In this paper, we study a Riemann-Hilbert problem related to inhomogeneous complex partial differential operators of higher order on the unit disk. Applying the Cauchy-Pompeiu formula, we find out the solvable conditions and obtain the representation of the solutions. Then, we investigate the boundary value problems for bi-polyanalytic functions with the Dirichlet and Riemann-Hilbert boundary conditions, obtain the specific solution and the solvable conditions, and extend the conclusion to the corresponding higher-order problems. Therefore, we obtain the solution to the half-Neumann problem of higher order for bi-polyanalytic functions.
Citation: Yanyan Cui, Chaojun Wang. The inhomogeneous complex partial differential equations for bi-polyanalytic functions[J]. AIMS Mathematics, 2024, 9(6): 16526-16543. doi: 10.3934/math.2024801
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In this paper, we study a Riemann-Hilbert problem related to inhomogeneous complex partial differential operators of higher order on the unit disk. Applying the Cauchy-Pompeiu formula, we find out the solvable conditions and obtain the representation of the solutions. Then, we investigate the boundary value problems for bi-polyanalytic functions with the Dirichlet and Riemann-Hilbert boundary conditions, obtain the specific solution and the solvable conditions, and extend the conclusion to the corresponding higher-order problems. Therefore, we obtain the solution to the half-Neumann problem of higher order for bi-polyanalytic functions.
The Cauchy integral representations play an important role in the function theory of one or several complex variables, among which the Cauchy integral formula and the Cauchy-Pompeiu formula [1] are the two most important categories. The classical Cauchy-Pompeiu formulas [2] in C are:
w(z)=12πi∫∂Gw(ζ)dζζ−z−1π∫Gwˉζ(ζ)dξdηζ−z,z∈G, | (1.1) |
w(z)=−12πi∫∂Gw(ζ)dˉζ¯ζ−z−1π∫Gwζ(ζ)dξdη¯ζ−z,z∈G, | (1.2) |
where G is a bounded smooth domain in the complex plane and w∈C1(G;C)⋂C(¯G;C).
Obviously, the classical Cauchy-Pompeiu formulas are closely related to complex partial differential operators. Thus, lots of boundary value problems related to complex partial differential equations were solved with the help of the Cauchy integral formula and the Cauchy-Pompeiu formula in the complex plane C, see [3,4,5,6]. By iterating the Cauchy-Pompeiu formula for one variable, the solutions to second order systems in polydomains composed by the Laplace and the Bitsadze operators were obtained [7]. The Riemann-Hilbert problems for generalized analytic vectors and complex elliptic partial differential equations of higher order were investigated in C [8]. By constructing a weighted Cauchy-type kernel, the Cauchy-Pompeiu integral representation in the constant weights case and orthogonal case were obtained [9]. The classical Cauchy-Pompeiu formula was generalized to different cases for different kinds of functions in C [7], among which a high order case is:
w(z)=m−1∑μ=012πi∫∂G1μ!(¯z−ζ)μζ−z∂μˉζw(ζ)dζ−1π∫G1(m−1)!(¯z−ζ)m−1ζ−z∂mˉζw(ζ)dξdη, | (1.3) |
where w∈Cm(G;C)⋂Cm−1(¯G;C) and 1≤m, see [10]. The first item in the right hand of (1.3) is the Cauchy integral expression for m-holomorphic functions on D, and the second item in (1.3) is a singular integral operator. Let
T0,mf(z)=−1π∫D1(m−1)!(¯z−ζ)m−1ζ−zf(ζ)dξdη, | (1.4) |
where m≥1, f∈C(∂G) with G being a bounded domain in the complex plane, then
∂mT0,mf(z)∂ˉzm=f(z), |
see [11]. Therefore, the solution to ∂mˉzw(z)=f(z) can be expressed as (1.3), and T0,mf(z) provides a particular solution to ∂mˉzw(z)=f(z). Thus it can be seen that, the Cauchy-Pompeiu formulas provide the integral representations of solutions to some partial differential equations.
The generalizations of the classical Cauchy-Pompeiu formula have contributed to the flourishing development of boundary value problems in C. Many types of boundary value problems for analytic functions or polyanalytic functions have emerged in C. Among them, Riemann boundary value problems and Dirichlet problems are the two major categories. By using the Cauchy-Pompeiu formula and its generalizations, the solvable conditions were found, and the integral expressions of the solutions to some problems were obtained. Some were discussed on different ranges of the unit disk, such as a generalized Cauchy-Riemann equation with super-singular points on a half-plane [12], the variable exponent Riemann boundary value problem for Liapunov open curves [13], Schwarz boundary value problems for polyanalytic equation in a sector ring [14], and so on [15,16,17]. Some were investigated for different equations, such as bi-harmonic equations with a p-Laplacian [18] and so on [12,19,20], which are different from the partial differential equations in [7].
As we all know, analytic functions are defined by a pair of Cauchy-Riemann equations. Bi-analytic functions [21] arise from the generalized system of Cauchy-Riemann 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, |
where k∈R,k≠−1, ϕ(z)=(k+1)θ−iω is called the associated function of f(z)=u+iv. A special case of bi-analytic functions is
∂ˉzf(z)=λ−14λϕ(z)+λ+14λ¯ϕ(z),∂ˉzϕ(z)=0. |
Bi-analytic functions are generalizations of analytic functions. They are important to studying elasticity problems and, therefore, have attracted the attention of many scholars. Begehr and Kumar [22] successfully obtained the solution to the Schwarz and Neumann boundary value problems for bi-polyanalytic functions. Lin and Xu [23] investigated the Riemann problem for (λ,k) bi-analytic functions. There are many other excellent conclusions as well [24,25].
So far, there have been few results for boundary value problems of bi-analytic functions with Riemann-Hilbert boundary conditions. Stimulated by this, we investigate this type of problems for bi-polyanalytic functions [26]
∂ˉzf(z)=λ−14λϕ(z)+λ+14λ¯ϕ(z),∂nˉzϕ(z)=0(n≥1), |
where λ∈R∖{−1,0,1}, which are the extensions of bi-analytic functions.
To solve the boundary value problems for bi-polyanalytic functions, in this paper, we first discuss a Riemann-Hilbert problem related to complex partial differential operators of higher order on the unit disk, and then we investigate the boundary value problems for bi-polyanalytic functions with the Dirichlet and Riemann-Hilbert boundary conditions, the half-Neumann boundary conditions and the mixed boundary conditions. Applying this method, we can also discuss other related systems of complex partial differential equations of higher order for bi-polyanalytic functions.
Let ℜz and ℑz represent the real and imaginary parts of the complex number z, respectively. Let Cm(G) represent the set of functions whose partial derivatives of order m are all continuous within G, and C(G) represent the set of continuous functions on G. To get the main results, we need the following lemmas:
Lemma 2.1. [2] Let G be a bounded smooth domain in the complex plane, f∈L1(G;C) and
Tf(z)=−1π∫Gf(ζ)ζ−zdξdη, |
then ∂ˉzTf(z)=f(z).
Lemma 2.2. Let D be the unit disk in C. For φ∈C(∂D,R) and f1∈Cm−1(ˉD,C) (m≥1) with ∂ˉzfκ−1=fκ(κ=2,⋯,m),fm=0, let
W(z)=12πi∫∂Dφ(ζ)ζ+zζ−zdζζ+12πi∫∂Dm−1∑μ=11μ!ℜ[(¯z−ζ)μfμ(ζ)]ζ+zζ−zdζζ+12π∫∂Dm−1∑μ=01μ!ℑ{[(¯z−ζ)μ−(−ˉζμ)]fμ(ζ)}dζζ, | (2.1) |
then ℜW=φ, ∂mˉzW(z)=0 and ∂sˉzW(z)=fs(z)(s=1,2,⋯,m−1).
Proof. ℜW=φ is due to the property of the Schwarz operator, as ζ+zζ−z is a pure imaginary number if z→∂D.
Let
{A=12πi∫∂Dm−1∑μ=11μ!ℜ[(¯z−ζ)μfμ(ζ)]ζ+zζ−zdζζ,B=12π∫∂Dm−1∑μ=01μ!ℑ{[(¯z−ζ)μ−(−ˉζμ)]fμ(ζ)}dζζ. |
Then
{Aˉz=m−1∑μ=11μ![12πi∫∂Dμ(¯z−ζ)μ−1fμ(ζ)dζζ−z−12πi∫∂Dμ(¯z−ζ)μ−1fμ(ζ)dζ2ζ],Asˉz=m−1∑μ=s1(μ−s),Amˉz=0, | (2.2) |
and
{Bˉz=m−1∑μ=114πiμ!∫∂Dμ(¯z−ζ)μ−1fμ(ζ)dζζ,Bsˉz=m−1∑μ=s14πi(μ−s)!∫∂D(¯z−ζ)μ−sfμ(ζ)dζζ(s=2,⋯,m−1),Bmˉz=0. | (2.3) |
By (2.2), (2.3), and the Cauchy-Pompeiu formula (1.1), we get
Wˉz=Aˉz+Bˉz=m−1∑μ=11μ!12πi∫∂Dμ(¯z−ζ)μ−1fμ(ζ)dζζ−z=12πi∫∂D[m−1∑μ=11(μ−1)!(¯z−ζ)μ−1fμ(ζ)]dζζ−z+−1π∫D∂∂ˉζ[m−1∑μ=11(μ−1)!(¯z−ζ)μ−1fμ(ζ)]dσζζ−z=[m−1∑μ=11(μ−1)!(¯z−ζ)μ−1fμ(ζ)]ζ=z=f1(z). |
Similarly,
∂sˉzW(z)=∂sˉzA+∂sˉzB=fs(z)(s=2,⋯,m−1),∂mˉzW(z)=∂mˉzA+∂mˉzB=0. |
Lemma 2.3. [11] Let G be a bounded smooth domain in the complex plane, and w∈ C1(G;C)⋂C(¯G;C), then
∫Gwˉz(z)dxdy=12i∫∂Gw(z)dz,∫Gwz(z)dxdy=−12i∫∂Gw(z)dˉz. |
Theorem 3.1. Let D be the unit disk in C. For γ,fk∈C(∂D) with ∂ˉzfk=fk+1 (k≥1,k∈Z), let
T0,kfk(z)=−1π∫D1(k−1)!(¯z−ζ)k−1ζ−zfk(ζ)dσζ, | (3.1) |
then the problem
{ℜ[¯ζpW(ζ)]=γ(ζ)(ζ∈∂D),∂kW(z)∂ˉzk=fk(z)(z∈D) |
is solvable on D.
(i) In the case of p≥0, the solution can be expressed as:
W(z)=zpφ1(z)+k−1∑s=0s∑l=0+∞∑v=0αsl(v−p)zvˉzl+T0,kfk(z), | (3.2) |
where
φ1(z)=12πi∫∂Dγ(ζ)ζ+zζ−zdζζ−1(k−1)!zpπ∫D¯fk(ζ)z(z−ζ)k−11−zˉζdσζ+12πi∫∂Dk−1∑μ=11μ!ℜ[¯(z−ζ)μμ∑m=0Cmμ(fμ−m(ζ)−T0,k−μ+mfk(ζ))∂mˉζˉζp]ζ+zζ−zdζζ+12π∫∂Dk−1∑μ=01μ!ℑ[(¯(z−ζ)μ−¯(−ζ)μ)μ∑m=0Cmμ(fμ−m(ζ)−T0,k−μ+mfk(ζ))∂mˉζˉζp]dζζ, | (3.3) |
αsl(v−p) are arbitrary complex constants satisfying
{k−1∑s=0s∑l=0αsl(v+l)=0(v≥p+k),k−1∑s=0s∑l=0[αsl(v+l)+¯αsl(l−v)]=0(−p≤v≤p),k−1∑s=0s∑l=0αsl(p+1+t+l)+k−1∑s=t+1s∑l=t+1¯αsl(−p−1−t+l)=0(t=0,1,⋯,k−2,k≥2). | (3.4) |
For k=1 the last equation in (3.4) is non-existent.
(ii) In the case of p<0, the solution can be expressed as:
W(z)=zpφ2(z)+k−1∑s=0s∑l=0+∞∑v=0αsl(v−p)zvˉzl+T0,kfk(z) | (3.5) |
on the condition that
12πi∫∂D{ζ−p[fk−1(ζ)−T0,1fk(ζ)]+ζp[¯fk−1(ζ)−T0,1fk(ζ)]}dζζl+1=0(l=0,1,⋯,−p−1), | (3.6) |
where αsl(v−p) is the same as in (i) and
φ2(z)=12πi∫∂Dγζ+zζ−zdζζ−1(k−1)!zpπ∫D¯fk(ζ)z(z−ζ)k−11−zˉζdσζ+12πi∫∂Dk−1∑μ=11μ!ℜ{¯(z−ζ)μζ−p[fμ(ζ)−T0,(k−μ)fk(ζ)]}ζ+zζ−zdζζ+12π∫∂Dk−1∑μ=01μ!ℑ{(¯(z−ζ)μ−¯(−ζ)μ)ζ−p[fμ(ζ)−T0,(k−μ)fk(ζ)]}dζζ. | (3.7) |
Proof. (1) For n=1, by Lemma 2.1, T0,kfk(z) is a particular solution to ∂kW(z)∂ˉzk=fk(z), then the corresponding general solution is W(z)=φ(z)+T0,kfk(z), where φ(z) is a k-holomorphic function with
ℜ[ζ−pφ(ζ)]=ℜ[ζ−pW(ζ)−ζ−pT0,kfk(ζ)]=γ(ζ)−ℜ[ζ−pT0,kfk(ζ)]≐γ0(ζ). |
(i) In the case of p≥0(p∈Z):
① Let φ(z)=zpφ1(z), then ℜ[ζ−pφ(ζ)]=γ0(ζ)⇔ℜ[φ1(ζ)]=γ0(ζ), and φ(z) is k-holomorphic if φ1(z) is k-holomorphic. As
∂μˉζ(ˉζpφ(ζ))=μ∑m=0Cmμ∂μ−mˉζφ(ζ)∂mˉζ(ˉζp)=μ∑m=0Cmμ∂μ−mˉζ[W(ζ)−T0,kfk(ζ)]∂mˉζ(ˉζp)=μ∑m=0Cmμ[fμ−m(ζ)−T0,k−μ+mfk(ζ)]∂mˉζ(ˉζp), |
by Lemma 2.2 we get that
φ1(z)=12πi∫∂Dk−1∑μ=01μ!ℜ[¯(z−ζ)μ∂μˉζ(ˉζpφ(ζ))]ζ+zζ−zdζζ+12π∫∂Dk−1∑μ=01μ!ℑ[(¯(z−ζ)μ−¯(−ζ)μ)∂μˉζ(ˉζpφ(ζ))]dζζ=12πi∫∂Dγ(ζ)ζ+zζ−zdζζ−12πi∫∂D[ζ−pT0,kfk(ζ)+ζp¯T0,kfk(ζ)](1ζ−z−12ζ)dζ+12πi∫∂Dk−1∑μ=11μ!ℜ[¯(z−ζ)μμ∑m=0Cmμ(fμ−m(ζ)−T0,k−μ+mfk(ζ))∂mˉζˉζp]ζ+zζ−zdζζ+12π∫∂Dk−1∑μ=01μ!ℑ[(¯(z−ζ)μ−¯(−ζ)μ)μ∑m=0Cmμ(fμ−m(ζ)−T0,k−μ+mfk(ζ))∂mˉζˉζp]dζζ | (3.8) |
satisfies ℜ[φ1(ζ)]=γ0(ζ) and ∂kˉzφ1(z)=0. Furthermore, by (3.1), we obtain that
12πi∫∂Dζ−pT0,kfk(ζ)dζζ−z=1(k−1)!−1π∫Dfk(ζ′)[12πi∫∂Dζ−p(¯ζ−ζ′)k−1ζ′−ζdζζ−z]dσζ′=1(k−1)!−1π∫Dfk(ζ′)[−12πi∫∂Dˉζp(¯ζ−ζ′)k−1ˉζζ′−1dˉζzˉζ−1]dσζ′=0, | (3.9) |
which follows
12πi∫∂Dζ−pT0,kfk(ζ)dζ2ζ=0. | (3.10) |
Meanwhile,
12πi∫∂Dζp¯T0,kfk(ζ)dζζ−z=1(k−1)!−1π∫D¯fk(ζ′)[12πi∫∂Dζp(ζ−ζ′)k−1¯ζ′−ˉζdζζ−z]dσζ′=1(k−1)!−1π∫D¯fk(ζ′)zp+1(z−ζ′)k−1z¯ζ′−1dσζ′=zp(k−1)!−1π∫D¯fk(ζ)z(z−ζ)k−1zˉζ−1dσζ, | (3.11) |
which follows
12πi∫∂Dζp¯T0,kfk(ζ)dζ2ζ=0. | (3.12) |
Plugging (3.9)–(3.12) into (3.8), we get (3.3). Therefore zpφ1(z) is a particular solution to ℜ[ζ−pφ(ζ)]=γ0(ζ) and ∂kˉzφ(z)=0.
② If φ0(z) is k-holomorphic on D with ℜ[ζ−pφ0(z)]=0, then zpφ1(z)+φ0(ζ) is the general solution to ℜ[ζ−pφ(ζ)]=γ0(ζ) and ∂kˉzφ(z)=0. In the following, we seek φ0(z).
As φ0(z) is k-holomorphic, it can be expressed as
φ0(z)=k−1∑s=0s∑l=0+∞∑v=0Cslvzvˉzl≐k−1∑s=0s∑l=0+∞∑v=0αsl(v−p)zvˉzl, | (3.13) |
where αsl(v−p)=Cslv are arbitrary complex constants. Let
αslv+¯αsl(2l−v)=Aslv,αslv=Bslv,¯αsl(2l−v)=Cslv, | (3.14) |
then
0=ℜ{ˉζpφ0(ζ)}=k−1∑s=0s∑l=0+∞∑v=−p(αslvζv−l+¯αslvζl−v)=k−1∑s=0s∑l=0p+2l∑v=−p(αslv+¯αsl(2l−v))ζv−l+k−1∑s=0s∑l=0+∞∑v=p+2l+1αslvζv−l+k−1∑s=0s∑l=0−p−1∑v=−∞¯αsl(2l−v)ζv−l=k−1∑s=0s∑l=0p+2l∑v=−pAslvζv−l+k−1∑s=0s∑l=0+∞∑v=p+2l+1Bslvζv−l+k−1∑s=0s∑l=0−p−1∑v=−∞Cslvζv−l={p∑v=−pA00vζv+(p∑v=−pA10vζv+p+2∑v=−pA11vζv−1)+(p∑v=−pA20vζv+p+2∑v=−pA21vζv−1+p+4∑v=−pA22vζv−2)+⋯+[p∑v=−pA(k−1)0vζv+p+2∑v=−pA(k−1)1vζv−1+⋯+p+2(k−1)∑v=−pA(k−1)(k−1)vζv−(k−1)]}+{+∞∑v=p+1B00vζv+(+∞∑v=p+1B10vζv++∞∑v=p+3B11vζv−1)+(+∞∑v=p+1B20vζv++∞∑v=p+3B21vζv−1++∞∑v=p+5B22vζv−2) | (3.15) |
+⋯+[+∞∑v=p+1B(k−1)0vζv++∞∑v=p+3B(k−1)1vζv−1+⋯++∞∑v=p+2k−1B(k−1)(k−1)vζv−(k−1)]}+−p−1∑v=−∞{C00vζv+(C10vζv+C11vζv−1)+(C20vζv+C21vζv−1+C22vζv−2)+⋯+[C(k−1)0vζv+C(k−1)1vζv−1+⋯+C(k−1)(k−1)vζv−(k−1)]}={p∑v=−pζv[k−1∑l=0k−1∑s=lAsl(v+l)]+ζ−p−1k−1∑l=1k−1∑s=lAsl(−p−1+l)+⋯+ζ−p−k+1A(k−1)(k−l)(−p)+ζp+1k−1∑l=1k−1∑s=lAsl(p+1+l)+ζp+2k−1∑l=2k−1∑s=lAsl(p+2+l)+⋯+ζp+k−1A(k−1)(k−1)(p+2(k−1))}+{ζp+1k−1∑s=0Bs0(p+1)+ζp+2[k−1∑s=0Bs0(p+2)+k−1∑s=1Bs1(p+3)]+⋯+ζp+k−1k−2∑l=0k−1∑s=lBsl(p+k−1+l)++∞∑v=p+kζv[k−1∑l=0k−1∑s=lBsl(v+l)]}+{ζ−p−1k−1∑s=0Cs0(−p−1)+ζ−p−2[k−1∑s=0Cs0(−p−2)+k−1∑s=1Cs1(−p−1)]+⋯+ζ−p−k+1k−2∑l=0k−1∑s=lCsl(−p−k+1+l)+−p−k∑v=−∞ζv[k−1∑l=0k−1∑s=lCsl(v+l)]}=−p−k∑v=−∞ζv[k−1∑l=0k−1∑s=lCsl(v+l)]+ζ−p−k+1[k−2∑l=0k−1∑s=lCsl(−p−k+1+l)+A(k−1)(k−l)(−p)]+⋯+ζ−p−1[k−1∑s=0Cs0(−p−1)+k−1∑l=1k−1∑s=lAsl(−p−1+l)]+p∑v=−pζv[k−1∑l=0k−1∑s=lAsl(v+l)]+ζp+1[k−1∑s=0Bs0(p+1)+k−1∑l=1k−1∑s=lAsl(p+1+l)]+ζp+k−1[k−2∑l=0k−1∑s=lBsl(p+k−1+l)+A(k−1)(k−1)(p+2(k−1))]++∞∑v=p+kζv[k−1∑l=0k−1∑s=lBsl(v+l)], | (3.16) |
which leads to (3.4) in consideration of (3.14) and
k−1∑l=0k−1∑s=lCslv=k−1∑s=0s∑l=0Cslv. |
By ① and ②, (3.2) is the solution to Problem RH.
(ii) In the case of p<0(p∈Z):
① Let φ(z)=zpφ2(z), then ℜ[ζ−pφ(ζ)]=γ0(ζ)⇔ℜ[φ2(ζ)]=γ0(ζ). Similar to the discussion of ① in (i), we get a particular solution φ2(z) for ℜ[φ2(ζ)]=γ0(ζ) where φ2(z) is k holomorphic:
φ2(z)=12πi∫∂Dγ0ζ+zζ−zdζζ+12πi∫∂Dk−1∑μ=11μ!ℜ[¯(z−ζ)μζ−p∂μˉζφ(ζ)]ζ+zζ−zdζζ+12π∫∂Dk−1∑μ=01μ!ℑ[(¯(z−ζ)μ−¯(−ζ)μ)ζ−p∂μˉζφ(ζ)]dζζ=12πi∫∂Dγζ+zζ−zdζζ−12πi∫∂D[ζ−pT0,kfk(ζ)+ζp¯T0,kfk(ζ)](1ζ−z−12ζ)dζ+12πi∫∂Dk−1∑μ=11μ!ℜ[¯(z−ζ)μζ−p∂μˉζφ(ζ)]ζ+zζ−zdζζ+12π∫∂Dk−1∑μ=01μ!ℑ[(¯(z−ζ)μ−¯(−ζ)μ)ζ−p∂μˉζφ(ζ)]dζζ. |
So we get (3.7) for (3.9)–(3.12) and
∂μˉζφ(ζ)=∂μˉζ[W(ζ)−T0,kfk(ζ)]=∂μˉζW(ζ)−∂μˉζT0,kfk(ζ)=fμ(ζ)−T0,(k−μ)fk(ζ). |
Therefore
φ(z)=zpφ2(z)+k−1∑s=0s∑l=0+∞∑v=0αsl(v−p)zvˉzl |
is the solution to ℜ[ζ−pφ(ζ)]=γ0(ζ) and ∂kˉzφ(z)=0, where αsl(v−p) are arbitrary complex constants satisfying (3.12).
② Secondly, we seek the condition to ensure that φ(z) is k-holomorphic from φ2(z)=z−pφ(z). As the k-holomorphism of φ2(z)=z−pφ(z) is equivalent to the holomorphism of ∂k−1ˉzφ2(z)=z−p∂k−1ˉzφ(z), applying the properties of the Schwarz operator for holomorphic functions and
1ζ−z=1ζ⋅11−zζ=∞∑l=0zlζl+1(|zζ|<1), |
we get that
z−p∂k−1ˉzφ(z)=12πi∫∂D[ζ−p∂k−1ˉζφ(ζ)+ζp¯∂k−1ˉζφ(ζ)][2ζζ−z−1]dζ2ζ=∞∑l=0{12πi∫∂D[ζ−p∂k−1ˉζφ(ζ)+ζp¯∂k−1ˉζφ(ζ)]dζζl+1}zl−12πi∫∂D[ζ−p∂k−1ˉζφ(ζ)+ζp¯∂k−1ˉζφ(ζ)]dζ2ζ. |
As p<0, then φ(z) is k-holomorphic (-i.e., ∂k−1ˉzφ(z) is holomorphic) if and only if z−p∂k−1ˉzφ(z) has a zero of order at least −p at z=0. Therefore,
12πi∫∂D[ζ−p∂k−1ˉζφ(ζ)+ζp¯∂k−1ˉζφ(ζ)]dζζl+1=0(l=0,1,⋯,−p−1), |
that is (3.6) as
∂k−1ˉζφ(ζ)=∂k−1ˉζ[W(ζ)−T0,kfk(ζ)]=∂k−1ˉζW(ζ)−∂k−1ˉζT0,kfk(ζ)=fk−1(ζ)−T0,1fk(ζ). |
By ① and ②, (3.5) is the solution of Problem RH on the condition of (3.6).
Theorem 3.1 extends the conclusions of Riemann-Hilbert boundary value problems for k-holomorphic functions. Given that k=1 in Theorem 3.1, we can get the following conclusion, which extends the existing results of the corresponding Riemann-Hilbert problems for analytic functions.
Corollary 3.2. Let D be the unit disk in C. For γ,f∈C(∂D), the problem
ℜ[ˉζpW(ζ)]=γ(ζ)(ζ∈∂D),∂W(z)∂ˉz=f(z)(z∈D) |
is solvable on D.
(i) In the case of p≥0, the solution can be expressed as:
W(z)=zp2πi∫∂Dγ(ζ)ζ+zζ−zdζζ−z2p+1π∫D¯f(ζ)1−zˉζdσζ+2p∑v=0αv−pzv−1π∫Df(ζ)ζ−zdσζ, |
where αv−p are arbitrary complex constants satisfying αv+¯α−v=0(−p≤v≤p);
(ii) In the case of p<0, the solution is the same as in (i) on the condition of
12πi∫∂D{γ(ζ)−ℜ[ζ−pTf(ζ)]}dζζl+1=0(l=0,1,⋯,−p−1). |
In this section we discuss several boundary value problems for bi-polyanalytic functions with the Dirichlet, Riemann-Hilbert boundary conditions or the mixed boundary conditions.
Theorem 4.1. Let D be the unit disk in C, and let λ∈R∖{−1,0,1}, φ,γ,fk∈C(∂D) with ∂ˉzfk=fk+1 (k=1,2,⋯,n−1,n≥2) and fn=0. Then the problem
{∂ˉzf(z)=λ−14λϕ(z)+λ+14λ¯ϕ(z),∂nˉzϕ(z)=0,∂kˉzϕ(z)=fk(z)(z∈D),f(ζ)=φ(ζ),ℜ[¯ζpϕ(ζ)]=γ(ζ)(ζ∈∂D), | (4.1) |
is solvable on D.
(i) In the case of p≥0, the solution can be expressed as:
f(z)=12πi∫∂Dφ(ζ)dζζ−z−1π∫D[λ−14λ(ζpφ1(ζ)+T0,kfk(ζ))+λ+14λ¯(ζpφ1(ζ)+T0,kfk(ζ))]dσζζ−z−λ−14λ{k−1∑s=0s∑l=0[l∑v=0−αsl(v−p)l+1zvˉzl+1++∞∑v=l+1αsl(v−p)l+1zv(z−l−1−ˉzl+1)]}−λ+14λ{k−1∑s=0s∑l=0[l−1∑v=0¯αsl(v−p)v+1zl(z−v−1−ˉzv+1)++∞∑v=l−¯αsl(v−p)v+1zlˉzv+1]} | (4.2) |
if and only if
12πi∫∂Dφ(ζ)dζ1−ˉzζ+1π∫D[λ−14λζpφ1(ζ)+λ+14λˉζp¯φ1(ζ)]dσζˉzζ−1+1π∫D{(¯z−ζ)kfk(ζ)k!(ˉzζ−1)−¯fk(ζ)(k−1)![k−2∑s=0Csk−1(ˉzζ−1)k−1−s−(−1)k−1−s(k−1−s)ˉzk(1−ˉzζ)s+(1−ˉzζ)k−1ln(1−ˉzζ)zk]}dσζ=λ−14λk−1∑s=0s∑l=0l∑v=0αsl(v−p)ˉzl−vl+1+λ+14λk−1∑s=0s∑l=0+∞∑v=l¯αsl(v−p)ˉzv−lv+1, | (4.3) |
where T0,kfk and φ1 are represented by (3.1) and (3.3), respectively, and αsl(v−p) are arbitrary complex constants satisfying (3.4);
(ii) In the case of p<0, the solution can be expressed as (4.2) on the condition of (3.6) and (4.3), where T0,kfk and αsl(v−p) are the same as in (i); however, φ1 is represented by (3.7).
Proof. As
∂ˉz[−1π∫Dg(ζ)ζ−zdσζ]=g(z), |
obviously,
−1π∫D[λ−14λϕ(ζ)+λ+14λ¯ϕ(ζ)]dσζζ−z |
is a special solution to ∂ˉzf(z)=λ−14λϕ(z)+λ+14λ¯ϕ(z), then the solution of the problem (4.1) can be represented as
f(z)=ψ(z)−1π∫D[λ−14λϕ(ζ)+λ+14λ¯ϕ(ζ)]dσζζ−z, | (4.4) |
where ψ(z) is analytic on D to be determined, and ϕ(ζ) is the solution to
ℜ[¯ζpϕ(ζ)]=γ(ζ)(ζ∈∂D),∂nˉzϕ(z)=0,∂kˉzϕ(z)=fk(z)(z∈D). |
Since ψ(z) is analytic, by (4.4),
ψ(z)=12πi∫∂Dψ(ζ)ζ−zdζ=12πi∫∂D{φ(ζ)+1π∫D[λ−14λϕ(˜ζ)+λ+14λ¯ϕ(˜ζ)]dσ˜ζ˜ζ−ζ}dζζ−z=12πi∫∂Dφ(ζ)dζζ−z+1π∫D[λ−14λϕ(˜ζ)+λ+14λ¯ϕ(˜ζ)][12πi∫∂D1˜ζ−ζdζζ−z]dσ˜ζ=12πi∫∂Dφ(ζ)ζ−zdζ | (4.5) |
if and only if
12πi∫∂Dˉzψ(ζ)1−ˉzζdζ=0(z∈D), |
that is,
12πi∫∂Dˉz1−ˉzζ{φ(ζ)+1π∫D[λ−14λϕ(˜ζ)+λ+14λ¯ϕ(˜ζ)]dσ˜ζ˜ζ−ζ}dζ=0⇔12πi∫∂Dˉzφ(ζ)1−ˉzζdζ+1π∫D[λ−14λϕ(˜ζ)+λ+14λ¯ϕ(˜ζ)][12πi∫∂Dˉz1−ˉzζdζ˜ζ−ζ]dσ˜ζ⇔12πi∫∂Dφ(ζ)1−ˉzζdζ+1π∫D[λ−14λϕ(ζ)+λ+14λ¯ϕ(ζ)]dσζˉzζ−1=0. | (4.6) |
Plugging (4.5) into (4.4),
f(z)=12πi∫∂Dφ(ζ)ζ−zdζ−1π∫D[λ−14λϕ(ζ)+λ+14λ¯ϕ(ζ)]dσζζ−z. | (4.7) |
(i) In the case of p≥0, by Theorem 3.1,
ϕ(z)=zpφ1(z)+k−1∑s=0s∑l=0+∞∑v=0αsl(v−p)zvˉzl+T0,kfk(z), | (4.8) |
where T0,kfk and φ1 are represented by (3.1) and (3.3), respectively, and αsl(v−p) are arbitrary complex constants satisfying (3.4).
Applying the Cauchy-Pompeiu formula (1.1),
1π∫Dζvˉζlζ−zdσζ=1π∫D∂ˉζ(ζvˉζl+1l+1)dσζζ−z=12πi∫∂Dζvˉζl+1l+1dζζ−z−zvˉzl+1l+1=1l+112πi∫∂Dζv−l−1ζ−zdζ−zvˉzl+1l+1={zvl+1(z−l−1−ˉzl+1),v≥l+1,−zvl+1ˉzl+1,v<l+1. | (4.9) |
Similarly,
1π∫Dˉζvζlζ−zdσζ={zlv+1(z−v−1−ˉzv+1),v≤l−1,−zlv+1ˉzv+1,v>l−1. | (4.10) |
Plugging (4.8)–(4.10) into (4.7), the solution (4.2) follows.
In order to obtain the solvable conditions, the following integrals need to be calculated: First, by Lemma 2.3,
1π∫Dζvˉζlˉzζ−1dσζ=1π∫D∂ˉζ(ˉζl+1l+1ζvˉzζ−1)dσζ=12πi∫∂Dˉζl+1l+1ζvdζˉzζ−1=1l+112πi∫∂Dζv−l−1ˉzζ−1dζ={0,v≥l+1,−ˉzl−vl+1,v<l+1. | (4.11) |
Similarly,
1π∫Dˉζvζlˉzζ−1dσζ={0,v≤l−1,−ˉzv−lv+1,v>l−1. | (4.12) |
Secondly, by the Cauchy-Pompeiu formula (1.1),
1π∫DT0,kfk(ζ)dσζˉzζ−1=1π∫D[−1π∫D1(k−1)!(¯ζ−˜ζ)k−1˜ζ−ζfk(˜ζ)dσ˜ζ]dσζˉzζ−1=−1π∫Dfk(˜ζ)(k−1)![1π∫D(¯ζ−˜ζ)k−1˜ζ−ζdσζˉzζ−1]dσ˜ζ=−1π∫Dfk(˜ζ)(k−1)![1π∫D∂ˉζ(¯ζ−˜ζ)kk(1−ˉzζ)dσζζ−˜ζ]dσ˜ζ=−1π∫Dfk(˜ζ)(k−1)![12πi∫∂D(¯ζ−˜ζ)kk(1−ˉzζ)dζζ−˜ζ−0]dσ˜ζ=−1π∫Dfk(˜ζ)(k−1)![1k¯12πi∫∂D(ζ−˜ζ)kdζ(ζ−z)(1−¯˜ζζ)]dσ˜ζ=1π∫D(¯z−ζ)kfk(ζ)k!(ˉzζ−1)dσζ. | (4.13) |
In addition, for z,˜ζ∈D, setting ˜z=1/ˉz, in view of
1π∫D(ζ−˜ζ)k−1¯˜ζ−ζdσζˉzζ−1=−˜zπ∫D(ζ−˜ζ)k−1ζ−˜zdσζ¯ζ−˜ζ=−˜zπ∫D[k−2∑s=0Csk−1(ζ−˜z)k−2−s(˜z−˜ζ)s+(˜z−˜ζ)k−1ζ−˜z]dσζ¯ζ−˜ζ=−˜zπ∫D∂ζ[k−2∑s=0Csk−1(ζ−˜z)k−1−sk−1−s(˜z−˜ζ)s+(˜z−˜ζ)k−1ln(ζ−˜z)]dσζ¯ζ−˜ζ=˜z[k−2∑s=0Csk−1(ζ−˜z)k−1−sk−1−s(˜z−˜ζ)s+(˜z−˜ζ)k−1ln(ζ−˜z)]ζ=˜ζ+˜z2πi∫∂D[k−2∑s=0Csk−1(ζ−˜z)k−1−sk−1−s(˜z−˜ζ)s+(˜z−˜ζ)k−1ln(ζ−˜z)]dˉζ¯ζ−˜ζ=˜z[k−2∑s=0Csk−1(˜ζ−˜z)k−1−sk−1−s(˜z−˜ζ)s+(˜z−˜ζ)k−1ln(˜ζ−˜z)]−˜z[k−2∑s=0Csk−1(−˜z)k−1−sk−1−s(˜z−˜ζ)s+(˜z−˜ζ)k−1ln(−˜z)]=˜z[k−2∑s=0Csk−1(˜ζ−˜z)k−1−s−(−˜z)k−1−sk−1−s(˜z−˜ζ)s+(˜z−˜ζ)k−1ln˜ζ−˜z−˜z]=k−2∑s=0Csk−1(ˉz˜ζ−1)k−1−s−(−1)k−1−s(k−1−s)ˉzk(1−ˉz˜ζ)s+(1−ˉz˜ζ)k−1ln(1−ˉz˜ζ)ˉzk, |
in which the logarithmic functions take the principal value, therefore,
1π∫D¯T0,kfk(ζ)dσζˉzζ−1=1π∫D[−1π∫D1(k−1)!(ζ−˜ζ)k−1¯˜ζ−ζ¯fk(˜ζ)dσ˜ζ]dσζˉzζ−1=−1π∫D¯fk(˜ζ)(k−1)![1π∫D(ζ−˜ζ)k−1¯˜ζ−ζdσζˉzζ−1]dσ˜ζ=−1π∫D[k−2∑s=0Csk−1(ˉzζ−1)k−1−s−(−1)k−1−s(k−1−s)ˉzk(1−ˉzζ)s+(1−ˉzζ)k−1ln(1−ˉzζ)ˉzk]¯fk(ζ)dσζ(k−1)!. | (4.14) |
Plugging (4.11)–(4.14) into (4.6), the condition (4.3) follows.
(ii) In the case of p<0, similar to (i), by Theorem 3.1, the result follows.
Given that k=1 in Theorem 4.1, we can get the solution to the following boundary value problem for bi-analytic functions.
Corollary 4.2. Let D be the unit disk in C, and let λ∈R∖{−1,0,1}. For φ,γ∈C(∂D), the problem
{∂ˉzf(z)=λ−14λϕ(z)+λ+14λ¯ϕ(z),∂ˉzϕ(z)=0(z∈D),f(ζ)=φ(ζ),ℜ[¯ζpϕ(ζ)]=γ(ζ)(ζ∈∂D) |
is solvable on D.
(i) In the case of p≥0, the solution can be expressed as:
f(z)=12πi∫∂Dφ(ζ)dζζ−z−1π∫D[λ−14λζpφ1(ζ)+λ+14λ¯ζpφ1(ζ)]dσζζ−z+λ−14λ[α−pˉz++∞∑v=1αv−pzv(ˉz−z−1)]+λ+14λ+∞∑v=0¯αv−pˉzv+1 |
if and only if
12πi∫∂Dφ(ζ)dζ1−ˉzζ+1π∫D[λ−14λζpφ1(ζ)+λ+14λˉζp¯φ1(ζ)]dσζˉzζ−1=λ−14λα(−p)+λ+14λ+∞∑v=0¯αv−pˉzv, |
where
φ1(z)=12πi∫∂Dγ(ζ)ζ+zζ−zdζζ |
and αv−p are arbitrary complex constants satisfying
αv=0(v≥p+1),αv+¯α−v=0(−p≤v≤p). |
(ii) In the case of p<0, the solution is the same as in (i) on the condition that
12πi∫∂Dγ(ζ)dζζl+1=0(l=0,1,⋯,−p−1). |
Applying Corollary 3.2 and Theorem 4.1 we can get the solution to the following boundary value problem of higher order for bi-polyanalytic functions.
Corollary 4.3. Let D be the unit disk in C, and let λ∈R∖{−1,0,1}, φ,˜γ,γ,fk∈C(∂D) with ∂ˉzfk=fk+1 (k=1,2,⋯,n−1,n≥2) and fn=0. Then the problem
{∂2ˉzW(z)=λ−14λϕ(z)+λ+14λ¯ϕ(z),∂nˉzϕ(z)=0,∂kˉzϕ(z)=fk(z)(z∈D),∂ˉζW(ζ)=φ(ζ),ℜ[¯ζqW(ζ)]=˜γ(ζ),ℜ[¯ζpϕ(ζ)]=γ(ζ)(ζ∈∂D) |
is solvable on D.
(i) In the case of q≥0, the solution can be expressed as:
W(z)=zq2πi∫∂D˜γ(ζ)ζ+zζ−zdζζ−z2q+1π∫D¯f(ζ)1−zˉζdσζ+2q∑v=0αv−qzv−1π∫Df(ζ)ζ−zdσζ, |
where αv−q are arbitrary complex constants satisfying αv+¯α−v=0(−q≤v≤q) and f(z) is the solution of the problem (4.1) in Theorem 4.1;
(ii) In the case of q<0, the solution is the same as in (i) on the condition of
12πi∫∂D{γ(ζ)−ℜ[ζ−qTf(ζ)]}dζζl+1=0(l=0,1,⋯,−q−1). |
ies
Remark 4.4. Applying Corollaries 3.2 and 4.3 we can get the solution of the problem:
{∂3ˉzW(z)=λ−14λϕ(z)+λ+14λ¯ϕ(z),∂nˉzϕ(z)=0,∂kˉzϕ(z)=fk(z)(z∈D),∂2ˉζW(ζ)=φ(ζ),ℜ[¯ζpϕ(ζ)]=γ(ζ)(ζ∈∂D),ℜ[¯ζq1W(ζ)]=~γ1(ζ),ℜ[¯ζq2W(ζ)]=~γ2(ζ)(ζ∈∂D), |
where λ∈R∖{−1,0,1}, φ,~γ1,~γ2,γ,fk∈C(∂D) with ∂ˉzfk=fk+1 (k=1,2,⋯,n−1,n≥2) and fn=0. Similarly, the corresponding higher-order problems can be solved.
Applying Theorem 4.1 and the results for the half-Neumann-n problem in [27], we can draw the following conclusion:
Corollary 4.5. Let D be the unit disk in C, and let λ∈R∖{−1,0,1}, cs∈C, γs∈C(∂D) (0≤s≤m−2,m≥2,m∈N), φ,γ,fk∈C(∂D) with ∂ˉzfk=fk+1 (k=1,2,⋯,n−1,n≥2) and fn=0. Then the problem
{∂mˉzW(z)=λ−14λϕ(z)+λ+14λ¯ϕ(z),∂nˉzϕ(z)=0,∂kˉzϕ(z)=fk(z)(z∈D),∂m−1ˉζW(ζ)=φ(ζ),ℜ[¯ζpϕ(ζ)]=γ(ζ)(ζ∈∂D),ζ∂sˉζWζ(ζ)=γs(ζ)(ζ∈∂D),∂sˉzW(0)=cs(0≤s≤m−2,m≥2,m∈N) |
is solvable on D, and the solution is given by
W(z)=m−2∑s=0csˉzs+m−2∑s=0(−1)s+12πi∫∂Dγs(ζ)[(1−|z|2)szslog(1−zˉζ)+s−1∑r=0ˉζs−m(s−m)zrr∑l=0Cls(−|z|2)l]dζ+(−1)m−1zπ∫Df(ζ)ζ(ζ−z)(¯ζ−z)m−2(m−2)!dσζ |
if and only if
12πi∫∂Dγm−2(ζ)(1−ˉzζ)ζdζ+ˉzπ∫Df(ζ)(1−ˉzζ)2dσζ=0 |
and
12πi∫∂Dγm−1−s(ζ)(1−ˉzζ)ζdζ+ˉz2πi∫∂Dγm−s(ζ)ζlog(1−ˉzζ)dζ=s−1∑r=2(−1)rr!(r−1)ˉz2πi∫∂Dγm−1−s+r(ζ)ζ[(¯ζ−z)r−1−(−ˉz)r−1]dζ+(−1)s(s−1)!ˉzπ∫Df(ζ)(¯ζ−z)s−1(1−ˉzζ)2dσζ,2≤s≤m−1, |
where f(z) is the solution of the problem (4.1) in Theorem 4.1.
Remark 4.6. By Theorem 4.1 and the mixed boundary value problems with combinations of Schwarz, Dirichlet, and Neumann conditions in [27], we can discuss the corresponding mixed boundary value problems for bi-polyanalytic functions, for example,
{∂mˉzW(z)=λ−14λϕ(z)+λ+14λ¯ϕ(z),∂nˉzϕ(z)=0,∂kˉzϕ(z)=fk(z)(z∈D),∂m−1ˉζW(ζ)=φ(ζ),ℜ[¯ζpϕ(ζ)]=γ(ζ)(ζ∈∂D),ζ∂sˉζWζ(ζ)=γs(ζ)(ζ∈∂D),∂sˉzW(0)=cs(0≤s≤m−3,m≥3,m∈N),∂m−2ˉζW(ζ)=γm−2(ζ)(ζ∈∂D). |
By using the Cauchy-Pompeiu formula in the complex plane, we first discuss a Riemann-Hilbert boundary value problem of higher order on the unit disk D in C and obtain the expression of the solution under different solvable conditions. On this basis, we get the specific solutions to the boundary value problems for bi-polyanalytic functions with the Dirichlet and Riemann-Hilbert boundary conditions and the corresponding higher-order problems. Therefore, we obtain the solution to the half-Neumann problem of higher order for bi-polyanalytic functions. The conclusions provide a favorable method for discussing other boundary value problems of bi-polyanalytic functions, such as mixed boundary value problems and the related systems of complex partial differential equations of higher order, and also provide a solid basis for future research on bi-polyanalytic functions.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
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 the Henan Provincial Education Department (No.19B110016).
The authors declare that they have no conflicts of interest.
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