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Research article

Systems of two-dimensional complex partial differential equations for bi-polyanalytic functions

  • Received: 27 June 2024 Revised: 21 August 2024 Accepted: 27 August 2024 Published: 06 September 2024
  • MSC : 30C45, 32A30

  • A class of Schwarz problems with the conditions concerning the real and imaginary parts of high-order partial differentiations for polyanalytic functions was discussed first on the bicylinder. Then, with the particular solution to the Schwarz problem for polyanalytic functions, a Dirichlet problem for bi-polyanalytic functions was investigated on the bicylinder. From the perspective of series, the specific representation of the solution was obtained. In this article, a novel and effective method for solving boundary value problems, with the help of series expansion, was provided. This method can also be used to solve other types of boundary value problems or complex partial differential equation problems of other functions in high-dimensional complex spaces.

    Citation: Yanyan Cui, Chaojun Wang. Systems of two-dimensional complex partial differential equations for bi-polyanalytic functions[J]. AIMS Mathematics, 2024, 9(9): 25908-25933. doi: 10.3934/math.20241265

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  • A class of Schwarz problems with the conditions concerning the real and imaginary parts of high-order partial differentiations for polyanalytic functions was discussed first on the bicylinder. Then, with the particular solution to the Schwarz problem for polyanalytic functions, a Dirichlet problem for bi-polyanalytic functions was investigated on the bicylinder. From the perspective of series, the specific representation of the solution was obtained. In this article, a novel and effective method for solving boundary value problems, with the help of series expansion, was provided. This method can also be used to solve other types of boundary value problems or complex partial differential equation problems of other functions in high-dimensional complex spaces.



    Partial differential equations are closely related to many physical problems in real life. For example, the ninth-order linear or non-linear boundary value problems are related to the laminar viscous flow in a semi-porous channel or the hydro-magnetic stability, and the telegraph equations are related to the vibrations within objects or the propagation of waves. There are also many different types of partial differential equations in engineering and other applied sciences. Zhang et al. [1] discussed cubic spline solutions of ninth-order linear and non-linear boundary value problems using a cubic B-spline. Shah et al. [2] proposed a new and efficient operational matrix method for solving time-fractional telegraph equations with Dirichlet boundary conditions. Nisar et al. [3] proposed a hybrid mesh free framework based on Padˊe approximation in order to solve the numerical solutions of nonlinear partial differential equations. There are also many different types of partial differential equations in chemistry, engineering, and other applied sciences. There have been many successful conclusions about these partial differential equations.

    Complex partial differential equations of analytic functions also have a wide range of applications. Bi-analytic functions, the generalizations of analytic functions, have important applications in elasticity. In 1961, Sander [4] studied the properties of pairs of functions {u(x,y),v(x,y)} with binary real variables (x,y), which satisfy the system of partial differential equations:

    {uxvy=θ,uy+vx=ω,(k+1)θx+ωy=0,(k+1)θyωx=0,

    for the real constant k(k1), where θ(x,y) and ω(x,y) are continuously differentiable functions of x and y. Sander provided the definition of bi-analytic functions of type k, which are of great significance for studying some physical problems for k>0, and extended some properties of analytic functions to bi-analytic functions.

    In 1965, Lin and Wu [5] introduced the function class that is more extensive than Sander's function class, i.e., bi-analytic functions of the type (λ,k) which are defined by the system of equations:

    {1kuxvy=θ,uy+1kvx=ω,kθx+λωy=0,kθyλωx=0,

    where θ(x,y) and ω(x,y) are continuously differentiable functions of x and y, and λ,k are real constants with λ0,1,k2, and 0<k<1. The complex form of the system is

    k+12fˉzk12fz=λk4λφ(z)+λ+k4λ¯φ(z),

    in which φ(z)=kϑiλω is analytic and is called the associate function of f(z)=u+iv. In [5], the general expression and the properties of bi-analytic functions of the type (λ,k) were researched in detail.

    Hua et al. [6] introduced a mechanical interpretation for bi-analytic functions and promoted the corresponding function theory. Thereafter, bi-analytic functions aroused widespread attention from many scholars [7,8,9]. In 1994, Kumar [10] discussed a broader class of functions, i.e., bi-polyanalytic functions, and investigated several Riemann-Hilbert problems for systems of n-order partial differential equations applying polyanalytic functions [11] and bi-polyanalytic functions on the unit disk. In 2005, Kumar and Prakash [12] investigated Dirichlet problems for the Poisson equation and some boundary value problems for bi-polyanalytic functions on the unit disk. They obtained the explicit representations of the solutions and the corresponding solvable conditions. In 2006, Begehr and Kumar [13] discussed some complex partial differential equations of higher order. Some boundary value problems for bi-polyanalytic functions were solved on different conditions on the unit disk.

    In recent years, some other boundary value problems for bi-analytic functions were solved [14,15,16,17]. With the gradual improvements of the theory for bi-analytic functions and polyanalytic functions [18,19,20,21] in the complex plane, some scholars attempted to generalize the relevant achievements to spaces of several complex variables [22,23].

    In this paper, based on the work of the former researchers, we study a class of Schwarz problems for polyanalytic functions on the bicylinder. Then, from the perspective of series and applying the particular solution to the Schwarz problem for polyanalytic functions, we discuss a Dirichlet problem for bi-polyanalytic functions on the bicylinder.

    In the following, let the bicylinder D2=D1×D2={(z1,z2):|z1|<1,|z2|<1}, and let 0D2 denote the characteristic boundary of D2. Let C(G) represent the set of continuous functions within G.

    To get the main results, we need to discuss the following Schwarz problem.

    Theorem 2.1. Let gμνC(0D2;R) for 1μ,νm1 (m2), and let

    ˜ϕ(z)=+m1,m2=0m1˜v1=μm1˜v2=νˉz˜v11ˉz˜v22˜v1!˜v2!um~v1,m~v2m1,m2zm11zm22, (2.1)

    where

    um~v1,m~v2m1,m2={1(2πi)20D2mμ1l1=0mν1l2=0g(μ+l1)(ν+l2)(ζ)l1!l2!˜Adζ1dζ2ζ1ζ2,{˜v1=μ˜v2=ν,1(2πi)20D2mμ1l1=0m1˜v2l2=0g(μ+l1)(˜v2+l2)(ζ)l1!l2!˜B|v2=˜v2νdζ1dζ2ζ1ζ2,{˜v1=μν<˜v2m1,1(2πi)20D2m1˜v1l1=0mν1l2=0g(˜v1+l1)(ν+l2)(ζ)l1!l2!˜C|v1=˜v1μdζ1dζ2ζ1ζ2,{μ<˜v1m1˜v2=ν,1(2πi)20D2m1˜v1l1=0m1˜v2l2=0g(˜v1+l1)(˜v2+l2)(ζ)l1!l2!˜D|v1=˜v1μv2=˜v2νdζ1dζ2ζ1ζ2,{μ<˜v1m1ν<˜v2m1, (2.2)

    and

    {˜A=2[m1j1=0m2j2=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22+D21|v1=0m2j2=0Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22+D31|v2=0m1j1=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11](D11D12+D22D23+D32D33)|v1=v2=0+2(ζ1ˉζ1)l1[m2j2=0Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22ˉζm11+B21D31D23+B222B21]|v2=0+2(ζ2ˉζ2)l2[m1j1=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11ˉζm22+D21C21D11+C222C21]|v1=0+2(¯ζ1m1C21+B21¯ζ2m2B21C21)(ζ1ˉζ1)l1(ζ2ˉζ2)l2,˜B=2{[m1j1=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11(ζ1ˉζ1)l1ˉζm11]m2j2=0Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22+D21|v1=0m2j2=0Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22+D31m1j1=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11B21D31}(D11D12+D22|v1=0D23+D32D33B21D23B21B22),
    {˜C=2{m1j1=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11[m2j2=0Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22(ζ2ˉζ2)l2ˉζm22]+D21m2j2=0Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22+D31|v2=0m1j1=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11D21C21}(D11D12+D22D23+D32D33|v2=0D11C21C22C21),˜D=2[m1j1=0m2j2=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22+D21m2j2=0Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22+D31m1j1=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11](D11D12+D22D23+D32D33),

    in which

    D11={Cm1l1(ζ1ˉζ1)l1m1,0m1l1,0,m1>l1,D12={Cm2l2(ζ2ˉζ2)l2m2,0m2l2,0,m2>l2,D21={0,0m1<l1,(1)l1+v1ˉζm1l11,m1l1,D22={(1)m1+v1,m1=l1,0,m1l1,D23={Cm2l2(ζ2ˉζ2)l2m2,0m2l2,0,m2>l2,D31={0,0m2<l2,(1)l2+v2ˉζm2l22,m2l2,D32={Cm1l1(ζ1ˉζ1)l1m1,0m1l1,0,m1>l1,D33={(1)m2+v2,m2=l2,0,m2l2,

    and

    C21={0,m21,1,m2=0,C22={(1)m1+v1,m1=l1,0,m1l1,
    B21={0,m11,1,m1=0,B22={(1)m2+v2,m2=l2,0,m2l2.

    Then, ˜ϕ(z) satisfies

    μˉz1νˉz2˜ϕ(z)=gμν(z)(z0D2),μˉz1νˉz2˜ϕ(0,z2)=0=μˉz1νˉz2˜ϕ(z1,0)(z1D1,z2D2).

    Proof: 1) Let

    ˜ϕ(z)=m1˜v1=μm1˜v2=νˉz˜v11ˉz˜v22˜v1!˜v2!u˜v1˜v2(z), (2.3)

    in which

    u˜v1˜v2(z)={1(2πi)20D2mμ1l1=0mν1l2=0g(μ+l1)(ν+l2)(ζ)l1!l2!(A1A2A3+A4)dζ1dζ2ζ1ζ2,{˜v1=μ˜v2=ν,1(2πi)20D2mμ1l1=0m1˜v2l2=0g(μ+l1)(˜v2+l2)(ζ)l1!l2!(B1B2)|v2=˜v2νdζ1dζ2ζ1ζ2,{˜v1=μν<˜v2m1,1(2πi)20D2m1˜v1l1=0mν1l2=0g(˜v1+l1)(ν+l2)(ζ)l1!l2!(C1C2)|v1=˜v1μdζ1dζ2ζ1ζ2,{μ<˜v1m1˜v2=ν,1(2πi)20D2m1˜v1l1=0m1˜v2l2=0g(˜v1+l1)(˜v2+l2)(ζ)l1!l2!D|v1=˜v1μv2=˜v2νdζ1dζ2ζ1ζ2,{μ<˜v1m1ν<˜v2m1, (2.4)

    is analytic on D2, and

    {A1=[(z1ζ1ˉζ1)l1(z2ζ2ˉζ2)l2+(z1)l1(z2ζ2ˉζ2)l2+(z1ζ1ˉζ1)l1(z2)l2][2ζ1ζ2(ζ1z1)(ζ2z2)1],A2=(ζ1ˉζ1)l1{(z2ζ2ˉζ2)l2[2ζ1ζ2(ζ1z1)(ζ2z2)1]+(z2)l2[2ζ2ζ2z21]},A3=(ζ2ˉζ2)l2{(z1ζ1ˉζ1)l2[2ζ1ζ2(ζ1z1)(ζ2z2)1]+(z1)l1[2ζ1ζ1z11]},A4=(ζ1ˉζ1)l1(ζ2ˉζ2)l2[2ζ1ζ1z1+2ζ2ζ2z22],B1=[(z1ζ1ˉζ1)l1(z2ζ2ˉζ2)l2+(z1)l1(z2ζ2ˉζ2)l2+(z1ζ1ˉζ1)l1(z2)l2(1)v2][2ζ1ζ2(ζ1z1)(ζ2z2)1],B2=(ζ1ˉζ1)l1{(z2ζ2ˉζ2)l2[2ζ1ζ2(ζ1z1)(ζ2z2)1]+(z2)l2(1)v2[2ζ2ζ2z21]},C1=[(z1ζ1ˉζ1)l1(z2ζ2ˉζ2)l2+(z1)l1(1)v1(z2ζ2ˉζ2)l2+(z1ζ1ˉζ1)l1(z2)l2][2ζ1ζ2(ζ1z1)(ζ2z2)1],C2=(ζ2ˉζ2)l2{(z1ζ1ˉζ1)l2[2ζ1ζ2(ζ1z1)(ζ2z2)1]+(z1)l1(1)v1[2ζ1ζ1z11]},D=[(z1ζ1ˉζ1)l1(z2ζ2ˉζ2)l2+(z1)l1(1)v1(z2ζ2ˉζ2)l2+(z1ζ1ˉζ1)l1(z2)l2(1)v2][2ζ1ζ2(ζ1z1)(ζ2z2)1].

    In the following, we will show that ˜ϕ(z) satisfies

    μˉz1νˉz2˜ϕ(z)=gμν(z)(z0D2),μˉz1νˉz2˜ϕ(0,z2)=0=μˉz1νˉz2˜ϕ(z1,0)(z1D1,z2D2).

    By (2.3), we get that

    μˉz1νˉz2˜ϕ(z)=m1˜v1=μm1˜v2=νˉz˜v1μ1ˉz˜v2ν2(˜v1μ)!(˜v2ν)!u˜v1˜v2(z)=m1μv1=0m1νv2=0ˉzv11ˉzv22v1!v2!u(v1+μ)(v2+ν)(z), (2.5)

    where

    u(v1+μ)(v2+ν)={1(2πi)20D2mμ1l1=0mν1l2=0g(μ+l1)(ν+l2)(ζ)l1!l2!(A1A2A3+A4)dζ1dζ2ζ1ζ2,{v1=0v2=0,1(2πi)20D2mμ1l1=0m1v2νl2=0g(μ+l1)(v2+ν+l2)(ζ)l1!l2!(B1B2)dζ1dζ2ζ1ζ2,{v1=00<v2m1ν,1(2πi)20D2m1v1μl1=0mν1l2=0g(v1+μ+l1)(ν+l2)(ζ)l1!l2!(C1C2)dζ1dζ2ζ1ζ2,{0<v1m1μv2=0,1(2πi)20D2m1v1μl1=0m1v2νl2=0g(v1+μ+l1)(v2+ν+l2)(ζ)l1!l2!Ddζ1dζ2ζ1ζ2,{0<v1m1μ0<v2m1ν.

    Therefore, for 1k1,k2m1,

    mk1ˉz1mk2ˉz2˜ϕ(z)=k11v1=0k21v2=0ˉzv11ˉzv22v1!v2!u(v1+mk1)(v2+mk2)(z), (2.6)

    in which

    u(v1+mk1)(v2+mk2)={0D2k11l1=0k21l2=0g(mk1+l1)(mk1+l2)(ζ)(2πi)2l1!l2!(A1A2A3+A4)dζ1dζ2ζ1ζ2,{v1=0v2=0,0D2k11l1=0k21v2l2=0g(mk1+l1)(mk2+v2+l2)(ζ)(2πi)2l1!l2!(B1B2)dζ1dζ2ζ1ζ2,{v1=00<v2k21,0D2k11v1l1=0k21l2=0g(mk1+v1+l1)(mk2+l2)(ζ)(2πi)2l1!l2!(C1C2)dζ1dζ2ζ1ζ2,{0<v1k11v2=0,0D2k11v1l1=0k21v2l2=0g(mk1+v1+l1)(mk2+v2+l2)(ζ)(2πi)2l1!l2!Ddζ1dζ2ζ1ζ2,{0<v1k110<v2k21.

    Let

    ϕ1(z)=ˉzv11ˉzv22v1!v2!1(2πi)20D2k11v1l1=0k21v2l2=0g(mk1+v1+l1)(mk2+v2+l2)(ζ)l1!l2!dζ1dζ2ζ1ζ2.

    For 1k1,k2m1, (2.6) follows that

    mk1ˉz1mk2ˉz2˜ϕ(z)=0v1=00v2=0ϕ1(z)(A1A2A3+A4)+0v1=0k21v2=1ϕ1(z)(B1B2)+k11v1=10v2=0ϕ1(z)(C1C2)+k11v1=1k21v2=1ϕ1(z)D=k11v1=0k21v2=0ϕ1(z)D0v1=0k21v2=0ϕ1(z)B2k11v1=00v2=0ϕ1(z)C2+0v1,v2=0ϕ1(z)A4=1(2πi)20D2k11v1=0k11v1l1=0k21v2=0k21v2l2=0ˉzv11ˉzv22v1!v2!g(mk1+v1+l1)(mk2+v2+l2)(ζ)l1!l2![(z1ζ1ˉζ1)l1(z2ζ2ˉζ2)l2+(z1)l1(1)v1(z2ζ2ˉζ2)l2+(z1ζ1ˉζ1)l1(z2)l2(1)v2][2ζ1ζ2(ζ1z1)(ζ2z2)1]dζζ1(2πi)20D2k11λ1=0k21v2=0k21v2l2=0ˉzv22v2!g(mk1+λ1)(mk2+v2+l2)(ζ)λ1!l2!(ζ1ˉζ1)λ1{(z2ζ2ˉζ2)l2[2ζ1ζ2(ζ1z1)(ζ2z2)1]+(z2)l2(1)v2[2ζ2ζ2z21]}dζ1dζ2ζ1ζ21(2πi)20D2k11v1=0k11v1l1=0k21λ2=0ˉzv11v1!g(mk1+v1+l1)(mk2+λ2)(ζ)l1!λ2!(ζ2ˉζ2)λ2{(z1ζ1ˉζ1)l1[2ζ1ζ2(ζ1z1)(ζ2z2)1]+(z1)l1(1)v1[2ζ1ζ1z11]}dζ1dζ2ζ1ζ2+0D2k11λ1=0k21λ2=0g(mk1+λ1)(mk2+λ2)(ζ)(2πi)2λ1!λ2!(ζ1ˉζ1)λ1(ζ2ˉζ2)λ2[2ζ1ζ1z1+2ζ2ζ2z22]dζζ=1(2πi)20D2k11λ1=0k21λ2=0λ1v1=0λ2v2=0ˉzv11ˉzv22λ1!λ2!Cv1λ1Cv2λ2g(mk1+λ1)(mk2+λ2)[(z1ζ1ˉζ1)λ1v1(z2ζ2ˉζ2)λ2v2+(z1)λ1v1(1)v1(z2ζ2ˉζ2)λ2v2+(z1ζ1ˉζ1)λ1v1(z2)λ2v2(1)v2][2ζ1ζ2(ζ1z1)(ζ2z2)1]dζζ1(2πi)20D2k11λ1=0k21λ2=0(ζ1ˉζ1)λ1λ1!λ2!λ2v2=0Cv2λ2ˉzv22g(mk1+λ1)(mk2+λ2)(ζ){(z2ζ2ˉζ2)λ2v2[2ζ1ζ2(ζ1z1)(ζ2z2)1]+(z2)λ2v2(1)v2[2ζ2ζ2z21]}dζ1dζ2ζ1ζ21(2πi)20D2k11λ1=0k21λ2=0(ζ2ˉζ2)λ2λ1!λ2!λ1v1=0Cv1λ1ˉzv11g(mk1+λ1)(mk2+λ2)(ζ){(z1ζ1ˉζ1)λ1v1[2ζ1ζ2(ζ1z1)(ζ2z2)1]+(z1)λ1v1(1)v1[2ζ1ζ1z11]}dζ1dζ2ζ1ζ2+0D2k11λ1=0k21λ2=0g(mk1+λ1)(mk2+λ2)(ζ)(2πi)2λ1!λ2!(ζ1ˉζ1)λ1(ζ2ˉζ2)λ2[2ζ1ζ1z1+2ζ2ζ2z22]dζ1dζ2ζ1ζ2=1(2πi)20D2k11λ1=0k21λ2=01λ1!λ2!{λ1v1=0Cv1λ1(z1ζ1ˉζ1)λ1v1ˉzv11λ2v2=0Cv2λ2(z2ζ2ˉζ2)λ2v2ˉzv22(ζ1ˉζ1)λ1λ2v2=0Cv2λ2(z2ζ2ˉζ2)λ2v2ˉzv22+λ1v1=0Cv1λ1(z1)λ1v1(ˉz1)v1λ2v2=0Cv2λ2(z2ζ2ˉζ2)λ2v2ˉzv22λ1v1=0Cv1λ1(z1ζ1ˉζ1)λ1v1ˉzv11(ζ2ˉζ2)λ2+λ1v1=0Cv1λ1(z1ζ1ˉζ1)λ1v1ˉzv11λ2v2=0Cv2λ2(z2)λ2v2(ˉz2)v2}g(mk1+λ1)(mk2+λ2)(ζ)[2ζ1ζ2(ζ1z1)(ζ2z2)1]dζ1dζ2ζ1ζ21(2πi)20D2k11λ1=0k21λ2=0g(mk1+λ1)(mk2+λ2)(ζ)λ1!λ2!{(ζ1ˉζ1)λ1λ2v2=0Cv2λ2(z2)λ2v2(ˉz2)v2[2ζ2ζ2z21]+λ1v1=0Cv1λ1(z1)λ1v1(ˉz1)v1(ζ2ˉζ2)λ2[2ζ1ζ1z11](ζ1ˉζ1)λ1(ζ2ˉζ2)λ2[2ζ1ζ1z1+2ζ2ζ2z22]}dζζ=0D2k11λ1=0k21λ2=0g(mk1+λ1)(mk2+λ2)(2πi)2λ1!λ2!{(z1+ˉz1ζ1ˉζ1)λ1(z2+ˉz2ζ2ˉζ2)λ2[(ζ1ˉζ1)λ1(z1ˉz1)λ1](z2+ˉz2ζ2ˉζ2)λ2(z1+ˉz1ζ1ˉζ1)λ1[(ζ2ˉζ2)λ2(z2ˉz2)λ2]}[2ζ1ζ2(ζ1z1)(ζ2z2)1]dζ1dζ2ζ1ζ2+0D2k11λ1=0k21λ2=0g(mk1+λ1)(mk2+λ2)(2πi)2λ1!λ2!(ζ1ˉζ1)λ1[(ζ2ˉζ2)λ2(z2ˉz2)λ2][2ζ2ζ2z21]dζ1dζ2ζ1ζ2+0D2k11λ1=0k21λ2=0g(mk1+λ1)(mk2+λ2)(2πi)2λ1!λ2![(ζ1ˉζ1)λ1(z1ˉz1)λ1](ζ2ˉζ2)λ2[2ζ1ζ1z11]dζ1dζ2ζ1ζ2. (2.7)

    Applying the properties of the Cauchy kernels on D2 and D, for z0D2, (2.7) leads to

    mk1ˉz1mk2ˉz2˜ϕ(z)=k11λ1=0k21λ2=0{g(mk1+λ1)(mk2+λ2)(ζ)λ1!λ2![(z1+ˉz1ζ1ˉζ1)λ1(z2+ˉz2ζ2ˉζ2)λ2[(ζ1ˉζ1)λ1(z1ˉz1)λ1](z2+ˉz2ζ2ˉζ2)λ2(z1+ˉz1ζ1ˉζ1)λ1[(ζ2ˉζ2)λ2(z2ˉz2)λ2]]}|ζ1=z1ζ2=z2+D1k11λ1=0k21λ2=0(ζ1ˉζ1)λ12πλ1!λ2!{g(mk1+λ1)(mk2+λ2)(ζ)[(ζ2ˉζ2)λ2(z2ˉz2)λ2]}|ζ2=z2dζ1iζ1+D2k11λ1=0k21λ2=0(ζ2ˉζ2)λ22πλ1!λ2!{g(mk1+λ1)(mk2+λ2)(ζ)[(ζ1ˉζ1)λ1(z1ˉz1)λ1]}|ζ1=z1dζ2iζ2=g(mk1)(mk2)(z),

    which means μˉz1νˉz2˜ϕ(z)=gμν(z) for z0D2.

    In addition, by (2.7), we get that

    mk1ˉz1mk2ˉz2˜ϕ(0,z2)={0D2k11λ1=0k21λ2=0g(mk1+λ1)(mk2+λ2)(ζ)(2πi)2λ1!λ2![(ζ1ˉζ1)λ1(z2+ˉz2ζ2ˉζ2)λ2(ζ1ˉζ1)λ1(z2+ˉz2ζ2ˉζ2)λ2(ζ1ˉζ1)λ1[(ζ2ˉζ2)λ2(z2ˉz2)λ2]](2ζ2ζ2z21)dζ1dζ2ζ1ζ2+0D2k11λ1=0k21λ2=0g(mk1+λ1)(mk2+λ2)(ζ)(2πi)2λ1!λ2!(ζ1ˉζ1)λ1[(ζ2ˉζ2)λ2(z2ˉz2)λ2][2ζ2ζ2z21]dζ1dζ2ζ1ζ2+0D2k11λ1=0k21λ2=0g(mk1+λ1)(mk2+λ2)(ζ)(2πi)2λ1!λ2!(ζ1ˉζ1)λ1(ζ2ˉζ2)λ2dζ1dζ2ζ1ζ2}={0D2k11λ1=0k21λ2=0g(mk1+λ1)(mk2+λ2)(ζ)(2πi)2λ1!λ2!(ζ1ˉζ1)λ1(ζ2ˉζ2)λ2dζ1dζ2ζ1ζ2}=0.

    Similarly, we have that

    mk1ˉz1mk2ˉz2˜ϕ(z1,0)={0D2k11λ1=0k21λ2=0g(mk1+λ1)(mk2+λ2)(ζ)(2πi)2λ1!λ2![(z1+ˉz1ζ1ˉζ1)λ1(ζ2ˉζ2)λ2[(ζ1ˉζ1)λ1(z1ˉz1)λ1](ζ2ˉζ2)λ2(z1+ˉz1ζ1ˉζ1)λ1(ζ2ˉζ2)λ2](2ζ1ζ1z11)dζ1dζ2ζ1ζ2+0D2k11λ1=0k21λ2=0g(mk1+λ1)(mk2+λ2)(ζ)(2πi)2λ1!λ2!(ζ1ˉζ1)λ1(ζ2ˉζ2)λ2dζ1dζ2ζ1ζ2+0D2k11λ1=0k21λ2=0g(mk1+λ1)(mk2+λ2)(ζ)(2πi)2λ1!λ2![(ζ1ˉζ1)λ1(z1ˉz1)λ1](ζ2ˉζ2)λ2[2ζ1ζ1z11]dζζ}={0D2k11λ1=0k21λ2=0g(mk1+λ1)(mk2+λ2)(ζ)(2πi)2λ1!λ2!(ζ1ˉζ1)λ1(ζ2ˉζ2)λ2dζ1dζ2ζ1ζ2}=0.

    Therefore,

    μˉz1νˉz2˜ϕ(0,z2)=0=μˉz1νˉz2˜ϕ(z1,0).

    2) In the expression of u~v1~v2(z) determined by (2.4),

    D=[(z1ζ1ˉζ1)l1(z2ζ2ˉζ2)l2+(z1)l1(1)v1(z2ζ2ˉζ2)l2+(z1ζ1ˉζ1)l1(z2)l2(1)v2][2ζ1ζ2(ζ1z1)(ζ2z2)1]=[l1p1=0Cp1l1zp11(ζ1ˉζ1)l1p1l2q1=0Cq1l2zq12(ζ2ˉζ2)l2q1+(z1)l1(1)v1l2q1=0Cq1l2zq12(ζ2ˉζ2)l2q1+l1p1=0Cp1l1zp11(ζ1ˉζ1)l1p1(z2)l2(1)v2][2j1=0zj11ζj11j2=0zj22ζj221]. (2.8)

    Moreover, we have that

    l1p1=0Cp1l1zp11(ζ1ˉζ1)l1p1l2q1=0Cq1l2zq12(ζ2ˉζ2)l2q1[2+j1=0zj11ζj11j2=0zj22ζj221]=2l1p1=0+j1=0Cp1l1(ζ1ˉζ1)l1p1ˉζj11zp1+j11l2q1=0+j2=0Cq1l2(ζ2ˉζ2)l2q1ˉζj22zq1+j22l1p1=0l2q1=0Cp1l1Cq1l2(ζ1ˉζ1)l1p1(ζ2ˉζ2)l2q1zp11zq12=2+j1=0l1+j1m1=j1Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11zm11+j2=0l2+j2m2=j2Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22zm22l1m1=0l2m2=0Cm1l1Cm2l2(ζ1ˉζ1)l1m1(ζ2ˉζ2)l2m2zm11zm22=2+m1=0m1j1=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11zm11+m2=0m2j2=0Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22zm22l1m1=0l2m2=0Cm1l1Cm2l2(ζ1ˉζ1)l1m1(ζ2ˉζ2)l2m2zm11zm22=+m1,m2=0[2m1j1=0m2j2=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22D11D12]zm11zm22, (2.9)

    in which

    D11={Cm1l1(ζ1ˉζ1)l1m1,0m1l1,0,m1>l1,D12={Cm2l2(ζ2ˉζ2)l2m2,0m2l2,0,m2>l2,

    and

    (z1)l1(1)v1l2q1=0Cq1l2zq12(ζ2ˉζ2)l2q1[2j1=0zj11ζj11+j2=0zj22ζj221]=2+j1=0(1)l1+v1ˉζj11zl1+j11+j2=0l2q1=0Cq1l2(ζ2ˉζ2)l2q1ˉζj22zq1+j22zl11(1)l1+v1l2q1=0Cq1l2(ζ2ˉζ2)l2q1zq12=2+m1=l1(1)l1+v1ˉζm1l11zm11+m2=0m2j2=0Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22zm22l2m2=0(1)l1+v1Cm2l2(ζ2ˉζ2)l2m2zl11zm22=+m1,m2=0[2D21m2j2=0Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22D22D23]zm11zm22, (2.10)

    where

    D21={0,0m1<l1,(1)l1+v1ˉζm1l11,m1l1,D22={(1)m1+v1,m1=l1,0,m1l1,D23={Cm2l2(ζ2ˉζ2)l2m2,0m2l2,0,m2>l2.

    Similarly, we get that

    l1p1=0Cp1l1zp11(ζ1ˉζ1)l1p1(z2)l2(1)v2[2j1=0zj11ζj11+j2=0zj22ζj221]=+m1,m2=0[2D31m1j1=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11D32D33]zm11zm22, (2.11)

    in which

    D31={0,0m2<l2,(1)l2+v2ˉζm2l22,m2l2,D32={Cm1l1(ζ1ˉζ1)l1m1,0m1l1,0,m1>l1,D33={(1)m2+v2,m2=l2,0,m2l2.

    Plugging (2.9)–(2.11) into (2.8) gives that

    D=2+m1,m2=0{[m1j1=0m2j2=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22+D21m2j2=0Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22+D31m1j1=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11]12(D11D12+D22D23+D32D33)}zm11zm22. (2.12)

    In addition, as the result of

    (z1ζ1ˉζ1)l2[2ζ1ζ2(ζ1z1)(ζ2z2)1]+(z1)l1(1)v1[2ζ1ζ1z11]=l1p1=0Cp1l1zp11(ζ1ˉζ1)l1p1[2+j1=0zj11ζj11+j2=0zj22ζj221]+zl11(1)l1+v1(2+j1=0zj11ζj111)=2+m1=0m1j1=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11zm11+m2=0ˉζm22zm22l1p1=0Cp1l1(ζ1ˉζ1)l1p1zp11+2+m1=l1(1)l1+v1ˉζm1l11zm11(1)l1+v1zl11=2+m1,m2=0[m1j1=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11ˉζm22+D21C2112(D11+C22)C21]zm11zm22,

    in which

    C21={0,m21,1,m2=0,C22={(1)m1+v1,m1=l1,0,m1l1,

    we have that

    C2=(ζ2ˉζ2)l2{(z1ζ1ˉζ1)l2[2ζ1ζ2(ζ1z1)(ζ2z2)1]+(z1)l1(1)v1[2ζ1ζ1z11]}=2(ζ2ˉζ2)l2+m1,m2=0[m1j1=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11ˉζm22+D21C21D11+C222C21]zm11zm22, (2.13)

    which follows that

    C1C2=D|v2=0C2=2+m1,m2=0{m1j1=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11[m2j2=0Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22(ζ2ˉζ2)l2ˉζm22]+D21m2j2=0Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22+D31|v2=0m1j1=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11D21C2112(D11D12+D22D23+D32D33|v2=0D11C21C22C21)}zm11zm22. (2.14)

    Similarly, we have that

    B2=(ζ1ˉζ1)l1{(z2ζ2ˉζ2)l2[2ζ1ζ2(ζ1z1)(ζ2z2)1]+(z2)l2(1)v2[2ζ2ζ2z21]}=2(ζ1ˉζ1)l1+m1,m2=0[m2j2=0Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22ˉζm11+B21D31D23+B222B21]zm11zm22, (2.15)

    and

    B1B2=D|v1=0B2=2+m1,m2=0{[m1j1=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11(ζ1ˉζ1)l1ˉζm11]m2j2=0Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22+D21|v1=0m2j2=0Cm2j2l2(ζ2ˉζ2)l2m2+j2ˉζj22+D31m1j1=0Cm1j1l1(ζ1ˉζ1)l1m1+j1ˉζj11B21D3112(D11D12+D22|v1=0D23+D32D33B21D23B21B22)}zm11zm22, (2.16)

    in which

    B21={0,m11,1,m1=0,B22={(1)m2+v2,m2=l2,0,m2l2.

    Therefore,

    A1A2A3+A4=D|v1=v2=0B2|v2=0C2|v1=0+A4=D|v1=v2=0B2|v2=0C2|v1=0+2+m1,m2=0(¯ζ1m1C21+B21¯ζ2m2B21C21)(ζ1ˉζ1)l1(ζ2ˉζ2)l2zm11zm22 (2.17)

    as the result of

    A4=(ζ1ˉζ1)l1(ζ2ˉζ2)l2[2ζ1ζ1z1+2ζ2ζ2z22]=2(ζ1ˉζ1)l1(ζ2ˉζ2)l2[+j1=0zj11ζj11++j2=0zj22ζj221]=2(ζ1ˉζ1)l1(ζ2ˉζ2)l2[+m1=0¯ζ1m1zm11+m2=0C21zm22++m1=0B21zm11+m2=0¯ζ2m2zm22+m1=0B21zm11+m2=0C21zm22]=2(ζ1ˉζ1)l1(ζ2ˉζ2)l2+m1,m2=0(¯ζ1m1C21+B21¯ζ2m2B21C21)zm11zm22.

    On the other hand, due to the analyticity of the function u~v1~v2(z), it can be expressed as

    u~v1~v2(z)=+m1,m2=0um~v1,m~v2m1,m2zm11zm22,μ~v1m1,ν~v2m1. (2.18)

    Plugging (2.12), (2.14), (2.16), and (2.17) into (2.4), and considering the Eq (2.18), we get (2.2). Moreover, (2.18) and (2.3) lead to (2.1). Therefore, from the result in the first part (1), it can be concluded that ˜ϕ(z) determined by (2.1) satisfies

    μˉz1νˉz2˜ϕ(z)=gμν(z)(z0D2),μˉz1νˉz2˜ϕ(0,z2)=0=μˉz1νˉz2˜ϕ(z1,0)(z1D1,z2D2).

    Theorem 3.1. Let φC(0D2;C), λR{1,0,1}, and let gμνC(0D2;R) for 1μ,νm1 (m2). Then, the problem

    ˉz1ˉz2f(z)=λ14λϕ(z)+λ+14λˉϕ(z),mˉz1mˉz2ϕ(z)=0(zD2)

    with the conditions

    f(z)=φ(z),μˉz1νˉz2ϕ(z)=gμν(z)(z0D2),μˉz1νˉz2ϕ(0,z2)=0=μˉz1νˉz2ϕ(z1,0)

    is solvable and the solution is

    f(z)=λ14λ[+m1,m2=0u0,0m1,m2zm11zm22ˉz1ˉz2++m1,m2=0m1v1=μm1v2=νumv1,mv2m1,m2zm11zm22ˉzv1+11ˉzv2+12(v1+1)!(v2+1)!]+λ+14λ[+m1,m2=0¯u0,0m1,m2ˉzm1+11m1+1ˉzm2+12m2+1++m1,m2=0m1v1=μm1v2=νzv11zv22v1!v2!¯umv1,mv2m1,m2ˉzm1+11m1+1ˉzm2+12m2+1]++m1,m2=0bm1,m2zm11zm22,

    where umv1,mv2m1,m2 is determined by (2.2) (in which ~v1 and ~v2 are replaced by v1 and v2, respectively), and u0,0m1,m2, bm1,m2 are determined by the following:

    (ⅰ) for 1m1m1 and 1m2m1,

    u0,0m1,m2=(m1+1)(m2+1){λλ+11π22π0eit1(m1+1)2π0eit2(m2+1)¯φ(eit1,eit2)dt2dt1λ1λ+1mμv1=1mνv2=1¯uv1,v2(mm1v1),(mm2v2)(mv1+1)!(mv2+1)!mμv1=1mνv2=11(mv1)!(mv2)!uv1,v2(m+m1v1),(m+m2v2)(m+m1v1+1)(m+m2v2+1)}; (3.1)

    (ⅱ) for m11 and m21,

    u0,00,0=λ+1(2π)22π0eit12π0eit2¯φ(eit1,eit2)dt2dt1λ1(2π)22π0eit12π0eit2φ(eit1,eit2)dt2dt1mμv1=1mνv2=1uv1,v2(mv1),(mv2)(mv1+1)!(mv2+1)!, (3.2)
    u0,00,m2=4λλ11(2π)22π0eit12π0eit2(1m2)φ(eit1,eit2)dt2dt1mμv1=1mνv2=1uv1,v2(mv1),(mv2+m2)(mv1+1)!(mv2+1)!λ+1λ1{mμv1=1mνv2=11(mv1+1)!(mv2)!¯uv1,v2(mv1),(mv2m2)mv2m2+1,1m2ν,mμv1=1mm2v2=11(mv1+1)!(mv2)!¯uv1,v2(mv1),(mv2m2)mv2m2+1,ν<m2m1,0,m2>m1, (3.3)
    u0,0m1,0=4λλ11(2π)22π0ei(1m1)t12π0eit2φ(eit1,eit2)dt2dt1mμv1=1mνv2=1uv1,v2(mv1+m1),(mv2)(mv1+1)!(mv2+1)!λ+1λ1{mμv1=1mνv2=11(mv1)!(mv2+1)!¯uv1,v2(mv1m1,(mv2))mv1m1+1,1m1μ,mm1v1=1mνv2=11(mv1)!(mv2+1)!¯uv1,v2(mv1m1,(mv2))mv1m1+1,μ<m1m1,0,m1>m1; (3.4)

    (ⅲ) with u0,01,1 being determined by (3.1),

    b0,0=1(2π)22π02π0φ(eit1,eit2)dt2dt1λ14λ[u0,01,1+mμv1=1mνv2=1uv1,v2(mv1+1),(mv2+1)(mv1+1)!(mv2+1)!]λ+14λmμv1=1mνv2=11(mv1)!(mv2)!¯uv1,v2(mv11),(mv21)(mv1)(mv2); (3.5)

    (ⅳ) for {m1mm2m or {1m1m1m2m or {m1m1m2m1,

    u0,0m1,m2=4λλ+1(m1+1)(m2+1)(2π)22π0ei(m1+1)t12π0ei(m2+1)t2¯φ(eit1,eit2)dt2dt1mμv1=1mνv2=1(m1+1)(m2+1)(mv1)!(mv2)!uv1,v2(mv1+m1),(mv2+m2)(mv1+m1+1)(mv2+m2+1); (3.6)

    (ⅴ) for m1,m21,

    bm1,m2=1(2π)22π0eim1t12π0eim2t2φ(eit1,eit2)dt2dt1λ14λ[u0,0(m1+1),(m2+1)+mμv1=1mνv2=1uv1,v2(m1+mv1+1),(m2+mv2+1)(mv1+1)!(mv2+1)!]λ+14λ{m1m1v1=1m1m2v2=11(mv1)!(mv2)!¯uv1,v2(m1m1v1),(m1m2v2)(mv1m1)(mv2m2),1m1,m2<m1,0,m1,m2m1, (3.7)
    bm1,0=1(2π)22π02π0eim1t1φ(eit1,eit2)dt2dt1λ14λ[u0,0(m1+1),1+mμv1=1mνv2=1uv1,v2(m1+mv1+1),(mv2+1)(mv1+1)!(mv2+1)!]λ+14λ{m1m1v1=1mνv2=11(mv1)!(mv2)!¯uv1,v2(m1m1v1),(m1v2)(mv1m1)(mv2),1m1<m1,0,m1m1, (3.8)
    b0,m2=1(2π)22π02π0eim2t2φ(eit1,eit2)dt2dt1λ14λ[u0,01,(m2+1)+mμv1=1mνv2=1uv1,v2(mv1+1),(m2+mv2+1)(mv1+1)!(mv2+1)!]λ+14λ{mμv1=1m1m2v2=11(mv1)!(mv2)!¯uv1,v2(m1v1),(m1m2v2)(mv1)(mv2m2),1m2<m1,0,m2m1, (3.9)

    in which u0,0(m1+1),(m2+1), u0,0(m1+1),1, and u0,01,(m2+1) are determined by (3.6).

    Proof: 1) By Theorem 2.1,

    ϕ(z)=+m1,m2=0m1v1=μm1v2=νˉzv11ˉzv22v1!v2!umv1,mv2m1,m2zm11zm22+u0(z),

    where umv1,mv2m1,m2 is determined by (2.2) (in which ~v1 and ~v2 are replaced by v1 and v2, respectively), and u0(z) is analytic on D2. Thus, u0(z) can be represented as

    u0(z)=+m1,m2=0u0,0m1,m2zm11zm22,

    in which u0,0m1,m2 is to be determined.

    Let

    ϕ1(z)=u0(z)ˉz1ˉz2+m1v1=μm1v2=νuv1v2(z)(v1+1)!(v2+1)!ˉzv1+11ˉzv2+12,

    where

    uv1,v2(z)=+m1,m2=0umv1,mv2m1,m2zm11zm22,μv1m1,νv2m1.

    Thus, ˉz1ˉz2ϕ1(z)=ϕ(z). Let

    ~u0(z)=+m1,m2=0u0,0m1,m2zm1+11m1+1zm2+12m2+1,

    and let

    ˜uv1,v2(z)=+m1,m2=0umv1,mv2m1,m2zm1+11m1+1zm2+12m2+1,μv1m1,νv2m1.

    Then, we get that z1z2~u0(z)=u0(z) and z1z2˜uv1,v2(z)=uv1,v2(z). Let

    ϕ2(z)=¯~u0(z)+m1v1=μm1v2=νˉzv11ˉzv22v1!v2!˜uv1v2(z),

    which follows that

    ˉz1ˉz2ϕ2(z)=¯z1z2¯ϕ2=¯z1z2~u0(z)+m1v1=μm1v2=νˉzv11ˉzv22v1!v2!z1z2˜uv1v2(z)=¯u0(z)+m1v1=μm1v2=νˉzv11ˉzv22v1!v2!uv1v2(z)=¯ϕ(z).

    Therefore,

    ˉz1ˉz2[λ14λϕ1(z)+λ+14λϕ2(z)]=λ14λϕ(z)+λ+14λˉϕ(z),

    which means that

    λ14λϕ1(z)+λ+14λϕ2(z)

    is a special solution to

    ˉz1ˉz2f(z)=λ14λϕ(z)+λ+14λˉϕ(z).

    So, the solution of the problem is

    f(z)=[λ14λϕ1(z)+λ+14λϕ2(z)]+ψ(z)=λ14λ[u0(z)ˉz1ˉz2+m1v1=μm1v2=νuv1v2(z)(v1+1)!(v2+1)!ˉzv1+11ˉzv2+12]+λ+14λ[¯~u0(z)+m1v1=μm1v2=νzv11zv22v1!v2!¯˜uv1v2(z)]+ψ(z)=λ14λ[+m1,m2=0u0,0m1,m2zm11zm22ˉz1ˉz2++m1,m2=0m1v1=μm1v2=νumv1,mv2m1,m2zm11zm22ˉzv1+11ˉzv2+12(v1+1)!(v2+1)!]+λ+14λ[+m1,m2=0¯u0,0m1,m2ˉzm1+11m1+1ˉzm2+12m2+1++m1,m2=0m1v1=μm1v2=νzv11zv22v1!v2!¯umv1,mv2m1,m2ˉzm1+11m1+1ˉzm2+12m2+1]++m1,m2=0bm1,m2zm11zm22, (3.10)

    where ψ(z)=+m1,m2=0bm1,m2zm11zm22 is analytic on D2, and u0,0m1,m2 and bm1,m2 are to be determined.

    2) In this part, we seek the expressions of u0,0m1,m2 and bm1,m2.

    For z0D2, let z1=eit1 and z2=eit2 (t1,t2[0,2π]). Then, we get that

    φ(eit1,eit2)=f(eit1,eit2)=λ14λ[+m1,m2=0u0,0m1,m2ei(m11)t1ei(m21)t2++m1,m2=0mμ~v1=1mν~v2=1u~v1,~v2m1,m2eit1(m1m+~v11)eit2(m2m+~v21)(m~v1+1)!(m~v2+1)!]+λ+14λ[+m1,m2=0¯u0,0m1,m2eit1(m1+1)m1+1eit2(m2+1)m2+1++m1,m2=0mμ~v1=1mν~v2=1eit1(m1+1m+~v1)eit2(m2+1m+~v2)(m~v1)!(m~v2)!¯u~v1,~v2m1,m2(m1+1)(m2+1)]++m1,m2=0bm1,m2eim1t1eim2t2=λ14λ[0m1=0+m2=0u0,00,m2eit1ei(m21)t2++m1=10m2=0u0,0m1,0ei(m11)t1eit2++m1,m2=1u0,0m1,m2ei(m11)t1ei(m21)t2+mμ~v1=1mν~v2=1m~v1m1=0m~v2m2=0u~v1,~v2m1,m2eit1(m1(m~v1+1))eit2(m2(m~v2+1))(m~v1+1)!(m~v2+1)!+mμ~v1=1mν~v2=1+m1,m2=0u~v1,~v2(m1+(m~v1+1)),(m2+(m~v2+1))eim1t1eim2t2(m~v1+1)!(m~v2+1)!+mμ~v1=1mν~v2=1m~v1m1=0+m2=0u~v1,~v2m1,(m2+(m~v2+1))eit1(m1(m~v1+1))eim2t2(m~v1+1)!(m~v2+1)!+mμ~v1=1mν~v2=1+m1=0m~v2m2=0u~v1,~v2(m1+(m~v1+1)),m2eim1t1eit2(m2(m~v2+1))(m~v1+1)!(m~v2+1)!]+λ+14λ[+m1,m2=1¯u0,0(m11),(m21)eit1m1m1eit2m2m2++m1,m2=0mμ~v1=1mν~v2=1eit1(m1+1m+~v1)eit2(m2+1m+~v2)(m~v1)!(m~v2)!¯u~v1,~v2m1,m2(m1+1)(m2+1)]++m1,m2=0bm1,m2eim1t1eim2t2. (3.11)

    (ⅰ) In the case of 0r1,r2m2, multiplying both sides of the Eq (3.11) by eit1(mr1)eit2(mr2), and then integrating with respect to t1,t2[0,2π] yields that

    1(2π)22π0eit1(mr1)[2π0eit2(mr2)φ(eit1,eit2)dt2]dt1=λ14λ[+m2=0u0,00,m212π2π0eit1(mr11)dt112π2π0eit2(m21+mr2)dt2++m1=1u0,0m1,012π2π0eit1(m11+mr1)dt112π2π0eit2(1+mr2)dt2++m1,m2=1u0,0m1,m212π2π0eit1(m11+mr1)dt112π2π0eit2(m21+mr2)dt2+mμv1=1mνv2=1mv1m1=0mv2m2=0uv1,v2m1,m212π2π0eit1(m1+v11r1)dt112π2π0eit2(m2+v21r2)dt2(mv1+1)!(mv2+1)!+mμv1=1mνv2=1+m1,m2=0uv1,v2(m1+(mv1+1)),(m2+(mv2+1))12π2π0eit1(m1+mr1)dt112π2π0eit2(m2+mr2)dt2(mv1+1)!(mv2+1)!+mμv1=1mνv2=1mv1m1=0+m2=0uv1,v2m1,(m2+(mv2+1))12π2π0eit1(m1+v11r1)dt112π2π0eit2(m2+mr2)dt2(mv1+1)!(mv2+1)!+mμv1=1mνv2=1+m1=0mv2m2=0uv1,v2(m1+(mv1+1)),m212π2π0eit1(m1+mr1)dt112π2π0eit2(m2+v21r2)dt2(mv1+1)!(mv2+1)!]+λ+14λ[+m1,m2=1¯u0,0(m11),(m21)m1m212π2π0eit1(mr1m1)dt112π2π0eit2(mr2m2)dt2++m1,m2=0mμv1=1mνv2=112π2π0eit1(mr1(m1+1m+v1))dt112π2π0eit2(mr2(m2+1m+v2))dt2(mv1)!(mv2)!¯uv1,v2m1,m2(m1+1)(m2+1)]++m1,m2=0bm1,m212π2π0eit1(m1+mr1)dt112π2π0eit2(m2+mr2)dt2=λ14λ[mμv1=1mνv2=1uv1,v2(1+r1v1),(1+r2v2)(mv1+1)!(mv2+1)!]+λ+14λ[¯u0,0(mr11),(mr21)(mr1)(mr2)+mμv1=1mνv2=11(mv1)!(mv2)!¯uv1,v2(2mr11v1),(2mr21v2)(2mr1v1)(2mr2v2)],

    and the last equation is due to

    12π2π0eimtdt={0,m0(mZ),1,m=0.

    Therefore,

    u0,0(mr11),(mr21)=(mr1)(mr2){4λλ+11(2π)22π0eit1(mr1)2π0eit2(mr2)¯φ(eit1,eit2)dt2dt1λ1λ+1[mμv1=1mνv2=1¯uv1,v2(1+r1v1),(1+r2v2)(mv1+1)!(mv2+1)!]mμv1=1mνv2=11(mv1)!(mv2)!uv1,v2(2mr11v1),(2mr21v2)(2mr1v1)(2mr2v2)},

    which leads to (3.1) for 1m1,m2m1.

    (ⅱ) Multiplying both sides of the Eq (3.11) by eit1eit2, and then integrating with respect to t1,t2[0,2π] yields that

    1(2π)22π0eit1[2π0eit2φ(eit1,eit2)dt2]dt1=λ14λ[+m2=0u0,00,m212π2π0eim2t2dt2++m1=1u0,0m1,012π2π0eim1t1dt1++m1,m2=1u0,0m1,m212π2π0eim1t1dt112π2π0eim2t2dt2+mμv1=1mνv2=1mv1m1=0mv2m2=0uv1,v2m1,m212π2π0eit1(m1m+v1)dt112π2π0eit2(m2m+v2)dt2(mv1+1)!(mv2+1)!+mμv1=1mνv2=1+m1,m2=0uv1,v2(m1+(mv1+1)),(m2+(mv2+1))12π2π0eit1(m1+1)dt112π2π0eit2(m2+1)dt2(mv1+1)!(mv2+1)!+mμv1=1mνv2=1mv1m1=0+m2=0uv1,v2m1,(m2+(mv2+1))12π2π0eit1(m1m+v1)dt112π2π0eit2(m2+1)dt2(mv1+1)!(mv2+1)!+mμv1=1mνv2=1+m1=0mv2m2=0uv1,v2(m1+(mv1+1)),m212π2π0eit1(m1+1)dt112π2π0eit2(m2m+v2)dt2(mv1+1)!(mv2+1)!]+λ+14λ[+m1,m2=1¯u0,0(m11),(m21)m1m212π2π0eit1(1m1)dt112π2π0eit2(1m2)dt2++m1,m2=0mμv1=1mνv2=112π2π0eit1(m1m+v1)dt112π2π0eit2(m2m+v2)dt2(mv1)!(mv2)!¯uv1,v2m1,m2(m1+1)(m2+1)]++m1,m2=0bm1,m212π2π0eit1(m1+1)dt112π2π0eit2(m2+1)dt2=λ14λ[u0,00,0+mμv1=1mνv2=1uv1,v2(mv1),(mv2)(mv1+1)!(mv2+1)!]+λ+14λ[¯u0,00,0+mμv1=1mνv2=1¯uv1,v2(mv1),(mv2)(mv1+1)!(mv2+1)!],

    which leads to (3.2).

    Additionally, for r21, multiplying both sides of the Eq (3.11) by eit1ei(1r2)t2, and then integrating with respect to t1,t2[0,2π] yields that

    1(2π)22π0eit1[2π0eit2(1r2)φ(eit1,eit2)dt2]dt1=λ14λ[+m2=0u0,00,m212π2π0eit2(m2r2)dt2++m1=1u0,0m1,012π2π0eim1t1dt112π2π0eir2t2dt2++m1,m2=1u0,0m1,m212π2π0eim1t1dt112π2π0eit2(m2r2)dt2+mμv1=1mνv2=1mv1m1=0mv2m2=0uv1,v2m1,m212π2π0eit1(m1m+v1)dt112π2π0eit2(m2m+v2r2)dt2(mv1+1)!(mv2+1)!+mμv1=1mνv2=1+m1,m2=0uv1,v2(m1+(mv1+1)),(m2+(mv2+1))12π2π0eit1(m1+1)dt112π2π0eit2(m2+1r2)dt2(mv1+1)!(mv2+1)!+mμv1=1mνv2=1mv1m1=0+m2=0uv1,v2m1,(m2+(mv2+1))12π2π0eit1(m1m+v1)dt112π2π0eit2(m2+1r2)dt2(mv1+1)!(mv2+1)!+mμv1=1mνv2=1+m1=0mv2m2=0uv1,v2(m1+(mv1+1)),m212π2π0eit1(m1+1)dt112π2π0eit2(m2m+v2r2)dt2(mv1+1)!(mv2+1)!]+λ+14λ[+m1,m2=1¯u0,0(m11),(m21)m1m212π2π0eit1(1m1)dt112π2π0eit2(1r2m2)dt2++m1,m2=0mμv1=1mνv2=112π2π0eit1(m1m+v1)dt112π2π0eit2(m2m+v2+r2)dt2(mv1)!(mv2)!¯uv1,v2m1,m2(m1+1)(m2+1)]++m1,m2=0bm1,m212π2π0eit1(m1+1)dt112π2π0eit2(m2+1r2)dt2=λ14λ[u0,00,r2+mμv1=1mνv2=1uv1,v2(mv1),(mv2+r2)(mv1+1)!(mv2+1)!]+λ+14λ{mμv1=1mνv2=11(mv1+1)!(mv2)!¯uv1,v2(mv1),(mv2r2)mv2r2+1,1r2ν,mμv1=1mr2v2=11(mv1+1)!(mv2)!¯uv1,v2(mv1),(mv2r2)mv2r2+1,ν<r2m1,0,r2>m1,

    and the last equation is due to

    +m1,m2=0mμv1=1mνv2=112π2π0eit1(m1m+v1)dt112π2π0eit2(m2m+v2+r2)dt2(mv1)!(mv2)!¯uv1,v2m1,m2(m1+1)(m2+1)={mμv1=1mνv2=11(mv1+1)!(mv2)!¯uv1,v2(mv1),(mv2r2)mv2r2+1,1r2ν,mμv1=1mr2v2=11(mv1+1)!(mv2)!¯uv1,v2(mv1),(mv2r2)mv2r2+1,ν<r2m1,0,r2>m1.

    Therefore, for r21,

    u0,00,r2=4λλ11(2π)22π0eit12π0eit2(1r2)φ(eit1,eit2)dt2dt1mμv1=1mνv2=1uv1,v2(mv1),(mv2+r2)(mv1+1)!(mv2+1)!λ+1λ1{mμv1=1mνv2=11(mv1+1)!(mv2)!¯uv1,v2(mv1),(mv2r2)mv2r2+1,1r2ν,mμv1=1mr2v2=11(mv1+1)!(mv2)!¯uv1,v2(mv1),(mv2r2)mv2r2+1,ν<r2m1,0,r2>m1. (3.12)

    For r11, multiplying both sides of the equation (3.11) by ei(1r1)t1eit2 and then integrating with respect to t1,t2[0,2π], similar to (3.12), yields that

    u0,0r1,0=4λλ11(2π)22π0ei(1r1)t12π0eit2φ(eit1,eit2)dt2dt1mμv1=1mνv2=1uv1,v2(mv1+r1),(mv2)(mv1+1)!(mv2+1)!λ+1λ1{mμv1=1mνv2=11(mv1)!(mv2+1)!¯uv1,v2(mv1r1,(mv2))mv1r1+1,1r1μ,mr1v1=1mνv2=11(mv1)!(mv2+1)!¯uv1,v2(mv1r1,(mv2))mv1r1+1,μ<r1m1,0,r1>m1. (3.13)

    (ⅲ) In addition, integrating both sides of the Eq (3.11) with respect to t1,t2[0,2π] yields that

    1(2π)22π02π0φ(eit1,eit2)dt2dt1=λ14λ[+m2=0u0,00,m212π2π0eit1dt112π2π0eit2(m21)dt2++m1=1u0,0m1,012π2π0eit1(m11)dt112π2π0eit2dt2++m1,m2=1u0,0m1,m212π2π0eit1(m11)dt112π2π0eit2(m21)dt2+mμv1=1mνv2=1mv1m1=0mv2m2=0uv1,v2m1,m212π2π0eit1(m1+v11m)dt112π2π0eit2(m2+v21m)dt2(mv1+1)!(mv2+1)!+mμv1=1mνv2=1+m1,m2=0uv1,v2(m1+(mv1+1)),(m2+(mv2+1))12π2π0eim1t1dt112π2π0eim2t2dt2(mv1+1)!(mv2+1)!+mμv1=1mνv2=1mv1m1=0+m2=0uv1,v2m1,(m2+(mv2+1))12π2π0eit1(m1+v11m)dt112π2π0eim2t2dt2(mv1+1)!(mv2+1)!+mμv1=1mνv2=1+m1=0mv2m2=0uv1,v2(m1+(mv1+1)),m212π2π0eim1t1dt112π2π0eit2(m2+v21m)dt2(mv1+1)!(mv2+1)!]+λ+14λ[+m1,m2=1¯u0,0(m11),(m21)m1m212π2π0eim1t1dt112π2π0eim2t2dt2++m1,m2=0mμv1=1mνv2=112π2π0eit1(m1+1m+v1)dt112π2π0eit2(m2+1m+v2)dt2(mv1)!(mv2)!¯uv1,v2m1,m2(m1+1)(m2+1)]++m1,m2=0bm1,m212π2π0eim1t1dt112π2π0eim2t2dt2=λ14λ[u0,01,1+mμv1=1mνv2=1uv1,v2(mv1+1),(mv2+1)(mv1+1)!(mv2+1)!]+λ+14λmμv1=1mνv2=11(mv1)!(mv2)!¯uv1,v2(mv11),(mv21)(mv1)(mv2)+b0,0,

    which follows (3.5), where u0,01,1 is determined by (3.1).

    (ⅳ) In the case of r1,r2m+1, multiplying both sides of the Eq (3.11) by eir1t1eir2t2, and then integrating with respect to t1,t2[0,2π] yields that

    1(2π)22π0eir1t1[2π0eir2t2φ(eit1,eit2)dt2]dt1=λ14λ[+m2=0u0,00,m212π2π0eit1(r11)dt112π2π0eit2(m21+r2)dt2++m1=1u0,0m1,012π2π0eit1(m11+r1)dt112π2π0eit2(1+r2)dt2++m1,m2=1u0,0m1,m212π2π0eit1(m11+r1)dt112π2π0eit2(m21+r2)dt2+mμv1=1mνv2=1mv1m1=0mv2m2=0uv1,v2m1,m212π2π0eit1(m1(mv1+1)+r1)dt112π2π0eit2(m2(mv2+1)+r2)dt2(mv1+1)!(mv2+1)!+mμv1=1mνv2=1+m1,m2=0uv1,v2(m1+(mv1+1)),(m2+(mv2+1))12π2π0eit1(m1+r1)dt112π2π0eit2(m2+r2)dt2(mv1+1)!(mv2+1)!+mμv1=1mνv2=1mv1m1=0+m2=0uv1,v2m1,(m2+(mv2+1))12π2π0eit1(m1(mv1+1)+r1)dt112π2π0eit2(m2+r2)dt2(mv1+1)!(mv2+1)!+mμv1=1mνv2=1+m1=0mv2m2=0uv1,v2(m1+(mv1+1)),m212π2π0eit1(m1+r1)dt112π2π0eit2(m2(mv2+1)+r2)dt2(mv1+1)!(mv2+1)!]+λ+14λ[+m1,m2=1¯u0,0(m11),(m21)m1m212π2π0eit1(m1+r1)dt112π2π0eit2(m2+r2)dt2++m1,m2=0mμv1=1mνv2=112π2π0eit1(m1+1m+v1r1)dt112π2π0eit2(m2+1m+v2r2)dt2(mv1)!(mv2)!¯uv1,v2m1,m2(m1+1)(m2+1)]++m1,m2=0bm1,m212π2π0eit1(m1+r1)dt112π2π0eit2(m2+r2)dt2=λ+14λ[¯u0,0(r11),(r21)r1r2+mμv1=1mνv2=11(mv1)!(mv2)!¯uv1,v2(mv1+r11),(mv2+r21)(mv1+r1)(mv2+r2)].

    Therefore,

    u0,0r1,r2=4λλ+1(r1+1)(r2+1)(2π)22π0ei(r1+1)t12π0ei(r2+1)t2¯φ(eit1,eit2)dt2dt1mμv1=1mνv2=1(r1+1)(r2+1)(mv1)!(mv2)!uv1,v2(mv1+r1),(mv2+r2)(mv1+r1+1)(mv2+r2+1), (3.14)

    for r1,r2m.

    Similarly, in the case of r1m+1 and 0r2m2, multiplying both sides of the Eq (3.11) by eir1t1eit2(mr2), and then integrating with respect to t1,t2[0,2π] yields that u0,0r1,r2 (r1m,1r2m1) has the same representation as (3.14). In the case of 0r1m2 and r2m+1, multiplying both sides of the Eq (3.11) by eit1(mr1)eir2t2, and then integrating with respect to t1,t2[0,2π] yields that u0,0r1,r2 (1r1m1,r2m) has the same representation as (3.14).

    (ⅴ) Similarly, for r1,r21, multiplying both sides of the Eq (3.11) by eir1t1eir2t2, and then integrating with respect to t1,t2[0,2π], we can get the expression of br1r2 which leads to (3.7). For r11, multiplying both sides of the Eq (3.11) by eir1t1, and then integrating with respect to t1,t2[0,2π] yields the expression of br10 which leads to (3.8). For r21, we can get b0r2 which leads to (3.9).

    Remark 3.2. The results obtained in this article extend the existing conclusions about ployanalytic functions and bi-ployanalytic functions. On this basis, we can study other partial differential equation problems. For example, it would be interesting to discuss whether bi-polyanalytic or even ployanalytical function solutions exist for some nonlocal integrable partial differential equations (see, e.g., [24]), which need to explore the solutions to the corresponding Riemann-Hilbert problems. Additionally, the non-existence of solutions to Cauchy problems on the real line for first-order nonlocal differential equations (see, e.g., [25]) indicates that we can attempt to discuss the generalizations of analytical solutions for partial differential equation problems.

    With the help of the series expansion of polyanalytic functions, and applying the properties of Cauchy kernels on the bicylinder and the unit disk, we first discuss a class of Schwarz problems with the conditions concerning the real and imaginary parts of high-order partial differentiation for polyanalytic functions on the bicylinder. On this basis, we investigate a type of boundary value problem for bi-polyanalytic functions with Dirichlet boundary conditions on the bicylinder. From the perspective of series, we obtain the specific representation of the solution to the Dirichlet problem. The method used in this article, with the help of series expansion, is different from the previous methods for solving boundary value problems. It is a very effective method and can be used to solve other types of problems regarding complex partial differential equations of bi-polyanalytic functions in high-dimensional complex spaces. The conclusions of this article also lay a necessary foundation for further research on polyanalytic and bi-polyanalytic functions.

    Yanyan Cui: conceptualization, project administration, writing original, writing-review and editing; Chaojun Wang: investigation, writing original, writing-review and editing. All authors have read and approved the final version of the manuscript for publication.

    The authors are grateful to the anonymous referees for their valuable comments and suggestions which improved the quality of this article. This work was supported by the NSF of China (No. 11601543), the NSF of Henan Province (Nos. 222300420397 and 242300421394), and the Science and Technology Research Projects of Henan Provincial Education Department (No. 19B110016).

    The authors declare no conflict of interest.



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    1. Yanyan Cui, Chaojun Wang, Dirichlet and Neumann boundary value problems for bi-polyanalytic functions on the bicylinder, 2025, 10, 2473-6988, 4792, 10.3934/math.2025220
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