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Linear barycentric rational collocation method for solving generalized Poisson equations


  • We consider the Poisson equation by collocation method with linear barycentric rational function. The discrete form of the Poisson equation was changed to matrix form. For the basis of barycentric rational function, we present the convergence rate of the linear barycentric rational collocation method for the Poisson equation. Domain decomposition method of the barycentric rational collocation method (BRCM) is also presented. Several numerical examples are provided to validate the algorithm.

    Citation: Jin Li, Yongling Cheng, Zongcheng Li, Zhikang Tian. Linear barycentric rational collocation method for solving generalized Poisson equations[J]. Mathematical Biosciences and Engineering, 2023, 20(3): 4782-4797. doi: 10.3934/mbe.2023221

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  • We consider the Poisson equation by collocation method with linear barycentric rational function. The discrete form of the Poisson equation was changed to matrix form. For the basis of barycentric rational function, we present the convergence rate of the linear barycentric rational collocation method for the Poisson equation. Domain decomposition method of the barycentric rational collocation method (BRCM) is also presented. Several numerical examples are provided to validate the algorithm.



    Two-dimensional elliptic [1] boundary value problems

    t(a(t,s)u(t,s)t)+s(b(t,s)u(t,s)s)=f(t,s),(t,s)Ω (1.1)

    with

    u(t,s)=g(t,s),u(t,s)n=h(t,s),(t,s)Ω, (1.2)

    where Ω=[a,b]×[c,d] and f(t,s),g(t,s),h(t,s) on Ω, can be used in many scientific areas, such as electrostatics, mechanical engineering, magnetic fields, thermal fields and brain activity detection[2].

    Floater [3,4,5] proposed a rational interpolation scheme. In [6], a linear rational collocation method was presented for the lower regular function. Wang et al.[7,8,9,10] successfully solved initial value problems, plane elasticity problems, incompressible plane problems and non-linear problems by collocation method. The linear barycentric rational collocation method (LBRCM) to solve nonlinear parabolic partial differential equations[11], biharmonic equation [12], fractional differential equations[13], telegraph equation [14], Volterra integro-differential equation [15] and heat conduction equation[16] have been studied.

    In the following, we introduce the barycentric formula of a one dimensional function. Let

    p(t)=nj=1Lj(t)fj, (1.3)

    and

    Lj(t)=nk=1,kj(ttk)nk=1,kj(tjtk). (1.4)

    In order to get the barycentric formula, Eq (1.4) is changed into

    l(t)=(tt1)(tt2)(ttn) (1.5)

    and

    wj=1jk(tjtk),j=1,2,,n, (1.6)

    which means wj=1/l(tj).

    Lj(t)=l(t)wjttj,j=1,2,,n. (1.7)

    For Eq (1.2), we get

    p(t)=l(t)nj=1wjttjfj (1.8)

    which means

    p(t)=nj=1wjttjfjnj=1wjttj. (1.9)

    For the case uniform partition, we get

    wj=(1)njCjn=(1)njn!j!(nj)!. (1.10)

    For the case the nonuniform partition is chosen, we take the second Chebyshev point [7] a

    tj=cos(j1)πn1,j=1,,n (1.11)

    with

    wj=(1)jδj,δj={1/2,j=1,n1,otherwise (1.12)

    For the barycentric rational function, we first set

    r(t)=ndi=1λi(t)pi(t)ndi=1λi(t) (1.13)

    where

    λi(t)=(1)i(tti)(tti+d) (1.14)

    and

    pi(t)=i+dk=ii+dj=i,jkttjtktjfk. (1.15)

    Combining (1.14) and (1.15), we have

    ndi=1λi(t)pi(t)=ndi=1(1)ii+dk=i1ttki+di,jk1tktjfk=nk=1wkttkfk (1.16)

    where

    wk=iJk(1)ii+dj=i,jk1tktj (1.17)

    and Jk={iI;kdik}. By taking p(t)=1, we have

    1=i+dk=ii+dj=i,jkttjtktj (1.18)

    and then have

    ndi=1λi(t)=nk=1wkttk. (1.19)

    Combining (1.16), (1.19) and (1.13), we get

    r(t)=nj=1wjttjfjnj=1wjttj (1.20)

    where ωj is defined as in (1.17).

    In this paper, based on linear barycentric rational interpolation of one dimension, we construct a barycentric rational interpolation of a two-dimensional Poisson equation. In order to get the discrete linear equation of a two-dimensional Poisson equation, the equidistant nodes and second kind of Chebyshev points were chosen as collocation point. For the general area, a domain decomposition method of the barycentric rational collocation method is also presented.

    Let a=t1<<tm=b,h=bam and c=s1<<sn=d,τ=dcn with mesh point (ti,sj),i=1,2,,m;j=1,2,,n. Then, we have

    u(ti,s)=ui(s), (2.1)

    on [a,b], and

    u(t,s)=mi=1nj=1Li(t)Mj(s)uij (2.2)

    where

    Li(t)=wittinj=1wjttj (2.3)

    and

    Mj(s)=vjssjnj=1vjssj. (2.4)

    wi,vj is the weight function defined as (1.6) or (1.17); see [17].

    We have

    [nk=1Mk(s)u1knk=1Mk(s)umk]+[C(2)11C(2)1mC(2)m1C(2)mn][nk=1Mk(s)u1knk=1Mk(s)umk]=[f1(s)fm(s)].

    Then, we have

    [nk=1Mk(sj)u1knk=1Mk(sj)umk]+[C(2)11C(2)1mC(2)m1C(2)mn][nk=1Mk(sj)u1knk=1Mk(sj)umk]=[f1(sj)fm(sj)],

    where C(2)ij=Li(tj), and

    C(2)ij={2Li(tj)(Li(ti)1titj),jiijLi(tj),j=i. (2.5)

    ui=[ui1,ui2,uin]T,fi=[fi1,fi2,,fin]T=[fi(si),fi(s2),,fi(sn)]T. With the help of the matrix form, the linear equation systems can be written as

    (ImD(2))[u1um]+(C(2)In)[u1um]=[f1fm] (2.6)

    and D(2)ij=Mi(sj),

    D(2)ij={2Mi(sj)(Mi(si)1sisj),ji,ijMi(sj),j=i. (2.7)

    Then, we have

    [(C(2)In)+(ImD(2))]U=F (2.8)

    and

    LU=F (2.9)

    where

    L=C(2)In+ImD(2) (2.10)

    and is the Kronecker product of the matrices. The Kronecker product of A=(aij)m×n and B=(bij)k×l is defined as

    A×B=(aijB)mk×nl (2.11)

    where

    aijB=[aijb11aijb12aijb1laijb21aijb22aijb2laijbk1aijbk2aijbkl] (2.12)

    and the node of tensor is (ti,sj),i=1,2,,m;j=1,2,,n. Then, matrix A and B can be can be changed to (m×n) column vectors as

    t=[t1,,t1,t2,,t2,,tm,,tm]
    s=[s1,s2,,sn,s1,s2,,sn,,s1,s2,,sn],

    and then we get relationship of the partial differential equation and differential matrix as

    k+luktls=C(k)D(l),k,lN. (2.13)

    Consider the generalized elliptic boundary value problems as

    [β(t,s)u(t,s)]=f(t,s),(t,s)Ω (3.1)

    with boundary condition

    u(t,s)=u0(t,s),(t,s)Γ (3.2)

    where β(t,s) is the diffusion coefficient, and =(t,s) is the gradient operator.

    Taking the rectangle domain Ω into two sub-rectangle domains Ωi,i=1,2, the boundary of the domain is Γ, and the boundary of Ωi,i=1,2, is Γ0. Suppose β(t,s)CΩ and the interface conditions of Γ0 are

    [u]Γ=0,[(βu)n]Γ=0.

    Suppose β(t,s) is not continuous on Ω and the interface conditions of Γ0 are

    [u]Γ=δ(t,s),[(βu)n]Γ=γ(t,s).

    In the following, we take the two sub-domain Ωi,i=1,2, (t1,i,s1,j), the function u1,ij=u(t1,i,s1,j); (t2,i,s2,j),i=1,2,,m2;j=1,2,,n2 and u2,ij=u(t2,i,s2,j).

    On the sub-domain of Ω1, the barycentrix function is defined as

    u(t,s)=m1i=1n1j=1R1,i(t)R1,j(s)u1,ij (3.3)

    where R1,i(t),R1,j(s) are defined as (2.3) and (2.4).

    Equation (3.1) can be written as

    β(2ut2+2us2)+βtut+βsus=f(t,s),(t,s)Ω (3.4)

    Taking Eq (3.3) into Eq (3.4), we have

    β(t,s)(m1i=1n1j=1R1,i(t)R1,j(s)u1,ij+m1i=1n1j=1R1,i(t)R1,j(s)u1,ij)+β(t,s)tm1i=1n1j=1R1,i(t)R1,j(s)u1,ij+β(t,s)sm1i=1n1j=1R1,i(t)R1,j(s)u1,ij=f(t,s),(t,s)Ω (3.5)

    Taking (t1,i,s1,j) on the sub-domain of Ω1, we have

    β(t1,k,s1,l)(m1i=1n1j=1R1,i(t1,k)R1,j(s1,l)u1,ij+m1i=1n1j=1R1,i(t1,k)R1,j(s1,l)u1,ij)+β(t1,k,s1,l)tm1i=1n1j=1R1,i(t1,k)R1,j(s1,l)u1,ij+β(t1,k,s1,l)sm1i=1n1j=1R1,i(t1,k)R1,j(s1,l)u1,ij=f(t1,k,s1,l),(t,s)Ω. (3.6)

    As we have used

    R1,i(t1,k)=δki,R1,j(s1,l)=δljR1,i(t1,k)=C1(1)ki,R1,j(s1,l)=D1(1)ljR1,i(t1,k)=C1(2)ki,R1,j(s1,l)=D1(2)lj (3.7)

    Take the notation

    B=diag(β(t1,1,s1,1),β(t1,1,s1,2),,β(t1,1,s1,n1),,β(t1,m1,s1,1),β(t1,m1,s1,2),,β(t1,m1,s1,n1)), (3.8)
    B11=diag(βt(t1,1,s1,1),βt(t1,1,s1,2),,βt(t1,1,s1,n1),,βt(t1,m1,s1,1),βt(t1,m1,s1,2),,βt(t1,m1,s1,n1)), (3.9)
    B12=diag(βs(t1,1,s1,1),βs(t1,1,s1,2),,βs(t1,1,s1,n1),,βs(t1,m1,s1,1),βs(t1,m1,s1,2),,βs(t1,m1,s1,n1)), (3.10)
    F1=diag(f(t1,1,s1,1),f(t1,1,s1,2),,f(t1,1,s1,n1),,f(t1,m1,s1,1),f(t1,m1,s1,2),,f(t1,m1,s1,n1)), (3.11)
    U1=diag(u(t1,1,s1,1),u(t1,1,s1,2),,u(t1,1,s1,n1),,u(t1,m1,s1,1),u(t1,m1,s1,2),,u(t1,m1,s1,n1)). (3.12)

    The matrix equation of (3.5) can be written as

    [B1(C1(2)In1+Im1D1(2))+B11(C1(1)In1)+B12(Im1D1(2))]U1=F1 (3.13)

    where C1(1),C1(2),D1(1),D1(2) are the one order and two order differential matrices, and Im1,In1 are the identity matrices. Then, we write

    L1U1=F1. (3.14)

    Similarly in the sub-domain Ω2, we get the matrix equation

    [B2(C2(2)In2+Im2D2(2))+B21(C2(1)In2)+B22(Im2D2(2))]U2=F2, (3.15)

    and

    L2U2=F2. (3.16)

    Combining Eq (3.14) and Eq (3.16), we get

    [L200L2][U1U2]=[F1F2]. (3.17)

    Then, we have

    LU=F, (3.18)

    and

    L=[L200L2],U=[U1U2],F=[F1F2]. (3.19)

    Points of the boundary are 2(m1+m22)+n1+n2. Sb denotes the number of the domain, and boundary points are denoted as (tbk,sbk),kSb. The boundary condition can be discrete, as

    ekm1n1+m2n2U=u0(tbk,sbk)
    u(t1i,l,s1i,l)=m1i=1n1j=1R1,i(ti,l)R1,j(si,l)u1,ij (3.20)
    u(t2i,l,s2i,l)=m2i=1n2j=1R2,i(t2i,l)R2,j(s2i,l)u2,ij (3.21)
    ut(t1i,l,s1i,l)=m1i=1n1j=1R1,i(ti,l)R1,j(si,l)u1,ij (3.22)
    ut(t2i,l,s2i,l)=m2i=1n2j=1R2,i(t2i,l)R2,j(s2i,l)u2,ij (3.23)
    us(t1i,l,s1i,l)=m1i=1n1j=1R1,i(ti,l)R1,j(si,l)u1,ij (3.24)
    us(t2i,l,s2i,l)=m2i=1n2j=1R2,i(t2i,l)R2,j(s2i,l)u2,ij (3.25)

    where l=0,1,,m0, and

    (t1i,l,s1i,l)=(t2i,l,s2i,l)=(ti,l,si,l).

    Define

    R1,i(ti,l)=[R1,0(ti,l),R1,1(ti,l),,R1,m1(ti,l)] (3.26)
    R1,i(si,l)=[R1,0(si,l),R1,1(si,l),,R1,n1(si,l)] (3.27)
    R2,i(ti,l)=[R2,0(ti,l),R2,1(ti,l),,R2,m1(ti,l)] (3.28)
    R2,i(si,l)=[R2,0(si,l),R2,1(si,l),,R2,n1(si,l)] (3.29)
    Rt1,i(ti,l)=[R1,0(ti,l),R1,1(ti,l),,R1,m1(ti,l)] (3.30)
    Rs1,i(si,l)=[R1,0(si,l),R1,1(si,l),,R1,n1(si,l)] (3.31)
    Rt2,i(ti,l)=[R2,0(ti,l),R2,1(ti,l),,R2,m1(ti,l)] (3.32)
    Rs2,i(si,l)=[R2,0(si,l),R2,1(si,l),,R2,n1(si,l)]. (3.33)

    Take as the matrix equation

    u(t1i,l,s1i,l)=(L1M1)U (3.34)
    u(t2i,l,s2i,l)=(L2M2)U (3.35)
    ut(t1i,l,s1i,l)=(Lt1,lM1,1)U (3.36)
    ut(t2i,l,s2i,l)=(Lt2,lM2,l)U (3.37)
    us(t1i,l,s1i,l)=(L1,lMs1,1)U (3.38)
    us(t2i,l,s2i,l)=(L2,lMs2,1)U. (3.39)

    The discrete boundary condition condition can be given as

    [u](t1i,l,s1i,l)=u(t2i,l,s2i,l)u(t1i,l,s1i,l)=δ(ti,l,si,l) (3.40)
    [(βu)n](t1i,l,s1i,l)=[ut(t2i,l,s2i,l)ntlus(t2i,l,s2i,l)nsl] (3.41)
    [ut(t1i,l,s1i,l)ntlus(t1i,l,s1i,l)nsl]=γ(ti,l,si,l). (3.42)

    The matrix equations of (3.40) and (3.41) are

    [L2,lM2,lL1,lM1,l]U=δ(ti,l,si,l) (3.43)
    β2[(Lt2,lM2,l)ntl(L2,lMs2,l)nsl]Uβ1[(Lt1,lM1,l)ntl(L1,lMs1,l)nsl]U=γ(ti,l,si,l) (3.44)

    Example 1. Consider

    2u+u=f

    with f(t,s)=t22; its analytic solutions are

    u(t,s)=t2+es

    where Ω=[1,1]×[1,1].

    In Table 1 convergence rate is O(hd1+1) with d1=d2=2,3,4,5. In Table 2, for the Chebyshev nodes, the convergence rate is O(τd2+2) with d1=d2=2,3,4,5.

    Table 1.  Errors of equidistant nodes with d1=d2.
    m×n d1=d2=1 2 3 4 5
    8×8 2.6911e-02 1.6610e-03 7.1154e-04 7.2691e-05 4.0731e-05
    16×16 8.0799e-03 2.3171e-04 4.5299e-05 2.9892e-06 6.1418e-07
    32×32 2.4157e-03 2.9843e-05 2.8178e-06 1.0109e-07 9.4081e-09
    64×64 7.1483e-04 3.7641e-06 1.7520e-07 3.2724e-09 3.8841e-10
    hα 1.7448 2.9285 3.9959 4.8130 5.5594

     | Show Table
    DownLoad: CSV
    Table 2.  Errors of non-equidistant nodes with d1=d2.
    m×n d1=d2=1 2 3 4 5
    8×8 6.2451e-02 1.5475e-03 5.5808e-04 1.8735e-05 2.2295e-06
    16×16 1.5433e-02 1.1791e-04 9.1048e-06 4.3131e-07 7.8660e-08
    32×32 3.5514e-03 6.8688e-06 2.0908e-07 4.9400e-09 3.2320e-10
    64×64 8.4301e-04 3.9526e-07 4.5490e-09 4.9775e-09 4.3803e-09
    hα 2.0703 3.9783 5.6349 3.9593 2.9971

     | Show Table
    DownLoad: CSV

    Figure 1 shows the error estimate of equidistant nodes, and Figure 2 shows the error estimate of Chebyshev nodes.

    Figure 1.  Error estimate of equidistant nodes with m=20;n=20;d1=d2=9.
    Figure 2.  Error estimate of Chebyshev nodes with m=20;n=20;d1=d2=9.

    Example 2. Consider

    2u+u=f

    with f(t,s)=3sin(t+s). Its analytic solutions are

    u(t,s)=sin(t+s)

    where Ω=[1,1]×[1,1].

    Table 3 shows the convergence is O(hd1+1) with d1=d2=2,3,4,5. In Table 4, for the non-uniform partition with Chebyshev nodes for d1=d2=2,3,4,5, the convergence rate is O(τd2+2).

    Table 3.  Errors of equidistant nodes with d1=d2.
    m×n d1=d2=1 2 3 4 5
    8×8 5.0201e-03 1.0360e-03 3.7701e-04 4.8362e-05 2.5361e-05
    16×16 1.7960e-03 1.4558e-04 2.1324e-05 1.9874e-06 3.1977e-07
    32×32 6.0280e-04 1.8624e-05 1.2263e-06 6.5505e-08 4.2192e-09
    64×64 1.9404e-04 2.3331e-06 7.2859e-08 2.0682e-09 2.8110e-10
    hα 1.5644 2.9315 4.1124 4.8377 5.4871

     | Show Table
    DownLoad: CSV
    Table 4.  Errors of Chebyshev nodes with d1=d2.
    m×n d1=d2=1 2 3 4 5
    8×8 1.3288e-02 9.9717e-04 1.1162e-04 7.8012e-06 1.8963e-06
    16×16 2.8908e-03 6.3908e-05 4.7613e-06 3.3689e-07 4.0815e-08
    32×32 6.2445e-04 4.1969e-06 1.0285e-07 3.7866e-09 1.4989e-10
    64×64 1.4493e-04 2.4827e-07 2.4237e-09 1.4696e-09 1.8797e-09
    hα 2.1729 3.9906 5.1637 4.1247 3.3262

     | Show Table
    DownLoad: CSV

    We choose m=20;n=20;d1=9;d2=9 to test our algorithm.

    Figure 3 shows the error estimate of equidistant nodes, and Figure 4 shows the error estimate of Chebyshev nodes.

    Figure 3.  Error estimate of equidistant nodes with m=20;n=20;d1=d2=9.
    Figure 4.  Error estimate of Chebyshev nodes with m=20;n=20;d1=d2=9.

    Example 3. Consider the Poisson equation u=2sin(πt)cos(πs),(t,s)Ω and Ω=Ω1Ω2={t,s:1<t<1,1<s<1}{t,s:0<t<1,0<s<1}. Its analytic solutions are

    u(t,s)=sin(πt)cos(πs)

    with the boundary condition

    [u]Γ0=0,[ut]Γ0=0.

    We choose m1=m2=20;n1=n2=20;d1=9;d2=9 to test the domain decomposition method of the barycentric rational collocation method. Figure 5 shows the errors under equidistant nodes. From Figure 5 we know that the error can reach 107 with 21 collocation points.

    Figure 5.  Error estimate of equidistant nodes with m1=m2=20;n1=n2=20;d1=d2=9.

    Example 4. Consider u(t,s)=6ts(t2s22),(t,s)Ω and Ω=Ω1Ω2={t,s:1<t<1,1<s<1}{t,s:0.5<t<0.5,1<s<0}. Its analytic solutions are

    u(t,s)=ts(t21)(s21)

    with condition

    [u]Γ0=0,[ut]Γ0=0.

    We choose m1=m2=20;n1=n2=20;d1=9;d2=9 on each Ωi,i=1,2, to test the domain decomposition method of the barycentric rational collocation method. Figure 6 shows the error estimate of equidistant nodes, and Figure 7 shows the error estimate of Chebyshev nodes. The error of both equidistant nodes and Chebyshev nodes can reach 1011, which shows the accuracy of our algorithm.

    Figure 6.  Error estimate of equidistant nodes with m1=m2=20;n1=n2=20;d1=d2=9.
    Figure 7.  Error estimate of Chebyshev nodes with m1=m2=20;n1=n2=20;d1=d2=9.

    The work of Jin Li was supported by the Natural Science Foundation of Shandong Province (Grant No. ZR2022MA003) and Natural Science Foundation of Hebei Province (Grant No. A2019209533).

    The authors also gratefully acknowledge the helpful comments and suggestions of the reviewers, which have improved the presentation.

    The authors declare that there are no conflicts of interest.



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