LINEARIZATION METHOD FOR SOLVING NONLINEAR INTEGRAL EQUATIONS

The objective of this paper is to assess both the applicability and the accuracy of linearization method in several problems of general nonlinear integral equations. This method provides piecewise linear integral equations which can be easily integrated. It is shown that the accuracy of linearization method can be substantially improved by employing variable steps which adjust themselves to the solution. This approach can reveal that, under this method, the nonlinear integral equations can be transformed into the linear integral equations which may be integrated using classical methods. Numerical examples are used to illustrate the preciseness and effectiveness of the proposed method.


Introduction
Modeling and analysis of physical phenomena in applied sciences often generates nonlinear mathematical problems.Nonlinearity may be an inner feature of the model, that is, evolution equations with nonlinear terms, or of the problem, that is, nonlinear boundary conditions.The interplay between applied sciences and mathematics then leads to the development of initial and/or boundary value problems for nonlinear partial differential or integral or integrodifferential equations modeling real physical systems.The theory and application of integral equations is an important subject within applied mathematics.Integral equations are used as mathematical models for many and varied physical situations, and also occur as reformulations of other mathematical problems.Since many physical problems are modeled by integral equations, the numerical solutions of such integral equations have been highly studied by many authors.In recent years, numerous works have been focusing on the development of more advanced and efficient methods for integral equations such as implicitly linear collocation methods [1], product integration method [3], and Hermite-type collocation method [2] and semianalytical-numerical techniques such as Adomian's decomposition method [4].
The linearization method presented in this paper for integral equations is based on the piecewise linearization of the nonlinear integral equations and the analytical solution of the resulting linear integral equation.Thus, linearization methods provide closed-form solutions in a piecewise fashion and are not iterative; therefore, they do not need a judicious guess to the solution.Furthermore, since these methods are based on the linearization of the nonlinear integral equations, they also provide a means to adapt the step size to the solution.If one is only interested in the values of the dependent variables at discrete values of the independent variable, linearization methods provide explicit nonlinear mappings.

Model problems and method
We will consider the general nonlinear Volterra integral equation of the form where y(x) is an unknown function, a is a real constant, and the functions f (x) and K(x,t, y(t)) are analytical on R and R 3 , respectively.Note that K(x,t, y(t)) is a nonlinear function which may be singular at x = x 0 .Now, we are interested to find the numerical solution to (2.1) in domain T = [a,+∞).Consider the interval [a,b) ⊆ T, and divide it into a series of subintervals [x n ,x n+1 ) such that x 0 = a.In each subinterval K(x,t, y) may be linearized as follows.
(1) If K(x,t, y) is regular, K may be approximated by the first three terms of its Taylor series expansion around (x n ,t n , y n ) in the following form: where ) (2.5) P. Darania et al. 3 Since in the integral part of (2.3) t is an independent variable, y is a dependent variable, and x is a parameter, therefore by integrating it respect to t, we have From this equation and after some computations, we obtain (2.7) With differentiating (2.7), respect to x, we obtain It is clear that, this equation is a linear ordinary differential equation, whose analytical solution may be written as (2.9) Equation ( 2.9) provides the following nonlinear mapping (2.10) Note that, since the integral equation (2.1) is one-dimensional and x is independent variable, therefore x and t are equivalent.Now, we will use (2.10) and initial condition 4 On the numerical solutions of integral equations (2.4) for solving (2.1) by taking t n = x n .By this assumption, the numerical solution of (2.1) at the grid points x n+1 , for n = 0,1,2,..., can be obtained from where x n = x n+1 − x n is the local size step and y(x 0 ) = y 0 .
On the other hand, if we set x n = h for n = 0,1,2,..., namely, x n = x 0 + nh, then (2.11) will be reduced to the following equation: (2.12) (2) If K(x,t, y(t)) is singular at x n , the above derivation is not valid, but K(x,t, y) may be approximated by the first three terms of its Taylor series expansion around (x n+1 ,t n+1 , y n ), so (2.1) may be approximated by where P. Darania et al. 5 whose analytical solution may be expressed as (2.15) From (2.15), solution of (2.1) at the grid points x n+1 can be obtained from where x n , and t n+1 are as above.
In the similar manner, if we set x n = h, for n = 0,1,2,..., (x n = x 0 + nh), then (2.16) will be reduced to the following equation: (2.17) The nonlinear mappings or difference equations corresponding to (2.12) and (2.17) have been derived by considering that K is a function of three variables x, t, and y.Moreover, since these equations have been obtained by approximating K by the first three terms of its Taylor series expansion, while higher-order terms have been neglected, the local step size x n may be determined from the condition that higher-order terms be much smaller than the first-order ones, or from the condition that y n+1 does not differ significantly from y n .

Error analysis
In this section, we perform the estimating error for the integral equations.Since the truncated nonlinear mappings or difference equations corresponding to (2.12) and (2.17) are an approximate solution of (2.1), hence, the error function e(x) for (2.1) is defined as follows: where in the regular point x n , we have and in the singular point x n , we have In other words, by decreasing h the error function e(x r ) approaches to zero.

Presentation of results
In order to assess both the applicability and accuracy of the theoretical results of Section 2, we have applied them to a variety of nonlinear integral equations in the following examples.
has the following analytical solution y(x) = exp(x), and therefore, provides an example to verify the accuracy of this method.
Equation (4.1) by using (2.7) is reduced to the linear integral equation in the following form: and from (2.12), numerical solution of (4.2) at the grid points x n+1 , n = 0,1,2,..., can be obtained from The result for h = 0.1 and 0.01 and analytical solution are presented in Figure 4.1 which indicates that the numerical solutions obtained from (4.3) and step sizes equal to 0.0001, 0.001, 0.01, and 0.1 are nearly identical.The exact solution at x = 0.7 is 2.013752707000, whereas the numerical solutions corresponding to step sizes equal to 0.0001,0.001,0.01,and 0.1 are 2.0137523240, 2.0137511060, 2.0136061360, and 2.0005244930, respectively.These results indicate that, if we use the time steps smaller than about 0.01, then we obtain the numerical solutions of minimum error.In other words, by decreasing h the error function e(x r ) ≤ 10 −5 .Table 4.1 shows the errors-involved presented method with h = 0.0001,0.001,0.01,and 0.1 along with the exact solution.The absolute errors found by presented method are compared with the exact solution in Table 4.2.

Conclusion
By introducing the Linearization method, the nonlinear Volterra integral equations in applied sciences and physics can be transformed into the linear Volterra integral equations which may be integrated using classical methods.Both the applicability and the accuracy of linearization method for the solution of nonlinear integral equations have been examined by means of several problems.It has been shown that, for regular problems, linearization method is robust and accurate techniques whose accuracy is not a strong function of either the linearization point or the linearization with respect to the independent variable.
substituting the solutions y(x r ) (r = 0,1,...) in the error function e(x) we havee x r = y x r − Y x r ,(3.4)then our aim is e(x r ) ≤ 10 −kr (k r is any positive integer).If we prescribe max(10 −kr ) = 10 −k , then we decrease the step size h as far as the following inequality holds at each point x r : e x r ≤ 10 −k .(3.5)

Table 4 .
1. Comparison of numerical solution y i and exact solution y(x i ).

Table 4 .
2. Comparison of numerical solution y i and exact solution y(x i ).(2.17), we get the numerical solution of (4.4) at the grid points x n+1 for n = 0,1,2,..., in the following form: from