# Comment on “Nonlinear Response of Strong Nonlinear System Arisen in Polymer Cushion”

Authors: Jun Zhou
Journal: Abstract and Applied Analysis
 (2013)

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## 1. Introduction

In [1], the following nonlinear oscillator was studied:
(1)X¨+tanhX+λ2tanλ1X+β5tan3λ1X=0,X(0)=0,X˙(0)=V=T0L2gh=2β1ghβ3.
Similar nonlinear oscillators arising in packing systems and textile engineering are available in [2–7].In [1], (1) is approximated by Taylor’s series for the nonlinear terms:
(2)X¨+ω012X+(-13+λ3λ13+13λ2λ13)X3+(215+215λ2λ15+λ3λ15)X5+(1115λ3λ17)X7.
By using Taylor’s expansion function of MATLAB’s symbolic function, the nonlinear terms tanh(X), tan(λ1X), and tan3(λ1X) are expanded to 8th order. After careful calculation, (2) should be corrected as
(3)X¨+ω012X+(-13+λ3λ13+13λ2λ13)X3mm+(215+215λ2λ15+λ3λ15)X5mm+(-17315+17315λ2λ17+1115λ3λ17)X7,
where ω01=1+λ1λ2.


## 2. Variational Iteration Method

In [1], the following variational iteration algorithm was constructed:
(4)X1=X0+1ω01∫0tsin(ω01(s-t))mmmmmmmmm×{X¨0+ω012X0+(-13+λ3λ13+13λ2λ13)X03mmmmmmmmmmm+(215+215λ2λ15+λ3λ15)X05mmmmmmmmmmm+(1115λ3λ17)X07}ds,
which should be corrected as
(5)X1=X0+1ω01∫0tsin(ω01(s-t))mmmmmmm×{X¨0+ω012X0+(-13+λ3λ13+13λ2λ13)X03mmmmmmmmm+(215+215λ2λ15+λ3λ15)X05mmmmmmimmm+(-17315+17315λ2λ17+1115λ3λ17)X07}ds.
In order to simplify the solution process, we denote
(6)a=ω012b=-13+λ3λ13+13λ2λ13c=215+215λ2λ15+λ3λ15d=-17315+17315λ2λ17+1115λ3λ17.
Equation (5) becomes
(7)X1=X0+1ω01∫0tsin(ω01(s-t))mmmmmmmm×{X¨0(s)+aX0(s)+bX03(s)mmmmmmmmmm+cX05(s)+dX07(s)}ds.
We begin with
(8)X0=Asin(Ωt)
and use the following relationships:
(9)X03=A3sin3(Ωt)=-A34sin(3Ωt)+3A34sin(Ωt)X05=A5sin5(Ωt)=A516sin(5Ωt)-5A516sin(3Ωt)+10A516sin(Ωt)X07=A7sin7(Ωt)=-A764sin(7Ωt)+7A764sin(5Ωt)-21A764sin(3Ωt)+35A764sin(Ωt).
By (7), we have
(10)X1=X0+1ω01∫0tsin(ω01(s-t))mmmmmm×{A1sinΩs+A2sin(3Ωs)mmmmmmmm+A3sin(5Ωs)-A4sin(7Ωs)}ds,
where
(11)A1=-AΩ2+aA+3bA34+10cA516+35A7d64A2=-bA34-5cA516-21A7d64A3=cA516+7A7d64A4=-A7d64.
Consider the following identity:
(12)∫0tsin(ω01(s-t))sinkΩsds=mmmkΩsinω01t-ω01sinkΩtω012-k2Ω2,k=1,3,5,7.
From (10), we obtain
(13)X1=AsinΩt-1ω01(Ω2-ω012)A1(Ωsinω01t-ω01sinΩt)-1ω01(9Ω2-ω012)A2(3Ωsinω01t-ω01sin3Ωt)-1ω01(25Ω2-ω012)A3(5Ωsinω01t-ω01sin5Ωt)-1ω01(49Ω2-ω012)A4(7Ωsinω01t-ω01sin7Ωt),
where Aj (j = 1~4) are defined above.


## 3. Discussion and Conclusion

The result of(7) in [1] should be corrected as that given in (13). In [1], actually the following variational iteration algorithm is used:
(14)Xn+1=Xn+1ω01∫0tsin(ω01(s-t))mmmmmm×{X¨n(s)+aXn(s)+bXn3(s)mmmmmmmm+cXn5(s)+dXn7(s)}ds,
which is called the variational iteration algorithm I in [8, 9]; for the present problem, the variational iteration algorithm II in [10] is much more effective, which reads
(15)Xn+1=X0+1ω01∫0tsin(ω01(s-t))mmmmmm×{bXn3(s)+cXn5(s)mmmmmmmm+dXn7(s)}ds.
The variational iteration method is especially effective in solving nonlinear oscillators.



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*Source: 329638-2013-06-10.xml*