VARIATIONAL ITERATION METHOD FOR VIBRATION PROBLEMS Metin O. Kaya Istanbul Technical University, Faculty of Aeronautics and Astronautics, 34469, Maslak, Istanbul, Turkey Abstract In this study, linear/nonlinear, free/forced and damped/undamped vibrations of both one degree of freedom and continous systems are discussed by using the Variational Iteration Method. Additionally, common vibration problems are classified and Lagrange multipliers are derived for each type of problem. Keywords: Variational Iteration Method, VIM, Vibration, Lagrange Multiplier, Nonlinear Vibration 1. Introduction Vibration of dynamical systems can be divided into two main classes like discrete and distributed. The variables in discrete systems depend on time only, whereas in distributed systems such as beams, plates etc. variables depend on time and space. Therefore, equations of motion of discrete systems are described by ordinary differential equations, while equations of motion of distributed systems are described by partial differential equations ( Meirovitch [1] ). 1 A considerable amount of studies have been made in the area of vibration problems. Several techniques, such as finite element method, finite difference method, perturbation techniques, series techniques, etc. have been used to handle vibration problems. The Variational Iteration Method, VIM, was first proposed by He [2-11] and the method has been applied to investigate many nonlinear partial differential equations, autonomous and singular ordinary differential equations such that solitary wave solutions, rational solutions, compacton solutions and other types of solution were found by Abdou and Soliman [12, 13]. Additonally, He [4, 9] used VIM to solve linear/nonlinear vibration problems. In this study He’s studies are extended to cover vibration problems with damping, forced vibration and vibration of beams. Therefore, here VIM is applied to various vibration problems including vibration of linear/nonlinear, damped/undamped, free/forced vibrations of one degree of freedom systems and beams as an example of continuous systems. The procedure presented in this paper can be simply extended to solve more complex vibration problems; such as aeroelasticity, random vibrations etc. 2. Variational Iteration Method In order to illustrate the basic concepts of VIM, the following nonlinear partial differential equation can be considered Lu( x, t ) Ru( x, t ) Nu( x, t ) g ( x, t ) (1) 2 where R is a linear operator which has partial derivatives with respect to x , L is the linear time derivative operator, Nu ( x, t ) is a nonlinear term and g ( x, t ) is an inhomogeneous term. According to VIM, the following iteration formula can be constructed. t un1 ( x, t ) un ( x, t ) Lun Run Nun g d (2) 0 where is the general Lagrange multiplier which can be identified optimally via variational theory, Run and Nun are considered as restricted variations, i.e. Run 0 , Nun 0 . 3. Different Form of L Operator According to the L operator, a general classification of vibration problems can be made as follows 3.1. Case I: Here the following form of the L operator is used Lm 2 t 2 (3) Considering Eq. (3), Eq. (1) can be expressed as follows mu ( x, t ) Ru( x, t ) Nu( x, t ) g ( x, t ) (4) 3 The correction functional of Eq. (4) can be written as t un1 ( x, t ) un ( x, t ) mun Run Nun g d (5) 0 Making the above correction functional stationary, and noticing that u(0) 0 , the following iteration can be written t un1 ( x, t ) un ( x, t ) mun Run Nun g d (6a) 0 t un (t ) m ( ) un ( ) t m ( ) un ( ) t m un d 0 (6b) 0 which yields the following stationary conditions: m 0 (7a) 1 m t 0 (7b) t 0 (7c) Therefore, in this case the Lagrange multiplier can be identified as follows 1 ( t ) m (8) 4 3.2. Case II: Here the following form of the L operator is used Lm 2 c 2 t t (9) Considering Eq. (9), Eq. (1) can be expressed as follows mu ( x, t ) cu( x, t ) Ru( x, t ) Nu( x, t ) g ( x, t ) (10) Here Eq. (10) denotes vibration with damping where c is the damping coefficient and m is the mass. The correction functional of Eq. (10) can be written as t un1 ( x, t ) un ( x, t ) mun Run Nun g d (11) 0 Making the above correction functional stationary, and noticing that u(0) 0 , the following iteration can be written t un1 ( x, t ) un ( x, t ) mun cu Run Nun g d (12a) 0 t un (t ) m ( ) un ( ) t [c m ( )] un ( ) t (m c ) un d 0 (12b) 0 which yields the following stationary conditions 5 m c 0 1 m c t t (13a) 0 (13b) 0 (13c) Combining Eqs. (13b) and (13c), Eqs. (13a)-(13c) can be rewritten as follows m c 0 (14a) 1 m t 0 (14b) (14c) t 0 Therefore, in this case the Lagrange multiplier can be identified as follows 1 mc ( t ) [e 1] c (15) Approximate Lagrange multiplier can be obtained simply by expanding Eq. (15) as follows 1 mc ( t ) 1 c c2 2 [e 1] ( t ) ( t ) ( t )3 c m 2! m2 3! m3 (16) Hence, 1 c c2 2 ( t ) ( t ) ( t )3 2 3 m 2! m 3! m (17) 6 3.3. Case III: Here the following form of the L operator is used Lm 2 k t 2 (18) Considering Eq. (18), Eq. (1) can be expressed as follows mu ( x, t ) ku( x, t ) Ru( x, t ) Nu( x, t ) g ( x, t ) (19) whre k is the spring coefficient. The correction functional of Eq. (19) can be written as t un1 ( x, t ) un ( x, t ) mun kun Run Nun g d (20) 0 Making the above correction functional stationary, and noticing that u(0) 0 , the following iteration can be written t un1 ( x, t ) un ( x, t ) mun ku Run Nun g d (21a) 0 t un (t ) m ( ) un ( ) t m ( ) un ( ) t (m k ) un d 0 (21b) 0 which yields the following stationary conditions m k 0 (22a) 7 1 m t 0 (22b) (22c) t 0 Therefore, in this case the Lagrange multiplier can be identified as follows 1 sin ( t ) m (23) where the circular frequency, , is given by k . m (24) Approximate Lagrange multiplier can be obtained simply by expanding Eq. (23) as follows 1 ( t ) 2 ( t )3 sin ( t ) m m 3! m (25) Hence, ( t ) 2 ( t )3 m 3! m 3.4. (26) Case IV: Here the following form of the L operator is used Lm 2 c k 2 t t (27) 8 Considering Eq. (27), Eq. (1) can be expressed as follows mu ( x, t ) cu( x, t ) ku( x, t ) Ru( x, t ) Nu ( x, t ) g ( x, t ) (28) The correction functional of Eq. (28) can be written as t un1 ( x, t ) un ( x, t ) mun cu ku Run Nun g d (29) 0 Making the above correction functional stationary, and noticing that u(0) 0 , the following iteration can be written t un1 ( x, t ) un ( x, t ) mun cu ku Run Nun g d (30a) 0 t un (t ) m ( ) un ( ) t [c m ( )] un ( ) t (m c k ) un d 0 (30b) 0 which yields the following stationary conditions m c k 0 (31a) 1 m c (31b) t t 0 0 (31c) Combining Eqs. (31b) and (31c), Eqs. (31a)-(31c) can be rewritten as follows m c k 0 (32a) 9 1 m t 0 (32b) (32c) t 0 Therefore, in this case the Lagrange multiplier can be identified as follows Sin[ ( t )] 2cm( t ) e m (33) where 4mk c 2 2m (34) Approximate Lagrange multiplier can be obtained simply by expanding Eq. (34) as follows Sin[ ( t )] 2cm ( t ) ( t ) c( t ) 2 1 c2 k 3 e 2 ( t ) 2 m 2m 3! m m m m (35) Hence, ( t ) c( t ) 2 1 c2 k 3 2 ( t ) 2 2m 3! m m m m 10 (36) 4. Illustrative Examples EXAMPLE 1: In this example, a simple mass-spring system that undergoes forced vibration is examined. The differential equation of motion of this system is given by my ky f (t ) (37) where f (t ) F0 cos(t ) Dividing both sides by m , Eq.(37) can be rewritten as follows y 2 y F0 cos(t ) where F0 (38) F0 . m Here it is easily seen that the Lagrange multiplier of this problem is 1 sin ( t ) m (39) Additionally, the iteration formula of this problem is t yn1 (t ) yn (t ) 1 sin ( t ) [ my ( ) ky( ) f ( )]d m 0 The complementary solution of this problem is given by 11 (40) y0 A cos t B sin t (41) Using Eq. (41) as an initial approximation, we get y1 (t ) A cos t B sin t 1 t 0 sin ( t ) [ F0 cos( )]d (42) or y1 t A cost B sin t F0 cost F0 cost 2 2 2 2 (43) Since the last term in Eq. (43) automatically satisfies the complementary equation, this term will not be used. Thus, Eq. (43) can be simplified to y1 (t ) A cos t B sin t F0 cos(t ) 2 2 (44) which is the general solution of Eq. (37). In order to point out the importance of using the exact Lagrange multiplier instead of the approximate one, the following multiplier may be considered ( t ) m (45) Considering Eq. (45), the following iteration expression can be written t yn1 (t ) yn (t ) ( t )[ y ( ) 2 y( ) 0 f ( ) ]d m 12 (46) If we use the complementary solution given by Eq.(41) as an initial approximation, we get t y1 (t ) A cos t B sin t ( t )[ F0 cos( )]d (47a) 0 or in compact form y1 A cos t B sin t F0 [ 1 Cos(t ) ] 2 (47b) and t 2 2Cos( ) 1 Cos(t ) Cos ( ) F0 cos( ) d y2 A cos t B sin t F0 [ ] ( t ) F0 2 2 2 0 (48a) or in compact form 1 2 1 2 2t 2 y2 A cos t B sin t F0 2 4 F0Cos(t ) 2 4 F0 22 (48b) In the same way, y 4 can be written as follows 1 2 4 6 1 2 4 6 y4 A cos t B sin t F0 2 4 6 8 F0 cost 2 4 6 8 F0 2 t 2 2 1 2 4 F0 4 t 4 2 4 6 24 Since, 13 1 2 F0 6 t 6 2 4 720 1 2 (49) 1 1 2 4 6 2 2 2 4 6 8 (50) The expression for y n can be written as follows yn A cos t B sin t F0Cos(t ) F0 2t 2 4t 4 F0 6t 6 1 2 2 2 2 2 24 720 (51) As n , the compact form of solution becomes y(t ) A cos t B sin t F0 [cos(t ) cos(t )] 2 2 (52) Note that Eq. (52) is the same as Eq. (47b). If we use a different approximate form of the Lagrange multiplier, , such as ( t ) 2 ( t )3 m 3! m (53) the following iteration expression can be written t yn1 (t ) yn (t ) [( t ) 0 2 ( t )3 3! ][ y ( ) 2 y( ) f ( ) ]d m (54) Again using the complementary solution given by Eq.(41) as an initial approximation, we get t y1 (t ) A cos t B sin t [( t ) 0 2 ( t )3 3! 14 ][ F0 cos( )]d (55a) or in compact form 1 2 1 2 2t 2 y1 A cos t B sin t F0 2 4 F0Cos(t ) 2 4 F0 22 (55b) and 1 2 4 6 1 2 4 6 y2 A cos t B sin t F0 2 4 6 8 F0 cost 2 4 6 8 F0 2 t 2 2 1 2 4 F0 4 t 4 2 4 6 24 1 2 F0 6 t 6 2 4 720 1 2 (56) ( t ) 2 ( t )3 It can be easily seen that approximate Lagrange multiplier m 3! m converges faster than ( t ) . m EXAMPLE 2: In this example, a damped mass-spring system that undergoes forced vibration is examined. The differential equation of motion of this system is given by my cy ky F0 cos(t ) (57) The Lagrange multiplier of this problem is Sin[ ( t )] 2cm( t ) e m (58) 15 where 4mk c 2 2m (59) Hence using this Lagrange multiplier, the iteration formula can be written as t yn1 (t ) yn (t ) c ( t ) 1 2m e Sin[ ( t )][my ( ) cy ( ) ky( ) f ( )]d m 0 (60) Dividing both sides of Eq. (57) by m , the following equation of motion is obtained y 2 y 2 y F0 cos(t ) (61) where F0 F0 c . and m 2m Therefore, Eq. (60) becomes yn1 (t ) yn (t ) 1 t c e 2m ( t ) Sin[ ( t )] y ( ) 2 y ( ) 2 y ( ) F0 cos( ) d (62) 0 and 1 2 (63) The complementary solution of this problem is given by 16 y0 A cos t B sin t et (64) Taking this complementary solution as an initial approximation, the following expression is obtained y1 (t ) y0 (t ) 1 t e c ( t ) 2m Sin[ ( t )][ F0 cos( )]d (65) 0 Integrating the second term of Eq. (65), we get y1 (t ) y0 (t ) ( 2 2 )Cos[t ] 2Sin[t ] 4 2(2 2 1) 22 4 (2 2 )Cos[ t ]et (2 2 ) Sin[ t ]et 4 2(2 2 1) 22 4 4 2(2 2 1) 22 4 1 2 (66) Since the last two terms of Eq. (66) automatically satisfy the homogeneous equation, they will not be used. The second term of Eq. (66) can be written in a more compact form as follows ( 2 2 )Cos[t ] 2Sin[t ] 4 2(2 2 1) 22 4 Cos(t ) 1 2 / 2 / 2 2 (67) 2 where tan( ) 2 / 1 2 / 2 (68) Hence, the exact solution of Eq. (57) is obtained after one iteration 17 y1 A cos t B sin t et F0Cos(t ) 1 2 / 2 / 2 2 (69) 2 EXAMPLE 3: In this example, transverse vibration of a uniform beam with simply supported ends is examined. The differential equation of motion of this system is given by 4 2 y 2 y c 0, 0 x L, t > 0 t 2 x 4 (70) The initial and the boundary conditions for this problem are as follows: y ( x,0) sin x (71a) L y t ( x,0) 0 (71b) y (0, t ) 0 (71c) y ( L, t ) 0 (71d) yxx (0, t ) 0 (71e) yxx ( L, t ) 0 (71f) y t ( x,0) 0 , yxx (0, t ) 0 The Lagrange multiplier of this problem is 18 It is easily noticed that the Lagrange multiplier of this problem is 1 ( t ) m (72) Using this Lagrange multiplier, the iteration formula can be written as t yn1 (t ) yn (t ) 1 ( t )[ y c 2 yxxxx ]d m0 (73) The complementary solution of this problem is given by y0 (t ) sin x (74) L Using the iteration formula given by Eq. (73) and taking the complementary solution as an initial approximation, we get y1 (t ) sin x L t c2 4 x c 2 4t 2 x 1 x sin sin ( t ) sin d L4 L 2 L4 L m 0 L (75) and y2 (t ) sin t c2 4 x 2 4 x c2 4t 2 x 1 x c 2 8 2 x c sin sin sin ( t ) sin 4 d 8 4 4 L 2L L L 2L L m0 L L L sin x L c 2 4t 2 x c 4 8t 4 x sin sin 2! L4 L 4! L8 L Hence the expression for y n can be written as follows 19 (76) yn (t ) sin x c2 4t 2 c4 8t 4 1 L 2! L4 4! L8 (1) n c2 n 4 nt 2 n (2n)! L4 n (77) Thus we have, lim yn (t ) sin x n L cos 2 ct (78) L2 which is the exact solution. EXAMPLE 4: In this example, transverse vibration of a variable coefficient beam that was studied by Wazwaz [14] is considered. The differential equation of motion of this system is given by 2 y 4 y ( x ) f ( x, t ) , ( x ) 0 , 0 x L , t > 0 t 2 x 4 (79) Here it can be that the equation of motion of the beam is as follows: 2 y 4 y 6 (1 x ) ( x 4 x 3 )cos t 2 4 t x 7! (80) The initial and the boundary conditions for this problem are as follows: y ( x,0) 6 7 x , 7! y t ( x,0) 0 , 0 x 1 (81a) 0 x 1 (81b) 20 y (0, t ) 0 . y (1, t ) 6 cos t , 7! yxx (0, t ) 0 , y xx (1, t ) 1 cos t , 20 t>0 (81c) t>0 (81d) t>0 (81e) t>0 (81f) The Lagrange multiplier of this problem is 1 ( t ) m and m 1 for this problem. Hence the iteration formula can be written as t yn1 (t ) yn (t ) ( t )[ y (1 x) y xxxx ( x 4 x 3 0 6 )cos ]d 7! (82) The complementary solution of this problem that is used as an initial approximation is given by y0 ( t ) 6 7 x cos t ( x 3 x 4 )(1 cos t ) 7! (83) By using the iteration expression given by Eq. (82) and the initial approximation given by Eq. (83), we get y1 (t ) 6 7 x cos t ( x 3 x 4 )(1 cos t ) (1 x)(24 12t 2 x 3 (24 x 3 )cos t 7! 21 (84a) or in compact form y1 (t ) 12(t 2 2)(1 x) ( x7 24 x 24)cos t 7! (84b) and y2 (t ) 12(t 2 2)(1 x) ( x7 24 x 24)cos t 12(1 x)(t 2 2 2cos t ) 7! (85) Eq. (85) can be simplified to the following expression y2 ( t ) x7 cos t 7! (86) As it is seen, the exact solution of this problem is obtained quickly in two iteration. EXAMPLE 5: In this example, Mathematical Pendulum that was studied by He [4, 9] is considered. The differential equation of motion of the undamped mathematical pendulum is given by, y 2 sin y 0 (87) The initial conditions for this problem are as follows: y (0) A (88a) y (0) 0 (88b) 22 The Siny term in Eq. (87) is a nonlinear term and it can be expanded as sin y y 1 3 y 6 (89) Substituting Eq. (89) into Eq. (87) gives y y 2 2 6 y3 0 (90) A more detailed form of this mathematical pendulum was investigated by He [5, 7]. The Lagrange multiplier of this problem is 1 sin ( t ) (91) Hence the iteration formula is yn1 (t ) yn (t ) 1 t 2 sin ( t )[ y( ) y( ) 0 2 6 y 3 ( )]d (92) The complementary solution of this problem that is used as an initial approximation is given by y0 (t ) A cos(t ) (93) where is an unknown constant. Substituting the initial approximation into Eq. (90), the following residual is obtained 23 R0 (t ) y 2 y 2 1 1 3 2 y 3 A(1 A2 2 ) 2 cos(t ) A cos(3t ) 6 8 24 (94) The coefficient of the cos( t ) term is set to zero in order to eliminate the secular term which may occur in the next iteration. Doing so, the expression of is found as follows A2 1 8 (95) Hence, A3 y1 (t ) A cost (cos3t cost ) 24(9 2 1) 2 (96) with defined in Eq. (95). The period can be expressed as follows T 2 1 1 A2 8 If A 2 (97) , then T 1.20T0 . On the other hand He’s [4, 9] approximation gives T 1.17T0 , while the exact period is Tex 1.16T0 , where T0 2 / . 24 EXAMPLE 6: In this example, the problem that was studied by Nayfeh and Mook [15] is considered. The differential equation of motion is given by, u 2u u 2u 0 (98) The initial conditions for this problem are as follows: y (0) A (99a) y (0) 0 (99b) The Lagrange multiplier of this problem is 1 sin ( t ) (100) The iteration formula is given by un1 (t ) un (t ) 1 t sin ( t ) [u( ) u( ) u ( )u( )]d 2 2 (101) 0 The complementary solution of this problem that is used as an initial approximation is given by u0 (t ) A cos(t ) (102) where is an unknown constant. 25 Substituting the initial approximation given by Eq. (102), the following residual is obtained as follows R0 (t ) u 2u u 2u A 2 (1 2 3 2 1 A )cos(t ) A3 2 2 cos(3 t ) 4 4 (103) The coefficient of the cos( t ) term is set to zero in order to eliminate the secular term which may occur in the next iteration. Doing so, the expression for is obtained as follows 2 (104) 4 3 A2 Hence, y1 (t ) A cost A3 2 (cost cos3t ) 4(9 2 1) (105) with defined in Eq. (104). The new frequency is defined as follows 1 1 2 4 3 A2 (106) The frequency that is obtained by Nayfeh and Mook [15] using the perturbation method is 3 8 1 (1 A2 ) (107) 26 Note that Eq. (107) is valid only for small values. However, the frequency expression given by Eq. (106) is valid for all values and takes the following form for small values 3 8 1 1 A2 27 2 4 A 128 (108) EXAMPLE 7: In this example, the Duffing-harmonic oscillator that was studied by Lim and Wu [16] and Mickens [17] is considered. The differential equation of motion is given by, d2 y y3 0 dt 2 1 y 2 (109) The initial conditions for this problem are as follows: y (0) A (110a) y (0) 0 (110b) with initial conditions y (0) A and y (0) 0 . For small y values, Eq. (109) reduces to d2 y y3 dt 2 0 (111a) On the other hand, for large y values, Eq. (109) reduces to 27 d2 y y dt 2 (111b) 0 Considering Eqs. (111a) and (111b) respectively, it is noticed that for small y values, Eq. (109) reduces to the equation of motion of the Duffing-type nonlinear oscillator while for large y values, it reduces to the equation of motion of a linear harmonic oscillator. Therefore, Eq. (109) is called as Duffing-harmonic oscillator equation of motion. The following form of Eq. (109) is going to be studied in this example (1 y 2 ) d2 y y3 0 2 dt (112) He’s technique is going to be used to overcome seculer terms that appear in the iterations. The initial approximation is, y0 (t ) A cos( t ) (113) where is an unknown constant. Substituting the initial approximation into Eq. (112), the following residual is obtained 3 3 A3 R0 (t ) (1 y 2 ) y y 3 ( A2 2 A2 2 )cos( t ) (1 2 )cos(3 t ) 4 4 4 (114) In oreder to discard the seculer terms, the coefficient of cos( t ) is set to zero which gives the expression of as follows 28 3 2 A 4 3 1 A2 4 (115) Hence the new frequency is defined as follows 3 2 A 4 3 1 A2 4 (116) which is the same with the one found by Lim and Wu [16] and Mickens [17]. The iteration formula is given by t u1 (t ) u0 (t ) ( t )[ 0 A3 (1 2 )cos(3 )]d 4 (117) Hence, y1 (t ) cos t A (cos3t 1) 27 (118) with defined in Eq. (116). For small values of amplitude A, the frequency expression given in Eq. (116) is expressed as follows 3 A 4 (119a) 29 Additionally, for large values of amplitude A, the frequency expression given in Eq. (116) is expressed as follows 1 (119b) which agree with the approximations made for the equations of motion given in Eqs. (111a) and (111b). 5. Conclusion In thispaper, which extends He’s studies , various types of vibration problems including vibration of linear/nonlinear, damped/undamped, free/forced vibrations of one degree of freedom systems and beams as an example of continuous systems are solved using the Variational Iteration Method. Lagrange multipliers which arise from different types of vibration problems are presented and the solutions are made in a detailed way. Additionally, the procedure presented in this paper can be simply extended to solve more complex vibration problems; such as aeroelasticity, random vibrations etc. 30 REFERENCES [1] L. Meirovitch, Computational Methods in Structural Dynamics, Sijthoff & Noordhoff International Publishers, Alphen aan den Rijn, The Netherlands, 1980 [2] J. He, A New Approach to Nonlinear Partial Differential Equations, Communications in Nonlinear Science & Numerical Simulation, Vo1.2, No.4 (Dec. 1997), 230-235. [3] J.H. He, Approximate analytical solution for seepage flow with fractional derivatives in porous media, Comput. Methods Appl. Mech. Eng. 167 (1998) 57–68. [4] J.H. He, Approximate solution of nonlinear differential equations with convolution product nonlinearities, Comput. Methods Appl. Mech. Eng. 167 (1998) 69–73. [5] J.H. He, Variational iteration method—a kind of non-linear analytical technique: some examples, Internat. J. Nonlinear Mech. 34 (1999) 699–708. [6] J.H. He, Variational iteration method for autonomous ordinary differential systems, Appl. Math. Comput. 114 (2000) 115–123. [7] J. He, Homotopy perturbation method: a new nonlinear analytical technique, Applied Mathematics and Computation 135 (2003) 73–79. [8] J.H. He, The homotopy perturbation method for nonlinear oscillators with discontinuities, Appl. Math. Comput. 151 (2004) 287–292. [9] J.H. He, A generalized variational principle in micromorphic thermoelasticity, Mech. Res. Comm. 32 (1) (2005) 93–98. 31 [10] J.H. He, Some asymptotic methods for strongly nonlinearly equations, Internat. J. Modern Math. 20 (10) (2006) 1141–1199. [11] J.H. He, Variational iteration method—Some recent results and new interpretations, J. Comput. Appl. Math. in press doi:10.1016/j.cam.2006.07.009. [12] M.A. Abdou, A.A. Soliman, New applications of variational iteration method, Physica D 211 (2005) 1–8. [13] M.A. Abdou, A.A. Soliman, Variational iteration method for solving Burger’s and coupled Burger’s equations J. Comput. Appl. Math. 181 (2005) 245–251. [14] A. M. Wazwaz, Analytic treatment for variable coefficient fourth-order parabolic partial differential equation, Appl. Math. Comput. 123 (2001) 219-227. [15] A. H. Nayfeh and D. T. Mook, Nonlinear Oscillation, John Wiley & Sons, 1979 [16] C.W. Lim and B.S. Wu, A new analytical approach to the Duffing-harmonic oscillator, Physics Letters A 311 (2003) 365–373 [17] R. E. Mickens, Mathematical and Numerical Study of the Duffing Harmonic Oscillator, Journal of Sound and Vibration (2001) 244(3), 563-567. 32

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# Variational Iteration Method for Vibration Problems