Two Basic Boundary-Value Problems for inhomogeneous Cauchy–Riemann Equation in an infinite Sector Mohamed S. Akel and Hussein S. Hussein Dept. of Math., Faculty of Science, King Faisal University, Al-Ahsaa,31982, P.O. Box 380, Saudi Arabia *
Corresponding author E-mail adresses :
[email protected] (M. S. Akel),
[email protected] (H. S. Hussein)
Abstract The object of the present article is to investigate the Schwarz and Dirichlet boundary value problems for the inhomogeneous Cauchy–Riemann equation in an infinite sector. Firstly, we obtain the Schwarz–Poisson formula in a sector with angle 𝜋⁄𝛼 1
(𝛼 ≥ 2 ). Secondly, boundary values at the vertex point are proved to exist. Finally, the solutions and the conditions of solvability are explicitly obtained. The result we obtain is a generalization of recent results obtained by Begehr, H., Gaertner, E.A. (Georg. Math. J. 14(1), 33–52, 2007) and by Ying Wang, Yufeng Wang (Complex Anal. Oper. Theory, DOI 10.1007/s11785-010-0107-0).
Keywords Cauchy–Pompeiu formula, inhomogeneous Cauchy–Riemann equation , Schwarz –type operator, Pompeiu –type operator, Schwarz Problem, Dirichlet problem
Mathematics Subject Classification (2000) 30E25 - 30G30 - 45E05
1 Introduction Nowadays, complex analytic methods, e.g., the theory of boundary value problems (BVPs) for analytic and generalized analytic functions, are widely applied to solving physical problems in shell theory, fluid dynamics, elasticity theory and so on[1-4]. Kinds of BVPs have been investigated for complex partial differential equations (PDEs) [5-10]. On the other hand, BVPs of complex PDEs in some special domains are solved in a different way. Those special domains include the unit disc [11,12], half plane [13,14], circular ring [15–17], quarter plane [18–20], half disc and half ring [21], triangle[22-24]. In this paper we study the solvability of the Schwarz and Dirichlet problems for the inhomogeneous Cauchy–Riemann (a)
permanent address : Dept. of Math., Faculty of Science, South Valley University, Qena, Egypt 1
equation in an infinite sectoral domain in the plane. Firstly, we discuss the boundary behaviors of Schwarz-type operator and Pompeiu-type operator. Specially, boundary values at the vertex point of the sector are proved to exist. Finally, we obtain explicitly the expressions of the solutions and the solvability conditions. Let Ω be a infinite sector domain in the complex plane C defined by 𝜋
Ω = {𝑧 ∈ ℂ, 0 < arg 𝑧 < 𝛼}
(1.1) 1
Except for additional indication, we always assume that 𝛼(𝛼 ≥ ) 2
is a fixed real constant. The sector Ω is isomorphic with the upper half plane ℍ.
Indeed, the conformal mapping [25] 𝜁: Ω → ℍ 𝑧 ⟼ 𝑧𝛼
(1.2)
with the branch cut along (0, +∞), maps the boundary [0, +∞) onto itself and the boundary {𝑧 = 𝑡𝑒 𝑖𝜗 , 0 ≤ 𝑡 < +∞, 𝜗 = 𝜋/𝛼} onto (−∞, 0]. The inverse mapping is
𝜁: ℍ → Ω 𝑧 ⟼ 𝑧1/𝛼
(1.3)
which transforms ℝ onto the boundary 𝜕Ω of the sector Ω. Obviously, the above two conformal mappings can be firstly 1 − 1 extended to the boundary except for 0. Since the argument at 0 can be arbitrary and 𝛼 ≥ 1/2 , we can assume that the arguments of 𝑧 𝛼 and 𝑧 1/𝛼 at 𝑧 = 0 are 0. Hence, the above 1 − 1 mapping between the boundary is proper. In what follows, the main analytic branches of 𝑧 𝛼 and 𝑧 1/𝛼 are always chosen as above, respectively. Moreover, on the basis of the analytic branches of 𝑧 𝛼 and 𝑧 1/𝛼 , we define the analytic branch of several functions as follows, 𝑧 𝛼 1−𝛼 𝑧 𝑧1/𝛼 𝛼−1 1/𝛼−1 𝑧 = ,𝑧 = 𝛼, 𝑧 = . 𝑧 𝑧 𝑧
2
One of the fundamental tools for solving BPVs of complex PDEs is the Cauchy-Pompeiu formula which is valid for the upper half plane ℍ[14]. Theorem 1.1 If 𝑤 ∶ ℍ → ℂ satisfies 𝑤 ∈ 𝐿2 (ℝ; ℂ) ∩ 𝐶(ℝ, ℂ) and 𝑤𝑧 ∈ 𝐿𝑝,2 (ℍ; ℂ), 𝑝 > 2. Then, for 𝑧 ∈ ℍ, +∞
1 𝑑𝑡 1 𝑤(𝑧) = ∫ 𝑤(𝑡) − 2𝜋𝑖 𝑡−𝑧 𝜋 −∞
𝑤(𝑧) = −
+∞
∫ 𝑤𝜁 (𝜁) 0 2, can be represented as ∞ ∞ 1 𝑡 −1/2 𝑑𝑡 1 𝑡 −1/2 𝑑𝑡 ∫ 𝑤(𝑡) 1/2 − ∫ 𝑤(𝑡) 1/2 4𝜋𝑖 0 𝑡 − 𝑧 1/2 4𝜋𝑖 0 𝑡 + 𝑧 1/2
1 𝜁 −1/2 𝑤(𝑧), 𝑧 ∈ ℂ0 − ∫ 𝑤𝜁̅ (𝜁) 1/2 𝑑𝜉𝑑𝜂 = { 2𝜋 ℂ0 0, 𝑧 ∉ ℂ̅0 𝜁 − 𝑧 1/2
and for 𝑧 ∈ ℂ0
∞
1 1 𝑡 1/2 𝑤(𝑧) = 𝑖Im𝑤(−1) + ∫ Re𝑤(𝑡) ( 1/2 − ) 𝑡 −1/2 𝑑𝑡 1/2 2𝜋𝑖 𝑡 + 1 𝑡 −𝑧 0
∞
1 1 𝑡 1/2 − ∫ Re𝑤(𝑡) ( 1/2 − ) 𝑡 −1/2 𝑑𝑡 1/2 2𝜋𝑖 𝑡 + 1 𝑡 +𝑧 0
1 1 𝜁1/2 −1/2 − ∫ {𝜁 𝑤𝜁̅ (𝜁) ( 1/2 − ) 2𝜋 Ω 𝜁 − 𝑧 1/2 𝜁 + 1 –𝜁
−1/2
̅̅̅̅̅̅̅̅ 𝑤𝜁̅ (𝜁) (
1 𝜁
1/2
−
𝑧 1/2
−
𝜁
1/2
𝜁 + 1
)} 𝑑𝜉𝑑𝜂
3 Schwarz boundary value problem in the sector This Section is devoted to extending the Schwarz-Poisson representation formula to an infinite sectoral domain with angle 𝜋⁄𝛼 (𝛼 ≥ 1/2 ) by a proper conformal mapping to solve the Schwarz problem in the sector. Schwarz problem: Find a function 𝑤 satisfying the following conditions
6
𝑤𝑧̅ = 𝑓 in Ω , Re𝑤 = 𝛾 on 𝜕Ω Im𝑤(𝑖 1/𝛼 ) = 𝑐 , 𝑐 ∈ ℝ, with 𝑓 ∈ 𝐿𝑝 (Ω, ℂ) , 𝑝 > 2, 𝛾 ∈ 𝐿2 (𝜕Ω, ℝ) ∩ 𝐶(𝜕Ω, ℝ). Firstly, introducing a new kernel 1 1 𝐻(𝑧, 𝜁) = [ 𝛼 − ] 𝜁 𝛼−1 . 𝛼 𝛼 𝛼 ̅̅̅ 𝜁 −𝑧 𝜁 −𝑧 Then, the following lemma is valid.
(3.1)
(3.2)
Lemma 3.1. For 𝛾 ∈ 𝐶(𝜕Ω, ℝ), lim
𝑧∈Ω, 𝑧→0
𝛼 ∫ [𝛾(𝜁) − 𝛾(0)]𝐻(𝑧, 𝜁)𝑑𝜁 = 0. 2𝜋𝑖 𝜕Ω
Proof: For (𝑧, 𝜁) ∈ Ω × 𝜕Ω, we have 𝛼 ∫ [𝛾(𝜁) − 𝛾(0)]𝐻(𝑧, 𝜁)𝑑𝜁 𝑧∈Ω, 2𝜋𝑖 𝜕Ω lim
𝑧→0
∞
1 1 1 1 = lim ∫ [𝛾(𝜁 𝛼 ) − 𝛾(0)] [ − ] 𝑑𝜁 𝛼 𝑧∈Ω, 2𝜋𝑖 𝜁 − 𝑧 𝛼 𝜁 − 𝑧̅̅̅ 𝑧→0
−∞
∞
1 Im z α 1/𝛼 = lim ∫ [𝛾(𝜁 ) − 𝛾(0)] 𝑑𝜁 𝑧∈Ω, 𝜋 |𝜁 − 𝑧 𝛼 |2 𝑧→0
−∞
= lim [𝛾(𝑧) − 𝛾(0)] = 0, 𝑧∈Ω, 𝑧→0
by Poisson formula on ℝ.
Lemma 3.2. For 𝛾 ∈ 𝐶(𝜕Ω, ℝ), if 𝜁0 ∈ 𝜕Ω\{0} lim
𝑧∈Ω, 𝑧→𝜁0
𝛼 ∫ 𝛾(𝜁)𝐻(𝑧, 𝜁)𝑑𝜁 = 𝛾(𝜁0 ). 2𝜋𝑖 𝜕Ω
Proof: Since 𝐻(𝑧, 𝜁) = 0 for (𝑧, 𝜁) ∈ (0, ∞) × {t eiπ/α ; 0 ≤ 𝑡 < ∞}, then for 𝜁0 ∈ (0, ∞), 𝛼 ∫ 𝛾(𝜁)𝐻(𝑧, 𝜁)𝑑𝜁 𝑧→𝜁0 2𝜋𝑖 𝜕Ω lim
7
∞
𝛼 1 1 = lim ∫ 𝛾(𝜁) [ 𝛼 − ] 𝜁 𝛼−1 𝑑𝜁 𝛼 𝑧→𝜁0 2𝜋𝑖 𝜁 − 𝑧 𝛼 𝜁 𝛼 − 𝑧̅̅̅ 0
∞
𝛼 𝛼 𝑧 𝛼 − 𝑧̅̅̅ = lim ∫ 𝛾(𝜁) 𝛼 𝜁 𝛼−1 𝑑𝜁 𝑧→𝜁0 2𝜋𝑖 |𝜁 − 𝑧 𝛼 |2 0
∞
𝛼 ̅̅̅ 1 𝑧𝛼 − 𝑧 1 = lim ∫ 𝛾(𝜁 𝛼 ) 𝑑𝜁 = 𝛾(𝜁0 ) 𝑧→𝜁0 2𝜋𝑖 |𝜁 − 𝑧 𝛼 |2 0
Similarly, 𝐻(𝑧, 𝜁) = 0 for (𝑧, 𝜁) ∈ {t eiπ/α ; 0 < 𝑡 < ∞} × [0, ∞), then for 𝜁0 ∈ {t eiπ/α ; 0 < 𝑡 < ∞}, 𝛼 ∫ 𝛾(𝜁)𝐻(𝑧, 𝜁)𝑑𝜁 𝑧→𝜁0 2𝜋𝑖 𝜕Ω lim
∞
𝛼 1 1 = − lim ∫ 𝛾(𝑡 𝑒 𝑖𝜋/𝛼 ) [ 𝛼 − ] 𝑡 𝛼−1 𝑑𝑡 𝛼 𝛼 𝛼 𝑧→𝜁0 2𝜋𝑖 𝑡 +𝑧 𝑡 + ̅̅̅ 𝑧 0
0
𝛼 1 𝑧 𝛼 − 𝑧̅̅̅ 1/𝛼 = lim ∫ 𝛾(𝑡 ) 𝛼 𝑑𝑡 = 𝛾(𝜁0 ). 𝑧→𝜁0 2𝜋𝑖 |𝑡 − 𝑧 𝛼 |2 −∞
We introduce the Schwarz-type operator as follows ∞
∞
𝛼 𝑡 𝛼−1 𝛼 𝑡 𝛼−1 1/𝛼 𝑆[𝛾](𝑧) = ∫ γ(𝑡) 𝛼 𝑑𝑡 − ∫ γ((−1) 𝑡) 𝛼 𝑑𝑡 (3.3) 𝜋𝑖 𝑡 − 𝑧𝛼 𝜋𝑖 𝑡 + 𝑧𝛼 0
0
where 𝛾 ∈ 𝐿2 (𝜕Ω, ℝ) ∩ 𝐶(𝜕Ω, ℝ). Obviously, 𝑆[𝛾](𝑧) is analytic in the sector Ω. Further Re 𝑆[𝛾](𝑧) =
1 ∫ 𝛾(𝜁)𝐻(𝑧, 𝜁)𝑑𝜁 , 𝑧 ∈ Ω, 2𝜋𝑖 𝜕Ω
(3.4)
for 𝛾 ∈ 𝐿2 (𝜕Ω, ℝ) ∩ 𝐶(𝜕Ω, ℝ), with 𝐻(𝑧, 𝜁) is defined by (3.2). By Lemmas 3.1 and 3.2 the following result is true.
Theorem 3.1. If 𝛾 ∈ 𝐿2 (𝜕Ω, ℝ) ∩ 𝐶(𝜕Ω, ℝ), then 8
lim Re 𝑆[𝛾](𝑧) = γ(t), t ∈ ∂Ω,
z∈Ω,z→t
where 𝑆 is the Schwarz-type operator defined by (3.3). secondly, a Pompeiu-type operator for the sector Ω is introduced by 𝛼−1
𝛼 𝜁 𝛼−1 𝜁 ̅̅̅̅̅̅ ̂ 𝑇[𝑓](𝑧) = − ∫ {𝑓 (𝜁) 𝛼 − 𝑓(𝜁) } 𝑑𝜉𝑑𝜂 , 𝑧 ∈ Ω (3.5) 𝛼 𝜋 Ω 𝜁 − 𝑧𝛼 𝜁 − 𝑧𝛼 where 𝑓 ∈ 𝐿𝑝 (Ω, ℂ) , 𝑝 > 2. By simple computation, one obtains 𝛼 Re 𝑇̂[𝑓](𝑧) = − ∫ {𝑓(𝜁)𝐻(𝑧, 𝜁) − ̅̅̅̅̅̅ 𝑓(𝜁) 𝐻(𝑧, 𝜁)}𝑑𝜉𝑑𝜂 , 𝑧 2𝜋 Ω ∈ Ω (3.6) with 𝐻(𝑧, 𝜁) is defined by (3.2). the following theorem concludes basic properties of the Pompeiutype operator 𝑇̂ defined by (3.5).
Theorem 3.2. If 𝑓 ∈ 𝐿𝑝 (Ω, ℂ) , 𝑝 > 2, then 𝜕𝑧 𝑇̂[𝑓](𝑧) = 𝑓(𝑧), 𝑧 ∈ Ω in weak sense, and lim𝑧→𝑡 Re𝑇̂[𝑓](𝑧) = 0, 𝑡 ∈ 𝜕Ω.
According to the representation formula in Theorem 2.1, and based on the previous lemmas, we can combine both theorems 3.1 and 3.2 to solve Schwarz problem (3.1), therefore it is enough to prove the following theorem.
Theorem 3.3. The Schwarz problem (3.1) for the sector Ω is uniquely solvable by
∞
𝛼 1 𝑡𝛼 𝑤(𝑧) = 𝑖c + ∫ γ(𝑡) ( 𝛼 − 2𝛼 ) 𝑡 𝛼−1 𝑑𝑡 𝛼 𝜋𝑖 𝑡 −𝑧 𝑡 + 1 0
∞
𝛼 1 𝑡𝛼 1/𝛼 − ∫ γ((−1) 𝑡) ( 𝛼 − ) 𝑡 𝛼−1 𝑑𝑡 𝜋𝑖 𝑡 + 𝑧 𝛼 𝑡 2𝛼 + 1 0
𝛼 1 𝜁𝛼 − ∫ {𝜁 𝛼−1 𝑓(𝜁) ( 𝛼 − ) 𝜋 Ω 𝜁 − 𝑧 𝛼 𝜁 2𝛼 + 1 –𝜁
𝛼−1
1
𝜁
𝛼
̅̅̅̅̅̅ 𝑓(𝜁) ( 𝛼 − 2𝛼 )} 𝑑𝜉𝑑𝜂 𝜁 − 𝑧𝛼 𝜁 + 1
(3.8)
9
Proof: From Theorem 2.1, we only need to verify that (3.8) provides a solution. Since for 𝑧, 𝜁 ∈ Ω 𝛼𝜁 𝛼−1 lim(𝜁 − 𝑧) 𝛼 =1 𝑧→𝜁 𝜁 − 𝑧𝛼 then we know that z is a simple pole of
𝛼𝜁 𝛼−1
𝜁 𝛼 −𝑧 𝛼
in Ω, hence
𝛼𝜁 𝛼−1 1 (3.9) = + 𝑔(𝑧, 𝜁), 𝜁𝛼 − 𝑧𝛼 𝜁 − 𝑧 where for arbitrary 𝑧 ∈ Ω, 𝑔(𝑧, 𝜁) is analytic with respect to 𝑧. Thus, we obviously see that 1 𝑓(𝜁) 𝜕𝑧̅ 𝑤(𝑧) = 𝜕𝑧̅ {− ∫ 𝑑𝜉𝑑𝜂} = 𝑓(𝑧). 𝜋 Ω𝜁 −𝑧 Now, if we write (3.8) as 𝑤(𝑧) = 𝑖𝑐 + 𝑤1 (𝑧) + 𝑤2 (𝑧) with ∞
𝛼 1 𝑡𝛼 𝑤1 (𝑧) = ∫ γ(𝑡) ( 𝛼 − 2𝛼 ) 𝑡 𝛼−1 𝑑𝑡 𝛼 𝜋𝑖 𝑡 −𝑧 𝑡 + 1 0 ∞
𝛼 1 𝑡𝛼 1/𝛼 − ∫ γ((−1) 𝑡) ( 𝛼 − 2𝛼 ) 𝑡 𝛼−1 𝑑𝑡, 𝛼 𝜋𝑖 𝑡 +𝑧 𝑡 + 1 0
𝛼 1 𝜁𝛼 𝛼−1 𝑤2 (𝑧) = − ∫ {𝜁 𝑓(𝜁) ( 𝛼 − ) 𝜋 Ω 𝜁 − 𝑧 𝛼 𝜁 2𝛼 + 1 –𝜁
𝛼−1
1
𝜁
𝛼
̅̅̅̅̅̅ ( 𝛼 𝑓(𝜁) − 2𝛼 )} 𝑑𝜉𝑑𝜂. 𝛼 𝜁 −𝑧 𝜁 + 1
Then, we obtain 1. For 𝑧 ∈ 𝜕Ω, Re 𝑤(𝑧) = 𝛾(𝑧). Indeed, for (𝑧, 𝜁) ∈ 𝜕Ω × Ω, 𝐻(𝑧, 𝜁) = 0. Therefore, 𝛼 1 1 Re𝑤2 (𝑧) = − ∫ {𝜁 𝛼−1 𝑓(𝜁) ( 𝛼 − ) 𝜋 Ω 𝜁 − 𝑧 𝛼 𝜁 𝛼 − 𝑧𝛼 +𝜁
𝛼−1
1 1 ̅̅̅̅̅̅ 𝑓(𝜁) ( 𝛼 − )} 𝑑𝜉𝑑𝜂 𝜁 − 𝑧 𝛼 𝜁 𝛼 − 𝑧𝛼
𝛼 = − ∫ {𝑓 (𝜁)𝐻(𝑧, 𝜁) + ̅̅̅̅̅̅̅ 𝑓(𝜁) 𝐻(𝑧, 𝜁) }𝑑𝜉𝑑𝜂 = 0. 𝜋 Ω On the other hand, 𝛼 Re𝑤1 (𝑧) = ∫ 𝛾(𝜁)𝐻(𝑧, 𝜁)𝑑𝜁 . 2𝜋𝑖 𝜕Ω
10
In according to Lemma 3.2, we have lim𝑧→𝜁0 Re𝑤(𝑧) = 𝛾(𝜁0 ) for 𝜁0 ∈ 𝜕Ω\{0}. With respect to the vertex point 0, we have 𝛼 Re𝑤1 (𝑧) = ∫ [𝛾(𝜁) − 𝛾(0)]𝐻(𝑧, 𝜁)𝑑𝜁 2𝜋𝑖 𝜕Ω+ 𝛼𝛾(0) + ∫ 𝐻(𝑧, 𝜁)𝑑𝜁 2𝜋𝑖 𝜕Ω+ Applying the representation (2.3) to 𝑤(𝑧) ≡ 1 and then taking the real part on both sides, we obtain 𝛼 𝛼 1 1 ∫ 𝐻(𝑧, 𝜁)𝑑𝜁 = ∫ [ 𝛼 − ] 𝑑𝜁 = 1 𝛼 2𝜋𝑖 𝜕Ω 2𝜋𝑖 𝜕Ω 𝜁 − 𝑧 𝛼 𝜁 𝛼 − 𝑧̅̅̅ then, from Lemma 3.1, lim𝑧→0 Re𝑤1 (𝑧) = 𝛾(0). In a word, we have lim𝑧→𝜁0 Re𝑤(𝑧) = 𝛾(𝜁0 )) for 𝜁0 ∈ 𝜕Ω. 2. Because, 𝑤1 (𝑖
1/𝛼
𝛼 ∞ 1 𝑡𝛼 − 2𝛼 ) = ∫ 𝛾(𝑡) [ 𝛼 ] 𝑡 𝛼−1 𝑑𝑡 𝜋𝑖 0 𝑡 −𝑖 𝑡 +1 ∞
𝛼 1 𝑡𝛼 1/𝛼 − ∫ γ((−1) 𝑡) ( 𝛼 − 2𝛼 ) 𝑡 𝛼−1 𝑑𝑡 𝜋𝑖 𝑡 +𝑖 𝑡 + 1 0
𝛼 ∞ 𝑡 𝛼−1 1/𝛼 = ∫ {𝛾(𝑡) + γ((−1) 𝑡)} 2𝛼 𝑑𝑡, 𝜋 0 𝑡 +1 we have Im𝑤1 (𝑖 1/𝛼 ) = 0. Similarly, 𝑤2 (𝑖
1/𝛼
𝛼 1 𝜁𝛼 𝛼−1 − ) = − ∫ {𝜁 𝑓(𝜁) ( 𝛼 ) 𝜋 Ω 𝜁 − 𝑖 𝜁 2𝛼 + 1 𝛼−1
1
𝜁
𝛼
̅̅̅̅̅̅ 𝑓(𝜁) ( 𝛼 − 2𝛼 )} 𝑑𝜉𝑑𝜂 𝜁 −𝑖 𝜁 + 1 2𝛼 𝜁 𝛼−1 =− ∫ Im {𝑓(𝜁) 2𝛼 } 𝑑𝜉𝑑𝜂 , 𝜋 Ω 𝜁 + 1 –𝜁
so, Im𝑤2 (𝑖 1/𝛼 ) = 0. Therefore, we have shown that Im 𝑤(𝑖 1/𝛼 ) = 𝑐. Thus, the proof of Theorem 3.3 is completed. 11
4 Dirichlet boundary value problem in the sector In this section, we consider the Dirichlet boundary value problem an infinite sectoral domain with angle 𝜋⁄𝛼 (𝛼 ≥ 1/2 ). Dirichlet problem: Find a function 𝑤 satisfying the following conditions 𝑤𝑧̅ = 𝑓 in Ω , 𝑓 ∈ 𝐿𝑝 (Ω, ℂ) , 𝑝 > 2, (4.1) Re𝑤 = 𝛾 on 𝜕Ω, 𝛾 ∈ 𝐿2 (𝜕Ω, ℂ) ∩ 𝐶(𝜕Ω, ℂ). Firstly, we consider the Dirichlet problem for the homogeneous Cauchy–Riemann equation 𝑤𝑧̅ = 0 in Ω , Re𝑤 = 𝛾 on 𝜕Ω,
𝑓 ∈ 𝐿𝑝 (Ω, ℂ) , 𝑝 > 2, 𝛾 ∈ 𝐿2 (𝜕Ω, ℂ) ∩ 𝐶(𝜕Ω, ℂ).
(4.2)
Theorem 4.1. Dirichlet boundary value problem(4.2) is solvable if and only if for 𝑧 ∈ Ω, ∞
∞
𝛼 𝑡 𝛼−1 𝑑𝑡 𝛼 𝑡 𝛼−1 𝑑𝑡 1/𝛼 ∫ 𝛾(𝑡) 𝛼 − ∫ 𝛾((−) 𝑡) =0 2𝜋𝑖 𝑡 − 𝑧̅ 𝛼 2𝜋𝑖 𝑡 𝛼 + 𝑧̅ 𝛼 0
(4.3)
0
Then the solution can uniquely be given by +∞
+∞
𝛼 𝑡 𝛼−1 𝑑𝑡 𝛼 𝑡 𝛼−1 𝑑𝑡 1/𝛼 𝑤(𝑧) = ∫ 𝛾(𝑡) 𝛼 − ∫ 𝛾((−1) 𝑡) 𝛼 𝛼 2𝜋𝑖 𝑡 − 𝑧 𝛼 2𝜋𝑖 𝑡 +𝑧 for 𝑧 ∈ Ω.
0
(4.4)
0
Proof: Because of (3.9), it is obvious that (4.4) provides a weak solution to the equation 𝑤𝑧̅ = 0. Therefore, to complete the proof of this theorem it suffices to discuss the following two points. 1. The sufficiency of (4.3). If the condition (4.3) is satisfied, then we can rewrite 𝑤(𝑧) as 𝛼 𝑤(𝑧) = ∫ 𝛾(𝜁)𝐻(𝑧, 𝜁) 𝑑𝜁 , 2𝜋𝑖 𝜕Ω where 𝐻 is defined in (3.2). In according to Lemmas 3.1 and 3.2, one gets 𝛼 lim 𝑤(𝑧) = lim ∫ 𝛾(𝜁)𝐻(𝑧, 𝜁) 𝑑𝜁 , 𝑧→𝜁0 𝑧→𝜁0 2𝜋𝑖 𝜕Ω for any 𝜁0 ∈ 𝜕Ω. 12
Thus, the condition (4.3) suffices to solve the Dirichlet problem 𝜕𝑧̅ 𝑤(𝑧) = 0 in Ω and 𝑤 = 𝛾 on 𝜕Ω. 2. The necessity of (4.3). Define a new function ∞
∞
𝛼 𝑡 𝛼−1 𝑑𝑡 𝛼 𝑡 𝛼−1 𝑑𝑡 1/𝛼 𝑤 ̃(𝑧) = ∫ 𝛾(𝑡) 𝛼 − ∫ 𝛾((−) 𝑡) 𝛼 , 2𝜋𝑖 𝑡 − 𝑧̅ 𝛼 2𝜋𝑖 𝑡 + 𝑧̅ 𝛼 0
𝑧 ∈ Ω.
0
If 𝑤 be the solution to the Dirichlet problem (4.2), then lim 𝑤(𝑧) = 𝛾(𝜁),
for 𝜁 ∈ ∂Ω .
𝑧→𝜁
Thus, for the function
𝛼 ∫ 𝛾(𝜁)𝐻(𝑧, 𝜁) 𝑑𝜁 , 2𝜋𝑖 𝜕Ω we easily obtain, as above, that for any 𝜁0 ∈ 𝜕Ω, lim [𝑤(𝑧) − 𝑤 ̃(𝑧)] = 𝛾(𝜁0 ), 𝑤(𝑧) − 𝑤 ̃(𝑧) =
𝑧→𝜁0
which implies
lim 𝑤 ̃(𝑧) = 0, 𝜁 ∈ 𝜕Ω. 𝑧→𝜁
(4.5) (4.6)
On other hand, we observe that the function ∞
∞
1 𝛼 𝑡 𝛼−1 𝑑𝑡 𝛼 𝑡 𝛼−1 𝑑𝑡 ̅̅̅̅̅̅̅ 𝑤 ̃(𝑧) = − ∫ 𝛾(𝑡) 𝛼 + ∫ 𝛾((−)𝛼 𝑡) 𝛼 , 2𝜋𝑖 𝑡 − 𝑧 𝛼 2𝜋𝑖 𝑡 + 𝑧𝛼 0
𝑧 ∈ Ω.
0
is analytic on Ω, having zero boundary value. Then by the maximum principle for analytic functions, we have 𝑤 ̃(𝑧) = 0 on Ω, which is just the condition (4.3). This completes the proof.
Theorem 4.2. Dirichlet boundary value problem(4.1) is solvable if and only if for 𝑧 ∈ Ω,
∞
∞
𝛼 𝑡 𝛼−1 𝑑𝑡 𝛼 𝑡 𝛼−1 𝑑𝑡 1/𝛼 ∫ 𝛾(𝑡) 𝛼 − ∫ 𝛾((−) 𝑡) 𝛼 2𝜋𝑖 𝑡 − 𝑧̅ 𝛼 2𝜋𝑖 𝑡 + 𝑧̅ 𝛼 0
0
𝛼 𝜁 𝛼−1 − ∫ 𝑓(𝜁) 𝛼 𝑑𝜉𝜂 = 0 𝜋 Ω 𝜁 − 𝑧̅ 𝛼 Then the solution can uniquely be given by +∞
+∞
0
0
(4.7)
𝛼 𝑡 𝛼−1 𝑑𝑡 𝛼 𝑡 𝛼−1 𝑑𝑡 1/𝛼 𝑤(𝑧) = ∫ 𝛾(𝑡) 𝛼 − ∫ 𝛾((−1) 𝑡) 𝛼 𝛼 2𝜋𝑖 𝑡 − 𝑧 𝛼 2𝜋𝑖 𝑡 +𝑧 13
for 𝑧 ∈ Ω.
𝛼 𝜁 𝛼−1 𝑑𝜉𝑑𝜂 − ∫ 𝑓(𝜁) 𝛼 , 𝜋 Ω 𝜁 − 𝑧𝛼
(4.8)
Proof: Because of (3.9), it is obvious that (4.8) provides a weak solution to the equation 𝑤𝑧̅ = 𝑓. Thus, the proof of the theorem could be completed by discussing the following two points. 1. The sufficiency of (4.7). If the condition (4.7) is satisfied, then we can rewrite 𝑤(𝑧) as 𝛼 𝛼 𝑤(𝑧) = ∫ 𝛾(𝜁)𝐻(𝑧, 𝜁) 𝑑𝜁 − ∫ 𝑓(𝜁)𝐻(𝑧, 𝜁)𝑑𝜉𝑑𝜂 , 2𝜋𝑖 𝜕Ω 𝜋 Ω where 𝐻 is defined in (3.2). Since 𝐻(𝑧, 𝜁) = 0 for (𝑧, 𝜁) ∈ 𝜕Ω × Ω, then, for 𝑧 ∈ Ω and 𝜁0 ∈ 𝜕Ω 𝛼 lim 𝑤(𝑧) = lim ∫ 𝛾(𝜁)𝐻(𝑧, 𝜁) 𝑑𝜁 . 𝑧→𝜁0 𝑧→𝜁0 2𝜋𝑖 𝜕Ω In according to Lemmas 3.1 and 3.2, one gets 𝑤 = 𝛾 on 𝜕Ω. Thus, the condition (4.7) suffices to solve the Dirichlet problem (4.1). 2. The necessity of (4.7). Define a new function ∞
∞
𝛼 𝑡 𝛼−1 𝑑𝑡 𝛼 𝑡 𝛼−1 𝑑𝑡 1/𝛼 ̃ 𝑊 (𝑧) = ∫ 𝛾(𝑡) 𝛼 − ∫ 𝛾((−) 𝑡) 𝛼 2𝜋𝑖 𝑡 − 𝑧̅ 𝛼 2𝜋𝑖 𝑡 + 𝑧̅ 𝛼 0
0
𝛼 𝜁 𝛼−1 − ∫ 𝑓(𝜁) 𝛼 𝑑𝜉𝜂, 𝑧 ∈ Ω. 𝜋 Ω 𝜁 − 𝑧̅ 𝛼 If 𝑤 be the solution to the Dirichlet problem (4.1), then lim 𝑤(𝑧) = 𝛾(𝜁),
𝑧→𝜁
Thus, we have ̃ (𝑧) = 𝑤(𝑧) − 𝑊
for 𝜁 ∈ ∂Ω .
𝛼 𝛼 ∫ 𝛾(𝜁)𝐻(𝑧, 𝜁) 𝑑𝜁 − ∫ 𝑓(𝜁)𝐻(𝑧, 𝜁)𝑑𝜉𝑑𝜂 . (4.9) 2𝜋𝑖 𝜕Ω 𝜋 Ω
Since the area integral in (4.9) vanishes on 𝜕Ω, we easily obtain, as above, for 𝑧 ∈ Ω and 𝜁 ∈ 𝜕Ω, ̃ (𝑧)] = 𝛾(𝜁), lim[𝑤(𝑧) − 𝑊 (4.10) 𝑧→𝜁
which implies 14
̃ (𝑧) = 0, 𝜁 ∈ 𝜕Ω. lim 𝑊
(4.11)
𝑧→𝜁
On other hand, we observe that the function ∞
∞
𝛼 𝑡 𝛼−1 𝑑𝑡 𝛼 𝑡 𝛼−1 𝑑𝑡 ̅̅̅̅̅̅̅ 1/𝛼 ̃ 𝑊 (𝑧) = − ∫ 𝛾(𝑡) 𝛼 + ∫ 𝛾((−) 𝑡) 𝛼 2𝜋𝑖 𝑡 − 𝑧 𝛼 2𝜋𝑖 𝑡 + 𝑧𝛼 0
−
𝛼−1
𝛼 𝜁 ∫ 𝑓(𝜁) 𝛼 𝑑𝜉𝜂, 𝜋 Ω 𝜁 − 𝑧𝛼
0
𝑧∈Ω
is analytic for 𝑧 ∈ Ω, and then by the maximum principle for ̃ (𝑧) = 0 for 𝑧 ∈ Ω, which is just the analytic functions, we know 𝑊 condition (4.7). This completes the proof. Acknowledgment
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