# On Perturbative Cubic Nonlinear Schrodinger Equations under Complex Nonhomogeneities and Complex Initial Conditions

Authors: Magdy A. El-Tawil; Maha A. El-Hazmy
Journal: Differential Equations and Nonlinear Mechanics
 (2009)

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## Abstract

A perturbing nonlinear Schrodinger equation is studied under general complex nonhomogeneities and complex initial conditions for zero boundary conditions. The perturbation method together with the eigenfunction expansion and variational parameters methods are used to introduce an approximate solution for the perturbative nonlinear case for which a power series solution is proved to exist. Using Mathematica, the symbolic solution algorithm is tested through computing the possible approximations under truncation procedures. The method of solution is illustrated through case studies and figures.


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## Body

## 1. Introduction

The nonlinear Schrodinger (NLS) equation is the principal equation to be analyzed and solved in many fields, see [1–5], for examples. In the last two decades, there are a lot of  NLS problems depending on additive or multiplicative noise in the random case [6, 7] or a lot of solution methodologies in the deterministic case.Wang et al. [8] obtained the exact solutions to NLS using what they called the subequation method. They got four kinds of exact solutions of
(1.1)i∂u∂t+12∂2u∂x2+α|u|pu+β|u|2pu=0,
for which no sign to the initial or boundary conditions type is made. Xu and Zhang [9] followed the same previous technique in solving the higher-order NLS:
(1.2)i∂u∂x-12α∂2u∂t2+β|u|2u+iε∂3u∂t3+iδ|u|2∂u∂t+iγu2∂u*∂t=0.
Sweilam [10] solved
(1.3)i∂u∂t+∂2u∂x2+q|u|2u=0,t>0,L0<x<L1,with initial conditionu(x,0)=g(x) and boundary conditions ux(L0,t)=ux(L1,t)=0, which gives rise to solitary solutions using variational iteration method. Zhu [11] used the extended hyperbolic auxiliary equation method in getting the exact explicit solutions to the higher-order NLS:
(1.4)iqz-β12qtt+γ1|q|2q=iβ26qttt+β324qtttt-γ2|q|4q,
without any conditions. Sun et al. [12] solved the NLS:
(1.5)i∂ψ∂t+∂2ψ∂x2+a|ψ|2ψ=0,
with the initial condition  ψ(x,0)=ψ0(x) using Lie group method. By using coupled amplitude phase formulation, Porsezian and Kalithasan [13] constructed the quartic anharmonic oscillator equation from the coupled higher-order NLS. Two-dimensional grey solitons to the NLS were numerically analyzed by Sakaguchi and Higashiuchi [14]. The generalized derivative NLS was studied by Huang et al. [15] introducing a new auxiliary equation expansion method. Abou Salem and Sulem [16] studied the effective dynamics of solitons for the generalized Schrodinger equation in a random potential. El-Tawil [17] considered a nonlinear Schrodinger equation with random complex input and complex initial conditions. Colin et al. [18] considered three components of nonlinear Schrodinger equations related to the Raman amplification in a plasma. In [19], Jia-Min and Yu-Lu constructed an appropriate transformations and an extended elliptic subequation approach to find some exact solutions for variable coefficient cubic-quintic nonlinear Schrodinger equation with an external potential.In this paper, a straight forward solution algorithm is introduced using the transformation from a complex solution to a coupled equations in two real solutions, eliminating one of the solutions to get separate independent and higher-order equations, and finally introducing a perturbative approximate solution to the system.


## 2. The Linear Case

Consider the nonhomogeneous linear Schrodinger equation:(2.1)i∂u(t,z)∂z+α∂2u(t,z)∂t2=F1(t,z)+iF2(t,z),(t,z)∈(0,T)×(0,∞),
where u(t,z) is a complex valued function  which is subjected to
(2.2)I.C.:u(t,0)=f1(t)+if2(t),acomplexvaluedfunction,B.C.:uz(0,z)=0,uz(T,z)=0.
Let u(t,z)=ψ(t,z)+iϕ(t,z), ψ,ϕ: real valued functions. Substituting (2.2) in (2.1), the following coupled equations are got as follows:
(2.3)∂ϕ(t,z)∂z=α∂2ψ(t,z)∂t2+G1(t,z),∂ψ(t,z)∂z=α∂2ϕ(t,z)∂t2+G2(t,z),
where ψ(t,0)=f1(t), ϕ(t,0)=f2(t), G1(t,z)=-F1(t,z), G2(t,z)=F2(t,z), and all corresponding other I.C. and B.C. are zeros.Eliminating one of the variables in (2.3), one can get the following independent equations:
(2.4)∂4ψ(t,z)∂t4+1α2∂2ψ(t,z)∂z2=1α2ψ̃1(t,z),∂4ϕ(t,z)∂t4+1α2∂2ϕ(t,z)∂z2=1α2ψ̃2(t,z),
where
(2.5)ψ̃1(t,z)=∂G2∂z-α∂2G1∂t2,ψ̃2(t,z)=α∂G2∂t2+∂G1∂z.
Using the eigenfunction expansion technique [20],  the following solutions for (2.4) are obtained:
(2.6)ψ(t,z)=∑n=0∞Tn(z)sin(nπT)t,ϕ(t,z)=∑n=0∞τn(z)sin(nπT)t,
where  Tn(z) and  τn(z)  can be got through the applications of initial conditions and then solving the resultant second-order differential equations using the method of the variational parameter [21]. The final expressions can be got as follows
(2.7)Tn(z)=(C1+A1(z))sinβnz+(C2+B1(z))cosβnz,τn(z)=(C3+A2(z))sinβnz+(C4+B2(z))cosβnz,
where
(2.8)βn=α(nπT)2,A1(z)=1βn∫ψ̃1n(z;n)cos(βn)zdz,B1(z)=-1βn∫ψ̃1n(z;n)sin(βn)zdz,A2(z)=1βn∫ψ̃2n(z;n)cos(βn)zdz,B2(z)=-1βn∫ψ̃2n(z;n)sin(βn)zdz,
in which
(2.9)ψ̃1n(z;n)=2T∫0Tψ̃1(t,z)sin(nπTt)dt,ψ̃2n(z;n)=2T∫0Tψ̃2(t,z)sin(nπTt)dt.
The following conditions should also be satisfied:
(2.10)C2=2T∫0Tf1(t)sin(nπT)dt-B1(0),C4=2T∫0Tf2(t)sin(nπT)dt-B2(0).
Finally, the following solution is obtained:
(2.11)u(t,z)=(ψ(t,z)+iϕ(t,z)),
or
(2.12)|u(t,z)|2=(ψ2(t,z)+ϕ2(t,z)).


## 3. The Nonlinear Case

Consider the homogeneous nonlinear Schrodinger equation:(3.1)i∂u(t,z)∂z+α∂2u(t,z)∂t2+ε|u(t,z)|2u(t,z)+iγu(t,z)=F1(t,z)+iF2(t,z),(t,z)∈(0,T)×(0,∞),
where u(t,z) is a complex-valued function which is subjected to  the initial and boundary conditions (2.2).Lemma 3.1.
The solution of (3.1) with the constraints (2.2) is a power series in  ε if the solution exists.Proof.
Atε=0, the following linear equation is got:
(3.2)i∂u(t,z)∂z+α∂2u(t,z)∂t2+iγu(t,z)=F(t,z),(t,z)∈(0,T)×(0,∞),
which has the solution, see the previous section,
(3.3)u0(t,z)=e-γz(w0(t,z)+iv0(t,z)).
Following Pickard approximation, (3.13) can be rewritten as
(3.4)i∂un(t,z)∂z+α∂2un(t,z)∂t2+iγun(t,z)=F(t,z)-ε|un-1(t,z)|2un-1(t,z),n≥1.
At n=1, the iterative equation takes the following form:
(3.5)i∂u1(t,z)∂z+α∂2u1(t,z)∂t2+iγu1(t,z)=F(t,z)-ε|u0(t,z)|2u0(t,z)=F(t,z)+εh1(t,z),
which can be solved as a linear case with zero initial and boundary conditions. The following general solution can be obtained:
(3.6)w1(t,z)=∑n=0∞(T0n+εT1n)sin(nπT)t,v1(t,z)=∑n=0∞(τ0n+ετ1n)sin(nπT)t,u1(t,z)=e-γz(w1(t,z)+iv1(t,z))=u1(0)+εu1(1).
At n=2, the following equation is obtained:
(3.7)i∂u2(t,z)∂z+α∂2u2(t,z)∂t2+iγu2(t,z)=F(t,z)-ε|u1(t,z)|2u1(t,z)=F(t,z)+εh2(t,z),
which can be solved as a linear case with zero initial and boundary conditions. The following general solution can be obtained:
(3.8)u2(t,z)=u2(0)+εu2(1)+ε2u2(2)+ε3u2(3)+ε4u2(4).
Continuing like this, one can get
(3.9)un(t,z)=un(0)+εun(1)+ε2un(2)+ε3un(3)+⋯+ε(n+m)un(n+m).
As n→∞, the solution (if exists) can be reached as  u(t,z)=limn→∞un(t,z). Accordingly, the solution is a power series in ε.According to the previous lemma, one can assume the solution of (3.1) as the following:
(3.10)u(t,z)=∑n=0∞εnun.
Let u(t,z)=ψ(t,z)+iϕ(t,z), ψ,ϕ: real valued functions. The following coupled equations are got:
(3.11)∂ϕ(t,z)∂z=α∂2ψ(t,z)∂t2+ε(ψ2+ϕ2)ψ-γϕ-F1,∂ψ(t,z)∂z=-α∂2ϕ(t,z)∂t2-ε(ψ2+ϕ2)ϕ-γψ+F2,
where  ψ(t,0)=f1(t), ϕ(t,0)=f2(t), and all corresponding other I.C. and B.C. are zeros.As a perturbation solution, one can assume that(3.12)ψ(t,z)=ψ0+εψ1+ε2ψ2+⋯,ϕ(t,z)=ϕ0+εϕ1+ε2ϕ2+⋯,
where  ψ0(t,0)=f1(t), ϕ0(t,0)=f2(t), and all corresponding other I.C. and B.C. are zeros.Substituting (3.12) into (3.11) and then equating the equal powers of ε, one can get the following set of coupled equations:
(3.13)∂ϕ0(t,z)∂z=α∂2ψ0(t,z)∂t2-γϕ0-F1,(3.14)∂ψ0(t,z)∂z=-α∂2ϕ0(t,z)∂t2-γψ0+F2,(3.15)∂ϕ1(t,z)∂z=α∂2ψ1(t,z)∂t2-γϕ1+(ψ03+ψ0ϕ02),(3.16)∂ψ1(t,z)∂z=-α∂2ϕ1(t,z)∂t2-γψ1-(ϕ03+ϕ0ψ02),(3.17)∂ϕ2(t,z)∂z=α∂2ψ2(t,z)∂t2-γϕ2+(3ψ02ψ1+2ψ0ϕ0ϕ1+ψ1ϕ02),(3.18)∂ψ2(t,z)∂z=-α∂2ϕ2(t,z)∂t2-γψ2-(3ϕ02ϕ1+2ϕ0ψ0ψ1+ϕ1ψ02),
and so on. The prototype equations to be solved are
(3.19)∂ϕi(t,z)∂z=α∂2ψi(t,z)∂t2-γϕi+Gi(1),i≥1,∂ψi(t,z)∂z=-α∂2ϕi(t,z)∂t2-γψi+Gi(2),i≥1,
where ψi(t,0)=δi,0f1(t), ϕi(t,0)=δi,0f2(t) and all other corresponding conditions are zeros. The nonhomogeneity functions Gi(1) and Gi(2) are functions computed from previous steps.Following the solution algorithm described in the previous section for the linear case, the general symbolic algorithm in Figure39 can be simulated through the use of a symbolic package, mathematica-5 is used in this paper.


### 3.1. The Zero-Order Approximation

In this case,(3.20)u(0)(t,z)=(ψ0+iϕ0),
where
(3.21)ψ0(t,z)=e-γz∑n=0∞T0nsin(nπT)t,ϕ0(t,z)=e-γz∑n=0∞τ0nsin(nπT)t,
in which
(3.22)T0n(z)=A01(z)sinβnz+(C02+B01(z))cosβnz,τ0n(z)=A02(z)sinβnz+(C̃02+B02(z))cosβnz,
where the constants and variables A01(z),C02,B01(z), A02(z),C̃02, and B02(z) can be got by the aid of Section 2.The absolute value of the zero-order approximation is got from(3.23)|u(0)(t,z)|2=ψ02+ϕ02.


### 3.2. The First-Order Approximation

(3.24)u(1)(t,z)=u(0)+ε(ψ1+iϕ1),
where
(3.25)ψ1(t,z)=e-γz∑n=0∞T1n(z)sin(nπT)t,ϕ1(t,z)=e-γz∑n=0∞τ1n(z)sin(nπT)t,
in which
(3.26)T1n(z)=A11(z)sinβnz+(C12+B11(z))cosβnz,τ1n(z)=A12(z)sinβnz+(C̃12+B12(z))cosβnz,
where the constants and variables A11(z), B11(z), A12(z), and B12(z) can be evaluated in a similar manner as the zero-order approximation whereas C̃12=-B12(0) and C12=-B11(0).The absolute value of the first-order approximation can be got using(3.27)|u(1)(t,z)|2=|u(0)(t,z)|2+2ε(ψ0ψ1+ϕ0ϕ1)+ε2(ψ12+ϕ12).


### 3.3. The Second-Order Approximation

(3.28)u(2)(t,z)=u(1)(t,z)+ε2(ψ2+iϕ2),
where
(3.29)ψ2(t,z)=e-γz∑n=0∞T2n(z)sin(nπT)t,ϕ2(t,z)=e-γz∑n=0∞τ2n(z)sin(nπT)t,
in which
(3.30)T2n(z)=A21(z)sinβnz+(C22+B21(z))cosβnz,τ2n(z)=A22(z)sinβnz+(C̃22+B22(z))cosβnz,
where the constants and variables A21(z),C22,B21(z), A22(z),C̃22, and B22(z) can be evaluated similarly  as the previous approximation.The absolute value of the second-order approximation can be got using(3.31)|u(2)(t,z)|2=|u(1)(t,z)|2+2ε2(ψ0ψ2+ϕ0ϕ2)+2ε3(ψ1ψ2+ϕ1ϕ2)+ε4(ψ22+ϕ22).


## 3.1. The Zero-Order Approximation

In this case,(3.20)u(0)(t,z)=(ψ0+iϕ0),
where
(3.21)ψ0(t,z)=e-γz∑n=0∞T0nsin(nπT)t,ϕ0(t,z)=e-γz∑n=0∞τ0nsin(nπT)t,
in which
(3.22)T0n(z)=A01(z)sinβnz+(C02+B01(z))cosβnz,τ0n(z)=A02(z)sinβnz+(C̃02+B02(z))cosβnz,
where the constants and variables A01(z),C02,B01(z), A02(z),C̃02, and B02(z) can be got by the aid of Section 2.The absolute value of the zero-order approximation is got from(3.23)|u(0)(t,z)|2=ψ02+ϕ02.


## 3.2. The First-Order Approximation

(3.24)u(1)(t,z)=u(0)+ε(ψ1+iϕ1),
where
(3.25)ψ1(t,z)=e-γz∑n=0∞T1n(z)sin(nπT)t,ϕ1(t,z)=e-γz∑n=0∞τ1n(z)sin(nπT)t,
in which
(3.26)T1n(z)=A11(z)sinβnz+(C12+B11(z))cosβnz,τ1n(z)=A12(z)sinβnz+(C̃12+B12(z))cosβnz,
where the constants and variables A11(z), B11(z), A12(z), and B12(z) can be evaluated in a similar manner as the zero-order approximation whereas C̃12=-B12(0) and C12=-B11(0).The absolute value of the first-order approximation can be got using(3.27)|u(1)(t,z)|2=|u(0)(t,z)|2+2ε(ψ0ψ1+ϕ0ϕ1)+ε2(ψ12+ϕ12).


## 3.3. The Second-Order Approximation

(3.28)u(2)(t,z)=u(1)(t,z)+ε2(ψ2+iϕ2),
where
(3.29)ψ2(t,z)=e-γz∑n=0∞T2n(z)sin(nπT)t,ϕ2(t,z)=e-γz∑n=0∞τ2n(z)sin(nπT)t,
in which
(3.30)T2n(z)=A21(z)sinβnz+(C22+B21(z))cosβnz,τ2n(z)=A22(z)sinβnz+(C̃22+B22(z))cosβnz,
where the constants and variables A21(z),C22,B21(z), A22(z),C̃22, and B22(z) can be evaluated similarly  as the previous approximation.The absolute value of the second-order approximation can be got using(3.31)|u(2)(t,z)|2=|u(1)(t,z)|2+2ε2(ψ0ψ2+ϕ0ϕ2)+2ε3(ψ1ψ2+ϕ1ϕ2)+ε4(ψ22+ϕ22).


## 4. Case Studies

To examine the proposed solution algorithm, see  Figure39, some case studies are illustrated.


### 4.1. One Input Is On

Case Study 1
Takingf1(t)=0, f2(t)=0, F1(t,z)=1, F2(t,z)=0, and following the solution algorithm, the selective results for the zero-order approximation are got in Figures 1, 2, and 3.Figure 1
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 2
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 3
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for  different values of  t, considering only one term in the series (M=1).Case Study 2
Takingf1(t)=0, f2(t)=0, F1(t,z)=0, F2(t,z)=1 and following the solution algorithm, it has been noticed that the same results for the case study 1 are got.Case Study 3
Takingf1(t)=1, f2(t)=0, F1(t,z)=0, F2(t,z)=0 and following the solution algorithm, the selective results for the first-zero approximation are got in Figures 4, 5, and 6.Figure 4
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 5
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 6
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for  different values of  t, considering only one term in the series (M=1).One can notice the decrease of the solution level and its higher variability.Case Study 4
Takingf1(t)=0, f2(t)=1, F1(t,z)=0, F2(t,z)=0 and following the solution algorithm, it has been noticed that the same results for the case study 3 are got:


### 4.2. Two Inputs Are On

Case Study 5
Takingf1(t)=0, f2(t)=0, F1(t,z)=1, F2(t,z)=1 and following the solution algorithm, the selective results for the zero-order approximation are got in Figures 7, 8, and 9.Figure 7
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 8
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 9
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of  t, considering only one term in the series (M=1).One can notice that the solution level becomes a little bit higher than that of case study 2.Case Study 6
Takingf1(t)=1, f2(t)=1, F1(t,z)=0, F2(t,z)=0 and following the solution algorithm, the selective results for the zero-order approximation are got in Figures 10, 11, and 12.Figure 10
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 11
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 12
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of  t, considering only one term in the series (M=1).One can notice the little increase of the solution level than that of case studies 3 and 4.Case Study 7
Takingf1(t)=1, f2(t)=0, F1(t,z)=1, F2(t,z)=0 and following the solution algorithm, the selective results for the zero-order approximation are got in Figures 13, 14, and 15.Figure 13
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 14
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 15
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of t, considering only one term in the series (M=1).One can notice the small perturbations at small values ofz.


### 4.3. Three Inputs Are On

Case Study 8
Takingf1(t)=1, f2(t)=1, F1(t,z)=1, F2(t,z)=0 and following the solution algorithm, the selective results for the zero-order approximation are got in Figures 16, 17, and 18.Figure 16
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 17
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 18
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of  t, considering only one term in the series (M=1).One can notice the increase of the depth of the perturbations.Case Study 9
Takingf1(t)=1, f2(t)=0, F1(t,z)=1, F2(t,z)=1 and following the solution algorithm, the selective results for the zero-order approximation are got in Figures 19, 20, and 21.Figure 19
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 20
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 21
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of t, considering only one term in the series (M=1).One can notice that the perturbations become smaller than that of case study 8.


### 4.4. Four Inputs Are On

Case Study 10
Taking the case off1(t)=1, f2(t)=1, F1=1, F2=1, the following final results for the zero-order approximation are obtained in Figures 22, 23, and 24.Figure 22
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 23
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 24
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of  t, considering only one term in the series (M=1).One can notice the little increase in the solution level.


### 4.5. Exponential Nonhomogeneity

Case Study 11
Taking the case off1(t)=1, f2(t)=0, F1(t,z)=e-t, F2(t,z)=0, the following final results for the zero-order approximation are obtained in Figures 25, 26, and 27.Figure 25
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 26
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 27
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of t, considering only one term in the series (M=1).One can notice the low solution level and high perturbations.


### 4.6. Exponential Initial Condition

Case Study 12
Taking the case off1(t)=e-t, f2(t)=0, F1(t,z)=1, F2(t,z)=0, the following final results for the zero-order approximation are obtained in Figures 28, 29, and 30.Figure 28
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 29
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 30
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of t, considering only one term in the series (M=1).One can notice that a higher solution level is got compared with case study 11 and less perturbations are got at small values ofz.


### 4.7. First-Order Approximation

Case Study 13
Taking the case off1(t)=1, f2(t)=0, F1(t,z)=1, F2(t,z)=0, the following final results for the zero and first-order approximations are obtained in Figures 31, 32, and 33.Figure 31
The zero-order approximation of|u(0)| at α=1, T=10, γ=0 with considering only one term in the series (M=1).Figure 32
The zero-order approximation|u(0)| at α=1, T=10, γ=0 for different values of z, considering only one term in the series (M=1).Figure 33
The zero-order approximation|u(0)| at  α=1, T=10, γ=0 for different values of t, considering only one term in the series (M=1).Figure 34
The first-order approximation of|u(1)| at α=1, T=10, γ=0, ε=01 with considering only one term in the series (M=1).Figure 35
The first-order approximation of|u(1)| at α=1, T=10, γ=0, ε=05 with considering only one term in the series (M=1).Figure 36
The first-order approximation of|u(1)| at α=1, T=10, γ=0, ε=1 with  considering only one term in the series (M=1).Figure 37
The first-order approximation|u(1)| at  α=1, T=10, γ=0, ε=05 for different values of z, considering only one term in the series (M=1).Figure 38
The first-order approximation|u(1)| at α=1, T=10, γ=0, ε=05 for different values of t, considering only one term in the series (M=1).Figure 39
The general solution algorithm.One can notice the oscillations of the solution level compared with case 7.One can notice that the solution level increases with the increase ofε.


## 4.1. One Input Is On

Case Study 1
Takingf1(t)=0, f2(t)=0, F1(t,z)=1, F2(t,z)=0, and following the solution algorithm, the selective results for the zero-order approximation are got in Figures 1, 2, and 3.Figure 1
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 2
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 3
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for  different values of  t, considering only one term in the series (M=1).Case Study 2
Takingf1(t)=0, f2(t)=0, F1(t,z)=0, F2(t,z)=1 and following the solution algorithm, it has been noticed that the same results for the case study 1 are got.Case Study 3
Takingf1(t)=1, f2(t)=0, F1(t,z)=0, F2(t,z)=0 and following the solution algorithm, the selective results for the first-zero approximation are got in Figures 4, 5, and 6.Figure 4
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 5
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 6
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for  different values of  t, considering only one term in the series (M=1).One can notice the decrease of the solution level and its higher variability.Case Study 4
Takingf1(t)=0, f2(t)=1, F1(t,z)=0, F2(t,z)=0 and following the solution algorithm, it has been noticed that the same results for the case study 3 are got:


## 4.2. Two Inputs Are On

Case Study 5
Takingf1(t)=0, f2(t)=0, F1(t,z)=1, F2(t,z)=1 and following the solution algorithm, the selective results for the zero-order approximation are got in Figures 7, 8, and 9.Figure 7
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 8
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 9
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of  t, considering only one term in the series (M=1).One can notice that the solution level becomes a little bit higher than that of case study 2.Case Study 6
Takingf1(t)=1, f2(t)=1, F1(t,z)=0, F2(t,z)=0 and following the solution algorithm, the selective results for the zero-order approximation are got in Figures 10, 11, and 12.Figure 10
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 11
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 12
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of  t, considering only one term in the series (M=1).One can notice the little increase of the solution level than that of case studies 3 and 4.Case Study 7
Takingf1(t)=1, f2(t)=0, F1(t,z)=1, F2(t,z)=0 and following the solution algorithm, the selective results for the zero-order approximation are got in Figures 13, 14, and 15.Figure 13
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 14
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 15
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of t, considering only one term in the series (M=1).One can notice the small perturbations at small values ofz.


## 4.3. Three Inputs Are On

Case Study 8
Takingf1(t)=1, f2(t)=1, F1(t,z)=1, F2(t,z)=0 and following the solution algorithm, the selective results for the zero-order approximation are got in Figures 16, 17, and 18.Figure 16
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 17
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 18
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of  t, considering only one term in the series (M=1).One can notice the increase of the depth of the perturbations.Case Study 9
Takingf1(t)=1, f2(t)=0, F1(t,z)=1, F2(t,z)=1 and following the solution algorithm, the selective results for the zero-order approximation are got in Figures 19, 20, and 21.Figure 19
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 20
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 21
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of t, considering only one term in the series (M=1).One can notice that the perturbations become smaller than that of case study 8.


## 4.4. Four Inputs Are On

Case Study 10
Taking the case off1(t)=1, f2(t)=1, F1=1, F2=1, the following final results for the zero-order approximation are obtained in Figures 22, 23, and 24.Figure 22
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 23
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 24
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of  t, considering only one term in the series (M=1).One can notice the little increase in the solution level.


## 4.5. Exponential Nonhomogeneity

Case Study 11
Taking the case off1(t)=1, f2(t)=0, F1(t,z)=e-t, F2(t,z)=0, the following final results for the zero-order approximation are obtained in Figures 25, 26, and 27.Figure 25
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 26
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 27
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of t, considering only one term in the series (M=1).One can notice the low solution level and high perturbations.


## 4.6. Exponential Initial Condition

Case Study 12
Taking the case off1(t)=e-t, f2(t)=0, F1(t,z)=1, F2(t,z)=0, the following final results for the zero-order approximation are obtained in Figures 28, 29, and 30.Figure 28
The zero-order approximation of|u(0)| at α=1, T=10, γ=02 with considering only one term in the series (M=1).Figure 29
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of z, considering only one term in the series (M=1).Figure 30
The zero-order approximation|u(0)| at α=1, T=10, γ=02 for different values of t, considering only one term in the series (M=1).One can notice that a higher solution level is got compared with case study 11 and less perturbations are got at small values ofz.


## 4.7. First-Order Approximation

Case Study 13
Taking the case off1(t)=1, f2(t)=0, F1(t,z)=1, F2(t,z)=0, the following final results for the zero and first-order approximations are obtained in Figures 31, 32, and 33.Figure 31
The zero-order approximation of|u(0)| at α=1, T=10, γ=0 with considering only one term in the series (M=1).Figure 32
The zero-order approximation|u(0)| at α=1, T=10, γ=0 for different values of z, considering only one term in the series (M=1).Figure 33
The zero-order approximation|u(0)| at  α=1, T=10, γ=0 for different values of t, considering only one term in the series (M=1).Figure 34
The first-order approximation of|u(1)| at α=1, T=10, γ=0, ε=01 with considering only one term in the series (M=1).Figure 35
The first-order approximation of|u(1)| at α=1, T=10, γ=0, ε=05 with considering only one term in the series (M=1).Figure 36
The first-order approximation of|u(1)| at α=1, T=10, γ=0, ε=1 with  considering only one term in the series (M=1).Figure 37
The first-order approximation|u(1)| at  α=1, T=10, γ=0, ε=05 for different values of z, considering only one term in the series (M=1).Figure 38
The first-order approximation|u(1)| at α=1, T=10, γ=0, ε=05 for different values of t, considering only one term in the series (M=1).Figure 39
The general solution algorithm.One can notice the oscillations of the solution level compared with case 7.One can notice that the solution level increases with the increase ofε.


## 5. Conclusions

The perturbation technique introduces an approximate solution to the NLS equation with a perturbative nonlinear term for a finite time interval. Using mathematica, the difficult and huge computations problems were fronted to some extent for limited series terms. To get more improved orders, it is expected to face a problem of computation. With respect to the solution level, the effect of the nonhomogeneity is higher than the effect of the initial condition. The initial conditions also cause perturbations for the solution at small values of the space variable. The solution level increases with the increase ofε.



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*Source: 395894-2009-09-29.xml*