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IKONION JOURNAL OF MATHEMATICS
Year: 2020 Volume:2 Issue: 2
ISSN: 2687-6531
MULTIPLICATIVE VOLTERRA INTEGRAL EQUATIONS AND
THE RELATIONSHIP BETWEEN MULTIPLICATIVE
DIFFERENTIAL EQUATIONS
Nihan GรNGรR *1 and Hatice DURMAZ1
1 Gรผmรผลhane University, Faculty of Engineering and Naturel Sciences, Department of Mathematical
Engineering, Gรผmรผลhane, Turkey.
E-mail: [email protected] (corresponding author)
( Received:15.09.2020, Accepted: 01.11.2020, Published Online: 05.11.2020)
Abstract In this study, the multiplicative Volterra integral equation is defined by using the
concept of multiplicative integral. The solution of multiplicative Volterra integral
equation of the second kind is researched by using the successive approximations
method with respect to the multiplicative calculus and the necessary conditions for the
continuity and uniqueness of the solution are given. The main purpose of this study is
to investigate the relationship of the multiplicative integral equations with the
multiplicative differential equations. Keywords: Multiplicative calculus; Multiplicative differential equations;
Multiplicative Volterra integral equations; Successive approximations method.
1. Introduction
Grossman and Katz [10] have built non-Newtonian calculus between years 1967-1970 as an
alternative to classic calculus. They have set an infinite family of calculus, including classic, geometric,
harmonic, quadratic, bigeometric, biharmonic and biquadratic calculus. Also, they defined a new kind
of derivative and integral by using multiplication and division operations instead of addition and
subtraction operations. Later, the new calculus that establish in this way is named multiplicative calculus
by Stanley [16]. Multiplicative calculus provide different point of view for applications in science and
engineering. It is discussed and developed by many researchers. Stanley [16] developed multiplicative
calculus, gave some basic theorems about derivatives, integrals and proved infinite products in this
calculus. Aniszewska [1] used the multiplicative version of Runge-Kutta method for solving
multiplicative differential equations. Bashirov, Mฤฑsฤฑrlฤฑ and รzyapฤฑcฤฑ [2] demonstrated some applications
and usefulness of multiplicative calculus for the attention of researchers in the branch of analysis. Rฤฑza,
รzyapฤฑcฤฑ and Mฤฑsฤฑrlฤฑ [14] studied the finite difference methods for the numerical solutions of
multiplicative differential equations and Volterra integral equations. Mฤฑsฤฑrlฤฑ and Gurefe [13] developed
multiplicative Adams Bashforth-Moulton methods to obtain the numerical solution of multiplicative
differential equations. Bashirov, Rฤฑza [4] discussed multiplicative differentiation for complex valued
functions and Bashirov, Norozpour [6] extended the multiplicative integral to complex valued functions.
Bashirov [5] studied double integrals in the sense of multiplicative calculus. Bhat et al. [7] defined
multiplicative Fourier transform and found the solution of multiplicative differential equations by
applying multiplicative Fourier transform. Bhat et al. [8] defined multiplicative Sumudu transform and
solved some multiplicative differential equations by using multiplicative Sumudu transform. For more
details see in [1-10, 13, 14, 16-20].
10
Ikonion Journal of Mathematics 2020, 2 (2)
Integral equations have used for the solution of many problems in applied mathematics,
mathematical physics and engineering since the 18th century. The integral equations have begun to enter
the problems of engineering and other fields because of the relationship with differential equations and
so their importance has increased in recent years. The reader may refer for relevant terminology on the
integral equations to [11, 12, 15, 21, 22].
Now, we will give some necessary definitions and theorems in multiplicative calculus as follows:
Definition 1. Let ๐ be a function whose domain is โ the set of real numbers and whose range is a
subset of โ. The multiplicative derivative of the ๐ at ๐ฅ is defined as the limit
๐โ๐(๐ฅ)
๐๐ฅ= ๐โ(๐ฅ) = lim
โโ0(๐(๐ฅ+โ)
๐(๐ฅ))
1
โ.
The limit is also called โ-derivative of ๐ at ๐ฅ, briefly.
If ๐ is a positive function on an open set ๐ด โ โ and its classical derivative ๐โฒ(๐ฅ) exists, then its
multiplicative derivative also exists and
๐โ(๐ฅ) = ๐[๐โฒ(๐ฅ)
๐(๐ฅ)]= ๐(๐๐โ๐)
โฒ(๐ฅ)
where ๐๐ โ ๐(๐ฅ) = ๐๐๐(๐ฅ). Moreover, if ๐ is multiplicative differentiable and ๐โ(๐ฅ) โ 0, then its
classical derivative exists and
๐โฒ(๐ฅ) = ๐(๐ฅ) โ ๐๐๐โ(๐ฅ) [16].
The multiplicative derivative of ๐โ is called the second multiplicative derivative and it is denoted
by ๐โโ. Likewise, the ๐-th multiplicative derivative can be defined of ๐ and denoted by ๐โ(๐) for ๐ =
0,1,2, ... . If ๐-th derivative ๐(๐)(๐ฅ) exists, then its ๐-th multiplicative derivative ๐โ(๐)(๐ฅ) also exists
and
๐โ(๐)(๐ฅ) = ๐(๐๐โ๐)(๐)(๐ฅ), ๐ = 0,1,2,โฆ [2].
Definition 2. The multiplicative absolute value of ๐ฅ โ โ denoted with the symbol |๐ฅ|โ and defined
by
|๐ฅ|โ = ๐ฅ, ๐ฅ โฅ 1 1
๐ฅ, ๐ฅ < 1.
Theorem 1. Let ๐ and ๐ be multiplicative differentiable functions. Then the functions ๐. ๐, ๐. ๐, ๐ +
๐,๐๐โ , ๐๐ are multiplicative differentiable where ๐ is an arbitrary constant and their multiplicative
derivative can be shown as
(1) (๐๐)โ(๐ฅ) = ๐โ(๐ฅ)
(2) (๐๐)โ(๐ฅ) = ๐โ(๐ฅ)๐โ(๐ฅ)
(3) (๐ + ๐)โ(๐ฅ) = ๐โ(๐ฅ)๐(๐ฅ)
๐(๐ฅ)+๐(๐ฅ)๐โ(๐ฅ)๐(๐ฅ)
๐(๐ฅ)+๐(๐ฅ)
(4) (๐
๐)โ(๐ฅ) =
๐โ(๐ฅ)
๐โ(๐ฅ)
(5) (๐๐)โ(๐ฅ) = ๐โ(๐ฅ)๐(๐ฅ)๐(๐ฅ)๐โฒ(๐ฅ)
(6) [๐โ(๐ฅ)]๐ = [๐๐(๐ฅ)]โ for ๐ โ โ [16].
11
Ikonion Journal of Mathematics 2020, 2 (2)
Theorem 2. (Multiplicative Mean Value Theorem) If the function ๐ is continuous on [๐, ๐] and is
*-differentiable on (๐, ๐), then there exits ๐ < ๐ < ๐ such that
๐โ(๐) = (๐(๐)
๐(๐))
1
๐โ๐ [3].
Definition 3. Let ๐ be a function with two variables, then its multiplicative partial derivatives are
defined as
๐โ๐(๐ฅ,๐ฆ)
๐๐ฅ= ๐๐ฅ
โ(๐ฅ, ๐ฆ) = ๐๐
๐๐ฅ๐๐(๐(๐ฅ,๐ฆ))
and ๐โ๐(๐ฅ,๐ฆ)
๐๐ฆ= ๐๐ฆ
โ(๐ฅ, ๐ฆ) = ๐๐
๐๐ฆ๐๐(๐(๐ฅ,๐ฆ))
[5].
Theorem 3. (Multiplicative Chain Rule) Suppose that ๐ be a function of two variables ๐ฆ and ๐ง with
continuous multiplicative partial derivatives. If ๐ฆ and ๐ง are differentiable functions on (๐, ๐) such that
๐(๐ฆ(๐ฅ), ๐ง(๐ฅ)) is defined for every ๐ฅ โ (๐, ๐), then
๐โ๐(๐ฆ(๐ฅ), ๐ง(๐ฅ))
๐๐ฅ= ๐๐ฆ
โ(๐ฆ(๐ฅ), ๐ง(๐ฅ))๐ฆโฒ(๐ฅ)
๐๐งโ(๐ฆ(๐ฅ), ๐ง(๐ฅ))
๐งโฒ(๐ฅ) [2].
Definition 4. Let ๐ be a positive function and continuous on the interval [๐, ๐], then it is
multiplicative integrable or briefly โ-integrable on [๐, ๐] and
โ โซ ๐(๐ฅ)๐๐ฅ๐
๐
= ๐โซ ln(๐(๐ฅ))๐๐ฅ๐
๐ [16].
Theorem 4. If ๐ and ๐ are integrable functions on [๐, ๐] in the sense of multiplicative, then
(1) โ โซ (๐(๐ฅ)๐)๐๐ฅ๐
๐= (โ โซ (๐(๐ฅ))
๐๐ฅ๐
๐)๐
(2) โ โซ (๐(๐ฅ)๐(๐ฅ))๐๐ฅ๐
๐=โ โซ (๐(๐ฅ))
๐๐ฅ๐
๐โ โซ (๐(๐ฅ))
๐๐ฅ๐
๐
(3) โ โซ (๐(๐ฅ)
๐(๐ฅ))๐๐ฅ๐
๐=โโซ (๐(๐ฅ))
๐๐ฅ๐
๐
โโซ (๐(๐ฅ))๐๐ฅ๐
๐
(4) โ โซ (๐(๐ฅ))๐๐ฅ๐
๐=โ โซ (๐(๐ฅ))
๐๐ฅ๐
๐โ โซ (๐(๐ฅ))
๐๐ฅ๐
๐
where ๐ โ โ and ๐ โค ๐ โค ๐ [2,3].
Theorem 5. (Fundamental Theorem of Multiplicative Calculus) If the function ๐ has multiplicative
derivative on [๐, ๐] and ๐โ is multiplicative integrable on [๐, ๐] , then
โ โซ๐โ(๐ฅ)๐๐ฅ๐
๐
=๐(๐)
๐(๐) [2,3].
Definition 5. The equation of the form
๐ฆโ(๐ฅ) = ๐(๐ฅ, ๐ฆ(๐ฅ))
including the multiplicative derivative of ๐ฆ is called first order multiplicative differential equation. It is
equivalent to the ordinary differential equation ๐ฆโฒ(๐ฅ) = ๐ฆ(๐ฅ) ln ๐(๐ฅ, ๐ฆ(๐ฅ)). Similarly, ๐-th order
multiplicative differential equation is defined by ๐น (๐ฅ, ๐ฆ, ๐ฆโ, โฆ , ๐ฆโ(๐โ1), ๐ฆโ(๐)(๐ฅ)) = 1, (๐ฅ, ๐ฆ) โ โ ร
โ+ [2,3]. The equation of the form
12
Ikonion Journal of Mathematics 2020, 2 (2)
(๐ฆโ(๐))๐๐(๐ฅ)
(๐ฆโ(๐โ1))๐๐โ1(๐ฅ)
โฆ(๐ฆโโ)๐2(๐ฅ)(๐ฆโ)๐1(๐ฅ)๐ฆ๐0(๐ฅ) = ๐(๐ฅ)
that ๐ is a positive function, is called multiplicative linear differential equation. If the exponentials ๐๐(๐ฅ)
are constants, then the equation called as multiplicative linear differential equation with constant
exponentials; if not it is called as multiplicative linear differential equation with variable exponentials
[17].
2. Multiplicative Volterra Integral Equations
An equation in which an unknown function appears under one or more signs of multiplicative
integration is called a multiplicative integral equation (MIE), if the multiplicative integral exists. The
equation
๐ข(๐ฅ) = ๐(๐ฅ) โ โซ[๐ข(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก
๐ฅ
๐
where ๐(๐ฅ) and ๐พ(๐ฅ, ๐ก) are known functions, ๐ข(๐ฅ) is unknown function, is called linear multiplicative
Volterra integral equation (LMVIE) of the second kind. The function ๐พ(๐ฅ, ๐ก) is the kernel of
multiplicative Volterra integral equation. If ๐(๐ฅ) = 1, then the equation takes the form
๐ข(๐ฅ) = โ โซ[๐ข(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก
๐ฅ
๐
and it is called LMVIE of the first kind.
Example 1. Show that the function ๐ข(๐ฅ) = ๐2๐ฅ is a solution of the MVIE ๐ข(๐ฅ) = ๐๐ฅ โ โซ [(๐ข(๐ก))1
๐ฅ]๐๐ก
๐ฅ
0.
Solution. Substituting the function ๐2๐ฅ in place of ๐ข(๐ฅ) into the right side of the equation, we obtain
๐๐ฅ โ โซ [(๐ข(๐ก))1
๐ฅ]๐๐ก
๐ฅ
0= ๐๐ฅ โ โซ [(๐2๐ก)
1
๐ฅ]๐๐ก
๐ฅ
0 = ๐๐ฅ ๐โซ ln ๐
2๐ก๐ฅ ๐๐ก
๐ฅ
0 = ๐๐ฅ ๐โซ2๐ก
๐ฅ๐๐ก
๐ฅ
0 = ๐๐ฅ ๐2
๐ฅโ๐ก2
2|0
๐ฅ
= ๐2๐ฅ = ๐ข(๐ฅ)
So, this means that the function ๐ข(๐ฅ) = ๐2๐ฅ is a solution of the MVIE.
2.1. The Successive Approximation Method For Solving Multiplicative Volterra Integral
Equations
Theorem 6. Consider LMVIE of the second kind as
๐ข(๐ฅ) = ๐(๐ฅ) โ โซ [๐ข(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก๐ฅ
0. (1)
If ๐(๐ฅ) is positive and continuous on [0, ๐] and ๐พ(๐ฅ, ๐ก) is continuous on the rectangle 0 โค ๐ก โค ๐ฅ and
0 โค ๐ฅ โค ๐, then there exists an unique continuous solution of (1) as
๐ข(๐ฅ) = โ ๐๐(๐ฅ)โ๐=0 = ๐โ ๐๐๐๐(๐ฅ)
โ๐=0
such that the series โ ๐๐๐๐(๐ฅ)โ๐=0 is absolute and uniform convergent where
๐0(๐ฅ) = ๐(๐ฅ) , ๐๐(๐ฅ) =โ โซ [ ๐๐โ1(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก๐ฅ
0, ๐ = 1,2, โฆ .
Proof: Take the initial approximation as
๐ข0(๐ฅ) = ๐(๐ฅ) = ๐0(๐ฅ) . (2)
13
Ikonion Journal of Mathematics 2020, 2 (2)
If we write ๐ข0(๐ฅ) instead of ๐ข(๐ฅ) in equation (1), then we get the new function showed with ๐ข1(๐ฅ) as
๐ข1(๐ฅ) = ๐(๐ฅ) โ โซ[๐ข0(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก.
๐ฅ
0
(3)
Since the multiplicative integral which is in equation (3) depends on variable ๐ฅ, we can show it with
๐1(๐ฅ) =โ โซ[๐ข0(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก
๐ฅ
0
=โโซ[๐0(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก
๐ฅ
0
and write the equation (3) as follow
๐ข1(๐ฅ) = ๐(๐ฅ)๐1(๐ฅ) = ๐0(๐ฅ)๐1(๐ฅ) (4)
by using (2). Therefore the third approximation is obtained as
๐ข2(๐ฅ) = ๐(๐ฅ) โ โซ[๐ข1(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก.
๐ฅ
0
By the equation (4), we find
๐ข2(๐ฅ) = ๐(๐ฅ) โ โซ[๐0(๐ก). ๐1(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก
๐ฅ
0
= ๐(๐ฅ) โ โซ([๐0(๐ก)]๐พ(๐ฅ,๐ก)[๐1(๐ก)]
๐พ(๐ฅ,๐ก))๐๐ก
๐ฅ
0
= ๐(๐ฅ) โ โซ[๐0(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก
๐ฅ
0
โ โซ[๐1(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก
๐ฅ
0
.
If we set ๐2(๐ฅ) =โ โซ [๐1(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก๐ฅ
0, then ๐ข2(๐ฅ) = ๐0(๐ฅ) ๐1(๐ฅ) ๐2(๐ฅ). In a similar way, we get
๐ข๐(๐ฅ) = ๐0(๐ฅ) ๐1(๐ฅ) ๐2(๐ฅ) โฆ ๐๐(๐ฅ) (5)
where
๐0(๐ฅ) = ๐(๐ฅ) , ๐๐(๐ฅ) =โ โซ [ ๐๐โ1(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก๐ฅ
0, ๐ = 1,2, โฆ .
Continuing this process, we get the series
๐ข(๐ฅ) = ๐0(๐ฅ) ๐1(๐ฅ) ๐2(๐ฅ) โฆ ๐๐(๐ฅ). โฆ =โ๐๐(๐ฅ)
โ
๐=0
= ๐โ ๐๐๐๐(๐ฅ)โ๐=0 . (6)
From (5) and (6), it is clear that lim๐โโ
๐ข๐ (๐ฅ) = ๐ข(๐ฅ).
Assume that ๐น = ๐๐๐ฅ๐ฅโ[0,๐]
๐(๐ฅ) and ๐พ = ๐๐๐ฅ0โค๐กโค๐ฅโค๐
|๐พ(๐ฅ, ๐ก)|. Then we find
๐0(๐ฅ) = ๐(๐ฅ) โค ๐น
๐1(๐ฅ) =โ โซ[๐0(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก
๐ฅ
0
= ๐โซ ๐พ(๐ฅ,๐ก)๐๐๐0(๐ก)๐๐ก๐ฅ
0 โค ๐โซ |๐พ(๐ฅ,๐ก)| |๐๐๐0(๐ก)|๐๐ก๐ฅ
0 = ๐โซ |๐พ(๐ฅ,๐ก)| ๐๐|๐0(๐ก)|โ๐๐ก๐ฅ
0
โค ๐๐พ.๐๐๐น.๐ฅ
๐2(๐ฅ) =โ โซ[๐1(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก
๐ฅ
0
= ๐โซ ๐พ(๐ฅ,๐ก)๐๐๐1(๐ก)๐๐ก๐ฅ
0 โค ๐๐พ.๐๐๐น โซ |๐พ(๐ฅ,๐ก)|.๐ก๐๐ก๐ฅ
0 โค ๐๐พ2.๐๐๐น โซ ๐ก๐๐ก
๐ฅ
0 = ๐๐๐๐น.๐พ2.๐ฅ2
2!
โฎ
๐๐(๐ฅ) โค ๐๐๐๐น.
๐พ๐๐ฅ๐
๐! .
14
Ikonion Journal of Mathematics 2020, 2 (2)
Also, we can see that |๐๐๐๐(๐ฅ)| = ๐๐|๐๐(๐ฅ)|โ โค ๐๐๐น ๐พ๐๐๐
๐! for = 1,2,โฆ . Because of |๐๐๐๐(๐ฅ)| โค
๐๐๐น ๐พ๐๐๐
๐! , the series โ ๐๐๐๐(๐ฅ)
โ๐=0 is absolute and uniform convergence from the Weierstrass ๐-test.
Since each terms of this series are continuous, the function which this series convergences uniformly is
continuous. Hence ๐ข(๐ฅ) is continuous function. Now, we will show ๐ข(๐ฅ) is a solution of the equation
(1). Since ๐๐(๐ฅ) =โ โซ ๐๐โ1(๐ก)๐พ(๐ฅ,๐ก)๐๐ก๐ฅ
0 and ๐0(๐ฅ) = ๐(๐ฅ), we find
๐0(๐ฅ)โ๐๐(๐ฅ)
๐
๐=1
= ๐(๐ฅ)โ(โโซ๐๐โ1(๐ก)๐พ(๐ฅ,๐ก)๐๐ก
๐ฅ
0
)
๐
๐=1
โ๐๐(๐ฅ)
๐
๐=0
= ๐(๐ฅ) โ โซ(โ๐๐โ1(๐ก)
๐โ1
๐=1
)
๐พ(๐ฅ,๐ก)๐๐ก๐ฅ
0
โ๐๐(๐ฅ)
๐
๐=0
= ๐(๐ฅ) โ โซ(โ๐๐(๐ก)
๐
๐=0
)
๐พ(๐ฅ,๐ก)๐๐ก๐ฅ
0
(7)
From (6) and (7), we obtain
๐ข(๐ฅ) = lim๐โโ
โ๐๐(๐ฅ)
๐
๐=0
= lim๐โโ
๐(๐ฅ) ๐โซ ๐พ(๐ฅ,๐ก)โ ๐๐๐๐(๐ก)๐๐=0 ๐๐ก
๐ฅ
0 = ๐(๐ฅ) ๐โซ ๐พ(๐ฅ,๐ก) lim
๐โโ(โ ๐๐๐๐(๐ก)๐๐=0 )๐๐ก
๐ฅ
0
= ๐(๐ฅ) ๐โซ ๐พ(๐ฅ,๐ก)๐๐๐โ ๐๐๐๐(๐ก)โ๐=0 ๐๐ก
๐ฅ
0 = ๐(๐ฅ) ๐โซ ๐พ(๐ฅ,๐ก)๐๐๐ข(๐ก)๐๐ก๐ฅ
0 = ๐(๐ฅ) โ โซ๐ข(๐ก)๐พ(๐ฅ,๐ก)๐๐ก
๐ฅ
0
.
by using the uniform convergence of the series โ ๐๐๐๐(๐ฅ)โ๐=0 . This indicates that ๐ข(๐ฅ) is the solution
of the equation (1). Now, we will show the uniqueness of the solution. Assume that ๐ข(๐ฅ) and ๐ฃ(๐ฅ) are
different solutions of the equation (1). Since
๐ข(๐ฅ) = ๐(๐ฅ) โ โซ[๐ข(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก
๐ฅ
0
๐ฃ(๐ฅ) = ๐(๐ฅ) โ โซ[๐ฃ(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก
๐ฅ
0
we find
๐ข(๐ฅ)
๐ฃ(๐ฅ)= โ โซ [
๐ข(๐ก)
๐ฃ(๐ก)]
๐พ(๐ฅ,๐ก)๐๐ก
=
๐ฅ
0
๐โซ ๐พ(๐ฅ,๐ก)(๐๐๐ข(๐ก)โ๐๐๐ฃ(๐ก))๐๐ก๐ฅ
0 .
If we set ๐ข(๐ฅ)
๐ฃ(๐ฅ)= ๐(๐ฅ), we can write ๐(๐ฅ) = โ โซ [๐(๐ก)]๐พ(๐ฅ,๐ก)
๐๐ก=
๐ฅ
0๐โซ ๐พ(๐ฅ,๐ก)๐๐๐(๐ก)๐๐ก
๐ฅ
0 . Because of
๐๐๐(๐ฅ) = โซ ๐พ(๐ฅ, ๐ก)๐๐๐(๐ก)๐๐ก๐ฅ
0, we find
|๐๐๐(๐ฅ)| = |โซ ๐พ(๐ฅ, ๐ก)๐๐๐(๐ก)๐๐ก๐ฅ
0
| โค โซ |๐พ(๐ฅ, ๐ก)||๐๐๐(๐ก)|๐๐ก๐ฅ
0
โค ๐พโซ |๐๐๐(๐ก)|๐๐ก๐ฅ
0
.
It is taken as โ(๐ฅ) = โซ |๐๐๐(๐ก)|๐๐ก๐ฅ
0, then we write
|๐๐๐(๐ฅ)| โค ๐พโ(๐ฅ)
|๐๐๐(๐ฅ)| โ ๐พโ(๐ฅ) โค 0.
By multiplication with ๐โ๐พ๐ฅ both sides of the inequality, then
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Ikonion Journal of Mathematics 2020, 2 (2)
๐โ๐พ๐ฅ|๐๐๐(๐ฅ)| โ ๐โ๐พ๐ฅ๐พโ(๐ฅ) โค 0
๐
๐๐ฅ(๐โ๐พ๐ฅโ(๐ฅ)) โค 0
and by integration both sides of this inequality from 0 to ๐ฅ we find
๐โ๐พ๐ฅโ(๐ฅ) โ ๐โ๐พ0โ(0) โค 0
๐โ๐พ๐ฅโ(๐ฅ) โค 0.
Since โ(๐ฅ) โค 0 and โ(๐ฅ) โฅ 0, we find โ(๐ฅ) = 0. Therefore |๐๐๐(๐ฅ)| = 0 for every ๐ฅ โ [0, ๐], i.e.,
๐(๐ฅ) = 1 for every ๐ฅ โ [0, ๐]. Thus ๐(๐ฅ) =๐ข(๐ฅ)
๐ฃ(๐ฅ)= 1 and we obtain ๐ข(๐ฅ) = ๐ฃ(๐ฅ). This completes the
proof.
Remark 1. If the following iterations of method of successive approximations are set by
๐ข0(๐ฅ) = ๐(๐ฅ)
๐ข๐(๐ฅ) = ๐(๐ฅ) โ โซ[๐ข๐โ1(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก, ๐ = 1,2,3,โฆ
๐ฅ
0
for the multiplicative integral equation
๐ข(๐ฅ) = ๐(๐ฅ) โ โซ [๐ข(๐ก)]๐พ(๐ฅ,๐ก)๐๐ก๐ฅ
0
where ๐(๐ฅ) is positive and continuous on [0, ๐] and ๐พ(๐ฅ, ๐ก) is continuous for 0 โค ๐ฅ โค ๐, 0 โค ๐ก โค ๐ฅ,
then the sequence of successive approximations ๐ข๐(๐ฅ) converges to the solution ๐ข(๐ฅ).
Example 2. Solve the multiplicative Volterra integral equation
๐ข(๐ฅ) = ๐๐ฅ โ โซ [(๐ข(๐ก))(๐กโ๐ฅ)
]๐๐ก
๐ฅ
0
with using the successive approximations method.
Solution. Let taken ๐ข0(๐ฅ) = ๐๐ฅ, then the first approximation is obtained as
๐ข1(๐ฅ) = ๐๐ฅ โ โซ [(๐๐ก)(๐กโ๐ฅ)]
๐๐ก๐ฅ
0= ๐๐ฅ๐โซ ๐๐(๐๐ก.(๐กโ๐ฅ))๐๐ก
๐ฅ
0 = ๐๐ฅ๐โซ ๐ก.(๐กโ๐ฅ)๐๐ก๐ฅ
0 = ๐(๐ฅโ
๐ฅ3
3!)
and by using this approximation it can be obtained as
๐ข2(๐ฅ) = ๐๐ฅ โ โซ [(๐
(๐กโ๐ก3
3!))
(๐กโ๐ฅ)
]
๐๐ก
๐ฅ
0= ๐
(๐ฅโ๐ฅ3
3!+๐ฅ5
5!).
By proceeding similarly, the ๐๐กโ approximation is
๐ข๐(๐ฅ) = ๐(๐ฅโ
๐ฅ3
3!+๐ฅ5
5!โโฏ+(โ1)๐
๐ฅ2๐+1
(2๐+1)!).
Since the expression ๐ฅ โ๐ฅ3
3!+๐ฅ5
5!โโฏ+ (โ1)๐
๐ฅ2๐+1
(2๐+1)!+โฏ is the Maclaurin series of ๐ ๐๐๐ฅ,
lim๐โโ
๐ข๐(๐ฅ) = ๐๐ ๐๐๐ฅ. Therefore the solution of the equation ๐ข(๐ฅ) = ๐๐ ๐๐๐ฅ.
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Ikonion Journal of Mathematics 2020, 2 (2)
3. The Relationship Between Multiplicative Differential Equations
We will investigate the relationship of the multiplicative Volterra integral equations with the
multiplicative differential equations.
3.1. The Conversion of the Multiplicative Volterra Integral Equations to Multiplicative
Differential Equations
In this section, we demonstrate the method of converting a multiplicative Volterra integral
equation into a multiplicative differential equation. For this, we need the Leibniz Formula in the sense
of multiplicative calculus.
Firstly, we will give necessary lemma with using proof of multiplicative Leibniz formula.
Lemma 1. Let ฮฉ be an open set in โ2. Suppose that ๐:ฮฉ โ โ be a function such that the
multiplicative partial derivatives ๐๐ฅ๐ฆโโ(๐ฅ, ๐ฆ) , ๐๐ฆ๐ฅ
โโ(๐ฅ, ๐ฆ) exists in ฮฉ and are continuous, then we have
๐โ
๐๐ฅ(๐โ
๐๐ฆ๐(๐ฅ, ๐ฆ)) =
๐โ
๐๐ฆ(๐โ
๐๐ฅ๐(๐ฅ, ๐ฆ)).
Proof. Fix ๐ฅ and ๐ฆ. ๐น(โ, ๐) is taken as
๐น(โ, ๐) = (๐(๐ฅ + โ, ๐ฆ + ๐) ๐(๐ฅ, ๐ฆ)
๐(๐ฅ, ๐ฆ + ๐) ๐(๐ฅ + โ, ๐ฆ))
1โ๐
By using the multiplicative mean value theorem, we find
๐น(โ, ๐) = (๐(๐ฅ + โ, ๐ฆ + ๐) ๐(๐ฅ, ๐ฆ)
๐(๐ฅ, ๐ฆ + ๐) ๐(๐ฅ + โ, ๐ฆ))
1โ๐
=
(
(
๐(๐ฅ + โ, ๐ฆ + ๐)๐(๐ฅ, ๐ฆ + ๐)
๐(๐ฅ + โ, ๐ฆ)๐(๐ฅ, ๐ฆ)
)
1๐
)
1โ
= (๐โ
๐๐ฆ(๐(๐ฅ + โ, ๐ฆ+๐1๐)
๐(๐ฅ, ๐ฆ+๐1๐)))
1โ
=๐โ
๐๐ฆ((๐(๐ฅ + โ, ๐ฆ+๐1๐)
๐(๐ฅ, ๐ฆ+๐1๐))
1โ
) =๐โ
๐๐ฆ(๐โ
๐๐ฅ๐(๐ฅ+๐2โ, ๐ฆ+๐1๐))
and
๐น(โ, ๐) = (๐(๐ฅ + โ, ๐ฆ + ๐) ๐(๐ฅ, ๐ฆ)
๐(๐ฅ, ๐ฆ + ๐) ๐(๐ฅ + โ, ๐ฆ))
1โ๐
=
(
(
๐(๐ฅ + โ, ๐ฆ + ๐)๐(๐ฅ + โ, ๐ฆ)
๐(๐ฅ, ๐ฆ + ๐)๐(๐ฅ, ๐ฆ)
)
1โ
)
1๐
= (๐โ
๐๐ฅ(๐(๐ฅ+๐3โ, ๐ฆ + ๐)
๐(๐ฅ+๐3โ, ๐ฆ)))
1๐
=๐โ
๐๐ฅ((๐(๐ฅ+๐3โ, ๐ฆ + ๐)
๐(๐ฅ+๐3โ, ๐ฆ))
1๐
) =๐โ
๐๐ฆ(๐โ
๐๐ฆ๐(๐ฅ+๐3โ, ๐ฆ+๐4๐))
for some 0 < ๐1, ๐2, ๐3, ๐4 < 1 which all of them depend on ๐ฅ, ๐ฆ, โ, ๐. Therefore,
๐โ
๐๐ฆ(๐โ
๐๐ฅ๐(๐ฅ+๐2โ, ๐ฆ+๐1๐)) =
๐โ
๐๐ฆ(๐โ
๐๐ฆ๐(๐ฅ+๐3โ, ๐ฆ+๐4๐))
for all โ and ๐. Taking the limit โ, ๐ โ 0 and using the assumed continuity of both partial derivatives,
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Ikonion Journal of Mathematics 2020, 2 (2)
it gives
๐โ
๐๐ฆ(๐โ
๐๐ฅ๐(๐ฅ, ๐ฆ)) =
๐โ
๐๐ฆ(๐โ
๐๐ฆ๐(๐ฅ, ๐ฆ)).
Theorem 7. (Multiplicative Leibniz Formula) Let ๐ด, ๐ผ โ โ be open set and ๐ be a continuous
function on ๐ด ร ๐ผ into โ. If ๐๐ฅโ exists and is continuous on ๐ด ร ๐ผ, โ(๐ฅ), ๐ฃ(๐ฅ) are continuously
differentiable functions of ๐ด into ๐ผ, then we have
๐โ
๐๐ฅ(โ โซ ๐(๐ฅ, ๐ก)๐๐ก
๐ฃ(๐ฅ)
โ(๐ฅ)
) = โ โซ ๐๐ฅโ(๐ฅ, ๐ก)๐๐ก
๐ฃ(๐ฅ)
โ(๐ฅ)
๐(๐ฅ, ๐ฃ(๐ฅ))
๐ฃโฒ(๐ฅ)
๐(๐ฅ, โ(๐ฅ))โโฒ(๐ฅ)
Proof. Let ๐(๐ฅ, ๐ก) =๐โ
๐๐ก๐น(๐ฅ, ๐ก) = ๐น๐ก
โ(๐ฅ, ๐ก). Hence we can write โ โซ ๐(๐ฅ, ๐ก)๐๐ก๐ฃ(๐ฅ)
โ(๐ฅ)=โ โซ ๐น๐ก
โ(๐ฅ, ๐ก)๐๐ก๐ฃ(๐ฅ)
โ(๐ฅ).
Since โ โซ ๐(๐ฅ, ๐ก)๐๐ก๐ฃ(๐ฅ)
โ(๐ฅ)=๐น(๐ฅ,๐ฃ(๐ฅ))
๐น(๐ฅ,โ(๐ฅ)) , we find
๐โ
๐๐ฅ(โ โซ ๐(๐ฅ, ๐ก)๐๐ก
๐ฃ(๐ฅ)
โ(๐ฅ)
) =๐โ
๐๐ฅ(๐น(๐ฅ, ๐ฃ(๐ฅ))
๐น(๐ฅ, โ(๐ฅ))) =
๐โ
๐๐ฅ๐น(๐ฅ, ๐ฃ(๐ฅ))
๐โ
๐๐ฅ๐น(๐ฅ, โ(๐ฅ))
by using properties of multiplicative derivative. Therefore we get
๐โ
๐๐ฅ(โ โซ ๐(๐ฅ, ๐ก)๐๐ก
๐ฃ(๐ฅ)
โ(๐ฅ)
) =๐น๐ฅโ(๐ฅ, ๐ฃ(๐ฅ))
1 [๐น๐ฃ(๐ฅ)
โ (๐ฅ, ๐ฃ(๐ฅ))]๐ฃโฒ(๐ฅ)
๐น๐ฅโ(๐ฅ, โ(๐ฅ))
1 [๐นโ(๐ฅ)
โ (๐ฅ, โ(๐ฅ))]โโฒ(๐ฅ)
(8)
with multiplicative chain rule. By using Lemma 1, we obtain
๐โ
๐๐ฅโ(โ โซ ๐(๐ฅ, ๐ก)๐๐ก
๐ฃ(๐ฅ)
โ(๐ฅ)
) =โ โซ (๐โ
๐๐ก๐น๐ฅโ(๐ฅ, ๐ก))
๐๐ก๐ฃ(๐ฅ)
โ(๐ฅ)
๐(๐ฅ, ๐ฃ(๐ฅ))๐ฃโฒ(๐ฅ)
๐(๐ฅ, โ(๐ฅ))โโฒ(๐ฅ)
=โ โซ (๐โ
๐๐ก(๐โ
๐๐ฅ๐น(๐ฅ, ๐ก)))
๐๐ก
๐(๐ฅ, ๐ฃ(๐ฅ))๐ฃโฒ(๐ฅ)
๐(๐ฅ, โ(๐ฅ))โโฒ(๐ฅ)
๐ฃ(๐ฅ)
โ(๐ฅ)
=โ โซ๐โ
๐๐ฅ(๐โ
๐๐ก๐น(๐ฅ, ๐ก))
๐๐ก๐ฃ(๐ฅ)
โ(๐ฅ)
๐(๐ฅ, ๐ฃ(๐ฅ))๐ฃโฒ(๐ฅ)
๐(๐ฅ, โ(๐ฅ))โโฒ(๐ฅ)
=โ โซ๐โ
๐๐ฅ(๐น๐กโ(๐ฅ, ๐ก))
๐๐ก
๐ฃ(๐ฅ)
โ(๐ฅ)
๐(๐ฅ, ๐ฃ(๐ฅ))
๐ฃโฒ(๐ฅ)
๐(๐ฅ, โ(๐ฅ))โโฒ(๐ฅ)
=โ โซ ๐๐ฅโ(๐ฅ, ๐ก)๐๐ก
โ(๐ฅ)
๐ฃ(๐ฅ)
๐(๐ฅ, ๐ฃ(๐ฅ))
๐ฃโฒ(๐ฅ)
๐(๐ฅ, โ(๐ฅ))โโฒ(๐ฅ)
.
from the equality (8). This completes the proof.
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Ikonion Journal of Mathematics 2020, 2 (2)
Example 3. Show that the multiplicative integral equation ๐ข(๐ฅ) = sin๐ฅ โ โซ ([๐ข(๐ก)]๐ฅ tan ๐ก)๐๐ก๐ฅ
0
can be transformed to a multiplicative differential equation.
Solution. If we consider the equation ๐ข(๐ฅ) = sin๐ฅ โ โซ ([๐ข(๐ก)]๐ฅtan๐ก)๐๐ก๐ฅ
0 and differentiate it by
using multiplicative Leibniz formula, we write
๐ขโ(๐ฅ) =๐โ
๐๐ฅ(sin๐ฅ)
๐โ
๐๐ฅ(โ โซ([๐ข(๐ก)]๐ฅ tan๐ก)๐๐ก
๐ฅ
0
)
= ๐๐๐๐ ๐ฅ๐ ๐๐ ๐ฅ โ โซ [
๐โ
๐๐ฅ([๐ข(๐ก)]๐ฅ tan ๐ก)]
๐๐ก๐ฅ
0
(๐ข(๐ฅ)๐ฅ tan๐ฅ)๐ฅโฒ
(๐ข(0)๐ฅ tan0)0โฒ
= ๐cot๐ฅ โ โซ[๐ข(๐ก)tan ๐ก]๐๐ก๐ฅ
0
๐ข(๐ฅ)๐ฅ tan๐ฅ
To take derivative is continued until the expression gets rid of the integral sign. Hence, we obtain
๐ขโโ(๐ฅ) =๐โ
๐๐ฅ(๐cot๐ฅ)
๐โ
๐๐ฅ(โ โซ[๐ข(๐ก)tan ๐ก]๐๐ก
๐ฅ
0
)๐โ
๐๐ฅ(๐ข(๐ฅ)๐ฅ tan๐ฅ)
= ๐โ๐๐๐ ๐๐2๐ฅ โ โซ[1]๐๐ก
๐ฅ
0
(๐ข(๐ฅ)tan๐ฅ)๐ฅโฒ
(๐ข(0)tan0)0โฒ ๐((๐ฅ.๐ก๐๐๐ฅ)โฒ ๐๐๐ข(๐ฅ)+
๐ขโฒ(๐ฅ)๐ข(๐ฅ)
๐ฅ ๐ก๐๐๐ฅ)
= ๐โ๐๐๐ ๐๐2๐ฅ [๐ข(๐ฅ)]tan๐ฅ ๐
(๐๐๐ข(๐ฅ)(๐ฅ ๐ก๐๐๐ฅ)โฒ+๐ขโฒ(๐ฅ)
๐ข(๐ฅ) ๐ฅ ๐ก๐๐๐ฅ)
= ๐โ๐๐๐ ๐๐2๐ฅ [๐ข(๐ฅ)](2tan๐ฅ+๐ฅ ๐ ๐๐
2๐ฅ) (๐๐ขโฒ(๐ฅ)
๐ข(๐ฅ) )
๐ฅ๐ก๐๐๐ฅ
= ๐โ๐๐๐ ๐๐2๐ฅ [๐ข(๐ฅ)](2tan๐ฅ+๐ฅ ๐ ๐๐
2๐ฅ) [๐ขโ(๐ฅ)]๐ฅ tan๐ฅ .
Thus the multiplicative integral equation is equivalent to the multiplicative differential equation
๐ขโโ(๐ฅ) = ๐โ๐๐๐ ๐๐2๐ฅ ๐ข(๐ฅ)(2tan๐ฅ+๐ฅ ๐ ๐๐
2๐ฅ). [๐ขโ(๐ฅ)]๐ฅ tan๐ฅ .
3.2. The Conversion of the Multiplicative Linear Differential Equations to Multiplicative
Integral Equations
In this section, we prove that the multiplicative linear differential equation with constant or
variable exponentials is converted to MVIE. We need to following theorem for converting ๐๐กโ order
multiplicative differential equation to MVIE.
Theorem 8. If ๐ is a positive integer and ๐ is a constant with ๐ฅ โฅ ๐, then we have
โ โซโฆ(๐)โฆ
๐ฅ
๐
โ โซ๐ข(๐ก)๐๐กโฆ๐๐ก๐ฅ
๐
=โ โซ [(๐ข(๐ก))(๐ฅโ๐ก)(๐โ1)
(๐โ1)! ]
๐๐ก๐ฅ
๐
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Ikonion Journal of Mathematics 2020, 2 (2)
Proof. Let
๐ช๐ =โ โซ[๐ข(๐ก)](๐ฅโ๐ก)(๐โ1)
๐๐ก๐ฅ
๐
. (9)
If it is taken ๐น(๐ฅ, ๐ก) = [๐ข(๐ก)](๐ฅโ๐ก)(๐โ1)
, we can write that
๐โ๐ช๐๐๐ฅ
=โ โซ๐น๐ฅโ(๐ฅ, ๐ก)๐๐ก
๐ฅ
๐
[๐น(๐ฅ, ๐ฅ)]1
[๐น(๐ฅ, ๐)]0
=โ โซ๐น๐ฅโ(๐ฅ, ๐ก)๐๐ก
๐ฅ
๐
by using the multiplicative Leibniz formula to equation (9). Then we find
๐โ๐ช๐๐๐ฅ
=โ โซ (๐๐๐๐ฅ๐๐๐น(๐ฅ,๐ก)
)๐๐ก
๐ฅ
๐
=โ โซ (๐(๐โ1)(๐ฅโ๐ก)(๐โ2) ๐๐๐ข(๐ก))
๐๐ก๐ฅ
๐
=โ โซ (๐๐๐([๐ข(๐ก)](๐โ1)(๐ฅโ๐ก)
(๐โ2)))๐๐ก
๐ฅ
๐
=โ โซ ([๐ข(๐ก)](๐โ1)(๐ฅโ๐ก)(๐โ2)
)๐๐ก.
๐ฅ
๐
Hence we get
๐โ๐ช๐๐๐ฅ
= (โ โซ([๐ข(๐ก)](๐ฅโ๐ก)(๐โ2)
)๐๐ก
๐ฅ
๐
)
(๐โ1)
= (๐ช๐โ1)(๐โ1) (10)
where ๐ > 1 . Since ๐ช1(๐ฅ) =โ โซ ๐ข(๐ก)๐๐ก๐ฅ
๐ for ๐ = 1, then we can write
๐โ๐ช1๐๐ฅ
=๐โ
๐๐ฅ(โ โซ(๐ข(๐ก))
๐๐ก
๐ฅ
๐
) = ๐ข(๐ฅ). (11)
If it is taken multiplicative derivative of the equation (10) by using multiplicative Leibniz formula, then
๐โโ๐ช๐
๐๐ฅ(2)=๐โ
๐๐ฅ(โ โซ([๐ข(๐ก)](๐ฅโ๐ก)
(๐โ2))๐๐ก
๐ฅ
๐
)
(๐โ1)
= (๐โ
๐๐ฅ(โ โซ ([๐ข(๐ก)](๐ฅโ๐ก)
(๐โ2))๐๐ก
๐ฅ
๐
))
(๐โ1)
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Ikonion Journal of Mathematics 2020, 2 (2)
= (โ โซ [๐โ
๐๐ฅ ([๐ข(๐ก)](๐ฅโ๐ก)
(๐โ2))]๐๐ก [[๐ข(๐ฅ)](๐ฅโ๐ฅ)
(๐โ2)]1
[[๐ข(๐)](๐ฅโ๐)๐โ2]0
๐ฅ
๐
)
(๐โ1)
= (โ โซ(๐โ
๐๐ฅ ([๐ข(๐ก)](๐ฅโ๐ก)
(๐โ2)))
๐๐ก
๐ฅ
๐
)
(๐โ1)
= (โ โซ(๐
๐๐๐ฅ (๐๐([๐ข(๐ก)](๐ฅโ๐ก)
(๐โ2))))
๐๐ก
๐ฅ
๐
)
(๐โ1)
= (โ โซ [๐(๐๐((๐ข(๐ก))
(๐โ2)(๐ฅโ๐ก)(๐โ3)
))]
๐๐ก๐ฅ
๐
)
(๐โ1)
= (โ โซ [(๐ข(๐ก))(๐โ2)(๐ฅโ๐ก)(๐โ3)
]๐๐ก
๐ฅ
๐
)
(๐โ1)
= (โ โซ [(๐ข(๐ก))(๐ฅโ๐ก)(๐โ3)
]๐๐ก
๐ฅ
๐
)
(๐โ1).(๐โ2)
= (๐ช๐โ2)(๐โ1)(๐โ2) .
By proceeding similarly, we obtain
๐โ(๐โ1)๐ช๐
๐๐ฅ(๐โ1)= (๐ช1)
(๐โ1)!
Hence, we write
๐โ(๐)๐ช๐
๐๐ฅ(๐)= (๐โ๐ช1๐๐ฅ)
(๐โ1)!
= [๐ข(๐ฅ)](๐โ1)!
from the equation (11). Now, we will take multiplicative integral by considering the above relations.
From the equation (11), ๐ช1(๐ฅ) =โ โซ ๐ข(๐ก)๐๐ก๐ฅ
๐ . Also, we have
๐ช2(๐ฅ) =โ โซ ๐ช1(๐ฅ2)๐๐ฅ2
๐ฅ
๐
=โ โซโ โซ ๐ข(๐ฅ1)๐๐ฅ1๐๐ฅ2
๐ฅ2
๐
๐ฅ
๐
where ๐ฅ1 and ๐ฅ2 are parameters. By proceeding similarly, we obtain
๐ช๐(๐ฅ) = (โ โซโ โซ โฆโ โซ โ โซ ๐ข(๐ฅ1)๐๐ฅ1๐๐ฅ2โฆ๐๐ฅ๐
๐ฅ2
๐
๐ฅ3
๐
๐ฅ๐
๐
๐ฅ
๐
)
(๐โ1)!
where ๐ฅ1, ๐ฅ2, โฆ , ๐ฅ๐ are parameters. If we write the equation (9) instead of the statement ๐ช๐, then it is
find
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Ikonion Journal of Mathematics 2020, 2 (2)
โ โซ [(๐ข(๐ก))(๐ฅโ๐ก)๐โ1
]๐๐ก
๐ฅ
๐
= (โ โซโ โซ โฆโ โซ โ โซ ๐ข(๐ฅ1)๐๐ฅ1๐๐ฅ2โฆ๐๐ฅ๐
๐ฅ2
๐
๐ฅ3
๐
๐ฅ๐
๐
๐ฅ
๐
)
(๐โ1)!
Hence we can write
(โ โซ [(๐ข(๐ก))(๐ฅโ๐ก)๐โ1
]๐๐ก
๐ฅ
๐
)
1(๐โ1)!
=โ โซโ โซ โฆโ โซ โ โซ ๐ข(๐ฅ1)๐๐ฅ1๐๐ฅ2โฆ๐๐ฅ๐
๐ฅ2
๐
๐ฅ3
๐
๐ฅ๐
๐
๐ฅ
๐
If it is taken = ๐ฅ1 = ๐ฅ2 = โฏ = ๐ฅ๐ , therefore we obtain
โ โซโฆ(๐)โฆ
๐ฅ
๐
โ โซ๐ข(๐ก)๐๐กโฆ๐๐ก๐ฅ
๐
=โ โซ [(๐ข(๐ก))(๐ฅโ๐ก)๐โ1
(๐โ1)! ]
๐๐ก
.
๐ฅ
๐
This completes the proof.
Let the ๐๐กโ- order multiplicative linear differential equation
๐โ(๐)๐ฆ
๐๐ฅ(๐) (๐โ(๐โ1)๐ฆ
๐๐ฅ(๐โ1))
๐1(๐ฅ)
(๐โ(๐โ2)๐ฆ
๐๐ฅ(๐โ2))
๐2(๐ฅ)
โฆ(๐โ๐ฆ
๐๐ฅ)๐๐โ1(๐ฅ)
(๐ฆ)๐๐(๐ฅ) = ๐(๐ฅ) (12)
that given the initial conditions
๐ฆ(0) = ๐0 , ๐ฆโ(0) = ๐1 , ๐ฆ
โ(๐โ1)(0) = ๐๐โ1 (13)
It can be transformed the multiplicative Volterra integral equation. Hence the solution of (12)-(13)
may be reduced to a solution of some multiplicative Volterra integral equation.
Take ๐โ(๐)๐ฆ
๐๐ฅ(๐)= ๐ข(๐ฅ). By integrating both sides of the equality
๐โ
๐๐ฅ(๐โ(๐โ1)๐ฆ
๐๐ฅ(๐โ1)) = ๐ข(๐ฅ), we write
โ โซ๐โ (๐โ(๐โ1)๐ฆ
๐๐ฅ(๐โ1))
๐ฅ
0
= โ โซ๐ข(๐ก)๐๐ก๐ฅ
0
๐ฆโ(๐โ1)(๐ฅ)
๐ฆโ(๐โ1)(0) = โ โซ๐ข(๐ก)๐๐ก
๐ฅ
0
๐ฆโ(๐โ1)(๐ฅ) = ๐๐โ1 โ โซ๐ข(๐ก)๐๐ก
๐ฅ
0
By proceeding similarly, we find
โ โซ๐โ (๐โ(๐โ2)๐ฆ
๐๐ฅ(๐โ2))
๐ฅ
0
=โโซ(๐๐โ1 โ โซ๐ข(๐ก)๐๐ก
๐ฅ
0
)
๐๐ก๐ฅ
0
๐ฆโ(๐โ2)(๐ฅ)
๐ฆโ(๐โ2)(0)=โ โซ ๐๐โ1
๐๐ก โ โซโ โซ๐ข(๐ก)๐๐ก๐๐ก๐ฅ
0
๐ฅ
0
๐ฅ
0
๐ฆโ(๐โ2)(๐ฅ)
๐๐โ2= ๐โซ ๐๐ ๐๐โ1๐๐ก
๐ฅ
0 โ โซโ โซ๐ข(๐ก)๐๐ก๐๐ก๐ฅ
0
๐ฅ
0
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Ikonion Journal of Mathematics 2020, 2 (2)
๐ฆโ(๐โ2)(๐ฅ)
๐๐โ2= (๐๐โ1)
๐ฅ โ โซโ โซ๐ข(๐ก)๐๐ก๐๐ก๐ฅ
0
๐ฅ
0
๐ฆโ(๐โ2)(๐ฅ) = ๐๐โ2 (๐๐โ1)๐ฅ โ โซโ โซ๐ข(๐ก)๐๐ก๐๐ก
๐ฅ
0
๐ฅ
0
โ โซ๐โ (๐โ(๐โ3)๐ฆ
๐๐ฅ(๐โ3))
๐ฅ
0
=โโซ[(๐๐โ1)๐ฅ. ๐๐โ2]
๐๐ก
๐ฅ
0
โ โซโ โซโ โซ๐ข(๐ก)๐๐ก๐๐ก๐๐ก๐ฅ
0
๐ฅ
0
๐ฅ
0
๐ฆโ(๐โ3)(๐ฅ) = ๐๐โ3 (๐๐โ2)๐ฅ (๐๐โ1)
๐ฅ2 โ โซโ โซโ โซ๐ข(๐ก)๐๐ก๐๐ก๐๐ก๐ฅ
0
๐ฅ
0
๐ฅ
0
โฎ
๐ฆโ = ๐1 (๐2)๐ฅ (๐3)
๐ฅ2 โฆ(๐๐โ2)๐ฅ(๐โ3) (๐๐โ1)
๐ฅ(๐โ2) โ โซ. . . (๐ โ 1). . .โ โซ๐ข(๐ก)๐๐กโฆ๐๐ก๐ฅ
0
๐ฅ
0
Hence, we get
๐ฆ = ๐0 (๐1)๐ฅ (๐2)
๐ฅ2 โฆ(๐๐โ1)๐ฅ(๐โ1) โ โซโฆ(๐)โฆ โ โซ๐ข(๐ก)๐๐กโฆ๐๐ก
๐ฅ
0
๐ฅ
0
If we take into account the above expressions, the multiplicative linear differential equation (12) is
written as follows
๐ข(๐ฅ) [(๐๐โ1)๐1(๐ฅ)(โ โซ๐ข(๐ก)๐๐ก
๐ฅ
0
)
๐1(๐ฅ)
] [(๐๐โ2)๐2(๐ฅ) (๐๐โ1)
๐ฅ ๐2(๐ฅ)(โ โซโ โซ๐ข(๐ก)๐๐ก๐๐ก๐ฅ
0
๐ฅ
0
)
๐2(๐ฅ)
]โฏ
[(๐0)๐๐(๐ฅ) (๐1)
๐ฅ๐๐(๐ฅ) (๐2)๐ฅ2๐๐(๐ฅ). โฆ . (๐๐โ1)
๐ฅ๐โ1๐๐(๐ฅ) (โ โซโฆ(๐)โฆโ โซ๐ข(๐ก)๐๐กโฆ๐๐ก๐ฅ
0
๐ฅ
0
)
๐๐(๐ฅ)
] = ๐(๐ฅ)
๐ข(๐ฅ) (๐0)๐๐(๐ฅ) (๐1)
๐ฅ ๐๐(๐ฅ)+๐๐โ1(๐ฅ)โฆ(๐๐โ1)๐ฅ๐โ1๐๐(๐ฅ)+โฏ+๐1(๐ฅ)(โ โซ๐ข(๐ก)๐๐ก
๐ฅ
0
)
๐1(๐ฅ)
(โ โซโ โซ๐ข(๐ก)๐๐ก๐๐ก๐ฅ
0
๐ฅ
0
)
๐2(๐ฅ)
โฆ
(โ โซโฆ(๐)โฆโ โซ๐ข(๐ก)๐๐กโฆ๐๐ก๐ฅ
0
๐ฅ
0
)
๐๐(๐ฅ)
= ๐(๐ฅ) (14)
If we set
๐1(๐ฅ) + ๐2(๐ฅ) ๐ฅ + โฏ+ ๐๐(๐ฅ) ๐ฅ๐โ1 = ๐๐โ1(๐ฅ)
๐2(๐ฅ) + ๐3(๐ฅ) ๐ฅ +โฏ+ ๐๐(๐ฅ) ๐ฅ๐โ2 = ๐๐โ2(๐ฅ)
โฎ
๐๐โ1(๐ฅ) + ๐๐(๐ฅ) ๐ฅ = ๐1(๐ฅ)
๐๐(๐ฅ) = ๐0(๐ฅ)
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Ikonion Journal of Mathematics 2020, 2 (2)
and
๐น(๐ฅ) =๐(๐ฅ)
(๐0)๐0(๐ฅ) (๐1)
๐1(๐ฅ)โฆ(๐๐โ1)๐๐โ1(๐ฅ)
then we can edit the equation (14) in the form as follows
๐ข(๐ฅ) (โ โซ๐ข(๐ก)๐๐ก๐ฅ
0
)
๐1(๐ฅ)
(โ โซโ โซ๐ข(๐ก)๐๐ก๐๐ก๐ฅ
0
๐ฅ
0
)
๐2(๐ฅ)
โฆ(โ โซโฆ(๐)โฆโ โซ๐ข(๐ก)๐๐กโฆ๐๐ก๐ฅ
0
๐ฅ
0
)
๐๐(๐ฅ)
= ๐น(๐ฅ) .
By using Theorem 8, we get
๐ข(๐ฅ) (โ โซ๐ข(๐ก)๐๐ก๐ฅ
0
)
๐1(๐ฅ)
(โ โซ[๐ข(๐ก)๐ฅโ๐ก]๐๐ก๐ฅ
0
)
๐2(๐ฅ)
โฆ(โ โซ๐ข(๐ก)(๐ฅโ๐ก)๐โ1
(๐โ1)!๐๐ก
๐ฅ
0
)
๐๐(๐ฅ)
= ๐น(๐ฅ).
Then we find the equation
๐ข(๐ฅ) โ โซ(๐ข(๐ก)[๐1(๐ฅ)+(๐ฅโ๐ก)๐2(๐ฅ)+โฏ+๐๐(๐ฅ)
(๐ฅโ๐ก)๐โ1
(๐โ1)!])
๐๐ก๐ฅ
0
= ๐น(๐ฅ).
If we put ๐พ(๐ฅ, ๐ก) = ๐1(๐ฅ) + (๐ฅ โ ๐ก)๐2(๐ฅ) + โฏ+ ๐๐(๐ฅ)(๐ฅโ๐ก)๐โ1
(๐โ1)! as the kernel function, then the
equation (12) is turned into
๐ข(๐ฅ) โ โซ๐ข(๐ก)๐พ(๐ฅ,๐ก)๐๐ก
๐ฅ
0
= ๐น(๐ฅ)
which is a MVIE of the second kind.
Example 4. Form a multiplicative Volterra integral equation corresponding to the multiplicative
differential equation ๐โ2๐ฆ(๐ฅ)
๐๐ฅ(2)= ๐ฆ(๐ฅ)๐๐๐ ๐ฅ with the initial conditions ๐ฆ(0) = 1, ๐ฆโ(0) = 1.
Solution. Let ๐โ2๐ฆ(๐ฅ)
๐๐ฅ(2)= ๐ข(๐ฅ). Then we write
โ โซ๐โ๐ฆโ๐ฅ
0
=โโซ๐ข(๐ก)๐๐ก๐ฅ
0
๐ฆโ(๐ฅ)
๐ฆโ(0)=โ โซ๐ข(๐ก)๐๐ก
๐ฅ
0
๐ฆโ(๐ฅ) =โ โซ๐ข(๐ก)๐๐ก๐ฅ
0
.
Therefore we find
โ โซ๐ฆโ(๐ก)๐๐ก๐ฅ
0
=โโซโ โซ๐ข(๐ก)๐๐ก๐ฅ
0
๐ฅ
0
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Ikonion Journal of Mathematics 2020, 2 (2)
๐ฆ(๐ฅ)
๐ฆ(0)=โ โซโ โซ๐ข(๐ก)๐๐ก
๐ฅ
0
๐ฅ
0
๐ฆ(๐ฅ) =โ โซ[๐ข(๐ก)(๐ฅโ๐ก)]๐๐ก
๐ฅ
0
If we replace the equation ๐ฆ(๐ฅ) =โ โซ [๐ข(๐ก)(๐ฅโ๐ก)]๐๐ก๐ฅ
0 into the given multiplicative differential equation,
we obtain ๐ข(๐ฅ) =โ โซ [๐ข(๐ก)๐๐๐ ๐ฅ (๐ฅโ๐ก)]๐๐ก๐ฅ
0.
4. Conclusion
In this paper, the multiplicative Voltterra integral equation is defined by using the concept of
multiplicative integral. The solution of multiplicative Volterra integral equation is obtained with the
successive approximations method. The multiplicative Leibniz formula is proved and the multiplicative
Volterra integral equation is converted to a multiplicative differential equation by aid of multiplicative
Leibniz formula. The multiplicative linear differential equation with constant or variable exponentials
is converted to a multiplicative Volterra integral equation is proved.
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