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Research article

Fast finite difference/Legendre spectral collocation approximations for a tempered time-fractional diffusion equation

  • Received: 20 September 2024 Revised: 16 November 2024 Accepted: 20 November 2024 Published: 11 December 2024
  • MSC : 65M70, 65M06, 65M12

  • The present work is concerned with the efficient numerical schemes for a time-fractional diffusion equation with tempered memory kernel. The numerical schemes are established by using a L1 difference scheme for generalized Caputo fractional derivative in the temporal variable, and applying the Legendre spectral collocation method for the spatial variable. The sum-of-exponential technique developed in [Jiang et al., Commun. Comput. Phys., 21 (2017), 650-678] is used to discrete generalized fractional derivative with exponential kernel. The stability and convergence of the semi-discrete and fully discrete schemes are strictly proved. Some numerical examples are shown to illustrate the theoretical results and the efficiency of the present methods for two-dimensional problems.

    Citation: Zunyuan Hu, Can Li, Shimin Guo. Fast finite difference/Legendre spectral collocation approximations for a tempered time-fractional diffusion equation[J]. AIMS Mathematics, 2024, 9(12): 34647-34673. doi: 10.3934/math.20241650

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  • The present work is concerned with the efficient numerical schemes for a time-fractional diffusion equation with tempered memory kernel. The numerical schemes are established by using a L1 difference scheme for generalized Caputo fractional derivative in the temporal variable, and applying the Legendre spectral collocation method for the spatial variable. The sum-of-exponential technique developed in [Jiang et al., Commun. Comput. Phys., 21 (2017), 650-678] is used to discrete generalized fractional derivative with exponential kernel. The stability and convergence of the semi-discrete and fully discrete schemes are strictly proved. Some numerical examples are shown to illustrate the theoretical results and the efficiency of the present methods for two-dimensional problems.



    Fractional differential equations have attracted extensive attention for their ability to accurately describe complex physical phenomena in both micro and macro perspectives [1,2,3]. The differential equations with fractional derivative in time variable can better describe genetic and memory characteristics of some physical processes. For different physical situations, different memory function arise from various fractional derivative operators, such as the power memory kernel, exponential memory kernel, and Mittag-Leffler memory kernel functions [4].

    In this work, we consider the following initial boundary value problem:

    {α,λ(t)0,tu(x,t)=Δu(x,t)+f(x,t),xΩ,t(0,T],u(x,t)=0,xΩ,t(0,T],u(x,0)=u0(x),xΩ, (1.1)

    where Δ=2x21+2x22 denotes the Laplace operator, Ω=[1,1]2 denotes the computational domain with boundary Ω, x=(x1,x2)Ω stands for the space variables, and α,λ(t)0,t denotes the generalized Caputo time-fractional derivative [5]

    α,λ(t)0,tu(x,t)=1Γ(1α)t0λ(ts)(ts)αus(x,s)ds,0<α<1. (1.2)

    In fractional derivative (1.2), we assume that the kernel function satisfies λ(t)C2[0,T], λ(t)>0 and λ(t)0 for all t[0,T]. Particularly, if kernel takes λ(t)=ebt(b>0), fractional derivative (1.2) gives

    α,ebt0,tu(x,t)=1Γ(1α)t0eb(ts)(ts)αus(x,s)ds, (1.3)

    provides the tempered fractional derivative and integral

    α,ebt0,tu(x,t)=ebtΓ(1α)t0(ts)αs(u(x,s)ebs)dsbΓ(1α)t0eb(ts)(ts)αu(x,s)ds. (1.4)

    Obviously, from a mathematical perspective, the Caputo-type fractional derivative (1.2) can be interpreted as a kind of generalized fractional derivative [4,6,7]. From Figure 1, we can see the influence of parameters on the solution. More application (in macroscopic and microscopic scales) of the model given in problem (1.1) with tempered fractional derivative, see references [3,8,9]. There are many research works on the discretization of fractional derivative (1.2). By splitting into two parts, Alikhanov first discussed the L1 formula of fractional derivative (1.2) in the work [5], and proposed a new difference scheme for the time-fractional diffusion equation with the fractional derivatives (1.2). Recently, in view of relation (1.4), Chen et al. [10] established two different L1 discretizations for tempered Caputo fractional derivative on the graded meshes, which developed in reference [11].

    Figure 1.  The evolution of solution of problem (1.1) with λ(t)=ebt. Here, we take zero source term (i.e., f=0). The left figure illustrates behavior of solution with different α, the right figure illustrates behavior of solution with different b.

    The nonlocality of the fractional derivative will make storage and computation very expensive if we adopt a lower-order method. For example, if we use the L1 scheme to calculate (1.1), the required computational and storage amounts are O(NNT) and O(NN2T), respectively, where N is the number of spatial grid points and NT stands for the number of grid points in time. When the time step is small enough, it means that the numerical calculation needs expensive storage costs and computational costs, so it is particularly important to develop fast or high-order numerical schemes. The sum-of-exponential (SOE) technique is one of the effective techniques to approximate the fractional derivatives with the power kernel function [12,13]. In this topic, Jiang et al. [13] used the SOE method to approximate the kernel function t(α+1)(0<α<1) and solved fractional diffusion equations on unbounded domains. Yan et al. [14] combined the L2-1σ scheme with SOE approximation successfully solved an initial boundary value problem of the time fractional diffusion equation. Combining the SOE approximation, Xu et al. [15] proposed a fast L2 discretization of Caputo fractional derivatives, and they get the error estimates by rearranging the coefficients of the fast L2 difference approximation. With the help of the SOE technique, Gu et al. [16] proposed a fast difference algorithm to solve the generalized time-space fractional diffusion equation with fractional derivative (1.3). It seems that the authors of [16] fail to give any convergence analysis of the fast L1 difference scheme.

    As a higher-order numerical method, the spectral method has been widely used for solving many kinds of differential equations [17,18] including fractional differential equations [15]. Lin and Xu [19] solved the time-fractional diffusion equation based on the finite difference method in the time direction and the Legendre spectral method in space. Later, Li and Xu [20] improved their previous results and proposed a time-space spectral method for the time-fractional diffusion equation. Compared with one-dimensional problems, solving high-dimensional problems is more difficult due to the increase in storage and computational cost. Zeng et al. [21] proposed the Legendre Galerkin spectral method for two-dimensional Riesz space fractional nonlinear reaction-diffusion equations and obtained the optimal spatial error estimation. Guo et al. [22] used the Legendre spectral method to solve two-dimensional fractional nonlinear reaction-diffusion wave equations. Recently, combining the fast L2 discretization of Caputo fractional derivatives, Xu et al. [15] proposed a fast difference/spectral method for a time fractional equation in one dimension. In their numerical scheme, the convergence order in the temporal direction is (3α) and the convergence order in the spatial direction is O(N1m), where m is the regularity of the exact solution in the spatial direction. This work suggests that the fast difference approximation with high-order discretization schemes is an effective way to solve the time fractional differential equations. Following this line, in this paper, we try to develop a fast finite difference/Legendre spectral collocation method for problem (1.1) with smooth solution and discuss the error estimates for the present numerical schemes. First, the generalized fractional derivative is discretized by L1 discretization, and the time direction convergence order is analyzed. Secondly, by using SOE approximation of exponential kernel function, a fast L1 scheme with exponential memory kernel function is obtained. Therefore, it can be seen that the calculation and storage costs of the fast scheme are significantly reduced. With the similar technique given in [15], we will discuss the error estimates when the fast L1 method is used to solve problem (1.1).

    The rest of this paper is organized as follows: In Section 2, we present time discretizations for problem (1.1) and discuss the stability and convergence of the present semi-discrete scheme. Combing the Legendre spectral collocation method for spatial variables, in Section 3, we discuss the finite difference/spectral collocation method for problem (1.1). The error estimates of the fully discrete scheme are also established in this section. For the generalized fractional derivative with exponential kernel, we develop the sum-of-exponentials technique developed in [13] to accelerate the calculation process. The error estimates of the related numerical schemes are discussed in Section 4. In Section 5, we present two numerical examples to verify the efficiency of our numerical algorithm and theoretical results. Finally, some conclusions are provided in Section 6.

    First, we discretize the fractional derivative with a generalized memory kernel, assuming the solution u(t)C2[0,T]. The computational domain [0,T] is uniformly divided into NT intervals with the uniform grids tk=kτ, k=0,1,,NT. For λ(t)C2[0,T], the time fractional derivative with a generalized memory kernel is discretized as [5]

    α,λ(t)0,tu(t)|t=tk+1=1Γ(1α)tk+10λ(tk+1s)(tk+1s)αu(s)ds=τ1αΓ(2α)kl=0(λkl+1/2akl+(λklλkl+1)bkl)ut,l+Rk1+Rk2, (2.1)

    where

    λl=λ(tl), ut,l=u(tl+1)u(tl)τ, Π1,lu(s)=u(tl+1)stlτ+u(tl)tl+1sτ,al=(l+1)1αl1α, bl=12α[(l+1)2αl2α]12[(l+1)1α+l1α], l0,

    and the error

    Rk1=τ1αΓ(1α)kl=0ut,l10λkl+1zλkl+1/2(λklλkl+1)(z1/2)(kl+1z)αdz,Rk2=1Γ(1α)kl=0tl+1tlλ(tk+1s)(u(s)Π1,lu(s))(tk+1s)αds,

    give [5]

    |Rk1|t1αk+14Γ(2α)τ2max0ttk+1|u(t)|max0ttk+1|λ(t)|, |Rk2|˜cmax0ttk+1|u(t)|λ(0)τ2α,

    where ˜c depends on α.

    We rewrite (2.1) as follows:

    α,λ(t)0,tu(t)|t=tk+1=α10[k1j=0(cj+1cj)u(tkj)+c0u(tk+1)cku(t0)]+rk+1τ,

    where α0=ταΓ(2α),cl=(λl+1/2al+(λlλl+1)bl)(l0), and the truncation error gives rk+1τ=Rk1+Rk2. For the sake of simplification, we define the fractional differential operator Lα,λ(t)tu(x,tk+1) by

    Lα,λ(t)tu(x,tk+1)=α10[k1j=0(cj+1cj)u(x,tkj)+c0u(x,tk+1)cku(x,t0)]. (2.2)

    Using Lα,λ(t)tu(x,tk+1) to approximate α,λ(t)0,tu(x,t) in problem (1.1) at tk+1, and denoting uk+1 is an approximation of u(x,tk+1), fk+1=f(x,tk+1), we obtain the time semi-discrete scheme of problem (1.1), i.e.,

    c0u1α0Δu1=c0u0+α0f1, (2.3)

    for k=0.

    For 1kNT1,

    k1j=0(cjcj+1)ukj+cku0=c0uk+1α0Δuk+1α0fk+1. (2.4)

    After a simple calculation, the truncation error of the semi-discretized schemes (2.6) and (2.7) give rk+1=α0rk+1τ and satisfies

    |rk+1|=α0|rk+1τ|˜cmax0ttj+1|2tu(x,t)|λ(0)τ2. (2.5)

    Considering the variational formulation of (1.1), we define some functional spaces endowed with standard norms and inner products that will be used hereafter [17,18,26]

    Hm(Ω)={u|uL2(Ω),DβuL2(Ω) for 0|β|m},H10(Ω)={u|uH1(Ω),u|Ω=0},

    where β=(β1,β2), |β|=β1+β2,Dβ=|β|xβ11xβ22, L2(Ω) is the space of measurable functions whose square is Lebesgue integrable in Ω, which equip with the inner products

    (u,v)=Ωuvdx,  (u,v)1=(u,v)+α0c10(u,v),

    and the normH1 is defined by

    ||v||1=(v,v)1/21,  v1=(v2+α0c10v2)1/2.

    The weak formulation of the (2.3) and (2.4) reads: Find uk+1H10(Ω), such that

    (u1,v)+α0c10(u1,v)=(u0,v)+α0c10(f1,v), (2.6)

    for k=0, and

    (uk+1,v)+α0c10(uk+1,v)=c10k1j=0(cjcj+1)(ukj,v)+c10ck(u0,v)+α0c10(fk+1,v), (2.7)

    for all vH10(Ω) and 1kNT1.

    Lemma 1. [5] If λ(t)>0, λ(t)0, and λ(t)C2[0,T], the weighted coefficients al, bl and cl(l=0,1,) are given by (2.1), satisfy

    a0>a1>a2>>al>1α(l+1)α, b0>b1>bl>0,
    c0>c1>>cl>α0λ(tl+1/2)Γ(1α)tαl+1>α0λ(T)Γ(1α)Tα.

    Theorem 1. For all τ>0, the semi-discretized schemes (2.6) and (2.7) are unconditionally stable, i.e.,

    uk+11u0+˜cα0c10max0jk+1fj,k=0,1,,NT1. (2.8)

    Proof. For k=0, substituting v=u1 in (2.6) and using u1u11, we have

    u121u0u1+α0c10f1u1u0u11+α0c10max0j1fju11. (2.9)

    From (2.9), we have u11u0+˜cα0c10max0j1fj. For k>1, we prove (2.8) by mathematical induction. Assume that there holds

    uj1u0+˜cα0c10max0ljfl,j=1,2,,k. (2.10)

    Now prove (2.8) using inequality (2.10). Substituting v=uk+1 into (2.7), we obtain

    uk+12+α0c10uk+12=c10k1j=0(cjcj+1)(ukj,uk+1)+α0c10(fk+1,uk+1)+ckc10(u0,uk+1), (2.11)

    which implies

    uk+121c10k1j=0(cjcj+1)ukjuk+1+α0c10fk+1uk+1+ckc10u0uk+1.

    Using Lemma 1 and the inequality uk+1uk+11, we have

    uk+11c10k1j=0(cjcj+1)(u0+˜cα0c10max0jkjfj)+ckc10u0+α0c10fk+1=c10[k1j=0(cjcj+1)(u0+˜cα0c10max0jkjfj)+cku0+α0fk+1].

    Note that u0+˜cα0c10max0jkjfju0+˜cα0c10max0jkfj(j=0,1,,k1). Therefore

    uk+11c10[(k1j=0(cjcj+1)+ck)(u0+˜cα0c10max0jkfj)]+α0c10fk+1u0+˜cα0c10max0jk+1fj.

    Finally, we obtain (2.8).

    Here we provide some lemmas to analyze coefficients cl and al, which will help to study the convergence analysis of the solution. In the following sections, ˜c is used to represent a constant, which may be different under different situations. Here, we employ the technique used in reference [19] to do the error estimates of schemes (2.6) and (2.7).

    Theorem 2. Let u(x,tk) be the exact solution of problem (1.1), {uk}NTk=0 be the solution of schemes (2.6) and (2.7), there holds

    u(x,tk)uk1{˜cTαmaxt(0,T]2tu(x,t)τ2α,0<α<1,˜cTmaxt(0,T]2tu(x,t)τ,α1, (2.12)

    where ˜c is independent of u,τ and T.

    Proof. Firstly, we prove that the following inequality

    u(x,tj)uj1˜cc1j1maxt(0,T]|2tu(x,t)|λ(0)τ2,j=1,2,,NT. (2.13)

    Let ej=u(x,tj)uj, for j=1, we obtain

    (e1,v)+α0c10(e1,v)=(e0,v)+(r1,v). (2.14)

    If we assume the initial value is exact (i.e., e0=0), let v=e1 in (2.14), using (2.5), we obtain

    e11r1˜cc10maxt(0,T]|2tu(x,t)|λ(0)τ2.

    Suppose (2.13) holds for all j=1,2,,k, we need to prove it also holds for j=k+1. In view of

    (ek+1,v)+α0c10(ek+1,v)=c10k1j=0(cjcj+1)(ekj,v)+c10ck(e0,v)+(rk+1,v). (2.15)

    Let v=ek+1 in (2.15), we have

    ek+12+α0c10ek+12=c10k1j=0(cjcj+1)(ekj,ek+1)+c10ck(e0,ek+1)+(rk+1,ek+1). (2.16)

    We rewrite (2.16) as follows:

    ek+121c10[k1j=0(cjcj+1)˜cc1kjmaxt(0,T]|2tu(x,t)|λ(0)τ2+˜cmaxt(0,T]|2tu(x,t)|λ(0)τ2]ek+1.

    Using Lemma 1, we have c1kj<c1k, it follows that

    ek+11c10[k1j=0(cjcj+1)˜cc1kmaxt(0,T]|2tu(x,t)|λ(0)τ2+˜cmaxt(0,T]|2tu(x,t)|λ(0)τ2]˜cc1kmaxt(0,T]|2tu(x,t)|λ(0)τ2,

    which implies (2.13). Lemma 1 suggests that

    kαc1k1λ(tk+1/2)kαΓ(1α)tαkα0λ(tk1/2)λ(tk+1/2)Γ(1α)ταα0λ(tk+1/2)λ(tk1/2)Γ(1α)ταΓ(2α)τα=11α.

    Hence, we have

    u(x,tk)uk1˜cc1k1maxt(0,T]|2tu(x,t)|λ(0)τ2=˜ckαλ(tk+1/2)c1k11λ(tk+1/2)kαmaxt(0,T]|2tu(x,t)|λ(0)τ2˜c11α1λ(tk+1/2)(kτ)αmaxt(0,T]|2tu(x,t)|λ(0)τ2α˜cTαmaxt(0,T]|2tu(x,t)|τ2α.

    Next we consider the case α1, we firstly prove that

    u(tj)uj1˜cjmaxt(0,T]|2tu(x,t)|τ2,j=1,2,,NT. (2.17)

    Inequality (2.17) obviously holds for j=1. Suppose (2.17) holds for j=1,2,,k. Now we will prove that (2.17) still holds when j=k+1. In fact, with the similar method given in Theorem 1, we have

    u(tk+1)uk+11c10(c0c1)ek+c10k1j=1(cjcj+1)ekj+c10cke0+rk+1c10[(c0c1)˜ckmaxt(0,T]|2tu(x,t)|τ2+k1j=1(cjcj+1)(kj)˜cmaxt(0,T]|2tu(x,t)|τ2]+˜cmaxt(0,T]|2tu(x,t)|τ2=c10[(c0c1)kk+1+k1j=1(cjcj+1)kjk+1+c0k+1](k+1)˜cmaxt(0,T]|2tu(x,t)|τ2. (2.18)

    Simple calculations, we can check that

    (c0c1)kk+1+k1j=1(cjcj+1)kjk+1+c0k+1=(c0c1)+k1j=1(cjcj+1)(c0c1)1k+1k1j=1(cjcj+1)j+1k+1+c0k+1. (2.19)

    With the help of (2.18), Eq (2.19) provides that

    ek+11c10[(c0c1)+k1j=1(cjcj+1)(c0c1)1k+1k1j=1(cjcj+1)j+1k+1+c0k+1]˜c(k+1)maxt(0,T]|2tu(x,t)|τ2. (2.20)

    Since

    (c0c1)1k+1+k1j=1(cjcj+1)j+1k+1+ck1k+1[(c0c1)+k1j=1(cjcj+1)+ck]=c0k+1,

    we have

    (c0c1)1k+1k1j=1(cjcj+1)j+1k+1+c0k+1ck. (2.21)

    Substituting (2.21) into (2.20), we obtain

    ek+11c10[(c0c1)+k1j=1(cjcj+1)+ck](k+1)˜cmaxt(0,T]|2tu(x,t)|τ2˜cTmaxt(0,T]|2tu(x,t)|τ.

    We finally obtain our conclusion.

    Let PN(Ix) and PN(Iy) be the spaces of all polynomials of degree less than or equal to N defined on domains Ix=(1,1) and Iy=(1,1), respectively. Let VN(Ω)=PN(Ix)PN(Iy), {ξj}Nj=0, {ξi}Ni=0 are Legendre-Gauss-Labatoo (LGL) points, and {ωj}Nj=0 and {ωi}Ni=0 are the weights, which satisfy the following quadrature formula [25,26]

    1111φ(x1,x2)dx1x2=Nj1=0Nj2=0φ(ξj1,ξj2)ωj1ωj2,φV2N1(Ω).

    Define the inner product and norm as follows [18,25,26]

    (ϕ,ψ)N=Ni=0Nj=0ϕ(ξi,ξj)ψ(ξi,ξj)ωiωj,ψ0,N=(ϕ,ϕ)1/2N.

    We will use the H10-orthogonal projection operator πN such that [24]

    ((uπNu),v)=0,vV0N(Ω), (3.1)

    for all uH10(Ω). Here V0N(Ω) is defined as

    V0N(Ω)=(PN(Ix)PN(Iy))H10(Ω).

    For this projection, there have estimates [25,26]

    uπNul˜cNlmum,uHm(Ω)H10(Ω),m1,l=0,1. (3.2)

    Lemma 2. [25] For all φHm(Ω)( m1), vNPN(Ω), there holds

    (φ,vN)(φ,vN)N˜cNmφmvN0,N. (3.3)

    Finite difference/Legendre collocation approximation of problem (1.1) gives: Find ukNV0N(Ω),

    (uk+1N,vN)N+α0c10(uk+1N,vN)N=c10k1j=0(cjcj+1)(ukjN,vN)N+ckc10(u0N,vN)N+(fk+1,vN)N, (3.4)

    for all vNV0N(Ω). The discrete H1-norm is defined by

    ψN1,N=((ψN,ψN)N+α0c10(ψN,ψN)N)1/2,ψNVN(Ω).

    To do the error estimate, we split the error as u(tk)ukN=(u(tk)πNu(tk))(ukNπNu(tk)):=ϵkNekN. Without loss of generality, we only consider the homogeneous case of problem (1.1), i.e., f=0.

    Theorem 3. Let u(x,t) be the exact solution of problem (1.1), and ukN be the solution of fully discrete scheme (3.4) with the initial condition u0N=πNu0. Suppose 2tuL((0,T];Hm(Ω))(m1), then

    ||u(x,tk)ukN||1,N{˜cTα1α1λ(T)(Nm||α,λ(t)tu||L(Hm)+τ2αλ(0)(Nm||2tu||L(Hm)+2tuL(L2)))+˜cN1m||u||L(Hm),0<α<1,˜cTλ(T)(Nmα,λ(t)tuL(Hm)+τλ(0)(Nm2tuL(Hm)+2tuL(L2)))+˜cN1muL(Hm),α1, (3.5)

    where α,λ(t)tuL(Hm)=supt(0,T)α,λ(t)tu(x,t)m, uL(Hm)=supt(0,T)u(x,t)m, 2tuL(Hm)=supt(0,T)2tu(x,t)m, and 2tuL(L2)=supt(0,T)2tu(x,t).

    Proof. Subtracting ekN=ukNπNu(tk) on both sides of Eq (3.4), we obtain

    (ek+1N,vN)N+α0c10(ek+1N,vN)N=c10k1j=0(cjcj+1)(ukjN,vN)N+ckc10(u0N,vN)N(πNu(tk+1),vN)Nα0c10(πNu(tk+1),vN)N.

    Rewriting the above formula, we obtain

    (ek+1N,vN)N+α0c10(ek+1N,vN)N=c10k1j=0(cjcj+1)(ekjN,vN)N+ckc10(e0N,vN)N+(εk+11,vN)N+(εk+12,vN)N,

    where

    εk+11=(u(tk+1)πNu(tk+1))c10k1j=0(cjcj+1)(u(tkj)πNu(tkj))ckc10(u(t0)πNu(t0)),

    and

    εk+12=(u(tk+1)πNu(tk+1))+c10k1j=0(cjcj+1)u(tkj)+ckc10u(t0)πNu(tk+1)+α0c10ΔπNu(tk+1).

    For term (εk+11,vN)N, we have

    (εk+11,vN)N=(IdπN)(u(tk+1)c10k1j=0(cjcj+1)u(tkj)ckc10u(t0),vN)N=α0c10((IdπN)(α,λ(t)tu(tk+1)rk+1τ),vN)N,

    where Id is the identity operator. Using (3.3), we have

    |(εk+11,vN)N|α0c10[((IdπN)(α,λ(t)tu(tk+1)rk+1τ),vN)+˜cN1(IdπN)(αtu(tk+1)rk+1τ)1vN0,N]α0c10[(IdπN)(α,λ(t)tu(tk+1)rk+1τ)0,N+˜cN1(IdπN)(α,λ(t)tu(tk+1)rk+1τ)1]vN0,N.

    By using of (3.2), we obtain

    |(εk1,vN)N|˜cα0c10(Nmα,λ(t)tuL(Hm)+τ2αλ(0)Nm2tuL(Hm))vN0,N.

    Note that

    (εk+12,vN)N=(u(tk+1)c10k1j=0(cjcj+1)u(tkj)ckc10u(t0),vN)Nα0c10(πNu(tk+1),vN)N,

    the facts (πNu(tk),vN)N=(πNu(tk),vN) and (α,λ(t)tu(tk),vN)=(u(tk),vN), we obtain

    |(εk+12,vN)N|α0c10(Lα,λ(t)tu(tk+1),vN)α0c10(Lα,λ(t)tu(tk+1),vN)N+α0c10(α,λ(t)tu(tk+1)Lα,λ(t)tu(tk+1),vN).

    Furthermore, we obtain

    |(εk+12,vN)N|˜cα0c10[Nmα,λ(t)tuL(Hm)+λ(0)τ2α(2tuL(L2)+Nm||2tu||L(Hm))]vN0,N.

    Combining the estimates of (εk1,vN)N and (εk2,vN)N, we have

    ek+1N1,Nc10k1j=0(cjcj+1)ekjN0,N+ckc10e0N0,N+˜cα0c10[Nmα,λ(t)tuL(Hm)+λ(0)τ2α(2tuL(L2)+Nm||2tu||L(Hm))]. (3.6)

    Moreover, with the same technique used in Theorem 2, using inequality (3.6), we can prove that

    ekN1,Nc1k1˜cα0c10[Nmα,λ(t)tuL(Hm)+λ(0)τ2α(2tuL(L2)+Nm||2tu||L(Hm))]˜cc1k1kαλ(tk+1/2)Γ(2α)λ(tk+1/2)kατα(Nm||α,λ(t)tu||L(Hm)+τ2αλ(0)2tuL(L2)+τ2αλ(0)Nm2tuL(Hm))˜cTα(1α)(λ(T))[Nmα,λ(t)tuL(Hm)+λ(0)τ2α(2tuL(L2)+Nm||2tu||L(Hm))].

    Hence, applying triangle inequality, we obtain

    u(tk)ukN1,Nu(tk)πNu(tk)1,N+ukNπNu(tk)1,N˜cN1muL(Hm)+˜cTα(1α)λ(T)×[Nmα,λ(t)tuL(Hm)+λ(0)τ2α(2tuL(L2)+Nm||2tu||L(Hm))].

    For α1, similar to the case discussed in Theorem 2, we obtain

    ekN1,N˜cTλ(T)(Nm||α,λ(t)tu||L(Hm)+τλ(0)2tuL(L2)+τλ(0)Nm2tuL(Hm)).

    Applying triangle inequality once again, we have

    u(tk)ukN1,Nu(tk)πNu(tk)1,N+ukNπNu(tk)1,N,˜cN1muL(Hm)+˜cTλ(T)(Nm||α,λ(t)tu||L(Hm)+τλ(0)2tuL(L2)+τλ(0)Nm2tuL(Hm)),

    which implies the desired result.

    Applying the same technique given in reference [13], we use the sum-of-exponentials (SOE) approximation to approximate the power function tα(0<α<1) in fractional derivative (2.1). And for the approach error, we have the following results:

    Lemma 3. [13] For a given absolute error ε, the kernel functions 1tα(α(0,1)), there exist positive real numbers {ηαi}Nexpi=1 and {wαi}Nexpi=1 such that

    |tαNexpi=1wαieηαit|ε,t[τ,T], (4.1)

    where Nexp=O(log1ε(loglog1ε+logTτ)+log1τ(loglog1ε+log1τ)).

    For the details of the proof of Lemma 3, see [13]. Following the same idea given in references [13,16], we split the Caputo fractional derivative (1.3) into a local term L(tk) and a history term H(tk) in discretized fractional derivative (2.1) with kernel λ(t)=ebt(b0), i.e.,

    α,λ(t)0,tu(t)|t=tk=1Γ(1α)tk10eb(tks)(tks)αu(s)ds+1Γ(1α)tktk1eb(tks)(tks)αu(s)ds:=H(tk)+L(tk).

    For the local term, we discrete it as the follows

    L(tk):u(tk)u(tk1)τΓ(1α)tktk1eb(tks)(tks)αds,=u(tk)u(tk1)τΓ(2α)(ebττ1α+bτ0ebθθ1αdθ). (4.2)

    The history term, using SOE approximation, we obtain

    H(tk):=1Γ(1α)tk10eb(tks)(tks)αu(s)ds1Γ(1α)tk10Nexpi=1wαieηαi(tks)eb(tks)u(s)ds=1Γ(1α)Nexpi=1tk10wαie˜ηαi(tks)u(s)ds,

    where ˜ηαi=ηαi+b. For simplicity, we rewrite H(tk) as

    H(tk)1Γ(1α)Nexpi=1wαiUαh,i(tk),

    with

    Uαh,i(tk)=e˜ηαiτUαh,i(tk1)+tk1tk2e˜ηαi(tks)u(s)ds.

    Using linear interpolation operators to approximate u(s), we have

    tk1tk2e˜ηαi(tks)u(s)ds=tk1tk2e˜ηαi(tks)(Π1,k1u(s))ds=ˆaiu(tk1)+ˆbiu(tk2)+O(τ2α),

    where ˆai=1e(ηαi+b)ττ(ηαi+b)e(ηαi+b)τ, ˆbi=ˆai. Hence, history term has recurrence relation

    Uαh,i(tk)=e˜ηαiτUαh,i(tk1)+ˆaiu(tk1)+ˆbiu(tk2).

    Collecting the above formulas, we obtain the fast finite difference operator of fractional derivative (1.3)

    Fα,λ(t)tu1=Lα,λ(t)t(t1),k=1,Fα,λ(t)tuk=u(tk)u(tk1)τΓ(2α)(ebττ1α+bτ0ebθθ1αdθ)+1Γ(1α)Nexpi=1wαiUαh,i(tk),2kNT, (4.3)

    where Uαh,i(t1)=0.

    If we use the difference operator (4.3) to approximate fractional derivatives (1.3) at time level tk, we get the semi-discrete scheme

    Fα,λ(t)tukΔuk=f(tk),1kNT.

    The weak form in L2-inner product gives: Find ukH10(Ω)(1kNT), such that

    (Fα,λ(t)tuk,v)+(uk,v)=(fk,v),vH10(Ω). (4.4)

    For the sake of simplification, (2.2) can be rewritten as

    Lα,λ(t)tu(tk)=ταΓ(2α)c10[u(tk)kj=1ρk,αkju(tkj)],

    where ρk,αkj=cj1cjc0 as 1jk1, ρk,αkj=ck1c0 as j=k. By comparing the two operators Fα,λ(t)t and Lα,λ(t)t, we have

    Fα,λ(t)tu(tk)Lα,λ(t)tu(tk)=1Γ(1α)kj=1tjtj1[(tks)αNexpi=1wαieηαi(tks)]eb(tks)(Π1,ju(s))ds=1α0(1α)1kj=1[˜akju(tj1)+˜bkju(tj)],

    where Π1,ju(s)=u(tj1)tjsτ+u(tj)stj1τ, s[tj1,tj],

    ˜akj=τα1tjtj1[(tks)αNexpi=1wαieηαi(tks)]eb(tks)ds,

    and

    ˜bkj=τα1tjtj1[(tks)αNexpi=1wαieηαi(tks)]eb(tks)ds.

    Simple calculations, we can check that

    |˜akj|=|˜bkj|ταε.

    For the truncation error of the fast difference operator Fα,λ(t)t gives

    Rk,ατ:=α,λ(t)tu(tk)Fα,λ(t)tu(tk),2kNT.

    Lemma 4. Suppose that u(t)C2[0,T], for any 0<α<1, there holds

    |Rk,ατ|˜cτ2αmax0ttk|u(t)|+˜cεmax0ttk|u(t)|,2kNT.

    Proof. By the triangle inequality

    |Rk,ατ|=|α,λ(t)tu(tk)Fαtu(tk)||α,λ(t)tu(tk)Lαtu(tk)|+|Lαtu(tk)Fαtu(tk)|˜cτ2αmax0ttk|u(t)|+1α0(1α)1kj=1[˜akju(tj1)+˜bkju(tj)]˜cτ2αmax0ttk|u(t)|+˜cεmax0ttk|u(t)|.

    Furthermore, we can rewrite the fast difference operator as

    Fα,λ(t)tuk=1α0c10[ukkj=1ρk,αkjukj]+1α0(1α)1kj=1[˜akjuj1+˜bkjuj]=1α0c10[ukkj=1ρk,αkjukj]+1α0(1α)1kj=0ζk,αkjukj,

    where

    ζ1,α0=˜a0,ζ1,α1=˜b0;ζ2,α0=˜a1,ζ2,α1=˜a0+˜b1,ζ2,α2=˜b0;ζ3,α0=˜a2,ζ3,α1=˜a1+˜b2,ζ3,α2=˜a0+˜b1,ζ3,α3=˜b0;ζk,α0=˜ak1,ζ3,αkj=˜aj1+˜bj,ζk,αk1=˜b0,k4,j=2,3,,k1.

    The coefficients ρk,αkj have the following properties.

    Lemma 5. For k1, the coefficients ρk,αkj and c10 satisfy

    (1) ρk,αkj>0,j=1,2,k1; (2) kj=1ρk,αkj=1,j=1,2,k1;

    (3) ρk,α0>2(1α)λk1/23kα;    (4) 0<ρk,αk1<1;   (5) 1<c10<ebτ2.

    Proof. See the Appendix A.

    To do the error estimates of the fast difference scheme, we follow the technique used in reference [24]. Introducing a parameter σ, then ukkj=1ρk,αkjukj can be rewritten as following

    ukkj=1ρk,αkjukj=(ukσuk1)(ρk,αk1σ)(uk1σuk2)(ρk,α2+σρk,α3++σk3ρk,αk1σk2)(u2σu1)(ρk,α1+σρk,α2++σk2ρk,αk1σk1)(u1σu0)(ρk,α0+σρk,α1++σk1ρk,αk1σk)u0=ˉukkj=1ˉρk,αkjˉukj,

    where σ=12ρk,αk1, σj means power of σ, and

    ˉu0=u0,ˉuj=ujσuj1,ˉρk,αkj=jl=1σjlρk,αklσj,j=1,2,,k.

    Similarly, the sum-of-exponentials part can be rewritten as

    1α0(1α)1kj=0ζk,αkjukj=1α0(1α)1kj=0ˉζk,αkjˉukj,

    where ˉζk,αkj=jl=0σjlζk,αkl, j=0,1,,k. The above statements show that the fast difference operator can be rewritten as

    ˉFα,λ(t)tˉuk=Fα,λ(t)tuk=1α0c10[ˉukkj=1ˉρk,αkjˉukj]+1α0(1α)1kj=1ˉζk,αkjˉukj.

    Thus, the semi-discrete scheme (4.4) can be rewritten as

    (ˉFα,λ(t)tˉuk,v)+(uk,v)=(fk,v),vH10(Ω). (4.5)

    For the weighted coefficients ˉρk,αkj and ˉζk,αkj, there holds

    Lemma 6. For k2, the coefficients ˉρk,αkj and ˉζk,αkj satisfy

    (1) ˉρk,αkj>0, j=1,2,k; (2) 1ˉρk,α0<1ρk,α0<3kα2(1α)λk1/2;

    (3) kj=1ˉρk,αkj12σ(1α)λk1/23(1σ)kα; (4) |ˉζk,αkj|4ετα, j=1,2,k; (5) 0<σ<12.

    Proof. See the Appendix B.

    Theorem 4. Assume that the permissible error ε satisfies εmin{14Γ(1α)σλ(T)3ˆc(1σ)TαNTebτ/2,Γ(1α)4ˆc}, then the semi-discrete scheme (4.4) is stable, and its solution satisfies

    uk1(c10+˜c(α0c0)1/2)(u0+max0jkfj),1kNT. (4.6)

    Proof. For k=1, taking v=u1 in (4.4), there holds

    (u1,u1)+α0(u1,u1)=(u0,u1)+(f1,u1).

    Using Schwarz inequality, we arrive at u121u1u0+f1u1, which produces u11u0+f1. Using (5) of Lemma 5, we have

    u11c10(u0+f1)c10(1+˜c(α0c0)1/2)(u0+max0j1fj).

    For k2, using mathematical induction on the index k, we will prove the following result:

    α0c10uk2c20(u02+max0jkfj),2kNT. (4.7)

    Firstly, substituting v=2ˉuk into semi-discrete scheme (4.5), it yields

    2(ˉuk,ˉuk)+2α0c10(uk,ˉuk)+2c10(1α)kj=0ˉζk,αkj(ˉukj,ˉuk)=2kj=1ˉρk,αkj(ˉukj,ˉuk)+2(fk,ˉuk).

    Using the identity 2(uk,ˉuk)=uk2+ˉuk2σ2uk12 and (3) of Lemma 6, we have

    ˉuk2+α0c10uk2+α0c10ˉuk2σˉuk12+α0c10σuk12+kj=2ˉρk,αkjukj22c10(1α)kj=0ˉζk,αkj(ˉukj,ˉuk)+fk2+ˉuk2. (4.8)

    Furthermore, using Poincaré inequality [26]

    ˉukˆcˉuk, (4.9)

    and (5) of Lemma 5, if εΓ(1α)4ˆc, we obtain

    |2c10(1α)ˉζk,αk(ˉuk,ˉuk)|2ταεc10(1α)ˉuk2α0c102ˉuk2,

    where we used the relation ˉζk,αk=ζk,αk=˜b0ταε in above derivation. Setting θk=1kj=1ˉρk,αkj and using Young inequality, we obtain

    |2c10(1α)kj=1ˉζk,αkj(ˉukj,ˉuk)|θkkkj=1ˉukj2+kc20(1α)2θkkj=1(ˉζk,αkj)2ˉuk2.

    According to (3) and (4) of Lemma 6, we obtain θk=1kj=1ˉρk,αkj2σ(1α)λk1/23(1σ)kα. Moreover, with the help of Poincaré inequality, we obtain

    |2c10(1α)kj=1ˉζk,αkj(ˉukj,ˉuk)|θkkkj=1ˉukj2+3ˆcc20(1α)2(1σ)kα+12σ(1α)λ(T)(4ταε)2ˉuk2.

    If ε14Γ(1α)σλ(T)3ˆc(1σ)TαNTebτ/2, there holds

    |2c10(1α)kj=1ˉζk,αkj(ˉukj,ˉuk)|θkkkj=1ˉukj2+3ˆcc20(1α)2(1σ)kα+12σ(1α)λ(T)(4ταε)2ˉuk2θkkkj=1ˉukj2+α0c102ˉuk2.

    Rewriting (4.8) as follows:

    α0c10uk2(σ+θkk)uk12+α0c10σuk12+kj=2(ˉρk,αkj+θkk)ˉukj2+fk2.

    In view of relation σ+θ2+ˉρ2,α0=1, we obtain (4.7) for k=2. Assuming that (4.7) is already correct for all k=2,3,,n1, we can check that

    α0c10un2c20(σ+θnn+kj=2ˉρk,αkj+θnn(n1))u02+fn2=c20(u02+fn2)c20(u0+max0jnfj)2,n=2,,NT,

    which implies

    α0c10unc10(u0+max0jnfj). (4.10)

    Furthermore, using (4.10) and Poincaré inequality (4.9), we obtain

    un2ˆcun2˜c(α0c0)1/2(u0+max0jnfj). (4.11)

    Combining (4.10) and (4.11), we obtain the desired results.

    For the convergence of time semi-discrete scheme (4.4), we have

    Theorem 5. Let u(x,t) be the exact solution of problem (1.1), and {uk}NTk=1 be the semi-discrete solution of scheme (4.4) with the initial u0=u(0). Under the assumption given in Theorem 4 and 2tu(x,t)L((0,T];L2(Ω)), there holds

    u(tk)uk1˜cα,T(τ2α2tuL(L2)+εuL(L2)),2kNT, (4.12)

    where ˜cα,T depends only on α and T.

    Proof. Let ek=u(tk)uk. Subtracting (4.4) from problem (1.1) at time level tk, it follows that

    (Fα,λ(t)tek,v)+(ek,v)=(Rk,ατ,v),vH10(Ω).

    With the similar technique in Theorem 4, we obtain

    ek2+α0c10ek2˜cα,Tmax2iNTRk,ατ2,k2.

    Applying Lemma 4, we obtain (4.12).

    We consider the spectral collocation method in space as follows: Find ukNV0N(Ω), such that

    (Fα,λ(t)tukN,vN)N+(ukN,vN)N=0,vNV0N(Ω). (4.13)

    Theorem 6. Let u(x,t) be the exact solution of (1.1), and ukN be the solution of scheme (4.13) with the initial condition u0N=πNu0. Suppose 2tu(x,t)L((0,T];Hm(Ω))(m1), for k=2,,NT, there holds

    u(tk)ukN1˜cα,T(τ2α2tuL(L2)+Nmτ2α2tuL(Hm)+Nmα,λ(t)tuL(Hm)+N1muL(Hm)+εuL(L2)+εNmuL(Hm)). (4.14)

    Proof. Let ekN=ukNπNu(tk), using (4.13), we have

    (Fα,λ(t)tekN,vN)N+(ekN,vN)N=(Fα,λ(t)tukN,vN)N+(ukN,vN)N(Fα,λ(t)tπNu(tk),vN)N(πNu(tk),vN)N=(Fα,λ(t)tukNFα,λ(t)tπNu(tk),vN)N(Fα,λ(t)tu(tk),vN)N(πNu(tk),vN)N:=εk1(vN)+εk2(vN),vNV0N(Ω), (4.15)

    where

    εk1(vN)=(Fα,λ(t)tukNFα,λ(t)tπNu(tk),vN)N,εk2(vN)=(Fα,λ(t)tu(tk),vN)N(πNu(tk),vN)N. (4.16)

    Using equality ˉFα,λ(t)tˉuk=Fα,λ(t)tuk, scheme (4.15) can be rewritten as

    (ˉFα,λ(t)tˉekN,vN)N+(ekN,vN)N=εk1(vN)+εk2(vN),vNV0N(Ω),

    where ˉekN=ekNσek1N. Moreover, using Lemma 2, we have

    εk1(vN)=(Fα,λ(t)tukNFα,λ(t)tπNu(tk),vN)N=((IdπN)(α,λ(t)tu(tk)Rk,ατ),vN)N|((IdπN)(α,λ(t)tu(tk)Rk,ατ),vN)|+˜cN1(IdπN)(α,λ(t)tu(tk)Rk,ατ)1vN0,N[(IdπN)(α,λ(t)tu(tk)Rk,ατ)0,N+˜cN1(IdπN)(α,λ(t)tu(tk)Rk,ατ)1]vN0,N. (4.17)

    According to Lemma 4, for l=0,1, we have

    (IdπN)Rk,ατl˜cα,Tmax0ttk(IdπN)2tu(t)lτ2α+˜cα,Tεmax0ttk(IdπN)2tu(t)l.

    Combing (3.2), Eq (4.17) produces

    |εk1(vN)|˜c(Nmα,λ(t)tuL(Hm)+Nmτ2α2tuL(Hm)+εNmuL(Hm))vN0,N.

    According to the definition of the 0,N and projection operator πN, we have

    (πNu(tk),vN)N=(πNu(tk),vN)=(u(tk),vN),vNV0N(Ω). (4.18)

    Applying (1.1), (4.16), and (4.18), we obtain

    εk2(vN)=(Fα,λ(t)tu(tk),vN)N(πNu(tk),vN)N=(Fα,λ(t)tu(tk),vN)(Fα,λ(t)tu(tk),vN)N+(Rk,ατ,vN).

    Furthermore, using Lemmas 2 and 4, we obtain

    |εk2(vN)|(˜cNmα,λ(t)tu(tk)Rk,ατm+Rk,ατ0,N)vN0,N˜cα,T(Nmα,λ(t)tuL(Hm)+Nmτ2α2tuL(Hm)+εNmuL(Hm)+τ2α2tuL(L2)+εuL(L2))vN0,N.

    Combining the estimates of |εk1(vN)|, |εk2(vN)| and the Poincaré inequality, we have

    |εk1(vN)|+|εk2(vN)|˜cα,T(Nmα,λ(t)tuL(Hm)+Nmτ2α2tuL(Hm)+εNmuL(Hm)+τ2α2tuL(L2)+εuL(L2))vN0,N.

    With the similar argument given in Theorem 4, we obtain

    ekN+α0c10ekN˜cα,T(Nmα,λ(t)tuL(Hm)+Nmτ2α2tuL(Hm)+εNmuL(Hm)+τ2α2tuL(L2)+εuL(L2)).

    By the triangle inequality u(tk)ukN1,Nu(tk)πNu(tk)1,N+ukNπNu(tk)1,N, we finally obtain our conclusion.

    In this section, we will test our theory results given in previous sections. The error in time direction at terminal time T=1 is measured by the point-wise maximum norm. In the following numerical experiments, we used two examples with the exponential kernel function λ(t)=et, which modify the numerical examples given in reference [5]. In our numerical experiments, all the algorithms are implemented by MATLAB R2022b, which were run on a 2.90 GHz PC with 32 GB of RAM and the Windows 11 operating system.

    Example 1. First, we consider problem (1.1) with exact solution u(x,y,t)=(1+(6(6+6t+3t2+t3)et)sin(πx)sin(πy). The source term and initial values are given by the exact solution.

    In Table 1, we list the errors and convergence orders of the present numerical scheme (3.4). Here, we fix the degree of polynomial Nx=Ny=20. It can be observed that the time direction convergence order is O(τ2α), which is consistent with our theoretical analysis. To test the convergence order in spatial direction, we set τ=1/1000 for α=0.3,0.5, and 0.9. The relationship between the errors and polynomial degree N is shown in the semi-log graph in Figure 2. We find that the numerical solution has the exponential accuracy.

    Table 1.  Maximum errors and convergence orders at T=1 with Nx=Ny=20.
    α=0.3 α=0.5 α=0.9
    τ Error Order τ Error Order τ Error Order
    1/10 9.3238e-05 * 1/10 2.3796e-04 * 1/10 1.1903e-03 *
    1/20 2.9882e-05 1.6416 1/20 8.6493e-05 1.4601 1/20 5.6183e-04 1.0831
    1/40 9.4754e-06 1.6570 1/40 3.1118e-05 1.4748 1/40 2.6352e-04 1.0922
    1/80 2.9837e-06 1.6671 1/80 1.1129e-05 1.4834 1/80 1.2326e-04 1.0962

     | Show Table
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    Figure 2.  The numerical errors of the smooth solution with the polynomial degree Nx=Ny=20.

    Example 2. To test the effectiveness of the fast difference/spectral collocation scheme (4.13), we consider problem (1.1) with exact solution u(x,y,t)=(3(2+2t+t2)et)sin(πx)sin(2πy), and the source term and initial values are calculated by the exact solution.

    The numerical results of schemes (3.4) and (4.13) are listed in Tables 2 and 3. In these tables, we compared the errors, convergence orders, and CPU times of the two schemes in Tables 2 and 3 for α=0.5 and α=0.7, respectively. In the implementation of fast difference/spectral collocation, we take ε=1e9, and it can be seen that the time convergence order of fast solver is O(τ2α), which is consistent with our theoretical results. In this example, the CPU(s) are measured, and the total times of numerical errors for the time steps τ vary in [1/40,1/80,1/160,1/320,1/640,1/1280]. We also observe that the fast L1 scheme/spectral collocation has lower complexity and achieves the same accuracy as the L1 scheme ones.

    Table 2.  The comparison of maximum errors and convergence orders at T=1 for the fast scheme and the direct scheme with Nx=Ny=20, α = 0.5.
    Direct scheme (3.4) Fast scheme (4.13)
    τ Error Order τ Error Order
    1/40 6.7179e-06 * 1/40 6.5977e-06 *
    1/80 2.3550e-06 1.5123 1/80 2.3258e-06 1.5042
    1/160 8.2788e-07 1.5082 1/160 8.2074e-07 1.5027
    1/320 2.9156e-07 1.5056 1/320 2.8981e-07 1.5018
    1/640 1.0281e-07 1.5038 1/640 1.0237e-07 1.5013
    1/1280 3.6276e-08 1.5029 1/1280 3.6269e-08 1.5010
    CPU(s) 657.5 CPU(s) 42.6

     | Show Table
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    Table 3.  The comparison of maximum errors and convergence orders at T=1 for the fast scheme and the direct scheme with Nx=Ny=20, α = 0.7.
    Direct scheme (3.4) Fast scheme (4.13)
    τ Error Order1 τ Error Order1
    1/40 1.9589e-05 * 1/40 1.9405e-05 *
    1/80 7.8911e-06 1.3117 1/80 7.8473e-06 1.3061
    1/160 3.1894e-06 1.3069 1/160 3.1789e-06 1.3036
    1/320 1.2916e-06 1.3041 1/320 1.2891e-06 1.3021
    1/640 5.2369e-07 1.3023 1/640 5.2313e-07 1.3011
    1/1280 2.1246e-07 1.3015 1/1280 2.1237e-07 1.3005
    CPU(s) 659.0 CPU(s) 61.7

     | Show Table
    DownLoad: CSV

    In this paper, the two-dimensional fractional diffusion equation with generalized memory kernel is analyzed and approximated in time and space. We used the spectral collocation method in the spatial direction, and the L1 formula and the fast L1 formula are used in time to approximate the fractional derivative. Finally, we obtained estimates of the temporal and spatial errors of the present schemes, the time and space error estimates are O(τ2α+O(N1m). By comparing the L1 scheme with the fast L1 scheme, we found that the fast L1 scheme can significantly reduce storage costs and computational costs. There were also many improvements in this article, such as the use of graded grids instead of uniform grids (e.g., Stynes et al. [11]) for time grids when the initial singularity of solution is considered. Our future work will focus on this issue with the help of the existing techniques developed in references [10,27,28].

    Zunyuan Hu: Methodology, software, writing-original draft, writing-review & editing; Can Li: Conceptualization, methodology, writing-original draft, writing-review & editing; Shimin Guo: Methodology, writing-review & editing. All authors have read and approved the final version of the manuscript for publication.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    In this research, Can Li was partially supported by the Science Basic Research Plan in Shaanxi Province of China under Grant No. 2023-JC-YB-045. Shimin Guo was partially supported by the National Natural Science Foundation of China under Grant No. 12271427.

    It is declared that none of the authors have any competing interests in this manuscript.

    (1) According to the definitions of ρk,αkj and c0, we obtain

    ρk,αkj=(cj1cj)c10,ρk,α0=ck1c10,c10=1λ1/2a0+(λ0λ1)b0=1ebτ/2+(1ebτ)(12α12)>0.

    By simple calculation, and using Lemma 1, we have

    cj1cj=λj1/2aj1λj+1/2aj+(λj1λj)bj1(λjλj+1)bj>0.

    (2) After calculating term by term, we obtain

    kj=1ρk,αkj=ρk,α0+ρk,α1+ρk,α2++ρk,αk1=c10[(c0c1)+(c1c2)+(c2c3)++ck1]=c10c0=1.

    (3) Using Lemma 1 once again, we have

    ρk,α0=ck1c10=λk1/2ak1+(λk1λk)bk1λ1/2a0+(λ0λ1)b0>ταΓ(2α)λk1/2Γ(1α)tαk1λ1/2a0+(λ0λ1)b0>λk1/2(1α)kα1λ1/2a0+(λ0λ1)b0>2(1α)λk1/23kα.

    (4) Since (1) provides that ρk,αk1>0, then ρk,αk1=(c0c1)c10=1c1c0<1.

    (5) Note that

    c0=λ1/2a0+(λ0λ1)b0=e12bτ[1e12bτ(12α12)]+(12α12)=e12bτ[1e12bτZ]+Z,

    where Z=12α12, 0Z12. We may rewrite c0 as follows

    c0=e12bτ[1e12bτZ]+Z:=g(τ,Z).

    By checking the partial derivatives of g with respect to variable Z, we obtain g(τ,Z)Z=1ebτ0, which means g(τ,Z) is monotone increasing function with respect to variable Z, thus we get c0e12bτ[112e12bτ]+12, and c0e12bτ. Furthermore, we define ˜g(τ)=e12bτ[112e12bτ]+12 and take the derivative of ˜g(τ), we obtain

    ˜g(τ)τ=12be12bτ[112e12bτ]+14bebτ=14bebτ12be12bτ+14bebτ=be12bτ[12e12bτ12]0.

    Hence, ˜g(τ)1, 0<c0=g(τ,Z)˜g(τ)1, it provides that c10=1g(τ,Z)1˜g(τ)1. Finally we get 1c10e12bτ.

    (1) From Lemma 5, for j=1,2,...,k, we have ρk,αkj>0, thus σ=12ρk,αk1>0. By using ˉρk,αkj=jl=1σjlˉρk,αklσj, we further have

    ˉρk,αkj=jl=1σjlˉρk,αklσj=(2σj+σj2ˉρk,αk2+σj3ˉρk,αk3+...+σˉρk,αkj+l+ˉρk,αkj)σj=σj+σj2ˉρk,αk2+σj3ˉρk,αk3+...+σˉρk,αkj+l+ˉρk,αkj>0.

    (2) According to Lemma 5, we have 1ρk,α0<3kα2(1α)λk1/2. Moreover, note that ρk,α0<ˉρk,α0, we have 1ˉρk,α0<1ρk,α0<3kα2(1α)λk1/2.

    (3) Note that 1ˉρk,α0<3kα2(1α)λk1/2, we may check that

    kj=1ˉρk,αkj=ρk,αk1(1+σ+σ2+σ3+σk1)+ρk,αk2(1+σ+σ2+σ3+σk2)++ρk,α2(1+σ+σ2)+ρk,α1(1+σ)+ρk,α0(σ+σ2+σ3+σk)=11σ[(ρk,αk1+ρk,αk2+ρk,αk3++ρk,α1+ρk,α0)σ(1σk)σ(ρk,αk1σk1+ρk,αk2σk2++ρk,α1σ+ρk,α0)]=1σ1σˉρk,α012σ(1α)λk1/23(1σ)kα.

    (4) In view of |ζk,αkj|2ταε, we obtain |ˉζk,αkj|=|jl=0σjlζk,αkj|2ταεjl=0σl4ταε.

    (5) From Lemma 5, we have σ=12ρk,αk1<12.



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