The Use of Modiﬁed Fractional Differential Transform for Multi-term

Fractional Order Differential Equations

NGUYEN MINH TUAN1ID

1Department of Mathematics, Faculty of Applied Science

King Mongkut’s University of Technology North Bangkok, Bangkok 10800,

THAILAND

Abstract: The differential transform method has been prevalently expedited in the last decades by elucidating the

solutions of partial differential equations. In this paper, the multi-term fractional differential equations have been

solved by using the modiﬁed differential transform method combining the fractional integral operator to omit one

term consisting of fractional differential order. Compared to the previous research, the method is effective and

approached to approximate solutions that lead to exact solutions.

Key-Words: Multi-term fractional order; multi-term fractional order; dtm; modiﬁed differential transform

method

Received: March 5, 2024. Revised: October 6, 2024. Accepted: November 4, 2024. Published: December 2, 2024.

1 Introduction

A plethora of mathematical scientists have consid-

ered fractional differential equations to ﬁnd solutions

using different kinds of methods. The differential

transform method and its variations have been ex-

tended for fractional differential orders and used to

gain the solutions of fractional differential equations,

[1]. Modifying the fractional reduced differential

transform method effectively solves the multi-term

fractional order of the fractional differential equa-

tions, [2]. The high accuracy of the solution is crit-

ical to the method’s performance. The differential

transform method has been successfully applied to

solving fractional differential equations such as the

BagleyTorvik, Ricatti, and oscillation equation, [3].

The multi-term fractional differential equations have

been scrutinized by using the fractional Taylor vec-

tor method to attain the numerical solutions com-

pared to exact solutions, [4]. A series of multi-

term fractional differential equations in the Newto-

nian ﬂuid has been solved using Euler’s method, in-

cluding the trapezoidal quadrature formula and prod-

uct rectangle rule, [5]. Another effective method, the

pseudo-spectral method, solves the fractional differ-

ential equations and performs the beneﬁt outputs of

the extensive method, [6].

The attained solutions could be expressed as exact

or approximate solutions, [7]. The differential trans-

form method is extracted from the Taylor expansion

and is important in ﬁnding a series of solutions and

exact solutions, [8]. The extension of the differen-

tial transform method is a reduced differential trans-

form method that is applied to seek analytic solutions

and has been expanded to solve fractional differential

equations, [9]. The proposed method is for solving

one-term fractional differential derivatives and multi-

term fractional differential derivatives, [10] has de-

composed the fractional differential equation into a

system of differential equations, [11]. The output

of the differential transform method is usually per-

formed in inﬁnite power series or approximate solu-

tions, [12] and performs solutions using Matlab soft-

ware, [13].

In this paper, the modiﬁed fractional differential

transform method is used by solving multi-term frac-

tional derivative equations that have been taken into

account in the last few years. The application of

the differential transform method combining differ-

ential integral operator is the main aim of the paper,

[14], [15] that has been depicted in the research, [16].

Some practical related deﬁnitions, [17] are also used

in this paper to contribute to the performance of ﬁnd-

ing solutions.

2 Basic deﬁnitions and preliminaries

This section will perform some useful deﬁnitions ap-

plied in fractional differential equations and this pa-

per.

Deﬁnition 1. Gamma function (, [17])

Γ(p) = ∞

e−xxp−1dx

Theorem 2.1 (Caputo’s Fractional Derivatives).(,

[17]) Given a function y=f(t), and an arbitrary or-

der n−1<

≤n, the Caputo’s fractional derivative

of order

is given by:

International Journal of Computational and Applied Mathematics & Computer Science

DOI: 10.37394/232028.2024.4.12

Nguyen Minh Tuan

E-ISSN: 2769-2477

112

Volume 4, 2024

Table 1: Basic results using DTMT method , [18], [19, 20], [21].

Original Functions DTM for some fundamental functions

f(t) = ag(t)±bh(t)Fk=aGk±bHk

f(t) = g(t)h(t)Fk=∑k

r=0GrHk−r

f(t) = tpFk=

(k−

p) = 1 if k=

0 if k=

f(t) = C

tg(t)Fk=Γ(

(k+1)+1)

Γ(1+k

)Gk+1

f(t) = C

tg(t),

=pq Fk=Γ(q(k+p)+1)

Γ(kq+1)Gk+p

f(t) = atmFk=a

(k−m)

f(t) = C

0Dn

tg(t)Fk= (k+1)(k+2)···(k+n)Gk+n

f(t) = sin

t Fk=

k!sin k

f(t) = cos

t Fk=

k!cos k

f(t) = e

tFk=

xf(x) = 1

Γ(n−

)x

(x−s)n−

−1f(n)(s)ds

Based on the Theorem 2.1, we have derivative of

function f(x) = C, (C is a constant):

xf(x) = 0

The inverse operator C

x, called J

xis fractional

integral operator of order

is given as the following

f(x) = 1

Γ(

)x

(x−w)

−1f(w)dw

We have the Caputo fractional derivative property

xf(x)) = f(x)−

n−1

∑

k=0

f(k)(0)xk

k!,x>0 (1)

Suppose the

f(t) =

∞

∑

k=0

Fktkq

where

=pq is the fractional derivative order, p

is integer part, qis fractional part, and Fk,Gk,Hk

is the transformation of f(t),g(t),h(t)respectively.

For the initial integer-order derivative condition with

k=0,1,...,(pq −1), we could transform as follows:

Fk=





0 if kq /∈Z+

(kq)!dkq

dtkq f(t)t=t0

if kq ∈Z+

3 Methodology

Step 1: Integrating the equation (1) to transform one

of the fractional differential terms.

Step 2: Using differential transform method in Table

1 to transform the term of fractional order in the form

deﬁned by

Fk=

∞

∑

k=0

Step 3: Applying the inductive method to calculate

the terms of the transform and gain the solution satis-

ﬁes the condition:

f(t) = lim

k→∞Fk

4 Applications

Example 4.1.Consider the nonlinear fractional equa-

tion , [5]

ty(t) + y(t) = 2t2−

Γ(3−

)+t2−t,1<

<2 (2)

subjected to the condition y(0) = 0,y′(0) = −1.

Step 1:

Taking transform both sides Eq. (4.1) using Eq. (1):

y(t)−y(0)−ty′(0) + J

[y(t)] = J

2t2−

Γ(3−

)+t2−t

Step 2:

Taking DTM transform on both sides, we have

Hk+

(k−1) + J

[Hk] = J

2

Γ(3−

)

(k−2+

)

(k−2)−

(k−1)

Step 3:

Using the initial conditions, we have:

H0=0,H1=−1.

We choose k=2 then H2=1.

Using the inductive method, we have the following

terms:

H3=0,H4=0,···

The exact solution collected is y(t) = t2−t.

International Journal of Computational and Applied Mathematics & Computer Science

DOI: 10.37394/232028.2024.4.12

Nguyen Minh Tuan

E-ISSN: 2769-2477

113

Volume 4, 2024

The solution is compared in Table 3 and depicted

in Figure 2 (APPENDIX).

Example 4.2.Cogitate the fractional differential

equation, [4]

D2y(t) = −y(t) + 2y′(t)−D1

2y(t) + t7+2048

429√

t6.5

−14t6+42t5−t2−8

3√

t1.5+4t−2.(3)

satisfy the condition y(0) = y′(0) = 0,

Step 1:

Taking transform both sides Eq. (4.2) using Eq. (1):

D2y(t)=−y(t) + J

2y′(t)−y(t)

+t7+2048

429√

t6.5−14t6+42t5−t2

−8

3√

t1.5+4t−2

Step 2:

Taking DTM transform on both sides, we have

[(k+1)(k+2)Hk+2] = −Hk







2(k+1)Hk+1−Hk+

(k−7)

+2048

429√

πδ

(k−6.5)−14

(k−6)

+42

(k−5)−

(k−2)

−8

3√

πδ

(k−1.5) + 4

(k−1)−2

(k)







Step 3:

Using the initial condition, we have H0=H1=0. We

will calculate the terms of the DTM transform as the

following:

k=0 :H2=−1

k=1 :H3=0

k=2 :H4=0

k=3 :H5=0

k=4 :H6=0

k=5 :H7=1

k=6 :H8=0

k=7 :H9=0

k=8 :H10 =0

···

The exact solution obtained is y(t) = t7−t.

The solution is compared in Table 5 and depicted

in Figure 3 (APPENDIX).

Example 4.3.Taking into account the Bagley-Torvik

equation, [3]

aD2y(t) + bC

ty(t) + cy (t) = c(t+1)(4)

satisfy the condition y(0) = y′(0) = 1,

Step 1:

Taking transform both sides Eq. (4.3) using Eq. (1):

D2y(t)+by(t)−y(0)−ty′(0)

+cJ

[y(t)] = cJ

[t+1]

Step 2:

Taking the DTM transform of the equation, we have

[(k+1)(k+2)Hk+2] + b[Hk−

(k)−

(k−1)]

+cJ

[Hk] = cJ

[

(k−1) +

(k)].

Step 3:

We calculate the terms of the DTM transform:

H0=1;H1=1; We choose k=0 : then H2=0.

k=1 : then H3=1.

k=2 : then H4=0;

Using the inductive method, we have the following

terms of the transform:

H5=0;···

The exact solution is y(t) = 1+t.

The solution is compared in Table 4 and depicted

in Figure 4 (APPENDIX).

Example 4.4.Consider the two-term fractional order

initial value problem, [5]

ty(t) + 1.3C

ty(t) + 2.6y(t) = sin(2t)(5)

satisfy the condition y(0) = y′(0) = y′′ (0) = 0;

2.2,

=1.5.

Step 1:

Taking transform both sides Eq. (4.4) using Eq. (1):

y(t)−y(0)−ty′(0)−t2y′′ (0) + 1.3J

D

y(t)

[2.6y(t)] = J

[sin(2t)]

Using initial conditions, we simplify the equation as

follows

y(t) + 1.3J

C

ty(t)=J

[sin(2t)−2.6y(t)]

Step 2:

Using DTM to simplify the equation

Hk+p=Γ(kq +1)

1.3Γ(q(k+p) + 1)J

2k

k!sink

2

−2.6Hk−1.3Hk

where

=pq;p=3,q=0.5;

Step 3:

Using the initial conditions:

H0=0;H1=0;H2=0;

International Journal of Computational and Applied Mathematics & Computer Science

DOI: 10.37394/232028.2024.4.12

Nguyen Minh Tuan

E-ISSN: 2769-2477

114

Volume 4, 2024

We calculate the terms of DTM up to k=17 as fol-

lows:

k=0 :H3=0;

k=1 :H4=0.68171;

k=2 :H5=0;

k=3 :H6=−0.22724;

k=4 :H7=−0.23443;

k=5 :H8=2.8405 ×10−2;

k=6 :H9=5.2097 ×10−2;

k=7 :H10 =4.3553 ×10−2;

k=8 :H11 =−4.7361 ×10−3;

k=9 :H12 =−7.4958 ×10−3;

k=10 :H13 =−5.5859 ×10−3;

k=11 :H14 =6.049 ×10−4;

k=12 :H15 =7.6911 ×10−4;

k=13 :H16 =5.1853 ×10−4;

k=14 :H17 =−5.1113 ×10−5;

k=15 :H18 =−5.9492 ×10−5;

k=16 :H19 =−3.6897 ×10−5;

k=17 :H20 =3.3606 ×10−6,···

Solution is expressed in the form of y=∑∞

k=0Hktkq

y(t) = 0.68171t2−0.22724t3−0.23443t3.5

+0.028405t4+0.052097t4.5+0.043553t5

−0.0047361t5.5−0.0074958t6−5.5859t6.5

+0.0006049t7+0.00076911t7.5

+0.00051853t8−0.000051113t8.5

−0.000059492t9−0.000036897t9.5

+0.0000033606t10 ···

The solution is close to the solutions gained in, [4]

using the fractional Taylor method (FTM) performed

in Table 1 and depicted in Figure 5 (APPENDIX).

Example 4.5.Consider the general two-term frac-

tional order initial value problem

ty(t) + 1.3C

ty(t) + 2.6y(t) = sin(2t)(6)

satisfy the condition y(0) = y′(0) = y′′ (0) = 0.and

<3,0<

<1.

Step 1:

Taking transform both sides Eq. (4.5) using Eq. (1):

y(t)−y(0)−ty′(0)−t2y′′ (0)

+1.3J

C

ty(t)+J

[2.6y(t)] = J

[sin(2t)]

Using initial conditions and simplifying

y(t) + 1.3J

C

ty(t)=J

[sin(2t)−2.6y(t)]

Step 2:

Taking DTM on both sides, we have

[Hk+1] = Γ(k

+1)

1.3Γ(

(k+1) + 1)J

2k

k!sink

2

−2.6Hk−1.3Hk

Step 3:

We have H0=0 using initial conditions. We calculate

the terms of DTM: Taking

k=0 :H1=0

k=1 :H2=H2=1.5385Γ(

+1)

Γ(2

+1)

k=2 :H3=−3.077Γ(

+1)

Γ(3

+1)

k=3 :H4=−6.1540Γ(

+1)

Γ(4

+1)−1.0256Γ(3

+1)

Γ(4

+1)

k=4 :

H5=12.308Γ(

+1)

Γ(5

+1)+2.0512Γ(3

+1)

Γ(5

+1).

k=5 :

H6=Γ(5

+1)

1.3Γ(6

+1)4

15 −32.001Γ(

+1)

Γ(5

+1)

−5.3331Γ(3

+1)

Γ(5

+1)

k=6 :

H7=1.5385Γ(5

+1)

Γ(7

+1)32.001

Γ(5

+1)Γ(

+1)

+5.3331Γ(3

+1)

Γ(5

+1)−4

15

···

International Journal of Computational and Applied Mathematics & Computer Science

DOI: 10.37394/232028.2024.4.12

Nguyen Minh Tuan

E-ISSN: 2769-2477

115

Volume 4, 2024

The solution collected in the form as follows

y(t) = 1.5385Γ(

+1)

Γ(2

+1)t2−3.077Γ(

+1)

Γ(3

+1)t3

−6.1540Γ(

+1)

Γ(4

+1)+1.0256Γ(3

+1)

Γ(4

+1)t4

+12.308Γ(

+1)

Γ(5

+1)+2.0512Γ(3

+1)

Γ(5

+1)t5

+Γ(5

+1)

1.3Γ(6

+1)4

15 −32.001Γ(

+1)

Γ(5

+1)

−5.3331Γ(3

+1)

Γ(5

+1)t6

+1.5385Γ(5

+1)

Γ(7

+1)32.001Γ(

+1)

Γ(5

+1)

+5.3331Γ(3

+1)

Γ(5

+1)−4

15t7

+···

The solution is depicted in Figure 5 and Figure 6

(APPENDIX).

5 Conclusion

The modiﬁed differential transform method is the

combination of the fractional integral operator and

the differential transform method. The method has

been successfully applied to seek the solutions of

multi-fractional differential equations. The main pur-

pose of the method is to eliminate one of the frac-

tional differential orders and then use the differential

transform method for the other term. The solutions

have been compared to other methods by using the

Matlab application to illustrate the numerical results

and two-dimensional graph performance. The solu-

tions found could lead to exact solutions or be close

to the ones using other methods.

Acknowledgment:

The authors acknowledge the anonymous reviewer’s

comments, which improve the manuscript’s quality,

and kindly provide advice and valuable discussions.

References:

[1] Abuasad, S., Hashim, I., & Abdul Karim, S. A.

(2019). Modiﬁed Fractional Reduced

Differential Transform Method for the Solution

of Multiterm Time-Fractional Diffusion

Equations. Advances in Mathematical Physics,

2019, 114.

https://doi.org/10.1155/2019/5703916

[2] Al-rabtah, A., & Abuasad, S. (2023). Effective

Modiﬁed Fractional Reduced Differential

Transform Method for Solving Multi-Term

Time-Fractional Wave-Diffusion Equations.

Symmetry, 15(9), 1721.

https://doi.org/10.3390/sym15091721

[3] Arikoglu, A., & Ozkol, I. (2007). Solution of

fractional differential equations by using

differential transform method. Chaos, Solitons

& Fractals, 34(5), 14731481.

https://doi.org/10.1016/j.chaos.2006.09.004

[4] Avc, ., & Mahmudov, N. I. (2020). Numerical

Solutions for Multi-Term Fractional Order

Differential Equations with Fractional Taylor

Operational Matrix of Fractional Integration.

Mathematics, 8(1), 96.

https://doi.org/10.3390/math8010096

[5] El-Mesiry, A. E. M., El-Sayed, A. M. A., &

El-Saka, H. A. A. (2005). Numerical methods

for multi-term fractional (arbitrary) orders

differential equations. Applied Mathematics

and Computation, 160(3), 683699.

https://doi.org/10.1016/j.amc.2003.11.026

[6] Esmaeili, S., & Shamsi, M. (2011). A

pseudo-spectral scheme for the approximate

solution of a family of fractional differential

equations. Communications in Nonlinear

Science and Numerical Simulation, 16(9),

36463654.

https://doi.org/10.1016/j.cnsns.2010.12.008

[7] Tuan, N. M., Kooprasert, S., Sirisubtawee, S.,

& Meesad, P. (2024). The bilinear neural

network method for solving Benney-Luke

equation. Partial Differential Equations in

Applied Mathematics, 10, 100682.

https://doi.org/10.1016/j.padiff.2024.100682

[8] Idrees, M., & Muhammad, F. (2012).

Application of Differential Transform Method

Stiff Systems. 6.

[9] Keskin, Y., & Oturanç, G. (2009). Reduced

Differential Transform Method for Partial

Differential Equations. International Journal of

Nonlinear Sciences and Numerical Simulation,

10(6).

https://doi.org/10.1515/IJNSNS.2009.10.6.741

[10] Leszczyski, J., & Ciesielski, M. (2002). A

Numerical Method for Solution of Ordinary

Differential Equations of Fractional Order. In

R. Wyrzykowski, J. Dongarra, M. Paprzycki, &

J. Waniewski (Eds.), Parallel Processing and

Applied Mathematics (Vol. 2328, pp. 695702).

Springer Berlin Heidelberg.

https://doi.org/10.1007/3-540-48086-2_77

International Journal of Computational and Applied Mathematics & Computer Science

DOI: 10.37394/232028.2024.4.12

Nguyen Minh Tuan

E-ISSN: 2769-2477

116

Volume 4, 2024

[11] Noori, S. R. M., & Taghizadeh, N. (2020).

Study of convergence of reduced differential

transform method for different classes of

nonlinear differential equations [Preprint].

Preprints.

https://doi.org/10.22541/au.159285253.32751770

[12] Sohail, M., & Mohyud-Din, S. T. (2012).

Reduced Differential Transform Method for

Solving a System of Fractional PDEs. 10.

[13] Sunday, O. A., & Lois, J. F. (2021).

Construction of Functions for Fractional

Derivatives using Matlab. Journal of Advances

in Mathematics and Computer Science, 110.

https://doi.org/10.9734/jamcs/2021/v36i630368

[14] Tuan, N. M., Koonprasert, S., Sirisubtawee, S.,

Meesad, P., & Khansai, N. (2024). New

Solutions of Benney-Luke Equation Using The

(G’/G,1/G) Method. Wseas Transactions on

mathematics, 23, 267-275.

https://doi.org/10.37394/23206.2024.23.29

[15] Tuan, N. M. (2023). A Study of Applied

Reduced Differential Transform Method Using

Volterra Integral Equations in Solving Partial

Differential Equations. EQUATIONS, 3,

93103.

https://doi.org/10.37394/232021.2023.3.11

[16] Wang, Y. (2019). Positive solutions for a class

of two-term fractional differential equations

with multipoint boundary value conditions.

Advances in Difference Equations, 2019(1),

304.

https://doi.org/10.1186/s13662-019-2250-x

[17] Das, S. (2011). Functional Fractional Calculus.

Springer Berlin Heidelberg.

https://doi.org/10.1007/978-3-642-20545-3

[18] Nazari, D., & Shahmorad, S. (2010).

Application of the fractional differential

transform method to fractional-order

integro-differential equations with nonlocal

boundary conditions. Journal of Computational

and Applied Mathematics, 234(3), 883891.

https://doi.org/10.1016/j.cam.2010.01.053

[19] Benhammouda, B., Vazquez-Leal, H., &

Sarmiento-Reyes, A. (2014). Modiﬁed

Reduced Differential Transform Method for

Partial Differential-Algebraic Equations.

Journal of Applied Mathematics, 2014, 19.

https://doi.org/10.1155/2014/279481

[20] Chen, C.-K., & Ho, S.-H. (1996). Application

of differential transformation to eigenvalue

problems. Applied Mathematics and

Computation, 79(23), 173188.

https://doi.org/10.1016/0096-3003(95)00253-7

[21] Abuteen, E., Freihat, A., Al-Smadi, M., Khalil,

H., & Khan, R. A. (2016). Approximate Series

Solution of Nonlinear, Fractional Klein-Gordon

Equations Using Fractional Reduced

Differential Transform Method. Journal of

Mathematics and Statistics, 12(1), 2333.

https://doi.org/10.3844/jmssp.2016.23.33

Contribution of Individual Authors to the

Creation of a Scientiﬁc Article (Ghostwriting

Policy)

Nguyen Minh Tuan: Conceptualization, data cu-

ration, investigation, methodology, software, visu-

alization, writing-original draft and writing-review

and editing, validation, visualization, writing-original

draft and writing-review and editing.

Sources of Funding for Research Presented in a

Scientiﬁc Article or Scientiﬁc Article Itself No

funding was received for conducting this study.

Conﬂicts of Interest The authors

have no conﬂicts of interest to

declare that are relevant to the content of this

article.

Creative Commons Attribution License 4.0

(Attribution 4.0 International , CC BY 4.0)

This article is published under the terms of the

Creative Commons Attribution License 4.0

https://creativecommons.org/licenses/by/4.0/deed.en

_US

APPENDIX

Table 2: Solutions of example 4.1

Values DTM transform Exact solution errors

0.0 0.0000 0.0000 0.0000

0.1 -0.0900 -0.0900 0.0000

0.2 -0.1600 -0.1600 0.0000

0.3 -0.2100 -0.2100 0.0000

0.4 -0.2400 -0.2400 0.0000

0.5 -0.2500 -0.2500 0.0000

0.6 -0.2400 -0.2400 0.0000

0.7 -0.2100 -0.2100 0.0000

0.8 -0.1600 -0.1600 0.0000

0.9 -0.0900 -0.0900 0.0000

1.0 0.0000 0.0000 0.0000

International Journal of Computational and Applied Mathematics & Computer Science

DOI: 10.37394/232028.2024.4.12

Nguyen Minh Tuan

E-ISSN: 2769-2477

117

Volume 4, 2024

Table 3: Solutions of example 4.2

Values DTM transform Exact solution errors

0.0 0.0000 0.0000 0.0000

0.1 -0.1000 -0.1000 0.0000

0.2 -0.2000 -0.2000 0.0000

0.3 -0.2998 -0.2998 0.0000

0.4 -0.3984 -0.3984 0.0000

0.5 -0.4922 -0.4922 0.0000

0.6 -0.5720 -0.5720 0.0000

0.7 -0.6176 -0.6176 0.0000

0.8 -0.5903 -0.5903 0.0000

0.9 -0.4217 -0.4217 0.0000

1.0 0.0000 0.0000 0.0000

Table 4: Solutions of example 4.4

Values Fractional Taylor , [4] DTM errors

0.0 0.0000 0.0000 0.0000

0.1 -0.0000 0.0000 -0.0000

0.2 0.0001 0.0001 0.0000

0.3 0.0021 0.0014 0.0007

0.4 0.0137 0.0100 0.0037

0.5 0.0569 0.0473 0.0096

0.6 0.1791 0.1693 0.0098

0.7 0.4682 0.4978 -0.0296

0.8 1.0708 1.2675 -0.1967

0.9 2.2137 2.8907 -0.6770

1.0 4.2294 6.0437 -1.8144

Table 5: Solutions of example 4.3

Values DTM transform Exact solution errors

0.0 1.0000 1.0000 0.0000

0.1 1.1000 1.1000 0.0000

0.2 1.2000 1.2000 0.0000

0.3 1.3000 1.3000 0.0000

0.4 1.4000 1.4000 0.0000

0.5 1.5000 1.5000 0.0000

0.6 1.6000 1.6000 0.0000

0.7 1.7000 1.7000 0.0000

0.8 1.8000 1.8000 0.0000

0.9 1.9000 1.9000 0.0000

1.0 2.0000 2.0000 0.0000

Figure 1: 2D solution performance of exam-

ple 4.1

Figure 2: 2D solution of example 4.2

International Journal of Computational and Applied Mathematics & Computer Science

DOI: 10.37394/232028.2024.4.12

Nguyen Minh Tuan

E-ISSN: 2769-2477

118

Volume 4, 2024

Figure 3: 2D solution performance of exam-

ple 4.3

Figure 4: 2D solution of example 4.4,

1.5,

=2.2

Figure 5: 2D solution of example 4.5,

2.2,

=1.5,

=0.5

Figure 6: 2D solution of example 4.5,

2.2,

=1.5,

=1.115

International Journal of Computational and Applied Mathematics & Computer Science

DOI: 10.37394/232028.2024.4.12

Nguyen Minh Tuan

E-ISSN: 2769-2477

119

Volume 4, 2024