Boundary Value Problems for Complex Partial Differential Equations in Fan-shaped Domains

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1 Boundary Value Problems for Complex Partial Differential Equations in Fan-shaped Domains DISSERTATION des Fachbereichs Mathematik und Informatik der Freien Universität Berlin zur Erlangung des Grades eines Doktors der Naturwissenschaften Erster Gutachter : Prof. Dr. Heinrich Begehr Zweiter Gutachter : Prof. Dr. Jinyuan Du Dritter Gutachter : Prof. Dr. Alexander Schmitt Vorgelegt von Ying Wang October 200 Tag der Disputation:. February 20

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3 Contents Contents Abstract Acknowledgements i iii v Chapter Introduction Chapter 2 Boundary Value Problems for the Inhomogeneous Cauchy- Riemann Equation 5 2. Preliminaries Schwarz Problem with Angle /n n N Schwarz-Poisson Representation Schwarz Problem Dirichlet Problem with Angle /n n N Schwarz Problem with Angle /α α / Schwarz-Poisson Representation Schwarz Problem Dirichlet Problem with Angle /α α / Chapter 3 Harmonic Boundary Value Problems for the Poisson Equation Harmonic Dirichlet Problem Harmonic Neumann Problem Chapter 4 Boundary Value Problems for the Bi-Poisson Equation Biharmonic Green Function i

4 4.2 Biharmonic Neumann Function Dirichlet and Neumann Problems for the Bi-Poisson Equation.. 65 Chapter 5 Triharmonic Boundary Value Problems for the Tri-Poisson Equation Triharmonic Green Function Triharmonic Neumann Function Triharmonic Boundary Value Problems Chapter 6 Tetra-harmonic Boundary Value Problems Tetra-harmonic Dirichlet Problem Tetra-harmonic Neumann Function Chapter 7 Polyharmonic Dirichlet and Polyharmonic Neumann Problems 7. Polyharmonic Dirichlet Problem Polyharmonic Neumann Problem Appendix A: The Tetra-harmonic Green Function for the Unit Disc9 Appendix B: The Tri-harmonic Neumann Function for the Unit Disc 2 Bibliography 23 Zusammenfassung 29 Curriculum Vitae 3 ii

5 Abstract In this dissertation, we investigate some boundary value problems for complex partial differential equations in fan-shaped domains. First of all, we establish the Schwarz-Poisson representation in fan-shaped domains with angle /n n N by the reflection method, and study the corresponding Schwarz and Dirichlet problems respectively. Further, the Schwarz-Poisson formula is extended to the general fan-shaped domains with angle /α α /2 by proper conformal mappings, and then the Schwarz and Dirichlet problems for the Cauchy-Riemann equation are solved. Next, we also establish a bridge between the unit disc and the fan-shaped domain with α = /2, and the Schwarz-Poisson formula for the unit disc is derived from the Schwarz-Poisson formula for α = /2. Then, we firstly obtain a harmonic Green function and a harmonic Neumann function in the fan-shaped domain with angle /α α /2, and then investigate the Dirichlet and Neumann problems for the Poisson equation. In particular, the outward normal derivative at the three corner points is properly defined. Next, a biharmonic Green function, a biharmonic Neumann function, a triharmonic Green function, a triharmonic Neumann function and a tetra-harmonic Green function are constructed for the fan-shaped domain with angle /n n N in explicit form respectively. Moreover, we give the process of constructing a tetraharmonic Neumann function and the expression of the tetra-harmonic Neumann function with integral representation. Accordingly, the Dirichlet and Neumann problems are discussed. Finally, we establish the iterated expressions and the solvability conditions of polyharmonic Dirichlet and Neumann problems for the higher order Poisson equation in the fan-shaped domain with angle /n n N respectively. In the meantime, the boundary behavior of polyharmonic Green and polyharmonic Neumann functions by convolution are discussed in detail. Besides, in the Appendix, the tetra-harmonic Green function and the triharmonic Neumann function for iii

6 the unit disc are constructed in explicit form. Keywords: Schwarz-Poisson representation, polyharmonic Green function, polyharmonic Neumann function, Schwarz problem, Dirichlet problem, Neumann problem. iv

7 Acknowledgements I consider myself very fortunate that many people offer me help and give me much motivation, so that I can complete my thesis. First and foremost, I would like to express my deepest gratitude to Prof. Dr. Heinrich Begehr for giving me the chance to study at this institute. He is very kind and warm-hearted. He spent a lot of care in instructing and revising my papers, and also made great contribution to my extension and to the financial support for my Ph.D project. He always gave me much help with great patience whenever I needed help. For all this and much else besides, I offer him my deep gratitude. Next I am very grateful to Prof. Dr. Jinyuan Du for his encouragement and assistance in pursuing my study. He is strict and very kind to me. I appreciate him very much for all his instruction and help, which always encourages me to make progress in mathematics. I am also thankful to Prof. Dr. Alexander Schmitt for his kindness and spending time in assessing the thesis. Then, I owe deep thanks to Dr. Yufeng Wang for his many suggestions and aids both in my study and life in Berlin. I am very thankful to Dr. Zhongxiang Zhang for his much help, Dr. Zhihua Du for his kindness and the China Scholarship Council for the financial support. I also appreciate the help from Ms Caroline Neumann, the Center for International Cooperation in FU Berlin and the support from the STIBET-Program DAAD. Finally, I want to express my special thanks to my parents, sisters and friends for their selfless love and encouragement. They always gave me the biggest understanding and tolerance when I was stressful, which helps me to overcome many difficulties. v

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9 Chapter Introduction Complex analysis is a comparatively active branch in mathematics which has grown significantly. In particular, the investigation of boundary value problems possesses both theoretical and applicable values of importance to many fields, such as electricity and magnetism, hydrodynamics, elasticity theory, shell theory, quantum mechanics, medical imaging, etc. In recent years, many investigators have made great contribution to boundary value problems for complex partial differential equations. Numerous results are achieved, which rapidly enrich the development of generalized analytic functions, boundary value problems, Riemann-Hilbert analysis, mathematical physics and so on, reference to 6, 3, 32, 37, 45, 50, 5, 55, 56. The classical boundary value problems initiated by B.Riemann and D.Hilbert are the Riemann and the Riemann-Hilbert problems 46, 38. The theory of boundary value problems for analytic functions is extended to many branches. Analytic functions are in close connection with the Cauchy-Riemann operator z. Then one aspect is to investigate boundary value problems for different kinds of functions and the functions satisfying particular complex differential equations, e.g. generalized analytic functions, functions with several variables, functions in Hardy space, functions satisfying the Cauchy-Riemann equation, the Beltrami equation, the generalized Poisson equation, even the higher order complex differential equations, reference to 3, 4, 5, 6, 48, 50, 56. In particular, great interest has arisen for polyanalytic and polyharmonic equations, see 5, 22, 24, 39, 40. On the other hand, various types of conditions imposed on the boundary lead to different boundary value problems, such as the Riemann, the Hilbert, the Dirichlet, the Schwarz, the Neumann, the Hasemann, the Robin boundary value problems 8, 22, 33, 35, 39, 43. Moreover, besides the study in the classical

10 unit disc, much attention has been paid to boundary value problems in some particular domains, for example, a half unit disc, a triangle, a fan-shaped domain, the upper half plane, a quarter plane, a circular ring and a half circular ring 5, 9, 26, 36, 49, 54, 60. Also, some investigators have extended boundary value problems to higher dimensional spaces, such as a polydisc, a sphere and other torus related domains, reference to 2, 42, 44. Generally speaking, the fundamental tools for solving boundary value problems are the Gauss theorem and the Cauchy-Pompeiu formula. Besides, the higher order Cauchy-Pompeiu operators T m,n, due to H. Begehr and G. Hile 20, establish a bridge for boundary value problems between the homogeneous and the inhomogeneous complex partial differential equations. As is well known, Green, Neumann and Robin functions are three useful fundamental solutions for certain boundary value problems via integral representation formulas. Especially, in order to solve some polyharmonic Dirichlet and Neumann problems, certain polyharmonic Green and polyharmonic Neumann functions need to be studied. In fact, there are several different kinds of polyharmonic Green functions. Convoluting the harmonic Green function with itself consecutively leads to an iterated polyharmonic Green function, which can be used to solve an iterated Dirichlet problem for the higher order Poisson equation. In addition, different from the above polyharmonic Green functions, polyharmonic Green-Almansi functions are firstly introduced for the unit disc by Almansi, which also give rise to some particular polyharmonic Dirichlet problems. Similarly, convoluting the harmonic Neumann function with itself consecutively results in an iterated polyharmonic Neumann function. Besides, iteration of the harmonic Green, Neumann and Robin functions pairwise leads to different hybrid biharmonic Green functions due to H. Begehr 0,. Furthermore, convoluting the iterated polyharmonic Green functions with the polyharmonic Green-Almansi functions also gives a variety of hybrid polyharmonic Green functions 9, 2, 28. However, it should be noted that the expressions of the polyharmonic Green and Neumann functions by convolution are not easily constructed in explicit form even in the classical unit disc, although the iterated polyharmonic Dirichlet and 2

11 Neumann problems can be solved by iterated forms. Many results have been obtained for boundary value problems of complex partial differential equations in some particular domains. In the unit disc, basic boundary value problems Schwarz, Dirichlet, Neumann, Robin problems for the Cauchy-Riemann equation are studied and several hybrid biharmonic functions are given explicitly by convoluting the harmonic ones 7, 2, 25. Moreover, polyharmonic Poisson kernels for the higher order Poisson equation are constructed in a complicated form using vertical sums 6, 34, but the iterated polyharmonic m-harmonicgreen and Neumann functions by convolution are established only up to m = For the upper half plane 7, 36, polyharmonic Green-Almansi functions are given explicitly to solve the related polyharmonic Dirichlet problem. And in the circular ring 52, 53, 54, the existing results mainly include solving four fundamental boundary value problems for the Cauchy-Riemann equation, and then establishing the harmonic Green, Neumann, Robin functions as well as the biharmonic Green function in detail. Besides, the investigation for the quarter plane is just at the beginning with basic boundary value problems for the Cauchy-Riemann equation solved 2, 9. As to in the higher dimensional spaces, the related results can be viewed in 3, 4, 2, 4, 42, 44. Especially, the Schwarz problem for the Cauchy-Riemann equation, the harmonic Green and Neumann functions are studied in half disc and half ring 26. Also some results are achieved in fan-shaped domains 57, 58, 59. In this thesis, we systematically investigate some boundary value problems in fan-shaped domains. First of all, the Schwarz-Poisson representation formula is obtained in fan-shaped domains with angle /n n N by the reflection method, as well as the Schwarz and Dirichlet problems are discussed for the Cauchy-Riemann equation. For the general fan-shaped domains with angle /α α /2, the Schwarz-Poisson representation is established by a proper conformal mapping, and then some boundary value problems for the Cauchy- Riemann equation are investigated. In particular, the boundary behavior at the three corner points are discussed in detail for the above two kinds of domains. Next, the situation for α = /2 is especially investigated. Furthermore, we develop a bridge between the unit disc and the fan-shaped domain with α = /2, 3

12 and then the Schwarz-Poisson formula for the unit disc can be derived from the Schwarz-Poisson formula for the fan-shaped domain with angle 2 α = /2. Next, the harmonic Green and the harmonic Neumann functions are constructed in the fan-shaped domain with angle /α α /2. What is more, the outward normal derivatives at the three corner points are introduced properly and the corresponding Dirichlet and Neumann problems for the Poisson equation are studied. As we know, the construction of polyharmonic Green and polyharmonic Neumann functions of arbitrary order m in explicit form is a demanding and complicated procedure. Here the biharmonic Green, the biharmonic Neumann, the tri-harmonic Green, the tri-harmonic Neumann, and the tetraharmonic Green functions are established explicitly for the fan-shaped domain with angle /n n N, by means of a series of proper polyharmonic functions. Then we also give the construction process for the tetra-harmonic Neumann function in detail and obtain the expression of the tetra-harmonic Neumann function with integral representation. Accordingly, the Dirichlet and Neumann problems are investigated respectively. Finally, even though the explicit expressions for polyharmonic Green and Neumann functions are unknown except for the above lower order ones, we still establish the inductive expressions of solutions and solvability conditions for the iterated polyharmonic Dirichlet and polyharmonic Neumann problems in the fanshaped domain with angle /n n N. At the same time, the recursive expressions of polyharmonic Green and polyharmonic Neumann functions are given by convolution, and their boundary behaviors are investigated in detail. Besides, in the Appendix, the tetra-harmonic Green function and the tri-harmonic Neumann function for the unit disc are constructed in explicit forms respectively. 4

13 Chapter 2 Boundary Value Problems for the Inhomogeneous Cauchy-Riemann Equation In this Chapter, the Schwarz-Poisson representation formulas are obtained in fan-shaped domains with angle /n n N and /α α /2 respectively, and then the solutions and solvability conditions for the Schwarz and Dirichlet problems are given explicitly. 2. Preliminaries et C be the complex plane and the variable z = x i y, x, y R. Introducing the complex partial derivatives, z = 2 x i y, z = 2 x i. y If a continuously differentiable function wz satisfies the following homogeneous Cauchy-Riemann equation then wz is analytic. z wz = 0, The main tools for solving boundary value problems of complex differential equations are the Gauss theorem and the Cauchy-Pompeiu representation. Theorem , 7 Gauss Theorem et D C be a regular domain, w C D, C CD, C, z = x i y, then, w z zdxdy = 2i wzdz, 2. D w z zdxdy = 2i D wzdz. 2.2 D D 5

14 The above Gauss theorem leads to a generalization of the Cauchy representation for analytic functions, that is the so-called Cauchy-Pompeiu formula. Theorem , 7 Cauchy-Pompeiu representation Any w C D; C CD; C for a regular complex domain D C can be represented as and D D w d z D The Pompeiu operator w d z D T fz = D w dξdη z = w dξdη z = wz, z D, 0, z / D, wz, z D, 0, z / D f dξdη z, f D; C, z D 2.5 studied in detail by I.N.Vekua 50 plays a critical role in treating boundary value problems for the inhomogeneous Cauchy-Riemann equation. It has some important properties. Theorem , 50 et D C be a bounded domain, then for f p D; C, p > 2, T is a completely continuous linear operator from p D; C into C α 0 C with α 0 = p 2 p. Theorem , 50 If f D, then for all ϕ C0D T fzϕ z zdxdy fzϕzdxdy = 0. D D Remark 2... Theorem 2..4 implies that for f D, T f is differentiable T f with respect to z in weak sense with = f. z The Poisson kernels for the unit disc D = z C : z < and the upper half plane H = z : Imz > 0 are, respectively, z z, z D, D, 6

15 and z z, z H, H. Then for γ C D, R, γ 2 C H, R 47, 36, lim z t, z < t D = γ z d z = γ t. and lim z t, z H t H γ 2 z d = γ 2 t. z 2.2 Schwarz Problem with Angle /n n N et be a fan-shaped domain in the complex plane C defined by = z C : z <, 0 < arg z <. 2.6 n 0,, ω = e iθ are three corner points of the domain and the oriented circular arc is parameterized by : τ e iτ, τ 0, θ, where θ =. The boundary = 0, ω, 0 is oriented counter-clockwise. n ω = e iθ. In what follows, we always regard n as a fixed positive integer, θ = n and By rotations, we define some domains k ω 2k = ω 2k z : z, k = 0,,, n, 2.7 where 0 = is the sector defined by 2.6. By reflections on the real axis, we define E k = z : z k, k = 0,,, n. 2.8 Besides, by reflections on the unit circumference, define D k = z : z k, E k = z : z E k, k = 0,,, n

16 Obviously, k, D k, E k, E k, k = 0,,, n are disjoint domains and Moreover, C = n n k E k = z C : z, Obviously, the following lemma holds. k D k E k E k. 2.0 n D k E k = z C : z. emma If z, then zω 2k k, zω 2k E k, z ω 2k D k and z ω 2k E k for k = 0,,, n, where k, E k, D k, E k are defined by , respectively Schwarz-Poisson Representation To solve the Schwarz problem, the Schwarz-Poisson formula is derived from the Cauchy-Pompeiu formula. Theorem Any w C ; C C; C can be represented as wz = n w zω z d 2k ω 2k z n w zω z dξdη, z 2k ω 2k z and for z wz = i ω,0 0, n zω 2k zω 2k d Rew zω2k zω 2k n n Rew zω z d 2k ω 2k z n w n w zω 2k zω 2k 8 z ω 2k z z dξdη. ω 2k z 2. Imw d 2.2

17 Proof: From emma 2.2. and Theorem 2..2, and n wz = n 0 = n 0 = n 0 = w d zω2k zw d z ω2k w d zω 2k zw d z ω 2k ω,0 0, w dξdη zω2k ω,0 0, w zω zw z ω zw dξdη z ω2k ω,0 0, w zω w dξdη zω 2k ω,0 0, zw z ω zw dξdη z ω 2k 2k d, z, 2k d, z, 2k d, z, 2k d, z Clearly, adding 2.3 and 2.4 leads to the validity of 2.. gives Taking complex conjugation on both sides of 2.5 and 2.6, respectively, 0 = n ω,0 w d zω 2k w zω 2k2 d 9 0 w d zω2k w dξdη zω2k, z 2.7

18 and n 0 = ω,0 z w d z ω 2k zw z ω 2k2 d 0 zw d z ω2k zw dξdη z ω2k, z. 2.8 Obviously, 2.7 and 2.8 can be, respectively, rewritten as n w d 0 = zω 2k w d zω2k ω,0 0, w dξdη, z, zω2k and n 0 = z w d z ω 2k ω,0 0, zw d z ω2k zw dξdη, z. z ω2k Subtracting the sum of 2.9 and 2.20 from 2., we easily get wz = I I 2 I 3, z 2.2 with I = I 2 = n w zω z 2k z ω 2k w zω z 2k z ω 2k ω,0 0, n w zω 2k w z z ω 2k zω 2k d, z d z ω 2k

19 and I 3 = n w By simple computation, we have I = n zω 2k Rew w zω 2k zω 2k zω 2k zω2k zω 2k z ω 2k z z ω 2k z dξdη. d n 2.24 Imw d 2.25 and I 2 = i ω,0 0, This completes the proof. n Rew zω 2k z d ω 2k z Remark When n =, the sector is the upper half disc and Theorem 2.2. coincides with Theorem in Schwarz Problem Firstly, the boundary behavior of some linear integrals are investigated. et n Kz, = zω 2k zω 2k z z ω 2k z z ω 2k emma If γ C; C, then lim γ γ Kz, d = γt γ, t \ ω, z, z t lim γ γω Kz, d = γt γω, t \ z, z t 2.28 and lim z, z t where K is defined by γkz, d = 0, t ω, 0 0,, 2.29

20 Proof: Simple computation gives n lim z, z t k= zω 2k zω 2k z z ω 2k z = 0 z ω 2k for t \ ω and by emma Hence we see that lim γ γ Kz, d z, z t = lim γ γ z, z t = lim Λ z, z t z z z z d z, t \ ω, d z 2.30 where = τ : τ = e iϕ, n < ϕ < 0 is oriented counter-clockwise and Λ = γ γ,, γ γ,. Therefore, by the continuity of Λ on and the boundary property of the classical Poisson kernel on the unit circle, 2.30 implies that lim γ γ Kz, d = γt γ, t \ ω. z, z t Similarly, the other equality in 2.28 is valid. Finally, if and z ω, 0 0,, then Kz, 0, and hence 2.29 holds. emma If γ C0, ; C, then lim z, z t γ γ Kz, d = γt γ, t 0,, lim z, z t where K is given by γkz, d = 0, t \ 0,,

21 Proof: Similarly as before, we have with lim z, z t 0 = lim z, z t = lim z, z t γ γ Kz, d 0 0 γ γ z z z z Λ 2 z z d, t 0, γ γ, 0,, Λ 2 = γ γ,,. z d z Thus, from the continuity of Λ 2 on 0, and the property of the Poisson kernel on the real axis, 2.3 is obtained. If 0, and z \ 0,, then Kz, 0, and hence 2.32 is true. Similarly to emma 2.2.3, the following lemma is valid. emma If γ Cω, 0; C, then lim γ γω Kz, d = γt γω, t ω, 0, 2.33 z, z t ω,0 lim z, z t where K is given by ω,0 γkz, d = 0, t \ ω, 0, 2.34 It should be noted that the boundary behavior at the corner z = 0 needs to be especially investigated because 0 is the common point of the boundary of all the sectors k, E k, k = 0,,, n. emma If γ Cω, 0 0, ; C, then lim γ γ0 Kz, d = 0, 2.35 z, z 0 where K is defined by ω,0 0, 3

22 Proof: Firstly, we have lim z, z 0 = lim z, z 0 ω,0 0, ω,0 0, γ γ0 Kz, d n γ γ0 zω 2k zω 2k d Next we discuss the boundary behavior at z = 0 in two cases. Case : if n is an even number, by 2.36, lim z, z 0 0 = lim z, z 0 = n 2 lim z, z 0 γ γ0 Kz, d n 2 γ γ0 zω 2k zω 2k 0 zω d 2k zω 2k Λ 3 zω d 2k zω 2k 2.37 with γ γ0, 0,, Λ 3 = γ γ0,, 0. Since for z, all zω 2k, k = 0,, n 2 leads to Similarly, lim z, z 0 are in the upper half-plane, 2.37 γ γ0 Kz, d = n 2 Λ 30 = lim z, z 0 ω,0 γ γ0 Kz, d = Thus, the sum of 2.38 and 2.39 gives the desired conclusion

23 with and Case 2: if n is an odd number, then 2.36 equals lim z, z 0 0 n 2 γ γ0 zω d 2k zω 2k 0 n 2 γ γ0 zω d 2k zω 2k k= n 2 γω γ0 zω d 2k zω 2k k= n 2 γω γ0 zω d 2k zω 2k 0 0 = lim z, z 0 lim n 2 n 2 z, z 0 k= Λ 4 zω 2k zω 2k d Λ 5 zω 2k zω 2k γ γ0, 0,, Λ 4 = γ ω γ0,, 0, γω γ0, 0,, Λ 5 = γ γ0,, 0. Then the continuity of Λ 4, Λ 5 at = 0 implies that the limit in 2.40 equals n 2 Λ 4 0 n Λ 5 0 = 0. 2 d 2.40 Therefore 2.35 is valid in this case. In conclusion, the desired conclusion 2.35 is always true. emma If γ Cω, 0 0, ; C, then lim γ γ0 Kz, d = γt γ0, t ω, 0 0,, z, z t ω,0 0, where K is defined by

24 Proof: By emma 2.2.5, we only need to prove lim γ γ0 Kz, d = γt γ0, t ω, 0 0,. z, z t ω,0 0, 2.4 By emmas and 2.2.4, 2.4 is equivalent to lim γ γ0 Kz, d = γt γ0, t ω, z, z t and lim z, z t ω,0 γ γ0 Kz, d = γt γ0, t 0, The left-hand side of 2.42 equals lim γ γ0 z, z t z d zω 2 ω,0 = lim z, z t = γt γ0, t ω, 0. 0 γω γ0 Hence 2.42 is true. Similarly, 2.43 is also valid. zω zω d Theorem If γ C ; C, then lim γkz, d = γt, t, 2.44 z, z t where K is defined by Proof: et t ω, 0 0,, by emmas and 2.2.6, the left-hand side in 2.44 equals γt γ0 γ0 lim z, z t 6 ω,0 0, Kz, d. 2.45

25 On the other hand, by 2.2, = n zω 2k zω 2k zω2k zω 2k i ω,0 0, n d zω 2k z d. ω 2k z 2.46 Taking the real part on both sides of 2.46 gives = Kz, d ω,0 0, Kz, d Then, by 2.47 and 2.29, 2.45 equals γ0 γt lim z, z t Kz, d = γt. Therefore, 2.44 is valid for t ω, 0 0,. Similarly, 2.44 is also true for t \ 0, and t \ ω, 0, respectively. This completes the proof. We introduce the Schwarz-type operator as follows Sγz = i n zω 2k zω 2k d γ zω2k zω 2k n γ zω z d, z, 2k ω 2k z ω,0 0, 2.48 where γ C ; R. Obviously, Sγz is analytic in the domain, denoted as Sγ A. Further, ReSγz = γkz, d, z 2.49 for γ C ; R, where K is defined by Thus, by Theorem 2.2.2, the following result is valid. Theorem If γ C ; R, then ReSγ t = γt, t, where S is the Schwarz-type operator defined by

26 Finally, a Pompeiu-type operator for is introduced by T fz = n f zω 2k z ω 2k z f zω z dξdη, z, 2k ω 2k z 2.50 where f p ; C, p > 2. By simple computation, we have Re T fz = 2 where K is defined by fkz, fkz, dξdη, z, 2.5 Theorem If f p ; C, p > 2, then z T fz = fz, z in weak sense, and Re T f t = 0, t, where T is the Pompeiu-type operator defined by Proof: By 2.50, it is obvious that z T fz = z T fz = fz, z in weak sense, where T is defined by 2.5. On the other hand, Kz, = Kz, = 0, z, implies that Re T f t = 0, t. Therefore, we obtain the following Schwarz problem. Theorem The Schwarz problem z w = f in, n n 0 Imwe iϕ dϕ = c, Rew = γ on, c R 2.52 for f p ; C, p > 2, γ C ; R, is uniquely solvable by wz = Sγz T fz ic, 2.53 where S, T are defined by 2.48 and 2.50, respectively. 8

27 Proof: By Theorem 2.2., if there is a solution for the Schwarz problem 2.52, it must be of the form Thus we only need to verify that 2.53 provides a solution. If w is given by 2.53, then z w = f in since Sγ A and z T f = f by Theorem Also by Theorems and 2.2.4, Rew = γ on. Finally, we only need to prove that Actually, n = and = Hence z = n z = n n zω 2k zω 2k zω2k zω 2k z dz z z = zω 2k dz z z Similarly, for, which implies that 0 z = Imwe iϕ dϕ = c, c R dz z = z 2 z z dz ω 2k z z = = 0 for ω, 0 0,. = ImSγz dz z n zω 2k n zω 2k Im T fz dz z n dz = 0, n zω 2k zω2k dz zω2k zω 2k z zω 2k zω 2k dz z = z dz ω 2k z z z ω 2k z dz z = 0, = Thus, 2.53, 2.55 and 2.56 lead to Then the proof is completed. 9

28 2.3 Dirichlet Problem with Angle /n n N We firstly consider the Dirichlet problem for the homogeneous Cauchy-Riemann equation z wz = 0, z, 2.57 w t = γt, t, where the boundary data γ C ; C. Theorem The Dirichlet problem 2.57 is solvable if and only if n γ zω 2k and its solution is uniquely expressed as wz = n γ Proof: zω 2k z ω 2k z d = 0, z, 2.58 z ω 2k z d, z By 2. in Theorem 2.2., if there is a solution for the Dirichlet problem 2.57, then its solution can be represented as Now we prove that 2.58 is a necessary condition. If w is the solution to the Dirichlet problem 2.57 given by 2.59, then et Obviously, hz = w t = γt, t n γ zω 2k wz hz = where K is defined by By Theorem 2.2.2, z ω 2k z d, z. γkz, d, lim wz hz = γt, t. 2.6 z, z t 20

29 By 2.60 and 2.6, we have h t = 0, t. Since hz is analytic for z, then by the maximum principle of analytic functions, hz 0 for z, which is just the condition On the other hand, If the condition 2.58 is satisfied, then wz = n γ zω z d = γkz, d, 2k z ω 2k z. Hence, obviously z wz = 0, z, and w t = γt, t by Theorem Theorem The inhomogeneous Dirichlet problem z wz = fz, z, f p ; C, p > 2, w t = γt, t, γ C ; C is solvable if and only if for z, n γ zω 2k = n f zω 2k z d z ω 2k z z ω 2k dξdη, and its solution can be uniquely expressed as wz = n γ zω z d 2k z ω 2k n f zω z dξdη, z. 2k z ω 2k Proof: If the Dirichlet problem 2.62 is solvable, then its solution can be represented as 2.64 due to Theorem et φz = wz T fz, then z φ = 0 in, φ = γ T f on, 2

30 where T is defined by 2.5. From Theorem 2.3., the condition of solvability is n γ T f zω z d = 0, 2k ω 2k z which is just condition 2.63 by computation. Conversely, if the condition of solvability 2.63 is satisfied, 2.64 can be rewritten as wz = γkz, d fkz, dξdη, z Since fkz, dξdη tends to 0 as z t, 2.65 implies that w t = γt, t D by Theorem Obviously, we also have z wz = fz, z. This completes the proof. 2.4 Schwarz Problem with Angle /α α /2 In Section 2.2, we give the Schwarz-Poisson formula for the domain with angle /n n N explicitly by the reflection method. However, for a general angle /α α /2, it is difficult to bring this formula into effect likewise. This Section is devoted to extending the Schwarz-Poisson representation formula to a fan-shaped domain with angle /α by a proper conformal mapping and to solving the Schwarz problem in detail. et be a fan-shaped domain with angle ϑ = α = z <, 0 < arg z < α. α /2, that is, Its boundary = 0, Γ 0 ϖ, 0 is oriented counter-clockwise with ϖ = e iϑ. Further, 0,, ϖ are the three corner points and the oriented segment Γ 0 is parameterized by Γ 0 : τ e iτ, Obviously, the domain = for α = n. τ 0,. α Similarly, except for additional indication, we always assume that α α /2 is a fixed real constant, ϑ = /α and ϖ = e iϑ. 22

31 2.4. Schwarz-Poisson Representation From 26 and Theorem 2.2., when n =, the following lemma holds. emma Any w C D ; C CD ; C can be represented as w z z d z D w z z wz, z D, dξdη = z 0, z / D, D and for z D wz = 0 z Rew z z d z z Rew i w z D z d z 0 z w z z Imw d where D = z <, Imz > 0 and 0 = τ =, Imτ > 0. We consider the conformal mapping 30 : D z z dξdη, z z α 2.68 with the branch cut along 0,, which especially maps the boundary Γ 0 onto 0, ϖ, 0 onto, 0 and 0, onto 0,, respectively. The inverse mapping is ς : D z z /α 2.69 which transforms 0 onto Γ 0,, 0 onto ϖ, 0 and 0, onto 0,, respectively. Remark Obviously, the above two conformal mappings can be firstly - extended to the boundary except for 0. Since the argument at 0 can be arbitrary and α /2, we can assume that the argument of z α and z /α at z = 0 are 0. Hence, the above - mapping between the boundary is proper. 23

32 In what follows, the main analytic branches of z α and z /α are always chosen as above, respectively. Moreover, on the basis of the analytic branches of z α and z /α, we define the analytic branch of several functions as follows, z α = zα z, z α = z z α, z/α = z/α z. Then, based on emma 2.4. and the conformal mapping, we obtain the following Schwarz-Poisson representation for an arbitrary angle α α /2. Theorem Any w C ; C C ; C can be represented as α w α z z α α d α z α α α w α z z α wz, z, α dξdη = α z α α 0, z /, and for z wz = α α z α Rew α z α z α d α α z α α Imw d Γ 0 Γ 0 α Rew i α z z α α d α z α α ϖ,0 0, α w α z z α α α z α α w α z α z α dξdη. α z α α Proof: et z wz = fz and denote a new function W 0 z = wz /α, z D, then for z D, z W 0 z = α z α fz /α, z W 0 z = α z α z wz /α. 24

33 We observe that W 0 z C D ; C CD ; C, thus from emma 2.4., we have for z, W 0 z α = 0 z α ReW 0 z zα d α z α ReW 0 i W 0 D z zα d α z α z α zα W z α 0 0 ImW 0 d z α zα z α dξdη where W 0 = α α f /α and 0 is defined in emma Applying the transformation 2.68: = τ α with τ = τ iτ 2, τ, τ 2 R, thus, dξdη = J dτ dτ 2 with the Jacobi determinant J = α 2 τ 2α. Therefore, W 0 z α = α ReW 0 τ α τ α z α τ α z τ α z α α τ α z α Γ 0 α ReW 0 τ α i τ α z z α α z α τ α 0 α ReW 0 τ α i τ α z z α α z α τ α ϖ,0 τ α α fτ τ α z α τ α fτ τ α z α dτ τ α z α z α τ α τ α dτ Γ 0 τ α dτ ImW 0 τ α dτ τ z α α 2 τ dτ 2α z α τ α dτ 2. By simple computation, this is the desired conclusion 2.7. Similarly, expression 2.70 is also true. Remark We can verify that when α = n n N, the Schwarz-Poisson representation formula in Theorem 2.4. is just the result in Theorem

34 In particular, the investigation for α = /2 should be especially noted. For the case α = /2, the fan-shaped domain with angle 2 is just D 0 = z, z <, z / 0, with D 0 = 0, Γ, 0, where Γ = τ, τ =, τ, is oriented counter clockwise. et the unit disc Then we observe that D = z : z < with D = τ, τ =. D 0 = D\0,, Γ = D\ From Theorem 2.4., the result for α = /2 holds. Corollary Any w C D 0 ; C CD 0 ; C can be represented as w 4i /2 z z /2 /2 d /2 z /2 /2 D 0 w 2 /2 z z /2 wz, z D 0, /2 dξdη = /2 z /2 /2 D 0 0, z / D 0, 2.73 and for z D 0 wz = 4i 2 /2 z /2 Rew /2 z /2 z /2 /2 /2 z /2 Γ Rew /2 z z /2 /2 z /2 /2,0 0, D 0 w /2 z /2 z /2 z /2 /2 d 2 2 Γ /2 d Imw d w /2 z z /2 dξdη, /2 /2 z /2 / where for 0,,, 0 and =, we have Rew Rew, as well as w. 26

35 Actually, the classical Schwarz-Poisson formula for the unit disc see 7 can be derived from Corollary Proposition Any w C D; C CD; C is represented as wz = Rew z d z Imw d 2 = = w z zw dξdη, z <. z < 2.75 Proof: Obviously, when w C D; C CD; C, we have w C D 0 ; C CD 0 ; C. Moreover, w = w for 0,,, 0 and =. Thus the third integral on the right-hand side in 2.74 equals 0. Then from 2.72 and 2.74, wz is expressed as wz = Rew 4i D w 2 Since D /2 z /2 /2 z /2 z /2 /2 /2 z /2 /2 z /2 z /2 z /2 /2 d 2 2 D Imw d w /2 z z /2 dξdη. /2 /2 z /2 /2 z = 2 /2 /2 z /2 z z = 2 z/2 /2 /2 z /2, z /2, /2 z /2 /2 therefore, the area integral in 2.76 can be converted into w z zw dξdη /2 w z 2 /2 z /2 D D z/2 /2 w z /2 /2 /2 /2 z z/2 /2 dξdη. /2 z /2 /

36 et 0 be the second area integral in 2.77, then from the Gauss theorem, /2 0 = lim w ε 0 4i /2 z z/2 /2 /2 z /2 /2 = 4i = = Rew Thus, by , wz = 2 = 4i = =ε w /2 /2 z z/2 /2 /2 z /2 /2 /2 z /2 /2 z /2 z/2 /2 z /2 /2 Imw d Rew < d. d w z zw dξdη z /2 z /2 /2 z /2 z /2 /2 /2 z /2 /2 z /2 /2 z /2 z/2 /2 z /2 /2 d. By simple computation, 2.75 is valid. Then, the proof is completed Schwarz Problem In this part, we consider the Schwarz problem in the fan-shaped domain with angle /α α /2. Schwarz boundary value problem Find a function w satisfying the following conditions w z = f in, Rew = γ on, α α with f p ; C, p > 2, 0 Imwe iϕ dϕ = c, c R, γ C ; R Firstly, introducing a new kernel Hz, = α z α α z z α α z α z α α α z α α Then, the following lemmas are valid. 28

37 emma If γ C ; C, then lim z, z α γ γhz, d = 0. Γ 0 Proof: By simple computation, α γ γhz, d Γ 0 γ /α γ 0 /α γ γ z α 0 Γ z α lim z, z = lim z, z = lim z, z = z α d z α d z α d z α, where 0 = τ : τ =, Imτ < 0 is oriented counter-clockwise and Γ = γ /α γ, 0, γ /α γ, 0. Therefore, from the continuity of Γ at = and the property of the Poisson operator on the unit circle, lim z, z α γ γhz, d = Γ z α Γ 0 z= = 0. emma With γ C ; C, lim z, z α 0, ϖ,0 γ γhz, d = 0. 29

38 Proof: We observe that where lim z, z = lim z, z = lim z, z α γ γhz, d 0 γ /α γ 0 z α z α /α γ γ z α z α Γ 2 z d, α z α 0 γ /α γ, 0,, Γ 2 = γ /α γ,,. d d Hence, similarly from the continuity of Γ 2 at = and the property of the Poisson operator on the real axis, lim z, z α 0 γ γhz, d = Γ 2 = 0. Thus, the desired conclusion is obtained also from Hz, = 0 for z, ϖ, 0. emma With γ C ; C, α lim γ γϖhz, d = 0. z, z ϖ Γ 0 Proof: Similarly to emma 2.4.2, α lim γ γϖhz, d z, z ϖ Γ 0 = lim Γ 3 z, z ϖ z d α z α, = where γ /α γϖ, 0, Γ 3 = γ /α γϖ, 0. 30

39 Since when z ϖ, z α tends to and Γ 3 is continuous at = z α z = ϖ, thus, α lim γ γϖhz, d = Γ 3 z α z, z ϖ z=ϖ = 0. Γ 0 emma With γ C ; C, α lim γ γϖhz, d = 0. z, z ϖ 0, ϖ,0 Proof: Since Hz, = 0 for z, ϖ 0,, then α lim γ γϖhz, d z, z ϖ 0, ϖ,0 0 = lim Γ 4 z, z ϖ z d, α z α where γ /α γϖ,, 0, Γ 4 = /α γ γϖ,,. Therefore, similarly from the continuity of Γ 4 at = z α z = ϖ, we obtain, α lim γ γϖhz, d = Γ 4 z α z, z ϖ z=ϖ = 0. 0, ϖ,0 emma With γ C ; C, α lim γ γ0hz, d = 0. z, z 0 Proof: By Hz, = 0 for z, 0 Γ 0, we see that α lim γ γ0hz, d z, z 0 = lim γ /α γ0 z, z 0 0 z α z α 0 γ /α γ0 z d α z α = lim γ /α γ0 z, z 0 z α z α 3 d d = γz γ0 z=0 = 0.

40 Remark emmas suggest that the following boundary behavior is valid at the three corner points, that is, α lim γ γthz, d = 0, z, z t is true. t 0,, ϖ. According to the representation formula in Theorem 2.4., the next result Theorem The Schwarz problem 2.79 for the domain is uniquely solvable by wz = α α z α γ α z α z α α α z α Γ 0 α γ i α z α ϖ,0 0, α f α z α f α z α d z α z α α z α α z α α ic α d z α dξdη, α z. z α α 2.8 Proof: From Theorem 2.4., we only need to verify that 2.8 provides a solution. Since for z,, then we know that z is a simple pole of αα lim z z α z =, α α α α z α in and α α α z = gz,, α z where for arbitrary z, gz, is analytic with respect to z. obviously see that z wz = z f z dξdη = fz. Thus, we 32

41 When Γ 0, i.e, =, α α z α α z α z α dz α α z α z = Γ 0 = α z dz α z z = 0, z = and for 0, ϖ, 0, α α z z α dz α z α α z = Γ 0 0 = dz α z z = 0. z = Similarly, for, α Thus, Γ 0 0 α z α z α z dz α z z α z α z α z z α dz α z α α z = 0. α α Imwe iϕ dϕ = α 0 α 0 cdϕ = c. et w 0 be the area integral in 2.8, then Rew 0 z = α fhz, f Hz, dξdη. 2 dz z Since Hz, = 0 for z,, we obtain Rew 0 z = 0 for z. We can write Rewz as Rewz = α γhz, d Rew 0 z Since Hz, = 0 for z, Γ 0 \, ϖ ϖ, 0 0,, then for 0 Γ 0 \, ϖ, α α lim γhz, d = lim z 0 z 0 Γ 0 α z α 2 d = lim γ z 0 α z α 2 = lim z 0 Γ 0 z α 2 γ 0 α z α 2 zα 2 α z α 2 d z α 2 d γ α z α 2 = γ 0. 33

42 Similarly, Hz, = 0 for z, 0, ϖ, 0 Γ 0, then for 0 0,, lim z 0 = lim z 0 α α 0 γhz, d = lim z 0 γ zα z α α z α 2 α d = lim z 0 α z α z γ α α z α zα z α α d 2 z α α γ α z α z α z α 2 d = γ 0. Moreover, when z, ϖ, 0 0, Γ 0, Hz, = 0. Thus for 0 ϖ, 0, lim z 0 α γhz, d = lim z 0 0 γ α z α z α z α 2 d = γ 0. In a word, we have lim z 0 Rewz = γ 0 for 0 except for the three corner points. With respect to the three corner points, we can convert 2.82 into different representations. Firstly, Rewz can be expressed as Rewz = α αγ γ γ Hz, d Hz, d Rew 0 z Applying the representation 2.7 to wz and then taking the real part on both sides, we obtain α α Hz, d = α z α α z α zα z α z α α d =, α z α α then, also from emmas and 2.4.3, lim Rewz = γ. z, z 2.84 Similarly as before, Rewz = α γ γϖ Hz, d αγϖ Hz, d Rew 0 z

43 Hence, lim Rewz = γϖ follows from 2.84, emma and emma z, z ϖ Also obviously, we can rewrite Rewz as, Rewz = α γ γ0 Hz, d αγ0 Hz, d Rew 0 z Thus, the desired result lim Rewz = γ0 is true from 2.84 and emma z, z Therefore, the proof is completed. 2.5 Dirichlet Problem with Angle /α α /2 In this Section, a Dirichlet boundary value problem for the domain is discussed. Theorem Dirichlet boundary value problem w z = f in, f p ; C, p > 2, w = γ on, γ C ; C, is solvable if and only if for z, α = α γ α z z α α d α z α α f α z z α α dξdη. α z α α Then the solution can uniquely be expressed as wz = α γ α f α z α α z α z α α d z α α z α z α α α dξdη, z

44 Proof: If there is a solution, it must be of the form 2.88 from Theorem et w be the solution to the Dirichlet problem, then we know lim wz = γ for z We consider a new function Gz = α γ α z z α α d α z α α α f α z z α α dξdη, z. α z α α Thus, we have wz Gz = α α γhz, d fhz, dξdη Since the area integral in 2.90 vanishes on, then from the proof of Theorem 2.4.2, we easily obtain, for, which implies lim wz Gz = γ, 2.9 z, z lim Gz = 0. z, z We observe that Gz is analytic for z, and then by the maximum principle for analytic functions, we know Gz 0 for z, which is just the condition wz as On the other hand, if the condition 2.87 is satisfied, then we can rewrite wz = α α γhz, d fhz, dξdη. Obviously, z wz = f in and w = γ on. This completes the proof. 36

45 Chapter 3 Harmonic Boundary Value Problems for the Poisson Equation In this Chapter, we study a harmonic Green and a harmonic Neumann functions in the fan-shaped domain with angle /α α /2, and then solve the corresponding Dirichlet and Neumann problems for the Poisson equation. Here, Γ 0, ϖ, ϑ are given as in Section Harmonic Dirichlet Problem Definition A real-valued function Gz, = 2 G z, in a regular domain D C is called the Green function of D, more exactly the Green function of D for the aplace operator, if for any fixed D as a function of z, it possesses the following properties:. Gz, is harmonic in D\, 2. Gz, log z is harmonic in D, 3. lim Gz, = 0, t z t, z D D. The Green function is uniquely determined by -3. additional properties: 4. Gz, > 0, 5. Gz, = G, z. Actually, it also has the It must be noted that not any domain in the complex plane has a Green function. The existence of the Green function for a given domain D C can be proved when the Dirichlet problem for harmonic functions is solvable for D. The harmonic Green function G for the fan-shaped domain with angle 37

46 ϑ = /α is α z α z α α 2 G z, = log α z α z α α, 3. which can be verified easily. The outward normal derivative of the boundary is given by z z z z, z Γ 0 \, ϖ, νz = i z z, z 0,, ie iϑ z e iϑ z, z ϖ, Thus, for z Γ 0 \, ϖ, νz G z, = z z z z G z, z α = 2α α z α and for z 0,, zα α z α zα α z α νz G z, = i z z G z, = 2iαz α α z α α z α α α z α zα α z α, α z α α When z ϖ, 0, i.e. z = ρe iϑ, 0 < ρ <, we have z α = z α and e iϑ z α = e iϑ z α, then νz G z, = ie iϑ z e iϑ z G z, = 2iαe iϑ z α α z α α z α α α z α. α z α α From 27, we obtain the following representation formula. Theorem 3... Any w C 2 ; C C ; C can be represented as wz = w ν G z, ds w 4 G z, dξdη, where s is the arc length parameter on with respect to the variable and G z, is the harmonic Green function for. 38.

47 Based on Theorem 3.., the representation formula provides a solution for the related Dirichlet problem. Theorem The Dirichlet problem z z w = f in, f p ; C, p > 2, w = γ on, γ C ; C is uniquely solvable by wz = α γ α Γ 0 γ α α z α α α z α α z α α z α 0 α γ α z α α z α ϖ,0 fg z, dξdη, z, where G is given in 3.. α α z α α α z α d z α z α z α α d α z α α z α z α α z α z α α α d 3.3 Proof: Obviously, if the problem is solvable, it must be unique. By Theorem 3.., we only need to verify that 3.3 is a solution. Since z G z, = αz α α z α α z α α α z α, α z α α then similarly to the proof of Theorem 2.4.2, αz α α z = gz,, α z where for any z, gz, is analytic in z. αz α can be written as α zα Thus, we obviously see that expression 3.3 is a solution to the Poisson equation. Moreover, by the properties of the Green function, the area integral in 3.3 vanishes on. Actually, we can rewrite wz as wz = α γhz, d 39 fg z, dξdη,

48 with Hz, defined by Then, similarly from the proof of Theorem 2.4.2, we obtain This completes the proof. lim wz = γ for z. 3.2 Harmonic Neumann Problem Definition A real-valued function Nz, = 2 N z, in a regular domain D is called a Neumann function for the aplace operator, if as a function of z, it satisfies. Nz, is harmonic in D\, 2. Nz, log z is harmonic in D, 3. νz Nz, is constant on any boundary component of D for any D. Remark The Neumann function is not uniquely defined by -3 above. It is only given up to an arbitrary additive constant. Example A harmonic Neumann function for the ring domain R = z C : 0 < r < z < can be written as see 54 N z, = log z z z r 2k z r 2k r 2k z r 2k 2 z, z 2 2 k= and its boundary behavior is 2, z =, νz N z, = 0, z = r. A harmonic Neumann function N z, for the Poisson equation in can be expressed as N z, = log α z α α z α z α α z α α

49 We verify this in the following. For z Γ 0 \, ϖ, N z, = 2 log α z α α z α 2, νz N z, = z z z z N z, = αz α when z 0,, α z α α z α αz α α z α α z α α z α α α z α α α z α α α z α α N z, = 2 log α z α z α α 2, = 4α, νz N z, = i z z N z, = iαz α α z α α z α α z α α α z α α α z α α z α α z α α = 0, α z α α and for z ϖ, 0, z = ρe iϑ, 0 < ρ <, N z, = 2 log α z α z α α 2, νz N z, = ie iϑ z e iϑ z N z, = iαz α e iϑ α z α α z α α z α α α z α α iαz α e iϑ α z α α z α α z α α = 0. α z α α Moreover, the normalization condition holds, that is, dz N z, z = 2 log z α 2 dz α z = 0. Γ 0 Thus, N z, satisfies the properties:. N z, is harmonic in z \, z = 2. N z, log z 2 is harmonic in z for any, 3. νz N z, = 4α, z Γ0 \, ϖ, 0, z 0, ϖ, 0, for, 4

50 4. Γ 0 N z, dz z = 0. Then, we give a representation formula for the domain. Theorem Any w C 2 ; C C ; C can be represented as wz = α i Γ 0 d w 4 ν wn z, ds where N z, is the harmonic Neumann function for. w N z, dξdη. Proof: et z and ε > 0 such that B ε z with B ε z = C : z < ε. Suppose that ε = \B ε z, then w N z, dξdη = 2 w N z, w N z, ε ε w N z, w N z, dξdη = N z, w d w 4i d ε wn 2 z, wn z, 2 N z, w dξdη ε = N z, w d w 4i d wn z, d N 4i z, d ε ε = ν wn z, w ν N z, ds 4 d N z, zw zw 4i z z =ε zn z, zn 4i z, w d z. z =ε 42

51 Introducing the polar coordinates = z εe iψ gives rise to d N z, zw zw 4i z = ε 4 z =ε 2 0 N z, e iψ w e iψ w dψ, which tends to 0 when ε 0. Similarly, lim zn z, zn ε 0 4i z, w d z Then, wz = 4 z =ε 2 = lim ε wz εe iϕ dϕ = wz. ν wn z, ν N w ds w N z, dξdη. Hence, the desired result follows from 3.5 and the property 3 of N z,. 3.5 Next, the outward normal derivatives at the corner points are introduced. et the partial outward normal derivatives be ν z w0 = lim νz wt, ν z w0 = lim νz wt, 3.6 t 0, t 0, t 0, t ϖ,0 and ν z w = lim νz wt, ν z w = lim νz wt, 3.7 t, t Γ 0 \ t, t 0, ν z wϖ = lim νz wt, ν z wϖ = lim νz wt. 3.8 t ϖ, t ϖ,0 t ϖ, t Γ 0 \ϖ Definition If the partial outward normal derivatives exist, then the outward normal derivatives at the three corner points are, νz wt = 2 νz wt ν z wt, t 0,, ϖ. Finally, the Neumann problem for the Poisson equation is discussed. 43

52 Theorem The Neumann boundary value problem z z w = f in, ν w = γ on, α /α we iθ dθ = c 0 0 for f p ; C, p > 2, γ C ; C is solvable if and only if 2 γds = fdξdη Then the solution is uniquely expressed by wz = c 0 γ log α z α α z α 2 d Γ 0 γ log α z α z α α 2 d 2 0 e iϑ γ log α z α z α α 2 d 2 ϖ,0 fn z, dξdη, z, 3.0 where N z, is given in 3.4. Proof: From Theorem 3.2., if the Neumann problem is solvable, it should be of the form 3.0. Then we confirm that 3.0 is a solution. Similarly to the proof of Theorem 3..2, we easily get w zz = f in. Since when, α i Γ 0 log α z α z α α 2 dz z = i thus, the normalization condition z = α /α we iθ dθ = c 0 0 follows from 3. and the normalization of N z,. log z α 2 dz z = 0, 3. 44

53 In addition, we have z z z z wz = α γ Γ 0 γ α α z α z α α α z α zα α z α zα α z α α zα α z α α 2 α z α 0 α e iϑ z α γ 2 α z α ϖ,0 fz z z zn z, dξdη. Hence, for 0 Γ 0 \, ϖ, zα α z α α zα α z α zα d α z 2 α zα α z α α d zα α z α α νz w 0 = lim z z z z wz z,z 0 α α = lim γ z,z 0 α z α d α α z α Γ 0 α 4α γds fdξdη = γ 0 α 4α γds fdξdη, d which implies the sufficiency and necessary of 3.9. Further, from γ C, C, ν z w = lim νz w 0 = γ, 3.2 0, 0 Γ 0 \,ϖ Similarly, ν z wϖ = lim νz w 0 = γϖ ϖ, 0 Γ 0 \,ϖ i z z wz = α z α γ 2 α z zα α α z zα α α z zα d α α z α Γ 0 α z α γ α z zα α α z α zα α α z zα α d α z α α 0 45

54 α e iϑ z α γ α z zα α α z α zα α α z zα α d α z α α ϖ,0 f z zn z, dξdη. i Then for 0 0,, Thus, νz w 0 = lim i z z wz z,z 0 α z α γ α z α = lim z,z 0 = lim z z 0 α α 0 0 = lim z,z 0 0 γ γ 0 ν z w0 = ν z w = Finally, we also observe that, z α α α z α zα α z α α z α α z α γ /α d zα α α z α z α z α ie iϑ z e iϑ z wz = α e iϑ z α γ 2 α z eiϑ z α α α z e iϑ z α α α z α Γ 0 α e iϑ z α γ α z eiϑ z α α α z α e iϑ z α α α z α 0 α e iϑ γ ϖ,0 fe iϑ z e iϑ zn z, dξdη. i α d d d = γ 0. lim νz w 0 = γ0, , 0 0, lim νz w 0 = γ , 0 0, e iϑ z α d α z α e iϑ z α α z α α e iϑ z α α z eiϑ z α α α z α e iϑ z α α α z e iϑ z α α α z α α d d 46

55 Since for 0 ϖ, 0, that is, 0 = ρe iϑ, 0 < ρ <, we obtain α 0 = α 0, e iϑ 0 α = e iϑ α 0. Thus, νz w 0 = lim ie iϑ z e iϑ z wz z,z 0 α = lim e iϑ e iϑ z α γ z,z 0 α z e iϑ z α d α α z α ϖ,0 α = lim e iϑ γ z,z 0 ϖ,0 α e iϑ e iϑ α γ α z e iϑ α d α α z α ϖ,0 α = lim γ z,z 0 α z α α z α Furthermore, ϖ,0 ν z w0 = ν z wϖ = Hence, by and Definition 3.2.2, e iϑ z α α e iϑ z α α α z α α z α α d = γ 0. lim νz w 0 = γ0, , 0 ϖ,0 lim νz w 0 = γϖ ϖ, 0 ϖ,0 d νz wt = γt, t 0,, ϖ. Then the proof is completed. 47

56

57 Chapter 4 Boundary Value Problems for the Bi-Poisson Equation As stated before, convoluting the harmonic Green function with itself leads to a biharmonic Green function. Similarly, convoluting the harmonic Neumann function with itself also gives rise to a biharmonic Neumann function. This chapter is devoted to the construction of a biharmonic Green function and a biharmonic Neumann function in the fan-shaped domain with angle /n n N, by using the concept of convolution and some proper transformations, and finally give the solutions and solvability conditions for the corresponding Dirichlet and Neumann problems explicitly. Here,, ω = e iθ, θ are given as in Section Biharmonic Green Function In the last Chapter, we have studied the harmonic Green and the harmonic Neumann function explicitly in the fan-shaped domain with angle /α α /2. Thus, the harmonic Green function for is expressed as G z, = log n z n z n n 2 n z n z n n n ω 2k z zω 2k 2 = log, z,. ω 2k z zω 2k 4. Remark 4... Here the factorization is used. We only take n z n as an example. For k = 0,,, n and θ = /n, we have 0 2kθ < 2. Then = zω 2k are n different roots of n z n in the complex plane. Hence, n 2 n z n 2 = zω 2k. 49

58 What is more, the solution to the Dirichlet problem for α = n is given as follows. emma 4... The Dirichlet problem for the Poisson equation w z z = f in, f p ; C, p > 2, w = γ on, γ C ; C is uniquely solvable by wz = n γ zω 2k n γ 0, ω,0 zω 2k zω 2k zω 2k zω 2k z ω 2k z z ω 2k z fg z, dξdη, z. d d zω 2k et Ĝ 2 z, = G z, G, d ξd η, z,, 4.2 with G z, defined by 4.. Then, Ĝ 2., is the solution to the Dirichlet problem z z Ĝ 2 z, = G z, in, Ĝ 2 z, = 0 on. 4.3 Moreover, for any fixed, Ĝ2z, satisfies the properties,. Ĝ 2 z, is biharmonic in \, 2. Ĝ 2 z, z 2 log z 2 is biharmonic in, 3. Ĝ 2 z, = 0, z z Ĝ 2 z, = 0 for z, 4. Ĝ 2 z, = Ĝ2, z for z. Since the boundary consists of a circular arc and two lines, it is difficult to obtain Ĝ2z, explicitly by direct computation as in the unit disc. Therefore, we prefer to transform Ĝ2z, into a new unknown function. From the property 2 of Ĝ2z,, we see that Ĝ2z, can be represented as Ĝ 2 z, = z 2 G z, h 2 z,, z,,

59 where h 2 z, is a biharmonic function in. Thus from 4.3, z z h 2 z, = 2Re z z G z, in, h 2 z, = 0 on. Then from emma 4.., we obtain h 2 z, = F, G z, d ξd η, 4.5 where n F, = ω 2k ω 2k ω 2k ω 2 ω 2k 2k ω 2k 2 ω 2k ω 2k ω2k ω 2k ω 2k ω ω2k 2 2k ω 2k ω 2k 2 ω 2k Define a new biharmonic function n z H 0 z, = z zω 2k zω 2k log zω 2k 2 z ω2k z ω 2k z z log ω 2k z 2 zω 2k zω 2k z z log zω 2k 2 z z z ω2k z ω 2k log ω 2k z Then, the next result holds. Theorem 4... For z,, h 2 z, = H 0 z, J 0 z,, where H 0 z, is given by 4.6 and J 0 z, = 2 n ω2k ω 2k log ω 2k 2 ω2k ω 2k log ω 2k 2 n zω d 2k zω 2k zω 2k zω 2k

60 Proof: We can easily verify that z z H 0 z, = F z,, z,. What is more, for z, n z H 0 z, = ω2k z ω 2k zω 2k zω 2k log zω 2k 2 zω 2k zω 2k z ω2k z ω 2k log zω 2k 2, 4.8 and when z 0, ω, that is, z = ρe iθ, 0 ρ, n ρe H 0 z, = iθ ρe iθ ρω 2k ρω 2k log ρω 2k 2 ρω 2k ρω2k ρe iθ ρe iθ log ω 2k ρ 2 ρω 2k ρω 2k ρe iθ ρe iθ log ρω 2k 2 ρe iθ ρe iθ ρω2k ρω 2k log ω 2k ρ 2 = 0. Obviously, we also obtain H 0 z, = 0 for z 0,. Thus, again by emma 4.. and 4.8, the proof is completed. Now we only need to compute the boundary integral in Theorem 4... In fact, we observe that J 0 z, can be converted into an integral on the whole unit circle. That is J 0 z, = 2 n = 2 = 2Re zω 2k log 2 = n zω 2k log 2 zω 2k log 2 2 log zω 2k d. d