Some Basic Bounday Value Poblems fo Complex Patial Diffeential Equations in Quate Ring and Half Hexagon DISSERTATION des Fachbeeichs Mathematik und Infomatik de Feien Univesität Belin zu Elangung des Gades eines Doktos de Natuwissenschaften Este Gutachte: Pof.D.Heinich Begeh Zweite Gutachte: Pof.D.Alexande Schmitt Ditte Gutachte: Pof.D.Vladimi Mityushev Tag de Disputation: 6. Juni 03 Vogelegt von Bibinu Shupeyeva Apil 03
Este Beteue: Pof.D.Heinich Begeh Zweite Beteue: Pof.D.Alexande Schmitt
Contents Abstact Acknowledgments iii v Intoduction Peinaies 3 I Bounday Value Poblems in a Quate Ring Domain 7 Bounday Value Poblems fo the Inhomogeneous Cauchy-Riemann Equation 9. Schwaz poblem......................................... 9.. Schwaz-Poisson epesentation fomula........................ 9.. Schwaz poblem..................................... 4. Diichlet poblem......................................... 7.. Diichlet poblem fo analytic functions........................ 7.. Diichlet poblem fo the inhomogeneous Cauchy-Riemann equation........ 3.3 Neumann poblem fo the homogeneous Cauchy-Riemann equation............. 33 3 Bounday Value Poblems fo the Poisson Equation 39 3. Hamonic Diichlet Poblem................................... 39 3.. Hamonic Geen function and the Geen Repesentation fomula.......... 39 3.. Hamonic Diichlet Poblem............................... 4 3. Hamonic Neumann Poblem.................................. 45 3.. Hamonic Neumann function and the Neumann epesentation fomula....... 45 3.. Hamonic Neumann Poblem.............................. 5 II Bounday Value Poblems fo a Half Hexagon 57 4 Schwaz Poblem fo the Inhomogeneous Cauchy-Riemann Equation 59 4. Desciption of the domain.................................... 59 4. Schwaz-Poisson epesentation fomula............................ 6 4.3 Schwaz poblem fo the inhomogeneous Cauchy-Riemann equation............. 76 5 Hamonic Diichlet Poblem fo the Poisson equation 9 5. Geen epesentation fomula.................................. 9 5. Hamonic Diichlet poblem................................... 96 Bibliogaphy 09 Zusammenfassung 3 i
Abstact This dissetation is devoted to the investigation of some bounday value poblems fo complex patial diffeential equations in a quate ing and a half hexagon. The method of eflection is used among the main tools to obtain the Schwaz-Poisson epesentation fomula and the hamonic Geen function fo both domains. Fo the quate ing the elated Schwaz, Diichlet and Neumann poblems fo the Cauchy-Riemann equation ae solved explicitly. Fom the hamonic Geen function fo this domain the Neumann function is deived satisfying cetain pescibed popeties. By use of the Geen and Neumann functions the coesponding Diichlet and Neumann poblems fo the Poisson equation ae solved. Similaly, using the eflection points the Schwaz-Poisson epesentation fomula is found fo the half hexagon and the solution of the Schwaz poblem fo the Cauchy-Riemann equation is povided. The hamonic Geen function obtained fo this domain allows to solve the elated hamonic Diichlet poblem. Due to the fact that both domains ae non-egula, special attention is paid to the bounday behavio in the cone points. iii
Acknowledgments It was vey lucky to me to meet many good people who played a big ole in my life. I want to thank all of those who helped me in witing this pape. This thesis and my stay in Gemany would not be possible unless Pof.D. Heinich Begeh ageed to be my supeviso. Since the vey beginning he has been so kind and attentive to me. Giving numeous suggestions and sensible advice, he spent a lot of time fo diecting my wok in the ight diection. I appeciate his geat patience to check my papes again and again until the final esults wee composed in this thesis. It was my hono to wok with Pofesso Begeh, a peson with a heat of gold. I expess my deepest gatitude to Pof.D.Alexande Schmitt fo his suppot and contibution. He helped me a lot with my enollment as a PhD student and futhe extension of my scholaship. Thanks to him I had the oppotunity to have a woking place at an office and to do my eseach. I would also like to thank D.Tatyana Vaitekhovich fo he cae and being not only an adviso but a good fiend. My stay in Gemany and studying at Feie Univesität Belin was suppoted by DAAD Geman Academic Exchange Sevice in collaboation with UCA Univesity of Cental Asia and I am vey gateful to them fo the chance to accomplish my eseach aboad. Being fa fom my home, falling sometimes in despai, the thoughts on my paents, family and fiends helped me to ovecome all the difficulties. I also expess them my gatitude fo thei love and belief in me. v
Intoduction A method of solution is pefect if we can foesee fom the stat, and even pove, that following that method we shall attain ou aim Gotfied Wilhelm von Leibniz The theoy of complex bounday value poblems oiginating fom the wok of B.Riemann 35 and D.Hilbet 3 and developed by F.D.Gakhov 8, I.N.Muskhelishvili 33, I.N.Vekua 4, W.Haack and W.Wendland 30 and othes is still investigated. On one hand a theoy fo complex model equations of abitay ode is exploed afte the basic fundamental solutions wee appoached 5. On the othe hand explicit solutions ae found in many diffeent paticula domains. The explicit solutions ae impotant not only fo applications in engineeing, mathematical physics, fluid dynamics etc., but also influence the geneal theoy fo abitay domains. Complex model equations ae simple inhomogeneous equations with a diffeential opeato being the poduct of powes of the Cauchy-Riemann z and anti-cauchy- Riemann opeato z, i.e. z k z l w = f, k, l N. Thee ae thee diffeent basic model opeatos of this kind, the Cauchy-Riemann opeato z, the Laplace opeato z z and the Bitsadze opeato z. The basic bounday value poblems fo complex patial diffeential equations have been consideed fo diffeent paticula domains, see, 3, 4, 5, 6, 7, 8,, 3, 4, 7, 8, 9, 0,,, 5, 6, 7, 9, 3, 34, 4, 43, 44. They ae explicitly solved as well fo the inhomogeneous Cauchy-Riemann equation as fo the Poisson equation. Among these paticula domains the case of the unit disc is mostly consideed and explained in classical textbooks on complex analysis and patial diffeential equations. In paticula, the Schwaz and Poisson kenels fo the unit disc ae used to genealize the concept of the kenel functions and integal epesentations fo solutions. The Geen and Neumann functions fo the unit disc, see e.g 5, 6, ae given explicitly and thei geneal concepts have been developed. The confomal invaiance of these functions allows to calculate the hamonic Geen and Neumann functions fo othe domains, which ae confomal equivalent to the unit disc. The method of confomal invaiance is mainly of theoetical value. In ode to find the Geen and Neumann functions in explicit fom the confomal mapping has to be expessed. Its existence is assued by the Riemann mapping theoem, but a geneal method fo its computation is not available. Fo cetain polygonal domains the Schwaz-Chistoffel fomula povides a method to compute the confomal mapping to the unit disc. Howeve, this fomula is not pope fo pactical poblems since it involves elliptic functions. Polygonal domains which include cone points ae iegula. Bounday value poblems in such domains ae always delicate and the studying of the behavio of solutions in the neighbohood of the cone points is in geneal difficult. Among the iegula domains studied ecently ae those the boundaies of which consist of pieces of cicles and the staight lines. They ae elated to polygonal domains, whee some of the bounday sides ae eplaced by cicula acs e.g. half discs, quate discs, half ings, disc sectos, lenses and lunes, see 8,, 6, 43. Some polygonal domains ae studied in,, 4, 9, 9, 44 etc. Multiply connected domains ae also consideed, e.g. ing domains, in paticula a concentic ing domain 0, 4. Fo cetain of the domains mentioned above a method fo constucting the Schwaz kenel and the Geen and Neumann functions exists. It is well explained e.g. in 0,,. The method uses eflection of the domain at all pats of the bounday. It can be applied if the whole complex plane can
be coveed by continuously epeated eflections, see e.g., 8, 9, 0,, 4, 43, 44. This method is applicable fo a quate ing and a half hexagon and they ae subjects of investigation of the pesent thesis. The main tools fo teating the bounday value poblems fo the Cauchy-Riemann equation is the Cauchy-Pompeiu epesentation fomula. This fomula is to be adjusted to the bounday value conditions. Fo the bounday value poblems fo the Poisson equation the Geen and Neumann epesentation fomulas ae exploited. The method is based on the explicit fom of the Geen and Neumann functions. The solutions of bounday value poblems ae consideed in the distibutional sense. The eason fo this fact is that a paticula solution of the inhomogeneous Cauchy-Riemann equation is given by the Pompeiu opeato which has weak deivatives with espect to z and z in pope functions spaces, see 4 fo the extensive explanation of the opeato. This Pompeiu opeato is a kind of potential opeato of the Cauchy-Riemann opeato. The paticula solutions of the Poisson equation ae given by aea integal opeatos with the Geen and Neumann functions as kenels. These ae also potential opeatos. Othe potential opeatos of this kind fo highe ode model opeatos ae developed fo cetain simple domains and the bounday value poblems ae solved, see 0,, 3, 4. In the pesent thesis the Cauchy-Riemann equation z w = f and the Poisson equation z z w = f in two paticula domains ae studied, namely a quate ing and a half hexagon. The bounday of the quate ing consists of two staight segments and two cicula acs. Reflections at the segments gives the coveing of a ing domain and continued eflections at the bounday cicles of the ing poduces a coveing of the punctued complex plane. Similaly, the half hexagon is eflected to the whole hexagon which povides a paqueting of the whole plane as well. In this way fo both domains the equied kenel functions ae obtained and thus the Schwaz, Diichlet and Neumann poblems ae explicitly solved. Paticula attention is paid to the behavio of the solutions in the cone points. While the aea integal, the Pompeiu opeato, unde suitable assumptions is continuous in the entie plane, the bounday behavio of the bounday integal has to be investigated, especially in the cone points. Thus the Poisson kenel is needed. It tuns out to have diffeent foms on the diffeent pats of the bounday of these domains. It is shown that continuity of the bounday function poduces solutions which behave continuously in these points. In the fist pat of the thesis the bounday value poblems fo the quate ing ae consideed. At fist the Schwaz-Pompeiu epesentation fomula and the Schwaz kenel ae obtained. Fo the inhomogeneous Cauchy-Riemann equation the Schwaz and Diichlet poblems and also the Neumann poblem fo the homogeneous Cauchy-Riemann equation ae solved. Futhe, the hamonic Geen function is found and the solution of the elated Diichlet poblem fo the Poisson equation is pesented. The hamonic Neumann function is constucted by multiplying all the tems fom the Geen function adding some facto in ode to get convegence. By the Neumann epesentation fomula the elated Neumann poblem fo the quate ing is explicitly solved and the solubility condition is explicitly given. The second pat of the thesis is devoted to bounday value poblems fo the half hexagon. By use of the eflection points the Schwaz-Pompeiu epesentation fomula is eached and due to the diffeent behavio on the bounday pats, thee equivalent foms of the epesentation ae pesented and used to solve the Schwaz poblem fo the inhomogeneous Cauchy-Riemann equation. Similaly, the Geen function is constucted with espect to the diffeent bounday pats and the equality of the thee foms obtained is shown. The elated Diichlet poblem fo the Poisson equation is solved and the solution is given in explicit fom.
Chapte Peinaies Let C be the complex plane of the vaiable z = x iy. The eal numbe x is called the eal pat of z and is witten x = Re z. The eal numbe y is called the imaginay pat of z and is witten y = Im z. The complex numbe z = x iy is by definition the complex conjugate of z. A function of the complex vaiable z is a ule that assigns a complex numbe to each z within some specified set D, D is called the domain of definition of the function. The collection of all possible values of the function is called the ange of the function. A cuve γ is a continuous complex-valued function γt defined fo t in some inteval a, b in the eal axis. The cuve γ is simple if γt γt, a t < t b and it is closed if γa = γb. A cuve is smooth if fo γt exists γ t and it is continuous on a, b. A cuve is piecewise smooth if it consists of a finite numbe of smooth cuves and the end of one coincides with the beginning of the next. A domain D in C is called egula if it is bounded and its bounday D is being a smooth cuve. Let the complex-valued function w be defined in D C and let u and v denote its eal and imaginay pats: w = u iv, whee ux, y and vx, y ae eal-valued functions. The two patial diffeential equations u x = v y, u y = v x ae called the Cauchy-Riemann equations fo the pai of functions u, v. The function w is called diffeentiable at each point whee the patial deivatives of u, v ae continuous and satisfy the Cauchy-Riemann equations. A complex-valued function defined in D C diffeentiable at evey point of D is said to be analytic o holomophic in D. The patial diffeential opeatos x and y ae applied to the function w as w x = u x i v x, w y = u y i v y. Defining the complex patial diffeential opeatos z and z by z = x i, y z = x i,. y. then x = z z, y = i z z. The Cauchy-Riemann equation can be witten as w z = 0. As this is the condition fo w to be analytic, then w is independent of z in D C. The complex-valued function w, defined in D C, is said to be hamonic if it is of the class C and satisfies the Laplace equation w x w y = 0. An analytic holomophic function is hamonic. As it is fom C this follows fom u x = x v y = y v x = u y, 3
which poves that u is hamonic. Similaly can be poved fo v to be hamonic and thus w is hamonic. As z z = 4 x y.3 the Laplace equation can be witten as w z z = 0. The following fomulas ae used to solve the bounday value poblems. Definition.0.. 8 Let γ be a smooth closed contou in C and denote the domain within the contou as D, the inteio domain; the exteio domain denoted as D. If wz is an analytic function in D and continuous in D γ, then γ w d wz, z D z =, 0, z D..4 And if wz is analytic in D and continuous in D γ, then γ w d w, z D z =, wz w, z D..5 The integal pesented in the left-hand side of.4 and.5 is the Cauchy integal. Theoem.0.. Cauchy s Theoem 36 Let D be an open subset of C, and let Γ be a contou contained with its inteio in D. Then wzdz = 0 fo evey function w that is holomophic in D. This equality is valid as well fo simply connected domains. The definitions of some classes of functions ae needed 4. Γ Let a function fz and its patial deivatives up to the mth ode be continuous in a domain D. The set of these functions is denoted by C m D and C m D, whee D is the closue of the domain D. By C α D denote the set of all bounded functions fz satisfying the inequality fz fz Hf z z α, 0 < α,.6 fz fz Hf = sup z,z G z z α,.7 whee α is called the Hölde index of the function f. Let a function fz given in the domain D satisfy the inequality D fz p dxdy < M D, p, whee D is an abitay closed bounded subset of the domain D and M D is a constant depending on D. The set of such functions is denoted by L p D. The set of functions satisfying L p f L p f, D = D fz p dxdy /p < is denoted by L p D and is being the Banach type space. Let f C m G, and let thee exist a closed subset G f of the set G, such that f = 0 outside G f. The set of such functions is denoted by D 0 mg. 4
Definition.0.. 4 Let f, g L G. If f and g satisfy the elation g ϕ z dxdy fϕdxdy = 0 G G g ϕ z dxdy G G fϕdxdy = 0,.8 whee ϕ is an abitay function of the class D 0 G, f is said to be the genealized weak deivative of g with espect to z to z. This definition is a paticula case of the geneal one, which can be found in e.g. 38. We take the definition of the Sobolev space as Definition.0.3. 38 The linea manifold of all summable functions ϕx, x,..., x n having on a finite domain D all genealized deivatives of ode l summable to powe p >, ae called W l,p : l ϕ L x α p ; α i = l. xα... xαn n Theoem.0.. Gauss Theoem eal fom 5 Let f, g C D; R CD; R be a diffeentiable eal vecto field in a egula domain D R then f x x, y g y x, ydxdy = fx, ydy gx, ydx. D Theoem.0.3. Gauss theoem complex fom 5, 6 Let w C D; C CD; C in a egula domain D of the complex plane C then w z zdxdy = wzdz,.9 i D D D w z zdxdy = i D D wzdz..0 Theoem.0.4. Cauchy-Pompeiu epesentation fomula Let D C be a egula domain and w C D; C CD; C. Then using = ξ iη fo z D wz = w d z w π dξdη z,. hold. D wz = D D w d z π D w dξdη z. The Cauchy-Pompeiu epesentation fomula fo any function w C D; C CD; C in a bounded domain D of the entie complex plane C with piecewise smooth bounday is complemented with the elation 0 = w d z w π dξdη, z C\D..3 z D Let D be the unit disc defined as D = z C : z < }. The kenel 5 is called the Schwaz kenel and its eal pat D z, z D, D z z z 5
is the Poisson kenel fo the unit disc with a popety poved by H.A.Schwaz 37 γ z, z < z z d = γ, D.4 fo γ C D; R. = The Poisson kenel fo the uppe half plane H = z : Im z > 0} studied in 9 and it is t z, z H, H.5 t z with the popety z t,z H γ t z t z d = γt, z H, t H.6 fo pope γ C H; R. Fo the non-homogeneous Cauchy-Riemann equation w z 4. If f C D then wz = Φz T f, whee Φz = Γ T f = π D wd z, = f, w = u iv the solution is pesented in fdξdη..7 z The integal T f, which is called the Pompeiu opeato, exists fo all points z in C and is holomophic outside G with espect to z and vanishes at infinity. Theoem.0.5. 4 Let D be a bounded domain. If f L D, then T f, egaded as a function of a point z of the domain D, exists almost eveywhee and belongs to an abitay class L p D whee p is an abitay numbe satisfying the condition p < and D is an abitay bounded domain of the complex plane. Theoem.0.6. 4 If f L D, then T f ϕ z dxdy D D fϕdxdy = 0,.8 whee ϕ is an abitay function of the class D 0 G. By Theoem.0.6 the opeato T f D z G if f L G and in the weak sense. T f z = f.9 The method of eflection used in the pesent thesis allows to attain the Schwaz-Pompeiu epesentation fomula and constuct the hamonic Geen function fo a given domain D. As an example, let us conside the uppe half plane H, Im z > 0 with a fixed point, Im > 0. Reflection of at the bounday, i.e. eal axis, gives a point. The function z is analytic in z C and has a zeo at z =. The diffeence log z log z is a hamonic function in z H fo any, except fo the case z =, and it vanishes on the bounday. Then, this diffeence log z log z, satisfying the popeties of the Geen function, is the hamonic Geen function fo the uppe half plane. The method is applicable to cetain egula and non egula domains if the eflection povides a paqueting of the complex plane. 6
Pat I Bounday Value Poblems in a Quate Ring Domain 7
Chapte Bounday Value Poblems fo the Inhomogeneous Cauchy-Riemann Equation In this Chapte, the Schwaz-Poisson epesentation fomula is obtained in a quate ing domain and the Schwaz and Diichlet poblems as well fo the inhomogeneous as the Neumann poblem fo the homogeneous Cauchy-Riemann equation ae solved explicitly.. Schwaz poblem Let R be the uppe ight quate ing domain see Fig. in the complex plane C defined by R = z C : < z <, Re z > 0, Im z > 0}. The bounday R is piecewise smooth and oiented counte-clockwise. It is non-egula and contains fou cone points,, i, i. Thus the Cauchy-Pompeiu fomula holds fo pope functions. Fo attaining the Poisson kenel the quate ing is eflected acoss its bounday pats on the eal and the imaginay axes to the entie ing. The ing itself is epeatedly eflected acoss its two bounday cicles such that the entie punctued plane is eached. Applying the Cauchy-Pompeiu epesentation fomula in the points whee z fom the quate ing is mapped to, leads to the adjusted Schwaz-Poisson fomula obtained by some modifications. The latte enables to solve the elated Schwaz poblem... Schwaz-Poisson epesentation fomula To get the Cauchy-Pompeiu epesentation due to the bounday conditions a meomophic kenel function is constucted as follows. A point z R chosen to be a simple pole is eflected acoss the pats of the bounday: z =, Re z > 0, Im z > 0}; z =, Re z > 0, Im z > 0}; 0 < Re z <, Im z = 0}; 0 < Im z <, Re z = 0}. The diect eflection of the pole gives zeos. The eflected points ae, z, z, z z. These zeos in tun ae also eflected at the pats of the boundaies z =, z =, z =, z = and these eflection points z, z, z, z, z, z i i Z Fig.: Quate ing 9
as the diect eflection of the zeos become poles. Continuing this pocess, the new points can be expessed in geneal fom as ±z n, ±z n, ± z n, ± z, ±n n z, ±n z, ± z n, ±,. zn whee ±z n, ± z, ± n n z, ± z ae poles and ±z n, ± z n, ± n n z, ± z n 8 the Cauchy-Pompeiu fomula is to be modified. ae zeos. As it was done in Theoem... Any w C R ; C CR ; C fo the domain R C can be epesented as wz = } w z n n z z d n z π R R n= n= } w z n n z z n dξdη z. and wz = =, 0<Im, 0<Re =, 0<Im, 0<Re z Re w z z z 4n 4n z z 4n z z n= 4n z 4n z z Re w z z z 4n n= 4n z z 4n z z t t z tz t z 4n z 4n z } d } d Re wt } 4n t 4n t z z t 4n z t tz 4n z t t 4n z t dt n= t Re wit t z tz t z.3 } 4n t 4n t z z t 4n z t tz 4n z t t 4n z t dt n= Im w d π z w π z z =, 0<Im, 0<Re n= R 4n 4n z z 4n z z 4n z z 4n n= z w z z } 4n 4n z z 4n z z 4n z dξdη z 4n 0
Poof. Fom the Cauchy-Pompeiu epesentation fomulas.,.3 and the elation 0 = w n n z z d n z the equality π R wz = π R R R n= follows. Substituting the eflection points n= w n n z z n dξdη, z R z } w z n n z z d n z n= n= } w z n n z z n dξdη, z R z.4.5 z, z, z, z, z, z, z in.3 and.4 gives the following equalities 0 = } w z n n z z d n z π R R 0 = π R n= n= } w z n n z z n dξdη, z R. z R 0 = π R } z w z n z n z d z n n= } z w z n z n z z n dξdη, z R. R 0 = π R n= } z w z n z n z d z n n= } z w z n z n z z n dξdη, z R. R 0 = π R n= } w z n n z z d n z n= n= } w z n n z z n dξdη, z R. z R } w z n n z z d n z n= n= } w z n n z z n dξdη, z R. z.6.7.8.9.0
0 = π R R 0 = π R } z w z n z n z d z n n= } z w z n z n z z n dξdη, z R. R n= } z w z n z n z d z n n= } z w z n z n z z n dξdη, z R. n=.. To simplify the following calculations the evaluations above ae composed, namely.5 and.6,.7 and.8,.9 and.0,. and.. Then wz = } w z 4n 4n z z d 4n z R n=.3 } w π z 4n 4n z z 4n z dξdη, z R. R n= 0 = π R 0 = π R R z } w z 4n z 4n z d z 4n n= } z w z 4n z 4n z z 4n dξdη, z R. R w n= z n= 4n 4n z z 4n z } d } w z 4n 4n z z 4n z dξdη, z R. n=.4.5 0 = π R R z w z 4n n= z 4n z z 4n } d z } w z 4n z 4n z z dξdη, z R. 4n n=.6 Taking the complex conjugation of.5 and.6 and consideing the integals ove the diffeent bounday pats give thei new foms 0 = } w z 4n z 4n z d 4n z =, 0<Im, 0<Re =, 0<Im, 0<Re w n= 4 z 4 n= 4n 4 z z 4n 4n z 4 } d
π 0 = R wt t t z n= 4n t t 4n z z t 4n z } dt t } dt t } t wit t z 4n t t 4n z z t 4n z n= w z 4n 4n z z 4n z =, 0<Im, 0<Re =, 0<Im, 0<Re R w n= z z n= 4n z z 4n 4n z dξdη. } d 4 z w z 4 4n 4 z z 4n 4n z 4 n= t z wt t z 4n t z t 4n z t z 4n n= t z wit t z 4n t z t 4n z t z 4n n= z w π z 4n z n= } dt t } dt t 4n z z 4n } dξdη. } d.7.8 Then, subtacting.8 fom.3,.7 fom.4 and adding.8 and.3,.7 and.4 lead to wz = b b b 3 b 4 b 5 b 6, whee b b = =, 0<Im, 0<Re 4n n= = =, 0<Im, 0<Re 4n n= z Re w z z z 4n z z 4n z z z Re w z z z 4n z z 4n z z 4n z 4n z 4n z 4n z } d ; } d ; b 3 = t Re wt t z tz t z } 4n t 4n t z z t 4n z t tz 4n z t t 4n t z dt; n= 3
b 5 b 4 = = t Re wit t z tz t z } 4n t 4n t z z t 4n z t tz 4n z t t 4n t z dt; n= =, 0<Im, 0<Re w z z z 4n 4n z z 4n z z 4n z z 4n π 4n n= =, 0<Im, 0<Re 4n z z Im w z d =, 0<Im, 0<Re n= z w z z 4n z z 4n z z 4n z z z z d z = π } d = =, 0<Im, 0<Re Im w d ; b 6 = w π z z z R 4n 4n z z 4n z z 4n z z 4n n= w z z z } 4n z 4n z z 4n 4n z z dξdη. 4n z n= Hence, adding all of these fomulas fo b,..., b 6 the epesentation fomula.3 is obtained. Theoem.. enables to solve the elated Schwaz poblem... Schwaz poblem In ode to solve the Schwaz poblem on the basis of Theoem.., the bounday behavio of the bounday integal is studied. Let the pats of the bounday be denoted by R = =, Re 0, Im 0}; R = =, Re 0, Im 0}; 3 R = = t : t }; 4 R = = it : t } and new functions be expessed by K z, = z 4n n= z 4n z z 4n z z 4n z z z 4n z z 4n z z 4n z.9 4n z 4n z }, 4
K z, = z z 4n 4n z n= z 4n z z z 4n z z z z 4n z z z 4n z 4n z 4n z 4n z.0 }. Theoem... If γ C R ; C, then fo 0,, i, i γk z, d = γ 0, 0 R R,. z R,z 0 R R γk z, d = γ 0, 0 3 R 4 R,. z R,z 0 3R 4R whee K z, and K, z ae given in.9,.0 espectively. Poof. Studying the bounday behavio of the bounday integal implies computations on the diffeent pats of the bounday R. Taking the bounday elations into account, the following sum is obtainable easily γk z, d R R γ z =, 0<Im, 0<Re γk z, d = 3R 4R z z z 4n n= 4n z 4n z z 4 4n z 4 z z 4 4n z 4 z } d z 4 4n z 4 z z 4 4n z 4 z 4n z 4n z γ z z =, 0<Im, 0<Re z z 4n 4 n= 4n z 4 4n z z 4 4n z 4 z z 4 4n z 4 z } z 4 4n z 4 z z 4 4n z 4 z 4 4n z 4 d 4n z z z t γt t z t t z } 4n t dt n= z z 4n n= 4n t z t 4n z t t 4n z t t 4n t z t γit t z t t z } t 4n z t t 4n z t t 4n t z t 4n t z dt..3 5
Let 0 =, Re 0 > 0, Im 0 > 0, z R, then γk z, d z 0 R R γ z 0 z =, 0<Im, 0<Re =, 0<Im, 0<Re γ 4n n= 3R 4R z γk z, d } = } z z d 4n 0 4n 0 0 4 4n 0 4 0 0 4 4n 0 0 4 4n 0 4 0 0 4 4n 0 4 0 4n 0 4n 0 } γ 0 0 d 0 0 =, 0<Im, 0<Re z 0 4n =, 0<Im, 0<Re γ 4n z z z z 0 4n n= z z z 0 4n n= n= 4n 4n z z 4 4n z 4 4n z γt 4n 4n z z 4 4n z 4 4n z t t z t t z } d 4n 4n z z 44n z 4 4n z } z 4 4n d z 4 4n z } t 4n t z t 4n z t t 4n z t t 4n t z dt t γit t z t t z } t 4n z t t 4n z t t 4n t z t 4n t z dt. Since fo 0 =, Re 0 > 0, Im 0 > 0 the elations t 0 = t 0 0, 4n t 0 = 4n t 0 0, 4n 0 t = 4n t 0 0, t 0 = t 0 0,.4 4n t 0 = 4n t 0, 4n 0 t = 4n t 0 ae valid, then one has to calculate γk z, d = z 0 R R γ z 0 z =, 0<Im, 0<Re =, 0<Im, 0<Re γ 0 0 z 0 z 0 } z d 4n 4n 0 4 4 0 n= 6
4n 4n 0 4n 4 4 0 4n 4n 0 4n 0 γ z 0 z =, 0<Im, 0<Re 4n 4n 0 4 4 0 4n 4n 0 z 4n 4n 0 z z Finally, decomposing the integal so that the entie unit disc D is eached, γk z, d = γ z 0 z 0 z R =, 0<Im, 0<Re =, 0<Im, 0<Re z z z z γ z 0 z z d =, 0>Im, 0<Re γ z z d =, 0<Im, 0>Re =, 0>Im, 0>Re γ the popety.4 of the Poisson kenel fo D is used. Thus γk z, d = Γ z 0 z 0 R = z z γ z 4n 4n 0 4n 4n 0 } d. z } z z 4 4 0 } d = d = z z d, d z d = γ 0,.5 whee γ, Re 0, Im 0 γ, Re 0, Im 0 Γ = γ, Re 0, Im 0 γ, Re 0, Im 0.6 is a continuous function on =. Similaly, fo 0 =, Re 0 > 0, Im 0 > 0, z R z 0 z 0 γk z, d R R γ z =, 0<Im, 0<Re z 0 n= z 4n 4n z z 4n z z 4 z 3R 4R z γk z, d z z 4n z z 4 z 4n z z 4 z z 4n z z 4n z z 4n z } d } = 4n z z 4 7
then z 0 z 0 =, 0<Im, 0<Re γ z z z 4 4n 0 4 4n 0 4 4n 0 4 } z d 4n 4 n= 4n 0 4 4n 4 0 4n 4 0 4n 0 d z z z 0 4n n= z z z 0 4n n= R R γt t t 0 t t 0 4n 0 } t 4n t 0 t 4n 0 t t 4n 0 t t 4n t dt 0 t γit t t t } t 4n t t 4n t t 4n t t 4n t dt, γk z, d = z 0, z R whee Γ is defined as in.6 fo =. then then = Γ z z d = γ 0,.7 Fo Re 0 > 0, Im 0 = 0, 0 = t 0, < t 0 < afte simila computations one gets z z t γk z, d = γt z 0 z t0 t z t t z Hee z 0 z 0 z 0 n= 3R 4R } 4n t 4n t z t 4n z t t 4n z t t 4n t z dt. 4n t t 0 0, 4n t 0 t 0, 4n t 0t 0, 4n t t 0 0, 3R 4R z z z z γk z, d = z t0 z z γt t z t z dt z z dt γt t z dt γt t z z z dt γt z 0 t z dt γt t z z 0 3R 4R z z γk z, d = z 0 8 } γt t t z t t z dt = } γt tz tz dt = γ t dt t z γ t dt t z } γ t dt t z, } γ t dt t z = Γ t z z t z dt = γ 0,.8
whee γt, t γ t, t Γ t = 0, < t <, t > γt, t γ t, t.9 is piecewise continuous on R. Finally, fo Re 0 = 0, Im 0 > 0, 0 = it 0, < t 0 < } γk z, d γk z, d = z 0 R R 3R 4R z z t γit z 0 t z t t z } 4n t 4n z t t 4n z t t 4n t z t 4n t z dt. Since z 0 z 0 z 0 R ir z z π z z π n= t t 0 0, t t 0 0, 4n t t 0 0, 4n t 0 t 0, 4n t 0 t 0, 4n t t 0 0, γk z, d = z it 0 z z γit it z it z dt z z π dt γit it z dt γit it z z z dt γit z 0 π it z dt γit it z } γit t t z t t z dt = } γit itz itz dt = z z π γ i t dt t iz γ it dt it z } γ it dt it z, } γ i t dt t iz = then z 0 R ir γk z, d = z 0 π Γ it z z it z dt = γ 0.30 whee γit, t γ it, t Γ it = 0, < t >, t > γit, t γ it, t.3 is piecewise continuous on ir. Hence, the equalities.,. ae valid fo γ C R ; C. 9
The following lemma enables to see the bounday behavio at the cone points of R. Accoding to the epesentation fomula.3 the function wz can be pesented by = z z z z 4n 4n z 4n z 4n z 4n z 4n z 4n 4n z =, 0<Im, 0<Re =, 0<Im, 0<Re z z z z t t z tz t z n= t t z tz t z n= Lemma... If γ C, ; C, then z R, z t 0 z R, z t 0 whee K z, is given in.0. n= 4n 4n z 4n z 4n 4n z 4n 4n 4n z n= 4n t 4n t z 4n z t 4n z t 4n tz 4n z t 4n t 4n t z 4n z t 4n z t 4n tz 4n z t 4n t 4n z t } d } d.3 } dt } 4n t 4n z t dt. γ γk z, d = γt 0 γ, t 0 3 R \},.33 γ γk z, d = 0, t 0 R \,,.34 Poof. Taking the eal pat of.3 and multiplying the both sides of the equality by γ give γ = =, 0<Im, 0<Re γk z, d =, 0<Im, 0<Re γk z, d γk z, tdt Let t 0 be fom the bounday pat 3 R. We conside the diffeence γ γk z, d γ γk z, d z t 0 =, 0<Im, 0<Re γ γk z, tdt n= =, 0<Im, 0<Re γ γk z, itdt. It is seen that fo t 0 3 R we have to calculate only the following bounday integal t γ γk z, d = γt γ z t 0 z t0 t z t t z tz t z tz t z 4n t 4n t z t 4n t z z t 4n z t } z t 4n z t tz 4n z t tz 4n z t t 4n t z t 4n t z dt = z t 0 Λ t z z t z dt, t 0,, z R, 0 γk z, itdt..35
whee Λ t is defined on the basis of calculations.8 and Γ t given in.9 and it is γt γ, t γ t γ, t Λ t = 0, t, t γt γ, t γ t γ, t.36 Because Λ t defined on R is continuous in -;, the elation.33 holds. Thus, in paticula z γk z, d = γ.37 follows. If t 0 R \,, then the bounday integal tends to 0 fo z t 0 and then the equality.34 holds. Lemma... If γ Ci, ; C, then z R, z it 0 z R, z it 0 γ γik z, d = γit 0 γi, t 0 4 R \},.38 γ γik z, d = 0, t 0 R \ 4 R..39 Poof. Similaly as befoe, the bounday integal ove the inteval, of the imaginay axis emains z it 0 γ γik z, d = z it 0 π 4n it it γit γi t z it t z itz t z itz t z 4n t z it 4n t z z it 4n z t n=.40 } z it 4n z t itz 4n z t itz 4n z t it 4n t z it 4n t z dt = z it 0 π Λ it z z it z dt, t 0,, z R, whee Λ it is obtained on the basis of.30 and Γ it given in.3 and it is γit γi, t γ it γi, t Λ it = 0, < t <, t > γit γi, t γ it γi, t..4 Because Λ it defined fo t R is continuous in -;, the elation.38 holds. Thus, in paticula z i follows. Finally, if it 0 R \ 4 R, then K, z tends to 0 fo z it 0. γk z, d = γi.4
Theoem..3. The Schwaz poblem w z = f in R, f L p R ; C, p >, Re w = γ on R, γ C R ; C, π π 0 Im we iϕ dϕ = c, c R.43 is uniquely solvable in the space of functions with genealized deivatives with espect to z by wz = z γ z z z =, 0<Im, 0<Re 4n 4n z z 4n z z n= 4n z 4n z z γ z z z =, 0<Im, 0<Re } d } 4n 4n z z 4n z z n= 4n z d 4n z t γt t z tz t z } 4n t 4n t z z t 4n z t tz 4n z t t 4n z t dt n= t γit t z tz t z.44 } 4n t 4n t z z t 4n z t tz 4n z t t 4n z t dt ic n= f π z z z R 4n 4n z z 4n z z 4n z 4n z n= f z z z 4n 4n z n= z 4n z } z 4n z 4n dξdη z Poof. Theoem.. povides epesentation foms fo any smooth enough function. Theefoe if the Schwaz poblem has a solution, it has to be of the fom.44 with an analytic function on the ight-hand side up to the Pompeiu-type opeato 4, 5. Let us denote by T R fz = f π z z z R 4n 4n z z 4n z z 4n z 4n z.45 n=
f z z z 4n 4n z n= n= z 4n z } z 4n z 4n dξdη, z which can be easily epesented by T R fz = dξdη f π z f π z z z R R 4n 4n z z 4n z z 4n z 4n z f z z z 4n 4n z n= z 4n z } z 4n z 4n dξdη. z.46 Thus, by the popety of the Pompeiu opeato z T f = f, the opeato T R fz povides a solution of the Cauchy-Riemann equation w z = f in a weak sense. Then the side condition and bounday behavio ae to be checked. Conside the poof piece by piece. Pat. Side condition. At fist the bounday integal has to be studied. Fo =, Re > 0, Im > 0 z z z dz z z = z z z dz z z = The sum z =, 0<Im z, 0<Re z z =, 0<Im z, 0<Re z z =, 0<Im z, 0<Re z z =, 0<Im z, 0<Re z z z z z z =, 0<Im z, 0<Re z dz z z z dz z = z =, 0>Im z, 0>Re z 4n n= dz dz z z =, 0<Im z, 0<Re z z =, 0>Im z, 0<Re z z =, 0<Im z, 0<Re z 4n z z is divided and two integals ae calculated: n n z n= n= n z =, 0<Im z, 0<Re z z = n z dz n z z = n= z z dz z z =, 0<Im z, 0>Re z 4n z z dz z z = dz z =, 0<Im z, 0<Re z dz z = } dz = z 4n z 4n z z =, 0<Im z, 0<Re z z n z n z dz = dz z n z n= z = } dz = z dz z = 0. = 0 3
and n= n= n z =, 0<Im z, n 0<Re z z = z n z z dz n z z = 0. z dz n z z z =, 0<Im z, 0<Re z z n z z n z } dz = z Calculating the bounday integal on =, Re > 0, Im > 0 gives z z 4z 4 z 4z 4 z z =, 0<Im z, 0<Re z 4n z n= 4n z z 4 π z =, 0<Im z, 0<Re z n= z =, 0<Im z, 0<Re z z =, 0<Im z, 0<Re z 4n z 4n z Im n= Im n= Fo < t <, = t t t z z =, 0<Im z, 0<Re z z =, 0<Im z, 0<Re z dz t z 4nz 4n z 4n z 4n z z z z 4nz 4n z 4n z 4n z 4n z 4n z 4n z 4n z z n z z z n z } dz z tz dz t z z = z =, 0<Im z, 0<Re z z =, 0>Im, 0>Re dz t z z The tems in the sum ae ewitten as 4n n= n= n z =, 0>Im z, 0<Re z z =, 0<Im z, 0<Re z z =, 0<Im z, 0<Re z dz n t z z z =, 0>Im z, 0<Re z t 4n t z t 4n z t dz n t z z z =, 0>Im z, 0<Re z z =, 0>Im z, 0>Re z dz n t z z 4nz 4n z 4nz 4n z dz z = π = n= z =, 0<Im z, 0<Re z π n Im t dz t z z dz z t z dz z dz n t z z z =, 0>Im z, 0<Re z 4 z =, 0<Im z, 0>Re z z =, 0<Im z, 0<Re z z =, 0<Im, 0<Re } dz z = } dz z = Im n= z = z =, 0<Im z, 0<Re z z t z n z dz z n = 0. } t dz z = t z } dz = z z dz n z z z = t 4n t z t 4n z t dz t n tz z dz t n tz z z =, 0>Im z, 0<Re z z =, 0>Im z, 0>Re z t n tz dz z t z = 0. } dz = z dz t n tz z } dz = z
n n= Fo < t <, = it similaly t t z and the sum is z =, 0<Im z, 0<Re z z =, 0<Im z, 0<Re z 4n n= z = t dz t iz z z =, 0<Im z, 0<Re z t 4n t iz dz z n t z dz t n tz z = 0. tz dz t z z = z =, 0<Im z, 0<Re z z =, 0<Im z, 0<Re z t dz iz t t t iz tiz dz t iz z = z = π t 4n t iz t 4n iz t } = n= n z = Teating the side condition fo the aea integal, we obseve z π z dz z z = π z =, 0<Im z, 0<Re z π z =, 0<Im z, 0<Re z π π z =, 0<Im z, 0<Re z z = z z dz z z z dz z dz z π z =, 0>Im z, 0>Re z z =, 0<Im z, 0<Re z dz z z = 0 fo < <. z = dz z dz z it z = 0 z =, 0<Im z, 0<Re z iz n t t n itz z =, 0<Im z, 0<Re z z z z z =, 0>Im z, 0<Re z dz z z t 4n iz t dz z = 0. dz z z = dz z = z =, 0<Im z, 0>Re z z } dz = z Integating the tems of the fist sum unde the aea integal, it is similaly educed to the integal on the whole unit cicle 4n 4n z z 4n z z 4n z dz 4n z z = n= z =, 0<Im z, 0<Re z n= 4n n= n z =, 0<Im z, 0<Re z z =, 0<Im z, 0<Re z 4n z 4n z dz n z z z =, 0>Im z, 0>Re z dz n z dz z z z =, 0<Im z, 0<Re z z =, 0<Im, 0<Re 4n z 4n z dz n z z z =, 0>Im z, 0>Re z } dz = z dz n z z 5
z =, 0>Im z, 0<Re z n= n dz n z z z = Similaly, fo < < π π π z =, 0>Im z, 0<Re z dz n z z n z n z z =, 0<Im z, 0<Re z z =, 0>Im z, 0>Re z z =, 0>Im z, 0>Re z z z z z z =, 0>Im z, 0<Re z dz z = 0. dz z = π dz z z = π z dz z z = π z = dz n z z z =, 0>Im z, 0>Re z z =, 0>Im z, 0>Re z And the complex conjugation at sum unde the aea integal is 4n π 4n z z 4n z π π z =, 0<Im z, 0<Re z z =, 0>Im z, 0>Re z z =, 0>Im z, 0>Re z n= n= z =, 0>Im z, 0<Re z n z z z dz z z = dz z z = 0. z dz z z z 4n z 4n z 4n 4n z z 4n z z 4n z 4n z 4n 4n z 4n z n= 4n z dz z = π n= n z = Hence, fom Pat the following computations z =, 0<Im z, 0<Re z dz z π z =, 0>Im z, 0>Re z n z n z wz dz z = c π z =, 0<Im z, 0<Re z dz z, n= dz z = dz z = 4n 4n z dz z = 0. } dz = z follow. z =, 0<Im z, 0<Re z Im wz dz z = c, π π 0 Im we iϕ dϕ = c. Pat. Bounday behavio. Let s denote the aea integal, then Re s z = } fk z, fk z, dξdη, π R 6
whee K z, is given in.0. The eal pat of the aea integal tuns on to be zeo on the bounday R as can be veified by simple calculations on the bounday pats: z =, Re z > 0, Im z > 0} and z =, Re z > 0, Im z > 0}, z = z, Re z > 0, Im z = 0} and z = z, Re z = 0, Im z > 0}. Let s denote the sum of the bounday integals. Then Re s z = γk, z d =, 0<Im, 0<Re γtk t, zdt =, 0<Im, 0<Re γk, z d γitk it, zdt. To complete the poof one has to study the elations fo the Schwaz opeato fo the unit disc D and the half plane 5 Re Sϕ = ϕ on D, i.e. z Sϕz = ϕ, D. By the Theoem.., this bounday condition holds fo the domain R. The bounday behavio at the cone points is also studied in Lemmas.. -.... Diichlet poblem Using the Cauchy fomula fo the analytic functions in D and the esults of the pevious section, the Diichlet poblem is now solved as well fo the homogeneous as fo the inhomogeneous Cauchy-Riemann equations... Diichlet poblem fo analytic functions Theoem... The Diichlet poblem w z = 0 in R, w = γ on R, γ C R ; C.47 is uniquely solvable if and only if γ z z z R 4n z 4n z z 4n 4n z z n= and the solution can be pesented as wz = γ z z z R 4n n= 4n z 4n z z 4n z z 4n z z 4n } d = 0, } d, z R.48.49 Poof. At fist the condition in.48 is shown to be necessay. Let w defined by.49 be a solution to the Diichlet poblem. This fomula can be decomposed into the sum of Cauchy type integals. Then the equality wz = γ fo any z,z R R,.50 7
holds. Conside fo z < the function.49 at the eflected point z w z = γ z z z n= R 4n z 4n z z 4n 4n z z 4n z } d.5 and take the diffeence wz w z = γ z z z z z z R 4n 4n z z 4n z z 4n z z 4n.5 n= z 4n z z 4n 4n z z 4n z } d. Let L, z denote the sum of integands, then wz w z = γl, z R 4n n= z d = 4n z z 4n z z 4n γ z =, 0<Im, 0<Re 4n n= 4n z z =, 0<Im, 0<Re γ z 4n z z 4n z z 4n z z 4n z z 4n } d z z 4 z 4 z z 4 z 4 z z z 4n z z 4 4n z 4 z z z 4 4n z } d z 4 4n z 4 z z z 4 4n z 4n z z 4n z t γt t z 4n n= t 4n t z z t t z t z 4 t z 4 z t z 4 t z 4 z t 4n z t z 4 4n t z 4 z z t z 4 4n z } dt t t z 4 4n t z 4 z z t z 4 4n z t 4n t z z t 4n z t γit t z 4n n= t 4n t z z t t z t z 4 t z 4 z t z 4 t z 4 z t 4n z t z 4 4n t z 4 z z t z 4 4n z } dt t. t z 4 4n t z 4 z z t z 4 4n z t 4n t z z t 4n z z z 8
Calculating the bounday integals sepaately, letting z, z R, R, it is seen that solely the integal on the oute cicle is to be studied. wz w z, z < z = γ z, z < z z z z z z and 4n 4n z n= 4n z z 4n 4n z z 4n 4n z wz w z, z < z = =, 0<Im, =, 0>Im, 0>Re =, 0<Im, 0>Re γ γ z z whee Γ is given in.6. Then Conside fo z > the function w z = γ R 4n n= 0<Re z d =, 0<Im, 0<Re γ 4n z z4n 4n z 4n z z4n 4n z z z d =, 0>Im, 0<Re z d = z γ z w z = 0. } d z z d =.53 Γ z z d = γ, z z 4 4 4 z.54 z 4n z 4 4 z 4 4n 4n z z 4 4n z } d, which can be easily ewitten as z γ z 4 4 4 z R } 4n z 4n z z 4n z 4n z d 4n 4n z = n=0 n= z γ z 4 4 4 z z z z z 4 z 4 4 4 z R 4n z z 4n z z 4n z dz z 4n z = n= z γ z z z R } 4n 4n z z 4n z z 4n z d z 4n. n= 9
Thus w z = w z. Then the diffeence wz w z = R γ 4n n= z z z z z 4 4 4n z z 4 z 4n z z 4n z z 4n z 4n z 4 4 z 4 4n 4n z z 4 4n z is educed to the diffeence wz w z in.5. If z, =, Im > 0, Re > 0, then, similaly to.53 wz w z, z > z = z, z > = Γ whee Γ is defined as in.6 fo =. Then again z w z = 0. z z d } d = γ,.55 Conside next fo Re z > 0, Im z > 0, z < the function wz = γ z z z R 4n n= 4n z z 4n z z 4n z z 4n } d,.56 which is equal to w z. Then again and taking z, < t <, = t the equality n= wz wz = wz w z = γl, z R z z t wz wz = γt z t z t t z t t z } 4n t 4n t z t t 4n z t 4n t z t t z 4n dt.57 = z t γt follows, whee Γ t is detemined in.9. Thus γ t tz zz z t z dt = t z z t wz = γt, < t <. z t Conside now fo Re z > 0, Im z > 0, z < the function wz = γ z z z R 4n n= 4n z z d 4n z z 4n z z 4n 30 Γ t z z dt = γt t z } d = wz.58
then similaly wz wz = wz w z = γl, z d..59 R Letting z, = it, < t <, one gets z z t wz wz = γit z it z it t z t t z } 4n t 4n t z t 4n t z t t 4n z t t z 4n dt =.60 n= γit γ i z it t tz zz z it z dt = Γ it z it z z dt = γit, z it π it z whee Γ it is given in.3. Hence, wz = γit, < t <. z it Denote fo z R the function φz = w z = wz = wz = w z. Fom the fomulas.53,.55,.57,.60 the equality wz φz = γ, z R.6 follows and thus, because of.50, the elation φz = 0, R is valid. Because φz is an analytic function in R then fom the maximum pinciple fo analytic functions it follows that wz = 0 in R, which is given as condition.48. On the othe hand, if this solubility condition is valid, then, subtacting.48 fom.49, the elation w = γ on R follows fom.6... Diichlet poblem fo the inhomogeneous Cauchy-Riemann equation Theoem... The Diichlet poblem w z = f, z R, f L p R, C, < p, w = γ, γ C R, C.6 is solvable if and only if fo z R γ z z z R } 4n z 4n z z 4n 4n z z d 4n z = n=.63 f π R 4n n= z z z z 4n z z 4n The solution is unique and can be epesented as } 4n z z 4n z dξdη. 3
wz = γ z z z R } 4n 4n z z 4n z z 4n z d z 4n n=.64 f π z z z R } 4n 4n z z 4n z z 4n z z 4n dξdη, z R. n= Poof. By the Cauchy-Pompeiu epesentation fomula, if a solution exists, it must have the fom of.64. Let Then conside the homogeneous Diichlet poblem ϕ = w T f in R, ϕ = γ T f on R, then w z = f is ϕ T f z = ϕ z f and thus ϕ z = 0. ϕ z = 0 in R, ϕ = γ T f on R..65 By the solvability condition.48 the new condition fo.65 is γ T f z z z R } 4n z 4n z z 4n 4n z z d 4n z = 0. n=.66 Calculation of the successive integal T f z z z R 4n z 4n z z 4n 4n z z 4n z n= f π z z z R R } 4n z 4n z z 4n 4n z z 4n z n= f π z z z R 4n z 4n z z 4n n= } d = d d ξd η = } 4n z z 4n z dξdη educes.66 to the coesponding fomula.63. That.64 unde the condition.63 povides a solution to.6 follows because.64 can be witten as wz = d γl, z fl, z π dξdη, R R 3
whee L, z is defined as the tem in.5. Calculations of the integals on the diffeent pats of the bounday in the manne of the poblem.47 show that fl, z π dξdη R tends to 0 when z, R and wz γz if z 0 R, z R. The popety of the Pompeiu opeato gives the weak solution of the diffeential equation w z = f, z R..3 Neumann poblem fo the homogeneous Cauchy-Riemann equation The Neumann poblem fo an analytic function in R means to find a function with pescibed outwad nomal deivatives. In this section this Neumann condition is modified in a pope way in ode to adjust it to the concept of analyticity. Definition.3.. Let on R \, i, i, } denote z z z z, z =, Re z > 0, Im z > 0, ν = z z z z, z =, Re z > 0, Im z > 0, i z z, Re z > 0, Im z = 0, z z, Re z = 0, Im z > 0..67 Then the Neumann poblem is to find an analytic function w in R, such that ν wz = γz on R \, i, i, } fo given γ C R ; C, w = 0. Theoem.3.. The Neumann poblem whee w z = 0, z R, νz w = γ on R, γ C R ; C, w = 0 zw, z =, Re z > 0, Im z > 0, ν w = zw, z =, Re z > 0, Im z > 0, iw, Re z > 0, Im z = 0, w, Re z = 0, Im z > 0,.68.69 is solvable if and only if fo z R γ =, 0<Im, 0<Re z z z 4n z 4n z z 4n n= γ z z z =, 0<Im, 0<Re 4n n= z 4n z z 4n 33 } 4n z z d 4n z } 4n z z d 4n z
whee t γt π t z tz t z 4n tz 4n t z tt z 4n π n= n= γit it t z itz t z 4n itz 4n t z itt z 4n n= δ z z z R 4n z 4n z z 4n } t 4n t z z tt 4n z dt γ, z =, Re z > 0, Im z > 0, γ, z =, Re z > 0, Im z > 0, δ = γ, Re z > 0, Im z = 0, γ, Re z = 0, Im z > 0. } it 4n t z z itt 4n z dt = } 4n z z 4n z ds = 0,.70.7 Then the solution is wz = z π R γ ds π R γ n log n z n z n n z n n log z n n ds Poof. If w is a solution to the Neumann poblem, then ϕ = w is a solution to the Diichlet poblem ϕ z = 0 in R, ϕ = w on R,.7 whee fom.7 w is pesented by zγz, z =, Re z > 0, Im z > 0, w z γz, z =, Re z > 0, Im z > 0, z = iγz, Re z > 0, Im z = 0, γz, Re z = 0, Im z > 0. Then the solution of the Diichlet poblem is w z = γ =, 0<Im, 0<Re 4n n= =, 0<Im, 0<Re 4n n= z z z 4n z z 4n z z 4n z z 4n γ z z z 4n z z 4n z z 4n z z 4n } d } d.73.74 34
t γt π t z t z t z } 4n t 4n t z z t 4n z t z 4n t z dt t z 4n t n= t γit π t z t z t z } 4n t 4n t z z t 4n z t z 4n t z dt t z 4n t. n= Let wz = w z w z w 3 z w 4 z w, whee w = w w w 3 w 4 = 0, then w z w = z =, 0<Im, 0<Re γ z z z z 4n n z n z n n= z n z z n z n z γ z z n= =, 0<Im, 0<Re n n z n z z n z z z n z z n z n n z n n z n z n z n z n z n } d dz = n z n n n z n n z } d dz and w z = =, 0<Im, 0<Re γ log z z z log z z n log n z n n n z log n z n n n z n= n log n z n n n z log n z n n n z } d w..75 Similaly w z w = z n= =, 0<Im, 0<Re n n z zn n z γ z n n z zn n z z n z n z n z n z z n z n z n z n z z } d dz 35
and w z = =, 0<Im, 0<Re γ log z z z log z z n log n z n n n z log n z n n n z n= n log n z n n n z log n z n n n z } d w..76 Next z w 3 z w 3 = π n= γt n t n t z tzn t n z t t z t t z n t n t z tzn t n z tz tz n z t n z n tz n tz tz n z t n z n tz n } dt t dz, w 3 z = γt t log t z t π t t z zt t log t t zt z n t log n t z n t n t n t z n tz n t n log t t n tz n.77 n= } t n log t n z t n t n t n z t n log n zt n t dt n t n zt t w 3. Similaly fo the last integal z w 4 z w 4 = n= γit n it n it z itzn it n z it it z n it n it z itzn it n z it it z itz itz n z it n z n itz n itz itz n z it n z n itz n } dt t dz and finally w 4 z = γit it log it z it it it z zit it log it it zit z n it log n it z n it n it n it z n itz n it n log it it n itz n.78 n= } it n log it n z it n it n it n z it n log n zit n it dt n it n zit t w 4. Thus, composing.75-.78, the equality wz = z π R γ ds π R γ n log n z n z 36 n n z n n log z n n ds
follows. Obviously, wz is an analytic function in R. The bounday condition in.68 is satisfied due to the fact that the composition of.70 and.74 gives w z = π R γl, zds, whee L, z is the kenel in the integal.5, and, consequently, γ can be attained as its bounday value, i.e. as the bounday value of ϕ = w. 37
Chapte 3 Bounday Value Poblems fo the Poisson Equation 3. Hamonic Diichlet Poblem In this Chapte, the Geen and Neumann functions fo the quate ing ae constucted and the solutions to the Diichlet and the Neumann poblems fo the Poisson equation ae pesented. 3.. Hamonic Geen function and the Geen Repesentation fomula Definition 3... 6 A eal-valued function Gz, z 0 in a domain D of C having the popeties:. Gz, z 0 is hamonic in z D\z 0 },. log z z 0 Gz, z 0 is hamonic in the neighbohood of z 0, 3. z D Gz, z 0 = 0 is called the Geen function of D, moe exactly the Geen function of D fo the Laplace opeato. Theoem 3... 6 The Geen function of D has the following additional popeties:. 0 < Gz, z 0, z D\z 0 },. Gz, z 0 = Gz 0, z, z z 0, 3. it is uniquely given by the popeties of Definition 3... The hamonic Geen function fo the quate ing domain can be obtained on the basis of the Geen function Gz, = G z, fo the ing R 4 G R z, = log z log z z k k z log log z z k k z and the uppe half ing R 8 z z z n z n n z n z G R z, = log z z z n z n n z n z n= by obseving the additional eflection points and using the method of eflection. Thus, fo the quate ing domain R z z z G z, = log z n z n z z z z z n z n z n= z z n n n n z z n z n z z z n n n n z z n z n z, k= i.e. z z 4n z 4n z z 4n z 4n G z, = log z z 4n z 4n z z 4n z 4n. 3. n= G z, in 3. can be ewitten as z z 4n z 4n z z 4n z 4n G z, = log z z 4n z 4n z z 4n z 4n. 3. n= The popeties of the Geen function ae investigated in the following Lemmas. 39
Lemma 3... The infinite poduct 4n z 4n z z 4n z 4n 4n z 4n z z 4n z 4n n= conveges fo z R, R. 3.3 Poof. Let us conside the tems of this poduct sepaately. By the definition of the seies convegence, the sum 4n z 4n z must be obseved. Since then n= 4n z 4n z 4n z 4n z n= = 4 4n z 4n z z 4 8n 4 4n z 4n z z 4 8n, = n= Similaly fo the othe tems one gets 4n z 4n z = n= n= z 4n z 4n = n= n= z 4n z 4n = n= n= 4n z z 8n z 4 4 4n z z. 4n z z 8n 4 z 4 4n z z ; 4n z z 8n z 4 4n z z ; 4n z z 8n z 4 4n z z. Thus, the convegence of the paticula sums leads to the convegence of the whole poduct 3.3. Lemma 3... The function G z, has vanishing bounday values on R, i.e. G z, = 0. z z 0 R Poof. Conside the function in 3. on the diffeent pats of the bounday. Fo z R z z 4 z 4 z G z, = log z 4 z z n= then, taking z, yields z 8 z 4n z 4 4n z z 4 z 4 4n z z 4 4n z 4n z 4n z z 4n z 4n G z, = 0. z Fo z =, Re z > 0, Im z > 0 afte multiplying some tems of the numeato by z and by z in the denominato, the function becomes: G z 4 4 z z, = log z z 4 4 z z 4n 4n z z 4n 4n z z 4n 4n z z 4n 4n z. n= 40,
then Hee fo any M N M M M z 4n = z 4n = z 4 z 4n, n= n= n= M M M 4n z = 4n z = 4 z 4n z, n= n= M M M z 4n = z 4n = z z 4n, 4 n= n= M M M 4n z = 4n z = 4 z 4n z, n= G z, = 0. z Obviously, fo the eal z = z and imaginay z = z axes it follows that since z = z = z thee. G z, = 0 In the Geen epesentation fomula the hamonic Geen function and the Poisson kenel ae being used. The latte can be obtained as the outwad nomal deivative of the Geen function on the bounday. z z z z, z =, Re z > 0, Im z > 0 νz = z z z z, z =, Re z > 0, Im z > 0 3.4 i z z Re z > 0, Im z = 0 z z, Re z = 0, Im z > 0 So, diffeentiating G z, in 3., one gets n= n= n= n= z G z, = z z z4n 4n z n= z4n 4n z z z z z z 4n z z z z 4n z z z 4n z z 4n z 4n z 4n z 4n z 4n } 3.5 z z G z, = z z4n 4n z n= z4n 4n z z z z 4n z z z z z z 4n z z z 4n z z 4n z 4n z 4n z 4n z 4n } and consideing the outwad nomal deivatives on the diffeent bounday pats, one gets the following: fo z =, Re z > 0, Im z > 0 νz G z, = 8Re z z 4n n= 4n z 4n z 4n z 4n z 4 }.
Fo z =, Re z > 0, Im z > 0 νz G z, = 8 Re z z 4n 4n z n= 4n z z 4n z z 4n z Fo Re z > 0, Im z = 0 z νz G z, = 8Im z z z z } 4n z z 4n z z z 4n z z 4n. n= Fo Re z = 0, Im z > 0 z νz G z, = 8Re z z z z } 4n z z z 4n z 4n z z z 4n. n= Theoem 3... 6 Any w C D; C C D; C can be epesented as wz = w ν G z, ds w 4π π G z, dξdη, D whee s is the ac length paamete on D with espect to the vaiable and Gz, = G z, is the hamonic Geen function fo D. D }. 3.. Hamonic Diichlet Poblem The Geen epesentation fomula povides a solution to the Diichlet poblem. Theoem 3..3. The hamonic Diichlet poblem is uniquely solvable by w zz = f in R, w = γ on R 3.6 fo f L R ; C CR ; C, γ C R ; C wz = π =, 0<Im, 0<Re γk z, d γitk z, itdt π R =, 0<Im, 0<Re fg z, dξdη, γk z, d γtk z, tdt 3.7 whee K z, and K z, ae given in.9 and.0 espectively. Poof. We conside fist the bounday condition. Taking the popety of the Geen function to vanish on the bounday 6 into account, fo checking the bounday behavio only the bounday integals need to be consideed. 4
Let wz = w w w 3 w 4 π fg z, dξdη, then w z = =, 0<Im, 0<Re 4n n= z γ 4n z z 4n z z 4n z R z 4n z z 4n z z 4n z } d = z 4n z z =, 0<Im, 0<Re 4n z γk, zd, whee K, z is given in.9. Letting z 0, 0 =, Re 0 0, Im 0 0}, fom the calculations in Theoem.. and fomula.5 the equality w z = z 0,z R z 0,z R follows, whee Γ, z is defined in.6. Similaly, w z = =, 0<Im, 0<Re 4n n= z γ 4n z = z 4n z Γ z 4n z z 4n z z 4n z z z z d = γ 0 z z 4n z 4n z 4n z } d = γk, zd. =, 0<Im, 0<Re Letting z 0, 0 =, Re 0 0, Im 0 0}, and by Theoem.. one gets w z = Γ z 0,z R z 0,z R z z d = γ 0, = whee Γ is defined as in.6 fo =. Fo < Re, Im = 0 w 3 z = z z t γt t z t t z 4n t 4n t z t 4n z t } t 4n z t t 4n t z dt = n= γtk t, zdt, whee K, z is defined in.0. Taking z 0, < 0 <, 0 = t 0 on the basis of Theoem.. w 3z = z t 0,z R z t 0,z R γtk t, zdt = 43 z t 0,z R Γ t z z t z dt = γt 0,