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Numerical solution of coupled type Fractional order Burgers’ equation
using Finite Difference and Fibonacci Collocation Method
Mohd. Kashif, Kushal Dhar Dwivedi, T. Som
PII: S0577-9073(21)00353-1
DOI: https://doi.org/10.1016/j.cjph.2021.10.044
Reference: CJPH 1742
To appear in: Chinese Journal of Physics
Received date : 26 July 2021
Revised date : 16 October 2021
Accepted date : 22 October 2021
Please cite this article as: M. Kashif, K.D. Dwivedi and T. Som, Numerical solution of coupled
type Fractional order Burgers’ equation using Finite Difference and Fibonacci Collocation Method,
Chinese Journal of Physics (2022), doi: https://doi.org/10.1016/j.cjph.2021.10.044.
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©2022 The Physical Society of the Republic of China (Taiwan). Published by Elsevier B.V. All
rights reserved.
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Numerical solution of coupled type Fractional order Burgers’
equation using Finite Difference and Fibonacci Collocation Method
Mohd. Kashifa, Kushal Dhar Dwivedib, T. Soma
aDepartment of Mathematical Sciences, Indian Institute of Technology (BHU), Varanasi-221005, India
bDepartment of Mathematics, S. N. Government P. G. College Khandwa, M.P.-450001, India
In this article a non-standard finite difference collocation method is developed with the help of
Fibonacci polynomial to solve the coupled type fractional order Burgers’ equation. To show the
efficiency of the method, it is used to solve the coupled Burgers’ equation having exact solu-
tions and compared the obtained numerical results with the existing results through error analysis.
Through the tabular presentation of the results, it is shown that the proposed method is performing
much better as compared to the existing methods even for less degree of approximation and less
order of temporal discretization. After validation, the method is used for solving an unsolved non-
linear fractional order coupled Burgers’ equation and simulate the results for different fractional
order spatial derivative for different values of the parameters.
Keywords: Fractional coupled diffusion equation, Fibonacci Polynomial, Non-standard finite
difference method.
1. Introduction
The system of Burgers’ equation is the most common non-linear time-dependent partial differ-
ential equation as it consists of both non-linear propagation and diffusion effects. Things become
even more complicated when dealing with Burgers’ equations in coupled form as it consists of the
information of two different solutions depending on each other. Due to its wide range of applica-
bility in various filed of science, finding the best solutions of the Burgers’ equation always the hot
topic to the researchers [1]. Many researchers have solved the coupled Burgers’ equation (CBE)
of integer order analytically with particular initial and boundary conditions. Kaya [2] has given an
explicit solution of CBE with some particular initial conditions with the help of the decomposition
method. Soliman [3] has solved the CBE with the help of tanh-function. Mohamed and Torky
[4] have given an analytical solution of CBE with some particular initial and boundary conditions.
As we know that an analytical solution is not always possible when dealing with PDEs. Things
become more complex to deal with non-linear coupled PDEs. This is why researchers are seek-
ing their interest to develop an efficient method to solve this type of PDEs numerically. Many
researchers have solved the CBE of integer order numerically viz., Ahmed [5] has solved the CBE
numerically using variational iteration algorithm. Abazari and Borhanifar [6] have used the dif-
ferential transformation method to solve the CBEs. Bak et al. [7] have solved the CBEs with the
help of the semi-Lagrangian approach.
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The study of fractional order nonlinear diffusion equations is few in number. The growing
interest in fractional order diffusion equations (FDEs) is due of their numerous useful applications
in various engineering areas. It is used for an accurate description of the solute transportation in
complex media like porous aquifers. Due to the nonlocal property of fractional derivative, FDEs
have comprehensive applications in the fields of engineering, physics, economics, etc. Also due to
nonlocal property, FDEs have a greater memory effect than the standard order diffusion equation.
Time, space, and time-space FDEs are extensively used to describe physical and engineering prob-
lems like anomalous diffusion, solute transport, signal processing, control, etc. For the description
and understanding of dispersion phenomena, the FDEs have fundamental importance and have
received a lot of attention during last few decades [8–21]. Thus scientists and engineers are in-
volved to find the solutions of FDEs for their non-local behaviour, greater flexibility in models and
its convergence to the integer order systems. The current study is mainly focused on developing
an efficient numerical technique to deal with fractional order non-linear coupled type diffusion
equations. From the above discussions, we can say that this type of model contains almost every
type of complexity. The main goal of this present scientific contribution is not only developing the
numerical method but also making sure that the developed technique is performing far better than
previously existing numerical techniques that deal with these types of complex models.
In this article a numerical method has been developed to solve the following nonlinear
fractional-order coupled Burgers’ equation:
∂u
∂t−ε∂βu
∂xβ+a0u∂u
∂x+a1u∂v
∂x+v∂u
∂x=0,(1)
∂v
∂t−ε∂βv
∂xβ+b0v∂v
∂x+b1u∂v
∂x+v∂u
∂x=0,(2)
under the initial conditions
u(x,0) = f1(x),v(x,0) = f2(x),a≤x≤b,(3)
and the boundary conditions
u(a,t) = g1(t), , u(b,t) = g2(t),
v(a,t) = h1(t), , v(b,t) = h2(t),t∈[0,T],(4)
where, a0,b0,a1and b1are constants and εis positive kinematic viscous parameter depending on
Reynolds number ε=1
Reynolds Number .β(1<β≤2)is order of fractional spatial derivative.
Physically the above problem can be explained as two types of interacting pollutes (if a1=0
and b1=0) with concentrations uand vhave been taken place in the porous medium. This model
governs the effect on the solute concentrations of the pollutes due to mutual interaction, random
motion and flow of the medium. If we choose a1=b1=0, then there is no interaction taking
place between these two pollutants. Thus, the effect on their concentrations is caused by the
transportation of the particles into the medium individually.
In the present study a non-standard finite difference collocation (NSFDC) method is developed
with a Fibonacci polynomial, where the solution in series form is approximated with Fibonacci
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polynomial and consider the constants as function of t. After calculating the residue of the CBEs,
the collocation method is used to get the required number of a linear equation to obtain the un-
known constants.
The article is organized in the following manner. The section 1 consists of some introduction
about the model, previously existing methods, and a brief discussion of the proposed method.
In section 2, some basic definitions, some properties of the Fibonacci polynomial are given. In
section 3, the proposed method has applied to solve the considered model (1) with initial and
boundary conditions (3) and (4). Section 4 contains the validation of the proposed method, while
it is applied to integer order models having exact solution. After the validation, the method has
been applied to solve the unsolved fractional order non-linear CBE. In section 5, conclusion of
overall work is given.
2. Definitions
2.1. Caputo fractional order derivative
The fractional differential operator in Caputo sense is defined as [22]
(Dβf)(t) = 1
Γ(n−β)Zt
0
(t−τ)(n−β−1)f(n)(τ)dτ,β>0,τ>0,
where the operator Dβsatisfies the following properties for n−1<β<n,n∈N.
(DβIβf)(t) = f(t),
(IβDβf)(t) = f(t)−∑n−1
k=0
f(k)(0)
k!tk−β,t>0,
Dβ(λ1f(x) + λ2g(x)) = λ1Dβf(x) + λ2Dβg(x),(5)
Dβtn=(0n∈0,1,2, ..., ⌈β⌉,
Γ(n+1)
Γ(n+1−β)tt−β,n∈N,n≥ ⌈β⌉,(6)
where notation ⌈·⌉ is the ceiling function and Iβis the Riemann-Liouville integral operator.
2.2. NSFDC method
A NSFDC scheme is used to discretize the differential equation, which has constructed apply-
ing many rules [23–25].
In NSFDC scheme, the first-order derivative takes the form
du
dt =uk+1−ψ(h)uk
φ(h),
where ψ(h) = 1+o(h)and φ(h) = h+o(h2)depend on h=δt.
Functions ψ(h)and φ(h)also depend on different parameters appeared in the aforementioned
differential equations. φ(h)is a continuous function, which satisfies 0 <φ(h)<1 as h→0. The
details of the NSFD method can be found in [23–25].
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2.3. Properties of Fibonacci Polynomial
It is known that Fibonacci Polynomial can be constructed using recurrence relation:
Fm+2(x) = xFm+1(x) + Fm(x),m≥0,
with initial conditions as
F0(x) = 0,F1(x) = 1.
From above relation, the explicit form of the series is obtained as
Fm(x) =
⌊m−1
2⌋
∑
r=0m−r−1
rxm−2r−1,(7)
where ⌊·⌋ denotes the floor function, which is defined by ⌊α⌋=max{n∈Z|n≤α}.
Equation (7) is rewritten as
Fi(x) =
i
∑
j=0
(i+j−1
2)!
j!(i−j−1
2)!xj,(j+i) = odd,i≥0.(8)
The function f(x)which squared integrable is written in terms of Fibonacci polynomial as [26]
f(x) =
∞
∑
k=1
ckFk,(9)
where
ck=
∞
∑
j=0
k(−1)jf(2j+k−1)(0)
j!(j+k)!.
The series (9) is approximated by (n+1)termed finite sum as given below:
fn(x) =
n+1
∑
k=1
ckFk,(10)
where the unkowns ck’s have to be determined.
3. Numerical solution using NSFDC Scheme
The fractional order β-th derivative of the function fn(x)can be approximated in terms of
Fibonacci polynomial as
Dβ[fn(x)] =
n+1
∑
k=⌈β⌉+1
k
∑
i=⌈β⌉
(i+k)=odd
ck
(k+i−1
2)!
(k−i−1
2)!Γ(i+1−β)xi−β.(11)
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Let us consider the approximation of the solutions of the non-linear coupled Burgers’ equation (1)
with the help of Fibonacci polynomial as follows
u(x,t) =
n+1
∑
i=1
bi(t)Fi(x),
v(x,t) =
n+1
∑
j=1
cj(t)Fj(x).
(12)
Considering the expressions given in (12) at the time level m, one can obtain
u(x,tm) =
n+1
∑
i=1
bm
iFi(x),
v(x,tm) =
n+1
∑
j=1
cm
jFj(x).
(13)
Now writing the non-linear coupled Burgers’ equation at time level mwith the help of equation
(13), we obtain
∂u(x,tm)
∂t−ε∂βu(x,tm)
∂xβ+a0u(x,tm)∂u(x,tm)
∂x+a1u(x,tm)∂v(x,tm)
∂x+v(x,tm)∂u(x,tm)
∂x=0,
(14)
∂v(x,tm)
∂t−ε∂βv(x,tm)
∂xβ+b0v(x,tm)∂v(x,tm)
∂x+b1u(x,tm)∂v(x,tm)
∂x+v(x,tm)∂u(x,tm)
∂x=0.
(15)
As we all agree that dealing with non-linear equations is complicated in comparison to dealing with
linear equations. So the non-linear terms of the above equations (14) and (15) can be linearized
with the help of Taylor’s expansion as
u(x,tm)∂u(x,tm)
∂x=u(x,tm)∂u(x,tm−1)
∂x+u(x,tm−1)∂u(x,tm)
∂x−u(x,tm−1)∂u(x,tm−1)
∂x,
v(x,tm)∂v(x,tm)
∂x=v(x,tm)∂v(x,tm−1)
∂x+v(x,tm−1)∂v(x,tm)
∂x−v(x,tm−1)∂v(x,tm−1)
∂x,
u(x,tm)∂v(x,tm)
∂x=u(x,tm)∂v(x,tm−1)
∂x+u(x,tm−1)∂v(x,tm)
∂x−u(x,tm−1)∂v(x,tm−1)
∂x,
v(x,tm)∂u(x,tm)
∂x=v(x,tm)∂u(x,tm−1)
∂x+v(x,tm−1)∂u(x,tm)
∂x−v(x,tm−1)∂u(x,tm−1)
∂x,
(16)
Now on using above equations (16) in the equations (14) and (15), we have
∂u(x,tm)
∂t−ε∂βu(x,tm)
∂xβ+Rm
1(x),(17)
∂v(x,tm)
∂t−ε∂βv(x,tm)
∂xβ+Rm
2(x),(18)
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where
Rm
1=a0u(x,tm)∂u(x,tm−1)
∂x+u(x,tm−1)∂u(x,tm)
∂x−u(x,tm−1)∂u(x,tm−1)
∂x+a1u(x,tm)∂v(x,tm−1)
∂x
+u(x,tm−1)∂v(x,tm)
∂x−u(x,tm−1)∂v(x,tm−1)
∂x+v(x,tm)∂u(x,tm−1)
∂x+v(x,tm−1)∂u(x,tm)
∂x
−v(x,tm−1)∂u(x,tm−1)
∂x,
(19)
Rm
2=b0v(x,tm)∂v(x,tm−1)
∂x+v(x,tm−1)∂v(x,tm)
∂x−v(x,tm−1)∂v(x,tm−1)
∂x+b1u(x,tm)∂v(x,tm−1)
∂x
+u(x,tm−1)∂v(x,tm)
∂x−u(x,tm−1)∂v(x,tm−1)
∂x+v(x,tm)∂u(x,tm−1)
∂x+v(x,tm−1)∂u(x,tm)
∂x
−v(x,tm−1)∂u(x,tm−1)
∂x.
(20)
Now using the expressions (11) and (12) in the equations (17) and (18), we have
n+1
∑
i=1
dbm
i
dt Fi(x)−ε
n+1
∑
i=⌈β⌉+1
i
∑
k=⌈β⌉
(i+k)=odd
bi
(i+k−1
2)!
(i−k−1
2)!Γ(k+1−β)xk−β+Rm
1(x),(21)
n+1
∑
j=1
dcm
j
dt Fj(x)−ε
n+1
∑
j=⌈β⌉+1
j
∑
k=⌈β⌉
(j+k)=odd
cj
(j+k−1
2)!
(j−k−1
2)!Γ(k+1−β)xk−β+Rm
2(x).(22)
Rewriting the above two equations (21) and (22), we get
n+1
∑
i=1
bm
i−bm−1
i
φ(h)Fi(x)−ε
n+1
∑
i=⌈β⌉+1
i
∑
k=⌈β⌉
(i+k)=odd
bi
(i+k−1
2)!
(i−k−1
2)!Γ(k+1−β)xk−β+Rm
1(x),(23)
n+1
∑
j=1
cm
j−cm−1
j
φ(h)Fj(x)−ε
n+1
∑
j=⌈β⌉+1
j
∑
k=⌈β⌉
(j+k)=odd
cj
(j+k−1
2)!
(j−k−1
2)!Γ(k+1−β)xk−β+Rm
2(x).(24)
The above two equations (23) and (24) together with boundary conditions will help us of finding
the unknowns b′
ksand c′
ksat each time level other than zero time level. Initial conditions will help
us to find the values at zero time level. Hence rewriting initial conditions (4) with the help of (13),
we get
n+1
∑
i=1
b0
iFi(x) = f1(x),
n+1
∑
j=1
c0
iFi(x) = f2(x),a≤x≤b.(25)
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Now rewriting boundary conditions (4) with the help of expressions (13), we obtain
n+1
∑
i=1
bm
iFi(a) = g1(t), ,
n+1
∑
i=1
bm
iFi(b) = g2(t),t∈[0,T],
n+1
∑
j=1
cm
jFj(a) = h1(t), ,
n+1
∑
j=1
cm
jFj(b) = h2(t),t∈[0,T].
(26)
From the expressions (13) it is clear that total (2n+1)unknowns are obtained at each time level
m. So one can collocate equations (23) and (24) at (n−1)collocation points x=i
n+1,i=
1,2,3, ..., n−1. In this way togather with four boundary conditions given in (26), we will have
2n+1 algebraic linear equations in b′
ksand c′
ks, which we can solve easily with appropriate nu-
merical method. To start the process of calculation of unknowns, we need the b′
ksand b′
ksat zero
time level which can be calculated by collocating the initial conditions (25) at the set collocation
points and using the boundary conditions (26) at zero time level.
4. Numerical Applications
In this section, the proposed scheme has been applied on two problems having exact solutions
and compared the numerical results obtained by our method with the existing results through error
analysis. During comparison of the numerical error, the authors have calculated the L2and L∞
errors, which are defined as
L2=sh
m
∑
j=1
|uj−Uj|2,h=xi+1−xi,(27)
L∞=Max
j|uj−Uj|,(28)
where uand Uare numerical and exact solutions, respectively.
Example 1. Let us consider the following example
ut−uxx −2uux+uvx+vux=0,(29)
vt−vxx −2vvx+uvx+vux=0,(30)
with the initial and boundary conditions as
u(x,0) = v(x,0) = sin x,−π≤x≤π,(31)
u(−π,t) = u(π,t) = 0,(32)
v(−π,t) = v(π,t) = 0,t∈[0,T],(33)
which has the exact solution
u(x,t) = exp(−t)sin x =v(x,t).
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The problem is solved for N=17 and ∆t=0.01 with the proposed method and a comparison
of error analysis between the exact solution and the numerical results obtained by different tech-
niques is presented in Table 1. It is clearly observed from Table 1 that our proposed method
performs better as compared to previously solved methods even for very less order of approxi-
mation and discretization of time (∆t=0.01). It is also seen that in other methods viz., Onarcan
and Hepson [27] have used trigonometric B-spline algorithm to achieve the minimum error of
order L∞×106=392.13, Mittal and Jiwari [28] used differential quadrature method to get the
error of order L∞×106=74.601, Bhatt et al. used [29] fourth-order compact schemes to find the
minimum error L∞×106=1.193, while in our proposed scheme L∞×106=1.56073e−6.
Example 2. Let us consider the following example
ut−uxx +2uux+α(uvx+vux) = 0,(34)
vt−vxx +2vvx+β(uvx+vux) = 0,(35)
having the exact solutions as
u(x,t) = a0(1−tanh(k(x−2kt))),(36)
v(x,t) = a02β−1
2α−1−tanh(k(x−2kt)),(37)
where k=a04αβ −1
4α−2, and a0,αand βare constants. Here the initial and boundary conditions
are extracted from the exact solutions. In this problem a comparison of errors with other existing
methods are given in Table 2. It is very clear from the table that our method is performing well as
compared to an existing method [16] even for less temporal discretization. Bhatt et al. [29] have
also solved above problem with fourth-order scheme and obtained the maximum error of order
10−5for n=16 and n=20. Khater et al. [30] have used a Chebyshev spectral collocation method
to obtain the maximum absolute error of order 10−5for n=20. Rashid et al. [31] and Mittal and
Jiwari [28] have used different types of numerical schemes to obtain absolute error order 10−5
for n=20. But that much level of accuracy is obtained for a very less number of approximation
(n=3)and less order of temporal discretization while applying our proposed method on the
mentioned problem.
After the validation of the accuracy and efficiency of our proposed method while applying it
on the two existing integer order problems mentioned above, the authors have made an endeavour
to apply it on the following unsolved problem and to show the effect of spatial fractional order
derivative on the solution profiles for different cases.
Let us consider the following unsolved problem
∂u
∂t−ε∂βu
∂xβ−a0u∂u
∂x+a1u∂v
∂x+v∂u
∂x=0,(38)
∂u
∂t−ε∂βv
∂xβ−b0v∂v
∂x+b1u∂v
∂x+v∂u
∂x=0,(39)
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with the initial and boundary conditions as
u(x,0) = v(x,0) = sin πx,0≤x≤1,(40)
u(0,t) = u(1,t) = 0,(41)
v(0,t) = v(1,t) = 0,t∈[0,T],(42)
The unsolved problem has been tackled with the help of proposed method and observe the behavior
of u(x,t)and v(x,t)due to the change in spatial fractional order derivative for the different values
of parameters a0,a1,b0and b1. In the Figure 1, eight sub-figures have been plotted for different
values of a0=b0for a fixed value of a1=b1=10 at time t=0.5. From the sub-figures a, b, c and
d of the Figure 1, it is clear that both u(x,t)and v(x,t)are decreasing with increase in the order
of spatial derivative. But on increasing the values of a0=b0, both the solutions start behaving
irregularly in the sub-figures 1(e), 1(f), 1(g) and 1(h). In the Figure 2, eight sub-figures are plotted
for different values of a!=b1for a fixed value of a0=b0=1 at time t=0.5. From the sub-figures
2(a) and 2(b) of the Figure 2, it is clearly seen that at very low value of a1=b1=1, solutions
u(x,t)and v(x,t)are behaving irregularly but as the values of a1=b1are increased, the solutions
start behaving regularly. From the sub-figures 2(c)-2(h), it is also observed that for higher values
of a1=b1, the solutions start decreasing on increasing the values of spatial order derivative β.
5. Conclusion
It is well known that getting an exact solution of non-linear PDEs is not always possible, and
things become more complex when it comes in the form of non-linear coupled PDEs. So the only
way left is to simulate the solution numerically. In the present study, the authors have developed a
non-standard/standard numerical technique with the help of Fibonacci polynomial also have taken
care of its performance over the other existing methods. Two numerical examples of integer order
CBE are considered to validate the effectiveness of the proposed numerical scheme and then it
is applied to find the solution of an unsolved non-linear coupled spatial fractional order Burgers’
equations. The behavior of solutions of the unsolved coupled problem has been observed through
graphical representation by the change in various parameters presented in the unsolved model. The
authors are also optimist that the proposed scheme will help the engineers and scientists working
in the area of coupled non-linear PDEs in integer as well as fractional order systems.
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Table 1: Comparison of numerical results with different methods for Example 1.
Method N t L2×106L∞×106Method N t L2×106L∞×106
Our proposed 17 0.1 3.16812e-5 31173e-4 DQM [32] 131 0.1 2.26096 3.78768
scheme 0.5 9.84944e-5 6.83079e-4 0.5 2.70172 2.61424
1.0 1.49673e-5 1.00919e-4 1.0 2.48146 1.57097
2.0 7.20063e-6 3.68887e-5 2.0 1.61674 1.00892
3.0 3.31726e-6 1.54191e-5 3.0 0.86332 0.54705
4.0 4.69968e-6 4.21068e-5 4.0 0.42518 0.27676
5.0 2.67499e-6 1.63987e-5 5.0 0.20099 0.13728
10.0 3.49104e-7 1.56073e-6 10.0 0.01748 0.01017
LBM [33] 64 0.1 30.3 27.48 LB [34] 64 0.1 28.3 22.826
0.5 151.7 92.04 0.5 117.8 80.420
1.0 303.4 111.66 1.0 170.4 66.441
2.0 607.0 82.11 2.0 309.0 58.983
5.0 1518.1 10.22 5.0 1015.3 8.098
10.0 3038.6 0.13 10.0 2278.5 0.112
FDM [33] 64 0.1 80.2 72.68 Coll. [35] 200 0.1 8.2 7.4
0.5 401.5 243.54 0.5 24.9 41.0
1.0 803.2 295.46 1.0 30.0 82.1
2.0 1607.1 217.52 2.0 - -
5.0 4022.7 27.10 5.0 - -
10.0 8061.7 0.36 10.0 - -
Coll. [36] 100 0.1 32.9024 29.7713 Coll. [35] 400 0.1 2.0 1.86
0.5 164.5015 99.7752 0.5 10.2 6.22
1.0 328.9759 121.0234 1.0 20.4 7.56
Coll. [36] 128 0.1 30.1811 18.4560 Imp.FD [37] 200 0.1 58.6 53.0
0.5 39.4088 61.8548 0.5 294.0 179.0
1.0 28.0808 11.2589 1.0 591.0 217.0
Gal. [38] 64 0.1 1.3961 3.9846
0.5 2.4739 2.8698
1.0 3.5300 1.7864
2.0 5.5176 0.7300
3.0 7.8953 0.3903
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Table 2: Comparison of numerical results with different methods for Example 2.
Method N ∆tα β tL2(u)×103L∞(u)×103L2(v)×103L∞(v)×103
DQM [32] 3 0.01 0.1 0.3 0.1 0.213635 0.067569 0.109682 0.034684
0.3 0.639770 0.202348 0.328518 0.103886
0.5 1.064579 0.336650 0.546654 0.172867
0.7 1.487774 0.470476 0.764093 0.241627
1.0 2.119762 0.670328 1.088950 0.344356
Our proposed 3 0.1 0.1 0.3 0.1 0.00812135 0.00898543 0.00303799 0.00446763
scheme 0.3 0.0244485 0.0274415 0.00930321 0.012705
0.5 0.0407458 0.045736 0.0155668 0.0212789
0.7 0.0570096 0.0638709 0.0218201 0.0297751
1.0 0.0813408 0.0907786 0.0311794 0.0423763
DQM [32] 3 0.01 0.3 0.03 0.1 0.250138 0.079101 0.477433 0.150978
0.3 0.749491 0.237010 1.430031 0.452216
0.5 1.247615 0.394531 2.379619 0.752501
0.7 1.744516 0.551664 3.326207 1.051839
1.0 2.487583 0.786643 4.740498 1.499077
Our proposed 3 0.1 0.1 0.3 0.1 0.00507463 0.00738132 0.00848107 0.0122781
scheme 0.3 0.0151924 0.0220816 0.025398 0.0367252
0.5 0.0252694 0.036693 0.0422557 0.0610205
0.7 0.0353061 0.051216 0.0590544 0.0851642
1.0 0.0502862 0.0728356 0.084142 0.094347
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β=1.6
β=1.8
β=2.0
0.0 0.2 0.4 0.6 0.8 1.0
0.000
0.005
0.010
0.015
x
u(x, 0.5)
(a) for a0=b0=0.2
β=1.6
β=1.8
β=2.0
0.0 0.2 0.4 0.6 0.8 1.0
0.000
0.005
0.010
0.015
0.020
0.025
x
v(x, 0.5)
(b) for a0=b0=0.2
β=1.6
β=1.8
β=2.0
0.0 0.2 0.4 0.6 0.8 1.0
0.000
0.005
0.010
0.015
0.020
0.025
x
u(x, 0.5)
(c) for a0=b0=2.0
β=1.6
β=1.8
β=2.0
0.0 0.2 0.4 0.6 0.8 1.0
0.000
0.005
0.010
0.015
0.020
0.025
x
v(x, 0.5)
(d) for a0=b0=2.0
β=1.6
β=1.8
β=2.0
0.0 0.2 0.4 0.6 0.8 1.0
0.000
0.002
0.004
0.006
0.008
0.010
x
u(x, 0.5)
(e) for a0=b0=20
β=1.6
β=1.8
β=2.0
0.0 0.2 0.4 0.6 0.8 1.0
0.000
0.002
0.004
0.006
0.008
0.010
x
v(x, 0.5)
(f) for a0=b0=20
β=1.6
β=1.8
β=2.0
0.0 0.2 0.4 0.6 0.8 1.0
0.0000
0.0002
0.0004
0.0006
0.0008
x
u(x, 0.5)
(g) for a0=b0=200
β=1.6
β=1.8
β=2.0
0.0 0.2 0.4 0.6 0.8 1.0
0.0000
0.0002
0.0004
0.0006
x
v(x, 0.5)
(h) for a0=b0=200
Figure 1: Simulation of Example 3 for a1=b1=10 at t=0.5
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β=1.6
β=1.8
β=2.0
0.0 0.2 0.4 0.6 0.8 1.0
0.000
0.002
0.004
0.006
0.008
0.010
x
u(x, 0.5)
(a) for a1=b1=1
β=1.6
β=1.8
β=2.0
0.0 0.2 0.4 0.6 0.8 1.0
0.000
0.002
0.004
0.006
0.008
0.010
x
v(x, 0.5)
(b) for a1=b1=1
β=1.6
β=1.8
β=2.0
0.0 0.2 0.4 0.6 0.8 1.0
0.000
0.005
0.010
0.015
0.020
0.025
x
u(x, 0.5)
(c) for a1=b1=10
β=1.6
β=1.8
β=2.0
0.0 0.2 0.4 0.6 0.8 1.0
0.000
0.005
0.010
0.015
0.020
0.025
x
v(x, 0.5)
(d) for a1=b1=10
β=1.6
β=1.8
β=2.0
0.0 0.2 0.4 0.6 0.8 1.0
0.000
0.001
0.002
0.003
x
u(x, 0.5)
(e) for a1=b1=100
β=1.6
β=1.8
β=2.0
0.0 0.2 0.4 0.6 0.8 1.0
0.000
0.001
0.002
0.003
x
v(x, 0.5)
(f) for a1=b1=100
β=1.6
β=1.8
β=2.0
0.0 0.2 0.4 0.6 0.8 1.0
0.00000
0.00005
0.00010
0.00015
0.00020
0.00025
0.00030
x
u(x, 0.5)
(g) for a1=b1=1000
β=1.6
β=1.8
β=2.0
0.0 0.2 0.4 0.6 0.8 1.0
0.00000
0.00005
0.00010
0.00015
0.00020
0.00025
0.00030
x
v(x, 0.5)
(h) for a1=b1=1000
Figure 2: Simulation of Example 3 for a0=b0=2 at t=0.5
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Highlights
Numerical solution of coupled Burgers' equation using Non-standard finite difference
scheme.
Validation of efficiency and effectiveness of the proposed method by applying it to
numerical examples.
The damping nature of solute concentrations of the model shown through graphical
presentations.
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☒
☐
