# On a Solution of a Third Kind Mixed Integro-Differential Equation with Singular Kernel Using Orthogonal Polynomial Method

Authors: Ahmad Alalyani; M. A. Abdou; M. Basseem
Journal: Journal of Applied Mathematics
 (2023)

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## Abstract

This paper deals with the solution of a third kind mixed integro-differential equation (MIDE) in displacement type in spaceL2−1,1×C0,T,T<1. The singular kernel is modified to take a logarithmic form, while the kernels of time are continuous and positive functions. Using the separation of variables technique, we have a system of Fredholm integral equations (FIEs) that can be transformed into an algebraic system after using orthogonal polynomials. In all the previous researchers’ works, the time periods were divided, and the mixed equation transformed into an algebraic system of FIEs. While when using the separation method, we are able to obtain FIE with time coefficients, and these functions are described as an integral operator in time. Thus, we can study the behavior of the solution with the time dimension in a broader and deeper than the previous one. Some examples are given and discussed to show the performance and efficiency of the proposed methods.


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## Body

## 1. Introduction

Various problems in mathematical physics and contact problems in the theory of elasticity lead to an integral equation of the1st and 2nd kinds, (see Papov [1]). In thermoelasticity, Abdou and Basseem in [2] derived a closed form of Gaursat functions for first and second fundamental problems with variant time. Basseem and Alalyani in [3] used Toeplitz and Nystrom methods on the solution of quadratic nonlinear integral equation with different singular kernels. The theory of singular integral equations, developed by the scholar Muskhelishvili in [4], has assumed different techniques and increased important applications in different areas of science. Orthogonal polynomials are widely used in many areas such as airfoil (see [5]), polymer physics (see [6]), fracture mechanics (see [7]), contact radiation, and molecular conduction (see [8]). Abdou et al. in [9] used orthogonal polynomials method to discuss the solution of an MIE with singular kernel in some contact problems. Hafez and Youssri in [10] applied the Legendre-Chebyshev collocation method for solving two-dimensional linear and nonlinear mixed Volterra-Fredholm integral equation (MVFIE). Tohidi and Samadi in [11] used Legendre polynomial in optimal control of nonlinear Volterra integral equation (NLVIE). Nemati et al. in [12] applied Legendre polynomial in a class of two-dimensional nonlinear VIE. Khader et al. in [13, 14] used the Chybyshev polynomial method on the fractional diffusion equation, and they also used Legendre polynomials to solve Ricatii, logistic and delay DE with variable coefficients.For the third kind of MIDE problem, we will use two orthogonal polynomials for the first time to solve these types of problems. Then, we will compare the numerical results and errors between them.Consider the bounded differential operator in a general form(1)Λ=μ0+μ1∂∂t+μ2∂2∂t2+⋯+μn∂n∂tn.Consider MIDE in the form(2)LxΛφx,t−λ1t∫−11kx−yϖφy,tdy−λ2t∫0t∫−11Gt,τkx−yϖφy,τdydτ−λ3t∫0tVt,τφx,τdτ=fx,t,kν=12∫−∞∞tanhuuejuv,v=x−yϖ,j2=−1,ϖ∈0,∞.The functionsGt,τ and Vt,τ are considered as a continuous time functions in the space C0,T,T<1. λit,i=1,2,3 are time parameters. The kernel of position kx−y/ϖ is singular, while the unknown function φx,t behaves like fx,t; that is, the free term belongs to the class L2−1,1×C0,T. The function Lx determines the kind of Eq. (2). For the displacement and contact problems in the theory of elasticity, we assume the condition
(3)∫−11φx,tdx=Pt,where Pt is the external pressure that affects the surface of the elastic material under study, while in the nuclear and mathematical physics problems, we let
(4)φ±1,t=0.Many different and special cases with various applications can be derived from Eq. (2) as seen in some of previous references as an especial cases.The kernel of Eq. (2) can be written in the form (see [15])
(5)kν=12∫−∞∞tanhuuejuv=−lntanhπν4,ν=x−yϖ,ϖ∈0,∞.Sinceϖ⟶∞ and x−y are very small, ν=x−y/ϖ, d=ln4ϖ/π(6)lntanhπx−y4ϖ=lnπx−y4ϖ=lnx−y−d.Hence, Eq. (2) becomes
(7)LxΛφx,t+λ1t∫−11lnx−y−dφy,tdy+λ2t∫0t∫−11Gt,τlnx−y−dφy,τdydτ−λ3t∫0tVt,τφx,τdτ=fx,t.The importance of Eq. (7) comes from its special cases:
(i)
Whenλ2=λ3=0, a third kind FIE is obtained.(ii)
Whenλ2=0, we have an MFVIE of a third kind.(iii)
Whenλ1=λ3=0, a third kind VFIE is attained.The structure of this article is as follows: In Section2, a separation method is used to obtain a system of FIEs for the third kind of MIDE problem. In Section 3, we represent the unknown function in two different forms, Legendre polynomials and Chebyshev polynomials, and then, we use a technique of orthogonal polynomials to discuss the solution of each system. In Section 4, we illustrate two applications as numerical test examples to show the performance and efficiency of the proposed methods. In Section 5, we present our conclusion.


## 2. Separation Method

Consider(8)φx,t=ψxAt,A0≠0,fx,t=gxAt.Eq. (7) becomes
(9)LxψxΘAt+λ1tAt∫−11lnx−y−dψydy+λ2t∫0tGt,τAτdτ∫−11lnx−y−dψydy−λ3tψx∫0tVt,τAτdτ=gxAt,where
(10)Θ=μ0+μ1ddt+μ2d2dt2+⋯+μndndtn,where ΘAt<D1,D1 is a constant.In order to guarantee the existence of a unique solution of Eq. (9), we assume the following:
(i)
The singular kernel satisfies the discontinuous condition(11)∫−11∫−11k2x−yϖdxdy1/2<c,cisaconstant.(ii)
While the functionfx,t∈L2−1,1×C0,T and its normality are defined as fx,t=max0≤t≤T∫0t∫−11f2x,τdx1/2dτ, the function Lx belongs to the space L2−1,1 in which Lx<E, where E is a constant.(iii)
For all values oft,τ∈0,T, the given functions of time are bounded, in which Gt,τ<B1,Vt,τ<B2,λit<λi,i=1,2,3 and At=max0≤t≤T∫0tAτdτ<D2, where B1,B2,D2 and λi are constants.(iv)
The unknown functionφx,t satisfies Ho¨lder condition with respect to time and Lipchitz condition with respect to position.LetW be an operator defined as
(12)Wψx=1ηx,tgxAt−λ1tAt∫−11lnx−y−dψydy−λ2t∫0tAτGt,τdτ∫−11lnx−y−dψydy,where
(13)ηx,t=LxψxΘAt−λ3tψx∫0tVt,τAτdτ.Theorem 1.
Under the above assumptions, the solution of the system of IE (9) is unique under the condition
(14)cλ2D2B1+λ1ED1−λ3B2D2<1in L2−1,1 for every t∈C0,T,T<1.Proof.
Since(15)LxΘAt−λ3tψx∫0tVt,τAτdτ≤ηx,t≤LxΘAt+λ3tψx∫0tVt,τAτdτ.
Using Caushy Schwarz inequality and previous conditions, we get(16)ηx,t≥ED1−λ3B2D2.
Therefore,(17)Wψx≤1ED1−λ3B2D2gxD2+cψxλ2D2B1+λ1.It is obvious that the operatorW maps the ball Br∈L2−1,1 into itself where
(18)r=σ1−δ,σ=gxD2ED1−λ3B2D2,δ=cλ2D2B1+λ1ED1−λ3B2D2.The inequality (17) involves the boundedness of the operator W under the condition δ<1.Let two functionsψ1x and ψ2x be two solutions of (9), and then, the formula (12) leads to
(19)Wψ1x−Wψ2x≤1ηx,tλ1tAt∫−11lnx−y−dψ1y−ψ2ydy+λ2t∫0tGt,τAτdτ∫−11lnx−y−dψ1y−ψ2ydy.Using conditions (i)-(iv) and Cauchy Schwarz inequality, we deduce that(20)Wψ1x−Wψ2x≤cλ2D2B1+λ1ED1−λ3B2D2ψ1x−ψ2x.It follows that forδ<1,W is a contraction operator of a system (9). Hence, there exists a unique solution in L2−1,1 by a Banach fixed point theorem for every t∈C0,t. It is easy to show that the MIE (2) can be changed into the third kind FIE with time coefficients
(21)LxΘAtAt−Ntψx+Mt∫−11lnx−y−dψydy=gx,where
(22)Nt=λ3t∫0tVt,τAτdτAt,(23)Mt=λ1t+λ2tAt∫0tAτGt,τdτ.The formula (21) leads to discuss the following cases:AtLx=0, Eq. (2) is called MIE of the first kind, but using the separation technique, we get a system of Fredholm integral equation of the second kind depends on time. This system has a unique solution under the condition cλ1+λ2D2B1/λ3B2D2<1.IfLx is a constant function, Eq. (2) is called MIE of the second kind, while Lx is a function of x, and we have a system of Fredholm integral equation of the third kind, in which its uniqueness holds under the condition cλ1+λ2D2B1/ED1−λ3B2D2<1.In all previous research, the time periods in the mixed equation were divided, and this equation transformed into an algebraic system for FIEs (see [16, 17]). Here, we use the separation technique method to get FIEs with time parameters that are described as an integrated operator in time.


## 3. Method of Solution

### 3.1. Legendre Polynomial Form

To obtain the solution of Eq. (21), we assume the unknown function ψx in the Legendre polynomial form
(24)ψx=∑n=0∞CnPnx.Since it is difficult to discuss the numerical solution using formula (24), then it can be truncated to
(25)ψNx=∑n=0NCnPnx,limN⟶∞φNx=φx,where Cn are constants and Pnx are Legendre polynomials that satisfy the orthogonal relation (see [18])
(26)∫−11PnxPmxdx=22n+1,n=m,0,n≠m.By substituting (25) in Eq. (21), for n=0, we can easily show that
(27)C0=Atg0LxΘAt−NtAt−MtAt∫−11lnx−y−ddy.By differentiating Eq. (21) with respect to x, for n>0, we get
(28)LxΘAtAt−Ntψ′x+L′xΘAtAtψx+Mt∫−11ψyx−ydy=g′x.Eq. (24) through the use of the following relation
(29)Pnmx=1−x2m/2dmxmPnxleads to the following relation
(30)ψ′x=∑n=0∞Cn1−x2−1/2Pn1x.Here,Pnm,n,m≥0 are the associated Legendre polynomials of the first kind that satisfy the following orthogonal relations (see [18])
(31)∫−11PnlxPmlxdx=2n+l!n−l!2n+1,n=m,0,n≠m,(32)1−x2Pnlx=12n+1Pn+1l−1x−Pn+1l+1x.In sense of Eq. (24) and Eq. (30), the function g′x of Eq. (28) can be represented as (see [18])
(33)g′x=∑n=1∞g^n1−x2Pn1x.The constant coefficientsg^n can be determined using Eq. (31) in the following form
(34)g^n=n−1!2n+12n+1!∫−111−x2g′xPn1xdx,n≠0.Using the Legendre polynomial of the second kind with its relation(35)Qnx=12∫−11Pnyx−ydy,Eq. (28) through the use of (30) and (33) yields
(36)LxΘAtAt−Nt∑n=1∞Cn1−x2Pn1x+L′xΘAtAt∑n=1∞CnPnx+2Mt∑n=1∞CnQnx=∑n=1∞g^n1−x2Pn1x.Multiplying both sides of Eq. (36) by 1−x2Pn1x, integrating from −1 to 1, and through the use of Eq. (31), Eq. (32), and the following relation (see [18])
(37)∫−111−x2QnxPm1xdx=−2nm+11+−1nm−n−1m−n+1m+nm+n+2,n≠m±1,0,n=m±1,we get
(38)LxΘAtAt−NtCm−2Mt∑n=1∞χm,nCn=g^m,(39)χm,n=2m+1n+11+−1mn−m−1m+1n−m+1m+nm+n+2,where m=1,2,3,⋯


#### 3.1.1. Convergence of Algebraic System

The convergence of the algebraic system (38) can be derived from the following:
(40)∑n=1∞2m+1n+11+−1mn−m−1m+1n−m+1m+nm+n+2<∑n=1∞22m+1m+n+24.Since the series∑n=1∞22m+1/m+n+24 behaves like ∑n=1∞1/n4, χm,n⟶0, where n⟶∞ and m=1,2,3,⋯ The solution of this convergent linear algebraic system can be determined for Cm, and then, approximated solution φNx,t is obtained.


### 3.2. Chebyshev Polynomial Form

Recall Eq. (21) and consider ψx in Chebyshev form as
(41)ψx=∑n=0∞anTnx1−x2.To obtain the solution numerically, the formula (41) can be truncated to
(42)ψNx=∑n=0NanTnx1−x2,limN⟶∞ψNx=ψx,where an are constants and Tnx are Chebyshev polynomials that satisfy the following (see [18]):
(i)
Algebraic formula:(43)TnxTmx=12Tn+mx+Tn−mx.(ii)
Integral relations:(44)∫−11Tnxdx=21−n2,n=0,2,4,⋯,0,n=1,3,5,⋯,(45)∫−11lnx−y−dTny1−y2dy=πln2−d,n=0,πTnxn,n≥1.(iii)
Orthogonality rule:(46)∫−11TnxTmx1−x2dx=π,n=m=0,π2,n=m≠0,0,n≠m.In sense of Eq. (41), the two given functions Lx and gx are written as
(47)Lx=∑n=0∞LnTnx,Ln=2π∫−11LxTnx1−x2dx,n≠0,1π∫−11Lx1−x2dx,n=0,(48)gx=∑n=0∞gnTnx1−x2,gn=2π∫−11gxTnxdx,n≠0,1π∫−11gxdx,n=0.By substituting (41), (47), and (48), Eq. (21) through the use of (43) takes the form
(49)ΘAt2an∑m=0∞LmTn+m+Tn−m1−x2−NtanTnx1−x2+Mtπln2−da0,n=0πTnxn,n≥1=AtgnTnx1−x2.Multiplying byTlx, integrating with respect to x from -1 to 1, and using (43) and (46), we have the following cases:
(i)
Forl=n=0,m=0, we have
(50)a0=g0AtΘAtL0−Nt+Mtln2−d.(ii)
Forl=n=0,m≠0, we have
(51)a0=2g0AtΘAt∑m=0∞Lm−Nt+Mtln2−d.(iii)
Forl=n,m=0, we have
(52)an=gnAtΘAtL0−Nt+Mt41−n2/1−4n2.(iv)
Forn≠0,m≠0, we have
(53)ΘAtan∑m=0∞Lmδl,n+m+δl,n−m−2Ntanδl,n+8Mt℘n,lnan=2Atgnδl,n,where
(54)℘n,l=1−n2−l21−n2−l22−4n2l2,n+l=0,2,4,⋯,0,n+l=1,3,5,⋯The formula (53) represents LAS with time coefficients. The unknown constants an can be determine after the following cases:
(a)
Whenk=n+m, we have
(55)ΘAtan∑m=0MLm−2Ntan+8Mt∑m=0M℘1n,mnan=2Atgn,where
(56)℘1n,m=1−2n2−2nm−m21−m21−2n+m2,m=0,2,4,⋯,0,m=1,3,5,⋯(b)
Whenk=n−m, we have
(57)ΘAtan∑m=0MLm−2Ntan+8Mt∑m=0M℘2n,mnan=2Atgn,where
(58)℘2n,m=1−2n2+2nm−m21−m21−2n−m2,m=0,2,4,⋯,0,m=1,3,5,⋯The previous algebraic systems have a unique solution which can be obtained after determining the constants,an.


## 3.1. Legendre Polynomial Form

To obtain the solution of Eq. (21), we assume the unknown function ψx in the Legendre polynomial form
(24)ψx=∑n=0∞CnPnx.Since it is difficult to discuss the numerical solution using formula (24), then it can be truncated to
(25)ψNx=∑n=0NCnPnx,limN⟶∞φNx=φx,where Cn are constants and Pnx are Legendre polynomials that satisfy the orthogonal relation (see [18])
(26)∫−11PnxPmxdx=22n+1,n=m,0,n≠m.By substituting (25) in Eq. (21), for n=0, we can easily show that
(27)C0=Atg0LxΘAt−NtAt−MtAt∫−11lnx−y−ddy.By differentiating Eq. (21) with respect to x, for n>0, we get
(28)LxΘAtAt−Ntψ′x+L′xΘAtAtψx+Mt∫−11ψyx−ydy=g′x.Eq. (24) through the use of the following relation
(29)Pnmx=1−x2m/2dmxmPnxleads to the following relation
(30)ψ′x=∑n=0∞Cn1−x2−1/2Pn1x.Here,Pnm,n,m≥0 are the associated Legendre polynomials of the first kind that satisfy the following orthogonal relations (see [18])
(31)∫−11PnlxPmlxdx=2n+l!n−l!2n+1,n=m,0,n≠m,(32)1−x2Pnlx=12n+1Pn+1l−1x−Pn+1l+1x.In sense of Eq. (24) and Eq. (30), the function g′x of Eq. (28) can be represented as (see [18])
(33)g′x=∑n=1∞g^n1−x2Pn1x.The constant coefficientsg^n can be determined using Eq. (31) in the following form
(34)g^n=n−1!2n+12n+1!∫−111−x2g′xPn1xdx,n≠0.Using the Legendre polynomial of the second kind with its relation(35)Qnx=12∫−11Pnyx−ydy,Eq. (28) through the use of (30) and (33) yields
(36)LxΘAtAt−Nt∑n=1∞Cn1−x2Pn1x+L′xΘAtAt∑n=1∞CnPnx+2Mt∑n=1∞CnQnx=∑n=1∞g^n1−x2Pn1x.Multiplying both sides of Eq. (36) by 1−x2Pn1x, integrating from −1 to 1, and through the use of Eq. (31), Eq. (32), and the following relation (see [18])
(37)∫−111−x2QnxPm1xdx=−2nm+11+−1nm−n−1m−n+1m+nm+n+2,n≠m±1,0,n=m±1,we get
(38)LxΘAtAt−NtCm−2Mt∑n=1∞χm,nCn=g^m,(39)χm,n=2m+1n+11+−1mn−m−1m+1n−m+1m+nm+n+2,where m=1,2,3,⋯


### 3.1.1. Convergence of Algebraic System

The convergence of the algebraic system (38) can be derived from the following:
(40)∑n=1∞2m+1n+11+−1mn−m−1m+1n−m+1m+nm+n+2<∑n=1∞22m+1m+n+24.Since the series∑n=1∞22m+1/m+n+24 behaves like ∑n=1∞1/n4, χm,n⟶0, where n⟶∞ and m=1,2,3,⋯ The solution of this convergent linear algebraic system can be determined for Cm, and then, approximated solution φNx,t is obtained.


## 3.1.1. Convergence of Algebraic System

The convergence of the algebraic system (38) can be derived from the following:
(40)∑n=1∞2m+1n+11+−1mn−m−1m+1n−m+1m+nm+n+2<∑n=1∞22m+1m+n+24.Since the series∑n=1∞22m+1/m+n+24 behaves like ∑n=1∞1/n4, χm,n⟶0, where n⟶∞ and m=1,2,3,⋯ The solution of this convergent linear algebraic system can be determined for Cm, and then, approximated solution φNx,t is obtained.


## 3.2. Chebyshev Polynomial Form

Recall Eq. (21) and consider ψx in Chebyshev form as
(41)ψx=∑n=0∞anTnx1−x2.To obtain the solution numerically, the formula (41) can be truncated to
(42)ψNx=∑n=0NanTnx1−x2,limN⟶∞ψNx=ψx,where an are constants and Tnx are Chebyshev polynomials that satisfy the following (see [18]):
(i)
Algebraic formula:(43)TnxTmx=12Tn+mx+Tn−mx.(ii)
Integral relations:(44)∫−11Tnxdx=21−n2,n=0,2,4,⋯,0,n=1,3,5,⋯,(45)∫−11lnx−y−dTny1−y2dy=πln2−d,n=0,πTnxn,n≥1.(iii)
Orthogonality rule:(46)∫−11TnxTmx1−x2dx=π,n=m=0,π2,n=m≠0,0,n≠m.In sense of Eq. (41), the two given functions Lx and gx are written as
(47)Lx=∑n=0∞LnTnx,Ln=2π∫−11LxTnx1−x2dx,n≠0,1π∫−11Lx1−x2dx,n=0,(48)gx=∑n=0∞gnTnx1−x2,gn=2π∫−11gxTnxdx,n≠0,1π∫−11gxdx,n=0.By substituting (41), (47), and (48), Eq. (21) through the use of (43) takes the form
(49)ΘAt2an∑m=0∞LmTn+m+Tn−m1−x2−NtanTnx1−x2+Mtπln2−da0,n=0πTnxn,n≥1=AtgnTnx1−x2.Multiplying byTlx, integrating with respect to x from -1 to 1, and using (43) and (46), we have the following cases:
(i)
Forl=n=0,m=0, we have
(50)a0=g0AtΘAtL0−Nt+Mtln2−d.(ii)
Forl=n=0,m≠0, we have
(51)a0=2g0AtΘAt∑m=0∞Lm−Nt+Mtln2−d.(iii)
Forl=n,m=0, we have
(52)an=gnAtΘAtL0−Nt+Mt41−n2/1−4n2.(iv)
Forn≠0,m≠0, we have
(53)ΘAtan∑m=0∞Lmδl,n+m+δl,n−m−2Ntanδl,n+8Mt℘n,lnan=2Atgnδl,n,where
(54)℘n,l=1−n2−l21−n2−l22−4n2l2,n+l=0,2,4,⋯,0,n+l=1,3,5,⋯The formula (53) represents LAS with time coefficients. The unknown constants an can be determine after the following cases:
(a)
Whenk=n+m, we have
(55)ΘAtan∑m=0MLm−2Ntan+8Mt∑m=0M℘1n,mnan=2Atgn,where
(56)℘1n,m=1−2n2−2nm−m21−m21−2n+m2,m=0,2,4,⋯,0,m=1,3,5,⋯(b)
Whenk=n−m, we have
(57)ΘAtan∑m=0MLm−2Ntan+8Mt∑m=0M℘2n,mnan=2Atgn,where
(58)℘2n,m=1−2n2+2nm−m21−m21−2n−m2,m=0,2,4,⋯,0,m=1,3,5,⋯The previous algebraic systems have a unique solution which can be obtained after determining the constants,an.


## 4. Numerical Results

In this section, some numerical examples are considered to show the accuracy and efficiency of the proposed methods.


### 4.1. Application 1

Consider the MIE(59)Lx0.04+0.35d/dt+0.87d2/dt2AtAt−Ntψx+Mt∫−11lnx−y−0.01ψydy=11+x2,where
(60)Nt=0.3/1−tt∫0tτAτdτAt,Mt=0.3t+0.4t4∫0tτ2AτdτAt.In Tables1 and 2, the solution φ=ψxAt and its corresponding error are determined by using Legendre polynomials L and Chebyshev polynomials C.Table 1
N=17,Lx=1+x3.
At=ln1.3+tAt=etTimexφLELφCECφLELφCEC0.7−0.82.069011.3e−50.070842.9e−44.788601.0e−5−0.085122.9e−4−0.42.090681.0e−50.107891.5e−43.187223.6e−6−0.111141.4e−400.671091.1e−50.126673.2e−4−1.134953.5e−6−0.127403.1e−40.40.178181.4e−60.107891.5e−4−0.252901.0e−6−0.111141.4e−40.80.060371.3e−60.070842.9e−4−0.079681.1e−6−0.085122.9e−40.3−0.80.188365.3e−60.003421.3e−50.568754.3e−6−0.006692.5e−5−0.40.081501.0e−60.004896.8e−60.364648.0e−7−0.009221.2e−50.00.026572.7e−70.005691.4e−5−0.080084.2e−7−0.010672.7e−50.40.008885.5e−80.004896.8e−6−0.019779.6e−8−0.009221.2e−50.80.003015.6e−80.003421.3e−5−0.006241.0e−7−0.006692.5e−50.03−0.8−1.8e−43.9e−81.9e−67.3e−90.002993.0e−8−6.5e−62.5e−8−0.46.3e−53.8e−102.7e−63.7e−9−0.000261.2e−9−9.2e−61.2e−80.01.6e−51.4e−103.1e−67.9e−9−5.7e−54.7e−10−1.0e−52.7e−80.45.1e−62.9e−112.7e−63.7e−9−1.7e−51.0e−10−9.2e−61.2e−80.81.7e−63.0e−111.9e−67.3e−9−5.9e−61.0e−8−6.5e−62.5e−8Table 2
N=17,Lx=1+x3.
At=11+t4At=tcostTimexφLELφCECφLELφCEC0.7−0.84.042481.0e−5−0.023318.7e−52.048051.3e−5−0.063880.00022−0.4−1.666202.3e−6−0.032184.4e−51.960354.2e−6−0.085630.000110−0.217091.4e−6−0.037249.4e−5−0.887403.2e−6−0.098620.000240.4−0.064573.3e−7−0.032184.4e−5−0.193088.3e−7−0.085630.000110.8−0.021233.6e−7−0.023318.7e−5−0.060119.3e−7−0.063880.000220.3−0.81.140354.4e−6−0.007662.8e−50.200386.1e−60.001947.5e−6−0.40.293038.7e−7−0.010541.4e−50.050354.7e−70.002773.8e−60.0−0.098614.8e−7−0.012193.1e−50.015731.5e−70.003228.1e−60.4−0.023011.1e−7−0.010541.4e−50.005143.0e−80.002773.8e−60.8−0.007191.1e−7−0.007662.8e−50.001733.1e−80.001947.5e−60.03−0.80.001905.4e−88.4e−53.3e−76.8e−51.3e−91.1e−74.3e−10−0.4−0.001636.4e−81.2e−41.7e−73.8e−62.2e−111.5e−72.2e−100.05.9e−47.6e−91.4e−43.6e−79.6e−78.3e−121.8e−74.6e−100.42.0e−41.4e−91.2e−41.7e−73.0e−71.7e−121.5e−72.2e−100.87.1e−51.4e−98.4e−53.3e−71.0e−71.8e−121.1e−74.3e−10


### 4.2. Application 2

Letμ0=0.04,μ1=0.35,μ2=0.87, λ1=0.3t,λ2=0.4t2, and λ3=0.3/1−t in Eq. (21). In Table 3, the solution φ=ψxAt and its corresponding error are determined by using Legendre polynomials L and Chebyshev polynomials C, where gx=1+x.Table 3
Vt,τ=tτ,Gt,τ=t2τ2,Lx=1+x3,N=17,At=ln1.3+t.
TimexφLELφCEC0.7−0.83.421891.3e−30.047211.7e−4−0.43.166887.2e−30.094951.7e−40.00.498386.0e−50.125713.5e−40.40.248001.5e−50.149015.1e−40.80.142881.7e−50.165498.2e−40.3−0.8−5.054830.003810.002477.7e−6−0.40.051689.7e−50.004378.0e−60.00.025431.9e−60.005691.5e−50.40.012248.3e−70.006732.3e−50.80.006757.1e−70.007603.7e−50.03−0.8−0.000252.1e−61.4e−64.2e−9−0.45.5e−54.6e−82.4e−64.4e−90.01.6e−51.4e−93.1e−68.7e−90.47.0e−64.8e−103.7e−61.2e−80.83.7e−63.8e−104.2e−62.0e−8The following figures representφLx,t,φCx,t and its corresponding errors ELx,t,ECx,t.


## 4.1. Application 1

Consider the MIE(59)Lx0.04+0.35d/dt+0.87d2/dt2AtAt−Ntψx+Mt∫−11lnx−y−0.01ψydy=11+x2,where
(60)Nt=0.3/1−tt∫0tτAτdτAt,Mt=0.3t+0.4t4∫0tτ2AτdτAt.In Tables1 and 2, the solution φ=ψxAt and its corresponding error are determined by using Legendre polynomials L and Chebyshev polynomials C.Table 1
N=17,Lx=1+x3.
At=ln1.3+tAt=etTimexφLELφCECφLELφCEC0.7−0.82.069011.3e−50.070842.9e−44.788601.0e−5−0.085122.9e−4−0.42.090681.0e−50.107891.5e−43.187223.6e−6−0.111141.4e−400.671091.1e−50.126673.2e−4−1.134953.5e−6−0.127403.1e−40.40.178181.4e−60.107891.5e−4−0.252901.0e−6−0.111141.4e−40.80.060371.3e−60.070842.9e−4−0.079681.1e−6−0.085122.9e−40.3−0.80.188365.3e−60.003421.3e−50.568754.3e−6−0.006692.5e−5−0.40.081501.0e−60.004896.8e−60.364648.0e−7−0.009221.2e−50.00.026572.7e−70.005691.4e−5−0.080084.2e−7−0.010672.7e−50.40.008885.5e−80.004896.8e−6−0.019779.6e−8−0.009221.2e−50.80.003015.6e−80.003421.3e−5−0.006241.0e−7−0.006692.5e−50.03−0.8−1.8e−43.9e−81.9e−67.3e−90.002993.0e−8−6.5e−62.5e−8−0.46.3e−53.8e−102.7e−63.7e−9−0.000261.2e−9−9.2e−61.2e−80.01.6e−51.4e−103.1e−67.9e−9−5.7e−54.7e−10−1.0e−52.7e−80.45.1e−62.9e−112.7e−63.7e−9−1.7e−51.0e−10−9.2e−61.2e−80.81.7e−63.0e−111.9e−67.3e−9−5.9e−61.0e−8−6.5e−62.5e−8Table 2
N=17,Lx=1+x3.
At=11+t4At=tcostTimexφLELφCECφLELφCEC0.7−0.84.042481.0e−5−0.023318.7e−52.048051.3e−5−0.063880.00022−0.4−1.666202.3e−6−0.032184.4e−51.960354.2e−6−0.085630.000110−0.217091.4e−6−0.037249.4e−5−0.887403.2e−6−0.098620.000240.4−0.064573.3e−7−0.032184.4e−5−0.193088.3e−7−0.085630.000110.8−0.021233.6e−7−0.023318.7e−5−0.060119.3e−7−0.063880.000220.3−0.81.140354.4e−6−0.007662.8e−50.200386.1e−60.001947.5e−6−0.40.293038.7e−7−0.010541.4e−50.050354.7e−70.002773.8e−60.0−0.098614.8e−7−0.012193.1e−50.015731.5e−70.003228.1e−60.4−0.023011.1e−7−0.010541.4e−50.005143.0e−80.002773.8e−60.8−0.007191.1e−7−0.007662.8e−50.001733.1e−80.001947.5e−60.03−0.80.001905.4e−88.4e−53.3e−76.8e−51.3e−91.1e−74.3e−10−0.4−0.001636.4e−81.2e−41.7e−73.8e−62.2e−111.5e−72.2e−100.05.9e−47.6e−91.4e−43.6e−79.6e−78.3e−121.8e−74.6e−100.42.0e−41.4e−91.2e−41.7e−73.0e−71.7e−121.5e−72.2e−100.87.1e−51.4e−98.4e−53.3e−71.0e−71.8e−121.1e−74.3e−10


## 4.2. Application 2

Letμ0=0.04,μ1=0.35,μ2=0.87, λ1=0.3t,λ2=0.4t2, and λ3=0.3/1−t in Eq. (21). In Table 3, the solution φ=ψxAt and its corresponding error are determined by using Legendre polynomials L and Chebyshev polynomials C, where gx=1+x.Table 3
Vt,τ=tτ,Gt,τ=t2τ2,Lx=1+x3,N=17,At=ln1.3+t.
TimexφLELφCEC0.7−0.83.421891.3e−30.047211.7e−4−0.43.166887.2e−30.094951.7e−40.00.498386.0e−50.125713.5e−40.40.248001.5e−50.149015.1e−40.80.142881.7e−50.165498.2e−40.3−0.8−5.054830.003810.002477.7e−6−0.40.051689.7e−50.004378.0e−60.00.025431.9e−60.005691.5e−50.40.012248.3e−70.006732.3e−50.80.006757.1e−70.007603.7e−50.03−0.8−0.000252.1e−61.4e−64.2e−9−0.45.5e−54.6e−82.4e−64.4e−90.01.6e−51.4e−93.1e−68.7e−90.47.0e−64.8e−103.7e−61.2e−80.83.7e−63.8e−104.2e−62.0e−8The following figures representφLx,t,φCx,t and its corresponding errors ELx,t,ECx,t.


## 5. Conclusion

In this paper, we apply a separation of variable technique in an MIDE to have a system of FIE with time operator coefficients. The solution is then obtained after using the orthogonal polynomials method with two different polynomials (see Figures1–4). According to numerical results, the Legendre polynomial L has a higher efficiency than the Chebyshev polynomial C. From all Tables 1–3, corresponding errors show that the Legendre polynomial is a more powerful solver for these kinds of equations. We also observed that the solution behaves like its corresponding given function; that is, fx,t is even in the first example; using the Chebyshev orthogonal polynomial method, its solution is symmetric in the x domain.Figure 1
φC, where gx=1+x,At=ln1.3+t.Figure 2
EC, where gx=1+x, At=ln1.3+t.Figure 3
φL, where gx=1+x, At=ln1.3+t.Figure 4
EL, where gx=1+x, At=ln1.3+t.



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