# Approximate Solutions of Delay Parabolic Equations with the Dirichlet Condition

Authors: Deniz Agirseven
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

Finite difference and homotopy analysis methods are used for the approximate solution of the initial-boundary value problem for the delay parabolic partial differential equation with the Dirichlet condition. The convergence estimates for the solution of first and second orders of difference schemes in Hölder norms are obtained. A procedure of modified Gauss elimination method is used for the solution of these difference schemes. Homotopy analysis method is applied. Comparison of finite difference and homotopy analysis methods is given on the problem.


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## Body

## 1. Introduction

Increase in interest in the theoretical aspects of numerical methods for delay differential equations points out that delay differential equations are capable of generating extensive and conceivable models for phenomena in many branches of sciences. Numerical solutions of the delay ordinary differential equations have been studied mostly for ordinary differential equations (cf., e.g., [1–14] and the references therein). Nevertheless, delay partial differential equations are less in demand than delay ordinary differential equations. Different kinds of problems for delay partial differential equations are solved by using operator approach (see, e.g., [15–17]).In recent years, Ashyralyev and Sobolevskii considered the initial-value problem for linear delay partial differential equations of parabolic type in the spacesC(Eα) of functions defined on the segment [0,∞) with values in a Banach space Eα and the stability inequalities were established under stronger assumption than the necessary condition of the stability of the differential problem. The stability estimates for the solutions of difference schemes of the first- and second-order accuracy difference schemes for approximately solving this initial-value problem for delay differential equations of parabolic type were presented. They obtained the stability estimates in Hölder norms for solutions of the initial-value problem of the delay differential and difference equations of the parabolic type [15, 16]. Gabriella used extrapolation spaces to solve Banach spaces valued delay differential equations with unbounded delay operators. The author proved regularity properties of various types of solutions and investigated the existence of strong and weak solutions for a class of abstract semilinear delay equations [17].In this paper, finite difference (see, e.g., [18–28]) and homotopy analysis methods (HAM) (see, e.g., [29–37]) for the approximate solutions of the delay differential equation of the parabolic type(1.1)ut(t,x)+(-a(x)uxx(t,x)+b(x)ux(t,x)+c(x)u(t,x))=d(t)(-a(x)uxx(t-ω,x)+b(x)ux(t-ω,x)+c(x)u(t-ω,x)),0<t<∞,x∈(0,l),u(t,x)=g(t,x),-ω≤t≤0,x∈[0,l],u(t,0)=u(t,l)=0,t≥0,
are studied. Here g(t,x)(t∈(-∞,0),x∈[0,l]),a(x),b(x),c(x)(x∈(0,∞)) are given smooth bounded functions and a(x)≥a>0.Difference schemes which are accurate to first and second orders for the approximate solution of problem (1.1) are presented. The convergence estimates for the solution of these difference schemes are obtained. For the numerical study, procedure of modified Gauss elimination method is used to solve these difference schemes. Homotopy analysis method is applied to find the solution of problem (1.1). The numerical results are obtained at the same points for each method. Comparison of finite difference and homotopy analysis methods is given on the problem.


## 2. The Finite Difference Method

In this section, the first and second orders of accuracy int for the approximate solution of problem (1.1) are considered. The convergence estimates for the solution of these difference schemes are established. A procedure of modified Gauss elimination method is used to solve these difference schemes.


### 2.1. The Difference Scheme, Convergence Estimates

The discretization of problem (1.1) is carried out in two steps. In the first step, we define the grid space(2.1)[0,L]h={x=xn:xn=nh,0≤n≤M,Mh=L}.
To formulate our results, we introduce the Banach spaceC∘hα=C∘α[0,L]h, α∈[0,1), of all grid functions φh={φn}n=1M-1 defined on [0,L]h with φ0=φM=0 equipped with the norm(2.2)‖φh‖C∘hα=‖φh‖Ch+sup⁡1≤n<n+r≤M-1|φn+r-φn|(rh)-α,‖φh‖Ch=max⁡1≤n≤M-1|φn|.
Moreover, Cτ(E)=C([0,∞)τ,E) is the Banach space of all grid functions fτ={fk}k=1∞ defined on(2.3)[0,∞)τ={tk=kτ,k=0,1,…}
with values in E equipped with the norm(2.4)‖fτ‖Cτ(E)=sup⁡1≤k<∞‖fk‖E.
To the differential operator A generated by problem (1.1), we assign the difference operators Ahx,Bhx by the formulas(2.5)Ahxφh(x)={-a(xn)φn+1-2φn+φn-1h2+b(xn)φn+1-φn-12h+c(xn)φn}1M-1,Bhx(t)φh(x)=d(t)Ahxφh,
acting in the space of grid functions φh(x)={φn}1M-1 satisfying the conditions φ0=φM=0. It is well known that Ahx is a strongly positive operator in Ch. With the help of Ahx and d(t)Ahx, we arrive at the initial value problem(2.6)duh(t,x)dt+Ahxuh(t,x)=d(t)Ahxuh(t-w,x),0<t<∞,0<x<L,uh(t,x)=gh(t,x),-ω≤t≤0,0≤x≤L.
In the second step, we consider difference schemes of first and second orders of accuracy(2.7)1τ(ukh(x)-uk-1h(x))+Ahxukh(x)=d(tk)Ahxuk-Nh(x),tk=kτ,1≤k,Nτ=w,ukh(x)=gh(tk,x),tk=kτ,-N≤k≤0,(2.8)1τ(ukh(x)-uk-1h(x))+(Ahx+12τ(Ahx)2)ukh(x)=12(I+τ2Ahx)d(tk-τ2)Ahx(uk-Nh(x)+uk-N-1h(x)),tk=kτ,1≤k,ukh=gh(tk,x),tk=kτ,-N≤k≤0.Theorem 2.1.
Assume that(2.9)sup⁡0≤t<∞|d(t)|≤1-αM22-α.
Suppose that problem (1.1) has a smooth solution u(t,x) and
(2.10)∫0∞[max⁡0≤x≤L|uss(s,x)|+sup⁡0<x<x+y<L|uss(s,x+y)-uss(s,x)|y2α]ds<∞,∫0∞[max⁡0≤x≤L|uxxxx(s,x)|+sup⁡0<x<x+y<L|uxxxx(s,x+y)-uxxxx(s,x)|y2α]ds<∞.
Then, for the solution of difference scheme (2.7), the following convergence estimate holds:
(2.11)sup⁡k‖ukh-uh(tk,⋅)‖C∘h2α≤M1(τ+h2)
with M1 being a real number independent of τ, α, and h.Proof.
Using notations ofAhx and Bhx, we can obtain the following formula for the solution:
(2.12)ukh(x)=Rkgh(0,x)+∑j=1kRk-j+1Bjxgh(tj-N,x)τ,1≤k≤N,ukh(x)=Rk-nNunNh(x)+∑j=nN+1kRk-j+1Bjxuj-Nh(x)τ,nN≤k≤(n+1)N,
where R=(I+τAhx)-1. The proof of Theorem 2.1 is based on the formulas (2.12), on the convergence theorem, on the difference schemes in Cτ(Eαh) (see [38]), on the estimate
(2.13)∥exp⁡{-tkAhx}∥Ch→Ch≤M,k≥0,
and on the fact that in Eαh=Eα(Ahx,Ch) the norms are equivalent to the norms in C∘h2α uniformly in h for 0<α<1/2 (see, [18]).Theorem 2.2.
Assume that assumption (2.9) of Theorem 2.1 and the following conditions hold:
(2.14)∫0∞[max⁡0≤x≤L|usss(s,x)|+sup⁡0<x<x+y<L|usss(s,x+y)-usss(s,x)|y2α]ds<∞,∫0∞[max⁡0≤x≤L|uxxss(s,x)|+sup⁡0<x<x+y<L|uxxss(s,x+y)-uxxss(s,x)|y2α]ds<∞,∫0∞[max⁡0≤x≤L|uxxxxs(s,x)|+sup⁡0<x<x+y<L|uxxxxs(s,x+y)-uxxxxs(s,x)|y2α]ds<∞.
Then for the solution of difference scheme (2.8), the following convergence estimate is satisfied:
(2.15)sup⁡k‖ukh-uh(tk,⋅)‖C∘h2α≤M2(τ2+h2)
with M2 being a real number independent of τ, α, and h.Proof.
Using notations ofAhx and Bhx again, we can obtain the following formula for the solution:
(2.16)ukh(x)=Rkgh(0,x)+∑j=1kRk-j+1(I+τAhx2)(gh(tj-N,x)-gh(tj-N-1,x))τ,1≤k≤N,ukh(x)=Rk-nNunNh(x)+∑j=nN+1kRk-j+1(I+τAhx2)Bjx12(uj-Nh(x)+uj-N-1h(x))τ,nN≤k≤(n+1)N,
where R=(I+τAhx+(τAhx)2/2)-1. The proof of Theorem 2.2 is based on the formulas (2.16), on the convergence theorem, on the difference schemes in Cτ(Eαh) (see, [38]), on the estimate (2.13), and on the equivalence of the norms as in Theorem 2.1.
Finally, the numerical methods are given in the following section for the solution of delay parabolic differential equation with the Dirichlet condition. The method is illustrated by numerical examples.


### 2.2. Numerical Results

We consider the initial-boundary-value problem(2.17)∂u(t,x)∂t-∂2u(t,x)∂x2+(0.1)∂2u(t-1,x)∂x2=0,t>0,0<x<π,u(t,x)=e-tsin⁡x,-1≤t≤0,0≤x≤π,u(t,0)=u(t,π)=0,t≥0,
for the delay parabolic differential equation.The exact solution of this problem fort∈[n-1,n], n=0,1,2,…,x∈[0,π] is(2.18)u(t,x)={e-tsin⁡x,-1≤t≤0,e-t{1+(0.1)et}sin⁡x,0≤t≤1,e-t{1+(0.1)et+[(0.1)e(t-1)]22!}sin⁡x,1≤t≤2,⋮e-t{1+(0.1)et+[(0.1)e(t-1)]22!+⋯+[(0.1)e(t-n)](n+1)(n+1)!}sin⁡x,n≤t≤n+1,⋮
For the approximate solution of delay parabolic equation (2.17), consider the set of grid points(2.19)[-1,∞]τ×[0,π]h={(tk,xn):tk=kτ,-N≤k≤∞,xn=nh,0≤n≤M,Mh=π}.
Using difference scheme accurate to first order for the approximate solutions of the initial-boundary-value problem for the delay parabolic equation (2.17), we get the following system of equations:(2.20)unk-unk-1τ-un+1k-2unk+un-1kh2+(0.1)un+1k-N-2unk-N+un-1k-Nh2=0,mN+1≤k≤(m+1)N,m=0,1,…,1≤n≤M-1,unk=e-tksin⁡xn,-N≤k≤0,0≤n≤M,u0k=uMk=0,k≥0.
In this first step, applying difference scheme accurate to first order, we obtain a system of equations in matrix form(2.21)AUn+1m+BUnm+CUn-1m=Rφnm,1≤n≤M-1,m=0,1,…,U0m=0̃,UMm=0̃,
where A, B, C are (N+1)×(N+1) matrices defined by(2.22)A=[00000⋅⋅⋅00000a000⋅⋅⋅000000a00⋅⋅⋅0000000a0⋅⋅⋅00000000a⋅⋅⋅0000⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅00000⋅⋅⋅a00000000⋅⋅⋅0a0000000⋅⋅⋅00a000000⋅⋅⋅000a](N+1)×(N+1),B=[100⋅⋅⋅0⋅⋅⋅0⋅⋅⋅00bc0⋅⋅⋅0⋅⋅⋅0⋅⋅⋅000bc⋅⋅⋅0⋅⋅⋅0⋅⋅⋅0000b⋅⋅⋅0⋅⋅⋅0⋅⋅⋅00000⋅⋅⋅0⋅⋅⋅0⋅⋅⋅00000⋅⋅⋅0⋅⋅⋅0⋅⋅⋅00000⋅⋅⋅0⋅⋅⋅0⋅⋅⋅00⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅000⋅⋅⋅0⋅⋅⋅0⋅⋅⋅00000⋅⋅⋅0⋅⋅⋅0⋅⋅⋅c0000⋅⋅⋅0⋅⋅⋅0⋅⋅⋅bc](N+1)×(N+1).C=A, R is (N+1)×(N+1) identity matrix and φnm, Usm are (N+1)×1 column vectors as(2.23)φnm=[φnmNφnmN+1⋮φn(m+1)N](N+1)×(1),Usm=[UsmNUsmN+1⋮Us(m+1)N](N+1)×(1)fors=n±1,n,
where unmN is given for any m=0,1,…,(2.24)φnk=-(0.1)un+1k-N-2unk-N+un-1k-Nh2,mN+1≤k≤(m+1)N,m=0,1,…,1≤n≤M-1,unk=e-tksin⁡xn,-N≤k≤0.
Here, we denote(2.25)a=-1h2,b=-1τ,c=1τ+2h2.
So, we have second-order difference equation (2.21) with matrix coefficients. To solve this difference equation, we have applied a procedure of modified Gauss elimination method. Hence, we obtain a solution of the matrix equation in the following form:(2.26)Ujm=αj+1Uj+1m+βj+1m,j=M-1,…,2,1,UMm=0,
where αj(j=1,…,M) are (N+1)×(N+1) square matrices and βjm(j=1,…,M) are (N+1)×1 column matrices defined by(2.27)αj+1=-(B+Cαj)-1A,βj+1m=(B+Cαj)-1(Rφjm-Cβj),
where j=1,…,M-1, α1 is the (N+1)×(N+1) zero matrix, and β1m is the (N+1)×1 zero matrix.Second, using the second order of accuracy difference scheme for the approximate solutions of problem (2.17) and applying formulae(2.28)2u(0)-5u(h)+4u(2h)-u(3h)h2-u′′(0)=O(h2),2u(1)-5u(1-h)+4u(1-2h)-u(1-3h)h2-u′′(1)=O(h2),
we obtain the following system of equations:(2.29)unk-unk-1τ-un+1k-2unk+un-1kh2+τ2(un+2k-4un+1k+6unk-4un-1k+un-2kh4)+(0.1){un+1k-N-2unk-N+un-1k-N2h2+un+1k-1-N-2unk-1-N+un-1k-1-N2h2-τ2[un+2k-N-4un+1k-N+6unk-N-4un-1k-N+un-2k-N2h4+un+2k-1-N-4un+1k-1-N+6unk-1-N-4un-1k-1-N+un-2k-1-N2h4]}=0mN+1≤k≤(m+1)N,m=0,1,…,2≤n≤M-2,unk=e-tksin⁡xn,-N≤k≤0,0≤n≤M,u1k=45u2k-15u3k,k≥0,uM-1k=45uM-2k-15uM-3k,k≥0,u0k=uMk=0,k≥0.
In the second step, we apply second-order difference scheme to get the system of linear equations in matrix form(2.30)AUn+2m+BUn+1m+CUnm+DUn-1m+EUn-2m=Rφnm,m=0,1,…,2≤n≤M-2,U0m=0̃,UMm=0̃,U1m=45U2m-15U3m,UM-1m=45UM-2m-15UM-3m,
where A, B, C are (N+1)×(N+1) matrices defined by(2.31)A=[00000⋅⋅⋅00000x000⋅⋅⋅000000x00⋅⋅⋅0000000x0⋅⋅⋅00000000x⋅⋅⋅0000⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅00000⋅⋅⋅x00000000⋅⋅⋅0x0000000⋅⋅⋅00x000000⋅⋅⋅000x](N+1)×(N+1),B=[00000⋅⋅⋅00000y000⋅⋅⋅000000y00⋅⋅⋅0000000y0⋅⋅⋅00000000y⋅⋅⋅0000⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅00000⋅⋅⋅y00000000⋅⋅⋅0y0000000⋅⋅⋅00y000000⋅⋅⋅000y](N+1)×(N+1),C=[10000⋅⋅⋅0000zt000⋅⋅⋅00000zt00⋅⋅⋅000000zt0⋅⋅⋅0000000zt⋅⋅⋅0000⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅00000⋅⋅⋅zt0000000⋅⋅⋅0zt000000⋅⋅⋅00zt](N+1)×(N+1),B=D, E=A, R is (N+1)×(N+1) identity matrix, and φnm, Usm are (N+1)×1 column vectors as(2.32)φnm=[φnmNφnmN+1⋮φn(m+1)N](N+1)×(1),Usm=[UsmNUsmN+1⋮Us(m+1)N](N+1)×(1)fors=n±1,n±2,n,
where unmN is given for any m=0,1,…,(2.33)φnk=-(0.1){un+1k-N-2unk-N+un-1k-N2h2+un+1k-1-N-2unk-1-N+un-1k-1-N2h2-τ2[un+2k-N-4un+1k-N+6unk-N-4un-1k-N+un-2k-N2h4+un+2k-1-N-4un+1k-1-N+6unk-1-N-4un-1k-1-N+un-2k-1-N2h4]},mN+1≤k≤(m+1)N,m=0,1,…,1≤n≤M-1,unk=e-tksin⁡xn,-N≤k≤0.Here, we denote(2.34)x=τ2h4,y=-1h2-2τh4,z=-1τ,t=1τ+2h2+3τh4,
Hence, we have second-order difference equation (2.30) with matrix coefficients. For the solution of this matrix equation, we use the modified Gauss elimination method. We seek a solution of the matrix equation by the following form:(2.35)Ujm=αj+1Uj+1m+βj+1Uj+2m+γj+1m,j=M-2,…,2,1,0UMm=0,UM-1m=[(βM-2+5I)-4(I-αM-2)αM-1]-1[4(I-αM-2)γM-1m-γM-2m],
where αj(j=2,…,M-2) and βj(j=2,…,M-2) are (N+1)×(N+1) square matrices and γjm(j=2,…,M-2) are column matrices defined by(2.36)αj+1=-(C+Dαj+Eβj-1+Eαj-1αj)-1(B+Dβj+Eαj-1βj),βj+1=-(C+Dαj+Eβj-1+Eαj-1αj)-1(A),γj+1m=-(C+Dαj+Eβj-1+Eαj-1αj)-1(Rφjm-Dγjm-Eαj-1γjm-Eγj-1m),
where j=2,…,M-2, α1 is (N+1)×(N+1) zero matrix, and β1 is (N+1)×(N+1) zero matrix, γ1m and γ2m are (N+1)×1 zero matrices.We give the results of the numerical analysis. The numerical solutions are recorded for different values ofN and M and unk represent the numerical solutions of these difference schemes at (tk,xn). Tables 1, 2, 3, and 4 are constructed for N=M=20,40,80 in t∈[0,1], t∈[1,2], t∈[2,3], t∈[3,4], respectively, and the error is computed by the following formula:(2.37)EMN=max⁡-N≤k≤N1≤n≤M-1|u(tk,xn)-unk|.
Thus, by using the second order of accuracy difference scheme, the accuracy of solution increases faster than the first order of accuracy difference scheme.Table 1
Comparison of the errors of different difference schemes int∈[0,1].
MethodN=M=20N=M=40N=M=80Difference scheme (2.20)0.005990880.002860920.00139629Difference scheme (2.29)0.000762650.000203170.00005148Table 2
Comparison of the errors of different difference schemes int∈[1,2].
MethodN=M=20N=M=40N=M=80Difference scheme (2.20)0.073241510.036930610.01845008Difference scheme (2.29)0.000770550.000205410.00005206Table 3
Comparison of the errors of different difference schemes int∈[2,3].
MethodN=M=20N=M=40N=M=80Difference scheme (2.20)0.035200810.017498620.00872690Difference scheme (2.29)0.000670840.000178450.00004521Table 4
Comparison of the errors of different difference schemes int∈[3,4].
MethodN=M=20N=M=40N=M=80Difference scheme (2.20)0.016871660.008445770.00421761Difference scheme (2.29)0.000455270.000121070.00003067


## 2.1. The Difference Scheme, Convergence Estimates

The discretization of problem (1.1) is carried out in two steps. In the first step, we define the grid space(2.1)[0,L]h={x=xn:xn=nh,0≤n≤M,Mh=L}.
To formulate our results, we introduce the Banach spaceC∘hα=C∘α[0,L]h, α∈[0,1), of all grid functions φh={φn}n=1M-1 defined on [0,L]h with φ0=φM=0 equipped with the norm(2.2)‖φh‖C∘hα=‖φh‖Ch+sup⁡1≤n<n+r≤M-1|φn+r-φn|(rh)-α,‖φh‖Ch=max⁡1≤n≤M-1|φn|.
Moreover, Cτ(E)=C([0,∞)τ,E) is the Banach space of all grid functions fτ={fk}k=1∞ defined on(2.3)[0,∞)τ={tk=kτ,k=0,1,…}
with values in E equipped with the norm(2.4)‖fτ‖Cτ(E)=sup⁡1≤k<∞‖fk‖E.
To the differential operator A generated by problem (1.1), we assign the difference operators Ahx,Bhx by the formulas(2.5)Ahxφh(x)={-a(xn)φn+1-2φn+φn-1h2+b(xn)φn+1-φn-12h+c(xn)φn}1M-1,Bhx(t)φh(x)=d(t)Ahxφh,
acting in the space of grid functions φh(x)={φn}1M-1 satisfying the conditions φ0=φM=0. It is well known that Ahx is a strongly positive operator in Ch. With the help of Ahx and d(t)Ahx, we arrive at the initial value problem(2.6)duh(t,x)dt+Ahxuh(t,x)=d(t)Ahxuh(t-w,x),0<t<∞,0<x<L,uh(t,x)=gh(t,x),-ω≤t≤0,0≤x≤L.
In the second step, we consider difference schemes of first and second orders of accuracy(2.7)1τ(ukh(x)-uk-1h(x))+Ahxukh(x)=d(tk)Ahxuk-Nh(x),tk=kτ,1≤k,Nτ=w,ukh(x)=gh(tk,x),tk=kτ,-N≤k≤0,(2.8)1τ(ukh(x)-uk-1h(x))+(Ahx+12τ(Ahx)2)ukh(x)=12(I+τ2Ahx)d(tk-τ2)Ahx(uk-Nh(x)+uk-N-1h(x)),tk=kτ,1≤k,ukh=gh(tk,x),tk=kτ,-N≤k≤0.Theorem 2.1.
Assume that(2.9)sup⁡0≤t<∞|d(t)|≤1-αM22-α.
Suppose that problem (1.1) has a smooth solution u(t,x) and
(2.10)∫0∞[max⁡0≤x≤L|uss(s,x)|+sup⁡0<x<x+y<L|uss(s,x+y)-uss(s,x)|y2α]ds<∞,∫0∞[max⁡0≤x≤L|uxxxx(s,x)|+sup⁡0<x<x+y<L|uxxxx(s,x+y)-uxxxx(s,x)|y2α]ds<∞.
Then, for the solution of difference scheme (2.7), the following convergence estimate holds:
(2.11)sup⁡k‖ukh-uh(tk,⋅)‖C∘h2α≤M1(τ+h2)
with M1 being a real number independent of τ, α, and h.Proof.
Using notations ofAhx and Bhx, we can obtain the following formula for the solution:
(2.12)ukh(x)=Rkgh(0,x)+∑j=1kRk-j+1Bjxgh(tj-N,x)τ,1≤k≤N,ukh(x)=Rk-nNunNh(x)+∑j=nN+1kRk-j+1Bjxuj-Nh(x)τ,nN≤k≤(n+1)N,
where R=(I+τAhx)-1. The proof of Theorem 2.1 is based on the formulas (2.12), on the convergence theorem, on the difference schemes in Cτ(Eαh) (see [38]), on the estimate
(2.13)∥exp⁡{-tkAhx}∥Ch→Ch≤M,k≥0,
and on the fact that in Eαh=Eα(Ahx,Ch) the norms are equivalent to the norms in C∘h2α uniformly in h for 0<α<1/2 (see, [18]).Theorem 2.2.
Assume that assumption (2.9) of Theorem 2.1 and the following conditions hold:
(2.14)∫0∞[max⁡0≤x≤L|usss(s,x)|+sup⁡0<x<x+y<L|usss(s,x+y)-usss(s,x)|y2α]ds<∞,∫0∞[max⁡0≤x≤L|uxxss(s,x)|+sup⁡0<x<x+y<L|uxxss(s,x+y)-uxxss(s,x)|y2α]ds<∞,∫0∞[max⁡0≤x≤L|uxxxxs(s,x)|+sup⁡0<x<x+y<L|uxxxxs(s,x+y)-uxxxxs(s,x)|y2α]ds<∞.
Then for the solution of difference scheme (2.8), the following convergence estimate is satisfied:
(2.15)sup⁡k‖ukh-uh(tk,⋅)‖C∘h2α≤M2(τ2+h2)
with M2 being a real number independent of τ, α, and h.Proof.
Using notations ofAhx and Bhx again, we can obtain the following formula for the solution:
(2.16)ukh(x)=Rkgh(0,x)+∑j=1kRk-j+1(I+τAhx2)(gh(tj-N,x)-gh(tj-N-1,x))τ,1≤k≤N,ukh(x)=Rk-nNunNh(x)+∑j=nN+1kRk-j+1(I+τAhx2)Bjx12(uj-Nh(x)+uj-N-1h(x))τ,nN≤k≤(n+1)N,
where R=(I+τAhx+(τAhx)2/2)-1. The proof of Theorem 2.2 is based on the formulas (2.16), on the convergence theorem, on the difference schemes in Cτ(Eαh) (see, [38]), on the estimate (2.13), and on the equivalence of the norms as in Theorem 2.1.
Finally, the numerical methods are given in the following section for the solution of delay parabolic differential equation with the Dirichlet condition. The method is illustrated by numerical examples.


## 2.2. Numerical Results

We consider the initial-boundary-value problem(2.17)∂u(t,x)∂t-∂2u(t,x)∂x2+(0.1)∂2u(t-1,x)∂x2=0,t>0,0<x<π,u(t,x)=e-tsin⁡x,-1≤t≤0,0≤x≤π,u(t,0)=u(t,π)=0,t≥0,
for the delay parabolic differential equation.The exact solution of this problem fort∈[n-1,n], n=0,1,2,…,x∈[0,π] is(2.18)u(t,x)={e-tsin⁡x,-1≤t≤0,e-t{1+(0.1)et}sin⁡x,0≤t≤1,e-t{1+(0.1)et+[(0.1)e(t-1)]22!}sin⁡x,1≤t≤2,⋮e-t{1+(0.1)et+[(0.1)e(t-1)]22!+⋯+[(0.1)e(t-n)](n+1)(n+1)!}sin⁡x,n≤t≤n+1,⋮
For the approximate solution of delay parabolic equation (2.17), consider the set of grid points(2.19)[-1,∞]τ×[0,π]h={(tk,xn):tk=kτ,-N≤k≤∞,xn=nh,0≤n≤M,Mh=π}.
Using difference scheme accurate to first order for the approximate solutions of the initial-boundary-value problem for the delay parabolic equation (2.17), we get the following system of equations:(2.20)unk-unk-1τ-un+1k-2unk+un-1kh2+(0.1)un+1k-N-2unk-N+un-1k-Nh2=0,mN+1≤k≤(m+1)N,m=0,1,…,1≤n≤M-1,unk=e-tksin⁡xn,-N≤k≤0,0≤n≤M,u0k=uMk=0,k≥0.
In this first step, applying difference scheme accurate to first order, we obtain a system of equations in matrix form(2.21)AUn+1m+BUnm+CUn-1m=Rφnm,1≤n≤M-1,m=0,1,…,U0m=0̃,UMm=0̃,
where A, B, C are (N+1)×(N+1) matrices defined by(2.22)A=[00000⋅⋅⋅00000a000⋅⋅⋅000000a00⋅⋅⋅0000000a0⋅⋅⋅00000000a⋅⋅⋅0000⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅00000⋅⋅⋅a00000000⋅⋅⋅0a0000000⋅⋅⋅00a000000⋅⋅⋅000a](N+1)×(N+1),B=[100⋅⋅⋅0⋅⋅⋅0⋅⋅⋅00bc0⋅⋅⋅0⋅⋅⋅0⋅⋅⋅000bc⋅⋅⋅0⋅⋅⋅0⋅⋅⋅0000b⋅⋅⋅0⋅⋅⋅0⋅⋅⋅00000⋅⋅⋅0⋅⋅⋅0⋅⋅⋅00000⋅⋅⋅0⋅⋅⋅0⋅⋅⋅00000⋅⋅⋅0⋅⋅⋅0⋅⋅⋅00⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅000⋅⋅⋅0⋅⋅⋅0⋅⋅⋅00000⋅⋅⋅0⋅⋅⋅0⋅⋅⋅c0000⋅⋅⋅0⋅⋅⋅0⋅⋅⋅bc](N+1)×(N+1).C=A, R is (N+1)×(N+1) identity matrix and φnm, Usm are (N+1)×1 column vectors as(2.23)φnm=[φnmNφnmN+1⋮φn(m+1)N](N+1)×(1),Usm=[UsmNUsmN+1⋮Us(m+1)N](N+1)×(1)fors=n±1,n,
where unmN is given for any m=0,1,…,(2.24)φnk=-(0.1)un+1k-N-2unk-N+un-1k-Nh2,mN+1≤k≤(m+1)N,m=0,1,…,1≤n≤M-1,unk=e-tksin⁡xn,-N≤k≤0.
Here, we denote(2.25)a=-1h2,b=-1τ,c=1τ+2h2.
So, we have second-order difference equation (2.21) with matrix coefficients. To solve this difference equation, we have applied a procedure of modified Gauss elimination method. Hence, we obtain a solution of the matrix equation in the following form:(2.26)Ujm=αj+1Uj+1m+βj+1m,j=M-1,…,2,1,UMm=0,
where αj(j=1,…,M) are (N+1)×(N+1) square matrices and βjm(j=1,…,M) are (N+1)×1 column matrices defined by(2.27)αj+1=-(B+Cαj)-1A,βj+1m=(B+Cαj)-1(Rφjm-Cβj),
where j=1,…,M-1, α1 is the (N+1)×(N+1) zero matrix, and β1m is the (N+1)×1 zero matrix.Second, using the second order of accuracy difference scheme for the approximate solutions of problem (2.17) and applying formulae(2.28)2u(0)-5u(h)+4u(2h)-u(3h)h2-u′′(0)=O(h2),2u(1)-5u(1-h)+4u(1-2h)-u(1-3h)h2-u′′(1)=O(h2),
we obtain the following system of equations:(2.29)unk-unk-1τ-un+1k-2unk+un-1kh2+τ2(un+2k-4un+1k+6unk-4un-1k+un-2kh4)+(0.1){un+1k-N-2unk-N+un-1k-N2h2+un+1k-1-N-2unk-1-N+un-1k-1-N2h2-τ2[un+2k-N-4un+1k-N+6unk-N-4un-1k-N+un-2k-N2h4+un+2k-1-N-4un+1k-1-N+6unk-1-N-4un-1k-1-N+un-2k-1-N2h4]}=0mN+1≤k≤(m+1)N,m=0,1,…,2≤n≤M-2,unk=e-tksin⁡xn,-N≤k≤0,0≤n≤M,u1k=45u2k-15u3k,k≥0,uM-1k=45uM-2k-15uM-3k,k≥0,u0k=uMk=0,k≥0.
In the second step, we apply second-order difference scheme to get the system of linear equations in matrix form(2.30)AUn+2m+BUn+1m+CUnm+DUn-1m+EUn-2m=Rφnm,m=0,1,…,2≤n≤M-2,U0m=0̃,UMm=0̃,U1m=45U2m-15U3m,UM-1m=45UM-2m-15UM-3m,
where A, B, C are (N+1)×(N+1) matrices defined by(2.31)A=[00000⋅⋅⋅00000x000⋅⋅⋅000000x00⋅⋅⋅0000000x0⋅⋅⋅00000000x⋅⋅⋅0000⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅00000⋅⋅⋅x00000000⋅⋅⋅0x0000000⋅⋅⋅00x000000⋅⋅⋅000x](N+1)×(N+1),B=[00000⋅⋅⋅00000y000⋅⋅⋅000000y00⋅⋅⋅0000000y0⋅⋅⋅00000000y⋅⋅⋅0000⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅00000⋅⋅⋅y00000000⋅⋅⋅0y0000000⋅⋅⋅00y000000⋅⋅⋅000y](N+1)×(N+1),C=[10000⋅⋅⋅0000zt000⋅⋅⋅00000zt00⋅⋅⋅000000zt0⋅⋅⋅0000000zt⋅⋅⋅0000⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅00000⋅⋅⋅zt0000000⋅⋅⋅0zt000000⋅⋅⋅00zt](N+1)×(N+1),B=D, E=A, R is (N+1)×(N+1) identity matrix, and φnm, Usm are (N+1)×1 column vectors as(2.32)φnm=[φnmNφnmN+1⋮φn(m+1)N](N+1)×(1),Usm=[UsmNUsmN+1⋮Us(m+1)N](N+1)×(1)fors=n±1,n±2,n,
where unmN is given for any m=0,1,…,(2.33)φnk=-(0.1){un+1k-N-2unk-N+un-1k-N2h2+un+1k-1-N-2unk-1-N+un-1k-1-N2h2-τ2[un+2k-N-4un+1k-N+6unk-N-4un-1k-N+un-2k-N2h4+un+2k-1-N-4un+1k-1-N+6unk-1-N-4un-1k-1-N+un-2k-1-N2h4]},mN+1≤k≤(m+1)N,m=0,1,…,1≤n≤M-1,unk=e-tksin⁡xn,-N≤k≤0.Here, we denote(2.34)x=τ2h4,y=-1h2-2τh4,z=-1τ,t=1τ+2h2+3τh4,
Hence, we have second-order difference equation (2.30) with matrix coefficients. For the solution of this matrix equation, we use the modified Gauss elimination method. We seek a solution of the matrix equation by the following form:(2.35)Ujm=αj+1Uj+1m+βj+1Uj+2m+γj+1m,j=M-2,…,2,1,0UMm=0,UM-1m=[(βM-2+5I)-4(I-αM-2)αM-1]-1[4(I-αM-2)γM-1m-γM-2m],
where αj(j=2,…,M-2) and βj(j=2,…,M-2) are (N+1)×(N+1) square matrices and γjm(j=2,…,M-2) are column matrices defined by(2.36)αj+1=-(C+Dαj+Eβj-1+Eαj-1αj)-1(B+Dβj+Eαj-1βj),βj+1=-(C+Dαj+Eβj-1+Eαj-1αj)-1(A),γj+1m=-(C+Dαj+Eβj-1+Eαj-1αj)-1(Rφjm-Dγjm-Eαj-1γjm-Eγj-1m),
where j=2,…,M-2, α1 is (N+1)×(N+1) zero matrix, and β1 is (N+1)×(N+1) zero matrix, γ1m and γ2m are (N+1)×1 zero matrices.We give the results of the numerical analysis. The numerical solutions are recorded for different values ofN and M and unk represent the numerical solutions of these difference schemes at (tk,xn). Tables 1, 2, 3, and 4 are constructed for N=M=20,40,80 in t∈[0,1], t∈[1,2], t∈[2,3], t∈[3,4], respectively, and the error is computed by the following formula:(2.37)EMN=max⁡-N≤k≤N1≤n≤M-1|u(tk,xn)-unk|.
Thus, by using the second order of accuracy difference scheme, the accuracy of solution increases faster than the first order of accuracy difference scheme.Table 1
Comparison of the errors of different difference schemes int∈[0,1].
MethodN=M=20N=M=40N=M=80Difference scheme (2.20)0.005990880.002860920.00139629Difference scheme (2.29)0.000762650.000203170.00005148Table 2
Comparison of the errors of different difference schemes int∈[1,2].
MethodN=M=20N=M=40N=M=80Difference scheme (2.20)0.073241510.036930610.01845008Difference scheme (2.29)0.000770550.000205410.00005206Table 3
Comparison of the errors of different difference schemes int∈[2,3].
MethodN=M=20N=M=40N=M=80Difference scheme (2.20)0.035200810.017498620.00872690Difference scheme (2.29)0.000670840.000178450.00004521Table 4
Comparison of the errors of different difference schemes int∈[3,4].
MethodN=M=20N=M=40N=M=80Difference scheme (2.20)0.016871660.008445770.00421761Difference scheme (2.29)0.000455270.000121070.00003067


## 3. Homotopy Analysis Method

In this section, we consider homotopy analysis method for the solution of problem (1.1). We study the initial-boundary-value problem for the delay parabolic equation (1.1). To illustrate the basic idea of homotopy analysis method (HAM) developed by Liao (see, e.g., [29–35]), the following differential equation is considered:(3.1)N[u(t,x)]=f(t,x),
where N is a linear operator for problem (1.1), t and x denote independent variables, u(t,x) is an unknown function, and f(t,x) is a known analytical function. Liao constructs the so-called zero-order deformation equation(3.2)(1-q)L[ϕ(t,x;q)-u0(t,x)]=qℏ{N[ϕ(t,x;q)]-f(t,x)},
where q∈[0,1] is an embedding parameter, ℏ is a nonzero auxiliary parameter, L is an auxiliary linear operator, u0(t,x) is an initial guess of u(t,x), and ϕ(t,x;q) is an unknown function. When q=0 and q=1, it holds(3.3)ϕ(t,x;0)=u0(t,x),ϕ(t,x;1)=u(t,x),
respectively. As q increases from 0 to 1, the solution ϕ(t,x;q) varies from the initial guess u0(t,x) to the solution u(t,x). Expanding ϕ(t,x;q) in Taylor series with respect to q, we get(3.4)ϕ(t,x;q)=u0(t,x)+∑m=1+∞um(t,x)qm,
where(3.5)um(t,x)=1m!∂mϕ(t,x;q)∂qm|q=0,
when the initial guess u0(t,x), the auxiliary linear operator L and the auxiliary parameter ℏ are chosen properly, the series (3.4) converges at q=1. We get(3.6)u(t,x)=u0(t,x)+∑m=1+∞um(t,x).
Then define the vectors(3.7)un⃗={u0(t,x),u1(t,x),…,un(t,x)}.
Differentiating the zero-order deformation equation (3.2) m times with respect to the embedding parameter q and dividing them by m!, we obtain the mth-order deformation equation(3.8)L[um(t,x)-χmum-1(t,x)]=ℏRm(u⃗m-1),
where(3.9)Rm(u⃗m-1)=1(m-1)!∂m-1(N[ϕ(t,x;q)]-f(t,x))∂qm-1|q=0,χm={0,m≤1,1,m>1.
with the initial condition(3.10)um(0,x)=0,m≥1.
High-order deformation equation (3.8) is governed by the linear operator L. ℜm(u⃗m-1) can be represented by u1(t,x),um(t,x),u2(t,x),…,um-1(t,x) and high-order deformation equation can be solved consecutively. The Nth-order approximation of u(t,x) is given by(3.11)u(t,x)≈u0(t,x)+∑m=1Num(t,x).


### 3.1. Homotopy Analysis Solution

For the approximate solution of the delay parabolic differential equation with the Dirichlet condition, we consider the delay parabolic equation (2.17) and rewrite the equation for t∈[0,1] in the following form:(3.12)∂u(t,x)∂t-∂2u(t,x)∂x2=(0.1)e-(t-1)sin⁡x,0<t≤1,0<x<π,u(0,x)=sin⁡x,0≤x≤πu(t,0)=u(t,π)=0,0≤t≤1.
To solve the initial-boundary-value problem (3.12) by means of HAM, we choose the initial approximation(3.13)u0(t,x)=sin⁡x,
and the linear operator(3.14)L[ϕ(t,x;q)]=∂ϕ(t,x;q)∂t,
with the property(3.15)L[c]=0,
where c is constant of integration. From (3.12), we define a linear operator as(3.16)N[ϕ(t,x;q)]=∂ϕ(t,x;q)∂t-∂2ϕ(t,x;q)∂x2.
Firstly, we construct the zero-order deformation equation(3.17)(1-q)L[ϕ(t,x;q)-u0(t,x)]=qℏ{N[ϕ(t,x;q)]-f(t,x)},
when q=0 and q=1,(3.18)ϕ(t,x;0)=u0(t,x)=u(0,x),ϕ(t,x;1)=u(t,x).
Then, we get mth-order deformation equations (3.8) for m≥1 with the initial conditions(3.19)um(0,x)=0,
where(3.20)Rm(u⃗m-1)=∂um-1(t,x)∂t-∂2um-1(t,x)∂x2-(1-χm)(0.1)e-(t-1)sin⁡x.
The solution of the mth-order deformation equations (3.20) for m≥1 is(3.21)um(t,x)=χmum-1(t,x)+ℏL-1[Rm(u⃗m-1)].
From (3.12) and (3.21), we obtain(3.22)u0(t,x)=sin⁡x,u1(t,x)=-ℏ(0.1)e∑k=1∞(-1)k+1tkk!sin⁡x+ℏtsin⁡x,u2(t,x)=ℏ(ℏ+1)tsin⁡x-ℏ(ℏ+1)∑k=2∞(-1)ktk-1(k-1)!sin⁡x+ℏ2(0.1)e∑k=2∞(-1)k+1tkk!sin⁡x+ℏ2t22!sin⁡x,u3(t,x)=ℏ(ℏ+1)tsin⁡x+3ℏ2(ℏ+1)t22!sin⁡x-ℏ(ℏ+1)2∑k=3∞(-1)k+1tk-2(k-2)!sin⁡x-2ℏ2(ℏ+1)∑k=3∞(-1)k+1tk-1(k-1)!sin⁡x-ℏ3(0.1)e∑k=3∞(-1)k+1tkk!sin⁡x+ℏ3t33!sin⁡x,⋮un(t,x)=f1(ℏ+1,t,x)+(-1)nℏn(0.1)e∑k=n∞(-1)k+1tkk!sin⁡x+ℏntnn!sin⁡x,⋮
and so on. Then for ℏ=-1, we get(3.23)u0(t,x)=sin⁡x,u1(t,x)=(0.1)e∑k=1∞(-1)k+1tkk!sin⁡x-tsin⁡x,u2(t,x)=(0.1)e∑k=2∞(-1)k+1tkk!sin⁡x+t22!sin⁡x,u3(t,x)=(0.1)e∑k=3∞(-1)k+1tkk!sin⁡x-t33!sin⁡x,⋮un(t,x)=(0.1)e∑k=n∞(-1)k+1tkk!+(-1)ntnn!sin⁡x,⋮
and so on.From (3.6), when we take ℏ=-1, the solution of (3.12) can be obtained as(3.24)u(t,x)=((0.1)et+1)∑k=0∞(-1)ktkk!sin⁡x.
Equation (3.24) has the closed form(3.25)u(t,x)=((0.1)et+1)e-tsin⁡x,
which is the exact solution of (3.12).Second, we consider the solution of the delay parabolic equation (2.17) for t∈[1,2] and rewrite this equation in the following form:(3.26)∂u(t,x)∂t-∂2u(t,x)∂x2=(0.1)e-(t-1)[(0.1)e(t-1)+1]sin⁡x,1<t≤2,0<x<π,u(1,x)=e-1[(0.1)e+1]sin⁡x,0≤x≤πu(t,0)=u(t,π)=0,1≤t≤2.
Now, we choose the initial approximation(3.27)u0(t,x)=e-1[(0.1)e+1]sin⁡x.
We take the linear operator (3.14) with the property (3.15), and we define the operator (3.16) from (3.26).Firstly, we construct the zero-order deformation equation (3.2) and then, we obtain mth-order deformation equations (3.8) for m≥1 with the initial conditions(3.28)um(1,x)=0,
where(3.29)Rm(u⃗m-1)=∂um-1(t,x)∂t-∂2um-1(t,x)∂x2-(1-χm)(0.1)e-(t-1)[(0.1)e(t-1)+1]sin⁡x.
The solution of the mth-order deformation equations (3.29) for m≥1 is(3.30)um(t,x)=χmum-1(t,x)+ℏL-1[Rm(u⃗m-1)].
From (3.26) and the mth-order deformation equations (3.30), we get
(3.31)u0(t,x)=e-1[(0.1)e+1]sin⁡x,u1(t,x)=ℏe-1[(0.1)e+1]sin⁡x(t-1)-ℏ(0.1)∑k=1∞(-1)k+1(t-1)kk!-ℏ(0.1)2e∑k=1∞(-1)k+1(t-1)k+1(k+1)(k-1)!sin⁡x,u2(t,x)=ℏ(ℏ+1)e-1[(0.1)e+1](t-1)sin⁡x-ℏ(ℏ+1)(0.1)∑k=2∞(-1)k(t-1)k-1(k-1)!sin⁡x-ℏ(ℏ+1)(0.1)2e∑k=2∞(-1)k(t-1)kk(k-2)!sin⁡x+ℏ2e-1[(0.1)e+1]sin⁡x(t-1)22!+ℏ2(0.1)∑k=2∞(-1)k+1(t-1)kk!sin⁡x+ℏ2(0.1)2e∑k=2∞(-1)k+1(t-1)k+1(k+1)k(k-2)!sin⁡x,u3(t,x)=ℏ(ℏ+1)2e-1[(0.1)e+1]sin⁡x(t-1)+2ℏ2(ℏ+1)e-1[(0.1)e+1]sin⁡x(t-1)22!+ℏ(ℏ+1)2(0.1)∑k=3∞(-1)k+1(t-1)k-2(k-2)!sin⁡x-ℏ(ℏ+1)2(0.1)2e∑k=3∞(-1)k+1(t-1)k(k-1)(k-3)!sin⁡x-2ℏ2(ℏ+1)(0.1)∑k=3∞(-1)k(t-1)k-1(k-1)!sin⁡x-ℏ(ℏ+1)(0.1)2e∑k=3∞(-1)k(t-1)kk(k-1)(k-3)!sin⁡x+ℏ2(ℏ+1)(0.1)2e∑k=3∞(-1)k(t-1)kk(k-1)(k-2)!sin⁡x+ℏ3e-1[(0.1)e+1]sin⁡x(t-1)33!-ℏ3(0.1)∑k=3∞(-1)k+1(t-1)kk!sin⁡x-ℏ3(0.1)2e∑k=3∞(-1)k+1(t-1)k+1(k+1)k(k-1)(k-3)!sin⁡x,⋮un(t,x)=f2(ℏ+1,t,x)+ℏne-1[(0.1)e+1]sin⁡x(t-1)nn!-ℏn(0.1)∑k=n∞(-1)k+1(t-1)kk!sin⁡x+(-1)nℏn(0.1)2e∑k=n∞(-1)k+1(t-1)k+1(k+1)k(k-1)⋯(k-(n-2))(k-n)!sin⁡x,⋮
and so on. When we choose ℏ=-1, we obtain
(3.32)u0(t,x)=e-1[(0.1)e+1]sin⁡x,u1(t,x)=-e-1[(0.1)e+1]sin⁡x(t-1)+(0.1)∑k=1∞(-1)k+1(t-1)kk!sin⁡x+(0.1)2e∑k=1∞(-1)k+1(t-1)k+1(k+1)(k-1)!sin⁡x,u2(t,x)=e-1[(0.1)e+1]sin⁡x(t-1)22!+(0.1)∑k=2∞(-1)k+1(t-1)kk!sin⁡x+(0.1)2e∑k=2∞(-1)k+1(t-1)k+1(k+1)k(k-2)!sin⁡x,u3(t,x)=-e-1[(0.1)e+1]sin⁡x(t-1)33!+(0.1)∑k=3∞(-1)k+1(t-1)kk!sin⁡x+(0.1)2e∑k=3∞(-1)k+1(t-1)k+1(k+1)k(k-1)(k-3)!sin⁡x,⋮un(t,x)=(-1)ne-1[(0.1)e+1]sin⁡x(t-1)nn!+(0.1)∑k=n∞(-1)k+1(t-1)kk!sin⁡x+(0.1)2e∑k=n∞(-1)k+1(t-1)k+1(k+1)k(k-1)⋯(k-(n-2))(k-n)!sin⁡x,⋮
and so on.From (3.6), the solution of (3.26) for ℏ=-1 can be obtained as(3.33)u(t,x)=([(0.1)e(t-1)]22!+(0.1)et+1)∑k=0∞(-1)ktkk!sin⁡x.
Equation (3.33) has the closed form(3.34)u(t,x)=([(0.1)e(t-1)]22!+(0.1)et+1)e-tsin⁡x,
which is the exact solution of the (3.26).Now, we consider the solution of the delay parabolic equation (2.17) for t∈[2,3] and rewrite the equation in the following form:(3.35)∂u(t,x)∂t-∂2u(t,x)∂x2=(0.1)e-(t-1)[[(0.1)e(t-2)]22!+(0.1)e(t-1)+1]sin⁡x,2<t≤3,0<x<π,u(2,x)=e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x,0≤x≤πu(t,0)=u(t,π)=0,2≤t≤3.
The initial approximation is(3.36)u0(t,x)=e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x.We take the linear operator (3.14) with the property (3.15). From parabolic equation (3.35), we define a linear operator (3.16) and obtain the zero-order deformation equation (3.2). Thus, we get the mth-order deformation equations (3.8) for m≥1 with the initial conditions(3.37)um(2,x)=0,
where(3.38)Rm(u⃗m-1)=∂um-1(t,x)∂t-∂2um-1(t,x)∂x2-(1-χm)×(0.1)e-(t-1)[[(0.1)e(t-2)]22!+(0.1)e(t-1)+1]sin⁡x.
The solution of the mth-order deformation equations (3.38) for m≥1 is(3.39)um(t,x)=χmum-1(t,x)+ℏL-1[Rm(u⃗m-1)].
From (3.35) and (3.39), we obtain
(3.40)u0(t,x)=e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x,u1(t,x)=ℏe-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)-ℏe-1(0.1)∑k=1∞(-1)k+1(t-2)kk!sin⁡x-ℏ(0.1)2∑k=1∞(-1)k+1(t-2)k+1(k+1)(k-1)!sin⁡x-ℏ(0.1)2∑k=1∞(-1)k+1(t-2)kk!sin⁡x-ℏe(0.1)32!∑k=1∞(-1)k+1(t-2)k+2(k+2)(k-1)!sin⁡x,u2(t,x)=ℏ(ℏ+1)e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)-ℏ(ℏ+1)e-1(0.1)∑k=2∞(-1)k(t-2)k-1(k-1)!sin⁡x-ℏ(ℏ+1)(0.1)2∑k=2∞(-1)k(t-2)k-1(k-1)!sin⁡x-ℏ(ℏ+1)(0.1)2∑k=2∞(-1)k(t-2)kk(k-2)!sin⁡x-ℏ(ℏ+1)∑k=2∞(-1)k(t-2)k+1(k+1)(k-2)!sin⁡x+ℏ2e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)22!+ℏ2e-1(0.1)∑k=2∞(-1)k+1(t-2)kk!sin⁡x+ℏ2(0.1)2∑k=2∞(-1)k+1(t-2)k+1(k+1)k(k-2)!sin⁡x+ℏ2(0.1)2∑k=2∞(-1)k+1(t-2)kk!sin⁡x+ℏ2e(0.1)32!∑k=2∞(-1)k+1(t-2)k+2(k+2)(k+1)(k-2)!sin⁡x,u3(t,x)=ℏ(ℏ+1)2e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)+2ℏ2(ℏ+1)e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)22!-ℏ(ℏ+1)2(0.1)e-1∑k=3∞(-1)k+1(t-2)k-2(k-2)!sin⁡x-ℏ(ℏ+1)2(0.1)2∑k=3∞(-1)k+1(t-2)k-2(k-2)!sin⁡x-3ℏ(ℏ+1)(0.1)2∑k=3∞(-1)k+1(t-2)k-1(k-1)!sin⁡x-ℏ2(ℏ+1)(0.1)2∑k=3∞(-1)k+1(t-2)k-1(k-1)!sin⁡x-2ℏ2(ℏ+1)e-1(0.1)∑k=3∞(-1)k+1(t-2)k-1(k-1)!sin⁡x-ℏ(ℏ+1)2(0.1)32!e∑k=3∞(-1)k+1(t-2)kk(k-3)!sin⁡x-2ℏ2(ℏ+1)(0.1)2∑k=3∞(-1)k+1(t-2)kk(k-1)(k-3)!sin⁡x-2ℏ2(ℏ+1)(0.1)32!e∑k=3∞(-1)k+1(t-2)k+1(k+1)k(k-3)!sin⁡x+ℏ3e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)33!-ℏ3e-1(0.1)∑k=3∞(-1)k+1(t-2)kk!sin⁡x-ℏ3(0.1)2∑k=3∞(-1)k+1(t-2)k+1(k+1)k(k-1)(k-3)!sin⁡x-ℏ3(0.1)2∑k=3∞(-1)k+1(t-2)kk!sin⁡x-ℏ3e(0.1)32!∑k=3∞(-1)k+1(t-2)k+2(k+2)(k+1)k(k-3)!sin⁡x,⋮un(t,x)=f3(ℏ+1,t,x)+ℏne-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)nn!+(-1)nℏne-1(0.1)∑k=n∞(-1)k+1(t-2)kk!sin⁡x+(-1)nℏn(0.1)2∑k=n∞(-1)k+1(t-2)k+1(k+1)k(k-1)⋯(k-(n-2))(k-n)!sin⁡x+(-1)nℏn(0.1)2∑k=n∞(-1)k+1(t-2)kk!sin⁡x+(-1)nℏne(0.1)32!∑k=n∞(-1)k+1(t-2)k+2(k+2)(k+1)k⋯(k-(n-3))(k-n)!sin⁡x,⋮
and so on. When we choose ℏ=-1, we obtain
(3.41)u0(t,x)=e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x,u1(t,x)=-e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)+e-1(0.1)∑k=1∞(-1)k+1(t-2)kk!sin⁡x+(0.1)2∑k=1∞(-1)k+1(t-2)k+1(k+1)(k-1)!sin⁡x+(0.1)2∑k=1∞(-1)k+1(t-2)kk!sin⁡x+e(0.1)32!∑k=1∞(-1)k+1(t-2)k+2(k+2)(k-1)!sin⁡x,u2(t,x)=e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)22!+e-1(0.1)∑k=2∞(-1)k+1(t-2)kk!sin⁡x+(0.1)2∑k=2∞(-1)k+1(t-2)k+1(k+1)k(k-2)!sin⁡x+(0.1)2∑k=2∞(-1)k+1(t-2)kk!sin⁡x+e(0.1)32!∑k=2∞(-1)k+1(t-2)k+2(k+2)(k+1)(k-2)!sin⁡x,u3(t,x)=-e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)33!+e-1(0.1)∑k=3∞(-1)k+1(t-2)kk!sin⁡x+(0.1)2∑k=3∞(-1)k+1(t-2)k+1(k+1)k(k-1)(k-3)!sin⁡x+(0.1)2∑k=3∞(-1)k+1(t-2)kk!sin⁡x+e(0.1)32!∑k=3∞(-1)k+1(t-2)k+2(k+2)(k+1)k(k-3)!sin⁡x,⋮un(t,x)=(-1)ne-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)nn!+e-1(0.1)∑k=n∞(-1)k+1(t-2)kk!sin⁡x+(0.1)2∑k=n∞(-1)k+1(t-2)k+1(k+1)k(k-1)⋯(k-(n-2))(k-n)!sin⁡x+(0.1)2∑k=n∞(-1)k+1(t-2)kk!sin⁡x+e(0.1)32!∑k=n∞(-1)k+1(t-2)k+2(k+2)(k+1)k⋯(k-(n-3))(k-n)!sin⁡x,⋮
and so on.From (3.6), the solution of (3.35) for ℏ=-1 is(3.42)u(t,x)=([(0.1)e(t-2)]33!+[(0.1)e(t-1)]22!+(0.1)et+1)∑k=0∞(-1)ktkk!sin⁡x.
This series has the closed form(3.43)u(t,x)=([(0.1)e(t-2)]33!+[(0.1)e(t-1)]22!+(0.1)et+1)e-tsin⁡x,
which is the exact solution of the (3.35).Finally, we consider the solution of the delay parabolic equation (2.17) for t∈[3,4] and rewrite this equation in the following form:(3.44)∂u(t,x)∂t-∂2u(t,x)∂x2=(0.1)e-(t-1)[[(0.1)e(t-3)]33!+[(0.1)e(t-2)]22!+(0.1)e(t-1)+1]sin⁡x,3<t≤4,0<x<π,u(3,x)=e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x,0≤x≤π,u(t,0)=u(t,π)=0,3≤t≤4.
We take the initial approximation(3.45)u0(t,x)=e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x
and the linear operator (3.14) with the property (3.15). From (3.44), we define linear operator (3.16).We construct the zero-order deformation equation (3.2) and the mth-order deformation equations (3.8) for m≥1 with the initial conditions(3.46)um(3,x)=0,
where(3.47)Rm(u⃗m-1)=∂um-1(t,x)∂t-∂2um-1(t,x)∂x2-(1-χm)×(0.1)e-(t-1)([(0.1)e(t-3)]33!+[(0.1)e(t-2)]22!+(0.1)e(t-1)+1)sin⁡x.The solution of themth-order deformation equations (3.47) for m≥1 is(3.48)um(t,x)=χmum-1(t,x)+ℏL-1[Rm(u⃗m-1)].From (3.44) and (3.48), we get
(3.49)u0(t,x)=e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x,u1(t,x)=ℏe-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)-ℏe-2(0.1)∑k=1∞(-1)k+1(t-3)kk!sin⁡x-ℏe-1(0.1)2∑k=1∞(-1)k+1(t-3)k+1(k+1)(k-1)!sin⁡x-2ℏe-1(0.1)2∑k=1∞(-1)k+1(t-3)kk!sin⁡x-ℏ(0.1)32!∑k=1∞(-1)k+1(t-3)k+2(k+2)(k-1)!sin⁡x-ℏ(0.1)3∑k=1∞(-1)k+1(t-3)k+1(k+1)(k-1)!sin⁡x-ℏ(0.1)32!∑k=1∞(-1)k+1(t-3)kk!sin⁡x-ℏe(0.1)43!∑k=1∞(-1)k+1(t-3)k+3(k+3)(k-1)!sin⁡x,u2(t,x)=ℏ(ℏ+1)e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)-ℏ(ℏ+1)(0.1)∑k=2∞(-1)k(t-3)k-1(k-2)!sin⁡x-ℏ(ℏ+1)(0.1)2e-1∑k=2∞(-1)k(t-3)kk!sin⁡x-2ℏ(ℏ+1)(0.1)2e-1∑k=2∞(-1)k(t-3)k-1(k-1)!sin⁡x-ℏ(ℏ+1)(0.1)32!∑k=2∞(-1)k(t-3)k-1(k-1)!sin⁡x-ℏ(ℏ+1)(0.1)3∑k=2∞(-1)k(t-3)kk(k-2)!sin⁡x-ℏ(ℏ+1)(0.1)3∑k=2∞(-1)k(t-3)k+1(k+1)(k-2)!sin⁡x-ℏ(ℏ+1)(0.1)43!e∑k=2∞(-1)k(t-3)k+2(k+2)(k-2)!sin⁡x+ℏ2e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)22!+ℏ2e-2(0.1)∑k=2∞(-1)k+1(t-3)kk!sin⁡x+ℏ2e-1(0.1)2∑k=2∞(-1)k+1(t-3)k+1(k+1)k(k-2)!sin⁡x+2ℏ2e-1(0.1)2∑k=2∞(-1)k+1(t-3)kk!sin⁡x+ℏ2(0.1)32!∑k=2∞(-1)k+1(t-3)k+2(k+2)(k+1)(k-2)!sin⁡x+ℏ2(0.1)3∑k=2∞(-1)k+1(t-3)k+1(k+1)k(k-2)!sin⁡x+ℏ2(0.1)32!∑k=2∞(-1)k+1(t-3)kk!sin⁡x+ℏ2e(0.1)43!∑k=2∞(-1)k+1(t-3)k+3(k+3)(k+2)(k-2)!sin⁡x,u3(t,x)=ℏ(ℏ+1)2e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)-ℏ(ℏ+1)2(0.1)e-2∑k=3∞(-1)k+1(t-3)k-2(k-2)!sin⁡x-ℏ(ℏ+1)2(0.1)2e-1∑k=3∞(-1)k+1(t-3)k-1(k-1)(k-3)!sin⁡x-2ℏ(ℏ+1)2(0.1)2e-1∑k=3∞(-1)k+1(t-3)k-2(k-2)!sin⁡x-ℏ(ℏ+1)2(0.1)32!∑k=3∞(-1)k+1(t-3)kk(k-3)!sin⁡x-ℏ(ℏ+1)2(0.1)3∑k=3∞(-1)k+1(t-3)k-1(k-1)(k-3)!sin⁡x-ℏ(ℏ+1)2(0.1)32!∑k=3∞(-1)k+1(t-3)k-2(k-2)!sin⁡x-ℏ(ℏ+1)2e(0.1)43!∑k=3∞(-1)k+1(t-3)k+1(k+1)(k-3)!sin⁡x+2ℏ2(ℏ+1)e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)22!-2ℏ2(ℏ+1)e-2(0.1)∑k=3∞(-1)k+1(t-3)k-1(k-1)!sin⁡x-ℏ2(ℏ+1)e-1(0.1)2∑k=3∞(-1)k+1(t-3)kk(k-1)(k-3)!sin⁡x-4ℏ2(ℏ+1)e-1(0.1)2∑k=3∞(-1)k+1(t-3)k-1(k-1)!sin⁡x-2ℏ2(ℏ+1)(0.1)32!∑k=3∞(-1)k+1(t-3)k+1(k+1)k(k-3)!sin⁡x-2ℏ2(ℏ+1)(0.1)3∑k=3∞(-1)k+1(t-3)kk(k-1)(k-3)!sin⁡x-2ℏ2(ℏ+1)(0.1)32!∑k=3∞(-1)k+1(t-3)k-1(k-1)!sin⁡x-2ℏ2(ℏ+1)e(0.1)43!∑k=3∞(-1)k+1(t-3)k+2(k+2)(k+1)(k-3)!sin⁡x+ℏ3e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)33!-ℏ3e-2(0.1)∑k=3∞(-1)k+1(t-3)kk!sin⁡x-ℏ3e-1(0.1)2∑k=3∞(-1)k+1(t-3)k+1(k+1)k(k-1)(k-3)!sin⁡x-2ℏ3e-1(0.1)2∑k=3∞(-1)k+1(t-3)kk!sin⁡x-ℏ3(0.1)32!∑k=3∞(-1)k+1(t-3)k+2(k+2)(k+1)k(k-3)!sin⁡x-ℏ3(0.1)3∑k=3∞(-1)k+1(t-3)k+1(k+1)k(k-1)(k-3)!sin⁡x-ℏ3(0.1)32!∑k=3∞(-1)k+1(t-3)kk!sin⁡x-ℏ3e(0.1)43!∑k=3∞(-1)k+1(t-3)k+3(k+3)(k+2)(k+1)(k-3)!sin⁡x,⋮un(t,x)=f4(ℏ+1,t,x)+ℏne-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)nn!+(-1)nℏne-2(0.1)∑k=n∞(-1)k+1(t-3)kk!sin⁡x+(-1)nℏne-1(0.1)2∑k=n∞(-1)k+1(t-3)k+1(k+1)k(k-1)⋯(k-(n-2))(k-n)!sin⁡x+(-1)nℏn2e-1(0.1)2∑k=n∞(-1)k+1(t-3)kk!sin⁡x+(-1)nℏn(0.1)32!∑k=n∞(-1)k+1(t-3)k+2(k+2)(k+1)k(k-3)⋯(k-(n-3))(k-n)!sin⁡x+(-1)nℏn(0.1)3∑k=n∞(-1)k+1(t-3)k+1(k+1)k(k-1)⋯(k-(n-2))(k-n)!sin⁡x+(-1)nℏn(0.1)32!∑k=n∞(-1)k+1(t-3)kk!sin⁡x+(-1)nℏne(0.1)43!∑k=n∞(-1)k+1(t-3)k+3(k+2)(k+1)k⋯(k-(n-4))(k-n)!sin⁡x,⋮
and so on. For ℏ=-1, we get
(3.50)u0(t,x)=e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x,u1(t,x)=-e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)+e-2(0.1)∑k=1∞(-1)k+1(t-3)kk!sin⁡x+e-1(0.1)2∑k=1∞(-1)k+1(t-3)k+1(k+1)(k-1)!sin⁡x+2e-1(0.1)2∑k=1∞(-1)k+1(t-3)kk!sin⁡x+(0.1)32!∑k=1∞(-1)k+1(t-3)k+2(k+2)(k-1)!sin⁡x+(0.1)3∑k=1∞(-1)k+1(t-3)k+1(k+1)(k-1)!sin⁡x+(0.1)32!∑k=1∞(-1)k+1(t-3)kk!sin⁡x+e(0.1)43!∑k=1∞(-1)k+1(t-3)k+3(k+3)(k-1)!sin⁡x,u2(t,x)=e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)22!+e-2(0.1)∑k=2∞(-1)k+1(t-3)kk!sin⁡x+e-1(0.1)2∑k=2∞(-1)k+1(t-3)k+1(k+1)k(k-2)!sin⁡x+2e-1(0.1)2∑k=2∞(-1)k+1(t-3)kk!sin⁡x+(0.1)32!∑k=2∞(-1)k+1(t-3)k+2(k+2)(k+1)(k-2)!sin⁡x+(0.1)3∑k=2∞(-1)k+1(t-3)k+1(k+1)k(k-2)!sin⁡x+(0.1)32!∑k=2∞(-1)k+1(t-3)kk!sin⁡x+e(0.1)43!∑k=2∞(-1)k+1(t-3)k+3(k+3)(k+2)(k-2)!sin⁡x,u3(t,x)=-e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)33!+e-2(0.1)∑k=3∞(-1)k+1(t-3)kk!sin⁡x+e-1(0.1)2∑k=3∞(-1)k+1(t-3)k+1(k+1)k(k-1)(k-3)!sin⁡x+2e-1(0.1)2∑k=3∞(-1)k+1(t-3)kk!sin⁡x+(0.1)32!∑k=3∞(-1)k+1(t-3)k+2(k+2)(k+1)k(k-3)!sin⁡x+(0.1)3∑k=3∞(-1)k+1(t-3)k+1(k+1)k(k-1)(k-3)!sin⁡x+(0.1)32!∑k=3∞(-1)k+1(t-3)kk!sin⁡x+e(0.1)43!∑k=3∞(-1)k+1(t-3)k+3(k+3)(k+2)(k+1)(k-3)!sin⁡x,⋮un(t,x)=(-1)ne-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)nn!+e-2(0.1)∑k=n∞(-1)k+1(t-3)kk!sin⁡x+e-1(0.1)2∑k=n∞(-1)k+1(t-3)k+1(k+1)k(k-1)⋯(k-(n-2))(k-n)!sin⁡x+2e-1(0.1)2∑k=n∞(-1)k+1(t-3)kk!sin⁡x+(0.1)32!∑k=n∞(-1)k+1(t-3)k+2(k+2)(k+1)k(k-3)⋯(k-(n-3))(k-n)!sin⁡x+(0.1)3∑k=n∞(-1)k+1(t-3)k+1(k+1)k(k-1)⋯(k-(n-2))(k-n)!sin⁡x+(0.1)32!∑k=n∞(-1)k+1(t-3)kk!sin⁡x+e(0.1)43!∑k=n∞(-1)k+1(t-3)k+3(k+2)(k+1)k⋯(k-(n-4))(k-n)!sin⁡x,⋮
and so on.From (3.6), the solution of (3.44) is obtained as(3.51)u(t,x)=([(0.1)e(t-3)]44!+[(0.1)e(t-2)]33!+[(0.1)e(t-1)]22!+(0.1)et+1)∑k=0∞(-1)ktkk!sin⁡x.
This series has the closed form(3.52)u(t,x)=([(0.1)e(t-3)]44!+[(0.1)e(t-2)]33!+[(0.1)e(t-1)]22!+(0.1)et+1)e-tsin⁡x,
which is the exact solution of (3.44).We give the HAM solutions int∈[0,1], t∈[1,2], t∈[2,3], t∈[3,4]. We use four terms for evaluating the approximate solution uapp=∑k=03uk(t,x). According to the ℏ-curve of uxt(0,0), the solution series is convergent when -1.48≤ℏ≤0.48, -1.41≤ℏ≤2.10, -1.19≤ℏ≤0.12, and -1.02≤ℏ≤0, respectively, in t∈[0,1], t∈[1,2], t∈[2,3], t∈[3,4]. We take ℏ=-1 to determine how much the approximate solution is accurate and compute the absolute errors uerr=|uexact-uapp| at the points (0.5,π/2), (1.5,π/2), (2.5,π/2), (3.5,π/2) in Table 5.Table 5
The absolute error atx=π/2 when ℏ=-1.
tuexactuappuerr0.50.688966723247640.687185174004180.001781549243461.50.316170660693700.315284387317410.000886273376282.50.144725356942580.144321999823470.000403357119103.50.066241553295910.066192994488770.00004855880714


## 3.1. Homotopy Analysis Solution

For the approximate solution of the delay parabolic differential equation with the Dirichlet condition, we consider the delay parabolic equation (2.17) and rewrite the equation for t∈[0,1] in the following form:(3.12)∂u(t,x)∂t-∂2u(t,x)∂x2=(0.1)e-(t-1)sin⁡x,0<t≤1,0<x<π,u(0,x)=sin⁡x,0≤x≤πu(t,0)=u(t,π)=0,0≤t≤1.
To solve the initial-boundary-value problem (3.12) by means of HAM, we choose the initial approximation(3.13)u0(t,x)=sin⁡x,
and the linear operator(3.14)L[ϕ(t,x;q)]=∂ϕ(t,x;q)∂t,
with the property(3.15)L[c]=0,
where c is constant of integration. From (3.12), we define a linear operator as(3.16)N[ϕ(t,x;q)]=∂ϕ(t,x;q)∂t-∂2ϕ(t,x;q)∂x2.
Firstly, we construct the zero-order deformation equation(3.17)(1-q)L[ϕ(t,x;q)-u0(t,x)]=qℏ{N[ϕ(t,x;q)]-f(t,x)},
when q=0 and q=1,(3.18)ϕ(t,x;0)=u0(t,x)=u(0,x),ϕ(t,x;1)=u(t,x).
Then, we get mth-order deformation equations (3.8) for m≥1 with the initial conditions(3.19)um(0,x)=0,
where(3.20)Rm(u⃗m-1)=∂um-1(t,x)∂t-∂2um-1(t,x)∂x2-(1-χm)(0.1)e-(t-1)sin⁡x.
The solution of the mth-order deformation equations (3.20) for m≥1 is(3.21)um(t,x)=χmum-1(t,x)+ℏL-1[Rm(u⃗m-1)].
From (3.12) and (3.21), we obtain(3.22)u0(t,x)=sin⁡x,u1(t,x)=-ℏ(0.1)e∑k=1∞(-1)k+1tkk!sin⁡x+ℏtsin⁡x,u2(t,x)=ℏ(ℏ+1)tsin⁡x-ℏ(ℏ+1)∑k=2∞(-1)ktk-1(k-1)!sin⁡x+ℏ2(0.1)e∑k=2∞(-1)k+1tkk!sin⁡x+ℏ2t22!sin⁡x,u3(t,x)=ℏ(ℏ+1)tsin⁡x+3ℏ2(ℏ+1)t22!sin⁡x-ℏ(ℏ+1)2∑k=3∞(-1)k+1tk-2(k-2)!sin⁡x-2ℏ2(ℏ+1)∑k=3∞(-1)k+1tk-1(k-1)!sin⁡x-ℏ3(0.1)e∑k=3∞(-1)k+1tkk!sin⁡x+ℏ3t33!sin⁡x,⋮un(t,x)=f1(ℏ+1,t,x)+(-1)nℏn(0.1)e∑k=n∞(-1)k+1tkk!sin⁡x+ℏntnn!sin⁡x,⋮
and so on. Then for ℏ=-1, we get(3.23)u0(t,x)=sin⁡x,u1(t,x)=(0.1)e∑k=1∞(-1)k+1tkk!sin⁡x-tsin⁡x,u2(t,x)=(0.1)e∑k=2∞(-1)k+1tkk!sin⁡x+t22!sin⁡x,u3(t,x)=(0.1)e∑k=3∞(-1)k+1tkk!sin⁡x-t33!sin⁡x,⋮un(t,x)=(0.1)e∑k=n∞(-1)k+1tkk!+(-1)ntnn!sin⁡x,⋮
and so on.From (3.6), when we take ℏ=-1, the solution of (3.12) can be obtained as(3.24)u(t,x)=((0.1)et+1)∑k=0∞(-1)ktkk!sin⁡x.
Equation (3.24) has the closed form(3.25)u(t,x)=((0.1)et+1)e-tsin⁡x,
which is the exact solution of (3.12).Second, we consider the solution of the delay parabolic equation (2.17) for t∈[1,2] and rewrite this equation in the following form:(3.26)∂u(t,x)∂t-∂2u(t,x)∂x2=(0.1)e-(t-1)[(0.1)e(t-1)+1]sin⁡x,1<t≤2,0<x<π,u(1,x)=e-1[(0.1)e+1]sin⁡x,0≤x≤πu(t,0)=u(t,π)=0,1≤t≤2.
Now, we choose the initial approximation(3.27)u0(t,x)=e-1[(0.1)e+1]sin⁡x.
We take the linear operator (3.14) with the property (3.15), and we define the operator (3.16) from (3.26).Firstly, we construct the zero-order deformation equation (3.2) and then, we obtain mth-order deformation equations (3.8) for m≥1 with the initial conditions(3.28)um(1,x)=0,
where(3.29)Rm(u⃗m-1)=∂um-1(t,x)∂t-∂2um-1(t,x)∂x2-(1-χm)(0.1)e-(t-1)[(0.1)e(t-1)+1]sin⁡x.
The solution of the mth-order deformation equations (3.29) for m≥1 is(3.30)um(t,x)=χmum-1(t,x)+ℏL-1[Rm(u⃗m-1)].
From (3.26) and the mth-order deformation equations (3.30), we get
(3.31)u0(t,x)=e-1[(0.1)e+1]sin⁡x,u1(t,x)=ℏe-1[(0.1)e+1]sin⁡x(t-1)-ℏ(0.1)∑k=1∞(-1)k+1(t-1)kk!-ℏ(0.1)2e∑k=1∞(-1)k+1(t-1)k+1(k+1)(k-1)!sin⁡x,u2(t,x)=ℏ(ℏ+1)e-1[(0.1)e+1](t-1)sin⁡x-ℏ(ℏ+1)(0.1)∑k=2∞(-1)k(t-1)k-1(k-1)!sin⁡x-ℏ(ℏ+1)(0.1)2e∑k=2∞(-1)k(t-1)kk(k-2)!sin⁡x+ℏ2e-1[(0.1)e+1]sin⁡x(t-1)22!+ℏ2(0.1)∑k=2∞(-1)k+1(t-1)kk!sin⁡x+ℏ2(0.1)2e∑k=2∞(-1)k+1(t-1)k+1(k+1)k(k-2)!sin⁡x,u3(t,x)=ℏ(ℏ+1)2e-1[(0.1)e+1]sin⁡x(t-1)+2ℏ2(ℏ+1)e-1[(0.1)e+1]sin⁡x(t-1)22!+ℏ(ℏ+1)2(0.1)∑k=3∞(-1)k+1(t-1)k-2(k-2)!sin⁡x-ℏ(ℏ+1)2(0.1)2e∑k=3∞(-1)k+1(t-1)k(k-1)(k-3)!sin⁡x-2ℏ2(ℏ+1)(0.1)∑k=3∞(-1)k(t-1)k-1(k-1)!sin⁡x-ℏ(ℏ+1)(0.1)2e∑k=3∞(-1)k(t-1)kk(k-1)(k-3)!sin⁡x+ℏ2(ℏ+1)(0.1)2e∑k=3∞(-1)k(t-1)kk(k-1)(k-2)!sin⁡x+ℏ3e-1[(0.1)e+1]sin⁡x(t-1)33!-ℏ3(0.1)∑k=3∞(-1)k+1(t-1)kk!sin⁡x-ℏ3(0.1)2e∑k=3∞(-1)k+1(t-1)k+1(k+1)k(k-1)(k-3)!sin⁡x,⋮un(t,x)=f2(ℏ+1,t,x)+ℏne-1[(0.1)e+1]sin⁡x(t-1)nn!-ℏn(0.1)∑k=n∞(-1)k+1(t-1)kk!sin⁡x+(-1)nℏn(0.1)2e∑k=n∞(-1)k+1(t-1)k+1(k+1)k(k-1)⋯(k-(n-2))(k-n)!sin⁡x,⋮
and so on. When we choose ℏ=-1, we obtain
(3.32)u0(t,x)=e-1[(0.1)e+1]sin⁡x,u1(t,x)=-e-1[(0.1)e+1]sin⁡x(t-1)+(0.1)∑k=1∞(-1)k+1(t-1)kk!sin⁡x+(0.1)2e∑k=1∞(-1)k+1(t-1)k+1(k+1)(k-1)!sin⁡x,u2(t,x)=e-1[(0.1)e+1]sin⁡x(t-1)22!+(0.1)∑k=2∞(-1)k+1(t-1)kk!sin⁡x+(0.1)2e∑k=2∞(-1)k+1(t-1)k+1(k+1)k(k-2)!sin⁡x,u3(t,x)=-e-1[(0.1)e+1]sin⁡x(t-1)33!+(0.1)∑k=3∞(-1)k+1(t-1)kk!sin⁡x+(0.1)2e∑k=3∞(-1)k+1(t-1)k+1(k+1)k(k-1)(k-3)!sin⁡x,⋮un(t,x)=(-1)ne-1[(0.1)e+1]sin⁡x(t-1)nn!+(0.1)∑k=n∞(-1)k+1(t-1)kk!sin⁡x+(0.1)2e∑k=n∞(-1)k+1(t-1)k+1(k+1)k(k-1)⋯(k-(n-2))(k-n)!sin⁡x,⋮
and so on.From (3.6), the solution of (3.26) for ℏ=-1 can be obtained as(3.33)u(t,x)=([(0.1)e(t-1)]22!+(0.1)et+1)∑k=0∞(-1)ktkk!sin⁡x.
Equation (3.33) has the closed form(3.34)u(t,x)=([(0.1)e(t-1)]22!+(0.1)et+1)e-tsin⁡x,
which is the exact solution of the (3.26).Now, we consider the solution of the delay parabolic equation (2.17) for t∈[2,3] and rewrite the equation in the following form:(3.35)∂u(t,x)∂t-∂2u(t,x)∂x2=(0.1)e-(t-1)[[(0.1)e(t-2)]22!+(0.1)e(t-1)+1]sin⁡x,2<t≤3,0<x<π,u(2,x)=e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x,0≤x≤πu(t,0)=u(t,π)=0,2≤t≤3.
The initial approximation is(3.36)u0(t,x)=e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x.We take the linear operator (3.14) with the property (3.15). From parabolic equation (3.35), we define a linear operator (3.16) and obtain the zero-order deformation equation (3.2). Thus, we get the mth-order deformation equations (3.8) for m≥1 with the initial conditions(3.37)um(2,x)=0,
where(3.38)Rm(u⃗m-1)=∂um-1(t,x)∂t-∂2um-1(t,x)∂x2-(1-χm)×(0.1)e-(t-1)[[(0.1)e(t-2)]22!+(0.1)e(t-1)+1]sin⁡x.
The solution of the mth-order deformation equations (3.38) for m≥1 is(3.39)um(t,x)=χmum-1(t,x)+ℏL-1[Rm(u⃗m-1)].
From (3.35) and (3.39), we obtain
(3.40)u0(t,x)=e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x,u1(t,x)=ℏe-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)-ℏe-1(0.1)∑k=1∞(-1)k+1(t-2)kk!sin⁡x-ℏ(0.1)2∑k=1∞(-1)k+1(t-2)k+1(k+1)(k-1)!sin⁡x-ℏ(0.1)2∑k=1∞(-1)k+1(t-2)kk!sin⁡x-ℏe(0.1)32!∑k=1∞(-1)k+1(t-2)k+2(k+2)(k-1)!sin⁡x,u2(t,x)=ℏ(ℏ+1)e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)-ℏ(ℏ+1)e-1(0.1)∑k=2∞(-1)k(t-2)k-1(k-1)!sin⁡x-ℏ(ℏ+1)(0.1)2∑k=2∞(-1)k(t-2)k-1(k-1)!sin⁡x-ℏ(ℏ+1)(0.1)2∑k=2∞(-1)k(t-2)kk(k-2)!sin⁡x-ℏ(ℏ+1)∑k=2∞(-1)k(t-2)k+1(k+1)(k-2)!sin⁡x+ℏ2e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)22!+ℏ2e-1(0.1)∑k=2∞(-1)k+1(t-2)kk!sin⁡x+ℏ2(0.1)2∑k=2∞(-1)k+1(t-2)k+1(k+1)k(k-2)!sin⁡x+ℏ2(0.1)2∑k=2∞(-1)k+1(t-2)kk!sin⁡x+ℏ2e(0.1)32!∑k=2∞(-1)k+1(t-2)k+2(k+2)(k+1)(k-2)!sin⁡x,u3(t,x)=ℏ(ℏ+1)2e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)+2ℏ2(ℏ+1)e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)22!-ℏ(ℏ+1)2(0.1)e-1∑k=3∞(-1)k+1(t-2)k-2(k-2)!sin⁡x-ℏ(ℏ+1)2(0.1)2∑k=3∞(-1)k+1(t-2)k-2(k-2)!sin⁡x-3ℏ(ℏ+1)(0.1)2∑k=3∞(-1)k+1(t-2)k-1(k-1)!sin⁡x-ℏ2(ℏ+1)(0.1)2∑k=3∞(-1)k+1(t-2)k-1(k-1)!sin⁡x-2ℏ2(ℏ+1)e-1(0.1)∑k=3∞(-1)k+1(t-2)k-1(k-1)!sin⁡x-ℏ(ℏ+1)2(0.1)32!e∑k=3∞(-1)k+1(t-2)kk(k-3)!sin⁡x-2ℏ2(ℏ+1)(0.1)2∑k=3∞(-1)k+1(t-2)kk(k-1)(k-3)!sin⁡x-2ℏ2(ℏ+1)(0.1)32!e∑k=3∞(-1)k+1(t-2)k+1(k+1)k(k-3)!sin⁡x+ℏ3e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)33!-ℏ3e-1(0.1)∑k=3∞(-1)k+1(t-2)kk!sin⁡x-ℏ3(0.1)2∑k=3∞(-1)k+1(t-2)k+1(k+1)k(k-1)(k-3)!sin⁡x-ℏ3(0.1)2∑k=3∞(-1)k+1(t-2)kk!sin⁡x-ℏ3e(0.1)32!∑k=3∞(-1)k+1(t-2)k+2(k+2)(k+1)k(k-3)!sin⁡x,⋮un(t,x)=f3(ℏ+1,t,x)+ℏne-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)nn!+(-1)nℏne-1(0.1)∑k=n∞(-1)k+1(t-2)kk!sin⁡x+(-1)nℏn(0.1)2∑k=n∞(-1)k+1(t-2)k+1(k+1)k(k-1)⋯(k-(n-2))(k-n)!sin⁡x+(-1)nℏn(0.1)2∑k=n∞(-1)k+1(t-2)kk!sin⁡x+(-1)nℏne(0.1)32!∑k=n∞(-1)k+1(t-2)k+2(k+2)(k+1)k⋯(k-(n-3))(k-n)!sin⁡x,⋮
and so on. When we choose ℏ=-1, we obtain
(3.41)u0(t,x)=e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x,u1(t,x)=-e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)+e-1(0.1)∑k=1∞(-1)k+1(t-2)kk!sin⁡x+(0.1)2∑k=1∞(-1)k+1(t-2)k+1(k+1)(k-1)!sin⁡x+(0.1)2∑k=1∞(-1)k+1(t-2)kk!sin⁡x+e(0.1)32!∑k=1∞(-1)k+1(t-2)k+2(k+2)(k-1)!sin⁡x,u2(t,x)=e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)22!+e-1(0.1)∑k=2∞(-1)k+1(t-2)kk!sin⁡x+(0.1)2∑k=2∞(-1)k+1(t-2)k+1(k+1)k(k-2)!sin⁡x+(0.1)2∑k=2∞(-1)k+1(t-2)kk!sin⁡x+e(0.1)32!∑k=2∞(-1)k+1(t-2)k+2(k+2)(k+1)(k-2)!sin⁡x,u3(t,x)=-e-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)33!+e-1(0.1)∑k=3∞(-1)k+1(t-2)kk!sin⁡x+(0.1)2∑k=3∞(-1)k+1(t-2)k+1(k+1)k(k-1)(k-3)!sin⁡x+(0.1)2∑k=3∞(-1)k+1(t-2)kk!sin⁡x+e(0.1)32!∑k=3∞(-1)k+1(t-2)k+2(k+2)(k+1)k(k-3)!sin⁡x,⋮un(t,x)=(-1)ne-2[[(0.1)e]22!+2(0.1)e+1]sin⁡x(t-2)nn!+e-1(0.1)∑k=n∞(-1)k+1(t-2)kk!sin⁡x+(0.1)2∑k=n∞(-1)k+1(t-2)k+1(k+1)k(k-1)⋯(k-(n-2))(k-n)!sin⁡x+(0.1)2∑k=n∞(-1)k+1(t-2)kk!sin⁡x+e(0.1)32!∑k=n∞(-1)k+1(t-2)k+2(k+2)(k+1)k⋯(k-(n-3))(k-n)!sin⁡x,⋮
and so on.From (3.6), the solution of (3.35) for ℏ=-1 is(3.42)u(t,x)=([(0.1)e(t-2)]33!+[(0.1)e(t-1)]22!+(0.1)et+1)∑k=0∞(-1)ktkk!sin⁡x.
This series has the closed form(3.43)u(t,x)=([(0.1)e(t-2)]33!+[(0.1)e(t-1)]22!+(0.1)et+1)e-tsin⁡x,
which is the exact solution of the (3.35).Finally, we consider the solution of the delay parabolic equation (2.17) for t∈[3,4] and rewrite this equation in the following form:(3.44)∂u(t,x)∂t-∂2u(t,x)∂x2=(0.1)e-(t-1)[[(0.1)e(t-3)]33!+[(0.1)e(t-2)]22!+(0.1)e(t-1)+1]sin⁡x,3<t≤4,0<x<π,u(3,x)=e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x,0≤x≤π,u(t,0)=u(t,π)=0,3≤t≤4.
We take the initial approximation(3.45)u0(t,x)=e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x
and the linear operator (3.14) with the property (3.15). From (3.44), we define linear operator (3.16).We construct the zero-order deformation equation (3.2) and the mth-order deformation equations (3.8) for m≥1 with the initial conditions(3.46)um(3,x)=0,
where(3.47)Rm(u⃗m-1)=∂um-1(t,x)∂t-∂2um-1(t,x)∂x2-(1-χm)×(0.1)e-(t-1)([(0.1)e(t-3)]33!+[(0.1)e(t-2)]22!+(0.1)e(t-1)+1)sin⁡x.The solution of themth-order deformation equations (3.47) for m≥1 is(3.48)um(t,x)=χmum-1(t,x)+ℏL-1[Rm(u⃗m-1)].From (3.44) and (3.48), we get
(3.49)u0(t,x)=e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x,u1(t,x)=ℏe-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)-ℏe-2(0.1)∑k=1∞(-1)k+1(t-3)kk!sin⁡x-ℏe-1(0.1)2∑k=1∞(-1)k+1(t-3)k+1(k+1)(k-1)!sin⁡x-2ℏe-1(0.1)2∑k=1∞(-1)k+1(t-3)kk!sin⁡x-ℏ(0.1)32!∑k=1∞(-1)k+1(t-3)k+2(k+2)(k-1)!sin⁡x-ℏ(0.1)3∑k=1∞(-1)k+1(t-3)k+1(k+1)(k-1)!sin⁡x-ℏ(0.1)32!∑k=1∞(-1)k+1(t-3)kk!sin⁡x-ℏe(0.1)43!∑k=1∞(-1)k+1(t-3)k+3(k+3)(k-1)!sin⁡x,u2(t,x)=ℏ(ℏ+1)e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)-ℏ(ℏ+1)(0.1)∑k=2∞(-1)k(t-3)k-1(k-2)!sin⁡x-ℏ(ℏ+1)(0.1)2e-1∑k=2∞(-1)k(t-3)kk!sin⁡x-2ℏ(ℏ+1)(0.1)2e-1∑k=2∞(-1)k(t-3)k-1(k-1)!sin⁡x-ℏ(ℏ+1)(0.1)32!∑k=2∞(-1)k(t-3)k-1(k-1)!sin⁡x-ℏ(ℏ+1)(0.1)3∑k=2∞(-1)k(t-3)kk(k-2)!sin⁡x-ℏ(ℏ+1)(0.1)3∑k=2∞(-1)k(t-3)k+1(k+1)(k-2)!sin⁡x-ℏ(ℏ+1)(0.1)43!e∑k=2∞(-1)k(t-3)k+2(k+2)(k-2)!sin⁡x+ℏ2e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)22!+ℏ2e-2(0.1)∑k=2∞(-1)k+1(t-3)kk!sin⁡x+ℏ2e-1(0.1)2∑k=2∞(-1)k+1(t-3)k+1(k+1)k(k-2)!sin⁡x+2ℏ2e-1(0.1)2∑k=2∞(-1)k+1(t-3)kk!sin⁡x+ℏ2(0.1)32!∑k=2∞(-1)k+1(t-3)k+2(k+2)(k+1)(k-2)!sin⁡x+ℏ2(0.1)3∑k=2∞(-1)k+1(t-3)k+1(k+1)k(k-2)!sin⁡x+ℏ2(0.1)32!∑k=2∞(-1)k+1(t-3)kk!sin⁡x+ℏ2e(0.1)43!∑k=2∞(-1)k+1(t-3)k+3(k+3)(k+2)(k-2)!sin⁡x,u3(t,x)=ℏ(ℏ+1)2e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)-ℏ(ℏ+1)2(0.1)e-2∑k=3∞(-1)k+1(t-3)k-2(k-2)!sin⁡x-ℏ(ℏ+1)2(0.1)2e-1∑k=3∞(-1)k+1(t-3)k-1(k-1)(k-3)!sin⁡x-2ℏ(ℏ+1)2(0.1)2e-1∑k=3∞(-1)k+1(t-3)k-2(k-2)!sin⁡x-ℏ(ℏ+1)2(0.1)32!∑k=3∞(-1)k+1(t-3)kk(k-3)!sin⁡x-ℏ(ℏ+1)2(0.1)3∑k=3∞(-1)k+1(t-3)k-1(k-1)(k-3)!sin⁡x-ℏ(ℏ+1)2(0.1)32!∑k=3∞(-1)k+1(t-3)k-2(k-2)!sin⁡x-ℏ(ℏ+1)2e(0.1)43!∑k=3∞(-1)k+1(t-3)k+1(k+1)(k-3)!sin⁡x+2ℏ2(ℏ+1)e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)22!-2ℏ2(ℏ+1)e-2(0.1)∑k=3∞(-1)k+1(t-3)k-1(k-1)!sin⁡x-ℏ2(ℏ+1)e-1(0.1)2∑k=3∞(-1)k+1(t-3)kk(k-1)(k-3)!sin⁡x-4ℏ2(ℏ+1)e-1(0.1)2∑k=3∞(-1)k+1(t-3)k-1(k-1)!sin⁡x-2ℏ2(ℏ+1)(0.1)32!∑k=3∞(-1)k+1(t-3)k+1(k+1)k(k-3)!sin⁡x-2ℏ2(ℏ+1)(0.1)3∑k=3∞(-1)k+1(t-3)kk(k-1)(k-3)!sin⁡x-2ℏ2(ℏ+1)(0.1)32!∑k=3∞(-1)k+1(t-3)k-1(k-1)!sin⁡x-2ℏ2(ℏ+1)e(0.1)43!∑k=3∞(-1)k+1(t-3)k+2(k+2)(k+1)(k-3)!sin⁡x+ℏ3e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)33!-ℏ3e-2(0.1)∑k=3∞(-1)k+1(t-3)kk!sin⁡x-ℏ3e-1(0.1)2∑k=3∞(-1)k+1(t-3)k+1(k+1)k(k-1)(k-3)!sin⁡x-2ℏ3e-1(0.1)2∑k=3∞(-1)k+1(t-3)kk!sin⁡x-ℏ3(0.1)32!∑k=3∞(-1)k+1(t-3)k+2(k+2)(k+1)k(k-3)!sin⁡x-ℏ3(0.1)3∑k=3∞(-1)k+1(t-3)k+1(k+1)k(k-1)(k-3)!sin⁡x-ℏ3(0.1)32!∑k=3∞(-1)k+1(t-3)kk!sin⁡x-ℏ3e(0.1)43!∑k=3∞(-1)k+1(t-3)k+3(k+3)(k+2)(k+1)(k-3)!sin⁡x,⋮un(t,x)=f4(ℏ+1,t,x)+ℏne-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)nn!+(-1)nℏne-2(0.1)∑k=n∞(-1)k+1(t-3)kk!sin⁡x+(-1)nℏne-1(0.1)2∑k=n∞(-1)k+1(t-3)k+1(k+1)k(k-1)⋯(k-(n-2))(k-n)!sin⁡x+(-1)nℏn2e-1(0.1)2∑k=n∞(-1)k+1(t-3)kk!sin⁡x+(-1)nℏn(0.1)32!∑k=n∞(-1)k+1(t-3)k+2(k+2)(k+1)k(k-3)⋯(k-(n-3))(k-n)!sin⁡x+(-1)nℏn(0.1)3∑k=n∞(-1)k+1(t-3)k+1(k+1)k(k-1)⋯(k-(n-2))(k-n)!sin⁡x+(-1)nℏn(0.1)32!∑k=n∞(-1)k+1(t-3)kk!sin⁡x+(-1)nℏne(0.1)43!∑k=n∞(-1)k+1(t-3)k+3(k+2)(k+1)k⋯(k-(n-4))(k-n)!sin⁡x,⋮
and so on. For ℏ=-1, we get
(3.50)u0(t,x)=e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x,u1(t,x)=-e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)+e-2(0.1)∑k=1∞(-1)k+1(t-3)kk!sin⁡x+e-1(0.1)2∑k=1∞(-1)k+1(t-3)k+1(k+1)(k-1)!sin⁡x+2e-1(0.1)2∑k=1∞(-1)k+1(t-3)kk!sin⁡x+(0.1)32!∑k=1∞(-1)k+1(t-3)k+2(k+2)(k-1)!sin⁡x+(0.1)3∑k=1∞(-1)k+1(t-3)k+1(k+1)(k-1)!sin⁡x+(0.1)32!∑k=1∞(-1)k+1(t-3)kk!sin⁡x+e(0.1)43!∑k=1∞(-1)k+1(t-3)k+3(k+3)(k-1)!sin⁡x,u2(t,x)=e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)22!+e-2(0.1)∑k=2∞(-1)k+1(t-3)kk!sin⁡x+e-1(0.1)2∑k=2∞(-1)k+1(t-3)k+1(k+1)k(k-2)!sin⁡x+2e-1(0.1)2∑k=2∞(-1)k+1(t-3)kk!sin⁡x+(0.1)32!∑k=2∞(-1)k+1(t-3)k+2(k+2)(k+1)(k-2)!sin⁡x+(0.1)3∑k=2∞(-1)k+1(t-3)k+1(k+1)k(k-2)!sin⁡x+(0.1)32!∑k=2∞(-1)k+1(t-3)kk!sin⁡x+e(0.1)43!∑k=2∞(-1)k+1(t-3)k+3(k+3)(k+2)(k-2)!sin⁡x,u3(t,x)=-e-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)33!+e-2(0.1)∑k=3∞(-1)k+1(t-3)kk!sin⁡x+e-1(0.1)2∑k=3∞(-1)k+1(t-3)k+1(k+1)k(k-1)(k-3)!sin⁡x+2e-1(0.1)2∑k=3∞(-1)k+1(t-3)kk!sin⁡x+(0.1)32!∑k=3∞(-1)k+1(t-3)k+2(k+2)(k+1)k(k-3)!sin⁡x+(0.1)3∑k=3∞(-1)k+1(t-3)k+1(k+1)k(k-1)(k-3)!sin⁡x+(0.1)32!∑k=3∞(-1)k+1(t-3)kk!sin⁡x+e(0.1)43!∑k=3∞(-1)k+1(t-3)k+3(k+3)(k+2)(k+1)(k-3)!sin⁡x,⋮un(t,x)=(-1)ne-3[[(0.1)e]33!+[2(0.1)e]22!+3(0.1)e+1]sin⁡x(t-3)nn!+e-2(0.1)∑k=n∞(-1)k+1(t-3)kk!sin⁡x+e-1(0.1)2∑k=n∞(-1)k+1(t-3)k+1(k+1)k(k-1)⋯(k-(n-2))(k-n)!sin⁡x+2e-1(0.1)2∑k=n∞(-1)k+1(t-3)kk!sin⁡x+(0.1)32!∑k=n∞(-1)k+1(t-3)k+2(k+2)(k+1)k(k-3)⋯(k-(n-3))(k-n)!sin⁡x+(0.1)3∑k=n∞(-1)k+1(t-3)k+1(k+1)k(k-1)⋯(k-(n-2))(k-n)!sin⁡x+(0.1)32!∑k=n∞(-1)k+1(t-3)kk!sin⁡x+e(0.1)43!∑k=n∞(-1)k+1(t-3)k+3(k+2)(k+1)k⋯(k-(n-4))(k-n)!sin⁡x,⋮
and so on.From (3.6), the solution of (3.44) is obtained as(3.51)u(t,x)=([(0.1)e(t-3)]44!+[(0.1)e(t-2)]33!+[(0.1)e(t-1)]22!+(0.1)et+1)∑k=0∞(-1)ktkk!sin⁡x.
This series has the closed form(3.52)u(t,x)=([(0.1)e(t-3)]44!+[(0.1)e(t-2)]33!+[(0.1)e(t-1)]22!+(0.1)et+1)e-tsin⁡x,
which is the exact solution of (3.44).We give the HAM solutions int∈[0,1], t∈[1,2], t∈[2,3], t∈[3,4]. We use four terms for evaluating the approximate solution uapp=∑k=03uk(t,x). According to the ℏ-curve of uxt(0,0), the solution series is convergent when -1.48≤ℏ≤0.48, -1.41≤ℏ≤2.10, -1.19≤ℏ≤0.12, and -1.02≤ℏ≤0, respectively, in t∈[0,1], t∈[1,2], t∈[2,3], t∈[3,4]. We take ℏ=-1 to determine how much the approximate solution is accurate and compute the absolute errors uerr=|uexact-uapp| at the points (0.5,π/2), (1.5,π/2), (2.5,π/2), (3.5,π/2) in Table 5.Table 5
The absolute error atx=π/2 when ℏ=-1.
tuexactuappuerr0.50.688966723247640.687185174004180.001781549243461.50.316170660693700.315284387317410.000886273376282.50.144725356942580.144321999823470.000403357119103.50.066241553295910.066192994488770.00004855880714


## 4. Conclusion

The numerical solutions of first order of difference scheme (2.20) and second order of difference scheme (2.29) for different values of N and M and the approximate solutions obtained by HAM for N=3 in (3.11) when ℏ=-1 are given at the same points (0.5,π/2), (1.5,π/2), (2.5,π/2), (3.5,π/2) in Tables 6, 7, 8, and 9, respectively. The absolute errors computed show that, with homotopy analysis method, the results are more accurate for the parabolic delay equation (2.17).Table 6
Comparison of the absolute error in HAM whenℏ=-1 and the errors of different difference schemes at (0.5,π/2).
MethoduappuerrHAM forN=30.687185174004180.00178154924346Difference scheme (2.20) for N=M=480.690692365741470.00172564249383Difference scheme (2.29) for N=M=100.690678050573830.00171132732619Table 7
Comparison of the absolute error in HAM whenℏ=-1 and the errors of different difference schemes at (1.5,π/2).
MethoduappuerrHAM forN=30.315284387317410.00088627337628Difference scheme (2.20) for N=M=1480.317053893170700.00088323247700Difference scheme (2.29) for N=M=180.317079574674490.00090891398079Table 8
Comparison of the absolute error in HAM whenℏ=-1 and the errors of different difference schemes at (2.5,π/2).
MethoduappuerrHAM forN=30.144321999823470.00040335711910Difference scheme (2.20) for N=M=2900.145127314595680.00040195765309Difference scheme (2.29) for N=M=240.145127239122490.00047857754134Table 9
Comparison of the absolute error in HAM whenℏ=-1 and the errors of different difference schemes at (3.5,π/2).
MethoduappuerrHAM forN=30.066192994488770.00004855880714Difference scheme (2.20) for N=M=4800.066406948789630.00016953954937Difference scheme (2.29) for N=M=580.066287275636900.00004572234990Although HAM seems to be more rapid than finite difference method, the series solutions obtained by HAM are convergence only for the regions determined by convergence control parameterℏ. So, convergence region is limited for HAM. The comparison of two methods of finite difference and homotopy analysis shows that latter is more rapid and more accurate in the cases that series solutions are convergence. When we take ℏ out of the convergence region determined by ℏ curves, it is shown that finite difference method is faster and more accurate than HAM. The approximate solutions obtained by HAM for different values of ℏ chosen from out of the convergence region of the series solutions and the numerical solutions of first and second order of difference schemes (2.20) and (2.29) for N=M=4 are given in Tables 10, 11, 12, and 13, respectively at the same points (0.5,π/2), (1.5,π/2), (2.5,π/2), (3.5,π/2).Table 10
Comparison of the absolute error in HAM whenℏ=-2 and the errors of different difference schemes at (0.5,π/2).
MethoduappuerrHAM forN=31.823821039402681.13485431615504Difference scheme (2.20) for N=M=40.717545844299230.02857912105159Difference scheme (2.29) for N=M=40.643968291648380.04499843159926Table 11
Comparison of the absolute error in HAM whenℏ=-2.1 and the errors of different difference schemes at (1.5,π/2).
MethoduappuerrHAM forN=3−0.716406886517041.0325774721074Difference scheme (2.20) for N=M=40.266254680316870.04991598037683Difference scheme (2.29) for N=M=40.269080217975660.04709044271804Table 12
Comparison of the absolute error in HAM whenℏ=1.5 and the errors of different difference schemes at (2.5,π/2).
MethoduappuerrHAM forN=31.857439682841201.71271432589862Difference scheme (2.20) for N=M=40.092504552941440.05222080400114Difference scheme (2.29) for N=M=40.114183784830600.03054157211198Table 13
Comparison of the absolute error in HAM whenℏ=2 and the errors of different difference schemes at (3.5,π/2).
MethoduappuerrHAM forN=31.339546030117371.27330447682146Difference scheme (2.20) for N=M=40.029041277169670.03720027612623Difference scheme (2.29) for N=M=40.048428042022390.01781351127352Despite HAM, by finite difference method, we can guarantee the convergence in the whole domain that (2.17) is defined in. Therefore finite difference method is more efficient than HAM.



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*Source: 682752-2012-06-03.xml*