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The nonlinear Schrödinger equations (NLSEs) are extensively used to describe numerous crucial phenomena and dynamic processes in various fields such as fluid dynamics, plasma, chemistry, biology, optical fibers[1,2,3,4], nuclear physics, stochastic mechanics, biomolecule dynamics, dynamics of accelerators, and Bose-Einstein condensates [5,6,7,8,9,10]. The last few decades have seen significant advancements in the field of nonlinear optics [11,12,13]. These equations also appear in other forms of nonlinearities, such as cubic-quintic (CQ), cubic-quintic-septic (CQS), power-law, logarithmic nonlinearities, and various other forms. There is a growing interest in studying soliton pulses that can propagate without changing their shape in optical fibers [14,15].
Therefore, obtaining the exact soliton solutions of these NLSEs can help us to understand these phenomena better. In recent years, several effective approaches have been developed to construct accurate solutions of these equations. For example, the simple equation method [16], Kudryashov's method [17], the Jacobi elliptic function method [18], the inverse scattering method [19], the extended trial function method [20], the tanh method [21], the F-expansion method [22], generalized extended tanh-function method, the sine–cosine method [23], the generalized Riccati equation method [24], the new ϕ6-model expansion method [25], Hirota bilinear method [26], the exp-function method [27], the Darboux transformation [28], the auxiliary equation method [29], the Binary Bell polynomials [30], the extended hyperbolic function method [31], the homogeneous balance Method [32], the (G′G)-expansion method [33,34], the exponential rational function method [35], the Bäcklund transformation [36], the homotopy perturbation method [37], the modified Kudryashov method, the sine-Gordon expansion approach [38], the Riccati-Bernoulli sub-ODE method [39], the modified extended direct algebraic method [40], the truncated Painlevé expansion method [41], the (G′/G,1/G)-expansion method [42], the tan(ϕ/2)-expansion method [43,44], the soliton ansatz method [45,46], and so on.
Researches have been conducted on the study of NLSEs with the polynomial law of nonlinearity. For instance, Seadawy et al. [47] established soliton solutions using the extended simplest equation method, and they also described this model [48], which includes the conservation principles for optical soliton (OS) with polynomial and triple power law nonlinearities, while Aziz et al. [49] discovered chirped soliton solutions by utilizing Jacobi elliptic functions. Furthermore, this model was discussed by Dieu-donne et al. [50], who examined the optical soliton (OS) solutions with two types of nonlinearities, which are triple power and polynomial laws. Sugati et al.[51] also examined this model, obtaining the traveling pulse solutions to the NLSE when it is treated with both spatio-temporal dispersion (STD) and group velocity dispersion (GVD).
In this paper, our aim is to investigate soliton solutions for two models of the NLSE that carry the polynomial law of nonlinearity (cubic-quintic-septic). We will use the variational principle based on finding Lagrangian, which we will then apply it with different trial functions that have one or two nontrivial variational parameters. Furthermore, we will employ another technique called the amplitude ansatz method to extract new solitary wave solutions.
The paper is categorized as follows. In Section 2, we discuss model-I of NLSE with polynomial law nonlinearity in terms of formulation of the variational principle and finding solitary wave solutions. In Section 3, we apply the same steps on model-II of NLSE with polynomial law nonlinearity. Finally, the work concludes in Section 4.
Nonlinear Schrödinger equation (NLSE) with the polynomial nonlinear law is [47]:
iΓt+aΓxx+b1Γ|Γ|2+b2Γ|Γ|4+b3Γ|Γ|6=0. | (2.1) |
Such that Γ is a complex function on the form: Γ(x,t)=Θ(x,t)+iΨ(x,t). Since Θ and Ψ are real functions of x and t, furthermore |Γ|2=(Θ+iΨ)(Θ−iΨ). Using the variational approach, we will search for solutions to NLSE with the polynomial nonlinear law. As a result, we derive Γ with respect to x and t to investigate the existence of a Lagrangian and the invariant variational principle for this equation, which are expressed in the following way:
Let M and N are functionals in Θ and Ψ:
M(Θ,Ψ)=∂Θ∂t+a∂2Ψ∂x2+b1ΨΘ2+b1Ψ3+b2ΨΘ4+2b2Θ2Ψ3+b2Ψ5+b3ΨΘ6+3b3Θ4Ψ3+3b3Θ2Ψ5+b3Ψ7, | (2.2) |
N(Θ,Ψ)=−∂Ψ∂t+a∂2Θ∂x2+b1Θ3+b1ΘΨ2+b2Θ5+2b2Ψ2Θ3+b2ΘΨ4+b3Θ7+3b3Ψ2Θ5+3b3Θ3Ψ4+b3ΘΨ6. | (2.3) |
Put Θ=λΘ and Ψ=λΨ,
∫10M(λΘ,λΨ)dλ=12∂Θ∂t+12a∂2Ψ∂x2+14b1ΨΘ2+14b1Ψ3+16b2ΨΘ4+13b2Θ2Ψ3+16b2Ψ5+18b3ΨΘ6+38b3Θ4Ψ3+38b3Θ2Ψ5+18b3Ψ7, |
∫10N(λΘ,λΨ)dλ=−12∂Ψ∂t+12a∂2Θ∂x2+14b1Θ3+14b1ΘΨ2+16b2Θ5+13b2Ψ2Θ3+16b2ΘΨ4+18b3Θ7+38b3Ψ2Θ5+38b3Θ3Ψ4+18b3ΘΨ6. |
The consistency conditions are expressed as follows [52,53], where Θx, Ψx, Θt and Ψt stand for the partial derivatives of Θ and Ψ with respect to variables x and t. If the system of Eqs (2.2) and (2.3) satisfies the the previously mentioned conditions, then a functional integral J(Θ,Ψ) can be written down using the formula given by Tonti [53]:
J(Θ,Ψ)=∫ω[14b1Θ4+16b2Θ6+18b3Θ8+12b1Θ2Ψ2+12b2Θ4Ψ2+12b3Θ6Ψ2+14b1Ψ4+12b2Θ2Ψ4+34b3Θ4Ψ4+16b2Ψ6+12b3Θ2Ψ6+18b3Ψ8+12ΨΘt−12ΘΨt+12aΘΘxx+12aΨΨxx]dω, |
where dω=dxdt.
The terms ΘΘxx and ΨΨxx solve integration by parts and choosing the boundary on Θx and Ψx to be such that the boundary terms vanish, we get
J(Θ,Ψ)=∫ω[14b1Θ4+16b2Θ6+18b3Θ8+12b1Θ2Ψ2+12b2Θ4Ψ2+12b3Θ6Ψ2+14b1Ψ4+12b2Θ2Ψ4+34b3Θ4Ψ4+16b2Ψ6+12b3Θ2Ψ6+18b3Ψ8+12ΨΘt−12ΘΨt−12aΘ2x−12aΨ2x]dω, | (2.4) |
where dω=dxdt.
We obtain the Lagrangian L as
L(Θ,Ψ)=14b1Θ4+16b2Θ6+18b3Θ8+12b1Θ2Ψ2+12b2Θ4Ψ2+12b3Θ6Ψ2+14b1Ψ4+12b2Θ2Ψ4+34b3Θ4Ψ4+16b2Ψ6+12b3Θ2Ψ6+18b3Ψ8+12ΨΘt−12ΘΨt−12aΘ2x−12aΨ2x. | (2.5) |
We find the value of L in the Euler-Lagrange equations to validate our interpretations:
∂L∂Θ−∂∂t(∂L∂Θt)−∂∂x(∂L∂Θx)=0, |
∂L∂Ψ−∂∂t(∂L∂Ψt)−∂∂x(∂L∂Ψx)=0. |
The resulting derivatives give us the system of Eqs (2.2) and (2.3).
We demonstrate the simplest example of the application of this technique, by taking the box-shaped initial pulse and an ansatz based on linear Jost functions in a single nontrivial variational parameter in three cases.
Case 1: We use the following ansatz for Θ(x,t) and Ψ(x,t) functions:
Θ(x,t)={100exp(−μ(t−5)(x−5)),atx>5,t>5,(t+5)(x+5),at|x|<5,|t|<5,0,atx<−5,t<−5, | (2.6) |
Ψ(x,t)={0,atx>5,t>5,(5−t)(5−x),at|x|<5,|t|<5,100exp(μ(t+5)(x+5)),atx<−5,t<−5. | (2.7) |
We found the values of the Lagrangian L from substituting Eqs (2.6) and (2.7) into Eq (2.5) (see Figure 1):
J(Θ,Ψ)=∫−5−∞∫−5−∞Ldxdt+∫5−5∫5−5Ldxdt+∫∞5∫∞5Ldxdt, |
where
∫−5−∞∫−5−∞Ldxdt=∫∞5∫∞5Ldxdt=0. |
By using Mathematica program we get the value of J(Θ,Ψ) at interval −5<x<5,−5<t<5 as
J(Θ,Ψ)=−10000a3+1850000000b19+304000000000000b2441+1760000000000000000b3567+250003. | (2.8) |
Case 2: Assume Θ(x,t) and Ψ(x,t) for the Jost function as the following:
Θ(x,t)={20exp(−μ(t−5)(x−5)),atx>5,t>5,t+x+10,at|x|<5,|t|<5,0,atx<−5,t<−5, | (2.9) |
Ψ(x,t)={0,atx>5,t>5,10−t−x,at|x|<5,|t|<5,20exp(μ(t+5)(x+5)),atx<−5,t<−5. | (2.10) |
We found the values of the Lagrangian L from substituting by Eqs (2.9) and (2.10) into Eq (2.5), then we get
J(Θ,Ψ)=10063(−63a+882000b1+145800000b2+28120000000b3+630). | (2.11) |
Case 3:
Θ(x,t)={12(exp(6π−t−x)−exp(−2π−t−x)),atx>π,t>π,sinh(t+x+2π),at|x|<π,|t|<π,0,atx<−π,t<−π, | (2.12) |
ψ(x,t)={0,atx>π,t>π,sinh(2π−t−x),at|x|<π,|t|<π,12(exp(6π+t+x)−exp(−2π+t+x)),atx<−π,t<−π. | (2.13) |
We found the values of the Lagrangian L from substituting by Eqs (2.12) and (2.13) into Eq (2.5), then we have
J(Θ,Ψ)=−1.02783×1010a+2.64108×1019b1+1.60864×1029b2+1.39506×1039b3+2.83012×106. | (2.14) |
Case 1: We try the following Jost functions:
Θ(x,t)={12(exp(π2(3+2α+α2)−πt−πx)−exp(π2(1−2α−α2)−πt−πx)),atx>π,t>π,sinh((π+αt)(π+αx)),at0<x<π,0<t<π,sinh((t+π)(x+π)),at−π<x<0,−π<t<0,0,atx<−π,t<−π, | (2.15) |
Ψ(x,t)={0,atx>π,t>π,sinh((π−t)(π−x)),at0<x<π,0<t<π,sinh((π−αt)(π−αx)),at−π<x<0,−π<t<0,12(exp(π2(3+2α+α2)+πt+πx)−exp(π2(1−2α−α2)+πt+πx)),atx<−π,t<−π. | (2.16) |
This ansatz now contains nontrivial variational parameter α. Substituting Eqs (2.15) and (2.16) into Eq (2.5) (see Figure 2), then we obtain the functional integral at α=−1 as follows:
J(Θ,Ψ)=−4.43569×107a+1.13554×1014b1+6.23167×1021b2+4.89178×1029b3. | (2.17) |
Also, we can use α=2, then we get
∫0−π∫0−πLdxdt=∫π0∫π0Ldxdt=−4.42653×1075a+5.59314×10148b1+5.89626×10224b2+8.85443×10300b3−2.66975×1036. |
The functional integral at α=2 becomes
J(Θ,Ψ)=−1.77562×1076a+4.12756×10150b1+4.35525×10226b2+6.5433×10302b3−5.33949×1036. | (2.18) |
Case 2: We now try the following Jost function:
Θ(x,t)={12(exp(4π−t−x+2πα)−exp(−t−x−2πα)),atx>π,t>π,sinh(2π+αt+αx),at0<x<π,0<t<π,sinh(2π+t+x),at−π<t<0,−π<x<0,0,atx<−π,t<−π, | (2.19) |
Ψ(x,t)={0,atx>π,t>π,sinh(2π−t−x),at0<x<π,0<t<π,sinh(2π−αt−αx),at−π<x<0,−π<t<0,12(exp(4π+t+x+2πα)−exp(t+x−2πα)),atx<−π,t<−π. | (2.20) |
This ansatz contains nontrivial variational parameter α. Substituting Eqs (2.19) and (2.20) into Eq (2.5), then we get the functional integral at α=0.5 as follows:
J(Θ,Ψ)=−1.84004×107a+2.29576×1014b1+2.6187×1021b2+4.24151×1028b3+270561. | (2.21) |
Also, we can take α=−0.2, then we obtain
∫0−π∫0−πLdxdt=∫π0∫π0Ldxdt=−13516.2a+2.22124×109b1+7.54018×1013b2+3.79568×1018b3+19035.7. |
The functional integral at α=−0.2 becomes
J(Θ,Ψ)=−28484a+4.44353×109b1+1.50805×1014b2+7.59137×1018b3+38071.3. | (2.22) |
We assume the Jost function by quadratic polynomials at interval |x|<5,|t|<5:
Case 1: We use the following Jost functions:
Θ(x,t)={(100+10000α)exp(−μ(t−5)(x−5)),atx>5,t>5,(t+5)(x+5)+α(5+t)2(5+x)2,at|x|<5,|t|<5,0,atx<−5,t<−5, | (2.23) |
Ψ(x,t)={0,atx>5,t>5,(5−t)(5−x)+α(5−t)2(5−x)2,at|x|<5,|t|<5,(100+10000α)exp(μ(t+5)(x+5)),atx<−5,t<−5. | (2.24) |
We found the values of Lagrangian calculation from Eqs (2.23) and (2.24) into Eq (2.5), and we get
J(Θ,Ψ)=5.55556×106α2−2.66667×107aα2−500000aα+555556α−3333.33a+8.65053×1030α8b3+7.81255×1029α7b3+1.97241×1023α6b2+3.11121×1028α6b3+1.38897×1022α5b2+7.14358×1026α5b3+6.1741×1015α4b1+4.13321×1020α4b2+1.03576×1025α4b3+3.12755×1014α3b1+6.67171×1018α3b2+9.72777×1022α3b3+6.14172×1012α2b1+6.18569×1016α2b2+5.79228×1020α2b3+5.61111×1010αb1+3.14172×1014αb2+2.00529×1018αb3+2.05556×108b1+6.89342×1011b2+3.10406×1015b3+8333.33. | (2.25) |
We choose a=3, b1=2, b2=12, and b3=16, then the functional integral J(Θ,Ψ) becomes
J(Θ,Ψ)=1.44175×1030α8+1.30209×1029α7+5.18545×1027α6+1.19067×1026α5+1.72648×1024α4+1.62163×1022α3+9.65689×1019α2+3.34372×1017α+5.17688×1014. | (2.26) |
Derive Eq (2.26) with respect to α, then values of α are:
α=−0.0136245,α=−0.012952−0.00218808i,α=−0.012952+0.00218808i,α=−0.0110959−0.00364364i,α=−0.0110959+0.00364364i,α=−0.0086518−0.00394467i,α=−0.0086518+0.00394467i. |
We substitute the exact roots of α into Eq (2.26) give the following analytical equations:
J(Θ,Ψ)=4.83907×1010,J(Θ,Ψ)=3.84513×1010−4.55557×1010i,J(Θ,Ψ)=3.84513×1010+4.55557×1010i,J(Θ,Ψ)=−3.14336×109−1.06115×1011i,J(Θ,Ψ)=−3.14336×109+1.06115×1011i,J(Θ,Ψ)=−1.8701×1011−2.04551×1011i,J(Θ,Ψ)=−1.8701×1011+2.04551×1011i. | (2.27) |
Case 2: We try the following Jost functions:
Θ(x,t)={(20+200α)exp(−μ(t−5)(x−5)),atx>5,t>5,(10+t+x)+α(t+5)2+α(x+5)2,at|x|<5,|t|<5,0,atx<−5,t<−5, | (2.28) |
Ψ(x,t)={0,atx>5,t>5,(10−t−x)+α(5−t)2+α(5−x)2,at|x|<5,|t|<5,(20+200α)exp(μ(t+5)(x+5)),atx<−5,t<−5. | (2.29) |
We found the values of Lagrangian calculation from Eqs (2.28) and (2.29) into Eq (2.5) (see Figure 3), and we have
J(Θ,Ψ)=50000α2−13333.3aα2−2000aα+15000α−100a+9.90698×1017α8b3+9.04639×1017α7b3+5.97273×1013α6b2+3.65565×1017α6b3+4.25597×1013α5b2+8.55415×1016α5b3+4.78095×109α4b1+1.28615×1013α4b2+1.27048×1016α4b3+2.39683×109α3b1+2.11556×1012α3b2+1.22953×1015α3b3+4.62698×108α2b1+2.00368×1011α2b2+7.59375×1013α2b3+4.08889×107αb1+1.03937×1010αb2+2.7454×1012αb3+1.4×106b1+2.31429×108b2+4.46349×1010b3+1000. | (2.30) |
We put a=6, b1=3, b2=14, and b3=13, then the functional integral J(Θ,Ψ) becomes
J(Θ,Ψ)=3.30233×1017α8+3.01546×1017α7+1.2187×1017α6+2.85245×1016α5+4.23816×1015α4+4.10379×1014α3+2.5364×1013α2+9.17853×1011α+1.49404×1010. | (2.31) |
Derive Eq (2.31) with respect to α, then values of α are:
α=−0.131984,α=−0.129184−0.0176052i,α=−0.129184+0.0176052i,α=−0.117682−0.0366915i,α=−0.117682+0.0366915i,α=−0.0866387−0.051817i,α=−0.0866387+0.051817i. |
We substitute the exact roots of α into Eq (2.31) give the following analytical equations:
J(Θ,Ψ)=207360,J(Θ,Ψ)=−102453−293854i,J(Θ,Ψ)=−102453+293854i,J(Θ,Ψ)=−1.38256×106+542198i,J(Θ,Ψ)=−1.38256×106−542198i,J(Θ,Ψ)=−1.31177×107+1.75734×107i,J(Θ,Ψ)=−1.31177×107−1.75734×107i. | (2.32) |
We inspect here the exact solutions of the nonlinear Schrödinger equation with the polynomial nonlinear law. This equation is defined as Eq (2.1).
Case 1: We suppose the ansatz function of the NLSE with the polynomial nonlinear law is in the form of a bright solitary wave solution
h1(x,t)=Asech(w(x−tv)),Γ(x,t)=Asech(w(x−tv))ei(kx−ωt), | (2.33) |
where A,w and v are the amplitude, the pulse width, and velocity of soliton in normalized unites. Substituting from Eq (2.33) in Eq (2.1), and separating the real and imaginary parts, we obtain
−ak2v+avw2+vω+(A2b1v−2avw2)sech2(w(x−tv))+A4b2vsech4(w(x−tv))+A6b3vsech6(w(x−tv))=0, | (2.34) |
(w−2akvw)tanh(w(x−tv))=0. | (2.35) |
Equating the coefficients of the linearly independent terms to zero, we obtain the dynamical system in A,w,v,k,ω,a,b1,b2,b3 by solving this system we get:
Family I:
A=±w√b1kv,a=12kv,ω=k2−w22kv. | (2.36) |
The sufficient conditions for solitary wave solution stability are
b1kv>0,kv≠0. | (2.37) |
Family II:
A=±√k−2vωb1v,a=12kv,w=±√k(k−2vω), | (2.38) |
provided that
k−2vωb1v>0,kv≠0,k(k−2vω)>0. | (2.39) |
Family III:
A=±w√2ωb1k2−b1w2,a=ωk2−w2,v=k2−w22kω, | (2.40) |
whenever
ωb1(k2−w2)>0,kω≠0,k2−w2≠0. | (2.41) |
Family IV:
A=±√1b1(√v4ω2+v2w2v2−ω),a=√v4ω2+v2w2−v2ω2v2w2,k=√v4ω2+v2w2v+vω, | (2.42) |
provided that
v4ω2+v2w2>0,1b1(√v4ω2+v2w2v2−ω)>0,v2w2≠0. | (2.43) |
Then, the solutions of the NLSE with the polynomial nonlinear law as bright solitary wave solutions are (see Figures 4 and 5):
Γ11(x,t)=±w√b1kvsech(w(x−tv))ei(kx−ωt), | (2.44) |
Γ12(x,t)=±√k−2vωb1vsech(w(x−tv))ei(kx−ωt), | (2.45) |
Γ13(x,t)=±w√2ωb1k2−b1w2sech(w(x−tv))ei(kx−ωt), | (2.46) |
Γ14(x,t)=±√1b1(√v4ω2+v2w2v2−ω)sech(w(x−tv))ei(kx−ωt). | (2.47) |
Case 2: Another choice of the dark solitary wave solution of the NLSE with the polynomial nonlinear law is
h2(x,t)=A+Btanh(w(x−tv)),Γ(x,t)=(A+Btanh(w(x−tv)))ei(kx−ωt). | (2.48) |
By replacement from Eq (2.48) in Eq (2.1) and separating the real and imaginary parts
−aAk2+A7b3+21A5b3B2+A5b2+35A3b3B4+10A3b2B2+A3b1+7Ab3B6+5Ab2B4+3Ab1B2+Aω+(−21A5b3B2−70A3b3B4−10A3b2B2−21Ab3B6−10Ab2B4−3Ab1B2)×sech2(w(x−tv))+(35A3b3B4+21Ab3B6+5Ab2B4)×sech4(w(x−tv))−7Ab3B6sech6(w(x−tv))+(−aBk2+7A6b3B+35A4b3B3+5A4b2B+21A2b3B5+10A2b2B3+3A2b1B+b3B7+b2B5+b1B3+Bω)×tanh(w(x−tv))+(−2aBw2−35A4b3B3−42A2b3B5−10A2b2B3−3b3B7−2b2B5−b1B3)×tanh(w(x−tv))sech2(w(x−tv))+(21A2b3B5+3b3B7+b2B5)tanh(w(x−tv))sech4(w(x−tv))−B7b3tanh(w(x−tv))sech6(w(x−tv))=0, | (2.49) |
(2aBkw−Bwv)sech2(w(x−tv))=0 | (2.50) |
Equating the coefficients of the linearly independent terms to zero, we deduce the coefficients A,B,w,v,k,ω,a,b1,b2,b3 in the form:
Family I:
A=0,B=±i√b23b3,b1=27b23ω+2b32−27ab23k29b2b3,w=±13b3√27b23ω−b32−27ab23k26a,v=12ak. | (2.51) |
The sufficient conditions for dark solitary wave solution stability are
b2b3<0,b2b3≠0,ak≠0,27b23ω−b32−27ab23k2a>0. | (2.52) |
Family II:
A=0,B=±i√b23b3,w=±13b3√3b1b2b3−b322a,v=12ak,ω=27ab23k2−2b32+9b1b3b227b23, | (2.53) |
provided that
b2b3<0,3b1b2b3−b32a>0,ak≠0,b3≠0. | (2.54) |
Family III:
A=0,B=±6√k2−2kvω+2w22b3kv,a=12kv,b1=3√b3(3k2−6kvω+4w2)3√4k2v2(k2−2kvω+2w2),b2=−33√b23(k2−2kvω+2w2)2kv, | (2.55) |
such that
k2−2kvω+2w2b3kv>0,kv≠0,4k2v2(k2−2kvω+2w2)≠0. | (2.56) |
Family IV:
A=0,B=±6√ak2+2aw2−ωb3,b1=(3ak2+4aw2−3ω)3√b3ak2+2aw2−ω,b2=−33√b23(ak2+2aw2−ω),v=k2+2w22k(ak2+2aw2), | (2.57) |
whenever
ak2+2aw2−ωb3>0,ak2+2aw2−ω≠0,k(ak2+2aw2)≠0. | (2.58) |
Then, the dark soliton solutions of the NLSE with the polynomial nonlinear law Eq (2.1) are (see Figure 6):
Γ21(x,t)=±i√b23b3tanh(w(x−tv))ei(kx−ωt), | (2.59) |
Γ23(x,t)=±6√k2−2kvω+2w22b3kvtanh(w(x−tv))ei(kx−ωt), | (2.60) |
Γ24(x,t)=±6√ak2+2aw2−ωb3tanh(w(x−tv))ei(kx−ωt). | (2.61) |
The model II of the NLSE with polynomial nonlinearity is given by [49]:
iΨt+aΨxx+bΨxt+(k1|Ψ|2+k2|Ψ|4+k3|Ψ|6)Ψ=0, | (3.1) |
where Ψ(x,t) denotes the complex valued function. The coefficients a and b indicate group velocity dispersion (GVD) and spatio-temporal dispersion (STD), respectively. The first term represents the linear evolution of pulses in nonlinear optical fibers. The final term represents the nonlinearity of the non-Kerr law. Since Ψ(x,t) is a complex function on the form Ψ(x,t)=u(x,t)+iv(x,t) such that u and v are real functions of x and t, also |Ψ|2=(u+iv)(u−iv). We will use the variational technique to find solutions for the Eq (3.1), where we derive Ψ with respect to x and t to obtain the Lagrangian of this equation, which is expressed in the following way:
Let M and N are functionals in u and v,
M(u,v)=∂u∂t+a∂2v∂x2+b∂2v∂x∂t+k1vu2+k1v3+k2vu4+2k2u2v3+k2v5+k3vu6+3k3v3u4+3k3u2v5+k3v7, | (3.2) |
N(u,v)=−∂v∂t+a∂2u∂x2+b∂2u∂x∂t+k1u3+k1uv2+k2u5+2k2v2u3+k2uv4+k3u7+3k3v2u5+3k3u3v4+k3uv6. | (3.3) |
The system of Eqs (3.2) and (3.3) satisfies the conditions mentioned in [52,53], then a functional integral J(u,v) can be written down using the formula given by Tonti [53]:
J(u,v)=∫ω[12auuxx+12avvxx+12buuxt+12bvvxt+12k3v2u6+34k3u4v4+12k2v2u4+12k3u2v6+12k2u2v4+12k1u2v2+18k3u8+16k2u6+14k1u4+18k3v8+16k2v6+14k1v4+12vut−12uvt]dω, | (3.4) |
where dω=dxdt.
Then the Lagrangian L is given by
L(u,v)=12auuxx+12avvxx+12buuxt+12bvvxt+12k3v2u6+34k3u4v4+12k2v2u4+12k3u2v6+12k2u2v4+12k1u2v2+18k3u8+16k2u6+14k1u4+18k3v8+16k2v6+14k1v4+12vut−12uvt. | (3.5) |
As a necessary check to our calculations, we use the value of L in the Euler-Lagrange equations:
∂L∂u−∂∂t(∂L∂ut)+∂2∂x2(∂L∂uxx)+∂2∂x∂t(∂L∂uxt)=0, |
∂L∂v−∂∂t(∂L∂vt)+∂2∂x2(∂L∂vxx)+∂2∂x∂t(∂L∂vxt)=0, |
which yields the system of Eqs (3.2) and (3.3).
We provide an example of applying this technique by using a box-shaped initial pulse and a single nontrivial variational parameter based on linear Jost functions in three cases.
Case 1: We use the following ansatz for u(x,t) and v(x,t) functions:
u(x,t)={100exp(−μ(t−5)(x−5)),atx>5,t>5,(t+5)(x+5),at|x|<5,|t|<5,0,atx<−5,t<−5, | (3.6) |
v(x,t)={0,atx>5,t>5,(5−t)(5−x),at|x|<5,|t|<5,100exp(μ(t+5)(x+5)),atx<−5,t<−5. | (3.7) |
We found the values of the Lagrangian L from substituting Eqs (3.6) and (3.7) into Eq (3.5), and by using Mathematica program we get the value of J(u,v) at interval −5<x<5,−5<t<5 as
J(u,v)=2500b+2.05556×108k1+6.89342×1011k2+3.10406×1015k3+8333.33. | (3.8) |
Case 2: Assume u(x,t) and v(x,t) for the Jost function as the following:
u(x,t)={20exp(−μ(t−5)(x−5)),atx>5,t>5,t+x+10,at|x|<5,|t|<5,0,atx<−5,t<−5, | (3.9) |
v(x,t)={0,atx>5,t>5,10−t−x,at|x|<5,|t|<5,20exp(μ(t+5)(x+5)),atx<−5,t<−5. | (3.10) |
We found the values of the Lagrangian L from substituting by Eqs (3.9) and (3.10) into Eq (3.5) (see Figure 7), then we have
J(u,v)=100063(88200k1+14580000k2+2812000000k3+63). | (3.11) |
Case 3:
u(x,t)={12(exp(6π−t−x)−exp(−2π−t−x)),atx>π,t>π,sinh(t+x+2π),at|x|<π,|t|<π,0,atx<−π,t<−π, | (3.12) |
v(x,t)={0,atx>π,t>π,sinh(2π−t−x),at|x|<π,|t|<π,12(exp(6π+t+x)−exp(−2π+t+x)),atx<−π,t<−π. | (3.13) |
We found the values of the Lagrangian L from substituting by Eqs (3.12) and (3.13) into Eq (3.5), then we get
J(u,v)=1.02783×1010a+1.02783×1010b+2.64108×1019k1+1.60864×1029k2+1.39506×1039k3+2.83012×106. | (3.14) |
Case 1: We try the following Jost functions:
u(x,t)={12(exp(π2(3+2α+α2)−πt−πx)−exp(π2(1−2α−α2)−πt−πx)),atx>π,t>π,sinh((π+αt)(π+αx)),at0<x<π,0<t<π,sinh((t+π)(x+π)),at−π<x<0,−π<t<0,0,atx<−π,t<−π, | (3.15) |
v(x,t)={0,atx>π,t>π,sinh((π−t)(π−x)),at0<x<π,0<t<π,sinh((π−αt)(π−αx)),at−π<x<0,−π<t<0,12(exp(π2(3+2α+α2)+πt+πx)−exp(π2(1−2α−α2)+πt+πx)),atx<−π,t<−π. | (3.16) |
This ansatz now contains nontrivial variational parameter α. Substituting Eqs (3.15) and (3.16) into Eq (3.5), then we obtain the functional integral at α=−1 as follows:
J(u,v)=4.43568×107a+4.92254×107b+1.13554×1014k1+6.23167×1021k2+4.89178×1029k3. | (3.17) |
Also, we can use α=2, then we get
∫0−π∫0−πLdxdt=∫π0∫π0Ldxdt=4.42653×1075a+4.47679×1075b+5.59314×10148k1+5.89626×10224k2+8.85443×10300k3−2.66975×1036. |
The functional integral at α=2 becomes
J(u,v)=1.77562×1076a+1.78568×1076b+4.12756×10150k1+4.35525×10226k2+6.5433×10302k3−5.33949×1036. | (3.18) |
Case 2: We now try the following Jost function:
u(x,t)={12(exp(4π−t−x+2πα)−exp(−t−x−2πα)),atx>π,t>π,sinh(2π+αt+αx),at0<x<π,0<t<π,sinh(2π+t+x),at−π<t<0,−π<x<0,0,atx<−π,t<−π, | (3.19) |
v(x,t)={0,atx>π,t>π,sinh(2π−t−x),at0<x<π, 0<t<π,sinh(2π−αt−αx),at−π<x<0,−π<t<0,12(exp(4π+t+x+2πα)−exp(t+x−2πα)),atx<−π,t<−π. | (3.20) |
This ansatz contains nontrivial variational parameter α. Substituting Eqs (3.19) and (3.20) into Eq (3.5) (see Figure 8), then we get the functional integral at α=−0.2 as follows:
J(u,v)=28473.8a+28473.8b+4.44353×109k1+1.50805×1014k2+7.59137×1018k3+38071.3. | (3.21) |
Also, we can take α=0.5 then we obtain
∫0−π∫0−πLdxdt=∫π0∫π0Ldxdt=4.40169×106a+4.40169×106b+9.17623×1013k1+1.04745×1021k2+1.6966×1028k3+135281. |
The functional integral at α=0.5 becomes
J(u,v)=1.84004×107a+1.84004×107b+2.29576×1014k1+2.6187×1021k2+4.24151×1028k3+270561. | (3.22) |
We assume the Jost function by quadratic polynomials at interval |x|<5,|t|<5:
Case 1: We use the following Jost functions:
u(x,t)={(100+10000α)exp(−μ(t−5)(x−5)),atx>5,t>5,(t+5)(x+5)+α(5+t)2(5+x)2,at|x|<5,|t|<5,0,atx<−5,t<−5, | (3.23) |
v(x,t)={0,atx>5,t>5,(5−t)(5−x)+α(5−t)2(5−x)2,at|x|<5,|t|<5,(100+10000α)exp(μ(t+5)(x+5)),atx<−5,t<−5. | (3.24) |
We found the values of Lagrangian calculation from Eqs (3.23) and (3.24) into Eq (3.5) (see Figure 9), and we get
J(u,v)=5.55556×106α2+1.33333×107aα2+250000aα+555556α+2.5×107α2b+555556αb+2500b+8.65053×1030α8k3+7.81255×1029α7k3+1.97241×1023α6k2+3.11121×1028α6k3+1.38897×1022α5k2+7.14358×1026α5k3+6.1741×1015α4k1+4.13321×1020α4k2+1.03576×1025α4k3+3.12755×1014α3k1+6.67171×1018α3k2+9.72777×1022α3k3+6.14172×1012α2k1+6.18569×1016α2k2+5.79228×1020α2k3+5.61111×1010αk1+3.14172×1014αk2+2.00529×1018αk3+2.05556×108k1+6.89342×1011k2+3.10406×1015k3+8333.33. | (3.25) |
We put a=2, b=6, k1=2, k2=3, and k3=4, then the functional integral J(u,v) becomes
J(u,v)=3.46021×1031α8+3.12502×1030α7+1.24449×1029α6+2.85747×1027α5+4.14318×1025α4+3.89131×1023α3+2.3171×1021α2+8.02211×1018α+1.24183×1016. | (3.26) |
Derive Eq (3.26) with respect to α, then values of α are:
α=−0.0136235,α=−0.0129514−0.00218597i,α=−0.0129514+0.00218597i,α=−0.0110962−0.00364064i,α=−0.0110962+0.00364064i,α=−0.00865263−0.00394263i,α=−0.00865263+0.00394263i. |
We substitute the exact roots of α into Eq (3.26) give the following analytical equations:
J(u,v)=1.15381×1012,J(u,v)=9.16876×1011−1.08685×1012i,J(u,v)=9.16876×1011+1.08685×1012i,J(u,v)=−7.64569×1010−2.53327×1012i,J(u,v)=−7.64569×1010+2.53327×1012i,J(u,v)=−4.47645×1012−4.88381×1012i,J(u,v)=−4.47645×1012+4.88381×1012i. | (3.27) |
Case 2: We try the following Jost functions:
u(x,t)={(20+200α)exp(−μ(t−5)(x−5)),atx>5,t>5,(10+t+x)+α(t+5)2+α(x+5)2,at|x|<5,|t|<5,0,atx<−5,t<−5, | (3.28) |
v(x,t)={0,atx>5,t>5,(10−t−x)+α(5−t)2+α(5−x)2,at|x|<5,|t|<5,(20+200α)exp(μ(t+5)(x+5)),atx<−5,t<−5. | (3.29) |
We found the values of Lagrangian calculation from Eqs (3.28) and (3.29) into Eq (3.5), and we have
J(u,v)=50000α2+13333.3aα2+2000aα+15000α+9.90698×1017α8k3+9.04639×1017α7k3+5.97273×1013α6k2+3.65565×1017α6k3+4.25597×1013α5k2+8.55415×1016α5k3+4.78095×109α4k1+1.28615×1013α4k2+1.27048×1016α4k3+2.39683×109α3k1+2.11556×1012α3k2+1.22953×1015α3k3+4.62698×108α2k1+2.00368×1011α2k2+7.59375×1013α2k3+4.08889×107αk1+1.03937×1010αk2+2.7454×1012αk3+1.4×106k1+2.31429×108k2+4.46349×1010k3+1000. | (3.30) |
We put a=3, b=−2, k1=2, k2=4, and k3=6, then the functional integral J(u,v) becomes
J(u,v)=5.94419×1018α8+5.42784×1018α7+2.19363×1018α6+5.13419×1017α5+7.62802×1016α4+7.38564×1015α3+4.56427×1014α2+1.6514×1013α+2.68738×1011. | (3.31) |
Derive Eq (3.31) with respect to α, then values of α are:
α=−0.131961,α=−0.129142−0.0174743i,α=−0.129142+0.0174743i,α=−0.117682−0.0366515i,α=−0.117682+0.0366515i,α=−0.0866909−0.0518096i,α=−0.0866909+0.0518096i. |
We substitute the exact roots of α into Eq (3.31) give the following analytical equations:
J(u,v)=3.53554×106,J(u,v)=−1.84827×106−5.13825×106i,J(u,v)=−1.84827×106+5.13825×106i,J(u,v)=−2.46359×107+1.0125×107i,J(u,v)=−2.46359×107−1.0125×107i,J(u,v)=−2.31965×108+3.16423×108i,J(u,v)=−2.31965×108−3.16423×108i. | (3.32) |
We survey here the exact solutions of the nonlinear Schrödinger equation with polynomial law nonlinearity Eq (3.1).
Case 1: We suppose the ansatz function of the NLSE with polynomial law nonlinearity is in the form of a bright solitary wave solution.
h1(x,t)=Asech(w(x−tv)),Ψ(x,t)=Asech(w(x−tv))ei(μx−ωt), | (3.33) |
where A,w and v are the amplitude, the pulse width, and velocity of soliton in normalized unites. Substituting from Eq (3.33) in Eq (3.1), and separating the real and imaginary parts, we obtain
avw2+bμvω−aμ2v−bw2+vω+(−2avw2+A2k1v+2bw2)×sech2(w(x−tv))+A4k2vsech4(w(x−tv))+A6k3vsech6(w(x−tv))=0, | (3.34) |
(−2aμvw+bvwω+bμw+w)tanh(w(x−tv))=0. | (3.35) |
Equating the coefficients of the linearly independent terms to zero, we obtain the dynamical system in A,w,v,μ,ω,a,b,k1,k2,k3 by solving this system we get:
Family I:
A=±√2(μw2−vw2ω)k1μ2v+k1vw2,a=−v2ω2−w2v(μ2+w2)(vω−μ),b=2μvω+w2−μ2(μ2+w2)(μ−vω). | (3.36) |
The sufficient conditions for solitary wave solution stability are:
μw2−vw2ωk1μ2v+k1vw2>0,v(μ2+w2)(vω−μ)≠0,(μ2+w2)(μ−vω)≠0. | (3.37) |
Family II:
A=±√bμ2−bμvω+μ−2vωk1v,a=bμ+bvω+12μv,w=±√bμ3−bμ2vω+μ2−2μvωbvω+1−bμ, | (3.38) |
provided that
bμ2−bμvω+μ−2vωk1v>0,bμ3−bμ2vω+μ2−2μvωbvω+1−bμ>0,μv≠0. | (3.39) |
Family III:
A=±w√2ωbk1μ3+bk1μw2+k1μ2−k1w2,a=ω(b2μ2+b2w2+2bμ+1)bμ3+bμw2+μ2−w2,v=bμ3+bμw2+μ2−w2ω(bμ2+bw2+2μ), | (3.40) |
whenever
ωbk1μ3+bk1μw2+k1μ2−k1w2>0,bμ3+bμw2+μ2−w2≠0,ω(bμ2+bw2+2μ)≠0. | (3.41) |
Then, the solutions of the NLSE with polynomial law nonlinearity as bright solitary wave solutions are (see Figures 10 and 11):
Ψ11(x,t)=±√2(μw2−vw2ω)k1μ2v+k1vw2sech(w(x−tv))ei(μx−ωt), | (3.42) |
Ψ12(x,t)=±√bμ2−bμvω+μ−2vωk1vsech(w(x−tv))ei(μx−ωt), | (3.43) |
Ψ13(x,t)=±w√2ωbk1μ3+bk1μw2+k1μ2−k1w2sech(w(x−tv))ei(μx−ωt). | (3.44) |
Case 2: Another choice of the dark solitary wave solution of the NLSE with polynomial law nonlinearity is
Ψ(x,t)=(A+Btanh(w(x−tv)))ei(μx−ωt). | (3.45) |
By replacement from Eq (3.45) in Eq (3.1) and separating the real and imaginary parts
A7k3−aAμ2+21A5B2k3+A5k2+35A3B4k3+10A3B2k2+A3k1+Abμω+7AB6k3+5AB4k2+3AB2k1+Aω+(−21A5B2k3−70A3B4k3−10A3B2k2−21AB6k3−10AB4k2−3AB2k1)×sech2(w(x−tv))+(35A3B4k3+21AB6k3+5AB4k2)×sech4(w(x−tv))−7AB6k3sech6(w(x−tv))+(−aBμ2+7A6Bk3+35A4B3k3+5A4Bk2+21A2B5k3+10A2B3k2+3A2Bk1+bBμω+B7k3+B5k2+B3k1+Bω)tanh(w(x−tv))+(−2aBw2−35A4B3k3−42A2B5k3−10A2B3k2+2bBw2v−3B7k3−2B5k2−B3k1)tanh(w(x−tv))sech2(w(x−tv))+(21A2B5k3+3B7k3+B5k2)tanh(w(x−tv))sech4(w(x−tv))−B7k3tanh(w(x−tv))sech6(w(x−tv))=0, | (3.46) |
(2aBμw−bBμwv−bBwω−Bwv)sech2(w(x−tv))=0. | (3.47) |
Equating the coefficients of the linearly independent terms to zero, we deduce the coefficients A,B,w,v,μ,ω,a,b,k1,k2,k3 in the form:
Family I:
A=0,B=±i√k23k3,a=k32v2ω+k32μv+54k23w2−27k23v2ω227k23v(μ2−2w2)(vω−μ),b=54k23μvω−27k23μ2−2k32μv−54k23w227k23(μ2−2w2)(μ−vω),k1=3k32μ2v−54k23vw2ω−4k32vw2+54k23μw29k2k3v(μ2−2w2). | (3.48) |
The sufficient conditions for dark solitary wave solution stability are
k2k3<0,k23v(μ2−2w2)(vω−μ)≠0,k2k3v(μ2−2w2)≠0. | (3.49) |
Family II:
A=0,B=±i√k23k3,a=27b2k23μ2−54b2k23w2+54bk23μ+bk32v+27k2327k23v(bμ2−2bw2+2μ),k1=3bk32μ2v−4bk32vw2+6k32μv+54k23w29k2k3v(bμ2−2bw2+2μ),ω=27bk23μ3−54bk23μw2+27k23μ2+2k32μv+54k23w227k23v(bμ2−2bw2+2μ), | (3.50) |
provided that
k2k3<0,k2k3v(bμ2−2bw2+2μ)≠0,k23v(bμ2−2bw2+2μ)≠0. | (3.51) |
Family III:
A=0,B=±i√k23k3,a=bμ+bvω+12μv,k1=27bk23μvω−27bk23μ2−27k23μ+54k23vω+4k32v18k2k3v,w=−√μ(27bk23μvω−27k23μ+54k23vω−2k32v−27bk23μ2)54bk23vω+54k23−54bk23μ, | (3.52) |
whenever
k2k3<0,μv≠0,k2k3v≠0,μ(27bk23μvω−27k23μ+54k23vω−2k32v−27bk23μ2)54bk23vω+54k23−54bk23μ>0. | (3.53) |
Family IV:
B=±6√aμ2v2ω−2av2w2ω−aμ3v+2aμvw2+v2ω2−2w2v(2w2−μ2)(vω−μ)(k3(μ+vω)(2w2−μ2)(μ−vω)),A=0,b=2aμv−1μ+vω,k1=1v2/3(μ+vω)3√2w2(av(vω−μ)+1)−v(aμ2(vω−μ)+vω2)×(3√(k3(μ+vω)(2w2−μ2)(μ−vω))(2w2−μ2)(vω−μ)×(3v(aμ2(vω−μ)+vω2)−4w2(av(vω−μ)+1))),k2=3k33√2w2(av(vω−μ)+1)−v(aμ2(vω−μ)+vω2)3√v(2w2−μ2)(vω−μ)(k3(μ+vω)(2w2−μ2)(μ−vω)), | (3.54) |
such that
aμ2v2ω−2av2w2ω−aμ3v+2aμvw2+v2ω2−2w2v(2w2−μ2)(vω−μ)(k3(μ+vω)(2w2−μ2)(μ−vω))>0,μ+vω≠0,v2/3(μ+vω)3√2w2(av(vω−μ)+1)−v(aμ2(vω−μ)+vω2)≠0,3√v(2w2−μ2)(vω−μ)(k3(μ+vω)(2w2−μ2)(μ−vω))≠0. | (3.55) |
Then, the dark soliton solutions of the NLSE with polynomial law nonlinearity Eq (3.1) are (see Figure 12)
Ψ21(x,t)=±i√k23k3tanh(w(x−tv))ei(μx−ωt), | (3.56) |
Ψ24(x,t)=±6√aμ2v2ω−2av2w2ω−aμ3v+2aμvw2+v2ω2−2w2v(2w2−μ2)(vω−μ)(k3(μ+vω)(2w2−μ2)(μ−vω))×tanh(w(x−tv))ei(μx−ωt). | (3.57) |
In this paper, we discussed the two models of the nonlinear Schrödinger equation (NLSE) with polynomial law nonlinearity by effective and understandable techniques, such as the variational principle method based on the Lagrangian and the amplitude ansatz method. We found the functional integral and the Lagrangian of these models. Meanwhile, the solutions of the proposed equations were extracted by choice of different ansatz functions based on the Jost function, and they are continuous at all intervals. Firstly, the Jost function was approximated by piecewise linear ansatz function with a single nontrivial variational parameter in three cases from a region of a rectangular box. Then, the Jost function was approximated by the piecewise ansatz function containing the hyperbolic function in two cases of the two-box potential and was also approximated by quadratic polynomials with two free parameters rather than a piecewise linear ansatz function. Finally, this trial function had been approximated by the tanh function. Besides, we applied the amplitude ansatz method to obtain the new soliton solutions of the offered equations that contain bright soliton, dark soliton, bright-dark solitary wave solutions, rational dark-bright solutions, and periodic solitary wave solutions. The conditions for the stability of solutions were conducted. Graphical models, such as 2D, 3D, and contour plots, were induced using appropriate parameter values. These solutions have vital applications in applied sciences and might provide valuable support for investigators and physicists to solve more complex physical phenomena.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
The authors extend their appreciation to the Deputyship for Research and Innovation, Ministry of Education in Saudi Arabia for funding this research work through project number 445-9-693. Furthermore, the authors would like to extend their appreciation to Taibah University for its supervision support.
The authors declare that they have no competing interests.
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