# Analytical Solution for the Cubic-Quintic Duffing Oscillator Equation with Physics Applications

Authors: Alvaro H. Salas; Lorenzo J. Martínez H; David L. Ocampo R
Journal: Complexity
 (2022)

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## Abstract

The nonlinear differential equation governing the periodic motion of the one-dimensional, undamped, and unforced cubic-quintic Duffing oscillator is solved exactly by obtaining the period and the solution. The period is given in terms of the complete elliptic integral of the first kind and the solution involves Jacobian elliptic functions. We solve the cubic-quintic Duffing equation under arbitrary initial conditions. Physical applications are provided. The solution to the mixed parity Duffing oscillator is also formally derived. We illustrate the obtained results with concrete examples. We give high accurate trigonometric approximations to the Jacobian function cn.


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## Body

## 1. Introduction

It is well known that many engineering problems are not linear and their analytical solutions are not easy to obtain. Disturbance methods are among the known methods for solving nonlinear problems, which are based on the existence of small/large parameters, the so-called disturbance parameters. Our approach is different from known solutions to this problem [1–5]. On the contrary, here, we express the solutions without imaginary quantities: both frequency and modulus are real numbers. The quintic term appearing in the cubic-quintic Duffing equation makes this nonlinear oscillator not only more complex but also more interesting to study.


## 2. The Analytical Solution to the Cubic-Quintic Duffing Equation

Letp, q, r, y0, and y˙0 be arbitrary real numbers. We will solve the initial value problem:(1)y′′t+pyt+qyt3+ryt5=0given thaty0=y0andy′0=y˙0.We will assume thaty02+y˙02>0. Multiplying (1) by y′t and integrating it with respect to t gives(2)12y′t2+p2yt2+q4yt4+r6yt6=12y˙02+p2y02+q4y04+r6y06.Let(3)yt=c0cnωt+c1|m1+λ⋅cnωt+c1|m2+μ⋅cnωt+c1|m4,where the parameter values c0, c1, m, ω, λ, and μ are to be determined. If the solution in (3) is periodic, it will have the same period as the function cn ωt+c1|m and this period may be evaluated by means of the formula(4)T=4ωKm.In the case, when−1≤m≤1/2, we may approximate the value of Km using the formula(5)Km≈πm409m−3984+486450m41m−208+9728.The error for this approximation is given by(6)Error=max−1≤m≤1/2Km−Km<0.000314.On the contrary, we may obtain approximate trigonometric solution making use of the following approximation formula:(7)cnt,m≈cosmt=1+κcos1+κt1+κcos21+κt,whereκ=114m2−144m+144−m+12.See Table1, for the error = max−2Km≤t≤2Kmcosmt−cnt,m.Table 1
Error of approximating the Jacobian cn function by means of the cosine function (7) with T = 4K(m).
mErrormError−10.006800−0.950.006240.050.0000332−0.90.0057420.10.00014−0.850.00530.150.00033−0.80.00450.20.00061−0.750.00430.250.0010−0.70.00390.30.00153−0.650.003440.350.00212−0.60.003010.40.003−0.550.002610.450.0041−0.50.002220.50.0054−0.450.001860.550.007−0.40.0015170.60.009−0.350.00120.650.0116−0.30.00090.70.0145−0.250.000660.750.0188−0.20.000440.80.0241−0.150.000250.850.0314−0.10.000120.90.042−0.050.0000310.950.059A more accurate trigonometric approximation may be obtained using the formula(8)cnt,m≈cosmt≔1+ρ+κcoswt1+ρcos2wt+κcos4wt,being(9)w=κ−1m−2κ+2,ρ=κ−7κ−κ−12mκ+2,andκ=8m22409m4−29600m3+111520m2−163840m+8192035767m6−831840m5+6197600m4−21217280m3+36823040m2−31457280m+10485760.See Table2, for the error = max−2Km≤t≤2Kmcosmt−cnt,m.Table 2
Error of approximating the Jacobian cn function by means of the cosine function (8) with T = 4K(m).
mErrormError−10.00007400−0.950.0000630.052.002e −9−0.90.0000540.13.52e −8−0.850.00004520.151.98e −7−0.80.00003780.26.93e −7−0.750.000030.251.89e −6−0.70.0000250.34.40e −6−0.650.0000120.359.22e −6−0.60.0000150.40.000018−0.550.0000110.450.000033−0.58.37e −60.50.00006−0.455.88e −60.550.0000991−0.43.94e −60.60.00017−0.352.48e −60.650.00023−0.31.44e −60.70.00045−0.257.5e −70.750.00074−0.23.322e −70.80.0012−0.151.145e −70.850.0021−0.12.44e −80.90.0036−0.051.67e −90.950.015Let(10)Rt=12y′t2+p2yt2+q4yt4+r6yt6−12y˙02+p2y02+q4y04+r6y06:Residual.Introduce the notation:(11)cn=cnωt|mand cn=ζ.Definition 1.
The discriminant to the i.v.p. (1) is defined as(12)Δ=p+qy02+ry0423q2−16pr−4qry02−4r2y04+6q+2ry02q2−6pr−2qry02−2r2y04y˙02−36r2y˙04.


### 2.1. First Case:Δ>0

We defineμ=0. Inserting the ansatz (3) into (10), we obtain(13)Rt=1121+λζ3−6c02mω+6c02ω−6py02−3qy04−2ry0−6y0.2+34c02mω+2c02p−2c02ω−6λpy02−3λqy04−2λry06−6λy0.2ζ,3−2c02mω+4c02λp+c04q−6λ2py0−3λ2qy0−2λ2ry06−6λ2y.02ζ26c02λ2p+3c04λq+2c06r−6λ3py02−3λ3qy02−2λ3ry02−6λ3y0.2ζ3.Equating to zero the coefficients ofζjj=0,1,2,3 in the numerator of the last expression gives an algebraic system. Solving it, we obtain(14)m=3λω−p+ω3λ+2ω.ω=3λ+26py02+3qy04+2ry06+6y˙026c02−p.For these choices, we obtain the following system:(15)−18py02−9qy04−6ry06−18y˙02λ2+12c02p−18py02−9qy04−6ry06−18y˙02λ+6c02p+3c04q−6py02−3qy04−2ry06−6y˙02=0.−6py02−3qy04−2ry06−6y˙02λ3+6c02pλ2+3c04qλ+2c06r=0.Eliminatingc0 from system (15) gives the sextic(16)d0+d1λ+d2λ2+d3λ3+d4λ4+d5λ5+d6λ6=0,where(17)d0=4r26py02+3qy04+2ry06+6y˙022,d1=36r6py02+3qy04+2ry06+6y˙02pq+6pry02+3qry04+2r2y06+6ry˙02,d2=948p3r+240p2qry02+576p2r2y04−18pq3y02+120pq2ry04+656pqr2y06+240pqry˙02+384pr3y08+1152pr2y˙02y02−9q4y04−6q3ry06−18q3y˙02+144q2r2y08+192qr3y010+576qr2y˙02y04+64r4y012+384r3y˙02y06+576r2y˙04,d3=16216p3r−2p2q2+48p2qry02+72p2r2y04−6pq3y02+24pq2ry04+88pqr2y06+48pqry˙02+48pr3y08+144pr2y˙02y02−3q4y04−2q3ry06−6q3y˙02+18q2r2y08+24qr3y010+72qr2y˙02y04+8r4y012+48r3y˙02y06+72r2y˙04,d4=27208p3r−36p2q2+504p2qry02+576p2r2y04−78pq3y02+252pq2ry04+744pqr2y06+504pqry˙02+384pr3y08+1152pr2y˙02y02−39q4y04−26q3ry06−78q3y˙02+144q2r2y08+192qr3y010+576qr2y˙02y04+64r4y012+384r3y˙02y06+576r2y˙04,d5=32416p3r−3p2q2+36p2qry02+36p2r2y04−6pq3y02+18pq2ry04+48pqr2y06+36pqry˙02+24pr3y08+72pr2y˙02y02−3q4y04−2q3ry06−6q3y˙02+9q2r2y08+12qr3y010+36qr2y˙02y04+4r4y012+24r3y˙02y06+36r2y˙04,d6=10816p3r−3p2q2+36p2qry02+36p2r2y04−6pq3y02+18pq2ry04+48pqr2y06+36pqry˙02+24pr3y08+72pr2y˙02y02−3q4y04−2q3ry06−6q3y˙02+9q2r2y08+12qr3y010+36qr2y˙02y04+4r4y012+24r3y˙02y06+36r2y˙04.Sextic (16) is solvable by radicals. Indeed, let(18)λ=z−12.Then,(19)s0+s1z+s2z2+s3z3=0,where(20)s0=9pq+6pry02+3qry04+2r2y06+6ry˙022,s1=948p3r−27p2q2−12p2qry02+36p2r2y04−18pq3y02−6pq2ry04+32pqr2y06−12pqry˙02+24pr3y08+72pr2y˙02y02−9q4y04−6q3ry06−18q3y˙02+9q2r2y08+12qr3y010+36qr2y˙02y04+4r4y012+24r3y˙02y06+36r2y˙04,s2=−2732p3r−9p2q2+36p2qry02−36p2r2y04−12pq3y02+18pq2ry04−24pqr2y06+36pqry˙02−24pr3y08−72pr2y˙02y02−6q4y04−4q3ry06−12q3y˙02−9q2r2y08−12qr3y010−36qr2y˙02y04−4r4y012−24r3y˙02y06−36r2y˙04,s3=2716p3r−3p2q2+36p2qry02+36p2r2y04−6pq3y02+18pq2ry04+48pqr2y06+36pqry˙02+24pr3y08+72pr2y˙02y02−3q4y04−2q3ry06−6q3y˙02+9q2r2y08+12qr3y010+36qr2y˙02y04+4r4y012+24r3y˙02y06+36r2y˙04.The discriminant to cubic (19) is(21)δcubic=20639121408r26py02+3qy04+2ry06+6y˙02216p3r+6pq3y02+3q4y04+2q3ry06+6q3y˙022Δ>0.Cubic (19) has three real roots and at least one of them must be positive. Indeed, let z1, z2, and z3 be the roots to this cubic. Then,(22)z1z2z3=−s0s3=9pq+6pry02+3qry04+2r2y06+6ry˙0221728Δ>0.Sincez1z2z3>0, at least one of the numbers z1, z2, and z3 must be positive. We choose the closest to 1/4 positive root to cubic (19) so that, in view of (17), the number λ will be the closest to zero real root to the sextic in (15). Observe also that the condition Δ>0 implies that q2−4pr>0. Thus, if q2−4pr≤0, then Δ≤0. Moreover, the discriminant to the cubic (17) and Δ have the same sign.The numbersc0 and c1 are determined from the initial conditions:(23)c1=cn−1±y0c02−λy02|m.The numberc0 is a solution to the sextic:(24)λλ+12λ+1y026py02+3qy04+2ry06+6y˙02+−24λ2py02−24λpy02−6py02−9λ2qy04−9λqy04−3qy04−6λ2ry06−6λry06−2ry06−18λ2y˙02−18λy˙02−6y˙02c02+62λ+1pc04+3q+2ry02c06=0.Example 1.
Let(25)p=1,q=5,r=1,y0=1andy˙0=1.
Sextic (15) reads(26)213084λ6+639252λ5+598887λ4+132354λ3−75861λ2−35496λ−3364=0.
The roots to this sextic are(27)λ1=−1.39266,λ2=−0.855911,λ3=−0.626144.λ4=−0.373856,λ5=−0.144089,λ6=0.392663.
We choose the valueλ=λ5=−0.144089. The values of c0 and c1 are determined from the initial conditions. They read(28)c0=0.985102andc1=−0.327845.
The exact solution to the i.v.p.,(29)y′′t+yt+5yt3+yt5=0given thaty0=1andy′0=1,reads(30)yt=0.985102cn2.60927t−0.327845,0.2684471−0.144089cn2.60927t−0.327845,0.2684472.
The solution is periodic with periodT=4Km/ω=2.5997790766024407. The approximate trigonometric solution is given by(31)ytrigot=3.41275cos2.41528t−0.12581814−3.72995cos22.41528t−0.125818.
The error of this trigonometric approximant compared with the exact solution on0≤t≤T equals 0.00352816, see Figure 1.Figure 1
Comparison between the exact and the approximate numerical solution.


### 2.2. Second Case:Δ<0

Letμ≠0. Inserting the ansatz (3) into (10), we obtain(32)12ζ6μ3+36ζ5λμ2+12ζ43λ2μ+3μ2+12ζ3λ3+6λμ+12ζ23λ2+3μ+36ζλ+12Rt=−6c02mω+6c02ω−6py02−3qy04−2ry06−6y˙02+34c02μω−4c02μmω+2c02mω−4c02λp−c04q+6λ2py02+6μpy02+3λ2qy04+3μqy04+2λ2ry06+2μry06+6λ2y˙02+6μy˙02ζ2+3ζ4c02mω+2c02p−2c02ω−6λpy02−3λqy04−2λry06−6λy˙02−12c02μω−24c02μmω+6c02λ2p+12c02μp+3c04λq+2c06r−6λ3py02−36λμpy02−3λ3qy04−18λμqy04−2λ3ry06−12λμry06−6λ3y˙02−36λμy˙02ζ3−3μ−2c02μω+2c02μmω−4c02mω−4c02λp−c04q+6λ2py02+6μpy02+3λ2qy04+3μqy04+2λ2ry06+2μry06+6λ2y˙02+6μy˙02ζ4+3μ24c02mω+2c02p−2c02ω−6λpy02−3λqy04−2λry06−6λy˙02ζ5−μ26c02mω+6μpy02+3μqy04+2μry06+6μy˙02ζ6.Equating to zero the coefficients ofζjj=0,1,2,3,4,5,6 gives an algebraic system. Solving it gives(33)ω=3λ+26py02+3qy04+2ry06+6y˙026c02−p,m=μμ−1,μ=6c02p6py02+3qy04+2ry06+6y˙02−3λ−1.The system reduces to(34)−24pc02+3qc04+24py02+12qy04+8ry06+24y˙02+12λpc02+6py02+3qy04+2ry06+6y˙02−3λ26py02+3qy04+2ry06+6y˙02=0,2c0272p2c02−72p2y02+6prc04y02−36pqy04+3qrc04y04−24pry06+2r2c04y06−72py˙02+6rc04y˙02−348pc02−qc04−24py02−12qy04−8ry06−24y˙026py02+3qy04+2ry06+6y˙02λ+66py02+3qy04+2ry06+6y˙02pc02+36py02+18qy04+12ry06+36y˙02λ2−6py02+3qy04+2ry06+6y˙022λ3=0.Eliminatingc0 from this system gives the sextic(35)D0+D1λ+D2λ2+D3λ3+D4λ4+D5λ5+D6λ6=0,where(36)D0=642ry06+3qy04+6py02+6y˙022r2y06+3qry04+6pry02+6ry˙02+9pq2,D1=5762r2y06+3qry04+6pry02+6ry˙02+9pq4r3y012+12qr2y010+24pr2y08+9q2ry08+24r2y˙02y06+50pqry06+21pq2y04+36qry˙02y04+36p2ry04+72pry˙02y02+42p2qy02+36ry˙04+24p3+42pqy˙02,D2=14488r5y018+396qr4y016+792pr4y014+594q2r3y014+3056pqr3y012+309q3r2y012+792r4y˙02y012+2376p2r3y010+3822pq2r2y010+2376qr3y˙02y010+36q4ry010+27q5y08+7644p2qr2y08+4752pr3y˙02y08+1782q2r2y˙02y08+1602pq3ry08+108pq4y06+2376r3y˙04y06+2664p3r2y06+11208pqr2y˙02y06+72q3ry˙02y06+7254p2q2ry06+3564qr2y˙04y04+1809p2q3y04+108q4y˙02y04+7128p2r2y˙02y04+6120pq2ry˙02y04+6552p3qry04+7128pr2y˙04y02+3402p3q2y02+216pq3y˙02y02+12240p2qry˙02y02+864p4ry02+2376r2y˙06+108q3y˙04+6120pqry˙04+3402p2q2y˙02+864p3ry˙02+1728p4q,D3=8648r5y018+36qr4y016+72pr4y014+54q2r3y014+312pqr3y012+39q3r2y012+72r4y˙02y012+216p2r3y010+450pq2r2y010+216qr3y˙02y010+36q4ry010+27q5y08+900p2qr2y08+432pr3y˙02y08+162q2r2y˙02y08+288pq3ry08+108pq4y06+216r3y˙04y06+232p3r2y06+1224pqr2y˙02y06+72q3ry˙02y06+1038p2q2ry06+324qr2y˙04y04+369p2q3y04+108q4y˙02y04+648p2r2y˙02y04+864pq2ry˙02y04+888p3qry04+648pr2y˙04y02+522p3q2y02+216pq3y˙02y02+1728p2qry˙02y02+48p4ry02+216r2y˙06+108q3y˙04+864pqry˙04+522p2q2y˙02+48p3ry˙02+216p4q,D4=108−88r5y018−396qr4y016−792pr4y014−594q2r3y014−2496pqr3y012−145q3r2y012−792r4y˙02y012−2376p2r3y010−2142pq2r2y010−2376qr3y˙02y010+456q4ry010+342q5y08−4284p2qr2y08−4752pr3y˙02y08−1782q2r2y˙02y08+642pq3ry08+1368pq4y06−2376r3y˙04y06−2440p3r2y06−7848pqr2y˙02y06+912q3ry˙02y06−510p2q2ry06−3564qr2y˙04y04+2223p2q3y04+1368q4y˙02y04−7128p2r2y˙02y04−1080pq2ry˙02y04−1176p3qry04−7128pr2y˙04y02+1710p3q2y02+2736pq3y˙02y02−2160p2qry˙02y02−192p4ry02−2376r2y˙06+1368q3y˙04−1080pqry˙04+1710p2q2y˙02−192p3ry˙02+576p4q,D5=3248r5y018+36qr4y016+72pr4y014+54q2r3y014+216pqr3y012+35q3r2y012+72r4y˙02y012+216p2r3y010+162pq2r2y010+216qr3y˙02y010+24q4ry010+18q5y08+324p2qr2y08+432pr3y˙02y08+162q2r2y˙02y08+48pq3ry08+72pq4y06+216r3y˙04y06+200p3r2y06+648pqr2y˙02y06+48q3ry˙02y06+30p2q2ry06+324qr2y˙04y04+117p2q3y04+72q4y˙02y04+648p2r2y˙02y04−24p3qry04+648pr2y˙04y02+90p3q2y02+144pq3y˙02y02−48p4ry02+216r2y˙06+72q3y˙04+90p2q2y˙02−48p3ry˙02+24p4q,D6=−272ry06+3qy04+6py02+6y˙024r4y012+12qr3y010+24pr3y08+9q2r2y08+48pqr2y06+24r3y˙02y06−2q3ry06−3q4y04+36p2r2y04+36qr2y˙02y04+,D6=−272ry06+3qy04+6py02+6y˙02.Sextic (34) has at least one real root. Indeed, let zj (j=1,2,3,4,5,6) be its roots. Then,(37)z1z2z3z4z5z6=D0D6=649pq+6pry02+3qry04+2r2y06+6ry˙02227Δ<0.We will choose the closest to zero real root to sextic (34). The values for c0 and c1 are determined from the initial conditions:(38)c1=cn−1c02−λy02±c02−λy022−4μy042μy02|m.The numberc0 is found from the algebraic equation:(39)6λy06−15λy04+15λy02+3λy˙02+2y06−5y04+5y02+y˙02−5c022y06−5y04+y˙02c02+4λy08−10λy06+10λy04+2λy˙02y022y06−5y04+5y02+y˙02c04+−4λy08+10λy06−10λy04−2λy˙02y02−20y04c02+2λ2y010−5λ2y08+5λ2y06+λ2y˙02y04+24λy010−60λy08+60λy06+12λy˙02y04+8y010−20y08+20y06+4y˙02y04=0.Example 2.
Letp=q=r=y0=y˙0=1. The i.v.p. problem to be solved is(40)y′′t+yt+yt3+yt5=0given thaty0=1andy′0=1.
This problem has a negative discriminantΔ=−387. Sextic (35) reads(41)177633λ6−1825416λ5+5060232λ4−12568608λ3−16971120λ2−6469632λ−735488=0.
The roots are(42)λ1=−0.404933−0.145779i,λ2=−0.404933+0.145779i,λ3=−0.208601,λ4=1.63821−3.26816i,λ5=1.63821+3.26816i,λ6=8.01838.
We chooseλ=λ3=−0.208601. The values for c0 and c1 are(43)c0−0.993235,c1=2.72596.
The exact solution is given by(44)yt=−0.993235cn1.71662t+2.72596|0.0253557−0.0260153cn1.71662t+2.72596|0.02535574−0.208601cn1.71662t+2.72596|0.02535572+1.
See Figure2, for a comparison with the numerical solution.Figure 2
Comparison between the numerical and the exact solution for the i.v.p. (40).


### 2.3. Third Case:Δ=0

In this case, we have two roots to the cubic:(45)z1=198p24p2+q6py02+3qy04+2ry06+6y˙02+4pr+q2q2−4pr,andz2=198p24p2+q6py02+3qy04+2ry06+6y˙02+4pr+q2q2−4pr.At least one of these two numbers must be positive. So, we proceed the same way as we did for a positive discriminant.


### 2.4. Two Particular Cases

#### 2.4.1. First Particular Case:y˙0=0

Let(46)y′′t+pyt+qyt3+ryt5=0given thaty0=y0andy′0=0.The discriminant to the i.v.p. (46) equals(47)Δ1=p+qy02+ry0423q2−4qry02−4r2y04−16pr.IfΔ1>0, then the exact solution is given by(48)yt=λ+1y0cntω|m1+λcntω|m2,where(49)m=2λ3λ+2p+λ+13λ+1qy0λ+16λ+2p+3λ+2qy0,ω=λ+16λ+2p+3λ+2qy06λλ+1+2,λ=−y03q+6ry0±33q2−16pr−4qry0−4r2y012p+qy0+ry0.Assume now a negative discriminant. The exact solution reads(50)ω=2pλ3λ+4−5μ+1+3λ+2qy02λ+μ+16λλ+1+10μ+2,m=2pλ3λ+2−5μ+3λ+1qy02λ+μ+12pλ3λ+4−5μ+1+3λ+2qy02λ+μ+1,λ=23q+2ry02−26Δ2+12p+9qy02+6ry043y0216pr−3q2+4ry02q+ry02,μ=−96p2+16p6Δ2−9qy02−7ry04+y023q+2ry0246Δ2−17qy02−14ry04y0416pr−3q2+4ry02q+ry02,whereΔ2=p+qy02+ry046p+3qy02+2ry04.We claim thatΔ2>0. Indeed, let(51)δ=3q2−4qry02−4r2y04−16pr.Then,δ<0. However,(52)Δ2=3q+2ry022−δ3q+2ry022−3δ128r2.From the last identity, it is clear thatΔ2>0.Assume now thatΔ1=0. In this case, we have the following solutions:(53)yt=y03q+2ry02cos1/4q+2ry023q+2ry02/rt3q+4ry02−2ry02cos1/4q+2ry023q+2ry02/rtforq+2ry023q+2ry02r>0,yt=y03q+2ry02cosh1/4−q+2ry023q+2ry02/rt3q+4ry02−2ry02cosh1/4−q+2ry023q+2ry02/rtforq+2ry023q+2ry02r<0.Solutions (53) are called solitons. They arise in soliton theory.


#### 2.4.2. Second Particular Case:y0=0

Let(54)y′′t+pyt+qyt3+ryt5=0given thaty0=0andy′0=v˙0.The solution has the form(55)yt=v˙0λ+μ+1⋅snx−pλ+μ+1/2λ+4μ−1|μ/λ+μ+1−pλ+μ+1/2λ+4μ−11+μ⋅cnx−pλ+μ+1/2λ+4μ−1|μ/λ+μ+14+λ⋅cnx−pλ+μ+1/2λ+4μ−1|μ/λ+μ+12.The values ofλ and μ are obtained by solving the two sextics:(56)Pλ=0andQμ=0,where(57)Pλ=v˙029q3−48pqr−64r2v˙02+12−3p2q2+16p3r−3q3v˙02+20pqrv˙02+16r2v˙04λ+6−3p2q2+16p3r+3q3v˙02−12pqrv˙02−16r2v˙04λ2+4v˙029q3−48pqr−32r2v˙02λ3+3v˙023q3−16pqr+16r2v˙02λ4+48r2v˙04λ5+8r2v˙04λ6,and(58)Qμ=3v˙043p2q2−16p3r+6q3v˙02−36pqrv˙02−36r2v˙04−1444p6+17p4qv˙02+21p2q2v˙04+14p3rv˙04+6q3v˙06+27pqrv˙06+18r2v˙08μ+966p6+48p4qv˙02+129p2q2v˙04−124p3rv˙04+114q3v˙06−252pqrv˙06−198r2v˙08μ2−768v˙023p4q+15p2q2v˙02−10p3rv˙02+18q3v˙04−18pqrv˙04+36r2v˙06μ3+768v˙043p2q2+8p3r+6q3v˙02+28pqrv˙02+132r2v˙04μ4−12288rv˙06pq+6rv˙02μ5+16384r2v˙08μ6.


## 2.1. First Case:Δ>0

We defineμ=0. Inserting the ansatz (3) into (10), we obtain(13)Rt=1121+λζ3−6c02mω+6c02ω−6py02−3qy04−2ry0−6y0.2+34c02mω+2c02p−2c02ω−6λpy02−3λqy04−2λry06−6λy0.2ζ,3−2c02mω+4c02λp+c04q−6λ2py0−3λ2qy0−2λ2ry06−6λ2y.02ζ26c02λ2p+3c04λq+2c06r−6λ3py02−3λ3qy02−2λ3ry02−6λ3y0.2ζ3.Equating to zero the coefficients ofζjj=0,1,2,3 in the numerator of the last expression gives an algebraic system. Solving it, we obtain(14)m=3λω−p+ω3λ+2ω.ω=3λ+26py02+3qy04+2ry06+6y˙026c02−p.For these choices, we obtain the following system:(15)−18py02−9qy04−6ry06−18y˙02λ2+12c02p−18py02−9qy04−6ry06−18y˙02λ+6c02p+3c04q−6py02−3qy04−2ry06−6y˙02=0.−6py02−3qy04−2ry06−6y˙02λ3+6c02pλ2+3c04qλ+2c06r=0.Eliminatingc0 from system (15) gives the sextic(16)d0+d1λ+d2λ2+d3λ3+d4λ4+d5λ5+d6λ6=0,where(17)d0=4r26py02+3qy04+2ry06+6y˙022,d1=36r6py02+3qy04+2ry06+6y˙02pq+6pry02+3qry04+2r2y06+6ry˙02,d2=948p3r+240p2qry02+576p2r2y04−18pq3y02+120pq2ry04+656pqr2y06+240pqry˙02+384pr3y08+1152pr2y˙02y02−9q4y04−6q3ry06−18q3y˙02+144q2r2y08+192qr3y010+576qr2y˙02y04+64r4y012+384r3y˙02y06+576r2y˙04,d3=16216p3r−2p2q2+48p2qry02+72p2r2y04−6pq3y02+24pq2ry04+88pqr2y06+48pqry˙02+48pr3y08+144pr2y˙02y02−3q4y04−2q3ry06−6q3y˙02+18q2r2y08+24qr3y010+72qr2y˙02y04+8r4y012+48r3y˙02y06+72r2y˙04,d4=27208p3r−36p2q2+504p2qry02+576p2r2y04−78pq3y02+252pq2ry04+744pqr2y06+504pqry˙02+384pr3y08+1152pr2y˙02y02−39q4y04−26q3ry06−78q3y˙02+144q2r2y08+192qr3y010+576qr2y˙02y04+64r4y012+384r3y˙02y06+576r2y˙04,d5=32416p3r−3p2q2+36p2qry02+36p2r2y04−6pq3y02+18pq2ry04+48pqr2y06+36pqry˙02+24pr3y08+72pr2y˙02y02−3q4y04−2q3ry06−6q3y˙02+9q2r2y08+12qr3y010+36qr2y˙02y04+4r4y012+24r3y˙02y06+36r2y˙04,d6=10816p3r−3p2q2+36p2qry02+36p2r2y04−6pq3y02+18pq2ry04+48pqr2y06+36pqry˙02+24pr3y08+72pr2y˙02y02−3q4y04−2q3ry06−6q3y˙02+9q2r2y08+12qr3y010+36qr2y˙02y04+4r4y012+24r3y˙02y06+36r2y˙04.Sextic (16) is solvable by radicals. Indeed, let(18)λ=z−12.Then,(19)s0+s1z+s2z2+s3z3=0,where(20)s0=9pq+6pry02+3qry04+2r2y06+6ry˙022,s1=948p3r−27p2q2−12p2qry02+36p2r2y04−18pq3y02−6pq2ry04+32pqr2y06−12pqry˙02+24pr3y08+72pr2y˙02y02−9q4y04−6q3ry06−18q3y˙02+9q2r2y08+12qr3y010+36qr2y˙02y04+4r4y012+24r3y˙02y06+36r2y˙04,s2=−2732p3r−9p2q2+36p2qry02−36p2r2y04−12pq3y02+18pq2ry04−24pqr2y06+36pqry˙02−24pr3y08−72pr2y˙02y02−6q4y04−4q3ry06−12q3y˙02−9q2r2y08−12qr3y010−36qr2y˙02y04−4r4y012−24r3y˙02y06−36r2y˙04,s3=2716p3r−3p2q2+36p2qry02+36p2r2y04−6pq3y02+18pq2ry04+48pqr2y06+36pqry˙02+24pr3y08+72pr2y˙02y02−3q4y04−2q3ry06−6q3y˙02+9q2r2y08+12qr3y010+36qr2y˙02y04+4r4y012+24r3y˙02y06+36r2y˙04.The discriminant to cubic (19) is(21)δcubic=20639121408r26py02+3qy04+2ry06+6y˙02216p3r+6pq3y02+3q4y04+2q3ry06+6q3y˙022Δ>0.Cubic (19) has three real roots and at least one of them must be positive. Indeed, let z1, z2, and z3 be the roots to this cubic. Then,(22)z1z2z3=−s0s3=9pq+6pry02+3qry04+2r2y06+6ry˙0221728Δ>0.Sincez1z2z3>0, at least one of the numbers z1, z2, and z3 must be positive. We choose the closest to 1/4 positive root to cubic (19) so that, in view of (17), the number λ will be the closest to zero real root to the sextic in (15). Observe also that the condition Δ>0 implies that q2−4pr>0. Thus, if q2−4pr≤0, then Δ≤0. Moreover, the discriminant to the cubic (17) and Δ have the same sign.The numbersc0 and c1 are determined from the initial conditions:(23)c1=cn−1±y0c02−λy02|m.The numberc0 is a solution to the sextic:(24)λλ+12λ+1y026py02+3qy04+2ry06+6y˙02+−24λ2py02−24λpy02−6py02−9λ2qy04−9λqy04−3qy04−6λ2ry06−6λry06−2ry06−18λ2y˙02−18λy˙02−6y˙02c02+62λ+1pc04+3q+2ry02c06=0.Example 1.
Let(25)p=1,q=5,r=1,y0=1andy˙0=1.
Sextic (15) reads(26)213084λ6+639252λ5+598887λ4+132354λ3−75861λ2−35496λ−3364=0.
The roots to this sextic are(27)λ1=−1.39266,λ2=−0.855911,λ3=−0.626144.λ4=−0.373856,λ5=−0.144089,λ6=0.392663.
We choose the valueλ=λ5=−0.144089. The values of c0 and c1 are determined from the initial conditions. They read(28)c0=0.985102andc1=−0.327845.
The exact solution to the i.v.p.,(29)y′′t+yt+5yt3+yt5=0given thaty0=1andy′0=1,reads(30)yt=0.985102cn2.60927t−0.327845,0.2684471−0.144089cn2.60927t−0.327845,0.2684472.
The solution is periodic with periodT=4Km/ω=2.5997790766024407. The approximate trigonometric solution is given by(31)ytrigot=3.41275cos2.41528t−0.12581814−3.72995cos22.41528t−0.125818.
The error of this trigonometric approximant compared with the exact solution on0≤t≤T equals 0.00352816, see Figure 1.Figure 1
Comparison between the exact and the approximate numerical solution.


## 2.2. Second Case:Δ<0

Letμ≠0. Inserting the ansatz (3) into (10), we obtain(32)12ζ6μ3+36ζ5λμ2+12ζ43λ2μ+3μ2+12ζ3λ3+6λμ+12ζ23λ2+3μ+36ζλ+12Rt=−6c02mω+6c02ω−6py02−3qy04−2ry06−6y˙02+34c02μω−4c02μmω+2c02mω−4c02λp−c04q+6λ2py02+6μpy02+3λ2qy04+3μqy04+2λ2ry06+2μry06+6λ2y˙02+6μy˙02ζ2+3ζ4c02mω+2c02p−2c02ω−6λpy02−3λqy04−2λry06−6λy˙02−12c02μω−24c02μmω+6c02λ2p+12c02μp+3c04λq+2c06r−6λ3py02−36λμpy02−3λ3qy04−18λμqy04−2λ3ry06−12λμry06−6λ3y˙02−36λμy˙02ζ3−3μ−2c02μω+2c02μmω−4c02mω−4c02λp−c04q+6λ2py02+6μpy02+3λ2qy04+3μqy04+2λ2ry06+2μry06+6λ2y˙02+6μy˙02ζ4+3μ24c02mω+2c02p−2c02ω−6λpy02−3λqy04−2λry06−6λy˙02ζ5−μ26c02mω+6μpy02+3μqy04+2μry06+6μy˙02ζ6.Equating to zero the coefficients ofζjj=0,1,2,3,4,5,6 gives an algebraic system. Solving it gives(33)ω=3λ+26py02+3qy04+2ry06+6y˙026c02−p,m=μμ−1,μ=6c02p6py02+3qy04+2ry06+6y˙02−3λ−1.The system reduces to(34)−24pc02+3qc04+24py02+12qy04+8ry06+24y˙02+12λpc02+6py02+3qy04+2ry06+6y˙02−3λ26py02+3qy04+2ry06+6y˙02=0,2c0272p2c02−72p2y02+6prc04y02−36pqy04+3qrc04y04−24pry06+2r2c04y06−72py˙02+6rc04y˙02−348pc02−qc04−24py02−12qy04−8ry06−24y˙026py02+3qy04+2ry06+6y˙02λ+66py02+3qy04+2ry06+6y˙02pc02+36py02+18qy04+12ry06+36y˙02λ2−6py02+3qy04+2ry06+6y˙022λ3=0.Eliminatingc0 from this system gives the sextic(35)D0+D1λ+D2λ2+D3λ3+D4λ4+D5λ5+D6λ6=0,where(36)D0=642ry06+3qy04+6py02+6y˙022r2y06+3qry04+6pry02+6ry˙02+9pq2,D1=5762r2y06+3qry04+6pry02+6ry˙02+9pq4r3y012+12qr2y010+24pr2y08+9q2ry08+24r2y˙02y06+50pqry06+21pq2y04+36qry˙02y04+36p2ry04+72pry˙02y02+42p2qy02+36ry˙04+24p3+42pqy˙02,D2=14488r5y018+396qr4y016+792pr4y014+594q2r3y014+3056pqr3y012+309q3r2y012+792r4y˙02y012+2376p2r3y010+3822pq2r2y010+2376qr3y˙02y010+36q4ry010+27q5y08+7644p2qr2y08+4752pr3y˙02y08+1782q2r2y˙02y08+1602pq3ry08+108pq4y06+2376r3y˙04y06+2664p3r2y06+11208pqr2y˙02y06+72q3ry˙02y06+7254p2q2ry06+3564qr2y˙04y04+1809p2q3y04+108q4y˙02y04+7128p2r2y˙02y04+6120pq2ry˙02y04+6552p3qry04+7128pr2y˙04y02+3402p3q2y02+216pq3y˙02y02+12240p2qry˙02y02+864p4ry02+2376r2y˙06+108q3y˙04+6120pqry˙04+3402p2q2y˙02+864p3ry˙02+1728p4q,D3=8648r5y018+36qr4y016+72pr4y014+54q2r3y014+312pqr3y012+39q3r2y012+72r4y˙02y012+216p2r3y010+450pq2r2y010+216qr3y˙02y010+36q4ry010+27q5y08+900p2qr2y08+432pr3y˙02y08+162q2r2y˙02y08+288pq3ry08+108pq4y06+216r3y˙04y06+232p3r2y06+1224pqr2y˙02y06+72q3ry˙02y06+1038p2q2ry06+324qr2y˙04y04+369p2q3y04+108q4y˙02y04+648p2r2y˙02y04+864pq2ry˙02y04+888p3qry04+648pr2y˙04y02+522p3q2y02+216pq3y˙02y02+1728p2qry˙02y02+48p4ry02+216r2y˙06+108q3y˙04+864pqry˙04+522p2q2y˙02+48p3ry˙02+216p4q,D4=108−88r5y018−396qr4y016−792pr4y014−594q2r3y014−2496pqr3y012−145q3r2y012−792r4y˙02y012−2376p2r3y010−2142pq2r2y010−2376qr3y˙02y010+456q4ry010+342q5y08−4284p2qr2y08−4752pr3y˙02y08−1782q2r2y˙02y08+642pq3ry08+1368pq4y06−2376r3y˙04y06−2440p3r2y06−7848pqr2y˙02y06+912q3ry˙02y06−510p2q2ry06−3564qr2y˙04y04+2223p2q3y04+1368q4y˙02y04−7128p2r2y˙02y04−1080pq2ry˙02y04−1176p3qry04−7128pr2y˙04y02+1710p3q2y02+2736pq3y˙02y02−2160p2qry˙02y02−192p4ry02−2376r2y˙06+1368q3y˙04−1080pqry˙04+1710p2q2y˙02−192p3ry˙02+576p4q,D5=3248r5y018+36qr4y016+72pr4y014+54q2r3y014+216pqr3y012+35q3r2y012+72r4y˙02y012+216p2r3y010+162pq2r2y010+216qr3y˙02y010+24q4ry010+18q5y08+324p2qr2y08+432pr3y˙02y08+162q2r2y˙02y08+48pq3ry08+72pq4y06+216r3y˙04y06+200p3r2y06+648pqr2y˙02y06+48q3ry˙02y06+30p2q2ry06+324qr2y˙04y04+117p2q3y04+72q4y˙02y04+648p2r2y˙02y04−24p3qry04+648pr2y˙04y02+90p3q2y02+144pq3y˙02y02−48p4ry02+216r2y˙06+72q3y˙04+90p2q2y˙02−48p3ry˙02+24p4q,D6=−272ry06+3qy04+6py02+6y˙024r4y012+12qr3y010+24pr3y08+9q2r2y08+48pqr2y06+24r3y˙02y06−2q3ry06−3q4y04+36p2r2y04+36qr2y˙02y04+,D6=−272ry06+3qy04+6py02+6y˙02.Sextic (34) has at least one real root. Indeed, let zj (j=1,2,3,4,5,6) be its roots. Then,(37)z1z2z3z4z5z6=D0D6=649pq+6pry02+3qry04+2r2y06+6ry˙02227Δ<0.We will choose the closest to zero real root to sextic (34). The values for c0 and c1 are determined from the initial conditions:(38)c1=cn−1c02−λy02±c02−λy022−4μy042μy02|m.The numberc0 is found from the algebraic equation:(39)6λy06−15λy04+15λy02+3λy˙02+2y06−5y04+5y02+y˙02−5c022y06−5y04+y˙02c02+4λy08−10λy06+10λy04+2λy˙02y022y06−5y04+5y02+y˙02c04+−4λy08+10λy06−10λy04−2λy˙02y02−20y04c02+2λ2y010−5λ2y08+5λ2y06+λ2y˙02y04+24λy010−60λy08+60λy06+12λy˙02y04+8y010−20y08+20y06+4y˙02y04=0.Example 2.
Letp=q=r=y0=y˙0=1. The i.v.p. problem to be solved is(40)y′′t+yt+yt3+yt5=0given thaty0=1andy′0=1.
This problem has a negative discriminantΔ=−387. Sextic (35) reads(41)177633λ6−1825416λ5+5060232λ4−12568608λ3−16971120λ2−6469632λ−735488=0.
The roots are(42)λ1=−0.404933−0.145779i,λ2=−0.404933+0.145779i,λ3=−0.208601,λ4=1.63821−3.26816i,λ5=1.63821+3.26816i,λ6=8.01838.
We chooseλ=λ3=−0.208601. The values for c0 and c1 are(43)c0−0.993235,c1=2.72596.
The exact solution is given by(44)yt=−0.993235cn1.71662t+2.72596|0.0253557−0.0260153cn1.71662t+2.72596|0.02535574−0.208601cn1.71662t+2.72596|0.02535572+1.
See Figure2, for a comparison with the numerical solution.Figure 2
Comparison between the numerical and the exact solution for the i.v.p. (40).


## 2.3. Third Case:Δ=0

In this case, we have two roots to the cubic:(45)z1=198p24p2+q6py02+3qy04+2ry06+6y˙02+4pr+q2q2−4pr,andz2=198p24p2+q6py02+3qy04+2ry06+6y˙02+4pr+q2q2−4pr.At least one of these two numbers must be positive. So, we proceed the same way as we did for a positive discriminant.


## 2.4. Two Particular Cases

### 2.4.1. First Particular Case:y˙0=0

Let(46)y′′t+pyt+qyt3+ryt5=0given thaty0=y0andy′0=0.The discriminant to the i.v.p. (46) equals(47)Δ1=p+qy02+ry0423q2−4qry02−4r2y04−16pr.IfΔ1>0, then the exact solution is given by(48)yt=λ+1y0cntω|m1+λcntω|m2,where(49)m=2λ3λ+2p+λ+13λ+1qy0λ+16λ+2p+3λ+2qy0,ω=λ+16λ+2p+3λ+2qy06λλ+1+2,λ=−y03q+6ry0±33q2−16pr−4qry0−4r2y012p+qy0+ry0.Assume now a negative discriminant. The exact solution reads(50)ω=2pλ3λ+4−5μ+1+3λ+2qy02λ+μ+16λλ+1+10μ+2,m=2pλ3λ+2−5μ+3λ+1qy02λ+μ+12pλ3λ+4−5μ+1+3λ+2qy02λ+μ+1,λ=23q+2ry02−26Δ2+12p+9qy02+6ry043y0216pr−3q2+4ry02q+ry02,μ=−96p2+16p6Δ2−9qy02−7ry04+y023q+2ry0246Δ2−17qy02−14ry04y0416pr−3q2+4ry02q+ry02,whereΔ2=p+qy02+ry046p+3qy02+2ry04.We claim thatΔ2>0. Indeed, let(51)δ=3q2−4qry02−4r2y04−16pr.Then,δ<0. However,(52)Δ2=3q+2ry022−δ3q+2ry022−3δ128r2.From the last identity, it is clear thatΔ2>0.Assume now thatΔ1=0. In this case, we have the following solutions:(53)yt=y03q+2ry02cos1/4q+2ry023q+2ry02/rt3q+4ry02−2ry02cos1/4q+2ry023q+2ry02/rtforq+2ry023q+2ry02r>0,yt=y03q+2ry02cosh1/4−q+2ry023q+2ry02/rt3q+4ry02−2ry02cosh1/4−q+2ry023q+2ry02/rtforq+2ry023q+2ry02r<0.Solutions (53) are called solitons. They arise in soliton theory.


### 2.4.2. Second Particular Case:y0=0

Let(54)y′′t+pyt+qyt3+ryt5=0given thaty0=0andy′0=v˙0.The solution has the form(55)yt=v˙0λ+μ+1⋅snx−pλ+μ+1/2λ+4μ−1|μ/λ+μ+1−pλ+μ+1/2λ+4μ−11+μ⋅cnx−pλ+μ+1/2λ+4μ−1|μ/λ+μ+14+λ⋅cnx−pλ+μ+1/2λ+4μ−1|μ/λ+μ+12.The values ofλ and μ are obtained by solving the two sextics:(56)Pλ=0andQμ=0,where(57)Pλ=v˙029q3−48pqr−64r2v˙02+12−3p2q2+16p3r−3q3v˙02+20pqrv˙02+16r2v˙04λ+6−3p2q2+16p3r+3q3v˙02−12pqrv˙02−16r2v˙04λ2+4v˙029q3−48pqr−32r2v˙02λ3+3v˙023q3−16pqr+16r2v˙02λ4+48r2v˙04λ5+8r2v˙04λ6,and(58)Qμ=3v˙043p2q2−16p3r+6q3v˙02−36pqrv˙02−36r2v˙04−1444p6+17p4qv˙02+21p2q2v˙04+14p3rv˙04+6q3v˙06+27pqrv˙06+18r2v˙08μ+966p6+48p4qv˙02+129p2q2v˙04−124p3rv˙04+114q3v˙06−252pqrv˙06−198r2v˙08μ2−768v˙023p4q+15p2q2v˙02−10p3rv˙02+18q3v˙04−18pqrv˙04+36r2v˙06μ3+768v˙043p2q2+8p3r+6q3v˙02+28pqrv˙02+132r2v˙04μ4−12288rv˙06pq+6rv˙02μ5+16384r2v˙08μ6.


## 2.4.1. First Particular Case:y˙0=0

Let(46)y′′t+pyt+qyt3+ryt5=0given thaty0=y0andy′0=0.The discriminant to the i.v.p. (46) equals(47)Δ1=p+qy02+ry0423q2−4qry02−4r2y04−16pr.IfΔ1>0, then the exact solution is given by(48)yt=λ+1y0cntω|m1+λcntω|m2,where(49)m=2λ3λ+2p+λ+13λ+1qy0λ+16λ+2p+3λ+2qy0,ω=λ+16λ+2p+3λ+2qy06λλ+1+2,λ=−y03q+6ry0±33q2−16pr−4qry0−4r2y012p+qy0+ry0.Assume now a negative discriminant. The exact solution reads(50)ω=2pλ3λ+4−5μ+1+3λ+2qy02λ+μ+16λλ+1+10μ+2,m=2pλ3λ+2−5μ+3λ+1qy02λ+μ+12pλ3λ+4−5μ+1+3λ+2qy02λ+μ+1,λ=23q+2ry02−26Δ2+12p+9qy02+6ry043y0216pr−3q2+4ry02q+ry02,μ=−96p2+16p6Δ2−9qy02−7ry04+y023q+2ry0246Δ2−17qy02−14ry04y0416pr−3q2+4ry02q+ry02,whereΔ2=p+qy02+ry046p+3qy02+2ry04.We claim thatΔ2>0. Indeed, let(51)δ=3q2−4qry02−4r2y04−16pr.Then,δ<0. However,(52)Δ2=3q+2ry022−δ3q+2ry022−3δ128r2.From the last identity, it is clear thatΔ2>0.Assume now thatΔ1=0. In this case, we have the following solutions:(53)yt=y03q+2ry02cos1/4q+2ry023q+2ry02/rt3q+4ry02−2ry02cos1/4q+2ry023q+2ry02/rtforq+2ry023q+2ry02r>0,yt=y03q+2ry02cosh1/4−q+2ry023q+2ry02/rt3q+4ry02−2ry02cosh1/4−q+2ry023q+2ry02/rtforq+2ry023q+2ry02r<0.Solutions (53) are called solitons. They arise in soliton theory.


## 2.4.2. Second Particular Case:y0=0

Let(54)y′′t+pyt+qyt3+ryt5=0given thaty0=0andy′0=v˙0.The solution has the form(55)yt=v˙0λ+μ+1⋅snx−pλ+μ+1/2λ+4μ−1|μ/λ+μ+1−pλ+μ+1/2λ+4μ−11+μ⋅cnx−pλ+μ+1/2λ+4μ−1|μ/λ+μ+14+λ⋅cnx−pλ+μ+1/2λ+4μ−1|μ/λ+μ+12.The values ofλ and μ are obtained by solving the two sextics:(56)Pλ=0andQμ=0,where(57)Pλ=v˙029q3−48pqr−64r2v˙02+12−3p2q2+16p3r−3q3v˙02+20pqrv˙02+16r2v˙04λ+6−3p2q2+16p3r+3q3v˙02−12pqrv˙02−16r2v˙04λ2+4v˙029q3−48pqr−32r2v˙02λ3+3v˙023q3−16pqr+16r2v˙02λ4+48r2v˙04λ5+8r2v˙04λ6,and(58)Qμ=3v˙043p2q2−16p3r+6q3v˙02−36pqrv˙02−36r2v˙04−1444p6+17p4qv˙02+21p2q2v˙04+14p3rv˙04+6q3v˙06+27pqrv˙06+18r2v˙08μ+966p6+48p4qv˙02+129p2q2v˙04−124p3rv˙04+114q3v˙06−252pqrv˙06−198r2v˙08μ2−768v˙023p4q+15p2q2v˙02−10p3rv˙02+18q3v˙04−18pqrv˙04+36r2v˙06μ3+768v˙043p2q2+8p3r+6q3v˙02+28pqrv˙02+132r2v˙04μ4−12288rv˙06pq+6rv˙02μ5+16384r2v˙08μ6.


## 3. Applications

The cubic-quintic Duffing oscillator has many interesting applications in soliton theory, optics, nonlinear circuits, plasma physics, and other areas of science and engineering.


### 3.1. Nonlinear Odd Parity Oscillators

Let us consider the nonlinear oscillator:(59)u′′t+fut=0subjected tou0=u0andu′0=0,where fx is and odd function:(60)f−x=−fx.We may approximate the functionf=fx by means of a Chebyshev polynomial on some interval −A,A (A>0) as follows:(61)fx≈px+qx3+rx5,where(62)p=f−A2/2−fA2/2−5+33f−1/22−3A+5+33f1/22−3A+5−33f−1/22+3A−f1/22+3A32A,q=2−8f−A2/2+8fA2/2+7+53f−1/22−3A−7+53f1/22−3A+53−7f−1/22+3A−f1/22+3A3A3,r=421+33−1f−A2/2−3−1fA2/2−f−1/22−3A+f1/22−3A+3−2f−1/22+3A−f1/22+3A3A5.Then, we may obtain an approximated analytic solution to oscillator (59) by solving the cubic-quintic Duffing oscillator:(63)y′′t+pyt+qyt3+ryt5=0subjected toy0=u0andy′0=0.Example 3.
Let us obtain an approximate analytical solution to the pendulum equation:(64)θ′′t+k2sinθt=0subjected toθ0=θ0andθ′0=0,having the analytical solution [9](65)θt=2tan−1tanθ02cnk⋅t|sin2θ02.
The Chebyshev approximantPx of fx=sinx on −π/2,π/2 reads(66)Px=162sinπ2−3sin1232πcosπ22+sinπ22cos1232π−3π5x5+424sinπ2−53sin1232πcosπ22+7sinπ22cos1232π3π3x3−2xsinπ2−63sin1232πcosπ22+10sinπ22cos1232π3π.
We may rationalize this expression up to10−7 to obtain(67)Px=17x52926−2461x315601+1641x1649.
The square mean error is(68)∫−ππsinx−Px2=0.0141773.
Thus, our aim is to solve the cubic-quintic Duffing oscillator:(69)y′′t+1641k21649yt−2461k215601y3t+17k22926y5t=0subjected toy0=θ0andy′0=0.
Letk=1 and θ0=5π/6. For these data, we have a positive discriminant and the solution to the initial value problem,(70)y′′t+16411649yt−246115601y3t+172926y5t=0subjected toy0=5π6andy′0=0,may be obtained making use of formula (48).


### 3.2. The Duffing–Helmholtz Oscillator

Let us consider the i.v.p.:(71)x′′t+αxt+βx2t+γx3t=0given thatx0=x0andx′0=x˙0.Suppose that the functiony=yt is the solution to the cubic-quintic Duffing equation:(72)y′′t+14α+2Aβ+3A2γyt−13A−x0β+3Aγyt3+38A−x02γyt5=0given thaty0=1andy′0=x˙02x0−A.Then, the function,(73)xt=A+x0−Ay2t,is the solution to the i.v.p. (71) provided that A is a solution to the quartic:(74)3γA4+4βA3+6αA2−6αx02−4βx03−3γx04−6x˙02=0.


### 3.3. Nonlinear Conservative Oscillators

Suppose we are given to sole the i.v.p.:(75)u′′t+gut=0subjected tou0=u0andu′0=u˙0.Assume thatu≤A. We approximate the function g=gu by means of a cubic polynomial on −A≤u≤A using Chebyshev approach so that(76)gu≈a0+a1u+a2u2+a3u3,−A≤u≤A,where(77)a0=cos2π/8g−Asinπ/8+gAsinπ/8−sin2π/8g−Acosπ/8+gAcosπ/82,a1=2−18secπ/8g−Acosπ/8−gAcosπ/8+csc5π/8gAsinπ/8−g−Asinπ/832A,a2=−g−Asinπ/8−gAsinπ/8+g−Acosπ/8+gAcosπ/82A2,a3=2−1csc2π/8secπ/8−g−Acosπ/8+gAcosπ/8+cotπ/8g−Asinπ/8−gAsinπ/84A3.We now replace the i.v.p. (75) with the i.v.p.:(78)u′′t+a0+a1ut+a2u2t+a3u3t=0subjected tou0=u0andu′0=u˙0.Let(79)ut=ρ+xt,wherea0+a1ρ+a2ρ2+aρ3.The problem reduces to the i.v.p.:(80)x′′t+a1+2ρa2+3ρ2a3xt+a2+3ρa3x2t+a3x3t=0given thatx0=u0−ρandx′0=u˙0.This is a Duffing–Helmholtz oscillator (71) with(81)α=a1+2ρa2+3ρ2a3,β=a2+3ρa3andγ=a3.


## 3.1. Nonlinear Odd Parity Oscillators

Let us consider the nonlinear oscillator:(59)u′′t+fut=0subjected tou0=u0andu′0=0,where fx is and odd function:(60)f−x=−fx.We may approximate the functionf=fx by means of a Chebyshev polynomial on some interval −A,A (A>0) as follows:(61)fx≈px+qx3+rx5,where(62)p=f−A2/2−fA2/2−5+33f−1/22−3A+5+33f1/22−3A+5−33f−1/22+3A−f1/22+3A32A,q=2−8f−A2/2+8fA2/2+7+53f−1/22−3A−7+53f1/22−3A+53−7f−1/22+3A−f1/22+3A3A3,r=421+33−1f−A2/2−3−1fA2/2−f−1/22−3A+f1/22−3A+3−2f−1/22+3A−f1/22+3A3A5.Then, we may obtain an approximated analytic solution to oscillator (59) by solving the cubic-quintic Duffing oscillator:(63)y′′t+pyt+qyt3+ryt5=0subjected toy0=u0andy′0=0.Example 3.
Let us obtain an approximate analytical solution to the pendulum equation:(64)θ′′t+k2sinθt=0subjected toθ0=θ0andθ′0=0,having the analytical solution [9](65)θt=2tan−1tanθ02cnk⋅t|sin2θ02.
The Chebyshev approximantPx of fx=sinx on −π/2,π/2 reads(66)Px=162sinπ2−3sin1232πcosπ22+sinπ22cos1232π−3π5x5+424sinπ2−53sin1232πcosπ22+7sinπ22cos1232π3π3x3−2xsinπ2−63sin1232πcosπ22+10sinπ22cos1232π3π.
We may rationalize this expression up to10−7 to obtain(67)Px=17x52926−2461x315601+1641x1649.
The square mean error is(68)∫−ππsinx−Px2=0.0141773.
Thus, our aim is to solve the cubic-quintic Duffing oscillator:(69)y′′t+1641k21649yt−2461k215601y3t+17k22926y5t=0subjected toy0=θ0andy′0=0.
Letk=1 and θ0=5π/6. For these data, we have a positive discriminant and the solution to the initial value problem,(70)y′′t+16411649yt−246115601y3t+172926y5t=0subjected toy0=5π6andy′0=0,may be obtained making use of formula (48).


## 3.2. The Duffing–Helmholtz Oscillator

Let us consider the i.v.p.:(71)x′′t+αxt+βx2t+γx3t=0given thatx0=x0andx′0=x˙0.Suppose that the functiony=yt is the solution to the cubic-quintic Duffing equation:(72)y′′t+14α+2Aβ+3A2γyt−13A−x0β+3Aγyt3+38A−x02γyt5=0given thaty0=1andy′0=x˙02x0−A.Then, the function,(73)xt=A+x0−Ay2t,is the solution to the i.v.p. (71) provided that A is a solution to the quartic:(74)3γA4+4βA3+6αA2−6αx02−4βx03−3γx04−6x˙02=0.


## 3.3. Nonlinear Conservative Oscillators

Suppose we are given to sole the i.v.p.:(75)u′′t+gut=0subjected tou0=u0andu′0=u˙0.Assume thatu≤A. We approximate the function g=gu by means of a cubic polynomial on −A≤u≤A using Chebyshev approach so that(76)gu≈a0+a1u+a2u2+a3u3,−A≤u≤A,where(77)a0=cos2π/8g−Asinπ/8+gAsinπ/8−sin2π/8g−Acosπ/8+gAcosπ/82,a1=2−18secπ/8g−Acosπ/8−gAcosπ/8+csc5π/8gAsinπ/8−g−Asinπ/832A,a2=−g−Asinπ/8−gAsinπ/8+g−Acosπ/8+gAcosπ/82A2,a3=2−1csc2π/8secπ/8−g−Acosπ/8+gAcosπ/8+cotπ/8g−Asinπ/8−gAsinπ/84A3.We now replace the i.v.p. (75) with the i.v.p.:(78)u′′t+a0+a1ut+a2u2t+a3u3t=0subjected tou0=u0andu′0=u˙0.Let(79)ut=ρ+xt,wherea0+a1ρ+a2ρ2+aρ3.The problem reduces to the i.v.p.:(80)x′′t+a1+2ρa2+3ρ2a3xt+a2+3ρa3x2t+a3x3t=0given thatx0=u0−ρandx′0=u˙0.This is a Duffing–Helmholtz oscillator (71) with(81)α=a1+2ρa2+3ρ2a3,β=a2+3ρa3andγ=a3.


## 4. Analysis and Discussion

We have solved the cubic-quintic Duffing oscillator equation for any given arbitrary initial conditions. In [1], authors considered the particular case:(82)d2xdt2+a1x+a3x3+a5x5=0given thatx0=A>0anddxdt0=0,under the restrictions,(83)a1≥0,a3≥0anda5≥0.These conditions, however, are too restrictive. In [6], author considered the ansatz:(84)y2t=1a+bcn2ωt+ϕ,k2.This approach does not allow to determine the sign ofyt. On the contrary, this ansatz sometimes gives complex values for ω or k, which makes it difficult to interpret the obtained solution physically. Our approach avoids obtaining such complex values.Other authors solved this equation using perturbative methods [2, 4, 7–9]. In [5], author studied the stability analysis to a cubic-quintic Duffing equation. There are other numerical and analytical methods that allow to solve this oscillator equation.The cubic-quintic Duffing oscillator may also be solved making use of perturbative methods. One of them is the famous Krylov–Bogoliubov–Mitropolsky method (KBM). For example, let us consider the following oscillator:(85)y′′t+ω02yt+αyt3+βyt5=0given thaty0=y0andy′0=y˙0.Using the Krylov–Bogoliubov–Mitropolsky method gives the following approximate analytical solution:(86)yt=Acosω−123αA4β+63α2A2−96αω02+55A6β2−120A2βω023072ω04A3cos3ω+9α2+5A4β2+9αA2β+24βω029216ω04A5cos5ω+β72α+95A2β294912ω04A7cos7ω+β298304ω04A9cos9ω,where(87)ω=ωt=ω0+3α8ω0A2−53α2−16βω02256ω03A4−5αβ64ω03A6−55β23072ω03A8t+B.For the damped oscillator,(88)y′′t+2εy′t+ω02yt+qyt3+ryt5=0given thaty0=y0andy′0=y˙0.The KBM gives(89)yt=atcosψt+4q+5rat2128ω02at3cos3ψt+r384ω02at5cos5ψt,where(90)at=Aexp−εtandψt=ω0t+B+31−e−2tε16εω0qA2+51−e−4tε64εω0rA4.The constantsA and B are determined from the initial conditions. These results are also valid for the cubic Duffing oscillator r=0:(91)y″t+2εy′t+ω02yt+qyt3=0given thaty0=y0andy′0=y˙0.


## 5. Conclusions

The cubic-quintic Duffing oscillator has been solved exactly for arbitrary initial conditions. The obtained results may be applied to solve strongly nonlinear conservative oscillators like the pendulum oscillator equation. We may go further by considering a damped oscillator of the formx¨+2εx˙+Fx=0. In the case when F−x=−Fx, we may approximate the function Fx by means of some cubic-quintic polynomial using Chebyshev approximation formulas. The solution is then assumed in the form xt=exp−ρtyt, where ρ is some parameter having a value near ρ=ε and y=yt is the exact solution to some cubic-quintic Duffing oscillator equation.



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*Source: 9269957-2022-06-15.xml*