# A more accurate solution of nonlinear conservative oscillator by energy balance method

Md Helal Uddin Molla (Department of Mathematics, Rajshahi University of Engineering and Technology, Rajshahi, Bangladesh)
Md Abdur Razzak (Department of Mathematics, Rajshahi University of Engineering and Technology, Rajshahi, Bangladesh)
M.S. Alam (Department of Mathematics, Rajshahi University of Engineering and Technology, Rajshahi, Bangladesh)

ISSN: 1573-6105

Publication date: 3 December 2018

## Abstract

### Purpose

The purpose of this paper is to present an analytical approximate technique to solve nonlinear conservative oscillator based on the He’s energy balance method (improved version recently presented by Khan and Mirzabeigy). The method is illustrated by solving double well Duffing oscillator.

### Design/methodology/approach

The Duffing equation with a double-well potential (with a negative linear stiffness) is an important model of a mass particle moving in a symmetric double well potential. This form of the equation also appears in the transverse vibrations of a beam when the transverse and longitudinal deflections are coupled (Thompsen, 2003).

### Findings

The approximate solutions obtained by the present technique have good agreement with the numerical solution and also provide better results than other existing methods.

### Originality/value

The results are more accurate than those obtained by other existing methods. The relative errors obtained by the present paper are less than those obtained by other existing methods. Therefore, the present technique is very effective and convenient for solving nonlinear conservative oscillator.

## Keywords

#### Citation

Molla, M., Razzak, M. and Alam, M. (2018), "A more accurate solution of nonlinear conservative oscillator by energy balance method", Multidiscipline Modeling in Materials and Structures, Vol. 14 No. 4, pp. 634-646. https://doi.org/10.1108/MMMS-10-2017-0130

### Publisher

:

Emerald Publishing Limited

## 1. Introduction

Nonlinear oscillators often arise in the exact modeling of phenomena in physical sciences, mechanical structures, nonlinear circuits, chemical oscillation and other engineering field. The study of them is of interest to many researchers. Especially the Duffing equation with a double-well potential (with a negative linear stiffness) is an important model. Physical realization of such a Duffing oscillator model is a mass particle moving in a symmetric double-well potential. This form of the equation also appears in the transverse vibrations of a beam when the transverse and longitudinal deflections are coupled (Thompsen, 2003). Many analytical methods are available to solve nonlinear oscillators, such as the classical perturbation methods (Bogoliubov and Mitropolskii, 1961; Nayfeh and Mook, 1979; Nayfeh, 1973) which can be used to solve weak nonlinearity problems. Homotopy perturbation method (He, 1999), harmonic balance method (Mickens, 1996; Hosen et al., 2012; Alam et al., 2016; Molla et al., 2016; Mondal et al., 2017), Hamiltonian approach (He, 2010), Variational approach (Yazdi, 2016), Rational variational approaches (Yazdi and Tehrani, 2017), homotopy analysis method (Liao and Cheung, 1998), max-min approach (He, 2008; Yazdi et al. 2012), coupling method (Akbarzede et al., 2011; Razzak and Alam, 2016), energy balance method (He, 2002; Durmaz et al., 2010; Yazdi et al., 2010; Momeni et al., 2011; Durmaz and Kaya, 2012; Khan and Mirzabeigy, 2014; Yazdi and Tehrani, 2015; Razzak and Rahman, 2015; Molla et al., 2017; Molla and Alam, 2017) have been developed for solving strong nonlinear oscillators. The three analytical techniques, such as the max-min, second-order Hamiltonian, and the global error minimization approaches, are applied to achieve natural frequencies of conservative Helmholtz-Duffing oscillator by Mirzabeigy et al. (2014). Energy balance method was first introduced by He (2002). JH He provided that the first-order approximations of energy balance method give a good accurate solution in comparison with first-order approximation obtained by other techniques. Khan and Mirzabeigy (2014) presented approximate solutions by adding second harmonic term (i.e. cos3ωt), and the solution gave a better result than that obtained by other energy balance methods (Durmaz et al., 2010; Durmaz and Kaya, 2012; Momeni et al., 2011). Recently, Yazdi and Tehrani (2015) presented approximate solutions by applying the energy balance to nonlinear oscillations via Jacobi collocation method.

In the present article, an analytical approximate technique (proposed by Molla et al., 2017) has been presented to obtain approximate periods and periodic solutions to free oscillation of the undamped double-well Duffing oscillator based on the improved energy balance method (Khan and Mirzabeigy, 2014). The results are more accurate than those obtained by other existing methods.

## 2. The basic idea of He’s energy balance method

Consider a general form of a nonlinear oscillator with initial conditions as:

(1) x ¨ + f ( x ) = 0 , x ( 0 ) = A , x ˙ ( 0 ) = 0 ,
where over dot denotes the derivative with respect to time t. According to the variational principle, Equation (1) can be written as:
(2) J ( x ) = 0 t [ 12 x ˙ 2 + F ( x ) ] d t ,
where T = (2π)/(ω) is a period of the nonlinear oscillator and F(x) = 2∫f (x)dx.

The Hamiltonian can be written in the form:

(3) H ( t ) = 12 x ˙ 2 + F ( x ) = F ( A )

From Equation (3), the following residual is obtained:

(4) R ( t ) = 12 x ˙ 2 + F ( x ) F ( A ) = 0.

Consider the first approximate solution in the form:

(5) x ( t ) = A cos ω t ,
substituting Equation (5) into Equation (4) yields the following residual:
(6) R ( t ) = 12 A 2 ω 2 sin 2 ω t + F ( A cos ω t ) F ( A ) = 0.

And finally collocation at ωt = (π)/(4) gives:

(7) ω = 2 A F ( A ) F ( 2 2 A ) .

## 3. Improved energy balance method

According to Khan and Mirzabeigy’s (2014) presented method, the solution of Equation (1) is of the following form:

(8) x ( t ) = b cos ω t + b 1 cos3 ω t ,
using initial conditions it becomes:
(9) A = b + b 1 .

Eliminating b1 from Equations (8) and (9), the solution takes the form:

(10) x ( t ) = b cos ω t + ( A b ) cos3 ω t .

By substituting Equation (10) into Equation (4), residual is obtained which contains two unknown parameters, ω and b. For determining these parameters, two equations are essential; the first equation obtained by collocation method as:

(11) Lim ω t π4 R ( t ) = 0.

Then, the second equation is obtained using Galerkin-Petrov method as:

(12) 0 T / 4 R ( t ) cos ω t d t = 0 .

By solving Equations (11) and (12) simultaneously, ω and b can be determined.

## 4. More accurate solution

Let us consider b = A(1−u) and b1 = Au, then Equation (8) becomes:

(13) x ( t ) = A ( ( 1 u ) cos ω t + u cos3 ω t ) .

Substituting Equation (13) into Equation (4) residual is obtained as:

(14) R ( t ) = 12 ( A ω ( ( 1 u ) sin ω t + 3 u sin3 ω t ) ) 2 + F ( A ( ( 1 u ) cos ω t + u cos3 ω t ) ) F ( A ) = 0.

This residual contains two unknown constants ω and u. In order to determine these constants, we need two algebraic equations which are obtained from:

(15) 0 T / 4 R ( t ) cos ( 2 n 2 ) ω t d t sin 2 ω t = 0
for n = 1 and n = 2, respectively. Solving these two equations simultaneously, we obtain ω and u.

In a similar way, we consider the third approximate solution as in the form:

(16) x ( t ) = A ( ( 1 u v ) cos ω t + u cos3 ω t + v cos5 ω t )
where ω, u and v are unknown constants to be determined. Substituting Equation (16) into Equation (4) residual is obtained as:
(17) R ( t ) = ( A ω ( ( 1 u v ) sin ω t + 3 u sin3 ω t + 5 v sin5 ω t ) ) 2 / 2 + F ( A ( ( 1 u v ) cos ω t + u cos3 ω t + v cos5 ω t ) ) F ( A ) = 0.

This residual contains three unknown constants ω, u and v. To determine these unknown constants, we need three algebraic equations which are obtained from Equation (15) for n = 1, 2, 3, respectively. Solving these equations simultaneously, we obtain ω, u and v.

## 5. Example

### 5.1 Duffing oscillator with double-well potential

Double-well Duffing equation appears in the transverse vibrations of a beam when the transverse and longitudinal deflections are coupled (Wu et al., 2007; Momeni et al., 2011). Double-well Duffing equation is in the form:

(18) x ¨ x + x 3 = 0 , x ( 0 ) = A , x ˙ ( 0 ) = 0.

When Α > 2 , oscillation occurs between symmetric limit [−A, A]. For the case 1 < A < 2 , the oscillation occurs around stable equilibrium points x = ±1 and is asymmetric about it. In the present study, we consider the first case with A > 2 . The variational of Equation (18) is given as:

(19) J ( x ) = 0 t [ 12 x ˙ 2 x 2 2 + x 4 4 ] d t .

Its Hamiltonian can be written in the form:

(20) H = x ˙ 2 2 x 2 2 + x 4 4 = A 2 2 + A 4 4 .

The residual of Equation (20) becomes:

(21) R ( t ) = x ˙ 2 2 x 2 2 + x 4 4 + A 2 2 A 4 4 = 0.

Substituting Equation (13) into Equation (21), we obtain:

(22) R ( t ) = A 2 ( ( ω ( 1 u ) sin ω t + 3 u sin3 ω t ) ) 2 + ( ( 1 u ) cos ω t + u cos3 ω t ) 2 1 ) / 2 + A 4 ( ( ( 1 u ) cos ω t + u cos3 ω t ) 4 1 ) / 4 = 0.

Using Equation (22) into Equation (15), we obtain the first two equations, respectively, for n = 1, 2:

(23) 4 + 16 u 8 u 2 A 2 ( 3 + 12 u 12 u 2 + 12 u 3 6 u 4 ) + ( 4 + 16 u + 88 u 2 ) ω 2 = 0 ,
(24) 16 u A 2 ( 1 + 16 u 16 u 2 + 2 u 4 ) + ( 48 u + 96 u 2 ) ω 2 = 0.

Eliminating ω2 from Equations (23) and (24) and neglecting the higher order terms more than u2, we obtain the following equation as:

(25) A 2 32 u + 16 A 2 u 224 u 2 + 136 A 2 u 2 = 0.

Thus, Equation (25) becomes a quadratic equation, the smallest value of |u| is the required value of u. Substituting the value of u into Equation (23), we obtain the frequency ω = ω2 as well as the period T2 = 2π/ω2.

For the third approximate solution, substituting Equation (16) into Equation (21), residual is obtained as:

(26) R ( t ) = A 2 ( ( ω ( ( 1 u v ) sin ω t + 3 u sin3 ω t + 5 v sin5 ω t ) 2 + ( ( 1 u v ) cos ω t + u cos3 ω t + v cos5 ω t ) 2 1 ) / 2 + A 4 ( ( ( 1 u v ) cos ω t + u cos3 ω t + v cos5 ω t ) 4 1 ) / 4 = 0.

Substituting Equation (26) into Equation (15), we obtain, respectively, for n = 1, 2, 3:

(27) 4 ( 1 + 4 u 2 u 2 + 8 v 4 u v 4 v 2 ) + A 2 ( 3 12 u + 12 u 2 12 u 3 + 6 u 4 28 v + 36 u v 36 u 2 v + 16 u 3 v + 36 v 2 60 u v 2 + 36 u 2 v 2 36 v 3 + 36 u v 3 + 16 v 4 ) + ( 4 + 16 u + 88 u 2 + 32 v + 304 u v + 464 v 2 ) ω 2 = 0 ,
(28) 16 ( u + 3 v u v v 2 ) + A 2 ( 1 16 u + 6 u 2 2 u 4 44 v + 36 u v 24 u 2 v + 8 u 3 v + 42 v 2 48 u v 2 + 18 u 2 v 2 36 v 3 + 28 u v 3 + 14 v 4 ) + 16 ( 3 u + 6 u 2 + 5 v + 37 u v + 45 v 2 ) ω 2 = 0.
(29) 2 ( u 2 + 2 v + 2 u v + v 2 ) A 2 ( u + 3 u 2 4 u 3 + 2 u 4 + 5 v + 6 u v 12 u 2 v + 7 u 3 v + 3 v 2 15 u v 2 + 12 u 2 v 2 7 v 3 + 11 u v 3 + 4 v 4 ) + ( 18 u 2 + 20 v + 100 u v + 130 v 2 ) ω 2 = 0.

Eliminating ω2 from Equations (27) and (28), then from Equations (27) and (29), we obtain:

(30) A 2 + 32 u 16 A 2 u + 224 u 2 136 A 2 u 2 64 u 3 192 u 4 + v ( 32 + 992 u + 576 u 2 1 , 216 u 3 ) + v 2 ( 992 + 1 , 792 u 2 , 624 u 2 2 , 176 u v 1 , 024 v 2 ) + A 2 ( 184 u 3 + 14 u 4 208 u 5 + 188 u 6 v ( 8 + 676 u 240 u 2 808 u 3 + 1 , 816 u 4 1 , 248 u 5 ) v 2 ( 674 + 360 u 2 , 922 u 2 + 5 , 432 u 3 3 , 540 u 4 ) + v 3 ( 484 + 3 , 308 u 8 , 104 u 2 + 6 , 160 u 3 ) + v 4 ( 1 , 162 7 , 192 u + 7 , 668 u 2 ) 2 , 096 v 5 + 4 , 536 u v 5 + 1 , 256 v 6 ) = 0.
(31) 32 u + 128 u 3 160 u 4 + v ( 32 + 320 u + 768 u 2 1 , 024 u 3 ) + v 2 ( 512 + 1 , 792 u 2 , 624 u 2 ) + 960 v 3 3 , 072 u v 3 1 , 504 v 4 + A 2 ( 2 u 13 u 2 48 u 3 + 212 u 4 268 u 5 + 142 u 6 v ( 20 + 202 u + 288 u 2 1 , 378 u 3 + 1 , 912 u 4 1 , 056 u 5 ) v 2 ( 389 + 738 u 3 , 996 u 2 + 6 , 164 u 3 3 , 570 u 4 ) v 3 ( 266 5 , 114 u + 10 , 396 u 2 7 , 096 u 3 ) + v 4 ( 2 , 572 9 , 676 u + 8 , 916 u 2 ) 3 , 740 v 5 + 6 , 300 u v 5 + 1 , 968 v 6 ) = 0.

By eliminating A from Equations (30) and (31), it provides:

(32) u 2 + 13 u 3 + 62 u 4 + 170 u 5 40 u 6 v 12 u v 13 u 2 v + 223 u 3 v + 1 , 622 u 4 v 28 v 2 193 u v 2 122 u 2 v 2 237 v 3 + = 0.

From the Equation (32), we can expand v in a power series of u as:

(33) v = m 2 u 2 + m 3 u 3 + ,

Substituting Equation (33) into Equation (32) and equating the equal powers of u on both sides, we obtain a set of algebraic equations which provide the values of m2, m3, … . Using these values in Equation (33), we obtain:

(34) v = u 2 + u 3 + 9 u 4 + 23 u 5 + 135 u 6 .

Substituting the value of v in Equation (30) and ignoring higher order terms with u7,…,we obtain:

(35) 32 u + 256 u 2 + 960 u 3 + 265 u 4 + 12 , 800 u 5 + 50 , 912 u 6 + A 2 ( 1 16 u 144 u 2 5 00 u 3 1 , 168 u 4 7 , 136 u 5 25 , 408 u 6 ) = 0.

It is laborious to solve Equation (35). On the other hand, the numerical solution suggests that u is positive. To determine an approximate solution near A = 1.42, we may consider an iterative formula. At first, we ignore terms with u4, u5, u6 of Equation (35). Thus, it becomes a cubical equation that has only one real solution, denoted by u0. Then, the terms with u4, u5, u6 can be replaced by u0 u3, u 0 2 u 3 , u 0 3 u 3 . Therefore, Equation (35) again becomes a cubical equation whose solution gives the desired result of u (by using the value of u0). After taking two or three iterations, we obtain desired results. Substituting the value of u and v into Equation (27), we obtain the frequency ω = ω3 as well as period T3 = 2π/ω3.

To obtain more accurate approximate solution for A > 1.43, we divide Equation (35) by: 1 + 8u + 30u2 + 83u3 + 400u4 + 1,591u5 + 7,073u6 and expand u in a power series of λ as:

(36) u = λ 18 λ 3 7 λ 4 588 λ 5 561 λ 6 23 , 159 λ 7 35 , 994 λ 8 1 , 008 , 384 λ 9 .
Where λ = A2/(32 − 24A2).

Substituting the value of u and v from Equation (36) and Equation (34) into Equation (27), we obtain the frequency ω = ω3 as well as period T3 = 2π/ω3.

Now, we consider the case with 1 < A < (2)1/2. We introduce a new variable y as:

(37) y = x 1.

Substituting Equation (37) into Equation (18), we obtain:

(38) y ¨ + 2 y + 3 y 2 + y 3 = 0 ,
with initial conditions y ( 0 ) = α and y ˙ ( 0 ) = 0 where α = A − 1 and over dot denotes the derivative with respect to time t. This is a quadratic-cubic nonlinear oscillator. So it has two separate solutions, respectively, for two regions y >0 and y <0.

The variational of Equation (38) is given by:

(39) J ( x ) = 0 t [ y ˙ 2 + 2 y 2 + 2 y 3 + y 4 2 ] d t .

Its Hamiltonian can be written in the form:

(40) H = y ˙ 2 + 2 y 2 + 2 y 3 + y 4 2 = 2 α 2 + 2 α 3 + α 4 2 .

The residual of Equation (40) becomes:

(41) R ( t ) = y ˙ 2 + 2 y 2 + 2 y 3 + y 4 2 2 α 2 2 α 3 α 4 2 = 0.

Substituting Equation (16) into Equation (41), we obtain:

(42) R ( t ) = α 2 ( ( ω ( ( 1 u v ) sin ω t + 3 u sin3 ω t + 5 v sin5 ω t ) ) 2 + 2 ( ( 1 u v ) cos ω t + u cos3 ω t + v cos5 ω t ) 2 1 ) ) + 2 α 3 ( ( ( 1 u v ) cos ω t + u cos3 ω t + v cos5 ω t ) 3 1 ) + α 4 ( ( ( 1 u v ) cos ω t + u cos3 ω t + v cos5 ω t ) 4 1 ) / 2 = 0.

Substituting Equation (42) into Equation (15), we obtain, respectively, for n = 1, 2, 3:

(43) 2 8 u + 4 u 2 16 v + 8 u v + 8 v 2 8 α ( 15 , 015 + 60 , 060 u 48 , 048 u 2 + 27 , 456 u 3 + 132 , 132 v 123 , 552 u v + 64 , 064 u 2 v 121 , 264 v 2 + 108 , 992 u v 2 + 61 , 632 v 3 ) / ( 15 , 015 π ) + ( 3 / 4 3 u + 3 u 2 3 u 3 + 3 u 4 / 2 7 v + 9 u v 9 u 2 v + 4 u 3 v + 9 v 2 15 u v 2 + 9 u 2 v 2 9 v 3 + 9 u v 3 + 4 v 4 ) α 2 + ( 1 + 4 u + 22 u 2 + 8 v + 76 u v + 116 v 2 ) ω 2 = 0.
(44) 2 + 4 ( u u 2 + v u v v 2 ) + 4 α ( 60 , 060 + 45 , 045 π + 144 , 144 u 247 , 104 u 2 + 183 , 040 u 3 + 185 , 328 v 384 , 384 u v + 249 , 600 u 2 v 326 , 976 v 2 + 426 , 240 u v 2 + 189 , 696 v 3 ) / ( 45 , 045 π ) + α 2 ( 5 + 8 u 18 u 2 + 24 u 3 14 u 4 + 12 v 36 u v + 48 u 2 v 24 u 3 v 30 v 2 + 72 u v 2 54 u 2 v 2 + 36 v 3 44 u v 3 18 v 4 ) / 8 + ( 1 + 2 u 10 u 2 + 2 v 2 u v 26 v 2 ) ω 2 = 0 .
(45) 2 + 8 u 6 u 2 + 12 v 12 u v 10 v 2 + 8 α ( 1 , 633 , 632 + 765 , 765 π + 2 , 800 , 512 u 3 , 734 , 016 u 2 + 2 , 263 , 040 u 3 + 4 , 667 , 520 v 882 , 556 u v + 5 , 744 , 640 u 2 v 7 , 494 , 144 v 2 + 8 , 460 , 288 u v 2 + 4 , 503 , 552 v 3 ) / ( 765 , 765 π ) + α 2 ( 3 + 10 u 18 u 2 + 20 u 3 10 u 4 + 18 v 48 u v + 60 u 2 v 30 u 3 v 42 v 2 + 90 u v 2 60 u 2 v 2 + 50 v 3 58 u v 3 24 v 4 ) / 4 ( 1 + 4 u + 13 u 2 2 v + 26 u v 51 v 2 ) ω 2 = 0.

Eliminating ω2 from Equations (43) and (44), then from Equations (43) and (45), we obtain:

(46) m 0 + m 1 α + m 2 α 2 = 0 ,
(47) n 0 + n 1 α + n 2 α 2 = 0 ,
where m0, m1, m2, n0, n1, and n2 are functions of u and v and presented by:
(48) m 0 = u + 7 u 2 u 3 + v + 31 u v + 18 u 2 v + 31 v 2 + 56 u v 2 + ,
(49) m 1 = ( 150 , 150 + 45 , 045 π 276 , 276 u + 180 , 180 π u 883 , 740 u 2 + 990 , 990 π u 2 1 , 979 , 120 u 3 907 , 764 v + 360 , 360 π v 187 , 044 u v + 3 , 423 , 420 π u v + 675 , 168 u 2 v 5 , 840 , 484 v 2 + 5 , 225 , 220 π v 2 + 12 , 302 , 208 u v 2 3 , 365 , 616 v 3 + ) / ( 90 , 090 π ) ,
(50) m 2 = ( 1 + 16 u + 136 u 2 184 u 3 + 8 v + 676 u v 240 u 2 v + 674 v 2 + 360 u v 2 484 v 3 + ) / 64 ,
(51) n 0 = u 2 + 4 u 3 + v + 10 u v + 24 u 2 v + 16 v 2 + 56 u v 2 + 30 v 3 + ,
(52) n 1 = ( 2 , 399 , 397 + 765 , 765 π 9 , 860 , 136 u + 3 , 063 , 060 π u 48 , 228 , 609 u 2 + 16 , 846 , 830 π u 2 + 17 , 519 , 996 u 3 13 , 608 , 738 v + 6 , 126 , 120 π v 126 , 345 , 258 u v+ 58 , 198 , 140 π u v + 105 , 886 , 812 u 2 v 179 , 047 , 383 v 2 + 88 , 828 , 740 π v 2 + 262 , 621 , 644 u v 2 + 126 , 795 , 228 v 3 + ) / ( 1 , 531 , 530 π ) ,
(53) n 2 = ( 2 u + 13 u 2 + 48 u 3 + 20 v + 202 u v + 288 u 2 v + 389 v 2 + 738 u v 2 + 266 v 3 + ) / 64 ,

Eliminating α from Equations (46) and (47), we obtain:

(54) ( m 2 n 1 m 1 n 2 ) ( m 1 n 0 m 0 n 1 ) ( m 2 n 0 m 0 n 2 ) 2 = 0.

Substituting Equations (48)-(53) into Equation (54) and ignoring terms of more than the second order of u and v, we obtain:

(55) ( 247 , 408 157 , 920 π + 25 , 200 π 2 ) u ( 982 , 864 629 , 520 π + 100 , 800 π 2 ) u 2 ( 15 , 792 5 , 040 π ) v + ( 11 , 882 , 640 u 7 , 661 , 280 π u + 1 , 234 , 800 π 2 u ) v + ( 2 , 971 , 216 2 , 127 , 360 π + 376 , 425 π 2 ) v 2 = 0.

From Equation (55), we can expand v in a power series of u as:

(56) v = v 1 u + v 2 u 2 +

Substituting Equation (56) into Equation (55) and equating the equal powers of u on both sides, we obtain a set of algebraic equations which provide the values of v1, v2, …. Using these values in Equation (56), we obtain:

(57) v = ( 47 15 π ) u / 3 + ( 175 , 106 , 864 167 , 477 , 520 π + 53 , 306 , 775 π 2 5 , 646 , 375 π 3 ) u 2 / 3 , 024 .

Substituting Equation (57) into Equation (46), we obtain a polynomial equation of u and find the value of u from this equation. Then, substituting the value of u and v into Equation (43), we obtain the approximate frequency ω3(α) as well as period T3(α) = π/ω3(α) for 1 < A < 2 :

i .e .0 < α < 0.414 .

For the region y < 0, Equation (38) becomes:

(58) y ¨ + 2 y 3 y 2 + y 3 = 0 ,
with initial conditions y ( 0 ) = β and y ˙ ( 0 ) = 0 .

In a similar way, we can obtain:

(59) m ¯ 0 + m ¯ 1 β + m ¯ 2 β 2 = 0 ,
(60) n ¯ 0 + n ¯ 1 β + n ¯ 2 β 2 = 0 ,
where m ¯ 0 , m ¯ 1 , m ¯ 2 , n ¯ 0 , n ¯ 1 , are n ¯ 2 are functions of u and v and presented by:
(61) m ¯ 0 = u + 7 u 2 u 3 + v + 31 u v + 18 u 2 v + 31 v 2 + 56 u v 2 + ,
(62) m ¯ 1 = ( 150 , 150 + 45 , 045 π 276 , 276 u + 180 , 180 π u 883 , 740 u 2 + 990 , 990 π u 2 1 , 979 , 120 u 3 907 , 764 v + 360 , 360 π v 187 , 044 u v + 3 , 423 , 420 π u v + 675 , 168 u 2 v 5 , 840 , 484 v 2 + 5 , 225 , 220 π v 2 + 12 , 302 , 208 u v 2 3 , 365 , 616 v 3 + ) / ( 90 , 090 π ) ,
(63) m ¯ 2 = ( 1 + 16 u + 136 u 2 184 u 3 + 8 v + 676 u v 240 u 2 v + 674 v 2 + 360 u v 2 484 v 3 + ) / 64 ,
(64) n ¯ 0 = u 2 + 4 u 3 + v + 10 u v + 24 u 2 v + 16 v 2 + 56 u v 2 + 30 v 3 + ,
(65) n ¯ 1 = ( 2 , 399 , 397 + 765 , 765 π 9 , 860 , 136 u + 3 , 063 , 060 π u 48 , 228 , 609 u 2 + 16 , 846 , 830 π u 2 + 17 , 519 , 996 u 3 13 , 608 , 738 v + 6 , 126 , 120 π v 126 , 345 , 258 u v+ 58 , 198 , 140 π u v + 105 , 886 , 812 u 2 v 179 , 047 , 383 v 2 + 88 , 828 , 740 π v 2 + 262 , 621 , 644 u v 2 + 126 , 795 , 228 v 3 + ) / ( 1 , 531 , 530 π ) ,
(66) n ¯ 2 = ( 2 u + 13 u 2 + 48 u 3 + 20 v + 202 u v + 288 u 2 v + 389 v 2 + 738 u v 2 + 266 v 3 + ) / 64 ,
and:
(67) v = ( 47 15 π ) u / 3 + ( 175 , 106 , 864 167 , 477 , 520 π + 53 , 306 , 775 π 2 5 , 646 , 375 π 3 ) u 2 / 3 , 024 .

Substituting Equation (67) into Equation (59), we obtain a polynomial equation of u and find the value of u from this equation.

Here, β (see Wu et al., 2007, for details) can be provided as:

(68) β = 1 1 2 α α 2 .

Also, in this case, the corresponding equation of Equation (43) is:

(69) 2 8 u + 4 u 2 16 v + 8 u v + 8 v 2 + 8 β ( 15 , 015 + 60 , 060 u 48 , 048 u 2 + 27 , 456 u 3 + 132 , 132 v 123 , 552 u v + 64 , 064 u 2 v 121 , 264 v 2 + 108 , 992 u v 2 + 61 , 632 v 3 ) / ( 15 , 015 π ) + ( 3 / 4 3 u + 3 u 2 3 u 3 + 3 u 4 / 2 7 v + 9 u v 9 u 2 v + 4 u 3 v + 9 v 2 15 u v 2 + 9 u 2 v 2 9 v 3 + 9 u v 3 + 4 v 4 ) β 2 + ( 1 + 4 u + 22 u 2 + 8 v + 76 u v + 116 v 2 ) ω 2 = 0

Then, using the value of u and v into Equation (69), we obtain approximate frequency ω3(β) as well as period T3(β) = π/ω3(β).

Finally, we have determined the approximate period (by averaging above two periods) T3 for different values of A as well as α in the following way:

(70) T 3 = T 3 ( α ) + T 3 ( β ) 2 = π ω 3 ( α ) + π ω 3 ( β ) .

## 6. Results and discussions

A simple analytical technique based on the energy balance method (proposed by Molla et al., 2017) has been presented to determine the approximate solution of strong nonlinear oscillators with a double-well potential. The determination of the solution is quite easy. It has been mentioned that the procedures of Khan and Mirzabeigy (2014), Momeni et al. (2011), Wu et al. (2007) and harmonic balance method are laborious, especially for obtaining the higher approximations. We have determined the second and third approximate frequencies and periods, respectively, of Equations (18) and (38) and the results have been presented in Tables I-II with corresponding exact results and other existing results. Then, the solutions of Equation (18) have presented in Figures 1 and 2, respectively, for A = 1.42 and A = 1.43. The numerical solutions have been obtained by Runge-Kutta fourth-order method which are presented in all figures. Comparing all the results, we can say that the present results are better than those obtained by other existing results. Therefore, the present approximate solutions are in nice agreement with the numerical solution.

## 7. Conclusion

Based on the energy balance method (proposed by Molla et al., 2017), a new analytical technique has been presented to obtain analytical approximate solutions of the nonlinear conservative oscillator. The results obtained in the present method provide high accuracy in comparison with other existing methods. Therefore, the present technique is very effective and convenient for solving of the nonlinear conservative oscillator.

## Figures

#### Figure 1

Comparison of 3rd approximate periodic solution of Equation (18) obtained by present method (EBM) (denoting by circle line) with EBM (Khan and Mirzabeigy, 2014) (denoting by dashes line) and numerical results (denoting by solid line) for A=1.42

#### Figure 2

Comparison of 3rd approximate periodic solution of Equation (18) obtained by present method (EBM) (denoting by circle line) with EBM (Khan and Mirzabeigy, 2014) (denoting by dashes line) and numerical results (denoting by solid line) for A=1.43

## Table I

Comparison of the approximate frequencies of Equation (18) with exact and other existing frequencies (Momeni et al., 2011; Khan and Mirzabeigy, 2014) for A > 2

A ωExact ωM(EBM) Er(%) ωKM(EBM) Er(%) ω3(HBM) Er(%) ω2 Er(%) ω3 Er(%)
1.42 0.4165 0.7158 0.6019 0.5111 0.5497 0.4380
71.860 44.514 22.723 31.981 5.162
1.45 0.5603 0.7595 0.6522 0.5873 0.6124 0.5741
35.552 16.402 4.818 9.299 2.462
1.5 0.6812 0.8292 0.7301 0.6908 0.7039 0.6842
21.726 7.179 1.409 3.332 0.440
1.7 0.9890 1.0805 0.9981 0.9908 0.9931 0.9892
9.252 0.920 0.182 0.415 0.020
2 1.3409 1.4142 1.3386 1.3417 1.3421 1.3409
5.466 0.172 0.059 0.089 0.000
10 8.4102 8.6024 8.3687 8.4109 8.4102 8.4102
2.285 0.493 0.008 0.000 0.000
100 84.7151 86.5453 84.3381 84.7226 84.7149 84.7153
2.160 0.445 0.008 0.000 0.000

Notes: Where A is amplitude, ω2,ω3 second and third approximate frequency obtained by present method,ωExact, ωM(EBM),ωKM(EBM) are, respectively, exact frequency, Momeni et al. (2011) presented, Khan and Mirzabeigy (2014) presented second approximate frequencies, ω3(HBM) presented 3rd approximate frequencies obtained by harmonic balance method. Er(%) denotes the absolute percentage error

## Table II

Comparison of the approximate periods of Equation (18) with exact and other existing periods (Wu et al., 2007) for 1 < A < 2

A TExact T2(Wu) Er(%) T3(Wu) Er(%) T3(HBM) Er(%) T3 Er(%)
1.05 4.45169 4.45156 4.45169 4.45168 4.45169
0.0029 0.0000 0.0002 0.0000
1.10 4.48053 4.47993 4.48057 4.48053 4.48053
0.0134 0.0009 0.0000 0.0000
1.15 4.53484 4.53325 4.53498 4.53481 4.53483
0.0351 0.0031 0.0007 0.0002
1.20 4.62391 4.62035 4.62423 4.61939 4.62390
0.0770 0.0069 0.0978 0.0002
1.25 4.76522 4.75760 4.76589 4.76508 4.76522
0.1599 0.0141 0.0029 0.0000
1.30 4.99674 4.97967 4.99834 4.99635 4.99675
0.3416 0.0320 0.0078 0.0002
1.35 5.42749 5.38170 5.43194 5.42587 5.42755
0.8437 0.0820 0.0298 0.0011
1.40 6.75637 6.49434 6.77454 6.73249 6.75880
3.8783 0.2689 0.3534 0.0359
1.41 7.92344 7.26163 7.92748 7.81444 7.95334
8.3526 0.0510 1.3757 0.3773
1.412 8.55534 7.58661 8.51545 8.34831 8.65382
11.3231 0.4663 2.4199 1.1510

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#### Supplementary materials

MMMS_14_4.pdf (20.3 MB)

## Acknowledgements

The authors are grateful to the honorable reviewers for their helpful comments and suggestions in improving the manuscript.

## Corresponding author

Md Helal Uddin Molla can be contacted at: helal.mathru@yahoo.com