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Linear barycentric rational collocation method to solve plane elasticity problems


  • A linear barycentric rational collocation method for equilibrium equations with polar coordinates is considered. The discrete linear equations is changed into the matrix forms. With the help of error of barycentrix polar coordinate interpolation, the convergence rate of the linear barycentric rational collocation method for equilibrium equations can be obtained. At last, some numerical examples are given to valid the proposed theorem.

    Citation: Jin Li. Linear barycentric rational collocation method to solve plane elasticity problems[J]. Mathematical Biosciences and Engineering, 2023, 20(5): 8337-8357. doi: 10.3934/mbe.2023365

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  • A linear barycentric rational collocation method for equilibrium equations with polar coordinates is considered. The discrete linear equations is changed into the matrix forms. With the help of error of barycentrix polar coordinate interpolation, the convergence rate of the linear barycentric rational collocation method for equilibrium equations can be obtained. At last, some numerical examples are given to valid the proposed theorem.



    Equilibrium equations in elasticity are classical equations to solve plane problems. There are displacement and stress methods that are used to solve equilibrium equations. The displacement method takes displacement as an unknown quantity, as there is only the displacement component to deduce the equations and boundary conditions. For the stress method, there is only the stress component to deduce the equations and boundary conditions as the unknown quantities.

    In the area of in-plane crack problems, heat transfer, nuclear reactor dynamics and so on, the impact of system memory is often dependent on the nonlinear fraction equation and nonlinear time-dependent Burgers' equations. These problems have been studied by using the Galerkin finite element method [1], localized collocation schemes [2] and singular boundary method (SBM) [3]. Lots of numerical methods such as finite element methods (FEM)[4,5,6,7], finite difference methods, spectral method [8,9] and the differential quadrature method and so on are developed to solve plane elastic problems [10,11].

    In what follows, we consider the equilibrium equations

    {σxx+τyxy+fx=0,σyy+τxyx+fy=0 (1.1)

    where σx,σy and τyx are the stress components.

    Geometric equations:

    {ϵx=ux,ϵy=vy,γyz=wy+vz,ϵij=12(ui,j+uj,i) (1.2)

    where ϵx,ϵx and τyx are the strain components and u and v are displacement variables.

    Constitutive relations of plane stress problem:

    {ϵx=1E(σxμσy),ϵy=1E(σyμσx),γxy=2(1+μ)Eτxy. (1.3)

    Constitutive relations of the plane strain problem:

    {ϵx=1μ2E(σxμ1μσy),ϵy=1μ2E(σyμ1μσx),γxy=2(1+μ)Eτxy. (1.4)

    Displacement boundary equation:

    u|Γu=ˉu,v|Γu=ˉv,ui|Γu=ˉui (1.5)

    Stress boundary equations:

    {(n1σx+n2τyx)Γσ=ˉtx,(n2σy+n1τxy)Γσ=ˉty,njσij=ˉti. (1.6)

    Combining Eqs (1.4) and (1.2), we have the stress components of displacement:

    {σx=E1μ2(ux+μvy),σy=E1μ2(vy+μux),τxy=E2(1+μ)(uy+vx) (1.7)

    and the equilibrium equations:

    {E1μ2(2ux2+1μ22vy2+1+μ22vxy)+fx=0,E1μ2(2vy2+1μ22ux2+1+μ22vxy)+fy=0 (1.8)

    and the displacement boundary equations:

    {E1μ2[(ux+μvy)n1+1μ2(uy+vx)n2]Γσ=ˉtx,E1μ2[(vy+μux)n2+1μ2(uy+vx)n1]Γσ=ˉty. (1.9)

    Barycentric formulae have been studied by [12,13,14,15,16,17,18] to avoid the Runge phenomenon. Volterra equations (VE) and Volterra Integro-Differential equations (VIDE) [19,20,21,22,23] have been investigated through the use of linear barycentric collocation methods (LBCMs). The LBCM types include the linear barycentric Lagrange collocation method (LBLCM) and linear barycentric rational collocation method (LBRCM). By comparing the with LBLCM and LBRCM, we can get the error estimate of linear rational barycentric interpolation; then, the convergence rate of the LBRCM can be obtained. Initial value and boundary value problems[24], plane elasticity problems [25], incompressible plane problems [26] and non linear problems [27] have been the focus of barycentric interpolation and rational collocation method in recent years. In previous studies [28,29], heat conduction and telegram equations were solved by LBRCM. In other studies [30,31], biharmonic equation and fractional differential equations were solved by using the LBRCM.

    In this paper, first, the polar coordinates of the equilibrium equations are obtained via the transformation of x=ρcosθ,y=ρsinθ. Second, the LBRCM for equilibrium equations is constructed and the matrix equation of the LBRCM is also presented. Third, the convergence rate of LBRCM is proved for the equilibrium equations. At last, some numerical examples are given to validate the proposed theorem.

    In order to get the polar coordinates of the equilibrium equations, let us take x=ρcosθ,y=ρsinθ and (ρ,θ) at some point P(ρ,θ); the displacement components are uρ and uθ, stress components are σρ, σθ and τθr and the physical components are fρ and fθ. The equilibrium equations of the polar coordinates can be represented as:

    {σρρ+σρσθρ+1ρτρθθ+fρ=0,1ρσθθ+τρθρ+2τρθρ+fθ=0. (2.1)

    Geometric equations:

    {ϵρ=uρρ,ϵθ=uρρ+1ρuθθ,γρθ=γθρ=1ρuρθ+uθρuθρ. (2.2)

    Constitutive relations of plane stress problem:

    {ϵρ=1E(σρμσθ),ϵθ=1E(σθμσρ),γρθ=2(1+μ)Eτρθ. (2.3)

    Combining Eqs (2.1)–(2.3), the displacement of the equilibrium equations for the plane stress problem is expressed as:

    {2uρρ2+1ρuρρuρρ2+1+μ2ρ2uθρθ3μ2ρ2uθθ+1μ2ρ22uρθ2+fρ=0,1+μ2ρ2uρρθ+3μ2ρ2uρθ+1μ2(2uθρ2+1ρuθρuθρ2)+1ρ22uθθ2+fθ=0. (2.4)

    The displacement of th stress components is expressed as follows:

    {σρ=E1μ2[uρρ+μ(uρρ+1ρuθθ)],σθ=E1μ2(μuρρ+uρρ+1ρuθθ),τρθ=E2(1+μ)(uθρuθρ+1ρuρθ) (2.5)

    where we have used

    x=ρcosθ,y=ρsinθ; (2.6)

    then we have

    {σρ=σxcos2θ+σysin2θ+2τxysinθcosθ,σθ=σxsin2θ+σycos2θ2τxysinθcosθ,τρθ=(σxσy)cosθsinθ+τxy(cos2θsin2θ) (2.7)

    and

    {σx=σρcos2θ+σθsin2θ+2τρθsinθcosθ,σy=σρsin2θ+σθcos2θ2τρθsinθcosθ,τxy=(σρσθ)cosθsinθ+τρθ(cos2θsin2θ) (2.8)

    with the equilibrium condition, we get ϕ(r,θ) below:

    {σρ=1ρϕρ+1ρ22ϕθ2,σθ=2ϕρ2,τρθ=1ρ2ϕθ1ρ2ϕρθ (2.9)

    and

    22ϕ=(2ρ2+1ρρ+1ρ22θ2)2ϕ=0 (2.10)

    where 2ϕ=(2ρ2+1ρρ+1ρ22θ2) is the Laplace operator.

    We partition the area [ρa,ρb]×[θ0,θ2π] into ρa=ρ0<ρ1<<ρm=ρb,h=ρbρam and [θa,θb] into θ0=θ0<θ1<<θn=θ2π,τ=θ2πθ0n with [ρa,ρb]×[θ0,θ2π] and (ρi,θj),i=0,1,,m;j=0,1,,,n.

    ϕ(ρ,θ):=rm,n(ρ,θ)=mi=0nj=0ri(ρ)rj(θ)ϕij (3.1)

    where ri(ρ) and rj(θ) are the barycentric rational interpolation basis functions [24] for ρ and θ, respectively, which can be given as

    ri(ρ)=wiρρinj=0wjρρj, wi=kJi(1)kk+d1j=k,ji1ρkρj,i=0,1,2,,n (3.2)

    Ji={kIm:id1ki},Im={0,,md1}, and

    rj(θ)=wjθθjnj=0wjθθj, wj=kJj(1)kk+d2i=k,ji1θkθi,j=0,1,2,,n, (3.3)

    Jj={kIn: jd2kj},In={0,,nd2}.

    Combining Eqs (3.1), (2.4) and (2.5), we get the discrete equilibrium equations, which be expressed as

    {mi=0nj=0ri(ρ)rj(θ)ϕρij+1ρmi=0nj=0ri(ρ)rj(θ)ϕρij1ρ2mi=0nj=0ri(ρ)rj(θ)ϕρij+1+μ2rmi=0nj=0ri(ρ)rj(θ)ϕθij+3μ2ρ2mi=0nj=0ri(ρ)rj(θ)ϕθij+1μ2ρ2mi=0nj=0ri(ρ)rj(θ)ϕθij+fρij(ρi,θj)=0,1+μ2ρmi=0nj=0ri(ρ)rj(θ)ϕρij+3μ2ρ2mi=0nj=0ri(ρ)rj(θ)ϕρij+1μ2(mi=0nj=0ri(ρ)rj(θ)ϕθij+1ρmi=0nj=0ri(ρ)rj(θ)ϕθij1ρ2mi=0nj=0ri(ρ)rj(θ)ϕθij)+1ρ2mi=0nj=0ri(ρ)rj(θ)ϕθij+fθij(ρi,θj)=0 (3.4)

    and

    {σρ=E1μ2[mi=0nj=0ri(ρ)rj(θ)ϕρij+μ(1ρmi=0nj=0ri(ρ)rj(θ)ϕρij+1ρmi=0nj=0ri(ρ)rj(θ)ϕθij)],σθ=E1μ2(μmi=0nj=0ri(ρ)rj(θ)ϕρij+1ρmi=0nj=0ri(ρ)rj(θ)ϕθij+1ρmi=0nj=0ri(ρ)rj(θ)ϕρij),τρθ=E2(1+μ)(mi=0nj=0ri(ρ)rj(θ)ϕρij1ρmi=0nj=0ri(ρ)rj(θ)ϕθij+1ρmi=0nj=0ri(ρ)rj(θ)ϕθij). (3.5)

    Equations (3.4) and (3.5) can be written in matrix form

    {[(R(2,0)In)+diag(1ρ)(R(1,0)In)diag(1ρ2)(ImIn)]Uρ+[diag(1+μ2r)(R(1,0)R(0,1))+diag(3μ2ρ2)(ImR(0,1))+diag(1μ2ρ2)(InR(0,2))]Uθ+Fρ=0,[diag(1+μ2ρ)(R(1,0)R(0,1))+diag(3μ2ρ2)(ImR(0,1))]Uρ+[1μ2((R(2,0)In)+diag1ρ(R(1,0)In)diag1ρ2(ImIn))+diag1ρ2(R(2,0)In)]Uθ+Fθ=0 (3.6)

    and

    {σσρ=E1μ2[{(R(1,0)In)+μdiag(1ρ)(ImIn)}Uρ+μdiag1ρ(ImR(0,1))Uθ],σσθ=E1μ2[{μ(R(1,0)In)+diag(1ρ)(ImIn)}Uρ+diag1ρ(ImR(0,1))Uθ],ττρθ=E2(1+μ)[diag(1ρ)(ImR(0,1))Uρ+[diag(1ρ)(ImIn)+(R(1,0)In)]Uθ] (3.7)

    where is the Kronecher product of the matrix and R(0,k)=(R(0,k)ij)m×m, R(k,0)=(R(k,0)ij)n×n,k=1,2, U=[u00,u01,,u0n,u10,u11,,u1n,,um0,um1,,umn]T, Fρ=[f00,f01,,f0n,f10,f11,,f1n,,fm0,fm1,,fmn]T, fij=ρ2if(ρi,θj) and

    R(0,1)ij=ri(θj),R(0,2)ij=ri(θj),R(1,0)ij=ri(ρj),R(2,0)ij=ri(ρj). (3.8)

    Taking the notations as

    {A11=(R(2,0)In)+diag(1ρ)(R(1,0)In)diag(1ρ2)(ImIn)A12=diag(1+μ2r)(R(1,0)R(0,1))+diag(3μ2ρ2)(ImR(0,1))+diag(1μ2ρ2)(InR(0,2)),A21=diag(1+μ2ρ)(R(1,0)R(0,1))+diag(3μ2ρ2)(ImR(0,1))A22=1μ2((R(2,0)In)+diag1ρ(R(1,0)In)diag1ρ2(ImIn))+diag1ρ2(R(2,0)In), (3.9)

    then we have

    {A11Uρ+A21Uθ+Fρ=0,A21Uρ+A22Uθ+Fρ=0. (3.10)

    Replacing the barycentric rational interpolants of the function u(ρ,θ) with rm,n(ρ,θ) in Eq (3.1), we have

    rm,n(ρ,θ)=mi=0nj=0wi,j(ρρi)(θθj)ui,jmi=0nj=0wi,j(ρρi)(θθj), (4.1)

    where

    wi,j=(1)id1+jd2k1Jik2Jjk1+d1h1=k1,h1j1|ρiρh1|k2+d2h2=k2,h2j1|θjθh2|. (4.2)

    Then the error function is defined as

    e(ρ,θ):=u(ρ,θ)rm,n(ρ,θ)=(ρρi)(ρρi+d1)u[ρi,ρi+1,,ρi+d1,ρ]+(θθj)(θθj+d2)u[θj,θj+1,,θj+d2,θ]. (4.3)

    Now we give the theorem as below

    Theorem 1. For e(ρ,θ) defined in Eq(4.3) and u(ρ,θ)Cd1+2[0,ρ]×Cd2+2[0,θ], we have

    |e(ρ,θ)|C(hd1+1+τd2+1). (4.4)

    {Proof}. For (ρ,θ), the function wi,j(ρ,θ) is defined as Eq (4.2); then, we get

    u(ρ,θ)rm,n(ρ,θ)=md1i=0nd2j=0λi(ρ)λj(θ)(u(ρ,θ)rn(ρ,θ))md1i=0nd2j=0λi(ρ)λj(θ), (4.5)

    where

    λi(ρ)=(1)i(ρρi)(ρρi+d1),λj(θ)=(1)j(θθj)(θθj+d2)

    see [24].

    By the error formula

    u(ρ,θ)rm,n(ρ,θ)=u(ρ,θ)u1(ρ,θ)+u1(ρ,θ)rm,n(ρ,θ)=(ρρi)(ρρi+d1)u[ρi,ρi+1,,ρi+d1,ρ,θ]+(θθj)(θθj+d2)u[θj,θj+1,,θj+d2,ρ,θ]. (4.6)

    it follows that

    u(ρ,θ)rm,n(ρ,θ)=md1i=0(1)iu[ρi,ρi+1,,ρi+d1,ρ,θ]md1i=0λi(ρ)+nd2j=0(1)ju[θj,θj+1,,θj+d2,ρ,θ]nd2j=0λj(θ). (4.7)

    By a similar method of analysis as that of Floater and Kai [15], we have

    |md1i=0λi(ρ)|1d1!hd1+1 (4.8)

    and

    |nd2j=0λj(θ)|1d2!τd2+1. (4.9)

    Combining Eqs (4.7)–(4.9) together, the proof of Theorem 1 is completed.

    Corollary 1. For e(ρ,θ) defined in (4.3), we have

    {|eρ(ρ,θ)|C(hd1+τd2+1),u(ρ,θ)Cd1+3[ρa,ρb]×Cd2+2[θ0,θ2π],|eθ(ρ,θ)|C(hd1+1+τd2),u(ρ,θ)Cd1+2[ρa,ρb]×Cd2+3[θ0,θ2π],|eρρ(ρ,θ)|C(hd11+τd2+1),u(ρ,θ)Cd1+4[ρa,ρb]×Cd2+2[θ0,θ2π],d12.|eθθ(ρ,θ)|C(hd1+1+τd21),u(ρ,θ)Cd1+2[ρa,ρb]×Cd2+4[θ0,θ2π],d12. (4.10)

    This corollary can be obtained similarly as Theorem 1, so it is omitted here.

    Theorem 2. Let

    2ϕρ2+1ρϕρ+1ρ22ϕθ2=0,(ρ,θ)Ω (4.11)

    and

    ϕ(ρ,θ)=g(ρ,θ),(ρ,θ)Ω (4.12)

    where Ω=[ρa,ρb]×[θ0,θ2π] and g(ρ,θ) is consistent. Then we get

    max (4.13)

    where u(\rho, \theta) \in C^{d_{1}+4}[\rho_a, \rho_b]\times C^{d_{2}+4}[\theta_{0}, \theta_{2\pi}], d_{1} \ge 2, d_{2} \ge 2.

    Theorem 3. Let

    \begin{equation} \nabla^2\nabla^2\phi = \left(\frac{\partial^2 }{\partial \rho^2} +\frac{1}{\rho}\frac{\partial }{\partial \rho} +\frac{1}{\rho^2}\frac{\partial^2 }{\partial \theta^2}\right)^2\phi = 0,(\rho,\theta) \in \Omega \end{equation} (4.14)

    and

    \begin{equation} \phi(\rho,\theta) = g(\rho,\theta), (\rho,\theta) \in \partial \Omega \end{equation} (4.15)

    where \Omega = [\rho_a, \rho_b] \times[\theta_{0}, \theta_{2\pi}] g(\rho, \theta) is consistent and

    \begin{equation} \max \limits_{\Omega_{kl}}|\phi_{i,j}-\phi(\rho_{i},\theta_{j})| \le C (h^{d_{1}-3}+\tau^{d_{2}-3}) \end{equation} (4.16)

    also, \Omega_{kl} = [\rho_k, \rho_{k+1}] \times[\theta_{l} and \theta_{l+1}], \phi(\rho, \theta) \in C^{d_{1}+6}[\rho_a, \rho_b]\times C^{d_{2}+6}[\theta_{0}, \theta_{2\pi}], d_{1} \ge 4, d_{2} \ge 4.

    Proof. Let \phi(\rho, \theta) and \phi_{i, j} be the analysis solution and numerical solution of Eq (4.14) respectively:

    \begin{eqnarray} \begin{array}{ll} &\nabla^2\nabla^2\phi(\rho,\theta)- \nabla^2\nabla^2\phi(\rho_{i},\theta_{j}) \\& = \frac{\partial^4 \phi(\rho,\theta)}{\partial \rho^4} +\frac{\partial^2 \phi(\rho,\theta)}{\rho^2 \partial \rho^2} + \frac{2}{\rho^2}\frac{\partial^3 \phi(\rho,\theta)}{ \partial \rho^3}+ \frac{1}{\rho^4}\frac{\partial^4 \phi(\rho,\theta)}{ \partial \theta^4}+ \frac{2}{\rho^2}\frac{\partial^4 \phi(\rho,\theta)}{ \partial \rho^2\partial \theta^2} + \frac{2}{\rho^3}\frac{\partial^3 \phi(\rho,\theta)}{ \partial \rho \partial \theta^2} \\& -\left[\frac{\partial^4 \phi(\rho_{i},\theta_{j})}{\partial \rho^4} +\frac{\partial^2 \phi(\rho_{i},\theta_{j})}{\rho^2 \partial \rho^2} + \frac{2}{\rho^2}\frac{\partial^3 \phi(\rho_{i},\theta_{j})}{ \partial \rho^3}+ \frac{1}{\rho^4}\frac{\partial^4 \phi(\rho_{i},\theta_{j})}{ \partial \theta^4}+ \frac{2}{\rho^2}\frac{\partial^4 \phi(\rho_{i},\theta_{j})}{ \partial \rho^2\partial \theta^2} + \frac{2}{\rho^3}\frac{\partial^3 \phi(\rho_{i},\theta_{j})}{ \partial \rho \partial \theta^2}\right] \\& = \left[\frac{\partial^4 \phi(\rho,\theta)}{\partial \rho^4}-\frac{\partial^4 \phi(\rho_{i},\theta_{j})}{\partial \rho^4}\right]+ \left[\frac{\partial^2 \phi(\rho,\theta)}{\rho^2 \partial \rho^2}-\frac{\partial^2 \phi(\rho_{i},\theta_{j})}{\rho^2 \partial \rho^2} \right] \\&+ \left[\frac{1}{\rho^4}\frac{\partial^4 \phi(\rho,\theta)}{\partial \rho^4}-\frac{1}{\rho^4}\frac{\partial^4 \phi(\rho_{i},\theta_{j})}{\partial \rho^4}\right]+ \left[\frac{2}{\rho^2}\frac{\partial^3 \phi(\rho,\theta)}{ \partial \rho^3}-\frac{2}{\rho^2}\frac{\partial^3 \phi(\rho_{i},\theta_{j})}{ \partial \rho^3} \right] \\&+ \left[\frac{2}{\rho^2}\frac{\partial^4 \phi(\rho,\theta)}{ \partial \rho^2\partial \theta^2}- \frac{2}{\rho^2}\frac{\partial^4 \phi(\rho_{i},\theta_{j})}{ \partial \rho^2\partial \theta^2}\right]+ \left[\frac{2}{\rho^3}\frac{\partial^3 \phi(\rho,\theta)}{ \partial \rho \partial \theta^2}- \frac{2}{\rho^3}\frac{\partial^3 \phi(\rho_{i},\theta_{j})}{ \partial \rho \partial \theta^2} \right] \\&: = R_{1}(\rho,\theta)+R_{2}(\rho,\theta)+R_{3}(\rho,\theta) \\&+R_{4}(\rho,\theta)+R_{5}(\rho,\theta)+R_{6}(\rho,\theta) \end{array} \end{eqnarray} (4.17)

    where

    R_{1}(\rho,\theta) = \frac{\partial^4 \phi(\rho,\theta)}{\partial \rho^4}-\frac{\partial^4 \phi(\rho_{i},\theta_{j})}{\partial \rho^4},
    R_{2}(\rho,\theta) = \frac{\partial^2 \phi(\rho,\theta)}{\rho^2 \partial \rho^2}-\frac{\partial^2 \phi(\rho_{i},\theta_{j})}{\rho^2 \partial \rho^2},
    R_{3}(\rho,\theta) = \frac{1}{\rho^4}\frac{\partial^4 \phi(\rho,\theta)}{\partial \theta^4}-\frac{1}{\rho^4}\frac{\partial^4 \phi(\rho_{i},\theta_{j})}{\partial \theta^4}
    R_{4}(\rho,\theta) = \frac{2}{\rho^2}\frac{\partial^3 \phi(\rho,\theta)}{ \partial \rho^3}-\frac{2}{\rho^2}\frac{\partial^3 \phi(\rho_{i},\theta_{j})}{ \partial \rho^3} ,
    R_{5}(\rho,\theta) = \frac{2}{\rho^2}\frac{\partial^4 \phi(\rho,\theta)}{ \partial \rho^2\partial \theta^2}- \frac{2}{\rho^2}\frac{\partial^4 \phi(\rho_{i},\theta_{j})}{ \partial \rho^2\partial \theta^2},
    R_{6}(\rho,\theta) = \frac{2}{\rho^3}\frac{\partial^3 \phi(\rho,\theta)}{ \partial \rho \partial \theta^2}- \frac{2}{\rho^3}\frac{\partial^3 \phi(\rho_{i},\theta_{j})}{ \partial \rho \partial \theta^2}

    for R_{1}(\rho, \theta) , we have

    \begin{equation} \label{eq3-13} \begin{array}{ll} R_{1}(\rho,\theta) = \frac{\partial^4 \phi(\rho,\theta)}{\partial \rho^4}-\frac{\partial^4 \phi(\rho_{i},\theta_{j})}{\partial \rho^4} \\ = \frac{\partial^4 \phi(\rho,\theta)}{\partial \rho^4}-\frac{\partial^4 \phi(\rho,\theta_{j})}{\partial \rho^4}+\frac{\partial^4 \phi(\rho,\theta_{j})}{\partial \rho^4}-\frac{\partial^4 \phi(\rho_{i},\theta_{j})}{\partial \rho^4} \\ = \frac{ \sum\limits_{i = 0}^{m-d_{1}} (-1)^{i}\frac{\partial^4 \phi}{\partial \rho^4}[\rho_{i}, \rho_{i+1}, \ldots, \rho_{i+d_1}, \rho,\theta]} { \sum\limits_{i = 0}^{m-d_{1}} \lambda_{i}(\rho)} \\ \nonumber+ \frac{ \sum\limits_{j = 0}^{n-d_{2}}(-1)^{j}\frac{\partial^4 \phi}{\partial \rho^4}[ \theta_{j}, \theta_{j+1}, \ldots, \theta_{j+d_2}, \rho_{i},\theta]} { \sum\limits_{j = 0}^{n-d_{2}}\lambda_{j}(\theta)} \\ = \frac{\partial^4 e (\rho,\theta_{j})}{\partial \rho^4}+\frac{\partial^4 e(\rho_{i},\theta_{j})}{\partial \rho^4}, \end{array} \end{equation}

    where

    \begin{equation} |R_{1}(\rho,\theta)|\le\left|\frac{\partial^4 e (\rho,\theta_{j})}{\partial \rho^4}+\frac{\partial^4 e(\rho_{i},\theta_{j})}{\partial \rho^4}\right|\le C (h^{d_{1}-3}+\tau^{d_{2}-3}). \end{equation} (4.18)

    For R_{2}(\rho, \theta) , we have

    \begin{equation} \begin{array}{ll} R_{2}(\rho,\theta) = \frac{\partial^2 \phi(\rho,\theta)}{\rho^2 \partial \rho^2}-\frac{\partial^2 \phi(\rho_{i},\theta_{j})}{\rho^2 \partial \rho^2} \\ = \frac{\partial^2 \phi(\rho,\theta)}{\rho^2 \partial \rho^2}-\frac{\partial^2 \phi(\rho,\theta_{j})}{\rho^2 \partial \rho^2} +\frac{\partial^2 \phi(\rho,\theta_{j})}{\rho^2 \partial \rho^2}-\frac{\partial^2 \phi(\rho,\theta_{j})}{\rho^2 \partial \rho^2} \\ = e_{\rho\rho}(\theta,\theta_{n})+e_{\rho\rho}(\rho_{m},\theta_{n}) \end{array} \end{equation} (4.19)

    and

    \begin{equation} |R_{2}(\rho,\theta)|\le|e_{\rho\rho}(\theta,\theta_{i})+e_{\rho\rho}(\rho_{i},\theta_{j})|\le C (h^{d_{1}-1}+\tau^{d_{2}-1}). \end{equation} (4.20)

    For R_{3}(\rho, \theta) we have

    \begin{equation} \begin{array}{ll} R_{3}(\rho,\theta) = \frac{1}{\rho^4}\frac{\partial^4 \phi(\rho,\theta)}{\partial \theta^4}-\frac{1}{\rho^4}\frac{\partial^4 \phi(\rho_{i},\theta_{j})}{\partial \theta^4} = \frac{\partial^4 e (\rho,\theta_{i})}{\partial \theta^4}+\frac{\partial^4 e(\rho_{i},\theta_{j})}{\partial \theta^4} \end{array} \end{equation} (4.21)

    and

    \begin{equation} |R_{3}(\rho,\theta)|\le \left|\frac{\partial^4 e (\rho,\theta_{i})}{\partial \theta^4}+\frac{\partial^4 e(\rho_{i},\theta_{j})}{\partial \theta^4} \right|\le C (h^{d_{1}-3}+\tau^{d_{2}-3}). \end{equation} (4.22)

    Similarly, for R_{4}(\rho, \theta) , R_{5}(\rho, \theta) and R_{6}(\rho, \theta) , we also get

    \begin{equation} |R_{4}(\rho,\theta)|\le \left|\frac{2}{\rho^2}\frac{\partial^3 \phi(\rho,\theta)}{ \partial \rho^3}-\frac{2}{\rho^2}\frac{\partial^3 \phi(\rho_{i},\theta_{j})}{ \partial \rho^3} \right|\le C (h^{d_{1}-2}+\tau^{d_{2}-2}), \end{equation} (4.23)
    \begin{equation} |R_{5}(\rho,\theta)|\le \left|\frac{2}{\rho^2}\frac{\partial^4 \phi(\rho,\theta)}{ \partial \rho^2\partial \theta^2}- \frac{2}{\rho^2}\frac{\partial^4 \phi(\rho_{i},\theta_{j})}{ \partial \rho^2\partial \theta^2} \right|\le C (h^{d_{1}-3}+\tau^{d_{2}-3}), \end{equation} (4.24)
    \begin{equation} |R_{6}(\rho,\theta)|\le \left|\frac{2}{\rho^3}\frac{\partial^3 \phi(\rho,\theta)}{ \partial \rho \partial \theta^2}- \frac{2}{\rho^3}\frac{\partial^3 \phi(\rho_{i},\theta_{j})}{ \partial \rho \partial \theta^2} \right|\le C (h^{d_{1}-2}+\tau^{d_{2}-2}). \end{equation} (4.25)

    Combining Eqs (4.18) and (4.19)–(4.25), the proof of Theorem 4.3 is completed.

    In the following part, we present some examples to illustrate our numerical scheme analysis.

    We define the absolute error estimate and relative error estimate as

    Er = \max \{u_e-u_a\}

    and

    Err = \frac{\max \{u_e-u_a\}}{u_e} .

    Example 1. Consider the following elastic polar curved bar bending

    \begin{equation} \left \lbrace \begin{array}{lll} u_{\rho} = \frac{\sin \theta}{E}\left[D(1-\mu)\ln \rho+A(1-3\mu)\rho^2+\frac{B(1+\mu)}{\rho^2} \right] \\ -\frac{2D}{E}\theta \cos\theta+K\sin \theta+L\cos \theta, \\ u_{\theta} = -\frac{\cos \theta}{E}\left[-D(1-\mu)\ln \rho+A(5+\mu)+\frac{B(1+\mu)}{\rho^2} \right] \\ +\frac{2D}{E}\theta \sin\theta+\left[\frac{D(1+\mu)}{E}+K\right]\cos \theta-L\sin \theta \end{array}\right. \end{equation} (5.1)

    and

    \begin{equation} \left \lbrace \begin{array}{lll} \sigma_{\rho} = \left( 2A\rho-\frac{2B}{\rho^2}+\frac{D}{\rho}\right)\sin \theta, \\ \sigma_{\theta} = \left( 6A\rho+\frac{2B}{\rho^2}+\frac{D}{\rho}\right)\sin \theta, \\ \tau_{\rho\theta} = -\left( -2A\rho-\frac{2B}{\rho^2}+\frac{D}{\rho}\right)\cos \theta, \end{array}\right. \end{equation} (5.2)

    where A = \frac{P}{2N}, B = -\frac{Pa^2b^2}{2N}, D = -\frac{P}{N}(a^2+b^2), L = \frac{D\pi}{E}, N = a^2-b^2+(a^2+b^2)\ln\frac{a}{n} , K = -\frac{1}{E}\left[D(1-\mu)\ln \rho_{0}+A(1-3\mu)\rho^2_{0}+\frac{B(1+\mu)}{\rho^2_{0}} \right], \rho_{0} = \frac{a+b}{2}, a < \theta < b and 0 < \theta < \frac{\pi}{2} and the boundary conditions are given as \sigma_{\rho}|_{\rho = a} = 0, \sigma_{\rho}|_{\rho = b} = 0, \tau_{\rho \theta}|_{\rho = a} = 0, \tau_{\rho \theta}|_{\rho = b} = 0, \sigma_{ \theta} = 0, \int_{a}^{b}\tau_{\rho \theta}d\rho = P, \theta = 0, u_{\rho} = 0, u_{\theta} = 0, \theta = \pi/2 and

    \sigma_{\theta} = 0,\tau_{\rho \theta} = -\left( -2A\rho-\frac{2B}{\rho^2}+\frac{D}{\rho}\right),\theta = 0,
    u_{\rho} = 0,\tau_{\rho \theta} = \frac{1}{E}\left[D(1-\mu)\ln \rho+A(1-3\mu)\rho^2+\frac{B(1+\mu)}{\rho^2} \right]+K,\theta = \frac{\pi}{2}
    u_{\theta} = \frac{2D}{E} \sin\theta+L.

    In Tables 1 and 2, the error estimates of displacement and stress are presented for the barycentric rational interpolation collocation methods (BRICMs) with d = 5 and barycentric Lagrange interpolation collocation methods with n = 11 and n = 19 . From the table, the displacement and stress have higher accuracy for the Lagrange interpolation collocation methods than for the BRICMs.

    Table 1.  Error estimate of barycentric rational interpolation collocation methods with d = 5 .
    n Function Equidistant nodes Quasi-equidistant nodes
    Absolute error Relative error Absolute error Relative error
    u_{\rho} 1.0843e-07 5.2465e-04 3.9398e-10 1.9063e-06
    11 \sigma_{\rho} 4.9419e-01 2.4710e-04 1.0401e-01 5.2007e-05
    \sigma_{\theta} 2.1686e+00 5.9143e-04 2.9761e-02 8.1168e-06
    u_{\rho} 1.0468e-08 5.0653e-05 5.5667e-12 2.6935e-08
    19 \sigma_{\rho} 5.5156e-02 2.7578e-05 1.0824e-02 5.4119e-06
    \sigma_{\theta} 2.0936e-01 5.7099e-05 3.2952e-03 8.9868e-07

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    Table 2.  Error estimate of Lagrange interpolation collocation methods.
    n Function Equidistant nodes Quasi-equidistant nodes
    Absolute error Relative error Absolute error Relative error
    u_{\rho} 1.4060e-09 6.8033e-06 2.0507e-11 9.9229e-08
    11 \sigma_{\rho} 6.1776e-03 3.0888e-06 2.4225e-04 1.2112e-07
    \sigma_{\theta} 2.8120e-02 7.6692e-06 4.1015e-04 1.1186e-07
    u_{\rho} 2.3788e-13 1.1510e-09 1.4206e-15 6.8738e-12
    19 \sigma_{\rho} 3.2498e-06 1.6249e-09 1.8951e-08 9.4753e-12
    \sigma_{\theta} 3.5000e-06 9.5454e-10 2.8475e-08 7.7659e-12

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    In Tables 3 and 4, the error estimates of \sigma_{\rho} and \sigma_{\theta} are presented for barycentric rational interpolation with d_{1} = d_{2} = 2, 3, 4, 5 for equidistant nodes.

    Table 3.  Errors of equidistant nodes with d_{1} for \sigma_{\rho} .
    n d_{1}=2 d_{1}=3 d_{1}=4 d_{1}=5
    8 3.8722e+00 1.8512e+00 6.7278e-02 3.9574e-02
    16 1.4909e+00 1.3770 2.5799e-01 2.8430 6.2873e-03 3.4196 1.3624e-03 4.8604
    32 6.0715e-01 1.2961 3.3272e-02 2.9549 4.5096e-04 3.8014 4.4118e-05 4.9486
    64 2.0902e-01 1.5384 4.2086e-03 2.9829 2.9866e-05 3.9164 1.3797e-06 4.9990

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    Table 4.  Errors of equidistant nodes with d_{2} for \sigma_{\theta} .
    n d_{2}=2 d_{2}=3 d_{2}=4 d_{2}=5
    8 2.4659e+02 1.2345e+02 5.0487e+00 3.1449e+00
    16 1.4900e+02 7.2685e-01 3.1157e+01 1.9863 8.4074e-01 2.5862 1.9203e-01 4.0336
    32 8.3557e+01 8.3445e-01 7.7866e+00 2.0005 1.1597e-01 2.8578 1.1921e-02 4.0097
    64 4.4684e+01 9.0301e-01 1.9452e+00 2.0011 1.5115e-02 2.9398 7.4421e-04 4.0016

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    In Tables 5 and 6, the error estimates of \sigma_{\rho} and \sigma_{\theta} are presented for barycentric rational interpolation with d_{1} = d_{2} = 2, 3, 4, 5 for quasi-equidistant nodes.

    Table 5.  Errors of quasi-equidistant nodes with \sigma_{\rho} .
    n d_{1}=2 d_{1}=3 d_{1}=4 d_{1}=5
    8 7.6984e+00 3.0925e+00 2.7111e-02 7.8893e-03
    16 2.5097e+00 1.6170 1.2113e-01 4.6742 1.6515e-03 4.0370 3.3223e-04 4.5696
    32 7.5343e-01 1.7360 1.3255e-02 3.1919 7.3991e-05 4.4803 4.6671e-06 6.1535
    64 2.0590e-01 1.8715 1.0018e-03 3.7259 6.6256e-06 3.4812 3.8948e-06

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    Table 6.  Errors of quasi-equidistant nodes with d_{2} for \sigma_{\theta} .
    n d_{2}=2 d_{2}=3 d_{2}=4 d_{2}=5
    8 1.6510e+02 1.1570e+02 2.1174e+00 1.0292e+00
    16 1.0058e+02 7.1503e-01 3.4378e+00 5.0727 5.6098e-02 5.2382 1.3486e-02 6.2539
    32 5.1539e+01 9.6454e-01 5.7939e-01 2.5689 1.6043e-03 5.1280 8.7592e-05 7.2665
    64 2.0298e+01 1.3443e+00 6.4441e-02 3.1685 1.2028e-04 3.7375 8.0433e-05

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    In Tables 711, the errors of u_{\rho} , u_{\theta} , \sigma_{\rho} , \sigma_{\theta} and \tau_{r\theta} are shown for barycentric rational interpolation with d = 2, 3, 4, 5 for equidistant nodes. The convergence rate is O(h^{d}) for u_{\rho} , u_{\theta} , \sigma_{\rho} and \sigma_{\theta} , and O(h^{d-1}) for \tau_{\rho \theta} which agrees with our theorem analysis.

    Table 7.  Errors of equidistant nodes with d_{1} = d_{2} for u_{\rho} .
    m\times n d_{1}=d_{2}=2 d_{1}=d_{2}=3 d_{1}=d_{2}=4 d_{1}=d_{2}=5
    8\times8 4.6386e-01 2.3538e-01 7.3939e-02 4.3353e-02
    16\times16 2.2488e-01 1.0445 3.7954e-02 2.6326 5.0343e-03 3.8765 8.9588e-04 5.5967
    32\times32 7.1140e-02 1.6604 4.5791e-03 3.0511 3.0493e-04 4.0452 2.1490e-05 5.3816
    64\times64 1.8606e-02 1.9349 5.4216e-04 3.0783 1.8514e-05 4.0418 5.8452e-07 5.2003

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    Table 8.  Errors of equidistant nodes with d_{1} = d_{2} for u_{\theta} .
    m\times n d_{1}=d_{2}=2 d_{1}=d_{2}=3 d_{1}=d_{2}=4 d_{1}=d_{2}=5
    8\times8 3.8459e-01 2.0080e-01 6.4882e-02 3.7789e-02
    16\times16 1.9199e-01 1.0023 3.3038e-02 2.6036 4.3719e-03 3.8915 7.7682e-04 5.6042
    32\times32 6.1740e-02 1.6368 4.0051e-03 3.0442 2.6542e-04 4.0419 1.8712e-05 5.3756
    64\times64 1.6255e-02 1.9253 4.7537e-04 3.0747 1.6149e-05 4.0387 5.1045e-07 5.1960

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    Table 9.  Errors of equidistant nodes with d for \sigma_{\rho} .
    m\times n d_{1}=d_{2}=2 d_{1}=d_{2}=3 d_{1}=d_{2}=4 d_{1}=d_{2}=5
    8\times8 2.4305e+04 9.1855e+03 3.3907e+03 1.9465e+03
    16\times16 9.0087e+03 1.4319 1.6141e+03 2.5087 2.2601e+02 3.9071 4.0143e+01 5.5996
    32\times32 3.1667e+03 1.5084 2.0801e+02 2.9560 1.4637e+01 3.9487 1.3872e+00 4.8549
    64\times64 9.2345e+02 1.7778 2.8337e+01 2.8759 1.0162e+00 3.8484 7.2075e-02 4.2665

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    Table 10.  Errors of equidistant nodes with d for \sigma_{\theta} .
    m\times n d_{1}=d_{2}=2 d_{1}=d_{2}=3 d_{1}=d_{2}=4 d_{1}=d_{2}=5
    8\times8 6.7329e+04 3.6992e+04 1.3543e+04 7.7757e+03
    16\times16 3.6203e+04 0.8951 6.4833e+03 2.5124 9.0279e+02 3.9070 1.6079e+02 5.5957
    32\times32 1.2708e+04 1.5104 8.3482e+02 2.9572 5.8440e+01 3.9494 4.1882e+00 5.2627
    64\times64 3.7039e+03 1.7786 1.1076e+02 2.9140 3.9820e+00 3.8754 1.4185e-01 4.8839

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    Table 11.  Errors of equidistant nodes with d for \tau_{\rho\theta} .
    m\times n d_{1}=d_{2}=2 d_{1}=d_{2}=3 d_{1}=d_{2}=4 d_{1}=d_{2}=5
    8\times8 4.7320e+03 2.8512e+03 1.0721e+03 6.4853e+02
    16\times16 3.8711e+03 0.2897 7.7859e+02 1.8726 1.1139e+02 3.2667 2.0975e+01 4.9504
    32\times32 1.8997e+03 1.0270 1.3801e+02 2.4961 9.7341e+00 3.5165 7.2237e-01 4.8598
    64\times64 6.7700e+02 1.4886 2.1544e+01 2.6795 7.6752e-01 3.6648 2.5826e-02 4.8058

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    Example 2. Consider the the following elastic thick circular:

    \begin{equation} \left \lbrace \begin{array}{lll} \sigma_{\rho} = \frac{a^2P_{a}-b^2P_{b}}{b^2-a^2}+\frac{a^2b^2(P_{b}-P_{a})}{b^2-a^2}\frac{1}{\rho^2} \\ \sigma_{\rho} = \frac{a^2P_{a}-b^2P_{b}}{b^2-a^2}-\frac{a^2b^2(P_{b}-P_{a})}{b^2-a^2}\frac{1}{\rho^2} \end{array}\right. \end{equation} (5.3)

    with

    \begin{equation} u_{\rho} = \frac{(1-\mu)(a^2P_{a}-b^2P_{b})}{E(b^2-a^2)}+\frac{(1+\mu)a^2b^2(P_{b}-P_{a})}{E(b^2-a^2)\rho}. \end{equation} (5.4)

    Then we get the displacement equation as

    \begin{equation} \frac{\texttt{d}^2u_{\rho}}{\texttt{d}\rho^2}+\frac{1}{\rho}\frac{\texttt{d}u_{\rho}}{\texttt{d}\rho}-\frac{1}{\rho^2}u_{\rho} = 0,a < \rho < b \end{equation} (5.5)

    and the boundary conditions can be given as

    \sigma_{\rho}(a) = -P_{a},\sigma_{\rho}(b) = -P_{b},

    which means that

    \begin{equation} \frac{E}{1-\mu^2}\left(\frac{\partial u_{\rho} }{\partial {\rho}}+\mu\frac{ u_{\rho} }{{\rho}} \right)_{\rho = a} = -P_{a}, \frac{E}{1-\mu^2}\left(\frac{\partial u_{\rho} }{\mu\partial {\rho}}+\frac{ u_{\rho} }{{\rho}} \right)_{\rho = b} = -P_{b}, \end{equation} (5.6)

    and the matrix equations can be given as

    \begin{eqnarray} \left[{R}^{(2,0)} +{\mathrm{diag}}(\frac{1}{\rho}) {R}^{(1,0)}+{\mathrm{\mathrm{diag}}}(\frac{1}{\rho^2})\right] U_{\rho} = {0} \end{eqnarray} (5.7)

    and

    \begin{equation} \left \lbrace \begin{array}{lll} \sigma_{\rho} = \frac{E}{1-\mu^2}\left( {R}^{(1,0)}+\mu {\mathrm{diag}}(\frac{1}{\rho}) \right)U_{\rho}, \\ \sigma_{\theta} = \frac{E}{1-\mu^2} \left( \mu{R}^{(1,0)}+ {\mathrm{diag}}(\frac{1}{\rho} \right)U_{\rho} \end{array}\right. \end{equation} (5.8)

    with a = 0.5 m, b = 1 m, P_{a} = 1000 Pa, P_{b} = 2000 Pa, E = 10^7 Pa, \mu = 0.3.

    In Tables 12 and 13, the error estimates of the BRICM with d = 5 and Lagrange interpolation collocation methods with n = 11 and n = 19 for displacement and stress are given.

    Table 12.  Error estimates of the BRICM with different d with d = 5 .
    n Function Equidistant nodes Quasi-equidistant nodes
    Absolute error Relative error Absolute error Relative error
    u_{\rho} 2.9925e-02 9.2382e-03 1.7356e-04 4.9117e-05
    u_{\theta} 1.7418e-02 9.9419e-03 9.7870e-05 4.7756e-05
    11 \sigma_{\rho} 4.7749e+02 2.1424e-02 8.8496e+00 4.7244e-04
    \sigma_{\theta} 2.8427e+03 9.4039e-03 2.3377e+01 6.5748e-05
    \tau_{{\rho}\theta} 2.7594e+02 1.2381e-02 4.3209e+00 2.3067e-04
    u_{\rho} 2.9642e-03 5.5736e-04 6.5569e-07 1.1227e-07
    u_{\theta} 1.6923e-03 6.0049e-04 2.1714e-07 6.4625e-08
    19 \sigma_{\rho} 3.8746e+01 1.0246e-03 2.2016e-01 6.9270e-06
    \sigma_{\theta} 2.6988e+02 5.5473e-04 1.7981e+00 3.0837e-06
    \tau_{{\rho}\theta} 2.8528e+01 7.5436e-04 4.6880e-01 1.4750e-05

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    Table 13.  Error estimates of Lagrange interpolation collocation methods.
    n Function Equidistant nodes Quasi-equidistant nodes
    Absolute error Relative error Absolute error Relative error
    u_{\rho} 5.7368e-05 1.7710e-05 2.1383e-09 6.0515e-10
    u_{\theta} 3.1838e-05 1.8173e-05 1.3071e-09 6.3780e-10
    11 \sigma_{\rho} 4.4327e+00 1.9889e-04 3.2005e-03 1.7086e-07
    \sigma_{\theta} 5.4486e+00 1.8024e-05 3.7800e-04 1.0631e-09
    \tau_{{\rho}\theta} 4.0339e-01 1.8099e-05 5.6771e-05 3.0307e-09
    u_{\rho} 5.3104e+00 9.9853e-01 9.0233e-09 1.5450e-09
    u_{\theta} 2.8335e+00 1.0054e+00 5.5352e-09 1.6474e-09
    19 \sigma_{\rho} 6.3820e+05 1.6876e+01 5.9085e-03 1.8591e-07
    \sigma_{\theta} 4.8026e+05 9.8716e-01 1.9612e-03 3.3635e-09
    \tau_{{\rho}\theta} 4.7011e+04 1.2431e+00 1.2233e-03 3.8490e-08

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    In Tables 1416, the errors of barycentric rational interpolation, i.e., u_{\rho} , \sigma_{\rho} and \sigma_{\theta} are shown with d = 2, 3, 4, 5 for equidistant nodes and O(h^{d}) for u_{\rho} , \sigma_{\rho} and \sigma_{\theta} which agrees with our theorem analysis.

    Table 14.  Errors of equidistant nodes with d_{1} for u_{\rho} .
    n d_{1}=2 d_{1}=3 d_{1}=4 d_{1}=5
    8 6.9993e-06 3.3694e-06 1.0401e-06 6.6611e-07
    16 1.7434e-06 2.0053 4.4148e-07 2.9320 9.2244e-08 3.4952 2.9354e-08 4.5041
    32 4.3685e-07 1.9967 5.7223e-08 2.9477 6.7371e-09 3.7753 1.1405e-09 4.6858
    64 1.0946e-07 1.9967 7.2943e-09 2.9717 4.5402e-10 3.8913 4.0016e-11 4.8330

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    Table 15.  Errors of equidistant nodes with d_{1} for \sigma_{\rho} .
    n d_{1}=2 d_{1}=3 d_{1}=4 d_{1}=5
    8 2.8959e+01 1.3156e+01 3.9232e+00 2.4593e+00
    16 9.5402e+00 1.6019 2.3041e+00 2.5135 4.6863e-01 3.0655 1.4678e-01 4.0665
    32 2.6910e+00 1.8258 3.3753e-01 2.7711 3.8783e-02 3.5949 6.4731e-03 4.5031
    64 7.1232e-01 1.9176 4.5531e-02 2.8901 2.7689e-03 3.8080 2.4068e-04 4.7492

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    Table 16.  Errors of equidistant nodes with d_{2} for \sigma_{\theta} .
    n d_{2}=2 d_{2}=3 d_{2}=4 d_{2}=5
    8 6.7387e+01 1.3999e+02 2.0802e+01 1.3322e+01
    16 8.8297e+00 2.0053 3.4869e+01 2.9320 1.8449e+00 3.4952 5.8709e-01 4.5041
    32 1.1445e+00 1.9967 8.7371e+00 2.9477 1.3474e-01 3.7753 2.2810e-02 4.6858
    64 1.4589e-01 1.9967 2.1892e+00 2.9717 9.0804e-03 3.8913 8.0031e-04 4.8330

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    In Tables 1719, for quasi-equidistant nodes, the errors of u_{\rho} , \sigma_{\rho} and \sigma_{\theta} are given for the barycentric rational interpolation with d_{1} = d_{2} = 2, 3, 4, 5 . The convergence rate is O(h^{2d-2}) for u_{\rho} and O(h^{d}) for \sigma_{\rho} and \sigma_{\theta} which coincides with our theorem analysis.

    Table 17.  Errors of quasi-equidistant nodes with d_{1} for u_{\rho} .
    n d_{1}=2 d_{1}=3 d_{1}=4 d_{1}=5
    8 5.1242e-07 4.7200e-08 5.6060e-08 4.4966e-08
    16 5.7993e-08 3.1434 4.6281e-09 3.3503 5.2101e-10 6.7495 5.7807e-11 9.6034
    32 7.9035e-09 2.8753 1.8128e-10 4.6741 2.8497e-12 7.5144 1.5231e-13 8.5681
    64 7.2603e-10 3.4444 5.1410e-13 8.4620 1.3194e-12 1.1110 6.6684e-12 -

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    Table 18.  Errors of quasi-equidistant nodes with d_{1} for \sigma_{\rho} .
    n d_{1}=2 d_{1}=3 d_{1}=4 d_{1}=5
    8 2.3800e+01 6.5238e+00 1.0394e+00 5.2459e-01
    16 4.1561e+00 2.5176 7.9468e-01 3.0373 1.3660e-01 2.9276 3.1504e-02 4.0576
    32 8.5118e-01 2.2877 8.1116e-02 3.2923 7.5261e-03 4.1819 9.8814e-04 4.9947
    64 1.9227e-01 2.1464 8.8700e-03 3.1930 5.2740e-04 3.8349 5.4116e-04 0.8686

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    Table 19.  Errors of quasi-equidistant nodes with d_{2} for \sigma_{\theta} .
    n d_{2}=2 d_{2}=3 d_{2}=4 d_{2}=5
    8 1.0685e+01 2.6150e+00 1.1212e+00 8.9933e-01
    16 1.9955e+00 2.4207 2.8961e-01 3.1746 4.3436e-02 4.6900 1.0093e-02 6.4774
    32 3.5851e-01 2.4767 2.6615e-02 3.4438 2.2886e-03 4.2463 2.9545e-04 5.0943
    64 6.6816e-02 2.4237 2.6669e-03 3.3190 1.5324e-04 3.9007 2.3041e-04 -

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    In this paper, first, the equilibrium equations were transformed into polar coordinates; then, the LBRCM was presented to solve the equilibrium equations. Third, the matrix equations of the equilibrium equations were obtained and the convergence rate of the LBRCM was also proved. At last, some numerical examples were given to validate the proposed theorem. The plane elasticity problems under the irregular domain can also be solved by using the LBRCM, as will be discussed it in the near future.

    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 author declares that there are no conflicts of interest.



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