# Homotopy Analysis Method for Second-Order Boundary Value Problems of Integrodifferential Equations

Authors: Ahmad El-Ajou; Omar Abu Arqub; Shaher Momani
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

In this paper, series solution of second-order integrodifferential equations with boundary conditions of the Fredholm and Volterra types by means of the homotopy analysis method is considered. The new approach provides the solution in the form of a rapidly convergent series with easily computable components using symbolic computation software. The homotopy analysis method provides us with a simple way to adjust and control the convergence region of the infinite series solution by introducing an auxiliary parameter. The proposed technique is applied to a few test examples to illustrate the accuracy, efficiency, and applicability of the method. The results reveal that the method is very effective, straightforward, and simple.


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## Body

## 1. Introduction

Integrodifferential equations (IDEs) are often involved in the mathematical formulation of physical and engineering phenomena. IDEs can be encountered in various fields of science such as physics, chemistry, biology, and engineering. These kinds of equations can also be found in numerous applications, such as electromagnetic, plasma physics, elasticity, fluid dynamics, oscillation theory, polymer rheology, chemical kinetics, biomechanics, and control theory [1–5]. Since it is usually impossible to obtain the closed-form solutions to second-order boundary value problems of IDEs met in practice, these problems must be attacked by various approximate and numerical methods.The purpose of this paper is to extend the application of the homotopy analysis method (HAM) to provide symbolic approximate solution for the second-order boundary value problems of IDEs of the following Fredholm type:(1.1)u′′(x)+∫abK(x,s)f(u(s),u′(s))ds+g(u(x),u′(x))+h(x)=0,a≤s,x≤b,
and the following Volterra type:
(1.2)u′′(x)+∫axK(x,s)f(u(s),u′(s))ds+g(u(x),u′(x))+h(x)=0,a≤s,x≤b,
subject to the boundary conditions
(1.3)u(a)=κ1,u(b)=κ2,
where a,b,κ1,κ2 are real finite constants, u is an unknown function to be determined, K and h are continuous functions on [a,b]×[a,b] and [a,b], respectively, and f,g are nonlinear continuous functions of u,u′. For details about the existence and uniqueness of solutions for such problems, see [6–8]. However, we assume that (1.1) and (1.2) subject to the boundary conditions (1.3) have a unique analytic solution on the given interval.The numerical solvability of second-order IDEs with boundary conditions of the Fredholm and Volterra types and other related equations has been pursued by several authors. To mention a few, in [9], the   authors have discussed the Legendre polynomials method for solving Fredholm equation u′′(x)+∫abK(x,s)u(s)ds+f(x)+p(x)u(x)+q(x)u′(x)=0. Furthermore, the compact finite difference method is carried out in [10] for the Volterra IDE u′′(x)+∫axK(x,s)u(s)ds+g(x,u(x))=0 and the Fredholm IDE u′′(x)+∫abK(x,s)u(s)ds+g(x,u(x))=0. Recently, the monotone iterative sequences method for solving Volterra equation u′′(x)+∫0xK(x,s)f(u(s))ds+g(x)=0 is proposed in [11].The HAM, which is proposed by Liao [12–17], is effectively and easily used to solve some classes of linear and nonlinear problems without linearization, perturbation, or discretization. The HAM is based on the homotopy, a basic concept in topology. The auxiliary parameter ℏ is introduced to construct the so-called zero-order deformation equation. Thus, unlike all previous analytic techniques, the HAM provides us with a family of solution expressions in auxiliary parameter ℏ. As a result, the convergence region and rate of solution series are dependent upon the auxiliary parameter ℏ and thus can be greatly enlarged by means of choosing a proper value of ℏ. This provides us with a convenient way to adjust and control convergence region and rate of solution series given by the HAM.In the recent years, extensive work has been done using HAM, which provides analytical approximations for linear and nonlinear equations. This method has been implemented in several functional equations, such as nonlinear water waves [15], unsteady boundary-layer flows [16], solitary waves with discontinuity [17], Klein-Gordon equation [18], projectile motion with the quadratic resistance law [19], systems of fractional differential equations [20], nonlinear fractional differential equations [21, 22], systems of fractional algebraic-differential equations [23], fractional SIR model [24], singular Volterra integral equation [25], charged particle motion for certain configurations of oscillating magnetic fields [26], Volterra population growth model [27], systems of fractional IDEs [28], and high-order IDEs [29].The outline of the paper is as follows. In the next section, the basic idea of the HAM is introduced. In Section3, we utilize the statement of the HAM for solving second-order IDEs with boundary conditions of the Fredholm and Volterra types. In Section 4, numerical examples are given to illustrate the capability of HAM. This paper ends in Section 5 with some concluding remarks.


## 2. Homotopy Analysis Method

The principles of the HAM and its applicability for various kinds of functional equations are given in [12–28]. For convenience of the reader, we will present a review of the HAM [12–17] then we will extend the HAM to construct a symbolic approximate solution to second-order IDEs with boundary conditions of the Fredholm and Volterra types.  To achieve our goal, we consider the following nonlinear functional equation:
(2.1)N[u(x)]=0,x≥a,
where N is a nonlinear operator and u(x) is an unknown function of independent variable x.Letu0(x) denote an initial guess of the exact solution of (2.1), ℏ≠0 an auxiliary parameter, H(x)≠0 an auxiliary function, and Λ an auxiliary linear operator with the property Λ[f(x)]=0 when f(x)=0. The auxiliary parameter ℏ, the auxiliary function H(x), and the auxiliary linear operator Λ play important roles within the HAM [12]. As we will see later, choosing suitable values of the auxiliary parameter ℏwill help us to adjust and control the convergence region of the series solution.Liao [12–17] constructs, using q∈[0,1] as an embedding parameter, the so-called zero-order deformation equation as
(2.2)(1-q)Λ[ϕ(x;q)-u0(x)]=qℏH(x)N[ϕ(x;q)],
where ϕ(x;q) is the solution of  (2.1) which depends on ℏ, H(x), Λ, u0(x), and q. When q=0, the zero-order deformation equation (2.2) becomes
(2.3)ϕ(x;0)=u0(x),
and when q=1, since ℏ≠0 and H(x)≠0, the zero-order deformation equation (2.2) reduces to
(2.4)N[ϕ(x;1)]=0.So,ϕ(x;1) is exactly the solution of the nonlinear equation (2.1). Thus, according to (2.3) and (2.4), as q increases from 0 to 1, the solution ϕ(x;q) varies continuously from the initial approximation u0(x) to the exact solution u(x).Define the so-calledmth-order deformation derivatives
(2.5)um(x)=1m!∂mϕ(x;q)∂qm|q=0,
expanding ϕ(x;q) in Taylor series with respect to the embedding parameter q, using (2.3) and (2.5), we obtain
(2.6)ϕ(x;q)=u0(x)+∑m=1∞um(x)qm.Assume that the auxiliary parameterℏ, the auxiliary function H(x), the initial approximation u0(x), and the auxiliary linear operator Λ are properly chosen so that the power series (2.6) of ϕ(x;q) converges at q=1. Then, we have under these assumptions the series solution u(x)=u0(x)+∑m=1∞um(x).Define the vectoru→n={u0(x),u1(x),u2(x),…,un(x)}. Differentiating equation (2.2) m-times with respect to embedding parameter q, then setting q=0 and dividing them by m!. Using (2.5), we have the so-called mth-order deformation equation as
(2.7)Λ[um(x)-χmum-1(x)]=ℏH(x)Rm(u→m-1(x)),m=1,2,…,n,
where
(2.8)Rm(u→m-1(x))=1(m-1)!∂m-1N(ϕ(x;q))∂qm-1|q=0,(2.9)χm={0,m≤1,1,m>1.For any given nonlinear operatorΛ, the term Rm(u→m-1(x)) can be easily expressed by  (2.8). Thus, we can gain u0(x),u1(x),u2(x),… by means of solving the linear high-order deformation equation (2.7) one after the other in order. The mth-order approximation of u(x) is given by u(x)=∑k=0m-1uk(x).


## 3. Solving Second-Order Boundary Value Problems of IDEs by HAM

Throughout this section, we will utilize the construction for the IDEs of the Fredholm type in order not to increase the length of the paper without the loss of generality for the remaining type. However, similar construction can be implemented for the Volterra type.Accordingly, we extend the application of the HAM to solve the second-order IDEs of the Fredholm type(3.1)N[u(x)]:=u′′(x)+∫abK(x,s)f(u(s),u′(s))ds+g(u(x),u′(x))+h(x)=0,a≤s,x≤b,
subject to the boundary conditions (1.3).First of all, we assume that (3.1) satisfies the initial condition u′(a)=κ,  where the unknown constant κ can be later determined by substituting the boundary condition u(b)=κ2 into the obtained solution. After that, we chose the initial guess u0 such that the initial conditions u0(a)=κ1 and u0′(a)=κ are satisfied. That is, the initial guess will be of the form u0=u0(x;κ). Finally, we take the auxiliary linear operator Λ=d2/dx2 which satisfies the property Λ[c1+c2t]=0, wherec1,c2 are constants of integrations.The so-called zero-order deformation equation will be defined as(3.2)(1-q)d2dx2[ϕ(x;q,κ)-u0(x;κ)]=qℏH(x)N[ϕ(x;q,κ)],
where
(3.3)N[ϕ(x;q,κ)]=d2ϕ(x;q,κ)dx2+∫abK(x,s)f(ϕ(s;q,κ),ϕ′(s;q,κ))ds+g(ϕ(x;q,κ),ϕ′(x;q,κ))+h(x),
when q=0, (3.2) reduces to ϕ(x;q,κ)=u0(x;κ). In this case, the so-called mth-order deformation equation can be constructed as
(3.4)d2dx2[um(x;κ)-χmum-1(x;κ)]=ℏH(x)Rm(u→m-1(x;κ)).
Operating the operator J2, the inverse operator of d2/dx2 to both sides of (3.4), then, the mth-order deformation equation will have the following form:
(3.5)um(x;κ)=χmum-1(x;κ)+ℏ∫ax[(x-τ)H(τ)Rm(u→m-1(τ;κ))]dτ,
where
(3.6)Rm(u→m-1(x;κ))=d2um-1(x;κ)dx2+∫abK(x,s)1(m-1)!∂m-1(f(ϕ(s;q,κ),ϕ′(s;q,κ)))∂qm-1|q→0ds+1(m-1)!∂m-1(g(ϕ(x;q,κ),ϕ′(x;q,κ)))∂qm-1|q→0+h(x)(1-χm),
and the mth-order approximation of u(x) can be given as
(3.7)u(x)=∑k=0m-1uk(x;κ).If we substitute the boundary conditionu(b)=κ2 into (3.7), then we obtain an equation of the variable κ. And so, the approximate solution for (3.1) and (1.3) will be completely constructed. For simplicity, in this paper, we assume that the value of the auxiliary function is kept at H(x)=1.


## 4. Numerical Results and Discussion

HAM provides an analytical approximate solution in terms of an infinite power series. However, there is a practical need to evaluate this solution and to obtain numerical values from the infinite power series. The consequent series truncation and the practical procedure are conducted to accomplish this task. In this section, we consider four examples to demonstrate the performance and efficiency of the present technique. Through this paper all the symbolic and numerical computations are performed by using Mathematica 7.0 software package.To show the accuracy of the present method for our problems, we report three types of error. The first one is the residual error,Re, defined as
(4.1)Re(x)≔|d2dx2uHAMm(x)+∫abK(x,s)f(uHAMm(s),ddsuHAMm(s))ds+g(uHAMm(x),ddxuHAMm(x))+h(x)|,
for the Fredholm type and similarly for the Volterra type, while the exact, Ex, and relative, Rl, errors are defined, respectively, by
(4.2)Ex(x):=|uExact(x)-uHAMm(x)|,Rl(x):=|uExact(x)-uHAMm(x)||uExact(x)|,
where s,x∈[a,b], uHAMm is the mth-order approximation of u(x) obtained by the HAM, and uExact is the exact solution.Example 4.1.
Consider the following linear Fredholm IDE:(4.3)u′′(x)-∫-11se-scosxu(s)ds-2u(x)-x=0,-1≤s,x≤1,
subject to the boundary conditions
(4.4)u(-1)=-1,u(1)=1.
The exact solution is u(x)=c1e2x+c2e-2x+c3cosx-0.5x, where c1=0.4206…, c2=-0.3545…, and c3=-0.2662….
According to (3.6) and (2.5), we have
(4.5)Rm(u→m-1(x;κ))=d2um-1(x;κ)dx2-∫-11(se-scosx)um-1(s;κ)ds-2um-1(x;κ)-x(1-χm).
Assuming the initial approximation of (4.3) has the form u0(x)=κ(x+1)-1, which satisfies the initial conditions u0(-1)=-1 and u0′(-1)=κ. Consequently, the first few terms of the HAM series solution for (4.3) according to these initial conditions are as follows:
(4.6)u1(x)=ℏ[0.31668-0.5311κ+1.88088x-1.12044κx+(1-κ)x2-(0.16667+0.33333κ)x3+(0.73576+0.14313κ)cosx],(4.7)u2(x)=u1(x)-0.033333ℏ2[∫-11(60.81144+6.54379κ)-(21.76335-17.39749κ)x-0.033333ℏ2-(20.49956-14.06698κ)x2+(23.8088-1.20436κ)x3+5(1-κ)x4-0.033333ℏ2-(0.5+κ)x5-(81.0036+0.77298κ)cosx∫-11].
To determine the value of κ, we must chose a value to the auxiliary parameter ℏ. It was proved that if we set ℏ=-1, then we have the Adomian decomposition solution which is a special case of the HAM solution [30, 31]. Therefore, -1 is available value forℏ. Now, substitute ℏ=-1and the boundary condition at x=1 into the 10th order approximation of u(x) to get κ=1.4818348333.
Figure1 shows the 10th-order approximate solution of (4.3) and (4.4) obtained by HAM at ℏ=-1 together with the exact solution. It is clear from the figure that the approximate solution is in high agreement with the exact solution.
Since the value ofκ has been identified previously, the valid region for the values of ℏ can be obtained by the so-called ℏ-curve which correspond to the horizontal line as shown in Figure 2. Thus, the valid region of ℏ is -1.5<ℏ<-0.8.Figure 1
Solutions of (4.3) and (4.4). Dotted line: the 10th-order approximate HAM solution at ℏ=-1 and solid line: the exact solution.Figure 2
Theℏ-curves of u′(0) and u′′(0) which are corresponding to the 10th-order approximate HAM solution of (4.3) and (4.4). Dotted line: u′(0) and solid line: u′′(0).Figure3 shows a comparison between the exact and residual errors obtained from the 10th-order approximation of u(x) for (4.3) and (4.4) at ℏ=-1, while Figures 4 and 5 show respectively how the exact and residual errors change with varying the values of ℏ.Figure 3
Errors of (4.3) and (4.4) at ℏ=-1. Dotted line: the exact error and solid line: the residual error.Figure 4
Exact error of  (4.3) and (4.4) at different values of ℏ. Dash-dotted line: ℏ=-1.1, dotted line: ℏ=-1, and solid line: ℏ=-0.9.Figure 5
Residual error of (4.3) and (4.4) at different values of ℏ. Dash-dotted line: ℏ=-1.1, dotted line: ℏ=-1, and solid line: ℏ=-0.9.Example 4.2.
Consider the following nonlinear Fredholm IDE:(4.8)u′′(x)-∫01(x-s)2eu(s)ds+(u′(x))2+(32x2-53x+712)=0,0≤s,x≤1,
subject to the boundary conditions
(4.9)u(0)=0,u(1)=ln2.
The exact solution is u(x)=ln(x+1).
According to (3.6) and (2.5), we have
(4.10)Rm(u→m-1(x;κ))=d2dx2um-1(x;κ)+(32x2-53x+712)(1-χm)-∫01(x-s)21(m-1)!∂m-1(exp(ϕ(s;q,κ)))∂qm-1|q→0ds+∑i=0m-1ddxui(x;κ)ddxum-1-i(x;κ).
Choose the initial guess approximation as u0(x)=kx, which satisfies the initial conditions u0(0)=0 and u0′(0)=κ. Then, we have the following approximations terms of the HAM series solution for (4.8):
(4.11)u1(x)=hx272κ3(72-72ⅇκ+72ⅇκκ+24xκ-24ⅇκxκ-36ⅇκκ2hx272κ3+24ⅇκxκ2+6x2κ2-6ⅇκx2κ2+21κ3-20xκ3+9x2κ3+36κ5),(4.12)u2(x)=u1(x)+h2x22160κ8(47520-95040ⅇκ+47520ⅇ2κ+97200κ-66960ⅇκκ-30240ⅇ2κκ+5760xκ-11520ⅇκxκ+5760ⅇ2κxκ+36000κ2+24480ⅇκκ2+19440ⅇ2κκ2+10800xκ2-6480ⅇκxκ2-4320ⅇ2κxκ2+360x2κ2+360ⅇ2κx2κ2-720ⅇκx2κ2+7560κ3-12240ⅇκκ3-12240ⅇ2κκ3+4800xκ3+960ⅇκxκ3+3600ⅇ2κxκ3+540x2κ3-180ⅇκx2κ3-360ⅇ2κx2κ3+3780ⅇκκ4+5580ⅇ2κκ4+1260xκ4-900ⅇκxκ4-2400ⅇ2κxκ4+300x2κ4-120ⅇκx2κ4+360ⅇ2κx2κ4+15120κ5-16770ⅇκκ5-1620ⅇ2κκ5+600ⅇκxκ5+900ⅇ2κxκ5+105x2κ5-75ⅇκx2κ5-180ⅇ2κx2κ5+15600ⅇκκ6+270ⅇ2κκ6+4320xκ6-4540ⅇκxκ6-180ⅇ2κxκ6+30ⅇκx2κ6+45ⅇ2κx2κ6-7710ⅇκκ7+4420ⅇκxκ7+720x2κ7-745ⅇκx2κ7+630κ8+2160ⅇκκ8-600xκ8-1800ⅇκxκ8+270x2κ8+540ⅇκx2κ8+72x3κ8-72ⅇκx3κ8-540ⅇκκ9+420xκ9+360ⅇκxκ9-300x2κ9-90ⅇκx2κ9+108x3κ9+1080κ10+720xκ11).
As in the previous example, we determine an introductory value of the constant κ by substituting ℏ=-1 and the boundary condition at x=1 into the obtained HAM solution. So, the 5th-order approximation of u(x) gives κ=0.9507338904. Figure 6 shows the 5th-order approximate solution of (4.8) and (4.9) obtained by HAM at ℏ=-1 together with the exact solution, while Figure 7 shows the ℏ-curve corresponding to the 5th-order approximate HAM solution. It is clear from Figure 7 that the valid region of ℏ is -0.7<ℏ<-0.4.
Our next goal is to show how the auxiliary parameterℏ affects the approximate solutions. However, it is evident from Figure 6 that both curves corresponding to the 5th-order approximate HAM solution and the exact solution are not coinciding in general at  ℏ=-1. To improve the solution, let us compute other values of κ at different values ofℏ. Table 1 shows the values of the constantκ at different values of the auxiliary parameterℏ.
Representation the exact solution and the 5th-order approximate HAM solution of (4.8) and (4.9) at different values of ℏ mentioned in Table 1 shows that all these solutions are approximately identical. However, Figure 8 shows the correspondence between the exact solution and the 5th-order approximate HAM solution of (4.8) and (4.9) at ℏ=-0.6. From the last mentioned figure, we see that we can achieve a good approximation with the exact solution by using a few terms in HAM.
Figure9 shows a comparison between the exact errors of (4.8) and (4.9) at different values ofℏ corresponding to the 5th-order approximate HAM solution. It is to be noted that the accuracy of certain node at ℏ=-0.6 is more convincing than the other values of ℏ.Table 1
The values ofκ at different values of ℏ.
ℏ
κ
-
0.8
0
.
9973778496
-
0.6
0.9987971735
-
0.4
0.9834454136Figure 6
Solutions of (4.8) and (4.9). Dotted line: the 5th-order approximate HAM solution at  ℏ=-1 and solid line: the exact solution.Figure 7
Theℏ-curves of u′(1) and u′′(1) which are corresponding to the 5th-order approximate HAM solution of (4.8) and (4.9). Dotted line:  u′(1) and solid line: u′′(1).Figure 8
Solutions of (4.8) and (4.9). Dotted line: the 5th-order approximate HAM solution at ℏ=-0.6 and solid line: the exact solution.Figure 9
Exact error of (4.8) and (4.9) at different values of ℏ. Dash-dotted line: ℏ=-0.4, dotted line: ℏ=-0.6, and solid line: ℏ=-0.8.Example 4.3.
Consider the following linear Volterra IDE:(4.13)u′′(x)-∫0xe-ssinxu′(s)ds+u(x)-(12e-xsin2x-sinx)=0,0≤s,x≤π2,
subject to the boundary conditions
(4.14)u(0)=-1,u(π2)=1.
The exact solution is u(x)=sinx-cosx.
According to (3.6) and (2.5), we have
(4.15)Rm(u→m-1(x;κ))=d2dx2um-1(x;κ)-∫0xsinxe-s(ddsum-1(s;κ))ds+um-1(x;κ)-(12e-xsin2x-sinx)(1-χm).
Choose the initial guess approximation as u0(x)=κx-1, which satisfies the initial conditions u0(0)=-1 and u0′(0)=κ. Using the iteration formula (3.5), we can directly obtain the following approximations terms of the HAM series solution for (4.13):
(4.16)u1(x)=1150ℏ(12-75κ+120x-75κx-75x2+25κx3+75κⅇ-xcosx1150ℏ1150ℏ-12ⅇ-xcos2x-150sinx+150κsinx+9ⅇ-xsin2x),(4.17)u2(x)=u1(x)+11200ℏ2(21598564225+300ⅇ-x-5376x65-552x2+160x3-50x4-2349κ2u1(x)+11200ℏ2-15ⅇ-2xκ-300ⅇ-xκ+1080xκ-300x2κ+100x3κ+10x5κu1(x)+11200ℏ2+350ⅇ-2x(-375κ+8ⅇx(25κ-177))cos2x+465169ⅇ-2xcos3x-750sinxu1(x)+11200ℏ2-3ⅇ-2xsinx+600ⅇ-xsinx+840κsinx-1800ⅇ-xκsinx-600ⅇ-xxκsinxu1(x)+11200ℏ2-3ⅇ-2xcosx(3-20ⅇx(-12-10x+25κ+20xκ+5x2κ)+5κsinx)u1(x)+11200ℏ2+406825ⅇ-xsin2x-84ⅇ-xκsin2x-405169ⅇ-2xsin3x).
The rest of components of the iteration formula (3.5) can be obtained in a similar way.
To find the value ofκ, substitute ℏ=-1and the boundary condition  at x=π/2 into the 5th-order approximation of u(x) to get κ=1.0000000668. After that, substituting the value of κ in the 5th-order approximation of  u(x)  and plotting the ℏ-curve, we can obtain the valid region of ℏ which is  -1.3<ℏ<-0.7  as shown in Figure 10.Figure 10
Theℏ-curves of u′(0) and u′′(0) which are corresponding to the 5th-order approximate HAM solution of (4.13) and (4.14). Dotted line: u′(0) and solid line: u′′(0).The graph of the 5th-order approximate HAM solution of (4.13) and (4.14) at ℏ=-1 together with the exact solution are depicted in Figure 11. It is evident from the figure that the 5th-order approximate HAM solution agrees with the exact solution in the interval [0,π/2].Figure 11
Solutions of (4.13) and (4.14). Dotted line: the 5th-order approximate HAM solution at ℏ=-1 and solid line: the exact solution.In Table2, the exact, relative, and residual errors have been calculated for various x in [0,π/2] to measure the extent of agreement between the 5th-order approximate HAM solution at ℏ=-1 and the exact solution. From the table, it can be seen that the HAM provides us with the accurate approximate solution of (4.13) and (4.14).Table 2
Exact, relative, and residual errors of (4.13) and (4.14) at ℏ=-1 andκ=1.0000000668.
x
Exact error
Relative error
Residual Error
0
.
1
4.973784
×
10
-
10
5.556241
×
10
-
10
1.112207
×
10
-
9
0.2
2.426173
×
10
-
11
3.104917
×
10
-
11
5.782878
×
10
-
9
0.3
3.202272
×
10
-
12
4.853278
×
10
-
12
5.440901
×
10
-
7
0.4
1.244961
×
10
-
11
2.341725
×
10
-
11
2.572090
×
10
-
6
0.5
1.030532
×
10
-
10
2.588255
×
10
-
10
8.420042
×
10
-
6
0.6
5.499366
×
10
-
10
2.109517
×
10
-
9
2.202234
×
10
-
5
0.7
2.259143
×
10
-
9
1.872873
×
10
-
8
4.964877
×
10
-
5
0.8
7.674839
×
10
-
9
3.716740
×
10
-
7
1.008886
×
10
-
4
0.9
2.263322
×
10
-
8
1.399558
×
10
-
7
1.900279
×
10
-
4
1.0
5.985469
×
10
-
8
1.987414
×
10
-
7
3.378840
×
10
-
4
⋮
⋮
⋮
⋮
π
/
2
4.370067
×
10
-
6
0.370067
×
10
-
6
4.598268
×
10
-
3Example 4.4.
Consider the following nonlinear Volterra IDE:(4.18)u′′(x)+∫0x(u(s))2ds+(x2-sinhx-14sinh2x)=0,0≤s,x≤1,
subject to the boundary conditions
(4.19)u(0)=0,u(1)=sinh(1).
The exact solution is u(x)=sinhx.
According to (3.6) and (2.5), we have
(4.20)Rm(u→m-1(x;κ))=d2dx2um-1(x;κ)+(x2-sinhx-14sinh2x)(1-χm)+∫0x(∑j=0m-1uj(s;κ)um-1-j(s;κ))ds.
Assume that the initial approximation of (4.18) has the form u0(x)=κx, which satisfies the initial conditions u0(0)=0 and u0′(0)=κ. Using the iteration formula (3.5), we can directly obtain the following approximations terms of the HAM series solution for (4.18) subject to these initial conditions:
(4.21)u1(x)=1120ℏ(x(135+10x2+2x4κ2)-15(8+coshx)sinhx),(4.22)u2(x)=u1(x)+13360ℏ2(3780x+280x3-13545xκ+126x5κ+83x7κ+56x5κ2+29x9κ3-6720xκcoshx-1052xκcosh2x-3360sinhx+20160κsinhx-210sinh2x+3154κsinh2x).
Similarly to the previous discussion, one can substitute ℏ=-1and the boundary condition at x=1 into the 5th-order approximation of u(x) to get κ=1.0000000063. Figure 12 shows the agreement between the 5th-order approximate HAM solution of (4.18) and (4.19) at ℏ=-1 and the exact solution.Figure 12
Solutions of (4.18) and (4.19). Dotted line: the 5th-order approximate HAM solution at ℏ=-1 and solid line: the exact solution.Since the value ofκ has been identified previously, the valid region for the values of ℏ can be obtained by the so-called ℏ-curve which correspond to the horizontal line as shown in Figure 13. Thus, it is clear from the figure that the valid region of ℏ is -1.2<ℏ<-0.7.Figure 13
Theℏ-curves of u′(1) and u′′(1) which are corresponding to the 5th-order approximate HAM solution of (4.18) and (4.19). Dotted line:u′(1) and solid line: u′′(1).The detailed data ofu(x) for (4.18) and (4.19) that includes the exact nodal values, the 5th-order approximate HAM solution nodal values, and the exact error are given in Table 3. It is clear that the accuracy obtained using HAM is in advanced by using only five terms approximations.Table 3
Numerical results of (4.18) and (4.19) at ℏ=-1 and κ=1.0000000063.
x
Exact solution
HAM solution
Exact error
0
.
1
0.1001667500
0.1001667500
5.760464
×
10
-
14
0.2
0.2013360025
0.2013360025
2.119789
×
10
-
13
0.3
0.3045202934
0.3045202934
5.767938
×
10
-
13
0.4
0.4107523258
0.4107523258
8.997664
×
10
-
14
0.5
0.5210953055
0.5210953055
2.401551
×
10
-
13
0.6
0.6366535821
0.6366535821
1.919631
×
10
-
12
0.7
0.7585837018
0.7585837018
3.099931
×
10
-
11
0.8
0.8881059822
0.8881059820
2.174120
×
10
-
10
0.9
1.0265167257
1.0265167244
1.282535
×
10
-
9
1.0
1.1752011936
1.1752011876
6.091291
×
10
-
9


## 5. Conclusion

The main concern of this work has been to propose an efficient algorithm for the solution of second-order IDEs with boundary conditions of the Fredholm and Volterra types. The goal has been achieved by extending the HAM to solve this class of boundary value problems. We can conclude that the HAM is a powerful and efficient technique in finding approximate solutions for linear and nonlinear boundary value problems of second-order IDEs of different types. The proposed algorithm produced a rapidly convergent series by choosing suitable values of the auxiliary parameterℏ. After computing several approximants and using the boundary conditions at the boundary points, we can easily determine the approximate solution.There are two important points to make here. First, HAM provides us with a simple way to adjust and control the convergence region of the series solution by introducing the auxiliary parameterℏ. Second, the results obtained by HAM are very effective and convenient in linear and nonlinear cases with less computational work. This confirms our belief that the efficiency of our technique gives it much wider applicability for general classes of linear and nonlinear problems.



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*Source: 365792-2012-09-19.xml*