .
2.5 2 1.5
1 X 0.5 0 t -0.5 0 2 4 6 8 10 12 -1 -1.5 Numerical results -2 Rahman et al [10] results -2.5
Figure 2. Comparison of Rahman et al [10] results with numerical results for ε = 1 . Science Journal of Applied Mathematics and Statistics 2019; 7(4): 51-55 55
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