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Abstract

In this paper, a semi-analytical method based on variational iteration formulae is proposed. Second-order nonlinear oscillator equations are numerically investigated by this method. A matrix Lagrange multiplier is given to derive analytical solutions and compared with the old one by use of the same order approximation. The initial value is updated successively and numerical solutions are obtained.

Keywords

Initial value problems nonlinear oscillator variational iteration method semi-analytical solutions

Introduction

Variational iteration method^1,2 is one of the famous analytical methods for nonlinear equations. It has many applications in nonlinear problems; for example, fractional differential and difference equations, delay differential equations,³ heat equations⁴ and some other modified versions with applications.^5–14

However, approximate analytical solutions are not satisfied for the need of high accuracies. If differential equations are in chaotic status or the trivial solution is non-stable, first order approximation are not enough at all since the exact solution holds sensitivity and truncation of approximations cause great numerical errors. Hence, new high-accuracy methods are needed and this is the motivation of this paper. On the other hand, numerical methods can be adopted to solve nonlinear oscillators.

We propose a new semi-analytical method based on the variational iteration method which combines the two features of numerical and analytical solutions.

Matrix Lagrange multiplier

New Lagrange multiplier

Let us consider the strong nonlinear oscillator $\frac{d^{2} x}{d t^{2}} + x + ε x^{2} = 0, x (t_{0}) = 0.1, x' (t_{0}) = 0, ε = 1$ (1)

We rewrite it as ${\begin{array}{l} \frac{d x}{d t} = y, x (t_{0}) = 0.1, \\ \frac{d y}{d t} = - x - ε x^{2}, y (t_{0}) = 0 \end{array}$ (2)and ${(\begin{array}{l} x \\ y \end{array})}^{'} = (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) - (\begin{matrix} 0 \\ x^{3} \end{matrix}), (\begin{matrix} x (t_{0}) \\ y (t_{0}) \end{matrix}) = (\begin{array}{l} 0.1 \\ 0 \end{array})$ (3)

According to the method,⁹ we give a Lagrange multiplier and variational iteration formula in the form of matrix $λ (t, τ) = e^{A (t - τ)}, A = (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix})$ (4)such that the variational iteration reads $(\begin{array}{l} x_{n + 1} \\ y_{n + 1} \end{array}) = (\begin{array}{l} x_{0} \\ y_{0} \end{array}) + \int_{t_{0}}^{t} e^{A (t - τ)} (\begin{array}{l} 0 \\ - x_{n}^{3} \end{array}) d τ, 0 \leq n$ (5)

The initial iteration is given by $(\begin{array}{l} x_{0} \\ y_{0} \end{array}) = e^{A (t - t_{0})} (\begin{array}{l} x (t_{0}) \\ y (t_{0}) \end{array})$ (6)

We can calculate the exponential matrix $λ (t, τ) = e^{A (t - τ)} = (\begin{matrix} \cos (t - τ) & \sin (t - τ) \\ - \sin (t - τ) & \cos (t - τ) \end{matrix})$ (7)and $e^{A (t - t_{0})} = (\begin{array}{l} \cos (t - t_{0}) \sin (t - t_{0}) \\ - \sin (t - t_{0}) \cos (t - t_{0}) \end{array})$ (8)

To the best of our knowledge, the Lagrange multiplier (7) is new and was not found in any other existing works. Hence, we obtain some new variational iteration solutions $(\begin{array}{l} x_{n + 1} \\ y_{n + 1} \end{array}) = (\begin{array}{l} x_{0} \\ y_{0} \end{array}) + \int_{t_{0}}^{t} (\begin{matrix} \cos (t - τ) & \sin (t - τ) \\ - \sin (t - τ) & \cos (t - τ) \end{matrix}) (\begin{array}{l} 0 \\ - x_{n}^{3} \end{array}) d τ, 0 \leq n$ and $\begin{matrix} (\begin{array}{l} x_{0} \\ y_{0} \end{array}) = e^{A (t - t_{0})} (\begin{array}{l} x (t_{0}) \\ y (t_{0}) \end{array}) = (\begin{matrix} \cos (t - t_{0}) & \sin (t - t_{0}) \\ - \sin (t - t_{0}) & \cos (t - t_{0}) \end{matrix}) (\begin{array}{l} x (t_{0}) \\ y (t_{0}) \end{array}) = (\begin{array}{l} \cos (t - t_{0}) x (t_{0}) + \sin (t - t_{0}) y (t_{0}) \\ - \sin (t - t_{0}) x (t_{0}) + \cos (t - t_{0}) y (t_{0}) \end{array}) \end{matrix}$ which finally yields $(\begin{array}{l} x_{n + 1} \\ y_{n + 1} \end{array}) = (\begin{array}{l} \cos (t - t_{0}) x (t_{0}) + \sin (t - t_{0}) y (t_{0}) - \int_{t_{0}}^{t} \sin (t - τ) x_{n}^{3} d τ \\ - \sin (t - t_{0}) x (t_{0}) + \cos (t - t_{0}) y (t_{0}) - \int_{t_{0}}^{t} \cos (t - τ) x_{n}^{3} d τ \end{array}), 0 \leq n$ (9)

Comparison with old Lagrange multiplier and error analysis

If we consider the system (2) separately and use the Lagrange multiplier for single equation, we directly use the one as ${\begin{array}{l} x_{n + 1} = x_{0} + \int_{t_{0}}^{t} λ_{1} (t, s) y_{n} d s, x_{0} = x (t_{0}) = 0.1, \\ y_{n + 1} = y_{0} + \int_{t_{0}}^{t} λ_{2} (t, s) (x_{n} + ε x_{n}^{3}) d s, y_{0} = y (t_{0}) = 0 \end{array}$ (10)where $λ_{1} (t, s) = λ_{2} (t, s) = 1$ .

Define the residual function $e_{n} = \ln | \frac{d^{2} x_{n}}{d t^{2}} + x_{n} + ε x_{n}^{3} |$ (11)

We assume $n = 3.$ We can see that more terms are used to identify the Lagrange multipliers, and higher accuracy will be obtained concerning the iteration formulae. Even for the first order approximation, we can obtain relative high-accuracy solution from Figure 1. We also notice that our variational iteration formula is valid in Figure 2 which gives the values of the residual function (11).

Figure 1.

Comparasion between variational iteration formulae (9) (red) and (10) (black).

Figure 2.

Variational iteration formula (9): first order approximation.

Semi-analytical solutions

We give the following scheme for numerical solutions of equation (1)

Step I: To obtain approximate solution $x_{n + 1}$ and $y_{n + 1}$ ${\begin{matrix} x_{n + 1} = ϕ_{1} (t, t_{0}, C) \\ y_{n + 1} = ϕ_{2} (t, t_{0}, C) \end{matrix}$ (12)

Step II: Let the spatial step size and the node number to be $h$ and $N$ . By substituting $t = t_{1}$ into equation (9), one can obtain the values $x_{1}^{*}$ and $y_{1}^{*}$ ${\begin{matrix} x_{1}^{*} = ϕ_{1} (t_{1}, t_{0}, C) \\ y_{1}^{*} = ϕ_{2} (t_{1}, t_{0}, C) \end{matrix}$ (13)

Step III: Replace $t_{0}$ and $C$ with $t_{1}$ , $x_{1}^{*}$ and $y_{1}^{*}$ in equation (13), respectively, which lead to ${\begin{matrix} x_{n + 1} = ϕ_{1} (t, t_{1}, x_{1}^{*}) \\ y_{n + 1} = ϕ_{2} (t, t_{1}, x_{1}^{*}) \end{matrix}$ (14)

Similarly, one can have the value $x_{2}^{*}$ and $y_{2}^{*}$ ${\begin{matrix} x_{2}^{*} = ϕ_{1} (t_{2}, t_{1}, x_{1}^{*}) \\ y_{1}^{*} = ϕ_{2} (t_{2}, t_{1}, x_{1}^{*}) \end{matrix}$ (15)

As a result, we can obtain all the information of the numerical solutions ${\begin{matrix} x_{0}^{*}, x_{1}^{*}, …, x_{N}^{*} \\ y_{0}^{*}, y_{1}^{*}, …, y_{N}^{*} \end{matrix}$ (16)where $x_{0}^{*}$ is the initial value $x_{0}^{*} (0) = C$ (17)

We use the above steps to derive the numerical solutions of equation (1). Their behaviors are illustrated in Figures 3 and 4 where we adopt third approximation, $N = 1000$ and $h = \frac{2 π}{N}$ .

Figure 3.

Semi-analytical solution x.

Figure 4.

Semi-analytical solution y.

Conclusions

This paper suggests a semi-analytical method for nonlinear oscillators. The method is proposed based on the famous variational iteration method. We newly use a Matrix Lagrange multiplier which yields analytical solutions of higher accuracy. The Matrix Lagrange multiplier is given in an exponential function. In view of this point, we use the analytical solution to derive numerical solutions. The result shows the method’s efficiency. Our method can be further used in fractional differential equation or other models. We will consider these applications in our future study.

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) disclosed receipt of the following financial support for the research,authorship,and/or publication of this article: This study is financially supported by Key program of Sichuan Province Educational Department (16ZA0311).

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