# KdV Equation Solution by Double Laplace Adomian Decomposition Method and Its Convergence Analysis

Authors: Umar Habib; Salman Zeb; Kamal Shah; Sardar Muhammad Hussain
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

In this paper, the double Laplace Adomian decomposition method (DLADM) is utilized for the solution of the nonlinear Korteweg–de Vries (KdV) equation. The convergence analysis of the suggested approach has been carried out and some examples of the KdV equation are solved for demonstrating the proposed approach. We also compared the obtained series solutions with existing method solutions which show close agreement with each other.


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## Body

## 1. Introduction

A differential equation (DE) is one which involves the derivative of one or more dependent variables in terms of one or more independent variables [1]. DE has a long history that started in 17th century when Newton and Leibniz developed calculus [2]. In 18th century the major work is done in the Leibnizian convention and its multi-component formation is expanded which resulted in PDEs. New masters emerged such as Lagrange, Laplace, and Euler who made significant contributions to astronomy, mechanics, and continuous media. In the 19th century various types of equations, methods, and their solutions were developed such as special functions and Fourier analysis. Also in this century, with the influence of Maxwell, most developments were made in the field of electricity and magnetism, heat theory, and optics. Relativity theory was enriched and most applications were made to the field of dynamical systems and quantum mathematics in the 20th century [3]. Differential equations are used in a variety of recent research work in scientific fields and technology including physics, engineering, and biology. For example, Goodwine [4] studied the applications of DE in aeronautical and mechanical engineering. Moreover, Sumithra [5] investigated DE and modeled the variation of physical quantities such as velocity, displacement, pressure, temperature, current, and voltage with respect to time and position. Applications of DEs in physics such as Newton’s second law of motion, the Euler–Lagrange, Hamilton’s, and Einstein’s field equations has also been discussed. DEs also has a wide range of applications in biology for example Sumithra [5] explored the Lotka–Volterra and Predator–Prey equations which shows the importance of DEs in biology. He also discussed other significant DEs like equations of population growth and radioactive decay.Nonlinear partial differential equations (NLPDEs) are found in various physical models including plasma physics and fluid dynamics. Nonlinear system of PDEs have also received great attention and they have been occurred in a variety of processes of biology and chemistry and it has various uses in these disciplines [6]. Sine–Gordon, Klein–Gordon, KdV, and Boussinesq equations are some of the most significant examples of NLPDEs [7]. As the superposition principle does not apply to NLPDEs, theretofore numerical solutions are usually employed to solve to such kinds of equations. The KdV problem is an NLPDE of the third order idealized in 1877 by Boussinesq [8]. The equation is named after two Dutchmen Korteweg and Devries, who originally obtained the problem in 1895 [9]. The KdV problem was derived essentially from a hydromagnetic model of cold plasma [10, 11]. It is also been implied to depict numerous essential physical processes in different fields like in plasma physics, harmonic crystals [6], and shallow water waves [12] among others. It depicts shallow sea waves with a long wavelength and a tiny amplitude. Some approaches like numerical strategies, soliton solution, inverse scattering transform, and least action principle [13] are found in the literature to solve the KdV problem.The Laplace transform is one of the approaches for investigating the analytical solution of linear differential equations. Various famous practitioners have used the aforementioned method extensively [14, 15]. Kilicman and Gadain [16] expanded the idea of the Laplace transform to double Laplace transform (DLT). In comparison to the single Laplace transform, the DLT is more useful and appropriate [17]. However, there is a scarcity of work on the DLT [18]. There are certain uses of the DLT in combination with decomposition methods to handle NLPDEs [19–23]. It is thought that a universal strategy for discovering accurate solutions to NLPDEs has not yet been discovered [6]. For approaches related to fuzzy systems we refer the readers to [24–26]. The ADM on the other hand is thought to be capable of handling NLPDEs. The ADM is used to solve linear as well as nonlinear equations. This method has various advantages over other strategies, most notably are as it avoids linearization and perturbation when solving nonlinear equations. It provides a precise and explicit answer with fewer calculations and avoids physically implausible assumptions. Several scholars have looked into the convergence of Adomian’s approach [27–30]. The method generates a series solution and converges to its exact solution if it exists.In this work, we consider a non-linear KdV problem of third order with initial and boundary conditions in the following form:(1)wtx,t+awx,twxx,t+bwxxxx,t=fx,t,wx,0=fx,w0,t=gt,wx0,t=ht,wxx0,t=kt,where t is any positive real number, x∈−∞,∞, a,b are real numbers [31]. The solution of (1) has been obtained using DLADM and its convergence analysis has also been studied. The suggested approach has been employed on some examples for its applications and the obtained solutions are compared with existing method solutions in the literature.


## 2. Preliminaries

In this section, some basic concepts are described which are used in our main work.


### 2.1. Double Laplace Transform

DLT of a functionwx,t is defined as follows:(2)w¯p,s=L2wx,t=∫0∞e−px∫0∞e−stwx,tdtdx,where L2 denotes DLT and w¯p,s is the DLT of wx,t.


### 2.2. Differentiation Property

Letw(x, t) be continuous and wtx,t be piecewise continuous functions in 0≤x≤X and 0≤t≤Tfor allX>0, T>0. Let wx,t be of exponential order as x⟶∞, t⟶∞. Then, the DLT of wtx,t and wxx,t exists and are defined as follows:(3)LxLtwtx,t=sW¯p,s−W¯p,0,LxLtwxx,t=pW¯p,s−W¯0,s.We will utilize the following formulae for DLT of derivatives of functions:(4)L2∂∂twx,t=sL2wx,t−Lxwx,0,(5)L2∂3∂x3wx,t=p3L2wx,t−p2Ltw0,t−pLt∂∂xw0,t−Lt∂2∂x2w0,t,where ∂/∂twx,t and ∂3/∂x3wx,t are first and third order partial derivatives w.r.t. t and x respectively, L2 denotes double Laplace transform w.r.t. x and t.


### 2.3. Double Inverse Laplace Transform

For an analytical functionW¯¯p,s where s,p are complex numbers and such that Res≥a, Rep≥b where a and b are real constants, then the inverse double Laplace transform Lx−1Lt−1  =  wx,y is defined as follows:(6)wx,y=Lx−1Lt−1W¯¯p,s=12πι∫a−ι∞a+ι∞epxdp12πι∫b−ι∞b+ι∞esyw¯¯p,sds,a>0,b>0.The following Table1 contains some important functions and their Laplace transform used in this work.Table 1
Double Laplace transform of some functions.
ux,tL2ux,t=Up,s11/pst1/ps2x1/sp2tnn!/psn+1xmtnm!n!/pm+1sn+1utx,tsUp,s−Up,0uxx,tpUp,s−U0,suxxx,tp2Up,s−pU0,s−Ux0,suxxxx,tp3Up,s−p2U0,s−pUx0,s−Uxx0,seax+bt1/p−as−bFor further study about DLT, see [16, 18, 32, 33].


## 2.1. Double Laplace Transform

DLT of a functionwx,t is defined as follows:(2)w¯p,s=L2wx,t=∫0∞e−px∫0∞e−stwx,tdtdx,where L2 denotes DLT and w¯p,s is the DLT of wx,t.


## 2.2. Differentiation Property

Letw(x, t) be continuous and wtx,t be piecewise continuous functions in 0≤x≤X and 0≤t≤Tfor allX>0, T>0. Let wx,t be of exponential order as x⟶∞, t⟶∞. Then, the DLT of wtx,t and wxx,t exists and are defined as follows:(3)LxLtwtx,t=sW¯p,s−W¯p,0,LxLtwxx,t=pW¯p,s−W¯0,s.We will utilize the following formulae for DLT of derivatives of functions:(4)L2∂∂twx,t=sL2wx,t−Lxwx,0,(5)L2∂3∂x3wx,t=p3L2wx,t−p2Ltw0,t−pLt∂∂xw0,t−Lt∂2∂x2w0,t,where ∂/∂twx,t and ∂3/∂x3wx,t are first and third order partial derivatives w.r.t. t and x respectively, L2 denotes double Laplace transform w.r.t. x and t.


## 2.3. Double Inverse Laplace Transform

For an analytical functionW¯¯p,s where s,p are complex numbers and such that Res≥a, Rep≥b where a and b are real constants, then the inverse double Laplace transform Lx−1Lt−1  =  wx,y is defined as follows:(6)wx,y=Lx−1Lt−1W¯¯p,s=12πι∫a−ι∞a+ι∞epxdp12πι∫b−ι∞b+ι∞esyw¯¯p,sds,a>0,b>0.The following Table1 contains some important functions and their Laplace transform used in this work.Table 1
Double Laplace transform of some functions.
ux,tL2ux,t=Up,s11/pst1/ps2x1/sp2tnn!/psn+1xmtnm!n!/pm+1sn+1utx,tsUp,s−Up,0uxx,tpUp,s−U0,suxxx,tp2Up,s−pU0,s−Ux0,suxxxx,tp3Up,s−p2U0,s−pUx0,s−Uxx0,seax+bt1/p−as−bFor further study about DLT, see [16, 18, 32, 33].


## 3. Solution of KdV Equation Using DLADM

Considering (1) and utilizing the DLT L2 on it where x⟶p, t⟶s while making use of (4) and (5), we obtain the following equation:(7)s+bp3w¯p,s−Fp+aL2wx,twxx,t−p2bGs−pbHs−bKs=Fp,s,where w¯p,s and Fp,s are the DLTs of wx,t and fx,t respectively, and Fp, Gs, Hs and Ks are the LT of wx,0,w0,t,wx0,t and wxx0,t, respectively.(7) has been simplified and rearranged, and we obtain the following equation:(8)w¯p,s=1s+bp3Fp+bp2Gs+bpHs+bKs+Fp,s−as+bp3L2wx,twxx,t.Using the inverse LTL2−1 in (8), we obtain the following equation:(9)L2−1wp,s=wx,t=L2−1Fp+bp2Gs+bpHs+bKs+Fp,ss+bp3−L2−1as+bp3L2wx,twxx,t.Using the ADM, the solution function and the non-linear terms are treated as follows:(10)wx,t=∑m=0∞wmx,t,(11)wx,twxx,t=∑m=0∞Am=Nwx,t.The Adomian polynomialAm is given by the following equation:(12)Am=1m!dmdλmN∑k=0mλkwkx,tλ=0,m=0,1,2,….Using (10) and (11), (9) becomes the following equation:(13)∑m=0∞wmx,t=L2−1Fp+bp2Gs+bpHs+bKs+Fp,ss+bp3−L2−1as+bp3L2∑m=0∞Am.Utilizing the Adomian decomposition formula (9) and (12) for m=0,1,2,…, we have the following equation:(14)A0=w0w0x,A1=w0w1x+w1w0x,A2=w0w2x+w1w1x+w2w0x.(13) implies that(15)w0+w1+w2+…=L2−1Fp+bp2Gs+bpHs+bKs+Fp,ss+bp3−L2−1−as+bp3L2A0+A1+A2+….Comparing the terms we have the following equation:(16)w0=L2−1Fp+bp2Gs+bpHs+bKs+Fp,ss+bp3,w1=L2−1−as+bp3L2A0=L2−1−as+bp3L2w0∂∂xw0,w2=L2−1−as+bp3L2A1=L2−1−as+bp3L2w0∂∂xw1+w1∂∂xw0,⋮wm+1x,t=L2−1−as+bp3L2Am.The series solution is(17)wx,t=w0+w1+w2+⋯+wm+1.


## 4. Convergence Analysis

This section is devoted to the convergence analysis of DLADM for KdV equation. The general KdV equation is(18)wtx,t+awx,twxx,t+bwxxxx,t=fx,t.Forw,v∈H (Hilbert Space) define by the following equation:(19)H=L2α,β×0,T,whereα≪0,where w:α,β×0,T⟶R×R with ‖w‖2=∫α,β×0,Tw2x,tdxdt<∞, such that(20)H=L2−1−1s+p3L2wx,tp,sx,t<∞.Theorem 1.
DLADM is used for the nonlinear KdV equation approaches to the exact solution if the following hypotheses holds:
∀w,v∈H and hemi-continuous operator “K”(21)H1:Kw−Kv,w−v≥M∗w−v2,M∗>0.
ForM∗>0, there existCM∗>0withw≤M∗,v≤M∗(22)H2:Kw−Kv,w1≤CM∗w−vw1,∀w1∈H,Proof.
The operator form of (18) for all w∈H is given by the following equation:(23)wtx,t=Kw=−awx,twxx,t−bwxxxx,t+fx,t,Similarly for all v∈H we have the following equation:(24)vtx,t=Kv=−avx,tvxx,t−bvxxxx,t+fx,t.
Subtracting (24) from (23) and putting wt=∂w/∂t,vt=∂v/∂t where w⟶wx,t then simplifying, we obtain the following equation:(25)Kw−Kv=−a2∂∂xw2−v2−b∂3∂x3w−v.
The inner product form of (25) is(26)Kw−Kv,w−v=a2−∂∂xw2−v2,w−v+b−∂3∂x3w−v,w−v,as ∂/∂x is a differential operator in H therefore utilizing the Schwartz inequality for C1>0, we obtain the following equation:(27)∂∂xw2−v2,w−v≤C1w2−v2w−v.
Sincew2−v2=w−vw+v, hence(28)∂∂xw2−v2,w−v≤C1w−v2w+v,utilizing the triangle inequality w+v≤w+v, (28) implies that(29)∂∂xw2−v2,w−v≤C1w−v2w+v.
Using the fact that for allw,v∈H there must exists constants say C2 and C3 such that C2>0, C3>0 with w≤C2, v≤C3, (29) becomes(30)∂∂xw2−v2,w−v≤C1C2+C3w−v2,which implies(31)−∂∂xw2−v2,w−v≥C1C2+C3w−v2.
Similarly for differential∂3/∂x3 in H and for constant C4>0, we obtain the following equation:(32)∂3∂x3w−v,w−v≤C4w−v2,which implies(33)−∂3∂x3w−v,w−v≥C4w−v2.
Making use of (31) and (33), (26) becomes(34)Kw−Kv,w−v≥a2C1C2+C3+bC4‖w−v‖2.
TakingM∗=a/2C1C2+C3+bC4, we obtain the following equation:(35)Kw−Kv,w−v≥M∗w−v2,hence H1 holds.
To verifyH2, for all w1∈H the inner product form of (25) is(36)Kw−Kv,w1=a2−∂∂xw2−v2,w1+b−∂3∂x3w−v,w1.
For differential operator and anyM∗>0, we have the following equation:(37)−∂∂xw2−v2,w1≤M∗w2−v2w1‖,utilizing w2−v2=w+vw−v and making use of triangle inequality, (37) becomes(38)−∂∂xw2−v2,w1≤M∗w+vw−vw1.
Using thatw≤M∗, v≤M∗, we obtain the following equation:(39)−∂∂xw2−v2,w1≤2M∗2w−vw1.
Similarly for differential operator−∂3/∂x3 we have the following equation:(40)−∂3∂x3w−v,w1≤M∗w−vw1,and making use of (39) and (40) in (36) we obtain the following equation:(41)Kw−Kv,w1≤aM∗2+bM∗w−vw1.
TakingaM∗2+bM∗=CM∗, (41) implies(42)Kw−Kv,w1≤CM∗w−vw1,which shows that H2 holds and the proof is completed.


## 5. Applications

In this section, we solved some examples of the KdV problem to explain the applications of the proposed method for nonlinear PDEs. Comparison of DLADM solutions with existing method solutions are provided in Tables2 and 3.Table 2
Comparison of the approximate solutions obtained via DLADM with Picard’s successive approximation method (SAM) for Example1 when t=0.1
xSAM solution [34]DLADM solution00.0000000000000000.0100000000000000.10.1010000000000000.1199916667000000.20.2020000000000000.2298666718000000.30.3030000000000000.3393250868000000.40.4040000000000000.4478673169000000.50.5050000000000000.5547947669000000.60.60600000000000000.6592111086000000.70.7070000000000000.7600243470000000.80.8080000000000000.8559498870000000.90.9090000000000000.94551480040000011.0100000000000001.027063492000000Table 3
Comparison of an approximate solution by DLADM with SAM solution fort=0.1 Example 2.
xSAM solution [34]DLADM solution01.4000000000000001.7333625570000000.10.9384480644000001.1501937020000000.20.6290605497000000.8393359610000000.30.4216718967000000.6635765631000000.40.2826551252000000.5570123291000000.50.1894693965000000.4869200210000000.60.1270051346000000.4363423751000000.70.0851340876800000.3960086214000000.80.0570670855700000.3604972474000000.90.0382532114300000.32635234410000010.0256418944500000.291116367000000Example 1.
We consider the following problem with conditions:(43)wtx,t−wx,twxx,t+wxxxx,t=−δ′xt+t2,w0,t=wxx0,t=0,wx,0=x,wx0,t=1+t+t2,where δx is the Dirac Delta function.Applying DLT to (43) and then utilizing (4) and (5), we obtain the following equation:(44)p3+sw¯p,s=w¯p,0+p2w¯0,s+pwx¯0,s+w¯xx0,s−L2δ′xt+t2+L2wx,twxx,t.Utilizing the initial and boundary conditions and simplifying we obtain the following equation:(45)w¯p,s=1p2s+1p3+sL2wx,twxx,t,where we have used the following equation:(46)w¯p,0=1p2,w¯0,s=0,wx¯0,s=1s+1s2+2s3,w¯xx0,s=0.Taking inverse DLT of (45), we have the following equation:(47)L2−1w¯p,s=wx,t=L2−11p2s+L2−11p3+sL2wx,twxx,t.Using the ADM, from (47) we have the following equation:(48)∑m=0∞wmx,t=x+L2−11p3+sL2∑n=0∞Am.By comparing terms we obtainw0=x,(49)w1=L2−11p3+sL2A0,evaluating (12) for n=0 we obtain A0=x, putting this value in (49) and simplifying, we obtain the following equation:(50)w1=xt−12δ′xt2,Similarly,(51)w2=xt2−2x44!+2x77!.As the series solution is given by the following equation:(52)wx,t=w0+w1+w2+…,Putting values we obtain the following solution:(53)wx,t=x+xt−12δ′xt2+xt2+….


### 5.1. Convergence Analysis

For allw,v∈H and a=−1, b=1, (34) implies that(54)Kw−Kv,w−v≥−12C1C2+C3+C4w−v2.TakingM∗=−1/2C1C2+C3+C4, (54) implies that(55)Kw−Kv,w−v≥M∗w−v2.HenceH1 holds.To verifyH2, for all w,v,w1∈H and a=−1, b=1, (41) implies(56)Kw−Kv,w1≤−M∗2+M∗w−vw1,taking −M∗2+M∗=CM∗, we ge the following from (56):(57)Kw−Kv,w1≤CM∗w−vw1.ThereforeH2 also holds.Example 2.
Considering the following problem(58)wtx,t+wx,twxx,t+wxxxx,t=−68e−4x−4t,wx,0=1e4x,w0,t=1e4t,wx0,t=−4e4t,wxx0,t=16e4t,and applying DLT on the problem while using the formulae (4) and (5) and utilizing initial and boundary conditions. Simplifying, we obtain the following equation:(59)wx,t=e−4x−4t−L2−11p3+sL2wwx.Using the ADM, we have the following equation:(60)∑n=0∞wnx,t=e−4x−4t−L2−11p3+sL2∑n=0∞Am.By comparison, we have the following equation:(61)w0=e−4x−4t,and(62)w1=4Lx−11p+8Lt−11p3+ss+8.Using the convolution theorem forLt−11/p3+ss+8, (62) becomes the following equation:(63)w1=e2x−8t5−cos3x39e8t+x−2sin3x133e8t+x−1130e8t+x−e2x5+cos3x39ex+2sin3x133ex+1130e8x+256te−8x65+4te2x15−8te−xcos3x39−16tsin3x133−4t.In the same way we can findw2,w3,…,wn., and the solution in series form is then given by the following equation:(64)wx,t=e−4x−4t+e2x−8t5−cos3x39e8t+x−2sin3x133e8t+x−1130e8t+x−e2x5+cos3x39ex+2sin3x133ex+1130e8x+256te−8x65+4te2x15−8te−xcos3x39−16tsin3x133−4t.


### 5.2. Convergence Analysis

For allw,v∈H and a=1, b=1, the result specified in (20) gives the following equation:(65)Kw−Kv,w−v≥12C1C2+C3+C4w−v2.TakingM∗=1/2C1C2+C3+C4, (65) implies(66)Kw−Kv,w−v≥M∗w−v2,hence H1 holds.To verifyH2, for all w,v,w1∈H and a=1, b=1, the result given in (41) implies(67)Kw−Kv,w1≤M∗2+M∗w−vw1.TakingM∗2+M∗=CM∗, we have from (67), the following:(68)Kw−Kv,w1≤CM∗w−vw1,which implies H2 holds.


## 5.1. Convergence Analysis

For allw,v∈H and a=−1, b=1, (34) implies that(54)Kw−Kv,w−v≥−12C1C2+C3+C4w−v2.TakingM∗=−1/2C1C2+C3+C4, (54) implies that(55)Kw−Kv,w−v≥M∗w−v2.HenceH1 holds.To verifyH2, for all w,v,w1∈H and a=−1, b=1, (41) implies(56)Kw−Kv,w1≤−M∗2+M∗w−vw1,taking −M∗2+M∗=CM∗, we ge the following from (56):(57)Kw−Kv,w1≤CM∗w−vw1.ThereforeH2 also holds.Example 2.
Considering the following problem(58)wtx,t+wx,twxx,t+wxxxx,t=−68e−4x−4t,wx,0=1e4x,w0,t=1e4t,wx0,t=−4e4t,wxx0,t=16e4t,and applying DLT on the problem while using the formulae (4) and (5) and utilizing initial and boundary conditions. Simplifying, we obtain the following equation:(59)wx,t=e−4x−4t−L2−11p3+sL2wwx.Using the ADM, we have the following equation:(60)∑n=0∞wnx,t=e−4x−4t−L2−11p3+sL2∑n=0∞Am.By comparison, we have the following equation:(61)w0=e−4x−4t,and(62)w1=4Lx−11p+8Lt−11p3+ss+8.Using the convolution theorem forLt−11/p3+ss+8, (62) becomes the following equation:(63)w1=e2x−8t5−cos3x39e8t+x−2sin3x133e8t+x−1130e8t+x−e2x5+cos3x39ex+2sin3x133ex+1130e8x+256te−8x65+4te2x15−8te−xcos3x39−16tsin3x133−4t.In the same way we can findw2,w3,…,wn., and the solution in series form is then given by the following equation:(64)wx,t=e−4x−4t+e2x−8t5−cos3x39e8t+x−2sin3x133e8t+x−1130e8t+x−e2x5+cos3x39ex+2sin3x133ex+1130e8x+256te−8x65+4te2x15−8te−xcos3x39−16tsin3x133−4t.


## 5.2. Convergence Analysis

For allw,v∈H and a=1, b=1, the result specified in (20) gives the following equation:(65)Kw−Kv,w−v≥12C1C2+C3+C4w−v2.TakingM∗=1/2C1C2+C3+C4, (65) implies(66)Kw−Kv,w−v≥M∗w−v2,hence H1 holds.To verifyH2, for all w,v,w1∈H and a=1, b=1, the result given in (41) implies(67)Kw−Kv,w1≤M∗2+M∗w−vw1.TakingM∗2+M∗=CM∗, we have from (67), the following:(68)Kw−Kv,w1≤CM∗w−vw1,which implies H2 holds.


## 6. Conclusion

In this paper, the KdV problem solution has been obtained via DLADM. A convergence analysis of the proposed strategy is also carried out for studying its convergence analysis in order to show the accuracy of the resulting series solution. Furthermore, examples of the KdV equation were solved using DLADM and the results are compared with the SAM procedure which agree with each other. In the future, the DLADM procedure can be studied for finding solutions of nonlinear PDEs arising in different fields of science and engineering.



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