# Approximation of Elliptic Functions by Means of Trigonometric Functions with Applications

Authors: Alvaro H. Salas; Lorenzo J. H. Martinez; David L. R. Ocampo R.
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

In this work, we give approximate expressions for Jacobian and elliptic Weierstrass functions and their inverses by means of the elementary trigonometric functions, sine and cosine. Results are reasonably accurate. We show the way the obtained results may be applied to solve nonlinear ODEs and other problems arising in nonlinear physics. The importance of the results in this work consists on giving easy and accurate way to evaluate the main elliptic functions cn, sn, and dn, as well as the Weierstrass elliptic function and their inverses. A general principle for solving some nonlinear problems through elementary functions is stated. No similar approach has been found in the existing literature.


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## Body

## 1. Introduction

Historically, elliptic functions were first discovered by Niels Henrik Abel [2] as inverse functions of elliptic integrals, and their theory was improved by Carl Gustav Jacobi; these in turn were studied in connection with the problem of the arc length of an ellipse, whence the name derives. Jacobi’s elliptic functions have found numerous applications in physics and were used by Jacobi to prove some results in elementary number theory. The solution of the quintic algebraic equation was provided by Hermite. His famous solution of the general quintic equation appeared in Sur la ré solution de l’équation du cinquième degré (1858; “On the Solution of the Equation of the Fifth Degree”). Many late 19th century mathematicians first gained recognition for their work largely through the encouragement and publicity supplied by Hermite. In general, any entire algebraic equation of nth degree may be expressed through elliptic theta functions [3]. The elliptic functions have a variety of applications in science and technology. These functions are the main tool for solving nonlinear differential equations. However, these functions, as well as elliptic integrals, are difficult to evaluate. Sometimes, a nonlinear problem is replaced by linear one. However, this approach reduces the accuracy of the desired solution. The main Jacobian elliptic functions may be expressed by means of trigonometric series. Truncating these series, we get approximate analytic expressions in terms of the elementary trigonometric function. Thus, we have the following series representations [1]:(1)cnz,m=2πmKm∑n=0∞qmn+1/21+qm2n+1cos2n+1πz2Km,snz,m=2πmKm∑n=0∞qmn+1/21−qm2n+1sin2n+1πz2Km,dnz,m=π2Km+2πKm∑n=1∞qmn1+qm2ncosnπzKm.There are also infinite product representations. They read as [1](2)cnz,m=2qm41−m4m4cosπz2Km∏n=1∞1+2qm2ncosπz/Km+qm4n1−2qm2n−1cosπz/Km+qm4n−2,snz,m=2qm4m4sinπz2Km∏n=1∞1−2qm2ncosπz/Km+qm4n1−2qm2n−1cosπz/Km+qm4n−2,dnz,m=1−m4∏n=1∞1+2qm2n−1cosπz/Km+qm4n−21−2qm2n−1cosπz/Km+qm4n−2.Here,(3)qm=explogm∑k=0∞1/2k2mk/k!2−2∑k=0∞1/2k2−ψk+1/2+ψk+1mk/k!2∑k=0∞1/2k1/2kmk/k!2.On the contrary, there is an important elliptic function, the℘-function introduced by Karl Weierstrass. Applications of Weierstrass functions include integrable nonlinear differential equations, motion in cubic and quartic potentials, description of the movement of a spherical pendulum, and construction of minimal surfaces. This function allows to solve the following nonlinear ode called Duffing–Helmholtz equation:(4)x¨+n+px+qx2+rx3=0,x0=x0,x′0=x˙0.The Elliptic expansion function method [5, 6] is based on the solution of (4). The Weierstrass elliptic function is related to Jacobian elliptic function sn and cn and vice versa. However, sometimes, it is more convenient using Weierstrass elliptic function rather than Jacobian ones. For example, it is easier to solve the Duffing–Helmholtz equation (4) by means of the ℘-function. We have the following trigonometric representation [1]:(5)℘z;g2,g3=−η1ω1+π2ω12csc2πz2ω1−2π2ω12∑k=1∞kq2k1−q2kcoskπzω1.However, even if we truncate the series or the product, the approximation will involve other quantities (Km, qm, and η1 ) that are not elementary quantities. Many identities related to elliptic functions and elliptic integrals may be found in [1, 4].Taking into account these difficulties, there is a need to obtain elementary and reasonable accurate approximate analytical expressions for these functions and their inverses.In this paper, our aim consists on providing simple tools for evaluating such functions through elementary trigonometric functions. These representations will allow to estimate the main period in terms of either the modulus or elliptic invariants. We also illustrate the obtained results in the solution of some nonlinear equations arising in mathematical physics. Our main idea is to choose some ansatz that will satisfy the nonlinear ODE associated to a given elliptic function in the best possible way.


## 2. Trigonometric Approximation of the Main Elliptic Jacobian Functions

The three main Jacobian elliptic functions are cn = cnt,m, sn = sn t,m, and dn = dn t,m. Functions cn and sn are called elliptic cosine (generalized cosine) and ellipic sine (generalized sine), respectively. Their associated odes are(6)cn″t,m+2mcn3t,m+1−2mcnt,m=0,cn0,m=1,cn′0,m=0,sn″t−2msnt3+m+1snt=0,sn0,m=0,sn′0,m=1.The functions sn and cn have period:(7)T=4Km.We have the following approximation formulas:(8)Km≈5π18−32π99m−16.The error for−1≤m≤1/4 is given by(9)max−1≤m≤1/4Km−5π18−32π99m−16=0.00844.This formula allows us to solve the equationKm=μ for 1.31≤μ≤1.68 as follows:(10)m=169−64π918μ−5π.A better approximation is obtained by means of the formula(11)Km≈540191π3416182−6470621467805πm2−271301025968πm+224946883200π17080911708091m3−16837184m2+38748416m−24653824.The error for−1≤m≤0.8 is given by(12)max−1≤m≤1/4540191π3416182−6470621467805πm2−271301025968πm+224946883200π17080911708091m3−16837184m2+38748416m−24653824−Km=0.00275195.This formula allows us to solve the equationKm=μ for 1.31≤μ≤2.25721 by solving the following cubic:(13)540191π−3416182μm3−64165891π−526162μm2+256127285π−302722μm−24653824π−2μ=0.The function dn = dnt,m is a solution to the ODE:(14)dn″t+m−2dnm−2+2dnt3=0,dn0,m=1,dn′0,m=0.On the contrary, the elliptic Weierstrass function℘=℘t;g2,g3 with elliptic invariants g2 and g3 obeys the ODE:(15)℘′2=4℘3−g2℘−g3.The main period of the function℘ is given by(16)T=2∫a+∞dx4x3−g2x−g3,where a is the greatest real root to the cubic:(17)4x3−g2x−g3=0.Suppose thata is the greatest real root of cubic (17) and let b and −a+b be the other two roots. Then,(18)T=2∫a+∞dx4x3−g2x−g3=2∫a+∞dx4x−ax−bx+a+b=2a−bK3ab−a+2.In general, ifRea>maxReb,Rec, then(19)∫audx4x−ax−bx−c=2iFisinh−1a−b/u−a|a−c/a−b+2K−b+c/a−ba−b,foru>Rea.


### 2.1. Approximation for Elliptic Cosine and Elliptic Sine

Define the generalized cosine and sine functions as follows:(20)cosλt=1+λcos1+λt1+λcos21+λt,sinλt=sin1+λt1+λcos21+λt.Our aim is to find someλ so that(21)cnt,m≈cosλt.Define(22)Rt=12mxt4+121−2mxt2+12x′t2+12m−1.Observe thatRt=0 when xt=cnt,m. Let xt=cosλt. We have(23)Rt=12λcos2λ+1t+1313214λ2+2m+12λ+12m132−17λ2−32λ−λm−16mcos2λ+1t+1322λ2+8λ−2λm+4mcos4λ+1t+132λ2+λmcos6λ+1t.We will chooseλ so that(24)14λ2+2m+12λ+12m=0.Define(25)λ=114m2−144m+144−m−12.The obtained approximations are good. This is seen from Table1.Table 1
Errors for approximations of Jacobian elliptic functions by means of trigonometric functions.
mmax−T≤t≤Tcnt,m−cosmtmax−T≤t≤Tsnt,m−sinmt−10.006636680.00397374−0.90.005615630.00337397−0.80.004638480.00280023−0.70.003715820.0022567−0.60.002859370.00174918−0.50.00208230.00128479−0.40.001399520.000872175−0.30.0008281080.00052203−0.20.0003879270.000247775−0.10.0001024610.00006643330000.10.0001154010.00007756370.20.000492760.0003383990.30.001189670.0008372070.40.002284330.00165310.50.003888360.002900540.60.00617190.004748480.70.009419550.007482090.80.01418140.01165060.90.02184770.018634We thus have(26)cnt,m≈cosmt≔1+λcos1+λt1+λcos21+λt,(27)snt,m≈sinmt≔sin1+λt1+λcos21+λt,(28)dnt,m≈dnmt≔1−msinm2t.The numberλ is given by (25). We now give approximate expressions for inverse elliptic cosine and inverse elliptic cosine as follows:(29)cn−1t,m≈cosm−1t=1λ+1cos−1tλ−λt2+1,−1≤t≤1,(30)sn−1t,m≈sinm−1t=sgntλ+1cos−11−t21+λt2,−1≤t≤1.For the error concerning these approximations, see Table2.Table 2
Errors for approximations of inverse Jacobian elliptic functions by means of inverse trigonometric functions.
mmax−1≤t≤1cn−1t,m−cosm−1tmax−1≤t≤1sn−1t,m−sinm−1t0000.10.0001863390.0001566730.20.0008584470.0007228530.30.002255280.001903140.40.004762050.004038920.50.00903320.007715150.60.01626030.01407410.70.02882840.02551690.80.05218580.04828570.90.1059810.105797In the case whenm>1, we use the approximations(31)cnt,m≈dn1/mmt,snt,m≈1msin1/mmt,dnt,m≈cos1/mmt.Example 1.
Suppose we are given to find the least positive numbert such that(32)cnt,14=13.
The exact solution is(33)t=cn−113,14=1.2957603108659352.
On the contrary, using formula (29), we obtain(34)cos1/4−113=1.296968374157.
The error is 0.00121.


## 2.1. Approximation for Elliptic Cosine and Elliptic Sine

Define the generalized cosine and sine functions as follows:(20)cosλt=1+λcos1+λt1+λcos21+λt,sinλt=sin1+λt1+λcos21+λt.Our aim is to find someλ so that(21)cnt,m≈cosλt.Define(22)Rt=12mxt4+121−2mxt2+12x′t2+12m−1.Observe thatRt=0 when xt=cnt,m. Let xt=cosλt. We have(23)Rt=12λcos2λ+1t+1313214λ2+2m+12λ+12m132−17λ2−32λ−λm−16mcos2λ+1t+1322λ2+8λ−2λm+4mcos4λ+1t+132λ2+λmcos6λ+1t.We will chooseλ so that(24)14λ2+2m+12λ+12m=0.Define(25)λ=114m2−144m+144−m−12.The obtained approximations are good. This is seen from Table1.Table 1
Errors for approximations of Jacobian elliptic functions by means of trigonometric functions.
mmax−T≤t≤Tcnt,m−cosmtmax−T≤t≤Tsnt,m−sinmt−10.006636680.00397374−0.90.005615630.00337397−0.80.004638480.00280023−0.70.003715820.0022567−0.60.002859370.00174918−0.50.00208230.00128479−0.40.001399520.000872175−0.30.0008281080.00052203−0.20.0003879270.000247775−0.10.0001024610.00006643330000.10.0001154010.00007756370.20.000492760.0003383990.30.001189670.0008372070.40.002284330.00165310.50.003888360.002900540.60.00617190.004748480.70.009419550.007482090.80.01418140.01165060.90.02184770.018634We thus have(26)cnt,m≈cosmt≔1+λcos1+λt1+λcos21+λt,(27)snt,m≈sinmt≔sin1+λt1+λcos21+λt,(28)dnt,m≈dnmt≔1−msinm2t.The numberλ is given by (25). We now give approximate expressions for inverse elliptic cosine and inverse elliptic cosine as follows:(29)cn−1t,m≈cosm−1t=1λ+1cos−1tλ−λt2+1,−1≤t≤1,(30)sn−1t,m≈sinm−1t=sgntλ+1cos−11−t21+λt2,−1≤t≤1.For the error concerning these approximations, see Table2.Table 2
Errors for approximations of inverse Jacobian elliptic functions by means of inverse trigonometric functions.
mmax−1≤t≤1cn−1t,m−cosm−1tmax−1≤t≤1sn−1t,m−sinm−1t0000.10.0001863390.0001566730.20.0008584470.0007228530.30.002255280.001903140.40.004762050.004038920.50.00903320.007715150.60.01626030.01407410.70.02882840.02551690.80.05218580.04828570.90.1059810.105797In the case whenm>1, we use the approximations(31)cnt,m≈dn1/mmt,snt,m≈1msin1/mmt,dnt,m≈cos1/mmt.Example 1.
Suppose we are given to find the least positive numbert such that(32)cnt,14=13.
The exact solution is(33)t=cn−113,14=1.2957603108659352.
On the contrary, using formula (29), we obtain(34)cos1/4−113=1.296968374157.
The error is 0.00121.


## 3. Approximation for Elliptic Weierstrass Function

It is known that(35)cnωt,m=1−64m+11+12/4m+1ω℘t;1/1216m2−16m+1ω2,1/2162m−132m2−32m−1ω3.On the contrary,(36)℘t;g2,g3=−g24m+1248m2−48m+3+3g216m2−16m+1⋅11−cn23g2/16m2−16m+14t,m,where(37)m=142−ζ+3.Here,ζ is the least in magnitude root of the cubic:(38)4g23−27g32z3−27g23z+23g23+112g32=0.In the case wheng23−27g32=0, we will have only one real root:(39)z=733g3227g23.Then,(40)m=12−154730≈−3.4282×10−4.Letg23−27g32≠0. The discriminant of cubic (38) reads(41)Δ=432g23−27g32200g29+9131g32g26+126560g34g23+338688g36.


### 3.1. First Case:Δ>0

The cubic equation (38) has three real roots. They are(42)zk=3g23/2g23−27g32cos132πk+cos−1−g23−27g3223g23+112g3227g29/2,k=−1,0,1.


### 3.2. Second Case:Δ<0

The cubic equation (38) has only one real root. It is given by(43)ζ=Δ−g23−27g32223g23+112g322/3+9g26−243g32g232g23−27g32Δ−g23−27g32223g23+112g323.


### 3.3. Third Case:Δ=0

Suppose thatg23−27g32=0. We have two real roots for cubic equation (38):(44)z1=1223g23+112g32g23−27g323,z2=−23g23+112g32g23−27g323.It follows from (36) that the period of Weierstrass elliptic function reads(45)T=2∫a+∞dx4x3−g2x−g3=22ζ3g24K142−ζ+3,where a is the greatest real root to the cubic:(46)4x3−g2x−g3=0.Using identity (36) and approximation (26), we have the following approximation:(47)℘t;g2,g3≈−g24m+1248m2−48m+3+3g216m2−16m+11−cosm23g2/16m2−16m+14t,where m is given by (37).Example 2.
Letg2=2 and g3=1. For this choice, m=0.0119056. We have(48)℘t;2,1≈−0.474691+2.718671−3.73048cos2.32484t/14.−0.0835474cos22.32484t.
The period isT=2.70262 and the error on the interval −T/2≤t≤T/2 in the sup norm is E=8.52×10−7, see Figure 1. The reciprocals of the two functions are plotted on the interval −T/2≤t≤T/2.
We now give approximate expression for inverse elliptic Weierstrass function as follows. Let(49)a=−g24m+1248m2−48m+3,b=3g216m2−16m+1.
We know that(50)℘≈a+b1−cosm23g2/16m2−16m+14t.
Now, we solve this equation fort to obtain(51)t≈±16m2−16m+14234g24λ+1cos−1±a+b−℘a2−2abλ−b2λ−2a−bλ℘+℘2,so(52)℘−1t;g2,g3≈±16m2−16m+14234g24λ+1cos−1±a+b−ta2−2abλ−b2λ−2a−bλt+℘2.Figure 1
Comparison between the approximate trigonometric solution and the Runge–Kutta numerical solution for Example2.Example 3.
Suppose we are given that(53)℘t;−2,−5=15.
The exact solution is(54)t=℘−115;−2,−5=−1.210122529012178.
Using (52), we obtain four values:(55)t1=−1.20908,t2=1.20908,t3=−0.763507,t4=0.763507.
We chooset=−1.20908.
From all previous facts, we may establish the following.


#### 3.3.1. General Principle

If the solution to some problem may be written in terms of either Jacobian elliptic functions or in terms of the Weierstrass elliptic function, then such solution may be approximated by means of trigonometric functions.


## 3.1. First Case:Δ>0

The cubic equation (38) has three real roots. They are(42)zk=3g23/2g23−27g32cos132πk+cos−1−g23−27g3223g23+112g3227g29/2,k=−1,0,1.


## 3.2. Second Case:Δ<0

The cubic equation (38) has only one real root. It is given by(43)ζ=Δ−g23−27g32223g23+112g322/3+9g26−243g32g232g23−27g32Δ−g23−27g32223g23+112g323.


## 3.3. Third Case:Δ=0

Suppose thatg23−27g32=0. We have two real roots for cubic equation (38):(44)z1=1223g23+112g32g23−27g323,z2=−23g23+112g32g23−27g323.It follows from (36) that the period of Weierstrass elliptic function reads(45)T=2∫a+∞dx4x3−g2x−g3=22ζ3g24K142−ζ+3,where a is the greatest real root to the cubic:(46)4x3−g2x−g3=0.Using identity (36) and approximation (26), we have the following approximation:(47)℘t;g2,g3≈−g24m+1248m2−48m+3+3g216m2−16m+11−cosm23g2/16m2−16m+14t,where m is given by (37).Example 2.
Letg2=2 and g3=1. For this choice, m=0.0119056. We have(48)℘t;2,1≈−0.474691+2.718671−3.73048cos2.32484t/14.−0.0835474cos22.32484t.
The period isT=2.70262 and the error on the interval −T/2≤t≤T/2 in the sup norm is E=8.52×10−7, see Figure 1. The reciprocals of the two functions are plotted on the interval −T/2≤t≤T/2.
We now give approximate expression for inverse elliptic Weierstrass function as follows. Let(49)a=−g24m+1248m2−48m+3,b=3g216m2−16m+1.
We know that(50)℘≈a+b1−cosm23g2/16m2−16m+14t.
Now, we solve this equation fort to obtain(51)t≈±16m2−16m+14234g24λ+1cos−1±a+b−℘a2−2abλ−b2λ−2a−bλ℘+℘2,so(52)℘−1t;g2,g3≈±16m2−16m+14234g24λ+1cos−1±a+b−ta2−2abλ−b2λ−2a−bλt+℘2.Figure 1
Comparison between the approximate trigonometric solution and the Runge–Kutta numerical solution for Example2.Example 3.
Suppose we are given that(53)℘t;−2,−5=15.
The exact solution is(54)t=℘−115;−2,−5=−1.210122529012178.
Using (52), we obtain four values:(55)t1=−1.20908,t2=1.20908,t3=−0.763507,t4=0.763507.
We chooset=−1.20908.
From all previous facts, we may establish the following.


### 3.3.1. General Principle

If the solution to some problem may be written in terms of either Jacobian elliptic functions or in terms of the Weierstrass elliptic function, then such solution may be approximated by means of trigonometric functions.


## 3.3.1. General Principle

If the solution to some problem may be written in terms of either Jacobian elliptic functions or in terms of the Weierstrass elliptic function, then such solution may be approximated by means of trigonometric functions.


## 4. Applications in Physics

### 4.1. Undamped and Unforced Duffing Equation

This is the equation,(56)x¨+px+qx3=0,x=xt,given under the initial conditions,(57)x0=x0,x′0=x˙0.The general solution to equation (56) may be written in terms of any of the twelve Jacobian elliptic functions. Let, for example,(58)xt=c1cnωt+c2,m.Then,(59)x¨+px+qx3=c13q−2c1mωcn3+2c1mω+c1p−c1ωcn,where cn=cnωt+c2,m.Equating to zero, the coefficients ofcnj gives an algebraic system whose solution is(60)ω=p+qc12,m=qc122p+qc12.Thus, the general solution to the Duffing equation is(61)xt=cnp+qc12t+c2,qc122p+qc12.The values for the constantsc1 and c2 are determined under the initial conditions.Definition 1.
The numberΔ=p+qx022+2qx˙02 is called the discriminant for the Duffing equations (56) and (57). We will distinguish three cases depending on the sign of the discriminant.


### 4.2. First Case:Δ>0

The solution to the IVP (56) and (57) is given by(62)xt=Δ−pqcnΔ4t−signx˙0cn−1qΔ−px0,12−p2Δ,12−p2Δ.Making use of the addition formula,(63)cnx+y,m=cnx,mcny,m+snx,mdnx,msny,mdny,m1−msnx,msny,m,solution (62) may be expressed as(64)xt=x0cnΔ4t|m+x˙0/ωdnΔ4t|msnΔ4t|m1+p+qx02−Δ/2ΔsnΔ4t|m2,where(65)m=121−pΔ,ω=Δ.Solution (64) is a periodic solution with period:(66)T=4Kmω.Using our general principle, we may give approximate analytical solution to the IVP (56) and (57) using the functions sinmt, cosmt, and dnmt defined by (26)–(28) as follows:(67)xappt=x0cosmωt+x˙0/ωdnmωtsinmωt1+p+qx02−ω/2Δsinm2ωt.The numbersm and ω are the same as in (65). However, we also may use the following formula, obtained from (62), replacing cn with cosm and cn−1 with cosm−1. The resulting expression reads(68)xappt=Δ−pqcosmΔ4t−signx˙0cosm−1qΔ−px0,where m is given by (65).Example 4.
Let us consider the IVP(69)x″t+xt+x3t=0,x0=1&x′0=−1.
Using formula (62), the exact solution to (69) is(70)xt=−6−1cn64t+cn−1−16−1,12−126,1126−6.
According to relations (64) and (65), the exact solution to the IVP (69) is also written as(71)xt=264dn64t|1/126−6sn64t|1/126−6−26cn64t|1/126−66−2sn64t|1/126−62−26.
The period is given by(72)T=4K1/126−664≈3.27458.
In Figure2, the comparison between the exact analytical solution (70) and the approximate numerical RK4 solution is presented. Full compatibility between the two analytical and numerical solutions is observed.
The approximate trigonometric solution from (68) equals(73)xtrigt=1.10428cos1.43552t+0.551331−0.15872cos21.43552t+0.55133.
The error of the trigonometric solution compared with the exact analytical solution is(74)max0≤t≤Txtrigt−xexactt=0.0033048.
For a graphical comparison, see Figure3.Figure 2
Comparison between the exact analytical solution and the Runge–Kutta numerical solution for Example3.Figure 3
Comparison between the exact and the approximate trigonometric solution for Example3.


### 4.3. Second Case:Δ<0

In this case,q<0, and then, since δ=p2−Δ/−q>0, δ=def2p+qx02x02+2x˙02>0. Let(75)xt=A−2A1+yt,where y=yt is a solution to some Duffing equation:(76)y″t+myt+ny3t,under initial conditions,(77)y0=y0=2Ax˙0A−x02,y′0=y˙0=A+x0A−x0.Inserting ansatz (76) into x″t+pxt+qx3t=0 and taking into account the relation,(78)y′t2=y˙0+my02+n2y04−my2t−n2y4t,we obtain(79)−AA2q+4my02+2ny04+p+4y˙02+A3A2q−2m−pyt−A3A2q−2m−pyt2+AA2q−2n+pyt3=0.Equating to zero, the coefficients ofyjt gives an algebraic system. A solution to this system is(80)m=12−p+3A2q,n=12p+A2q,A=2p+qx02x02+2x˙02−q4=δ−q4.Observe that the Duffing equations (76) and (77) has a positive discriminant given by(81)m+ny022+2ny˙02=δA−x042A4x˙02+δA2+x0224A8x02.Then, the problem reduces to the first case. The solution to the IVP (56) and (57) may be written in the form,(82)xt=A−2A1+Bb0cnωt|m+b1snωt|mdnωt|m/1+b2sn2ωt|m,where(83)m=B2A2q+p2A2B2+3q+2B2−1p,ω=12A2B2+3q+B2−1p,b0=A+x0AB−Bx0,b1=2Ax˙0BωA−x02,b2=−2Ax0x0−Ap+qx02+ωA+x0A−x02+4Ax˙022ωA−x02A+x0,A=2p+qx02x02+2x˙02−q4=δ−q4,B=p+A22qA2q−p−3AqA2q+p.The solution is unbounded and periodic with period:(84)T=4Kmω=4K1−mmω.Solution (82) is unbounded.Example 5.
Let us assume the following IVP:(85)x″t+xt−x3t=0,x0=−1,x′0=−1.
The solution of IVP (84) according to relation (82) reads(86)xt=1.31607−2.632151+0.13647cn1.75396t|1.00353−0.27976dn1.75396t|1.00353sn1.75396t|1.00353/1.−1.00463sn1.75396t|1.003532.
The period of solution (86) is given by(87)T=9.57783,see Figure 4.Figure 4
Plot of the unbounded periodic function for Example5.


### 4.4. Third Case:Δ=0

If the discriminant vanishesΔ=0, then q<0, and the only solution to problem (56) and (57) with(88)x′02=x˙02=p+qy022−2q,reads(89)xt=−pqtanhp2t±tanh−1x0−qp,which may be verified by direct computation.Remark 1.
The solution to the IVP,(90)x¨+px+qx3=0,x0=x0,x′0=0,is(91)xt=x0cnp+qx02t,qx022p+qx02,
An approximate trigonometric solution is(92)xtrigot=x0cosmp+qx02t,m=qx022p+qx02.Remark 2.
Letp+p2+2qx˙02>0. Then, the solution to the IVP(93)x¨+px+qx3=0,x0=0,x′0=x˙0,is(94)xt=2x˙0p2+2qx˙02+psnp+p2+2qx˙022t,−p2+qx˙02−p2+2qx˙02pqx˙02.
An approximate trigonometric solution is(95)xtrigot=2x˙0p2+2qx˙02+psinmp+p2+2qx˙022t,where(96)m=−p2+qx˙02−p2+2qx˙02pqx˙02.


## 4.1. Undamped and Unforced Duffing Equation

This is the equation,(56)x¨+px+qx3=0,x=xt,given under the initial conditions,(57)x0=x0,x′0=x˙0.The general solution to equation (56) may be written in terms of any of the twelve Jacobian elliptic functions. Let, for example,(58)xt=c1cnωt+c2,m.Then,(59)x¨+px+qx3=c13q−2c1mωcn3+2c1mω+c1p−c1ωcn,where cn=cnωt+c2,m.Equating to zero, the coefficients ofcnj gives an algebraic system whose solution is(60)ω=p+qc12,m=qc122p+qc12.Thus, the general solution to the Duffing equation is(61)xt=cnp+qc12t+c2,qc122p+qc12.The values for the constantsc1 and c2 are determined under the initial conditions.Definition 1.
The numberΔ=p+qx022+2qx˙02 is called the discriminant for the Duffing equations (56) and (57). We will distinguish three cases depending on the sign of the discriminant.


## 4.2. First Case:Δ>0

The solution to the IVP (56) and (57) is given by(62)xt=Δ−pqcnΔ4t−signx˙0cn−1qΔ−px0,12−p2Δ,12−p2Δ.Making use of the addition formula,(63)cnx+y,m=cnx,mcny,m+snx,mdnx,msny,mdny,m1−msnx,msny,m,solution (62) may be expressed as(64)xt=x0cnΔ4t|m+x˙0/ωdnΔ4t|msnΔ4t|m1+p+qx02−Δ/2ΔsnΔ4t|m2,where(65)m=121−pΔ,ω=Δ.Solution (64) is a periodic solution with period:(66)T=4Kmω.Using our general principle, we may give approximate analytical solution to the IVP (56) and (57) using the functions sinmt, cosmt, and dnmt defined by (26)–(28) as follows:(67)xappt=x0cosmωt+x˙0/ωdnmωtsinmωt1+p+qx02−ω/2Δsinm2ωt.The numbersm and ω are the same as in (65). However, we also may use the following formula, obtained from (62), replacing cn with cosm and cn−1 with cosm−1. The resulting expression reads(68)xappt=Δ−pqcosmΔ4t−signx˙0cosm−1qΔ−px0,where m is given by (65).Example 4.
Let us consider the IVP(69)x″t+xt+x3t=0,x0=1&x′0=−1.
Using formula (62), the exact solution to (69) is(70)xt=−6−1cn64t+cn−1−16−1,12−126,1126−6.
According to relations (64) and (65), the exact solution to the IVP (69) is also written as(71)xt=264dn64t|1/126−6sn64t|1/126−6−26cn64t|1/126−66−2sn64t|1/126−62−26.
The period is given by(72)T=4K1/126−664≈3.27458.
In Figure2, the comparison between the exact analytical solution (70) and the approximate numerical RK4 solution is presented. Full compatibility between the two analytical and numerical solutions is observed.
The approximate trigonometric solution from (68) equals(73)xtrigt=1.10428cos1.43552t+0.551331−0.15872cos21.43552t+0.55133.
The error of the trigonometric solution compared with the exact analytical solution is(74)max0≤t≤Txtrigt−xexactt=0.0033048.
For a graphical comparison, see Figure3.Figure 2
Comparison between the exact analytical solution and the Runge–Kutta numerical solution for Example3.Figure 3
Comparison between the exact and the approximate trigonometric solution for Example3.


## 4.3. Second Case:Δ<0

In this case,q<0, and then, since δ=p2−Δ/−q>0, δ=def2p+qx02x02+2x˙02>0. Let(75)xt=A−2A1+yt,where y=yt is a solution to some Duffing equation:(76)y″t+myt+ny3t,under initial conditions,(77)y0=y0=2Ax˙0A−x02,y′0=y˙0=A+x0A−x0.Inserting ansatz (76) into x″t+pxt+qx3t=0 and taking into account the relation,(78)y′t2=y˙0+my02+n2y04−my2t−n2y4t,we obtain(79)−AA2q+4my02+2ny04+p+4y˙02+A3A2q−2m−pyt−A3A2q−2m−pyt2+AA2q−2n+pyt3=0.Equating to zero, the coefficients ofyjt gives an algebraic system. A solution to this system is(80)m=12−p+3A2q,n=12p+A2q,A=2p+qx02x02+2x˙02−q4=δ−q4.Observe that the Duffing equations (76) and (77) has a positive discriminant given by(81)m+ny022+2ny˙02=δA−x042A4x˙02+δA2+x0224A8x02.Then, the problem reduces to the first case. The solution to the IVP (56) and (57) may be written in the form,(82)xt=A−2A1+Bb0cnωt|m+b1snωt|mdnωt|m/1+b2sn2ωt|m,where(83)m=B2A2q+p2A2B2+3q+2B2−1p,ω=12A2B2+3q+B2−1p,b0=A+x0AB−Bx0,b1=2Ax˙0BωA−x02,b2=−2Ax0x0−Ap+qx02+ωA+x0A−x02+4Ax˙022ωA−x02A+x0,A=2p+qx02x02+2x˙02−q4=δ−q4,B=p+A22qA2q−p−3AqA2q+p.The solution is unbounded and periodic with period:(84)T=4Kmω=4K1−mmω.Solution (82) is unbounded.Example 5.
Let us assume the following IVP:(85)x″t+xt−x3t=0,x0=−1,x′0=−1.
The solution of IVP (84) according to relation (82) reads(86)xt=1.31607−2.632151+0.13647cn1.75396t|1.00353−0.27976dn1.75396t|1.00353sn1.75396t|1.00353/1.−1.00463sn1.75396t|1.003532.
The period of solution (86) is given by(87)T=9.57783,see Figure 4.Figure 4
Plot of the unbounded periodic function for Example5.


## 4.4. Third Case:Δ=0

If the discriminant vanishesΔ=0, then q<0, and the only solution to problem (56) and (57) with(88)x′02=x˙02=p+qy022−2q,reads(89)xt=−pqtanhp2t±tanh−1x0−qp,which may be verified by direct computation.Remark 1.
The solution to the IVP,(90)x¨+px+qx3=0,x0=x0,x′0=0,is(91)xt=x0cnp+qx02t,qx022p+qx02,
An approximate trigonometric solution is(92)xtrigot=x0cosmp+qx02t,m=qx022p+qx02.Remark 2.
Letp+p2+2qx˙02>0. Then, the solution to the IVP(93)x¨+px+qx3=0,x0=0,x′0=x˙0,is(94)xt=2x˙0p2+2qx˙02+psnp+p2+2qx˙022t,−p2+qx˙02−p2+2qx˙02pqx˙02.
An approximate trigonometric solution is(95)xtrigot=2x˙0p2+2qx˙02+psinmp+p2+2qx˙022t,where(96)m=−p2+qx˙02−p2+2qx˙02pqx˙02.


## 5. Undamped Duffing–Helmholtz Equation

Let(97)x¨+px+qx2+rx3=0,x0=x0,x′0=x˙0.Equation (97) is called Duffing–Helmholtz equation. We will give a solution in terms of Jacobian and Weierstrass elliptic function. We will solve this problem using the ansatz:(98)xt=A+B1+C℘t+t0;g2,g3,BC≠0.Plugging ansatz (98) into the ode x¨+px+qx2+rx3=0 gives(99)x¨+px+qx2+rx3=121+C℘32C2A3Cr+A2Cq+ACp+2B℘3+2C3A3Cr+3A2BCr+3A2Cq+2ABCq+3ACp+BCp−6B℘2+C6A3r+12A2Br+6A2q+6AB2r+8ABq+6Ap+2B2q−3BCg2+4Bp℘+A3r+6A2Br+2A2q+6AB2r+4ABq+2Ap+2B3r+2B2q−4BC2g3+BCg2+2Bp.Equating to zero, the coefficients of℘j of the numerator of last expression in (99) give an algebraic system. Solving it, we obtain the following solution:(100)g3=12169pr2−3q2rA4+12pqr−4q3A3+18p2r−6pq2A2+p3,g2=−1123r2A4+4qrA3+6prA2−p2,B=−6AA2r+Aq+p3A2r+2Aq+p,C=123A2r+2Aq+p.The values oft0 and A are determined under the initial conditions x0=x0 and x′0=x˙0 and(101)x¨0+px0+qx20+rx30=0.We have(102)t0=±℘−1x0−A−BCA−x0;g2,g3.The numberA is a solution to the quartic:(103)3rA4+4qA2+6pA−3rx04+4qx03+6px02+6x˙02=0.Example 6.
Let(104)x¨+x+2x2+3x3=0,x0=1,x′0=0.
The solution is(105)1.07627−2.720781+0.762858℘t−0.148317;−7.16667,0.675926.
See Figure5, for a comparison with Runge–Kutta numerical solution.
Solution (105) is periodic. Its period equals(106)T=2∫0.0938538∞14x3+7.16667x−0.675926dx=3.12129.
We also may give approximate analytical trigonometric solution to this problem by means of formulas (47) and (52).Figure 5
Comparison between the exact solution and the Runge–Kutta numerical solution for Example6.


## 6. The Elliptic Function Expansion Method

Many important models in soliton theory are solved by means of a traveling wave transformation. Given a pdf,(107)Pu,ut,ux,uxx,…=0,we seek a solution in the ansatz form u=uξ, ξ−x+λt. This transformation gives a nonlinear ODE:(108)Qλ,uξ,u′ξ,u″ξ,…=0.The solution to this ODE in many cases may be found in the form(109)uξ=a0+a1vξ+⋯+anvnξ,where the function v=vξ is the solution to some elliptic equation:(110)dφdξ=c0+c1φξ+c2φ2ξ+c3φ3ξ+c4φ4ξ,φ0=φ0.This equation may be reduced to an equation of form (97) by differentiating it with respect to ξ:(111)φ″ξ=c12φξ+c2φ′ξ+3c32φ2ξ+2c4φ3ξ.The exact solution to equation (110) reads(112)φξ=A+B1+C℘ξ+ξ0;g2,g3.Here,(113)B=−34A3c4+3A2c3+2Ac2+c16A2c4+3Ac3+c2,C=126A2c4+3Ac3+c2,g2=112−12A4c42−12A3c3c4−12A2c2c4−12Ac1c4+c22−3c1c3,g3=143272A4c2c42−27A4c32c4−27A3c33+72A3c2c3c4−27A2c2c32+72A2c22c4−27Ac1c32+72Ac1c2c4+2c23−9c1c2c3+27c12c4,ξ0=℘−1−A+B−φ0CA−φ0;g2,g3.The numberA must obey the quartic:(114)c0+c1A+c2A2+c3A3+c4A4=0.On the contrary, the nonlinear ODE,(115)dφdξ2=c0+c1φξ+c2φ2ξ+c3φ3ξ+c4φ4ξ,may admit other solutions that are expressed in terms of elementary functions. Some of them are as follows.Forc0=c34−8c2c4c32+16c22c42/64c43,c1=−c3c32−4c2c4/8c42,anda=−c3/4c4,(116)φξ=−c34c4±3c32/c4−8c24c4tanh122c2−3c324c4ξ.Forc0=c34−8c2c4c32+16c22c42/64c43andc1=4c2c3c4−c33/8c42,(117)φξ=−c3+3c32−8c2c44c4±3c32−8c2c42c41+exp±1/28c2−3c32/c4ξ.Forc0=−c325c32−16c2c4/256c43andc1=−c3c32−4c2c4/8c42,(118)φξ=−c34c4±3c32−8c2c4/c422c4sech123c322c4−4c2ξ.Forc0=1/27−2c23/c32−2c22−3c1c3c22/c32+9c1c2/c3+6c1c22−3c1c3/c3,c4=0,andb=c22−3c1c3/c3,(119)φξ=−c2−c22−3c1c33c3+c22−3c1c3c3sech212c22−3c1c34t.Forc0=c1c2−c22−2c1c3/4c3,c4=c323c2−5c22−2c1c3/8c2−c22−2c1c32,(120)φξ=c22−2c1c3−c2+c3+6c1c3+4c2c22−2c1c3−c2/c3c3/c1sech1/23c22−2c1c3−c2t.Forc0=1/2a2ac3+c2,c1=−1/3a3ac3+4c2,andc4=−3ac3+c2/6a2,(121)φξ=a+12a3ac3+2c2−63ac3+c2+3ac3+2c22t2.Application of the elliptic functions’ expansion method may be found in [5, 6]. For applications to other physical problems, see [7–36].


## 7. Conclusions

We have obtained reasonable accurate approximations for the main Jacobian elliptic functions and Weierstrass elliptic function and their inverses through elementary functions. We also gave reasonable approximations for the elliptic integralKm. The accuracy of these approximations is seen in (11) and (12) and tables. We established a general principle that allows to solve some nonlinear problems in elementary way. To the best of the authors’ knowledge, there is no similar approach in the literature.



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