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Research article

Error estimate and superconvergence of a high-accuracy difference scheme for 2D heat equation with nonlocal boundary conditions

  • Received: 29 July 2024 Revised: 05 September 2024 Accepted: 11 September 2024 Published: 26 September 2024
  • MSC : 65M06, 65M12, 65T50

  • In this work, we initially construct an implicit Euler difference scheme for a two-dimensional heat problem, incorporating both local and nonlocal boundary conditions. Subsequently, we harness the power of the discrete Fourier transform and develop an innovative transformation technique to rigorously demonstrate that our scheme attains the asymptotic optimal error estimate in the maximum norm. Furthermore, we derive a series of approximation formulas for the partial derivatives of the solution along the two spatial dimensions, meticulously proving that each of these formulations possesses superconvergence properties. Lastly, to validate our theoretical findings, we present two comprehensive numerical experiments, showcasing the efficiency and accuracy of our approach.

    Citation: Liping Zhou, Yumei Yan, Ying Liu. Error estimate and superconvergence of a high-accuracy difference scheme for 2D heat equation with nonlocal boundary conditions[J]. AIMS Mathematics, 2024, 9(10): 27848-27870. doi: 10.3934/math.20241352

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  • In this work, we initially construct an implicit Euler difference scheme for a two-dimensional heat problem, incorporating both local and nonlocal boundary conditions. Subsequently, we harness the power of the discrete Fourier transform and develop an innovative transformation technique to rigorously demonstrate that our scheme attains the asymptotic optimal error estimate in the maximum norm. Furthermore, we derive a series of approximation formulas for the partial derivatives of the solution along the two spatial dimensions, meticulously proving that each of these formulations possesses superconvergence properties. Lastly, to validate our theoretical findings, we present two comprehensive numerical experiments, showcasing the efficiency and accuracy of our approach.



    In recent years, nonclassical boundary and initial-boundary value problems have garnered significant attention across diverse disciplines such as physics, biology, ecology, chemistry, and beyond. Among these, parabolic partial differential equations (PDEs) with nonlocal initial and/or boundary conditions have emerged as powerful tools for modeling a wide array of phenomena. These include, but are not limited to, heat conduction[1], thermoelasticity[2], biotechnology[3], electrochemistry[4], population dynamics [5], and petroleum exploration [6]. The incorporation of nonlocal conditions into these PDEs allows for a more nuanced and realistic representation of the complex interactions and dynamics at play within these systems.

    Let QT=Ω×I be the computational domain, where Ω=(0,1)2 and I=(0,T) represent the spatial domain and the time domain, respectively, and T is a positive constant. Here, we consider the following 2D parabolic problem to find a high-accuracy numerical scheme and obtain its theoretical error estimates:

    ut=a2Δu+f(x,y,t),(x,y)Ω,t(0,T], (1.1)

    which is subject to the initial conditions

    u|t=0=g(x,y),(x,y)Ω, (1.2)

    the Dirichlet boundary conditions

    u|x=0=μ1(y,t),y(0,1),t(0,T], (1.3)
    u|x=1=μ2(y,t),y(0,1),t(0,T], (1.4)

    and the nonlocal boundary conditions

    u|y=0=u|y=1+μ3(x,t),x(0,1),t(0,T], (1.5)
    uy|y=0=μ4(x,t),x(0,1),t(0,T], (1.6)

    where u(x,y,t) is the unknown function, g(x,y), μi(y,t) (i=1,2) and μj(x,t) (j=3,4) are known functions, and a is a positive constant.

    These two nonlocal boundary conditions (1.5) and (1.6) are often be used to describe the correlation of a physical quantity across two parallel boundaries in a physical system, as well as the situation where the normal derivative at the boundaries is controlled by external factors, which is commonly used to simulate the interactions between boundaries and boundary effects in processes such as heat conduction and fluid flow.

    If the exact solution u of problems (1.1)–(1.6) satisfies certain smootheness conditions, then the compatible condition is deduced as follows: (x,y)Ω, the following relations hold:

    g(0,y)=μ1(y,0),g(1,y)=μ2(y,0),g(x,0)=g(x,1)+μ3(x,0),gy(x,0)=μ4(x,0).

    The analytical frameworks and numerical techniques employed in tackling parabolic problems with nonlocal conditions have aroused the concern of many scholars. Pertaining to the crucial aspects of convergence and stability for such problems, we acknowledge the foundational work presented in [7,8,9], as well as the extensive references cited therein. Among the prevalent numerical methodologies, finite difference methods (FDM) stand out prominently, with notable contributions from studies such as [7,10,11,12,13]. Additionally, finite element methods (FEM) have garnered substantial attention, exemplified by works cited in [14,15]. Furthermore, the realm of numerical solutions encompasses innovative approaches like Adomian expansions [16], the local coordinates method [17], and the utilization of reproducing kernel spaces [18], each offering unique insights and advancements in this field.

    It is widely acknowledged that two-dimensional parabolic partial differential equations (PDEs), characterized by their two spatial variables, pose significant challenges for theoretical analysis, particularly in the realms of convergence analysis and error estimation. The dimensionality of these variables often complicates the mathematical treatment, necessitating innovative strategies. One promising approach to mitigate these difficulties is the utilization of the discrete fourier transform (DFT) method, which offers advantages in reducing the complexity of self-variables during convergence analysis. In this study, we build upon our previous work [19,20] by extending the numerical schemes and integrating the DFT method on the spatial variable x for error estimation within the context of a two-dimensional parabolic PDE subject to a nonlocal boundary condition.

    However, a major obstacle arises from the complex boundary condition imposed on the spatial variable y. This condition presents a challenge to traditional DFT methods, which are inherently designed to preserve some boundary conditions. To overcome this limitation, we propose a novel transformation tailored specifically to handle this periodic boundary scenario. Furthermore, we contribute by deriving formulas for the solution derivatives and rigorously proving that these formulas enable us to achieve optimal asymptotic error estimates in the maximum norm. This achievement underscores the effectiveness and applicability of our proposed methodology in accurately approximating and analyzing solutions to two-dimensional parabolic PDEs with intricate boundary conditions.

    This paper is organized as follows. In Section 2, the backward Euler difference scheme for the solution of problems (1.1)–(1.6) is presented. Then, in Section 3, we utilize the DFT and develop a new transformation to analyze the error estimate for the corresponding difference equation. The superconvergence for the derivative and its theoretical results are also considered. Finally, some numerical experiments are presented in Section 5.

    Now, we use the FDM to discretize problems (1.1)–(1.6). The domain ¯QT is discretized by the uniformly distributed grid points (xi,yj,tn), where

    xi=ih,i=0,1,,2N,2Nh=1,yj=jh,j=0,1,,2N,2Nh=1,tn=nτ,n=0,1,,M,Mτ=T,

    where τ is time stepsize, and h is space stepsize along both x and y directions.

    Define a function space by

    Cm(¯QT)={s1+s2+s3uxs1ys2ts3C(¯QT)|s1+s2+s3m},

    and its norm by

    um,=maxs1+s2+s3m{|s1+s2+s3uxs1ys2ts3|},(x,y,t)¯QT,

    where m and si (i=1,2,3) are given nonnegative integers.

    The key to seeking a numerical solution for problems (1.1)–(1.6) lies in how to discretize the nonlocal boundary conditions (1.6). Suppose uC4(¯QT), using the Taylor formula, we have

    u(x,h,t)=u(x,0,t)+huy(x,0,t)+h22uyy(x,0,t)+h33!uyyy(x,0,t)+O(h4). (2.1)

    Using (1.1), we have

    uyy(x,0,t)=1a2ut(x,0,t)uxx(x,0)1a2f(x,0,t). (2.2)

    Moreover, we obtain

    uyyy(x,0,t)=1a2uty(x,0,t)uxxy(x,0,t)1a2fy(x,0,t).

    Therefore, with (1.6), we obtain

    uyyy(x,0,t)=1a2(μ4)t(x,t)(μ4)xx(x,t)1a2fy(x,0,t). (2.3)

    Substituting (1.6), (2.2), and (2.3) into (2.1), we have

    u(x,h,t)=u(x,0,t)+hμ4(x,t)+h22(1a2ut(x,0,t)uxx(x,0,t)1a2f(x,0))+h33!(1a2(μ4)t(x,t)(μ4)xx(x,t)1a2fy(x,0,t))+O(h4),

    i.e.

    ut(x,0,t)=2a2h2(u(x,h,t)u(x,0,t))+a2uxx(x,0,t)+˜μ4(x,t)+O(h2), (2.4)

    where

    ˜μ4(x,t)=f(x,0,t)2a2hμ4(x,t)h3((μ4)t(x,t)a2(μ4)xx(x,t)fy(x,0,t)). (2.5)

    From the derivation process described above, the discretization of (1.6) is converted to discretizing (2.4).

    Let uni,j and Uni,j be the exact value and the approximation of u(x,y,t) at grid point (xi,yj,tn), respectively. Let fni,j=f(xi,yj,tn), gi,j=g(xi,yj), (μm)nj=μm(yj,tn) (m=1,2), (μ3)ni=μ3(xi,tn) and (˜μ4)ni=˜μ4(xi,tn).

    Then, (2.4) is approximated by the following difference equations:

    Uni,0Un1i,0τ=2a2h2(Uni,1Uni,0)+a2Uni+1,02Uni,0+Uni1,0h2+2(μ4)nih+(˜μ4)ni,i=1,2,,2N1. (2.6)

    Also, we obtain the difference equations of (1.1)

    Uni,jUn1i,jτ=a2(Uni1,j2Uni,j+Uni+1,jh2+Uni,j12Uni,j+Uni,j+1h2)+fni,j,i,j=1,2,,2N1,n=1,2,,M. (2.7)

    Let αni,0 be the local truncature error of (2.6). When uC4(¯QT), using the Taylor formula, we can easily deduce that

    |αni,0|u4,(τ+h2)τ+h2,i=1,2,,2N1.

    Similarly, when uC4(¯QT), it holds that

    |αni,j|u4,(τ+h2)τ+h2,i=1,2,,2N1,j=1,2,,2N1,

    where αni,j​ is the local truncation error of (2.7).

    Moreover, we obtain

    |αni,j|τ+h2,i=1,2,,2N1,j=0,1,,2N1. (2.8)

    From the above, we obtain the backward Euler difference scheme of problems (1.1)–(1.6).

    Uni,jUn1i,jτ=a2(Uni1,j2Uni,j+Uni+1,jh2+Uni,j12Uni,j+Uni,j+1h2)+fni,j,i,j=1,2,,2N1,n=1,2,,M, (2.9a)
    U0i,j=gi,j,i,j=0,1,,2N, (2.9b)
    Un0,j=(μ1)nj,j=0,1,,2N,n=1,2,,M, (2.9c)
    Un2N,j=(μ2)nj,j=0,1,,2N,n=1,2,,M, (2.9d)
    Uni,0=Uni,2N+(μ3)ni,i=1,2,,2N1,n=1,2,,M, (2.9e)
    Uni,0Un1i,0τ=2a2h2(Uni,1Uni,0)+a2h2(Uni1,02Uni,0+Uni+1,0)+(˜μ4)ni,i=1,2,,2N1,n=1,2,,M. (2.9f)

    Let eni,j=uni,jUni,j be the error of the approximation solution U at the grid point (xi,yj,tn), and μ=τh2 be the grid ratio. Then, the error equations of (2.9a)–(2.9f) are

    eni,jen1i,j=a2μ(eni1,j+eni+1,j+eni,j1+eni,j+14eni,j)+ταni,j,i,j=1,,2N1,n=1,2,,M, (3.1a)
    e0i,j=0,i,j=0,1,,2N, (3.1b)
    en0,j=0,j=0,1,,2N,n=1,2,,M, (3.1c)
    en2N,j=0,j=0,1,,2N,n=1,2,,M, (3.1d)
    eni,0=eni,2N,i=1,2,,2N1,n=1,2,,M, (3.1e)
    eni,0en1i,0=2a2μ(eni,1eni,0)+a2μ(eni1,02eni,0+eni+1,0)+ταni,0,i=1,2,,2N1,n=1,2,,M. (3.1f)

    Given the complexity of the above error equations, the key to obtaining an error estimate lies in finding transformations that separate the index variables i, j, and n.

    Since the error sequence {eni,j} satisfies (3.1c) and (3.1d), applying the DFT to {eni,j} with respect to i, we obtain

    eni,j=2h2N1k=1ˆenk,jsin(kπxi),i,j=0,1,,2N. (3.2)

    Similarly, applying the DFT to {αni,j} with respect to i, we obtain

    αni,j=2h2N1k=1ˆαnk,jsin(kπxi),i=1,2,,2N1,j=0,1,,2N1. (3.3)

    It follows from (2.8) and (3.3) that

    |ˆαnk,j|τ+h2h12,k=1,2,,2N1,j=0,1,,2N1. (3.4)

    Substituting (3.2) and (3.3) into (3.1a), we obtain

    2h2N1k=1(ˆenk,jˆen1k,j)sin(kπxi)=2ha2μ2N1k=1(ˆenk,j(sin(kπxi1)2sin(kπxi)+sin(kπxi+1))+(ˆenk,j12ˆenk,j+ˆenk,j+1)sin(kπxi))+2hτ2N1k=1ˆαnk,jsin(kπxi),i,j=1,2,,2N1. (3.5)

    Utilizing the properties of the DFT and (3.5), we obtain

    ˆenk,jˆen1k,j=a2μ(ˆenk,j1(2+4sin2kπh2)ˆenk,j+ˆenk,j+1)+τˆαnk,j,j=1,2,,2N1. (3.6)

    Similarly, substituting (3.2) and (3.3) into (3.1f), we have

    ˆenk,0ˆen1k,0=2a2μ(ˆenk,1ˆenk,0)4a2μsin2kπh2ˆenk,0+τˆαnk,0. (3.7)

    Substituting (3.2) into (3.1b) and (3.1e), we deduce that

    ˆe0k,j=0,j=0,1,,2N, (3.8)

    and

    ˆenk,0=ˆenk,2N. (3.9)

    Given that the sequence {ˆenk,j} adheres to the condition specified in (3.9), the conventional DFT is found to be inadequate for our analytical needs. In pursuit of a suitable tool for analysis, we aspire for a novel transformation that not only fulfills the criteria outlined in (3.8) but also possesses the property of invertibility. Drawing inspiration from the formulation of the DFT, we introduce a fresh transformation tailored specifically for the sequence {ˆenk,j} with respect to j in the following way, aiming to address the aforementioned limitations and meet our analytical needs.

    ˆenk,j=2N1l=0˜enk,lTl(yj),j=0,1,,2N1, (3.10)

    where

    Tl(y)={cos(2lπy),l=0,1,,N,ysin(2lπy),l=N+1,N+2,,2N1. (3.11)

    It is straightforward to verify Lemma 3.1.

    Lemma 3.1. The sequence {Tl(yj)} has the following properties.

    (1) Tl(y0)={1,l=0,1,,N,0,l=N+1,N+2,,2N1.

    (2) Tl(y1)Tl(y0)={(cos(2lπh)1)Tl(y0),l=0,1,,N,hsin(2lπh),l=N+1,N+2,,2N1.

    (3) For any 0l,j2N1,

    Tl(yj1)2Tl(yj)+Tl(yj+1)={2(cos(2lπh)1)Tl(yj),l=0,1,,N,2(cos(2lπh)1)Tl(yj)+2hT2Nl(yj)sin(2lπh),l=N+1,N+2,,2N1.

    (4) Tl(y2Nj)={cos(2lπyj),l=0,1,,N,(yj1)sin(2lπyj),l=N+1,N+2,,2N1.

    For the sake of simplicity in the subsequent analysis, we introduce

    Pl(y):={cos(2lπy),l=0,1,,N,sin(2lπy),l=N+1,N+2,,2N1, (3.12)

    and consider the orthogonality relation of the polynomials Pl(yi) and Pl(yj).

    Lemma 3.2. Given that i,j=0,,N, we have the following identity:

    Nl=0σlPl(yi)Pl(yj)={N2σi,i=j,0,ij, (3.13)

    where

    σl={12,l=0,N,1,otherwise. (3.14)

    Proof. Using (3.14) and (3.12), and noticing 2Nh=1 and yi=ih, we have

    Nl=0σlPl(yi)Pl(yj)=N1l=1Pl(yi)Pl(yj)+12l=0,NPl(yi)Pl(yj)=N1l=1cos(2lπyi)cos(2lπyj)+12(1+cos(2Nπyi)cos(2Nπyj))=12N1l=1cos(2lπyi+j)+12N1l=1cos(2lπyij)+1+(1)i+j2. (3.15)

    For 0m2N, we have

    2N1l=1cos(2lπym)=N1l=1(cos(2(l1)πym)+cos(2(l+1)πym))1+cos(2(N1)πym)+cos(2πym)cos(2Nπym)=2cos(2πym)N1l=1cos(2lπym)+(1+(1)m)(cos(2πym)1),

    i.e.,

    2(1cos(2lπym))N1l=1cos(2lπym)=(1+(1)m)(cos(2πym)1).

    If cos(2lπym)1, then

    N1l=1cos(2lπym)=1+(1)m2. (3.16)

    Now we focus on the case ij. Since 0i,jN, it follows that 0<yi+j<1 and 0<|yij|<1. Furthermore, when l ranges from 1 to N1, we observe that cos(2lπyij)1 and cos(2lπyi+j)1.

    Thus, with (3.16), and observing the same parity of i+j and ij, we obtain

    N1l=1cos(2lπyi+j)=N1l=1cos(2lπyij)=1+(1)i+j2.

    Substituting the above equality into (3.15), we deduce that

    Nl=0σlPl(yi)Pl(yj)=0,ij. (3.17)

    In the next, we consider the case i=j. Noting that when 1lN1, cos(2lπy2i) is equal to 1 only in i=0 and i=N. Therefore, by utilizing (3.16), we obtain

    N1l=1cos(2lπy2i)={1,i=1,2,,N1,N1,i=0,N. (3.18)

    Substituting (3.18) into (3.15), we deduce that

    Nl=0σlPl(yi)Pl(yj)={N2,i=1,2,,N1,N,i=0,N. (3.19)

    Furthermore, with (3.14), (3.17), and (3.19), we arrive at the conclusion stated in (3.13).

    Similar to Lemma 3.2, we can derive the subsequent lemma as well.

    Lemma 3.3. Given that i,j=N+1,,2N1, we have the following identity

    2N1l=N+1Pl(yi)Pl(yj)={N2,i=j,0,ij. (3.20)

    Therefore, we can conclude the following lemma.

    Lemma 3.4. Suppose

    ai=2N1l=N+1ˆalPl(yi),i=N+1,N+2,,2N1. (3.21)

    Then

    ˆal=2N2N1i=N+1aiPl(yi),l=N+1,N+2,,2N1. (3.22)

    Proof. Using (3.21), (3.12), and Lemma 3.3, we obtain

    2N1i=N+1aiPl(yi)=2N1i=N+12N1m=N+1ˆamPl(yi)Pm(yi)=2N1m=N+1ˆam2N1i=N+1Pi(yl)Pi(ym)=N2ˆal.

    The proof is finished.

    Similar to Lemma 3.4, we obtain

    Lemma 3.5. Suppose

    ai=Nl=0ˆalPl(yi),i=0,1,,N. (3.23)

    Then

    ˆal=2σlNNi=0σiaiPl(yi),l=0,1,,N. (3.24)

    Based on Lemmas 3.4 and 3.5, we obtain the invertible transformation of (3.25).

    Lemma 3.6. Suppose

    ˆaj=2N1l=0˜alTl(yj),j=0,1,,2N1. (3.25)

    Then,

    ˜al={2σlNNj=0σj((1yj)ˆaj+yjˆan2Nj)cos(2lπyj),l=0,1,,N,2NNj=0(ˆajˆa2Nj)sin(2lπyj),l=N+1,N+2,,2N1, (3.26)

    where

    σj={12,j=0,N,1,otherwise. (3.27)

    Proof. Using (3.25), (3.12), and Lemma 3.1, we have

    ˆa2Nj=Nl=0˜alPl(yj)+(yj1)2N1l=N+1˜alPl(yj),

    and

    ˆaj=Nl=0˜alPl(yj)+yj2N1l=N+1˜alPl(yj).

    From the two equalities above, it follows that

    (1yj)ˆaj+yjˆa2Nj=Nl=0˜alPl(yj)

    and

    ˆajˆa2Nj=2N1l=N+1˜alPl(yj).

    Moreover, using Lemmas 3.5 and 3.4, we arrive at the conclusion stated in (3.26).

    Similar to (3.10), we use the same transformation to {ˆαnk,j} with respect to j in the following way:

    ˆαnk,j=2N1l=0˜αnk,lTl(yj),j=0,1,,2N1. (3.28)

    Using (3.4), (3.27), and Lemma 3.6, and noting that 0yj1 (j=0,1,,2N), we can deduce

    |˜αnk,l|τ+h2h12,l=0,1,,2N1. (3.29)

    Substituting (3.10) and (3.28) into (3.6), we obtain

    2N1l=0(˜enk,l˜en1k,l)Tl(yj)=a2μ2N1l=0˜enk,l(Tl(yj1)(2+4sin2kπh2)Tl(yj)+Tl(yj+1))+τ2N1l=0˜αnk,lTl(yj),j=1,2,,2N1. (3.30)

    Using Lemma 1, (3.30) can be rewritten as

    2N1l=0(˜enk,l˜en1k,l)Tl(yj)=a2μ(2N1l=0(2(cos(2lπh)1)4sin2kπh2)˜enk,lTl(yj)+2h2N1l=N+1˜enk,lT2Nl(yj)sin(2lπh))+τ2N1l=0˜αnk,lTl(yj),j=1,2,,2N1.

    Let l:=2Nl in 2N1l=N+1˜enk,lT2Nl(yj)sin(2lπh), the above equalities have the following form:

    2N1l=0(˜enk,l˜en1k,l)Tl(yj)=2a2μ(22N1l=0(sin2(lπh)+sin2kπh2)˜enk,lTl(yj)+hN1l=1˜enk,2NlTl(yj)sin(2lπh))+τ2N1l=0˜αnk,lTl(yj),j=1,2,,2N1. (3.31)

    Substitute (3.10) into (3.7), then

    2N1l=0(˜enk,0˜en1k,0)Tl(y0)=2a2μ2N1l=0˜enk,l(Tl(y1)Tl(y0))4a2μsin2kπh22N1l=0˜enk,lTl(y0)+τ2N1l=0˜αnk,lTl(y0).

    Moreover, using Lemma 3.1, we obtain

    2N1l=0(˜enk,0˜en1k,0)Tl(y0)=2a2μ(22N1l=0(sin2(lπh)+sin2kπh2)˜enk,lTl(y0)+hN1l=1˜enk,2NlTl(y0)sin(2lπh))+τ2N1l=0˜αnk,lTl(y0).

    Through comparing with the above equality and (3.31), we find that (3.31) also holds for j=0. Therefore,

    2N1l=0(˜enk,l˜en1k,l)Tl(yj)=2a2μ(22N1l=0(sin2(lπh)+sin2kπh2)˜enk,lTl(yj)+hN1l=1˜enk,2NlTl(yj)sin(2lπh))+τ2N1l=0˜αnk,lTl(yj),j=0,1,,2N1. (3.32)

    Using Lemma 3.6 to perform an invertible transformation on (3.32), we obtain

    ˜enk,l˜en1k,l={4a2μ(sin2(lπh)+sin2kπh2)˜enk,l2a2μh˜enk,2Nlsin(2lπh)+τ˜αnk,l,l=1,2,,N1,4a2μ(sin2(lπh)+sin2kπh2)˜enk,l+τ˜αnk,l,l=0,N,N+1,,2N1. (3.33)

    Let

    ωk,l=11+4a2μ(sin2(lπh)+sin2kπh2). (3.34)

    Obviously,

    0<ωk,l<1. (3.35)

    Using (3.34), (3.33) can be rewritten as

    ˜enk,l={ωk,l˜en1k,l2a2μhωk,l˜enk,2Nlsin(2lπh)+τωk,l˜αnk,l,l=1,2,,N1,ωk,l˜en1k,l+τωk,l˜αnk,l,l=0,N,N+1,,2N1. (3.36)

    Substituting (3.10) into (3.8), and using Lemma 3.6, we can easily deduce that

    ˜enk,l=0,l=0,1,,2N1. (3.37)

    Using (3.36) and (3.37), we obtain the following recursive formula for {˜enk,l}:

    ˜enk,l={2a2μhsin(2lπh)nm=1(ωk,l)nm+1˜emk,2Nl+τnm=1(ωk,l)nm+1˜αmk,l,l=1,2,,N1,τnm=1(ωk,l)nm+1˜αmk,l,l=0,N,N+1,,2N1. (3.38)

    In order to estimate {˜enk,l}, we first prove the following estimation

    nm=1(ωk,l)nm+1{1τ(l2+k2),l=0,1,,N,1τ((2Nl)2+k2),l=N+1,N+2,,2N1. (3.39)

    In fact, from (3.34) and (3.35), we can derive that

    nm=1(ωk,l)nm+1=ωk,l(ωk,l)n+11ωk,l=1(ωk,l)nμ(sin2(lπh)+sin2kπh2)1μ(sin2(lπh)+sin2kπh2). (3.40)

    For 0lN, we have lπh[0,π2]. Observe that kπh2(0,π2)(1k2N1). Therefore, (3.40) can be rewritten as

    nm=1(ωk,l)nm+11μh2(4l2+k2)1τ(l2+k2). (3.41)

    For Nl2N1, from 2Nh=1 and 0<(2Nl)πhπ2, we have

    sin(lπh)=sin(2(2Nl)πh)2(2Nl)h.

    Moreover, (3.40) can be written as

    nm=1(ωk,l)nm+11μh2(4(2Nl)2+k2)1τ((2Nl)2+k2). (3.42)

    Therefore, combining (3.41) with (3.42), (3.39) holds.

    Now, we give the estimation of {˜enk,l} in three cases.

    Case 1. Nl2N1

    With (3.38), (3.35), (3.29), and (3.39), we have

    |˜enk,l|τnm=1(ωk,l)nm+1|˜αnk,l|τ(τ+h2)h12nm=1(ωk,l)nm+1τ+h2h12((2Nl)2+k2). (3.43)

    Case 2. 1lN1

    Using (3.43), we obtain

    |˜enk,2Nl|τ+h2h12(l2+k2). (3.44)

    From the above inequality, and using (3.38), (3.35), (3.39), (3.29), and μ=τh2, we obtain

    |˜enk,l|2a2μhsin(2lπh)nm=1(ωk,l)nm+1|˜emk,2Nl|+τnm=1(ωk,l)nm+1|˜αmk,l|(μhsin(2lπh)τ+h2h12(l2+k2)+τ(τ+h2)h12)nm=1(ωk,l)nm+1τ+h2h12(l2+k2)(sin(lπh)h(l2+k2)+1)τ+h2h12(l2+k2). (3.45)

    Case 3. l=0

    Observing that ωk,0=11+4a2μsin2kπh2, similar to deduce (3.43), we obtain

    |˜enk,0|τ+h2k2h12. (3.45 = 6)

    From (3.10), (3.43), (3.45), and (3.46), and noticing that Tl(yj) is bounded for any 0l,j2N1, we have

    |ˆenk,j|2N1l=0|˜enk,l||Tl(yj)|τ+h2k2h12(1k2+N1l=11l2+k2+2N1l=N1(2Nl)2+k2)τ+h2h12N1l=01l2+k2. (3.47)

    Furthermore, from (3.2), we obtain

    |eni,j|2h2N1k=1|ˆenk,j|(τ+h2)2N1k=1N1l=01l2+k2(τ+h2)(32Nk=11k2+2Nk=22Nl=21l2+k2)(τ+h2)2Nk=22Nl=21l2+k2. (3.48)

    Since 1x2+y2 is increasing monotonically with respect to variables x and y for x,y>0, respectively, it follows that

    2Nk=22Nl=21l2+k2=h22Nk=22Nl=21(lh)2+(kh)2Ωh1x2+y2dxdy<π20dθ2hdrrπ2|lnh|, (3.49)

    where Ωh=[h,1]×[h,1].

    Using (3.49) and (3.48), and noticing (3.1c)–(3.1e), we can obtain the following error estimation theorem.

    Theorem 3.1. Suppose uC4(¯QT). For any postive integer 1nM, the following estimates for (2.9a)–(2.9f)

    |eni,j|(τ+h2)|lnh|,i,j=0,1,,2N.

    hold.

    In this section, we present the approximation formulas for the partial derivatives of u with respect to two spatial variables, which exhibit superconvergence under certain smooth conditions.

    Let Ux and Uy be the approximation functions for the partial derivatives ux and uy, respectively. For any tn (1nM), we introduce the following approximation formulas for ux and uy at the grid point (xi,yj,tn), respectively:

    Ux(xi,yj,tn)=Uni+1,jUni1,j2h,1i2N1,0j2N, (4.1)

    and

    Uy(xi,yj,tn)=Uni,j+1Uni,j12h,0i2N,1j2N1. (4.2)

    Before exploring the superconvergence of (4.1) and (4.2), we first present the following lemma.

    Lemma 4.1. Suppose that the function p(x)C1[0,1] satisfies

    maxx[0,1]{|p(x)|,|p(x)|}M, (4.3)

    where M is a positive constant. If

    ˆpk=2h2N1i=1pisin(iπxk),i=1,2,,2N1, (4.4)

    then

    |ˆpk|Mπkh121kh12. (4.5)

    Proof. Let θk=kπh2. Obviously, θk(0,π2). We can easily verify the following equality:

    2N1i=1(pi12pi+pi+1)sin(iπxk)=2N1i=1pi(sin((i+1)πxk)2sin(iπxk)+sin((i1)πxk))+p0sin(πxk)+p2Nsin((2N1)πxk)=4sin2θk2N1i=1pisin(iπxk)+(p0+(1)kp2N)sin(2θk). (4.6)

    Noting that

    2N1i=1(pipi1)sin(iπxk)=2N1i=1(xixi1p(x)sin(kπx)dx+xixi1p(x)(sin(kπxi)sin(kπx))dx)=x2N10p(x)sinkπxdx+22N1i=1xixi1p(x)coskπ(xi+x)2sinkπ(xix)2dx, (4.7)

    the following equality is also verified:

    2N1i=1(pi+1pi)sin(iπxk)=1x1p(x)sin(kπx)dx+22N1i=1xi+1xip(x)coskπ(xi+1+x)2sinkπ(xi+1x)2dx. (4.8)

    Subtracting (4.7) from (4.8), we have

    2N1i=1(pi12pi+pi+1)sin(iπxk)=1x2N1p(x)sin(kπx)dxx10p(x)sin(kπx)dx+21x2N1p(x)coskπ(1+x)2sinkπ(1x)2dx2x10p(x)coskπ(x1+x)2sinkπ(x1x)2dx.

    Thus, using (4.3), we obtain

    |2N1i=1(pi12pi+pi+1)sin(iπxk)|6Mh. (4.9)

    Using (4.6), (4.3), and (4.9), and noting θk(0,π2), we obtain

    |2N1i=1pisin(iπxk)|=14sin2θk|2N1i=1(pi12pi+pi+1)sin(iπxk)(p0+(1)kp2N)sin(2θk)|3Mh+2Msinθk2sin2θkMπkh.

    Therefore, the lemma is proved with (4.4).

    Next, we study the superconvergence of (4.1).

    Theorem 4.1. Suppose uC5(¯QT). Then, for any integer 1nM,

    |Ux(xi,yj,tn)ux(xi,yj,tn)|(τ+h2)|lnh|,i=1,2,,2N1,j=0,1,,2N. (4.10)

    Proof. From (4.1) and uC5(¯QT), we obtain

    Ux(xi,yj,tn)=ux(xi,yj,tn)+eni+1,jeni1,j2h+O(h2).

    Thus, in order to prove this theorem, it suffices to prove the following inequality, i.e., for any given integer 1nM,

    |eni+1,jeni1,j2h|(τ+h2)|lnh|,i=1,2,,2N1,j=0,1,,2N. (4.11)

    From (3.2) and (3.10), we have

    eni+1,jeni1,j2h=22h2N1k=1ˆenk,jsin(kπh)cos(kπxi)=22h2N1k=12N1l=0˜enk,lTl(yj)sin(kπh)cos(kπxi). (4.12)

    Since uC5(¯QT), using Lemma 4.1, (2.8), and (3.3), we obtain

    |ˆαnk,j|τ+h2kh12. (4.13)

    Correspondingly, (3.29) is written as

    |˜αnk,l|τ+h2kh12. (4.14)

    For this, by modifying (3.43), (3.45), and (3.46), we obtain

    |˜enk,l|{τ+h2kh12(l2+k2),0lN,τ+h2kh12((2Nl)2+k2),N+1l2N1. (4.15)

    Using (4.12) and (4.15), and given |Tl(y)|1 for y[0,1], we obtain

    |eni+1,jeni1,j2h|h122N1k=1k2N1l=0|˜enk,l|(τ+h2)2N1k=1(N1l=01l2+k2+2N1l=N1(2Nl)2+k2)(τ+h2)2Nk=22Nl=21l2+k2. (4.16)

    From this, using (3.49), we prove (4.11). Therefore, (4.10) holds.

    In the following, we discuss the superconvergence properties of (4.2).

    Theorem 4.2. Suppose uC5(¯QT). Then, for any integer 1nM,

    |Uy(xi,yj,tn)uy(xi,yj,tn)|(τ+h2)ln2h,i=0,1,,2N,j=1,2,,2N1, (4.17)

    hold.

    Proof. From (4.2) and uC5(¯QT), we obtain

    Ux(xi,yj,tn)=ux(xi,yj,tn)+eni,j+1eni1,j12h+O(h2). (4.18)

    From (4.18), in order to prove this theorem, we only need to prove that for any integer 1nM,

    |eni,j+1eni,j12h|(τ+h2)ln2h,i=0,1,,2N,j=1,2,,2N1. (4.19)

    Using (3.2) and (3.10), and noting that T0(yj+1)T0(yj1)=0, we find that

    eni,j+1eni,j12h=22h2N1k=1(ˆenk,j+1ˆenk,j1)sin(kπxi)=22h2N1k=12N1l=0˜enk,l(Tl(yj+1)Tl(yj1))sin(kπxi)=22h2N1k=12N1l=1˜enk,l(Tl(yj+1)Tl(yj1))sin(kπxi). (4.20)

    Using (3.11), when Nl2N1, we have

    |Tl(yj+1)Tl(yj)|2|yjcos((yj+1+yj)lπ)sin(2lπh)|+h|sin(2lπyj+1)||sin((2Nl)πh)|+h(2Nl)h. (4.21)

    Furthermore, we also deduce that

    |Tl(yj+1)Tl(yj)|=2|sin((yj+1+yj)lπ)sin(2lπh)|lh. (4.22)

    Using (4.20), (4.15), (4.21), and (4.22), we obtain

    |eni,j+1eni,j12h|1h122N1k=12N1l=1|˜enk,l||Tl(yj+1)Tl(yj1)|τ+h2h2N1k=1(Nl=1lhk(k2+l2)+2N1l=N+1(2Nl)hk(k2+(2Nl)2))(τ+h2)2N1k=1Nl=1lk(k2+l2)(τ+h2)(2Nl=1l1+l2+2Nk=11k(k2+1)+2Nk=22Nl=2lk(k2+l2))(τ+h2)(|lnh|+2Nk=22Nl=2lk(k2+l2)). (4.23)

    Upon observing

    22Nk=22Nl=2lk(k2+l2)=2Nk=22Nl=2(lk(k2+l2)+kl(k2+l2))=2Nk=22Nl=21klln2h,

    by substituting this result into (4.23), we obtain

    |eni,j+1eni,j12h|(τ+h2)ln2h.

    This result confirms (4.19), thereby establishing the validity of (4.17).

    In this section, we present two numerical examples to validate the theoretical results and investigate the efficiency and the superconvergence properties of the numerical schemes. Our aim is to demonstrate the practical implications of the theoretical findings and assess the performance of the proposed methods. Let

    Uu:=max1nM0i,j2N|Uni,juni,j|,Uxux:=max1nM1i2N10j2N|(Ux)ni,j(ux)ni,j|,Uyuy:=max1nM0i2N1j2N1|(Uy)ni,j(uy)ni,j|.

    Example 5.1. In (1.1)–(1.6), take

    a=1,T=1,f(x,y,t)=0,g(x,y)=ex+y,μ1(y,t)=ey+2t,μ2(y,t)=e1+y+2t,μ3(x,t)=ex+2t(1e),μ4(x,t)=ex+2t.

    The exact solution is u=ex+y+2t which can be easily verified.

    The results are reported in Tables 13. From Table 1, we can observe that in the cases of τ=h2 and τ=h, the error Uu​ is approximately of the order O(h2) and O(h), respectively. This observation verifies the correctness of Theorem 3.1.

    Table 1.  Error with respect to u in τ=h and τ=h2 for Example 5.1.
    h τ=h2 τ=h
    Uu ratio Uu ratio
    1/32 4.4377e-003 - 1.2941e-001 -
    1/64 1.1104e-003 4.00 6.5173e-002 1.99
    1/128 2.7768e-004 4.00 3.2701e-002 1.99
    1/256 6.9427e-005 4.00 1.6379e-002 2.00

     | Show Table
    DownLoad: CSV
    Table 2.  Error of ux and uy in τ=h2 for Example 5.1.
    h τ Uxux ratio Uyuy ratio
    1/32 1/322 1.7481e-002 - 7.0681e-003 -
    1/64 1/642 4.4901e-003 3.89 1.8960e-003 3.73
    1/128 1/1282 1.1379e-003 3.95 5.0110e-004 3.78
    1/256 1/2562 2.8640e-004 3.97 1.3025e-004 3.85

     | Show Table
    DownLoad: CSV
    Table 3.  Error of ux and uy in τ=h for Example 5.1.
    h τ Uxux ratio Uyuy ratio
    1/32 1/32 6.1798e-001 - 1.7722e-001 -
    1/64 1/64 3.3247e-001 1.86 1.0037e-001 1.77
    1/128 1/128 1.7242e-001 1.93 5.3329e-002 1.88
    1/256 1/256 8.7795e-002 1.96 2.7478e-002 1.94

     | Show Table
    DownLoad: CSV

    Furthermore, from Tables 2 and 3, it is evident that when τ=h2, both Uxux and Uyuy are close to the order O(h2). On the other hand, when τ=h, Uxux and Uyuy approach the order O(h). These findings support the theoretical expectations regarding the convergence rates of the spatial derivatives. Therefore, the correctness of Theorems 4.1 and 4.2 is verified.

    Example 5.2. In problems (1.1)–(1.6), take

    a=1,T=1,f(x,y,t)=0,g(x,y)=(1+y)ex,μ1(y,t)=(1+y)et,μ2(y,t)=(1+y)e1+t,μ3(x,t)=ex+t,μ4(x,t)=ex+t.

    It is easily verified that its exact solution is u=(1+y)ex+t.

    Numerical results for Example 5.2 are reported in Tables 46. These results verify the correctness of Theorems 3.1, 4.1 and 4.2 again.

    Table 4.  Error with respect to u in τ=h and τ=h2 for Example 5.2.
    h τ=h2 τ=h
    Uu ratio Uu ratio
    1/32 4.2726e-004 - 1.1677e-002 -
    1/64 1.1104e-003 4.00 5.8545e-003 1.99
    1/128 1.0692e-004 4.00 2.9303e-003 2.00
    1/256 6.6840e-006 4.00 1.4660e-003 2.00

     | Show Table
    DownLoad: CSV
    Table 5.  Error of ux and uy in τ=h2 for Example 5.2.
    h τ Uxux ratio Uyuy ratio
    1/32 1/322 2.2208e-003 - 3.8223e-004 -
    1/64 1/642 5.6284e-004 3.95 1.0418e-004 3.67
    1/128 1/1282 1.4169e-004 3.97 2.7176e-005 3.83
    1/256 1/2562 3.5544e-005 3.99 6.9401e-006 3.92

     | Show Table
    DownLoad: CSV
    Table 6.  Error of ux and uy in τ=h for Example 5.2.
    h τ Uxux ratio Uyuy ratio
    1/32 1/32 5.1907e-002 - 1.0436e-002 -
    1/64 1/64 2.7983e-002 1.85 5.7010e-003 1.83
    1/128 1/128 1.4518e-002 1.93 2.9780e-003 1.91
    1/256 1/256 7.3924e-003 1.96 1.5219e-003 1.96

     | Show Table
    DownLoad: CSV

    This work focuses on a heat conduction problem with nonlocal boundary conditions. We develop an implicit Euler scheme and demonstrate that it achieves asymptotic optimal order with the DFT. Furthermore, we introduce two approximation formulas that exhibit superapproximation for first-order partial derivatives along the x and y directions of the exact solution, respectively. In the future, we plan to extend this work to other difference schemes for parabolic problems with nonlocal boundary conditions, such as the explicit Euler scheme, the Crank-Nicolson scheme, and other schemes. Additionally, we aim to consider heat conduction problems with different nonlocal boundary conditions.

    Liping Zhou: Conceptualization, methodology, formal analysis, writing-original draft, validation; Yumei Yan: Editing, software; Ying Liu: Writing-review and editing. All authors contributed equally to the manuscript. All authors have read and approved the final version of the manuscript for publication.

    This work is partially supported by the National Natural Science Foundation of China (No.12101224), the Hunan Provincial Natural Science Foundation of China (No. 2022JJ30271, 2024JJ7203) and the Key Project of Hunan Provincial Education Department of China (No. 23A0577).

    The authors declare no conflicts of interest.



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