# An Asymptotic Model for Solving Mixed Integral Equation in Position and Time

Authors: A. R. Jan
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this paper, we considered a mixed integral equation(MIE) of the second kind in the space L2−b,b×C0,T,T<1. The kernel of position has a singularity and takes some different famous forms, while the kernels of time are positive and continuous. Using an asymptotic method of separating the variables, we have a Fredholm integral equation (FIE) in position with variable parameters in time. Then, using the Toeplitz matrix method (TMM), we obtain a linear algebraic system (LAS) that can be solved numerically. Some applications with the aid of the maple 18 program are discussed when the kernel takes Coleman function, Cauchy kernel, Hilbert kernel, and a generalized logarithmic function. Also the error estimate, in each case, is computed.


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## Body

## 1. Introduction

The theory ofIEs has many wide applications in different sciences, for example, in the contact displacement problems, in the theory of elasticity (see Abdou [1]), in the spectrum of mathematical physics problems (see Abdou and Alharbi [2]), in the thermomechanics (see Chirita et al., [3]) and in models (see Chirita [4]). For this aim, many authors have been interested to establish different kinds of analytical and numerical methods for solving the IEs of different types and different kinds with the continuous or discontinuous kernel. In [5], Ladopulas used the collocation evaluation method to discuss the solution of a singular integro-differential equation in Banach space. In [6], Mirzaee and Hoseini used the Fibonacci collocation method for solving Volterra–Fredholm integral equations with continuous kernels. In [7], Khairullina and Makletsov used the wavelet-collocation method to solve an integral equation with a singular kernel. Mirzaee and Hadadiyan[8] discussed the solution of the nonlinear class of mixed Volterra–Fredholm integral equations using an operational matrix. In addition, the same authors Mirzaee and Hadadiyan in [9], used three-dimensional triangular functions and their applications for solving nonlinear mixed Volterra–Fredholm integral equations. Mosa, et al. [10] applied the Adomian decomposition method to discuss the solution of a mixed integral equation in position and time with a phase-lag term in the position. In [11], Zaheer, et al. applied the meshless method to discuss the solution of the Volterra integral equation in one dimensional with an oscillatory kernel. Abd El-Rahman [12] considered a mixed integral equation with a weak singular kernel.In this paper, we review a mixed integral equation in position and time in a general form, so that all previously solved equations of this type are considered special cases of this work. In addition, we consider the kernel of position as a singular term. Numerical applications have been made when the kernel of the position takes many forms of singular form, which are famous in different sciences. Due to the difficulty of solving the singular integral equations, the Toeplitz matrix method will be used. This method is one of the best methods to solve the singular integral equations, where the singular of integral term disappeared, and we have directed the numerical results.Consider, in the space of position and timeL2−b,b×C0,T,T<1, the following MIE:(1)μ1Φx;t=fx;t+λ1∫0t∫−bbkx−yζt,τΦy;τdydτ+λ2∫−bbkx−yΦy;tdy+λ3∫0tξt,τΦx;τdτ;x=x¯x1,x2,...,xn,y=y¯y1,y2,…,yn,where f,k,ζ,ξ are known functions, while Φ is an unknown function. The constant μ1 defines the kind of the IE, while the constant λi,i=1,2,3,, may be complex and have several physical meanings.We can adapt equation (1) in the integral operator form(2)W¯Φx;t=1μ1fx;t+1μ1WΦx;t,μ1≠0,(3)WΦx;t=λ1∫0t∫−bbζt,τkx−yΦy;τdydτ+λ2∫−bbkx−yΦy;tdy+λ3∫0tξt,τΦx;τdτ.In the remaining part of the paper, using the separation of variables method, we have quadratic SFIEs. Then, after using the TMM, as a famous numerical method for solving singular integral equations, see Abdou, et al. [13], we have LAS which can be solved numerically. Some numerical results are calculated, and the error in each case is computed.


## 2. Existence of a Unique Solution

To prove the existence of a unique solution of equation (1), we assume the following:(i)
The kernel of position satisfies the discontinuity condition(4)∫−bb∫−bbkx−y2dxdy1/2≤cc−constant.(ii)
The two kernels of Volterraζt,τ,ξt,τ belong to the class C0,T, 0≤τ≤t≤T<1, respectively, and satisfy for two constants N and M, and the conditions are as follows:(5)ζt,τ≤N,ξt,τ≤M.(iii)
The given functionfx,t is continuous in L2−b,b×C0,T with its partial derivatives with respect to position x and time t, and its norm is defined as(6)fx;tL2a,b×C0,T=max0≤t≤T∫0t∫−bbfx;τ2dx1/2dτ=H,His constant.Whenμ1≠0, the existence of a unique solution of equation (1) can be proved using two different methods: Banach fixed point theorem or Picard method while Picard’s method fails when μ1=0. In this case, we use Banach fixed point theorem. In Banach’s fixed point theorem, we must prove that the integral operator of equation (2) is bounded and continuous. And then, if the constant of continuity is less than one, the integral operator is contracting. Hence, we have a unique solution.Theorem 1.
TheMIE (1) has a unique solution in the space L2−b,b×C0,T, under the condition(7)μ1≥Nc2bλ1T+λ2+λ3TM;T=maxt0≤t≤τ.
The idea is to construct the solution as a sequence of functionsΦnx;t as n⟶∞,i.e, and we assume the solution Φx,t in the form(8)Φx;t=limn⟶∞Φnx;t,∀t∈0,T,T<1,where(9)Φnx;t=∑i=0nVix;t,and(10)Vnx;t=Φnx;t−Φnx;t;n≥1,V0x;t=fx;t.
Hence, the proof of the theorem comes as a result of the following two lemmas.Lemma 1.
Besides the conditions (i)–(iii), the infinite series∑i=0nVix;tasn⟶∞ is uniformly convergent to a continuous solution function Φx,t.Proof.
From (8), we pick up two sequences Φnx;t,Φn−1x;t, to construct the following:(11)μ1Φnx,t=fx,t+λ1∫0t∫abkx−yζt,τΦn−1y,τdydτ+λ2∫abkx−yΦn−1y,tdy+λ3∫0tξt,τΦn−1x,τdτ,n≥1,(12)μ1Φn−1x,t=fx,t+λ1∫0t∫abkx−yζt,τΦn−2y,τdydτ+λ2∫abkx−yΦn−2y,tdy+λ3∫0tξt,τΦn−2x,τdτ.
Adapting (12), with the aid of (9), in the form(13)μ1Vnx,t=λ1∫0t∫abζt,τkx−yVn−1y,τdydτ+λ2∫abkx−yVn−1y,tdy+λ3∫0tξt,τVn−1x,τdτ.
Forn=1, after applying Cauchy–Schwarz inequality and using the conditions (i)-(iii), formula (13) yields(14)V1x,t≤α;α=Nc2bλ1T+λ2+λ3TMμ1,T=maxt0≤t≤τ.
By induction, we have(15)Vnx,t≤Hαn,α<1.
This bound, according to condition (7), makes the sequence Vnx;t converges uniformly. In addition, the result of inequality (15) leads us to decide that the sequence Φnx;t has a convergent solution. So, we have(16)Φx;t=limn⟶∞Φnx;t=limn⟶∞∑i=0nVix;t.
Since eachVix;t is continuous, therefore, Φx;t is also continuous, and the infinite series of equation (16) is uniformly continuous. Hence, the lemma is proved.Lemma 2.
The continuous solutionΦx;t of formula (16) represents a unique solution of equation (1).Proof.
AssumeΦx;t and Φ¯x;t are two different solutions of (1). Hence, we have(17)μ1Φx;t−Φ¯x;t=λ1∫0t∫−bbkx−yζt,τΦy;τ−Φ¯y;τdydτ+λ2∫−bbkx−yΦy;t−Φ¯y;tdy+λ3∫0tξt,τΦx;τ−Φ¯x;τdτ,after applying Cauchy–Schwarz inequality, and using the conditions (i)–(iii), formula (17) yields(18)Φx;t−Φ¯x;t≤αΦx;t−Φ¯x;t.
Sinceα<1, we deduce that the solution of (1) must be unique.


## 3. Separation of Variables Method

One of the most important ways to solve mathematical physics problems is the method of separating variables. The importance of this method comes when the unknown function is related to the position and time variables. This helps the researcher to determine the appropriate form to choose the time function and the appropriate time to be used, and accordingly, we study the behavior of the position function.For this, consider the following approximations:(19)Φx,t=ΨxAt,fx,t=FxAt,A0≠0.It is preferable to choose the time function in the form of a polynomial. From the constants of polynomials, the function of time can be classified in the form of an exponential, trigonometric, hyperbolic, and so on. This choice depends on the basis that the time period of the experiment (solving the problem) is always less than one and that the time, in the beginning, is not equal to zero. Based on that, we assume that(20)At=∑n=0∞antn,a0≠0.From equation (20), we can deduce the following:(1)
A0=a0, A∞=0. These results lead us to say that the function of time is bounded.(2)
Ifan=1/n!, the function of time becomes At=et. While At=e−tifan=−1n/n!.As for different values ofan,∀n≥0, we have different time functions.Using (19) in (1), we can arrive at the following equation:(21)βtΨx−γt∫−bbkx−yΨydy=Fx,βt=μ1−λ3At∫0tξt,τAτdτ,γt=λ2+λ1At∫0tζt,τAτdτ.It is clear by separating the variables that the mixed integral equation is transformed directly into equation (21) and has become a Fredholm equation in position with constants related to time t;t∈0,T,T<1.The existence of a unique solution of equation (21) is satisfied under the condition:(22)kx−y≤βtγt.


### 3.1. Convergence Analysis of Integral equation (21)

To discuss the convergence analysis of equation (21), assume the sequence of solutions in the following form:(23)Ψx=limj⟶∞ψjx.Letψn and ψm are two arbitrary distinct partial sums in the sequence of solution ψn;n>m. Using the definition of metric space,(24)dψn,ψm=maxn,m∈jψn−ψm≤δdψn,ψm−1:δ=cγtβt.Hence, we have(25)dψn,ψm≤δm1−δdψ1,ψ0;0<δ<1.AsFt,τ∈C0,T,0≤τ≤t≤T<∞, and dψ1,ψ0 is fixed so, dψn,ψm⟶0, we conclude that the sequence ψn is a Cauchy sequence in a complete metric space. Then, the series is convergent.


## 3.1. Convergence Analysis of Integral equation (21)

To discuss the convergence analysis of equation (21), assume the sequence of solutions in the following form:(23)Ψx=limj⟶∞ψjx.Letψn and ψm are two arbitrary distinct partial sums in the sequence of solution ψn;n>m. Using the definition of metric space,(24)dψn,ψm=maxn,m∈jψn−ψm≤δdψn,ψm−1:δ=cγtβt.Hence, we have(25)dψn,ψm≤δm1−δdψ1,ψ0;0<δ<1.AsFt,τ∈C0,T,0≤τ≤t≤T<∞, and dψ1,ψ0 is fixed so, dψn,ψm⟶0, we conclude that the sequence ψn is a Cauchy sequence in a complete metric space. Then, the series is convergent.


## 4. Toeplitz Matrix Method

Toeplitz matrix method is considered one of the best methods for solving the singular integral equations, where the singular term directly disappears, and we have a linear system of algebraic equations. For this, rewrite the integral term of equation (21) in the following form:(26)∫−bbkx−yψydy=∑ℓ=−NN−1∫b=ℓhℓ+1hkx−yψydy,=∑ℓ=−NN−1Gℓxψb+Hℓxψb+h+Rℓ;h=bN.Here in (26), Gℓx and Hℓx are arbitrary functions, and Rℓ is the estimated error.In the light ofTMM, we can determine the values of Gℓx and Hℓx, respectively, in the following forms:(27)Gℓx=1hb+hSℓx−Zℓx,Hℓx=1hZℓx−bSℓx,where(28)Sℓx=∫ℓhℓ+1hkx−ydy,Zℓx=∫ℓhℓ+1hkx−yydy.Assumingx=jh,−N≤j≤N0≤ti≤T,i=0,1,…,M and then using the following notations:(29)ψjh=ψj,Dℓjh=Dℓ,j,Fjh=Fj,At=Ati=Ai,βt=βti=βi,δt=δti=δi.Hence, the integral equation (21) yields the following LAS of (2N + 1) equations:(30)βiψj−γi∑ℓ=−NNDℓ,jψℓ=Fj,−N≤j≤N,i=0,1,…,M.The matricesDn,ℓ can be written in the following Toeplitz matrix form:(31)Dℓ,j=Pn−ℓ−Ln,ℓ,Pℓ−j=Gℓ,j+Hℓ−1,j,−N≤j,ℓ≤N,Lℓ,j=H−N−1,j,ℓ=−N,0,−N<ℓ<N,GN,j,ℓ=N.It is clear from (31) that we have two kinds of matrices: the first Pℓ−j is called the Toeplitz matrix of order 2N+1 while Lℓ,j represents a matrix of the order 2N+1 whose elements are zeros except the first and the last rows (columns). Moreover, it is worth mentioning that such a Toeplitz-like linear system (31) can be solved by using the preconditioned Krylov subspace methods of ΟMlogM operations and ΟM memory cost, where M = 2N + 1; see [14,15] for details.The error termRn,ℓ can be determined from the following formula:(32)Rℓ=max0≤n≤p∫ℓhℓh+hy2kx−ydy−Gℓxℓh2+Hℓxℓh+h2=Oh3.


## 5. Convergence Analysis of Linear Algebraic System (30)

To discuss and prove the convergence ofLAS (30) in the Banach space ℓ∞×C0,T, we write it in the operator form(33)T¯ψj=Tψj+1βiFj,Tψj=δi∑ℓ=−NNDℓ,jψℓ;δi=γiβi;−N≤j≤N,βi=μ1−λ3Ati∫0tiξti,τAτdτ,γi=λ2+λ1Ati∫0tiζti,τAτdτ.Then, consider the following.Lemma 3.
Forx,x′∈−b,b if the kernel of equation (1) satisfies the conditions(34)1kx−y∈L2−b,b×−b,b,2limx′⟶xkx′−y−kx−y=0,then, we have(i)
supN∑ℓ=−NNDℓ,j≤X,X=maxjXj, (X is constant)(ii)
limn′⟶nsupN∑ℓ=−NNDn′,ℓ−Dn,ℓ=0Proof.
Applying Cauchy–Schwarz inequality for the first formula of (27) and summing from ℓ=−N to ℓ=N and then using the first condition of (34), we have ∑ℓ=−NNGℓx≤ρ1. Since each term of ∑ℓ=−NNGℓx is bounded above, hence, for x=jh, we get(35)supN∑ℓ=−NNGℓjh≤ρ1.
Similarly, we consider a small constantρ2, such that(36)supN∑ℓ=−NNHℓjh≤ρ2.
From (35) and (36), we deduce that(37)supN∑ℓ=−NNDℓ,j≤supN∑ℓ=−NNGℓ,j+supN∑ℓ=−NNHℓ,j≤ρ;ρ=ρ1+ρ2.
Hence, case (i) holds.
For the second case of (26), consider x,x′∈−b,b. After applying Hölder inequality, and summing from ℓ=−N to ℓ=N, we have(38)1limj′⟶jsupN∑ℓ=−NNGℓj′h−Gℓjh=0;2limj′⟶jsupN∑ℓ=−NNHℓj′h−Hℓjh=0.
Finally, case (ii) of Lemma3 is held.


## 6. The Existence and Uniqueness Solution of the LAS (21) of Toeplitz Matrix

To discuss the unique solution of theLAS (30), we consider the following assumptions:visupNFℓ≤η,η−constantvsupN∑ℓ=−NNDℓ,j≤X (X is a constant)iv For the constants Q>Q1,Q>P1, the points ψℓjh satisfy(39)supjψℓjh≤Q1Ψℓ∞,supjψℓjh−φℓjh≤P1Ψ−Φℓ∞.Theorem 2.
For all values of timeti;ti∈0,T,T<1, the LAS of equation (30), in the Banach space ℓ∞, has a unique solution under the following condition:(40)QXγ<β.
To prove this theorem, we must consider the following lemmas.Lemma 4.
If the conditions (vi)–(iv) are verified, then the operatorT¯ defined by equation (33) maps the space ℓ∞ into itself.Proof.
Let U be the set of all functionsΦ=ψmm=0∞ in ℓ∞ such that Φℓ∞≤β, βis a constant. Define the norm of the operator T¯Φ in the Banach space ℓ∞ by(41)T¯Φℓ∞=supmT¯ψm.
From formula (43), and with the aid of the conditions (vi)–(iv), we have(42)T¯φm≤γβQΦℓ∞supN∑n=−NNDmn+ηβ.
The above inequality can be adapted in the following form:(43)supmT¯φm≤σ1Φℓ∞+Hβ,σ1=γβQX.
The inequality (43) shows that the operator T¯ maps the set U into itself, where(44)δ=ηβ−γQX.
Sinceδ>0,η>0, therefore, we have σ1<1. Also, the inequality (43) involves that the operator T is bounded, where TΦℓ∞≤σ1Φℓ∞. Hence, the operator T¯ is bounded.Lemma 5.
Under the two conditions (vi) and (iv),T¯ is a contraction operator in the space ℓ∞.Proof.
For the two functionsΦ and Ψ in ℓ∞, formula (33) leads to(45)T¯φm−T¯ψm≤γβ∑n=−NNDmnsupnφnh−ψnh.
In the light of conditions (vi) and (iv), and in view of (41), we obtain(46)T¯Φ−T¯Ψℓ∞≤σ1Φ−Ψℓ∞.
The inequality (46) shows that the operator T¯ is continuous in the space ℓ∞, and then, T¯ is a contraction operator under the condition σ1<1.
In the light of Lemmas 4 and 5, the operatorT¯ defined by (33) is contractive in the Banach space ℓ∞. Hence, T¯ has a unique fixed point which is the unique solution of the linear algebraic system in ℓ∞.Definition 1.
TheTMM is said to be convergent of order r in the interval −b,b, if and only if for sufficiently large N, there exists a constant ϑ>0 independent of N such that(47)ψx−ψNx≤ϑN−r,∀t∈0,T.


## 7. Applications

We solve equation (1) numerically using TMM with the aid of the maple 18 program, when the kernel of Fredholm term takes some weakly singular and singular forms, and at different times, the approximate solution and the estimated error, in each case, are calculated. In general, consider the following general form of a mixed integral equation with its exact solution(48)0.9Φx,t=fx,t+0.32∫0t∫−11kx−yt2τ2⋅Φy,τdydτ+0.25∫−11kx−yΦy,τdy+0.3∫0tt2τΦx,τdτ;At=0.25+t2Exact solutionΦx,t=0.25+t2x2.Example 1.
When the kernel of (26) takes the Carleman kernel,(49)kx−y=x−y−υ,0<υ<1.υis called Poisson ratio.
The Carleman kernel is called in the sense of contact problem weakly kernel. The importance of the Carleman kernel came from the works of Russian scholes in the nonlinear theory of plasticity; see Popov [16] and Aleksandrovsk and Covalence [17]. In this case, the elements of the Toeplitz coefficient matrix of (31) take the following form:(50)Pℓ−j=h2−υυ−2ℓ+1−j2−υ−ℓ−1−j2−υ−2ℓ−j2−υ+h1−υυ−1×ℓ+1−jℓ+1−j1−υ+ℓ−1−jℓ−1−j1−υ−2ℓ−jℓ−j1−υ.
The estimated error becomes(51)Rj≃h3−υj1−j1−υ+1−2j2−υ−13−υ.
In Table1, we calculate numerically, the numerical solution of (35), using TMM for two different values of Poisson’s coefficient ʋ at time t=0.3 and the estimating error in each case is computing.
Figures1(a) and 1(b) describe the numerical solution of (41) with the Carleman kernel at T = 0.3.Table 1
Describes the numerical solution of (41) with Carleman kernel and the corresponding error.
xExtractυ=0.12υ=0.73Num. TMMErr. TMMNum. TMMErr. TMM-14.00000E-043.63253E-043.67468E-053.08591E-049.14093E-05-0.82.56000E-042.73278E-041.72775E-052.24550E-043.14502E-05-0.61.44000E-041.61339E-041.73390E-051.10959E-043.30406E-05-0.46.40000E-058.14052E-051.74052E-053.00103E-053.39897E-05-0.21.60000E-053.34485E-051.74485E-05-1.85140E-053.45140E-0500.00000E+001.74632E-051.74632E-05-3.46828E-053.46828E-050.21.60000E-053.34485E-051.74485E-05-1.85140E-053.45140E-050.46.40000E-058.14052E-051.74052E-053.00103E-053.39897E-050.61.44000E-041.61339E-041.73390E-051.10959E-043.30406E-05.82.56000E-042.73278E-041.72775E-052.24550E-043.14502E-0514.00000E-043.63253E-043.67468E-053.08591E-049.14093E-05Figure 1
The error of Caleman kernel at time. (a) The error of Caleman kernel whenυ=0.12,T=0.3 (b) The error of the Caleman kernel when. υ=0.72,T=0.3
(a)(b)Example 2.
With the Cauchy kernel,(52)kx−y=1x−y.
The integral equations with Cauchy kernels have wide applications in mathematical physics problems and in the different sciences, see Abdou [1], Constanda [18], and Frankel [19].
In this case, the elements of the Toeplitz matrix of (31), at time T=0.3, take the following form:(53)Pℓ−j=ℓ+1−jlnℓ+1−j−2ℓ−jlnℓ−j+ℓ−1−jlnℓ−1−j.
The estimated error, at timeT=0.3, takes the following form:(54)Rj≃h2σ,σ=16j+112j2+120j3+130j4+142j5+⋯,
See Table2 and Figure 2.Table 2
Describes the numerical solution of equation (1) with the Cauchy kernel and the corresponding error.
XExactNum. TMMErr. TMM-14.00000E-044.25183E-045.51830E-07-0.82.56000E-042.54476E-042.52438E-07-0.61.44000E-041.43139E-041.60871E-07-0.46.40000E-056.33420E-056.58005E-07-0.21.60000E-051.54227E-055.77310E-0800.00000E+005.54788E-075.54788E-080.21.60000E-051.54227E-055.77310E-070.46.40000E-056.33420E-056.58005E-070.61.44000E-041.43139E-041.60871E-070.82.56000E-042.54476E-042.52438E-0714.00000E-044.25183E-045.51830E-07Figure 2
The error of Cauchy kernel at time.T=0.3Example 3.
Consider the Hilbert kernel(55)kx−y=cotx−y2,x∈−π,π;φ±π,t=0,Exact solutionΦx,t=0.25+t2sinx.
The importance of the integral equation with the Hilbert kernel with its application in the crack problem in the theory of elasticity can be found in Kamalyan, et al., [20].
To obtain the elements of the Toeplitz matrix, we must use the famous integral formula, see Gradstein and Ryzhik [21].(56)∫nhn+1hxmcotxdx=∑s=0∞−1s22sB2sm+2s2s!xm+2sm≥1,x<π,where B2s are Bernoulli numbers.
After using the famous integral of Bernoulli, we have the elements of the Toeplitz coefficient matrix in the following form:(57)Pℓ−j=2ℓ−j+1lnsinhℓ−j+12−2ℓ−jlnsinhℓ−j2ℓ−j−1lnsinhℓ−j−12−∑s=0∞−1s22sB2sm+2s2s!×ℓ−j+11+2s−2ℓ−j1+2s+ℓ−j−11+2s,
See Table3 and Figure 3.Table 3
Describes the numerical solution of equation (1) with the Hilbert kernel and the corresponding error.
xExactNum. TMMErr. TMM-2.51327-2.35114E-04-2.35361E-042.46666E-07-1.88496-3.80423E-04-3.80880E-044.56924E-07-1.25664-3.80423E-04-3.76145E-044.27804E-06-0.62832-2.35114E-04-2.35259E-041.45217E-070.0E+000.00000E+00-2.21595E-092.21595E-090.6283192.35114E-042.35254E-041.39694E-071.2566373.80423E-043.80824E-044.01774E-071.8849563.80423E-043.76176E-044.24685E-062.5132742.35114E-042.35439E-043.24793E-07Figure 3
The error of the Hilbert kernel at time.T=0.3Example 4.
When the kernel takes the following generalized logarithmic function form:(58)kx−y=lnx−yq,q=1,2,…,nExact solutionφx,t=x20.25+t2.
Several problems have been considered when the kernel takes the logarithmic form.kx−y=lnx−y, for example, see Abdou, et al., [22,23].
The importance of logarithmic function came from its form and its derivatives form, where we can derive the following cases:(i)
lnx−y=lnx−yx−yυ︸ux,yx−y−υ (u (x, y) is continuous function)(ii)
∂/∂xkx−y|=1/x−y Cauchy kernel
In addition, we can establish a strong singular kernel and so on. In our application, we use the result of the following integral, see Gradstein and Ryzhik [21](59)∫xℓlnxqdx=xq+1q+1∑n=0q−1jq+1qq−1…q−n+1lnxq−nℓ+1n+1,ℓ≠−1.
In the generalized logarithmic kernel, the elements of the Toeplitz matrix become(60)Pℓ−j=h∑n=0q−1nqq−1q−2…q−n+11−2−n−1×ℓ−j+12lnℓ−j+1hq−n+ℓ−j−12lnℓ−j−1hq−n−2ℓ−j2lnℓ−jhq−n.
The result of estimate error is(61)Rℓ,j=h3∑n=0q−1nqq−1q−2…q−n+1×j2lnjhq−1j−1+3−n+1j−2−n+12j−1+1−j2ln1−jhq−1−j+3−n+11−j+2−n+12j−3.(4-1):kx−y|=lnx−y;T=0.3We have the logarithmic kernel. We useTMM to solve numerically equation (21) at the time interval t∈0,0.3.See Table4 and Figure 4.(4-2):kx−y|=lnx−y5,0<t≤0.3We have a strong singular logarithmic kernel of power five at the interval of timet∈0,0.3 with the exact solution φx,t=x20.25+t2See Table 5 and Figure 5.Table 4
Represents the numerical solution of equation (1) with logarithmic kernel and the error of TMM.
xExactApp TMM.Error TMM-14.00E-044.14364E-041.43636E-09-0.82.56E-042.53924E-042.07618E-09-0.61.44E-041.43201E-047.98768E-09-0.46.40E-056.37610E-052.38987E-10-0.21.60E-051.60212E-052.12470E-1000.00E+009.86122E-089.86122E-110.21.60E-051.60212E-052.12433E-100.46.40E-056.37610E-052.38987E-100.61.44E-041.43201E-047.98770E-090.82.56E-042.53924E-042.07618E-0914.00E-044.14364E-041.43636E-09Figure 4
The error of logarithmic kernel at.k=1,T=0.3Table 5
Represents the numerical solution of (26) with generalized Logarithmic kernel and the error of TMM.
xExactApp TMM.Error TMM-10.640.6353448840.00465512-0.80.40960.4064219430.00317806-0.60.23040.2286578590.00174214-0.40.10240.1017988210.00060118-0.20.02560.0257243910.00012439000.0003726950.000372690.20.02560.0257243910.000124390.40.10240.1017988210.000601180.60.23040.2286578590.001742140.80.40960.4064219430.00317806Figure 5
The error of super strong singular logarithmic kernel at.k=k,T=0.3


## 8. Conclusions

From the above work, we can deduce the following:(1)
From Equation (1), we can establish the following famous special cases:(i)
Let in (1), ζt,τ=0 to obtain(62)μ1Φx,t=fx,t+λ2∫abkx−yΦy,tdy+λ3∫0tξt,τΦx,τdτ;x=x¯x1,x2,…,xn,y=yy1,y2,…,yn.The above formula (62) represents the Fredholm–Volterra integral equation of the second kind with discontinuous kernels. Many authors established the solution of (62), using different numerical methods, see Khairullina, et al., [7] and Bazm [24](ii)
Fredholm–Volterra integral equations of the first kind, with different singular kernels, are obtained directly from (62), when μ1=0, for the solution of these cases with some different methods, see Abdou et al., [2,25].(iii)
Let in (1) ξt,τ=0, to have(63)μ1Φx,t=fx,t+λ1∫0t∫abkx−yζt,τΦy,τdydτ+λ2∫abkx−yΦy,tdy.Formula (63) represents the Volterra–Fredholm integral equation of the second kind. The solution of (63), using some different numerical methods, is discussed by many authors, for example, see Abdou, et al., [22,23], Mirzaee and Hoseini [6], and Mirzaee and Samadyar [26](iv)
If in (3) μ=0, we have V-FIE of the first kind. Many spectral relationships, when the kernel of position has a singular term and the kernel of time is continuous, have been obtained and discussed in Mirzaee and Hoseini [6].(2)
We notice the following from the results of the programs in all previous cases.(i)
In Table1, the error increases with increasing of υ, where υ is called the Poisson ratio, and in the theory of elasticity and when 0≤υ≤0.5, the atomic bond between the particles of the material is normal, while υ≥0.5; the atomic bond becomes strong, and for this, the error may be large. Also, it was found that the maximum error value is obtained when υ=0.73. Moreover, the error decreases gradually and has less value when υ=0.07. The maximum error is at x=±1, when υ=0.73,T=0.8. The minimum error is at x=±0.0, when υ=0.07,T=0.004. In all the studied situations, the error value increases when it gets closer to the end points x=±1. It also decreases at the middle when it gets closer to zero.(ii)
In Table2, the maximum error is at x=±1, and when the values of T are increasing, the error values increase slowly.(3)
TheTMM is considered as one of the best methods for solving the singular integral equation, where the singularity disappears and the solution can be obtained directly.



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