The spread of rumors has an important impact on the production and life of human society. Moreover, in the process of rumor propagation, individuals with different educational levels show different degrees of trust and ability to spread rumors. Therefore, a new rumor propagation model was established, which considers the influence of education level on rumor propagation. Initially, the basic reproduction number of the model was calculated. Then, we analyzed the existence and stability of the rumor equilibrium point. Next, based on the principle of Pontryagin's maximum value, we obtained a control strategy, which effectively reduced the spread of rumors. Numerical simulations verified the results of theoretical analysis. The results showed that the higher the education level of the population, the slower the spread of rumors to a certain extent, but it could not prevent the spread of rumors. In addition, through the support of the government and the propaganda of the official media, strengthening education can improve people's education level to a certain extent, and then minimize the speed of rumor propagation.
Citation: Hongshuang Wang, Sida Kang, Yuhan Hu. Dynamic analysis and optimal control of rumor propagation models considering different education levels and hesitation mechanisms[J]. AIMS Mathematics, 2024, 9(8): 20089-20117. doi: 10.3934/math.2024979
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The spread of rumors has an important impact on the production and life of human society. Moreover, in the process of rumor propagation, individuals with different educational levels show different degrees of trust and ability to spread rumors. Therefore, a new rumor propagation model was established, which considers the influence of education level on rumor propagation. Initially, the basic reproduction number of the model was calculated. Then, we analyzed the existence and stability of the rumor equilibrium point. Next, based on the principle of Pontryagin's maximum value, we obtained a control strategy, which effectively reduced the spread of rumors. Numerical simulations verified the results of theoretical analysis. The results showed that the higher the education level of the population, the slower the spread of rumors to a certain extent, but it could not prevent the spread of rumors. In addition, through the support of the government and the propaganda of the official media, strengthening education can improve people's education level to a certain extent, and then minimize the speed of rumor propagation.
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:
{∂u∂x−∂v∂y=θ,∂u∂y+∂v∂x=ω,(k+1)∂θ∂x+∂ω∂y=0,(k+1)∂θ∂y−∂ω∂x=0, |
for the real constant k(k≠−1), 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:
{1k∂u∂x−∂v∂y=θ,∂u∂y+1k∂v∂x=ω,k∂θ∂x+λ∂ω∂y=0,k∂θ∂y−λ∂ω∂x=0, |
where θ(x,y) and ω(x,y) are continuously differentiable functions of x and y, and λ,k are real constants with λ≠0,1,k2, and 0<k<1. The complex form of the system is
k+12∂f∂ˉz−k−12∂f∂z=λ−k4λφ(z)+λ+k4λ¯φ(z), |
in which φ(z)=kϑ−iλω 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 gμν∈C(∂0D2;R) for 1≤μ,ν≤m−1 (m≥2), and let
˜ϕ(z)=+∞∑m1,m2=0m−1∑˜v1=μm−1∑˜v2=νˉz˜v11ˉz˜v22˜v1!˜v2!um−~v1,m−~v2m1,m2zm11zm22, | (2.1) |
where
um−~v1,m−~v2m1,m2={1(2πi)2∫∂0D2m−μ−1∑l1=0m−ν−1∑l2=0g(μ+l1)(ν+l2)(ζ)l1!l2!˜Adζ1dζ2ζ1ζ2,{˜v1=μ˜v2=ν,1(2πi)2∫∂0D2m−μ−1∑l1=0m−1−˜v2∑l2=0g(μ+l1)(˜v2+l2)(ζ)l1!l2!˜B|v2=˜v2−νdζ1dζ2ζ1ζ2,{˜v1=μν<˜v2≤m−1,1(2πi)2∫∂0D2m−1−˜v1∑l1=0m−ν−1∑l2=0g(˜v1+l1)(ν+l2)(ζ)l1!l2!˜C|v1=˜v1−μdζ1dζ2ζ1ζ2,{μ<˜v1≤m−1˜v2=ν,1(2πi)2∫∂0D2m−1−˜v1∑l1=0m−1−˜v2∑l2=0g(˜v1+l1)(˜v2+l2)(ζ)l1!l2!˜D|v1=˜v1−μv2=˜v2−νdζ1dζ2ζ1ζ2,{μ<˜v1≤m−1ν<˜v2≤m−1, | (2.2) |
and
{˜A=2[m1∑j1=0m2∑j2=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22+D21|v1=0m2∑j2=0Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22+D31|v2=0m1∑j1=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11]−(D11D12+D22D23+D32D33)|v1=v2=0+2(−ζ1−ˉζ1)l1[m2∑j2=0Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22ˉζm11+B21D31−D23+B222B21]|v2=0+2(−ζ2−ˉζ2)l2[m1∑j1=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11ˉζm22+D21C21−D11+C222C21]|v1=0+2(¯ζ1m1C21+B21¯ζ2m2−B21C21)(−ζ1−ˉζ1)l1(−ζ2−ˉζ2)l2,˜B=2{[m1∑j1=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11−(−ζ1−ˉζ1)l1ˉζm11]m2∑j2=0Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22+D21|v1=0m2∑j2=0Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22+D31m1∑j1=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11−B21D31}−(D11D12+D22|v1=0D23+D32D33−B21D23−B21B22), |
{˜C=2{m1∑j1=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11[m2∑j2=0Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22−(−ζ2−ˉζ2)l2ˉζm22]+D21m2∑j2=0Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22+D31|v2=0m1∑j1=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11−D21C21}−(D11D12+D22D23+D32D33|v2=0−D11C21−C22C21),˜D=2[m1∑j1=0m2∑j2=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22+D21m2∑j2=0Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22+D31m1∑j1=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11]−(D11D12+D22D23+D32D33), |
in which
D11={Cm1l1(−ζ1−ˉζ1)l1−m1,0≤m1≤l1,0,m1>l1,D12={Cm2l2(−ζ2−ˉζ2)l2−m2,0≤m2≤l2,0,m2>l2,D21={0,0≤m1<l1,(−1)l1+v1ˉζm1−l11,m1≥l1,D22={(−1)m1+v1,m1=l1,0,m1≠l1,D23={Cm2l2(−ζ2−ˉζ2)l2−m2,0≤m2≤l2,0,m2>l2,D31={0,0≤m2<l2,(−1)l2+v2ˉζm2−l22,m2≥l2,D32={Cm1l1(−ζ1−ˉζ1)l1−m1,0≤m1≤l1,0,m1>l1,D33={(−1)m2+v2,m2=l2,0,m2≠l2, |
and
C21={0,m2≥1,1,m2=0,C22={(−1)m1+v1,m1=l1,0,m1≠l1, |
B21={0,m1≥1,1,m1=0,B22={(−1)m2+v2,m2=l2,0,m2≠l2. |
Then, ˜ϕ(z) satisfies
ℜ∂μˉz1∂νˉz2˜ϕ(z)=gμν(z)(z∈∂0D2),ℑ∂μˉz1∂νˉz2˜ϕ(0,z2)=0=ℑ∂μˉz1∂νˉz2˜ϕ(z1,0)(z1∈D1,z2∈D2). |
Proof: 1) Let
˜ϕ(z)=m−1∑˜v1=μm−1∑˜v2=νˉz˜v11ˉz˜v22˜v1!˜v2!u˜v1˜v2(z), | (2.3) |
in which
u˜v1˜v2(z)={1(2πi)2∫∂0D2m−μ−1∑l1=0m−ν−1∑l2=0g(μ+l1)(ν+l2)(ζ)l1!l2!(A1−A2−A3+A4)dζ1dζ2ζ1ζ2,{˜v1=μ˜v2=ν,1(2πi)2∫∂0D2m−μ−1∑l1=0m−1−˜v2∑l2=0g(μ+l1)(˜v2+l2)(ζ)l1!l2!(B1−B2)|v2=˜v2−νdζ1dζ2ζ1ζ2,{˜v1=μν<˜v2≤m−1,1(2πi)2∫∂0D2m−1−˜v1∑l1=0m−ν−1∑l2=0g(˜v1+l1)(ν+l2)(ζ)l1!l2!(C1−C2)|v1=˜v1−μdζ1dζ2ζ1ζ2,{μ<˜v1≤m−1˜v2=ν,1(2πi)2∫∂0D2m−1−˜v1∑l1=0m−1−˜v2∑l2=0g(˜v1+l1)(˜v2+l2)(ζ)l1!l2!D|v1=˜v1−μv2=˜v2−νdζ1dζ2ζ1ζ2,{μ<˜v1≤m−1ν<˜v2≤m−1, | (2.4) |
is analytic on D2, and
{A1=[(z1−ζ1−ˉζ1)l1(z2−ζ2−ˉζ2)l2+(−z1)l1(z2−ζ2−ˉζ2)l2+(z1−ζ1−ˉζ1)l1(−z2)l2][2ζ1ζ2(ζ1−z1)(ζ2−z2)−1],A2=(−ζ1−ˉζ1)l1{(z2−ζ2−ˉζ2)l2[2ζ1ζ2(ζ1−z1)(ζ2−z2)−1]+(−z2)l2[2ζ2ζ2−z2−1]},A3=(−ζ2−ˉζ2)l2{(z1−ζ1−ˉζ1)l2[2ζ1ζ2(ζ1−z1)(ζ2−z2)−1]+(−z1)l1[2ζ1ζ1−z1−1]},A4=(−ζ1−ˉζ1)l1(−ζ2−ˉζ2)l2[2ζ1ζ1−z1+2ζ2ζ2−z2−2],B1=[(z1−ζ1−ˉζ1)l1(z2−ζ2−ˉζ2)l2+(−z1)l1(z2−ζ2−ˉζ2)l2+(z1−ζ1−ˉζ1)l1(−z2)l2(−1)v2]⋅[2ζ1ζ2(ζ1−z1)(ζ2−z2)−1],B2=(−ζ1−ˉζ1)l1{(z2−ζ2−ˉζ2)l2[2ζ1ζ2(ζ1−z1)(ζ2−z2)−1]+(−z2)l2(−1)v2[2ζ2ζ2−z2−1]},C1=[(z1−ζ1−ˉζ1)l1(z2−ζ2−ˉζ2)l2+(−z1)l1(−1)v1(z2−ζ2−ˉζ2)l2+(z1−ζ1−ˉζ1)l1(−z2)l2]⋅[2ζ1ζ2(ζ1−z1)(ζ2−z2)−1],C2=(−ζ2−ˉζ2)l2{(z1−ζ1−ˉζ1)l2[2ζ1ζ2(ζ1−z1)(ζ2−z2)−1]+(−z1)l1(−1)v1[2ζ1ζ1−z1−1]},D=[(z1−ζ1−ˉζ1)l1(z2−ζ2−ˉζ2)l2+(−z1)l1(−1)v1(z2−ζ2−ˉζ2)l2+(z1−ζ1−ˉζ1)l1(−z2)l2(−1)v2]⋅[2ζ1ζ2(ζ1−z1)(ζ2−z2)−1]. |
In the following, we will show that ˜ϕ(z) satisfies
ℜ∂μˉz1∂νˉz2˜ϕ(z)=gμν(z)(z∈∂0D2),ℑ∂μˉz1∂νˉz2˜ϕ(0,z2)=0=ℑ∂μˉz1∂νˉz2˜ϕ(z1,0)(z1∈D1,z2∈D2). |
By (2.3), we get that
∂μˉz1∂νˉz2˜ϕ(z)=m−1∑˜v1=μm−1∑˜v2=νˉz˜v1−μ1ˉz˜v2−ν2(˜v1−μ)!(˜v2−ν)!u˜v1˜v2(z)=m−1−μ∑v1=0m−1−ν∑v2=0ˉzv11ˉzv22v1!v2!u(v1+μ)(v2+ν)(z), | (2.5) |
where
u(v1+μ)(v2+ν)={1(2πi)2∫∂0D2m−μ−1∑l1=0m−ν−1∑l2=0g(μ+l1)(ν+l2)(ζ)l1!l2!(A1−A2−A3+A4)dζ1dζ2ζ1ζ2,{v1=0v2=0,1(2πi)2∫∂0D2m−μ−1∑l1=0m−1−v2−ν∑l2=0g(μ+l1)(v2+ν+l2)(ζ)l1!l2!(B1−B2)dζ1dζ2ζ1ζ2,{v1=00<v2≤m−1−ν,1(2πi)2∫∂0D2m−1−v1−μ∑l1=0m−ν−1∑l2=0g(v1+μ+l1)(ν+l2)(ζ)l1!l2!(C1−C2)dζ1dζ2ζ1ζ2,{0<v1≤m−1−μv2=0,1(2πi)2∫∂0D2m−1−v1−μ∑l1=0m−1−v2−ν∑l2=0g(v1+μ+l1)(v2+ν+l2)(ζ)l1!l2!Ddζ1dζ2ζ1ζ2,{0<v1≤m−1−μ0<v2≤m−1−ν. |
Therefore, for 1≤k1,k2≤m−1,
∂m−k1ˉz1∂m−k2ˉz2˜ϕ(z)=k1−1∑v1=0k2−1∑v2=0ˉzv11ˉzv22v1!v2!u(v1+m−k1)(v2+m−k2)(z), | (2.6) |
in which
u(v1+m−k1)(v2+m−k2)={∫∂0D2k1−1∑l1=0k2−1∑l2=0g(m−k1+l1)(m−k1+l2)(ζ)(2πi)2l1!l2!(A1−A2−A3+A4)dζ1dζ2ζ1ζ2,{v1=0v2=0,∫∂0D2k1−1∑l1=0k2−1−v2∑l2=0g(m−k1+l1)(m−k2+v2+l2)(ζ)(2πi)2l1!l2!(B1−B2)dζ1dζ2ζ1ζ2,{v1=00<v2≤k2−1,∫∂0D2k1−1−v1∑l1=0k2−1∑l2=0g(m−k1+v1+l1)(m−k2+l2)(ζ)(2πi)2l1!l2!(C1−C2)dζ1dζ2ζ1ζ2,{0<v1≤k1−1v2=0,∫∂0D2k1−1−v1∑l1=0k2−1−v2∑l2=0g(m−k1+v1+l1)(m−k2+v2+l2)(ζ)(2πi)2l1!l2!Ddζ1dζ2ζ1ζ2,{0<v1≤k1−10<v2≤k2−1. |
Let
ϕ1(z)=ˉzv11ˉzv22v1!v2!1(2πi)2∫∂0D2k1−1−v1∑l1=0k2−1−v2∑l2=0g(m−k1+v1+l1)(m−k2+v2+l2)(ζ)l1!l2!dζ1dζ2ζ1ζ2. |
For 1≤k1,k2≤m−1, (2.6) follows that
∂m−k1ˉz1∂m−k2ˉz2˜ϕ(z)=0∑v1=00∑v2=0ϕ1(z)(A1−A2−A3+A4)+0∑v1=0k2−1∑v2=1ϕ1(z)(B1−B2)+k1−1∑v1=10∑v2=0ϕ1(z)(C1−C2)+k1−1∑v1=1k2−1∑v2=1ϕ1(z)D=k1−1∑v1=0k2−1∑v2=0ϕ1(z)D−0∑v1=0k2−1∑v2=0ϕ1(z)B2−k1−1∑v1=00∑v2=0ϕ1(z)C2+0∑v1,v2=0ϕ1(z)A4=1(2πi)2∫∂0D2k1−1∑v1=0k1−1−v1∑l1=0k2−1∑v2=0k2−1−v2∑l2=0ˉzv11ˉzv22v1!v2!g(m−k1+v1+l1)(m−k2+v2+l2)(ζ)l1!l2![(z1−ζ1−ˉζ1)l1(z2−ζ2−ˉζ2)l2+(−z1)l1(−1)v1(z2−ζ2−ˉζ2)l2+(z1−ζ1−ˉζ1)l1(−z2)l2(−1)v2][2ζ1ζ2(ζ1−z1)(ζ2−z2)−1]dζζ−1(2πi)2∫∂0D2k1−1∑λ1=0k2−1∑v2=0k2−1−v2∑l2=0ˉzv22v2!g(m−k1+λ1)(m−k2+v2+l2)(ζ)λ1!l2!(−ζ1−ˉζ1)λ1⋅{(z2−ζ2−ˉζ2)l2[2ζ1ζ2(ζ1−z1)(ζ2−z2)−1]+(−z2)l2(−1)v2[2ζ2ζ2−z2−1]}dζ1dζ2ζ1ζ2−1(2πi)2∫∂0D2k1−1∑v1=0k1−1−v1∑l1=0k2−1∑λ2=0ˉzv11v1!g(m−k1+v1+l1)(m−k2+λ2)(ζ)l1!λ2!(−ζ2−ˉζ2)λ2⋅{(z1−ζ1−ˉζ1)l1[2ζ1ζ2(ζ1−z1)(ζ2−z2)−1]+(−z1)l1(−1)v1[2ζ1ζ1−z1−1]}dζ1dζ2ζ1ζ2+∫∂0D2k1−1∑λ1=0k2−1∑λ2=0g(m−k1+λ1)(m−k2+λ2)(ζ)(2πi)2λ1!λ2!(−ζ1−ˉζ1)λ1(−ζ2−ˉζ2)λ2[2ζ1ζ1−z1+2ζ2ζ2−z2−2]dζζ=1(2πi)2∫∂0D2k1−1∑λ1=0k2−1∑λ2=0λ1∑v1=0λ2∑v2=0ˉzv11ˉzv22λ1!λ2!Cv1λ1Cv2λ2g(m−k1+λ1)(m−k2+λ2)[(z1−ζ1−ˉζ1)λ1−v1(z2−ζ2−ˉζ2)λ2−v2+(−z1)λ1−v1(−1)v1(z2−ζ2−ˉζ2)λ2−v2+(z1−ζ1−ˉζ1)λ1−v1(−z2)λ2−v2(−1)v2][2ζ1ζ2(ζ1−z1)(ζ2−z2)−1]dζζ−1(2πi)2∫∂0D2k1−1∑λ1=0k2−1∑λ2=0(−ζ1−ˉζ1)λ1λ1!λ2!λ2∑v2=0Cv2λ2ˉzv22g(m−k1+λ1)(m−k2+λ2)(ζ)⋅{(z2−ζ2−ˉζ2)λ2−v2[2ζ1ζ2(ζ1−z1)(ζ2−z2)−1]+(−z2)λ2−v2(−1)v2[2ζ2ζ2−z2−1]}dζ1dζ2ζ1ζ2−1(2πi)2∫∂0D2k1−1∑λ1=0k2−1∑λ2=0(−ζ2−ˉζ2)λ2λ1!λ2!λ1∑v1=0Cv1λ1ˉzv11g(m−k1+λ1)(m−k2+λ2)(ζ)⋅{(z1−ζ1−ˉζ1)λ1−v1[2ζ1ζ2(ζ1−z1)(ζ2−z2)−1]+(−z1)λ1−v1(−1)v1[2ζ1ζ1−z1−1]}dζ1dζ2ζ1ζ2+∫∂0D2k1−1∑λ1=0k2−1∑λ2=0g(m−k1+λ1)(m−k2+λ2)(ζ)(2πi)2λ1!λ2!(−ζ1−ˉζ1)λ1(−ζ2−ˉζ2)λ2[2ζ1ζ1−z1+2ζ2ζ2−z2−2]dζ1dζ2ζ1ζ2=1(2πi)2∫∂0D2k1−1∑λ1=0k2−1∑λ2=01λ1!λ2!{λ1∑v1=0Cv1λ1(z1−ζ1−ˉζ1)λ1−v1ˉzv11λ2∑v2=0Cv2λ2(z2−ζ2−ˉζ2)λ2−v2ˉzv22−(−ζ1−ˉζ1)λ1λ2∑v2=0Cv2λ2(z2−ζ2−ˉζ2)λ2−v2ˉzv22+λ1∑v1=0Cv1λ1(−z1)λ1−v1(−ˉz1)v1λ2∑v2=0Cv2λ2(z2−ζ2−ˉζ2)λ2−v2ˉzv22−λ1∑v1=0Cv1λ1(z1−ζ1−ˉζ1)λ1−v1ˉzv11(−ζ2−ˉζ2)λ2+λ1∑v1=0Cv1λ1(z1−ζ1−ˉζ1)λ1−v1ˉzv11λ2∑v2=0Cv2λ2(−z2)λ2−v2(−ˉz2)v2}⋅g(m−k1+λ1)(m−k2+λ2)(ζ)[2ζ1ζ2(ζ1−z1)(ζ2−z2)−1]dζ1dζ2ζ1ζ2−1(2πi)2∫∂0D2k1−1∑λ1=0k2−1∑λ2=0g(m−k1+λ1)(m−k2+λ2)(ζ)λ1!λ2!{(−ζ1−ˉζ1)λ1λ2∑v2=0Cv2λ2(−z2)λ2−v2(−ˉz2)v2[2ζ2ζ2−z2−1]+λ1∑v1=0Cv1λ1(−z1)λ1−v1(−ˉz1)v1(−ζ2−ˉζ2)λ2[2ζ1ζ1−z1−1]−(−ζ1−ˉζ1)λ1(−ζ2−ˉζ2)λ2[2ζ1ζ1−z1+2ζ2ζ2−z2−2]}dζζ=∫∂0D2k1−1∑λ1=0k2−1∑λ2=0g(m−k1+λ1)(m−k2+λ2)(2πi)2λ1!λ2!{(z1+ˉz1−ζ1−ˉζ1)λ1(z2+ˉz2−ζ2−ˉζ2)λ2−[(−ζ1−ˉζ1)λ1−(−z1−ˉz1)λ1]⋅(z2+ˉz2−ζ2−ˉζ2)λ2−(z1+ˉz1−ζ1−ˉζ1)λ1[(−ζ2−ˉζ2)λ2−(−z2−ˉz2)λ2]}[2ζ1ζ2(ζ1−z1)(ζ2−z2)−1]dζ1dζ2ζ1ζ2+∫∂0D2k1−1∑λ1=0k2−1∑λ2=0g(m−k1+λ1)(m−k2+λ2)(2πi)2λ1!λ2!(−ζ1−ˉζ1)λ1[(−ζ2−ˉζ2)λ2−(−z2−ˉz2)λ2][2ζ2ζ2−z2−1]dζ1dζ2ζ1ζ2+∫∂0D2k1−1∑λ1=0k2−1∑λ2=0g(m−k1+λ1)(m−k2+λ2)(2πi)2λ1!λ2λ2[2ζ1ζ1−z1−1]dζ1dζ2ζ1ζ2. | (2.7) |
Applying the properties of the Cauchy kernels on D2 and D, for z∈∂0D2, (2.7) leads to
ℜ∂m−k1ˉz1∂m−k2ˉz2˜ϕ(z)=k1−1∑λ1=0k2−1∑λ2=0{g(m−k1+λ1)(m−k2+λ2)(ζ)λ1!λ2λ2−(z1+ˉz1−ζ1−ˉζ1)λ1[(−ζ2−ˉζ2)λ2−(−z2−ˉz2)λ2]]}|ζ1=z1ζ2=z2+∫∂D1k1−1∑λ1=0k2−1∑λ2=0(−ζ1−ˉζ1)λ12πλ1!λ2!{g(m−k1+λ1)(m−k2+λ2)(ζ)[(−ζ2−ˉζ2)λ2−(−z2−ˉz2)λ2]}|ζ2=z2dζ1iζ1+∫∂D2k1−1∑λ1=0k2−1∑λ2=0(−ζ2−ˉζ2)λ22πλ1!λ2!{g(m−k1+λ1)(m−k2+λ2)(ζ)[(−ζ1−ˉζ1)λ1−(−z1−ˉz1)λ1]}|ζ1=z1dζ2iζ2=g(m−k1)(m−k2)(z), |
which means ℜ∂μˉz1∂νˉz2˜ϕ(z)=gμν(z) for z∈∂0D2.
In addition, by (2.7), we get that
ℑ∂m−k1ˉz1∂m−k2ˉz2˜ϕ(0,z2)=ℑ{∫∂0D2k1−1∑λ1=0k2−1∑λ2=0g(m−k1+λ1)(m−k2+λ2)(ζ)(2πi)2λ1!λ2![(−ζ1−ˉζ1)λ1(z2+ˉz2−ζ2−ˉζ2)λ2−(−ζ1−ˉζ1)λ1⋅(z2+ˉz2−ζ2−ˉζ2)λ2−(−ζ1−ˉζ1)λ1[(−ζ2−ˉζ2)λ2−(−z2−ˉz2)λ2]](2ζ2ζ2−z2−1)dζ1dζ2ζ1ζ2+∫∂0D2k1−1∑λ1=0k2−1∑λ2=0g(m−k1+λ1)(m−k2+λ2)(ζ)(2πi)2λ1!λ2!(−ζ1−ˉζ1)λ1[(−ζ2−ˉζ2)λ2−(−z2−ˉz2)λ2][2ζ2ζ2−z2−1]dζ1dζ2ζ1ζ2+∫∂0D2k1−1∑λ1=0k2−1∑λ2=0g(m−k1+λ1)(m−k2+λ2)(ζ)(2πi)2λ1!λ2!(−ζ1−ˉζ1)λ1(−ζ2−ˉζ2)λ2dζ1dζ2ζ1ζ2}=ℑ{∫∂0D2k1−1∑λ1=0k2−1∑λ2=0g(m−k1+λ1)(m−k2+λ2)(ζ)(2πi)2λ1!λ2!(−ζ1−ˉζ1)λ1(−ζ2−ˉζ2)λ2dζ1dζ2ζ1ζ2}=0. |
Similarly, we have that
ℑ∂m−k1ˉz1∂m−k2ˉz2˜ϕ(z1,0)=ℑ{∫∂0D2k1−1∑λ1=0k2−1∑λ2=0g(m−k1+λ1)(m−k2+λ2)(ζ)(2πi)2λ1!λ2λ2−(z1+ˉz1−ζ1−ˉζ1)λ1(−ζ2−ˉζ2)λ2](2ζ1ζ1−z1−1)dζ1dζ2ζ1ζ2+∫∂0D2k1−1∑λ1=0k2−1∑λ2=0g(m−k1+λ1)(m−k2+λ2)(ζ)(2πi)2λ1!λ2!(−ζ1−ˉζ1)λ1(−ζ2−ˉζ2)λ2dζ1dζ2ζ1ζ2+∫∂0D2k1−1∑λ1=0k2−1∑λ2=0g(m−k1+λ1)(m−k2+λ2)(ζ)(2πi)2λ1!λ2λ2[2ζ1ζ1−z1−1]dζζ}=ℑ{∫∂0D2k1−1∑λ1=0k2−1∑λ2=0g(m−k1+λ1)(m−k2+λ2)(ζ)(2πi)2λ1!λ2!(−ζ1−ˉζ1)λ1(−ζ2−ˉζ2)λ2dζ1dζ2ζ1ζ2}=0. |
Therefore,
ℑ∂μˉz1∂νˉz2˜ϕ(0,z2)=0=ℑ∂μˉz1∂νˉz2˜ϕ(z1,0). |
2) In the expression of u~v1~v2(z) determined by (2.4),
D=[(z1−ζ1−ˉζ1)l1(z2−ζ2−ˉζ2)l2+(−z1)l1(−1)v1(z2−ζ2−ˉζ2)l2+(z1−ζ1−ˉζ1)l1(−z2)l2(−1)v2]⋅[2ζ1ζ2(ζ1−z1)(ζ2−z2)−1]=[l1∑p1=0Cp1l1zp11(−ζ1−ˉζ1)l1−p1l2∑q1=0Cq1l2zq12(−ζ2−ˉζ2)l2−q1+(−z1)l1(−1)v1l2∑q1=0Cq1l2zq12(−ζ2−ˉζ2)l2−q1+l1∑p1=0Cp1l1zp11(−ζ1−ˉζ1)l1−p1(−z2)l2(−1)v2][2∞∑j1=0zj11ζj11∞∑j2=0zj22ζj22−1]. | (2.8) |
Moreover, we have that
l1∑p1=0Cp1l1zp11(−ζ1−ˉζ1)l1−p1l2∑q1=0Cq1l2zq12(−ζ2−ˉζ2)l2−q1[2+∞∑j1=0zj11ζj11∞∑j2=0zj22ζj22−1]=2l1∑p1=0+∞∑j1=0Cp1l1(−ζ1−ˉζ1)l1−p1ˉζj11zp1+j11l2∑q1=0+∞∑j2=0Cq1l2(−ζ2−ˉζ2)l2−q1ˉζj22zq1+j22−l1∑p1=0l2∑q1=0Cp1l1Cq1l2(−ζ1−ˉζ1)l1−p1(−ζ2−ˉζ2)l2−q1zp11zq12=2+∞∑j1=0l1+j1∑m1=j1Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11zm11+∞∑j2=0l2+j2∑m2=j2Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22zm22−l1∑m1=0l2∑m2=0Cm1l1Cm2l2(−ζ1−ˉζ1)l1−m1(−ζ2−ˉζ2)l2−m2zm11zm22=2+∞∑m1=0m1∑j1=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11zm11+∞∑m2=0m2∑j2=0Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22zm22−l1∑m1=0l2∑m2=0Cm1l1Cm2l2(−ζ1−ˉζ1)l1−m1(−ζ2−ˉζ2)l2−m2zm11zm22=+∞∑m1,m2=0[2m1∑j1=0m2∑j2=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22−D11D12]zm11zm22, | (2.9) |
in which
D11={Cm1l1(−ζ1−ˉζ1)l1−m1,0≤m1≤l1,0,m1>l1,D12={Cm2l2(−ζ2−ˉζ2)l2−m2,0≤m2≤l2,0,m2>l2, |
and
(−z1)l1(−1)v1l2∑q1=0Cq1l2zq12(−ζ2−ˉζ2)l2−q1[2∞∑j1=0zj11ζj11+∞∑j2=0zj22ζj22−1]=2+∞∑j1=0(−1)l1+v1ˉζj11zl1+j11+∞∑j2=0l2∑q1=0Cq1l2(−ζ2−ˉζ2)l2−q1ˉζj22zq1+j22−zl11(−1)l1+v1l2∑q1=0Cq1l2(−ζ2−ˉζ2)l2−q1zq12=2+∞∑m1=l1(−1)l1+v1ˉζm1−l11zm11+∞∑m2=0m2∑j2=0Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22zm22−l2∑m2=0(−1)l1+v1Cm2l2(−ζ2−ˉζ2)l2−m2zl11zm22=+∞∑m1,m2=0[2D21m2∑j2=0Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22−D22D23]zm11zm22, | (2.10) |
where
D21={0,0≤m1<l1,(−1)l1+v1ˉζm1−l11,m1≥l1,D22={(−1)m1+v1,m1=l1,0,m1≠l1,D23={Cm2l2(−ζ2−ˉζ2)l2−m2,0≤m2≤l2,0,m2>l2. |
Similarly, we get that
l1∑p1=0Cp1l1zp11(−ζ1−ˉζ1)l1−p1(−z2)l2(−1)v2[2∞∑j1=0zj11ζj11+∞∑j2=0zj22ζj22−1]=+∞∑m1,m2=0[2D31m1∑j1=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11−D32D33]zm11zm22, | (2.11) |
in which
D31={0,0≤m2<l2,(−1)l2+v2ˉζm2−l22,m2≥l2,D32={Cm1l1(−ζ1−ˉζ1)l1−m1,0≤m1≤l1,0,m1>l1,D33={(−1)m2+v2,m2=l2,0,m2≠l2. |
Plugging (2.9)–(2.11) into (2.8) gives that
D=2+∞∑m1,m2=0{[m1∑j1=0m2∑j2=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22+D21m2∑j2=0Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22+D31m1∑j1=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11]−12(D11D12+D22D23+D32D33)}zm11zm22. | (2.12) |
In addition, as the result of
(z1−ζ1−ˉζ1)l2[2ζ1ζ2(ζ1−z1)(ζ2−z2)−1]+(−z1)l1(−1)v1[2ζ1ζ1−z1−1]=l1∑p1=0Cp1l1zp11(−ζ1−ˉζ1)l1−p1[2+∞∑j1=0zj11ζj11+∞∑j2=0zj22ζj22−1]+zl11(−1)l1+v1(2+∞∑j1=0zj11ζj11−1)=2+∞∑m1=0m1∑j1=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11zm11+∞∑m2=0ˉζm22zm22−l1∑p1=0Cp1l1(−ζ1−ˉζ1)l1−p1zp11+2+∞∑m1=l1(−1)l1+v1ˉζm1−l11zm11−(−1)l1+v1zl11=2+∞∑m1,m2=0[m1∑j1=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11ˉζm22+D21C21−12(D11+C22)C21]zm11zm22, |
in which
C21={0,m2≥1,1,m2=0,C22={(−1)m1+v1,m1=l1,0,m1≠l1, |
we have that
C2=(−ζ2−ˉζ2)l2{(z1−ζ1−ˉζ1)l2[2ζ1ζ2(ζ1−z1)(ζ2−z2)−1]+(−z1)l1(−1)v1[2ζ1ζ1−z1−1]}=2(−ζ2−ˉζ2)l2+∞∑m1,m2=0[m1∑j1=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11ˉζm22+D21C21−D11+C222C21]zm11zm22, | (2.13) |
which follows that
C1−C2=D|v2=0−C2=2+∞∑m1,m2=0{m1∑j1=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11[m2∑j2=0Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22−(−ζ2−ˉζ2)l2ˉζm22]+D21m2∑j2=0Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22+D31|v2=0m1∑j1=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11−D21C21−12(D11D12+D22D23+D32D33|v2=0−D11C21−C22C21)}zm11zm22. | (2.14) |
Similarly, we have that
B2=(−ζ1−ˉζ1)l1{(z2−ζ2−ˉζ2)l2[2ζ1ζ2(ζ1−z1)(ζ2−z2)−1]+(−z2)l2(−1)v2[2ζ2ζ2−z2−1]}=2(−ζ1−ˉζ1)l1+∞∑m1,m2=0[m2∑j2=0Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22ˉζm11+B21D31−D23+B222B21]zm11zm22, | (2.15) |
and
B1−B2=D|v1=0−B2=2+∞∑m1,m2=0{[m1∑j1=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11−(−ζ1−ˉζ1)l1ˉζm11]m2∑j2=0Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22+D21|v1=0m2∑j2=0Cm2−j2l2(−ζ2−ˉζ2)l2−m2+j2ˉζj22+D31m1∑j1=0Cm1−j1l1(−ζ1−ˉζ1)l1−m1+j1ˉζj11−B21D31−12(D11D12+D22|v1=0D23+D32D33−B21D23−B21B22)}zm11zm22, | (2.16) |
in which
B21={0,m1≥1,1,m1=0,B22={(−1)m2+v2,m2=l2,0,m2≠l2. |
Therefore,
A1−A2−A3+A4=D|v1=v2=0−B2|v2=0−C2|v1=0+A4=D|v1=v2=0−B2|v2=0−C2|v1=0+2+∞∑m1,m2=0(¯ζ1m1C21+B21¯ζ2m2−B21C21)(−ζ1−ˉζ1)l1(−ζ2−ˉζ2)l2zm11zm22 | (2.17) |
as the result of
A4=(−ζ1−ˉζ1)l1(−ζ2−ˉζ2)l2[2ζ1ζ1−z1+2ζ2ζ2−z2−2]=2(−ζ1−ˉζ1)l1(−ζ2−ˉζ2)l2[+∞∑j1=0zj11ζj11++∞∑j2=0zj22ζj22−1]=2(−ζ1−ˉζ1)l1(−ζ2−ˉζ2)l2[+∞∑m1=0¯ζ1m1zm11+∞∑m2=0C21zm22++∞∑m1=0B21zm11+∞∑m2=0¯ζ2m2zm22−+∞∑m1=0B21zm11+∞∑m2=0C21zm22]=2(−ζ1−ˉζ1)l1(−ζ2−ˉζ2)l2+∞∑m1,m2=0(¯ζ1m1C21+B21¯ζ2m2−B21C21)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≤~v1≤m−1,ν≤~v2≤m−1. | (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
ℜ∂μˉz1∂νˉz2˜ϕ(z)=gμν(z)(z∈∂0D2),ℑ∂μˉz1∂νˉz2˜ϕ(0,z2)=0=ℑ∂μˉz1∂νˉz2˜ϕ(z1,0)(z1∈D1,z2∈D2). |
Theorem 3.1. Let φ∈C(∂0D2;C), λ∈R∖{−1,0,1}, and let gμν∈C(∂0D2;R) for 1≤μ,ν≤m−1 (m≥2). Then, the problem
∂ˉz1∂ˉz2f(z)=λ−14λϕ(z)+λ+14λˉϕ(z),∂mˉz1∂mˉz2ϕ(z)=0(z∈D2) |
with the conditions
f(z)=φ(z),ℜ∂μˉz1∂νˉz2ϕ(z)=gμν(z)(z∈∂0D2),ℑ∂μˉz1∂νˉz2ϕ(0,z2)=0=ℑ∂μˉz1∂νˉz2ϕ(z1,0) |
is solvable and the solution is
f(z)=λ−14λ[+∞∑m1,m2=0u0,0m1,m2zm11zm22ˉz1ˉz2++∞∑m1,m2=0m−1∑v1=μm−1∑v2=νum−v1,m−v2m1,m2zm11zm22⋅ˉzv1+11ˉzv2+12(v1+1)!(v2+1)!]+λ+14λ[+∞∑m1,m2=0¯u0,0m1,m2ˉzm1+11m1+1ˉzm2+12m2+1++∞∑m1,m2=0m−1∑v1=μm−1∑v2=νzv11zv22v1!v2!⋅¯um−v1,m−v2m1,m2ˉzm1+11m1+1ˉzm2+12m2+1]++∞∑m1,m2=0bm1,m2zm11zm22, |
where um−v1,m−v2m1,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 1≤m1≤m−1 and 1≤m2≤m−1,
u0,0m1,m2=(m1+1)(m2+1){λλ+11π2∫2π0e−it1(m1+1)∫2π0e−it2(m2+1)¯φ(eit1,eit2)dt2dt1−λ−1λ+1m−μ∑v1=1m−ν∑v2=1¯uv1,v2(m−m1−v1),(m−m2−v2)(m−v1+1)!(m−v2+1)!−m−μ∑v1=1m−ν∑v2=11(m−v1)!(m−v2)!uv1,v2(m+m1−v1),(m+m2−v2)(m+m1−v1+1)(m+m2−v2+1)}; | (3.1) |
(ⅱ) for m1≥1 and m2≥1,
u0,00,0=λ+1(2π)2∫2π0e−it1∫2π0e−it2¯φ(eit1,eit2)dt2dt1−λ−1(2π)2∫2π0eit1∫2π0eit2φ(eit1,eit2)dt2dt1−m−μ∑v1=1m−ν∑v2=1uv1,v2(m−v1),(m−v2)(m−v1+1)!(m−v2+1)!, | (3.2) |
u0,00,m2=4λλ−11(2π)2∫2π0eit1∫2π0eit2(1−m2)φ(eit1,eit2)dt2dt1−m−μ∑v1=1m−ν∑v2=1uv1,v2(m−v1),(m−v2+m2)(m−v1+1)!(m−v2+1)!−λ+1λ−1{m−μ∑v1=1m−ν∑v2=11(m−v1+1)!(m−v2)!¯uv1,v2(m−v1),(m−v2−m2)m−v2−m2+1,1≤m2≤ν,m−μ∑v1=1m−m2∑v2=11(m−v1+1)!(m−v2)!¯uv1,v2(m−v1),(m−v2−m2)m−v2−m2+1,ν<m2≤m−1,0,m2>m−1, | (3.3) |
u0,0m1,0=4λλ−11(2π)2∫2π0ei(1−m1)t1∫2π0eit2φ(eit1,eit2)dt2dt1−m−μ∑v1=1m−ν∑v2=1uv1,v2(m−v1+m1),(m−v2)(m−v1+1)!(m−v2+1)!−λ+1λ−1{m−μ∑v1=1m−ν∑v2=11(m−v1)!(m−v2+1)!¯uv1,v2(m−v1−m1,(m−v2))m−v1−m1+1,1≤m1≤μ,m−m1∑v1=1m−ν∑v2=11(m−v1)!(m−v2+1)!¯uv1,v2(m−v1−m1,(m−v2))m−v1−m1+1,μ<m1≤m−1,0,m1>m−1; | (3.4) |
(ⅲ) with u0,01,1 being determined by (3.1),
b0,0=1(2π)2∫2π0∫2π0φ(eit1,eit2)dt2dt1−λ−14λ[u0,01,1+m−μ∑v1=1m−ν∑v2=1uv1,v2(m−v1+1),(m−v2+1)(m−v1+1)!(m−v2+1)!]−λ+14λm−μ∑v1=1m−ν∑v2=11(m−v1)!(m−v2)!¯uv1,v2(m−v1−1),(m−v2−1)(m−v1)(m−v2); | (3.5) |
(ⅳ) for {m1≥mm2≥m or {1≤m1≤m−1m2≥m or {m1≥m1≤m2≤m−1,
u0,0m1,m2=4λλ+1(m1+1)(m2+1)(2π)2∫2π0e−i(m1+1)t1∫2π0e−i(m2+1)t2¯φ(eit1,eit2)dt2dt1−m−μ∑v1=1m−ν∑v2=1(m1+1)(m2+1)(m−v1)!(m−v2)!uv1,v2(m−v1+m1),(m−v2+m2)(m−v1+m1+1)(m−v2+m2+1); | (3.6) |
(ⅴ) for m1,m2≥1,
bm1,m2=1(2π)2∫2π0e−im1t1∫2π0e−im2t2φ(eit1,eit2)dt2dt1−λ−14λ[u0,0(m1+1),(m2+1)+m−μ∑v1=1m−ν∑v2=1uv1,v2(m1+m−v1+1),(m2+m−v2+1)(m−v1+1)!(m−v2+1)!]−λ+14λ{m−1−m1∑v1=1m−1−m2∑v2=11(m−v1)!(m−v2)!¯uv1,v2(m−1−m1−v1),(m−1−m2−v2)(m−v1−m1)(m−v2−m2),1≤m1,m2<m−1,0,m1,m2≥m−1, | (3.7) |
bm1,0=1(2π)2∫2π0∫2π0e−im1t1φ(eit1,eit2)dt2dt1−λ−14λ[u0,0(m1+1),1+m−μ∑v1=1m−ν∑v2=1uv1,v2(m1+m−v1+1),(m−v2+1)(m−v1+1)!(m−v2+1)!]−λ+14λ{m−1−m1∑v1=1m−ν∑v2=11(m−v1)!(m−v2)!¯uv1,v2(m−1−m1−v1),(m−1−v2)(m−v1−m1)(m−v2),1≤m1<m−1,0,m1≥m−1, | (3.8) |
b0,m2=1(2π)2∫2π0∫2π0e−im2t2φ(eit1,eit2)dt2dt1−λ−14λ[u0,01,(m2+1)+m−μ∑v1=1m−ν∑v2=1uv1,v2(m−v1+1),(m2+m−v2+1)(m−v1+1)!(m−v2+1)!]−λ+14λ{m−μ∑v1=1m−1−m2∑v2=11(m−v1)!(m−v2)!¯uv1,v2(m−1−v1),(m−1−m2−v2)(m−v1)(m−v2−m2),1≤m2<m−1,0,m2≥m−1, | (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=0m−1∑v1=μm−1∑v2=νˉzv11ˉzv22v1!v2!um−v1,m−v2m1,m2zm11zm22+u0(z), |
where um−v1,m−v2m1,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+m−1∑v1=μm−1∑v2=νuv1v2(z)(v1+1)!(v2+1)!ˉzv1+11ˉzv2+12, |
where
uv1,v2(z)=+∞∑m1,m2=0um−v1,m−v2m1,m2zm11zm22,μ≤v1≤m−1,ν≤v2≤m−1. |
Thus, ∂ˉz1∂ˉz2ϕ1(z)=ϕ(z). Let
~u0(z)=+∞∑m1,m2=0u0,0m1,m2zm1+11m1+1zm2+12m2+1, |
and let
˜uv1,v2(z)=+∞∑m1,m2=0um−v1,m−v2m1,m2zm1+11m1+1zm2+12m2+1,μ≤v1≤m−1,ν≤v2≤m−1. |
Then, we get that ∂z1∂z2~u0(z)=u0(z) and ∂z1∂z2˜uv1,v2(z)=uv1,v2(z). Let
ϕ2(z)=¯~u0(z)+m−1∑v1=μm−1∑v2=νˉzv11ˉzv22v1!v2!˜uv1v2(z), |
which follows that
∂ˉz1∂ˉz2ϕ2(z)=¯∂z1∂z2¯ϕ2=¯∂z1∂z2~u0(z)+m−1∑v1=μm−1∑v2=νˉzv11ˉzv22v1!v2!∂z1∂z2˜uv1v2(z)=¯u0(z)+m−1∑v1=μm−1∑v2=νˉzv11ˉzv22v1!v2!uv1v2(z)=¯ϕ(z). |
Therefore,
∂ˉz1∂ˉz2[λ−14λϕ1(z)+λ+14λϕ2(z)]=λ−14λϕ(z)+λ+14λˉϕ(z), |
which means that
λ−14λϕ1(z)+λ+14λϕ2(z) |
is a special solution to
∂ˉz1∂ˉz2f(z)=λ−14λϕ(z)+λ+14λˉϕ(z). |
So, the solution of the problem is
f(z)=[λ−14λϕ1(z)+λ+14λϕ2(z)]+ψ(z)=λ−14λ[u0(z)ˉz1ˉz2+m−1∑v1=μm−1∑v2=νuv1v2(z)(v1+1)!(v2+1)!ˉzv1+11ˉzv2+12]+λ+14λ[¯~u0(z)+m−1∑v1=μm−1∑v2=νzv11zv22v1!v2!¯˜uv1v2(z)]+ψ(z)=λ−14λ[+∞∑m1,m2=0u0,0m1,m2zm11zm22ˉz1ˉz2++∞∑m1,m2=0m−1∑v1=μm−1∑v2=νum−v1,m−v2m1,m2zm11zm22⋅ˉzv1+11ˉzv2+12(v1+1)!(v2+1)!]+λ+14λ[+∞∑m1,m2=0¯u0,0m1,m2ˉzm1+11m1+1ˉzm2+12m2+1++∞∑m1,m2=0m−1∑v1=μm−1∑v2=νzv11zv22v1!v2!⋅¯um−v1,m−v2m1,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 z∈∂0D2, let z1=eit1 and z2=eit2 (t1,t2∈[0,2π]). Then, we get that
φ(eit1,eit2)=f(eit1,eit2)=λ−14λ[+∞∑m1,m2=0u0,0m1,m2ei(m1−1)t1ei(m2−1)t2++∞∑m1,m2=0m−μ∑~v1=1m−ν∑~v2=1u~v1,~v2m1,m2⋅eit1(m1−m+~v1−1)eit2(m2−m+~v2−1)(m−~v1+1)!(m−~v2+1)!]+λ+14λ[+∞∑m1,m2=0¯u0,0m1,m2e−it1(m1+1)m1+1e−it2(m2+1)m2+1++∞∑m1,m2=0m−μ∑~v1=1m−ν∑~v2=1e−it1(m1+1−m+~v1)e−it2(m2+1−m+~v2)(m−~v1)!(m−~v2)!¯u~v1,~v2m1,m2(m1+1)(m2+1)]++∞∑m1,m2=0bm1,m2eim1t1eim2t2=λ−14λ[0∑m1=0+∞∑m2=0u0,00,m2e−it1ei(m2−1)t2++∞∑m1=10∑m2=0u0,0m1,0ei(m1−1)t1e−it2++∞∑m1,m2=1u0,0m1,m2ei(m1−1)t1ei(m2−1)t2+m−μ∑~v1=1m−ν∑~v2=1m−~v1∑m1=0m−~v2∑m2=0u~v1,~v2m1,m2eit1(m1−(m−~v1+1))eit2(m2−(m−~v2+1))(m−~v1+1)!(m−~v2+1)!+m−μ∑~v1=1m−ν∑~v2=1+∞∑m1,m2=0u~v1,~v2(m1+(m−~v1+1)),(m2+(m−~v2+1))eim1t1eim2t2(m−~v1+1)!(m−~v2+1)!+m−μ∑~v1=1m−ν∑~v2=1m−~v1∑m1=0+∞∑m2=0u~v1,~v2m1,(m2+(m−~v2+1))eit1(m1−(m−~v1+1))eim2t2(m−~v1+1)!(m−~v2+1)!+m−μ∑~v1=1m−ν∑~v2=1+∞∑m1=0m−~v2∑m2=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(m1−1),(m2−1)e−it1m1m1e−it2m2m2++∞∑m1,m2=0m−μ∑~v1=1m−ν∑~v2=1e−it1(m1+1−m+~v1)e−it2(m2+1−m+~v2)(m−~v1)!(m−~v2)!¯u~v1,~v2m1,m2(m1+1)(m2+1)]++∞∑m1,m2=0bm1,m2eim1t1eim2t2. | (3.11) |
(ⅰ) In the case of 0\leq r_1, r_2\leq m-2 , multiplying both sides of the Eq (3.11) by e^{it_1(m-r_1)}e^{it_2(m-r_2)} , and then integrating with respect to t_1, t_2\in[0, 2\pi] yields that
\begin{aligned} &\frac{1}{(2\pi)^2}\int^{2\pi}_0e^{it_1(m-r_1)}\Big[\int^{2\pi}_0e^{it_2(m-r_2)}\varphi(e^{it_1}, e^{it_2})dt_2\Big]dt_1\\ & = \!\frac{\lambda-1}{4\lambda} \Big[\!\sum\limits_{m_2 = 0}^{+\infty}u_{0, m_2}^{0, 0}\frac{1}{2\pi}\int^{2\pi}_0e^{it_1(m-r_1-1)}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(m_2-1+m-r_2)}dt_2\\ &\; \; +\!\sum\limits_{m_1 = 1}^{+\infty}u_{m_1, 0}^{0, 0}\frac{1}{2\pi}\int^{2\pi}_0e^{it_1(m_1-1+m-r_1)}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(-1+m-r_2)}dt_2\\ &\; \; +\!\sum\limits_{m_1, m_2 = 1}^{+\infty}\!\!\!u_{m_1, m_2}^{0, 0}\frac{1}{2\pi}\int^{2\pi}_0e^{it_1(m_1-1+m-r_1)}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(m_2-1+m-r_2)}dt_2\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \sum\limits_{m_1 = 0}^{m-v_1}\sum\limits_{m_2 = 0}^{m-v_2} \!\frac{u_{m_1, m_2}^{v_1, v_2}\frac{1}{2\pi}\int^{2\pi}_0 e^{it_1(m_1+v_1-1-r_1)}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(m_2+v_2-1-r_2)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \!\sum\limits_{m_1, m_2 = 0}^{+\infty} \frac{u_{(m_1+(m-v_1+1)), (m_2+(m-v_2+1))}^{v_1, v_2} \frac{1}{2\pi}\!\!\int^{2\pi}_0 \!\!\!e^{it_1(m_1+m-r_1)}dt_1\!\cdot\! \frac{1}{2\pi}\!\!\int^{2\pi}_0\!\!\!e^{it_2(m_2+m-r_2)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \sum\limits_{m_1 = 0}^{m-v_1}\sum\limits_{m_2 = 0}^{+\infty} \frac{u_{m_1, (m_2+(m-v_2+1))}^{v_1, v_2} \frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{it_1(m_1+v_1-1-r_1)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{it_2(m_2+m-r_2)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \sum\limits_{m_1 = 0}^{+\infty}\sum\limits_{m_2 = 0}^{m-v_2} \frac{u_{(m_1+(m-v_1+1)), m_2}^{v_1, v_2} \frac{1}{2\pi}\!\!\int^{2\pi}_0 \!\!e^{it_1(m_1+m-r_1)}dt_1\cdot \frac{1}{2\pi}\!\!\int^{2\pi}_0\!\!e^{it_2(m_2+v_2-1-r_2)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\Big]\\ &\; \; +\frac{\lambda+1}{4\lambda} \Big[\sum\limits_{m_1, m_2 = 1}^{+\infty}\frac{\overline{u_{(m_1-1), (m_2-1)}^{0, 0}}}{m_1m_2} \frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{it_1(m-r_1-m_1)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{it_2(m-r_2-m_2)}dt_2\\ &\; \; +\sum\limits_{m_1, m_2 = 0}^{+\infty}\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \frac{\frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{it_1(m-r_1-(m_1+1-m+v_1))}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{it_2(m-r_2-(m_2+1-m+v_2))}dt_2} {(m-v_1)!(m-v_2)!}\\ &\; \; \cdot\frac{\overline{u_{m_1, m_2}^{v_1, v_2}}}{(m_1+1)(m_2+1)} \Big]+ \sum\limits_{m_1, m_2 = 0}^{+\infty}b_{m_1, m_2}\frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{it_1(m_1+m-r_1)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{it_2(m_2+m-r_2)}dt_2\\ & = \!\frac{\lambda-1}{4\lambda} \Big[\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \!\frac{u_{(1+r_1-v_1), (1+r_2-v_2)}^{v_1, v_2}} {(m-v_1+1)!(m-v_2+1)!}\Big]\\ &\; \; +\!\frac{\lambda+1}{4\lambda} \Big[\frac{\overline{u_{(m-r_1-1), (m-r_2-1)}^{0, 0}}}{(m-r_1)(m-r_2)} +\!\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \frac{1}{(m-v_1)!(m-v_2)!} \frac{\overline{u_{(2m-r_1-1-v_1), (2m-r_2-1-v_2)}^{v_1, v_2}}}{(2m\!-\!r_1\!-\!v_1)(2m\!-\!r_2\!-\!v_2)} \Big], \end{aligned} |
and the last equation is due to
\frac{1}{2\pi}\!\int^{2\pi}_0e^{imt}dt = \left\{ \begin{array}{ll} 0, \quad m\neq 0 \; (m\in\mathbb{Z}), \\ 1, \quad m = 0. \end{array} \right. |
Therefore,
\begin{aligned} u_{(m-r_1-1), (m-r_2-1)}^{0, 0}&\! = \!(m\!-\!r_1)(m\!-\!r_2)\Big\{\frac{4\lambda}{\lambda\!+\!1} \frac{1}{(2\pi)^2}\!\!\int^{2\pi}_0\!\!\!\!\!e^{-it_1(m-r_1)}\!\!\!\int^{2\pi}_0\!\!\!\!e^{-it_2(m-r_2)}\overline{\varphi(e^{it_1}\!, \! e^{it_2})}dt_2dt_1\\ &\; \; -\frac{\lambda-1}{\lambda+1}\Big[\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \!\frac{\overline{u_{(1+r_1-v_1), (1+r_2-v_2)}^{v_1, v_2}}}{(m-v_1+1)!(m-v_2+1)!}\Big]\\ &\; \; -\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \frac{1}{(m-v_1)!(m-v_2)!} \frac{u_{(2m-r_1-1-v_1), (2m-r_2-1-v_2)}^{v_1, v_2}}{(2m-r_1-v_1)(2m-r_2-v_2)} \Big\}, \end{aligned} |
which leads to (3.1) for 1\leq m_1, \; m_2\leq m-1 .
(ⅱ) Multiplying both sides of the Eq (3.11) by e^{it_1}e^{it_2} , and then integrating with respect to t_1, t_2\in[0, 2\pi] yields that
\begin{aligned} &\frac{1}{(2\pi)^2}\int^{2\pi}_0e^{it_1}\Big[\int^{2\pi}_0e^{it_2}\varphi(e^{it_1}, e^{it_2})dt_2\Big]dt_1\\ & = \!\frac{\lambda-1}{4\lambda} \Big[\sum\limits_{m_2 = 0}^{+\infty}u_{0, m_2}^{0, 0}\frac{1}{2\pi}\int^{2\pi}_0e^{im_2t_2}dt_2 +\sum\limits_{m_1 = 1}^{+\infty}u_{m_1, 0}^{0, 0}\frac{1}{2\pi}\int^{2\pi}_0e^{im_1t_1}dt_1\\ &\; \; +\!\sum\limits_{m_1, m_2 = 1}^{+\infty}\!\!\!u_{m_1, m_2}^{0, 0}\frac{1}{2\pi}\int^{2\pi}_0e^{im_1t_1}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{im_2t_2}dt_2\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \sum\limits_{m_1 = 0}^{m-v_1}\sum\limits_{m_2 = 0}^{m-v_2} \!\frac{u_{m_1, m_2}^{v_1, v_2}\frac{1}{2\pi}\int^{2\pi}_0 e^{it_1(m_1-m+v_1)}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(m_2-m+v_2)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\\ &\; \; +\!\!\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \!\sum\limits_{m_1, m_2 = 0}^{+\infty} \!\!\!\!\!\frac{u_{(m_1+(m-v_1+1)), (m_2+(m-v_2+1))}^{v_1, v_2} \frac{1}{2\pi}\!\!\int^{2\pi}_0 \!\!\!e^{it_1(m_1+1)}dt_1\!\cdot\! \frac{1}{2\pi}\!\!\int^{2\pi}_0\!\!\!e^{it_2(m_2+1)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \sum\limits_{m_1 = 0}^{m-v_1}\sum\limits_{m_2 = 0}^{+\infty} \!\frac{u_{m_1, (m_2+(m-v_2+1))}^{v_1, v_2} \frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{it_1(m_1-m+v_1)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{it_2(m_2+1)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \sum\limits_{m_1 = 0}^{+\infty}\sum\limits_{m_2 = 0}^{m-v_2} \!\frac{u_{(m_1+(m-v_1+1)), m_2}^{v_1, v_2} \frac{1}{2\pi}\int^{2\pi}_0 e^{it_1(m_1+1)}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(m_2-m+v_2)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\Big]\\ &\; \; +\frac{\lambda+1}{4\lambda} \Big[\sum\limits_{m_1, m_2 = 1}^{+\infty}\frac{\overline{u_{(m_1-1), (m_2-1)}^{0, 0}}}{m_1m_2} \frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{it_1(1-m_1)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{it_2(1-m_2)}dt_2\\ &\; \; +\sum\limits_{m_1, m_2 = 0}^{+\infty}\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \frac{\frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{-it_1(m_1-m+v_1)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{-it_2(m_2-m+v_2)}dt_2} {(m-v_1)!(m-v_2)!}\\ &\; \; \cdot\frac{\overline{u_{m_1, m_2}^{v_1, v_2}}}{(m_1+1)(m_2+1)} \Big]+ \sum\limits_{m_1, m_2 = 0}^{+\infty}b_{m_1, m_2}\frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{it_1(m_1+1)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{it_2(m_2+1)}dt_2\\ & = \!\frac{\lambda-1}{4\lambda} \Big[u_{0, 0}^{0, 0}+\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \!\frac{u_{(m-v_1), (m-v_2)}^{v_1, v_2}} {(m-v_1+1)!(m-v_2+1)!}\Big]\\ &\; \; +\frac{\lambda+1}{4\lambda} \Big[\overline{u_{0, 0}^{0, 0}} +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \frac{\overline{u_{(m-v_1), (m-v_2)}^{v_1, v_2}}}{(m-v_1+1)!(m-v_2+1)!} \Big], \end{aligned} |
which leads to (3.2).
Additionally, for r_2\geq 1 , multiplying both sides of the Eq (3.11) by e^{it_1}e^{i(1-r_2)t_2} , and then integrating with respect to t_1, t_2\in[0, 2\pi] yields that
\begin{aligned} &\frac{1}{(2\pi)^2}\int^{2\pi}_0e^{it_1}\Big[\int^{2\pi}_0e^{it_2(1-r_2)}\varphi(e^{it_1}, e^{it_2})dt_2\Big]dt_1\\ & = \!\frac{\lambda-1}{4\lambda} \Big[\!\sum\limits_{m_2 = 0}^{+\infty}u_{0, m_2}^{0, 0} \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(m_2-r_2)}dt_2 +\!\sum\limits_{m_1 = 1}^{+\infty}u_{m_1, 0}^{0, 0}\frac{1}{2\pi}\int^{2\pi}_0e^{im_1t_1}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{-ir_2t_2}dt_2\\ &\; \; +\!\sum\limits_{m_1, m_2 = 1}^{+\infty}\!\!\!u_{m_1, m_2}^{0, 0}\frac{1}{2\pi}\int^{2\pi}_0e^{im_1t_1}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(m_2-r_2)}dt_2\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \sum\limits_{m_1 = 0}^{m-v_1}\sum\limits_{m_2 = 0}^{m-v_2} \!\frac{u_{m_1, m_2}^{v_1, v_2}\frac{1}{2\pi}\int^{2\pi}_0 e^{it_1(m_1-m+v_1)}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(m_2-m+v_2-r_2)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \!\sum\limits_{m_1, m_2 = 0}^{+\infty} \!\!\!\!\!\frac{u_{(m_1+(m-v_1+1)), (m_2+(m-v_2+1))}^{v_1, v_2} \frac{1}{2\pi}\!\!\int^{2\pi}_0 \!\!\!e^{it_1(m_1+1)}dt_1\!\cdot\! \frac{1}{2\pi}\!\!\int^{2\pi}_0\!\!\!e^{it_2(m_2+1-r_2)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \sum\limits_{m_1 = 0}^{m-v_1}\sum\limits_{m_2 = 0}^{+\infty} \!\frac{u_{m_1, (m_2+(m-v_2+1))}^{v_1, v_2} \frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{it_1(m_1-m+v_1)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{it_2(m_2+1-r_2)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \sum\limits_{m_1 = 0}^{+\infty}\sum\limits_{m_2 = 0}^{m-v_2} \!\frac{u_{(m_1+(m-v_1+1)), m_2}^{v_1, v_2} \frac{1}{2\pi}\int^{2\pi}_0 e^{it_1(m_1+1)}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(m_2-m+v_2-r_2)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\Big]\\ &\; \; +\frac{\lambda+1}{4\lambda} \Big[\sum\limits_{m_1, m_2 = 1}^{+\infty}\frac{\overline{u_{(m_1-1), (m_2-1)}^{0, 0}}}{m_1m_2} \frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{it_1(1-m_1)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{it_2(1-r_2-m_2)}dt_2\\ &\; \; +\sum\limits_{m_1, m_2 = 0}^{+\infty}\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \frac{\frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{-it_1(m_1-m+v_1)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{-it_2(m_2-m+v_2+r_2)}dt_2} {(m-v_1)!(m-v_2)!}\\ &\; \; \cdot\frac{\overline{u_{m_1, m_2}^{v_1, v_2}}}{(m_1+1)(m_2+1)} \Big]+ \sum\limits_{m_1, m_2 = 0}^{+\infty}b_{m_1, m_2}\frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{it_1(m_1+1)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{it_2(m_2+1-r_2)}dt_2\\ & = \!\frac{\lambda-1}{4\lambda} \Big[u^{0, 0}_{0, r_2}+\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \!\frac{u_{(m-v_1), (m-v_2+r_2)}^{v_1, v_2}} {(m-v_1+1)!(m-v_2+1)!}\Big]\\ &\; \; +\frac{\lambda+1}{4\lambda}\left\{ \begin{array}{ll} \sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \frac{1}{(m-v_1+1)!(m-v_2)!} \frac{\overline{u_{(m-v_1), (m-v_2-r_2)}^{v_1, v_2}}}{m-v_2-r_2+1}, \quad 1\leq r_2\leq \nu, \\ \sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-r_2}_{v_2 = 1} \frac{1}{(m-v_1+1)!(m-v_2)!} \frac{\overline{u_{(m-v_1), (m-v_2-r_2)}^{v_1, v_2}}}{m-v_2-r_2+1}, \quad \nu < r_2\leq m-1, \\ 0, \quad r_2 > m-1, \end{array} \right. \end{aligned} |
and the last equation is due to
\begin{aligned} &\sum\limits_{m_1, m_2 = 0}^{+\infty}\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \frac{\frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{-it_1(m_1-m+v_1)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{-it_2(m_2-m+v_2+r_2)}dt_2} {(m-v_1)!(m-v_2)!} \frac{\overline{u_{m_1, m_2}^{v_1, v_2}}}{(m_1+1)(m_2+1)}\\ & = \left\{ \begin{array}{ll} \sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \frac{1}{(m-v_1+1)!(m-v_2)!} \frac{\overline{u_{(m-v_1), (m-v_2-r_2)}^{v_1, v_2}}}{m-v_2-r_2+1}, \quad 1\leq r_2\leq \nu, \\ \sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-r_2}_{v_2 = 1} \frac{1}{(m-v_1+1)!(m-v_2)!} \frac{\overline{u_{(m-v_1), (m-v_2-r_2)}^{v_1, v_2}}}{m-v_2-r_2+1}, \quad \nu < r_2\leq m-1, \\ 0, \quad r_2 > m-1. \end{array} \right. \end{aligned} |
Therefore, for r_2\geq 1 ,
\begin{equation} \begin{aligned} u^{0, 0}_{0, r_2}& = \!\frac{4\lambda}{\lambda-1} \frac{1}{(2\pi)^2}\!\!\int^{2\pi}_0\!\!e^{it_1}\!\!\int^{2\pi}_0\!\!e^{it_2(1-r_2)}\varphi(e^{it_1}, e^{it_2})dt_2dt_1 \!-\!\!\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \!\frac{u_{(m-v_1), (m-v_2+r_2)}^{v_1, v_2}} {(m\!-\!v_1\!+\!1)!(m\!-\!v_2\!+\!1)!}\\ &\; \; -\frac{\lambda+1}{\lambda-1}\left\{ \begin{array}{ll} \sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \frac{1}{(m\!-\!v_1\!+\!1)!(m\!-\!v_2)!} \frac{\overline{u_{(m-v_1), (m-v_2-r_2)}^{v_1, v_2}}}{m\!-\!v_2\!-\!r_2\!+\!1}, \quad 1\leq r_2\leq \nu, \\ \sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-r_2}_{v_2 = 1} \frac{1}{(m\!-\!v_1\!+\!1)!(m\!-\!v_2)!} \frac{\overline{u_{(m-v_1), (m-v_2-r_2)}^{v_1, v_2}}}{m-v_2-r_2+1}, \quad \nu < r_2\leq m-1, \\ 0, \quad r_2 > m-1. \end{array} \right. \end{aligned} \end{equation} | (3.12) |
For r_1\geq 1 , multiplying both sides of the equation (3.11) by e^{i(1-r_1)t_1}e^{it_2} and then integrating with respect to t_1, t_2\in[0, 2\pi] , similar to (3.12), yields that
\begin{equation} \begin{aligned} u^{0, 0}_{r_1, 0}& = \!\frac{4\lambda}{\lambda-1} \frac{1}{(2\pi)^2}\!\!\int^{2\pi}_0\!\!e^{i(1-r_1)t_1}\!\!\int^{2\pi}_0\!\!e^{it_2}\varphi(e^{it_1}, e^{it_2})dt_2dt_1 \!-\!\!\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \!\frac{u_{(m-v_1+r_1), (m-v_2)}^{v_1, v_2}} {(m\!-\!v_1\!+\!1)!(m\!-\!v_2\!+\!1)!}\\ &\; \; -\frac{\lambda+1}{\lambda-1}\left\{ \begin{array}{ll} \!\! \sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \frac{1}{(m\!-\!v_1)!(m\!-\!v_2\!+\!1)!} \frac{\overline{u_{(m-v_1-r_1, (m-v_2))}^{v_1, v_2}}}{m-v_1-r_1+1}, \quad 1\!\leq\! r_1\!\leq\! \mu, \\ \!\! \sum\limits^{m-r_1}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \frac{1}{(m\!-\!v_1)!(m\!-\!v_2\!+\!1)!} \frac{\overline{u_{(m-v_1-r_1, (m-v_2))}^{v_1, v_2}}}{m-v_1-r_1+1}, \quad \mu\! < \! r_1\!\leq\! m\!-\!1, \\ \!0, \quad r_1 > m-1. \end{array} \right. \end{aligned} \end{equation} | (3.13) |
(ⅲ) In addition, integrating both sides of the Eq (3.11) with respect to t_1, t_2\in[0, 2\pi] yields that
\begin{aligned} &\frac{1}{(2\pi)^2}\int^{2\pi}_0\int^{2\pi}_0\varphi(e^{it_1}, e^{it_2})dt_2dt_1\\ & = \!\frac{\lambda-1}{4\lambda} \Big[\!\sum\limits_{m_2 = 0}^{+\infty}u_{0, m_2}^{0, 0}\frac{1}{2\pi}\int^{2\pi}_0e^{it_1}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(m_2-1)}dt_2\\ &\; \; +\!\sum\limits_{m_1 = 1}^{+\infty}u_{m_1, 0}^{0, 0}\frac{1}{2\pi}\int^{2\pi}_0e^{it_1(m_1-1)}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{-it_2}dt_2\\ &\; \; +\!\sum\limits_{m_1, m_2 = 1}^{+\infty}u_{m_1, m_2}^{0, 0}\frac{1}{2\pi}\int^{2\pi}_0e^{it_1(m_1-1)}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(m_2-1)}dt_2\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \sum\limits_{m_1 = 0}^{m-v_1}\sum\limits_{m_2 = 0}^{m-v_2} \!\frac{u_{m_1, m_2}^{v_1, v_2}\frac{1}{2\pi}\int^{2\pi}_0 e^{it_1(m_1+v_1-1-m)}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(m_2+v_2-1-m)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \!\sum\limits_{m_1, m_2 = 0}^{+\infty} \frac{u_{(m_1+(m-v_1+1)), (m_2+(m-v_2+1))}^{v_1, v_2} \frac{1}{2\pi}\!\!\int^{2\pi}_0 \!\!\!e^{im_1t_1}dt_1\!\cdot\! \frac{1}{2\pi}\!\!\int^{2\pi}_0\!\!\!e^{im_2t_2}dt_2} {(m-v_1+1)!(m-v_2+1)!}\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \sum\limits_{m_1 = 0}^{m-v_1}\sum\limits_{m_2 = 0}^{+\infty} \frac{u_{m_1, (m_2+(m-v_2+1))}^{v_1, v_2} \frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{it_1(m_1+v_1-1-m)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{im_2t_2}dt_2} {(m-v_1+1)!(m-v_2+1)!}\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \sum\limits_{m_1 = 0}^{+\infty}\sum\limits_{m_2 = 0}^{m-v_2} \frac{u_{(m_1+(m-v_1+1)), m_2}^{v_1, v_2} \frac{1}{2\pi}\int^{2\pi}_0 e^{im_1t_1}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(m_2+v_2-1-m)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\Big]\\ &\; \; +\frac{\lambda+1}{4\lambda} \Big[\sum\limits_{m_1, m_2 = 1}^{+\infty}\frac{\overline{u_{(m_1-1), (m_2-1)}^{0, 0}}}{m_1m_2} \frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{-im_1t_1}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{-im_2t_2}dt_2\\ &\; \; +\sum\limits_{m_1, m_2 = 0}^{+\infty}\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \frac{\frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{-it_1(m_1+1-m+v_1)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{-it_2(m_2+1-m+v_2)}dt_2} {(m-v_1)!(m-v_2)!}\\ &\; \; \cdot\frac{\overline{u_{m_1, m_2}^{v_1, v_2}}}{(m_1+1)(m_2+1)} \Big]+ \sum\limits_{m_1, m_2 = 0}^{+\infty}b_{m_1, m_2}\frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{im_1t_1}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{im_2t_2}dt_2\\ & = \!\frac{\lambda-1}{4\lambda} \Big[u^{0, 0}_{1, 1}+\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \!\frac{u_{(m-v_1+1), (m-v_2+1)}^{v_1, v_2}} {(m-v_1+1)!(m-v_2+1)!}\Big]\\ &\; \; +\frac{\lambda+1}{4\lambda} \sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \frac{1}{(m-v_1)!(m-v_2)!} \frac{\overline{u_{(m-v_1-1), (m-v_2-1)}^{v_1, v_2}}}{(m-v_1)(m-v_2)}+b_{0, 0}, \end{aligned} |
which follows (3.5), where u^{0, 0}_{1, 1} is determined by (3.1).
(ⅳ) In the case of r_1, r_2\geq m+1 , multiplying both sides of the Eq (3.11) by e^{ir_1t_1}e^{ir_2t_2} , and then integrating with respect to t_1, t_2\in[0, 2\pi] yields that
\begin{aligned} &\frac{1}{(2\pi)^2}\int^{2\pi}_0e^{ir_1t_1}\Big[\int^{2\pi}_0e^{ir_2t_2}\varphi(e^{it_1}, e^{it_2})dt_2\Big]dt_1\\ & = \!\frac{\lambda-1}{4\lambda} \Big[\!\sum\limits_{m_2 = 0}^{+\infty}u_{0, m_2}^{0, 0}\frac{1}{2\pi}\int^{2\pi}_0e^{it_1(r_1-1)}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(m_2-1+r_2)}dt_2\\ &\; \; +\!\sum\limits_{m_1 = 1}^{+\infty}u_{m_1, 0}^{0, 0}\frac{1}{2\pi}\int^{2\pi}_0e^{it_1(m_1-1+r_1)}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(-1+r_2)}dt_2\\ &\; \; +\!\sum\limits_{m_1, m_2 = 1}^{+\infty}\!\!\!u_{m_1, m_2}^{0, 0}\frac{1}{2\pi}\int^{2\pi}_0e^{it_1(m_1-1+r_1)}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(m_2-1+r_2)}dt_2\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \sum\limits_{m_1 = 0}^{m-v_1}\sum\limits_{m_2 = 0}^{m-v_2} \frac{u_{m_1, m_2}^{v_1, v_2}\frac{1}{2\pi}\int^{2\pi}_0 e^{it_1(m_1-(m-v_1+1)+r_1)}dt_1\cdot \frac{1}{2\pi}\int^{2\pi}_0e^{it_2(m_2-(m-v_2+1)+r_2)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \!\sum\limits_{m_1, m_2 = 0}^{+\infty} \frac{u_{(m_1+(m-v_1+1)), (m_2+(m-v_2+1))}^{v_1, v_2} \frac{1}{2\pi}\!\!\int^{2\pi}_0 \!\!\!e^{it_1(m_1+r_1)}dt_1\!\cdot\! \frac{1}{2\pi}\!\!\int^{2\pi}_0\!\!\!e^{it_2(m_2+r_2)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \sum\limits_{m_1 = 0}^{m-v_1}\sum\limits_{m_2 = 0}^{+\infty} \frac{u_{m_1, (m_2+(m-v_2+1))}^{v_1, v_2} \frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{it_1(m_1-(m-v_1+1)+r_1)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{it_2(m_2+r_2)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\\ &\; \; +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \sum\limits_{m_1 = 0}^{+\infty}\sum\limits_{m_2 = 0}^{m-v_2} \frac{u_{(m_1+(m-v_1+1)), m_2}^{v_1, v_2} \frac{1}{2\pi}\!\!\int^{2\pi}_0 \!\!e^{it_1(m_1+r_1)}dt_1\cdot \frac{1}{2\pi}\!\!\int^{2\pi}_0\!\!e^{it_2(m_2-(m-v_2+1)+r_2)}dt_2} {(m-v_1+1)!(m-v_2+1)!}\Big]\\ &\; \; +\frac{\lambda+1}{4\lambda} \Big[\sum\limits_{m_1, m_2 = 1}^{+\infty}\frac{\overline{u_{(m_1-1), (m_2-1)}^{0, 0}}}{m_1m_2} \frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{it_1(-m_1+r_1)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{it_2(-m_2+r_2)}dt_2\\ &\; \; +\sum\limits_{m_1, m_2 = 0}^{+\infty}\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \frac{\frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{-it_1(m_1+1-m+v_1-r_1)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{-it_2(m_2+1-m+v_2-r_2)}dt_2} {(m-v_1)!(m-v_2)!}\\ &\; \; \cdot\frac{\overline{u_{m_1, m_2}^{v_1, v_2}}}{(m_1+1)(m_2+1)} \Big]+ \sum\limits_{m_1, m_2 = 0}^{+\infty}b_{m_1, m_2}\frac{1}{2\pi}\!\int^{2\pi}_0 \!e^{it_1(m_1+r_1)}dt_1\cdot \frac{1}{2\pi}\!\int^{2\pi}_0\!e^{it_2(m_2+r_2)}dt_2\\ & = \frac{\lambda+1}{4\lambda} \Big[\frac{\overline{u_{(r_1-1), (r_2-1)}^{0, 0}}}{r_1r_2} +\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \frac{1}{(m-v_1)!(m-v_2)!} \frac{\overline{u_{(m-v_1+r_1-1), (m-v_2+r_2-1)}^{v_1, v_2}}}{(m-v_1+r_1)(m-v_2+r_2)} \Big]. \end{aligned} |
Therefore,
\begin{equation} \begin{aligned} u_{r_1, r_2}^{0, 0}& = \frac{4\lambda}{\lambda+1} \frac{(r_1+1)(r_2+1)}{(2\pi)^2}\int^{2\pi}_0e^{-i(r_1+1)t_1}\int^{2\pi}_0e^{-i(r_2+1)t_2} \overline{\varphi(e^{it_1}, e^{it_2})}dt_2dt_1\\ &\; \; -\sum\limits^{m-\mu}_{v_1 = 1}\sum\limits^{m-\nu}_{v_2 = 1} \frac{(r_1+1)(r_2+1)}{(m-v_1)!(m-v_2)!} \frac{u_{(m-v_1+r_1), (m-v_2+r_2)}^{v_1, v_2}}{(m-v_1+r_1+1)(m-v_2+r_2+1)}, \end{aligned} \end{equation} | (3.14) |
for r_1, r_2\geq m .
Similarly, in the case of r_1\geq m+1 and 0\leq r_2\leq m-2 , multiplying both sides of the Eq (3.11) by e^{ir_1t_1}e^{it_2(m-r_2)} , and then integrating with respect to t_1, t_2\in[0, 2\pi] yields that u_{r_1, r_2}^{0, 0} ( r_1\geq m, \; 1\leq r_2\leq m-1 ) has the same representation as (3.14). In the case of 0\leq r_1\leq m-2 and r_2\geq m+1 , multiplying both sides of the Eq (3.11) by e^{it_1(m-r_1)}e^{ir_2t_2} , and then integrating with respect to t_1, t_2\in[0, 2\pi] yields that u_{r_1, r_2}^{0, 0} ( 1\leq r_1\leq m-1, \; r_2\geq m ) has the same representation as (3.14).
(ⅴ) Similarly, for r_1, r_2\geq 1 , multiplying both sides of the Eq (3.11) by e^{-ir_1t_1}e^{-ir_2t_2} , and then integrating with respect to t_1, t_2\in[0, 2\pi] , we can get the expression of b_{r_1r_2} which leads to (3.7). For r_1\geq 1 , multiplying both sides of the Eq (3.11) by e^{-ir_1t_1} , and then integrating with respect to t_1, t_2\in[0, 2\pi] yields the expression of b_{r_10} which leads to (3.8). For r_2\geq 1 , we can get b_{0r_2} 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 |