In this study, we used a natural resource, yarosite minerals, as a Fe2O3 precursor. Yarosite minerals were used for the synthesis of LaFeO3/Fe2O3 doped with ZnO via a co-precipitation method using ammonium hydroxide, which produced a light brown powder. Then, an ethanol gas sensor was prepared using a screen-printing technique and characterized using gas chamber tools at 100,200, and 300 ppm of ethanol gas to investigate the sensor's performance. Several factors that substantiate electrical properties such as crystal and morphological structures were also studied using X-Ray Diffraction (XRD) and Scanning Electron Microscopy (SEM), respectively. The crystallite size decreased from about 61.4 nm to 28.8 nm after 0.5 mol% ZnO was added. The SEM characterization images informed that the modified LaFeO3 was relatively the same but not uniform. Lastly, the sensor's electrical properties exhibited a high response of about 257% to 309% at an operating temperature that decreased from 205 ℃ to 180 ℃. This finding showed that these natural resources have the potential to be applied in the development of ethanol gas sensors in the future. Hence, yarosite minerals can be considered a good natural resource that can be further explored to produce an ethanol gas sensor with more sensitive response. In addition, this method reduces the cost of material purchase.
Citation: Endi Suhendi, Andini Eka Putri, Muhamad Taufik Ulhakim, Andhy Setiawan, Syarif Dani Gustaman. Investigation of ZnO doping on LaFeO3/Fe2O3 prepared from yarosite mineral extraction for ethanol gas sensor applications[J]. AIMS Materials Science, 2022, 9(1): 105-118. doi: 10.3934/matersci.2022007
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In this study, we used a natural resource, yarosite minerals, as a Fe2O3 precursor. Yarosite minerals were used for the synthesis of LaFeO3/Fe2O3 doped with ZnO via a co-precipitation method using ammonium hydroxide, which produced a light brown powder. Then, an ethanol gas sensor was prepared using a screen-printing technique and characterized using gas chamber tools at 100,200, and 300 ppm of ethanol gas to investigate the sensor's performance. Several factors that substantiate electrical properties such as crystal and morphological structures were also studied using X-Ray Diffraction (XRD) and Scanning Electron Microscopy (SEM), respectively. The crystallite size decreased from about 61.4 nm to 28.8 nm after 0.5 mol% ZnO was added. The SEM characterization images informed that the modified LaFeO3 was relatively the same but not uniform. Lastly, the sensor's electrical properties exhibited a high response of about 257% to 309% at an operating temperature that decreased from 205 ℃ to 180 ℃. This finding showed that these natural resources have the potential to be applied in the development of ethanol gas sensors in the future. Hence, yarosite minerals can be considered a good natural resource that can be further explored to produce an ethanol gas sensor with more sensitive response. In addition, this method reduces the cost of material purchase.
Bi-analytic functions arise from the research on systems of some partial differential equations and play an important role in studying elasticity problems. In 1961, Sander [1] studied a system of partial differential equations of first order:
{∂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 k∈R,k≠−1, and introduced the concept of bi-analytic functions of type k. He extended the elementary properties of analytic functions to bi-analytic functions and extended the related problems of the plane strain, the generalized plane stress, and the flow of viscous fluids to the theory of bi-analytic functions.
Later, Lin and Wu [2] introduced a more extensive class of functions, i.e., bi-analytic functions of 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 f(z)=u+iv, λ,k are real constants with λ≠0,1,k2 and 0<k<1. The complex form of the system of equations 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 associated function of f(z). Lin and Wu [2] obtained the general expression and some properties of bi-analytic functions of type (λ,k), including the Cauchy integral theorem and formula, the Morera theorem, the Weierstrass theorem, and the power series expansions.
Hua et al. [3,4] investigated the systems of partial differential equations of second order, which have close association with (λ,k) bi-analytic functions, and showed that (λ,k) bi-analytic functions provide a powerful tool to deal with problems of plane elasticity.
Mu [5] obtained the Cauchy integral representation, the removable singularity theorem, the Weierstrass theorem, and the mean value theorem of bi-analytic functions of type (λ,k) in the complex plane. Wen et al. [6,7] studied various types of boundary value problems for elliptic complex equations and systems, including boundary problems of (λ,k) bi-analytic functions. Begehr and Lin [8] studied a mix-contact problem by applying (λ,k) bi-analytic functions and singular integral operators. In addition, Lin and Zhao [9] reviewed the work on the applications of bi-analytic functions in elasticity, especially for solving the basic boundary value problems of plane elasticity (in isotropic or orthotropic cases). In [10], Lai investigated the properties of the vector-valued bi-analytic functions of type (λ,k) in a complex couple Banach space and obtained the solution of the corresponding Dirichlet problem.
The special case of bi-analytic functions of type (λ,k) for k=1 is
∂f∂ˉz=λ+14λϕ(z)+λ−14λ¯ϕ(z),∂ϕ∂ˉz=0, |
which means
(1−λ)∂2f∂ˉz2+(1+λ)¯∂2f∂z∂ˉz=0. |
Thus, bi-analytic functions are different from bianalytic functions (which only satisfy ∂2f∂ˉz2=0) in the sense of Bitsadze [11]. The generalization of a bianalytic function f is ∂nf∂ˉzn=0(n≥1), which is called a polyanalytic function [12]. Obviously, bi-analytic functions can be similarly generalized.
In [13], Kumar extended bi-analytic functions on bounded simply connected domains. He considered the system of complex equations:
∂f∂ˉz=λ+14λϕ(z)+λ−14λ¯ϕ(z),∂nϕ∂ˉzn=0(n≥1), |
where f is called bi-polyanalytic functions, and obtained the expressions of solutions to the corresponding boundary value problems on the unit disc. In [13], the systems
∂f∂ˉz=λ+14λϕ(z)+λ−14λ¯ϕ(z),∂nϕ∂ˉzn=h(z,ϕ,f,fz),∂f∂ˉz=λ+14λϕ(z)+λ−14λ¯ϕ(z),∂nϕ∂ˉzn=h(z,ϕ,ϕz,f), |
and the associated boundary problems were also discussed.
Begehr et al. [14,15] investigated boundary value problems for bi-polyanalytic functions in the upper half plane and on the unit disc. Similar researches can also be found in the reference [16].
In 2022, Lin and Xu [17] successfully obtained the solutions of Riemann problems of (λ,k) bi-analytic functions. In 2023, Lin [18] discussed a type of inverse boundary value problems for (λ,1) bi-analytic functions. Nowadays, bi-analytic functions and polyanalytic functions have made great progress [19,20,21,22].
The above conclusions are all in the complex plane. If these results can be extended to spaces of several complex variables, there will be more applications. There have been some conclusions about bi-analytic functions or polyanalytic functions with several complex variables. For example, in [23], Begehr and Kumar studied complex bi-analytic functions of n variables. They successfully obtained the corresponding Cauchy integral formula, Taylor series and Poisson integral formula on the polydisc, and discussed the Dirichlet problem of the systems of partial differential equations on the polydisc. In [24], Kumar discussed a generalized Riemann boundary value problem by the Cauchy integral of bi-analytic functions with two complex variables. In 2023, Vasilevski [25] studied polyanalytic functions with several complex variables in detail.
However, there is relatively little research on bi-polyanalytic functions in spaces of several complex variables. Inspired by these, and on the basis of the previous works of the former researchers, we first investigate a kind of boundary value problem for polyanalytic functions with Schwarz conditions on the bicylinder, then we discuss boundary value problems with the Dirichlet, Neumann boundary conditions and mixed type boundary conditions for bi-polyanalytic functions on the bicylinder.
Throughout this paper, let the bicylinder D2=D1×D2={(z1,z2):|z1|<1,|z2|<1}, whose characteristic boundary is denoted as ∂0D2, where z1 and z2 are on the complex plane. Let Cm(G) represent the set of functions whose partial derivatives of order m are all continuous within a bounded smooth domain G.
Lemma 2.1. [26] Let G be a bounded smooth domain in the complex plane, f∈L1(G;C) and
Tf(z)=−1π∫Gf(ζ)ζ−zdξdη,ζ=ξ+iη. |
Then, ∂ˉzTf(z)=f(z).
Lemma 2.2. [26] Let w∈C1(G;C)⋂C(¯G;C), where m≥1 and G is a bounded smooth domain in the complex plane, then
w(z)=12πi∫∂Gw(ζ)dζζ−z−1π∫Gwˉζ(ζ)dσζζ−z,∫Gwˉz(z)dσz=12i∫∂Gw(z)dz,∫Gwz(z)dσz=−12i∫∂Gw(z)dˉz, |
where dσζ=dξdη(ζ=ξ+iη) and dσz=dxdy(z=x+iy).
Lemma 2.3. [27] For n∈N, f∈Lp(D),p≥1 with D being the unit disc, let
Tnf(z)=(−1)n2π(n−1)!∫D(¯ζ−z+ζ−z)n−1[f(ζ)ζζ+zζ−z+¯f(ζ)ˉζ1+zˉζ1−zˉζ]dξdη, |
and let T0f=f. Then
∂l∂ˉzlTnf=Tn−lf,1≤l≤n,ℜ∂l∂ˉzlTnf=0(z∈∂D),ℑ∂l∂ˉzlTnf(0)=0,0≤l≤n−1. |
Lemma 2.4. Let gμν∈C(∂0D2;R) for 1≤μ,ν≤m−1 (m≥2), and let
˜ϕ(z)=m−1∑˜v1=μm−1∑˜v2=νˉz˜v11ˉz˜v22˜v1!˜v2!u˜v1˜v2(z), | (2.1) |
where
u˜v1˜v2={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=μ,ν<˜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−1,ν<˜v2≤m−1, |
in which
{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]. |
Then, ˜ϕ(z) is a specific solution to the problem
{∂mˉz1∂mˉz2˜ϕ(z)=0,z∈D2,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. Obviously, u˜v1˜v2(z) is analytic on D2, therefore, ∂mˉz1∂mˉz2˜ϕ(z)=0. In addition, from the proof of Theorem 2.1 in [28]: We obtain that ℜ∂μˉz1∂νˉz2˜ϕ(z)=gμν(z) for z∈∂0D2, and
ℑ∂μˉz1∂νˉz2˜ϕ(0,z2)=0=ℑ∂μˉz1∂νˉz2˜ϕ(z1,0) |
for z1∈D1,z2∈D2. So, ˜ϕ(z) is a specific solution to the problem.
Lemma 2.5. [15] Let D be the unit disc. For 1≤k≤n−1, let gk∈C(∂D) and ck∈C be given. Then, there exists an analytic function uk on D satisfying
∂νk−1∑μ=01μ!un+μ−k(z)ˉzμ=gn−k(z)(z∈∂D),un−k(0)=cn−k |
if and only if for ∀z∈D
{12πi∫∂Dgn−1(ζ)dζ(1−¯zζ)ζ=0,k−1∑v=0(−1)vv!12πi∫∂D¯ζvgn+v−k(ζ)dζ1−¯zζ=cn−k+1,2≤k≤n−1, |
and uk is uniquely represented as
un−k(z)=cn−k−12πi∫∂Dk−1∑v=0(−2)vv!ˉζvgn+v−k(ζ)log(1−zˉζ)dζζ−k−1∑λ=112πi∫∂Dk−1−λ∑μ=0(−2)μμ!ˉζμgn+λ+μ−k(ζ)1λ!zλ[log(1−zˉζ)+λ∑σ=1zσˉζσσ]dζζ. |
Lemma 2.6. [15] Let f∈L1(D), φ∈C(∂D) and c∈C be given; then the problem
wˉz=f(z∈D),∂νw(z)=φ(z)(z∈∂D),w(0)=c |
has a unique solution
w(z)=c−12πi∫∂Dφ(ζ)log(1−zˉζ)dζζ−12πi∫∂Df(ζ)log(1−zˉζ)dˉζ−1π∫Dzf(ζ)ζ(ζ−z)dσζ, |
if and only if
12πi∫∂Dφ(ζ)dζ(1−ˉzζ)ζ+12πi∫∂Df(ζ)dˉζ1−ˉzζ=−1π∫Df(ζ)ˉz(1−ˉzζ)2dσζ. |
Lemma 2.7. Let u0 be an analytic function on D2. Then
1π2∫D2u0(ζ)dσζ1ζ1−z1dσζ2ζ2−z2=(ˉz2−1z2)[ˉz1u0(z)−1z1(u0(z)−u0(0,z2))]+1z2[ˉz1u0(z1,0)−1z1(u0(z1,0)−u0(0,0))], | (2.2) |
and
1π2∫D2¯u0(ζ)dσζ1ζ1−z1dσζ2ζ2−z2=¯u1(z)−¯u1(0,z2)−(¯u1(z1,0)−¯u1(0,0)), | (2.3) |
where ∂ζ1∂ζ2u1(ζ)=u0(ζ).
Proof. By Lemma 2.2,
1π2∫D2u0(ζ)dσζ1ζ1−z1dσζ2ζ2−z2=−1π∫D1[−1π∫D2∂ˉζ2(ˉζ2u0(ζ))dσζ2ζ2−z2]dσζ1ζ1−z1=−1π∫D1[ˉz2u0(ζ1,z2)−12πi∫∂D2ˉζ2u0(ζ)dζ2ζ2−z2]dσζ1ζ1−z1=−1π∫D1[ˉz2u0(ζ1,z2)−1z2(u0(ζ1,z2)−u0(ζ1,0))]dσζ1ζ1−z1=(ˉz2−1z2)−1π∫D1u0(ζ1,z2)dσζ1ζ1−z1+1z2−1π∫D1u0(ζ1,0)dσζ1ζ1−z1=(ˉz2−1z2)−1π∫D1∂ˉζ1(ˉζ1u0(ζ1,z2))dσζ1ζ1−z1+1z2−1π∫D1∂ˉζ1(ˉζ1u0(ζ1,0))dσζ1ζ1−z1=(ˉz2−1z2)[ˉz1u0(z1,z2)−12πi∫∂D1ˉζ1u0(ζ1,z2)dζ1ζ1−z1]+1z2[ˉz1u0(z1,0)−12πi∫∂D1ˉζ1u0(ζ1,0)dζ1ζ1−z1]=(ˉz2−1z2)[ˉz1u0(z)−1z1(u0(z)−u0(0,z2))]+1z2[ˉz1u0(z1,0)−1z1(u0(z1,0)−u0(0,0))]. |
So we obtain (2.2). Let ∂ζ1u1(ζ)=u2(ζ). Since ∂ζ1∂ζ2u1(ζ)=u0(ζ), then we have ∂ζ2u2(ζ)=u0(ζ), which follows ∂ˉζ2¯u2(ζ)=¯u0(ζ). Therefore, by Lemma 2.2,
−1π∫D2¯u0(ζ)dσζ2ζ2−z2=−1π∫D2∂ˉζ2¯u2(ζ)dσζ2ζ2−z2=¯u2(ζ1,z2)−12πi∫∂D2¯u2(ζ)dζ2ζ2−z2=¯u2(ζ1,z2)−¯12πi∫∂D2u2(ζ)dζ2ζ2(1−ˉz2ζ2)=¯u2(ζ1,z2)−¯u2(ζ1,0). | (2.4) |
In addition, ∂ζ1u1(ζ)=u2(ζ) follows ∂ˉζ1¯u1(ζ)=¯u2(ζ). Similarly, we have
−1π∫D1¯u2(ζ1,z2)dσζ1ζ1−z1=¯u1(z)−¯u1(0,z2). | (2.5) |
Therefore, by (2.4) and (2.5), we obtain
1π2∫D2¯u0(ζ)dσζ1ζ1−z1dσζ2ζ2−z2=−1π∫D1[−1π∫D2dσζ2ζ2−z2]dσζ1ζ1−z1=−1π∫D1¯u2(ζ1,z2)dσζ1ζ1−z1−−1π∫D1¯u2(ζ1,0)dσζ1ζ1−z1=¯u1(z)−¯u1(0,z2)−(¯u1(z1,0)−¯u1(0,0)). |
Lemma 2.8. Let φ∈C(∂0D2;C) and gμν∈C(∂0D2;R) for 1≤μ,ν≤m−1 (m≥2), and let λ∈R∖{−1,0,1}. Let W(z) and u0(z) be analytic functions on D2 and
W(τ)=φ(τ)−1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1ζ1−τ1dσζ2ζ2−τ2−λ−14λ{(ˉτ2−1τ2)[ˉτ1u0(τ)−u0(τ)−u0(0,τ2)τ1]+1τ2[ˉτ1u0(τ1,0)−u0(τ1,0)−u0(0)τ1]}−λ+14λ[¯u1(τ)−¯u1(0,τ2)−¯u1(τ1,0)+¯u1(0,0)] | (2.6) |
for τ∈∂D2, where ˜ϕ is determined in Lemma 2.4 and ∂ζ1∂ζ2u1(ζ)=u0(ζ). Then
W(z)=1(2πi)2∫∂0D2φ(ζ)dζ1dζ2(ζ1−z1)(ζ2−z2), | (2.7) |
and
u0(z)=4λλ+1{¯1(2πi)2∫∂0D2φ(ζ)dζ1(1−ˉz1ζ1)2dζ2(1−ˉz2ζ2)2−¯1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2(1−ˉz1ζ1)2(1−ˉz2ζ2)2}−(λ−1){1(2πi)2∫∂0D2φ(ζ)dζ−1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2}+(λ−1)2λ+1{¯1(2πi)2∫∂0D2φ(ζ)dζ−¯1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2}. | (2.8) |
Proof. As W(z) is analytic, applying the properties of the Poisson kernel on D2, W(z) can be expressed as
W(z)=1(2πi)2∫∂0D2W(ζ)(ζ1−z1)(ζ2−z2)dζ1dζ2 | (2.9) |
if and only if
1(2πi)2∫∂0D2W(ζ)[ˉz11−ˉz1ζ1ˉz21−ˉz2ζ2+ˉz21−ˉz2ζ21ζ1−z1+ˉz11−ˉz1ζ11ζ2−z2]dζ1dζ2=0. | (2.10) |
(2.6) and (2.9) derive the expression of W:
W(z)=1(2πi)2∫∂0D2W(ζ)(ζ1−z1)(ζ2−z2)dζ1dζ2=1(2πi)2∫∂0D2{φ(ζ)−1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)]dσ~ζ1~ζ1−ζ1dσ~ζ2~ζ2−ζ2−λ−14λ[(ˉζ2−1ζ2)(ˉζ1u0(ζ)−u0(ζ)−u0(0,ζ2)ζ1)+1ζ2(ˉζ1u0(ζ1,0)−u0(ζ1,0)−u0(0)ζ1)]−λ+14λ[¯u1(ζ)−¯u1(0,ζ2)−¯u1(ζ1,0)+¯u1(0,0)]}dζ1dζ2(ζ1−z1)(ζ2−z2)=1(2πi)2∫∂0D2{φ(ζ)−1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)]dσ~ζ1~ζ1−ζ1dσ~ζ2~ζ2−ζ2−λ−14λu0(0)ζ1ζ2−λ+14λ[¯u1(ζ)−¯u1(0,ζ2)−¯u1(ζ1,0)+¯u1(0,0)]}dζ1dζ2(ζ1−z1)(ζ2−z2)=1(2πi)2∫∂0D2{φ(ζ)−1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)]dσ~ζ1~ζ1−ζ1dσ~ζ2~ζ2−ζ2}dζ1dζ2(ζ1−z1)(ζ2−z2)−λ−14λu0(0)12πi∫∂D2[12πi∫∂D1dζ1ζ1(ζ1−z1)]dζ2ζ2(ζ2−z2)−λ+14λ¯12πi∫∂D2{12πi∫∂D1[u1(ζ)−u1(0,ζ2)−u1(ζ1,0)+u1(0)]dζ1ζ1(1−ˉz1ζ1)}dζ2ζ2(1−ˉz2ζ2)=1(2πi)2∫∂0D2{φ(ζ)−1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)]dσ~ζ1~ζ1−ζ1dσ~ζ2~ζ2−ζ2}dζ1dζ2(ζ1−z1)(ζ2−z2)=1(2πi)2∫∂0D2φ(ζ)dζ1dζ2(ζ1−z1)(ζ2−z2)−12πi∫∂D1{12πi∫∂D2[1π2∫D2(λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ))dσ~ζ1~ζ1−ζ1dσ~ζ2~ζ2−ζ2]dζ2ζ2−z2}dζ1ζ1−z1=1(2πi)2∫∂0D2φ(ζ)dζ1dζ2(ζ1−z1)(ζ2−z2)−12πi∫∂D1{1π2∫D2(λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ))[12πi∫∂D21~ζ2−ζ2dζ2ζ2−z2]dσ~ζ1~ζ1−ζ1dσ~ζ2=1(2πi)2∫∂0D2φ(ζ)dζ1dζ2(ζ1−z1)(ζ2−z2), |
which is due to
12πi∫∂D1dζ1ζ1(ζ1−z1)=0,12πi∫∂D21~ζ2−ζ2dζ2ζ2−z2=0 |
and
12πi∫∂D1[u1(ζ1,ζ2)−u1(0,ζ2)−u1(ζ1,0)+u1(0,0)]dζ1ζ1(1−ˉz1ζ1)=0 |
for z∈D2.
Similarly, by (2.6) and (2.10), we obtain
1(2πi)2∫∂0D2W(ζ)[ˉz11−ˉz1ζ1ˉz21−ˉz2ζ2+ˉz21−ˉz2ζ21ζ1−z1+ˉz11−ˉz1ζ11ζ2−z2]dζ1dζ2=1(2πi)2∫∂0D2{φ(ζ)−1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)]dσ~ζ1~ζ1−ζ1dσ~ζ2~ζ2−ζ2−λ−14λ[(ˉζ2−1ζ2)(ˉζ1u0(ζ)−u0(ζ)−u0(0,ζ2)ζ1)+1ζ2(ˉζ1u0(ζ1,0)−u0(ζ1,0)−u0(0)ζ1)]−λ+14λ[¯u1(ζ)−¯u1(0,ζ2)−¯u1(ζ1,0)+¯u1(0,0)]}⋅[ˉz11−ˉz1ζ1ˉz21−ˉz2ζ2+ˉz21−ˉz2ζ21ζ1−z1+ˉz11−ˉz1ζ11ζ2−z2]dζ1dζ2=1(2πi)2∫∂0D2{φ(ζ)−1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)]dσ~ζ1~ζ1−ζ1dσ~ζ2~ζ2−ζ2−λ−14λ1ζ1ζ2u0(0)−λ+14λ[¯u1(ζ)−¯u1(0,ζ2)−¯u1(ζ1,0)+¯u1(0,0)]}⋅[ˉz11−ˉz1ζ1ˉz21−ˉz2ζ2+ˉz21−ˉz2ζ21ζ1−z1+ˉz11−ˉz1ζ11ζ2−z2]dζ1dζ2=0. |
Therefore,
1(2πi)2∫∂0D2{φ(ζ)−1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)]dσ~ζ1~ζ1−ζ1dσ~ζ2~ζ2−ζ2}⋅[ˉz11−ˉz1ζ1ˉz21−ˉz2ζ2+ˉz21−ˉz2ζ21ζ1−z1+ˉz11−ˉz1ζ11ζ2−z2]dζ1dζ2=λ−14λu0(0)1(2πi)2∫∂0D2[ˉz11−ˉz1ζ1ˉz21−ˉz2ζ2+ˉz21−ˉz2ζ21ζ1−z1+ˉz11−ˉz1ζ11ζ2−z2]dζ1dζ2ζ1ζ2+λ+14λ1(2πi)2∫∂0D2[¯u1(ζ)−¯u1(0,ζ2)−¯u1(ζ1,0)+¯u1(0,0)]}⋅[ˉz11−ˉz1ζ1ˉz21−ˉz2ζ2+ˉz21−ˉz2ζ21ζ1−z1+ˉz11−ˉz1ζ11ζ2−z2]dζ1dζ2=λ−14λu0(0){12πi∫∂D1ˉz11−ˉz1ζ1dζ1ζ1[12πi∫∂D2ˉz21−ˉz2ζ2dζ2ζ2+12πi∫∂D2dζ2(ζ2−z2)ζ2]+12πi∫∂D1dζ1(ζ1−z1)ζ1⋅12πi∫∂D2ˉz21−ˉz2ζ2dζ2ζ2}+λ+14λ{¯12πi∫∂D2[12πi∫∂D1[u1(ζ)−u1(0,ζ2)−u1(ζ1,0)+u1(0)]z1dζ1ζ1(ζ1−z1)]z2dζ2ζ2(ζ2−z2)+¯12πi∫∂D2[12πi∫∂D1[u1(ζ)−u1(0,ζ2)−u1(ζ1,0)+u1(0)]dζ1ζ1(1−ˉz1ζ1)]z2dζ2ζ2(ζ2−z2)+¯12πi∫∂D1[12πi∫∂D2[u1(ζ)−u1(0,ζ2)−u1(ζ1,0)+u1(0)]dζ2ζ2(1−ˉz2ζ2)]z1dζ1ζ1(ζ1−z1)} |
=λ−14λu0(0)ˉz1ˉz2+λ+14λ{¯12πi∫∂D2[u1(z1,ζ2)−u1(0,ζ2)−u1(z1,0)+u1(0)]z2dζ2ζ2(ζ2−z2)+¯12πi∫∂D2u1(ζ)−u1(0,ζ2)−u1(ζ1,0)+u1(0)1−ˉz1ζ1|ζ1=0z2dζ2ζ2(ζ2−z2)+¯12πi∫∂D1u1(ζ)−u1(0,ζ2)−u1(ζ1,0)+u1(0)1−ˉz2ζ2|ζ2=0z1dζ1ζ1(ζ1−z1)}=λ−14λu0(0)ˉz1ˉz2+λ+14λ{¯[u1(z1,ζ2)−u1(0,ζ2)−u1(z1,0)+u1(0)]ζ2=z2−¯[u1(z1,ζ2)−u1(0,ζ2)−u1(z1,0)+u1(0)]ζ2=0}=λ−14λu0(0)ˉz1ˉz2+λ+14λ¯[u1(z)−u1(0,z2)−u1(z1,0)+u1(0)], | (2.11) |
which is in virtue of
12πi∫∂D1ˉz11−ˉz1ζ1dζ1ζ1=ˉz1,12πi∫∂D2ˉz21−ˉz2ζ2dζ2ζ2=ˉz2 |
and
12πi∫∂D1dζ1ζ1(ζ1−z1)=0=12πi∫∂D2dζ2ζ2(ζ2−z2). |
Taking the partial derivative on both sides of Eq (2.11) with respect to ˉz1ˉz2 gives
1(2πi)2∫∂0D2{φ(ζ)−1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)]dσ~ζ1~ζ1−ζ1dσ~ζ2~ζ2−ζ2}dζ1(1−ˉz1ζ1)2dζ2(1−ˉz2ζ2)2=λ−14λu0(0)+λ+14λ¯u0(z). |
Thus,
λ−14λu0(0)+λ+14λ¯u0(z)=1(2πi)2∫∂0D2φ(ζ)dζ1(1−ˉz1ζ1)2dζ2(1−ˉz2ζ2)2−12πi∫∂D2{1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)]⋅[12πi∫∂D11~ζ1−ζ1dζ1(1−ˉz1ζ1)2]dσ~ζ1dσ~ζ2~ζ2−ζ2}dζ2(1−ˉz2ζ2)2=1(2πi)2∫∂0D2φ(ζ)dζ1(1−ˉz1ζ1)2dζ2(1−ˉz2ζ2)2−1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)]−dσ~ζ1(1−ˉz1~ζ1)2[12πi∫∂D21~ζ2−ζ2dζ2(1−ˉz2ζ2)2]dσ~ζ2=1(2πi)2∫∂0D2φ(ζ)dζ1(1−ˉz1ζ1)2dζ2(1−ˉz2ζ2)2−1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2(1−ˉz1ζ1)2(1−ˉz2ζ2)2. | (2.12) |
Particularly,
λ−14λu0(0)+λ+14λ¯u0(0)=1(2πi)2∫∂0D2φ(ζ)dζ−1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2, | (2.13) |
which follows
λ−14λ¯u0(0)+λ+14λu0(0)=¯1(2πi)2∫∂0D2φ(ζ)dζ−¯1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2, | (2.14) |
(2.13) and (2.14) derive
u0(0)=(λ+1){¯1(2πi)2∫∂0D2φ(ζ)dζ−¯1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2}−(λ−1){1(2πi)2∫∂0D2φ(ζ)dζ−1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2}. | (2.15) |
Plugging (2.15) into (2.12), we obtain the expression of u0(z), i.e., (2.8).
Theorem 3.1. Let φ∈C(∂0D2;C) and gμν∈C(∂0D2;R) for 1≤μ,ν≤m−1 (m≥2), and let λ∈R∖{−1,0,1}. 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) |
for 1≤μ,ν≤m−1 has a unique solution
f(z)=1(2πi)2∫∂0D2φ(ζ)(ζ1ζ1−z1+ˉζ1¯ζ1−z1−1)(ζ2ζ2−z2+ˉζ2¯ζ2−z2−1)dζζ+λ−14λ[(|z1|2−1)(|z2|2−1)z1z2u0(z)+|z2|2−1z1z2u0(0,z2)+|z1|2−1z1z2u0(z1,0)+1−|z1|2|z2|2z1z2u0(0)]+1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)][1(ζ1−z1)(ζ2−z2)−ˉz1ˉz2(ˉz1ζ1−1)(ˉz2ζ2−1)]dσζ1dσζ2, | (3.1) |
where ˜ϕ is determined by (2.1) and
u0(z)=4λλ+1{¯1(2πi)2∫∂0D2φ(ζ)dζ1(1−ˉz1ζ1)2dζ2(1−ˉz2ζ2)2−¯1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2(1−ˉz1ζ1)2(1−ˉz2ζ2)2}−(λ−1){1(2πi)2∫∂0D2φ(ζ)dζ−1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2}+(λ−1)2λ+1{¯1(2πi)2∫∂0D2φ(ζ)dζ−¯1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2}. | (3.2) |
Proof. (1) Applying Lemma 2.1,
∂ˉz1∂ˉz2{1π2∫D2[λ−14λϕ(ζ)+λ+14λˉϕ(ζ)]dσζ1ζ1−z1dσζ2ζ2−z2}=λ−14λϕ(z)+λ+14λˉϕ(z), |
which means
1π2∫D2[λ−14λϕ(ζ)+λ+14λˉϕ(ζ)]dσζ1ζ1−z1dσζ2ζ2−z2 |
is a special solution to
∂ˉz1∂ˉz2f(z)=λ−14λϕ(z)+λ+14λˉϕ(z). |
Therefore, by Lemma 2.4, the solution of the problem is
f(z)=W(z)+1π2∫D2{λ−14λ[u0(ζ)+˜ϕ(ζ)]+λ+14λ[¯u0(ζ)+˜ϕ(ζ)]}dσζ1ζ1−z1dσζ2ζ2−z2, | (3.3) |
where ˜ϕ is determined in Lemma 2.4, W and u0 are analytic functions on D2 to be determined, and f=φ.
Applying Lemma 2.7 and plugging (2.2) and (2.3) into (3.3), we obtain
f(z)=W(z)+1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1ζ1−z1dσζ2ζ2−z2+λ−14λ{(ˉz2−1z2)[ˉz1u0(z)−u0(z)−u0(0,z2)z1]+1z2[ˉz1u0(z1,0)−u0(z1,0)−u0(0)z1]}+λ+14λ[¯u1(z)−¯u1(0,z2)−¯u1(z1,0)+¯u1(0,0)], | (3.4) |
which leads to
W(z)=f(z)−1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1ζ1−z1dσζ2ζ2−z2−λ−14λ{(ˉz2−1z2)[ˉz1u0(z)−u0(z)−u0(0,z2)z1]+1z2[ˉz1u0(z1,0)−u0(z1,0)−u0(0)z1]}−λ+14λ[¯u1(z)−¯u1(0,z2)−¯u1(z1,0)+¯u1(0,0)]. | (3.5) |
Considering the boundary condition f=φ and applying Lemma 2.8, we obtain that W(z) and u0(z) are determined by (2.7) and (2.8), respectively. Therefore, we have
u0(0,z2)=4λλ+1{¯1(2πi)2∫∂0D2φ(ζ)dζ(1−ˉz2ζ2)2−¯1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2(1−ˉz2ζ2)2}−(λ−1){1(2πi)2∫∂0D2φ(ζ)dζ−1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2}+(λ−1)2λ+1{¯1(2πi)2∫∂0D2φ(ζ)dζ−¯1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2}, | (3.6) |
and
u0(z1,0)=4λλ+1{¯1(2πi)2∫∂0D2φ(ζ)dζ(1−ˉz1ζ1)2−¯1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2(1−ˉz1ζ1)2}−(λ−1){1(2πi)2∫∂0D2φ(ζ)dζ−1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2}+(λ−1)2λ+1{¯1(2πi)2∫∂0D2φ(ζ)dζ−¯1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2}. | (3.7) |
In addition, (2.11) gives
λ+14λ¯[u1(z)−u1(0,z2)−u1(z1,0)+u1(0)]=1(2πi)2∫∂0D2{φ(ζ)−1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)]dσ~ζ1~ζ1−ζ1dσ~ζ2~ζ2−ζ2}⋅[ˉz11−ˉz1ζ1ˉz21−ˉz2ζ2+ˉz21−ˉz2ζ21ζ1−z1+ˉz11−ˉz1ζ11ζ2−z2]dζ1dζ2−λ−14λu0(0)ˉz1ˉz2=1(2πi)2∫∂0D2φ(ζ)[ˉz11−ˉz1ζ1ˉz21−ˉz2ζ2+ˉz21−ˉz2ζ21ζ1−z1+ˉz11−ˉz1ζ11ζ2−z2]dζ1dζ2−12πi∫∂D2{1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)][12πi∫∂D11~ζ1−ζ1ˉz1dζ11−ˉz1ζ1]dσ~ζ1dσ~ζ2~ζ2−ζ2}ˉz2dζ21−ˉz2ζ2−12πi∫∂D2{1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)][12πi∫∂D11~ζ1−ζ1dζ1ζ1−z1]dσ~ζ1dσ~ζ2~ζ2−ζ2}ˉz2dζ21−ˉz2ζ2−12πi∫∂D1{1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)][12πi∫∂D21~ζ2−ζ2dζ2ζ2−z2]dσ~ζ1dσ~ζ2~ζ1−ζ1}ˉz1dζ11−ˉz1ζ1−λ−14λu0(0)ˉz1ˉz2=1(2πi)2∫∂0D2φ(ζ)[ˉz11−ˉz1ζ1ˉz21−ˉz2ζ2+ˉz21−ˉz2ζ21ζ1−z1+ˉz11−ˉz1ζ11ζ2−z2]dζ1dζ2−1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)]ˉz1dσ~ζ1ˉz1~ζ1−1[12πi∫∂D21~ζ2−ζ2ˉz2dζ21−ˉz2ζ2]dσ~ζ2−λ−14λu0(0)ˉz1ˉz2=1(2πi)2∫∂0D2φ(ζ)[ˉz11−ˉz1ζ1ˉz21−ˉz2ζ2+ˉz21−ˉz2ζ21ζ1−z1+ˉz11−ˉz1ζ11ζ2−z2]dζ1dζ2−1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)]ˉz1dσ~ζ1ˉz1~ζ1−1ˉz2dσ~ζ2ˉz2~ζ2−1−λ−14λu0(0)ˉz1ˉz2, | (3.8) |
in which
12πi∫∂D11~ζ1−ζ1ˉz1dζ11−ˉz1ζ1=ˉz1ˉz1~ζ1−1,12πi∫∂D11~ζ1−ζ1dζ1ζ1−z1=0=12πi∫∂D21~ζ2−ζ2dζ2ζ2−z2 |
are used.
Plugging (2.7) and (3.8) into (3.4), f(z) is determined as
f(z)=1(2πi)2∫∂0D2φ(ζ)dζ1dζ2(ζ1−z1)(ζ2−z2)+1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1ζ1−z1dσζ2ζ2−z2+λ−14λ[(|z1|2−1)(|z2|2−1)z1z2u0(z)+|z2|2−1z1z2u0(0,z2)+|z1|2−1z1z2u0(z1,0)+u0(0)z1z2]+1(2πi)2∫∂0D2φ(ζ)[ˉz11−ˉz1ζ1ˉz21−ˉz2ζ2+ˉz21−ˉz2ζ21ζ1−z1+ˉz11−ˉz1ζ11ζ2−z2]dζ1dζ2−1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)]ˉz1dσ~ζ1ˉz1~ζ1−1ˉz2dσ~ζ2ˉz2~ζ2−1−λ−14λu0(0)ˉz1ˉz2=1(2πi)2∫∂0D2φ(ζ)dζ1dζ2(ζ1−z1)(ζ2−z2)+1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1ζ1−z1dσζ2ζ2−z2+λ−14λ[(|z1|2−1)(|z2|2−1)z1z2u0(z)+|z2|2−1z1z2u0(0,z2)+|z1|2−1z1z2u0(z1,0)+1−|z1|2|z2|2z1z2u0(0)]+1(2πi)2∫∂0D2φ(ζ)[ˉz11−ˉz1ζ1ˉz21−ˉz2ζ2+ˉz21−ˉz2ζ21ζ1−z1+ˉz11−ˉz1ζ11ζ2−z2]dζ1dζ2−1π2∫D2[λ−14λ˜ϕ(˜ζ)+λ+14λ¯˜ϕ(˜ζ)]ˉz1dσ~ζ1ˉz1~ζ1−1ˉz2dσ~ζ2ˉz2~ζ2−1=1(2πi)2∫∂0D2φ(ζ)(ζ1ζ1−z1+ˉζ1¯ζ1−z1−1)(ζ2ζ2−z2+ˉζ2¯ζ2−z2−1)dζζ+λ−14λ[(|z1|2−1)(|z2|2−1)z1z2u0(z)+|z2|2−1z1z2u0(0,z2)+|z1|2−1z1z2u0(z1,0)+1−|z1|2|z2|2z1z2u0(0)]+1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)][1(ζ1−z1)(ζ2−z2)−ˉz1ˉz2(ˉz1ζ1−1)(ˉz2ζ2−1)]dσζ1dσζ2, | (3.9) |
where u0(z),u0(0,z2),u0(z1,0),u0(0) are defined in (2.8), (2.15), (3.6), and (3.7), and the last equation is due to
1(2πi)2∫∂0D2φ(ζ)[1(ζ1−z1)(ζ2−z2)+ˉz11−ˉz1ζ1ˉz21−ˉz2ζ2+ˉz21−ˉz2ζ21ζ1−z1+ˉz11−ˉz1ζ11ζ2−z2]dζ=1(2πi)2∫∂0D2φ(ζ)[1(ζ1−z1)(ζ2−z2)+ˉz11−ˉz1ζ1ˉz21−ˉz2ζ2+(z1ζ1(ζ1−z1)ˉz21−ˉz2ζ2+ˉζ1ˉz21−ˉz2ζ2)+(ˉz11−ˉz1ζ1z2ζ2(ζ2−z2)+ˉz1ˉζ21−ˉz1ζ1)]dζ=1(2πi)2∫∂0D2φ(ζ)[1(ζ1−z1)(ζ2−z2)+ˉζ1ˉz2+ˉz1ˉζ2−ˉz1ˉz2(1−ˉz1ζ1)(1−ˉz2ζ2)+ˉζ1z1ζ1−z1ˉz21−ˉz2ζ2+ˉz11−ˉz1ζ1ˉζ2z2ζ2−z2]dζ=1(2πi)2∫∂0D2φ(ζ)[1(ζ1−z1)(ζ2−z2)+ˉζ1ˉζ2(1−ˉz1ζ1)(1−ˉz2ζ2)−1ζ1ζ2+ˉζ1z1ζ1−z1ˉz21−ˉz2ζ2+ˉz11−ˉz1ζ1ˉζ2z2ζ2−z2]dζ=1(2πi)2∫∂0D2φ(ζ)[ζ1ζ2(ζ1−z1)(ζ2−z2)+ˉζ1ˉζ2(¯ζ1−z1)(¯ζ2−z2)−1+z1ζ1−z1ˉz2¯ζ2−z2+ˉz1¯ζ1−z1z2ζ2−z2]dζζ=1(2πi)2∫∂0D2φ(ζ)(ζ1ζ1−z1+ˉζ1¯ζ1−z1−1)(ζ2ζ2−z2+ˉζ2¯ζ2−z2−1)dζζ. |
(2) In the following, we verify that (3.1) is the solution to the problem.
(ⅰ) For z∈∂0D2 (i.e., |z1|=|z2|=1), applying the properties of the Poisson kernel on D2, (3.1) satisfies the boundary condition f=φ obviously. In addition, applying Lemma 2.1, by (2.12) and (3.1), we obtain
∂ˉz1∂ˉz2f(z)=1(2πi)2∫∂0D2φ(ζ)dζ1(1−ˉz1ζ1)2dζ2(1−ˉz2ζ2)2+λ−14λ[u0(z)−u0(0)]+λ−14λ˜ϕ(z)+λ+14λ¯˜ϕ(z)−1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2(1−ˉz1ζ1)2(1−ˉz2ζ2)2=λ−14λ˜ϕ(z)+λ+14λ¯˜ϕ(z)+λ−14λu0(z)+λ+14λ{−λ−1λ+1u0(0)+4λλ+1⋅[1(2πi)2∫∂0D2φ(ζ)dζ1(1−ˉz1ζ1)2dζ2(1−ˉz2ζ2)2−1π2∫D2[λ−14λ˜ϕ(ζ)+λ+14λ¯˜ϕ(ζ)]dσζ1dσζ2(1−ˉz1ζ1)2(1−ˉz2ζ2)2]=λ−14λ˜ϕ(z)+λ+14λ¯˜ϕ(z)+λ−14λu0(z)+λ+14λ¯u0(z)=λ−14λ[˜ϕ(z)+u0(z)]+λ+14λ[¯˜ϕ(z)+¯u0(z)]=λ−14λϕ(z)+λ+14λ¯ϕ(z). |
(ⅱ) By Lemma 2.4, ϕ(z)=˜ϕ(z)+u0(z) satisfies ∂mˉz1∂mˉz2ϕ(z)=0(z∈D2) and
ℜ∂μˉz1∂νˉz2ϕ(z)=gμν(z)(z∈∂0D2),ℑ∂μˉz1∂νˉz2ϕ(0,z2)=0=ℑ∂μˉz1∂νˉz2ϕ(z1,0) |
for 1≤μ,ν≤m−1.
From the above analysis in (1) and (2), and by the expression of f in (3.1) and the uniqueness of u0(z) in (3.2), it is shown that (3.1) is the unique solution of the problem.
Theorem 3.2. Let φ∈C(∂0D2;C) and hμ1ν1,gμ2ν2∈C(∂0D2;R) for 0≤μ1,ν1≤n−2, 1≤μ2,ν2≤m−1 (m,n≥2), and let λ∈R∖{−1,0,1}. Then
f(z)=Tn−1,D1(Tn−1,D2F)(z)+˜ϕ(z) |
is the solution to the problem
∂nˉz1∂nˉz2f(z)=λ−14λϕ(z)+λ+14λˉϕ(z),∂mˉz1∂mˉz2ϕ(z)=0,z∈D2, |
with the conditions
{∂n−1ˉz1∂n−1ˉz2f=φ,ℜ∂μ1ˉz1∂ν1ˉz2f=hμ1ν1(z∈∂0D2),ℑ∂μ1ˉz1∂ν1ˉz2f(0,z2)=0=ℑ∂μ1ˉz1∂ν1ˉz2f(z1,0),ℜ∂μ2ˉz1∂ν2ˉz2ϕ(z)=gμ2ν2(z)(z∈∂0D2),ℑ∂μ2ˉz1∂ν2ˉz2ϕ(0,z2)=0=ℑ∂μ2ˉz1∂ν2ˉz2ϕ(z1,0), |
where F(z) is determined by (3.1) (in which gμν is replaced by gμ2ν2), ˜ϕ(z) is determined by (2.1) (in which gμν is replaced by hμ1ν1 and m is replaced by n−1), and
Tn,DiF(zi)=(−1)n2π(n−1)!∫Di(¯ζi−zi+ζi−zi)n−1[F(ζi)ζiζi+ziζi−zi+¯F(ζi)¯ζi1+ziˉζi1−ziˉζi]dξidηi, | (3.10) |
with T0,DiF=F and n∈N,i=1,2.
Proof. Let F(z)=∂n−1ˉz1∂n−1ˉz2f(z). Then, the problem is transformed to be
∂ˉz1∂ˉz2F(z)=λ−14λϕ(z)+λ+14λˉϕ(z),∂mˉz1∂mˉz2ϕ(z)=0,z∈D2, |
with the conditions
{∂n−1ˉz1∂n−1ˉz2f=F,ℜ∂μ1ˉz1∂ν1ˉz2f=hμ1ν1(z∈∂0D2),ℑ∂μ1ˉz1∂ν1ˉz2f(0,z2)=0=ℑ∂μ1ˉz1∂ν1ˉz2f(z1,0),F=φ,ℜ∂μ2ˉz1∂ν2ˉz2ϕ(z)=gμ2ν2(z)(z∈∂0D2),ℑ∂μ2ˉz1∂ν2ˉz2ϕ(0,z2)=0=ℑ∂μ2ˉz1∂ν2ˉz2ϕ(z1,0). |
By Theorem 3.1, F(z) determined by (3.1) (where gμν is replaced by gμ2ν2) is the unique solution to
{∂ˉz1∂ˉz2F(z)=λ−14λϕ(z)+λ+14λˉϕ(z),∂mˉz1∂mˉz2ϕ(z)=0(z∈D2),F=φ,ℜ∂μ2ˉz1∂ν2ˉz2ϕ(z)=gμ2ν2(z)(z∈∂0D2),ℑ∂μ2ˉz1∂ν2ˉz2ϕ(0,z2)=0=ℑ∂μ2ˉz1∂ν2ˉz2ϕ(z1,0). |
For Tn,DiF defined by (3.10); applying Lemma 2.3,
∂n−1ˉz1∂n−1ˉz2Tn−1,D1(Tn−1,D2F)=∂n−1ˉz1(Tn−1,D1F)=F,ℜ∂μ1ˉz1∂ν1ˉz2Tn−1,D1(Tn−1,D2F)=ℜ∂μ1ˉz1Tn−1,D1(Tn−1−ν1,D2F)=0(z∈∂D),ℑ∂μ1ˉz1∂ν1ˉz2Tn−1,D1(Tn−1,D2F)(z1,0)=ℑ∂μ1ˉz1Tn−1,D1(Tn−1−ν1,D2F)(z1,0)=0,ℑ∂μ1ˉz1∂ν1ˉz2Tn−1,D1(Tn−1,D2F)(0,z2)=ℑ∂μ1ˉz1Tn−1,D1(Tn−1−ν1,D2F)(0,z2)=0. |
Therefore, Tn−1,D1(Tn−1,D2F) is a special solution to ∂n−1ˉz1∂n−1ˉz2f=F, and the solution to
∂n−1ˉz1∂n−1ˉz2f=F,ℜ∂μ1ˉz1∂ν1ˉz2f=hμ1ν1(z∈∂0D2),ℑ∂μ1ˉz1∂ν1ˉz2f(0,z2)=0=ℑ∂μ1ˉz1∂ν1ˉz2f(z1,0) |
is
f(z)=Tn−1,D1(Tn−1,D2F)(z)+˜ϕ(z), |
where ˜ϕ(z) is a n−1-holomorphic function on D2 satisfying
ℜ∂μ1ˉz1∂ν1ˉz2˜ϕ=hμ1ν1(z∈∂0D2),ℑ∂μ1ˉz1∂ν1ˉz2˜ϕ(0,z2)=0=ℑ∂μ1ˉz1∂ν1ˉz2˜ϕ(z1,0). |
By Lemma 2.4, ˜ϕ(z) is determined by (2.1), in which gμν and m are replaced by hμ1ν1 and n−1, respectively.
In this section, we discuss systems of complex partial differential equations with Neumann boundary conditions on the bicylinder. Let ∂νif(z)=zi∂zif(z)+¯zi∂¯zif(z) denote the directional derivative of f(z) in relation to the outer normal vector, where i=1,2.
Theorem 4.1. Let gk1k2∈C(∂0D2) and ck1k2∈C. Let bk1k2(z1) and dk1k2(z2) be analytic functions about z1 and z2, respectively, with bk1k2(0)=dk1k2(0)=ck1k2 (1≤k1,k2≤m−1, m≥2). Then, there exists an analytic function u(m−k1)(m−k2)(z) on D2:
u(m−k1)(m−k2)(z)=d(m−k1)(m−k2)(z2)+b(m−k1)(m−k2)(z1)−c(m−k1)(m−k2)+1(2πi)2⋅∫∂D1∫∂D2k1−1∑t1=0k2−1∑t2=0(−2)t1+t2t1!t2!¯ζ1t1¯ζ2t2g(t1+m−k1)(t2+m−k2)(ζ)log(1−z1¯ζ1)log(1−z2¯ζ2)dζ1dζ2ζ1ζ2+k1−1∑λ1=112πi∫∂D1k1−1−λ1∑s1=0(−2)s1s1!¯ζ1s1[12πi∫∂D2k2−1∑t2=0(−2)t2t2!¯ζ2t2g(λ1+s1+m−k1)(t2+m−k2)(ζ)⋅log(1−z2¯ζ2)dζ2ζ2]1λ1!zλ11[log(1−z1¯ζ1)+λ1∑σ1=1zσ11¯ζ1σ1σ1]dζ1ζ1+k2−1∑λ2=112πi∫∂D2k2−1−λ2∑s2=0(−2)s2s2!¯ζ2s2[12πi∫∂D1k1−1∑t1=0(−2)t1t1!¯ζ1t1g(t1+m−k1)(λ2+s2+m−k2)(ζ)⋅log(1−z1¯ζ1)dζ1ζ1]1λ2!zλ22[log(1−z2¯ζ2)+λ2∑σ2=1zσ22¯ζ2σ2σ2]dζ2ζ2+k1−1∑λ1=1k2−1∑λ2=11(2πi)2∫∂D1∫∂D2k1−1−λ1∑s1=0k2−1−λ2∑s2=0(−2)s1+s2s1!s2!¯ζ1s1¯ζ2s2g(λ1+s1+m−k1)(λ2+s2+m−k2)(ζ)⋅1λ1!zλ111λ2!zλ22[log(1−z1¯ζ1)+λ1∑σ1=1zσ11¯ζ1σ1σ1][log(1−z2¯ζ2)+λ2∑σ2=1zσ22¯ζ2σ2σ2]dζ1dζ2ζ1ζ2 | (4.1) |
that satisfies
{∂ν1∂ν2k1−1∑l1=0k2−1∑l2=0ˉzl11ˉzl22l1!l2!u(l1+m−k1)(l2+m−k2)(z)=g(m−k1)(m−k2)(z),z∈∂0D2,u(m−k1)(m−k2)(z1,0)=b(m−k1)(m−k2)(z1),z1∈D1,u(m−k1)(m−k2)(0,z2)=d(m−k1)(m−k2)(z2),z2∈D2,u(m−k1)(m−k2)(0)=c(m−k1)(m−k2) | (4.2) |
if and only if
{12πi∫∂D1g(m−1)(m−k2)(ζ1,z2)dζ1(1−¯z1ζ1)ζ1=0,k1−1∑τ1=0(−1)τ1τ1!12πi∫∂D1¯ζ1τ1g(τ1+m−k1)(m−k2)(ζ1,z2)dζ11−¯z1ζ1=∂ν2[k2−1∑l2=01l2!d(m−k1+1)(l2+m−k2)(z2)¯z2l2],2≤k1≤m−1,1≤k2≤m−1 | (4.3) |
for ∀z1∈D1 (z2∈∂D2), and
{∫∂D2{∫∂D1k1−1∑t1=0(−2)t1t1!¯ζ1t1g(t1+m−k1)(m−1)(ζ)log(1−z1¯ζ1)dζ1ζ1+k1−1∑λ1=1∫∂D1k1−1−λ1∑s1=0(−2)s1s1!¯ζ1s1⋅g(λ1+s1+m−k1)(m−1)(ζ)1λ1!zλ11[log(1−z1¯ζ1)+λ1∑σ1=1zσ11¯ζ1σ1σ1]dζ1ζ1}dζ2(1−¯z2ζ2)ζ2=0,k2−1∑τ2=0(−1)τ2τ2!12πi∫∂D2¯ζ2τ2{−12πi∫∂D1k1−1∑t1=0(−2)t1t1!¯ζ1t1g(t1+m−k1)(τ2+m−k2)(ζ)log(1−z1¯ζ1)dζ1ζ1−k1−1∑λ1=112πi∫∂D1k1−1−λ1∑s1=0(−2)s1s1!¯ζ1s1g(λ1+s1+m−k1)(τ2+m−k2)(ζ)λ1!zλ11[log(1−z1¯ζ1)+λ1∑σ1=1zσ11¯ζ1σ1σ1]dζ1ζ1}dζ21−¯z2ζ2=b(m−k1)(m−k2+1)(z1)−c(m−k1)(m−k2+1),1≤k1≤m−1,2≤k2≤m−1. | (4.4) |
for ∀z2∈D2, z1∈D1.
Proof. From (4.2), we obtain
{∂ν1k1−1∑l1=01l1!{∂ν2[k2−1∑l2=01l2!u(l1+m−k1)(l2+m−k2)(z)ˉzl22]}ˉzl11=g(m−k1)(m−k2)(z),∂ν2[k2−1∑l2=01l2!u(m−k1)(l2+m−k2)(0,z2)ˉzl22]=∂ν2[k2−1∑l2=01l2!d(m−k1)(l2+m−k2)(z2)ˉzl22] | (4.5) |
for z1∈∂D1 (at the same time z2∈∂D2). Applying Lemma 2.5 on (4.5) for z1∈∂D1, there exists an analytic function about z1, i.e., ∂ν2[k2−1∑l2=0ˉzl22l2!u(m−k1)(l2+m−k2)(z)], satisfying (4.5) if and only if (4.3) is satisfied for ∀z1∈D1 (z2∈∂D2). Besides that, ∂ν2[k2−1∑l2=0ˉzl22l2!u(m−k1)(l2+m−k2)(z)] is determined by
∂ν2[k2−1∑l2=01l2!u(m−k1)(l2+m−k2)(z)ˉzl22]=∂ν2[k2−1∑l2=01l2!d(m−k1)(l2+m−k2)(z2)ˉzl22]−12πi∫∂D1k1−1∑t1=0(−2)t1t1!¯ζ1t1g(t1+m−k1)(m−k2)(ζ1,z2)log(1−z1¯ζ1)dζ1ζ1−k1−1∑λ1=112πi∫∂D1k1−1−λ1∑s1=0(−2)s1s1!¯ζ1s1g(λ1+s1+m−k1)(m−k2)(ζ1,z2)⋅1λ1!zλ11[log(1−z1¯ζ1)+λ1∑σ1=1zσ11¯ζ1σ1σ1]dζ1ζ1, |
that is,
∂ν2{k2−1∑l2=01l2![u(m−k1)(l2+m−k2)(z)−d(m−k1)(l2+m−k2)(z2)]ˉzl22}=−12πi∫∂D1k1−1∑t1=0(−2)t1t1!¯ζ1t1g(t1+m−k1)(m−k2)(ζ1,z2)log(1−z1¯ζ1)dζ1ζ1−k1−1∑λ1=112πi∫∂D1k1−1−λ1∑s1=0(−2)s1s1!¯ζ1s1g(λ1+s1+m−k1)(m−k2)(ζ1,z2)1λ1!zλ11[log(1−z1¯ζ1)+λ1∑σ1=1zσ11¯ζ1σ1σ1]dζ1ζ1. | (4.6) |
Further, applying Lemma 2.5 on (4.6) for z2∈∂D2, there exists an analytic function about z2, i.e., u(m−k1)(m−k2)(z), satisfying (4.6) and
u(m−k1)(m−k2)(z1,0)−d(m−k1)(m−k2)(0)=b(m−k1)(m−k2)(z1)−d(m−k1)(m−k2)(0), |
if and only if (4.4) is satisfied for ∀z2∈D2 (at the same time z1∈D1). Moreover, u(m−k1)(m−k2)(z) is determined by
u(m−k1)(m−k2)(z)−d(m−k1)(m−k2)(z2)=b(m−k1)(m−k2)(z1)−d(m−k1)(m−k2)(0)−12πi∫∂D2k2−1∑t2=0(−2)t2t2!¯ζ2t2{−12πi∫∂D1k1−1∑t1=0(−2)t1t1!¯ζ1t1g(t1+m−k1)(t2+m−k2)(ζ)log(1−z1¯ζ1)dζ1ζ1−k1−1∑λ1=112πi∫∂D1k1−1−λ1∑s1=0(−2)s1s1!¯ζ1s1g(λ1+s1+m−k1)(t2+m−k2)(ζ)⋅1λ1!zλ11[log(1−z1¯ζ1)+λ1∑σ1=1zσ11¯ζ1σ1σ1]dζ1ζ1}log(1−z2¯ζ2)dζ2ζ2−k2−1∑λ2=112πi∫∂D2k2−1−λ2∑s2=0(−2)s2s2!¯ζ2s2{−12πi∫∂D1k1−1∑t1=0(−2)t1t1!¯ζ1t1g(t1+m−k1)(λ2+s2+m−k2)(ζ)⋅log(1−z1¯ζ1)dζ1ζ1−k1−1∑λ1=112πi∫∂D1k1−1−λ1∑s1=0(−2)s1s1!¯ζ1s1g(λ1+s1+m−k1)(λ2+s2+m−k2)(ζ)⋅1λ1!zλ11[log(1−z1¯ζ1)+λ1∑σ1=1zσ11¯ζ1σ1σ1]dζ1ζ1}1λ2!zλ22[log(1−z2¯ζ2)+λ2∑σ2=1zσ22¯ζ2σ2σ2]dζ2ζ2, |
which leads to (4.1).
Since \partial_{\nu_2}\Big[\sum\limits^{k_2-1}_{l_2 = 0}\frac{\bar{z}_2^{l_2}}{l_2!} u_{(m-k_1)(l_2+m-k_2)}(z)\Big] is analytic about z_1 and u_{(m-k_1)(m-k_2)}(z) is analytic about z_2 , then u_{(m-k_1)(m-k_2)}(z) is analytic on D^2 . Therefore, there exists an analytic function u_{(m-k_1)(m-k_2)}(z) on D^2 , determined by (4.1), satisfying (4.2) on the conditions of (4.3) and (4.4).
Theorem 4.2. Let \varphi, g_{\mu\nu}\in C(\partial_0D^2) and c_{\mu\nu}\in\mathbb{C} . Let b_{\mu\nu}(z_1) and d_{\mu\nu}(z_2) be analytic functions about z_1 and z_2 , respectively, with b_{\mu\nu}(0) = d_{\mu\nu}(0) = c_{\mu\nu} ( 1\leq\mu, \nu\leq m-1, \; \; m\geq2 ). Let \lambda\in\mathbb{R}\setminus\{-1, 0, 1\} . Then the problem
\partial_{\bar{z}_1}\partial_{\bar{z}_2}f(z) = \frac{\lambda-1}{4\lambda}\phi(z)+\frac{\lambda+1}{4\lambda}\bar{\phi}(z), \; \partial^{m}_{\bar{z}_1}\partial^{m}_{\bar{z}_2}\phi(z) = 0, \; z\in D^2, |
with the conditions
\begin{equation} \left\{ \begin{array}{ll} f(z) = \varphi(z), \; \partial\nu_1\partial\nu_2(\partial^{\mu}_{\bar{z}_1}\partial^{\nu}_{\bar{z}_2}\phi(z)) = g_{\mu\nu}(z), \; z\in \partial_0D^2, \\ \partial^{\mu}_{\bar{z}_1}\partial^{\nu}_{\bar{z}_2}\phi(z_1, 0) = \sum\limits^{k_1-1}_{l_1 = 0}\frac{1}{l_1!}b_{(l_1+m-k_1)(m-k_2)}(z_1)\overline{z_1}^{l_1}, \; z_1\in D_1, \\ \partial^{\mu}_{\bar{z}_1}\partial^{\nu}_{\bar{z}_2}\phi(0, z_2) = \sum\limits^{k_2-1}_{l_2 = 0}\frac{1}{l_2!}d_{(m-k_1)(l_2+m-k_2)}(z_2)\overline{z_2}^{l_2}, \; z_2\in D_2, \\ \partial^{\mu}_{\bar{z}_1}\partial^{\nu}_{\bar{z}_2}\phi(0, 0) = c_{(m-k_1)(m-k_2)} \end{array} \right. \end{equation} | (4.7) |
for 1\leq \mu, \nu, k_1, k_2\leq m-1 , has a unique solution if and only if (4.3) and (4.4) are satisfied. The solution is given by (3.1), where u_0 is determined by (3.2), and
\begin{equation} \widetilde{\phi}(z) = \sum\limits^{m-1}_{l_1, l_2 = 1}\frac{u_{l_1l_2}(z)}{l_1!l_2!}\overline{z_1}^{l_1}\overline{z_2}^{l_2}, \end{equation} | (4.8) |
in which u_{l_1l_2}(z) is determined by (4.1).
Proof. (4.8) follows that
\left\{ \begin{array}{ll} \partial\nu_1\partial\nu_2(\partial^{\mu}_{\bar{z}_1}\partial^{\nu}_{\bar{z}_2}\phi(z)) = g_{\mu\nu}(z), \; z\in \partial_0D^2, \\ \partial^{\mu}_{\bar{z}_1}\partial^{\nu}_{\bar{z}_2}\phi(z_1, 0) = \sum\limits^{k_1-1}_{l_1 = 0}\frac{1}{l_1!}b_{(l_1+m-k_1)(m-k_2)}(z_1)\overline{z_1}^{l_1}, \; z_1\in D_1, \\ \partial^{\mu}_{\bar{z}_1}\partial^{\nu}_{\bar{z}_2}\phi(0, z_2) = \sum\limits^{k_2-1}_{l_2 = 0}\frac{1}{l_2!}d_{(m-k_1)(l_2+m-k_2)}(z_2)\overline{z_2}^{l_2}, \; z_2\in D_2, \\ \partial^{\mu}_{\bar{z}_1}\partial^{\nu}_{\bar{z}_2}\phi(0, 0) = c_{(m-k_1)(m-k_2)} \end{array} \right. |
is equivalent to (4.2). By Theorem 4.1 and similar to the proof of Theorem 3.1, we obtain the desired conclusion.
Theorem 4.3. Let g\in L_1(D^2) and \varphi_1, \varphi_2\in C(\partial D^2) . Then the problem
\begin{equation} \left\{ \begin{array}{ll} \partial_{\bar{z}_1}\partial_{\bar{z}_2}f(z) = g(z), f(z_1, 0) = \alpha_1(z_1), f(0, z_2) = \alpha_2(z_2), \; z_1\in D_1, \; z_2\in D_2, \\ \partial_{\nu_1}f(z) = \varphi_1(z), \partial_{\nu_2}f(z) = \varphi_2(z), \; z\in \partial_0D^2 \end{array} \right. \end{equation} | (4.9) |
with the compatibility condition
\alpha_1(0) = \alpha_2(0) = \alpha, \; \partial_{\nu_1}\varphi_2(z) = \partial_{\nu_2}\varphi_1(z) = \varphi(z) |
has a unique solution
\begin{equation} \begin{aligned} f(z)& = \alpha_1(z_1)+\alpha_2(z_2)-\alpha -\frac{1}{2\pi i}\int_{\partial D_2}[\varphi_2(z_1, \zeta_2)-\partial_{\nu_2}\alpha_2(\zeta_2)]\log(1-z_2\overline{\zeta_2})\frac{d\zeta_2}{\zeta_2}\\ &\quad+\frac{1}{2\pi i}\int_{\partial D_2}\Big\{\frac{1}{2\pi i}\int_{\partial D_1}[\zeta_2\varphi(\zeta)-\zeta_2^2\partial_{\zeta_2}\varphi_1(\zeta)]\log(1-z_1\overline{\zeta_1})\frac{d\zeta_1}{\zeta_1}\\ &\quad+\frac{1}{2\pi i}\int_{\partial D_1}g(\zeta)\log(1-z_1\overline{\zeta_1})d\overline{\zeta_1}+\frac{1}{\pi}\int_{D_1}g(\zeta) \frac{z_1}{\zeta_1(\zeta_1-z_1)}d\sigma_{\zeta_1}\Big\}\log(1-z_2\overline{\zeta_2})d\overline{\zeta_2}\\ &\quad+\frac{1}{\pi}\int_{D_2}\Big\{\frac{1}{2\pi i}\int_{\partial D_1}[\zeta_2\varphi(\zeta)-\zeta_2^2\partial_{\zeta_2}\varphi_1(\zeta)]\log(1-z_1\overline{\zeta_1})\frac{d\zeta_1}{\zeta_1}\\ &\quad+\frac{1}{2\pi i}\int_{\partial D_1}g(\zeta)\log(1-z_1\overline{\zeta_1})d\overline{\zeta_1}+\frac{1}{\pi}\int_{D_1}g(\zeta) \frac{z_1}{\zeta_1(\zeta_1-z_1)}d\sigma_{\zeta_1}\Big\}\frac{z_2}{\zeta_2(\zeta_2-z_2)}d\sigma_{\zeta_2}, \end{aligned} \end{equation} | (4.10) |
if and only if
\begin{equation} \begin{aligned} &\quad\frac{1}{2\pi i}\int_{\partial D_1}[z_2\varphi(\zeta_1, z_2)-z_2^2\partial_{z_2}\varphi_1(\zeta_1, z_2)]\frac{d\zeta_1}{(1-\bar{z}_1\zeta_1)\zeta_1} +\frac{1}{2\pi i}\int_{\partial D_1}g(\zeta_1, z_2)\frac{d\overline{\zeta_1}}{1-\bar{z}_1\zeta_1}\\ & = \frac{-1}{\pi}\int_{D_1}g(\zeta_1, z_2)\frac{\bar{z}_1}{(1-\bar{z}_1\zeta_1)^2}d\sigma_{\zeta_1} \end{aligned} \end{equation} | (4.11) |
and
\begin{equation} \begin{aligned} &\quad\frac{1}{2\pi i}\int_{\partial D_2}[\varphi_2(z_1, \zeta_2)-\partial_{\nu_2}\alpha_2(\zeta_2)]\frac{d\zeta_2}{(1-\bar{z}_2\zeta_2)\zeta_2}\\ & = \frac{1}{2\pi i}\int_{\partial D_2}\Big\{\frac{1}{2\pi i}\int_{\partial D_1}[\zeta_2\varphi(\zeta)-\zeta_2^2\partial_{\zeta_2}\varphi_1(\zeta)]\log(1-z_1\overline{\zeta_1})\frac{d\zeta_1}{\zeta_1}\\ &\quad+\frac{1}{2\pi i}\int_{\partial D_1}g(\zeta)\log(1-z_1\overline{\zeta_1})d\overline{\zeta_1}+\frac{1}{\pi}\int_{D_1}g(\zeta) \frac{z_1}{\zeta_1(\zeta_1-z_1)}d\sigma_{\zeta_1}\Big\}\frac{d\overline{\zeta_2}}{1-\bar{z}_2\zeta_2}\\ &\quad+\frac{1}{\pi}\int_{D_2}\Big\{\frac{1}{2\pi i}\int_{\partial D_1}[\zeta_2\varphi(\zeta)-\zeta_2^2\partial_{\zeta_2}\varphi_1(\zeta)]\log(1-z_1\overline{\zeta_1})\frac{d\zeta_1}{\zeta_1}\\ &\quad+\frac{1}{2\pi i}\int_{\partial D_1}g(\zeta)\log(1-z_1\overline{\zeta_1})d\overline{\zeta_1}+\frac{1}{\pi}\int_{D_1}g(\zeta) \frac{z_1}{\zeta_1(\zeta_1-z_1)}d\sigma_{\zeta_1}\Big\}\frac{\bar{z}_2}{(1-\bar{z}_2\zeta_2)^2}d\sigma_{\zeta_2}. \end{aligned} \end{equation} | (4.12) |
Proof. (4.9) follows \partial_{\nu_1}(\partial_{\nu_2}f(z)) = \varphi(z) , that is
\partial_{\nu_1}[z_2\partial_{z_2}f(z)+\bar{z}_2\partial_{\bar{z}_2}f(z)] = \varphi(z), |
which is equivalent to
\partial_{\nu_1}[\partial_{\bar{z}_2}f(z)] = z_2\varphi(z)-z_2^2\partial_{z_2}\partial_{\nu_1}f(z) = z_2\varphi(z)-z_2^2\partial_{z_2}\varphi_1(z). |
Applying Lemma 2.6 on the problem
\partial_{\bar{z}_1}[\partial_{\bar{z}_2}f(z)] = g(z), \; \partial_{\nu_1}[\partial_{\bar{z}_2}f(z)] = z_2\varphi(z)-z_2^2\partial_{z_2}\varphi_1(z), \; \partial_{\bar{z}_2}f(0, z_2) = \partial_{\bar{z}_2}\alpha_2(z_2), |
we get that the unique solution \partial_{\bar{z}_2}f(z) is determined by
\begin{equation} \begin{aligned} \partial_{\bar{z}_2}f(z)& = \partial_{\bar{z}_2}\alpha_2(z_2) -\frac{1}{2\pi i}\int_{\partial D_1}[z_2\varphi(\zeta_1, z_2)-z_2^2\partial_{z_2}\varphi_1(\zeta_1, z_2)]\log(1-z_1\overline{\zeta_1})\frac{d\zeta_1}{\zeta_1}\\ &\quad-\frac{1}{2\pi i}\int_{\partial D_1}g(\zeta_1, z_2)\log(1-z_1\overline{\zeta_1})d\overline{\zeta_1} -\frac{1}{\pi}\int_{D_1}g(\zeta_1, z_2)\frac{z_1}{\zeta_1(\zeta_1-z_1)}d\sigma_{\zeta_1} \end{aligned} \end{equation} | (4.13) |
if and only if (4.11) is satisfied.
In addition, (4.13) is equivalent to
\begin{equation} \begin{aligned} \partial_{\bar{z}_2}[f(z)-\alpha_2(z_2)]& = -\frac{1}{2\pi i}\int_{\partial D_1}[z_2\varphi(\zeta_1, z_2)-z_2^2\partial_{z_2}\varphi_1(\zeta_1, z_2)]\log(1-z_1\overline{\zeta_1})\frac{d\zeta_1}{\zeta_1}\\ &\quad-\frac{1}{2\pi i}\int_{\partial D_1}g(\zeta_1, z_2)\log(1-z_1\overline{\zeta_1})d\overline{\zeta_1} -\frac{1}{\pi}\int_{D_1}g(\zeta_1, z_2)\frac{z_1d\sigma_{\zeta_1}}{\zeta_1(\zeta_1-z_1)}. \end{aligned} \end{equation} | (4.14) |
In view of
\begin{equation} \partial_{\nu_2}[f(z)-\alpha_2(z_2)] = \varphi_2(z)-\partial_{\nu_2}\alpha_2(z_2), \; \; f(z_1, 0)-\alpha_2(0) = \alpha_1(z_1)-\alpha, \end{equation} | (4.15) |
applying Lemma 2.6, we obtain the unique solution of the problem (4.14) with the conditions (4.15) is (4.10) if and only if (4.12) is satisfied. So we get the desired conclusion.
Theorem 4.4. Let \varphi, g_{\mu\nu}\in C(\partial_0D^2) and c_{\mu\nu}\in\mathbb{C} . Let b_{\mu\nu}(z_1) and d_{\mu\nu}(z_2) be analytic functions about z_1 and z_2 , respectively, with b_{\mu\nu}(0) = d_{\mu\nu}(0) = c_{\mu\nu} ( 1\leq\mu, \nu\leq m-1, \; \; m\geq2 ). Let \lambda\in\mathbb{R}\setminus\{-1, 0, 1\} . Then the problem
\begin{equation} \partial_{\bar{z}_1}\partial_{\bar{z}_2}f(z) = \frac{\lambda-1}{4\lambda}\phi(z)+\frac{\lambda+1}{4\lambda}\bar{\phi}(z), \; \partial^{m}_{\bar{z}_1}\partial^{m}_{\bar{z}_2}\phi(z) = 0, \; z\in D^2, \end{equation} | (4.16) |
under the conditions
\begin{equation} \left\{ \begin{array}{ll} \partial_{\nu_1}f(z) = \varphi_1(z), \; \; \partial_{\nu_2}f(z) = \varphi_2(z), \; \; \partial\nu_1\partial\nu_2(\partial^{\mu}_{\bar{z}_1}\partial^{\nu}_{\bar{z}_2}\phi(z)) = g_{\mu\nu}(z), \; \; z\in \partial_0D^2, \\ f(z_1, 0) = \alpha_1(z_1), \; \; \partial^{\mu}_{\bar{z}_1}\partial^{\nu}_{\bar{z}_2}\phi(z_1, 0) = \sum\limits^{k_1-1}_{l_1 = 0}\frac{1}{l_1!}b_{(l_1+m-k_1)(m-k_2)}(z_1)\overline{z_1}^{l_1}, \; \; z_1\in D_1, \\ f(0, z_2) = \alpha_2(z_2), \; \; \partial^{\mu}_{\bar{z}_1}\partial^{\nu}_{\bar{z}_2}\phi(0, z_2) = \sum\limits^{k_2-1}_{l_2 = 0}\frac{1}{l_2!}d_{(m-k_1)(l_2+m-k_2)}(z_2)\overline{z_2}^{l_2}, \; \; z_2\in D_2, \\ \partial^{\mu}_{\bar{z}_1}\partial^{\nu}_{\bar{z}_2}\phi(0, 0) = c_{(m-k_1)(m-k_2)} \end{array} \right. \end{equation} | (4.17) |
with 1\leq \mu, \nu, k_1, k_2\leq m-1 and
\alpha_1(0) = \alpha_2(0) = \alpha, \quad \partial_{\nu_1}\varphi_2(z) = \partial_{\nu_2}\varphi_1(z) = \varphi(z) |
has a unique solution if and only if (4.3), (4.4) and
\begin{equation} \begin{aligned} &\quad\frac{1}{2\pi i}\int_{\partial D_1}[z_2\varphi(\zeta_1, z_2)-z_2^2\partial_{z_2}\varphi_1(\zeta_1, z_2)]\frac{d\zeta_1}{(1-\bar{z}_1\zeta_1)\zeta_1}\\ & = \frac{1}{\pi}\int_{D_1} \Big[\frac{\lambda-1}{4\lambda}\phi_{\zeta_1}(\zeta_1, z_2)+\frac{\lambda+1}{4\lambda} \overline{\phi_{\overline{\zeta_1}}(\zeta_1, z_2)}\Big] \frac{d\sigma_{\zeta_1}}{1-\bar{z}_1\zeta_1}, \end{aligned} \end{equation} | (4.18) |
and
\begin{equation} \begin{aligned} &\quad\frac{1}{2\pi i}\int_{\partial D_2}[\varphi_2(z_1, \zeta_2)-\partial_{\nu_2}\alpha_2(\zeta_2)]\frac{d\zeta_2}{(1-\bar{z}_2\zeta_2)\zeta_2}\\ & = \frac{1}{2\pi^2 i}\int_{\partial D_1}\int_{D_2}\log(1-z_1\overline{\zeta_1}) [\varphi(\zeta)+\zeta_2\varphi_{\zeta_2}(\zeta)-2\zeta_2\partial_{\zeta_2}\varphi_1(\zeta) -\zeta_2^2\partial_{\zeta_2}^2\varphi_1(\zeta)] \frac{d\sigma_{\zeta_2}}{\bar{z}_2\zeta_2-1}\frac{d\zeta_1}{\zeta_1}\\ &\quad+\frac{1}{2\pi^2 i}\int_{\partial D_1}\int_{D_2}\log(1-z_1\overline{\zeta_1}) \Big[\frac{\lambda-1}{4\lambda}\phi_{\zeta_2}(\zeta)+\frac{\lambda+1}{4\lambda}\overline{\phi_{\overline{\zeta_2}}(\zeta)}\Big] \frac{d\sigma_{\zeta_2}}{\bar{z}_2\zeta_2-1}d\overline{\zeta_1}\\ &\quad+\frac{1}{\pi^2}\int_{D_1}\int_{D_2} \Big[\frac{\lambda-1}{4\lambda}\phi_{\zeta_2}(\zeta)+\frac{\lambda+1}{4\lambda}\overline{\phi_{\overline{\zeta_2}}(\zeta)}\Big] \frac{d\sigma_{\zeta_2}}{\bar{z}_2\zeta_2-1} \frac{z_1d\sigma_{\zeta_1}}{\zeta_1(\zeta_1-z_1)} \end{aligned} \end{equation} | (4.19) |
are satisfied. The solution is given by (4.10) where g(\zeta) is replaced by \frac{\lambda-1}{4\lambda}\phi(\zeta)+\frac{\lambda+1}{4\lambda}\bar{\phi}(\zeta) , and
\phi(z) = \sum\limits^{m-1}_{l_1, l_2 = 1}\frac{u_{l_1l_2}(z)}{l_1!l_2!}\overline{z_1}^{l_1}\overline{z_2}^{l_2}, |
in which u_{l_1l_2}(z) is determined by (4.1).
Proof. Applying Theorems 4.1 and 4.3, the problem (4.16) with conditions (4.17) has a unique solution (4.10) if and only if (4.3), (4.4), (4.11), and (4.12) are satisfied, where g(\zeta) is replaced by \frac{\lambda-1}{4\lambda}\phi(\zeta)+\frac{\lambda+1}{4\lambda}\bar{\phi}(\zeta) . To obtain the specific representations of the solvable conditions, we need the following equations.
By the Gauss formula, we have
\begin{aligned} &\quad\frac{1}{2\pi i}\int_{\partial D_2}\frac{[\zeta_2\varphi(\zeta)-\zeta_2^2\partial_{\zeta_2}\varphi_1(\zeta)]}{1-\bar{z}_2\zeta_2}d\overline{\zeta_2} = \frac{-1}{\pi}\int_{D_2} \partial_{\zeta_2}\frac{[\zeta_2\varphi(\zeta)-\zeta_2^2\partial_{\zeta_2}\varphi_1(\zeta)]}{1-\bar{z}_2\zeta_2}d\sigma_{\zeta_2}\\ & = \frac{-1}{\pi}\int_{D_2} \Big\{[\zeta_2\varphi(\zeta)-\zeta_2^2\partial_{\zeta_2}\varphi_1(\zeta)]\frac{\bar{z}_2}{(1-\bar{z}_2\zeta_2)^2}\\ &\quad+\frac{1}{1-\bar{z}_2\zeta_2}[\varphi(\zeta)+\zeta_2\varphi_{\zeta_2}(\zeta)-2\zeta_2\partial_{\zeta_2}\varphi_1(\zeta) -\zeta_2^2\partial_{\zeta_2}^2\varphi_1(\zeta)]\Big\}d\sigma_{\zeta_2}, \end{aligned} |
which follows
\begin{equation} \begin{aligned} &\quad\frac{1}{2\pi i}\int_{\partial D_2}\frac{[\zeta_2\varphi(\zeta)-\zeta_2^2\partial_{\zeta_2}\varphi_1(\zeta)]}{1-\bar{z}_2\zeta_2}d\overline{\zeta_2} +\frac{1}{\pi}\int_{D_2} [\zeta_2\varphi(\zeta)-\zeta_2^2\partial_{\zeta_2}\varphi_1(\zeta)]\frac{\bar{z}_2}{(1-\bar{z}_2\zeta_2)^2}d\sigma_{\zeta_2}\\ & = \frac{1}{\pi}\int_{D_2} \frac{1}{\bar{z}_2\zeta_2-1}[\varphi(\zeta)+\zeta_2\varphi_{\zeta_2}(\zeta)-2\zeta_2\partial_{\zeta_2}\varphi_1(\zeta) -\zeta_2^2\partial_{\zeta_2}^2\varphi_1(\zeta)]d\sigma_{\zeta_2}. \end{aligned} \end{equation} | (4.20) |
In addition,
\frac{1}{2\pi i}\int_{\partial D_2}\frac{\phi(\zeta)}{1-\bar{z}_2\zeta_2}d\overline{\zeta_2} = \frac{-1}{\pi}\int_{D_2} \Big[\phi(\zeta)\frac{\bar{z}_2}{(1-\bar{z}_2\zeta_2)^2}+\frac{\phi_{\zeta_2}(\zeta)}{1-\bar{z}_2\zeta_2}\Big]d\sigma_{\zeta_2} |
follows
\begin{equation} \frac{1}{2\pi i}\int_{\partial D_2}\frac{\phi(\zeta)}{1-\bar{z}_2\zeta_2}d\overline{\zeta_2}+\frac{1}{\pi}\int_{D_2} \Big[\phi(\zeta)\frac{\bar{z}_2}{(1-\bar{z}_2\zeta_2)^2}d\sigma_{\zeta_2} = \frac{1}{\pi}\int_{D_2}\frac{\phi_{\zeta_2}(\zeta)}{\bar{z}_2\zeta_2-1}d\sigma_{\zeta_2}. \end{equation} | (4.21) |
Similarly, we obtain
\begin{equation} \frac{1}{2\pi i}\int_{\partial D_2}\frac{\overline{\phi(\zeta)}}{1-\bar{z}_2\zeta_2}d\overline{\zeta_2}+\frac{1}{\pi}\int_{D_2} \Big[\overline{\phi(\zeta)}\frac{\bar{z}_2}{(1-\bar{z}_2\zeta_2)^2}d\sigma_{\zeta_2} = \frac{1}{\pi}\int_{D_2}\frac{\overline{\phi_{\overline{\zeta_2}}(\zeta)}}{\bar{z}_2\zeta_2-1}d\sigma_{\zeta_2}.\ \end{equation} | (4.22) |
Using (4.21) and replacing g(\zeta_1, z_2) in (4.11) by \frac{\lambda-1}{4\lambda}\phi(\zeta_1, z_2)+\frac{\lambda+1}{4\lambda}\bar{\phi}(\zeta_1, z_2) , we get (4.18). Using (4.20)–(4.22) and replacing g(\zeta) in (4.12) by \frac{\lambda-1}{4\lambda}\phi(\zeta)+\frac{\lambda+1}{4\lambda}\bar{\phi}(\zeta) , we get
\begin{aligned} &\quad\frac{1}{2\pi i}\int_{\partial D_2}[\varphi_2(z_1, \zeta_2)-\partial_{\nu_2}\alpha_2(\zeta_2)]\frac{d\zeta_2}{(1-\bar{z}_2\zeta_2)\zeta_2}\\ & = \frac{1}{2\pi i}\int_{\partial D_2}\Big\{\frac{1}{2\pi i}\int_{\partial D_1}[\zeta_2\varphi(\zeta)-\zeta_2^2\partial_{\zeta_2}\varphi_1(\zeta)]\log(1-z_1\overline{\zeta_1})\frac{d\zeta_1}{\zeta_1}\\ &\quad+\frac{1}{2\pi i}\int_{\partial D_1}g(\zeta)\log(1-z_1\overline{\zeta_1})d\overline{\zeta_1}+\frac{1}{\pi}\int_{D_1}g(\zeta) \frac{z_1}{\zeta_1(\zeta_1-z_1)}d\sigma_{\zeta_1}\Big\}\frac{d\overline{\zeta_2}}{1-\bar{z}_2\zeta_2}\\ &\quad+\frac{1}{\pi}\int_{D_2}\Big\{\frac{1}{2\pi i}\int_{\partial D_1}[\zeta_2\varphi(\zeta)-\zeta_2^2\partial_{\zeta_2}\varphi_1(\zeta)]\log(1-z_1\overline{\zeta_1})\frac{d\zeta_1}{\zeta_1}\\ &\quad+\frac{1}{2\pi i}\int_{\partial D_1}g(\zeta)\log(1-z_1\overline{\zeta_1})d\overline{\zeta_1}+\frac{1}{\pi}\int_{D_1}g(\zeta) \frac{z_1}{\zeta_1(\zeta_1-z_1)}d\sigma_{\zeta_1}\Big\}\frac{\bar{z}_2}{(1-\bar{z}_2\zeta_2)^2}d\sigma_{\zeta_2}\\ & = \frac{1}{2\pi i}\int_{\partial D_1}\log(1-z_1\overline{\zeta_1}) \Big\{\frac{1}{2\pi i}\int_{\partial D_2}[\zeta_2\varphi(\zeta)-\zeta_2^2\partial_{\zeta_2}\varphi_1(\zeta)]\frac{d\overline{\zeta_2}}{1-\bar{z}_2\zeta_2}\\ &\quad+\frac{1}{\pi}\int_{D_2}[\zeta_2\varphi(\zeta)-\zeta_2^2\partial_{\zeta_2}\varphi_1(\zeta)] \frac{\bar{z}_2d\sigma_{\zeta_2}}{(1-\bar{z}_2\zeta_2)^2}\Big\}\frac{d\zeta_1}{\zeta_1}\\ &\quad+\frac{1}{2\pi i}\int_{\partial D_1}\log(1-z_1\overline{\zeta_1}) \Big\{\frac{1}{2\pi i}\int_{\partial D_2}g(\zeta)\frac{d\overline{\zeta_2}}{1-\bar{z}_2\zeta_2} +\frac{1}{\pi}\int_{D_2}g(\zeta) \frac{\bar{z}_2d\sigma_{\zeta_2}}{(1-\bar{z}_2\zeta_2)^2}\Big\}d\overline{\zeta_1}\\ &\quad+\frac{1}{\pi}\int_{D_1}\Big\{\frac{1}{2\pi i}\int_{\partial D_2}g(\zeta)\frac{d\overline{\zeta_2}}{1-\bar{z}_2\zeta_2} +\frac{1}{\pi}\int_{D_2}g(\zeta) \frac{\bar{z}_2d\sigma_{\zeta_2}}{(1-\bar{z}_2\zeta_2)^2}\Big\} \frac{z_1d\sigma_{\zeta_1}}{\zeta_1(\zeta_1-z_1)}\\ & = \frac{1}{2\pi i}\int_{\partial D_1}\log(1-z_1\overline{\zeta_1}) \Big\{\frac{1}{\pi}\int_{D_2}[\varphi(\zeta)+\zeta_2\varphi_{\zeta_2}(\zeta)-2\zeta_2\partial_{\zeta_2}\varphi_1(\zeta) -\zeta_2^2\partial_{\zeta_2}^2\varphi_1(\zeta)] \frac{d\sigma_{\zeta_2}}{\bar{z}_2\zeta_2-1}\Big\}\frac{d\zeta_1}{\zeta_1}\\ &\quad+\frac{1}{2\pi i}\int_{\partial D_1}\log(1-z_1\overline{\zeta_1}) \Big\{\frac{1}{\pi}\int_{D_2} \Big[\frac{\lambda-1}{4\lambda}\phi_{\zeta_2}(\zeta)+\frac{\lambda+1}{4\lambda}\overline{\phi_{\overline{\zeta_2}}(\zeta)}\Big] \frac{d\sigma_{\zeta_2}}{\bar{z}_2\zeta_2-1}\Big\}d\overline{\zeta_1}\\ &\quad+\frac{1}{\pi}\int_{D_1}\Big\{\frac{1}{\pi}\int_{D_2} \Big[\frac{\lambda-1}{4\lambda}\phi_{\zeta_2}(\zeta)+\frac{\lambda+1}{4\lambda}\overline{\phi_{\overline{\zeta_2}}(\zeta)}\Big] \frac{d\sigma_{\zeta_2}}{\bar{z}_2\zeta_2-1}\Big\} \frac{z_1d\sigma_{\zeta_1}}{\zeta_1(\zeta_1-z_1)}, \end{aligned} |
which leads to (4.19).
Remark 4.1. Dirichlet problems and Neumann problems are two typical types of boundary value problems. The conclusions obtained in this paper have enriched the research on boundary value problems for bi-polyanalytic functions. With the methods used in this paper, we can discuss other complex partial differential equation problems for bi-polyanalytic functions. For example, it would be interesting to discuss more complex mixed boundary value problems for bi-polyanalytic functions that simultaneously satisfy multiple boundary conditions, such as Schwarz boundary conditions, Riemann-Hilbert boundary conditions, Neumann boundary conditions and other boundary conditions. However, the corresponding boundary value problems of polyanalytic functions need to be investigated first. Besides that, with the methods used in this paper, we can also solve some complex partial differential equation problems (homogeneous or non-homogeneous) in higher-dimensional complex spaces.
We first discuss a kind of boundary value problem for polyanalytic functions with Schwarz conditions on the bicylinder. On this basis, with the help of the properties of the singular integral operators as well as the Cauchy-Pompeiu formula on the unit disc, we investigate a type of boundary value problem with Dirichlet boundary conditions and a type of mixed boundary value problems of higher order for bi-polyanalytic functions on the bicylinder and obtain the specific representations of the solutions. In addition, we discuss a system of complex partial differential equations with respect to polyanalytic functions with Neumann boundary conditions. On this foundation, we obtain the solutions to Neumann boundary value problems for bi-polyanalytic functions on the bicylinder. The conclusions in this paper provide effective methods for discussing other boundary value problems of inhomogeneous complex partial differential equations of higher order in spaces of several complex variables.
Yanyan Cui: Conceptualization, Project administration, Writing original draft, Writing–review and editing; Chaojun Wang: Investigation, Writing original draft, Writing–review and editing. All authors have read and agreed to the published version of the manuscript.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
This work was supported by the NSF of China (No. 11601543), the NSF of Henan Province (No. 222300420397), and the Science and Technology Research Projects of Henan Provincial Education Department (No. 19B110016).
The authors declare that they have no conflicts of interest.
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