General Integral Transform Performance for Space-Time Fractional
Telegraph Equations
NGUYEN MINH TUAN1ID , SANOE KOONPRASERT1,3ID ,
PHAYUNG MEESAD2ID
1Department of Mathematics, Faculty of Applied Science,
King Mongkut’s University of Technology North Bangkok,
1518 Pracharat 1 Road, Wongsawang, Bangsue, Bangkok 10800,
THAILAND
2Information Technology and Management Department,
King Mongkut’s University of Technology North Bangkok,
1518 Pracharat 1 Road, Wongsawang, Bangsue, Bangkok 10800,
THAILAND
3Centre of Excellence in Mathematics,
CHE, Si Ayutthaya Road, Bangkok 10400,
THAILAND
Abstract: The development of technology has supported effective tools in industrial machines and set up the re-
markable phase that serves well-being such as kinetic energy, kinetic movement, and nuclear energy. Applied
mathematics has also contributed valuable procedures in various fields of these sciences, especially the creation
of transformation. With practical relevance, a new general integral (NGI) transform has also shown a crucial role
in the same pragmatic methods. In this paper, the NGI transform using the combination of Padé approximation
including continued fraction expansions (CFE) has been used to attain approximate solutions of space-time frac-
tional telegraph equations by directly getting the inverse transform.
Key-Words: New general integral transform; Padé approximation; Continued fraction expansions (CFE)
Received: March 11, 2023. Revised: January 3, 2024. Accepted: March 5, 2024. Published: April 11, 2024.
1 Introduction
Transformation is an increasingly astounding varia-
tion created and applied in solving partial differential
equations, [1]. In recent years, transform has been
shown not only in different fields of mathematics but
also in other majors such as physics, chemistry, and
technology, [2]. The following decades produced a
lot of new transformations, for example, the Shehu
transformation was also created and applied to find
the approximate solutions, [3]; the Sumudu trans-
formation is used to solve delay fractional Bagley-
Torvik equations shown combining efficient method,
[4]; the Elzaki transformation is created from Laplace
transform and applied in finding solutions of partial
differential equations, [5], [6]; the natural transform
(N-transform) was invented by, [7], and was per-
formed in solving unstable fluid flow problems; the
Aboodh transforms, [8], and new
α
−Integral Laplace
transform, [9], derived from Laplace transform are
useful tools combining the Homotopy method for
finding the exact solution of various differential equa-
tions. Following this stream, Pourreza transforms,
[10], and Mohand transforms, [11], set a new expan-
sion to solve higher differential equations with con-
stant coefficients. Sawi transforms, [12], and Kamal
transforms, [13], are new structures of Fourier inte-
gration establishing new methods to find the zeros of
partial differential equations. G_transform was in-
troduced, [14], [15], as a state-of-the-art method to
evaluate the improper integral. However, it also has
some limitations as shown in, [16]. Almost all trans-
forms were introduced in L1or L2space. H-transform
followed by the sequences of relatives was written
in, [17], and has been generalized with hypergeo-
metric and Bessels-typed kernels related to fractional
differential equations. H-Transform, a whole trans-
formation motivated by H-functions kernels in class
Lebesgue functions, has set up an expansion on the
Space Lv,rfor 1 ≤r≤∞on the upper half-space with
the types of Bessel Integral Transforms and Hyper-
geometric Integral Transforms. The role of the H-
function, a form banked on Mellin-Barnes integrals,
has been illustrated in, [18], and applied in partial dif-
ferential equations with rational coefficients.
In this paper, a new general integral (NGI) has been
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performed to find solutions of fractional differential
equations as shown in equation (1).
∂
k
α
∂
t
α
f(x,t) =
∂β
∂
t
β
f(x,t) +
∂
∂
tf(x,t) + f(x,t),
t≥0,0<
α
≤2.(1)
satisfy the initial conditions f(x,0) = h(x),ft(x,0) =
0, and
∂α
∂
t
α
is fractional Caputo derivative with or-
der
α
,
∂β
∂
t
β
represents the integer or fractional Caputo
derivative of the function.
The new general integral (NGI) transform, [19], is
defined as the following:
T[f(t)] = F(s) = p(s)Z∞
0
e−q(s)tf(t)dt.(2)
The function f(t)is defined in the set
A={f(t):∃N,n1,n2>0,|f(t)|<Ne|t|
ni,
if t∈(−1)i×[0,∞)]}.(3)
The inverse new general integral transform of the
function f(t)is defined by
T−1F(s) = f(t),f or t ≥0.(4)
Another form is
u(t) = T−1[F(s)] = 1
2
π
iZc+i∞
c−i∞
eq(s)t
p(s)F(s)ds.(5)
where sis the new general integral transform variable,
and cis a real constant and the integral has got along
the s=cin the complex plane s=x+iy.
2 Fundamental Definitions
This section will perform some useful definitions ap-
plied in fractional differential equations and this pa-
per.
Definition 1. Gamma function, [20]:
Γ(p) = Z∞
0
e−xxp−1dx.(6)
Definition 2. Beta function, [20]:
B(p,q) = Z1
0
xp−1(1−x)q−1dx (7)
where Re(p),Re(q)>0.
Definition 3. Mittag-Leffler Function, [20]:
E
α
(z) = ∞
∑
k=0
zk
Γ(
α
k+1),Re(
α
)>0.(8)
E
α
,
β
(z) = ∞
∑
k=0
zk
Γ(
α
k+
β
),Re(
α
),Re(
β
)>0,
β
∈C.
(9)
E(t,
α
,a) = ∞
∑
k=0
(at)k
Γ(k+
α
+1)=t
α
E1,
α
+1(at).
(10)
We have the following special values
E1,1(z) = ez;E1,2(z) = ez−1
z.(11)
E2,1(z2) = cosh(z);E2,2(z2) = sinhz
z.(12)
Definition 4. Hypergeometric functions, [20]:
pFq(a1, .., ap,b1, .., bq;z) = Γ(b1)...Γ(bq)
Γ(a1)...Γ(aq)×
∞
∑
k=0
Γ(a1+k)...Γ(ap+k)zk
Γ(b1+k)...Γ(aq).
(13)
In particular, Rec>Re (a+b)and cis not a nonpos-
itive integer
2F1(a,b,c; 1) = Γ(c)Γ(c−a−b)
Γ(c−a)Γ(c−b).(14)
1F1(a,b,c;z) = Γ(c)
Γ(a)
∞
∑
k=0
Γ(a+k)
Γ(c+k)
zk
k!.(15)
Definition 5. The incomplete gamma function, [20]:
The incomplete gamma function
γ
∗(v,z)is de-
fined
γ
∗(v,z) = e−z∞
∑
k=0
zk
Γ(v+k+1).(16)
If Rez>0,then
γ
∗(v,z)has the integral presentation
γ
∗(v,z) = 1
Γ(v)zvZz
0
tv−1e−tdt.(17)
The special forms of the confluent hypergeometric
functions:
γ
∗(v,t) = e−t
Γ(v+1)1F1(1,v+1;t).(18)
An alternative form
γ
∗(v,t) = 1
Γ(v+1)1F1(1,v+1; −t).(19)
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when t=0 then
γ
∗(v,0) = 1
Γ(v+1).(20)
Considering pis a non-negative integer, we can read-
ily deduce in the form
γ
∗(p,at) = e−at ∞
∑
k=p
(at)k−p
k!.(21)
and
γ
∗(−p,at)=(at)p.
In particular, we have
γ
∗(1,at) = 1−e−at
at ;
γ
∗(0,at) = 1;
γ
∗(−1,at) = at.
(22)
Now we define the function
Et(v,a) = zveaz
γ
∗(v,z).(23)
The properties of the Et(v,a)may be determined by
Et(v,a) = tv∞
∑
k=0
(at)k
Γ(v+k+1).(24)
We have the following special values
Et(0,a) = eat ;E0(v,a) = 0;Et(−1,a) = aEt(0,a);
Et(1,a) = Et(0,a)−1
a;Et(v,0) = tv
Γ(v+1).(25)
3 Transformation of Basic Functions
Some fundamental functions using new general inte-
gral transformation, [19], are shown in Table 1, and
some transcendental functions are summarized in the
form of propositions derived from, [20], as the fol-
lowing:
Proposition 1. The inverse NGT derived from, [21],
of special functions as follows
T−1"p(s)q(s)
α
−
β
q(s)
α
−
λ
#=t
β
−1E
α
,
β
(
λ
t
α
),
for |
λ
|<q(s)
α
.(26)
Proposition 2. The inverse NGT for specific case is
derived from, [20]:
T−1"p(s)
q(s)q(s)
α
−a2#
=
q
∑
j=1
q
∑
k=1
aj+k−2ntEt[( j+k)
α
−1,aq]
−[( j+k)
α
−1]Et[( j+k)
α
,aq]o,
α
=1
q,q∈Z+.
(27)
Proposition 3. The inverse NGT for the general case
is derived from, [20]:
T−1p(s)
qu(s)[q(s)
α
−a]n
=
q
∑
j1=1···
q
∑
jn=1
aJ−n
(n−1)!Γ(u−n+J
α
)
×
n−1
∑
k=0
(−1)kCn−1
kΓ(u−n+J
α
+k)
×tn−1−kEt(u−n+J
α
+k,aq).(28)
where J=∑n
i=1ji;
α
=1
q,q∈Z+,Cn
k=n!
k!(n−k)!.
Table 1: Special definition of the newfangled
Laplace-typed NGLT transform.
Cases Transform
1p(s)
q(s)
tp(s)
q(s)2
t
α
Γ(
α
+1)p(s)
q(s)
α
+1
sintp(s)
q(s)2+1
sinat ap(s)
q(s)2+a2
costq(s)p(s)
q(s)2+1
cosat aq(s)p(s)
q(s)2+a2
etp(s)
q(s)−1
t f (t−1)e−q(s)(q(s)+1)p(s)
q(s)2
teat p(s)
(p−a)2
f′(t)q(s)F(s)−p(s)f(0)
f(
α
)(t)q
α
(s)F(s)−p(s)∑n−1
k=0q
α
−(k+1)(s)f(k)(0)
Theorem 3.1.Caputo’s Fractional Derivatives, [21],
[22]: Considering a continuous function y=f(t),
and an arbitrary order n−1<
α
≤n, the Caputo’s
fractional derivative of order
α
is given by:
C
aD
α
xf(x) = 1
Γ(n−
α
)Zx
a
(x−s)n−
α
−1f(n)(s)ds.
(29)
Theorem 3.2.Residue theorem (Cauchy’s residue
theorem), [23], [24], [25]. Let Cbe a simple closed
contour, described positively. If a function fis ana-
lytic, then
f(x,t) = 1
2
π
iZc+i∞
c−i∞
e−q(s)t
p(s)F(x,s)ds
=
m
∑
j=1
Res[f(x,zj),zj]in u.h.p.(30)
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where zjis a finite number of isolated singularities in
the upper half-plane (u.h.p) in the complex plane z=
x+iy, and f(x,z)is the function under the integration,
and if f(z)has a pole of order Nat z=z0,N≥m, we
use the particular expression as follows:
Res[f(z),z0] = lim
z→z0
1
(N−1)!
dN−1
dzN−1[(z−z0)Nf(z)] in u.h.p.
(31)
Theorem 3.3.CFE (Continued Fraction Expansions)
approximation method, [26], [27]:
Continued Fraction Expansions is one of the presti-
gious theories applied to find the approximate expan-
sion, especially in fractional order derivatives. The
main idea is to express the infinitive expansion of
truncated form as the following:
(1+x)
α
=1+
α
x
1+(1−
α
)x
2+(1+
α
)x
3+(2−
α
)x
2+(2+
α
)x
5+(3−
α
)x
2+(3+
α
)x
7+···
=b0+a1
b1+a2
b2+a3
b3+(a4
b4+a5
b5+(a6
b6+a7
b7+···
(32)
(1+x)
α
=b0+a1
b1+
a2
b2+
a3
b3+···
,(33)
where b0=1,b2=2,bi+1=i+1, a1=
α
x,ai= (k−
α
)x,ai+1= (k+
α
)xfor i=2k,k≥1.
Following this, we have the approximate for (1+x)
α
:
(1+x)
α
≈1+
α
x
1+(1−
α
)x
2≈2+ (1+
α
)x
2+ (1−
α
)x.(34)
By taking s=x−1, we have the first-order and the
second-order CFE approximate
s
α
≈(1+
α
)s+ (1−
α
)
(1−
α
)s+ (1+
α
).(35)
s
α
≡
α
2+3
α
+2s2+−2
α
2+8s+
α
2−3
α
+2
(
α
2−3
α
+2)s2+ (−2
α
2+8)s+ (
α
2+3
α
+2).
(36)
Theorem 3.4.Padé-approximation, [28], [29]: A
Padé approximation with real coefficients whose nu-
merator and denominator have degrees Land Mis
performed
[L
M] = a0+a1s+a2s+···+aLsL
b0+b1s+b2s+···+bLsM=
∞
∑
i=0
cisi.(37)
where cisatisfies the following simultaneous equa-
tions
a0=c0;
a1=c1+c0;
a2=c2+b1c1+b2c0;
···
al=cl+
min{L,M}
∑
i=0
bicL−i.(38)
For example, the second-order approximation is trun-
cated as
a0+a1s+a2x2
1+b1s+b2s2≈c0+c1s+c2s2,(39)
where c0=a0,a1=c1+c0,a2=c2+b1c1+b2c0.
Lemma 3.5.Using Theorem 3.3 and Theorem 3.4
for the second-order expression with combination of
equation (36) and equation (39), we derived the ap-
proximation
s
α
≈
α
2−3
α
+2
(
α
2+3
α
+2)−3(
α
−2)
α
+2s
−6
α
4−3
α
3−6
α
2+8
(
α
2+3
α
+2)2s2.(40)
4 Applications
Example 4.1.Consider the space-time fractional tele-
graph equation
∂α
∂
t
α
f(x,t) = 2f(x,t),t≥0,1≤
α
≤2,(41)
f(x,0) = ex,ft(x,0) = 0,t≥0,x>0.(42)
Solution
Taking the NGI transform on both sides equation (41)
using Table 1, using initial value conditions (42), and
simplify as follows
[q
α
(s)−2]F(x,s) = p(s)q(s)
α
−1F(x,0).
+p(s)q(s)
α
−2F
t(x,0)(43)
Simplify on both sides of equation (43), we have
F(x,s) = p(s)q(s)
α
−1ex
q
α
(s)−2.(44)
Taking inverse NGI transform equation (44) and us-
ing propositional Proposition 1, we obtain the solu-
tion
f(x,t) = E
α
,1(2t
α
)ex.(45)
When
α
=2 the exact solution becomes
f(x,t) = cosh(√2t)ex.(46)
The solution (46) is depicted in Figure 1, and Figure
2 (Appendix).
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Example 4.2.Consider the following space-time
fractional telegraph equation
∂
2
α
∂
t2
α
f(x,t) = 2
∂α
∂
t
α
f(x,t)−f(x,t),0<
α
≤1,
(47)
f(x,0) = ex,ft(x,0) = 0,t≥0,0<x≤1.
(48)
Solution
Taking the NGI transform equation (47) in Table 1,
using initial value condition (48), we have
h(q
α
(s)−1)2iF(x,s) = p(s)hq(s)2
α
−1−2q(s)
α
−1iex.
(49)
Simplify the equation (49), we have
F(x,s) =
p(s)hq(s)2
α
−1−2q(s)
α
−1iex
(q
α
(s)−1)2
=p(s)
q(s)ex−p(s)ex
(q
α
(s)−1)2.(50)
Taking inverse NGI transform equation (50) and ap-
ply the Proposition 2:
f(x,t) = ex−
q
∑
j=1
q
∑
k=1ntEt[( j+k)
α
−1,1]
−[( j+k)
α
−1]Et[( j+k)
α
,1]oex,(51)
where
α
=1
q,q∈Z+.
When
α
=1, the exact solution attained is
f(x,t) = ex−ex[tEt(1,1)−Et(2,2)].(52)
or in the form of f(x,t) = ex+t(1−t). The solu-
tion (52) is performed in Figure 3, and Figure 4 (Ap-
pendix).
Example 4.3.Consider the space-time fractional tele-
graph equation
∂
3
α
∂
t3
α
f(x,t) = 6
∂
2
α
∂
t2
α
f(x,t)−12
∂α
∂
t
α
f(x,t) + 8f(x,t),
0<
α
≤1.(53)
f(x,0) = ex,ft(x,0) = 0,ftt (x,0) = 0,t≥0,0<x≤1.
(54)
Solution
Taking NGT on both sides equation (53) using condi-
tion (54), we have
T
∂
3
α
∂
t3
α
f(x,t)=T6
∂
2
α
∂
t2
α
f(x,t)−12
∂α
∂
t
α
f(x,t)
+8f(x,t).(55)
Simplify the equation (55), we have
h(q
α
(s)−2)3iF(x,s) = p(s)q(s)3
α
−1−6q(s)2
α
−1
+12q(s)
α
−1ex.(56)
Equation (56) can be reduced as follows
F(x,s) =
p(s)hq(s)3
α
−6q(s)2
α
+12q(s)
α
iex
q(s)(q
α
(s)−2)3
=p(s)
q(s)ex+8p(s)ex
q(s)(q
α
(s)−2)3.(57)
Using Proposition 3, from equation (57), we get the
solution:
f(x,t) = ex+
q
∑
j1=1
q
∑
j2=1
q
∑
j3=1
2J−3
2Γ(J
α
−2)×
2
∑
k=0
(−1)kCk
2Γ(J
α
+k−2)
×t2−kEt(J
α
+k−2,2q)ex.(58)
where J=∑3
i=1ji;
α
=1
q,q∈Z+,Ck
2=2!
k!(2−k)!,0≤
k≤2.
When
α
=1 the exact solution becomes
f(x,t) = ex+4ex[t2Et(1,2)−2tEt(2,2) + 2Et(3,2)].
(59)
The solution (59) is portrayed in Figure 5, and Figure
6 (Appendix).
Example 4.4.Consider the approximate solution of
the space-fractional telegraph equations , [30], [31],
[32], [33]:
∂α
∂
t
α
f(x,t) =
∂
2
∂
t2f(x,t) +
∂
∂
tf(x,t) + f(x,t),
t≥0,1<
α
≤2,(60)
subjected to the initial condition
f(x,0) = ex,ft(x,0) = −e−t,0<x≤1.(61)
Solution
Taking NGI transform on both sides equation (60),
and condition (61), the equation is turned into
[q(s)
α
−q(s)2−q(s)−1]F(x,s) = p(s)[q(s)
α
−1
−q(s)
α
−2−s]f(x,0).(62)
F(x,s) = p(s)[q(s)
α
−1−q(s)
α
−2−s]
q(s)
α
−q(s)2−q(s)−1ex.(63)
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Taking inverse NGI transformation on both sides
equation (63), and applying the Theorem 3.2 to find
the solution for the general value
α
:
f(x,t)
=1
2
π
iZc+i∞
c−i∞
eq(s)t+x
p(s)
p(s)[q(s)
α
−1−q(s)
α
−2−s]
[q(s)
α
−q(s)2−q(s)−1]exds
(64)
=
m
∑
j=1
Res[f(x,q(s)),q(s)j].(65)
where q(sj),j=1,··· ,mbe the isolated singularities
in the domain and f(x,q(s)) is the function under the
integration.
We find the solution of the equation q(s)
α
−q(s)2−
q(s)−1 by apply Lemma 3.5, and getting two solu-
tions q(s1),q(s2)from the approximate expression:
h[q(s)] = −(
α
−2)27
α
2+16
α
+13
(
α
2+3
α
+2)2s2
−4
α
−1
α
+2s−6
α
α
2+3
α
+2.(66)
Finally, the approximate solution is gathered from
the equation (64) shown in Table 2 (Appendix). The
expressions of q(s1),q(s2)is shown in Table 3 (Ap-
pendix).
When
α
=2, we gain the exact solution:
f(x,t) = ex−t.(67)
The solution (73) is demonstrated in Figure 7, and
Figure 8 (Appendix).
Example 4.5.Considering the approximate solution
of the space-fractional telegraph equations, [30],
[31], [32], [33]:
∂α
∂
t
α
f(x,t) =
∂
2
∂
t2u(x,t) +
∂
∂
tf(x,t) + f(x,t)
−x2−t+1,(68)
subjected to the initial condition
f(x,0) = x2,ft(x,0) = 0,0<x≤1,t≥0,0<
α
≤1.
(69)
Solution
Taking NGI transform on both sides of the equation
(68), using equation (69), the equation becomes
[q(s)
α
−q(s)2−q(s)−1]F(x,s) = p(s)[q(s)
α
−1
−q(s)−1]f(x,0)−p(s)
q(s)2+ (1−x2)p(s)
q(s).
(70)
Simplify on both sides of equation (70), we attain
F(x,s) = p(s)
q(s)x2+p(s)[q(s)−1]
[q(s)
α
−q(s)2−q(s)−1]q(s)2.
(71)
We find the solution of the equation q(s)
α
−q(s)2−
q(s)−1 by apply Lemma 3.5 to get the solutions
q(s1),q(s2), two simple simple poles, given by equa-
tions (74) and equation (75), and q(s) = 0 is the pole
of order 2 of the denominator of equation (71):
f(x,t) = x2
+−
α
2+3
α
+22(q(s1)−1)eq(s1)t
(
α
−2)2(7
α
2+16
α
+13)[q(s1)−q(s2)]q(s1)2
+−
α
2+3
α
+22(q(s2)−1)eq(s2)t
(
α
−2)2(7
α
2+16
α
+13)[q(s2)−q(s1)]q(s2)2
+
∂
∂
q(s)q(s)−1
h[q(s)] q(s)=0
.(72)
where h[q(s)] given by equation (66).
When
α
=2, from equation (71) we have an exact
solution f(x,t) = x2+t2, [30], [31], [32], [33].
The solution (72) is shown in Figure 9, and Figure 10
(Appendix).
5 Conclusion
In this paper, a new general integral transform (NGI)
for solving space-time fractional telegraph equations
has been constructed by taking an inverse NGI trans-
form based on the present formulas. Some particu-
lar functions are cumbersome to use by inverse NGI
transform, so we apply the Padé approximate proce-
dure including continued fraction expansion to evalu-
ate the polynomial expansions. The limitation of the
Laplace-typed transform is the polynomial expres-
sion and it is hard to take inverse transform directly.
However, they are also useful in the case of using in-
verse integral transform by establishing the sequence
formula of the transform. The results showed the ap-
proximate solutions approaching exact solutions in-
cluding the order CFE model of s
α
and the [L/M]
Padé approximation.
Acknowledgment:
The authors acknowledge the anonymous reviewer’s
comments, which improve the manuscript’s quality,
and kindly provide advice and valuable discussions.
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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.
Sanoe Koonprasert: Conceptualization, data cura-
tion, formal analysis, investigation, methodology,
project administration, resources, supervision, vali-
dation, visualization, writing-original draft, writing-
review and editing.
Phayung Meesad: Conceptualization, data curation,
formal analysis, investigation, resources, supervision,
validation, visualization, writing-original draft, 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
Figure 1: 2D solution performance of example 4.1.
Figure 2: 3D solution of example 4.1 when
α
=2.
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Table 2: Solution of example 4.4.
f(x,t) =
α
2+3
α
+22eq(s1)t+x[q(s1)
α
−1+q(s1)
α
−2−q(s1)]
(
α
−2)2(7
α
2+16
α
+13)[q(s1)−q(s2)] +
α
2+3
α
+22eq(s2)t+x[q(s2)
α
−1+q(s2)
α
−2−q(s2)]
(
α
−2)2(7
α
2+16
α
+13)[q(s2)−q(s1)] .(73)
Table 3: Form of q(s1),q(s2).
q(s1) = 3
α
+
α
2+22
(
α
−2)2(
α
+2)(16
α
+7
α
2+13)×
−2
α
+8√2q1
(
α
+1)3(
α
+2)7(−150
α
−42
α
2+57
α
3+36
α
4−15
α
5+2
α
6+4)
+12√2
α
q1
(
α
+1)3(
α
+2)7(−150
α
−42
α
2+57
α
3+36
α
4−15
α
5+2
α
6+4)
+6√2
α
2q1
(
α
+1)3(
α
+2)7(−150
α
−42
α
2+57
α
3+36
α
4−15
α
5+2
α
6+4)
+√2
α
3q1
(
α
+1)3(
α
+2)7(−150
α
−42
α
2+57
α
3+36
α
4−15
α
5+2
α
6+4) + 2
(74)
and
q(s2) = −3
α
+
α
2+22
(
α
−2)2(
α
+2)(16
α
+7
α
2+13)×
2
α
+8√2q1
(
α
+1)3(
α
+2)7(−150
α
−42
α
2+57
α
3+36
α
4−15
α
5+2
α
6+4)
+12√2
α
q1
(
α
+1)3(
α
+2)7(−150
α
−42
α
2+57
α
3+36
α
4−15
α
5+2
α
6+4)
+6√2
α
2q1
(
α
+1)3(
α
+2)7(−150
α
−42
α
2+57
α
3+36
α
4−15
α
5+2
α
6+4)
+√2
α
3q1
(
α
+1)3(
α
+2)7(−150
α
−42
α
2+57
α
3+36
α
4−15
α
5+2
α
6+4)−2
(75)
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Figure 3: 2D solution performance of example 4.2.
Figure 4: 3D solution of example 4.2 when
α
=1.
Figure 5: 2D solution of example 4.3.
Figure 6: 3D solution of example 4.3 when
α
=1.
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Figure 7: 2D solution performance for example 4.4.
Figure 8: 3D solution for example 4.4 when
α
=2.
Figure 9: 2D solution performance for example 4.5.
Figure 10: 3D solution for example 4.5 when
α
=2.
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Volume 19, 2024
