Oblique Derivative Problems for Elliptic Equations in Conical Domains 9783031283802, 9783031283819

Table of contents :
Contents
List of Symbols
1 Introduction
2 Preliminaries
2.1 Elementary Inequalities
2.2 Domains with a Conical Point
2.3 The Quasi-Distance Function r and Its Properties
2.4 Function Spaces
2.4.1 Lebesgue Spaces
2.4.2 Space M(G)
2.4.3 Regularization and Approximation by Smooth Functions
2.5 Hölder and Sobolev Spaces
2.5.1 Notations and Definitions
2.5.2 Sobolev Embedding Theorems
2.6 Weighted Sobolev Spaces
2.7 Spaces of Dini Continuous Functions
2.8 Variable Exponent Spaces
2.9 The Nemyckij Operator and Its Properties
2.10 Some Functional Analysis
2.11 The Cauchy Problem for Differential Inequalities
2.12 The Dependence of the Eigenvalues on the Coefficients of the Differential Equation
2.13 Basic Properties of the Gamma and Gegenbauer Functions
2.14 Additional Auxiliary Results
2.14.1 The Stampacchia Lemma
2.14.2 Other Assertions
2.15 Notes
3 Eigenvalue Problems
3.1 The Linear Eigenvalue Problem
3.1.1 The Eigenvalue Problem for n=2
3.1.2 The Eigenvalue Problem for n≥3
3.1.3 On Properties of Eigenvalues
3.2 The Nonlinear Eigenvalue Problem
4 Integral Inequalities
4.1 Classical Hardy Inequalities
4.2 The Friedrichs-Wirtinger Type Inequality
5 The Linear Oblique Derivative Problem for Elliptic Second Order Equation in a Domain with Conical Boundary Point
5.1 Preliminaries
5.2 Setting of the Problem
5.3 The Global Integral Weighted Estimate
5.4 Local Integral Weighted Estimates
5.5 The Power Modulus of Continuity
5.6 Examples
5.7 Notes
6 The Oblique Derivative Problem for Elliptic Second Order Semi-linear Equations in a Domain with a Conical Boundary Point
6.1 Setting of the Problem
6.2 Main Results
6.3 Global Integral Weighted Estimate
6.4 Local Integral Weighted Estimates
6.5 Power Modulus of Continuity
7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak Quasi-Linear Equations in a Neighborhood of a Conical Boundary Point
7.1 Setting of the Problem
7.2 The Maximum Principle
7.3 The Comparison Principle
7.4 The Barrier Function. The Preliminary Estimate of the Solution Modulus
7.5 Local Estimate at the Boundary
7.6 Global Integral Estimate
7.7 Local Integral Weighted Estimates
7.8 The Power Modulus of Continuity at the Conical Point for Weak Solutions
7.9 Example
7.10 Notes
8 Behavior of Strong Solutions to the Degenerate Oblique Derivative Problem for Elliptic Quasi-linear Equations in a Neighborhood of a Boundary Conical Point
8.1 Setting of the Problem
8.2 The Barrier Function. The Preliminary Estimate of the Solution Modulus
8.3 Integral Weighted Estimates
8.4 The Power Modulus of the Continuity at the Conical Point
8.5 Notes
9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order Equation with Perturbed p(x)-Laplacian
9.1 Setting of the Problem
9.2 Preliminary
9.3 The Maximum Principle
9.4 The Comparison Principle
9.5 The Barrier Function. Estimation of the Solution Modulus
9.6 Proof of the Main Theorem 9.4
10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for Strong Quasi-Linear Elliptic Second Order Equation with Perturbed p(x)-Laplacian
10.1 Setting of the Problem
10.2 Preliminary
10.3 The Maximum Principle
10.4 The Comparison Principle
10.5 The Barrier Function
10.6 Estimation of the Solution Modulus. The Proof of the Main Theorem 10.3
11 Existence of Bounded Weak Solutions
11.1 Setting of the Problem
11.2 Proof of the Existence Theorem
Bibliography
Index
Notation Index

Citation preview

Frontiers in Elliptic and Parabolic Problems

Mikhail Borsuk

Oblique Derivative Problems for Elliptic Equations in Conical Domains

Frontiers in Mathematics

Frontiers in Elliptic and Parabolic Problems Series Editor Michel Chipot, Institute of Mathematics, Zürich, Switzerland

The goal of this series is to reflect the impressive and ongoing evolution of dealing with initial and boundary value problems in elliptic and parabolic PDEs. Recent developments include fully nonlinear elliptic equations, viscosity solutions, maximal regularity, and applications in finance, fluid mechanics, and biology, to name a few. Many very classical notions have been revisited, such as degree theory or Sobolev spaces. Books in this series present the state of the art keeping applications in mind wherever possible. The series is curated by the Series Editor.

Mikhail Borsuk

Oblique Derivative Problems for Elliptic Equations in Conical Domains

Mikhail Borsuk Mathematics and Informatics University of Warmia and Mazury in Olsztyn Olsztyn, Poland

ISSN 1660-8046 ISSN 1660-8054 (electronic) Frontiers in Mathematics ISSN 2730-549X ISSN 2730-5503 (electronic) Frontiers in Elliptic and Parabolic Problems ISBN 978-3-031-28380-2 ISBN 978-3-031-28381-9 (eBook) https://doi.org/10.1007/978-3-031-28381-9 Mathematics Subject Classification: 35J20, 35J25, 35J92, 35J60, 35J65, 35J70, 35J85, 35B05, 35B45, 35B65 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This book is published under the imprint Birkhäuser, www.birkhauser-science.com by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Contents

1

Introduction .. . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . .

1

2

Preliminaries . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.1 Elementary Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.2 Domains with a Conical Point .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.3 The Quasi-Distance Function rε and Its Properties .. . . . . . . .. . . . . . . . . . . . . . . 2.4 Function Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.4.1 Lebesgue Spaces .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.4.2 Space M(G) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.4.3 Regularization and Approximation by Smooth Functions .. . . . . . 2.5 Hölder and Sobolev Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.5.1 Notations and Definitions . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.5.2 Sobolev Embedding Theorems . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.6 Weighted Sobolev Spaces .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.7 Spaces of Dini Continuous Functions .. . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.8 Variable Exponent Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.9 The Nemyckij Operator and Its Properties .. . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.10 Some Functional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.11 The Cauchy Problem for Differential Inequalities .. . . . . . . . .. . . . . . . . . . . . . . . 2.12 The Dependence of the Eigenvalues on the Coefficients of the Differential Equation .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.13 Basic Properties of the Gamma and Gegenbauer Functions . . . . . . . . . . . . . . 2.14 Additional Auxiliary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.14.1 The Stampacchia Lemma . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.14.2 Other Assertions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 2.15 Notes . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . .

9 9 10 14 15 15 17 17 19 19 23 24 27 28 30 31 33 38 38 40 40 41 43

v

vi

Contents

3

Eigenvalue Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.1 The Linear Eigenvalue Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.1.1 The Eigenvalue Problem for n = 2 . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.1.2 The Eigenvalue Problem for n ≥ 3 . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.1.3 On Properties of Eigenvalues . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 3.2 The Nonlinear Eigenvalue Problem . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . .

45 45 45 48 52 55

4

Integral Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 4.1 Classical Hardy Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 4.2 The Friedrichs-Wirtinger Type Inequality . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . .

65 65 68

5

The Linear Oblique Derivative Problem for Elliptic Second Order Equation in a Domain with Conical Boundary Point . . . . . . . . . . .. . . . . . . . . . . . . . . 73 5.1 Preliminaries.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 73 5.2 Setting of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 74 5.3 The Global Integral Weighted Estimate .. . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 77 5.4 Local Integral Weighted Estimates . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 88 5.5 The Power Modulus of Continuity . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 101 5.6 Examples . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 103 5.7 Notes . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 104

6

The Oblique Derivative Problem for Elliptic Second Order Semi-linear Equations in a Domain with a Conical Boundary Point . . . . . . . . 6.1 Setting of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 6.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 6.3 Global Integral Weighted Estimate .. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 6.4 Local Integral Weighted Estimates . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 6.5 Power Modulus of Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . .

7

Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak Quasi-Linear Equations in a Neighborhood of a Conical Boundary Point . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 7.1 Setting of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 7.2 The Maximum Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 7.3 The Comparison Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 7.4 The Barrier Function. The Preliminary Estimate of the Solution Modulus . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 7.5 Local Estimate at the Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 7.6 Global Integral Estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 7.7 Local Integral Weighted Estimates . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 7.8 The Power Modulus of Continuity at the Conical Point for Weak Solutions . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . .

107 107 108 110 119 122

129 129 134 139 142 150 157 163 178

Contents

vii

7.9 Example.. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 180 7.10 Notes . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 182 8

9

Behavior of Strong Solutions to the Degenerate Oblique Derivative Problem for Elliptic Quasi-linear Equations in a Neighborhood of a Boundary Conical Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 8.1 Setting of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 8.2 The Barrier Function. The Preliminary Estimate of the Solution Modulus . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 8.3 Integral Weighted Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 8.4 The Power Modulus of the Continuity at the Conical Point.. . . . . . . . . . . . . . 8.5 Notes . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order Equation with Perturbed .p(x)-Laplacian . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 9.1 Setting of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 9.2 Preliminary . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 9.3 The Maximum Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 9.4 The Comparison Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 9.5 The Barrier Function. Estimation of the Solution Modulus . . . . . . . . . . . . . . 9.6 Proof of the Main Theorem 9.4 .. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . .

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for Strong Quasi-Linear Elliptic Second Order Equation with Perturbed p(x)-Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 10.1 Setting of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 10.2 Preliminary . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 10.3 The Maximum Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 10.4 The Comparison Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 10.5 The Barrier Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 10.6 Estimation of the Solution Modulus. The Proof of the Main Theorem 10.3 .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . .

183 183 186 196 214 230

231 231 235 241 252 255 258

271 271 274 276 287 291 294

11 Existence of Bounded Weak Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 307 11.1 Setting of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 307 11.2 Proof of the Existence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 308 Bibliography . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 319 Index . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 325 Notation Index . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 327

List of Symbols

Let us fix some notations used in the whole book: • • • • • • • • •

[l] : the integral part of l (if l is not integer) R—the set of real numbers R+ —the set of positive numbers Rn —the n–dimensional Euclidean space, n ≥ 2 N—the set of natural numbers N0 = N ∪ {0}—the set of nonnegative integers x = (x1 , . . . , xn )—an element of Rn O = (0, . . . , 0) (r, ω) = (r, ω1 , ω ), ω = (ω2 , . . . , ωn−1 )—spherical coordinates in Rn with pole O defined by x1 = r cos ω1 ,

.

x2 = r sin ω1 cos ω2 , ...

............

xn−1 = r sin ω1 sin ω2 · · · sin ωn−2 cos ωn−1 , xn = r sin ω1 sin ω2 · · · sin ωn−2 sin ωn−1 • • • • • • • •

S n−1 —the unit sphere in Rn Br (x0 )—the open ball with radius r centred at x0 B r (x0 )—the closed ball with radius r centred at x0 2π n/2 ωn = n(n/2) —the volume of the unit ball in Rn σn = nωn —the area of the n-dimensional unit sphere Rn+ —the half-space {x : xn > 0} —the hyperplane {x : xn = 0} G—a bounded domain in Rn ix

x

• • • • • • • • • • • • • • • • • •

List of Symbols

G ⊂⊂ G—G has compact closure in G; G is strictly contained in G dx—volume element in Rn ds—area element in Rn−1 dσ —area element in Rn−2 ∂G—the boundary of G, in what follows we shall assume that O ∈ ∂G d(x) := dist(x, ∂G) → − n = (n1 , . . . , nn ) —exterior unit normal vector on ∂G G = G ∪ ∂G—the closure of G meas G—the Lebesgue measure of G diam G—the diameter of G K—an open cone with vertex in O := K ∩ S n−1 ; C : the rotational cone {x1 > r cos ω20 } ω0 ∂C  : the lateral surface of C : {x1 = r cos 2 } ·, · —the scalar product of two vectors ∂u Di u := ∂x i ∇u := (D1 u, . . . , Dn u) 2u Dij u := ∂x∂i ∂x j

• D 2 u—the Hessian of u n  • |∇u| := ( (Di u)2 )1/2 i=1 n 

• |D 2 u| := (

i,j =1

(Dij u)2 )1/2

 

p(x)u = div |∇u|p(x)−2∇u denotes the p(x)-Laplacian; β = (β1 , . . . , βn ), βi ∈ N0 —an n-dimensional multi-index |β| := β1 + · · · + βn —the length of the multi-index β |β| β Dx = D β := β1 ∂β2 β —a partial derivative of order |β| ∂x1 ∂x2 ...∂xn n   → − • ∂u ∂ n = ∇u, n —the exterior normal derivative of u on ∂G • • • •

j

• δi —Kronecker’s delta • supp u : the support of u, the closure of the set on which u = 0 • c = c(∗, . . . , ∗)—a constant depending only on the quantities appearing in the parentheses. The same letter c will sometimes be used to denote different constants depending on the same set of arguments

1

Introduction

The aim of our book is the investigation of the behavior of strong and weak solutions of regular oblique derivative problems for second order elliptic equations (linear and quasilinear) in the neighborhood of the boundary singularities. The oblique derivative problem is regular, if it is assumed that unit vector fields that define the oblique derivative boundary conditions is nowhere tangential to the boundaries considered. The author’s main goal is to establish the precise exponent of the solution decrease rate and under the best possible conditions. The question on the behavior of solutions of elliptic boundary value problems near boundary singularities is of great importance for many applications, e.g., in hydrodynamics, aerodynamics, fracture mechanics. Problems in which the boundary value condition has the form .B(x, u, Du) .= 0, where .B depends on the gradient .∇u of the unknown function u in a suitable way, are called oblique derivative problems. Oblique derivative problems play a major role in the study of reflected shocks in transonic flow [37] and in the theory of the capillary problem (see, e.g., [57]). In geodesy [44], the most fundamental problems of the gravity field determination from boundary observations are translated into exterior boundary value problems for the Laplace or Poisson equations. Motivated by the results of the above cited papers, we shall give the asymptotic behavior of solutions near a boundary corner point to the singular p-Laplacian oblique problem. We shall investigate how the behavior of weak solution depends on the first eigenvalue of the eigenvalue problem. The systematic development of the theory of oblique derivative problems for elliptic equations in smooth domains was presented recently by Lieberman in [89]. G. Lieberman [86,88,89] derived some pointwise estimates and regularity results for solution to the linear

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Borsuk, Oblique Derivative Problems for Elliptic Equations in Conical Domains, Frontiers in Elliptic and Parabolic Problems, https://doi.org/10.1007/978-3-031-28381-9_1

1

2

1 Introduction

and quasi-linear uniformly elliptic second order oblique derivative problem in smooth domains, a priori estimates for weak solutions in the Sobolev–Kondratiev weighted spaces. It should be noted that investigations in the aforementioned works refer to linear boundary value problems in sufficiently smooth domains. The two-dimensional basic theory of linear oblique derivative problems is quite old. For two-dimensional domains Talenti [120] established .W 2,2 -solvability under the assumption that .a ij are measurable functions only. For higher-dimensional case the .W 2,2 -regularity and invertibility properties for some linear oblique derivative problems are obtained if .a ij ∈ W 1,n ( ) (see Miranda [96], Chicco [40]) or if .a ij are measurable functions satisfying the Cordes condition (Chicco [40]). Agmon, Douglis and Nirenberg in [2], using explicit representations for solutions to derive suitable .Lp -estimates, established that the condition .a ij ∈ C 0 ( ) is sufficient to .W 2,p -regularity of solutions for all values of .p ∈ (0, ∞). An exact a priori estimate of Schauder type in . C 2+α (G) for solutions of the linear oblique derivative problem for elliptic second order equations with smooth coefficients and smooth boundary . ∂G was derived by R. Fiorenza [58] and with discontinuous coefficients (for the transmission oblique derivative problem) by the author: Theorem 2.1 [21] for elliptic equation and Theorem 1.5 [24] for parabolic equation. These estimates were used for the proof of conditional existence theorems for nonlinear oblique derivative problems. However, many problems of physics and technology lead to boundary value problems in domains with a non-smooth boundary, in particular, in domains which have a finite number of angular (.n = 2) or conical (.n ≥ 3) points on the boundary. The theory of linear boundary value problems in non-smooth domains was described in well known survey of Kondrat’ev and Oleinik [74] and in the work of Kufner and Sändig [81], as well as in the monograph of Kozlov et al. [78]. In non-smooth domains, linear oblique derivative problems were studied earlier by Faierman [51], Garroni et al. [59], Grisvard [63], Lieberman [85, 88], Reisman [109], and others. Lieberman considered the oblique derivative problem in Lipschitz domains. His results concern elliptic equations with Hölder-continuous coefficients. The local and global maximum principle for general second order linear and quasi-linear elliptic oblique derivative problems were established by him. Grisvard in his work investigated the properties of the second weak derivatives of the oblique problem for the Laplace operator in a plane domain with a polygonal boundary. Solonnikov et al. proved the uniqueness of solutions and obtained a priori estimates for weak solutions of the Laplace operator in the Sobolev-Kondrat’ev weighted spaces. We study the behavior of strong solutions to the degenerate oblique derivative problem for general linear second order elliptic equation in a neighborhood of the boundary conical point of a n-dimensional, .n ≥ 2, bounded domain G (see Figure 2.1 on page 12). The degeneration of the problem occurs in the boundary condition. We derive a priori estimates in the Kondratiev spaces and an estimate of the type .u(x) = O (|x|α ) with exact exponent .α. V. Solonnikov et. al. in [59,118] have studied similar problem for the Laplace operator in a plane angle. They proved the unique existence of solutions and obtained a priori estimates for weak solutions to the Laplace operator in the Sobolev–Kondrat’ev weighted spaces.

1 Introduction

3

We study also the asymptotic behavior of strong solutions to the oblique problem for general semi-linear second order elliptic equations near the boundary conical point, i.e., we obtain the estimation of the type .|u(x)| = O(|x|α ) with the sharp exponent .α. In our previous papers [15–18, 27, 33] we obtained similar results for linear and quasi-linear oblique problems. The result presented in this paper extends our previous results to a wider class of elliptic equations. New regularity theorems were established. Our results refer to general equations of second order. It should be pointed out that assumptions concerning of the equation coefficients are the least restrictive possible, i.e., the leading coefficients of the equation must be Dini continuous at the conical point and the lower coefficients can grow in a particular way. Some properties of solutions of the semi-linear problem in a smooth domain and in a neighborhood of an isolated singular point were studied by Kondrat’ev et al., see, e.g., [72,73]. Other problems for elliptic semi-linear equations were considered by Veron et al., see, e.g., [14, 50, 107]. Here we will generalize results of [16, 17, 19, 26, 32, 33] to the p-Laplacian case with oblique boundary condition. In [32] the p-Laplacian problem with Robin condition was investigated, in [26, 33] more general equations were investigated but only with zero Dirichlet and Robin conditions, respectively, and in [16, 17, 19] oblique problems for second order linear, quasi-linear, and semi-linear equations in a corner domains were studied. A number of variations of problems involving the p-Laplacian equation have been investigated in the literature. See, for example, [20,124]. For other results of singular elliptic equations, see [62, 98, 123] and references therein. The growing attention in the study of the p-Laplace operator is motivated by the fact that it arises in various applications, e.g., in the generalized reaction-diffusion theory [69], non-Newtonian fluid theory [5] both for the case .p > 2 (dilatant fluids) and the case .1 < p < 2 (pseudo-plastic fluids), non-Newtonian filtration [68], quantum physics [12], the turbulent flow of a gas in porous medium [48], glacial sliding, theory of superconductors, biology, etc. We consider also bounded weak solutions to the oblique derivative problem for quasilinear elliptic second order equations with perturbed .p(x)-Laplacian and with strong nonlinearities singular absorption term in the neighborhood of a corner boundary point of the bounded plane sector. Such problem appears under the study of initial—boundary value problems for the parabolic equation in a dihedral angle. The degeneracy of the problem stems from the term in the boundary condition. We establish a priori estimates for the power modulus of the continuity of solutions at the angle vertex. Analogous estimates have been obtained in [28,29,32] to the Robin problem for singular and degenerate .p−, p(x)− Laplacian equation with strong nonlinearities singular absorption term. This book consists of ten chapters and is organized as follows. In Chap. 2 “Preliminaries ” we introduce notations for a domain with a conical boundary point and function spaces that are used in the following chapters. We formulate some theorems of the functional analysis, give solutions of the Cauchy problem for differential first order inequalities, and present some information about the special functions: gamma and Gegenbauer functions.

4

1 Introduction

In Chap. 3 “Eigenvalue problems” we consider linear  .

ω∈

ω ψ + λ(λ + n − 2)ψ(ω) = 0, , ∂ψ λχ(ω + ) + γ (ω ) ψ(ω) = 0, ω ∈ ∂ 1 1 ∂ ν

and nonlinear ⎧   ⎪ ⎪ −divω (λ2 ψ 2 + |∇ω ψ|2 )(p+ −2)/2 ∇ω ψ = ⎪ ⎪ ⎪ ⎪ ⎪ ⎪λ (λ(p+ − 1) + n − p+ ) (λ2 ψ 2 + |∇ω ψ|2 )(p+ −2)/2 ψ, ⎪ ⎨ ω ∈ , .

⎪ ⎪ ∂ψ 2 2 2 (p −2)/2 ⎪ + ⎪ (λ ψ + |∇ω ψ| ) + λχ(ω1 )ψ + → ⎪ ∂− ν ⎪ ⎪

p+ −1 ⎪ ⎪ ⎩ γ (ω1 ) p+p−1+μ · ψ|ψ|p+ −2 = 0, ω ∈ ∂ , + −1

(EVP)

(NEVP)

eigenvalue problems. Here we study properties of eigenvalues and eigenfunctions. In Chap. 4 “Integral inequalities” we present well known classical Hardy inequalities. Next we prove new Friedrichs-Wirtinger type inequality, adapting to the oblique derivative problems. Chapter 5 “Linear oblique derivative problem for elliptic second order equation in a domain with boundary conical point” deals with linear elliptic oblique derivative problem for the nondivergent form equation in non-smooth bounded domain with boundary conical point:  .

L[u] ≡ a ij (x)uxi xj + a i (x)uxi + a(x)u = f (x), x ∈ G . ∂u 1 x ∈ ∂G\O B[u] ≡ ∂u ∂ n + χ(ω) ∂r + |x| γ (ω)u = g(x),

(L)

Our assumptions concerning the smoothness of the equation coefficients are the least restrictive possible; leading coefficients of the equation must be Dini continuous at the conical point .O, whereas lower coefficients can grow and we indicate the exact power growth order. In Sect. 5.6 we construct the examples which show that the Dini condition for leading coefficients of the the equation at the conical point as well as the assumption concerning the lower equation coefficients are essential for the validity of the estimates derived in the Chap. 4. Otherwise in these estimates the exponent .λ should be changed .λ − ε with any .ε > 0. The fact that the exponent .λ in these estimates cannot be increased is shown by the partial solutions of the Laplace equation in the domain with the angular or conical point. In this sense the estimates of chapter 4 are the best possible.

1 Introduction

5

In Chap. 6 we study the boundary value problem with oblique derivative for semi-linear elliptic equations in a domain with a conical boundary point ⎧ ⎪ ⎪ a ij (x)uxi xj + a i (x)uxi + a(x)u(x) = h(u) + f (x), x ∈ G, ⎨ .

⎪ ⎪ ⎩

h(u) = a0 (x)u(x)|u(x)|q−1,

∂u ∂ n

+ χ(ω) ∂u ∂r +

1 |x| γ (ω)u(x)

q ∈ (0, 1),

= g(x),

(SL) x ∈ ∂G\O.

In Sect. 6.2 we describe the main results, i.e., Theorems 6.3–6.5. In Theorem 6.3 we assume that the leading coefficients are Dini continuous at zero. Then, we generalize the result assuming that the condition of Dini-continuity is not satisfied. It gives less accurate regularity of solutions. In Theorem 6.5, we obtain estimates for a particular function describing the growth of the leading coefficients. In the next two sections, we derive global and local weighted estimates that are used in the last section to prove the main theorems. Chapter 7 is devoted to investigating of the behavior of weak solutions to the conormal problem for weak quasi-linear with the linear uniformly elliptic principal part second order equations of divergence form in the neighborhood of a corner boundary point of the bounded n-dimensional domain .G : ⎧  ij  d −2 ⎪ ⎨ ℵ[u] ≡ − dxi a (x)uxj + a0 r u(x) + b(x, u, ux ) = 0, . a ij = a j i , x ∈ G, ⎪ ⎩ 1 x ∈ ∂G \ O. B[u] ≡ ∂u ∂ ν + |x| γ (ω)u(x) = g(x, u),

(WQL)

The main results are described in Sect. 6.1. Just as in chapters 5 and 6 we study the cases both of Dini continuous at zero leading coefficients and Dini continuous not at zero leading coefficients. In Sect. 7.9 we consider the example of the elliptic oblique derivative problem with Laplacian in the principal part for 3-dimensional domain with conical boundary point. We construct the explicit solution of such problem. Chapter 8 is devoted to an investigation of the behavior of strong solutions to the quasilinear elliptic oblique derivative problem in the neighborhood of a conical boundary point ⎧ ⎨ .

a ij (x, u, ux )uxi xj + a(x, u, ux ) = 0, a ij = a j i ,

⎩B[u] ≡

∂u ∂ n

+

χ(x) ∂u ∂r

+

1 |x| γ (x)u

= g(x),

x ∈ G, x ∈ ∂G\O.

(QL)

This chapter essentially generalizes and develops the results of section 10.2 in Chapter 10 [33] for the Robin boundary value problem.

6

1 Introduction

In Chap. 9 we consider the following oblique derivative problem involving the perturbed .p(x)-Laplacian with singular nonlinearities and absorption term in the neighborhood of an angular boundary point of the bounded 2-dimensional (plane) sector: ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ .

− p(x)u −

⎪ |∇u|p(x)−2 ⎪ ⎪ ⎪ ⎩

d dx2

 ∂u



χ(ω) p(x)−2 ∂u cos ω |∇u| ∂r

+ a0 (x)u|u|p(x)−1+

b(u, ∇u) = f (x), x ∈ GR 0,  γ (ω) p(x)−2 + |x|p(x)−1 u|u| = 0,

∂u ∂ n + χ(ω)sign ω ∂r

(OP2 )

x ∈ 0R ∪ R .

In Chap. 10 we investigate the behavior of bounded weak solutions to the oblique derivative problem for quasi-linear elliptic second order equations with perturbed .p(x)Laplacian and with strong nonlinearities singular absorption term in the neighborhood of a corner boundary point of the bounded n-dimensional cone .(n ≥ 3) : ⎧   p(x)−1+ ⎪ ⎪ − p(x)u + 1ω0 dxd 1 χ(ω1 )|∇u|p(x)−2 ∂u ⎪ ∂r + a0 (x)u|u| sin ⎪ 2 ⎪ ⎪ ⎨ b(u, ∇u) = f (x), x ∈ GR 0, .   ⎪|∇u|p(x)−2 ∂u + χ(ω ) ∂u + γ (ω1 ) u|u|p(x)−2 = 0, ⎪ 1 ∂r ⎪ ∂ n

|x|p(x)−1 ⎪ ⎪ ⎪ ⎩ x ∈ R ∪ . 0

(OPn )

R

In Chaps. 9–10 we give the asymptotic behavior of solutions near a boundary angular or corner point to the perturbed .p(x)-Laplacian oblique problem, precisely we investigate how the behavior of weak solution depends on the first eigenvalue of the corresponding eigenvalue problems. Our approach in this chapter follows the technique of [26, 33]. However, the problem in this chapter is nonlinear with oblique boundary condition. The feature that needs to be highlighted in the boundary value problems .(OP2 ), .(OPn ) is the possible singularity, which the nonlinearity b could exhibit when .u → 0. To investigate the behavior of weak solutions to the problems .(OP2 ), .(OPn ) near a boundary corner point, we use a technique of barrier function. Chapter 11 is devoted to the existence of bounded weak solutions to the oblique derivative problem .(OPn ) for an elliptic quasi-linear second order equation with the variable perturbed .p(x)-Laplacian in the Lipschitz bounded n-dimensional domain. Boundary value problems for elliptic second order equations with a non-standard growth in function spaces with variable exponents have been an active investigation in recent years. We refer to [66] for an overview. There are many essential differences between the variable exponent problems and the constant exponent problems. In the variable exponent problems, many singular phenomena occurred and many special questions were raised. V. Zhikov [125] gave examples of the Lavrentiev phenomenon for the variational problems with variable exponent. The proof the existence of bounded weak solutions is based on the application of the Leray-Lions Theorem 2.59.

1 Introduction

7

Remark 1.1 The term . χ(ω) ∂u ∂r in the boundary operator . B[u] of problems . (L), (SL), (QL) has geophysical interpretation (see [44]) and value . χ = χ(ω) generally is small on average (apart from mountains areas) and . χ(ω) ∂u ∂r can be considered as a perturbation . In one important case, when gravity anomalies data are with respect to the operator . ∂u ∂ n

reduced to the ellipsoid for the determination of the geoid, we have that . χ(ω) ≡ 0, i.e., the oblique derivative problem is the Robin problem. Such problem is a modification of the crucial problem of physical geodesy which is known as the field gravimetric BVP. The Additional Notes The mixed boundary value problem (BVP) of Dirichlet oblique derivative type in plane domains with piecewise differentiable boundary was considered by J. Banasiak [6, 7], as well as by J. Banasiak and G. Roach [8, 9]. They proved decomposition formulas for solutions of mixed BVP of Dirichlet oblique derivative type for general elliptic second order equations; they investigated the behavior of variational solutions of elliptic second order equations, as well as they obtained some results of .L2 -solvability for mixed problem in plane and n-dimensional domains with curvilinear polygons, edges, and points where the edges are vanishing. Oblique derivative and interface problems on polygonal domains were studied by M. Dauge and S. Nicaise [43]. The behavior of weak solutions to BVP of Dirichlet-Robin type for divergent elliptic quasi-linear second order equations with triple degeneration in a neighborhood of boundary edge was investigated by author [25].

2

Preliminaries

2.1

Elementary Inequalities

In this section we review some elementary inequalities (see, e.g., [11, 65]) which will be frequently used throughout this book. Lemma 2.1 (Cauchy’s Inequality) For .a, b ≥ 0 and .ε > 0, we have ab ≤

.

ε 2 1 a + b2 . 2 2ε

(2.1.1)

Lemma 2.2 (Young’s Inequality) For .a, b ≥ 0, ε > 0 and .p, q > 1 with . p1 + we have 1 1 (εa)p + .ab ≤ p q

 q b . ε

1 q

= 1,

(2.1.2)

Lemma 2.3 (Hölder’s Inequality) Let .ai , bi , i = 1, . . . , n, be nonnegative real numbers and .p, q ∈ R with . p1 + q1 = 1, we have n  .

i=1

 ai bi ≤

n  i=1

1/p  p ai

n 

1/q q bi

.

(2.1.3)

i=1

Lemma 2.4 (Theorem 41 [65]) Let .a, b be nonnegative real numbers and .m ≥ 1. Then ma m−1 (a − b) ≥ a m − b m ≥ mbm−1 (a − b).

.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Borsuk, Oblique Derivative Problems for Elliptic Equations in Conical Domains, Frontiers in Elliptic and Parabolic Problems, https://doi.org/10.1007/978-3-031-28381-9_2

(2.1.4) 9

10

2 Preliminaries

Lemma 2.5 ((Jensen’s Inequality) (Theorem 65 [65], Lemma 1 [114])) Let .ai , i = 1, . . . , n, be nonnegative real numbers and .p > 0. Then n 

λ

.

i=1

 p ai

≤

n 

p ai

i=1

≤

n 

p

(2.1.5)

ai ,

i=1

where .λ = min(1, np−1 ) and . = max(1, np−1 ). Lemma 2.6 Let .a, b ∈ R, m > 1. Then the familiar inequality |b|m ≥ |a|m + m|a|m−2 a(b − a)

(2.1.6)

.

is valid. Proof By Young’s inequality (2.1.2) with .ε = 1, p = m, q =

m m−1 ,

we have

m|a|m−2 ab ≤ m|b| · |a|m−1 ≤ |b|m + (m − 1)|a|m ⇒ (2.1.6).

 

.

Lemma 2.7 (See, E.g., Proof of Theorem 3.1 (i) [52], Inequality (4.8) [3]) For any n . ξ, η ∈ R the following inequalities hold ⎧  p−2  ⎪ ⎨ |ξ |p−2 ξ − |η|p−2 η (ξ − η) ≥ (p − 1)(|ξ |p + |η|p ) p |ξ − η|2 , . if 1 < p < 2; ⎪  p−2  ⎩ 1 p−2 p |ξ | ξ − |η| η (ξ − η) ≥ 2p |ξ − η| , if p ≥ 2.

2.2

(2.1.7)

Domains with a Conical Point

Definition 2.8 Let G ⊂ Rn be a bounded domain. We say that G has a conical point in O if • O ∈ ∂G. • ∂G \ O is smooth. • G coincides in some neighborhood of O with an open cone K, with vertex in O. •  π ω0 ∈ (0, ), ωi ∈ (0, π), i = 2, . . . , n − 2; ∂K = (r, ω) : r > 0, w1 = 2 2

.

 ωn−1 ∈ (0, 2π) .

• ∂K ∩ S n−1 is smooth.

2.2 Domains with a Conical Point

11

For a domain G which has a conical point at O ∈ ∂G we introduce the notations: • • • • • •

:= K ∩ S n−1 . d := area element of . Gba := G ∩ {(r, ω) : 0 ≤ a < r < b, ω ∈ }− a layer in Rn . ab := ∂G ∩ {(r, ω) : 0 ≤ a < r < b, ω ∈ ∂ }− the lateral surface of the layer Gba . Gd := G \ Gd0 . d := ∂G \ 0d .

• ρ := Gd0 ∩ ∂B (0),  ≤ d. −k d • G(k) := G22−(k+1) , k = 0, 1, 2, . . . . d Without loss of generality we assume that there exists R > 0 such that GR 0 is a rotational cone with the vertex at O and the aperture ω0 ∈ (0, π), thus 

 n  ω ω 0 0 R .0 = (r, ω) : x12 = cot2 xi2 , r ∈ (0, R),  ω1 = ± , ω0 ∈ (0, π) . 2 2 i=2

Let us recall some well known formulae related to spherical coordinates (r, ω1 , . . . , ωN−1 ) centered at the conical point O : dx = r n−1 drd , .

(2.2.1)

d ρ = ρ n−1 d , .

(2.2.2)

d = J (ω)dω

(2.2.3)

.

denotes the (n − 1)−dimensional area element of the unit sphere; J (ω) = sinn−2 ω1 sinn−3 ω2 · · · sin ωn−2 , .

(2.2.4)

dω = dω1 · · · dωn−1 , .

(2.2.5)

ds = r n−2 drdσ

(2.2.6)

.

denotessurface of the cone K, where dσ denotes the (n − 2)−dimensional area element on ∂ : dσ = sinn−2

.

ω0 sinn−3 ω2 · · · sin ωn−2 dω2 dω3 . . . dωn−1 . 2  2 ∂u 1 2 . |∇u| = + 2 |∇ω u|2 , ∂r r

(2.2.7)

12

2 Preliminaries

where |∇ω u| denotes the projection of the vector ∇u onto the tangent plane to the unit sphere at the point ω  ∇ω u =

.

 1 ∂u 1 ∂u ,..., √ ,. √ q1 ∂ω1 qn−1 ∂ωn−1

(2.2.8)

  n−1  1 ∂u 2 |∇ω u| = , qi ∂ωi 2

(2.2.9)

i=1

where q1 = 1, qi = (sin ω1 · · · sin ωi−1 )2 , i ≥ 2,

u =

.

1 ∂ 2 u n − 1 ∂u + 2 ω u, . + 2 r ∂r ∂r r

(2.2.10)

1  ∂  J (ω) ∂u  = · J (ω) ∂ωi qi ∂ωi

(2.2.11)

n−1

ω u =

i=1

=

n−1  i=1

1

∂ n−i−1 qi sin ωi ∂ωi

  ∂u n−i−1 ωi sin ∂ωi

denotes the Beltrami-Laplace operator; → u = divω −

.

1  ∂ J (ω) ∂ωi n−1 i=1



 J (ω) √ ui . qi

(2.2.12)

 

p(x)u = div |∇u|p(x)−2∇u denotes the p(x)-Laplacian. Fig. 2.1 n-dimensional bounded domain with a boundary conical point

x1

∂G

G Ωd

Γ0d

Γ0d

O x

2.2 Domains with a Conical Point

13

Lemma 2.9 Let α ∈ R and v(x) = r α u(x). Then Di v = αr α−2 xi u + r α Di u,

|∇v|2 ≤ c1 α 2 r 2α−2u2 + r 2α |∇u|2 ,   Dij v = r α Dij u + αr α−2 xi Dj u + xj Di u .

j

+(α 2 − 2α)r α−4 xi xj u + αr α−2 uδi ,

|D 2 v|2 ≤ c2 r 2α |D 2 u|2 + r 2α−2 |∇u|2 + r 2α−4 u2 with constants c1 , c2 > 0 depending only on α and n. Lemma 2.10 Let Gd0 be the convex rotational cone with the vertex at O and the aperture ω0 , Then xi cos(

n, xi )| d = 0, and cos(

n, x1 )| d = − sin

.

0

  cos(n, x1 )

d

= cos ω1 ,

0

ω0 , 2

  ω1 ∈ (−ω0 /2, ω0 /2); xi cos(

n, xi )

(2.2.13) d

= d.

Proof We can rewrite the equation of 0d in this way F (x) ≡ x12 − cot2

.

n ω0  2 xi = 0. 2 i=2

We use the formula cos(

n , xi ) = .

∂F ∂xi

|∇F | ,

∀i = 1, . . . , n. Because of

∂F ∂F ω0 xi , ∀i = 2, . . . , n, = 2x1 and = −2 cot2 ∂x1 ∂xi 2

then  .xi cos(

n, xi ) d = 0

  N   1 2 ∂F  ω 0 xi2  = 0. xi x 2 − cot2  = |∇F | ∂xi 0d |∇F | 1 2 d i=2

0

14

2 Preliminaries

Because of  |∇F | =

.

2

∂F ∂x1

2

  |∇F |2 

+

 n   ∂F 2 i=2

0d

∂xi

  n  2 4 ω0 2 = 4 x1 + cot xi ⇒ 2 i=2

  cos2 ω20 4x12 2 = = 4x1 1 + , sin2 ω20 sin2 ω20

we have .

  cos(

n, x1 )

0d

= −2x1

sin ω20 ω0 = − sin , 2x1 2

since



n, x1 ) > (

π . 2  

The last equality of (2.2.13) is trivial.

The Quasi-Distance Function rε and Its Properties

2.3

Let us assume as in Definition 2.8 that the cone K is contained in a circular cone .K˜ with opening angle .ω0 . Furthermore, let us suppose that the axis of .K˜ coincides with .{(x1 , 0, . . . , 0) : x1 > 0}. In this case we define the quasi-distance .rε (x) as follows. ¯ and consider the unit radius-vector We fix the point .Q = (−1, 0, . . . , 0) ∈ S n−1 \ ¯ and .l = OQ = {−1, 0, . . . , 0}. We denote by .r

the radius-vector of the point .x ∈ G / Gd0 for all .ε ∈]0, d[, it follows introduce the vector .r ε = r − εl for each .ε > 0. Since .εl ∈

= 0 for all .x ∈ G. ¯ It is easy to see that .rε (x) has the following that .rε (x) = | r − εl| properties: (1) Lemma 2.11 There exists an .h > 0 such that .rε (x) ≥ hr and .rε (x) ≥ hε, where ⎧ ⎨1, if 0 < ω0 ≤ π, .h = ⎩sin ω0 , if π < ω0 < 2π.

∀x ∈ G,

2

Proof From the definition of .rε (x) we know that .rε2 = (x1 + ε)2 +

n  i=2

xi2 = (x1 +

ε)2 + r 2 − x12 = r 2 + 2εx1 + ε2 . If .0 < ω0 ≤ π, we have .x1 ≥ 0 and therefore we obtain either .rε2 ≥ r 2 ⇒ rε ≥ r or 2 2 .rε ≥ ε ⇒ rε ≥ ε. If .x1 = r cos ω ≤ 0 and .|ω| ∈ [ π2 , ω20 ], we obtain, by the Cauchy inequality either |2εr cos ω| ≤ r 2 cos2 ω + ε2 ⇒ 2εr cos ω ≥ −r 2 cos2 ω − ε2 ⇒ rε ≥ r · sin

.

ω0 2

2.4 Function Spaces

15

or |2εr cos ω| ≤ ε2 cos2 ω + r 2 ⇒ 2εr cos ω ≥ −r 2 − ε2 cos2 ω ⇒ rε ≥ ε · sin

.

ω0 . 2  

(2) Corollary 2.12 hr ≤ rε (x) ≤ r + ε ≤

.

2 rε (x); ∀x ∈ G, ∀ε > 0. h

(3) If .x ∈ Gd , then .rε (x) ≥ d2 for all .ε ∈]0, d2 [. ¯ (4) . lim rε (x) = r, for all .x ∈ G. ε→0+

(5) .|∇rε |2 = 1, and .rε =

n−1 rε .

xi ∂rε ε Proof Because . ∂x = x1r+ε and . ∂r ∂xi = rε (i ≥ 2), then ε 1 n   2 (x1 +ε)2 + xi2 n  ∂rε i=2 2 .|∇rε | = = = 1 and ∂xi rε2 i=1   n 2 n   xi2 (x1 +ε)2 ∂ 2 rε ∂ rε 1 1 .rε = + = − + − = rε rε r3 r3 ∂x 2 ∂x 2 1

.

=

i=2

ε

i

ε

i=2

N rε

−

rε2 rε3

=  

n−1 rε .

2.4

Function Spaces

2.4.1

Lebesgue Spaces

Let G be a domain in .Rn . For .p ≥ 1 we denote by .Lp (G) be the space of Lebesgue integrable functions equipped with the norm ⎛ uLp (G) = ⎝



.

⎞1/p |u|p dx ⎠

.

G

Theorem 2.13 ((Fubini’s Theorem), See Theorem 9 §11, Chapter III [46]) Let .G1 ⊂ Rm1 , G2 ⊂ Rm2 , and .f ∈ L1 (G1 × G2 ). Then for almost all .x ∈ G1 and .y ∈ G2 the integrals 

 .

G1

f (x, y)dx

and

f (x, y)dy G2

16

2 Preliminaries

exist. Moreover, 

 .

f (x, y)dxdy =

G1 ×G2

⎛ ⎜ ⎝

G1



⎞ ⎟ f (x, y)dy ⎠ dx =

G2



⎛ ⎞  ⎜ ⎟ ⎝ f (x, y)dx ⎠ dy.

G2

G1

Theorem 2.14 (Hölder’s Inequality, See Theorem 189 [65]) Let .p, q > 1 with . p1 + q1 = 1 and .u ∈ Lp (G), v ∈ Lq (G). Then  |uv|dx ≤ uLp (G) vLq (G) .

.

(2.4.1)

G

If .p = 1, then (2.4.1) is valid with .q = ∞. Corollary 2.15 Let .1 ≤ p ≤ q and .u ∈ Lq (G). Then uLp (G) ≤ (meas G)1/p−1/q uLq (G) .

.

(2.4.2)

Corollary 2.16 ((Interpolation Inequality), See Inequality (7.9) Section 7.1 of Chapter 7 [61]) Let .1 < p ≤ q ≤ r and .1/q = λ/p + (1 − λ)/r. Then the inequality uLq (G) ≤ uλLp (G) u1−λ Lr (G)

.

holds for all .u ∈ Lr (G). Theorem 2.17 ((Fatou’s Theorem), (See Theorem 3 §44, Chapter VII [70])) If a sequence of measurable nonnegative functions .{fn (x)} converges a.e. on G to .f (x) and  fn (x)dμ ≤ K,

.

G

then .f (x) is integrable on G and  f (x)dμ ≤ K.

.

G

Lemma 2.18 ([103, Lemma 1.3.8]) Let .G1 ⊂ Rm1 , G2 ⊂ Rm2 and .f, fk ∈ Lp (G1 × G2 ), k = 1, 2, . . . , with .1 ≤ p ≤ ∞ and .

lim f − fk Lp (G1 ×G2 ) = 0.

k→∞

2.4 Function Spaces

17

Then there is a subsequence .{fkl } of .{fk } such that .

lim f (y, z) − fkl (y, z)Lp (G2 ) = 0

kl →∞

holds for almost every .y ∈ G1 .

2.4.2

Space M(G)

The space .M(G) : it is the set of all measurable and bounded almost everywhere in .G functions .u(x) with the norm  u = vrai max |u(x)| =

.

x∈G

inf

meas E=0

 sup |u(x)| . x∈G\E

The convergence in .M(G) is the uniform convergence almost everywhere.

2.4.3

Regularization and Approximation by Smooth Functions p

Let us denote by .Lloc (G) the linear space of all measurable functions which are locally p p–integrable in .G, that is p–integrable on every compact subset of .G. Although .Lloc (G) is not a normed space, it can be readily topologized. p

Definition 2.19 We say that a sequence .{um } converges to u in the sense of .Lloc (G) if p   .{um } converges to u in .L (G ) for each .G ⊂⊂ G. Let .r = |x − y| for all .x, y ∈ Rn and h be any positive number. Furthermore, let .ψh (r) be a nonnegative function in .C ∞ (Rn ) vanishing outside the ball .Bh (0) and satisfying . ψh (r)dx = 1. Such a function is often called a mollifier. A typical example is the

Rn

function .ψh (r) given by ψh (r) =

.

⎧ ⎨ch−n · exp ⎩0

h2 |r|2 −h2

for r < h, c = const > 0; for |r| ≥ h,

where c is chosen so that . ψh (r)dx = 1 and whose graph has the familiar bell shape.

18

2 Preliminaries

Definition 2.20 For .L1loc (G) and .h > 0, the regularization of u, denoted by .uh is then defined by the convolution  uh (x) =

(2.4.3)

ψh (r)u(y)dy

.

G

provided that .h < dist (x, ∂G). It is clear that .uh belongs to .C ∞ (G ) for any .G ⊂⊂ G provided that .h < dist(G , ∂G). Furthermore, if u belongs to .L1 (G) and G is bounded, then .uh belongs to .C ∞ (Rn ) for arbitrary .h > 0. As h tends to zero, the function .ψh (r) tends to the Dirac delta distribution at the point x. The significant feature of regularization, which we partly explore now, is the sense in which .uh approximates u as h tends to zero. It turns out, roughly stated, that if u lies in a local space, then .uh approximates u in the natural topology of that space. Lemma 2.21 Let .u ∈ C 0 (G). Then .uh converges to u uniformly on any subdomain  .G ⊂⊂ G. Proof We have 



uh (x) =

ψh (r)u(y)dy =

.

|x−y|≤h

putting z =

.

x−y h

ψ1 (|z|)|u(x − hz)|dz |z|≤1

. Hence if .G ⊂⊂ G and .2h < dist(G , ∂G), 

.

sup |u − uh | ≤ sup G

x∈G |z|≤1

ψ1 (|z|)|u(x) − u(x − hz)|dz ≤ ≤ sup sup |u(x) − u(x − hz)|. x∈G |z|≤1

Since u is uniformly continuous over the set .Bh (G ) = {x | dist(x, G ) < h}, the sequence  .uh tends to u uniformly on .G .    p  p Lemma 2.22 Let .u ∈ Lloc (G) . L (G) , where .1 ≤ p < ∞. Then .uh converges to u in   p the sense of .Lloc (G) . Lp (G) . Proof Using Hölder’s inequality, we obtain from (2.4.3)  |uh (x)| ≤

.

ψ1 (|z|)|u(x − hz)|p dz

p

|z|≤1

2.5 Hölder and Sobolev Spaces

19

so that if .G ⊂⊂ G and .2h < dist(G , ∂G), then  

 |uh (x)|p dx ≤

.

G

ψ1 (|z|)|u(x − hz)|p dzdx = G |z|≤1

 =



 |u(x − hz)|p dx ≤

ψ1 (|z|)dz G

|z|≤1

|u|p dx, Bh (G )

where .Bh (G ) = {x : dist(x, G ) < h}. Consequently uh Lp (G ) ≤ uLp (Bh (G )) .

.

(2.4.4)

The proof can now be completed by an approximation based on Lemma 2.21. We choose ε > 0 together with a .C 0 (G) function w satisfying

.

u − wLp (Bh (G )) ≤ ε,

.

where .2h < dist(G , ∂G). By virtue of Lemma 2.21, we have for sufficiently small h that .w − wh Lp (G ) ≤ ε. Applying the estimate (2.4.4) to the difference .u − w we obtain u − uh Lp (G ) ≤ u − wLp (G ) + w − wh Lp (G ) + uh − wh Lp (G ) ≤

.

≤ 2ε + u − wLp (Bh (G )) ≤ 3ε for small enough .h ≤ h . Hence .uh converges to u in .Lloc (G). The result for .u ∈ Lp (G) can then be obtained by extending u to be zero outside G and applying the result for p n .L   loc (R ). p

Lemma 2.23 (On the Passage to the Limit Under the Integral Symbol [115, Theorem III.10]) Let .χ(x) ∈ L∞ (G) and let .χh (x) be the regularization of .χ. Then for any .u ∈ L1 (G) 

 .

χh (x)u(x)dx =

lim

h→0 G

2.5

Hölder and Sobolev Spaces

2.5.1

Notations and Definitions

χ(x)u(x)dx. G

In this section .G ⊂ Rn is a bounded domain of the class .C 0,1 . Let .x0 ∈ Rn be a point and f a function defined on .G  x0 . The function f is Hölder continuous with exponent

20

2 Preliminaries

α ∈ (0, 1) at .x0 if the quantity

.

|f (x) − f (x0 )| |x − x0 |α x∈G

[f ]α;x0 = sup

.

is finite. .[f ]α;x0 is said to be the .α−Hölder coefficient of f at .x0 with respect to .G. The function f is uniformly Hölder continuous with exponent .α ∈ (0, 1) in G if the quantity |f (x) − f (y)| |x − y|α x,y∈G

[f ]α;G = sup

.

x=y

is finite. We consider the following spaces. • .C l (G) : the Banach space of functions having all the derivatives of order at most l (if l is a nonnegative integer) and of order .[l] (if l is non-integer) continuous in .G and whose .[l]-th order partial derivatives are uniformly Hölder continuous with exponent .l − [l] in l .G. .|u|l;G is the norm of the element .u ∈ C (G); if .l = [l] then [l] 

|u|l;G =

.

|D α u(x) − D α u(y)| . |x − y|l−[l] |α|=[l] x,y∈G

sup |D j u| + sup sup

j =0 G

x=y

• .C0l (G) : the set of functions in .C l (G) with compact support in .G. • .W k,p (G), 1 ≤ p < ∞ : the Sobolev space equipped with the norm

uW k,p (G)

.

⎛ ⎞1/p   =⎝ |D β u|p dx ⎠ , G |β|≤k

• .W0 (G) is the closure of .C0∞ (G) with respect to the norm . · W k,p (G) . • .W k,p (G \ O) = W k,p (G \ Bε (0)), ∀ε > 0. • For .p = 2 we use the notation k,p

W k (G) ≡ W k,2 (G),

.

W0k (G) ≡ W0k,2 (G).

Definition 2.24 Let us say that .u ∈ W k,p (G) satisfies .u ≤ 0 on .∂G in the sense of traces, k,p if its positive part .u+ = max{u, 0} ∈ W0 (G). If u is continuous in a neighborhood of .∂G, then u satisfies .u ≤ 0 on .∂G, if the inequality holds in the classical pointwise sense. Other definitions of the inequality at .∂G follow naturally. For example: .u ≥ 0 on .∂G, if

2.5 Hölder and Sobolev Spaces

21

−u ≤ 0 on .∂G; .u = v on .∂G, if both .u − v ≤ 0 and .u − v ≥ 0 on .∂G;

.

.

sup u = inf{k| u ≤ k on ∂G, k ∈ R};

inf u = − sup(−u). ∂G

∂G

∂G

k− 1 ,p

• For . ⊆ ∂G and .k ∈ 1, 2, . . . , the space .W p () consists of traces on . of functions from .W k,p (G) and is equipped with the norm ϕ

.

W

1 ,p k− p

()

= inf W k,p (G) ,

where the infimum is taken over the set of all functions . ∈ W k,p (G) such that . = ϕ on . in the sense of traces. For .p = 2 we use the notation 1

W k−1/2 () ≡ W k− 2 ,2 ().

.

Theorem 2.25 ([61, Theorem 7.28] (Interpolation Inequality)) Let G be a .C 1,1 domain in .Rn and let .u ∈ W 2,p (G) with .p ≥ 1. Then for all .ε > 0 ∇uLp (G) ≤ εuW 2,p (G) + cε−1 uLp (G)

.

with a constant c depending only on the domain .G. Theorem 2.26 ([49, Section 4.3] (Trace Theorem)) Let .1 ≤ p < ∞. There exists a bounded linear operator T : W 1,p (G) → Lp (∂G)

.

such that Tu = u

.

on ∂G

for all .u ∈ W 1,p (G) ∩ C 0 (G). Henceforth, we will write simply u instead of .T u. Theorem 2.27 (See, E.g., (6.23), (6.24) Chapter I [82] or Lemma 6.36 [86]) Let .∂G be piecewise smooth and .u ∈ W 1,1 (G). Then there is a constant .c > 0 which depends only on G such that   . |u|ds ≤ c (|u| + |∇u|) dx, ∀ ⊆ ∂ . (2.5.1) 

G

22

2 Preliminaries

If .u ∈ W 1,2 (G), then 

 .

2

(δ|∇v|2 + cδ v 2 )dx, ∀v(x) ∈ W 1,2 (G), ∀δ > 0.

v ds ≤

∂G

(2.5.2)

G

If .u ∈ W 2,2 (G), then   .

∂u ∂n

2 ds ≤ c

∂G



 |f (x)| · |u|p dx ≤ θ ε

(2.5.3)

G

Lemma 2.28 Let .f (x) ∈ Ls (G), s > inequality .

2|∇u||D 2 u| + |∇u|2 dx.

θ −1 θ

1

n p

> 1, u(x) ∈ Lp# (G), p# =

Then the



f (x)Lθ s (G)

G

pn n−p .

|u|p dx+ G

⎛ ⎞p  p# # p + (1 − θ )ε · ⎝ |u| dx ⎠ ,

θ =1−

n , ∀ε > 0 ps

(2.5.4)

G

holds true. Proof By Hölder’s inequality with exponents s and .s  , where . 1s +  .

1 s

=1

⎛ ⎞ 1 s   |f (x)| · |u|p dx ≤ f (x)Ls (G) · ⎝ |u|ps dx ⎠ .

G

(2.5.5)

G

From the inequality . 1s < pn it follows that .ps  < p# = inequality (see Corollary 2.16) gives

pn n−p

and then the interpolation

⎞ 1 ⎛ ⎞θ ⎛ ⎞ (1−θ )p ⎛ s    p#  # ps p p ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ . |u| dx ≤ |u| dx · |u| dx G

G

G

with .θ ∈ (0, 1), which is defined by the equality . ps1  = Therefore, by using the Young inequality with obtain the required inequality (2.5.4).

θ 1−θ n ⇒ θ = 1 − ps . p + p# 1 1 exponents . θ and . (1−θ) , from (2.5.5) we

 

2.5 Hölder and Sobolev Spaces

2.5.2

23

Sobolev Embedding Theorems

We give the well known Sobo- lev inequalities and Kondrashov compactness results also called the embedding theorems (see [117], §§1.4.5–1.4.6 [93], §7.7[61]). Theorem 2.29 ([126, Theorem 2.4.1], [61, Theorem 7.10] (Sobolev Inequalities)) Let G be a bounded open domain in .Rn and .p > 1. Then 1,p .W 0 (G)

→

⎧ Np ⎨L n−p (G)

for p < n,

⎩C 0 (G)

for p > n.

(2.5.6)

1,p

Furthermore, there exists a constant .c = c(n, p) such that for all .u ∈ W0 (G) we have uLNp/(n−p) (G) ≤ c∇uLp (G)

(2.5.7)

sup |u| ≤ c(meas G)1/n−1/p ∇uLp (G)

(2.5.8)

.

for .p < n and .

G

for .p > n. The following Embedding Theorems 2.30–2.33 first were proved by Sobolev [117] and can be found with complete proofs in [99], [80, Sections 5.7,5.8] and [93, Section 1.4]. Let G be a .C 0,1 bounded domain in .Rn . Theorem 2.30 Let .k ∈ N and .p ∈ R with .p ≥ 1, kp < n. Then the embedding W k,p (G) → Lq (G)

.

(2.5.9)

is continuous for .1 ≤ q ≤ Np/(n − kp) and compact for .1 ≤ q < Np/(n − kp). If kp = n, then the embedding (2.5.9) is continuous and compact for any .q ≥ 1.

.

Theorem 2.31 Let .k ∈ N0 , let .m ∈ N and let .p, q ∈ R with .p, q ≥ 1. If .kp < n, then the embedding W m+k,p (G) → W m,q (G)

.

(2.5.10)

is continuous for any .q ∈ R satisfying .1 ≤ q ≤ np/(n − kp). If .k = np, then the embedding (2.5.10) is continuous for any .q ≥ 1.

24

2 Preliminaries

Theorem 2.32 Let .k, m ∈ ∇0 and .p > 1. Then the embedding W k,p (G) → C m+β (G)

.

is continuous if (k − m − 1)p < n < (k − m)p

.

and 0 < β ≤ k − m − n/p,

(2.5.11)

and compact if the inequality in (2.5.11) is sharp. If .(k −m−1)p = n, then the embedding is continuous for any .β ∈ (0, 1). Theorem 2.33 Let .u ∈ W k,p (G) with .k ∈ N, p ∈ R, kp > n and .p > 1. Then m .u ∈ C (G) for .0 ≤ m < k − n/p and there exists a constant .c, independent of u, such that .

sup |D α u(x)| ≤ cuW k,p (G)

x∈G

for all .|α| < k − n/p. Theorem 2.34 Let G be a lipschitzian domain and let .Ts ⊂ G be piecewise .C k -smooth s-dimensional manifold. Let .k ≥ 1, p > 1, kp < n, n − kp < s ≤ n and .1 ≤ q ≤ q∗ = sp/(n − kp). Then .W k,p (G) → Lq (Ts ) and the inequality uLq (Ts ) ≤ cuW k,p (G)

.

(2.5.12)

holds. If .q < q∗, then this embedding is compact.

2.6

Weighted Sobolev Spaces

Definition 2.35 For k ∈ N0 , 1 < p < ∞ and α ∈ R we define the weighted Sobolev k (G) as the closure of C ∞ (G \ 0) with respect to the norm space Vp,α 0

uVp,α k (G)

.

⎛ ⎞1/p     p =⎝ r α+p(|β|−k) D β u dx ⎠ . G |β|≤k

2.6 Weighted Sobolev Spaces

25 k−1/p

For  ⊆ ∂G and k ∈ 1, 2, . . . , the space Vp,α k (G) and is equipped with the norm from Vp,α

() consists of traces on  of functions

uV k−1/p () = inf vVp,α k (G) ,

.

p,α

k (G) such that v = u on . where the infimum is taken over the set of all functions v ∈ Vp,α For p = 2 we use the notations ◦k

.W α

◦ k−1/2

k (G) = V2,α (G) and W α

k−1/2

() = V2,α

().

Lemma 2.36 ([71, 78]) Let k  , k ∈ N with k  ≤ k and α − pk ≤ α  − pk  .

.



k (G) is continuously imbedded in V k (G). Furthermore, the embedding is Then Vp,α p,α  compact if k  < k and α − pk < α  − pk  . k (G) the following Lemma 2.37 ([102]) Let (k − |γ |)p > n, then for every u ∈ Vp,α inequality is valid

|D γ u(x)| ≤ c|x|k−|γ |−(α+n)/p uVp,α k (G) ,

.

∀x ∈ Gd0

with a constant c independent of u and some d > 0 depending only on G. In particular k Vp,α (G) → C m (G)

.

for m < k − (α + n)/p. Proof Without loss of generality we can assume that G is a cone. We introduce new variables y = (y1 , . . . , yn ) by x = yt with t > 0 and set v(y) := u(x). By the Sobolev Embedding Theorem 2.33, we have γ

|Dy v(y)| ≤ c

.

 |δ|≤k

Dyδ vLp (G2 ) ,

∀y ∈ G21 .

1

Returning back to the variables x t |γ | |Dx u(x)| ≤ c

.

γ

 |δ|≤k

t |δ|−n/p Dxδ uLp (G2t ) , t

∀x ∈ G2t t .

26

2 Preliminaries

Multiplying both sides of this inequality by t n/p−k+α/p we obtain t |γ |−k+(α+n)/p |Dx u(x)| ≤ c γ

.



t |δ|−k+α/p Dxδ uLp (G2t ) , t

|δ|≤k

∀x ∈ G2t t .

Therefore, because of t ≤ |x| ≤ 2t in G2t t , we have |x||γ |−k+(α+n)/p |Dx u(x)| ≤ c γ

.



|x||δ|−k+α/p Dxδ uLp (G2t ) ,

∀x ∈ G2t t

t

|δ|≤k

 

with a constant c independent of t. Thus the assertion holds. Lemma 2.38 Let k, m ∈ N0 , β ∈ R with (k − m − 1)p < n < (k − m)p

and 0 < β ≤ k − m − n/p.

.

k (G) Then for any u ∈ Vp,α

 .

|γ |=m

|D γ u(x) − D γ u(y)| ≤ c|x|k−m−β−(α+n)/p uVp,α k (G) , β |x − y| x,y∈G,x=y sup

∀x ∈ Gd0 with a constant c and some d > 0. Proof The proof is completely analogous to the proof of Lemma 2.37. ◦ k−1/2

Lemma 2.39 ([71, Lemma 1.1]) Let u ∈ W α

(0d ). Then

 r α−2k+1 u2 (x)ds ≤ cu2◦ k−1/2

.

Wα

0d

. (0d )

Lemma 2.40 Let d > 0 and ρ ∈ (0, d). Then the inequality 

 r

.

3−n

ρ

∂u ∂n

0

2

 r 1−n u2 ds + c2 u2◦ 2

ds ≤ c1 ρ

0 ◦2

ρ

ρ

W 4−n (G0 )

is valid for all u ∈ W 4−n (G0 ) with constants c1 , c2 independent of u.

 

2.7 Spaces of Dini Continuous Functions

27

Proof Let us first recall that due to Theorem 2.27 we have   . ρ

∂v ∂n

2 ds ≤ c3

   1 r|D 2 v|2 + |∇v|2 dx r ρ

0

G0

with a constant c3 > 0 depending only on Gd0 . Setting v = r (3−n)/2u we have    ∂v  3 − n (1−n)/2 (3−n)/2 ∂u  r . = r + u·  ∂n  ρ \O ∂n  ρ \O 2 0

n 

Since

xi ni

i=1

r



 r

. ρ

3−n

0

n 

xi ni

i=1

r

.

≤ 1, therefore ∂u ∂n

 

2 ds ≤ 2

ρ

0

0

≤ c5

∂v ∂n

2

 (3 − n)2 1−n 2  x 2 ds ≤ r + u n, 4 r

 

(3 − n)2 r|D 2 v|2 + r −1 |∇v|2 ds + r 1−n u2 ds. 2 ρ

ρ

G0

0

The assertion then follows by Lemma 2.9.

2.7

 

Spaces of Dini Continuous Functions

Definition 2.41 The function A is called Dini continuous at zero if the integral d .

A(t) dt t

0

is finite for some d > 0. Definition 2.42 The Banach space C 0,A (G) is the set of all bounded and continuous functions u on G ⊂ Rn for which [u]A;G =

.

|u(x) − u(y)| < ∞. x,y∈G,x=y A(|x − y|) sup

28

2 Preliminaries

It is equipped with the norm uC 0,A (G) = uC 0 (G) + [u]A;G .

.

If k ≥ 1, then we denote by C k,A (G) the subspace of C k (G) consisting of all functions whose (k −1)-th order partial derivatives are uniformly Lipschitz continuous and each k-th order derivative belongs to C 0,A (G). It is a Banach space equipped with the norm uC k,A (G) = uC k (G) +



.

[D β u]A;G .

|β|=k

Furthermore, let us introduce the following notation [u]A,x =

.

|u(x) − u(y)| . y∈G\{x} A(|x − y|) sup

Definition 2.43 We shall say that the boundary portion T ⊂ ∂G is of class C 1,A if for each point x0 ∈ T there are a ball B = B(x0 ), a one-to-one mapping ψ of B onto a ball B  , and a constant K > 0 such that (i) (ii) (iii) (iv)

B ∩ ∂G ⊂ T , ψ(B ∩ G) ⊂ Rn+ . ψ(B ∩ ∂G) ⊂ . ψ ∈ C 1,A (B), ψ −1 ∈ C 1,A (B  ). ψC 1,A (B) ≤ K, ψ −1 C 1,A (B  ) ≤ K.

It is not difficult to see that for the diffeomorphism ψ one has K −1 |ψ(x) − ψ(x  )| ≤ |x − x  | ≤ K|ψ(x) − ψ(x  )|,

∀x, x  ∈ B.

.

2.8

(2.7.1)

Variable Exponent Spaces

At first, we remind some theories on the variable exponent Sobolev space .W 1,p(x) (G) (we refer to [4, 53, 77]). Let G be an open subset of .Rn and let .p : G → R be a measurable function satisfying condition (i). The variable exponent Lebesgue space .Lp(x) (G) is defined by Lp(x) (G) =

.

⎧ ⎨ ⎩

 u : G → R is measurable, Ap(·)(u) :=

|u(x)|p(x)dx < ∞ G

⎫ ⎬ ⎭

2.8 Variable Exponent Spaces

29

with the norm u % $ ≤1 . |u|p(x) = inf λ > 0 : Ap(·) λ

.

Proposition 2.44 The following inequalities hold (see, e.g., (15) [4]):

p− p+ p− p+ ≤ Ap(·) (u) ≤ max |u|p(·) ; min |u|p(·) , |u|p(·) , |u|p(·)

(2.8.1)

 1   1  1 1 p− p+ p− p+ min Ap(·) (u), Ap(·) (u) ≤ |u|p(·) ≤ max Ap(·) (u), Ap(·) (u) .

(2.8.2)

.

.

Proposition 2.45 ((Generalized Hölder Inequality) (See, E.g., (16) [4])) The inequality  |f (x)g(x)|dx ≤ 2|f |p(x)|g|p (x)

.

G

holds for every .f ∈ Lp(x)(G) and .g ∈ Lp (x)(G), where .

1 p  (x)

+

1 p(x)

= 1.

The variable exponent Sobolev space .W 1,p(x) (G) is defined by   W 1,p(x)(G) = u ∈ Lp(x)(G) : |∇u| ∈ Lp(x) (G)

.

with the norm u1,p(x) = |u|p(x) + |∇u|p(x).

.

We need some properties on spaces .W 1,p(x)(G). Proposition 2.46 If . 1 < p(x) < n the spaces .Lp(x)(G), W 1,p(x) (G) are separable, uniform convex and reflexive Banach spaces (see, e.g., Theorem 2.5, Corollary 2.7, Corollary 2.12, Theorem 3.1 [77], Theorems 1.10 and 2.1 [53]). Proposition 2.47 (See Theorem 2.3[53]) Let .p, q ∈ C(G) and .p, q ∈ L∞ + (G) =   t ∈ L∞ (G) : ess inf t ≥ 1 . Assume that G

p(x) < n, q(x)
0 such that |h(x, ξ )| ≤ g(x) + c

n 

.

i=1

|ξi |pi (x)/r(x)

2.10 Some Functional Analysis

31

for every ξ ∈ Rn and a.e. x ∈ G, then the Nemyckij operator maps the space Lp1 (x) (G) × · · · × Lpn (x) (G) in Lr(x) (G) and it is continuous and bounded. Corollary 2.53 (See Theorems 4.3 [77]) Let p(x) ∈ [1, +∞) a.e. in G and let h ∈ CAR(G) be such that the inequality |h(x, ξ )| ≤ g(x) + c|ξ |p(x)−1

.

p(x)

for some c > 0 and g ∈ L p(x)−1 (G) holds for every ξ ∈ Rn and a.e. x ∈ G. Then the p(x)

operator H : u → h(x, ∇u(x)) maps the space W 1,p(x)(G) in L p(x)−1 (G) and is continuous and bounded.

2.10

Some Functional Analysis

Proposition 2.54 (See, E.g., Proposition 2.1.22 (iv) [45]) If X is a uniformly convex Banach space, xn  x and xn  → x, then {xn }∞ n=1 converges to x in the norm topology. Definition 2.55 Let X, Y be Banach spaces. Then we denote by L(X, Y ) the linear space of all continuous linear mappings L : X → Y. Theorem 2.56 ([61, Theorem 5.2] (The Method of Continuity)) Let X, Y be Banach spaces and L0 , L1 ∈ L(X, Y ). Furthermore, let Lt := (1 − t)L0 + tL1

.

∀t ∈ [0, 1]

and suppose that there exists a constant c such that uX ≤ cLt uY

.

∀t ∈ [0, 1].

Then L1 maps X onto Y if and only if L0 maps X onto Y. Theorem 2.57 ([111] (Variational Principle for the Least Positive Eigenvalue)) Let H, V be Hilbert spaces with dense and compact embedding V ⊂ H ⊂ V

.

and let A : V → V  be a continuous operator. We assume that the bilinear form a(u, v) = (Au, v)H

.

32

2 Preliminaries

is continuous and V-coercive, that is there are constants c1 and c2 such that |a(u, v)| ≤ c1 uV vV ,

.

a(u, u) ≥ c2 u2V for all u, v ∈ V . Then the smallest eigenvalue ϑ of the eigenvalue problem Au + ϑu = 0

.

satisfies ϑ = inf

.

v∈V

a(v, v) . v2H

Theorem 2.58 ([61, Theorem 11.3] (The Leray-Schauder Theorem)) Let T be a compact mapping of a Banach space B into itself, and suppose that there exists a constant M such that xB < M

.

for all x ∈ B and σ ∈ [0, 1] satisfying x = σ T x. Then T has a fixed point. Theorem 2.59 ((Leray-Lions) (See Theorem 5.3.23[45])) Let X be a reflexive real Banach space. Let T : X → X∗ be an operator satisfying the conditions • (a) T is bounded. • (b) T is demicontinuous. • (c) T is coercive. Moreover, let there exists a bounded mapping  : X × X → X∗ such that • (d) (u, u) = T (u) for every u ∈ X. • (e) For all u, w, h ∈ X and any sequence {tn }∞ n=1 of real numbers such that tn → 0 we have (u + tn h, w))  (u, w).

.

• (f) For all u, w ∈ X we have .

(u, u) − (w, u), u − w ≥ 0

(the so-called condition of monotonicity in the principal part).

2.11 The Cauchy Problem for Differential Inequalities

33

• (g) If un  u and .

lim (un , un ) − (u, un ), un − u = 0,

n→∞

then we have (w, un )  (w, u)

.

∀w ∈ X.

• (h) If w ∈ X, un  u, (w, un )  z, then .

lim (w, un ), un  =< z, u > .

n→∞

Then the equation T (u) = f ∗ has at least one solution u ∈ X for every f ∗ ∈ X∗ .

2.11

The Cauchy Problem for Differential Inequalities

Theorem 2.60 (The Comparison Principle (See [106], [38] or [39])) • (1) Let y(x), z(x) w(x) ∈ C 1 (x > x0 ) and satisfy the Cauchy problem: ⎧  ⎪ x > x0 , ⎪ ⎨y = f (x, y),  . z > f (x, z) x > x0 , ⎪ ⎪ ⎩ z(x0 ) = y(x0) = a;

⎧  ⎪ x > x0 , ⎪ ⎨y = f (x, y),  w < f (x, w) x > x0 , ⎪ ⎪ ⎩ w(x0 ) = y(x0 ) = b.

Then w(x) ≤ y(x) ≤ z(x), ∀x ≥ x0 .

.

• (2) Let y(x), z(x) w(x) ∈ C 1 (x < x0 ) and satisfy the Cauchy problem: ⎧  ⎪ x < x0 , ⎪ ⎨y = f (x, y),  . z > f (x, z) x < x0 , ⎪ ⎪ ⎩ z(x0 ) = y(x0) = a;

⎧  ⎪ x < x0 , ⎪ ⎨y = f (x, y),  w < f (x, w) x < x0 , ⎪ ⎪ ⎩ w(x0 ) = y(x0 ) = b.

Then z(x) ≤ y(x) ≤ w(x), ∀x ≤ x0 .

.

34

2 Preliminaries

Proof We shall consider (1). Setting q(x) ≡

.

f (x, z(x)) − f (x, y(x)) , ∀x ≥ x0 , z(x) − y(x)

we have z (x) − y  (x) > q(x) · (z(x) − y(x)), ∀x ≥ x0 .

.

 Now we multiply this inequality by exp −



x

and next integrate by parts:

q(t)dt x0

x .

⎛ ξ ⎞     z (ξ ) − y  (ξ ) · exp ⎝− q(t)dt ⎠ dξ >

x0

x0

x

⎛ (z(ξ ) − y(ξ )) · exp ⎝−

ξ

⎞

⎛ q(ξ ) (z(ξ ) − y(ξ )) · exp ⎝−

x0

q(t)dt ⎠ dξ ⇒

x0

ξ =x0

x

⎛ q(ξ ) (z(ξ ) − y(ξ )) · exp ⎝−

x0

⎞

ξ =x  ⎠ q(t)dt  +

x0

x

ξ

⎛ q(ξ ) (z(ξ ) − y(ξ )) · exp ⎝−

x0

ξ x0

ξ

⎞ q(t)dt ⎠ dξ >

x0

⎞

q(t)dt ⎠ dξ ⇒ ⎛

(z(x) − y(x)) · exp ⎝−

x

⎞ q(t)dt ⎠ > z(x0 ) − y(x0) = 0 ⇒

x0

z(x) ≥ y(x), ∀x ≥ x0 .  

The rest of the inequalities are proved in the same way.

Theorem 2.61 Let V () be a monotonically increasing, nonnegative differentiable function defined on [0, 2d] that satisfies the problem

.

⎧ ⎨ V  (ρ) − P()V () + N (ρ)V (2ρ) + Q(ρ) ≥ 0, ⎩ V (d) ≤ V0 ,

0 < ρ < d,

(CP)

2.11 The Cauchy Problem for Differential Inequalities

35

where P(), N (), Q() are nonnegative continuous functions defined on [0, 2d] and V0 is a constant. Then  d V () ≤ exp

 B(τ )dτ

.



d −

V0 exp



 P(τ )dτ +





d +

Q(τ ) exp 

  P(σ )dσ dτ

τ −

(2.11.1)



with  2 B() = N () exp

.

 P(σ )dσ .

(2.11.2)



Proof We define functions  d w() = V () exp

 P(σ )dσ

.

(2.11.3)

.



 d

d R() = V0 +

Q(τ ) exp 

 P(σ )dσ dτ.

(2.11.4)

τ



d

Multiplying the differential inequality (CP ) by the integrating factor exp

 P(s)ds



and integrating from ρ to d we get  d V (d) − V () exp

 P(s)ds +

.



 d

d N (τ ) exp 

 P(s)ds V (2τ )dτ +

τ

 d

d Q(τ ) exp

+ 

 P(s)ds dτ ≥ 0.

τ

Hence it follows that d w() ≤ R() +

B(τ )w(2τ )dτ.

.



(2.11.5)

36

2 Preliminaries

Now we have d

w() ≤1+ . R()

B(τ ) 

w(2τ ) R(2τ ) dτ. R(2τ ) R()

(2.11.6)

Since R(2τ ) ≤ R() for τ > , setting z() =

.

w() R()

(2.11.7)

B(τ )z(2τ )dτ.

(2.11.8)

we get d z() ≤ 1 +

.



Let us define the function d Z() = 1 +

B(τ )z(2τ )dτ.

.



From (2.11.8) we have z() ≤ Z()

(2.11.9)

.

and Z  () = −B()z(2) ≥ −B()Z(2).

.

Multiplying the obtained differential inequality by the integrating factor   d exp − B(s)ds and using the equality 

d & Z() exp . d



d −

B(s)ds 

'



= Z () exp



d −

 B(s)ds +



 + B() exp

d − 

 B(s)ds Z(),

2.11 The Cauchy Problem for Differential Inequalities

37

we have d & Z() exp . d



d −

B(s)ds

'

 ≥ B() exp

d −



 ( ) B(s)ds Z() − Z(2) .



But d Z(2) = 1 +

d B(s)z(2s)ds ≤ 1 +

.

B(s)z(2s)ds = Z(), 

2

therefore d & Z() exp .Z() − Z(2) ≥ 0 ⇒ d



d −

B(s)ds

'

≥ 0.



Integrating from  to d we have  Z() exp

.



d −

 d

B(s)ds ≤ Z(d) = 1 ⇒ Z() ≤ exp 

 B(s)ds .



Hence, by (2.11.9), we get  d z() ≤ exp

.

 B(s)ds .

(2.11.10)



Now, in virtue of (2.11.3), (2.11.7), and (2.11.10), we finally obtain  V () ≤ exp

.

d −

  d  P(σ )dσ R() exp B(σ )dσ



or with regard to (2.11.4) the desired estimate (2.11.1).



 

38

2 Preliminaries

2.12

The Dependence of the Eigenvalues on the Coefficients of the Differential Equation

We consider the ordinary second order differential equation ⎧ ⎨p(x)y   − q(x)y + θ r(x)y = 0, . ⎩p(x) > 0. q(x) > 0, r(x) > 0.

x ∈ (a, b);

(2.12.1)

Theorem 2.62 (See, E.g., Theorem 1 §26 [121] as Well as Theorem 7 §2, chapter VI [42]) If function . q(x) decreases, leaving functions . p(x) and . r(x) unchanged, then eigenvalue . θ decreases.

2.13

Basic Properties of the Gamma and Gegenbauer Functions

We will need some information about the Gegenbauer function (see, e.g., §3.15 [10]). The Gegenbauer function is function .Cαν (z) that is a solution of the differential equation (z2 − 1)w + (2ν + 1)zw − α(α + 2ν)w = 0, ∀α.

(2.13.1)

.

Let .(z), Re z > 0 be the gamma function. The following formulae hold: •

(α + 2ν) ν + 12

1 × Cαν (z) = √ · π (ν)(2ν)(α + 1)

.

π (z +

*

z2 − 1 cos t)α (sin t)2ν−1 dt,

Re ν > 0;

(2.13.2)

0

hence it follows that •

π (α + 2ν) ν + 12 ν .Cα (1) = √ · (sin t)2ν−1 dt, π(ν)(2ν)(α + 1)

Re ν > 0.

(2.13.3)

0

• π (sin t)α dt =

.

0

π (1 + α)   , Re α > −1 · 2α  2 1 + α2

(2.13.4)

2.13 Basic Properties of the Gamma and Gegenbauer Functions

39

(see for .β = 0 (29) §1.5.1 [10]). From (2.13.3)–(2.13.4) we get • √

Cαν (1) =

.

π (α + 2ν)

· 22ν−1 (ν) ν + 1 (α + 1) 2

and hence, by formula . (1 + z) = z(z), we derive • n .

2 Cλ−1 (1) n−2 2

Cλ

(1)

=

λ(λ + n − 2) , n ≥ 3. (n − 1)(n − 2)

(2.13.5)

• We use the Legendre duplication formula (see, e.g., (15) §1.2 [10])   22z−1 1 .(2z) = √ (z) z + 2 π and obtain from (2.13.2), (2.13.4)

1 π (2ν) ν + 2 1 ν · (sin t)2ν−1 dt = .C (z) = √ · 0 π (ν)(2ν)(1) 0

1 (2ν) π 1

= 1, Re ν > 0, √ · · 2ν−1 π (ν) 2  ν+1

(2.13.6)

2

in virtue of .(1) = 1. • Other representation of the Gegenbauer function (see (3) §3.15 [10]) Cαν (x) =

.

  1 1−x (α + 2ν) F α + 2ν; −α; ν + ; ; x ∈ (−1, 1), (α + 1)(2ν) 2 2

(2.13.7)

where .F (· · · ) is the hypergeometric function (see, e.g., (15.4.19) [1])     1−a−b a+b+1 a+b+1 ,x =  · (x − x 2 ) 4 × .F a, b, 2 2 1−a−b

2 (1 − 2x); P a−b−1 2

but .Pαν (x) is the Legendre function (see, e.g., chapter 3 [10]).

x ∈ (0, 1),

(2.13.8)

40

2 Preliminaries

• From (2.13.7) and (2.13.8) we derive

1 (α + 2ν) ν + 1 2

Cαν (x) = 2ν− 2

.

(α + 1)(2ν)

(1 − x 2 )

1−2ν 4

1

P2

×

−ν

α+ν− 12

(x);

x ∈ (0, 1).

(2.13.9)

• The following formula holds (see, e.g., (30) §3.15.2 [10]) .

d ν ν+1 (z). C (z) = 2νCα−1 dz α

(2.13.10)

• The asymptotic expansion (see (2) §3.9.1 [10], (8.10.7) [1]; (4) §1.18 [10]).

− 1 (ν + μ + 1) π 2

sin θ × 3 2  ν+2  ,  +  π μπ 1 −1 θ− + + O(ν ) , cos ν + 2 4 2

Pνμ (cos θ ) =

.

ε ≤ θ ≤ π − ε, ∀ε > 0; (2.13.11)

+ , 1 (z + α) α−β −2 =z 1 + (α − β)(α + β − 1) + O(z ) . . (z + β) 2z

2.14

(2.13.12)

Additional Auxiliary Results

2.14.1 The Stampacchia Lemma Lemma 2.63 (See Lemma 3.11 of [101], [119]) Let ϕ : [k0 , ∞) → R be a nonnegative and non-increasing function which satisfies ϕ(h) ≤

.

C [ϕ(k)]β (h − k)α

for h > k > k0 ,

where C, α, β are positive constants with β > 1. Then ϕ(k0 + d) = 0,

.

where d α = C |ϕ(k0 )|β−1 2αβ/(β−1).

.

(2.14.1)

2.14 Additional Auxiliary Results

41

Proof Define the sequence ks = k0 + d −

.

d , 2s

s = 1, 2, . . . .

From (2.14.1) it follows that ϕ(ks+1 ) ≤

.

C2(s+1)α [ϕ(ks )]β , dα

s = 1, 2, . . . .

(2.14.2)

Let us prove by induction that ϕ(ks ) ≤

.

ϕ(k0 ) α < 0. , where μ = 2−sμ 1−β

(2.14.3)

For s = 0 the claim is trivial. Let us suppose that (2.14.3) is valid up to s. By (2.14.2) and the definition of d α it follows that ϕ(ks+1) ≤ C

.

2(s+1)α [ϕ(k0]β ϕ(k0 ) ≤ −(s+1)μ . α −sβμ d 2 2

(2.14.4)

Since the right hand side of (2.14.4) tends to zero as s → ∞, we obtain 0 ≤ ϕ(k0 + d) ≤ ϕ(ks ) → 0.

.

 

2.14.2 Other Assertions Lemma 2.64 (See Lemma 2.1 [41]) Let us consider the function ⎧ ⎨eςx − 1, .η(x) = ⎩−e−ςx + 1,

x ≥ 0, x ≤ 0,

(2.14.5)

where ς > 0. Let a, b be positive constants, p > 1. If ς > (2b/a) + p, then we have aη (x) − bη(x) ≥

.

a ςx e , 2

∀x ≥ 0

(2.14.6)

and +  ,p x .η(x) ≥ η , p

∀x ≥ 0.

(2.14.7)

42

2 Preliminaries

Moreover, there exists a q ≥ 0 and an M > 0 such that +  ,p +  ,p x x  .η(x) ≤ M η and η (x) ≤ M η , p p |η(x)| ≥ x,

∀x ≥ q; .

∀x ∈ R.

(2.14.8) (2.14.9)

Proof Formula (2.14.6) is easily verified by direct calculation. By definition, inequality (2.14.7) can be stated as .

x

p ς e p − 1 ≤ eςx − 1,

∀x ≥ 0.

(2.14.10)

We set for x ≥ 0 y=e

.

ς px

≥ 1 and f (y) = (y − 1)p + 1 − y p .

Then f  (y) = p(y − 1)p−1 − py p−1 < 0,

.

hence it follows that f (y) is decreasing function, that is f (y) ≤ f (1), ∀y ≥ 1. Because of f (1) = 0, we get (2.14.10). Further, the first inequality from (2.14.8) has the form y p − 1 ≤ M(y − 1)p .

(1.11.10)

.

We consider the function g(y) = M(y − 1)p − y p + 1. Then we have 1

g(y) > M(y − 1) − y ≥ 0, if y ≥ y0 = p

.

Mp

p

1

Mp −1

> 1,

if we choose M > 1. Therefore 1

g(y) > 0, ∀y ≥ y0

.

or for e

ς px

≥

Mp

1

Mp p x ≥ q1 = ln 1 , ς Mp −1 that is the first inequality from (2.14.8) is proved.

1

Mp −1

⇒

2.15 Notes

43

Let us now prove the second inequality from (2.14.8). We rewrite it in the form M(y − 1)p ≥ ςy p .

.

Hence it follows: 1

1 p

1 p

M (y − 1) ≥ ς y

.

⇒ y ≥

Mp 1

1

Mp −ςp

,

if M > ς. The last inequality means that 1

e

.

ς px

≥

1

Mp 1

1

Mp −ςp

⇒ x ≥ q2 =

Mp p ln 1 . ς M p − ς p1

Thus, inequalities (2.14.8) are proved, if we take 1

Mp p ln 1 .M > ς ; q = max(q1 , q2 ) = , ς M p − ς p1 (since ς > 1). Finally, we prove the inequality (2.14.9). From the definition we have ⎧ ⎨eςx − 1, .|η(x)| = ⎩e−ςx − 1,

x ≥ 0, x ≤ 0.

It is sufficient prove the inequality eςx − 1 ≥ x,

.

x ≥ 0.

Since ς > 1, this result follows from the Taylor formula,

2.15

 

Notes

The proof of the Cauchy, Young, and Hölder inequalities §1.2 can be found in Chapter 1 [11] or in Chapter II [65]. The formulae (2.2.1)–(2.2.12) are proved in §2, Chapter 1 [95]. For the proof of the Fubini and Fatou Theorems see, for example, Theorem 9 §11 and Theorem 19 §6, Chapter III [46]. The proof of the integral inequalities §1.5 can be found in Chapter VI [65]. The Clarkson inequality is proved in Subsection 2 §3, Chapter I [117].

44

2 Preliminaries

The material in §1.8 is due to [36]. The simplest version of Theorem 2.61 in §1.10 goes back to G. Peano [104]; the special case was formulated and proved by T. Gronwall [64] and S. Chaplygin [39]. The case .N () ≡ 0 of this theorem was considered in [75,76]. The general case belongs [22, 23].

3

Eigenvalue Problems

3.1

The Linear Eigenvalue Problem

In this section we investigate the eigenvalue problem for the Laplace–Beltrami operator . ω on the unit sphere  .

ω∈

ω ψ + λ(λ + n − 2)ψ(ω) = 0, , ∂ψ λχ(ω + ) + γ (ω ) ψ(ω) = 0, ω ∈ ∂ 1 1 ∂ ν

(EVP)

which consists of the determination of all values .λ (eigenvalues), for which (EVP) has a non-zero weak solutions .ψ(ω) (eigenfunctions). Here we suppose: .ν is the unite exterior normal vector to .∂K at the points of .∂ ; .χ(ω1 ) ≥ 0, .γ (ω1 ) > 0 be .C 0 (∂ )—functions.

3.1.1

The Eigenvalue Problem for n = 2

  Let . = − ω20 ; ω20 be an open interval. Let .γ± > 0, χ± ≥ 0, μ ≥ 0. We consider the eigenvalue problem: ⎧  2 ⎪ ⎪ ⎨ ψ (ω) + λ ψ(ω) = 0,

. ψ  ( ω20 ) + λχ+ + (1 + μ)γ+  ψ( ω20 ) = 0, ⎪ ⎪ ⎩ −ψ  (− ω20 ) + λχ− + (1 + μ)γ−  ψ(− ω20 ) =

ω ∈ , (EVP2 ) 0.

Lemma 3.1 There exists the least positive eigenvalue .λ∗ ∈



π π 2ω0 , ω0

.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Borsuk, Oblique Derivative Problems for Elliptic Equations in Conical Domains, Frontiers in Elliptic and Parabolic Problems, https://doi.org/10.1007/978-3-031-28381-9_3

45

46

3 Eigenvalue Problems

Proof By direct calculations, we find that .λ is defined from the transcendence equation

.

tan(λω0 ) =

(χ+ + χ− )λ2 + (1 + μ)(γ+ + γ− )λ (1 − χ+ χ− )λ2 − (1 + μ)(γ+ χ− + γ− χ+ )λ − (1 + μ)2 γ+ γ−

(3.1.1)

and the eigenfunction: (   (  ω0 )  ω0 ) ψ(ω) = λ cos λ ω − − (1 + μ)γ+ + λχ+ sin λ ω − . 2 2

.

(3.1.2)

By method, we deduce that there exists the least positive eigenvalue .λ∗ ∈

the graphic π π 2ω0 , ω0 (see figures below). Indeed, putting .λω0 = and .(1 + μ)γ± ω0 = γ± equation (3.1.1) takes the form .

tan( ) =

(χ+ + χ− ) 2 + (γ+ + γ− ) . (1 − χ+ χ− ) 2 − (γ+ χ− + γ− χ+ ) − γ+ γ−

We consider three cases. Case 1 . χ+ χ− < 1. We see that . π2 < ∗ < π or .

π 2ω0

< λ∗
0, q ≡ μ2 sinn−4 ω,  ≡ sinn−2 ω; ω ∈ 0, 2

.

By (3.1.8) and Theorem 2.62, we know that if the coefficient q changes everywhere in the same sense, every eigenvalue θ of (3.1.8) changes in this same sense. Thus, we get the smallest positive eigenvalue of (3.1.8) for μ = 0. Therefore we shall consider (EVP∗n ) with μ = 0 and θ = λ(λ + n − 2). From (3.1.6) and μ = 0, we obtain the equation for v(ω )   n−1  J∂ ∂v = 0. . · ∂ωi (sin ω2 · · · sin ωi−1 )2 ∂wi i=2

We see that v = const = 0 is a solution of above equation. In this way we can choose function u(ω, ω ) = ψ(ω) as a solution of (3.1.5). Now we shall to find eigenvalue λ and eigenfunction ψ(ω) for (EVP∗n ). To this let us denote ξ = cos ω, ψ(ω) = η(ξ ).

(3.1.9)

.

Then the equation of problem (EVP∗n ) takes the form ω0 (ξ 2 − 1)η (ξ ) + (n − 1)ξ η (ξ ) − λ(λ + n − 2)η(ξ ) = 0, ξ ∈ cos , 1 . 2

.

n−2

The above equation solution is the Gegenbauer function: η(ξ ) = Cλ 2 (ξ ) for n ≥ 3 (see Sect. 2.13). Therefore we obtain n−2

ψ(ω) = Cλ 2 (cos ω) , n ≥ 3

.

(3.1.10)

3.1 The Linear Eigenvalue Problem

51

as well as, by (2.13.10), ψ  (ω) = − sin ω ·

.

 n d n−2  2 Cλ 2 (ξ ) = −(n − 2) sin ω · Cλ−1 (cos ω). ξ =cos ω dξ

(3.1.11)

Hence we have that ψ(−ω) = ψ(ω) and ψ  (0) = 0. Now from the second boundary condition we obtain the equation for λ with n ≥ 3, ω0 ∈ (0, π): .

(n − 2) sin

n n−2 ω0 ω0 ω0 2 · Cλ−1 cos = (λχ+ + γ+ ) · Cλ 2 cos . 2 2 2

(3.1.12)

Let us define the following function F (λ) = (n − 2) sin

.

n ω0 ω0 2 cos · Cλ−1 − 2 2 n−2

(λχ+ + γ+ ) · Cλ 2

ω0 cos . 2

(3.1.13)

Now, using formulae (2.13.10), (2.13.6), we calculate n n−2 ω0 ω0 ω0 2 cos · C−1 − γ+ · C0 2 cos = 2 2 2  ω0 d n−2  sin · C0 2 (x) ω0 − γ+ = −γ+ < 0. x= cos 2 2 dx

F (0) =(n − 2) sin .

(3.1.14)

Next, by (2.13.9), from (3.1.13) we derive ω 3−n (λ + n − 2)( n−1 2 0 2 ) sin × (3.1.15) (λ + 1)(n − 3)! 2 + , 1−n 3−n ω0 ω0 2 2 − (λχ+ + γ+ ) · P n−3 cos λ(λ + n − 2) · P n−3 cos . λ+ 2 λ+ 2 2 2 F (λ) = 2

.

n−3 2

Now, we apply the asymptotic expansion (2.13.11) and obtain 1−n 2 λ+ n−3 2

P

.

ω0 cos = 2 0   & n ω0 nπ ' 1 (λ) 2 λ−1+ − +O ; ω0 · n cos π sin 2 (λ + 2 ) 2 2 4 λ

52

3 Eigenvalue Problems

ω0 = 2   & n ω0 nπ ' 1 (λ + 1) λ−1+ − +O · . n sin (λ + 2 ) 2 2 4 λ 3−n 2 λ+ n−3 2

P

0 −

2 π sin ω20



cos

Choosing n 2π + .λ = λm ≡ 1 − 2 ω0



 1 n + + 2m > 0, since ω0 < π); 3 4 m ∈ N, m

1,

from (3.1.15) together with above asymptotic expansions we derive: ( n−1 1 ω0 2−n 2 2 ) × √ · (3.1.16) sin 2 (n − 2) π 1 2 √   (λm + n − 2) 1 3 1 (λ (λ + n − 2) + χ + γ ) + O . m m + + (λm + n2 ) 2 2 λm F (λm ) = 2

.

n−2 2

Further, using asymptotic formula (2.13.12) and taking into account that χ+ ≥ 0, γ+ > 0 , hence it follows +  , n−2 1 F (λm ) = C(n, ω0 , χ+ , γ+ )λm2 · 1 + O ; n ≥ 3. λm

.

Further, we have

(3.1.17)

lim λm = +∞, therefore, by (3.1.17),

m→+∞

F (+∞) ≡

.

lim F (λm ) = +∞.

m→+∞

Hence and from (3.1.14) we obtained that the continuous function F (λ) takes different sings at the ends of the interval [0, +∞) and therefore it must have the first positive zero.  

3.1.3

On Properties of Eigenvalues

Further we shall derive the estimate of .λ. Let us set .y(ω) = (3.1.11) it follows

ψ  (ω) ψ(ω) .

From above (3.1.10),

n

y(ω) = −(n − 2) sin ω ·

.

2 (cos ω) Cλ−1 n−2

Cλ 2 (cos ω)

, n ≥ 3.

(3.1.18)

3.1

53

PROPERTIES OF EIGENVALUES

Therefore, passing to the limit .ω → 0 in (3.1.18), by (2.13.5),we obtain n

y(0) = −(n − 2)

.

2 (1) Cλ−1 n−2

Cλ 2 (1)

· lim sin ω = 0. ω→0

Now problem .(EVP∗n ) takes the form  .

  y  + y 2 + (n − 2)y cot ω + λ(λ + n − 2) = 0, ω ∈ 0, ω20 ,   y(0) = 0; y ω20 = −(λχ+ + γ+ ).

(3.1.19)

Hence  .

  y  + (n − 2)y cot ω < 0, ω ∈ 0, ω20 , y(0) = 0.

Let us consider the following Cauchy problem  .

  z + (n − 2)z cot ω = 0, ω ∈ 0, ω20 ⇒ z(ω) ≡ 0. z(0) = 0

(3.1.20)

By the Chaplygin comparison principle (see Theorem 2.60) we obtain .y(ω) ≤ z(ω) ≡ 0. Hence and from (3.1.19) it follows  .

  y  ≥ −y 2 − λ(λ + n − 2), ω ∈ 0, ω20 , y(0) = 0.

(3.1.21)

Solving this problem we obtain .

√

  y 1 arctan √ ≥ −ω λ(λ + n − 2) λ(λ + n − 2)

⇒

*

& ω ' * 0 . − λ(λ + n − 2) tan ω λ(λ + n − 2) ≤ y ≤ 0, ω ∈ 0, 2

(3.1.22)

Let us denote

.

ω0 * = λ(λ + n − 2) 2

3 4 4 n − 2 2 4 2 n − 2 . ⇒ λ = 5 + 2 − 2 2 ω0

(3.1.23)

54

3 Eigenvalue Problems

Then from (3.1.22), in virtue of the second boundary condition of problem (3.1.19), it follows .

tan

λχ0 + γ0 ω0 or with regard to (3.1.23) 2 ⎧ ⎤⎫ ⎡3 4 2 ⎨ 2 4 ω0 4 n − 2 ⎦⎬ n−2 tan ≥ + 2 − γ 0 + χ0 · ⎣ 5 , ⎭ 2 ⎩ 2 2 ω0

≥

ω0 ∈ (0, π), n ≥ 3.

(3.1.24)

Since for .n ≥ 3 3 4 4 n − 2 2 4 2 4 2 n−2 1 5 = 2 · : . + 2 −

2 2 2 ω0 ω0 n−2 + 2

≤ 4 2 ω02

+

n−2 2 2 4 2 4 · 2 ≤ 2, n − 2 ω0 ω0

it is sufficient to consider the inequality .

tan

≥

2χ0 ω0

+

γ0 ω0 1 · , 2

ω0 ∈ (0, π), n ≥ 3.

(3.1.25)

γ0 ω0 0 By the graphic method (see Fig. 3.1, here: .a = 2χ ω0 and .b = 2 ), we obtain that π ∗ ∗ .0 < < 2 , where . is the smallest positive solution of (3.1.25). Because of (3.1.23), we obtain

0 ∗

0 γ− ≥ 1.

.

(3.2.1)

Note that if any two eigenfunctions solve the problem for the same value of .λ, then they  scalar multiplies of each other. Without loss of generality we can assume that  ω0are .ψ 2 = 1. Moreover, note that there are two possible cases: either .ψ(−ω) = ψ(ω) or .ψ(−ω) = −ψ(ω). Let .n = p+ = 2. Then, problem .(OEVP) takes the form ⎧   ⎨ψ  + λ2 ψ = 0, ω ∈ − ω20 , ω20 , . ⎩±ψ  + λχ± + γ± (1 + μ)ψ = 0, ω = ± ω0 . 2

(3.2.2)

Let us satisfy the equality λ(χ− − χ+ ) = (1 + μ)(γ+ − γ− ).

.

(3.2.3)

Solving the problem we get: • For .ψ(−ω) = ψ(ω) we have .ψ(ω) =

cos(λω)

.

Thus .ψ(0) = 0 and .ψ  (0) = 0, and by

= χ+ +

(1 + μ)γ+ . λ

cos

λω0 2

the boundary condition  .

tan

λω0 2



By the graphic method, we establish that the least positive root .λ∗ of the above equation satisfies the condition .0 < λ∗ < ωπ0 (see Fig. 3.2).

57

y = tg

λω0 2

3.2 The Nonlinear Eigenvalue Problem

y=

λ∗

π ω0

2π ω0

β λ λ

Fig. 3.2

• For .ψ(−ω) = −ψ(ω) we have .ψ(ω) =

sin(λω)

. λω sin 2 0

Thus .ψ(0) = 0 and .ψ  (0) = 0, and

by the boundary condition 

λω0 . tan 2

 =−

λ < 0. λχ+ + (1 + μ)γ+

Hence it follows that the least positive root .λ∗ of the above equation satisfies the condition .λ∗ > ωπ0 . We are interested in the least positive eigenvalue and therefore we consider the case ψ(−ω) = ψ(ω) ⇒ ψ  (−ω) = −ψ  (ω), ψ  (−ω) = ψ  (ω). It is obviously the (OEVP) equation is satisfies.

.

58

3 Eigenvalue Problems

  Thus, we can rewrite problem (OEVP) as a problem for .ψ(ω) ∈ C 2 0, ω20 ∩ ( ) C 1 0, ω20 : ⎧  ⎪ λ2 ψ 2 + (p+ − 1)ψ 2 ψ  (ω)+ ⎪ ⎪ ⎪   ⎪ ⎪ ⎪ (n − 2) cot ω λ2 ψ 2 + ψ 2 ψ  (ω)+ ⎪ ⎨ . λ (λ(2p+ − 3) + n − p+ ) ψ 2 ψ(ω)+ ⎪ ⎪   ⎪ ⎪ ⎪λ3 (λ(p+ − 1) + n − p+ ) ψ 3 (ω) = 0, ω ∈ 0, ω20 , ⎪ ⎪ ⎪ ⎩ψ  ω0  = 1, ψ  (0) = 0

(3.2.4)

2

and the equation for .λ :   (λ2 + ψ 2 (ω0 /2))(p+ −2)/2 ψ  (ω0 /2) + λχ+ +

.



p+ − 1 + μ p+ − 1

p+ −1

= 0.

(3.2.5)

& ω ω ' 0 0 ψ  (ω) ≤ 0, ∀ω ∈ − , ; 2 2

(3.2.6)

γ+

Lemma 3.6 The eigenpair .(λ, ψ(ω)) in (OEVP) satisfies conditions λ > 0;

.

λ (λ(p+ − 1) + n − p+ ) > 0;

1 ≤ ψ(ω) ≤ ψ0 = const (p, λ, ω0 ),

moreover, if .n = 2 and . 2 − p+ ≤ μ < 1 then is true inequality .

λψ 2 (ω) − 1 −

λ

ψ 2 (ω) + ψ(ω)ψ  (ω) ≤ 0,

=

p+ − 1 λ. p+ − 1 + μ

(3.2.7)

Proof For the (OEVP) boundary conditions in the point .+ ω20 must be: ψ

.

(λ2 ψ 2

+

ψ 2 )

2−p+ 2

ψ (λ2 ψ 2 + ψ 2 )

2−p+ 2

+

(λ2 ψ 2

−

or because of, by (3.2.4), .ψ

.



λχ+ ψ +

ψ 2 )

2−p+ 2

 ω0  2

2−p+ 2

+ γ−

p+ −1

p+ − 1 + μ p+ − 1

2−p+ 2



p+ − 1 + μ = (γ+ − γ− ) p+ − 1

ψ|ψ|p+ −2 =

p+ −1

= 1 hence it follows

λ(χ− − χ+ ) (λ2 + ψ 2 )

p+ − 1 + μ p+ − 1



λχ− ψ (λ2 ψ 2 + ψ 2 )

+ γ+

p+ −1 .

ψ|ψ|p+ −2

3.2 The Nonlinear Eigenvalue Problem

59

Therefore we obtain: • (1) If .p+ = 2 we obtain (3.2.3). • (2) If .χ+ = χ− then we have the Robin problem (see [32]). • (3) By . μ ≥ 0. γ+ > γ− > 0 and let .χ− > 0; then we derive ⎧ ⎪ ⎨ .

λ≥

⎪ ⎩0 < λ ≤

p+ −1+μ p+ −1 p+ −1+μ p+ −1



γ+ −γ− χ− −χ+

γ+ −γ− χ− −χ+



1 p+ −1

1 p+ −1

> 0, if if

,

1 < p+ < 2;

(3.2.8)

p+ > 2.

So we have proved .λ > 0. We multiply the (NEVP) equation by .ψ(ω) and integrate over . :  .

−



ψ(ω)divω (λ2 ψ 2 + |∇ω ψ|2 )(p+ −2)/2 ∇ω ψ d =



 λ (λ(p+ − 1) + n − p+ )

(λ2 ψ 2 + |∇ω ψ|2 )(p+ −2)/2 ψ 2 (ω)d .



Integrating by parts the left integral we obtain:  .

−



ψ(ω)divω (λ2 ψ 2 + |∇ω ψ|2 )(p+ −2)/2 ∇ω ψ d =



 − 

ψ(ω)

  n  p+ −2 ∂ψ J (ω) 2 2 ∂ dω = (λ ψ + |∇ω ψ|2 ) 2 ∂ωi qi ∂ωi 1

(λ2 ψ 2 + |∇ω ψ|2 )(p+ −2)/2|∇ω ψ|2 d −



 ψ(ω)

∂ψ 2 2 (λ ψ + |∇ω ψ|2 )(p+ −2)/2 dσ > 0, ∂ν

∂

because of, by the boundary condition of (NEVP), we have on .∂ : .

− ψ(ω)

∂ψ 2 2 (λ ψ + |∇ω ψ|2 )(p+ −2)/2 = ∂ν

λχ(ω1 )ψ 2 (ω)(λ2 ψ 2 + |∇ω ψ|2 )(p+ −2)/2 +   p+ − 1 + μ p= −1 γ (ω1 ) |ψ(ω)|p+ > 0. p+ − 1

60

3 Eigenvalue Problems

Thus, from above it follows that λ (λ(p+ − 1) + n − p+ ) > 0.

.

Now, we consider the problem (3.2.4) for the function .y(ω) = the following Cauchy problem

ψ  (ω) ψ(ω) ,

ψ(0) = 0. We get

.

⎧     2 2 2 2 ⎪ ⎪ (p+ − 1)y + λ y (ω) + (n − 2) cot ω y + λ y(ω)+ ⎪ ⎪ ⎪ (p+ − 1)y 4 + λ (2λ(p+ − 1) + n − p+ ) y 2 + ⎨ . λ3 (λ(p+ − 1) + n − p+ ) = 0, ⎪   ⎪ ⎪ ω ∈ 0, ω20 , ⎪ ⎪ ⎩ y(0) = 0.

(CP)

and the equation for .λ : .

 

p+ −2 p+ − 1 + μ p+ −1 2 λ2 + y02 = 0, (y0 + λχ+ ) + γ+ p+ − 1 where y0 = y

ω 0

2

< 0.

(3.2.9)

Since . λ > 0, 1 < p+ < n , from (CP) it follows that ⎧     2 2 2 2 ⎪ ⎪ ⎨ (p+ − 1)y + λ y (ω) + (n − 2) cot ω y + λ y(ω) ≤ 0, . ω ∈ 0, ω20 , ⎪ ⎪ ⎩ y(0) = 0.

(CI)

Solving the Cauchy problem ⎧ ⎨(p − 1)Y 2 + λ2  Y  (ω) + (n − 2) cot ω Y 2 + λ2  Y (ω) = 0, + . ⎩ Y (0) = 0,

  ω ∈ 0, ω20 ,

we derive .Y (ω) ≡ 0 and applying the comparison principle for ordinary differential equations (see Theorem 2.60) we obtain & ω ' 0 . y(ω) ≤ 0, ω ∈ 0, 2

.

(3.2.10)

3.2 The Nonlinear Eigenvalue Problem

61

Next from (CP) it follows ⎧    2 2 4 2 ⎪ ⎪ ⎨− (p+ − 1)y + λ y (ω) ≤ (p+ − 1)y + λ (2λ(p+ − 1) + n − p+ ) y +  . λ3 (λ(p+ − 1) + n − p+ ) , ω ∈ 0, ω20 , ⎪ ⎪ ⎩ y(0) = 0. Now we solve the Cauchy problem ⎧   ⎪ − (p+ − 1)z2 + λ2 z (ω) = (p+ − 1)z4 + ⎪ ⎪ ⎪ ⎪ ⎨ λ (2λ(p+ − 1) + n − p+ ) z2 + λ3 (λ(p+ − 1) + n − p+ ) , .   ⎪ ⎪ ω ∈ 0, ω20 , ⎪ ⎪ ⎪ ⎩ z(0) = 0.

(3.2.11)

By integrating of this problem, we obtain |z(ω)| = λ (p+ − 2)λ + n − p+ |z(ω)| ; arctan ; − (n − p+ )ω, n−p + + λ2 + λ p+ −1 λ2 + λ n−p p+ −1

(p+ − 2) arctan

.

1 < p+ < n.

(3.2.12)

Applying once more the comparison principle for ordinary differential equations we have & ω ' 0 |y(ω)| ≤ |z(ω)|, ∀ω ∈ 0, (3.2.13) . 2 ) ( From the definition of .y(ω) and because .y(ω) ≤ 0 ∀ω ∈ 0, ω20 it follows that .

⎛ ⎜ ψ(ω) = exp ⎝−

ω0

2

.

⎞ ⎟ y(ξ )dξ ⎠ ≥ 1,

& ω ' 0 ∀ω ∈ 0, . 2

ω

( )  (ω) Now, since . ψψ(ω) = y(ω) ≤ 0 and . ψ(ω) ≥ 1, ∀ω ∈ 0, ω20 we have that . ψ  (ω) ≤ ( )   0, ∀ω ∈ 0, ω20 . Moreover, since .z(0) = 0 and .z (ω) ≤ 0∀ω ∈ 0, ω20 (see (3.2.11)) we have that  ω  ω ω 0 0 0   ≤ z(ω) ≤ z(0) = 0 ∀ω ∈ 0, ⇒ |z(ω)| ≤ z .z  = z0 2 2 2

62

3 Eigenvalue Problems

and, by (3.2.13),  ω  0   z0 ≥ y  = |y0| 2

.

(3.2.14)

as well as ⎛ ⎜ ψ(ω) = exp ⎝

ω0

2

.

⎛ ω0 ⎞ 2 ω ⎜ ⎟ ⎟ 0 z0 ≡ ψ0 . |y(ξ )|dξ ⎠ ≤ exp ⎝ |z(ξ )|dξ ⎠ ≤ exp 2 ⎞

ω

0

Now, from (3.2.12) we have (p+ − 2) arctan

.

(p+ − 2)λ + n − p+ z0 z0 ω0 ; = arctan ; − (n − p+ ) . n−p n−p λ 2 + + λ2 + λ p+ −1 λ2 + λ p+ −1 (3.2.15)

Equation (3.2.15) was studied in [32] (see Lemma 2.2, equation (14)) and there was proven that z0 ≤ const (n, p+ , λ, ω0 ).

.

(3.2.16)

Thus, (3.2.6) is proven. For the proof of (3.2.7) we calculate

.

1−

λ

ψ 2 (ω) − ψ(ω)ψ  (ω) − λψ 2 (ω) =

μψ 2 (ω) − (p+ − 1)λ2 ψ 2 (ω) + p+ − 1 + μ

λ λ(2p+ − 3) + 2 − p+  ψ 2 (ω)ψ 2 (ω) + λ3 λ(p+ − 1) + 2 − p+  ψ 4 (ω) = λ2 ψ 2 (ω) + (p+ − 1)ψ 2 (ω)   (p+ − 1 + μ)λ λ(2p+ − 3) + 2 − p+  + μλ2 − (p+ − 1)2 λ2 ψ 2 (ω)ψ 2 (ω)   + (p+ − 1 + μ) λ2 ψ 2 (ω) + (p+ − 1)ψ 2 (ω) μ(p+ − 1)ψ 4 (ω)  + (p+ − 1 + μ) λ2 ψ 2 (ω) + (p+ − 1)ψ 2 (ω) (p+ − 1 + μ)λ3 λ(p+ − 1) + 2 − p+  − (p+ − 1)λ4 4   ψ (ω) ≥ (p+ − 1 + μ) λ2 ψ 2 (ω) + (p+ − 1)ψ 2 (ω) (p+ − 1)(p+ − 2 + 2μ)λ2 ψ 2 (ω)ψ 2 (ω) + (p+ − 1)(p+ − 2 + μ)λ4 ψ 4 (ω)   ≥ 0, (p+ − 1 + μ) λ2 ψ 2 (ω) + (p+ − 1)ψ 2 (ω) by our assumptions.

 

3.2 The Nonlinear Eigenvalue Problem

63

Proposition 3.7 If assumption (i) satisfies and .γ+ ≥ 1 (see assumption (iii1)), then  ; .

λ

λ2 + y02

p(x)−p+

≤ 1,

∀x ∈ 0R ,

(3.2.17)

where . is defined by (10.1.3). Proof . Case p+ = 2. The inequality (3.2.17) is true if .p(x) ≡ p+ = 2. Now, let . 1 < p(x) ≤ p+ = 2, ∀x ∈ 0R . From (3.2.9) with regard to (10.1.3) we have ; γ+ λ ⇒ |y0 | = γ+ (1 + μ) + λχ+ ≥ λ2 + y02 ≥ |y0| ≥ γ+ ≥ 1 λ λ   ; λ2 + y02 ≤ 0 ⇒ (3.2.17) is true. (p(x) − p+ ) ln λ

.

Case 1 < p+ < 2. From (3.2.9) we obtain that

.

 p+ −1 ; 2−p+  p+ −1 λ λ · λ2 + y02 + λχ+ ≥ γ+ |y0 |2−p+ ⇒

|y0 | = γ+

.

|y0 | ≥

;

λ

and

λ

λ2 + y02 ≥

λ

|y0 | ≥ 1

⇒

(3.2.17) is true.

Case p+ > 2. From (3.2.9) it follows that

.

|y0 |(λ + 2

.

γ+

 p+ −1 λ ;

2

(λ

y02 )

p+ −2 2

⇒

= λχ+ (λ + 2

|y0 | ≥ γ+

λ2 + y02 ≥ |y0| ≥ γ+

+ y02 )

p+ −1 2

(p(x) − p+ ) ln

≥ γ+

y02 )

 p+ −1 λ

 p+ −1 λ

 p+ −1 λ

p+ −2 2

(λ2 + y02 )

(λ2 + y02 )

; and

+ γ+

2−p+ 2

2−p+ 2

λ2 + y02 ≥

 p(x) − p+ ln γ ≤ 0 λ2 + y02 ≤ λ p+ − 1

 ;

 p+ −1 λ

λ

≥ ⇒

⇒ 1

· γ p+ −1

⇒

⇒

(3.2.17) is true.  

64

3 Eigenvalue Problems

Corollary 3.8 Here we consider the case . n = 2. From the Cauchy problem equation (CP) we have . y  (ω) < 0, therefore

.

y0 ≤ y(ω) ≤ 0,

.

⇒ |y(ω)| ≤ |y0 |.

(3.2.18)

From Eq. (3.2.9) for .p+ = 2 we have |y0 | = (1 + μ)γ+ + λχ+ .

.

(3.2.19)

Let now . 1 < p+ < 2. By solving the Cauchy problem .(CP ), we obtain .

arctan

1−λ |y(ω)| |y(ω)| +; arctan ; =ω 2−p λ + + λ2 + λ p+ −1 λ2 + λ 2−p p+ −1

and consequently .

Since . arctan ;

arctan

1−λ |y0 | |y0 )| ω0 +; . arctan ; = 2−p 2−p λ 2 + + λ2 + λ p+ −1 λ2 + λ p+ −1

|y0 | 2−p+ + −1

λ2 +λ p

0 |y0 |
1, s = 1 and

F (x) =

.

⎧x ⎪ ⎪ f (ξ )dξ, ⎪ ⎪ ⎪ ⎨0

if s > 1;

⎪ ⎪ ∞ ⎪ ⎪ ⎪ ⎩ f (ξ )dξ, if s < 1; x

then 

∞

x

.

−s

 F (x)dx ≤ p

0

p |s − 1|

p 

∞

x −s (xf )p dx.

(4.1.1)

0

The constant is the best. We prove the partial case p = 2. Theorem 4.2 Let f ∈ L2 (0, d), d, β > 0 and F (x) =

x

1

y β− 2 f (y)dy. Then

0

d x

.

0

−2β−1

1 F (x)dx ≤ 2 β

d

2

f 2 (x)dx.

(4.1.2)

0

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Borsuk, Oblique Derivative Problems for Elliptic Equations in Conical Domains, Frontiers in Elliptic and Parabolic Problems, https://doi.org/10.1007/978-3-031-28381-9_4

65

66

4 Integral Inequalities

Proof Let 0 < δ < β. Then by Hölder’s inequality (2.14)  x 2     2 δ β−δ− 21  .|F (x)| = dy  ≤  y f (y)y   0

x

x 2δ

≤

2

y 2β−2δ−1 dy =

y f (y)dy 0

0

1 = x 2(β−δ) 2(β − δ)

x y 2δ f 2 (y)dy. 0

Therefore, by the Fubini Theorem 2.13, d

x −2β−1 F 2 (x)dx ≤

.

0

1 2(β − δ)

1 = 2(β − δ)

1 . = 2(β − δ)

d 0

1 4δ(β−δ)

0

d

⎛ d ⎞  y 2δ f 2 (y) ⎝ x −2δ−1 dx ⎠ dy = y

0

d y 2δ f 2 (y)

y −2δ − d −2δ dy ≤ 2δ

0

1 ≤ 4δ(β − δ) Noting that assertion.

⎞ ⎛ x  x −2δ−1 ⎝ y 2δ f 2 (y)dy ⎠ dx =

d f 2 (x)dx. 0

becomes minimal for δ =

1 2 β,

we choose δ :=

1 2β

and obtain the  

Corollary 4.3 Let v ∈ C 0 [0, d] ∩ W 1,2 (0, d), d > 0 with v(0) = 0. Then d r N−5+α v 2 (r)dr ≤

.

0

4 (4 − N − α)2



d r N−3+α

∂v ∂r

2 dr

0

for α < 4 − N provided that the integral on the right hand side is finite.

(4.1.3)

4.1 Classical Hardy Inequalities

67

Proof We apply Hardy’s inequality (4.1.2) with F = v, β :=  2 1 F  (r) = r β− 2 f (r) and therefore f 2 (r) = r 1−2β ∂v . ∂r

4−N−α , 2

noting that  

Remark 4.4 The constant in (4.1.3) is the best possible. Corollary 4.5 If u ∈ C0∞ (Rn ), α < 4 − N and u(0) = 0, then  r α−4 u2 (x)dx ≤

.

RN

4 (4 − N − α)2

 r α−2 |∇u(x)|2 dx RN

provided that the integral on the right hand side is finite. Proof The assertion follows by integrating both sides of (4.1.3) over for large enough d and applying (2.2.7).   Corollary 4.6 If u ∈ W01,2 (G), α < 4 − N, then  r

.

4 u (x)dx ≤ (4 − N − α)2

 r α−2 |∇u(x)|2dx,

α−4 2

G

(4.1.4)

G

provided that the integral on the right hand side of (4.1.4) is finite. Proof The claim follows from Corollary 4.5 because C0∞ (G) is dense in W01,2 (G).

 

Corollary 4.7 Let v ∈ C 0 [ε, d] ∩ W 1,2 (ε, d), d > 0 with v(ε) = 0. Then d r N−5+α v 2 (r)dr ≤

.

ε

4 (4 − N − α)2



d r N−3+α

∂v ∂r

2 dr

(4.1.5)

ε

for α < 4 − N. Proof We apply the inequality (4.1.3) to the function v(r) extended by zero into [0, ε).   Note also another generalization of the Hardy inequality: Theorem 4.8 The inequality ∞ x

.

0

α−p

pp |f (x)| dx ≤ |α + 1 − p|p

∞

p

0

x α |f  (x)|p dx

(4.1.6)

68

4 Integral Inequalities

is true if p > 1, α = p − 1 and f (x) is absolutely continuous on [0, ∞) and satisfies the following boundary condition

.

⎧ ⎨f (0) = 0

when α < p − 1,

⎩ lim f (x) = 0 when α > p − 1. x→+∞

 

Proof For the proof of Theorem 4.8 we refer to [47, 116]

4.2

The Friedrichs-Wirtinger Type Inequality

Theorem 4.9 Let λ be the smallest positive eigenvalue of problem (EVP) and let χ(ω1 ), γ (ω1 ) be positive bounded piecewise smooth functions on ∂ . For any u ∈ W 1 ( ) ∩ C 0 ( ) holds the inequality  .

⎤ ⎡   1 u2 d ≤ ⎣ |∇ω u|2 d + λχ(ω1 ) + γ (ω1 ) u2 dσ ⎦ , θ





(F W )

∂

where θ = λ(λ + n − 2). Proof At first, let ψ(ω) ∈ C 1 ( ) ∩ C 2 ( ) be the eigenfunction corresponding to the eigenvalue λ and u(ω), v(ω) ∈ C 1 ( ) ∩ C 0 ( ). Setting u(ω) = ψ(ω)v(ω) we obtain 1  2   2 n−1 n−1  ∂u 2  J J ∂ψ ∂v 2 ∂ψ v J |∇ω u| = ≥ + 2ψv qi ∂ωi qi ∂ωi ∂ωi ∂ωi 2

i=1

.

=

i=1

   , n−1 +  J ∂ψ ∂ ∂ J ∂ψ ψv 2 − ψv 2 . ∂ωi qi ∂ωi ∂ωi qi ∂ωi i=1

Therefore 

       n−1 n−1  J ∂ψ ∂ ∂ 2 J ∂ψ 2 |∇ω u| d ≥ ψv dω − ψv dω ∂ωi qi ∂ωi ∂ωi qi ∂ωi 2

.

i=1



=

  n−1 ∂ i=1

i=1



ψv 2

1 ∂ψ cos ( ν , ωi ) dσ − qi ∂ωi

 ψv 2 ω ψd .

4.2 The Friedrichs-Wirtinger Type Inequality

69

Taking into account that cos ( ν , ω1 ) = 1, cos ( ν , ωi ) = 0 for all 2 ≤ i < n − 1 and that ψ(ω is the solution of (EVP), we get 

 |∇ω u| d ≥ θ

.



 λχ(ω1 ) + γ (ω1 ) ψ 2 v 2 dσ.

ψ v d −

2

2 2



∂

Finally, returning to u = ψv, we have the required inequality (F W ). The extension to   u ∈ W 1 ( ) ∩ C 0 ( ) follows directly by the approximation arguments.   ◦1 Theorem 4.10 Let Gd0 be a conical domain, u ∈ W 2−n Gd0 , λ > 0 be the smallest positive eigenvalue of problem (EVP) and let χ(ω1 ), γ (ω1 ) be positive bounded piecewise smooth functions on ∂ . Then  .

r −n u2 dx

Gd0

≤

⎡ ⎢ 1 ⎢ λ(λ + n − 2) ⎣



⎤



⎥ λχ(ω1 ) + γ (ω1 ) r 1−n u2 ds ⎥ ⎦.

r 2−n |∇u|2 dx +

Gd0

(4.2.1)

0d

Proof Multiplying the inequality (F W ) by r −1 and integrating over r ∈ (0, d), we have  λ(λ + n − 2)

.

r

d 

−n 2

u dx = λ(λ + n − 2) 0

Gd0

d  ≤

r −1 u2 drd

r

−1

d  |∇ω u| drd + 2

0

 =

λχ(ω1 ) + γ (ω1 ) r −1 u2 drdσ

0 ∂

r −n |∇ω u|2 dx +

Gd0

Hence we get the desired inequality (4.2.1).

 λχ(ω1 ) + γ (ω1 ) r 1−n u2 ds. 0d

 

70

4 Integral Inequalities

Lemma 4.11 Let Gd0 be a conical domain, ∇u(, ω) ∈ L2 ( ) for almost everywhere  ∈ (0, d) and χ(ω1 ), γ (ω1 ) ∈ C 0 (∂ ). Let λ > 0 be the smallest positive eigenvalue of problem (EVP) and -() = U



 r 2−n |∇u|2 dx +

.



G0

γ (ω1 )r 1−n u2 ds.

(4.2.2)



0

Then   .



   ∂u   - n − 2 2  1 u  d ≤ u  + χ(ω1 )u2 dσ. U () + r= ∂r r= 2 2λ 2 ∂

-() in spherical coordinates. As a result we obtain Proof Let us write U 2   1 2   ∂u 1 1 -() = r .U + 2 |∇ω u|2 d dr + γ (ω1 )u2 dσ dr. ∂r r r 0

0



∂

Differentiating the last equality by  we have 2  1  2    1 ∂u  1 2 |∇ .U () = + u| γ (ω1 )u2 (, ω)dσ.  d +  ω  r= ∂r   r= -



∂

Moreover, by the Cauchy inequality, we have for any ε > 0     ∂u ∂u 1 2 ∂u 2 ε 2 =u  .u . ≤ u +  ∂r ∂r 2 2ε ∂r Applying the above inequality, the Friedrichs-Wirtinger type inequality (F W ) and choosing ε = λ, we obtain   .

 1



   n − 2 2  ∂u  u  + u  d ≤ ∂r r= 2 r=

    2 ε + n − 2 2  2 ∂u 2  d ≤ u  +  2 2ε ∂r r= r=

4.2 The Friedrichs-Wirtinger Type Inequality

 1

71

  2 2  ∂u 2    ε+n−2  |∇ω u|2  + d +  2λ(λ + n − 2) 2ε ∂r r= r=

ε+n−2 2λ(λ + n − 2)

 λχ(ω1 ) + γ (ω1 ) u2 (, ω)dσ = ∂

⎧ 1 ⎫  2  2    ⎬  ∂u  1  ⎨ 1 2 2 |∇  d + + u| γ (ω )u (, ω)dσ + ω 1   ⎭ 2λ ⎩ ∂r   r= r=



1 2

 - 1 χ(ω1 )u2 (, ω)dσ = U () + 2λ 2

∂



∂

χ(ω1 )u2 (, ω)dσ. ∂

  Lemma 4.12 Let v ∈ C 0 (Gdε ) ∩ W 1 (Gdε ) for all ε > 0, γ (ω1 ) > 0, χ(ω1 ) ≥ 0 and λ be the smallest positive eigenvalue of (EVP). Then  λ(λ + n − 2)

.

r −n v 2 dx ≤

Gdε



 r 2−n |∇v|2 dx + Gdε

λχ(ω1 ) + γ (ω1 ) r 1−n v 2 ds.

(4.2.3)

εd

Proof By Theorem 4.9, the inequality (F W ) holds. Multiplying it by r −1 and integrating   over r ∈ (ε, d) we obtain the required (4.2.3). Lemma 4.13 Let v ∈ C0 (G) ∩ W1 (G) and v(·, ω) satisfies the boundary conditions of (EVP). Let χ(ω) ≥ 0, γ (ω) ≥ γ0 > 0. Then for any ε > 0  .

Gd0

rεα−2 r −2 v 2 dx

1 ≤ λ(λ + n − 2)

 rεα−2 |∇v|2 dx+ Gd0

 0d

 r −1 rεα−2 (λχ(ω) + γ (ω)) v 2 ds .

(4.2.4)

72

4 Integral Inequalities

Proof Multiplying both sides of the Friedrichs-Wirtinger inequality (W) by ( + ε)α−2 r n−3 and integrating over r ∈ ( 2 , ) we obtain  .

( + ε)α−2 r −2 v 2 dx ≤



G/2

1 λ(λ + n − 2) 

  ( + ε)α−2 |∇v|2 dx+ 

G/2

 r −1 ( + ε)α−2 (λχ(ω) + γ (ω)) v 2 ds , ∀ε > 0



/2

or since  + ε ∼ rε  . 

G/2

rεα−2 r −2 v 2 dx ≤

1 λ(λ + n − 2)

  rεα−2 |∇v|2 dx+ 

G/2



 r −1 rεα−2 (λχ(ω) + γ (ω)) v 2 ds , ∀ε > 0.



/2

Letting ρ = 2−k d, (k = 0, 1, 2, . . .) and summing the obtained inequalities over all k we   get the desired inequality (4.2.4).

The Linear Oblique Derivative Problem for Elliptic Second Order Equation in a Domain with Conical Boundary Point

5.1

Preliminaries

At first, we formulate well known statements. Proposition 5.1 (The Local Maximum Principle (See Theorem 3.3 [85], Theorem 4.3 [88]; See as Well [87])) Let G be a bounded domain in .Rn with the .C 1 -boundary .∂G\0d and .Gd0 be a convex rotational cone with vertex at .O and the aperture .ω0 ∈ ( π2 , π). Let .u(x) be a strong solution of the problem ⎧ ij i ⎪ ⎪ ⎨L[u] ≡ a (x)uxi xj + a (x)uxi + a(x)u = f (x), .

⎪ ⎪ ⎩

a ij = a j i ,

B0 [u] ≡

∂u β i (x) ∂x i

+ γ (x)u = g(x),

x ∈ G,

(OP )

x ∈ ∂G.

Suppose the operator .L is uniformly elliptic with the ellipticity constants .0 < ν ≤ μ, a i (x) ∈ Lp (G), p > n, a(x) ∈ Ln (G), a(x) ≤ 0, x ∈ G; f (x) ∈ Ln (G), g(x) ∈ L∞ (∂G); .β i (x), γ (x) ∈ C 1 (∂G), γ (x) ≥ γ0 > 0 on .∂G. Then for any .q > 0 and .σ ∈ (0, 1), we have .

⎧⎛ ⎫ ⎞1/q ⎪ ⎪ |u|q dx ⎪ ⎪ ⎪ ⎨⎜ GR ⎬

⎪ ⎟ 0 ⎜ ⎟ + R f  n R + gL∞ (∂G) , . sup |u(x)| ≤ C L (G ) ⎝ measGR ⎠ 0 ⎪ ⎪ ⎪ ⎪ Gσ0 R 0 ⎪ ⎪ ⎩ ⎭ where .C = C(ν, μ, γ0 , n, p, R, G, a i Lp (G) , aLn (G) ). © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Borsuk, Oblique Derivative Problems for Elliptic Equations in Conical Domains, Frontiers in Elliptic and Parabolic Problems, https://doi.org/10.1007/978-3-031-28381-9_5

73

5

74

5 The Linear Oblique Derivative Problem for Elliptic Second Order Equation. . .

Theorem 5.2 (.Lp -Estimate of Solutions of the Elliptic Oblique Problem in the Smooth Domain (See Theorem 15.3 of [2])) Let G be a domain in .Rn with a .C 2 boundary portion .T ⊂ ∂G. Let .L be uniformly elliptic in G with the ellipticity constants .0 < ν ≤ μ and .u ∈ W 2,p (G), p > 1 be a strong solution of the problem ⎧ ⎨L[u] ≡ a ij (x)u i ij ji xi xj + a (x)uxi + a(x)u = f (x), a = a , . ⎩B0 [u] ≡ β i (x) ∂u + γ (x)u = g(x),

x ∈ G, x ∈ T,

∂xi

(OP )

where (i) .a ij (x), a i (x), a(x) ∈ C 0 (G); β i (x), γ (x) ∈ C 1 (T ) (ii) .f (x) ∈ Lp (G), g(x) ∈ W

1− p1 ,p

(T ).

Then, for any domain .G ⊂⊂ G ∪ T we have  uW 2,p (G ) ≤ C uLp (G) + f Lp (G) + g

.

1− 1 ,p W p (T )

 ,

(5.1.1)

where the constant C is independent of u and depends only on .N, p, ν, μ, T , G , G, a i (x)C 0 (G) , a(x)C 0 (G) , β i (x)C 1 (T ) , γ (x)C 1 (T ) and the moduli of continuity of the coefficients .a ij (x) on .G .

5.2

Setting of the Problem

Let .G ⊂ Rn be a bounded domain with boundary .∂G that is a smooth surface everywhere except at the origin .O ∈ ∂G and near .O it is a conical surface. We consider the following problem  .

L[u] ≡ a ij (x)uxi xj + a i (x)uxi + a(x)u = f (x), x ∈ G , ∂u 1 B[u] ≡ ∂u x ∈ ∂G\O ∂ n + χ(ω) ∂r + |x| γ (ω)u = g(x),

(L)

where .n denotes the unite exterior normal vector to .∂G\O, .(r, ω) are spherical coordinates in .Rn with pole .O; summation over repeated indices from 1 to n is understood. Definition 5.3 A function .u(x) is called a strong solution of problem (L) provided that   2,n (G) ∩ W 2 (Gε ) ∩ C 0 G and satisfies the equation for any .ε > 0 function .u(x) ∈ Wloc .Lu = f for almost all .x ∈ Gε as well as the boundary condition .Bu = g in the sense of traces on .ε .

5.2 Setting of the Problem

75

Regarding the problem we assume that following conditions are satisfied: (a) The uniformly ellipticity condition ν|ξ |2 ≤

n 

.

a ij (x)ξi ξj ≤ μ|ξ |2 , ∀ξ ∈ Rn , x ∈ G

i,j =1 j

ij ji ij with the ellipticity .ν, μ > 0; .a (x) = a (x); a (0) = δ . i   constants ij 0 i p (b) .a (x) ∈ C G , .a (x) ∈ L (G), .p > n; .a(x) and ◦0

◦ 1/2

f (x) ∈ Ln (G) ∩ W 4−n (G), .g(x) ∈ W 4−n (∂G); there exists a monotonically increasing nonnegative function .A, continuous at zero, .A(0) = 0, such that for .x ∈ G .

⎛

⎞1 1  n n  2 2 2     j .⎝ |a i (x)|2 + |x|2 |a(x)| ≤ A(|x|). a ij (x) − δi  ⎠ + |x| i,j =1

i=1

  (c) .γ (ω), χ(ω) ∈ C 1 and there exist numbers .χ0 ≥ 0, .γ0 > 0 such that .γ (ω) ≥ γ0 > 0, .0 ≤ χ(ω) ≤ χ0 . (d) There exist numbers .f1 ≥ 0, .g1 ≥ 0, .g0 ≥ 0, .s > 0 such that  r 4−n |∇g|2 dx ≤ g02 2s ,  ∈ (0, 1).

|f (x)| ≤ f1 |x|s−2, |g(x)| ≤ g1 |x|s−1 ,

.



G0

Let .λ be the smallest positive eigenvalue of .(EVP) and : ks =

.

g02 +

 1  2 f1 + g12 . 2s

(5.2.1)

Main results of this article are the following statements about power modulus of continuity: Theorem 5.4 Let .u(x) be a strong solution of problem (L) and assumptions (a)–(d) are satisfied with .A(r) being Dini continuous at zero. Then there are .d ∈ (0, 1) and constant .C > 0 depending only on .ν, .μ, s, .λ, .γ0 , .χ0 , .γ C 1 (∂G) , .χC 1 (∂G) , .diam G, .meas G, on

76

5 The Linear Oblique Derivative Problem for Elliptic Second Order Equation. . . 1

the modulus of continuity of leading coefficients and on the quantity . Ar(r) dr, such that 0

for all .x ∈ Gd0 |u(x)| ≤

.

 C |u(x)|0,G + ks + f (x) ◦ 0



W 4−n (G)

+ g(x) ◦ 1/2

×

W 4−n (∂G)

⎧ λ ⎪ if s > λ ⎨ |x| , 1 λ |x| ln |x| , if s = λ . ⎪ ⎩ s if s < λ |x| ,

(5.2.2)

Theorem 5.5 Let .u(x) be a strong solution of problem (L) and assumptions (a)–(d) are satisfied with .A(r), which is a continuous at zero function, but not Dini continuous. Then for any .ε > 0 there are .d ∈ (0, 1) and a constant .Cε > 0 depending only on .ν, .μ, s, .λ, .ε, d .γ0 , .χ0 , .γ C 1 (∂G) , .χC 1 (∂G) , .meas G and on .A (diam G) such that for all .x ∈ G 0 |u(x)| ≤

.

 Cε |u(x)|0,G + ks + f (x) ◦ 0



W 4−n (G)

+ g(x) ◦ 1/2

W 4−n (∂G)



×

|x|λ−ε , if s ≥ λ . |x|s−ε , if s < λ

(5.2.3)

Theorem 5.6 Let .u(x) be a strong solution of problem (L) and assumptions (a)–(d) are satisfied with .A(r) ∼ 11 . Then there are .d ∈ (0, 1) and constants .C > 0, .cs > 0 ln

r

depending only on .ν, .μ, s, .λ, .γ0 , .χ0 , .γ C 1 (∂G) , .χC 1 (∂G) , .meas G and on .A (diam G) such that for all .x ∈ Gd0 |u(x)| ≤

.

 C |u(x)|0,G + ks + f (x) ◦ 0



W 4−n (G)

+ g(x) ◦ 1/2

W 4−n (∂G)

 ln

cs (λ)

1 |x|



×

|x|λ , if s ≥ λ . |x|s , if s < λ

(5.2.4)

5.3 The Global Integral Weighted Estimate

5.3

77

The Global Integral Weighted Estimate

Theorem 5.7 Let u(x) be a strong solution of problem (L) and assumptions (a)–(c) are ◦2

satisfied. Then u(x) ∈ W 4−n (G) and ⎛ u(x) ◦ 2

.

W 4−n (G)

+⎝

⎞1



2

r 1−n γ (ω)u2 (x)ds ⎠ ≤

∂G

 C |u(x)|0,G + f (x) ◦ 0



W 4−n (G)

+ g(x) ◦ 1/2

,

(5.3.1)

W 4−n (∂G)

where C > 0 depends on ν, μ, diam G, χC 1 (∂G) , γ C 1 (∂G) and on the modulus of continuity of leading coefficients. Proof Rewriting the equation of (L) in the form

u = f (x) −

.

& ' j a ij (x) − δi uxi xj + a i (x)uxi + a(x)u ,

(5.3.2)

multiplying both sides by r 2−n u(x) and integrating over Gε we obtain 

 r

.

2−n

u udx =

Gε

 r

2−n

uf dx −

Gε

r 2−n u

& j a ij (x) − δi uxi xj +

Gε

' a i (x)uxi + a(x)u dx.

(5.3.3)

Calculating the integral from the left by parts and using the boundary condition, we obtain 

 .

r 2−n u udx =

r 2−n u ε

Gε

 ε

∂u ds − ε2−n ∂ n

 u ε



∂u d ε − ∂r

+ , ∂r uxi r 2−n uxi + (2 − n)r 1−n u dx = ∂xi

Gε

   ∂u 1 ∂u 2−n r u g(x) − γ (ω)u − χ(ω) ds − ε2−n u d ε − r ∂r ∂r 

 r 2−n |∇u|2 dx + (n − 2)

Gε

Gε

ε

r −n uuxi xi dx.

(5.3.4)

78

5 The Linear Oblique Derivative Problem for Elliptic Second Order Equation. . .

We integrate by parts the last integral of the previous equality  r

.

−n



1 uuxi xi dx = 2

Gε

r

−n

Gε

1 2





∂u2 1 xi dx = ∂xi 2

r −n u2 xi cos(

n, xi )ds−

ε

r −n u2 xi cos(

n, xi )d ε −

1 2

ε

 u2

∂  −n  r xi dx. ∂xi

(5.3.5)

Gε

But  n n   ∂  −n   −n ∂r −n−1 r xi = . xi r − nr = 0. ∂xi ∂xi i=1

(5.3.6)

i=1

Hence, because of   xi cos(

n , x i )

.

  = ε, xi cos (

n, xi ) 

ε

0d

= 0,

equality (5.3.5) takes the form  .

r −n uuxi xi dx = −

1 2

 u2 d +

Gε

1 2



r −n u2 xi cos(

n, xi )ds.

(5.3.7)

d

Therefore from (5.3.4), (5.3.7) we have   ∂u 1 r 2−n u g(x) − γ (ω)u − χ(ω) ds− r ∂r



 r 2−n u udx =

.

G!

ε



ε

∂u u d ε − ∂r

2−n ε

 r

2−n

n−2 |∇u| dx − 2



2

Gε

n−2 2



u2 d +

r −n u2 xi cos(

n, xi )ds.

(5.3.8)

d

Now, from (5.3.3) and (5.3.8) we get 

 .

r 2−n |∇u|2 dx +

Gε

 −

γ (ω)r 1−n u2 ds + ε

χ(ω)r 2−n u ε

∂u ds − ε2−n ∂r

 u2 d =

 u ε

n−2 2

∂u n−2 d ε + ∂r 2

 d

r −n u2 xi cos(

n, xi )ds

5.3 The Global Integral Weighted Estimate

 −

 r

2−n

Gε

& ' j a ij (x) − δi uxi xj + a i (x)uxi + a(x)u dx+

r 2−n u

uf dx +

79

Gε

 r 2−n ugds.

(5.3.9)

ε

Now, we estimate the integral over ε in the above equation. To this we introduce the function M(ε) = max |u(x)|

(5.3.10)

lim M(ε) = |u(0)|,

(5.3.11)

.

x∈ ε

and then .

ε→+0

because of u ∈ C 0 (G). Lemma 5.8 There exists a positive constant c0 depending only on ν, μ, G, max A(|x − x,y∈G

y|), χC 1 (∂G) , γ C 1 (∂G) such that       ∂u  2−n  . lim ε  u d ε  ≤ c0 |u(0)|2 .   ε→+0 ∂r  ε 

(5.3.12)

2ε Proof Let us consider the set G2ε ε , so that ε ⊂ ∂Gε . We use the inequality (2.5.1) of Theorem 2.27   . |w|d ε ≤ c (|w| + |∇w|) dx, ε

G2ε ε

where c depends only on the domain G. We set w = u ∂u ∂r and we find    

 ∂u  2 2 2 −2 2   d ε ≤ c r dx. . u + |∇u| + r u u xx  ∂r  ε

(5.3.13)

G2ε ε

 Considering sets Gε/2 and G2ε ε ⊂ Gε/2 and introducing new variables x defined by 5ε/2

5ε/2

x = εx  , we have that function w(x  ) = u(εx  ) satisfies in G1/2 the following problem for 5/2

80

5 The Linear Oblique Derivative Problem for Elliptic Second Order Equation. . .

the uniformly elliptic equation ⎧ ⎪ a ij (εx  )wxi xj + εa i (εx  )wxi + ε2 a(εx  )w = ε2 f (εx  ), ⎪ ⎪ ⎪ ⎪ 5/2 ⎨ x  ∈ G1/2 ; . ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ∂w 5/2 ∂w 1  x  ∈ 1/2 . ∂ n  + χ(ω) ∂r  + |x  | γ (ω)w = εg(εx ),

(5.3.14)

We use L2 -estimate for the solution of problem (5.3.14) inside the domain and near a smooth portion of the boundary (see Theorem 5.2): 

. wx2 x  + |∇  w|2 + w2 dx  ≤ G21

c1

 

ε4 f 2 + w2 dx  + c2 ε2 inf |∇  G|2 + |G|2 dx  ; 5/2

5/2

G1/2

G1/2



 5/2 here infimum is taken over all G ∈ W 1 G1/2 such that G  5/2 = g and constants c1 , c2 > A(|x  − y  |), χ

1/2



5/2 , 1/2



5/2 . C 1 1/2

γ 

Let us return



r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx ≤

(5.3.15)

0 depend only on ν, μ, G,

max

5/2

x  ,y  ∈G1/2

C1

to the variable x. As a result we get .

 c3

G2ε ε



r 4−n f 2 + r −n u2 dx + c4 inf

5ε/2



r 4−n |∇G|2 + r 2−n |G|2 dx.

5ε/2

Gε/2

Gε/2

Further,  r

. 5ε/2

Gε/2

−n 2

5ε/2 

r −1

u dx = ε/2

 u2 (r, ω)d dr ≤

2ε(θ1 ε)−1



u2 (θ1 ε, ω)d ≤ 2θ1 −1 M 2 (θ1 ε) · meas

(5.3.16)

5.3 The Global Integral Weighted Estimate

81

for some 12 < θ1 < 52 , by the mean value theorem with regard to u ∈ C 0 (G) and (5.3.10). From (5.3.13), (5.3.15), (5.3.16) it follows that       ∂u  2−n  .ε c3 f 2◦ 0 5ε/2 + c4 g2◦ 1/2 5ε/2 .  u d ε  ≤ c0 M 2 (θ1 ε) +   ∂r W 4−n Gε/2 W 4−n ε/2  ε  Hence, by (5.3.11) and assumptions about functions f , g, we finally obtain (5.3.12).

 

Estimating integrals from the right side of equality (5.3.9), by the Cauchy inequality, we get: •  r 2−n uf dx ≤

.

Gε

 



n n δ r −n u2 dx+ r − 2 |u| r 2− 2 |f | dx ≤ 2 Gε

1 2δ



Gε

r 4−n f 2 dx, ∀δ > 0.

(5.3.17)

Gε

•  r 2−n ugds ≤

.

ε

 ε

δ1 2

1

n

r 2−2

*

γ (ω)|u|

 γ (ω)r



 3 n 1 r 2−2 √ |g| ds ≤ γ (ω)

1 u ds + 2δ1 γ0

ε

•



1−n 2

r 3−n g 2 ds, ∀δ1 > 0,

(5.3.18)



|∇u|2 + u2 dx,

(5.3.19)

ε

because of γ (ω) ≥ γ0 .  .

d

r −n u2 xi cos(

n, xi )ds ≤ d 1−n

 u2 ds ≤ d

cd 1−n

Gd

by Theorem 2.27 (inequality (2.5.2).

82

5 The Linear Oblique Derivative Problem for Elliptic Second Order Equation. . .

•  .

  

     j r 2−n |u| a ij (x) − δi  uxi xj  + a i (x) uxi  + |a(x)| |u| dx ≤

Gε



A(r) r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx,

(5.3.20)

Gε

by assumption (b). • Further,  .

−

χ(ω)r 2−n u ε

∂u ds = − ∂r

 χ(ω)r 2−n u

∂u ds− ∂r

εd

 χ(ω)r 2−n u

∂u ds; ∂r

(5.3.21)

d

 .

−

χ(ω)r

2−n

∂u u ds ≤ d 2−n χ0 ∂r

d

    ∂u  u  ds ≤  ∂r  d



u2xx + |∇u|2 + u2 dx, C(d, χ0 )

(5.3.22)

Gd

because of 0 ≤ χ(ω) ≤ χ0 and Theorem 2.27 (inequality (2.5.2);  .

−

χ(ω)r 2−n u εd



1 ∂u ds = − ∂r 2

χ(ω)

∂u2 drdσ = ∂r

εd

 ω   d  1 ω0  ∂u2 r, 20 , ω  − J (ω) , w drdω ≤ χ ω w1 = 20 2 2 ∂r ∂

1 2

ε

 χ ∂

ω

0 , ω u2 ε, , ω dω 2 2

ω

0

(5.3.23)

5.3 The Global Integral Weighted Estimate

83

in virtue of ωi ∈ (0, π) for i = 2, 3, . . . , n − 2 and χ(ω) ≥ 0, where w = (w2 , . . . , wn−1 ); therefor from (5.3.21)–(5.3.23) it follows  .

−

χ(ω)r

2−n

∂u u ds ≤ C (χ0 , d) ∂r

ε



u2xx + |∇u|2 + u2 dx+

Gd

1 2



χ

ω

0 , ω u2 ε, , ω dω . 2 2

ω

0

(5.3.24)

∂

Now, using the inequality  .

r 3−n g 2 ds ≤ Cg2◦ 1/2

0d

(5.3.25)

W 4−n (0d )

(see Lemma 2.39) and choosing δ1 = 1, from (5.3.9), (5.3.17)–(5.3.24) we obtain 

 r 2−n |∇u|2 dx +

.

Gε

γ (ω)r 1−n u2 ds ≤ ε

      1  ω

ω

∂u 0 0  2−n  ε , ω u2 ε, , ω dω + χ  u d ε  +   2 ∂r 2 2  ε  ∂  

δ 2 A(r) r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx + r −n u2 dx+ 2 Gε

Gε



1 c u2xx + |∇u|2 + u2 dx + f 2◦ 0 C(χ0 , d) + g2◦ 1/2 . 2δ γ 0 W 4−n (G) W 4−n (∂G) Gd

By assumption (b), we have that A(r) is continuous in zero and A(0) = 0, thus ∀δ > 0 ∃d > 0 such that A(r) < δ for all 0 < r < d.

.

Hence, by (5.3.15) and (5.3.16), assuming that 2ε < d, we obtain  .

Gε

A(r) r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx =  G2ε ε

A(r) r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx+

(5.3.26)

84

5 The Linear Oblique Derivative Problem for Elliptic Second Order Equation. . .



A(r) r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx+

Gd2ε





A(r) r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx ≤

Gd





CA(2ε) M 2 (ε) + f 2◦ 0



5ε/2 W 4−n Gε/2

δ

+ g2◦ 1/2

5ε/2

+

W 4−n ε/2



r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx+

Gd2ε



C1 (d, diam G) u2xx + |∇u|2 + u2 dx,

(5.3.27)

Gd

for all δ > 0 and 0 < ε < d/2. Substituting ε = 2−k−1 d to (5.3.15), we obtain .



r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx ≤

G(k)





C3

r −n u2 + r 4−n f 2 dx+

G(k−1) ∪G(k) ∪G(k+1)





r 4−n |∇G|2 + r 2−n G 2 dx;

C4 inf G(k−1) ∪G(k) ∪G(k+1) ◦1

here the infimum is taken over all functions G ∈W 4−n (G) such that G = g on ∂G. Summing these inequalities over k = 0, 1, .., [log2 (d/4ε)], for any ε ∈ (0, d/2) we obtain

.



r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx ≤

Gd2ε



C3 r −n u2 + r 4−n f 2 dx + C4 g2◦ 1/2 G2d ε

W 4−n (ε2d )

.

(5.3.28)

5.3 The Global Integral Weighted Estimate

85

Now, from (5.3.27)–(5.3.28) it follows that  .

A(r) r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx ≤

Gε





CA(2ε) M (ε) + f  ◦ 0 2

+ g ◦ 1/2

2

2



5ε/2 W 4−n Gε/2

5ε/2



 

2 2 2 uxx + |∇u| + u + c f 2◦ 0 C1 (d, diam G)

G2d ε



+ g ◦ 1/2 2

W 4−n (G)

Gd

r −n u2 dx+

+δ

W 4−n ε/2

(5.3.29)

.

W 4−n (∂G)

Thus, from (5.3.26) and (5.3.29), we obtain  r

.

2−n

1 |∇u| dx + 2

Gε



2

γ (ω)r 1−n u2 ds ≤ ε

      

ω

ω ∂u   0 0 ε2−n  u d ε  + χ , ω u2 ε, , ω dω +   ∂r 2 2  ε  ∂    A(2ε) M 2 (ε) + f 2◦ 0



5ε/2 W 4−n Gε/2

+ g2◦ 1/2

 C(diam G, χ0 , d) Gd

5ε/2

r −n u2 dx+

+δ

W 4−n ε/2

Gε

u2xx + |∇u|2 + u2 dx+ 



- f 2◦ 0 C

W 4−n (G)

+ g2◦ 1/2

(5.3.30)

W 4−n (∂G)

- > 0 depends on γ0 and not depends on ε. for any δ > 0, where C By Lemma 4.12,  .

Gε

r

1 u dx ≤ λ(λ + n − 2)

−n 2

 r 2−n |∇u|2 dx+ Gdε

 εd

  λχ(ω) + γ (ω) r 1−n u2 ds + d −n u2 dx Gd

86

5 The Linear Oblique Derivative Problem for Elliptic Second Order Equation. . .

or, because of χ(ω) ≤ χ0 , γ (ω) ≥ γ0 > 0 ,  .

r −n u2 dx ≤

Gε

1 λ(λ + n − 2)

 r 2−n |∇u|2 dx+ Gε

   λχ0 1 γ (ω)r 1−n u2 ds + d −n u2 dx. 1+ λ(λ + n − 2) γ0 ε

(5.3.31)

G

By Theorem 5.2 about L2 —estimate for solutions of problem (L) inside the domain and near a smooth portion of the boundary, we have  

2 2 2 . uxx + |∇u| + u dx ≤ c1 f 2 + u2 dx + c2 g2W 1/2 ( ) ≤ d/2 Gd

Gd/2

 c

u2L2 (G)

+ f  ◦ 0 2

W 4−n (G)

 + g ◦ 1/2 2

(5.3.32)

,

W 4−n (∂G)

where constant c > 0 depends only on ν, μ, d, G, max A(|x −y|), χC 1 (∂G) , γ C 1 (∂G) x,y∈G % $ γ0 (λ+n−2) λ(λ+n−2) , from (5.3.30)–(5.3.32) and is independent of u. Choosing δ ≤ min 4(γ0 +λχ0 ) ; 2 it follows   . r 2−n |∇u|2 dx + γ (ω)r 1−n u2 ds ≤ Gε

ε

      

ω

ω ∂u  0 0 2−n  ε , ω u2 ε, , ω dω +  u d ε  + χ   ∂r 2 2  ε  ∂   A(2ε) M 2 (ε) + f 2◦ 0



5ε/2 W 4−n Gε/2

C(diam G, χ0 , d) 

+ g2◦ 1/2

5ε/2

+

W 4−n ε/2



u2xx + |∇u|2 dx+

Gd



-1 |u|20,G + f 2◦ 0 C

W 4−n (G)

+ g ◦ 1/2 2

W 4−n (∂G)

.

(5.3.33)

5.3 The Global Integral Weighted Estimate

87

In virtue of Lemma 5.8, (5.3.31), (5.3.32) as well as u ∈ C 0 (G), we can pass in (5.3.33) to the limit ε → +0, using the Fatou Theorem. As a result we get .

 

r 2−n |∇u|2 + r −n u2 dx + γ (ω)r 1−n u2 ds ≤ G

∂G

-3 C



 |u|20,G

+ f  ◦ 0 2

g ◦ 1/2 2

W 4−n (G)

.

(5.3.34)

.

(5.3.35)

W 4−n (∂G)

Passing in (5.3.28) to the limit ε → +0, by (5.3.34), we get  r 4−n u2xx dx ≤

.

Gd0



 r

C3

−n 2



u dx + C4 f  ◦ 0 2

W 4−n (G)

G

+ g ◦ 1/2 2

W 4−n (∂G)

 

In virtue of (5.3.34), (5.3.35) we have the required estimate (5.3.1).

Theorem 5.9 Let u(x) be a strong solution of problem (L) and assumptions (a)–(c) are ◦ 2

satisfied. Then (u(x) − u(0)) ∈ W 4−n (G) and ⎛ u(x) − u(0) ◦ 2

.

W 4−n (G)



+⎝



⎞1 2

1 γ (ω)|u(x) − u(0)|2 ds ⎠ ≤ r

∂G



C |u|0,G + |u(0)| · a ◦ 0

W 4−n (G)

+ f  ◦ 0

W 4−n (G)

+ g ◦

1 2

,

(5.3.36)

W 4−n (∂G)

where C > 0 depends on ν, μ, diam G, χC 1 (∂G) , γ C 1 (∂G) and on the modulus of continuity of leading coefficients. Proof Setting v(x) = u(x) − u(0) we have v ∈ C 0 (G), v(0) = 0 and v is a strong solution of the problem

.

⎧ ⎨a ij (x)v

xi xj

⎩ ∂v ∂ n

+

+ a i (x)vxi + a(x)v = f (x) − a(x)u(0) ≡ f0 (x),

χ(ω) ∂v ∂r

+

1 |x| γ (x)v

= g(x) −

1 |x| γ (x)u(0)

≡ g0 (x),

x ∈ G, x ∈ ∂G \ O.

Without loss of generality we can suppose u(0) ≥ 0 and then g0 (x) ≤ g(x), since γ (ω) >   0. We repeat verbatim the arguments of the Theorem 5.7 proof.

88

5 The Linear Oblique Derivative Problem for Elliptic Second Order Equation. . .

Corollary 5.10 Let u(x) be a strong solution of problem (L) and satisfy assumptions (a)– (c). Then u(0) = 0. Proof By the Cauchy inequality, we have .

1 |u(0)|2 ≤ |u(x)|2 + |u(x) − u(0)|2 . 2

Thus .

1 |u(0)|2 2



r −n dx ≤

Gd0



r −n |u(x)|2 dx +

Gd0



r −n |u(x) − u(0)|2dx.

(5.3.37)

Gd0

Both integrals from the right side are finite by Theorems 5.7, 5.9. Hence it follows that the right side of inequality (5.3.37) is finite. However, the left side of this inequality is infinite if u(0) = 0, because of  .

r

−n

d dx ∼

dr = ∞. r

0

Gd0

It leads to a contradiction. Therefore must be u(0) = 0.

5.4

 

Local Integral Weighted Estimates

Theorem 5.11 Let u(x) be a strong solution of problem (L) and assumptions (a)–(d) are satisfied with A(r) that is Dini continuous at zero. Then there are d ∈ (0, 1) and a constant C > 0 depending only on ν, μ, d, s, λ, γ0 , χ0 , meas G, diam G, χC 1 (∂G) , γ C 1 (∂G) d

and on the quantity 0

u(x) ◦ 2

.

  W 4−n G0

A(τ ) τ dτ ,

such that for all  ∈ (0, d)

≤

 C |u(x)|0,G + f (x) ◦ 0

W 4−n (G)

where ks is defined by (5.2.1).

+ g(x) ◦ 1/2

W 4−n (∂G)

 + ks ×

⎧ λ ⎪ if s > λ ⎨ , λ ln 1 , if s = λ , ⎪ ⎩ s  , if s < λ

(5.4.1)

5.4 Local Integral Weighted Estimates

89 ◦2

Proof By Theorem 5.7 we have that u(x) ∈W 4−n (G). Let us consider the equation of problem (L) in the form (5.3.2). Multiplying both side of (5.3.2) by r 2−n u(x) and  integrating over the domain G0 , 0 <  < d, we obtain  r 2−n u udx =

. 

G0



$ & '% j r 2−n u f − a ij (x) − δi uxi xj + a i (x)uxi + a(x)u dx.

(5.4.2)



G0

We integrate by parts 

 r 2−n u udx = −

. 

ux i 

G0



G0

 −

 ∂u ∂ 2−n r u dx + r 2−n u ds = ∂xi ∂ n

r 2−n |∇u|2 dx +

 G0

n−2 2



∂G0

r −n xi

 G0

∂u2 dx + ∂xi

 r 2−n u

 ∂G0

∂u ds. ∂ n

(5.4.3)

Again, integrating by parts  .

r −n xi



∂u2 dx = − ∂xi

G0

 u2 

∂  −n  xi r dx + ∂xi

G0



r −n u2 xi cos(

n, xi )ds. 

∂G0

    Because of ∂G0 = 0 ∪  , (5.3.6) and xi cos(

n, xi ) = ,    xi cos(

n, xi )  = 0 we finally obtain 0



r −n xi

.  G0

∂u2 dx = 1−n ∂xi



 u2 d  =



u2 d .

(5.4.4)



By the boundary condition of (L), we have  r 2−n u

. 

∂G0



∂u ds = ∂ n

r 2−n u 

0



 0

∂u ds + ∂ n

 2−n u

∂u d  = ∂r



   ∂u 1 ∂u 2−n − γ (ω)u ds +  u d . r u g − χ(ω) ∂r r ∂r

(5.4.5)

90

5 The Linear Oblique Derivative Problem for Elliptic Second Order Equation. . .

From (5.4.3)–(5.4.5) it follows    ∂u n − 2 2 + u d + u ∂r 2



 r 2−n u udx = − 

r 2−n |∇u|2 dx + 

G0



G0

  ∂u 1 − γ (ω)u ds. r 2−n u g − χ(ω) ∂r r



.



0

Now, hence and from (5.4.2) 

 r

.

2−n

|∇u| dx + 2



χ(ω)r

2−n



G0

∂u u ds + ∂r

0

 γ (ω)r 1−n u2 ds = 

0

     ∂u n − 2 2 + u d + r 2−n ugds − r 2−n uf dx+ u ∂r 2 



 r 2−n u

&



0

G0

'

j a ij (x) − δi uxi xj + a i (x)uxi + a(x)u dx.

(5.4.6)

 G0

Let us estimate the right side terms of (5.4.6): • By assumption (b) and the Cauchy inequality  .

    j r 2−n |u| a ij (x) − δi  |uxi xj |dx ≤ A() r 2−n |u||uxx |dx =





G0

G0

A()

 

n n 1 r 2− 2 |uxx | r − 2 |u| dx ≤ A() r 4−n u2xx + r −n u2 dx. 2 



G0

G0

•  . 

G0

  r 2−n |u||a i (x)| uxi  dx ≤ A()

 r 1−n |u||∇u|dx ≤ 

G0

1 A() 2



r −n u2 + r 2−n |∇u|2 dx. 

G0

5.4 Local Integral Weighted Estimates

91

• 

 r

.

2−n

r −n u2 dx.

|a(x)|u dx ≤ A() 2





G0

G0

•  r

.

2−n



δ |u||f |dx ≤ 2

G0



r −n u2 dx +



G0

1 f 2◦ 0    , ∀δ > 0. 2δ W 4−n G0

•  r 2−n |u||g|ds ≤

. 

0

δ1 2

 r 1−n u2 ds + 

1 2δ1

0

 r 3−n g 2 ds ≤ 

0

δ1 2

 r 1−n u2 ds + 

0

c g2◦ 1/2    , δ1 W 4−n 0

by (5.3.25), for all δ1 > 0. • Similar to (5.3.28)  r 4−n u2xx dx

.  G0

≤ C1



r −n u2 + r 4−n f 2 dx + C2 g2◦ 1/2

2

.

(5.4.7)

W 4−n 0

2 G0

• Further,  χ(ω)r 2−n u

. 

0

1 ∂u ds = ∂r 2

 χ(ω) 

∂u2 drdσ = ∂r

0

1 n−2 ω0 sin sinn−3 ω2 · · · sin ωn−2 2 2

 χ ∂

 ∂u2 r, ω0 , ω  2 ,w drdω = 2 ∂r

ω

0





0

1 2

 χ (ω) u2 (, ω) dσ ∂

in virtue of ωi ∈ (0, π) for i = 2, 3, . . . , n − 2 and χ(ω) ≥ 0, u(0) = 0, where w = (w2 , . . . , wn−1 ).

92

5 The Linear Oblique Derivative Problem for Elliptic Second Order Equation. . .

• Because of γ (ω) ≥ γ0 , 0 ≤ χ(ω) ≤ χ0 , by (4.2.2), we have 

 χ(ω)r 1−n u2 ds =

. 



0

χ(ω) γ (ω)r 1−n u2 ds ≤ γ (ω)

0

χ0 γ0

 γ (ω)r 1−n u2 ds ≤  0

χ0 U (), γ0

(5.4.8)

as well as  r 1−n u2 ds ≤

. 

0

1 γ0

 γ (ω)r 1−n u2 ds ≤ 

1 U (). γ0

0

• Applying inequality (4.2.1), by (4.2.2) and (5.4.8),  .

r −n u2 dx ≤



G0

⎡ ⎢ 1 ⎢ λ(λ + n − 2) ⎣



 r 2−n |∇u|2 dx + 

⎤ ⎥ (χλ + γ ) r 1−n u2 ds ⎥ ⎦=



G0

1 1 -() + U λ(λ + n − 2) λ+n−2



0

-(), χ(ω)r 1−n u2 ds ≤ C(λ, χ0 , γ0 )U



0

1 C(λ, χ0 , γ0 ) = λ(λ + n − 2)



γ0 + λχ0 γ0

 .

(5.4.9)

By Lemma 4.11 and above estimates from inequality (5.4.6) we get .

-() ≤ 1 − (A() + δ) U

   - −1 2 2 -(2) + c1 δ f  ◦ 0 2 + g ◦ 1/2 2 , ∀δ > 0, U () + A()U 2λ W 4−n G0 W 4−n 0 (5.4.10)

5.4 Local Integral Weighted Estimates

93

where constant c1 > 0 depends on γ0 , χ0 , λ. Using assumption (d), inequality (5.4.10) takes the form .

-() ≤ 1 − (A() + δ) U  - -(2) + c2 ks2 δ −1 2s , ∀δ > 0. U () + A()U 2λ

(5.4.11)

By Theorem 5.7 we have 

 -(d) ≤ C |u|20,G + f 2◦ 0 U

.

W 4−n (G)

≡ U0 .

+ g2◦ 1/2

(5.4.12)

W 4−n (∂G)

Equations (5.4.11) and (5.4.12) are the Cauchy problem (CP ) of Theorem 2.61 [33] with P() =

.

2λ 2λ 1 − (A() + δ) , N () = A(),   Q() = 2λc2 ks2 δ −1 2s−1 ,

(5.4.13)

for all δ > 0. The solution to this problem, by Theorem 2.61 [33], is the following ⎛

⎡

-() ≤ ⎣U0 exp ⎝− U

d

⎞ P(ς )dς ⎠ +



⎛ d ⎞  × exp ⎝ B(ς )dς ⎠ ,

.



d

⎛ Q(ς ) exp ⎝−



ς

⎞

⎤

P(σ )dσ ⎠ dς ⎦



⎛ ⎞ 2 ⎜ ⎟ B() = N () exp ⎝ P(σ )dσ ⎠ .

(5.4.14)



There are three possible cases: s > λ, s = λ and s < λ. (1) Case s > λ. P() =

.

Choosing δ = ε , for any ε > 0, we obtain from (5.4.13)

A() 2λ A() − 2λ − 2λε−1 , N () = 2λ , Q() = 2λc2 ks2 2s−1−ε .   

94

5 The Linear Oblique Derivative Problem for Elliptic Second Order Equation. . .

Let us calculate ς .

−

 2λ ς  ς ε − ε A(σ ) + 2λ dσ, ς ∈ (, d); P(σ )dσ = ln + 2λ ς ε σ



⎛ .

exp ⎝−



ς

⎞ P(σ )dσ ⎠ ≤ C1

 2λ  , ς



⎞ ⎛  d A(σ ) 2λ ε d exp ⎝2λ dσ ⎠ C1 = exp ε σ 

0

and ⎛ .

⎜ exp ⎝

2

⎞ ⎟ P(σ )dσ ⎠ ≤ 22λ .



Therefore d

d B(ς )dς ≤ λ2

.



2λ+1

A(ς ) dς = C2 0



and d .



⎛ Q(ς ) exp ⎝−

ς

⎞ P(σ )dσ ⎠ dς ≤ 2λC1 c2 ks2 2λ



d ς 2s−1−ε−2λ dς = 

2λC1 c2 ks2 2λ

d 2(s−λ)−ε − 2(s−λ)−ε ≤ C3 ks2 2λ , 2(s − λ) − ε

if we choose ε = s − λ > 0. From above inequalities, by (5.4.14), we obtain

-() ≤ C -1 U0 + ks2 2λ ; U

.

-1 > 0 depends only on λ, d, s and on here constant C

d 0

A(ς) ς dς .

(5.4.15)

5.4 Local Integral Weighted Estimates

95

(2) Case s = λ. In this case we can choose in (5.4.13), instead of δ > 0, function δ() = 1 1 ,  ∈ (0, d), where e is the Euler number. Then we obtain the Cauchy 2λ ln 

problem (5.4.14), with P() =

.

2λ 1 − (A() + δ() , 

N () = 2λ

A() , 

Q() = 2λc2 kλ2 δ −1 ()2λ−1 . We calculate ς .

− 

 2λ ς d dσ A(σ )  dσ = P(σ )dσ ≤ ln + + 2λ 1 ς σ σ ln σ 

0

 1  2λ d ln   A(σ ) + 2λ dσ, ln + ln 1 ς σ ln ς 0

which yields ⎛ .

exp ⎝−

ς 

⎞ ⎛  2λ ln 1 d A(σ ) ⎠   dσ , ς ∈ (, d) P(σ )dσ ⎠ ≤ exp ⎝2λ ς σ ln ς1 ⎞

0

and 2 .

⎛ P(σ )dσ ≤ ln 22λ + ln ⎝

ln



because ln

1 2

ln

1 2

⎞

⎠ ≤ ln 22λ , 1 

1 

< ln

. Therefore

d

d B(ς )dς ≤ λ2

.



2λ+1 0

A(ς ) dς = C2 . ς

96

5 The Linear Oblique Derivative Problem for Elliptic Second Order Equation. . .

Moreover ⎛

d

Q(ς ) exp ⎝−

.



ς

⎞ P(σ )dσ ⎠ dς +



⎛ 2

4λ

c2 kλ2 2λ ln

1 exp ⎝2λ 

d

⎞ A(σ ) ⎠ dσ σ

d

dς 1 ≤ C4 kλ2 2λ ln2 , ς 



0

⎛ C4 = 4λ2 c2 exp ⎝2λ

d

⎞ A(σ ) ⎠ dσ . σ

0

From above inequalities, by (5.4.14), we obtain

-() ≤ C -2 U0 + kλ2 2λ ln2 1 , U 

.

-2 > 0 depends only on λ, d and on where constant C

d 0

(5.4.16)

A(ς) ς dς .

(3) Case s < λ. We consider (5.4.13) with any δ > 0, then we obtain the Cauchy problem (5.4.14). Let us calculate ς .

−

 2λ(1−δ) d A(σ )  dσ, P(σ )dσ ≤ ln + 2λ ς σ



0

hence ⎛ .

exp ⎝−

ς

⎞ ⎛  2λ(1−δ) d A(σ ) ⎠  dσ , ς ∈ (, d). P(σ )dσ ⎠ ≤ exp ⎝2λ ς σ ⎞

 d

Therefore

0

B(ς )dς ≤ C2 with constant C2 as above in the cases (1)–(2). Moreover



d .



⎛ Q(ς ) exp ⎝−

ς

⎞ P(σ )dσ ⎠ dς ≤



⎛ 2λc2 ks2 δ −1 2λ(1−δ) exp ⎝2λ

d 0

⎞ d  A(σ ) ⎠ dσ ς 2(s−λ+λδ)−1 dς = σ 

5.4 Local Integral Weighted Estimates

⎛ 2λc2 ks2 δ −1 2λ(1−δ) exp ⎝2λ

97

d 0

⎞ A(σ ) d 2s−2λ+2λδ − 2s−2λ+2λδ dσ ⎠ ≤ σ 2s − 2λ + 2λδ C5 ks2 2s ,

if we choose δ =

λ−s 2λ

> 0. From above inequalities, by (5.4.14), we obtain

-() ≤ C -3 U0 + ks2 2s , U

(5.4.17)

.

-3 > 0 depends only on λ, d, s and on where constant C

d 0

A(ς) ς dς .

Finally, taking into account of (5.4.7), (5.4.9), (5.4.12), by (5.4.15)–(5.4.17), we obtain the   required estimate (5.4.1). Theorem 5.12 Let u(x) be a strong solution of problem (L) and assumptions (a)–(d) are satisfied with A(r), being continuous at zero function, but not Dini continuous. Then for all ε > 0 there are d ∈ (0, 1) and a constant Cε > 0 depending only on ν, μ, d, s, λ, ε, γ0 , χ0 , meas G, diam G, χC 1 (∂G) , γ C 1 (∂G) , such that for all  ∈ (0, d) u(x) ◦ 2

.

  W 4−n G0

≤  Cε |u(x)|0,G + f (x) ◦ 0

 + g(x) ◦ 1/2

+ ks ×

λ−ε , if s > λ , s−ε , if s ≤ λ

(5.4.18)

W 4−n(G)



W 4−n (∂G)

where ks is defined by (5.2.1). Proof Similar to Theorem 5.11 we derive the Cauchy problem (5.4.11)–(5.4.13). Then we get ς .

− 

 P(σ )dσ = 2λ(1 − δ) ln + 2λ ς

ς 

A(σ ) dσ, ς ∈ (, d). σ

98

5 The Linear Oblique Derivative Problem for Elliptic Second Order Equation. . .

Now, by properties of A(r) ς .

ς A(σ ) dσ ≤ A(d) ln σ 



and for any δ > 0 we choose d > 0, by continuity of A(r), so that A(d) < δ. Thus we obtain ⎛ ς ⎞  2λ(1−2δ)   ⎝ ⎠ . exp − P(σ )dσ ≤ , ∀δ > 0. ς 

Further we calculate ⎛ ⎞ ⎛ 2ς ⎞ 2 d d  ⎜ ⎟ . exp ⎝ P(σ )dσ ⎠ ≤ 22λ ⇒ B(ς )dς = N (ς ) exp ⎝ P(σ )dσ ⎠ dς ≤ 



d λ22λ+1



ς

⎛ A(ς ) d dς ≤ λ22λ+1 δ ln ⇒ exp ⎝ ς 



d .

⎛ Q(ς ) exp ⎝−



d

⎞ B(ς )dς ⎠ ≤

 −λ22λ+1 δ d

;



ς

⎞ P(σ )dσ ⎠ dς ≤ 2λks2 c2 δ −1 2λ(1−2δ)×



d

ς 2s−2λ(1−2δ)−1dς ≤ 2λks2 c2 δ −1 2λ(1−2δ)



 c4 ks2

d 2s−2λ(1−2δ) − 2s−2λ(1−2δ) ≤ 2s − 2λ(1 − 2δ)

2λ(1−2δ), if s ≥ λ , ∀δ > 0 if s < λ 2s ,

  (in this connection we choose any δ > 0 for s ≥ λ and δ ∈ 0, λ−s 2λ for s < λ). Thus, applying (5.4.14), we obtain

-() ≤ C U0 + ks2 .U



2(λ−ε), if s ≥ λ , ∀ε > 0. 2(s−ε), if s < λ

From above and (5.4.7), (5.4.9) with regard to (5.4.12) we get the required estimation (5.4.18).  

5.4 Local Integral Weighted Estimates

99

Now, we can make a correction of Theorem 5.12 in the case A(r) ∼

1 ln 1r

.

Theorem 5.13 Let u(x) be a strong solution of problem (L) and assumptions (a)–(d) are satisfied with A(r) ∼ 11 . Then there are d ∈ (0, 1) and constants C > 0, cs > 0 ln

r

depending only on ν, μ, d, s, λ, γ0 , χ0 , meas G, diam G, χC 1 (∂G) , γ C 1 (∂G) , such that for all  ∈ (0, d) u(x) ◦ 2

  W 4−n G0

.

≤  C |u(x)|0,G + f (x) ◦ 0

W 4−n(G)

+ g(x) ◦ 1/2

W 4−n (∂G)

   λ  , if s ≥ λ cs (λ) 1 ln · ,  s , if s < λ

 + ks × (5.4.19)

where ks is defined by (5.2.1). Proof Similar to Theorem 5.11 we derive the Cauchy problem (5.4.11)–(5.4.13). In this case we choose δ=

.

1 , 0 <  < d. 2λ ln 1

Then (5.4.13) takes the form P() =

.

  1 + 2λ 2λ 2λ 1 , N () = , Q() = 4λ2 c2 ks2 2s−1 ln − .    ln 1  ln 1

Because of A() ∼ δ(), for suitable small d > 0 we have ς .

− 

⎛ ⎞1+2λ  2λ ln 1  ⎝ ⎠ P(σ )dσ = ln + ln ⇒ ς ln 1 ς

⎛ exp ⎝−

ς 

⎛ ⎞2λ+1  2λ ln 1  ⎝  ⎠ P(σ )dσ ⎠ = . ς ln 1 ⎞

ς

100

5 The Linear Oblique Derivative Problem for Elliptic Second Order Equation. . .

Further ⎛ .

⎜ exp ⎝

2

⎞ ⎟ P(σ )dσ ⎠ ≤ 22λ

⎛ d ⎞    1 ⇒ exp ⎝ B(ς )dς ⎠ ≤ C(d, λ) lnc(λ) 





and d .

⎛ Q(ς ) exp ⎝−



2

4λ

c2 ks2 2λ ln2λ+1

2

4λ

c2 ks2 2λ ln2λ+1

ς

⎞ P(σ )dσ ⎠ dς =



  d 2(s−λ)−1 ς 1 dς ≤ 2λ 1  ln ς    d 1 ς 2(s−λ)−1dς =  

⎧   ⎨ d 2(s−λ) −2(s−λ) , if s = λ, 1

s−λ 4λ2 c2 ks2 2λ ln2λ+1 ≤ d  ⎩ln if s = λ  , ⎧ 2λ ln2λ+1 1 , if s > λ, ⎪  ⎪ ⎪ ⎨  - s2 2λ ln2λ+2 1 , if s = λ, Ck ⎪  ⎪ ⎪ ⎩2s ln2λ+1 1 , if s < λ.  Thus, applying (5.4.14), we obtain ⎧   ⎨ 2λ

 , if s ≥ λ, 1 -() ≤ C U0 + ks2 ln2cs (λ) .U  ⎩2s if s < λ. From above and (5.4.7), (5.4.9) with regard to (5.4.12) we get the required estimation   (5.4.19).

5.5 The Power Modulus of Continuity

5.5

101

The Power Modulus of Continuity

Proof of Theorem 5.4 To prove Theorem 5.4 we introduce the function ⎧ λ ⎪ s>λ ⎨ , 1 λ  ln  , s = λ .ψ() = ⎪ ⎩ s  , s 0. Making the following transformation: x = x  , u(x  ) = ψ()w(x  ), we get that the function w(x  ) satisfies the problem  .

2  ψ() f (x ), 2 x  ∈ 1/4

a ij (x  )wxi xj + a i (x  )wxi + 2 a(x  )w = ∂w ∂ n 

+

1 |x  | γ (ω)w

+ χ(ω) ∂w ∂r  =

  ψ() g(x ),

x  ∈ G21/4

.

Based on the local maximum principle (see Proposition 5.1), we have ⎡    12 ⎢       2  sup g(x  ) . sup w(x ) ≤ C ⎢ w dx + ⎣ ψ() G2 G1 1/2

1/4

G21/4

+

2

 

ψ()

  f (x  )n dx 

 n1

⎤ ⎥ ⎥, ⎦

(5.5.1)

G21/4

where constant C > 0 depends only on ν, μ, γ0 , χ0 , n, p, G, ω0 , max γ (ω), a n 2 , L G1/4 ω∈∂G > 1>  >  > > n i 2 2> |a | . > > > i=1 > p 2 L

G1/2

Now, we return to variable x and to function u(x). Then  .

G21/4

w2 dx  =

1 2 ψ ()

 G21/4

u2 (x  )dx  ≤

2n ψ 2 ()

 2

G/4

r −n u2 dx.

102

5 The Linear Oblique Derivative Problem for Elliptic Second Order Equation. . .

Hence, by Theorem 5.11, we obtain 

w2 dx  ≤

.

G21/4

 C |u(x)|0,G + f (x) ◦ 0

2

W 4−n (G)

+ g(x) ◦ 1/2

W 4−n (∂G)

+ ks

,

(5.5.2)

where ks is defined by (5.2.1). By assumption (d),

.

  sup |g(x)| ≤ g1 s−1 ψ() G2 ψ() /4

⎧ s−λ < 1, s > λ ⎪ ⎪ ⎨ 1 = g1 ln 1 < 1, s = λ ⇒  ⎪ ⎪ ⎩ 1, s λ ⎪ ⎪  n1   ⎨ 2 1 n < 1, s = λ |f (x)| dx ⇒ .f1 ≤ f-1 . ln 1 ⎪ ψ() ⎪ ⎩ 1, 2 s λ, s−ε , if s ≤ λ,

and applying Theorem 5.12 we prove Theorem 5.5.

 

Proof of Theorem 5.6 Repeating verbatim the proof of Theorem 5.4 with ψ() = ln

.

cs (λ)

   λ 1  , if s ≥ λ, ·  s , if s < λ,

and applying Theorem 5.13 we prove Theorem 5.6.

5.6

 

Examples

Let the domain .G ⊂ R3 lies inside the cone $ ω π % 0 ⊂ 0, , ω2 ∈ (0, 2π) , .G0 = (r, ω1 , ω2 ) : r > 0, ω1 ∈ 0, 2 2 O ∈ ∂G and in a some neighborhood of .O the boundary .∂G coincides with the lateral surface of the cone .G0 . Denote

.

$ % ω0 , ω2 ∈ (0, 2π); ω0 ∈ (0, π) . 0 = (r, ω1 , ω2 ) : r > 0, ω1 = 2

.

Let .χ ≥ 0, .γ > 0 be constants. We present two examples which demonstrate that the coefficients of operator .L assumptions are essential for validity of Theorems 5.4 and 5.6. Example 5.14 The function 1 u(r, ω1 , ω2 ) = r λ ln ψ(ω1 ), r

.

where .λ > 0 and .ψ(ω1 ) are defined by (3.1.12), (3.1.10), is the solution of the following problem ⎧ ⎨ u = −(2λ + 1)r λ−2 ψ(ω1 ), x ∈ G0

    . ∂u ∂u 1 λ−1 ψ ω0 . + χ + γ u = −χr  ⎩ ∂ n

ω0 ∂r r 2 ω1 =

2

104

5 The Linear Oblique Derivative Problem for Elliptic Second Order Equation. . .

Hence we have



f (x) = O |x|λ−2 , g(x) = O |x|λ−1 ,

.

therefore in this case .s = λ. Thus, this example confirms the validity (5.2.2) of Theorem 5.4 for .s = λ. Example 5.15 The function 1 u(r, ω1 , ω2 ) = r λ ln ψ(ω1 ), r

.

where .λ > 0 and .ψ(ω1 ) are defined by (3.1.12), (3.1.10), is the solution of problem ⎧ τ −2 ⎪ ⎨ r τ u + (2λ+1)r u = 0, x ∈ G0 , ln 1r

 .    ∂u ∂u 1 λ−1 ψ ω0 . ⎪ ⎩ ∂ n + χ ∂r + r γ u  ω0 = −χr 2 ω1 =

2

In this case we have A(r) =

.

2λ + 1 ln

1 r

1 ⇒

A(r) dr = +∞. r

0

Therefore the assumption of Dini-continuity is not satisfied. Thus, this example confirms the validity (5.2.4) of Theorem 5.6 for .s = λ.

5.7

Notes

The systematic theory of oblique derivative problems for linear second order elliptic equations in smooth domains was developed by G. Lieberman [89] (Chapters 1–6). There, classical Schauder theory is exhaustively studied (that is, estimates are obtained and solvability in space .C 2+β (G) is proved), and the theory of strong solutions from 2,n 0 .W loc (G) ∩ C (G) is also investigated. Ibid there is also an extensive bibliography with a thorough discussion of works related to the problem considered. Let us take a look at some other work. In [55] a priori estimates and strong solvability results in Sobolev space .W 2,p (G), p ∈ (1, ∞) are proved for the regular oblique derivative problem for linear uniformly elliptic equations with discontinuous coefficients: ⎧ n  ij ⎪ ⎨ a (x)uxi xj = f (x) a.e. G ⊂ Rn , n ≥ 3, .

⎪ ⎩

i,j =1 ∂u → + γ (x)u − ∂ l

= g(x)

on ∂G ∈ C 1, 1,

5.7 Notes

105

→ − where . a ij ∈ VMO ∩ L∞ ; it is assumed that unit vector field . l (x) is nowhere tangential to the boundary . ∂G, i.e., the oblique derivative problem is regular (see also [108]). In [113] is studied the oblique derivative problems for linear second order elliptic equations on conical domains under the assumption of axi-symmetry of the solutions. In this work are found sufficient conditions on the angle of the oblique vector for Hölder regularity of the gradient of such solutions to hold up to the vertex of the cone. Let us call more attention to the work [105]. It investigates the smoothness in the Hölder spaces of solutions to the linear oblique derivative problem.

The Oblique Derivative Problem for Elliptic Second Order Semi-linear Equations in a Domain with a Conical Boundary Point

6.1

Setting of the Problem

Let .G ⊂ Rn be a bounded domain with the boundary .∂G that is a smooth surface everywhere except at the origin .O ∈ ∂G and near .O it is a conical surface. We consider the semi-linear problem ⎧ ij i ⎪ ⎪ ⎨ a (x)uxi xj + a (x)uxi + a(x)u(x) = h(u) + f (x), x ∈ G, .

⎪ ⎪ ⎩

h(u) = a0 (x)u(x)|u(x)|q−1,

∂u ∂ n

+ χ(ω) ∂u ∂r

+

1 |x| γ (ω)u(x)

q ∈ (0, 1),

= g(x),

(SL) x ∈ ∂G\O.

where .n denotes the unite exterior normal vector to .∂G\O; .(r, ω) are spherical coordinates in .Rn with the pole .O. Remark 6.1 For .q = 1, problem (SL) takes the form of linear problem .(L) with .a(x) −→ a(x) − a0 (x). Definition 6.2 A function u is called a strong solution of problem .(SL) provided that for 2,n (G) ∩ W 2 (G! ) ∩ C 0 (G) and satisfies the equation of .(SL) any .! > 0 function .u ∈ Wloc for almost all .x ∈ G! as well as the boundary condition in the sense of traces on .! .

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Borsuk, Oblique Derivative Problems for Elliptic Equations in Conical Domains, Frontiers in Elliptic and Parabolic Problems, https://doi.org/10.1007/978-3-031-28381-9_6

107

6

108

6 The Oblique Derivative Problem for Elliptic Second Order Semi-linear. . .

Regarding the problem we assume that the following conditions are satisfied: (A1) the uniform ellipticity condition n 

ν|ξ |2 ≤

a ij (x)ξi ξj ≤ μ|ξ |2 ,

.

∀ξ ∈ Rn , x ∈ G

(6.1.1)

i,j =1 j

with the ellipticity constants .ν, μ > 0; .a ij = a j i ; .a ij (0) = δi . ˚ 0 (G), .a(x) ≤ 0, .a0 ∈ (A2) .a ij ∈ C 0 (G), .a i ∈ Lp (G), .p > n; .a ∈ Ln (G) ∩ W 4−n n

L 1−q (G) ∩ V 02

4 1−q , 1−q −n

(G); there exists a monotonically increasing nonnegative

continuous function .A, .A(0) = 0, such that for .x, y ∈ G n  .

1/2

j

|a ij (x) − δi |2

+ |x|

i,j =1

n 

|a i (x)|2

1/2

+ |x|2 |a(x)| ≤ A(|x|).

(6.1.2)

i=1

(A3) .γ (ω), χ(ω) ∈ C 1 ( ) and there exist numbers .χ0 ≥ 0, .γ0 > 0, such that .γ (ω) ≥ γ0 > 0, .χ0 ≥ χ(ω) > 0. 1 ˚ 0 (G), .g ∈ V 1− n (∂G) ∩ W ˚ 1/2 (∂G), there exist nonnegative (A4) .f ∈ Ln (G) ∩ W n,0 4−n 4−n numbers .f1 , g0 , g1 , a1 and .s > 0, such that |f (x)| ≤ f1 |x|s−2,

|g(x)| ≤ g0 |x|s−1, 1

a0  1−q 2

.

(A5) .a0 ∈ V 0n

2qn 1−q , 1−q

6.2



4 1−q , 1−q −n(G0 )

|∇g| ≤ g1 |x|s−2, ≤ a1 s ,  ∈ (0, 1);

(6.1.3)

(G).

Main Results

Our main results are the following statements about power modulus of continuity, whose proofs are Sect. 6.5. Theorem 6.3 Let u be a strong solution of problem (SL) and .λ > 1 be the smallest positive eigenvalue of problem .(EVP). Suppose that assumptions (A1)–(A5) with .A(r) being Dini continuous at zero are satisfied. Suppose in addition that there exists a nonnegative constant .k0 , such that a0 

.

n

2

L 1−q (G/4 )

≤ k0 1−2q ψ q (),

(6.2.1)

6.2 Main Results

109

where ⎧ λ ⎪ ⎪ ⎨ .ψ() = λ ln 1 ⎪ ⎪ ⎩ s 

if s > λ, if s = λ,

(6.2.2)

if s < λ,

with .0 <  < b. Then there are positive constants .d ∈ (0, b) and .c1 , .c2 , which depend only on .ν, .μ, s, b, .λ, .γ0 , .χ0 , .k0 , .f1 , .g0 , .g1 , .a1 , .γ C 1 (∂ ) , .χC 1 (∂ ) , .diam G, .meas G, on 1

the modulus of continuity of the leading coefficients and on the quantity . Ar(r) dr, and do 0

not depend on u, such that for all .x ∈ Gd0 : • If .0 < q < 1 −

2 λ

and .λ > s, then 2

|u(x)| ≤ c1 |x| 1−q .

(6.2.3)

.

• If .1 −

2 λ

≤ q ≤ 1, then ⎧ λ ⎪ ⎪ ⎨|x| 1 .|u(x)| ≤ c2 |x|λ ln |x| ⎪ ⎪ ⎩ s |x|

if λ < s, if λ = s,

(6.2.4)

if λ > s.

Theorem 6.4 Let u be a strong solution of problem (SL) and .λ > 1 be the smallest positive eigenvalue of problem .(EVP). Suppose that assumptions (A1)–(A5) are satisfied with .A(r), which is a continuous at zero function, but not Dini continuous. Suppose in addition that there exists a nonnegative constant .k0 , such that (6.2.1) holds with ψ() =

.

⎧ ⎨λ−!

if s > λ,

⎩s−!

if s ≤ λ,

(6.2.5)

and .0 <  < b. Then, for any .! > 0, there are positive constants .d ∈ (0, b) and .c1 , c2 , which depend only on .ν, .μ, s, b, .λ, .γ0 , .χ0 , .k0 , .f1 , .g0 , .g1 , .a1 , .γ C 1 (∂ ) , .χC 1 (∂ ) , .diam G, .meas G, on the modulus of continuity of the leading coefficients and do not depend on u, such that for all .x ∈ Gd0 .

• If .0 < q < 1 −

2 λ

and .λ > s, then 2

|u(x)| ≤ c1 |x| 1−q

.

+!

.

110

6 The Oblique Derivative Problem for Elliptic Second Order Semi-linear. . .

• If .1 −

2 λ

≤ q ≤ 1, then

|u(x)| ≤ c2

.

⎧ ⎨|x|λ−!

if λ ≤ s,

⎩|x|s−!

if λ > s.

Theorem 6.5 Let u be a strong solution of problem (SL) and .λ > 1 be the smallest positive eigenvalue of problem .(EVP). Suppose that assumptions (A1)–(A5) are satisfied with .A(r) ∼ 11 . Suppose in addition that there exists a nonnegative constant .k0 , such ln

r

that (6.2.1) holds with ⎧  ⎨ λ  cs (λ) 1 .ψ() = ln  ⎩ s

if s > λ, if s ≤ λ,

(6.2.6)

and .0 <  < b. Then there are positive constants .d ∈ (0, b) and .c1 , .c2 , .cs , which depend only on .ν, .μ, s, b, .λ, .γ0 , .χ0 , .k0 , .f1 , .g0 , .g1 , .a1 , .γ C 1 (∂ ) , .χC 1 (∂ ) , .diam G, .meas G, on the modulus of continuity of the leading coefficients and do not depend on u, such that for all .x ∈ Gd0 • If .0 < q < 1 −

2 λ

and .λ > s, then  |u(x)| ≤ c1 ln

cs (λ)

.

• If .1 −

2 λ

≤ q ≤ 1, then  |u(x)| ≤ c2 lncs (λ)

.

6.3

 2 1 |x| 1−q . |x|

⎧ ⎨

|x|λ 1 |x| ⎩|x|s

if λ ≤ s, if λ > s.

Global Integral Weighted Estimate

Let us introduce the function M(!) = max |u(x)|

(6.3.1)

lim M(!) = |u(0)|.

(6.3.2)

.

x∈ !

then .

!→0+

6.3 Global Integral Weighted Estimate

111

Lemma 6.6 Let u be a strong solution of problem (SL) and assumptions (A1)–(A3) be satisfied. Then there exists a positive constant .c0 depending only on .ν, .μ, G, .maxx,y∈G A(|x − y|), .χC 1 (∂ ) , .γ C 1 (∂ ) , such that .

  ∂u    lim ! 2−n  u d !  ≤ c0 |u(0)|2 . ∂r !→0+

(6.3.3)

! 2! Proof Let us consider the set .G2! ! , . ! ⊂ ∂G! . By Theorem 2.27 we have



 .

|w|d ! ≤ c

!

(|w| + |∇w|) dx,

G2! !

where c depends only on the domain G. Setting .w = u ∂u ∂r and using the Cauchy inequality, we obtain  

 ∂u   r 2 u2xx + |∇u|2 + r −2 u2 dx. . u d ! ≤ c (6.3.4) ∂r !

G2! !

  Let us consider two sets .G!/2 and .G2! ! ⊂ G!/2 , and a new variable .x defined by .x = !x . 5!/2

5!/2

Thus, the function .w(x  ) = u(!x  ) satisfies in .G1/2 the problem 5/2

a ij (!x  )wxi xj + !a i (!x  )wxi + ! 2 a(!x  )w(x  ) = ! 2 h(w(x  )) + ! 2 f (!x  ),

.

x  ∈ G1/2 , 5/2

h(w(x  )) = a0 (!x  )w(x  )|w(x  )|q−1 ,

q ∈ (0, 1),

∂w ∂w 1 + χ(ω)  +  γ (ω)w(x  ) = !g(!x  ),  ∂ n

∂r |x |

x  ∈ 1/2 . 5/2

(6.3.5)

We apply .L2 -estimate for the solution of problem (6.3.5) (see Theorem 5.2). As a result we obtain the following estimation 

. wx2 x  + |∇  w|2 + w2 dx  ≤ G21



c1 5/2

G1/2

[! 4 f 2 + ! 4 h2 (w) + w2 ]dx  + c2 ! 2 inf

 5/2

G1/2

|∇  G|2 + |G|2 dx  ,

112

6 The Oblique Derivative Problem for Elliptic Second Order Semi-linear. . .

  5/2  here infimum is taken over all .G ∈ W 1 G1/2 , such that .G  5/2 = g, and constants .c1 and 1/2

c2 are positive and depend only on . max A(|x − y|), .ν, .μ, G, .χC 1 ( 5/2 ) , .γ C 1 ( 5/2 ) .

.

5/2 x,y∈G1/2

1/2

1/2

Returning to the variable x we obtain 

r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx ≤ G2! !





c3 .

 r 4−n f 2 + r 4−n h2 + r −n u2 dx+

(6.3.6)

5!/2

G!/2





c4 inf

r 4−n |∇G|2 + r 2−n |G|2 dx.

5!/2

G!/2

Now, by the mean value theorem with regard to .u ∈ C 0 (G) and (6.3.1), we obtain  r

5!/2 

−n 2

r −1

u dx =

5!/2

u2 (r, ω)d dr ≤

!/2

G . !/2



2!(θ1!)−1



(6.3.7) u2 (θ1 !, ω)d ≤ 2θ1−1 M 2 (θ1 !) meas



for some . 12 < θ1 < 52 . By (6.3.4), (6.3.6) and (6.3.7), it follows that !



2−n 

 u

 ∂u d !  ≤ ∂r

!



.

c5 M 2 (θ1 !) + c6

(6.3.8) (r 4−n h2 + r 4−n f 2 )dx + c7 g2˚1/2

5!/2

W4−n (!/2 )

.

5!/2 G!/2

1 Taking into account that .q < 1, by the Young inequality with . q1 and . 1−q , we deduce



r 4−n h2 = r 4−n a02 |u|2q = r −nq |u|2q r 4−n+nq a02 ≤ .

δr

−n 2

u + c(δ, q)r

4 1−q −n

2 1−q

a0

(6.3.9) ,

∀δ > 0.

6.3 Global Integral Weighted Estimate

113

By assumptions about functions f , g, and .a0 , from (6.3.7) and (6.3.9), we obtain that $ 

&

lim

.

' r 4−n h2 + r 4−n f 2 dx + g2˚1/2

%

5!/2 W4−n (!/2 )

!→0+

=0

(6.3.10)

5!/2

G!/2

 

and therefore, by (6.3.2), (6.3.8), and (6.3.10), we finally obtain (6.3.3).

Theorem 6.7 Let u be a strong solution of problem (SL) and assumptions (A1)–(A3) be ˚ 2 (G) and satisfied. Then .u ∈ W 4−n uW ˚2

.

4−n

(G) +



a0 r 2−n |u|1+q dx

1/2

+



G

r 1−n γ (ω)u2 ds

1/2

≤

∂G

(6.3.11) 1 C |u|2,G + a0  1−q 0 V

2 , 4 −n (G) 1−q 1−q

+ f W ˚0

4−n (G)

+ gW ˚1/2 (∂G) , 4−n

where .C > 0 depends on .ν, .μ, b, q, n, .meas G, .χC 1 (∂ ) , .γ C 1 (∂ ) and on modulus of continuity of the leading coefficients. Proof Let us rewrite the equation of (SL) in the form

u = h(u) + f (x) −

.

& ' j a ij (x) − δi uxi xj + a i (x)uxi + a(x)u(x) ,

(6.3.12)

multiply both sides by .r 2−n u(x) and integrate over .G! . As a result we obtain 

 r

2−n

r 2−n u (h + f ) dx−

u udx =

G!

G!



.

( ) j r 2−n u a ij − δi uxi xj + a i uxi + au dx.

(6.3.13)

G!

Calculating the integral from the left side by parts, using the boundary condition, the representation .∂G! = ! ∪ ! , .d ! = ! n−1 d , and the fact that   xi cos(

n, xi )

.

!

= !,

  xi cos (

n, xi ) 

0d

= 0,

114

6 The Oblique Derivative Problem for Elliptic Second Order Semi-linear. . .

(see Lemma 2.10), we obtain 

 r 2−n |∇u|2 dx +

γ (ω)r 1−n u2 ds +

n−2 2

!

G!



 u2 d +

r 2−n a0 (x)|u|q+1 dx = G!



 r 2−n ug ds −

.

!

n−2 2



χ(ω)r 2−n u

∂u ds − ! 2−n ∂r

!

r −n u2 xi cos(

n, xi )ds −

d



 u

∂u d ! + ∂r

(6.3.14)

!



r 2−n uf dx+ G!

( ) j r 2−n u a ij − δi uxi xj + a i uxi + au dx

G!

(see formulas (5.3.4)–(5.3.7). By estimations (5.3.17)–(5.3.24), taking .δ1 = 1, identity (6.3.14) takes the form 

 .

G!

r

2−n

|∇u| dx + 2

γ (ω)r 1−n u2 ds+ !



r 2−n a0 (x)|u|q+1 dx ≤ G!

1 2

 χ

ω

ω

0 , ω u2 !, , ω dω + 2 2 0

∂

   ∂u 

  ! 2−n  u d !  + 2 A(r) r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx+ ∂r !

δ 2

G!

 r G!

−n 2

u dx + C(χ0 , d)



u2xx + |∇u|2 + u2 dx+

Gd

1 c f 2W g2˚1/2 , 0 (G) + ˚ W4−n (∂G) 2δ γ0 4−n

(6.3.15)

for all .δ > 0. By assumption (6.1.2), function .A(r) is continuous at zero and .A(0) = 0. Therefore for all .δ > 0 there exists .d > 0 such that .A(r) < δ for all .0 < r < d < b.

6.3 Global Integral Weighted Estimate

115

Hence, by (6.3.6), (6.3.7), and (6.3.9), assuming that .2! < d, we obtain for all .δ > 0 

A(r) r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx =

.

G!



A(r) r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx+

G2! !



A(r) r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx+

Gd2!



A(r) r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx =

Gd

 CA(2!) M 2 (!) +

4

r 1−q

−n

2

a01−q (x)dx + f 2˚0

5!/2

W4−n (G!/2 )

+

5!/2

G!/2

g2˚1/2

5!/2

W4−n !/2

+δ

 C1 (d, diam G)



r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx+

Gd2!

u2xx + |∇u|2 + u2 dx.

Gd

Substituting .! = 2−k−1 d to (6.3.6), we obtain .



r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx ≤

G(k)





C2

r −n u2 + r 4−n f 2 + r 4−n h2 dx+

G(k−1) ∪G(k) ∪G(k+1)



C3 inf G(k−1) ∪G(k) ∪G(k+1)



r 4−n |∇G|2 + r 2−n |G|2 dx,

116

6 The Oblique Derivative Problem for Elliptic Second Order Semi-linear. . .

 ˚ 1 (G) such that .G  here infimum is taken over all .G ∈ W = g. Summing these 4−n  d  ∂G inequalities over .k = 0, 1, . . . , $log2 4! %, for any .! ∈ 0, d2 , we obtain 

r 4−n u2xx + r 2−n |∇u|2 + r −n u2 dx ≤ .

Gd2!



C2 r −n u2 + r 4−n f 2 + r 4−n h2 dx + C3 g2˚1/2

(6.3.16)

W4−n (!2d )

.

G2d !

By (6.3.15) and (6.3.16), we have 

 r

.

2−n

|∇u| dx + 2

G!



r 1−n γ (ω)u2 ds+ !

 r 2−n a0 (x)|u|q+1 dx ≤ ! 2−n 

G!

 δ

 u

 ∂u d !  + ∂r

!

 χ

ω

0 , ω u2 !, , ω dω + 2 2

ω

0

∂

 2 4 −n r 1−q a01−q (x)dx + A(2!) M 2 (!) +

G!

4

r 1−q

−n

2

a01−q (x)dx+

5!/2

G!/2

f 2˚0



5!/2 W4−n G!/2

+ g2˚1/2

5!/2

W4−n !/2





+δ

r −n u2 dx+

G!



u2xx + |∇u|2 + u2 dx+ C(χ0 , d, diam G) Gd

C4 f 2W ˚0

4−n

+ g2˚1/2 (G)

W4−n (∂G)

,

∀ δ > 0,

(6.3.17)

where .C4 > 0 does not depend on .!. By Lemma 4.12, taking into account that .χ(ω) ≤ χ0 and .0 < γ0 ≤ γ (ω), we find  r G! .

1 u dx ≤ λ(λ + n − 2)

−n 2

 r 2−n |∇u|2 dx+ G!

  λχ0 1 1+ r 1−n γ (ω)u2 ds + d −n u2 dx. λ(λ + n − 2) γ0 !

G

(6.3.18)

6.3 Global Integral Weighted Estimate

117

We apply .L2 -estimate for the solution of problem (SL) (see Theorem 5.2). As a result we obtain the estimate 

u2xx + |∇u|2 + u2 dx ≤ Gd

.



f 2 + h2 + u2 dx + C6 g2W 1/2 (

C5

d/2 )

≤

(6.3.19)

Gd/2



4

C7

r 1−q

−n

2

a01−q dx + |u|22,G + f 2W ˚0

4−n (G)

+ g2˚1/2

W4−n (∂G)

,

Gd/2

where the constant .C7 > 0 depends only on .ν, .μ, d, G, . max A(|x − y|), .χC 1 (∂G) , x,y∈G

γ C 1 (∂G) . Now, let us choose

.

δ ≤ min

.

 γ0 (λ + n − 2) λ(λ + n − 2)  , . 4(γ0 + λχ0 ) 2

Thus, by (6.3.17)–(6.3.19), we obtain 





r 2−n |∇u|2 dx + G!

!



 ! 2−n 

r 1−n γ (ω)u2 ds +

u

 ∂u d !  + ∂r

!



.

C8 

r 2−n a0 |u|q+1 dx ≤ G!

 χ

ω

0 0 , ω u2 !, , ω dω + 2 2

ω

∂ 4

r 1−q

−n

2 1−q

a0

(x)dx + A(2!) M 2 (!)+

(6.3.20)

G 4

r 1−q

−n

2

a01−q dx + f 2˚0



5!/2 W4−n G!/2

5!/2

+ g2˚1/2

G!/2

C9 |u|22,G + f 2W ˚0

4−n

+ g2˚1/2 (G)

W4−n (∂G)

5!/2 W4−n !/2

.



+

118

6 The Oblique Derivative Problem for Elliptic Second Order Semi-linear. . .

By Lemma 6.6, (6.3.18), as well as .u ∈ C 0 (G), using the Fatou lemma, we can pass in (6.3.20) to the limit .! → +0. As a result we obtain .

  

r 2−n |∇u|2 + r −n u2 dx + r 1−n γ (ω)u2 ds + r 2−n a0 |u|q+1 dx ≤ G

∂G

G

(6.3.21)  C8

4

r 1−q

−n

2 a01−q dx + C10 |u|22,G + f 2W ˚0

4−n

+ g2˚1/2 (G)

W4−n (∂G)

.

G

Passing to the limit .! → +0 in (6.3.16) and taking into account (6.3.21), we obtain 

 r 4−n u2xx dx

≤ C11

r



−n 2

u dx + C2

G

d .G0

4

r 1−q

−n

2

a01−q dx+

G

C2 f 2W ˚0

4−n (G)

C3 g2˚1/2

W4−n (∂G)

(6.3.22) .

Finally, by (6.3.21) and (6.3.22), we obtain the required estimation (6.3.11).

 

Theorem 6.8 Let u be a strong solution of problem (SL) and assumptions (A1)–(A3) be ˚ 2 (G) and satisfied. Then .(u(x) − u(0)) ∈ W 4−n u(x) − u(0)W ˚2

4−n

 .

(G) +



a0 r 2−n |u(x) − u(0)|1+q dx

G

r 1−n γ (ω)|u(x) − u(0)|2 ds

1/2

1/2

+

≤ (6.3.23)

∂G



C |u(0)| · aW ˚0

4−n (G)

f W ˚0

4−n

1 1−p V 02 (G) 4 1−q , 1−q −n

+ |u|2,G + a0 

+ g 1/2 ˚ (∂G) , (G) W

+

4−n

where .C > 0 depends on .ν, .μ, b, q, n, .diam G, .χC 1 (∂ ) , .γ C 1 (∂ ) and on modulus of continuity of the leading coefficients.

6.4 Local Integral Weighted Estimates

119

Proof Setting .v(x) = u(x) − u(0), we have .v(x) ∈ C 0 (G), .v(0) = 0 and v is a strong solution of the problem ⎧ ij i ⎪ x ∈ G, ⎪ ⎪ a (x)vxi xj + a (x)vxi + a(x)v(x) = h(v) + f0 (x), ⎪ ⎪ ⎪ q−1 ⎪ h(v) = a0 (x) (v + u(0)) |v + u(0)| , q ∈ (0, 1), ⎪ ⎨ .

f0 (x) = f (x) − a(x)u(0); ⎪ ⎪ ⎪ ∂v ∂v 1 ⎪ ⎪ ∂ n + χ(ω) ∂r + |x| γ (ω)v(x) = g0 (x), ⎪ ⎪ ⎪ ⎩ g (x) = g(x) − 1 γ (ω)u(0). 0

x ∈ ∂G\O,

|x|

Without loss of generality we can suppose that .u(0) ≥ 0. Then .g0 (x) ≤ g(x), since ˚ 0 (G). Proceeding step by step the arguments of the γ (ω) > 0. We have that .f0 (x) ∈ W 4−n proof of Theorem 6.7 for the function v we obtain the required estimation (6.3.23).  

.

Corollary 6.9 Let u be a strong solution of problem (SL) and assumptions (A1)–(A3) be satisfied. Then .u(0) = 0. Proof By the Cauchy inequality we have .

1 |u(0)|2 ≤ |u(x)|2 + |u(x) − u(0)|2 . 2

Thus .

1 |u(0)|2 2



Gd0

r −n dx ≤



r −n |u(x)|2 dx +

Gd0



r −n |u(x) − u(0)|2dx.

(6.3.24)

Gd0

By Theorems 6.7 and 6.8 both integrals from the right side are finite, i.e., the right side of (6.3.24) is finite. On the other hand, because of .

r −n dx ∼

Gd0

d 0

dr r

= ∞, the left side of this

 inequality is infinite if .u(0) = 0. It leads to a contradiction. Thus, it must be .u(0) = 0. 

6.4

Local Integral Weighted Estimates

Theorem 6.10 Let u be a strong solution of problem (SL) and assumptions (A1)–(A4) be satisfied with A(r) that is Dini continuous at zero. Then there are d ∈ (0, b) and a constant c > 0 depending only on ν, μ, b, s, λ, q, γ0 , χ0 , meas G, diam G, χC 1 (∂ ) ,

120

6 The Oblique Derivative Problem for Elliptic Second Order Semi-linear. . .

γ C 1 (∂ ) and on the quantity

d 0

uW ˚2



4−n (G0 )

A(τ ) τ dτ ,

such that for all  ∈ (0, d),

≤ c uL2 (G) + a0 V 0

(G) 4 2 1−q , 1−q −n

⎧ ⎪λ ⎨

⎪ gW + k 1/2 λ ln 1 s ˚ (∂G) ⎪ 4−n ⎪ ⎩ s 

.

+ f W ˚0

4−n (G)

+

if s > λ,

(6.4.1)

if s = λ, if s < λ,

where : ks =

.

g02 + a12 +

1 2 (f + g12 ). 2s 1

(6.4.2)

˚ 2 (G). Let us now introduce the function Proof By Theorem 6.7 we have that u ∈ W 4−n -() = U



 r 2−n |∇u|2 dx +

.



r 1−n γ (ω)u2 ds,

0 <  < d < b.



G0

0



Multiplying both sides of (6.3.12) by r 2−n u(x) and integrating it over the domain G0 , 0 <  < d, we obtain 

 r 2−n u udx = 

r 2−n u(h + f )dx− 

G0

G0



.

(6.4.3) j

r 2−n u[(a ij − δi )uxi xj + a i uxi + au]dx. 

G0

Calculating the integral from the left side by parts, similarly to (6.3.14), we obtain 

 r 2−n |∇u|2 dx + 



G0

r 2−n uh dx = 

r 

 u 



2−n

uf dx +

 γ (ω)r 1−n u2 ds+ 

0

∂u n − 2 2 + u d + ∂r 2



G0

G0

∂u ds + ∂r

0



.

χ(ω)r 2−n u

 r 2−n ug ds− 

0 j

r 2−n u[(a ij − δi )uxi xj + a i uxi + au]dx. 

G0

(6.4.4)

6.4 Local Integral Weighted Estimates

121

Similar to (6.3.16), we have  .



r −n u2 + r 4−n h2 + r 4−n f 2 dx + C2 g2˚1/2

r 4−n u2xx dx ≤ C1

 G0

2

W4−n (0 )

.

2

G0

Repeating verbatim estimations of formulas (5.4.6)–(5.4.9), identity (6.4.4) together with the above inequality, takes the form -() + .[1 − (A() + δ)]U

 r 2−n a0 |u|1+q dx ≤ 

G0

 - -(2) + c1 δ −1 f 2 2 + U () + A()U ˚0 G W 2λ 0 4−n 2

a0  1−q 0



2 V 2 G0 4 1−q , 1−q −n

+ g2˚1/2

2 W4−n (0 )

,

∀δ > 0,

where constant c1 > 0 depends on γ0 , χ0 , λ. We used inequalities (6.3.9) for δ = 1 and (6.3.16) for ! = 0. Now, using assumption (A4), we finally obtain -() ≤ [1 − (A() + δ)]U

.

 - -(2) + c2 ks2 δ −1 2s , U () + A()U 2λ

∀δ > 0.

(6.4.5)

Moreover, by Theorem 6.7, we have the initial condition -(d) ≤ C u2 2 + f 2W U ˚0 L (G)

4−n

.

g2˚1/2 W4−n (∂G)

2

+ a0  1−q 0 (G) V

(G) 4 2 1−q , 1−q −n

+ (6.4.6)

≡ U0 .

The differential inequality (6.4.5) with the initial condition (6.4.6) is the Cauchy problem (CP) of Theorem 2.61. By the statement of this Theorem (see inequality (2.11.1)), we derive ⎧ ⎪2λ ⎨

⎪ -() ≤ c U0 + ks2 .U 2λ ln 1 ⎪ ⎪ ⎩ 2s 

d

here constant c > 0 depends only on λ, d, s and on 0

if s > λ, if s = λ,

(6.4.7)

if s < λ,

A(ς) ς dς . Finally, taking into account

(6.3.9), (6.3.16), (6.3.18), (6.4.6), and (6.4.7), we obtain the required estimate (6.4.1).

 

122

6 The Oblique Derivative Problem for Elliptic Second Order Semi-linear. . .

Theorem 6.11 Let u be a strong solution of problem (SL) and assumptions (A1)–(A4) be satisfied with function A(r) continuous at zero, but not Dini continuous. Then for all ! > 0 there are d ∈ (0, b) and a constant c! > 0 depending only on ν, μ, b, s, λ, q, !, γ0 , χ0 , meas G, diam G, χC 1 (∂ ) , γ C 1 (∂ ) , such that for all  ∈ (0, d) uW ˚2

 4−n (G0 )

≤ c uL2 (G) + a0 V 0

(G) 4 2 1−q , 1−q −n

.

gW ˚1/2 (∂G) + ks

+ f W ˚0

4−n (G)

⎧

⎨λ−!

if s > λ,

⎩s−!

if s ≤ λ,

4−n

+ (6.4.8)

where ks is defined by (6.4.2). Proof Similar to the proof of Theorem 6.10, we obtain the Cauchy problem (6.4.5)– (6.4.6). Repeating verbatim the proof of Theorem 5.12 and taking into account Theorem 6.7 we obtain estimation (6.4.8).   Theorem 6.12 Let u be a strong solution of problem (SL) and assumptions (A1)–(A4) be satisfied with function A(r) ∼ 11 . Then there are d ∈ (0, b) and constants c > 0 ln

r

and cs > 0 depending only on ν, μ, b, s, λ, q, !, γ0 , χ0 , meas G, diam G, χC 1 (∂ ) , γ C 1 (∂ ) , such that for all  ∈ (0, d), uW ˚2

 4−n (G0 )

≤ c uL2 (G) + a0 V 0

2 , 4 −n (G) 1−q 1−q

.

gW ˚1/2 (∂G) + ks 4−n

+ f W ˚0

4−n (G)

⎧  ⎨ λ  1 lncs (λ)  ⎩s

+ (6.4.9)

if s > λ, if s ≤ λ,

where ks is defined by (6.4.2). Proof Similar to the proof of Theorem 6.10, we obtain the Cauchy problem (6.4.5)– (6.4.6). Repeating verbatim the proof of Theorem 5.13 and taking into account Theo  rem 6.7 we obtain estimation (6.4.9).

6.5

Power Modulus of Continuity

Theorem 6.13 Let u ∈ W 2,n (G) be a strong solution of (SL). Suppose that assumptions (A1)–(A5) are satisfied. Then there is a positive constant c1 , such that uV 2

.

n,0 (G)

1 ≤ c1 f Ln (G) + a0  1−q 0 V

n , 2qn 1−q 1−q

(G)

+ g

1− 1 Vn,0 n (∂G)

.

(6.5.1)

6.5 Power Modulus of Continuity

123

2 (G) of the problem Proof By Kozlov et al. [79, Theorem 1.4.1], for a solution u ∈ Vn,0

.

⎧ ⎨a ij (x)u

xi xj

+ a i (x)uxi + a(x)u(x) = F (x), x ∈ G,

⎩ ∂u + χ(ω) ∂u + ∂r ∂ n

1 |x| γ (ω)u(x)

= g(x),

x ∈ ∂G\O,

the estimate uV 2

.

n,0 (G)

≤ c0 F Ln (G) + g

1− 1 Vn,0 n (∂G)

+ uLn (G )

holds for any nonempty open set G ⊂⊂ G, provided that λ > 1 and F ∈ Ln (G), where constant c0 depends only on ν, μ, n, χ0 , γ0 , max A(|x|), a i p,G , a p ,G , p > n, and 2

x∈G

the domain G. Thus, using the Jensen inequality with F (x) = f (x) + a0 (x)u|u|q−1 , we obtain 

r −2n |u|n + r −n |∇u|n + |uxx |n dx G .

≤C

 $   |a0 |n |u|qn + |f |n dx + |u|n dx + gn 1− 1

Vn,0 n (∂G)

G

G

(6.5.2)

% .

Using the Young inequality and taking into account q ∈ (0, 1), we deduce



|a0 |n |u|qn = r −2qn |u|qn r 2qn |a0 (x)|n ≤

.

qn

2qn

n

! q−1 r 1−q |a0 | 1−q + ! n r −2n |u|n ,

(6.5.3)

for all ! > 0. Choosing ! = 2−n , from (6.5.2) and (6.5.3), we obtain the estimation uV 2

.

n,0 (G)

1 ≤ c f Ln (G) + a0  1−q 0 V

n , 2qn 1−q 1−q

(G)

+ g

1− 1 Vn,0 n (∂G)

+ uLn (G ) ,

for an arbitrary nonempty open set G ⊂⊂ G. Choosing the domain G such that (diam G )2n < 1/2, we have 

 |u|n dx ≤

.

G

G

r 2n r −2n |u|n dx ≤ (diam G )2n



r −2n |u|n dx ≤

G

1 2

 G

r −2n |u|n dx.

124

6 The Oblique Derivative Problem for Elliptic Second Order Semi-linear. . .

Thus, formula (6.5.2) takes the form  .

r −2n |u|n + r −n |∇u|n + |uxx |n dx ≤

G

$   C1 |a0 |n |u|qn + |f |n dx + gn 1− 1

% ,

Vn,0 n (∂G)

G

 

and we derive the required estimation (6.5.1). Now we will prove the main results. 2



2

Proof of Theorem 6.3 We consider two sets G/4 and G/2 ⊂ G/4 ,  > 0. We make the transformation: x = x  , u(x  ) = ψ()v(x  ), where function ψ() is defined by (6.2.2). The function v(x  ) satisfies the problem ⎧ ⎪ ⎪ a ij (x  )vxi xj + a i (x  )vxi + 2 a(x  )v = ⎪ ⎨ 2   2 q−1 ()v|v|q−1 , . ψ() f (x ) + a0 (x ) ψ ⎪ ⎪ ⎪ ⎩ ∂v + χ(ω) ∂v + 1 γ (ω)v =  g(x  ), ∂ n 

|x |

∂r

ψ()

x  ∈ G21/4 , 2 . x  ∈ 1/4

Based on the local maximum principle (see Proposition 5.1) we conclude that .

sup |v(x  )| ≤ C

& 

x  ∈G11/2

v 2 dx 

1/2

+

2 f Ln (G2 ) + 1/4 ψ()

G21/4

2

 ψ

q−1

 ()

|a0 (x  )|n |v|qn dx 

1/n

+

'  sup |g(x  )| , ψ() x  ∈G2

(6.5.4)

1/4

G21/4

where constant C > 0 depends only on ν, μ, b, γ0 , χ0 , n, G, ω0 , max γ (ω), g1 , ω∈∂  aLn (G4 ) , ( ni=1 |a i |2 )1/2 Lp (G2 ) , p > n. Now, we use the Young inequality. Taking 1/4 1/2 into account that q ∈ (0, 1), we deduce



|a0 (x  )|n |v|qn = |x  |−2qn |v|qn |x  |2qn |a0 (x  )|n ≤

.

qn

2qn

n

! q−1 |x  | 1−q |a0 (x  )| 1−q + ! n |x  |−2n |v|n ,

∀! > 0.

(6.5.5)

6.5 Power Modulus of Continuity

125

By (6.5.4) and (6.5.5), we obtain 

sup |v(x )| ≤ C

.

x  ∈G11/2



v 2 dx 

1/2

+C

2 f  n 2 + L G1/4 ψ()

G21/4



&  C sup |g(x  )| + C2 ψ q−1 () ! ψ() x  ∈G2 1/4

q

|x  |−2n |v|n dx 

1/n

+

G21/4



2qn

n

|x  | 1−q |a0 (x  )| 1−q dx 

! q−1

1/n ' ,

(6.5.6)

G21/4

for all ! > 0. Returning to the variable x and to the function u, we have 

v 2 dx 

1/2

=

G21/4

.

1 2 ψ ()

 2



u (x )dx



1/2

22 ≤ ψ()

G21/4

n



r −n u2 (x)dx

1/2

≤ (6.5.7)

2 G/4

c uL2 (G) + a0 V 0

(G) 2 4 1−q , 1−q −n

+ f W ˚0

4−n (G)

+

= const., + k gW 1/2 s ˚ (∂G) 4−n

by the definition of the function ψ() and Theorem 6.10. By assumption (A4), .

2 2 f  n 2 = L G1/4 ψ() ψ()



|f (x  )|n dx 

1/n

≤

 ψ()

G21/4

 f1 ψ()

|f (x)|n dx 2

G/4

2 r n(s−2)r n−1 dr meas

1/n

≤

/4

⎧ ⎪ s−λ < 1 ⎪ ⎪ ⎨ s 1 = c1 f1 ln 1 < 1 c1 f1 ⎪  ψ() ⎪ ⎪ ⎩1



if s > λ, if s = λ, if s < λ.

1/n

≤

126

6 The Oblique Derivative Problem for Elliptic Second Order Semi-linear. . .

Hence .

2 f  n 2 ≤ c1 f1 = const. L G1/4 ψ()

(6.5.8)

Similarly

  sup |g(x  )| ≤ g1 s−1 . ψ() x  ∈G2 ψ() 1/4

⎧ ⎪ s−λ < 1 ⎪ ⎪ ⎨ 1 = g1 ln 1 < 1 ⎪  ⎪ ⎪ ⎩ 1

if s > λ, if s = λ, if s < λ.

Thus .

 sup |g(x  )| ≤ g1 = const. ψ() x  ∈G2

(6.5.9)

1/4

We calculate 

|x  |−2n |v|n dx 

1/n

=

G21/4



r −2n |v|n dx

1/n

=

2

G/4

 ψ()

.



r −2n |u|n dx

(6.5.10)

1/n

2

G/4

and  .

2qn

n

|x  | 1−q |a0 (x  )| 1−q dx 

1/n

≤

G21/4

c(q, n)−1

 2

G/4

n

|a0 | 1−q dx

1/n .

(6.5.11)

6.5 Power Modulus of Continuity

127

Choosing ! = ψ()/ in (6.5.6), because of (6.5.2) and (6.5.3), by (6.5.10) and (6.5.11), we obtain

1/n  .! |x  |−2n |v|n dx  ≤ c(n, q, b)gn 1− 1 + G21/4

Vn,0 n ()

'1/n & n |a0 (x)| 1−q + |f (x)|n dx c(n, q, b) ≤ c(n, q, b, s, f1 , k0 ) = const. G

(6.5.12) and ! .

q q−1



2qn

n

|x  | 1−q |a0 (x  )| 1−q dx 

1/n

≤

G21/4

(6.5.13)

 q 1 

1/n n 1−q c2 |a0 | 1−q dx ≤ c2 k0 = const., ψ()  G

by (6.2.1) and assumption (A4). From (6.5.7)–(6.5.9), (6.5.12) and (6.5.13), with regard to (6.5.6), we have .

sup |v(x  )| ≤ c3 (1 + 2 ψ q−1 ()).

(6.5.14)

x  ∈G11/2

We need to show that for all  > 0, 2 ψ q−1 () < ∞.

(6.5.15)

.

Let us assume 0 < q < 1 −

2 λ

and λ > s. In this case we have that if s ≤

2 1−q ,

then

2 ψ q−1 () = s(q−1)+2 < ∞

.

2 holds for all  > 0. Choosing the best exponent s = 1−q < λ, we obtain the required estimation (6.2.3). In fact, by (6.5.14) and (6.5.15), it follows

|v(x  )| ≤ M0 = const.

.

128

6 The Oblique Derivative Problem for Elliptic Second Order Semi-linear. . .

for all x  ∈ G11/2 . Returning to the variable x, we obtain 2

|u(x)| ≤ M0 ψ() = M0  1−q

.



for all x ∈ G/2 , 0 <  < b. Setting |x| = 2/3, we obtain (6.2.3). Let us assume that 1 −

2

 ψ

.

q−1

2 λ

() =

≤ q ≤ 1. Thus for all  > 0 ⎧ 2+λ(q−1) < ∞ ⎪ ⎪ ⎨

3 2+λ(q−1) ln 2 (q−1) 1 < ∞ ⎪ ⎪ ⎩ 2+s(q−1) ≤ 2q (λ−2)(q−1) 

if s > λ, if s = λ, 0 such that .γ (ω) ≥ γ0 > 0. Theorem 7.7 Let operator Q satisfy assumptions (i)–(iii) and functions .u, v ◦1   C 0 (Gd0 )∩ W 0 Gd0 (.d & 1) satisfy inequality Q(u, η) ≤ Q(v, η)

.

∈

(7.3.2)

140

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

◦1   for all nonnegative .η ∈ C 0 (Gd0 )∩ W 0 Gd0 and also inequality

u(x) ≤ v(x),

.

x ∈ ∂Gd0 .

(7.3.3)

Then .u(x) ≤ v(x) in .Gd0 . Proof We define function .z = u − v and .uτ = τ u + (1 − τ )v for .τ ∈ [0, 1]. Then for all ◦1   nonnegative .η ∈ C 0 (Gd0 )∩ W 0 Gd0  .

0 ≥ Q(u, η) − Q(v, η) =

.

  a ij (x) uxj − vxj ηxi +

Gd0

/ [b(x, u, ux ) − b(x, v, vx )] η(x) + a0 r −2 [u(x) − v(x)] η(x) dx−    a ij (x) uxj − vxj cos(r, xi )η(x)d d + d



γ (ω) (u − v)η(x)ds − r

0d

 [g(x, u) − g(x, v)] η(x)ds.

(7.3.4)

0d

We calculate a ij (x)(uxj − vxj ) = a ij (x)zxj .

.

1 b(x, u, ux ) − b(x, v, vx ) =

.

(7.3.5)

db(x, uτ , uτx ) dτ = dτ

0

1 1 0

2 ∂b ∂b · z(x) + τ · zxj dτ. ∂uτ ∂uxj

Let .k ≥ 1 be any odd number. Let us choose function $ % $ % η = max (u − v)k , 0 = max zk , 0

.

as a test function in the integral inequality (7.3.4) and introduce the following set: $ % (Gd0 )+ := x ∈ Gd0 : u(x) > v(x) ⊂ Gd0 .

.

(7.3.6)

7.3 The Comparison Principle

141

By assumptions (i)–(iii) and (7.3.6), (7.3.5), we obtain 

.

.

  a ij (x) uxj − vxj ηxi + a0 r −2 [u(x) − v(x)] η(x)+

(Gd0 )+

 / [b(x, u, ux ) − b(x, v, vx )] η(x) dx ≥

& kνzk−1 |∇z|2 +

(Gd0 )+

1 z

τ −2

|u |

k+1

1 |∇u | dτ − z zxj τ 2

k

0

' |uτ |−1 |∇uτ |dτ dx.

(7.3.7)

0

 According to (7.3.7) and the fact that .η(x) d ∪ = 0, inequality (7.3.4) takes the d 0 following form:  kν

.

 zk−1 |∇z|2 dx +

(Gd0 )+

⎞ ⎛ 1  zk+1 ⎝ |uτ |−2 |∇uτ |2 dτ ⎠ dx ≤

(Gd0 )+

0



⎛ 1 ⎞  τ −1 τ z |∇z| ⎝ |u | |∇u |dτ ⎠ dx. k

(Gd0 )+

(7.3.8)

0

By the Cauchy inequality,

k−1

k+1 zk |∇z||uτ |−1 |∇uτ | = |uτ |−1 z 2 |∇uτ | z 2 |∇z| ≤

.

1 ε τ −2 k+1 |u | z |∇uτ |2 + zk−1 |∇z|2 , ∀ε > 0. 2 2ε Taking .ε = 2, from (7.3.8), it follows .

   1 kν − 4

zk−1 |∇z|2 dx ≤ 0.

(7.3.9)

(Gd0 )+

% $ 1 Finally, choosing the odd number .k ≥ max 1, 4ν , (n − 2) χγ00 and taking into account that .z(x) ≡ 0 on .∂(Gd0 )+ , we get that .z(x) ≡ 0 in .(Gd0 )+ . It leads to contradiction with   the definitions of set .(Gd0 )+ . By this, the theorem was proved.

142

7.4

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

The Barrier Function. The Preliminary Estimate of the Solution Modulus

We shall derive preliminary estimates of .|u(x)| and .|∇u(x)| for a weak solution of problem (WQL) near a conical boundary point. Theorem 7.8 Let u be a weak solution of problem (WQL), assumptions .(A)–.(E) be satisfied. Then there exist . ∈ (0, 1), .d ∈ (0, 1) and positive constants .C0 , C1 that do not depend on u, such that |u(x)| ≤ C0 |x|

+1

,

(7.4.1)

|∇u(x)| ≤ C1 |x| ,

(7.4.2)

.

.

for all .x ∈ Gd0 . Proof Let .Gd0 be a convex rotational cone with a solid angle .ω0 ∈ (0, π) and the lateral surface .0d such that .Gd0 ⊂ {x1 ≥ 0}. Let us consider the operator Q defined by (7.3.1) and integrate by parts the first integral, and taking into account that .∂Gd0 = 0d ∪ d , we get   Q(Aw, η) ≡

−

.

' d & ij a (x)Awxj + a0 r −2 Aw(x)+ dxi

Gd0

  b(x, Aw, Awx ) η(x)dx + a ij (x)Awxj cos(

n, xi )η(x)ds− ∂Gd0





a (x)Awxj cos(r, xi )η(x)d d + ij

d

g(x, Aw)η(x)ds = 0d

0d

 ?



γ (ω) Aw(x)η(x)ds− r

−

' d & ij a (x)Awxj + a0 r −2 Aw(x)η(x)+ dxi

Gd0

@  b(x, Aw, Awx ) η(x)dx + a ij (x)Awxj cos(

n, xi )η(x)ds+  0d

0d

γ (ω) Aw(x)η(x)ds − r

 g(x, Aw)η(x)ds, 0d

◦1   for all nonnegative .η ∈ C 0 (Gd0 )∩ W 0 Gd0 and some .A > 0.

(7.4.3)

7.4 The Barrier Function. The Preliminary Estimate of the Solution Modulus

143

Lemma 7.9 (Existence of the Barrier Function) Let assumptions .(A)–.(D) satisfy, and we set .h = cot ω20 , 0 < ω0 < π. There exist the number . 0 > 0 depending only on .ν, μ, h and the barrier function w(x; y) ≡ x

.

−1

(x 2 − h2 y 2 ) + Bx

+1

,

0
0 the following inequality holds: ℵ[Aw] ≥ 0,

.

x ∈ Gd0 ; 0 < d < 1.

(7.4.5)

Proof Let us define the following linear elliptic operator: ∂2 ; a ij (x) = a j i (x), x ∈ Gd0 , ∂xi ∂xj

L0 ≡ a ij (x)

.

where νξ 2 ≤ a ij (x)ξi ξj ≤ μξ 2 , ∀x ∈ Gd0 , ∀ξ ∈ Rn and ν, μ = const > 0.

.

Let .x = (x, y, x  ) ∈ Rn , where .x = x1 , .y = x2 , .x  = (x3 , . . . , xn ). In .{x1 ≥ 0}, we consider the cone K with the vertex .O such that .K ⊃ Gd0 . We recall that .Gd0 ⊂ {x1 ≥ 0}. Let .∂K be the lateral surface of K and let on .∂K ∩ yOx = ± be .x = ±hy, such that in the interior of K the inequality .x > h|y| holds. Let the coefficients of the operator .L0 be .a 2,2 = a, a 1,2 = b, a 1,1 = c. Then we have L0 w = awyy + 2bwxy + cwxx ,

(7.4.6)

.

where νη2 ≤ aη12 + 2bη1η2 + cη22 ≤ μη2

.

and (7.4.7)

η2 = η12 + η22 ; ∀η1 , η2 ∈ R. Let us calculate the operator .L0 on the function (7.4.4). For .t = yx , |t| < h1 , we obtain L0 w = −h2 x

.

−1

φ( ),

144

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

where φ( ) = 2a − 4bt + 4bt − ch−2 (1 + B)(

.

= c(t 2 − h−2 (1 + B))

2

2

+ ) + ct 2

2

− 3ct 2 + 2ct 2 =

+ (4bt − ch−2 (1 + B) − 3ct 2 ) + 2(ct 2 − 2bt + a)

with φ(0) = 2(ct 2 − 2bt + a) ≥ 2ν, because of (7.4.7),  2  y 1+B ν B c(t 2 − h−2 (1 + B)) = c 2 − ≤ −c 2 ≤ − 2 < 0 2 x h h h .

and discriminant  =(4bt − ch−2 (1 + B) − 3ct 2 )2 − 8c(t 2 − h−2 (1 + B))(ct 2 − 2bt + a) ≥ .

(3ct 2 + 2ch−2 − 4bt)2 + 8c(2h−2 − t 2 )(ct 2 − 2bt + a) > (3ct 2 + 2ch−2 − 4bt)2 + 8t 2 ν(ct 2 − 2bt + a) > (5νt 2 − 4bt)2 + 4t 2 ν 2 > 0.

φ( ) is a square polynomial with the positive discriminant; therefore, there exist two roots 1 , 2 of this square polynomial and, by the Viète formula,

. .

.

1

·

2

ct 2 − 2bt + a  < 0. =2  2 c t − h−2 (1 + B)

Hence, since . φ(0) ≥ 2ν, it follows that there exists . 0 > 0 depending only on . ν, μ, h, B such that .φ( ) ≥ ν for . ∈ (0; 0 ]. Therefore, we obtain L0 [w(x, y)] ≤ −νh2 x

.

−1

; 0
2(1 + h), A> 2μ(1 + h)

√

1 + h2 h

 (7.4.21)

.

In this way, we get the required estimation (7.4.16). Now, we consider the barrier function .w(x) defined by (7.4.4) and function .u(x) that satisfies the integral identity ((I I )loc ). We verify that these functions satisfy the assumptions of Theorem 7.7. By (7.4.3), (7.4.9), and (7.4.16), it follows that Q(Aw, η) ≥ 0 = Q(u, η),

(7.4.22)

.

◦1

for all nonnegative functions .η(x) ∈ C 0 (G)∩ W 0 (G). Since .x 2 ≥ h2 y 2 in .K and by the definition of the barrier function (7.4.4), we get   Aw(x)

.

+1

≥ ABd

∂Gd0

 √



h

+1

.

1 + h2

On the other hand, we have   u(x)

.

∂Gd0

  ≤ |u(x)|

∂Gd0

≤ M0 .

Choosing M0 .A ≥ 2(1 + h)d

√ +1

1 + h2 h



+1

(7.4.23)

and .B > 2(1 + h), we obtain   Aw(x)

.

∂Gd0

≥ ABd

+1



h √ 1 + h2



+1

  ≥ M0 ≥ u(x)

∂Gd0

.

Writing barrier function . w(x, y, x  ) (7.4.4) in the spherical coordinates and recalling that .h = cot ω20 , we get  

w(x, y, x ) = r

.

+1

cos

−1

ω1 B cos ω1 + 2

ϕ(ω1 ) sin2

ω0 2



& ω ω ' 0 0 , ∀ω1 ∈ − , , 2 2 (7.4.24)

148

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

where ϕ(ω1 ) = sin

.

ω

0 − ω1 sin + ω1 . 2 2

ω

0

Because of .ϕ  (ω) = − sin 2ω and .ϕ  (ω1 ) = 0 for .ω1 = 0, we can see that .ϕ  (0) = −2 cos 0 = −2 < 0, and we have .

& max ' ϕ(ω1 ) ω ω ω∈ − 20 , 20

= ϕ(0) = sin2

ω0 . 2

(7.4.25)

Finally, from (7.4.24) and (7.4.25), we obtain

ω1 B cos2 ω1 + 1   1 ≤ r +1 cos +1 ω1 B + cos2 ω1   1 ≤ x +1 B + . cos2 ω1

w(x, y, x  ) ≤ r .

+1

cos

−1

(7.4.26)

Now, if we choose number .B > 2(1 + h), fixed number d ∈ (0, 1), . ∈ (0, 0 ] according to (7.4.15) and number A according to (7.4.15), (7.4.21), and (7.4.23), we assure the validity of Theorem 7.7. Therefore, we have u(x) ≤ Aw(x), x ∈ Gd0 .

.

(7.4.27)

Replacing .u(x) by .−u(x), we can derive the estimation .u(x) ≥ −Aw(x). In this way, because of (7.4.26), we derived (7.4.1). To estimate the modulus of gradient of the problem (WQL) solution near a conical  point, we consider a set .G/2 ⊂ G, .0 <  < d, and make the following transformation:   − −1 u(x  ). Then the function .z(x  ) satisfies the problem .x = x , .z(x ) =  ⎧ + , ⎪ d ∂z ij   −2 ⎪ ⎪ ⎨ − dx  a (x ) ∂xj + a0 (r ) z+ .

i

1− b(x  ,  +1 z,  zx  ) = 0, x  ∈ G11/2 , ⎪ ⎪ ⎪ ⎩ ∂z + γ (ω) z = − g( x  ,  +1 z), x  ∈  1 . 1/2 |x  |

 ∂ν

(7.4.28)

7.4 The Barrier Function. The Preliminary Estimate of the Solution Modulus

149

Applying assumption (E) about a priori estimate of the modulus of gradient of the problem (7.4.28) solution .

max |∇  z(x  )| ≤ M1

x  ∈G11/2

and returning to the variable x and to the function .u(x), we obtain |∇u(x)| ≤ M1 

.



for .x ∈ G/2 , .0 <  < d. Setting .|x| = 23 , we obtain the required estimation (7.4.2).   Corollary 7.10 Let .u(x) be a weak solution of problem (WQL) and assumptions of Theorem 7.8 be satisfied. Let us also suppose that    ∂g(x, u)  s−1 α   .  ∂u  ≤ g1 |x| |u(x)| ,

s > 1, α ≥ −1.

Then .u(0) = 0. Proof From the boundary condition of problem (WQL), .

∂u γ (ω) + u(x) = g(x, u) ∂ ν

|x|

& ' ⇒ γ (ω)u(x) = x g(x, u) − a ij (x)uxj cos(

n, xi ) , x ∈ ∂G\{O}.

By assumptions (A), (D), (E) and estimates (7.4.1)–(7.4.2) and (7.4.19), we get γ0 |u(x)| ≤ γ (ω)|u(x)| ≤ |x||g(x, u)| + μ|x||∇u| ≤    ∂g(x, u)   + μ|x||∇u| ≤  |x||g(x, 0)| + |x||u(x)|  ∂u 

.

g0 |x|s + g1 |x|s |u(x)|1+α + C1 μ|x|

+1

g0 |x|s + g1 C01+α |x|s+(

≤ +1)(1+α)

+ C1 μ|x|

+1

.

Because of continuity of .u(x), letting .|x| tend to 0, we get that .γ0 |u(0)| ≤ 0, and taking   into account that .γ0 > 0, we finally get .u(0) = 0.

150

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

7.5

Local Estimate at the Boundary

Theorem 7.11 Let u(x) be a weak solution of (WQL) and assumptions (A1), (C)–(D) be satisfied. Suppose, in addition, that g(x, 0) ∈ L∞ (∂G). Then inequality .

& n 2− n(p+2) 2p b (x) sup |u(x)| ≤ C − 2 u(x)2,G +  0 G0

0



p > m# > 2 with

holds for

m# =

 2p p+2 ,G0

+ ' g(x, 0)∞,  , 0

⎧ ⎨

2n n−2 , ⎩ 2m , m−2

if n > 2; m > 2, if n = 2;

(7.5.1)

∈ (0, 1) and  ∈ (0, d), where d ∈ (0, 1).

Proof Let us consider the integral identity (I I ) and let us make the following transformation: x = x  , z(x  ) = u(x  ), ξ(x  ) = η(x  ). We have: dx = n dx  , ds = n−1 ds  . We denote by G and ∂G the images of G and ∂G, respectively. Thus the integral identity takes the form  & a ij (x  )zxj ξxi + a0 (r  )−2 z(x  )ξ(x  )+ . G

 ' γ (ω) 2 b(x  , z, −1 zx  )ξ(x  ) dx  + z(x  )ξ(x  )ds  = r ∂G

 ·

g(x  , z)ξ(x  )ds 

(7.5.2)

∂G ◦1

for all ξ(x  ) ∈ C 0 (G )∩ W 0 (G ). Let us choose ξ(x  ) = z(x  )ζ 2 (|x  |) as a test function in the above integral inequality, where ζ (|x  |) ∈ C0∞ ([0, 1]) is a nonnegative and non-increasing function that will be further specified. We calculate ξxi (x  ) = zxi ζ 2 (|x  |) + 2z(x  )ζ(|x  |)ζxi (|x  |),

.

7.5 Local Estimate at the Boundary

151

and then (7.5.2) takes the form  .

a ij (x  )zxi zxj ζ 2 (|x  |)dx  +

G10













a (x )z(x )zxj ζ(|x |)ζxi dx + ij

2 G10

a0 (r  )−2 z2 (x  )ζ 2 (|x  |)ds  +

G10



b(x  , z, −1 zx  )z(x  )ζ 2 (|x  |)dx  +

2 · G10

 ·



γ (ω) 2  2  z (x )ζ (|x |)ds  = r

01

g(x  , z)z(x  )ζ 2 (|x  |)ds  .

(7.5.3)

01

Now, we introduce z(x  ) = |z(x  )| + k,

(7.5.4)

.

where k = k() = g(x  , 0)∞, 1 + 2 b0 (x  )

.

0

2p 1 p+2 ,G0

.

(7.5.5)

Let us estimate some integrals from the right side of (7.5.3): • Because of, by (A4), a0 > 0, we have  .

a0 (r  )−2 z2 (x  )ζ 2 (|x  |)ds  ≥ 0.

(7.5.6)

G10

• By the ellipticity condition (A1),  .

G10

a ij (x  )zxi zxj ζ 2 (|x  |)dx  ≥ ν

 G10

|∇  z|2 ζ 2 (|x  |)dx  ,

(7.5.7)

152

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

and the Cauchy inequality 

a ij (x  )|z(x  )||zxj |ζ(|x  |)|ζxi |dx  ≤

2

.

G10



2μ  μδ

|∇  z||z(x  )||∇  ζ |ζ(|x  |)dx  ≤

G10

|∇  z|2 ζ 2 dx  +

G10

μ δ



(7.5.8)

z2 |∇  ζ |2 dx  , ∀δ > 0.

G10

• By assumption (B1) and (7.5.4), & ' 2 b(x  , z, −1 z)zζ 2 ≤ 2 |z|ζ 2 β|z|−1 −2 |∇  z|2 + b0 (x  )

.

≤ β|∇  z|2 ζ 2 + 2

z2 2 2 b0 (x  ) 2 2 ζ b0 (x  ) ≤ β|∇  z|2 ζ 2 + z ζ . z k

(7.5.9)

• By assumption (D), we get zg(x  , z) = zg(x  , 0) + z

1

.

dg(x  , τ z) dτ = zg(x  , 0) + z2 dτ

0

1

∂g(x  , τ z) dτ ∂(τ z)

0

≤ z|g(x  , 0)|. Thus, using the Cauchy inequality, 







g(x , z)zζ ds ≤



.

01

1 2

 01

2

|g(x  , 0)|zζ 2 ds  ≤

01

  

1 2 g(x  , 0)2∞ ζ 2 z2 + 2 |g(x  , 0)|2 ds  ≤ z2 ζ 2 1 + ds  . 2 k2 01

(7.5.10)

7.5 Local Estimate at the Boundary

153

By virtue of (7.5.6)–(7.5.11), inequality (7.5.3) takes the form 



|∇  z|2 ζ 2 (|x  |)dx  +

ν

.

G10



μ δ

|∇  z|2 ζ 2 dx  +

μδ G10



γ (ω) 2  2  z (x )ζ (|x |)ds  ≤ r

01

z2 |∇  ζ |2dx  + β

G10

 G10

|∇  z|2 ζ 2 dx  +



2 b0 (x  ) 2 2  z ζ dx + k

G10

 1+

2 g(x  , 0)2∞ k2



z2 ζ 2 ds  .

01

Further, by inequality (2.5.2) of Theorem 2.27, we have  .

z2 ζ 2 ds  ≤ c

01

 &

 '   z2 ζ 2 + ∇  (z2 ζ 2 ) dx  .

G10

By the Cauchy inequality, taking into account (7.5.4)   1   2 2  ∇ (z ζ ) ≤ 2z|∇  z|ζ 2 + 2z2 ζ |∇  ζ | ≤ δ1 |∇  z|2 ζ 2 + z2 ζ 2 + 2z2 |∇  ζ |ζ ≤ δ1   1 1 z2 ζ 2 + z2 |∇  ζ |2 , δ1 |∇  z|2 ζ 2 + z2 ζ 2 + z2 ζ 2 + z2 |∇  ζ |2 = δ1 |∇  z|2 ζ 2 + 1 + δ1 δ1 .

for all δ1 > 0. Thus  .

  ,  + 1 z2 ζ 2 ds  ≤ c z2 ζ 2 dx  . δ1 |∇  z|2 ζ 2 + z2 |∇  ζ |2 + 2 + δ1

01

G10

Thus, choosing δ =

.

(7.5.11)

ν −β 2



ν−β 2μ ,

we obtain

|∇  z|2 ζ 2 dx  ≤

G10

2μ2 ν −β

 G10

z2 |∇  ζ |2 dx  +

 G10

2 b0 (x  ) 2 2  z ζ dx + k

   +  , 1 2 g(x  , 0)2∞  2 2 2  2 2 2 |∇ z| ζ + z |∇ ζ | + 2 + z ζ c 1+ δ dx  ≤ 1 k2 δ1 G10

154

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

2μ2 ν−β

  ,  + 1 z2 |∇  ζ |2 dx  + 2c z2 ζ 2 dx  + δ1 |∇  z|2 ζ 2 + z2 |∇  ζ |2 + 2 + δ1

 G10

G10



2 b0 (x  ) 2 2  z ζ dx , k

(7.5.12)

G10

in virtue of, by the definition (7.5.5), k ≥ g(x  , 0). Now we choose δ1 = then inequality(7.5.12) takes the form 



2 2





G10



 2 2  z |∇ ζ | + ζ dx + c2 F (x  )z2 ζ 2 dx  , 2

|∇ z| ζ dx ≤ c1

.

ν−β 8c ,

G10

and

(7.5.13)

G10

where   2μ2 16c2 4 max + 2c; 4c+ .c1 = ; ν−β ν −β ν −β

c2 =

4 ; ν−β

β 2,

(7.5.15)

p > m# > 2,

(7.5.16)

G10

the interpolation inequality for Lp -norms (see Corollary 2.16): zζ 

.

2p 1 p−2 ,G0

≤ εzζ 

m#

2m# ,G10 m# −2

m# =

+ ε m# −p zζ 2,G1 , 0

⎧ ⎨

2n n−2 , ⎩ 2m , m−2

∀ε > 0,

if n > 2; m > 2, if n = 2;

and the Sobolev embedding theorem (see (7.2.11)) zζ 22m#

.

m# −2

,G10

≤ c2

 &

' |∇  ζ |2 + ζ 2 |z|2 + |∇  z|2 ζ 2 dx  , n ≥ 2,

G10

(7.5.17)

7.5 Local Estimate at the Boundary

155

where c2 depends only on n and the domain G. By (7.5.13)–(7.5.17), we obtain zζ 

.

≤ c3 (ζ + |∇  ζ |)z2,G1 +

2m# ,G10 m# −2

0

c4 F 

1 2 2p 1 p+2 ,G0

 εzζ 

2m# ,G10 m# −2

+ε

m# m# −p

 zζ 2,G1 , 0

1

for all ε > 0, where p > m# ≥ 2. From (7.5.5) and (7.5.14), it follows F  22p

1 p+2 ,G0

Thus, choosing ε = zζ 

.

− 12 1 2c4 F  2p ,G1 , 0 p+2

2m# ,G10 m# −2

≤ c(p).

we get

≤ c(ζ + |∇  ζ |)z2,G1 , m# > 2.

(7.5.18)

0

Now, we apply the Moser iteration method for inequality (7.5.18). To this, we define sets +(1− )2−j , j = 0, 1, 2, . . ., for all ∈ (0, 1). It is easy to see that G0 ≡ G(j ) ≡ G0 G(∞) ⊂ . . . ⊂ G(j +1) ⊂ G(j ) ⊂ . . . ⊂ G(0) ≡ G10 . We introduce the sequence of cut-off functions ζj (x  ) ∈ C ∞ (G(j ) ) such that

.

⎧ ⎪ ⎪0 ≤ ζj (x  ) ≤ 1 ⎪ ⎪ ⎪ ⎨ζ (x  ) ≡ 1 j

⎪ ⎪ ζj (x  ) ≡ 0 ⎪ ⎪ ⎪ j+1 ⎩  |∇ ζj | ≤ 21−

in G(j ) , in G(j +1) , for |x  | > for

+ (1 − )2−j ,

+ (1 − )2−j −1 < |x  |
t > > 2 > >z ζ >

2m# ,G10 m# −2

> > t > > ≤ c >(ζ + |∇  ζ |)z 2 >

2,G10

(7.5.19)

,

where t = 2. We replace in (7.5.19) ζ(|x  |) by ζj (x  ) and t by tj . We have > t > > j > . > z 2 ζj > > >

2m# ,G(0) m# −2

> > tj > >  > 2 ≤ c >(ζj + |∇ ζj |)z > >

2,G(0)

.

Then taking tj -th root, by direct calculation, we obtain  ztj+1 ,G

.

(j+1)

≤

C 1−

2/tj

2j

· 2 tj · ztj ,G . (j)

156

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

After iteration, we find that  ztj+1 ,G

.

(j+1)

Notice that the series

∞ 

j tj

j =0

≤

∞ 2 

1 t j=0 j

C 1−

·2

2

∞ 

j t j=0 j

· z2,G1 . 0

is convergent by the d’Alembert ratio test, but the series

j ∞ ∞   1  m# − 2 m# 1 . = = tj 2 m# 4 j =0

j =0

as a geometric series. Therefore, we get  ztj+1 ,G

.

(j+1)

≤

m# /2

C 1−

· z2,G1 . 0

Consequently, letting j → ∞, we have

.

sup z(x  ) ≤

x  ∈G0



m# /2

C 1−

· z2,G1 . 0

Thus, from the above and taking into account (7.5.4) and (7.5.5), we obtain

.

sup |z(x  )| ≤



x  ∈G0

m# /2 &

C 1−

z2,G1 + k()12,G1 0

'

0

and, returning to the variables x and function u, .

& n ' - − 2 u  + K() , sup |u(x)| ≤ C 2,G x∈G0

0



where K() = g(x, 0)∞,  + 

.

2− n(p+2) 2p

0

-= C



C 1−

Thus, Theorem 7.11 was proved.

m# /2

b0 (x) 

· max 1;



2p p+2

)

, G0 , p > m# > 2;

meas n

1/2 .  

7.6 Global Integral Estimate

7.6

157

Global Integral Estimate

Theorem 7.12 Let u(x) be a weak solution of problem (WQL) and assumptions (A)–(D) ◦1

be satisfied with function A(r) being continuous at zero, A(0) = 0. Then u(x) ∈W α−2 (G), 4 − n ≤ α ≤ 2, and there exists positive constant C, which depends only on ν, μ, β, λ, n, d, χ0 , γ0 such that inequality .

 

r α−4 u2 + r α−2 |∇u|2 dx + r α−3 γ (ω)u2 ds G

∂G

⎤   & ' u2 + (1 + r α )b02 dx + r α−1 g 2 (x, 0)ds ⎦ ≤ C1 ⎣ ⎡

G

(7.6.1)

∂G

holds. Proof Let rε (x) be the function defined in Sect. 2.3 of Chap. 2. We consider the integral identity (I I ). Putting η(x) = rεα−2 u(x) and calculating ηxi = rεα−2 uxi + (α − 2)(xi − εli )r α−4 u(x),

.

we obtain  rεα−2 a ij (x)uxi uxj dx +

.

G



r −2 rεα−2 u2 (x)dx

a0 G

α−2 2

a ij (x)(xi − εli )rεα−4 G

 +



∂u2 dx+ ∂xj



rεα−2 u(x)b(x, u, ux )dx G

+ ∂G

rεα−2 u2 (x)

γ (ω) ds = r

 rεα−2 u(x)g(x, u)ds ∂G

and, because of assumption (A1), 

 a ij (x)rεα−2 |∇u|2 dx +

.

 a0 G

r −2 rεα−2 u2 (x)dx = (2 − α)

 G

G

&

∂G

rεα−2 u2 (x)

γ (ω) ds+ r

' (xi − εli ) a ij (x) − a ij (0) rεα−4 uxj u(x)dx+

158

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

2−α 2

 (xi − εli )rεα−4

∂u2 dx − ∂xi

G

 b(x, u, ux )rεα−2 u(x)dx+ G

 rεα−2 u(x)g(x, u)ds.

(7.6.2)

∂G

Now we calculate: • Integrating by parts, we get  (xi − εli )rεα−4

.

∂u2 dx = − ∂xi

G

 u2 (x) G

' ∂ & α−4 rε (xi − εli ) dx+ ∂xi

 u2 (x)rεα−4 (xi − εli ) cos(

n, xi )ds.

(7.6.3)

∂G

We calculate n 

(xi − εli )2 ' ∂ & α−4 α−5 i=1 r . (xi − εli ) = (α − 4)rε + nrεα−4 = (n + α − 4)rεα−4 . ∂xi ε rε By Lemma 2.10,   xi cos(

n, xi )

.

0d

  = 0 and cos(

n , x i )

0d

= − sin

ω0 , 2

we obtain   (xi − εli ) cos(

n, xi )

.

0d

= −ε sin

ω0 . 2

Thus and from the representation ∂G = 0d ∪ d ,  .

u2 (x)rεα−4 (xi − εli ) cos(

n, xi )ds =

∂G

− ε sin

ω0 2



 u2 (x)rεα−4 ds + 0d

u2 (x)rεα−4 (xi − εli ) cos(

n, xi )ds. d

(7.6.4)

7.6 Global Integral Estimate

159

From (7.6.3) and (7.6.4), it follows 

2−α . 2

(xi − εli )rεα−4 G

2−α ω0 −ε · sin · 2 2

∂u2 dx = ∂xi

 rεα−4 u2 (x)ds +

(2 − α)(4 − n − α) 2

 rεα−4 u2 (x)dx+ G

0d

2−α 2

 (xi − εli )rεα−4 u2 (x) cos(

n, xi )ds.

(7.6.5)

d

• Because of α ≤ 2, on d , we have (see Corollary 2.12)   rε ≥ hr ≥ hd ⇒ (α − 3) ln rε ≤ (α − 3) ln(hd) ⇒ rεα−3 

.

d

≤ (hd)α−3 .

Therefore, 

2−α . 2

(xi − εli )rεα−4 u2 (x) cos(

n, xi )ds d

≤

2−α (hd)α−3 2



2−α ≤ 2

 rεα−3 u2 (x)ds d



u2 + |∇u|2 dx u2 (x)ds ≤ c

d

(7.6.6)

Gd

because of (2.5.2) in Theorem 2.27. • By assumption (B1) and the Cauchy inequality, 

 rεα−2 u(x)b(x, u, ux )dx ≤ β

.

G

 ≤β

rεα−2 |∇u|2 dx G

δ + 2

 G

 rεα−2 |∇u|2 dx +

G

r −2 rεα−2 u2 dx

rεα−2 |u|b0(x)dx G

1 + 2δ

 r 2 rεα−2 b02 (x)dx, ∀δ > 0.

G

(7.6.7)

160

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

• To estimate integrals over G in (7.6.2), we use the representation G = Gd0 ∪ Gd . To estimate integral over Gd0 , we use assumption (A2) and the Cauchy inequality (2 − α)

 &

.

 c1 A(d)

' a ij (x) − a ij (0) (xi − εli )rεα−4 u(x)uxj dx ≤

Gd0

rεα−3 |∇u||u|dx

  α−4    α−2 2 = c1 A(d) rε 2 |u| dx ≤ rε |∇u|

Gd0

 c2 A(d)

Gd0

rεα−4 u2 dx.

(7.6.8)

Gd0

To estimate integrals over Gd , we use assumption (A1), the Cauchy inequality, and α ≤ 2, as well as the fact that for r ≥ d we have rε ≥ hd. Thus (2 − α)

.

 & ' a ij (0) − a ij (x) (xi − εli )rεα−4 u(x)uxj dx ≤

Gd

 c3

|∇u|2 + u2 dx.

(7.6.9)

Gd

• By assumption (D), we obtain 1 ug(x, u) = ug(x, 0) + u

.

dg(x, τ u) dτ = ug(x, 0)+ dτ

0

1 u2

 1*

 1 ∂g(x, τ u) 1 dτ ≤ |u||g(x, 0)| = r − 2 γ (ω)|u| · r 2 √ |g(x, 0)| ≤ ∂(τ u) γ (ω)

0

1 δ1 −1 r γ (ω)u2 + rg 2 (x, 0), ∀δ1 > 0. 2 2δ1 γ0

(7.6.10)

7.6 Global Integral Estimate

161

From (7.6.5)–(7.6.10), identity (7.6.2) takes the following form:  rεα−2 |∇u|2 dx + ε sin

ν

.

G

 a0

rεα−4 u2 ds +

r −2 rεα−2 u2 dx  c2 A(d) δ 2

(2 − α)(4 − n − α) ≤ 2



 +c

rεα−4 u2 dx G

(u2 + |∇u|2 )dx+

Gd

 

u2 + |∇u|2 dx + β rεα−2 |∇u|2 dx rεα−4 u2 dx + c3 Gd

Gd0



γ (ω) 2 u ds+ r

rεα−2

∂G

0d

G

+





ω0 2 − α 2 2

r −2 rεα−2 u2 dx +

G



1 2δ

δ1 2

r 2 rεα−2 b02 (x)dx +

G



G

rεα−2 r −1 γ (ω)u2 ds+

∂G

1 2δ1γ0

 rεα−2 rg 2 (x, 0)ds, ∂G

and, by inequalities rε ≥ hr, 4 − n ≤ α ≤ 2, and assumption (C),  (ν − β)

.

  γ (ω) 2 δ1 u ds rεα−2 |∇u|2 dx + 1 − rεα−2 2 r

G



+ a0  + c2 A(d)

r −2 rεα−2 u2 dx

G

rεα−4 u2 dx +

Gd0

1 2δ

δ ≤ 2

∂G



r −2 rεα−2 u2 dx

 + c4

G



r α b02 (x)dx +

1 2δ1 γ0

G

(u2 + |∇u|2 )dx

Gd



r α−1 g 2 (x, 0)ds.

(7.6.11)

∂G

Using Lemma 4.13 and rεα−4 ≤ r −2 rεα−2 , we obtain  c2 A(d)

.

Gd0

 rεα−4 u2 dx

r −2 rεα−2 u2 dx ≤

≤ c2 A(d) Gd0

⎡

⎢ cA(d) ⎢ ⎣



Gd0

⎤

 rεα−2 |∇u|2 dx +

rεα−2 0d

γ (ω) 2 ⎥ u ds ⎥ ⎦ r

(7.6.12)

162

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

and δ . 2



δ = 2

r −2 rεα−2 u2 dx

G

⎡ ⎢ c-1 δ ⎢ ⎣



r −2 rεα−2 u2 dx

δ + 2



r −2 rεα−2 u2 dx ≤

Gd

Gd0

⎤



 rεα−2 |∇u|2 dx +

Gd0

rεα−2

γ (ω) 2 ⎥ u ds ⎥ ⎦ + c-2 δ r

 u2 dx.

(7.6.13)

G

0d

Hence, from (7.6.11), it follows  .

cA(d)] [ν − β − c-1 δ − -

rεα−2 |∇u|2 dx G

  γ (ω) 2 δ1 − c-1 δ − u ds cA(d) rεα−2 + 1− 2 r  ≤ c5

1 (u2 + |∇u|2 )dx + 2δ

G



∂G

r α b02(x)dx + G

1 2δ1 γ0

 r α−1 g 2 (x, 0)ds. ∂G

ν−β 1 1 Now we choose δ1 = 1 and δ ≤ -c+min ; c1 2 4 . By the continuity of function A(r), we can choose d > 0 such that A(d) ≤ δ. Thus we get 

 rεα−2 |∇u|2 dx

.

G

⎡

+

rεα−2

γ (ω) 2 u ds r

∂G



≤ C ⎣ (u2 + |∇u|2 )dx + G



 r α b02 (x)dx +

G

⎤ r α−1 g 2 (x, 0)ds ⎦ , ∀ε > 0.

∂G

(7.6.14) The right side of (7.6.14) does not depend on ε. Therefore, using the Fatou theorem, we can pass in (7.6.14) to the limit ε → +0. Thus, 

 r α−2 |∇u|2 dx +

.

G

r α−3 γ (ω)u2 ds

∂G

⎡ ⎤    ≤ C ⎣ (u2 + |∇u|2 )dx + r α b02 (x)dx + r α−1 g 2 (x, 0)ds ⎦ . G

G

∂G

(7.6.15)

7.7 Local Integral Weighted Estimates

163

Setting η(x) = u(x) in the integral identity (I I ), by (7.6.10) with δ1 = 1, assumptions (A1), (B1), (C) and the Cauchy inequality, we obtain 

1 |∇u| dx + 2



2

(ν − β)

.

G

γ (ω) 2 u ds ≤ r

∂G

1 2

 (u2 + b02 )dx +



1 2γ0

G

rg 2 (x, 0)ds.

(7.6.16)

∂G

Now we use Theorem 4.9 with χ(ω) ≡ 0. For any u ∈ W 1 ( ), we have  .



⎡ ⎤   1 ⎣ |∇ω u|2 d + γ (ω)u2 dσ ⎦ , u2 d ≤ λ(λ + n − 2)

∂

where λ is the smallest positive eigenvalue of (EVP). Multiplying this inequality by r α−5+n and integrating over r ∈ (0, d), we obtain  .

Gd0

r α−4 u2 dx ≤ ⎡ ⎢ 1 ⎢ λ(λ + n − 2) ⎣



 r α−2 |∇u|2 dx +

Gd0

⎤ ⎥ γ (ω)r α−3 u2 ds ⎥ ⎦.

(7.6.17)

0d ◦1

Finally, from (7.6.15)–(7.6.17), we derive that u(x) ∈W α−2 (G) and the desired estimation (7.6.1).  

7.7

Local Integral Weighted Estimates

Theorem 7.13 Let u be a weak solution of problem (WQL), λ be the smallest positive eigenvalue of problem (EVP), and assumptions (A)–(D)

be satisfied with function A(r) being Dini continuous at zero. Then there exist d ∈ 0, 1e and a constant C > 0, which

164

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . . d

depends only on n, λ, s, β, γ0 , and on 0

.

A(r) r dr

such that the inequality

 

r −n u2 + r 2−n |∇u|2 dx + r 1−n γ (ω)u2 ds ≤ 



G0

0

⎧ 2λ(1−β) ⎨

⎪ C u22,G + b02 + g02 · 2λ(1−β) ln2 ⎪ ⎩ 2s 

1 

if s > λ(1 − β), if s = λ(1 − β),

(7.7.1)

if s < λ(1 − β)

holds for all  ∈ (0, d). ◦1

Proof According to Theorem 7.12, we have that u(x) ∈W 2−n (G). We consider the integral identity ((I I )loc ). Putting η(x) = r 2−n u(x) in it, we obtain 

 r 2−n a ij (x)uxi uxj dx + (2 − n)

. 



G0



r

a0

r −n a ij (x)xi u(x)uxj dx+

G0



−n 2

u (x)dx +



r

2−n



u(x)b(x, u, ux )dx +



G0



G0



r 1−n γ (ω)u2 (x)ds = 0



r 2−n a ij (x)u(x)uxj cos(r, xi )d  +

r 2−n u(x)g(x, u)ds,

(7.7.2)

 0



and because of assumption (A1), we get 

 r

.

2−n

2

|∇u| dx + a0





G0

G0

 (n − 2)



 

r

G0

−n 2

 r 1−n γ (ω)u2 (x)ds =

u (x)dx + 

0

r −n xi u(x)uxi dx −

 r 2−n u(x)b(x, u, ux )dx+ 

G0

 & ' r 2−n a ij (x) − a ij (0) u(x)uxj cos(r, xi )d  + r 2−n u(x)uxi cos(r, xi )d  + 

7.7 Local Integral Weighted Estimates

165



 r

2−n

u(x)g(x, u)ds −



& ' r 2−n a ij (x) − a ij (0) uxi uxj dx+



0

G0



(n − 2)

& ' r −n a ij (x) − a ij (0) xi u(x)uxj dx.

(7.7.3)

 G0

We calculate: • 

 .

r 2−n u(x)uxi cos(r, xi )d  = 



u(x)

 ∂u  d . ∂r r=



(7.7.4)



• Integrating by parts, using the representation G0 =  ∪ 0 and equalities      n, xi )   = 0, xi cos(r, xi ) =  (see (2.2.13), ∂x∂ i r −n xi = 0, we xi cos(

 0 obtain  (n − 2)

r

.



−

n−2 2

−n

G0

 u2 

n−2 ∂  −n  r xi dx + ∂xi 2

G0

n − 2 −n  2





n−2 u(x)xi uxi dx = 2



∂u2 dx = ∂xi

G0



r −n u2 (x)xi cos(

n, xi )dx = 

∂G0

n−2 u (x)xi cos(r, xi )d  + 2



2



r −n xi

r −n u2 (x)xi cos(

n, xi )ds =



0

n−2 2



  u2 (x)

r=

d .

(7.7.5)



• According to (7.6.10), we have  r

. 

0

2−n

δ1 u(x)g(x, u)ds ≤ 2

 r 1−n γ (ω)u2 ds+ 

0

1 2δ1 γ0

 r 3−n |g(x, 0)|2 ds, ∀δ1 > 0.  0

(7.7.6)

166

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

• By assumption (B1), we obtain  .



−

r 2−n u(x)b(x, u, ux )dx ≤ β 

r 2−n |∇u|2 dx+ 

G0

G0

 r

2−n



b0 (x)|u(x)|dx ≤ β



r

2−n



G0

G0

1 2δ2

δ2 |∇u| dx + 2



2

r −n u2 dx+



G0



r 4−n |b0 (x)|2dx, ∀δ2 > 0.

(7.7.7)

 G0

• By assumption (B4), we obtain 

 r

.

4−n

|b0 (x)| dx ≤ 2

b02 meas ·

 G0

r 2s−1 dr =

b02 meas 2s  2s

(7.7.8)

0

and 1 . γ0

 r

3−n

 0

g2 |g(x, 0)| ds ≤ 0 meas ∂ · γ0



2

r 2s−1dr =

g02 2s meas ∂ . γ0 2s

(7.7.9)

0

• By assumption (A2) and the Cauchy inequality, we have  .

& ' r 2−n u(x) a ij (x) − a ij (0) uxj cos(r, xi )d  ≤



 A()

  |u(x)||∇u|

A() d ≤ r= 2



 

  |∇u|  2

2

  + u (x) 2

r=

r=

 d ,



(7.7.10)  (n − 2)

.

& ' r −n u(x) a ij (x) − a ij (0) xi uxj dx−





G0

 & ' r 2−n a ij (x) − a ij (0) uxi uxj dx ≤ (n − 2)A() r 1−n |u(x)||∇u|dx+





G0

 



A()  G0

r

2−n

|∇u| dx ≤ A() 2

 G0

G0

 n 2−n n − 2 −n 2 2 r u (x) + r |∇u| dx. 2 2

(7.7.11)

7.7 Local Integral Weighted Estimates

167

In virtue of (7.7.4)–(7.7.11), from (7.7.3), we obtain 

 r 2−n |∇u|2 dx + a0

. 

G0



r −n u2 (x)dx +



r 1−n γ (ω)u2 (x)ds ≤ 

G0

0

   ' & ∂u  n n − 2 2  u (x)   d + β + A() + r 2−n |∇u|2 dx+ r= ∂r r= 2 2 



A() 2

    2 |∇u|2 

r=

  + u2 (x)

r=



 d +

G0

δ2 n−2 A() + 2 2



δ1 2

 r 1−n γ (ω)u2 ds +  0

1 4s







r −n u2 dx+



G0

g2 b02 meas + 0 meas ∂ 2s ; δ2 δ1 γ0

∀δ1 , δ2 > 0. (7.7.12)

Let us denote -() = U



 r 2−n |∇u|2 dx +

.



G0

r 1−n γ (ω)u2 ds.

(7.7.13)



0

Then we have 2   1  2   ∂u 1 1 -() = r γ (ω)u2 (r, ω)dσ dr; .U + |∇ω u|2 d dr + ∂r r r 0

0 ∂

(7.7.14) 2  1  2    1 ∂u  1 2 - () = |∇ + u| γ (ω)u2 (, ω)dσ.  d + U  ω  r= ∂r   r=

∂

Applying Lemma 4.11 with χ(ω) ≡ 0, we have    n − 2 2  ∂u   - u (x) . +   d ≤ U (). r= ∂r r= 2 2λ

(7.7.15)

168

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

Now from (7.7.14), we derive 

  |∇u| 

.

2 1     ∂u 2  1 2 d ≤   + |∇ω u|   r= ∂r  r=



2

2

r=

d =





- (). U

(7.7.16)

By the Friedrichs–Wirtinger type inequality (see Theorem F W ) with χ(ω) ≡ 0, we obtain 

A

1 u d ≤ λ(λ + n − 2)

.



 λ(λ + n − 2)

2



A

B

 |∇ω u| d +

2

γ (ω)u dσ = 2

∂

B    1 1 2 2 |∇ u|  d + γ (ω)u  dσ ≤ r= r=  



∂

 - (). U λ(λ + n − 2)

(7.7.17)

, + A() 1 - (). + 1 U 2 λ(λ + n − 2)

(7.7.18)

From (7.7.16) and (7.7.17), we obtain .

A() 2

    2 |∇u|2 

r=

  + u2 (x)



 r=

d ≤

Using inequality (7.6.17) for α = 4 − n and function (7.7.13), we have  .  G0

r −n u2 dx ≤

1 -(). U λ(λ + n − 2)

(7.7.19)

Now, from (7.7.13), (7.7.15), (7.7.18), and (7.7.19), inequality (7.7.12) takes the following form: , + δ2 δ1 -() ≤ − ) U . (1 − β) − c1 (n, λ)A() − 2 2λ(λ + n − 2)   g02  1 b02  meas + meas ∂ 2s ; [1 + c2 (n, λ)A()] U () + 2λ 4s δ2 δ1 γ0 ∀δ1 , δ2 > 0,

(7.7.20)

7.7 Local Integral Weighted Estimates

169

where c1 (n, λ) =

.

n−2 n + ; 2 2λ(λ + n − 2)

c2 (n, λ) = λ +

1 . λ+n−2

Now, we put δ1 = δ2 = q(n, λ)δ, ∀δ > 0

.

with

⎧ ⎨n − 2, if q(n, λ) = ⎩ 2λ2 , if 1+λ2

n > 2; n = 2.

Then, (7.7.20) takes the form -() ≤ [(1 − β) − c1 (n, λ) (A() + δ)] U   g2  c0 - () + b02 + 0 δ −1 2s ; ∀δ > 0; [1 + c2 (n, λ)A()] U 2λ 4sq(n, λ) γ0

.

c0 = max (meas , meas ∂ ) .

(7.7.21)

By Theorem 7.12 with α = 4 − n, .

⎡ C1 ⎣

-(d) ≤ U  &

'



u2 + (1 + r 4−n )b02 dx +

G

⎤ -0 . r 3−n g 2 (x, 0)ds ⎦ ≡ U

(7.7.22)

∂G

Thus, from (7.7.21)–(7.7.22), we obtain the Cauchy problem (CP ) for the differential inequality  .

-() + N ()U -(2) + Q() ≥ 0, 0 <  < d, - () − P()U U U (d) ≤ U0 ,

with 2λ (1 − β) − c1 (A() + δ)) · , N () ≡ 0,  1 + c2 (n, λ)A()   2 g λc0 1 Q() = · δ −1 2s−1 , ∀δ > 0. b2 + 0 · 2sq(n, λ) 0 γ0 1 + c2 (n, λ)A() P() =

.

(CP )

170

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

(1) Case s > λ(1 − β). We choose δ = ε for all ε > 0. We can rewrite the function P() in the following form: P() =

.

K() = 2λ ·

2λ(1 − β) K() − ;  

[(1 − β)c2 (n, λ) + c1 (n, λ)]A() + c1 (n, λ)ε ; 1 + c2 (n, λ)A()

moreover, function K() satisfies the Dini condition at zero: d .

K() d ≤ 2λ[(1 − β)c2 (n, λ) + c1 (n, λ)] · 

0

d

A() d+ 

0

2λc1 (n, λ)

dε < ∞, ∀ε > 0. ε

We calculate τ .

−

  τ  2λ(1−β) K(σ ) τ + dσ ≤ ln P(σ )dσ = −2λ(1 − β) ln +  σ τ





d

K(σ ) dσ ⇒ σ

0

⎞ ⎛ d τ   2λ(1−β)  2λ(1−β) K(σ ) dσ ⎠ = K0 exp ⎝− P(σ )dσ ⎠ ≤ · exp ⎝ ; τ σ τ ⎛

⎞



d

⎛ Q(τ ) exp ⎝−



τ 



0

dτ ·

 ×

  g02 τ 2s−1−ε−2λ(1−β) λc0 K0 2λ(1−β) 2 × dτ ·  b + ≤ 1 + c2 (n, λ)A(τ ) 2sq(n, λ) 0 γ0

d τ



g2 λc0 K0 b02 + 0 P(σ )dσ ⎠ dτ = 2sq(n, λ) γ0 d

2s−1−ε−2λ(1−β)

0

⎞

2λ(1−β)

  g02 d s−λ(1−β) λc0 K0 2 · b0 + ·2λ(1−β), = 2sq(n, λ) γ0 s − λ(1 − β)

if we choose ε = s − λ(1 − β) > 0.

7.7 Local Integral Weighted Estimates

171

Thus, according to Theorem 2.61, we obtain ⎛

d

-() ≤ U -0 exp ⎝− U

.

-0 + U

P(τ )dτ ⎠ +







⎞

g2 λc0 b02 + 0 2sq(n, λ) γ0

d

⎛ Q(τ ) exp ⎝−



 ·

d s+λ(1−β) s − λ(1 − β)

τ

⎞ P(σ )dσ ⎠ dτ ≤





 2λ(1−β)

· K0

d

(7.7.23)

.

From (7.7.23) and (7.7.19), taking into account (7.7.13), (7.7.22), and by assumption (D), we derived the statement of (7.7.1) for s > λ(1 − β). 1 (2) Case s = λ(1 − β). In this case, as δ > 0, we take δ() = 1 . Then we 2λc1 (n,λ) ln 

obtain the Cauchy problem (CP ), with P() =

.

2λ(1 − β) L() − ;  

1 1 [(1 − β)c2 (n, λ) + c1 (n, λ)]A() + · 1; 1 + c2 (n, λ)A() 1 + c2 (n, λ)A() ln    g2 1 1 λ2 c0 c1 b02 + 0 · · 2s−1 ln . Q() = sq(n, λ) γ0 1 + c2 (n, λ)A() 

L() = 2λ ·

We calculate τ .

−

P(σ )dσ ≤ ln

 2λ(1−β) τ

d + 2λ[(1 − β)c2 (n, λ) + c1 (n, λ)]



A(τ ) dτ + τ

0

τ

⎛ exp ⎝−

τ 



⎞

 2λ(1−β) ln 1  ⎠ P(σ )dσ ≤ K1 , τ ln τ1

⎧ ⎨

d

K1 = exp 2λ[(1 − β)c2 (n, λ) + c1 (n, λ)] ⎩ ⎛ exp ⎝−

d 

⎞ P(σ )dσ ⎠ ≤ K1

dσ σ ln σ1

; ⇒

where ⎫ ⎬

A(τ ) dτ ; ⎭ τ

0

 2λ(1−β) d

1 1 ln , 0 <  < d <  e

172

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

and d .

⎛ Q(τ ) exp ⎝−



τ

⎞ λ2 c0 c1 K1 P(σ )dσ ⎠ dτ ≤ sq(n, λ)

 b02



d

g2 + 0 γ0



1 · 2λ(1−β) ln × 

τ 2λ(1−β)−1τ −2λ(1−β)





λ2 c0 c1 K1 sq(n, λ)

b02

g2 + 0 γ0

 ·

2λ(1−β)

1 ln 

d

1 ln τ1

1 ln dτ = τ

dτ ≤ τ



λ2 c0 c1 K1 sq(n, λ)

 b02

g2 + 0 γ0

 · 2λ(1−β) ln2

1 . 

Thus, according to Theorem 2.61, we obtain ⎛ -() ≤ U -0 exp ⎝− U

d

.

⎞ P(τ )dτ ⎠ +







2 g2 -0 + λ c0 c1 b02 + 0 U sq(n, λ) γ0



d

⎛ Q(τ ) exp ⎝−



τ

⎞ P(σ )dσ ⎠ dτ ≤



· K1

 2λ(1−β) d

ln2

1 . 

(7.7.24)

From (7.7.24) and (7.7.19), taking into account (7.7.13), (7.7.22) and by assumption (D), we derived the statement of (7.7.1) for s = λ(1 − β). (3) Case s < λ(1 − β). Similarly to case (1), we get the Cauchy problem (CP ) with any δ > 0. We calculate ⎛ .

exp ⎝−

τ 

⎞ P(σ )dσ ⎠ ≤

 2λ(1−β−c1 δ) τ

⎛ · exp ⎝2λc1

d

⎞ A(σ ) ⎠ dσ = σ

0

K2

 2λ(1−β−c1 δ) τ

,

7.7 Local Integral Weighted Estimates

d .

⎛ Q(τ ) exp ⎝−



τ 

d τ

173

⎞

  2 g λc0 K2 P(σ )dσ ⎠ dτ ≤ b2 + 0 δ −1 · 2λ(1−β−c1 δ) × 2sq(n, λ) 0 γ0

2s−1 −2λ(1−β−c1 δ)

τ



  g02 λc0 K2 2 dτ ≤ δ −1 · 2λ(1−β−c1 δ) × b + 2sq(n, λ) 0 γ0 d 2s−2λ(1−β−c1δ) − 2s−2λ(1−β−c1δ) . 2s − 2λ(1 − β − c1 δ)

Now we choose δ = d .

⎛ Q(τ ) exp ⎝−



λ(1−β)−s 2c1 (n,λ)

τ

> 0. Then we derive ⎞

P(σ )dσ ⎠ dτ ≤



λ2 c0 c1 K2 sq(n, λ)[λ(1 − β) − s]2

 b02

g2 + 0 γ0

 2s

as well as, since in our case 2λ(1 − β − c1 δ) = λ(1 − β) + s > 2s and 0 <  < d, ⎛ .

exp ⎝−

d

⎞ P(σ )dσ ⎠ ≤ K2

 2s d

.



Thus, according to Theorem 2.61, we obtain ⎛ -() ≤ U -0 exp ⎝− U

d

.

 -0 + U

⎞

d

P(τ )dτ ⎠ +



λ2 c0 c1 K2 sq(n, λ)[λ(1 − β) − s]2



⎛ Q(τ ) exp ⎝−



b02 +

g02 γ0

τ

⎞ P(σ )dσ ⎠ dτ ≤



 · K2

 2s d

.

(7.7.25)

From (7.7.25) and (7.7.19), taking into account (7.7.13), (7.7.22) and by assumption (D), we derived the statement of (7.7.1) for s < λ(1 − β).   Theorem 7.14 Let u be a weak solution of problem (WQL), λ be the smallest positive eigenvalue of problem (EVP), and assumptions (A)–(D) be satisfied with function A(r) being continuous at zero function, but not Dini continuous. Then for all ε > 0, there are

174

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

d ∈ (0, 1) and a constant Cε > 0, which depends only on n, λ, γ0 such that the inequality .

 

r −n u2 + r 2−n |∇u|2 dx + r 1−n γ (ω)u2 ds 



G0

0



2λ(1−β−ε), if s ≥ λ(1 − β), ≤ Cε u22,G + b02 + g02 if s < λ(1 − β), 2s ,

(7.7.26)

holds for all  ∈ (0, d). Proof Similarly to Theorem 7.13, we derive the Cauchy problem (CP ). By the continuity of A(r), we can choose d > 0 such that A(d) < δ for any δ > 0. Thus τ .

τ τ A(σ ) dσ ≤ A(d) ln ≤ δ ln , 0 <  < τ < d < 1. σ  



Rewrite P(σ ) =

2λ σ

·

P(σ ) =

.

(1−β)−c1 (A(σ )+δ)) 1+c2 (n,λ)A(σ )

so

2λ(1 − β − c1 δ) 2λ c1 + c2 (1 − β − c1 δ) − · · A(σ ), σ σ 1 + c2 A(σ )

and assuming c1 δ < 1 − β, we calculate τ .

−

τ P(σ )dσ = −2λ(1 − β − c1 δ) ln + 2λ[c1 + c2 (1 − β − c1 δ)] 



τ

A(σ ) dσ ≤ σ



 2λ(1 − β − ε) ln , τ

where ε = [2c1 + c2 (1 − β − c1 δ)]δ > 2c1 δ > 0,   1−β . ∀δ ∈ 0, c1

Hence, it follows that ⎛ .

exp ⎝−

τ 

⎞ P(σ )dσ ⎠ ≤

 2λ(1−β−ε) τ

, ∀ε > 0,

7.7 Local Integral Weighted Estimates

175

and d .

⎛ Q(τ ) exp ⎝−

τ



⎞ P(σ )dσ ⎠ dτ ≤



  g02 λc0 2 · 2λ(1−β−ε)× b + 2sδq(n, λ) 0 γ0 d τ 2s−1−2λ(1−β−ε)dτ ≤



g2 λc0 b02 + 0 2sδq(n, λ) γ0



 · 2λ(1−β−ε) ·

d 2s−2λ(1−β−ε) − 2s−2λ(1−β−ε) . 2s − 2λ(1 − β − ε)

Thus, according to Theorem 2.61, we obtain 2λ(1−β−ε) -() ≤ U -0  + U d   g02 d 2s−2λ(1−β−ε) − 2s−2λ(1−β−ε) λc0 2 . · 2λ(1−β−ε) · b0 + 2sδq(n, λ) γ0 2s − 2λ(1 − β − ε)

.

Now, if s ≥ λ(1 − β) , then from above we have  -() ≤ U

.

-0 U d 2λ(1−β−ε)

   g02 λc0 d 2s−2λ(1−β−ε) 2 + · b + × 2sδq(n, λ) 0 γ0 2s − 2λ(1 − β − ε) 2λ(1−β−ε).

(7.7.27)

1−β− s

But, if s < λ(1 − β) , then we choose 0 < δ < 2c1 λ , and therefore, we can take   ε ∈ 2c1 δ, 1 − β − λs ⇒ 2λ(1 − β − ε) > 2s, and from above, we get  -() ≤ .U

-0 + U

    2s g02 λc0 d 2s 1 2 . · b0 + · 2sδq(n, λ) γ0 2λ(1 − β − ε) − 2s d

(7.7.28)

From (7.7.27)–(7.7.28), taking into account (7.7.13), (7.7.22) and by assumption (D), we derived the statement of our theorem.   Now, we can make a correction of Theorem 7.14 in the case A(r) ∼

1 ln 1r

.

Theorem 7.15 Let u be a weak solution of problem (WQL), λ be the smallest positive eigenvalue of problem (EVP),

assumptions (A)–(D) be satisfied with function A(r) ∼ and 1 1 . Then there are d ∈ 0, 1 e and a constant C > 0, which depends only on n, λ, γ0 ln

r

176

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

such that the inequality .

 

r −n u2 + r 2−n |∇u|2 dx + r 1−n γ (ω)u2 ds ≤ 



G0

0

⎧

 1 ξ(λ) ⎨2λ(1−β), C u22,G + b02 + g02 ln · ⎩2s , 

if s > λ(1 − β), if s ≤ λ(1 − β)

(7.7.29)

holds for all  ∈ (0, d). Proof Choosing as δ > 0 function δ() =

1 2λc1 ln

1 

0 <  < d, similarly to

,

Theorem 7.13, we derive the Cauchy problem (CP ) with P() =

.

2λ(1 − β) T () − ;  

1 + 2λ[(1 − β)c2 (n, λ) + c1 (n, λ)] 1 · 1; N () ≡ 0, 1 + c2 (n, λ)A() ln    2 g 1 1 λ2 c0 c1 b2 + 0 · · 2s−1 ln . Q() = sq(n, λ) 0 γ0 1 + c2 (n, λ)A() 

T () =

We calculate τ .

−

P(σ )dσ ≤ ln

 2λ(1−β) τ

+



 {1 + 2λ[(1 − β)c2 (n, λ) + c1 (n, λ)]} ln ⎛ exp ⎝−

τ

⎞ P(σ )dσ ⎠ ≤



 2λ(1−β) τ



ln 1

 ; ⇒

ln τ1 1 ξ(λ)

ln  ln τ1

,

where

ξ(λ) = 1 + 2λ[(1 − β)c2 (n, λ) + c1 (n, λ)]; ⎞ ⎛ d     2λ(1−β) ln 1 ξ(λ) 1  , 0 λ(1 − β), if s ≤ λ(1 − β).  

Thus we derived the statements of (7.7.29).

7.8

The Power Modulus of Continuity at the Conical Point for Weak Solutions

Proof of Theorem 7.3 Let us introduce the function ⎧ λ(1−β) , ⎪ if s > λ(1 − β), ⎨ 1 λ(1−β) .ψ() = ln  , if s = λ(1 − β),  ⎪ ⎩ s  , if s < λ(1 − β).

(7.8.1)

By Theorem 7.11, we have the inequality .

& n 2− n(p+2) 2p b (x) sup |u(x)| ≤ C − 2 u(x)2,G +  0 0

/2

G0

p > m# > 2

with

m# =

 2p p+2 ,G0

+ ' g(x, 0)∞,  , 0

⎧ ⎨

2n n−2 , ⎩ 2m , m−2

if n > 2; m > 2, if n = 2;

 ∈ (0, d),

d ∈ (0, 1). (7.8.2)

Further, by Theorem 7.13, we have



.

− n2

u(x)2,G = 

− n2

 2

u (x)dx

0

 G0

 12

 ≤

r

−n 2

u (x)dx

 12

≤

 G0

C (M0 + b0 + g0 ) ψ().

(7.8.3)

7.8 The Power Modulus of Continuity at the Conical Point for Weak. . .

179

By assumptions (B4), (D), 

.

2− n(p+2) 2p

b0 (x)

+ g(x, 0)∞,  ≤ C(b0 + g0 )ψ().

 2p p+2 ,G0

0

(7.8.4)

Thus, from (7.8.2)–(7.8.4), it follows .

sup |u(x)| ≤ C (M0 + b0 + g0 ) ψ() /2

G/4

for  ∈ (0, d). Finally, setting |x| = 13 , we obtain the desired estimation (7.1.1). To estimate the modulus of gradient of the problem (WQL) solution near a conical 2  2 point, we introduce two sets G/4 and G/2 ⊂ G/4 ,  > 0. We make the following transformation: x = x  , u(x  ) = v(x  )ψ(). Then the function v(x  ) satisfies in the weak sense problem ⎧

d ij (x  )v  + a r  −2 v(x  )+ ⎪ − a ⎪  0 x ⎪ dxi j ⎨

2 . x ∈ G21/4 , b x  , v(x  )ψ(), ψ() vx  = 0, ψ()  ⎪ ⎪ ⎪ ⎩ ∂v + 1 γ (ω)v(x  ) =  g(x  , ψ()v(x  )), x ∈  2 . 1/4 ψ()

 |x  |

(7.8.5)

∂ν

Applying assumption about a priori estimate of the modulus of gradient of the problem (7.8.5) solution .

max |∇  v(x  )| ≤ M1 .

x  ∈G11/2

(7.8.6)

Returning to the variable x and the function u(x), we obtain from the above |∇u(x)| ≤ M1 −1 ψ(),

.

for  ∈ (0, d). Setting |x| = 23 , we obtain the desired estimation (7.1.2).

 

Proof of Theorem 7.4 Repeating verbatim the proof of Theorem 7.3 with function  ψ() =

.

λ(1−β−ε), if s ≥ λ(1 − β), s , if s < λ(1 − β),

and applying Theorem 7.14, we prove Theorem 7.4.

 

180

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

Proof of Theorem 7.5 Repeating verbatim the proof of Theorem 7.3 with function    1 ξ(λ) λ(1−β) , if s > λ(1 − β), .ψ() = ln ·  if s ≤ λ(1 − β), s ,  

and applying Theorem 7.15, we prove Theorem 7.5.

7.9

Example

In the cone % $ ω π 0 ⊂ 0, , ϕ ∈ (0, 2π) , G0 = (r, ω, ϕ) : r > 0, ω ∈ 0, 2 2

.

with the lateral surface $ % ω0 , ϕ ∈ (0, 2π); ω0 ∈ (0, π) .0 = (r, ω, ϕ) : r > 0, ω = 2 and .O ∈ ∂G0 , we consider the three-dimensional problem for the Laplace operator with absorption:  .

u + βu−1 |∇u|2 = a0 r −2 u, x ∈ G0 , ∂u ∂u 1 x ∈ 0 , ∂ n + χ ∂r + r γ u = 0,

(7.9.1)

where .χ ≥ 0, .γ > 0 are constants and 0 ≤ a0 < (1 + β)

.

2

+ ,

0 ≤ β < 1,

and . is defined by (7.9.6)–(7.9.7). Now, we rewrite the equation of problem (7.9.1) using formulas, which are related to the spherical coordinates .(r, ω, ϕ) centered at the point .O: .

1 ∂u 1 ∂ 2u ∂ 2 u 2 ∂u 1 ∂ 2u · + cot ω + + · · + · ∂r 2 r ∂r r 2 ∂ω r 2 ∂ω2 r 2 sin2 ω ∂ϕ 2 1     2 2 ∂u 2 1 ∂u 2 1 ∂u −1 = a0 r −2 u. + βu + 2 + ∂r r ∂ω r 2 sin2 ω ∂ϕ

Using the separation of variables method, we want to find the exact solution of this problem in the form .u(r, ω, ϕ) = r ψ(ω)z(ϕ). Then, the equation of the problem for the function

7.9 Example

181

u takes the form 1 .

2

2

sin ω (1 + β)

+

ψ ψ  + cot ω + +β ψ ψ



ψ ψ

2

2

− a0 =

z −β − z

  2 z = μ2 . z

(7.9.2)

By (7.9.2) and the boundary condition of (7.9.1), we obtain the following problem for .ψ(ω): ⎧   2 ⎪ ⎨ ψ ψ +&ψ ψ cot ω + βψ + '   2 (1 + β) 2 + − a0 − μ2 ψ 2 = 0, ω ∈ 0, ω20 , . sin ω ⎪     ⎩  ψ (0) = 0; ψ  ω20 + [ χ + γ ] ψ ω20 = 0.

(7.9.3)

1

Making the transformation .ψ(ω) = y 1+β (ω), the equation of problem (7.9.3) takes the form + y  + y  cot ω + (1 + β)

.

2

+

− a0 −

,

μ2

(1 + β)y = 0.

sin2 ω

We multiply this equation by .sin ω1 and write it in the following form: (py  ) − qy + $ · sin ω · y = 0,

(7.9.4)

.

p ≡ sin ω > 0, q ≡

& μ2 (1 + β) , $ = (1 + β) sin ω ω π 0 ⊂ 0, . ω ∈ 0, 2 2

2

+

' − a0 (1 + β).

By (7.9.4) and the proof of Theorem 3.5, Chap. 2, Sect. 3.1.2, we know that if the coefficient q changes everywhere in the same sense, every eigenvalue of (7.9.4) changes in this same sense. Thus, for .μ = 0, we get the equation for the smallest positive eigenvalue of (7.9.4). Finally, putting & .

(1 + β)

2

+

' − a0 (1 + β) = τ (τ + 1), τ > 0,

problem (7.9.3) takes the following form:  .

  ω ∈ 0, ω20 , y  + y  cot ω + τ (τ + 1)y = 0,     y  (0) = 0; y  ω20 + ( χ + γ )(1 + β)y ω20 = 0.

(7.9.5)

182

7 Behavior of Weak Solutions to the Conormal Problem for Elliptic Weak. . .

We obtained problem .(EVP∗3 ) with . θ = τ (τ + 1) and . μ = 0, which was studied in Chap. 3, Sect. 3.1.2. Therefore, we have: y(ω) = Cτ1/2 (cos ω)

.

and the equation for . τ,

.

sin

:

3 1 ω0 ω0 ω0 · Cτ2−1 cos = ( χ + γ ) · Cτ2 cos , 2 2 2

(7.9.6)

where τ=

.

1 2

;

1 + 4(1 + β) + 4(1 + β)2

2

 − 4(1 + β)a0 − 1 .

(7.9.7)

By (7.9.2) and .μ = 0, we have the equation for .z(ϕ) : .

z +β z

  2 z = 0, z

⇒

z(ϕ) ≡ C = const.

Finally, we have found the solution of our problem in the form u(r, ω, ϕ) = Cr

.

7.10



Cτ1/2(cos ω)

1 1+β

.

Notes

The conormal problem for elliptic quasi-linear equations in smooth domains has been intensely studied recently (see Section 8.3–8.4, 11.9 and Notes [89] as well as Notes to Chapter 9 [33]). We investigated the conormal problems for elliptic quasi-linear equations with triple degeneration in a domain with boundary edge in Chapter 9 [33]. In our chapter here, we used other methods and obtained the best sharp estimates of solutions because of the leading part to the .(WQL) equation is linear.

8

Behavior of Strong Solutions to the Degenerate Oblique Derivative Problem for Elliptic Quasi-linear Equations in a Neighborhood of a Boundary Conical Point

8.1

Setting of the Problem

Let .G ⊂ Rn be a bounded domain with boundary .∂G that is a smooth surface everywhere except at the origin .O ∈ ∂G, and near .O, it is a conical surface. We consider the following quasi-linear elliptic value problem: ⎧ ⎨ .

a ij (x, u, ux )uxi xj + a(x, u, ux ) = 0, a ij = a j i , x ∈ G,

⎩B[u] ≡

∂u ∂ n

+ χ(x) ∂u ∂r +

1 |x| γ (x)u

= g(x),

x ∈ ∂G\O,

(QL)

where .n denotes the unite exterior normal vector to .∂G\O, .(r, ω) are spherical coordinates in .Rn with pole .O, and repeated indices are understood as summation from 1 to n. Definition 8.1 A function .u(x) is called a strong solution of problem .(QL) if .u(x) ∈   2,q Wloc (G) ∩ W 1 (G) ∩ C 0 G , .q > n, and satisfies the equation for almost all .x ∈ Gε for all .ε > 0 as well as the boundary condition in the sense of traces on .∂G\O. We assume that .M0 = max |u(x)| is known (see, e.g., Corollary 8 [122], Lemma 8.1 [89], Theorem 13.1 x∈G

[86]). Let .M = {(x, u, z) : x ∈ G, u ∈ R, z ∈ Rn }. With regard to problem (QL), we assume that the following conditions are satisfied on .M: (A) .a ij (x, u, z) ∈ W 1,q (M), .q > n; .a(x, u, z), ∂a(x,u,z) are Caratheodory functions. ∂u

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Borsuk, Oblique Derivative Problems for Elliptic Equations in Conical Domains, Frontiers in Elliptic and Parabolic Problems, https://doi.org/10.1007/978-3-031-28381-9_8

183

184

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

(B) The uniform ellipticity ν|ξ |2 ≤

n 

.

a ij (x, u, z)ξi ξj ≤ μ|ξ |2 ,

∀ξ ∈ Rn ,

i,j =1 j

with the ellipticity constants .μ ≥ ν > 0; .a ij (0, 0, 0) = δi , .i, j = 1, . . . , n—the Kronecker symbol. (C) . ∂a(x,u,z) ≤ 0. ∂u   (D) .γ (ω), χ(ω) ∈ C 1 ; there exist numbers .s > 1, .χ0 ≥ 0, .γ0 > tan ω20 ≥ 0 and .γ1 ≥ γ (ω) ≥ γ0 > 0, .0 ≤ χ(ω) ≤ χ0 as well as nonnegative constants .μ1 , .k1 , .g0 , .g1 q and functions .b(x), f (x) ∈ Lloc (G), .q ≥ n such that the inequalities    ∂a(x, u, z)   ≤ μ1 |z|2 + b(x)|z| + f (x), |a(x, u, z)| +   ∂u

.

0 ≤ b(x), f (x) ≤ k1 |x|s−2, |g(x)| ≤ g0 |x|s−1 , |∇g| ≤ g1 |x|s−2 hold. (E) The problem (QL) coefficients satisfy such conditions that guarantee .u ∈ C 1+-(G ) and the existence of the local a priori estimate |u|1+-,G ≤ M1 , - ∈ (0, 1),

.

for any smooth .G ⊂⊂ G\O (see Theorems 13.13 and 13.14 [86], Lemma 2.3 [84, 90]). Proposition 8.2 (The Lieberman Local Maximum Principle (See Theorem 3.3 [89], Theorem 4.3 [88]; See As Well [87, 90])) Let G be a bounded domain in .RN with the d d 1 .C -boundary .∂G \  and .G be a convex rotational cone with the vertex at .O and the 0 0 π aperture .ω0 ∈ ( 2 , π). Let .u(x) be a strong solution of the problem .(QL) with .|u| ≤ M0 . Suppose the conditions .(A), (B), (C) are satisfied. In addition, suppose that there are nonnegative number .μ1 and nonnegative functions .b(x) ∈ Ls (G), s > n, f (x) ∈ Ln (G), such that |a(x, u, z)| ≤ μ1 |z|2 + b(x)|z| + f (x).

.

Suppose finally that .g ∈ L∞ (∂G), γ (ω) ≥ γ0 > 0, 0 < χ(ω) ≤ χ0 .

8.1 Setting of the Problem

185

Then for any .q > 0 and .σ ∈ (0, 1), there is a constant C = C(ν, μ, μ1 , M0 , γ0 , χ0 , ω0 , n, p, G, bLs (G) , f Ln (G) ) such that

.



⎛

⎜ 1 sup |u(x)| ≤ C ⎜ ⎝ meas GR σR

.

G0

0



⎞1/q ⎟ |u|q dx ⎟ ⎠

GR 0

+

 + R f Ln (GR ) + gL∞ ( R ) , R ∈ (0, d). 0

0

The main result is the following statement about power modulus of continuity: Theorem 8.3 Let .u(x) be a strong solution of problem (QL) and .λ > 1 be the smallest positive eigenvalue of eigenvalue problem .(EVP). Suppose that assumptions .(A)–.(E) are -0 , .C -1 that do not depend on u, but satisfied. Then there exist numbers .d ∈ (0, e−1 ), .C depend only on n, .γ0 , .χ0 , .λ, .ν, .μ, .μ1 , s, .k1 , q, .g0 , .g1 , .M0 , .M1 and domain G, such that: • ⎧ ⎪ |x|λ , if s > λ, ⎪ ⎨ 3 1 λ -0 |x| ln 2 .|u(x)| ≤ C |x| , if s = λ, ⎪ ⎪ ⎩ s |x| if, s < λ, ⎧ ⎪ |x|λ−1, ⎪ ⎨ -1 |x|λ−1 ln 32 1 , .|∇u(x)| ≤ C |x| ⎪ ⎪ ⎩ s−1 |x| , ⎧ ⎨4 + q(λ − 2) > 0 for s > λ, • If . ⎩4 + q(s − 2) > 0 for s < λ,

d/2

x ∈ G0 ;

(8.1.1)

if s > λ, if s = λ,

d/2

x ∈ G0 .

(8.1.2)

if s < λ,

  2 G0 , and there exist numbers then .u(x) ∈ Vq,4−n

-2 , which do not depend on u, but depend only on n, .γ0 , .χ0 , .λ, .ν, .μ, .μ1 , s, .k1 , d > 0, .C q, .g0 , .g1 , .M0 , .M1 and domain G, such that

.

  .u(x) 2 Vq,4−n G0

⎧ λ−2+ 4 q, ⎪  ⎪ ⎨ 4 -2 λ−2+ q ln3 1 , ≤C  ⎪ ⎪ ⎩ s−2+ q4  ,

if s > λ, if s = λ, if s < λ,

0 <  < d/2.

(8.1.3)

186

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . . n 2−λ ,

• If . 1 < λ < 2, .q >

then

 d/2  u(x) ∈ C λ G0 , if s > λ, .

 d/2 u(x) ∈ C λ−ε G0 , ∀ε > 0, if s = λ,

(8.1.4)

 d/2 u(x) ∈ C s G0 , if s < λ.

8.2

The Barrier Function. The Preliminary Estimate of the Solution Modulus

Let us define the following linear elliptic operator: L0 ≡ a ij (x)

.

∂2 ; a ij (x) = a j i (x), x ∈ Gd0 , ∂xi ∂xj

where νξ 2 ≤ a ij (x)ξi ξj ≤ μξ 2 , ∀x ∈ Gd0 , ∀ξ ∈ Rn and ν, μ = const > 0

.

and the boundary operator 1 ∂ ∂ γ (x), B≡ → + χ(x) + ∂r |x| ∂− n

.

0 ≤ χ(x) ≤ χ0 ,

γ (x) ≥ γ0 > 0, x ∈ 0d \O.

Lemma 8.4 (Existence of the Barrier Function) Let .0 ≤ χ(ω) ≤ χ0 ; γ1 ≥ γ (ω) ≥ γ0 > tan ω20 > 0. Fix the numbers .g0 ≥ 0, d ∈ (0, 1). There exist .h > 0 depending only on .ω0 , the number . 0 ∈ (0, γ0 cot ω20 − 1), a number .B > 0, and a function .w(x) ∈ C 1 (G0 ) ∩ C 2 (G0 ) that depend only on .ω0 , the ellipticity constants .ν, μ of the operator .L0 , and the quantities .χ0 , γ0 , g1 , such that for any . ∈ (0; 0 ) the following inequalities hold: L0 [w(x)] ≤ −νh2 |x|

.

−1

; x ∈ Gd0 ;

B[w(x)] ≥ g0 |x| ; x ∈ 0d \ O;

.

0 ≤ w(x) ≤ C0 ( 0 , B, ω0 )|x|

.

+1

; x ∈ Gd0 ;

|∇w(x)| ≤ C1 ( 0 , B, ω0 )|x| ; x ∈ Gd0 .

.

(8.2.1) (8.2.2) (8.2.3) (8.2.4)

8.2 The Barrier Function. The Preliminary Estimate of the Solution Modulus

187

Proof Let .(x, y, x  ) ∈ Rn , where .x = x1 , y = x2 , x  = (x3 , . . . , xn ). In .{x1 ≥ 0}, we consider the cone K with the vertex in .O, such that .K ⊃ Gd0 . (We recall that .Gd0 ⊂ {x1 ≥ 0}.) Let .∂K be the lateral surface of K and let .∂K ∩ yOx = ± be .x = ±hy, where ω0 .h = cot 2 , 0 < ω0 < π, such that in the interior of K the inequality .x > h|y| holds. We shall consider the following function: w(x; y, x  ) ≡ x

−1

.

+1

(x 2 − h2 y 2 ) + Bx

, (8.2.5)

∈ (0; 1), B > 0.

with some

Let the coefficients of the operator .L0 be .a 2,2 = a, a 1,2 = b, a 1,1 = c. Then we have L0 w = awyy + 2bwxy + cwxx ,

(8.2.6)

.

where νη2 ≤ aη12 + 2bη1η2 + cη22 ≤ μη2

and (8.2.7)

.

=

η2

η12

+ η22 ;

∀η1 , η2 ∈ R.

Let us calculate the operator .L0 on the function (8.2.5). For .t = yx , |t| < h1 , we obtain L0 w = −h2 x

.

−1

φ( ),

where φ( ) = 2a − 4bt + 4bt − ch−2 (1 + B)(

.

= c(t 2 − h−2 (1 + B))

2

2

+ ) + ct 2

2

− 3ct 2 + 2ct 2 =

+ (4bt − ch−2 (1 + B) − 3ct 2 ) + 2(ct 2 − 2bt + a)

and −2

c(t − h

.

2



y2 1+B (1 + B)) = c 2 − x h2

 ≤ −c

B < 0. h2

Because of (8.2.7), we have .φ(0) = 2(ct 2 − 2bt + a) ≥ 2ν, and since .φ( ) is a square function, there exists the number . 0 > 0 depending only on .ν, μ, h such that .φ( ) ≥ ν for . ∈ [0; 0 ]. Therefore, we obtain (8.2.1). Now, let us notice that ± : x = ±hy, h = cot

.

ω0 , 0 < ω0 < π. 2

(8.2.8)

188

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

Then we have ⎧ ⎨x = r cos ω0 , 2 .on + : ⎩y = r sin ω0 2

⎧ ⎨x = r cos ω0 , 2 on − : ⎩y = −r sin ω0 2 sin

⎧ ⎨ (

n, x) = ⎩ (

n, y) =

ω0 π 2 + 2 , ω0 2

⎧ ⎨ (

n, x) = π2 + ω20 , ⎩ (

n, y) = π + ω20

1 ω0 =√ , 2 1 + h2

cos

⎧ ⎨ ( r , x) = ⎩ ( r , y) =

⎧ ⎨ ( r , x) = ⎩ ( r , y) =

ω0 2 , ω0 π 2 − 2 ω0 2 , ω0 π 2 + 2

h ω0 =√ . 2 1 + h2

Therefore, we obtain wx = (1 + )x (1 + B) − ( − 1)h2 y 2 x

.

−2

 ⇒ wx ± = [2 + B(1 + )]x , (8.2.9)

wy = −2h2 yx

−1

 ⇒ wy  = ∓2hx . ±

Because of      ∂w     . = wx cos (

n, x) + wy cos (

n, y)  ∂ n ± ± ±      ∂w     . = wx cos ( r , x) + wy cos ( r , y)  ∂ r ± ± ± and (8.2.9), we get  ∂w  h . = −r  ∂ n ± (1 + h2 )

+1 2

[2(1 + h2 ) + B(1 + )].

 ∂w  . = (1 + )Br h1+ . ∂ r ± Hence, it follows that   B[w]

.

±

h ≥   1 + h2

+1 2

( ) r Bhγ0 − B(1 + ) − 2(1 + h2 ) .

8.2 The Barrier Function. The Preliminary Estimate of the Solution Modulus

Since .h >

1 γ0

  B[w]

.

for . ≤

≥

d ±

0,

we obtain (

h 0r  1 + h2

189

0 +1 2

B(hγ0 − 1 −

0)

) − 2(1 + h2 ) ≥ g0 r , 0 < r < d < 1

if we choose B≥

.

$ g0 1 + h2  h

0 +1 2

0

% + 2(1 + h2 ) ·

1 hγ0 − 1 −

(8.2.10)

. 0

(It should be pointed out that we can choose, if necessary, . 0 so small that . 0 < hγ0 − 1.) Now will show (8.2.3). Let us rewrite the function (8.2.5) in spherical coordinates. Recalling that .h = cot ω20 , we obtain w(x; y, x  ) = (1 + B)(r cos ω)1+ − h2 r 2 sin2 ω(r cos ω) −1 =   ( ω0 ω0 ) σ (ω) 1+ −1 2 =r cos ω B cos ω + , , ∀ω ∈ − ω0 , 2 2 2 sin 2

.

where σ (ω) = sin

.

ω

0 − ω · sin +ω . 2 2

ω

0

We find . σ  (ω) = − sin 2ω and . σ  (ω) = 0 for ω = 0. Now we see that .σ  (0) = −2 cos 0 = −2 < 0. In this way, we have .

max

ω∈[−ω0 /2,ω0 /2]

σ (ω) = σ (0) = sin2

ω0 , 2

and therefore, 

w(x; y, x ) ≤ r

1+

.

≤ r 1+

cos

−1

 B+

2

ω(B cos ω + 1) ≤ r

1+

cos

+1

 1 . cos2 ω

Hence, (8.2.3) follows. Finally, (8.2.4) follows in virtue of (8.2.9).

 ω B+

1 cos2 ω



 

Now we shall derive preliminary estimates of .|u(x)| and .|∇u(x)| for a strong solution of problem (QL) near a conical boundary point.

190

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

Theorem 8.5 Let .u(x) be a strong solution of problem (QL) and assumptions .(A)−(D) be satisfied. Then there exist number .d ∈ (0, 1) and . 0 > 0 depending only on ellipticity constants .ν, .μ and .μ1 , n, .ω0 , .k1 , s, .γ0 , .χ0 , .g0 , .M0 and domain G such that .

|u(x) − u(0)| ≤ C0 |x|

+1

,

0
0 depends only on .ν, .μ, .μ1 , n, .k1 , s, .γ0 , .g0 , .M0 , and domain G but does not depend on u. Proof Let us take the linear elliptic operator - ≡ a ij (x) L

.

∂2 ∂ + a i (x) , x ∈ G, ∂xi ∂xj ∂xi

where a ij (x) = aij (x, u(x), ux (x)); a i (x) = b(x)|∇u(x)|−1uxi (x).

.

(8.2.12)

Here we suppose that .a i (x) = 0, i = 1, . . . , n, in such points x, where .|∇u(x)| = 0. Let us take the barrier function (8.2.5) and define the auxiliary function .v(x) as follows: v(x) = −1 + exp(ν −1 μ1 (u(x) − u(0))).

.

(8.2.13)

For those functions, we shall show that ⎧ ⎪ ⎪ ⎨ L(Aw(x)) ≤ Lv(x), .

⎪ ⎪ ⎩

x ∈ Gd0 ;

B[Aw(x)] ≥ B[v(x)], x ∈ 0d \ O; Aw(x) ≥ v(x),

(8.2.14)

x ∈ d ∪ O

and further use: Proposition 8.6 (The Comparison Principle (See Proposition 10.16 [33])) Let .Gd0 be a convex rotational cone with vertex at .O and the aperture .ω0 ∈ ( π2 , π). Let - be uniformly elliptic in .Gd with the ellipticity constants .0 < ν ≤ μ, .a i (x) ∈ L∞ (Gd ), .L loc 0 0 in .Gd0 . Let .0 ≤ χ(x) ≤ χ0 and .γ (x) ∈ L∞ (0d ), γ (x) ≥ γ0 > 0 on .0d . Suppose that v 2,N and w are functions in .Wloc (Gd0 ) ∩ C 0 (Gd0 ) satisfying (8.2.14). Then .v(x) ≤ Aw(x) in

Gd0 .

.

8.2 The Barrier Function. The Preliminary Estimate of the Solution Modulus

191

- on the function (8.2.13). We obtain Let us calculate the operator .L Lv(x) ≡ ν −1 μ1 (a ij (x)uxi xj + ν −1 μ1 a ij (x)uxi uxj + b(x)|∇u(x)|)× ( × exp(ν −1 μ1 (u(x) − u(0))) = ν −1 μ1 b(x)|∇u(x)| − a(x, u(x), ux (x)) . ) + a ij (x)uxi uxj exp(ν −1 μ1 (u(x) − u(0))) ≥ − ν −1 μ1 f (x) exp(2ν −1 μ1 M0 ) in virtue of assumptions .(C) and .(D). Since .f (x) ≤ k1 r s−2 , we obtain Lv(x) ≥ −ν −1 μ1 k1 r s−2 exp(2ν −1 μ1 M0 ), x ∈ Gd0 .

(8.2.15)

.

- on the barrier function (8.2.5). Let the number . Let us calculate the operator .L such that Lemma 8.4 holds, and suppose . satisfies the inequality

0

be

≤ min( 0 , s − 1).

0
−1,

(8.2.21)

8.2 The Barrier Function. The Preliminary Estimate of the Solution Modulus

193

Let us define the following function . f (v) = v − (1 + v) ln(1 + v), v > −1; we get f  (v) = − ln(1 + v), f  (v) = −

.

1 . 1+v

We see that .f  (v) = 0 ⇔ v = 0 and .f  (0) = −1 < 0. Then we obtain . max f (v) = v>−1

f (0) = 0 ⇒   B[v]

.

d \O ±

≤ ν −1 μ1 exp(2ν −1 μ1 M0 )g0 r s−1 .

(8.2.22)

Taking into account (8.2.18) and (8.2.22), we choose A ≥ ν −1 μ1 g0−1 exp(2ν −1 μ1 M0 ),

(8.2.23)

.

and we obtain d B[Aw] ≥ B[v] on ± \ O,

.

r ∈ (0, d],

∈ (0, s − 1].

If .u(0) < 0, instead of the function .v(x), defined by (8.2.13), we have to take the function z(x) := 1 − exp(−ν −1 μ1 (u(x) − u(0))).

.

(8.2.24)

Now we compare .v(x) and .w(x) on . d . Since .x 2 ≥ h2 y 2 in .K, from (8.2.5), we have   w(x)

.

r=d

  ≥ B|x|1+ 

r=d

+1

= Bd 1+ cos

ω0 . 2

(8.2.25)

On the other hand,   v(x) .

d

  = −1 + exp(ν −1 μ1 (u(x) − u(0))) − 1 + exp(2ν

−1

d

≤ (8.2.26)

μ1 M0 ),

and, therefore, from (8.2.25) and (8.2.26), in virtue of (8.2.10), we obtain   Aw(x)

d

.

≥ ABd

1+

1 hγ0 − 1 −

cos

0

+1

$ g0 1 + h2  ω0 ≥A 2 h 0

d 1+ 0 h1+ 0 (1 + h2 )−

  exp(2ν −1 μ1 M0 ) − 1 ≥ v 

d

,

1+ 0 2

0 +1 2

≥

% + 2(1 + h2 ) ×

194

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

if we choose A enough great A≥

.

[exp(2ν −1 μ1 M0 ) − 1](hγ0 − 1 − 0 ) . &   1− 0 ' hd 1+ 0 g0 + 2h 0 1 + h2 2

(8.2.27)

Thus, if we choose a small number .d > 0 according to (8.2.16) and large numbers .A, B according to (8.2.10), (8.2.17), (8.2.23) and (8.2.27), we provide the validity of (8.2.14). Therefore, the functions (8.2.5) and (8.2.13) satisfy Proposition 8.6—the comparison principle and, by it, we have v(x) ≤ Aw(x), x ∈ Gd0 .

(8.2.28)

.

Returning to the function .u(x) from (8.2.13), on the basis of (8.2.28), we have −1 −1 u(x) − u(0) = νμ−1 1 ln(v(x) + 1) ≤ νμ1 ln(Aw(x) + 1) ≤ νμ1 Aw(x).

.

Similarly, we derive the estimate u(x) − u(0) ≥ −νμ−1 1 Aw(x),

.

if we consider an auxiliary function (8.2.24). In virtue of (8.2.3), the theorem is proved.   Theorem 8.7 Let .u(x) be a strong solution of the problem (QL) and assumptions .(A)– (E) be satisfied. Let . > 0 be a number defined by Theorem 8.5. Then there exists the number .d ∈ (0, 1) such that

.

|∇u(x)| ≤ C1 |x| , x ∈ Gd0 ,

(8.2.29)

.

where constant .C1 depends only on .ν, .μ, n, .k1 , s, .g0 , .γ0 , .χ0 , .M1 , and domain G but does not depend on u. ρ

Proof Let us consider the set .Gρ/2 ⊂ G, 0 < ρ < d. We make the transformation   −1− u(ρx  ). The function .v(x  ) satisfies the problem .x = ρx ; v(x ) = ρ ⎧ ⎨ a ij (x  )v .

⎩

xi xj

∂v ∂ n 

= F (x  ),

∂v + χ(ω) ∂r  +

1  |x  | γ (ω)v(x )

x  ∈ G11/2 , = ρ − g(ρx  ),

1 , x ∈ 1/2

(QL)

8.2 The Barrier Function. The Preliminary Estimate of the Solution Modulus

195

where a ij (x  ) ≡ aij (ρx  , ρ 1+ v(x  ), ρ vx  (x  ))

.

and F (x  ) ≡ −ρ 1− a(ρx  , ρ 1+ v(x  ), ρ vx  (x  )).

.

Now we apply the assumption .(E) .

max |∇  v(x  )| ≤ M1 .

(8.2.30)

x  ∈G11/2

Returning to the variable x and the function .u(x), we obtain from (8.2.30) ρ

|∇u(x)| ≤ M1 ρ , x ∈ Gρ/2 , 0 < ρ < d.

.

Putting now .|x| = 23 ρ, we obtain the desired estimate (8.2.29).

 

Corollary 8.8 Let .u(x) be a strong solution of the problem (QL) and assumptions .(A)– (E) be satisfied. Then .u(0) = 0, and therefore the inequality (8.2.11) takes a form

.

|u(x)| ≤ C0 |x|

.

+1

, x ∈ Gd0 .

(8.2.31)

Proof From the problem boundary condition, it follows that γ (x)u(x) = |x|g(x) − |x|

.

∂u ∂u − |x|χ(ω) , ∂n ∂r

x ∈ ∂G \ O.

By the assumption (D) and the estimate (8.2.29), we obtain γ0 |u(x)| ≤ γ (x)|u(x)| ≤ |x||g(x)| + (1 + χ0 )|x||∇u| ≤ g0 |x|s + C2 |x|

.

+1

.

By letting .|x| tend to 0, we get, because of the continuity of .u(x), that .γ0 |u(0)| = 0, and   taking into account .γ0 > 0, we find .u(0) = 0.

196

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

8.3

Integral Weighted Estimates

Theorem 8.9 Let u(x) be a strong solution of problem (QL) and assumptions (A)–(E) be satisfied. Then there exist numbers d ∈ (0, 1) and C > 0 not depending on u, such that

◦2 d/2 u(x) ∈W 4−n G0 and .



r 4−n u2xx + r 2−n |∇u|2 + r −n |u(x)|2 dx ≤ d/2

G0

⎛ ⎜ C⎜ ⎝

⎞



⎟ |∇u|2 dx + |u|20 + g02 + g12 + k12 + 1⎟ ⎠.

(8.3.1)

G2d 0

Proof Because of (8.2.29) and (8.2.31), we obtain   2−n  . |∇u|2 + r −n |u(x)|2 dx ≤ r Gd0

d r 2−n+2 · r n−1 dr ≤ Cd 2+2 ≤ const.

≤ Cmeas

(8.3.2)

0 ◦1

Hence, it follows that u ∈ W 2−n (Gd0 ). By assumption (A), we have aij (x, u, z) ∈ W 1,q (M), q > N, and, by the Sobolev embedding theorem, aij (x, u, z), i, j = 1, . . . , N, are uniformly continuous on M. Therefore, for ∀δ > 0, there exists dδ > 0 such that |aij (x, u(x), ux (x)) − aij (y, u(y), ux (y))| < δ,

(8.3.3)

|x − y| + |u(x) − u(y)| + |ux (x) − ux (y)| < dδ .

(8.3.4)

.

if only .

By (8.2.29) and (8.2.31), we get |x − y| + |u(x) − u(y)| + |ux (x) − ux (y)| < d + C0 d 1+ + C1 d ,

.

∀x ∈ Gd0 .

(8.3.5)

8.3 Integral Weighted Estimates

197

Now we choose d > 0 such that the inequality d + C0 d 1+ + C1 d ≤ dδ

(8.3.6)

.

holds. For such d, we may guarantee (8.3.3) in Gd0 . Now we shall estimate the second derivatives of the problem (QL) solution. We make  the transformation x = x  , u(x  ) = v(x  ). Then (x1 , . . . , xN ) ∈ G/2 → G11/2   ), and the function v(x  ) satisfies the problem (x1 , . . . , xN ⎧ ⎨ a ij (x  )v .

⎩

xi xj

∂v ∂ n 

= F (x  ),

∂v + χ(ω) ∂r 

+

x  ∈ G11/2 ,

1  |x  | γ (ω)v(x )

=

ρg(ρx  ),

x∈

1 , 1/2

(QL)

where a ij (x  ) ≡ aij (x  , v(x  ), −1 vx  (x  )),

.

(8.3.7) F (x  ) ≡ −2 a(x  , v(x  ), −1 vx  (x  )). Because of (8.3.3), we can apply Theorem 5.2 about interior and near a smooth portion of the boundary L2 -estimate to the solution of the (QL) equation:   v 2 (x  ) + F 2 (x  )+ . (vx2 x  + |∇  v|2 )dx  ≤ C4 G11/2

G21/4

C4 2 inf(|∇  G|2 + G 2 )dx  ,

(8.3.8)

where the constant C4 does not depend on v, F, g, and it is defined by ν, μ, γ (ω)C 1 ( 2 ) , χ(ω)C 1 ( 2 ) , the continuity moduli of a ij (x  ) and G21/4 . Returning to 1/4 1/4 the variable x and the function u(x) in (8.3.8), we obtain  

4−n 2 r −n u2 + r 4−n a 2(x, u, ux ) dx+ . r uxx dx ≤ C4 

G/2

2

G/4

C4 inf



r 4−n |∇G|2 + +r 2−n G 2 dx.

2 G/4

(8.3.9)

198

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

Putting in (8.3.9)  = 2−k d and summing up over k = 0, 1, . . . , log2 (d/ε) ∀ε ∈ (0, d), we get  

r −n u2 + r 4−n a 2 (x, u, ux ) dx+ . r 4−n u2xx dx ≤ C5 Gdε

G2d ε/4

C5 inf



r 4−n |∇G|2 + +r 2−n G 2 dx.

(8.3.10)

G2d ε/4

By the assumption (D) and (8.2.29) with regard to (8.3.2), we have   . r 4−n u2xx dx ≤ C6 r −n u2 + r 4−n f 2 (x) + r 4−n b2(x)+ G2d 0

Gdε

r

2−n



r 4−n |∇G|2 + +r 2−n G 2 dx., ∀ε > 0, |∇u| dx + C5 inf 2

(8.3.11)

G2d 0

where the constants C5 , C6 do not depend on ε. Therefore, we can perform the passage to the limit as ε → +0, by the Fatou theorem, and we get  $  4−n 2 r −n u2 + r 4−n f 2 (x) + r 4−n b2 (x)+ . r uxx dx ≤ C6 G2d 0

Gd0

 r 2−n |∇u|2 dx + g2◦ 1/2

% .

(8.3.12)

◦2 W α (Gd0 ).

Now we

W 4−n (02d )

On the basis of the inequalities (8.3.2) and (8.3.12), we have u(x) ∈ shall prove (8.3.1). Let ζ(r) ∈ C 2 [0, d] be a cut-off function such that ζ (r) ≡ 1 f or r ∈ [0, d/2],

.

0 ≤ ζ(r) ≤ 1 f or r ∈ [d/2, d],

ζ(r) ≡ 0 f or r > d, ζ  (r) ≤ 0 and |ζ  (r)| ≤

ζ (d) = ζ  (d) = 0, C f or r ∈ [d/2, d]. r

We multiply both sides of the (QL) equation by ζ 2 (r)r 2−n u(x) and integrate over Gd0 . We get   2 2−n . ζ (r)r uudx = − ζ 2 (r)r 2−n u{(aij (x, u, ux )− Gd0

Gd0

aij (0, 0, 0))uxi xj + a(x, u, ux )}dx.

(8.3.13)

8.3 Integral Weighted Estimates

199

We apply the Gauss–Ostrogradsky formula:   ∂u 2 2−n . ζ (r)r uudx = ζ 2 (r)r 2−n u ds− ∂ n

Gd0

∂Gd0

 2

ζ (r)r



2−n

ζ(r)ζ  (r)xi r 1−n

2

|∇u| dx −

Gd0

Gd0

n−2 2



∂u2 dx+ ∂xi

ζ 2 (r)xi r −n

∂u2 dx. ∂xi

(8.3.14)

Gd0

Because of the (QL) boundary condition and by the properties of ζ(r), we obtain   . ζ 2 (r)r 2−n uudx = − ζ 2 (r)r 2−n |∇u|2 dx− Gd0

Gd0



ζ(r)ζ  (r)xi r 1−n

Gd0



∂u2 2−α dx + ∂xi 2



ζ 2 (r)xi r −n

∂u2 dx+ ∂xi

Gd0

  ∂u 1 2 2−n − γ (ω)u ds. ζ (r)r u g(x) − χ(ω) ∂r r

(8.3.15)

0d

Now we calculate the second and third integrals from the right in (8.3.15). For this, we use the Gauss–Ostrogradsky formula once more  .

 −

ζ (r)ζ  (r)xi r 1−n

Gd0

∂u2 dx = ∂xi

 ∂Gd0

 ∂  ζ(r)ζ  (r)xi r 1−n dx = u ∂xi

ζ(r)ζ  (r)r 1−n u2 xi cos(

n, xi )ds− 

2

Gd0

 − Gd0

ζ(r)ζ  (r)r 1−n u2 xi cos(

n, xi )ds−

d ∪0d

(8.3.16) .

/

u2 ζ  (r)r 2−n + ζ(r)ζ  (r)r 2−n + ζ(r)ζ  (r)r 1−n dx 2

200

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

and 

ζ 2 (r)xi r −n

.





∂u2 dx = ∂xi

Gd0

∂Gd0

∂ u ζ 2 (r)xi r −n dx = ∂xi

−

ζ 2 (r)r −n u2 xi cos(

n, xi )ds− 

2

Gd0

 −2

ζ 2 (r)r −n u2 xi cos(

n, xi )ds−

d ∪0d

u2 ζ(r)ζ  (r)r 1−n dx.

(8.3.17)

Gd0

  Since ζ (r)

d

  = 0, ζ  (r)

it follows that

  = 0 and xi cos(

n, xi )

d/2 0

 .





ζ 2 (r)r 2−n ug(x)ds −

=

= 0, from (8.3.13)–(8.3.17),

 ζ 2 (r)r 2−n |∇u|2 dx +

Gd0



0d

0d

ζ 2 (r)r 1−n γ (ω)u2 ds 0d

ζ 2 (r)r 2−n χ(ω)u

∂u ds+ ∂r

(8.3.18)

0d

. / u2 (3 − n)ζ (r)ζ  (r)r 1−n + ζ(r)ζ  (r)r 2−n + (ζ  (r))2 r 2−n dx+

Gdd/2



+

ζ 2 (r)r 2−n u{(aij (x, u, ux ) − aij (0, 0, 0))uxi xj + a(x, u, ux )}dx.

Gd0

Now, by the Cauchy inequality, we estimate the first integral from the right 

 .

ζ 2 (r)r 2−n |u||g|ds =

0d

1 ≤ 2

0d

 2

ζ (r)r 0d

3−n

1−n * 1 ζ 2 (r) r 2 √ |g| r 2 γ (ω)|u| ds ≤ γ (ω)

1−n

1 γ (ω)u ds + 2γ0



2

r 3−n g 2 ds, ∂G

(8.3.19)

8.3 Integral Weighted Estimates

201

because of 0 < γ0 ≤ γ (ω). Now from (8.3.18) and (8.3.19), we obtain  ζ 2 (r)r 2−n |∇u|2 dx +

.

Gd0

1 2γ0

Gdd/2

 ζ 2 (r)r 1−n γ (ω)u2 ds ≤ 0d



 r 3−n g 2 ds − ∂G



1 2

ζ 2 (r)r 2−n χ(ω)u

∂u ds+ ∂r

0d

. / u2 (3 − n)ζ (r)ζ  (r)r 1−n + ζ(r)ζ  (r)r 2−n + (ζ  (r))2 r 2−n dx+  Gd0

  ζ 2 (r)r 2−n uuxi xj aij (x, u, ux ) − aij (0, 0, 0) dx+  ζ 2 (r)r 2−n ua(x, u, ux )dx.

(8.3.20)

Gd0

Using the Cauchy inequality, (8.3.3) and (8.3.12), we obtain  . ζ 2 (r)r 2−n uuxi xj (aij (x, u, ux ) − aij (0, 0, 0))dx ≤ Gd0

 δ Gd0

r 2−n |uxx ||u|dx ≤

δ 2





 r 4−n |uxx |2 + r −n u2 dx ≤

Gd0

 . / r −n u2 + r 4−n f 2 (x) + r 4−n b2 (x) + r 2−n |∇u|2 dx+ δC6 G2d 0

C7 |g2◦ 1/2

W 4−n (02d )

, ∀δ > 0.

(8.3.21)

From the assumption (D) and by (8.2.29), we get   r 2−n ua(x, u, ux ) ≤ r 2−n |u| μ1 |∇u|2 + b(x)|∇u| + f (x) ≤   |∇u| μ1 r 2−n |u||∇u| + r 2−n b(x)|u| + r 2−n |u||f | ≤ .

  δ ≤ C1 d r 2−n |∇u|2 + r −n u2 + r 4−n b2 (x) + r −n u2 + 2 1 4−n 2 r f , ∀δ > 0 2δ

202

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

and therefore,   2 2−n . ζ (r)r ua(x, u, ux )dx ≤ Cd ζ 2 (r)r 2−n |∇u|2 dx+ Gd0

Gd0

  δ Cd + ζ 2 (r)r −n u2 (x)dx + Cd ζ 2 (r)r 4−n b2 (x)dx+ 2 Gd0

1 2δ

Gd0



ζ 2 (r)r 4−n f 2 (x)dx, ∀δ > 0.

(8.3.22)

Gd0

In virtue of χ(ω) ≥ 0, u(0) = 0, ζ(d) = 0, ζ  (r) ≤ 0, r ∈ [0, d] and ω = (ω1 , ω ), ω = (ω2 , . . . ωn−1 ) , we have  .

−

2

ζ (r)χ(ω)r

2−n



1 ∂u u ds = − ∂r 2

0d

∂u2 drdσ = ∂r

ζ 2 (r)χ(ω) 0d

1 − 2



ω

0

J ∂

−

1 2

2





ω

0

2



0

,ω χ

ω

J

d ∂u2 r, ω0 , ω  2 ,ω ζ 2 (r)drdω = 2 ∂r



, ω χ

ω

0

2

0

, ω

$ ω

0 u2 d, , ω ζ 2 (d)− 2

∂

1 2

 J

ω

0

2



ω

% 0 u2 0, , ω ζ 2 (0) dω + 2

ω

0

,ω χ

2

,ω



d

ω

0 u2 r, , ω 2ζ(r)ζ  (r)drdω ≤ 0. 2

(8.3.23)

0

∂

From (8.3.20)–(8.3.23), it follows that   1 2−n 2 . r |∇u| dx + r 1−n γ (ω)u2 ds ≤ 2 d/2

d/2

G0

 C5 (δ + d )

0

r 2−n |∇u|2 + r −n u2 dx + C6  C7 G2d d/2

r 4−n (b2 (x) + f 2 (x))dx+

G2d 0

d/2

G0



(|∇u|2 + u2 )dx +

1 2γ0



r 3−n g 2 ds, ∀δ > 0. ∂G

(8.3.24)

8.3 Integral Weighted Estimates

203

Now, from (8.3.12), by (8.3.24) with the choosing d, δ appropriately positive small, it   follows the required assertion (8.3.1). Theorem 8.10 Let u(x) be a strong solution of (QL) problem

and assumptions (A)–(E) be satisfied. Then there exist positive numbers C and d ∈ 0, 1e not depending on u, such that u(x) ◦ 2

.



W 4−n (G0 )

≤

 C uW 1 (G) + g0 + g1 + k1

⎧ ⎪ ⎨

λ

3 λ ln 2 1 ⎪ ⎩ s

if s > λ,   d . if s = λ,  ∈ 0, 2 if s < λ,



◦2 d/2 Proof Because of Theorem 8.9, we have that u(x) ∈W 4−n G0 . Now we set   2−n 2 .U () = r |∇u| dx + r 1−n γ (ω)u2 ds.  G0

(8.3.25)

(8.3.26)

 0

Similarly (8.3.13), we have  . r 2−n u udx = 

G0

 −

r 2−n u

&

' a ij (x, u, ux ) − a ij (0, 0, 0) uxi xj + a(x, u, ux ) dx.

(8.3.27)



G0





On the other hand, using the Gauss–Ostrogradsky formula, the fact that ∂G0 = 0 ∪  , and the boundary condition of problem (QL), we have     ∂u 1 2−n 2−n − γ (ω)u ds+ . r u udx = r u g(x) − χ(ω) ∂r r 

G0





0

2−n u 

∂u d  − ∂r



 G0

 r 2−n |∇u|2 dx + (n − 2)  G0

r −n uuxi xi dx.

(8.3.28)

204

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

Now we calculate the last integral of the right side of (8.3.28)    1 1 . r −n uuxi xi dx = r −n u2 xi cos(

n, xi )ds + −n u2 xi cos(

n, xi )d  − 2 2 



G0



0

1 2

 u2 

∂  −n  r xi dx. ∂xi

G0

Because of .

 n n   ∂  −n   ∂r nr −n − nr −n−1 xi =0 r xi = ∂xi ∂xi i=1

  and xi cos(

n, xi )



 .

i=1

  = , xi cos(

n , x i )

r −n uuxi xi dx =



1 2



0

= 0, d  = n−1 d , we obtain

 1−n u2 d  =

1 2



G0

 u2 d .

(8.3.29)



From (8.3.28) and (8.3.29), it follows that     ∂u 1 2−n 2−n − γ (ω)u ds+ . r u udx = r u g(x) − χ(ω) ∂r r 



G0

0

    ∂u n − 2 2 + u d − r 2−n |∇u|2 dx. u ∂r 2

(8.3.30)





G0

Comparing (8.3.27) and (8.3.30), we get    ∂u 2−n 2 1−n 2 . r |∇u| dx + γ (ω)r u ds + χ(ω)r 2−n u ds = ∂r 



G0

0



0

    ∂u n − 2 2 + u d + r 2−n ugds+ u ∂r 2



 r 2−n u  G0



0

&

' a ij (x, u, ux ) − a ij (0, 0, 0) uxi xj + a(x, u, ux ) dx.

(8.3.31)

8.3 Integral Weighted Estimates

205

Now, we estimate integrals from the right side of (8.3.31): • By Lemma 4.11, we have .

    1 ∂u n − 2  - + U () + χ(ω)u2 d . u d ≤ ∂r 2 2λ 2

(8.3.32)

∂

• By the Cauchy inequality, we obtain  r 2−n |u||g|ds = 

 * 

0

0

.

δ1 2

γ (ω)r

1−n 2

|u|



 3−n 1 r 2 |g| ds ≤ √ γ (ω)

 γ (ω)r 1−n u2 ds + 

0

γ (ω)r 1−n u2 ds + 

γ (ω)r 3−n g 2 ds ≤

(8.3.33)



0



δ1 2



1 2δ1

c 2 2s g  , ∀δ1 > 0, δ1 γ0 0

0

in virtue of assumption (D) and 0 < γ0 ≤ γ (ω). • Analogously to (8.3.23), we obtain  ∂u . χ(ω)r 2−n u ds = ∂r 

0

1 n−2 ω0 sin 2 2

 sinn−3 ω2 · · · sin ωn−2 · χ

ω

0 , ω u2 , , ω dω = 2 2

ω

0

∂

1 2

 χ(ω)u2 (, ω)d ,

(8.3.34)

∂

in virtue of ωi ∈ (0, π) for i = 2, 3, . . . , n − 2 and χ(ω) ≥ 0, u(0) = 0; here w = (w2 , . . . , wn−1 ). • 

 .  0

χ(ω)r

u ds =

1−n 2

 0

because of 0 ≤ χ(ω) ≤ χ0 .

χ(ω) χ0 γ (ω)r 1−n u2 ds ≤ γ (ω) γ0

 γ (ω)r 1−n u2 ds,  0

(8.3.35)

206

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

• By assumption (A) and the embedding theorem, we have a ij (x, u, z) ∈ W 1,q (M), q > n ⇒ a ij (x, u, z) ∈ C δ (M),

.

(8.3.36)

0 < δ < 1 − qn , i, j = 1, . . . , n. Hence, together with (8.2.29) and (8.2.31), it follows that |a ij (x, u, ux ) − a ij (0, 0, 0)| ≤ C (|x| + |u| + |ux |)δ ≤

δ ≤ C0 δ , C1  +  +1 + 

.

(8.3.37)



δ ∈ 0; 1 − qn , for all |x| ≤  < d. Therefore, using the Cauchy inequality and the Friedrichs–Wirtinger type inequality (see Theorem 4.2.1), we obtain  r 2−n |u||uxx ||a ij (x, u, ux ) − a ij (0, 0, 0)|dx ≤

. 

G0

C0 

δ



4−n

n r − 2 |u| r 2 |uxx | dx ≤ 

G0

C0 δ 2

 r 

G0

δ

C0 2

C0 δ u dx + 2

−n 2

 r 4−n |uxx |2 dx ≤ 

G0

 r 4−n |uxx |2 dx + C1 δ  G0



C1 δ λ  0

⎧ ⎪ ⎪ ⎨



⎪ ⎪ ⎩G 0

r 2−n |∇u|2 dx +

  n 1−n 2 χ(ω)r u ds, δ ∈ 0; 1 − . q

γ (ω)r 1−n u2 ds  0

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

+

(8.3.38)

By (8.3.26) and (8.3.35), estimation (8.3.38) takes the following form:  . r 2−n |u||uxx ||a ij (x, u, ux ) − a ij (0, 0, 0)|dx ≤ 

G0

 C2 δ 

G0

-(). r 4−n |uxx |2 dx + C2 δ U

(8.3.39)

8.3 Integral Weighted Estimates

207

• By assumption (D) and the inequality (8.2.31), using Theorem 4.2.1 analogously to (8.3.38) and using (8.3.35), we get  r 2−n ua(x, u, ux )dx ≤

. 

G0

C3 

+1

ε1 2



ε2 2

r 2−n |∇u|2 dx + 

G0



1 2ε2

+



r −n u2 dx+



G0

r 4−n f 2 dx + C4 ε1−1 2s+2 ≤



G0

C5 

+1

-() + 1 f 2◦ 0   + k12 C4 ε−1 2s+2 , + ε1 + ε2 U 1  2ε2 W 4−n G0

(8.3.40)

∀ε1 , ε2 > 0, because of (8.3.26). • By inequality (8.3.12) and Theorem 4.2.1 with regard to notation (8.3.26), we obtain  . r 4−n u2xx dx ≤ 

G0

C6



r −n u2 + r 4−n f 2 + r 4−n b2 + r 2−n |∇u|2 dx + C7 g2◦ 1/2

2

≤

W 4−n 0

2

G0

-(2) + C6 f 2◦ 0 C8 U



2 W 4−n G0

+ C6 b2◦ 0



2 W 4−n G0

+ C7 g2◦ 1/2

2

W 4−n 0

.

(8.3.41)

• By (8.3.39) and (8.3.41), we get  . r 2−n |u||uxx ||a ij (x, u, ux ) − a ij (0, 0, 0)|dx ≤ 

G0

 C9 δ



-() + U -(2) + f 2◦ 0 U



2 W 4−n G0

+ b2◦ 0



2 W 4−n G0

+ g2◦ 1/2

2

.

W 4−n 0

(8.3.42)

208

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

From (8.3.31), in virtue of (8.3.32)–(8.3.34), (8.3.40), and (8.3.42), it follows that  1 .U () + χ(ω)u2 d ≤ 2 ∂

 - 1 U () + 2λ 2

and & .

1−  

C

+1

χ(ω)u2 d + 

+1

-() + δ U -(2)+ +  δ + ε3 U

∂

 C





ε3−1 f 2◦ 0 2 W 4−n G0

+  δ + ε3

ε3−1 f 2◦ 0 2 W 4−n G0

'

+ b ◦ 0 2 + g ◦ 1/2 2 + k12 ε3−1 2s+2 W 4−n G0 W 4−n 0 2

2

-() ≤  U - () + δ U -(2)+ U 2λ

+ b ◦ 0 2



2 W 4−n G0

+ g ◦ 1/2 2

2

+

W 4−n 0



k12 ε3−1 2s+2

(8.3.43)

for all ε3 > 0 and δ ∈ 0; 1 − qn . By Theorem 8.9, we have that 



-(d) = .U

r

2−n

|∇u| dx +

γ (ω)r 1−n u2 ds ≤

2

Gd0

0d

-0 . C u2W 1 (G) + g02 + g12 + k12 ≡ U

(8.3.44)

From (8.3.43) and (8.3.44), we obtain the following Cauchy problem for the differential inequality  .

-() + N ()U -(2) + Q() ≥ 0, 0 <  < d, - () − P()U U U (d) ≤ U0 ,

(CP )

where P() =

.

2λ & 1−  

+1

N () = 2λδ

.

+  δ + ε3 −1

,

' ,

(8.3.45)

(8.3.46)

8.3 Integral Weighted Estimates

C .Q() = 

209

 ε3−1 f 2◦ 0



2 W 4−n G0

+ b2◦ 0



2 W 4−n G0

+ g2◦ 1/2

2

+

W 4−n 0

k12 ε3−1 2s+2

,

(8.3.47)

for all ε3 > 0. Then, in virtue of Theorem 2.61, ⎞ ⎛ d  -() ≤ exp ⎝ B(τ )dτ ⎠ × .U 

⎛

⎡

⎣U0 exp ⎝−

d

⎞ P(τ )dτ ⎠ +



d

⎛ Q(τ ) exp ⎝−



τ

⎞

⎤

P(σ )dσ ⎠ dτ ⎦

(8.3.48)



with ⎛ ⎞ 2 ⎜ ⎟ .B() = N () exp ⎝ P(σ )dσ ⎠ . 

There are three possible cases: s > λ, s = λ, and s < λ. .

s > λ.

Choosing ε3 = ε ∀ε > 0, we obtain from (8.3.45)–(8.3.47) P(ρ) =

.

2λ − 2λ ρ δ ρ

−1

+ ρ + ρ ε−1 ; . −1

N (ρ) = 2λρ δ

(8.3.49)

;

(8.3.50)

and $ Q(ρ) = C b2◦ 0

.

2

W 4−N (G0 )

+ −ε f 2◦ 0

2

W 4−N (G0 )

+

+ g2◦ 1/2

2ρ

W 4−N (0 )

+ k12 2s+2

−ε

%

ρ −1 .

(8.3.51)

210

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

Now, by means of simple calculations, from (8.3.49) and (8.3.50), we have d

ρ 2λ .exp − P(τ )dτ ≤ C d

d B(τ )dτ ≤ C = const.

and

ρ

(8.3.52)

ρ

In addition, τ



Q(τ ) exp −

.

$

P(σ )dσ ≤ C2λ τ −2λ−1 b2◦ 0

W 4−N (G2τ 0 )

ρ

+ τ −ε f 2◦ 0

W 4−N (G2τ 0 )

+ g2◦ 1/2

W 4−N (02τ )

+

+ k12 τ 2s+2

−ε

%

, ∀ε > 0.

(8.3.53)

Let us recall the assumption (D): b2◦ 0

.

W 4−N (G2τ 0 )

τ −ε f 2◦ 0

≤ ck12 τ 2s ,

W 4−N (G2τ 0 )

g2◦ 1/2

W 4−N (02τ )

≤ ck12 τ 2s−ε

≤ c(g02 + g12 )τ 2s .

Since s > λ, we can put s = λ + ε, ∀ε > 0. Therefore, we get b2◦ 0

.

W 4−N (G2τ 0 )

+ τ −ε f 2◦ 0

W 4−N (G2τ 0 )

xi12τ 2s+2

+ g2◦ 1/2

W 4−N (02τ )

−ε

+

≤ c(k12 + g02 + g12 )τ 2λ+ε , ∀ε > 0.

(8.3.54)

From (8.3.53) and (8.3.54), it follows that d



τ

Q(τ ) exp −

.



P(σ )dσ dτ ≤ C(k12 + g02 + g12 )2λ .

(8.3.55)

ρ

Finally, by (8.3.44), (8.3.48)–(8.3.55), we derive / . 2 2 2 2λ -(ρ) ≤ C(n, λ, d, ) u2 1  . + k + g + g U 1 0 1 W (G)

.

At last, from (8.3.41) and (8.3.56), we deduce the validity of (8.3.25) for s > λ. .

s = λ.

(8.3.56)

8.3 Integral Weighted Estimates

Choosing ε3 =

1 2λ ln2

1 

211

, we obtain the Cauchy problem (CP ) with functions

 2λ − 2λ  + δ .P() =  C .Q() = 

 2λ ln2

−1

+



1 2λ ln2

1 

−1

, N () = 2λδ

,

1 f 2◦ 0 2 + b2◦ 0 2 + g2◦ 1/2 2  W 4−n G0 W 4−n G0 W 4−n 0 1 2 2s+2 k   1

+2λ ln2

 .

Now we calculate ⎛ d ⎞  . exp ⎝− P(τ )dτ ⎠ = 

⎧ d1  ⎨  2λ − 2λ τ + τ δ exp − ⎩ τ

−1

+



2λτ ln2

⎫ ⎬

2

1 1 τ

dτ

⎭

=

, +  δ d +1 −  +1 d − δ + × · exp 2λ δ +1 ⎛ d ⎞     2λ dτ 1 ⎠≤C , exp ⎝ , d ∈ 0, d e τ ln2 τ1

 2λ d

(8.3.57)



because of d .



dτ τ ln2

1 τ

=

1 ln d1

−

⎛ .

exp ⎝

2τ

1 ln 1


0. We calculate ⎛ .

exp ⎝−

d 

 2λ d

⎧ d ⎨  + 2λ − 2λ τ + τ δ P(τ )dτ ⎠ = exp − ⎩ τ ⎞



+ · exp 2λ



dδ

− δ

δ

+

d

+1

− +1

+1

− ε3 ln

 d

⎫

, ⎬ ε 3 −1 + dτ = ⎭ τ

, ≤C

 2λ(1−ε3) d (8.3.61)

8.3 Integral Weighted Estimates

213

and ⎛ 2τ ⎞  . exp ⎝ P(σ )dσ ⎠ ≤ C22λ(1−ε3) ⇒ τ

d

d B(τ )dτ =



⎛

N (τ ) exp ⎝



2τ

⎞ P(σ )dσ ⎠ dτ ≤ λC2

d 2λ(1−ε3 )+1

τ

−1

dτ



 = λC22λ(1−ε3)+1

τδ

dδ

− δ

δ

 ≤ const.

(8.3.62)

By assumption (D), we have ⎛ τ ⎞   2λ(1−ε ) −2λ(1−ε )−1 3 3 .Q(τ ) exp ⎝− P(σ )dσ ⎠ ≤ C τ ε−1 f 2◦ 0 3



  W 4−n G2τ 0

 +b2◦ 0

  W 4−n G2τ 0

+ g2◦ 1/2 

W 4−n 02τ



+ k12 ε3−1 τ 2s+2

≤ C2λ(1−ε3 ) τ 2s−2λ(1−ε3)−1 (g02 + g12 + k12 ), and therefore, d .

⎛ Q(τ ) exp ⎝−



τ

⎞ P(σ )dσ ⎠ dτ



≤ C2λ(1−ε3) (g02 + g12 + k12 )

if we choose ε3 =

λ−s 2λ

d 2s−2λ(1−ε3) − 2s−2λ(1−ε3) ≤ C2s (g02 + g12 + k12 ), 2s − 2λ(1 − ε3 ) (8.3.63)

> 0. Finally, from (8.3.44), (8.3.48), (8.3)–(8.3.63), it follows that

2 2 2 2s -() ≤ C(n, λ, d, ) u2 1 + g + g + k U 0 1 1  . W (G)

.

From (8.3.41) and (8.3.64), we obtain the validity of (8.3.25) for s < λ.

(8.3.64)  

214

8.4

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

The Power Modulus of the Continuity at the Conical Point

Now we shall make the exponent . in the estimates (8.2.31) and (8.2.29) more precise and prove the Hölder continuity of the first derivatives of the strong solutions in the neighborhood of a conical point. 2ρ

ρ

ρ

Proof of Theorem 8.3 Let us consider the sets .Gρ/2 and .Gρ/4 ⊃ Gρ/2 , 0 < ρ < d/2. We need to consider three cases: .

s > λ, s < λ, s = λ. .

s > λ.

We make the transformation .x = ρx  ; w(x  ) = ρ −λ u(ρx  ). The function .w(x  ) satisfies the problem ⎧ ⎨ .

⎩

a ij (x  )wxi xj = F (x  ), x  ∈ G11/2, ∂w ∂ n 

+ χ(ω) ∂w ∂r  +

1  |x  | γ (ω)w(x )

1 , = 1−λ g(ρx  ), x ∈ 1/2

(QL)0

where a ij (x  ) ≡ aij (ρx  , ρ λ w, ρ λ−1 wx  (x  ))

.

and F (x  ) ≡ −ρ 2−λ a(ρx  , ρ λ w(x  ), ρ λ−1 wx  (x  )).

.

The .Lq -estimate (5.1.1) is satisfied for the function .w(x  ) (see Theorem 5.2 about .Lq estimate of solutions of the elliptic oblique problem in the smooth domain), that is,   q q |w| + ρ qλ |∇  w|2q + q |b|q |∇  w|q + .w 2,q ≤ C3 W (G1 ) 1/2

G21/4

 ρ q(2−λ)|f |q dx  + C4 ρ q(1−λ)



G21/4

where the constants .C3 , C4 do not depend on .w.

(|∇  g|q + |g|q )dx  ,

(8.4.1)

8.4 The Power Modulus of the Continuity at the Conical Point

215

At first, we consider the case .2 ≤ n < 4. By the Sobolev Imbedding Theorem, we have sup |w(x  )| ≤ Cw2,2;G1 .

.

(8.4.2)

1/2

x  ∈G11/2

Returning to the variable x and because of the estimate (8.3.25) of Theorem 8.10, we get 

w22,2;G1 = 1/2

|wx  x  |2 + |∇  w|2 + w2 dx  ≤

G11/2 .

C(n)ρ

−2λ



r 4−n |uxx |2 + r 2−n |∇u|2 +

(8.4.3)

G11/2

r −n u2 dx ≤ C. From (8.4.2) and (8.4.3), it follows that sup |w(x  )| ≤ C0 ,

.

x  ∈G11/2

and returning to the variable x, we get ρ

|u(x)| ≤ C0 ρ λ , x ∈ Gρ/2 .

.

Putting now .|x| = 23 ρ, we obtain the first estimate of (8.1.1) for our theorem in our case. Let us now .n ≥ 4. We apply the Lieberman local maximum principle, Proposition 8.2. Then, by the condition .(D), we have .



sup w(x ) ≤ C

x  ∈G11/2

 

w2 dx 

1 2

+ ρ s−λ g0 +

G21/4

+ρ

2−λ







|a(ρx , ρ w(x ), ρ λ

λ−1

1 n . wx  )| dx n



(8.4.4)

G21/4

We shall estimate each integral from the right hand side of (8.4.4). We estimate the first integral:  .

G21/4

w2 dx  ≤ ρ −2λ

 2ρ

Gρ/4

r −n u2 dx ≤ C,

(8.4.5)

216

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

by Theorem 8.10. Because of the assumption .(D) and the estimate (8.2.29), we have    λ  λ−1 n  . |a(ρx , ρ w(x ), ρ wx  )| dx ≤ c(n) μn1 |∇u|2n + G21/4

2ρ

Gρ/4



−n + b (x)|∇u| + f (x) r dx ≤ c(n) μn1 (r 2−n |∇u|2 )× n

n

n

2ρ

Gρ/4

× (r −2 |∇u|2n−2 ) + (r 2−n |∇u|2 ) · (k1n r (s−2)n−2 |∇u|n−2 ) + k1n r (s−2)n−n dx ≤    ≤ c(n) μn1 C12n−2 ρ 2 (n−1)−2 + k1n C1n−2 ρ (n−2)+(s−2)n−2 r 2−N |∇u|2 dx 2ρ

Gρ/4

  + c(n)(s − 2)−1 (k1 )n meas 2(s−2)n − 2−2(s−2)n ρ (s−2)n , 0 < ρ < d/2.

(8.4.6)

Because of (8.3.25), hence, we obtain   ρ

.

2−λ





|a(ρx , ρ w(x ), ρ λ

λ−1

n

wx  )| dx



 n1

≤

G21/4

2(λ−1) 2 ≤ C ρ 2−λ+ n +

(n−1) n

+ ρ s−λ+

2(λ−1) (n−2) n + n

d + ρ s−λ , ∀ρ ∈ (0, ). 2

(8.4.7)

We recall that .s > λ > 1 and . > 0. Hence and from (8.4.4),(8.4.5) and (8.4.7), it follows that .

sup w(x  ) ≤ C1 + C2 ρ 2−λ+

2(λ−1) 2 (n−1) n + n

.

(8.4.8)

x  ∈G11/2

To prove the validity of theorem in our first case, it is enough to obtain the following estimate: sup w(x  ) ≤ const.

.

(8.4.9)

x  ∈G11/2

We shall show that the repetition by the finite time of the procedure of (8.4.8) receiving for various . can lead to the estimate (8.4.9). Let the exponent of .ρ in (8.4.8) be negative (otherwise (8.4.8) means (8.4.9)). Returning to the function .u(x) in (8.4.8) and putting .|x| = 23 ρ, we obtain u(x) ≤ C|x|2+

.

2(λ−1) n

,

(8.4.10)

8.4 The Power Modulus of the Continuity at the Conical Point

and hence, by Theorem 8.7 for . =

217

1,

.

1

=1+

2(λ − 1) , n

(8.4.11)

we get |∇u(x)| ≤ C|x| 1 .

(8.4.12)

.

Let us repeat the procedure of receiving the inequalities (8.4.7) and (8.4.8), applying the estimate (8.4.12) instead of (8.2.29) (i.e., changing . for . 1 ). As a result, we obtain .

sup w(x  ) ≤ C1 + C2 ρ 2−λ+

2(λ−1) 2 1 (n−1) n + n

.

(8.4.13)

x  ∈G11/2

If the exponent of .ρ in (8.4.13) is negative, then letting .

by Theorem 8.7 for . =

=1+

2

2,

2(λ − 1) 2(n − 1) + n n

(8.4.14)

1,

we get |∇u(x)| ≤ C19 |x| 2 ,

(8.4.15)

.

and repeating the above procedure, we get the estimate .

sup w(x  ) ≤ C1 + C2 ρ 2−λ+

2(λ−1) 2 2 (n−1) n + n

.

(8.4.16)

x  ∈G11/2

Letting t=

.

3 2(n − 1) ≥ n 2

∀n ≥ 4,

(8.4.17)

we consider the following number sequence .{ k } .

1

defined by (8.4.11), 2

3

=

=

1 (1 + t),

1 (1 + t

+ t 2 ),

·········································· k+1

=

k 1 (1 + t + . . . + t ) =

t k+1 − 1 , 1 t −1

k = 0, 1, . . . .

(8.4.18)

218

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

Repeating the stated process k times, we obtain the estimates .

sup w(x  ) ≤ C1 + C2 ρ 1−λ+

k+1

,

0 < ρ < d/2;

x  ∈G11/2

k = 0, 1, . . . .

(8.4.19)

Now we shall show that for .∀n ≥ 4 exists integer k such that 1−λ+

.

k+1

≥ 0.

(8.4.20)

From (8.4.11) and (8.4.18), we have 1−λ+

.

k+1

=

λ−1 t k+1 − 1 + (2t k+1 − 2 − nt + n). t −1 n(t − 1)

The first term on the right is positive, by (8.4.17). For the second term from (8.4.17), it follows that 2t k+1 − 2 − nt + n = 2k+2 (1 − 1/n)k+1 − n ≥ 0

.

if .{(2n − 2)/n}k+1 ≥ n/2. Hence, we get that (8.4.20) holds if k+1≥

.

+ Choosing .k =

ln n2 ln 2n−2 n

ln n2 ln 2n−2 n

.

, , where .[a] is an integer part of .a, we guarantee (8.4.20) .∀n ≥ 4.

By this, the validity of (8.1.1) for .s > λ is proved. The validity of (8.1.2) for .s > λ we get from Theorem 8.7 for . = λ − 1. Now we shall prove (8.1.3) for .s > λ. Returning to the variable x and the function .u(x) in (8.4.1), we have  

. |uxx |q + −q |∇u|q + −2q |u|q dx ≤ C4 −2q |u|q + |∇u|2q + 

G/2

2

G

|b|q |∇u|q + |f |q + |∇g|q + −q |g|q dx.

8.4 The Power Modulus of the Continuity at the Conical Point

219

Multiplying this inequality by .4−n , replacing . by .2−k , and summing up over all .k = 0, 1, . . ., we obtain  q .u 2 ≤ C r 4−n−2q |u|q + r 4−n |∇u|2q + r 4−n |b|q |∇u|q +  4 Vq,4−n (G0 )

2

G0

r 4−n |f |q + r 4−n |∇g|q + r 4−n−q |g|q dx, ∀q > 1.

Using the estimates (8.1.1)–(8.1.2), by the assumption .(D), taking into consideration .s > λ > 1, we get uV 2

.

ρ q,4−n (G0 )

≤ C

λ−2+ q4

(8.4.21)

If only .4 + (λ − 2)q > 0. From (8.4.21), we obtain the validity of (8.1.3) for .s > λ. Finally, we shall prove (8.1.4) of our theorem for .s > λ. The function .w(x  ) satisfies the problem .(QL)0 . By the Sobolev Embedding Theorem, we have .

sup x  , y  ∈ G11/2 x  = y 

|∇  w(x  ) − ∇  w(y  )| |x  − y  |

-3 C

 

1− nq



≤ C(n, q, G)wW 2,q (G1

1/2 )

≤

|w|q + ρ qλ |∇  w|2q + q |b|q |∇  w|q +

G21/4

ρ

q(2−λ)

 |f | dx  q

 q1

  q1  (1−λ)  q q  inf + C4 ρ (|∇ G| + |G| )dx ,

q > n,

(8.4.22)

G21/4

-4 do not depend on .w. Returning to the variable x -3 , C by (8.4.1), where the constants .C and using the estimate (8.1.1), by the assumption .(D), we obtain 





|w| dx = q

G21/4

ρ −qλ |u(x)|q ρ −n dx ≤

2ρ

Gρ

4 .

q C0 meas · C(q, λ)

2ρ ρ 4

dr q = C0 meas · C(q, λ) ln 8. r

220

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

Similarly, by the assumption .(D) and the estimate (8.1.2): •  .

ρ qλ |∇  w(x  )|2q dx  ≤

G21/4

 r qλ r 2q−n |∇u(x)|2q dx ≤ 2

G/4

2ρ r 3qλ−1 dr ≤ const (d, q, λ, meas ),

meas

 ∈ (0, d/2).

ρ 4

• 



 q



 |b| |∇ w(x )| dx ≤ q

.

q

q k1 meas

2ρ r q(s+λ−1)−1dr ≤ ρ 4

G21/4

const (d, k1 , q, λ, meas ),

 ∈ (0, d/2).

•  ρ

.

q(2−λ)



|f | dx ≤ q

q k1 meas

2ρ r q(s−λ)−1dr ≤ ρ 4

G21/4

const (d, k1 , q, λ, meas ),

 ∈ (0, d/2).

•  .



(|∇  G|q + |G|q )dx  · ρ q(1−λ) =

G21/4

(ρ q |∇G|q + |G|q )ρ q(1−λ)−n dx ≤ 2ρ

Gρ

4

q

2ρ

q

(g1 + g0 )meas

r q(s−λ)−1dr ≤ ρ 4

const (d, g0 , g1 , q, λ, meas ),

 ∈ (0, d/2).

From (8.4.22), by above estimates, it follows that .

sup x  , y  ∈ G11/2 x  = y 

|∇  w(x  ) − ∇  w(y  )| |x  − y  |

1− qn

≤ C5 , q > n.

(8.4.23)

8.4 The Power Modulus of the Continuity at the Conical Point

221

Returning to the variable x and the function u, we have .

sup ρ x, y ∈ Gρ/2

|∇u(x) − ∇u(y)| |x − y|

1− nq

≤ C5 ρ

λ−2+ qn

,

x = y q > n, ρ ∈ (0, d/2).

(8.4.24)

n 2−λ .

We take .τ =

|∇u(x) − ∇u(y)| ≤ C5 ρ τ |x − y|λ−1−τ , ∀x, y ∈ Gρ/2 , ρ ∈ (0, d/2).

(8.4.25)

Let us recall that from the assumptions of our theorem, we have .q > λ − 2 + qn < 0. Then from (8.4.24), it follows that ρ

.

ρ

Because .x, y ∈ Gρ/2 , then .|x − y| ≤ 2ρ, and because .τ < 0, .|x − y|τ ≥ (2ρ)τ . That is the way we obtain |∇u(x) − ∇u(y)| ≤ C5 2−τ |x − y|λ−1 , ∀x, y ∈ Gρ/2 , ρ ∈ (0, d/2) ⇒ ρ

.

.

|∇u(x) − ∇u(y)| ≤ C5 2−τ , ρ ∈ (0, d/2). |x − y|λ−1

sup ρ x, y ∈ Gρ/2 x = y d/2

(8.4.26)

ρ

Let now .x, y ∈ G0 . If .x, y ∈ Gρ/2 , ∀ρ ∈ (0, d/2), we have (8.4.26). If .|x − y| > ρ = |x|, then, because of (8.1.2), we have .

|∇u(x) − ∇u(y)| ≤ 2ρ 1−λ |∇u(x)| ≤ 2C1 . |x − y|λ−1

From this inequality and (8.4.26), it follows that .

sup d/2 x, y ∈ G0 x = y

|∇u(x) − ∇u(y)| ≤ const. |x − y|λ−1

d/2

Because of (8.4.27), (8.1.1), and (8.1.2), we get that . u ∈ C λ (G0 ). .

s < λ.

(8.4.27)

222

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

We make the transformation .x = ρx  ; w(x  ) = ρ −s u(ρx  ). The function .w(x  ) satisfies the problem ⎧ ⎨ .

⎩

a ij (x  )wxi xj = F (x  ), x  ∈ G11/2 , ∂w ∂ n 

+ χ(ω) ∂w ∂r  +

1  |x  | γ (ω)w(x )

1 , = 1−s g(ρx  ), x ∈ 1/2

(QL)0

where a ij (x  ) ≡ aij (ρx  , ρ s w, ρ s−1 wx  (x  ))

.

and F (x  ) ≡ −ρ 2−s a(ρx  , ρ s w(x  ), ρ s−1 wx  (x  )).

.

The .Lq -estimate (5.1.1) is satisfied for the function .w(x  ) (see Theorem 5.2 about .Lq estimate of solutions of the elliptic oblique problem in the smooth domain), that is, we have 

q q qs  2q q q  q q(2−s) q

≤ C1 .w +  |∇ w| +  |b| |∇ w| +  |f | |w| dx  1 2,q W

G1/2

G21/4

 + C2 q(1−s)



 |∇  G|q + |G|q dx  ,

G21/4

where constants .C1 , .C2 do not depend on w but depend on .|χ|1,∂G and .|γ |1,∂G . Now, repeating verbatim the above proof for .s > λ but taking s instead of .λ, we show the validity of our theorem in the case .s < λ. .

s = λ.

We make the transformation 3

x = x  , w(x  ) = −λ ln− 2

.

  1 u(x  ). 

Thus the function .w(x  ) satisfies the problem ⎧ ⎪ a ij (x  )wxi xj = F (x  ), ⎪ ⎪ ⎨ .

∂w

∂ n  ⎪ ⎪ ⎪ ⎩

+

1 ∂w  |x  | γ (ω)w(x ) + χ(ω) ∂r  = 3 1−λ g(x  ) ln− 2 1 ,

x ∈ G11/2 , (QL)∗0 1 , x  ∈ 1/2

8.4 The Power Modulus of the Continuity at the Conical Point

223

where       3 3 1 1 w(x  ), λ−1 ln 2 wx  (x  ) a ij (x  ) = a ij x  , λ ln 2  

.

and 



F (x ) = −

.

2−λ

       3 1 1 1  λ−1 32  − 2 a x ,  ln ln w(x ),  wx  (x ) · ln .    

λ

3 2

Let .2 ≤ n < 4. By the Sobolev Imbedding Theorem, we have .

sup |w(x  )| ≤ Cw2,2;G1 .

(8.4.28)

1/2

x  ∈G11/2

Let us return to the variable x. Because of estimation (8.3.25) of Theorem 8.10 for .s = λ, we have 

|wx  x  |2 + |∇  w|2 + |w|2 dx  ≤ w22,2;G1 = 1/2

G11/2

      1 1 r 4−2λ ln−3 |uxx |2 + r 2−2λ ln−3 |∇u|2 + r r 

G/2 .

   1 |u|2 r −n dx ≤ r   

1 r 4−n |uxx |2 + r 2−n |∇u|2 + r −n |u|2 dx ≤ −2λ ln−3  r −2λ ln−3



G/2

C

−2λ

−3

ln

    1 2λ 3 1 ·  ln = C.   (8.4.29)

Because of (8.4.28) and (8.4.29), we get .

-0 . sup |w(x  )| ≤ C

x  ∈G11/2

224

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

Therefore, returning to the variable x, we obtain -0 λ ln 2 .|u(x)| ≤ C 3

  1  , x ∈ G/2. 

We get the desired estimation (8.1.1) for .s = λ and .2 ≤ n < 4 if we put .|x| = 23 . Let .n ≥ 4. We apply the Lieberman local maximum principle, Proposition 8.2. By assumption .(D), we get .

sup |w(x  )| ≤

x  ∈G11/2

+  2

C

w dx



 12

A

 

+ g0 + 

 n

2−λ

G21/2

|a(x )| dx



 n1 B

− 32

ln

 , 1 . 

(8.4.30)

G21/4

Let us estimate integrals from the right side of (8.4.30). For the first integral, we use estimation (8.3.25) for .s = λ of Theorem 8.10. We have  .

w2 dx  ≤ −2λ ln−3

   1 r −n u2 dx ≤ C. 

G21/4

(8.4.31)

2

G/4

Now, we estimate the second integral from the right side of (8.4.30). Because of assumption .(D) and (8.2.29), we obtain  

μn1 |∇u|2n + bn |∇u|n + f n r −n dx ≤ . |a(x  )|n dx  ≤ c(n) G21/4

2

G/4

 & c(n)





μn1 r 2−n |∇u|2 r −2 |∇u|2n−2 +

2

G/4





' r 2−n |∇u|2 k1n r (λ−2)n−2 |∇u|n−2 + k1n r (λ−2)n−n dx ≤

 n 2n−1 (2n−2) −2 n 2n−2 (n−2)+(λ−2)n−2 c(n) μ1 C1  + k1 C1  r 2−n |∇u|2 dx+ 2

 c(n, ω0 )k1n

G/4 1 (λ−2)n , (λ−2)n 

if λ = 2,

ln 8,

if λ = 2,

0 λ), we derive sup |w(x  )| ≤ const.

.

(8.4.34)

x  ∈G11/2

By this, putting .|x| = 23 , we proved the validity of (8.1.1) for .s = λ. 3 Further, the function .w(x  ) = −λ ln− 2 1 u(x  ) satisfies the problem .(QL)∗0 . We apply again the assumption .(E): .

max |∇  w(x  )| ≤ M1 .

x  ∈G11/2

(8.4.35)

Hence, it follows that |∇u(x)| ≤ M1 

.

λ−1

  1  ln , x ∈ G/2, 0 <  < d.  3 2

Putting .|x| = 23 , we get the validity of (8.1.2) for .s = λ. Now, we prove the validity of (8.1.3). By Theorem 5.2, we have q

w

.



W 2,q G11/2

≤



C

q w q 2 L G1/4

+

q F  q 2 L G1/4

+

(1−λ)q

ln

− 3q 2

   1 q g 1− 1 ,q ≤ 2  W q 1/4

226

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

 & C1

|w| +  q

(2−λ)q

ln

− 3q 2

    1  2q (2λ−2)q 3q 1 ln · |∇ w|  +  

G21/4 

|b| |∇ w|  q

q (λ−1)q

    q  1 · |G| + |∇  G|q dx  = 

3q

C2 (1−λ)q ln− 2

ln

3q 2

 

' 1 + |f |q dx  + 

G21/4

    3q 1 q λq  2q 2 C1 |w| +  |∇ w| ln + |b|q q |∇  w|q +  G21/4



(2−λ)q

− 3q 2

|f | ln q

      q  1 1  (1−λ)q − 3q 2 dx + C2  · |G| + |∇  G|q dx  ≤ ln   G21/4

3q

- (2−λ)q−n ln− 2 C

   $ 1 r −2q |u|q + |∇u|2q + |b|q |∇u|q + |f |q + |∇G|q + ·  2

G/4

% r −q |G|q dx,

(8.4.36)

- do not depend on u. However, where constants .C1 , C2 , C   q  q

= |w| + |∇  w|q + |wx  x  |q dx  = .w 1 2,q W

G1/2

G11/2 3q

(2−λ)q−n ln− 2

  

1 r −2q |u|q + r (−q |∇u|q + |uxx |q dx. · 

(8.4.37)

 G/2

From the above (8.4.36)–(8.4.37), it follows that 

r −2q |u|q + r −q |∇u|q + |uxx |q dx ≤ . 

G/2

-· C

 $ 2

G/4

% r −2q |u|q + |∇u|2q + |b|q |∇u|q + |f |q + |∇G|q + r −q |G|q dx.

8.4 The Power Modulus of the Continuity at the Conical Point

227

We multiply this inequality by .r 4−n , replace . by .2−k , and sum up over all .k = 0, 1, 2, . . ., and we obtain 

q    ≤ C4 .u 2 r 4−n |∇G|q + r 4−n−q |G|q dx+ Vq,4−n G0

2

G0

 C3

r 4−n−2q |u|q + r 4−n |∇u|2q + r 4−n |b|q |∇u|q + r 4−n |f |q dx.

2

G0

By (8.1.1), (8.1.2), and assumption .(D), taking into account that .s = λ > 1, q > n, we obtain q

uV 2

.

q,4−n





G0

≤     2  1 3+q(λ−2) 3q 3+2q(λ−1) 3q 1 2 +r + r ln ln C6 r r 0

r



3+2q λ− 23

ln

3q 2

   1 + r 3+q(λ−2) dr ≤ r 2 r 3+q(λ−2) ln3q

C7

  d 1 1 dr, 0 <  < < . r 2 2e

(8.4.38)

0

By the assumption of our Theorem . 4 + q(λ − 2) > 0. Then there exists a value .α, such that .−3 − q(λ − 2) < α < 1. Let us consider the following function: 1 h(r) = r α+3+q(λ−2) ln3q . r

.

We calculate: h (r) = r α+2+q(λ−2) ln3q−1

.

  1 1 [α + 3 + q(λ − 2)] ln − 3q . r r

228

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

Then, for all .r ∈ (0, 2], where .0 <  < 0 ⇒ h(r) ≤ h(2) and 2 3+q(λ−2)

r

.

3q

ln

1 dr = r

0

2 r

d 2,

3q d ≤ exp − α+3+q(λ−2) , we have .h (r) >

−α α+3+q(λ−2)

r

ln

3q

1 dr = r

2

0

r −α h(r)dr ≤

0

2 h(2)

r −α dr ≤ c(α, q)4+q(λ−2) ln3q

1 , c(α, q) > 0. 

0

From here and (8.4.38), we get the validity of (8.1.3).  Finally, we

shall prove (8.1.4) of our theorem for .s = λ. The function .w(x ) = 3 −λ ln− 2 1 u(x  ) satisfies the problem .(QL)∗0 . By the Sobolev Embedding Theorem, .

|∇  w(x  ) − ∇  w(y  )|

sup x  , y  ∈ G11/2 x  = y 

|x  − y  |

≤

1− nq

C(n, q, G).wW 2,q (G1

1/2 )

(8.4.39)

From (8.4.36), we have q

w

.



W 2,q G11/2

3q

- (2−λ)q−n ln− 2 ≤ C

 $

  1 × 

% r −2q |u|q + |∇u|2q + |b|q |∇u|q + |f |q + |∇G|q + r −q |G|q dx.

2

G/4

Returning to the variable x and the function u, by (8.1.1), (8.1.2), and assumption .(D), taking into account that .s = λ > 1, q > n, we obtain from above .

sup ρ x, y ∈ Gρ/2 x = y

|∇u(x) − ∇u(y)| |x − y|

3q

r 2(λ−1)q ln 2

1− nq

1

-q ≤C

+  

3q

r (λ−2)q ln 2

  1 + r

2

G/4

     , q1 3q 1 1 q ≤ + k1 r (2λ−3)q ln 2 + (k1 + g0 + g1 )q r (λ−2)q dx r r

8.4 The Power Modulus of the Continuity at the Conical Point

-1 (q, k1 , g0 , g1 , meas ) C

229

+ 2 r

(λ−2)q+n−1

3q 2

  , q1 1 dr , r

2 . e

(8.4.40)

ln

 4

q > n, ρ ∈ (0, d/2), d < Since 3q

.

lim r qε ln 2

r→+0

    3q 1 1 = 0 ⇒ ln 2 ≤ const · r −qε , r ∈ (0, d/2), r r

from (8.4.40), it follows that .

sup ρ x, y ∈ Gρ/2 x = y

|∇u(x) − ∇u(y)| |x − y|

1− nq

-2  ≤C

λ−2+ nq −ε

, ∀ε > 0.

Let us recall that from the assumptions of our theorem, we have .q > λ − 2 + qn − ε < 0. Then from above, we obtain

n 2−λ .

We take .τ =

-2 ρ τ |x − y|λ−1−τ −ε , ∀x, y ∈ G , ρ ∈ (0, d/2), ∀ε > 0. |∇u(x) − ∇u(y)| ≤ C ρ/2 ρ

.

ρ

Because .x, y ∈ Gρ/2 , then .|x − y| ≤ 2ρ, and because .τ < 0, .|x − y|τ ≥ (2ρ)τ . That is the way we get -2 2−τ |x − y|λ−1−ε , ∀ε > 0 ⇒ |∇u(x) − ∇u(y)| ≤ C

.

.

sup ρ x, y ∈ Gρ/2 x = y

|∇u(x) − ∇u(y)| - −τ ≤ C2 2 , ρ ∈ (0, d/2), ∀ε > 0. |x − y|λ−1−ε

d/2

ρ

(8.4.41)

Let now .x, y ∈ G0 . If .x, y ∈ Gρ/2 , ∀ρ ∈ (0, d/2), we have (8.4.41). If .|x − y| > ρ = |x|, then, because of (8.1.2), we have 3q |∇u(x) − ∇u(y)| . ≤ 2ρ 1−λ+ε |∇u(x)| ≤ 2C1 ε ln 2 λ−1−ε |x − y|

  1 ≤ 2const. r

230

8 Behavior of Strong Solutions to the Degenerate Oblique Derivative. . .

From this inequality and (8.4.41), it follows that .

sup d/2 x, y ∈ G0 x = y

|∇u(x) − ∇u(y)| ≤ const. |x − y|λ−1−ε

d/2

Because of (8.4.42), (8.1.1), and (8.1.2), we get that . u ∈ C λ−ε (G0 ), ∀ε > 0.

8.5

(8.4.42)

 

Notes

The systematic theory of oblique derivative problems for quasi-linear second order elliptic equations in smooth domains was developed by G. Lieberman [89] (Chapters 6–11). There, maximum and Hölder estimates, gradient and higher order estimates as well as existence theorems for classical solutions are proved. Ibid there is also an extensive bibliography with a thorough discussion of works related to the problem considered. Let us take a look at some other work. In [56], a priori estimates and strong solvability 2,n (G) ∩ results in Sobolev space . W 2,q (G), q > n, as well as uniqueness result in . Wloc 1 C (G) are proved for the regular oblique derivative problem for quasi-linear uniformly elliptic equations: ⎧ n  ij ⎪ ⎨ a (x, u)uxi xj + b(x, u, Du) = 0 a.e. G ⊂ Rn , n ≥ 3, .

⎪ ⎩

i,j =1 ∂u → + γ (x)u − ∂ l

= g(x)

on ∂G ∈ C 1, 1,

where . a ij ∈ VMO ∩ L∞ with respect to x, and b to grow at most quadratically with the → − gradient; it is assumed that unit vector field . l (x) is nowhere tangential to the boundary . ∂G, i.e., the oblique derivative problem is regular.

The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order Equation with Perturbed p(x)-Laplacian

.

9.1

Setting of the Problem

We use the following notations: • .(r, ω): Polar coordinates of a point .x = (x1 , x2 ) in .R2 with the pole .O = (0, 0) defined by .

• • • • • •

x1 = r cos ω, x2 = r sin ω

S 1 : The unit circle in .R2 centered at .O   K: An open angle .{r > 0, ω ∈ − ω20 , ω20 } with vertex .O ω0 ω0 .∂K: The lateral sides of .K : x1 = r cos 2 , x2 = ±r sin 2  ω0 ω0  1 . = − 2 , 2 An arc obtained by intersecting the angle K with .S R .G := {(r, ω) : 0 < r < R, ω ∈ } ⊂ {x1 ≥ 0}; be a sector with angle .ω0 ∈ (0, π) 0 R R R R := {(r, ω) : 0 < r < R, ω = ± ω0 } be lateral sides of this . ± 0 = + ∪ − , 2 angle, thus .

R ± =

.

  ω0  ω0 r, ± x1 = ±x2 cot ; r ∈ (0, R) 2 2

(9.1.1)

• . R := K ∩ {|x| = R}

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Borsuk, Oblique Derivative Problems for Elliptic Equations in Conical Domains, Frontiers in Elliptic and Parabolic Problems, https://doi.org/10.1007/978-3-031-28381-9_9

231

9

232

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

Let us fix .R > 0. We consider the following oblique derivative problem involving the perturbed .p(x)—Laplacian with singular nonlinearities and absorption term:

(OP2)

.



⎧ χ(ω) d p(x)−2 ∂u + a (x)u|u|p(x)−1+ ⎪ −

u − |∇u| 0 p(x) ⎪ dx2 cos ω ∂r ⎪ ⎪ ⎪ ⎨ b(u, ∇u) = f (x), x ∈ GR 0,   γ (ω) ∂u ∂u p(x)−2 p(x)−2 = 0, ⎪ ⎪ |∇u| ⎪ ∂ n + χ(ω)sign ω ∂r + |x|p(x)−1 u|u| ⎪ ⎪ ⎩ x ∈ 0R ∪ R ,

where .n denotes the unit outward normal vector on .0R ∪ R . We will work under the following assumptions: (i) .1 < p− ≤ p(x) ≤ p+ = p(0) ≤ 2, ∀x ∈ GR 0. (ii) The Lipschitz condition: .p(x) ∈ C 0,1 (GR 0 ) ⇒ 0 ≤ p+ − p(x) ≤ L|x|, ∀x ∈ GR 0 , where L is the Lipschitz constant for .p(x).

(iii) There exist a positive constant .a0 such that .a0 (x) ≥ a0 > 0 and a nonnegative + −1 constant .f0 ≥ 0 such that .|f (x)| ≤ f0 |x|β(x), β(x) > p+p−1+μ (p(x) − 1) λ− 2s ; s > 2 p− ;

∀x ∈ GR μ ∈ [0, 1) and .λ is the least positive eigenvalue of .(OEV P ) (see 0; below Sect. 9.4).

9.1 Setting of the Problem

233

Let ω ω 0 0 ; χ± = χ ± . γ± = γ ± 2 2

(9.1.2)

.

) ( (iii1) .γ (ω) ≥ γ0 ≥ 1, ∀ω ∈ − ω20 , + ω20 ; and .γ− > γ+ ≥ 1. (iii2) .χ(ω) ∈ C 1 ( ), χ  (ω) < 0, ⇒ 0 ≤ χ+ ≤ χ(ω) ≤ χ− , ∀ω ∈ ;

χ− ≤

.

 ⎧ ⎪ p− −1 ⎪ min ⎪ ω ; p ⎪ 2− 2+ +2 tan 20 ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

 ( (μ−δ) cos(ω0 /2) 6(3+4| −1|)

(p− −1) cos(ω0 /2)  p  6(3+4| −1|)+ 2+2 tan(ω0 /2)− 2+ cos(ω0 /2)

for μ > 0, ∀δ ∈ (0, μ);

for μ = 0, (9.1.3)

+ −1 where . = p+p−1+μ λ. (iv) The function .b(u, ξ ) is differentiable with respect to the .u, ξ variables in .M = R × Rn and satisfies in .M the following inequalities:

(iv)a

|b(u, ξ )| ≤ δ|u|−1 |ξ |p(x) + b0 |u|p(x)−1,

(iv)b

b(u, ξ ) ≥ ν|u|−1 |ξ |p(x) − b0 |u|p(x)−1,

.

(iv)c

.

3 4 n   4 ∂b(u, ξ ) 2 5   ≤ b1 |u|−1 |ξ |p(x)−1 ;  ∂ξ  i=1

i

0 < δ < μ; if μ > 0; if μ = 0;

ν > 0;

∂b(u, ξ ) ≥ b2 |u|−2 |ξ |p(x); ∂u

b0 ≥ 0, b1 ≥ 0, b2 ≥ 0. Definition 9.1 A function u is called a bounded weak solution of problem .(OP2) provided 1,p(x) that .u ∈ N−1,∞ (GR 0 ) and satisfies the integral identity .

  |∇u|p(x)−2 ∇u, ∇η + χ(ω)|∇u|p(x)−2(ux1 + ux2 tan ω)ηx2 +

GR 0

(I I )

  a0 (x)u|u|p(x)−1η + b(u, ∇u)η dx + γ (ω)r 1−p(x)u|u|p(x)−2ηds+ 0R

 R 1,p(x)

for all .η ∈ N−1,∞ (GR 0 ).

2−p(x)

1 + χ(ω) tan ω γ (ω)u|u|p(x)−2ηdω = 1 + χ(ω)sign ω

 f ηdx GR 0

234

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

Remark 9.2 For the justification of this definition, see below (9.2.3) in Sect. 9.3. Lemma 9.3 Let u be a weak bounded solution of problem .(OP2) . For any function .η ∈ 1,p(x) N−1,∞ (GR 0 ), the equality Q[u, η] ≡

.

  |∇u|p(x)−2 ∇u, ∇η + χ(ω)|∇u|p(x)−2(ux1 + ux2 tan ω)ηx2 + Gd0

 a0 (x)u|u|p(x)−1η + b(u, ∇u)η dx+

(I I )loc  d 2−p(x)

 d·

|∇u|p(x)−2

1 + χ(ω) tan ω γ (ω)u|u|p(x)−2ηdω− 1 + χ(ω)sign ω

∂u (1 + χ(ω) tan ω) ηdω + ∂r



 γ (ω)r 1−p(x)u|u|p(x)−2ηds = 0d

 f ηdx Gd0

holds for a.e. .d ∈ (0, R). Proof Let us consider the integral identity .(I I ) in which we replace .η(x) by .η(x)χd (x), where .χd (x) is the characteristic function of .Gd0 . We have .

∂ ∂χd (x) ; i = 1, 2. (η(x)χd (x)) = ηxi χd (x) + η(x) ∂xi ∂xi

According to formula (7’) of Subsection 3 §1, Chapter 3 [60], .

∂χd (x) xi = − δ(d − r), i = 1, 2, ∂xi r

where . δ(d − r) denotes the Dirac delta distribution lumped on the circle .r = d. Thus,   .

|∇u|

p(x)−2

∇u, ∇χd (x) + χ(ω)|∇u|

p(x)−2

(ux1

 ∂χd (x) + ux2 tan ω) × ∂x2

Gd0

η(x)dx =

9.2 Preliminary

235



 −

|∇u|p(x)−2

 x2 ∂u xi + χ(ω)(ux1 + ux2 tan ω) · · η(x)δ(d − r)dx = ∂xi r r

Gd0

 −

(by below formulas (9.2.1)–(9.2.2);) |∇u|p(x)−2

∂u (1 + χ(ω) tan ω) ηd d = ∂r

d

 −d ·

|∇u|p(x)−2

∂u (1 + χ(ω) tan ω) ηdω. ∂r



 

Hence, we derive the required statement. The main result about power modulus of continuity is as follows:

Theorem 9.4 Let u be a weak bounded solution of problem .(OP2) , let .M0 = sup |u(x)| x∈GR 0

(see Theorem 9.8 Sect. 9.3), and let .λ be the least positive eigenvalue of problem .(OEVP) (see Sect. 9.4). Suppose that assumptions (i)–(iv) hold. Then there exist .d- ∈ (0, 1) and a a0 , γ0 , χ− , b0 , f0 constant .C0 > 0 depending only on .λ, d, M0 , p+ , p− , L, n, (μ − δ), ν,and such that |u(x)| ≤ C0 |x| ,

.

9.2

=

p+ − 1 λ; p+ − 1 + μ

-

∀x ∈ Gd0 .

(9.1.4)

Preliminary

We will need the following obvious (see figure) equalities: .

  cos(n, x1 )   cos(n, x2 )

d +

d +

  cos(n, x1 )

.

d

= − sin = cos

ω0 ; 2

ω0 ; 2

= cos ω;

d d d = cos ω + sin ω ; dr dx1 dx2

  cos(n, x1 )   cos(n, x2 )

d −

d −

  cos(n, x2 )

d

ω0 ; 2 ω0 = − cos ; 2 = − sin

(9.2.1)

= sin ω;

cos ω d d d + . = sin ω dx2 dr r dω

(9.2.2)

236

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

By these equalities, we shall justify integral identity .(I I ) for definition of the week solution of our problem. For this, we calculate via the integration by parts the integral  @  ? d χ(ω) p(x)−2 ∂u |∇u| .−

p(x)u + η(x)dx = dx2 cos ω ∂r 

GR 0



|∇u|p(x)−2 ∇u, ∇η + χ(ω)(ux1 + ux2 tan ω)ηx2 dx−

GR 0



 |∇u|

R ∪ R ∪ + R −

p(x)−2

 ∂u χ(ω) ∂u + cos(

n, x2 ) η(x)ds. ∂ n

cos ω ∂r

From the boundary condition of problem .(P ) with regard to (9.1.2) and (9.2.1), we have: R: • On .±

 |∇u|

.

p(x)−2

   ∂u ∂u χ(ω) ∂u p(x)−2 ∂u + cos(

n, x2 ) = |∇u| ± χ± = ∂ n

cos ω ∂r ∂ n

∂r − γ± r 1−p(x)u|u|p(x)−2.

• On . R :  ∂u χ(ω) ∂u ∂u + cos(

n, x2 ) = |∇u|p(x)−2 (1 + χ(ω) tan ω) = .|∇u| ∂ n

cos ω ∂r ∂r 1 + χ(ω) tan ω ∂u  ∂u  − γ (ω)d 1−p(x)u|u|p(x)−2, because of  =  . 1 + χ(ω)sign ω ∂ n R ∂r R 

p(x)−2

Thus, from above equalities, we obtain @   ? χ(ω) d p(x)−2 ∂u |∇u| η(x)dx =

p(x)u + .− dx2 cos ω ∂r  GR 0

GR 0



|∇u|p(x)−2 ∇u, ∇η + χ(ω)(ux1 + ux2 tan ω)ηx2 dx+

(9.2.3)

9.2 Preliminary

237

 γ (ω)|x|1−p(x)u|u|p(x)−2η(x)ds+ 0R



R 2−p(x)

1 + χ(ω) tan ω γ (ω)u|u|p(x)−2η(x)dω. 1 + χ(ω)sign ω

By this is justified integral identity .(I I ). Now we define functions: A1 (ω, ξ ) = |ξ |p(x)−2ξ1 ; A2 (ω, ξ ) = |ξ |p(x)−2ξ2 + χ(ω)|ξ |p(x)−2(ξ1 + ξ2 tan ω); ' ∂A1 (ω, ξ ) & . = (p(x) − 1)ξ12 + ξ22 · |ξ |p(x)−4; a11 (ω, ξ ) ≡ ∂ξ1 a12 (ω, ξ ) ≡

.

a21 (ω, ξ ) ≡

∂A1 (ω, ξ ) = (p(x) − 2)ξ1 ξ2 |ξ |p(x)−4; ∂ξ2

∂A2 (ω, ξ ) & = (p(x) − 1)χ(ω)ξ12 + ∂ξ1

' (p(x) − 2)(1 + χ(ω) tan ω)ξ1 ξ2 + χ(ω)ξ22 |ξ |p(x)−4;

a22 (ω, ξ ) ≡

∂A2 (ω, ξ ) = (1 + χ(ω) tan ω)|ξ |p(x)−2+ ∂ξ2 ' & (p(x) − 2) χ(ω)ξ1 ξ2 + (1 + χ(ω) tan ω)ξ22 · |ξ |p(x)−4.

We shall study the quadratic form F (ζ ) = a11 (ω, ξ )ζ12 + a12(ω, ξ )ζ1 ζ2 + a21 (ω, ξ )ζ1 ζ2 + a22 (ω, ξ )ζ22 ,

.

∀ζ = (ζ1 , ζ2 ) ∈ R2 \ {0}. Proposition 9.5 The quadratic form .F (ζ ) is positive definite, if 0 ≤ χ(ω) < cot

.

ω0 2

& ω ω ' 0 0 ∀ω ∈ − , . 2 2

Proof It is well known that for this, the inequalities a11 > 0,

.

    a11 a12  >0  a21 a22 

(9.2.4)

238

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

must be satisfied. The first inequality is obvious. Next we calculate a11 a22 − a12 a21 = (p(x) − 1)(1 + χ(ω) tan ω)|ξ |2(p(x)−2) ≥ ω0 2(p(x)−2) |ξ | > 0, (p(x) − 1) 1 − χ(ω) tan 2

.

by (9.2.4).  

Proposition 9.6 Suppose that (9.1.3) holds. form .F (ζ ) is positive   Then the quadratic definite, and there exists positive .ν = const p± , ω0 , χ− , such that F (ζ ) ≥ ν|ξ |p(x)−2 |ζ |2.

.

(9.2.5)

Proof We compose the characteristic equation of the quadratic form .F (ζ ) :    a +a  a11 − λ 12 2 21  .  = 0. 21  a12 +a a22 − λ 2 Solving this equation, we find characteristic numbers λ1 =

.

* 1 a11 + a22 −  , 2

λ2 =

* 1 a11 + a22 +  , . 2

 = (a22 − a11)2 + (a12 + a21 )2 .

(9.2.6) (9.2.7)

It is well known that the inequality λ1 |ζ |2 ≤ F (ζ ) ≤ λ2 |ζ |2

.

(9.2.8)

holds. We are interested in the lower bound; therefore, we shall calculate .λ1 . ) ( Using the Cauchy inequality and that . tan ω ≥ − tan ω20 , ∀ω ∈ − ω20 , ω20 , we obtain . a11 + a22 = (p(x) + χ(ω) tan ω) ξ12 − (2 − p(x))χ(ω)ξ1 ξ2 + & ' / p(x) + (p(x) − 1) χ(ω) tan ω ξ22 |ξ |p(x)−4 ≥

.

.

2 − p(x) χ(ω) ξ12 + ξ22 + (p(x) + χ(ω) tan ω) ξ12 − 2 & ' / p(x) + (p(x) − 1) χ(ω) tan ω ξ22 |ξ |p(x)−4 ≥

9.2 Preliminary

239

 , .+ 2 − p(x) ω0 + p(x) − χ(ω) tan ξ12 + 2 2 +  , / ω0 2 − p(x) p(x) − χ(ω) (p(x) − 1) tan + ξ22 |ξ |p(x)−4 ≥ 2 2 ?  @ 2 − p(x) ω0 + p(x) − χ(ω) tan |ξ |p(x)−2 ≥ 2 2 ? @  2 − p+ ω0 p− − χ− tan + |ξ |p(x)−2 > 0, (9.2.9) 2 2 if χ−
0, 2 2 2

λ1 >

.

(9.2.11)

by (9.1.3) (see assumption (iii2)). Thus, from (9.2.11), we get ν=

.

p+ ' 1& ω0 − p− − 1 − χ− 2 + 2 tan > 0. 2 2 2

(9.2.12)  

9.3 The Maximum Principle

241

Remark 9.7 Proved above propositions mean that our problem equation is uniformly elliptic.

9.3

The Maximum Principle

Theorem 9.8 Let u be a weak solution to the problem (OP2 ) and assumptions (i)–(iv) be satisfied. Then, there exists a positive constant M0 depending only on p, s, μ,a0 , γ0 , χ− , f Ls (GR ) , such that 0

uL∞ (GR ) ≤ M0 .

.

0

$

% Proof Let us define the set A(k) = x ∈ GR : |u(x)| > k with χA(k) being the 0 characteristic function of the set A(k). Then, for all q > 0, we have that A(k + q) ⊆ A(k). We take η((|u| − k)+)χA(k) · sign u as a test function in the integral identity (I I ), where η is defined by (2.14.5) and k ≥ k0 ≥ 1. Note that η((|u|−k)+ ) ≥ 0 and η ((|u|−k)+ ) ≥ 0 on A(k). By assumption (iv)a, (iv)b , we have b(u, ∇u)η((|u| − k)+ ) · sign u ≥

.

− δ|u|−1 |∇u|p(x) + b0 |u|p(x)−1 η((|u| − k)+ ). p −1

But, since |u| > k ≥ k0 ≥ 1, p(x) ≥ p− > 1, we have |u|p(x)−1 ≥ k0 − from above and from the integral identity (I I ), it follows:  $. .

A(k)

/ |∇u|p(x) + χ(ω)|∇u|p(x)−2(ux1 + ux2 tan ω)ux2 η ((|u| − k)+ )+

% p p −1 a0 k0 − η((|u| − k)+ ) − δk0−1 |∇u|p(x) + b0 |k0 − η((|u| − k)+ ) dx+  p− −1 γ0 k 0 · r 1−p(x)η((|u| − k)+ )ds+ 0R ∩A(k)

 R 2−p(x) ∩A(k)

, and therefore

1 + χ(ω) tan ω γ (ω)|u|p(x)−1η((|u| − k)+ )dω ≤ 1 + χ(ω)sign ω  |f |η((|u| − k)+ )dx. A(k)

242

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

  ) ( Setting − = − ω20 , 0 ; + = 0, + ω20 , we have = − ∪ + . Therefore, = + . Now we obtain − ∩A(k)

∩A(k)

+ ∩A(k)

 R 2−p(x)

.

+ ∩A(k)

1 + χ(ω) tan ω γ (ω)|u|p(x)−1η((|u| − k)+ )dω = 1 + χ(ω)sign ω  R 2−p(x)

+ ∩A(k)

1 + χ(ω) tan ω γ (ω)|u|p(x)−1η((|u| − k)+ )dω ≥ 0, 1 + χ(ω)

by assumptions (iii1)−(iii2). Next, by (9.1.3), we verify that .

1 + χ(ω) tan ω > 0, ∀ω ∈ − 1 − χ(ω)

(9.3.1)

and therefore again, by assumptions (iii1)−(iii2),  R 2−p(x)

.

− ∩A(k)

1 + χ(ω) tan ω γ (ω)|u|p(x)−1η((|u| − k)+ )dω = 1 + χ(ω)sign ω  R 2−p(x)

− ∩A(k)

1 + χ(ω) tan ω γ (ω)|u|p(x)−1η((|u| − k)+ )dω ≥ 0. 1 − χ(ω)

At finally, from above inequalities, we get  $. .

/ |∇u|p(x) + χ(ω)|∇u|p(x)−2(ux1 + ux2 tan ω)ux2 η ((|u| − k)+ )+

A(k)

%

p p −1 a0 k0 − η((|u| − k)+ ) − δk0−1 |∇u|p(x) + b0 |k0 − η((|u| − k)+ ) dx ≤  |f |η((|u| − k)+ )dx. (9.3.2) A(k)

9.3 The Maximum Principle

243

Next, since |ux1 u x2 | ≤ 12 |∇u|2 and | tan ω| ≤ tan ω20 , then χ(ω)(ux1 + ux2 tan ω)ux2 ≥ −χ− 12 + tan ω20 |∇u|2 ; therefore, from (9.3.2), it follows .

 @   ? ω0 1 p + tan χ− |∇u|p(x)η ((|u| − k)+ ) + 1− a0 k0 − η((|u| − k)+ )− 2 2

A(k)



% p −1 δk0−1 |∇u|p(x) + b0 |k0 − η((|u| − k)+ ) dx ≤ |f |η((|u| − k)+ )dx.

(9.3.3)

A(k)

From assumption (iii2) (see (9.1.3)), it follows that χ− ≤

.

1 ⇒ 1 − 1 + 2 tan ω20



ω0 1 + tan 2 2

 χ− ≥

1 ; 2

then we obtain .

 

p −1 η((|u| − k)+ )+ |∇u|p(x)η ((|u| − k)+ ) − 2 δk0−1 |∇u|p(x) + b0 k0 −

A(k)

 % p 2a0 k0 − η((|u| − k)+ ) dxdx ≤ 2 |f |η((|u| − k)+ )dx, A(k)

or .

 

|∇u|p(x) η ((|u| − k)+ ) − 2δk0−1 η((|u| − k)+ ) +

A(k) p −1

2k0 −

 % (a0 k0 − b0 )η((|u| − k)+ ) dxdx ≤ 2 |f |η((|u| − k)+ )dx. A(k)

Choosing k0 >

.

b0 , a0

(9.3.4)

244

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

we set aˆ 0 = 2(a0k0 − b0 ) > 0 , and from above, it follows .

 

|∇u|p(x) η ((|u| − k)+ ) − 2δk0−1 η((|u| − k)+ ) +

A(k) p −1

aˆ 0 k0 −

 % η((|u| − k)+ ) dxdx ≤ 2 |f |η((|u| − k)+ )dx.

(9.3.5)

A(k)

Additionally, let us define the sets .

A− (k) = A(k) ∩ {|∇u| ≤ 1}, ⇒ A+ (k) = A(k) ∩ {|∇u| ≥ 1}

A(k) = A− (k) ∪ A+ (k)

(9.3.6)

and the functions  vk (x) := η

.

((|u| − k)+ )+ p−



 ,

wk (x) := η

((|u| − k)+ )+ p+

 .

(9.3.7)

We note that the inequalities |∇u|p+ ≤ |∇u|p(x) ≤ |∇u|p−

on A− (k); .

(9.3.8)

|∇u|p− ≤ |∇u|p(x) ≤ |∇u|p+

on A+ (k)

(9.3.9)

.

hold by (i). Direct calculations give 1 .|∇vk | = |∇u| · η p−



((|u| − k)+ )+ p− 

|∇vk |p− = Choosing ς > p− +

4δ k0



  ς ((|u| − k)+ )+ = , |∇u| · exp ς p− p−

ς > 0; ⇒  ς p− |∇u|p− · eς((|u|−k)+ )+ . p−

according to (2.14.6), we have

η (((|u| − k)+ )+ ) − 2δk0−1 η(((|u| − k)+ )+ ) ≥

.

(9.3.10)

1 ς((|u|−k)+)+ e . 2

(9.3.11)

9.3 The Maximum Principle

245

From (9.3.10), (9.3.11), it follows that / . |∇u|p− η (((|u| − k)+ )+ ) − 2δk0−1 η(((|u| − k)+ )+ ) ≥   1 p− p− |∇vk |p− ⇒ (by (9.3.9)) 2 ς  . / |∇u|p(x) η (((|u| − k)+ )+ ) − 2δk0−1 η(((|u| − k)+ )+ ) dx ≥

.

A+ (k)



. / |∇u|p− η (((|u| − k)+ )+ ) − 2δk0−1 η(((|u| − k)+ )+ ) dx ≥

A+ (k)



1 2

p− ς

p−

 ·

|∇vk |p− dx.

(9.3.12)

A+ (k)

Similarly, choosing ς > p+ + 

4δ k0

and taking into account (9.3.8), we obtain

/ . |∇u|p(x) η (((|u| − k)+ )+ ) − 2δk0−1 η(((|u| − k)+ )+ ) dx ≥

.

A− (k)

1 2



p+ ς

p+

 ·

|∇wk |p+ dx.

(9.3.13)

A− (k)

Since p+ ≥ p− , the inequalities (9.3.12) and (9.3.13) hold for ς > p+ + adding the inequalities (9.3.12) and (9.3.13), we get 1 . 2



p− ς

p−

 ·

|∇vk |

p−

1 dx + 2



p+ ς

p+

A+ (k)



4δ k0 .

Therefore,

 ·

|∇wk |p+ dx ≤

A− (k)

. / |∇u|p(x) η (((|u| − k)+ )+ ) − 2δk0−1 η(((|u| − k)+ )+ ) dx

(9.3.14)

A(k)

by (9.3.6). Finally, from (9.3.5), (9.3.14), we derive 1 . 2



p− ς

p−

 ·

|∇vk |

p−

A+ (k) p −1

aˆ 0 k0 −

1 dx + 2



p+ ς

p+

 ·

|∇wk |p+ dx+

A− (k)

 · A(k)

 η(((|u| − k)+ )+ )dx ≤

|f (x)| · η(((|u| − k)+ )+ )dx. A(k)

246

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

=

Since A(k)

+ A+ (k)

.

1 2



p −1

aˆ 0 k0 −

p− ς

, by (9.3.6), we have A− (k)

p−

 ·

|∇vk |p− dx +

1 2



p+ ς

p+

A+ (k)



p −1

η(((|u| − k)+ )+ )dx + aˆ 0 k0 −

· A+ (k)

 ·



|∇wk |p+ dx+

A− (k)

·

η(((|u| − k)+ )+ )dx ≤

A− (k)

(9.3.15)



 |f (x)| · η(((|u| − k)+ )+ )dx + A+ (k)

|f (x)| · η(((|u| − k)+ )+ )dx.

A− (k)

Now, by (2.14.7) and (9.3.7), we derive p −1

aˆ 0 k0 −

.



p −1

η(((|u| − k)+ )+ )dx + aˆ 0 k0 −

· A+ (k)

⎛ p −1

aˆ 0 k0 −

⎜ ·⎝

 ·

η(((|u| − k)+ )+ )dx ≥

A− (k)





p

⎟ p wk + dx ⎠ .

vk − dx +

A+ (k)

⎞ (9.3.16)

A− (k)

From (9.3.15)–(9.3.16), it follows that .

1 2



p− ς

p−

 ·

|∇vk |p− dx +

A+ (k) p −1

aˆ 0 k0 −

⎛ ⎜ ·⎝

1 2

p+ ς

p+

 ·

|∇wk |p+ dx+

A− (k)





p

vk − dx +

A+ (k)







⎟ p wk + dx ⎠ ≤

(9.3.17)

A− (k)

|f (x)| · η(((|u| − k)+ )+ )dx + A+ (k)

⎞

|f (x)| · η(((|u| − k)+ )+ )dx.

A− (k)

Next, we have 

 .

A± (k)

|f (x)| · η(((|u| − k)+ )+ )dx = 

|f (x)| · η(((|u| − k)+ )+ )dx+ A± (k+q)

|f (x)| · η(((|u| − k)+ )+ )dx, A± (k)\A± (k+q)

∀q > 0.

(9.3.18)

9.3 The Maximum Principle

247

By (2.14.8), we obtain   .η(((|u| − k)+ )+ ) 

A± (k+q)

+  , (|u| − k)+ p∓ ≤M η . p∓

Then (9.3.7) implies that 



p

|f (x)| · vk − dx;

|f (x)| · η(((|u| − k)+ )+ )dx ≤ M ·

.

A+ (k+q)

A+ (k+q)





(9.3.19) p

|f (x)| · wk + dx.

|f (x)| · η(((|u| − k)+ )+ )dx ≤ M · A− (k+q)

A− (k+q)

Using the definition of η by (2.14.5), we come to   .η(((|u| − k)+ )+ )  

≤ eςq ,

∀q > 0

⇒

A± (k)\A± (k+q)

|f (x)| · η(((|u| − k)+ )+ )dx ≤ A± (k)\A± (k+q)



e

ςq

·

|f (x)| dx,

∀q > 0.

(9.3.20)

A± (k)\A± (k+q)

By assumption (iii), we get that f (x) ∈ Ls (G), where s > inequality with exponents s and

s,

where

1 s

+

1 s

= 1:

⎛



⎜ f (x)|z| dx ≤ f (x)Ls (G) · ⎝

.

A(k+d)

From the inequality inequality gives ⎛ ⎜ .⎝

s

|z|

ps 

⎟ dx ⎠ .

(9.3.21)

A(k) 1 s


1. By Hölder’s

⎞ 1



p

2 p±

s

|z|

ps 

⎟ dx ⎠

⎛ ⎜ ≤⎝



A(k)

⎞θ ⎛ ⎟ ⎜ |z| dx ⎠ · ⎝

2p 2−p ,



p

A(k)

and then the interpolation ⎞ (1−θ )p

⎟ |z| dx ⎠ p#

p#

248

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

with θ ∈ (0, 1), which is defined by the equality

=

1 ps 

Therefore by using the Young inequality with exponents obtain  f (x)|z|p dx ≤ θ ε

.

θ −1 θ

θ p 1 θ

+

⇒ θ = 1 −

1−θ p#

and

1 (1−θ)

2 ps .

from (9.3.21), we



1

f (x)Lθ s (G)

A(k+d)

|z|p dx+

A(k)

⎛ ⎜ + (1 − θ )ε · ⎝



⎞p ⎟ |z| dx ⎠

p#

p#

,

∀ε > 0.

(9.3.22)

A(k)

Therefore, we derive inequalities: ⎛



p |f (x)| · vk − dx

.

⎜ ≤ ε(1 − θ− ) ⎝

A+ (k+q)

θ− ε

θ− −1 θ−



f 

1 θ−

⎞ p− p# ⎟ vk − dx ⎠

# p−

+

A+ (k)



p

Ls (GR 0) A+ (k)

vk − dx; (9.3.23)

⎛





⎜ p |f (x)| · wk + dx ≤ ε(1 − θ+ ) ⎝

A− (k+q)

θ+ ε

# ∀ε > 0, p∓ =

θ+ −1 θ+

f 

1 θ+

⎞ ⎟ wk dx ⎠ # p+

p+ # p+

+

A− (k)



p

Ls (GR 0) A− (k)

wk + dx,

  2 2p∓ 2 2 2 = , θ∓ = 1 − ; s > max ; > 1. 2 − p∓ sp∓ p− p+ p−

Then applying (9.3.23) to (9.3.18)–(9.3.20), we get that  .

⎛ ⎜ |f (x)| · η(((|u| − k)+ )+ )dx ≤ Mε(1 − θ− ) ⎝

A+ (k)

A+ (k)

⎞ p− p# ⎟ vk − dx ⎠

# p−

+

A+ (k)

 eςq ·



|f (x)| dx + Mθ− ε

θ− −1 θ−

f 

1 θ−



Ls (GR 0) A+ (k)

p

vk − dx (9.3.24)

9.3 The Maximum Principle

249

⎛



⎜ |f (x)| · η(((|u| − k)+ )+ )dx ≤ Mε(1 − θ+ ) ⎝

A− (k)

p# ⎟ wk + dx ⎠

# p+

+

A− (k)

 eςq ·

⎞ p+



|f (x)| dx + Mθ+ ε

θ+ −1 θ+



1 θ+

f L

p

wk + dx.

R s (G0 )

A− (k)

A− (k)

By the well known Sobolev embedding theorem and taking into account (9.3.23), we obtain ⎛ ⎜ ⎝ .

⎞ p−



p# ⎟ vk − dx ⎠

≤ c− ·

A+ (k)

⎛ ⎜ ⎝



# p−

p# ⎟ wk + dx ⎠

A− (k)

 p vk − + |∇vk |p− dx;

A+ (k)

⎞ p+







# p+

≤ c+ ·

(9.3.25) 

 p wk + + |∇wk |p+ dx,

A− (k)

where c∓ = const > 0. Finally, (9.3.17)–(9.3.25) imply that +  p− ,  1 p− . − Mc− (1 − θ− )ε · 2 ς

|∇vk |p− dx+

A+ (k)

,  +  p+ 1 p+ − Mc+ (1 − θ+ )ε · 2 ς

|∇wk |p+ dx+

A− (k)

+ ,  1 θ− −1 θ− p− −1 θ− aˆ 0 k0 − Mc− (1 − θ− )ε − Mθ− ε f L (GR ) · s

+

p −1 aˆ 0 k0 −

− Mc+ (1 − θ+ )ε − Mθ+ ε

θ+ −1 θ+

f 

0

,

1 θ+

Ls (GR 0)

 eςq ·

p

vk − dx +

A+ (k)



·

p

wk + dx ≤

A− (k)

|f (x)| dx,

∀ε > 0.

A(k)

(9.3.26)

250

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

Further, at first, we choose     p+  1 1 1 p− p− p+ min .ε = ; 4M c− (1 − θ− ) ς c+ (1 − θ+ ) ς

(9.3.27)

and next  k0 ≥

.

2MF aˆ 0



1 p− −1

where

,

(9.3.28)

1 $ θ− −1 θ F = max c− (1 − θ− )ε + θ− ε θ− f L−(GR ) ; s

c+ (1 − θ+ )ε + θ+ ε

θ+ −1 θ+

0

% f L (GR ) ; . 1 θ+ s

0

Thus, by the above arguments, we derive 

 p  |∇vk |p− + vk − dx +

.

A+ (k)





p  |∇wk |p+ + wk + dx ≤

A− (k)

 |f (x)| dx,

C

(9.3.29)

A(k)

where C = const p− , p+ , aˆ 0 , k0 , δ, ν, s, f Ls (GR ) > 0. The inequalities (9.3.25) 0 together with (9.3.29) give ⎛ ⎜ .⎝



⎞ p− p# ⎟ vk − dx ⎠

A+ (k)

# p−

⎛ ⎜ +⎝



⎞ p+ p# ⎟ wk + dx ⎠

# p+

≤

A− (k)

 max{c− , c+ } · C

|f (x)| dx, ∀k ≥ k0 .

A(k)

At last, by the Hölder inequality, we have  .

A(k)

1

|f (x)|dx ≤ f (x)Ls (GR ) meas 1− s A(k); 0

s>

2 > 1. p−

(9.3.30)

9.3 The Maximum Principle

251

Then from (9.3.30), it follows that ⎛ ⎜ .⎝

⎞ p−



# p−

vk

⎟ dx ⎠

# p−

⎛ ⎜ +⎝

A+ (k)

⎞ p+



# p+

wk

⎟ dx ⎠

# p+

≤

A− (k) 1

max{c− , c+ } · C f (x)Ls (GR ) meas 1− s A(k);

2 > 1, ∀k ≥ k0 . p−

s>

0

(9.3.31)

Now, let l > k > k0 . By (2.14.9) and the definition of the functions vk (x), wk (x), we have 1 1 ((|u| − k)+ )+ , wk ≥ ((|u| − k)+ )+ . p− p+

vk ≥

.

Therefore, 



p# vk − dx

.

≥

A+ (l)

l−k p−

p−#

 · measA+ (l);

p# wk + dx

 ≥

A− (l)

l−k p+

p+#

· measA− (l).

Hence, (9.3.31) together with A± (l) ⊆ A± (k) implies that measA(l) = meas (A+ (l) ∪ A− (l)) ≤ measA+ (l) + measA− (l) ≤

.



p− l−k

p−#

p− l−k

 C+

·

# p− p−

≤

# p+ p+

 +

· f (x)

p+ l−k

# p− p−

Ls (GR 0)

p+#

· f (x)

p+#

 A− (k)

# p−

meas p−

# p+ p+

(1− 1s )

# p+

Ls (GR 0)

p#

wk + dx ≤

·

meas p+

A(k)+

(1− 1s )

A(k)

# p∓

where C∓ = (C max{c− , c+ }) p∓ .

(see (9.3.23)), we have # p−

meas p−

.

p−#

p+ l−k

∀l > k ≥ k0 , Since

p# vk − dx

A+ (k)

 C−



(1− 1s )

# p+

A(k) ≥ meas p+

(1− 1s )

A(k), if measA(k) ≤ 1.

(9.3.32)

252

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

Moreover, # p+ . p+

   #  p− 2 1 1 ≥ > 1 for s > 1− 1− > 1. s p− s p−

Let us introduce ψ(k) = meas A(k). Then from (9.3.32), it follows that

ψ(l) ≤ 2Cψ

.

ζ

⎧ ⎪ ⎨

1 #

p− (k) (l−k) ⎪ 1

⎩

# (l−k)p+

if

l − k ≥ 1;

if

0 < l − k < 1,

∀l > k ≥ k0 ,

where

2 - = const (p− , p+ , aˆ 0 , k0 , δ, ν, , s, f  > 1, C ) > 0. ζ = 1 − 1s 2−p Ls (GR − 0) By the Stampacchia Lemma, we have that ψ(k0 +ϑ) = 0 with ϑ depending only on the quantities given in being proved Theorem. This fact means that |u(x)| ≤ k0 + ϑ for almost all x ∈ GR 0 . Thus, we derive M0 = k0 + ϑ, where k0 is defined by (9.3.4), (9.3.28) with (9.3.23) and (9.3.27) being proved theorem. Then, by the Stampacchia lemma, we have ψ(k0 + ϑ) = 0,

.

p# β

#

ϑ p = C6 |ψ(k0 )|β−1 2 β−1 .

It means that |u(x)| ≤ k0 + ϑ for almost all x ∈ GR 0 , where k0 > 1 is sufficiently large,   such that inequality (9.3.28) holds.

9.4

The Comparison Principle

In .Gd0 , we consider the second order quasi-linear degenerate operator .Q of the form Q(u, η) ≡

 .

/   Ai (x, ux )ηxi + a0 (x)u|u|p(x)−1 + b(u, ∇u) η(x) dx−

Gd0





Ai (x, ux ) cos(r, xi )ηd d +

.

d

0d

 d 1−p(x) d 1,p(x)

γ (ω)r 1−p(x)u|u|p(x)−2ηds+

1 + χ(ω) tan ω γ (ω)u|u|p(x)−2ηd d 1 + χ(ω)sign ω 1,p(x)

for .u(x) ∈ N−1,∞ (Gd0 ) and for all nonnegative .η(x) belonging to .N−1,∞ (Gd0 ). We assume that conditions .(i), (iii), (iii1), (iii2), (iv) are satisfied as well as that functions

9.4 The Comparison Principle

253

Ai (x, ξ ) are Caratheodory, continuously differentiable with respect to the .ξ variables in d 2 .M = G × R × R , and satisfy in .M the following inequality: 0 .

(vii) . ∂ A∂ξi j(ξ ) ζi ζj ≥ ν|ξ |p−2 ζ 2 for all .ζ ∈ R2 , .ν > 0. We get the following comparison principle that plays an important role in the proof of Theorem 9.4. Theorem 9.9 Let operator .Q satisfy above assumptions and functions .u, w 1,p N−1,∞ (Gd0 ), .d & 1, satisfy the following inequality: Q(u, η) ≤ Q(w, η)

∈

(9.4.1)

.

1,p

for all nonnegative functions .η ∈ N−1,∞ (Gd0 ), as soon as u(x) ≤ w(x), a.e on d .

.

Then .u(x) ≤ v(x) in .Gd0 . Proof Let us define . z = u − w and uτ = τ u + (1 − τ )w, τ ∈ [0, 1]. Then we have  ?  .0 ≥ Q(u, η) − Q(w, η) = ηxi zxj Gd0

1 0

∂Ai (x, uτx ) dτ + ∂uτxj 

1

a0 (x)z(x)η(x)  η(x)zxi

1

0

∂b(x, uτ , uτx ) dτ ∂uτxi

0



@ ∂b(x, uτ , uτx ) dτ dx− ∂uτ

1

+ η(x)z(x) 0



 zxj

 d

1 + χ(ω) tan ω γ (ω) d 1−p(x) 1 + χ(ω)sign ω 

γ (ω) r p(x)−1

0d 1,p(x)

for all nonnegative .η ∈ N−1,∞ (Gd0 ).

d



1 0



1 0

1 0

∂(uτ |uτ |p(x)−1) dτ + ∂uτ

∂Ai (x, uτx ) dτ · cos(r, xi )ηd d + ∂uτxj

 ∂(uτ |uτ |p(x)−2) dτ z(x)η(x)d d + ∂uτ

 ∂(uτ |uτ |p(x)−2) dτ z(x)η(x)ds ∂uτ

(9.4.2)

254

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

Now, we introduce the sets (Gd0 )+ := {x ∈ Gd0 | u(x) > w(x)} ⊂ Gd0 ,

.

(0d )+ := {x ∈ 0d | u(x) > w(x)} ⊂ 0d and assume that .(Gd0 )+ =  ∅, (0d )+ = ∅. Let .k ≥ 1 be any odd number. We choose k . η = max{(u − w) , 0} as a test function in the integral inequality (9.4.2). We have 

1

.

0



1 0

∂(uτ |uτ |p(x)−1) dτ = p(x) ∂uτ



1 0

∂(uτ |uτ |p(x)−2) dτ = (p(x) − 1) ∂uτ

  Then, by our assumptions and .η 

$

.

|uτ |p(x)−1dτ > 0;



1

|uτ |p(x)−2dτ > 0.

0

= 0, we obtain from (9.4.2) that

d

kνzk−1



1

|∇uτ |p(x)−2dτ |∇z|2 +

0

(Gd0 )+

b2 zk+1



1

|uτ |−2 |∇uτ |p(x)dτ

% dx ≤

0

 b1 ·

zk



1

|uτ |−1 |∇uτ |p(x)−1dτ |∇z|dx.

(9.4.3)

0

(Gd0 )+

By the Cauchy inequality, b1 zk |∇z||uτ |−1 |∇uτ |p(x)−1 =     p(x) p(x) k+1 k−1 · b1 z 2 |∇z||∇uτ | 2 −1 ≤ |uτ |−1 z 2 |∇uτ | 2 .

b2 ε τ −2 k+1 |u | z |∇uτ |p(x) + 1 zk−1 |∇z|2 |∇uτ |p(x)−2, ∀ε > 0. 2 2ε Hence, taking .ε = 2b2 , we obtain from (9.4.3) the inequality  .

b2 kν − 1 4b2

 

 z

(Gd0 )+

k−1

|∇z|

1

2 0

 |∇u |

τ p(x)−2

dτ dx ≤ 0.

(9.4.4)

9.5 The Barrier Function. Estimation of the Solution Modulus

255

  b12 Choosing the odd number .k ≥ max 1; 2b2 ν , in view of .z(x) ≡ 0 on .∂(Gd0 )+ , we get from (9.4.4) that .z(x) ≡ 0 in .(Gd0 )+ . We got a contradiction to our definition of the set d + .(G ) , and this completes the proof.   0

9.5

The Barrier Function. Estimation of the Solution Modulus

Let us consider the following auxiliary problem for a function .w(r, ω) ≡ 0  (BFP)

.

− p+ w = μ|w|−1 |∇w|p+ , x ∈ Gd0 ,   γ ∂w ∂w ± d. |∇w|p+ −2 ∂ n ± χ± ∂r + |x|p+ −1 w|w|p+ −2 = 0, x ∈ ±

By direct calculations, we can show that (BF) w = w(r, ω) = r ψ λ (ω),

.

=

p+ − 1 λ, p+ − 1 + μ

is a solution to problem (BFP), where .(λ, ψ(ω)) is an eigenpair of the eigenvalue problem (OEVP)2 . For this function, we calculate

.

∂w ∂w = r −1 ψ /λ (ω); = r ψ λ −1 (ω)ψ  (ω); ∂r ∂ω λ . 2 / 2 |∇w|2 = r 2( −1) ψ 2( λ −1) (ω) λ2 ψ 2 (ω) + ψ  (ω) ; ⇒ λ p+

p+ /2 2 |∇w|p+ = r p+ ( −1) ψ p+ ( λ −1) (ω) λ2 ψ 2 (ω) + ψ  (ω) ; ⇒ λ p+

p+ /2 2 r (p+ −1) −p+ ψ (p+ −1) λ −p+ (ω) λ2 ψ 2 (ω) + ψ  (ω) ; |w|−1 |∇w|p+ = λ .

|w|p+ −1 = r (p+ −1) ψ (p+ −1) λ ; 2 ∂ 2w ∂ 2w = r −1 ψ λ −1 (ω)ψ  (ω); = ( − 1)r −2 ψ /λ (ω); 2 ∂r ∂r∂ω λ

∂ 2w − 1 r ψ λ −2 (ω)ψ 2 (ω) + r ψ λ −1 (ω)ψ  (ω). = ∂ω2 λ λ λ 1,p(x)

Proposition 9.10 .w ∈ N−1,∞ (Gd0 ).

(9.5.1)

256

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

Proof From .(BF ) and (3.2.6), it follows that .w ∈ L∞ (Gd0 ). Next,  .

r −p(x)wp(x) dx =

Gd0

 r(

−1)p(x)

ψ λ p(x)(ω)dx.

Gd0

By .r ≤ d & 1 and assumption (i), we have

.

r(

−1)p(x)

≤ r(

−1)p−

, if

≥ 1;

r(

−1)p(x)

≤ r(

−1)p+

, if

≤ 1;

p

ψ λ p(x)(ω) ≤ ψ0 + in virtue of (3.2.6) and

≤ λ.

Hence, it follows that  .

r

−p(x)

w

p(x)

dx ≤

⎧ ( −1)p− +n d ⎪ ⎪ ⎨ ( −1)p− +n ,

p ψ0 + meas · ⎪

⎪ ⎩ d(

Gd0

if

≥ 1; (9.5.2)

−1)p+ +n

( −1)p+ +n ,

if

≤ 1.

From (9.5.1) and (CP) with regard to (3.2.9) (see Sect. 3.2), we obtain that  .

 p(x) λ

w−1 |∇w|p(x)dx =

Gd0 −p(x)

r (p(x)−1)

Gd0 p −1



p(x)/2 2 ψ (p(x)−1) λ −p(x)(ω) λ2 ψ 2 (ω) + ψ  (ω) dx ≤



ψ0 +

r (p(x)−1)

Gd0 p −1

−p(x)



ψ0 +

Gd0

r (p(x)−1)



p(x)/2 λ2 + y 2 (ω) dx ≤

−p(x)



p(x)/2 λ2 + y02 dx,

9.5 The Barrier Function. Estimation of the Solution Modulus

where . y(ω) = .

ψ  (ω) ψ(ω) ;

y0 = y

257

 ω0  2 ) . Next, by (3.2.19), (3.2.20)



p(x)/2 λ2 + y02 (ω) ≤ C1 = const (μ, λ, ω0 , p+ , χ+ , γ+ ).

From the above inequality, we obtain that  .

p −1

w−1 |∇w|p(x)dx ≤ C1 ψ0 +

Gd0

 r (p(x)−1)

Gd0

−p(x)



p −1 C1 ψ0 +

dx =

−1)(p(x)−p+ )

r(

· r(

−1)p+ −

dx.

(9.5.3)

Gd0

Now, by assumptions (i)–(ii) and .r & 1, we derive −1)(p(x)−p+ )

r(

.

≤

⎧ ⎨r (1− ⎩1,

)Lr ,

if

> 1,

if

≤ 1.

Using the well known inequality r α | ln r| ≤

.

1 , ∀α > 0, 0 < r < 1, αe

(9.5.4)

where e is the Euler number, we establish for . > 1 the inequality r (1−

.

)Lr

≤e

L( −1) e

,

0 < r < 1.

Thus, from (9.5.3), it follows that  .

Gd0

p −1

w−1 |∇w|p(x) dx ≤ C1 ψ0 +

e

L( −1) e

meas ·

d (p+ −1)+2−p+ . (p+ − 1) + 2 − p+  

258

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

9.6

Proof of the Main Theorem 9.4

Let .A > 1, and let .w(r, ω) ≥ 0 be the barrier function defined above. By the definition of the operator Q in .(I I )loc , we have   Q[Aw, η] ≡ Ap(x)−1|∇w|p(x)−2 ∇w, ∇η + Gd0

χ(ω)Ap(x)−1|∇w|p(x)−2(wx1 + wx2 tan ω)ηx2 +  p(x) p(x) w η + b(Aw, ∇Aw)η dx− a0 (x)A 

.

d· 

Ap(x)−1|∇w|p(x)−2

∂w (1 + χ(ω) tan ω) ηdω+ ∂r

(9.6.1)



γ (ω)

1 + χ(ω) tan ω p(x)−1 2−p(x) p(x)−1 A d w ηdω+ 1 + χ(ω)sign ω



γ (ω)Ap(x)−1r 1−p(x)wp(x)−1ηds 0d 1,p(x)

for all .η ∈ N−1,∞ (Gd0 ). Integrating by parts, we obtain 

 Ap(x)−1|∇w|p(x)−2wxi ηxi dx = −

.

Gd0

/ d . p(x)−1 A |∇w|p(x)−2wxi dx+ dxi

Gd0

 A

0d

η(x)

p(x)−1

|∇w|

p(x)−2 dw

dn



η(x)dS +

Ap(x)−1|∇w|p(x)−2

∂w η(x)d d = ∂r

d

 ?

μAp(x)−1w−1 |∇w|p(x) − Ap(x)−1|∇w|p+ −2 wxi

d|∇w|p(x)−p+ − dxi

Gd0

@  ∂w ∂Ap(x)−1 p(x)−2 η(x)ds+ wxi |∇w| η(x)dx + Ap(x)−1|∇w|p(x)−2 ∂xi ∂ n



0d

Ap(x)−1|∇w|p(x)−2 d

∂w η(x)d d , ∂r

(9.6.2)

9.6

Proof of the Main Theorem 9.4

259

by the (BF P ) equation. In the same way, we have  χ(ω)Ap(x)−1|∇w|p(x)−2(wx1 + wx2 tan ω)ηx2 dx =

.

Gd0



 −

χ(ω) p(x)−1 ∂w A ηx dx = |∇w|p(x)−2 cos ω ∂r 2

Gd0

η(x)

@ ? ∂w d χ(ω) p(x)−1 A dx+ |∇w|p(x)−2 dx2 cos ω ∂r

Gd0



χ(ω) p(x)−1 ∂w → A cos(− n , x2 )η(x)ds, by (9.2.2). |∇w|p(x)−2 cos ω ∂r

0d ∪ d

Adding (9.6.2) with (9.6.3), we obtain .

  Ap(x)−1|∇w|p(x)−2 ∇w, ∇η + χ(ω)Ap(x)−1|∇w|p(x)−2(wx1 +

Gd0

 ?

 wx2 tan ω)ηx2 = μAp(x)−1w−1 |∇w|p(x) − Ap(x)−1|∇w|p+ −2 wxi

d|∇w|p(x)−p+ − dxi

Gd0

? @@ d χ(ω) p(x)−1 ∂w ∂Ap(x)−1 A wxi |∇w|p(x)−2 − |∇w|p(x)−2 η(x)dx+ ∂xi dx2 cos ω ∂r    χ(ω) ∂w ∂w → + cos(− n , x2 ) η(x)ds+ Ap(x)−1|∇w|p(x)−2 ∂ n

cos (ω0 /2) ∂r 0d

 (1 + χ(ω) tan ω) Ap(x)−1|∇w|p(x)−2

d

∂w η(x)d d , ∂r

by (9.2.1).

(9.6.3)

260

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . . 1,p(x)

Hence and from (9.6.1), it follows that for all .η ∈ N−1,∞ (Gd0 ) Q[Aw, η] ≡

.

 ? d|∇w|p(x)−p+ μAp(x)−1w−1 |∇w|p(x) − Ap(x)−1|∇w|p+ −2 wxi − dxi Gd0

 0d



 χ(ω) p(x)−1 ∂w A |∇w|p(x)−2 + cos ω ∂r @ a0 (x)Ap(x)wp(x) + b(Aw, ∇Aw) η(x)dx+

d ∂Ap(x)−1 wxi |∇w|p(x)−2 − ∂xi dx2

?   χ(ω) ∂w ∂w → + cos(− n , x2 ) + Ap(x)−1 |∇w|p(x)−2 ∂ n

cos (ω0 /2) ∂r @ γ (ω)r 1−p(x)wp(x)−1 η(x)ds+

 (Aw)p(x)−1

(1 + χ(ω) tan ω)γ (ω)d 2−p(x) ηdω ≡ JGd + J d + J . 0 0 1 + χ(ω)sign ω

(9.6.4)

Now, with regard to (9.2.1) and the (BF P ) boundary condition, we calculate ?

 J d ≡

.

A

0

p(x)−1

0d

 |∇w|

p(x)−2

 χ(ω) ∂w ∂w → − + cos( n , x2 ) + ∂ n

cos (ω0 /2) ∂r

γ (ω)r 

1−p(x)

w

p(x)−1

@ η(x)ds =



γ (ω)Ap(x)−1r 1−p+ wp+ −1 r p+ −p(x) wp(x)−p+ − |∇w|p(x)−p+ ≥ 0,

0d

because of, by (BF ) and (9.5.1), for .∀x ∈ 0d : r p+ −p(x) wp(x)−p+ − |∇w|p(x)−p+ = ? p(x)−p+ ψ p+ −p(x (ω) λ2 ψ 2 + r ( −1)(p(x)−p+) ψ (p(x)−p+ ) λ (ω) 1 − λ .

(9.6.5)

9.6

Proof of the Main Theorem 9.4

261

ψ 2

r

( −1)(p(x)−p+ )

? ψ

(p(x)−p+ ) λ

(ω) 1 −

 ;

+ y02

p(x)−p+ @

p(x)−p+ @

2

=

≥ 0,

(9.6.6)

γ (ω)d 2−p(x) wp(x)−1ηdω ≥ 0. 1 + χ(ω)sign ω

(9.6.7)

λ

λ2

in virtue of Proposition 3.7. Further, by assumption .(iii2), it is obvious that  J ≡

(1 + χ(ω) tan ω)Ap(x)−1

.



Thus, from (9.6.4)–(9.6.7), it follows Q[Aw, η] ≥

.

 ? d|∇w|p(x)−p+ − μAp(x)−1w−1 |∇w|p(x) − Ap(x)−1|∇w|p+ −2 wxi dxi Gd0

 χ(ω) p(x)−1 ∂w A + |∇w|p(x)−2 cos ω ∂r @ p(x) p(x) a0 (x)A w + b(Aw, ∇Aw) η(x)dx ≡ JGd .

∂Ap(x)−1 d wxi |∇w|p(x)−2 − ∂xi dx2



0

(9.6.8)

Further, we proceed to the estimating of integral .JGd . Setting .W (x) = |∇w|p(x)−p+ , we 0 calculate .

ln W (x) = (p(x) − p+ ) ln |∇w|,

⇒

∂W ∂p p(x) − p+ d|∇w| 1 · · = ln |∇w| + ⇒ W (x) ∂xi ∂xi |∇w| dxi ? @

p(x) − p+ d|∇w| d p(x)−p+ p(x)−p+ ∂p |∇w| = |∇w| · ln |∇w| + . dxi ∂xi |∇w| dxi Similarly, .

∂ p(x)−1 ∂p A = Ap(x)−1 ln A. ∂xi ∂xi

262

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

By (9.6.8), we obtain that JGd

.

0

?   p(x)−1 p(x)−2 ≥ |∇w| A μw−1 |∇w|2 − (∇p · ∇w) · ln(A|∇w|)− Gd0

 @  p(x) − p+ d χ(ω) p(x)−1 d|∇w| p(x)−2 ∂w · wxi A + − |∇w| |∇w| dxi dx2 cos ω ∂r  p(x) p(x) a0 (x)A w + b (Aw, A∇w) η(x)dx.

(9.6.9)

Further, by direct calculation,  χ(ω) p(x)−1 ∂w A = |∇w|p(x)−2 cos ω ∂r    χ(ω) ∂w ∂p p(x) − 2 ∂|∇w| Ap(x)−1|∇w|p(x)−2 · + ln(A|∇w|) + cos ω ∂r ∂x2 |∇w| ∂x2  2   1 ∂ 2w ∂ w 1 ∂w 1 ∂w + χ(ω) + χ  (ω) + χ(ω) tan ω . r ∂r r ∂r∂ω ∂r 2 r ∂r .

d dx2



(9.6.10)

Passing to polar coordinates with regard to (9.2.2), we calculate wxi

.

1 ∂w ∂|∇w| d|∇w| ∂w ∂|∇w| · + 2 · ; = dxi ∂r ∂r r ∂ω ∂ω

1 ∂|∇w| ∂|∇w| ∂|∇w| + cos ω . = sin ω ∂x2 ∂r r ∂ω Thus, hence and from (9.6.8)–(9.6.10) with regard to assumptions .a0 > 0, .(iv)a −(iv)b , we obtain that  Q[Aw, η] ≥

A

.

Gd0

p(x)−1

  ? p(x)−2 −1 2 |∇w| σ w |∇w| − (∇p · ∇w)+

  χ(ω) ∂w ∂p ∂w ∂|∇w| p+ − p(x) · · + ln(A|∇w|) + cos ω ∂r ∂x2 |∇w| ∂r ∂r    ∂w 1 ∂w ∂|∇w| 2 − p(x) ∂|∇w| 1 ∂|∇w| · · χ(ω) + · + tan ω · − r 2 ∂ω ∂ω |∇w| ∂r ∂r r ∂ω

9.6

Proof of the Main Theorem 9.4

χ  (ω)

263



 @ ∂ 2 w 1 ∂w + χ(ω) tan ω − ∂r 2 r ∂r  b0 wp(x)−1 η(x)dx,

1 ∂ 2w 1 ∂w − χ(ω) − r ∂r r ∂r∂ω

(9.6.11)

⎧ ⎨μ − δ, if μ > 0; with . σ = ⎩ν, if μ = 0. 

(ω) Taking into account (9.5.1), the Lipschitz condition of .p(x) (ii), and . ψψ(ω) = y(ω), we direct calculate:

(1) . |w|−1 |∇w|2 = (2)

 2 λ

r

−2 ψ

λ

  (ω) λ2 + y 2 (ω) .

|(∇p · ∇w)(ln A + ln |∇w|)| ≤ |∇p| · |∇w| · (ln A + | ln |∇w||) ≤

.

L|∇w| · (ln A + | ln |∇w||); in virtue of (9.5.1), (3.2.6), (3.2.18)–3.2.20), we derive:   | ln |∇w|| ≤ ln

.

λ

ln

   + | − 1| · | ln r| +

1 ln ψ + | ln(λ2 + y 2 (ω))| ≤ λ 2

λ 1 + ln ψ0 + | ln(λ2 + y02 )| + | − 1| · | ln r| = 2

ln C1 (μ, p+ , λ, γ+ , χ+ , p+ , ω0 ) + | − 1| · | ln r| (note that . C1 ≥ 1 : indeed, by virtue of .(BF ) and (3.2.6), . λ ψ0 ≥ 1; from (3.2.17), ; it follows that . λ2 + y02 ≥ λ ≥ 1 ⇒ ; C1 = λ ψ0 λ2 + y02 ≥ 1); using inequality (9.5.4) with .α = 12 , we get that ; r −1 ψ λ (ω) λ2 + y 2 (ω) · (ln C1 + | − 1| · | ln r|) = λ ;   √ √ r −2 ψ λ (ω) λ2 + y 2 (ω) · r ln C1 + (| − 1| r) · r| ln r| ≤ λ ; √ r −2 ψ λ (ω) λ2 + y 2 (ω) · (r ln C1 + | − 1| r), λ

|∇w| · | ln |∇w|| ≤

.

264

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

from which we obtain that |(∇p · ∇w)(ln A + ln |∇w|)| ≤

.

λ

−2

r

; √ ψ λ (ω) λ2 + y 2 (ω) · (r ln(AC1 ) + | − 1| r)

and therefore, hence, we get    χ(ω) ∂w ∂p   · ln(A|∇w|) ≤ . (∇p · ∇w) +  cos ω ∂r ∂x2   1+  1+

χ− cos ω20



−2

L r λ

χ− cos ω20

 |∇p| · |∇w| · ln(A|∇w|) ≤

; √ ψ λ (ω) λ2 + y 2 (ω) · (r ln(AC1 ) + | − 1| r).

(3) By (9.5.1), .

∂|∇w| = ∂r 

−1 |∇w|; r 

(9.6.12)

y  (ω) ∂|∇w| = |∇w| + 2 · y(ω) = ∂ω λ λ + y 2 (ω)   y  (ω) p+ − 1 + 2 · y(ω), |∇w| p+ − 1 + μ λ + y 2 (ω) and therefore, we get .

 ∂w ∂|∇w| 1 ∂w ∂|∇w| · + 2 · = ∂r ∂r r ∂ω ∂ω  @ ? p+ − 1 y  (ω) y2 w + 2 ; −1+ (p+ − p(x)) 2 r λ p+ − 1 + μ λ + y 2 (ω)

p+ − p(x) · |∇w|



(9.6.13)

now, since (p+ − 1)y 4 + λ (2λ(p+ − 1) + 2 − p+ ) y 2 + λ3 (λ(p+ − 1) + 2 − p+ ) =   2 − p+ 2 2 2 2 λ , (p+ − 1)(y + λ ) y + λ + p+ − 1

.

9.6

Proof of the Main Theorem 9.4

265

the (CP ) equation can be rewritten as follows: .

y  (ω) (p+ − 1)(y 2 + λ2 ) + (2 − p+ )λ = − , y 2 + λ2 (p+ − 1)y 2 + λ2

ω 0 ; ω ∈ 0, 2

(9.6.14)

by (9.6.14) and .p+ ≤ 2, μ ≥ 0, we establish that .

μ

y  (ω) (2 − p+ ) p+ − 1 + 2 ≥ −λ − 2 p+ − 1 + μ λ + y (ω) (p+ − 1)y 2 + λ2 y 2 + λ2 (2 − p+ ) y 2 + λ2 p+ − 1 · ≥ −λ − μ ; p+ − 1 + μ (p+ − 1)y 2 + λ2 (p+ − 1)y 2 + λ2 (p+ − 1)y 2 + λ2

hence, it follows that .

y2 λ



p+ − 1 y  (ω) + 2 p+ − 1 + μ λ + y 2 (ω)

 ≥−

. / y2 1 · · (2 − p+ )λ + μ(y 2 + λ2 ) ≥ 2 2 λ (p+ − 1)y + λ −

μ(y 2 + λ2 ) 2 − p+ − ; p+ − 1 λ(p+ − 1)

(9.6.15)

finally, from (9.6.13), (9.6.15), and the Lipschitz condition of .p(x) (see assumption (ii)), we derive

.

.

p+ − p(x) · |∇w|



 ∂w ∂|∇w| 1 ∂w ∂|∇w| · + 2 · ≥ ∂r ∂r r ∂ω ∂ω ? @ μ(y 2 + λ2 ) 2 − p+ + − L | − 1| + r p+ − 1 λ(p+ − 1)

−1

ψ λ (ω).

(4) In virtue of (9.5.1), (9.6.13), (3.2.18)–(3.2.20),   ∂w 2 − p(x) ∂|∇w| 1 ∂|∇w| χ(ω) + . tan ω · = |∇w| ∂r ∂r r ∂ω ?   @ y  (ω) p+ − 1 w + · y(ω) ≥ (2 − p(x)) χ(ω) 2 ( − 1) tan ω + r p+ − 1 + μ λ2 + y 2 (ω)   1 −(2 − p− ) χ− | − 1| tan(ω0 /2) + |y(ω)| r −2 ψ λ (ω). λ

266

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

(5) Because of . χ  (ω) < 0 (see assumption .(iii2)), . ψ  (ω) ≤ 0 (see (3.2.6)), and by (9.5.1), .

− χ  (ω)



1 ∂ 2w 1 ∂w − χ(ω) ≥ 0; r ∂r r ∂r∂ω

 ∂ 2 w 1 ∂w − + χ(ω) tan ω ≥ − χ− (1 + | − 1|) tan(ω0 /2) · r ∂r 2 r ∂r

−2

ψ λ (ω).

From (1)–(5), it follows that in .Gd0 :   χ(ω) ∂w ∂p .σ w |∇w| − (∇p · ∇w) + ln(A|∇w|)+ cos ω ∂r ∂x2   ∂w ∂|∇w| 1 ∂w ∂|∇w| p+ − p(x) · · + 2 · + |∇w| ∂r ∂r r ∂ω ∂ω   ∂w ∂|∇w| 1 ∂|∇w| 2 − p(x) · χ(ω) tan ω · + · − |∇w| ∂r ∂r r ∂ω  2  1 ∂ 2w ∂ w 1 ∂w 1 ∂w − χ(ω) − χ(ω) tan ω ≥ χ  (ω) + r ∂r r ∂r∂ω r ∂r ∂r 2  2

? λμL d − r −2 ψ λ (ω) λ2 + y 2 (ω) σ− λ (p+ − 1)    χ− Ld Ld L| − 1| √ 1+ ln A + ln C1 + d − cos(ω0 /2)   Ld 1 | − 1| + − p+ − 1 −1

χ−

2

| − 1| tan(ω0 /2) −

χ−

(9.6.16)

@ λ|y(ω)| χ− · 2 (1 + | − 1|) tan(ω0 /2) . − y (ω) + λ2

Now, in virtue of (9.6.11) and assumption .(iv)a , we note that

.

 σ−

λμL d (p+ − 1)

⎧ ⎨ = ν > 0, if μ = 0; ⎩≥ 1 σ > 0, if μ > 0, d ≤ 2

σ (p+ −1)2 2μL(p+ −1+μ) .

(9.6.17)

9.6

Proof of the Main Theorem 9.4

267

Further, we choose .d > 0 and .χ− ≥ 0 such small that are fulfilled the following inequalities: ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ . ⎪ ⎪ L ⎪ 1+ ⎪ ⎪ ⎨ .

Ld

χ− cos(ω0 /2)

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

1

1 2

χ− cos(ω0 /2)

ln C1 + | − 1| + 1+

.

1+

χ− cos(ω0 /2)

1 p+ −1



ln A /

·d

L| −1|

·

≤ ≤

√ d

/

+ (1 + 2| − 1|) tan(ω0 /2) · χ−

≤ ≤

⎧ ⎨

1 12 ν, ⎩ 1 σ, , ⎧ 24

for μ = 0,

1 12 ν, ⎩ 1 σ, , ⎧ 24

for μ = 0,

⎨

⎨ 1 ν,

6 ⎩ 1 σ, , ⎧ 12

⎨ 1 ν,

6 ⎩ 1 σ, , 12

for μ > 0;

for μ > 0; for μ = 0, for μ > 0; for μ = 0, for μ > 0. (9.6.18)

We introduce quantities

.

0

⎧ ⎨1 = ⎩ p+

for for

≥ 1, 0,

(9.6.19)

as well as we note that . p(x) ≥ 0 . Then from (9.6.16), (9.6.17), (9.6.18), (9.6.19), we obtain  ?  χ(ω) ∂w ∂p |∇w|p(x)−2 σ w−1 |∇w|2 − (∇p · ∇w) + ln(A|∇w|)+ cos ω ∂r ∂x2   ∂w ∂|∇w| 1 ∂w ∂|∇w| p+ − p(x) · · + 2 · + |∇w| ∂r ∂r r ∂ω ∂ω   ∂w ∂|∇w| 1 ∂|∇w| 2 − p(x) · χ(ω) tan ω · + · − |∇w| ∂r ∂r r ∂ω  2  @ 1 ∂ 2w ∂ w 1 ∂w 1 ∂w − χ(ω) − χ(ω) tan ω − b0 wp(x)−1 ≥ χ  (ω) + r ∂r r ∂r∂ω ∂r 2 r ∂r p(x) p(x) r (p(x)−1)−p(x)ψ λ (p(x)−1)(ω)(λ2 + y 2 (ω)) 2 − μ∗ λ

(p(x)−1) λ (p(x)−1) ψ (ω) ≥ r (p(x)−1)−p(x) μ∗ p(x) − b0 r p(x) ≥ b0 r

.

 ∗ μ

0

 − b0 d r (

−1)p(x)−

≥

μ∗ 2

0r

( −1)p(x)−

,

(9.6.20)

268

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

if we add to inequalities (9.6.18) yet the requirement 1 2

b0 d ≤

.

0μ

∗

(9.6.21)

.

From (9.6.11), (9.6.20), it follows that 

μ∗ .Q[Aw, η] ≥ 2

Ap(x)−1r (

0

−1)p(x)−

η(x)dx.

Gd0

Since .p(x) ≥ p− > 1 and .A > 1, we have .Ap(x)−1 ≥ Ap− −1 . Therefore, taking into consideration assumption (iii), the last inequality takes the form Q[Aw, η] ≥

.

μ∗ 2

μ∗ 2

0A

p− −1 0A

p− −1



 r(

Gd0

r

η(x)dx ≥ 

β(x)

Gd0



−1)p(x)−

η(x)dx ≥ Gd0



|f (x)|η(x)dx ≥

f0 r β(x) η(x)dx ≥

f (x)η(x)dx = Q[u, η], by (I I )loc

Gd0

Gd0 1,p(x)

for all nonnegative .η ∈ N−1,∞ (Gd0 ), if .A > 1 satisfies  A≥

.

2f0 μ∗ 0



1 p− −1

.

(9.6.22)

Further, we show that .u(x) ≤ Aw(x) on . d . By (BF ) and (3.2.6), w(x)| d = d ψ

.

/λ

(ω) ≥ d .

In virtue of .|u(x)| ≤ M0 , ∀x ∈ Gd0 , we can choose A such that A≥

.

M0 d

and therefore Aw(x)| d ≥ Ad ≥ M0 ≥ u(x)| d .

.

(9.6.23)

9.6

Proof of the Main Theorem 9.4

269

Thus, if we choose a small .χ− ≥ 0 according to (9.1.3), . d > 0 according to (9.6.17), (9.6.18), (9.6.21), and a large .A > 1 according to (9.6.21)–(9.6.23):  1   M0 2f0 p− −1 , .A ≥ max , d μ∗ 0 

then we come to the Comparison Principle Q(u, η) ≤ Q(Aw, η)

in Gd0 ;

.

u(x) ≤ Aw(x)

on d .

For this purpose, we need to check the consistency for the system of two inequalities ⎧ ⎨ Ld 1 + . ⎩

χ− cos(ω0 /2)

ln A

≤

A

≥

μ∗ 6 , M0 d .

At first, for the last inequality of (9.6.18), we have .

1 cos(ω0 /2) + (1 + 2| − 1|) sin(ω0 /2) + (1 + 2| − 1|) tan(ω0 /2) = ≤ 2 2 cos(ω0 /2) 3 + 4| − 1| . 2 cos(ω0 /2)

Hence, it follows that the last inequality of (9.6.18) is fulfilled if χ− ≤

.

2 μ∗ cos(ω0 /2) , 3(3 + 4| − 1|)

(see (9.1.3)).

Now, our above system can be rewritten in the form ⎧ ⎨ Ld 1 + . ⎩

2 μ∗ 3(3+4| −1|)

ln A

≤

A

≥

μ∗ 6 , M0 d .

From this system, it follows that .

  μ∗ 1

ln M0 d − ≤ ln A ≤ · 2 μ∗ d 6L 1 + 3(3+4| −1|)   d · ln M0 d − ≤

6L 1 +

μ∗ 2 μ∗ 3(3+4| −1|)

.

⇒

270

9 The Oblique Derivative Problem in a Plane Sector for Elliptic Second Order. . .

Now, in virtue of .

  ln M0 − lim d · ln M0 d − = lim 1

d→+0

d→+0

ln d

=−

d

lim d ln d =

d→+0

lim d = 0,

d→+0

consequently, there is .d- > 0 such that for .d ∈ (0, d) the desired inequality is true. Thus, the Comparison Principle implies that u(x) ≤ Aw(x)

.

-

in Gd0 . -

Similarly, we derive the estimate .u(x) ≥ −Aw(x) in Gd0 if we replace .u(x) with .−u(x). By this and (BF ), (3.2.6), we get the required estimate |u(x)| ≤ Aw(x) ≤ C|x| ,

.

-

in Gd0 .

The Oblique Derivative Problem in a Bounded n-Dimensional Cone for Strong Quasi-Linear Elliptic Second Order Equation with Perturbed p(x)-Laplacian

10.1

10

Setting of the Problem

In this chapter, we investigate the behavior of bounded weak solutions to the oblique derivative problem for quasi-linear elliptic second order equations with perturbed .p(x)Laplacian and with strong nonlinearities singular absorption term in the neighborhood of a corner boundary point of the bounded .n- dimensional cone .(n ≥ 3): ⎧   p(x)−1+ ⎪ − p(x)u + 1ω0 dxd 1 χ(ω1 )|∇u|p(x)−2 ∂u ⎪ ∂r + a0 (x)u|u| sin 2 ⎪ ⎪ ⎨ b(u, ∇u) = f (x), x ∈ GR 0, .(OPn )   γ (ω1 ) ∂u ∂u p(x)−2 p(x)−2 ⎪ = 0, |∇u| ⎪ ∂ n + χ(ω1 ) ∂r + |x|p(x)−1 u|u| ⎪ ⎪ ⎩ x ∈ 0R ∪ R , where .R > 0 is fix.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Borsuk, Oblique Derivative Problems for Elliptic Equations in Conical Domains, Frontiers in Elliptic and Parabolic Problems, https://doi.org/10.1007/978-3-031-28381-9_10

271

272

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for. . .

We will work under the following assumptions: (i) .1 < p− ≤ p(x) ≤ p+ = p(0) < n, ∀x ∈ GR 0 . (ii) The Lipschitz condition: .p(x) ∈ C 0,1 (GR 0 ) ⇒ where L is the Lipschitz constant for 0 ≤ p+ − p(x) ≤ L|x|, ∀x ∈ GR 0, .p(x), 0 ≤ μ < 1, and .λ is the least positive eigenvalue of problem .(NEV P ) (see below Sect. 10.4). (iii) There exist a positive constant .a0 such that .a0 (x) ≥ a0 > 0 and a nonnegative constant .f0 ≥ 0 such that .|f (x)| ≤ f0 |x|β(x), + −1 β(x) > p+p−1+μ (p(x) − 1) λ − ns ; s > pn− ≥ pn+ > 1; ∀x ∈ GR 0; μ ∈ [0, 1), and .λ is the least positive eigenvalue of .(OEV P ) (see below Sect. 9.5). Let ω ω 0 0 ; χ± = χ ± . γ± = γ ± 2 2

.

(10.1.1)

( ) (iii1) .γ (ω1 ) ∈ C 0 − ω20 , + ω20 ; there exist numbers .γ0 > 0, such that .γ (ω1 ) ≥ ) ( ω0 γ0 , ∀ω1 ∈ − 2 , + ω20 and .γ− > γ+ ≥ 1.

10.1 Setting of the Problem

273

( ) (iii2) .χ(ω1 ) ∈ C 1 − ω20 , + ω20 , χ  (ω1 ) < 0, ∀ω1 ) ( .0 ≤ χ+ ≤ χ(ω1 ) ≤ χ− , ∀ω1 ∈ − ω20 , + ω20 ; ω0 0 sin , where .0 ≤ χ− < n−1 2

0

=

∈

⎧ ⎨1,

( ω0 ) − 2 , + ω20

⇒

if p(x) ≥ 2;

⎩p− − 1, if 1 < p(x) ≤ 2. (10.1.2)

(iv) The function .b(u, ξ ) is differentiable with respect to the .u, ξ variables in .M = R × Rn and satisfies in .M the following inequalities: (iv)a

|b(u, ξ )| ≤ δ|u|−1 |ξ |p(x) + b0 |u|p(x)−1,

(iv)b

b(u, ξ ) ≥ ν|u|−1 |ξ |p(x) − b0 |u|p(x)−1,

.

.

(iv)c

3 4 n   4  ∂b(u, ξ ) 2 −1 p(x)−1 5   ;  ∂ξ  ≤ b1 |u| |ξ | i=1

0 < δ < μ; if μ > 0. ν > 0;

if μ = 0.

∂b(u, ξ ) ≥ b2 |u|−2 |ξ |p(x); ∂u

i

b0 ≥ 0, b1 ≥ 0, b2 ≥ 0. (v) The spherical region . ⊂ S n−1 is invariant with respect to rotations in .S n−2 . Definition 10.1 A function u is called a bounded weak solution of problem .(OPn ) 1,p(x) provided that .u ∈ N−1,∞ (GR 0 ) and satisfies the integral identity .

   p(x)−2 ∇u, ∇η − |∇u| GR 0

  b(u, ∇u)η dx +

 1 ∂u χ(ω1 ) ηx1 + a0 (x)u|u|p(x)−1η+ sin ω20 ∂r

  γ (ω1 ) cos ω1 R n−p(x) u|u|p(x)−2η(x)d + 1 − χ(ω1 ) 1 + χ(ω1 ) sin ω20 

 γ (ω1 )r 1−p(x)u|u|p(x)−2η(x)ds =

0R

f ηdx GR 0

1,p(x)

for all .η ∈ N−1,∞ (GR 0 ). Remark 10.2 For the justification of this definition, see below (10.2.1) in Sect. 9.2.

(II)

274

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for. . .

The main results about power modulus of continuity are as follows: Theorem 10.3 Let u be a weak bounded solution of problem .(OPn ), .M0 = sup |u(x)| x∈GR 0

(see Sect. 9.3), and let .λ be the least positive eigenvalue of problem .(NEV P ) (see Sect. 9.5). Suppose that (i)–(iiv) hold. Then there exist .d- ∈ (0, R) and a constant .C0 > 0 depending only on .λ, R, M0 , p+ , p− , L, n, (μ − δ), ν,a0 , γ0 , χ− , b0 , f0 and such that |u(x)| ≤ C0 |x| ,

=

.

10.2

p+ − 1 λ; p+ − 1 + μ

-

∀x ∈ Gd0 .

(10.1.3)

Preliminary

By equalities (9.2.1), we shall justify integral identity .(I I ) for definition of the week solution of our problem. For this, we calculate via the integration by parts the integral  ? . − p(x)u + GR 0





 @ 1 d p(x)−2 ∂u χ(ω1 )|∇u| η(x)dx = sin ω20 dx1 ∂r

|∇u|p(x)−2 ∇u, ∇η −

GR 0

 |∇u|

0R ∪ R

p(x)−2

1 ∂u χ(ω ) ηx dx− 1 sin ω20 ∂r 1

 1 ∂u ∂u − χ(ω1 ) cos(

n, x1 ) η(x)ds. ∂ n

sin ω20 ∂r

From the boundary condition of problem .(P ) with regard to (9.2.1), we have: • On .0R :   1 ∂u ∂u + χ(ω1 ) cos(

n , x |∇u|p(x)−2 − ) = 1 ∂ n

sin ω20 ∂r   ∂u ∂u + χ(ω1 ) − |∇u|p(x)−2 = γ (ω1 )r 1−p(x)u|u|p(x)−2. ∂ n

∂r

.

• On . R :   1 ∂u ∂u + χ(ω1 ) cos(

n , x |∇u|p(x)−2 − ) = 1 ∂ n

sin ω20 ∂r   cos ω1 ∂u |∇u|p(x)−2 − 1 = χ(ω1 ) ∂r sin ω20

.

10.2 Preliminary

275

  γ (ω1 ) cos ω1 R 1−p(x) u|u|p(x)−2, 1 − χ(ω1 ) 1 + χ(ω1 ) sin ω20 ∂u  ∂u  because of  =  . ∂ n R ∂r R Thus, from above equalities, we obtain .

 ? − p(x)u +

GR 0



 @ d 1 p(x)−2 ∂u )|∇u| χ(ω η(x)dx = 1 sin ω20 dx1 ∂r

|∇u|p(x)−2 ∇u, ∇η −

GR 0

1 ∂u ηx dx+ ω0 χ(ω1 ) sin 2 ∂r 1

 γ (ω1 )r 1−p(x)u|u|p(x)−2η(x)ds+ 0R

  cos ω1 γ (ω1 ) R n−p(x) u|u|p(x)−2η(x)d . 1 − χ(ω1 ) 1 + χ(ω1 ) sin ω20



(10.2.1)

By this is justified integral identity (II). Now we define functions:  A1 (ω, ξ ) = |ξ |

p(x)−2

.

χ(ω1 )  xk ξk ξ1 − sin ω20 r n

 ;

k=1

Ai (ω, ξ ) = |ξ |p(x)−2ξi ;

i ≥ 2.

Proposition 10.4 Let assumption (10.1.2) (see (iii2)) be satisfied. Then the equation of (P ) is uniformly elliptic, i.e.,

.

.

∂Ai (x, ξ ) ζi ζj ≥ ν|ξ |p(x)−2ζ 2 , ∂ξj  ν=

0

−

∀ζ ∈ Rn \ {0};

(n − 1)χ− sin ω20

p(x) ≥ p− > 1; (10.2.2)

 > 0.

Proof It is well known that   ∂ |ξ |p(x)−2ξi . ζi ζj ≥ ∂ξj

0 |ξ |

ζ , ∀ζ ∈ Rn \ {0}.

p(x)−2 2

276

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for. . .

Therefore, .

∂Ai (x, ξ ) ζi ζj ≥ ∂ξj

0 |ξ |

p(x)−2 2

ζ −

n  χ(ω1 ) xk ∂ p(x)−2 |ξ | ζ ζ ξk . 1 j sin ω20 r ∂ξj k=1

Next we calculate A B n n   xj ∂ xk xk ξk p(x)−2 |ξ | ξk = |ξ |p(x)−2 + (p(x) − 2) |ξ |p(x)−4ξj . ∂ξj r r r k=1 k=1 A B n n n    xj ∂ xk ξk = ζ1 |ξ |p(x)−2 ζj + ⇒ ζ1 ζj |ξ |p(x)−2 ∂ξj r r j =1

j =1

k=1

(p(x) − 2) ζ1 |ξ |p(x)−4

n 

ξj ζj ·

j =1

⎛

n  k=1

xk ξk ≤ (by the Hölder inequality) r ⎞

⎝1 + max |p(x) − 2|⎠ |ξ |p(x)−2|ζ |2 < (n − 1)|ξ |p(x)−2|ζ |2 , GR 0

by assumption (i). Thus we obtain ∂Ai (x, ξ ) . ζi ζj ≥ ∂ξj



(n − 1)χ− 0− sin ω20



 

by (10.1.2).

10.3

|ξ |p(x)−2ζ 2 > 0,

The Maximum Principle

Theorem 10.5 Let u be a weak solution to the problem (OPn ) and assumptions (i)–(iv) be satisfied. Then, there exists a positive constant M0 depending only on p, s, μ,a0 , γ0 , χ− , f Ls (GR ) , such that 0

uL∞ (GR ) ≤ M0 .

.

0

$ % Proof Let us define the set A(k) = x ∈ GR 0 : |u(x)| > k with χA(k) being the characteristic function of the set A(k). Then, for all q > 0, we have that A(k + q) ⊆ A(k). We take η((|u| − k)+)χA(k) · sign u as a test function in the integral identity (I I ), where η is defined by (2.14.5) and k ≥ k0 ≥ 1. Note that η((|u|−k)+ ) ≥ 0 and η ((|u|−k)+ ) ≥ 0

10.3 The Maximum Principle

277

on A(k). From the integral identity (I I ), it follows: .

   p(x)−2 |∇u| |∇u|2 −

A(k)

 1 ∂u ∂u χ(ω1 ) η ((|u| − k)+ )+ sin ω20 ∂r ∂x1 

a0 (x)|u|

p(x)

η((|u| − k)+ ) + b(u, ∇u)η((|u| − k)+ ) · sign u dx+

 γ (ω1 )r 1−p(x)|u|p(x)−1η((|u| − k)+ )ds+ 0R ∩A(k)

 ∩A(k)

  γ (ω1 ) cos ω1 1 − χ(ω1 ) R n−p(x) |u|p(x)−1η((|u| − k)+ )d = 1 + χ(ω1 ) sin ω20  f (x)η((|u| − k)+ ) · sign udx. A(k)

By assumption (iv)a , (iv)b , we have b(u, ∇u)η((|u| − k)+ ) · sign u ≥

σ |u|−1 |∇u|p(x) · sign u − b0 |u|p(x)−1 η((|u| − k)+ ) ≥

− σ |u|−1 |∇u|p(x) + b0 |u|p(x)−1 η((|u| − k)+ ); σ = min(δ, ν) > 0.

.

p −1

But, since |u| > k ≥ k0 ≥ 1, p(x) ≥ p− > 1, we have |u|p(x)−1 ≥ k0 − from above, it follows: .

, and therefore

   χ(ω1 ) p(x) p(x)−2 ∂u ∂u |∇u| η ((|u| − k)+ )− − |∇u| sin ω20 ∂r ∂x1

A(k)

σ k0−1 |∇u|p(x)η((|u| − k)+ ) p −1 γ0 k 0 −

 ∩A(k)

p −1 + (a0 k0 − b0 )k0 − η((|u| −

 k)+ ) dx+

 ·

r 1−p(x)η((|u| − k)+ )ds+

0R ∩A(k)

  cos ω1 γ (ω1 ) R n−p(x) |u|p(x)−1η((|u| − k)+ )d ≤ 1 − χ(ω1 ) 1 + χ(ω1 ) sin ω20  |f |η((|u| − k)+ )dx. A(k)

278

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for. . .

Hence, by (10.1.2) and χ(ω1 ) ≥ 0 , we obtain .

   χ(ω1 ) p(x) p(x)−2 ∂u ∂u − |∇u| |∇u| η ((|u| − k)+ )− sin ω20 ∂r ∂x1

A(k)

σ k0−1 |∇u|p(x)η((|u| −

p −1 k)+ ) + (a0 k0 − b0 )k0 − η((|u| − k)+ )

 dx ≤

 |f |η((|u| − k)+ )dx.

(10.3.1)

A(k)

Next, since

∂u ∂u ∂r ∂x1

≤ |∇u|2 and

χ(ω1 ) ω sin 20

≤

χ− ω sin 20

1 n−1 ,




.

b0 , a0

(10.3.2)

we set aˆ 0 = (n − 1)(a0k0 − b0 ) > 0 , and from above, we have .

 

|∇u|p(x) η ((|u| − k)+ ) − (n − 1)σ k0−1 η((|u| − k)+ ) +

A(k) p −1

aˆ 0 k0 −

  η((|u| − k)+ ) dx ≤ (n − 1) |f |η((|u| − k)+ )dx. A(k)

(10.3.3)

10.3 The Maximum Principle

279

Additionally, let us define the sets

.

A− (k) = A(k) ∩ {|∇u| ≤ 1}, A+ (k) = A(k) ∩ {|∇u| ≥ 1}

⇒

A(k) = A− (k) ∪ A+ (k)

(10.3.4)

and the functions  vk (x) := η

.

((|u| − k)+ )+ p−



 ,

wk (x) := η

((|u| − k)+ )+ p+

 .

(10.3.5)

We note that the inequalities |∇u|p+ ≤ |∇u|p(x) ≤ |∇u|p−

on A− (k); .

(10.3.6)

|∇u|p− ≤ |∇u|p(x) ≤ |∇u|p+

on A+ (k)

(10.3.7)

.

hold by (i). Direct calculations give    ((|u| − k)+ )+ ((|u| − k)+ )+ ς |∇u| · exp ς = , p− p− p−   ς p− p− |∇u|p− · eς((|u|−k)+ )+ . (10.3.8) ⇒ |∇vk | = p−

1 .|∇vk | = |∇u| · η p− ς > 0;

Choosing ς > p− +



2(n−1)σ k0

according to (10.3.6), we have

η (((|u| − k)+ )+ ) − (n − 1)σ k0−1 η(((|u| − k)+ )+ ) ≥

.

1 ς((|u|−k)+ )+ e . 2

From (10.3.8), (10.3.9), it follows that . / |∇u|p− η (((|u| − k)+ )+ ) − (n − 1)σ k0−1 η(((|u| − k)+ )+ ) ≥   1 p− p− |∇vk |p− ⇒ (by (9.3.9)) 2 ς  . / |∇u|p(x) η (((|u| − k)+ )+ ) − (n − 1)σ k0−1 η(((|u| − k)+ )+ ) dx ≥

.

A+ (k)

(10.3.9)

280

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for. . .



. / |∇u|p− η (((|u| − k)+ )+ ) − (n − 1)σ k0−1 η(((|u| − k)+ )+ ) dx ≥

A+ (k)

1 2



p− ς

p−

 ·

|∇vk |p− dx.

(10.3.10)

A+ (k)

Similarly, choosing ς > p+ + 

2(n−1)σ k0

and taking into account (10.3.7), we obtain

. / |∇u|p(x) η (((|u| − k)+ )+ ) − (n − 1)σ k0−1 η(((|u| − k)+ )+ ) dx ≥

.

A− (k)

1 2



p+ ς

p+

 ·

|∇wk |p+ dx.

(10.3.11)

A− (k)

Since p+ ≥ p− , the inequalities (10.3.10) and (10.3.11) hold for ς > p+ + Therefore, adding the inequalities (10.3.10) and (10.3.11), we get 1 . 2



p− ς

p−

 ·

|∇vk |

p−

1 dx + 2



p+ ς

p+

 ·

A+ (k)



2(n−1)σ . k0

|∇wk |p+ dx ≤

A− (k)

. / |∇u|p(x) η (((|u| − k)+ )+ ) − (n − 1)σ k0−1 η(((|u| − k)+ )+ ) dx,

(10.3.12)

A(k)

by (10.3.4). Finally, from (10.3.3), (10.3.12), we derive .

1 2



p− ς

p−

 ·

|∇vk |p− dx +

1 2



p+ ς

p+

 ·

A+ (k) p −1 aˆ 0 k0 −

|∇wk |p+ dx+

A− (k)

 ·

 η(((|u| − k)+ )+ )dx ≤ (n − 1)

A(k)

A(k)

=

Since, by (10.3.4),

1 . 2



p −1

aˆ 0 k0 −

p− ς

+

A(k)

A+ (k)

p−



 · A+ (k)

|f (x)| · η(((|u| − k)+ )+ )dx.

·

, we have A− (k)

|∇vk |

p−

1 dx + 2



p+ ς

p+

A+ (k) p −1

η(((|u| − k)+ )+ )dx + aˆ 0 k0 −

 ·



|∇wk |p+ dx+

A− (k)

· A− (k)

η(((|u| − k)+ )+ )dx ≤

10.3 The Maximum Principle

281

 (n − 1)

|f (x)| · η(((|u| − k)+ )+ )dx+

A+ (k)



(n − 1)

|f (x)| · η(((|u| − k)+ )+ )dx.

(10.3.13)

A− (k)

Now, by (2.14.7) and (10.3.5), we derive p −1

aˆ 0 k0 −

.



p −1

η(((|u| − k)+ )+ )dx + aˆ 0 k0 −

· A+ (k)

⎛

p −1

aˆ 0 k0 −

⎜ ·⎝

 ·

η(((|u| − k)+ )+ )dx ≥

A− (k)



⎞



p

⎟ p wk + dx ⎠ .

vk − dx +

A+ (k)

(10.3.14)

A− (k)

From (10.3.13)–(10.3.14), it follows that .

1 2



p− ς

p−

 ·

|∇vk |p− dx +

A+ (k)

⎛ ⎜ ·⎝

p −1

aˆ 0 k0 −

1 2



p+ ς

p+

 · A− (k)





p

vk − dx +

A+ (k)

⎞

⎟ p wk + dx ⎠ ≤

A− (k)

 (n − 1)

|∇wk |p+ dx+

|f (x)| · η(((|u| − k)+ )+ )dx+

A+ (k)



(n − 1)

|f (x)| · η(((|u| − k)+ )+ )dx.

(10.3.15)

A− (k)

Next, we have 

 |f (x)| · η(((|u| − k)+ )+ )dx =

.

A± (k)



|f (x)| · η(((|u| − k)+ )+ )dx+ A± (k+q)

|f (x)| · η(((|u| − k)+ )+ )dx, A± (k)\A± (k+q)

∀q > 0.

(10.3.16)

282

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for. . .

By (2.14.8), we obtain   .η(((|u| − k)+ )+ ) 

A± (k+q)

+  , (|u| − k)+ p∓ ≤M η . p∓

Then (10.3.5) implies that 

 A+ (k+q)

p

(10.3.17)

p

(10.3.18)

|f (x)| · vk − dx; .

|f (x)| · η(((|u| − k)+ )+ )dx ≤ M ·

.

A+ (k+q)





|f (x)| · wk + dx.

|f (x)| · η(((|u| − k)+ )+ )dx ≤ M · A− (k+q)

A− (k+q)

Using the definition of η by (2.14.5) from Lemma 2.64, we come to   .η(((|u| − k)+ )+ )  

≤ eςq ,

∀q > 0

⇒

A± (k)\A± (k+q)

|f (x)| · η(((|u| − k)+ )+ )dx ≤ A± (k)\A± (k+q)



e

ςq

·

|f (x)| dx,

∀q > 0.

(10.3.19)

A± (k)\A± (k+q)

By assumption (iii), we get that f (x) ∈ Ls (G), where s > inequality with exponents s and s  , where

1 s

+

1 s

=1:

⎛



⎜ f (x)|z| dx ≤ f (x)Ls (G) · ⎝

.

A(k+d)

⎜ .⎝

s

|z|

ps 

⎟ dx ⎠ .

(10.3.20)

A(k)

From the inequality inequality gives ⎛

> 1. By Hölder’s

⎞ 1



p

n p±

1 s


0.

,

(10.3.21)

A(k)

Therefore, we derive inequalities: 

p

|f (x)| · vk − dx ≤

.

A+ (k+q)

⎛

⎞ p−



⎜ ε(1 − θ− ) ⎝

p# ⎟ vk − dx ⎠

A+ (k)

# p−

+ θ− ε

θ− −1 θ−



1 θ

p

vk − dx;

f L−(GR ) s

0

A+ (k)



p

|f (x)| · wk + dx ≤

(10.3.22)

A− (k+q)

⎛ ⎜ ε(1 − θ+ ) ⎝

⎞ p+



p# ⎟ wk + dx ⎠

# p+

+ θ+ ε

θ+ −1 θ+

s

A− (k) # ∀ε > 0, p∓ =



1 θ

p

wk + dx,

f L+(GR ) 0

A− (k)



np∓ n n n , θ∓ = 1 − ; s > max ; n − p∓ sp∓ p− p+

 =

n > 1. p−

Then applying (10.3.22) to (10.3.16)–(10.3.19), we get that  .

⎛ ⎜ |f (x)| · η(((|u| − k)+ )+ )dx ≤ Mε(1 − θ− ) ⎝

A+ (k)

e

· A+ (k)

p# ⎟ vk − dx ⎠

A+ (k)

 ςq

⎞ p−



|f (x)| dx + Mθ− ε

θ− −1 θ−



1 θ−

f L

R s (G0 )

A+ (k)

# p−

+

p

vk − dx

(10.3.23)

284

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for. . .

 .

⎛ ⎜ |f (x)| · η(((|u| − k)+ )+ )dx ≤ Mε(1 − θ+ ) ⎝

A− (k)

p# ⎟ wk + dx ⎠

A− (k)

 eςq ·

⎞ p+



|f (x)| dx + Mθ+ ε

θ+ −1 θ+



1 θ

p

0

A− (k)

+

wk + dx.

f L+(GR ) s

# p+

(10.3.24)

A− (k)

By the well known Sobolev embedding theorem and taking into account (10.3.22), we obtain ⎛ ⎜ ⎝ .



⎞ p− p# ⎟ vk − dx ⎠

⎜ ⎝



A− (k)

 p−  vk + |∇vk |p− dx;

≤ c− ·

A+ (k)

⎛



# p−

A+ (k)

⎞ p+ p# ⎟ wk + dx ⎠



# p+

(10.3.25)  p+  wk + |∇wk |p+ dx,

≤ c+ · A− (k)

where c∓ = const > 0. Finally, (10.3.15)–(10.3.25) imply that .

,  +   1 p− p− − M(n − 1)c− (1 − θ− )ε · 2 ς

|∇vk |p− dx+

A+ (k)

 ,  +  1 p+ p+ − M(n − 1)c+ (1 − θ+ )ε · 2 ς

|∇wk |p+ dx+

A− (k)

+ 1 θ− −1 θ p −1 aˆ 0 k0 − − M(n − 1)c− (1 − θ− )ε − M(n − 1)θ− ε θ− f  − 

,

Ls (GR 0)

×

p

vk − dx + A+ (k)

+ , 1 θ+ −1 θ p −1 aˆ 0 k0 − − M(n − 1)c+ (1 − θ+ )ε − M(n − 1)θ+ ε θ+ f L+(GR ) × 

s

0

p

wk + dx ≤  (n − 1)eςq · A(k)

A− (k)

|f (x)| dx,

∀ε > 0.

(10.3.26)

10.3 The Maximum Principle

285

Further, at first, we choose ε=

.

  p− 1 1 p− min ; 4M(n − 1) c− (1 − θ− ) ς 1 c+ (1 − θ+ )



p+ ς

p+  (10.3.27)

and next  k0 ≥

.

2(n − 1)MF aˆ 0



1 p− −1

where

,

 1 θ− −1 θ F = max c− (1 − θ− )ε + θ− ε θ− f L−(GR ) ; s

c+ (1 − θ+ )ε + θ+ ε

θ+ −1 θ+

0

 ; .

(10.3.28)

|f (x)| dx,

(10.3.29)

f 

1 θ+

Ls (GR 0)

Thus, by the above arguments, we derive 

 p  |∇vk |p− + vk − dx +

.

A+ (k)





p  |∇wk |p+ + wk + dx ≤

A− (k)

 C A(k)

where C = const n, p− , p+ , aˆ 0 , k0 , δ, ν, s, f Ls (GR ) > 0. The inequalities (10.3.25) 0 together with (10.3.29) give ⎛ ⎜ .⎝



⎞ p− p# ⎟ vk − dx ⎠

A+ (k)

# p−

⎛ ⎜ +⎝



⎞ p+ p# ⎟ wk + dx ⎠

# p+

≤

A− (k)

 max{c− , c+ } · C

|f (x)| dx, ∀k ≥ k0 .

A(k)

At last, by the Hölder inequality, we have  .

A(k)

1

|f (x)|dx ≤ f (x)Ls (GR ) meas 1− s A(k); 0

s>

2 > 1. p−

(10.3.30)

286

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for. . .

Then from (10.3.30), it follows that ⎛ ⎜ .⎝

⎞ p−



p# ⎟ vk − dx ⎠

# p−

⎛ ⎜ +⎝

A+ (k)

⎞ p+



p# ⎟ wk + dx ⎠

# p+

≤

A− (k) 1

max{c− , c+ } · C f (x)Ls (GR ) meas 1− s A(k);

2 > 1, ∀k ≥ k0 . p−

s>

0

(10.3.31)

Now, let l > k > k0 . By (2.14.9) and the definition of the functions vk (x), wk (x), we have vk ≥ p1− ((|u| − k)+ )+ , wk ≥ p1+ ((|u| − k)+ )+ . Therefore, 



p#

vk − dx ≥

.

A+ (l)





p#

wk + dx ≥ A− (l)

l−k p− l−k p+

p−#

p+#

· measA+ (l);

· measA− (l).

Hence, (10.3.31) together with A± (l) ⊆ A± (k) implies that measA(l) = meas (A+ (l) ∪ A− (l)) ≤ measA+ (l) + measA− (l) ≤

.



p− l−k

p−#

 ·

p# vk − dx

 +

A+ (k)

p+ l−k

p+#



p#

wk + dx ≤

· A− (k)

(10.3.32)  C−

p− l−k

 C+

p−#

p+ l−k

Since

# p− p−

≤

# p+ p+

s

# p−

meas p− (GR )

A(k)+

0

# p+ p+

· f (x)L

(1− 1s )

s

# p+

meas p+ (GR )

(1− 1s )

A(k)

0

# p∓

where C∓ = (C max{c− , c+ }) p∓ .

(see (10.3.22)), we have # p−

meas p−

.

· f (x)L

p+#

∀l > k ≥ k0 ,

# p− p−

(1− 1s )

# p+

A(k) ≥ meas p+

(1− 1s )

A(k), if measA(k) ≤ 1.

10.4 The Comparison Principle

287

Moreover, # p+ . p+

   #  p− 2 1 1 ≥ > 1 for s > 1− 1− > 1. s p− s p−

Let us introduce ψ(k) = meas A(k). Then from (10.3.32), it follows that

ψ(l) ≤ 2Cψ

.

ζ

⎧ ⎪ ⎨

1 #

p− (k) (l−k) ⎪ 1

⎩

# (l−k)p+

if

l − k ≥ 1;

if

0 < l − k < 1,

∀l > k ≥ k0 ,



2 - = const (n, p− , p+ , aˆ 0 , k0 , δ, ν, , s, f  where ζ = 1 − 1s 2−p > 1, C )> Ls (GR − 0) 0. By the Stampacchia Lemma, we have that ψ(k0 +ϑ) = 0 with ϑ depending only on the quantities given in being proved theorem. This fact means that |u(x)| ≤ k0 + ϑ for almost all x ∈ GR 0 . Thus, we derive M0 = k0 + ϑ, where k0 is defined by (10.3.2), (10.3.28) with (10.3.22) and (10.3.27) being proved theorem. Then, by the Stampacchia lemma, we have ψ(k0 + ϑ) = 0,

.

p# β

#

ϑ p = C6 |ψ(k0 )|β−1 2 β−1 .

It means that |u(x)| ≤ k0 + ϑ for almost all x ∈ GR 0 , where k0 > 1 is sufficiently large,   such that inequalities (10.3.2), (10.3.28) hold.

10.4

The Comparison Principle

Lemma 10.6 Let u be a weak bounded solution of problem (OPn ). For any function η ∈ 1,p(x) N−1,∞ (GR 0 ), the equality    p(x)−2 ∇u, ∇η − .Q[u, η] ≡ |∇u| Gd0

a0 (x)u|u|

p(x)−1

 1 ∂u χ(ω1 ) ηx1 + sin ω20 ∂r

  η + b(u, ∇u)η dx + γ (ω1 )r 1−p(x)u|u|p(x)−2η(x)ds− 0d

288

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for. . .

 d

n−1



·

|∇u|

  cos ω1 ηd + 1 − χ(ω1 ) ∂r sin ω20

p(x)−2 ∂u

(I I )loc

   cos ω1 γ (ω1 ) n−p(x) p(x)−2 u|u| η(x)d = f ηdx d 1 − χ(ω1 ) 1 + χ(ω1 ) sin ω20 Gd0

holds for a.e. d ∈ (0, R). Proof Let us consider the integral identity (I I ) in which we replace η(x) by η(x)χd (x), where χd (x) is the characteristic function of Gd0 . We have .

∂ ∂χd (x) ; i = 1, 2. (η(x)χd (x)) = ηxi χd (x) + η(x) ∂xi ∂xi

According to formula (7’) of Subsection 3 §1, Chapter 3 [60], .

∂χd (x) xi = − δ(d − r); i = 1, 2, ∂xi r

where δ(d − r) denotes the Dirac delta distribution lumped on the circle r = d. Thus, .

  |∇u|p(x)−2 ∇u, ∇χd (x) −

Gd0



 .

− Gd0

|∇u|p(x)−2

 1 ∂u ∂χd (x) · η(x)dx = χ(ω ) 1 sin ω20 ∂r ∂x1

 1 ∂u x1 ∂u xi − · · χ(ω ) η(x)δ(d − r)dx = 1 ∂xi r sin ω20 ∂r r

 − d

(by formulas (9.2.1)–(9.2.2) )   cos ω1 p(x)−2 ∂u 1 − χ(ω1 ) ηd d = |∇u| ∂r sin ω20  −d

n−1

·

|∇u|

Hence, we derive the required statement.

  cos ω1 ηd . 1 − χ(ω1 ) ∂r sin ω20

p(x)−2 ∂u

 

10.4 The Comparison Principle

289

In Gd0 , we consider the second order quasi-linear degenerate operator Q of the form Q(u, η) ≡

 . /   Ai (x, ux )ηxi + a0 (x)u|u|p(x)−1 + b(u, ∇u) η(x) dx+

Gd0



.

d

  γ (ω1 ) cos ω1 1 − χ(ω1 ) d 1−p(x)u|u|p(x)−2η(x)d d − 1 + χ(ω1 ) sin ω20



 γ (ω)r 1−p(x)u|u|p(x)−2ηds

Ai (x, ux ) cos(r, xi )ηd d + d

0d

1,p(x)

1,p(x)

for u(x) ∈ N−1,∞ (Gd0 ) and for all nonnegative η(x) belonging to N−1,∞ (Gd0 ). We assume that conditions (i), (iii), (iii1), (iii2), (iv), (iv)c are satisfied as well as that functions Ai (x, ξ ) are Caratheodory, continuously differentiable with respect to the u, ξ variables in M = Gd0 × R × R2 , and satisfy in M the following inequality: (vii) ∂ A∂ξi j(ξ ) ζi ζj ≥ ν|ξ |p−2 ζ 2 for all ζ ∈ R2 , ν > 0. We get the following comparison principle that plays an important role in the proof of Theorem 10.3. Theorem 10.7 Let operator Q satisfy above assumptions and functions u, w 1,p N−1,∞ (Gd0 ), d & 1, satisfy the following inequality: Q(u, η) ≤ Q(w, η)

(10.4.1)

.

1,p

for all nonnegative functions η ∈ N−1,∞ (Gd0 ), as soon as u(x) ≤ w(x), a.e on d .

.

Then u(x) ≤ v(x) in Gd0 . Proof Let us define z = u − w and uτ = τ u + (1 − τ )w, τ ∈ [0, 1]. Then we have  ?  .0 ≥ Q(u, η) − Q(w, η) = ηxi zxj  a0 (x)z(x)η(x) 0

Gd0 1

∈

1 0

∂(uτ |uτ |p(x)−1) dτ + η(x)zxi ∂uτ

∂Ai (x, uτx ) dτ + ∂uτxj 

1 0

∂b(x, uτ , uτx ) dτ + ∂uτxi

290

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for. . .



@   1 ∂b(x, uτ , uτx ) ∂Ai (x, uτx ) dτ dx − z dτ · cos(r, xi )ηd d + x j ∂uτ ∂uτxj 0

1

η(x)z(x) 0

 d

d

    1 τ |uτ |p(x)−2) γ (ω ) ∂(u cos ω 1 1 1 − χ(ω1 ) d 1−p(x) dτ z(x)η(x)d d + 1 + χ(ω1 ) sin ω20 ∂uτ 0 

γ (ω)



r p(x)−1

1 0

0d

 ∂(uτ |uτ |p(x)−2) dτ z(x)η(x)ds ∂uτ

(10.4.2)

1,p(x)

for all nonnegative η ∈ N−1,∞ (Gd0 ). Now, we introduce the sets (Gd0 )+ := {x ∈ Gd0 | u(x) > w(x)} ⊂ Gd0 ,

.

(0d )+ := {x ∈ 0d | u(x) > w(x)} ⊂ 0d and assume that (Gd0 )+ = ∅, (0d )+ = ∅. Let k ≥ 1 be any odd number. We choose η = max{(u − w)k , 0} as a test function in the integral inequality (10.4.2). We have 

1

.

0



1 0

∂(uτ |uτ |p(x)−1) dτ = p(x) ∂uτ



1 0

∂(uτ |uτ |p(x)−2) dτ = (p(x) − 1) ∂uτ

  Then, by our assumptions and η

.



  kνzk−1

1

d



1

|uτ |p(x)−2dτ > 0.

0

= 0, we obtain from (10.4.2) that

 |∇uτ |p(x)−2dτ |∇z|2 + b2 zk+1

0

(Gd0 )+

|uτ |p(x)−1dτ > 0;

1

|uτ |−2 |∇uτ |p(x)dτ

 dx

0

 ≤ b1 · (Gd0 )+

zk



1 0

|uτ |−1 |∇uτ |p(x)−1dτ |∇z|dx.

(10.4.3)

10.5 The Barrier Function

291

By the Cauchy inequality, b1 zk |∇z||uτ |−1 |∇uτ |p(x)−1 =     k−1 τ −1 k+1 τ p(x) τ p(x) −1 2 2 2 2 · b1 z |∇z||∇u | ≤ |u | z |∇u |

.

b2 ε τ −2 k+1 |u | z |∇uτ |p(x) + 1 zk−1 |∇z|2 |∇uτ |p(x)−2, ∀ε > 0. 2 2ε Hence, taking ε = 2b2 , we obtain from (10.4.3) the inequality  .

b2 kν − 1 4b2

 



1

zk−1 |∇z|2 (Gd0 )+

 |∇uτ |p(x)−2dτ dx ≤ 0.

(10.4.4)

0

  b12 Choosing the odd number k ≥ max 1; 2b2 ν , in view of z(x) ≡ 0 on ∂(Gd0 )+ , we get from (10.4.4) that z(x) ≡ 0 in (Gd0 )+ . We got a contradiction to our definition of the set   (Gd0 )+ , and this completes the proof.

10.5

The Barrier Function

Let us consider the following auxiliary problem for a function .w(r, ω) ≡ 0  (BFP)

.

x ∈ Gd0 , − p+ w = μ|w|−1 |∇w|p+ ,   γ ∂w ∂w ± d. |∇w|p+ −2 ∂ n ± χ± ∂r + |x|p+ −1 w|w|p+ −2 = 0, x ∈ ±

By direct calculations, we can show that (BF)

.

w = w(r, ω) = r ψ λ (ω),

=

p+ − 1 λ, p+ − 1 + μ

is a solution to problem (.BFP), where .(λ, ψ(ω)) is an eigenpair of the eigenvalue problem (OEV P ). For this function, we calculate ∂w ∂w = r −1 ψ /λ (ω); = r ψ λ −1 (ω)ψ  (ω); ∂r ∂ω λ 2 / . 2 |∇w|2 = r 2( −1) ψ 2( λ −1) (ω) λ2 ψ 2 (ω) + ψ  (ω) ; ⇒ λ p+

p+ /2 2 |∇w|p+ = r p+ ( −1) ψ p+ ( λ −1) (ω) λ2 ψ 2 (ω) + ψ  (ω) ; λ .

292

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for. . .

p+ λ

r (p+ −1)

|w|−1 |∇w|p+ =

p+ /2 2 −p+ (p+ −1) λ −p+ ψ (ω) λ2 ψ 2 (ω) + ψ  (ω) ; |w|p+ −1 = r (p+ −1) ψ (p+ −1) λ ;

(10.5.1)

2 ∂ 2w ∂ 2w = r −1 ψ λ −1 (ω)ψ  (ω); = ( − 1)r −2 ψ /λ (ω); 2 ∂r ∂r∂ω λ

∂ 2w − 1 r ψ λ −2 (ω)ψ 2 (ω) + r ψ λ −1 (ω)ψ  (ω). = ∂ω2 λ λ λ 1,p(x)

Proposition 10.8 .w ∈ N−1,∞ (Gd0 ). Proof From .(BF ) and (3.2.6), it follows that .w ∈ L∞ (Gd0 ). Next,  .

r −p(x)wp(x) dx =

Gd0

 r(

−1)p(x)

ψ λ p(x)(ω)dx.

Gd0

By .r ≤ d & 1 and assumption (i), we have

.

r(

−1)p(x)

≤ r(

−1)p−

, if

≥ 1;

r(

−1)p(x)

≤ r(

−1)p+

, if

≤ 1;

p

ψ λ p(x)(ω) ≤ ψ0 + in virtue of (3.2.6) and

≤ λ.

Hence, it follows that  .

r

−p(x)

w

p(x)

dx ≤

⎧ ( −1)p−+n d ⎪ ⎪ ( −1)p− +n , ⎨

p ψ0 + meas · ⎪

⎪ ⎩ d(

Gd0

if

≥ 1; (10.5.2)

−1)p+ +n

( −1)p+ +n ,

if

≤ 1.

From (10.5.1) with regard to (3.2.13), (3.2.14), we obtain that  .

 p(x) λ Gd0

w−1 |∇w|p(x)dx =

Gd0

r (p(x)−1)

−p(x)



p(x)/2 2 ψ (p(x)−1) λ −p(x)(ω) λ2 ψ 2 (ω) + ψ  (ω) dx ≤

10.5 The Barrier Function

293



p −1 ψ0 +

−p(x)

r (p(x)−1)

Gd0 p −1



ψ0 +



p(x)/2 λ2 + y 2 (ω) dx ≤

−p(x)

r (p(x)−1)



λ2 + z02

p(x)/2 dx,

Gd0

where . y(ω) =

ψ  (ω) ψ(ω) .

Next, by (3.2.16),



p(x)/2

.

λ2 + y02 (ω)

≤ C1 = const (n, p+ , λ, ω0 ).

From the above inequality, we obtain that  .

w

−1

|∇w|

p(x)

dx ≤

p −1 C1 ψ0 +

Gd0



−p(x)

r (p(x)−1)

Gd0



p −1

mathbbC1ψ0 +

dx =

−1)(p(x)−p+)

r(

· r(

−1)p+ −

dx.

(10.5.3)

Gd0

Now, by assumptions (i)–(ii) and .r & 1, we derive −1)(p(x)−p+ )

r(

.

≤

⎧ ⎨r (1−

)Lr ,

⎩1,

if

> 1,

if

≤ 1.

Using the well known inequality, r α | ln r| ≤

.

1 , ∀α > 0, 0 < r < 1, αe

where e is the Euler number, we establish for .α = 1, r (1−

.

)Lr

≤e

L( −1) e

,

(10.5.4)

> 1 the inequality

0 < r < 1.

Thus, from (10.5.3), it follows that  .

Gd0

p −1

w−1 |∇w|p(x)dx ≤ C1 ψ0 +

e

L( −1) e

meas ·

d (p+ −1)+n−p+ . (p+ − 1) + n − p+  

294

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for. . .

10.6

Estimation of the Solution Modulus. The Proof of the Main Theorem 10.3

Let .A > 1, and let .w(r, ω) ≥ 0 be the barrier function defined above. By the definition of the operator Q in .(I I )loc , we have  Q[Aw, η] ≡

A

p(x)−1

  p(x)−2 ∇w, ∇η − |∇w|

Gd0

a0 (x)Aw

p(x)

 1 ∂w ηx + χ(ω) sin ω20 ∂r 1

  η dx + b(Aw, ∇Aw)ηdx+ Gd0



γ (ω)Ap(x)−1r 1−p(x)wp(x)−1 η(x)ds−

.

0d

d



n−1

·

A



Ap(x)−1

p(x)−1

|∇w|

  cos ω ηd + 1 − χ(ω) ∂r sin ω20

p(x)−2 ∂w

  cos ω γ (ω) 1 − χ(ω) d n−p(x) wp(x)−1η(x)d 1 + χ(ω) sin ω20

1,p(x)

for all .η ∈ N−1,∞ (Gd0 ) and for a.e. .d ∈ (0, R). Integrating by parts, we obtain 

 Ap(x)−1|∇w|p(x)−2wxi ηxi dx = −

.

Gd0

/ d . p(x)−1 A |∇w|p(x)−2wxi dx+ dxi

Gd0

 A

0d

η(x)

p(x)−1

|∇w|

p(x)−2 ∂w

∂n



η(x)dS +

Ap(x)−1|∇w|p(x)−2

∂w η(x)d d = ∂r

d

 ?

μAp(x)−1w−1 |∇w|p(x) − Ap(x)−1|∇w|p+ −2 wxi

d|∇w|p(x)−p+ − dxi

Gd0

@  ∂Ap(x)−1 ∂w η(x)ds+ wxi |∇w|p(x)−2 η(x)dx + Ap(x)−1|∇w|p(x)−2 ∂xi ∂ n



0d

Ap(x)−1|∇w|p(x)−2 d

∂w η(x)d d , ∂r

(10.6.1)

10.6 Estimation of the Solution Modulus. The Proof of the Main. . .

295

by the (BF P ) equation. In the same way, we have .

−

1 sin ω20

1 sin ω20 1 sin ω20





 χ(ω1 )Ap(x)−1|∇w|p(x)−2

∂w ηx dx = ∂r 1

Gd0

? @ d p(x)−1 p(x)−2 ∂w η(x) |∇w| χ(ω1 )A dx− dx1 ∂r

Gd0

χ(ω1 )Ap(x)−1|∇w|p(x)−2

∂w → cos(− n , x1 )η(x)ds. ∂r

(10.6.2)

0d ∪ d

Adding (10.6.1) with (10.6.2) and taking into account (9.2.1), we obtain for all .η ∈ 1,p(x) N−1,∞ (Gd0 ): Q[Aw, η] ≡

.

 ?

μAp(x)−1w−1 |∇w|p(x) − Ap(x)−1|∇w|p+ −2 wxi

d|∇w|p(x)−p+ − dxi

Gd0

∂Ap(x)−1 wxi |∇w|p(x)−2 + a0 (x)Ap(x)wp(x) + b(Aw, ∇Aw)+ ∂xi  @ 1 d p(x)−1 p(x)−2 ∂w |∇w| χ(ω)A η(x)dx+ sin ω20 dx1 ∂r  ?  @  ∂w p(x)−1 p(x)−2 ∂w 1−p(x) p(x)−1 + χ(ω) + γ (ω)r |∇w| η(x)ds+ A w ∂ n

∂r

0d

 d

  cos ω . ∂w + |∇w|p(x)−2 Ap(x)−1 1 − χ(ω) ω0 sin 2 ∂r γ (ω) w p(x)−1 / η(x)d d ≡ 1 + χ(ω) d JGd + J d + J d . 0

0

(10.6.3)

296

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for. . .

Now, with regard to the (BF P ) boundary condition, we calculate J d ≡ 0  ?  @  ∂w ∂w + χ(ω) + γ (ω)r 1−p(x)wp(x)−1 η(x)ds = Ap(x)−1 |∇w|p(x)−2 ∂ n

∂r .



0d



γ (ω)Ap(x)−1r 1−p+ wp+ −1 r p+ −p(x)wp(x)−p+ − |∇w|p(x)−p+ ≥ 0,

(10.6.4)

0d

because of, by (BF ) and (10.5.1), for .∀x ∈ 0d : r p+ −p(x) wp(x)−p+ − |∇w|p(x)−p+ =

.

r( ? 1− r

−1)(p(x)−p+ )

p(x)−p+ λ

( −1)(p(x)−p+ )

ψ (p(x)−p+) λ (ω)×

p(x)−p+ @ 2 ψ p+ −p(x (ω) λ2 ψ 2 + ψ 2 = ?

ψ

(p(x)−p+ ) λ

(ω) 1 −

 ; λ

λ2

+

y02

p(x)−p+ @

≥ 0,

in virtue of Proposition 3.7. Further, by (10.1.2) (see assumption .(iii2)), it is obvious that



 A d

p(x)−1

J d ≡

.

cos ω 1 − χ(ω) sin ω20

. ∂w |∇w|p(x)−2 + ∂r

γ (ω) w p(x)−1 / η(x)d d ≥ 0. 1 + χ(ω) d Thus, from (10.6.3), (10.6.4), (10.6.5), it follows Q[Aw, η] ≥

.

 ? Gd0

μAp(x)−1w−1 |∇w|p(x) − Ap(x)−1|∇w|p+ −2 wxi

d|∇w|p(x)−p+ − dxi

(10.6.5)

10.6 Estimation of the Solution Modulus. The Proof of the Main. . .

297

  ∂Ap(x)−1 d 1 p(x)−1 p(x)−2 ∂w + χ(ω)A wxi |∇w|p(x)−2 + |∇w| ∂xi sin ω20 dx1 ∂r @ a0 (x)Ap(x)wp(x) + b(Aw, ∇Aw) η(x)dx ≡ JGd . (10.6.6) 0

Further, we proceed to the estimating of integral .JGd . Setting .W (x) = |∇w|p(x)−p+ , we 0 calculate .

ln W (x) = (p(x) − p+ ) ln |∇w|,

⇒

∂W 1 ∂p p(x) − p+ d|∇w| · · = ln |∇w| + ⇒ W (x) ∂xi ∂xi |∇w| dxi ? @

p(x) − p+ d|∇w| ∂p d |∇w|p(x)−p+ = |∇w|p(x)−p+ · ln |∇w| + . dxi ∂xi |∇w| dxi Similarly, .

∂ p(x)−1 ∂p A = Ap(x)−1 ln A. ∂xi ∂xi

By (10.6.6), we obtain that JGd ≥

.

0

?   Ap(x)−1|∇w|p(x)−2 μw−1 |∇w|2 − (∇p · ∇w) · ln(A|∇w|)− Gd0

 @  p(x) − p+ 1 d|∇w| d p(x)−1 p(x)−2 ∂w |∇w| · wxi + + χ(ω)A |∇w| dxi sin ω20 dx1 ∂r  a0 (x)Ap(x)wp(x) + b (Aw, A∇w) η(x)dx. (10.6.7) Further, by direct calculation,   d ∂w = χ(ω)Ap(x)−1|∇w|p(x)−2 dx1 ∂r    ∂w ∂p p(x) − 2 ∂|∇w| Ap(x)−1|∇w|p(x)−2 χ(ω) · − ln(A|∇w|) + ∂r ∂x1 |∇w| ∂x1   ∂ 2 w sin ω ∂ 2 w sin ω ∂w  χ (ω) + χ(ω) cos ω 2 − (10.6.8) . r ∂r ∂r r ∂r∂ω .

298

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for. . .

Passing to polar coordinates with regard to (9.2.2), we calculate wxi

.

1 ∂w ∂|∇w| d|∇w| ∂w ∂|∇w| · + 2 · ; = dxi ∂r ∂r r ∂ω ∂ω ∂|∇w| ∂|∇w| sin ω ∂|∇w| − . = cos ω ∂x1 ∂r r ∂ω

Thus, hence and from (10.6.6)–(10.6.8) with regard to assumptions .a0 > 0, .(iv)a −(iv)b , we obtain that 

.Q[Aw, η] ≥  ? Ap(x)−1 |∇w|p(x)−2 σ w−1 |∇w|2 − (∇p · ∇w) ln(A|∇w|)

Gd0

  χ(ω) ∂w ∂p ∂w ∂|∇w| 1 ∂w ∂|∇w| p+ − p(x) · · + · ln(A|∇w|) + − sin ω20 ∂r ∂x1 |∇w| ∂r ∂r r 2 ∂ω ∂ω   2 − p(x) χ(ω) ∂w ∂|∇w| sin ω ∂|∇w| · − · (10.6.9) cos ω · − |∇w| sin ω20 ∂r ∂r r ∂ω @   χ(ω) ∂ 2 w sin ω ∂ 2 w sin ω 1 ∂w  p(x)−1 + − b0 w cos ω 2 − η(x)dx, χ (ω) sin ω20 r ∂r sin ω20 ∂r r ∂r∂ω

+

⎧ ⎨μ − δ, if μ > 0; with . σ = ⎩ν, if μ = 0. Taking into account assumption (ii)—the Lipschitz condition of .p(x) and y(ω), we direct calculate: (1) . |w|−1 |∇w|2 = (2)

 2 λ

r

−2 ψ

λ

.

ψ  (ω) ψ(ω)

  (ω) λ2 + y 2 (ω) .

|(∇p · ∇w)(ln A + ln |∇w|)| ≤ |∇p| · |∇w| · (ln A + | ln |∇w||) ≤

.

L|∇w| · (ln A + | ln |∇w||); in virtue of (10.5.1), (3.2.6), (3.2.18)–3.2.20), we derive   | ln |∇w|| ≤ ln

.

   + | − 1| · | ln r| +

1 ln ψ + | ln(λ2 + y 2 (ω))| ≤ λ 2

λ λ 1 ln + ln ψ0 + | ln(λ2 + y02 )| + | − 1| · | ln r| = 2

ln C1 (μ, p+ , λ, γ+ , χ+ , p+ , ω0 ) + | − 1| · | ln r|

=

10.6 Estimation of the Solution Modulus. The Proof of the Main. . .

299

(note that . C1 ≥ 1 : indeed, by virtue of .(BF ) and (3.2.6), . λ ψ0 ≥ 1; from (3.2.17), ; ; it follows that . λ2 + y02 ≥ λ ≥ 1 ⇒ C1 = λ ψ0 λ2 + y02 ≥ 1); using inequality (10.5.4) with .α = 12 , we get that ; r −1 ψ λ (ω) λ2 + y 2 (ω) · (ln C1 + | − 1| · | ln r|) = λ ;   √ √ r −2 ψ λ (ω) λ2 + y 2 (ω) · r ln C1 + (| − 1| r) · r| ln r| ≤ λ ; √ r −2 ψ λ (ω) λ2 + y 2 (ω) · (r ln C1 + | − 1| r), λ

|∇w| · | ln |∇w|| ≤

.

from which we obtain that |(∇p · ∇w)(ln A + ln |∇w|)| ≤

.

−2

L r λ

; √ ψ λ (ω) λ2 + y 2 (ω) · (r ln(AC1 ) + | − 1| r)

and therefore, hence, we get

 χ− 1+ sin ω20

   χ(ω) ∂w ∂p   · ln(A|∇w|) ≤ . (∇p · ∇w) −  sin ω20 ∂r ∂x1    χ− 1+ |∇p| · |∇w| · ln(A|∇w|) ≤ sin ω20  ; √ −2 λ L r ψ (ω) λ2 + y 2 (ω) · (r ln(AC1 ) + | − 1| r). λ

(3) −1 ∂|∇w| = |∇w|; ∂r r   ∂|∇w| y  (ω) = |∇w| + 2 · y(ω) = ∂ω λ λ + y 2 (ω)   y  (ω) p+ − 1 + |∇w| · y(ω), p+ − 1 + μ λ2 + y 2 (ω) .

(10.6.10)

300

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for. . .

and by .(BF ), .

p+ − p(x) · |∇w|



 ∂w ∂|∇w| 1 ∂w ∂|∇w| · + 2 · = ∂r ∂r r ∂ω ∂ω ?  @ y  (ω) w y2 p+ − 1 + 2 −1+ ; (p+ − p(x)) 2 r λ p+ − 1 + μ λ + y 2 (ω)

now, since (p+ − 1)y 4 + λ (2λ(p+ − 1) + n − p+ ) y 2 + λ3 (λ(p+ − 1) + n − p+ ) =   n − p+ λ , (p+ − 1)(y 2 + λ2 ) y 2 + λ2 + p+ − 1

.

the (CP ) equation can be rewritten as follows: .

y  (ω) (n − 2)y(ω) cot ω + (p+ − 1)(y 2 + λ2 ) + (n − p+ )λ = − ≥ y 2 + λ2 (p+ − 1)y 2 + λ2 ω (p+ − 1)(y 2 + λ2 ) + (n − p+ )λ 0 , ω ∈ 0, (10.6.11) − , (p+ − 1)y 2 + λ2 2

in virtue of . y(ω) ≤ 0 (see (3.2.10)); next from (10.6.11), by and .1 < p+ < n, μ ≥ 0, we establish that .

y  (ω) p+ − 1 ≥ + 2 p+ − 1 + μ λ + y 2 (ω) −λ

but, since . (p y 2 (ω) . λ



(n − p+ ) (n − 2)λ2 y 2 + λ2 − − μ ; (p+ − 1)y 2 + λ2 (p+ − 1)y 2 + λ2 (p+ − 1)y 2 + λ2

y2 2 +λ2 −1)y +


0, if μ = 0; λμL d σ− 3 ⎩≥ σ > 0, if μ > 0 and d ≤ (p+ − 1) 4

σ (p+ −1)2 4μL(p+ −1+μ) .

(10.6.14)

From assumption (9.1.3), we have ⎧ ⎨ν, 2 + p − 1) λ(p+ if μ = 0; + 2 . χ < − ω0 ⎩ 1 σ > 0, if μ > 0. 2 sin 2 4

(10.6.15)

From (10.6.14)–(10.6.15), it follows ⎧ ⎨> 1 ν > 0, if μ = 0; 2 + p − 1) λ(p λμL + + 2 σ− d− χ − ⎩> 1 σ > 0, if μ > 0. (p+ − 1) 2 sin ω20 2

 .

(10.6.16)

Further, we choose .d > 0 and .χ− ≥ 0 such small that are fulfilled the following inequalities: ⎧ ⎧   ⎨ 1 ν, ⎪ ⎪ for μ = 0, ⎪ L ⎪ 1 + χ−ω0 · d ln A ≤ 20 ⎪ ⎪ sin 2 1 ⎩ ⎪ for μ > 0; ⎪ ⎪ 20 σ, ,  ⎪ . ⎪ ⎪ L χ − ⎪ ⎪ ln C1 + 1+ ⎪ ⎪ ⎪ sin ω20 ⎪ ⎧ ⎪ ⎪ ⎪ ⎨ 1 ν, / ⎪ for μ = 0, ⎪ n−p +λ(n−2) + 20 ⎪ ⎪| − 1| + · d ≤ ⎪ p+ −1 1 ⎪ ⎩ ⎪ σ, , for μ > 0; ⎪ ⎪ ⎧20 ⎪ ⎪   ⎨ ⎨ 1 ν, √ for μ = 0, . 1 + χ−ω0 L| −1| · d ≤ 20 sin 1 ⎪ ⎩ 2 ⎪ ⎪ 20 σ, , @ for μ > 0; ⎪ ? ⎪ ⎪ 2 ⎪ p+ (n−p+ ) 1 ⎪ ⎪ (p+ − 1)| − 1| + 2(p · χ− ≤ ω ⎪ + −1) sin 20 ⎪ ⎪ ⎧ ⎪ ⎪ ⎪ ⎨ 1 ν, ⎪ for μ = 0, ⎪ 20 ⎪ ⎪ ⎪ 1 ⎪ ⎩ ⎪ for μ > 0. ⎪ 20 σ, , ⎪ ⎧ ⎪ ⎪ ⎪ ⎨ 1 ν, ⎪ for μ = 0, ⎪ 1 20 ⎪ ⎪ ω0 · χ1 ≤ ⎪ sin 2 ⎩ ⎩ 1 σ, , for μ > 0. 20

(10.6.17)

304

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for. . .

We introduce quantities

.

∗ 0

=

⎧ ⎨1 ⎩

p+

for

≥ 1,

for

0,

as well as we note that . p(x) ≥ 0∗ . Then from (10.6.13), (10.6.14), (10.6.17), (10.6.18), we obtain  ?  χ(ω) ∂w ∂p |∇w|p(x)−2 σ w−1 |∇w|2 − (∇p · ∇w) − ln(A|∇w|)+ sin ω20 ∂r ∂x1   ∂w ∂|∇w| 1 ∂w ∂|∇w| p+ − p(x) · · + 2 · − |∇w| ∂r ∂r r ∂ω ∂ω   ∂|∇w| sin ω ∂|∇w| 2 − p(x) χ(ω) ∂w · − · cos ω · − |∇w| sin ω20 ∂r ∂r r ∂ω @  χ(ω) ∂ 2 w sin ω ∂ 2 w sin ω 1 ∂w + − b0 wp(x)−1 ≥ cos ω − χ  (ω) sin ω20 r ∂r sin ω20 r ∂r∂ω ∂r 2 p(x) p(x) μ∗ r (p(x)−1)−p(x)ψ λ (p(x)−1)(ω)(λ2 + y 2 (ω)) 2 − λ

.

r

(p(x)−1)−p(x)



ψ λ (p(x)−1)(ω) ≥

  p(x) − b0 r p(x) ≥ μ∗ 0∗ − b0 d r (

b0 r μ∗

(p(x)−1)

−1)p(x)−

μ∗ 2

≥

∗ ( −1)p(x)− 0r

,

if we add to inequalities (10.6.17) yet the requirement d≤

.

∗ ∗ 0μ

2b0

(10.6.19)

.

From (10.6.9), (10.3.18), it follows that Q[Aw, η] ≥

.

μ∗ 2

∗ 0



Gd0

Ap(x)−1r (

−1)p(x)−

η(x)dx.

10.6 Estimation of the Solution Modulus. The Proof of the Main. . .

305

Since .p(x) ≥ p− > 1 and .A > 1, we have .Ap(x)−1 ≥ Ap− −1 . Therefore, taking into consideration assumption (iii), the last inequality takes the form

μ∗ 2

∗ p− −1 0A

Gd0



Q[Aw, η] ≥

.

 r(

μ∗ 2

η(x)dx ≥



f0 r β(x) η(x)dx ≥ Gd0

−1)p(x)−

∗ p− −1 0A

r β(x) η(x)dx ≥

Gd0

 |f (x)|η(x)dx ≥

Gd0



f (x)η(x)dx = Q[u, η], by (I I )loc Gd0

1,p(x)

for all nonnegative .η ∈ N−1,∞ (Gd0 ), if .A > 1 satisfies  A≥

.

2f0 μ∗ 0∗



1 p− −1

(10.6.20)

.

Further, we show that .u(x) ≤ Aw(x) on . d . By (BF ) and (3.2.6), w(x)| d = d ψ

.

/λ

(ω) ≥ d .

In virtue of .|u(x)| ≤ M0 , ∀x ∈ Gd0 , we can choose A such that A≥

.

M0 , d

(10.6.21)

and therefore, Aw(x)| d ≥ Ad ≥ M0 ≥ u(x)| d .

.

Thus, if we choose a small .χ− ≥ 0 according to (9.1.3), . d > 0 according to (10.6.14), (10.6.17), (10.6.19) and a large .A > 1 according to (10.6.20)–(10.6.21): 

  1  M0 2f0 p− −1 .A ≥ max , , d μ∗ 0∗ then we come to the Comparison Principle Q(u, η) ≤ Q(Aw, η)

.

in Gd0 ;

u(x) ≤ Aw(x)

on d .

306

10 The Oblique Derivative Problem in a Bounded n-Dimensional Cone for. . .

For this purpose, we need to check the consistency for the system of two inequalities

.

⎧  ⎪ ⎨L 1 + ⎪ ⎩

 χ− ω sin 20

· d ln A

≤

μ∗ 5 ,

A

≥

M0 d .

At first, for the penultimate inequality of (10.6.17), we have 1+

.

2 μ∗ (p+ − 1) χ− 1 · ≡ ϒ. ω0 ≤ 1 + 2 sin 2 5 2| − 1|(p+ − 1)2 + (n − p+ )p+

Now, our above system can be rewritten in the form ⎧ ⎨ Lϒ · d ln A ≤ . ⎩ A ≥

μ∗ 5 , M0 d .

From this system, it follows that .

  μ∗ 1 ln M0 d − ≤ ln A ≤ · d 5Lϒ

⇒

  d · ln M0 d − ≤

μ∗ . 5Lϒ

Now, in virtue of .

  ln M0 − lim d · ln M0 d − = lim 1

d→+0

d→+0

ln d

=−

lim d ln d =

d→+0

d

lim d = 0,

d→+0

consequently, there is .d- > 0 such that for .d ∈ (0, d) the desired inequality is true. Thus, the Comparison Principle implies that u(x) ≤ Aw(x)

.

-

in Gd0 . -

Similarly, we derive the estimate .u(x) ≥ −Aw(x) in Gd0 if we replace .u(x) with .−u(x). By this and (BF ), (3.2.6), we get the required estimate |u(x)| ≤ Aw(x) ≤ C|x| ,

.

-

in Gd0 .

11

Existence of Bounded Weak Solutions

11.1

Setting of the Problem

Let us fix .R > 1. We consider the oblique derivative problem .(OPn ). Remark 11.1 We observe that if . χ(ω1 ) ≡ 0 then we have the Robin problem. This problem was investigated in our works [28–33]. We will work under the following assumptions: (i) .1 < p− ≤ p(x) ≤ p+ < n, ∀x ∈ GR 0;

p(x) ∈ C 0,1 (GR 0 ).

(ii) .a0 (x) is measurable and .0 < a0 ≤ a0 (x) ≤ a1 ; a0 , a1 − constants. R n+1 (iii) .b : G0 × R ⇒ R is a Carathéodory function (.b ∈ CAR) satisfying for almost all R n+1 , the following inequalities: .x ∈ G and for all .(u, ξ ) ∈ R 0 on .GR 0

(iii)a



|b(x, u, ξ )| ≤ b1 b0 (x) + |u(x)|q0(x) + |ξ |q1 (x) , where b1 = const ≥ 0, b0 ∈ Lq ∗ (x) (G),

.

1 q ∗ (x)

q0 (x) < p∗ (x) − 1, (iii)b

+

1 p∗ (x)

p∗ (x) =

= 1,

q1 (x) < p(x) − 1 +

ub(x, u, ξ ) ≥ b2 |u|p(x)

for |u| > 1;

np(x) ; n − p(x)

p(x) ; n

where b2 = const ≥ 0

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Borsuk, Oblique Derivative Problems for Elliptic Equations in Conical Domains, Frontiers in Elliptic and Parabolic Problems, https://doi.org/10.1007/978-3-031-28381-9_11

307

308

11 Existence of Bounded Weak Solutions

( ) (iv) .γ (ω1 ), χ(ω1 ) ∈ C 0 − ω20 , + ω20 ; with the norm . | · |0 = & max ' | · |; there exist numbers .γ1 , γ0 > 0, such that .γ1 ≥ γ (ω1 ) ≥ −

ω0 ω0 2 ,+ 2

) ( γ0 ; χ(ω1 ) ≥ 0, ∀ω1 ∈ − ω20 , + ω20 (iw) .f ∈ Lp (x) (G) ∩ Ls (G), p1(x) +

1 p ( x)

= 1, s >

2 p− .

By Proposition proposition 10.4 the equation of .(OPn ) is uniformly elliptic, Definition 11.2 The function u is called a bounded weak solution of problem .(OPn ) 1,p(x) (GR ) ∩ M(GR ) and satisfies the integral identity provided that .u ∈ Vp(x)(GR 0)≡W 0 0 (II) (see Sect. 9.1) for all .η ∈ Vp(x)(GR ). 0 Main result is the following statement: Theorem 11.3 Let the assumptions .(i)−(iv) be satisfied. Then problem .(OPn ) has at least one bounded weak solution.

11.2

Proof of the Existence Theorem

∗ R Proof We define nonlinear operators J, I, A, B, , : Vp(x)(GR 0 ) → Vp(x) (G0 ) and an ∗ (GR ) by element f ∗ ∈ Vp(x) 0

 < J (u), η > =

|∇u|p(x)−2∇u∇ηdx, GR 0



< I (u), η > = GR 0

χ(ω1 ) p(x)−2 ∂u ηx dx, ω0 |∇u| sin 2 ∂r 1



.

< A(u), η > =

a0 (x)u|u|p(x)−1η(x)dx, GR 0



< B(u), η > =

b(x, u(x), ∇u(x)η(x)dx, GR 0



< (u), η > =

γ (ω1 )|x|1−p(x)u|u|p(x)−2η(x)ds, 0R

11.2 Proof of the Existence Theorem

 < (u), η > = R

< f ∗, η > =

309

  γ (ω1 ) cos ω1 R 1−p(x) u|u|p(x)−2η(x)d R , 1 − χ(ω1 ) 1 + χ(ω1 ) sin ω20



f (x)η(x)dx GR 0

for all u, η ∈ Vp(x)(GR 0 ). ∗ By the definition of Vp(x)(GR 0 ), it is obvious that f is well defined.

 

Lemma 11.4 J (u), I (u), A(u), B(u), (u), (u) are bounded and continuous operators. Proof By Hölder inequality with Proposition 3.2 we have:  .

|∇u|p(x)−1|∇η|dx ≤ 2u1,p(x) · η1,p(x) .

< J (u), η >≤

(11.2.1)

GR 0

Thus, < J (u), η > is well defined and operator J (u) is bounded. Therefore, by Corollary 2.53, operator J (u) is continuous. In just the same way .

|χ|0 < I (u), η >≤ sin ω20

 |∇u|p(x)−1|∇η|dx ≤ GR 0

2

|χ|0 u1,p(x) · η1,p(x) . sin ω20

(11.2.2)

Thus, < I (u), η > is well defined and operator I (u) is bounded and continuous. Next, we denote M0 = u(x)1 (here normal is in M(GR 0 )) and estimate | < A(u), η > p |. For this it is sufficient assume 1 < |u(x)| ≤ M0 . Then we get u(x)p(x) ≤ M0 + and therefore         p 1 p(x)  .| < A(u), η > | ≤ a1 η(x)dx  ≤ sup |η(x)| · M0 + R n meas ,  |u(x)| n R  GR0 G0  (11.2.3)

1 See Theorem 10.5.

310

11 Existence of Bounded Weak Solutions

where is a domain on the unit sphere with smooth boundary ∂ , obtained by the intersection of the cone with the unit sphere. The estimate (11.2.3) means the boundedness R R of A(u). Let {un }∞ n=1 ⊂ Vp(x) (G0 ) be any sequence and let for u, un ∈ M(G0 ) : R un − u → 0. By the property of M(G0 ), we get that un (x) → u(x) uniformly almost everywhere in GR 0 . Moreover, p 1 a0 (x)|un (x)|p(x)η(x)dx ≤ a1 sup |η(x)| · M0 + R n meas , n R G

.

x ∈ GR 0.

0

Therefore, we can pass to the limit under the symbol of integral over GR 0 and we obtain .

lim | < A(un ) − A(u), η > | = 0,

n→∞

i.e., operator A(u) is continuous. Similarly, since R > 1, we derive the boundedness and the continuity of operator :   |χ|0 ) p −1 .| < (u), η > | ≤ γ1 1 + sup |η(x)| · R n−p− M0 + meas , sin ω20

(11.2.4)

i.e., operator (u) is bounded as well as continuous. Now we consider operator B(u). According to (iii)a , it is clear that  | B(u), η | ≤

|b (x, u(x), ∇u(x)) | · |η(x)|dx GR 0

.

≤ b1 sup |η(x)| GR 0



|b0 (x)| + |u(x)|q0(x) + |∇u(x)|q1 (x) dx.

GR 0

Next, we derive using the Hölder inequality  .

|b0 (x)|dx ≤ 2|b0 (x)|q ∗ (x) · |1|(q ∗ (x)) ≤ const (n, p+ , p− , meas G) · |b0 (x)|q ∗ (x).

GR 0

Further, it is clear that q1 (x) < p(x),

.

q0 (x)
≥ 0.

(11.2.7)

Moreover, .

< J (u) − J (η), u − η > ≥

1 2 p+

$ % p− p+ min |∇(u − η)|p(x) ; |∇(u − η)|p(x) , if p(x) ≥ 2;

.

(11.2.8)

< J (u) − J (η), u − η > ≥

%C $ p− p+ (p− − 1) min |∇(u − η)|p(x); |∇(u − η)|p(x) 

⎛

⎜ 2 max ⎜ ⎝



⎞ 2−p− 2

⎟ (|∇u(x)|p(x) + |∇η(x)|p(x))dx ⎟ ⎠

;

GR 0

⎛ ⎜ ⎜ ⎝



⎞ 2−p+ 2

⎟ (|∇u(x)|p(x) + |∇η(x)|p(x))dx ⎟ ⎠

 ,

GR 0

if 1 < p(x) < 2.

(11.2.9)

Proof By direct calculation, we have

 GR 0

J (u) − J (η), u − η =

|∇u(x)|p(x)−2∇u(x) − |∇η(x)|p(x)−2∇η(x) (∇u(x) − ∇η(x))dx. .

(11.2.10)

11.2 Proof of the Existence Theorem

313

Now, we use the inequalities (2.1.7) (see Lemma 2.7). Then for p(x) ≥ 2 we obtain . J (u) − J (η), u − η = 

|∇u(x)|p(x)−2∇u(x) − |∇η(x)|p(x)−2∇η(x) (∇u(x) − ∇η(x))dx ≥

GR 0



1

|∇u − ∇η|p(x)dx ≥

2 p+

1 2 p+

% $ p− p+ , min |∇(u − η)|p(x) ; |∇(u − η)|p(x)

GR 0

by the inequality (2.8.1). Now we consider the case 1 < p(x) < 2. For this case let s(x) = 2 2−p(x) .

2 p(x) ,

s  (x) =

Then we have

s− =

.

2 2 ≤ s(x) ≤ = s+ ; p+ p−

 s− =

2 2  ≤ s  (x) ≤ = s+ ; ⇒ 2 − p− 2 − p+ (11.2.11)

p− 1 = , s+ 2

p+ 1 = ; s− 2

2 − p+ 1 = ,  s+ 2

2 − p− 1 = .  s− 2

By (2.1.7) for 1 < p(x) < 2, we obtain . J (u) − J (η), u − η = 

|∇u(x)|p(x)−2∇u(x) − |∇η(x)|p(x)−2∇η(x) (∇u(x) − ∇η(x))dx ≥

GR 0

 (p− − 1)



GR 0

|∇u(x) − ∇η(x)|2 |∇u(x)|p(x) + |∇η(x)|p(x)

 2−p(x) dx. p(x)

Next we consider the integral 

 .

GR 0

|∇u(x) − ∇η(x)|

p(x)

dx = GR 0

|∇u(x) − ∇η(x)|p(x)   2−p(x) × |∇u(x)|p(x) + |∇η(x)|p(x) 2



2−p(x) 2 |∇u(x)|p(x) + |∇η(x)|p(x) dx ≤     p(x)   |∇u(x) − ∇η(x)|  2   ×  2−p(x)   |∇u(x)|p(x) + |∇η(x)|p(x) 2  s(x)

314

11 Existence of Bounded Weak Solutions

  

2−p(x)  2    |∇u(x)|p(x) + |∇η(x)|p(x)    

⎛

⎜ 2 max ⎜ ⎝ ⎛ ⎜ ⎜ ⎝

⎞



GR 0

|∇u(x) − ∇η(x)|2

  |∇u(x)|p(x) + |∇η(x)|p(x)

2−p(x) p(x)

⎞



GR 0

≤ s  (x)

|∇u(x) − ∇η(x)|2

 |∇u(x)|p(x) + |∇η(x)|p(x) 

⎛

⎜ max ⎜ ⎝

⎟ ⎟  2−p(x) dx ⎠

⎟ dx ⎟ ⎠ 1 s+

1 s−

;

 ×

p(x)

⎞



⎟ |∇u(x)|p(x) + |∇η(x)|p(x) dx ⎟ ⎠

1  s−

;

GR 0

⎞ 1 s+ 

⎟  ⎜ ⎜ |∇u(x)|p(x) + |∇η(x)|p(x) dx ⎟ . ⎠ ⎝ ⎛

GR 0

Hence, by (11.2.11) and by the inequality (2.8.1), it follows that ⎛   ⎝ . max G

⎞ p+ |∇u(x) − ∇η(x)|2

 |∇u(x)|p(x) + |∇η(x)|p(x) ⎛ ⎝

.

;

p(x)

⎞ p−

 

G



2

⎠  2−p(x) dx

|∇u(x) − ∇η(x)|2 |∇u(x)|p(x) + |∇η(x)|p(x)

2

⎠  2−p(x) dx

 ≥

p(x)

⎞ 2−p− ⎛ 2 C  

p(x) p(x) p(x) ⎠ ⎝ |∇u(x)| dx |∇u(x) − ∇η(x)| dx 2 max + |∇η(x)| ;

G

G

⎞ 2−p+ ⎛ 2  

p(x) p(x) ⎠ ⎝ |∇u(x)| dx + |∇η(x)| ≥ G

11.2 Proof of the Existence Theorem

.

315

%C $ p− p+ min |∇(u − η)|p(x) ; |∇(u − η)|p(x) ⎛ ⎞ 2−p− 2  

p(x) p(x) ⎝ ⎠ |∇u(x)| dx 2 max + |∇η(x)| ; G

⎞ 2−p+ ⎛ 2  

p(x) p(x) ⎝ |∇u(x)| dx ⎠ + |∇η(x)| . G

  Lemma 11.6 If un  u in Vp(x)(GR 0 ) (weak convergence) and |∇un |p(x) → |∇u|p(x),

.

(11.2.12)

∗ then B(un ) → B(u) in (Vp(x)(GR 0 )) .

Proof By the assumption (i), the space W 1,p(x) (GR 0 ) is Banach space, which is separable and reflexive (see Proposition 2.46). Therefore from the weak convergence un → u follows the boundedness of the set {un 1,p(x)}. In addition, by proposition 2.47, there is the operator of the compact embedding that maps this set to the compact set in Lp( x) (GR 0 ), i.e., this operator is compact operator. But then there is subsequence {unk } → u in Lp( x) (GR {unk } → u in Vp( x) (GR 0 ). Together with (11.2.12) we have that 0 ). By Lemma 11.4 operator B is continuous operator. Therefore, we can perform the passage ∗ to the limit under the integral symbol and thus B(un ) → B(u) in (Vp(x)(GR 0 )) . Set T := J − I + A + B +  + . Then the operator equation T (u) = f ∗

.

(11.2.13)

is equivalent to validity of the integral identity (I I ). This fact means that the solutions of (11.2.13) correspond one-to-one to the weak solutions of (P ). Now, we shall verify the assumptions (a)–(h) of the Leray-Lions Theorem 2.59 to prove that there is a solution of (11.2.13). Assumptions (a)–(b) follow directly from Lemma 11.4. The coercivity of T (assumption (c)) is a direct consequence of our assumptions (ii), (iii)b , (iii1), (iii2, ) and (10.1.2):

  GR 0

. < T (u), u >=    χ(ω1 ) ∂u ∂u p(x)+1 (x)|u| + ub(u, ∇u) dx+ |∇u|p(x)−2 |∇u|2 − + a 0 sin ω20 ∂r ∂x1

316

11 Existence of Bounded Weak Solutions

  γ (ω1 ) cos ω1 R n−p(x) |u|p(x)d + 1 − χ(ω1 ) 1 + χ(ω1 ) sin ω20

 .



 γ (ω1 )|x|1−p(x)|u|p(x)ds ≥

a0 , b2 ) · const (p− , n,-



0R

|∇u(x)|p(x) + |u(x)|p(x) dx

for |u| > 1.

GR 0

Now, we use the inequality (2.8.1) .





p− p+ |∇u(x)|p(x) + |u(x)|p(x) ≥ min |u|1,p(x) . , |u|1,p(x)

GR 0

Then we obtain .

lim

u1,p(x) →∞

< T (u), u > p± −1 ≥ const (p− , n,a0 , b2 ) · lim u1,p(x) = ∞. u1,p(x) →∞ u1,p(x)

Let us define an operator  : Vp(x)(GR ) × Vp(x)(GR ) → (Vp(x)(GR ) )∗ by 0

.

0

0

< (u, w), η >:=< J (u), η > − < I (w), η > + < A(w), η > + < B(w), η > + < (w), η > + < (w), η >

for all u, w, η ∈ Vp(x)(GR ) . The assumption (d) is obviously. 0

Next, let u, w, h ∈ Vp(x)(GR 0 ) and tn → 0. Then, by continuity of operator J, we have (u + tn h, w) = J (u + tn h) − I (w) + A(w) + B(w) + (w) + (w) →

.

J (u) − I (w) + A(w) + B(w) + (w) + (w) = (u, w). Thus, the assumption (e) satisfies. The assumption (f) satisfies by Lemma 11.5, because of (u, u) − (w, u) = J (u) − J (w).

.

11.2 Proof of the Existence Theorem

317

Now, we shall verify the assumption (g). Let un → u in Vp(x)(GR 0 ) and .

lim < (un , un ) − (u, un ), un − u >= 0

n→∞

⇒

lim < J (un ) − J (u), un − u >= 0.

(11.2.14)

n→∞

From (11.2.14) and (11.2.8)–(11.2.9) it follows that |∇un |p(x) → |∇u|p(x), i.e., (11.2.12) is satisfied. Moreover, un → u in M. The last facts and that W 1,p(x) is a uniformly convex Banach space together with the weak convergence imply un → u

.

in

V(p(x)(GR 0)

(see, e.g., Proposition 2.54). Further, by Lemmas 11.4 and 11.6, we have (w, un ) = J (w) − I (un ) + A(un ) + B(un ) + (un ) + (un ) →

.

J (w) − I (u) + A(u) + B(u) + (u) + (u) = (w, u) for arbitrary w ∈ V(p(x)(GR 0 ). Finally, we verify the assumption (h). Let w ∈ ). Then, in virtue of B(u ) → B(u) and un  u in V(p(x)(GR n 0 I (un ) → I (u), A(u ) → A(u), (un ) n  ∗ ) (see Lemmas 11.4, 11.6), we obtain (un ) → (u) in V(p(x)(GR 0 .

V(p(x)(GR 0 ), →

(u),

< (w, un ), un >=

< J (w) − I (un ) + A(un ) + B(un ) + (un ) + (un ), un > → < J (w), u > − < I (u), u > + < A(u), u > + < B(u), u > + < (u), u > + < (u), u >= < (w, u), u > . Hence we have: un  u (w, un )

.

→

in

V(p(x)(G) implies that

J (w) − I (u) + A(u) + B(u) + (u) + (u)) = (w, u).

Thus, all assumptions of the Leray-Lions Theorem are satisfied.

 

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Index

B Barrier function, 143, 186, 187, 255, 291 Beltrami-Laplace operator, 12

C Caratheodory condition, 30 Cauchy’s inequality, 9 Comparison principle mixed problem, 190 Conical point, 10 Convex rotational cone, 13

D Differential inequality .(CP ), 34 Dini continuity, 27

F Fatou’s theorem, 16 Fixed point Leray–Schauder theorem, 32 Fubini’s theorem, 15

G Generalized Hölder inequality, 29

I Inequality for boundary and domain integrals, 21, 22, 24, 26 Interpolation inequality, 16, 21

J Jensen’s inequality, 10

L Leray-Lions theorem, 32, 317 p .L –estimates oblique problem local boundary, 74

M Maximum principle oblique problem local, 184 Robin problem local, 73 Method of continuity, 31

N n-dimensional bounded cone, 271 Nemyckij operator, 30

H Hölder continuity, 19 Hölder’s inequality, 9, 16

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Borsuk, Oblique Derivative Problems for Elliptic Equations in Conical Domains, Frontiers in Elliptic and Parabolic Problems, https://doi.org/10.1007/978-3-031-28381-9

325

326 P Problem (EVP), 4, 45 (L), 4, 74 (NEVP), 4, 55 (OEVP), 56 (OP2 ), 6, 232 (OPn ), 6, 271 (QL), 5, 183 SL, 5, 107 WQL, 5, 129 .p(x)-Laplacian, x, 12

Index S Sobolev embedding theorems, 23 Space of Dini continuous functions, 27 Stampacchia’s Lemma, 40

V Variable exponent Lebesgue space, 28 Variable exponent Sobolev space, 28 Variational principle, 31

W Weighted Sobolev space, 24 Q Quasi-distance function, 14 Y Young’s inequality, 9 R Regularization of a function, 18

Notation Index

A(k), 134, 244, 279 A− (k), 244, 279 A+ (k), 244, 279 C l C l (G), 20 C 0,A C 0,A , 27 C k,A (G), 28 C0l (G), 20 G(k) , 11 Gba , 11 Gd , 11 Lp (G), 15 p Lloc (G), 17 Lp(x) (G), 28 M(ε), 79, 110 M(G), 17 k (G), 24 Vp,α k−1/p

Vp,α (), 25 W 1,p(x) (G), 28, 29 W k,p (G), 20 W k,p (G \ O), 20 k,p W0 (G), 20 W k−1/2 (), 21 k− 1 ,p

W p (), 21 W k (G), 20 W0k (G), 20 [u]A;G , 28 [f ]α;G , 20

ω u, 12 divω u , 12

p(x) , 12

ab , 11 0R , 11 d , 11 , 11 ρ , 11

◦k W α (G), 25 ◦ k−1/2 Wα (), 25

λ, 45 λ∗ , 47, 54, 57 ψ(ω), 45 y(ω), 52 M, 183 ∇ω u, 12 ω0 , 13 θ, 68 d , 11 J (ω), 11 ds, 11 dσ , 11 rε (x), 14 uh , 18 1,p(x) N−1,∞ (GR 0 ), 30 (z), 38 Cαν (z), 38 Pαν (x), 39 μ Pν (cos θ), 40 |u|C k,A (G) , 28 U(), 70 - (), 70 U

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Borsuk, Oblique Derivative Problems for Elliptic Equations in Conical Domains, Frontiers in Elliptic and Parabolic Problems, https://doi.org/10.1007/978-3-031-28381-9

327