Differential transform method for special systems of integral equations

1 Introduction

The differential transform scheme is a method for solving a wide range of problems whose mathematical models yield equations or systems of equations involving algebraic, differential, integral and integro-differential (e.g. see Chen and Ho, 1996; Jang et al., 1997; Chen and Ho, 1999; Arikhoglu and Ozkol, 2006; Arikhoglu and Ozkol, 2007; Ayaz, 2003; Ayaz, 2004; Hassan, 2008; Biazar and Eslami, 2010; Biazar and Eslami, 2010). The concept of the differential transform was first proposed by Zhou (1986), and its main applications therein is solved both linear and non-linear initial value problems in electric circuit analysis. This method constructs an analytical solution in the form of polynomials. It is different from the high-order Taylor series method, which requires symbolic computation of the necessary derivatives of the data functions. The Taylor series method is computationally expansive for large orders. The differential transform is an iterative procedure for obtaining analytic Taylor series solutions of differential equations. In recent years the application of differential transform theory has been appeared in many researches.

The theory and application of integral equation is an important subject within applied mathematics. Integral equations are used as mathematical models for many physical situations, and integral equations also occur as reformulations of other mathematical problems such as partial differential equations and ordinary differential equations. These are motivations to solve this kind of equations.

In this paper, we present differential transform method for solving systems of integral equations:

2 Basic idea of differential transform method

The basic definitions and fundamental operations of two-dimensional differential transform are defined in Chen and Ho (1999). The differential transform of the function $u(x\text{,}y)$ is defined as the following

The inverse differential transform of $U\left(k\text{,}h\right)$ is defined as follows:

From the definitions of Eqs. (2) and (3), it is readily proved that the transformed functions comply with the following basic mathematical operations.

The following theorems can be proved easily.

1. If $u(x\text{,}y)=g(x\text{,}y)\pm h(x\text{,}y)\text{,}$ then $U(k\text{,}h)=G(k\text{,}h)\pm H(k\text{,}h)$ .

2. If $u(x\text{,}y)=\lambda g(x\text{,}y)$ , then $U(k\text{,}h)=\lambda G(k\text{,}h)$ , where, $\lambda$ is a constant.

3. If $u(x\text{,}y)=\frac{\partial g\left(x\text{,}y\right)}{\partial x}\text{,}$ then $U(k\text{,}h)=(k+1)G(k+1\text{,}h)$ .

4. If $u(x\text{,}y)=\frac{\partial g\left(x\text{,}y\right)}{\partial y}\text{,}$ then $U(k\text{,}h)=(h+1)G(k\text{,}h+1)$ .

5. If $u(x\text{,}y)=\frac{{\partial }^{r+s}g\left(x\text{,}y\right)}{\partial x^r\partial y^s}$ , then $$U(k\text{,}h)=(k+1)(k+2)+\cdots (k+r)(h+1)(h+2)\cdots (h+s)G(k+r\text{,}h+s)\text{.}$$

6. If $u(x\text{,}y)=x^m{y}^n$ , then $U(k\text{,}h)=\delta (k-m\text{,}h-n)=\delta (k-m)\delta (h-n)\text{,}$ where $$\delta (k-m\text{,}h-n)=\left\{\begin{array}{cc}1\text{,}\hfill & k=m\text{,}h=n\text{.}\hfill \\ o\hfill & \mathit{otherwise}\text{.}\hfill \end{array}\right.$$

7. If $u(x\text{,}y)=g(x\text{,}y)h(x\text{,}y)\text{,}$ then $U(k\text{,}h)={\sum }_{r=0}^k{\sum }_{s=0}^{h}G(r\text{,}h-s)H(k-r\text{,}s)$ .

8. If $u(x\text{,}y)=f(x\text{,}y)g(x\text{,}y)h(x\text{,}y)\text{,}$ then $$U(k\text{,}h)={\sum }_{r=0}^k{\sum }_{t=0}^{k-r}{\sum }_{s=0}^h{\sum }_{p=0}^{h-s}F(r\text{,}h-s-p)G(t\text{,}s)H(k-r-t\text{,}p)\text{.}$$

9. If $u(x\text{,}y)=f(x\text{,}y)g(x\text{,}y)h(x\text{,}y)v(x\text{,}y)$ , then $$U(k\text{,}h)={\sum }_{r=0}^k{\sum }_{t=0}^{k-r}{\sum }_{d=0}^{k-r-t}{\sum }_{s=0}^h{\sum }_{p=0}^{h-s}{\sum }_{q=0}^{h-s-p}F(r\text{,}h-s-p-q)G(t\text{,}s)H(d\text{,}q)\times V(k-r-t-d\text{,}q)\text{.}$$

10. If $u(x\text{,}y)={\int }_0^{x}g(x\text{,}y)h(x\text{,}y)\mathit{dx}\text{,}$ then $U(k\text{,}h)=\frac{1}{k}{\sum }_{r=0}^{k-1}{\sum }_{s=0}^{h}G(r\text{,}h-s)H(k-r-1\text{,}s)$ , where $k⩾1$ and $U(0\text{,}h)=0$ .

11. If $u(x\text{,}y)={\int }_0^{x}f(x\text{,}y)g(x\text{,}y)h(x\text{,}y)\mathit{dx}\text{,}$ then $$U(k\text{,}h)=\frac{1}{k}{\sum }_{r=0}^{k-1}{\sum }_{t=0}^{k-r-1}{\sum }_{s=0}^h{\sum }_{p=0}^{h-s}F(r\text{,}h-s-p)G(t\text{,}s)H(k-r-t-1\text{,}p)\text{,}$$ where $k⩾1$ and $U(0\text{,}h)=0$ .

12. If $u(x\text{,}y)={\int }_0^{x}f(x\text{,}y)g(x\text{,}y)h(x\text{,}y)v(x\text{,}y)\mathit{dx}\text{,}$ then $$U(k\text{,}h)=\frac{1}{k}{\sum }_{r=0}^{k-1}{\sum }_{t=0}^{k-r-1}{\sum }_{d=0}^{k-r-t-1}{\sum }_{s=0}^h{\sum }_{p=0}^{h-s}{\sum }_{q=0}^{h-s-p}F(r\text{,}h-s-p-q)G(t\text{,}s)H(d\text{,}q)V(k-r-t-d-1\text{,}q)\text{,}$$ where $k⩾1$ and $U(0\text{,}h)=0$ .

3 Numerical case studies

Three examples are presented to demonstrate the procedure of solving system (1) by differential transform method. The results reveal that the proposed technique is very effective and simple and leads to an exact solutions.

4 Conclusion

In this work differential transform method, has been used successfully for solving special systems of integral equations. The results reveal the efficiency of this method for solving these systems. The present method reduces the computational difficulties of other usual methods and the calculations are manipulated simply. In this paper, the Maple 13 Package, is used to perform calculations.