# Closed-Form Exact Solutions for the Unforced Quintic Nonlinear Oscillator

Authors: Augusto Beléndez; Enrique Arribas; Tarsicio Beléndez; Carolina Pascual; Encarnación Gimeno; Mariela L. Álvarez
Journal: Advances in Mathematical Physics
 (2017)

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## Abstract

Closed-form exact solutions for the periodic motion of the one-dimensional, undamped, quintic oscillator are derived from the first integral of the nonlinear differential equation which governs the behaviour of this oscillator. Two parameters characterize this oscillator: one is the coefficient of the linear term and the other is the coefficient of the quintic term. Not only the common case in which both coefficients are positive but also all possible combinations of positive and negative values of these coefficients which provide periodic motions are considered. The set of possible combinations of signs of these coefficients provides four different cases but only three different pairs of period-solution. The periods are given in terms of the complete elliptic integral of the first kind and the solutions involve Jacobi elliptic function. Some particular cases obtained varying the parameters that characterize this oscillator are presented and discussed. The behaviour of the periods as a function of the initial amplitude is analysed and the exact solutions for several values of the parameters involved are plotted. An interesting feature is that oscillatory motions around the equilibrium point that is not atx=0 are also considered.


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## Body

## 1. Introduction

Mathematical models based on nonlinear oscillators have been widely used in physics, engineering, applied mathematics, and related fields [1, 2]. These nonlinear systems have been the focus of attention for many years and several methods have been used to find approximate solutions to them [1, 3]. In conservative nonlinear oscillators the restoring force is not dependent on time, the total energy is constant [1], and any oscillation is stationary. The aim of this paper is to obtain closed-form exact periodic solutions to the quintic equation corresponding to a nonlinear oscillator described by a differential equation with fifth-power nonlinearity. Due to the presence of the fifth-power nonlinearity, this oscillator is difficult to handle and has not been studied as extensively as the Duffing oscillator with cubic nonlinearity. For this reason, several techniques have been used to obtain analytical approximate expressions for the period and the solution of the quintic oscillator. Lin [4] proposed a new parameter iteration technique to solve the Duffing equation with strong and higher-order nonlinearity. Ramos [5] approximately solved the quintic equation using some Lindstedt-Poincaré techniques. Pirbodaghi et al. [6] obtained an accurate analytical approximate solution to Duffing equations with cubic and quintic nonlinearities using the homotopy analysis method and homotopy Padé technique. Wu et al. [7] approximately solved this nonlinear oscillator using a method that incorporates salient features of both Newton’s method and the harmonic balance method. Later, Lai et al. [8] used a Newton-harmonic balancing approach to obtain accurate approximate analytical higher-order solutions for strong nonlinear Duffing oscillators with cubic-quintic restoring force. They also discussed the effect of strong quantic nonlinearity on accuracy as compared to cubic nonlinearity. Beléndez et al. [9] approximately solved the quintic oscillator using a cubication method which allowed them to obtain approximate analytical expressions for the period and the solution in terms of elementary functions. Scarpello and Ritelli [10] exactly solved the quintic oscillator, but only when the coefficient for the linear term is equal to one and the coefficient for the nonlinear term is positive. Elías-Zúñiga [11] derived the exact solution of the cubic-quintic Duffing equation based on the use of Jacobi elliptic functions by the same method that he used to obtain the exact solution of the mixed-parity Helmholtz-Duffing oscillator [12]. However, in both cases he did not solve the nonlinear differential equation but assumed that its exact solution is given by a rational equation which includes the cn Jacobian elliptic function and five unknown parameters that need to be determined. Based on his pervious results, Elías-Zúñiga obtained the analytical approximate solution of the damped cubic-quintic Duffing oscillator [13] and also developed a “quintication” method [14] to obtain approximate analytical solutions of conservative nonlinear oscillators. Recently, Beléndez et al. [15] have exactly solved the unforced cubic-quintic Duffing oscillator, providing exact expressions for the period and the solution, but only for oscillations around x=0 and taken into account that the coefficients for the linear and the nonlinear terns are positive.In this paper we obtain the closed-form exact expressions for the period and the solution of the quintic Duffing nonlinear oscillator modelled by the second-order differential equationd2x/dt2+a1x+a5x5=0, where a1 and a5 are the coefficients of the linear and the nonlinear terms, respectively, considering all possible combinations of signs of a1 and a5 that provide oscillatory motions. Unlike the procedure considered by Elías-Zúñiga [11, 12], we do not assume an expression for the solution but solve the nonlinear differential equation exactly as was done in [15]. This is done using elliptic functions so that, after inversion, the solution x is provided as an explicit function of timet. When a1=0 and a5>0, the system becomes a truly nonlinear oscillator [16] for which the exact expressions for the period and the solution have been already obtained [17]. The particular situation in which coefficients a1 and a5 are both positive is the most common case analysed. However, we obtain closed-form exact solutions not only for the case in which both coefficients are positive, but also for all possible combinations of positive and negative values of these coefficients which provide periodic motions. The set of possible combinations of signs of these coefficients gives rise to four different cases. In three of these combinations ((a) a1≥0, a5>0, and x0>0, (b) a1<0, a5>0 and x0>-3a1/a51/4, and (d) a1>0, a5<0 and 0<x0<-a1/a51/4) the system oscillates around the equilibrium position x=0 with x∈-x0,x0, where x0>0 is the oscillation amplitude. However, there is still one more case ((c) a1>0, a5<0, and 0<x0<-3a1/a51/4 or --3a1/a51/4<x0<0) in which the system does not oscillate around the positionx = 0 with x∈[-x0,x0], but around the equilibrium position x=(-a1/a5)1/4 with x∈[x1,x0]. Three different sets of closed-form expressions for the exact period and solution were obtained. Following the procedure considered by Lai and Chow [18], some examples are analysed and plots including periods, solutions, or phase-diagrams are presented and discussed.


## 2. Formulation and Solution Procedure

A quintic oscillator is an example of a conservative autonomous oscillatory system, which is modelled by the following second-order differential equation:(1)d2xdt2+a1x+a5x5=0with initial conditions(2)x0=x0>0,dxdt0=0.In (1) x and t are generalized dimensionless displacement and time variables, and we assume that the coefficients for the linear and the nonlinear terms satisfy. In order to obtain the exact period and periodic solution for (1), we take into account that this is a conservative system and has the following first integral:(3)dxdt2+a1x2+13a5x6=a1x02+13x06which can be written as follows:(4)v2=dxdt2=a1x02-x2+13a5x06-x6≥0.The dynamical study [19] of the nonlinear differential equation given in (1) showed that its motion is periodic in the following situations:(a)
a1≥0, a5>0, and x0>0: the system oscillates around the equilibrium position x=0 and the periodic solution x satisfies x∈-x0,x0.(b)
a1<0, a5>0, and x0>-3a1/a51/4: the system oscillates around the equilibrium position x=0 and the periodic solution x satisfies x∈-x0,x0.(c)
a1<0, a5>0, and 0<x0<(-3a1/a5)1/4: the system oscillates around the equilibrium position x=(-3a1/a5)1/4 and the periodic solution satisfies x∈x1,x0 with 0<x1≤x≤x0 (or around the equilibrium position x=-(-3a1/a5)1/4 with x∈x0,x1 when --3a1/a51/4<x0<0).(d)
a1>0, a5<0, and 0<x0<-a1/a51/4: the system oscillates around the equilibrium position x=0 and the periodic solution x satisfies x∈-x0,x0.The phase plots in Figure 1 illustrate four examples of these situations: (a) a1=1, a5=3, and x0=1, (b) a1=-1, a5=3 with x0=1.1 (x0 must be >1), (c) a1=-1, a5=-3, and x0=0.95 (x0 must be <1), and (d) a1=1, a5=-3, and x0=0.5 (x0 must be <3-1/4≈0.7598). As can be seen in these figures, for cases (a), (b), and (d) the system oscillates around the equilibrium position x=0, whereas for case (c) the system oscillates around the equilibrium position x=-a1/a51/4. In the following sections we obtain the exact expressions for the period and the periodic solution for each of these four cases.Figure 1
Phase space portraits for parameters values of (a)a1=1, a5=3, and x0=1; (b) a1=-1, a5=3, and x0=1.1; (c) a1=-1, a5=3, and x0=0.95; and (d) a1=1, a5=-3, and x0=0.5.
(a) (b) (c) (d)


## 3. Exact Solution Whena1≥0, a5>0, and x0>0

In this situation all the solutions are periodical and the phase space diagram is made up of an infinite number of closed orbits, each of them for each value of the initial amplitudex0 (as can be seen in Figure 1(a)). This system oscillates around the equilibrium position x=0 with x∈-x0,x0, where x0>0 is the initial amplitude and the period,T, and periodic solution, x, are dependent on the oscillation amplitude x0. This system corresponds to a nonlinear oscillator for which the nonlinear function fx=a1x+a3x3+a5x5 is odd; that is, f-x=-fx. From (4) we obtain(5)dtdx2=1a1x02-x2+1/3a5x06-x6and then(6)dt=±dxa1x02-x2+1/3a5x06-x6,where the sign (±) is chosen taking into account the sign of dx/dt in each quadrant.Taking into account thata5>0, from (6) we can write(7)a53dt=±dx3a1/a5x02-x2+x06-x6and after some mathematical operations we obtain(8)a53dt=±dxx02-x2x4+x02x2+3a1/a5+x02.Integrating (8) we obtain(9)a53t=±∫xx0dxx02-x2x4+x02x2+3a1/a5+x02.The change of variable x2=u gives(10)a53t=±∫x2x02dux02-uuu2+x02u+3a1/a5+x02.This is an improper integral which contains a square root of a four-degree polynomial in the denominator and so its solution can be expressed as a function of an elliptic integral.


### 3.1. Calculation of the Exact Period

The symmetry of the problem indicates that the period of the oscillationT is four times the time taken by the oscillator to go from u=0 to u=x02. Therefore, from (10) it follows that(11)T=23a5∫x2x02dux02-uuu2+x02u+3a1/a5+x02.We consider the definite integral [20, section 3.145, formula 2, pages 270-271](12)∫βyduα-uu-βu-σ2+ρ2=1pqF2arccotqα-ypy-β,α-β2-p-q24pq,where β<u<α, Fϕ,m is the incomplete elliptic integral of the first kind [20](13)Fϕ,m=∫0ϕdθ1-msin2θ,and p,q, and m are defined as(14)p=σ-α2+ρ2,q=σ-β2+ρ2,m=α-β2-p-q24pq.By comparing the integrals in (11) and (12) we obtain y=x02, α=x02, and =0, as well as the following values for the different parameters which appear in (12)(15)σ=-12x02,ρ=1212a1a5+3x02,p=3a5a1+a5x04,q=1a53a1+a5x04,m=12-32a1+a5x044a1+a5x043a1+a5x04.As y=α=x02, then F2arccot0,m=Fπ,m=2Km, where Km is the complete elliptic integral of the first kind defined as [20](16)Km=∫0π/2dθ1-msin2θ.From (11) to (16) we conclude that the exact period of the quintic nonlinear oscillator can be written in the compact form(17)T=43q1q31/4Kma,where ma takes the form(18)ma=12-q243q1q31/2and coefficients qn—which depend on a1, a5, and x0—are defined as follows:(19)qn=na1+a5x04.Figure 2 shows the periodT as a function of the initial amplitude x0 (17) when a1=1 and (a) a5=0.1, (b) a5=0.5, (c) a5=5, and (d) a5=50.Figure 2
Exact periodT in (17) as a function of the initial amplitude x0 when a1=1 and (a) a5=0.1, (b) a5=0.5, (c) a5=5, and (d) a5=50.


### 3.2. Calculation of the Exact Solution

From (6) we obtaint as a function ofx for the following cases:Trajectory1→2 (0≤t≤T/4 and x0≥x≥0), x is positive and dx/dt is negative.Trajectory2→3 (T/4≤t≤T/2 and 0≥x≥-x0), x is negative and dx/dt is negative.Trajectory3→4 (T/2≤t≤3T/4 and -x0≤x≤0), x is negative and dx/dt is positive.Trajectory4→1 (3T/4≤t≤T and 0≤x≤x0),x is positive and dx/dt is positive.From (10), it follows that for trajectory 1→2 (0≤t≤T/4 and x0≥x≥0) we have(20)2a53t=∫x2x02dux02-uuu2+x02u+3a1/a5+x02.The definite integral in (20) can be split as follows: (21)2a53t=∫0x02dux02-uuu2+x02u+3a1/a5+x02-∫0x2dux02-uuu2+x02u+3a1/a5+x02.The values of the two integrals on the right-hand side of (21) can be calculated taking into account (11) and (12) and their values are(22)∫0x02dux02-uuu2+x02u+3a1/a5+x02=2a53T4,∫0x2dux02-uuu2+x02u+3a1/a5+x02=1pqF2arccotqA2-x2px2,ma.Substituting (22) into (21) gives (23)2pqa53T4-t=F2arccotqA2-x2px2,ma,and using (15)–(19), (23) can be written as(24)2Kma-q1q331/4t=F2arccotq33q11/4x02-x2x21/2,ma.The inverse function of Fφ,m is the Jacobi amplitude φ [21, 22](25)F-1u,m=φ=amu,mwhose cosine is the Jacobi cosine function, cn [21](26)cosφ=cosamu,m=cnu,m.In order to introduce an “arccos” function in (24) we take into account that(27)2arccot⁡z=φ,where (28)z=q33q11/4x02-x2x21/2.From (28) we obtain(29)tan2φ2=1z2.Taking into account the relation(30)tan2φ2=1-cosφ1+cosφwe can finally write(31)cosφ=z2-1z2+1=q3x02-x2-3q1x2q3x02-x2+3q1x2which allows us to write (24) as follows:(32)2Kma-2q1q331/4t=Farccosq3x02-x2-3q1x2q3x02-x2+3q1x2,maand from (25), (26), and (32) we can write, after some simplifications,(33)q3x02-q3+3q1x2q3x02-q3-3q1x2=cn2Kma-213q1q31/4t,ma=-cn213q1q31/4t,ma,where the relation cn2Km-u,m=-cnu,m [22] has been taken into account.Finally we can write(34)xat=x01+3q1q31-cn1/28/3q1q31/4t,ma1+cn1/28/3q1q21/4t,ma-1/2which is valid for 0≤t≤T/4.From (7), it follows for trajectory 2→3 that(35)2a53∫T/4tdt=2a53t-T4=-∫0x2dux02-uuu2+x02u+3a1/a5+x02and for trajectory 3→4T/4≤t≤T/2(36)2a53∫T/2tdt=2a53t-T2=∫x02x2dux02-uuu2+x02u+3a1/a5+x02.Proceeding in the same manner as for trajectory 1→2, it is follows that xbt=-xat which is valid for T/4≤t≤3T/4, because x<0 for these values of time.Finally, for trajectory4→1 we have(37)2a53∫3T/4tdt=2a53t-3T4=∫0x2dux02-uuu2+x02u+3a1/a5+x02and we obtain the same value for the solution as that given in (34).The exact solution of the quintic oscillator can be written as follows:(38)
x
t
=
x
a
t
0
≤
t
≤
T
4
-
x
a
t
T
4
≤
t
≤
3
T
4
x
a
t
3
T
4
≤
t
≤
TTaking into account the relation [22, formula 16.18.4, page 574](39)1-cn2u1+cn2u=sn2udn2ucn2uand (34) and (38), the exact periodic solution of the quintic oscillator can be written in compact form as follows:(40)xt=x0cn1/3q1q31/4t,macn21/3q1q31/4t,ma+3q1/q31/2sn21/3q1q31/4t,madn21/3q1q31/4t,ma1/2which is valid for all value of t and where cn, sn, and dn are the basic Jacobi elliptic functions [20]. Figure 3 shows the plot of the displacement x as a function of time t when a1=1, a5=3, and x0=1. This displacement was obtained using (40). This figure corresponds to the phase space portrait shown in Figure 1(a).Figure 3
Exact solutionx in (40) as a function of time t when a1=1, a5=3, and x0=1.


## 3.1. Calculation of the Exact Period

The symmetry of the problem indicates that the period of the oscillationT is four times the time taken by the oscillator to go from u=0 to u=x02. Therefore, from (10) it follows that(11)T=23a5∫x2x02dux02-uuu2+x02u+3a1/a5+x02.We consider the definite integral [20, section 3.145, formula 2, pages 270-271](12)∫βyduα-uu-βu-σ2+ρ2=1pqF2arccotqα-ypy-β,α-β2-p-q24pq,where β<u<α, Fϕ,m is the incomplete elliptic integral of the first kind [20](13)Fϕ,m=∫0ϕdθ1-msin2θ,and p,q, and m are defined as(14)p=σ-α2+ρ2,q=σ-β2+ρ2,m=α-β2-p-q24pq.By comparing the integrals in (11) and (12) we obtain y=x02, α=x02, and =0, as well as the following values for the different parameters which appear in (12)(15)σ=-12x02,ρ=1212a1a5+3x02,p=3a5a1+a5x04,q=1a53a1+a5x04,m=12-32a1+a5x044a1+a5x043a1+a5x04.As y=α=x02, then F2arccot0,m=Fπ,m=2Km, where Km is the complete elliptic integral of the first kind defined as [20](16)Km=∫0π/2dθ1-msin2θ.From (11) to (16) we conclude that the exact period of the quintic nonlinear oscillator can be written in the compact form(17)T=43q1q31/4Kma,where ma takes the form(18)ma=12-q243q1q31/2and coefficients qn—which depend on a1, a5, and x0—are defined as follows:(19)qn=na1+a5x04.Figure 2 shows the periodT as a function of the initial amplitude x0 (17) when a1=1 and (a) a5=0.1, (b) a5=0.5, (c) a5=5, and (d) a5=50.Figure 2
Exact periodT in (17) as a function of the initial amplitude x0 when a1=1 and (a) a5=0.1, (b) a5=0.5, (c) a5=5, and (d) a5=50.


## 3.2. Calculation of the Exact Solution

From (6) we obtaint as a function ofx for the following cases:Trajectory1→2 (0≤t≤T/4 and x0≥x≥0), x is positive and dx/dt is negative.Trajectory2→3 (T/4≤t≤T/2 and 0≥x≥-x0), x is negative and dx/dt is negative.Trajectory3→4 (T/2≤t≤3T/4 and -x0≤x≤0), x is negative and dx/dt is positive.Trajectory4→1 (3T/4≤t≤T and 0≤x≤x0),x is positive and dx/dt is positive.From (10), it follows that for trajectory 1→2 (0≤t≤T/4 and x0≥x≥0) we have(20)2a53t=∫x2x02dux02-uuu2+x02u+3a1/a5+x02.The definite integral in (20) can be split as follows: (21)2a53t=∫0x02dux02-uuu2+x02u+3a1/a5+x02-∫0x2dux02-uuu2+x02u+3a1/a5+x02.The values of the two integrals on the right-hand side of (21) can be calculated taking into account (11) and (12) and their values are(22)∫0x02dux02-uuu2+x02u+3a1/a5+x02=2a53T4,∫0x2dux02-uuu2+x02u+3a1/a5+x02=1pqF2arccotqA2-x2px2,ma.Substituting (22) into (21) gives (23)2pqa53T4-t=F2arccotqA2-x2px2,ma,and using (15)–(19), (23) can be written as(24)2Kma-q1q331/4t=F2arccotq33q11/4x02-x2x21/2,ma.The inverse function of Fφ,m is the Jacobi amplitude φ [21, 22](25)F-1u,m=φ=amu,mwhose cosine is the Jacobi cosine function, cn [21](26)cosφ=cosamu,m=cnu,m.In order to introduce an “arccos” function in (24) we take into account that(27)2arccot⁡z=φ,where (28)z=q33q11/4x02-x2x21/2.From (28) we obtain(29)tan2φ2=1z2.Taking into account the relation(30)tan2φ2=1-cosφ1+cosφwe can finally write(31)cosφ=z2-1z2+1=q3x02-x2-3q1x2q3x02-x2+3q1x2which allows us to write (24) as follows:(32)2Kma-2q1q331/4t=Farccosq3x02-x2-3q1x2q3x02-x2+3q1x2,maand from (25), (26), and (32) we can write, after some simplifications,(33)q3x02-q3+3q1x2q3x02-q3-3q1x2=cn2Kma-213q1q31/4t,ma=-cn213q1q31/4t,ma,where the relation cn2Km-u,m=-cnu,m [22] has been taken into account.Finally we can write(34)xat=x01+3q1q31-cn1/28/3q1q31/4t,ma1+cn1/28/3q1q21/4t,ma-1/2which is valid for 0≤t≤T/4.From (7), it follows for trajectory 2→3 that(35)2a53∫T/4tdt=2a53t-T4=-∫0x2dux02-uuu2+x02u+3a1/a5+x02and for trajectory 3→4T/4≤t≤T/2(36)2a53∫T/2tdt=2a53t-T2=∫x02x2dux02-uuu2+x02u+3a1/a5+x02.Proceeding in the same manner as for trajectory 1→2, it is follows that xbt=-xat which is valid for T/4≤t≤3T/4, because x<0 for these values of time.Finally, for trajectory4→1 we have(37)2a53∫3T/4tdt=2a53t-3T4=∫0x2dux02-uuu2+x02u+3a1/a5+x02and we obtain the same value for the solution as that given in (34).The exact solution of the quintic oscillator can be written as follows:(38)
x
t
=
x
a
t
0
≤
t
≤
T
4
-
x
a
t
T
4
≤
t
≤
3
T
4
x
a
t
3
T
4
≤
t
≤
TTaking into account the relation [22, formula 16.18.4, page 574](39)1-cn2u1+cn2u=sn2udn2ucn2uand (34) and (38), the exact periodic solution of the quintic oscillator can be written in compact form as follows:(40)xt=x0cn1/3q1q31/4t,macn21/3q1q31/4t,ma+3q1/q31/2sn21/3q1q31/4t,madn21/3q1q31/4t,ma1/2which is valid for all value of t and where cn, sn, and dn are the basic Jacobi elliptic functions [20]. Figure 3 shows the plot of the displacement x as a function of time t when a1=1, a5=3, and x0=1. This displacement was obtained using (40). This figure corresponds to the phase space portrait shown in Figure 1(a).Figure 3
Exact solutionx in (40) as a function of time t when a1=1, a5=3, and x0=1.


## 4. Exact Solution Whena1<0, a5>0, and x0>-3a1/a51/4

Equations (17) and (40) can be used to obtain the period and the solution in case (b), a1<0, a5>0, and x0>-3a1/a51/4. Figure 4 shows the period T as a function of the initial amplitude x0 (17) when a1=-1 and (a) a5=1, (b) a5=3, (c) a5=10, and (d) a5=50. Figure 5 shows the plot of the displacement x as a function of time t when a1=-1, a5=3, and x0=1.1. This displacement was obtained using (40). This figure corresponds to the phase space portrait shown in Figure 1(b).Figure 4
Exact periodT in (17) as a function of the initial amplitude x0 when a1=-1, x0>-3a1/a51/4 and (a) a5=1, (b) a5=3, (c) a5=10, and (d) a5=50.Figure 5
Exact solutionx in (40) as a function of timet when a1=-1, a5=3, and x0=1.1.


## 5. Exact Solution Whena1<0, a5>0, and 0<x0<-3a1/a51/4

The phase portrait in Figure1(c) shows the behaviour of the oscillator when 0<x0<-3a1/a51/4 (a1=-1, a5=3, and x0=19/20=0.95<1=-3a1/a51/4). As was previously mentioned, now the system oscillates around the equilibrium position x=-a1/a51/4 and the periodic solution satisfies x∈x1,x0 with 0<x1≤x≤x0.


### 5.1. Calculation of the Exact Period

We shall now obtain the period and the solution for the right orbit in Figure1(c) for which x>0 taking into account that x0 is the highest value for x. From (7) and considering that x>0, we obtain t as a function of x as follows:(41)2a53t=∫x2x02dux02-uuu2+x02u+3a1/a5+x02which can be written as(42)2a53t=∫x2x02dux02-uu-u1uu-u2,where u1 and u2 are defined by the following equations: (43)u1=12a5-34a1+a5x04-a5x04=12a5-3q4-q0,(44)u2=-12a5-34a1+a5x04+a5x04=-12a5-3q4+q0,where q2 and q4 are defined in (19).For the right orbit in Figure1(c) it is easy to verify that it satisfies (45)x02≥u≥u1>0>u2.Now the period,T, and the periodic solution, x, are dependent on x0 and x∈[x1,x0], where x1=u1 and u1 is given in (43). The following integral is valid for a≥u≥b>c>d [20, 6th edition, Formula 3.147, Integral 7, page 272]:(46)∫uadya-yy-by-cy-d=2a-cb-dFμ,r,where Fµ,r is the incomplete elliptic integral of the first kind defined in (13) and µ and r are given by the following equations:(47)μ=arcsinb-da-ua-bu-d,r=a-bc-da-cb-d.We have a=x02, b=u1, c=0, and d=u2.Then the value of the integral in (42) is(48)2a53t=2x02u1-u2Farcsinu1-u2x02-x2x02-u1x2-u2,u2u1-x02x02u1-u2.As can be seen in Figure 1(c), the period of the oscillation is twice the time taken by the oscillator to go from u=x02 to u=u1. Therefore,(49)T=2-3q0q41/4Kmc,where mc is given by (50)mc=12+q22-3q0q41/2.q0, q2, and q4 are defined in (19) and Km is the complete elliptic integral of the first kind defined in (16). Figure 6 shows the periodT as a function of the initial amplitude x0 (49) when a1=-1, 0<x0<-3a1/a51/4 and (a) a5=1, (b) a5=3, (c) a5=10, and (d) a5=50. This figure corresponds to the right orbit in the phase space portrait shown in Figure 1(c). Figure 7 shows the variation in the period as a function of the initial position when a1=-1 and a5=3, (a) x0>-3a1/a51/4=1, Section 4 (17), and (b) 0<x0<-3a1/a51/4=1, Section 3 (49). As can be seen the motion is not periodic when the initial position is x=-3a1/a51/4.Figure 6
Exact periodT in (49) as a function of the initial amplitude x0 when a1=-1, 0<x0<-3a1/a51/4 and (a) a5=1, (b) a5=3, (c) a5=10, and (d) a5=50.Figure 7
Exact periodT as a function of the initial position x0 when a1=-1 and a5=3, (a) x0>-3a1/a51/4=1, Section 4 (17), and (b) x0<(-3a1/a5)1/4=1, Section 3 (49). As can be seen the motion is not periodic when the initial position is x=-3a1/a51/4.


### 5.2. Calculation of the Exact Solution

In order to obtain the solutionx as a function of time t, from (48) it follows that(51)u1-u2x02-x2x02-u1x2-u2=sn213a5x02u1-u2t,u2u1-x02x02u1-u2and taking into account (19), (43), (44), and (50), (51) allows us to obtain the solution as follows:(52)xt=x01-1/2+q2/2-3/q0q41/2sn2-1/3q0q41/4t,mc1-1/2-1/2-3q0/q41/2sn2-1/3q0q41/4t,mc1/2.When a1<0 and a5>0, (49) and (52) are valid provided that the initial position x0 satisfies the condition 0<x0<-3a1/a51/4, except for the equilibrium point (when x0=(-a1/a5)1/4).The expressions for the period and the exact solution for the left orbit in Figure1(c), that is, for x<0, can be obtained following the same procedure as that used for x>0 and are not included here.Figure8(a) shows the plot of the displacement x as a function of time t when a1=-1, a5=3, and x0=19/20=0.95. This displacement was obtained using (52). In this example, and from (43), we obtain (53)x1=u1=12012-361+249037≈0.415385.As can easily be verified in Figure 8(a), the oscillatory motion of this system is bounded between x1≈0.415385 and x0=0.95 and the equilibrium point is located at x=-a1/a51/4=3-1/4≈0.759839. Figure 8(b) shows the displacement x as a function of time t when a1=-1, a5=3, and (54)x0=12012-361+249037≈0.415385.In this situation we obtained x1=u1=19/20=0.95.Figure 8
(a) Exact periodic solutionx in (52) for a1=-1, a5=3, and x0=19/20=0.95 and (b) a1=-1, a5=3, and x0=1/201/2-361+249037. In this situation x1=19/20=0.95 is obtained.
(a) (b)It is easy to verify that in this case the system is equivalent to a mixed-parity nonlinear oscillator. The change of variable(55)y=x--a1a51/4gives(56)d2ydt2-4a1y-10-a1a51/33/4y2+10-a1a51/2y3-5-a1a51/31/4y4+a5y5=0.The system modelled by (56) oscillates around the equilibrium position y=0 and the periodic solution y satisfies y∈y1,y0 with y1=x1--a1/a51/4 and y0=x0--a1/a51/4. Figure 9 shows the phase portrait when a1=-1, a5=3, and y0=x0--a1/a51/4=0.95-3-1/4≈0.190164. As can be seen the system oscillates around the equilibrium point y=0.Figure 9
Phase space portrait for the mixed-parity nonlinear oscillator in (56) when a1=-1, a1=3 and y0=x0--a1/a51/4=0.95-3-1/4≈0.190164. As can be seen the system oscillates around the equilibrium point y=0.


## 5.1. Calculation of the Exact Period

We shall now obtain the period and the solution for the right orbit in Figure1(c) for which x>0 taking into account that x0 is the highest value for x. From (7) and considering that x>0, we obtain t as a function of x as follows:(41)2a53t=∫x2x02dux02-uuu2+x02u+3a1/a5+x02which can be written as(42)2a53t=∫x2x02dux02-uu-u1uu-u2,where u1 and u2 are defined by the following equations: (43)u1=12a5-34a1+a5x04-a5x04=12a5-3q4-q0,(44)u2=-12a5-34a1+a5x04+a5x04=-12a5-3q4+q0,where q2 and q4 are defined in (19).For the right orbit in Figure1(c) it is easy to verify that it satisfies (45)x02≥u≥u1>0>u2.Now the period,T, and the periodic solution, x, are dependent on x0 and x∈[x1,x0], where x1=u1 and u1 is given in (43). The following integral is valid for a≥u≥b>c>d [20, 6th edition, Formula 3.147, Integral 7, page 272]:(46)∫uadya-yy-by-cy-d=2a-cb-dFμ,r,where Fµ,r is the incomplete elliptic integral of the first kind defined in (13) and µ and r are given by the following equations:(47)μ=arcsinb-da-ua-bu-d,r=a-bc-da-cb-d.We have a=x02, b=u1, c=0, and d=u2.Then the value of the integral in (42) is(48)2a53t=2x02u1-u2Farcsinu1-u2x02-x2x02-u1x2-u2,u2u1-x02x02u1-u2.As can be seen in Figure 1(c), the period of the oscillation is twice the time taken by the oscillator to go from u=x02 to u=u1. Therefore,(49)T=2-3q0q41/4Kmc,where mc is given by (50)mc=12+q22-3q0q41/2.q0, q2, and q4 are defined in (19) and Km is the complete elliptic integral of the first kind defined in (16). Figure 6 shows the periodT as a function of the initial amplitude x0 (49) when a1=-1, 0<x0<-3a1/a51/4 and (a) a5=1, (b) a5=3, (c) a5=10, and (d) a5=50. This figure corresponds to the right orbit in the phase space portrait shown in Figure 1(c). Figure 7 shows the variation in the period as a function of the initial position when a1=-1 and a5=3, (a) x0>-3a1/a51/4=1, Section 4 (17), and (b) 0<x0<-3a1/a51/4=1, Section 3 (49). As can be seen the motion is not periodic when the initial position is x=-3a1/a51/4.Figure 6
Exact periodT in (49) as a function of the initial amplitude x0 when a1=-1, 0<x0<-3a1/a51/4 and (a) a5=1, (b) a5=3, (c) a5=10, and (d) a5=50.Figure 7
Exact periodT as a function of the initial position x0 when a1=-1 and a5=3, (a) x0>-3a1/a51/4=1, Section 4 (17), and (b) x0<(-3a1/a5)1/4=1, Section 3 (49). As can be seen the motion is not periodic when the initial position is x=-3a1/a51/4.


## 5.2. Calculation of the Exact Solution

In order to obtain the solutionx as a function of time t, from (48) it follows that(51)u1-u2x02-x2x02-u1x2-u2=sn213a5x02u1-u2t,u2u1-x02x02u1-u2and taking into account (19), (43), (44), and (50), (51) allows us to obtain the solution as follows:(52)xt=x01-1/2+q2/2-3/q0q41/2sn2-1/3q0q41/4t,mc1-1/2-1/2-3q0/q41/2sn2-1/3q0q41/4t,mc1/2.When a1<0 and a5>0, (49) and (52) are valid provided that the initial position x0 satisfies the condition 0<x0<-3a1/a51/4, except for the equilibrium point (when x0=(-a1/a5)1/4).The expressions for the period and the exact solution for the left orbit in Figure1(c), that is, for x<0, can be obtained following the same procedure as that used for x>0 and are not included here.Figure8(a) shows the plot of the displacement x as a function of time t when a1=-1, a5=3, and x0=19/20=0.95. This displacement was obtained using (52). In this example, and from (43), we obtain (53)x1=u1=12012-361+249037≈0.415385.As can easily be verified in Figure 8(a), the oscillatory motion of this system is bounded between x1≈0.415385 and x0=0.95 and the equilibrium point is located at x=-a1/a51/4=3-1/4≈0.759839. Figure 8(b) shows the displacement x as a function of time t when a1=-1, a5=3, and (54)x0=12012-361+249037≈0.415385.In this situation we obtained x1=u1=19/20=0.95.Figure 8
(a) Exact periodic solutionx in (52) for a1=-1, a5=3, and x0=19/20=0.95 and (b) a1=-1, a5=3, and x0=1/201/2-361+249037. In this situation x1=19/20=0.95 is obtained.
(a) (b)It is easy to verify that in this case the system is equivalent to a mixed-parity nonlinear oscillator. The change of variable(55)y=x--a1a51/4gives(56)d2ydt2-4a1y-10-a1a51/33/4y2+10-a1a51/2y3-5-a1a51/31/4y4+a5y5=0.The system modelled by (56) oscillates around the equilibrium position y=0 and the periodic solution y satisfies y∈y1,y0 with y1=x1--a1/a51/4 and y0=x0--a1/a51/4. Figure 9 shows the phase portrait when a1=-1, a5=3, and y0=x0--a1/a51/4=0.95-3-1/4≈0.190164. As can be seen the system oscillates around the equilibrium point y=0.Figure 9
Phase space portrait for the mixed-parity nonlinear oscillator in (56) when a1=-1, a1=3 and y0=x0--a1/a51/4=0.95-3-1/4≈0.190164. As can be seen the system oscillates around the equilibrium point y=0.


## 6. Exact Solution Whena1>0, a5<0, and 0<x0<-a1/a51/4

For case (d) it is necessary to obtain the equation for the period again since ifa5<0, the root a5/3 is not a real number and so (7) cannot be used. In order to obtain the exact solution when a1>0, a5<0, and 0<x0<-a1/a51/4 in which case the system oscillates around the equilibrium position x=0 and the periodic solution x satisfies x∈-x0,x0, we proceed as follows. From (6) we can write(57)-a53dt=±dx-3a1/a5x02-x2+x06-x6.


### 6.1. Calculation of the Exact Period

We can consider the same four trajectories that we analysed at the beginning of Section3.2. Doing the change of variable x2=u, it follows that for trajectory 1→2 (0≤t≤T/4 and x0≥x≥0) we have(58)2-a53t=∫x2x02dux02-uu-u2-x02u-3a1/a5-x02which can be written as(59)2-a53t=∫x2x02duw1-ux02-uuu-w2,where w1 and w2 are defined by the following equations: (60)w1=12-a534a1+a5x04--a5x04=12-a53q4--q0,w2=-12-a534a1+a5x04+-a5x04=-12a53q4+-q0.It is easy to verify that it satisfies (61)w1>x02≥u>0>w2.Now the period,T, and the periodic solution, x, are dependent on x0 and x∈[-x0,x0]. The following integral is valid for a>b≥u>c>d [20, 6th edition, Formula 3.147, Integral 5, page 272]:(62)∫ubdya-yb-yy-cy-d=2a-cb-dFκ,q,where F(κ,q) is the incomplete elliptic integral of the first kind defined in (13) and κ and q are given by the following equations:(63)κ=arcsina-cb-ub-ca-u,q=b-ca-da-cb-d.We have a=w1, b=x02, c=0, and d=w2. Then the value of the integral in (59) is(64)2-a53t=2w1x02-w2Farcsinw1x02-x2x02w1-x2,x02w1-w2w1x02-w2.As can be seen in Figure 1(d), the period of the oscillation is four times the time taken by the oscillator to go from u=0 and u=x02. Therefore,(65)T=463q2--3q0q41/2Kmd,where md is given by (66)md=12-3q22-3q0q4-1.q0, q2, and q4 are defined in (19) and K(m) is the complete elliptic integral of the first kind defined in (16). Figure 10 shows the periodT as a function of the initial amplitude x0 (65) when a1=1, 0<x0<(-a1/a5)1/4 and (a) a5=-1, (b) a5=-3, (c) a5=-10, and (d) a5=-50. This figure corresponds to the right orbit in the phase space portrait shown in Figure 1(d).Figure 10
Exact periodT in (65) as a function of the initial amplitude x0 when a1=1, x0<-a1/a51/4 and (a) a5=-1, (b) a5=-3, (c) a5=-10, and (d) a5=-50.


### 6.2. Calculation of the Exact Solution

In order to obtain the solutionx as a function of timet, from (64) it follows that(67)Farcsinw1x02-x2x02w1-x2,x02w1-w2w1x02-w2=-13a5w1x02-w2tand then(68)w1x02-x2x02w1-x2=sn2-13a5w1x02-w2t,x02w1-w2w1x02-w2and taking into account (19), (60), (66), and (68) we finally obtain(69)xt=x0cn1/2q2-1/6-3q0q41/2t,md1+2/1+-3q4/q01/2sn21/2q2-1/6-3q0q41/2t,md1/2.It is easy to verify that (69) is also valid for trajectories 2→3 (T/4≤t≤T/2 and 0≥x≥-x0), 3→4 (T/2≤t≤3T/4 and -x0≤x≤0), and 4→1 (3T/4≤t≤T and 0≤x≤x0), where the exact periodT is given in (65). It is important to point out that when a1>0 and a5<0, (65) and (69) are valid provided that the initial amplitude x0 satisfies the condition 0<x0<-a1/a51/4. Figure 11 shows the plot of the displacement x as a function of timet when a1=1, a5=-3, and x0=0.5. This displacement was obtained using (69). This figure corresponds to the phase space portrait shown in Figure 1(d).Figure 11
Exact solutionx in (69) as a function of timet when a1=1, a5=-3, and x0=0.5.


## 6.1. Calculation of the Exact Period

We can consider the same four trajectories that we analysed at the beginning of Section3.2. Doing the change of variable x2=u, it follows that for trajectory 1→2 (0≤t≤T/4 and x0≥x≥0) we have(58)2-a53t=∫x2x02dux02-uu-u2-x02u-3a1/a5-x02which can be written as(59)2-a53t=∫x2x02duw1-ux02-uuu-w2,where w1 and w2 are defined by the following equations: (60)w1=12-a534a1+a5x04--a5x04=12-a53q4--q0,w2=-12-a534a1+a5x04+-a5x04=-12a53q4+-q0.It is easy to verify that it satisfies (61)w1>x02≥u>0>w2.Now the period,T, and the periodic solution, x, are dependent on x0 and x∈[-x0,x0]. The following integral is valid for a>b≥u>c>d [20, 6th edition, Formula 3.147, Integral 5, page 272]:(62)∫ubdya-yb-yy-cy-d=2a-cb-dFκ,q,where F(κ,q) is the incomplete elliptic integral of the first kind defined in (13) and κ and q are given by the following equations:(63)κ=arcsina-cb-ub-ca-u,q=b-ca-da-cb-d.We have a=w1, b=x02, c=0, and d=w2. Then the value of the integral in (59) is(64)2-a53t=2w1x02-w2Farcsinw1x02-x2x02w1-x2,x02w1-w2w1x02-w2.As can be seen in Figure 1(d), the period of the oscillation is four times the time taken by the oscillator to go from u=0 and u=x02. Therefore,(65)T=463q2--3q0q41/2Kmd,where md is given by (66)md=12-3q22-3q0q4-1.q0, q2, and q4 are defined in (19) and K(m) is the complete elliptic integral of the first kind defined in (16). Figure 10 shows the periodT as a function of the initial amplitude x0 (65) when a1=1, 0<x0<(-a1/a5)1/4 and (a) a5=-1, (b) a5=-3, (c) a5=-10, and (d) a5=-50. This figure corresponds to the right orbit in the phase space portrait shown in Figure 1(d).Figure 10
Exact periodT in (65) as a function of the initial amplitude x0 when a1=1, x0<-a1/a51/4 and (a) a5=-1, (b) a5=-3, (c) a5=-10, and (d) a5=-50.


## 6.2. Calculation of the Exact Solution

In order to obtain the solutionx as a function of timet, from (64) it follows that(67)Farcsinw1x02-x2x02w1-x2,x02w1-w2w1x02-w2=-13a5w1x02-w2tand then(68)w1x02-x2x02w1-x2=sn2-13a5w1x02-w2t,x02w1-w2w1x02-w2and taking into account (19), (60), (66), and (68) we finally obtain(69)xt=x0cn1/2q2-1/6-3q0q41/2t,md1+2/1+-3q4/q01/2sn21/2q2-1/6-3q0q41/2t,md1/2.It is easy to verify that (69) is also valid for trajectories 2→3 (T/4≤t≤T/2 and 0≥x≥-x0), 3→4 (T/2≤t≤3T/4 and -x0≤x≤0), and 4→1 (3T/4≤t≤T and 0≤x≤x0), where the exact periodT is given in (65). It is important to point out that when a1>0 and a5<0, (65) and (69) are valid provided that the initial amplitude x0 satisfies the condition 0<x0<-a1/a51/4. Figure 11 shows the plot of the displacement x as a function of timet when a1=1, a5=-3, and x0=0.5. This displacement was obtained using (69). This figure corresponds to the phase space portrait shown in Figure 1(d).Figure 11
Exact solutionx in (69) as a function of timet when a1=1, a5=-3, and x0=0.5.


## 7. Discussion

In this section we briefly discuss the derived solutions presented in this paper compared to the exact one derived by Elías-Zúñiga [11], providing a discussion in which both solutions are compared for all the cases discussed in the manuscript. As it was pointed out in the introduction, Elías-Zúñiga does not solve the nonlinear differential equation exactly as we have done here, but he assumes an expression for the solution. He proposed a solution for (1) with the initial conditions given in (2) which can be written as follows:(70)x2t=1a+bcn2ωt,m.Substituting (70) into (1) it is possible to obtain the values for a,b,ω, and m, whose expressions in terms of a1, a5, and x0 are(71)a=-2a5a5x02±3-a54a1a5+a5x04,b=1-ax02x02,m=1-a2x041+2ax02,ω=-a1+2aa1x021+ax02+a2x04.Elías-Zúñiga also concluded that the corresponding exact period of oscillation T of this nonlinear oscillator is given by [11, Eq. (15), page 2576](72)T=4Kmωand pointed out that, depending on the system parameter values of a1, a5, and x0, we can have real, complex, or imaginary values for a,b,ω, and m. Now we are going to compare the exact period and solutions presented in this paper with Elías-Zúñiga’s paper for all the cases discussed in the manuscript.


### 7.1. Exact Solution Whena1≥0, a5>0, and x0>0

We have obtained the exact period given in (17) and we have written the exact periodic solution of the quintic oscillator in the compact form given in (40). For this case, Elías-Zúñiga exact period is given in (72) and the exact solution has to be written as a piecewise function as follows:(73)
x
t
=
a
+
b
c
n
2
ω
t
,
m
-
1
/
2
0
≤
t
≤
T
4
-
a
+
b
c
n
2
ω
t
,
m
-
1
/
2
T
4
≤
t
≤
3
T
4
a
+
b
c
n
2
ω
t
,
m
-
1
/
2
3
T
4
≤
t
≤
T
.When a1=1, a5=3, and x0=1 we obtain(74)a=-0.25±0.661438i,b=1.25∓0.661438i,ω=0.405233∓1.63224i,m=0.5625±0.826797iand (72) gives for the period of this nonlinear oscillator the following value:(75)T=1.89604±3.37321iwhich is not a real number and cannot correspond to a period of an oscillatory system. The period we have obtained in this paper by using (17) isT = 3.79208 s, which is the correct period. Substituting the values for a,b,ω, and m given in (74) into (73) and plotting x as a function of time, we obtain Figure 3, and it is possible to conclude that the solution given in (40) and the piecewise solution given in (73) are the same exact solution. However, the equation for the exact period given in (72) [11, Eq. (15), page 2576] does not give the exact period.


### 7.2. Exact Solution Whena1<0, a5>0, and x0>(-3a1/a5)1/4

The exact period is given in (17) and the exact periodic solution is given in (40). For this case, Elías-Zúñiga exact period is given in (72) and the exact solution is written as a piecewise function given in (73).Whena1=-1, a5=3, and x0=1.1 we obtain(76)a=-1.3036±0.674788i,b=2.13004∓0.674788i,ω=0.171154∓1.107i,m=0.817588±0.575804iand (72) gives for the period of this nonlinear oscillator the following value:(77)T=2.8212±5.85337iwhich is not a real number and cannot correspond to a period of a physical oscillatory system. The period we have obtained in this paper by using (17) isT = 5.64241 s, which is the correct period. Substituting the values for a,b,ω, and m given in (76) into (73) and plotting x as a function of time, we obtain Figure 5, and it is possible to conclude that the solution given in (40) and the piecewise solution given in (73) are the same exact solution. However, the equation for the exact period given in (72) [11, Eq. (15), page 2576] does not give the exact period.


### 7.3. Exact Solution Whena1<0, a5>0, and 0<x0<-3a1/a51/4

We have obtained the exact period given in (49) and the exact periodic solution given in (52). For this case, Elías-Zúñiga exact period is given in (72) and the exact solution can be written as follows:(78)xt=1a+bcn2ωt,m.When a1=-1, a5=3, and x0=0.95 we obtain(79)a=-0.930194,b=2.03823,ω=0.885853i,m=-0.434817and (72) gives for the period of this nonlinear oscillator the following value:(80)T=-6.46749iwhich is an imaginary number and cannot correspond to a period of an oscillatory system. The period we have obtained in this paper by using (49) isT = 3.90327 s, which is the correct period. Substituting the values for a,b,ω, and m given in (79) into (78) and plotting x as a function of time, we obtain Figure 8(a), and it is possible to conclude that the solution given in (52) and the solution given in (78) are the same exact solution. However, the equation for the exact period given in (72) [11, Eq. (15), page 2576] does not give the exact period.


### 7.4. Exact Solution Whena1>0, a5<0, and 0<x0<-a1/a51/4

We have obtained the exact period given in (65) and the exact periodic solution given in (69). For this case, Elías-Zúñiga exact period is given in (72) and the exact solution is written as a piecewise function given in (73).Whena1=-1, a5=3, and x0=0.5 we obtain(81)a=1.1747,b=2.8253,ω=1.07253i,m=0.575648and (72) gives for the period of this nonlinear oscillator the following value:(82)T=-7.80654iwhich is an imaginary number and cannot correspond to a period of an oscillatory system. The period we have obtained in this paper by using (65) isT = 6.69369 s, which is the correct period. Substituting the values for a,b,ω, and m given in (81) into (73) and plotting x as a function of time, we obtain Figure 11, and it is possible to conclude that the solution we obtained (69) and the solution given in (73) are the same exact solution. However, the equation for the exact period given in (72) [11, Eq. (15), page 2576] does not give the exact period.


### 7.5. Why the Period Given in [11, Eq. (15), page 2576] Does Not Give the Exact Period?

As we have previously pointed out, Elías-Zúñiga concludes that “since the Jacobian elliptic function cn(ωt,m) has a period in ωt equal to 4Km, we may see that the corresponding exact period of oscillationT is given by T=4Km/ω” [11, page 2576]. As we have just proved, this equation for the period gives complex and imaginary values for the period of a physical oscillator but not real numbers. However, if Elías-Zúñiga’s equations for the exact solution have been proven correct, where is the error in the expression for the period? The answer of this question is that not only the Jacobian elliptic function cn(ωt,m) has a period in ωt equal to 4Km, but also periods 4iK′m and 2Km+2iK′m, where K′m=K1-m [22].Whena1=1, a5=3, and x0=1, Elías-Zúñiga’s period would have to be(83)T=-4iK′mω=-4iK1-mω=3.79208swhich gives the correct period. When a1=-1, a5=3, and x0=1.1, Elías-Zúñiga’s period would have to be(84)T=-4iK′mω=-4iK1-mω=5.64241swhich corresponds to the correct period for these numerical parameters. When a1=-1, a5=3, and x0=0.95, the correct expression for Elías-Zúñiga’s period would have to be(85)T=-2Km+2iK′mω=-2Km+2iK1-mω=3.90327swhich is the correct period. Finally, when a1=1, a5=-3, and x0=0.5, the correct expression for Elías-Zúñiga’s period would have to be(86)T=4iK′mω=4iK1-mω=6.69369swhich corresponds to the correct period for these parameters.


## 7.1. Exact Solution Whena1≥0, a5>0, and x0>0

We have obtained the exact period given in (17) and we have written the exact periodic solution of the quintic oscillator in the compact form given in (40). For this case, Elías-Zúñiga exact period is given in (72) and the exact solution has to be written as a piecewise function as follows:(73)
x
t
=
a
+
b
c
n
2
ω
t
,
m
-
1
/
2
0
≤
t
≤
T
4
-
a
+
b
c
n
2
ω
t
,
m
-
1
/
2
T
4
≤
t
≤
3
T
4
a
+
b
c
n
2
ω
t
,
m
-
1
/
2
3
T
4
≤
t
≤
T
.When a1=1, a5=3, and x0=1 we obtain(74)a=-0.25±0.661438i,b=1.25∓0.661438i,ω=0.405233∓1.63224i,m=0.5625±0.826797iand (72) gives for the period of this nonlinear oscillator the following value:(75)T=1.89604±3.37321iwhich is not a real number and cannot correspond to a period of an oscillatory system. The period we have obtained in this paper by using (17) isT = 3.79208 s, which is the correct period. Substituting the values for a,b,ω, and m given in (74) into (73) and plotting x as a function of time, we obtain Figure 3, and it is possible to conclude that the solution given in (40) and the piecewise solution given in (73) are the same exact solution. However, the equation for the exact period given in (72) [11, Eq. (15), page 2576] does not give the exact period.


## 7.2. Exact Solution Whena1<0, a5>0, and x0>(-3a1/a5)1/4

The exact period is given in (17) and the exact periodic solution is given in (40). For this case, Elías-Zúñiga exact period is given in (72) and the exact solution is written as a piecewise function given in (73).Whena1=-1, a5=3, and x0=1.1 we obtain(76)a=-1.3036±0.674788i,b=2.13004∓0.674788i,ω=0.171154∓1.107i,m=0.817588±0.575804iand (72) gives for the period of this nonlinear oscillator the following value:(77)T=2.8212±5.85337iwhich is not a real number and cannot correspond to a period of a physical oscillatory system. The period we have obtained in this paper by using (17) isT = 5.64241 s, which is the correct period. Substituting the values for a,b,ω, and m given in (76) into (73) and plotting x as a function of time, we obtain Figure 5, and it is possible to conclude that the solution given in (40) and the piecewise solution given in (73) are the same exact solution. However, the equation for the exact period given in (72) [11, Eq. (15), page 2576] does not give the exact period.


## 7.3. Exact Solution Whena1<0, a5>0, and 0<x0<-3a1/a51/4

We have obtained the exact period given in (49) and the exact periodic solution given in (52). For this case, Elías-Zúñiga exact period is given in (72) and the exact solution can be written as follows:(78)xt=1a+bcn2ωt,m.When a1=-1, a5=3, and x0=0.95 we obtain(79)a=-0.930194,b=2.03823,ω=0.885853i,m=-0.434817and (72) gives for the period of this nonlinear oscillator the following value:(80)T=-6.46749iwhich is an imaginary number and cannot correspond to a period of an oscillatory system. The period we have obtained in this paper by using (49) isT = 3.90327 s, which is the correct period. Substituting the values for a,b,ω, and m given in (79) into (78) and plotting x as a function of time, we obtain Figure 8(a), and it is possible to conclude that the solution given in (52) and the solution given in (78) are the same exact solution. However, the equation for the exact period given in (72) [11, Eq. (15), page 2576] does not give the exact period.


## 7.4. Exact Solution Whena1>0, a5<0, and 0<x0<-a1/a51/4

We have obtained the exact period given in (65) and the exact periodic solution given in (69). For this case, Elías-Zúñiga exact period is given in (72) and the exact solution is written as a piecewise function given in (73).Whena1=-1, a5=3, and x0=0.5 we obtain(81)a=1.1747,b=2.8253,ω=1.07253i,m=0.575648and (72) gives for the period of this nonlinear oscillator the following value:(82)T=-7.80654iwhich is an imaginary number and cannot correspond to a period of an oscillatory system. The period we have obtained in this paper by using (65) isT = 6.69369 s, which is the correct period. Substituting the values for a,b,ω, and m given in (81) into (73) and plotting x as a function of time, we obtain Figure 11, and it is possible to conclude that the solution we obtained (69) and the solution given in (73) are the same exact solution. However, the equation for the exact period given in (72) [11, Eq. (15), page 2576] does not give the exact period.


## 7.5. Why the Period Given in [11, Eq. (15), page 2576] Does Not Give the Exact Period?

As we have previously pointed out, Elías-Zúñiga concludes that “since the Jacobian elliptic function cn(ωt,m) has a period in ωt equal to 4Km, we may see that the corresponding exact period of oscillationT is given by T=4Km/ω” [11, page 2576]. As we have just proved, this equation for the period gives complex and imaginary values for the period of a physical oscillator but not real numbers. However, if Elías-Zúñiga’s equations for the exact solution have been proven correct, where is the error in the expression for the period? The answer of this question is that not only the Jacobian elliptic function cn(ωt,m) has a period in ωt equal to 4Km, but also periods 4iK′m and 2Km+2iK′m, where K′m=K1-m [22].Whena1=1, a5=3, and x0=1, Elías-Zúñiga’s period would have to be(83)T=-4iK′mω=-4iK1-mω=3.79208swhich gives the correct period. When a1=-1, a5=3, and x0=1.1, Elías-Zúñiga’s period would have to be(84)T=-4iK′mω=-4iK1-mω=5.64241swhich corresponds to the correct period for these numerical parameters. When a1=-1, a5=3, and x0=0.95, the correct expression for Elías-Zúñiga’s period would have to be(85)T=-2Km+2iK′mω=-2Km+2iK1-mω=3.90327swhich is the correct period. Finally, when a1=1, a5=-3, and x0=0.5, the correct expression for Elías-Zúñiga’s period would have to be(86)T=4iK′mω=4iK1-mω=6.69369swhich corresponds to the correct period for these parameters.


## 8. Conclusions

Closed-form expressions for the exact periods and solutions of the nonlinear quintic oscillator have been obtained for all possible oscillatory motions of this nonlinear system. Unlike Elías-Zúñiga’s procedure [11, 12], we do not assume any expression for the solution but exactly solve the nonlinear differential equation, which allows us to obtain the period and, after inversion, the solution for this conservative nonlinear oscillator. As shown, there are four possible combinations of coefficients a1 and a5 which provide periodic motions. In three of them ((a) a1≥0, a5>0, and x0>0, (b) a1<0, a5>0, and x0>-3a1/a51/4>0, and (d) a1>0, a5<0, and 0<x0<-a1/a51/4) the system oscillates around the equilibrium positionx = 0 with x∈-x0,x0. However, there is one more case ((c) a1>0, a5<0, and 0<x0<-3a1/a51/4) in which the system oscillates around the equilibrium position x=-a1/a51/4 with x∈x1,x0. In all cases the exact periods are given in terms of a complete elliptic integral of the first kind and the exact solutions are expressed in terms of Jacobi elliptic functions. We also showed that making a convenient change of variable to a new variable y=x--a1/a51/4 it is possible to verify that case (c) is equivalent to a mixed-parity nonlinear oscillator oscillating around the equilibrium position y=0. Finally, our exact solutions are compared with the exact ones obtained by Elías-Zúñiga [11].



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*Source: 7396063-2017-02-02.xml*