# Impact of Multiplicative Noise on the Exact Solutions of the Fractional-Stochastic Boussinesq-Burger System

Authors: Wael W. Mohammed; Farah M. Al-Askar; M. El-Morshedy
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this paper, we consider the fractional-stochastic Boussinesq-Burger system (FSBBS) generated by the multiplicative Brownian motion. The Jacobi elliptic function techniques are used to create creative elliptic, hyperbolic, and rational fractional-stochastic solutions for FSBBS. Furthermore, we draw 2D and 3D graphs by using the MATLAB Package for some obtained solutions of FSBBS to discuss the influence of the Brownian motion on these solutions. Finally, we indicate that the Brownian motion stabilizes the solutions of FSBBS around zero.


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## Body

## 1. Introduction

Nonlinear partial differential equations (NLPDEs) have grown in popularity in the area of nonlinear science, owing to their large variety of uses in economics [1], engineering [2], civil engineering [3], soil mechanics [4], physics [5], quantum mechanics [6], statistical mechanics [7], solid-state physics [8], population ecology [9], etc. Solitons are among the most common in the setting of NLPDE solutions, and they are essential for understanding nonlinear physical phenomena. Solitons are utilized to understand the properties of nonlinear media in various areas including quantum electronics, plasma physics, nonlinear optics, and fluid dynamics [10–13]. Recently, the searching of exact soliton solutions to NLPDEs has become an enthralling research topic in engineering and applied sciences. Many techniques have been used to determine exact solutions for NLPDE including tanh-sech [14, 15], Darboux transformation [16], sine-cosine [17, 18], exp−ϕς-expansion [19], G′/G-expansion [20–22], Lie symmetry analysis method [23], improved F-expansion method [24, 25], Hirota’s function [26], the Jacobi elliptic function [27, 28], and perturbation [29, 30].The fractional differential equation is extensively used in fluid mechanics, solid state physics, optical fibers, neural physics, quantum field theory, mathematical biology, plasma physics, and other areas [31–34]. Researchers recommend fractional-order derivative over ordinary order derivative because integer-order derivative is essentially a local operator, but fractional-order derivative is so much more. Also, they explain physical phenomena such as quantum mechanics, diffusion, gravity, heat, elasticity, fluid dynamics, electrodynamics, electrostatics, and sound. Recently, the exact solutions with conformable derivative have been obtained in many papers for instance [35–40].On the other hand, a wide variety of complex nonlinear physical phenomena can be represented using stochastic partial differential equations (SPDEs). These kind of equations can be found in many fields, such as physics and finance.On the other side, stochastic partial differential equations (SPDEs) can be used to represent a wide range of complicated nonlinear physical processes. These kind of equations appear in a variety of areas including engineering, geophysical, biology, climate dynamics, finance, and physics [41–43].To realize a higher level of qualitative accord, we take the following fractional-stochastic Boussinesq-Burger system (FSBBS) perturbed in the itô sense by multiplicative noise:(1)dΦ+2ΦDxαΦ−12DxαΨdt=σΦdB,(2)dΨ+2DxαΦΨ−12Dx3αΦdt=σΨdB,where Φx,t denotes the horizontal velocity field. Dα is the conformable derivative (CD) [44]. Ψx,t is the height of the water surface above the bottom horizontal level. Bt is a Brownian motion (BM) and σ is the noise strength.The Boussinesq–Burgers system (BBS), withα=1 and σ=0, appears in fluid flow research and explains how shallow water waves spread. Due to the importance of BBS, many researchers have created its exact solutions by using various methods such as Hirota method [45], Lie symmetry method [46], sine-Gordon expansion method [47], Jacobi elliptic function method [48], extended homogeneous balance [49], Darboux transformation [50], The modified exp−ϕζ-expansion function method [51], and Exp-function method [52]. On the other side, many techniques have been documented for fractional BBS, including a domain decomposition method [53] and generalized Kudryashov method [54]. The exact solutions of the FSBBS (1-2) have not yet been studied.Our novelty of this paper is to find the exact fractional stochastic solutions of FSBBS (1-2). In the presence of a stochastic term and the fractional space, this study is the first to obtain analytical solutions to the FSBBS (1-2). Numerous solutions, including those involving elliptic, trigonometric, rational, and hyperbolic functions, can be obtained using the Jacobi elliptic function technique. Moreover, we utilize MATLAB to build 2D and 3D figures for some of the obtained solutions in this study to display how the BM influences on the solutions of FSBBS (1-2).The layout of the document is as follows: in Sec.2, we define and give some properties of the CD and BM. In Sec. 3, we use an effective wave transformation to establish the FSBBS (1-2) wave equation. In Sec.4, we use the Jacobi elliptic function method to generate the analytic of FSBBS (1-2). While, in Sec.5, the effect of the BM on the solutions obtained is studied. In Sec. 6, the document’s conclusion is shown.


## 2. Preliminaries

In this section, we define and clarify some characteristics of the BM and CD. In the following, we define BMBt as:Definition 1 (see [55]).
Stochastic processBtt≥0 is said a BM if the following conditions satisfy: Bt is continuous function of t≥0; B0=0; for τ1<τ2,Bτ2−Bτ1 is independent; and Bτ2−Bτ1 has a Gaussian distribution ℵ0,τ2−τ1.Lemma 2 (see [55]).
EeσBt=e1/2σ2t for σ≥0.Definition 3 (see [44]).
Letϕ:0,∞⟶ℝ, then the CD of ϕ of order α∈0,1 is defined as
(3)Dxαϕx=limκ⟶0ϕx+κx1−α−ϕxκ.Let us go through some of the CD’s features. Ifa,b are constant, then
(1)
Dxαa=0,(2)
Dxαxb=bxb−α,(3)
DxαaΘ1x+bΘ2x=aDxαΘ1x+bDxαΘ2x,(4)
DxαΘx=x1−αdΘ/dx,(5)
DxαΘ1∘Θ2x=x1−αΘ2′xΘ1′Θ2x.


## 3. Wave Equation of FSBBS

The next wave transformation is used(4)Φx,t=φξeσBt−1/2σ2t,Ψx,t=ψξeσBt−1/2σ2t,ξ=1αxα+ωt,in order to attain the wave equation of FSBBS (1-2). Where ω is a constant, φ and ψ are deterministic functions. Plugging Equation (4) into Equations (1) and (2) and utilizing
(5)dΦ=ωφ′dt+σφdBeσBt−1/2σ2t,dΨ=ωψ′dt+σψdBeσBt−1/2σ2t,DxαΦ=φ′eσBt−1/2σ2t,DxαΨ=ψ′eσBt−1/2σ2t,Dx3αΦ=φ′′′eσBt−1/2σ2t,DxαΦΨ=φψ′eσBt−1/2σ2t,we get(6)ωφ′+2φφ′eσBt−1/2σ2t−12ψ′=0,(7)ωψ′+2φψ′eσBt−1/2σ2t−12φ′′′=0.Taking expectationE· for Equations (6) and (7), we attain
(8)ωφ′+2φφ′e−1/2σ2tEeσBt−12ψ′=0,(9)ωψ′+2φψ′e−1/2σ2tEeσBt−12φ′′′=0.SinceBt is a Gaussian distribution, then EeσBt=eσ2/2t. Now Equations (8) and (9) become
(10)ωφ′+2φφ′−12ψ′=0,(11)ωψ′+φψ′−12φ′′′=0.Integrating Equations (10) and (11) and putting the constants of integration equal zero, we have
(12)ψ=2ωφ+2φ2,(13)ωψ+2φψ−12φ′′=0.Plugging Equation (12) into (13), we attain
(14)φ′′−8φ3−12ωφ2−4ω2φ=0.


## 4. Exact Solutions of FSBBS

We use the Jacobi elliptic functions approach described by Peng [56] to find the solutions of Equation (14). Consequently, we can therefore derive the exact solutions of FSBBS (1-2).


### 4.1. Jacobi Elliptic Functions Method

First, we suppose the solutions of Equation (14) are
(15)φξ=∑i=1Naiχi,where χ is the solution of
(16)χ′=12pχ4+qχ2+r,where r,q and p are real parameters.We note from the next Table1 that Equation (16) has different types of solutions relying on r,q and p:Table 1
All possible solutions for Equation (16).
Casepqrχξ12 m2-(1 + m2)1snξ222 m2-1-m2(1-m2)dsξ322-m2(1-m2)csξ4-2 m22 m2-1(1-m2)cnξ5-22-m2(m2-1)dnξ6m22m2−2214snξ1±dnξ7m22m2−22m24snξ1±dnξ8−12m2+12−1−m224mcnξ±dnξ9m2−12m2+12m2−14dnξ1±snξ101−m221−m221−m24cnξ1±snξ111−m2221−m22214snξdn±cnξ12200cξ13010ceξWherednξ,m=dnξ,m,cnξ=cnξ,m,snξ=snξ,m, for 0<m<1 are the Jacobi elliptic functions (JEFs). If m⟶1, then JEFs are converted into the hyperbolic functions as follows:
(17)dnξ⟶sechξsnξ⟶tanhξ,cnξ⟶sechξ,csξ⟶cschξ,ds⟶cschξ.


### 4.2. Solutions of FSBBS

By balancingφ′′ with φ3 in Equation (14), we can calculate the parameter N as
(18)N+2=3N⇒N=1.Hence, Equation (15) with N=1 becomes
(19)φ=a0+a1χ.Differentiating Equation (19) twice, we have, by using (16),
(20)φ′′=a1qχ+a1pχ3.Substituting Equation (19) and Equation (20) into Equation (14) we obtain
(21)a1p−8a13χ3−24a0a12+12ωa12χ2+a1q−24a02a1−24ωa0a1−4ω2a1χ−8a03+12ωa02+4ω2a0=0.Equating each coefficient ofχk, for k=0,1,2,3, to zero, we have
(22)a1p−8a13=0,24a0a12+12ωa12=0,a1q−24a02a1−24ωa0a1−4ω2a1=0,and
(23)8a03+12ωa02+4ω2a0=0.We get by solving these equations:(24)a0=12−12q,a1=±p8,ω=±−12q,for p>0 and q≤0. Then, the Equation (14) has the solutions:
(25)φξ=−18q±p8χξ.Therefore, by utilizing (4) and (12), the solution of the FSBBS (2-1) are
(26)Φx,t=−18q+p8χξeσBt−1/2σ2t,Ψx,t=−3q4+−pqχξ+p4χ2eσBt−1/2σ2t.There are numerous cases, by using the previous Table1, for q≤0, p>0 and r as follows:Case 1.
Ifq=−1+m2,p=2m2 and r=1, then χξ=snξ. Therefore, the FSBBS (1-2) has the solution
(27)Φx,t=−18q+p8sn1αxα+−q2teσBt−1/2σ2t,(28)Ψx,t=−3q4+−pqsnxαα+−q2t+p4sn2xαα+−q2teσBt−1/2σ2t.If m⟶1, then Equations (27) and (28) degenerates to
(29)Φx,t=−18q+p8tanh1αxα+−q2teσBt−1/2σ2t,Ψx,t=−3q4+−pqtanh1αxα+−q2t+p4tanh21αxα+−q2teσBt−1/2σ2t.Case 2.
Ifq=2m2−1 for m≤1/√2,p=2 and r=−m21−m2, then χξ=dsξ. So, the FSBBS (1-2) has the solution:
(30)Φx,t=−18q+p8ds1αxα+−q2teσBt−1/2σ2t,(31)Ψx,t=−3q4+−pqdsxαα+−q2t+p4ds2xαα+−q2teσBt−1/2σ2t.If m⟶1, then Equations (30) and (31) degenerates to
(32)Φx,t=−18q+p8cschxαα+−q2teσBt−1/2σ2t,Ψx,t=−3q4+−pqcschxαα+−q2t+p4csch2xαα+−q2teσBt−1/2σ2t.Case 3.
Ifq=m2−2/2,p=m2/2 and r=1/4orr=m2/4, then χξ=snξ/1±dnξ. Therefore, the FSBBS (1-2) has the solution:
(33)Φx,t=−18q+p8sn1/αxα+−q/2t1±dn1/αxα+−q/2teσBt−1/2σ2t,(34)Ψx,t=−3q4+−pqsn1/αxα+−q/2t1±dn1/αxα+−q/2t++p4sn21/αxα+−q/2t1±dn1/αxα+−q/2t2eσBt−1/2σ2t.Ifm⟶1, then Equations (33) and (34) degenerates to
(35)Φx,t=−18q+p8tanh1/αxα+−q/2t1±sech1/αxα+−q/2teσBt−1/2σ2t,(36)Ψx,t=−3q4+−pqtanh1/αxα+−q/2t1±sech1/αxα+−q/2t++p4tanh2xα/α+−q/2t1±sechxα/α+−q/2t2eσBt−1/2σ2t.Case 4.
Ifq=0,p=2 and r=0, then χξ=C/ξ. Hence, the FSBBS (1-2) has the solution:
(37)Φx,t=αC2x−αeσBt−1/2σ2t,Ψx,t=αC2x−2αeσBt−1/2σ2t.Remark 4.
If we setσ=0 and α=1 in Equations (27) and (36), then we get the same results as reported in [48].


## 4.1. Jacobi Elliptic Functions Method

First, we suppose the solutions of Equation (14) are
(15)φξ=∑i=1Naiχi,where χ is the solution of
(16)χ′=12pχ4+qχ2+r,where r,q and p are real parameters.We note from the next Table1 that Equation (16) has different types of solutions relying on r,q and p:Table 1
All possible solutions for Equation (16).
Casepqrχξ12 m2-(1 + m2)1snξ222 m2-1-m2(1-m2)dsξ322-m2(1-m2)csξ4-2 m22 m2-1(1-m2)cnξ5-22-m2(m2-1)dnξ6m22m2−2214snξ1±dnξ7m22m2−22m24snξ1±dnξ8−12m2+12−1−m224mcnξ±dnξ9m2−12m2+12m2−14dnξ1±snξ101−m221−m221−m24cnξ1±snξ111−m2221−m22214snξdn±cnξ12200cξ13010ceξWherednξ,m=dnξ,m,cnξ=cnξ,m,snξ=snξ,m, for 0<m<1 are the Jacobi elliptic functions (JEFs). If m⟶1, then JEFs are converted into the hyperbolic functions as follows:
(17)dnξ⟶sechξsnξ⟶tanhξ,cnξ⟶sechξ,csξ⟶cschξ,ds⟶cschξ.


## 4.2. Solutions of FSBBS

By balancingφ′′ with φ3 in Equation (14), we can calculate the parameter N as
(18)N+2=3N⇒N=1.Hence, Equation (15) with N=1 becomes
(19)φ=a0+a1χ.Differentiating Equation (19) twice, we have, by using (16),
(20)φ′′=a1qχ+a1pχ3.Substituting Equation (19) and Equation (20) into Equation (14) we obtain
(21)a1p−8a13χ3−24a0a12+12ωa12χ2+a1q−24a02a1−24ωa0a1−4ω2a1χ−8a03+12ωa02+4ω2a0=0.Equating each coefficient ofχk, for k=0,1,2,3, to zero, we have
(22)a1p−8a13=0,24a0a12+12ωa12=0,a1q−24a02a1−24ωa0a1−4ω2a1=0,and
(23)8a03+12ωa02+4ω2a0=0.We get by solving these equations:(24)a0=12−12q,a1=±p8,ω=±−12q,for p>0 and q≤0. Then, the Equation (14) has the solutions:
(25)φξ=−18q±p8χξ.Therefore, by utilizing (4) and (12), the solution of the FSBBS (2-1) are
(26)Φx,t=−18q+p8χξeσBt−1/2σ2t,Ψx,t=−3q4+−pqχξ+p4χ2eσBt−1/2σ2t.There are numerous cases, by using the previous Table1, for q≤0, p>0 and r as follows:Case 1.
Ifq=−1+m2,p=2m2 and r=1, then χξ=snξ. Therefore, the FSBBS (1-2) has the solution
(27)Φx,t=−18q+p8sn1αxα+−q2teσBt−1/2σ2t,(28)Ψx,t=−3q4+−pqsnxαα+−q2t+p4sn2xαα+−q2teσBt−1/2σ2t.If m⟶1, then Equations (27) and (28) degenerates to
(29)Φx,t=−18q+p8tanh1αxα+−q2teσBt−1/2σ2t,Ψx,t=−3q4+−pqtanh1αxα+−q2t+p4tanh21αxα+−q2teσBt−1/2σ2t.Case 2.
Ifq=2m2−1 for m≤1/√2,p=2 and r=−m21−m2, then χξ=dsξ. So, the FSBBS (1-2) has the solution:
(30)Φx,t=−18q+p8ds1αxα+−q2teσBt−1/2σ2t,(31)Ψx,t=−3q4+−pqdsxαα+−q2t+p4ds2xαα+−q2teσBt−1/2σ2t.If m⟶1, then Equations (30) and (31) degenerates to
(32)Φx,t=−18q+p8cschxαα+−q2teσBt−1/2σ2t,Ψx,t=−3q4+−pqcschxαα+−q2t+p4csch2xαα+−q2teσBt−1/2σ2t.Case 3.
Ifq=m2−2/2,p=m2/2 and r=1/4orr=m2/4, then χξ=snξ/1±dnξ. Therefore, the FSBBS (1-2) has the solution:
(33)Φx,t=−18q+p8sn1/αxα+−q/2t1±dn1/αxα+−q/2teσBt−1/2σ2t,(34)Ψx,t=−3q4+−pqsn1/αxα+−q/2t1±dn1/αxα+−q/2t++p4sn21/αxα+−q/2t1±dn1/αxα+−q/2t2eσBt−1/2σ2t.Ifm⟶1, then Equations (33) and (34) degenerates to
(35)Φx,t=−18q+p8tanh1/αxα+−q/2t1±sech1/αxα+−q/2teσBt−1/2σ2t,(36)Ψx,t=−3q4+−pqtanh1/αxα+−q/2t1±sech1/αxα+−q/2t++p4tanh2xα/α+−q/2t1±sechxα/α+−q/2t2eσBt−1/2σ2t.Case 4.
Ifq=0,p=2 and r=0, then χξ=C/ξ. Hence, the FSBBS (1-2) has the solution:
(37)Φx,t=αC2x−αeσBt−1/2σ2t,Ψx,t=αC2x−2αeσBt−1/2σ2t.Remark 4.
If we setσ=0 and α=1 in Equations (27) and (36), then we get the same results as reported in [48].


## 5. The Influence of Fractional Derivative and Noise

Here, the influence of noise on the achieved solutions of FSBBS (1-2) is explained. For various values ofα (the fractional derivative order) and σ (noise strength), some graphs are provided using the MATLAB tools.Firstly the Fractional Derivative Influence. In Figures 1 and 2, if σ=0 and m=0.4, we can observe that the surface shrinks when α is decreasing:Figure 1
3D-graph of Equation (33) with σ=0 and various values of α=1,0.5.Figure 2
3D-graph of Equation (34) with σ=0 and various values of α=1,0.5.Secondly the Noise Influence. In Figures 3 and 4, when noise is introduced, the surface flattens significantly if its strength is increased σ=1,2Figure 3
3D-graph of Equation (33) with σ=1,2 and α=1,0.5.Figure 4
3D-graph of Equation (34) with σ=1,2 and α=1,0.5.In Figure5, we introduce 2D-graph of the Φx,t in (33) with α=1 and with σ=0,0.5,1,2, which highlights the previous outcomes:Figure 5
2D graph of Equation (33)We may deduct from Figures1–5 that:
(1)
When the fractional-orderα increases, the surface expands,(2)
The multiplicative noise stabilizes the solutions of FSBBS at zero.This results show that it is important to add the stochastic term into the Boussinesq-Burger equation in order to obtain accurate solutions.


## 6. Conclusions

In this article, the exact fractional-stochastic solutions of the fractional-stochastic Boussinesq-Burger system (1-2) driven by multiplicative noise were successfully obtained by using the Jacobi elliptic function method. Numerous analytical solutions for FSBBS (1-2) including elliptic, trigonometric, rational, and hyperbolic functions can be determined using the Jacobi elliptic function method. Because of the importance of FSBBS in fluid flow research and in explaining the propagation of shallow water waves, the acquired solutions are far more beneficial and efficient in understanding several critical complicated physical phenomena. In addition, we utilized the MATLAB package to demonstrate how multiplicative noise and fractional derivative influenced the solutions of FSBBS. As a result, we concluded that the stabilization of the solutions of the FSBBS (1-2) is affected by the multiplicative noise.



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