Two Basic Boundary-Value Problems for inhomogeneous CauchyβRiemann 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 CauchyβRiemann equation in an infinite sector. Firstly, we obtain the SchwarzβPoisson formula in a sector with angle πβπΌ 1
(πΌ β₯ 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), 33β52, 2007) and by Ying Wang, Yufeng Wang (Complex Anal. Oper. Theory, DOI 10.1007/s11785-010-0107-0).
Keywords CauchyβPompeiu formula, inhomogeneous CauchyβRiemann equation , Schwarz βtype operator, Pompeiu βtype 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 [15β17], quarter plane [18β20], 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 CauchyβRiemann (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 Ξ© be a infinite sector domain in the complex plane C defined by π
Ξ© = {π§ β β, 0 < arg π§ < πΌ}
(1.1) 1
Except for additional indication, we always assume that πΌ(πΌ β₯ ) 2
is a fixed real constant. The sector Ξ© is isomorphic with the upper half plane β.
Indeed, the conformal mapping [25] π: Ξ© β β π§ βΌ π§πΌ
(1.2)
with the branch cut along (0, +β), maps the boundary [0, +β) onto itself and the boundary {π§ = π‘π ππ , 0 β€ π‘ < +β, π = π/πΌ} onto (ββ, 0]. The inverse mapping is
π: β β Ξ© π§ βΌ π§1/πΌ
(1.3)
which transforms β onto the boundary πΞ© of the sector Ξ©. Obviously, the above two conformal mappings can be firstly 1 β 1 extended to the boundary except for 0. Since the argument at 0 can be arbitrary and πΌ β₯ 1/2 , we can assume that the arguments of π§ πΌ and π§ 1/πΌ at π§ = 0 are 0. Hence, the above 1 β 1 mapping between the boundary is proper. In what follows, the main analytic branches of π§ πΌ and π§ 1/πΌ are always chosen as above, respectively. Moreover, on the basis of the analytic branches of π§ πΌ and π§ 1/πΌ , we define the analytic branch of several functions as follows, π§ πΌ 1βπΌ π§ π§1/πΌ πΌβ1 1/πΌβ1 π§ = ,π§ = πΌ, π§ = . π§ π§ π§
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 β[14]. Theorem 1.1 If π€ βΆ β β β satisfies π€ β πΏ2 (β; β) β© πΆ(β, β) and π€π§ β πΏπ,2 (β; β), π > 2. Then, for π§ β β, +β
1 ππ‘ 1 π€(π§) = β« π€(π‘) β 2ππ π‘βπ§ π ββ
π€(π§) = β
+β
β« π€π (π) 0 2, can be represented as β β 1 π‘ β1/2 ππ‘ 1 π‘ β1/2 ππ‘ β« π€(π‘) 1/2 β β« π€(π‘) 1/2 4ππ 0 π‘ β π§ 1/2 4ππ 0 π‘ + π§ 1/2
1 π β1/2 π€(π§), π§ β β0 β β« π€πΜ
(π) 1/2 ππππ = { 2π β0 0, π§ β βΜ
0 π β π§ 1/2
and for π§ β β0
β
1 1 π‘ 1/2 π€(π§) = πImπ€(β1) + β« Reπ€(π‘) ( 1/2 β ) π‘ β1/2 ππ‘ 1/2 2ππ π‘ + 1 π‘ βπ§ 0
β
1 1 π‘ 1/2 β β« Reπ€(π‘) ( 1/2 β ) π‘ β1/2 ππ‘ 1/2 2ππ π‘ + 1 π‘ +π§ 0
1 1 π1/2 β1/2 β β« {π π€πΜ
(π) ( 1/2 β ) 2π Ξ© π β π§ 1/2 π + 1 βπ
β1/2
Μ
Μ
Μ
Μ
Μ
Μ
Μ
Μ
π€πΜ
(π) (
1 π
1/2
β
π§ 1/2
β
π
1/2
π + 1
)} ππππ
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 πβπΌ (πΌ β₯ 1/2 ) by a proper conformal mapping to solve the Schwarz problem in the sector. Schwarz problem: Find a function π€ satisfying the following conditions
6
π€π§Μ
= π in Ξ© , Reπ€ = πΎ on πΞ© Imπ€(π 1/πΌ ) = π , π β β, with π β πΏπ (Ξ©, β) , π > 2, πΎ β πΏ2 (πΞ©, β) β© πΆ(πΞ©, β). Firstly, introducing a new kernel 1 1 π»(π§, π) = [ πΌ β ] π πΌβ1 . πΌ πΌ πΌ Μ
Μ
Μ
π βπ§ π βπ§ Then, the following lemma is valid.
(3.1)
(3.2)
Lemma 3.1. For πΎ β πΆ(πΞ©, β), lim
π§βΞ©, π§β0
πΌ β« [πΎ(π) β πΎ(0)]π»(π§, π)ππ = 0. 2ππ πΞ©
Proof: For (π§, π) β Ξ© Γ πΞ©, we have πΌ β« [πΎ(π) β πΎ(0)]π»(π§, π)ππ π§βΞ©, 2ππ πΞ© lim
π§β0
β
1 1 1 1 = lim β« [πΎ(π πΌ ) β πΎ(0)] [ β ] ππ πΌ π§βΞ©, 2ππ π β π§ πΌ π β π§Μ
Μ
Μ
π§β0
ββ
β
1 Im z Ξ± 1/πΌ = lim β« [πΎ(π ) β πΎ(0)] ππ π§βΞ©, π |π β π§ πΌ |2 π§β0
ββ
= lim [πΎ(π§) β πΎ(0)] = 0, π§βΞ©, π§β0
by Poisson formula on β.
Lemma 3.2. For πΎ β πΆ(πΞ©, β), if π0 β πΞ©\{0} lim
π§βΞ©, π§βπ0
πΌ β« πΎ(π)π»(π§, π)ππ = πΎ(π0 ). 2ππ πΞ©
Proof: Since π»(π§, π) = 0 for (π§, π) β (0, β) Γ {t eiΟ/Ξ± ; 0 β€ π‘ < β}, then for π0 β (0, β), πΌ β« πΎ(π)π»(π§, π)ππ π§βπ0 2ππ πΞ© lim
7
β
πΌ 1 1 = lim β« πΎ(π) [ πΌ β ] π πΌβ1 ππ πΌ π§βπ0 2ππ π β π§ πΌ π πΌ β π§Μ
Μ
Μ
0
β
πΌ πΌ π§ πΌ β π§Μ
Μ
Μ
= lim β« πΎ(π) πΌ π πΌβ1 ππ π§βπ0 2ππ |π β π§ πΌ |2 0
β
πΌ Μ
Μ
Μ
1 π§πΌ β π§ 1 = lim β« πΎ(π πΌ ) ππ = πΎ(π0 ) π§βπ0 2ππ |π β π§ πΌ |2 0
Similarly, π»(π§, π) = 0 for (π§, π) β {t eiΟ/Ξ± ; 0 < π‘ < β} Γ [0, β), then for π0 β {t eiΟ/Ξ± ; 0 < π‘ < β}, πΌ β« πΎ(π)π»(π§, π)ππ π§βπ0 2ππ πΞ© lim
β
πΌ 1 1 = β lim β« πΎ(π‘ π ππ/πΌ ) [ πΌ β ] π‘ πΌβ1 ππ‘ πΌ πΌ πΌ π§βπ0 2ππ π‘ +π§ π‘ + Μ
Μ
Μ
π§ 0
0
πΌ 1 π§ πΌ β π§Μ
Μ
Μ
1/πΌ = lim β« πΎ(π‘ ) πΌ ππ‘ = πΎ(π0 ). π§βπ0 2ππ |π‘ β π§ πΌ |2 ββ
We introduce the Schwarz-type operator as follows β
β
πΌ π‘ πΌβ1 πΌ π‘ πΌβ1 1/πΌ π[πΎ](π§) = β« Ξ³(π‘) πΌ ππ‘ β β« Ξ³((β1) π‘) πΌ ππ‘ (3.3) ππ π‘ β π§πΌ ππ π‘ + π§πΌ 0
0
where πΎ β πΏ2 (πΞ©, β) β© πΆ(πΞ©, β). Obviously, π[πΎ](π§) is analytic in the sector Ξ©. Further Re π[πΎ](π§) =
1 β« πΎ(π)π»(π§, π)ππ , π§ β Ξ©, 2ππ πΞ©
(3.4)
for πΎ β πΏ2 (πΞ©, β) β© πΆ(πΞ©, β), with π»(π§, π) is defined by (3.2). By Lemmas 3.1 and 3.2 the following result is true.
Theorem 3.1. If πΎ β πΏ2 (πΞ©, β) β© πΆ(πΞ©, β), then 8
lim Re π[πΎ](π§) = Ξ³(t), t β βΞ©,
zβΞ©,zβt
where π is the Schwarz-type operator defined by (3.3). secondly, a Pompeiu-type operator for the sector Ξ© is introduced by πΌβ1
πΌ π πΌβ1 π Μ
Μ
Μ
Μ
Μ
Μ
Μ π[π](π§) = β β« {π (π) πΌ β π(π) } ππππ , π§ β Ξ© (3.5) πΌ π Ξ© π β π§πΌ π β π§πΌ where π β πΏπ (Ξ©, β) , π > 2. By simple computation, one obtains πΌ Re πΜ[π](π§) = β β« {π(π)π»(π§, π) β Μ
Μ
Μ
Μ
Μ
Μ
π(π) π»(π§, π)}ππππ , π§ 2π Ξ© β Ξ© (3.6) with π»(π§, π) is defined by (3.2). the following theorem concludes basic properties of the Pompeiutype operator πΜ defined by (3.5).
Theorem 3.2. If π β πΏπ (Ξ©, β) , π > 2, then ππ§ πΜ[π](π§) = π(π§), π§ β Ξ© in weak sense, and limπ§βπ‘ ReπΜ[π](π§) = 0, π‘ β πΞ©.
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 Ξ© is uniquely solvable by
β
πΌ 1 π‘πΌ π€(π§) = πc + β« Ξ³(π‘) ( πΌ β 2πΌ ) π‘ πΌβ1 ππ‘ πΌ ππ π‘ βπ§ π‘ + 1 0
β
πΌ 1 π‘πΌ 1/πΌ β β« Ξ³((β1) π‘) ( πΌ β ) π‘ πΌβ1 ππ‘ ππ π‘ + π§ πΌ π‘ 2πΌ + 1 0
πΌ 1 ππΌ β β« {π πΌβ1 π(π) ( πΌ β ) π Ξ© π β π§ πΌ π 2πΌ + 1 βπ
πΌβ1
1
π
πΌ
Μ
Μ
Μ
Μ
Μ
Μ
π(π) ( πΌ β 2πΌ )} ππππ π β π§πΌ π + 1
(3.8)
9
Proof: From Theorem 2.1, we only need to verify that (3.8) provides a solution. Since for π§, π β Ξ© πΌπ πΌβ1 lim(π β π§) πΌ =1 π§βπ π β π§πΌ then we know that z is a simple pole of
πΌπ πΌβ1
π πΌ βπ§ πΌ
in Ξ©, hence
πΌπ πΌβ1 1 (3.9) = + π(π§, π), ππΌ β π§πΌ π β π§ where for arbitrary π§ β Ξ©, π(π§, π) is analytic with respect to π§. Thus, we obviously see that 1 π(π) ππ§Μ
π€(π§) = ππ§Μ
{β β« ππππ} = π(π§). π Ξ©π βπ§ Now, if we write (3.8) as π€(π§) = ππ + π€1 (π§) + π€2 (π§) with β
πΌ 1 π‘πΌ π€1 (π§) = β« Ξ³(π‘) ( πΌ β 2πΌ ) π‘ πΌβ1 ππ‘ πΌ ππ π‘ βπ§ π‘ + 1 0 β
πΌ 1 π‘πΌ 1/πΌ β β« Ξ³((β1) π‘) ( πΌ β 2πΌ ) π‘ πΌβ1 ππ‘, πΌ ππ π‘ +π§ π‘ + 1 0
πΌ 1 ππΌ πΌβ1 π€2 (π§) = β β« {π π(π) ( πΌ β ) π Ξ© π β π§ πΌ π 2πΌ + 1 βπ
πΌβ1
1
π
πΌ
Μ
Μ
Μ
Μ
Μ
Μ
( πΌ π(π) β 2πΌ )} ππππ. πΌ π βπ§ π + 1
Then, we obtain 1. For π§ β πΞ©, Re π€(π§) = πΎ(π§). Indeed, for (π§, π) β πΞ© Γ Ξ©, π»(π§, π) = 0. Therefore, πΌ 1 1 Reπ€2 (π§) = β β« {π πΌβ1 π(π) ( πΌ β ) π Ξ© π β π§ πΌ π πΌ β π§πΌ +π
πΌβ1
1 1 Μ
Μ
Μ
Μ
Μ
Μ
π(π) ( πΌ β )} ππππ π β π§ πΌ π πΌ β π§πΌ
πΌ = β β« {π (π)π»(π§, π) + Μ
Μ
Μ
Μ
Μ
Μ
Μ
π(π) π»(π§, π) }ππππ = 0. π Ξ© On the other hand, πΌ Reπ€1 (π§) = β« πΎ(π)π»(π§, π)ππ . 2ππ πΞ©
10
In according to Lemma 3.2, we have limπ§βπ0 Reπ€(π§) = πΎ(π0 ) for π0 β πΞ©\{0}. With respect to the vertex point 0, we have πΌ Reπ€1 (π§) = β« [πΎ(π) β πΎ(0)]π»(π§, π)ππ 2ππ πΞ©+ πΌπΎ(0) + β« π»(π§, π)ππ 2ππ πΞ©+ Applying the representation (2.3) to π€(π§) β‘ 1 and then taking the real part on both sides, we obtain πΌ πΌ 1 1 β« π»(π§, π)ππ = β« [ πΌ β ] ππ = 1 πΌ 2ππ πΞ© 2ππ πΞ© π β π§ πΌ π πΌ β π§Μ
Μ
Μ
then, from Lemma 3.1, limπ§β0 Reπ€1 (π§) = πΎ(0). In a word, we have limπ§βπ0 Reπ€(π§) = πΎ(π0 )) for π0 β πΞ©. 2. Because, π€1 (π
1/πΌ
πΌ β 1 π‘πΌ β 2πΌ ) = β« πΎ(π‘) [ πΌ ] π‘ πΌβ1 ππ‘ ππ 0 π‘ βπ π‘ +1 β
πΌ 1 π‘πΌ 1/πΌ β β« Ξ³((β1) π‘) ( πΌ β 2πΌ ) π‘ πΌβ1 ππ‘ ππ π‘ +π π‘ + 1 0
πΌ β π‘ πΌβ1 1/πΌ = β« {πΎ(π‘) + Ξ³((β1) π‘)} 2πΌ ππ‘, π 0 π‘ +1 we have Imπ€1 (π 1/πΌ ) = 0. Similarly, π€2 (π
1/πΌ
πΌ 1 ππΌ πΌβ1 β ) = β β« {π π(π) ( πΌ ) π Ξ© π β π π 2πΌ + 1 πΌβ1
1
π
πΌ
Μ
Μ
Μ
Μ
Μ
Μ
π(π) ( πΌ β 2πΌ )} ππππ π βπ π + 1 2πΌ π πΌβ1 =β β« Im {π(π) 2πΌ } ππππ , π Ξ© π + 1 βπ
so, Imπ€2 (π 1/πΌ ) = 0. Therefore, we have shown that Im π€(π 1/πΌ ) = π. 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 πβπΌ (πΌ β₯ 1/2 ). Dirichlet problem: Find a function π€ satisfying the following conditions π€π§Μ
= π in Ξ© , π β πΏπ (Ξ©, β) , π > 2, (4.1) Reπ€ = πΎ on πΞ©, πΎ β πΏ2 (πΞ©, β) β© πΆ(πΞ©, β). Firstly, we consider the Dirichlet problem for the homogeneous CauchyβRiemann equation π€π§Μ
= 0 in Ξ© , Reπ€ = πΎ on πΞ©,
π β πΏπ (Ξ©, β) , π > 2, πΎ β πΏ2 (πΞ©, β) β© πΆ(πΞ©, β).
(4.2)
Theorem 4.1. Dirichlet boundary value problem(4.2) is solvable if and only if for π§ β Ξ©, β
β
πΌ π‘ πΌβ1 ππ‘ πΌ π‘ πΌβ1 ππ‘ 1/πΌ β« πΎ(π‘) πΌ β β« πΎ((β) π‘) =0 2ππ π‘ β π§Μ
πΌ 2ππ π‘ πΌ + π§Μ
πΌ 0
(4.3)
0
Then the solution can uniquely be given by +β
+β
πΌ π‘ πΌβ1 ππ‘ πΌ π‘ πΌβ1 ππ‘ 1/πΌ π€(π§) = β« πΎ(π‘) πΌ β β« πΎ((β1) π‘) πΌ πΌ 2ππ π‘ β π§ πΌ 2ππ π‘ +π§ for π§ β Ξ©.
0
(4.4)
0
Proof: Because of (3.9), it is obvious that (4.4) provides a weak solution to the equation π€π§Μ
= 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 π€(π§) as πΌ π€(π§) = β« πΎ(π)π»(π§, π) ππ , 2ππ πΞ© where π» is defined in (3.2). In according to Lemmas 3.1 and 3.2, one gets πΌ lim π€(π§) = lim β« πΎ(π)π»(π§, π) ππ , π§βπ0 π§βπ0 2ππ πΞ© for any π0 β πΞ©. 12
Thus, the condition (4.3) suffices to solve the Dirichlet problem ππ§Μ
π€(π§) = 0 in Ξ© and π€ = πΎ on πΞ©. 2. The necessity of (4.3). Define a new function β
β
πΌ π‘ πΌβ1 ππ‘ πΌ π‘ πΌβ1 ππ‘ 1/πΌ π€ Μ(π§) = β« πΎ(π‘) πΌ β β« πΎ((β) π‘) πΌ , 2ππ π‘ β π§Μ
πΌ 2ππ π‘ + π§Μ
πΌ 0
π§ β Ξ©.
0
If π€ be the solution to the Dirichlet problem (4.2), then lim π€(π§) = πΎ(π),
for π β βΞ© .
π§βπ
Thus, for the function
πΌ β« πΎ(π)π»(π§, π) ππ , 2ππ πΞ© we easily obtain, as above, that for any π0 β πΞ©, lim [π€(π§) β π€ Μ(π§)] = πΎ(π0 ), π€(π§) β π€ Μ(π§) =
π§βπ0
which implies
lim π€ Μ(π§) = 0, π β πΞ©. π§βπ
(4.5) (4.6)
On other hand, we observe that the function β
β
1 πΌ π‘ πΌβ1 ππ‘ πΌ π‘ πΌβ1 ππ‘ Μ
Μ
Μ
Μ
Μ
Μ
Μ
π€ Μ(π§) = β β« πΎ(π‘) πΌ + β« πΎ((β)πΌ π‘) πΌ , 2ππ π‘ β π§ πΌ 2ππ π‘ + π§πΌ 0
π§ β Ξ©.
0
is analytic on Ξ©, having zero boundary value. Then by the maximum principle for analytic functions, we have π€ Μ(π§) = 0 on Ξ©, 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 π§ β Ξ©,
β
β
πΌ π‘ πΌβ1 ππ‘ πΌ π‘ πΌβ1 ππ‘ 1/πΌ β« πΎ(π‘) πΌ β β« πΎ((β) π‘) πΌ 2ππ π‘ β π§Μ
πΌ 2ππ π‘ + π§Μ
πΌ 0
0
πΌ π πΌβ1 β β« π(π) πΌ πππ = 0 π Ξ© π β π§Μ
πΌ Then the solution can uniquely be given by +β
+β
0
0
(4.7)
πΌ π‘ πΌβ1 ππ‘ πΌ π‘ πΌβ1 ππ‘ 1/πΌ π€(π§) = β« πΎ(π‘) πΌ β β« πΎ((β1) π‘) πΌ πΌ 2ππ π‘ β π§ πΌ 2ππ π‘ +π§ 13
for π§ β Ξ©.
πΌ π πΌβ1 ππππ β β« π(π) πΌ , π Ξ© π β π§πΌ
(4.8)
Proof: Because of (3.9), it is obvious that (4.8) provides a weak solution to the equation π€π§Μ
= π. 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 π€(π§) as πΌ πΌ π€(π§) = β« πΎ(π)π»(π§, π) ππ β β« π(π)π»(π§, π)ππππ , 2ππ πΞ© π Ξ© where π» is defined in (3.2). Since π»(π§, π) = 0 for (π§, π) β πΞ© Γ Ξ©, then, for π§ β Ξ© and π0 β πΞ© πΌ lim π€(π§) = lim β« πΎ(π)π»(π§, π) ππ . π§βπ0 π§βπ0 2ππ πΞ© In according to Lemmas 3.1 and 3.2, one gets π€ = πΎ on πΞ©. Thus, the condition (4.7) suffices to solve the Dirichlet problem (4.1). 2. The necessity of (4.7). Define a new function β
β
πΌ π‘ πΌβ1 ππ‘ πΌ π‘ πΌβ1 ππ‘ 1/πΌ Μ π (π§) = β« πΎ(π‘) πΌ β β« πΎ((β) π‘) πΌ 2ππ π‘ β π§Μ
πΌ 2ππ π‘ + π§Μ
πΌ 0
0
πΌ π πΌβ1 β β« π(π) πΌ πππ, π§ β Ξ©. π Ξ© π β π§Μ
πΌ If π€ be the solution to the Dirichlet problem (4.1), then lim π€(π§) = πΎ(π),
π§βπ
Thus, we have Μ (π§) = π€(π§) β π
for π β βΞ© .
πΌ πΌ β« πΎ(π)π»(π§, π) ππ β β« π(π)π»(π§, π)ππππ . (4.9) 2ππ πΞ© π Ξ©
Since the area integral in (4.9) vanishes on πΞ©, we easily obtain, as above, for π§ β Ξ© and π β πΞ©, Μ (π§)] = πΎ(π), lim[π€(π§) β π (4.10) π§βπ
which implies 14
Μ (π§) = 0, π β πΞ©. lim π
(4.11)
π§βπ
On other hand, we observe that the function β
β
πΌ π‘ πΌβ1 ππ‘ πΌ π‘ πΌβ1 ππ‘ Μ
Μ
Μ
Μ
Μ
Μ
Μ
1/πΌ Μ π (π§) = β β« πΎ(π‘) πΌ + β« πΎ((β) π‘) πΌ 2ππ π‘ β π§ πΌ 2ππ π‘ + π§πΌ 0
β
πΌβ1
πΌ π β« π(π) πΌ πππ, π Ξ© π β π§πΌ
0
π§βΞ©
is analytic for π§ β Ξ©, and then by the maximum principle for Μ (π§) = 0 for π§ β Ξ©, which is just the analytic functions, we know π condition (4.7). This completes the proof. Acknowledgment
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