# Numerical Treating of Mixed Integral Equation Two-Dimensional in Surface Cracks in Finite Layers of Materials

Authors: A. M. Al-Bugami
Journal: Advances in Mathematical Physics
 (2022)

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## Abstract

The goal of this paper is study the mixed integral equation with singular kernel in two-dimensional adding to the time in the Volterra integral term numerically. We established the problem from the plane strain problem for the bounded layer medium composed of different materials that contains a crack on one of the interface. Also, the existence of a unique solution of the equation proved. Therefore, a numerical method is used to translate our problem to a system of two-dimensional Fredholm integral equations (STDFIEs). Then, Toeplitz matrix (TMM) and the Nystrom product methods (NPM) are used to solve the STDFIEs with Cauchy kernel. Numerical examples are presented, and their results are compared with the analytical solution to demonstrate the validity and applicability of the methods. The codes were written in Maple.


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## Body

## 1. Introduction

Many problems of engineering, mathematical physical, and contact problems in the theory of elasticity lead to singular integral equations. Integral equations provide an important tool for solving the ordinary and partial deferential equations. Therefore, many different methods are used to obtain the solution of the linear and nonlinear integral equations. Brunner and Kauthen [1] introduced collocation and iterated collocation methods for solving the two-dimensional Volterra integral equation (T-DVIE). In [2], authors proposed a class of explicit Runge-Kutta-type methods of order 3 for solving nonlinear T-DVIE. In [3], authors studied the approximate solution of T-DVIEs by the two-dimensional differential transform method. Abdou, in [4, 5], used different methods to obtain the solution of F-VIE of the first and second kinds in which the Fredholm integral term is considered in position while the Volterra integral term is considered in time. EL-Borai et al., in [6], studied the numerical solution for the T-DFIE with weak singular kernel, but they have studied the problem on a rectangular path of the parties only. AL-Bugami in [7] studied and discussed the solution of the T-DFIE with applications in contact problems. AL-Bugami in [8] studied and discussed the solution of the two-dimensional singular Fredholm integral equation (T-DFIE) with time. The solution of a large of mixed boundary value problem of a great variety of contact and crack problems in solid mechanics, physical, and engineering can be related to a system of the singular IEs have a simple Cauchy-type singularity (Ref. [9]). In [10], the authors studied the linear two-dimensional Volterra integral equation with continuous kernel numerically. In [11], the authors discussed continuous Fredholm-Volterra integral equation. Al-Bugami, in [12], studied the singular Hammerstein-Volterra integral equation and solved numerically. In [13], the author studied the surface cracks of finite layers of fractional materials.In this work, we consider a mixed integral equation with singular kernel in two-dimensional (MIE)(1)ηwx,t+λπ∫−11∫−11p1x−up2y−vwu,v;tdudv+λ∫−11kx,ywy,tdy+∫0tζt,τwx,τdτ=fx,t.Formula (1) is called the MIE with singular kernel in two-dimensional of the second kind with Cauchy kernel in L2−1,1×L2−1,1×C0;Y;Y<1, where the FI term is considered in position with singular kernel, and the VI term is considered in time with a positive and continuous kernelζt,τ. fx,t is known function, while wx,t is unknown function to be determined. The numerical coefficient λ is called the parameter of the IE.


## 2. The Basic Formulas of the Problem

Consider the plane strain problem for the bounded layer medium (Figure1), composed of three different materials. Let the medium material contain a crack on one of the interface. Without any loss in generality, the half length of the crack is assumed to be unity. We will consider with the effect of the ratio of the layer thickness to the crack length on the stress, intensity factors, and the strain energy release rate. For interesting the disturbed stress state, while is variable also with time, caused by the crack.Figure 1
The layers of materials.We assume that the overall stress distributionσij0x,y,t, in the imperfection free medium, is known. The stress state σij1x,y,t, in the cracked medium, may be expressed as
(2)σij1x,y,t=σij0x,y,t+σijx,y,t,i,j=x,y,z,whereσij is the disturbed state, which may be obtained by using the tractions
(3)P1x,t=−σyy0x,0;t;P2x,t=−σxy0x,0;t,x<1,t∈0,T,which are the only external loads applied to the medium. The problem can always be expressed as the sum of a symmetric component and an antisymmetric component. The tractions are Pix,t,i=1,2, where
(4)P1x,t=P1−x,t,P2x,t=−P2−x,t,x<1,t∈0,T.The solution of the antisymmetric problem requires only a slight modification. Letui,vi be the x and y components of the displacement vector in the i-th materials and satisfy the filed equations in the form:
(5)μi∇2ui+λi+μi∂∂x∂ui∂x+∂vi∂y=ρd2uidt2,(6)μi∇2vi+λi+μi∂∂x∂ui∂x+∂vi∂y=ρd2uidt2.Then, assume the displacement functions in the following:(7)ix,y,t=Uix,y+Ft,(8)vix,y,t=Vix,y+Ft.Ft is known function of t. Hence, using (7) and (8) in (5) and (6), we get
(9)λi+2μi∂2Ui∂x2+μi∂2Ui∂y2+λi+μi∂2Vi∂x∂y=0,(10)λi+2μi∂2Vi∂y2+μi∂2Vi∂x2+λi+μi∂2Ui∂x∂y=0,(11)d2Ftdt2=μiρFt.Formula (11) has a solution
(12)Ft=Be−μiρ,F∞⟶0.For solving the two formulas (9) and (10), we use the Fourier integral transform:
(13)Uix,y=2π∫0∞wiα,ysinαxdα,i=1,2,3,(14)Vix,y=2π∫0∞ψiα,ycosαxdα.Then, we have(15)−α2λi+2μiwi+μid2widy−αλi+μidψidy=0,(16)−α2λi+2μiψi+μid2ψidy+αλi+μidwidy=0.After solving the system of Eqs. (15) and (16), and then using the two formulas (13) and (14), we get
(17)Uix,y=2π∫0∞Ai1+Ai2ye−αy+Ai3+Ai4yeαysinαxdα,Vix,y=2π∫0∞Ai1+Kiα+yAi1e−αy+−Ai3+Kiα−yAi4eαycosαxdα,whereKi has physical meaning and Ki=3−4υi for plane strain and Ki=3−υi/1+υi for generalized plane stress, υi is Poisson’s coefficient for each materials, and Ai,j,j=1,2,3,4 are functions of α which can be determined from the boundary conditions. After obtaining, the values of Ui,Vi, the stresses may be evaluated by Hooks law.The components of the stress vector at the interfaces and boundaries may be expressed as(18)12μiσyyi=2π∫0∞−αAi1+Ai2y+21−υiAi2e−αy+−αAi3+Ai4y+21−υiAi4eαycosαxdα,12μiσxyi=2π∫0∞−αAi1+Ai2y+1−υiAi2e−αy+αAi3+Ai4y−1−2υiAi4eαysinαxdα.On the boundaries, the medium may have formally any one of the following four groups of homogeneous boundary conditions(19)σyyi=0=σxyi,ui=0=vi,σxyi=0=vi,σyyi=0=ui,i=1,2,3.The continuity requires that on the interfaces, the stress and displacement vectors in the adjacent layers be equal, i.e.,(20)ui+1−ui=0,vi+1−vi=0,σyyi+1−σyyi=0,σxyi+1−σxyi=0,i=1,2,3.Now, to obtain the integral equation, we first assume that aty=0, the bond between the two adjacent layers is perfect except for the dislocations at y=0 and x=y defied by
(21)∂∂xu2+−u3−=f1x,t,∂∂xv2+−v3−=f2x,t,where the superscripts + and – refer to the limiting values of the displacement asy approaches zero from + and – sides, respectively. In addition to (21), on the interface y=0, we have the following conditions
(22)σyy2−σyy3=0,σxy2−σxy3=0,0≤x<∞,y=0.The components of the stress vector aty=0 and x>0 may be expresses as
(23)1+K3μ3σyy3x,0,t=limy⟶0−2π∫0∞eαxα11A1α,t+α12A2α,tcosαxdα+2π∫0∞H11αA1α,t+H12αA2α,tcosαxdα+∫0tFτfix,0,τdτ,(24)−1+K3μ3σxy3x,0,t=limy⟶0−2π∫0∞eαxα21A1α,t+α22A2α,tsinαxdα+2π∫0∞H21αA1α,t+H22αA2α,tsinαxdα+∫0tFτfix,0,τdτ,whereHα is the Heaviside functions, and Ai is the Fourier transforms of fi defined as follows:
(25)A1α,t=∫0∞f1z,tcosαzdz,A2α,t=∫0∞f2z,tsinαzdz.The constantsaij depend on the elastic properties of the materials adjacent to the crack only and are given by
(26)a11=−a22=1+λ2λ4/λ4,a12=−a21=−1+2λ4−λ2λ4/λ4,λ2=K2μ3−K3μ2/μ2+K2μ3,λ4=μ3+μ2K3/μ2+μ3,whereμi is the shear modules, and λ’s is Lame’s constants.Note that once the dislocationsfix on the interface are specified, formulas (23)and (24) give the stresses for all values of x. The crack problem under consideration fix is zero for ∣x∣>1 and is unknown for ∣x∣<1. On the other hand, the stress vector on the interface y=0 is unknown for ∣x∣>1 that is given by the following known functions for ∣x∣<1, i.e.,
(27)f1x,t=f1−x,t,f2x,t=−f2−x,t.Then,(28)∫0∞Hαcosαxdα∫0tf1z,tcosαzdz=12∫−11∫−11f1∗z,tf1∗∗z,tdxdx∫0∞Hαcosαz−xdα.Hence, we obtain(29)−1+K3μ3P1x,t=limy⟶0−a11π∫−11∫−11f1∗z,tf1∗∗z,tdzdz∫0∞eαycosαz−xdα+a12π∫−11∫−11f2∗z,tf2∗∗z,tdzdz×∫0∞eαysinαz−xdα+1π∫−11∫−11∑j=12k1j∗x,zk1j∗∗x,zfj∗z,tfj∗∗z,tdzdz+∫0tFτf1x,τdτ,(30)−1+K3μ3P3x,t=limy⟶0−a21π∫−11∫−11f1∗z,tf1∗∗z,tdzdz∫0∞eαysinαz−xdα+a22π∫−11∫−11f2∗z,tf2∗∗z,tdzdz×∫0∞eαycosαz−xdα+1π∫−11∑j=12k2jx,zfjz,tdz+∫0tFτf2x,τdτ.Evaluating the infinite integrals in (30), passing to the Cauchy theorems, we have
(31)−1+K3a12μ3P1x,t=γw1x,t+1π∫−11∫−11w2u,v;tx−uy−vdudv−1a12π∫−11∑j=12k1jx,ywjy,tdy+1a12∫0tFτw1x,τdt,−1+K3a21μ3P2x,t=1π∫−11∫−11w1u,v;tx−uy−vdudv−γw2x,t−1a21π∫−11∑j=12k2jx,ywjy,tdy+1a21∫0tFτw2x,τdt,where
(32)γ=a11a12=a22a21=μ2+K3μ2−μ2+K2μ3μ2+K2μ3+μ3+K3μ2.The two formulas of (31) represent a system of MIE with Cauchy kernel. For one layer, we can have the following MIE, on noting the difference notations.
(33)ηwx,t−1π∫−11∫−11wu,v;tx−uy−vdudv−1πa12∫−11kx,ywy,tdy+1a12∫0tFt,τwx,τdτ=fx,t.In general, we can write Eq. (33) in the form:
(34)ηwx,t+λπ∫−11∫−11p1x−up2y−vwu,v;tdudv+λ∫−11kx,ywy,tdy+λ∫0tζt,τwx,τdτ=fx,t,wherep1x−up2y−v=1x−uy−v


## 3. The Existence and Uniqueness of the Solution

We write this formula in the integral operator form(35)W¯wx,t=1ηfx,t−Wwx,t,η≠0,(36)Ww=Hw+Dw+ζw,Hw=ληπ∫−11∫−11p1x−up2y−vwu,v;tdudv,Dw=λη∫−11kx,ywy,tdy,ζw=λη∫0tζt,τwx,τdτ.We assume the following conditions:(1)
The singular kernel of FI term satisfies inL2−1,1×L2−1,1 the discontinuity condition(37)∫−11∫−11∫−11∫−11p1x−up2y−v2dxdudtdv12=M,Misaconstant.(2)
The kernel of VI termζ(t, τ) is continuous in the Banach space C0,T,0≤τ≤t≤T<∞ and satisfies(38)ζt,τ≤N,∀t,τ∈0,T.(3)
The continuous kernelkx,y≤L(39)fx,t=max0≤t≤T∫0t∫−11f2x,τdx12dτ=R,Risaconstant.(4)
wx,t in the space, L2−1,1×L2−1,1×C0,T, behaves as the known function fx,tTheorem 1.
Eq. (34) has an exact unique solution in L2−1,1×L2−1,1×C0,T, under the condition
(40)η>λMπ+2TL+N,T=max0≤t≤Tt.Lemma 1.
The integral operatorW¯ maps L2−1,1×L2−1,1×C0,T into itself.Proof.
From (35) and (40), the normality of the integral operator takes the forms
(41)W¯wx,t≤1ηfx,t+λπη∫−11∫−11p1x−up2y−vwu,v;tdudv+λη∫−11∣kx,ywy,t∣dy+λη∫0t∣ζt,τwx,τ∣dτ.
Applying Cauchy-Schwarz inequality, we have(42)Hw≤λπη∫−11∫−11p12x−up22y−vdudv12∫−11∫−11w2u,v;tdudv12.
Using the definition of the norm in the spaceL2−1,1×L2−1,1×C0,T, we get
(43)Hw≤λπη∫−11∫−11∫−11∫−11p12x−up22y−vdudvdudv12max0≤t≤T∫0t∫−11∫−11w2u,v;tdudv12dτ.
Then, using condition (1), we obtain
(44)Hw≤λπηMw.
Also, the termDw takes the form
(45)Dw≤λη∫−11kx,ywy,tdy.
Using condition (3), we get
(46)Dw≤ληLmax0≤t≤T∫0t∫−11w2y,τ∣dy12dτ.
Hence,(47)Dw≤λη2LTw.
In the same manner, we can write(48)ζw≤λη∫0t∣ζt,τwx,τ∣dτ.
Using condition (2), we obtain
(49)ζw≤ληNmax0≤t≤T∫0t∫−11w2y,τdy12dτ.
Thus, one has(50)ζw≤λη2NTw,T=max0≤t≤Tt.
Hence, with the aid of conditions (5), (44), (47), and (50), Eq. (41) takes the form
(51)W¯w≤Rη+αw,α=η−1λMπ+2TL+N,T=max0≤t≤Tt.
The inequality (51) involves the boundedness of the operators W and W¯.☐Lemma 2.
The integral operator (35) under the condition (40) is continuous and contraction operator.Proof.
For the functionsw1x,t,w2x,t in the space L2−1,1×L2−1,1×C0,T, formula (35) yields
(52)W¯w1−W¯w2x,t=Ww1−w2x,t.
Hence, we have(53)W¯w1−W¯w2x,t≤ληπ∫‐11∫‐11∣p1x−up2y−vw1u,v;t−w2u,v;t∣dudv+ληπ∫‐11∣k∣x,y∣∣w1y,t−w2y,t∣dy+λη∫0t∣ζt,τ∣w1x,τ−w2x,τ∣dτ.
Using formula (53) with the conditions (1), (2), and (3), then applying Cauchy-Schwarz inequality, we obtain
(54)W¯w1−W¯w2≤αw1−w2,α=η−1λMπ+2TL+N.
Hence,W¯ is a continuous operator in the space L2−1,1×L2−1,1×C0,T, and under the condition α<1, W¯ is a contraction operator.☐


## 4. THE STDFIEs

Consider Eq. (34). In this section, we divide the interval 0,T,0≤t≤T<∞,0=t0<t1<⋯<tn=T, where t=ti,i=0,1,2,..,n,, to get
(55)ηwx,ti+λπ∫−11∫−11wu,v,tix−uy−vdudv+λ∫−11kx,ywy,tidy+λ∫0tiζti,τwx,τdτ=fx,ti.Using the quadrature formula, the Volterra term becomes(56)∫0tiζti,τwx,τdτ=∑j=0iκjζti,tjwx,tj+Rx,ti.κj is the weight, where κ0=1/2h0,κi=1/2hi,κi=hj,0<j<i,h denotes the constant step size for integration. Using (56) in (55), we have
(57)ηwix+λπ∫−11∫−11wiu,vx−uy−vdudv+λ∫−11kx,ywiydy+λ∑j=0iκjζi,jwjx=fix+Rix.Formula (57) can be adapted in the form
(58)η+λκiζi,jwix=fix−λπ∫−11∫−11p1∣x−u∣p2∣y−v∣wiu,vdudv−λ∫−11kx,ywiydy‐λ∑j=0iκjζi,jwjx,0≤j≤i−1.Then, the general form of Eq. (58) can be represented as
(59)μiwix+λπ∫−11∫−11p1∣x−u∣p2∣y−v∣wiu,vdudv+λ∫−11kx,ywiydy=ψix,where
(60)μi=η+λκiζi,j,ψi=fix−λ∑j=0i−1κjζi,jwjx,i=0,1,⋯,n.Formula (59) represents a linear system of TDFIEs of the second kind, which contains n+1 equation of n+1 unknown functions of wix corresponding to the time interval [0,T].


## 5. Some Numerical Methods

### 5.1. The TMM

We present the TMM to obtain numerical solution of TDFIE of the second kind with Cauchy form, which it expresses in the form(61)μiwix+λπ∫−11∫−11p1∣x−u∣p2∣y−v∣wiu,vdudv+λ∫−11kx,ywiydy=ψix,which it may be adapted as
(62)μwx=ψx−λπ∫−11∫−11Fx−u,y−vwu,v;tdudv,where
(63)Fx−u,y−v=1+πkx,yy−xx−uy−v=1x−uy−v1+πkx,yy−x.Then, write the integral term in Eq. (62) as the form
(64)∫−11∫−11Fx−u,y−vwu,v;tdudv=∑n=−NN∫nhnh+hFx−u,y−vwu,v;tdudv,h=1N.Formula (64) reduces as
(65)∫−11∫−11Fx−u,y−vwu,v;tdudv=∑n=−NN∑m=−MMFx−u,y−vwu,v;tdudvh=1N.Then,(66)∫nhnh+h∫mhmh+hFx−u,y−vwu,v;tdudv=An,mx,ywnh,mh+Bn,mx,ywnh+h,mh+h+R.Then, we putwu,v=1.1,uv in Eq. (66), and then we obtain
(67)An,mx,y=1hnh+hmh+hInh+mh+h−Jnh+mh+h,Bn,mx,y=1hJnh+mh+h−nhmhInh+mh+h,where
(68)Ix,y=∫nhnh+h∫mhmh+hkt−τ,x−ydudv,Jx,y=∫nhnh+h∫mhmh+huv⋅kx−u,y−vdudv.Eq. (65) becomes
(69)∫−11∫−11Fx−u,y−vwu,v;tdudv=∑n=−NN−1∑m=−MM−1An,mx,ywnh,mh;t+Bn,mx,ywnh+h,mh+h;t=∑n=−NN‐1∑m=−MM−1An,mx,ywnh,mh;t+∑n=−NN∑m=−MMBn−1m−1x,ywnh,mh;t=∑n=−NN∑m=−MMDn,mx,ywnh,mh;t,where
(70)Dn,mx,y=A−Nx,yn=m=−N,Anx,y+Bn−1x,y−N<n=m<N,BN−1x,yn=m=N.Thus, the IE (62) becomes
(71)μwx,y+λπ∑n=−NN∑m=−MMDn,mx,ywnh,mh;t=fx,y.If we putx=kh,y=lh, then we get
(72)μwk,l+λπ∑n=−NN∑m=−MMDkln,mwnm=fkl−N≤k≤N,−M≤l≤M,where
(73)Dkln,m=A−Nkh,lhn=m=−N,Ankh,lh+Bn−1kh,lh−N<n=m<N,BN−1kh,lhn=m=N.The matrixDkln,m may be written as Dkln,m=Gkln,m−Ekln,m, where
(74)Gkln,m=Ankh,lh+Bn−1kh,lh,−N≤k,l,n,≤Nis the TM of order 2N+1, and the matrix
(75)Ekln,m=B−N−1kh,lhn=m=−N,0−N<n=m<N,ANkh,lhn=m=N.However, the solution of the system can be obtained in the form(76)wk,l=μI−λGkln−Ekln−1fkl,where I is the identity matrix and μI−λGkln−Ekln≠0..


### 5.2. The PNM

Consider(77)μwx=ψx−λπ∫−11∫−11px−u;y−vk¯x−u;y−vwu,v;tdudv,where
(78)px−u;y−vk¯x−u;y−v=1x−uy−v1+πkx,yy−x,wherepand k¯ are badly behaved and well-behaved functions of their arguments, respectively. We approximate the integral term in (77) when ψsxi,yi by
(79)∫−11∫−11pxi−u;ys−vk¯xi−u;ys−vwu,v;tdudv≈∑j=0N∑i=0Mκijslk¯xi−u;ys−vwuj,vl,where κijsl is the weights. Also, we approximate the integral term in (77) in the form:
(80)∫−11∫−11pxi−u;ys−vk¯xi−u;ys−vwu,v;tdudv=∑j=0N∑i=0M∫u2ju2j+2∫v2jv2j+1pxi−u;ys−vk¯xi−u;ys−vwu,v;t,wherexi=ui=yi=vi=a+ih,i=0,1,⋯,N with h=b−a/N and N even. Now, if we approximate the nonsingular part of the integrand over each interval u2j,u2j+2, V2l,v2l+2, by the second degree Lagrange interpolation polynomial that interpolates, we find
(81)∫−11∫−11pui−u,vi−vk~ui−u,vi−vwu,v;tdudv=∑j=0N−22∑l=0M−22∫u2ju2j+2∫v2lv2l+2pui−u,vi−v×u2j+1−uv2l+1−vu2j+2−uv2l+2−v2h22h2k~ui−u2j,vi−v2lwu2j,v2l;t+u−u2jv−v2lu2j+2−uv2l+2−vh2h2k~ui−u2j+1,vi−v2l+1wu2j+1,v2l+1;t+u−u2jv−v2lu−u2j+1v−v2l+12h22h2k~ui−u2j+2,vi−v2l+2wu2j+2,v2l+2;tdudv=∑j=0N∑l=0Mwijwilk~ui−uj,vi−vlwui,vl;t,where uj=jh,uj+1=j+1h,uj−uj+1=vl−vl+1=−h.If we define(82)αijui,vs=14h4∫u2j−2u2j∫v2j−2v2jpui,u;vs,vu−u2j−2v−v2j−2u−u2j−1v−v2j−1dudv,βijui,vs=14h4∫u2j−2u2j∫v2j−2v2jpui,u;vs,vu2j−2−uv2j−2−vu2j−uv2j−vdudv,γijui,vs=14h4∫u2j−2u2j∫v2j−2v2jpui,u;vs,vu−u2j−2v−v2j−2u−u2j−1v−v2j−1dudv.In general, assumeK0 thus (82) become
(83)αijui,vs=h24∫02∫02ζδζ−1ζ−1pui,u2j−2+ζh;vs,v2l−2+δhdζdδ,βijui,vs=h24∫02∫02ζ−1ζ−2ζ−1ζ−2pui,u2j−2+ζh;vs,v2l−2+δhdζdδ,γijui,vs=h24∫02∫02ζδ2−ζ2−ζpui,u2j−2+ζh;vs,v2l−2+δhdζdδ.If we defineψk=∫02∫02ξkδkpui−u2j−2+ξh,vi−v2l−2+δhdξdδ,k=0,1,2 and let ui−u2j−2=i−2j+2h,vi−v2l−2=i−2l+2h, we have ψk=∫02∫02ξkδkpz−ζh,g−δhdζdδ,k=0,1,2,z=i−2j+2,g=s−2l+2.


## 5.1. The TMM

We present the TMM to obtain numerical solution of TDFIE of the second kind with Cauchy form, which it expresses in the form(61)μiwix+λπ∫−11∫−11p1∣x−u∣p2∣y−v∣wiu,vdudv+λ∫−11kx,ywiydy=ψix,which it may be adapted as
(62)μwx=ψx−λπ∫−11∫−11Fx−u,y−vwu,v;tdudv,where
(63)Fx−u,y−v=1+πkx,yy−xx−uy−v=1x−uy−v1+πkx,yy−x.Then, write the integral term in Eq. (62) as the form
(64)∫−11∫−11Fx−u,y−vwu,v;tdudv=∑n=−NN∫nhnh+hFx−u,y−vwu,v;tdudv,h=1N.Formula (64) reduces as
(65)∫−11∫−11Fx−u,y−vwu,v;tdudv=∑n=−NN∑m=−MMFx−u,y−vwu,v;tdudvh=1N.Then,(66)∫nhnh+h∫mhmh+hFx−u,y−vwu,v;tdudv=An,mx,ywnh,mh+Bn,mx,ywnh+h,mh+h+R.Then, we putwu,v=1.1,uv in Eq. (66), and then we obtain
(67)An,mx,y=1hnh+hmh+hInh+mh+h−Jnh+mh+h,Bn,mx,y=1hJnh+mh+h−nhmhInh+mh+h,where
(68)Ix,y=∫nhnh+h∫mhmh+hkt−τ,x−ydudv,Jx,y=∫nhnh+h∫mhmh+huv⋅kx−u,y−vdudv.Eq. (65) becomes
(69)∫−11∫−11Fx−u,y−vwu,v;tdudv=∑n=−NN−1∑m=−MM−1An,mx,ywnh,mh;t+Bn,mx,ywnh+h,mh+h;t=∑n=−NN‐1∑m=−MM−1An,mx,ywnh,mh;t+∑n=−NN∑m=−MMBn−1m−1x,ywnh,mh;t=∑n=−NN∑m=−MMDn,mx,ywnh,mh;t,where
(70)Dn,mx,y=A−Nx,yn=m=−N,Anx,y+Bn−1x,y−N<n=m<N,BN−1x,yn=m=N.Thus, the IE (62) becomes
(71)μwx,y+λπ∑n=−NN∑m=−MMDn,mx,ywnh,mh;t=fx,y.If we putx=kh,y=lh, then we get
(72)μwk,l+λπ∑n=−NN∑m=−MMDkln,mwnm=fkl−N≤k≤N,−M≤l≤M,where
(73)Dkln,m=A−Nkh,lhn=m=−N,Ankh,lh+Bn−1kh,lh−N<n=m<N,BN−1kh,lhn=m=N.The matrixDkln,m may be written as Dkln,m=Gkln,m−Ekln,m, where
(74)Gkln,m=Ankh,lh+Bn−1kh,lh,−N≤k,l,n,≤Nis the TM of order 2N+1, and the matrix
(75)Ekln,m=B−N−1kh,lhn=m=−N,0−N<n=m<N,ANkh,lhn=m=N.However, the solution of the system can be obtained in the form(76)wk,l=μI−λGkln−Ekln−1fkl,where I is the identity matrix and μI−λGkln−Ekln≠0..


## 5.2. The PNM

Consider(77)μwx=ψx−λπ∫−11∫−11px−u;y−vk¯x−u;y−vwu,v;tdudv,where
(78)px−u;y−vk¯x−u;y−v=1x−uy−v1+πkx,yy−x,wherepand k¯ are badly behaved and well-behaved functions of their arguments, respectively. We approximate the integral term in (77) when ψsxi,yi by
(79)∫−11∫−11pxi−u;ys−vk¯xi−u;ys−vwu,v;tdudv≈∑j=0N∑i=0Mκijslk¯xi−u;ys−vwuj,vl,where κijsl is the weights. Also, we approximate the integral term in (77) in the form:
(80)∫−11∫−11pxi−u;ys−vk¯xi−u;ys−vwu,v;tdudv=∑j=0N∑i=0M∫u2ju2j+2∫v2jv2j+1pxi−u;ys−vk¯xi−u;ys−vwu,v;t,wherexi=ui=yi=vi=a+ih,i=0,1,⋯,N with h=b−a/N and N even. Now, if we approximate the nonsingular part of the integrand over each interval u2j,u2j+2, V2l,v2l+2, by the second degree Lagrange interpolation polynomial that interpolates, we find
(81)∫−11∫−11pui−u,vi−vk~ui−u,vi−vwu,v;tdudv=∑j=0N−22∑l=0M−22∫u2ju2j+2∫v2lv2l+2pui−u,vi−v×u2j+1−uv2l+1−vu2j+2−uv2l+2−v2h22h2k~ui−u2j,vi−v2lwu2j,v2l;t+u−u2jv−v2lu2j+2−uv2l+2−vh2h2k~ui−u2j+1,vi−v2l+1wu2j+1,v2l+1;t+u−u2jv−v2lu−u2j+1v−v2l+12h22h2k~ui−u2j+2,vi−v2l+2wu2j+2,v2l+2;tdudv=∑j=0N∑l=0Mwijwilk~ui−uj,vi−vlwui,vl;t,where uj=jh,uj+1=j+1h,uj−uj+1=vl−vl+1=−h.If we define(82)αijui,vs=14h4∫u2j−2u2j∫v2j−2v2jpui,u;vs,vu−u2j−2v−v2j−2u−u2j−1v−v2j−1dudv,βijui,vs=14h4∫u2j−2u2j∫v2j−2v2jpui,u;vs,vu2j−2−uv2j−2−vu2j−uv2j−vdudv,γijui,vs=14h4∫u2j−2u2j∫v2j−2v2jpui,u;vs,vu−u2j−2v−v2j−2u−u2j−1v−v2j−1dudv.In general, assumeK0 thus (82) become
(83)αijui,vs=h24∫02∫02ζδζ−1ζ−1pui,u2j−2+ζh;vs,v2l−2+δhdζdδ,βijui,vs=h24∫02∫02ζ−1ζ−2ζ−1ζ−2pui,u2j−2+ζh;vs,v2l−2+δhdζdδ,γijui,vs=h24∫02∫02ζδ2−ζ2−ζpui,u2j−2+ζh;vs,v2l−2+δhdζdδ.If we defineψk=∫02∫02ξkδkpui−u2j−2+ξh,vi−v2l−2+δhdξdδ,k=0,1,2 and let ui−u2j−2=i−2j+2h,vi−v2l−2=i−2l+2h, we have ψk=∫02∫02ξkδkpz−ζh,g−δhdζdδ,k=0,1,2,z=i−2j+2,g=s−2l+2.


## 6. Numerical Applications and Discussions

In this section, we state some applications and numerical results to discuss the approximate solution (i.e., the treat of the surface cracks that occurs in the materials as a result, the constant in finite of materials, which the deformation increases as well as the time increases in the interval0,T. The TMM and PNM are used to get numerical solution for values of μ=1, and for different materials, plutonium υ=0.21 and fiber υ=0.22, where the Poisson ratio is 0≤υ<1,λ=λ¯/2G,λ¯=2Gυ/1−2υ,Gshearmodules.We divided the position interval by N=10, 20 units. Since 0≤t≤T<∞, we choose the time T=0.02, 0.2. Tables 1 and 2 are as follows: exact sol. → the exact solution, Approx.T. →approximate solution of TMM, error.T. → the absolute error of TMM, Approx.N. →approximate solution of PNM, and error.N. → the absolute error of PNM. Figures 2–17 show the value absolute error by TMM and PNM at N=10, 20 and the time T=0.02, 0.2, for the materials plutonium υ=0.21 and fiber υ=0.22.Table 1
The values of exact, approximate, and absolute error values by TMM and PNM for (fiberυ=0.22).
TNxExact sol.Approx.T.Error.T.Approx.N.Error.N.0.0210-1.000.004000000.000452990.00005299010.004459210.00045921-0.600.000144000.000140290.370969×10−50.002028740.00188474-0.200.000016000.000020220.422877×10−50.002649460.002633460.200.000016000.000020220.422877×10−50.004832890.004816890.600.000144000.000140290.37096×10−50.007972190.007828191.000.004000000.000452990.00005299010.002468120.0015318720-1.000.004000000.000479520.00007952000.004432680.00043268-0.600.000144000.000144640.6491943×10−60.002024390.00188039-0.200.000016000.000023480.7485778×10−50.002646200.002630200.200.000016000.000023480.748577×10−50.004829630.004813630.600.000144000.000144640.6491944×10−60.007967480.007823841.000.004000000.000479520.00007952000.002441590.001558400.210-1.000.040000000.087091720.0470917250.082179520.04217952-0.600.014400000.024304910.0099049130.022135870.00773587-0.200.001600000.002384840.0007848410.000284840.001315150.200.001600000.002384840.0007848410.002468270.000868270.600.014400000.024304910.0099049130.016192420.001792421.000.040000000.087091720.0470917250.084170600.0441706020-1.000.040000000.092517550.052517550.087605350.04760535-0.600.014400000.025180750.010780750.023011710.00861171-0.200.001600000.003028130.001428130.000358440.001241550.200.001600000.003028130.001428130.001824980.000224980.600.014400000.025180750.010780750.017068260.002668261.000.040000000.092517550.052517550.089596430.04959643Table 2
The values of exact, approximate, and absolute error values by TMM and PNM for (Plutoniumυ=0.21).
TNxExact sol.Approx.T.ErrorT.Approx.N.ErrorN.0.0210-1.000.004000000.0004430780.0000430780.004469120.00046912-0.600.000144000.0001408360.316366×10−50.002028190.00188419-0.200.000016000.0000200810.408164×10−50.002649600.002633600.200.000016000.0000200810.408164×10−50.004833030.004817030.600.000144000.0001408360.316366×10−50.007971640.007827641.000.004000000.0004430780.00004307840.002478040.0015219620-1.000.004000000.000463330.0000633390.004448870.00044887-0.600.000144000.000144840.843735×10−60.002024190.00188019-0.200.000016000.000023090.7095783×10−50.002646590.002630590.200.000016000.000023090.7095783×10−50.004830020.004814020.600.000144000.000144840.843735×10−60.007967640.007823641.000.004000000.000463330.0000633390.002457780.001542210.210-1.000.040000000.078639960.0386399640.073727760.03372776-0.600.014400000.023896630.0094663790.021727590.00732759-0.200.001600000.002588110.0009881110.000081570.001518420.200.001600000.002588110.0009881110.002265000.000665000.600.014400000.023896630.0094966370.015784140.001384141.000.040000000.078639960.0386399640.075718840.0357188420-1.000.040000000.082322990.042322990.077410790.03741079-0.600.014400000.024620940.010220940.022451900.00805190-0.200.001600000.003128930.0015289330.000459240.001140750.200.001600000.003128940.0015289330.001724170.000124170.600.014400000.024620940.0102209430.016508450.002108451.000.040000000.082322990.0423229940.079401870.03940187Figure 2
The value absolute error by TMM atN=10 and T=0.02.Figure 3
The value absolute error by PNM atN=10 and T=0.02.Figure 4
The value absolute error by TMM atN=20 and T=0.02.Figure 5
The value absolute error by PNM atN=20 and T=0.02.Figure 6
The value absolute error by TMM atN=10 and T=0.2.Figure 7
The value absolute error by PNM atN=10 and T=0.2.Figure 8
The value absolute error by TMM atN=20 and T=0.2.Figure 9
The value absolute error by PNM atN=20 and T=0.2.Figure 10
The value absolute error by TMM atN=10 and T=0.02.Figure 11
The value absolute error by PNM atN=10 and T=0.02.Figure 12
The value absolute error by TMM atN=20 and T=0.02.Figure 13
The value absolute error by PNM atN=20 and T=0.02.Figure 14
The value absolute error by TMM atN=10 and T=0.2.Figure 15
The value absolute error by PNM atN=10 and T=0.2.Figure 16
The value absolute error by TMM atN=20 and T=0.2.Figure 17
The value absolute error by PNM atN=20 and T=0.2.Consider(84)μwx,t=fx,t−λμ∫−11∫−111x−uy−vwu,v;tdudv−λ∫−11∣x−y∣wy,tdy−∫0ttτwx,τdτ,μ=1.The exact solutionϕx,y=x2y2.


## 7. The Conclusion

We have presented a successful technique for the numerical solution of MIE with singular kernel in two-dimensional by using TMM and PNM which is established from the plane strain problem for the bounded layer medium composed of different materials. From Tables1 and 2 and Figures 2–17, we note that the errors due to the TMM are less than the errors due to PNM. In addition, we note that N increases for the two different materials (fiber υ=0.22) (plutonium υ=0.21), the values of kgx−gy=cotgy−gx/2 and x∈−π,π,ϕ±π,t=0. are fixed, and the error values increase. The approximate solution is nearly coincident with the exact solution for t>0 at each value of x∈−1,1.



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