A short remark on Ren–Hu’s modification of He’s frequency–amplitude formulation and the temperature oscillation in a polar bear hair

Abstract

Recently Zhong-Fu Ren and Gui-Fang Hu suggested a modification of He’s frequency–amplitude formulation for nonlinear oscillators. This paper shows that this modification results in a lower accuracy than that by the original one. The temperature oscillation in a polar bear hair is used as an example to illustrate the effectiveness of He’s frequency–amplitude formulation.

Introduction

Recently He’s frequency–amplitude formulation^1–7 has been caught much attention due its extremely simple calculation with high accuracy for the whole solution domain for nonlinear oscillators, it is the simplest method among all analytical methods for nonlinear oscillators, for example, the homotopy perturbation method^8,9 and the variational iteration.^10–12 It is also very effective for oscillators with fractal derivatives^13–18 which arise in oscillations in discontinuous media, e.g. a nanofiber membrane.^19–21 Tian and Liu gave a complete review on various period/frequency estimation of a nonlinear oscillator, showing that He’s frequency–amplitude formulation is the simplest method. ²² Ren and Hu ²³ suggested a modification of He’s frequency–amplitude formulation; however, their modification cannot lead to a higher accuracy than that by the original one.

He’s frequency–amplitude formulation and Ren–Hu modification

Consider a nonlinear oscillator in the form

$$\ddot{u}+{\alpha }_{0}u+{\alpha }_1{u}^3+{\alpha }_2{u}^5=0,u(0)=A,\dot{u}\left(0\right)=0$$

(1)

He’s frequency–amplitude formulation reads

$${\omega }_{\mathit{He}}^2=\frac{{\omega }_1^2{\tilde{R}}_2-{\omega }_2^2{\tilde{R}}_1}{{\tilde{R}}_2-{\tilde{R}}_1}$$

(2)

where ${\omega }_1$ and ${\omega }_2$ are freely chosen frequencies, for example ${\omega }_1=1$ and ${\omega }_2=2$ for simple calculation, ${\tilde{R}}_1$ and ${\tilde{R}}_2$ are the residuals for two trial solutions with frequencies of ${\omega }_1$ and ${\omega }_2$ , and there some different approaches to calculation of the residuals.^1–7 The frequency is estimated as^1–7

$$\omega \mathit{He}=\sqrt{{\alpha }_0+\frac{3}{4}{\alpha }_1{A}^2+\frac{5}{8}{\alpha }_2{A}^4}$$

(3)

While the Ren-Hu’s modification results in ²³

$${\omega }_{\mathit{Ren}-\mathit{Hu}}=\sqrt{{\alpha }_0+\frac{2}{3}{\alpha }_1{A}^2+\frac{8}{15}{\alpha }_2{A}^4}$$

(4)

Ren and Hu showed that equation (4) gives a good accuracy when ${\alpha }_1$ and ${\alpha }_2$ tends to infinity. We compare equations (3) and (4) for two special cases

Case 1: ${\alpha }_0=4$ , ${\alpha }_1=1$ , ${\alpha }_2=0$ , and $A=\pi /3$

The numerical method by the variational iteration method gives the following result ²⁴

$$\omega \mathit{VIM}=2.199549$$

(5)

While He’s formulation and Ren–Hu’s modification give, respectively

$${\omega }_{\mathit{He}}=\sqrt{4+\frac{3}{4}{\left(\frac{\pi }{3}\right)}^2}=2.196011$$

(6)

$${\omega }_{\mathit{Ren}-\mathit{Hu}}=\sqrt{4+\frac{2}{3}{\left(\frac{\pi }{3}\right)}^2}=2.175105$$

(7)

Case 2 ${\alpha }_0=2$ , ${\alpha }_1=5/6$ , ${\alpha }_2=0$ and $A=\pi /6$

The numerical method by the variational iteration method gives the following result ²⁴

$$\omega \mathit{VIM}=1.4928$$

(8)

While He’s formulation and Ren–Hu’s modification give, respectively

$$\omega \mathit{He}=\sqrt{2+\frac{3}{4}\times \frac{5}{6}{\left(\frac{\pi }{6}\right)}^2}=1.473549$$

(9)

$${\omega }_{\mathit{Ren}-\mathit{Hu}}=\sqrt{2+\frac{2}{3}\times \frac{5}{6}{\left(\frac{\pi }{6}\right)}^2}=1.4670748$$

(10)

Both cases show that He’s formula give a higher accuracy than that by Ren–Hu’s modification.

Case 3 ${\alpha }_0=0.01$ , ${\alpha }_1=1$ , ${\alpha }_2=0.01$ and $A=1$

The exact frequency is

$$\omega \mathit{exact}=0.85668$$

(11)

He’s formulation and Ren-Hu’s modification predict, respectively

$${\omega }_{\mathit{He}}=0.87536$$

(12)

$${\omega }_{\mathit{Ren}-\mathit{Hu}}=0.82583$$

(13)

The relative errors are, respectively, 2.18% and 3.60%, showing He’s formulation gives a higher accuracy.

Temperature oscillation in a hollow hair of polar bear

Temperature along a polar bear changes periodically. ¹⁵ The heat conduction equation can be written in the form

$$\frac{d}{\mathit{dx}}\left(k\frac{\mathit{dT}}{\mathit{dx}}\right)=Q,T\left(0\right)=T_0,\frac{\mathit{dT}}{\mathit{dx}}\left(0\right)=0$$

(14)

where k is the heat conduction coefficient of the polar bear hair, Q is the heat source, and it can be expressed as

$$Q=-\frac{Q_0}{1+\mathit{cx}}$$

(15)

where

$$c=\frac{Q_0-Q_{\infty }}{L{Q}_{\infty }}$$

(16)

where L is the length of the polar hair, $Q_0$ is the heat source due to body temperature, $Q_{\infty }$ is the heat source from the environment.

Equation (16) becomes

$$\frac{d^{2}T}{d{x}^2}+\frac{b}{1+\mathit{cx}}=0$$

(17)

where $b=Q_0/k$ .

By He’s frequency formulation, we have

$$\omega =\sqrt{\frac{b}{\frac{\sqrt{3}}{2}{T}_0\left(1+\frac{\sqrt{3}}{2}c{T}_0\right)}}=\sqrt{\frac{1}{\frac{\sqrt{3}}{2}k\left(1+\frac{\sqrt{3}}{2}\frac{Q_0-Q_{\infty }}{L{Q}_{\infty }}{T}_0\right)}}$$

(18)

Conclusion

This paper shows that He’s frequency–amplitude formulation gives much better estimation of the frequency–amplitude relationship for a nonlinear oscillator than Ren–Hu’s modification.