Optimal Homotopy Asymptotic Method for Solving the Linear Fredholm Integral Equations of the First Kind

and Applied Analysis 3 By substituting g0(s), g1(s), and g3(s) solutions in (6), we obtain g (s) = − 1 2 sin (s) + 1 4 c1 sin (s) + 1 4 (1 + c1) c1 sin (s) − 1 8 c 1 sin (s) + 1 4 c2 sin (s) . (18) For the calculations of the constants c1 and c2, the use of the technique mentioned in (8)–(12) yields c1 = 6.000000004, c2 = −24.00000002. (19) Substituting values in (18), the final solution becomes g (s) = sin (s) . (20) This is the exact solution. Table 1 shows some numerical results of these solutions calculated according to the present method. The exact solution, OHAM solution and absolute error of this example are shown in Figure 1. Example 2. We consider the following equation [10]:


Introduction
Integral equations of the first kind arise in several applications. These include applications in biology, chemistry, physics, and engineering. In recent years, much work has been carried out by researchers in mathematics and engineering in applying and analyzing novel numerical and semi analytical methods for obtaining solutions of integral equations of the first kind. Among these are the homotopy analysis method [1], operational Tau method [2], homotopy perturbation method [3], Adomian decomposition [3], quadrature rule [4], and automatic augmented Galerkin algorithms [5].
In this study, we develop the optimal homotopy asymptotic method (OHAM), which was proposed by Marinca et al. [6,7], for solving the linear Fredholm integral equations of the first kind. This method is characterized by it is convergence criteria which are more flexible than other methods.
The general form of the linear Fredholm integral equations of the first kind is where and are constant and the functions ( , ) and ( ) are known. It should be noted that OHAM has been applied to the nonlinear Fredholm integral equations of the second kind by [8].

Application of OHAM to the Linear Fredholm Integral Equations of the First Kind
In this section, we formulate the optimal homotopy asymptotic method (OHAM) for solving the linear Fredholm integral equations of the first kind following the procedure as outlined in [6,7] and other papers. Let us consider a form of the linear Fredholm integral equation of the first kind: Using OHAM, we can obtain a family of equations as follows: where ∈ [0, 1] is an embedding parameter, ( , ) is unknown function, and ( ) is an (nonzero) auxiliary function for ̸ = 0 and (0) = 0 and given as ( ) = ∑ =1 where , = 1, 2, . . ., are auxiliary constants, and when = 0 and = 1 it holds that ( , ) , = 1, 2, . . . .

(5)
If the series (5) convergence occurs when = 1, one has Substituting (5) in (3) and equating the coefficients of like powers of , we get as follows: For finding the constants 1 , 2 , 3 , . . ., we can get the result of the th-order approximations as follows: If we substitute (8) into (1) we obtain the residual equation If ( , ) = 0, then ( , ) will be the exact solution. The least squares method can be used to determine 1 , 2 , 3 , . . . . At first we consider the functional By using Galerkin's method we get the following system: and then minimizing it to obtain the values of 1 , 2 , . . . , , we have With these constants, the approximate solution is determined.

Numerical Examples and Discussion
In this section, three examples of the linear Fredholm integral equations of the first kind were solved to show the efficiency of the present method. Maple software with long format and double accuracy was used to carry out the computations.
Example 1. We consider the following equation [9]: for which the exact solution is ( ) = sin( ). Applying OHAM to the linear Fredholm integral equation of first kind yields ( ( , )) = ( ) , which satisfies Now we use (7) to obtain a series of problems: Hence the solutions are Substituting values in (18), the final solution becomes This is the exact solution. Table 1 shows some numerical results of these solutions calculated according to the present method.
The exact solution, OHAM solution and absolute error of this example are shown in Figure 1.

Example 2.
We consider the following equation [10]:  Now we use (7) to obtain a series of problems: Hence the solutions are By substituting 0 ( ), 1 ( ), and 3 ( ) solutions in (6), we obtain For the calculations of the constants 1 and 2 , the use of the technique mentioned in (8)-(12) yields Substituting values in (26), the final solution becomes This is the exact solution. Table 2 shows some numerical results of these solutions calculated according to the present method.
The exact solution, OHAM solution and absolute error of this example are shown in Figure 2.  Example 3. We consider the following equation [9]: Now we use (7) to obtain a series of problems: Hence the solutions are By substituting 0 ( ), 1 ( ), and 3 ( ) solutions in (6) For the calculations of the constants 1 and 2 , the use of the technique mentioned in (8)-(12) yields 1 = 6, 2 = −24.
Substituting values in (34), the final solution becomes This is the exact solution.   Table 3 shows some numerical results of these solutions calculated according to the present method.
The exact solution, OHAM solution and absolute error of this example are shown in Figure 3.