linear and quasilinear complex equations of hyperbolic and...
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Linear and Quasilinear Complex Equations of Hyperbolic and Mixed Type
Asian Mathematics Series A Series edited by Chung-Chun YangDepartment of Mathematics The Hong Kong University ofScience and Technology Hong Kong
Volume 1Dynamics of transcendental functionsXin-Hou Hua and Chung-Chun Yang
Volume 2Approximate methods and numerical analysis for elliptic complex equationsGuo Chun Wen
Volume 3Introduction to statistical methods in modern geneticsMark CK Yang
Volume 4Mathematical theory in periodic plane elasticityHai-Tao Cai and Jian-ke Lu
Volume 5Gamma lines On the geometry of real and complex functionsGrigor A Barsegian
Volume 6Linear and quasilinear complex equations of hyperbolic and mixed typeGuo Chun Wen
Linear and Quasilinear Complex Equations of Hyperbolic and Mixed Type
Guo Chun Wen School of Mathematical Sciences Peking University Beijing China
London and New York
First published 2002 by Taylor amp Francis 11 New Fetter Lane London EC4P 4EE
Simultaneously published in the USA and Canada by Taylor amp Francis Inc 29 West 35th Street New York NY 10001
Taylor amp Francis is an imprint of the Taylor amp Francis Group
copy 2002 Guo Chun Wen
All rights reserved No part of this book may be reprinted or reproducedor utilized in any form or by any electronicmechanical or other meansnow known or hereafter invented including photocopying and recording or in any information storage or retrieval system without permission inwriting from the publishers
Every effort has been made to ensure that the advice and information in this book is true and accurate at the time of going to press However neither the publisher nor the authors can accept any legal responsibility or liability for any errors or omissions that may be made In the case of drug administration any medical procedure or the use of technical equipment mentioned within this book you are strongly advised to consult the manufacturerrsquos guidelines
British Library Cataloguing in Publication DataA catalogue record for this book is available from the British Library
Library of Congress Cataloging in Publication DataA catalog record for this book has been requested
ISBN 0ndash415ndash26971ndash7
This edition published in the Taylor amp Francis e-Library 2006
ldquoTo purchase your own copy of this or any of Taylor amp Francis or Routledgersquoscollection of thousands of eBooks please go to wwweBookstoretandfcoukrdquo
ISBN 0-203-16658-2 Master e-book ISBN
ISBN 0-203-26135-6 (Adobe eReader Format)
(Print Edition)
Contents
Introduction to the series viiiPreface ix
Chapter IHyperbolic complex equations of first order 1
1 Hyperbolic complex functions and hyperbolic pseudoregular functions 1
2 Complex forms of linear and nonlinear hyperbolicsystems of first order equations 10
3 Boundary value problems of linear hyperbolic complex equations of first order 18
4 Boundary value problems of quasilinear hyperboliccomplex equations of first order 25
5 Hyperbolic mappings and quasi-hyperbolic mappings 35
Chapter IIHyperbolic complex equations of second order 39
1 Complex form of hyperbolic equations of second order 392 Oblique derivative problems for quasilinear hyperbolic
equations of second order 433 Oblique derivative problems for general quasilinear
hyperbolic equations of second order 504 Other oblique derivative problems for quasilinear
hyperbolic equations of second order 595 Oblique derivative problems for degenerate hyperbolic
equations of second order 66
vi Contents
Chapter IIINonlinear elliptic complex equations of first and second order 79
1 Generalizations of KeldychndashSedov formula for analytic functions 79
2 Representation and existence of solutions for elliptic complex equations of first order 90
3 Discontinuous oblique derivative problems for quasilinear elliptic equations of second order 95
4 Boundary value problems for degenerate ellipticequations of second order in a simply connected domain 108
Chapter IVFirst order complex equations of mixed type 119
1 The RiemannndashHilbert problem for simplest first order complex equation of mixed type 119
2 The RiemannndashHilbert problem for first order linear complex equations of mixed type 126
3 The RiemannndashHilbert problem for first orderquasilinear complex equations of mixed type 134
4 The RiemannndashHilbert problem for first order quasilinearequations of mixed type in general domains 138
5 The discontinuous RiemannndashHilbert problem for quasilinear mixed equations of first order 143
Chapter VSecond order linear equations of mixed type 157
1 Oblique derivative problems for simplest second order equation of mixed type 157
2 Oblique derivative problems for second order linear equations of mixed type 162
3 Discontinuous oblique derivative problems for secondorder linear equations of mixed type 171
4 The Frankl boundary value problem for second order linear equations of mixed type 177
5 Oblique derivative problems for second order degenerate equations of mixed type 194
Contents vii
Chapter VISecond order quasilinear equations of mixed type 200
1 Oblique derivative problems for second order quasilinear equations of mixed type 200
2 Oblique derivative problems for second order equations of mixed type in general domains 209
3 Discontinuous oblique derivative problems for second order quasilinear equations of mixed type 218
4 Oblique derivative problems for quasilinear equationsof mixed type in multiply connected domains 227
References 240Index 250
Introduction to the Series
The Asian Mathematics Series provides a forum to promote and reflect timelymathematical research and development from the Asian region and to providesuitable and pertinent reference on text books for researchers academics andgraduate students in Asian universities and research institutes as well as in theWest With the growing strength of Asian economic scientific and technologicaldevelopment there is a need more than ever before for teaching and researchmaterials written by leading Asian researchers or those who have worked in orvisited the Asian region particularly tailored to meet the growing demands ofstudents and researchers in that region Many leading mathematicians in Asiawere themselves trained in the West and their experience with Western methodswill make these books suitable not only for an Asian audience but also for theinternational mathematics community
The Asian Mathematics Series is founded with the aim to present significantcontributions from mathematicians written with an Asian audience in mind tothe mathematics community The series will cover all mathematical fields andtheir applications with volumes contributed to by international experts who havetaught or performed research in Asia The material will be at graduate level orabove The book series will consist mainly of monographs and lecture notes butconference proceedings of meetings and workshops held in the Asian region willalso be considered
Preface
In this book we mainly introduce first and second order complex equations ofhyperbolic and mixed (elliptic-hyperbolic) type in which various boundaryvalue problems for first and second order linear and quasilinear complex equationsof hyperbolic and mixed type are considered In order to obtain the results oncomplex equations of mixed type we need to first discuss some boundary valueproblems for elliptic and hyperbolic complex equations
In Chapters I and II the hyperbolic pseudoregular functions and quasi-hyperbolicmappings are introduced which are corresponding to pseudoanalytic functionsand quasiconformal mappings in the theory of elliptic complex equations On thebasis of hyperbolic notations the hyperbolic systems of first order equations andhyperbolic equations of second order with some conditions can be reduced tocomplex forms In addition several boundary value problems mainly the RiemannndashHilbert problem oblique derivative problems for some hyperbolic complex equationsof first and second order are discussed in detail
In Chapter III firstly the generalizations of the KeldychndashSedov formula foranalytic functions are given Moreover discontinuous boundary value problemsfor nonlinear elliptic complex equations of first and second order are discussedBesides some oblique derivative problems for degenerate elliptic equations ofsecond order are also introduced
In Chapter IV we mainly consider the discontinuous boundary value problemsfor first order linear and quasilinear complex equations of mixed type whichinclude the discontinuous Dirichlet problem and discontinuous RiemannndashHilbertproblem In the meantime we give some a priori estimates of solutions for theabove boundary value problems
For the classical dynamical equation of mixed type due to S A Chaplygin[17] the first really deep results were published by F Tricomi [77] 1) In ChaptersV and VI we consider oblique derivative boundary value problems for secondorder linear and quasilinear complex equations of mixed type by using a complexanalytic method in a special domain and in general domains which include theDirichlet problem (Tricomi problem) as a special case We mention that in thebooks [12] 1) 3) the author investigated the Dirichlet problem (Tricomi problem)for the simplest second order equation of mixed type ie uxx+sgnyuyy=0 in
x Preface
general domains by using the method of integral equations and a complicatedfunctional relation In the present book we use the uniqueness and existence ofsolutions of discontinuous RiemannndashHilbert boundary value problem for ellipticcomplex equations and other methods to obtain the solvability result of obliquederivative problems for more general equations and domains which includes theresults in [12] 1) 3) as special cases
Similarly to the book [86] 1) the considered complex equations and boundaryconditions in this volume are rather general and several methods are used Thereare two characteristics of this book one is that mixed complex equations are includedin the quasilinear case and boundary value conditions are almost considered inthe general oblique derivative case especially multiply connected domains areconsidered Another one is that complex analytic methods are used to investigatevarious problems about complex equations of hyperbolic and mixed type Wemention that some free boundary problems in gas dynamics and some problem inelasticity can be handled by using the results stated in this book
The great majority of the contents originates in investigations of the author andhis cooperative colleagues and many results are published here for the first timeAfter reading the book it can be seen that many questions about complexequations of mixed type remain for further investigations
The preparation of this book was supported by the National Natural ScienceFoundation of China The author would like to thank Prof H Begehr ProfW Tutschke and Mr Pi Wen Yang because they proposed some beneficialimproving opinions to the manuscript of this book
Beijing Guo Chun WenAugust 2001 Peking University
CHAPTER I
HYPERBOLIC COMPLEX EQUATIONS OFFIRST ORDER
In this chapter we first introduce hyperbolic numbers hyperbolic regular functionsand hyperbolic pseudoregular functions Next we transform the linear and non-linear hyperbolic systems of first order equations into complex forms Moreoverwe discuss boundary value problems for some hyperbolic complex equations of firstorder Finally we introduce the so-called hyperbolic mappings and quasihyperbolicmappings
1 Hyperbolic Complex Functions and HyperbolicPseudoregular Functions
11 Hyperbolic numbers and hyperbolic regular functions
First of all we introduce hyperbolic numbers and hyperbolic complex functions Theso-called hyperbolic number is z = x + jy where x y are two real numbers and j iscalled the hyperbolic unit such that j2 = 1 Denote
e1 = (1 + j)2 e2 = (1minus j)2 (11)
it is easy to see that
e1 + e2 = 1 ekel =
ek if k = l
0 if k = lk l = 1 2 (12)
and (e1 e2) will be called the hyperbolic element Moreover w = f(z) = u(x y) +jv(x y) is called a hyperbolic complex function where u(x y) v(x y) are two realfunctions of two real variables x y which are called the real part and imaginary partof w = f(z) and denote Rew = u(z) = u(x y) Imw = v(z) = v(x y) Obviously
z = x+ jy = microe1 + νe2 w = f(z) = u+ jv = ξe1 + ηe2 (13)
in which
micro = x+ y ν = x minus y x = (micro+ ν)2 y =(micro minus ν)2
ξ = u+ v η = u minus v u = (ξ + η)2 v = (ξ minus η)2
2 I Hyperbolic Equations of First Order
z = xminusjy will be called the conjugate number of z The absolute value of z is definedby |z| =
radic|x2 minus y2| and the hyperbolic model of z is defined by z= radic
x2 + y2The operations of addition subtraction and multiplication are the same with the realnumbers but j2 = 1 There exists the divisor of zero and denote by O = z |x2 = y2the set of divisors of zero and zero It is clear that z isin O if and only if |z| = 0 andz has an inversion
1z=
z
zz=
1x+ y
e1 +1
x minus ye2 =
1micro
e1 +1νe2
if and only if x + jy isin O and if the hyperbolic numbers z1 = micro1e1 + ν1e2 z2 =micro2e1 + ν2e2 isin O then
z1
z2= (micro1e1 + ν1e2)
(1micro2
e1 +1ν2
e2
)=
micro1
micro2e1 +
ν1
ν2e2
It is clear that |z1z2| = |z1||z2| but the triangle inequality is not true As for thehyperbolic model of z we have the triangle inequality z1+z2 le z1 + z2 and z1z2 le radic
2 z1 z2 In the following the limits of the hyperbolic number aredefined by the hyperbolic model The derivatives of a hyperbolic complex functionw = f(z) with respect to z and z are defined by
wz = (wx + jwy)2 wz = (wx minus jwy)2 (14)
respectively and then we have
wz = (wx minus jwy)2=[(ux minus vy) + j(vx minus uy)]2
= [(wx minus wy)e1 +(wx + wy)e2]2=wνe1 + wmicroe2
= [ξνe1 + ηνe2]e1+[ξmicroe1 + ηmicroe2]e2=ξνe1 + ηmicroe2
wz = [(ux + vy) + j(vx + uy)]2 = wmicroe1 + wνe2
= (ξe1 + ηe2)microe1 + (ξe1 + ηe2)νe2=ξmicroe1 + ηνe2
(15)
LetD be a domain in the (x y)-plane If u(x y) v(x y) are continuously differentiablein D then we say that the function w = f(z) is continuously differentiable in D andwe have the following result
Theorem 11 Suppose that the hyperbolic complex function w = f(z) is continu-ously differentiable Then the following three conditions are equivalent
(1) wz = 0 (16)
(2) ξν = 0 ηmicro = 0 (17)
(3) ux = vy vx = uy (18)
Proof From (15) it is easy to see that the conditions (1)(2) and (3) in Theorem11 are equivalent
1 Hyperbolic Complex Functions 3
The system of equations (18) is the simplest hyperbolic system of first orderequations which corresponds to the CauchyndashRiemann system in the theory of ellipticequations The continuously differentiable solution w = f(z) of the complex equation(16) in D is called a hyperbolic regular function in D
If the function w(z) is defined and continuous in the neighborhood of a point z0and the following limit exists and is finite
wprime(z0)= limzrarrz0
w(z)minusw(z0)z minus z0
= limmicrorarrmicro0νrarrν0
[ξ(z)minusξ(z0)
micro minus micro0e1 +
η(z)minusη(z0)ν minus ν0
e2
]
= [ξmicroe1 + ηνe2]|micro=micro0ν=ν0 = wz(z0)
then we say that w(z) possesses the derivative wprime(z0) at z0 From the above for-mula we see that w(z) possesses a derivative at z0 if and only if ξ(z) = Rew(z) +Imw(z) η(z) = Rew(z) minus Imw(z) possess derivatives at micro0 = x0 + y0 ν0 = x0 minus y0
respectively
Now we can define some elementary hyperbolic regular functions according toseries representations in their convergent domains as follows
zn = [microe1 + νe2]n = microne1 + νne2 =(x+ y)n + (x minus y)n
2+j(x+ y)n minus(x minus y)n
2
ez = 1 + z +z2
2+ middot middot middot+ zn
n+ middot middot middot = emicro e1 + eν e2 =
ex+y + exminusy
2+ j
ex+y minus exminusy
2
ln z = lnmicro e1 + ln ν e2 =ln(x+ y) + ln(x minus y)
2+ j
ln(x+ y)minus ln(x minus y)2
sin z = z minus z3
3+ middot middot middot+ (minus1)n z2n+1
(2n+ 1)+ middot middot middot = sinmicro e1 + sin ν e2
cos z = 1minus z2
2+ middot middot middot+ (minus1)n z2n
(2n)+ middot middot middot = cosmicro e1 + cos ν e2
tgz=sin z
cos z= (sinmicroe1 + sin νe2)
(1
cosmicroe1 +
1cos ν
e2
)= tgmicro e1 + tgν e2
ctgz=cos zsin z
=(cosmicroe1 + cos νe2)(
1sinmicro
e1+1
sin νe2
)=ctgmicro e1+ctgν e2
(1+z)α=1+αz+middot middot middot+α(αminus1) (αminusn+1)n
zn+middot middot middot=(1+micro)αe1+(1+ν)αe2
where n is a positive integer and α is a positive number Moreover we can define theseries expansion of hyperbolic regular functions and discuss its convergency
4 I Hyperbolic Equations of First Order
12 Hyperbolic continuous functions and their integrals
Suppose that w = f(z) = u(x y) + jv(x y) is any hyperbolic complex function in adomain D and possesses continuous partial derivatives of first order in D Then forany point z0 isin D we have
∆w = f primex(z0)∆x+ f prime
y(z0)∆y + ε(∆z)
where
∆w = f(z)minus f(z0) f primex(z0) = ux(z0) + jvx(z0) f prime
y(z0) = uy(z0) + jvy(z0)
and z = z0 +∆z ε is a function of ∆z and
lim∆zrarr0
ε(∆z) rarr 0
Suppose that C is a piecewise smooth curve in the domain D and w = f(z) =u+ jv = ξe1 + ηe2 is a continuous function in D Then the integral of f(z) along Cand D are defined byint
Cf(z)dz =
intC
udx+ vdy + j[int
Cvdx+ udy] =
intC[ξdmicroe1 + ηdνe2]int int
Df(z)dxdy =
int intD
udxdy + jint int
Dvdxdy
We easily obtain some properties of integrals of f(z) as follows
Theorem 12 (1) If f(z) g(z) are continuous functions in D and C is a piecewisesmooth curve in D thenint
C[f(z) + g(z)]dz =
intC
f(z)dz +int
Cg(z)dzint int
D[f(z) + g(z)]dxdy =
int intD
f(z)dxdy +int int
Dg(z)dxdy
(2) Under the conditions in (1) and denoting M1 = maxzisinC f(z) M2 = supzisinD
f(z) the length of C by l and the area of D by S then
int
Cf(z)dz le
radic2M1l
int int
Df(z)dxdy le
radic2M2S
(3) If C is a piecewise smooth closed curve and G is the finite domain bounded byC f(z) is continuously differentiable in G then we have Greenrsquos formulasint int
G[f(z)]zdxdy =
j
2
intC
f(z)dz
int intG[f(z)]zdxdy = minusj
2
intC
f(z)dz
1 Hyperbolic Complex Functions 5
(4) Under the conditions as in (3) of Theorem 11 and w = f(z) is a hyperbolicregular function in G then int
Cf(z)dz = 0
In the following we introduce the definition of hyperbolic pseudoregular functionsand prove some properties of hyperbolic pseudoregular functions
13 Hyperbolic pseudoregular functions and their properties
Let w(z) F (z) G(z) be continuous functions in a domain D and G(z) F (z) satisfythe conditions
ImF (z)G(z) = 0 in D (19)
Then for every point z0 isin D we can obtain a unique pair of real numbers δ0 and γ0such that
w(z0) = δ0F (z0) + γ0G(z0) (110)
SettingW (z) = w(z)minus δ0F (z)minus γ0G(z) (111)
it is easy to see thatW (z0) = 0 (112)
If the following limit exists and is finite
w(z0) = limzrarrz0
w(z)minus δ0F (z)minus γ0G(z)z minus z0
= limzrarrz0
W (z)minus W (z0)z minus z0
ie
w(z0)= [ReW (z0)+ImW (z0)]microe1+[ReW (z0)minusImW (z0)]νe2=W prime(z0)(113)
where micro = x + y ν = x minus y then we say that w(z) is a (FG)-derivative of w(z) atz0 In order to express the existence of (113) by partial differential equations wesuppose again that
Fz(z) Fz(z) Gz(z) and Gz(z) exist and are continuous (114)
in a neighborhood of z0 According to the definition of W (z) if Wz Wz exist then
Wz(z) = wz(z)minus δ0Fz(z)minus γ0Gz(z)
Wz(z) = wz(z)minus δ0Fz(z)minus γ0Gz(z)(115)
From (113) (115) and Theorem 11 we see that if w(z0) exists then Wz(z0) exists
Wz(z0) = w(z0) (116)
andWz(z0) = 0 (117)
6 I Hyperbolic Equations of First Order
and if wz(z) wz(z) are continuous in a neighborhood of z0 and (117) holds then wehave (113) and (116) Since
W (z) =
∣∣∣∣∣∣∣∣∣w(z) w(z0) w(z0)
F (z) F (z0) F (z0)
G(z) G(z0) G(z0)
∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣F (z0) F (z0)
G(z0) G(z0)
∣∣∣∣∣∣ (118)
(117) can be rewritten as ∣∣∣∣∣∣∣∣∣wz(z0) w(z0) w(z0)
Fz(z0) F (z0) F (z0)
Gz(z0) G(z0) G(z0)
∣∣∣∣∣∣∣∣∣ = 0 (119)
If (113) exists then (116) can be written as
w(z0) =
∣∣∣∣∣∣∣∣∣wz(z0) w(z0) w(z0)
Fz(z0) F (z0) F (z0)
Gz(z0) G(z0) G(z0)
∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣F (z0) F (z0)
G(z0) G(z0)
∣∣∣∣∣∣ (120)
Unfolding (119) and (120) respectively and arranging them we obtain
wz = aw + bw (121)
w = wz minus Aw minus Bw (122)
where
a = minus FGz minus FzG
F G minus FG b =
FGz minus FzG
FG minus FG
A = minus FGz minus FzG
F G minus FG B =
FGz minus FzG
FG minus FG
(123)
here a(z) b(z) A(z) and B(z) are called the characteristic coefficients of the gener-ating pair (FG) Obviously F = G = 0 and
Fz = aF + bF Gz = aG+ bG
Fz = AF +BF Gz = AG+BG
uniquely determine a b A and B Denote them by a(FG) b(FG) A(FG) and B(FG)
respectively
1 Hyperbolic Complex Functions 7
From the above discussion we see that if w(z0) exists then wz at z0 exists and(121) (122) are true If wz and wz(z) exist and are continuous in a neighborhoodz0 isin D and (121) holds at z0 then w(z0) exists and (122) is true
For any function w(z) if w(z) exists and is continuous in the domain D thenw(z) is called the first-class (FG) hyperbolic pseudoregular function or hyperbolicpseudoregular function for short It is clear that the following theorem holds
Theorem 13 w(z) is a hyperbolic pseudoregular function if and only if wz(z) andwz(z) exist and are continuous and (121) holds
By (19) it is easy to see that every function w(z) has a unique expression
w(z) = φ(z)F (z) + ψ(z)G(z) (124)
where φ(z) and ψ(z) are two real-valued functions Let
K(z) = φ(z) + jψ(z) (125)
Then we can give the following definition
If w(z) is the first-class (FG) hyperbolic pseudoregular complex function thenK(z) = φ(z) + jψ(z) is called the second-class (FG) hyperbolic pseudoregular func-tion
Theorem 14 K(z) = φ(z)+jψ(z) is a second-class (FG) hyperbolic pseudoregularfunction if and only if φ and ψ have continuous partial derivatives and
Fφz +Gψz = 0 (126)
Under this conditionw(z) = Fφz +Gψz (127)
holds where
φz = (e1φmicro + e2φν) = [(φx + φy)e1 + (φx minus φy)e2]2
φz = (e1φν + e2φmicro) = [(φx minus φy)e1 + (φx + φy)e2]2
ψz = (e1ψmicro + e2ψν) = [(ψx + ψy)e1 + (ψx minus ψy)e2]2
ψz = (e1ψν + e2ψmicro) = [(ψx minus ψy)e1 + (ψx + ψy)e2]2
(128)
Proof From
W (z) = [φ(z)minus φ(z0)]F (z) + [ψ(z)minus ψ(z0)]G(z)
it follows thatWz(z0) = F (z0)φz(z0) +G(z0)ψz(z0)
Wz(z0) = F (z0)φz(z0) +G(z0)ψz(z0)
8 I Hyperbolic Equations of First Order
Thus the proof can be immediately obtained
Setting
minusG
F= σ + jτ = (σ + τ)e1 + (σ minus τ)e2
minusF
G= σ + jτ = (σ + τ)e1 + (σ minus τ)e2
(129)
where σ τ σ and τ are real-valued functions and τ = 0 τ = 0 Hence (126) isequivalent to the system of equations
φx = σψx minus τψy φy = minusτψx + σψy (130)
If φ and ψ have continuous partial derivatives up to second order we find the deriva-tives with respect to x and y in (130) and then obtain
φxx minus φyy + δφx + γφy = 0
ψxx minus ψyy + δψx + γψy = 0(131)
where
δ =σy + τx
τ γ = minus σx + τy
τ
δ =σy + τx
τ γ = minusσx + τy
τ
(132)
In accordance with the following theorem the elimination is reasonable
Theorem 15 Let δ γ δ and γ be determined by (132) Then the real-valued func-tion φ (ψ) is the real (imaginary) part of a second-class hyperbolic pseudoregular func-tion if and only if it has continuous partial derivatives up to second order and satisfiesthe first (second) equation in (131)
Proof From the second formula in (131) we see that the function
φ(z) =int z
z0
[(σψx minus τψy)dx+ (minusτψx + σψy)dy] z0 z isin D
is single-valued and φ(z) ψ(z) satisfy system (130) The part of necessity can bederived from Theorem 23 Chapter II below
14 Existence of a generating pair (FG)
Theorem 16 Let a(z) and b(z) be two continuous functions in a bounded andclosed domain D = micro0 le micro le micro0+R1 ν0 le ν le ν0+R2 where R1 R2 are positiveconstants and denote z0 = micro0e1 + ν0e2 Then there exists a unique continuouslydifferentiable hyperbolic pseudoregular function w(z) satisfying the complex equation
wz = a(z)w(z) + b(z)w(z) (133)
1 Hyperbolic Complex Functions 9
and the boundary conditions
w(z) = c1(micro)e1 + c2(ν0)e2 when z isin L1
w(z) = c1(micro0)e1 + c2(ν)e2 when z isin L2(134)
where c1(micro) and c2(ν) are two real continuous functions on L1 L2 respectively L1 =micro0 le micro le micro0 +R1 ν = ν0 and L2 = micro = micro0 ν0 le ν le ν0 +R2The theorem is a special case of Theorems 33 and 34 below
Theorem 17 Let a(z) and b(z) be two continuous complex functions in the domainD as stated in Theorem 16 Then there exists a generating pair (FG) in D suchthat
a = a(FG) and b = b(FG) (135)
Proof Denote by F (z) and G(z) two solutions of the complex equation (133) sat-isfying the boundary conditions
w(z) = e1 + e2 = 1 when z isin L1 cup L2
andw(z) = e1 minus e2 = j when z isin L1 cup L2
respectively Then by Theorem 16 F (z) and G(z) have continuous partial deriva-tives and
Fz = aF + bF and Gz = aG+ bG
F = 1 G = j when z isin L1 cup L2
Hence a = a(FG) and b = b(FG) Whether ImF (z)G(z) = 0 in D remains to bediscussed
Theorem 18 Under the same conditions as in Theorem 17 and letting
b = minusa or b = a z isin D (136)
there exists a generating pair (FG) in D satisfying the complex equation (133) and
F (z) = 1 or G(z) = j in D (137)
Proof By the hypotheses in Theorem 17 and equation (136) there exists a uniquegenerating pair (FG) in D satisfying the conditions
Fz = a(F minus F ) or Gz = a(G+ G) z isin D
F (z) = 1 or G(z) = j when z isin L1 cup L2
Hence we have (137) The above results are similar to those in [9]1) (see [89])
10 I Hyperbolic Equations of First Order
2 Complex Forms of Linear and Nonlinear HyperbolicSystems of First Order Equations
In this section we transform linear and nonlinear hyperbolic systems of first orderequations into complex forms
21 Complex forms of linear hyperbolic systems of first order equations
We consider the linear hyperbolic system of first order partial differential equations⎧⎨⎩a11ux + a12uy + b11vx + b12vy = a1u+ b1v + c1
a21ux + a22uy + b21vx + b22vy = a2u+ b2v + c2(21)
where the coefficients akl bkl ak bk ck (k l = 1 2) are known functions in D in whichD is a bounded domain System (21) is called hyperbolic at a point in D if at thepoint the inequality
I = (K2 +K3)2 minus 4K1K4 gt 0 (22)
holds in which
K1 =∣∣∣∣∣a11 b11
a21 b21
∣∣∣∣∣ K2 =∣∣∣∣∣a11 b12
a21 b22
∣∣∣∣∣ K3 =∣∣∣∣∣a12 b11
a22 b21
∣∣∣∣∣ K4 =∣∣∣∣∣a12 b12
a22 b22
∣∣∣∣∣ If the inequality (22) at every point (x y) in D holds then (21) is called a hyperbolicsystem in D We can verify that (22) can be rewritten as
I = (K2 minus K3)2 minus 4K5K6 gt 0 (23)
where
K5 =∣∣∣∣∣a11 a12
a21 a22
∣∣∣∣∣ K6 =∣∣∣∣∣ b11 b12
b21 b22
∣∣∣∣∣ If the coefficients akl bkl (k l = 1 2) in D are bounded and the condition
I = (K2 +K3)2 minus 4K1K4 = (K2 minus K3)2 minus 4K5K6 ge I0 gt 0 (24)
holds in which I0 is a positive constant then (21) is called uniformly hyperbolic inD In the following we reduce system (21) to complex form
1) If K2 K3 are of same signs and K6 = 0 at the point (x y) isin D then we cansolve (21) for vy minusvx and obtain the system of first order equations⎧⎨⎩ vy = aux + buy + a0u+ b0v + f0
minusvx = dux + cuy + c0u+ d0v + g0(25)
where a b c d are known functions of akl bkl(k l = 1 2) and a0 b0 c0 d0 f0 g0 areknown functions of bkl ak bk ck(k l = 1 2) and
a = K1K6 b = K3K6 c = K4K6 d = K2K6
2 Complex Forms of Hyperbolic Systems 11
The condition (22) of hyperbolic type is reduced to the condition
∆ = I4K26 =
(b+ d)2
4minus ac gt 0 (26)
There is no harm in assuming that a minus c ge 0 because otherwise let y be replacedby minusy this requirement can be realized If a c are not of the same sign or one ofthem is equal to zero then minusac ge 0 bd ge 0 and may be such that a ge 0 minusc ge 0or a c are of same signs then we may assume that a gt 0 c gt 0 because otherwiseif v is replaced by minusv this requirement can be realized Moreover we can assumethat 0 lt c lt 1 otherwise setting v = hv herein h is a positive constant such thath ge c + 1 we have K4 = hK4 K6 = h2K6 and c = K4K6 = ch lt 1 bd ge 0Multiply the first formula of (25) by minusj and then subtract the second formula of(25) This gives
vx minus jvy = minus j(aux + buy + a0u+ b0v + f0)
minus dux minus cuy minus c0u minus d0v minus g0
Noting z = x+ jy w = u+ jv and using the relations⎧⎨⎩ux = (wz + wz + wz + wz)2 uy = j(minuswz + wz + wz minus wz)2
vx = j(wz minus wz + wz minus wz)2 vy = (minuswz minus wz + wz + wz)2
we getj(wz minus wz) = minus(aj + d)(wz + wz + wz + wz)2
minus(c+ bj)j(minuswz + wz + wz minus wz)2
+lower order terms
namely
(1 + q1)wz + q2wz = minusq2wz + (1minus q1)wz + lower order terms (27)
in whichq1 = [a minus c+ (d minus b)j]2 q2 = [a+ c+ (d+ b)j]2
Notingq0 = (1 + q1)(1 + q1)minus q2q2
= [(2 + a minus c)2 minus (d minus b)2 minus (a+ c)2 + (d+ b)2]4
= 1 + a minus c minus (d minus b)24 + (d+ b)24minus ac
= 1 + a minus c minus (d minus b)24 + ∆
= 1 + a minus c+ σ = 1 + a minus c+ bd minus ac
= (1 + a)(1minus c) + bd gt 0
12 I Hyperbolic Equations of First Order
where σ = ∆minus (b minus d)24 = bd minus ac ge 0 thus we can solve (27) for wz giving
wz minus Q1(z)wz minus Q2(z)wz = A1(z)w + A2(z)w + A3(z) (28)
in which
Q1(z) =minus2q2(z)q0(z)
Q2(z) =[q2q2 minus (q1 minus 1)(q1 + 1)]
q0
For the complex equation (28) if (a minus c)2 minus 4∆ ge 0 (1 + σ)2 minus 4∆ ge 0 ie
(K1 minus K4)2 minus (K2 minus K3)2 + 4K5K6 ge 0
(K26 +K2K3 minus K1K4)2 minus (K2 minus K3)2 + 4K5K6 ge 0
(29)
then we can prove
|Q1|+ |Q2| = |Q1Q1|12 + |Q2Q2|12 lt 1 (210)
where |Q1| = |Q1Q1|12 is the absolute value of Q1 In fact
|2q2| = |(a+ c)2 minus (d+ b)2|12 = |(a minus c)2 minus 4∆|12
|q2q2 minus (q1 minus 1)(q1 + 1)|= |(a minus c)24minus∆minus [a minus c+ (b minus d)j minus 2]
times[a minus c minus (d minus b)j + 2]4| = |(1 + σ)2 minus 4∆|12
(1 + σ)2 + (a minus c)2 + 2(1 + σ)(a minus c) = [1 + σ + (a minus c)]2 gt 0
(1 + σ)2(a minus c)2 + (4∆)2 minus 4∆(1 + σ)2 minus 4∆(a minus c)2
lt (4∆)2 + (1 + σ)2(a minus c)2 + 8∆(1 + σ)(a minus c)
[(a minus c)2 minus 4∆][(1 + σ)2 minus 4∆]
lt (4∆)2 + (1 + σ)2(a minus c)2 + 8∆(1 + σ)(a minus c)
2[(a minus c)2 minus 4∆][(1 + σ)2 minus 4∆]12 lt 8∆ + 2(1 + σ)(a minus c)
(a minus c)2 minus 4∆ + (1 + σ)2 minus 4∆
+2[(a minus c)2 minus 4∆][(1 + σ)2 minus 4∆]12
lt (1 + σ + a minus c)2
and then|(a minus c)2 minus 4∆|12 + |(1 + σ)2 minus 4∆|12 lt 1 + σ + a minus c
thus we can derive (210)
2)K2 K3 at (x y) isin D have same signs K6 = 0 K5 = 0 by using similar methodswe can transform (21) into a complex equation in the form (28)
2 Complex Forms of Hyperbolic Systems 13
Now we discuss the case
3) K2 K3 are not of same signs K4 = 0 at the point (x y) isin D then we can solve(21) for uy vy and obtain the system of first order equations⎧⎨⎩ vy = aux + bvx + a0u+ b0v + f0
minusuy = dux + cvx + c0u+ d0v + g0(211)
where a b c d are known functions of akl bkl (k l = 1 2) and a0 b0 c0 d0 f0 g0 areknown functions of ak2 bk2 ak bk ck (k l = 1 2) and
a = K5K4 b = minusK3K4 c = K6K4 d = K2K4
The condition (22) of hyperbolic type is reduced to the condition
∆ = I4K24 = (b+ d)24minus ac gt 0 (212)
Similarly to (25) multiply the second formula of (211) by j and then subtract thefirst formula of (211) we get
minusvy minus juy = wz minus wz = minus(a minus d j)ux minus (b minus c j)vx + lower order terms
= minus(a minus d j)(wz + wz + wz + wz)2
+(c minus b j)(wz minus wz + wz minus wz)2 + lower order terms
namely(1 + q1)wz + q2wz = (1minus q1)wz minus q2wz + lower order terms (213)
whereq1 =
[a minus c minus (d minus b)j]2
q2 =[a+ c minus (d+ b)j]
2
It is clear that
q0 = (1 + q1)(1 + q1)minus q2q2
=[(2 + a minus c)2 minus (d minus b)2 minus (a+ c)2 + (d+ b)2]
4
= (1 + a)(1minus c) + bd gt 0
thus we can solve (213) for wz ie
wz minus Q1(z)wz minus Q2(z)wz = A1(z)w + A2w + A3(z) (214)
in which
Q1(z) =[(1minus q1)(1 + q1) + |q2|2]
q0 Q2 =
minus2q2(z)q0
4) K2 K3 are not of same signs K4 = 0 K1 = 0 by using similar methods as in3) we can transform (21) into the complex equation of the form (214)
14 I Hyperbolic Equations of First Order
22 Complex forms of nonlinear hyperbolic systems of first orderequations
Next we consider the general nonlinear hyperbolic system of first order partial dif-ferential equations
Fk(x y u v ux uy vx vy) = 0 k = 1 2 (215)
where the real functions Fk(k = 1 2) are defined and continuous at every point (x y)in D and possess continuous partial derivatives in ux uy vx vy For system (215)its condition of hyperbolic type can be defined by the inequality (22) or (23) butin which
K1 =D(F1 F2)D(ux vx)
K2 =D(F1 F2)D(ux vy)
K3 =D(F1 F2)D(uy vx)
K4 =D(F1 F2)D(uy vy)
K5 =D(F1 F2)D(ux uy)
K6 =D(F1 F2)D(vx vy)
(216)
where Fkux Fkuy Fkvx Fkvy(k = 1 2) can be found as follows
Fkux =int 1
0Fktux(x y u v tux tuy tvx tvy)dt
Fkuy =int 1
0Fktuy(x y u v tux tuy tvx tvy)dt
Fkvx =int 1
0Fktvx(x y u v tux tuy tvx tvy)dt
Fkvy =int 1
0Fktvy(x y u v tux tuy tvx tvy)dt
(217)
By using the method in Subsection 21 for cases 1) K2 K3 are of same signs K5 orK6 = 0 2) K2 K3 are not of same signs K1 or K4 = 0 then system (215) can bereduced to the complex form
wz minus Q1wz minus Q2wz = A1w + A2w + A3 (218)
where z = x+ jy w = u+ jv and
Qk = Qk(z w wz wz) k = 1 2 Ak = Ak(z w wz wz) k = 1 2 3
In particular if (29) holds from the condition of hyperbolic type in (22) it followsthat (210) holds
Theorem 21 Let system (215) satisfy the condition of hyperbolic type as in (22)and the conditions from the existence theorem for implicit functions Then (215) issolvable with respect to wz and the corresponding hyperbolic complex equation of firstorder (218) can be obtained
2 Complex Forms of Hyperbolic Systems 15
As for the cases 3) K1 = K4 = 0 K2 K3 are not of same signs K5 = 0 or K6 = 0and 4) K5 = K6 = 0 K2 K3 are of same signs K1 = 0 or K4 = 0 we can transformthe quasilinear case of hyperbolic system (215) into the complex forms by using asimilar method in the next subsection
23 Complex forms of quasilinear hyperbolic systems of first orderequations
Finally we discuss the quasilinear hyperbolic system of first order partial differentialequations ⎧⎨⎩a11ux + a12uy + b11vx + b12vy = a1u+ b1v + c1
a21ux + a22uy + b21vx + b22vy = a2u+ b2v + c2(219)
where the coefficients akl bkl(k l = 1 2) are known functions in (x y) isin D andak bk ck(k = 1 2) are known functions of (x y) isin D and u v isin IR The hyperbolicitycondition of (219) is the same as for system (21) ie for any point (x y) isin D theinequality
I = (K2 +K3)2 minus 4K1K4 = (K2 minus K3)2 minus 4K5K6 gt 0 (220)
holds in which
K1 =∣∣∣∣∣a11 b11
a21 b21
∣∣∣∣∣ K2 =∣∣∣∣∣a11 b12
a21 b22
∣∣∣∣∣ K3 =∣∣∣∣∣a12 b11
a22 b21
∣∣∣∣∣ K4 =
∣∣∣∣∣a12 b12
a22 b22
∣∣∣∣∣ K5 =∣∣∣∣∣a11 a12
a21 a22
∣∣∣∣∣ K6 =∣∣∣∣∣ b11 b12
b21 b22
∣∣∣∣∣ We first consider the case 1) K1 = K4 = 0 K2 K3 are not of same signs K6 = 0
at the point (x y) isin D From K1 = K4 = 0 there exist real constants λ micro such that
a11 = λb11 a21 = λb21 a12 = microb12 a22 = microb22
thusK2 = λK6 K3 = minusmicroK6 K5 = λmicroK6
and thenI = (K2 minus K3)2 minus 4K5K6 = [(λ+ micro)2 minus 4λmicro]K2
6
= (K2 +K3)2 minus 4K1K4 = (λ minus micro)2K26 gt 0
It is easy to see that λ = micro ie K2 = minusK3 in this case system (219) becomes theform ⎧⎨⎩ b11(λu+ v)x + b12(microu+ v)y = a1u+ b1v + c1
b21(λu+ v)x + b22(microu+ v)y = a2u+ b2v + c2(221)
Setting U = λu+ v V = microu+ v and noting∣∣∣∣∣∣Uu Uv
Vu Vv
∣∣∣∣∣∣ =∣∣∣∣∣∣λ 1
micro 1
∣∣∣∣∣∣ = λ minus micro = 0
16 I Hyperbolic Equations of First Order
andu =
U minus V
λ minus micro v =
microU minus λV
micro minus λ
system (221) can be written in the form⎧⎨⎩ b11Ux + b12Vy = aprime1U + bprime
1V + c1
b21Ux + b22Vy = aprime2U + bprime
2V + c2(222)
where
aprime1 =
a1 minus microb1
λ minus micro bprime
1 =minusa1 + λb1
λ minus micro aprime
2 =a2 minus microb2
λ minus micro bprime
2 =minusa2 + λb2
λ minus micro
thus ⎧⎨⎩Ux = [(aprime1b22 minus aprime
2b12)U + (bprime1b22 minus bprime
2b12)V + (c1b22 minus c2b12)]K6
Vy = [(aprime2b11 minus aprime
1b21)U + (bprime2b11 minus bprime
1b21)V + (c2b11 minus c1b21)]K6(223)
Subtracting the first equation from the second equation the complex equation ofW = U + jV
Wz +W z = A1(z W )W + A2(z W )W + A3(z W ) (224)
can be derived where A1 A2 A3 are known functions of bkl ak bk ck(k l = 1 2)
Moreover we consider system (219) with the condition 2) K5 = K6 = 0 K2 K3
are of same signs and K4 = 0 at the point (x y) isin D In this case due to K5 =K6 = 0 at the point (x y) isin D there exist real constants λ micro such that
a11 = λa12 a21 = λa22 b11 = microb12 b21 = microb22
thusK1 = λmicroK4 K2 = λK4 K3 = microK4
and then
I = (K2 +K3)2 minus 4K1K4 = [(λ+ micro)2 minus 4λmicro]K24 = (λ minus micro)2K2
4 gt 0
It is clear that if λ = micro ie K2 = K3 then system (219) can become the form⎧⎨⎩a12(λux + uy) + b12(microvx + vy) = a1u+ b1v + c1
a22(λux + uy) + b22(microvx + vy) = a2u+ b2v + c2(225)
Letξ =
x minus microy
λ minus micro η =
minusx+ λy
λ minus micro
it is easy to see that ∣∣∣∣∣∣ξx ξy
ηx ηy
∣∣∣∣∣∣ = 1(λ minus micro)2
∣∣∣∣∣∣1 minusmicro
minus1 λ
∣∣∣∣∣∣ = 1λ minus micro
= 0
2 Complex Forms of Hyperbolic Systems 17
and x = λξ + microη y = ξ + η Thus system (225) can be rewritten in the form⎧⎨⎩a12uξ + b12vη = a1u+ b1v + c1
a22uξ + b22vη = a2u+ b2v + c2(226)
This system can be solved for uξ vη namely⎧⎨⎩uξ = aprime1u+ bprime
1v + cprime1
vη = aprime2u+ bprime
2v + cprime2
(227)
whereaprime
1 = (a1b22 minus a2b12)K4 aprime2 = (a2a12 minus a1a22)K4
bprime1 = (b1b22 minus b2b12)K4 bprime
2 = (b2a12 minus b1a22)K4
cprime1 = (c1b22 minus c2b12)K4 cprime
2 = (c2a12 minus c1a22)K4
Denoting ζ = ξ + jη then system (227) can be written in the complex form
wζ + wζ = Aprime1(ζ w)w + Aprime
2(ζ w)w + Aprime3(ζ w) (228)
in which Aprime1 A
prime2 A
prime3 are known functions of ak2 bk2 ak bk ck(k = 1 2)
For 3) K1 = K4 = 0 K2 K3 are not of same signs and K5 = 0 and 4) K5 = K6 =0 K2 K3 are of same signs and K1 = 0 by using a similar method we can transform(219) into the complex equations (224) and (228) respectively We mention that itis possible that the case
b11 = λa11 b21 = λa21 b12 = microa12 b22 = microa22
occurs for 1) and 2) and
a12 = λa11 a22 = λa21 b12 = microb11 b22 = microb21
occurs for 3) and 4) then we can similarly discuss In addition if λ(x y) micro(x y) areknown functions of (x y) in D then in the left-hand sides of the two equations in(221) should be added b11λxu+ b12microyu and b21λxu+ b22microyu It is sufficient to modifythe coefficient of u and the system is still hyperbolic For 2)ndash4) we can similarlyhandle
The complex equations as stated in (224) and (228) can be written in the form
wz + wz = A(z w)w +B(z w)w + C(z w) (229)
which is equivalent to the system of first order equations
ux = au+ bv + f vy = cu+ dv + g (230)
where z = x + jy w = u + jv A = (a + jb minus c minus jd)2 B = (a minus jb minus c + jd)2C = f minus g Let
W (z) = ve1 + ue2 Z = xe1 + ye2 (231)where e1 = (1+j)2 e2 = (1minusj)2 From (230) we can obtain the complex equation
WZ = vye1 + uxe2 = A1W + A2W + A3 = F (ZW ) (232)
in which A1(ZW ) = de1 + ae2 A2(ZW ) = ce1 + be2 A3(ZW ) = ge1 + fe2 [92]
18 I Hyperbolic Equations of First Order
3 Boundary Value Problems of Linear Hyperbolic ComplexEquations of First Order
In this section we mainly discuss the RiemannndashHilbert boundary value problem forlinear hyperbolic complex equations of first order in a simply connected domain Wefirst give a representation of solutions for the above boundary value problem andthen prove the uniqueness and existence of solutions for the above problem by usingthe successive iteration
31 Formulation of the RiemannndashHilbert problem and uniqueness of itssolutions for simplest hyperbolic complex equations
Let D be a simply connected bounded domain in the x+ jy-plane with the boundaryΓ = L1 cup L2 cup L3 cup L4 where L1 = x = minusy 0 le x le R1 L2 = x = y + 2R1 R1 lex le R2 L3 = x = minusy minus 2R1 + 2R2 R = R2 minus R1 le x le R2 L4 = x = y 0 lex le R2 minus R1 and denote z0 = 0 z1 = (1 minus j)R1 z2 = R2 + j(R2 minus 2R1) z3 =(1 + j)(R2 minus R1) = (1 + j)R and L = L1 cup L4 where j is the hyperbolic unit Forconvenience we only discuss the case R2 ge 2R1 the other case can be discussed by asimilar method We consider the simplest hyperbolic complex equation of first order
wz = 0 in D (31)
The RiemannndashHilbert boundary value problem for the complex equation (31) maybe formulated as follows
Problem A Find a continuous solution w(z) of (31) in D satisfying the boundaryconditions
Re [λ(z)w(z)] = r(z) z isin L Im [λ(z0)w(z0)] = b1 (32)
where λ(z) = a(z) + jb(z) = 0 z isin L and λ(z) r(z) b1 satisfy the conditions
Cα[λ(z) L] = Cα[Reλ L] + Cα[Imλ L] le k0 Cα[r(z) L] le k2 (33)
|b1| le k2 maxzisinL1
1|a(z)minus b(z)| max
zisinL4
1|a(z) + b(z)| le k0 (34)
in which b1 is a real constant and α (0 lt α lt 1) k0 k2 are non-negative constants Inparticular when a(z) = 1 b(z) = 0 ie λ(z) = 1 z isin L Problem A is the Dirichletproblem (Problem D) whose boundary condition is
Re [w(z)] = r(z) z isin L Im [w(z0)] = b1 (35)
Problem A with the conditions r(z) = 0 z isin L b1 = 0 is called Problem A0
On the basis of Theorem 11 it is clear that the complex equation (31) can bereduced to the form
ξν = 0 ηmicro = 0 (micro ν) isin Q = 0 le micro le 2R 0 le ν le 2R1 (36)
3 Linear Hyperbolic Equations 19
where micro = x + y ν = x minus y ξ = u + v η = u minus v Hence the general solution ofsystem (36) can be expressed as
ξ = u+ v = f(micro) = f(x+ y) η = u minus v = g(ν) = g(x minus y) ie
u = [f(x+ y) + g(x minus y)]2 v = [f(x+ y)minus g(x minus y)]2(37)
in which f(t) g(t) are two arbitrary real continuous functions on [0 2R] [0 2R1]respectively From the boundary condition (32) we have
a(z)u(z) + b(z)v(z)=r(z) on L λ(z0)w(z0)=r(z0) + jb1 ie
[a((1minus j)x) + b((1minus j)x)]f(0) + [a((1minus j)x)minus b((1minus j)x)]
times g(2x) = 2r((1minus j)x) on [0 R1]
[a((1 + j)x) + b((1 + j)x)]f(2x) + [a((1 + j)x)minus b((1 + j)x)]
times g(0) = 2r((1 + j)x) on [0 R]
f(0)=u(0)+v(0)=r(0)+b1
a(0)+b(0) g(0)=u(0)minusv(0)=
r(0)minusb1
a(0)minusb(0)
(38)
The above formulas can be rewritten as
[a((1minus j)t2)+b((1minus j)t2)]f(0) + [a((1minus j)t2)minus b((1minus j)t2)]times g(t) = 2r((1minus j)t2) t isin [0 2R1]
[a((1 + j)t2) + b((1 + j)t2)]f(t) +[a((1 + j)t2)minus b((1 + j)t2)]times g(0) = 2r((1 + j)t2) t isin [0 2R] ie
f(x+ y) =2r((1 + j)(x+ y)2)
a((1 + j)(x+ y)2)+b((1 + j)(x+ y)2)(39)
minus [a((1+j)(x+y)2)minus b((1+j)(x+y)2)]g(0)a((1+j)(x+y)2) + b((1+j)(x+y)2)
0lex+yle2R
g(x minus y) =2r((1minus j)(x minus y)2)
a((1minus j)(x minus y)2)minus b((1minus j)(x minus y)2)
minus [a((1minusj)(xminusy)2)+b((1minusj)(xminusy)2)]f(0)a((1minusj)(xminusy)2)minusb((1minusj)(xminusy)2)
0lexminusyle2R1
Thus the solution w(z) of (31) can be expressed as
w(z) = f(x+ y)e1 + g(x minus y)e2
=12f(x+ y) + g(x minus y) + j[f(x+ y)minus g(x minus y)]
(310)
20 I Hyperbolic Equations of First Order
where f(x + y) g(x minus y) are as stated in (39) and f(0) g(0) are as stated in (38)It is not difficult to see that w(z) satisfies the estimate
Cα[w(z) D]leM1=M1(α k0 k2 D) Cα[w(z) D]leM2k2=M2(α k0 D)k2 (311)
where M1 M2 are two non-negative constants only dependent on α k0 k2 D andα k0 D respectively The above results can be written as a theorem
Theorem 31 Any solution w(z) of Problem A for the complex equation (31) pos-sesses the representation (310) which satisfies the estimate (311)
32 Uniqueness of solutions of the RiemannndashHilbert problem for linearhyperbolic complex equations
Now we discuss the linear case of the complex equation (232) namely
wz = A1(z)w + A2(z)w + A3(z) (312)
and suppose that the complex equation (312) satisfies the following conditions
Condition C Al(z) (l = 1 2 3) are continuous in D and satisfy
C[Al D] = C[ReAl D] + C[ImAl D] le k0 l = 1 2 C[A3 D] le k1 (313)
Due to w = u + jv = ξe1 + ηe2 wz = ξmicroe1 + ηνe2 wz = ξνe1 + ηmicroe2 from theformulas in Section 1 equation (312) can be rewritten in the form
ξνe1 + ηmicroe2 = [A(z)ξ +B(z)η + E(z)]e1
+[C(z)ξ +D(z)η + F (z)]e2 z isin D ie⎧⎨⎩ ξν = A(z)ξ +B(z)η + E(z)
ηmicro = C(z)ξ +D(z)η + F (z)z isin D
(314)
in which
A = ReA1 + ImA1 B = ReA2 + ImA2 C = ReA2 minus ImA2
D = ReA1 minus ImA1 E = ReA3 + ImA3 F = ReA3 minus ImA3
The boundary condition (32) can be reduced to
Re [λ(ξe1 + ηe2)] = r(z) Im [λ(ξe1 + ηe2)]|z=z0 = b1 (315)
where λ = (a + b)e1 + (a minus b)e2 Moreover the domain D is transformed into Q =0 le micro le 2R 0 le ν le 2R1 R = R2 minus R1 which is a rectangle and A B C D E F
3 Linear Hyperbolic Equations 21
are known functions of (micro ν) isin Q There is no harm in assuming that w(z0) = 0otherwise through the transformation
W (z) = w(z)minus [a(z0)minus jb(z0)][r(z0) + jb1]
[a2(z0)minus b2(z0)] (316)
the requirement can be realized For convenience sometimes we write z isin D or z isin Qand denote L1 = micro = 0 0 le ν le 2R1 L4 = 0 le micro le 2R ν = 0
Now we give a representation of solutions of Problem A for equation (312)
Theorem 32 If equation (312) satisfies Condition C then any solution w(z) ofProblem A for (312) can be expressed as
w(z) = w0(z) + Φ(z) + Ψ(z) in D
w0(z) = f(x+ y)e1 + g(x minus y)e2 Φ(z) = f(x+ y)e1 + g(x minus y)e2
Ψ(z) =int xminusy
0[Aξ +Bη + E]d(x minus y)e1 +
int x+y
0[Cξ +Dη + F ]d(x+ y)e2
(317)
where f(x+ y) g(x minus y) are as stated in (39) and f(x+ y) g(x minus y) are similar tof(x+ y) g(x minus y) in (39) but Φ(z) satisfies the boundary condition
Re [λ(z)Φ(z)]=minusRe [λ(z)Ψ(z)] z isin L Im [λ(z0)Φ(z0)]=minusIm [λ(z0)Ψ(z0)](318)
Proof It is not difficult to see that the functions w0(z) Φ(z) are solutions of thecomplex equation (31) in D which satisfy the boundary conditions (32) and (318)respectively and Ψ(z) satisfies the complex equation
[Ψ]z = [Aξ +Bη + E]e1 + [Cξ +Dη + F ]e2 (319)
and Φ(z) + Ψ(z) satisfies the boundary condition of Problem A0 Hence w(z) =w0(z)+Φ(z)+Ψ(z) satisfies the boundary condition (32) and is a solution of ProblemA for (312)
Theorem 33 Suppose that Condition C holds Then Problem A for the complexequation (312) has at most one solution
Proof Let w1(z) w2(z) be any two solutions of Problem A for (312) and substitutethem into equation (312) and the boundary condition (32) It is clear that w(z) =w1(z)minus w2(z) satisfies the homogeneous complex equation and boundary conditions
wz = A1w + A2w in D (320)
Re [λ(z)w(z)] = 0 if (x y) isin L w(z0) = 0 (321)
On the basis of Theorem 32 the function w(z) can be expressed in the form
w(z) = Φ(z) + Ψ(z)
Ψ(z) =int xminusy
0[Aξ +Bη]e1d(x minus y) +
int x+y
0[Cξ +Dη]e2d(x+ y)
(322)
22 I Hyperbolic Equations of First Order
Suppose w(z) equiv 0 in the neighborhood (sub D) of the point z0 = 0 We maychoose a sufficiently small positive number R0 lt 1 such that 8M3MR0 lt 1 whereM3 = maxC[A Q0] C[B Q0] C[CQ0] C[DQ0] M = 1+4k2
0(1+k20) is a positive
constant and m = C[w(z) Q0] gt 0 herein Q0 = 0 le micro le R0 cap 0 le ν le R0From (39)(310)(317)(318)(322) and Condition C we have
Ψ(z) le 8M3mR0 Φ(z) le 32M3k20(1 + k2
0)mR0
thus an absurd inequalitym le 8M3MmR0 lt m is derived It shows w(z) = 0 (x y) isinQ0 Moreover we extend along the positive directions of micro = x + y and ν = x minus ysuccessively and finally obtain w(z) = 0 for (x y) isin D ie w1(z)minus w2(z) = 0 in DThis proves the uniqueness of solutions of Problem A for (312)
33 Solvability of Problem A for linear hyperbolic complex equations offirst order
Theorem 34 If the complex equation (312) satisfies Condition C then ProblemA for (312) has a solution
Proof In order to find a solution w(z) of Problem A in D we can express w(z)in the form (317) In the following by using successive iteration we can find asolution of Problem A for the complex equation (312) First of all substitutingw0(z) = ξ0e1 + η0e2 of Problem A for (31) into the position of w = ξe1 + ηe2 in theright-hand side of (312) the function
w1(z) = w0(z) + Φ1(z) + Ψ1(z)
Ψ1(z) =int ν
0[Aξ0 +Bη0 + E]e1dν +
int micro
0[Cξ0 +Dη0 + F ]e2dmicro
(323)
can be determined where micro = x + y ν = x minus y Φ1(z) is a solution of (31) in Dsatisfying the boundary conditions
Re [λ(z)Φ1(z)] = minusRe [λ(z)Ψ1(z)] z isin L
Im [λ(z0)Φ1(z0)] = minusIm [λ(z0)Ψ1(z0)](324)
Thus from (323) (324) we have
w1(z)minus w0(z) = C[w1(z)minus w0(z) D] le 2M4M(4m+ 1)Rprime (325)
where M4 = maxzisinD(|A| |B| |C| |D| |E| |F |) m = w0 C(D) Rprime = max(2R1 2R)M = 1+4k2
0(1+ k20) is a positive constant as in the proof of Theorem 33 Moreover
we substitute w1(z) = w0(z)+Φ1(z)+Ψ1(z) and the corresponding functions ξ1(z) =Rew1(z)+Imw1(z) η1(z) = Rew1(z)minusImw1(z) into the positions of w(z) ξ(z) η(z)in (317) and similarly to (323)ndash(325) we can find the corresponding functionsΨ2(z)Φ2(z) in D and the function
w2(z) = w0(z) + Φ2(z) + Ψ2(z) in D
3 Linear Hyperbolic Equations 23
It is clear that the function w2(z)minus w1(z) satisfies the equality
w2(z)minus w1(z) = Φ2(z)minus Φ1(z) + Ψ2(z)minusΨ1(z)
= Φ2(z)minus Φ1(z) +int ν
0[A(ξ1 minus ξ0) +B(η1 minus η0)]e1dν
+int micro
0[C(ξ1 minus ξ0) +D(η1 minus η0)]e2dmicro
and
w2 minus w1 le [2M3M(4m+ 1)]2int Rprime
0RprimedRprime le [2M3M(4m+ 1)Rprime]2
2
where M3 is a constant as stated in the proof of Theorem 33 Thus we can find asequence of functions wn(z) satisfying
wn(z) = w0(z) + Φn(z) + Ψn(z)
Ψn(z) = +int ν
0[Aξn +Bηn + E]e1dν +
int micro
0[Cξn +Dηn + F ]e2dmicro
(326)
and wn(z)minus wnminus1(z) satisfies
wn(z)minuswnminus1(z)=Φn(z)minusΦnminus1(z) + Ψn(z)minusΨnminus1(z)
=Φn(z)minusΦnminus1(z) +int ν
0[A(ξnminus1minusξnminus2)
+B(ηnminus1minusηnminus2)]e1dν
+int micro
0[C(ξnminus1minusξnminus2)+D(ηnminus1minusηnminus2)]e2dmicro (327)
and then
wn minus wnminus1 le [2M3M(4m+ 1)]nint Rprime
0
Rprimenminus1
(n minus 1) dRprime le [2M3M(4m+ 1)Rprime]n
n (328)
From the above inequality it is seen that the sequence of functions wn(z) ie
wn(z) = w0(z) + [w1(z)minus w0(z)] + middot middot middot+ [wn(z)minus wnminus1(z)](n = 1 2 ) (329)
uniformly converges to a function wlowast(z) and wlowast(z) satisfies the equality
wlowast(z) = w0(z) + Φlowast(z) + Ψlowast(z)
Ψlowast(z) =int ν
0[Aξlowast +Bηlowast + E]e1dν +
int micro
0[Cξlowast +Dηlowast + F ]e2dmicro
(330)
It is easy to see that wlowast(z) satisfies equation (312) and the boundary condition (32)hence it is just a solution of Problem A for the complex equation (312) in the closeddomain D ([87]2)
24 I Hyperbolic Equations of First Order
34 Another boundary value problem for linear hyperbolic complexequations of first order
Now we introduce another boundary value problem for equation (312) in D with theboundary conditions
Re [λ(z)w(z)] = r(z) on L1 cup L5 Im [λ(z1)w(z1)] = b1 (331)
where L1=y=minusx0lexleRL5=y=(R+R1)[x(RminusR1)minus2R1R(R2minusR21)] R1
lexleR=R2minusR1R2 ge2R1 λ(z)=a(z)+jb(z)zisinL1λ(z)=a(z) +jb(z)=1+jzisinL5 and λ(z) r(z) b1 satisfy the conditions
Cα[λ(z) L1] le k0 Cα[r(z) L1cupL5] le k2 |b1| le k2maxzisinL1
1|a(z)minusb(z)| le k0 (332)
in which b1 is a real constant and α (0 lt α lt 1) k0 k2 are non-negative constantsThe boundary value problem is called Problem A1
On the basis of Theorem 11 it is clear that the complex equation (31) can bereduced to the form (36) in D The general solution of system (36) can be expressedas
w(z) = u(z) + jv(z)
= [u(z) + v(z)]e1 + [u(z)minus v(z)]e2
=f(x+y)e1+g(xminus y)e2
=12f(x+y)+g(xminusy)+j[f(x+y)minusg(xminusy)]
(333)
where f(t) (0 le t le 2R) g(t) (0 le t le 2R1) are two arbitrary real continuousfunctions Noting that the boundary condition (331) namely
a(z)u(z) + b(z)v(z)=r(z) on L1 cup L5 λ(z1)w(z1) =r(z1) + jb1 ie
[a((1minus j)x) + b((1minus j)x)]f(0) + [a((1minus j)x)minus b((1minus j)x)]g(2x)
= 2r((1minus j)x) on [0 R1] f(z1)=u(z1)+v(z1)=r(z1)+b1
a(z1)+b(z1)
Re [λ(z)w(z)]=u(z)+v(z)=r[(1+
R+R1
RminusR1j)
xminusj2RR1
RminusR1
]on [R1 R]
(334)
It is easy to see that the above formulas can be rewritten as
[a((1minusj)t2)+b((1minusj)t2)]f(0)+[a((1minusj)t2)minusb((1minusj)t2)]
timesg(t)=2r((1minusj)t2) tisin [0 2R1] f(x+y)=f[( 2R
RminusR1
)xminus 2RR1
RminusR1
]
f(t) = r[((1 + j)R minus (1minus j)R1)
t
2R+ (1minus j)R1
] t isin [0 2R]
4 Quasilinear Hyperbolic Equations 25
and then
g(x minus y) =2r((1minus j)(x minus y)2)
a((1minus j)(x minus y)2)minus b((1minus j)(x minus y)2)
minus [a((1minusj)(xminusy)2)+b((1minusj)(xminusy)2)]f(0)a((1minusj)(xminusy)2)minusb((1minusj)(xminusy)2)
0lexminusyle2R1
f(x+y) = r[((1+j)Rminus(1minusj)R1)x+y
2R+(1minusj)R1] 0lex+yle2R
(335)
Substitute the above function f(x+y) g(xminusy) into (333) the solution w(z) of (36)is obtained We are not difficult to see that w(z) satisfies the estimate
Cα[w(z) D]le M1 Cα[w(z) D] le M2k2 (336)
where M1 = M1(α k0 k2 D) M2 = M2(α k0 D) are two non-negative constants
Next we consider Problem A1 for equation (312) Similarly to before we canderive the representation of solutions w(z) of Problem A1 for (312) as stated in(317) where f(x + y) g(x minus y) possess the form (335) and L = L1 cup L2 z0 in theformula (318) should be replaced by L1 cup L5 z1 Moreover applying the successiveiteration the uniqueness and existence of solutions of Problem A1 for equation (312)can be proved but L z0 in the formulas (321) and (324) are replaced by L1 cupL5 z1We write the results as a theorem
Theorem 35 Suppose that equation (312) satisfies Condition C Then ProblemA1 for (312) has a unique solution w(z) which can be expressed in the form (317)where f(x+ y) g(x minus y) possess the form (335)
4 Boundary Value Problems of Quasilinear HyperbolicComplex Equations of First Order
In this section we mainly discuss the RiemannndashHilbert boundary value problem forquasilinear hyperbolic complex equations of first order in a simply connected domainWe first prove the uniqueness of solutions for the above boundary value problemand then give a priori estimates of solutions of the problem moreover by usingthe successive iteration the existence of solutions for the above problem is provedFinally we also discuss the solvability of the above boundary value problem in generaldomains
26 I Hyperbolic Equations of First Order
41 Uniqueness of solutions of the RiemannndashHilbert problem forquasilinear hyperbolic complex equations
In the subsection we first discuss the quasilinear hyperbolic complex equation
wz = F (z w) F = A1(z w)w + A2(z w)w + A3(z w) in D (41)
whereD is a simply connected bounded domain in the x+jy-plane with the boundaryΓ = L1 cup L2 cup L3 cup L4 as stated in Section 3
Suppose that the complex equation (41) satisfies the following conditions
Condition C
1) Al(z w) (l = 1 2 3) are continuous in z isin D for any continuous complexfunction w(z) and satisfy
C[Al D] = C[ReAl D] + C[ImAl D] le k0 l = 1 2 C[A3 D] le k1 (42)
2) For any continuous complex functions w1(z) w2(z) in D the equality
F (z w1)minusF (z w2)= A1(z w1 w2)(w1minusw2)+A2(z w1 w2)(w1minusw2) in D (43)
holds where C[Al D] le k0 l = 1 2 and k0 k1 are non-negative constants In particu-lar when (41) is a linear equation the condition (43) obviously holds
In order to give an a priori Cα(D)-estimate of solutions for Problem A we needthe following conditions For any hyperbolic numbers z1 z2(isin D) w1 w2 the abovefunctions satisfy
Al(z1 w1)minus Al(z2 w2) le k0[ z1 minus z2 α + w1 minus w2 ] l = 1 2
A3(z1 w1)minus A3(z2 w2) le k2[ z1 minus z2 α + w1 minus w2 ](44)
in which α(0 lt α lt 1) k0 k2 are non-negative constants
Similarly to (312) and (314) due to w = u + jv = ξe1 + ηe2 = ζ wz = ξmicroe1 +ηνe2 wz = ξνe1 + ηmicroe2 the quasilinear hyperbolic complex equation (41) can berewritten in the form
ξνe1 + ηmicroe2 = [A(z ζ)ξ +B(z ζ)η + E(z ζ)]e1
+[C(z ζ)ξ +D(z ζ)η + F (z ζ)]e2 z isin D ie⎧⎨⎩ ξν = A(z ζ)ξ +B(z ζ)η + E(z ζ)
ηmicro = C(z ζ)ξ +D(z ζ)η + F (z ζ)z isin D
(45)
in which
A = ReA1 + ImA1 B = ReA2 + ImA2 C = ReA2 minus ImA2
D = ReA1 minus ImA1 E = ReA3 + ImA3 F = ReA3 minus ImA3
4 Quasilinear Hyperbolic Equations 27
Obviously any solution of Problem A for equation (41) possesses the same rep-resentation (317) as stated in Theorem 32 In the following we prove the existenceand uniqueness of solutions for Problem A for (41) with Condition C
Theorem 41 If Condition C holds then Problem A for the quasilinear complexequation (41) has at most one solution
Proof Let w1(z) w2(z) be any two solutions of Problem A for (41) and substitutethem into equation (41) and boundary condition (32) By Condition C we see thatw(z) = w1(z) minus w2(z) satisfies the homogeneous complex equation and boundaryconditions
wz = A1w + A2w in D (46)
Re [λ(z)w(z)] = 0 z isin L Im [λ(z0)w(z0)] = 0 (47)
On the basis of Theorem 32 the function w(z) can be expressed in the form
w(z) = Φ(z) + Ψ(z)
Ψ(z) =int xminusy
0[Aξ + Bη]e1d(x minus y) +
int x+y
0[Cξ + Dη]e2d(x+ y)
(48)
where the relation between the coefficients A B C D and A1 A2 is the same with thatbetween A B C D and A1 A2 in (45) Suppose w(z) equiv 0 in the neighborhood Q0(subD) of the point z0 = 0 we may choose a sufficiently small positive number R0 lt 1such that 8M5MR0 lt 1 where M5 = maxC[A Q0] C[B Q0] C[C Q0] C[DQ0]Similarly to the proof of Theorem 33 we can derive a contradiction Hence w1(z) =w2(z) in D
42 Solvability of Problem A for quasilinear hyperbolic complexequations
Theorem 42 If the quasilinear complex equation (41) satisfies Condition C thenProblem A for (41) has a solution
Proof Similarly to the proof of Theorem 34 we use the successive iteration tofind a solution of Problem A for the complex equation (41) Firstly substitutingw0(z) = ξ0e1 + η0e2 of Problem A for (31) into the position of w = ξe1 + ηe2 in theright-hand side of (41) the function
w1(z) = w0(z) + Φ1(z) + Ψ1(z)
Ψ1(z) =int ν
0[Aξ0 +Bη0 + E]e1dν +
int micro
0[Cξ0 +Dη0 + F ]e2dmicro
(49)
can be determined where micro = x + y ν = x minus y Φ1(z) is a solution of (31) in Dsatisfying the boundary conditions
Re [λ(z)Φ1(z)] = minusRe [λ(z)Ψ1(z)] z isin L
Im [λ(z0)Φ1(z0)] = minusIm [λ(z0)Ψ1(z0)](410)
28 I Hyperbolic Equations of First Order
Moreover we can find a sequence of functions wn(z) satisfyingwn(z) = w0(z) + Φn(z) + Ψn(z)
Ψn(z) =int ν
0[Aξnminus1 +Bηnminus1 + E]e1dν +
int micro
0[Cξnminus1 +Dηnminus1 + F ]e2dmicro
(411)
and wn(z)minus wnminus1(z) satisfies
wn(z)minuswnminus1(z)=Φn(z)minusΦnminus1(z)+Ψn(z)minusΨnminus1(z)=Φn(z)minus Φnminus1(z)
+int ν
0[A(ξnminus1minusξnminus2)+B(ηnminus1minusηnminus2)]e1dν (412)
+int micro
0[C(ξnminus1minusξnminus2)+D(ηnminus1minusηnminus2)]e2dmicro
Denoting M5 = maxD(|A| |B| |C| |D|) we can obtain
wn minus wnminus1 le [2M5M(4m+ 1)]nint Rprime
0
Rprimenminus1
(n minus 1) dRprime le [2M5M(4m+ 1)Rprime]n
n (413)
in which m = w0 C(Q) Rprime = max(2R1 2R) M = 1 + 4k20(1 + k2
0) is a positiveconstant as in the proof of Theorem 33 The remained proof is identical with theproof of Theorem 34
43 A priori estimates of solutions of the RiemannndashHilbert problem forhyperbolic complex equations
We first give the boundedness estimate of solutions for Problem A
Theorem 43 If Condition C holds then any solution u(z) of Problem A for thehyperbolic equation (41) satisfies the estimates
C[w(z) D] le M6 C[w(z) D] le M7k (414)
in which M6 = M6(α k0 k1 k2 D) k = k1 + k2 M7 = M7(α k0 D) are non-negativeconstants
Proof On the basis of Theorems 41 and 42 we see that under Condition CProblem A for equation (41) has a unique solution w(z) which can be found by usingsuccessive iteration Due to the functions wn+1(z) minus wn(z) (n = 1 2 ) in D arecontinuous the limit function w(z) of the sequence wn(z) in D is also continuousand satisfies the estimate
C[w(z) D] leinfinsum
n=0
[2M5M(4m+ 1)Rprime]n
n= e2M5M(4m+1)Rprime
= M6 (415)
where Rprime = max(2R1 2R) This is the first estimate in (414) As for the second esti-mate in (414) if k = k1+k2 = 0 then it is true from Theorem 41 If k = k1+ k2 gt 0
4 Quasilinear Hyperbolic Equations 29
let the solution w(z) of Problem A be substituted into (41) and (32) and dividingthem by k we obtain the equation and boundary conditions for w(z) = w(z)k
wz = A1w + A2w + A3k z isin D
Re [λ(z)w(z)] = r(z)k z isin D Im [λ(z0)w(z0)] = b1k(416)
Noting that A3k rk b1k satisfy the conditions
C[A3k D] le 1 C[rk L] le 1 |b1k| le 1
by using the method of deriving C[w(z) D] in (415) we can obtain the estimate
C[w(z) D] le M7 = M7(α k0 D)
From the above estimate the second estimate in (414) is immediately derived
Here we mention that in the proof of the estimate (414) we have not requiredthat the coefficients λ(z) r(z) of (32) satisfy a Holder (continuous) condition andonly require that they are continuous on L
Next we shall give the Cα(D)-estimates of solutions of Problem A for (41) andfirst discuss the linear hyperbolic complex equation (312) or (314)
Theorem 44 Suppose that the linear complex equation (312) satisfies the condi-tions (313) and (44) ie the coefficents of (312) satisfies the conditions
Cα[Al D] le k0 l = 1 2 Cα[A3 D] le k2 (417)
in which α(0 lt α lt 1) k0 k2 are non-negative constants Then the solution w(z) =w0(z) + Φ(z) + Ψ(z) satisfies the following estimates
Cα[w0(z) D] le M8 Cα[Ψ(z) D] le M8
Cα[Φ(z) D] le M8 Cα[w(z) D] le M8(418)
where w0(z) is a solution of (31) as stated in (310) M8 = M8(α k0 k1 k2 D) is anon-negative constant
Proof As stated before w0(z) is the function as in (310) which satisfies the esti-mate (311) namely the first estimate in (418) In order to prove that Ψ(z) =Ψ1(z) = Ψ1
1(z)e1 +Ψ21(z)e2 satisfies the second estimate in (418) from
Ψ11(z) =
int xminusy
0G1(z)d(x minus y) Ψ2
1(z) =int x+y
0G2(z)d(x+ y)
G1(z) = A(z)ξ +B(z)η + E(z) G2(z) = C(z)ξ +D(z)η + F (z)(419)
and (417) we see that Ψ11(z) = Ψ1
1(micro ν) Ψ21(z) = Ψ2
1(micro ν) in D with respect toν = x minus y micro = x+ y satisfy the estimates
Cα[Ψ11(middot ν) D] le M9R
prime Cα[Ψ21(micro middot) D] le M9R
prime (420)
30 I Hyperbolic Equations of First Order
respectively where Rprime = max(2R 2R1) M9 = M9(α k0 k1 k2 D) is a non-negativeconstant If we substitute the solution w0 = w0(z) = ξ0e1 + η0e2 of Problem A of(31) into the position of w = ξe1 + ηe2 in (419) and ξ0 = Rew0 + Imw0 η0 =Rew0 minus Imw0 from (417) and (311) we obtain
Cα[G1(micro middot) D] le M10 Cα[G2(middot ν) D] le M10
Cα[Ψ11(micro middot) D] le M10R
prime Cα[Ψ21(middot ν) D] le M10R
prime(421)
in which M10 = M10(α k0 k1 k2 D) is a non-negative constant Due to Φ(z) =Φ1(z) satisfies the complex equation (31) and boundary condition (318) and Φ1(z)possesses a representation similar to that in (317) the estimate
Cα[Φ1(z) D] le M11Rprime = RprimeM11(α k0 k1 k2 D) (422)
can be derived Thus setting w1(z) = w0(z)+Φ1(z)+Ψ1(z) w1(z) = w1(z)minusw0(z) itis clear that the functions w1
1(z) = Re w1(z)+ Im w1(z) w21(z) = Re w1(z) minusIm w1(z)
satisfy as functions of micro = x+ y ν = x minus y respectively the estimates
Cα[w11(middot ν) D] le M12R
prime Cα[w11(micro middot) D] le M12R
prime
Cα[w21(micro middot) D] le M12R
prime Cα[w21(middot ν) D] le M12R
prime(423)
where M12 = 2M13M(4m + 1) M = 1 + 4k20(1 + k0) m = Cα[w0 D] M13 =
maxD[|A| |B| |C| |D| |E| |F |] By using successive iteration we obtain thesequence of functions wn(z) (n = 1 2 ) and the corresponding functions w1
n =Re wn + Im wn w2
n = Re wn minus Im wn satisfying the estimates
Cα[w1n(middot ν) D] le
(M12Rprime)n
n Cα[w1
n(micro middot) D] le (M12Rprime)n
n
Cα[w2n(micro middot) D] le (M12R
prime)n
n Cα[w2
n(middot ν) D] le(M12R
prime)n
n
(424)
and denote by w(z) the limit function of wn(z) =sumn
m=0 wn(z) in D the correspondingfunctions w1 = Rew(z) + Imw(z) w2 = Rew(z)minus Imw(z) satisfy the estimates
Cα[w1(middot ν) D] le eM12Rprime Cα[w1(micro middot) D] le eM12Rprime
Cα[w2(micro middot) D] le eM12Rprime Cα[w2(middot ν) D] le eM12Rprime
Combining the first formula in (418) (420)ndash(424) and the above formulas the lastthree estimates in (418) are derived
Theorem 45 Let the quasilinear complex equation (41) satisfy Condition C and(44) Then the solution w(z) of Problem A for (41) satisfies the following estimates
Cα[w(z) D] le M14 Cα[w(z) D] le M15k (425)
where k = k1+k2 M14 = M14(α k0 k1 k2 D) M15 = M15(α k0 D) are non-negativeconstants
4 Quasilinear Hyperbolic Equations 31
Proof According to the proof of Theorem 43 from the first formula in (425) thesecond formula in (425) is easily derived Hence we only prove the first formulain (425) Similarly to the proof of Theorem 44 we see that the function Ψ1(z) =Ψ1
1(micro ν)e1+Ψ21(micro ν)e2 still possesses the estimate (420) Noting that the coefficients
are the functions of z isin D and w and applying the condition (44) we can derivesimilar estimates as in (421) Hence we also obtain estimates similar to (422) and(423) and the constant M12 in (423) can be chosen as M12 = 2M13M(4m+1) m =Cα[w0(z) D] Thus the first estimate in (425) can be derived
Moreover according to the above methods we can obtain estimates for [Rew +Imw]ν [Rew minus Imw]micro analogous to those in (414) and (425)
44 The boundary value problem for quasilinear hyperbolic equations offirst order in general domains
In this subsection we shall generalize the domain D to general cases
1 The boundary L1 of the domain D is replaced by a curve Lprime1 and the boundary
of the domain Dprime is Lprime1 cup Lprime
2 cup L3 cup L4 where the parameter equations of the curvesLprime
1 Lprime2 are as follows
Lprime1 = γ1(x) + y = 0 0 le x le l Lprime
2 = x minus y = 2R1 l le x le R2 (426)
in which γ1(x) on 0 le x le l = 2R1 minus γ1(l) is continuous and γ1(0) = 0 γ1(x) gt 0on 0 lt x le l and γ1(x) is differentiable except at isolated points on 0 le x le l and1 + γprime
1(x) gt 0 By this condition the inverse function x = σ(ν) of x + γ1(x) = νcan be found and σprime(ν) = 1[1 + γprime
1(x)] hence the curve Lprime1 can be expressed by
x = σ(ν) = (micro+ ν)2 ie micro = 2σ(ν)minus ν 0 le ν le 2R1 We make a transformation
micro =2R[micro minus 2σ(ν) + ν]2R minus 2σ(ν) + ν
ν = ν 2σ(ν)minus ν le micro le 2R 0 le ν le 2R1 (427)
its inverse transformation is
micro =12R[2R minus 2σ(ν) + ν]micro+ 2σ(ν)minus ν ν = ν 0 le micro le 2R 0 le ν le 2R1 (428)
The transformation (427) can be expressed by⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
x =12(micro+ ν)
=2R(x+ y) + 2R(x minus y)minus (2R + x minus y)[2σ(x+ γ1(x))minus x minus γ1(x)]
4R minus 4σ(x+ γ1(x)) + 2x+ 2γ1(x)
y =12(micro minus ν)
=2R(x+ y)minus 2R(x minus y)minus (2R minus x+ y)[2σ(x+ γ1(x))minus x minus γ1(x)]
4R minus 4σ(x+ γ1(x)) + 2x+ 2γ1(x)
(429)
32 I Hyperbolic Equations of First Order
where γ1(x) = minusy and its inverse transformation is⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
x =12(micro+ ν) =
[2R minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+ y)4R
+σ(x+ γ1(x))minus x+ γ1(x)minus x+ y
2
y =12(micro minus ν) =
[2R minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+ y)4R
+σ(x+ γ1(x))minus x+ γ1(x) + x minus y
2
(430)
Denote by z = x+ jy = f(z) z = x+ jy = fminus1(z) the transformation (429) and theinverse transformation (430) respectively In this case the system of equations andboundary conditions are
ξν = Aξ +Bη + E ηmicro = Cξ +Dη + F z isin Dprime (431)
Re [λ(z)w(z)] = r(z) z isin Lprime1 cup Lprime
2 Im [λ(z0)w(z0)] = b1 (432)
in which zprime0 = l minus jγ1(l) λ(z) r(z) b1 on Lprime
1 cup Lprime2 satisfy the conditions (33)(34)
Suppose system (431) in Dprime satisfies Condition C through the transformation (428)and ξν = ξν ηmicro = [2R minus 2σ(ν) + ν]ηmicro2R system (431) is reduced to
ξν = Aξ +Bη + E ηmicro = [2R minus 2σ(ν) + ν][Cξ +Dη + F ]
2R (433)
Moreover through the transformation (430) ie z = fminus1(z) the boundary condition(432) is reduced to
Re [λ(fminus1(z))w(fminus1(z))] = r(fminus1(z)) z isin L1 cup L2
Im [λ(fminus1(z0))w(fminus1(z0)] = b1(434)
in which z0 = f(z0) = 0 Therefore the boundary value problem (431)(432) istransformed into the boundary value problem (433)(434) On the basis of Theorems41 and 42 we see that the boundary value problem (433) (434) has a uniquesolution w(z) and then w[f(z)] is just a solution of the boundary value problem(431)(432) in Dprime
Theorem 46 If the complex equation (41) satisfies Condition C in the domainDprime with the boundary Lprime
1 cup Lprime2 cup L3 cup L4 where Lprime
1 Lprime2 are as stated in (426) then
Problem A with the boundary conditions
Re [λ(z)w(z)] = r(z) z isin Lprime1 cup Lprime
2 Im [λ(z0)w(z0)] = b1
has a unique solution w(z)
4 Quasilinear Hyperbolic Equations 33
2 The boundary L1 L4 of the domain D are replaced by the two curves Lprimeprime1 Lprimeprime
4respectively and the boundary of the domain Dprimeprime is Lprimeprime
1 cup Lprimeprime2 cup Lprimeprime
3 cup Lprimeprime4 where the
parameter equations of the curves Lprimeprime1 Lprimeprime
2 Lprimeprime3 Lprimeprime
4 are as follows
Lprimeprime1 = γ1(x) + y = 0 0 le x le l1 Lprimeprime
2 = x minus y = 2R1 l1 le x le R2
Lprimeprime3 = x+ y = 2R l2 le x le R2 Lprimeprime
4 = minusγ4(x) + y = 0 0 le x le l2(435)
in which and γ1(0) = 0 γ4(R2) = 2R minus R2 γ1(x) gt 0 0 le x le l1 γ4(x) gt 0 0 lex le l2 γ1(x) on 0 le x le l1 γ4(x) on 0 le x le l2 are continuous and γ1(x) γ4(x)possess derivatives except at finite points on 0 le x le l1 0 le x le l2 respectively and1 + γprime
1(x) gt 0 1 + γprime4(x) gt 0 zprimeprime
1 = x minus jγ1(l1) isin L2 zprimeprime3 = x + jγ2(l2) isin L3 By the
conditions the inverse functions x = σ(ν) x = τ(micro) of x+ γ1(x) = ν x+ γ4(x) = microcan be found respectively namely
micro = 2σ(ν)minus ν 0 le ν le l1 + γ1(l1) ν = 2τ(micro)minus micro l2 + γ4(l2) le micro le 2R1 (436)
We make a transformation
micro = micro ν =2R1(ν minus 2τ(micro) + micro)2R1 minus 2τ(micro) + micro
0 le micro le 2R 2τ(micro)minus micro le micro le 2R1 (437)
its inverse transformation is
micro = micro ν =2R1 minus 2τ(micro) + micro
2R1ν + 2τ(micro)minus ν 0 le micro le 2R 0 le ν le 2R1 (438)
Hence we have
x =12(micro+ ν) =
4R1x minus [2τ(x+ γ1(x))minus x minus γ1(x)](2R1 + x+ y)4R1 minus 4τ(x+ γ1(x)) + 2x+ 2γ1(x)
y =12(micro minus ν) =
2R1y + [2τ(x+ γ1(x))minus x minus γ1(x)](2R1 minus x minus y)4R1 minus 4τ(x+ γ1(x)) + 2x+ 2γ1(x)
(439)
and its inverse transformation is
x =12(micro+ ν) =
4R1x+ [2τ(x+ γ1(x))minus x minus γ1(x)](2R1 minus x+ y)4R1
y =12(micro minus ν) =
4R1y minus [2τ(x+ γ1(x))minus x minus γ1(x)](2R1 + x minus y)4R1
(440)
Denote by z = x + jy = g(z) z = x + jy = gminus1(z) the transformation (439) andits inverse transformation (440) respectively Through the transformation (438) wehave
(u+ v)ν =2R1 minus 2τ(micro) + micro
2R1(u+ v)ν (u minus v)micro = (u minus v)micro
system (431) in Dprimeprime is reduced to
ξν =2R1 minus 2τ(micro) + micro
2R1[Aξ +Bη + E] ηmicro = Cξ +Dη + F z isin Dprime (441)
34 I Hyperbolic Equations of First Order
where Dprime is a bounded domain with boundary Lprime1 cup Lprime
2 cup L3 cup L4 and Lprime1 = Lprimeprime
1 More-over through the transformation (440) the boundary condition (432) on Lprimeprime
1 cup Lprimeprime4
is reduced to
Re [λ(gminus1(z))w(gminus1(z))] = r(gminus1(z)) z isin Lprime1 cup L4
Im [λ(gminus1(z0))w(gminus1(z0)] = b1(442)
in which z0 = g(z0) = 0 Therefore the boundary value problem (431)(432) in Dprimeprime istransformed into the boundary value problem (441)(442) On the basis of Theorem46 we see that the boundary value problem (441)(442) has a unique solution w(z)and then w[g(z)] is just a solution of the boundary value problem (431)(432)
Theorem 47 If the complex equation (41) satisfies Condition C in the domainDprimeprime with the boundary Lprimeprime
1 cupLprimeprime2 cupLprimeprime
3 cupLprimeprime4 then Problem A with the boundary conditions
Re [λ(z)w(z)] = r(z) z isin Lprimeprime1 cup Lprimeprime
4 Im [λ(z0)w(z0)] = b1
has a unique solution w(z) where zprimeprime1 = x minus jγ1(l1) isin L2 z
primeprime3 = x+ jγ4(l2) isin L3
Now we give an example to illustrate the above results When R2 = 2R1 theboundary of the domain D is L1 = x = minusy 0 le x le R1 L2 = x = y + 2R1 R1 lex le R1 L3 = x = minusy + 2R1 R1 le x le 2R1 L4 = x = y 0 le x le R1 Wereplace L1 cup L4 by a left semi-circumference Lprimeprime
1 cup Lprimeprime4 with the center R1 and the
radius R1 namely
Lprimeprime1 = x minus y = ν y = minusγ1(x) = minus
radicR2
1 minus (x minus R1)2 0 le x le R1
Lprimeprime4 = x+ y = micro y = γ4(x) =
radicR2
1 minus (x minus R1)2 0 le x le R1
where 1 + γprime1(x) gt 0 0 lt x le R1 1 + γprime
4(x) gt 0 0 le x lt R1 and
x = σ(ν) =R1 + ν minus
radicR2
1 + 2R1ν minus ν2
2
x = τ(micro) =R1 + micro minus
radicR2
1 + 2R1micro minus micro2
2
It is clear that according to the above method the domain D can be generalized toa general domainDprimeprime namely its boundary consists of the general curves Lprimeprime
1 Lprimeprime2 L
primeprime3 L
primeprime4
with some conditions which includes the circumference L = |z minus R1| = R1Finally we mention that some boundary value problems for equations (312) and
(41) with one of the boundary conditions
Re [λ(z)w(z)] = r(z) z isin L1 cup L2 Im [λ(z1)w(z1)] = b1
Re [λ(z)w(z)] = r(z) z isin L3 cup L4 Im [λ(z3)w(z3)] = b1
Re [λ(z)w(z)] = r(z) z isin L1 cup L5 Im [λ(z0)w(z0)] = b1
Re [λ(z)w(z)] = r(z) z isin L1 cup L6 Im [λ(z0)w(z0)] = b1
5 Quasi-hyperbolic Mappings 35
can be discussed where λ(z) r(z) b1 satisfy the conditions similar to those in(33)(34)(331) and Lj(j = 1 5) zj(j = 0 1 3) are as stated in Section 3
L6 = y =R2 minus 2R1
R2x 0 le x le R2 and λ(z) = 1 + j z isin L5 λ(z) = 1minus j z isin L6
For corresponding boundary value problems of hyperbolic systems of first ordercomplex equations whether there are similar results as before The problem needsto be investigated
5 Hyperbolic Mappings and Quasi-hyperbolic Mappings
Now we introduce the definitions of hyperbolic mappings and quasi-hyperbolic map-pings and prove some properties of quasi-hyperbolic mappings
51 Hyperbolic mappings
A so-called hyperbolic mapping in a domain D is a univalent mapping given by ahyperbolic continuously differentiable function w = f(z) = u + jv satisfying thesimplest hyperbolic system of first order equations
ux = vy vx = uy (51)
which maps D onto a domain G in the w-plane By Theorem 11 system (51) isequivalent to the system
ξν = 0 ηmicro = 0 (52)
where ξ = u+ v η = u minus v micro = x+ y ν = x minus y Noting∣∣∣∣∣microx νx
microy νy
∣∣∣∣∣ =∣∣∣∣∣ 1 1
1 minus1
∣∣∣∣∣ = minus2∣∣∣∣∣ ξu ηu
ξv ηv
∣∣∣∣∣ =∣∣∣∣∣ 1 1
1 minus1
∣∣∣∣∣ = minus2 (53)
if we find a homeomorphic solution of (52) then the solution
w = u+ jv = ξ[(micro+ ν)2 (micro minus ν)2]e1 + η[(micro+ ν)2 (micro minus ν)2]e2 (54)
of the corresponding system (51) is also homeomorphic In fact
ξ = ξ(micro) η = η(ν) (55)
is a homeomorphic solution of (52) if ξ(micro) and η(ν) are strictly monotonous continu-ous functions of micro(micro0 le micro le micro1) and ν(ν0 le ν le ν1) respectively When [ξ(micro) η(ν)]is univalent continuous in ∆ = micro0 le micro le micro1 ν0 le ν le ν1 and D(∆) is the closeddomain in the z = x + jy-plane corresponding to ∆ it is easy to see that[u(x y) v(x y)] is a homeomorphic solution of (51) in D(∆)
36 I Hyperbolic Equations of First Order
52 Quasi-hyperbolic mappings
In this subsection we first discuss the uniformly hyperbolic system in the complexform
wz = Q(z)wz Q(z) = a(z) + jb(z) (56)
where Q(z) is a continuous function satisfying the condition |Q(z)| le q0 lt 1 here q0
is a non-negative constant On the basis of the representation
wz = ξνe1 + ηmicroe2 wz = ξmicroe1 + ηνe2 Q = q1e1 + q2e2
from (56) it follows thatξν = q1ξmicro ηmicro = q2ην (57)
Due to Q = a + jb = q1e1 + q2e2 here q1 = a + b q2 = a minus b thus |Q|2 = |QQ| =|a2 minus b2| = |q1q2| le q2
0 lt 1 and a representation theorem of solutions for (56) can beobtained
Theorem 51 Let χ(z) be a homeomorphic solution of (56) in a domain D andw(z) be a solution of (56) in D Then w(z) can be expressed as
w(z) = Φ[χ(z)] (58)
where Φ(χ) is a hyperbolic regular function in the domain G = χ(D)
Proof Suppose that z(χ) is the inverse function of χ(z) we can find
w[z(χ)]χ = wzzχ + wz zχ = wz[zχ +Qzχ]
andχχ = 1 = χzzχ + χz zχ = χz[zχ +Qzχ]
χχ = 0 = χzzχ + χz zχ = χz[zχ +Qzχ]
From the above equalities we see χz = 0 zχ+Qzχ = 0 consequently Φ(χ) = w[z(χ)]satisfies
[Φ(χ)]χ = 0 χ isin G = χ(D)
This shows that Φ(χ) is a hyperbolic regular function in G = χ(D) therefore therepresentation (58) holds
Next we prove the existence of a homeomorphic solution of equation (56) withsome conditions for the coefficient of (56)
From (57) we see that the complex equation (56) can be written in the form
ξν = (a+ b)ξmicro ηmicro = (a minus b)ην (59)
Let ∆ = micro0 le micro le micro0 + R1 ν0 le ν le ν0 + R2 in which micro0 ν0 are two real num-bers and R1 R2 are two positive numbers if a b possess continuously differentiable
5 Quasi-hyperbolic Mappings 37
derivatives with respect to micro ν in ∆ then the solution w = ξe1+ ηe2 of (59) in ∆ isa homeomorphism provided that one of the following sets of conditions
b gt 0 minusb lt a lt b (510)
b lt 0 b lt a lt minusb (511)
holds and ξ and η are strictly monotonous continuous functions of micro (micro0 le micro lemicro0 +R1) and ν (ν0 le ν le ν0 +R2) respectively
Thus we have the following theorem
Theorem 52 Denote by D the corresponding domain of ∆ in the (x + jy)-planeand let w(z) be a continuous solution of (56) in D If (510) or (511) in D holdsξmicro(ξν) gt 0 and ηmicro (ην) gt 0 except some possible isolated points on micro (micro0 le micro lemicro0 +R1) (ν (ν0 le ν le ν0 +R2)) then the solution w(z) in D is a homeomorphism
In particular if the coefficient Q(z) = a + jb of the complex equation (56) is ahyperbolic constant which satisfies the condition
|Q(z)|2 = |QQ| = |a2 minus b2| = |q1q2| le q20 lt 1 z isin D
we make a nonsingular transformation
micro = minus(a+ b)σ + τ ν = σ minus (a minus b)τ (512)
Thus system (59) can be transformed into the system
ξσ = 0 ητ = 0 (σ τ) isin G (513)
where the domain G is the corresponding domain of ∆ under the transformation(512) According to the discussion of hyperbolic mappings we see that system (513)in G possesses a homeomorphic solution hence system (59) in ∆ has a homeomorphicsolution and then the complex equation (56) in D has a homeomorphic solution Theabove result can be written as a theorem
Theorem 53 Suppose that Q(z) = a+ jb is a hyperbolic constant and |Q(z)| lt 1Then the complex equation (56) in D has a homeomorphic solution
53 Other Quasi-hyperbolic mappings
Now we consider the hyperbolic complex equation
wz = Q(z)wz (514)
where Q(z) is a continuous function in D = 0 le x le R1 0 le y le R2 satisfyingthe condition |Q(z)| le q0 lt 1 If Q(z) in D is a hyperbolic regular function of z weintroduce a transformation of functions
W = w minus Qw ie w =W +Q(z)W1minus |Q(z)|2 (515)
38 I Hyperbolic Equations of First Order
Then (514) is reduced to the complex equation
Wz = 0 (516)
The solution W (z) of (516) in D is a hyperbolic regular function of z As statedbefore the complex equation (516) possesses a homeomorphic solution which realizesa hyperbolic mapping in D Moreover if Q(z) is a hyperbolic regular function in Dwe find the partial derivative with respect to z in (514) and obtain
wzz = Q(z)wzz ie wzz = Q(z)wzz (517)
from |Q(z)| lt 1 it follows that
(1minus |Q(z)|2)wzz = 0 ie wzz = 0 (518)
the solution of the above complex equation (518) is called a hyperbolic harmoniccomplex function
A hyperbolic harmonic complex function w(z) can be expressed as
w(z) = u(z) + jv(z) = φ(z) + φ(z) + ψ(z)minus ψ(z)
= φ(z) + ψ(z) + φ(z)minus ψ(z) = f(z) + g(z)
in which φ(z) ψ(z) are hyperbolic regular functions hence f(z) = φ(z)+ψ(z) g(z) =φ(z) minusψ(z) are hyperbolic regular functions This is a representation of hyperbolicharmonic functions through hyperbolic regular functions Hence in order to find ahyperbolic harmonic function it is sufficient to find solutions of the following twoboundary value problems with the boundary conditions
Re f(x) = Reφ0(x) Re f(jy) = Reφ1(y)
andIm g(x) = Imφ0(x) Im g(jy) = Imφ1(y)
respectively where φ0(x) φ1(y) are given hyperbolic complex functions on0 le x le R1 0 le y le R2 respectively and R1 R2 are two positive constants
At last we mention that the notations of hyperbolic numbers and hyperboliccomplex functions are mainly used in this and next chapters From Chapter III toChapter VI except in Section 5 Chapter V we do not use them
The references for this chapter are [5][9][12][19][26][29][32][34][38][44][51][59][68][74][80][83][85][87][89][92][97]
CHAPTER II
HYPERBOLIC COMPLEX EQUATIONS OFSECOND ORDER
In this chapter we mainly discuss oblique derivative boundary value problems forlinear and quasilinear hyperbolic equations of second order in a simply connecteddomain Firstly we transform some linear and nonlinear uniformly hyperbolic equa-tions of second order with certain conditions into complex forms give the uniquenesstheorem of solutions for the above boundary value problems Moreover by usingthe successive iteration the existence of solutions for several oblique derivative prob-lems is proved Finally we introduce some boundary value problems for degeneratehyperbolic equations of second order with certain conditions
1 Complex Form of Hyperbolic Equations of Second Order
This section deals with hyperbolic equations of second order in the plane domains wefirst transform some linear and nonlinear uniformly hyperbolic equations of secondorder with certain conditions into complex forms and then we state the conditionsof some hyperbolic complex equations of second order
11 Reduction of linear and nonlinear hyperbolic equations of secondorder
Let D be a bounded domain we consider the linear hyperbolic partial differentialequation of second order
auxx + 2buxy + cuyy + dux + euy + fu = g (11)
where the coefficients a b c d e f g are known continuous functions of (x y) isin Din which D is a bounded domain The condition of hyperbolic type for (11) is thatfor any point (x y) in D the inequality
I = ac minus b2 lt 0 a gt 0 (12)
holds If a b c are bounded in D and
I = ac minus b2 le I0 lt 0 a gt 0 (13)
40 II Hyperbolic Equations of Second Order
in D where I0 is a negative constant then equation (11) is called uniformly hyper-bolic in D Introduce the notations as follows
( )z =( )x + j( )y
2 ( )z =
( )x minus j( )y2
( )zz =( )xx minus ( )yy
4
( )zz =( )xx + ( )yy + 2j( )xy
4 ( )zz =
( )xx + ( )yy minus 2j( )xy
4
( )x = ( )z + ( )z ( )y = j[( )z minus ( )z] ( )xy = j[( )zz minus ( )zz]
( )xx = ( )zz + ( )zz + 2( )zz ( )yy = ( )zz + ( )zz minus 2( )zz
(14)
equation (11) can be written in the form
2(a minus c)uzz + (a+ c+ 2bj)uzz + (a+ c minus 2bj)uzz
+(d+ ej)uz + (d minus ej)uz + fu = g in D(15)
If a = c in D then equation (15) can be reduced to the complex form
uzz minus Re [Q(z)uzz + A1(z)uz]minus A2(z)u = A3(z) in D (16)
in which
Q =a+ c+ 2bj
a minus c A1 =
d+ ej
a minus c A2 =
f
2(a minus c) A3 =
g
2(a minus c)
If (a+ c)2 ge 4b2 then the conditions of hyperbolic type and uniformly hyperbolic aretransformed into
|Q(z)| lt 1 in D (17)
and|Q(z)| le q0 lt 1 in D (18)
respectively
For the nonlinear hyperbolic equation of second order
Φ(x y u ux uy uxx uxy uyy) = 0 in D (19)
from (14) we have Φ = F (z u uz uzz uzz) Under certain conditions equation (19)can be reduced to the real form
auxx + 2buxy + cuyy + dux + euy + fu = g in D (110)
and its complex form is as follows
a0uzz minus Re [quzz + a1uz]minus a2u = a3 in D (111)
1 Complex Form of Hyperbolic Equations 41
in which
a =int 1
0Φτuxx(x y u ux uy τuxx τuxy τuyy)dτ = a(x y u ux uy uxx uxy uyy)
2b =int 1
0Φτuxy(x y u ux uy τuxx τuxy τuyy)dτ = 2b(x y u ux uy uxx uxy uyy)
c =int 1
0Φτuyy(x y u ux uy τuxx τuxy τuyy)dτ = c(x y u ux uy uxx uxy uyy)
d =int 1
0Φτux(x y u τux τuy 0 0 0)dτ = d(x y u ux uy)
e =int 1
0Φτuy(x y u τux τuy 0 0 0)dτ = e(x y u ux uy)
f =int 1
0Φτu(x y τu 0 0 0 0 0)dτ = f(x y u)
g = minusΦ(x y 0 0 0 0 0 0) = g(x y)
and
a0 = 2(a minus c) =int 1
0Fτuzz(z u uz τuzz τuzz)dτ = a0(z u uz uzz uzz)
q=2(a+c+2bj)=minus2int 1
0Fτuzz(z u uz τuzz τuzz)dτ=q(z u uz uzz uzz)
a1 = 2(d+ ej) = minus2int 1
0Fτuz(z u τuz 0 0 0)dτ = a1(z u uz)
a2 = f = minusint 1
0Fτu(z τu 0 0 0 0)dτ = a2(z u)
a3 = minusF (z 0 0 0 0) = a3(z)
(112)
The condition of uniformly hyperbolic type for equation (110) is the same with (13)If a = c in D the complex equation (111) can be rewritten in the form
uzz minus Re [Quzz + A1uz]minus A2u = A3 in D (113)
whereQ = qa0 A1 = a1a0 A2 = a2a0 A3 = a3a0
are functions of z isin D u uz uzz uzz the condition of uniformly hyperbolic type for(113) is as stated in the form (18)
As stated in [12] 3) for the linear hyperbolic equation (11) or its complex form(16) if the coefficients a b c are sufficiently smooth through a nonsingular transfor-mation of z equation (11) can be reduced to the standard form
uxx minus uyy + dux + euy + fu = g (114)
or its complex form
uzz minus Re [A1(z)uz]minus A2(z)u = A3(z) (115)
42 II Hyperbolic Equations of Second Order
12 Conditions of some hyperbolic equations of second order
Let D be a simply connected bounded domain with the boundary Γ = L1 cup L2 cupL3 cup L4 as stated in Chapter I where L1 = x = minusy 0 le x le R1 L2 = x =y + 2R1 R1 le x le R2 L3 = x = minusy minus 2R1 + 2R2 R2 minus R1 le x le R2 L4 = x =y 0 le x le R2 minus R1 and denote z0 = 0 z1 = (1minus j)R1 z2 = R2+ j(R2 minus 2R1) z3 =(1 + j)(R2 minus R1) L = L3 cup L4 here there is no harm in assuming that R2 ge 2R1In the following we mainly consider second order quasilinear hyperbolic equation inthe form
uzz minus Re [A1(z u uz)uz]minus A2(z u uz)u = A3(z u uz) (116)
whose coefficients satisfy the following conditions Condition C
1) Al(z u uz)(l = 1 2 3) are continuous in z isin D for all continuously differentiablefunctions u(z) in D and satisfy
C[Al D] = C[ReAl D] + C[ImAl D] le k0 l = 1 2 C[A3 D] le k1 (117)
2) For any continuously differentiable functions u1(z) u2(z) in D the equality
F (z u1 u1z)minus F (z u2 u2z) = Re [A1(u1 minus u2)z] + A2(u1 minus u2) in D (118)
holds whereC[Al(z u1 u2) D] le k0 l = 1 2 (119)
in (117)(119) k0 k1 are non-negative constants In particular when (116) is alinear equation from (117) it follows that the conditions (118) (119) hold
In order to give a priori estimates in Cα(D) of solutions for some boundary valueproblems we need to add the following conditions For any two real numbers u1 u2
and hyperbolic numbers z1 z2 isin D w1 w2 the above functions satisfy
Al(z1 u1 w1)minus Al(z2 u2 w2) le k0[ z1 minus z2 α + u1 minus u2 α + w1 minus w2 ] l = 1 2 A3(z1 u1 w1)minus A3(z2 u2 w2) le k1[ z1 minus z2 α + u1 minus u2 α + w1 minus w2 ]
(120)
where α (0 lt α lt 1) k0 k1 are non-negative constants
It is clear that (116) is the complex form of the following real equation of secondorder
uxx minus uyy = aux + buy + cu+ d in D (121)
in which a b c d are functions of (x y)(isin D) u ux uy(isin IR) and
A1 =a+ jb
2 A2 =
c
4 A3 =
d
4in D
2 Quasilinear Hyperbolic Equations 43
Due to z = x+ jy = microe1 + νe2 w = uz = ξe1 + ηe2 and
wz =wx + jwy
2= ξmicroe1 + ηνe2 wz =
wx minus jwy
2= ξνe1 + ηmicroe2
the quasilinear hyperbolic equation (116) can be rewritten in the form
ξνe1 + ηmicroe2 = [A(z u w)ξ +B(z u w)η + C(z u w)u+D(z u w)]e1
+[A(z u w)ξ +B(z u w)η + C(z u w)u+D(z u w)]e2 ie⎧⎨⎩ ξν = A(z u w)ξ +B(z u w)η + C(z u w)u+D(z u w)
ηmicro = A(z u w)ξ + A(z u w)η + C(z u w)u+D(z u w)in D
(122)
in which
A =a+ b
4 B =
a minus b
4 C =
c
4 D =
d
4
In the following we mainly discuss the oblique derivative problem for linearhyperbolic equation (16) and quasilinear hyperbolic equation (116) in thesimply connected domain We first prove that there exists a unique solution of theboundary value problem and give a priori estimates of their solutions and then provethe solvability of the boundary value problem for general hyperbolic equations
2 Oblique Derivative Problems for Quasilinear HyperbolicEquations of Second Order
Here we first introduce the oblique derivative problem for quasilinear hyperbolic equa-tions of second order in a simply connected domain and give the representationtheorem of solutions for hyperbolic equations of second order
21 Formulation of the oblique derivative problem and the representationof solutions for hyperbolic equations
The oblique derivative problem for equation (116) may be formulated as follows
Problem P Find a continuously differentiable solution u(z) of (116) in D satisfyingthe boundary conditions
12
partu
partl= Re [λ(z)uz] = r(z) z isin L = L3 cup L4
u(0) = b0 Im [λ(z)uz]|z=z3 = b1
(21)
44 II Hyperbolic Equations of Second Order
where l is a given vector at every point on L λ(z) = a(z) + jb(z) = cos(l x) +j cos(l y) z isin L b0 b1 are real constants and λ(z) r(z) b0 b1 satisfy the conditions
Cα[λ(z) L] le k0 Cα[r(z) L] le k2 |b0| |b1| le k2
maxzisinL3
1|a(z)minus b(z)| maxzisinL4
1|a(z) + b(z)| le k0
(22)
in which α(0 lt α lt 1) k0 k2 are non-negative constants The above boundary valueproblem for (116) with A3(z u w) = 0 z isin D u isin IR w isin CI and r(z) = b0 = b1 =0 z isin L will be called Problem P0
By z = x+ jy = microe1+ νe2 w = uz = ξe1+ ηe2 the boundary condition (21) canbe reduced to
Re [λ(z)(ξe1 + ηe2)] = r(z) u(0) = b0 Im [λ(z)(ξe1 + ηe2)]|z=z3 = b1 (23)
where λ(z) = (a + b)e1 + (a minus b)e2 Moreover the domain D is transformed intoQ = 0 le micro le 2R 0 le ν le 2R1 R = R2 minus R1 which is a rectangle and A B C Dare known functions of (micro ν) and unknown continuous functions u w and they satisfythe condition
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
λ(z3)w(z3) = λ(z)[ξe1 + ηe2]|z=z3 = r(z3) + jb1 u(0) = b0
Re [λ(z)w(2Re1 + νe2)] = r(z)
if (x y) isin L3 = micro = 2R 0 le ν le 2R1
Re [λ(z)w(microe1 + 0e2)] = r(z)
if (x y) isin L4 = 0 le micro le 2R ν = 0
(24)
where λ(z) r(z) b0 b1 are as stated in (21) We can assume that w(z3) = 0otherwise through the transformation W (z) = w(z) minus [a(z3) minus jb(z3)][r(z3) + jb1][a2(z3)minus b2(z3)] the requirement can be realized
It is not difficult to see that the oblique derivative boundary value problem (Prob-lem P ) includes the Dirichlet boundary value problem (Problem D) as a special caseIn fact the boundary condition of Dirichlet problem (Problem D) for equation (121)is as follows
u(z) = φ(z) on L = L3 cup L4 (25)
We find the derivative with respect to the tangent direction s = (x∓jy)radic2 for (25)
in which ∓ are determined by L3 and L4 respectively it is clear that the followingequalities hold
Re [λ(z)uz] = r(z) z isin L Im [λ(z)uz]|z=z3 = b1 (26)
2 Quasilinear Hyperbolic Equations 45
in which
λ(z) = a+ jb =
⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩1minus jradic2on L3
1 + jradic2on L4
r(z) =φxradic2on L = L3 cup L4
b+1 = Im
[1minus jradic2
uz(z3)]= minusφx + φx
2radic2
|z=z3minus0 = minusradic2φx|z=z3minus0
bminus1 = Im
[1 + jradic2
uz(z3)]=
radic2φx|z=z3+0 b0 = φ(0)
(27)
in which a = 1radic2 = b = minus1radic2 on L3 and a = 1
radic2 = minusb = minus1radic2 on L4
Noting that Problem P for (116) is equivalent to the RiemannndashHilbert problem(Problem A) for the complex equation of first order and boundary conditions
wz = Re [A1w] + A2u+ A3 in D (28)
Re [λ(z)w(z)] = r(z) z isin L Im [λ(z)w(z)]|z=z3 = b1 (29)
and the relationu(z) = 2Re
int z
0w(z)dz + b0 in D (210)
from Theorem 12 (3) Chapter I and equation (28) we see that
2ReintΓw(z)dz =
intΓw(z)dz +
intΓw(z)dz
= 2jint int
D[wz minus wz]dxdy = 4
int intDIm [wz]dxdy = 0
the above equality for any subdomain in D is also true hence the function determinedby the integral in (210) is independent of integral paths in D In this case wemay choose that the integral path is along two family of characteristic lines namelyfirst along one of characteristic lines x + y = micro (0 le micro le 2R) and then alongone of characteristic lines x minus y = ν (0 le ν le 2R1) for instance the value ofu(zlowast)(zlowast = xlowast + jylowast isin D ylowast le 0) can be obtained by the integral
u(zlowast) = 2Re[int
s1
w(z)dz +int
s2
w(z)dz]+ b0
where s1 = x+y = 0 0 le x le (xlowast minusylowast)2 s2 = xminusy = xlowast minusylowast (xlowast minusylowast)2 le x lexlowastminusylowast in which xlowastminusylowast is the intersection of the characteristic line xminusy = xlowastminusylowastthrough the point zlowast and real axis In particular when Aj = 0 j = 1 2 3 equation(116) becomes the simplest hyperbolic complex equation
uzz = 0 (211)
46 II Hyperbolic Equations of Second Order
Problem P for (211) is equivalent to Problem A for the simplest hyperbolic complexequation of first order
wz = 0 in D (212)
with the boundary condition (29) and the relation (210) Hence similarly to Theorem31 Chapter I we can derive the representation and existence theorem of solutionsof Problem A for the simplest equation (212) namely
Theorem 21 Any solution u(z) of Problem P for the hyperbolic equation (211)can be expressed as (210) where w(z) is as follows
w(z) = f(x+ y)e1 + g(x minus y)e2
=12f(x+ y) + g(x minus y) + j[f(x+ y)minus g(x minus y)]
f(2R) = u(z3) + v(z3) =r((1 + j)R) + b1
a((1 + j)R) + b((1 + j)R)
g(0) = u(z3)minus v(z3) =r((1 + j)R)minus b1
a((1 + j)R)minus b((1 + j)R)
(213)
here f(x+ y) g(x minus y) possess the forms
g(x minus y)=2r((1minus j)(x minus y)2 + (1 + j)R)
a((1minus j)(x minus y)2 + (1 + j)R)minus b((1minus j)(x minus y)2 + (1 + j)R)
minus [a((1minusj)(xminusy)2+(1 + j)R)+b((1minus j)(x minus y)2 + (1 + j)R)]f(2R)a((1minus j)(x minus y)2 + (1 + j)R)minusb((1minus j)(x minus y)2 + (1 + j)R)
0 le x minus y le 2R1 (214)
f(x+ y)=2r((1+j)(x+y)2)minus[a((1 + j)(x+ y)2)minusb((1 + j)(x+ y)2)]g(0)
a((1 + j)(x+ y)2) + b((1 + j)(x+ y)2)
0 le x+ y le 2R
Moreover u(z) satisfies the estimate
C1α[u(z) D]leM1=M1(α k0 k2 D) C1
α[u(z) D]leM2k2=M2(α k0 D)k2 (215)
where M1 M2 are two non-negative constants only dependent on α k0 k2 D and αk0 D respectively
Proof Let the general solution w(z) = uz = 12f(x+ y) + g(x minus y) + j[f(x+ y)minus
g(x minus y)] of (212) be substituted in the boundary condition (21) we obtain
a(z)u(z) + b(z)v(z) = r(z) on L λ(z3)w(z3) = r(z3) + jb1 ie
[a((1minus j)x+ 2jR) + b((1minus j)x+ 2jR)]f(2R)+[a((1minus j)x+ 2jR)
minusb((1minus j)x+ 2jR)]g(2x minus 2R) = 2r((1minus j)x+ 2jR) on L3
2 Quasilinear Hyperbolic Equations 47
[a((1 + j)x) + b((1 + j)x)]f(2x) + [a((1 + j)x)
minusb((1 + j)x)]g(0) = 2r((1 + j)x) on L4
f(2R)=u(z3)+v(z3)=r((1 + j)R)+b1
a((1 + j)R)+b((1 + j)R)
g(0)=u(z3)minusv(z3)=r((1 + j)R)minusb1
a((1 + j)R)minusb((1 + j)R)
and the above formulas can be rewritten as
[a((1minus j)t2 + (1 + j)R) + b((1minus j)t2 + (1 + j)R)]f(2R)
+ [a((1minus j)t2 + (1 + j)R)minus b((1minus j)t2 + (1 + j)R)]g(t)
= 2r((1minus j)t2 + (1 + j)R) t isin [0 2R1]
[a((1 + j)t2) + b((1 + j)t2)]f(t) + [a((1 + j)t2)
minusb((1 + j)t2)]g(0) = 2r((1 + j)t2) t isin [0 2R]thus the solution w(z) can be expressed as (213)(214) From the condition (22)and the relation (210) we see that the estimate (215) of u(z) for (211) is obviouslytrue
Next we give the representation of Problem P for the quasilinear equation (116)
Theorem 22 Under Condition C any solution u(z) of Problem P for the hyper-bolic equation (116) can be expressed as
u(z) = 2Reint z
0w(z)dz + b0 in D
w(z) = W (z) + Φ(z) + Ψ(z) in D
W (z) = f(micro)e1 + g(ν)e2 Φ(z) = f(micro)e1 + g(ν)e2
Ψ(z)=int ν
0[Aξ+Bη+Cu+D]e1dν+
int micro
2R[Aξ+Bη+Cu+D]e2dmicro
(216)
where f(micro) g(ν) are as stated in (214) and f(micro) g(ν) are similar to f(micro) g(ν) in(214) but r(z) b1 are replaced by minusRe [λ(z)Ψ(z)] minusIm [λ(z3)Ψ(z3)] namely
Re [λ(z)Φ(z)]=minusRe [λ(x)Ψ(x)] z isin L Im [λ(z3)Φ(z3)]=minusIm [λ(z3)Ψ(z3)](217)
Proof Let the solution u(z) of Problem P be substituted into the coefficients ofequation (116) Then the equation in this case can be seen as a linear hyperbolicequation (115) Due to Problem P is equivalent to the Problem A for the complexequation (28) with the relation (210) from Theorem 32 Chapter I it is not difficultto see that the function Ψ(z) satisfies the complex equation
[Ψ]z = [Aξ +Bη + Cu+D]e1 + [Aξ +Bη + Cu+D]e2 in D (218)
48 II Hyperbolic Equations of Second Order
and Φ(z) = w(z) minus W (z) minus Ψ(z) satisfies the complex equation and the boundaryconditions
ξνe1 + ηmicroe2 = 0 (219)
Re [λ(z)(ξe1 + ηe2)] = minusRe [λ(z)Ψ(z)] on L
Im [λ(z)(ξe1 + ηe2)]|z=z3 = minusIm [λ(z3)Ψ(z3)](220)
By the representation of solutions of Problem A for (116) as stated in (317) Chap-ter I we can obtain the representation (216) of Problem P for (116)
22 Existence and uniqueness of solutions of Problem P for hyperbolicequations of second order
Theorem 23 If the complex equation (116) satisfies Condition C then ProblemP for (116) has a solution
Proof We consider the expression of u(z) in the form (216) In the following byusing successive iteration we shall find a solution of Problem P for equation (116)Firstly substitute
u0(z) = 2Reint z
0w0(z)dz + b0 w0(z) = W (z) = ξ0e1 + η0e2 (221)
into the position of u w = ξe1 + ηe2 in the right-hand side of (116) where w0(z) isthe same function with W (z) in (216) and satisfies the estimate (215) Moreoverwe have
u1(z) = 2Reint z
0w1(z)dz + b0 w1(z) = w0(z) + Φ1(z) + Ψ1(z)
Ψ1(z) =int xminusy
0[Aξ0 +Bη0 + Cu0 +D]e1d(x minus y)
+int x+y
2R[Aξ0 +Bη0 + Cu0 +D]e2d(x+ y)
(222)
from the first equality in (222) the estimate
C1[u0(z) D] le M3C[w0(z) D] + k2 (223)
can be derived where M3 = M3(D) From the third and second equalities in (222)we can obtain
C[Ψ1(z) D] le 2M4[(4 +M3)m+ k2 + 1]Rprime
C[Φ1(z) D] le 8M4k20(1 + k2
0)[(4 +M3)m+ k2 + 1]Rprime
C[w1(z)minus w0(z) D] le 2M4M [(4 +M3)m+ k2 + 1]Rprime
(224)
2 Quasilinear Hyperbolic Equations 49
where M4 = maxD(|A| |B| |C| |D|) Rprime = max(2R1 2R) m = w0(z) C(D) M =1 + 4k2
0(1 + k20) is a positive constant Thus we can find a sequence of functions
wn(z) satisfying
un+1(z) = 2Reint z
0wn+1(z)dz + b0
wn+1(z) = w0(z) + Φn(z) +int ν
0[Aξn +Bηn + Cun +D]e1dν
+int micro
2R[Bηn + Aηn + Cun +D]e2dmicro
(225)
and then
wn minus wnminus1 le 2M4M [(4 +M3)m+ 1]n timesint Rprime
0
Rprimenminus1
(n minus 1) dRprime
le 2M4M [(4 +M3)m+ 1)]Rprimen
n
(226)
From the above inequality it is seen that the sequence of functions wn(z) ie
wn(z) = w0(z) + [w1(z)minus w0(z)] + middot middot middot+ [wn(z)minus wnminus1(z)](n = 1 2 ) (227)
in D uniformly converges a function wlowast(z) and wlowast(z) satisfies the equality
wlowast(z) = ξlowaste1 + ηlowaste2
= w0(z) + Φlowast(z) +int ν
0[Aξlowast +Bηlowast + Culowast +D]e1dν
+int micro
2R[Aξlowast +Bηlowast + Culowast +D]e2dmicro
(228)
and the functionulowast(z) = 2Re
int z
0wlowast(z)dz + b0 (229)
is just a solution of Problem P for equation (116) in the closed domain D
Theorem 24 Suppose that Condition C holds Then Problem P for the complexequation (116) has at most one solution in D
Proof Let u1(z) u2(z) be any two solutions of Problem P for (116) we see thatu(z) = u1(z)minus u2(z) and w(z) = u1z(z)minus u2z(z) satisfies the homogeneous complexequation and boundary conditions
wz = Re [A1w] + A2u in D (230)
Re [λ(z)w(z)] = 0 on L Im [λ(z3)w(z3)] = 0 (231)
and the relationu(z) = 2Re
int z
0w(z)dz z isin D (232)
50 II Hyperbolic Equations of Second Order
From Theorem 22 we see that the function w(z) can be expressed in the form
w(z) = Φ(z) + Ψ(z)
Ψ(z) =int ν
0[Aξ + Bη + Cu]e1dν +
int micro
2R[Aξ + Bη + Cu]e2dmicro
(233)
moreover from (232)C1[u(z) D] le M3C[w(z) D] (234)
can be obtained in which M3 = M3(D) is a non-negative constant By using theprocess of iteration similar to the proof of Theorem 23 we can get
w(z) = w1 minus w2 le 2M5M [(4 +M3)m+ 1]Rprimen
n
where M5 = maxD(|A| |B| |C|) Let n rarr infin it can be seen w(z) = w1(z)minus w2(z) =0 Ψ(z) = Φ(z) = 0 in D This proves the uniqueness of solutions of Problem P for(116)
3 Oblique Derivative Problems for General QuasilinearHyperbolic Equations of Second Order
In this section we first give a priori estimates in C1(D) of solutions for the obliquederivative problem moreover by using the estimates of solutions the existence ofsolutions for the above problem for general quasilinear equation is proved Finallywe discuss the oblique derivative problem for hyperbolic equations of second order ingeneral domains
31 A priori estimates of solutions of Problem P for hyperbolic equationsof second order
From Theorems 23 and 24 we see that under Condition C Problem P for equation(116) has a unique solution u(z) which can be found by using successive iterationNoting that wn+1(z)minuswn(z) satisfy the estimate (226) the limit w(z) of the sequenceof functions wn(z) satisfies the estimate
maxzisinD
w(z) = C[w(z) D] le e2M5M [(4+M3)m+1]Rprime= M6 (31)
and the solution u(z) of Problem P is as stated in (210) which satisfies the estimate
C1[u(z) D] le RlowastM6 + k2 = M7 (32)
where Rlowast = 2R1 + 2R Thus we have
3 General Hyperbolic Equations 51
Theorem 31 If Condition C holds then any solution u(z) of Problem P for thehyperbolic equation (116) satisfies the estimates
C1[u D] le M7 C1[u D] le M8k (33)
in which M7 = M7(α k0 k1 k2 D) k = k1 + k2 M8 = M8(α k0 D) are non-negativeconstants
In the following we give the C1α(D)-estimates of solution u(z) for Problem P for
(116)
Theorem 32 If Condition C and (120) hold then any solution u(z) of ProblemP for the hyperbolic equation (116) satisfies the estimates
Cα[uz D] le M9 C1α[u D] le M10 C1
α[u D] le M11k (34)
in which k = k1 + k2 Mj = Mj(α k0 k1 k2 D) j = 9 10 M11 = M11(α k0 D) arenon-negative constants
Proof Similarly to Theorem 43 Chapter I it suffices to prove the first estimatein (34) Due to the solution u(z) of Problem P for (116) is found by the successiveiteration through the integral expression (216) we first choose the solution
u0(z) = 2Reint z
0w0(z)dz + b0 w0(z) = W (z) = ξ0e1 + η0e2 (35)
of Problem P for the equationuzz = 0 in D (36)
and substitute u0 w0 into the position of u w = ξe1 + ηe2 on the right-hand side of(116) where w0(z) is the same function with W (z) in (216) and w0(z) u0(z) satisfythe first estimates
Cα[w0 D] = Cα[Rew0D] + Cα[Imw0 D] le M12k2 C1α[u0 D] le M13k2 = M14 (37)
where Mj = Mj(α k0 D) j = 12 13 and then we have
u1(z) = 2Reint z
0w1(z)dz + b0 w1(z) = w0(z) + Φ1(z) + Ψ1(z)
Ψ1(z) = Ψ11(z)e1 +Ψ2
1(z)e2 Ψ11(z) =
int ν
0G1(z)dν
Ψ21(z) =
int micro
2RG2(z)dmicro G1(z) = G2(z) = Aξ0 +Bη0 + Cu0 +D
(38)
From (37) and the last three equalities in (38) it is not difficult to see thatΨ1
1(z) = Ψ11(micro ν) Ψ2
1(z) = Ψ21(micro ν) satisfy the Holder continuous estimates about
ν micro respectively namely
Cα[Ψ11(middot ν) D] le M15R
prime Cα[Ψ21(micro middot) D] le M15R
prime (39)
52 II Hyperbolic Equations of Second Order
in which M15 = M15(α k0 k1 k2 D) there is no harm assuming that Rprime = max(2R12R) ge 1 By Condition C and (120) we can see that G1(z) Ψ1
1(z) G2(z) Ψ21(z)
about micro ν satisfy the the Holder conditions respectively ie
Cα[G1(micro middot) D] le M16Rprime Cα[Ψ1
1(micro middot) D] le M16Rprime
Cα[G2(middot ν) D] le M16Rprime Cα[Ψ2
1(middot ν) D] le M16Rprime
(310)
where M16 = M16(α k0 k1 k2 D) Moreover we can obtain the estimates of Ψ1(z)Φ1(z) as follows
Cα[Ψ1(z) D] le M17Rprime Cα[Φ1(z) D] le M17R
prime (311)
in which M17 = M17(α k0 k1 k2 D) Setting w11(z) = Re w1(z) + Im w1(z) w1
2(z) =Re w1(z) minus Im w1(z) w1(z) = w1(z) minus w0(z) and u1(z) = u1(z) minus u0(z) from (38)ndash(311) it follows that
Cα[w11(z) D] le M18R
prime Cα[w21(z) D] le M18R
prime
Cα[w1(z) D] le M18Rprime C1
α[u1(z) D] le M18Rprime
(312)
where M18 = M18(α k0 k1 k2 D) According to the successive iteration the esti-mates of functions w1
n(z) = Re wn(z) + Im wn(z) w2n(z) = Re wn(z) minus Im wn(z)
wn(z) = wn(z) minus wnminus1(z) and the corresponding function un(z) = un(z) minus unminus1(z)can be obtained namely
Cα[w1n(z) D] le
(M18Rprime)n
n Cα[w2
n(z) D] le(M18R
prime)n
n
Cα[wn(z) D] le (M18Rprime)n
n C1
α[un(z) D] le (M18Rprime)n
n
(313)
Therefore the sequences of functions
wn(z) = w0(z) +nsum
m=1wm(z) un(z) = u0(z) +
nsumm=1
um(z) (n = 1 2 )
uniformly converge to w(z) u(z) in D respectively and w(z) u(z) satisfy the esti-mates
Cα[w(z) D] le M9 = eM18Rprime C1
α[u(z) D] le M10 (314)
this is just the first estimate in (34)
32 The existence of solutions of Problem P for general hyperbolicequations of second order
Now we consider the general quasilinear equation of second order
uzz = F (z u uz) +G(z u uz)
F = Re [A1uz] + A2u+ A3
G = A4 uz σ +A5|u |τ z isin D
(315)
3 General Hyperbolic Equations 53
where F (z u uz) satisfies Condition C σ τ are positive constants and Aj(z u uz)(j = 4 5) satisfy the conditions in Condition C ie
C[Aj(z u uz) D] le k0 j = 4 5
and denote by Condition C prime the above conditions
Theorem 33 Let the complex equation (315) satisfy Condition C prime
(1) When 0 lt max(σ τ) lt 1 Problem P for (315) has a solution u(z) isin C11(D)
(2) When min(σ τ) gt 1 Problem P for (315) has a solution u(z) isin C11(D)
provided thatM19 = k1 + k2 + |b0|+ |b1| (316)
is sufficiently small
(3) In general the above solution of Problem P is not unique if 0ltmax(σ τ)lt1
Proof (1) Consider the algebraic equation for t
M8k1 + k2 + 2k0tσ + 2k0t
τ + |b1|+ |b0| = t (317)
where M8 is a non-negative constant in (33) it is not difficult to see that equation(317) has a unique solution t = M20 ge 0 Now we introduce a closed and convexsubsetBlowast in the Banach space C1(D) whose elements are the functions u(z) satisfyingthe conditions
u(z) isin C1(D) C1[u(z) D] le M20 (318)
We arbitrarily choose a function u0(z) isin Blowast for instance u0(z) = 0 and substitute itinto the position of u in coefficients of (315) and G(z u uz) from Theorems 23 and24 it is clear that problem P for
uzzminusRe [A1(z u0 u0z)uz]minusA2(z u0 u0z)uminusA3(z u0 u0z)=G(z u0 u0z) (319)
has a unique solution u1(z) isin Blowast By Theorem 31 we see that the solution u1(z)satisfies the estimate in (318) By using the successive iteration we obtain a sequenceof solutions um(z)(m = 1 2 ) isin Blowast of Problem P which satisfy the equations
um+1zz minus Re [A1(z um umz)um+1z]minus A2(z um umz)um+1
minusA3(z um umz) = G(z um umz) in D m = 1 2 (320)
and um+1(z) isin Blowast From (320) we see that um+1(z) = um+1(z)minus um(z) satisfies theequations and boundary conditions
um+1zz minus Re[A1um+1z]minus A2um+1
= G(z um umz)minus G(z umminus1 umminus1z) in D m = 1 2
Re [λ(z)(um+1(z)] = 0 on L Im [λ(z3)(um+1(z3)] = 0
(321)
54 II Hyperbolic Equations of Second Order
Noting that C[G(z um umz)minus G(z umminus1 umminus1z) D] le 2k0M20 M20 is a solution ofthe algebraic equation (317) and according to Theorem 31
um+1 = C1[um+1 D] le M20 (322)
can be obtained Due to um+1 can be expressed as
um+1(z) = 2Reint z
0wm+1(z)dz wm+1(z) = w0(z) + Φm+1(z) + Ψm+1(z)
Ψm+1(z) =int xminusy
0[Aξm + Bηm + Cum + G]e1d(x minus y)
+int x+y
2R[Aξm + Bηm + Cum + G]e2d(x+ y)
(323)
in which the relation between A1 A2 G and A B C G is the same as that ofA1 A2 A3 and A B C D in Section 1 and G = G(z um umz) minus G(z umminus1 umminus1z)By using the method in the proof of Theorem 23 we can obtain
um+1 minus um = C1[um+1 D] le (M21Rprime)n
n
where M21 = 4M22M(M3 + 2)(2m0 + 1) Rprime = max(2R1 2R) m0 = w0(z) C(D)herein M22 = maxC[A Q] C[B Q] C[C Q] C[G Q] M = 1+ 4k2
0(1 + k20) From
the above inequality it is seen that the sequences of functions um(z) wm(z)ie
um(z)=u0(z)+[u1(z)minusu0(z)]+middot middot middot+[um(z)minusumminus1(z)](m=1 2 )
wm(z)=w0(z)+[w1(z)minusw0(z)]+middot middot middot+[wm(z)minuswmminus1(z)](m=1 2 )(324)
uniformly converge to the functions ulowast(z) wlowast(z) respectively and wlowast(z) satisfies theequality
wlowast(z) = w0(z) + Φlowast(z)
+int xminusy
0[Aσlowast +Bηlowast + Culowast +D]e1d(x minus y)
+int x+y
2R[Bσlowast + Aηlowast + Culowast +D]e2d(x+ y)
(325)
and the functionulowast(z) = 2Re
int z
0wlowast(z)dz + b0 (326)
is just a solution of Problem P for the nonlinear equation (315) in the closeddomain D
(2) Consider the algebraic equation for t
M8k1 + k2 + 2k0tσ + 2k0t
τ + |b0|+ |b1| = t (327)
3 General Hyperbolic Equations 55
it is not difficult to see that equation (327) has a solution t = M20 ge 0 providedthat M19 in (316) is small enough If there exist two solutions then we choosethe minimum of both as M20 Now we introduce a closed and convex subset Blowastof the Banach space C1(D) whose elements are of the functions u(z) satisfying theconditions
u(z) isin C1(D) C1[u(z) D] le M20 (328)
By using the same method in (1) we can find a solution u(z) isin Blowast of Problem P forequation (315) with min(σ τ) gt 1
(3) We can give an example to explain that there exist two solutions for equation(315) with σ = 0 τ = 12 namely the equation
uxx minus uyy = Au12 A = 8sgn(x2 minus y2) in D (329)
has two solutions u1(x y) = 0 and u2(x y) = (x2 minus y2)24 and they satisfy theboundary conditions
Re [λ(z)uz] = r(z) z isin L u(0) = b0 Im [λ(z)uz]|z=0 = b1 (330)
whereλ(z) = 1minus i z isin L1 λ(x) = 1 + i z isin L4
b0 = 0 r(z) = 0 z isin L = L1 cup L4 b1 = 0
33 The existence of solutions of Problem P for hyperbolic equations ofsecond order in general domains
In this subsection we shall generalize the domain D to general cases
1 The boundaries L3 L4 of the domain D are replaced by the curves Lprime3 L
prime4 and
the boundary of the domain Dprime is L1 cup L2 cup Lprime3 cup Lprime
4 where the parameter equationsof the curves Lprime
3 Lprime4 are as follows
Lprime3 = x+ y = 2R l le x le R2 Lprime
4 = x+ y = micro y = γ1(x) 0 le x le l (331)
in which γ1(x) on 0 le x le l = γ1(l) is continuous and γ1(0) = 0 γ1(x) gt 0 on0 lt x le l and γ1(x) possesses the derivatives on 0 le x le l except some isolatedpoints and 1 + γprime
1(x) gt 0 By the condition we can find the inverse function x =τ(micro) = (micro+ ν)2 of x+ γ1(x) = micro and then ν = 2τ(micro)minus micro 0 le micro le 2R We make atransformation
micro = micro ν =2R1(ν minus 2τ(micro) + micro)2R1 minus 2τ(micro) + micro
0 le micro le 2R 2τ(micro)minus micro le ν le 2R1 (332)
its inverse transformation is
micro = micro ν =2R1 minus 2τ(micro) + micro
2R1ν + 2τ(micro)minus micro 0 le micro le 2R 0 le ν le 2R1 (333)
56 II Hyperbolic Equations of Second Order
Hence we have
x =12(micro+ ν) =
4R1x minus [2τ(x+ γ1(x))minus x minus γ1(x)](2R1 + x+ y)4R1 minus 4τ(x+ γ1(x)) + 2x+ 2γ1(x)
y =12(micro minus ν) =
4R1y + [2τ(x+ γ1(x))minus x minus γ1(x)](2R1 minus x minus y)4R1 minus 4τ(x+ γ1(x)) + 2x+ 2γ1(x)
(334)
and
x =12(micro+ ν) =
4R1x+ [2τ(x+ γ1(x))minus x minus γ1(x)](2R1 minus x+ y)4R1
y =12(micro minus ν) =
4R1y minus [2τ(x+ γ1(x))minus x minus γ1(x)](2R1 + x minus y)4R1
(335)
Denote by z = x + jy = f(z) z = x + jy = fminus1(z) the transformations (334)and (335) respectively In this case setting w(z) = uz equation (116) in Dprime andboundary condition (21) on Lprime
3 cup Lprime4 can be rewritten in the form
ξν = Aξ +Bη + Cu+D ηmicro = Aξ +Bη + Cu+D z isin Dprime (336)
Re [λ(z)w(z)] = r(z) z isin Lprime3 cup Lprime
4 u(0) = b0 Im [λ(z3)w(z3)] = b1 (337)
in which z3 = l+ jγ1(l) λ(z) r(z) b1 on Lprime3 cupLprime
4 satisfy the condition (22) and u(z)and w(z) satisfy the relation
u(z) = 2Reint z
0w(z)dz + b0 in Dprime (338)
Suppose equation (116) in Dprime satisfies Condition C through the transformation(333) we have ξν = [2R1 minus 2τ(micro) + micro]ξν2R1 ηmicro = ηmicro system (336) is reduced to
ξν =[2R1 minus 2τ(micro) + micro][Aξ +Bη + Cu+D]
2R1 ηmicro = Aξ +Bη + Cu+D (339)
Moreover through the transformation (335) the boundary condition (337) on Lprime3cupLprime
4is reduced to
Re [λ(fminus1(z))w(fminus1(z))]=r(fminus1(z)) z isin L3 cup L4
Im [λ(fminus1(z3))w(fminus1(z3)] = b1(340)
in which z3 = f(z3) Therefore the boundary value problem (336)(337) is trans-formed into the boundary value problem (339)(340) On the basis of Theorems 22and 23 we see that the boundary value problem (339)(340) has a unique solutionw(z) and then w[f(z)] is just a solution of the boundary value problem (336)(337)
Theorem 34 If equation (116) satisfies Condition C in the domain Dprime with theboundary L1cupL2cupLprime
3cupLprime4 then Problem P with the boundary condition (337)(w = uz)
has a unique solution u(z) as stated in (338)
3 General Hyperbolic Equations 57
2 The boundaries L3 L4 of the domain D are replaced by two curves Lprimeprime3 Lprimeprime
4respectively and the boundary of the domain Dprimeprime is L1 cup L2 cup Lprimeprime
3 cup Lprimeprime4 where the
parameter equations of the curves Lprimeprime3 Lprimeprime
4 are as follows
Lprimeprime3 = x minus y = ν y = γ2(x) l le x le R2
Lprimeprime4 = x+ y = micro y = γ1(x) 0 le x le l
(341)
in which γ1(0) = 0 γ2(R2) = R2 minus 2R1 γ1(x) gt 0 0 le x le l γ2(x) gt 0 l le x leR2 γ1(x) on 0 le x le l γ2(x) on l le x le R2 are continuous and γ1(x) γ2(x) possessthe derivatives on 0 le x le l l le x le R2 except isolated points respectively and1+γprime
1(x) gt 0 1minusγprime2(x) gt 0 zprime
3 = l+jγ1(l) = l+jγ2(l) isin L3 (or L4) By the conditionswe can find the inverse functions x = τ(micro) x = σ(ν) of x+ γ1(x) = micro x minus γ2(x) = νrespectively namely
ν = 2τ(micro)minus micro 0 le micro le 2R micro = 2σ(ν)minus ν 0 le ν le l minus γ2(l) (342)
We first make a transformation
micro =2Rmicro
2σ(ν)minus ν ν = ν 0 le micro le 2σ(ν)minus ν 0 le ν le 2R1 (343)
its inverse transformation is
micro =(2σ(ν)minus ν)micro
2R ν = ν 0 le micro le 2R 0 le ν le 2R1 (344)
The above transformation can be expressed by
x =12(micro+ ν) =
2R(x+ y) + (x minus y)[2σ(x minus γ2(x))minus x+ γ2(x)]2[2σ(x minus γ2(x))minus x+ γ2(x)]
y =12(micro minus ν) =
2R(x+ y)minus (x minus y)[2σ(x minus γ2(x))minus x+ γ2(x)]2[2σ(x minus γ2(x))minus x+ γ2(x)]
(345)
and its inverse transformation is
x =12(micro+ ν) =
[2σ(x minus γ2(x))minus x+ γ2(x)](x+ y) + 2R(x minus y)4R
y =12(micro minus ν) =
[2σ(x minus γ2(x))minus x+ γ2(x)](x+ y)minus 2R(x minus y)4R
(346)
Denote by z = x + jy = g(z) z = x + jy = gminus1(z) the transformation (345) andthe inverse transformation (346) respectively Through the transformation (344)we have
(u+ v)ν = (u+ v)ν (u minus v)micro =2σ(ν)minus ν
2R(u minus v)micro (347)
system (336) in Dprimeprime is reduced to
ξν = Aξ +Bη + Cu+D ηmicro =2σ(ν)minus ν
2R[Aξ +Bη + Cu+D] in Dprime (348)
58 II Hyperbolic Equations of Second Order
Moreover through the transformation (346) the boundary condition (337) on Lprimeprime3cupLprimeprime
4is reduced to
Re [λ(gminus1(z))w(gminus1(z))] = r(gminus1(z)) z isin Lprime3 cup Lprime
4
Im [λ(gminus1(z3))w(gminus1(z3)] = b1(349)
in which z3 = g(zprime3) Besides the relation (338) is valid Therefore the boundary
value problem (336)(337) in Dprimeprime is transformed into the boundary value problem(348)(349) On the basis of Theorem 34 we see that the boundary value problem(348)(349) has a unique solution w(z) and then w[g(z)] is just a solution of theboundary value problem (336)(337) in Dprimeprime but we mention that the conditions ofcurve Lprime
3 Lprime4 through the transformation z = gminus1(z) must satisfy the conditions of
the curves in (331) For instance if z3 isin L3 γ1(x) ge x+2lminus2R on 2Rminus2l le x le lthen the above condition holds If z3 isin L4 γ2(x) ge 2l minusx on l le x le 2l then we cansimilarly discuss For other case it can be discussed by using the method as statedin Section 2 Chapter VI below
Theorem 35 If equation (116) satisfies Condition C in the domain Dprimeprime with theboundary L1 cup L2 cup Lprimeprime
3 cup Lprimeprime4 then Problem P with the boundary conditions
Re[λ(z)uz] = r(z) z isin Lprimeprime3 cup Lprimeprime
4 u(0) = b0 Im[λ(zprimeprime3 )uz(zprimeprime
3 )] = b1 (350)
has a unique solution u(z) as stated in (338) in Dprimeprime
By using the above method we can generalize the domain D to more generaldomain including the disk Dprimeprime = ||z minus R1|| lt R1 For the domain Dprimeprime we chooseR2 = 2R1 the boundary Lprimeprime of the domain Dprimeprime consists of Lprimeprime
1 Lprimeprime2 L
primeprime3 L
primeprime4 namely
Lprimeprime1 =
y = minusγ1(x) = minus
radicR2
1 minus (x minus R1)2 0 le x le R1
Lprimeprime2 =
y = minusγ2(x) = minus
radicR2
1 minus (x minus R1)2 R1 le x le 2R1
Lprimeprime3 =
y = γ3(x) =
radicR2
1 minus (x minus R1)2 R1 le x le 2R1
Lprimeprime4 =
y = γ4(x) =
radicR2
1 minus (x minus R1)2 0 le x le R1
where 1 + γprime1(x) gt 0 1 + γprime
4(x) gt 0 on 0 lt x le R1 1 minus γprime2(x) gt 0 1 minus γprime
3(x) gt 0 onR1 le x lt 2R1 The above curves can be rewritten as
Lprimeprime1 =
⎧⎨⎩x = σ1(ν) =R1 + ν minus
radicR2
1 + 2R1ν minus ν2
2
⎫⎬⎭
Lprimeprime2 =
⎧⎨⎩x = τ1(micro) =R1 + micro minus
radicR2
1 + 2R1micro minus micro2
2
⎫⎬⎭
4 Other Oblique Derivative Problems 59
Lprimeprime3 =
⎧⎨⎩x = σ2(ν) =R1 + ν minus
radicR2
1 + 2R1ν minus ν2
2
⎫⎬⎭
Lprimeprime4 =
⎧⎨⎩x = τ2(micro) =R1 + micro minus
radicR2
1 + 2R1micro minus micro2
2
⎫⎬⎭
where σ1(ν) σ2(ν) are the inverse functions of x + γ1(x) = ν x minus γ3(x) = ν on0 le ν le 2R1 and x = τ1(micro) x = τ2(micro) are the inverse functions of x minus γ2(x) =micro x+ γ4(x) = micro on 0 le micro le 2R1 respectively Through a translation we can discussthe unique solvability of corresponding boundary value problem for equation (116) inany disk ||z minus z0|| lt R where z0 is a hyperbolic number and R is a positive number
4 Other Oblique Derivative Problems for QuasilinearHyperbolic Equations of Second Order
In this section we discuss other oblique derivative problems for quasilinear hyperbolicequations Firstly the representation theorem of solutions for the above boundaryvalue problems is given moreover the uniqueness and existence of solutions for theabove problem are proved The results obtained include the corresponding result ofthe Dirichlet boundary value problem or the Darboux problem([12]3) as a specialcase
41 Formulation of other oblique derivative problems for quasilinearhyperbolic equations
We first state four other oblique derivative problems for equation (116) here thedomain D is the same as that in Section 1 but R2 = 2R1
Problem P1 Find a continuously differentiable solution u(z) of (116) in D satisfy-ing the boundary conditions
Re [λ(z)uz] = r(z) z isin L1 cup L2
u(0) = b0 Im [λ(z)uz]|z=z1 = b1(41)
where b0 b1 are real constants λ(z) = a(z) + jb(z) z isin L1 cup L2 and λ(z) r(z) b0 b1
satisfy the conditions
Cα[λ(z) L1 cup L2] le k0 Cα[r(z) L1 cup L2] le k2
|b0| |b1| le k2 maxzisinL1
1|a(z)minus b(z)| max
zisinL2
1|a(z) + b(z)| le k0
(42)
in which α(0 lt α lt 1) k0 k2 are non-negative constants
60 II Hyperbolic Equations of Second Order
If the boundary condition (41) is replaced by
Re [λ(z)uz] = r(z) z isin L1 Re [Λ(x)uz(x)] = R(x) x isin L0 = (0 R2)
u(0) = b0 Im [λ(z)uz]|z=z1 = b1(43)
where b0 b1 are real constants λ(z) = a(z) + jb(z) z isin L1 Λ(x) = 1 + j x isin L0and λ(z) r(z) R(x) b0 b1 satisfy the conditions
Cα[λ(z) L1] le k0 Cα[r(z) L1] le k2 |b0| |b1| le k2
Cα[R(x) L0] le k2 maxzisinL1
1|a(z)minus b(z)| le k0
(44)
in which α(0 lt α lt 1) k0 k2 are non-negative constants then the boundary valueproblem for (116) will be called Problem P2
If the boundary condition in (41) is replaced by
Re [λ(z)uz] = r(z) z isin L2 Re [Λ(x)uz(x)] = R(x) x isin L0 = (0 R2)
u(0) = b0 Im [λ(z)uz]|z=z1 = b1(45)
where b0 b1 are real constants λ(z) = a(z) + jb(z) z isin L2 Λ(x) = 1 minus j x isin L0and λ(z) r(z) R(x) b0 b1 satisfy the conditions
Cα[λ(z) L2] le k0 Cα[r(z) L2] le k2 |b0| |b1| le k2
Cα[R(x) L0] le k2 maxzisinL2
1|a(z) + b(z)| le k0
(46)
then the boundary value problem for (116) is called Problem P3 For Problem P2
and Problem P3 there is no harm in assuming that w(z1) = 0 otherwise throughthe transformation W (z) = w(z)minus [a(z1)minus jb(z1)][r(z1)+ jb1][a2(z1)minus b2(z1)] therequirement can be realized
If the boundary condition in (41) is replaced by
u(x) = s(x) uy = R(x) x isin L0 = (0 R2) (47)
where s(z) R(x) satisfy the conditions
C1α[s(x) L0] le k2 Cα[R(x) L0] le k2 (48)
then the boundary value problem for (116) is called Problem P4
In the following we first discuss Problem P2 and Problem P3 for equation (211)
4 Other Oblique Derivative Problems 61
42 Representations of solutions and unique solvability of Problem P2
and Problem P3 for quasilinear hyperbolic equations
Similarly to Theorem 21 we can give the representation of solutions of Problem P2
and Problem P3 for equation (211) namely
Theorem 41 Any solution u(z) of Problem P2 for the hyperbolic equation (211)can be expressed as
u(z) = 2Reint z
0w(z)dz + b0 in D (49)
where w(z) is as follows
w(z) = f(x+ y)e1 + g(x minus y)e2
=12f(x+ y) + g(x minus y) + j[f(x+ y)minus g(x minus y)]
(410)
here f(x+ y) g(x minus y) possess the forms
f(x+ y) = Re [(1 + j)uz(x+ y)] = R(x+ y) 0 le x+ y le 2R
g(xminusy)=2r((1minusj)(xminusy)2)minus[a((1minusj)(xminusy)2)+b((1minusj)(xminusy)2)]f(0)
a((1minus j)(x minus y)2)minus b((1minus j)(x minus y)2)
0 le x minus y le 2R1
f(0) = u(z1) + v(z1) =r((1minus j)R1) + b1
a((1minus j)R1) + b((1minus j)R1)
(411)
As for Problem P3 for (211) its solution can be expressed as the forms (49) (410)but where f(x+ y) g(x minus y) possess the forms
f(x+ y) =2r((1+j)(x+y)2+(1minusj)R1)
a((1+j)(x+y)2+(1minusj)R1)minusb((1+j)(x+y)2+(1minusj)R1)
minus [a((1+j)(x+y)2+(1minusj)R1)minusb((1+j)(x+y)2+(1minusj)R1)]g(2R1)a((1 + j)(x+ y)2+(1minus j)R1)+b((1 + j)(x+ y)2+(1minus j)R1)
0 le x+ y le 2R (412)
g(2R1) = u(z1)minus v(z1) =r((1minus j)R1)minus b1
a((1minus j)R1)minus b((1minus j)R1)
g(x minus y) = Re [(1minus j)uz(x minus y)] = R(x minus y) 0 le x minus y le 2R1
Moreover the solution u(z) of Problem P2 and Problem P3 satisfies the estimate
C1α[u(z) D] le M23 C1
α[u(z) D] le M24k2 (413)
where M23 = M23(α k0 k2 D) M24 =M24(α k0 D) are two non-negative constants
Next we give the representation of solutions of Problem P2 and Problem P3 forthe quasilinear hyperbolic equation (116)
62 II Hyperbolic Equations of Second Order
Theorem 42 Under Condition C any solution u(z) of Problem P2 for the hyper-bolic equation (116) can be expressed as
u(z) = 2Reint z
0w(z)dz + b0 w(z) = W (z) + Φ(z) + Ψ(z) in D
W (z) = f(micro)e1 + g(ν)e2 Φ(z) = f(micro)e1 + g(ν)e2
Ψ(z) =int ν
2R1
[Aξ +Bη + Cu+D]e1dν +int micro
0[Aξ +Bη + Cu+D]e2dmicro
(414)
where f(micro) g(ν) are as stated in (411) and f(micro) g(ν) are similar to f(micro) g(ν) in(411) but the functions r(z) R(x) b1 are replaced by the corresponding functionsminusRe [λ(z)Ψ(z)] minusRe [Λ(x)Ψ(x)] minusIm [λ(z1)Ψ(z1)] namely
Re [λ(z)Φ(z)] = minusRe [λ(z)Ψ(z)] z isin L1
Re [Λ(x)Φ(x)] = minusRe [Λ(x)Ψ(x)] x isin L0
Im [λ(z1)Φ(z1)] = minusIm [λ(z1)Ψ(z1)]
(415)
As to Problem P3 for (116) its solution u(z) possesses the expression (414) whereW (z) is a solution of Problem P3 for (211) and Φ(z) is also a solution of (211)satisfying the boundary conditions
Re [λ(z)Φ(z)] = minusRe [λ(z)Ψ(z)] z isin L2
Re [Λ(x)Φ(x)] = minusRe [Λ(x)Ψ(x)] x isin L0
Im [λ(z1)Φ(z1)] = minusIm [λ(z1)Ψ(z1)]
(416)
Proof Let the solution u(z) of Problem P2 be substituted into the coefficients ofequation (116) Due to Problem P2 is equivalent to the Problem A2 for the complexequation (28) with the relation (210) where the boundary conditions are as follows
Re [λ(z)w(z)]=r(z) on L1 Re [Λ(x)w(x)]=R(x) on L0
Im [λ(z1)w(z1)] = b1(417)
According to Theorem 32 in Chapter I it is not difficult to see that the functionΨ(z) satisfies the complex equation
[Ψ]z = [Aξ +Bη + Cu+D]e1 + [Aξ +Bη + Cu+D]e2 in D (418)
and noting that W (z) is the solution of Problem A2 for the complex equation (212)hence Φ(z) = w(z)minusW (z)minusΨ(z) = ξe1+ ηe2 satisfies the complex equation and theboundary conditions
ξmicroe1 + ηλe2 = 0 (419)
4 Other Oblique Derivative Problems 63
Re [λ(z)(ξe1 + ηe2)] = minusRe [λ(z)Ψ(z)] on L1
Re [Λ(x)(ξe1 + ηe2)] = minusRe [Λ(x)Ψ(x)] on L0
Im [λ(z)(ξe1 + ηe2)]|z=z1 = minusIm [λ(z1)Ψ(z1)]
(420)
The representation of solutions of Problem A2 for (116) is similar to (317) in ChapterI we can obtain the representation (414) of Problem P2 for (116) Similarly we canverify that the solution of Problem P3 for (116) possesses the representation (414)with the boundary condition (416)
By using the same method as stated in the proofs of Theorems 23 and 24 wecan prove the following theorem
Theorem 43 If the complex equation (116) satisfies Condition C then ProblemP2 (Problem P3) for (116) has a unique solution
43 Representations of solutions and unique solvability of Problem P4 forquasilinear hyperbolic equations
It is clear that the boundary condition (47) is equivalent to the following boundarycondition
ux = sprime(x) uy = R(x) x isin L0 = (0 R2) u(0) = s(0) ie
Re [(1 + j)uz(x)] = σ(x) Re [(1minus j)uz(x)] = τ(x) u(0) = r(0)(421)
in which
σ(x) =sprime(x) +R(x)
2 τ(x) =
sprime(x)minus R(x)2
(422)
Similarly to Theorem 41 we can give the representation of solutions of Problem P4
for equation (211)
Theorem 44 Any solution u(z) of Problem P4 for the hyperbolic equation (211)can be expressed as (49) (410) where b0 = s(0) f(x+y) g(xminusy) possess the forms
f(x+ y) = σ(x+ y) =sprime(x+ y) +R(x+ y)
2 0 le x+ y le 2R
g(x minus y) = τ(x minus y) =sprime(x minus y)minus R(x minus y)
2 0 le x minus y le 2R1
(423)
and
f(0) =sprime(0) +R(0)
2 g(2R1) =
sprime(2R1)minus R(2R1)2
(424)
Moreover u(z) of Problem P4 satisfies the estimate (413)
Next we give the representation of Problem P4 for the quasilinear hyperbolicequation (116)
64 II Hyperbolic Equations of Second Order
Theorem 45 Under Condition C any solution u(z) of Problem P4 for the hyper-bolic equation (116) can be expressed as (414) where b0 = s(0) W (z) is a solutionof Problem A4 for (212) satisfying the boundary condition (421) (W = uz) and Φ(z)is also a solution of (212) satisfying the boundary conditions
Re [(1+ j)Φ(x)] = minusRe [(1+ j)Ψ(x)] Re [(1minus j)Φ(x)] = minusRe [(1minus j)Ψ(x)] (425)
Proof Let the solution u(z) of Problem P4 be substituted into the coefficients ofequation (116) Due to Problem P4 is equivalent to the Problem A4 for the complexequation (28) with the relation (210) and the boundary conditions
Re [(1 + j)w(x)] = σ(z) Re [(1minus j)w(x)] = τ(x) x isin L0 (426)
according to Theorem 32 in Chapter I it can be seen that the function Ψ(z) satisfiesthe complex equation (418) and noting that W (z) is the solutions of Problem A4 forthe complex equation (212) ie (419) hence Φ(z) = w(z)minus W (z)minus Ψ(z) satisfiesequation (419) and boundary conditions
Re [(1 + j)(ξe1 + ηe2)] = minusRe [(1 + j)Ψ(x)]
Re [(1minus j)(ξe1 + ηe2)] = minusRe [(1minus j)Ψ(x)]x isin L0 (427)
Similarly to Theorem 42 the representation of solutions of Problem A4 for (28) issimilar to (317) in Chapter I we can obtain the representation (414) of Problem P4
for (116)
By using the same method as stated in the proofs of Theorems 23 and 24 wecan prove the following theorem
Theorem 46 If the complex equation (116) satisfies Condition C then ProblemP4 for (116) has a unique solution
Besides we can discuss the unique solvability of Problem P2 Problem P3 and Prob-lem P4 for general quasilinear hyperbolic equation (315) and generalize the aboveresults to the general domains Dprime with the conditions (331) and (341) respectively
Similarly to Problem P as in Section 2 we can discuss Problem P1 for equation(211) here the solution w(z) of equation (212) is as follows
w(z) = f(x+ y)e1 + g(x minus y)e2
=12f(x+ y) + g(x minus y) + j[f(x+ y)minus g(x minus y)]
f(0) = u(z1) + v(z1) =r((1minus j)R1) + b1
a((1minus j)R1) + b((1minus j)R1)
g(2R1) = u(z1)minus v(z1) =r((1minus j)R1)minus b1
a((1minus j)R1)minus b((1minus j)R1)
(428)
4 Other Oblique Derivative Problems 65
here f(x+ y) g(x minus y) possess the forms
g(x minus y) =2r((1minusj)(xminusy)2)minus[a((1minusj)(xminusy)2)+b((1minusj)(xminusy)2)]f(0)
a((1minus j)(x minus y)2)minus b((1minus j)(x minus y)2)
0 le x minus y le 2R1 (429)
f(x+y)=2r((1+j)(x+y)2+(1minusj)R1)
a((1+j)(x+y)2+(1minusj)R1)minusb((1+j)(x+y)2+(1minusj)R1)
minus [a((1+j)(x+y)2+(1minusj)R1)minusb((1+j)(x+y)2+(1minusj)R1)]g(2R1)a((1+j)(x+y)2+(1minusj)R1)+b((1+j)(x+y)2+(1minusj)R1)
0 le x+ y le 2R1
Moreover when we prove that Problem P1 of equation (116) has a unique solutionu(z) the integrals in (216)(222)(225)(228) and (233) possess the similar formsand the integral path in (221) can be chosen for instance the integral Ψ(z) in (216)is replaced by
Ψ(z) =int ν
2R1
[Aξ +Bη + Cu+D]e1dν +int micro
0[Aξ +Bη + Cu+D]e1dmicro (430)
Now we explain that the Darboux problem is a special case of Problem P1 Theso-called Darboux problem is to find a solution u(z) for (116) such that u(z) satisfiesthe boundary conditions
u(x) = s(x) x isin L0 u(z) = φ(x) z = x+ jy isin L1 (431)
where s(x) φ(x) satisfy the conditions
C1[s(x) L0] le k2 C1[φ(x) L1] le k2 (432)
herein k2 is a non-negative constant [12]3) From (431) we have
u(x) = s(x) x isin L0 Re [(1minusj)uz]=φx(x) = φprime(x) z isin L1
u(0) = s(0) Im [(1minusj)uz]|z=z1minus0 = minusφprime(1) ie
u(x)=s(x) x isin L0 Re [λ(z)uz] = r(z) z isin L1
u(0) = b0 Im [λ(z)uz]|z=z1=b1
(433)
where we choose
λ(z)=(1minusj)radic2 r(z)=φprime(x)
radic2 z isin L1 b0 = s(0) b1=minusφprime(1)
radic2 (434)
it is easy to see that the boundary conditions (433)(434) possess the form of theboundary condition (41) This shows that the above Darboux problem is a specialcase of Problem P2
For more general hyperbolic equations of second order the corresponding boundaryvalue problems remain to be discussed
66 II Hyperbolic Equations of Second Order
5 Oblique Derivative Problems for Degenerate HyperbolicEquations of Second Order
This section deals with the oblique derivative problem for the degenerate hyperbolicequation in a simply connected domain We first give the representation theorem ofsolutions of the oblique derivative problem for the hyperbolic equation and then byusing the method of successive iteration the existence and uniqueness of solutionsfor the above oblique derivative problem are proved
51 Formulation of the oblique derivative problem for degenerate hyper-bolic equations
It is known that the Chaplygin equation in the hyperbolic domain D possesses theform
K(y)uxx + uyy = 0 in D (51)
where K(y) possesses the first order continuous derivative K prime(y) and K prime(y) gt 0 ony1 lt y lt 0 K(0) = 0 and the domain D is a simply connected domain with theboundary L = L0 cup L1 cup L2 herein L0 = (0 2)
L1=x+int y
0
radicminusK(t)dt=0 xisin(0 1)
L2=
xminusint y
0
radicminusK(t)dt=2 xisin(1 2)
are two characteristic lines and z1 = x1 + jy1 = 1 + jy1 is the intersection point ofL1 and L2 In particular if K(y) = minus|y|m m is a positive constant thenint y
0
radicminusK(t)dt =
int y
0|t|m2dt = minus
int |y|
0d
2m+ 2
|t|(m+2)2 = minus 2m+ 2
|y|(m+2)2
In this case the two characteristic lines L1 L2 are as follows
L1 x minus 2m+ 2
|y|(m+2)2 = 0 L2 x+2
m+ 2|y|(m+2)2 = 2 ie
L1 y = minus(m+ 22
x)2(m+2) L2 y = minus[m+ 22
(2minus x)]2(m+2)
In this section we mainly consider the general Chaplygin equation of second order
K(y)uxx + uyy = dux + euy + fu+ g in D (52)
where DK(y) are as stated in (51) its complex form is the following equation ofsecond order
uzz = Re [Quzz + A1uz] + A2u+ A3 z isin D (53)
where
Q =K(y) + 1K(y)minus 1
A1 =d+ je
K(y)minus 1 A2 =
f
2(K(y)minus 1) A3 =
g
2(K(y)minus 1)
5 Degenerate Hyperbolic Equations 67
and assume that the coefficients Aj(z)(j = 1 2 3) satisfy Condition C It is clearthat equation (52) is a degenerate hyperbolic equation
The oblique derivative boundary value problem for equation (52) may be formu-lated as follows
Problem P1 Find a continuous solution u(z) of (53) in D which satisfies theboundary conditions
12
partu
partl= Re [λ(z)uz] = r(z) z isin L = L1 cup L2
u(0) = b0 Im [λ(z)uz]|z=z1 = b1
(54)
where l is a given vector at every point z isin L uz = [radic
minusK(y)ux + juy]2 umacrz =
[radic
minusK(y)ux minus juy]2 b0 b1 are real constants λ(z) = a(x) + jb(x) = cos(l x) +j cos(l y) z isin L and λ(z) r(z) b0 b1 satisfy the conditions
C1α[λ(z) Lj] le k0 C1
α[r(z) Lj] le k2 j = 1 2
|b0| |b1| le k2 maxzisinL1
1|a(z)minus b(z)| max
zisinL2
1|a(z) + b(z)| le k0
(55)
in which α (12 lt α lt 1) k0 k2 are non-negative constants For convenience we canassume that uz(z1) = 0 ie r(z1) = 0 b1 = 0 otherwise we make a transformationuz minus [a(z1) minus jb(z1)][r(z1) + jb1][a2(z1) minus b2(z1)] the requirement can be realizedProblem P1 with the conditions A3(z) = 0 z isin D r(z) = 0 z isin L and b0 = b1 = 0will be called Problem P0
For the Dirichlet problem (Tricomi problem D) with the boundary condition
u(x) = φ(x) on L1 = AC =x = minus
int y
0
radicminusK(t)dt 0 le x le 1
L2 = BC =x = 2 +
int y
0
radicminusK(t)dt 1 le x le 2
(56)
we find the derivative for (56) according to s = x on L = L1 cup L2 and obtain
us = ux + uyyx = ux minus (minusK(y))minus12uy = φprime(x) on L1
us = ux + uyyx = ux + (minusK(y))minus12uy = φprime(x) on L2 ie
(minusK(y))12U + V = (minusK(y))12φprime(x)2 = r(x) on L1
(minusK(y))12U minus V = (minusK(y))12φprime(x)2 = r(x) on L2 ie
Re [(1minus j)(U + jV )] = U minus V = r(x) on L1
Im [(1minus j)(U + jV )] = [minusU + V ]|z=z1minus0 = minusr(1minus 0)
Re [(1 + j)(U + jV )] = U + V = r(x) on L2
Im [(1 + j)(U + jV )] = [U + V ]|z=z1+0 = r(1 + 0)
68 II Hyperbolic Equations of Second Order
whereU =
radicminusK(y)ux2 = U V = minusuy2 = minusV
a+ jb = 1minus j a = 1 = b = minus1 on L1
a+ jb = 1 + j a = 1 = minusb = minus1 on L2
From the above formulas we can write the complex forms of boundary conditions ofU + jV
Re [λ(z)(U + jV )] = r(z) z isin Lj (j = 1 2)
λ(z)=
⎧⎨⎩1minus j = a+ jb
1 + j = a minus jbr(x)=
⎧⎨⎩(minusK(y))12φprime(x)2 on L1
(minusK(y))12φprime(x)2 on L2
andu(z) = 2Re
int z
0(U minus jV )dz + φ(0) in D (57)
Hence Problem D is a special case of Problem P1
52 Unique solvability of Problem P for Chaplygin equation (51) in thehyperbolic domain D
In the subsection we discuss the Chaplygin equation (51) in the hyperbolic domainD where the arcs L1 = AC L2 = BC are the characteristics of (51) ie
x+int y
0
radicminusK(t)dt = 0 0 le x le 1 x minus
int y
0
radicminusK(t)dt = 2 1 le x le 2 (58)
Setting thatmicro = x+
int y
0
radicminusK(t)dt ν = x minus
int y
0
radicminusK(t)dt (59)
and thenmicro+ ν = 2x micro minus ν = 2
int y
0
radicminusK(t)dt
(micro minus ν)y = 2radic
minusK(y)radic
minusK(y) = (micro minus ν)y2
xmicro = 12 = xν ymicro = 12radic
minusK(y) = minusyν
(510)
hence we haveUx = Umicro + Uν Vy =
radicminusK(y)(Vmicro minus Vν)
Vx = Vmicro + Vν Uy =radic
minusK(y)(Umicro minus Uν)(511)
andK(y)Ux minus Vy = K(y)(Umicro + Uν)minus
radicminusK(y)(Vmicro minus Vν) = 0
Vx + Uy = Vmicro + Vν +radic
minusK(y)(Umicro minus Uν) = 0 in D(512)
5 Degenerate Hyperbolic Equations 69
where U = ux2 V = minusuy2 and U =radic
minusK(y)U V = minusV Noting that
(radicminusK(y)
)micro= minus1
2(minusK)minus12K prime(y)ymicro =
K prime
4K(radic
minusK(y))
ν= minusK prime4K (513)
we have(U minus V )micro =
radicminusK(y)Umicro + Vmicro +
K primeU4K
(U + V )ν =radic
minusK(y)Uν minus Vν minus K primeU4K
(514)
Moreover by (511) and (514) we obtain
minus2radic
minusK(y)(U + V )ν
=radic
minusK(y)[(U + V )micro minus (U + V )ν ]minusradic
minusK(y)[(U + V )micro + (U + V )ν ]
= minusradic
minusK(y)(U + V )x + (U + V )y
= minus12(minusK(y))minus12K prime(y)U =
K prime(y)2K(y)
U 2radic
minusK(y)(U minus V )micro
=radic
minusK(y)[(U minus V )micro + (U minus V )ν ]
+radic
minusK(y)[(U minus V )micro minus (U minus V )ν)]
=radic
minusK(y)(U minus V )x + (U minus V )y
= minus12(minusK(y))minus12K prime(y)U =
K prime(y)2K(y)
U
(515)
Thus equation (515) can be written as a system of equations
(U+V )ν=minus(minusK)minus12 K prime(y)4K(y)
U (UminusV )micro=(minusK)minus12 K prime(y)4K(y)
U ie
Wmacrz=12[(minusK(y))12WxminusjWy]=A1W (z)+A2W (z) in ∆
(516)
in which W = U + jV A1(z)=A2(z)=minusjK prime8K in ∆ = 0 le micro le 2 0 le ν le 2and
u(z) = 2Reint z
0(U minus jV )dz + b0 in D (517)
where U minus jV = (minusK)minus12U + jV
In the following we first give the representation of solutions for the oblique deriva-tive problem (Problem P1) for system (516) inD For this we first discuss the systemof equations
(U + V )ν = 0 (U minus V )micro = 0 in D (518)
70 II Hyperbolic Equations of Second Order
with the boundary condition12
partu
partl= Re [λ(z)(U + jV )] = r(z) z isin L
u(0) = b0 Im [λ(z)(U + iV )]|z=z1 = b1
(519)
in which λ(z) = a(z) + jb(z) on L1 cup L2 Similarly to Chapter I the solution ofProblem P1 for (518) can be expressed as
ξ = U + V = f(micro) η = U minus V = g(ν)
U(x y) =f(micro) + g(ν)
2 V (x y) =
f(micro)minus g(ν)2
ie W (z) =(1 + j)f(micro) + (1minus j)g(ν)
2
(520)
in which f(t) g(t) are two arbitrary real continuous functions on [0 2] For conve-nience denote by the functions a(x) b(x) r(x) of x the functions a(z) b(z) r(z) of zin (519) thus the first formula in (519) can be rewritten as
a(x)U(x y) + b(x)V (x y) = r(x) on L ie
[a(x)minusb(x)]f(x+y)+[a(x)+b(x)]g(xminusy)=2r(x) on L ie
[a(t2) + b(t2)]f(0) + [a(t2)minus b(t2)]g(t) = 2r(t2) t isin [0 2][a(t2+1)minusb(t2+1)]f(t)+[a(t2+1)+b(t2+1)]g(2)=2r(t2+1) t isin [0 2]
where
f(0) = U(0) + V (0) =r(1) + b1
a(1) + b(1) g(2) = U(2)minus V (2) =
r(1)minus b1
a(1)minus b(1)
Noting that the boundary conditions in (519) we can derive
U =12
f(micro) +
2r(ν2)minus (a(ν2) + b(ν2))f(0)a(ν2)minus b(ν2)
V =12
f(micro)minus 2r(ν2)minus (a(ν2) + b(ν2))f(0)
a(ν2)minus b(ν2)
or
U =12
g(ν) +
2r(micro2 + 1)minus (a(micro2 + 1)minus b(micro2 + 1))g(2)a(micro2 + 1) + b(micro2 + 1)
V =12
minusg(ν) +
2r(micro2 + 1)minus (a(micro2 + 1)minus b(micro2 + 1))g(2)a(micro2 + 1) + b(micro2 + 1)
(521)
if a(x)minus b(x) = 0 on [01] and a(x) + b(x) = 0 on [12] respectively From the aboveformulas it follows that
Re[(1+j)W (x)]= U+V =2r(x2+1)minus(a(x2+1)+b(x2+1))f(0)
a(x2+1)minusb(x2+1)
Re[(1minusj)W (x)]= UminusV =2r(x2)minus(a(x2)minusb(x2))g(2)
a(x2) + b(x2) x isin [0 2]
(522)
5 Degenerate Hyperbolic Equations 71
if a(x)minus b(x) = 0 on [01] and a(x) + b(x) = 0 on [12] respectively From (522)
W (z) =
⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
12
[(1 + j)
2r(micro2 + 1)minus (a(micro2 + 1)minus b(micro2 + 1))g(2)a(micro2 + 1) + b(micro2 + 1)
+(1minus j)2r(ν2)minus (a(ν2) + b(ν2))f(0)
a(ν2)minus b(ν2)
] (523)
can be derived
Next we find the solution of Problem P1 for system (516) From (516) we have
U + V = minusint ν
2(minusK)minus12 K prime(y)
4K(y)Udν U minus V =
int micro
0(minusK)minus12 K prime(y)
4K(y)Udmicro
W = U+jV =minus1+j
2
int ν
2(minusK)minus12 K prime(y)
4K(y)Udν+
1minusj
2
int micro
0(minusK)minus12 K prime(y)
4K(y)Udmicro
= minus1 + j
2
int ν
2
K prime(y)4K(y)
Udν +1minus j
2
int micro
0
K prime(y)4K(y)
Udmicro
the above last two integrals are along two characteristic lines s2 and s1 respectivelyBut according to the method in [66]1) if we denote by s1 the member of the familyof characteristic lines dxdy = minus
radicminusK(y) and by s2 the member of the family of
characteristic lines dxdy =radic
minusK(y) passing through the point P isin D and
ds1 =radic(dx)2 + (dy)2 =
radic1 +
(dy
dx
)2dx =
radic1minus K
minusKdx = minusradic
1minus Kdy
ds2 =radic(dx)2 + (dy)2 =
radic1 +
(dy
dx
)2dx =
radic1minus K
minusKdx =
radic1minus Kdy
(524)
then system (51) can be rewritten in the form
(U+V )s1 = (U+V )xxs1+(U+V )yys1
=1radic1minusK
[radicminusK(y)(U+V )xminus(U+V )y
]
=1
2radic1minusK
(minusK(y))minus12K prime(y)U=minus 12radic1minusK
K prime(y)K(y)
U
(U minusV )s2 = (U minusV )xxs2+(U minusV )yys2
=1radic1minusK
[radicminusK(y)(U minusV )x+(U minusV )y
]
= minus 12radic1minusK
(minusK(y))minus12K prime(y)U=1
2radic1minusK
K prime(y)K(y)
U
(525)
72 II Hyperbolic Equations of Second Order
thus we obtain the system of integral equations
ξ = U + V = minusint s1
0
1
2radic(1minus K)
K prime(y)K(y)
Uds1 =int s1
0
12
K prime(y)K(y)
Udy
η = U minus V =int s2
0
1
2radic(1minus K)
K prime(y)K(y)
Uds2 =int s2
0
12
K prime(y)K(y)
Udy
(526)
where the integrals are along the two families of characteristics s1 and s2 respectivelySimilarly to next subsection the solution U+V UminusV can be obtained by the methodof successive iteration which can be expressed as
W (z) = U + jV = W (z) + Φ(z) + Ψ(z)
U+V =minusint s1
0
1
2radic(1minusK)
K prime(y)K(y)
Uds1 UminusV =int s2
0
1
2radic(1minusK)
K prime(y)K(y)
Uds2(527)
whereW (z) Φ(z) are the solutions of system (518) satisfying the boundary condition(519) respectively but the function r(z) b1 in the boundary condition of Φ(z) shouldbe replaced by minusRe [λ(z)Ψ(z)] on L1cupL2 and minusIm [λ(z1)Ψ(z1)] and then the function
u(z) = 2Reint z
0(U minus jV )dz + b0 = 2Re
int z
0[(minusK(y))minus12U + jV ]dz + b0 in D (528)
is just the solution of Problem P1 for (516) and the solution is unique Here wemention that firstly it suffices to consider the case of y1 le y le minusδ where δ is asufficiently small positive number and when we find out the solution of Problem P1
for equation (51) with the condition y1 le y le minusδ and then let δ rarr 0
Theorem 51 If the Chaplygin equation (51) in D satisfies the above conditionsthen Problem P1 for (51) in D has a unique solution
Finally we mention that the boundary condition of Problem P1
Re [λ(z)(U + jV )] = r(z) z isin L1 cup L2
can be replaced by
2U(x)=S(x)=radic
minusK[y(x)]ux ie u(x)=int x
0
S(x)dxradicminusK[y(x)]
+u(0)=s(x)
2V (x) = R(x) = uy on AB = L0 = (0 2)
(529)
where Cα[R(x) AB] le k2 lt infin C1α[s(x) AB] le k2
5 Degenerate Hyperbolic Equations 73
53 Unique solvability of the oblique derivative problem for degeneratehyperbolic equations
In this subsection we prove the uniqueness and existence of solutions of Problem P4
for the degenerate hyperbolic equation (52) the boundary condition of Problem P4
is as followsu(x) = s(x) uy(x) = R(x) on L0 = (0 2) (530)
where s(x) R(x) satisfy the condition C2α[s(x) L0] C2
α[R(x) L0] le k2 the aboveboundary value problem is also called the Cauchy problem for (52) Making a trans-formation of function
v(z) = u(z)minus yR(x)minus s(x) in D (531)
equation (52) and boundary condition (530) are reduced to the form
K(y)vxx + vyy = dvx + evy + fv +G
G = g + f(yR + s) + eR + d[yRprime(x) + sprime(x)]
minusK(y)[yRprimeprime(x) + sprimeprime(x)] in D
(532)
v(x) = 0 vy(x) = 0 on L0 (533)
Hence we may only discuss Cauchy problem (532)(533) and denote it by ProblemP4 again According to Subsection 43 Problem P4 for (532) is equivalent to theboundary value problem A for the hyperbolic system of first order equations therelation and the boundary conditions
ξs1 =2radicminusKradic1minus K
ξν =1
2radic1minus K
[( minusdradicminusKminus e minus 1
2K prime(y)K(y)
)ξ
+( minusdradicminusK
+ e minus 12
K prime(y)K(y)
)η minus fv minus G
]
ηs2 =2radicminusKradic1minus K
ηmicro =1
2radic1minus K
[( minusdradicminusKminus e+
12
K prime(y)K(y)
)ξ
+( minusdradicminusK
+ e+12
K prime(y)K(y)
)η minus fv minus G
]
ξ = U + V η = U minus V vy = ξ minus η v(x) = 0
(534)
In particular if K(y) = minus|y|mh(y) m is a positive constant then
K prime(y)K(y)
=m|y|mminus1h(y)
K(y)minus |y|mhy
K(y)=
m
y+
hy
h
74 II Hyperbolic Equations of Second Order
Integrating the hyperbolic system in (534) along the characteristics s1 s2 we obtainthe system of integral equations as follows
v(z) =int y
0(ξ minus η)dy in D
ξ(z) = minusint y
0[A1ξ +B1η + C1(ξ + η) +Dv + E]dy z isin s1
η(z) =int y
0[A2ξ +B2η + C2(ξ + η) +Dv + E]dy z isin s2
(535)
in this case
A1 = minuse
2minus hy
4h B1 =
e
2minus hy
4h A2 = minuse
2+
hy
4h B2 =
e
2+
hy
4h
C1 = minus12
dradicminusKminus m
4y C2 = minus1
2dradicminusK
+m
4y D = minusf
2 E = minusG
2
In the following we may only discuss the case ofK(y) = minus|y|mh(y) because otherwiseit can be similarly discussed In order to find a solution of the system of integralequations (535) we need to add the condition
limyrarr0
|y|d(x y)|y|m2 = 0 ie d(x y) asymp ε(y)|y|m2minus1 (536)
where ε(y) rarr 0 as y rarr 0 It is clear that for two characteristics s01 x = x1(y z0)
s02 x = x2(y z0) passing through P0 = z0 = x0 + jy0 isin D we have
|x1 minus x2| le 2|int y
0
radicminusKdy| le M |y|m2+1 for yprime lt y lt 0 (537)
for any z1 = x1 + jy isin s01 z2 = x2 + jy isin s0
2 where M(gt max[4radic
h(y)(m + 2) 1])is a positive constant Suppose that the coefficients of (535) possess continuouslydifferentiable with respect to x isin L0 and satisfy the condition
|Aj| |Ajx| |Bj| |Bjx| |yCj| |yCjx| |D| |Dx| |E||Ex| |1radich| |hyh| le M z isin D j = 1 2
(538)
According to the proof of Theorem 51 it is sufficient to find a solution of ProblemP4 for arbitrary segment minusδ le y le 0 where δ is a sufficiently small positive numberand choose a positive constant γ(lt 1) close to 1 such that the following inequalitieshold
3Mδ
2+[ε(y)M +m2]δm2
m+ 2lt γ
6δ2M2
m+ 6+8δMm+ 2
+2ε(y)M +m
m+ 2lt γ
(539)
Similar to [66]1) a solution of Problem P4 for (535) on minusδ lt y lt 0 can be foundFirstly let y0 isin (minusδ 0) and D0 be a domain bounded by the boundary y = 0 s0
1 s02
5 Degenerate Hyperbolic Equations 75
we choose v0 = 0 ξ0 = 0 η0 = 0 and substitute them into the corresponding positionsof v ξ η in the right-hand sides of (535) and obtain
ξ1(z)=minusint y
0[A1ξ0+B1η0+C1(ξ0+η0)+Dv0 + E]dy=minus
int y
0Edy z isin s0
1
η1(z)=int y
0[A2ξ0+B2η0+C2(ξ0+η0)+Dv0+E]dy=
int y
0Edy zisin s0
2
v1(z) = Reint y
0(ξ0 minus η0)dy = 0 in D0
(540)
By the successive iteration we find the sequences of functions vk ξk ηk whichsatisfy the relations
ξk+1(z) = minusint y
0[A1ξk +B1ηk + C1(ξk + ηk) +Dvk + E]dy z isin s0
1
ηk+1(z) =int y
0[A2ξk +B2ηk + C2(ξk + ηk) +Dvk + E]dy z isin s0
2
vk+1(z) =int y
0(ξk minus ηk)dy in D0 k = 0 1 2
(541)
We can prove that vk ξk ηk in D0 satisfy the estimates
|vk(z)| |ξk(z)| |ηk(z)| le Mksum
j=0γj|y| |ξk(z) + ηk(z)|
|vk(z1)minusvk(z2)| |ξk(z1)minusξk(z2)| |ηk(z1)minusηk(z2)|leMksum
j=0γj|y|m2+1
|vk+1(z)minus vk(z)| |ξk+1(z)minus ξk(z)| |ηk+1(z)minus ηk(z)| le Mγk|y||ξk+1(z) + ηk+1(z)minus ξk(z)minus ηk(z)| |vk+1(z1)minus vk+1(z2)
minusvk(z1)minus vk(z2)| |ξk+1(z1)minus ξk+1(z2)minus ξk(z1) + ξk(z2)||ηk+1(z1)minus ηk+1(z2)minus ηk(z1) + ηk(z2)| le Mγk|y|m2+1
(542)
In fact from (540) it follows that the first formula with k = 1 holds namely
|v1(z)|=0 le M |y| |ξ1(z)| le M |y| |η1(z)| le M |y|=Mγ0|y| le M1sum
j=0γj|y|
Moreover we get
|v1(z1)minusv1(z2)|=0|ξ1(z)+η1(z)|le|
int y
0[E(z1)minusE(z2)]dy| le2|
int y
0Ex[x1minusx2]dy|
le 4m+4
M2|y|m2+2 leMγ|y|m2+1 leM1sum
j=0γj|y|m2+1
76 II Hyperbolic Equations of Second Order
|ξ1(z1)minusξ1(z2)| le |int y
0[E(xj(t z1)+jt)minusE(xj(t z2)+jt)]dt|
le |int y
0|Ex||xj(t z1)minus xj(t z2)|dy| le M |
int y
0|x1 minus x2|dy|
le M |int y
0M |y|m2+1dy| le Mγ|y|m2+1 le M
1sumj=0
γj|y|m2+1
|η1(z1)minusη1(z2)|= |int y
0[E(xj(t z1)+jt)minusE(xj(t z2)+jt)]dt| leM
1sumj=0
γj|y|m2+1
|v1(z)minus v0(z)| = |v1(z)| le Mγ|y| |ξ1(z)minus ξ0(z)| = |ξ1(z)| le Mγ|y||η1(z)minusη0(z)|= |η1(z)|leMγ|y| |v1(z1)minusv1(z2)minusv0(z1)minusv0(z2)|leMγ|y|m2+1
|ξ1(z)+η1(z)minusξ0(z)minusη0(z)|= |ξ1(z)+η1(z)|leγ|y|m2+1 leMγ|y|m2+1
|ξ1(z1)minusξ1(z2)minusξ0(z1)+ξ0(z2)|= |ξ1(z1)minusξ1(z2)| le Mγ|y|m2+1
|η1(z1)minusη1(z2)minusη0(z1)+η0(z2)|= |η1(z1)minusη1(z2)|leMγ|y|m2+1
In addition we use the inductive method namely suppose the estimates in (542) fork = n are valid then they are also valid for k = n + 1 In the following we onlygive the estimates of |vn+1(z)| |ξn+1(z)| |ξn+1 + ηn+1(z)| the other estimates can besimilarly given From (541) we have
|vn+1(z)| le |int y
0[ξnminusηn]dy|le2M |
int y
0
nsumj=0
γjydy|leMnsum
j=0γj|y|2 leM
n+1sumj=0
γj|y|
|ξn+1(z)| le |int y
0
⎡⎣(|A1|+|B1|+|D|)Mnsum
j=0γj|y|+|C1|M
nsumj=0
γj|y|m2+1+|E|⎤⎦ dy|
le M |int y
0
⎡⎣3M nsumj=0
γj|y|+(
ε(y)2|y|radich
+m
4|y|)
nsumj=0
γj|y|m2+1 + 1
⎤⎦ dy|
le M |y|⎧⎨⎩[32M |y|+
(|ε(y)|M +
m
2
) |y|m2
m+ 2
]nsum
j=0γj + 1
⎫⎬⎭ le Mn+1sumj=0
γj|y|
and
|ξn+1(z)+ηn+1(z)| le |int y
0
2sumj=1[A2(z2)ξn(z2)minusA1(z1)ξn(z1)+B2(z2)ηn(z2)
minusB1(z1)ηn(z1)] + C2(z2)(ξn(z2) + ηn(z2))
minusC1(z1)(ξn(z1) + ηn(z1))
+D(z2)vn(z2)minus D(z1)vn(z1) + E(z2)minus E(z1)]
dy|
5 Degenerate Hyperbolic Equations 77
Noting that
|D(z2)vn(z2)minus D(z1)vn(z1)| = |[D(z2)minus D(z1)]vn(z2) +D(z1)
times[vn(z2)minus vn(z1)]|
lensum
j=0γj|y||D(z2)minus D(z1)|+M2
nsumj=0
γj|y|m2+1
le M2nsum
j=0γj|y||x2minusx1|+M2
nsumj=0
γj|y|m2+1
le (M |y|+1)M2nsum
j=0γj|y|m2+1
and
|A2(z2)ξn(z2)minus A1(z1)ξn(z1) +B2(z2)ηn(z2)minus B1(z1)ηn(z1)|le|[A2(z2)minusA2(z1)]ξn(z2)+[A2(z1)minusA1(z1)]ξn(z2)+A1(z1)[ξn(z2)minusξn(z1)]
+[B2(z2)minusB2(z1)]ηn(z2)+B1(z1)[ηn(z2)minusηn(z1)]|+[B2(z1)minusB1(z1)]ηn(z2)
le 2M |y|[M |x1 minus x2|+ |y|m2]nsum
j=0γj +
∣∣∣∣∣hy
2h
∣∣∣∣∣ |ξn(z2)minus ηn(z2)|
le (2M |y|+ 3)M2nsum
j=0γj|y|m2+1
we get
|ξn+1(z) + ηn+1(z)| le |int y
0[(3M |y|+ 4)M2
nsumj=0
γj|y|m2+1
+(|C1|+ |C2|)Mnsum
j=0γj|y|m2+1 +M2|y|m2+1]dy|
le M |y|m2+1
[6M2y2
m+ 6+8M |y|m+ 2
+(|ε(y)|M +
m
2
)
times 2m+ 2
nsumj=0
γj +2M
m+ 4|y|⎤⎦
le Mn+1sumj=0
γj|y|m2+1
On the basis of the estimates (542) we can derive that vk ξk ηk in D0
uniformly converge to vlowast ξlowast ηlowast satisfying the system of integral equations
ξlowast(z) = minusint y
0[A1ξlowast +B1ηlowast + C1(ξlowast + ηlowast) +Dvlowast + E]dy z isin s1
78 II Hyperbolic Equations of Second Order
ηlowast(z) =int y
0[A2ξlowast +B2ηlowast + C2(ξlowast + ηlowast) +Dvlowast + E]dy z isin s2
vlowast(z) =int y
0(ξlowast minus ηlowast)dy in D0
and the function vlowast(z) satisfies equation (532) and boundary condition (533) henceulowast(z) = vlowast(z) + yR(x) + s(x) is a solution of Problem P4 for (52) From the abovediscussion we can see that the solution of Problem P4 for (52) in D is unique
Theorem 52 If the equation (52) in D satisfies the above conditions then ProblemP4 for (52) in D has a unique solution
Now we mention that if we denote
W (z)= U+jV = |y|m2UminusjV =12[|y|m2ux+juy]
W (z)= UminusjV = |y|m2U+jV =12[|y|m2uxminusjuy]
then W (z) = |y|m2UminusjV is a solution of the first order hyperbolic complex equation
Wmacrz = A1(z)W + A2(z)W + A3(z)u+ A4(z) in D
A1 = minus d
4|y|m2 + j
(m
8|y| minus e
4
) A3=minusf
4
A2 = minus d
4|y|m2 + j
(m
8|y| +e
4
) A4=minusg
4
(543)
and
u(z)=2Reint z
0uzdz=2Re
int z
0(UminusjV )d(x+jy) =2Re
int z
0(U+jV )d(xminusjy)
is a solution of equation (52) with K(y) = minus|y|mBy using the similar method we can prove the solvability of Problem P1 Problem
P2 and Problem P3 for equation (52) Moreover for general domain Dprime with non-characteristics boundary we can also discuss the solvability of Problem P1 ProblemP2 Problem P3 and Problem P4 for equation (52) Besides we can discuss thesolvability of corresponding boundary value problems for the hyperbolic equation inthe form
uxx +K(y)uyy = dux + euy + fu+ g in D (544)
under certain conditions where K(y) is as stated in (52)
The references for the chapter are [2][7][12][13][24][25][34][41][44][47][54][60][66][70][79][85][87][89][95]
CHAPTER III
NONLINEAR ELLIPTIC COMPLEXEQUATIONS OF FIRST ANDSECOND ORDER
In this chapter we discuss the representation and existence of solutions of discontinu-ous boundary value problems for nonlinear elliptic complex equations of first andsecond order which will be used in latter chapters
1 Generalizations of KeldychndashSedov Formula for AnalyticFunctions
It is known that the KeldychndashSedov formula gives the representation of solutions ofthe mixed boundary value problem for analytic functions in the upper half-plane (see[53]) But for many problems in mechanics and physics one needs a more generalformulas of solutions of the discontinuous RiemannndashHilbert boundary value problemfor analytic functions in the upper half-plane and other special domains In thissection we shall establish the representations of solutions of the general discontinuousboundary value problem for analytic functions in the upper half-plane and upper half-unit disk In the following sections and chapters we shall give applications to somenonlinear elliptic complex equations and quasilinear equations of mixed type
11 General discontinuous boundary value problem for analytic functionsin the upper half-plane
Let D be the upper half-plane and a(x) b(x) c(x) be known real functions on L =minusinfin lt x lt infin y = 0 where a(x) b(x) possess discontinuities of first kind at mdistinct points xj(j = 1 m minusinfin lt x1 lt middot middot middot lt xm lt infin) m is a positive integerand c(x) = O(|xminusxj|minusβj) in the neighborhood of xj(j = 1 2 m) on L herein βj(lt1 j = 1 2 m) are non-negative constants such that βj+γj lt 1 γj(j = 1 m)are as stated in (13) below Denote λ(x) = a(x)minus ib(x) and |a(x)|+ |b(x)| = 0 thereis no harm in assuming that |λ(x)| = 1 x isin Llowast = Lx1 xm Suppose thatλ(x) c(x) satisfy the conditions
λ(x) isin Cα(Lj) |x minus xj|βjc(x) isin Cα(Lj) j = 1 2 m (11)
80 III Elliptic Complex Equations
where Lj is the line segment from the point xjminus1 to xj on L x0 = xm Lj(j =1 2 m) do not include the end points L1 = x lt x1 cup x gt xm α(0 lt α lt 1)is a constant and the function λ(x) isin Cα(Linfin)(Linfin is a neighborhood of the point infin)is indicated as λ(1x) isin Cα(Llowast) here Llowast(sub L) is a neighborhood of the point x = 0
The discontinuous RiemannndashHilbert boundary value problem for analytic functionsin D may be formulated as follows
Problem A Find an analytic function Φ(z) = u(z)+iv(z) in D which is continuousin Dlowast = Dx1 x2 xm satisfying the boundary condition
Re [λ(x)Φ(x)] = au minus bv = c(x) z isin Llowast (12)
Problem A with the condition c(x) = 0 on Llowast is called Problem A0
Denote by λ(xj minus 0) and λ(xj + 0) the left limit and right limit of λ(x) as x rarrxj(j = 1 2 m) on L and
eiφj =λ(xjminus0)λ(xj+0)
γj=1πiln[λ(xjminus0)λ(xj+0)
]=
φj
πminusKj
Kj =[φj
π
]+ Jj Jj = 0 or 1 j = 1 m
(13)
in which 0 le γj lt 1 when Jj = 0 and minus1 lt γj lt 0 when Jj = 1 j = 1 m and
K =12(K1 + middot middot middot+Km) =
12
msumj=1
[φj
πminus γj
](14)
is called the index of Problem A and Problem A0 If λ(x) on L is continuous thenK = ∆Γ arg λ(x)2π is a unique integer If the function λ(x) on L is not continuouswe can choose Jj = 0 or 1 hence the index K is not unique We can require that thesolution Φ(z) satisfy the condition
Φ(z) = O(|z minus xj|minusδ) δ =
⎧⎨⎩βj + τ for γj ge 0 and γj lt 0 βj gt |γj||γj|+ τ for γj lt 0 βj le |γj| j = 1 m
(15)
in the neighborhood (sub D) of xj where τ (lt α) is an arbitrary small positive number
In order to find the solution of Problem A for analytic functions we first considerProblem A0 Making a transformation
Ψ(z) =Φ(z)Π(z)
Π(z) =mprod
j=1
(z minus xj
z + i
)γj
(16)
in which γj (j = 1 m) are as stated in (13) the boundary condition
Re [λ(x)Φ0(x)] = 0 x isin Llowast (17)
1 Generalizations of KeldychndashSedov Formula 81
of Problem A0 for analytic functions Φ0(z) is reduced to the boundary condition
Re [Λ(x)Ψ0(x)] = 0 Λ(x) = λ(x)Π(x)|Π(x)| x isin Llowast (18)
of Problem Alowast0 for analytic functions Ψ0(z) = Φ0(z)Π(z) Noting that
Λ(xj minus 0)Λ(xj + 0)
=λ(xj minus 0)λ(xj + 0)
(Π(xj minus 0)Π(xj + 0)
)=
λ(xj minus 0)λ(xj + 0)
eminusiπγj = plusmn1 (19)
the index of Λ(x) on L is
K =12π∆L arg Λ(x) =
12
msumj=1
[φj
πminus γj
]=12
msumj=1
Kj (110)
which is the same as the index of λ(x) on L If 2K is even provided that we changethe signs of Λ(x) on some line segments of Lj (j = 1 m) then the new functionΛlowast(x) on L is continuous its index is K too When 2K is odd we rewrite theboundary condition (18) in the form
Re[Λ(x)
x minus x0
x+ i
x+ i
x minus x0Ψ0(x)
]= 0 x isin Llowast (111)
where x0(isin L) is a real number and x0 isin x1 xm thus similarly to before wechange the signs of Λ(x)(x minus x0)|x + i|(x + i)|x minus x0| on some line segments of Lsuch that the new function Λlowast(x) on L is continuous its index is Klowast = K minus 12Next we find an analytic function
Ψlowast(z) = i(
z minus i
z + i
)[K] (z minus x0
z + i
)eiS(z) in D (112)
which satisfies the homogeneous boundary condition
Re [Λlowast(x)Ψlowast(x)] = 0 x isin Llowast (113)
where [K] is the integer part of K S(z) is an analytic function in D satisfying theboundary condition
Re [S(x)] = arg
⎡⎣Λlowast(x)(
x minus i
x+ i
)[K](x minus x0
x+ i
)⎤⎦ x isin L Im [S(i)] = 0 (114)
Hence Problem Alowast0 for analytic functions possesses the solution
Ψ0(z) =
⎧⎨⎩Ψlowast(z) when 2K is even
(z minus x0)Ψlowast(z)(z + i) when 2K is odd(115)
82 III Elliptic Complex Equations
and then Problem A0 for analytic functions has a non-trivial solution in the form
X(z) = Π(z)Ψ0(z) =
⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩i(
z minus i
z + i
)K
Π(z)eiS(z) when 2K is even
i(
zminusi
z+i
)[K] zminusx0
z + iΠ(z)eiS(z) when 2K is odd
(116)
Take into account that X(z) has a zero of order [K] at the point z = i for K ge 0and a pole of order |[K]| at the point z = i for K lt 0 and a zero of order 1 at thepoint z = x0 when 2K is an odd integer moreover X(z) satisfies the homogeneousboundary condition (17) it is clear that iλ(x)X(x) is a real-valued function on L Letus divide the nonhomogeneous boundary condition (12) by iλ(x)X(x) and obtain
Re[Φ(x)iX(x)
]=
c(x)iλ(x)X(x)
=λ(x)c(x)iX(x)
x isin Llowast (117)
By using the Schwarz formula we get
Φ(z)iX(z)
=1πi
[int infin
minusinfinλ(t)c(t)
(t minus z)iX(t)dt+
Q(z)i
]
Φ(z) =X(z)πi
[int infin
minusinfinλ(t)c(t)
(t minus z)X(t)dt+Q(z)
]
(118)
If K ge 0 the function Q(z) possesses the form
Q(z) = i[K]sumj=0
[cj
(z minus i
z + i
)j
+ cj
(z minus i
z + i
)minusj]+
⎧⎪⎪⎨⎪⎪⎩0 when 2K is even
iclowastx0 + z
x0 minus z when 2K is odd
(119)
where clowast c0 are arbitrary real constants and cj (j=1 [K]) are arbitrary complexconstants from this we can see that the general solution Φ(z) includes 2K+1 arbitraryreal constants If 2K is odd we note (zminusx0)[(tminusz)(tminusx0)] = 1(tminusz)minus1(tminusx0)then the integral in (118) is understood as the difference of two integrals of Cauchytype If K lt 0 we have to take
Q(z) =
⎧⎪⎪⎨⎪⎪⎩iclowast = 0 when 2K is even
iclowastx0 + z
x0 minus z when 2K is odd
(120)
and require that the function in the square bracket of (118) has at least a zero point
1 Generalizations of KeldychndashSedov Formula 83
of order |[K]| at z = i From
int infin
minusinfinλ(t)c(t)
(t minus z)X(t)dt+ iclowast
x0 + z
x0 minus z
=int infin
minusinfinλ(t)c(t)
(1minus(zminusi)(tminusi))(tminusi)X(t)dt+iclowast
1+(zminusi)(x0+i)1minus(zminusi)(x0minusi)
x0+i
x0minusi
=infinsum
j=0
int infin
minusinfinλ(t)c(t)(z minus i)j
(t minus i)j+1X(t)dt+ iclowast
[1 +
z minus i
x0 + i
]x0 + i
x0 minus i
infinsumj=0
(z minus i
x0 minus i
)j
=int infin
minusinfinλ(t)c(t)(tminusi)X(t)
dt+iclowastx0+i
x0minusi+
infinsumj=1
[int infin
minusinfinλ(t)c(t)
(tminusi)j+1X(t)dt+
2iclowastx0
(x0minusi)j+1
](zminusi)j
(121)
in the neighborhood of z = i this shows thatint infin
minusinfinλ(t)c(t)
X(t)(t minus i)jdt = 0 j = 1 minusK(= |[K]|) when 2K is even
int infin
minusinfinλ(t)c(t)
X(t)(t minus i)jdt+
2iclowastx0
(x0 minus i)j=0 j=2 [minusK]+1(= |[K]|)
when 2K is odd
(122)
then the function in the square bracket of (118) has a zero point of order |[K]| atz = i hence the function Φ(z) is analytic at z = i Besides when 2K is odd
clowast = ix0 minus i
x0 + i
int infin
minusinfinλ(t)c(t)
X(t)(t minus i)dt (123)
is a determined constant Therefore when K lt 0 Problem A has minus2K minus1 solvabilityconditions Thus we have the following theorem
Theorem 11 Problem A for analytic functions in D = Im z gt 0 has the follow-ing solvability result
(1) If the index K ge 0 the general solution Φ(z) of Problem A possesses the form(118) (119) which includes 2K + 1 arbitrary real constants
(2) If the index K lt 0 Problem A has minus2K minus 1 solvability conditions as stated in(122) When the conditions hold the solution of Problem A is given by the secondformula in (118) in particular
Φ(z)=X(z)(z minus i)|[K]|
πi
[int infin
minusinfinλ(t)c(t)
(t minus z)(t minus i)|[K]|X(t)dt+
2iclowastx0
(x0 minus z)(x0 minus i)|[K]|
](124)
in |z minus i| lt 1 where the constant clowast is determined as stated in (123)
Finally we mention that if x1 xm are first kind of discontinuities of c(x)and if γj gt 0 j = 1 2 m then the solution Φ(z) of Problem A is bounded in
84 III Elliptic Complex Equations
Dlowast = Dx1 xm In general if γj le 0 (1 le j le m) the solution Φ(z) ofProblem A may not be bounded in the neighborhood of xj in Dlowast = Dx1 xmWe have
Φ(z) =
⎧⎨⎩O(|z minus xj|minusγj) if γj lt 0 Jj = 1
O(ln |z minus xj|) if γj = 0 Jj = 0(125)
in the neighborhood of xj on Dlowast but the integral
int z
iΦ(z)dz in D
is bounded In particular if m = 2n and
λ(x) =
⎧⎨⎩ 1 x isin (x2jminus1 x2j)
i x isin (x2j x2j+1)j = 1 n
and xj(j = 1 m = 2n) are first kind of discontinuous points of c(x) we canchoose γ2jminus1 = 12 K2jminus1 = 0 γ2j = minus12 K2j = 0 j = 1 n and thenthe index of the mixed boundary value problem is K = 0 In this case one canchoose Π(z) =
radicΠn
j=1(z minus x2jminus1)(z minus x2j) From the formula (118) with K = 0the KeldychndashSedov formula of the mixed boundary value problem for analytic func-tions in the upper half-plane is derived [53] If we chose γ2jminus1 = minus12 K2jminus1 =1 γ2j = minus12 K2j = 0 j = 1 n and the index of the mixed boundary valueproblem is K = n = m2 then the representation of solutions of the mixed bound-ary value problem for analytic functions can be written from (118) with K = nwhich includes 2K + 1 = m + 1 arbitrary real constants where the functionΠ(z) = 1
radicΠn
j=1(z minus x2jminus1)(z minus x2j)
12 The general discontinuous boundary value problem for analyticfunctions in the upper half-disk
Now we first introduce the general discontinuous RiemannndashHilbert problem (ProblemB) for analytic functions in the unit disk D = |z| lt 1 with the boundary conditions
Re [λ(z)Φ(z)] = au minus bv = c(z) Γ = |z| = 1 (126)
where λ(z) = a(z)minus ib(z) |λ(z)| = 1 on Γ and Z = z1 z2 zm are first kind ofdiscontinuous points of λ(z) on Γ and λ(z) c(z) satisfy the conditions
λ(z) isin Cα(Γj) |z minus zj|βjc(z) isin Cα(Γj) j = 1 2 m (127)
herein Γj is the arc from the point zjminus1 to zj on Γ and z0 = zm and Γj(j = 1 2 m)does not include the end points α(0 lt α lt 1) is a constant
1 Generalizations of KeldychndashSedov Formula 85
Denote by λ(zj minus 0) and λ(zj + 0) the left limit and right limit of λ(z) as z rarrzj(j = 1 2 m) on Γ and
eiφj =λ(zj minus 0)λ(zj + 0)
γj =1πiln[λ(zj minus 0)λ(zj + 0)
]=
φj
πminus Kj
Kj =[φj
π
]+ Jj Jj = 0 or 1 j = 1 m
(128)
in which 0 le γj lt 1 when Jj = 0 and minus1 lt γj lt 0 when Jj = 1 j = 1 m Theindex K of Problem B is defined by (14) Let βj + γj lt 1 j = 1 m we requirethat the solution Φ(z) possesses the property
Φ(z) = O(|z minus zj|minusδ) δ =
⎧⎨⎩βj + τ for γj ge 0 and γj lt 0 βj gt |γj||γj|+ τ for γj lt 0 βj le |γj| j = 1 m
(129)
in the neighborhood (sub Dlowast) of zj where τ(lt α) is an arbitrary small positive numberBy using a similar method as stated in Subsection 1 we can obtain the formula forsolutions of the boundary value problem
Theorem 12 Problem B for analytic functions in D = |z| lt 1 has the followingsolvability result
(1) If the index K ge 0 the general solution Φ(z) of Problem B possesses the form
Φ(z) =X(z)2πi
[intΓ
(t+ z)λ(t)c(t)(t minus z)tX(t)
dt+Q(z)] (130)
with
Q(z) = i[K]sumj=0(cjz
j + cjzminusj) +
⎧⎪⎪⎨⎪⎪⎩0 when 2K is even
iclowastz0 + z
z0 minus z when 2K is odd
(131)
where the constant clowast c0 are arbitrary real constants and cj (j = 1 [K]) arearbitrary complex constants which includes 2K + 1 arbitrary real constants
(2) If the index K lt 0 Problem B has minus2K minus 1 solvability conditions given by
intΓ
λ(t)c(t)X(t)tj
dt=0 j=1 minusK(= |[K]|) when 2K is even
intΓ
λ(t)c(t)X(t)tj
dt+iclowastzminusj+10 =0 j=1 [minusK]+1(= |[K]|) when 2K is odd
(132)When the conditions hold the solution of Problem B possesses the form
Φ(z) =X(z)z[K]
πi
[intΓ
λ(t)c(t)(t minus z)X(t)t|[K]| dt+
iclowast(z0 minus z)z|[K]|minus1
0
] (133)
86 III Elliptic Complex Equations
where the constant clowast is determined via (132) as
clowast = iintΓ
λ(t)c(t)X(t)t
dt
In the above formula X(z) is a non-trivial solution of the homogeneous boundaryvalue problem (Problem B0) for analytic functions in the form
X(z) =
⎧⎨⎩ izKΠ(z)eiS(z) when 2K is even
iz[K](z minus z0)Π(z)eiS(z) when 2K is oddΠ(z) =
mprodj=1(z minus zj)γj (134)
in which S(z) is an analytic function in D satisfying the boundary conditions
Re [S(z)] = arg[Λlowast(z)z[K]] z isin Γ Im [S(0)] = 0
the function Λlowast(z) is similar to that in (113) [85]11)[86]1)
In addition through the conformal mapping from the upper half-unit disk D =|z| lt 1 Im z gt 0 onto the unit disk G = |ζ| lt 1 namely
ζ(z) = minusiz2 + 2iz + 1z2 minus 2iz + 1
z(ζ) =1
ζ + i
[1 + iζ minus
radic2(1minus ζ2)
]
we can obtain the result of the general discontinuous RiemannndashHilbert problem(Problem C) for analytic functions in the upper half-unit disk D = |z| lt 1 Im z gt0 namely
w(z) = Φ[ζ(z)] in D = |z| lt 1 Im z gt 0 (135)
is the solution of Problem C for analytic functions
In order to the requirement in latter chapters we give a well posed version (Prob-lem Bprime) of Problem B for analytic functions in D = |z| lt 1 namely we findan analytic function Φ(z) which is continuous in DZ and satisfies the boundarycondition (126) and the point conditions
Im [λ(zprimej)Φ(z
primej)] = bj j = 1 m (136)
where zprime1 z
primem (isin Z) are distinct points on Γ and bj (j = 1 m) are real con-
stants and we choose the index K = (m minus 1)2 of λ(z) on Γ = |z| = 1The homogeneous problem of Problem Bprime with the conditions c(z) = 0 on Γ andbj = 0 (j = 1 m) will be called Problem Bprime
0
Theorem 13 Problem Bprime for analytic functions in D = |z| lt 1 has a uniquesolution
Proof First of all we verify the uniqueness of solutions of Problem Bprime LetΦ1(z)Φ2(z) be two solutions of Problem Bprime for analytic functions Then the functionΦ(z) = Φ1(z) minus Φ2(z) is a solution of Problem Bprime
0 with the homogeneous boundaryconditions
Re [λ(z)Φ(z)]=0 on Γ=|z| = 1 Im [λ(zprimej)Φ(z
primej)]=0 j = 1 m (137)
1 Generalizations of KeldychndashSedov Formula 87
According to the method in the proof of Theorem 11 or [8]2)[80]1) and [85]11) wesee that if Φ(z) equiv 0 in D then
m = 2K + 1 le 2ND +NΓ le 2K (138)
where ND NΓ are numbers of zero points in D and Γlowast = ΓZ respectively Thiscontradiction proves that Φ(z) equiv 0 ie Φ1(z) = Φ2(z) in D
Now we prove the existence of solutions of Problem Bprime for analytic functions Bythe representation (130) of the general solution of Problem B for analytic functionsit is easy to see that the general solution Φ(z) can be written as
Φ(z) = Φ0(z) +msum
j=1djΦj(z) (139)
where Φ0(z) is a special solution of Problem Bprime and Φj(z) (j = 1 m) are acomplete system of linearly independent solutions of Problem Bprime
0 and dj(j = 1 m)are arbitrary real constants In the following we prove that there exists a uniquesystem of real constants dprime
j(j = 1 m) such that |dprime1| + middot middot middot + |dprime
m| = 0 satisfyingthe equalities
msumj=1
dprimejΦj(zprime
j) = λ(zprimej)[c(z
primej) + ibj]minus Φ0(zprime
j) j = 1 m (140)
Then the analytic function Φ(z) = Φ0(z) +summ
j=1 dprimejΦj(z) satisfies the boundary con-
ditions (126) and (136) and thus is a solution of Problem Bprime According to thealgebraic theory it suffices to verify that the homogeneous system of algebraic sys-tem of equations (140) ie
Φlowast(zprimej) =
msumj=1
dprimejΦj(zprime
j) = 0 j = 1 m (141)
has no non-trivial solution Noting that the analytic function Φlowast(z) =summ
j=1 dprimejΦj(z)
is a solution of Problem Bprime0 from the uniqueness of solutions of Problem Bprime we see
that Φlowast(z) = 0 This proves the existence of solutions of Problem Bprime for analyticfunctions
Next we consider that D is the upper half-unit disk a(z) b(z) possess discon-tinuities of first kind at m distinct points z1 zm isin Γ cup L0 = Γprime = partD which arearranged according to the positive direction of partD Here Γ = |z| = 1 Im z gt 0L0 = minus1 le x le 1 y = 0 and z1 znminus1 isin Γ = |z| = 1 Im z gt 0 xn =minus1 xm = x0 = 1 isin L0 where n (lt m) m are positive integers and c(z) =O(|z minus zj|minusβj) in the neighborhood of zj (j = 1 2 m) on Γ in which βj(lt 1j = 1 2 m) are non-negative constants such that βj + γj lt 1 γj(j = 1 m)are as stated in (129) Denote λ(z) = a(z)minus ib(z) and |a(z)|+ |b(z)| = 0 there is noharm in assuming that |λ(z)| = 1 z isin Γprime = Γ cup L0 Suppose that λ(z) c(z) satisfyconditions again (127)
88 III Elliptic Complex Equations
Problem C Find an analytic function Φ(z) = u(z)+iv(z) inD which is continuouson Dlowast = DZ satisfying the boundary condition
Re [λ(z)Φ(z)] = au minus bv = c(z) z isin Γlowast = ΓprimeZ (142)
here Z = z1 zm Problem C with the condition r(z) = 0 on Γlowast is calledProblem C0
The index K of Problem C is the same as stated in (14) We can require thatthe solution Φ(z) satisfies the condition (129)
In order to find the solution of Problem C for analytic functions it suffices tochoose a conformal mapping from the upper half-unit disk onto the upper half planeor the unit disk In the following we shall use the other method namely first find asolution of Problem A for analytic functions in D+ = Im z gt 0 with the boundarycondition
Re [λ(x)Φ(x)]=r(x) on L=(minusinfin infin)
r(x)=
⎧⎨⎩r(x) on L0=(minus1 1)c(x) on L1=(minusinfinltxltminus1) cup (1ltxltinfin)
(143)
in which λ(x) c(x) on L2 = (minusinfin lt x le minus1) cup (1 le x lt infin) are appropriatefunctions such that λ(x) |x minus xj|βjc(x) are piecewise Holder continuous functionsand continuous at the points x = minus1 1 and the index of λ(x) on L is K = 0 Forinstance setting
λ(x) =
⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩λ(minus1 + 0) on (minusinfin minus1]λ(x) on (minus1 1)λ(16(x+ 3)minus 3) on [1 infin)
(144)
and denoting x2mminusj = 16(xj + 3)minus 3 j = n+ 1 m minus 1 we can determine thatthe index of above function λ(x) on L is K = 0 On the basis of Theorem 11 thesolution Ψ(z) of Problem A can be expressed in the form (118)(119) with K = 0and λ(z) c(z) are as stated in (127) Thus the function Φ(z) = Φ(z) minus Ψ(z) isanalytic in D and satisfies the boundary condition
Re [λ(z)Φ(z)] = r(z) =
⎧⎨⎩ r(z)minus Re [λ(z)Ψ(z)] z isin Γ
0 z isin L0(145)
Next similarly to Section 1 we make a transformation
Ψ(z) =Φ(z)Π(z)
Π(z) =2mminusnminus1prod
j=n+1j =m
(z minus xj
z + i
)γj
(146)
in which γj(j = n+1 mminus1 m+1 2mminusnminus1) are similar to those in (128)the boundary condition
Re [λ(z)Φ(z)] = 0 z isin L (147)
1 Generalizations of KeldychndashSedov Formula 89
of Problem C0 for the analytic function Φ(z) is reduced to the boundary condition
Re [Λ(z)Ψ(z)] = 0 Λ(z) = λ(z)Π(z)|Π(z)| z isin L (148)
for the analytic function Ψ(z) = Φ(z)Π(z) Noting that
Λ(zj minus 0)Λ(zj + 0)
=λ(zj minus 0)λ(zj + 0)
(Π(zj minus 0)Π(zj + 0)
)=
λ(zj minus 0)λ(zj + 0)
eminusiπγj = plusmn1
j = n+ 1 m minus 1 m+ 1 2m minus n minus 1
(149)
the index of Λ(z) on L is the same as the index of λ(z) on L Due to 2K = 0 iseven provided that we change the sign of Λ(z) on some arcs Lj = (xjminus1 xj) (j =n + 1 2m minus n minus 1) Ln+1 = (minusinfin xn+1) cup (x2mminusnminus1 infin) then the new functionΛlowast(z) on Γ is continuous the index of λ(z) on L has not been changed Moreover wefind a solution of Problem A for analytic functions in Im z gt 0 with the boundaryconditions
Re [S(z)] = arg Λlowast(z) on L = (minusinfin infin) ImS(i) = 0 (150)
and denote Ψ(z) = Ψ(z)eminusiS(z)
Now we extend the analytic function Ψ(z) as follows
Φ(z) =
⎧⎪⎨⎪⎩Ψ(z) in D = |z| lt 1 Im z gt 0
minusΨ(z) in D = |z| lt 1 Im z lt 0(151)
It can be seen that the analytic function Φ(z) in |z| lt 1 satisfies the boundarycondition
Re [Λ(z)Φ(z)] = R(z) on |z| = 1 (152)
where
Λ(z)=
⎧⎨⎩λ(z)
λ(z)R(z)=
⎧⎨⎩r(z)eIm S(z) on Γ0=|z|=1 Im zgt0
minusr(z)eIm S(z) on Γ0=|z|=1 Im zlt0(153)
We can find the indexK prime of Λ(z) on |z| = 1 and by Theorem 12 the analytic functionΦ(z) in D with the boundary condition (152) can be found ie
Φ(z) = Φ(i1 + ζ
1minus ζ
)in D (154)
where Φ(z) is an analytic function as the function Φ(z) in (118) but in whichλ(z) c(z) K are replaced by λ[i(1 + ζ)(1minus ζ)] R[i(1 + ζ)(1minus ζ)] K prime respectivelyherein λ(z) R(z) are as stated in (153) It is clear that Φ(z) includes 2K prime + 1 arbi-trary real constants when K prime ge 0 and minus2K prime minus 1 solvability conditions when K prime lt 0
90 III Elliptic Complex Equations
Thus the solution of Problem C for analytic functions in the upper half-unit disk Dis obtained ie
w(z) = Ψ(z) + Φ(z)Π(z)eiS(z) in D (155)
Theorem 14 When the index K ge 0 Problem C for analytic functions in D hasa solution in the form (155) including 2K + 1 arbitrary real constants and whenK lt 0 under minus2K minus 1 conditions Problem C for analytic functions possesses thesolution as stated in (155) Moreover the above solution of Problem C for analyticfunctions can be expressed by (135)
The KeldychndashSedov formula for analytic functions in the upper half-plane possessesimportant applications to the Tricomi problem for some equations of mixed type (see[12]1)3)) But more general boundary value problems for equations of mixed typecannot be solved by this formula Due to we have Theorems 11ndash14 such that theabove general problems can be solved In addition we can give the representation ofsolutions to the discontinuous RiemannndashHilbert boundary value problem for analyticfunctions in the zone domain D = 0 lt Im z lt 1 which can be used to solve someboundary value problems for nonlinear problems in mechanics
2 Representation and Existence of Solutions for EllipticComplex Equations of First Order
In this section we shall establish the representations for solutions of the generaldiscontinuous boundary value problem for elliptic complex equations of first orderin the upper half-unit disk Moreover we shall prove the existence of solutions fornonlinear elliptic complex equations of first order
21 Representation of solutions of the discontinuous RiemannndashHilbertproblem for elliptic complex equations in the upper half-unit disk
Let D be an upper half-unit disk with the boundary Γprime = Γcup L0 as stated in Section1 We consider the nonlinear uniformly elliptic systems of first order equations
Fj(x y u v ux vx uy vy) = 0 in D j = 1 2
Under certain conditions the system can be transformed into the complex form
wz = F (z w wz) F = Q1wz +Q2wz + A1w + A2w + A3 z isin D (21)
(see [86]1)) in which F (z w U) satisfy the following conditions
Condition C
1) Qj(z w U) Aj(z w) (j = 1 2) A3(z) are measurable in z isin D for all continu-ous functions w(z) in Dlowast = DZ and all measurable functions U(z) isin Lp0(Dlowast) and
2 Elliptic Equations of First Order 91
satisfyLp[Aj D] le k0 j = 1 2 Lp[A3 D] le k1 (22)
where Z = z1 zm Dlowast is any closed subset in D p0 p (2 lt p0 le p) k0 k1 arenon-negative constants
2) The above functions are continuous in w isin CI for almost every point z isin DU isin CI and Qj = 0 (j = 1 2) Aj = 0 (j = 1 2 3) for z isin D
3) The complex equation (21) satisfies the uniform ellipticity condition
|F (z w U1)minus F (z w U2)| le q0|U1 minus U2| (23)
for almost every point z isin D in which w U1 U2 isin CI and q0(lt 1) is a non-negativeconstant
Problem A The discontinuous RiemannndashHilbert boundary value problem for (21)is to find a continuous solution w(z) in Dlowast satisfying the boundary condition
Re [λ(z)w(z)] = c(z) z isin Γlowast = ΓprimeZ (24)
where λ(z) c(z) are as stated in Section 1 satisfying
Cα[λ(z)Γj] le k0 Cα[|z minus zj|βjc(z)Γj] le k2 j = 1 m (25)
herein α (12ltαlt1) k0 k2 are non-negative constants Assume that (βj + γj)β lt1 β = min(α 1 minus 2p0)2 γj βj(j = 1 m) are as stated in (128) (129)Problem A with A3(z) = 0 in D c(z) = 0 on Γlowast is called Problem A0 Theindex K of Problem A and Problem A0 is defined as in (14)
In order to prove the solvability of Problem A for the complex equation (21) weneed to give a representation theorem for Problem A
Theorem 21 Suppose that the complex equation (21) satisfies Condition C andw(z) is a solution of Problem A for (21) Then w(z) is representable by
w(z) = Φ[ζ(z)]eφ(z) + ψ(z) (26)
where ζ(z) is a homeomorphism in D which quasiconformally maps D onto the unitdisk G= |ζ| lt 1 with boundary L = |ζ| = 1 where ζ(minus1) = minus1 ζ(i) = iζ(1) = 1 Φ(ζ) is an analytic function in G ψ(z) φ(z) ζ(z) and its inverse functionz(ζ) satisfy the estimates
Cβ[ψ D] le k3 Cβ[φ D] le k3 Cβ[ζ(z) D] le k3 Cβ[z(ζ) G] le k3 (27)
Lp0 [|ψz|+ |ψz| D] le k3 Lp0 [|φz|+ |φz| D] le k3 (28)
Cβ[z(ζ) G] le k3 Lp0 [|χz|+ |χz| D] le k4 (29)
in which χ(z) is as stated in (214) below β = min(α 1 minus 2p0)2 p0 (2 lt p0 le p)kj = kj(q0 p0 k0 k1 D) (j = 3 4) are non-negative constantsMoreover if the
92 III Elliptic Complex Equations
coefficients Qj(z) = 0 (j = 1 2) of the complex equation (21) in D then the rep-resentation (26) becomes the form
w(z) = Φ(z)eφ(z) + ψ(z) (210)
and if K lt 0 Φ(z) satisfies the estimate
Cδ[X(z)Φ(z) D] le M1 = M1(p0 β δ k D) (211)
in which
X(z)=mprod
j=1j =nm
|zminuszj|ηj |zminuszn|2ηn|zminuszm|2ηm ηj= |γj|+τ γj lt0 βj le|γj|
|βj|+τ other case(212)
Here γj (j = 1 m) are real constants as stated in (128) and δ τ (0 lt δ ltmin(β τ)) are sufficiently small positive constants and M1 is a non-negative con-stant
Proof We substitute the solution w(z) of ProblemA into the coefficients of equation(21) and consider the following system
ψz = Qψz + A1ψ + A2ψ + A3 Q =
Q1 +Q2wzwz for wz = 00 for wz = 0 or z isin D
φz = Qφz + A A =
⎧⎨⎩A1 + A2ww for w(z) = 00 for w(z) = 0 or z isin D
Wz = QWz W (z) = Φ[ζ(z)]
(213)
By using the continuity method and the principle of contracting mappings we canfind the solution
ψ(z) = Tf = minus 1π
int intD
f(ζ)ζ minus z
dσζ
φ(z) = Tg ζ(z) = Ψ[χ(z)] χ(z) = z + Th
(214)
of (213) where f(z) g(z) h(z) isin Lp02(D) 2 lt p0 le p χ(z) is a homeomorphism inD Ψ(χ) is a univalent analytic function which conformally maps E = χ(D) onto theunit disk G(see [85]11)) and Φ(ζ) is an analytic function in G We can verify thatψ(z) φ(z) ζ(z) satisfy the estimates (27) and (28) It remains to prove that z = z(ζ)satisfies the estimate (29) In fact we can find a homeomorphic solution of the lastequation in (213) in the form χ(z) = z + Th such that [χ(z)]z [χ(z)]z isin Lp0(D)[80]1)[85]9) Next we find a univalent analytic function ζ = Ψ(χ) which maps χ(D)onto G hence ζ = ζ(z) = Ψ[χ(z)] By the result on conformal mappings applyingthe method of Lemma 21 Chapter II in [86]1) we can prove that (29) is trueWhen Qj(z) = 0 in D j = 1 2 then we can choose χ(z) = z in (214) in this caseΦ[ζ(z)] can be replaced by the analytic function Φ(z) herein ζ(z)Ψ(z) are as statedin (214) it is clear that the representation (26) becomes the form (210) Thus theanalytic function Φ(z) satisfies the boundary conditions
Re [λ(z)eφ(z)Φ(z)] = c(z)minus Re [λ(z)ψ(z)] z isin Γlowast (215)
On the basis of Theorem 12 and the estimate (27) Φ(z) satisfies the estimate (211)
2 Elliptic Equations of First Order 93
22 Existence of solutions of the discontinuous RiemannndashHilbert problemfor nonlinear complex equations in the upper half-unit disk
Theorem 22 Under the same conditions as in Theorem 21 the following state-ments hold
(1) If the index K ge 0 then Problem A for (21) is solvable and the generalsolution includes 2K + 1 arbitrary real constants
(2) If K lt 0 then Problem A has minus2K minus 1 solvability conditions
Proof Let us introduce a closed convex and bounded subset B1 in the Banachspace B = Lp0(D) times Lp0(D) times Lp0(D) whose elements are systems of functionsq = [Q(z) f(z) g(z)] with norms q = Lp0(Q D) + Lp0(f D) + Lp0(g D) whichsatisfy the condition
|Q(z)| le q0 lt 1 (z isin D) Lp0 [f(z) D] le k3 Lp0 [g(z) D] le k3 (216)
where q0 k3 are non-negative constants as stated in (23) and (27) Moreoverintroduce a closed and bounded subset B2 in B the elements of which are systemsof functions ω = [f(z) g(z) h(z)] satisfying the condition
Lp0 [f(z) D] le k4 Lp0 [g(z) D] le k4 |h(z)| le q0|1 + Πh| (217)
where Πh = minus 1π
intintD[h(ζ)(ζ minus z)2]dσζ
We arbitrarily select q = [Q(z) f(z) g(z)] isin B1 and using the principle of con-tracting mappings a unique solution h(z) isin Lp0(D) of the integral equation
h(z) = Q(z)[1 + Πh] (218)
can be found which satisfies the third inequality in (217) Moreover χ(z) = z+ This a homeomorphism in D Now we find a univalent analytic function ζ = Φ(χ)which maps χ(D) onto the unit disk G as stated in Theorem 21 Moreover we findan analytic function Ψ(ζ) in G satisfying the boundary condition in the form
Re [Λ(ζ)Φ(ζ)] = R(ζ) ζ isin L = ζ(Γ) (219)
in which ζ(z) = Ψ[χ(z)] z(ζ) is its inverse function ψ(z) = Tf φ(z) = Tg Λ(ζ) =λ[z(ζ)] exp[φ(z(ζ))] R(ζ) = r[z(ζ)] minus Re [λ[z(ζ)]ψ(z(ζ))] where Λ(ζ) R(ζ) on Lsatisfy conditions similar to λ(z) c(z) in (25) and the index of Λ(ζ) on L is K In thefollowing we first consider the case ofK ge 0On the basis of Theorem 12 we can findthe analytic function Φ(ζ) in the form (130) here 2K+1 arbitrary real constants canbe chosen Thus the function w(z) = Φ[ζ(z)]eφ(z) + ψ(z) is determined Afterwardswe find out the solution [f lowast(z) glowast(z) hlowast(z) Qlowast(z)] of the system of integral equations
f lowast(z)=F (z wΠf lowast)minusF (z w 0)+A1(z w)Tflowast+A2(z w)Tf lowast+A3(z w) (220)
Wglowast(z)=F (z w WΠglowast+Πf lowast)minusF (z wΠf lowast)+A1(z w)W+A2(z w)W (221)
94 III Elliptic Complex Equations
S prime(χ)hlowast(z)eφ(z) = F [z w S prime(χ)(1 + Πhlowast)eφ(z) +WΠglowast +Πf lowast]
minusF (z wWΠglowast +Πf lowast)(222)
Qlowast(z) =hlowast(z)
[1 + Πhlowast] S prime(χ) = [Φ(Ψ(χ))]χ (223)
and denote by qlowast = E(q) the mapping from q = (Q f g) to qlowast = (Qlowast f lowast glowast)According to Theorem 21 from Chapter IV in [86]1) we can prove that qlowast = E(q)continuously maps B1 onto a compact subset in B1 On the basis of the Schauderfixed-point theorem there exists a system q = (Q f g) isin B1 such that q = E(q)Applying the above method from q = (Q f g) we can construct a functionw(z) = Φ[ζ(z)]eφ(z) + ψ(z) which is just a solution of Problem A for (21) Asfor the case of K lt 0 it can be similarly discussed but we first permit that thefunction Φ(ζ) satisfying the boundary condition (215) has a pole of order |[K]| atζ = 0 and find the solution of the nonlinear complex equation (21) in the formw(z) = Φ[ζ(z)]eφ(z) + ψ(z) From the representation we can derive the minus2K minus 1solvability conditions of Problem A for (21)
Besides we can discuss the solvability of the discontinuous RiemannndashHilbertboundary value problem for the complex equation (21) in the upper half-plane andthe zone domain For some problems in nonlinear mechanics as stated in [90] it canbe solved by the results in Theorem 22
23 The discontinuous RiemannndashHilbert problem for nonlinear complexequations in general domains
In this subsection let Dprime be a general simply connected domain with the boundaryΓprime = Γprime
1 cup Γprime2 herein Γ
prime1Γ
prime2 isin C1
α (0 lt α lt 1) and their intersection points zprime zprimeprime withthe inner angles α1π α2π(0 lt α1 α2 lt 1) respectively We discuss the nonlinearuniformly elliptic complex equation
wz = F (z w wz) F = Q1wz +Q2wz + A1w + A2w + A3 z isin Dprime (224)
in which F (z w U) satisfies Condition C in Dprime There exist m point Z = z1 =zprime znminus1 zn = zprimeprime zm on Γprime arranged according to the positive directionsuccessively Denote by Γprime
j the curve on Γprime from zjminus1 to zj j = 1 2 m and
Γprimej(j = 1 m) does not include the end points
Problem Aprime The discontinuous RiemannndashHilbert boundary value problem for (21)is to find a continuous solution w(z) in Dlowast = DprimeZ satisfying the boundary condition
Re [λ(z)w(z)] = c(z) x isin Γlowast = ΓprimeZ
Im [λ(zprimej)w(zprime
j)] = bj j = 1 m(225)
where zprimej bj(j = 1 m) are similar to those in (136) λ(z) c(z) bj(j = 1 m)
are given functions satisfying
Cα[λ(z)Γj] le k0 Cα[|z minus zj|βjc(z)Γj] le k2 |bj| le k2 j = 1 m (226)
3 Elliptic Equations of Second Order 95
herein α (12 lt α lt 1) k0 k2 are non-negative constants and assume that (βj +γj)β lt 1 β = min(α 1 minus 2p0)α0 βj(j = 1 m) are similar to those in (129)α0 = max(1α1 1α2 1) Problem A with A3(z) = 0 in D r(z) = 0 on Γprime andbj = 0 (j = 1 m) is called Problem Aprime
0 The index K = (m minus 1)2 of Problem Aand Problem A0 is defined as in (14)
In order to give the unique result of solutions of Problem Aprime for equation (224)we need to add one condition For any complex functions wj(z) isin C(Dlowast) Uj(z) isinLp0(Dprime)(j = 1 2 2 lt p0 le p) the following equality holds
F (z w1 U1)minus F (z w1 U2) = Q(U1 minus U2) + A(w1 minus w2) in Dprime (227)
in which |Q(z w1 w2 U1 U2)|le q0 A(z w1 w2 U1 U2)isinLp0(D) Especially if (224)is a linear equation then the condition (227) obviously is true
Applying a similar method as before we can prove the following theorem
Theorem 23 If the complex equation (224) in Dprime satisfies Condition C thenProblem Aprime for (224) is solvable If Condition C and the condition (227) hold thenthe solution of Problem Aprime is unique Moreover the solution w(z) can be expressed as(26)ndash(29) where β = min(α 1 minus 2p0)α0 If Qj(z) = 0 in D j = 1 2 in (224)then the representation (26) becomes the form
w(z) = Φ(z)eφ(z) + ψ(z) (228)
and w(z) satisfies the estimate
Cδ[X(z)w(z) D] le M1 = M1(p0 β δ k D) (229)
in which
X(z) =mprod
j=2j =1n
|z minus zj|ηj |z minus z1|max(1α11)η1 |z minus zn|max(1α21)ηn
ηj = |γj|+ τ if γj lt 0 βj le |γj|
|βj|+ τ if γj ge 0 and γj lt 0 βj lt |γj|
(230)
here γj(j = 1 m) are real constants as stated in (128) δ τ (0 lt δ lt min(β τ))are sufficiently small positive constants and M1 is a non-negative constant
3 Discontinuous Oblique Derivative Problems for Quasilinear Elliptic Equations of Second Order
This section deals with the oblique derivative boundary value problems for quasi-linear elliptic equations of second order We first give the extremum principle andrepresentation of solutions for the above boundary value problem and then obtaina priori estimates of solutions of the above problem finally we prove the uniquenessand existence of solutions of the above problem
96 III Elliptic Complex Equations
31 Formulation of the discontinuous oblique derivative problem forelliptic equations of second order
Let D be the upper half-unit disk as stated in Section 1 and Γprime = Γ cup L0 of D bethe boundary where Γ = |z| = 1 Im z ge 0 and L0 = (minus1 1) We consider thequasilinear uniformly elliptic equation of second order
auxx + 2buxy + cuyy + dux + euy + fu = g in D (31)
where a b c d e f g are given functions of (x y) isin D and u ux uy isin IR Undercertain conditions equation (31) can be reduced to the the complex form
uzz=F (z u uz uzz) F =Re [Quzz+A1uz]+A2u+A3 in D (32)
where Q = Q(z u uz) Aj = Aj(z u uz) and
z = x+ iy uz =12[ux minus iuy] uz =
12[ux + iuy] uzz =
14[uxx + uyy]
Q(z) =minusa+ c minus 2bi
a+ c A1(z) =
minusd minus ei
a+ c A2(z) =
minusf
2(a+ c) A3(z) =
g
2(a+ c)
Suppose that equation (32) satisfies the following conditions
Condition C
1) Q(z u w) Aj(z u w) (j = 1 2 3) are continuous in u isin IR w isin CI for almostevery point z isin D u isin IRw isin CI and Q = 0 Aj = 0 (j = 1 2 3) for z isin D
2) The above functions are measurable in z isin D for all continuous functionsu(z) w(z) on Dlowast = DZ and satisfy
Lp[Aj(z u w) D]lek0 j=1 2 Lp[A3(z u w) D]lek1 A2(z u w)ge0 in D (33)
in which p0 p (2 lt p0 le p) k0 k1 are non-negative constants Z = minus1 13) Equation (32) satisfies the uniform ellipticity condition namely for any number
u isin IRw isin CI the inequality
|Q(z u w)| le q0 lt 1 (34)
for almost every point z isin D holds where q0 is a non-negative constant
The discontinuous oblique derivative boundary value problem for equation (32)may be formulated as follows
Problem P Find a continuously differentiable solution u(z) of (32) in Dlowast = DZwhich is continuous in D and satisfies the boundary conditions
12
partu
partν= Re [λ(z)uz] = r(z) z isin Γlowast = ΓprimeZ u(minus1) = b0 u(1) = b1 (35)
3 Elliptic Equations of Second Order 97
where Z = minus1 1 is the set of discontinuous points of λ(z) on Γlowast ν is a given vectorat every point on Γlowast λ(z) = a(x)+ ib(x) = cos(ν x)minus i cos(ν y) cos(ν n) ge 0 on ΓlowastIf cos(ν n) equiv 0 on Γlowast = ΓprimeZ then the condition u(1) = b1 can be canceled Heren is the outward normal vector at every point on Γlowast δ0(lt 1) is a constant b0 b1 arereal constants and λ(z) r(z) b0 b1 satisfy the conditions
Cα[λ(z)Γj] le k0 Cα[|z minus zj|βjr(z)Γj] le k2 j = 1 2 |b0| |b1| le k2 (36)
Herein α (12 lt α lt 1) k0 k2 are non-negative constants We assume that (βj +γj)β lt 1 β = min(α 1minus 2p0)2 βj(j = 1 m) are as stated in (127) ProblemP with A3(z) = 0 in D r(z) = 0 on Γ b0 = b1 = 0 is called Problem P0 Theindex of Problem P is K where K is defined as in (14) here we choose K = 0 andK = minus12 if cos(ν n) equiv 0 on Γlowast If A2(z) = 0 in D the last point condition in (35)can be replaced by
Im [λ(z)uz]|z=0 = b2 (37)
and we do not need the assumption cos(ν n) ge 0 on Γ where b2 is a real constantsatisfying the condition |b2| le k2 Then the boundary value problem for (32) will becalled Problem Q In the following we only discuss the case of K = 0 and the caseof K = minus12 can be similarly discussed
32 The representation theorem of Problem P for equation (32)
We first introduce a theorem
Theorem 31 Suppose that equation (32) satisfies Condition C Then there existtwo solutions ψ(z)Ψ(z) of the Dirichlet problem (Problem D) of (32) and its relatedhomogeneous equation
uzz minus Re [Q(z u uz)uzz + A1(z u uz)uz]minus A2(z u uz)u = 0 in D (38)
satisfying the boundary conditions
ψ(z) = 0 Ψ(z) = 1 on Γ (39)
respectively and ψ(z)Ψ(z) satisfy the estimates
C1β[ψ(z) D] le M2 C1
β[Ψ(z) D] le M2
Lp0 [ψzz D] le M3 Lp0 [Ψzz D] le M3 Ψ ge M4 gt 0 in D(310)
where β (0 lt β le α) Mj = Mj(q0 p0 β k0 k1 D) (j = 2 3 4) are non-negativeconstants
Proof We first assume that the coefficients Q = Aj = 0 (j = 1 2 3) of (32) inthe ε-neighborhood of z = minus1 1 ie Dε = |z plusmn 1| le ε Im z ge 0 ε gt 0 whereε = 1m (m is a positive integer) Introduce the transformation and its inversion
ζ(z) = minusiz2 + 2iz + 1z2 minus 2iz + 1
z(ζ) =1
ζ + i
[1 + iζ minus
radic2(1minus ζ2)
] (311)
98 III Elliptic Complex Equations
The function ζ(z) maps D onto G = |ζ| lt 1 such that the boundary points minus1 0 1are mapped onto the points minus1 minusi 1 respectively Through the transformation equa-tion (32) is reduced to the equation
uζζ= |zprime(ζ)|2Re [Quζζ(zprime(ζ))2+(A1zprime(ζ)minusQzprimeprime(ζ)(zprime(ζ))3)uζ ]+A2u+A3 (312)
in G It is clear that equation (312) in G satisfies conditions similar to Condition CHence equation (312) and its related homogeneous equation
uζζ= |zprime(ζ)|2Re [Quζζ(zprime(ζ))2+(A1zprime(ζ)minusQzprimeprime(ζ)(zprime(ζ))3)uζ ]+A2u in G (313)
possess the solutions ψ(ζ) Ψ(ζ) satisfying the boundary conditions
ψ(ζ) = 0 Ψ(ζ) = 1 on L = ζ(Γ)
and ψ[ζ(z)]Ψ[ζ(z)] in D are the solutions of Problem D of (32)(38) satisfying theboundary condition (39) respectively and ψ(z)Ψ(z) satisfy the estimate (310) butthe constants Mj = Mj(q0 p0 β k0 k1 D ε) (j = 2 3 4) Now we consider
ψ(z) =
⎧⎨⎩ψ(z) in D
minusψ(z) in D = |z| lt 1 Im z lt 0(314)
It is not difficult to see that ψ(z) in ∆ = |z| lt 1 is a solution of the elliptic equationuzz minus Re [Quzz + A1uz]minus A2u = A3 in ∆ (315)
where the coefficients
Q=
⎧⎨⎩Q(z)
Q(z)A1=
⎧⎨⎩A1(z)
A1(z)A2=
⎧⎨⎩A2(z)
A2(z)A3=
⎧⎨⎩A3(z)
minusA3(z)in
⎧⎨⎩D
D
⎫⎬⎭
where D is the symmetrical domian of D with respect to the real axis It is clear thatthe coefficients in ∆ satisfy conditions similar to those from Condition C Obviouslythe solution ψ(z) satisfies the boundary condition ψ(z) = 0 on part∆ = |z| = 1Denote by ψm(z) the solution of equation (32) with Q = Aj = 0(j = 1 2 3) in theε = 1m-neighborhood of z = minus1 1 we can derive that the function ψm(z) in ∆satisfies estimates similar to ψ(z) in (310) where the constants Mj(j = 2 3) areindependent of ε = 1m Thus we can choose a subsequence of ψm(z) whichuniformly converges to ψlowast(z) and ψlowast(z) is just a solution of Problem D for theoriginal equation (32) in D Noting that the solution Ψ(z) = ψ(z)+ 1 of Problem Dfor equation (38) is equivalent to the solution ψ(z) of Problem D for the equation
uzz minus Re [Quzz + A1uz]minus A2u = A2 in D (316)
with the boundary condition ψ(z) = 0 on Γ by using the same method we can provethat there exists a solution Ψ(z) of Problem D for (38) with the boundary conditionΨ(z) = 1 on Γ and the solution satisfies the estimates in (310)
3 Elliptic Equations of Second Order 99
Theorem 32 Suppose that equation (32) satisfies Condition C and u(z) is asolution of Problem P for (32) Then u(z) can be expressed as
u(z)=U(z)Ψ(z)+ψ(z) U(z)=2Reint z
0w(z)dz+b0 w(z)=Φ[ζ(z)]eφ(z) (317)
where ψ(z)Ψ(z) are as stated in Theorem 31 satisfying the estimate (310) ζ(z)is a homeomorphism in D which quasiconformally maps D onto the unit disk G=|ζ| lt 1 with boundary L where ζ(minus1) = minus1 ζ(1) = 1 ζ(i) = i Φ(ζ) is an analyticfunction in G φ(z) ζ(z) and its inverse function z(ζ) satisfy the estimates
Cβ[φ(z) D] le k3 Cβ[ζ(z) D] le k3
Cβ[z(ζ) G] le k3 Lp0 [|φz|+ |φz| D] le k3
Cβ[z(ζ) G] le k3 Lp0 [|χz|+ |χz| D] le k4
(318)
in which χ(z) is as stated in (214) β = min(α 1 minus 2p0)2 p0(2 lt p0 le p) kj =kj(q0 p0 k0 k1 D)(j = 3 4) are non-negative constants
Proof We substitute the solution u(z) of Problem P into the coefficients of equation(32) It is clear that (32) in this case can be seen as a linear equation Firstly onthe basis of Theorem 31 there exist two solutions ψ(z)Ψ(z) of Problem D of (32)and its homogeneous equation (38) satisfying the estimate (310) Thus the function
U(z) =u(z)minus ψ(z)
Ψ(z)in D (319)
is a solution of the equation
Uzz minus Re [QUzz + AUz] = 0 A = A1 minus 2(lnψ)z + 2Q(lnΨ)z in D (320)
and w(z) = Uz is a solution of the first order equation
wz =12[Qwz +Qwz + Aw + Aw] in D (321)
satisfying the boundary condition
12[partU
partν+ (lnΨ)νU ] = r(z)minus Re [λ(z)ψz] on Γlowast ie
Re [λ(z)Uz + (lnΨ)νU2] = r(z)minus Re [λ(z)ψz] on Γlowast(322)
By the following Lemma 33 we see that (lnΨ)ν gt 0 on Γlowast and similarly to Theorem21 the last formula in (317) can be derived and φ(z) ζ(z) and its inverse functionz(ζ) χ(z) satisfy the estimates (27)ndash(29)
Now we consider the linear homogeneous equation
uzz minus Re [Quzz + A1(z)uz]minus A2(z)u = 0 in D (323)
and give a lemma
100 III Elliptic Complex Equations
Lemma 33 Let the equation (323) in D satisfy Condition C and u(z) be a con-tinuously differentiable solution of (323) in D If M = maxzisinD u(z) ge 0 then thereexists a point z0 isin partD such that u(z0) = M If z0 = x0 isin (minus1 1) and u(z) lt u(z0)in Dz0 then
partu
partl= lim
z(isinl)rarrz0
u(z0)minus u(z)|z minus z0| gt 0 (324)
where z (isin D) approaches z0 along a direction l such that cos(l y) gt 0
Proof From the result in Section 2 Chapter III [86]1) we see that the solutionu(z) in D attains its non-negative maximumM at a point z0 isin partD There is no harmin assuming that z0 is a boundary point of ∆ = |z| lt R because we can choose asubdomain(isin D) with smooth boundary and the boundary point z0 and then makea conformal mapping Thus this requirement can be realized By Theorem 31 wefind a continuously differentiable solution Ψ(z) of (323) in ∆ satisfying the boundarycondition Ψ(z) = 1 z isin part∆ = |z| = R and can derive that 0 lt Ψ(z) le 1 z isin ∆Due to V (z) = u(z)Ψ(z) is a solution of the following equation
LV = Vzz minus Re [A(z)Vz] = 0 A(z) = minus2(lnΨ)z + A1(z) in ∆ (325)
it is clear that V (z) lt V (z0) z isin ∆ and V (z) attains the maximum at the pointz0 Afterwards we find a continuously differentiable solution V (z) of (325) in ∆ =R2 le |z| le R satisfying the boundary condition
V (z) = 0 z isin part∆ V (z) = 1 |z| = R
2
It is easy to see that partV parts = 2Re [izVz] z isin part∆ and
partV
partn= 2Re
zVz
R z isin part∆
partV
partn= minus4Re zVz
R |z| = R
2
where s n are the tangent vector and outward normal vector on the boundary part∆Noting that W (z) = Vz satisfies the equation
Wz minus Re [A(z)W ] = 0 z isin ∆
and the boundary condition Re [izW (z)] = 0 z isin part∆ and the index of iz on theboundary part∆ equals to 0 hence W (z) has no zero point on part∆ thus partV partn =2Re [zW (z)R] lt 0 z isin part∆ The auxiliary function
V (z) = V (z)minus V (z0) + εV (z) z isin ∆
by selecting a sufficiently small positive number ε such that V (z) lt 0 on |z| = R2obviously satisfies V (z) le 0 z isin part∆ Due to LV = 0 z isin ∆ on the basis of themaximum principle we have
V (z) le 0 z isin part∆ ie V (z0)minus V (z) ge minusε[V (z0)minus V (z)] z isin ∆
3 Elliptic Equations of Second Order 101
Thus at the point z = z0 we have
partV
partnge minusε
partV
partngt 0
partu
partn= Ψ
partV
partn+ V
partΨpartn
ge minusεpartV
partn+ V
partΨpartn
gt 0
Moreover noting the condition cos(l n) gt 0 cos(l s) gt 0 partUparts = 0 at the point z0where s is the tangent vector at z0 it follows the inequality
partu
partl= cos(l n)
partu
partn+ cos(l s)
partu
partsgt 0 (326)
Theorem 34 If equation (32) satisfies Condition C and for any uj(z) isinC1(Dlowast) j = 1 2 uzz isin CI the following equality holds
F (z u1 u1z u1zz)minus F (z u2 u2z u2zz) = minusRe [Quzz+A1uz]minusA2u
where Lp[Aj D] lt infin j = 1 2 then the solution u(z) of Problem P is unique
Proof Suppose that there exist two solutions u1(z) u2(z) of Problem P for (32)it can be seen that u(z) = u1(z) minus u2(z) satisfies the homogeneous equation andboundary conditions
uzz = Re [Quzz + A1uz] + A2u in D
12
partu
partν= 0 z isin Γlowast u(minus1) = 0 u(1) = 0
(327)
If the maximum M = maxD u(z) gt 0 it is clear that the maximum point zlowast = minus1and 1 On the basis of Lemma 33 the maximum of u(z) cannot attain on (minus1 1)hence its maximum M attains at a point zlowast isin Γlowast If cos(ν n) gt 0 at zlowast from Lemma33 we get partupartν gt 0 at zlowast this contradicts the boundary condition in (327) ifcos(ν n) = 0 at zlowast denote by Γprime the longest curve of Γ including the point zlowast so thatcos(ν n) = 0 and u(z) = M on Γprime then there exists a point zprime isin ΓΓprime such that atzprime cos(ν n) gt 0 partupartn gt 0 cos(ν s) gt 0 (lt 0) partuparts ge 0 (le 0) hence (326) at zprime
holds it is impossible This shows zlowast isin Γ Hence maxD+ u(z) = 0 By the similarmethod we can prove minD+ u(z) = 0 Therefore u(z) = 0 u1(z) = u2(z) in D
Theorem 35 Suppose that equation (32) satisfies Condition C then the solutionu(z) of Problem P for (32) satisfies the estimates
C1δ [u(z) D]=Cβ[u(z) D]+Cδ[|X(z)|1βuz D]leM5
C1δ [u(z) D] le M6(k1 + k2)
(328)
in which β = min(α 1 minus 2p0) X(z) = |z + 1|2η1|z minus 1|2η2 M5 = M5(p0 β δ k D)M6 = M6(p0 β δ k0 D) are two non-negative constants
Proof We first verify that any solution u(z) of Problem P for (32) satisfies theestimate
S(u) = C[u(z) D] + C[|X(z)|1βuz D] le M7 = M7(p0 α k D) (329)
102 III Elliptic Complex Equations
Otherwise if the above inequality is not true there exist sequences of coefficientsQm Am
j (j = 1 2 3) λm rm bmj (j = 1 2) satisfying the same conditions
of Q Aj(j = 1 2 3) λ r bj(j = 0 1) and Qm Amj (j = 1 2 3) weakly con-
verge in D to Q0 A0j(j = 1 2 3) and λm rm bm
j (j = 0 1) uniformly convergeon Γlowast to λ0 r0 b0
j(j = 0 1) respectively Let um is a solution of Problem P for(32) corresponding to Qm Am
j (j = 1 2 3) λm rm bmj (j = 0 1) but
maxD |um(z)| = Hm rarr infin as m rarr infin There is no harm in assuming that Hm ge 1Let Um = umHm It is clear that Um(z) is a solution of the boundary value problem
Umzz minus Re [QmUm
zz + Am1 Um
z ]minus Am2 Um =
Am3
Hm
partUm
partνm
=rm(z)Hm
z isin Γlowast Um(minus1) = bm0
Hm
Um(1) =bm1
Hm
From the conditions in the theorem we have
Lp[Am2 Um +
Am3
Hm
D] le M7 C[λjΓj] le M7
Cα
[|z minus zj|βj
rm(z)Hm
Γlowast]
le M7 j = 1 m
∣∣∣∣∣ bmj
Hm
∣∣∣∣∣ le M7 j = 0 1
whereM7=M7(q0 p0 α K D) is a non-negative constant According to the methodin the proof of Theorem 23 we denote
wm = Umz Um(z) = 2Re
int z
minus1wm(z)dz +
bm0
Hm
and can obtain that Um(z) satisfies the estimate
Cβ[Um(z) D] + Cδ[|X(z)|1βUmz D] le M8 (330)
in which M8 = M8 (q0 p0 δ α K D) δ (gt 0) are non-negative constants Hencefrom Um(z) and |X(z)|1βUm
z we can choose subsequences Umk(z) and|X(z)|1βUmk
z which uniformly converge to U0(z) and |X(z)|1βU0z in D respect-
ively and U0(z) is a solution of the following boundary value problem
U0zz = Re [Q
0U0zz + A0
1uz] + A02U
0 = 0 in D
partU0
partν= 0 on Γlowast U0(minus1) = 0 U0(1) = 0
By the result as stated before we see that the solution U0(z) = 0 However fromS(Um) = 1 the inequality S(U0) gt 0 can be derived Hence the estimate (329) istrue Moreover by using the method from S(Um) = 1 to (330) we can prove thefirst estimate in (328) The second estimate in (328) can be derived from the firstone
3 Elliptic Equations of Second Order 103
33 Existence of solutions of the discontinuous oblique derivative problemfor elliptic equations in the upper half-unit disk
Theorem 36 If equation (32) satisfies Condition C then Problem P for (32) issolvable
Proof Noting that the index K = 0 we introduce the boundary value problem Pt
for the linear elliptic equation with a parameter t(0 le t le 1)
Lu = uzz minus Re [Quzz + A1(z)uz] = G(z u) G = tA2(z)u+ A(z) (331)
for any A(z) isin Lp0(D) and the boundary condition (35) It is evident that whent = 1 A(z) = A3(z) Problem Pt is just Problem P When t = 0 the equation in(331) is
Lu=uzzminusRe [Quzz+A1uz]=A(z) ie wzminusRe [Qwz+A1w]=A(z) (332)
where w = uz By Theorem 37 below we see that Problem P for the first equation in(332) has a unique solution u0(z) which is just a solution of Problem P for equation(331) with t = 0 Suppose that when t = t0 (0 le t0 lt 1) Problem Pt0 is solvableie Problem Pt for (331) has a unique solution u(z) such that |X(z)|1βuz isin Cδ(D)We can find a neighborhood Tε = |t minus t0| lt ε 0 le t le 1 ε gt 0 of t0 such that forevery t isin Tε Problem Pt is solvable In fact Problem Pt can be written in the form
Luminust0[G(z u)minusG(z 0)]=(tminust0)[G(z u)minusG(z 0)]+A(z) z isin D (333)
and (35) Replacing u(z) in the right-hand side of (333) by a function u0(z) withthe condition |X(z)|1βu0z isin Cδ(D) especially by u0(z) = 0 it is obvious that theboundary value problem (333)(35) then has a unique solution u1(z) satisfying theconditions |X(z)|1βu1z isin Cδ(D) Using successive iteration we obtain a sequence ofsolutions un(z) satisfying the conditions |X(z)|1βunz isin Cδ(D)(n = 1 2 ) and
Lun+1minust0[G(z un+1)minusG(z 0)]=(tminust0)[G(z un)minusG(z 0)]+A(z) z isin D
Re [λ(z)un+1z] = r(z) z isin Γ un+1(minus1) = b0 un+1(1) = b1 n = 1 2
From the above formulas it follows that
L(un+1 minus un)z minus t0[G(z un+1)minus G(z un)]
= (t minus t0)[G(z un)minus G(z unminus1)] z isin D
Re [λ(z)(un+1z minus unz)] = 0 z isin Γ
un+1(minus1)minus un(minus1) = 0 un+1(1)minus un(1) = 0
(334)
Noting that
Lp[(t minus t0)(G(z un)minus G(z unminus1)) D] le |t minus t0|k0C1δ [un minus unminus1 D] (335)
104 III Elliptic Complex Equations
where C1δ [un minus unminus1 D] = Cβ[un minus unminus1 D] + Cδ[|X(z)|1β(unz minus unminus1z) D] and
applying Theorem 35 we get
C1δ [un+1 minus un D] le |t minus t0|M6C
1δ [un minus unminus1 D] (336)
Choosing the constant ε so small that 2εM6 lt 1 it follows that
C1δ [un+1 minus un D] le C1
δ
un minus unminus1 D
2 (337)
and when n m ge N0 + 1 (N0 is a positive integer)
C1δ [un+1minusun D]le2minusN0
infinsumj=02minusjC1
δ [u1minusu0 D]le2minusN0+1C1δ [u1minusu0 D] (338)
Hence un(z) is a Cauchy sequence According to the completeness of the Banachspace C1
δ (D) there exists a function ulowast(z) isin C1δ (D) so that C1
δ [un minus ulowast D] rarr 0 forn rarr infin From (338) we can see that ulowast(z) is a solution of Problem Pt for everyt isin Tε = |t minus t0| le ε Because the constant ε is independent of t0 (0 le t0 lt 1)therefore from the solvability of Problem Pt when t = 0 we can derive the solvabilityof Problem Pt when t = ε 2ε [1ε] ε 1 In particular when t = 1 and A(z) =A3(z) Problem P1 for the linear case of equation (32) is solvable
Next we discuss the quasilinear equation (32) satisfying Condition C but we firstassume that the coefficients Q = 0 Aj(j = 1 2 3) = 0 in Dm = z isin D dist(zΓ) lt1m here m(ge 2) is a positive integer namely consider
uzz = Re [Qmuzz + Am1 uz] + Am
2 u+ Am3 in D (339)
where
Qm =
⎧⎨⎩Q(z u uz)
0Am
j =
⎧⎨⎩Aj(z u uz)
0in
⎧⎨⎩Dm = DDm
Dm
⎫⎬⎭ j = 1 2 3
Now we introduce a bounded closed and convex set BM in the Banach space B =C1
δ (D) any element of which satisfies the inequality
C1δ [u(z) D] le M5 (340)
where M5 is a non-negative constant as stated in (328) We are free to choose anarbitrary function U(z) isin BM and insert it into the coefficients of equation (339)It is clear that the equation can be seen as a linear equation hence there exists aunique solution u(z) of Problem P and by Theorem 35 we see u(z) isin BM Denoteby u(z) = S[U(z)] the mapping from U(z) isin BM to u(z) obviously u(z) = S[U(z)]maps BM onto a compact subset of itself It remains to verify that u(z) = S[U(z)]continuously maps the set BM onto a compact subset In fact we arbitrarily selecta sequence of functions Un(z) such that C1
δ [Un(z) minus U0(z) D] rarr 0 as n rarr infin
3 Elliptic Equations of Second Order 105
Setting un(z) = S[Un(z)] and subtracting u0(z) = S[U0(z)] from un(z) = S[Un(z)]we obtain the equation for un = un(z)minus u0(z)
unzminusRe [Qm(z Un Unz)unzz+Am1 (z Un Unz)unz]minusAm
2 (z Un Unz)un=Cn
Cn = Cn(z Un U0 u0) = Am3 minus Re [Qmu0zz + Am
1 u0z]minus Am2 u0
(341)
in which Qm = Qm(z Un Unz)minusQm(z U0 U0z) Amj = Am
j (z Un Unz)minusAmj (z U0 U0z)
j = 1 2 3 and the solution un(z) satisfies the homogeneous boundary conditions
Re [λ(z)uz] = 0 z isin Γlowast = ΓZ u(minus1) = 0 u(1) = 0 (342)
Noting that the function Cn = 0 in Dm according to the method in the formula(243) Chapter II [86]1) we can prove that
Lp[Cn D] rarr 0 as n rarr infin
On the basis of the second estimate in (328) we obtain
C1δ [un(z)minus u0(z) D] le M6Lp[Cn D] (343)
thus C1δ [un(z) minus u0(z) D] rarr 0 as n rarr infin This shows that u(z) = S[U(z)] in the
set BM is a continuous mapping Hence by the Schauder fixed-point theorem thereexists a function u(z) isin BM such that u(z) = S[u(z)] and the function u(z) is justa solution of Problem P for the quasilinear equation (339)
Finally we cancel the conditions the coefficients Q = 0 Aj (j = 1 2 3) = 0 inDm = z dist(zΓ) lt 1m Denote by um(z) a solution of Problem P for equation(339) By Theorem 35 we see that the solution satisfies the estimate (328) Hencefrom the sequence of solutions um(z) m = 2 3 we can choose a subsequenceumk(z) for convenience denote umk(z) by um(z) again which uniformly con-verges to a function u0(z) in D and u0(z) satisfies the boundary condition (35) ofProblem P At last we need to verify that the function u0(z) is a solution of equation(32) Construct a twice continuously differentiable function gn(z) as follows
gn(z) =
⎧⎨⎩ 1 z isin Dn = DDn
0 z isin D2n0 le gn(z) le 1 in DnD2n (344)
where n(ge 2) is a positive integer It is not difficult to see that the function umn (z) =
gn(z)um(z) is a solution of the following Dirichlet boundary value problem
umnzz minus Re [Qmum
nzz] = Cmn in D (345)
umn (z) = 0 on Γ (346)
where
Cmn =gn[Re (Am
1 umz )+Am
2 um]+um[gnzzminusRe (Qmgnzz)]+2Re [gnzumz minusQmgnzu
mz ] (347)
106 III Elliptic Complex Equations
By using the method from the proof of Theorem 35 we can obtain the estimates ofum
n (z t) = um(z t) in Dn namely
C1β[u
mn Dn] le M9 um
n W 2p0
(Dn)le M10 (348)
where β = min(α 1 minus 2p0) 2 lt p0 le p Mj = Mj(q0 p0 α k0 k1 Mprimen gn Dn) j =
9 10 hereM primen = max1lemltinfin C10[um D2n] Hence from um
n (z) we can choose a sub-sequence unm(z) such that unm(z) unmz(z) uniformly converge to u0(z) u0z(z)and unmzz(z) unmzz(z) weakly converge to u0zz(z) u0zz(z) in Dn respectivelyFor instance we take n = 2 um
2 (z) = um(z) in D2 um2 (z) has a subsequence
um2(z) in D2 the limit function of which is u0(z) in D2 Next we take n = 3from um
3 (z) we can select a subsequence um3(z) in D3 the limit function is u0(z)in D3 Similarly from um
n (z)(n gt 3) we can choose a subsequence umn(z) inDn and the limit of which is u0(z) in Dn Finally from umn(z) in Dn we choosethe diagonal sequence umm(z) (m = 2 3 4 ) such that umm(z) ummz(z)uniformly converge to u0(z) u0z(z) and ummzz(z) ummzz(z) weakly converge tou0zz(z) u0zz(z) in any closed subset of D respectively the limit function u(z) = u0(z)is just a solution of equation (32) in D This completes the proof
Theorem 37 If equation (32) with A2(z) = 0 satisfies Condition C then ProblemQ for (32) has a unique solution
Proof By Theorem 23 we choose Dprime = Dn = m = 2 z1 = minus1 z2 = 1 and K = 0the second linear equation in (332) with A(z) = A3(z) has a unique solution w0(z)and the function
u0(z) = 2Reint z
minus1w0(z)dz + b0 (349)
is a solution of Problem Q for the first linear equation in (332) If u0(1) = bprime = b1then the solution is just a solution of Problem P for the linear equation (32) withA2(z) = 0 Otherwise u0(1) = bprime = b1 we find a solution u1(z) of Problem Q withthe boundary conditions
Re [λ(z)u1z] = 0 on Γ Im [λ(z)u1z]|z=0 = 1 u1(minus1) = 0
On the basis of Theorem 34 it is clear that u1(1) = 0 hence there exists a realconstant d = 0 such that b1 = bprime + du1(1) thus u(z) = u0(z) + du1(z) is just asolution of Problem P for the linear equation (32) with A2(z) = 0 As for thequasilinear equation (32) with A2 = 0 the existence of solutions of Problem Q andProblem P can be proved by the method as stated in the proof of last theorem
34 The discontinuous oblique derivative problem for elliptic equationsin general domains
In this subsection let Dprime be a general simply connected domain whose boundaryΓprime = Γprime
1 cupΓprime2 herein Γ
prime1Γ
prime2 isin C2
α(12 lt α lt 1) have two intersection points zprime zprimeprime with
3 Elliptic Equations of Second Order 107
the inner angles αprimeπ αprimeprimeπ (0 lt αprime αprimeprime lt 1) respectively We discuss the quasilinearuniformly elliptic equation
uzz = F (z u uz uzz) F = Re [Quzz + A1uz] + A1u+ A3 z isin Dprime (350)
in which F (z u uz uzz) satisfy Condition C in Dprime There are m points Z = z1 = zprime
znminus1 zn = zprimeprime zm on Γprime arranged according to the positive direction succes-sively Denote by Γprime
j the curve on Γprime from zjminus1 to zj j = 1 2 m z0 = zm and
Γprimej(j = 1 2 m) does not include the end points
Problem P prime The discontinuous oblique derivative boundary value problem for(350) is to find a continuous solution w(z) in Dlowast = DprimeZ satisfying the boundarycondition
12
partu
partν= Re [λ(z)uz] = c(z) z isin Γlowast = ΓprimeZ u(zj) = bj j = 1 m (351)
where cos(ν n) ge 0 λ(z) c(z) are given functions satisfying
Cα[λ(z)Γj] le k0 Cα[|z minus zj|βjc(z)Γj] le k2 |bj| le k2 j = 1 m (352)
herein α (12 lt α lt 1) k0 k2 are non-negative constants and assume that (βj +γj)β lt 1 β = min(α 1 minus 2p0)α0 βj(j = 1 m) are as stated in (127) α0 =max(1α1 1α2 1) Problem P prime with A3(z) = 0 inD r(z) = 0 on Γprime is called ProblemP prime
0 If cos(ν n) equiv 0 on each of Γj(j = 1 m) we choose the index K = m2 minus 1of Problem P prime which is defined as that in Subsection 23 If A2 = 0 in D the lastpoint conditions in (351) can be replaced by
u(zn) = bn Im [λ(z)uz]|zprimej= bj j = 1 n minus 1 n+ 1 m (353)
Here zprimej(isin Z j = 1 nminus 1 n+1 m) isin Γprime are distinct points and the condition
cos(ν n) ge 0 on Γprime can be canceled This boundary value problem is called ProblemQprime
Applying a similar method as before we can prove the following theorem
Theorem 38 Let equation (350) in Dprime satisfy Condition C similar to before ThenProblem P prime and Problem Qprime for (350) are solvable and the solution u(z) can beexpressed by (317) but where β = min(α 1 minus 2p0)α0 Moreover if Q(z) = 0in D then the solution u(z) of equation (350) possesses the form in (317) wherew(z) = Φ(z)eφ(z) + ψ(z) and u(z) satisfies the estimate
C1δ [u D] = Cδ[u(z) D] + Cδ[X(z)w(z) D] le M11 = M11(p0 β δ k D) (354)
in which X(z) is given as
X(z) =mprod
j=2j =n
|z minus zj|ηj |z minus z1|η1αprime|z minus zn|ηnαprimeprime (355)
108 III Elliptic Complex Equations
where ηj(j = 1 m) are as stated in (230) Besides the solution of Problem P prime
and Problem Qprime for (350) are unique if the following condition holds For any realfunctions uj(z) isin C1(Dlowast) Vj(z) isin Lp0(D)(j = 1 2) the equality
F (z u1 u1z V1)minusF (z u1 u1z V2) = Re [Q(V1minusV2)+A1(u1minusu2)z]+A2(u1minusu2) in Dprime
holds where |Q| le q0 in Dprime A1 A2 isin Lp0(Dprime)
Finally we mention that the above results can be generalized to the case of secondorder nonlinear elliptic equation in the form
uzz = F (z u uz uzz)
F = Re [Q(z u uz uzz) + A1(z u uz)uz] + A2(z u uz)u+ A3(z u uz) in D
satisfying the conditions similar to Condition C which is the complex form of thesecond order nonlinear elliptic equation
Φ(x y u ux uy uxx uxy uyy) = 0 in D
with certain conditions (see [86]1))
4 Boundary Value Problems for Degenerate EllipticEquations of Second Order in a Simply Connected Domain
This section deals with the oblique derivative problem for the degenerate ellipticequation of second order in a simply connected domain We first give a boundednessestimate of solutions of the oblique derivative problem for the equation and then byusing the principle of compactness the existence of solutions for the above obliquederivative problem is proved
41 Formulation of boundary value problems for degenerate ellipticequations
Let D be a simply connected domain with the boundary Γprime = Γ cup L0 where Γ isinC2
α (0 lt α lt 1) in the upper half plane with the end points minus1 1 and L0 = [minus1 1]and the inner angles of D at minus1 1 equal α1π α2π herein 0 lt α1 α2 lt 1 We considerthe elliptic equation of second order
Lu = ymuxx + uyy + a(x y)ux + b(x y)uy + c(x y)u = d(x y) in D (41)
here m is a positive number Its complex form is as follows
uzz minus Re [Q(z)uzz + A1(z)uz] + A2u = A3 in D (42)
4 Degenerate Elliptic Equations 109
where
Q(z) =1minus ym
1 + ym A1(z) = minus a+ bi
1 + ym A2(z) = minus c
2(1 + ym) A3(z) =
d
2(1 + ym)
Suppose that equation (42) satisfies the following conditions
Condition C
The coefficients Aj(z)(j = 1 2) are continuously differentiable in D and satisfy
C1α[Aj(z) D] le k0 j = 1 2 3 A2 = minus c
2(1 + ym)ge minus c
4ge 0 on D (43)
in which α(0 lt α lt 1) k0 are non-negative constants
The oblique derivative boundary value problem is as follows
Problem P In the domain D find a solution u(z) of equation (41) which iscontinuously differentiable in D and satisfies the boundary condition
lu=12
partu
partν+σ(z)u=φ(z) zisinΓ
partu
party=ψ(x) xisinL0
u(minus1)=b0 u(1)=b1
(44)
where ν is any unit vector at every point on Γ cos(ν n)ge 0 σ(z)ge σ0 gt 0 n is theunit outer normal at every point on Γ λ(z) = cos(ν x)minus i cos(ν y) and λ(z) φ(z)ψ(x) are known functions and b0 b1 are known constants satisfying the conditions
C1α[ηΓ] le k0 η = λ σ C1
α[φ L0] le k0 C1α[ψ L0] le k0 |b0| |b1| le k0 (45)
in which α (12 lt α lt 1) k0 σ0 are non-negative constants Problem P with theconditions A3(z) = 0 in D φ(z) = 0 on Γ ψ(z) = 0 on L0 and b0 = b1 = 0 is calledProblem P0 If cos(ν n) = 1 here n is a outward normal vector on Γ then Problem Pis the Neumann boundary value problem (Problem N) and if cos(ν n) gt 0 σ(z) = 0on Γ then Problem P is the regular oblique derivative problem ie third boundaryvalue problem (Problem O) in this case we choose σ(z) gt 0 on Γ If cos(ν n) = 0and σ(z) = 0 on Γ then from (44) we can derive
u(z) = 2Reint z
minus1uzdz + b0 = r(z) on Γ u(1) = b1 = 2Re
int 1
minus1uzdz + b0 (46)
In this case Problem P is called Problem D In the following there is no harm inassuming d(z) = 0 in (41)
42 A priori estimates of solutions for Problem P for (41)
First of all we give a lemma and then give a priori estimate of boundedness ofsolutions of Problem P for (41)
110 III Elliptic Complex Equations
Lemma 41 Suppose that equation (41) or (42) satisfies Condition C and Lu ge0 (or Lu le 0) in D if the solution u(z) isin C2(D) cap C(D) of (41) attains its positivemaximum (or negative minimum ) at a point x0 isin (minus1 1) and maxΓ u(z) lt u(x0) (orminΓ u(z) gt u(x0)) on Γ then
limyrarr0
partu(x0 y)party
lt 0 (or limyrarr0
partu(x0 y)party
gt 0) (47)
if the limit exists
Proof Assume that the first inequality is not true namely
limyrarr0
partu(x0 y)party
= M prime ge 0 (48)
Obviously M prime = 0 Denote M = u(x0) B = maxD |b(z)| and by d the diameter ofD Thus there exists a small positive constant ε lt M such that maxΓ u(z) le M minus εMaking a function
v(z) =εu(z)
(MeBd minus εeBy)
we have
v(z)le ε(M minus ε)MeBdminusεeBd
ltεM
MeBdminusεon Γ v(x)lev(x0)=
εM
MeBdminusεon L0 (49)
Noting that Lu ge 0 the function v(x y) satisfies the inequality
ymvxx + vyy + a(x y)vx + b(x y)vy + c(x y)v ge 0 in D
where b = bminus2εBeBy(MeBd minusεeBy) c(x y) = cminusε(B+ b)BeBy(MeBd minusεeBy) le 0in D According to the above assumption we get
limyrarr0
partv(x0 y)party
=ε2BM
(MeBd minus ε)2gt 0
Hence v(x y) attains its maximum in D but from (49) it is impossible This provesthe first inequality in (47) Similarly we can prove the second inequality in (47)
Now we choose a positive constant η lt 1 and consider the equation
Lηu=(y+η)muxx+uyy+a(x y)ux+b(x y)uy+c(x y)u=d in D (410)
It is easy to see that (410) is a uniformly elliptic equation in D From Theorem36 we can derive that for every one of η = 1n gt 0 (n = 2 3 ) there exists asolution un(z) of Problem D for equation (410) In the following we shall give someestimates of the solution un(z)
Lemma 42 If Condition C holds then any solution un(z) of Problem P for (410)with d = 0 satisfies the estimate
C[un(z) D] le M12 = M12(α k0 D) (411)
4 Degenerate Elliptic Equations 111
where M12 is a non-negative constant
Proof We first discuss Problem D and choose two positive constants c1 c2 suchthat
c1 gec2+maxΓ
|r(z)|+maxD
ec2y c2 gtmaxL0
|ψ(x)|+maxD
|b|+2maxD
|d|+1
and make a transformation of function v(z) = c1 minus ec2y plusmn un(z) thus we have
Lηv le minusc2(c2 + b)ec2y + c(c1 minus ec2y) + 2maxD
|d| lt 0 z = x+ iy isin D
v gt 0 on Γ vy =partv
party= minusc2e
c2y plusmn ψ(z) lt 0 on L0(412)
by the extremum principle for elliptic equations the function v(z) cannot take thenegative minimum in D hence
v(z) = c1 minus ec2y plusmn un(z) ge 0 ie c1 ge ec2y ∓ un(z) in D (413)
hence |un(z)| le c1 minus ec2y le c1 = M12
For other case we introduce an auxiliary function v(z) = c1 minus ec2y plusmn un(z) wherec1 c2 are two positive constants satisfying the conditions
c2 gt maxD
|b(z)|+maxL0
|ψ(x)|+maxD
ec2y + 2maxD
|d|
c1 gt c2 +maxD
ec2y(1 +
c2
σ0
)+max
Γ
|φ(z)|σ0
(414)
We can verify that the function v(z) satisfies the conditions
Lηv lt 0 in D lv gt 0 on Γ vy lt 0 on L0 (415)
hence v(z) cannot attain the negative minimum in D Thus |un(z)| le c1 minus ec2y lec1 = M12 This completes the proof
Secondly from the sequence of solutions un(z) of Problem P for equation (410)we can choose a subsequence unk
(z) which uniformly converges to a solution ulowast(z)of (41) in any closed subset of DL In fact by Condition C and the estimate (411)we can derive the estimate of the solution un(z) as follows
C1β[un(z) D] le M13 = M13(β k0 D η) (416)
where η = 1n gt 0 and β (0 lt β le α) is a constant
Lemma 43 If Condition C holds then any solution un(z) of Problem P for (410)satisfies the estimate (416)
From the above lemma we can derive that the limit function ulowast(z) of unk(z)
satisfies the first boundary condition in (44) In order to prove that ulowast(z) satisfies
112 III Elliptic Complex Equations
the second boundary condition in (44) we write the similar results in [24]1) as alemma
Lemma 44 Suppose that Condition C holds and 0 lt m lt 2 or m ge 2
a(x y) = O(ym2minus1+ε) ay = O(ym2minus2+ε) (417)
where ε is sufficiently small positive number Then any solution um(z) of Problem Pfor (410) with d = 0 satisfies the estimate
|uny| |((y + η)m+ε minus ηm+ε)u2nx| le M14 = M14(α k0 D) in Rn2δ0 (418)
where Rnδ = |x minus x0| lt ρ minus δ 0 lt y lt δl(Rnδ sub D) x0 isin (minus1 1) δ0 δ1 δ ρ(0 lt δ le 2δ0 lt ρ δ1 lt 1n) are small positive constants and M14 is a non-negativeconstant
Proof (1) First of all we prove the estimate
(uy)2 le [(y + η)m+ε1 minus ηm+ε1 ](ux)2 +M15 in Rnδ0 = Dlowast (419)
in which ε1(lt ε) M15 are non-negative constants f = f(x) = X4 = [ρ2 minus (x minusx0)2]4 g = g(x) = X2 = [ρ2 minus (x minus x0)2]2 are functions of x and F = ηm+ε1 minusY m+ε1 G = 1 minus Y ε1 H = minusY ε1 are functions of Y = y + η and introducing anauxiliary function
v(z) = f [F (ux)2 +G(uy)2] + gu2 +H in Dlowast (420)
if v(z) attains a positive maximum value at a point zlowast isin Dlowast then
v(z) gt 0 vx = vy = 0 Lη(v) = Lη(v) + cv le 0 at zlowast (421)
From (420) we get
vx = 2f [Fuxuxx +Guyuxy] + 2guux + f prime[F (ux)2 +G(uy)2] + gprimeu2 = 0
vy = 2f [Fuxuxy +Guyuyy] + 2guuy + f [F prime(ux)2 +Gprime(uy)2] +H prime = 0
Lη(v)=2f [FuxLη(ux)+GuyLη(uy)]+2guLη(u)+2fF [Y m(uxx)2+(uxy)2]
+2fG[Y m(uxy)2+(uyy)2]+2g[Y m(ux)2+(uy)2]+4Y mf prime[Fuxuxx+Guyuxy]
+4f [F primeuxuxy+Gprimeuyuyy]+Y mf primeprime[F (ux)2+G(uy)2]+f [F primeprime(ux)2+Gprimeprime(uy)2]
(422)
+4Y mgprimeuux+Y mgprimeprimeu2+H primeprime+af prime[F (ux)2+G(uy)2]
+agprimeu2+bf [F prime(ux)2+Gprime(uy)2]+bH prime+2cH
in which Y = y + η and from (410) we obtain
Lη(ux) = minus(axux + bxuy + cxu)
Lη(uy) = minus(mY mminus1uxx + ayux + byuy + cyu)
2fFY m(uxx)2 = 2fFY minusm(uyy + aux + buy + cu)2
(423)
4 Degenerate Elliptic Equations 113
and then we have
2fGuyLη(uy) =minusm
Y[2guuy + f [F prime(ux)2 +Gprime(uy)2] +H prime]minus 2mfF
Y(424)
timesuxuxy +2mfG
Yuy(aux + buy + cu)minus 2fGuy(ayux + byuy + cyu)
Substituting (423)(424) into Lη(v) it is not difficult to derive
1f
Lη(v) = 2(FY minusm +G)(Y m(uxy)2 + (uyy)2) + 2(2F prime minus mF
Y)uxuxy
+4[Gprimeuy + FY minusm(aux + buy + cu)]uyy minus 2Fux(axux + bxuy + cxu)
minus2Guy(ayux + byuy + cyu) +2mG
Yuy(aux + buy + cu)
+2FY minusm(aux + buy + cu)2 +[Y m
(f primeprime
fminus 2
f prime2
f 2
)+ a
f prime
f
](425)
times[F (ux)2 +G(uy)2] +(minusm
Y+ b)[F prime(ux)2 +Gprime(uy)2] + F primeprime(ux)2
+Gprimeprime(uy)2+2g
f[Y m(ux)2+(uy)2]+4Y m
(gprime
fminus f primeg
f 2
)uuxminus2mg
fYuuy
+[Y m
(gprimeprime
fminus 2
f primegprime
f 2
)+ a
gprime
f
]u2 +
H primeprime + (minusmY + b)H prime + 2cHf
Moreover by (411)(417)(420) and (425) we obtain
1f
Lη(v) = (2+o(1))[Y m(uxy)2+(uyy)2]+O
Y m+ε1minus1|uxuxy|
+Y mminus2+2ε+ε1|ux|2+(Y m2minus1+ε+ε1|ux|+Y ε1minus1|uy|+Y ε1)|uyy|+Y m2minus2+ε|uxuy|+Y minus1(uy)2+Y m2minus1+ε+ε1|ux|+Y minus1|uy|+Y ε1
+(
Y m
X2 +Y m2minus1+ε
X
)[Y m+ε1(ux)2+(uy)2]
+(1minusε1)(m+ε1)2(1+o(1))Y mminus2+ε1(ux)2
+ε1(m+1minusε1)(1+o(1))Y ε1minus2(uy)2
+2Y m
X2 (ux)2+2(uy)2
X2 +O
(Y m
X3 |ux|+Y minus1
X2 |uy|)
+ε1(m+1minusε1)(1+o(1))
X4 Y ε1minus2
(426)
When 0ltε1 ltmin(ε1) it is easy to see that the right-hand side of (426) is positivewhich contradicts (421) hence v(z) cannot have a positive maximum in Dlowast
On the basis of the estimate (411) we see that v(z) on the upper boundary |xminusx0|ltρy=1n of Dlowast is bounded and v(x)lefGψ2+gM2
12 on the lower boundary|xminusx0|ltρy=0 of Dlowast moreover v(x)lt0 on the left-hand side and right-hand side|xminusx0|=ρ0ltylt1n of Dlowast Thus the estimate (419) is derived
114 III Elliptic Complex Equations
(2) Now we give the estimate
|((y+η)m+εminusηm+ε)u2nx|leM16 isin Rnδ0=Dlowast (427)
in which M16 is independent of η In fact we introduce the auxiliary function(420) where we choose that f=f(x)=X4=[(ρminusδ0)2minus(xminusx0)2]4g=g(x)=X2=[(ρminusδ0)2minus(xminusx0)2]2 and F =Y m+ε1 minusηm+ε1 G=Y ε2 H=Y ε3 herein Y =y+ηε2ε3
are positive constants satisfying 0lt2ε3 ltε2 leε12 If v(z) attains a positive maxi-mum value at a point zlowast isinDlowast then we have (421) Substituting (410)(420)(423)into (422) we get
Lη(v)f
= 2F [Y m(uxx)2+(uxy)2]+2G[Y m(uxy)2
+(uyy)2]+2g
f[Y m(ux)2+(uy)2]+Σ
(428)
in which
Σ = 4F primeuxuxy+2(2Gprime+
mG
Y
)uyuyy minus2Fux(axux+bxuy+cxu)
minus2Guy(ayux+byuy+cyu)+2mG
Yuy(aux+buy+cu)
+[Y m
(f primeprime
fminus2f
prime2
f 2
)+a
f prime
f
][F (ux)2+G(uy)2]+b[F prime(ux)2
+Gprime(uy)2]+F primeprime(ux)2+Gprimeprime(uy)2+4Y m
(gprime
fminus f primeg
f 2
)uux
+[Y m
(gprimeprime
fminus2f
primegprime
f 2
)+a
gprime
f
]u2+
H primeprime+bH prime+2cHf
(429)
From (411)(417)(420) and (429) it follows that
Σ = O
Y m+ε1minus1|uxuxy|+Y ε2minus1|uyuyy|+Y m+ε1minus2|ux|2+Y m2minus2+ε+ε2
times|uxuy|+Y ε2minus2|uy|2+Y m+ε1|ux|+Y ε2minus1|uy|+(
Y m
X2 +Y m2minus1+ε
X
)
times(Y m+ε1|ux|2+Y ε2|uy|2)+Y m
X3 |ux|+Y ε3minus2
X4
geminusG(Y m|uxy|2+|uyy|2)
minus2 g
f(Y m|ux|2+|uy|2)+O
(Y m+ε1minus2|ux|2+Y ε2minus2|uy|2+ Y ε3minus2
X4
)
(430)
By (428)(430) if we can verify the following inequality
G[Y m(uxy)2+(uyy)2]+O
(Y m+ε1minus2|ux|2+Y ε2minus2|uy|2+ Y ε3minus2
X4
)gt0 (431)
4 Degenerate Elliptic Equations 115
then the inequality Lη(v)gt0 Noting that F primeGprimeH prime are positive and from (411)(420) (422) we have
2|Fuxuxy+Guyuyy|geF prime|ux|2+Gprime|uy|2minus2 g
f|uuy|+H prime
f
geF prime|ux|2+ 1f(H primeminus u2
Gprime2 )geF prime|ux|2+1+o(1)f
H prime
Hence(F 2|ux|2+Y mG2|uy|2)(Y m|uxy|2+|uyy|2)=Y m(Fuxuxy+Guyuyy)2+(Y mGuyuxy minusFuxuyy)2
ge Y m
4(F prime|ux|2+1+o(1)
fH prime)2
(432)
By (419)(420)(432) we obtain
(F 2|ux|2+Y mG2|uy|2)[Y m|uxy|2+|uyy|2
+1G
O
(Y m+ε1minus2|ux|2+Y ε2minus2|uy|2+ Y ε3minus2
X4
)]
ge Y m
4(mY m+ε1minus1|ux|2+ ε3+o(1)
X4 Y ε3minus1)2
+Y m+ε2(Y m+ε1+ε2|ux|2+Y ε2M15)Y minusε2O
(Y m+ε1minus2|ux|2+ Y ε2minus2
X4
)gt0
(433)
From (422) we see that uxuy cannot simultaneously be zero By (433) we have(431) such that Lη(v)gt0 holds This contradicts (421) Therefore v(z) cannotattain a positive maximum in Dlowast
On the basis of (411)(420) and the boundary condition (44) we see that vlefGψ2+gM2
12+H on the lower boundary of Dlowast is uniformly bounded Moreover thefunction v(z) is uniformly bounded on the upper left-hand and right-hand boundariesof Dlowast Thus the estimates in (418) are derived
Now we prove a theorem as follows
Theorem 45 Suppose that Condition C(417) hold and ax+cle0 in D Then anysolution un(z) of Problem P for (410) satisfies the estimate
|ux|leM17=M17(αk0D) in D (434)
where we assume that the inner angles αjπ(j=12) of D at z=minus11 satisfy the con-ditions 0ltαj(=12)lt1j=12 and M17 is a non-negative constant
116 III Elliptic Complex Equations
Proof We find the derivative with respect to x to equation (410) and obtain
(y+η)muxxx+uxyy+a(xy)uxx+b(xy)uxy+[ax+c(xy)]ux=F (xy)
F =f(xy)minusbxuy minuscxu in D(435)
On the basis of Lemmas 41 and 44 we have
|F (xy)|= |f(xy)minusbxuy minuscxu|leM18 ltinfin in D
and equation (435) can be seen as a elliptic equation of ux and the solution ux
satisfies the boundary conditions
partu
parts=cos(sx)ux+cos(sy)uy=
partr(z)parts
on Γ (ux)y=ψprime(x) on L0 (436)
in which s is the tangent vector at every point Γ Noting that the angles αjπ(j=12)satisfy the conditions 0ltαj (=12)lt1j=12 it is easy to see that cos(sx) =0 atz=minus11 Thus the first boundary condition in (46) can be rewritten in the form
ux=R(z)=minuscos(sy)cos(sx)
uy+1
cos(sx)partr(z)parts
on Γ (437)
here R(z) is a bounded function in the neighborhood (subΓ) of z=minus11 hence by themethod in the proof of Lemma 42 we can prove that the estimate (434) holds Asfor cos(sx)=0 at z=minus1 or z=1 the problem remains to be solved
Theorem 46 Suppose that Condition C and (417) hold Then Problem P for(41) or (42) has a unique solution
Proof As stated before for a sequence of positive numbers η=1nn=23 wehave a sequence of solutions un(z) of the corresponding equations (410) withη=1n(n=23) which satisfy the estimate (416) hence from un(z) we canchoose a subsequence unk
(z) which converges to a solution u0(z) of (42) in DcupΓsatisfying the first boundary condition in (44) It remains to prove that u0(z) satisfiesthe other boundary condition in (44) For convenience we denote unk
(z) by u(z)x0 isany point in minus1ltx0 lt1 and give a small positive number β there exists a sufficientlysmall positive number δ such that |ψ(x)minusψ(x0)|ltβ when |xminusx0|ltδ Moreover weconsider an auxiliary function
v(z)=F (ux)2plusmnuy+G+f G=minusCyε2 minusτ ∓ψ(x0) f=minusC(xminusx0)2
F =
⎧⎨⎩Y m+1+ε2 minusηm+1+ε2 0ltmlt1
Y m+ε1 minus(m+ε1)ηmminus1+ε1Y +(m+ε1minus1)ηm+ε1 mge1(438)
where Y =y+ηη=1nε2(0ltε2 leε13) are positive constants and C is an undeter-mined positive constants We first prove that v(z) cannot attain its positive max-imum in Dlowast=|xminusx0|2+y2 ltσ2ygt0 Otherwise there exists a point zlowast such that
4 Degenerate Elliptic Equations 117
v(zlowast)=maxDlowast v(z)gt0 and then
vx=2Fuxuxxplusmnuxy+f prime=0 vy=2Fuxuxy plusmnuyy+F prime(ux)2+Gprime=0
Lη(v)=2FuxLη(ux)plusmnLη(uy)+2F [Y m(uxx)2+(uxy)2]+4F primeuxuxy
+[F primeprime+bF prime+cF ](ux)2plusmncuy+Y mf primeprime
+af prime+2cf+Gprimeprime+bGprime+2cG
(439)
and from (423) and (439) we obtain
plusmnLη(uy) = minusm
Y[2Fuxuxy+F prime(ux)2+Gprime]
plusmnm
Y(aux+buy+cu)∓(ayux+byuy+cyu)
(440)
Moreover by (423) and (439)ndash(440) we have
Lη(v) = minus2Fux(axux+bxuy+cxu)minus m
Y[2Fuxuxy+F prime(ux)2+Gprime
∓(aux+buy+cu)]∓(ayux+byuy+cyu)
+2F [Y m(uxx)2+(uxy)2]∓4F primeux(2Fuxuxx+f prime)
+[F primeprime+bF prime+cF ](ux)2plusmncuy+Y mf primeprime+af prime+2cf+Gprimeprime+bGprime+2cG
= 2FY m[uxx∓2F primeY minusm(ux)2]2+2F(uxy minus m
2Yux
)2
+[F primeprimeminus m
YF prime+bF prime+cF minus8F prime2Y minusmF (ux)2](ux)2
+[minus2F (bxuy+cxu)plusmn ma
Y∓ay ∓4F primef prime
]ux+Gprimeprimeminus m
YGprime+bGprime
+2cGminus(
m2
2Y 2+2ax
)F (ux)2plusmn m
Y(buy+cu)∓(byuy+cyu)
plusmncuy+Y mf primeprime+af prime+2cf
(441)
Choosing a sufficiently small positive number σ such that the domain Dlowast=|xminusx0|2+y2 ltσ2capygt0subDσltmin[ρminus2δ0δ1] where δ0δ1 are constants as stated inLemma 44 and |ψ(z)minusψ(x0)|ltτ we can obtain
Lη(v) ge ε2(m+1+ε2)(1+o(1))Y m+ε2minus1(ux)2+O( 1
Y
)|ux|
+Cε2(m+1minusε2)(1+o(1))yε2minus2 gt0 if 0ltmlt1(442)
and
Lη(v) ge (m+ε1)(m+ε1minus1)Y m+ε1minus2(ux)2+O(Y m2+εminus2)|ux|+Cε2(m+1minusε2)(1+o(1))Y ε2minus2 gt0 if mge1 ε=ε1
(443)
in which we use Lemmas 42 and 44 the conditions (417) (438) (441) and
F (ux)2=O(Y 2ε2) F prime(ux)2=O(Y minus1+2ε2) if mge1
118 III Elliptic Complex Equations
It is clear that (442)(443) contradict (421) hence v(z) cannot attain a positivemaximum in Dlowast From (438) we get
v(z)=F (ux)2plusmn [uy minusψ(x0)]minusτ minusC[(xminusx0)2+yε2 ] (444)
Moreover it is easy to see that v(z)lt0 on the boundary of Dlowast provided that theconstant C is large enough Therefore v(z)le0 in Dlowast From F ge0 and (442)ndash(444)the inequality
plusmn[uyminusψ(x0)]minusτminusC[(xminusx0)2+Y ε2 ]le0 ie|uy minusψ(x0)|leτ+C[(xminusx0)2+Y ε2 ] in Dlowast
(445)
is derived Firstly let ηrarr0 and then let zrarrx0 τ rarr0 we obtain limpartupartyrarrψ(x0)Similarly we can verify limz(isinDlowast)rarrx0 uy=ψ(x0) when x0=minus11 Besides we can alsoprove that u(z)rarru(x0) as z(isinDlowast)rarrx0 when x0=minus11 This shows that the limitfunction u(z) of un(z) is a solution of Problem P for (41)
Now we prove the uniqueness of solutions of Problem P for (41) it suffices toverify that Problem P0 has no non-trivial solution Let u(z) be a solution of ProblemP0 for (41) with d=0 and u(z) equiv0 in D Similarly to the proof of Theorem 34we see that its maximum and minimum cannot attain in DcupΓ Moreover by usingLemma 41 we can prove that the maximum and minimum cannot attain at a pointin (minus11) Hence u(z)equiv0 in D
Finally we mention that for the degenerate elliptic equation
K(y)uxx+uyy=0 K(0)=0 K prime(y)gt0 in D (446)
which is similar to equation (41) satisfying Condition C and other conditions asbefore hence any solution u(z) of Problem P0 for (446) satisfies the estimates(411)(418) and (434) in D provided that the inner angles αjπ(j=12) of D atz=minus11 satisfy the conditions 0ltαj (=12)lt1j=12 Equation (446) is the Chap-lygin equation in elliptic domain Besides oblique derivative problems for the degen-erate elliptic equations of second order
uxx+ymuyy+a(xy)ux+b(xy)uy+c(xy)u=f(xy) in D
ym1uxx+ym2uyy+a(xy)ux+b(xy)uy+c(xy)u=f(xy) in D
needs to be considered where mm1m2 are non-negative constants
The references for this chapter are [3][6][11][15][18][23][24][30][33][38][39][40][46][48][50][53][58][60][65][67][76][78][80][81][82] [85][86][94][96][99]
CHAPTER IV
FIRST ORDER COMPLEX EQUATIONS OFMIXED TYPE
In this chapter we introduce the RiemannndashHilbert boundary value problem for firstorder complex equation of mixed (elliptic-hyperbolic) type in a simply connecteddomain We first prove uniqueness and existence of solutions for the above boundaryvalue problem and then give a priori estimates of solutions for the problem finallydiscuss the solvability of the above problem in general domains The results in thischapter will be used in the following chapters
1 The RiemannndashHilbert Problem for Simplest First OrderComplex Equation of Mixed Type
In this section we discuss the RiemannndashHilbert boundary value problem for the sim-plest mixed complex equation of first order in a simply connected domain Firstly weverify a unique theorem of solutions for the above boundary value problem Moreoverthe existence of solutions for the above problem is proved
11 Formulation of the RiemannndashHilbert problem for the simplestcomplex equation of mixed type
Let D be a simply connected bounded domain in the complex plane CI with theboundary partD = Γ cup L where Γ(sub y gt 0) isin Cα(0 lt α lt 1) with the end pointsz =0 2 and L=L1cupL2 L1=x=minusy 0 le x le 1
L2 = x = y + 2 1 le x le 2 Denote D+ =D cap y gt 0 Dminus = D cap y lt 0 and z1 = 1 minus iWithout loss of generality we may assume that Γ=|z minus 1|=1 yge0 otherwise through a conformalmapping this requirement can be realizedWe discuss the mixed system of first order equa-
tions
uxminusvy=0 vx+sgny uy=0 in D (11)
120 IV First Order Mixed Complex Equations
Its complex form is the following complex equation of first orderwz
wzlowast
= 0 in
D+
Dminus
(12)
where
w = u+ iv z = x+ iy wz =12[wx + iwy] wzlowast =
12[wx minus iwy]
The RiemannndashHilbert boundary value problem for the complex equation (12) maybe formulated as follows
Problem A Find a continuous solution w(z) of (12) in Dlowast = D(0 2cupxplusmn y =2 Im z le 0) or Dlowast = D(0 2 cup x plusmn y = 0 Im z le 0) satisfying the boundaryconditions
Re [λ(z)w(z)] = r(z) z isin Γ (13)
Re [λ(z)w(z)] = r(z) z isin Lj(j = 1 or 2) Im [λ(z1)w(z1)] = b1 (14)
where λ(z) = a(z) + ib(z) |λ(z)| = 1 z isin Γ cup Lj(j = 1 or 2) b1 is a real constantand λ(z) r(z) b1 satisfy the conditions
Cα[λ(z)Γ] le k0 Cα[r(z)Γ] le k2 |b1| le k2
Cα[λ(z) Lj] le k0 Cα[r(z) Lj] le k2 j = 1 or 2
a(z)minus b(z) = 0 on L1 or a(z) + b(z) = 0 on L2
(15)
in which α(0 lt α lt 1) k0 k2 are non-negative constants For convenience we mayassume that w(z1) = 0 otherwise through a transformation of the function w(z) minusλ(z1)[r(z1) + ib1] the requirement can be realized
This RiemannndashHilbert problem (Problem A) for (12) with r(z) = 0 z isin Γ cupL1 (or L2) and b1 = 0 will be called Problem A0 The number
K =12(K1 +K2) (16)
is called the index of Problem A and Problem A0 where
Kj =[φj
π
]+ Jj Jj = 0 or 1 eiφj =
λ(tj minus 0)λ(tj + 0)
γj =φj
πminus Kj j = 1 2 (17)
in which t1 = 2 t2 = 0 λ(t) = (1 minus i)radic2 on L0 = [0 2] or λ(t) = (1 + i)
radic2 on
L0 = [0 2] and λ(t1 minus0) = λ(t2+0) = exp(7πi4) or exp(πi4) Here we only discussthe case of K = (K1+K2)2 = minus12 on the boundary partD+ of D+ In order to ensurethat the solution w(z) of Problem A is continuous in the neighborhood(sub Dminus) of thepoint z = 0 or z = 2 we need to choose γ1 gt 0 or γ2 gt 0 respectively
1 Simplest Mixed Complex Equation 121
12 Uniqueness of solutions of the RiemannndashHilbert problem for thesimplest complex equation of mixed type
Theorem 11 Problem A for (12) has at most one solution
Proof Let w1(z) w2(z) be any two solutions of Problem A for (12) It is clear thatw(z) = w1(z)minus w2(z) is a solution of Problem A0 for (12) with boundary conditions
Re [λ(z)w(z)] = 0 z isin Γ (18)
Re [λ(z)w(z)] = 0 z isin Lj(j = 1 or 2) Im [λ(z1)w(z1)] = 0 (19)
Due to the complex equation (12) in Dminus can be reduced to the form
ξν = 0 ηmicro = 0 in Dminus (110)
where micro = x + y ν = x minus y ξ = u + v η = u minus v the general solution of system(110) can be expressed as
ξ = u+ v = f(micro) = f(x+ y) η = u minus v = g(ν) = g(x minus y) ie
u(z) =f(x+ y) + g(x minus y)
2 v(z) =
f(x+ y)minus g(x minus y)2
(111)
in which f(t) g(t) are two arbitrary real continuously differentiable functions on [0 2]Noting the boundary condition (19) we have
au+bv = 0 on L1 or L2 [av minus bu]|z=z1=0 ie
[a((1minusi)x)+b((1minusi)x)]f(0)
+[a((1minusi)x)minusb((1minusi)x)]g(2x)=0 on [0 1] or
[a((1+i)xminus2i)+b((1+i)xminus2i)]f(2xminus2)+[a((1+i)xminus2i)minusb((1+i)xminus2i)]g(2)=0 on [1 2]
w(z1)=0 [u+v]|z1=f(0)=0 or [uminusv]|z1=g(2)=0
(112)
The second formula in (112) can be rewritten as
[a((1minusi)t2)+b((1minusi)t2)]f(0) +[a((1minusi)t2)minus b((1minus i)t2)]g(t) = 0
f(0) = g(t) = 0 or [a((1+i)t2+1minusi)+b((1minusi)t2+1minusi)]f(t)
+[a((1minusi)t2+1minusi)minusb((1minusi)t2+1minusi)]g(2)=0
g(t) = f(t) = 0 t isin [0 2]
(113)
Thus the solution (111) becomes
u(z) = v(z) =12f(x+ y) g(x minus y) = 0 or
u(z) = minusv(z) =12g(x minus y) f(x+ y) = 0
(114)
122 IV First Order Mixed Complex Equations
if a(z)minus b(z) = 0 on L1 or a(z) + b(z) = 0 on L2 respectively In particular we have
u(x) = v(x) =12f(x) x isin [0 2] or
u(x) = minusv(x) =12g(x) x isin [0 2]
(115)
Next due to f(0) = 0 or g(2) = 0 from (115) we can derive that
u(x)minus v(x) = 0 ie Re [(1 + i)w(x)] = 0 x isin [0 2] oru(x) + v(x) = 0 ie Re [(1minus i)w(x)] = 0 x isin [0 2]
(116)
Noting the index K = minus12 of Problem A for (12) in D+ and according to the resultin Section 1 Chapter III and [85]11)[86]1) we know that w(z) = 0 in D+ Thus
u(z) + v(z) = Re [(1minus i)w(z)] = f(x+ y) = 0 g(x minus y) = 0 or
u(z)minus v(z) = Re [(1 + i)w(z)] = g(x minus y) = 0 f(x+ y) = 0(117)
obviously
w(z) = u(z) + iv(z) = w1(z)minus w2(z) = 0 on Dminus (118)
This proves the uniqueness of solutions of Problem A for (12)
13 Existence of solutions of the RiemannndashHilbert problem for thesimplest complex equation of mixed type
Now we prove the existence of solutions of the RiemannndashHilbert problem (ProblemA) for (12)
Theorem 12 Problem A for (12) has a solution
Proof As stated before the general solution of (12) in Dminus can be expressed as
u(z) =f(x+ y) + g(x minus y)
2 v(z) =
f(x+ y)minus g(x minus y)2
ie w(z) =(1 + i)f(x+ y) + (1minus i)g(x minus y)
2
(119)
in which f(t) g(t) are two arbitrary real continuously differentiable functions on [0 2]Taking into account the boundary condition (14) we have
1 Simplest Mixed Complex Equation 123
au+ bv = r(x) on L1 or L2 ie
[a((1minusi)x)+b((1minusi)x)]f(0)
+ [a((1minusi)x)minusb((1minusi)x)]g(2x)
=2r((1minusi)x) on [0 1] f(0)
= [a(z1) + b(z1)]r(z1) + [a(z1)minus b(z1)]b1 or
[a((1+i)xminus2i)+b((1+i)xminus2i)]f(2xminus2)+[a((1+i)xminus2i)minusb((1+i)xminus2i)]g(2)
=2r((1+i)xminus2i) on [1 2] g(2)= [a(z1)minus b(z1)]r(z1)minus [a(z1) + b(z1)]b1
(120)
The second and third formulas in (120) can be rewritten as
[a((1minus i)t2)minus b((1minus i)t2)]g(t)
= 2r((1minus i)t2)minus [a((1minus i)t2) + b((1minus i)t2)]f(0) t isin [0 2] or[a((1+i)t2+1minusi)+b((1+i)t2+1minusi)]f(t)
=2r((1+i)t2+1minusi)
minus[a((1+i)t2+1minusi)minusb((1+i)t2+1minusi)]g(2) tisin [0 2]
(121)
thus the solution (119) possesses the form
u(z) =12f(x+ y) + g(x minus y) v(z) =
12f(x+ y)minus g(x minus y)
g(xminusy)=2r((1minusi)(xminusy)2)minus[a((1minusi)(xminusy)2)+b((1minusi)(xminusy)2)]f(0)
a((1minusi)(xminusy)2)minusb((1minusi)(xminusy)2)
u(z)=12g(xminusy)+f(x+ y) v(z)=
12minusg(xminusy)+f(x+ y) (122)
f(x+ y) =2r((1 + i)(x+ y)2 + 1minus i)
a((1 + i)(x+ y)2 + 1minus i) + b((1 + i)(x+ y)2 + 1minus i)
minus [a((1 + i)(x+ y)2 + 1minus i)minus b((1 + i)(x+ y)2 + 1minus i)]g(2)a((1 + i)(x+ y)2 + 1minus i) + b((1 + i)(x+ y)2 + 1minus i)
if a(z)minus b(z) = 0 on L1 or a(z) + b(z) = 0 on L2 respectively In particular we get
u(x) =12f(x) + g(x) v(x) =
12f(x)minus g(x)
g(x) =2r((1minus i)x2)minus [a((1minus i)x2) + b((1minus i)x2)]f(0)
a((1minus i)x2)minus b((1minus i)x2)
u(x) =12g(x) + f(x) v(x) =
12minusg(x) + f(x)
124 IV First Order Mixed Complex Equations
f(x)=2r((1+i)x2+1minusi)
a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)
minus [a((1+i)x2+1minusi)minusb((1+i)x2+1minusi)]g(2)a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)
x isin [0 2](123)
From the above formulas it follows that
u(x)minusv(x)=2r((1minusi)x2)minus[a((1minusi)x2)+b((1minusi)x2)]f(0)
a((1minusi)x2)minusb((1minusi)x2) or
u(x)+v(x)=2r((1+i)x2+1minusi)
a((1+i)x2+1minusi)+b((1+i)x2+1minusi)
minus [a((1+i)x2+1minusi)minusb((1+i)x2+1minusi)]g(2)a((1+i)x2+1minusi)+b((1+i)x2+1minusi)
xisin [02]
f(0)=[a(1minusi)+b(1minusi)]r(1minusi)+[a(1minusi)minusb(1minusi)]b1 or
g(2)=[a(1minusi)minusb(1minusi)]r(1minusi)minus [a(1minusi)+b(1minusi)]b1
(124)
ieRe [(1+i)w(x)]=s(x)
s(x)=2r((1minusi)x2)minus[a((1minusi)x2)+b((1minusi)x2)]f(0)
a((1minus i)x2)minus b((1minus i)x2)
xisin [0 2] or
Re [(1minusi)w(x)]=s(x)=2r((1+i)x2+1minusi)
a((1+i)x2+1minus i)minusb((1+i)x2+1minus i)
minus [a((1+i)x2+1minusi)minusb((1+i)x2+1minusi)]g(2)a((1+i)x2+1minus i)minusb((1+i)x2+1minus i)
xisin [0 2]
(125)
if a((1minus i)x)minus b((1minus i)x) = 0 on [01] or a((1 + i)x minus 2i) + b((1 + i)x minus 2i) = 0 on[12] respectively We introduce a conformal mapping ζ = ζ(z) from the domain D+
onto the upper half-plane G = Im ζ gt 0 such that the three points z = 0 1 2 mapto ζ = minus1 0 1 respectively it is not difficult to derive that the conformal mappingand its inverse mapping can be expressed by the elementary functions namely
ζ(z) =5(z minus 1)
(z minus 1)2 + 4 z(ζ) = 1 +
52ζ(1minus
radic1minus 16ζ225)
Denoting W (ζ) = w[z(ζ)] and
Λ(ζ) =
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩
λ[z(ζ)] ζ isin Γ1 = ζ(Γ)
1minus iradic2
ζ isin Γ2 = minus1 le Re ζ le 1 Im ζ = 0 or
1 + iradic2
ζ isin Γ2 = minus1 le Re ζ le 1 Im ζ = 0
(126)
1 Simplest Mixed Complex Equation 125
in which the points ζ1 = 1 ζ2 = minus1 are the discontinuous points of Λ(ζ) on partG =Im ζ = 0 from (16)(17) it can be seen that the index of Λ(ζ) on partG = Im ζ =0 is K = minus12 Hence according to the result of Theorem 11 Chapter III weknow that the discontinuous RiemannndashHilbert boundary value problem for analyticfunctions W (ζ) in G with the boundary condition
Re [Λ(z)W (ζ)] = R(ζ) =
⎧⎨⎩ r[z(ζ)] ζ isin Γ1
s[z(ζ)]radic2 ζ isin Γ2
(127)
has a unique solution W (ζ) in G as follows
W (ζ) =X(ζ)πi
[int infin
minusinfinΛ(t)R(t)(t minus ζ)X(t)
dt+ iclowast2 + ζ
2minus ζ
]in G (128)
and
X(ζ) = iζ minus 2ζ minus i
Π(ζ)eiS(ζ) Π(ζ) =(
ζ minus 1ζ + i
)γ1(
ζ + 1ζ + i
)γ2
clowast =2i+ 12 + i
int infin
minusinfinΛ(t)R(t)
X(t)(t minus i)dt
and S(ζ) is an analytic function in Im ζ gt 0 with the boundary condition
Re [S(t)] = arg[Λ1(t)(
t minus 2t+ i
)] on Im t = 0 Im [S(i)] = 0 (129)
where γj(j = 1 2) are as stated in (17) and Λ1(t) = λ(t)Π(t)(t minus 2)|x + i|[|Π(t)|times|t minus 2|(x + i)] and the boundedness of w(z) or boundedness of integral of thesolution w(z) in the neighborhood sub D0 2 of t1 = 2 and t2 = 0 is determined byJj = 0 γj gt 0 or Jj = 0 γj = 0 and Jj = 1(j = 1 2) respectively Hence Problem Afor (12) has a solution w(z) in the form
w(z)=
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
W [ζ(z)] z isin D+0 2
12
(1 + i)f(x+ y) + (1minus i)
times2r((1minusi)(xminusy)2)minus[a((1minusi)(xminusy)2)+b ((1minusi)(xminusy)2)]f(0)a((1minus i)(x minus y)2)minus b ((1minus i)(x minus y)2)
or
12
(1minusi)g(xminusy)+
2(1+i)r((1+i)(x+y)2+1minusi)a((1+i)(x+y)2+1minusi)+b ((1+i)(x+y)2+1minusi)
minus [a((1 + i)(x+ y)2 + 1minus i) + b ((1 + i)(x+ y)2 + 1minus i)]g(2)a((1+i)(x+y)2+1minusi)+b ((1+i)(x+y)2+1minusi)
z = x+ iy isin Dminus0 2(130)
126 IV First Order Mixed Complex Equations
in which f(0) g(2) are as stated in (124) W (ζ) in D+0 2 is as stated in (128)and from (123) we derive that
f(x+ y) = u(x+ y) + v(x+ y) = Re [(1minus i)W (ζ(x+ y))]
g(x minus y) = u(x minus y)minus v(x minus y) = Re [(1 + i)W (ζ(x minus y))](131)
where W [ζ(x+ y)] and W [ζ(x minus y)] are the values of W [ζ(z)] on 0 le z = x+ y le 2and 0 le z = x minus y le 2 respectively
From the foregoing representation of the solution w(z) of Problem A for (12) andthe mapping ζ(z) we can derive that w(z) satisfies the estimate
Cβ[w(z)X(z) D+] + Cβ[wplusmn(z)Y plusmn(z) Dminus] le M1 (132)
in which X(z) = Π2j=1|z minus tj|2|γj |+δ Y plusmn(z) = |xplusmn y minus tj|2|γj |+δ wplusmn(z) = Rew plusmn Imw
β(0 lt β lt δ) δ are sufficiently small positive constants and M1 = M1(β k0 k2 D)is a non-negative constant [85]15)
Finally we mention that if the index K is an arbitrary even integer or 2K isan arbitrary odd integer the above RiemannndashHilbert problem for (12) can be con-sidered but in general the boundary value problem for K le minus1 have some solvabilityconditions or its solution for K ge 0 is not unique
2 The RiemannndashHilbert Problem for First Order LinearComplex Equations of Mixed Type
In this section we discuss the RiemannndashHilbert boundary value problem for first orderlinear complex equations of mixed (elliptic-hyperbolic) type in a simply connecteddomain Firstly we give the representation theorem and prove the uniqueness ofsolutions for the above boundary value problem secondly by using the method ofsuccessive iteration the existence of solutions for the above problem is proved
21 Formulation of RiemannndashHilbert problem of first order complexequations of mixed type
Let D be a simply connected bounded domain in the complex plane CI with theboundary partD = Γcup L where Γ L = L1 cup L2 D
+ = D cap y gt 0 Dminus = D cap y lt 0and z1 = 1minus i are as stated in Section 1
We discuss the first order linear system of mixed (elliptic-hyperbolic) type equa-tions ⎧⎨⎩ux minus vy = au+ bv + f
vx + sgny uy = cu+ dv + gz = x+ iy isin D (21)
2 Linear Mixed Complex Equations 127
in which a b c d f g are functions of (x y)(isin D) its complex form is the followingcomplex equation of first order
wz
wzlowast
= F (z w) F = A1(z)w + A2(z)w + A3(z) in
D+
Dminus
(22)
where
w = u+ iv z = x+ iy wz =12[wx + iwy] wzlowast =
12[wx minus iwy]
A1 =a minus ib+ ic+ d
4 A2 =
a+ ib+ ic minus d
4 A3 =
f + ig
2
Suppose that the complex equation (22) satisfies the following conditions
Condition C
Aj(z) (j = 1 2 3) are measurable in z isin D+ and continuous in Dminus in Dlowast =D(0 2 cup x plusmn y = 2 Im z le 0) or Dlowast = D(0 2 cup x plusmn y = 0 Im z le 0) andsatisfy
Lp[Aj D+] le k0 j = 1 2 Lp[A3 D+] le k1 (23)
C[Aj Dminus] le k0 j = 1 2 C[A3 Dminus] le k1 (24)
where p (gt 2) k0 k1 are non-negative constants
22 The representation and uniqueness of solutions of the RiemannndashHilbert problem for mixed complex equations
We first introduce a lemma which is a special case of Theorem 21 Chapter III
Lemma 21 Suppose that the complex equation (22) satisfies Condition C Thenany solution of Problem A for (22) in D+ with the boundary conditions (13) and
Re [λ(x)w(x)] = s(x) λ(x) = 1minus i or 1 + i x isin L0 Cα[s(x) L0] le k3 (25)
can be expressed asw(z) = Φ(z)eφ(z) + ψ(z) z isin D+ (26)
where Im [φ(z)] = 0 z isin L0 = (0 2) and φ(z) ψ(z) satisfies the estimates
Cβ[φD+] + Lp0 [φz D+] le M2 Cβ[ψD+] + Lp0 [ψz D+] le M2 (27)
in which k3 β (0 lt β le α) p0 (2 lt p0 le 2) M2 = M2(p0 β k D+) are non-negativeconstants k = (k0 k1 k2 k3) Φ(z) is analytic in D+ and w(z) satisfies the estimate
Cβ[w(z)X(z) D+] le M3(k1 + k2 + k3) (28)
128 IV First Order Mixed Complex Equations
in which
X(z) = |z minus t1|η1|z minus t2|η2 ηj =2|γj|+ δ if γj lt 0
δ γj ge 0j = 1 2 (29)
here γj(j = 1 2) are real constants as stated in (17) and δ is a sufficiently smallpositive constant and M3 = M3(p0 β k0 D
+) is a non-negative constant
Theorem 22 If the complex equation (22) satisfies Condition C in D then anysolution of Problem A with the boundary conditions (13) (14) for (22) can beexpressed as
w(z) = w0(z) +W (z) (210)
where w0(z) is a solution of Problem A for the complex equation (12) and W (z)possesses the form
W (z) = w(z)minus w0(z) w(z) = Φ(z)eφ(z) + ψ(z) in D+
φ(z) = φ0(z) + Tg = φ0(z)minus 1π
int intD+
g(ζ)ζ minus z
dσζ ψ(z) = Tf in D+
W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν
2g1(z)dνe1 +
int micro
0g2(z)dmicroe2 in z isin Dminus
(211)
in which φ(z) = 0 on L0 e1 = (1 + i)2 e2 = (1minus i)2 micro = x+ y ν = x minus y φ0(z)is an analytic function in D+ and
g(z) =
A1 + A2ww w(z) = 00 w(z) = 0
f = A1ψ + A2ψ + A3 in D+
g1(z) = Aξ +Bη + E g2(z) = Cξ +Dη + F in Dminus
(212)
where ξ = Rew+Imw η = RewminusImw A = ReA1+ImA1 B = ReA2+ImA2 C =ReA2 minus ImA2 D = ReA1 minus ImA1 E = ReA3 + ImA3 F = ReA3 minus ImA3 andφ(z) ψ(z) satisfy the estimates
Cβ[φ(z) D+] + Lp0 [φz D+] le M4 Cβ[ψ(z) D+] + Lp0 [ψz D+] le M4 (213)
where M4 = M4(p0 β k D+) is a non-negative constant Φ(z) is analytic in D+ andΦ(z) is a solution of equation (12) in Dminus satisfying the boundary conditions
Re [λ(z)(eφ(z)Φ(z) + ψ(z))] = r(z) z isin Γ
Re [λ(x)(Φ(x)eφ(x) + ψ(x))] = s(x) x isin L0
Re [λ(x)Φ(x)] = Re [λ(x)(W (x)minusΨ(x))] z isin L0
Re [λ(z)Φ(z)] = minusRe [λ(z)Ψ(z)] z isin L1 or L2
Im [λ(z1)Φ(z1)] = minusIm [λ(z1)Ψ(z1)]
(214)
2 Linear Mixed Complex Equations 129
Moreover the solution w0(z) of Problem A for (12) satisfies the estimate (132)namely
Cβ[w0(z)X(z) D+] + Cβ[wplusmn0 (z)Y
plusmn(z) Dminus] le M5(k1 + k2) (215)
where wplusmn0 (z) = Rew0(z)plusmn Imw0(z) Y plusmn(z) =
prod2j=1 |x plusmn y minus tj|ηj j = 1 2 X(z) ηj =
2|γj| + δ (j = 1 2) β are as stated in (132) and M5 = M5(p0 β k0 D) is a non-negative constant
Proof Let the solution w(z) be substituted in the position of w in the complex equa-tion (22) and (212) thus the functions g1(z) g2(z) and Ψ(z) in Dminus in (211)(212)can be determined Moreover we can find the solution Φ(z) of (12) with the boundarycondition (214) where
s(x)=
⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
2r((1minus i)x2)minus 2R(1minus i)x2)a((1minus i)x2)minus b((1minus i)x2)
+Re [λ(x)Ψ(x)] x isin L0 or
2r((1+i)x2+1minusi)minus2R((1+i)x2+1minusi)a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)
+ Re [λ(x)Ψ(x)] x isin L0
(216)here and later R(z) = Re [λ(z)Ψ(z)] on L1 or L2 thus
w(z) = w0(z) +W (z) =
⎧⎨⎩ Φ(z)eφ(z) + ψ(z) in D+
w0(z) + Φ(z) + Ψ(z) in Dminus(217)
is the solution of Problem A for the complex equation
wz
wzlowast
= A1w + A2w + A3 in
⎧⎨⎩D+
Dminus
⎫⎬⎭ (218)
which can be expressed as in (210) and (211)
23 The unique solvability of the RiemannndashHilbert problem for firstorder complex equations of mixed type
Theorem 23 Let the mixed complex equation (22) satisfy Condition C ThenProblem A for (22) has a solution in D
Proof In order to find a solution w(z) of Problem A in D we express w(z) in theform (210)ndash(212) In the following we shall find a solution of Problem A by usingthe successive iteration First of all denoting the solution w0(z) = (ξ0e1 + η0e2) ofProblem A for (12) and substituting them into the positions of w = (ξe1 + ηe2)in the right-hand side of (22) similarly to (210)ndash(212) we have the correspondingfunctions g0(z) f0(z) in D+ and the functions
130 IV First Order Mixed Complex Equations
W1(z) = w1(z)minus w0(z) w1(z) = Φ1(z)eφ1(z) + ψ1(z)
φ1(z) = φ0(z)minus 1π
int intD+
g0(ζ)ζ minus z
dσζ ψ1(z) = Tf0 in D+
w1(z) = w0(z) +W1(z) W1(z) = Φ1(z) + Ψ1(z)
Ψ1(z)=int ν
2[Aξ0+Bη0+E]e1dν+
int micro
0[Cξ0+Dη0+F ]e2dmicro in Dminus
(219)
can be determined where micro = x + y ν = x minus y and the solution w0(z) satisfies theestimate (215) ie
Cβ[w0(z)X(z) D+] + Cβ[wplusmn0 (z)Y
plusmn(z) Dminus] le M6 = M5(k1 + k2) (220)
where βX(z) Y plusmn(z) are as stated in (215) Moreover we find an analytic functionΦ1(z) in D+ and a solution Φ1(z) of (12) in Dminus satisfying the boundary conditions
Re [λ(z)(eφ1(z)Φ1(z) + ψ1(z))] = r(z) z isin Γ
Re [λ(x)(Φ1(x)eφ1(x) + ψ1(x))] = s1(x) x isin L0
Re [λ(x)Φ1(x)] = minusRe [λ(x)Ψ1(x)] z isin L0
Re [λ(z)Φ1(z)] = minusRe [λ(z)Ψ1(z)] z isin L1 or L2
Im [λ(z1)Φ1(z1)] = minusIm [λ(z1)Ψ1(z1)]
(221)
in which
s1(x)=
⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
2r((1minusi)x2)minus2R1((1minusi)x2)a((1minusi)x2)minusb((1minusi)x2)
+Re [λ(x)Ψ1(x)] x isin L0 or
2r((1+i)x2+1minusi)minus2R1((1+i)x2+1minusi)a((1+i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)
+Re [λ(x)Ψ1(x)] xisinL0
here and later R1(z) = Re [λ(z)Ψ1(z)] on L1 or L2 and
w1(z) = w0(z) +W1(z) =
⎧⎨⎩ Φ1(z)eφ1(z) + ψ1(z) in D+
w0(z) + Φ1(z) + Ψ1(z) in Dminus(222)
satisfies the estimate
Cβ[w1(z)X(z) D+] + C[wplusmn1 (z)Y
plusmn(z) Dminus] le M7 = M7(p0 β k D) (223)
where φ1(z) ψ1(z) Φ1(z) are similar to the functions in Theorem 22 Furthermore wesubstitute w1(z) = w0(z) +W1(z) and the corresponding functions w+
1 (z) = ξ1(z) =Rew1(z)+Imw(z) wminus
1 (z) = η1(z) = Rew1(z)minusImw(z) into the positions of w ξ η in(211)(212) and similarly to (219)ndash(222) we can find the corresponding functions
2 Linear Mixed Complex Equations 131
φ2(z) ψ2(z) Φ2(z) in D+ and Ψ2(z)Φ2(z) and W2(z) = Φ2(z) + Ψ2(z) in Dminus andthe function
w2(z) = w0(z) +W2(z) =
⎧⎨⎩ Φ2(z)eφ2(z) + ψ2(z) in D+
w0(z) + Φ2(z) + Ψ2(z) in Dminus(224)
satisfies a similar estimate of the form (223) Thus there exists a sequence of functionswn(z) as follows
wn(z) = w0(z) +Wn(z) =
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
Φn(z)eφn(z) + ψn(z) in D+
w0(z) + Φn(z) + Ψn(z) in Dminus
Ψn(z)=int ν
2[Aξnminus1+Bηnminus1+E]e1dν
+int micro
0[Cξnminus1+Dηnminus1+F ]e2dmicro in Dminus
(225)
and then
|[wplusmn1 (z)minuswplusmn
0 (z)]Yplusmn(z)|le|Φplusmn
1 (z)Yplusmn(z)|+
radic2[|Y +(z)
int ν
2[Aξ0+Bη0+E]e1dν|
+|Y minus(z)int micro
0[Cξ0 +Dη0 + F ]e2dmicro|
]le 2M8M(4m+ 1)Rprime in Dminus
(226)where m = maxC[w+
0 (z)Y +(z) Dminus] + C[wminus0 (z)Y minus(z) Dminus] M8 = maxzisinDminus(|A|
|B| |C| |D| |E| |F |) Rprime = 2 M = 1 + 4k20(1 + k2
0) M5 is a constant as stated in(220) It is clear that wn(z)minus wnminus1(z) satisfies
wn(z)minuswnminus1(z)=Φn(z)minusΦnminus1(z)+int ν
2[A(ξnminusξnminus1)+B(ηnminusηnminus1)]e1dν
+int micro
0[C(ξn minus ξnminus1) +D(ηn minus ηnminus1)]e2dmicro in Dminus
(227)
where n = 1 2 From the above equality we can obtain
|[wplusmnn minus wplusmn
nminus1]Yplusmn(z)| le [2M8M(4m+ 1)]n
timesint Rprime
0
Rprimenminus1
(n minus 1) dRprime le [2M8M(4m+ 1)Rprime]n
n in Dminus
(228)
and then we can see that the sequence of functions wplusmnn (z)Y
plusmn(z) ie
wplusmnn (z)Y
plusmn(z)=wplusmn0 (z)+[w
plusmn1 (z)minuswplusmn
0 (z)]+middot middot middot+[wplusmnn (z)minuswplusmn
nminus1(z)]Y plusmn(z) (229)
(n=1 2 ) in Dminus uniformly converge to functions wplusmnlowast (z)Y
plusmn(z) and wlowast(z) satisfiesthe equality
wlowast(z) = w0(z) + Φlowast(z) + Ψlowast(z)
Ψlowast(z) =int ν
2[Aξlowast +Bηlowast + E]e1dν
+int micro
0[Cξlowast +Dηlowast + F ]e2dmicro in Dminus
(230)
132 IV First Order Mixed Complex Equations
where ξlowast = Rewlowast + Imwlowast η = Rewlowast minus Imwlowast and wlowast(z) satisfies the estimate
C[wplusmnlowast (z)Y
plusmn(z) Dminus] le e2M8M(4m+1)Rprime (231)
Moreover we can find a sequence of functions wn(z)(wn(z) = Φn(z)eφn(z) + ψn(z))in D+ and Φn(z) is an analytic function in D+ satisfying the boundary conditions
Re [λ(z)(Φn(z)eφn(z) + ψn(z))] = r(z) z isin Γ
Re [λ(x)(Φn(x)eφn(x) + ψn(x))] = s(x) x isin L0(232)
in which
sn(x)=
⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
2r((1minusi)x2)minus2Rn((1minusi)x2)a((1minusi)x2)minusb((1minusi)x2)
+Re [λ(x)Ψn(x))] x isin L0 or
2r((1+i)x2+1minusi)minus2Rn((1minusi)x2 +1minusi)a((1+i)x2+1minusi)+b((1+i)x2+1minusi)
+Re [λ(x)Ψn(x)] xisinL0
(233)here and later Rn(z) = Re [λ(z)Ψn(z)] on L1 or L2 From (231) it follows that
Cβ[sn(x)X(x) L0] le 2k2k0 +[2M8M(4m+ 1)Rprime]n
n = M9 (234)
and then the estimate
Cβ[wn(z)X(z) D+] le M3(k1 + k2 +M9) (235)
thus from wn(z)X(x) we can choose a subsequence which uniformly converge afunction wlowast(z)X(z) in D+ Combining (231) and (235) it is obvious that thesolution wlowast(z) of Problem A for (22) in D satisfies the estimate
Cβ[wlowast(z)X(z) D+] + C[wplusmnlowast (z)Y
plusmn(z) Dminus] le M10 = M10(p0 β k D) (236)
where M10 is a non-negative constant
Theorem 24 Suppose that the complex equation (22) satisfies Condition C ThenProblem A for (22) has at most one solution in D
Proof Let w1(z) w2(z) be any two solutions of Problem A for (22) By ConditionC we see that w(z) = w1(z)minusw2(z) satisfies the homogeneous complex equation andboundary conditions
Lw = A1w + A2w in D (237)
Re [λ(z)w(z)] = 0 z isin Γ Re [λ(x)w(x)] = s(x) x isin L0
Re [λ(z)w(z)] = 0 z isin L1 or L2 Re [λ(z1)w(z1)] = 0(238)
2 Linear Mixed Complex Equations 133
From Theorem 22 the solution w(z) can be expressed in the form
w(z) =
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
Φ(z)eφ(z) φ(z) = T g in D+
g(z) =
⎧⎨⎩A1 + A2ww w(z) = 0 z isin D+
0 w(z) = 0 z isin D+
Φ(z) + Ψ(z)
Ψ(z) =int ν
2[Aξ +Bη]e1dν +
int micro
0[Cξ +Dη]e2dmicro in Dminus
(239)
where Φ(z) is analytic in D+ and Φ(z) is a solution of (12) in Dminus satisfying theboundary condition (214) but ψ(z) = 0 z isin D+ r(z) = 0 z isin Γ and
s(x)=
⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
minus2R((1minusi)x2)a((1minusi)x2)minus b((1minusi)x2)
+Re [λ(x)Ψ(x)] x isin L0 or
minus2R((1minus i)x2 + 1minus i)a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)
+Re [λ(x)Ψ(x)] x isin L0
By using the method in the proof of Theorem 23 we can derive that
|wplusmn(z)Y plusmn(z)| le [2M8M(4m+ 1)Rprime]n
n in Dminus (240)
Let n rarr infin we get wplusmn(z) = 0 ie w(z) = w1(z) minus w2(z) = 0 Ψ(z) = Φ(z) = 0 inDminus Noting that w(z) = Φ(z)eφ(z) satisfies the boundary conditions in (238) we seethat the analytic function Φ(z) in D+ satisfies the boundary conditions
Re [λ(z)eφ(z)Φ(z)] = 0 z isin Γ Re [λ(x)Φ(x)] = 0 x isin L0 (241)
and the index of the boundary value problem (241) is K = minus12 hence Φ(z) = 0 inD+ and then w(z) = Φ(z)eφ(z) = 0 in D+ namely w(z) = w1(z)minus w2(z) = 0 in D+This proves the uniqueness of solutions of Problem A for (22)
From Theorems 23 and 24 we see that under Condition C Problem A forequation (22) has a unique solution w(z) which can be found by using successiveiteration and w(z) of Problem A satisfies the estimates
Cβ[w(z)X(z) D+] le M11 C[wplusmn(z)Y plusmn(z) Dminus] le M12 (242)
where wplusmn(z) = Rew(z)plusmn Imw(z) X(z) Y plusmn(z) are as stated in (132) and β(0 lt βlt δ) Mj = Mj(p0 β k D) (j = 11 12) are non-negative constants k = (k0 k1 k2)Moreover we can derive the following theorem
Theorem 25 Suppose that equation (22) satisfies Condition C Then any solutionw(z) of Problem A for (22) satisfies the estimates
Cβ[w(z)X(z) D+] le M13(k1 + k2) C[wplusmn(z)Y plusmn(z) Dminus] le M14(k1 + k2) (243)
134 IV First Order Mixed Complex Equations
in which Mj = Mj(p0 β k0 D)(j = 13 14) are non-negative constants
Proof When k1 + k2 = 0 from Theorem 23 it is easy to see that (243) holds Ifk1 + k2 gt 0 then it is seen that the function W (z) = w(z)(k1 + k2) is a solution ofthe homogeneous boundary value problem
Lw = F (z w)k Fk = A1W + A2W + A3k in D
Re [λ(z)W (z)] = r(z)k z isin Γ Im [λ(z1)W (z1)] = b1k
Re [λ(z)W (z)] = r(z)k z isin Lj j = 1 or 2
where Lp[A3kD+] le 1 C[A3kDminus] le 1 Cα[r(z)kΓ] le 1 Cα[r(z)k Lj] le1 j = 1 or 2 |b1k| le 1 On the basic of the estimate (242) we can obtain theestimates
Cβ[w(z)X(z) D+] le M13 C[wplusmn(z)Y plusmn(z) Dminus] le M14 (244)
where Mj = Mj(p0 β k0 D) (j = 13 14) are non-negative constants From (244) itfollows the estimate (243)
From the estimates (243)(244) we can see the regularity of solutions of ProblemA for (22) In the next section we shall give the Holder estimate of solutions ofProblem A for first order quasilinear complex equation of mixed type with the morerestrictive conditions than Condition C which includes the linear complex equation(22) as a special case
3 The RiemannndashHilbert Problem for First Order QuasilinearComplex Equations of Mixed Type
In this section we discuss the RiemannndashHilbert boundary value problem for firstorder quasilinear complex equations of mixed (elliptic-hyperbolic) type in a simplyconnected domain We first give the representation theorem and prove the uniquenessof solutions for the above boundary value problem and then by using the successiveiteration the existence of solutions for the above problem is proved
31 Representation and uniqueness of solutions of RiemannndashHilbertproblem for first order quasilinear complex equations of mixed type
Let D be a simply connected bounded domain as stated in Subsection 21 We discussthe quasilinear mixed (elliptic-hyperbolic) system of first order equations
⎧⎨⎩ux minus vy = au+ bv + f
vx + sgny uy = cu+ dv + gz = x+ iy isin D (31)
3 Quasilinear Mixed Complex Equations 135
in which a b c d f g are functions of (x y) (isin D) u v (isin IR) its complex form isthe following complex equation of first order
wz
wzlowast
= F (z w) F = A1w + A2w + A3 in
D+
Dminus
(32)
where Aj = Aj(z w) j = 1 2 3 and the relations between Aj (j = 1 2 3) anda b c d f g are the same as those in (22)
Suppose that the complex equation (32) satisfies the following conditions
Condition C
1) Aj(z w) (j = 1 2 3) are continuous in w isin CI for almost every point z isin D+and are measurable in z isin D+ and continuous on Dminus for all continuous functionsw(z) in Dlowast = D(0 2cupxplusmny = 2 y le 0) or Dlowast = D(0 2cupxplusmny = 0 y le 0)and satisfy
Lp[Aj D+] le k0 j = 1 2 Lp[A3 D+] le k1
C[Aj Dminus] le k0 j = 1 2 C[A3 Dminus] le k1(33)
2) For any continuous functions w1(z) w2(z) on Dlowast the following equality holds
F (z w1)minus F (z w2) = A1(z w1 w2)(w1 minus w2) + A2(z w1 w2)(w1 minus w2) in D (34)
whereLp[Aj D+] le k0 C[Aj Dminus] le k0 j = 1 2 (35)
in (33)(35) p (gt 2) k0 k1 are non-negative constants In particular when (32) isa linear equation (22) the condition (34) is obviously valid
The boundary conditions of RiemannndashHilbert problem for the complex equation(32) are as stated in (13)(14) Let the solution w(z) of Problem A be substitutedin the coefficients of (32) Then the equation can be viewed as a linear equation(22) Hence we have the same representation theorems as Lemma 21 and Theorem22
Theorem 31 Suppose that the quasilinear complex equation (32) satisfies Condi-tion C Then Problem A for (32) has a unique solution in D
Proof We first prove the uniqueness of the solution of Problem A for (32) Letw1(z) w2(z) be any two solutions of Problem A for (32) By Condition C we seethat w(z) = w1(z)minusw2(z) satisfies the homogeneous complex equation and boundaryconditions
wz = A1w + A2w in D (36)
Re [λ(z)w(z)] = 0 z isin L1 or L2 Re [λ(z1)w(z1)] = 0 (37)
where the conditions on the coefficients Aj(j = 1 2) are the same as in the proof ofTheorem 24 for the linear equation (22) Besides the remaining proof is the samein the proof of Theorems 23 and 24
136 IV First Order Mixed Complex Equations
Next noting the conditions (33)(34) by using the same method the existenceof solutions of Problem A for (32) can be proved and any solution w(z) of ProblemA for (32) satisfies the estimate (243)
In order to give the Holder estimate of solutions for (32) we need to add thefollowing condition
3) For any complex numbers z1 z2(isin D) w1 w2 the above functions satisfy
|Aj(z1 w1)minus Aj(z2 w2)| le k0[|z1 minus z2|α + |w1 minus w2|] j = 1 2
|A3(z1 w1)minus A3(z2 w2)| le k1[|z1 minus z2|α + |w1 minus w2|] z isin Dminus(38)
in which α(12 lt α lt 1) k0 k1 are non-negative constants
On the basis of the results of Theorem 44 in Chapter I and Theorem 23 inChapter III we can derive the following theorem
Theorem 32 Let the quasilinear complex equation (32) satisfy Condition C and(38) Then any solution w(z) of Problem A for (32) satisfies the following estimates
Cδ[X(z)w(z) D+] le M15 Cδ[Y plusmn(z)wplusmn(z) Dminus] le M16 (39)
in which wplusmn(z) = Rew(z)plusmn Imw(z) and
X(z)=2prod
j=1|z minus tj|ηj Y plusmn(x)=
2prodj=1
|xplusmn y minus tj|ηj ηj=2|γj|+2δ if γj lt0
2δ γj ge 0(310)
here γj(j = 1 2) are real constants as stated in (17) and δ is a sufficiently smallpositive constant and Mj = Mj(p0 β k D) (j = 15 16) are non-negative constantsk = (k0 k1 k2)
32 Existence of solutions of Problem A for general first order complexequations of mixed type
Now we consider the general quasilinear mixed complex equation of first order
Lw =
wz
wzlowast
= F (z wz) +G(z w) z isin
D+
Dminus
F = A1w + A2w + A3 G = A4 |w |σ z isin D
(311)
in which F (z w) satisfies Condition C σ is a positive constant and A4(z w) satisfiesthe same conditions as Aj(j = 1 2) where the main condition is
C[A4(z w) D] le k0 (312)
and denote the above conditions by Condition C prime
3 Quasilinear Mixed Complex Equations 137
Theorem 33 Let the mixed complex equation (311) satisfy Condition C prime
(1) When 0 lt σ lt 1 Problem A for (311) has a solution w(z)
(2) When σ gt 1 Problem A for (311) has a solution w(z) provided that
M17 = k1 + k2 + |b1| (313)
is sufficiently small
Proof (1) Consider the algebraic equation for t
(M13 +M14)k1 + k2 + 2k0tσ + |b1| = t (314)
in which M13 M14 are constants stated in (243) It is not difficult to see that theequation (314) has a unique solution t = M18 ge 0 Now we introduce a closed andconvex subset Blowast of the Banach space C(D) whose elements are the function w(z)satisfying the condition
C[w(z)X(z) D+] + C[wplusmn(z)Y plusmn(z) Dminus] le M18 (315)
We arbitrarily choose a function w0(z) isin Blowast for instance w0(z) = 0 and substitute itinto the position of w in the coefficients of (311) and G(z w) From Theorem 31 itis clear that problem A for
Lw minus A1(z w0)w minus A2(z w0)w minus A3(z w0) = G(z w0) (316)
has a unique solution w1(z) By (243) we see that the solution w1(z) satisfies theestimate in (315) By using successive iteration we obtain a sequence of solutionswm(z)(m = 1 2 ) of Problem A which satisfy the equations
Lwm+1 minus A1(z wm)wm+1z minus A2(z wm)wm+1
+A3(z wm) = G(z wm) in D m = 1 2 (317)
and wm+1(z)X(z) isin Blowast m = 1 2 From (317) we see that wm+1(z) = wm+1(z)minuswm(z) satisfies the complex equation and boundary conditions
Lwm+1minusA1wm+1minusA2wm+1=G G(z)=G(z wm)minusG(z wmminus1) in D
Re [λ(z)wm+1(z)]=0 on Γ cup Lj j=1 or 2 Im [λ(z1)wm+1(z1)]=0(318)
where m=1 2 Noting that C[X(z)G(z) D] le 2k0M18 M18 is a solution of thealgebraic equation (314) and according to the proof of Theorem 23
C[wm+1X(z) D+] + C[wplusmnm+1(z)Y
plusmn(z) Dminus] le M18 (319)
can be obtained The function wm+1 can be expressed as
wm+1(z) = w0(z) + Φm+1(z) + Ψm+1(z)
Ψm+1(z) =int xminusy
2[Aξm+1 + Bηm+1 + E]e1d(x minus y)
+int x+y
0[Cξm+1 + Dηm+1 + F ]e2d(x+ y) in Dminus
(320)
138 IV First Order Mixed Complex Equations
in which the relation between A B C D E F and A1 A2 G is the same as that ofA B C D E F and A1 A2 A3 in (212) By using the method from the proof ofTheorem 25 we can obtain
C[wm+1X(z) D+] + C[wplusmnm+1(z)Y
plusmn(z) Dminus] le (M20Rprime)m
m
where M20 = 2M19M(M5+1)(4m+1) Rprime = 2 m = C[wplusmn0 (z)Y plusmn(z) D] herein M19 =
maxC[AQ] C[BQ] C[CQ] C[DQ] C[EQ] C[F Q] M =1 + 4k20(1 + k2
0) Fromthe above inequality it is seen that the sequence of functions wm(z)X(z) ie
wplusmnm(z)Y
plusmn(z)=wplusmn0 (z)+[w
plusmn1 (z)minuswplusmn
0 (z)]+middot middot middot+[wplusmnm(z)minuswplusmn
mminus1(z)]Y plusmn(z) (321)
(m = 1 2 ) uniformly converge to wplusmnlowast (z)Y
plusmn(z) and similarly to (230) the corres-ponding function wlowast(z) satisfies the equality
wlowast(z) = w0(z) + Φlowast(z) + Ψlowast(z)
Ψlowast(z) =int xminusy
2[Aξlowast +Bηlowast + E]e1d(x minus y)
+int x+y
0[Cξlowast +Dηlowast + F ]e2d(x+ y) in Dminus
(322)
and the function wlowast(z) is just a solution of Problem A for the quasilinear equation(311) in the closure of the domain D
(2) Consider the algebraic equation
(M13 +M14)k1 + k2 + 2k0tσ + |b1| = t (323)
for t It is not difficult to see that equation (323) has a solution t = M18 ge 0provided that M17 in (313) is small enough Now we introduce a closed and convexsubset Blowast of the Banach space C(D) whose elements are the functions w(z) satisfyingthe conditions
C[w(z)X(z) D+] + C[wplusmn(z)Y plusmn(z) Dminus] le M18 (324)
By using the same method as in (1) we can find a solution u(z) isin Blowast of Problem Afor equation (311) with σ gt 1
4 The RiemannndashHilbert Problem for First Order QuasilinearEquations of Mixed type in General Domains
This section deals with the RiemannndashHilbert boundary value problem for quasilinearfirst order equations of mixed (elliptic-hyperbolic) type in general domains
4 Mixed Equations in General Domains 139
41 Formulation of the oblique derivative problem for second orderequations of mixed type in general domains
Let D be a simply connected bounded domain D in the complex plane CI with theboundary partD = Γ cup L where Γ L are as stated in Section 1 Now we consider thedomain Dprime with the boundary Γ cup Lprime
1 cup Lprime2 where the parameter equations of the
curves Lprime1 Lprime
2 are as follows
Lprime1 = γ1(x) + y = 0 0 le x le l Lprime
2 = x minus y = 2 l le x le 2 (41)
in which γ1(x) on 0 le x le l = γ1(l) + 2 is continuous and γ1(0) = 0 γ1(x) gt 0on 0 le x le l and γ1(x) is differentiable on 0 le x le l except finitely many pointsand 1 + γprime
1(x) gt 0 Denote Dprime+ = Dprime cap y gt 0 = D+ Dprimeminus = Dprime cap y lt 0and zprime
1 = l minus iγ1(l) Here we mention that in [12]1)3) the author assumes that thederivative of γ(x) satisfies γprime
1(x) gt 0 on 0 le x le l and other conditions
We consider the first order quasilinear complex equation of mixed type as statedin (32) in Dprime and assume that (32) satisfies Condition C in Dprime
The oblique derivative boundary value problemfor equation (32) may be formulated as follows
Problem Aprime Find a continuous solution w(z)of (32) in Dlowast = D0 Lprime
2 which satisfies theboundary conditions
Re [λ(z)w(z)] = r(z) z isin Γ (42)
Re [λ(z)w(z)] = r(z) z isin Lprime1
Im [λ(z)uz]|z=zprime1= b1
(43)
where λ(z) = a(x)+ ib(x) and |λ(z)| = 1 on ΓcupLprime1 and b0 b1 are real constants and
λ(z) r(z) b0 b1 satisfy the conditions
Cα[λ(z)Γ] le k0 Cα[r(z)Γ] le k2 |b1| le k2
Cα[λ(z) Lprime1] le k0 Cα[r(z) Lprime
1] le k2 maxzisinLprime
1
1|a(x)minus b(x)| le k0
(44)
in which α (12 lt α lt 1) k0 k2 are non-negative constants The boundary valueproblem for equation (32) with A3(z u uz) = 0 z isin D u isin IR uz isin CI r(z) =0 z isin Γ cup Lprime
1 and b0 = b1 = 0 will be called Problem Aprime0 The number
K =12(K1 +K2) (45)
is called the index of Problem Aprime and Problem Aprime0 as stated in Section 1 Similarly
we only discuss the case of K = minus12 on partD+ because in this case the solution ofProblem Aprime is unique Besides we choose γ1 gt 0 In the following we first discussthe domain Dprime and then discuss another general domain Dprimeprime
140 IV First Order Mixed Complex Equations
42 The existence of solutions of Problem A for first order equations ofmixed type in general domains
1 By the conditions in (41) the inverse function x = σ(ν) of x + γ1(x) = ν =x minus y can be found and σprime(ν) = 1[1 + γprime
1(x)] Hence the curve Lprime1 can be expressed
by x = σ(ν) = (micro + ν)2 ie micro = 2σ(ν) minus ν 0 le ν le l + γ1(l) We make atransformation
micro=2[microminus2σ(ν)+ν
2minus2σ(ν)+ν
] ν=ν 2σ(ν)minusν lemicrole2 0leν le2 (46)
where micro ν are real variables its inverse transformation is
micro = [2minus 2σ(ν) + ν]micro2 + 2σ(ν)minus ν ν = ν 0 le micro le 2 0 le ν le 2 (47)
It is not difficult to see that the transformation in (46) maps the domain Dprimeminus ontoDminus The transformation (46) and its inverse transformation (47) can be rewrittenas ⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
x =12(micro+ ν) =
4x minus (2 + x minus y)[2σ(x+ γ1(x))minus x minus γ1(x)]4minus 4σ(x+ γ1(x)) + 2x+ 2γ1(x)
y =12(micro minus ν) =
4y minus (2minus x+ y)[2σ(x+ γ1(x))minus x minus γ1(x)]4minus 4σ(x+ γ1(x)) + 2x+ 2γ1(x)
(48)
and ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
x =12(micro+ ν) =
[2minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+ y)4
+σ(x+ γ1(x))minus x+ γ1(x)minus x+ y
2
y =12(micro minus ν) =
[2minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+ y)4
+σ(x+ γ1(x))minus x+ γ1(x) + x minus y
2
(49)
Denote by z = x+ jy = f(z) z = x+ jy = fminus1(z) the transformation (48) and theinverse transformation (49) respectively In this case the system of equations is
ξν = Aξ +Bη + E ηmicro = Cξ +Dη + F z isin Dprimeminus (410)
which is another form of (32) in Dprimeminus Suppose that (32) in Dprime satisfies ConditionC through the transformation (47) we obtain ξν = ξν ηmicro = [2minus 2σ(ν) + ν]ηmicro2 inDprimeminus and then
ξν = Aξ +Bη + E
ηmicro =[2minus 2σ(ν) + ν][Cξ +Dη + F ]
2
in Dminus (411)
and through the transformation (48) the boundary condition (43) is reduced to
Re [λ(fminus1(z))w(fminus1(z))] = r(fminus1(z)) z isin L1
Im [λ(fminus1(z1))w(fminus1(z1)] = b1(412)
4 Mixed Equations in General Domains 141
in which z1 = f(zprime1) Therefore the boundary value problem (410)(43) is trans-
formed into the boundary value problem (411)(412) ie the corresponding Prob-lem A in D On the basis of Theorem 31 we see that the boundary value problem(32)(in D+)(411)(42)(412) has a unique solution w(z) and w(z) is just a solutionof Problem A for (32) in Dprime with the boundary conditions (42)(43)
Theorem 41 If the mixed equation (32) in Dprime satisfies Condition C in the domainDprime with the boundary Γ cup Lprime
1 cup Lprime2 where Lprime
1 Lprime2 are as stated in (41) then Problem
Aprime for (32) with the boundary conditions (42) (43) has a unique solution w(z)
2 Next let the domain Dprimeprime be a simply connected domain with the boundaryΓ cup Lprimeprime
1 cup Lprimeprime2 where Γ is as stated before and
Lprimeprime1 = γ1(x) + y = 0 0 le x le l Lprimeprime
2 = γ2(x) + y = 0 l le x le 2 (413)
in which γ1(0) = 0 γ2(2) = 0 γ1(x) gt 0 0 le x le l γ2(x) gt 0 l le x le 2 γ1(x) on0 le x le l γ2(x) on l le x le 2 are continuous and differentiable except at isolatedpoints and 1 + γprime
1(x) gt 0 1 minus γprime2(x) gt 0 Denote Dprimeprime+ = Dprimeprime cap y gt 0 = D+ and
Dprimeprimeminus = Dprimeprime cap y lt 0 and zprimeprime1 = l minus iγ1(l) = l minus iγ2(l) We consider the Riemannndash
Hilbert problem (Problem Aprime) for equation (32) in Dprimeprime with the boundary conditions(42) and
Re [λ(z)w(z)] = r(z) z isin Lprimeprime2 Im [λ(zprimeprime
1 )w(zprimeprime1 )] = b1 (414)
where zprimeprime1 = l minus iγ1(l) = l minus iγ2(l) and λ(z) r(z) satisfy the corresponding condition
Cα[λ(z)Γ] le k0 Cα[r(z)Γ] le k2 |b1| le k2 Cα[λ(z) Lprimeprime2] le k0
Cα[r(z) Lprimeprime2] le k2 max
zisinLprimeprime1
1|a(x)minus b(x)| max
zisinLprimeprime2
1|a(x) + b(x)| le k0
(415)
in which α (12 lt α lt 1) k0 k2 are non-negative constants By the conditions in(413) the inverse function x = τ(micro) of x minus γ2(x) = micro can be found namely
ν = 2τ(micro)minus micro 0 le micro le 2 (416)
We make a transformation
micro = micro ν =2ν
2τ(micro)minus micro 0 le micro le 2 0 le ν le 2τ(micro)minus micro (417)
where micro ν are real variables its inverse transformation is
micro = micro = x+ y
ν =[2τ(micro)minus micro]ν
2
=[2τ(x minus γ2(x))minus x+ γ2(x)](x minus y)
2 0 le micro le 2 0 le ν le 2
(418)
142 IV First Order Mixed Complex Equations
Hence we have
x =12(micro+ ν) =
2(x minus y) + (x+ y)[2τ(x minus γ2(x))minus x+ γ2(x)]2[2τ(x minus γ2(x))minus x+ γ2(x)]
y =12(micro minus ν) =
minus2(x minus y) + (x+ y)[2τ(x minus γ2(x))minus x+ γ2(x)]2[2τ(x minus γ2(x))minus x+ γ2(x)]
x =12(micro+ ν) =
14[(2τ(x minus γ2(x))minus x+ γ2(x))(x minus y) + 2(x+ y)]
y =12(micro minus ν) =
14[(minus2τ(x minus γ2(x)) + x minus γ2(x))(x minus y) + 2(x+ y)]
(419)
Denote by z = x+ jy = g(z) z = x+ jy = gminus1(z) the transformation (418) and itsinverse transformation in (419) respectively Through the transformation (418) weobtain
(u+ v)ν = [τ(micro)minus micro2](u+ v)ν (u minus v)micro = (u minus v)micro in Dprimeminus (420)
System (410) in Dprimeprimeminus is reduced to
ξν = [τ(micro)minus micro2][Aξ +Bη + E]
ηmicro = Cξ +Dη + Fin Dprimeminus (421)
Moreover through the transformation (419) the boundary condition (414) on Lprimeprime2 is
reduced to
Re [λ(gminus1(z))w(gminus1(z))] = r[gminus1(z)] z isin Lprime2
Im [λ(gminus1(zprime1))w(gminus1(zprime
1)] = b1(422)
in which zprime1 = zprimeprime
1 = g(zprimeprime1 ) Therefore the boundary value problem (410)(414) is
transformed into the boundary value problem (421)(422) According to the methodin the proof of Theorem 41 we can see that the boundary value problem (32) (inD+) (421) (42) (422) has a unique solution w(z) and then w(z) is a solution of theboundary value problem (32)(42)(414) But we mention that through the trans-formation (417) or (419) the boundaries Lprimeprime
1 Lprimeprime2 are reduced to Lprime
1 Lprime2 respectively
such that Lprime1 L
prime2 satisfy the condition as stated in (41) In fact if the intersection zprimeprime
1of Lprimeprime
1 and Lprimeprime2 belongs to L2 and γ1(x) ge 2(1minus l) + x 2minus 2l le x le l then the above
requirement can be satisfied If zprimeprime1 isin L1 = x + y = 0 γ2(x) ge 2l minus x l le x le 2l
then we can proceed similarly
Theorem 42 If the mixed equation (32) satisfies Condition C in the domain Dprimeprime
with the boundary Γcup Lprimeprime1 cup Lprimeprime
2 where Lprimeprime1 Lprimeprime
2 are as stated in (413) then Problem Aprime
for (32) (42) (414) in Dprimeprime has a unique solution w(z)
5 Discontinuous RiemannndashHilbert Problem 143
5 The Discontinuous RiemannndashHilbert Problem forQuasilinear Mixed Equations of First Order
This section deals with the discontinuous RiemannndashHilbert problem for quasilinearmixed (elliptic-hyperbolic) complex equations of first order in a simply connecteddomain Firstly we give the representation theorem and prove the uniqueness ofsolutions for the above boundary value problem Afterwards by using the method ofsuccessive iteration the existence of solutions for the above problem is proved
51 Formulation of the discontinuous RiemannndashHilbert problem forcomplex equations of mixed type
Let D be a simply connected domain with the boundary Γ cup L1 cup L2 as stated asbefore where D+ = |z minus 1| lt 1 Im z gt 0 We discuss the first order quasilinearcomplex equations of mixed type as stated in (32) with Condition C
In order to introduce the discontinuous Riemann-Hilbert boundary value problemfor the complex equation (32) let the functions a(z) b(z) possess discontinuities offirst kind at m minus 1 distinct points z1 z2 zmminus1 isin Γ which are arranged accordingto the positive direction of Γ and Z = z0 = 2 z1 zm = 0 cup x plusmn y = 0 x plusmn y =2 y le 0 wherem is a positive integer and r(z) = O(|zminuszj|minusβj) in the neighborhoodof zj(z = 0 1 m) on Γ in which βj(j = 0 1 m) are sufficiently small positivenumbers Denote λ(z) = a(z) + ib(z) and |a(z)| + |b(z)| = 0 there is no harmin assuming that |λ(z)| = 1 z isin Γlowast = ΓZ Suppose that λ(z) r(z) satisfy theconditions
λ(z) isin Cα(Γj) |z minus zj|βjr(z) isin Cα(Γj) j = 0 1 m (51)
herein Γj is an arc from the point zjminus1 to zj on Γ and Γj(j = 1 m) does notinclude the end points α(0 lt α lt 1) is a constant
Problem Alowast Find a continuous solution w(z) of (32) in Dlowast = DZ and w(z) onZ maybe become infinite of an order lower than unity which satisfies the boundaryconditions
Re [λ(z)w(z)] = r(z) z isin Γ (52)
Re [λ(z)w(z)] = r(z) z isin Lj (j = 1 or 2) Im [λ(z)w(z)]|z=z1 = b1 (53)
where b1 are real constants λ(z) = a(x)+ ib(x)(|λ(z)| = 1) z isin Γcup Lj(j = 1 or 2)and λ(z) r(z) b1 satisfy the conditions
Cα[λ(z)Γj] le k0 Cα[r(z)Γj] le k2 |b1| le k2
Cα[λ(z) Lj] le k0 Cα[r(z) Lj] le k2 j = 1 or 2
maxzisinL1
1|a(z)minus b(z)| le k0 or max
zisinL2
1|a(z) + b(z)| le k0
(54)
144 IV First Order Mixed Complex Equations
in which α (12 lt α lt 1) k0 k2 are non-negative constants The above discontinuousRiemannndashHilbert boundary value problem for (32) is called Problem Alowast
Denote by λ(zjminus0) and λ(zj+0) the left limit and right limit of λ(z) as z rarr zj (j =0 1 m) on Γ and
eiφj =λ(zj minus 0)λ(zj + 0)
γj =1πiln[λ(zj minus 0)λ(zj + 0)
]=
φj
πminus Kj
Kj =[φj
π
]+ Jj Jj = 0 or 1 j = 0 1 m
(55)
in which zm = 0 z0 = 2 λ(z) = exp(minusiπ4) on L0 = (0 2) and λ(z0 minus 0) = λ(zn +0) = exp(minusiπ4) or λ(z) = exp(iπ4) on L0 and λ(z0 minus 0) = λ(zn + 0) = exp(iπ4)and 0 le γj lt 1 when Jj = 0 and minus1 lt Jj lt 0 when Jj = 1 j = 0 1 m and
K =12(K0 +K2 + middot middot middot+Km) =
msumj=0
[φj
2πminus γj
2
](56)
is called the index of Problem Alowast Now the function λ(z) on ΓcupL0 is not continuouswe can choose Jj = 0 or 1 (0 le j le m) hence the index K is not unique Here wechoose the index K = minus12 Let βj + γj lt 1 j = 0 1 m We can require thatthe solution u(z) satisfy the conditions
uz=O(|zminuszj|minusδ) δ=
⎧⎨⎩βj+τ for γj ge0 and γj lt0 βj gt |γj||γj|+τ for γj lt0 βj le|γj| j=1 m
(57)
in the neighborhood of zj in Dlowast where τ(lt α) is a small positive number
For Problem Alowast of the quasilinear complex equation (32) we can prove that thereexists a unique solution by using a similar method as stated in the last section
Next we discuss the more general discontinuous RiemannndashHilbert problem Asstated before denote L = L1 cup L2 L1 = x = minusy 0 le x le 1 L2 = x = y + 2 1 lex le 2 and D+ = D cap y gt 0 Dminus = D cap y lt 0 Here there are n pointsE1 = a1 E2 = a2 En = an on the segment AB = (0 2) = L0 where a0 = 0 lta1 lt a2 lt middot middot middot lt an lt an+1 = 2 and denote by A = A0 = 0 A1 = (1minus i)a12 A2 =(1minusi)a22 An = (1minusi)an2 An+1 = C = 1minusi and B1 = 1minusi+(1+i)a12 B2 =1 minus i + (1 + i)a22 Bn = 1 minus i + (1 + i)an2 B = Bn+1 = 2 on the segmentsAC CB respectively Moreover we denote Dminus
1 = Dminus capcup[n2]j=0 (a2j le xminusy le a2j+1)
Dminus2 = Dminuscapcup[(n+1)2]
j=1 (a2jminus1 le x+y le a2j) and Dminus2j+1 = Dminuscapa2j le xminusy le a2j+1
j = 0 1 [n2] Dminus2j = Dminus cap a2jminus1 le x + y le a2j j = 1 [(n + 1)2] and
Dminuslowast = DminusZ Z = cupn+1
j=0 (x plusmn y = aj y le 0) Dlowast = D+ cup Dminuslowast
Problem Alowast Find a continuous solution w(z) of (32) in Dlowast = DZ whereZ = z0 z1 zm a1 an cup x plusmn y = aj y le 0 j = 1 n and the abovesolution w(z) satisfies the boundary conditions (52) and
5 Discontinuous RiemannndashHilbert Problem 145
Re [λ(z)w(z)]=r(z)
zisinL3=sum[n2]
j=0 A2jA2j+1
Re [λ(z)w(z)]=r(z)
zisinL4=sum[(n+1)2]
j=1 B2jminus1B2j
Im [λ(z)w(z)]|z=A2j+1=c2j+1
j=0 1 [n2]
Im [λ(z)w(z)]|z=B2jminus1=c2j
j=1 [(n+ 1)2]
(58)
where cj(j = 1 n+ 1) are real constants λ(z) = a(x) + ib(x) |λ(z)| = 1 z isin Γand λ(z) r(z) cj(j = 1 n+ 1) satisfy the conditions
Cα[λ(z)Γ] le k0 Cα[r(z)Γ] le k2 |cj| le k2 j = 0 1 n+ 1
Cα[λ(z) Lj] le k0 Cα[r(z) Lj] le k2 j = 3 4
maxzisinL3
1|a(x)minus b(x)| and max
zisinL4
1|a(x) + b(x)| le k0
(59)
where α (12 lt α lt 1) k0 k2 are non-negative constants The above discontinuousRiemannndashHilbert boundary value problem for (32) is called Problem Alowast
Denote by λ(tj minus 0) and λ(tj +0) the left limit and right limit of λ(z) as z rarr tj =zj(j = 0 1 m zm+k = ak k = 1 n zn+m+1 = 2) on Γ cup L0 (L0 = (0 2)) and
eiφj =λ(tj minus 0)λ(tj + 0)
γj =1πiln(
λ(tj minus 0)λ(tj + 0)
)=
φj
πminus Kj
Kj =[φj
π
]+ Jj Jj = 0 or 1 j = 0 1 m+ n
(510)
in which [a] is the largest integer not exceeding the real number a λ(z) = exp(minusiπ4)on Lprime
1 = AB cap Dminus1 and λ(a2j + 0) = λ(a2j+1 minus 0) = exp(minusiπ4) j = 0 1 [n2]
and λ(z) = exp(iπ4) on Lprime2 = AB cap Dminus
2 and λ(a2jminus1 + 0) = λ(a2j minus 0) = exp(iπ4)j = 1 [(n + 1)2] and 0 le γj lt 1 when Jj = 0 and minus1 lt γj lt 0 whenJj = 1 j = 0 1 m+ n and
K =12(K0 +K1 + middot middot middot+Km+n) =
m+nsumj=0
(φj
2πminus γj
2
)(511)
is called the index of Problem Alowast and Problem Alowast0 We can require that the solution
w(z) in D+ satisfy the conditions
146 IV First Order Mixed Complex Equations
w(z) = O(|z minus zj|minusτ ) τ = γprimej + δ j = 0 1 m+ n (512)
in the neighborhood of zj(0 le j le m + n) in D+ where γprimej = max(0 minusγj) (j =
1 m minus 1 m + 1 m + n) γprimem = max(0 minus2γm) γprime
0 = max(0 minus2γ0) and γj(j =0 1 m+n) are real constants in (510) δ is a sufficiently small positive numberand choose the index K = minus12 Now we explain that in the closed domain Dminusthe functions u + v u minus v corresponding to the solution w(z) in the neighborhoodsof the 2n + 2 characteristic lines Z0 = x + y = 0 x minus y = 2 x plusmn y = aj(j =m + 1 m + n) y le 0 may be not bounded if γj le 0(j = m m + n + 1)Hence if we require that u + v u minus v in DminusZ0 is bounded then it needs to chooseγj gt 0 (j = 0 1 m+ n+ 1)
52 Representation of solutions for the discontinuous RiemannndashHilbertproblem
We first introduce a lemma
Lemma 51 Suppose that the complex equation (32) satisfies Condition C Thenthere exists a solution of Problem Alowast for (32) in D+ with the boundary conditions(52) and
Re [λ(z)w(z)]|z=x = s(x) Cβ[s(x) Lprimej] le k3 j = 1 2
λ(x) =
⎧⎨⎩ 1minus i on Lprime1 = Dminus
1 cap AB
1 + i on Lprime2 = Dminus
2 cap AB
(513)
and w(z) satisfies the estimate
Cβ[w(z)X(z) D+] le M21(k1 + k2 + k3) (514)
in which k3 is a non-negative constant s(x) is as stated in the form (525) belowX(z) = Πm+n
j=0 |z minus zj|γprimej+δ herein γprime
j = max(0 minusγj)(j = 1 m minus 1 m + 1 m +n) γprime
0 = max(0 minus2γ0) γprimem = max(0 minus2γm) and γj(j = 0 1 m + n) are real
constants in (510) β (0 lt β lt δ) δ are sufficiently small positive numbers andM21 = M21(p0 β k0 D
+) is a non-negative constant
By using the method as in the proofs of Theorems 21ndash23 Chapter III Theorem12 and Lemma 21 we can prove the lemma
Theorem 52 Problem Alowast for equation (12) in D has a unique solution w(z)
Proof First of all similarly to Theorem 12 the solution w(z) = u(z) + iv(z) ofequation (12) in Dminus can be expressed as (119) According to the proof of Theorem12 we can obtain f(x + y) on Lprime
1 = Dminus1 cap AB and g(x minus y) on Lprime
2 = Dminus2 cap AB in
the form
5 Discontinuous RiemannndashHilbert Problem 147
g(x)=k(x)=2r((1minusi)x2)minus[a((1minusi)x2)+b ((1minusi)x2)]h2j
a((1minus i)x2)minus b ((1minus i)x2)
on L2j+1 = Dminus2j+1 cap AB
f(x)=h(x)=2r((1 + i)x2 + 1minus i)
a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)
minus [a((1+i)x2+1minusi)minusb ((1+i)x2+1minusi)]k2jminus1
a((1+i)x2+1minusi)+b ((1+i)x2+1minusi)
on L2j = Dminus2j cap AB
(515)
where Dminusj (j = 1 2 2n+ 1) are as stated in Subsection 51 and
h2j = Re [λ(A2j+1)(r(A2j+1) + ic2j+1)] + Im [λ(A2j+1)(r(A2j+1) + ic2j+1)]
on L2j+1 j = 0 1 [n2]
k2jminus1 = Re [λ(B2jminus1)(r(B2jminus1) + ic2j)]minus Im [λ(B2jminus1)(r(B2jminus1) + ic2j)]
on L2j j = 1 [(n+ 1)2]
where L2j+1 = Dminus2j+1 cap AB j = 0 1 [n2] L2j = Dminus
2j cap AB j = 1 [(n + 1)2]By using Theorem 22 Chapter III choosing an appropriate index K = minus12 thereexists a unique solution w(z) of Problem Alowast in D+ with the boundary conditions(52) and
Re [λ(x)w(x)] =
⎧⎨⎩k(x)
h(x)λ(x) =
⎧⎪⎨⎪⎩1minus i on Lprime
1 = Dminus1 cap AB
1 + i on Lprime2 = Dminus
2 cap AB
and denote
Re [λ(x)w(x)] =
⎧⎨⎩h(x) on Lprime1
k(x) on Lprime2
Cβ[X(x)k(x) Lprime1] le k2 Cβ[X(x)h(x) Lprime
2] le k2
(516)
herein β(0 lt β le α lt 1) k2 are non-negative constants
Next we find a solution w(z) of Problem Alowast for (12) in Dminus with the boundaryconditions
Re [λ(z)w(z)] = r(z) on L3 cup L4
Im [λ(z)w(z)]|z=A2j+1 = c2j+1 j = 0 1 [n2]
Im [λ(z)w(z)]|z=B2jminus1 = c2j j = 1 [(n+ 1)2]
(517)
and (516) where cj(j = 1 n + 1) are as stated in (58) By the result and themethod in Chapters I and II we can find the solution of Problem Alowast for (12) in Dminus
1in the form
148 IV First Order Mixed Complex Equations
w(z) = w(z) + λ(A2j+1)[r(A2j+1) + ic2j+1]
w(z) =12[(1 + i)f2j+1(x+ y) + (1minus i)g2j+1(x minus y)]
f2j+1(x+ y) = Re [λ(x+ y)w(x+ y)] g2j+1(x minus y)
=2r((1minus i)(x minus y)2)
a((1minus i)(x minus y)2)minus b((1minus i)(x minus y)2)in Dminus
2j+1
j = 0 1 [n2]
w(z) = w(z) + λ(B2jminus1)[r(B2jminus1) + ic2j]
w(z)=12[(1 + i)f2j(x+ y)+(1minus i)g2j(x minus y)] f2j(x+ y)
= h(x+ y)=2r((1+i)(x+y)2+1minusi)
a((1+i)(x+y)2+1minusi)+b ((1+i)(x+y)2+1minusi)
g2j(x minus y) = Re [λ(x minus y)w(x minus y)] in Dminus2j j = 1
[n+ 12
]
Furthermore from the above solution we can find the solution of Problem Alowast for (12)in DminusDminus
1 cup Dminus2 and the solution w(z) of Problem Alowast for (12) in Dminus possesses
the form
w(z) =12[(1 + i)f(x+ y) + (1minus i)g(x minus y)]
f(x+ y) = Re [(1minus i)w(x+ y)] in DminusDminus2
g(x minus y) = k(x minus y) in Dminus1
f(x+ y) = h(x+ y) in Dminus2
g(x minus y) = Re [(1 + i)w(x minus y)] in DminusDminus1
(518)
where k(x) h(x) are as stated in (515) and w(z) is the solution of Problem Alowast for(12) with the boundary conditions (52)(516)
Theorem 53 Let the complex equation (32) satisfy Condition C Then any solu-tion of Problem Alowast for (32) can be expressed as
w(z) = w0(z) +W (z) in D (519)
where w0(z) is a solution of Problem Alowast for equation (12) and W (z) possesses theform
W (z) = w(z)minus w0(z) in D w(z) = Φ(z)eφ(z) + ψ(z)
φ(z) = φ0(z) + Tg = φ0(z)minus 1π
int intD+
g(ζ)ζ minus z
dσζ ψ(z) = Tf in D+
5 Discontinuous RiemannndashHilbert Problem 149
W (z) = Φ(z) + Ψ(z) in Dminus
Ψ(z) =
⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩
int ν
a2j+1
g1(z)dνe1+int micro
0g2(z)dmicroe2 in Dminus
2j+1 j=0 1 [n2]
int ν
2g1(z)dνe1+
int micro
a2jminus1
g2(z)dmicroe2 in Dminus2j j=1 [(n+1)2]
(520)
in which e1 = (1 + i)2 e2 = (1 minus i)2 micro = x + y ν = x minus y φ0(z) is an analyticfunction in D+ Im [φ(z)] = 0 on L0 and
g(z) =
⎧⎨⎩A1 + A2w(w) w(z) = 00 w(z) = 0 z isin D+
f = A1Tf + A2Tf + A3 in D+
g1(z) = Aξ +Bη + E g2(z) = Cξ +Dη + F in Dminus
(521)
where ξ = Rew + Imw η = Rew minus Imw and φ(z) ψ(z) satisfy the estimates
Cβ[φ(z) D+] + Lp0 [φz D+] le M22
Cβ[ψ(z) D+] + Lp0 [ψz D+] le M22(522)
where p0 (0 lt p0 le p) M22 = M22 (p0 α k D+) are non-negative constants Φ(z) isanalytic in D+ and Φ(z) is a solution of equation (12) in Dminus satisfying the boundaryconditions
Re [λ(z)eφ(z)Φ(z)] = r(z)minus Re [λ(z)ψ(z)] z isin Γ
Re [λ(x)(Φ(x)eφ(x) + ψ(x))] = s(x) x isin L0
Re [λ(x)Φ(x)] = Re [λ(x)(W (x)minusΨ(x))] z isin L0
Re [λ(z)Φ(z)] = minusRe [λ(z)Ψ(z)] z isin L3 cup L4
Im [λ(z)(Φ(z) + Ψ(z))]|z=A2j+1 = 0 j = 0 1 [n2]
Im [λ(z)(Φ(z) + Ψ(z))]|z=B2jminus1 = 0 j = 1 [(n+ 1)2]
(523)
Moreover the solution w0(z) of Problem Alowast for (12) satisfies the estimate in the form
Cβ[X(z)w0(z) D+] + C[wplusmn0 (micro ν)Y plusmn(micro ν) Dminus] le M23(k1 + k2) (524)
in which X(z) = Πm+nj=0 |z minus zj|γprime
j+δ Y plusmn(z) = Y plusmn(micro ν) = Πm+nj=0 |x plusmn y minus zj|γprime
j+δwplusmn
0 (micro ν) = Rew0(z)plusmn Imw0(z) w0(z) = w0(micro ν) micro = x + y ν = x minus yγprime
j(j = 0 1 m + n) are as stated in (514) M23 = M23 (p0 β k0 D) is a non-negative constant
150 IV First Order Mixed Complex Equations
Proof Let the solution w(z) of Problem Alowast be substituted into the complex equa-tion (32) and the solution w0(z) = ξ0e1 + η0e2 of Problem Alowast for equation (12)be substituted in the position of w in (521) Thus the functions f(z) g(z) in D+
and g1(z) g2(z) and Ψ(z) in Dminus in (520)(521) can be determined Moreover byTheorem 52 we can find an analytic function Φ(z) in D+ and a solution Φ(z) of(12) in Dminus with the boundary conditions (523) where
s(x) =
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
2r((1minus i)x2)minus 2R((1minus i)x2)a((1minus i)x2)minus b((1minus i)x2)
minus [a((1minus i)x2) + b((1minus i)x2)]h2j
a((1minus i)x2)minus b((1minus i)x2)+ Re [λ(x)Ψ(x)]
x isin (a2j a2j+1) j = 0 1 [n2]
2r((1+i)x2+1minusi)minus2R((1+i)x2+1minusi)a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)
minus [a((1 + i)x2 + 1minus i)minus b((1 + i)x2 + 1minus i)]k2jminus1
a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)
+Re [λ(x)Ψ(x)] x isin (a2jminus1 a2j) j = 1 [(n+ 1)2]
(525)
in which the real constants h2j(j = 0 1 [n2]) k2jminus1(j = 1 [(n + 1)2]) are asstated in (515) thus
w(z) = w0(z) +W (z) =
⎧⎨⎩ Φ(z)φ(z) + ψ(z) in D+
w0(z) + Φ(z) + Ψ(z) in Dminus
is the solution of Problem Alowast for the complex equation
wz
wzlowast
= A1w + A2w + A3 in
⎧⎨⎩D+
Dminus
⎫⎬⎭ (526)
and the solution w0(z) of Problem Alowast for (12) satisfies the estimate (524)
53 Existence and uniqueness of solutions of the RiemannndashHilbertproblem for (32)
Theorem 54 Suppose that the complex equation (32) satisfies Condition C ThenProblem Alowast for (32) is solvable
Proof In order to find a solution w(z) of Problem Alowast in D we express w(z) inthe form (519)ndash(521) On the basis of Theorem 52 we see that Problem Alowast for(12) has a unique solution w0(z)(= ξ0e1+η0e2) and substitute it into the position ofw = ξe1 + ηe2 in the right-hand side of (32) Similarly to (219) from (519)ndash(521)
5 Discontinuous RiemannndashHilbert Problem 151
we obtain the corresponding functions g0(z) f0(z) in D+ g10(z) g
20(z) in Dminus and the
functions
φ1(z)= φ0(z)minus 1π
int intD+
g0(ζ)ζ minus z
dσζ ψ(z)=Tf0 in D+
Ψ1(z)=
⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩
int ν
a2j+1
g10(z)dνe1+
int micro
0g20(z)dmicroe2 in Dminus
2j+1 j=0 1 [n2]
int ν
2g10(z)dνe1+
int micro
a2jminus1
g20(z)dmicroe2 in Dminus
2j j=1 [(n+1)2]
(527)
can be determined where micro = x + y ν = x minus y and the solution w0(z) satisfies theestimate (524) ie
Cβ[w0(z)X(z) D+] + Cβ[wplusmn0 (z)Y
plusmn(z) Dminus] le M23(k1 + k2) (528)
Moreover by Theorem 52 we can find an analytic function Φ(z) inD+ and a solutionΦ1(z) of (12) in Dminus satisfying the boundary conditions
Re [λ(z)(Φ1(z)eφ1(z) + ψ1(z))] = r(z) z isin Γ
Re [λ(x)(Φ1(x) + ψ1(x))] = s(z) x isin L0
Re [λ(x)Φ1(x)] = Re [λ(x)(W1(x)minusΨ1(x))] z isin L0
Re [λ(z)Φ1(z)] = minusRe [λ(z)Ψ1(z)] z isin L3 cup L4
Im [λ(z)(Φ1(z) + Ψ1(z))]|z=A2j+1 = 0 j = 0 1 [n2]
Im [λ(z)(Φ1(z) + Ψ1(z))]|z=B2jminus1 = 0 j = 1 [(n+ 1)2]
(529)
where
s1(x) =
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
2r((1minus i)x2)minus 2R1((1minus i)x2)a((1minus i)x2)minus b ((1minus i)x2)
minus [a((1minus i)x2) + b ((1minus i)x2)]h2j
a((1minus i)x2)minus b ((1minus i)x2)+ Re [λ(x)Ψ1(x)]
x isin (a2j a2j+1) j = 0 1 [n2]
2r((1+i)x2+1minusi)minus2R1((1+i)x2+1minusi)a((1 + i)x2 + 1minus i) + b ((1 + i)x2 + 1minus i)
minus [a((1 + i)x2 + 1minus i)minus b ((1 + i)x2 + 1minus i)]k2jminus1
a((1 + i)x2 + 1minus i) + b ((1 + i)x2 + 1minus i)
+Re [λ(x)Ψ1(x)] x isin (a2jminus1 a2j) j = 1 [(n+ 1)2]
in which the real constants h2j k2jminus1 are as stated in (515) and
w1(z) = w0(z) +W1(z) = w0(z) + Φ1(z) + Ψ1(z) in Dminus (530)
152 IV First Order Mixed Complex Equations
satisfies the estimate
Cβ[w1(z)X(z) D+]+C[wplusmn1 (micro ν)Y plusmn(micro ν) Dminus]leM24=M24(p0 β k Dminus) (531)
here wplusmn1 (micro ν) = Rew1(micro ν) plusmn Imw1(micro ν) Y plusmn(micro ν) X(z) Y plusmn(z) are as stated in
(524) Furthermore we substitute w1(z) = w0(z)+W1(z) and corresponding functionsw1(z) ξ1(z) = Rew1(z) + Imw1(z) η1(z) = Rew1(z) minus Imw1(z) into the positionsw(z) ξ(z) η(z) in (520) (521) and similarly to (527)ndash(530) we can find thecorresponding functions φ2(z) ψ2(z) Φ2(z) in D+ and Ψ2(z)Φ2(z) and W2(z) =Φ2(z) + Ψ2(z) in Dminus The function
w2(z) = w0(z) +W2(z) =
⎧⎨⎩ Φ2(z)eφ2(z) + ψ2(z) in D+
w0(z) + Φ2(z) + Ψ2(z) in Dminus(532)
satisfies the similar estimate in the form (531) Thus there exists a sequence offunctions wn(z) as follows
wn(z) = w0(z) +Wn(z) =
⎧⎨⎩ Φn(z)eφn(z) + ψn(z) in D+
w0(z) + Φn(z) + Ψn(z) in Dminus
Ψn(z)=
⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩
int ν
a2j+1
g1nminus1(z)e1dν+
int micro
0g2
nminus1(z)e2dmicro in Dminus2j+1 j=0 1 [n2]
int ν
2g1
nminus1(z)e1dν+int micro
a2jminus1
g2nminus1(z)e2dmicro in Dminus
2j j=1 [(n+1)2]
g1nminus1(z)=Aξnminus1+Bηnminus1+E g2
nminus1(z)=Cξnminus1+Dηnminus1 + F in Dminus
(533)
and then
|[wplusmn1 (micro ν)minus wplusmn
0 (micro ν)]Y plusmn(micro ν)| le |Φplusmn1 (micro ν)Y plusmn(micro ν)|
+radic2
|Y minus(micro ν)|[max
1lejlen+1|int ν
a2j+1
g10(z)e1dν|+ |
int ν
2g20(z)e1dν|
]
+|Y +(micro ν)|[|int micro
0g10(z)e2dmicro|+ max
1lejlen+1|int micro
a2jminus1
g20(z)e2dmicro|
]
le 2M25M(4m+ 1)Rprime in Dminus
(534)
wherem = C[w+0 (micro ν)Y +(micro ν) Dminus]+C[wminus
0 (micro ν)Y minus(micro ν) Dminus] M = 1+ 4k20(1+k2
0)
5 Discontinuous RiemannndashHilbert Problem 153
Rprime = 2 M25 = maxzisinDminus(|A| |B| |C| |D|) It is clear that wn(z)minus wnminus1(z) satisfies
wn(z)minus wnminus1(z)
= Φn(z)minus Φnminus1(z) +
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
int ν
a2j+1
[g1nminus1minusg1
nminus2]e1dν +int micro
0[g2
nminus1minusg2nminus2]e2dmicro in Dminus
2j+1
j=0 1 [n2]int ν
2[g1
nminus1minusg1nminus2]e1dν+
int micro
a2jminus1
[g2nminus1minusg2
nminus2]e2dmicro inDminus2j
j=1 [(n+1)2]
(535)
Moreover we can find the solution w(z) of Problem Alowast for (32) in the setDminus(cup[n2]
j=0 Dminus2j+1) cup (cup[(n+1)2]
j=1 Dminus2j) = DminusDminus
1 cup Dminus2 From the above result
|[wplusmnn minus wplusmn
nminus1]Yplusmn| le [2M25M(4m+ 1)]n
int Rprime
0
Rprimenminus1
(n minus 1) dRprime
le [2M25M(4m+ 1)Rprime]n
n in Dminus
(536)
can be obtained and then we see that the sequence of functions wplusmnn (micro ν)Y plusmn(micro ν)
ie
wplusmnn (micro ν)Y plusmn(micro ν)=wplusmn
0 +[wplusmn1 minuswplusmn
0 ]+ +[wplusmnn minuswplusmn
nminus1]Y plusmn(micro ν)(n=1 2 ) (537)
in Dminus uniformly converge to wplusmnlowast (micro ν)Y plusmn(micro ν) and wlowast(z) = [w+(micro ν) + wminus(micro ν)
minusi(w+(micro ν)minus wminus(micro ν))]2 satisfies the equality
wlowast(z)=w0(z)+Φlowast(z)+Ψlowast(z)
Ψlowast(z)=
⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩
int ν
a2j+1
g1lowast(z)e1dν+
int micro
0g2
lowast(z)e2dmicro inDminus2j+1 j=01[n2]
int ν
2g1
lowast(z)e1dν+int micro
a2jminus1
g2lowast(z)e2dmicro inDminus
2j j=1[(n+1)2]
g1lowast(z)=Aξlowast+Bηlowast+Eg2
lowast(z)=Cξlowast+Dηlowast+F inDminus
(538)
and the corresponding function ulowast(z) is just a solution of Problem Alowast for equation(32) in the domain Dminus and wlowast(z) satisfies the estimate
C[wplusmnlowast (micro ν)Y plusmn(micro ν) Dminus] le M26 = e4M25M(2m+1)Rprime
(539)
In addition we can find a sequence of functions wn(z)(wn(z) = Φn(z)eφn(z)+ψn(z))in D+ and Φn(z) is an analytic function in D+ satisfying the boundary conditions
Re [λ(z)(Φn(z)eφn(z) + ψn(z))] = r(z) z isin Γ
Re [λ(x)(Φn(x)eφn(x) + ψn(x))] = s(x) x isin L0(540)
154 IV First Order Mixed Complex Equations
in which
sn(x) =
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
2r((1minus i)x2)minus 2Rn((1minus i)x2)a((1minus i)x2)minus b((1minus i)x2)
minus [a((1minus i)x2) + b((1minus i)x2)]h2jradic2[a((1minus i)x2)minus b((1minus i)x2)]
+ Re [λ(x)Ψn(x)]
x isin (a2j a2j+1) j = 0 1 [n2]
2r((1+i)x2+1minusi)minus2Rn((1+i)x2+1minusi)a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)
minus [a((1 + i)x2 + 1minus i)minus b((1 + i)x2 + 1minus i)]k2jminus1
a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)
+Re [λ(x)Ψn(x)] x isin (a2jminus1 a2j) j = 1 [(n+ 1)2]
(541)
in which the real constants h2j k2jminus1 are as stated in (515) From (531)(539) itfollows that
Cβ[X(x)sn(x) L0] le M27 = M27(p0 β k D) (542)
and then the estimate
Cβ[wn(z)X(z) D+] le M21(k1 + k2 +M27) (543)
Thus from wn(z)X(x) we can choose a subsequence which uniformly converges toa function wlowast(z)X(z) in D+ Combining (543) and (539) it is obvious that thesolution wlowast(z) of Problem Alowast for (32) in D satisfies the estimate
Cβ[wlowast(z)X(z) D+] + C[wplusmnlowast (z)Y
plusmn(z) Dminus] le M28 = M28(p0 β k D) (544)
where M28 is a non-negative constant
Theorem 55 If the complex equation (32) satisfies Condition C then ProblemAlowast for (32) has at most one solution in D
Proof Let w1(z) w2(z) be any two solutions of Problem Alowast for (32) By ConditionC we see that w(z) = w1(z)minus w2(z) satisfies the homogeneous complex equation
wz
wzlowast
= A1w + A2w in
⎧⎨⎩D+
Dminus
⎫⎬⎭ (545)
and boundary conditions
Re [λ(z)w(z)] = 0 z isin Γ
Re [λ(x)w(x)] = s(x) x isin L0
Re [λ(z)w(z)] = 0 z isin L3 cup L4
Im [λ(z)w(z)]|z=A2j+1 = 0 j = 0 1 [n2]
Im [λ(z)w(z)]|z=B2jminus1 = 0 j = 1 [(n+ 1)2]
(546)
5 Discontinuous RiemannndashHilbert Problem 155
in which
s(x) =
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
2r((1minus i)x2)minus 2R((1minus i)x2)a((1minus i)x2)minus b((1minus i)x2)
+Re [λ(x)Ψ(x)] x isin (a2j a2j+1) j = 0 1 [n2]
2r((1+i)x2+1minusi)minus2R((1+i)x2+1minusi)a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)
+Re [λ(x)Ψ(x)] x isin (a2jminus1 a2j) j = 1 [(n+ 1)2]
(547)
From Theorem 53 the solution w(z) can be expressed in the form
w(z) =
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
Φ(z)eφ(z) φ(z) = T g in D+
g(z) =
⎧⎨⎩A1 + A2ww w(z) = 0 z isin D+
0 w(z) = 0 z isin D+
Φ(z) + Ψ(z)
Ψ(z) =
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
int ν
a2j+1
[Aξ + Bη]e1dν +int micro
0[Cξ + Dη]e2dmicro
in Dminus2j+1 j = 0 1 [n2]int ν
2[Aξ + Bη]e1dν +
int micro
a2jminus1
[Cξ + Dη]e2dmicro
in Dminus2j j = 1 [(n+ 1)2]
(548)
By using the method in the proof of Theorem 54 we can get that
|wplusmn(z)Y plusmn(z)| le [2M25M(4m+ 1)Rprime]n
n in Dminus (549)
Let n rarr infin we get wplusmn(z) = 0 ie w(z) = w1(z) minus w2(z) = 0 and then Φ(z) =Ψ(z) = 0 in Dminus thus s(x) = 0 on L0 Noting that w(z) = Φ(z)eφ(z) satisfies theboundary conditions in (546) we see that the analytic function Φ(z) in D+ satisfiesthe boundary conditions
Re [λ(z)eφ(z)Φ(z)] = 0 z isin Γ Re [λ(x)Φ(x)] = 0 x isin L0 (550)
and the index of the boundary value problem (550) is K = minus12 hence Φ(z) = 0 inD+ and then w(z) = Φ(z)eφ(z) = 0 in D+ namely w(z) = w1(z)minus w2(z) = 0 in D+This proves the uniqueness of solutions of Problem Alowast for (32)
From Theorems 54 and 55 we see that under Condition C Problem Alowast forequation (32) has a unique solution w(z) which can be found by using successiveiteration and the solution w(z) satisfies the estimate (544) ie
Cβ[w(z)X(z) D+] + C[wplusmn(micro ν)Y plusmn(micro ν) Dminus] le M29 (551)
156 IV First Order Mixed Complex Equations
where k = (k0 k1 k2) M29 = M29 (p0 β k D) is a non-negative constant Moreoverwe have
Theorem 56 Suppose that equation (32) satisfies Condition C Then any solutionw(z) of Problem Alowast for (32) satisfies the estimates (544) and
Cβ[w(z)X(z) D+] + C[w(z) Dminus] le M30 (k1 + k2) (552)
where X(z) Y (z) are as stated in (524) respectively and M30 = M30 (p0 β δ k0 D)is a non-negative constant
From the estimates (551) and (552) we can see the regularity of solutions ofProblem Alowast for (32)
Finally we mention that if the index K is an arbitrary even integer or 2K is anarbitrary odd integer the above Problem Alowast for (32) can be considered But ingeneral Problem Alowast for (32) with K le minus1 has minus2K minus 1 solvability conditions orwhen K ge 0 its general solution includes 2K + 1 arbitrary real conditions
For more general first order complex equations of mixed type the discontinuousRiemannndashHilbert boundary value problem remains to be discussed
The references for this chapter are [3][8][12][16][20][25][35][36][42][44][52][55][60][63][73][75][83][85][95][98]
CHAPTER V
SECOND ORDER LINEAR EQUATIONS OFMIXED TYPE
In this chapter we discuss the oblique derivative boundary value problem for sec-ond order linear equation of mixed (elliptic-hyperbolic) type in a simply connecteddomain We first prove uniqueness and existence of solutions for the above bound-ary value problem and then give a priori estimates of solutions for the problemfinally discuss the existence of solutions for the above problem in general domainsIn books [12]1)3) the author investigated the Dirichlet problem (Tricomi problem)for the mixed equation of second order ie uxx + sgny uyy = 0 In [69] theauthor discussed the Tricomi problem for the generalized Lavrentprimeev-Bitsadze equa-tion uxx + sgny uyy + Aux + Buy + Cu = 0 ie uξη + auξ + buη + cu = 0 with theconditions a ge 0 aξ+abminus c ge 0 c ge 0 in the hyperbolic domain In this section wecancel the above assumption in [69] and obtain the solvability result on the discon-tinuous Poincare problem which includes the corresponding results in [12]1)3)[69]as special cases
1 Oblique Derivative Problems for Simplest Second OrderEquation of Mixed Type
In this section we introduce the oblique derivative boundary value problem for sim-plest mixed equation of second order in a simply connected domain and verify theuniqueness and existence of solutions for the above boundary value problem
Let D be a simply connected bounded domain in the complex plane CI with theboundary partD = Γ cup L where Γ(sub y gt 0) isin C2
α(0 lt α lt 1) with the end pointsz = 0 2 and L = L1 cup L2 L1 = x = minusy 0 le x le 1 L2 = x = y + 2 1 le x le 2and denote D+ = D cap y gt 0 Dminus = D cap y lt 0 and z1 = 1 minus i We mayassume that Γ = |z minus 1| = 1 y ge 0 otherwise through a conformal mapping therequirement can be realized
11 The oblique derivative problem for simplest second order equationof mixed type
In A V Bitsadzersquos books [12]1)3) the author discussed the solvability of sev-eral boundary value problems including the Dirichlet problem or Tricomi problem
158 V Second Order Linear Mixed Equations
(Problem D or Problem T ) for the second order equation of mixed type
uxx + sgny uyy = 0 in D (11)
the equation is sondashcalled Lavrentprimeev-Bitsadze equation its complex form is as followsuzz
uzlowastzlowast
= 0 in
D+
Dminus
(12)
whereuzlowast = uz wzlowast =
12[wx minus iwy]
Now we formulate the oblique derivative boundary value problem as follows
Problem P Find a continuously differentiable solution u(z) of (12) in Dlowast =D0 x minus y = 2 or Dlowast = Dx + y = 0 2 which is continuous in D and sat-isfies the boundary conditions
12
partu
partl= Re [λ(z)uz] = r(z) z isin Γ u(0) = b0 u(2) = b2 (13)
12
partu
partl= Re [λ(z)uz] = r(z) z isin Lj (j = 1 or 2) Im [λ(z)uz]|z=z1 = b1 (14)
where l is a given vector at every point on Γcup Lj (j = 1 or 2) λ(z) = a(x) + ib(x) =cos(l x) minus i cos(l y) if z isin Γ and λ(z) = a(z) + ib(z) = cos(l x) + i cos(l y) ifz isin Lj (j = 1 or 2) b0 b1 are real constants and λ(z) r(z) b0 b1 b2 satisfy theconditions
Cα[λ(z)Γ]lek0 Cα[r(z)Γ]lek2 Cα[λ(z)Lj]lek0 Cα[r(z)Lj]lek2 j=1 or 2
cos(l n)ge0 on Γ |bj|lek2 j=0 1 2 maxzisinL1
1|a(z)minusb(z)| or maxzisinL2
1|a(z)+b(z)| lek0
(15)
in which n is the outward normal vector at every point on Γ α (12 lt α lt 1) k0 k2
are non-negative constants For convenience we may assume that uz(z1) = 0 other-wise through a transformation of function Uz(z) = uz(z) minus λ(z1)[r(z1) + ib1] therequirement can be realized
The boundary value problem for (12) with r(z) = 0 z isin Γ cup Lj (j = 1 or 2) andb0 = b1 = b2 = 0 will be called Problem P0 The number
K =12(K1 +K2) (16)
is called the index of Problem P and Problem P0 where
Kj =[φj
π
]+Jj Jj = 0 or 1 eiφj =
λ(tj minus 0)λ(tj + 0)
γj =φj
πminusKj j = 1 2 (17)
1 Simplest Mixed Equation of Second Order 159
in which t1 = 2 t2 = 0 λ(t) = eiπ4 on L0 = (0 2) and λ(t1 minus 0) = λ(t2 + 0) =exp(iπ4) or λ(t) = ei7π4 on L0 and λ(t1 minus 0) = λ(t2 + 0) = exp(i7π4) Here wechoose K = 0 or K = minus12 on the boundary partD+ of D+ if cos(ν n) equiv 0 on Γand the condition u(2) = b2 can be canceled In this case the solution of ProblemP for (12) is unique In order to ensure that the solution u(z) of Problem P iscontinuously differentiable in Dlowast we need to choose γ1 gt 0 If we require that thesolution u(z) in D is only continuous it is suffices to choose minus2γ1 lt 1 minus2γ2 lt 1Problem P in this case still includes the Dirichlet problem as a special case If theboundary condition Re [λ(z)uz] = r(z) on Lj(j = 1 or 2) in (14) is replaced byRe [λ(z)uz] = r(z) on Lj(j = 1 or 2) then Problem P does not include the aboveDirichlet problem (Problem D) as a special case
Setting that uz = w(z) it is clear that Problem P for (12) is equivalent to theRiemannndashHilbert boundary value problem (Problem A) for the first order complexequation of mixed type
wz
wzlowast
= 0 in
D+
Dminus
(18)
with the boundary conditions
Re [λ(z)w(z)] = r(z) z isin Γ u(2) = b2
Re [λ(z)w(z)] = r(z) z isin Lj (j = 1 or 2) Im[λ(z1)w(z1)] = b1(19)
and the relationu(z) = 2Re
int z
0w(z)dz + b0 in D (110)
in which the integral path in Dminus is as stated in Chapter II On the basis of the resultin Section 1 Chapter IV we can find a solution w(z) of Problem A for the mixedcomplex equation (18) as stated in (130) Chapter IV but the function λ(x) in theintegral formula in D+ should be replaced by λ(x) on L0 = (0 2) the function w(z)in Dminus should be replaced by w(z) in the second formula in (130) Chapter IV Hencewe have the following theorem
Theorem 11 Problem P for the mixed equation (12) has a unique solution in theform (110) where
w(z) = w(z) + λ(z1)[r(z1)minus ib1]
w(z) =
⎧⎪⎪⎨⎪⎪⎩W [ζ(z)] z isin D+0 2
12(1minus i)f(x+ y) +
12(1 + i)g(x minus y)
g(x minus y) =2r((1minus i)(x minus y)2)
a((1minus i)(x minus y)2)minus b((1minus i)(x minus y)2)
minus [a((1minusi)(xminusy)2)+b((1minusi)(xminusy)2)]f(0)a((1minus i)(x minus y)2)minus b((1minus i)(x minus y)2)
or
160 V Second Order Linear Mixed Equations
w(z) =12(1 + i)g(x minus y) +
12(1minus i)f(x+ y)
f(x+ y) =2r((1 + i)(x+ y)2 + 1minus i)
a((1 + i)(x+ y)2 + 1minus i) + b((1 + i)(x+y)2+1minusi)
minus [a((1+i)(x+y)2+1minusi)minusb((1+i)(x+y)2+1minusi)]g(2)a((1+i)(x+y)2+1minusi)+b((1+i)(x+y)2+1minusi)
z = x+ iy isin Dminus0 2
(111)
in which f(0) = [a(z1) + b(z1)]r(z1) + [a(z1) minus b(z1)]b1 g(2) = [a(z1) minus b(z1)]r(z1) minus[a(z1)+ b(z1)]b1 W (ζ) on D+0 2 is as stated in (128) Chapter IV but where thefunction λ(z) on L0 is as stated in (17) and λ(z) r(z) b1 are as stated in (13) (14)Moreover the functions
f(x+ y) = U(x+ y 0)minus V (x+ y 0) = Re [(1 + i)W (ζ(x+ y))]
g(x minus y) = U(x minus y 0) + V (x minus y 0) = Re [(1minus i)W (ζ(x minus y))](112)
where U = ux2 V = minusuy2 W [ζ(x+ y)] and W [ζ(xminus y)] are the values of W [ζ(z)]on 0 le z = x+ y le 2 and 0 le z = x minus y le 2 respectively
From the above representation of the solution u(z) of Problem P for (12) we canderive that u(z) satisfies the estimate
Cβ[u(z) D] + Cβ[w(z)X(z) D+] + C[wplusmn(z)Y plusmn(z) Dminus] le M1k2 (113)
in which k2 are as stated in (15) wplusmn(z) = Rew ∓ Imw and
X(z) =2prod
j=1|z minus tj|ηj Y plusmn(z) =
2prodj=1
|x plusmn y minus tj|ηj
ηj =2|γj|+ δ if γj lt 0
δ γj ge 0j = 1 2
(114)
here γj (j = 1 2) are real constants as stated in (17) and β δ (β lt δ) are sufficientlysmall positive constant andM1 = M1(p0 β k0 D
+) is a non-negative constant Fromthe estimate (113) we can see the regularity of solutions of Problem P for (12)
Finally we mention that if the index K is an arbitrary even integer or 2K is anarbitrary odd integer the above oblique derivative problem for (11) or (12) can beconsidered but in general these boundary value problems for K le minus1 have somesolvability conditions or their solutions for K ge 0 are not unique
12 The Dirichlet boundary value problem for simplest second orderequation of mixed type
The Dirichlet problem (Problem D or Problem T ) for (12) is to find a solution of(12) with the boundary conditions
u(z) = φ(z) on Γ cup Lj(j = 1 or 2) (115)
1 Simplest Mixed Equation of Second Order 161
where φ(z) satisfies the condition
C1[φ(z)Γ cup Lj] le k2 j = 1 or 2 (116)
In the following we shall explain that Problem D is a special case of Problem P Infact denote w = uz in D Problem D for the mixed equation (12) is equivalent toProblem A for the mixed equation (18) with the boundary condition (19) and therelation (110) in which
λ(z) = a+ ib =
⎧⎪⎪⎨⎪⎪⎩i(z minus 1) θ = arg(z minus 1) on Γ
1minus iradic2on L1 or
1 + iradic2on L2
r(z) =
⎧⎪⎪⎨⎪⎪⎩φθ on Γ
φxradic2on L1 or
φxradic2on L2
b1 = Im[1 + iradic2
uz(z1 minus 0)]=
φx minus φxradic2
|z=z1minus0 = 0 or
b1 = Im[1minus iradic2
uz(z1 + 0)]= 0 b0 = φ(0)
(117)
in which a = 1radic2 = b = minus1radic2 on L1 or a = 1
radic2 = minusb = minus1radic2 on L2
As for the index K = minus12 of Problem D on partD+ we can derive as followsAccording to (117) the boundary conditions of Problem D in D+ possess the form
Re [i(z minus 1)w(z)] = r(z) = φθ on Γ
Re[1minus iradic2
w(x)]= s(x) =
φprime((1minus i)x2)radic2
x isin [0 2] or
Re[1 + iradic2
w(x)]= s(x) =
φprime((1 + i)x2 + 1minus i)radic2
x isin [0 2]
it is clear that the possible discontinuous points of λ(z) on partD+ are t1 = 2 t2 = 0and
λ(t1 + 0) = e3πi2 λ(t2 minus 0) = eπi2
λ(t1 minus 0) = λ(t2 + 0) = eπi4 or λ(t1 minus 0) = λ(t2 + 0) = e7πi4
λ(t1 minus 0)λ(t1 + 0)
= eminus5πi4 = eiφ1 λ(t2 minus 0)λ(t2 + 0)
= eπi4 = eiφ2 or
λ(t1 minus 0)λ(t1 + 0)
= eπi4 = eiφ1 λ(t2 minus 0)λ(t2 + 0)
= eminus5πi4 = eiφ2
162 V Second Order Linear Mixed Equations
In order to insure the uniqueness of solutions of Problem D we choose that
minus1 lt γ1 =φ1
πminus K1 = minus5
4minus (minus1) = minus1
4lt 0
0 le γ2 =φ2
πminus K2 =
14
lt 1 or
0 le γ1 =φ1
πminus K1 =
14
lt 1
minus1 lt γ2 =φ2
πminus K2 = minus5
4minus (minus1) = minus1
4lt 0
thusK1 = minus1 K2 = 0 K =
K1 +K2
2= minus1
2 or
K1 = 0 K2 = minus1 K =K1 +K2
2= minus1
2
In this case the unique solution w(z) is continuous in Dlowast = D0 x minus y = 2 orDlowast = Dx + y = 0 2 for the first case w(z) in the neighborhood of t2 = 0 isbounded and w(z) in the neighborhood of t1 = 2 possesses the singularity in theform |z minus 2|minus12 and its integral (110) is bounded for the second case w(z) in theneighborhood of t1 = 2 is bounded and w(z) in the neighborhood of t2 = 0 possessesthe singularity of |z|minus12 and its integral is bounded If we require that the solutionw(z) = uz is bounded in D+0 2 then it suffices to choose the index K = minus1 inthis case the problem has one solvability condition
From Theorem 11 it follows that the following theorem holds
Theorem 12 Problem D for the mixed equation (12) has a unique continuoussolution in D as stated in (110) where λ(z) r(z) b1 are as stated in (117) and W (ζ)in D+0 2 is as stated in (128) Chapter IV but in which λ(x) = (1 + i)
radic2 or
(1minus i)radic2 on L0 and f(x+ y) g(x minus y) are as stated in (112) [85]15)
2 Oblique Derivative Problems for Second Order LinearEquations of Mixed Type
In this section we mainly discuss the uniqueness and existence of solutions for secondorder linear equations of mixed type
21 Formulation of the oblique derivative problem for mixed equationsof second order
Let D be a simply connected bounded domain D in the complex plane CI with theboundary partD = Γcup L as stated in Section 1 We consider the linear mixed equation
2 Oblique Derivative Problems 163
of second orderuxx + sgny uyy = aux + buy + cu+ d in D (21)
where a b c d are functions of z(isin D) its complex form is the following equation ofsecond order ⎧⎨⎩uzz = Re [A1(z)uz] + A2(z)u+ A3(z) in D+
uzlowastzlowast = Re [A1(z)uz] + A2(z)u+ A3(z) in Dminus(22)
where
z = x+ iy uz =12[ux minus iuy] uz =
12[ux + iuy] uzz =
14[uxx + uyy]
uzlowast =12[ux + iuy] = uz uzzlowast =
12[(uz)x minus i(uz)y] =
14[uxx minus uyy]
A1 =a+ ib
2 A2 =
c
4 A3 =
d
4in D
Suppose that equation (22) satisfies the following conditions Condition C
The coefficients Aj(z) (j = 1 2 3) in (22) are measurable in z isin D+ and continu-ous in Dminus and satisfy
Lp[Aj D+] le k0 j = 1 2 Lp[A3 D+] le k1 A2 ge 0 in D+ (23)
C[Aj Dminus] le k0 j = 1 2 C[A3 Dminus] le k1 (24)
where p (gt 2) k0 k1 are non-negative constants If the condition (24) is replaced by
Cα[Aj Dminus] le k0 j = 1 2 Cα[A3 Dminus] le k1
in which α(0 lt α lt 1) is a real constant then the conditions will be called ConditionC prime
The oblique derivative boundary value problem (Problem P ) for equation (22) isto find a continuously differentiable solution u(z) of (22) in Dlowast = D0 x minus y = 2or Dlowast = Dx + y = 0 2 which is continuous in D and satisfies the boundaryconditions (13) and (14) in which b0 b2 is a real constant satisfying the condition|b0| |b2| le k2 The index K is defined as stated in Section 1 now we discuss the case
K =12(K1 +K2) = 0 (25)
where
Kj =[φj
π
]+ Jj Jj = 0 or 1 eiφj =
λ(tj minus 0)λ(tj + 0)
γj =φj
πminus Kj j = 1 2 (26)
in which t1 = 2 t2 = 0 λ(t) = eiπ4 on L0 = (0 2) and λ(t1 minus 0) = λ(t2 + 0) =exp(iπ4) or λ(t) = ei7π4 on L0 and λ(t1 minus0) = λ(t2+0) = exp(i7π4) We mention
164 V Second Order Linear Mixed Equations
that if A2 = 0 in D or cos(l n) equiv 0 on Γ then we do not need the point conditionu(2) = b2 in (13) and only choose the index K = minus12 Because if cos(l n) equiv 0 onΓ from the boundary condition (13) we can determine the value u(2) by the valueu(0) namely
u(2)=2Reint 2
0uzdz+u(0)=2
int 2
0Re [i(z minus 1)uz]dθ+b0=2
int 0
πr(z)dθ+b0
in which λ(z) = i(z minus 1) θ = arg(z minus 1) on Γ In brief we choose that
K=
⎧⎪⎪⎨⎪⎪⎩0
minus12the point conditions are
⎧⎨⎩u(0)=b0 u(2)=b2
u(0) = b0
⎫⎬⎭ if
⎧⎨⎩ cos(l n) equiv0cos(l n)equiv0on Γ
In order to ensure that the solution u(z) of Problem P is continuously differentiablein Dlowast we need to choose γ1 gt 0 or γ2 gt 0 If we only require that the solution iscontinuous it suffices to choose minus2γ2 lt 1 minus2γ1 lt 1 respectively In the followingwe shall only discuss the case K = 0 and the case K = minus12 can be similarlydiscussed
22 The representation and uniqueness of solutions for the obliquederivative problem for (22)
Now we give the representation theorems of solutions for equation (22)
Theorem 21 Let equation (22) satisfy Condition C in D+ u(z) be a continuoussolution of (22) in D+ and continuously differentiable in D+
lowast = D+0 2 Thenu(z) can be expressed as
u(z) = U(z)Ψ(z) + ψ(z) in D+
U(z) = 2Reint z
0w(z)dz + b0 w(z) = Φ(z)eφ(z) in D+
(27)
where ψ(z) Ψ(z) are the solutions of equation (22) in D+ and
uzz minus Re [A1uz]minus A2u = 0 in D+ (28)
respectively and satisfy the boundary conditions
ψ(z) = 0 Ψ(z) = 1 on Γ cup L0 (29)
where ψ(z) Ψ(z) satisfies the estimates
C1β[ψD+] le M2 ψ W 2
p0(D+)le M2 (210)
C1β[Ψ D+] le M3 Ψ W 2
p0(D+)le M3Ψ(z) ge M4 gt 0 z isin D+ (211)
2 Oblique Derivative Problems 165
in which β (0 lt β le α) p0 (2 lt p0 le p) Mj = Mj(p0 β k D) (j = 2 3 4) arenon-negative constants k = (k0 k1 k2) Moreover U(z) is a solution of the equation
Uzz minus Re [AUz] = 0 A = minus2(lnΨ)z + A1 in D+ (212)
where Im [φ(z)] = 0 z isin L0 = (0 2) and φ(z) satisfies the estimate
Cβ[φD+] + Lp0 [φz D+] le M5 (213)
in which β(0 lt β le α) M5 = M5 (p0 β k0 D) are two non-negative constantsΦ(z) is analytic in D+ If u(z) is a solution of (22) in D+ satisfying the boundaryconditions (13) and
Re [λ(z)uz]|z=x = s(x) λ(x) = 1 + i or 1minus i x isin L0 Cβ[s(x) L0] le k3 (214)
then the following estimate holds
Cβ[u(z) D+] + Cβ[uzX(z) D+] le M6(k1 + k2 + k3) (215)
in which k3 is a non-negative constant s(x) can be seen as stated in the form (223)below X(z) is as stated in (114) and M6 = M6(p0 β k0 D
+) is a non-negativeconstant
Proof According to the method in the proof of Theorem 31 Chapter III theequations (22)(28) in D+ have the solutions ψ(z) Ψ(z) respectively which satisfythe boundary condition (29) and the estimates (210)(211) Setting that
U(z) =u(z)minus ψ(z)
Ψ(z) (216)
it is clear that U(z) is a solution of equation (212) which can be expressed the secondformula in (27) where φ(z) satisfies the estimate (213) and Φ(z) is an analyticfunction in D+ If s(x) in (214) is a known function then the boundary valueproblem (22)(13)(214) has a unique solution u(z) as stated in the form (27)which satisfies the estimate (215)
Theorem 22 Suppose that the equation (22) satisfies Condition C Then anysolution of Problem P for (22) can be expressed as
u(z) = 2Reint z
0w(z)dz + b0 w(z) = w0(z) +W (z) (217)
where w0(z) is a solution of Problem A for the complex equation (18) with the bound-ary conditions (13) (14)(w0(z) = u0z) and W (z) possesses the form
W (z) = w(z)minus w0(z) in D w(z) = Φ(z)eφ(z) + ψ(z) in D+
φ(z) = φ0(z) + Tg = φ0(z)minus 1π
int intD+
g(ζ)ζ minus z
dσζ ψ(z) = Tf in D+
W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν
2g1(z)dνe1 +
int micro
0g2(z)dmicroe2 in Dminus
(218)
166 V Second Order Linear Mixed Equations
in which e1 = (1 + i)2 e2 = (1 minus i)2 micro = x + y l = x minus y φ0(z) is an analyticfunction in D+ such that Im [φ(x)] = 0 on L0 and
g(z)=
A12+A1w(2w) w(z) =00 w(z)=0 zisinD+
f(z)=Re[A1φz]+A2u+A3 in Dminus
g1(z)=g2(z)=Aξ+Bη+Cu+D ξ=Rew+Imw η=RewminusImw
A=ReA1+ImA1
2 B=
ReA1minusImA1
2 C=A2D=A3 in Dminus
(219)
where Φ(z) and Φ(z) are the solutions of equation (18) in D+ and Dminus respectivelysatisfying the boundary conditions
Re [λ(z)(Φ(z)eφ(z) + ψ(z))] = r(z) z isin Γ
Re [λ(x)(Φ(x)eφ(x) + ψ(x))] = s(x) x isin L0
Re [λ(x)Φ(x)] = Re [λ(x)(W (x)minusΨ(x))] z isin L0
Re [λ(z)Φ(z)] = minusRe [λ(z)Ψ(z)] z isin L1 or L2
Im [λ(z1)Φ(z1)] = minusIm [λ(z1)Ψ(z1)]
(220)
where λ(x) = 1 + i or 1 minus i x isin L0 Moreover by Theorem 12 Chapter IV thesolution w0(z) of Problem A for (18) and u0(z) satisfy the estimate in the form
Cβ[u0(z) D]+Cβ[w0(z)X(z) D+]+Cβ[wplusmn0 (z)Y
plusmn(z) Dminus]leM7(k1+k2) (221)
in which wplusmn(z) = Rew(z)∓ Imw(z) X(z) Y plusmn(z) are as stated in (114)
u0(z) = 2Reint z
0w0(z)dz + b0 (222)
and M7 = M7(p0 β k0 D) is a non-negative constant From (222) it follows that
Cβ[u0(z) D] le M8Cβ[w0(z)X(z) D+] + Cβ[wplusmn0 (z)Y
plusmn(z) Dminus+ k2
where M8 = M8(D) is a non-negative constant
Proof Let u(z) be a solution of Problem P for equation (22) and w(z) = uzu(z) be substituted in the positions of w u in (219) thus the functions g(z)f(z) g1(z) g2(z) and ψ(z) φ(z) in D+ and Ψ(z) in Dminus in (218)(219) can bedetermined Moreover we can find the solution Φ(z) in D+ and Φ(z) in Dminus of (18)with the boundary conditions (220) where
s(x)=
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
2r((1minusi)x2)minus2R((1minusi)x2)a((1minus i)x2)minus b((1minus i)x2)
+Re [λ(x)Ψ(x)] or
2r((1+i)x2+1minusi)minus2R((1+i)x2+1minusi)a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)
+Re [λ(x)Ψ(x)] on L0
(223)
2 Oblique Derivative Problems 167
here and later R(z) = Re [λ(z)Ψ(z)] on L1 or L2 thus
w(z) = w0(z) +W (z) =
⎧⎪⎨⎪⎩Φ(z)φ(z) + ψ(z) in D+
w0(z) + Φ(z) + Ψ(z) in Dminus
is the solution of Problem A for the complex equation
wz
wzlowast
= Re [A1w] + A2u+ A3 in
⎧⎨⎩D+
Dminus
⎫⎬⎭ (224)
which can be expressed as the second formula in (217) and u(z) is a solution ofProblem P for (22) as stated in the first formula in (217)
Theorem 23 If equation (22) satisfies Condition C then Problem P for (22) hasat most one solution in D
Proof Let u1(z) u2(z) be any two solutions of Problem P for (22) By ConditionC we see that u(z) = u1(z)minusu2(z) and w(z) = uz satisfies the homogeneous equationand boundary condition
wz
wzlowast
= Re [A1w] + A2u in
D+
Dminus
(225)
Re [λ(z)w(z)] = 0 z isin Γ u(0) = 0 u(2) = 0
Re [λ(z)w(z)] = 0 z isin Lj(j = 1 or 2) Im [λ(z1)w(z1)] = 0(226)
From Theorem 22 the solution w(z) can be expressed in the form
w(z) =
⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
Φ(z)eφ(z) + ψ(z) ψ(z) = Tf φ(z) = φ0(z) + T g in D+
w0(z) + Φ(z) + Ψ(z)
Ψ(z) =int ν
2[Aξ+Bη+Cu]e1dν+
int micro
0[Aξ+Bη+Cu]e2dmicro in Dminus
(227)
where g(z) is as stated in (219) Φ(z) in D+ is an analytic function and Φ(z) is asolution of (18) in Dminus satisfying the boundary condition (220) φ(z) ψ(z) possessthe similar properties as φ(z) ψz(z) in Theorem 21 If A2 = 0 in D+ then ψ(z) = 0Besides the functions Φ(z) Φ(z) satisfy the boundary conditions⎧⎨⎩Re [λ(x)Φ(x)] = s(x)
Re [λ(x)Φ(x)] = Re [λ(x)(W (x)minusΨ(x))]on L0 (228)
where s(x) is as stated in (223) From (217) with b0 = 0 we can obtain
C[u(z) D] le M8C[w(z)X(z) D+] + C[wplusmn0 (z)Y
plusmn(z) Dminus (229)
168 V Second Order Linear Mixed Equations
By using the method of iteration the estimate
C[w(z) Dminus] le [2M9M(4m+ 1)Rprime]n
n(230)
can be derived where M9 = maxC[A Dminus] C[BDminus] C[CDminus] M = 1+ 4k20(1 +
k20) and m = C[w(z) Dminus] gt 0 Let n rarr infin from (229) it follows that w(z) = 0 in
Dminus and Ψ(z) = 0 Φ(z) = 0 z isin Dminus Thus the solution u(z) = 2Reint z0 w(z)dz is
the solution of equation (28) with the boundary conditions
Re [λ(z)uz]=0 on Γ Re [λ(x)uz(x)]=0 on L0=(0 2) u(0)=0 u(2)=0(231)
in which λ(x) = 1 + i or 1 minus i x isin L0 Similarly to the proof of Theorem 34Chapter III we can obtain u(z) = 0 on D+ This shows the uniqueness of solutionsof Problem P for (22)
23 The solvability of the oblique derivative problem for (22)
Theorem 24 Suppose that the mixed equation (22) satisfies Condition C ThenProblem P for (22) has a solution in D
Proof It is clear that Problem P for (22) is equivalent to Problem A for thecomplex equation of first order and boundary conditions
wz
wzlowast
= F F = Re [A1w] + A2u+ A3 in
⎧⎨⎩D+
Dminus
⎫⎬⎭ (232)
Re [λ(z)w(z)] = r(z) z isin Γ
Re [λ(z)w(z)] = r(z) z isin Lj(j = 1 or 2) Im [λ(z1)w(z1)] = b1(233)
and the relation (217) From (217) it follows that
C[u(z) D] le M8[C(w(z)X(z) D+) + C(wplusmn(z)Y plusmn(z) Dminus)] + k2 (234)
where X(z) Y plusmn(z) wplusmn(z) are as stated in (114) respectively M8 = M8(D) is anon-negative constant In the following by using successive iteration we shall find asolution of Problem A for the complex equation (232) inD Firstly denoting the solu-tion w0(z)(= ξ0e1+η0e2) of Problem A for (18) and u0(z) in (217) and substitutingthem into the position of w = (ξe1+ ηe2) u(z) in the right-hand side of (232) simi-larly to (218)(219) we have the corresponding functions f1(z) g1(z) g1
2(z) g12(z)
and
w1(z) = Φ1(z)eφ1(z) + ψ1(z) in D+
φ1(z)= φ0(z)+Tg1= φ0(z)minus 1π
int intD+
g1(ζ)ζminusz
dσζ ψ1(z)=Tf1 in D+
W1(z)=Φ(z)+Ψ(z) Ψ(z)=int ν
2g11(z)dνe1+
int micro
0g21(z)dmicroe2 in Dminus
(235)
2 Oblique Derivative Problems 169
where micro = x + y ν = x minus y where Φ1(z) is a solution of (18) in Dminus satisfying theboundary conditions
Re [λ(x)Φ1(x)] = Re [λ(x)(W1(z)minusΨ1(x))] z isin L0
Re [λ(z)Φ1(z)] = minusRe [λ(z)Ψ1(z)] z isin L1 or L2
Im [λ(z1)Φ1(z1)] = minusIm [λ(z1)Ψ1(z1)]
(236)
andw1(z) = w0(z) +W1(z) = w0(z) + Φ1(z) + Ψ1(z) in Dminus (237)
satisfies the estimate
Cβ[w(z)X(z) D+] + C[wplusmn1 (z)Y
plusmn(z) Dminus] le M10 = M10(p0 β k Dminus) (238)
Furthermore we substitute w1(z) = w0(z)+W1(z) and corresponding functions w1(z)ξ1(z) = w+(z) = Rew1(z)minusImw1(z) η1(z) = wminus(z) = Rew1(z)+Imw1(z) u1(z) intothe positions w(z) ξ(z) η(z) u(z) in (218)(219) and similarly to (235)ndash(237)we can find the corresponding functions ψ2(z) φ2(z) Φ2(z) in D+ Ψ2(z)Φ2(z) andW2(z) = Φ2(z) + Ψ2(z) in Dminus and the function
w2(z) = Φ2(z)eφ2(z) + ψ2(z) in D+
w2(z) = w0(z) +W2(z) = w0(z) + Φ2(z) + Ψ2(z) in Dminus(239)
satisfies the similar estimate in the form (238) Thus there exists a sequence offunctions wn(z) and
wn(z) = Φn(z)eφn(z) + ψn(z) in D+
wn(z) = w0(z) +Wn(z) = w0(z) + Φn(z) + Ψn(z)
Ψn(z)=int ν
2g1
n(z)e1dν +int micro
0g2
n(z)e2dmicro in Dminus
(240)
and then
|[wplusmn1 (z)minus wplusmn
0 (z)]Y plusmn(z)| le |Φplusmn1 (z)Y plusmn(z)|
+radic2[|Y +(z)
int ν
2[Aξ0 +Bη0 + Cu0 +D]e1dν|
+|Y minus(z)int micro
0[Aξ0 +Bη0 + Cu0 +D]e2dmicro|] le 2M11M(4m+ 1)Rprime in Dminus
(241)
where M11 = maxzisinDminus(|A| |B| |C| |D|) m = C[w0(z)X(z) Dminus] Rprime = 2 M = 1 +4k2
0(1 + k20) It is clear that wn(z)minus wnminus1(z) satisfies
wn(z)minus wnminus1(z)
= Φn(z)minus Φnminus1(z) +int ν
2[A(ξn minus ξnminus1) +B(ηn minus ηnminus1) + C(un minus unminus1)]e1dν
+int micro
0[A(ξn minus ξnminus1) +B(ηn minus ηnminus1) + C(un minus unminus1)]e2dmicro in Dminus
(242)
170 V Second Order Linear Mixed Equations
where n = 1 2 From the above equality the estimate
|[wplusmnn minus wplusmn
nminus1]Y plusmn(z)| le [2M11M(4m+ 1)]n timesint Rprime
0
Rprimenminus1
(n minus 1) dRprime
le [2M11M(4m+ 1)Rprime]n
n in Dminus
(243)
can be obtained and then we can see that the sequence of functions wplusmnn (z)Y
plusmn(z)ie
wplusmnn (z)Y
plusmn(z) = wplusmn0 (z) + [w
plusmn1 (z)minus wplusmn
0 (z)] + middot middot middot+ [wplusmnn (z)minus wplusmn
nminus1(z)]Y plusmn(z) (244)
(n = 1 2 ) in Dminus uniformly converge to wplusmnlowast (z)Y
plusmn(z) and wlowast(z) = [w+lowast (z) +
wminuslowast (z)minus i(w+
lowast (z)minus wminuslowast (z))]2 satisfies the equality
wlowast(z) = w0(z) + Φlowast(z) + Ψlowast(z)
Ψlowast(z) =int ν
2[Aξlowast+Bηlowast+Culowast+D]e1dν+
int micro
0[Aξlowast+Bηlowast+Culowast+D]e2dmicro in Dminus
(245)
and the corresponding function ulowast(z) is just a solution of Problem P for equation(22) in the domain Dminus and wlowast(z) satisfies the estimate
C[wplusmnlowast (z)Y
plusmn(z) Dminus] le e2M11M(4m+1)Rprime (246)
In the meantime we can obtain the estimate
Cβ[wn(z)X(z) D+] le M12 = M12(p0 β k D) (247)
hence from the sequence wn(z) we can choose a subsequence which uniformlyconverges to wlowast(z) in D+ and wlowast(z) satisfies the same estimate (247) Combining(246) and (247) it is obvious that the solution wlowast(z) = uz of Problem A for (22)in D satisfies the estimate
Cβ[wlowast(z)X(z) D+] + C[wplusmnlowast (z)Y
plusmn(z) Dminus] le M13 = M13(p0 β k D)
where M13 is a non-negative constant Moreover the function u(z) in (217) is asolution of Problem P for (22) where w(z) = wlowast(z)
From Theorems 23 and 24 we see that under Condition C Problem A forequation (232) has a unique solution w(z) which can be found by using successiveiteration and the corresponding solution u(z) of Problem P satisfies the estimates
Cβ[u(z) D+] + Cβ[uzX(z) D+] le M14
C[u(z) D] + C[uplusmnz Y plusmn(z) D] le M15
(248)
where X(z) Y plusmn(z) is as stated in (114) and Mj = Mj (p0 β k D) (j = 14 15)are non-negative constants k = (k0 k1 k2) Moreover we can derive the followingtheorem
3 Discontinuous Oblique Derivative Problems 171
Theorem 25 Suppose that equation (22) satisfies Condition C Then any solutionu(z) of Problem P for (22) satisfies the estimates
Cβ[u(z) D+] + Cβ[uzX(z) D+] le M16(k1 + k2)
C[u(z) Dminus] + C[uplusmnz Y plusmn(z) Dminus] le M17(k1 + k2)
(249)
in which Mj = Mj(p0 β k0 D) (j = 16 17) are non-negative constants
From the estimates (248)(249) we can see that the regularity of solutions ofProblem P for (22) (see [85]15))
3 Discontinuous Oblique Derivative Problems for SecondOrder Linear Equations of Mixed Type
This section deals with an application of method of integral equations to second orderequations of mixed type We mainly discuss the discontinuous Poincare boundaryvalue problem for second order linear equation of mixed (elliptic-hyperbolic) type iethe generalized Lavrentprimeev-Bitsadze equation with weak conditions by the methodof integral equations We first give the representation of solutions for the aboveboundary value problem and then give the solvability conditions of the above problemby the Fredholm theorem for integral equations
31 Formulation of the discontinuous Poincare problem for mixedequations of second order
Let D be a simply connected bounded domain D in the complex plane CI with theboundary partD = Γ cup L as stated in Section 1 We consider the second order linearequation of mixed type (21) and its complex form (22) with Condition C prime
In order to introduce the discontinuous Poincare boundary value problem forequation (22) let the functions a(z) b(z) possess the discontinuities of first kind atm+ 2 distinct points z0 = 2 z1 zm+1 = 0 isin Γ and Z = z0 z1 zm+1 whichare arranged according to the positive direction of Γ where m is a positive integerand r(z) = O(|z minus zj|minusβj) in the neighborhood of zj(j = 0 1 m + 1) on Γ inwhich βj(j = 0 1 m+1) are small positive numbers Denote λ(z) = a(x)+ ib(x)and |a(x)|+ |b(x)| = 0 there is no harm in assuming that |λ(z)| = 1 z isin Γlowast = ΓZSuppose that λ(z) r(z) satisfy the conditions
λ(z) isin Cα(Γj) |z minus zj|βjr(z) isin Cα(Γj) j = 0 1 m+ 1 (31)
herein Γj is an arc from the point zjminus1 to zj on Γ and zm+1 = 0 and Γj(j =0 1 m+ 1) does not include the end points and α (0 lt α lt 1) is a constant
Problem Q Find a continuously differentiable solution u(z) of (22) in Dlowast =DZ(Z = 0 x minus y = 2 y le 0 or Z = x+ y = 0 y le 0 2) which is continuous in
172 V Second Order Linear Mixed Equations
D and satisfies the boundary conditions
12
partu
partl+εσ(z)u=Re [λ(z)uz]+εσ(z)u=r(z)+Y (z)h(z) zisinΓ u(0)=b0 (32)
12
partu
partl= Re [λ(z)uz] = r(z) z isin L1 or L2 Im [λ(z)uz]|z=z1 = b1 (33)
where l is a vector at every point on Γ cup Lj (j = 1 or 2) z1 = 1 minus i b0 b1 are realconstants λ(z) = a(x)+ib(x) = cos(l x)minusi cos(l y) z isin Γ and λ(z) = a(x)+ib(x) =cos(l x) + i cos(l y) z isin Lj (j = 1 or 2) and λ(z) r(z) b0 b1 satisfy the conditions
Cα[λ(z)Γ] le k0 Cα[σ(z)Γ] le k0 Cα[r(z)Γ] le k2 |b0| |b1| le k2
Cα[λ(z) Lj] le k0 Cα[σ(z) Lj] le k0 Cα[r(z) Lj] le k2 j = 1 or 2
maxzisinL1
1|a(x)minus b(x)| or max
zisinL2
1|a(x) + b(x)| le k0
(34)
in which α (12 lt α lt 1) k0 k2 are non-negative constants ε is a real parameterBesides the functions Y (z) h(z) are as follows
Y (z)=ηm+1prodj=0
|z minus zj|γj |zminuszlowast|l z isin Γlowast h(z)=
⎧⎨⎩ 0zisinΓ ifK geminus12hjηj(z) zisinΓj ifKltminus12
(35)
in which Γj(j = 0 1 m) are arcs on Γlowast = ΓZ and ΓjcapΓk = φ j = k hj isin J (J =φ if K ge minus12 J = 1 2K prime minus 1 if K lt minus12 K prime = [|K| + 12]) are unknownreal constants to be determined appropriately herein h1 = 0 l = 1 if 2K is oddzlowast(isin Z) isin Γlowast is any fixed point and l = 0 if 2K is even Γj(j = 1 2K prime minus 1) arenon-degenerate mutually disjointed arcs on Γ and Γj cap Z = φ j = 1 2K prime minus 1ηj(z) is a positive continuous function on the interior point set of Γj such thatηj(z) = 0 on ΓΓj and
Cα[ηj(z)Γ] le k0 j = 1 2K prime minus 1 (36)
and η = 1 or minus1 on Γj (0 le j le m + 1Γm+1 = (0 2)) as stated in [93] Theabove discontinuous Poincare boundary value problem for (22) is called Problem QProblem Q for (22) with A3(z) = 0 z isin D r(z) = 0 z isin Γ cup Lj (j = 1 or 2) andb0 = b1 = 0 will be called Problem Q0
Denote by λ(zj minus 0) and λ(zj + 0) the left limit and right limit of λ(z) as z rarrzj (0 le j le m+ 1) on Γ cup L0 and
eiφj =λ(zj minus 0)λ(zj + 0)
γj =1πiln
λ(zj minus 0)λ(zj + 0)
=φj
πminus Kj
Kj =[φj
π
]+ Jj Jj = 0 or 1 j = 0 1 m+ 1
(37)
3 Discontinuous Oblique Derivative Problems 173
in which zm+1 = 0 z0 = 2 λ(z) = eiπ4 on L0 = (0 2) and λ(z0 minus0) = λ(zm+1+0) =exp(iπ4) or λ(z) = eminusiπ4 on L0 and λ(z0 minus 0) = λ(zm+1 + 0) = exp(minusiπ4) and0 le γj lt 1 when Jj = 0 and minus1 lt Jj lt 0 when Jj = 1 j = 0 1 m+ 1 and
K =12(K0 +K2 + middot middot middot+Km+1) =
m+1sumj=0
(φj
2πminus γj
2
)(38)
is called the index of Problem Q and Problem Q0 Let βj+γj lt 1 j = 0 1 m+1we can require that the solution u(z) satisfy the condition uz = O(|z minus zj|minusδj) in theneighborhood of zj (j = 0 1 m+ 1) in Dlowast where
τj=
⎧⎨⎩βj+τ for γj ge0 and γj lt0 βj gt |γj||γj|+τ for γj lt0 βj le|γj|
δj=
⎧⎨⎩ 2τj j=0 m+1
τj j=1 m(39)
and τ δ(lt τ) are small positive numbers In order to ensure that the solution u(z)of Problem Q is continuously differentiable in Dlowast we need to choose γ1 gt 0 or γ2 gt 0respectively
32 The representation and solvability of the oblique derivative problemfor (22)
Now we write a representation theorem of solutions for equation (22) which is similarto Theorem 22
Theorem 31 If equation (22) satisfies Condition C prime and ε = 0 A2 ge 0 in D+then any solution of Problem Q for (22) can be expressed as
u(z) = 2Reint z
0w(z)dz + c0 w(z) = w0(z) +W (z) (310)
where w0(z) is a solution of Problem A for equation (18) with the boundary conditions
Re [λ(z)w(z)] = r(z) + Y (z)h(z) z isin Γ
Re [λ(z)w(z)] = r(z) z isin Lj (j = 1 or 2) Im [λ(z)w(z)]|z=z1 = b1(311)
and W (z) possesses the form
W (z) = w(z)minus w0(z) W (z) = Φ(z)eφ(z) + ψ(z)
φ(z) = φ0(z) + Tg = φ0(z)minus 1π
int intD+
g(ζ)ζ minus z
dσζ ψ(z) = Tf in D+
W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν
2g1(z)dνe1 +
int micro
0g2(z)dmicroe2 in Dminus
(312)
174 V Second Order Linear Mixed Equations
in which e1 =1 + i
2 e2 =
1minus i
2 micro = x+ y ν = x minus y and
g(z)=A12+A1w(2w) w(z) = 00 w(z) = 0 z isin D+
f(z)=Re [A1φz]+A2u+A3 in D+
g1(z)=g2(z)=Aξ+Bη+Cu+D ξ=RewminusImw η=Rew+Imw
A=ReA1+ImA1
2 B=
ReA1minusImA1
2 C=A2 D=A3 in Dminus
(313)
where φ0(z) is an analytic function in D+ such that Im [φ(x)] = 0 on L0 = (0 2) andΦ(z)Φ(z) are the solutions of the equation (18) in D+ Dminus respectively satisfyingthe boundary conditions
Re [λ(z)eφ(z)Φ(z)] = r(z)minus Re [λ(z)ψ(z)] z isin Γ
Re [λ(x)(Φ(x)eφ(x) + ψ(x))] = s(x) x isin L0
Re [λ(x)Φ(x)] = Re [λ(x)(W (x)minusΨ(x))] z isin L0
Re [λ(z)Φ(z)] = minusRe [λ(z)Ψ(z)] z isin L1 or L2
Im [λ(z1)Φ(z1)] = minusIm [λ(z1)Ψ(z1)]
(314)
where
s(x)=
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
2r((1minusi)x2)minus2R((1minusi)x2)a((1minus i)x2)minus b((1minus i)x2)
+Re [λ(x)Ψ(x)] or
2r((1+i)x2+1minusi)minus2R((1+i)x2+1minusi)a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)
+Re [λ(x)Ψ(x)] on L0
(315)
in which s(x) can be written similar to (29) Moreover from Theorem 11 if theindex K le minus12 the solution u0(z)(w0(z) = u0z(z)) of Problem Q for (12) satisfiesthe estimate in the form
Cδ[u0(z) D] + Cδ[w0(z)X(z) D+] + C[wplusmn0 (z)Y
plusmn(z) Dminus] le M18(k1 + k2) (316)
in which δ is a small positive constant p0 (2 lt p0 le p) M18 = M18 (p0 δ k0 D) aretwo non-negative constants
wplusmn0 (z)=Rew0(z)∓ Imw0(z) X(z)=Πm+1
j=0 |zminustj|ηj Y plusmn(z)=prod2
j=1 |x plusmn yminustj|ηj
ηj = 2|γj|+ δ j = 0 m+ 1 ηj = |γj|+ δ j = 1 m
andu0(z) = 2Re
int z
0w0(z)dz + c0 (317)
3 Discontinuous Oblique Derivative Problems 175
In order to prove the solvability of ProblemQ for (22) denote w = uz and considerthe equivalent boundary value problem (Problem B) for the mixed complex equation⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩
wz minus Re [A1(z)w] = εA2(z)u+ A3(z) z isin D+
wzlowast minus Re [A1(z)w] = A3(z) z isin Dminus
u(z) = 2Reint z
0w(z)dz + b0
(318)
with the boundary conditions
Re [λ(z)w] = r(z)minus εσ(z)u+ Y (z)h(z) z isin Γ
Re [λ(z)uz] = r(z) z isin Lj (j = 1 or 2) Im [λ(z)uz]|z=z1 = b1(319)
where b0 b1 are real constants are as stated in (32)(33) According to the method inSection 5 Chapter IV we can find the general solution of Problem B1 for the mixedcomplex equation ⎧⎨⎩wz minus Re [A1(z)w] = A3(z) z isin D+
wzlowast minus Re [A1(z)w] = A3(z) z isin Dminus(320)
with the boundary conditions
Re [λ(z)w(z)] = r(z) + Y (z)h(z) z isin Γ
Re [λ(z)w(z)] = r(z) z isin Lj (j = 1 or 2) Im [λ(z)w(z)]|z=z1 = b1(321)
which can be expressed as
w(z) = w0(z) +2K+1sumk=1
ckwk(z) (322)
in which w0(z) is a special solution of Problem B1 and wk(z)(k = 1 2K + 1K ge 0) is the complete system of linear independent solutions for the homogeneousproblem of Problem B1 Moreover denote by H2u the solution of Problem B2 for thecomplex equation ⎧⎨⎩wz minus Re [A1(z)w] = A2(z)u z isin D+
wzlowast minus Re [A1(z)w] = A2(z)u z isin Dminus(323)
with the boundary conditions
Re [λ(z)w(z)] = minusσ(z)u+ Y (z)h(z) z isin Γ
Re [λ(z)w(z)] = 0 z isin Lj (j = 1 or 2) Im [λ(z)w(z)]|z=z1 = 0(324)
176 V Second Order Linear Mixed Equations
and the point conditions
Im [λ(aj)w(aj)] = 0 j isin J =
⎧⎨⎩ 1 2K + 1 K ge 0
φ K lt 0(325)
where aj isin ΓZ are the distinct points It is easy to see that H2 is a boundedoperator from u(z) isin C1(D) (ie C(u D)+C(X(z)uz D+) + C(Y plusmn(z)uplusmn
z Dminus) lt infin)to w(z) isin Cδ(D) (ie Cδ(u D) + Cδ(X(z)w(z) D+) + Cδ(Y plusmn(z)wplusmn(z) Dminus) lt infin)herein X(z) Y plusmn(z) are functions as stated in (316) Furthermore denote
u(z) = H1w + c0 = 2Reint z
0w(z)dz + c0 (326)
where c0 is arbitrary real constant It is clear that H1 is a bounded operator fromX(z)w(z) isin Cδ(D) to u(z) isin C1(D) On the basis of Theorem 31 the function w(z)can be expressed as an integral From (326) and w(z) = w(z)+ εH2u we can obtaina nonhomogeneous integral equation (K ge 0)
u minus εH1H2u = H1w(z) + c0 +2K+1sumk=1
ckH1wk(z) (327)
Due to H1H2 is a completely continuous operator in C1(D) we can use the Fredholmtheorem for the integral equation (327) Denote by
εj(j = 1 2 ) 0 lt |ε1| le |ε2| le middot middot middot le |εn| le |εn+1| le middot middot middot (328)
are the discrete eigenvalues for the homogeneous integral equation
u minus εH1H2u = 0 (329)
Noting that Problem Q for the complex equation (22) with ε = 0 is solvable hence|ε1| gt 0 In the following we first discuss the case of K ge 0 If ε = εj(j =1 2 ) ie it is not an eigenvalue of the homogeneous integral equation (329)then the nonhomogeneous integral equation (327) has a solution u(z) and the generalsolution of Problem Q includes 2K + 2 arbitrary real constants If ε is an eigenvalueof rank q as stated in (328) applying the Fredholm theorem we obtain the solv-ability conditions for nonhomogeneous integral equation (327) there is a system ofq algebraic equations to determine the 2K + 2 arbitrary real constants setting thats is the rank of the corresponding coefficients matrix and s le min(q 2K +2) we candetermine s equalities in the q algebraic equations hence Problem Q for (22) hasq minus s solvability conditions When these conditions hold then the general solutionof Problem Q includes 2K + 2 + q minus s arbitrary real constants As for the case ofK lt 0 it can be similarly discussed Thus we can write the above result as in thefollowing theorem
Theorem 32 Suppose that the linear mixed equation (22) satisfies Condition C primeIf ε = εj (j = 1 2 ) where εj(j = 1 2 ) are the eigenvalues of the homogeneousintegral equation (329) Then
4 Frankl Boundary Value problem 177
(1) When K ge 0 Problem Q for (22) is solvable and the general solution u(z)of Problem Q for (22) includes 2K + 2 arbitrary real constants
(2) When K lt0 Problem Q for (22) has minus2Kminus1minuss solvability conditions sle1If ε is an eigenvalue of homogeneous integral equation (329) with the rank q
(3) When K ge 0 Problem Q for (22) has q minus s solvability conditions and s lemin (q 2K + 2)
(4) When K lt 0 Problem Q for (22) has minus2K minus 1 + q minus s solvability conditionsand s le min (minus2K minus 1 + q 1 + q)
Moreover we can derive the solvability result of Problem P for equation (22) withthe boundary condition (32) in which h(z) = 0
4 The Frankl Boundary Value Problem for Second OrderLinear Equations of Mixed Type
This section deals with the Frankl boundary value problem for linear second orderequations of mixed (elliptic-hyperbolic) type ie for generalized Lavrentprimeev-Bitsadzeequations We first give representation formula and prove uniqueness of solutions forthe above boundary value problem moreover we obtain a priori estimates of solutionsfinally by the method of parameter extension the existence of solutions is proved Inthe books [12]1)3) the Frankl problem was discussed for the special mixed equationsof second order uxx+sgny uyy = 0 In the book [73] the Frankl problem was discussedfor the mixed equation with parabolic degeneracy sgny|y|muxx + uyy = 0 which is amathematical model of problem of gas dynamics There the existence of solutions ofFrankl problem was proved by using the method of integral equations In this sectionwe will not use this method We are proving the solvability of the Frankl problemfor generalized linear Lavrentprimeev-Bitsadze equations generalizing the correspondingresult from [12]1)3)
41 Formulation of the Frankl problem for second order equations ofmixed type
Let D be a simply connected bounded domain in the complex plane CI with theboundary partD = Γ cup AprimeA cup AprimeC cup CB where Γ(sub x gt 0 y gt 0) isin C2
micro(0 lt micro lt 1)with the end points A = i and B = a AprimeA = x = 0 minus1 le y le 1 AprimeC = x minus y =1 x gt 0 y lt 0 is the characteristic line and CB = 1 le x le a y = 0 and denoteD+ = D cap y gt 0 Dminus = D cap y lt 0 Without loss of generality we may assumethat Γ = x2a2 + y2 = 1 x gt 0 y gt 0 otherwise through a conformal mappingfrom D+ onto the domain Dprime+ = x2a2 + y2 lt 1 x gt 0 y gt 0 such that threeboundary points i 0 1 are not changed then the above requirement can be realized
178 V Second Order Linear Mixed Equations
Frankl Problem Find a continuously differentiable solution u(z) of equation(22) in Dlowast = D1 a i minusi x + y = 0 which is continuous in D and satisfies theboundary conditions
u = ψ1(s) on Γ (41)
u = ψ2(x) on CB (42)
partu
partx= 0 on AprimeA (43)
u(iy)minus u(minusiy) = φ(y) minus1 le y le 1 (44)
Here ψ1(s) ψ2(x) φ(y) are given real-valuedfunctions satisfying the conditions
C1α[ψ1(s)S]lek2 C1
α[ψ2(x)CB]lek2
C1α[φ(y)A
primeA]lek2 ψ1(0)=ψ2(a)(45)
in which S = 0 le s le l s is the arc length parameter on Γ normalized suchthat s = 0 at the point B l is the length of Γ and α (0 lt α lt 1) k2 are non-negative constants The above boundary value problem is called Problem F and thecorresponding homogeneous problem is called Problem F0
LetU =
12ux V = minus1
2uy W = U + iV in D (46)
then equation (22) can be written as the complex equationWz
W zlowast
= Re [A1W ] + A2u+ A3 in
⎧⎨⎩D+
Dminus
⎫⎬⎭
u(z) = 2Reint z
aW (z)dz + ψ1(0)
(47)
If A1 = A2 = A3 = 0 in D then it is clear that
U(x y) =12ux =
12[f(x+ y) + g(x minus y)]
minusV (x y) =12uy =
12[f(x+ y)minus g(x minus y)]
(48)
in Dminus From the boundary conditions (41)ndash(43) it follows that
U(0 y) =12
partu
partx= 0 minusV (0 y) =
12[u(0 y)]y =
12[u(0 minusy)]y +
12φprime(y)
= V (0 minusy) +12φprime(y) = minusF (y) +
12φprime(y)
F (y) = minusV (0 minusy) minus1 le y le 0
(49)
4 Frankl Boundary Value problem 179
and then
U(0 y) =12[f(y) + g(minusy)] = 0 minus1 le y le 0
minusV (0 y) =12[f(y)minus g(minusy)] = minusF (y) +
12φprime(y) minus1 le y le 0
U(0 y) + V (0 y) = g(minusy) = F (y)minus 12φprime(y) minus1 le y le 0
U(0 y)minus V (0 y) = f(y) = minusF (y) +12φprime(y) minus1 le y le 0
f(y) = minusg(minusy) f(y) = g(minusy)minus 2F (y) + φprime(y) minus1 le y le 0 ie
f(y minus x) = minusg(x minus y) f(y minus x) = g(x minus y)minus 2F (y minus x) + φprime(y minus x)
U(x y) + V (x y) = g(x minus y) = F (y minus x)minus 12φprime(y minus x) 0 le x minus y le 1
U(x y)minus V (x y) = f(x+ y) = minusg(minusx minus y)
= minusF (x+ y) +12φprime(x+ y) 0 le minusx minus y le 1
(410)
Hence
U(x y) =12[f(x+ y)minus f(y minus x)] 0 le x minus y le 1
minusV (x y)=12[f(x+y)minusf(yminusx)]minusF (y minus x)+
12φprime(yminusx) 0lexminusyle1
U(x 0) + V (x 0) = g(x) = F (minusx)minus 12φprime(minusx) 0 le x le 1
(411)
In particular we have
U(x 0) =12[f(x)minus f(minusx)] =
12[f(x) + F (minusx)minus 1
2φprime(minusx)]
minusV (x 0) =12[f(x)minus f(minusx)]minus F (minusx) +
12φprime(minusx)
=12[f(x)minus F (minusx) +
12φprime(minusx)] on OC
(412)
The boundary conditions of the Frankl problem are
partu
partl=2Re [λ(z)W (z)] = r(z) z isin Γ cup CB u(a) = b0 = ψ1(0)
U(0 y) =12
partu
partx= r(0 y) = Re [λ(iy)W (iy)] = 0 minus1 le y le 1
(413)
Re [λ(x)W (x)] = r(x)=1radic2[F (minusx)minus 1
2φprime(minusx)] xisinL0=(0 1) (414)
180 V Second Order Linear Mixed Equations
in which l is the tangent vector on the boundary Γ and
λ(z) =
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
cos(l x)minus i cos(l y)
1
1 + iradic2
1
r(z) =
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
ψprime1(s) on Γ = BA
0 on AO
1radic2[F (minusx)minus 1
2φprime(minusx)] on OC
ψprime2(x) on CB
We shall prove the solvability of the Frankl problem for equation (22) by using themethods of parameter extension and symmetry extension
We can choose the index K = minus12 of λ(z) on the boundary partD+ of D+ In factdue to the boundary condition
Re [λ(z)W (z)] =12Re [λ(z)(ux minus iuy)] = r(z) on partD+ = AO cup OB cup BA (415)
and λ(z) = 1 on AO cup CB λ(z) = exp(iπ4) on OC λ(z) = cos(l x) minus i cos(l y)on Γ denote t1 = 0 t2 = 1 t3 = a t4 = i it is seen λ(a + 0) = exp(i3π2) andλ(i minus 0) = exp(iπ) we have
Kj=[φj
π
]+Jj Jj = 0 or 1 eiφj =
λ(tjminus0)λ(tj+0)
γj=φj
πminusKj j=1 4
eiφ1=λ(t1minus0)λ(t1+0)
=ei0
eiπ4 = eminusiπ4 0ltγ1=φ1
πminusK1=minus1
4minusK1 lt
34
lt1
eiφ2=λ(t2minus0)λ(t2+0)
=eiπ4
ei0 = eiπ4 minus1ltγ2=φ2
πminusK2=
14minusK2=minus3
4lt0
eiφ3 =λ(t3 minus 0)λ(t3 + 0)
=ei0
ei3π2 0 le γ3 =φ3
πminus K3 = minus3
2minus K3 =
12
lt 1
eiφ4 =λ(t4 minus 0)λ(t4 + 0)
=eiπ
ei0 = eiπ 0 le γ4 =φ4
πminus K4 = 1minus K4 = 0 lt 1
(416)
here [b] is the largest integer not exceeding the real number b we choose K1 = minus1K3 = minus2 K2 = K4 = 1 Under these conditions the index K of λ(z) on theboundary partD+ of D+ is just as follows
K =12(K1 +K2 +K3 +K4) = minus1
2 (417)
Noting that U(0 y) = 0 on AprimeA we can extend W (z) onto the reflected domainD of D about the segment AprimeA In fact we introduce the function
W (z) =
⎧⎨⎩W (z) in D
minusW (minusz) on D(418)
4 Frankl Boundary Value problem 181
this function W (z) is a solution of the equation
⎧⎨⎩ Wz
W zlowast
⎫⎬⎭ = Re [A1W ] + A2u+ A3 in
⎧⎨⎩D
D
⎫⎬⎭u(z) = 2Re
int z
1W (z)dz + ψ1(0)
(419)
with the boundary conditions
2Re[λ(z)W (z)]=r(z) zisinΓcupCBcupΓcupBC u(a)=b0=ψ1(0)=u(minusa)
Re[λ(x)W (x)]=r(x) xisinL2=(01)cup(minus10)(420)
in which
A1 =
⎧⎨⎩A1(z)
minusA1(minusz)A2 =
⎧⎨⎩A2(z)
A2(minusz)A3 =
⎧⎨⎩A3(z) in D
A3(minusz) in D+ cup Dminus(421)
and
λ(z) =
⎧⎨⎩λ(z)
λ(minusz)r(z) =
⎧⎨⎩ r(z) Γ cup CB
minusr(minusz) Γ cup BC
λ(z) =
⎧⎪⎪⎪⎨⎪⎪⎪⎩1 + iradic2
1minus iradic2
r(z) =
⎧⎨⎩r(z) on OC = (0 1)
minusr(minusz) on CO = (minus1 0)
(422)
herein Γ BC = (minusa minus1) CO and AB are the reflected curves of Γ CB OC and BAabout the imaginary axis respectively We choose the index of the function λ(z) onthe boundary part(D+cupD+cupAO) of the elliptic domain D+cupD+cupAO asK = minus12 Infact noting that λ(z) = 1 on CBcupBC λ(z) = exp(iπ4) on OC λ(z) = exp(minusiπ4)on CO we denote t1 = 0 t2 = 1 t3 = a t4 = i t5 = minusa t6 = minus1 it is seenλ(a+ 0) = exp(i3π2) λ(i minus 0) = λ(i+ 0) = exp(iπ) λ(minusa minus 0) = exp(iπ2) hencewe have
Kj=[φj
π
]+Jj Jj=0 or 1 ejφj =
λ(tjminus0)λ(tj+0)
γj=φj
πminusKj j=1 6
eiφ1=λ(t1minus0)λ(t1+0)
=eminusiπ4
eiπ4 =eminusiπ2 0ltγ1=φ1
πminusK1=minus1
2minusK1=
12
lt1
eiφ2=λ(t2minus0)λ(t2+0)
=eiπ4
ei0 =eiπ4 minus1ltγ2=φ2
πminusK2=
14minusK2=minus3
4lt0
182 V Second Order Linear Mixed Equations
eiφ3=λ(t3minus0)λ(t3+0)
=ei0
ei3π2 =eminusi3π2 0ltγ3=φ3
πminusK3=minus3
2minusK3=
12
lt1
eiφ4 =λ(t4 minus 0)λ(t4 + 0)
=eiπ
eiπ= ei0 0 le γ4 =
φ4
πminus K4 = 0minus K4 = 0 lt 1
eiφ5=λ(t5 minus 0)λ(t5 + 0)
=eiπ2
ei0 =eiπ2 0 lt γ5=φ5
πminus K5=
12
minus K5 =12
lt 1
eiφ6=λ(t6minus0)λ(t6+0)
=ei0
eminusiπ4 =eiπ4 minus1ltγ6=φ6
πminusK6=
14minusK6=minus3
4lt0
(423)
If we choose K1 = minus1 K2 = K6 = 1 K3 = minus2 K4 = K5 = 0 the index K of λ(z) isjust
K =12(K1 +K2 + middot middot middot+K6) = minus1
2 (424)
We can discuss the solvability of the corresponding boundary value problem(419) (420) and then derive the existence of solutions of the Frankl problemfor equation (22)
42 Representation and a priori estimates of solutions to the Franklproblem for (22)
First of all similarly to Lemma 21 we can prove the following theorem
Theorem 41 Let equation (22) satisfy Condition C in D+ and u(z) be a continu-ous solution of (22) in D+
lowast = D+0 1 a i Then u(z) can be expressed as
u(z) = U(z)Ψ(z) + ψ(z) in D+
U(z) = 2Reint z
aw(z)dz + b0 w(z) = Φ(z)eφ(z) in D+
(425)
Here ψ(z) Ψ(z) are the solutions of equation (22) in D+ and
uzz minus Re [A1uz]minus A2u = 0 in D+ (426)
respectively and satisfy the boundary conditions
ψ(z) = 0 Ψ(z) = 1 on Γ cup L
partψ(z)partx
= 0partΨ(z)
partx= 0 on AO
(427)
where L = (0 a) They satisfy the estimates
C1γ [X(z)ψ(z) D+] le M19 X(z)ψ(z) W 2
p0(D+)le M19 (428)
4 Frankl Boundary Value problem 183
C1γ [X(z)Ψ(z) D+]leM20X(z)Ψ(z)W 2
p0(D+)leM20Ψ(z)geM21 gt0 zisinD+ (429)
in which X(z) = |x + y minus t1|η1prod4
j=2 |z minus tj|ηj ηj = maxminusγj + δ δ j = 1 2 3 4herein tj γj(j = 1 2 3 4) are as stated in (416) δ γ(γ lt δ) are small positiveconstants p0 (2 lt p0 le p) Mj = Mj (p0 γ k D) (j = 19 20 21) are non-negativeconstants k = (k0 k1 k2) Moreover U(z) is a solution of the equation
Uzz minus Re [AUz] = 0 A = minus2(lnΨ)z + A1 in D+ (430)
where Im [φ(z)] = 0 z isin partD+ Re [φ(0)] = 0 and φ(z) satisfies the estimate
Cβ[φ(z) D+] + Lp0 [φz D+] le M22 (431)
in which β (0 lt β le α) M22 = M22 (p0 α k0 D+) are two non-negative constants
Φ(z) is analytic in D+ If u(z) is a solution of Problem F then W (z) = uz satisfiesthe boundary conditions
Re [λ(z)W (z)] = r(z) on Γ cup AO u(a) = b0 = ψ1(0) (432)
Re [λ(x)W (x)] = r(x) λ(x) =
⎧⎪⎨⎪⎩1 + iradic2on L0 = (0 1)
1 on L1 = (1 a)(433)
Theorem 42 Suppose that equation (22) satisfies Condition C Then any solutionof the Frankl problem for (22) can be expressed as
u(z) = 2Reint z
aW (z)dz + b0 b0 = ψ1(0) (434)
Here W (z) is a solution of the equation
Wz
Wzlowast
= Re [A1W ] + A2u+ A3 in
D+
Dminus
(435)
satisfying the boundary conditions (413) minus (414) (W (z) = uz) and W (z) possessesthe form
W (z) = Φ(z)eφ(z) + ψ(z)
φ(z) = φ0(z) + Tg = φ0(z)minus 1π
int intD+
g(ζ)ζ minus z
dσζ ψ(z) = Tf in D+
W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν
1g1(z)dνe1 +
int micro
0g2(z)dmicroe2 in Dminus
(436)
184 V Second Order Linear Mixed Equations
in which e1 = (1 + i)2 e2 = (1 minus i)2 micro = x + y ν = x minus y φ0(z) is an analyticfunction in D+ such that Im φ(x) = 0 on Γ cup AO cup L and
g(z)=A12+A1W(2W ) W (z) = 00 W (z)=0
in D+
f(z) = Re [A1uz] + A2u+ A3 f(z) = Re [A1φz] + A2u+ A3 in D+
g1(z)=g2(z)=Aξ+Bη+Cu+D ξ=ReW+ImW η=ReW minusImW
A=ReA1+ImA1
2 B=
ReA1minusImA1
2 C=A2 D= A3 in Dminus
(437)
where Φ(z) in D+ and Φ(z) in Dminus are solutions of the equationWz
W zlowast
= 0 in
D+
Dminus
(438)
satisfying the boundary conditions
Re [λ(z)eφ(z)Φ(z)] = r(z)minus Re [λ(z)ψ(z)] z isin Γ
Re [eφ(z)Φ(z)] = minusRe [ψ(z)] z = iy isin AO
Re [λ(x)Φ(x)eφ(x)] = r(x)minus Re [λ(x)ψ(x)] x isin L0 = (0 1)
Re [λ(x)(Φ(x) + Ψ(x))] = r(x) = Re [λ(x)W (x)] x isin L0 = (0 1)
Re [Φ(x)] = minusRe [Ψ(x)] x isin OAprime u(a) = b0 = ψ1(0)
(439)
where λ(x) on L = (0 a) is as stated in (415)
Proof Let u(z) be a solution of the Frankl problem for equation (22) andW (z) = uz u(z) be substituted in the positions of w u in (437) Thus the functionsg(z) f(z) g1(z) g2(z) and ψ(z) φ(z) in D+ and Ψ(z) in Dminus in (436)(437) aredetermined Moreover we can find the solution Φ(z) in D+ and Φ(z) in Dminus of (438)with the boundary condition (439) where r(z) as stated in (415) namely
r(z) = H(F φ) z isin Γ cup AO cup L (440)
thus
W (z) =
⎧⎪⎨⎪⎩Φ(z)eφ(z) + ψ(z) in D+
Φ(z) + Ψ(z) in Dminus(441)
is the solution of Problem A for the complex equation (435) with the boundaryconditions (413)(414) which can be expressed as in (436) and u(z) is a solutionof the Frankl problem for (22) as stated in (434)
Next we discuss the uniqueness of solutions of the Frankl problem for (22)
4 Frankl Boundary Value problem 185
Theorem 43 Suppose that the mixed equation (22) satisfies Condition C Thenthe Frankl problem for (22) has at most one solution u(z) isin C(D) cap C1(D)
Proof We consider equation (22) in D+ As stated before if u1(z) u2(z) are twosolutions of the Frankl problem for (22) then u(z) = u1(z) minus u2(z) is a solution ofthe homogeneous equation⎧⎨⎩uzz
uzzlowast
⎫⎬⎭=Re [A1uz]+A2u in
⎧⎨⎩D+
Dminus
⎫⎬⎭
u(z) = 2Reint z
aW (z)dz W (z) = uz in D
u(z)=U(z)Ψ(z) Uz=Φ(z)eφ(z) in D+
W (z) = Φ0(z) + Φ(z) + Ψ(z) in Dminus
(442)
in which
Ψ(z) =int ν
1g1(z)dνe1 +
int micro
0g2(z)dmicroe2 g1(z) = g2(z) = Aξ +Bη + Cu in Dminus
and Ψ(z) Φ(z) φ(z)Φ(z) are similar to those in Theorem 42 and Φ(z)Φ(z) Φ0(z)are solutions of equation (438) in D+ and Dminus respectively satisfying the conditions
2Re [λ(z)Φ(z)]=r(z)=0 on ΓcupAOcupCB U(a)=0
Re [λ(x)Φ(x)] =12[f(x) + F (minusx)] on L0 = (0 1)
Φ(x) = Φ(x) Φ0(x) = uz(x)minusΨ(x)minus Uz(x)eminusφ(x) on L0
Re [Φ0(z)]=Re [Ψ(z)] Im [Φ0(z)]=Im [uzminusΨ(z)minusUz(z)eminusφ(z)] on AprimeO
(443)
According to Theorem 41 the solution U(z) = 2Reint za Φ(z)dz of equation (12)
satisfies the boundary conditions
U(z) = S(z) = 0 on Γ cup CB
U(x) = 2int x
aΦ(x)dx = 2
int x
aUz(x)eminusφ(x)dx =
int x
0Φ(x)dx minus
int a
0Φ(x)dx
=S(x)=int x
0[f(x)+F (minusx)]dx=g(x)+
12U(ix) on (0 1)
(444)
where g(x) =int x0 f(x)dx
int x0 F (minusx)dx = U(ix)2 Besides the harmonic function U(z)
in D+ satisfies the boundary condition
partU(z)partx
= 0 on AO (445)
Moreover there exists a conjugate harmonic function V (z) in D+ such that V (0) = 0From the above last formula we can derive that V (iy) =
int y0 Vydy =
int y0 Uxdy = 0 on
186 V Second Order Linear Mixed Equations
AO By the Cauchy theorem we haveintpartD+
[U(z) + iV (z)]2dz = 0
= minusint a
1[V (x)]2dx minus
intΓ[V (z)]2
(dx
ds+ i
dy
ds
)ds (446)
+iint 0
1[U(iy)]2dy +
int 1
0[U2(x)minus V 2(x) + 2i U(x)V (x)]dx
Due to the continuity of Uy on (01) V (x) =int x0 Vxdx = minus int x
0 Uydx = 2int x0 V (x)dx =
minus int x0 [f(x) minus F (minusx)]dx = minusg(x) + U(ix)2 is obtained From the imaginary part in
(446) and the above formula it is clear thatintΓ[V (z)]2
party
partsds+
int 1
0[U(iy)]2dy + 2
int 1
0[g(x)]2 minus 1
4[U(ix)]2dx = 0 (447)
Hence we getU(iy) = 0 on AO g(x) = 0 on OC
and then f(x) = gprime(x) = 0 F (minusx) = [U(ix)]x2 = 0 on OC Due to the functionr(z) = S(z) = 0 on partD+ in (443)(444) and the index K = minus12 hence Φ(z) = 0 inD+ and then the solution u(z) of the homogeneous Frankl problem for (442) in D+
satisfies u(z) = u1(z)minus u2(z) = 0 Moreover we can derive u(z) = u1(z)minus u2(z) = 0in Dminus This proves the uniqueness of solutions for the Frankl problem for (22) in D
Finally we give an a priori estimate of solutions to the Frankl problem for equation(22) From the estimate we can see the singular behavior of uz at the discontinuityset Z = 1 a i minusi x + y = 0 It becomes infinity of an order not exceeding 34 atz = 1 infinite of order not exceeding a small positive number δ at the points i minusiand uz is bounded at the point set a x+y = 0 In fact we can prove that z = i minusiare removable singular points In [12]3) the author pointed out that uz can becomeinfinity of an order less than 1
Theorem 44 Suppose that equation (22) satisfies Condition C in D and the func-tion r(z) in (414) is H(F φ) especially
r(x) = H(F φ) =1radic2[F (minusx)minus 1
2φprime(minusx)] x isin L0 = (0 1) (448)
Then any solution u(z) of the Frankl problem for equation (22) in D+ satisfies theestimate
C1γ [u D+] = Cγ[u(z) D+] + C[uzX(z) D+] le M23(k1 + k2) (449)
where X(z) is as stated in (429) ie
X(z) = |x+ y minus t1|η1
4prodj=2
|z minus tj|ηj ηj = maxminusγj + δ δ j = 1 2 3 4 (450)
4 Frankl Boundary Value problem 187
and M23 = M23(p0 γ δ k0 D) is a non-negative constant
Proof On the basis of the uniqueness of solutions of the Frankl problem for (22)in Theorem 43 and the results in [12]3) by using reductio ad absurdum we canderive the estimate (449) In fact from (483)(484) in the proof of Theorem 46below we see that the function [Wn+1 un+1] (Wn+1(z) = Wn+1(z)minusWn(z) un+1(z) =un+1(z)minus un(z)) is a solution of the boundary value problem
[Wn+1]z minus t0G(z un+1 Wn+1) = (t minus t0)G(z un Wn) z isin D+
Re [λ(z)Wn+1(z)] = 0 z isin Γ cup AO cup CB
Re [λ(z)Wn+1(z)]minus t0H(Fn+1 0) = (t minus t0)H(Fn 0)] on L0 = (0 1)
un+1(z) =int z
aWn+1(z)dz z isin D+
(451)
where G(z u W ) = Re [A1W ] + A2u + A3 G(z un wn) = G(z un+1 wn+1) minus G(zun wn) Fn+1 = Fn+1 minus Fn On the basis of Theorem 41 the solution Wn+1 of theboundary value problem (451) can be expressed as
un+1(z) = Un+1(z)Ψn+1(z) + ψn+1(z) in D+
Un+1(z)=2Reint z
1Φn+1(z)dz Un+1z=Φn+1(z)eφn+1(z) in D+
Wn+1(z) = Φ0n+1(z) + Φn+1(z) + Ψn+1(z) in Dminus
(452)
where Φn+1(z)Φ0n+1(z) are the solutions of equation (438) in Dminus Ψn+1(z) is a
solution of the equation in (451) in Dminus ψn+1(z) Ψn+1(z) are the solutions ofthe equation in (451) and its homogeneous complex equation in D+ satisfying theboundary conditions
ψn+1(z) = 1 Ψn+1(z) = 1 on Γ cup L
partψn+1(z)partx
= 0partΨn+1(z)
partx= 0 on AO
(453)
According to the proof of Theorem 43 we see that the function Un+1(z) satisfies theboundary conditions
Un+1(z) = S(z) = 0 on Γ cup CB Un+1(a) = 0
Un+1(x)=Reint x
aΦn+1xdx=S(x)
S(x)=int x
0[f(x)+Fn+1(minusx)]dx minus 1
2
int x
0φprime
n+1(minusx)dx
= g(x) +12Un+1(ix) +
12φn+1(minusx) on (0 1)
(454)
188 V Second Order Linear Mixed Equations
where
g(x)=int x
0f(x)dx
int x
0Fn+1(minusx)dx=
Un+1(ix)2
Un+1(minusx)=minusint x
0φprime
n+1(minusx)dx
Besides we can see that the harmonic function Un+1(z) in D+ satisfies the boundarycondition
partUn+1(z)partx
= 0 on AO (455)
Moreover there exists a conjugate harmonic function Vn+1(z) in D+ such thatVn+1(0) = 0 We shall verify that
limnrarrinfinmax
D+
|X(x)Un+1x| = 0 limnrarrinfin
int 1
0[Un+1(iy)]2dy = 0 (456)
Suppose that limnrarrinfinint 10 |Un+1(iy)|dy = C gt 0 due to
intΓ[Vn+1(z)]2
party
partsds+
int 1
0[Un+1(iy)]2dy
+2int 1
0
[g(x)]2 minus 1
4[t0Un+1(ix) + (t minus t0)Un(ix)]2
dx = 0
(457)
provided that |t minus t0| is sufficently small such that |t minus t0|2 int 10 |Un(iy)|2dy leint 1
0 |Un+1(iy)|2dy2 for n = nk rarr infin then similarly to (447) from (457) we canderive that
Un+1(iy) = 0 on AO g(x) = 0 on OC (458)
This contradiction proves thatint 10 [Un+1(ix)]2dx = 0
int 10 [Un+1(ix)]2dx = 0 and Un+1 =
Un+1(z) minus Un(z) = 0 in D+ for n ge N0 where N0 is a sufficiently large positivenumber Hence un+1 = un+1(z) minus un(z) = ψn+1(z) minus ψn(z) in D+ for n ge N0Similarly to the proof of the first estimate in (428) we can obtain
C1γ [X(z)un+1(z) D+] le M24|t minus t0|C1
γ [X(z)un(z) D+] (459)
in whichM24 = M24(p γ δ k0 D+) is a non-negative constant Choosing the constant
ε so small that εM24 le 12 and |t minus t0| le ε it follows that
C1γ [un+1 D+] le εM24|t minus t0|C1
γ [un D+] le 12C1
γ [un D+]
C1γ [un+1 D+]le2minusn+N0
infinsumj=N0
2minusjC1γ [u1 minus u0 D+]le2minusn+N0+1C1
γ [u1minusu0 D+]
for n gt N0 Therefore there exists a continuous function ulowast(z) on D+ such that
ulowast(z) =infinsum
j=0un+1 =
infinsumj=0[un+1 minus un(z)]
4 Frankl Boundary Value problem 189
From the estimate ofsumn
j=0[uj+1(z)minus uj(z)] = un+1(z)minus u0(z) in D+ the estimate
C1γ [un+1 D] = Cγ[un+1 D+] + C[un+1zXD+] le M25 (460)
can be derived where M25 = M25(p0 γ δ k0 D+) is a non-negative constant More-
over we can derive a similar estimate of ulowast(z) in D+ and Dminus which gives the estimate(449)
43 The solvability of the Frankl problem for (22)
Theorem 45 Suppose that the mixed equation (22) satisfies Condition C andA1(z) = A2(z) = 0 in D ie⎧⎨⎩uzz = A3(z) z isin D+
uzzlowast = A3(z) z isin Dminus(461)
Then the Frankl problem for (461) has a solution in D
Proof It is clear that the Frankl problem for (461) is equivalent to the followingProblem A for the complex equation of first order and boundary conditions
Wz
W zlowast
= A3(z) in
⎧⎨⎩D+
Dminus
⎫⎬⎭ (462)
Re [λ(z)W (z)] = r(z) = H(F φ) z isin Γ cup AO cup L
Re [λ(z)W (z)] = r(z) = 0 λ(z) = 1 z isin OAprime(463)
and the relationu(z) = 2Re
int z
aW (z)dz + b0 in D (464)
in which λ(z) r(z) are as stated in (415) and (440)
In order to find a solution W (z) of Problem A in D we can express W (z) in theform (434)ndash(437) In the following by using the method of parameter extensionwe shall find a solution of Problem A for the complex equation (462) We considerequation (462) and the boundary conditions with the parameter t isin [0 1]
Re [λ(z)W (z)] = tH(F φ) +R(z) on partD+ = Γ cup AO cup L (465)
in which H(F φ) on partD+ = Γ cup AO cup L is as stated in (448) and R(z)X(z) isinCγ(partD+) this problem is called Problem Ft
When t = 0 the unique solution of Problem F0 for the complex equation (461)can be found by a method given in Section 1 and its solution [W0(z) u0(z)] can be
190 V Second Order Linear Mixed Equations
expressed as
u0(z) = 2Reint z
aW0(z)dz + b0 W0(z) = W (z) in D b0 = ψ1(0)
W (z) = Φ(z) + ψ(z) ψ(z) = TA3 = minus 1π
int intD+
A3(ζ)ζminusz
dσζ in D+
W (z)=Φ(z)+Ψ(z) Ψ(z)=int ν
1A3(z)e1dν+
int micro
0A3(z)e2dmicro in Dminus
(466)
where Φ(z) is an analytic function in D+ and Φ(z) is a solution of (438) in Dminussatisfying the boundary conditions
Re [λ(z)W (z)] = R(z) z isin Γ cup L
Re [λ(z)(Φ(z) + Ψ(z))] = R(z) λ(z) = 1 z = iy isin AO
Re [λ(z)(Φ(z) + Ψ(z))] = R(z) λ(z) = 1 z = iy isin OAprime
Re [λ(x)(Φ(x)+Ψ(x))]=R(x)=Re [λ(x)W (x)] z=x isin OC u(a)=b0
(467)
Suppose that when t = t0 (0 le t0 lt 1) Problem Ft0 is solvable ie Problem Ft0
for (462) has a solution [W0(z) u0(z)] (u0(z) isin C1γ(D)) We can find a neighborhood
Tε = |t minus t0| le ε 0 le t le 1(0 lt ε lt 1) of t0 such that for every t isin Tε Problem Ft
is solvable In fact Problem Ft can be written in the formWz
Wzlowast
= A3(z) in
⎧⎨⎩D+
Dminus
⎫⎬⎭
Re [λ(z)W (z)]minus t0H(F φ) = (t minus t0)H(F φ) +R(z) on partD+
(468)
ReplacingW (z) u(z) in the right-hand sides of (468) by a functionW0(z) isin Cγ(partD+)and the corresponding function u0(z) in (466) ie u0(z) isin C1
γ(D) especially by thesolution [W0(z) u0(z)] of Problem F0 it is obvious that the boundary value problemfor such an equation in (468) then has a solution [W1(z) u1(z)] u1(z) isin C1
γ(partD)Using successive iteration we obtain a sequence of solutions [Wn(z) un(z)] un(z) isinC1
γ(D) n = 1 2 which satisfy the equations and boundary conditionsWn+1z
Wn+1zlowast
= A3(z) in
⎧⎨⎩D+
Dminus
⎫⎬⎭
Re [λ(z)Wn+1(z)]minus t0H(Fn+1 φ) = (t minus t0)H(Fn φ) +R(z) on partD+
Re [λ(z)Wn+1(z)] = 0 z isin OAprime
(469)
From the above formulas it follows that
[Wn+1 minus Wn]z = 0 z isin D
Re [λ(z)(Wn+1(z)minusWn(z))]minust0[H(Fn+1minusFn 0)]=(tminust0)[H(FnminusFnminus10)](470)
4 Frankl Boundary Value problem 191
Noting that
|t minus t0|Cγ[XH(Φn minus Φnminus1 0) L0] le |t minus t0|Cγ[X(Φn minus Φnminus1) L0] (471)
and applying Theorem 44 we have
C1γ [un+1 minus un D+] le M26C
1γ [Φn minus Φnminus1 D+] (472)
where M26 = M26(p0 γ δ k0 D+) Choosing the constant ε so small that εM24 le 12
and |t minus t0| le ε it follows that
C1γ [un+1 minus un D+] le εM26C
1γ [un minus unminus1 D+] le 1
2C1
γ [un minus unminus1 D+] (473)
and when n m ge N0 + 1 (N0 is a positive integer)
C1γ [un+1 minus un D+] le 2minusN0
infinsumj=02minusjC1
γ [u1 minus u0 D+] le 2minusN0+1C1γ [u1 minus u0 D+] (474)
Hence un(z) is a Cauchy sequence According to the completeness of the Banachspace C1
γ(D+) there exists a function ulowast(z) isin C1γ(D+) and Wlowast(z) = ulowastz(z) so that
C1γ [un minus ulowast D+] = Cγ[un minus ulowast D+] + C[X(Wn minus Wlowast) D+] rarr 0 as n rarr infin We cansee that [Wlowast(z) ulowast(z)] is a solution of Problem Ft for every t isin Tε = |t minus t0| leε Because the constant ε is independent of t0 (0 le t0 lt 1) therefore from thesolvability of Problem F0 when t0 = 0 we can derive the solvability of Problem Ft
when t = ε 2ε [1ε]ε 1 In particular when t = 1 and R(z) = 0 Problem F1 iethe Frankl problem for (461) in D+ is solvable
As for the solution [W (z) u(z)] in Dminus it can be obtained by (410)(411) and themethod in Chapters I and II namely
u(z) = 2Reint z
aW (z)dz + b0 on Dminus b0 = ψ1(0)
W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν
1A3(z)e1dν +
int micro
0A3(z)e2dmicro
Φ(z) =12[(1 + i)f(x+ y) + (1minus i)g(x minus y)]
f(x+ y) = Re [(1minus i)(W (x+ y)minusΨ(x+ y))]
g(x minus y) = Re [(1 + i)(W (x minus y)minusΨ(x minus y))] z isin Dminus cap x+ y ge 0
(475)
where W (x+ y) W (x minus y) are the values on 0 le z = x+ y le 1 0 le x minus y le 1 of thesolution W (z) of Problem F for (461) in D+ and Ψ(x+ y) Ψ(x minus y) are the valueson 0 le z = x + y le 1 0 le x minus y le 1 of Ψ(z) respectively Moreover the functionW (z) in Dminus cap x+ y le 0 can be obtained by (475)(418) In fact from (475) wehave found the function W (z) on OC prime = x + y = 0 0 le x le 12 by (418) weobtain the function W (z) = minusW (minusz) on OC primeprime = x minus y = 0 minus12 le x le 0 and
192 V Second Order Linear Mixed Equations
denote σ(x) = Re [(1 minus i)Ψ(z)] on OC prime τ(x) = Re [(1 + i)Ψ(z)] on OC primeprime Hence thesolution u(z) in Dminus cap x+ y le 0 is as follows
u(z) = 2Reint z
0W (z)dz + u(0) Ψ(z) =
int ν
0A3(z)e1dν +
int micro
0A3(z)e2dmicro
W (z) =12[(1minus i)f(x+ y) + (1 + i)g(x minus y)] + Ψ(z)
f(x+ y) = τ((x+ y)2) + ReW (0) + ImW (0)
g(x minus y) = σ((x minus y)2) + ReW (0)minus ImW (0)z isin Dminus cap x+ y le 0
(476)
in which Φ(z) and Ψ(z) are the functions from (475) Furthermore we can provethat the solution u(z) satisfies the boundary conditions (41)ndash(44) This completesthe proof
Theorem 46 Suppose that the mixed equation (22) satisfies Condition C Thenthe Frankl problem for (22) has a solution in D
Proof Similarly to the proof of Theorem 45 we see that the Frankl problem for(22) is equivalent to Problem A for first order complex equation and boundary con-ditions
Wz
Wzlowast
= G G = G(z u W ) = Re [A1W ] + A2u+ A3 in
⎧⎨⎩D+
Dminus
⎫⎬⎭ (477)
Re [λ(z)W (z)] = r(z) = H(F φ) z isin Γ cup AO cup L (478)
and the relation (464) in which r(z) = H(F φ) on z isin partD+ = Γ cup AO cup L is asstated in (463)
In order to find a solution W (z) of Problem A in D we can express W (z) in theform (434)ndash(437) In the following by using the method of parameter extension asolution of Problem A for the complex equation (477) will be found We considerthe equation and boundary conditions with the parameter t isin [0 1]
Wz = tG+K(z) G = G(z u W ) = Re [A1W ] + A2u+ A3 in D+ (479)
Re [λ(z)W (z)] = tH(F φ) +R(z) on partD+ = Γ cup AO cup L (480)
where K(z) isin Lp(D+) and R(z)X(z) isin Cγ(partD+) This problem is called ProblemFt
When t = 0 the complex equation (479) becomes the equation
Wz = K(z) in D+ (481)
From Theorem 45 we can find the unique solution of Problem F0 for (479) Supposethat when t = t0 (0 le t0 lt 1) Problem Ft0 is solvable ie Problem Ft0 for (479) hasa solution [W0(z) u0(z)] (u0 isin C1
γ(D)) We can find a neighborhood Tε = |t minus t0| le
4 Frankl Boundary Value problem 193
ε 0 le t le 1(0 lt ε lt 1) of t0 such that for every t isin Tε Problem Ft is solvable Infact Problem Ft can be written in the form
Wz minus t0G(z u W ) = (t minus t0)G(z u W ) +K(z) in D+
Re [λ(z)W (z)]minus t0H(F φ) = (t minus t0)H(F φ) +R(z) on partD+(482)
ReplacingW (z) u(z) in the right-hand sides of (482) by a functionW0(z) isin Cγ(partD+)and the corresponding function u0(z) in (466) ie u0(z) isin C1
γ(D) especially by thesolution [W0(z) u0(z)] of Problem F0 it is obvious that the boundary value problemfor such equation in (482) then has a solution [W1(z) u1(z)] u(z) isin C1
γ(partD) Usingsuccessive iteration we obtain a sequence of solutions [Wn(z) un(z)] un(z) isin C1
γ(D)n = 1 2 which satisfy the equations and boundary conditions
Wn+1zminust0G(z un+1 Wn+1) = (t minus t0)G(z un Wn) +K(z) in D+ (483)
Re [λ(z)Wn+1(z)]minust0H(Fn+1 φ)=(t minus t0)H(Fn φ)+R(z) on partD+ (484)
From the above formulas it follows that
[Wn+1 minus Wn]z minus t0[G(z un+1 Wn+1)minus G(z un Wn)]
= (t minus t0)[G(z un Wn)minus G(z unminus1 Wnminus1)] z isin D+
Re [λ(z)(Wn+1(z)minus Wn(z))]minus t0[H(Fn+1 minus Fn 0)]
= (t minus t0)[H(Fn minus Fnminus1 0)] z isin L0
(485)
Noting that
Lp[(tminust0)(G(z un Wn)minusG(z unminus1 Wnminus1)) D+] le 2k0|tminust0|C1γ [unminusunminus1 D+]
|t minus t0|Cγ[XH(Fn minus Fnminus1 0) L0] le |t minus t0|Cγ[X(Fn minus Fnminus1) L0](486)
and according to the method in the proof of Theorem 44 we can obtain
C1γ [un+1 minus un D+] le M27[2k0 + 1]C1
γ [un minus unminus1 D+] (487)
whereM27 = M27(p0 γ δ k0 D+) Choosing the constant ε so small that εM27
(2k0 + 1) le 12 and |t minus t0| le ε it follows that
C1γ [un+1 minus un D+] le εM27(2k0+1)C1
γ [un minus unminus1 D+] le 12C1
γ [un minus unminus1 D+] (488)
and when n m ge N0 + 1 (N0 is a positive integer)
C1γ [un+1 minus un D+] le 2minusN0
infinsumj=02minusjC1
γ [u1 minus u0 D+] le 2minusN0+1C1γ [u1 minus u0 D+]
Hence un(z) is a Cauchy sequence According to the completeness of the Banachspace C1
γ(D+) there exists a function ulowast(z) isin C1γ(D+) and Wlowast(z) = ulowastz(z) so that
194 V Second Order Linear Mixed Equations
C1γ [un minus ulowast D+] = Cγ[un minus ulowast D+] + C[X(Wn minus Wlowast) D+] rarr 0 as n rarr infin We cansee that [Wlowast(z) ulowast(z)] is a solution of Problem Ft for every t isin Tε = |t minus t0| le εBecause the constant ε is independent of t0 (0 le t0 lt 1) therefore from the solvabilityof Problem Ft0 when t0 = 0 we can derive the solvability of Problem Ft whent = ε 2ε [1ε] ε 1 In particular when t = 1 and K(z) = 0 R(z) = 0 ProblemF1 ie the Frankl problem for (22) in D+ is solvable
The existence of the solution [W (z) u(z)] of Problem F for (22) in Dminus can beobtained by the method in Chapters I and II
5 Oblique Derivative Problems for Second Order DegenerateEquations of Mixed Type
In this section we discuss the oblique derivative problem for second order degenerateequations of mixed type in a simply connected domain We first give the represen-tation of solutions of the boundary value problem for the equations and then provethe uniqueness of solutions for the problem Moreover we introduce the possibilityto prove the existence of the above oblique derivative problem
51 Formulation of oblique derivative problems for degenerate equationsof mixed type
Let D be a simply connected bounded domain in the complex plane CI with theboundary partD = Γ cup L where Γ(sub y gt 0) isin C2
α(0 lt α lt 1) with the endpoints z = 0 2 and L = L1 cup L2 L1 = x +
int y0
radicminusK(t)dt = 0 x isin (0 1) L2 =
x minus int y0
radicminusK(t)dt = 2 x isin (1 2) and z1 = x1 + jy1 = 1 + jy1 is the intersection
point of L1 and L2 In this section we use the hyperbolic numbers Denote D+ =D cap y gt 0 Dminus = D cap y lt 0 We may assume that Γ = |z minus 1| = 1 y ge 0 andconsider the linear degenerate mixed equation of second order
Lu = K(y)uxx + uyy = dux + euy + fu+ g in D (51)
where K(y) possesses the first order continuous derivatives K prime(y) and K prime(y) gt 0 ony = 0 K(0) = 0 The following degenerate mixed equation is a special case
Lu = sgny|y|m uxx + uyy = dux + euy + fu+ g in D (52)
where m is a positive constant d e f g are functions of z(isin D) Similarly to (543)Chapter II we denote W (z)= UminusiV =ym2U+iV =[ym2uxminusiuy]2 Wmacrz=[ym2Wx+iWy]2 in D+ and W (z)= U+jV = |y|m2UminusjV =[|y|m2ux+juy]2 Wmacrz=[|y|m2WxminusjWy]2 in Dminus then equation (52) in D can be reduced to the form
5 Degenerate Mixed Equations 195
⎧⎨⎩ Wmacrz
Wmacrz
⎫⎬⎭ = A1(z)W + A2(z)W + A3(z)u+ A4(z) in
D+
Dminus
A1 =
⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩im
8y+
d
4ym2 +ie
4=
d
4ym2 + i
(m
8y+
e
4
)
jm
8|y| +minusd
4|y|m2 minus je
4=
minusd
4|y|m2 + j
(m
8|y|minuse
4
) in
⎧⎨⎩D+
Dminus
⎫⎬⎭
A2 =
⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩d
4ym2+ i
(m
8yminus e
4
)
minusd
4|y|m2+j
(m
8|y|+e
4
)
A3=
⎧⎪⎪⎪⎨⎪⎪⎪⎩f
4
minusf
4
A4=
⎧⎪⎪⎨⎪⎪⎩g
4
minusg
4
in
⎧⎨⎩D+
Dminus
⎫⎬⎭
(53)
and
u(z)=
⎧⎪⎪⎨⎪⎪⎩2Re
int z
0uzdz+u(0) in D+
2Reint z
0(UminusjV )d(x+jy)+u(0) in Dminus
is a solution of equation (52)
Suppose that equation (52) satisfies the following conditions Condition C
The coefficients Aj(z) (j = 1 2 3) in (52) are continuous in D+ and continuousin Dminus and satisfy
C[Aj D+] le k0 j = 1 2 C[A3 D+] le k1 A2 ge 0 in D+
C[Aj Dminus] le k0 j = 1 2 C[A3 Dminus] le k1(54)
where p (gt 2) k0 k1 are non-negative constants If the above conditions is replaced by
C1α[Aj Dplusmn] le k0 j = 1 2 C1
α[A3 Dplusmn] le k1 (55)
in which α (0 lt α lt 1) is a real constant then the conditions will be called ConditionC prime
Now we formulate the oblique derivative boundary value problem as follows
Problem P Find a continuously differentiable solution u(z) of (52) in Dlowast =D0 L2 which is continuous in D and satisfies the boundary conditions
lu=partu
partl=2Re [λ(z)uz]=r(z) zisinΓ u(0)=b0 u(2)=b2 (56)
Re [λ(z)umacrz] = r(z) z isin L1 Im [λ(z)umacrz]|z=z1 = b1 (57)
where uz = [radicminusKux + iuy]2 λ(z) = a(x) + ib(x) = cos(l x) minus i cos(l y) if z isin Γ
and λ(z) = a(z) + jb(z) if z isin L1 b0 b1 b2 are real constants and λ(z)(|λ(z)| =1) r(z) b0 b1 b2 satisfy the conditions
196 V Second Order Linear Mixed Equations
Cα[λ(z)Γ]lek0 Cα[r(z)Γ]lek2 Cα[λ(z)L1]lek0 Cα[r(z)L1]lek2
cos(l n) ge 0 on Γ |b0| |b1| |b2|lek2 maxzisinL1
1|a(z)minus b(z)| le k0
(58)
in which n is the outward normal vector at every point on Γ α (12 lt α lt 1) k0 k2
are non-negative constants For convenience we may assume that uz(z1) = 0 other-wise through a transformation of function Uz(z) = uz(z)minusλ(z1) [r(z1)+jb1][a2(z1)+b2(z1)] the requirement can be realized If cos(l n) = 0 on Γ where n is theoutward normal vector on Γ then the problem is called Problem D in whichu(z) = 2Re
int z0 uzdz + b0 = φ(z) on Γ
Problem P for (52) with A3(z) = 0 z isin D r(z) = 0 z isin Γcup Lj (j = 1 or 2) andb0 = b1 = 0 will be called Problem P0 The number
K =12(K1 +K2)
is called the index of Problem P and Problem P0 where
Kj =[φj
π
]+ Jj Jj = 0or 1 eiφj =
λ(tj minus 0)λ(tj + 0)
γj =φj
πminus Kj j = 1 2 (59)
in which t1 = 2 t2 = 0 λ(t) = eiπ4 on L0 = (0 2) and λ(t1 minus 0) = λ(t2 + 0) =exp(iπ4) Here we choose K = 0 or K = minus12 on the boundary partD+ of D+ ifcos(ν n) equiv 0 on Γ and the condition u(2) = b2 can be canceled In fact if cos(l n) equiv 0on Γ from the boundary condition (56) we can determine the value u(2) by the valueu(0) namely
u(2)=2Reint 2
0uzdz+u(0)=2
int 2
0Re [i(zminus1)uz]dθ+b0=2
int 0
πr(z)dθ+b0 (510)
in which λ(z) = i(zminus1) θ = arg(zminus1) on Γ In order to ensure that the solution u(z)of Problem P is continuously differentiable in Dlowast we need to choose γ1 gt 0 If werequire that the solution is only continuous it suffices to choose minus2γ2 lt 1 minus2γ1 lt 1respectively In the following we shall only discuss the case K = 0 and the caseK = minus12 can be similarly discussed Problem P in this case still includes theDirichlet problem (Problem D) as a special case
52 Representation and uniqueness of solutions of oblique derivativeproblem for degenerate equations of mixed type
Now we give the representation theorem of solutions for equation (52)
Theorem 51 Suppose that the equation (52) satisfies Condition C prime Then anysolution of Problem P for (52) can be expressed as
u(z) = 2Reint z
0w(z)dz + b0 w(z) = w0(z) +W (z) (511)
5 Degenerate Mixed Equations 197
where w0(z) is a solution of Problem A for the complex equation
Wmacrz = 0 in D (512)
with the boundary conditions (56) (57) (w0(z) = u0z on Γ w0(z) = u0z on L1) andW (z) in Dminus possesses the form
W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν
2g1(z)dνe1 +
int micro
0g2(z)dmicroe2 in Dminus (513)
in which e1 = (1 + j)2 e2 = (1 minus j)2 micro = x minus 2|y|m2+1(m + 2) ν = x +2|y|m2+1(m+ 2)
g1(z)= A1ξ+B1η+Cu+D ξ=Rew+Imw η=RewminusImw
g2(z)= A2ξ+B2η+Cu+D C= minus f
4|y|m2 D= minus g
4|y|m2
A1=1
4|y|m2
[m
2yminus d
|y|m2 minuse
] B1=
14|y|m2
[m
2yminus d
|y|m2+e
]
A2=1
4|y|m2
[m
2|y|minusd
|y|m2 minuse
] B2=
14|y|m2
[m
2|y|minusd
|y|m2+e
]in Dminus
(514)
Φ(z) is the solutions of equation (52) and w(z) in D+ and Φ(z) in Dminus satisfy theboundary conditions
Re [λ(z)w(z)] = r(z) z isin Γ Re [λ(x)w(x)] = s(x) x isin L0
Re [λ(x)Φ(x)] = Re [λ(x)(W (z)minus Φ(x))] x isin L0 = (0 2)
Re [λ(z)(Φ(z) + Ψ(z))] = 0 z isin L1 Im [λ(z1)(Φ(z1) + Ψ(z1))] = 0
(515)
where λ(x) = 1 + i x isin L0 Moreover by Section 5 Chapter II we see that w0(z) isa solution of Problem A for equation (512) and
u0(z) = 2Reint z
0w0(z)dz + b0 in D (516)
Proof Let u(z) be a solution of Problem P for equation (52) and w(z) = uz u(z)be substituted in the positions of w u in (513) (514) thus the functions g1(z) g2(z)and Ψ(z) in Dminus in (513)(514) can be determined Moreover we can find the solutionΦ(z) in Dminus of (512) with the boundary condition (515) where s(x) on L0 is afunction of λ(z) r(z)Ψ(z) thus
w(z) = w0(z) + Φ(z) + Ψ(z) in Dminus (517)
is the solution of Problem A in Dminus for equation (52) which can be expressed as thesecond formula in (511) and u(z) is a solution of Problem P for (52) as stated inthe first formula in (511)
198 V Second Order Linear Mixed Equations
Theorem 52 Suppose that equation (52) satisfies Condition C prime Then ProblemP for (52) has at most one solution in D
Proof Let u1(z) u2(z) be any two solutions of Problem P for (52) It is easy tosee that u(z) = u1(z)minus u2(z) and W (z) = uz satisfy the homogeneous equation andboundary conditions
Wz
Wz
= A1W + A2W + A3u in
D+
Dminus
(518)
Re [λ(z)W (z)] = 0 z isin Γ u(0) = 0 u(2) = 0
Re [λ(z)W (z)] = 0 z isin L1 Im [λ(z1)W (z1)] = 0(519)
where W (z) = uz in D+ According to the method as stated in Section 5 ChapterII the solution W (z) in the hyperbolic domain Dminus can be expressed in the form
W (z) = Φ(z) + Ψ(z)
Ψ(z) =int ν
2[Aξ+Bη+Cu]e1dν+
int micro
0[Aξ+Bη+Cu]e2dmicro in Dminus
(520)
where Φ(z) is a solution of (512) in Dminus satisfying the boundary condition (515)Similarly to the method in Section 5 Chapter II Ψ(z) = 0 Φ(z) = 0 W (z) = 0z isin Dminus can be derived Thus the solution u(z) = 2Re
int z0 w(z)dz is the solution of
the homogeneous equation of (52) with homogeneous boundary conditions of (56)and (57)
2Re [λ(z)uz]=0 on Γ Re [λ(x)uz(x)]=0 on L0 u(0)=0 u(2)=0 (521)
in which λ(x) = 1 + i x isin L0 = (0 2)
Now we verify that the above solution u(z) equiv 0 in D+ If u(z) equiv 0 in D+noting that u(z) satisfies the boundary condition (521) and similarly to the proof ofTheorem 34 Chapter III we see that its maximum and minimum cannot attain inD+ cup Γ Hence u(z) attains its maximum and minimum at a point zlowast = xlowast isin L0 =(0 2) By using Lemma 41 Chapter III we can derive that ux(xlowast) = 0 uy(xlowast) = 0and then
Re [λ(x)uz(xlowast)] =12
[radicminusK(y)ux(xlowast) + uy(xlowast)
]= 0
this contradicts the second equality in (521) Thus u(z) equiv 0 in D+ This completesthe proof
53 Solvabilty problem of oblique derivative problems for degenerateequations of mixed type
From the above discussion we see in order to prove the existence of solutions ofProblem P for equation (52) the main problem is to find a solution of the oblique
5 Degenerate Mixed Equations 199
derivative problem for the degenerate elliptic equation of second order ie equation(52) in elliptic domain D+ and the oblique derivative boundary conditions is (56)and
Re [λ(x)uz(x)]=s(x) on L0=(0 2) ie
12
[radicminusK(y)ux + uy
]= s(x) on L0
(522)
which is more general than the case as stated in Section 4 Chapter III We try to solvethe problem by using the method of integral equations or the method of auxiliaryfunctions which will be discussed in detail in our other publishers
The references for this chapter are [1][10][12][17][21][22][28][37][43][47][49][57][62][66][69][70][73][77][85][91][93]
CHAPTER VI
SECOND ORDER QUASILINEAR EQUATIONSOF MIXED TYPE
This chapter deals with several oblique derivative boundary value problems for sec-ond order quasilinear equations of mixed (elliptic-hyperbolic) type We shall dis-cuss oblique derivative boundary value problems and discontinuous oblique derivativeboundary value problems for second order quasilinear equations of mixed (elliptic-hyperbolic) type Moreover we shall discuss oblique derivative boundary value prob-lems for general second order quasilinear equations of mixed (elliptic-hyperbolic) typeand the boundary value problems in multiply connected domains In the meantimewe shall give a priori estimates of solutions for above oblique derivative boundaryvalue problems
1 Oblique Derivative Problems for Second Order QuasilinearEquations of Mixed Type
In this section we first give the representation of solutions for the oblique derivativeboundary value problem and then prove the uniqueness and existence of solutionsof the problem and give a priori estimates of solutions of the above problem Finallywe prove the solvability of oblique derivative problems for general quasilinear secondorder equations of mixed type
11 Formulation of the oblique derivative problem for second orderequations of mixed type
Let D be a simply connected bounded domain D in the complex plane CI as statedin Chapter V We consider the second order quasilinear equation of mixed type
uxx + sgny uyy = aux + buy + cu+ d in D (11)
where a b c d are functions of z(isin D) u ux uy (isin IR) its complex form is the fol-lowing complex equation of second order
Luz=
uzz
uzzlowast
=F (z u uz) F =Re [A1uz]+A2u+A3 in
D+
Dminus
(12)
1 Oblique Derivative Problems 201
where Aj = Aj(z u uz) j = 1 2 3 and
uzz =14[uxx + uyy] uzzlowast =
12[(uz)x minus i(uz)y] =
14[uxx minus uyy]
A1 =a+ ib
2 A2 =
c
4 A3 =
d
4in D
Suppose that the equation (12) satisfies the following conditions namely
Condition C
1) Aj(z u uz) (j = 1 2 3) are continuous in u isin IR uz isin CI for almost everypoint z isin D+ and measurable in z isin D+ and continuous in Dminus for all continuouslydifferentiable functions u(z) in Dlowast = D0 xminusy = 2 or Dlowast = Dx+y = 0 2 andsatisfy
Lp[Aj D+] le k0 j = 1 2 Lp[A3 D+] le k1 A2 ge 0 in D+
C[Aj Dminus] le k0 j = 1 2 C[A3 Dminus] le k1(13)
2) For any continuously differentiable functions u1(z) u2(z) in Dlowast the equality
F (z u1 u1z)minus F (z u2 u2z) = Re [A1(u1 minus u2)z] + A2(u1 minus u2) in D (14)
holds where Aj = Aj(z u1 u2) (j = 1 2) satisfy the conditions
Lp[Aj D+] le k0 C[Aj Dminus] le k0 j = 1 2 (15)
in (13)(15) p (gt 2) k0 k1 are non-negative constants In particular when (12) isa linear equation the condition (14) obviously holds
Problem P The oblique derivative boundary value problem for equation (12) isto find a continuously differentiable solution u(z) of (12) in Dlowast = D0 x minus y = 2which is continuous in D and satisfies the boundary conditions
12
partu
partl= Re [λ(z)uz] = r(z) z isin Γ
12
partu
partl= Re [λ(z)uz] = r(z) z isin L1
Im [λ(z)uz]z=z1 = b1 u(0) = b0 u(2) = b2
(16)
where l is a given vector at every point on Γ cup L1 λ(z) = a(x) + ib(x) = cos(l x)∓i cos(l y) and ∓ are determined by z isin Γ and z isin L1 respectively b0 b1 b2 are realconstants and λ(z) r(z) b0 b1 b2 satisfy the conditions
Cα[λ(z)Γ]lek0 Cα[r(z)Γ]lek2 cos(ln)ge0onΓ |b0||b1||b2|lek2
Cα[λ(z)L1]lek0 Cα[r(z)L1]lek2 maxzisinL1 [1|a(x)minusb(x)|]lek0(17)
in which n is the outward normal vector at every point on Γ α (12 lt α lt 1) k0 k2
are non-negative constants For convenience we may assume that w(z1) = 0
202 VI Second Order Quasilinear Mixed Equations
otherwise through a transformation of function W (z) = w(z) minus λ(z1)[r(z1) minus ib1]the requirement can be realized Here we mention that if A2(z) = 0 in D1we can cancel the assumption cos(l n) ge 0 on Γ and if the boundary conditionRe [λ(z)uz] = r(z) z isin L1 is replaced by Re [λ(z)uz] = r(z) z isin L1 then ProblemP does not include the Dirichlet problem (Tricomi problem) as a special case
The boundary value problem for (12) with A3(z u uz) = 0 z isin D u isin IR uz isinCIr(z) = 0 z isin partD and b0 = b2 = b1 = 0 will be called Problem P0 The number
K =12(K1 +K2) (18)
is called the index of Problem P and Problem P0 where
Kj =[φj
π
]+ Jj Jj = 0 or 1 eiφj =
λ(tj minus 0)λ(tj + 0)
γj =φj
πminus Kj j = 1 2 (19)
in which [a] is the largest integer not exceeding the real number a t1 = 2 t2 = 0λ(t) = exp(iπ4) on L0 and λ(t1minus0) = λ(t2+0) = exp(iπ4) here we only discuss thecase of K=0 on partD+ if cos(l n) equiv 0 on Γ or K=minus12 if cos(l n) equiv 0 on Γ becausein this case the last point condition in (16) can be eliminated and the solution ofProblem P is unique In order to ensure that the solution u(z) of Problem P in Dlowast iscontinuously differentiable we need to choose γ1 gt 0 If we require that the solutionof Problem P in D is only continuous it suffices to choose minus2γ1 lt 1 minus2γ2 lt 1
Besides if A2 = 0 in D the last condition in (16) is replaced by
Im [λ(z)uz]|z=z2 = b2 (110)
where the integral path is along two family of characteristic lines similar to thatin (210) Chapter II z2(= 0 2) isin Γ and b2 is a real constant with the condition|b2| le k2 and here the condition cos(l n) ge 0 is canceled then the boundary valueproblem for (12) will be called Problem Q
12 The existence and uniqueness of solutions for the oblique derivativeproblem for (12)
Similarly to Section 2 Chapter V we can prove the following results
Lemma 11 Let equation (12) satisfy Condition C Then any solution of ProblemP for (12) can be expressed as
u(z) = 2Reint z
0w(z)dz + b0 w(z) = w0(z) +W (z) in D (111)
where the integral path in Dminus is the same as in Chapter II and w0(z) is a solutionof Problem A for the equation
Lw =
wz
wzlowast
= 0 in
D+
Dminus
(112)
1 Oblique Derivative Problems 203
with the boundary condition (16) (w0(z) = u0z) and W (z) possesses the form
W (z) = w(z)minus w0(z) in D w(z) = Φ(z)eφ(z) + ψ(z) in D+
φ(z) = φ0(z) + Tg = φ0(z)minus 1π
int intD+
g(ζ)ζ minus z
dσζ ψ(z) = Tf in D+
W (z)=Φ(z) + Ψ(z) Ψ(z)=int ν
2g1(z)dνe1 +
int micro
0g2(z)dmicroe2 in Dminus
(113)
in which Im [φ(z)] = 0 on L0 = (0 2) e1 = (1 + i)2 e2 = (1 minus i)2 micro = x + yν = x minus y φ0(z) is an analytic function in D+ and continuous in D+ such thatIm [φ(x)] = 0 on L0 and
g(z)=
A12+A1w(2w)w(z) =00w(z)=0 zisinD+
f(z)=Re[A1φz]+A2u+A3 inD+
g1(z)=g2(z)=Aξ+Bη+Cu+D ξ=Rew+Imw η=RewminusImw
A=ReA1+ImA1
2 B=
ReA1minusImA1
2 C=A2 D=A3 inDminus
(114)
where Φ(z) is an analytic function in D+ and Φ(z) is a solution of equation (112)in Dminus satisfying the boundary conditions
Re [λ(z)(Φ(z)eφ(z) + ψ(z))] = r(z) z isin Γ
Re [λ(x)(Φ(x)eφ(x) + ψ(x))] = s(x) x isin L0
Re [λ(x)Φ(x)] = Re [λ(x)(W (x)minusΨ(x))] z isin L0
Re [λ(z)Φ(z)] = minusRe [λ(z)Ψ(z)] z isin L1
Im [λ(z1)Φ(z1)] = minusIm [λ(z1)Ψ(z1)]
(115)
in which λ(x) = 1 + i x isin L0 = (0 2) and s(x) is as stated in (223) Chapter VMoreover by Theorem 11 Chapter V the solution w0(z) of Problem A for (112) andu0(z) satisfy the estimate in the form
Cβ[u0(z) D]+Cβ[w0(z)X(z) D+]+Cβ[wplusmn0 (micro ν)Y plusmn(micro ν) Dminus]leM1(k1+k2) (116)
where
X(z) =2prod
j=1
|z minus tj|ηj Y plusmn(z) = Y plusmn(micro ν) = [|ν minus 2||micro minus 2|]ηj
ηj =
2|γj|+ δ γj lt 0
δ γj ge 0j = 1 2
(117)
herein wplusmn0 (micro ν) = Rew0(z) ∓ Imw0(z) w0(z) = w0(micro ν) micro = x + y ν = x minus y
and γ1 γ2 are the real constants in (19) β(lt δ) δ are sufficiently small positiveconstants and
u0(z) = 2Reint z
0w0(z)dz + b0 in D (118)
where p0(2 lt p0 le p) M1 = M1(p0 β k0 D) are non-negative constants
204 VI Second Order Quasilinear Mixed Equations
Theorem 12 Suppose that equation (12) satisfies Condition C Then Problem Pfor (12) has a unique solution u(z) in D
Theorem 13 Suppose that the equation (12) satisfies Condition C Then anysolution u(z) of Problem P for (12) satisfies the estimates
C1β[u D+] = Cβ[u(z) D+] + Cβ[uzX(z) D+] le M2
C1[u Dminus] = Cβ[u(z) Dminus] + C[uplusmnz (micro ν)Y plusmn(micro ν) Dminus] le M3
C1β[u D+] le M4(k1 + k2) C1[u Dminus] le M4(k1 + k2)
(119)
where X(z) Y plusmn(micro ν) are stated in (117) and Mj = Mj (p0 β k0 D) (j = 2 3 4) arenon-negative constants
13 C1α(D)-estimate of solutions of Problem P for second order equations
of mixed type
Now we give the C1α(D)-estimate of solutions u(z) for Problem P for (12) but it
needs to assume the following conditions For any real numbers u1 u2 and complexnumbers w1 w2 we have
|Aj(z1 u1 w1)minusAj(z2 u2 w2)|lek0[|z1minusz2|α+|u1minusu2|α+|w1minusw2|] j=1 2
|A3(z1 u1 w1)minusA3(z2 u2 w2)|lek1[|z1minusz2|α+|u1minusu2|α+|w1minusw2|] z1z2 isinDminus(120)
where α (0 lt α lt 1) k0 k1 are non-negative constants
Theorem 14 If Condition C and (120) hold then any solution u(z) of ProblemP for equation (12) in Dminus satisfies the estimates
C1β[u Dminus]=Cβ[u Dminus]+Cβ[uplusmn
z (micro ν)Y plusmn(micro ν) Dminus]
le M5 C1β[u Dminus] le M6(k1 + k2)
(121)
in which uplusmnz (micro ν) = Reuz ∓ Imuz β (0 lt β le α) M5 = M5(p0 β k D) M6 =
M6(p0 β k0 D) are non-negative constants k = (k0 k1k2)
Proof Similarly to Theorem 13 it suffices to prove the first estimate in (121)Due to the solution u(z) of Problem P for (12) is found by the successive iterationthrough the integral expressions (111) (113) and (114) we first choose the solutionof Problem A of (112) in the form (118) ie
u0(z) = 2Reint z
0w0(z)dz + b0 w0(z) = ξ0(z)e1 + η0(z)e2 in D (122)
and substitute them into the positions of u0 w0 in the right-hand side of (114) wecan obtain Ψ1(z) w1(z) u1(z) as stated in (111)ndash(114) Denote
1 Oblique Derivative Problems 205
u1(z) = 2Reint z
0w1(z)dz + b0 w1(z) = w0(z) + Φ1(z) + Ψ1(z)
Ψ11(z) =
int ν
2G1(z)dν G1(z) = Aξ0 +Bη0 + Cu0 +D
Ψ21(z) =
int micro
0G2(z)dmicro G2(z) = Aξ0 +Bη0 + Cu0 +D
(123)
from the last two equalities in (123) it is not difficult to see that G1(z) =G1(micro ν) Ψ1
1(z) = Ψ11(micro ν) andG2(z) = G2(micro ν) Ψ2
1(z) = Ψ21(micro ν) satisfy the Holder
estimates about ν micro respectively namely
Cβ[G1(middot ν) Dminus] le M7 Cβ[Ψ11(middot ν) Dminus] le M7R
Cβ[G2(micro middot) Dminus] le M8 Cβ[Ψ21(micro middot) Dminus] le M8R
(124)
where Mj = Mj(p0 β k D) (j = 7 8) and R = 2 Moreover from (123) we canderive that Ψ1
1(micro ν) Ψ21(micro ν) about micro ν satisfy the Holder conditions respectively
namelyCβ[Ψ1
1(micro middot) Dminus] le M9R Cβ[Ψ21(middot ν) Dminus] le M9R (125)
where M9 = M9(p0 β k D) Besides we can obtain the estimate of Φ1(z) ie
Cβ[Φ1(z) Dminus] le M10R = M10(p0 β k D)R (126)
in which Φ1(z) satisfies equation (112) and boundary condition of Problem P butin which the function Ψ(z) is replaced by Ψ1(z) Setting w1(z) = w0(z) + Φ1(z) +Ψ1(z) and by the first formula in (123) we can find the function u1(z) from w1(z)Furthermore from (125)(126) we can derive that the functions wplusmn
1 (z) = wplusmn1 (micro ν) =
Re w1(z) ∓ Im w1(z) (w1(z) = w1(z) minus w0(z)) and u1(z) = u1(z) minus u0(z) satisfy theestimates
Cβ[wplusmn1 (micro ν)Y plusmn(micro ν) Dminus] le M11RCβ[u1(z) Dminus] le M11R (127)
where M11 = M11(p0 β k D) Thus according to the successive iteration we canobtain the estimates of functions wplusmn
n (z) = wplusmnn (micro ν) = Re wn(z)∓ Im wn(z) (wn(z) =
wn(z)minuswnminus1(z)) and the corresponding function un(z) = un(z)minusunminus1(z) satisfy theestimates
Cβ[wplusmnn (micro ν)Y plusmn(micro ν) Dminus] le (M11R)n
n Cβ[un(z) Dminus] le (M11R)n
n (128)
Therefore the sequences of functions
wn(z) =nsum
m=1
wm(z) + w0(z) un(z) =nsum
m=1
um(z) + u0(z) n = 1 2 (129)
uniformly converge to w(z) u(z) in any close subset ofDlowast respectively and w(z) u(z)satisfy the estimates
206 VI Second Order Quasilinear Mixed Equations
Cβ[wplusmn(micro ν)Y plusmn(micro ν) Dminus] le eM11R Cβ[u(z) Dminus] le M5 (130)
this is just the first estimate in (121)
From the estimates (119) and (121) we can see the regularity of solutions of Prob-lem P for (12) Moreover it is easy to see that the derivatives [w+(micro ν)]ν [wminus(micro ν)]microsatisfy the estimates similar to those in (130)
As for Problem Q for (12) we can similarly discuss its unique solvability
14 The solvability for the oblique derivative problem for general secondorder quasilinear equations of mixed type
Now we consider the general quasilinear equation of second order
Luz =
uzz
uzzlowast
= F (z u uz) +G(z u uz) z isin
D+
Dminus
F = Re [A1uz] + A2u+ A3 G = A4|uz|σ + A5|u|τ z isin D
(131)
where F (z u uz) satisfies Condition C σ τ are positive constants and Aj(z u uz)(j = 4 5) satisfy the conditions in Condition C the main conditions of which are
Lp[Aj(z u uz) D+] le k0 C[Aj(z u uz) Dminus] le k0 j = 4 5
and denote the above conditions by Condition C prime
Theorem 15 Let the complex equation (131) satisfy Condition C prime
(1) When 0 lt max(σ τ) lt 1 Problem P for (131) has a solution u(z) isin C(D)
(2) When min(σ τ) gt 1 Problem P for (131) has a solution u(z) isin C(D)provided that
M12 = k1 + k2 + |b0|+ |b1| (132)
is sufficiently small
(3) When min(σ τ) gt 1 Problem P for the equation
Luz =
uzz
uzzlowast
= F (z u uz) + εG(z u uz) z isin
D+
Dminus
(133)
has a solution u(z) isin C(D) provided that the positive number ε in (133) is appro-priately small where the functions F (z u uz) G(z u uz) are as stated in (131)
Proof (1) Consider the algebraic equation
M4k1 + k2 + 2k0tσ + 2k0t
τ + |b0|+ |b1| = t (134)
1 Oblique Derivative Problems 207
for t where M4 is the constant stated in (119) It is not difficult to see that equation(134) has a unique solution t = M13 ge 0 Now we introduce a bounded closed andconvex subset Blowast of the Banach space B = C1(D) whose elements are the functionsu(z) satisfying the condition
C1[u(z) D]=Cβ[u(z) D]+Cβ[uzX(z) D+]+C[uplusmnz (micro ν)Y plusmn(micro ν) Dminus]leM13 (135)
We arbitrarily choose a function u0(z) isin B for instance u0(z) = 0 and substitute itinto the position of u in the coefficients of (131) and G(z u uz) From Theorem 12it is clear that problem P for
Luz minus Re [A1(z u0 u0z)uz]minus A2(z u0 u0z)u minus A3(z u0 u0z) = G(z u0 u0z) (136)
has a unique solution u1(z) From Theorem 14 we see that the solution u1(z) satisfiesthe estimate in (135) By using successive iteration we obtain a sequence of solutionsum(z) (m = 1 2 ) isin Blowast of Problem P which satisfy the equations
Lum+1z minus Re [A1(z um umz)um+1z]minus A2(z um umz)um+1
+A3(z um umz) = G(z um umz) in D m = 1 2 (137)
and um+1(z) isin Blowast From (137) we see that and um+1(z) = um+1(z)minusum(z) satisfiesthe equations and boundary conditions
Lum+1zminusRe[A1um+1z]minusA2um+1=G(zumumz)minusG(zumminus1umminus1z) in D
Re[λ(z)um+1z]=0 on Γ Re[λ(z)um+1z]=0 on L1 Im[λ(z)um+1z]|z=z1=0(138)
in which m = 1 2 Noting that C[G(z um umz) minus G(z umminus1 umminus1z) D] le2k0M13 M13 is a solution of the algebraic equation (134) and according to The-orem 13 the estimate
um+1 = C1[um+1 D] le M14 = M14(p0 β k0 D) (139)
can be obtained Due to um+1 can be expressed as
um+1(z) = 2Reint z
0wm+1(z)dz in D wm+1(z) = Φm+1(z) + Ψm+1(z)
Ψm+1(z) =int xminusy
2[Aξm+1 + Bηm+1 + Cum+1 + D]e1d(x minus y)
+int x+y
0[Aξm+1 + Bηm+1 + Cum+1 + D]e2d(x+ y) in Dminus
(140)
in which Φm+1(z) Ψm+1(z) are similar to the functions Φ(z) Ψ(z) in (113) the rela-tion between A1 A2 G and A B C D is the same as that of A1 A2 A3 and A B C Din (114) and G = G(z um umz)minus G(z umminus1 umminus1z) By using the method from theproof of Theorem 13 we can obtain
um+1 minus um = C1[um+1 D] le (M14Rprime)m
m
208 VI Second Order Quasilinear Mixed Equations
where M14 = 2M4(M15 + 1)MR(4m0 + 1) R = 2 m0 = w0(z)X(z) C(D) hereinM15 = maxC[A Q] C[B Q] C[C Q] C[DQ] M = 1 + 4k2
0(1 + k20) From the
above inequality we see that the sequence of functions um(z) ieum(z) = u0(z) + [u1(z)minus u0(z)] + middot middot middot+ [um(z)minus umminus1(z)] m = 1 2 (141)
uniformly converges to a function ulowast(z) and wlowast(z) = ulowastz satisfies the equality
wlowast(z) = w0(z) + Φlowast(z) + Ψlowast(z) in Dminus
Ψlowast(z) = +int xminusy
2[Aξlowast +Bηlowast + Culowast +D]e1d(x minus y)
+int x+y
0[Aξlowast +Bηlowast + Culowast +D]e2d(x+ y) in Dminus
(142)
and the functionulowast(z) = 2Re
int z
0wlowast(z)dz + b0 in D (143)
is just a solution of Problem P for the general quasilinear equation (131) in D
(2) Consider the algebraic equation
M4k1 + k2 + 2k0tσ + 2k0t
τ + |b0|+ |b1| = t (144)
for t It is not difficult to see that the equation (144) has a solution t = M13 ge 0provided that the positive constant M12 in (132) is small enough Now we introducea bounded closed and convex subset Blowast of the Banach space C1(D) whose elementsare the functions u(z) satisfying the condition
C1[u(z) D]=Cβ[u(z) D]+Cβ[uzX(z) D+]+C[uplusmnz (micro ν)Y plusmn(micro ν) Dminus]leM13 (145)
By using the same method as in (1) we can find a solution u(z) isin Blowast of Problem Pfor equation (131) with min(σ τ) gt 1
(3) There is no harm in assuming that k1 k2 in (13)(17) are positive constantswe introduce a bounded closed and convex subset Bprime of the Banach space C1(D)whose elements are the functions u(z) satisfying the condition
C1[u(z) D] le (M4 + 1)(2k1 + k2) (146)
where M4 is a constant as stated in (119) and we can choose an appropriately smallpositive number ε such that C[εG(z u uz) D] le k1 Moreover we are free to choosea function u0(z) isin Bprime for instance u0(z) = 0 and substitute it into the position of uin the coefficients of (133) and G(z u uz) From Theorem 12 it is seen that thereexists a unique solution of Problem P for
Lu minus Re [A1(z u0 u0z)uz]minus A2(z u0 u0z)u minus A3(z u0 u0z) = G(z u0 u0z)
and u1(z) isin Bprime Thus similarly to the proof in (1) by the successive iteration asolution of Problem P for equation (133) can be obtained
2 Mixed Equations in General Domains 209
By using a similar method as before we can discuss the solvability of ProblemP and the corresponding Problem Q for equation (12) or (131) with the boundaryconditions
Re [λ(z)uz] = r(z) z isin Γ u(0) = b0 u(2) = b2
Re [λ(z)uz] = r(z) z isin L2 Im [λ(z)uz]|z=z1 = b1
in which the coefficients λ(z) r(z) b0 b1 b2 satisfy the condition (17) but where theconditions Cα[λ(z) L1] le k0 Cα[r(z) L1] le k2 maxzisinL1 [1|a(x) minus b(x)|] le k0 arereplaced by Cα[λ(z) L2] le k0 Cα[r(z) L2] le k2 maxzisinL2 [1|a(x) + b(x)|] le k0 andin (19) the condition λ(t) = eiπ4 on L0 = (0 2) and λ(t1minus0) = λ(t2+0) = exp(iπ4)is replaced by λ(t) = eminusiπ4 on L0 = (0 2) and λ(t1 minus 0) = λ(t2 + 0) = exp(minusiπ4)Besides the setDlowast = D0 xminusy = 2 in Condition C is replaced byDlowast = Dx+y =0 2 if the constant γ2 gt 0 in (19)
2 Oblique Derivative Problems for Second Order Equationsof Mixed Type in General Domains
This section deals with oblique derivative boundary value problem for secondorder quasilinear equations of mixed (elliptic-hyperbolic) type in general domainsWe prove the uniqueness and existence of solutions of the above problem In refs[12]1)3) the author discussed the Dirichlet problem (Tricomi problem) for secondorder equations of mixed type uxx + sgny uyy = 0 by using the method of integralequations and a complicated functional relation In the present section by usinga new method the solvability result of oblique derivative problem for more generaldomains is obtained
21 Oblique derivative problem for second order equations of mixed typein another domain
Let D be a simply connected bounded domain D in the complex plane CI with theboundary partD = Γ cup L where Γ(sub y gt 0) isin C2
micro (0 lt micro lt 1) with the end pointsz = 0 2 and L = L1 cup L2 cup L3 cup L4 and
L1 = x+ y = 0 0 le x le a2 L2 = x minus y = a a2 le x le a
L3 = x+ y = a a le x le 1 + a2 L4 = x minus y = 2 1 + a2 le x le 2(21)
where a (0 lt a lt 2) is a constant Denote D+ = D cap y gt 0 and Dminus = D cap y lt0 Dminus
1 = Dminuscapxminusy lt a andDminus2 = Dminuscapx+y gt a Without loss of generality we
may assume that Γ = |z minus 1| = 1 y ge 0 otherwise through a conformal mappingthis requirement can be realized
We consider the quasilinear second order mixed equation (12) and assume that(12) satisfies Condition C in D here Dminus is as stated before Problem P for(12)
210 VI Second Order Quasilinear Mixed Equations
inD is to find a continuously differentiablesolution of (12) in Dlowast = D0 a 2 sat-isfying the boundary conditions
12
partu
partl=Re[λ(z)uz]=r(z) zisinΓ
u(0)=b0 u(a)=b1 u(2)=b2 (22)12
partu
partl=Re[λ(z)uz]=r(z) zisinL1cupL4
Im[λ(z)uz]|z=zj=bj+2 j=12
where l is a given vector at every point on Γ cup L1 cup L4 z1 = (1 minus i)a2 z2 =(1 + a2) + i(1 minus a2) λ(z) = a(x) + ib(x) = cos(l x) minus i cos(l y) z isin Γ andλ(z) = a(x) + ib(x) = cos(l x) + i cos(l y) z isin L1 cup L4 bj(j = 0 1 4) are realconstants and λ(z) r(z) bj(j = 0 1 4) satisfy the conditions
Cα[λ(z)Γ] le k0 Cα[r(z)Γ] le k2 Cα[λ(z) Lj] le k0
Cα[r(z) Lj] le k2 j = 1 4 |bj| le k2 j = 0 1 4
cos(l n)ge0 on Γ maxzisinL1
1|a(x)minusb(x)| max
zisinL4
1|a(x)+b(x)| lek0
(23)
in which n is the outward normal vector at every point on Γ α (12 lt α lt 1) k0 k2
are non-negative constants The number
K =12(K1 +K2 +K3) (24)
is called the index of Problem P on the boundary partD+ of D+ where
Kj=[φj
π
]+Jj Jj=0or1 eiφj =
λ(tj minus0)λ(tj+0)
γj=φj
πminusKj j=123 (25)
in which [b] is the largest integer not exceeding the real number b t1 = 2 t2 = 0t3 = a λ(t) = eiπ4 on (0 a) and λ(t3 minus 0) = λ(t2+0) = exp(iπ4) and λ(t) = eminusiπ4
on (a 2) and λ(t1 minus 0) = λ(t3 + 0) = exp(minusiπ4) Here we only discuss the caseK = 12 or K = 0 if cos(l n) = 0 on Γ because in this case the solution of ProblemP is unique and includes the Dirichlet problem (Tricomi problem) as a special caseWe mention that if the boundary condition Re [λ(z)uz] = r(z) z isin Lj (j = 1 4) isreplaced by
Re [λ(z)uz] = r(z) z isin Lj (j = 1 4)
then Problem P does not include the Dirichlet problem (Tricomi problem) as a spe-cial case In order to ensure that the solution u(z) of Problem P is continuouslydifferentiable in Dlowast we need to choose γ1 gt 0 γ2 gt 0 and can select γ3 = 12 Ifwe only require that the solution u(z) in D is continuous it is sufficient to chooseminus2γ1 lt 1 minus2γ2 lt 1 minusγ3 lt 1
2 Mixed Equations in General Domains 211
Besides we consider the oblique derivative problem (Problem Q) for equation(12) with A2 = 0 and the boundary condition (22) but the last point conditionsu(a) = b1 u(2) = b2 in (22) is replaced by
Im [λ(z)uz]|zprimej= cj j = 1 2 (26)
in which zprimej(j = 1 2) isin Γlowast = Γ0 2 are two points and c1 c2 are real constants and
|c1| |c2| le k2
Similarly to Section 1 we can give a representation theorem of solutions of ProblemP for equation (12) in which the functions Ψ(z) in (113) λ(x) s(x) on L0 in (115)X(z) Y plusmn(micro ν) in (117) are replaced by
Ψ(z) =
⎧⎪⎪⎨⎪⎪⎩int ν
a
g1(z)dνe1 +int micro
0g2(z)dmicroe2 z isin Dminus
1 int ν
2g1(z)dνe1 +
int micro
a
g2(z)dmicroe2 z isin Dminus2
(27)
λ(x) =
1 + i x isin Lprime
0 = (0 a)
1minus i x isin Lprimeprime0 = (a 2)
(28)
s(x)=
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
2r((1minusi)x2)minus2Re[λ((1minusi)x2)Ψ((1minusi)x2)a((1minusi)x2)minusb((1minusi)x2)
+Re[λ(x)Ψ(x)]
minus [a((1minusi)x2)+b((1minusi)x2)]f(0)a((1minusi)x2)minusb((1minusi)x2)
xisin(0a)
2r((1+i)x2+1minusi)minus2Re[λ((1+j)x2+1minusi)Φ((1+j)x2+1minusi)]a((1+i)x2+1minusi)+b((1+i)x2+1minusi)
minusa((1+i)x2+1minusi)minusb((1+i)x2+1minusi)g(2)minush(x)a((1+i)x2+1minusi)+b((1+i)x2+1minusi)
h(x)=Re[λ(x)Ψ(x)]
times[a((1+i)x2+1minusi)+b((1+i)x2+1minusi)]2xisin(a2)
f(0)=[a(z1)+b(z1)]r(z1)+[a(z1)minusb(z1)]b3
g(2)=[a(z2)minusb(z2)]r(z2)minus [a(z2)+b(z2)]b4
(29)
X(z) =3prod
j=1
|z minus tj|ηj Y plusmn(z) = Y plusmn(micro ν) =3prod
j=1
[|micro minus tj||ν minus tj|]ηj
ηj =
2max(minusγj 0) + δ j = 1 2
max(minusγj 0) + δ j = 3
(210)
respectively δ is a sufficiently small positive constant besides L1 and the point z1 in(115) should be replaced by L1 cup L4 and z1 = (1minus i)a2 z2 = 1 + a2 + (1minus a2)i
Now we first prove the unique solvability of Problem Q for equation (12)
212 VI Second Order Quasilinear Mixed Equations
Theorem 21 If the mixed equation (12) in the domain D satisfies Condition Cthen Problem Q for (12) has a unique solution u(z) as stated in the form
u(z) = 2Reint z
0w(z)dz + b0 in D (211)
wherew(z) = Φ(z)eφ(z) + ψ(z) in D+
φ(z) = φ0(z) + Tg = φ0(z)minus 1π
int intD+
g(ζ)ζ minus z
dσζ ψ(z) = Tf in D+
w(z) = w0(z) + Φ(z) + Ψ(z) in Dminus
Ψ(z) =
⎧⎪⎪⎪⎨⎪⎪⎪⎩int ν
a
g1(z)dνe1 +int micro
0g2(z)dmicroe2 z isin Dminus
1 int ν
2g1(z)dνe1 +
int micro
a
g2(z)dmicroe2 z isin Dminus2
(212)
where Im [φ(z)] = 0 on L0 = (0 2) e1 = (1 + i)2 e2 = (1 minus i)2 micro = x + yν = x minus y φ0(z) is an analytic function in D+ and continuous in D+ such thatIm [φ(z)] = 0 on L0 and f(z) g(z) g1(z) g2(z) are as stated in (114) and Φ(z) isan analytic function in D+ and Φ(z) is a solution of equation (112) satisfying theboundary conditions the first conditions and
Re [λ(z)Φ(z)] = minusRe [λ(z)Ψ(z)] z isin L1 cup L4
Im [λ(zj)Φ(zj)] = minusIm [λ(zj)Ψ(zj)] j = 1 2
Im [λ(zprimej)Φ(z
primej)] = cj j = 1 2
(213)
in which λ(x) s(x) are as stated in (28)(29)
Proof By using a similar method as stated in Section 2 Chapter V we can proveTheorem 21 provided that L1 or L2 in the boundary conditions Section 2 ChapterV is replaced by L1 cup L4 the point conditions Im [λ(z1)w(z1)] = b1 in Section 2Chapter V is replaced by Im [λ(zj)w(zj)] = bj j = 1 2 and so on the formula (214)Chapter V is replaced by
Re [λ(z)uz] = Re [λ(z)w(z)] = s(x)
λ(x) =
1 + i x isin Lprime
0 = (0 a)
1minus i x isin Lprimeprime0 = (a 2)
Cβ[s(x) Lprime0] + Cβ[s(x) Lprimeprime
0] le k3
Theorem 22 Suppose that the equation (12) satisfies Condition C Then ProblemP for (12) has a unique solution u(z) and the solution u(z) satisfies the estimates
C1β[u(z) D+] = Cβ[u(z) D+] + Cβ[uzX(z) D+] le M15
C1[u(z) Dminus] = Cβ[u(z) Dminus] + C[uplusmnz Y plusmn(z) Dminus] le M15
C1β[u(z) D+]leM16(k1+k2) C1[u(z) Dminus]leM16(k1+k2)
(214)
2 Mixed Equations in General Domains 213
where
X(z)=3prod
j=1
|zminustj|ηj Y plusmn(z)=3prod
j=1
|xplusmnyminustj|ηj ηj=
2max(minusγj0)+δ j=12
max(minusγj0)+δ j=3(215)
herein t1 = 2 t2 = 0 t3 = a γ1 γ2 γ3 are real constants in (25) β(lt δ) δ aresufficiently small positive constants and M15 = M15(p0 β k D) M16 = M16(p0 βk0 D) are non-negative constants k = (k0 k1 k2)
Proof First of all we prove the uniqueness of solutions of Problem P for (12)Suppose that there exist two solutions u1(z) u2(z) of Problem P for (12) By Condi-tion C we can see that u(z) = u1(z)minusu2(z) and w(z) = uz satisfies the homogeneousequation and boundary conditions
wz
wzlowast
= f f = Re [A1w] + A2u in
D+
Dminus
(216)
Re [λ(z)w(z)] = r(z) z isin Γ u(0) = 0 u(a) = 0
u(2) = 0 Re [λ(z)w(z)] = 0 z isin L1 cup L4(217)
and (211) By using the method of proofs in Theorems 23 and 24 Chapter Vw(z) = uz = 0 u(z) = 0 in Dminus can be derived Thus we have
2Re[1minus iradic2
uz
]=
partu
partl= 0 on (0 a) 2Re
[1 + iradic2
uz
]=
partu
partl= 0 on (a 2) (218)
it is clear that (1minus i)radic2 = cos(l x) minus i cos(l y) = exp(minusiπ4) on (0 a) and
(1 + i)radic2 = cos(l x) + i cos(l y) = exp(iπ4) on (a 2) On the basis of the max-
imum principle of solutions for (12) with A3 = 0 in D+ if maxD+ u(z) gt 0 then itsmaximum M attains at a point zlowast isin Γ cup L0 obviously zlowast = 0 a and 2 and we canprove zlowast isin Γ by the method as stated in the proof of Theorem 23 Chapter V More-over it is not difficult to prove that if zlowast isin L0 then partupartl = 0 at zlowast This contradicts(217) Thus maxD+ u(z) = 0 By the similar method we can prove minD+ u(z) = 0Hence u(z) = 0 u1(z) = u2(z) in D+
Secondly we first prove the existence of solutions of Problem P for the linearequation (12) with A2 = 0 ie
uzz = Re [A1uz] + A3 z isin D+
uzzlowast = Re [A1uz] + A3 z isin Dminus(219)
By Theorem 21 we can prove the solvability of Problem P for (219) In fact ifu0(a) = b1 u0(2) = b2 then the solution u0(z) is just a solution of Problem P for(12) Otherwise u0(a) = cprime
1 = b1 or u0(2) = cprime2 = b2 we find a solution u2(z) of
214 VI Second Order Quasilinear Mixed Equations
Problem P for (219) with the boundary conditions
Re [λ(z)ukz] = 0 z isin Γ uk(0) = 0
Re [λ(z)ukz] = 0 z isin L1 cup L4 k = 1 2
Im [λ(z)ukz]|z=zprimej= δkj k j = 1 2
(220)
in which δ11 = δ22 = 1 δ12 = δ21 = 0 It is clear that
J =
∣∣∣∣∣u1(a) u2(a)
u1(2) u2(2)
∣∣∣∣∣ = 0 (221)
Because otherwise there exist two real constants d1 d2 (|d1| + |d2| = 0) such thatd1u1(z) + d2u2(z) equiv 0 in D and
d1u1(a) + d2u2(a) = 0 d1u1(2) + d2u2(2) = 0 (222)
According to the proof of uniqueness as before we can derive d1u1(z) + d2u2(z) equiv 0in D the contradiction proves J = 0 Hence there exist two real constants d1 d2such that
d1u1(a) + d2u2(a) = cprime1 minus b1 d1u1(2) + d2u2(2) = cprime
2 minus b2 (223)
thus the functionu(z) = u0(z)minus d1u1(z)minus d2u2(z) in D (224)
is just a solution of Problem P for equation (12) in the linear case Moreover we canobtain that the solution u(z) of Problem P for (12) satisfies the estimates in (214)we can rewrite in the form
C1[u D] = Cβ[u(z) D] + Cβ[uzX(z) D+]
+C[uplusmnz (micro ν)Y plusmn(micro ν) Dminus] le M17
C1[u D] le M18(k1 + k2)
(225)
where X(z) Y plusmn(micro ν) are as stated in (210) and M17 = M17(p0 β k D) M18 =M18(p0 β k0 D) are non-negative constants k = (k0 k1 k2) By using the estimateand method of parameter extension the existence of solutions of Problem P forquasilinear equation (12) can be proved
22 Oblique derivative problem for second order equations of mixed typein general domains
Now we consider the domain Dprime with the boundary partDprime = Γ cup Lprime where Lprime =Lprime
1 cup Lprime2 cup Lprime
3 cup Lprime4 and the parameter equations of the four curves Lprime
1 Lprime2 Lprime
3 Lprime4 are
Lprime1 = γ1(x) + y = 0 0 le x le l1 Lprime
2 = x minus y = a l1 le x le a
Lprime3 = x+ y = a a le x le l2 Lprime
4 = γ2(x) + y = 0 l2 le x le 2(226)
2 Mixed Equations in General Domains 215
in which γ1(0) = 0 γ2(2) = 0 γ1(x)gt 0 on 0le x le l1 = γ(l1)+a γ2(x)gt 0 on l2 =minusγ(l2)+a le x le 2 γ1(x) on 0 le x le l1 γ2(x) on l2 le x le 2 are continuous andγ1(x) γ2(x) are differentiable on 0le x le l1 l2 le x le 2 except some isolated pointsand 1 + γprime
1(x)gt 0 on 0 le x le l1 1 minus γprime2(x)gt 0 on l2 le x le 2 Denote Dprime+ =Dprimecap
y gt 0=D+ Dprimeminus=Dprime cap y lt 0 Dprimeminus1 =Dprimeminus cap x lt a and Dprimeminus
2 =Dprimeminus cap x gt aHere we mention that in [12]1)3) the author assumed 0 lt minusγprime
1(x) lt 1 on 0 le x le l1and some other conditions
We consider the quasilinear second order equation of mixed (elliptic-hyperbolic)type (12) in Dprime Assume that equation (12) satisfies Condition C but the hyper-bolic domain Dminus is replaced by Dprimeminus
Problem P prime The oblique derivative problem for equation (12) is to find a con-tinuously differentiable solution of (12) in Dlowast = Dprime0 a 2 for (12) satisfying theboundary conditions
12
partu
partl= Re [λ(z)uz] = r(z) z isin Γ u(0) = b0 u(a) = b1 u(2) = b2
12
partu
partl=Re [λ(z)uz]=r(z) zisinLprime
1cupLprime4 Im [λ(z)uz]|z=zj
=bj+2 j=12(227)
Here l is a given vector at every point on Γ cup Lprime1 cup Lprime
4 z1 = l1 minus iγ1(l1) z2 =l2 minus iγ2(l2) λ(z) = a(x) + ib(x) = cos(l x) minus i cos(l y) z isin Γ and λ(z) = a(x) +ib(x) = cos(l x) + i cos(l y) z isin Lprime
1 cup Lprime4 bj(j = 0 1 4) are real constants and
λ(z) r(z) bj(j = 0 1 4) satisfy the conditions
Cα[λ(z)Γ] le k0 Cα[r(z)Γ] le k2
Cα[λ(z) Lprimej] le k0 Cα[r(z) Lprime
j] le k2 j = 1 4
cos(l n) ge 0 on Γ |bj| le k2 0 le j le 4
maxzisinLprime
1
1|a(x)minus b(x)| max
zisinLprime4
1|a(x) + b(x)| le k0
(228)
in which n is the outward normal vector at every point on Γ α (12 lt α lt 1) k0 k2
are non-negative constants In particular if Lprimej = Lj(j = 1 2 3 4) then Problem P prime
in this case is called Problem P
In the following we discuss the domain Dprime with the boundary ΓcupLprime1cupLprime
2cupLprime3cupLprime
4where Lprime
1 Lprime2 Lprime
3 Lprime4 are as stated in (226) and γ1(x) γ2(x) satisfy the conditions
1 + γprime1(x) gt 0 on 0 le x le l1 and 1minus γprime
2(x) gt 0 on l2 le x le 2 By the conditions the
216 VI Second Order Quasilinear Mixed Equations
inverse functions x = σ(ν) x = τ(micro) of x+ γ1(x) = ν = xminus y xminus γ2(x) = micro = x+ ycan be found respectively namely
micro = 2σ(ν)minus ν 0 le ν le a ν = 2τ(micro)minus micro a le micro le 2 (229)
They are other expressions for the curves Lprime1 Lprime
4 Now we make a transformationin Dprimeminus
micro =a[micro minus 2σ(ν) + ν]a minus 2σ(ν) + ν
ν = ν 2σ(ν)minus ν le micro le a 0 le ν le a
micro=micro ν=a[2τ(micro)minusmicrominus2]+(2minusa)ν
2τ(micro)minus micro minus a alemicrole2 aleν le2τ(micro)minusmicro
(230)
in which micro ν are variables If (micro ν) isin Lprime1 Lprime
2 Lprime3 Lprime
4 then (micro ν) isin L1 L2 L3 L4
respectively The inverse transformation of (230) is
micro =1a[a minus 2σ(ν) + ν]micro+ 2σ(ν)minus ν
=1a[a minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+y)+2σ(x+γ1(x))minusxminusγ1(x)
ν= ν= xminusy 0 le micro le a 0 le ν le a micro = micro = x+ y
ν =1
2minus a[(2τ(micro)minus micro)(ν minus a)minus a(ν minus 2)]
=12minusa
[(2τ(xminusγ2(x))minusx+γ2(x))(xminusyminusa)minusa(xminusyminus2)]
a le microle2 ale ν le2
(231)
It is not difficult to see that the transformations in (231) map the domains Dprimeminus1 Dprimeminus
2onto the domains Dminus
1 Dminus2 respectively Moreover we have
x =12(micro+ ν) =
2ax minus (a+ x minus y)[2σ(x+ γ1(x))minus x minus γ1(x)]2a minus 4σ(x+ γ1(x)) + 2x+ 2γ1(x)
y =12(micro minus ν) =
2ay minus (a minus x+ y)[2σ(x+ γ1(x))minus x minus γ1(x)]2a minus 4σ(x+ γ1(x)) + 2x+ 2γ1(x)
x =12(micro+ ν) =
12a[a minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+ y)
+σ(x+ γ1(x))minus 12(x+ γ1(x)minus x+ y)
y =12(micro minus ν) =
12a[a minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+ y)
+σ(x+ γ1(x))minus 12(x+ γ1(x) + x minus y)
(232)
2 Mixed Equations in General Domains 217
and
x=12(micro+ν) =
[2τ(xminusγ2(x))minusx+γ2(x)](x+y+a)+2(xminusy)minus2a(1+x)2[2τ(xminusγ2(x))minusx+γ2(x)minusa]
y=12(microminusν) =
[2τ(xminusγ2(x))minusx+γ2(x)](x+yminusa)minus2(xminusy)+2a(1minusy)2[2τ(xminusγ2(x))minusx+γ2(x)minusa]
x=12(micro+ν) =
12(2minusa)
[2τ(xminusγ2(x))minusx+γ2(x))(xminus yminusa)
+2(x+ y+a)minus2ax]
y=12(microminusν) =
12(2minusa)
[2τ(xminusγ2(x))minusx+γ2(x))(minusx+ y+a)
+2(x+ yminusa)minus2ay]
(233)
Denote by z = x + iy = f(z) z = x + iy = g(z) and z = x + iy = fminus1(z) z =x+iy = gminus1(z) the transformations and their inverse transformations in (232) (233)respectively Through the transformation (230) we have
(U+V )ν=(U+V )ν (UminusV )micro=1a[aminus2σ(ν)+ν](UminusV )micro in Dprimeminus
1
(U + V )ν=2τ(micro)minus micro minus a
2minus a(U + V )ν (U minus V )micro=(U minus V )micro in Dprimeminus
2
(234)
Equation (12) in Dprimeminus can be rewritten in the form
ξν = Aξ +Bη + Cu+D ηmicro = Aξ +Bη + Cu+D z isin Dprimeminus (235)
where ξ = U + V = (ux minus uy)2 η = U minus V = (ux+ uy)2 under the transformation(234) it is clear that system (235) in Dprimeminus is reduced to
ξν=Aξ+Bη+ Cu+D ηmicro=1a[aminus2σ(ν)+ν][Aξ+Bη+Cu+D] in Dminus
1
ξν =2τ(micro)minusmicrominusa
2minus a[Aξ+Bη+Cu+D] ηmicro = Aξ+Bη+Cu+D in Dminus
2
(236)
Moreover through the transformations (232)(233) the boundary condition (223)on Lprime
1 cup Lprime4 is reduced to
Re [λ(fminus1(z))w(fminus1(z))]=r[fminus1(z)] zisinL1 Im [λ(fminus1(z3))w(fminus1(z3))]=b1
Re [λ(gminus1(z))w(gminus1(z))]=r[gminus1(z)] z isin L4 Im [λ(gminus1(z4))w(gminus1(z4))]=b2(237)
in which z3 = f(z3) z4 = g(z4) Therefore the boundary value problem (12) (inD+) (235) (227) (26) is transformed into the boundary value problem (12) (236)(227) (237) According to the proof of Theorem 21 we see that the boundary value
218 VI Second Order Quasilinear Mixed Equations
problem (12) (236) (227) (237) has a unique solution w(z) and then w[z(z)] is asolution of the boundary value problem (12)(22) (w = uz) in Dprimeminus and the function
u(z) =
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩
2Reint z
0w(z)dz + b0 in D+
2Reint z
a
w[f(z)]dz + u(a) in Dprimeminus1
2Reint z
2w[g(z)]dz + u(2) in Dprimeminus
2
(238)
is just a solution of Problem P for (12) in Dprime where u(a) = b1 u(2) = b2
Theorem 23 If the mixed equation (12) in the domain Dprime satisfies Condition Cthen Problem P prime for (12) with the boundary condition (22) has a unique solutionu(z) as stated in (238) where z1 = l1 minus iγ1(l1) z2 = l2 minus iγ2(l2)
By using the above method and the method in Section 4 Chapter IV we candiscuss the unique solvability of Problem P prime for equation (12) in some more generaldomains Dprimeprime including the domain Dprimeprime = |z minus 1| lt 1 Im z ge 0 cup |z minus a2| lta24 Im z lt 0 cup |z minus 1minus a2| lt (2minus a)24 Im z lt 0
3 Discontinuous Oblique Derivative Problems for SecondOrder Quasilinear Equations of Mixed Type
This section deals with discontinuous oblique derivative problems for quasilinear sec-ond order equations of mixed (elliptic-hyperbolic) type in a simply connected domainFirstly we give a representation theorem and prove the uniqueness of the solutionfor the above boundary value problem and then by using the method of successiveiteration the existence of solutions for the above problem is proved
31 Formulation of discontinuous oblique derivative problems for secondorder equations of mixed type
Let D be a simply connected domain with the boundary ΓcupL1 cupL2 as stated beforewhere D+ = |zminus1| lt 1 Im z gt 0 We discuss the second order quasilinear complexequations of mixed type as stated in (12) with Condition C In order to introducethe discontinuous oblique derivative boundary value problem for equation (12) letthe functions a(z) b(z) possess discontinuities of first kind at m minus 1 distinct pointsz1 z2 zmminus1 isin Γ which are arranged according to the positive direction of Γ andZ = z0 = 2 z1 zm = 0 cup x + y = 0 x minus y = 2 Im z le 0 where m is apositive integer and r(z) = O(|z minus zj|minusβj) in the neighborhood of zj(j = 0 1 m)on Γ in which βj(j = 0 1 m) are sufficiently small positive numbers Denoteλ(z) = a(z) + ib(z) and |a(z)| + |b(z)| = 0 There is no harm in assuming that
3 Discontinuous Oblique Derivative Problems 219
|λ(z)| = 1 z isin Γlowast = ΓZ Suppose that λ(z) r(z) satisfy the conditions
λ(z) isin Cα(Γj) |z minus zj|βjr(z) isin Cα(Γj) j = 1 m (31)
herein Γj is the arc from the point zjminus1 to zj on Γ and z0 = 2 and Γj(j = 1 m)does not include the end points α (0 lt α lt 1) is a constant Besides there existn points E1 = a1 E2 = a2 En = an on the segment AB = L0 = (0 2) andE0 = 0 En+1 = 2 where a0 = 0 lt a1 lt a2 lt middot middot middot lt an lt an+1 = 2 Denote byA = A0 = 0 A1 = (1minus i)a12 A2 = (1minus i)a22 An = (1minus i)an2 An+1 = C =1minus i and B1 = 1minus i + (1 + i)a12 B2 = 1minus i + (1 + i)a22 Bn = 1minus i + (1 +i)an2 Bn+1 = B = 2 on the segments AC CB respectively Moreover we denoteDminus
1 = Dminuscapcup[n2]j=0 (a2j le xminusy le a2j+1) Dminus
2 = Dminuscapcup[(n+1)2]j=1 (a2jminus1 le x+y le a2j)
and Dminus2j+1 = Dminus cap a2j le x minus y le a2j+1 j = 0 1 [n2] Dminus
2j = Dminus cap a2jminus1 lex + y le a2j j = 1 [(n + 1)2] and Dminus
lowast = Dminuscupn+1j=0 (x plusmn y = aj y le 0)
Dlowast = D+ cup Dminuslowast
The discontinuous oblique derivative boundary value problem for equation (12)may be formulated as follows
Problem P prime Find a continuous solution u(z) of (12) in D which is continuouslydifferentiable in Dlowast = D+ cup Dminus
lowast and satisfies the boundary conditions
12
partu
partl= Re [λ(z)uz] = r(z) z isin Γ
12
partu
partl= 2Re [λ(z)uz] = r(z) z isin L3 =
[n2]sumj=0
A2jA2j+1
12
partu
partl= 2Re [λ(z)uz] = r(z) z isin L4 =
[(n+1)2]sumj=1
B2jminus1B2j
(32)
Im [λ(z)uz]|z=A2j+1 = c2j+1 j = 0 1 [n2]
Im [λ(z)uz]|z=B2jminus1 = c2j j = 1 [(n+ 1)2]
u(zj) = bj j = 0 1 m u(aj) = bm+j j = 1 n
(33)
where l is a vector at every point on Γ cup L3 cup L4 bj(j = 0 1 m + n)cj(j = 0 1 n + 1 c0 = b0) are real constants λ(z) = a(x) + ib(x) =cos(l x)minus i cos(l y) z isin Γ λ(z) = a(x) + ib(x) = cos(l x) + i cos(l y) z isin L3 cup L4and λ(z) r(z) cj(j = 0 1 n+ 1) satisfy the conditions
Cα[λ(z)Γ] le k0 Cα[r(z)Γ] le k2 |cj| le k2 j = 0 1 n+ 1
Cα[λ(z) Lj]lek0 Cα[r(z) Lj]lek2 j=34 |bj|lek2 j=0 1 m+n
cos(l n) ge 0 on Γ maxzisinL3
1|a(x)minus b(x)| le k0 max
zisinL4
1|a(x) + b(x)| le k0
(34)
220 VI Second Order Quasilinear Mixed Equations
where n is the outward normal vector at every point on Γ α (12 lt α lt 1) k0 k2 arenon-negative constants The above discontinuous oblique derivative boundary valueproblem for (12) is called Problem P prime Problem P prime for (12) with A3(z u uz) = 0z isin D r(z) = 0 z isin Γ cup L3 cup L4 bj = 0(j = 0 1 m + n) and cj = 0(j = 0 1 n+1) will be called Problem P prime
0 Moreover we give the same definitionsas in (510) (511) Chapter IV but choose K = (m+nminus1)2 or K = (m+n)2minus1if cos(ν n) equiv 0 on Γ and the condition u(zm) = bm can be canceled Besides werequire that the solution u(z) in D+ satisfies the conditions
uz = O(|z minus zj|minusδ) δ =
βj + τ for γj ge 0 and γj lt 0 βj gt |γj||γj|+ τ for γj lt 0 βj le |γj| j = 0 1 m+ n
(35)in the neighborhood of zj (0 le j le m) aj (1 le j le n) in D+ where γprime
j =max (0 minusγj) (j = 0 1 m + n) γprime
0 = max (0 minus2γ0) γprimem = max (0 minus2γm) and
γj (j = 0 1 m + n) are real constants as stated in (510) Chapter IV δ is asufficiently small positive number Now we explain that in the closed domain Dminusthe derivatives ux+uy ux minusuy of the solution u(z) in the neighborhoods of the 2n+2characteristic lines Z prime = x+ y = 0 x minus y = 2 x plusmn y = aj (j = 1 n) y le 0 maybe not bounded if γj le 0(j = 0 1 n+1) Hence if we require that the derivativeuz of u(z) in DminusZ prime is bounded then we need to choose γj gt 0 (j = 0 1 n+ 1)If we only require that the solution u(z) is continuous in D it suffices to chooseminus2γ0 lt 1 minus2γm lt 1 minusγj lt 1 (j = 1 m minus 1 m+ 1 m+ n)
Furthermore we need to introduce another oblique derivative boundary valueproblem
Problem Qprime If A2(z) = 0 in D we find a continuously differentiable solution u(z) of(12) inDlowast which is continuous in D and satisfies the boundary conditions (32)(33)but the point conditions in (33) are replaced by
u(2) = b0 = d0 Im [λ(z)uz]|z=zprimej= dj j = 1 m+ n (36)
where zprimej(isin Z) isin Γ(j = 0 1 m+n) are distinct points dj(j=0 1 m+n) are
real constants satisfying the conditions |dj| le k2 j = 0 1 m + n but we do notassume cos(ν n) ge 0 on each Γj(j = 1 m)
32 Representations of solutions for the oblique derivative problem for(12)
First of all we give the representation of solutions of Problem Qprime for the equationuzz
uzzlowast
= 0 in
D+
Dminus
(37)
3 Discontinuous Oblique Derivative Problems 221
It is clear that Problem Qprime for (37) is equivalent to the following boundary valueproblem (Problem Aprime) for the first order complex equation
Lw =
wz
wzlowast
= 0 in
D+
Dminus
(38)
with the boundary conditions
Re [λ(z)w(z)] = r(z) on Γ
Re [λ(z)w(z)] = r(z) on L3 cup L4
Im [λ(z)w(z)]|z=zprimej= bj j = 1 m+ n
Im [λ(z)w(z)]|z=A2j+1 = c2j+1 j = 0 1 [n2]
Im [λ(z)w(z)]|z=B2jminus1 = c2j j = 1 [(n+ 1)2]
(39)
and the relation
u(z) = 2Reint z
2w(z)dz + b0 (310)
where the integral path is appropriately chosen Thus from Theorem 52 ChapterIV we can derive the following theorem
Theorem 31 The boundary value problem Qprime for (37) in D has a unique continu-ous solution u(z) as stated in (310) where the solution w(z) of Problem Aprime for (38)in Dminus possesses the form
w(z) =12[(1minus i)f(x+ y) + (1 + i)g(x minus y)]
f(x+ y) = Re [(1 + i)w(x+ y)] in DminusDminus2
g(x minus y) = k(x minus y) in Dminus1
f(x+ y) = h(x+ y) in Dminus2
g(x minus y) = Re [(1minus i)w(x minus y)] in DminusDminus1
(311)
herein w(x + y)(0 le x + y le 2) w(x minus y)(0 le x minus y le 2) are values of the solutionw(z) of Problem Aprime for (38) in D+ with the first boundary condition in (39) and theboundary condition
Re [λ(x)w(x)] =
⎧⎨⎩k(x) on Lprime1 = Dminus
1 cap AB
h(x) on Lprime2 = Dminus
2 cap AB(312)
222 VI Second Order Quasilinear Mixed Equations
in which k(x) h(x) can be expressed as
k(x)=2r((1minusi)x2)minus[a((1minusi)x2)+b((1minusi)x2)]h2j
a((1minus i)x2)minus b((1minus i)x2)
on L2j+1 = Dminus2j+1 cap AB
h(x)=2r((1 + i)x2 + 1minus i)
a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)
minus [a((1+i)x2+1minusi)minusb((1+i)x2+1minusi)]k2jminus1
a((1+i)x2+1minusi)+b((1+i)x2+1minusi)
on L2j = Dminus2j cap AB
(313)
where Dminusj (j = 1 2 2n+ 1) are as stated before and
h2j = Re [λ(A2j+1)(r(A2j+1) + ic2j+1)] + Im [λ(A2j+1)(r(A2j+1) + ic2j+1)]
on L2j+1 j = 0 1 [n2]
k2jminus1 = Re [λ(B2jminus1)(r(B2jminus1) + ic2j)]minus Im [λ(B2jminus1)(r(B2jminus1) + ic2j)]
on L2j j = 1 [(n+ 1)2]
where L2j+1 = Dminus2j+1 cap AB j = 0 1 [n2] L2j = Dminus
2j cap AB j = 1 [(n+ 1)2]
Next we give the representation theorem of solutions of Problem Qprime for equation(12)
Theorem 32 Suppose that equation (12) satisfies Condition C Then any solutionof Problem Qprime for (12) can be expressed as
u(z) = 2Reint z
2w(z)dz + c0 w(z) = w0(z) +W (z) in D (314)
where w0(z) is a solution of Problem Aprime for the complex equation (38) with the bound-ary condition (32) (36) (w0(z) = u0z) and w(z) possesses the form
w(z)=W (z)+w0(z) inD w(z)=Φ(z)eφ(z)+ψ(z)
φ(z)=φ0(z)+Tg=φ0(z)minus 1π
int intD+
g(ζ)ζminusz
dσζ ψ(z)=Tf in D+
W (z)=Φ(z)+Ψ(z) in Dminus
Ψ(z)=
⎧⎪⎪⎪⎨⎪⎪⎪⎩
int ν
a2j+1
g1(z)dνe1+int micro
0g2(z)dmicroe2 in Dminus
2j+1 j=01[n2]
int ν
2g1(z)dνe1+
int micro
a2jminus1
g2(z)dmicroe2 in Dminus2j j=1[(n+1)2]
(315)
3 Discontinuous Oblique Derivative Problems 223
in which e1 = (1 + i)2 e2 = (1 minus i)2 micro = x + y ν = x minus y φ0(z) is an analyticfunction in D+ Im [φ(z)] = 0 on Lprime = (0 2) and
g(z)=
A12+A1w(2w) w(z) =00 w(z)=0 zisinD+
f(z)=Re[A1φz]+A2u+A3
g1(z)=g2(z)=Aξ+Bη+Cu+D ξ=Rew+Imw η=RewminusImw
A=ReA1+ImA1
2 B=
ReA1minusImA1
2 C=A2D=A3 in D+
(316)
where Φ(z) is an analytic function in D+ and Φ(z) is a solution of the equation (38)in Dminus satisfying the boundary conditions
Re [λ(z)(Φ(z)eφ(z) + ψ(z))] = r(z) z isin Γ
Re [λ(x)(Φ(x)eφ(x) + ψ(x))] = s(x) x isin L0 = (0 2)
Im [λ(z)(Φ(z)eφ(z) + ψ(z))|z=zprimej= bj j = 1 m+ n
Re [λ(x)Φ(x)] = Re [λ(x)(W (x)minusΨ(x))] z isin L0
Re [λ(z)Φ(z)] = minusRe [λ(z)Ψ(z)] z isin L3 cup L4
Im [λ(z)(Φ(z) + Ψ(z))]|z=A2j+1 = 0 j = 0 1 [n2]
Im [λ(z)(Φ(z) + Ψ(z))]|z=B2jminus1 = 0 j = 1 [(n+ 1)2]
(317)
where s(x) can be written as in (319) below Moreover the solution u0(z) of ProblemQprime for (37) the estimate
C1[u0(z) Dminus] = Cβ[u0(z) Dminus] + C[wplusmn0 (micro ν)Y plusmn(micro ν) Dminus] le M19(k1 + k2) (318)
in which Y plusmn(z) = Y plusmn(micro ν) = Πn+1j=0 |x plusmn y minus aj|γprime
j+δ wplusmn0 (micro ν) = Rew0(z)∓ Imw0(z)
w0(z) = w0(micro ν) micro = x+y ν = xminusy are as stated in (524) Chapter IV and u0(z) isas stated in the first formula of (314) where w(z) = w0(z) M19 = M19(p0 β k0 D)is a non-negative constant
Proof Let u(z) be a solution of Problem Qprime for equation (12) and w(z) = uzu(z) be substituted in the positions of w u in (316) Thus the functions g(z) f(z)ψ(z) φ(z) in D+ and g1(z) g2(z)Ψ(z) in Dminus in (315)(316) can be determinedMoreover we can find the solution Φ(z) in D+ and Φ(z) in Dminus of (38) with theboundary condition (317) where
s(x)=2r((1minusi)x2)minus [a((1minusi)x2)+b((1minusi)x2)]h2j
a((1minusi)x2)minusb((1minusi)x2)
xisin(a2ja2j+1) j=01[n2]
s(x)=2r((1+i)x2+1minusi)minus[a((1+i)x2+1minusi)minusb((1+i)x2+1minusi)]k2jminus1
a((1+i)x2+1minusi)+b((1+i)x2+1minusi)
xisin(a2jminus1a2j) j=1[(n+1)2]
(319)
224 VI Second Order Quasilinear Mixed Equations
in which the real constants hj(j = 0 1 n) are of the form
h2j = Re [λ(A2j+1)(r(A2j+1) + ic2j+1) + Im [λ(A2j+1)(r(A2j+1) + ic2j+1)]
on L2j+1 j = 0 1 [n2]
k2jminus1 = Re [λ(B2jminus1)(r(B2jminus1) + ic2j)minus Im [λ(B2jminus1)(r(B2jminus1) + ic2j)]
on L2j j = 1 [(n+ 1)2]
where L2j+1 = Dminus2j+1capAB j = 0 1 [n2] L2j = Dminus
2jcapAB j = 1 [(n+1)2]Thus
w(z) = w0(z) +W (z) =
⎧⎨⎩ Φ(z)φ(z) + ψ(z) in D+
w0(z) + Φ(z) + Ψ(z) in Dminus
is the solution of Problem Aprime for the complex equationwz
wzlowast
= Re [A1w] + A2u+ A3 in
D+
Dminus
(320)
and u(z) is a solution of Problem Qprime as stated in (314)
33 Unique solvability for the discontinuous oblique derivative problemfor (12)
Theorem 33 Suppose that equation (12) satisfies Condition C Then Problem Qprime
for (12) has a unique solution in D
Proof The proof is similar to the proof of Theorems 23 and 24 Chapter V but theboundary condition on Lj(j = 1 or 2) and the point condition in which are modifiedFor instance the boundary condition on Lj(j = 1 or 2) and the point condition in(14) Chapter V are replaced by that on L3 cup L4 and Im [λ(z)w(z)]|z=A2j+1 = 0j = 0 1 [n2] Im [λ(z)w(z)]|z=B2jminus1 = 0 j = 1 [(n + 1)2] respectively theintegral in (218) Chapter V is replaced by
Ψ1(z)=
⎧⎪⎪⎪⎨⎪⎪⎪⎩int ν
a2j+1
g0(z)e1dν+int micro
0g0(z)e2dmicro in Dminus
2j+1 j=0 1 [n2]int ν
2g0(z)e1dν+
int micro
a2jminus1
g0(z)e2dmicro in Dminus2j j=1 [(n+1)2]
g0(z) = Aξ0 +Bη0 + Cu0 +D in Dminus
(321)
and so on and the characteristic lines through the points z1 = 1 minus i are re-placed by the characteristic lines through the points A2j+1(j = 0 1 [n2]) B2jminus1
(j = 1 [(n+ 1)2]
Moreover we can obtain the estimates of solutions of Problem Qprime for (12)
3 Discontinuous Oblique Derivative Problems 225
Theorem 34 Suppose that equation (12) satisfies Condition C Then any solutionu(z) of Problem Qprime for (12) satisfies the estimates
C1[u(z) Dminus] = Cβ[u(z) Dminus] + C[wplusmn(micro ν)Y plusmn(micro ν) Dminus] le M20
C1β[u(z) D+] = Cβ[u(z) D+] + Cβ[uzX(z) D+] le M21(k1 + k2)
C1[u(z) Dminus] le M21(k1 + k2) = M22
(322)
where
X(z) =m+nprodj=0
|z minus aj|γprimej+δ Y plusmn(z) = Y plusmn(micro ν) =
n+1prodj=0
|x plusmn y minus aj|γprimej+δ
wplusmn0 (micro ν)=Rew0(z)∓Imw0(z) w0(z)=w0(micro ν) micro=x+y ν=xminusy
(323)
in which γprimej = max (0 minusγj) (j = 1 mminus1 m+1 m+n) γprime
0 = max (0 minus2γ0) γprimem =
max (0 minus2γm) and γj (j = 0 1 m + n) are real constants as stated beforeβ (0 lt β lt δ) δ are sufficiently small positive numbers and k = (k0 k1 k2) M20 =M20(p0 β k D) M21 = M21(p0 β δ k0 D) are two non-negative constants
From the estimate (322) we can see the regularity of solutions of Problem Qprime for(12)
Next we consider the oblique derivative problem(Problem P prime) for the equation(12)
Theorem 35 Suppose that the mixed equation (12) satisfies Condition C ThenProblem P prime for (12) has a solution in D
Proof First of all we prove the uniqueness of solutions of Problem P prime for (12)Suppose that there exist two solutions of Problem P prime for (12) By Condition Cit can be seen that u(z) = u1(z) minus u2(z) and w(z) = uz satisfy the homogeneousequation and boundary conditions
wz
wzlowast
= f f = Re [A1w] + A2u in
D+
Dminus
u(zj) = 0 j = 0 1 m u(aj) = 0 j = 1 n
Re [λ(z)w(z)]=0 z isin Γ Re [λ(z)w(z)]=0 zisinL3cupL4
Im [λ(z)w(z)]|z=A2j+1 = 0 j = 0 1 [n2]
Im [λ(z)w(z)]|z=B2jminus1 = 0 j = 1 [(n+ 1)2]
(324)
By using the method of proof in Theorem 33 w(z) = uz = 0 u(z) = 0 in Dminus canbe verified Thus we have
2Re [λ(x)uz] =partu
partl= 0 on Lprime = (0 2)
226 VI Second Order Quasilinear Mixed Equations
it is clear that λ(x) = cos(l x)minusi cos(l y) = exp(minusiπ4) on Lprime1 and λ(x) = cos(l x)minus
i cos(l y) = exp(iπ4) on Lprime2 On the basis of the maximum principle of solutions for
the equationuzz = Re [A1uz] + A2u z isin D+ (325)
if maxD+ u(z) gt 0 then its maximum attains at a point zlowast isin Γ cup Lprime obviouslyzlowast = zj (j = 0 1 m) aj (j = 1 n) and we can prove zlowast isin Γ by the methodas stated in the proof of Theorem 34 Chapter III Moreover it is not difficult toprove that if zlowast isin Lprime then partupartl = 0 at zlowast Hence maxD+ u(z) = 0 By the similarmethod we can prove minD+ u(z) = 0 Hence u(z) = 0 u1(z) = u2(z) in D
Secondly we first prove the existence of solutions of Problem P prime for equation(12) in the linear case From Theorem 33 it can be seen that Problem Qprime for(12) has a solution ulowast(z) in D if ulowast(zj) = bj j = 0 1 m ulowast(aj) = bm+j j =1 n then the solution ulowast(z) is just a solution of Problem P prime for (12) Otherwise[ulowast(aprime
1) ulowast(aprime
m+n)] = [dlowast1 d
lowastn+m] in which aprime
j = zj j = 1 m aprimej = ajminusm j =
m+1 m+n we find m+n solutions u1(z) um+n(z) of Problem Qprime for (325)with the boundary conditions
Re [λ(z)ukz] = 0 z isin Γ Re [λ(z)ukz] = 0 z isin L3 cup L4
uk(2) = 0 Im [λ(z)ukz]|z=zprimej= δjk j k = 1 m+ n
Im [λ(z)ukz]|z=A2j+1 = 0 j = 0 1 [n2]
Im [λ(z)ukz]|z=B2jminus1 = 0 j = 1 [(n+ 1)2]
(326)
It is obvious that U(z) =summ+n
k=1 uk(z) equiv 0 in D moreover we can verify that
J =
∣∣∣∣∣∣∣∣∣u1(aprime
1) middot middot middot um+n(aprime1)
u1(aprimem+n) middot middot middot um+n(aprime
m+n)
∣∣∣∣∣∣∣∣∣ = 0thus there exist m + n real constants cprime
1 cprime2 middot middot middot cprime
m+n which are not equal to zerosuch that
cprime1u1(aprime
k) + cprime2u2(aprime
k) + middot middot middot+ cprimem+num+n(aprime
k) = dlowastk minus bk k = 1 middot middot middot m+ n
thus the function
u(z) = ulowast(z)minusm+nsumk=1
cprimekuk(z) in D
is just a solution of Problem P prime for the linear equation (12) with A2 = 0 in D Inaddition we can derive that the solution u(z) of Problem P prime for (12) satisfies theestimates similar to (322) Afterwards we consider the equation with the parametert isin [0 1]
Luz=
uzz
uzzlowast
==Re [A1uz]+ t[A2u+A3] + A(z) in
D+
Dminus
(327)
4 Problems in Multiply Connected Domains 227
where A(z) is any function in D satisfying the condition C[A(z)X(z) D+] +C[Aplusmn(micro ν)Y plusmn(micro ν) Dminus] lt infin By using the method of parameter extension namelywhen t = 0 we see that Problem P prime for such equation has a unique solution bythe above discussion Moreover assuming that when t = t0 isin (0 1] Problem P prime forequation (327) has a solution then we can prove that there exists a small positiveconstant ε such that for any t isin |tminust0| le ε t isin [0 1] Problem P prime for such equation(327) has a solution Thus we can derive that there exists a solution u(z) of ProblemP prime for equation (327) with t = 1 especially when A(z) = 0 in D ie Problem P prime forequation (12) has a solution u(z) This completes the proof
4 Oblique Derivative Problems for Quasilinear Equations ofMixed Type in Multiply Connected Domains
In this section we discuss the oblique derivative boundary value problems for quasilin-ear second order equations of mixed (elliptic-hyperbolic) type in multiply connecteddomains We first give a representation of solutions for the above boundary valueproblem and then prove the uniqueness and existence of solutions of the above prob-lem and give a priori estimates of solutions of the above problem In the book [9]2)the author proposed the Dirichlet boundary value problem (Tricomi problem) for sec-ond order equations of mixed type in multiply connected domains In [12] 1)3) theauthor only discussed the Dirichlet problem (Problem T2) for the Lavrentprimeev-Bitsadzeequation of mixed (elliptic-hyperbolic) type uxx + sgny uyy = 0 in a special doublyconnected domain Up to now we have not seen that other authors have solved itin multiply connected domains In this section we try to discuss the oblique deriva-tive problem for quasilinear equations of mixed type in multiply connected domainswhich includes the Dirichlet problem (Problem T2) as a special case
41 Formulation of the oblique derivative problem for second order equa-tions of mixed type
Let D be an N -connected bounded domain D in the complex plane CI with theboundary partD = Γ cup L where Γ =
sumNj=1 Γj isin C2
α(0 lt α lt 1) in y gt 0 withthe end points z = a1 = 0 b1 a2 b2 aN bN = 2 and L = cup2N
j=1Lj L1 = x =minusy 0 le x le 1 L2 = x = minusy + b1 b1 le x le b1 + (a2 minus b1)2 L3 = x =y+ a2 b1+(a2 minus b1)2 le x le a2 L4 = x = minusy+ b2 b2 le x le b2+(a3 minus b2)2 L2Nminus1 = x = y + aN bNminus1 + (aN minus bNminus1)2 le x le aN L2N = x = y + 2 1 lex le 2 in which a1 = 0 lt b1 lt a2 lt b2 lt middot middot middot lt aN lt bN = 2 and denoteD+ = D cap y gt 0 Dminus = D cap y lt 0 Dminus
1 = Dminus capx+ y lt b1 Dminus2 = Dminus cap b1 lt
x + y lt a2 Dminus3 = Dminus cap a2 lt x + y lt b2 Dminus
2Nminus2 = Dminus cap bNminus1 lt x + y ltaN Dminus
2Nminus1 = DminuscapaN lt x+y and z1 = 1minusi z2 = b1+(a2minusb1)(1minusi)2 zN =bNminus1+ (aN minus bNminus1)(1minus i)2 We assume that the inner angles πα2jminus1 πα2j of D+
at the points z = aj bj(j = 1 N) are greater than zero and less than π
228 VI Second Order Quasilinear Mixed Equations
We consider the quasilinear second orderequation of mixed type (11) and its com-plex form (12) with Condition C
The oblique derivative boundary valueproblem for equation (12) may be formu-lated as follows
Problem P primeprime Find a continuous solutionu(z) of equation (12) in D which is con-tinuously differentiable in Dlowast = DZ andsatisfies the boundary conditions
12
partu
partl= Re [λ(z)uz] = r(z) z isin Γ
12
partu
partl= Re [λ(z)uz] = r(z) z isin Lprime (41)
Im [λ(z)uz]|z=zj=cj j=1 N u(aj)=dj u(bj)=dN+j j=1 N
(42)where Z = xplusmny = aj xplusmny = bj j = 1 N y le 0 Lprime = cupN
j=1L2jminus1 l is a vectorat every point on ΓcupLprime λ(z) = a(x)+ ib(x) = cos(l x)∓ i cos(l y) z isin ΓcupLprime ∓ aredetermined by z isin Γ and Lprime respectively cj dj dN+j(j = 1 N) are real constantsand λ(z) r(z) cj dj dN+j(j = 1 N) satisfy the conditions
Cα[λ(z)Γ]lek0 Cα[r(z)Γ]lek2 Cα[λ(z) Lprime]lek0 Cα[r(z) Lprime]lek2
cos(l n) ge 0 on Γ |cj| |dj| |dN+j| le k2 j = 1 N
maxzisinL1
1|a(x)minus b(x)| max
zisinLprimeprime
1|a(x) + b(x)| le k0
(43)
in which n is the outward normal vector on Γ Lprimeprime = cupNj=2L2jminus1 α (12 lt α lt 1)
k0 k2 are non-negative constants Here we mention that if A2 = 0 in D+ then we cancancel the condition cos(l n) ge 0 on Γ and if the boundary condition Re [λ(z)uz] =r(z) z isin Lprime is replaced by Re [λ(z)uz] = r(z) z isin Lprime then Problem P primeprime does notinclude the Dirichlet problem (Tricomi problem)
The boundary value problem for (12) with A3(z u uz) = 0 z isin D u isin IR uz isin CIr(z) = 0 z isin Γ cup Lprime and cj = 0 (j = 0 1 N) and dj = 0 (j = 1 2N) will becalled Problem P primeprime
0 The number
K =12(K1 +K2 + middot middot middot+K2N) (44)
is called the index of Problem P primeprime and Problem P primeprime0 on the boundary partD+ of D+
where
Kj=[φj
π
]+Jj Jj=0 or 1 ejφj =
λ(tj minus0)λ(tj+0)
γj=φj
πminusKj j=1N (45)
in which [a] is the largest integer not exceeding the real number a and t1 = a1 = 0t2 = b1 t3 = a2 t4 = b2 t2Nminus1 = aN t2N = bN = 2 and
λ(t) = eiπ4 on lj = (aj bj) λ(t2jminus1 + 0) = λ(t2j minus 0) = eiπ4 j = 1 N
4 Problems in Multiply Connected Domains 229
If cos(l n) equiv 0 on each of Γj (j = 1 N) then we select the index K = N minus 1 onpartD+ If cos(l n) equiv 0 on Γj (j = 1 N) then we select the index K = N2 minus 1on partD+ and the last N point conditions in (42) can be eliminated In this caseProblem P includes the Dirichlet problem (Tricomi problem) as a special case Nowwe explain that in the closed domain Dminus the derivative ux plusmn uy of the solution u(z)in the neighborhoods of 4N characteristic lines x plusmn y = aj x plusmn y = bj(j = 1 N)may not be bounded if γjαj le 0(j = 1 2N) Hence if we require that the solutionu(z) in DminusZ is bounded where Z = x + y = aj x + y = bj x minus y = aj x minus y =bj y le 0 (j = 1 N) then it needs to choose γj gt 0 (j = 1 2N) hereinγj (j = 1 2N) are as stated in (45) If we require that solution u(z) is onlycontinuous in D it suffices to choose minusγjαj lt 1 (j = 1 2N)
Moreover we need to introduce another oblique derivative boundary value prob-lem
Problem Qprimeprime If A2 = 0 in D one has to find a continuously differentiable solutionu(z) of (12) in Dlowast which is continuous in D and satisfies the boundary conditions(41)(42) but the last N point conditions are replaced by
Im [λ(z)uz]|z=zprimej= dprime
j j = 1 N (46)
where zprimej(j = 1 N minus 1) are distinct points such that zprime
j isin Γ zprimej isin Z (j =
1 N minus 1) and dprimej (j = 1 N) are real constants satisfying the conditions
|dprimej| le k2 j = 1 N In the case the condition cos(l n) ge 0 on Γ can be can-
celed and we choose the index K = N minus 1
42 Representation and uniqueness of solutions for the oblique derivativeproblem for (12)
Now we give representation theorems of solutions for equation (12)
Theorem 41 Suppose that equation (12) satisfies Condition C Then any solutionof Problem P primeprime for (12) can be expressed as
u(z) = 2Reint z
0w(z)dz + d1 w(z) = w0(z) +W (z) (47)
where w0(z) is a solution of Problem A for the equation
wz
wzlowast
= 0 in
D+
Dminus
(48)
230 VI Second Order Quasilinear Mixed Equations
with the boundary conditions (41) and (42)(w0(z) = u0z) and W (z) possesses theform
W (z) = w(z)minus w0(z) w(z) = Φ(z)eφ(z) + ψ(z)
φ(z) = φ0(z) + Tg = φ0(z)minus 1π
int intD+
g(ζ)ζ minus z
dσζ ψ(z) = Tf in D+
W (z) = Φ(z) + Ψ(z) in Dminus
Ψ(z) =
⎧⎪⎪⎨⎪⎪⎩int micro
0g2(z)dmicroe2 +
int ν
2g1(z)dνe1 in Dminus
2jminus1 j = 1 2 Nint micro
0g2(z)dmicroe2 +
int ν
aj+1
g1(z)dνe1 in Dminus2j j = 1 2 N minus 1
(49)
in which Im [φ(z)] = 0 on L0 = cupNj=1lj lj = (aj bj) j = 1 N e1 = (1 + i)2
e2 = (1minus i)2 micro = x+ y ν = x minus y φ0(z) is an analytic function in D+ and
g(z)=
A12+A1w(2w) w(z) =00 w(z) = 0 z isin D+
f(z)=Re [A1φz]+A2u+A3 in D+
g1(z) = g2(z) = Aξ +Bη + Cu+D ξ = Rew + Imw η = Rew minus Imw
A =ReA1 + ImA1
2 B =
ReA1 minus ImA1
2 C = A2 D = A3 in Dminus
(410)where Φ(z) is analytic in D+ and Φ(z) is a solution of equation (48) in Dminus satisfyingthe boundary conditions
Re [λ(z)(Φ(z)eφ(z) + ψ(z))] = r(z) z isin Γ
Re [λ(x)(Φ(x)eφ(x) + ψ(x))] = s(x) x isin L0
Re [λ(x)Φ(x)] = Re [λ(x)(W (x)minusΨ(x))] z isin L0
Re [λ(z)Φ(z)] = minusRe [λ(z)Ψ(z)] z isin Lprime
Im [λ(zj)Φ(zj)] = minusIm [λ(zj)Ψ(zj)] j = 1 N
(411)
where λ(x) = 1+i on L0 and s(x) is as stated in (414) below Moreover the solutionu0(z) of Problem P primeprime for (48) in Dminus satisfies the estimate in the form
Cβ[u0(z) D] + Cβ[X(z)w(z) D] + C[wplusmn0 (micro ν)Y plusmn(micro ν) Dminus] le M23(k1 + k2) (412)
in which X(z) = Π2Nj=1[|zminus tj||γj |αj+δ Y plusmn(z) = Y plusmn(micro ν) = Π2N
j=1[|microminus tj||ν minus tj|]|γj |αj+δδ β (0 lt β lt δ) γj (j = 1 2N) are as stated in (45) wplusmn
0 (micro ν) = Rew0(z) ∓Imw0(z) w0(z) = w0(micro ν) micro = x+ y ν = x minus y and
u0(z) = 2Reint z
0w0(z)dz + d1 (413)
4 Problems in Multiply Connected Domains 231
and M23 = M23(p0 β k0 D) is a non-negative constant
Proof Let u(z) be a solution of Problem P primeprime for equation (12) and w(z) = uzu(z) be substituted in the positions of w u in (410) thus the functions g(z) f(z)ψ(z) φ(z) in D+ and g1(z) g2(z)Ψ(z) in Dminus in (49)(410) can be determinedMoreover we can find the solution Φ(z) in D+ and Φ(z) in Dminus of (48) with theboundary conditions (411) where
s(x)=2r((1minusi)x2)minus2Re [λ((1minusi)x2)Ψ((1minusi)x2)]minush(x)
a((1minus i)x2)minus b((1minus i)x2)+Re [λ(x)Ψ(x)] (414)
on L0 in which
h(x)=[a
((1minusi)x2
)+b
((1minusi)x2
)][Re (λ(z1)(r(z1)+ic1))+Im (λ(z1)(r(z1)+ic1))]
and
Ψ(x) =
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
int ν
2g1(z)dνe1 =
int ν
2g1((1minus i)ν2)dνe1 z = x+ iy = (1minus i)x isin L1int micro
0g2(z)dmicroe2 =
int micro
0g2((1 + i)micro2 + (1minus i)aj2))dmicroe2
z = x+ iy = (1 + i)x minus aji isin L2jminus1 j = 2 Nint micro
0g2(z)dmicroe2=
int micro
0g2((1+i)micro2+1minusi))dmicroe2 z=(1+i)xminus2iisinL2N
Thus
w(z) = w0(z) +W (z) =
⎧⎨⎩ Φ(z)φ(z) + ψ(z) in D+
w0(z) + Φ(z) + Ψ(z) in Dminus
is the solution of Problem A for the complex equationwz
wzlowast
= Re [A1w] + A2u+ A3 in
D+
Dminus
(415)
and u(z) is a solution of Problem P for (12) as stated in the first formula in (47)
Theorem 42 Suppose that equation (12) satisfies Condition C Then ProblemP primeprime for (12) has at most one solution in D
Proof Let u1(z) u2(z) be any two solutions of Problem P primeprime for (12) It is clearthat u(z) = u1(z) minus u2(z) and w(z) = uz satisfies the homogeneous equation andboundary conditions
wz
wzlowast
= Re [A1w] + A2u in
D+
Dminus
Re [λ(z)w(z)] = 0 z isin Γ Re [λ(x)w(x)] = s(x) x isin L0 u(0) = 0
Re [λ(z)w(z)] = 0 z isin Lprime Im [λ(zj)w(zj)] = 0 j = 1 N
(416)
232 VI Second Order Quasilinear Mixed Equations
in which
s(x)=minus2Re [λ((1minus i)x2)Ψ((1minus i)x2)]
a((1minus i)x2)minus b((1minus i)x2)+ Re [λ(x)Ψ(x)] on L0 (417)
From Theorem 41 the solution w(z) can be expressed in the form
w(z)=
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
Φeφ(z)+ψ(z) ψ(z)=Tf φ(z)= φ0(z)+T g in D+
Φ(z) + Ψ(z) in Dminus
Ψ(z)=
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
int micro
0[Aξ+Bη+Cu]edmicroe2+
int ν
2[Aξ+Bη+Cu]dνe1 in Dminus
2jminus1
j = 1 2 Nint micro
0[Aξ+Bη+Cu]edmicroe2 +
int ν
aj+1
[Aξ+Bη+Cu]dνe1 in Dminus2j
j = 1 2 N minus 1
(418)
where g(z) is as stated in (410) Φ(z) in D+ is an analytic function and Φ(z) is asolution of (48) in Dminus satisfying the boundary condition (411) in which W (z) =w(z) If A2 = 0 in D+ then ψ(z) = 0 besides the functions Φ(z) Φ(z) satisfy theboundary conditions⎧⎨⎩Re [λ(x)Φ(x)] = s(x)
Re [λ(x)Φ(x)] = Re [λ(x)(W (x)minusΨ(x))]x isin L0 (419)
where s(x) is as stated before Noting that
C[u(z) Dminus] le M24C[X(z)w(z) D+] + C[wplusmn(z)Y plusmn(z) Dminus]and applying the method of iteration we can get
|wplusmn(z)Y plusmn(z)| le [2NM25M((4 +M24)m+ 1)Rprime]n
n in Dminus
where M24 = M24(D) M25 = maxDminus [|A| |B| |C|] M = 1 + 4k20(1 + k2
0) m =C[w(z) Dminus] Rprime = 2 Let n rarr infin we can derive wplusmn(z) = 0 ie w(z) =w1(z) minus w2(z) = 0 u(z) = 0Ψ(z) = Φ(z) = 0 in Dminus and s(x) = 0 on L0Besides noting that the solution u(z) of the equation
uzz = Re [A1uz] + A2u in D+ (420)
12
partu
partl=Re[λ(z)uz]=0 zisinΓcupL0 u(aj)=0 u(bj)=0 j=1N (421)
and the index of the above boundary value problem is K = N minus 1 on the basis ofTheorem 37 Chapter III we see that u(z) = 0 in D+ This proves the uniquenessof solutions of Problem P primeprime for (12) in D As for the general equation (12) we canprove the uniqueness of solutions of Problem P primeprime by the extremum principle for ellipticequations of second order by the method in the proof of Theorem 34 Chapter III
4 Problems in Multiply Connected Domains 233
43 The solvability for the oblique derivative problem for (12)
First of all we prove the existence of solutions of Problem P primeprime for equation (37) inD It is obvious that Problem P primeprime for equation (37) is equivalent to the followingboundary value problem (Problem Aprimeprime) for (38) with the boundary conditions
Re [λ(z)w(z)] = r(z) on Γ Re [λ(z)w(z)] = r(z) on Lprime
Im [λ(z)w(z)]|z=zj= cj j = 1 N
u(aj) = dj u(bj) = dN+j j = 1 N
(422)
and the relation
u(z) = 2Reint z
0w(z)dz + d1 (423)
Similarly to the method in the proof of Theorem 31 we can get the following theorem
Theorem 43 Problem P primeprime for (37) in D has a unique continuous solution u(z)
Proof From the second and third boundary conditions in (422) we can obtain thefollowing conditions
Re [λ(x)w(x)] = k(x) on L0
k(x) =2r((1minus i)x2)minus [a((1minusi)x2)+b((1minusi)x2)]h
a((1minus i)x2)minus b((1minus i)x2)on L0
h = [Re (λ(z1)(r(z1)+ic1))+Im (λ(z1)(r(z1)+ic1))]
(424)
According to the method in the proof of Theorem 22 Chapter III we can find asolution w(z) of (38) in D+ with the first boundary condition in (422) and (424)Thus we can find the solution of Problem P primeprime for (37) in D as stated in (423) andthe solution w(z) of Problem Aprimeprime for (38) in Dminus
1 possesses the form
w(z) = w(z) + λ(z1)[r(z1)minus ic1]
w(z) =12[(1minus i)f(x+ y) + (1 + i)g(x minus y)]
f(x+ y) = Re [(1 + i)w(x+ y)]
g(x minus y) =2r((1minus i)(x minus y)2)
a((1minusi)(xminusy)2)minus b((1minusi)(xminusy)2)in Dminus
1 0 b1
234 VI Second Order Quasilinear Mixed Equations
Similarly we can write the solution of Problem P primeprime in Dminusj (j = 2 3 2N minus 1) as
w(z)= w(z)+λ(zj)[r(zj)minusicj]
w(z)=12[(1minusi)fj(x+y)+(1+i)gj(xminusy)] in Dminus
j j=2N
f2j(x+y)=2r((1+i)(x+y)2+(1minusi)aj+12)
a((1+i)(x+y)+(1minusi)aj+12)minusb((1+i)(x+y)2+(1minusi)aj+12)
g2j(xminusy)=2r((1minusi)(xminusy)2)
a((1minusi)(xminusy)2)+b((1minusi)(xminusy)2)in Dminus
2j j=12Nminus1
f2jminus1(x+y)=Re[(1minusi)w(x+y)]
g2jminus1(xminusy)=2r((1minusi)(xminusy)2)
a((1minusi)(xminusy)2)minusb((1minusi)(xminusy)2)in Dminus
2jminus1 j=2N
(425)in which Dminus
j (j = 2 3 2N minus 1) are as stated before
Theorem 44 Suppose that the mixed equation (12) satisfies Condition C ThenProblem P primeprime for (12) in D has a solution
Proof It is clear that Problem P primeprime for (12) is equivalent to Problem Aprimeprime for thecomplex equation of first order and boundary conditions
wz
wzlowast
= Re [A1w] + A2u+ A3 in
D+
Dminus
(426)
Re [λ(z)w(z)] = r(z) z isin Γ Re [λ(z)w(z)] = r(z) z isin Lprime
Im [λ(zj)w(zj)] = cj u(aj)=dj u(bj)=dN+j j=1 N(427)
and the relation (423) In order to find a solution w(z) of Problem Aprimeprime for (426)in D we express w(z) in the form (49)ndash(410) and use the successive iterationFirst of all denoting the solution w0(z) of Problem Aprimeprime for (426) and substitutingw0(z)(= ξ0e1 + η0e2) and the corresponding function u0(z) into the positions of w(z)(= ξe1 + ηe2) u(z) in the right hand side of (426)(49) and (410) thus the corres-ponding functions g0(z) f0(z) and the functions
W1(z) = w1(z)minus w0(z) w1(z) = Φ1(z)eφ1(z) + ψ1(z)
φ1(z) = φ0(z)minus 1π
int intD+
g0(ζ)ζ minus z
dσζ ψ1(z) = Tf0 in D+
w1(z) = w0(z) +W1(z) W1(z) = Φ1(z) + Ψ1(z) (428)g0(z) = Aξ0 +Bη0 + Cu0 +D in Dminus
Ψ1(z) =
⎧⎪⎪⎨⎪⎪⎩int micro
0g0(z)dmicroe2 +
int ν
2g0(z)dνe1 in Dminus
2jminus1 j = 1 2 Nint micro
0g0(z)dmicroe2 +
int ν
aj+1
g0(z)dνe1 in Dminus2j j = 1 2 N minus 1