Differential Transform Method for Solving Volterra q -Integral Equations

: In this paper, Volterra q -integral equations are solved by using the method of q -differential transformation. Exact solutions of linear and nonlinear q -integral equations have been investigated. To illustrate the method, several problems are discussed for the effectiveness and performance of the proposed method


Introduction
In q-calculus, we are looking for q-analogues of mathematical functions that have the original function as limits when q tends to unity [1].The history of q-calculus (and q-hypergeometric functions) dates back to Kac et al. [2] and Jackson [3,4], who first introduced the q in their introduction [5] in the tracks of Newton's infinite series.In recent years, interest in the subject has exploded [6].This is, of course, due to the fact that q-analysis has proven itself extremely fruitful in quantum theory, mechanics, number theory, and the theory of relativity [7], has wideranging applications in areas like computer science and particle physics, and also acts as an important research tool for researchers working with analytic number theory or theoretical physics.
The differential transform scheme is a numerical method for solving systems of equations involving algebraic, differential, integral, and integro-differential differential equations.The concept of differential transformation was first proposed by Zhou [8].The main applications of this method are to solve both linear and nonlinear initial value problems in electric circuit analysis and to give exact values of the nth derivative of an analytical function at a point in terms of known and unknown boundary conditions in a fast manner [9].
The main motive for solving integral equations is their use in mathematical models for many physical situations.Integral equations also occur as reformulations of other mathematical problems, such as partial differential equations and ordinary differential equations [10].In this study, the q-differential transform is introduced to solve Volterra q-integral equations.The concept of one-dimensional q-differential was first proposed and applied to solve linear and nonlinear initial value problems by Jafari et al. [11].
A general integral equation for an unknown function y(x) can be written [12] as where f (x), a(x), and K(x, t) are given functions.The function K(x, t) is called the kernel.
There are different types of integral equations: The equation ( 1) is said to be of the first kind if the unknown function only appears under the integral sign, if a(x) = 0, otherwise of the second kind.
The equation ( 1) is said to be a Fredholm equation if the integration limits a and b are constants and Volterra if a and b are functions of x.
The equation ( 1) is said to be homogeneous if f (x) = 0, otherwise it is nonhomogeneous.Moreover, the q-analogue of integral equations has been defined by Mansour [13] where 0 <| p |< 1, and Fredhlom q-integral equation is defined as In this paper, the q-differential transform has been employed to solve some q-integral equations of the Volterra type.

Preliminaries and q-notations
Some basic concepts of q-calculus are given in [3,4,14].The q-derivative of a real continuous function ζ(x) is defined as where qa ≤ c ≤ qb, this result is an analogue of du Bois-Reymond's theorem.Definition 1.The series representation of the q-exponential function is defined as [16] 0 q q e e x δ δ = If all the q-differentials of a function ζ(x) exist in some neighbourhood of a, then q-Taylor formula [17] is given as For further details in q-calculus go through [3,7,14,18].

Analysis of q-differential transform method
For the function f (y) the differential transform of k th derivative is defined by Zhou [8] as The differential inverse transform of F(k) is defined as From equations ( 3) and (4), we get However, the basic definitions and results of the differential transform method are given in [8,[19][20][21][22][23][24][25][26].
Jafari et al. [11] has defined the q-analogue of differential transform.Definition 2. The q-differential transform of the function f (z, t) is defined as where all the q-differentials of f (z, t) exist in some neighbourhood of t = a, and F(k) is the transformed function.Definition 3. The inverse q-differential transform of the function F(k) is defined as From the definitions ( 5) and ( 6), we get The fundamental operations of q-differential transform can readily be obtained and are very useful on our approach for solving q-integral equations listed below in Table 1 [11].
Table 1.Operations of q-differential transformations

Original function
Transformed function

Main result
Theorem 1. Suppose that F(k) and G(k) are the q-differential transformations of the function f (y) and g(y), respectively, then we have with the help of (2).( , ) 0, 0,1, 2...

y y h y t g y z f y z d z y y h y t m y
Now, with the help of ( 5) and ( 6), we have This can be proved on the same lines as above in Proof (a).

Applications and numerical results
In this section, we apply the method to solve some Volterra q-integral equations [13].These results show that the method is simple and effective and leads to an exact solution.
Example 1.We consider the following Volterra q-integral equation ∫ Sol: By using operations from Table 1 and according to theorem as stated above, we have the following relation: Consequently, we find , (4) 0, [3] ! 1 (5) ... (6), the solution of the above integral equation is given by which is the solution of the above q-integral equation.The exact solution for different values of q and y are given in Figure 1 and exact solution for fractional order 0 < q < 1 is given in Figure 2.   1, the relation is given by , 1, (0) 0. 1 , (4) 0, (6), the solution of the above integral equation is given by which is an exact solution.The exact solution for different values of q and y are given in Figure 3 and exact solution for fractional order 0 < q < 1 is given in Figure 4.  which is an exact solution.The exact solution for different values of q and y are given in Figure 5 and exact solution for q = 0.2 q = 0.4 q = 0.6 q = 0.

Figure 1 . 2 . 1 Example 2 .
Figure 1.Exact solution of Example 1, for Figure 2. Exact solution of Example 1, different values of q and y for fractional order 0 < q < 1 From the Theorem 1 above and from Table

Figure 3 . 4 . 1 Example 3 .
Figure 3. Exact solution of Example 2, for Figure 4. Exact solution of Example 2, different values of q and y for fractional order 0 < q < 1