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Abstract

The variational iteration method is a famous mathematical tool to nonlinear oscillator, this paper shows that the method can be further improved by using matrix Lagrange multipliers. Four examples are given to show the effectiveness of the modified variational iteration method.

Keywords

Variational iteration method nonlinear oscillator Lagrange multiplier matrix

Introduction

Ji-Huan He¹ gave a very lucid as well as elementary introduction to the variational iteration method and its applications. The method is based on a generalized Lagrange multiplier which is also called He multiplier in some literatures.² Identification of He multiplier is of great importance in practical applications, an accurate and suitable identification of He multiplier always leads to an ideal result. In order to make the identification as simple as possible so that all non-mathematicians can follow the method for various practical applications, we find in this paper that the matrix Lagrange multipliers method^3,4 is a promising technology, which is extremely suitable for nonlinear vibration problems. Integration of the variational iteration method and the matrix Lagrange multipliers method provides a universal approach to nonlinear oscillators.

In this paper, we will apply the modified variational iteration method to four problems: a singular oscillator,^5,6 the Duffing-harmonic oscillator,^7–9 the Lienard problem,^10–12 and the nonlinear oscillator with coordinate-dependent mass.^13,14

A singular oscillator

Consider an oscillator

\ddot{x} + \frac{d E}{d x} = 0, x (t_{0}) = α, \dot{x} (t_{0}) = 0

(1)

where E is the potential. For the well-known Duffing oscillator, the potential can be written as

E (x) = \frac{1}{2} x^{2} + \frac{1}{4} b x^{4}

(2)

In this paper, we first study the potential, which can be written in the form

E (x) = ln x

(3)

This potential arises in nanotechnology, including the bubble vibration and nano-fiber membrane vibration.^15–21

The oscillator under the potential of equation (3) can be written as

\ddot{x} + x^{- 1} = 0, x (t_{0}) = α, \dot{x} (t_{0}) = 0

(4)

This is a singular oscillator. In order to use the matrix Lagrange multipliers method, we re-write equation (4) in the form

{\begin{array}{l} \frac{d x}{d t} = y, x (t_{0}) = α, \\ \frac{d y}{d t} = - x^{- 1}, y (t_{0}) = 0 \end{array}

(5)

{(\begin{matrix} x \\ y \end{matrix})}^{′} = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) - (\begin{matrix} 0 \\ x^{- 1} \end{matrix}), (\begin{matrix} x (t_{0}) \\ y (t_{0}) \end{matrix}) = (\begin{matrix} α \\ 0 \end{matrix})

(6)

According to the variational iteration method, there is the Lagrange multiplier in the form of matrix

λ (t, τ) = e^{A (t - τ)}, A = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})

(7)

such that the variation iteration reads

(\begin{matrix} x_{n + 1} \\ y_{n + 1} \end{matrix}) = (\begin{matrix} x_{0} \\ y_{0} \end{matrix}) + \int_{t_{0}}^{t} e^{A (t - τ)} (\begin{matrix} 0 \\ - x_{n}^{- 1} \end{matrix}) d τ, n \geq 0

(8)

The initial iteration is given by

(\begin{matrix} x_{0} \\ y_{0} \end{matrix}) = e^{A (t - t_{0})} (\begin{matrix} x (t_{0}) \\ y (t_{0}) \end{matrix})

(9)

Calculating the exponential matrix, there comes

λ (t, τ) = e^{A (t - τ)} = (\begin{matrix} 1 & t - τ \\ 0 & 1 \end{matrix})

(10)

and

e^{A (t - t_{0})} = (\begin{matrix} 1 & t - t_{0} \\ 0 & 1 \end{matrix})

(11)

Therefore, there are the variational iteration solutions

(\begin{matrix} x_{n + 1} \\ y_{n + 1} \end{matrix}) = (\begin{matrix} x_{0} \\ y_{0} \end{matrix}) + \int_{t_{0}}^{t} (\begin{matrix} 1 & t - τ \\ 0 & 1 \end{matrix}) (\begin{matrix} 0 \\ - x_{n}^{- 1} \end{matrix}) d τ, n \geq 0

and

(\begin{matrix} x_{0} \\ y_{0} \end{matrix}) = (\begin{matrix} 1 & t - t_{0} \\ 0 & 1 \end{matrix}) (\begin{matrix} x (t_{0}) \\ y (t_{0}) \end{matrix}) = (\begin{matrix} x (t_{0}) + (t - t_{0}) y (t_{0}) \\ y (t_{0}) \end{matrix}) = (\begin{matrix} α \\ 0 \end{matrix})

That is

(\begin{matrix} x_{n + 1} \\ y_{n + 1} \end{matrix}) = (\begin{matrix} x (t_{0}) + (t - t_{0}) y (t_{0}) - \int_{t_{0}}^{t} (t - τ) x_{n}^{- 1} d τ \\ y (t_{0}) - \int_{t_{0}}^{t} x_{n}^{- 1} d τ \end{matrix}), n \geq 0

(\begin{matrix} x_{n + 1} \\ y_{n + 1} \end{matrix}) = (\begin{matrix} α - \int_{t_{0}}^{t} (t - τ) x_{n}^{- 1} d τ \\ - \int_{t_{0}}^{t} x_{n}^{- 1} d τ \end{matrix}), n \geq 0

(12)

For given $α$ , results comparison with those from classical variational iteration method and/or results in references shows computation results from equation (12) have no obvious improvement on solutions.

Duffing-harmonic oscillator

We consider the Duffing-harmonic oscillator

\ddot{x} + \frac{x^{3}}{1 + x^{2}} = 0, x (t_{0}) = α, \dot{x} (t_{0}) = 0

(13)

Rewrite this as

{\begin{array}{l} \frac{d x}{d t} = y, x (t_{0}) = α, \\ \frac{d y}{d t} = - x + \frac{x}{1 + x^{2}}, y (t_{0}) = 0 \end{array}

(14)

{(\begin{matrix} x \\ y \end{matrix})}^{′} = (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) + (\begin{matrix} 0 \\ \frac{x}{1 + x^{2}} \end{matrix}), (\begin{matrix} x (t_{0}) \\ y (t_{0}) \end{matrix}) = (\begin{matrix} α \\ 0 \end{matrix})

(15)

According to the variational iteration method, there is the Lagrange multiplier in the form of matrix

λ (t, τ) = e^{A (t - τ)}, A = (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix})

(16)

such that the variation iteration reads

(\begin{matrix} x_{n + 1} \\ y_{n + 1} \end{matrix}) = (\begin{matrix} x_{0} \\ y_{0} \end{matrix}) + \int_{t_{0}}^{t} e^{A (t - τ)} (\begin{matrix} 0 \\ \frac{x_{n}}{1 + x_{n}^{2}} \end{matrix}) d τ, n \geq 0

(17)

The initial iteration is given by

(\begin{matrix} x_{0} \\ y_{0} \end{matrix}) = e^{A (t - t_{0})} (\begin{matrix} x (t_{0}) \\ y (t_{0}) \end{matrix})

(18)

Calculating the exponential matrix, there comes

λ (t, τ) = e^{A (t - τ)} = (\begin{matrix} \cos (t - τ) & \sin (t - τ) \\ - \sin (t - τ) & \cos (t - τ) \end{matrix})

(19)

and

e^{A (t - t_{0})} = (\begin{matrix} \cos (t - t_{0}) & \sin (t - t_{0}) \\ - \sin (t - t_{0}) & \cos (t - t_{0}) \end{matrix})

(20)

Therefore, there are the variational iteration solutions

(\begin{matrix} x_{n + 1} \\ y_{n + 1} \end{matrix}) = (\begin{matrix} x_{0} \\ y_{0} \end{matrix}) + \int_{t_{0}}^{t} (\begin{matrix} \cos (t - τ) & \sin (t - τ) \\ - \sin (t - τ) & \cos (t - τ) \end{matrix}) (\begin{matrix} 0 \\ \frac{x_{n}}{1 + x_{n}^{2}} \end{matrix}) d τ, n \geq 0

and

\begin{matrix} (\begin{matrix} x_{0} \\ y_{0} \end{matrix}) = (\begin{array}{l} \cos (t - t_{0}) & \sin (t - t_{0}) \\ - \sin (t - t_{0}) & \cos (t - t_{0}) \end{array}) (\begin{matrix} x (t_{0}) \\ y (t_{0}) \end{matrix}) \\ = (\begin{array}{l} \cos (t - t_{0}) x (t_{0}) + \sin (t - t_{0}) y (t_{0}) \\ - \sin (t - t_{0}) x (t_{0}) + \cos (t - t_{0}) y (t_{0}) \end{array}) = (\begin{array}{l} α \cos (t - t_{0}) \\ - α \sin (t - t_{0}) \end{array}) \end{matrix}

That is

\begin{array}{l} (\begin{matrix} x_{n + 1} \\ y_{n + 1} \end{matrix}) = (\begin{matrix} \cos (t - t_{0}) x (t_{0}) + \sin (t - t_{0}) y (t_{0}) + \int_{t_{0}}^{t} \sin (t - τ) \frac{x_{n}}{1 + x_{n}^{2}} d τ \\ - \sin (t - t_{0}) x (t_{0}) + \cos (t - t_{0}) y (t_{0}) + \int_{t_{0}}^{t} \cos (t - τ) \frac{x_{n}}{1 + x_{n}^{2}} d τ \end{matrix}), n \geq 0 \\ = (\begin{matrix} α \cos (t - t_{0}) + \int_{t_{0}}^{t} \sin (t - τ) \frac{x_{n}}{1 + x_{n}^{2}} d τ \\ - α \sin (t - t_{0}) + \int_{t_{0}}^{t} \cos (t - τ) \frac{x_{n}}{1 + x_{n}^{2}} d τ \end{matrix}) \end{array}

(21)

For given initial values, results comparison with those from classical variational iteration method shows computation results from equation (21) have improvement on solutions. This agrees with the corresponding conclusion in Hua and Yong-Yan.⁴

The Lienard equation

Consider the conservative Lienard-type problem

\ddot{x} + (1 + {\dot{x}}^{2}) x = 0, x (t_{0}) = α, \dot{x} (t_{0}) = 0

(22)

Rewrite this as

{\begin{array}{l} \frac{d x}{d t} = y, x (t_{0}) = α, \\ \frac{d y}{d t} = - x - x y^{2}, y (t_{0}) = 0 \end{array}

(23)

{(\begin{matrix} x \\ y \end{matrix})}^{′} = (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) + (\begin{matrix} 0 \\ - x y^{2} \end{matrix}), (\begin{matrix} x (t_{0}) \\ y (t_{0}) \end{matrix}) = (\begin{matrix} α \\ 0 \end{matrix})

(24)

According to the variational iteration method, there is the Lagrange multiplier in the form of matrix

λ (t, τ) = e^{A (t - τ)}, A = (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix})

(25)

such that the variation iteration reads

(\begin{matrix} x_{n + 1} \\ y_{n + 1} \end{matrix}) = (\begin{matrix} x_{0} \\ y_{0} \end{matrix}) + \int_{t_{0}}^{t} e^{A (t - τ)} (\begin{matrix} 0 \\ - x_{n} y_{n}^{2} \end{matrix}) d τ, n \geq 0

(26)

The initial iteration is

(\begin{matrix} x_{0} \\ y_{0} \end{matrix}) = e^{A (t - t_{0})} (\begin{matrix} x (t_{0}) \\ y (t_{0}) \end{matrix})

(27)

Calculating the exponential matrix, there comes

λ (t, τ) = e^{A (t - τ)} = (\begin{matrix} \cos (t - τ) & \sin (t - τ) \\ - \sin (t - τ) & \cos (t - τ) \end{matrix})

(28)

and

e^{A (t - t_{0})} = (\begin{matrix} \cos (t - t_{0}) & \sin (t - t_{0}) \\ - \sin (t - t_{0}) & \cos (t - t_{0}) \end{matrix})

(29)

Therefore, there are the variational iteration solutions

(\begin{matrix} x_{n + 1} \\ y_{n + 1} \end{matrix}) = (\begin{matrix} x_{0} \\ y_{0} \end{matrix}) + \int_{t_{0}}^{t} (\begin{matrix} \cos (t - τ) & \sin (t - τ) \\ - \sin (t - τ) & \cos (t - τ) \end{matrix}) (\begin{matrix} 0 \\ - x_{n} y_{n}^{2} \end{matrix}) d τ, n \geq 0

and

\begin{matrix} (\begin{matrix} x_{0} \\ y_{0} \end{matrix}) = (\begin{array}{l} \cos (t - t_{0}) & \sin (t - t_{0}) \\ - \sin (t - t_{0}) & \cos (t - t_{0}) \end{array}) (\begin{matrix} x (t_{0}) \\ y (t_{0}) \end{matrix}) \\ = (\begin{array}{l} \cos (t - t_{0}) x (t_{0}) + \sin (t - t_{0}) y (t_{0}) \\ - \sin (t - t_{0}) x (t_{0}) + \cos (t - t_{0}) y (t_{0}) \end{array}) = (\begin{array}{l} α \cos (t - t_{0}) \\ - α \sin (t - t_{0}) \end{array}) \end{matrix}

That is

\begin{matrix} (\begin{matrix} x_{n + 1} \\ y_{n + 1} \end{matrix}) = (\begin{array}{l} α \cos (t - t_{0}) - \int_{t_{0}}^{t} x_{n} y_{n}^{2} \sin (t - τ) d τ \\ - α \sin (t - t_{0}) - \int_{t_{0}}^{t} x_{n} y_{n}^{2} \cos (t - τ) d τ \end{array}) \\ = (\begin{array}{l} α \cos (t - t_{0}) - \int_{t_{0}}^{t} x_{n} y_{n}^{2} \sin (t - τ) d τ \\ - α \sin (t - t_{0}) - \int_{t_{0}}^{t} x_{n} y_{n}^{2} \cos (t - τ) d τ \end{array}), n \geq 0 \end{matrix}

(30)

For given initial values, results comparison with those from classical variational iteration method shows computation results from equation (30) have certain improvement on solutions. This agrees with the conclusion in Hua and Yong-Yan.⁴

The oscillator with coordinate-dependent mass

Consider the nonlinear oscillator with coordinate-dependent mass

(1 + β x^{2}) \ddot{x} + β x {\dot{x}}^{2} - x (1 - x^{2}) = 0, x (t_{0}) = α, \dot{x} (t_{0}) = 0, β > 0

(31)

Rewrite this as

{\begin{array}{l} \frac{d x}{d t} = y, x (t_{0}) = α, \\ \frac{d y}{d t} = - \frac{x}{β} - \frac{β x}{1 + β x^{2}} y^{2} + \frac{(β + 1) x}{β (1 + β x^{2})}, y (t_{0}) = 0 \end{array}

(32)

{(\begin{matrix} x \\ y \end{matrix})}^{′} = (\begin{matrix} 0 & 1 \\ - \frac{1}{β} & 0 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) + (\begin{matrix} 0 \\ \frac{(β + 1) x}{β (1 + β x^{2})} - \frac{β x}{1 + β x^{2}} y^{2} \end{matrix}), (\begin{matrix} x (t_{0}) \\ y (t_{0}) \end{matrix}) = (\begin{matrix} α \\ 0 \end{matrix})

(33)

According to the variational iteration method, there is the Lagrange multiplier in the form of matrix

λ (t, τ) = e^{A (t - τ)}, A = (\begin{matrix} 0 & 1 \\ - \frac{1}{β} & 0 \end{matrix})

(34)

such that the variation iteration reads

(\begin{matrix} x_{n + 1} \\ y_{n + 1} \end{matrix}) = (\begin{matrix} x_{0} \\ y_{0} \end{matrix}) + \int_{t_{0}}^{t} e^{A (t - τ)} (\begin{matrix} 0 \\ \frac{(β + 1) x_{n}}{β (1 + β x_{n}^{2})} - \frac{β x_{n}}{1 + β x_{n}^{2}} y_{n}^{2} \end{matrix}) d τ, n \geq 0

(35)

The initial iteration is given by

(\begin{matrix} x_{0} \\ y_{0} \end{matrix}) = e^{A (t - t_{0})} (\begin{matrix} x (t_{0}) \\ y (t_{0}) \end{matrix})

(36)

Calculating the exponential matrix, there comes

λ (t, τ) = e^{A (t - τ)} = (\begin{array}{l} \cos \frac{t - τ}{\sqrt{β}} & \sin \frac{t - τ}{\sqrt{β}} \\ - \sin \frac{t - τ}{\sqrt{β}} & \cos \frac{t - τ}{\sqrt{β}} \end{array})

(37)

and

e^{A (t - t_{0})} = (\begin{array}{l} \cos \frac{t - t_{0}}{\sqrt{β}} & \sin \frac{t - t_{0}}{\sqrt{β}} \\ - \sin \frac{t - t_{0}}{\sqrt{β}} & \cos \frac{t - t_{0}}{\sqrt{β}} \end{array})

(38)

Therefore, there are the variational iteration solutions

(\begin{matrix} x_{n + 1} \\ y_{n + 1} \end{matrix}) = (\begin{matrix} x_{0} \\ y_{0} \end{matrix}) + \int_{t_{0}}^{t} (\begin{matrix} \cos \frac{t - τ}{\sqrt{β}} & \sin \frac{t - τ}{\sqrt{β}} \\ - \sin \frac{t - τ}{\sqrt{β}} & \cos \frac{t - τ}{\sqrt{β}} \end{matrix}) (\begin{matrix} 0 \\ \frac{(β + 1) x_{n}}{β (1 + β x_{n}^{2})} - \frac{β x_{n}}{1 + β x_{n}^{2}} y_{n}^{2} \end{matrix}) d τ

and

(\begin{matrix} x_{0} \\ y_{0} \end{matrix}) = (\begin{matrix} \cos \frac{t - t_{0}}{\sqrt{β}} & \sin \frac{t - t_{0}}{\sqrt{β}} \\ - \sin \frac{t - t_{0}}{\sqrt{β}} & \cos \frac{t - t_{0}}{\sqrt{β}} \end{matrix}) (\begin{matrix} x (t_{0}) \\ y (t_{0}) \end{matrix}) = (\begin{matrix} \cos \frac{t - t_{0}}{\sqrt{β}} x (t_{0}) + \sin \frac{t - t_{0}}{\sqrt{β}} y (t_{0}) \\ - \sin \frac{t - t_{0}}{\sqrt{β}} x (t_{0}) + \cos \frac{t - t_{0}}{\sqrt{β}} y (t_{0}) \end{matrix}) = (\begin{matrix} α \cos \frac{t - t_{0}}{\sqrt{β}} \\ - α \sin \frac{t - t_{0}}{\sqrt{β}} \end{matrix})

That is

(\begin{matrix} x_{n + 1} \\ y_{n + 1} \end{matrix}) = (\begin{array}{l} α \cos \frac{t - t_{0}}{\sqrt{β}} + \int_{t_{0}}^{t} (\frac{(β + 1) x_{n}}{β (1 + β x_{n}^{2})} - \frac{β x_{n}}{1 + β x_{n}^{2}} y_{n}^{2}) \sin \frac{t - τ}{\sqrt{β}} d τ \\ - α \sin \frac{t - t_{0}}{\sqrt{β}} + \int_{t_{0}}^{t} (\frac{(β + 1) x_{n}}{β (1 + β x_{n}^{2})} - \frac{β x_{n}}{1 + β x_{n}^{2}} y_{n}^{2}) \cos \frac{t - τ}{\sqrt{β}} d τ \end{array}), n \geq 0

(39)

Given initial values, results comparison with those from classical variational iteration method shows computation results from equation (39) have improvement on solutions. This agrees with the conclusion in Hua and Yong-Yan.⁴

Remark

Variational iteration method is a powerful approach. We discover that the matrix Lagrange multipliers method is a promising one, especially suitable for nonlinear vibration problems. The variational iteration method combined with the matrix Lagrange multipliers method gives a universal approach to nonlinear oscillators, though it has no obvious merits for the singular oscillator here. Maybe, singularity is a non-ignorance factor. However, for some famous oscillators, such as Duffing-harmonic oscillator, it shows the advantage.

Using analytical idea to derive numerical solutions is a good way, we are confident that the modified variational iteration method has broader application range for other models.

Footnotes

Declaration of conflicting interests

The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.

Funding

The author(s) received no financial support for the research, authorship, and/or publication of this article.

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Variational iteration method with matrix Lagrange multiplier for nonlinear oscillators

Abstract

Keywords

Introduction

A singular oscillator

Duffing-harmonic oscillator

The Lienard equation

The oscillator with coordinate-dependent mass

Remark

Footnotes

Declaration of conflicting interests

Funding

References