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Superclose analysis of a two-grid finite element scheme for semilinear parabolic integro-differential equations

  • In this paper, a two-grid finite element scheme for semilinear parabolic integro-differential equations is proposed. In the two-grid scheme, continuous linear element is used for spatial discretization, while Crank-Nicolson scheme and Leap-Frog scheme are ultilized for temporal discretization. Based on the combination of the interpolation and Ritz projection technique, some superclose estimates between the interpolation and the numerical solution in the H1-norm are derived. Notice that we only need to solve nonlinear problem once in the two-grid scheme, namely, the first time step on the coarse-grid space. A numerical example is presented to verify the effectiveness of the proposed two-grid scheme.

    Citation: Changling Xu, Tianliang Hou. Superclose analysis of a two-grid finite element scheme for semilinear parabolic integro-differential equations[J]. Electronic Research Archive, 2020, 28(2): 897-910. doi: 10.3934/era.2020047

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  • In this paper, a two-grid finite element scheme for semilinear parabolic integro-differential equations is proposed. In the two-grid scheme, continuous linear element is used for spatial discretization, while Crank-Nicolson scheme and Leap-Frog scheme are ultilized for temporal discretization. Based on the combination of the interpolation and Ritz projection technique, some superclose estimates between the interpolation and the numerical solution in the H1-norm are derived. Notice that we only need to solve nonlinear problem once in the two-grid scheme, namely, the first time step on the coarse-grid space. A numerical example is presented to verify the effectiveness of the proposed two-grid scheme.



    In this paper, we consider the following semilinear parabolic integro-differential equations:

    utu+t0u(s)ds=f(u), XΩ, tJ, (1)
    u(X,t)=0, XΩ, tJ, (2)
    u(X,0)=u0(X), XΩ, (3)

    where ΩR2 is a rectangle with boundary Ω, X=(x,y), J=(0,T], f() is twice continuously differentiable and u0 is a given function. We assume that

    |f(y)|+|f(y)|M, yR.

    There exists a lot of numerical methods for solving nonlinear partial differential equations in the literature. For example, Cannon and Lin [1] derived a priori error estimates of semidiscrete and Crank-Nicolson finite element approximations to the solution of the nonlinear diffusion equations with memory. Eriksson and Johnson [3] used adaptive finite element method to solve nonlinear parabolic problems. Moore [9] considered a posteriori error estimates for semi- and fully discrete finite element methods using a degree polynomial basis for solving nonlinear parabolic equations. Garcia [4] discussed a priori error estimates of fully discrete Raviart-Thomas mixed finite element scheme for nonlinear parabolic equations.

    Two-grid method was first proposed by Xu [11,12] as an efficient discretization technique for solving the nonlinear and the nonsymmetric problems. Liu et al. exploited the two kinds of two-grid algorithms for finite difference solutions of semilinear parabolic equations in [8]. Chen et al. [2] presented a two-grid scheme of mixed finite element method for fully nonlinear reaction-diffusion equations. Hou et al. [5] presented a two-grid method of P20-P1 mixed finite element method combined with Crank-Nicolson scheme for a class of nonlinear parabolic equations. Shi and Mu [10] discussed some superclose results of a two-grid finite element method for semilinear parabolic equations. Yang and Xing [13] discussed the convergence of two-grid discontinuous Galerkin scheme for a kind of nonlinear parabolic problems.

    This paper is motivated by the ideas of the works [10,11], we present a two-grid scheme for semilinear parabolic integro-differential equations discretized by finite element method combined with Crank-Nicolson scheme. We mainly discuss the superclose estimates between the numerical solution and the interpolation.

    The plan of this paper is as follows. In Section 2, we give the Crank-Nicolson scheme and deduce the superclose result of order O(h2+(t)2) in the H1-norm. In Section 3, we present the two-grid method and derive the superclose estimates of order O(H2+(t)2) and order O(h2+H4+(t)2), respectively. In Section 4, we present a numerical example to demonstrate the effectiveness of our method.

    We adopt the standard notation Wm,p(Ω) for Sobolev spaces on Ω with a norm m,p given by vpm,p=|α|mDαvpLp(Ω), a semi-norm ||m,p given by |v|pm,p=|α|=mDαvpLp(Ω). We set Wm,p0(Ω)={vWm,p(Ω):v|Ω=0}. For p=2, we denote Hm(Ω)=Wm,2(Ω), Hm0(Ω)=Wm,20(Ω), and m=m,2, =0,2.

    We denote by Ls(J;Wm,p(Ω)) the Banach space of all Ls integrable functions from J into Wm,p(Ω) with norm vLs(J;Wm,p(Ω))=(T0||v||sWm,p(Ω)dt)1s for s[1,), and the standard modification for s=. For simplicity of presentation, we denote vLs(J;Wm,p(Ω)) by vLs(Wm,p). Similarly, one can define the spaces H1(J;Wm,p(Ω)). In addition C denotes a general positive constant independent of h and t, where h is the spatial mesh-size, and t is the time step.

    Let Th be a uniform rectangular partition of Ω with mesh size h. Vh be the bilinear finite element space with vanishes on Ω. Let Ih and Rh be the associated interpolation and Ritz projection operators on Vh, respectively(see[10]).

    Then, for uH10(Ω)H3(Ω), from [10] we know that

    uRhu+h(uRhu)Ch2u2, (4)
    IhuRhu1Ch2u3, (5)
    ((uIhu),vh)Ch2u3vh, (6)
    ((uRhu),vh)=0,  vhVh. (7)

    The weak formulation of (1) is to find u:JH10(Ω), such that

    (ut,v)+(u,v)t0(u(s),v)ds=(f(u),v),  vH10(Ω). (8)

    Let {tntn=nt;0nN} be a uniform partition in time with time step t, un=u(X,tn) and tn1/2=(tn1+tn)/2. For a sequence of functions {ϕn}Nn=0, we denote dtϕn=(ϕnϕn1)/t, then the Crank-Nicolson scheme of (1) is to find unhVh for n=1,2,...,N, such that

    (dtunh,vh)(n2j=0t2(ujh+uj+1h)+t8(3un1h+unh),vh)=((unh+un1h)2,vh)+(f(unh)+f(un1h)2,vh),  vhVh, (9)
    u0h=Rhu0(X), XΩ. (10)

    For the proof of existence and uniqueness of the solution for the nonlinear algebraic problem (9)-(10), please refer to [6].

    Theorem 2.1. Let u and unh be the solutions of (8) and (9), respectively. Assume that uL(H3), utL2(H2), uttL2(L2), utttL2(L2), uttL2(H2)L(L2) and utttL2(L2), then, for n=1,2,,N, we have

    unhIhun1C(h2+(t)2). (11)

    Proof. Letting t=tn1/2 and v=vh in (8), we get

    (dtun,vh)+((un+un1)2,vh)tn1/20(u(s),vh)ds=(f(un1/2),vh)+(Rn1,vh)+(Rn2,vh),  vhVh, (12)

    where Rn1=dtunun1/2t, Rn2=un+un12un1/2.

    Setting ununh=unRhun+Rhununh:=ηn+ξn. Subtracting (9) from (12), with the help of (7), we have

    (dtξn,vh)+((ξn+ξn1)2,vh)=(dtηn,vh)+(tn1/20u(s)dsn2j=0t2(ujh+uj+1h)t8(3un1h+unh),vh)+(Rn1,vh)+(Rn2,vh)+(f(un1/2)f(unh)+f(un1h)2,vh),  vhVh. (13)

    Selecting vh=dtξn in (13), noting that

    ((ξn+ξn1)2,dtξn)=12t(ξn2ξn12), (14)

    then multiplying (13) by 2t and summing from n=1,...,l(1lN), we conclude that

    2ln=1dtξn2t+ξl2=2ln=1(dtηn,dtξn)t+2ln=1(tn1/20u(s)dsn2j=0t2(uj+uj+1)t4(un1+un1/2),dtξn)t+2ln=1(n2j=0t2(ξj+ξj+1)+t4(ξn1+Rhun1/2unh+un1h2),dtξn)t+2ln=1(f(un1/2)f(unh)+f(un1h)2,dtξn)t+2ln=1(Rn1,dtξn)t+2ln=1(Rn2,ξnξn1)=:6i=1Ii, (15)

    where we used (7) and ξ0=0.

    Now, we estimate the right-hand terms of (15). For I1, it is easy to check that

    dtηn21ttntn1ηt2ds,

    which together with Cauchy inequality, Young's inequality and (4) yields

    I1Ch4ut2L2(H2)+18ln=1dtξn2t. (16)

    For I2, we decompose it as

    I2=2ln=1(n2j=0tj+1tj(u(s)(uj+uj+1)2)ds+tn1/2tn1(u(s)(un1+un1/2)2)ds,ξnξn1)=2ln=2(n2j=0tj+1tj(u(s)(uj+uj+1)2)ds,ξnξn1)+2ln=2(tn1/2tn1(u(s)(un1+un1/2)2)ds,ξnξn1)+2(t1/20(u(s)(u0+u1/2)2)ds,ξ1)=:3i=1Ai. (17)

    Using Cauchy inequality and Young's inequality, we see that

    A1=2(l2j=0tj+1tj(u(s)(uj+uj+1)2)ds,ξl)2(t10(u(s)(u0+u1)2)ds,ξ1)2l1n=2(tntn1(u(s)(un+un1)2)ds,ξn)C(t)4utt2L2(L2)+Cln=1ξn2t+18ξl2, (18)
    A2=2(tl1/2tl1(u(s)(ul1+ul1/2)2)ds,ξl)2(t3/2t1(u(s)(u1+u3/2)2)ds,ξ1)2l1n=2(tn+1/2tn(u(s)(un+un+1/2)2)dstn1/2tn1(u(s)(un1+un1/2)2)ds,ξn)C(t)4utt2L2(L2)+Cln=1ξn2t+18ξl2, (19)

    and

    A3C(t)4utt2L2(L2)+Cξ12t. (20)

    Thus, we get

    I2C(t)4utt2L2(L2)+Cln=1ξn2t+14ξl2. (21)

    For I3, by virtue of Cauchy inequality and Young's inequality, we have

    I3=2ln=2(n2j=0t2(ξj+ξj+1),ξnξn1)+ln=1(t2(ξn1+(Rhun1/2unh+un1h2)),ξnξn1)=2(l2j=0t2(ξj+ξj+1),ξl)2(l1n=1t2(ξn1+ξn),ξn)+2(t4(ξl1+(Rhul1/2ulh+ul1h2)),ξl)2l1n=1(t4(ξn+(Rhun+1/2un+1h+unh2))t4(ξn1+(Rhun1/2unh+un1h2)),ξn)C(t)4utt2L2(H3)+Cln=1ξn2t, (22)

    where we used the following estimate

    (Rhun1/2unh+un1h2)=(Rhun1/2Rhun1+Rhun2+Rhun1+Rhun2unh+un1h2)=12(Rhun1Rhun1/2)+(RhunRhun1/2)+ξn1+ξnCRhut(ϕn)(tntn1/2)+Rhut(ψn)(tn1tn1/2)+Cξn+Cξn1CtRhutt(λn)(ϕnψn)+Cξn+Cξn1C(t)2Rhutt(λn)+Cξn+Cξn1C(t)2(utt(λn)+utt(λn)Rhutt(λn))+Cξn+Cξn1C(t)2utt(λn)2+Cξn+Cξn1, (23)

    where ϕn is located between tn and tn1/2, ψn is located between tn1/2 and tn1, λn is located between ϕn and ψn, and

    ϕnψntntn1=t.

    Next, we estimate I4. By use of mean value theorem and the assumption on f, we conclude that

    f(un1/2)f(un)+f(un1)212f(λn2)(un1/2un)+f(λn1)(un1/2un1)=t4f(λn1)ut(θn1)f(λn2)ut(θn2)=t4f(λn1)ut(θn1)f(λn1)ut(θn2)+f(λn1)ut(θn2)f(λn2)ut(θn2)=t4f(λn1)utt(θn3)(θn2θn1)+f(λn3)(λn1λn2)ut(θn2)C(t)2(utt(θn3)+ut(θn2)), (24)

    where we also used

    θn2θn1tntn1=t

    and

    |λn2λn1||unλn1|+|un1/2λn1||unun1|+|unun1/2|+|un1/2un1|2(|unun1/2|+|un1/2un1|)=2(|ut(θn1)|+|ut(θn2)|)Δt,

    where θn1 is located between tn1 and tn1/2, θn2 is located between tn1/2 and tn, θn3 is located between θn1 and θn2, λn1 is located between un1 and un1/2, λn2 is located between un1/2 and un, λn3 is located between λn1 and λn2.

    Using (4), mean value theorem and the assumption on f, we easily get

    f(un)+f(un1)2f(unh)+f(un1h)2C(ununh+un1un1h)Ch2(un2+un12)+C(ξn+ξn1). (25)

    By use of Cauchy inequality, Young's inequality and (24)-(25), we derive

    I4C(t)4(utt2L2(L2)+ut2L2(L2))+Ch4u2L2(H2)+Cln=1ξn2t+18ln=1dtξn2t. (26)

    For I5, from Cauchy inequality, Young's inequality and

    Rn1C(t)2uttt(ρn),

    we have

    I5C(t)4uttt2L2(L2)+18ln=1dtξn2t, (27)

    where ρn is located between tn1 and tn.

    Notice that

    ln=1(Rn2,dtξn)t=(Rl2,ξl)l1n=1(Rn+12Rn2,ξn). (28)

    Using Taylor expansion, we know that

    un=un+1/2un+1/2tt2+un+1/2tt(t)28148uttt(βn)(t)3, (29)
    un+1=un+1/2+un+1/2tt2+un+1/2tt(t)28+148uttt(γn)(t)3, (30)

    where tn<βn<tn+1/2<γn<tn+1.

    Using Cauchy inequality, Young's inequality, (29), (30) and mean value theorem, we get

    (Rl2,ξl)C(t)4utt(λl)2+12ξl2, (31)
    l1n=1(Rn+12Rn2,ξn)=196l1n=1(uttt(γn)(t)3uttt(βn)(t)3,ξn)196l1n=1(uttt(γn1)(t)3uttt(βn1)(t)3,ξn)+l1n=1((un+1/2ttun1/2tt)(t)28,ξn)=196l1n=1(uttt(γn)(t)3uttt(βn)(t)3,ξn)196l1n=1(uttt(γn1)(t)3uttt(βn1)(t)3,ξn)+l1n=1((t)28uttt(δn)t,ξn)C(t)4uttt2L2(L2)+Cln=1ξn2t, (32)

    where λl is located between tl1 and tl, δn is located between tn1/2 and tn+1/2.

    For I6, combining Cauchy inequality, Young's inequality, (28) with (31)-(32), we derive

    I6C(t)4(uttt2L2(L2)+utt2L(L2))+Cln=1ξn2t+12ξl2. (33)

    Now, substituting the estimates for I1-I6 into (15), then applying discrete Gronwall's lemma, for sufficiently small t, we have

    ξlCh2(u2L2(H2)+ut2L2(H2))1/2+C(t)2(ut2L2(L2)+utt2L2(L2)+uttt2L2(L2)+utt2L2(H3)+utt2L(L2)+uttt2L2(L2))1/2, (34)

    which together with (5), Poincare's inequality and triangle inequality yields (11). We complete the proof of the theorem.

    In this section, we present the main algorithm of the paper, which has the following two steps:

    Step 1. On the coarse grid TH, compute u1HVH to satisfy the following original nonlinear system:

    (dtu1H,vH)(t8(3u0H+u1H),vH)=((u1H+u0H)2,vH)+(f(u1H)+f(u0H)2,vH),  vHVH, (35)
    u0H=RHu0(X), XΩ. (36)

    For n=1,...,N1, compute un+1HVH to satisfy the following linear system:

    (un+1Hun1H2t,vH)(nj=1t2(ujH+uj1H),vH)=((un+1H+un1H)2,vH)+(f(unH),vH),  vHVH. (37)

    Step 2. On the fine grid Th, for n=1,...,N, compute ˜unhVh to satisfy the following linear system:

    (dt˜unh,vh)(n2j=0t2(˜ujh+˜uj+1h)+t8(3˜un1h+˜unh),vh)+((˜unh+˜un1h)2,vh)=(f(unH)+f(unH)(˜unhunH)2,vh)+(f(un1H)+f(un1H)(˜un1hun1H)2,vh),  vhVh, (38)
    ˜u0h=Rhu0(X), XΩ. (39)

    Now, we shall discuss the superclose estimates of the above two-grid algorithm in the following theorem.

    Theorem 3.1. Let u, unH and ˜unh be the solutions of (8), (35)-(37) and (35)-(39), respectively. Then under the conditions of Theorem 2.1, for n=1,2,,N, we have

    unHIHun1C(H2+(t)2), (40)
    ˜unhIhun1C(h2+H4+(t)2). (41)

    Proof. Setting ununH=unRHun+RHununH:=ωn+φn. From Theorem 2.1, (40) is obvious for n=1.

    For n=1,...,N1, letting t=tn and v=vH in (8), we get

    (un+1un12t,vH)+((un+1+un1)2,vH)tn0(u(s),vH)ds=(f(un),vH)+(Rn3,vH)+(Rn4,vH),  vHVH, (42)

    where Rn3=un+1un12tunt, Rn4=un+1+un12un.

    Subtracting (37) from (42), with the help of (7), we have

    (φn+1φn12t,vH)+((φn+1+φn1)2,vH)=(ωn+1ωn12t,vH)+(tn0u(s)dsnj=1t2(ujH+uj1H),vH)+(f(un)f(unH),vH)+(Rn3,vH)+(Rn4,vH),  vHVH. (43)

    Selecting vH=φn+1φn1t in (43), using φ0=0 and the equality

    ((φn+1+φn1)2,(φn+1φn1)t)=12t(φn+12+φn2φn2φn12), (44)

    then multiplying the resulting equation by 2t and summing from n=1,...,l(1lN1), we conclude that

    ln=1φn+1φn1t2t+φl+12+φl2=ln=1(ωn+1ωn1t,φn+1φn1t)t+2ln=1(tn0u(s)dsnj=1t2(uj+uj+1),φn+1φn1)+2ln=1(nj=1t2(φj+φj1),φn+1φn1)+2ln=1(f(un)f(unH),φn+1φn1t)t+2ln=1(Rn3,φn+1φn1t)t+2ln=1(Rn4,φn+1φn1)+φ12=:7i=1Di, (45)

    where we used (7) and φ0=0.

    Similar to the estimates of I1-I6, we can estimate D1-D6 as

    D1CH4ut2L2(H2)+18ln=1φn+1φn1t2t, (46)
    D2C(t)4utt2L2(L2)+Cln=0φn+12t+14φl+12, (47)
    D3Cln=0φn+12t, (48)
    D4CH4u2L2(H2)+Cln=1φn2t+18ln=1φn+1φn1t2t, (49)
    D5C(t)4uttt2L2(L2)+18ln=1φn+1φn1t2t, (50)
    D6C(t)4(uttt2L2(L2)+utt2L(L2))+Cln=0φn+12t+14φl+12. (51)

    At last, for D7, using (5), (40) and triangle inequality, we see that

    D72(IHu1RHu1)2+2(IHu1u1H)2C(H4+(t)4). (52)

    Now, substituting the estimates for D1-D7 into (45), then applying discrete Gronwall's lemma, for sufficiently small t, we have

    φl+1CH2(u2L2(H2)+ut2L2(H2))1/2+C(t)2(uttt2L2(L2)+utt2L2(H3)+utt2L(L2)+uttt2L2(L2))1/2, (53)

    which together with (5), Poincare's inequality and triangle inequality yields

    unHIHun1unHRHun1+RHunIHun1CH2(u2L(H3)+ut2L2(H2))1/2+C(t)2(uttt2L2(L2)+utt2L(L2)+utt2L2(H3)+uttt2L2(L2))1/2, n2. (54)

    Using Taylor expansion, we have

    f(un)=f(unH)+f(unH)(ununH)+f(αn)(ununH)22, (55)

    where αn is located between un and unH.

    Setting Rhun˜unh:=˜ξn. Subtracting (38) from (12), similar to (15), we conclude that

    2ln=1dt˜ξn2t+˜ξl2=2ln=1(dtηn,dt˜ξn)t+2ln=1(tn1/20u(s)dsn2j=0t2(˜ujh+˜uj+1h)t8(3˜un1h+˜unh),˜ξn˜ξn1)+2ln=1(f(un1/2)f(un)+f(un1)2,dt˜ξn)t+2ln=1(f(unH)(un˜unh)2+f(αn)(ununH)24,dt˜ξn)t+2ln=1(f(un1H)(un1˜un1h)2+f(αn1)(un1un1H)24,dt˜ξn)t+2ln=1(Rn1,dt˜ξn)t+2ln=1(Rn2,˜ξn˜ξn1)=:7i=1Bi. (56)

    Now, we estimate B1-B7, respectively. For B3, similar to (24), we know that

    B3C(t)4utt2L2(L2)+18ln=1dt˜ξn2t. (57)

    For B4, we find from Cauchy inequality and the assumption on f that

    B412ln=1(2f(unH)(un˜unh)+f(αn)(ununH)2)dt˜ξntCln=1un˜unhdt˜ξnt+Cln=1(ununH)2dt˜ξnt=:G1+G2. (58)

    Using Young's inequality and (4), we know that

    G1Ch4u2L2(H2)+Cln=1˜ξn2t+18ln=1dt˜ξn2t. (59)

    Combining (40), Cauchy inequality, Young's inequality and interpolation theory with H1L4, we derive

    G2ln=1(unIHun20,4+IHununH20,4)dt˜ξntCln=1(H4un22,4+IHununH21)dt˜ξntC((t)4+H8)+18ln=1dt˜ξn2t. (60)

    Now, from (59)-(60), we find that

    B4Cln=1˜ξn2t+C(h4+(t)4+H8)+14ln=1dt˜ξn2t. (61)

    Similarly, we can estimate B5 as

    B5Cln=1˜ξn12t+C(h4+(t)4+H8)+14ln=1dt˜ξn2t. (62)

    Similar to (16), (21), (27) and (33), we easily have

    |B1+B2+B6+B7|Ch4ut2L2(H2)+Cln=1˜ξn2t+12˜ξn2+18ln=1dt˜ξn2t+C(t)4(uttt2L2(L2)+utt2L2(L2)+uttt2L2(L2)). (63)

    It follows from (56)-(57), (61)-(63) and Poincare's inequality that

    ˜ξl2C(h4+H8+(t)4)+Cln=1˜ξn2t. (64)

    Thus, for sufficiently small t, using discrete Gronwall's lemma and Poincare's inequality, we arrive at

    ˜ξl1C(h2+H4+(t)2), (65)

    which together with (5) and triangle inequality yields

    ˜unhIhun1˜unhRhun1+RhunIhun1C(h2+H4+(t)2). (66)

    The proof is complete.

    In this section, we are going to validate the superclose estimates for two-grid discretization method for semilinear parabolic integro-differential equations by a concrete numerical example.

    We consider the following semi-linear parabolic integro-differential equations

    utu+t0u(s)ds=u3+g(X,t), XΩ, tJ, (67)
    u(X,t)=0, XΩ, tJ, (68)
    u(X,0)=u0(X), XΩ, (69)

    where Ω=(0,1)2 and J=(0,1]. We choose u(X,t)=sin(πt)sin(πx1)sin(πx2) as the exact solution. Then, the explicit formulation of g(X,t) is

    g(X,t)=(πcos(πt)+2π2sin(πt)+2π(cos(πt)1))sin(πx1)sin(πx2)+(u(X,t))3.

    We first test the example for the Crank-Nicolson scheme. The error and the convergence order of unhIhun1 at t=0.125 with h=Δt are presented in Table 1. Obviously, it is the same with the result in Theorem 2.1. Next, the two-grid scheme is tested. The error and the convergence order of unHIHun1 and ˜unhIhun1 are provided in Table 2 and Table 3. We find from these two tables that the result coincides with that in Theorem 3.1. Finally, we show the efficiency of the two-grid method by comparing the cpu time in Table 4.

    Table 1.  The error and the convergence order of unhIhun1 at t=0.125 with h=Δt.
    h unhIhun1 order
    1/32 9.6461e-04 -
    1/64 2.4062e-04 2.00
    1/128 6.0130e-05 2.00
    1/256 1.5037e-05 2.00

     | Show Table
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    Table 2.  The error and the convergence order of unHIHun1 at t=0.0625 with H=Δt.
    H unHIHun1 order
    1/16 1.9302e-02 -
    1/32 4.8349e-03 2.00
    1/64 1.2096e-03 2.00
    1/128 3.0244e-04 2.00

     | Show Table
    DownLoad: CSV
    Table 3.  The error and the convergence order of ˜unhIhun1 at t=0.001 with Δt=0.0001 and h=H2.
    H ˜unhIhun1 order
    1/2 8.7973e-04 -
    1/4 7.1420e-05 3.51
    1/8 4.6798e-06 3.91
    1/16 2.9328e-07 3.99

     | Show Table
    DownLoad: CSV
    Table 4.  The cpu time of two-grid scheme and Crank-Nicolson scheme for each time step (h=Δt).
    (H,h) two-grid time (s) Crank-Nicolson time (s)
    (1/4,1/16) 0.0998 0.1164
    (1/8,1/64) 0.9118 1.2019
    (1/16,1/256) 13.6126 17.9624

     | Show Table
    DownLoad: CSV


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