EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 17, No. 2, 2024, 1046-1069
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

A force function formula for solutions of nonlinear
weakly singular Volterra integral equations (WSVIE)

Kwasi Frempong Sarfo1,2,∗, William Obeng Denteh1, Ishmael Takyi1,
Kwaku Forkuoh Darkwah1

1 Department of Mathematics, Kwame Nkrumah University of Science and Technology,
Kumasi, Ashanti Region, Ghana
2 Department of Mathematical Sciences, Kumasi Technical University, Kumasi,
Ashanti Region, Ghana

Abstract. In this paper, we examine the nonlinear Weakly Singular Volterra Integral Equation

(WSVIE), u(x) = f(x)+
∫ x

0
tµ−1

xµ [u(t)]βdt. Al-Jawary and Shehan used the Daftardar-Jafari Method
(DJM) and solved the above integral equation for the investigation parameter µ > 1 using specific
force functions with µ and β values and obtained unique solutions. We have discovered a force

function f(x) = xk1 − xγk1

γk1+µ that allows the introduction of noise terms phenomena discovered
by Wazwaz that cancel out the terms of the power series in the successive solution terms um,
m = 0, 1, 2, ..., n: we thus obtain a maximum finite power series terms for each solution term
called truncation point and denoted by xg(n). Such that the integral solution can be written as
u(x) = u0+

∑n
m=1 um, where n is finite. Simplifying the solution terms, we get the unique solution

u(x) = xk1 , irrespective of the n−value in the truncation point. We have discovered a formula
relation between the last solution term un and the truncation point as un = anx

g(n). Our results
confirm the results of the two solution examples of AL-Jawary and Shehan for the investigation
parameter µ > 1. We extend the parameter range to include µ > 1 and 0 < µ ≤ 1 for our solution.
In addition, for any chosen rational parameter k1, the solution u(x) = xk1 is extrapolated to be
valid for all integer parameter values β ≥ 2 and positive rational parameter values µ > 0 and for
any finite value of n ≥ 2.

2020 Mathematics Subject Classification: 45A05, 45D05, 45G05

Key Words and Phrases: Volterra Integral Equation, Force function, Weakly singular kernel,
Nonlinear operator, Series solutions, Truncation point formula, Unique solution.

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v17i2.5071

Email addresses: kwasifrempongsarfo@gmail.com (K.F. Sarfo), wobeng-denteh.cos@knust.edu.gh (W.
Obeng-Denteh), ismael.takyi@knust.edu.gh (I. Takyi), kfdarkwah.cos@knust.edu.gh (K.F. Darkwah)

https://www.ejpam.com 1046 © 2023 EJPAM All rights reserved.



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1047

1. Introduction

Mathematical problems can be formulated using integral equations. Mathematical for-
mulations of physical problems as differential equations are converted to integral equations[12]
or physical problems are formulated as integral transforms using appropriate kernels[29].
Some of the kernels of integral equations found in the literature are the logarithmic
kernel, K(x, t) = 1√

π
1√
ln( t

s
)
( st )

µ 1
s , Abel’s kernel, K(x, t) = 1

(x−t)
1
2
, difference kernel,

K(x, t) = (x − t), and the reproducing kernel, K(x, t) = tµ−1

xµ . While a singular inte-
gral equation has infinite limits as well as the kernel being undefined at one or two points
in the range of integration, a weaker singularity occurs when only the kernel is undefined
at some points[17].
The weakly singular Volterra integral equation,

u(x) = f(x) +

∫ x

0
K(x, t)[u(t)]βdt (1)

is linear if β = 1 and nonlinear otherwise, has various applications of scientific problems
including stereology[23], heat conduction with mixed boundary conditions[13], crystal for-
mation, electrochemistry, superfluidity, and the radiation of heat from a semi-infinite solid
state [11].
Various numerical and analytic methods have been used to solve both linear and non-
linear Volterra integral equations. These include the interpolation approach[8, 9], the
optimal homotopy asymptotic method[19], the Riemann-Liouville fractional operator[27],
the extrapolation technique[22], the Adomian Decomposition Method(ADM)[2] and the
Variational Iteration Method(VIM)[24].
In [14], the authors stated the existence, uniqueness, and singularity properties of linear
WSVIE for the case when

1. 0 < µ ≤ 1, if 0 < µ < 1, the kernel is singular at x = 0 and t = 0 for all
values of t > 0. Equation(1) has an infinite set of solutions, but if µ = 1, then
the kernel has a singularity only at x = 0, and in this case, when fϵC1[0, X] with
f(0) = 1, equation(1) has an infinite set of solutions in C[0,X], which contains only
one particular solution belonging to C1[0, X].

2. When µ > 1, the kernel has a singularity only at x = 0, and equation (1) is said to
have a unique solution in Cm[0, X], fϵCm[0, X][17]

In solving the WSVIE, authors mostly use a specific force function or force function
formula in their solution. In[2, 15, 18, 30], the authors applied the Adomian Decomposition
Method (ADM) for solving linear and nonlinear weakly singular Volterra integral equations
with the reproducing kernel. The Homotopy method was used in[7, 16, 20, 21, 26], to
solve linear and nolinear Volterra integral equations using specific force functions. In
[5, 28, 29, 31], the authors used the Variational Iteration Method (VIM) to solve linear and
nonlinear Volterra integral equations using specific force functions. Applying the Series
Solution Method (SSM) given in [18, 29, 31], the authors solved the linear WSVIE and



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1048

obtained an exact solution. In[6], the Modified Adomian Decomposition Method (MADM)
introduced by Wazwaz[29] to accelerate the convergence of the ADM was implemented.
Daftardar-Jafari discovered the iterative method[12], popularly known as DJM, for solving
general functional equations, including nonlinear Volterra integral equations, algebraic
equations and systems of ordinary differential equations, nonlinear algebraic equations,
and fractional differential equations. In [1, 3, 4, 25], the authors solved various problems
using DJM.
In [18], the authors used a force function formula instead of a specific force function
to obtain unique solutions for linear WSVIE. In [29],the authors discovered noise term
phenomena such that terms in series solutions cancelled out to give an exact solution in a
finite number of solution terms. The noise term phenomenon was reinforced in [32].
To the best of our knowledge, only AL-Jawary and Shehan [4] have implemented the DJM
to solve both linear and nonlinear WSVIE while using the reproducing kernel K(x, t) =
tµ−1

xµ in

u(x) = f(x) +

∫ x

0

tµ−1

xµ
[u(t)]βdt. (2)

In [4], the authors provided a limited solution to the WSVIE of equation (1) using two

specific force functions, f(x) = x
1
2 − 5

11x and f(x) = x − 2
9x

3 with specific parameter
values of β =2 and 3.
In this paper, our solution is also based on the method of DJM[12], wherein we introduce
a force function formula in line with Hasan and Mohammed[18]. We use the force function
formula to expand the specific integral values of β =2 and 3 in AL-Jawary and Shehan[4]
to β ≥ 2. The force function formula introduces cancellation of terms in the integral series
solution to facilitate a unique solution, as discussed in [32] as a noise term phenomenon.
We have derived a truncation point formula to augment the force function to minimise
length computation. In addition, we have provided solution models to facilitate solution
examples. Finally, we have extended the investigation parameter µ > 1 of [4] to 0 < µ ≤ 1,
which the existing literature has not considered.
The paper is organised as follows: In Section 2, the authors provided the Banach space
assumptions for the solutions of the nonlinear WSVIE using DJM. In the same section,
the authors introduced a force function formula to expand the DJM for solutions of the
nonlinear WSVIE. The authors then introduced a truncation point formula that relates
the last solution term and derived solution models in sections 2.2.1, 2.2.2, 2.2.3, and 2.2.4.
In Sections 3 and 4, the solution models were used to compute solution examples for
various parameter values of β ≥ 2, k1, and µ being rational. In Section 5, results were
displayed using tables and summarized. Discussion of the results was done in Section 6
and ended with a conclusion in Section 7.



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1049

2. Daftardar-Jafari Method(DJM) for nonlinear WSVIE with
reproducing kernel

Following the Daftardar-Jafari Method given in [12], let f, u be in Banach Space B,
then the nonlinear WSVIE of equation (1), represented in operator form, is expressed as:

u = f +N(u), (3)

is in Banach Space B, such that B 7−→ B with the operator N being,

u = N(u) =

∫ x

0
K(x, t)u[(t)]βdt. (4)

The solution of equation (3) can be represented in series form:

u =
∞∑
n=0

un. (5)

The decomposition of the nonlinear operator N yields

N

( ∞∑
n=0

un

)
= N(u0) +

∞∑
n=1

N

( n∑
j=0

uj

)
−N

( n−1∑
j=0

uj

) . (6)

From eqns. (5) and (6), eqn. (3) is equivalent to

∞∑
n=0

un = f +N(u0) +

∞∑
n=1

N

( n∑
j=0

uj

)
−N

( n−1∑
j=0

uj

) . (7)

The recurrence relation is defined as:

u0 = f,

u1 = N(u0),

un+1 = N{(u0 + . . .+ un)} − (u1 + . . .+ un−1), n = 1, 2, . . . (8)

u = f +
∞∑
n=1

un. (9)

2.1. Implementation of the DJM and the force function formula

In this section, we present a solution approach using our new force function formula
for the nonlinear WSVIE, leading to a unique solution.
Let us consider the general form of the weakly singular Volterra integral equation in[10]

u(x) = f(x) +

∫ x

0

tµ−1

xµ
[u(t)]βdt (10)



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1050

where

u0(x) = f(x) = xk1 − xγk1

µ+ γk1
, (11)

is the force function formula.

u1(x) = N [u0(t)]

=

∫ x

0

tµ−1

xµ

[
tk1 − tγk1

µ+ γk1

]β
dt. (12)

u2(x) = N [u0(x) + u1(x)]−N [u0(x)]

=

∫ x

0

tµ−1

xµ

[
tk1 − tγk1

µ+ γk1
+ u1(t)

]β
dt− u1. (13)

u3(x) = N [u0(x) + u1(x) + u2(x)]−N [u0(x) + u1(x)]

=

∫ x

0

tµ−1

xµ

[
tk1 − tγk1

µ+ γk1
+ (u1 + u2)(t)

]β
dt−

∫ x

0

tµ−1

xµ

[
tk1 − tγk1

µ+ γk1
+ u1(t)

]β
dt.

(14)

In continuing from equation(14), we generate successive solution terms as:

u0(x) = f(x),

u1(x) =

∫ x

0

tµ−1

xµ

[
tk1 − tγk1

µ+ γk1

]β
dt,

u2(x) =

∫ x

0

tµ−1

xµ

[
tk1 − tγk1

µ+ γk1
+ u1(t)

]β
dt−

∫ x

0

tµ−1

xµ

[
tk1 − tγk1

µ+ γk1

]β
dt

u3(x) =

∫ x

0

tµ−1

xµ

[
tk1 − tγk1

µ+ γk1
+ (u1 + u2)t

]β
dt−

∫ x

0

tµ−1

xµ

[
tk1 − tγk1

µ+ γk1
+ u1(t)

]β
dt

...

um(x) =

∫ x

0

tµ−1

xµ

[
tk1 − tγk1

µ+ γk1
+ ...+ um−1(t)

]β
dt−

∫ x

0

tµ−1

xµ

[
tk1 − tγk1

µ+ γk1
+ ...+ um−2(t)

]β
dt,

... (15)

un(x) =

∫ x

0

tµ−1

xµ

[
tk1 − tγk1

µ+ γk1
+ ...+ un−1(t)

]β
dt−

∫ x

0

tµ−1

xµ

[
tk1 − tγk1

µ+ γk1
+ ...+ un−2(t)

]β
dt,

u(x) = xk1 − xγk1

µ+ γk1
+

n∑
m=1

um. (16)

which reduces to a unique solution, u(x) = xk1 , for every integer value γ = β, (β ≥ 2),
positive rational values of k1 and µ > 0, and un is a finite solution term and is related to
the truncation point by the relation un = anx

g(n).



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1051

Thus, the force function formula, f(x) = xk1 − xγk1

µ+γk1
, generates noise term cancella-

tion to obtain a unique solution when the truncation point is introduced. The relation
between the truncation point xg(n) and un is given by:

un(x) = anx
[n(γ−1)+1]k1 , n ≥ 2, (17)

where,

an =
βn−1

(γk1 + µ)
∏n

m=2

[
[m(γ − 1) + 1]k1 + µ

] . (18)

For example, when β = γ = 2, k1 = 3, and µ = 3
2 , if n = 2, then the final series solution

term is u2(x) =
8

315x
9. If n = 3, then the final series solution term is u3(x) =

16
8505x

12, and
so on. Equation (17) was discovered through the solution process.

2.2. Solution Models for the nonlinear WSVIE

Based on the solution series of equation (15) in section (2.1) and the subsequent trun-
cation point formula in equations (17) and (18), we provide explicit algebraic solutions
for the nonlinear WSVIE for values of β = 2, 3, 4, and 5 and show that for each case we
obtain the unique solution for our force function.

2.2.1. Solution model for 2nd -order nonlinear WSVIE (γ = β = 2)

Consider the 2nd order WSVIE given by:

u(x) = a1(x)
k1 − a2(x)

γk1 +

∫ x

0

tµ−1

xµ
u(t)βdt, γ = β = 2, µ > 0 and k1 ∈ Q+

Truncation point is given by

un(x) = anx
[n(γ−1)+1]k1 , n ≥ 2

u0(x) = f(x) = a1(x)
k1 − a2(x)

γk1 = xk1 − xγk1

µ+ γk1
where

a1 = µ− (µ− 1) = 1, a2 =
1

µ+ γk1

u1 =

∫ x

0

tµ−1

xµ
(u0)(t)

βdt

=
xβk1

βk1 + µ
− βa2

xk1+γk1

k1 + γk1 + µ
+ aβ2

xβγk1

βγk1 + µ

u2 =

∫ x

0

tµ−1

xµ
(u0 + u1)(t)

βdt− u1



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1052

= βa2
xk1+γk1

(k1 + γk1 + µ)
− aβ2

xβγk1

βγk1 + µ
− 2βa2

x2k1+γk1

(k1 + γk1 + µ)(2k1 + γk1 + µ)

+2aβ2
xk1+βγk1

(βγk1 + µ)(k1 + βγk1 + µ)
+ β2a22

x2k1+2γk1

(k1 + γk1 + µ)2(2k1 + 2γk1 + µ)

−2βaβ+1
2

xk1+3γk1

((k1 + γk1 + µ)(βγk1 + µ)(k1 + 3γk1 + µ)
+ ...

u3(x) =

∫ x

0

tµ−1

xµ
(u0 + u1 + u2)(t)

βdt−
2∑

i=0

ui

= 2βa2
x2k1+γk1

(k1 + γk1 + µ)(2k1 + γk1 + µ)
− 2aβ2

xk1+βγk1

(βγk1 + µ)(k1 + βγk1 + µ)

−4βa2
x3k1+γk1

(k1 + γk1 + µ)(2k1 + γk1 + µ)(3k1 + γk1 + µ)

−β2a22
x2k1+2γk1

(k1 + γk1 + µ)2(2k1 + 2γk1 + µ)

+4aβ2
x2k1+βγk1

(βγk1 + µ)(k1 + βγk1 + µ)(2k1 + βγk1 + µ

+2βaβ+1
2

x2k1+γk1+βγk1

(k1 + γk1 + µ)(βγk1 + µ)(k1 + γk1 + βγ + µ)

+2β2a22
x3k1+2γk1

(k1 + γk1 + µ)2(2k1 + βγk1 + µ)(3k1 + 2γk1 + µ)

u4(x) =

∫ x

0

tµ−1

xµ
(u0 + u1 + u2 + u3)(t)

βdt−
3∑

i=1

ui

= 4βa2
x3k1+γk1

(k1 + γk1 + µ)(2k1 + γk1 + µ)(3k1 + γk1 + µ)

−4aβ2
x2k1+βγk1

(βγk1 + µ)(k1 + βγk1 + µ)(2k1 + βγk1 + µ

−8βa2
x4k1+γk1

(k1 + γk1 + µ)(2k1 + γk1 + µ)(3γk1 + γk1 + µ)(4k1 + γk1 + µ)

−2β2a22
x3k1+2γk1

(k1 + γk1 + µ)2(2k1 + βγk1 + µ)(3k1 + 2γk1 + µ)

+8aβ2
x3k1+βγk1

(βγk1 + µ)(k1 + βγk1 + µ)(2k1 + βγk1 + µ) + (3k1 + βγk1 + µ)

u5(x) =

∫ x

0

tµ−1

xµ
(u0 + u1 + u2 + u3 + u4)(t)

βdt−
4∑

i=1

ui

= 8βa2
x4k1+γk1

(k1 + γk1 + µ)(2k1 + γk1 + µ)(3γk1 + γk1 + µ)(4k1 + γk1 + µ)



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1053

−8aβ2
x3k1+βγk1

(βγk1 + µ)(k1 + βγk1 + µ)(2k1 + βγk1 + µ)(3k1 + βγk1 + µ)

−16βa2

[(
x5k1+γk1

(k1 + γk1 + µ)(2k1 + γk1 + µ)(3k1 + γk1 + µ)
×

1

(4k1 + γk1 + µ)(5k1 + γk1 + µ)

)]
The series solutions reduces to

u6(x) =

∫ x

0

tµ−1

xµ
(u0 + u1 + u2 + u3 + u4 + u5(x))(t)

βdt−
5∑

i=1

ui

= 16βa2

[(
x5k1+γk1

(k1 + γk1 + µ)(2k1 + γk1 + µ)(3k1 + γk1 + µ)
×

1

(4k1 + γk1 + µ)(5k1 + γk1 + µ)

)]
+ ...

u(x) =
6∑

i=0

ui

∴ u(x) = xk1

2.2.2. Solution model for 3rd -order nonlinear WSVIE(γ = β = 3)

Consider the 3rd order WSVIE of the form:

u(x) = a1(x)
k1 − a2(x)

γk1 +

∫ x

0
− tµ−1

xµ
u(t)βdt, γ = β = 3, µ > 0 and k1 ∈ Q+ (19)

Truncation point is given by

un(x) = anx
[n(γ−1)+1]k1 , n ≥ 2

u0(x) = f(x) = a1(x)
k1 − a2(x)

γk1 = xk1 − xγk1

µ+ γk1
where,

a1 = µ− (µ− 1), a2 =
1

µ+ γk1

u1(x) =

∫ x

0

tµ−1

xµ
(u0)(t)

βdt

=
x3k1

3k1 + µ
− 3a2

x2k1+γk1

2k1 + γk1 + µ
+ 3a22

xk1+2γk1

k1 + 2γk1 + µ
− a32

x3γk1

3γk1 + µ

u2(x) =

∫ x

0

tµ−1

xµ
(u0 + u1)(t)

βdt− u1

= 3a2
x2k1+γk1

2k1 + γk1 + µ
+ 3a22

xk1+2γk1

k1 + 2γk1 + µ
+ 9a2

x4k1+γk1

(2k1 + γk1 + µ)(4k1 + γk1 + µ)



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1054

−9a22
x3k1+2γk1

(k1 + 2γk1 + µ)(3k1 + 2γk1 + µ)
+ a32

x3γk1

3γk1 + µ

u3(x) =

∫ x

0

tµ−1

xµ
(u0 + u1 + u2)(t)

βdt−
2∑

i=0

ui

= 9a2
x4k1+γk1

(2k1 + γk1 + µ)(4k1 + γk1 + µ)
− 9a22

x3k1+2γk1

(k1 + 2γk1 + µ)(3k1 + 2γk1 + µ)

−27a2
x6k1+γk1

(2k1 + γk1 + µ)(4k1 + γk1 + µ)(6k1 + γk1 + µ)

u4(x) =

∫ x

0

tµ−1

xµ
(u0 + u1 + u2 + u3)(t)

βdt−
3∑

i=1

ui

= 27a2
x6k1+γk1

(2k1 + γk1 + µ)(4k1 + γk1 + µ)(6k1 + γk1 + µ)

u(x) =
4∑

i=0

ui

∴ u(x) = xk1

2.2.3. Solution model for 4th -order nonlinear WSVIE(γ = β = 4)

Consider the 4th order WSVIE of the form:

u(x) = a1(x)
k1 + a2(x)

γk1 +

∫ x

0
− tµ−1

xµ
u(t)βdt, γ = β = 3, µ > 0 and k1 ∈ Q+(20)

Truncation point is given by,

un(x) = anx
[n(γ−1)+1]k1 , n ≥ 2

u0(x) = f(x) = a1(x)
k1 − a2(x)

γk1 = xk1 − xγk1

µ+ γk1
where,

a1 = µ− (µ− 1), a2 =
1

µ+ γk1

u1 =

∫ x

0

tµ−1

xµ
(u0)(t)

βdt

=
x4k1

4k1 + µ
− 4a2

x3k1+γk1

3k1 + γk1 + µ
+ 6a22

x2k1+2γk1(10)

2k1 + 2γk1 + µ
− 4a32

xk1+3γk1

k1 + 3γk1 + µ

+a42
x4γk1

4γk1 + µ

u2 =

∫ x

0

tµ−1

xµ
(u0 + u1)(t)

βdt− u1

= 4a2
x3k1+γk1

3k1 + γk1 + µ
− 6a22

x2k1+2γk1

2k1 + 2γk1 + µ



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1055

−16a2
x6k1+γk1

(3k1 + γk1 + µ)(6k1 + γk1 + µ)
+ 4a32

xk1+3γk1

(k1 + 3γk1 + µ)

+24a32
x5k1+2γk1

(2k1 + 2γk1 + µ)(5k1 + 2γk1 + µ)
− a42

x4γk1

4γk1 + µ

−16a32
x4k1+3γk1

(k1 + 3γk1 + µ)(4k1 + 3γk1 + µ)
+ 96a22

x8k1+2γk1

(3k1 + γk1 + µ)2(8k1 + 2γk1 + µ)
(21)

u3(x) =

∫ x

0

tµ−1

xµ
(u0 + u1 + u2)(t)

βdt−
2∑

i=0

ui

= 16a2
x6k1+γk1

(3k1 + γk1 + µ)(6k1 + γk1 + µ)
− 24a32

x5k1+2γk1

(2k1 + 2γk1 + µ)(5k1 + 2γk1 + µ)

−64a2
x9k1+γk1

(3k1 + γk1 + µ)(6k1 + γk1 + µ)(9k1 + γk1 + µ)

+16a32
x4k1+3γk1

(k1 + 3γk1 + µ)(4k1 + 3γk1 + µ)

+96a22
x8k1+2γk1

(2k1 + 2γk1 + µ)(5k1 + 2γk1 + µ)(8k1 + 2γk1 + µ)

−96a22
x8k1+2γk1

(3k1 + γk1 + µ)2(8k1 + 2γk1 + µ)

u4(x) =

∫ x

0

tµ−1

xµ
(u0 + u1 + u2 + u3)(t)

βdt−
3∑

i=1

ui

= 64a2
x9k1+γk1

(3k1 + γk1 + µ)(6k1 + γk1 + µ)(9k1 + γk1 + µ)

−96a22
x8k1+2γk1

(2k1 + 2γk1 + µ)(5k1 + 2γk1 + µ)(8k1 + 2γk1 + µ)

−256a2
x12k1+γk1

(3k1 + γk1 + µ)(6k1 + γk1 + µ)(9k1 + γk1 + µ)(12k1 + γk1 + µ)

u5(x) =

∫ x

0

tµ−1

xµ
(u0 + u1 + u2 + u3 + u4)(t)

βdt−
4∑

i=1

ui

−256a2
x12k1+γk1

(3k1 + γk1 + µ)(6k1 + γk1 + µ)(9k1 + γk1 + µ)(12k1 + γk1 + µ)

u(x) =

5∑
i=0

ui

∴ u(x) = xk1

(22)



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1056

2.2.4. Solution model for 5th -order nonlinear WSVIE(γ = β = 5)

Consider the general forcing function for a unique solution of WSVIE given by:

u(x) = xk1 − xγk1

µ+ γk1
+

∫ x

0

tµ−1

xµ
uβ(t)dtγ = β = 3, µ > 0 and k1 ∈ Q+ (23)

Truncation point is given by,

un(x) = anx
[n(γ−1)+1]k1 , n ≥ 2

u0(x) = f(x) = a1(x)
k1 + a2(x)

γk1 = xk1 − xγk1

µ+ γk1
where,

a1 = µ− (µ− 1), a2 =
1

µ+ γk1

u1(x) =

∫ x

0

tµ−1

xµ
(u0)

β(t)dt

=
x5k1

5k1 + µ
− 5a2

x4k1+γk1

4k1 + γk1 + µ
+ 10a22

x3k1+2γk1

3k1 + 2γk1 + µ

−a32
x2k1+3γk1

2k1 + 3γk1 + µ
+ 5a42

xk1+4γk1

k1 + 4γk1 + µ
+ ...

u2(x) =

∫ x

0

tµ−1

xµ
(u0 + u1)(t)

βdt− u1

= 5a2
x4k1+γk1

4k1 + γk1 + µ
− 10a22

x3k1+2γk1

3k1 + 2γk1 + µ

−25a2
x8k1+γk1

(4k1 + γk1 + µ)(8k1 + γk1 + µ)
+ 10a32

x2k1+3γk1

2k1 + 3γk1 + µ

+50a22
x7k1+2γk1

(3k1 + 2γk1 + µ)(7k1 + 2γk1 + µ)
− 5a42

xk1+4γk1

k1 + 4γk1 + µ

−50a32
x6k1+3γk1

(2k1 + 3γk1 + µ)(6k1 + 3γk1 + µ)

+250a22
x11k1+2γk1

(4k1 + γk1 + µ)2(11k1 + 2γk1 + µ)

u3(x) =

∫ x

0

tµ−1

xµ
(u0 + u1 + u2)(t)

βdt−
2∑

i=1

ui

= 25a2
x8k1+γk1

(4k1 + γk1 + µ)(8k1 + γk1 + µ)

−50a22
x7k1+2γk1

(3k1 + 2γk1 + µ)(7k1 + 2γk1 + µ)

−125a2
x12k1+γk1

(4k1 + γk1 + µ)(8k1 + γk1 + µ)(12k1 + γk1 + µ)



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1057

+50a32
x6k1+3γk1

(2k1 + 3γk1 + µ)(6k1 + 3γk1 + µ)

−250a22
x11k1+2γk1

(4k1 + γk1 + µ)2(11k1 + 2γk1 + µ)

+250a22
x11k1+2γk1

(3k1 + 2γk1 + µ)(7k1 + 2γk1 + µ)(11k1 + 2γk1 + µ)

u4(x) =

∫ x

0

tµ−1

xµ
(u0 + u1 + u2 + u3)(t)

βdt−
3∑

i=1

ui

= 125a2
x12k1+γk1

(4k1 + γk1 + µ)(8k1 + γk1 + µ)(12k1 + γk1 + µ)

−250a22
x11k1+2γk1

(3k1 + 2γk1 + µ)(7k1 + 2γk1 + µ)(11k1 + 2γk1 + µ)

−625a2
x16k1+γk1

(4k1 + γk1 + µ)(8k1 + γk1 + µ)(12k1 + γk1 + µ)(16k1 + γk1 + µ)

u5(x) =

∫ x

0

tµ−1

xµ
(u0 + u1 + u2 + u3 + u4)(t)

βdt−
4∑

i=1

ui

= 625a2
x16k1+γk1

(4k1 + γk1 + µ)(8k1 + γk1 + µ)(12k1 + γk1 + µ)(16k1 + γk1 + µ)

u(x) =

5∑
i=0

ui

∴ u(x) = xk1

3. Examples for β solution models using the investigation parameter
µ > 1

In this section, we implement the DJM and the force function formula for solutions of
nonlinear WSVIE.
Example 1(a). Consider the 2nd order nonlinear WSVIE given by

u(x) = xk1 − xγk1

γk1 + µ
+

∫ x

0

tµ−1

xµ
u(t)βdt (24)

Following the algorithm in equation (15), the relation between the final series solution
term and the truncation point is determined using equation (17).
Solution,

u0(x) = f(x) = x
1
2 − 5

11
x,

u1(x) =
5

11
x− 100

297
x

3
2 +

125

1936
x2



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1058

u2(x) =
100

297
x

3
2 − 125

594
x2 − 125

1936
x2 +

625

17908
x

5
2 +

50000

1852389
x3 + ...

u3(x) =
125

594
x2 − 625

17908
x

5
2 − 1250

10989
x

5
2 − 50000

1852389
x3 +

3125

188034
x3 + ...

u4(x) =
1250

10989
x

5
2 − 3125

188034
x3 − 12500

230769
x3 + ...

u5(x) =
12500

230769
x3 + ...

u(x) = u0 + ...+ u5 = x
1
2

Example 1(b). Consider the 3rd order nonlinear WSVIE given by

u(x) = xk1 − xγk1

γk1 + µ
+

∫ x

0

tµ−1

xµ
u(t)βdt (25)

For third order nonlinear parameter, γ = β = 3, k1 =
1
2 , µ = 2.

Following the algorithm in equation (15), the relation between the final series solution
term and the truncation point is determined using equation (17).

u0(x) = x
1
2 − 2

7
x

3
2

u1(x) =
2

7
x

3
2 − 4

21
x

5
2 +

24

539
x

7
2 − 16

4459
x

9
2

u2(x) =
4

21
x

5
2 − 24

539
x

7
2 − 8

77
x

7
2 +

144

7007
x

9
2 +

16

4459
x

9
2 + ...

u3(x) =
8

77
x

7
2 − 144

7007
x

9
2 − 48

1001
x

9
2 + ...

u4(x) =
48

1001
x

9
2 + ...

u(x) = u0 + ...+ u6 = x
1
2

Example 1(c). For k1 =
1
2 , µ = 3

2 ,γ = β = 4,
we substitute the parameter values in the 4th-order solution model to obtain the solution
u(x) = x

1
2

Example 1(d). For k1 =
1
2 , µ = 3,γ = β = 5,

we substitute the parameter values in the 5th-order solution model to obtain the unique
solution, u(x) = x

1
2

Example 2(a). Consider the nonlinear WSVIE of the form

u(x) = xk1 − xγk1

γk1 + µ
+

∫ x

0

tµ−1

xµ
u(t)3dt (26)



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1059

For 3rd order nonlinear parameter, γ = β = 3, µ = 3
2 and k1 = 1.

Following the algorithm in equation (15), the relation between the final series solution
term and the truncation point is determined using equation (17).

u0(x) = f(x)x− 2

9
x3

u1(x) =
2

9
x3 − 4

39
x5 +

8

459
x7 − 16

15309
x9

u2(x) =
4

39
x5 − 8

459
x7 − 8

221
x7 + ...

u3(x) =
8

221
x7 − 16

3216
x9 − 16

1547
x9 + ...

u4(x) =
16

1547
x9 + ...

u = u0 + ...+ u4 = x.

Example 2(b). Consider the fourth order nonlinear WSVIE given by

u(x) = xk1 − xγk1

γk1 + µ
+

∫ x

0

tµ−1

xµ
u(t)βdt (27)

For fourth order nonlinear parameter , γ = β = 4, k1 = 1, µ = 7
2 .

Following the algorithm in equation (15), the relation between the final series solution
term and the truncation point is determined using equation (17).

u0(x) = f(x) = x− 2

15
x4

u1(x) =
2

15
x4 − 16

315
x7 +

16

2025
x10 − 64

111375
x13 + ...

u2(x) =
16

315
x7 − 16

2025
x10 − 128

8505
x10 +

64

111375
x13 +

256

1002375
x13 + ...

u3(x) =
128

8505
x10 − 256

1002375
x13 − 1024

280665
x13 + ...

u4(x) =
1024

280665
x13 + ...

u = u0 + ...+ u4 = x.

Example 2(c).For k1 = 1, µ = 3,γ = β = 2,
we substitute the parameter values in the 2nd-order solution model to obtain the solution
u(x) = x
Example 2(d). For k1 = 1, µ = 4

3 ,γ = β = 5,
we substitute the parameter values in the 5th-order solution model to obtain the solution



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1060

u(x) = x
Example 3(a). Consider the 3rd order nonlinear WSVIE given by

u(x) = xk1 − xγk1

γk1 + µ
+

∫ x

0

tµ−1

xµ
u(t)βdt (28)

For the 3rd order nonlinear parameter, γ = β = 3, k1 =
1
4 , µ = 5

2 .
Following the algorithm in equation (15), the relation between the final series solution
term and the truncation point is determined using equation (17).

u0(x) = f(x) = x
1
4 − 4

13
x

3
4

u1(x) =
4

13
x

3
4 − 16

65
x

5
4 +

192

2873
x

7
4 − 256

41743
x

9
4

u2(x) =
16

65
x

5
4 − 192

2873
x

7
4 − 192

1105
x

7
4 +

2304

54587
x

9
4 +

256

41304
x

9
4 + ...

u3(x) =
192

1105
x

7
4 − 2304

54587
x

9
4 − 2304

20995
x

9
4 + ...

u4(x) =
2304

20995
x

9
4 + ...

u = u0 + ...+ u4 = x
1
4 .

Example 3(b). Consider the nonlinear WSVIE of the form,

u(x) = xk1 − xγk1

γk1 + µ
+

∫ x

0

tµ−1

xµ
u(t)βdt (29)

For fifth order nonlinear parameter, γ = β = 5, k1 =
1
4 and µ = 5

4 . Following the algorithm
in equation (15), the relation between the final series solution term and the truncation
point is determined using equation (17).

u0(x) = f(x) = x
1
4 − 2

5
x

5
4

u1(x) =
2

5
x

5
4 − 4

7
x

9
4 +

16

45
x

13
4 − 32

275
x

17
4 +

32

1625
x

21
4 + ...

u2(x) =
4

7
x

9
4 − 16

45
x

13
4 − 40

63
x

13
4 +

32

275
x

17
4 +

32

99
x

17
4 − 32

1625
x

21
4 +

64

715
x

21
4 +

320

637
x

21
4 + ...

u3(x) =
40

63
x

13
4 − 32

99
x

17
4 − 400

693
x

17
4 +

64

715
x

21
4 − 320

637
x

21
4 +

320

1287
x

21
4 + ...

u4(x) =
400

693
x

17
4 − 320

1287
x

21
4 − 4000

9009
x

21
4 + ...

u5(x) =
4000

9009
x

21
4 + ...



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1061

u = u0 + ...+ u5 = x
1
4 .

Example 3(c). For k1 =
1
4 , µ = 4,γ = β = 4,

we substitute the parameter values in the 4th-order solution model to obtain the solution,
u(x) = x

1
4

Example 3(d). For k1 =
1
4 , µ = 5

2 ,γ = β = 2,
we substitute the parameter values in the 2nd-order solution model to obtain the solution,
u(x) = x

1
2

Remark 1. The solution for the above six examples is obtained as u(x) = xk1, irrespective
of the values of the assigned parameters defined for k1, β and µ. See verification of the
series solution for various β solution models in Section 2.2.

4. Examples for various β solution models using the investigation
parameter 0 < µ ≤ 1

In this section, we examine the solutions to nonlinear WSVIE problems using the
investigation parameter µ being 0 < µ ≤ 1 and the nonlinear integer parameter β ≥ 2
Example 4(a). Consider the nonlinear WSVIE of the form

u(x) = xk1 − xγk1

µ+ γk1
+

∫ x

0

tµ−1

xµ
u(t)βdt (30)

For the 2nd-order nonlinear parameter, β = γ = 2, k1 = 3
2 and µ = 1

3 . Following the
algorithm in equation (15), the relation between the final series solution term and the
truncation point is determined using equation (17).

u0(x) = f(x) = x
3
2 − 3

10
x3

u1(x) =
3

10
x3 − 18

145
x

9
2 +

27

1900
x6

u2(x) =
18

145
x

9
2 − 27

1900
x6 − 108

2755
x6 +

81

22325
x

15
2 − 243

147175
x9 − 1458

4476875
x

21
2 + ...

u3(x) =
2916

148555
x6 − 81

22325
x

15
2 − 1296

129485
x

15
2 +

243

147175
x9 +

243

312550
x9 +

1458

4476875
x12 +

2916

9566375
x

21
2 + ...

u4(x) =
1296

129485
x

15
2 − 243

312550
x9 − 972

713545
x9 − 2916

9566375
x

21
2 +

1458

10157875
x

21
2 + ...

u5(x) =
972

713545
x9 − 1458

10157875
x

21
2 − 23328

58915675
x

21
2 + ...

u6(x) =
23328

58915675
x

21
2 + ...

u(x) =

6∑
n=0

un = x
3
2



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1062

Example 4(b). Consider the nonlinear WSVIE of the form

u(x) = xk1 − xγk1

µ+ γk1
+

∫ x

0

tµ−1

xµ
u(t)βdt (31)

For 3rd order nonlinear parameter, β = γ = 3, k1 =
3
2 , µ = 1.

Following the algorithm in equation (15), the relation between the final series solution
term and the truncation point is determined using equation (17).

u0(x) = f(x) = x
3
2 − 2

11
x

9
2

u1(x) =
2

11
x

9
2 − 12

187
x

15
2 +

24

2783
x

21
2 − 16

38599
x

27
2

u2(x) =
12

187
x

15
2 − 24

2783
x

21
2 − 72

4301
x

21
2 +

144

80707
x

27
2 +

16

38599
x

27
2 + ...

u3(x) =
72

4301
x

21
2 − 144

80707
x

27
2 − 432

124729
x

27
2 + ...

u4(x) =
432

124729
x

27
2 + ...

u(x) =
4∑

n=0

un = x
3
2 .

Example 5(a). Consider the nonlinear WSVIE of the form,

u(x) = xk1 − xγk1

µ+ γk1
+

∫ x

0

tµ−1

xµ
u(t)βdt

For the 3rd order nonlinear parameter, β = γ = 3, k1 =
1
4 , µ = 1

2 .
Following the algorithm in equation (15), the relation between the final series solution
term and the truncation point is determined using equation (17).

u0(x) = f(x) = x
1
4 − 4

5
x

3
4

u1(x) =
4

5
x

3
4 − 48

65
x

5
4 +

64

75
x

7
4 − 256

1375
x

9
4

u2(x) =
48

65
x

5
4 − 64

75
x

7
4 − 64

35
x

7
4 +

256

1375
x

9
4 +

256

275
x

9
4 + ...

u3(x) =
64

35
x

7
4 − 256

275
x

9
4 − 6912

3465
x

9
4 + ...

u4(x) =
6912

3465
x

9
4 + ...

u(x) =

4∑
n=0

un = x
1
4 .



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1063

Example 5(b). Consider the nonlinear WSVIE of the form,

u(x) = xk1 − xγk1

µ+ γk1
+

∫ x

0

tµ−1

xµ
u(t)βdt (32)

5th-order nonlinear parameter with β = γ = 5, k1 = 1 and µ = 3
4 .

Following the algorithm in equation (15), the relation between the final series solution
term and the truncation point is determined using equation (17).

u0(x) = f(x) = x− 4

23
x5

u1(x) =
4

23
x5 − 80

897
x9 +

128

5819
x13 − 2560

863857
x17 + ...

u2(x) =
80

897
x9 − 128

5819
x13 − 320

9867
x13 +

2560

863857
x17 − 12800

2065745
x17 + ...

u3(x) =
320

9867
x13 +

12800

2065745
x17 − 6400

700557
x17 + ...

u4(x) =
6400

700557
x17 + ...

u(x) =
4∑

i=0

un = x.

Example 6(a). Consider the nonlinear WSVIE of the form,

u(x) = xk1 − xγk1

µ+ γk1
+

∫ x

0

tµ−1

xµ
u(t)βdt (33)

4th order nonlinear parameter with β = γ = 4, k1 =
1
4 and µ = 4

5 .
Following the algorithm in equation (15), the relation between the final series solution
term and the truncation point is determined using equation (17).

u0(x) = f(x) = x
1
4 − 5

9
x

u1(x) =
5

9
x− 400

459
x

7
4 +

500

891
x

5
2 − 10000

59049
x

13
4 + ...

u2(x) =
400

459
x

7
4 − 500

891
x

5
2 − 16000

15147
x

5
2 +

10000

59049
x

13
4 − 200000

649539
x

13
4 + ...

u3(x) =
16000

15147
x

5
2 − 200000

649539
x

13
4 − 1280000

1226907
x

13
4 + ...

u4(x) =
1280000

1226907
x

13
4 + ...

u(x) =

4∑
n=0

un = x
1
4 .



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1064

Example 6(b). Consider the nonlinear WSVIE of the form

u(x) = xk1 − xγk1

µ+ γk1
+

∫ x

0

tµ−1

xµ
u(t)βdt (34)

5th-order nonlinear parameter β = 5 with β = γ = 5, k1 = 2, µ = 1
4 .

Following the algorithm in equation (15), we obtain the solution in series, and the trun-
cation point with the highest power is determined using equation (17).
Solution in series,

u0(x) = f(x) = x2 − 4

41
x10

u1(x) =
4

41
x10 − 80

2993
x18 +

128

35301
x26 + ...

u2(x) =
80

2993
x18 − 128

35301
x26 − 320

62853
x26 + ...

u3(x) =
320

62853
x26 + ...

u(x) =
∞∑
n=0

un = x2.

Remark 2. The solution for the above six examples is obtained as u(x) = xk1, irrespective
of the values of the assigned parameters defined for k1, β and µ. See verification of the
series solution for various β solution models in Section 3.

5. Summary of results of solved examples

The tables 1 and 2 below show row 1 as example of numbering; the second, third,
and fourth rows are parameter values in the nonlinear WSVIE; and the fifth row is the
solution.
Tables 1 and 2 show results for µ > 1 and 0 < µ ≤ 1, respectively.
Table 1: Solutions of worked examples for the case of µ > 1.

Example 1(a) 1(b) 1(c) 1(d) 2(a) 2(b) 2(c) 2(d) 3(a) 3(b) 3(c) 3(d)

k1
1
2

1
2

1
2

1
2 1 1 1 1 1

4
1
4

1
4

1
4

β 2 3 4 5 3 4 2 5 3 5 4 2

µ 6
5 2 3

2 3 3
2

7
2 3 4

3
5
2

5
4 4 5

2

u(x) x
1
2 x

1
2 x

1
2 x

1
2 x x x x x

1
4 x

1
4 x

1
4 x

1
4



K. F. Sarfo et al. / Eur. J. Pure Appl. Math, 17 (2) (2024), 1046-1069 1065

Table 2: Solutions of worked examples for the case of 0 < µ ≤ 1.
Example 4(a) 4(b) 5(a) 5(b) 6(a) 6(b)

k1
3
2

3
2

1
4 1 1

4 2

β 2 3 3 5 4 5

µ 1
3 1 1

2
3
4

4
5

1
4

u(x) x
3
2 x

3
2 x

1
4 x x

1
4 x2

From Tables 1 and 2 above, the solution is obtained as u(x) = xk1 for the indicated integer
values of β ≥ 2, rational values of µ being µ > 1 and 0 < µ ≤ 1

6. Discussion

In this paper, we used DJM and the force function formula in the solution process.
In line with [18], the force function used is f(x) = xk1 − xγk1

µ+γk1
. As discussed in [32], we

obtained the relation between the last solution term and the truncation point to be

un(x) =
βn−1x[n(γ−1)+1]k1

(γk1 + µ)
∏n

m=2

[
[m(γ − 1)]k1 + µ

] .
The relation between the last term in the solution series and the truncation point was
discovered through the solution process in the appendix, and the relation holds for all
integer values of β ≥ 2 and rational values of k1 and µ > 0. Due to noise term cancellation,
our solution becomes

u(x) = u0(x) +
n∑

m=1

um = xk1 .

We extended the range of investigation parameter values from µ > 1 to 0 < µ ≤ 1.
From Table 1, our solution examples 1(a) and 2(a) confirm the solutions of Al-Jawary and

Shehan[4], who respectively used specific force functions, f(x) = x
1
2 − 5

11x with β = 2 and
µ = 6

5 ,
and f(x) = x− 2

9x
3 with β = 3 and µ = 3

2 for their nonlinear WSVIE solutions.
Our solutions depend on the force function formula parameter k1 and produce the unique
solution u = xk1 irrespective of the nonlinear integer parameter value of β ≥ 2, positive
rational values of k1 and µ > 0, and for any finite value of n ≥ 2.

7. Conclusion

In this paper, we have solved the nonlinear WSVIE of equation (2) with the reproducing

kernel, K(x, t) = tµ−1

xµ , by extending the range of the investigation parameter µ > 1 to

0 < µ ≤ 1. We have discovered a force function formula, f(x) = xk1 − xγk1

γk1+µ , used
in equation (2). We are able to determine a formula relation between the final series
solution term and the truncation point. For the purpose of verifying different β solutions,
the authors have derived solution models in Section 3 that facilitate the computation of



REFERENCES 1066

solution examples. The examples of solutions validate the outcomes found in [4]. From
the various β solution models and table of solutions, we extrapolate that for the force
function, f(x) = xk1 − xγk1

γk1+µ where k1 is rational, we shall always get the unique solution

u(x) = xk1 , irrespective of any chosen parameter values of the integer β ≥ 2, positive
rational parameter value µ > 0, and for any finite value of n ≥ 2. The authors are
working on extending the range of µ values from µ > 0 to µ ≤ 0

Acknowledgements

The authors are grateful to the reviewers for their insightful remarks that helped us
improve this article.

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