analytic methods in partial differential equations

Analytic Methods in Partial DifferentialEquations

A collection of notes and insights into the world

of differential equations.

Gregory T. von Nessi

October 31, 2011

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Where I’m Coming From viii

I Classical Approaches 1

Chapter 1: Physical Motivations 2

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 The Heat Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2.1 A Note on Physical Assumptions . . . . . . . . . . . . . . . . . . . . 5

1.3 Classical Model for Traffic Flow . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4 The Transport Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.4.1 Method of Characteristics: A First Look . . . . . . . . . . . . . . . . 7

1.4.2 Duhamel’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.5 Minimal Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.5.1 Divergence Theorem and Integration by Parts . . . . . . . . . . . . . 10

1.6 The Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.7 Classification of PDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

Chapter 2: Laplace’s Equation ∆u = 0 and Poisson’s Equation −∆u = f 18

2.1 Laplace’s Eqn. in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . 18

2.2 The Fundamental Solution of the Laplacian . . . . . . . . . . . . . . . . . . 21

2.3 Properties of Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . 26

2.3.1 WMP for Laplace’s Equation . . . . . . . . . . . . . . . . . . . . . . 26

2.3.2 Poisson Integral Formula . . . . . . . . . . . . . . . . . . . . . . . . 27

2.3.3 Strong Maximum Principle . . . . . . . . . . . . . . . . . . . . . . . 32

2.4 Estimates for Derivatives of Harmonic Functions . . . . . . . . . . . . . . . 32Gregory T. von Nessi

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2.4.1 Mean-Value and Maximum Principles . . . . . . . . . . . . . . . . . 34

2.4.2 Regularity of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 36

2.4.3 Estimates on Harmonic Functions . . . . . . . . . . . . . . . . . . . 39

2.4.4 Analyticity of Harmonic Functions . . . . . . . . . . . . . . . . . . . 42

2.4.5 Harnack’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2.5 Existence of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

2.5.1 Green’s Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

2.5.2 Green’s Functions by Reflection Methods . . . . . . . . . . . . . . . 51

2.5.2.1 Green’s Function for the Halfspace . . . . . . . . . . . . . 51

2.5.2.2 Green’s Function for Balls . . . . . . . . . . . . . . . . . . 55

2.5.3 Energy Methods for Poisson’s Equation . . . . . . . . . . . . . . . . 57

2.5.4 Perron’s Method of Subharmonic Functions . . . . . . . . . . . . . . 58

2.5.4.1 Boundary behavior . . . . . . . . . . . . . . . . . . . . . . 60

2.6 Solution of the Dirichlet Problem by the Perron Process . . . . . . . . . . . 62

2.6.1 Convergence Theorems . . . . . . . . . . . . . . . . . . . . . . . . . 62

2.6.2 Generalized Definitions & Harmonic Lifting . . . . . . . . . . . . . . 62

2.6.3 Perron’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

2.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

Chapter 3: The Heat Equation 70

3.1 Fundamental Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

3.2 The Non-Homogeneous Heat Equation . . . . . . . . . . . . . . . . . . . . 74


3.3 Mean-Value Formula for the Heat Equation . . . . . . . . . . . . . . . . . . 76

3.4 Regularity of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

3.4.1 Application of the Representation to Local Estimates . . . . . . . . . 84

3.5 Energy Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

3.6 Backward Heat Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

3.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

Chapter 4: The Wave Equation 88

4.1 D’Alembert’s Representation of Solution in 1D . . . . . . . . . . . . . . . . 88

4.1.1 Reflection Argument . . . . . . . . . . . . . . . . . . . . . . . . . . 90

4.1.2 Geometric Interpretation (h ≡ 0) . . . . . . . . . . . . . . . . . . . 92

4.2 Representations of Solutions for n = 2 and n = 3 . . . . . . . . . . . . . . . 93

4.2.1 Spherical Averages and the Euler-Poisson-Darboux Equation . . . . . 93

4.2.2 Kirchoff’s Formula in 3D . . . . . . . . . . . . . . . . . . . . . . . . 94

4.2.3 Poisson’s Formula in 2D . . . . . . . . . . . . . . . . . . . . . . . . 95

4.2.4 Representation Formulas in 2D/3D . . . . . . . . . . . . . . . . . . 97

4.3 Solutions of the Non-Homogeneous Wave Equation . . . . . . . . . . . . . . 98


4.3.2 Solution in 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

4.4 Energy methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

4.4.1 Finite Speed of Propagation . . . . . . . . . . . . . . . . . . . . . . 102

4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

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Chapter 5: Fully Non-Linear First Order PDEs 106

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

5.2 Method of Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

5.2.1 Strategy to Prove that the Method of Characteristics Gives a Solution

of F (Du, u, x) = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

5.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

Chapter 6: Conservation Laws 124

6.1 The Rankine-Hugoniot Condition . . . . . . . . . . . . . . . . . . . . . . . . 124

6.1.1 Information about Discontinuous Solutions . . . . . . . . . . . . . . 128

6.2 Shocks, Rarefaction Waves and Entropy Conditions . . . . . . . . . . . . . . 129

6.2.1 Burger’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

6.3 Viscous Approximation of Shocks . . . . . . . . . . . . . . . . . . . . . . . . 135

6.4 Geometric Solution of the Riemann Problem for General Fluxes . . . . . . . 138

6.5 A Uniqueness Theorem by Lax . . . . . . . . . . . . . . . . . . . . . . . . . 139

6.6 Entropy Function for Conservation Laws . . . . . . . . . . . . . . . . . . . . 141

6.7 Conservation Laws in nD and Kruzkov’s Stability Result . . . . . . . . . . . 143

6.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

Chapter 7: Elliptic, Parabolic and Hyperbolic Equations 152

7.1 Classification of 2nd Order PDE in Two Variables . . . . . . . . . . . . . . . 153

7.2 Maximum Principles for Elliptic Equations of 2nd Order . . . . . . . . . . . . 158

7.2.1 Weak Maximum Principles . . . . . . . . . . . . . . . . . . . . . . . 160

7.2.1.1 WMP for Purely 2nd Order Elliptic Operators . . . . . . . 161

7.2.1.2 WMP for General Elliptic Equations . . . . . . . . . . . . 163

7.2.2 Strong Maximum Principles . . . . . . . . . . . . . . . . . . . . . . 169

7.3 Maximum Principles for Parabolic Equations . . . . . . . . . . . . . . . . . . 172

7.3.1 Weak Maximum Principles . . . . . . . . . . . . . . . . . . . . . . . 173

7.3.2 Strong Maximum Principles . . . . . . . . . . . . . . . . . . . . . . 174

7.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

II Functional Analytic Methods in PDE 181

Chapter 8: Facts from Functional Analysis and Measure Theory 182

8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

8.2 Banach Spaces and Linear Operators . . . . . . . . . . . . . . . . . . . . . . 183

8.3 Fredholm Alternative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

8.4 Spectral Theorem for Compact Operators . . . . . . . . . . . . . . . . . . . 191

8.5 Dual Spaces and Adjoint Operators . . . . . . . . . . . . . . . . . . . . . . 195

8.6 Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196

8.6.1 Fredholm Alternative in Hilbert Spaces . . . . . . . . . . . . . . . . 200

8.7 Weak Convergence and Compactness; the Banach-Alaoglu Theorem . . . . . 201

8.8 Key Words from Measure Theory . . . . . . . . . . . . . . . . . . . . . . . 204

8.8.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

8.8.2 Integrability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206

8.8.3 Calderon-Zygmund Decomposition . . . . . . . . . . . . . . . . . . . 207

8.8.4 Distribution functions . . . . . . . . . . . . . . . . . . . . . . . . . . 209

8.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209Gregory T. von Nessi

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Chapter 9: Function Spaces 216

9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216

9.2 Holder Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217

9.3 Lebesgue Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

9.3.1 Convolutions and Mollifiers . . . . . . . . . . . . . . . . . . . . . . . 221

9.3.2 The Dual Space of Lp . . . . . . . . . . . . . . . . . . . . . . . . . 224

9.3.3 Weak Convergence in Lp . . . . . . . . . . . . . . . . . . . . . . . . 225

9.4 Morrey and Companato Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 225

9.4.1 L∞ Companato Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 233

9.5 The Spaces of Bounded Mean Oscillation . . . . . . . . . . . . . . . . . . . 238

9.6 Sobolev Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240

9.6.1 Calculus with Sobolev Functions . . . . . . . . . . . . . . . . . . . . 244

9.6.2 Boundary Values, Boundary Integrals, Traces . . . . . . . . . . . . . 247

9.6.3 Extension of Sobolev Functions . . . . . . . . . . . . . . . . . . . . 249

9.6.4 Traces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253

9.6.5 Sobolev Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . 257

9.6.6 The Dual Space of H10 = W 1,2

0 . . . . . . . . . . . . . . . . . . . . . 260

9.7 Sobolev Imbeddings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261

9.7.1 Campanato Imbeddings . . . . . . . . . . . . . . . . . . . . . . . . . 264

9.7.2 Compactness of Sobolev/Morrey Imbeddings . . . . . . . . . . . . . 266

9.7.3 General Sobolev Inequalities . . . . . . . . . . . . . . . . . . . . . . 270

9.8 Difference Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271

9.9 A Note about Banach Algebras . . . . . . . . . . . . . . . . . . . . . . . . . 274

9.10 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274

Chapter 10: Linear Elliptic PDE 282

10.1 Existence of Weak Solutions for Laplace’s Equation . . . . . . . . . . . . . . 282

10.2 Existence for General Elliptic Equations of 2nd Order . . . . . . . . . . . . . 286

10.2.1 Application of the Fredholm Alternative . . . . . . . . . . . . . . . . 288

10.2.2 Spectrum of Elliptic Operators . . . . . . . . . . . . . . . . . . . . . 290

10.3 Elliptic Regularity I: Sobolev Estimates . . . . . . . . . . . . . . . . . . . . . 291

10.3.1 Caccioppoli Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . 292

10.3.2 Interior Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298

10.3.3 Global Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301

10.4 Elliptic Regularity II: Schauder Estimates . . . . . . . . . . . . . . . . . . . 303

10.4.1 Systems with Constant Coefficients . . . . . . . . . . . . . . . . . . 304

10.4.2 Systems with Variable Coefficients . . . . . . . . . . . . . . . . . . . 308

10.4.3 Interior Schauder Estimates for Elliptic Systems in Non-divergence

Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310

10.4.4 Regularity Up to the Boundary: Dirichlet BVP . . . . . . . . . . . . 311

10.4.4.1 The nonvariational case . . . . . . . . . . . . . . . . . . . 315

10.4.5 Existence and Uniqueness of Elliptic Systems . . . . . . . . . . . . . 315

10.5 Estimates for Higher Derivatives . . . . . . . . . . . . . . . . . . . . . . . . 318

10.6 Eigenfunction Expansions for Elliptic Operators . . . . . . . . . . . . . . . . 320

10.7 Applications to Parabolic Equations of 2nd Order . . . . . . . . . . . . . . . 324

10.7.1 Outline of Evan’s Approach . . . . . . . . . . . . . . . . . . . . . . 326

10.7.2 Evan’s Approach to Existence . . . . . . . . . . . . . . . . . . . . . 327

10.8 Neumann Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . 330

10.8.1 Fredholm’s Alternative and Weak Solutions of the Neumann Problem 332

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10.8.2 Fredholm Alternative Review . . . . . . . . . . . . . . . . . . . . . . 333

10.9 Semigroup Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334

10.10Applications to 2nd Order Parabolic PDE . . . . . . . . . . . . . . . . . . . 342

10.11Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344

Chapter 11: Distributions and Fourier Transforms 348

11.1 Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348

11.1.1 Distributions on C∞ Functions . . . . . . . . . . . . . . . . . . . . . 348

11.1.2 Tempered Distributions . . . . . . . . . . . . . . . . . . . . . . . . . 351

11.1.3 Distributional Derivatives . . . . . . . . . . . . . . . . . . . . . . . . 351

11.2 Products and Convolutions of Distributions; Fundamental Solutions . . . . . 354

11.2.1 Convolutions of Distributions . . . . . . . . . . . . . . . . . . . . . . 355

11.2.2 Derivatives of Convolutions . . . . . . . . . . . . . . . . . . . . . . 356

11.2.3 Distributions as Fundemental Solutions . . . . . . . . . . . . . . . . 356

11.3 Fourier Transform of S(Rn) . . . . . . . . . . . . . . . . . . . . . . . . . . 357

11.4 Definition of Fourier Transform on L2(Rn) . . . . . . . . . . . . . . . . . . . 361

11.5 Applications of FTs: Solutions of PDEs . . . . . . . . . . . . . . . . . . . . 364

11.6 Fourier Transform of Tempered Distributions . . . . . . . . . . . . . . . . . 365

11.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366

Chapter 12: Variational Methods 368

12.1 Direct Method of Calculus of Variations . . . . . . . . . . . . . . . . . . . . 368

12.2 Integral Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377

12.2.1 The Implicit Function Theorem . . . . . . . . . . . . . . . . . . . . 377

12.2.2 Nonlinear Eigenvalue Problems . . . . . . . . . . . . . . . . . . . . . 380

12.3 Uniqueness of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384

12.4 Energies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384

12.5 Weak Formulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385

12.6 A General Existence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 387

12.7 A Few Applications to Semilinear Equations . . . . . . . . . . . . . . . . . . 390

12.8 Regularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392

12.8.1 DeGiorgi-Nash Theorem . . . . . . . . . . . . . . . . . . . . . . . . 393

12.8.2 Second Derivative Estimates . . . . . . . . . . . . . . . . . . . . . . 396

12.8.3 Remarks on Higher Regularity . . . . . . . . . . . . . . . . . . . . . 399

12.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400

III Modern Methods for Fully Nonlinear Elliptic PDE 401

Chapter 13: Non-Variational Methods in Nonlinear PDE 402

13.1 Viscosity Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402

13.2 Alexandrov-Bakel’man Maximum Principle . . . . . . . . . . . . . . . . . . . 406

13.2.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408

13.2.1.1 Linear equations . . . . . . . . . . . . . . . . . . . . . . . 408

13.2.1.2 Monge-Ampere Equation . . . . . . . . . . . . . . . . . . 409

13.2.1.3 General linear equations . . . . . . . . . . . . . . . . . . . 410

13.2.1.4 Oblique boundary value problems . . . . . . . . . . . . . . 410

13.3 Local Estimates for Linear Equations . . . . . . . . . . . . . . . . . . . . . . 411

13.3.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411

13.3.2 Local Maximum Principle . . . . . . . . . . . . . . . . . . . . . . . . 411Gregory T. von Nessi

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13.3.3 Weak Harnack inequality . . . . . . . . . . . . . . . . . . . . . . . . 412

13.3.4 Holder Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415

13.3.5 Boundary Gradient Estimates . . . . . . . . . . . . . . . . . . . . . 417

13.4 Holder Estimates for Second Derivatives . . . . . . . . . . . . . . . . . . . . 419

13.4.1 Interior Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419

13.4.2 Function Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422

13.4.3 Interpolation inequalities . . . . . . . . . . . . . . . . . . . . . . . . 423

13.4.4 Global Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425

13.4.5 Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427

13.5 Existence Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428

13.5.1 Method of Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . 428

13.5.2 Uniformly Elliptic Case . . . . . . . . . . . . . . . . . . . . . . . . . 431

13.5.3 Monge-Ampere equation . . . . . . . . . . . . . . . . . . . . . . . . 433

13.5.4 Degenerate Monge-Ampere Equation . . . . . . . . . . . . . . . . . 436

13.5.5 Bondary Curvatures . . . . . . . . . . . . . . . . . . . . . . . . . . . 437

13.5.6 General Boundary Values . . . . . . . . . . . . . . . . . . . . . . . . 438

13.6 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438

13.6.1 Isoperimetric inequalities . . . . . . . . . . . . . . . . . . . . . . . . 438

13.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441

Appendices 443

Appendix A: Notation 444

Bibliography 446

Index 450

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viii Where I’m Coming From

Where I’m Coming From

You may think me a freak or just completely insane, but I love partial differential equations

(PDE). It is important that I get this out quickly because I don’t want you to get all com-

mitted to reading this book and then start thinking I’ve just started going all weird on you

half-way through. I believe in being honest in relationships and writing a textbook on PDE

should be no different... right? So, if you are going to have issues with for me being...

excessively demonstrative... with this subject, it is probably best to put the book down now

and walk away... slowly.

When I was going through school, I was extremely fortunate to have all of my PDE

courses (I took many) taught by really awesome professors; and it’s probably fair to say that

this continues to influence my thoughts and feelings toward the subject to this very day.

My professors were not the type to drone out solution estimates or regurgitate whatever

mathematical progression was necessary to complete a proof. Instead, I was taught PDE

from various intuitive perspectives. It’s not to say that there was no rigour in my education

(there was quite a bit); but my teachers never let me lose sight of the bigger picture, be it

mathematical, physical or computational in nature.

This book embodies my own endeavour to share my enthusiasm and intuition that I have

gained in this area of maths with you. I am no stranger to the idea that PDE is generally

viewed as a terrifying subject, that only masochists or the clinically insane truly enjoy. While

I am going to refrain from commenting on my own mental stability (or lack thereof), I will

say that I believe that PDE (and mathematics in general) can be understood by anyone. It’s

not to say that I believe everyone should be a mathematician; but I think if someone likes

the subject, they can learn it. This is where I am coming from.


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Where I’m Coming From ix

As much as I would like to present a 10,000 page self-contained treatise on the subject

of PDE, I am told that I will be dead before such a manuscript would get through peer

review. Given that I am writing this book for fame and money, having this book published

post-mortem is really not a viable option. So, I have decided to focus on the functional-

analytic foundation of PDE in this book; but this will require you to know some things about

real analysis and measure theory to suck out all of the information contained in these pages.

However, I still encourage anyone to at least flip through this book (or download/steal it...

Go ahead be a pirate! ARGHHH!) because my hope is that you will still gain some insight

into this really awesome and beautiful subject. Even if you have no understanding of maths,

I still encourage you to read the words and try to absorb the maths by some osmotic means.

Just keep in mind that no matter how scary the maths looks, the underlying concepts are

always within your grasp to understand. I believe in you!

If you get through this book, will you love PDE? That I can not honestly say. To me PDE

is like Opera (the music not the web browser), in the sense that (to rip off the sentiment

from Pretty Woman) some fall in love with Opera the first time they hear it, while others

can grow to appreciate it over time but never truly love it. Again, I believe you have all

the abilities to grasp every aspect of PDE, but whether or not you love or appreciate it is

intrinsic to your personality (I feel). My hope is that by reading this book that you are able

to figure out whether you are a lover of PDE or lean more toward appreciating the subject

(both are great). As for myself, PDE represents a new perspective by which I can see not

only abstract ideas but also every aspect of the universe that surrounds me. If you can at

least gain a glimpse of that perspective by reading this book, I will be happy.

As I said earlier, my enthusiasm for this subject has been cultivated by the teachings of

great professors from my past. I just want to thank Kiwi (James Graham-Eagle), Stephen

(Pennell), (Louis) Rossi, Georg (Dolzmann) and Neil (Trudinger) for robbing me of my sanity

and showing me that I am a lover of PDE.

Finally, I am initially distributing this work under an international Creative Commons

license that allows for the free distribution of these notes (but prevents commercial us-

ages and modifications). This license enables me to make these notes available, as I

continue the slow process of their revision. The license mark is at the footer of each

page, and you can find more information out by visiting the Creative Commons website

at http://creativecommons.org.

Gregory T. von Nessi Jr.

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Part I

Classical Approaches

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2Chapter 1: Physical Motivations

1 Physical Motivations

“. . . because it’s dull you TWIT; it’ll ’urt more!”

— Alan Rickman

Contents

1.1 Introduction 2

1.2 The Heat Equation 3

1.3 Classical Model for Traffic Flow 5

1.4 The Transport Equation 6

1.5 Minimal Surfaces 9

1.6 The Wave Equation 13

1.7 Classification of PDE 13

1.8 Exercises 16

1.1 Introduction

As the chapter title indicates, the following material is meant to simply acquaint the reader

with the concept of Partial Differential Equations (PDE). The chapter starts out with a few

simple physical derivations that motivate some of the most fundamental and famous PDE. In

some of these examples analytic techniques, such as Duhamel’s Principle, are introduced to

whet the reader’s appetite for the material forthcoming. After this initial motivation, basic

definitions and classifications of PDE are presented before moving on to the next chapter.Gregory T. von Nessi

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1.2 The Heat Equation3

1.2 The Heat Equation

One of the most, if not THE most, famous PDE is the Heat Equation. Obviously, this

equation is somehow connected with the physical concept of heat energy, as the name not-

so subtly suggests. The natural questions now are: what is the heat equation and where

does it come from? To answer these, we will first construct a physical model; but before we

proceed with this, let us define some notation.

B

Ω

In our model, take Ω to be a heat conducting body, ρ(x) as the mass density at x ∈ Ω,

c(x) is the specific heat at x ∈ Ω, and u(x, t) is the temperature at point x and time t. Also

take B to be an arbitrary ball in Ω.

In order to construct any physical model, one has to make some assumption about the

reality they are trying to replicate. In our model we assume that heat energy is conserved in

any subset of Ω; and in particular, is conserved in B.

With this in mind, we first consider the total heat energy in B, which is∫

B

ρ(x)c(x)u(x, t) dx ;

and the rate of change of the heat energy in B is

d

dt

∫

B

ρ(x)c(x)u(x, t) dx.

If we know the heat flux rate, ~F (x, t) (a vector-valued function), across a surface element

A centered at x with normal ~ν(x), then we know the flux across a surface is given by:

|A| · ~F (x, t) · ~ν(x).

Thus, using the assumption that heat energy is conserved in B, we calculate

d

dt

∫

B

ρ(x)c(x)u(x, t) dx = −∫

∂B

~F (x, t) · ~ν(x) dS

= −∫

B

div ~F (x, t) dx.

Upon rewriting this equality, we see that∫

B

(ρ(x)c(x)ut(x, t) + div ~F (x, t)︸︷︷︸turns out to be smooth

) dx = 0.

Since B is arbitrary, this last equation must be true for all balls B ⊂ Ω. We can now say,

ρ(x)c(x)ut(x, t) + div ~F (x, t) = 0. in Ω (1.1)

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We seek u, but we have only one equation that involves the unknown heat flux F ; in order

to proceed with our model, we must make another assumption based on what (we think) we

know of reality. So we make the following constitutive assumption as to what the heat flux

“looks like” mathematically:

~F (x, t) = −k(x) · ~Du(x, t).

(See Appendix A for notation). Plugging this constitutive assumption into (1.1), we get

Fourier’s Law of Cooling:

ρ(x)c(x)ut(x, t)− div (k(x) · ~Dxu(x, t)) = 0. (1.2)

To further simplify the problem, we consider the special case that ρ(x), c(x) and k(x)

are all constant; take them to be ρ1, c and k respectively. Now, we can further manipulate

the second term of the LHS of (1.2):

div Du = div

(∂

∂x1u, . . . ,

∂

∂xnu

)

=∂2

∂x21

u + . . .+∂2

∂x2n

u

=

(∂2

∂x21

+ . . .+∂2

∂x2n

)u

= ∆u (≡ ∇2u).

Thus we conclude that (1.2) can be written as

ρc · ut(x, t) = k · ∆uut(x, t) = K · ∆u, (1.3)

where K = kρc is the diffusion constant. (1.3) is the legendary Heat Equation in all its glory.

We shall see later that the solutions of the Heat Equation will evolve to a steady state

which does not change in time. These steady state solutions thus solve

∆u = 0. (1.4)

This is the most ubiquitous equation in all of physics. This is Laplace’s Equation.Gregory T. von Nessi

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1.3 Classical Model for Traffic Flow5

1.2.1 A Note on Physical Assumptions

[The reader may skip this section if they feel comfortable with the concepts of conservation

laws and constitutive assumptions.]

In our derivation of the Heat Equation, we used two physical observations to construct

a model of heat flow. Specifically, we first assumed that heat energy was conserved; this is

particular manifestation of what is known as a conservation law this was followed by a consti-

tutive assumption based on the actual observed behavior of heat. Considering conservation

laws, it is observed that many quantities are conserved nature; examples are conservation of

mass/energy, momentum, angular momentum etc. It turns out that these conservation laws

correspond to specific symmetries (via Noether’s Theorem). Physics, as we know it, is con-

structed off of the assumed validity of these conservation laws; and it is through these laws

that one finds that conclusions in physics manifest in the form of PDE. Some examples are

Schrodinger’s Equation, the Heat Equation (via the argument above), Einstein’s Equations,

and Maxwell’s Equation’s, to name a few.

In the above derivation, the use of the conservation of energy did not depend on the

fact that we were trying to model heat energy. Indeed, are model lost it’s generality (i.e.

pertaining to the general flow of energy) via the constitutive assumption, which is based on

the observations pertaining specifically to heat flow. In the above derivation, the constitutive

assumption was simply a mathematical way of saying that heat energy is observed to move

from a hot to a cold object, and that the rate of this flow will increase with the temperature

differential present. It is worth trying to understand this physically, through the example of

your own sense of “temperature”. To do this, one first should realize that when one feels

“temperature”, they are really feeling the heat energy flux on the surface of their skin. This

is why people with hypothermia cease to feel “cold”. In this situation their body’s internal

temperature has dropped below the normal 98.6F. As a result, the heat energy flux across

the surface of the skin has decreased as the temperature differential between the person’s

body and outside air has decreased.

Now, let us move on to another model to try and solidify the ideas of conservation laws

and constitutive assumptions.

1.3 Classical Model for Traffic Flow

We will now proceed to construct a model for (very) simple traffic flow in the same spirit as

the heat equation was derived; i.e. our model will be born out of a conservation law and a

constitutive assumption.

First, we define ρ(x, t) to represent the density of cars on an ideal high (one that is

straight, one lane with no ramps). Here, we take the highway to be represented by the

x-axis. Since we do not expect cars to be created or destroyed (hopefully) on our highway,

it is reasonable to assume that ρ(x, t) represents a conserved quantity. The corresponding

conservation law is written as

ρt(x, t) + div ~F (x, t) = 0.

Again, ~F represents the flux of car density at any given point in space-time:

~F (x, t) = ρ(x, t) · ~v(x, t),

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where ~v(x, t) is the velocity of cars at a given point in space-time. With that, we rewrite

our conservation law as

ρt + (ρ · v)x = 0. (1.5)

Now it is time to bring in a constitutive assumption based on traffic flow observations. The

assumption we will use comes from the Lighthill-Whitham-Richards model of traffic flow,

and this corresponds to ~v(x, t) assumed to have the form

~v(x, t) = vmax

(1− ρ(x, t)

ρmax

)x ,

where vmax is the maximum car velocity (speed limit) and ρmax is the maximum density of

cars (parked bumper-to-bumper). Looking at this assumption, we see that this model rather

naively indicates that car velocity is directly proportional to car density; cars approach the

speed-limit as their density goes to zero and, conversely, slow to a stop as they get bumper-

to-bumper. Using this constitutive assumption with our conservation law (1.5) yields

ρt + vmax ·(ρ− ρ2

ρmax

)

x

= 0

=⇒ ρt + vmax ·(

1− 2ρ

ρmax

)ρx = 0. (1.6)

It turns out that this PDE predicts that a continuous density profile can develop a dis-

continuity in finite time, i.e. a car pile-up is an inevitability on an ideal highway!

1.4 The Transport Equation

Here we will again present another PDE via a physical derivation. In this model we seek

to understand the dispersion of an air-born pollutant in the presence of a constant and uni-

directional wind. For this, we take u(x, t) to be the pollutant density, and ~b to be the velocity

of the wind, which we are assuming to be constant in both space and time.

Notation: From this point forward, the vector notation over the symbols will be dropped, as

the context will indicate whether or not a quantity is a vector. With this in mind, we acknowl-

edge that x may indeed represent a spatial variable in any number of dimenions (although

for physical relevance, x ∈ R3; but this restriction is not necessary in our mathematical

derivation).

If we assume that the pollutant doesn’t react chemically with the surrounding air, then

we its total mass is conserved:

ut + div F (x, t) = 0,

where

F (x, t) = u(x, t)b.

Again, the form of the pollutant density flux is born out of a constitutive assuumption

regarding the dispersion of pollutant. It is easy to see that this assumption corresponds to

the pollutant simply being carried along by the wind. Combining the above conservation law

and constitutive assumption yields

ut +Du · b = 0,Gregory T. von Nessi

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1.4 The Transport Equation7

which is called the Transport Equation. Viewed in a space-time setting, this equation corre-

sponds to the directional derivative of u (as a function of x and t) being zero in the direction

(b, 1):

(Du, ut) · (b, 1)︸︷︷︸Directional derivative

= 0. (1.7)

Before moving on to another PDE, let us see if we can make any progress in terms of

solving (1.7).

1.4.1 Method of Characteristics: A First Look

Expect u = constant on the integral curves. The curve through (x, t): γ(s) = (x+sb, t+s)

z(s) = u(γ(s)) = u(x + sb, t + s)

z(s) = Du(x + sb, t + s) · b + ut(x + sb, t + s)

= 0

If we know the concentration of the pollutant at t = 0, i.e. we are solving the initial value

problem

ut +Du · b = 0

u(x, 0) = g(x)

then u(x, t) = u(γ(0)) by definition of γ

= u(γ(s)) ∀s. (because u is constant on curve γ(s))

Thus u(x, t) is the value of g at the intersection of the line γ(s) with the plane Rn×t = 0for s = −t.

u(x, t) = u(x − tb, 0) = g(x − tb)

Remarks: If you want to check that u(x, t) = g(x − t, b)

1. is a solution of ut +Du · b = 0, then we need that g is differentiable.

The formula also makes sense for g not differentiable, and u(x, t) = g(x − tb) would

be a so-called generalized or weak solution.

2. We were solving the PDE by solving the ODE

z(s) = 0

z(−t) = u(γ(−t)) = u(x − tb, 0)

= g(x − tb)

This is a general concept which is called the method of characteristics.

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1.4.2 Duhamel’s Principle

If you can solve the homogeneous equation ut +Du ·b = 0 then you can solve the inhomoge-

neous equation ut+Du ·b = f (x, t). f (x, t) is the source of the pollutant (e.g. power plant).

Idea: Suppose the pollutant were released only once per hour, then you could trace the

evolution of the pollutant by homogeneous equation, and you could add this evolution to the

evolution of the initial data. Equation we want to solveut +Du · b = f

u(x, 0) = g(x)

Since the sum of two solutions is a solution (since the equation is linear) we can write

u = v + w , wherevt +Dv · b = 0

v(x, 0) = g(x)

wt +Dw · b = f

w(x, 0) = 0

Check: ut +Du · b = (v + w)t +D(v + w) · b= (vt +Dv · b)︸︷︷︸

=0

+ (wt +Dw · b)︸︷︷︸f

= f

u(x, 0) = v(x, 0) + w(x, 0)

= g(x) + 0

= g(x)

Suspect: w(x, t) =

∫ t

0

w(x, t; s) ds

wherewt(x, t; s) +Dw(x, t; s) · b = 0 for x ∈ Rn, t > s

w(x, s; s) = f (x, s)

Solutions: v(x, t) = g(x − tb)

w(x, t; s) = f (x + (s − t)b, s)

Expectation: u(x, t) = g(x − tb) +

∫ t

0

f (x + (s − t)b, s) ds

Theorem 1.4.1 (): Let f and g be smooth. Then the solution of the initial value problem

ut +Du · b = f

u(x, 0) = g

is given by

u(x, t) = g(x − tb) +

∫ t

0

f (x + (s − t)b, s) ds

Proof: Differentiate explicitly. Gregory T. von Nessi

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1.5 Minimal Surfaces9

1.5 Minimal Surfaces

D = unit disk in the plane.

All surfaces that are graphs of functions on D with a given fixed boundary curve

u(x, t)

D

Question: Find a surface with minimal surface are A(u)

A(u) =

∫∫

D

√1 + |Du|2

Idea: Find a PDE that is a necessary condition for A(u) to be minimal. If you take any φ

which is zero on ∂D, then A(u + φ) ≥ A(u). Define

g(ε) = A(u + εφ) ≥ A(u)

∀ε ∈ R, and we have a minimum (as a function of ε!) if ε = 0. Thus g′(ε) = 0

0 ǫ

A(u + ǫφ)

d

dε

∣∣∣∣∣ε=0

g(ε) = 0 =d

dε

∣∣∣∣∣ε=0

∫∫

D

√1 + |D(u + εφ)|2

=1

2

∫∫

D

2Du ·Dφ+ 2ε|Dφ|2√1 + |Du + εDφ|2

dx

∣∣∣∣∣ε=0

=

∫∫

D

Du ·Dφ√1 + |Du|2

dx

= −∫∫

D

div

(Du√

1 + |Du|2

)φ dx

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This holds ∀φ

div

(Du√

1 + |Du|2

)= 0 Minimal Surface Equation

1.5.1 Divergence Theorem and Integration by Parts

Theorem 1.5.1 (Divergence Theorem): ~F vector field domain Ω, smooth boundary ∂Ω =

S, outward unit normal ~ν(x)

∂Ω

~ν(x)

Ω

∫

Ω

div ~F dx =

∫

∂Ω

~F · ~ν dS Divergence Theorem

Example: Take ~F (x) = w(x)~v(x)

~v(x) = vector field

w(x) = scalar

LHS =

∫

Ω

div (w(x)~v(x)) dx

=

∫

Ω

n∑

i=1

∂i(w(x)vi(x)) dx

=

∫

Ω

n∑

i=1

[(∂iw(x))vi(x) + w(x)(∂ivi(x))] dx

=

∫

Ω

[Dw · ~v + w(x) · div v(x)] dx

RHS =

∫

∂Ω

(w(x)~v(x)) · ~ν(x) dS

=

∫

∂Ω

w(x)~v(x) · ~ν(x) dS

Thus, we have∫

Ω

Dw · ~v dx =

∫

∂Ω

w~v · ~ν dS −∫

Ω

w · div ~v dx (1.8)

Integration by parts

1 dimension:

∫ b

a

w ′v dx = wv

∣∣∣∣∣

b

a

−∫ b

a

wv ′ dx

Applications:Gregory T. von Nessi

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1.5 Minimal Surfaces11

1. ~v(x) = Du(x)

w(x) = φ(x)

∫

Ω

Dφ ·Du dx =

∫

∂Ω

φ Du · ~ν dS −∫

Ω

φ div Du dx

i.e.

∫

Ω

Dφ ·Du dx =

∫

∂Ω

φ ∂νu dS −∫

Ω

φ ∆u dx

2. Soap bubbles = minimal surfaces

u(x, t)

D

surface area

∫

D

√1 + |Du|2 dx.

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Minimize.

Assume that u is a minimizer. i.e. that A(u) ≤ A(v) for all functions v with the same

boundary values. Family of functions u + εφ where φ has zero boundary values. Look

at g(ε) = A(u + εφ)

g(ǫ)

ǫ

d

dε=

∫

Ω

d

dε

√1 + |Du + εDφ|2 dx

=

∫

Ω

d

dε

√1 + |Du|2 + 2εDu ·Dφ+ ε2|Dφ|2 dx

=1

2

∫

Ω

2Du ·Dφ+ 2ε|Dφ|2√1 + |Du|2 + 2εDu ·Dφ+ ε2|Dφ|2

dx

Now, since u is a minimum,

d

dε

∣∣∣∣∣ε=0

∫. . . dx = 0

=⇒∫

Ω

Du ·Dφ√1 + |Du|2

dx = 0

If we identify

w = φ ~v(x) =Du√

1 + |Du|2,

we can apply (1.8):∫

Ω

Dφ · Du√1 + |Du|2

dx =

∫

∂Ω

φDu√

1 + |Du|2· ~ν dS −

∫

Ω

φ div

(Du√

1 + |Du|2

)dx

= 0

The first term on the RHS is 0 since φ = 0 on ∂Ω. Thus,∫

Ω

φ div

(Du√

1 + |Du|2

)dx = 0

for all φ that are zero on ∂Ω. If u is smooth, then the integrand must be zero pointwise,

i.e.

div

(Du√

1 + |Du|2

)= 0 Minimal surface equation


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1.6 The Wave Equation13

1.6 The Wave Equation

[Ant80] Comments on ad hoc derivations (e.g. wave equation).

Outcome of this “incomprehensible” derivation is the wave equation

utt = c2uxx

Newton’s law f = ma.

1.7 Classification of PDE

1. Diffusion equation (2nd order):

ut = k∆u

2. Traffic flow (1st order):

ρt + vmax∂x

(ρ

(1− ρ

ρmax

))= 0

3. Transport equation (1st order):

ut +Du · b = 0

4. Minimal surfaces (2nd order):

div

(Du√

1 + |Du|2

)= 0

5. Wave equation (2nd order):

utt = c2uxx

Definition 1.7.1: A PDE is an expression of the form

F (Dku,Dk−1u, . . . , Du, u, x) = 0,

where Du = gradient of u

D2u = all 2nd order derivatives of u...

Dku = all kth order derivatives of u

k is called the order of the PDE.

Definition 1.7.2: A PDE is linear if it’s of the form

∑

|α|≤kaα(x) Dαu(x) = 0

where α is a multi-index.

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Notation: α = (α1, . . . , αn) with αi non-negative integers. We define

|α| =

n∑

i=1

αi

So,

Dα =∂α1

∂xα11

· ∂α2

∂xα22

· · · ∂αn

∂xαnn

Example:

• n = 2. α = (0, 0) corresponds to Dαu = u.

|α| = 1 is associated with α = (1, 0) or α = (0, 1).

α = (1, 0) : Dαu =∂u

∂x1

α = (0, 1) : Dαu =∂u

∂x2

|α| = 2 is associated with α = (2, 0), α = (1, 1) or α = (0, 2).

α = (2, 0) : Dαu =∂2u

∂x21

α = (1, 1) : Dαu =∂2u

∂x1∂x2

α = (0, 2) : Dαu =∂2u

∂x22

• n = 3. Just look at α = (1, 1, 2) with u = u(x1, x2, x3)

Dαu =∂u

∂x1

∂u

∂x2

∂2u

∂x23

u =∂4u

∂x1∂x2∂x23

u

Example: 2nd order linear equation in 2D]

∑

|α|≤2

aα(x) Dαu(x)

= a(2,0)(x)uxx + a(1,1)(x)ux,y + a(0,2)(x)uyy |α| = 2

+a(1,0)(x)ux + a(0,1)(x)uy |α| = 1

+a(0,0)(x)u |α| = 0

= f

Crucial term is the highest order derivative. Generalizations:

• a PDE is semilinear if it is linear in the highest order derivatives, i.e. it is of the form∑

|α|=kaα(x) Dαu(x) + a0(Dk−1u, . . . , Du, u, x)

• a PDE is quasilinear if it is of the form

∑

|α|=kaα(Dk−1u, . . . , Du, u, x) Dαu + a0(Dk−1u, . . . , Du, u, x)


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1.8 Exercises15

Example:

• diffusion equation:

ut = k∆u linear

• traffic flow:

ρt + vmax∂x

(ρ

(1− ρ

ρmax

))= 0 quasilinear

• transport equation:

ut +Du · b = 0 linear

• minimal surface equation:

difDu√

1 + |Du|2= 0

=⇒∑ ∂

∂xi

∂xiu√1 + |Du|2

= 0 quasilinear

• wave equation:

utt = c2uxx linear

1.8 Exercises

1.1: Classify each of the following PDEs as follows:

a) Is the PDE linear, semilinear, quasilinear or fully nonlinear?

b) What is the order of the PDE?

Here is the list of the PDEs:

i) Linear transport equation ut +∑ni=1 biDiu = 0.

ii) Schrodinger’s equation iut + ∆u = 0.

iii) Beam equation ut + uxxxx = 0.

iv) Eikonal equation |Du| = 1.

v) p-Laplacian div(|Du|p−2Du) = 0.

vi) Monge-Ampere equation det(D2u) = f .

1.2: Let Ω ⊂ R2 be an open and bounded domain representing a heat conducting plate with

density ρ(x, t) and specific heat c(x). Find an equation for the evolution of the temperature

u(x, t) at the point x at the time t under the assumption that there is a heat source f (x) in

Ω. Use Fourier’s law of cooling as the constitutive assumption and assume that all functions

are sufficiently smooth.

1.3:

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a) Verify that the function u(x, t) = e−kt sin x satisfies the heat equation ut = kuxx on

−∞ < x <∞ and t > 0. Here k > 0 is the diffusion constant.

b) Verify that the function v(x, t) = sin(x − ct) satisfies the wave equation utt = c2uxxon −∞ < x <∞ and t > 0. Here c > 0 is the wave speed.

c) Sketch u and v . What is the most striking difference between the evolution of the

temperature u and the wave v?

1.4: Suppose that u is a smooth function that minimizes the Dirichlet integral

I(u) =1

2

∫

Ω

|Du|2 dx

subject to given boundary conditions u(x) = g(x) on ∂Ω. Show that u satisfies Laplace’s

equation ∆u = 0 in Ω.

1.5: [Evans 2.5 #1] Suppose g is smooth. Write down an explicit formula for the func-

tion u solving the initial value problemut + b ·Du + cu = 0 in Rn × (0,∞),

u = g on Rn × t = 0.

Here c ∈ R and b ∈ Rn are constants.

1.6: Show that the minimal surface equation

div

(Du√

1 + |Du|2

)= 0

can in two dimensions be written as

uxx(1 + u2y ) + uyy (1 + u2

x )− 2uxuyuxy = 0.

1.7: Suppose that u = u(x, y) = v(r) is a radially symmetric solution of the minimal surface

equation

uxx(1 + u2y ) + uyy (1 + u2

x )− 2uxuyuxy = 0.

Show that v satisfies

vr r +vrr

(1 + v2r ) = 0.

Use this result to show that u(x, y) = ρ log(r+√r2 − ρ2) is a solution of the minimal surface

equation for r =√x2 + y2 ≥ ρ > 0.

Hint: You can verify this by differentiating the given solution or by solving the ordinary

differential equation for v .

1.8: [Evans 2.5 #2] Prove that Laplace’s equation ∆u = 0 is rotation invariant; that is,

if R is an orthogonal n × n matrix and we define

v(x) = u(Rx), x ∈ Rn,

then ∆v = 0.Gregory T. von Nessi

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18Chapter 2: Laplace’s Equation ∆u = 0 and Poisson’s Equation −∆u = f

2Laplace’s Equation ∆u = 0 and Poisson’sEquation −∆u = f

“. . . You are too young to be burdened by something as cool as sadness.”

–? Kurosaki Isshin

Contents

2.1 Laplace’s Eqn. in Polar Coordinates 18

2.2 The Fundamental Solution of the Laplacian 21

2.3 Properties of Harmonic Functions 26

2.4 Estimates for Derivatives of Harmonic Functions 32

2.5 Existence of Solutions 47

2.6 Solution of the Dirichlet Problem by the Perron Process 62

2.7 Exercises 66

2.1 Laplace’s Equation in Polar Coordinates and for Radially

Symmetric Functions

We say that u is harmonic if ∆u = 0.

Example:

The real and imaginary part of holomorphic functions are harmonic

f (z) = z2 = (x + iy)2 = x2 − y2 + 2ixy

u(x, y) = x2 − y2

u(x, y) = 2xy

are harmonic

r =√x2 + y2

θ = tan−1(yx

)

u(x, y) = v(r, θ)Gregory T. von Nessi

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2.1 Laplace’s Eqn. in Polar Coordinates19

x

y

r

θ

∂x f = ∂x√x2 + y2 =

1

2

2x√x2 + y2

=x

r

∂xx r = ∂x

(xr

)=

1

r+ x∂x

(1

r

)=

1

r− x

2

r3

∂x1

r= ∂x(x2 + y2)−1/2 = −1

2

2x√x2 + y2

3 =−xr3

∂xy r = ∂y

(xr

)=−xyr3

In Summary:

∂xy =x

r, ∂y r =

y

r

∂xx r =1

r− x2r3 =

y2

r3, ∂yy r =

x2

r3

∂xy r =−xyr3

Now, we look at derivatives with respect to θ.

∂xθ = ∂x tan−1 y

x=

1

1 + y2

x2

− y

x2=

−yx2 + y2

=−yr2

∂yθ =x

r2

∂xxθ = ∂x

(− yr2

)= ∂x(−y(x2 + y2)−1) =

2xy

(x2 + y2)2

=2xy

r4

∂xyθ = ∂y

(− yr2

)= − 1

r2− y∂y (x2 + y2)−1 = − 1

r2+

2y2

(x2 + y2)2

= − 1

r2+

2y2

r4

∂yyθ = ∂y

( xr2

)= x∂y (x2 + y2)−1

= −2xy

r4

In Summary:

∂xθ = − yr2, ∂yθ =

x

r2

∂xxθ =2xy

r4, ∂yyθ = −2xy

r4

∂xyθ =y2 − x2

r4

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Now, we consider

u(x, y) = v(r, θ)

∂xu = vr rx + vθθx

∂xxu = vr r (rx)2 + 2vrθrxθx + vθθ(θy )2 + vr ryy + vθθyy

∂yu = vr ry + vθθy

∂yyu = vr r (ry )2 + 2vrθryθy + vθθ(θy )2 + vr ryy + vθθyy

∆u = uxx + uyy

= vr r (r2x + r2

y ) + 2vrθ(rxθx + ryθy ) + vθθ(θ2x + θ2

y )

+vr (rxx + ryy ) + vθ(θxx + θyy )

= 0

= vr r

(x2 + y2

r2

)+ vθθ

(x2 + y2

r4

)

+vr

(1

r− x

2

r3+

1

r− y

2

r3

)

Result: If u(x, y) = v(r, θ) and u solves Laplace’s equation, then v solves

vr r +1

rvr +

1

r2vθθ = 0

Laplace’s equation in polar coordinates

Now we consider Laplace’s equation for radially symmetric functions:

u(x) = v(r), r =

√x2

1 + . . .+ x2n

∂u

∂xi= vr (r)

∂r

∂xi= vr (r)

xir

∂2u

∂x2i

=∂

∂xi

(vr (r)

xir

)

= vr rx2i

r2+ vr

(1

r+ xi∂xi

1

r

)

∂xi1

r= ∂xi (x

21 + . . .+ x2

n )−1/2 = −1

2(x2

1 + . . .+ x2n )−3/22xi = − xi

r3

= vr rx2i

r2+ vr

(1

r− x

2i

r3

)

Thus,

∆u =

n∑

i=1

∂2

∂x2i

u

=

n∑

i=1

vr rx2i

r2+ vr

(1

r− x

2i

r3

)

= vr r +

(n

r− 1

r

)vr

Result: If u(x) = v(r) is a radially symmetric solution of Laplace’s equation then

vr r +n − 1

rvr = 0 (2.1)


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2.2 The Fundamental Solution of the Laplacian21

2.2 The Fundamental Solution of the Laplacian

Special solution radially symmetric by solving (1) of the previous section. Assuming vr 6= 0

vr rvr

=1− nr

integrate in r

=⇒ ln vr = (1− n) ln r + C1

=⇒ vr = C2 · r1−n

Integrate again:

n = 2 : vr =C2

r=⇒ v = C2 · ln r + C3

n ≥ 3 : vr =C2

rn−1=⇒ v =

−1

n − 2

C2

rn−2+ C3

Find special solutions

u(x) = v(r) =

C2 · ln |x |+ C3 n = 2

− C2n−2

1|x |n−2 + C3 n ≥ 3

The fundamental solution is of this form with particular constants.

Definition 2.2.1:

Φ(x) =

− 1

2π · ln |x | n = 2

− 1n(n−2)α(n)

1|x |n−2 n ≥ 3

for x 6= 0, and α(n) is the volume of the unit ball in Rn.

Note: ∆Φ = 0 for x 6= 0 but not defined if x = 0. ∆Φ can be defined everywhere in the

sense of distributions,

−∆Φ = δ0 = “Dirac Mass”

placed at x = 0,

δ0(x) =

1 x = 0

0 x 6= 0

−∆Φ(y − x) =

0 x 6= y

1 x = y

Formally, this suggests a solution formula for Poisson’s equation,

−∆u = f in Rn

namely,

u(x) =

∫

RnΦ(x − y)f (y) dy

Why?

“− ∆u(x) =

∫

Rn−∆xΦ(x − y)f (y) dy

=

∫

Rnδx(y)f (y) dy = f (x)“

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Lemma 2.2.2 (): We have

|DΦ(x)| ≤ C

|x |n−1x 6= 0

|D2Φ(x)| ≤ C

|x |n x 6= 0

and∂Φ

∂ν(x) =

−1

nα(n)

1

rn−1∀x ∈ ∂B(0, r)

∂φ∂ν(x) = Dφ(x) · x|x |

ν(x) = x|x |

x

r

x

O

Proof:

n = 2 :∂

∂xiln |x | =

1

|x |∂

∂xi|x | =

1

|x |xi|x |

DΦ(x) =−2

2π

x

|x |2 , |DΦ| ≤ C 1

|x |√

∂

∂xj

∂

∂xiln |x | =

∂

∂xj

xi|x |2 =

δi j|x |2 + xi

∂

∂xj

1

|x |2

∂

∂xj

1

|x |2 =∂

∂xj

(∑x2k

)−1

= −(∑

x2k

)−22xj = − 2xj

|x |4

∂2Φ

∂xi∂xj= − 1

2π

(δi j|x |2 −

2xixj|x |4

)

∣∣∣∣∂2Φ

∂xi∂xj

∣∣∣∣ ≤ − 1

2π

(1

|x |2 −2|xi ||xj ||x |4

)

≤ 1

2π

3

|x |2

DΦ(x) = − 1

2πD ln |x | = − 1

2π

1

|x |x

|x |∂Φ

∂ν(x) = − 1

2π

x

|x |2x

|x |

= − 1

2π

|x |2|x |2 = − 1

2π

1

|x |Gregory T. von Nessi

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n ≥ 3 :∂

∂xi

1

|x |n−2=

∂

∂xi

(∑x2k

)− n−22

= −n − 2

2

(∑x2k

)−( n−22

+1)2xi

= −(n − 2)xi|x |n

DΦ(x) =−1

nα(n)

x

|x |n

|DΦ(x)| ≤ 1

nα(n)

|x ||x |n =

1

nα(n)

1

|x |n−1

Second derivatives are similar

∂Φ

∂ν(x) =

−1

nα(n)

x

|x |nx

|x |

=−1

nα(n)

|x |2|x |n+1

Remark: |D2Φ| ≤ C|x |n

i.e. D2Φ are not integrable about zero.

Theorem 2.2.3 (): Solution of Poisson’s equation in Rn. Suppose that f ∈ C2 and that f

has compact support, i.e. closure of (x ∈ Rn : f (x) 6= 0) ⊆ B(0, s) for some s > 0. Then

u(x) =

∫

RnΦ(x − y)f (y) dy

is twice differentiable and solves Poisson’s equations −∆u = f in Rn.

Proof: Change variables: z = x − y1. y = x − z and dy = dz , i.e.,

∫

RnΦ(x − y)f (y) dy =

∫

RnΦ(z)f (x − z) dz

=

∫

RnΦ(y)f (x − y) dy = u(x)

2. u is differentiable

limh→0

u(x + hei)− u(x)

h= limh→0

∫

RnΦ(y)

f (x + hei − y)− f (x − y)

hdy

Technical Step: If we can interchange limh→0 and integration, then (f ∈ C2):

∂u

∂xi=

∫

RnΦ(y)

∂f

∂xi(x − y) dy

and by the same arguments,

∂2u

∂xi∂xj=

∫

RnΦ(y)

∂2f

∂xi∂xj(x − y) dy

and ∆u =

∫

RnΦ(y)∆x f (x − y) dy

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So, what was the “black box” that enables use to switch the integration and limit?

Black Box: Lebesgue’s Dominated Convergence Theorem: fk be a sequence of integrable

functions such that |fk | ≤ g almost everywhere, g integrable fk → f pointwise almost every-

where. then

limk→∞

∫

Ω

fk =

∫

Ω

limk→∞

fk =

∫

Ω

f

Our application: Need

∣∣∣∣Φ(y)f (x + hei − y)− f (x − y)

h

∣∣∣∣ ≤ g

g(y) = |Φ(y)|maxz∈Rn

∣∣∣∣∂f

∂xi(z)

∣∣∣∣︸︷︷︸

≤M

is integrable since f and hence ∂f∂xi

have compact support.

∆u =

∫

RnΦ(y)∆x f (x − y) dy

=

∫

RnΦ(y)∆y f (x − y) dy

=

∫

Rn\B(0,ε)

Φ(y)∆y f (x − y) dy +

∫

B(0,ε)

Φ(y)∆y f (x − y) dy

= I1 + I2

Now, we have

|I2| ≤ maxRn|D2f |

∫

B(0,ε)

|Φ(y)| dy

n = 2 :

∫

B(0,ε)

ln |x | dx =

∫ 2π

0

∫ ε

0

ln r · rdrdθ

= 2π

∫ ε

0

r ln r dr

=

∫ ε

0

(1

2r ′)′

ln r dr

=1

2r2 ln r

∣∣∣∣∣

ε

0

−∫ ε

0

1

2r2 1

rdr

=1

2ε2 ln ε− 1

4r2

∣∣∣∣∣

ε

0

=1

2ε ln ε− 1

4ε2

∴ |I2| is O(ε ln ε)Gregory T. von Nessi

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n = 3 :

∫

B(0,ε)

1

|x |n−2dx =

∫ ε

0

∫

∂B(0,ε)

1

|r |n−2dSdr

=

∫ ε

0

1

rn−2Crn−1 dr

= C

∫ ε

0

r dr

=C

2ε2

∴ |I2| ≤ maxRn|D2f |

∫

B(0,ε)

|Φ(y)| dy

=

O(ε ln ε) n = 2

O(ε2) n ≥ 3

In particular: I2 → 0 as ε→ 0

I1 =

∫

Rn\B(0,ε)

Φ(y)∆y f (x − y) dy

=

∫

∂B(0,ε)

Φ(y)Df (x − y) · ν dS −∫

Rn\B(0,ε)

DΦ(y)Df (x − y) dy

= J1 + J2

Now,

|J1| =

∣∣∣∣∫

∂B(0,ε)

Φ(y)Df (x − y) · ν dS∣∣∣∣

≤ max |Df |∫

∂B(0,ε)

Φ(y) dS(y)

n = 2 : ε ln ε→ 0 as ε→ 0

n ≥ 3 :C

εn−2εn−1 = ε→ 0 as ε→ 0

So we are left to look at

J2 = −∫

Rn\B(0,ε)

DΦ(y)Df (x − y) dy

= −∫

∂B(0,ε)

DΦ(y) · νin(y)f (x − y) dS +

∫

Rn\B(0,ε)

∆Φ(y)︸︷︷︸=0

f (x − y) dy

=

∫

∂B(0,ε)

DΦ(y) · νout(y)f (x − y) dS

=

∫

∂B(0,ε)

−1

|∂B(0, ε)| f (x − y) dS

= −∫

∂B(0,ε)

−f (x − y) dS(y)→ −f (x) as ε→ 0

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Notation: −∫u f = |u|−1

∫u f , where |u| is the measure of u.

Finally,

∆u(x) =

∫

RnΦ(y)∆f (x − y) dy

= terms tending to zero as ε→ 0

−−∫

B(0,ε)

f (x − y) dS → −f (x) as ε→ 0

=⇒ −∆u(s) = f (x)

2.3 Properties of Harmonic Functions

2.3.1 WMP for Laplace’s Equation

To begin this section, we introduce the Laplacian differential operator:

∆u =

n∑

i=1

∂2u

∂x2i

, u ∈ C2(Ω).

Note that ∆u is sometimes written as div ·∇u in classical notation. Coupled to this we have

the following definition

Definition 2.3.1: u ∈ C2(Ω) is called harmonic in Ω if and only if ∆u = 0 in Ω.

Example 1: u(x1, x2) = x21 − x2

2 is a nonlinear harmonic function in R2.

In addition to the above, we have the next definition.

Definition 2.3.2: u ∈ C2(Ω) is called subharmonic( superharmonic) if and only if ∆u ≥ 0(≤0) in Ω.

Remark: It is obvious, due to the linearity of the Laplacian, that if u is subharmonic, then

−u is superharmonic and vice-versa.

Example 2: If x ∈ Rn, then ∆|x |2 = 2n. Thus, |x |2 is subharmonic in Rn.

Now, we come to our first theorem

Theorem 2.3.3 (Weak Maximum Principle for the Laplacian): Let u ∈ C2(Ω) ∩ C0(Ω)

with Ω bounded in Rn. If ∆u ≥ 0 in Ω, then

u ≤ max∂Ω

u

(⇐⇒ max

Ωu = max

∂Ωu

). (2.2)

Conversely, if ∆u ≤ 0 in Ω, then

u ≥ min∂Ω

u

(⇐⇒ min

Ωu = min

∂Ωu

). (2.3)

Consequently, if ∆u = 0 in Ω, then

min∂Ω

u ≤ u ≤ max∂Ω

u. (2.4)


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2.3 Properties of Harmonic Functions27

Proof: We will just prove the subharmonic result, as the superharmonic case is proved the

same way. Given u subharmonic, take ε > 0 and define v = u + ε|x |2, so that ∆v =

∆u + 2εn > 0 in Ω. If v takes attains a maximum in Ω, we have ∆v ≤ 0 at that point, a

contradiction. Thus, by the compactness of Ω, we have

v ≤ max∂Ω

v .

Finally, we take ε→ 0 in the above to get the result.

Corollary 2.3.4 (Uniqueness): Take u, v ∈ C2(Ω) ∩ C0(Ω). If ∆u = ∆v = 0 in Ω and

u = v on ∂Ω, then u = v in Ω.

Before moving onto more general elliptic equations, we briefly remind ourselves of the Dirich-

let Problem, which takes the following form.

∆u = 0 in Ω

u = φ on ∂Ω for some φ ∈ C0(∂Ω)

2.3.2 Poisson Integral Formula

The best possible scenario one can hope for when dealing in theoretical PDE is to actually

have an analytic solution to the equation. Obviously, it is easier to analyze a solution directly

rather than having to do so through its properties. Fortunately, this turns out to be possible

for the Dirichlet problem for Laplace’s equation in a ball.

To start off, let us consider some other examples of harmonic functions:

• Since Laplace’s equation is a constant coefficient equation, we know that if u ∈ C3(Ω)

is harmonic, then so is all its partial derivatives. Indeed,

Di(∆u) = 0 ⇐⇒ ∆(Diu) = 0.

• There are also radially symmetric solutions to Laplace’s equation. To derive these, let

r = |x | and u(x) = g(r). We calculate

Diu = g′ · xi|x | =g′

r· xi

∆u = Di iu = n · g′

r+g′′

r2· x2i −

g′

r3x2i

= n · g′

r+ g′′ − g

′

r

= (n − 1) · g′

r+ g′′

=(rn−1g′)′

rn−1= 0.

This implies that rn−1g′(r) = constant; integration then yields

g =

C1r

2−n + C2 n > 2

C1 ln r + C2 n = 2.

These are all the radial solution of the homogeneous Laplace’s equation outside of the

origin. For simplicity we will take C1 = 1 and C2 = 0 in the above.

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• It is easily verified that translating the above solutions so that y is taken to be the new

origin, are also harmonic functions:

|x − y |2−n n > 2

C1 ln |x − y | n = 2.

Since these functions are symmetric in x and y , it’s obvious the above are harmonic

with respect to either x or y provided x 6= y . Taking a derivative of the above indicates

that

xi − yi|xi − yi |n

is also harmonic. Next consider

V (x, y) =|y |2 − |x |2|x − y |n =

|y |2 + |x |2 − 2xy

|x − y |n − 2x(x − y)

|x − y |n

=1

|x − y |n−2− 2xi(xi − yi)|x − y |n .

The last term is harmonic since it is the derivative with respect to yi of |x − y |2−n.

Thus, v(x, y) is harmonic in both x and y variables.

It would be difficult to verify the above aforementioned functions solved Laplace’s equation,

but they are readily constructed from simpler solutions.

Now, let us consider |y | = R and define

v(x) =

∫

|y |=RV (x, y) dS(y).

First, it is clear that v(x) is harmonic for x ∈ BR(0). Indeed for any fixed x in the BR(0),

V (x, y) is bounded. So we may interchange the Laplacian with the integral to get that v(x)

is harmonic in BR(0). Next, we claim that v(x) is radial, i.e. only depends on |x |. To show

this we consider an arbitrary rotation transformation on v(x). Sometimes this is called an

orthonormal transformation as it corresponds to the transformation of x = Px ′, where P

is an orthonormal matrix (rows/columns are orthogonal basis in Rn, P T = P−1 is another

property). So, given the transformation, we have

v(Px) =

∫

|y |=R

|y |2 − |Px |2|Px − y |n dS(y)

=

∫

|y |=R

R2 − |Px |2|Px − y |n dS(y).

Now, we change variables with Pz = y :

v(Px) =

∫

|Pz |=R

R2 − |Px |2|Px − Pz |n dS(Pz)

=

∫

|z |=R

R2 − |x |2|P (x − z)|n det(P ) dS(z)

=

∫

|z |=R

R2 − |x |2|x − z |n dS(z).


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In the above, we have used the fact that rotations obviously do not change vector mag-

nitude. Also, we know det(P ) = 1 as 1 = det(I) = det(PP−1) = det(P ) det(P−1) =

det(P ) det(P T ) = det(P )2. So, v(x) is rotationally invariant. Thus, we calculate

v(0) =

∫

|y |=RR2−n dS(y) = R2−nA(SR(0)) = nωnR.

It turns out the above result is valid for any x ∈ BR(0), this can be verified by carrying out

the integration; but this is rather complicated.

Now, we can make the following definition

Definition 2.3.5: The Poisson Kernel:

K(x, y) =R2 − |x |2nωnR

1

|x − y |n ,

where |y | = R.

From the above, we already know that K(x, y) is harmonic in BR(0) with respect to x . From

the calculation for v(x), we also have

∫

|y |=RK(x, y) dS(y) = 1

by construction. Next, we have

Definition 2.3.6: The Poisson Integral for x ∈ BR(0) is

w(x) =

∫

|y |=RK(x, y)φ(y) dS(y).

Theorem 2.3.7 (): (Properties of the Poisson Integral) Given the above definition, w ∈C∞(BR(0)) ∩ C0(BR(0) with ∆w = 0 in BR(0). Also w(x)→ φ(y) as x → y ∈ ∂BR(0).

Remark: Basically one has that w solves the Dirichlet problem for Laplace’s equation:

∆w = 0 in BR(0)

w = φ on ∂BR(0)

Proof: w ∈ C∞(BR(0)) is obvious since for fixed x ∈ BR(0), K(x, y) is bounded and

harmonic. Hence on may take derivatives of any order inside the integral (the derivatives will

also be well behaved as x 6= y for our fixed x). So, all we need to show is that w(x)→ φ(y0)

as x → y0 ∈ ∂BR(0). Take an arbitrary ε > 0 and choose δ such that |φ(y) − φ(y0)| < ε

implies |y − y0| < δ. Keeping in mind the figure, we calculate:

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O

δ2

δ2

x

y0

δBR(0) ∩ |y − y0| > δ

|w(x)− φ(y0)| =

∣∣∣∣∫

∂BR(0)

K(x, y)(φ(y)− φ(y0)) dS(y)

∣∣∣∣

≤∫

∂BR(0)∩|y−y0|<δ|K(x, y)(φ(y)− φ(y0))| dS(y)

+

∫

∂BR(0)∩|y−y0|>δ|K(x, y)(φ(y)− φ(y0))| dS(y)

≤ ε

∫

∂BR(0)∩|y−y0|<δ|K(x, y)| dS(y)

+2 · max∂BR(0)

φ ·∫

∂BR(0)∩|y−y0|>δ

R2 − |x |2|x − y |n dS(y).


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Now, if one chooses x such that |x − y0| ≤ δ2 , this implies that |x − y | ≥ δ

2 for y ∈∂BR(0) ∩ |y − y0| > δ. Thus, we have

|w(x)− φ(y0)| ≤ ε+ 2 · max∂BR(0)

φ · R2 − |x |2δ2

n

≤ 2ε,

where we have chosen x so that ·max∂BR(0) φ · R2−|x |2δ2

n < ε2 . The last inequality comes from

simply picking x close enough to y0 (as this happens R2 − |x |2 clearly shrinks). So, we have

now shown that w(x) → φ(y0) as x → y0 ∈ ∂BR(0). This also implies the continuity of

w(x) on the closure of Ω.

A very important consequence to the last theorem is that if u ∈ C2(Ω) is harmonic, then for

any ball BR(z), one has

u(x) =R2 − |x − z |2

nωnR

∫

∂BR(z)

u(y)

|x − y |n dS(y).

In other words, the value of a harmonic function at any point is completely determined by

the values it takes on any given spherical shell surrounding that point! Taking, x = z in the

above, we get the following:

Theorem 2.3.8 (): (Mean Value Property) If u ∈ C2(Ω) and is harmonic on Ω, then one

has

u(z) =1

nωnRn−1

∫

∂BR(z)

u(y) dS(y)

=1

A(∂BR(z))

∫

∂BR(z)

u(y) dS(y), (2.5)

for any z ∈ Ω, BR(z) ⊂ Ω.

In addition, one has

Corollary 2.3.9 (): If u ∈ C2(Ω) and is subharmonic(superharmonic) on Ω, then one has

u(z) ≤ (≥)1

A(∂BR(z))

∫

∂BR(z)

u(y) dS(y), (2.6)

for any z ∈ Ω, BR(z) ⊂ Ω.

Proof: This is a simple matter of solving the following Dirichlet Problem:

∆v = 0 in BR(z)

v = u on ∂BR(z)

First, we know that ∆(u − v) ≥ (≤) 0. Thus, the weak maximum principle states that

u ≤ (≥) v in ∂BR(z). So, putting everything together, one has

u ≤ (≥) v =1

A(∂BR(z))

∫

∂BR(z)

v(y) dS(y)

=1

A(∂BR(z))

∫

∂BR(z)

u(y) dS(y).

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2.3.3 Strong Maximum Principle

Theorem 2.3.10 (Strong Maximum Principle): If u ∈ C2(Ω) with Ω connected and ∆u ≥(≤) 0, then u can not take on interior maximum(minimum) in Ω, unless u is constant.

Proof: Suppose u(y) = M := maxΩ u. Clearly, one has

∆(M − u) ≤ (≥) 0.

Thus, by the corollary to the mean value property,

(M − u)(y) ≥ (≤)1

nωnR

∫

∂BR(y)

M − u(x) dS(x)

for any BR(y) ⊂ Ω. This implies that u = M on BR(y), which in turn implies u = M on Ω

as Ω is connected.

Problem: Take Ω ⊂ Rn bounded and u ∈ C2(Ω) ∩ C0(Ω). If

∆u = u3 − u in Ω

u = 0 on ∂Ω

Prove −1 ≤ u ≤ 1. Can u take either value ±1?

2.4 Estimates for Derivatives of Harmonic Functions

Now, we will go through estimating the derivatives of harmonic functions. First, note that

the mean value property immediately implies that

u(x) =1

ωnRn

∫

BR(x)

u(y) dy

for any BR(x) ⊂ Ω and u harmonic. This can simply be ascertained from preforming a radial

integration on the statement of the mean value property. As we have shown derivatives to

be harmonic, we have that

Diu(x) =1

ωnRn

∫

BR(x)

div u,i dy,

where u,i is a vector whose ith place is u and whose other components are zero. So applying

the divergence theorem to the above, we have

Diu(x) =1

ωnRn

∫

∂BR(x)

u,i · ν dS(y)

≤ 1

ωnRnsupBR(x)

|u| ·∫

∂BR(x)

dS(y)

=n

Rsup

∂BR(x)|u|.

From this we see that any derivative is bounded by the final term in the above, thus

|Du(x)| ≤ n

RsupBR(x)

|u|.Gregory T. von Nessi

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2.4 Estimates for Derivatives of Harmonic Functions33

R|β|

R|β|

R|β|

R|β|

Applying this result iteratively over concentric balls whose radii increase by R/|β| for each

iteration, yields the following (refer to the figure):

|Dβu(x)| ≤(n|β|R

)|β|supBR(x)

|u|.

From this equation, the following is clear.

Theorem 2.4.1 (): Consider Ω bounded and u ∈ C2(Ω) ∩ C0(Ω). If u is harmonic in Ω,

then for any Ω′ ⊂⊂ Ω the following estimate holds

|Dβu(x)| ≤(n|β|dΩ′

)|β|sup

Ω

|u| ∀x ∈ Ω′, (2.7)

where dΩ′ = Dist (Ω′, ∂Ω).

2.4.1 Mean-Value and Maximum Principles

n = 1: u′′ = 0, u(x) = ax + b. u is affine.

u(x0) =1

2(u(x0 + r) + u(x0 − r))

= −∫

∂I

u(y) dS(y)

u(x0) = −∫

I

u(y) dy

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( )r rx0

ax+b

x

I

Theorem 2.4.2 (Mean value property for harmonic functions): Suppose that u is har-

monic on U, and that B(x, r) ⊂ U. Then

u(x) = −∫

∂B(x,r)

u(y) dS(y)

= −∫

B(x,r)

u(y) dy

Proof:

Idea: Φ(r) = −∫∂B(x,r) u(y) dS(y); Prove that Φ′(r) = 0.

d

drΦ(r) =

d

dr−∫

∂B(x,r)

u(y) dS(y)

=d

dr−∫

∂B(0,1)

u(x + rz) dS(z)

= −∫

∂B(0,1)

Du(x + rz) · z dS(z)

= −∫

∂B(x,r)

Du(y) · y − xr

dS(y)

= −∫

∂B(x,r)

Du(y) · ν(y) dS(y) (See below figure)

=1

|∂B(x, r)|

∫

B(x,r)

div Du(y)︸︷︷︸∆u=0

dy

∴ Φ′(r) = 0

Theorem 2.4.3 (Maximum principle for harmonic functions): u harmonic (u ∈ C2(Ω) ∩C(Ω),∆u = 0) in open bounded Ω. Then

maxu∈Ω

u = max∂Ω

u

Moreover, if Ω is connected and if the maximum of u is obtained in Ω, then u is constant.Gregory T. von Nessi

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x

y−xr= ν(y)

y

y − x

r

Remark:

• The second statement is also referred to as the strong maximum principle.

• The assertion holds also for the minimum of u, replace u by −u.

Proof: We prove the strong maximum principle.

Suppose that M = maxΩ u is attained at x0 ∈ Ω, then exists a ball B(x0, r) ⊂ Ω, r > 0.

Mean-value property:

M = u(x0) = −∫

B(x0,r)

u(y) dy

≤ −∫

B(x0,r)

M dy = M

=⇒ we have equality.

=⇒ u ≡ M on B(x0, r)

If u(z) < M for a point z ∈ B(x0, r), then we get a strict inequality.

M = x ∈ Ω, u(x) = MM is a closed set in Ω since u is constant; M is also open in Ω since it can be viewed as

the union of open balls.

=⇒ M = Ω, u is constant in Ω. Application of maximum principles: uniqueness of solutions of PDEs.

Theorem 2.4.4 (): There is at most one smooth solution of the boundary-value problem

−∆u = f in Ω

u = g on ∂Ω

Proof: Suppose u1 and u2 are solutions, take u1 = u2

maxΩ

(u1 − u2) = max∂Ω

(u1 − u2) = 0

=⇒ u1 − u2 = in Ω

Same argument for −(u1 − u2) gives

−(u1 − u2) ≤ in Ω

=⇒ |u1 − u2| ≤ 0in Ω

=⇒ u1 = u2 in Ω

=⇒ there exists at most one solution!

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0x

2.4.2 Regularity of Solutions

Goal: Harmonic functions are analytic (they can be expressed as convergent power series).

• Need that u is infinitely many times differentiable.

• Need estimates for the derivatives of u.

Next goal: Harmonic functions are smooth.

A way to smoothen functions is mollification.

Idea: (weighted) averages of functions are smoother.

Definition 2.4.5:

η(x) =

Ce

1

|x |2−1 |x | < 1

0 |x | ≥ 1

Fact: η is infinitely many times differentiable, η ≡ 0 outside B(0, 1).

We fix C such that∫Rn η(y) dy = 1.

ηε(x) =1

εnη(xε

)

• ηε = 0 outside B(0, ε)

•∫Rn ηε(y) dy = 1


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If u is integrable, then

uε(x) :=

∫

Rnηε(x − y)u(y) dy

=

∫

Rnηε(y)u(x − y) dy

=

∫

B(x,ε)

ηε(x − y)u(y) dy

is called the mollification of u.

Theorem 2.4.6 (Properties of uε): • uε is C∞ (infinitely many times differentiable)

• uε → u (pointwise) a.e. as ε→ 0

• If u is continuous then uε → u uniformly on compact subsets

Now since Φ′(r) = 0,

=⇒ Φ(r) = limρ→0

Φ(ρ)

= limρ→0−∫

∂B(x,ρ)

u(y) dS(y)

= u(x)

∴ Φ(r) = u(x) ∀r such that B(x, r) ⊂ U.

−∫

B(x,r)

u(y) dy =1

|B(x, r)|

∫ r

0

∫

∂B(x,ρ)

u(y) dS(y)dρ

=1

|B(x, r)|

∫ r

0

|∂B(x, ρ)| −∫

∂B(x,ρ)

u(y) dS(y)

︸︷︷︸u(x)

dρ

=u(x)

|B(x, r)|

∫ r

0

nα(n)ρn−1 dρ

=u(x)

rnα(n)nα(n)

1

nρn

∣∣∣∣∣

r

0

= u(x).

Note: nα(n) is the surface are of ∂B(0, 1) in Rn.

Complex differentiable =⇒ analytic

f (z) =

∞∑

n=1

ann!

(z − z0)n

an =n!

2πi

∮

γ

f (ω)

(ω − z0)n+1dω = f (n)(z)

Theorem 2.4.7 (): If u ∈ C(Ω), Ω bounded and open, satisfies the mean-value property,

then u ∈ C∞(Ω)

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ǫ

ǫ

Ωǫ

Ω

Proof: Take ε > 0

Ωε = x ∈ Ω : dist(x, ∂Ω) > ε

In Ωε : uε(x) =

∫

Rnηε(x − y)u(y) dy

η(y) =

Ce

1

|y |2−1 |y | ≤ 1

0 |y | > 1

uε(x) =1

εn

∫

B(x,ε)

η

( |x − y |ε

)u(y) dy

=1

εn

∫ ε

0

∫

∂B(x,ρ)

η(ρε

)u(y) dS(y)dρ

=1

εn

∫ ε

0

η(ρε

)|∂B(x, ρ)| −

∫

∂B(x,ρ)

u(y) dS

︸︷︷︸u(x)

dρ

=u(x)

εn

∫ ε

0

η(ρε

)nα(n)ρn−1 dρ

=u(x)

εn

∫ ε

0

∫

∂B(x,ρ)

η(ρε

)dSdρ

= u(x)

∫

Rnηε(y) dy

= u(x)

In Ωε, uε(x) = u(x)

=⇒ u ∈ C∞(Ωε)

ε > 0 arbitrary =⇒ u ∈ C∞(Ω).

Corollary 2.4.8 (): If u ∈ C(Ω) satisfies the mean-value property, then u is harmonic.Gregory T. von Nessi

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Proof: Mean-value property =⇒ u ∈ C∞(Ω). Our proof of harmonic =⇒ mean-value

property gave

Φ′(r) =r

n−∫

B(x,r)

∆u(y) dy,

where Φ(r) = −∫

∂B(x,r)

u(y) dy

Now Φ′(r) = 0 and thus

−∫

B(x,r)

∆u(y) dy = 0

Since ∆u is smooth, this is only possible if ∆u = 0 i.e. u is harmonic.

2.4.3 Estimates on Harmonic Functions

Theorem 2.4.9 (Local estimates for harmonic functions): Suppose u ∈ C2(Ω), u har-

monic. Then

|Dαu(x0)| ≤ Ckrn+k

∫

B(x0,r)

|u(y)| dy, |α| = k

where

Ck =

1

α(n) k = 0(2n+1nk)k

α(n) k = 1, 2, . . .

and B(x0, r) ⊂ Ω.

Remarks:

|Dα(x0)| ≤ C

r k−∫

B(x0,r)

|u(y)| dy |α| = k,

Proof: (by induction)

k = 0:

u(x0) = −∫

B(x0,r)

u(y) dy

∴ |u(x0)| ≤ 1

α(n)

1

rn

∫

B(x0,r)

|u(y)| dy

k = 1: u harmonic =⇒ u ∈ C∞(Ω)

∆uxi =∂

∂xi∆u = 0,

i.e., all derivatives of u are harmonic. Use mean value property for ∂∂xiu = uxi on B

(x0,

r2

)

uxi = −∫

B(xo , r2 )uxi︸︷︷︸

div u

dy

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r2

r

r2

x0

Note: uxi = div (0, . . . , 0, u, 0, . . . , 0) where u is in the ith component.

Use divergence theorem:

uxi =1∣∣B(x0,

r2

)∣∣∫

∂B(x0,r2 )u(y)νi dS(y)

≤ 2n

α(n)rn

∣∣∣∂B(x0,

r

2

)∣∣∣︸︷︷︸nα(n) r

n−1

2n−1

maxy∈∂B(x0,

r2 )|u(y)|

≤ 2n

r

1

α(n)

1(r2

)n maxy∈∂B(x0,

r2 )

∫

B(y, r2 )|u(z)| dz

≤ 2n+1n

α(n)rn+1

∫

B(x0,r)

|u(y)| dy

k ≥ 2:

|α| = k , then

Dαu(x) = Dβ∂

∂xiu(x),


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=⇒ |Dαu(x0)| ≤ nk

r

1

rn+k−12n

(2n+1nk)k−1

α(n)

∫

B(x0,r)

|u(t)| dt

≤ 1

rn+k

(2n+1nk)k

α(n)︸︷︷︸Ck

∫

B(x0,r)

|u(t)| dt

As a consequence of this, we get

Theorem 2.4.10 (Liouville’s Theorem): If u is harmonic on Rn and u is bounded, then u

is constant.

Proof:

|Du(x0)| ≤ C

rn+1

∫

B(x0,r)

|u(y)| dy

≤ C

rn+1M|B(x0, r)|,where M := sup

y∈Rn|u(y)|

≤ C

rM

where B(x0, r) ⊆ Rn.

As r →∞, Du(x0) = 0.

=⇒ u is constant.

2.4.4 Analyticity of Harmonic Functions

Theorem 2.4.11 (Analyticity of harmonic functions): Suppose that Ω is open (and bounded)

and that u ∈ C2(Ω), u harmonic in Ω. Then u is analytic i.e. u can be expanded locally into

a convergent power series.

Proof:

2r

r4r

Ω

Suppose that x0 ∈ Ω, show that u can be expanded into a power series at x0. i.e.

u(x) =∑

α

1

|α|Dαu(x0)(x − x0)α


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Let

r =1

4dist(x0, ∂Ω) and let

M =1

α(n)rn

∫

B(x0,2r)

|u(y)| dy

Local estimates: ∀x ∈ B(x0, r), we have B(x, r) ⊂ B(x0, 2r) ⊂ Ω, and thus

maxx∈B(x0,r)

|Dαu(x)| ≤ M1

r k(2n+1nk)k

= M

(2n+1nk

r

)k

= M

(2n+1n

r

)|α||α||α|

Stirling’s formula:

limk→∞

√kkk

k!ek=

1√2π

=⇒ kk ≤ C√2π

k!ek√k

≤ C1k!ek

Multinomial theorem:

nk =

(n∑

i=1

1

)k

=∑

|α|=k

|α|!α!

Multinomal Thm.

≥ |α|!α!

for every α with |α| = k

=⇒ |α|! ≤ nkα!

maxx∈B(x0,r)

|Dαu(x)| ≤ M

(2n+1n

r

)|α||α||α|

≤ CM

(2n+1n2

r

)|α||α|!e |α|

= CM

(2n+1en2

r

)|α||α|!

Taylor Series:

∞∑

N=0

∑

|α|=N

|Dαu(x0)|α!

(x − x0)α ≤∞∑

N=0

∑

|α|=NCM

(2n+1en2

r

)N|x − x0|α

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——

there are no more that nN multiindices α with |α| = N, thus

——

≤∞∑

N=0

CM

(2n+1en2

r

)NnN |x − x0|α

≤∞∑

N=0

CM

(2n+1en3

r

)N|x − x0|N

≤∞∑

N=0

CMqN |q| < 1

The last inequality is true only if the series is converging, i.e. for

2n+1en3

r|x − x0| < 1

=⇒ |x − x0| <r

2n+1en3

Result: The Taylor series converges in a ball with radius R = r2n+1en3 which only depends on

dist(x0, ∂Ω) and dimension.

2.4.5 Harnack’s Inequality

Theorem 2.4.12 (Harnack’s Inequality): Suppose that Ω is open and that Ω′ ⊂⊂ Ω (Ω′

is compactly contained in Ω), and Ω′ is connected. Suppose that u ∈ C2(Ω) and harmonic,

and nonnegative, then

supΩ′u ≤ C inf

Ω′u

and this constant does not depend on u.Gregory T. von Nessi

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x

2r

Ω′u(x0) = 0

Ω

4r

This means that non-negative harmonic functions can not have wild oscillations on Ω′.

Remark: infΩ′ u = 0, then supΩ′ u = 0 =⇒ u ≡ 0 on Ω′.

Ω

u(x0) = 0

Ω′

This follows also from the minimum principle for harmonic functions.

Proof of Harnack:

Let 4r = dist(Ω′, ∂Ω) and take x, y ∈ Ω′, |x−y | ≤ r =⇒ B(x, 2r) ⊂ Ω, B(y , r) ⊆ B(x, 2r).

u(x) = −∫

B(x,2r)

u(z) dz

≥ 1

|B(x, 2r)|

∫

B(y,r)

u(z) dz

=|B(y , r)||B(x, 2r)|−

∫

B(y,r)

u(z) dz

=1

2nu(y)

i.e.

u(x) ≥ 1

2nu(y)

Same argument interchanging x and y yields

u(y) ≥ 1

2nu(x)

=⇒ 2nu(x) ≥ u(y) ≥ 1

2nu(x) ∀x, y ∈ Ω′, |x − y | ≤ r

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y

Ω

Ω′

x

xk

x0

x

y0

x1x2

y

Ω′ is compact, cover Ω′ by at most N balls B(xi , r). Take x, y ∈ Ω′, take a curve γ joining

x and y .

Chain of balls covering γ, B(x0, r), . . . , B(xk , r), where k ≤ N. Now, |x − x0| ≤ r

u(x) ≥ 1

2nu(x0)

≥ 1

2n

(1

2nu(y0)

)

=1

22nu(y0)

≥ 1

23nu(x1)

...

≥ 1

2(2k+2)nu(y)

≥ 1

2(2N+2)nu(y)

∀x, y ∈ Ω′ we get

u(x) ≥ 1

2(2N+2)nu(y)

=⇒ Take xi such that u(xi)→ infx∈Ω′ u(x)

yi such that u(yi)→ supx∈Ω′ u(x)

=⇒ assertion.

Properties of harmonic functions

• mean-value propertyGregory T. von Nessi

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2.5 Existence of Solutions47

• maximum principle

• regularity

• local estimates

• analyticity

• Harnack’s inequality

2.5 Existence of Solutions

∆u = 0 in Ω

u = g on ∂ΩLaplace’s equation

−∆u = f in Ω

u = g on ∂ΩPoisson’s equation

• Green’s function → general representation formula

• Finding Green’s function for special domains, half-space and a ball

• Perron’s Method - get existence on fairly general domains from the existence on balls

2.5.1 Green’s Functions

We have in Rn

u(x) =

∫

RnΦ(y − x)f (y) dy

solves

−∆u = f in Rn

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Issue: Construct fundamental solutions with boundary conditions.

General representation formula for smooth functions u ∈ C2

Ω is a bounded, open domain. x ∈ Ω, ε > 0 such that B(x, ε) ⊂ Ω. Define Ωε = Ω\B(x, ε).

xǫ

Ω

0 =

∫

Ωε

u(y)∆Φ(y − x) dy

=

∫

∂Ωε

u(y)DΦ(y − x) · ν dS(y)

−∫

Ωε

Du(y) ·DΦ(y − x) dy

=

∫

∂Ωε

[u(y)DΦ(y − x) · ν −Du(y) · νΦ(y − x)] dS(y)

+

∫

Ωε

∆u(y)Φ(y − x) dy

Volume term for ε→ 0 gives

∫

Ω


singularity of Φ integrable.

boundary terms:

=

∫

∂Ω


+

∫

∂B(x,ε)

[u(y)DΦ(y − x) · ν︸︷︷︸I1

−Du(y) · νΦ(y − x)︸︷︷︸I2

] dS(y)

Remembering Du(y) ≤ Cεn−1 and

Φ ≤

Cεn−2 n ≥ 3

ln ε n = 2,

we see that

I2 ≤

ε n ≥ 3

ε ln ε n = 2

=⇒ I2 → 0 as ε→ 0.Gregory T. von Nessi

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Ω

x

ǫ

I1 =

∫

∂B(x,ε)

[u(y)DΦ(y − x) · νin dS(y)

DΦ(z) · νout =−1

nα(n)εn−1, z ∈ ∂B(0, ε)

DΦ(y − x) · νin =−1

nα(n)εn−1

I1 =

∫

∂B(x,ε)

u(y)1

nα(n)εn−1dS(y)

= −∫

∂B(x,ε)

u(y) dS(y)→ u(x) as ε→ 0

Putting things together:

0 =

∫

∂Ω


+u(x) +

∫

Ω


=⇒ u(x) =

∫

∂Ω

[Du(y) · νΦ(y − x)− u(y)DΦ(y − x) · ν] dS(y)

−∫

Ω


Recall: Existence of harmonic functions.

Green’s functions: Ω, u ∈ C2

0 =

∫

Ωε

u(y)∆Φ(y − x) dy

Representation of u(x)

u(x) = −∫

∂Ω

[u(y)

∂Φ

∂ν(y − x)− ∂u

∂ν(y)Φ(y − x)

]dS(y)

−∫

Ω


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Green’s function = fundamental solution with zero boundary data.

Idea: Correct the boundary values of Φ(y − x) by a harmonic corrector to be equal to

zero.

Green’s function G(x, y) = Φ(y − x)− φx(y),

where

∆yφ

x(y) = 0 in Ω

φx(y) = Φ(y − x) on ∂Ω

Same calculations as before:

u(x) = −∫

∂Ω

u(y)∂G

∂ν(x, y) dS(y)

−∫

Ω

G(x, y)∆u(y) dy

Expectation: The solution of

∆u = 0 in Ω

u = g on ∂Ω

is given by

u(x) = −∫

∂Ω

g(y)∂G

∂ν(x, y) dS(y)

Theorem 2.5.1 (Symmetry of Green’s Functions): ∀x, y ∈ Ω, x 6= y , G(x, y) = G(y , x).

Proof: Define

v(z) = G(x, z)

w(z) = G(y , z)Gregory T. von Nessi

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v harmonic for z 6= x , w harmonic for z 6= y .

v , w = 0 on ∂Ω.

xyǫ

Ω

ǫ

ε small enough such that B(x, ε) ⊂ Ω and B(y , ε) ⊂ Ω.

Ωε = Ω \ (B(x, ε) ∪ B(y , ε))

0 =

∫

Ωε

∆v · w dz

=

∫

∂Ωε

∂v

∂νw dS −

∫

Ωε

Dv ·Dw dz

=

∫

∂Ωε

(∂v

∂νw − v ∂w

∂ν

)dS

+

∫

Ωε

v · ∆w dz

Now, the last term on the RHS is 0. Thus

∫

∂B(x,ε)

(∂v

∂νw − v ∂w

∂ν

)dS =

∫

∂B(y,ε)

(v∂w

∂ν− ∂v∂νw

)dS

Now, since w is nice near x , we have∣∣∣∣∫

∂B(x,ε)

∂w

∂νv dS

∣∣∣∣ ≤ Cεn−1 sup∂B(x,ε)

|v | = o(1) as ε→ 0

On the other hand, v(z) = Φ(z − x)− φx(z), where φx is smooth in U. Thus

limε→0

∫

∂B(x,ε)

∂v

∂νw dS = lim

ε→0

∫

∂B(x,ε)

∂Φ

∂ν(x, z)w(z) dS = w(x)

by previous calculations. Thus, the LHS of (2.8) converges to w(x) as ε→ 0. Likewise the

RHS converges to v(y). Consequently

G(y , z) = w(x) = v(y) = G(x, y)

2.5.2 Green’s Functions by Reflection Methods

2.5.2.1 Green’s Function for the Halfspace

Rn+ = x ∈ Rn : xn ≥ 0

u(x) = −∫

∂Ω

u(y)∂v

∂ν(y − x) dy

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In this case: ∂Ω = ∂Rn+ = x ∈ Rn : xn = 0

Idea: reflect x into x = (x1, . . . , xn−1,−xn) and to take φx(y) = Φ(y − x)

G(x, y) = Φ(y − x)− φx(y)

= Φ(y − x)−Φ(y − x)

This works since x ∈ Rn+, x /∈ Rn+For y ∈ ∂Rn+, yn = 0, |y − x | = |y − x |. Thus, Φ(y − x) = Φ(y − x) for y ∈ ∂Rn+. Guessing

from what we showed on bounded domains:

u(x) = −∫

∂Ω

g(y)∂G

∂ν(x, y) dy,

where

∂G

∂ν= − ∂G

∂yn

Rn−1

R


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n ≥ 3:

Φ(z) =1

n(n − 2)α(n)

1

|z |n−2

=⇒ DΦ =−1

nα(n)

z

|z |n

− ∂G∂yn

(y − x) =1

nα(n)

yn − xn|y − x |n

− ∂G∂yn

(y − x) =1

nα(n)

yn + xn|y − x |n

So if y ∈ ∂Rn+,

∂G

∂ν(x, y) = − ∂G

∂yn(x, y) = − 2xn

nα(n)

1

|x − y |n

So our expectation is

u(x) = −∫

∂Rn+g(y)

∂G

∂ν(x, y) dy

=2xnnα(n)

∫

∂Rn+

g(y)

|x − y |n dy. (2.8)

This is Poisson’s formula on Rn+.

K(x, y) =2xn

nα(n)|x − y |n

is called Poisson’s kernel on Rn+.

Theorem 2.5.2 (Poisson’s formula on the half space): g ∈ C0(∂Rn+), g is bounded and

u(x) is given by (2.8), then

i.) u ∈ C∞(Rn+)

ii.) ∆u = 0 in Rn+

iii.) limx→x0 u(x) = g(x0) for x0 ∈ ∂Rn+Proof: y 7→ G(x, y) is harmonic x 6= y .

i.) symmetric G(x, y) = G(y , x)

=⇒ G is harmonic in its first argument x 6= y

=⇒ derivatives of G are harmonic for x 6= y

=⇒ our representation formula involves

∂νG(x, y) y ∈ ∂Rn+, x ∈ Rn+ =⇒ x 6= y

=⇒ ∂νG is harmonic for x ∈ Rn+.

ii.) u is differentiable

– integrate on n − 1 dimensional surface

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– decay at ∞ is of order 1|x |n =⇒ the function decays fast enough to be integrable,

and the same is true for its derivatives (since g is bounded). =⇒ differentiate

under the integral, =⇒ ∆u(x) = 0.

iii.) Boundary value for u. Fix x0 ∈ ∂Rn+, choose ε > 0, choose δ > 0 small enough such

that |g(x) − g(x0)| < ε if |x − x0| < δ, x ∈ ∂Rn+, by continuity of g. For x ∈ Rn+,

|x − x0| < δ2

|u(x)− g(x0)| =

∣∣∣∣∣2xn

mα(n)

∫

Rn+

g(y)− g(x0)

|x − y |n dy

∣∣∣∣∣ (2.9)

since∫

∂Rn+K(x, y) dy = 1 (HW# 4) (2.10)

(2.9) ≤∫

∂Rn+∩B(x0,δ)

K(x, y) |g(y)− g(x0)|︸︷︷︸≤ε

dy

︸︷︷︸≤ε by (2.10)

+

∫

∂Rn+\B(x0,δ)

K(x, y)|g(y)− g(x0)| d

R

y

x

x0

δ2

Rn−1

δ

|x − x0| <δ

2|x0 − y | > δ

|y − x0| ≤ |y − x |+ |x − x0|

≤ |y − x |+ δ

2

≤ |y − x |+ 1

2|x0 − y |

=⇒ 1

2|y − x0| ≤ |y − x |


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2nd integral

=

∫

∂Rn+\B(x0,δ)

2xnnα(n)

|g(y)− g(x0)||x − y |n dy

≤ 4x0n max |g|nα(n)

∫

∂Rn+\B(x0,δ)

(2

|x0 − y |

)ndy

≤ 4x0n max |g|nα(n)

C(δ) ≤ ε

If x0n is sufficiently small. Two estimates together: |u(x) − g(x0)| ≤ 2ε if f |x − x0| is

small enough. =⇒ continuity at x0 u(x0) = g(x0) as asserted.

2.5.2.2 Green’s Function for Balls

1. unit ball

2. scale the result to arbitrary balls

3. inversion in the unit sphere

O

x · x = |x |2|x |2 = 1

x = x|x |2 x 6= 0

x

1

Φ(y − x) =

− 1

2π ln |y − x | n = 21

n(n−2)α(n)1

|x−y |n−2 n ≥ 3

For x = 0 this simplifies to

− 1

2π ln |y |1

n(n−2)α(n)1

|y |n−2

=︸︷︷︸|y |=1

0

1n(n−2)α(n)

We choose Φ(y) to be this constant, and we assume x 6= 0 from now on. For x ∈ B(0, 1),

x /∈ B(0, 1), and Φ centered at x is a good function.

n ≥ 3 y 7→ Φ(y − x) is harmonic

y 7→ |x |2−nΦ(y − x) is harmonic

y 7→ Φ(|x |(y − x)) is harmonic

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Corrector φx(y) = Φ(|x |(y − x)) harmonic in B(0, 1)

|y | = 1⇐= y ∈ ∂B(0, 1) =⇒ |x |2|y − x |2

= |x |2∣∣∣∣y −

x

|x |2∣∣∣∣2

= |x |2∣∣∣∣|y |2 −

2(y , x)

|x |2 +|x |2|x |4

∣∣∣∣= |x |2 − 2(y , x) + |y |2

= |x − y |2

=⇒ Φ(|x |(y − x)) = Φ(y − x) if y ∈ ∂B(0, 1)

Definition 2.5.3: G(x, y) = Φ(y − x)−Φ(|x |(y − x)) is the Green’s function on B(0, 1).

Expectation: The solution of

∆u = 0 in B(0, 1)

u = g on ∂B(0, 1)

is given by

u(x) = −∫

∂B(0,1)

∂νG(x, y)g(y) dy

=1− |x |2nα(n)

∫

∂B(0,1)

g(y)

|x − y |n dy

Poisson’s formula on B(0, 1)

K(x, y) =1− |x |2nα(n)

1

|x − y |n

is called Poisson’s kernel on B(0, 1)

So now, we want to scale this result.

DΦ(z) =−1

nα(n)

z

|z |n∂Φ

∂y(y − x) =

−1

nα(n)

y − x|y − x |n

∂Φ

∂y(|x |(y − x)) =

−1

nα(n)

|x |(y − x)

||x |(y − x)|n |x |

on ∂B(0, 1). |y − x |2 = ||x |(y − x)|2

=⇒ ∂Φ

∂y(|x |(y − x)) =

−1

nα(n)

|x |2(y − x

|x |2)

|y − x |n∂G

∂y(y , x) =

−1

nα(n)

[y − x|y − x |n −

y |x |2 − x|y − x |n

]

=−1

nα(n)

[y(1− |x |2)

|y − x |n]

∂G

∂ν= y

∂G

∂y(x, y) =

−1

nα(n)

1− |x |2|x − y |n


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This gives for a solution of ∆u = 0 in B(0, 1), u(x) = g(x) on ∂B(0, 1)

u(x) =1− |x |2nα(n)

∫

∂B(0,1)

g(y)

|x − y |n DS(y)

Transform this to balls with radius r

Solve ∆u = 0 in B(0, r)

u = g on ∂B(0, r)

Define v(x) = u(rx) on B(0, 1)

=⇒ ∆v(x) = r2∆u(rx) = 0

v(x) = u(rx) = g(rx) for x ∈ ∂B(0, 1)

Poisson’s formula on B(0, 1)

v(x) =1− |x |2nα(n)

∫

∂B(0,1)

g(ry)

|x − y |n dS(y) for x ∈ B(0, 1)

u(x) = v(xr

)

=1−

∣∣ xr

∣∣2

nα(n)

∫

∂B(0,r)

g(y)∣∣ xr −

yr

∣∣ndS(y)

rn−1

=r2 − |x |2nα(n)r

∫

∂B(0,r)

g(y) dS(y)

u(x) =r2 − |x |2nα(n)r

∫

∂B(0,r)

g(y)

|x − y |n dS(y) (2.11)

Poisson’s formula on B(0, r)

Theorem 2.5.4 (): Suppose that g ∈ C(∂B(0, x)). then u(x) given by (2.11) has the

following properties:

i.) u ∈ C∞(B(0, r))

ii.) ∆u = 0

iii.) ∀x0 ∈ ∂B(0, r)

limx→x0

u(x) = g(x0), x ∈ B(0, r)

Proof: Exercise, similar to the proof in the half-space.

2.5.3 Energy Methods for Poisson’s Equation

Theorem 2.5.5 (): There exists at most one solution of

−∆u = f in Ω

u = g on ∂Ω

if ∂Ω is smooth, f , g continuous.

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Proof: Two solutions u1 and u2. Then define w = u1 − u2, thus

−∆w = 0 in Ω

w = 0 on ∂Ω

0 =

∫

Ω

−∆w · w dx

= −∫

∂Ω

Dw · ν w︸︷︷︸=0

dS(y) +

∫

Ω

|Dw |2 dx

=

∫

Ω

|Dw |2 dx

=⇒ Dw = 0 in Ω

=⇒ w = constant in Ω

=⇒ w = 0 on ∂Ω

=⇒ w = 0 in Ω

=⇒ u1 = u2 in Ω.

2.5.4 Perron’s Method of Subharmonic Functions

• Poisson’s formula on balls

• compactness argument

• construct a solution of ∆u = 0 in Ω

• Check boundary data, barriers.

Assumption: Ω open, bounded and connected.

Definition 2.5.6: A continuous function u is said to be subharmonic on Ω, if for all balls

B(x, r) ⊂⊂ Ω, and all harmonic functions h with u ≤ h on ∂B(x, r) we have u ≤ h on

B(x, r).

Definition 2.5.7: Superharmonic is analogous.

Proposition 2.5.8 (): Suppose u is subharmonic in Ω. Then u satisfies a strong maximum

principle. Moreover, if v is superharmonic on Ω with u ≤ v on ∂Ω, then either u < v on Ω,

or u = v on Ω.

Proof: u satisfies a mean-value property:

B(x, r) ⊂⊂ Ω. Let h be the solution of

∆h = 0 in B(x, r)

h = u on ∂B(x, r)

Then

u(x) ≤ h(x) = −∫

∂B(x,r)

h(y) dy = −∫

∂B(x,r)

u(y) dy

Idea: Local modification of subharmonic functions.Gregory T. von Nessi

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Definition 2.5.9: u subharmonic on Ω, B ⊂⊂ Ω. Let u be the solution of

∆u = 0 in B

u = u on ∂B

Define

U(x) =

u(x) in B

u(x) on Ω \ B

U is called the harmonic lifting of u on B. U subharmonic.

In 1D:

harmonic

u

Proposition 2.5.10 (): If u1, . . . , un are subharmonic, then u = maxu1, . . . , un is subhar-

monic.

Idea: Φ is bounded on ∂Ω. A continuous subharmonic function u is called a subfunction

relative to Φ if u ≤ Φ on ∂Ω.

Define superfunction analogously

inf∂Ω

Φ subfunction

sup∂Ω

Φ superfunction

Theorem 2.5.11 (): Φ bounded function on ∂Ω.

SΦ = v : v subfunction relative to Φ

and define u = supv∈SΦv . Then u is harmonic in Ω.

Remark: u is well-defined since w(x) = sup∂Ω Φ is a superfunction.

All subfunction ≤ all superfunctions on ∂Ω, and thus by Proposition 2.20, all subfunctions

≤ all superfunctions on Ω. Thus,

all subfunctions ≤ sup∂Ω

Φ <∞.

Proof: u is well-defined. Fix y ∈ Ω, there exists vk ∈ SΦ such that vk(y) → u(y). Replace

vk by maxvk , inf Φ if needed to ensure that vk is bounded from below. Choose

B(y , R) ⊂⊂ Ω, R > 0.

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vk = harmonic lifting of vk on B(y , R)

vk uniformly bounded

vk(y)→ u(y)

vk is harmonic in B(y , R).

HW1 (compactness result): ∃ subsequence vki that converges uniformly on all compact sub-

sets on B(y , R) to a harmonic function v . By construction, v(y) = u(y).

Assertion: u = v on B(y , R). =⇒ u harmonic on B(y , R) =⇒ u harmonic in Ω

Clearly: v ≤ u in B(y , R)

Suppose otherwise:

∃z ∈ B(y , R), v(z) < u(z)

=⇒ ∃u ∈ SΦ such that v(z) < u(z) ≤ u(z)

Take wL = maxu, vki

WL = harmonic lifting of wL on B.

As before, ∃ subsequence of WL that converges to a harmonic function w on B, by

construction:

v ≤ w ≤ u

v(y) = w(y) = u(y)

Maximum principle =⇒ v = w .

However, w(z) > v(z), contradiction.

=⇒ u = v .

2.5.4.1 Boundary behavior

Definition 2.5.12: Let ξ ∈ ∂Ω. A function ω ∈ C(Ω) is called a barrier at ξ relative to Ω if

i.) w is superharmonic in Ω

ii.) w > 0 in Ω \ ξ

Theorem 2.5.13 (): Φ bounded on ∂Ω. Let u be Perron’s solution of ∆u = 0 as constructed

before. Assume that ξ is a regular boundary point, i.e., there exists a barrier at ξ, and that

Φ is continuous at ξ. Then u is continuous at ξ and u(x)→ Φ(ξ) as x → ξ.Gregory T. von Nessi

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w(ξ) = 0

Ω

w > 0ξ

Proof: M = sup∂Ω |Φ|. w = barrier at ξ. Φ continuous at ξ. Fix ε > 0, ∃δ > 0 such that

|Φ(x)−Φ(ξ)| < ε if |x − ξ| < δ, x ∈ ∂Ω. Choose k > 0 such that kw(x) > 2M for x ∈ Ω,

and |x − ξ| > δ.

x 7→ Φ(ξ)− ε− kw(x) is a subfunction.

x 7→ Φ(ξ) + ε+ kw(x) is a superfunction.

w superharmonic =⇒ −w subharmonic.

By construction

Φ(ξ)− ε− kw(x) ≤ u(x)

≤ Φ(ξ) + ε+ kw(x)

⇐⇒ |u(x)−Φ(ξ)| ≤ ε+ kw(x)

and w(x)→ 0 as x → ξ

|u(x)−Φ(ξ)| < ε if |x − ξ| is small enough.

Theorem 2.5.14 (): Ω bounded, open, connected and all ξ ∈ ∂Ω regular. g continuous on

∂Ω =⇒ there exists a solution of

∆u = 0 in Ω

h = g on ∂Ω

Remark: A simple criterion for ξ ∈ ∂Ω to be regular is that Φ satisfies an exterior sphere

condition: there exists B(y , R), R > 0 such that B(y , R) ∩Ω = ξ

R

ξ

y

Barrier:

w(x) =

R2−n − |x − y |2−n n ≥ 3

ln |x−y |R n = 2

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2.6 Solution of the Dirichlet Problem by the Perron Process

2.6.1 Convergence Theorems

Important to the proof of existence of solutions to Laplace’s equation is the concept of

convergence of harmonic functions.

Theorem 2.6.1 (): A bounded sequence of harmonic functions on a domain Ω contains a

subsequence which converges uniformly, together with its derivatives, to a harmonic function.

Proof: Take Ω′ ⊂ Ω compact. Now, we calculate via the mean value theorem of calculus:

|um(x)− un(y)| ≤ maxtx + (1− t)y

0 ≤ t ≤ 1

|Dum| · |x − y |

≤ maxΩ′|Dum| · |x − y |

≤ C supΩ′|um| · |x − y |

≤ C′(K, n, L)|x − y |,where L is a constant which bounds the sequence elements um. From this, we see that

the sequence is equicontinuous. Next we recall that the Azalea-Ascoli theorem asserts that

such a sequence will have a subsequence converging uniformly to a limit. Now, one may

apply the same calculation above iteratively to any order of derivative (iteratively refining the

subsequence for each new derivative). From this we conclude that there is a subsequence

whose derivatives up to and including order |β| all converge uniformly to a limit for any |β|.It now follows that the limit is indeed a solution to Laplace’s equation.

2.6.2 Generalized Definitions & Harmonic Lifting

Before proving the existence of solution to the Dirichlet problem for Laplace’s equation, one

needs to extend the notions of subharmonic(superharmonic) functions to non-C2 functions.

Definition 2.6.2: u ∈ C0(Ω) is subharmonic(superharmonic) in Ω if for any ball B ⊂⊂ Ω

and harmonic function h ∈ C2(B) ∩ C0(B) with h ≥ (≤) u on ∂B, one has h ≥ (≤) u on B.

Note: If u is subharmonic and superharmonic, then it is harmonic.

From this new definition, some properties are immediate.

Properties:

i.) The Strong Maximum Principle still applies to this extended definition. Since u is

subharmonic, u ≤ h for h harmonic in B with the same boundary values, the Strong

Maximum Principle follows by applying the Poisson Integral formula to this situation.

ii.) w(x) = maxm(minm)u1(x), . . . , um(x) is subharmonic(superharmonic) if u1, . . . , umare subharmonic(superharmonic).

iii.) The Harmonic Lifting on a ball B for a subharmonic function u on Ω is defined as

U(x) =

u(x) for x /∈ Bh(x) for x ∈ B ,

where h ∈ C2(B) ∩ C0(B) is harmonic in B with h = u on ∂B.Gregory T. von Nessi

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2.6 Solution of the Dirichlet Problem by the Perron Process63

Note: The key thing to realize about the definition of harmonic lifting is that the “lifted”

part is harmonic in our classical C2 sense; it is simply the Poisson integral solution on the

ball with u taken as boundary data.

Proposition 2.6.3 (): U is subharmonic on Ω, when h is harmonic in B.

B′∩B

B′B

B′ \ B

Proof: Refer to the figure on the next page as you read this proof. Take another ball

B′ ⊂⊂ Ω along with another harmonic function h′ ∈ C2(B′) ∩ C0(B′) which is ≥ U on ∂B′.By definition of u(x), one has that h′ ≥ u on B′ \ B. This, in turn, leads to the assertion

that h′ ≥ U on ∂B ∩ B′, i.e. h′ ≥ U on ∂(B′ ∩ B). As, h′ − U is harmonic (in the classical

sense), we can apply the Weak Maximum Principle to ascertain that h′ ≥ U on B∩B′. Since

B′ is arbitrary, the proposition follows.

2.6.3 Perron’s Method

Now, we can start pursuit of the main objective of this section, namely to prove the existence

of the Dirichlet Problem:

∆u = 0 in Ω,Ω bounded domain in Rnu = φ on ∂Ω, φ ∈ C0(∂Ω)

The first step to doing this requires the following definition

Definition 2.6.4: A subfunction(superfunction) in Ω is subharmonic(superharmonic) in Ω

and is in C0(Ω) in addition to being ≤ φ on ∂Ω.

Now, we define

Sφ = all subfunctions,along with

u(x) = supv∈Sφ

v(x),

where the supremum is understood to be taken in the pointwise sense. u(x) is bounded as

v ∈ Sφ implies that v ≤ max∂Ω φ via the Weak Maximum Principle.

Remark: Essentially, the above definitions allow one to construct a weak formulation (or

weak solution) to the Dirichlet Problem via Maximum Principles. Indeed, at this point, it

has not been ascertained whether or not u is continuous.

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Theorem 2.6.5 (): u is harmonic in Ω.

Proof: First, fix y ∈ Ω. Since u(y) = supSφ v , there is a sequence vm ⊂ Sφ such that

vm(y) → u(y). Of course, we may assume that vm is bounded from above by max∂Ω φ.

Now, consider a ball B ⊂⊂ Ω and replace all vm by their harmonic liftings on B. Denoting

the resulting sequence as Vm, it is clear that Vm ⊂ Sφ and that Vm(y) → u(y). Since

Vm are all harmonic (in the C2 sense) in B, we apply our harmonic convergence theorem

to extract a subsequence (again denoted Vm after relabeling) which converges uniformly in

compact subset of B to a harmonic function v in B. Obviously, we have v(y) = u(y), but

we now make the following

Claim: v = u in B.

Proof of Claim: Consider z ∈ B such that v(z) < u(z). This implies that there exists

a function v ∈ Sφ such that v(z) > v(z). Replace Vm by vm = maxVm, v. Next, take the

harmonic liftings in B of these to get Vm. Again, utilizing the harmonic convergence theo-

rem, we extract a subsequence (still denoted Vm after relabeling) such that Vm converges

locally uniformly to a harmonic function w > v in B, but w(y) = v(y) = u(y). Thus, by

the Strong Maximum Principle w = v , a contradiction.

Thus, v = u in B which implies that u is indeed harmonic.

Now, to complete our existence proof, we need to see if the Perron solution of the Dirichlet

Problem attains the proper boundary data. To do this we make the next definition.

Definition 2.6.6: A Barrier at y ∈ ∂Ω is a function w ∈ C0(Ω) that has the following

properties:

i.) w is superharmonic,

ii.) w(y) = 0,

iii.) w > 0 on Ω \ y.

Theorem 2.6.7 (): Let u be a Perron solution. Then u ∈ C0(Ω) with u = φ on ∂Ω if and

only if there exists a barrier at each point y ∈ ∂Ω.

Proof:

(Only if) Solve with the BC φ = |x − y |.

(If) Take ε > 0 which implies there exists a δ > 0 such that |φ(x) − φ(y)| < ε implies

|x − y | < δ by continuity of φ. Now, choose a constant K(ε) large enough so that

Kw > 2 · max∂Ω |φ| if |x − y | ≥ δ. Next, define w± := φ(y) ± (ε + Kw). It is clear

that w− is a subfunction and that w+ is a superfunction with w− ≤ u ≤ w+ on ∂Ω.

Application of the Weak Maximum Principle now yields

|u(x)− φ(y)| ≤ ε+Kw.

Taking ε→ 0, one sees that u(x)→ φ(y) as x → y (since w(x)→ 0 as x → y by the

definition). Gregory T. von Nessi

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2.7 Exercises65

Ω

S

y

Now, that it has been shown the existence of Perron solutions is equivalent to the existence

of barriers on ∂Ω, we wish to ascertain precisely what criterion on ∂Ω implies the existence

of barriers. With that, we make the following definition.

Definition 2.6.8: A domain Ω satisfies an exterior sphere condition if at every point y ∈ ∂Ω,

there exists a sphere S such that S ∩Ω = y (refer to the above figure).

Correspondingly, we have the following theorem.

Theorem 2.6.9 (): If Ω satisfies an exterior sphere condition, then there exists a barrier at

each point y ∈ ∂Ω.

Proof: Simply take

w(x) =

R2−n − |x − y |2−n for n > 2

ln R|x−y | for n = 2

,

as the barrier at y ∈ ∂Ω. It is trivial to verify this function is indeed a barrier at y ∈ ∂Ω.

Remarks:

i.) It is easy to see that if the boundary ∂Ω is smooth, in a sense that locally it is the

graph of a C2 function, then it satisfies an exterior sphere condition. Also, all convex

domains satisfy an exterior sphere condition.

ii.) There are more general criterion for the existence of barriers, like the exterior cone

condition; but the proofs are harder.

2.7 Exercises

2.1:

a) Find all solutions of the equation ∆u+ cu = 0 with radial symmetry in R3. Here c ≥ 0

is a constant.

b) Prove that

Φ(x) = −cos(√c |x |)

4π|x |

is a fundamental solution for the equation, i.e., show that for all f ∈ C2(R3) with

compact support a solution of the equation ∆u + cu = f is given by

u(x) =

∫

R3

Φ(x − y)f (y) dy.

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Hint: You can solve the ODE you get in a) by defining w(r) = rv(r). Show that w solves

w ′′ + cw = 0.

2.2: [Evans 2.5 #4] We say that v ∈ C2(Ω) is subharmonic if

−∆v ≤ 0 in Ω.

a) Suppose that v is subharmonic. Show that

v(x) ≤ 1

|B(x, r)|

∫

B(x,r)

v(y) dy for all B(x, r) ⊂ Ω.

b) Prove that subharmonic functions satisfy the maximum principle,

maxΩv = max

∂Ωv .

c) Let φ : R → R be smooth and convex. Assume that u is harmonic and let v = φ(u).

Prove that v is subharmonic.

d) Let u be harmonic. Show that v = |Du|2 is subharmonic.

2.3: Let Rn+ be the upper half space, Rn+ = x ∈ Rn : xn > 0.

a) Suppose that u is a smooth solution of ∆u = 0 in Rn+ with u = 0 on ∂Rn+.

b) Use the result in a) to show that bounded solutions of the Dirichlet problem in the half

space Rn+,

∆u = 0 in Rn+,u = g on ∂Rn+,

are unique. Show by a counter example that the corresponding statement does not

hold for unbounded solutions.

2.4: [Qualifying exam 08/98] Suppose that u ∈ C2(R2+) ∩ C0(R2

+) is harmonic in R2+ and

bounded in R2+ ∪ ∂R2

+. Prove the maximum principle

supR2

+

u = sup∂R2

+

u

Hint: Consider ε > 0 the harmonic function

vε(x, y) = u(x, y)− ε ln(x2 + (y + 1)2).

2.5: Fundamental solutions and Green’s functions in one dimension.

a) Find a fundamental solution Φ of Laplace’s equation in one dimension,

−u′′ = 0 in (−∞,∞).

Show that

u(x) =

∫ ∞

−∞Φ(x − y)f (y) dy

is a solution of the equation −u′′ = f for all f ∈ C2(R) with compact support.Gregory T. von Nessi

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2.7 Exercises67

b) Find the Green’s function for Laplace’s equation on the interval (−1, 1) subject to the

Dirichlet boundary conditions u(−1) = u(1) = 0.

c) Find the Green’s function for Laplace’s equation on the interval (−1, 1) subject to the

mixed boundary conditions u′(−1) = 0 and u(1) = 0, i.e., find a fundamental solution

Φ with Φ′(−1) = 0 and Φ(1) = 0.

d) Can you find the Green’s function for the Neumann problem on the interval (−1, 1),

i.e., can you find a fundamental solution Φ with Φ′(−1) = Φ′(1) = 0?

2.6: Use the method of reflection to find the Green’s function for the first quadrant R2++ =

x ∈ R2 : x1 > 0, x2 > 0 in R2. Find a representation of a solution u of the boundary value

problem

∆u = 0 in R2

++,

u = g on ∂R2++,

where g is a continuous and bounded function on ∂R2++.

2.7: Let Ω1 and Ω2 be open and bounded sets in Rn (n ≥ 3) with smooth boundary

that are connected. Suppose that Ω1 is compactly contained in Ω2, i.e., Ω1 ⊂ Ω2. Let

Gi(x, y) be the Green’s functions for the Laplace operator on Ωi , i = 1, 2. Show that

G1(x, y) < G2(x, y) for x, y ∈ Ω1, x 6= y .

2.8: [Evans 2.5 #6] Use Poisson’s formula on the ball to prove that

rn−2 r − |x |(r + |x |)n−1

u(0) ≤ u(x) ≤ rn−2 r + |x |(r − |x |)n−1

u(0),

whenever u is positive and harmonic in B(0, r). This is an explicit form of Harnack’s inequality.

2.9: [Qualifying exam 08/97] Consider the elliptic problem in divergence form,

Lu = −n∑

i ,j=1

∂i(ai j(x)∂ju)

over a bounded and open domain Ω ⊂ Rn with smooth boundary. The coefficients satisfy

ai j ∈ C1(Ω) and the uniform ellipticity condition

n∑

i ,j=1

ai j(x)ξiξj ≥ λ0|ξ|2 ∀x ∈ Ω, ξ ∈ Rn

with a constant λ0 > 0. A barrier at x0 ∈ ∂Ω is a C2 function w satisfying

Lw ≥ 1 in Ω, w(x0) = 0, and w(x) ≥ 0 on ∂Ω.

a) Prove the comparison principle: If Lu ≥ Lv in Ω and u ≥ v on ∂Ω, then u ≥ v in Ω.

b) Let f ∈ C(Ω) and let u ∈ C2(Ω) ∩ C(Ω) be a solution of

Lu = f in Ω,

u = 0 on ∂Ω.

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Show that there exists a constant C that does not depend on w such that

∣∣∣∣∂u

∂ν(x0)

∣∣∣∣ ≤ C∣∣∣∣∂w

∂ν(x0)

∣∣∣∣

where ν is the outward normal to ∂Ω at x0.

c) Let Ω be convex. Show that there exists a barrier w for all x0 ∈ ∂Ω.


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70Chapter 3: The Heat Equation

3 The Heat Equation

“. . . it’s dull you TWIT; it’ll ’urt more!”

– Alan Rickman

Contents

3.1 Fundamental Solution 70

3.2 The Non-Homogeneous Heat Equation 74

3.3 Mean-Value Formula for the Heat Equation 76

3.4 Regularity of Solutions 82

3.5 Energy Methods 85

3.6 Backward Heat Equation 86

3.7 Exercises 86

3.1 Fundamental Solution

ut −4u = 0

Evan’s approach: find a solution that is invariant under dilation scaling u(x, t) 7→ λαu(λβx, λt).

Want u(x, t) = λαu(λβx, λt) for all λ > 0 and α, β constant that need to be determined.

Idea: λ = 1t , i.e. find

u(x, t) =1

tαu( xtα, 1)

=1

tαv( xtβ

)


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3.1 Fundamental Solution71

and let y = xtβ

. Find an equation for v .

∂

∂tu(x, t) =

−αtα+1

v( xtβ

)+

1

tα

n∑

j=1

∂v

∂xj

( xtβ

)· −βxjtβ+1

=−αtα+1

v( xtβ

)− β

tα+1Dxv

( xtβ

)· xtβ

∂

∂xiu(x, t) =

1

tα∂v

∂xi

1

tβ

∂2

∂x2i

u(x, t) =1

tα∂2v

∂xi

1

t2β

ut −4u = 0

=⇒ α

tα+1v(y) +

β

tα+1Dv(y) · y +

1

tα+2β4v = 0

λ = 1t , β = 1

2

αv(y) +1

2Dv · y +4u = 0

Assume: v is radial, v(y) = w(r) = w(|y |)

Diw(|y |) = w ′(|y |) yi|y |

Dv(y) · y = w ′(|y |) |y |2

|y | = w ′(r)r

Equation for w :

αw +1

2w ′(r)r + w ′′ +

n − 1

rw ′ = 0

α = n2 :

n

2w +

1

2w ′r + w ′′ +

n − 1

rw ′ = 0

xrn−1:

n

2rn−1w +

1

2rnw ′

︸︷︷︸12

(rnw)′

+ rn−1w ′′ + (n − 1)rn−2w ′︸︷︷︸(rn+1w ′)′

= 0

Integrate:

1

2rnw + rn−1w ′ = a = constant

Assume: rnw → 0, rn−1w ′ → 0 as r →∞, then a = 0 and

rn−1w ′ = −1

2rnw

w ′ = −1

2rw =⇒ w(r) = be−

14r2

v(y) = w(|y |) = be−|y |2

4 y =x√t

u(x, t) =1

tαv(y) =

b

tn2

e−|x |24t

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Definition 3.1.1: The function

Φ(x, t) =

1

(4πt)n2e−

|x |24t t > 0

0 t < 0

is called the fundamental solution of the heat equation.

Motivation for choice of b:

Lemma 3.1.2 ():

∫

RnΦ(y , t) dy = 1, t > 0

Proof:

(∫ ∞

−∞e−x

2

dx

)2

=

∫ ∞

−∞

∫ ∞

−∞e−x

2−y2

dydx

=

∫ 2π

0

∫ ∞

0

e−r2

r drdθ

= 2π

(−1

2e−r

2

) ∣∣∣∣∣

∞

0= π

∫

RnΦ(y , t) dy =

1

(4πt)n+2

∫

Rne−

|y |24t dy

Change coordinates: z = y√4t

, dz = 1

(4t)n2dy

∫

RnΦ(y , t) dy =

1

πn+2

∫

Rne−|z |

2

dz = 1.

Initial value problem for the heat equation in Rn: want to solve

ut −4u = 0 in Rn × 0,∞)

u = g on Rn × t = 0

Theorem 3.1.3 (): Suppose g is continuous and bounded on Rn, define

u(x, t) =

∫

RnΦ(x − y , t)g(y) dy

=1

(4πt)n2

∫

Rne−|x−y |2

4t g(y) dy

for x ∈ Rn, t > 0. Then

i.) u ∈ C∞(Rn \ (0,∞))

ii.) ut −4u = 0 in Rn × (0,∞)

iii.) lim(x,t)→(x0,0) u(x, t) = g(x0) if x0 ∈ Rn, x ∈ Rn, t > 0Gregory T. von Nessi

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3.2 The Non-Homogeneous Heat Equation73

Remark

i.) “smoothening”: the solution is C∞ for all positive times, even if g is only continuous.

ii.) Suppose that g > 0, g ≡ 0 outside B(0, 1) ∈ Rn, then u(x, t) > 0 in (Rn × (0,∞)),

“infinite speed of propagation”.

Proof: e−|x−y |2

4t is infinitely many times differentiable

i.) in x, t, the integral exists, and we can differentiate under the integral sign.

ii.) ut −4u = 0 since Φ(y , t) is a solution

iii.) Boundary data:

ε > 0, choose δ > 0 such that |g(x0)−g(x)| < ε for |x−x0| < δ, Then for |x−x0| < δ2

|u(x, t)− g(x0)| =

∣∣∣∣∫

RnΦ(x − y)[g(y)− g(x0)] dy

]

=

∫

B(x0,δ)

Φ(y + x, t)|g(y)− g(x0)| dy

+

∫

Rn\B(x0,δ)

Φ(x − y , t) |g(y)− g(x0)|︸︷︷︸≤C

dy

|x − x0| < δ2 , |y − x0| > δ

=⇒ (see Green’s function on half space)

12 |x0 − y | ≤ |x − y | Then

1

(4πt)n2

∫

Rn\B(x0,δ)

e−|x−y |2

4t dy ≤ 1

(4πt)n2

∫

Rn\B(x0,δ)

e−|x0−y |2

16t dy

≤ C

(4πt)n2

∫ ∞

0

e−r216t rn−1 dr → 0

as t → 0

|u(x, t) − g(x0)| < 2ε if |x − x0| < δ2 and t small enough, if |(x, t) − (x0, t)| is

small enough.

Formally: Φt −4Φ = 0 in Rn × (0,∞)

Φ as t → 0

∫

RnΦ(x, t) dx = 1 for t > 0.

Formally: Φ(x, 0) = δx , i.e., the total mass is concentrated at x = 0.

3.2 The Non-Homogeneous Heat Equation

Non-homogeneous equation:

ut − ∆u = f in Rn × t > 0u = 0 on Rn × t = 0

(3.1)

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Rn

Φ(x, t)


Φ(x − y , t − s) is a solution of the heat equation. For fixed x > 0, the function

u(x, t; s) =

∫

RnΦ(x − y , t − s)f (y , s) dy

Solves

ut(·; s)− ∆u(·; s) = 0 in Rn × s,∞u(·; s) = f (·, s) on Rn × t = s

Duhamel’s principle: the solution of (3.1) is given by

u(x, t) =

∫ t

0

u(x, t; s) ds ⇐⇒

u(x, t) =

∫ t

0

∫

RnΦ(x − y , t − s)f (y , s) dyds (3.2)

Compare to Duhamel’s principle applied to the transport equation.

Remark: The solution of

ut − ∆u = f in Rn × t > 0u = g on Rn × t = 0

(3.3)

is given by

u(x, t) =

∫

RnΦ(x − y , t)g(y) dy +

∫ t

0

∫

RnΦ(x − y , t − s)f (y , s) dyds

Proof: See proof of following theorem.

Theorem 3.2.1 (): Suppose that f ∈ C21 (Rn × (0,∞)) (i.e. f is twice continuously differ-

entiable in x , f is continuously differentiable in t), f has compact support, then the solution

of the non-homogeneous equation is given by (3.2).

Proof: Change coordinates

u(x, t) =

∫ t

0

∫

RnΦ(y , s)f (x − y , t − s) dyds


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3.3 Mean-Value Formula for the Heat Equation75

We can differentiate under the integral sign:

∂u

∂t(x, t) =

∫ t

0

∫

RnΦ(y , s)

∂f

∂t(x − y , t − s) dyds +

∫

RnΦ(y , t)f (x − y , 0) dy

∂2u

∂xi∂xj(x, t) =

∫ t

0

∫

RnΦ(y , s)

∂2f

∂xi∂xj(x − y , t − s) dyds

and all these derivatives are continuous.

ut − ∆u =

∫ t

0

∫

RnΦ(y , s)(∂t − ∆x)f (x − y , t − s) dyds

+

∫

RnΦ(y , t)f (x − y , 0) dy

︸︷︷︸K

=

∫ t

ε

∫

RnΦ(y , s)(−∂s − ∆y )f (x − y , t − s) dyds

+

∫ ε

0

∫

RnΦ(y , s)(−∂s − ∆y )f (x − y , t − s) dyds +K

= Iε + Jε +K

|Jε| ≤ (sup |∂t f |+ sup |D2f |) ·∫ ε

0

∫

RnΦ(y , s) dy

︸︷︷︸1

ds

= ε→ 0 as ε→ 0

Iε =

∫ t

ε

∫

Rn(∂s − ∆y )Φ(y , s)︸︷︷︸

=0

f (x − y , t − s) dyds

−∫

RnΦ(y , t)f (x − y , 0) dy

︸︷︷︸−K

+

∫

RnΦ(y , ε)f (x − y , t − ε) dy

Collecting the terms:

(ut − ∆u)(x, t) = limε→0

∫

RnΦ(y , ε)f (x − y , t − ε) dy

= f (x, t)

Calculation similar to last class.

3.3 Mean-Value Formula for the Heat Equation

u(x, t) =1

4rn

∫∫

E(x,t;r)

u(y , s)|x − y |2(t − s)2

dyds

Analogy: Laplace ∂B(x, r), B(x, r)

∂B(x, r) = level set of the fundamental solution.

Expectation: level sets of Φ(x, t) relevant.

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Definition 3.3.1: x ∈ Rn. t ∈ R, r > 0. Define the heat ball

Rn

t

(x, t)

x

E(x, t; r) =

(y , s) ∈ Rn × R

∣∣∣ s ≤ t, Φ(x − y , t − s) ≥ 1

rn

x, t is in the boundary of E(x, t; r)

Definition 3.3.2: 1. Parabolic cylinder: Ω open, bounded set, then

ΩT = Ω× (0, T ]

ΩT includes the top surface

Rn

t

ΩΓT

2. Parabolic boundary: ΓT := ΩT \ΩT = bottom surface + lateral boundary

(This body is the relevant boundary for the maximum principle).

Theorem 3.3.3 (Mean value formula): Ω bounded, open

u ∈ C21 (ΩT ) solution of the heat equation. Then

u(x, t) =1

4rn

∫∫

E(x,t,r)

u(y , s)|x − y |2(t − s)2

dyds


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∀E(x, t, r) ⊂ ΩT

Proof: Translate in space and time to assume x = 0, t = 0.

E(r) = E(0, 0, r)

φ(r) =1

4rn

∫∫

E(r)

u(y , s)|y |2|s|2 dyds

Idea: compute φ′(r).

E(r):

φ(−y − s) > 1

⇐⇒ 1

(4π(−s))n2

e−|y |2−4s ≥ 1

⇐⇒ 1(

4π−sr2

) n2

e−(y/r)2/(−4(s/r2)) ≥ 1

h(y , s) = (ry , r2s) maps E(1) to E(r).

Change coordinates:

1

4rn

∫∫

E(r)

u(y , s)|y |2s2

dyds =1

4rn

∫∫

h(E(1))

u(y , s)|y |2s2

dyds

[∫

h(Ω)

f (x) dx =

∫

Ω

(f h)(x) · |det(Dh)| dx]

=1

4rn

∫∫

E(1)

u(ry , r2s)|ry |2

(r2s)2rn+2 dyds

=1

4

∫∫

E(1)

u(ry , r2s)|y |2s2

dyds

φ′(r) =1

4

∫∫

E(1)

∑

i

∂u

∂yi(ryi , r

2s) · yi|y |2s2

+∂u

∂s(ry , r2s) · 2r s |y |

2

s2

dyds

=1

4

∫∫

E(r)

Du(y , s) · y

r

(y/r)2

(s/r2)2+∂u

∂s(y , s) · 2r |y/r |

2

s/r2

1

rn+2dyds

=1

4rn+1

∫∫

E(r)

Du(y , s) · y |y |

2

s2+ 2

∂u

∂s(y , s)

|y |2s

dyds

=: A+ B

Heat ball defined by φ(−y ,−s) = 1rn

⇐⇒ 1

(rπ(−s))n2

e−|y |2/(−4s) =

1

rn

Now, we define

ψ(y , s) =n

2ln(−4πs) +

|y |24s

+ n ln r

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It’s observed that ψ = 0 on ∂E(0, 0, r) which implies ψ(y , s) = 0 on ∂E(s)

Dψ(y) =y

2s

Rewrite B:

1

4rn+1

∫∫

E(r)

4∂u

∂s(y , s) · y ·Dψ dyds

= 0− 1

4rn+1

∫∫

E(r)

n∑

i=1

(4∂2u

∂s∂yiyiψ + 4

∂u

∂sψ

)dyds

= − 1

4rn+1

∫∫

E(r)

4

n∑

i=1

∂2u

∂s∂yiyiψ + 4n

∂u

∂sψ

dyds

=1

4rn+1

∫∫

E(r)

4

n∑

i=1

∂u

∂yiyi∂ψ

∂s− 4n

∂u

∂sψ

dyds

=1

4rn+1

∫∫

E(r)

4

n∑

i=1

[∂u

∂yiyi

(− n

2s− |y |

2

4s2

)− n∂u

∂sψ

]dyds

=1

4rn+1

∫∫

E(r)

−4n

∂u

∂s· ψ −

n∑

i=1

2n

s

∂u

∂yi· yidyds

Collecting our terms

φ′(r) = A+ B

=1

4rn+1

∫∫

E(r)

−4n∆u · ψ − 2n

s

n∑

i=1

∂u

∂yi· yidyds

=1

4rn+1

∫∫

E(r)

4nDu ·Dψ − 2n

s

n∑

i=1

∂u

∂yi· yidyds

= 0

=⇒ φ is constant

=⇒ φ(r) = limr→0

φ(r)

= u(0, 0) limr→0

1

4rn

∫∫

E(r)

|y |2s2

dyds

︸︷︷︸1

= u(0, 0)

Recall:

MVP :1

4rn

∫∫

E(x,t,r)

u(y , s)|x − y |2(t − s)2

dyds = u(x, t)

1

4rn

∫∫

E(x,t,r)

|x − y |2(t − s)2

dyds = 1

Theorem 3.3.4 (Maximum principle for the heat equation on bounded and open do-

mains): u ∈ C21 (ΩT ) ∩ C(ΩT ) solves the heat equation in ΩT . Then


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i.)

maxΩT

u = maxΓT

u max. principle

ii.) Ω connected, ∃(x0, t0) ∈ ΩT such that u(x0, t0) = maxΩTu then u is constant on ΩT

(strong max. principle).

Remarks:

1. In ii.) we can only conclude for t ≤ t0.

2. Since −u is also a solution, we have the same assertion with max replaced by min.

3. Heat conducting bar:

ΓT

T

Ω

ΩT

No heat concentration in the interior of the bar for t > 0.

Proof: Suppose u(x0, t0) = M = maxΩTu, with x0, t0 /∈ ΓT . For r > 0 small enough,

E(x0, t0, r) ⊂ ΩT . By the MVT:

M = u(x0, t0) =1

4rn

∫∫

E(x0,t0,r)

u(y , r)|x − y |2(t − s)2

dydS

≤ M

Suppose (y0, s0) ∈ ΩT , s0 < t0, line segment L connecting (y0, s0) and (x0, t0) in ΩT .

Rn

s0

t0

t

(y0, s0)

u(x0, t0) = M(x0, t0)

Define r0 as

mins ≥ s0

∣∣∣ u(x, t) = M, ∀(x, t) ∈ L, s ≤ t ≤ t0

= r0

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x0

x

Either r0 = s0 =⇒ u(y0, s0) = u(x0, t0) or r0 > s0. Then for r small enough, E(u0, s0, r) ⊆ΩT . Argument above =⇒ u ≡ constant on E(u0, s0, r). E(y0, s0, t) contains a part of L =⇒contradiction with definition of r0.

=⇒ u ≡ M for all points (y0, s0) connected to (x0, t0) by line segments.

General Case: (x, t) ∈ ΩT , t < t0, Ω connected.

Choose a polygonal path x0, x1, . . . , xn = x in Ω.

t0 > t1 > · · · > tn = t

=⇒ segments (xi , ti) + (xi+1, ti+1) in ΩT

=⇒ u(x, t) = u(x0, t0).

Theorem 3.3.5 (Uniqueness of solutions on bounded domains): If g ∈ C(ΓT ), f ∈ C(ΩT )

then there exists at most one solution to

ut − ∆u = f in ΩT

u = g on ΓT

Proof: If u1 and u2 are solutions then v = ±(u1 − u2) solve

vt − ∆v = 0 in ΩT

u = 0 on ΓT

Now we use the maximum principle.

Theorem 3.3.6 (Maximum principle for solutions of the Cauchy problem):

ut − ∆u = 0 in Rn × (0, T )

u = g on Rn × t = 0

Suppose u ∈ C21 (Rn × (0, T )) ∩ C(Rn × [0, T ]) is a solution of the heat equation with

u(x, t) ≤ Aea|x |2 for x ∈ Rn, t ∈ [0, T ], and A, a > 0 constants. Then

supRn×[0,T ]

u = supRnu = sup

Rng

Remark: Theorem not true without the exponential bound, [John].

Proof: Suppose that 4aT < 1, 4a(T + ε) < 1, ε > 0 small enough, a < 14(T+ε) , there

exists γ > 0 such that

a + γ =1

4(T + ε)Gregory T. von Nessi

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Fix y ∈ Rn, µ > 0 and define

v(x, t) = u(x, t)− µ

(T + ε− t) n2e|x−y |2

4(T+ε−t) x ∈ Rn, t > 0

Check: v is a solution of the heat equation since since Φ(x, t) is a solution.

|x − y | = r |x | ≤ |y |+ r

y rr

T

Ω = B(y , r) r > 0

ΩT = B(y , r)× (0, T ]

Use the maximum principle for v on the bounded set ΩT .

maxΩT

v ≤ maxΓT

v

on the bottom:

v(x, 0) = u(x, 0)− µ

(T + ε)n2

e|x−y |24(T+ε) x ∈ Rn

︸︷︷︸≥0

≤ u(x, 0)

≤ supRng

v(x, t) = u(x, t)− µ

(T + ε− t) n2e|x−y |2

4(T+ε−t) x ∈ Rn

≤ Aea|x |2 − µ

(T + ε)n2

er2

4(T+ε) x ∈ Rn

≤ Aea(|y |+r)2 − µ

(T + ε)n2

er2

4(T+ε) x ∈ Rn[a + γ =

1

4(T + ε)

1

T + ε= 4(a + γ)

]

= Aea(|y |+r)2 − µ(4(a + γ))n2 e(a+γ)r2

x ∈ Rn

≤ supRng if we choose r big enough.

Leading order term e(a+γ)r2dominates ear

2for r large enough. If r big enough, then

maxΩT

v ≤ supRng =⇒ max

Rn×[0,T ]v ≤ sup

Rng

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Now µ→ 0 and get

maxRn×[0,T ]

u ≤ supRng

If 4aT > 1, choose t0 > 0 such that 4at0 < 1, apply above argument on [0, t0], [t0, 2t0], . . .

Theorem 3.3.7 (Uniqueness of solutions for the Cauchy problem): There exists at most

one solution u ∈ C21 (Rn × (0, T ]) ∩ C(Rn × [0, T ]) of

ut − ∆u = f in Rn × (0, T ]

u = g on Rn × t = 0

which satisfies that

|u(x, t)| ≤ Aea|x |2 x ∈ Rn, t ∈ [0, T ]

Proof: Suppose u1 and u2 are solutions. Define w = ±(u1 − u2) solves

wt − ∆w = 0 in Rn × (0, T ]

w = 0 on Rn × t = 0

|w(x, t)| ≤ 2Aea|x |2

=⇒ max principle applies

=⇒ w = 0

=⇒ uniqueness.

3.4 Regularity of Solutions

Theorem 3.4.1 (): Ω open, u ∈ C21 (ΩT ), is a solution of the heat equation. Then u ∈

C∞(ΩT ).

Proof: Regularity is a local issue.

Parabolic cylinders

C(x, t, r) =

(y , s) ∈ Rn × R∣∣∣ |x − y | ≤ r, t − r2 ≤ s

Fix (x0, t0) ∈ ΩT

v(x, t) = u(x, t) · ξ(x, t)

ξ cut-off function.

0 ≤ ξ on ΩT

ξ ≡ 1 on C

(x0, t0,

3

4r

)

ξ ≡ 0 outside C, i.e. on ΩT \ Cξ ∈ C∞(ΩT )

Goal: Representation for u:

u(x, t) =

∫∫K(x, y , t, s)u(y , s) dyds


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3.4 Regularity of Solutions83

for r small enough

(x0, t0)

C(x0, t0,

r2

)= C ′′

C(x0, t0,

34r)= C ′

C (x0, t0, r) = C

C(x0, t0, r) ⊂ ΩT

with a smooth kernel =⇒ u as good as K.

vt = utξ + uξt

Dv = Du · ξ + uDξ

∆v = ∆u · ξ + sDu ·Dξ + u∆ξ

vt − ∆u = uξt − 2Du ·Dξ − u∆ξ = f

A solution v of this equation is given by Duhamel’s principle,

v(x, t) =

∫ t

0

∫


Uniqueness for the Cauchy problem

v(x, t) =

∫ t

0

∫


Suppose for the moment being that u is smooth.

(x, t) ∈ C′′, ξ(x, t) = 1

=⇒ v(x, t) = u(x, t)ξ(x, t) = u(x, t)

u(x, t) =

∫ t

0

∫

Rn︸︷︷︸C\C′

Φ(x − y , t − s)[uξs − 2Du ·Dξ − u∆ξ](y , s) dyds

=

∫∫

C\C′Φ(x − y , t − s)[uξs − u∆ξ + 2u∆ξ] dyds

+

∫∫

C\C′DΦ(x − y , t − s)2uDξ dyds

=

∫∫

C\C′K(x, y , t, s)u(y , s) dyds

with

K(x, y , t, s) = [ξs + ∆ξ](y , s)Φ(x − y , t − s) + 2Dξ ·DΦ(x − y , t − s)

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We don’t see the singularity of Φ, and K is C∞

=⇒ representation of u,

=⇒ u is as good as K

=⇒ u is C∞.

3.4.1 Application of the Representation to Local Estimates

Theorem 3.4.2 (local bounds on derivatives): u ∈ C21 (ΩT ) solution of the heat equation:

maxC(x,t, r2 )

|DkxDltu| ≤Ckl

r k+2l+n+2

∫

C(x,t,r)

|u(y , s)| dyds

Proof: Scaling argument. Calculation on a cylinder with r = 1.

v(x, t) = u(x0 + rx, t0 + r2t)

1

(x0, t0) (0, 0)

C (x0, t0, r)

r

r 2

C(0, 0, 1)

1

v solves the hear equations since

vt(x, t) = r2ut(x0 + rx, t0 + r2t)

∆v(x, t) = r2∆u(x0 + rx, t0 + r2t)

Representation of v :

v(x, t) =

∫∫

C(0,0,1)

K(x, y , t, s)v(y , s) dyds

x, t ∈ C(

0, 0, 12

)

|DkxDltu| ≤∫∫

C(0,0,1)

|DkxDltK(x, y , t, s)|︸︷︷︸Ckl

|v(y , s)| dyds

= Ckl

∫∫

C(0,0,1)

|v(y , s)| dyds

=⇒ max(x,t)∈C(0,0, 1

2 )|DkxDltv | ≤ Ckl

∫∫

C(0,0,1)

|v(y , s)| dyds

=⇒ r k+2l max(x,t)∈C(x0,t0,

r2 )|DkxDltu| ≤

Cklrn+2

∫∫

C(x0,t0,r)

|u(y , s)| dyds

=⇒ max(x,t)∈C(x0,t0,

r2 )|DkxDltu| ≤

Cklr k+2l+n+2

∫∫

C(x0,t0,r)

|u(y , s)| dyds


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3.5 Energy Methods85

Remark: This reflects again the different scaling in x and t.

3.5 Energy Methods

Theorem 3.5.1 (Uniqueness of solutions): Ω open and bounded, there exists at most one

solution u ∈ C21 (ΩT ) of

ut − ∆u = f in ΩT

u = g on Ω× t = 0u = h on ∂Ω× (0, T ]

Proof: Suppose you have two solutions u1, u2. Let w = u1 − u2. w solves the homoge-

neous heat equations.

Let

e(t) =

∫

Ω

w2(x, t) dx, 0 ≤ t ≤ T

d

dte(t) =

∫

Ω

2w(x, t) · wt(x, t) dx

= 2

∫

Ω

w(x, t) · ∆w(x, t) dx

I.P.= −2

∫

Ω

Dw(x, t) ·Dw(x, t) dx + 0

= −2

∫

Ω

|Dw(x, t)|2 dx ≤ 0

=⇒ e(t) is decreasing in t.

e(0) =

∫

Ω

w2(x, 0) dx = 0

=⇒ e(t) ≡ 0 on [0, T ]

=⇒ Dx(x, t) = 0 ∀x, t=⇒ w(x, t) = constant

∴ w(x, t) = 0 x ∈ ∂Ω

=⇒ w(x, t) = 0 ∀x ∈ Ω,∀t ∈ [0, T ]

=⇒ u1 = u2 in ΩT .

3.6 Backward Heat Equation

ut − ∆u = 0 in ΩT

u(x, 0) = g(x) in Ω

Backwards heat equation:

ut − ∆ = 0 in ΩT

u(x, T ) = g(x) in Ω

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Ω

T

Reverses the smoothening, expect bad solutions in finite time even if g(t) is smooth. Trans-

form: t 7→ T − t

v(x, t) = u(x, T − t)vt(x, t) = −ut(x, T − t)

∆v(x, t) = ∆u(x, T − t)

So, ut − ∆u = 0 transforms to

vt + ∆v = 0 Backwards heat equation

Theorem 3.6.1 (): If u1 and u2 are solutions of the heat equation with u1(x, T ) = u2(x, T )

with the same boundary values, then the solutions are identical.

3.7 Exercises

3.1: Apply the similarity method to the linear transport equation

ut + aux = 0

to obtain the special solutions of the form u(x, t) = c(at − x)−α.

3.2: [Evans 2.5 #10] Suppose that u is smooth and that u solves ut−∆u = 0 in Rn×(0,∞).

a) Show that uλ(x, t) = u(λx, λ2t) also solves the heat equation for each λ ∈ R.

b) Use a) to show that v(x, t) = x · Du(x, t) + 2tut(x, t) solves the heat equation as

well. The point of this problem is to use part a).

3.3: [Evans 2.5 #11] Assume that n = 1 and that u has the special form u(x, t) = v(x2/t).

a) Show that

ut = uxx

if and only if

4zv ′′(z) + (2 + z)v ′(z) = 0 for z > 0.

b) Show that the general solution of the ODE in a) is given by

v(z) = c

∫ z

0

e−s2/4s−1/2 ds + d, with c, d ∈ R.


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3.7 Exercises87

c) Differentiate the solution v(x2/t) with respect to x and select the constant c properly,

so as to obtain the fundamental solution Φ for n = 1.

3.4: [Evans 2.5 #12] Write down an explicit formula for a solution of the initial value problem

ut − ∆u + cu = f in Rn × (0,∞),

u = g on Rn × t = 0,

where c ∈ R.

Hint: Make the change of variables u(x, t) = eatv(x, t) with a suitable constant a.

3.5: [Evans 2.5 #13] Let g : [0,∞) → R be a smooth function with g(0) = 0. Show

that a solution of the initial-boundary value problem

ut − uxx = 0 in R+ × (0,∞),

u = 0 on R+ × t = 0,u = g on x = 0 × [0,∞),

is given by

u(x, t) =x√4π

∫ t

0

1

(t − s)3/2e−x

2/(t−s)g(s) ds.

Hint: Let v(x, t) = u(x, t)− g(t) and extend v to x < 0 by odd reflection.

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88Chapter 4: The Wave Equation

4 The Wave Equation


– Alan Rickman

Contents

4.1 D’Alembert’s Representation of Solution in 1D 88

4.2 Representations of Solutions for n = 2 and n = 3 93

4.3 Solutions of the Non-Homogeneous Wave Equation 98

4.4 Energy methods 101

4.5 Exercises 103

4.1 D’Alembert’s Representation of Solution in 1D

utt − ∆u = 0 in ΩT homogeneous

utt − ∆u = f in ΩT non-homogeneous

u = g on Ω× t = 0ut = h on Ω× t = 0

utt − uxx = 0

∂2t u − ∂2

xu = 0

(∂2t − ∂2

x

)u = 0

(∂t + ∂x) (∂t − ∂x) u︸︷︷︸v(x,t)

= 0


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4.1 D’Alembert’s Representation of Solution in 1D89

Solve (∂t + ∂x)v = vt + vx = 0

Transport equation for v : v(x, t) = a(x − t) an arbitrary function.

Thus ∂tu − ∂xu = v(x, t) = a(x − t).

General solution formula forut + bux = f (x, t) in Rn × (0,∞)

u = g on Rn × t = 0

Solution (Duhamel’s principle):

u(x, t) = g(x − tb) +

∫ t

0

f (x + (s − t)b, s) ds

Solution u (b = −1, f (x, t) = a(x − t))

u(x, t) = g(x − t(−1)) +

∫ t

0

a(x + (s − t)(−1)− s) ds

= g(x + t) +

∫ t

0

a (x + t − 2s)︸︷︷︸y

ds

a(x) = v(x, 0) = ∂tu(x, 0)− ∂xu(x, 0)

= h(x)− g′(x)

u(x, t) = g(x + t) +−1

2

∫ x−t

x+t

a(y) dy

= g(x + t) +1

2

∫ x+t

x−ta(y) dy

= g(x + t) +1

2

∫ x+t

x−t(h(y)− g′(y)) dy

=⇒ u(x, t) =1

2(g(x + t) + g(x − t)) +

1

2

∫ x+t

x−th(y) dy

D’Alembert’s formula.

Theorem 4.1.1 (D’Alembert’s solution for the wave equation in 1D): g ∈ C2(R), h ∈C1(R)

u(x, t) =1

2(g(x + t) + g(x − t)) +

1

2

∫ x+t

x−th(y) dy

Then

i.) u ∈ C2(R× (0,∞))

ii.)utt − uxx = 0

u(x, 0) = g(x)

ut(x, 0) = h(x)

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x − tt > 0

t = 0

information at x0travels with finitespeedc.f heat equationinfinite speed ofpropagation

x + t

x0

t > 0

t = 0

(x, t)

x − t x + t

Remark: We say u(x, 0) = g(x), if lim(x,t)→(x0,0),t>0 u(x, t) = g(x0) ∀x0 ∈ R. Remarks:

• finite speed of propagation.

• domain of influence

The values of u depend only on the value of g and h on [x − t, x + t]

4.1.1 Reflection Argument

u = 0

t

g, h

x = 0

Want to solve

utt − uxx = 0 in R+ × (0,∞)

u = g on R+ × t = 0ut = h on R+ × t = 0u = 0 on 0 × (0,∞)


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and suppose that g(0) = 0, h(0) = 0. Try to find a solution u on R of a wave equation that

is odd as a function of x , i.e.

u(−x, t) = −u(x, t)

=⇒ u(0, t) = −u(0, t)

=⇒ u(0, t) = 0

Extend g and h by odd reflection:

g(x) =

−g(−x) x < 0

g(x) x ≥ 0

h(x) =

−h(−x) x < 0

h(x) x ≥ 0

g

u solution of:utt − uxx = 0 in R× (0,∞)

u = g on R× t = 0ut = h on R× t = 0

D’Alembert’s formula:

u(x, t) =1

2(g(x − t) + g(x + t)) +

1

2

∫ x+t

x−th(y) dy

Assertion: u is odd, the restriction to [0,∞) is our solution u.

Proof:

u(−x, t) =1

2[g(−x − t) + g(−x + t)] +

1

2

∫ −x+t

−x−th(y) dy

= −1

2[g(x + t) + g(x − t)]− 1

2

∫ −x+t

−x−th(−y) dy

(sub. z = −y) = −1

2[g(x + t) + g(x − t)] +

1

2

∫ x−t

x+t

h(z) dz

= −1

2[g(x + t) + g(x − t)]− 1

2

∫ x+t

x−th(y) dy

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=⇒ u(−x, t) = −u(x, t)

=⇒ u is odd.

u(x, t) =

12 [g(x − t) + g(x + t)] + 1

2

∫ x+tx−t h(y) dy x ≥ t ≥ 0

12 [g(x + t)− g(t − x)] + 1

2

∫ t+xt−x h(y) dy t ≥ x ≥ 0

∫ x+t

x−th(y) dy =

∫ 0

x−th(y) dy +

∫ x+t

0

h(y) dy

=

∫ 0

x−t−h(−y) dy +

∫ x+t

x

h(y) dy

(sub. z = −y) =

∫ 0

−x+t

h(z) dz +

∫ x+t

x

h(y) dy

=

∫ x+t

t−xh(y) dy

4.1.2 Geometric Interpretation (h ≡ 0)

t > 0

0

0

0

g(x)

t = 0

Remark: D’Alembert:

u(x, t) =1

2[g(x + t) + g(x − t)] +

1

2

∫ x+t

x−th(y) dy

The solution is generally of the form u(x, t) = F (x − t) + G(x + t) with arbitrary functions

F and G.Gregory T. von Nessi

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4.2 Representations of Solutions for n = 2 and n = 393


4.2.1 Spherical Averages and the Euler-Poisson-Darboux Equation

U(x ; t, r) = −∫

∂B(x,r)

u(y , t) dS(y)

G(x, r) = −∫

∂B(x,r)

g(y) dS(y)

H(x, r) = −∫

∂B(x,r)

h(y) dS(y)

Recover u(x, t) = limr→0 U(x ; t, r)

x ∈ Rn fixed from now on.

Lemma 4.2.1 (): u a solution of the wave equation, then U satisfies

Utt − Ur r − n−1

r Ur = 0 in R+ × (0,∞)

U = G on R+ × t = 0Ut = H on R+ × t = 0

(4.1)

Once this is established, we transform (4.1) into Utt − Ur r = 0 and we use the represen-

tation on the half line to find U, U, u.

Calculation from MVP for harmonic functions:

Ur (x, t, r) =r

n−∫

B(x,r)

∆u(y , t) dy

=r

n

1

α(n)rn

∫

B(x,r)

∆u(y , t) dy

=1

nα(n)

1

rn−1

∫

B(x,r)

∆u(y , t) dy

∴ rn−1Ur =1

nα(n)

∫

B(x,r)

∆u(y , t) dy

(since u solves wave eq.) =1

nα(n)

∫

B(x,r)

utt(y , t) dy

(in polar) =1

nα(n)

∫ r

0

∫

∂B(x,ρ)

utt dSdρ

∴ [rn−1Ur ]r =1

nα(n)

∫

∂B(x,r)

utt dS

= rn−1−∫

∂B(x,r)

utt dS

= rn−1Utt

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∴ Utt −1

rn−1[rn−1Ur ]r = 0

⇐⇒ Utt −1

rn−1[(n − 1)rn−2Ur + rn−1Ur r ] = 0

⇐⇒ Utt −n − 1

rUr − Ur r = 0

Euler-Poisson-Darboux Equation.

4.2.2 Kirchoff’s Formula in 3D

Recall:

Representations forutt = ∆u in Rn × (0,∞)

u = g on Rn × t = 0ut = h on Rn × t = 0

n = 3 Kirchhoff’s formula

U(x, t, r) = −∫

∂B(x,r)

u(y , t) dS(y)

G(x, r), H(x, r)

U satisfies Euler-Poisson-Darboux equation

Utt − Ur r −n − 1

rUr = 0

U = G, Ut = H

Transformation: U = rU

Utt = rUtt = r [Ur r +2

rUr ]

= rUr r + 2Ur

= [U + rUr ]r

Ur r = [rU]r r = [U + rUr ]r

=⇒

Utt − Ur r = 0 in R+ × (0,∞)

U = G at t = 0

Ut = H at t = 0

U(0, t) = 0 t > 0

1D wave equation on the half line with Dirichlet condition at x = 0. From D’Alembert’s

solution in 1D:

u(x, t) = limr→0

U(x, t, r)

r

(formula with 0 ≤ r < t) = limr→0

1

2r[G(t + r) + G(t − r)] +

1

2r

∫ t+r

t−rH(y) dy

= G′(t) + H(t)Gregory T. von Nessi

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By definition

G(x ; t) = tG(x ; t)

= t−∫

∂B(x,t)

g(y) dS(y)

= t−∫

∂B(0,1)

g(x + tz) dS(z)

G′(x, t) = −∫

∂B(0,1)

g(x + tz) dS(z) + t−∫

∂B(0,1)

Dg(x + tz) · z dS(z)

= −∫

∂B(x,t)

[g(y) + t

∂y

∂ν(y)

]dS(y)

H(t) = t ·H(t) = t−∫

∂B(x,t)

h(y) dy

u(x, t) = −∫

∂B(x,t)

[g(y) + t

∂g

∂ν(y) + t · h(y)

]dS(y)

This is Kirchhoff’s Formula

Loss of regularity: need g ∈ C3, h ∈ C2, to get u in C2.

Finite speed of propagation:

B(x, t)

4.2.3 Poisson’s Formula in 2D

Method of descent: use the 3D solution to get the 2D solution by extending in 2D to 3D

u solves

utt − ∆u = 0 in R2 × (0,∞)

u = g on R2 × t = 0ut = h on R2 × t = 0

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Define

u(x1, x2, x3, t) = u(x1, x2, t)

g(x1, x2, x3) = g(x1, x2)

h(x1, x2, x3) = h(x1, x2)

Thenutt − ∆u = 0 in R3 × (0,∞)

u = g on R3 × t = 0ut = h on R3 × t = 0

Kirchhoff’s formula: x = (x1, x2, x3)

u(x, t) = −∫

∂B(x,t)

[g(y) + t

∂g

∂ν(y) + t · h(y)

]dS(y)

B(x, t) ⊂ R2x1

x2

x3

S−

S+B(x, t) ⊂ R3

tx

u(x1, x2) = u(x1, x2, 0)

Assume x = (x1, x2, 0)

Need to evaluate∫∂B(x,t)

|y − x |2 = t2 equation for the surface

⇐⇒ (y1 − x1)2 + (y2 − x2)2 + y23 = t2

Parametrization:

y23 = t2 − (y1 − x1)2 − (y2 − x2)2

= t2 − |y − x |2

Parametrization: f (y1, y2) =√t2 − |y − x |2

dS =√

1 + |Df |2 dy

∂f

∂yi=

1

2

1√t2 − |y − x |2

(−2)(yi − xi)

=−(yi − xi)√t2 − |y − x |2


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dS =

√1 +

|y − x |2t2 − |y − x |2 dy

=

(t2

t2 − |y − x |2) 1

2

dy

dS =t√

t2 − |y − x |2dy

4.2.4 Representation Formulas in 2D/3D

• Kirchoff’s formula

u(x, t) = −∫

∂B(x,t)

[g + t

∂y

∂ν+ t · h

]dS(y)

• Poisson’s formula u u

x = (x1, x2, 0)

u(x) = −∫

∂B(x,t)

[g + t

∂g

∂ν+ t · h

]dS(y)

B(x, t) ⊂ R2x1

x2

x3

S−

S+B(x, t) ⊂ R3

tx

Parameterize the surface integral

S+, f (y1, y2) =√t2 − |x − y |2

dS =√

1 + |Df |2 dy=

t√t2 − |y − x |2

∫

S+

[g + t

∂g

∂ν+ t · h

](y) dS(y)

=

∫

B(x,t)

[g + t

∂g

∂ν+ t · h

](y1, y2, f (y1, y2)) ·

√t + |Df |2 dy

=

∫

B(x,t)

[g(y1, y2) + t

Dg · (y − x)

t+ t · h(y1, y2)

]t dy√

t2 − |y − x |2

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∂g

∂ν= Dyg(y) · y − x

t

= Dyg ·y − xt

u(x) = −∫

∂B(x,t)

[g + t

∂g

∂ν+ t · h

]dS(y)

Since g, h do not appear on

u(x, t) = u(x, t)

=2

4πt2·∫

B(x,t)

[g(y1, y2) +Dg(y1, y2) · (y − x) + t · h(y1, y2)]t dy√

t2 − |x − y |2

=1

2−∫

B(x,t)

t · g(y) + t ·Dg(y) · (y − x) + t2h(y)√t2 − |x − y |2

dy x ∈ Rn, t > 0

Poisson’s formula in 2D.

u(x1, x2, x3, t) = u(x1, x2, t)

u(x1, x2, 0, t) = u(x1, x2, t)

u(x, t) = u(x, t)

Remarks

• Used method of descent extend 2D to 3D. The same idea works in n-dimensions,

extend from even to odd dimensions.

• Qualitative difference: integration is on the full ball in 2D.

• Loss of regularity:

need g ∈ C31 , h ∈ C2 to have u ∈ C2.

4.3 Solutions of the Non-Homogeneous Wave Equation

utt − ∆u = f (x, t) in Rn × (0,∞)

u = 0 on Rn × t = 0ut = 0 on Rn × t = 0

The fully non-homogeneous wave equationutt − ∆u = f

u = g

ut = h

can be solved by u = v + w , wherevtt − ∆v = f

v = 0

vt = 0vtt − ∆v = 0

w = f

wt = g


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4.3 Solutions of the Non-Homogeneous Wave Equation99


u(x, t; s)

utt(·; s)− ∆u(·; s) = 0 on Rn × (x,∞)

u(·; s) = 0 on Rn × t = s (u(x, t, 0) = 0)

ut(·; s) = f (·, s) on Rn × t = s (ut(x, t, 0) = f (·, t))

Then

u(x, t) =

∫ t

0

u(x, t; s) ds, x ∈ Rn, t > 0

Proof: We differentiate

ut(x, t) = u(x, t; t) +

∫ t

0

ut(x, t; s) ds

=

∫ t

0

ut(x, t; s) ds

utt(x, t) = ut(x, t; t) +

∫ t

0

utt(x, t; s) ds

= f (x, t) +

∫ t

0

utt(x, t; s) ds

∆u(x, t) =

∫ t

0

∆u(x, t; s) ds

=

∫ t

0

utt(x, t; s) ds

=⇒ utt(x, t)− ∆u(x, t) = f (x, t).

Examples: 1D: D’Alembert’s

Formula:

u(x, t) =1

2[g(x + t) + g(x − t)] +

1

2

∫ x+t

x−th(y) dy

To get the solution for u(x, t; s) translate the origin from 0 to s.

u(x, t; s) =1

2[g(x + t − s) + g(x − (t − s))] +

1

2

∫ x+t−s

x−(t−s)

h(y) dy

Recall that for u(x, t; s) we have g ≡ 0, h(y) = f (y , s).

Duhamel’s principle:

u(x, t) =

∫ t

0

1

2

∫ x+(t−s)

x−(t−s)

f (y , s) dyds

(sub. u = t − s) =1

2

∫ u

x−uf (y , t − u) dy(−du)

=1

2

∫ t

0

∫ x+s

x−sf (y , t − s) dyds

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s = 0

t

x

u(x, t)

x − t x + t

Which values of f influence u(x, t)

Kirchhoff’s formula (3D)

u(x, t) = −∫

∂B(x,t)

(g + t

∂g

∂ν+ t · h

)dS(y)

Poisson’s formula in 2D

u(x, t) =1

2−∫

B(x,t)

t · g(y) + t ·Dg · (y − x) + t2h(y)√t2 − |y − x |2

dy

Non-homogeneous problems

utt − ∆u = f

u = 0

ut = 0

u(x, t) =

∫ t

0

u(x, t; s) ds

u(·; s) solution on Rn × (s,∞)

u(·, s) = 0

ut(·, s) = f (·, s)

4.3.2 Solution in 3D

Initial data g(y) = s, h(y) = f (y , s). Translate the origin from t = 0 to t = s

u(x, t; s) = (t − s)−∫

∂B(x,t−s)

f (y , s) dS(y)

u(x, t) =

∫ t

0

(t − s)−∫

∂B(x,t−s)

f (y , s) dS(y)ds


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4.4 Energy methods101

(r = t − s)

=

∫ 0

t

r−∫

∂B(x,r)

f (y , t − r) dS(y)(−dr)

=

∫ t

0

r−∫

∂B(x,t)

f (y , t − r) dS(y)dr

=

∫ t

0

r

4πr2

∫

∂B(x,r)

f (y , t − r) dS(y)dr

=

∫

B(x,t)

1

4π|x − y | f (y , t − |x − y |) dy

= u(x, t) for x ∈ R3, t > 0

4.4 Energy methods

Ω ⊂ Rn bounded and open, ΩT = Ω× (0, T )

Theorem 4.4.1 (uniqueness): There is at most one smooth solution of

utt − ∆u = 0 in ΩT

u = g in Ω× t = 0ut = h in Ω× t = 0u = k on ∂Ω× (0, t)

Proof: Suppose u1, and u2 are solutions.

w = u1 − u2 satisfies

wtt − ∆w = 0 in ΩT

w = 0 in Ω× t = 0wt = 0 in Ω× t = 0w = 0 on ∂Ω× (0, t)

Multiply by wt :

wttwt − (∆w)wt = 0 inΩT∫

Ω

(wtt(x, t)wt(x, t) + ∆w(x, t) · wt(x, t)) dx = 0

=

∫

Ω

(wtt(x, t)wt(x, t) +Dw(x, t) ·Dwt(x, t)) dx

+

∫

∂Ω

(Dw · ν)wt dS

=

∫

Ω

(1

2∂t |wt |2 +

1

2∂t |Dw |2

)dx

Integrate from t = 0, to t = T

0 =

∫ T

0

∂t

∫

Ω

(1

2|wt |2 +

1

2|Dw |2

)dxdt

=1

2

∫

Ω

(|wt(x, T )|2 + |Dw(x, T )|2

)dx

−1

2

∫

Ω

(|wt(x, 0)|2 + |Dw(x, 0)|2

)dx = 0

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=⇒ wt , Dw = 0 in ΩT

=⇒ w = constant =⇒ w = 0

=⇒ u1 = u2 in ΩT .

Alternative: Define

e(t) =1

2

∫

Ω

(|wt(x, t)|+ |Dw(x, t)|) dx

and show that e(t) = 0, t ≥ 0.

4.4.1 Finite Speed of Propagation

(x0, t)

t

x0t0

B(x0, t0) ⊂ Rn

u(x, t0)

C = (x, t) : 0 ≤ t ≤ t0, |x − x0| ≤ t0 − t

If u is a solution of the wave equation with u ≡ 0, ut = 0 in B(x0, t0), then u ≡ 0 in C and

in particular u(x0, t0) = 0.

Define: e(t) =

∫

B(x0,t0−t)

(u2t (x, t) + |Du(x, t)|2

)dx

Compute: e(t) =

∫

B(x0,t0−t)(2ututt + 2Du ·Dut) dx

−∫

∂B(x0,t0−t)(u2t + |Du|2) dS

I.P.=

∫

B(x0,t0−t)(2ututt − 2∆u · ut) dx

+

∫

∂B(x0,t0−t)2(Du · ν)ut dS −

∫

∂B(x0,t0−t)(u2t + |Du|2) dS

= 0 +

∫

∂B(x0,t0−t)(2(Du · ν)ut − u2

t − |Du|2) dS


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4.5 Exercises103

|Du · ν|C-S≤ |Du| · |ν|︸︷︷︸=1

= |Du|

|2(Du · ν)ut | ≤ 2|Du| · |ut |Young’s≤ |Du|2 + |ut |2

Young’s inequality: 2ab ≤ a2 + b2, a, b ∈ R. Comes from the fact 0 ≤ a2 − 2ab + b2 =

(a − b)2.

∴ e(t) ≤∫

B(x0,t0−t)(|Du|2 + |ut |2 − u2

t − |Du|2) dx ≤ 0

=⇒ e(t) is decreasing in time.

But,

e(0) =

∫

B(x0,t0)

(|ut |2 + |Du|2) dx = 0

=⇒ e(t) = 0, t ≥ 0

=⇒ ut(x, t), Bu(x, t) = 0, (x, t) ∈ C=⇒ u ≡ 0 in C

=⇒ u(x0, t0) = 0.

4.5 Exercises

4.1: [Evans 2.6 #16] Suppose that E = (E1, E2, E3) and B = (B1, B2, B3) are a solution

of Maxwell’s equations

Et = curlB, Bt = − curlE, divE = 0, divB = 0.

Show that

utt − ∆u = 0

where u = E i or u = Bi .

4.2: [Evans 2.6 #17] (Equipartition of energy) Suppose that u ∈ C2(R × [0,∞)) is a

solution of the initial value problem for the wave equation in one dimension,utt − uxx = 0 in R× (0,∞),

u = g on R× t = 0,ut = h on R× t = 0.

Suppose that g and h have compact support. The kinetic energy is given by

k(t) =

∫ ∞

−∞u2t (x, t) dx

and the potential energy by

p(t) =

∫ ∞

−∞u2x (x, t) dx.

Prove that

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a) the total energy, e(t) = k(t) + p(t), is constant in t,

b) k(t) = p(t) for all large enough times t.

4.3: [Evans 2.6 #18] Let u be a solution ofutt − uxx = 0 in R3 × (0,∞),

u = g on R3 × t = 0,ut = h on R3 × t = 0.

where g and h are smooth and have compact support. Show that there exists a constant C

such that

|u(x, t)| ≤ C

t, x ∈ R3, t > 0

4.4: [Qualifying exam 01/01] Let Ω be a bounded domain in Rn with smooth boundary and

suppose that g : Ω → R is smooth and positive. Given f1, f2 : Ω → R, show that there can

be at most one C2(Ω) solution u : Ω× (0,∞)→ R to the initial value problem

utt = g(x)∆u in Ω× (0,∞),

u = f1 on Ω× t = 0,ut = f2 on Ω× t = 0,u = 0 on ∂Ω× (0,∞).

Hint: Find an energy E for which dE/dt ≤ CE and use Gronwall’s inequality which asserts

the following. If y(t) ≥ 0 satisfies dy/dt ≤ φy , where φ ≥ 0 and φ ∈ L1(0, T ), then

y(t) ≤ y(0)exp

(∫ t

0

φ(s) dx

).

4.5: [Qualifying exam 08/96] Let h ∈ C(R3) with h ≥ 0 and h = 0 for |x | ≥ a > 0. Let

u(x, t) be the solution of the initial-value problemutt − ∆u = 0 on R3 × (0,∞),

u(x, 0) = 0 on R3 × t = 0,ut(x, 0) = 0 on R3 × t = 0.

a) Show that u(x, t) = 0 in the set (x, t) : |x | ≤ t − a, t ≥ 0.b) Suppose that there exists a point x0 such that u(x0, t) = 0 for all t ≥ 0. Show that

u(x, t) ≡ 0.

4.6: [Qualifying exam 01/00] Suppose that q : R3 → R3 is continuous with q(x) = 0 for

|x | > a > 0. Let u(x, t) be the solution ofutt − ∆u = e iωtq(x) in R3 × (0,∞),

u(x, 0) = 0 on R3 × t = 0,ut(x, 0) = 0 on R3 × t = 0.

a) Show that e−iωtu(x, t)→ v(x) as t →∞ uniformly on compact sets, where

∆v + ω2v = q,

Hint: You may take the result of Exercise 2.1 for granted.

b) Show that v satisfies the “outgoing” radiation condition∣∣∣∣(∂

∂r+ iω

)v(x)

∣∣∣∣ ≤C

|x |2as r = |x | → ∞.


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106Chapter 5: Fully Non-Linear First Order PDEs

5 Fully Non-Linear First Order PDEs


– Alan Rickman

Contents

5.1 Introduction 106

5.2 Method of Characteristics 107

5.3 Exercises 121

5.1 Introduction

Solve F (Du(x), u(x), x) = 0 in Ω, Ω bounded and open in Rn.

• Method of characteristics

• Often time variable (conservation laws)

• existence of classical solution only for short times

• global existence for conservation laws via weak solutions (broaden the class of solutions,

solutions continuous and discontinuous).

Notation: Frequently, you set p = Du, z = u.

1st order PDE: F (p, z, x) = 0

F : Rn × R×Ω→ R

always f is smooth.

Example: Transport equation

ut(y , t) + a(y , t) ·Du(y , t) = 0 (5.1)Gregory T. von Nessi

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5.2 Method of Characteristics107

Set x = (y , t), p = (Du, ut). Then (5.1) is given by

F (p, z, x) = 0 with

F (p, z, x) =

(a(y , t)

1

)·(Du

ut

)

=

(a(x)

1

)·(Du(x)

ut(x)

)

F (p, z, x) =

(a(x)

1

)· ρ

5.2 Method of Characteristics

Idea: Reduce the PDE to a system of ODEs for x(s), z(s) = u(x(s)), p(s) = Du(x(s))

x(s)

Ω

Suppose u is a smooth solution of F (Du, u, x) = 0.

pi(s) =∂

∂suxi (x(s))

=

n∑

j=1

uxixj (x(s)) · xj(s)

1st order PDE, u ∈ C1 natural, but we get uxixj .

Eliminate them!

F (Du, u, x) = 0

∂

∂xiF (Du, u, x) = 0

=

n∑

j=1

∂F

∂pjuxixj +

∂F

∂zuxi +

∂F

∂xi= 0

To eliminate the second derivatives, we impose that

xj =∂F

∂pj(p(s), z(s), x(s))

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Then

pi(s) =

n∑

j=1

uxixj (x(s)) · xj(s)

=

n∑

j=1

uxixj (x(s)) · ∂F∂pj

(p(s), z(s), x(s))

= −∂F∂zuxi (x(s))− ∂F

∂xi

Thus,

pi(s) = −∂F∂zuxi (x(s))− ∂F

∂xi

Now find

z(s) =∂

∂su(x(s))

=

n∑

j=1

∂u

∂xj(x(s)) · xj(s)

=

n∑

j=1

pj · xj(s)

Thus,

z(s) =

n∑

j=1

pj · xj(s)

Result: If u is a solution of F (Du, u, x) = 0, and if

xj =∂F

∂pj(p(s), z(s), x(s)), then

p(s) = −∂F∂zp − ∂F

∂x

z(s) =∂F

∂p· p,

i.e., we get a system of 2n + 1 ODEs

x = Fpp = −Fzp − Fxz = Fp · p

Characteristics equations for the 1st order PDE. The ODE x = Fp is called the projected

characteristics equation. Issues:

• How do the boundary conditions relate to initial conditions for the ODEs?

• We can also have the following situation

Typically we can solve only locally, but not globally.Gregory T. von Nessi

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ΩCompatibility condition.

Where to prescribe the B.C.?

Ω

x

t

• If we construct u(x) from the values of u along the characteristic curves, do we get a

solution of the PDE?

Examples:

1. Transport equation

ut(y , t) + a(y , t) ·Du(y , t) = 0

F (p, z, x) =

(a(x)

1

)· p = 0

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Characteristic equations:

x = Fp =

(a(x)

1

)

x(s) = (y(s), t(s)) = (a(y , t), 1)

=⇒ t(s) = 1 =⇒ t = s

y(s) = a(y(s), s)

z(s) = Fp · p =

(a(s)

1

)· p = 0 (by the orig. PDE)

=⇒ z constant along characteristic curve

=⇒ u(y(s), s) = constant

=⇒ u determined from initial data.

(y(s), s)

t

RnΩ

2. Linear PDE: F (p, z, x) = b · p + cz

i.e. b(x) ·Du(x) + c(x)u(x) = 0

Characteristic equations:

x(s) = Fp = b(x(s))

z(s) = −c(x(s))u(x(s)) = −c(x(s))z(s)

x(s) = b(x(s))

z(s) = −c(x(s))z(s)

can solve without using equation for p.

Examples: x1ux2 − x2ux1 = u = 0 linear

F (p, z, x) = 0

F (p1, p2, z, x1, x2) = x1p2 − x2p1 − z = 0

= x⊥ · p − z,where

x⊥ =

(−x2

x1

)


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• x = Fp, x = x⊥

x1(s) = −x2(s), x1(0) = x0

x2(s) = x1(s), x2(0) = 0

I.C.

Solve subject to the initial data u = g on Γ = (x, 0) | x1 ≥ 0

(x0, 0)

Γ

• z = Fp · p = x⊥ · p = z by equation

z(0) = g(x0)

•

p = −Fx − Fzp =

(p2

−p1

)− (−1)p + suitable ICs, (see below)

˙(x1

x2

)=

(0 −1

1 0

)

︸︷︷︸A

(x1

x2

)

(x1

x2

)(0) =

(x0

0

)

(x1

x2

)(s) = eAs

(x0

0

)=

(cos s − sin s

sin s cos s

)(x0

0

)

x1(s) = x0 cos s

x2(s) = x0 sin scircular characteristics

z(s) = Ces = g(x0)es

Reconstruct u(x1, x2) from characteristics. Characteristic through (x1, x2) corresponds to

x0 =√x2

1 + x22 , (x1, x2) = x0e

iθ. u(x1, x2) = g(x0)e iθ

g

(√x2

1 + x22

)exp

(arctan

x2

x1

)= u(x1, x2)

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(x, y)

x(s)

(x0, 0)

Γ

5.2.1 Strategy to Prove that the Method of Characteristics Gives a

Solution of F (Du, u, x) = 0

i.) Reduce initial data on Γ to initial data on a flat part of the boundary, Rn−1

ii.) Check that the structure of the PDE does not change

iii.) Set up characteristic equations with the correct initial data

iv.) Solve the characteristic equations

v.) Reconstruct u from the solution

vi.) Check that u is a solution

Here we go!Gregory T. von Nessi

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i.) Flattening the boundary locally, in a neighborhood of x0 ∈ Γ

Suppose that Γ is of class C11 i.e. Γ is locally the graph of a C1 function of n − 1

variables

0

u

U

yn

Rn−1

xn Γ = γ(x) close to x0Γ

x0U

x1 ∈ Rn−1

Ψ(y)

Φ(x)

Define transformations (y = (y ′, yn))

y = Φ(x) :

y ′ = x ′

yn = xn − γ(x ′)

x = Ψ(y) :

x ′ = y ′

xn = yn + γ(x ′)

If u : U → R, then define

u : U → R

u(y) = u(x) = u(Ψ(y))

or vice versa u(x) = u(y) = u(Φ(x))

If u satisfies F (Du, u, x) = 0 in U, what is the equation satisfied by u?

ii.)

∂u

∂xi(x) =

∂

∂xiu(Φ(x))

=

n∑

j=1

∂u

∂yi

∂Φj

∂xi

Φ : Rn → Rn, DΦ =

Φ11 · · · Φ1

n...

. . ....

Φn1 · · · Φn

n

=(

(DΦ)TDu(Φ(x)))i

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In matrix notation

Du(x) = DΦT (x) ·Du(Φ(x))

=⇒ Du(x) = DΦT (Ψ(y) ·Du(y)

F (Du(x), u(x), x) = 0

F (DΦT (Ψ(y)) ·Du(y), u(y),Ψ(y)) = F (Du(y), u(y), y) = 0,

where F (p, z, y) is F (DΦT (Ψ(y))p, z,Φ(y)) = F .

In particular, F is a smooth function if the boundary of U is smooth.

From now on assume that x0 = 0 and that Γ ⊆ Rn−1

x0 Γ

Goal: construct u

Solution in a neighborhood of x0 = 0.Gregory T. von Nessi

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iii.) Find initial conditions for the ODEs.

0to 0 we want x(s)characteristic through y

∀y ∈ Rn close

=⇒ x(0) = y ∈ Rn−1

=⇒ z(0) = u(x(0)) = g(x(0)) = g(y)

=⇒ pi(0) = ∂g∂yi

i = 1, . . . , n − 1

Tangential derivatives, compatibility condition. One degree of freedom, pn(0). We

must choose pn(0) to satisfy

F

(∂g

∂yi(y), . . . ,

∂g

∂xi(y), pn(0), g(y), y

)= 0.

Definition 5.2.1: A triple (p0, z0, x0) of initial conditions is admissible at x0 if z0 =

g(x0), p0,i = ∂g∂yi

(x0), i = 1, . . . , n − 1, F (p0, z0, x0) = 0.

iii.) Initial conditions

x(0) = y (∈ Rn−1)

z(0) = g(y)

pi(0) = ∂g∂xi

(y) i ∈ 1, . . . , n − 1pn(0) chosen such that F (p1(0), . . . , pn(0), g(y), y) = 0

(5.2)

(x0, z0, p0) is called admissible if

z0 = g(x0)

p0,i = ∂g∂xi

(x0) i = 1, . . . , n − 1

F (p0, z0, x0) = 0

n equations for

the n ICs p01, . . . , p

0n

Question: We want to solve close to x0, can we find admissible initial data, y , g(y),

q(y) for y close to x . F (q(y), g(y), y) = 0?

Lemma 5.2.2 (): (p0, z0, x0) admissible, and suppose that Fp(p0, z0, x0) 6= 0. Then

there exists a unique solution of (5.2) for y sufficiently close to x . The triple (p0, z0, x0)

is said to be non-characteristic.

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Remark: If Γ is a smooth boundary then (p0, z0, x0) is non-characteristic if Fp(p0, z0, x0)·ν(x0) 6= 0, where ν(x0) is the normal to Γ at x0.

Proof: Use the implicit function theorem, G : Rn×Rn → Rn G(p, y) = (G1(p, y), . . . , Gn(p, y))

with G i(p, y) = pi − ∂g∂xi

(y). Gn(p, y) = F (p, g(y), y).

The triple (p0, z0, x0) admissible, G(p0, x0) = 0. Want to solve locally for p = q(y),

we can do this if det(Gp(p0, y0)) 6= 0.

Gp(p, y) =

1 0 · · · · · · 0 0

0. . .

. . .. . .

......

.... . .

. . .. . . 0

...

0 · · · · · · 0 1 0

Fp1 · · · · · · · · · · · · Fpn

det(Gp(p, y)) = Fpn(p, g(y), y)

=⇒ det(Gp(p0, x0) 6= 0

=⇒ there exists a unique solution p = q(y) close to x0

x0

(q(y), g(y), y) admissible initial data for the ODEs

iv.) Solution of the system of ODEs

x(s) = Fp(p(s), z(s), x(s))

z(s) = Fp(p(s), z(s), x(s)) · p(s)

p(s) = −Fx(p(s), z(s), x(s))− Fz(p(s), z(s), x(s))p(s)

Initial datax(0) = y

z(0) = g(y)

p(0) = q(y)

y ∈ Rn−1 parameter

Let

X(y , s) =

x(s)

z(s)

p(s)


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System:

˙X(y , s) = G(X(y , s))

X(y , 0) = (q(y), g(y), y)(5.3)

Lemma 5.2.3 (Existence of solution for (5.3)): If G ∈ Ck , X(y , 0) is Ck (as a

function of y) then there exists a neighborhood N of x0 ∈ Rn−1 and an s0 > 0 such

that (5.3) has a unique solution X : N × (−s0, s0)→ Rn × R×Ω and X is Ck .

( )

N

x0

v.) Reconstruction of u from the solutions of the characteristic equations

x

x0y

given x : find characteristic through x , i.e. find y = Y (x) and s = S(x) such that the

characteristic through y passes through x at time s.

Define: u(x) = z(Y (x), S(x))

Check: Invert x 7→ Y (x), S(x)

Lemma 5.2.4 (): Suppose that (p0, z0, x0) are non-characteristic initial data. Then

there exists a neighborhood V of x0 such that ∀x ∈ V there exists a unique Y and a

unique S with

x = X(Y (x), S(x))

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Proof: We need to invert X. Inverse functions. We can invert X(y , s) if DX(x0) 6= 0.

DX(x0) =

1 0 · · · · · · 0 Fp1

0. . .

. . .. . .

......

.... . .

. . .. . . 0

...

0 · · · · · · 0 1...

0 · · · · · · · · · 0 Fpn

To find ∂sX recall the characteristic equation, x = Fp =⇒ det(DX) = Fpn 6= 0 since

(p0, z0, x0) is non-characteristic.

vi.) prove that u is a solution of the PDE

u(x) = Z(Y (x), S(x))

p(x) = P (Y (x), S(x))

Crucial step Du(x) = p(x)

Step 1: f (y , s) := F (P (y , s), Z(y , s), X(y , s)) = 0

Proof:

f (y , 0) = F (P (y , 0), Z(y , 0), X(y , 0))

= F (q(y), g(y), y)

= 0

since (q(y), g(y), y) is admissible.

f (y , s) = FpP + Fz Z + Fx X

= Fp(−Fx − Fzp) + FzFpp + FxFp

= 0

=⇒ F (p(y , s), u(x), x) = 0.

Recall: Use capital letters to denote the dependence on y ,

P (y , s), Z(y , s), X(y , s)

solutions with initial data (q(y), g(y), y).

Invert the characteristics:

x Y (x), S(x) such that X(Y (x), S(x)) = x

Convention today: all gradients are rows.

Define:

u(x) := Z(Y (x), S(x))

p(x) := P (Y (x), S(x))

1. f (y , s) = F (P (y , s), Z(y , s), X(y , s)) ≡ 0 as a function of s.Gregory T. von Nessi

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2. We assert the following:

i.) Zs = P ·XTs = 〈P,Xs〉ii.) Zy = P ·Xy = P ·DyX

∂z

∂yi=

n−1∑

j=1

Pj∂X j

∂yii = 1, . . . , n − 1

Proof:

i.) Zs = Fp · p = Xs · p by the characteristic equations.

ii.) Before we prove the second identity we find

DyDsZ = DyZs

= P ·DyDsX +DsX ·DyP

Define

r(s) := DyZ − P ·DyX

and show that r(s) = 0.

Idea: find ODE for r and show that r = 0 is the unique solution.

ri(0) =∂g

∂xi(y)− q(y)

1. . .

1

for i = 1, . . . , n − 1

=⇒ ri(0) =∂g

∂xi(y)− qi(y) = 0

Since the triple (q(y), g(y), y) is admissible.

r(s) = DsDyZ −DsP ·DyX − P ·DsDyX= P ·DyDsX +DsX ·DyP−DsP ·DyX − P ·DsDyX

= DsX ·DyP −DsP ·DyXchar. eq. = Fp ·DyP − (−Fx − FzP )DyX − Fz ·DyZ

−(Fx − FzP )DyX

(Dy (F (P,Z,X)) = 0)

= FzP ·DyX − Fz ·DyZ= −Fz(DyZ − P ·DyX)

= −Fz · r(s)

∴

r(s) = −Fz · r(s)

r(0) = 0

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Linear ODE for r with r(0) = 0 =⇒ r = 0 for s > 0.

3. Prove that Du(x) = p(s)

=⇒ F (P,Z,X) = F (Du(x), u(x), x) = 0

=⇒ u is a solution of the PDE

u(x) = Z(Y (x), S(x))

=⇒ Dxu = DyZ ·DxY +DsZ ·DxS(by 2.) = P ·DyX ·DxY + P ·DsX ·DxS

= P (DyX ·DxY +DsX ·DxS)

= P ·DxX(Y, S)

= P · Id = p

=⇒ Dxu = P .

Examples:

1. Linear, homogeneous 1st order PDE:

b(x) ·Du(x) + c(x)u(x) = 0

Local existence of solutions:

Initial data non-characteristic:

x0

Fp · ν(x0) 6= 0

ν(x0)

F (p, z, x) = b(x) · p + c(x)z

Fp = b

Can solve if Fp(p0, z0, x0) · ν(x0) = b(x0) · ν(x0) 6= 0


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5.3 Exercises121

Local existence of b(x0) is not tangential

s

Characteristic equation:

x = Fp = b

(x(s) would be tangential to Γ if the initial conditions fail to be non-characteristic).

2. Quasilinear equation

b(x, u) ·Du(x) + c(x, u) = 0

Initial data non-characteristic Fp = b,

Fp(p0, z0, x0) · ν(x0) = b(x0, z0) · ν(x0) 6= 0.

Characteristics:

x = Fp, x(s) = b(x(s), z(s))

z = Fp · p= b(x(s), z(s)) · p(s)

= −c(x(s), z(s))

Closed system

x(s) = b(x(s), z(s)) x(0) = x0

z(s) = −c(x(s), z(s)) z(0) = g(x0)

for x and z , do not need the equation for p.

5.3 Exercises

5.1: [Evans 3.5 #2] Write down the characteristic equations for the linear equation

ut + b ·Du = f in Rn × (0,∞)

where b ∈ Rn and f = f (x, t). Then solve the characteristic equations with initial conditions

corresponding to the initial conditions

u = g on Rn × t = 0.

5.2: [Evans 3.5 #3] Solve the following first order equations using the method of charac-

teristics. Derive the full system of the ordinary differential equations including the correct

initial conditions before you solve the system!

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a) x1ux1 + x2ux2 = 2u with u(x1, 1) = g(x1).

b) uux1 + ux2 = 1 with u(x1, x2) = 12x1.

c) x1ux1 + 2x2ux2 + ux3 = 3u with u(x1, x2, 0) = g(x1, x2).

5.3: [Qualifying exam 08/00] Suppose that u : R × [0, T ] → R is a smooth solution of

ut + uux = 0 that is periodic in x with period L, i.e., u(x + L, t) = u(x, t). Show that

maxx∈R

u(x, 0)−minx∈R

u(x, 0) ≤ L

T.

5.4: Solve the nonlinear PDE of the first order

u3x1− ux2 = 0, u(x1, 0) = g(x1) = 2x

3/21 .

5.5: [Qualifying exam 01/90] Find a classical solution of the Cauchy problem

(ux)2 + ut = 0,

u(x, 0) = x2.

What is the domain of existence of the solution?


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124Chapter 6: Conservation Laws

6 Conservation Laws


– Alan Rickman

Contents

6.1 The Rankine-Hugoniot Condition 124

6.2 Shocks, Rarefaction Waves and Entropy Conditions 129

6.3 Viscous Approximation of Shocks 135

6.4 Geometric Solution of the Riemann Problem for General Fluxes 138

6.5 A Uniqueness Theorem by Lax 139

6.6 Entropy Function for Conservation Laws 141

6.7 Conservation Laws in nD and Kruzkov’s Stability Result 143

6.8 Exercises 148

6.1 The Rankine-Hugoniot Condition

Conservation law: ut + F (u)x = 0

initial data: u(x, 0) = u0(x) = g(x)

Example: Burger’s equation, F (u) = 12u

2

ut + F (u)x = ut +

(u2

2

)

x

= ut + u · ux = 0

Solutions via Method of Characteristics.

ut + F ′(u)ux = 0

Standard form: G(p, z, y) = 0, where p = (ux , ut) and y = (x, t).

G(p, z, y) = p2 + F ′(z)p1 = 0Gregory T. von Nessi

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6.1 The Rankine-Hugoniot Condition125

Characteristic equations: y = Gpz = Gp · p

y = (F ′(z), 1)

z = (F ′(z), 1) · p = p1F′(z) + p2 = 0 (equation)

Rewritten, we have

y1(s) = F ′(z(s)) y1(0) = x0 z(0) = g(x0)

y2(s) = 1 y2(0) = 0

=⇒ y2(s) = s (=⇒ identification t = s)

=⇒ z(s) = 0 =⇒ z(s) = constant = g(x0)

y1(s) = F ′(z(s)) = F ′(g(x0))

y1(s) = F ′(g(x0))s + x0

= F ′(g(x0))t + x0 (s = t)

=⇒ characteristic curves are straight lines.

Example: Burgers’ equation with g(x) = u(x, 0)

0 1

g(x) =

1 x ≤ 0

1− x 0 ≤ x < 1

0 x ≥ 0

Speed of the characteristic equations

F ′(g(x0)) = g(x0)

Characteristics:

Solution (local existence):

• Discontinuous “solutions” in finite time.

• Break down of classical solution.

• Same effect if you take smooth (C1, or more) initial data.

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x

1

t

0 1

0 112

t = 12

0 1

t = 1

• Equation ut + F (u)x = 0 involving derivatives is not defined.

• Weak solutions (or integral solutions).

• Go back from the PDE to an integrated version.

• v test function.

v smooth, with compact support in R× [0,∞).

• Multiply equation and integrate by parts.Gregory T. von Nessi

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6.1 The Rankine-Hugoniot Condition127

0 =

∫ ∞

0

∫

Rv(ut + F (u)x) dxdt

= −∫ ∞

0

∫

Rvt · u dxdt +

∫

Rv · u dx

∣∣∣∣∣

t=∞

t=0

−∫ ∞

0

∫

Rvx · F (u) dxdt +

∫ ∞

0

F (u) · v dt∣∣∣∣∣

x=∞

x=−∞

= −∫ ∞

0

∫

R(vt · u + vx · F (u)) dxdt −

∫

Rv(x, 0) · u(x, 0) dx

So, we have

∫ ∞

0

∫ ∞

−∞vt · u + vx · F (u) dxdt +

∫ ∞

−∞v(x, 0) · g(x) dx = 0 (6.1)

Definition 6.1.1: u is a weak solution of the conservation law ut +F (u)x = 0 if (6.1) holds

for all test functions v .

Question: What information is hidden in the weak formulation?

If u ∈ C1 is a weak solution, reverse the integration by parts and get

∫ ∞

0

∫

Rv(ut + F (u)x) dxdt = 0

for all smooth test functions.

=⇒ ut + F (u)x = 0

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6.1.1 Information about Discontinuous Solutions

t

VlVr

C

x

u is a solution in V = Vl ∪ Vr , u smooth in Vl and Vr and u can have a discontinuity across

C (e.g. u ≡ 1 on Vl and u ≡ 0 on Vr ).

Is this possible?

v test function. Support of v contained in Vl (Vr ) =⇒ ut + F (u)x = 0 in Vl (Vr ).

Support for v contains a part of C, and that v(x, 0) = 0.

(6.1)

∫∫

V

(vt · u + vx · F (u)) dxdt = 0

=

∫∫

Vl

(vt · u + vx · F (u)) dxdt +

∫∫

Vr

(vt · u + vx · F (u)) dxdt = 0

Now, let us look at∫∫

Vl

(vt · u + vx · F (u)) dxdtI.P.= −

∫∫

Vl

v (ut + F (u)x)︸︷︷︸0

dxdt −∫

C

(v · F (u)

v · u

)· ν dS

=

∫

C

(v · F (u) · ν1 + v · u · ν2) dS

Note that u here is the trace of u from Vl on the curve C, call this ul :∫

C

v(·F (ul) · ν1 + ul · ν2) dS

Same calculation for Vr with ν replaced by −ν∫∫

Vr

(vt · u + vx · F (u)) dxdt = −∫

C

v(F (ur ) · ν1 + ur · ν2) dS

Grand total:∫

C

v [F (ul)− F (ur )]ν1 + (ul − ur )ν2] dS = 0

=⇒ [F (ul)− F (ur )]ν1 + (ul − ur )ν2 = 0 along C.

Then ν = (1,−s) and

F (ul)− F (ur ) = s(ul − ur )Gregory T. von Nessi

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6.2 Shocks, Rarefaction Waves and Entropy Conditions129

ν = (1,−s)

C

(s , 1)

Suppose now that C is given by (x = s(t), t)

this is typically written as (adopting new notation)

[[F (u)]] = σ · [[u]] ,

where σ = s and [[w ]] = wl − wr = jump.

This is the Rankine-Hugoniot Condition

6.2 Shocks, Rarefaction Waves and Entropy Conditions

We return to our original example:

ut + u · ux = 0 g(x) =

1 x ≤ 0

1− x 0 ≤ x < 1

0 x ≥ 1

A discontinuity develops at t = 1, specifically, there is a jump for ul = 1 to ur = 0.

In the general situation, one solves the Riemann problem for the conservation laws

ut + F (u)x = 0 u(x, 0) =

ul x < 0

ur 0 ≤ x > 0

Example: The Riemann problem for Burger’s equation:

1. ul = 1, ur = 0. The discontinuity can propagate along a curve C with

s =[[F (u)]]

[[u]]=F (ul)− F (ur )

ul − ur=

12 − 0

1=

1

2

The characteristics are going into the curve C, a so-called shock wave.

2. ul = 0, ur = 1. Again, we have s = 12 ; but this time the characteristics are emerging

from the shock.

Is this the only solution? We can find another solution of Burger’s equation by the similarity

method, i.e. by selecting solutions of the form

u(x, t) =1

tαv( xtβ

)

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x

t

0

u = 0u = 1

x0

t

u = 1u = 0

u(x, t) =

0 x < 0xt0 ≤ x < t

1 x > t

x0

t

u = 0 u = 1

0

This leads to u(x, t) = xt . We know construct a different solution:

This solution is continuous and called a rarefaction wave. It is a weak solution.

Question: Which solution is the physically relevant solution?

No Ideas: The solution should be stable under perturbations, i.e. should look similar if we

smooth the initial data . The solution should be the limit of a viscous approximation of the

original equation (more later).

Recall that y = G(F ′(z), 1), i.e., F ′(ul) is the speed of the characteristics with initial data

ul .


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i.e., characteristics impinge into the shock.Lax entropy condition: F ′(ul) > s > F ′(ur)

s

F ′(ul)

F ′(ur)

Recall: Burger’s equation

ut +

(u2

2

)

x

= ut + u · ux = 0

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Rarefaction wave initial data:

0 x

t

0 x

t

Shock wave initial data:

s = 12

"non-physical" shock

x0

t

x0

t

x0

weak solutions not unique

ts = 12

Odd entropy conditions to select solutions:

Lax Characteristics going into the shock.

Viscous approximation: u as a limit of uε,

uεt + uε · uεx + ε · uεxx = 0

Hopf-Cole: Change of the dependent variables, w = Φ(u)

ut − a∆u + b|Du|2 = 0

u = g

wt = a∆w − |Du|2(aΦ′′(u)− bΦ′(u))

Choose Φ as a solution of aΦ′′ − bΦ′ = 0 =⇒ w solves the heat equation. One solution is

given by

Φ(t) = exp

(−bta

)


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Choosing this Φ, we havewt − a∆w = 0

w(x, 0) = exp(−b·g(x)

a

)

6.2.1 Burger’s Equation

Define

v(x, t) :=

∫ x

−∞u(y , t) dt

assuming that the integral exists and that u, ux → 0 as x → −∞.

The equation for v :

vt − ε · vxx +1

2v2x = 0

Proof: Compute∫ x

−∞ut dy − ε · ux +

1

2u2 =

∫ x

−∞(ε · uxx − u · ux︸︷︷︸

( 12u2)x

) dy − ε · ux +1

2u2

= ε · ux −1

2u2

∣∣∣∣∣

x

−∞− ε · ux +

1

2u2

= ε · ux −1

2u2 − ε · ux +

1

2u2

= 0

=⇒ v solves vt − ε · vxx + 12v

2x = 0

v(x, 0) = h(x) =

∫ x

−∞g(y) dy

This is of the form of the model equation with a = ε, b = 12 .

w = exp

(−b · v

a

)solves

wt − ε · wxx = 0

w(x, 0) = exp(−b·h(x)

a−ε

)

The unique bounded solution is

w(x, t) =1√

4πεt

∫

Rexp

(−(x − y)2

4εt

)exp

(−b · h(y)

ε

)dy

Now, v(x, t) = − εb lnw

v(x, t) = − εb

ln

1√

4πεt

∫

Rexp

(−(x − y)2

4εt

)exp

(−b · h(y)

ε

)dy

and u(x, t) = vx(x, t)

u(x, t) =1

b

∫Rx−y

4t exp(− (x−y)2

4εt

)exp

(−b·h(y)

ε

)dy

∫R exp

(− (x−y)2

4εt

)exp

(−b·h(y)

ε

)dy

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Formula for uε, let ε→ 0

Asymptotics of the formula for ε→ 0: Suppose that k and l are continuous functions, l

grows linearly, k grows at least quadratically. Suppose that k has a unique minimum point,

k(y0) = miny∈R

k(y)

Then

limε→0

∫R l(y) · exp

(−k(y)

ε

)dy

∫R exp

(−k(y)

ε

)dy

= l(y0)

Proof: See Evans

Evaluation for viscous approximation with

g(x) =

0 x < 0

1 x > 0

l(y) =x − yt

k(y) =(x − y)2

4t+h(y)

2

h(y) =

∫ y

−∞g(x) dx =

0 y < 0

y y ≥ 0=: y+

Find minimum of k :

Case 1: y < 0, k(y) = (x−y)2

4t

Minimum: y0 = x if x ≤ 0

y0 = 0 if x ≥ 0

Case 2: y ≥ 0, k(y) = (x−y)2

4t + y2

k ′(y) =x − y

2t+

1

2

k ′′(y) =1

2t> 0 =⇒ min

k ′(y) = 0, y = x − ty0 = x − t if x ≥ t

y0 = 0 if x ≤ t

Evaluation of the formula:

(1): x > t:

y ≤ 0: =⇒ y0 = 0, k(y0) = x2

4t

y ≥ 0: =⇒ u0 = x − t, k(x − t) = t4 + x−t

2 = −t4 + x

2

x2

4t>x

2⇐⇒ x2 > 2xt − t2 ⇐⇒ (x − t)2 ≥ 0

√

y0 = x − t, l(y) =x − yt

, l(x − t) =t

t= 1


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6.3 Viscous Approximation of Shocks135

Analogous arguments for:

(2): 0 ≤ x ≤ ty0 = 0, l(y) = x

t

(3): x < 0, y0 = x , l(x) = 0

u(x, t) = limε→0

uε(x, t) =

0 x < 0xt 0 ≤ x ≤ t1 x > t

Rarefaction wave!

6.3 Viscous Approximation of Shocks

Burger’s equation: Shocks (RH condition), rarefaction waves

x0

ulur

ut + F (u)x = ut + F ′(u) · ux = 0

Lax criterion: F ′(ul) > s > F ′(ur )

Since F ′(ul/r ) is the speed of the characteristics on the left/right of x0.

More generally, find shock solutions of the Riemann problem

ut + F (u)x = 0

u(x, 0) =

ul x < 0

ur x > 0

Building block for numerical schemes that approximate initial data by piecewise constant

functions.

Viscous approximation: ut + F (u)x = ε · uxx

Special solutions: traveling waves

u(x, t) = v(x − st), s = speed of profile.

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Find v = v(y) such that v(y)→ ul as y → −∞v(y)→ ur as y → +∞

Equation for v : ut = −s · v ′ux = v ′

uxx = v ′′

ut + F ′(u) · ux = ε · uxx−s · v ′ + F ′(v) · v ′ = ε · v ′′

Define w(y) = v(yε

), v(y) = w(ε · y),

=⇒

−s · w ′ + F ′(w) · w ′ = w ′

w(y)→ ul as y → −∞w(y)→ ur as y → +∞

Suppose for the moment being that ul > ur .

0

ul

ur

Suppose we have a unique solution

−s · w ′ + F ′(w) · w ′ = (−s · w + F (w))′

Integrate −s · w + F (w) + C0 = w ′

C0 = constant of integration

Let Φ(w) = −s · w + F (w) + C0

If w(y)→ ul , w′(y)→ 0 as y → −∞, then

−s · ul + F (ul) + C0 = 0

⇐⇒ C0 = s · ul − F (ul). Then

Φ(w) = −s(w − ul) + F (w)− F (ul)

If w → ur , w′ → 0 as y → +∞, =⇒ Φ(ur ) = 0

= −s(ur − ul) + F (ur )− F (ul) = 0

⇐⇒ s =F (ur )− F (ul)

ur − ulR.H.


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6.3 Viscous Approximation of Shocks137

w defined by the ODE w ′ = Φ(w)

Φ at least Lipschitz =⇒ local existence and uniqueness of solutions. Moreover, Φ(ur ) = 0,

Φ(ul) = 0 =⇒ w ≡ ur , w ≡ ul are solutions.

=⇒ ul > w > ur for all times.

can not happen

u

ur

ul

Analogously, there is no u with ul > u > ur with Φ(u) = 0. Φ has a sign on [ur , ul ] =⇒ w

decreasing =⇒ Φ(u) < 0 on (ur , ul)

Φ(u) = −s(u − ul) + F (u)− F (ul)

Φ(u) < 0 for u ∈ (ur , ul). So,

F (u) < s(u − ul) + F (ul)

F (u) < F (ul) +F (ur )− F (ul)

ur − ul︸︷︷︸σ

(u − ul)

F (ur)

ur ul

F (ul)

=⇒ F lies under the line segment connecting (ur , F (ur )) and (ul , F (ul)).

Remarks:

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1. This is always true (for all choices of ul and ur ) if F is convex.

2. Take the limit u ul

s(u − ul) > F (u)− F (ul)

⇐⇒ s <F (u)− F (ul)

u − ulIn the limit s ≤ F ′(ul)As u ur ,

F (u)− F (ur ) < F (ul)− F (ur ) + s(ur − ul + u − ur )= s(u − ur )

=⇒ s >F (u)− F (ur )

u − urand in the limit s ≥ F ′(ul)

Two conditions together yield: F ′(ul) ≥ s ≥ F ′(ur )

(Lax Shock admissibility criterion). So seeking traveling wave solution, we recover

the R.H. and Lax conditions.

3. Combining the inequalities

F (u)− F (ul)

u − ul> s >

F (u)− F (ur )

u − ur∀u ∈ (ur , ul)

Oleinik’s entropy criterion.

6.4 Geometric Solution of the Riemann Problem for General

Fluxes

K

u1 u2 ulur

slope s1slope s2 F

u


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6.5 A Uniqueness Theorem by Lax139

Let K = u, y : ur ≤ u ≤ ul , y ≤ F (u)The convex hull is bounded by line segments tangent to the graph of f and arcs on the

graph of f . The entropy solution consists of shocks (corresponding to the line segments)

and rarefaction waves (corresponding to the arcs).

s2(t)s1(t)

0

connects u1 and u2

u = uru = ul

6.5 A Uniqueness Theorem by Lax

Conservation Law ut + F (u)x = 0

A weak solution (say piecewise smooth with discontinuities) is an entropy solution if Oleinik’s

entropy criterion holds.

F (u)− F (ul)

u − ul> s >

F (u)− F (ur )

u − ur

for all discontinuities of u.

Theorem 6.5.1 (Lax): Suppose u, v are two entropy solutions. Then

µ(t) =

∫

R|u(x, t)− v(x, t)| dx

is decreasing in time.

Corollary 6.5.2 (Uniqueness): There exists at most on entropy solution with given initial

data, u(x, 0) = u0(x), x ∈ R

Proof: µ(0) = 0, µ decreasing =⇒ µ(t) = 0 ∀t =⇒ u(x, t) = v(x, t).

Proof of Theorem (6.5.1): Take u − vChoose xi(t) in such a way that u − v changes its sign at the points xi and has a sign on

xi , xi+1], namely (−1)i .

Now

µ(t) =∑

n

(−1)n∫ xn+1

xn

(u(x, t)− v(x, t)) dx

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xi+2(t)

u − v

xi(t)

i + 1 even

xi+1(t)

Find µ(t), check that the jumps in the intervals (xi , xi+1) make no difficulties

µ(t) =∑

n

(−1)n∫ xn+1

xn

(ut − vt) dx

+(u(x−n+1, t)− v(x−n−1, t)) · xn+1 − (u(x+n , t)− v(x+

n , t)) · xn (6.2)

∫ xn+1

xn

(u − v)t dx = −∫ xn+1

xn

(F (u)− F (v))x dx

= −F (u(x−n+1, t))± F (v(x−n+1, t))± F (u(x+n , t))± F (v(x+

n , t))

Rewrite (6.2) formally:

µ(t) =∑

n

(−1)n

−(F (u)− F (v)

∣∣∣∣∣

x−n+1

x+n

+ (u − v) · x∣∣∣∣∣

x−n+1

x+n

=∑

n

(−1)n

F (u)− F (v)− (u − v) · x

∣∣∣∣∣x−n

+ F (u)− F (v)− (u − v) · x∣∣∣∣∣x+n

(−1)n(−1)n−1

xn+1xnxn−1

Case 1: u − v continuous at xn. u(xn, t) = v(xn, t) =⇒ no contribution to the sum at xn

Case 2: u jumps from ul to ur with ur < ul and v continuous, u − v changes sign at xn =⇒ur < v = v(xn) < ul

u − v < 0 on [xn, xn+1] =⇒ n is odd.

Term at xn:

−F (ul)− F (v)− (ul − v)︸︷︷︸F (ul )−F (v)

ul−v>s≥0

·xn + F (ur )− F (v)− (ur − v)︸︷︷︸s>

F (ur )−F (v)ur−v ≥0

·xn


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6.6 Entropy Function for Conservation Laws141

=⇒ total contribution at xn is non-positive

=⇒ µ(t) ≤ 0

=⇒ µ is decreasing.

Analogous discussion for the remaining cases.

6.6 Entropy Function for Conservation Laws

Idea: Additional equations to be satisfied by smooth solutions, inequalities for non-smooth

solutions.

u smooth solution of ut + F (u)x = 0. Suppose you find a pair of functions (η,ψ)

η entropy, η convex, η′′ > 0

ψ entropy flux

such that

η(u)t + ψ(u)x = 0

η′(u) · ut + ψ′(u) · ux = 0

ut + F ′(u) · ux = 0

=⇒ η′(u) · ut + F ′(u) · η′(u) · ux = 0. This holds if

ψ′ = F ′ · η′

Suppose that uε is a solution of the viscous approximation

uεt + F (uε)x = ε · uεxx

suppose that (η,ψ) is an entropy-entropy flux pair with η convex.

=⇒ uεt · η′(uε) + F ′(uε) · η′(uε) · uεx = ε · η′(uε) · uεxx

⇐⇒ η(uε)t + ψ(uε) · uεx = ε(η′ · ux)x − ε · η′′(uε) · u2x

Integrate this on [x1, x2]× [t1, t2]

∫ t2

t1

∫ x2

x1

(η(uε)t + ψ(uε)x) dxdt = ε

∫ t2

t1

∫ x2

x1

(ε · η′(uε) · uεx)x dxdt

− ε∫ t2

t1

∫ x2

x1

ε · η′′(uε)uε 2x dxdt

︸︷︷︸≥0

=⇒∫ t2

t1

∫ x2

x1

(η(uε)t + ψ(uε)x) dxdt ≤ ε2

∫ t2

t1

−[η′(uε(x2, t)) · uεx(x2, t)

− η′(uε(x1, t)) · uεx(x1, t)]dxdt

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Suppose that uε → u, and let ε→ 0:

∫ t2

t1

∫ x2

x1

(η(u)t + ψ(u)x) dxdt ≤ 0 (6.3)

if uεx is uniformly integrable, i.e.

∣∣∣∣∫ t2

t1

∫ x2

x1

(η′(uε) · uε)x dxdt∣∣∣∣ ≤ C.

This means that a solution u of ut + F (ux) = 0 has to satisfy

η(u)t + ψ(u)x = 0

in a suitable weak sense.

Example: Burger’s equation.

η(u) = u2, F (u) = u2

2

ψ′(u) = 2u · u = 2u2

ψ(u) =2

3u2

Possible choice: (η,ψ) =(u2, 2

3u2)

Suppose u has a jump across a smooth curve.

x1 + ∆xt1x1 ∆x

C

t2 + ∆t

s = speed of the curve

s = ∆x∆t

(6.3) =

∫ x2

x1

η(u)

∣∣∣∣∣

t2

t1

dx +

∫ t2

t1

ψ(u)

∣∣∣∣∣

x2

x1

dt

≈ (x2 − x1)(η(ul)− η(ur )) + (t2 − t1)(ψ(ur − ul))

= s · ∆t · (u2l − u2

r ) + ∆t2

3(u3r − u3

l )

= . . . = −1

6∆t (ul − ur )3

︸︷︷︸≥0

≤ 0

=⇒ jump admissible only if ul > ur .Gregory T. von Nessi

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6.7 Conservation Laws in nD and Kruzkov’s Stability Result143

6.7 Conservation Laws in nD and Kruzkov’s Stability Result

Conservation laws in n dimensions

ut + divx ~F (u) = 0,

where ~F : R→ Rn vector value flux.

~F (u) =

F1(u)

...

Fn(u)

Definition 6.7.1: A pair (η,ψ) is called an entropy-entropy flux pair if ~ψ′(u) = ~F ′(u)η′(u),

where

~ψ : R→ Rn, ψ′i(u) = F ′i (u)η′(u)

η : R→ R

and if η is convex.

Take a viscous approximation:

uεt + divx ~F (uε) = ε · ∆uε

uεt · η′(uε) +

n∑

i=1

∂

∂xiFi(u

ε) · η′(uε) = ε · ∆uεη′ (6.4)

(∂

∂xiFi(u

ε)

)η′(uε) = F ′i (u

ε) · η′(uε) ∂∂xi

uε

= ψ′i(uε)∂

∂xiuε

=∂

∂xi~ψ(uε)

(6.4) is equivalent to

η(uε)t + div (~ψ(uε)) = ε · div (Duε · η′(uε))− ε ·Duε ·Dη′(uε)= ε · div (D(η(uε)))− ε ·Duε ·Duε · η′′(uε)

Multiply this by a test function φ ≥ 0 with compact support.∫ ∞

0

∫

Rnφ[η(uε)t + div ~ψ(uε)] dxdt =

∫ ∞

0

∫

Rnφ[ε · ∆η(uε)− ε · η′′(uε)|Duε|2] dxdt

∫ ∞

0

∫

Rnφ[η(uε)t + div ~ψ(uε)] dxdt ≤

∫ ∞

0

∫

Rnε · φ · ∆η(uε) dxdt

If uε → u as ε→ 0, then∫ ∞

0

∫

Rnφ[η(u)t + div ~ψ(u)] dxdt ≤ 0

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Definition 6.7.2: A bounded function u is a weak entropy solution of

ut + div ~F (u) = 0

u(x, 0) = u0(x)

if∫ ∞

0

∫

Rn(η(u) · φt + ~ψ(u) ·Dφ) dxdt ≤

∫

Rnφ(x, 0)η(u0(x)) dx

for all test functions φ ≥ 0 with compact support in Rn× [0,∞) and all entropy-entropy flux

pairs (η, ~ψ).

Additional conservation laws:

η(u)t + divx ~ψ(u) ≤ 0

Analogue of the Rankine-Hugoniot jump condition

n

v−

u−

v+

u+

Γ

Similar arguments to the 1D case show

νn+1 [[η]] +[[~ψ]]· ν0 ≤ 0

where ν = (ν0, νn+1) is the normal in space-time. Define n = ν0|ν0| , V = −νn+1

|ν0| . Then

V [[η]] ≥[[~ψ]]· n

n = 2

n

V = normal velocity of the interface.

By approximation, we can extend this setting to Lipschitz continuous η, and in particular to

Kruzkov’s entropy, η(u) = |u − k | k is constant.

~ψ(u) = sgn (u − k)(~F (u)− ~F (k))Gregory T. von Nessi

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Check: R.H. jump inequality s [[η]] ≥[[~ψ]]

with Kruzkov’s entropy in 1 dimension.

s(|ul − k | − |ur − k |) ≥ sgn (ul − k)(~F (ul)− ~F (k))− sgn (ur − k)(~F (ur )− ~F (k))

Suppose that ur < k < ul :

s(ul − k + ur − k) ≥ ~F (ul)− ~F (k) + ~F (ur )− ~F (k)

⇐⇒ s(ul + ur − 2k) ≥ ~F (ul) + ~F (ur )− 2~F (k)

⇐⇒ ~F (k) ≥ ~F (ul )+~F (ur )2 +

(k −

(ul+ur

2

))s

s =~F (ul)− ~F (ur )

ul − urSign error. Should recover Oleinik’s Entropy Criterion.

Correction: u+ = ur , u− = ul → Oleinik’s entropy criterion.

Theorem 6.7.3 (Kruzkov): Suppose that u1 and u2 are two weak entropy solutions of

ut + div ~F (u) = 0 which satisfies u1, u2 ∈ C0([0,∞); L1(Rn)) ∩ L∞((0,∞)× Rn). Then

||u1(·, t2)− u2(·, t2)||L1(Rn) ≤ ||u1(·, t1)− u2(,t1)||L1(Rn)

for all 0 ≤ t1 ≤ t2 <∞

Remark: This implies uniqueness if u1 and u2 have the same initial data.

Notation: L1(Rn) = space of all integrable functions with

||f ||L1(Rn) =

∫

Rn|f | dx

L∞ = space of all bounded measurable functions.

The first assumption means that the map

t 7→ Φ(t) = u(·, t) is continuous

i.e. for t0 fixed, ∀ε > 0, ∃δ > 0 such that ||Φ(t)−Φ(t0)||L1(Rn) ≤ ε if |t − t0| < δ. i.e.

∫

Rn|u(x, t0)− u(x, t)| dx < ε

Proof: (formal)

Kruzkov entropy-entropy flux:

η(u) = |u − k |, ~ψ(u) = sgn (u − k)(~F (u)− ~F (k))

∫

Rn

∫

R+

η(u)︸︷︷︸|u1−k|

φt + sgn (u1 − k)(~F (u1)− ~F (k)) ·Dφ dxdt ≥ 0

for all φ with compact support in Rn × (0,∞). Suppose you can choose k = u2(x, t) (it is

not clear you can do this since k is constant)∫

Rn

∫

R+

|u1(x, t)− u2(x, t)|φt + sgn (u1 − u2)[F (u1)− F (u2)](x, t) ·Dxφ dxdt ≥ 0

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By assumption, |u1|, |u2| ≤ N. Take M = sup|u|≤N |F ′(u)|.∫

|x |≤R−M(t2−t1)

|u1(x, t2)− u2(x, t2)| dx ≤∫

|x |≤R|u1(x, t)− u2(x, t2)| dx+ ≤ 0

Remark:

The proof shows more:∫

|x |≤R|u1(x, t2)− u2(x, t2)| dx ≤

∫

|x |≤R+M(t2−t1)

|u1(x, t1)− u2(x, t1)| dx

This shows again finite speed of propagation.

Fixing the proof:

• φ = χK . Fix by taking a regularization of χK , i.e. with standard mollifiers.

• Choice k = u2(x, t)

“Doubling of the variables”. Take u2(y , s) and integrate in y and s for the entropy

inequality for u1:

0 ≤∫

Rn×R+

∫

Rn×R+

|u1(x, t)− u2(y , s)|φt

+ sgn (u1 − u2)(~F (u1)− ~F (u2)) ·Dxφdxdtdyds

Entropy inequality for u2, written in y and s. Take k = u1(x, t), integrate in x and t:

0 ≤∫

Rn×R+

∫

Rn×R+

|u1(x, t)− u2(y , s)|φs

+ sgn (u1 − u2)(~F (u1)− ~F (u2)) ·Dyφdxdtdyds

Add the two inequalities:

0 ≤∫

Rn×R+

∫

Rn×R+

|u1 − u2|(φs + φt)

+ sgn (u1 − u2)(~F (u1)− ~F (u2)) · (Dxφ+Dyφ) dxdtdyds

Trick: Chose φε to enforce y = x as ε→ 0∫R ρε = 1

ρε(t) =1

ερ

(t

ε

)

ρε(x) =1

εn

n∏

i=1

ρ(xiε

)

Choose

φ(x, t, y , s) = ~ψ

(x + y

2,s + t

2

)· ρε(x − y

2

)· ρε(t − s

2

)

~ψ smooth, compact support in Rn × (0,∞).

Find

φt + φs = ~ψt(x , t) · ρε(y) · ρε(s)

Dxφ+Dyφ = Dx ~ψ(x , t) · ρε(y) · ρε(s)

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ρ1/4

ρ1

−1

ρ ∈ C∞0 (−1, 1)

1

ρ1/2

x =x + y

2, y =

x − y2

, t =t + s

2, s =

t − s2

→ weighted integral, as ε→ 0 the integration is on y = 0, s = 0 i.e. x = y and t = s.

You get exactly the inequality we get formally with k = u2(x, t).

6.8 Exercises

6.1: Let f (x) be a piecewise constant function

f (x) =

1 x < −1,

1/2 −1 < x < 1,

3/2 1 < x < 2,

1 x > 2.

Construct the entropy solution of the IVP, ut + uux = 0, u(x, 0) = f (x) for small values of

t > 0 using shock waves and rarefaction waves. Sketch the characteristics of the solution.

why does the structure of the solution change for large times?

6.2: Motivation of Lax’s entropy condition via stability. Consider Burgers’ equation

ut + uux = 0

subject to the initial conditions

u0(x) =

0 for x > 0

1 for x < 0, u1(x) =

1 for x > 0

0 for x < 0.

The physically relevant solution should be stable under small perturbations in the initial data.

Which solution (shock wave or rarefaction wave) do you obtain in the limit as n →∞ if you

approximate u0 and u1 by

u0,n(x) =

0 for x < 0

nx for 0 ≤ x ≤ 1n

1 for x > 1n

, u1,n(x) =

1 for x < 0

1− nx for 0 ≤ x ≤ 1n

0 for x > 1n

?

6.3: Let the initial data for Burgers’ equation ut + uux = 0 be given by

f (x) =

ul for x < 0,

ul − ul−urL x for 0 ≤ x ≤ L,ur for x > L,


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6.8 Exercises149

where 0 < ur < ul . Find the time t∗ when the profile of the solution becomes vertical. More

generally, suppose that the initial profile is smooth and has a point x0 where f ′(x0) < 0.

Find the time t∗ for which the solution of Burger’s equation will first break down (develop a

vertical tangent).

6.4: [Profile of rarefaction waves] Suppose that the conservation law

ut + f (u)x = 0

has a solution of the form u(x, t) = v(x/t). Show that the profile v(s) is given by

v(s) = (f ′)−1(s).

6.5: Verify that the viscous approximation of Burgers’ equation

ut + uux = εuxx

has a traveling wave solution of the form u(x, t) = v(x − st) with

v(y) = uR +1

2(uL − uR)

(1− tanh

((uL − uR)y

4ε

))

where s = (uL + uR)/2 is the shock speed given by the Rankine Hugoniot condition. Sketch

this solution. What happens as ε→ 0?

6.6: Consider the Cauchy problem for Burgers’ equation,

ut + uux = 0

with the initial data

u(x, 0) =

1 for |x | > 1,

|x | for |x | ≤ 1.

a) Sketch the characteristics in the (x, t) plane. Find a classical solution (continuous and

piecewise C1) and determine the time of breakdown (shock formation).

b) Find a weak solution for all t > 0 containing a shock curve. Note that the shock

does not move with constant speed. Therefore, find first the solution away from the

shock. Then use the Rankine-Hugoniot condition to find a differential equation for the

position of the shock given by (x = s(t), t) in the (x, t)-plane.

6.7: [Traffic flow] A simple model for traffic flow on a one lane highway without ramps leads

to the following first order equation which describes the conservation of mass,

ρt + vmax

(ρ

(1− ρ

ρmax

))

x

= 0, x ∈ R, t > 0.

In this model, the velocity of cars is related to the density by

v(x, t) = vmax

(1− ρ

ρmax

),

where we assume that the maximal velocity vmax is equation to one. Solve the Riemann

problem for this conservation law with

ρ(x, 0) =

ρL if x < 0,

ρR if x > 0.

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Under what conditions on ρL and ρR is the solution a shock wave and a rarefaction wave,

respectively? Sketch the characteristics. Does this reflect your daily driving experience? Find

the profile of the rarefaction wave.

6.8: [LeVeque] The Buckley-Leverett equations are a simple model for two-phase flow in

a porous medium with nonconvex flux

f (u) =u2

u2 + 12 (1− u)2

.

In secondary oil recovery, water is pumped into some wells to displace the oil remaining in the

underground rocks. Therefore u represents the saturation of water, namely the percentage

of water in the water-oil fluid, and varies between 0 and 1. Find the entropy solution to the

Riemann problem for the conservation law ut + f (u)x = 0 with initial states

u(x, 0) =

1 if x < 0,

0 if x > 0.

Hint: The line through the origin that is tangent to the graph of f on the interval [0, 1]

has slope 1/(√

3− 1) and touches at u = 1/√

3. If you are proficient with Mathematica or

Matlab you could try to plot the profile of the rarefaction wave.

6.9: Let u0(x) = sgn(x) be the sign function in R and let a be a constant. Let u(x, t)

be the solution of the transport equation

ut + aux = 0,

and let uε(x, t) be the solution of the so-called viscous approximation

uεt + auεx − εuεxx = 0,

both subject to the initial condition u(·, 0) = uε(·, 0) = u0.

a) Derive representation formulas for u and uε!

Hint: For the viscous approximation, consider v ε(x, t) = uε(x + at, t).

b) Prove the error estimate

∫ ∞

−∞|u(x, t)− uε(x, t)| dx ≤ C

√εt, for t > 0.


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152Chapter 7: Elliptic, Parabolic and Hyperbolic Equations

7 Elliptic, Parabolic and Hyperbolic Equations


– Alan Rickman

Contents

7.1 Classification of 2nd Order PDE in Two Variables 153

7.2 Maximum Principles for Elliptic Equations of 2nd Order 158

7.3 Maximum Principles for Parabolic Equations 172

7.4 Exercises 178


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7.1 Classification of 2nd Order PDE in Two Variables153

7.1 Classification of 2nd Order PDE in Two Variables

a(x, y) · uxx + 2b(x, y) · uxy + c(x, y) · uyy + d(x, y) · ux + e(x, y) · uy + g(x, y) · u = f (x, y)

General mth order PDE

L(x,D)u =∑

|α|≤maα(x) ·Dαu

where u : Rn → RPrinciple part:

Lp(x,D) =∑

|α|=maα(x)Dα

The symbol of the PDE is defined to be

L(x, iξ) =∑

|α|≤maα(x) · (iξ)α

(you take iξ because of Fourier transform)

ξ ∈ Rn, replace ∂∂xj

by iξj .

Examples: Laplace ∆ =∑j∂2

∂x2j

= L

L(x, iξ) =

n∑

j=1

(iξj)n = −|ξ|2

Heat Equation: t = n + 1.

Lu =∂u

∂xn+1− ∆u =

Symbol:

L(x, iξ) = iξn+1 −n∑

j=1

(iξj)2

Wave equation:

L =∂2

∂x2n+1

− ∆

L(x, iξ) = (iξn+1)2 −n∑

j=1

(iξj)2

Principle part of the symbol:

Lp(x, iξ) =∑

|α|=maα(x) · (iξ)α

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Examples:

∆ : Lp = −|ξ|2heat: Lp = ξ2

1 + · · ·+ ξ2n

wave: Lp = −ξ2n+1 + ξ2

1 + ·+ ξ2n

2nd order equation in two variables

Lp(x,D) = a(x, y)∂2

∂x2+ 2b(x, y)

∂2

∂x∂y+ c(x, y)

∂2

∂y2

Lp(x, iξ) = a(x, y)ξ21 + 2b(x, y)ξ1ξ2 + c(x, y)ξ2

2

=

(ξ1

ξ2

)T ( −a −b−b −c

)

︸︷︷︸A

(ξ1

ξ2

)

Definition 7.1.1: The PDE is

elliptic if detA > 0

parabolic if detA = 0

hyperbolic if detA < 0.

Typical situation:

(λ1 0

0 λ2

)

Equation: λ1uxx + λ2uyy+ lower order terms

elliptic: −A =

(1 0

0 1

)uxx + uyy = ∆

parabolic: −A =

(1 0

0 0

)uxx + lower order terms

hyperbolic: −A =

(1 0

0 −1

)uxx − uyy = = wave

Remarks: This definition even allows an ODE in the parabolic case (not good).

Better definition: parabolic is ut− elliptic = f .

ut − ∆u = f

Two reasons why this determinant is important:

(A) Given Cauchy data along a curve, C(x(s), y(s)) parametrized with respect to arc

length, that is

u(x(s), y(s)) = u0(s)

∂u

∂n(x(s), y(s)) = v0(s).

Question: Conditions of C that determine the second derivatives of a smooth solution.

Tangential derivative

Du · ~T = ux x + uy y = w0Gregory T. von Nessi

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n = (−y , x)(x , y)

Normal derivative Du · ~n = −ux y + uy x = v0

uxx x2 + 2uxy x y + uyy y

2 + ux x + uy y = w0(s)

−uxx x y − uxy y2 + uxy y2 + uxy x

2 + uyy x y − ux y + uy x = v0(s)

uxx x2 + 2uxy x y + uyy y

2 = p(s)

−uxx x y + (x2 − y2)uxy + uyy x y = r(s)

auxx + 2buxy + cuyy = q(s)

We can solve for the 2nd derivatives along the curve if

det

∣∣∣∣∣∣

x2 2x y y2

−x y x2 − y2 x y

a 2b c

∣∣∣∣∣∣6= 0

det

∣∣∣∣∣∣

x2 2x y y2

−x y x2 − y2 x y

a 2b c

∣∣∣∣∣∣= a(2x2y2 − x2y2 − x2y2 + y4)

−2b(x3y + xy3) + c(x4 − x2y2 + 2x2y2)

= a(y4 + x2y2)− 2b(x3y + x y3)

+c(x4 + x2y2) 6= 0

Divide by (x y)2:

a

(y2

x2+ 1

)− 2b

(y

x+x

y

)+ c

(x2

y2+ 1

)6= 0

⇐⇒ a

(dy2

dx2+ 1

)− 2b

(dy

dx+dx

dy

)+ c

(dx2

dy2+ 1

)6= 0

⇐⇒(

1 +

(dx

dy

)2)·(a

(dy

dx

)2

− 2bdy

dx+ c

)

︸︷︷︸6=0

6= 0

at2 − 2bt + c = 0

a

(t2 − 2b

at +

c

a

)= 0

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⇐⇒(t − b

a

)2

− b2

a2+ac

a2= 0

⇐⇒ t =b

a±√−ac + b2

a

Non-trivial solutions if

det

(a b

b c

)= ac − b2 ≤ 0

elliptic equation → 2nd derivative are determined.

Definition 7.1.2: A curve C is called a characteristic curve for L if

a

(dy

dx

)2

− 2bdy

dx+ c = 0.

(b) Curves of discontinuity

a, . . . , f , g continuous.

u solution of Lu = f , u, ux , uy continuous, uxx , uxy , uyy continuous except on a curve

C, jump along C. What condition has C to satisfy.

C = (x(s), y(s))

u l

ur

ulx(x(s), y(s)) = urx(x(s), y(s)).

Differentiate along the curve:

ulxx x + ulxy y = urxx x + urxy y

[[uxx ]] x + [[uxy ]] y = 0 (7.1)

Analogously: uly (x(s), y(s)) = ury (x(s), y(s))

=⇒ [[uxy ]] x + [[uyy ]] y = 0 (7.2)

Equation holds outside the curve, all 1st order derivatives and coefficients are contin-

uous =⇒ along the curve

a [[uxx ]] + 2b [[uxy ]] + c [[uyy ]] = 0 (7.3)

=⇒ linear system.

x y 0

0 x y

a 2b c

[[uxx ]]

[[uxy ]]

[[uyy ]]

= 0


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Non-trivial solution only if

det

x y 0

0 x y

a 2b c

= ay2 − 2bx y + cx2 = 0

⇐⇒ a

(dy

dx

)2

− 2bdy

dx+ c = 0

Jump discontinuity only possible if C is characteristic,

Elliptic: no characteristic curve

Parabolic: one characteristic

Hyperbolic: two characteristic

dy

dx=b

a±√b2 − aca

Laplace: a = c = 1, b = 0 dydx = ±

√−1. No solution

Parabolic: a = 1, c = 0, b = 0 dydx = 0 y = const. characteristic.

Hyperbolic: a = 1, c = −1, b = 0 dydx = ±1, y = ±x characteristic.

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7.2 Maximum Principles for Elliptic Equations of 2nd Order

Definition 7.2.1: L is an elliptic operator in divergence form if

L(x,D)u = −n∑

i ,j=1

∂i(ai j(x)∂ju) +

n∑

i=1

bi(x)∂iu + c(x)u(x)

and if the matrix (ai j) is positive definite, i.e.,

n∑

i ,j=1

ai j(x)ξiξj ≥ λ(x)|ξ|2

for all x ∈ Ω, all ξ ∈ Rn, λ(x) > 0

Remarks: Usually elliptic means uniformly elliptic, i.e.

(1) λ(x) ≥ λ0 > 0 ∀x ∈ Ω.

(2) Divergence form because L(x,D)u = −div (ADu) + ~b ·Du + cu where A = (ai j)

Definition 7.2.2: L is an elliptic operator in non-divergence form if

L(x,D) = −n∑

i ,j=1

ai j(x)∂i∂ju +

n∑

i=1

bi(x)∂iu + c(x)u(x)

and

n∑

i ,j=1

ai jξiξj > λ(x)|ξ|2

for all x ∈ Ω, ξ ∈ Rn

Remark: If ai j ∈ C1, then an operator in non-divergence form can be written as an operator

in divergence form. We may also assume that ai j = aj i .

Definition 7.2.3: A function u ∈ C2 is called a subsolution (supersolution) if L(x,D)u ≤ 0

(L(x,D)u ≥ 0)

Remark: n = 1, Lu = −uxxsubsolution −uxx ≤ 0 ⇐⇒ uxx ≥ 0 ⇐⇒ u convex =⇒ u ≤ v if vxx = 0 on v = u at the

endpoint of an interval.

Remark: We can not expect maximum principles to hold for general elliptic equations. Ob-

struction: the existence of eigenvalues and eigenfunctions.

Example: −u′′ = λu

λ = 1, u(x) = sin(x) on [0, 2π]

=⇒ sin x solves −u′′ = u

There is no maximum principle for −uxx − u = L.

Big difference between x ≥ 0 and c < 0.


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7.2 Maximum Principles for Elliptic Equations of 2nd Order159

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7.2.1 Weak Maximum Principles


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7.2.1.1 WMP for Purely 2nd Order Elliptic Operators

Again we take Ω ⊂ Rn. We now consider the following differential operator:

Lu :=

n∑

i ,j=1

ai jDi ju +

n∑

i=1

biDiu + cu,

where ai j , bi , c : Ω→ R. Now, we state the following.

Definition 7.2.4: The operator L is said to be elliptic(degenerate elliptic) if A := [ai j ] >

0(≥ 0) in Ω.

Remarks:

i.) In the above definition A > 0 means the minimum eigenvalue of the matrix A is > 0.

More explicitly, this means that

n∑

i ,j=1

ai jξiξj > 0, ∀ξ ∈ Rn.

ii.) The last definition indicates that parabolic PDE (such as the Heat Equation) are really

degenerate elliptic equations.

For convenience, we will be using the Einstein summation convention for repeated indicies.

This means that in any expression where letters of indicies appear more than once, a sum-

mation is applied to that index from 1 to n. With that in mind, we can write our general

operator as

Lu := ai jDi ju + biDiu + cu.

Now, we move onto the weak maximum principle for our special case operator:

Lu = ai jDi ju.

Theorem 7.2.5 (Weak Maximum Principle): Consider u ∈ C2(Ω)∩C0(Ω) and f : Ω→ R.

If Lu ≥ f , then

u ≤ max∂Ω

u +1

2

(sup

Ω

|f |T rA

)·Diam (Ω)2 , (7.4)

where T rA is the trace of the coefficient matrix A (i.e. T rA =∑ni=1 a

i i).

Note: If f ≡ 0, the above reduces to the result of the first weak maximum principle we

proved. Namely, u ≤ max∂Ω u.

Proof: Without loss of generality we may translate Ω so that it contains the origin. Again,

we consider an auxiliary function:

v = u + k |x |2,

where k is an arbitrary constant to be determined later. Since the Hessian Matrix D2u is

symmetric, we may choose a coordinate system such that D2u is diagonal. In this coordinate

system, we calculate

Lv = Lu + 2kai jSi j = Lu + 2k · T rA,

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where Si i = 1 and Si j = 0 when i 6= j . Now, suppose that v attains an interior maximum at

y ∈ Ω. From calculus, we then know that D2v(y) ≤ 0, which implies that ai jDi j ≤ 0 as Ais positive definite. Now, if we choose

k >1

2sup

Ω

|f |T rA ,

our above calculation indicates that Lv > 0; a contradiction. Thus,

v ≤ max∂Ω

v

which implies the result as Ω contains the origin (which indicates that |x |2 ≤ Diam (Ω)2).

Example 3: Now we will apply the weak maximum principle to the second most famous

elliptic PDE, the Minimal Surface Equation:

(1 + |Du|2)∆u −DiuDjuDi ju = 0.

This is a quasilinear PDE as only derivatives of the 1st order are multiplied against the 2nd

order ones (a fully nonlinear PDE would have products of second order derivatives). We will

now show that this is indeed an elliptic PDE. Recalling our special case Lu = ai jDi ju, we

can rewrite the minimal surface equation in this form with

ai j = (1 + |Du|2)Si j +DiuDju,

with S being as it was in the previous proof. Now, we calculate

ai jξiξj = (1 + |Du|2)Si jξiξj −DiuDjuξiξj= (1 + |Du|2)|ξ|2 −DiuDjuξiξj= (1 + |Du|2)|ξ|2 − (Diuξi)

2

≥ (1 + |Du|2)|ξ|2 − |Du|2|ξ|2

= |ξ|2.

To get the third equality in the above, we simply relabeled repeated indicies. Since one sums

over a repeated index, it’s representation is a dummy index, whose relabeling does not affect

the calculation.

From this calculation, we conclude this equation is indeed elliptic. Now, upon taking f = 0

and replacing u by −u, we can now apply the weak maximum principle to this equation.


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7.2.1.2 WMP for General Elliptic Equations

Now, we will go over the weak maximum principle for operators having all order of derivatives

≤ 2.

Theorem 7.2.6 (Weak Maximum Principle): Consider u ∈ C2(Ω) ∩ C0(Ω); ai j , bi , c :

Ω→ R; c ≤ 0; and Ω bounded. If u satisfies the elliptic equation (i.e. [ai j ] = A > 0)

Lu = ai jDi j + biDiu + cu ≥ f , (7.5)

then

u ≤ max∂Ω

u+ + C

(n,Diam (Ω) , sup

Ω

|b|λ

)· sup

Ω

|f |λ, (7.6)

where

λ = min|ξ|=1

ai jξiξj and u+ = maxu, 0.

Note: If ai j is not constant, then λ will almost always be non-constant on Ω as well.

Consequences:

• If f = 0, then u ≤ max∂Ω u+.

• If f = c = 0, then u ≤ max∂Ω u+.

• If Lu ≤ f one gains a lower bound

u ≥ max∂Ω

u− − C supΩ

|f |λ,

where u− = minu, 0.

• If we have the PDE Lu = f , then one has both an upper and lower bound:

|u| ≤ max∂Ω|u|+ C sup

Ω

|f |λ.

• Uniqueness of the Dirichlet Problem. Recall, that the problem is as follows

PDE: Lu = f in Ω

u = φ on ∂Ω.

So if one has u, v ∈ C2(Ω)∩C0(Ω) with Lu = Lv in Ω, and u = v on ∂Ω, then u ≡ vin Ω. The proof is the simple application of the weak maximum principle to w = u−v ,

which is obviously a solution of PDE as it’s linear.

Proof of Weak Max. Principle: As in previous proofs, we analyze a particular auxiliary function

to prove the result.

Aside: v = |x |2 will not work in this situation as Lv = 2 · T rA+ 2bixi + c |x |2; the bixi may

be a large negative number for large values of |x |.

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xi

x0

ξ

d

O

Ω− = x ∈ Ω | u < 0

Ω+ = x ∈ Ω | u > 0

Let us try

v = eα(x,ξ),

where ξ is some arbitrary vector. Also, we will consider Ω+ = x ∈ Ω | u > 0 instead of Ω,

for the rest of the proof. Clearly, this does not affect the result as we are seeking to bound

u by max∂Ω u+. With that, we have on Ω+

L0u := ai jDi ju + biDiu ≥ −cu + f

≥ f .

Now, we wish to specify which vector ξ is. For the following, please refer to the above figure.

We first choose our coordinates so that the origin corresponds to a boundary point of Ω such

that there exists a hyperplane through that point with Ω lying on one side of it. ξ is simply

taken to be the unit vector normal to this hyperplane. Next, we calculate

L0v = (α2ai jξiξj + αbiξi)eα(x,ξ)

= (α2ai jξiξj + αbiξi)eα(x,ξ)

≥ αλeα(x,ξ),Gregory T. von Nessi

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provided we choose α ≥(

supΩ|b|λ + 1

).

Aside: The calculation for the determination of α is as follows:

α2ai jξiξj + αbiξi ≥ α2λ+ αbiξi .

So, α2λ+ αbiξi ≥ αλ implies that

αλ ≥ λ− biξi=⇒ α ≥ sup

Ω

(1− b

iξiλ

)

So, clearly we have the desired inequality if we choose α ≥ b + 1, where b := supΩ|b|λ .

Given our choice of ξ, it is clear that (x, ξ) ≥ 0 for all x ∈ Ω. Thus, one sees that

L0v ≥ αλ. So, one can ascertain that

L0(u+ + kv) ≥ f + αλk > 0 in Ω+,

provided k is chosen so that

k >1

αsup

Ω

|f |λ.

Now, the proof proceeds as previous ones. Suppose w := u+ + kv has a positive maximum

in Ω+ at a point y . In this situation we have

[Di jw(y)] ≤ 0 =⇒ ai jDi jw(y) ≤ 0, biDiw(y) = 0,

which implies Lw ≤ 0; a contradiction. So, this implies the second inequality of the following:

u +e0

αsup

Ω

|f |λ≤ w

≤ max∂Ω+

w

≤ max∂Ω+

u+ +eαd

α· sup

Ω

|f |λ

≤ max∂Ω

u+ +eαd

α· sup

Ω

|f |λ,

again where α = b + 1 and d is the breadth of Ω in the direction of ξ. From the above, we

finally get

u ≤ max∂Ω

u+ +eαd − 1

α· sup

Ω

|f |λ.

Remark: Given the above, one needs |b|λ and |f |λ bounded for the weak maximum principle to

give a non-trivial result. Then one may apply the weak maximum principle to get uniqueness.

Theorem 7.2.7 (weak maximum principle, c = 0): Ω open, bounded, u ∈ C2(Ω)∩C(Ω).

L uniformly elliptic (for divergence form) Lu ≤ 0. Then

maxΩu = max

∂Ωu

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Analogously, if Lu ≥ 0, then

minΩu = min

∂Ωu

Proof:

(1) Lu < 0 and u has a maximum at x0 ∈ Ω. Du(x0) = 0 and

(∂2u

∂xi∂xj(x0)

)

i ,j

= D2u(x0)

is negative semidefinite.

Fact from linear algebra: A, B symmetric and positive definite, then A : B = tr (AB) ≥0.

Q orthogonal, QAQT = Λ,

λ = diag (λ1, . . . , λn), λi ≥ 0

tr (AB) = tr (QABQT )

= tr (QAQTQBQT )

= tr (ΛB)

B = QBQT . B symmetric and positive semidefinite =⇒ Bi i ≥ 0.

tr (ΛB) =

n∑

i=1

λi Bi i ≥ 0

tr (AB) =

n∑

i=1

(AB)i i

=

n∑

i=1

n∑

j=1

Ai jBj i

=

n∑

i ,j=1

Ai jBi j

= A : B

Lu = −n∑

i ,j=1

ai j∂i ju +

n∑

i=1

bi∂iu

Lu(x0) = −(ai j) : D2u(x0) + 0

(ai j) symmetric and positive definite. −D2u(x0) symmetric and positive semidefinite.

Conclusion:

0 >︸︷︷︸assump.

Lu(x0) ≥ 0

Contradiction, u can not have a maximum at x0.Gregory T. von Nessi

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(2) Lu(x0) = 0.

Consider uε(x) = u(x) + εerx1 ,

ε > 0, r > 0, r = r(x) chosen later.

Luε = Lu + εLerx1 ≤ εLerx1

= ε(−a11r2 + b1r)erx1

a11 > 0 (since∑ai jξiξj ≥ λ0|ξ|2, take ξ1 = e1 = (1, 0, . . . , 0), a11 ≥ λ0 > 0)

b = b(x) by assumption in C(Ω).

|b(x)| ≤ B = maxΩ |b|

Luε ≤ ε(−a11r2 + Br)erx1

= εr(−a11r + B)erx1 < 0

if r > 0 is big enough.

By the previous argument,

maxΩuε = max

∂Ωuε

As ε→ 0, uε → u and in the limit

maxΩu = max

∂Ωu

(In fact, r is an absolute constant, r = r(λ0, B)).

Corollary 7.2.8 (Uniqueness): There exists at most one solution of the BVP Lu − f in Ω,

u = u0 in ∂Ω with u ∈ C2(Ω) ∩ C(Ω).

Proof: By contradiction with the maximum and minimum principle.

Theorem 7.2.9 (weak maximum principle, c ≥ 0): Suppose u ∈ C2(Ω) ∩ C(Ω), and that

c ≥ 0.

i.) If Lu ≤ 0, then

maxΩu ≤ max

∂Ωu+

ii.) If Lu ≥ 0, then

minΩu ≥ −max

∂Ωu−

where u+ = max(u, 0), where u− = −min(u, 0),

that is, u = u+ − u−.

Proof: Proof for ii.) follows from i.) with −u.

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i.) Let Ω+ = Ω, u(x) > 0 and define Ku = Lu − cu. If Ω+ = ∅, then the assertion is

obviously true.

On Ω+, Ku = Lu − cu ≤ 0 − cu ≤ 0, i.e. on Ω+, u is a subsolution for K and

we may use the weak maximum principle for K.

If Ω+ 6= ∅, then use the weak maximum principle for K on Ω+

maxΩ

+u(x) ≤ max

∂Ω+u(x) = max

∂Ωu+ (7.7)

Ω

Ω+On ∂Ω+ ∩Ω we have u = 0.

Since u is not less that or equal to zero on Ω,

maxΩ

+u(x) = max

∂Ω+u(x) = max

∂Ωu+

where the second equality is due to (7.7).

Recall: Weak Maximum Principles Lu ≤ 0

c = 0 : maxΩu = max

∂Ωu

c ≥ 0 : maxΩu ≤ max

∂Ωu+

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7.2.2 Strong Maximum Principles

Theorem 7.2.10 ((Hopf’s maximum principle; Boundary point lemma)): Ω bounded,

connected with smooth boundary. u ∈ C2(Ω)∩C(Ω). Subsolution Lu ≤ 0 in Ω and suppose

that x0 ∈ ∂Ω such that u(x0) > u(x) for all x ∈ Ω. Moreover, assume that Ω satisfies an

interior ball condition at x0, i.e. there exists a ball

Ω

x0

yR

B(y , R) ⊂ Ω such that B(y , R) ∩Ω = x0

i.) If c = 0, then

∂u

∂ν(x0)

strict > 0 follows from assumption.

ii.) If c ≥ 0, then the same assertion holds if u(x0) ≥ 0

Proof:

Idea: Construct a barrier v ≥ 0, v > 0 in Ω,

w = u(x)− u(x0)∓ εv(x) ≤ 0 in B.

u(x)− u(x0) ≤ εv(x)

Construction of v(x): v(x) = e−αr2 − e−αR2

where r = |x − y |v(x) > 0 in B

v(x) = 0 on ∂B

∂v

∂xi= (−α)2(xi − yi) · e−αr

2

Lv = −∑

ai j∂i jv +∑

∂ivbi + cu

= −∑

4α2(xi − yi)(xj − yi)e−αr2

ai j

+∑(ai i2α− 2αbi(xi − yi)e−αr

2

+ c(e−αr2 − e−αR2

)

——

Ellipticity:∑

ai jξiξj ≥ λ0|ξ|2 ⇐⇒ −∑

ai jξiξj ≤ −λ0|ξ|2

——

≤ −λ04α2|x − y |2e−αr2

+ tr (ai j)2αe−αr2

+2α|~b| · |x − y |e−αr2

+ c(e−αr2 − e−αR2

)

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——[e−αr

2 − e−αR2

= e−αr2

(1− e−α(R2−r2))︸︷︷︸≤1

]

——

≤ e−αr2−λ04α2|x − y |2 + 2α tr (ai j) + 2α|b| · |x − y |+ |c |

If |x − y | bound from below, then Lv ≤ 0 if α is big enough.

∂Ω

x0

A

R2

B(y ,R)

A = B(y , R) ∩ B(x0,

R2

)

Then |x − y | ≥ R2 in A and Lv ≤ 0 for α big enough.

Define w(x) = u(x)− u(x0) + εv(x)

Lw = Lu︸︷︷︸≤0

−Lu(x0)︸︷︷︸c(x0)

+ε Lv︸︷︷︸≤0

≤ −cu(x0)

< 0

=⇒ w a subsolution.

w on ∂A: ∂A ∩ ∂B(y , R). w ≤ 0

∂A ∩ B(y , R): u(x0) > u(x) in Ω

=⇒ u(x)− u(x0) ≤ −δ, δ > 0 by compactness.

=⇒ w ≤ 0 if ε is small enough.

Lw ≤ 0 in B

w ≤ 0 on ∂B

Apply weak max principle to w .

maxBw ≤ max

∂Ωw+ = 0

=⇒ w(x) ≤ 0 for all x ∈ A.

w(x) = u(x)− u(x0) + εv(x)

w(x0) = 0 and u(x)− u(x0) ≤ −εv(x)Gregory T. von Nessi

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=⇒ ∂u

∂ν(x0) ≥ −ε∂v

∂ν(x0)

∂v

∂xi= −2α(xi − yi)e−αr

2

∂u

∂ν(x0) ≥ −ε(−2α)(x0 − y)e−αR

2 x0 − x|x0 − y |

= 2ε · αe−αR2 |x0 − y |2|x0 − y |︸︷︷︸

=R>0

Theorem 7.2.11 (Strong maximum principle, c = 0): Ω open and connected (not neces-

sarily bounded), c = 0

i.) If Lu ≤ 0, and u attains max at an interior point, then u ≡ constant.

ii.) Lu ≥ 0, u has a minimum at an interior point, the u is constant.

Proof:

M = maxΩu, u 6= M,

V = x ∈ Ω | u(x) < M 6= ∅ (contradiction assump.)

x

Ω

ν

x0

V

Choose x0 ∈ V such that dist(x0, ∂V ) <dist(x, ∂Ω)

=⇒ B(x0, |x − x0|) ⊂ V . We may assume that B ∩ ∂V = x=⇒ Satisfy assumptions in the boundary point lemma.

=⇒ ∂u∂ν (x) > 0

Contradiction, since ∂u∂ν (x) = Du(x) · ν = 0 since u(x) = M, i.e. x is a maximum point of

u.

Theorem 7.2.12 (Strong maximum principle, c ≥ 0): Ω open and connected (not neces-

sarily bounded). u ∈ C2(Ω), c ≥ 0.

i.) Lu ≤ 0 in Ω and u attains a non-negative maximum, then u = constant in Ω

ii.) Lu ≥ 0 u has non-positive maximum then u ≡ constant in Ω.

Proof: Hopf’s Lemma for non-negative maximum point at the boundary.

Example: c = 1, −u′′ + u = 0. u = sinh, cosh

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7.3 Maximum Principles for Parabolic Equations

Definition 7.3.1: L(x, t,D) = ut + Au,

Au = −n∑

i ,j=1

ai j(x)∂i ju +

n∑

i=1

bi∂iu + cu,

is said to be parabolic. A is (uniformly) elliptic

n∑

i ,j=1

ai jξiξj ≥ λ0|ξ|2 ∀ξ ∈ Rn, λ0 > 0.

Remark: Solutions of elliptic equations are stationary states of parabolic equations =⇒Same assumptions concerning c .

Ω open, bounded, ΩT = Ω× (0, T ]

ΓT = ∂ΩT = Ω× t = 0 ∪ ∂Ω× [0, T )

Ω

ΩTΓT


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7.3 Maximum Principles for Parabolic Equations173

7.3.1 Weak Maximum Principles

Theorem 7.3.2 (weak maximum principle, c = 0): u ∈ C21 (ΩT ) ∩ C(Ω), c = 0 in ΩT

i.) Lu = ut + Au ≤ 0 in ΩT , then

maxΩT

u = maxΓT

u

ii.) Lu ≥ 0,

minΩT

u = minΓTu

Proof:

i.) Suppose Lu < 0 and that the max is attained at x0, t0) in ΩT . 0 > Lu = ut + Au =

0 + (Au ≥ 0) ≥ 0

since Au ≥ 0 as in the elliptic case. contradiction.

If the max is attained at (x0, T ), 0 > Lu = ut + Au ≥ 0 since ut(x0, T ) ≥ 0 if u is

max at (x0, T )

If Lu ≤ 0, then take uε = u − εtLuε = (∂t + A)(u − εt) = Lu − ε ≤ −ε < 0

uε satisfies Luε < 0, and then

maxΩT

uε = maxΓT

uε,

∀ε > 0, and the assertion follows as ε 0.

Theorem 7.3.3 (weak maximum principle, c ≥ 0): u ∈ C21 (ΩT ) ∩ C(ΩT ), c ≥ 0 in ΩT .

u = u+ − u−, where u+ = max(u, 0)

u− = −min(u, 0)

i.) If Lu = ut + Au ≤ 0,

maxΩT

u ≤ maxΓT

u+

ii.) Lu ≥ 0, then

minΩT

u ≥ −maxΓT

u−.

Proof: Suppose first Lu < 0 and that u has a positive maximum at (x0, t0) in ΩT .

0 > Lu = (∂t + A)u

≥ c(x0, t0)u(x0, t0) > 0 contradiction

Similar for t0 = T

General case: uε = u − εtLuε = (∂t + A)(u − εt)

≤ 0− ε− ε c(x, t)︸︷︷︸≥0

t

≤ −ε < 0

Previous sept applies. If u has a positive max, then uε has a positive max in ΩT for ε small

enough.

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7.3.2 Strong Maximum Principles

Evans: Harnack inequality for parabolic equations

Protter, Weinberger: Maximum principles in differential equations, some form of Hopf’s

Lemma.

Theorem 7.3.4 (): Ω bounded, open, smooth boundary and u ∈ C21 (ΩT ) ∩ C(ΩT ) with

Lu = ut + Au ≤ 0. Suppose that either

i.) c = 0

ii.) c ≥ 0,

maxΩT

u = M ≥ 0

The the strong maximum principle holds.

Idea of Proof: Ω = (α, β). n = 1, ut − auxx + bux + cu = L, a = a(x, t) ≥ λ0 > 0

Step 1: B = Bε(ε, τ) open ball in ΩT , u < M in Bε, u(x0, t0) = M, (x0, t0) ∈ ∂Bε=⇒ x0 = ξ

ξ

T

x0(x0, t0)

x0

β

τ

α

The proof (by contradiction) uses the weak max principle for w = u − M + ηv , where

v(x, t) = e−Ar2 − e−Aε2

, A > 0. Compute

Lv = e−Ar2−α4A2(x − ξ)2 + . . .

∴ r2 = (x − ξ)2 + (t − τ)2

If the max is attained at a point (x, t0) with x0 6= ξ, choose δ small enough such that

|x − ξ| > c(δ) > 0 ∀x ∈ Bδ(x0)

w subsolution, w < 0 on ∂Bδ, w(x0, t0) = 0. Contradiction with weak max principle.

Step 2: Suppose u(x0, t0) < M with t0 ∈ (0, T ). Then u(x, t0) < M for all x ∈ (α, β).

Idea: Suppose ∃ x1 ∈ (α, β0) with u(x, t0) = M and u(x, t0) < M for x ∈ (x0, x1)

d(x) =distance to point (x, t0)

from the set u = M d(x1) = 0, d(x0) > 0

d(x + h) ≤√d2(x0) + h2

d(x0) ≤√d2(x0 + h) + h2


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7.3 Maximum Principles for Parabolic Equations175

(x0, t0)

βα x0 x1

u(x1, t0) = M

x

u(x0, t0 + d(x0)) = M

x0 x1

h

x

u(x0, t0 − d(x0)) = M

Suppose d is differentiable.

Rewrite these inequalities as

√d2(x)− h2 ≤ d(x + h) ≤

√d2(x) + h2

√d2(x)− h2 − d(x)

h≤ d(x + h)− d(x)

h

≤√d2(x) + h2 − d(x)

h

As h → 0, d ′(x) = 0 =⇒ d = constant

Contradiction, since d(x0) > 0, d(x1) = 0.

Step 3: Suppose u(x, t) < M for α < x < β, 0 < t < t1, t1 ≤ T , then u(x, t1) < M

for x ∈ (α, β)

Proof: Weak maximum principle for w(x, t) = u−M+ηv , where v(x, t) = exp(−(x−x1)2−A(t − t1))− 1

To conclude that ut > 0 Lu ≤ 0 subsolution.

Lu = ut︸︷︷︸>0

− auxx︸︷︷︸≥0

+ bux︸︷︷︸=0

+ cu︸︷︷︸c(x, t) ·M︸︷︷︸

≥0

at (xi , t1)

> 0. contradiction.

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u < M

T

βα

t1

(x1, t1)

(x − x1)2 − A(t − t1) = 0

(x0, t0)M

t1

L

Conclusion. Suppose that u(x0, t0) = M, (x0, t0) ∈ ΩT

To show if u(x0, t0) = M = max, then u ≡ M on Ωt0

We can not conclude that u ≡ M on ΩT .

Consider the line L(x0)− (x0, t), 0 ≤ t < t0Suppose there is a t1 ≤ 0 such that u(x0, t) < M for t < t1, u(x0, t1) = M

=⇒ u(x, t) < M for all (x, t) in (α, β)× (0, t1)

=⇒ Step 3: u(x0, t) < 0. Contradiction.

Modify the function v in n dimensions; the same argument works.


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7.4 Exercises177

7.4 Exercises

7.1: [Qualifying exam 08/90] Consider the hyperbolic equation

Lu = uyy − e2xuxx = 0 in R2.

a) Find the characteristic curves through the point(

0, 12

).

b) Suppose that u is a smooth of the initial value problem

uyy − e2xuxx = 0 in R2,

u(x, 0) = f (x) for x ∈ R,uy (x, 0) = g(x) for x ∈ R.

Show that the characteristic curves bound the domain of dependence for u at the point(0, 1

2

). To do this, prove that u

(0, 1

2

)= 0 provided f and g are zero on the intersection

of the domain of dependence with the x-axis. This can be done by defining the energy

e(y) =1

2

∫ h(y)

g(y)

(u2x + e−2xu2

y ) dx

where (g(y), h(y)) is the intersection of the domain of dependence with the parallel

to the x-axis through the point (0, y). Then show that e ′(y) ≤ 0. (See also proof for

the domain of dependence for the wave equation with energy methods.)

7.2: [Evans 7.5 #6] Suppose that g is smooth and that u is a smooth solution of

ut − ∆u + cu = 0 in Ω× (0,∞),

u = 0 on ∂Ω× [0,∞),

u = g on Ω× t = 0,

and that the function c satisfies c ≥ γ > 0. Prove the decay rate

|u(x, t)| ≤ Ce−γt for (x, t) ∈ Ω× (0,∞).

Hint: The solution of an ODE can also be viewed as a solution of a PDE!

7.3: [Evans 7.5 #7] Suppose that g is smooth and that u is a smooth solution of

ut − ∆u + cu = 0 in Ω× (0,∞),

u = 0 on ∂Ω× [0,∞),

u = g on Ω× t = 0.

Moreover, assume that c is bounded, but not necessarily nonnegative. Show that u ≥ 0.

Hint: What PDE does v(x, t) = eλtu(x, t) solve?

7.4: [Qualifying exam 08/01] Let Ω be a bounded domain in Rn with a smooth bound-

ary. Use

a) a sharp version of the Maximum principle

b) and an energy method

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to show that the only smooth solutions of the boundary-value problem

∆u = 0 in Ω,

∂nu = 0 on ∂Ω

(where ∂n is the derivative in the outward normal direction) have the form u = constant.

7.5: [Qualifying exam 01/02] Let B be the unit ball in Rn and let u be a smooth func-

tion satisfying

ut − ∆u + u1/2 = 0, 0 ≤ u ≤ M on B × (0,∞),

ur = 0, on ∂B × (0,∞).

Here M is a positive number and ur is the radial derivative on ∂B. Prove that there is a

number T depending only on M such that u = 0 on B × (T,∞).

Hint: Let v be the solution of the initial value problem

dv

dt+ v1/2 = 0, v(0) = M,

and consider the function w = v − u.


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Part II

Functional AnalyticMethods in PDE

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180Chapter 8: Facts from Functional Analysis and Measure Theory

8Facts from Functional Analysis and MeasureTheory


– Alan Rickman

Contents


8.2 Banach Spaces and Linear Operators 183

8.3 Fredholm Alternative 187

8.4 Spectral Theorem for Compact Operators 191

8.5 Dual Spaces and Adjoint Operators 195

8.6 Hilbert Spaces 196

8.7 Weak Convergence and Compactness; the Banach-Alaoglu Theorem 201

8.8 Key Words from Measure Theory 204

8.9 Exercises 209

8.1 Introduction

This chapter represents what, I believe, to be the most pedagogically challenging material

in this book. In order to break into modern theory of PDE, a certain amount of functional

analysis and measure theory must be understood; and that is what the material in this chap-

ter is meant to convey. Even though I’ve made every effort to keep the following grounded

in PDE via remarks and examples, it is understandable how the reader might regard some

parts of the chapter as unmotivated. Unfortunately, I believe this to be inevitable as there

is so much background material that needs to be “water over the back” to have a decent

understanding of modern methods in PDE. In the next chapter we will start using the tools

learned here to get some actual results in the context of linear elliptic and parabolic equations.


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8.2 Banach Spaces and Linear Operators181

In regard to organization, I have tried to order the material in this chapter as sequentially

as possible, i.e. material conveyed can be understood based off of material in the previous

section. In addition important theorems and their consequence are presented throughout

as soon as possible. I decided to focus on these two principles in hopes that the ordering

or material will directly indicate to what generality theorems and ideas can be applied. For

example, the validity of a given theorem in a Banach space as opposed to Hilbert spaces will

hopefully be made clear by the order in which the material is presented. Lastly, for the sake

of completeness, many proofs have been included in this chapter which can (and should) be

skipped on the first reading of this book, as many of these dive into rather pedantic points

of real and functional analysis.

8.2 Banach Spaces and Linear Operators

Philosophically, one of the major goals of modern PDE is to ascertain properties of solu-

tions when equations are not able to be solved analytically. These properties precipitate out

of classifying solutions by seeing whether or not they belong to certain abstract spaces of

functions. These abstractions of vector spaces are called Banach Spaces (defined below),

and they have many properties and characteristics that make them particularly useful for the

study of PDE.

To begin, take X to be a topological vector space (TVS) over R (C). Recalling that a

TVS is a linear space, this means that X satisfies the typical vector space axioms pertaining

to scalars in R (C) and has a topology corresponding to the continuity of vector addition and

scalar multiplication. If one is uncertain of this concept refer to [Rud91] for more details.

Definition 8.2.1: A norm is any mapping X → R, denoted ||x || for x ∈ X, having the

following properties:

i.) ||x || ≥ 0, ∀x ∈ X;

ii.) ||αx || = |α| · ||x ||, ∀α ∈ R, x ∈ X;

iii.) ||x + y || ≤ ||x ||+ ||y ||, ∀x, y ∈ X.

Remark: iii.) is usually referred to as the triangle inequality .

Definition 8.2.2: A TVS, X, coupled with a given norm defined on X, namely (X, || · ||), is

a normed linear space (NLS).

Definition 8.2.3: Given an NLS (X, || · ||), we define the following

• A metric is a mapping ρ : X ×X → R defined by ρ(x, y) = ||x − y ||.

• An open ball, denoted B(x, r), is a subset of X characterized by B(x, r) = y ∈ X :

ρ(x, y) < r.

• If U ⊂ X is such that ∀x ∈ U, ∃r > 0 with B(x, r) ⊂ U, then U is open in X.

Definition 8.2.4: A NLS X is said to be complete, if every Cauchy sequence in X has a limit

in X. Moreover,

• a complete NLS is called a Banach Space (BS);

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• A BS is separable if it contains a countable dense subset.

One can refer to either [Roy88] or [Rud91], if they are having a hard time understanding

any of the above definitions. To solidify these definitions, let us consider a few examples.

Example 4: i.) A classic example of a BS is C0([−1, 1]) = f : [−1, 1]→ R, continuouswith the norm ||f ||

C0[−1,1]:= maxx∈[−1,1] |f (x)|. (Showing completeness of this space

is a good exercise in real analysis. See [Roy88] if you need help.) Of course, C0 under

this norm is complete over any compact subset of Rn.

ii.) A non-example of a BS is X = f : [−1, 1] → R, f polynomial. Indeed X is not

complete as any function has an approximating sequence of Taylor polynomials that

converges (in the sense of Cauchy) in all norms, hence X is not complete. One may try

to argue that this statement is false considering the discrete map (||f || = 1 if f/≡ 0 and

||f || = 0 otherwise), but the discrete map is not a valid norm as it violates Definition

(8.2.1.ii).

iii.) Another classic example of a BS from real analysis is the space lp = a = (a1, a2, a3, . . .)∑∞i=1 |ai |p <

∞ 1 ≤ p <∞ with the norm

||a||lp

=

∞∑

i=1

|ai |p.

Again, showing this space is complete is another good exercise in real analysis.

Moving on, we now start to consider certain types of mapping on our newly defined

spaces.

Definition 8.2.5: Given X and Y are both NLS, T : X → Y is called a contraction if ∃θ < 1

such that

||Tx − Ty ||Y≤ θ||x − y ||

X, ∀x, y ∈ X.

A special case of the above definition is when X and Y are the same NLS. If we go further

and consider T : X → X, where X is a BS, then we have the following extremely powerful

theorem.

Theorem 8.2.6 (): [Banach’s fixed point theorem] If X is a BS and T a contraction on X,

then T has a unique fixed point in X; i.e. ∃!x ∈ X such that Tx = x . Moreover, for any

y ∈ X, l imi→∞T i(y) = x .Gregory T. von Nessi

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8.2 Banach Spaces and Linear Operators183

Proof : Consider a given point x0 ∈ X and define the sequence xi = T i(x0). Then, with

δ := ||x0 − Tx0||X = ||x0 − x1||X , we have

||x0 − xk ||X ≤k∑

i=1

||xi−1 − xi ||X

≤∞∑

i=1

θi−1||x0 − x1||X

= δ ·∞∑

i=1

θi−1

=δ

1− θ .

Also, for m ≥ n,

||xn − xm||X ≤ θ · ||xn−1 − xm−1||X ≤ θ2 · ||xn−2 − xm−2||X≤ · · · ≤ θn · ||x0 − xm−n||X ≤ δ ·

θn

1− θwhich tends to 0 as n → ∞, since θ < 1. Thus, xi is Cauchy. Put limi→∞(xi) =: x ∈ X,

then we see that Tx = limi→∞ Txi = limi→∞ xi+1 = x .

To finish the proof at hand, we need show uniqueness. Assume that w is another fixed

point of T , then

||x − w ||X = ||Tx − Tw ||X ≤ θ||x − w ||X .

Since θ < 1, this implies that ||x − w ||X

= 0 and hence x = w .

An immediate application of the above theorem is the proof of the Picard-Lindelof The-

orem for ODE, which asserts the existence of first order ODE under a Lipschitz condition

on any inhomogeneity. See [TP85] for the theorem and proof.

Continuing our exposition on properties of BS operators, we make the following definition.

Definition 8.2.7: Take X and Y as NLS. T : X → Y is said to be bounded if

||T || = supx∈X,x 6=0

||Tx ||Y

||x ||X

<∞. (8.1)

From (8.1), we readily verify that the space of all bounded linear mapping from X to Y ,

denoted L(X, Y ), forms a vector space. Also, concerning the boundedness of BS operators

we have the following theorem from functional analysis.

Theorem 8.2.8 (): Take X and Y to be BS. If X is infinite dimensional, then a map T is

linear and bounded if and only if it is continuous. On the other hand, if X is finite dimensional,

then T is continuous if and only if T is linear.

Proof : See [Rud91].

We now are in a position to state another very powerful idea that is useful for proving

existence of solutions for linear PDE.

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Theorem 8.2.9 (Method of Continuity): If X is a BS, Y a NLS, with L0 and L1 bounded

linear operators mapping X → Y such that exists a C > 0 satisfying the a priori estimate

||x ||X≤ C||Ltx ||Y ∀t ∈ [0, 1], (8.2)

where Lt := (1− t)L0 + tL1 and t ∈ [0, 1], then L1 maps X onto Y ⇐⇒ L0 map X onto

Y .

Proof : Suppose that there exists an s ∈ [0, 1] such that Ls is onto. (8.2) indicates that

Ls is also one-to-one; and hence, L−1s exists with ||L−1

s || ≤ C, where C is the same constant

in (8.2). Now fix an arbitrary y ∈ Y and t ∈ [0, 1]. We now seek to prove that exists x ∈ Xsuch that Lt(x) = y , i.e. Lt is onto Y . It is clear that Lt(x) = y if and only if

Ls(x) = Ltx + Lsx − Ltx= y + (1− s)L0 + sL1 − [(1− t)L0 + tL1](x)

= y + (t − s)[L0 − L1](x).

Upon rewriting this equality, we see

x = L−1s y + (t − s)L−1

s (L0 − L1)x︸︷︷︸Tx

. (8.3)

So, now we claim that ∃x ∈ X such that (8.3) is satisfied. This is equivalent to showing T

is a contraction mapping via Theorem 8.2.6. Taking x1, x2 ∈ X, we have

||Tx1 − Tx2||Y = ||(t − s)L−1s (L0 − L1)(x1 − x2)||

Y

≤ |t − s| · ||L−1s ||︸︷︷︸≤C

· ||L0 − L1||︸︷︷︸≤ 2C

·||x1 − x2||X .

Thus, if we consider t such that

|t − s| < 1

2C2,

we get that T is a contraction; and hence, Lt is onto. Now, we can cover [0, 1] with open

intervals of length C−2 to conclude that our original assumption (there existing an s ∈ [0, 1]

where Ls is onto) leads to Lt being onto ∀t ∈ [0, 1].

The method of continuity essentially falls under the general category of existence theorems.

While not particular good for explicit calculations, this theorem is very handy for proving,

well, existence. The following is a nice exercise that demonstrates this.

Exercise: Prove the existence of solution of the equation −(φu′)′ + u = f .

Hints: Consider the mappings

L0(u) = −u′′, L0(u) = f

L1(u) = −(φ · u′)′ + u, L1(u) = f ,

and note that L0 is onto (just integrate the equation for L0).Gregory T. von Nessi

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8.3 Fredholm Alternative185

8.3 Fredholm Alternative

In this section, one of the most important tools in the study of PDE will be presented, namely

the Fredholm Alternative. To begin the exposition, let us recall a fact from linear algebra.

Fact: If L : Rn → Rn is linear, the mapping L is onto and one-to-one (and hence, in-

vertible) if the kernel is trivial i.e. Ker(L) = x : Lx = 0 = 0.

It turns out in general that this is not true, unless you have a compact perturbation of

the identity map. The Fredholm Alternative can be viewed as a BS analogue of the above

fact; but before going on to that theorem, let us define what a compact operator is.

Definition 8.3.1: Consider X and Y NLS with a bounded linear map T : X → Y . T is said

to be compact if T maps bounded sets in X into relatively compact sets in Y .

Sequentially speaking, the above statement is equivalent to saying that T is compact if

it maps bounded sequences in X into sequences in Y that contain converging subsequences.

Not only is the next lemma (due to Riesz) used in the proof of the Fredholm Alternative,

but it is also a very important property of NLS in its own right.

Lemma 8.3.2 (Almost orthogonal element): If X is a NLS, Y a closed subspace of X and

Y 6= X. Then ∀θ < 1, ∃xθ ∈ X satisfying ||xθ|| = 1, such that dist(xθ, Y ) ≥ θ.

Proof: First, we fix arbitary x ∈ X \ Y . Since Y is closed,

dist(x, Y ) = infx∈Y||x − y || = d > 0.

Thus, ∃yθ ∈ Y , such that

||x − yθ|| ≤d

θ.

Now, we define

xθ =x − yθ||x − yθ||

;

it is clear that ||xθ|| = 1. Moreover, for any y ∈ Y ,

||xθ − y || =||x − [yθ − ||yθ − x ||y ]||

||yθ − x ||

≥ d

||yθ − x ||≥ θ,

where the second inequality is based on the fact that yθ − ||yθ − x ||y ∈ Y since Y is a proper

subspace of X.

Corollary 8.3.3 (): If X is an infinite dimensional NLS, then B1(0) is not precompact in X.

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Proof: The statement of the corollary is equivalent to saying there exists bounded se-

quences in B1(0) that have no converging subsequences. To see this, we argue directly by

making the following construction. For a given infinite dimensional BS X, pick an arbitrary

element x0 ∈ B1(0) ⊂ X. Since Span(x0) is a proper subspace of X, we know exists by

lemma 8.3.2 x1 ∈ X such that dist(x1,Span(x0)) ≥ 12 with ||x1|| = 1 =⇒ x1 ∈ B1(0). By

the same argument, we pick x2 ∈ X such that dist(x2,Span(x0, x1)) ≥ 12 with x2 ∈ B1(0).

Repeating iteratively, we now have constructed a sequence xi, for which there is clearly

no converging subsequence in X (the distance between any two elements is greater than12 ).

The above corollary indicates that compactness will be a big consideration in PDE anal-

ysis: many of the arguments forthcoming are based on limit principles, which the above

corollary presents a major obstacle. Thus, ideas of compactness of operators (and function

space imbeddings; more later) play a major role in PDE theory, as compactness “doesn’t

come for free” just because we work in a BS.

Now, we are finally in a position to present the objective of this section.

Theorem 8.3.4 (): [Fredholm Alternative] If X is a NLS and T : X → X is bounded, linear

and compact, then

i.) either the homogeneous equation has a non-trivial solution x−Tx = 0, i.e. Ker(I−T )

is non-trivial or

ii.) for each y ∈ X, the equation x − Tx = y has a unique solution x ∈ X. Moreover,

(I − T )−1 is bounded.

Proof 1: [GT01] For clarity the proof will be presented in four steps.

Step 1: Claim: Defining S := I − T , we claim there exists a constant C such that

dist(x,Ker(S)) ≤ C||S(x)|| ∀x ∈ X. (8.4)

Proof of Claim: We will argue indirectly by contradiction. Suppose the claim is

false, then there exists a sequence xn ⊂ X satisfying ||S(xn)|| = 1 and dn :=

dist(xn,Ker(S))→∞. Upon extracting a subsequence we can insure that dn ≤ dn+1.

Choosing a sequence yn ⊂ Ker(S) such that dn ≤ ||xn − yn|| ≤ 2dn, we see that if

we define

zn :=xn − yn||xn − yn||

,

then we have ||zn|| = 1. Moreover,

||S(zn)|| =

∣∣∣∣∣∣∣∣S(xn)− S(yn)

||xn − yn||

∣∣∣∣∣∣∣∣

=

∣∣∣∣∣∣∣∣

1− 0

||xn − yn||

∣∣∣∣∣∣∣∣

≤ 1

dn.

1Can be omitted on first reading.Gregory T. von Nessi

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8.3 Fredholm Alternative187

Thus, the sequence S(zn) → 0 as n →∞. Now acknowledging that T is compact,

by passing to a subsequence if necessary, we may assume that the sequence Tznconverges to an element w0 ∈ X. Since zn = (S + T )zn, we then also have

limn→∞

zn = limn→∞

(S + T )zn

= 0 + w0 = w0.

From this, we see that T (w0) = w0, i.e. zn → w0 ∈ Ker(S). However this leads to a

contradiction as

dist (zn,Ker(S)) = infy∈Ker(S)

||zn − y ||

= ||xn − yn||−1 infy∈Ker(S)

||xn − yn − ||xn − yn||y ||||xn − yn||

= ||xn − yn||−1dist (xn,Ker(S)) ≥ 1

2.

Step 2: We now wish to show that the range of S, R(S) := S(X), is closed in X. This is

equivalent to showing an arbitrary sequence in X, call it xn, is mapped to another

sequence with limit S(x) = y ∈ X, where x has to be shown to be in X.

Now consider the sequence dn defined by dn := dist(xn,Ker(S)). By (8.4) we

know that dn is bounded. Choosing yn ∈ Ker(S) such that dn ≤ ||xn − yn|| ≤ 2dn,

we see that the sequence defined by wn = xn − yn is also bounded. Thus, by the

compactness of T , T (wn) converges to some z0 ∈ X, upon passing to subsequence if

necessary. We also have S(wn) = S(xn) − S(yn) = S(xn) → y by definition. These

last two observations, combined with the fact that wn = (S+T )wn, leads us to notice

that wn → y + z0. Finally, we have that

limn→∞

S(wn) = limn→∞

S(xn − yn)

=⇒ S(y + z0) = y ,

where the limit on the RHS is justified by the continuity of S (see theorem (8.3.4)).

Since y + z0 ∈ X, we conclude that R(S) is indeed closed.

Step 3: We now will shot that if Ker(S) = 0 then R(S) = X, i.e. if case i.) of the theorem

does not hold then case ii.) is true.

To start we make the following. First we define the following series of spaces S(X)j :=

Sj(X). Now, it is easy to see that

S(X) ⊂ X =⇒ S2(X) ⊂ S(X) =⇒ · · · =⇒ Sn(X) ⊂ Sn−1(X).

This combined with the conclusion of Step 2, indicates that S(X)j constitute a set of

closed, non-expanding subspaces of X. Now, we make the following

Claim: There exists K such that for all i , j > K, S(X)i = S(X)j .

Proof of Claim: Suppose the claim is false, thus we can construct a sequence based on

the following rule that xn+1 ∈ Sn+1 with dist(xn+1, S(X)n) ≥ 12 , ||xn|| = 1 and x0 ∈ X

is chosen arbitrarily. So, let us now take n > m and consider

Txn − Txm = xn + (−xm + S(xm)− S(xn))

= xn + x,

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where x ∈ S(X)m+1. From this equation, we see that

||Txn − Txm|| ≥1

2(8.5)

by construction of our sequence. (8.5) contradicts the fact that T is a compact oper-

ator; our claim is proved.

As of yet, we still have not used the supposition that Ker(S) = 0. Utilizing our

claim, take k > K so that S(X)k = S(X)k+1. So, taking arbitrary y ∈ X, we see that

Sk(y) ∈ S(X)k = S(X)k+1. Thus, ∃x ∈ S(X)k+1 with Sk(y) = Sk+1(x). This leads

to

Sk(S(x)− y) = 0 =⇒ S(x)− y = S−k(0);

but since Ker(S) = 0, S−k(0) = 0. We now can conclude that S(x) = y , and since

y ∈ X was arbitrary we get the result: R(S) = X.

Step 4: We lastly seek to prove that if R(S) = X then Ker(S) = 0.

Proof of Step 4: This time we define a non-decreasing sequence of closed subspaces

Ker(S)j by setting Ker(S)j = S−j(0). This is indeed a non-decreasing sequence as

0 ∈ Ker(S) = S−1(0)

=⇒ S−1(0) ⊂ S−2(0)...

...

=⇒ S−k ⊂ S−(k+1)(0).

Also, Ker(S)j are closed due to the continuity of S. By employing an analogous

argument based on lemma 8.3.2 to that used in Step 3., we obtain that ∃l such that

Ker(S)j = Ker(S)l for all j ≥ l . Now, since R(S) = X =⇒ R(Sl) = X, ∃x ∈ Xsuch that for any arbitrary element y ∈ Ker(S)l , y = Slx is satisfied. Consequently

S2l(x) = Sl(y) = 0, which implies x ∈ Ker(S)2l = Ker(S)l whence

y = Slx = 0.

The result follows from seeing that Ker(S) ⊂ Ker(S)l = 0, which is indicated by

the preceding equation and the non-decreasing nature of Ker(S)l .

The boundedness of the operator S−1 = (I − T )−1 in case ii.) follows from (8.4) with

Ker(S) = 0. This completes the entire proof, whew!

To conclude this section, we will consider a few examples of compact and non-compact

operators.

Example 5: If T : X → Y is continuous and Y finite dimensional, then T is compact. It is

left to the reader to verify this, as it is a straight-forward calculation in real analysis.

Example 6: If X = C1([0, 1]), Y = C0([0, 1]), and T = ddx , then T not a compact operator.

If we look at

||Tu||C0([0,1])

≤ C||u||C0([0,1])


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8.4 Spectral Theorem for Compact Operators189

We see that this can’t hold ∀u ∈ X and ∀x ∈ [0, 1]. Just look at u =√x . Thus, T is not

even bounded; and hence, it is not compact. This indicates why taking derivatives is a “bad”

operation.

Example 7: Take X = C0([0, 1]), Y = C0([0, 1]), and Tu =∫ x

0 u(s) ds. Now, take uj to

be a bounded sequence in C0([0, 1]) (say it’s bounded by a constant C). Thus, we have that

vj = Tuj is also bounded in C0; in fact, it is easy to see that vj ≤ C. Now, we use the

following to show T compact:

Arzela-Ascoli Theorem: If uj is uniformly bounded and equicontinuous in a NLS X, then

exists a converging subsequence in X. (We will come back to this theorem later.)

So, we look at

|vj(x)− vj(y)| =

∣∣∣∣∫ y

x

v ′j (s) ds

∣∣∣∣ ≤ |x − y | · sup |uj | ≤ C|x − y |

Thus, we see that vj is equicontinuous. By applying Arzela-Ascoli, we find that vj does indeed

contain a converging subsequence in C0([0, 1]). We now conclude that T is compact.

8.4 Spectral Theorem for Compact Operators

In this section we will start investigations into the spectra of compact operators. First, for

completness, we recall the following elementary definitions.

Definition 8.4.1: Take X to be a NLS and T ∈ L(X,X). The resolvent set, ρ(T ), is the

set of all λ ∈ C such that (λI − T ) is one-to-one and onto. In addition, Rλ = (λI − T )−1 is

the resolvent operator assocaited with T .

From the Fredholm Alternative in the previous section, we see that Rλ is a bounded linear

map for λ ∈ ρ(T ).

Definition 8.4.2: Take X to be a NLS and T ∈ L(X,X). λ ∈ C is called an eigenvalue

of T if there exists a x 6= 0 such that T (xλ) = λxλ. The x in this equation is called the

eigenvector associated with λ. The set of eigenvalues of T is called the spectrum, denoted

σ(T ). Also, we define

• Nλ = x : Tx = λx is the eigenspace.

• mλ = dim(Nλ) is the multiplicity of λ.

Warning: It is not necessarily true that [ρ(T )]C = σ(T ).

The next lemma is a simple result from real analysis that will be important in the proof

of the main theorem of this section.

Lemma 8.4.3 (): If a set U ⊂ Rn (Cn) is bounded and has only a finite number of accumu-

lation points, then U is at most countable.

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are elements of U

Qr

O

x0r

R

Figure 8.1: Layout for lemma 8.4.3, note that U is only approximated and x0 is not necessarily

contained in U.

Proof 1: Without lose of generality we consider U ⊂ C with only one accumulation point

x0 (the argument doesn’t change for the other cases).

Since U is bounded, there exists 0 < R < ∞ such that U ⊂ BR(0). Since BR(0) is open,

there exists r > 0 such that Br (x0) ⊂ BR(0); define Qr := BR(0) \ Br (x0). To proceed, we

make the following

Claim: Qr contains only a finite number of points of U.

Suppose not, then Qr ∩ U is at least countably infinite. Since Qr is closed and bounded, it

is compact (Heine-Borel). Thus, we cover Qr with a finite number of balls with radius 1. At

least one of these balls, call it B∗1, is such that B∗1 ∩ U is at least countably infinite. Conse-

quently, we define V1 := B∗1 ∩Qr . In a similar way, we cover V1 with a finite number of balls

with radius 12 , where B∗1/2 is one of these balls with B∗1/2∩U being at least countably infinite.

In analogy with V1, we define V2 := B∗1/2∩ V1. We thus iterativly construct a series of closed

sets Vn := B∗1/n∩ Vn−1. By construction, we have Vn ⊂ Vm for m < n with diam(Vn) ≤ 1

n .

We now see that ∃x ∈ U such that

x ∈∞⋂

n=1

Vn,

by the compactness of Vn; but this implies that x is an accumulation point of V , since ev-

ery Vn ∩ U is at least countably infinite. Hence we have a contradiction, which proves the

claim.

Now, it is clear that

BR(0) = x0 ∪[ ∞⋃

n=1

Qr/n

].

Thus,

U ⊂ x0 ∪∞⋃

n=1

(U ∩Qr/n), (8.6)

1Can be omitted on first reading.Gregory T. von Nessi

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8.4 Spectral Theorem for Compact Operators191

since U ⊂ BR(0). Since the RHS of (8.6) is countable, so is U.

We now are in a position to state the main result of the section.

Theorem 8.4.4 (Spectral Theorem for Compact Operators): If X is a NLS with T ∈L(X,X) compact, then σ(T ) is at most countably infinite with λ = 0 as the only possible

accumulation point. Moreover, the multiplicity of each eigenvalue is finite.

Proof: It is easily verified that eigenvectors associated with different eigenvalues are

linearly independent. With that in mind, we can thus extract a series of λj (not necessarily

distinct) from σ(T ) with associated eigenvectors xj such that Txj = λjxj and x1, . . . , xn is

linearly independent ∀n. Now we define Mn := Spanx1, . . . , xn. Since Tx = λx , we see

that

||T || ≥ ||Tx ||||x || = λ;

hence σ(T ) is bounded.

To prove that λ = 0 is the only possible accumulation point, we will argue indirectly by as-

suming that λ 6= 0 is an accumulation point of λj. To do this, we first utilize lemma (8.4.3)

to construct a sequence yn ⊂ X such that yn ∈ Mn, ||yn|| = 1 with dist(yn,Mn−1) ≥ 12 .

We now wish to prove

∣∣∣∣∣∣∣∣

1

λnTyn −

1

λmTym

∣∣∣∣∣∣∣∣ ≥

1

2(8.7)

for n > m. If this is indeed true and the accumulation point of λj isn’t 0, then we get a

contradiction as ||yn|| is bounded, which implies Tyn contains a convergent subsequence as

T is compact (an impossibility in the face of (8.7)). So, let use prove (8.7).

Given yn ∈ Mn = Spanx1, . . . , xn, we can write

yn =

n∑

j=1

βjxj

for some βj ∈ C. Now, we calculate

yn −1

λnTyn =

n∑

j=1

(βjxj −

1

λnβjTxj

)

=

n∑

j=1

(βjxj −

βjλjλn

xj

)

=

n−1∑

j=1

(βjxj −

βjλjλn

xj

)∈ Mn−1.

Thus, 1λmTym ∈ Mn−1 since m < n. It is convenient to now define

n−1∑

j=1

(βjxj −

βjλjλn

xj

)− 1

λmTym =: z ∈ Mn−1.

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Finally, looking at (8.7), we see

∣∣∣∣∣∣∣∣

1

λnTyn −

1

λmTym

∣∣∣∣∣∣∣∣ = ||yn − z ||

≥ infω∈Mn−1

||yn − ω||

= dist(yn,Mn−1)

≥ 1

2.

This proves the claim; and hence, λj can only have 0 as an accumulation point by the

aforementioned contradiction. Applying this conclusion to lemma 8.4.3, we attain the full

theorem.

8.5 Dual Spaces and Adjoint Operators

In addition to the rich structure that we have already seen in BS, it turns out that bounded

linear mappings on BS also have some very nice properties that will be very useful in our

studies of linear PDE. First, we make a formal definition.

Definition 8.5.1: If X is a NLS, we define the dual space, X∗, as the set of all bounded

linear mappings taking X → R.

The following theorem is easily verified.

Theorem 8.5.2 (): If X is a NLS, X∗ is a BS with

||f ||X∗ = sup

x 6=0

|f (x)|||x ||

X

.

Notation: If x ∈ X and f ∈ X∗, then by definition f (x) ∈ R. Typically, since X∗ won’t

always be a space of functions, we write 〈f , x〉 := f (x) ∈ R.

Since X∗ is a BS in its own right, we state the following

Definition 8.5.3: If X is a NLS, we define the bidual space of X as X∗∗ := (X∗)∗. Moreover,

there is a canonical imbedding j : X → X∗∗ defined by 〈j(x), f 〉 = 〈f , x〉 for all x ∈ X and

f ∈ X∗. If the canonical imbedding is onto, then X is a reflexive BS.

Linear operators between BS also have an analogy in dual spaces.

Definition 8.5.4: If X, Y are BS and T ∈ L(X, Y ), the adjoint operator T ∗ : Y ∗ → X∗ is

defined by

〈T ∗g, x〉 = 〈g, T x〉 ∀x ∈ X, ∀g ∈ Y ∗

From this it is easy to see that T ∗ ∈ L(Y ∗, X∗).

Before we can elucidate some nice properties of these adjoints, we require a little more

“structure” beyond that stipulated in the definition of a BS. Thus, we move onto the next

section.Gregory T. von Nessi

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8.6 Hilbert Spaces193

8.6 Hilbert Spaces

Even though BS and their associated operators have a nice mathematical structure, they

lack constructions that would enable one to speak of “geometry”. As geometric structures

yield useful results in Euclidean space (inequalities, angle relations, etc.), we are motivated to

construct geometric analogies on the linear spaces (not necessarily BS) of previous sections.

This nebulous reference to “geometry” on a linear space (LS) will be clarified shortly.

Definition 8.6.1: Given X is an LS, a map (·, ·) : X ×X → R is called a scalar product if

i.) (x, y) = (y , x),

ii.) (λ1x1 + λ2x2, y) = λ1(x1, y) + λ2(x2, y), and

iii.) (x, x) > 0 ∀x 6= 0

for all x1, x2, y ∈ X and λ1, λ2 ∈ R.

Correspondingly, we now define

Definition 8.6.2: A LS with a scalar product is called a Scalar Product Space (SPS). It is

called a Hilbert Space (HS) if it is complete.

Warning: Do not assume that a scalar product is the same as taking an element of a

BS with it’s dual, i.e. “〈·, ·〉 6= (·, ·)”. Actually, this comparison doesn’t even make sense

as a scalar product has both arguments in a given space whereas the composition 〈·, ·〉 re-

quires an element from the space and another from the associated dual space. That aside,

the dual space of a BS may very well not be isomorphic to the BS itself. This will be the

case in a HS by the Riesz-representation theorem (stated below). With that theorem, one

may characterize any functional on a HS via the scalar product. This does not go the other

way around; i.e. on a BS, a dual space does not necessarily characterize a valid scalar product.

Also, one will notice that we only require that a SPS to be a LS (not a BS) with a

scalar product; but, it is easily verified that ||x || :=√

(x, x) satisfies the criteria stated in

the definition of a norm. Thus, any HS (SP’S) is a BS (NLS). This is not to say that the

only valid scalar product on a space is the one that generates the norm; case in point, Lp(Ω)

spaces are frequently considered with the norm

∫

Ω

f · g dx

where f , g ∈ Lp(Ω). It is easy to verify this satisfies definition 8.6.1 but does not generate

the Lp(Ω) norm.

Scalar products add a great deal of structure to the spaces we have been considering

previously. First, we can now start talking about the concept of “geometry” on a SPS.

We have seen that for any SPS, there exists a norm which can be analogized to length

in Euclidean space. Taking this analogy one step further, we can define the “angle”, φ,

between two elements x and y in a SPS, X, by cosφ := (x,y)||x ||·||y || . This leads directly to the

all-important orthogonality criterion x ⊥ y ⇐⇒ (x, y) = 0. In addition to geometry, there

are many useful inequalities that pertain to scalar products; here are a few.

• Cauchy Schwarz inequality: |(x, y)| ≤ ||x || · ||y ||

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• Triangle inequality: ||x + y || ≤ ||x ||+ ||y ||

• Parallelogram identity: ||x + y ||2 + ||x − y ||2 = 2(||x ||2 + ||y ||2)

Notation: From now on we will be referring to Hilbert Spaces, as it usually safe to assume

that we are working in a complete SPS.

With these concepts, we can now analyze some key theorems that pertain to Hilbert

Spaces.

Theorem 8.6.3 (Projection Theorem): If H is a HS and M is a closed subspace of H, then

each x ∈ H has a unique decomposition x = y +z with y ∈ M, z ∈ M⊥ (i.e. is perpendicular

to M, (z,M) = 0, ∀m ∈ M). Moreover dist(x,M) = ||z ||.

Proof: First, if x ∈ M then as x = x+0 we trivially get the conclusion. Thus, we consider

M 6= X with x /∈ M. Defining, d := dist(x,M), we select a minimizing sequence yn ⊂ Msuch that ||x − yn|| → d . Before proceeding, we need to show that yn is Cauchy. Utilizing

the parallelogram identity, we write

4

∣∣∣∣∣∣∣∣x −

1

2(yn + ym)

∣∣∣∣∣∣∣∣2

︸︷︷︸≥4d2

+||yn − ym||2 = 2(||x − yn||2 + ||x − ym||2)→ 4d2,

where the first term on the LHS is ≥ 4d2 as 12 (yn + ym) ∈ M. Thus, we see that

||yn − ym|| → 0; the sequence is Cauchy. Since X is complete and M is closed, we as-

certain that limn→∞ yn = y ∈ M.

Writing x = z + y , where z := x − y , the conclusion of the proof will follow provided

z ∈ M⊥. Consider any y∗ ∈ M and α ∈ R, as y + αy∗ ∈ M, it is calculated

d2 ≤ ||x − y − αy∗||2 = (z − αy∗, z − αy∗)= ||z ||2 − 2α(y∗, z) + α2||y∗||2

= d2 − 2α(y∗, z) + α2||y∗||2;

thus,

|(y∗, z)| ≤ α

2||y∗||2 ∀α ∈ R.

From this we conclude that |(y∗, z)| = 0; and since y∗ ∈ M was arbitrary, z ∈ M⊥.

One of the nicest properties of Hilbert Spaces, call it H, is that one can characterize

any bounded linear functional on H in terms of the scalar product. This is formalized by the

following theorem.

Theorem 8.6.4 (Riesz-representation Theorem): If F is a bounded linear function on a

Hilbert space H, then there exists a unique f ∈ H such that F (x) = (x, f ) and ||f || = ||F ||.Gregory T. von Nessi

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Proof: First, we consider F such that Ker(F ) = H; then, F = 0 and we correspondingly

choose f = 0. If, on the other hand, we have Ker(F ) 6= H, then ∃z ∈ H \ Ker(F ). By the

projection theorem, we may assume that z ∈ Ker(F )⊥ as it easily seen that Ker(F ) is a

proper subspace of H. Now, we see that

F

(x − F (x)

F (z)z

)= 0

=⇒ x − F (x)

F (z)z ∈ Ker(F ).

Since z ∈ Ker(F )⊥, we have by orthogonality

(x − F (x)

F (z)z, z

)= 0 ⇐⇒ (x, z) =

(F (x)

F (z)z, z

)=F (x)

F (z)||z ||2.

Solving for F (x) algebraically leads to

F (x) =

(x,F (z)

||z ||2 z)

= (x, f ), where f =F (z)

||z ||2 z.

Lastly, we need to show ||F || = ||f ||. So, on one hand, we have

||F || = supx∈H,x 6=0

|F (x)|||x || = sup

|(x, f )|||x ||

C-S≤ sup||x || · ||f ||||x || = ||f ||;

but on the other, its calculated that

||f ||2 = (f , f ) = F (f )C-S≤ ||F || · ||f || =⇒ ||f || ≤ ||F ||.

Thus, we conclude that ||f || = ||F ||.

Next, we will go over one of the key existence theorems pertaining to elliptic equations

but first a quick definition.

Definition 8.6.5: Give X a NLS, a mapping B : X × X → R is a bilinear form provided B

is linear in one argument when the other argument is fixed. B is bounded if there exists a

constant C0 > 0 such that

|B(x, y)| ≤ C0||x || · ||y ||

for all x, y ∈ X. Analogously, B is coercive if there exists a constant C1 > 0 such that

B(x, x) ≥ C1||x ||2

for all x ∈ X.

Warning: B is not necessarily symmetric in its arguments.

Theorem 8.6.6 (Lax-Milgram Theorem): If H is a HS and B : H × H → R is a bounded,

coercive bilinear form, then for all F ∈ H∗ there exists a unique f ∈ H such that

B(x, f ) = F (x).

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Proof: First, fix f ∈ H and define

Gf (x) := B(x, f ) = (x, T f ), (8.8)

where the last equality comes from the application of the Riesz representation theorem to

Gf (x). Given B is coercive, we calculate

C1||f ||2 ≤ B(f , f ) = (f , T f ) ≤ ||f || · ||T f ||,

where C1 is the coercivity constant associated with B. Analogously, via the boundedness of

B,

||T f ||2 = (T f , T f ) = B(T f , f ) ≤ C0||f || · ||T f ||,

where C0 is the boundedness constant of B. From these last two equations, we derive that

C1||f || ≤ ||T f || ≤ C0||f ||, (8.9)

which is tantamount to R(T ) being closed (right inequality) and T itself being one-to-one

(left inequality). Now, we wish to show that T is onto, since this along with (8.9) would

imply that T−1 is well-define.

If T is not onto, then by the projection theorem, there exists z ∈ R⊥(T ) (R(T ) is easily

noticed to be a proper subspace of H). Thus, B(z, f ) = (z, T f ) = 0 for all f ∈ H. Choosing

f = z , we calculate that 0 = B(z, z) ≥ C1||z ||2 which implies z = 0, i.e. R(T ) = H. So

T−1 is indeed well-defined as we have now shown T to be onto.

The last step of the proof is to apply the Riesz representation theorem: for any F ∈ H∗,there exists a unique g ∈ H such that F (x) = (x, g) for all x ∈ H. With that, we get

F (x) = (x, g) = (x, TT−1g) = B(x, T−1g).

From this applied to (8.8), we see that f = T−1g. The existence of f is thus guaranteed as

we have shown T−1 to be well defined.

As was mentioned above, the Lax-Milgram theorem turns out to be the right tool for

proving existence of solutions of many elliptic equation. That being said, one could be for-

given for thinking that the coercivity criterion is somewhat artificially induced, as it will be

shown to correspond to the ellipticity criterion of a differential operator. It is true that this

correspondence is very useful, but the Lax-Milgram theorem actually has a deeper implication

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Recalling the construction of Hilbert Spaces, one will remember that the properties of

such a space are all dependent on the definition of the associated scalar product. Indeed,

the norm and, hence, the topology of the space result from the particular definition of the

scalar product (on top of the assumed ambient properties of a linear space). Philosophically

speaking, Lax-Milgram indicates whether a given bilinear form acting as a scalar product

will “replicate” the topological structure an already-established HS. If this is found to be

the case, then one can view the conclusion of Lax-Milgram as a direct application of the

Riesz-representation theorem on the space with the bilinear form as the scalar product. Now

since topology is directly correlated to the norm of a HS, one can start to see the coercivity

condition (along with the boundedness of the bilinear form) as a stipulation on the bilinear

form that would guarantee the preservation of the topology of the HS if the scalar product

were replaced by the bilinear form itself. Of course the geometry, as described in the beginning

of the section, would be different; but this is not important, in this instance, as Lax-Milgram

only refers two what elements are contained in a space (a topological attribute). Hopefully,

this bit of insight will make the Lax-Milgram theorem (and existence theorems in general) a

little more intuitive for the reader.

8.6.1 Fredholm Alternative in Hilbert Spaces

With the additional structure of a HS, we are now able to realize more properties of adjoint

operators than was possible in BS.

Theorem 8.6.7 (): If H is a HS with T ∈ L(H,H), then T ∗ has the following properties:

i.) ||T ∗|| = ||T ||;

ii.) If T is compact, then T ∗ is also compact;

iii.) R(T ) = Ker(T ∗)⊥.

Proof:

i.) This is clear from the fact that

||T || = supx∈H

||Tx ||||x ||

(T ∗ has an analogous norm).

ii.) Let xn be a sequence in H satisfying ||xn|| ≤ M. Then

||T ∗xn||2 = (T ∗xn, T ∗xn) = (xn, TT∗xn)

≤ ||xn|| · ||TT ∗xn||≤ M||T || · ||T ∗xn||

so that ||T ∗xn|| ≤ M||T ||; that is, the sequence T ∗xn is also bounded. Hence,

since T is compact, by passing to a subsequence if necessary, we may assume that the

sequence TT ∗xn converges. But then

||T ∗(xn − xm)||2 = (T ∗(xn − xm), T ∗(xn − xm))

= (xn − xm, TT ∗(xn − xm))

≤ 2M||TT ∗(xn − xm)|| → 0 as m, n →∞.

SinceH is complete, the sequence T ∗xn is convergent and hence T ∗ is compact.

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iii.) Fix arbitrary x ∈ H and set Tx =: y ∈ R(T ). From this, it is seen that (y , f ) =

(Tx, f ) = (x, T ∗f ) = 0 for all f ∈ Ker(T ∗). Thus, R(T ) ⊂ Ker(T ∗)⊥; and since

Ker(T ∗) is closed, R(T ) ⊂ Ker(T ∗)⊥. Now suppose that y ∈ Ker(T ∗)⊥ \ R(T ).

By the projection theorem, y = y1 + y2 where y1 ∈ R(T ) and y2 ∈ R(T )⊥ \ 0

Consequently, 0 = (y2, T x) = (T ∗y2, x) for all x ∈ H; hence, y2 ∈ Ker(T ∗). Therefore,

(y2, y) = (y2, y1) + ||y2||2 = ||y2||2. Thus, we see that (y2, y) 6= 0, i.e. y /∈ Ker(T ∗)⊥.

This contradiction on the assumption of y ∈ Ker(T ∗)⊥ \R(T ) leads to the conclusion

that R(T ) = Ker(T ∗)⊥.

Now, we can can combine the above properties of adjoints, the results pertaining to the

spectra of compact operators and the Fredholm Alternative on BS to directly ascertain the

following generalization.

Theorem 8.6.8 (Fredholm Alternative in Hilbert Spaces): Consider a HS, H and a com-

pact operator T : H → H. In this case, there exists an at most countable set σ(T ) having no

accumulation points except possibly λ = 0 such that the following holds: if λ 6= 0, λ /∈ σ(T )

then the equations

λx − Tx = y λx − T ∗x = y (8.10)

have a unique solution x ∈ H, for all y ∈ H. Moreover, the inverse mappings (λI − T )−1,

(λI − T ∗)−1 are bounded. If λ ∈ σ(T ), then Ker(λI − T ) and Ker(λI − T ∗) have positive

and finite dimension, and we can solve the equation (8.10) if and only if y is orthogonal to

Ker(λI − T ∗) and Ker(λI − T ) respectively for both cases.

There is a lot of information in this theorem; it’s worth the effort to try and get one’s mind

around it. Also, keep in mind that Ker(λI − T ) and Ker(λI − T ∗) are just the eigenspaces

Tλ and T ∗λ respectively.

8.7 Weak Convergence and Compactness; the Banach-Alaoglu

Theorem

To begin this section we begin with some motivation. Consider a sequence of approx-

imate solutions of F (u,Du,D2u) = 0, i.e. a bounded sequence uk ∈ X such that

F (uk , Duk , D2uk) = εk → 0. Now, there are two important questions that arise:

• Does bounded imply there is a convergent subsequence i.e. ukj → u?

• Is this limit the solution?

As we have seen that bounded sequences in infinite dimensional BS do not necessarily contain

convergent subsequences (i.e. in the strong sense), these questions are not readily answered

with our current “set of tools”. Given that such approximations as the one stated are

frequently utilized in PDE, we now seek to define a new criterion for convergence so that we

can indeed answer both these questions in the affirmative.

Definition 8.7.1: Take X to be a BS, with X∗ as its dual.

i.) xj ⊂ X converges weakly to x ∈ X if

〈y , xj〉 → 〈y , x〉 ∀y ∈ X∗.

Notation: uk converging weakly to u is denoted by uk u.Gregory T. von Nessi

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8.7 Weak Convergence and Compactness; the Banach-Alaoglu Theorem199

ii.) yj ∈ X∗ converges weakly∗ to y ∈ X∗ if

〈yj , x〉 → 〈y , x〉 ∀x ∈ X

It can be readily seen that weak convergence implies strong convergence (that is conver-

gence in norm). Also, be the linearity of scalar products, it is also observed that weak limits

are unique.

Now, we present the key theorem which essential would enable us to answer “yes” to the

hypothetical situation presented at the beginning of the section.

Theorem 8.7.2 (Banach-Alaoglu): If X is a TVS, with Ω ⊂ X bounded, open set containing

the origin, then the unit ball in X∗ is sequentially weak∗ compact, i.e. define

K = F ∈ X∗ : |F (x)| ≤ 1 ∀x ∈ Ω,

then K is sequentially weak∗ compact.

Proof 2: First, since Ω is open, bounded and contains the origin, there corresponds to

each x ∈ X a number γx < ∞ such that x ∈ γxΩ, where aΩ :=x ∈ X : xa ∈ Ω

for any

fixed a ∈ R. Now for any x ∈ X, we know that xγx∈ Ω, hence

|F (x)| ≤ γ(x) (x ∈ X,Λ ∈ K).

Let Ax be the set of all scalars α such that |α| ≤ γx . Let τ be the product topology on

P , the Cartesian product of all Ax , one for each x ∈ X. Indeed, P will almost always be a

space with an uncountable dimension. By Tychonoff’s theorem, P is compact since each Exis compact. The elements of P are functions f on X (they may or may not be linear or even

continuous) that satisfy

|f (x)| ≤ γ(x) (x ∈ X).

Thus K ⊂ X∗ ∩ P , as K ⊂ X∗ and K ⊂ P . It follows that K inherits two topologies: one

from X∗ (its weak∗-topology, to which the conclusion of the theorem refers) and the other,

τ , from P . Again, see [Bre97] for the concept of inherited topologies.

We now claim that

i.) these two topologies coincide on K, and

ii.) K is closed subset of P .

If ii.) is proven, then we have that K is an uncountable Cartesian product of τ-compact sets

(each one contained in some Ax), hence it is compact by Tychonoff’s theorem. Moreover, if

i.) is shown to be true then K is concluded to be weak∗-compact from the previous statement.

So, we now see to prove our claims. To do this first fix F0 ∈ K; choose xi ∈ X, for

1 ≤ i ≤ n; and fix δ > 0. Defining

W1 := F ∈ X∗ : |Λxi − Λ0xi | < δ for 1 ≤ i ≤ n

and

W2 := f ∈ P : |f (xi)− Λ0xi | < δ for 1 ≤ i ≤ n,

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we let n, xi , and δ range over all admissible values. The resulting sets W1, then form a

local base for the weak∗-topology of X∗ at Λ0 and the sets W2 form a local base for the

product topology τ of P at Λ0; i.e. any open set in X∗ or P can be represented as a union

of translates of the sets contained in W1 or W2 respectively. Since K ⊂ P ∩X∗, we have

W1 ∩K = W2 ∩K.

This proves (i).

Now, suppose f0 is in the τ-closure of K. Next choose x, y ∈ X with scalars α, β, and

ε > 0. The set of all f ∈ P such that |f − f0| < ε at x , at y , and at αx + βy is a τ-

neighborhood of f0. Therefore K contains such an f . Since this f ∈ K ⊂ X∗, and hence,

linear, we have

f0(αx + βy)− αf0(x)− βf0(y) = (f0 − f )(αx + βy)

+α(f − f0)(x) + β(f − f0)(y).

so that

|f0(αx + βy)− αf0(x)− βf0(y)| < (1 + |α|+ |β|)ε.

Since ε was arbitrary, ,we see that f0 is linear. Finally, if x ∈ Ω and ε > 0, the same argument

shows that there is an g ∈ K such that |g(x)− f0(x)| < ε. Since |g(x)| ≤ 1, by the definition

of K, if follows that |f0(x)| ≤ 1. We conclude that f0 ∈ K. This proves ii.) and hence the

theorem.

As an immediate consequence we have

Corollary 8.7.3 (): If X is a reflexive BS, then the unit ball in X is sequentially weakly

compact.

From this corollary and the Riesz-representation theorem, we have that any bounded set

in any HS is sequentially weakly compact (the Riesz-representation theory proves that all

HS are reflexive BS). This is a key point that will be used frequently in the upcoming linear

theory, and gives us the criterion under which we can answer “yes” to the questions posed

at the beginning of the section.

Note: With this section I have broken the sequential structure of the chapter in that one

only needs to have a TVS to speak of weak convergence, etc. The reason for presenting

this material now, is that it can only be motivated once the student starts to realize how

restrictive the notion of strong convergence in a BS/HS is. Thus, for motivation’s sake I

have decided to present this material now, late in the chapter.

8.8 Key Words from Measure Theory

This section is intended to be a whirlwind review of basic measure theory, i.e. one that is not

already familiar with the following concepts should consult with [Roy88] for a proper treatise

of such. In reality, this section is a glorified appendix, in that most of the following results

are not proven but are explained in relation to what has already been explained.


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8.8 Key Words from Measure Theory201

8.8.1 Basic Definitions

Let us recall the “typical” definition of a measure:

Definition 8.8.1: Consider the set of all sets of real numbers, P(R). A mapping µ : P(R)→[0,∞) is a measure if the following properties hold:

i.) for any interval l , µl = the length of l ;

ii.) if En is a sequence of disjoint sets, then

µ

(⋃

n

En

)=∑

n

µEn;

iii.) µ is translationally invariant; i.e., if E is a set for which µ is defined and if E + y :=

x + y : x ∈ E, then µ(E + y) = µE for all y ∈ R.

Now, we say this is the “typical” definition as there are actually different ways to define

a measure. To elaborate, in addition to the above three criterion, it would be nice to have µ

defined for all subsets of R; but no such mapping exists. Thus, one of these four criterion

must be compromised to actually get a valid definition; obviously, this may be done in a

number of ways. We take defintion 8.8.1 as a proper definition as the measures we will

consider, Lebesgue (and more esoterically Hausdorff), obey the criterion contained in such.

The next definition is a necessary precursor to defining Lebesgue measure.

Definition 8.8.2: The outer measure, denoted µ∗ is defined for any A ⊆ R as

µ∗A = infA⊂⋃n In

∑

n

l(In),

where In are intervals in R and l(In) is the length of In.

Now, even though µ∗ is defined for all subsets of R it does not satisfy condition iii.) of

definition 8.8.1, which is undesirable for integration. Thus, we go about restricting the sets

of consideration to guarantee that iii.) is satisfied by the following definition.

Definition 8.8.3: A set E ⊂ R is Lebesgue measurable if

µ∗A = µ∗(A ∩ E) + µ∗(A ∩ Ec),

for all A ⊂ R. Moreover, the Lebesgue measure of E is just µ∗E.

It is left to the reader that definition does indeed imply condition iii.) of definition ; again,

see [Roy88] if you need help.

Denoting the set of all measurable sets B, let us review one of the properties of B. To

do this, we remember the following definition

Definition 8.8.4: A group of sets A is a σ-ring on a space X if the following are true:

• X ∈ A;

• if E ∈ A, then Ac := X \ A ∈ A;

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• and if Ei ∈ A then∞⋃

i=1

Ei ∈ A.

Canonically extending the above concept of Lebesgue measure on R to Rn by replacing

intervals with Cartesian products of intervals (and lengths with volumes), it is true that B is

a σ-ring over Rn.

Now, we move on to some important definitions pertaining to general measures.

Definition 8.8.5: N ∈ B is called a set of µ-measure zero if µ(N) = 0. This ie equivalent

to saying there exists open sets Ai ⊂ B with⋃Ai ⊂ N such that µ (

⋃Ai) < ε.

Definition 8.8.6: A property hold µ-a.e. (µ almost everywhere) if the set of all x for which

the property does not hold is a set of µ-measure zero.

Lastly, we go over what it means for a function to be differentiable.

Definition 8.8.7: A function f : Rn → R is B measurable if x : f (x) ∈ U ∈ B for all open

sets U in R.

8.8.2 Integrability

For integrability studies, we first define a special clase of functions.

Definition 8.8.8: A function f is simple if f is a finite non-zero constant on finitely many

disjoint sets Ai which are measurable and zero elsewhere.

A simple function f is said to be integrable if∑|fi | · µ(Ai) <∞;

and thus, ∫f dµ =

∑fi · χAi ,

where

χAi

=

1 x ∈ Ai0 x /∈ Ai

Note: Keep in mind that µ is a general measure; thus, the above criterion for integrability

of a simple function is dependent on our choice of µ.Still considering a general measure, we

now make the following generalization.

Definition 8.8.9: Consider a general function g and a sequence of simple functions fnsuch that g = lim fn a.e. g is said to be integrable (with respect to measure µ) if

limn,k→∞

∫|fn(x)− fk(x)| dλ = 0.

Moreover, since the limit

limk→∞

∫fk dµ

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Now, we specify µ to be henceforth the Lebesgue measure. The corresponding Lebesgue

integrals have nice convergence properties, i.e. there exists a good variety of conditions

when one can interchange limits and integrals. We now rapidly go over these convergence

theorems, referring the reader to [Roy88] for proofs.

Theorem 8.8.10 (Lebesgue’s Dominated Convergence): If g is integrable with some fi →f a.e., where f is measurable and |fi | ≤ g for all i , then f is integrable. Moreover,

∫fi dµ→

∫f dµ

[limi→∞

∫fi dµ =

∫limi→∞

fi dµ

].

Theorem 8.8.11 (Monotone Convergence (Beppo-Levi)): If fi is a sequence of integrable

functions such that 0 ≤ fi f monotonically and

∫fi dµ ≤ C <∞,

then f integrable and

∫f dµ = lim

i→∞

∫fi dµ.

Lemma 8.8.12 (Fatou’s): If fi is a sequence of integrable functions with f ≥ 0 and

∫fi dλ ≤ C <∞,

then lim inf fi is integrable and

∫limi→∞

fi dµ ≤ limi→∞

inf

∫fi dµ

Theorem 8.8.13 (Egoroff’s): Take fi to be a sequence of functions such that fi → f a.e.

on a set E, with f measurable, where E is measurable and µ(E) < ∞, then ∀ε > 0, ∃Eεmeasurable such that λ(E \ Eε) < ε and fi → f uniformly on Eε.

The next theorem technically states that multi-dimensional integrals can indeed be com-

puted via an iterative scheme.

Theorem 8.8.14 (Fubini’s): i.) If f : Rn × Rm → R is measurable such that

∫

Rn

∫

Rm|f (x, y)| dxdy

exists, then

∫

Rn

∫

Rmf (x, y) dxdy =

∫

Rng(y) dy,

where

g(y) =

∫

Rmf (x, y) dx

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ii.) On the otherhand, if

g(y) =

∫

Rm|f (x, y)| dx <∞

for almost every y with∫

Rng(y) dy <∞,

then∫

Rn

∫

Rm|f (x, y)| dxdy <∞

8.8.3 Calderon-Zygmund Decomposition

[Cube Decomposition]

A useful construction for some measure theoretic arguments in the forthcoming chapters is

that of cube decomposition. This is a recursive construction that enables one to ascertain

estimates for the measure of a set imbedded in Rn.

First, consider a cube K0 ∈ Rn, f a non-negative integrable function, and t > 0 all satisfying∫

K0

f dx ≤ t|K0|,

where | · | indicates the Lebesgue measure of the set. Now, by bisecting the edges of K0 we

divide K0 into 2n congruent subcubes whose interiors are disjoint. The subcubes K which

satisfy∫

K

f dx ≤ t|K| (8.11)

are similarly divided and the process continues indefinately. Set K∗ be the set of subcubes

K thus obtained that satisfy∫

K

f dx > t|K|,

and for each K ∈ K∗, denote by K the subcube whose subdivision gives K. It is clear that

|K||K| = 2n

and that for any K ∈ K∗, we have

t <1

|K|

∫

K

f dx ≤ 2nt.

Moreover, setting

F =⋃

K∈K∗K

and G = K0 \ F , we have

f ≤ t a.e. in G. (8.12)Gregory T. von Nessi

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This inequality is a consequence of Lebesgue’s differention theorem, as each point in G lies

in a nested sequence of cubes satisfying (8.11) with the diameters of the sequence of cubes

tending to zero.

For some pointwise estimates, we also need to consider the set

F =⋃

K∈K∗K

which satisfies by (8.11),

∫

F

f dx ≤ t|F |. (8.13)

In particular, when f is the characteristic function χΓ of a measurable subset Γ of K0, we

obtain from (8.12) and (8.13) that

|Γ | = |Γ ∩ F | ≤ t|F |

8.8.4 Distribution functions

Let f be a measurable function on a domain Ω (bounded or unbounded in Rn. The distribution

function µ = µf of f is defined by

µ(t) = µf (t) = |x ∈ Ω|f (x) > t|

for t > 0, and measures the relative size of f . Note that µ is a decreasing function on (0,∞).

The basic properities of the distribution function are embodied in the following lemma.

Lemma 8.8.15 (): For any p > 0 and |f |p ∈ L1(Ω), we have

µ(t) ≤ t−p∫

Ω

|f |p dx (8.14)

∫

Ω

|f |p dx = p

∫ ∞

0

tp−1µ(t) dt. (8.15)

Proof: Clearly

µ(t) · tp ≤∫

t≤|f ||f |p dx

≤∫

Ω

|f |p dx

for all t > 0 and hence (8.14) follows. Next, suppose that f ∈ L1(Ω). Then, by Fubini’s

Theorem

∫

Ω

|f | dx =

∫

Ω

∫ |f (x)|

0

dtdx

=

∫ ∞

0

µ(t) dt,

and the result (8.15) for general p follows by a change of variables. .

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8.9 Exercises

8.1: Let X = C0([−1, 1]) be the space of all continuous function on the interval [−1, 1], Y =

C1([−1, 1]) be the space of all continuously differentiable functions, and let Z = C10 ([−1, 1])

be the space of all function u ∈ Y with boundary values u(−1) = u(1) = 0.

a) Define

(u, v) =

∫ 1

−1

u(x)v(x) dx.

On which of the spaces X, Y , and Z does (·, ·) define a scalar product? Now consider

those spaces for which (·, ·) defines a scalar product and endow them with the norm

|| · || = (·, ·)1/2. Which of the spaces is a pre-Hilbert space and which is a Hilbert space

(if any)?

b) Define

(u, v) =

∫ 1

−1

u′(x)v ′(x) dx.




(if any)?

8.2: Suppose that Ω is an open and bounded domain in Rn. Let k ∈ N, k ≥ 0, and γ ∈ (0, 1].

We define the norm || · ||k,γ in the following way. The γth-Holder seminorm on Ω is given by

[u]γ;Ω = supx,y∈Ω,x 6=y

|u(x)− u(y)||x − y |γ ,

and we define the maximum norm by

|u|∞;Ω = supx∈Ω|u(x)|.

The Holder space Ck,γ(Ω) consists of all functions u ∈ Ck(Ω) for which the norm

||u||k,γ;Ω =∑

|α|≤k||Dαu||∞;Ω +

∑

|α|=k[Dαu]γ;Ω

is finite. Prove that Ck,α(Ω) endowed with the norm || · ||k,γ;Ω is a Banach space.

Hint: You may prove the result for C0,γ([0, 1]) if you indicate what would change for the

general result.

8.3: Consider the shift operator T : l2 → l2 given by

(x)i = xi+1.

Prove the following assertion. It is true that

x− Tx = 0 ⇐⇒ x = 0,Gregory T. von Nessi

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8.9 Exercises207

however, the equation

x− Tx = y

does not have a solution for all y ∈ l2. This implies that T is not compact and the Fredholm’s

alternative cannot be applied. Give an example of a bounded sequence xj such that Txj does

not have a convergent subsequence.

8.4: The space l2 is a Hilbert space with respect to the scalar product

(x, y) =

∞∑

i=1

xiyi .

Use general theorems to prove that the sequence of unit vectors in l2 given by (ei)j = δi jcontains a weakly convergent subsequence and characterize the weak limit.

8.5: Find an example of a sequence of integrable functions fi on the interval (0, 1) that

converge to an integrable function f a.e. such that the identity

limi→∞

∫ 1

0

fi(x) dx =

∫ 1

0

limi→∞

fi(x) dx

does not hold.

8.6: Let a ∈ C1([0, 1]), a ≥ 1. Show that the ODE

−(au′)′ + u = f

has a solution u ∈ C2([0, 1]) with u(0) = u(1) = 0 for all f ∈ C0([0, 1]).

Hint: Use the method of continuity and compare the equation with the problem −u′′ = f .

In order to prove the a-priori estimate derive first the estimate

sup |tut | ≤ sup |f |

where ut is the solution corresponding to the parameter t ∈ [0, 1].

8.7: Let λ ∈ R. Prove the following alternative: Either (i) the homogeneous equation

−u′′ − λu = 0

has a nontrivial solution u ∈ C([0, 1]) with u(0) = u(1) = 0 or (ii) for all f ∈ C0([0, 1]),

there exists a unique solution u ∈ C2([0, 1]) with u(0) = u(1) = 0 of the equation

−u′′ − λu = f .

Moreover, the mapping Rλ : f 7→ u is bounded as a mapping from C0([0, 1]) into C2([0, 1]).

Hint: Use Arzela-Ascoli.

8.8: Let 1 < p ≤ ∞ and α = 1 − 1p . Then there exists a constant C that depends

only on a, b, and p such that for all u ∈ C1([a, b]) and x0 ∈ [a, b] the inequality

||u||C0,α([a,b]) ≤ |u(x0)|+ C||u′||Lp([a,b])

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holds.

8.9: Let X = L2(0, 1) and

M =

f ∈ X :

∫ 1

0

f (x) dx = 0

.

Show that M is a closed subspace of X. Find M⊥ with respect to the notion of orthogonality

given by the L2 scalar product

(·, ·) : X ×X → R, (u, v) =

∫ 1

0

u(x)v(x) dx.

According to the projection theorem, every f ∈ X can be written as f = f1 + f2 with f1 ∈ Mand f2 ∈ M⊥. Characterize f1 and f2.

8.10: Suppose that pi ∈ (1,∞) for i = 1, . . . , n with

1

p1+ · · ·+ 1

pn= 1.

Prove the following generalization of Holder’s inequality: if fi ∈ Lpi (Ω) then

f =

n∏

i=1

fi ∈ L1(Ω) and

∣∣∣∣∣

∣∣∣∣∣n∏

i=1

fi

∣∣∣∣∣

∣∣∣∣∣L1(Ω)

≤n∏

i=1

||fi ||Lpi (Ω).

8.11: [Gilbarg-Trudinger, Exercise 7.1] Let Ω be a bounded domain in Rn. If u is a measurable

function on Ω such that |u|p ∈ L1(Ω) for some p ∈ R, we define

Φp(u) =

(1

|Ω|

∫

Ω

|u|p dx) 1

p

.

Show that

limp→∞

Φp(u) = supΩ|u|.

8.12: Suppose that the estimate

||u − uΩ||Lp(Ω) ≤ CΩ||Du||Lp(Ω)

holds for Ω = B(0, 1) ⊂ Rn with a constant C1 > 0 for all u ∈ W 1,p0 (B(0, 1)). Here uΩ

denotes the mean value of u on Ω, that is

uΩ =1

|Ω|

∫

Ω

u(z) dz.

Find the corresponding estimate for Ω = B(x0, R), R > 0, using a suitable change of coor-

dinates. What is CB(x0,R) in terms of C1 and R?

8.13: [Qualifying exam 08/99] Let f ∈ L1(0, 1) and suppose that

∫ 1

0

f φ′ dx = 0 ∀φ ∈ C∞0 (0, 1).


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8.9 Exercises209

Show that f is constant.

Hint: Use convolution, i.e., show the assertion first for fε = ρε ∗ f .

8.14: Let g ∈ L1(a, b) and define f by

f (x) =

∫ x

a

g(y) dy.

Prove that f ∈ W 1,1(a, b) with Df = g.

Hint: Use the fundamental theorem of Calculus for an approximation gk of g. Show that

the corresponding sequence fk is a Cauchy sequence in L1(a, b).

8.15: [Evans 5.10 #8,9] Use integration by parts to prove the following interpolation in-

equalities.

a) If u ∈ H2(Ω) ∩H10(Ω), then

∫

Ω

|Du|2 dx ≤ C(∫

Ω

u2 dx

) 12(∫

Ω

|D2u|2 dx) 1

2

b) If u ∈ W 2,p(Ω) ∩W 1,p0 (Ω) with p ∈ [2,∞), then

∫

Ω

|Du|p dx ≤ C(∫

Ω

|u|p dx) 1

2(∫

Ω

|D2u|p dx) 1

2

.

Hint: Assertion a) is a special case of b), but it might be useful to do a) first. For simplicity,

you may assume that u ∈ C∞c (Ω). Also note that

∫

Ω

|Du|p dx =

n∑

i=1

∫

Ω

|Diu|2|Du|p−2 dx.

8.16: [Qualifying exam 08/01] Suppose that u ∈ H1(R), and for simplicity suppose that u

is continuously differentiable. Prove that

∞∑

n=−∞|u(n)|2 <∞.

8.17: Suppose that n ≥ 2, that Ω = B(0, 1), and that x0 ∈ Ω. Let 1 ≤ p <∞. Show that

u ∈ C1(Ω\x0) belongs to W 1,p(Ω) if u and its classical derivative ∇u satisfy the following

inequality,∫

Ω\x0(|u|p + |∇u|p) dx <∞.

Note that the examples from the chapter show that the corresponding result does not hold

for n = 1. This is related to the question of how big a set in Rn has to be in order to be

“seen” by Sobolev functions.

Hint: Show the result first under the additional assumption that u ∈ L∞(Ω) and then

consider uk = φk(u) where φk is a smooth function which is close to

ψk(s) =

k if s ∈ (k,∞),

s if s ∈ [−k, k ],

−k if s ∈ (−∞,−k).

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You could choose φk to be the convolution of ψk with a fixed kernel.

8.18: Give an example of an open set Ω ⊂ Rn and a function u ∈ W 1,∞(Ω), such that

u is not Lipschitz continuous on Ω.

Hint: Take Ω to be the open unit disk in R2, with a slit removed.

8.19: Let Ω = B(

0, 12

)⊂ Rn, n ≥ 2. Show that log log 1

|x | ∈ W 1,n(Ω).

8.20: [Qualifying exam 01/00]

a) Show that the closed unit ball in the Hilbert space H = L2([0, 1]) is not compact.

b) Show that g : [0, 1]→ R is a continuous function and define the operator T : H → H

by

(Tu)(x) = g(x)u(x) for x ∈ [0, 1].

Prove that T is a compact operator if and only if the function g is identically zero.

8.21: [Qualifying exam 08/00] Let Ω = B(0, 1) be the unit ball in R3. For any function

u : Ω → R, define the trace Tu by restricting u to ∂Ω, i.e., Tu = u∣∣∂Ω

. Show that

T : L2(Ω)→ L2(∂Ω) is not bounded.

8.22: [Qualifying exam 01/00] Let H = H1([0, 1]) and let Tu = u(1).

a) Explain precisely how T is defined for all u ∈ H, and show that T is a bounded linear

functional on H.

b) Let (·, ·)H denote the standard inner product inH. By the Riesz representation theorem,

there exists a unique v ∈ H such that Tu = (u, v)H for all u ∈ H. Find v .

8.23: Let Ω = B(0, 1) ⊂ Rn with n ≥ 2, α ∈ R, and f (x) = |x |α. For fixed α, determine

for which values of p and q the function f belongs to Lp(Ω) and W 1,q(Ω), respectively. Is

there a relation between the values?

8.24: Let Ω ⊂ Rn be an open and bounded domain with smooth boundary. Prove that

for p > n, the space W 1,p(Ω) is a Banach algebra with respect to the usual multiplication

of functions, that is, if f , g ∈ W 1,p(Ω), then f g ∈ W 1,p(Ω).

8.25: [Evans 5.10 #11] Show by example that if we have ||Dhu||L1(V ) ≤ C for all 0 <

|h| < 12dist (V, ∂U), it does not necessarily follow that u ∈ W 1,1(Ω).

8.26: [Qualifying exam 01/01] Let A be a compact and self-adjoint operator in a Hilbert

space H. Let ukk∈N be an orthonormal basis of H consisting of eigenfunctions of A with

associated eigenvalues λkk∈N. Prove that if λ 6= 0 and λ is not an eigenvalue of A, then

A− λI has a bounded inverse and

||(A− λI)−1|| ≤ 1

infk∈N |λk − λ|<∞.

8.27: [Qualifying exam 01/03] Let Ω be a bounded domain in Rn with smooth boundary.Gregory T. von Nessi

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8.9 Exercises211

a) For f ∈ L2(Ω) show that there exists a uf ∈ H10(Ω) such that

||f ||2H−1(Ω) = ||uf ||2H10(Ω)

= (f , uf )L2(Ω).

b) Show that L2(Ω) ⊂⊂−→ H−1(Ω). You may use the fact that H10(Ω) ⊂⊂−→ L2(Ω).

8.28: [Qualifying exam 08/02] Let Ω be a bounded, connected and open set in Rn with

smooth boundary. Let V be a closed subspace of H1(Ω) that does not contain nonzero

constant functions. Using H1(Ω) ⊂⊂−→ L2(Ω), show that there exists a constant C < ∞,

independent of u, such that for u ∈ V ,

∫

Ω

|u|2 dx ≤ C∫

Ω

|Du|2 dx.

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212Chapter 9: Function Spaces

9 Function Spaces


– Alan Rickman

Contents


9.2 Holder Spaces 217

9.3 Lebesgue Spaces 218

9.4 Morrey and Companato Spaces 225

9.5 The Spaces of Bounded Mean Oscillation 238

9.6 Sobolev Spaces 240

9.7 Sobolev Imbeddings 261

9.8 Difference Quotients 271

9.9 A Note about Banach Algebras 274

9.10 Exercises 274

9.1 Introduction

A central goal to the theoretical study of PDE is to ascertain properties of solutions of

PDE that are not directly attainable by direct analytic means. To this end, one usually

seeks to classify solutions of PDE belonging to certain functions spaces whose elements

have certain known properties. This is, by no means, trivial, as the process requires not only

considering/defining appropriate function spaces but also elucidating the properties of such.

In this chapter we will consider some of the most prolific spaces in theoretical PDE along

with some of their respective properties.Gregory T. von Nessi

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9.2 Holder Spaces213

9.2 Holder Spaces

To start the discussion on function spaces, Holder spaces will first be considered as they are

easier to construct than Sobolev Spaces.

Consider open Ω ⊂ Rn and 0 < γ ≤ 1. Recall that a Lipschitz continuous function

u : Ω→ R satisfies

|u(x)− u(y)| ≤ C|x − y | (x, y ∈ Ω)

for some constant C. Of course, this estimates implies that u is continuous with a uniform

modulus of continuity. A generalization of the Lipschitz condition is

|u(x)− u(y)| ≤ C|x − y |γ (x, y ∈ Ω)

for some constant C. If this estimate holds for a function u, that function is said to be

Holder continuous with exponent γ.

Definition 9.2.1: (i) If u : Ω→ R is bounded and continuous, we write

||u||C(Ω) := supx∈Ω|u(x)|.

(ii) The γth-Holder seminorm of u : Ω→ R is

[u]C0,γ(Ω) := supx, y ∈ Ωx 6= y

|u(x)− u(y)||x − y |γ

,

and the γth-Holder norm is

||u||C0,γ(Ω) := ||u||C(Ω) + [u]C0,γ(Ω)

Definition 9.2.2: The Holder space

Ck,γ(Ω)

consists of all functions u ∈ Ck(Ω) for which the norm

||u||Ck,γ(Ω) :=∑

|α|≤k||Dαu||C(Ω) +

∑

|α|=k[Dαu]C0,γ(Ω)

is finite.

So basically, Ck,γ(Ω) consists of those functions u which are k-times continuous differen-

tiable with the kth-partial derivatives Holder continuous with exponent γ. As one can easily

ascertain, these functions are very well-behaved, but these spaces also have a very good

mathematical structure.

Theorem 9.2.3 (): The function spaces Ck,γ(Ω) are Banach spaces.

The proof of this is straight-forward and is left as an exercise for the reader.

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9.3 Lebesgue Spaces

Notation: E is a measurable set, and Ω is an open set ⊂ Rn.

Definition 9.3.1: For 1 ≤ p <∞ we define

Lp(E) = f : E → R, measurable,

∫

E

|f |p <∞

L∞(E) = f : E → R, ess sup |f | <∞where ess sup|f | = infK ≥ 0, |f | ≤ K a.e..

Two functions are said to be equivalent if f = g a.e. Then we define

Lp(E) = Lp(E)/ ∼Theorem 9.3.2 (): The spaces Lp(E) are BS with

||f ||p =

[∫

E

|f |p dλ]1/p

1 ≤ p <∞

||f ||∞ = ess sup |f |

• Lp(E) = f : E → R,∫|f |p dx < ∞. These are indeed equivalence classes of

functions (as stated in the last lecture). In general we can regard this as a space of

functions. We do however, need to be careful sometimes. The main such point is

that saying that f ∈ Lp is continuous means: there exists a representative of f that is

continuous. Sometimes referred to as “a version of f ”.

• Lploc(Ω) = f : Ω→ R measurable. ∀Ω′ ⊂⊂ Ω,∫

Ω′ |f |p dx <∞

Example: L1((0, 1))

f (x) = 1x /∈ L1((0, 1)), f ∈ L1

loc((0, 1))

Theorem 9.3.3 (Young’s Inequality): 1 < p <∞, a, b ≥ 0

ab ≤ ap

p+bp′

p′,

where p′ is the dual exponent to p defined by

1

p+

1

p′= 1

General convention (1)′ =∞, (∞)′ = 1.

Proof: a, b 6= 0, ln is convex downward (concave). Thus,

ln(λx + (1− λ)x) ≥ λ ln x + (1− λ) ln y

= ln xλ + ln y1−λ

take exponential of both sides

λx + (1− λ)y > xλy1−λ

Define λ = 1p , xλ = a, y1−λ = b =⇒ Young’s inequality.

1− λ = 1− 1

p=

1

p′


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9.3 Lebesgue Spaces215

Theorem 9.3.4 (Holder’s Inequality): f ∈ Lp(E), g ∈ Lp′(E), then f g ∈ L1(E) and

∣∣∣∣∫

E

f g

∣∣∣∣ ≤ ||f ||p||g||p′

Proof: Young’s inequality

|g|||g||p′

· |f |||f ||p≤ 1

p

|f |p||f ||pp

+1

p′|g|p′

||g||p′p′

Clear for p = 1 and p =∞. Assume 1 < p <∞ to use Young’s inequality. Integrate on E.

∫

E

|g|||g||p′

· |f |||f ||p≤ 1

p

∫

E

|f |p||f ||pp

+1

p′

∫

E

|g|p′

||g||p′p′=

1

p+

1

p′= 1

=⇒∫

E

|f | · |g| dx ≤ ||f ||p||g||p′

Remark: Sometimes f = 1 is a good choice

∫

E

1 · |g| dx ≤(∫

E

1p) 1

p(∫

E

gp′) 1

p′

= |E|1p ||g||p′

You gain |E|1p .

Lemma 9.3.5 (Minkowski’s inequality): Let 1 ≤ p ≤ ∞, then ||f + g||p ≤ ||f ||p + ||g||p.

This is clear for p = 1 and p =∞.

Assume 1 < p <∞. Convexity of xp implies that

|f + g|p ≤ 2p−1(|f |p + |g|p) =⇒ f + g ∈ Lp

|f + g|p = |f + g||f + g|p−1

≤ |f |︸︷︷︸Lp

|f + g|p−1

︸︷︷︸Lp′

+ |g|︸︷︷︸Lp

|f + g|p−1

︸︷︷︸Lp′

p′ = pp−1 , (|f + g|p−1)p

′= |f + g|p. Integrate and use Holder

∫|f + g|p dx ≤ ||f ||p

(∫|f + g|(p−1)p′

) 1p′

+ ||g||p(∫|f + g|(p−1)p′

) 1p′

= (||f ||p + ||g||p)

(∫|f + g|p

) 1p′

= (||f ||p + ||g||p) ||f + g||pp′p

Thus, we have

||f + g||p−pp′

p ≤ ||f ||p + ||g||p

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Now p − pp′ = p

(1− 1

p′

)= p 1

p = 1. Thus we have the result.

Proof of 9.3.2(Lp is complete) p =∞ (“simpler than 1 ≤ p <∞”).

If fn is Cauchy, then fn(x) is Cauchy for a.e. x and you can define the limit f (x) a.e.

Theorem (Riesz-Fischer) Lp complete for 1 ≤ p <∞

Proof: Suppose that fk is Cauchy, ∀ε > 0,∃k(ε) such that

||fk − fl || < ε if k, l ≥ k(ε)

Choose a subsequence that converges fast in the sense that

∞∑

i=1

||fki+1− fki ||p <∞

relabel the subsequence and call it again fk . Now define

gl =

l∑

k=1

|fk+1 − fk |

then gl ≥ 0 and by Minkowski’s inequality, ||gl ||p ≤ C. By Fatou’s Lemma∫

liml→∞

gpl dx ≤ liml→∞

inf

∫gpl dx

≤(

liml→∞

inf ||gl ||p)p

<∞

Thus liml→∞ gpl is finite for a.e. x , limk→∞ gl(x) exists =⇒ fk(x) is a Cauchy sequence for

a.e. x . Thus limk→∞ fk(x) = f (x) exists a.e. By Fatou∫|f − fl |p dx =

∫limk→∞

|fk(x)− fl(x)|p dx

≤ limk→∞

inf

∫|fk − fl |p dx

≤(

limk→∞

inf ||fk − fl ||p)p

≤(

limk→∞

inf

∞∑

k=l

||fk − fk+1||p)p→ 0 as l →∞

=⇒ ||fl − f ||p → 0

We know f is in Lp by a simple application of Minkowski’s inequality to the last result.

Sobolev Spaces ⇐⇒ f ∈ Lp with derivatives.

Calculus for smooth functions is linked to the calculus of Sobolev functions by ideas of

approximation via convolution methods.

Theorem 9.3.6 (): (see Royden for proof) Let 1 ≤ p < ∞. Then the space of continuous

functions with compact support is dense in in Lp, i.e.

C0c (Ω) ⊂ Lp(Ω) dense


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Idea of Proof:

1. Step functions

∞∑

i=1

aiχEi are dense

2. Approximate χEi by step functions on cubes.

3. Construct smooth approximation for χO, for cube O

Q

9.3.1 Convolutions and Mollifiers

• Sobolev functions = Lp functions with derivatives

• Convolution = approximation of Lp functions by smooth function approximation of the

Dirac δ

– φ ∈ C∞c (Rn), supp(φ) ∈ B(0, 1)

– φ ≥ 0,

∫

Rnφ(x) dx = 1

Rescaling: φε(x) = 1εnφ(xε

),

∫

Rnφε(x) dx = 1. φε(x) → 0 as ε → 0 pointwise a.e. and

∫

Rnφε(x) dx = 1.

B(0, 1)B(0, ǫ)

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Definition 9.3.7: The convolution,

fε(x) = (φε ∗ f )(x)

=

∫

Rnφε(x − y)f (y) dy

=

∫

B(x,ε)

φε(x − y)f (y) dy

is the weighted average of f on a ball of radius ε > 0.

Theorem 9.3.8 (): fε ∈ C∞

Proof: Difference Quotients and Lebesgue’s Dominated convergence theorem.

Now, we ask the question f ∈ Lp, fε → f in Lp. If p = ∞ fε → f implies f ∈ C0

which implies f ∈ Lp. We know f ∈ C0 since fε ∈ C∞ ⊂ C0.

Ω′′

Ω

Ω′

ǫ

d

Lemma 9.3.9 (): f ∈ C0(Ω), Ω′′ ⊂⊂ Ω compact subsets, then fε → f uniformly on Ω′′!

Proof:

d = dist(Ω′, ∂Ω)

= inf|x − y | : x ∈ Ω′, y ∈ ∂Ω

Ω′′ = x ∈ Ω,dist(x,Ω′) ≤ d2.

Let ε ≤ d2 , x ∈ Ω′ =⇒ B(x, ε) ⊂⊂ Ω′′, and fε(x) is well defined

|fε(x)− f (x)| =

∣∣∣∣∫

Rnφε(x − y)[f (y)− f (x)] dy

∣∣∣∣

≤ supy∈B(x,ε)

|f (y)− f (x)|∣∣∣∣∫

Rnφε(x − y) dy

∣∣∣∣︸︷︷︸

=1

——

f ∈ C0(Ω) =⇒ f uniformly continuous on Ω′′ ⊂⊂ Ω. Thus,

ω(ε) = sup|f (x)− f (y)|, x, y ∈ Ω′′, |x − y | < ε → 0

as ε→ 0Gregory T. von Nessi

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——

≤ ω(ε)→ 0 uniformly on Ω′

as ε→ 0

Proposition 9.3.10 (): If u, v ∈ L1(Rn), then u ∗ v ∈ L1(Rn) ∩ C0(Rn). This is a special

case of the following: given f ∈ L1(Rn) and g ∈ Lp(Rn) with p ∈ [1,∞], then f ∗ g ∈Lp(Rn) ∩ C0(Rn).

Proof: We will prove the specialized case (p = 1) first.

||f ∗ g||L1(Rn)| =

∫

Rn

∣∣∣∣∫

Rnf (x − y)g(y) dy

∣∣∣∣ dx

≤∫

Rn|g(y)|

∫

Rn|f (x − y)| dxdy

=

∫

Rn|g(y)|

∫

Rn|f (z)| dzdy = ||f ||L1(Rn)||g||L1(Rn)

There is nothing to show for p =∞. For p ∈ (1,∞), we use Holder’s inequality to compute

||f ∗ g||pLp(Rn)

=

∫

Rn

∣∣∣∣∫

Rnf (y)g(x − y) dy

∣∣∣∣p

dx

≤∫

Rn

(∫

Rn|f (y)|

1p′ |f (y)|

1p |g(x − y)| dy

)pdx

≤∫

Rn

(∫

Rn|f (y)| dy

) pp′(∫

Rn|f (y)| · |g(x − y)|p dy

)dx

= ||f ||p/p′L1(Rn)

∫

Rn

∫

Rn|f (y)| · |g(x − y)|p dydx

= ||f ||p/p′L1(Rn)

∫

Rn|f (y)|

∫

Rn|g(x − y)|p dydx

= ||f ||p/p′L1(Rn)

||f ||L1(Rn)||g||pLp(Rn)= ||f ||p

L1(Rn)||g||p

Lp(Rn)

which proves the assertion.

Lemma 9.3.11 (): f ∈ Lp(Ω), 1 ≤ p <∞. Then fε → f in Lp(Ω).

Remarks:

• The same assertion holds in Lploc(Ω).

• C∞c (Ω) is dense in Lp(Ω).

Proof: Extend f by 0 to Rn: assume Ω = Rn. By Proposition 8.31, we see that if f ∈ Lp(Rn),

g ∈ L1(Rn), then f ∗ g ∈ Lp(Rn) and ||f ∗ g||p ≤ ||f ||p||g||1.

So, take

g = φε : ||fε||p = ||f ∗ φε||p ≤ ||f ||p · ||φε||1 = ||f ||p (9.1)

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Need to show ∀δ > 0, ∃ε(δ) > 0, ||fε − f ||p < δ, ∀ε ≤ ε(δ).

By Theorem 8.29, ∃f1 ∈ C0c (Ω) such that

f = f1 + f2, ||f2||p = ||f − f1||p <δ

3

||fε − f || = ||(f1 + f2)ε + (f1 + f2)||p≤ ||f1ε − f1||p + ||f2ε − f2||p≤ ||f1ε − f1||p + ||f2ε||p + ||f ||p≤ ||f1ε − f1||p +

2δ

3by (9.1)

f1 has compact support in Ω, supp(f1) ⊂ Ω′ ⊂⊂ Ω, d = dist(Ω′, ∂Ω), Ω′′ =x ∈ Ω′,dist(x,Ω′) ≤ d

2

.

Now suppose that ε < d2 . By Lemma 8.30, f1ε → f1 uniformly in Ω′′ and thus in Ω (since

f1, f1ε = 0 in Ω \Ω′′). Now choose ε0 small enough such that ||f1ε − f ||p < δ3

=⇒ ||fε − f ||p ≤ δ ∀ε(δ) ≤ε0,

δ2

.

9.3.2 The Dual Space of Lp

Explicit example of T ∈ (Lp)∗:

Tg : Lp → R, Tgf =

∫f g dx

Tgf is defined by Holder, and

|Tgf | ≤ ||f ||p||g||p′

Thus,

||Tg|| = supf ∈Lp,f 6=0

|Tgf |||f ||p

≤ supf ∈Lp,f 6=0

||f ||p||g||p′||f ||p

= ||g||p′

In fact ||Tg|| = ||g||p′ . This follows by choosing f = |g|p′−2g.

Theorem 9.3.12 (): Let 1 ≤ p <∞,

Then J : Lp′ → (Lp)∗, J(g) = Tg is onto with ||J(g)|| = ||g||p′ . In particular, Lp is reflexive

for 1 < p <∞.

Remark: (L1)∗ = L∞, (L∞)∗ 6= L1.

9.3.3 Weak Convergence in Lp

• 1 ≤ p <∞, fj f weakly in Lp

∫fjg →

∫f g ∀g ∈ Lp′

• p =∞, fj * f weakly∗ in L∞

∫fjg →

∫f g ∀g ∈ L1


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9.4 Morrey and Companato Spaces221

Examples: Obstructions to strong convergence. (1) Oscillations. (2) Concentrations.

• Oscillations: Ω = (0, 1), fj(x) = sin(jx). fj * 0 in L∗∞, but fj 9 0 in L∞. (||0 −sin(jx)||∞ = 1).

• Concentration: g ∈ C∞c (Rn), fj(x) = j−n/pg(j−1x). So,∫|fj |p dx =

∫j−ngp(j−1x) dx =

∫|g|p dx 1 < p <∞

So we see that fj 0 in Lp, but fj 9 0 in Lp.

Last Lecture: We showed (Lp)∗ = Lp′, where 1

p + 1p′ = 1 and 1 ≤ p <∞. With this we can

talk about

• Weak-* convergence in L∞.

• Weak convergence in Lp, 1 ≤ p <∞.

So, as an example take weak convergence in L2(Ω), uj u in L2(Ω) if∫

Ω

ujv dx →∫

Ω

uv dx ∀v ∈ L2(Ω).

Obstructions to strong convergence which have weak convergence properties are oscillations

and concentrations.

9.4 Morrey and Companato Spaces

Loosely speaking, Morrey Spaces and Companato Spaces characterize functions based on

their “oscillations” in any given neighborhood in a set domain. This will be made clear by the

definitions. Philosophically, Morrey spaces are a precursor of Companato Spaces which in

turn are basically a modern re-characterization of the Holder Spaces introduced earlier in the

chapter. This new characterization is based on integration; and hence, is easier to utilize, in

certain proofs, when compared against the standard notion of Holder continuity.

It would have been possible to omit these spaces from this text altogether (and this is the

convention in books at this level); but I have elected to discuss these for the two following

reasons. First, the main theorem pertaining to the imbedding of Sobolev Spaces into Holder

spaces will be a natural consequence of the properties of Companato Spaces; this is much

more elegant, in the author’s opinion, than other proofs using standard properties of Holder

continuity. Second, we will utilize these spaces in the next chapter to derive the classical

Schauder regularity results for elliptic systems of equations. Again, it is the authors opin-

ion that these regularity results are proved much more intuitively via the use of Companato

spaces as opposed to the more standard proofs of yore. For the following introductory ex-

position, we essentially follow [Giu03, Gia93].

Per usual, we consider Ω to be open and bounded in Rn; but in addition to this, we will

also assume for all x ∈ Ω and for all ρ ≤ diam(Ω) that

|Bρ(x) ∩Ω| ≥ Aρn, (9.2)

where A > 0 is a fixed constant. This is, indeed, a condition on the boundary of Ω that pre-

cludes domains with “sharp outward cusps”; geometrically, it can be seen that any Lipschitz

domain will satisfy (9.2). With that we make our first definition.

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Definition 9.4.1: The Morrey Space Lp,λ(Ω) for all real p ≥ 1 and λ ≥ 0 is defined as

Lp,λ(Ω) :=

u ∈ Lp(Ω)

∣∣∣∣∣ ||u||Lp,λ(Ω) <∞,

where

||u||pLp,λ(Ω)

:= supx0 ∈ Ω

0 < ρ < diam(Ω)

(ρ−λ

∫

Ω(x0,ρ)

|u|p dx)

with Ω(x0, ρ) = B(x0, ρ) ∩Ω.

It is left up to the reader to verify that Lp,λ(Ω) is indeed a BS. One can easily see from

the definition that u ∈ Lp,λ(Ω) if and only if there exists a constant C > 0 such that

∫

Ω(x0,ρ)

|u|p dx < Cρλ, (9.3)

for all x0 ∈ Ω with 0 < ρ < ρ0 for some fixed ρ0 > 0. Sometimes, (9.3) is easier to use than

the original definition of a Morrey Space. Also, since x0 ∈ Ω is arbitrary in (9.3), it really

doesn’t matter what we set ρ0 > 0, as Ω is assumed to be bounded (i.e. it can be covered

by a finite number of balls of radius ρ0).

In addition to this, we get that Lp,λ(Ω) ∼= Lp(Ω) for free from the definition. Now, we

come to our first theorem concerning these spaces.

Theorem 9.4.2 (): Ifn − λp≥ n − µ

qand p ≤ q, then Lq,µ(Ω) ⊂−→ Lp,λ(Ω).

Proof: First note that the index condition above can be rewritten as

µp

q+ n − np

q≥ λ.

With this in mind, the rest of the proof is a straight-forward calculation using Holder’s

inequality:

∫

Ω(x0,ρ)

|u|p dx Holder≤(∫

Ω(x0,ρ)

|u|q dx)p/q

|Ω(x0, ρ)|1−p/q

≤ α(n)1−p/q ·(||u||q

Lq,µ(Ω)· ρµ)p/q

· ρn(1−p/q)

≤ α(n)1−p/q · ||u||qLq,µ(Ω)

· ρµp/q+n−np/q

≤ α(n)1−p/q · ||u||qLq,µ(Ω)

· ρλ.

Note in the third inequality utilizes the assumption p ≤ q. The above calculation implies

that

||u||pLp,λ(Ω)

≤ C||u||qLq,µ(Ω)

,

i.e. Lq,µ(Ω) ⊂−→ Lp,λ(Ω).

Corollary 9.4.3 (): Lp,λ(Ω) ⊂−→ Lp(Ω) for any λ > 0.Gregory T. von Nessi

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Proof: This is a direct consequence of the previous theorem.

The next two theorems are more specific index relations for Morrey Spaces, but are very

important properties nonetheless.

Theorem 9.4.4 (): If dim(Ω) = n, then Lp,n(Ω) ∼= L∞(Ω).

Proof: First consider u ∈ L∞(Ω). In this case,

∫

Ω(x0,ρ)

|u|p dx Holder≤ C · |Ω(x0, ρ)| · ||u||pL∞(Ω)

= C · α(n)ρn · ||u||pL∞(Ω)

<∞,

i.e. u ∈ Lp,n(Ω). If, on the other hand, u ∈ Lp,n(Ω), then, considering for a.e. x0 ∈ Ω we

have

u(x0) = limρ→0

1

|Ω(x0, ρ)|

∫

Ω(x0,ρ)

u dx

via Lebesgue’s Theorem, it is calculated that

|u(x0)| ≤ limρ→0−∫

Ω(x0,ρ)

|u| dx ≤ limρ→0

(−∫

Ω(x0,ρ)

|u|p dx)1/p

<∞.

Thus, we conclude u ∈ L∞(Ω).

Theorem 9.4.5 (): If dim(Ω) = n, then Lp,λ(Ω) = 0 for λ > n.

Proof: Fix an arbitrary x0 ∈ Ω, using Lebesgue’s Theorem, we calculate

|u(x0)| ≤ limρ→0

1

|Ω(x0, ρ)|

∫

Ω(x0,ρ)

|u|p dx

Holder≤ limρ→0

|Ω(x0, ρ)|1−1/p

|Ω(x0, ρ)|1/p(∫

Ω(x0,ρ)

|u|p dx)1/p

≤ limρ→0

diam(Ω)1−1/p

|Ω(x0, ρ)|1/p(∫

Ω(x0,ρ)

|u|p dx)1/p

≤ limρ→0

C · ρ−n/pρλ/p

= limρ→0

C · ρλ−n/p = 0,

since λ− n > 0. As x0 ∈ Ω is arbitrary, we see that u ≡ 0. .

Now, that a few of the basic properties of Morrey Spaces have been elucidated, we will

now consider a closely related group of spaces.

Definition 9.4.6: The Companato Spaces, Lp,λ(Ω), are defined as

Lp,λ(Ω) :=

u ∈ Lp(Ω)

∣∣∣∣∣ [u]Lp,λ(Ω) <∞,

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where

[u]pLp,λ(Ω):= sup

x0 ∈ Ω

0 < ρ < diam(Ω)

ρ−λ(∫

Ω(x0,ρ)

|u − ux0,ρ|p dx)

with

ux0,ρ := −∫

Ω(x0,ρ)

u dx.

First, one will notice that [u]Lp,λ(Ω) is a seminorm; this is seen from the fact that

[u]Lp,λ(Ω) = 0 if and only if u is a constant. Thus, we define

|| · ||Lp,λ(Ω) := || · ||Lp(Ω) + [ · ]pLp,λ(Ω)

as the norm on Lp,λ(Ω), making such a BS. The verification of this is left to the reader.

In analogy with theorem 9.4.2, we have the following.

Theorem 9.4.7 (): Ifn − λp≥ n − µ

qand p ≤ q, then Lq,µ(Ω) ⊂−→ Lp,λ(Ω)

Proof: See proof of theorem 9.4.2.

Just from looking at the definitions, naive intuition suggests that there should be some

correspondence between Morrey and Companato Spaces, but this correspondence is by no

means obvious. The next two theorems will actually elucidate the correspondence between

Morrey, Companato and Holder Spaces, but we first prove a lemma that will subsequently

be needed in the proofs of the aforementioned theorems.

Lemma 9.4.8 (): If Ω is bounded and satisfies (9.2), then for any R > 0,

|ux0,Rk − ux0,Rh | ≤ C · [u]Lp,λ(Ω)Rλ−n/pk , (9.4)

where Ri := 2−i · R and 0 < h < k .

Proof: First we fix an arbitrary R > 0 and define r such that 0 < r < R. Now, we

calculate that

rn|ux0,R − ux0,r |p

≤ |Ω(x0, r)|α(n)

|ux0,R − ux0,r |p

≤ 1

α(n)

∫

Ω(x0,r)

|ux0,R − ux0,r |p dx

≤ 2p−1

α(n)

[∫

Ω(x0,r)

|u(x)− ux0,R|p + |u(x)− ux0,r |p dx]

≤ 2p−1

α(n)

[∫

Ω(x0,R)

|u(x)− ux0,R|p dx +

∫

Ω(x0,r)

|u(x)− ux0,r |p dx]

≤ 2p−1

α(n)· [Rλ + rλ] · [u]pLp,λ(Ω)

≤ C · Rλ[u]pLp,λ(Ω).


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Dividing by rn and taking the p-th root of this result yields

|ux0,R − ux0,r | ≤ C · [u]Lp,λ(Ω) · Rλ/p · r−n/p. (9.5)

From here, we will utilize the definition of Ri and (9.5) to ascertain

|ux0,Ri− u

x0,Ri+1| ≤ C · [u]Lp,λ(Ω) · 2i(n−λ)/p+n/p · R(λ−n)/p

≤ C · [u]Lp,λ(Ω) · 2i(n−λ)/p · R(λ−n)/p

In the last equality, the 2n/p has been absorbed into the constant. Trudging on, we finish

the calculation to discover

|ux0,R− u

x0,Rh+1| ≤

h∑

i=0

|ux0,Ri− u

x0,Ri+1|

≤ C · [u]Lp,λ(Ω)R(λ−n)/p

h∑

i=0

2i(n−λ)/p

= C · [u]Lp,λ(Ω)R(λ−n)/p

(2(h+1)(n−λ)/p − 1

2(n−λ)/p − 1

)

≤ C · [u]Lp,λ(Ω)R(λ−n)/ph+1 . (9.6)

The conclusion follows as (9.6) is valid for any h ≥ 0 and R > 0 was chosen arbitrarily.

Theorem 9.4.9 (): If dim(Ω) = n and 0 ≤ λ ≤ n, then Lp,λ(Ω) ∼= Lp,λ(Ω).

Proof: First suppose u ∈ Lp,λ(Ω) and calculate

ρ−λ∫

Ω(x0,ρ)

|u − ux0,ρ|p dx

≤ 2p−1ρ−λ[∫

Ω(x0,ρ)

|u|p + |ux0,ρ|p dx]

≤ 2p−1ρ−λ[∫

Ω(x0,ρ)

|u|p dx + |Ω(x0, ρ)| · |ux0,ρ|p]

Holder≤ 2p−1ρ−λ[∫

Ω(x0,ρ)

|u|p dx + |Ω(x0, ρ)| · −∫

Ω(x0ρ)

|u|p dx]

≤ 2pρ−λ∫

Ω(x0ρ)

|u|p dx <∞.

Thus, we see that u ∈ Lp,λ(Ω).

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Now assume u ∈ Lp,λ(Ω). We find

ρ−λ∫

Ω(x0,ρ)

|u|p dx

= ρ−λ∫

Ω(x0,ρ)

|u − ux0,ρ + ux0,ρ|p dx

≤ 2p−1ρ−λ[∫

Ω(x0,ρ)

|u − ux0,ρ|p dx + |Ω(x0, ρ)| · |ux0,ρ|p]

≤ 2p−1

ρ−λ

∫

Ω(x0,ρ)

|u − ux0,ρ|p dx + C · ρn−λ|ux0,ρ|p.

(9.7)

To proceed with (9.7), we must endeavor to estimate estimate |ux0,ρ|. To do this, we consider

0 < r < R along with the calculation

|ux0,ρ|p ≤ 2p−1|ux0,R|p + |ux0,R − ux0,ρ|p. (9.8)

Thus, we now seek to estimate the RHS. To do this, we utilize (9.6) with diam(Ω) < R ≤2 · diam(Ω) chosen such that Rh = ρ for some h > 0. Inserting this result into (9.8) and

(9.8) in (9.7) yields

ρ−λ∫

Ω(x0,ρ)

|u|p dx ≤ 2p−1

[ρ−λ

∫

Ω(x0,ρ)

|u − ux0,ρ|p dx

+ C · ρn−λ2p−1 ·∣∣∣∣−∫

Ω(x0,R)

u dx

∣∣∣∣p

+ C · 2p−1 · [u]pLp,λ(Ω)

]<∞.

Thus, u ∈ Lp,λ(Ω).

Unlike Lp,λ(Ω), L∞(Ω) ( Lp,n(Ω). To show this we consider the following.

Example 8: Let us look at consider log x , where x ∈ (0, 1). On any given sub-interval

[a, b] ⊂ (0, 1), we have

∫ b

a

| log x | dx = a · log a − b · log b + (b − a).

Now, if we multiply by ρ−n = 2b−a , we have

2

b − a

∫ b

a

| log x | dx = 2

(a · log a − b · log b

b − a + 1

).

If we can show that the RHS is bounded for any 0 < a < b < 1, then we will have shown

that log x ∈ L1,1((0, 1)). But choosing b := e−i and a := e−(i+1), where i > 0 shows that

limi→∞

2e i

1− e−1

∫ e−i

e−(i+1)

| log x | dx = 2 · limi→∞

(i − e−1(i + 1)

1− e−1+ 1

)

= 2 · limi→∞

((1− e−1)i − e−1

1− e−1+ 1

)

= ∞.Gregory T. von Nessi

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Thus, log x /∈ L1,1(Ω) which is in agreement with theorem 9.4.4. Now considering the same

interval [a, b], it is also easily calculated that

2

b − a

∫ b

a

| log x − (log x)(b+a)/2,(b−a)/2| dx = 4,

i.e. log x ∈ L1,1(Ω). On the other hand, it is obvious that log x /∈ L∞((0, 1)). Thus,

L∞(Ω) ( Lp,n(Ω).

As we have seen Companato spaces can be identified with Morrey spaces in the interval

0 ≤ λ < n. For λ > n we have the following

Theorem 9.4.10 (): For n < λ ≤ n + p holds

Lp,λ(Ω) ∼= C0,α(Ω) with α =λ− np

whereas for λ > n + p

Lp,λ(Ω) = constants.

Proof: From lemma 9.4.8 and the assumption that λ > n we observe that ux0,Rk is a

Cauchy sequence for all x0 ∈ Ω. Thus by Lebesgue’s theorem

ux0,Rk → u(x0) (k →∞),

for all x ∈ Ω. Thus we know ux0,Rk <∞ exists. Next, we note that ux,Rk =: vk(x) ∈ C(Ω).

This can proved from a standard continuity argument noting that u ∈ L1(Ω). Taking h →∞in (9.6) yields

|ux0,Rk − u(x0)|C· [u]Lp,λ(Ω)R

λ−np

k . (9.9)

From this, we deduce that vk → u is of uniform convergence in Ω; hence u is continuous.

Taking u to represent u, we know go one to show that u is Holder continuous. First, take

x, y ∈ Ω and set R = |x − y |, and α =λ− np

.

Ω

x

R y

2R

,K$/ &'< 9 &$/\)GO-/8)N)9 &'< 8 X" "Q57&>-/-/ ',9h,/ |u(x)− u(y)|

|u(x)− u(y)| ≤ |ux,2R − u(x)|︸︷︷︸≤C·Rα

+|ux,2R − uy,2R|+ |uy,2R − u(y)|︸︷︷︸≤C·Rα

;

-/-/ ',9h "*,(2$/%/5/"*9;1)V9 &'< & < 'OM[ K$/

|Ω(x, 2R) ∩Ω(y, 2R)| · |ux,2R − uy,2R|

≤∫

Ω(x,2R)|u(z)− ux,2R| dz +

∫

Ω(y,2R)|u(z) − uy,2R| dz

≤(∫

Ω(x,2R)|u(z)− ux,2R|p dz

)1/p

|Ω(x, 2R)|1−1/p

+

(∫

Ω(y,2R)|u(z)− uy,2R|p dz

)1/p

|Ω(y, 2R)|1−1/p

≤ 21+n+(n−λ)/p · %'E9(Ω)n︸︷︷︸

C

·[u]Lp,λ(Ω)Rλ/pRn(1−1/p)

= C · [u]Lp,λ(Ω)Rα.

&'< 7 +*"

R = |x− y| ?'M 13P7$"\"!MO

[u]C0,α(Ω) = supx, y ∈ Ω

x 6= y

|u(x)− u(y)||x− y|α ≤ C · [u]Lp,λ(Ω).

&'< 77

Figure 9.1: Layout for proof of Theorem 9.4.10

We start first by approximating |u(x)− u(y)|:

|u(x)− u(y)| ≤ |ux,2R − u(x)|︸︷︷︸≤C·Rα

+|ux,2R − uy,2R|+ |uy,2R − u(y)|︸︷︷︸≤C·Rα

;

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the approximations in the underbraces come from (9.9). Next we figure

|Ω(x, 2R) ∩Ω(y , 2R)| · |ux,2R − uy,2R|

≤∫

Ω(x,2R)

|u(z)− ux,2R| dz +

∫

Ω(y,2R)

∣∣u(z)− uy,2R∣∣ dz

≤(∫

Ω(x,2R)

|u(z)− ux,2R|p dz)1/p

|Ω(x, 2R)|1−1/p

+

(∫

Ω(y,2R)

∣∣u(z)− uy,2R∣∣p dz

)1/p

|Ω(y , 2R)|1−1/p

≤ 21+n+(n−λ)/p · diam(Ω)n︸︷︷︸C

·[u]Lp,λ(Ω)Rλ/pRn(1−1/p)

= C · [u]Lp,λ(Ω)Rα.

(9.10)

As R = |x − y |, we have thus shown:

[u]C0,α(Ω) = supx, y ∈ Ω

x 6= y

|u(x)− u(y)||x − y |α ≤ C · [u]Lp,λ(Ω). (9.11)

We still aren’t quite done. Remember, we are seeking our conclusion in the BS definition

of C0,α(Ω) and Lp,λ(Ω); the last calculation only pertains to the seminorms. Let us pick

an arbitrary x ∈ Ω and define v(x) := |u(x)| − infΩ |u(x)|. Obviously, v(x) > 0 and exists

xn ⊂ Ω such that v(xn)→ 0 as n →∞. Thus, it is clear that

|v(x)| = limn→∞

|v(x)− v(xn)|

≤ (diam(Ω))α|v(x)− v(xn)||x − xn|α

≤ C(diam(Ω))α · [v ]Lp,λ(Ω)

= C(diam(Ω))α · [u]Lp,λ(Ω).

The last equality comes from the fact that |u(x)| and |v(x)| differ only by a constant. With

the last calculation in mind we finally realize

|u(x)| = |v(x)|+ infΩ|u(x)|

≤ C(diam(Ω))α · [u]Lp,λ(Ω) +1

|Ω|

∫

Ω

|u| dx

Holder≤ C(diam(Ω))α · [u]Lp,λ(Ω) + |Ω|−1/p

(∫

Ω

|u|p dx) 1

p

.

Combining this result with (9.11), we conclude

supx∈Ω|u(x)|+ [u]C0,α(Ω) ≤ C

(||u||Lp(Ω) + [u]Lp,λ(Ω)

),

completing the proof!

To conclude the section, two corollaries are put to the reader’s attention. Both of these

are proved by the exact same argument as the previous theorem.Gregory T. von Nessi

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Corollary 9.4.11 (): If

∫

Ω(x0,ρ)

|u − ux0,ρ|p dx ≤ Cρλ

for all ρ ≤ ρ0 (instead of ρ < diam(Ω)), then the conclusion of theorem (9.4.10) is true

with Ω replaced by Ω(x0, ρ0), i.e. local version of (9.4.10).

Corollary 9.4.12 (): If

∫

Bρ(x0)

|u − ux0,ρ|2 dx ≤ Cρλ

for all x0 ∈ Ω and ρ < dist(x0, ∂Ω), then for λ > n, we conclude that u is Holder continuous

in every subdomain of Ω.

9.4.1 L∞ Companato Spaces

Given the characterizations of Morrey and Companato spaces, one may ask if there is a

generalization of these spaces. It turns out there is, but only parts (corresponding to ranges

of indices) of these spaces are useful for our studies.

To construct our generalization, consider

Pk :=

T (x) |T (x) =

∑

|β|≤kCβ · xβ, Cβ = const.

,

i.e. the set of all polynomials of degree k or less. Also, let us define Tk,xu as the kth order

Taylor expansion of u about x :

Tk,xu(y) =∑

|β|≤k

Dβu(x)

β!· (y − x)β.

Now we define our generalization.

Definition 9.4.13: The Lp Companato Spaces, Lp,λk (Ω), are defined as

Lp,λk (Ω) :=

u ∈ Lp(Ω)

∣∣∣∣∣ [u]Lp,λk (Ω)<∞

,

where

[u]Lp,λk (Ω):= sup

x0 ∈ Ω

0 < ρ < diam(Ω)

ρ−(k+λ) · infT∈P‖

(∫

Ω(x0,ρ)

|u − T |p dx)

It is easy to notice that Lp,λ0 (Ω) is just Lp,λ0 (Ω). One could also fit Morrey spaces into

this picture by defining P∗ = 0 and denote the Morrey space Lp,λ∗ (Ω).

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Now since Pk ⊂ Pk+1, we readily ascertain that Lp,λk+1(Ω) ⊂ Lp,λk (Ω). At this point we

are inclined to then ask is it useful to consider k > 0 as Companato spaces can already char-

acterize Holder continuous functions very well. In addition, if we did need a weaker criterion,

we could always appeal to Sobolev spaces. We could counter act the weakening affect of in-

creasing k by increasing p, but such index balancing isn’t immediately useful within the scope

of this book. What does turn out to be useful, is to consider the “other end of the index

spectrum”, i.e. take k arbitrary with p = ∞ (as opposed to taking p arbitrary with k = 0:

the regular Companato space case). With p = ∞, it turns out that L∞,αk (Ω) ∼= Ck,α(Ω);

but before we prove this, we first need the following lemma.

Warning: The following proof is somewhat technically involved and can be skipped with-

out loosing continuity. That being said, the proof itself presents some clever tools that can

be employed to bounding Holder norms.

Lemma 9.4.14 (): If u ∈ Ck,α(Ω) with Ω bounded, then

ρkx · supΩ|Dβu| ≤ C(ε) · sup

Ω|u|+ ε · ρα+k

x · [u]Ck,α(Ω),

for any x ∈ Ω and R > 0, where ρx := dist(x, ∂Ω) and k = |β|.

Proof: First we fix an arbitrary x ∈ Ω and consider 12 ≥ λ, we define R := λ · ρx ; we

will fix λ later in the proof. Upon selecting a coordinate system, we select a group of n line

segments, denoted li , of length 2R with centers at x , which are parallel to the xi axis. Let

us denote the endpoints of each li by x ′i and x ′′i . With these preliminaries, we proceed in four

main steps.

!M ", Pk ⊂ Pk+1? M (!%',,&H",A3 Lp,λk+1(Ω) ⊂ Lp,λk (Ω)

<+T:/ "-J,7:M[+;,,,%=( "Bc "X,$" )V$/[`" %/

k > 0"

9R-332"-3"!L,!%'&:3! e %/7, $$")G$/ " P&M ,S< 4%/%', Y?F)NM %' %>%.LM[!BZ, Y?3M X$/ %4,MT&'"Z-/-J!:f85J PR"-3"!< Q$/ %;$/ TOM[!B/,/R J[)0,!",/

k5 &:,!",/p

?5/$/#"$R,%/653E,/X " N,9R9;%'E,&L$" )G$//MO,/,R"-JO)k/ "#508B0< 3#%/ "N$/;$/250T$" )G$/S?K "1" %/ # %W)TR,%/c"-J$/9%? S< K<RB

k5/,&cMO,

p = ∞ i"-/-0K"%LB8,/

p5/,&MO,

k = 0N2K$/E 9R-33h"-31!" < ,

p = ∞ ?k,\$/"2$/*3 L∞,αk (Ω) ∼= Ck,α(Ω)

'k5/$/Z5J )G:M[:-/!P6/ "!?0M["O%(2)G, MO,/; 9R9h'<

",& , )G, MO,/C-/8)# "\"9;MO3Z// !,,&(,7P,P%c%.!45J"B8,-/-J%:MO,$/ K",/X ,7$/, &< 3N5J,/X" %@?KO-/8)k,"F)J-/" ""9;1 P\ "O3*!(5019R-/ &%(R5J$/%',/ e %/Q9;"!<

u ∈ Ck,α(Ω)-/&'*)

Ω (,,% ( '*)+%,

ρkx · supΩ|Dβu| ≤ C(ε) · sup

Ω|u|+ ε · ρα+k

x · [u]Ck,α(Ω),

5",x ∈ Ω

" ,,R > 0 ( - )+% % ρx :=

& '(x, ∂Ω)

",,k = |β| -

,":M D 5/,&x ∈ Ω

%D" %/ 12 ≥ λ

? M +%/R := λ · ρx

'M[CMO,, λ

E:, C-/ )< -J=" ,/W4 %',3"&'"9(?3M[L" ZhK$/-g)

n,,:"K9;7"!?J%/%

li?0)# /K

2RMO,

"2x

?kMO/ c-3, xi

8 " < YZ$"*%/6%'-J,7")!

li57&

x′i%

x′′i< ,."-/,,9R,3 " ?JM[:-/ %.,4)G$/Z9h,

"-" <

Ω

ρy

x

ρx

yR

,K$/'&'< !&$/Q)iO-/ )N) 9R9h &'< 8 &

© "!$#&%(')+*,*.-0/&1&1&2Figure 9.2: Layout for proof of Lemma 9.4.14

Step 1: For the first step we claim that

supΩ|Du| ≤ 1

Rsup

Ω|u|. (9.12)

To prove this we first note that since u ∈ Ck,α(Ω), we can utilize the mean value

theorem to state ∃x∗i ∈ li such that

Diu(x∗) =u(x ′′i )− u(x ′i )x ′′ − x ′ ≤ 1

Rsup

Ω|u|.

Step 2: Next we prove the claim that

ρx · supΩ|Diu| ≤

1

λsup

Ω|u|+ 2232λ · ρ2

x · supBR(x)

|Di ju|.Gregory T. von Nessi

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We start with calculating

|Diu(x)| ≤ |Diu(x∗j )|+ |Diu(x)−Diu(x∗j )|

≤ |Diu(x∗j )|+∫ x

x∗j

|Di ju(t)| dt

Using (9.12) on the first term on the RHS and Holder’s inequality on the second we

get

|Diu(x)| ≤ 1

Rsup

Ω|u|+ R · sup

BR(x)|Di ju|

≤ 1

Rsup

Ω|u|+ R · ρ−2

y · ρ2y · sup

BR(x)|Di ju|.

We know that ρy + R ≥ ρx Longr ightar row ρy ≥ ρx − R = (1 − λ)ρx ≥ ρx2 . Using

this the above becomes

|Diu(x)| ≤ 1

λρx· sup

Ω|u|+ 22λ

ρx· ρ2y · sup

BR(x)|Di ju|.

Since we know that ρy ≤ R + ρx Longr ightar row ρy ≤ 3ρx . Thus,

supΩ|Diu| ≤

1

λρx· sup

Ω|u|+ 2232λ · ρx · sup

y∈BR(x)|Di ju|.

Multiplying by ρx gets the conclusion of this step.

Step 3: Our last claim is that

ρ2x · sup

Ω(|Di ju|) ≤

ρxλ· supBR(x)

|Diu|+ 223α+2 · λα · ρα+2x [u]C2,α(Ω). (9.13)

For this we start with

|Di ju| ≤ |Di ju(x∗j )|+ |Di ju(x)−Di ju(x∗j )|.

Again using (9.12) we get

|Di ju| ≤1

Rsup

Ω|Diu|+ Rα ·

|Di ju(x)−Di ju(x∗j )||x − x∗j |α

≤ 1

Rsup

Ω|Diu|+ Rα · ρ−α−2

y · ρα+2y · [u]C2,α(Ω)

≤ 1

λρxsup

Ω|Diu|+

22λα

ρ2x

ρα+2y · [u]C2,α(Ω)

≤ 1

λρxsup

Ω|Diu|+ 223α+2λα · ραx · [u]C2,α(Ω)

In the above, we again used the previously stated fact that ρx2 ≤ ρy ≤ 3ρx . As in the

previous step, we take the supremum of the LHS and multiply both sides by ρ2x to get

the claim for Step 3.

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Step 4: We are almost done. Next we realize that the proof in Step 3 can be easily adapted to

also state


1

λ· sup

Ω|u|+ 2 · 3α+1λα · ρα+1

x · [u]C1,α(Ω). (9.14)

Now λ ≤ 12 has not been fixed in any of these calculations. So for given ε > 0 let us

pick ε = 223α+2λα in (9.13) to get

ρ2x · sup

Ω(|Di ju|) ≤

22/α31+2/α

ε1/α· ρx · sup

BR(x)|Diu|+ ε · ρα+2

x · [u]C2,α(Ω).

For (9.13) we choose λ so that ε = 2232λ:


2232

εsup

Ω|u|+ ε · ρ2

x · supy∈BR(x)

|Di ju|.

Combining these last two equations we get

ρ2x · sup

Ω|Di ju| ≤

22+2/α33+2/α

ε1+1/αsup

Ω|u|

+22/α31+2/α

ε1/α−1· ρ2x supBR(x)

|Di ju(y)|+ ε · ρα+2x · [u]C2,α(Ω).

We can immediately algebraically deduce that

ρ2x · sup

Ω|Di ju| ≤

ε−222+2/α33+2/α

ε1/α−1 − 22/α31+2/αsup

Ω|u|

+ε1/α

ε1/α−1 − 22/α31+2/α· ρα+2x [u]C2,α(Ω).

Picking

ε > 22

1−α3α+21−α

guarantees everything stays positive. Taking this result with (9.14) we can apply an

induction argument to ascertain the conclusion.

Theorem 9.4.15 (): Given ∂Ω ∈ Ck,α and u ∈ Ck,α(Ω), there exists 0 ≤ C1 ≤ C2 < ∞such that

C1 · [u]Ck,α(Ω) ≤ [u]L∞,αk (Ω) ≤ C2 · [u]Ck,α(Ω),

i.e. Ck,α(Ω) ∼= L∞,αk (Ω).

Proof: First, we claim that

||u − Tx,ku||L∞(Ω(x,ρ)) ≤ C2 · ρk+α · [u]Ck,α(Ω),

for all x ∈ Ω. So we first fix y ∈ Ω(x, ρ) to find

u(y)− Tx,ku(y) =∑

|β|=k

Dβu(z)−Dβu(x)

β!· (z − x)β, (9.15)


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9.5 The Spaces of Bounded Mean Oscillation233

where z := tx + (1− t)y for some t ∈ (0, 1).

It needs to be noted that in order for the last equation to be valid, we may need to ex-

tend u to the convex hull of Ω(x, ρ). This is why we require the boundary regularity, to

guarantee that such a Holder continuous extension of u exists. We do not prove this here,

but the idea of the proof is similar to that of extending Sobolev functions.

Continuing on, given our definition of z , we readily see |z − x | ≤ |y − x | ≤ ρ. This combined

with (9.15) immediately indicates that

|u(y)− Tx,ku(y)| ≤ C2 · ρk+α · [u]Ck,α(Ω),

proving the RHS of (9.15).

For the other inequality, we first notice that ∀T ∈ Pk , [u − T ]Ck,α(Ω) = [u]Ck,α(Ω). We

know this since DβT = const. when |β| = k . Now, from lemma 9.4.14 we have

ρk · supΩ(x,ρ)

|Dβ(u − T )| ≤ C1

(ρα+k · [u]Ck,α(Ω) + sup

Ω(x,ρ)|Dβ(u − T )|

).

Taking the infimum of T ∈ Pk and dividing by ρk+α gives us

ρ−α supx,y∈Ω(x,ρ)

|Dβu(x)−Dβu(y)| ≤ C1

([u]Ck,α(Ω) + [u]L∞,αk (Ω)

),

proving the left inequality, concluding the proof.

9.5 The Spaces of Bounded Mean Oscillation

Note: These are also know as BMO Spaces or John-Nirenberg Spaces. Let Q0 be a cube in

Rn and u a measurable function on Q0. We define

u ∈ BMO(Q0)

if and only if

|u|BMO(Q0)

:= supQ

∫

Q∩Q0

−∫

Q∩Q0

|u − uQ∩Q0| dx <∞.

It is easy to see that we can also take as definition

|u|BMO(Q0)

= supQ⊂Q0

−∫

Q

|u − uQ| dx

and that u ∈ BMO(Q0), if and only if u ∈ L1,n(Q0).

Theorem 9.5.1 (John/Nirenberg): There exist constants C1 > 0 and C2 > 0 depending

only on n, such that the following holds: If u ∈ BMO(Q0), then for all Q ⊂ Q0

∣∣x ∈ Q0 | (u − uQ)(x) > t∣∣ ≤ C1 · exp

(−C2

|u|BMO(Q0)

)|Q|.

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Proof: It is sufficient to give the proof forQ = Q0 (because u ∈ BMO(Q0)Rightar rowu ∈BMO(Q) for Q ⊂ Q0). Moreover the inequality is homogenous and so we can assume that

|u|BMO(Q0)

= 1.

For α > 1 = supQ⊂Q0−∫Q |u − uQ| dx we use Calderon-Zygmund argument to get a

sequence Q(1)j with

i.) α < −∫Q

(1)j

|u − uQ0| dx ≤ 2nα

ii.) |u − uQ0| ≤ α on Q0 \

⋃j Q

(1)j .

Then we also have∣∣∣∣u

Q(1)j

− uQ0

∣∣∣∣ ≤ −∫

Q(1)j

|u − uQ0| dx ≤ 2nα

and

∑

j

|Q(1)j | ≤

1

α

∫

Q0

|u − uQ0| dx.

Take one of the Q(1)j . Then as |u|

BMO(Q0)= 1 we have

−∫

Q(1)j

∣∣∣∣u − uQ

(1)j

∣∣∣∣ dx ≤ 1 < α

and we can apply Calderon-Zygmund argument again. Thus, there exists a sequence Q(2)j

with

|u(x)− uQ0| ≤

∣∣∣∣u(x)− uQ

(1)j

∣∣∣∣+ |uQ

(1)j

− uQ0| ≤ 2 · 2nα

for a.e. x on Q0 \⋃Q

(2)k and

∑

j

|Q(2)j | ≤

1

α

∑

j

∫

Q(1)j

∣∣∣∣u − uQ

(1)j

∣∣∣∣ dx

≤ 1

α

∑

j

|Q(1)j | (u ∈ BMO and |u|

BMO= 1)

≤ 1

α2

∫

Q0

|u − uQ0| dx

≤ 1

α2|Q0|.

Repeating this argument one gets by induction: For all k ≥ 1 there exists sequence Q(k)j

such that

|u − uQ0| ≤ k2nα a.e. on Q0 \

⋃

j

Q(k)j

and

∑

j

|Q(k)j | ≤

1

αk

∫

Q0

|u − uQ0| dx.


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9.5 The Spaces of Bounded Mean Oscillation235

Now for t ∈ R+ either 0 < t < 2nα or there exists k ≥ 1 such that 2nαk < t < 2nα(k + 1).

In the first case

|x ∈ Q0 | |u(x)− uQ0| > t ≤ |Q0| ≤ |Q0|e−Ate2nAα

(o ≤ 2nα− tRightar row1 ≤ eA(2nα−t) for some A > 0); in the secont case

|x ∈ Q0 | |u(x)− uQ0| > t

≤ |x ∈ Q0 | |u(x)− uQ0| > 2nαk

≤∑

j

|Q(k)j | ≤

1

αk

∫

Q0

|u − uQ0| dx

≤ e(1− t2nα) lnα

∫

Q0

|u − uQ0|u − u

Q0| dx ≤ αe − lnα

2nα |Q0|

=: αe−At |Q0| (we need − k ≤ 1− t

2nαand α−k = e−k lnα).

Corollary 9.5.2 (): If u ∈ BMO(Q0), then u ∈ Lp(Q0) for all p > 1. Moreover

supQ⊂Q0

(∫

Q

|u − uQ|p dx

)1/p

≤ C(n, p) · |u|BMO(Q0)

.

Proof:∫

Q

|u − uQ|p dx = p

∫ ∞

0

tp−1|x ∈ Q | |u − uQ| > t| dt

≤ p · C1

∫ ∞

0

tp−1 exp

(− C2

|u|BMO(Q)

t

)|Q| dt

= p · C1

(C2

|u|BMO(Q)

)−p|Q|∫ ∞

0

e−ttp−1 dx ≤ C(n, p)|u|pBMO(Q)

.

Finally we have

Theorem 9.5.3 (Characterization of BMO): The following statements are equivalent:

i.) u ∈ BMO(Q0)

ii.) there exists constatns C1, C2 such that for all Q ⊂ Q0 and all s > 0

|x ∈ Q | |u(x)− uQ| > s| ≤ C1 exp(−C2s)|Q|

iii.) there exist constants C3, C4 such that for all Q ⊂ Q0,

∫

Q

(exp(C4|u − uQ |)− 1) dx ≤ C3|Q|

iv.) if v = exp(C4u), then

supQ⊂Q0

(−∫

Q

v dx

)(−∫

Q

1

vdx

)≤ C5.

(We can take C1 = C3 = C5 = C(n) and C2 = 2C4 = C(n)|u|−1BMO(Q0)

.)

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9.6 Sobolev Spaces

Take Ω ⊆ Rn

Definition 9.6.1: f ∈ L1(Ω) is said to be weakly differentiable if

∃g1, . . . , gn ∈ L1(Ω) such that

∫

Ω

f∂φ

∂xidx = −

∫

Ω

giφ dx ∀φ ∈ C∞c (Ω)

Notation: Df = (g1, . . . , gn), gi = Di f = ∂f∂xi

. If f ∈ C1(Ω), then Df = ∇f .

Definition 9.6.2: The Sobolev space W 1,p(Ω) is

W 1,p(Ω) = f ∈ Lp(Ω), Df exists, Di f ∈ Lp(Ω), i = 1, . . . , n

Proposition 9.6.3 (): The space W 1,p(Ω) is a Banach space with the norm

||f ||1,p =

(∫

Ω

|f |p + |Df |p dx) 1

p

1 ≤ p <∞,

where

|Df | =

(n∑

i=1

|Di f |2) 1

2

.

In the p =∞ case, we define

||f ||1,∞ = ||f ||∞ + ||Df ||∞

Proof: The triangle inequality follows from Minkowski’s inequality.

Completeness of W 1,p(Ω): Take fj Cauchy in W 1,p(Ω) =⇒ fj and Di fj are both Cauchy

in Lp(Ω). By the completeness of Lp, fj → f ∈ Lp(Ω) and Di fj → gi ∈ Lp(Ω).

Need: Df = g,

Call Di fj = (gj)i → gi ∈ Lp(Ω). By definition

∫

Ω

fj∂φ

∂xidx = −

∫

Ω

(gj)iφ dx ∀φ ∈ C∞c (Ω)

Cauchy ↓ ↓∫

Ω

f∂φ

∂xidx = −

∫

Ω

giφ dx

Remark: W 1,p(Ω) is a Hilbert space with

(u, v) =

∫

Ω

(uv +Du ·Dv) dx

Thus, we can use the Riesz-Representation Theorem and Lax-Milgram in this space.

Examples:Gregory T. von Nessi

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9.6 Sobolev Spaces237

|x|

−1 0 1 0

1

• f (x) = |x | on (−1, 1)

We will prove f ∈ W 1,1(−1, 1) and

g = Df =

−1 x ∈ (−1, 0)

1 x ∈ (0, 1)

To show this, look at∫ 1

−1

f φ′ dx = −∫ 1

−1

gφ dx = −[∫ 0

−1

φ dx +

∫ 1

0

φ dx

]

=

∫ 0

−1

φ dx −∫ 1

0

φ dx

But we also have∫ 1

−1

f φ′ dx = −∫ 0

−1

xφ′ dx +

∫ 1

0

xφ′ dx

=

∫ 0

−1

φ dx + 0−∫ 1

0

φ dx + 0√

Thus, g is the weak derivative of f .

•

f (x) = sgn(x) =

−1 x ∈ (−1, 0)

1 x ∈ (0, 1)

−1

0

1

−1 0 1

Our intuition says that f ′ = 2δ0 /∈ L2(−1, 1). Thus, we see to prove f is not weakly

integrable. So, suppose otherwise, g ∈ L1(−1, 1), Df = g.

∫ 1

−1

gφ dx = −∫ 1

−1

f φ′ dx =

∫ 0

−1

φ′ dx −∫ 1

0

φ′ dx

= φ(0)− φ(−1)− [φ(1)− φ(0)]

= 2φ(0)

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0

1

−1 0 1

Now choose φi such that φ ≥ 0, φ ≤ 1, φj(0) = 1, and φj(x)→ 0 a.e.

With this choice of φj we have

2φj(0) =

∫ 1

−1

gφj dx

=⇒ 2 = limj→∞

2φj(0) = limj→∞

∫ 1

−1

gφj dx

= 0 contradiction

Where the last equality is a result of Lebesgue’s dominated convergence theorem.

For higher derivatives, α is a multi-index. g = Dαf if

∫

Ω

f Dαφ dx = (−1)|α|∫

Ω

gφ dx ∀φ ∈ C∞c (Ω)

Definition 9.6.4:

W k,p(Ω) = f ∈ Lp(Ω), Dαf exists, Dαf ∈ Lp(Ω), |α| ≤ k

Sobolev functions are like classical functions. We will address the following in the next

few lectures

• Approximation

• Calculus. Specifically product and chain rules.

• Boundary values

• Embedding Theorems. In the HW we showed that

||u||C0,α(R) ≤ |u(x0)|+ C||u′||Lp(R)

which gives us the intuition that

W 1,p(R) ⊂−→ C0,α(R).

Recall: W 1,p(Ω) means we have weak derivatives in Lp(Ω), g = Dαf

∫

Ω

f Dαφ dx = (−1)|α|∫

Ω

gφ dx ∀φ ∈ C∞c (Ω)

• n = 1 W 1,p ⊂ C0,α with α = 1− 1p .


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• n ≥ 2 W n,p not holder continuous.

Lemma 9.6.5 (): f ∈ L1loc(Ω), α is a multi-index, and φε(x) = ε−nφ(ε−1x) with usual

mollification.

If Dαf exists, then Dα(φε ∗ f )(x) = (φε ∗Dαf )(x) ∀x with dist(x, ∂Ω) > ε.

Proof: Without loss of generality take |α| = 1, Dα = ∂∂xi

.

∂

∂xi(φε ∗ f )(x) =

∂

∂xi

∫


=

∫

Rn

∂

∂xiφε(x − y)f (y) dy

= −∫

Rn

∂

∂yiφε(x − y)f (y) dy

——

Now since φ(x − y) ∈ C∞c (Ω), we apply the definition of the weak partial derivative to get

——

=

∫

Rnφε(x − y)

∂

∂yif (y) dy

=

(φε ∗

∂

∂xif

)

Proposition 9.6.6 (): f , g ∈ L1loc(Ω). Then g = Dαf ⇐⇒ ∃fm ∈ C∞(Ω) such that fm → f

in L1loc(Ω) and Dαfm → g in L1

loc(Ω).

Proof:

(⇐=) fm → f and Dαfm → g∫

Ω

fm ·Dαφ dx = (−1)|α|∫

Ω

Dαfm · φ dx

(see following aside) ↓ ↓∫

Ω

f ·Dαφ dx = (−1)|α|∫

Ω

gφ dx

∣∣∣∣∫

Ω

fm ·Dαφ dx −∫

Ω

f ·Dαφ dx∣∣∣∣ ≤

∫

Ω

|fm − f | |Dαφ|︸︷︷︸C

dx

≤ C

∫

supp(φ)

|fm − f | dx → 0 given

(=⇒) Use convolution and lemma 8.37.

9.6.1 Calculus with Sobolev Functions

Proposition 9.6.7 (Product Rule): f ∈ W 1,p(Ω), g ∈ W 1,p′(Ω), then f g ∈ W 1,1(Ω) and

D(f g) = Df · g + f ·Dg.

Proof: By Holder, f g, Df · g, and f ·Dg are in L1(Ω). Explicitly, this is∫

Ω

|f g| dx ≤ ||f ||p||g||p′

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Now suppose p < ∞ (by symmetry) and fε is the convolution of f defined by fε = φε ∗ f .

fε → f in L1loc(Ω). Let Ω′ =supp(φ), ε <dist(Ω′, ∂Ω′).

∫

Ω

fεg ·Diφ dx =

∫

Ω

g ·Di(fεφ) dx −∫

Ω

g ·Di fε · φ dx

(def. of weak deriv. of g) = −∫

Ω

Dig · fεφ dx −∫

Ω

g ·Di fε · φ dx

=⇒∫

Ω

fεg ·Diφ dx = −∫

Ω

(Di fε · g + fε ·Dig)φ dx

Now fε → f in L1loc and by Lemma 8.37 Di fε → Di f in L1

loc as ε→ 0. Thus,∫

Ω

f g ·Diφ dx = −∫

Ω

(Di f · g + f ·Dig)φ dx

Now by applying the definition of weak derivatives, we get

Di(f g) = Di f · g + f ·Dig

Proposition 9.6.8 (Chain Rule): f ∈ C1, f bounded and g ∈ W 1,p, then (f g) ∈ W 1,p(Ω)

and

D(f g) = (f ′ g)Dg

Remark: The same assertion holds if f is continuous and piecewise C1 with bounded

derivative.

Proof: gε = φε ∗ g, gε → g in Lploc, Dgε → Dg in Lploc. (Assume p < ∞, even p = 1

would be sufficient).

D(f gε) = (f ′ gε)Dgε → (f ′ g)Dg in L1loc

Take Ω′ ⊂⊂ Ω, ε small enough.∫

Ω

|(f ′ gε)Dgε − (f ′ g)Dg| dx

≤∫

Ω

|(f ′ gε)(Dgε −Dg)| dx +

∫

Ω

|(f ′ g − f ′ gε)Dg| dx

Now f ′ gε is bounded since f ′ is bounded and Dgε −Dg → 0 in L1loc. Thus the first term

on the RHS goes to 0 as ε→ 0. Since gε → g in L1loc, gε → g in measure and thus exists a

subsequence such that gε → g a.e. Thus we can apply Lebesgue’s dominated convergence

theorem to see the second term in the RHS goes to 0 as ε→ 0. Thus we have∫

Ω

|(f ′ gε)Dgε − (f ′ g)Dg| dx = 0

and by the application of the definition of a weak derivative, we attain the result.

Lemma 9.6.9 (): Let f + = maxf , 0, f − = minf , 0, f = f + + f −, if f ∈ W 1,p, then f +

and f − are in W 1,p,

Df + =

Df on f > 0

0 on f ≤ 0

D|f | =

Df on f > 0

−Df on f < 0

0 f = 0


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Corollary 9.6.10 (): Let c ∈ R, E = x ∈ Ω, f (x) = c. Then Df = 0 a.e. in E.

Theorem 9.6.11 (): 1 ≤ p <∞, then W k,p(Ω) ∩ C∞(Ω) is dense in W k,p(Ω).

Proof: Choose Ω1 ⊂⊂ Ω2 ⊂⊂ Ω2 ⊂⊂ · · · ⊂⊂ Ω, e.g. Ωi =x ∈ Ω : dist(x, ∂Ω) < 1

i

.

Define the following open sets: Ui = Ωi+1 \Ωi−1

U2

Ω

· · ·· · ·

...

Ω1

Ω2

Ω3

U2

Then⋃i Ui = Ω, and thus Ui is an open cover of Ω. This cover is said to be locally finite,

i.e., ∀x ∈ Ω there exists ε > 0, such that B(x, ε) ∩ Ui 6= 0 only for finitely many i .

Theorem 9.6.12 (Partition of Unity): Ui is an open and locally finite cover of Ω. Then

there exists ηi ∈ C∞c (Ui), ηi ≥ 0, and

∞∑

i=1

ηi = 1 in Ω

Intuitive Picture in 1D:

1

Proof of 8.43 cont.:

ηi is a partition of unity corresponding to Ui , fi = ηi f ∈ W k,p(Ω).

supp(fi) ⊂ Ui = Ωi+1 \Ωi−1.

Choose εi > 0, such that φεi ∗ fi =: gi satisfies

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• supp(gi) ⊂ Ωi+2 \Ωi−2

• ||gi − fi ||W k,p(Ω) ≤ δ2i

, δ > 0 fixed.

• gi ∈ C∞c (Ω)

Now define

g(x) =

∞∑

i=1

gi(x)

(this sum is well-defined since gi(x) 6= 0 only for finitely many i).

||g − f ||W k,p(Ω) =

∣∣∣∣∣

∣∣∣∣∣∞∑

i=1

gi −∞∑

i=1

ηi f

∣∣∣∣∣

∣∣∣∣∣W k,p(Ω)

≤∞∑

i=1

||gi − ηi f ||W k,p(Ω)

≤∞∑

i=1

δ

2i= δ

Remark: The radius of the ball on which you average to get g(x) is proportional to dist(x, ∂Ω).

Theorem 9.6.13 (Fundamental Theorem of Calculus): i.) g ∈ L1(a, b), f (x) =

∫ x

a

g(t) dt

then f ∈ W 1,1(a, b) and Df = g.

ii.) If f ∈ W 1,1(a, b) and g = Df , then f (x) = C +

∫ x

a

g(t) dt for almost all x ∈ (a, b),

where C is a constant.

Remark: In particular, there exists a representative of f that is continuous.

Proof of ii.): Choose fε ∈ W 1,1(a, b)∩C∞(a, b) such that fε → f in W 1,1(Ω), i.e., fε → f in

L1(a, b) and Dfε → Df = g in L1(a, b). There exists a subsequence such that fε(x)→ f (x)

a.e.

Fundemental theorem for fε

fε(x)− fε(a) =

∫ x

a

f ′ε (t) dt

a.e. ↓ ↓ ↓

f (x)− L =

∫ x

a

g(t) dt

where L is finite and f (a) is chosen such that L = limx→a =: f (a) (Only changes f on a set

of measure zero).

9.6.2 Boundary Values, Boundary Integrals, Traces

Definition 9.6.14: 1 ≤ p <∞

1. W 1,p0 (Ω) = closure of C∞c (Ω) in W 1,p(Ω)


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2. p =∞: W 1,∞0 (Ω) is given by

W 1,∞0 (Ω) =

f ∈ W 1,∞(Ω) : ∃fi ∈ C∞c (Ω) s.t.

fj → f in W 1,1loc (Ω), sup ||fj ||W 1,∞(Ω) ≤ C <∞

3. If f0 ∈ W 1,p(Ω). Then f = f0 on ∂Ω if

f − f0 ∈ W 1,p0 (Ω)

Remark: It’s not always possible to define boundary values.

112

Γ

Ω ⊆ R2, Ω = B(0, 1) \ B(

0, 12

)\ Γ , where Γ =

x = (x1, 0), 1

2 < x1 < 1

. u(x1, x2) = φ =

polar angle. Ω not “on one side” of Γ .

Definition 9.6.15: Lipschitz boundary

f is Lipschitz continuous if f ∈ C0,1, i.e.

|f (x)− f (y)| ≤ L|x − y |,

for some constant L. ∂Ω is said to be Lipschitz, if ∃Ui , i = 1, . . . , m open such that

∂Ω ⊂ ⋃mi=1 Ui , where in any Ui , we can find a local coordinate system such that ∂Ω ∩ Ui is

the graph of a Lipschitz function and Ω ∩ Ui lies on one side of the graph.

Remark: Analogous definition with Ck,α boundary. ∂Ω∪Ui = graph of a Ck,α functions.

The following figure should make clear the concept of graphs of boundaries.

Definition 9.6.16: f : ∂Ω → R. f is measurable/integrable if f (g(y)) measurable inte-

grable.

Boundary integral: Choose U0 such that Ω ⊂ ⋃mi=0 Ui . Now pick a partition of unity, ηicorresponding to these Ui such that

∑mi=1 ηi = 1 on Ω

∫

∂Ω

f dHn−1 =

∫

∂Ω

f dS =

m∑

i=1

∫

∂Ω

ηi f dS

To make it clear what the integral in the last equality means, assume f has support in one

Ui , then in that Ui with coordinates properly redefined,∫

∂Ω

f dS =

∫

Rn−1

f (y , g(y))√

1 + |Dg(y)|2 dy .

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∂Ω

y1

y1

y2y2

g1(y) = |y |

U1

U4

U3

U2

R

Rn−1

Ω

y = (y1, . . . , yn−1)

∂Ω = graph

g(y)

U0

Ω

Ui

Ui

So, it’s clear how to extend this to general f since ηi f does have support in Ui .

Remark:√

1 + |Dg(y)|2 dy can be defined for g Lipschitz (derivatives well defined up except

for a countable number of “kinks”). We will assume ∂Ω is C1.

Definition 9.6.17:

Lp(∂Ω) =

f : ∂Ω→ R,

∫

∂Ω

|f |p dS <∞

L∞(∂Ω) =

f : ∂Ω→ R, ess sup

∂Ω|f | <∞


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9.6.3 Extension of Sobolev Functions

Theorem 9.6.18 (Extension): Ω ⊆ Rn, bounded, ∂Ω Lipschitz. Suppose Ω′ open neigh-

borhood of Ω, i.e. Ω ⊂⊂ Ω′, then exists a bounded and linear operator

E : W 1,p(Ω)→ W 1,p0 (Ω′) such that Ef

∣∣∣Ω

= f

Ω′Ω

reflection

cut-off function

Outline of the Proof: Reflect and multiply by cut-off functions.

C1 boundary: ∃ transformation that maps ∂Ω locally to a hyperplane “straightening or

flattening the boundary”.

x0

x-coordinate

y0

Φ

Ψ

y-coordinate

R

x0

Ω

∂Ω

Rn−1

Define Φ by

yi = xi = Φi(x), i = 1, . . . , n − 1

yn = xn − g(x) = Φn(x), x = (x1, . . . , xn−1)

=⇒ det(DΦ(x)) = 1

∂Ω flat near x0, ∂Ω ⊂ xn = 0B(x0, R) such that B(x0, R) ∩ ∂Ω ⊂ xn = 0

B+ = B ∩ xn ≥ 0 ⊂ Ω

B− = B ∩ xn ≤ 0 ⊂ Rn \Ω

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1.

B−

x0

B+

Rn−1

Ω

xn

2. Approximate: suppose first that u ∈ C∞(Ω) extend u by higher order reflection:

u(x) =

u(x) if x ∈ B+ := u+

−3u(x ,−xn) + 4u(x ,− xn2

)if x ∈ B− := u−

3. Check this is C1 across the boundary.

• u continuous xn = 0• all tangential derivatives are continuous across xn = 0

∂u

∂xn(x) = 3

∂u

∂xn(x − xn)− 2

∂u

∂xn

(x ,−xn

2

)

∂u

∂xn

∣∣∣∣∣xn=0

=∂u

∂xn(x , 0)

4. Calculate

||u||Lp(B) ≤ C||u||Lp(B+)

||Du||Lp(B) ≤ C||Du||Lp(B+)

5. Straightening out of the boundary:

Ψy-coordinate

u∗(y) = u (Ψ(y))

y y0

B

Φ(B)

Φ

x-coordinate

xu

x0

W = Ψ(B)

Choose B(y0, R0) ⊂ Φ(B). Construction for u∗ (see above figure). Now extend u∗

such that

||u∗||W 1,p(B) ≤ C||u∗||W 1,p(B+)

= C||u∗||W 1,p(B+)Gregory T. von Nessi

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Change coordinates from y back to x to get an extension u on W that satisfies

||u||W 1,p(W ) ≤ C||u||W 1,p(W∩Ω)

≤ ||u||W 1,p(W )

Remark: The constant C depends (by the chain rule) on ||Φ||∞, ||Ψ||∞, ||DΦ||∞, and

||DΨ||∞ but not on u.

6. ∀x0 ∈ ∂Ω, W (x0), u extension, ∂Ω is compact. Thus cover ∂Ω by finitely many sets

W1, . . . ,Wm (with corresponding ui extensions), choose W0 such that Ω ⊆ ⋃mi=0Wi ,

u0 = u and ηi the corresponding partition of unity.

u(x) =

m∑

i=0

ηiui

Making W (x0) smaller if necessary, we may assume that W (x0) ⊂ Ω′, thus u = 0

outside Ω′. By Minkowski

||u||W 1,p(Ω′) ≤ C||u||W 1,p(Ω)

7. Define Eu = u, bounded and linear on C∞(Ω)

8. Approximation: Refinement of approximation result.

Find um ∈ C∞(Ω) such that um → u in W 1,p(Ω)

||um − un||W 1,p(Ω′) = ||Eum − Eun||W 1,p(Ω′)

≤ C||um − un||W 1,p(Ω) → 0

=⇒ Eum is Cauchy in W 1,p(Ω′), and thus has a limit in this space, call it Eu.

9.6.4 Traces

Theorem 9.6.19 (): Ω bounded domain in Rn, Lipschitz boundary. 1 ≤ p <∞. Then exists

a unique linear bounded operator

T : W 1,p(Ω)→ Lp(∂Ω)

such that

Tu = u∣∣∣∂Ω

if u ∈ W 1,p(Ω) ∩ C0(Ω)

Proof: u ∈ C1(Ω), x0 ∈ ∂Ω, ∂Ω flat near x0. ∃B(x0, R) such that ∂Ω ∩ B(x0, R) ⊆xn = 0. Define B = B

(x0,

R2

)and let Γ = ∂Ω ∩ B.

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∂Ωx0

R2

R

Γ


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x = (x1, . . . , xn−1) ∈ xn = 0Choose ξ ∈ C∞c (B(x0, R)), ξ ≥ 0, ξ ≡ 1 on B

B+

xn

R

· · · · · ·

x x0

(x , x∗n )

x ∈ Rn−1

Note: From the above figure we see

ξ|u|p(x , 0) = (ξ|u|p)(x , 0)− (ξ|u|p)(x , x∗n )

=

∫ 0

x∗n

(ξ|u|p)xn dxn (9.16)

So we have for the main calculation:∫

Γ

|u(x , 0)|p dx ≤∫

Rn−1

ξ|u|p dx

= −∫

B+

(ξ|u|p)xn dx

= −∫

B+

|u|pξxn + ξ|u|p−1

︸︷︷︸Lp′

sgn(u) uxn︸︷︷︸Lp

dx

Holder≤ −∫

B+

|u|pξxn dx + C

(∫

B+

|u|p dx) 1

p′(∫

B+

|uxn |p dx) 1

p

Young’s≤ C

∫

B+

(|u|p + |Du|p) dx

General situation: non-linear change of coordinates just as before∫

Γ

|u|p dS ≤ C

∫

C

(|u|p + |Du|p) dx

where Γ ⊂ ∂Ω contains the points x0.

Now, ∂Ω compact, choose finitely many open sets Γi ⊂ ∂Ω such that ∂Ω ⊂ ⋃mi=1 Γi with∫

Γi

|u|p dS ≤ C

∫

C

(|u|p + |Du|p) dx

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=⇒ If we define for u ∈ C1(Ω), Tu = u∣∣∣∂Ω

=⇒ ||u||Lp(∂Ω) ≤ C||u||W 1,p(Ω)

Approximation: u ∈ W 1,p(Ω) choose uk ∈ C∞(Ω) such that

uk → u in W 1,p(Ω)

Estimate

||Tum − Tun||Lp(Ω) ≤ C||um − un||W 1,p(Ω)

=⇒ Tum is a Cauchy sequence in Lp(∂Ω).

Define

Tu = limk→∞

Tuk in Lp(∂Ω)

If u ∈ W 1,p(Ω) ∩ C0(Ω) then the approximation uk converges uniformly on Ω to u and

Tu = u∣∣∣∂Ω

.

Remark: W 1,∞(Ω) is space of all functions that have Lipschitz continuous representations.

Theorem 9.6.20 (Trace-zero functions in W 1,p): Assume Ω is bounded and ∂Ω is C1.

Suppose furthermore that u ∈ W 1,p(Ω). Then

u ∈ W 1,p0 (Ω) ⇐⇒ Tu = 0

Proof: (=⇒) We are given u ∈ W 1,p0 (Ω). Then by the definition, ∃um ∈ C∞c (Ω) such

that

um → u in W 1,p(Ω).

clearly Tum = 0 on ∂Ω. Thus,

sup∂Ω|Tu − Tum| ≤ C sup

∂Ω|u − um| = 0

Thus, Tu = limm→∞ Tum = 0.

(⇐=) Assume Tu = 0 on ∂Ω. As usual assume that ∂Ω is flat around a point x0 and

using standard partitions of unity we may assume for the following calculations that

u ∈ W 1,p(Rn+), supp(u) ⊂⊂ Rn+,T u = 0, on ∂Rn+ = Rn−1

Since Tu = 0 on Rn−1, there exists um ∈ C1(Rn+) such that

um → u in W 1,p(Rn+) (9.17)

and

Tum = um

∣∣∣Rn−1→ 0 in Lp(Rn−1) (9.18)


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Note: In deriving B, C is picked to bound |ξ′|p. Also the integral is restricted since ξ′ = 0 if

xn > 2/m by construction.

Now

limm→∞

A = 0

since ξm 6= 0 only if 0 ≤ xn ≤ 2/m. To estimate the term B, we use (9.19)

B ≤ C · 2p−1mp(

2

m

)p(∫ 2/m

0

∫

Rn−1

|Du|p dxdxn)

≤ C

∫ 2/m

0

∫

Rn−1

|Du|p dxdxn → 0 as m →∞

Thus, we deduce Dwm → Du in Lp(Rn+). Since clearly wm → u in Lp(R+n ) (is clear from

construction of wm). We can now conclude

wm → u in W 1,p(Rn+)

But wm = 0 if 0 < xn < 1/m. We can therefore mollify the wm to produce function

um ∈ C∞c (Rn+) such that um → u in W 1,p(Rn+). Hence w ∈ W 1,p0 (Rn+). .

9.6.5 Sobolev Inequalities

Theorem 9.6.21 (Sobolev-Gagliardo-Nirenberg Inequality): Ω is a bounded set in Rn. If

u ∈ W 1,1(Rn), then

||u||L

nn−1 (Rn)

≤ 1√n

∫

Rn|Du| dx.

Remark: In the above theorem means thatW 1,p0 (Ω) ⊆ Lp∗(Ω) and ||u||Lp∗(Ω) ≤ C||u||W 1,p

0 (Ω).

Proof: C∞c (Ω) dense in W 1,p0 (Ω).

First estimate for smooth functions, then pass to the limit in the estimate. It thus suffices

to prove

||f ||Lp∗(Ω) ≤ C||Df ||Lp(Ω) ∀f ∈ C∞c (Ω)

Special case: p = 1, p∗ = nn−1

n = 2: p∗ = 2

f (x) = f (x1, x2) =

∫ x1

−∞D1f (t, x2) dt

=⇒ |f (x)| ≤∫ x1

−∞|D1f (t, x2)| dt ≤

∫ ∞

−∞|D1f (t, x2)| dt = g1(x2)

Same idea

|f (x)| ≤∫ ∞

−∞|D2f | dx2 = g2(x1)

|f (x)|2 ≤ g1(x2)g2(x1)Gregory T. von Nessi

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||f ||22 =

∫

R2

|f (x)|2 dx =

∫

R2

g1(x2)g2(x1) dx1dx2

=

∫

Rg1(x2) dx2

∫

Rg2(x1) dx1

=

(∫

R

∫

R|D1f | dx1dx2

)(∫

R

∫

R|D2f | dx2dx1

)

≤(∫

R2

|Df | dx)2

=⇒ ||f ||L2(R2) ≤ ||Df ||L1(R2).

General Case:

|f (x)| ≤ gi(x) =

∫ ∞

−∞|Di f | dxi

We have

|f (x)| nn−1 ≤

(n∏

i=1

gi

) 1n−1

∫

R|f (x)| n

n−1 dx1 ≤∫

R

(n∏

i=1

gi

) 1n−1

dx1

Now taking Holder for n − 1 terms with pi = n − 1

[g1(x)]1n−1

∫

R

(n∏

i=2

gi

) 1n−1

dx1 ≤ [g1(x)]1n−1

n∏

i=2

(∫

Rgi(x) dx1

) 1n−1

integrate with respect to x2 to get

∫

R

∫

R|f (x)| n

n−1 dx1dx2

≤∫

R[g1(x)]

1n−1

n∏

i=2

(∫

Rgi(x) dx1

) 1n−1

dx2

=

(∫

Rg2(x) dx1

) 1n−1

[∫

R[g1(x)]

1n−1

n∏

i=3

(∫

Rgi(x) dx1

) 1n−1

dx2

]

Holder≤(∫

Rg2(x) dx1

) 1n−1(∫

Rg1(x) dx2

) 1n−1

n∏

i=3

(∫

R

∫

Rgi(x) dx1dx2

) 1n−1

Integrate∫dx3 · · ·

∫dxn + Holder

∫

Rn|f (x)| n

n−1 dx ≤n∏

i=1

(∫

Rn−1

gi(x) dx1 · · · dxi−1dxi+1 · · · dxn) 1

n−1

(def. of gi) =

n∏

i=1

(∫

Rn|Di f | dx

) 1n−1

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So this implies that

||f ||L

nn−1 (Rn)

≤n∏

i=1

(∫

Rn|Di f | dx

) 1n

=

∫

Rn|Df | dx = ||Df ||L1(Rn)

We can do a little better than this by observing

(n∏

i=1

ai

) 1n

≤ 1

n

n∑

i=1

ai

(which is just a generalized Young’s inequality). So we have

||f ||L

nn−1 (Rn)

≤ 1

n

∫

Rn

n∑

i=1

|Di f | dx ≤1√n

∫

Rn|Df | dx

Since

n∑

i=1

ai =√n

√√√√n∑

i=1

a2i

In the general case use: |f |γ with γ suitably chosen.

Theorem 9.6.22 (Poincare’s Inequality): Ω ⊆ Rn bounded, then

||u||Lp(Ω) ≤ C||Du||Lp(Ω) ∀v ∈ W 1,20 (Ω), 1 ≤ p ≤ ∞

If ∂Ω is Lipschitz, then

||u − u||Lp(Ω) ≤ C||Du||Lp(Ω) ∀v ∈ W 1,20 (Ω), 1 ≤ p ≤ ∞

where

u = uΩ =1

|Ω|

∫

Ω

u(x) dx = −∫

Ω

u(x) dx

Remark:

1. We need u − u to exclude constant functions

2. C depends on Ω, see HW for Ω = B(x0, R)

This will be an indirect proof with no explicit scaling given. Use additional scaling argument

to find scaling.

Proof: Sobolev-Gagliardo-Nirenberg:

||u||Lp∗(Ω) ≤ CS||Du||Lp(Ω)

Now we apply Holder under the assumption that Ω is bounded:

||u||Lp(Ω) ≤ C(Ω)||u||Lp∗(Ω) ≤ C(Ω)CS||Du||Lp(Ω)Gregory T. von Nessi

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Proof of ||u − u||Lp(Ω) ≤ C||Du||Lp(Ω). Without loss of generality take u = 0, other wise

apply inequality to v = u − u. Now we prove

Dv = Du.

Suppose otherwise ∃uj ∈ W 1,p(Ω), uj = 0, such that

||uj ||Lp(Ω) ≥ j ||Duj ||Lp(Ω)

May suppose, ||uj ||Lp(Ω) = 1

=⇒ 1

j≥ ||Duj ||Lp(Ω) =⇒ Duj → 0 in Lp(Ω)

and

||uj ||W 1,p(Ω) = ||uj ||Lp(Ω) + ||Duj ||Lp(Ω) ≤ 1 +1

j≤ 2

Compact Sobolev embedding (See Section 9.7): ∃ subsequence such that ujk ∈ u in Lp(Ω)

with Du = 0 =⇒ u is constant, 0 =∫

Ω uj dx →∫

Ω u dx =⇒ u = 0 =⇒ 1 = ||ujk ||W 1,p(Ω) →0. Contradiction.

9.6.6 The Dual Space of H10 = W 1,2

0

Definition 9.6.23: Ω ⊂ Rn bounded, H−1(Ω) = (H10)∗ = (W 1,2

0 )∗

Theorem 9.6.24 (): Suppose that T ∈ H−1(Ω). Then there exists functions f0, f1, . . . , fn ∈L2(Ω) such that

T (v) =

∫

Ω

(f0v +

n∑

i=1

fiDiv

)dx ∀v = H1

0(Ω) (9.20)

Moreover,

||T ||2H−1(Ω) = inf

∫

Ω

n∑

i=0

|fi |2 dx, fi ∈ L2(Ω), (9.20) holds

Remark: One frequently works formally with, T = f0 −∑ni=1Di fi

Proof: u, v ∈ H10(Ω),

(u, v) =

∫

Ω

(uv +Du ·Dv) dx

if T ∈ H−1(Ω), then by Riesz, ∃!u0 such that

T (u) = (v , u0) ∀v ∈ H10(Ω)

=

∫

Ω

(vu0 +Dv ·Du0) dx (9.21)

i.e. (9.20) with f0 = u0 and fi = Diu0.

Suppose g0, g1, . . . , gn satisfy

T (v) =

∫

Ω

(vg0 +Div · gi) dx

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take v = u0 in (9.21)

||u0||2H10(Ω)

= (u0, u0) = T (u0) =

∫

Ω

(u0g0 +

n∑

i=1

Diu0gi

)dx

Holder≤ ||u0||L2(Ω)||g0||L2(Ω) +

n∑

i=1

||Diu0||L2(Ω)||gi ||L2(Ω)

C-S≤ ||u0||H10(Ω)

(∫

Ω

|g0|2 + |g1|2 + · · ·+ |gn|2)dx

(9.20) =⇒∫

Ω

(u0v +Diu0 ·Div) dx ≤ ||v ||H10(Ω)||u0||H1

0(Ω)

Now, we see that

||T ||H−1(Ω) = supv∈H1

0(Ω)

|T (v)|||v ||H1

0(Ω)

≤ ||u0||H10(Ω)

On the other hand, choose v = u0||u0||H1

0(Ω)

T (v) = ||u0||H10(Ω) =⇒ ||T ||H−1(Ω) = ||u0||H1

0(Ω)

9.7 Sobolev Imbeddings

Theorem 9.7.1 (): Ω is a bounded set in Rn. Then the following embeddings are continuous

i.) 1 ≤ p < n, p∗ = npn−p , W 1,p

0 (Ω) ⊂−→ Lp∗(Ω)

ii.) p > n, α = 1− np , W 1,p

0 (Ω) ⊂−→ C0,α(Ω)

Remark:

1. i.) in the above theorem means that W 1,p0 (Ω) ⊆ Lp∗(Ω) and ||u||Lp∗(Ω) ≤ C||u||W 1,p

0 (Ω)

2. “Sobolev number”: Wm,p(Ω), Ω ⊆ Rn, then m − np is the Sobolev number.

Lq(Ω) = W 0,q(Ω), −nq is its Sobolev number.

Philosophy: W 1,p(Ω) ⊂−→ Lq(Ω) if 1 − np ≥ −nq , where maximal q corresponds to

equality.

1− np

= −nq⇐⇒ p − n

p= −n

q⇐⇒ n − p

p=n

q⇐⇒ q =

np

n − p = p∗

3. Analogously Ck,α(Ω) has a Sobolev number k + α

W 1,p(Ω) ⊂−→ C0,α(Ω) if 1− np ≥ 0 + α = α, optimal: α = 1− n

p .

4. More generally: Wm,p(Ω) ⊂−→ Lq(Ω) if m − np ≥ −nq and the optimal q satisfies

1q = 1

p − mn .

Wm,p(Ω) ⊂−→ C0,α(Ω) if m − np ≥ 0 + α = α


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9.7 Sobolev Imbeddings257

Proof: C∞c (Ω) dense in W 1,p0 (Ω).

First estimate for smooth functions, then pass to the limit in the estimate. It thus suffices

to prove

||f ||Lp∗(Ω) ≤ C||Df ||Lp(Ω) ∀f ∈ C∞c (Ω)

Special case: p = 1, p∗ = nn−1

n = 2: p∗ = 2

f (x) = f (x1, x2) =

∫ x1

−∞D1f (t, x2) dt

=⇒ |f (x)| ≤∫ x1

−∞|D1f (t, x2)| dt ≤

∫ ∞

−∞|D1f (t, x2)| dt = g1(x2)

Same idea

|f (x)| ≤∫ ∞

−∞|D2f | dx2 = g2(x1)

|f (x)|2 ≤ g1(x2)g2(x1)

||f ||22 =

∫

R2

|f (x)|2 dx =

∫

R2

g1(x2)g2(x1) dx1dx2

=

∫

Rg1(x2) dx2

∫

Rg2(x1) dx1

=

(∫

R

∫

R|D1f | dx1dx2

)(∫

R

∫

R|D2f | dx2dx1

)

≤(∫

R2

|Df | dx)2

=⇒ ||f ||L2(R2) ≤ ||Df ||L1(R2).

General Case:

|f (x)| ≤ gi(x) =

∫ ∞

−∞|Di f | dxi

We have

|f (x)| nn−1 ≤

(n∏

i=1

gi

) 1n−1

∫

R|f (x)| n

n−1 dx1 ≤∫

R

(n∏

i=1

gi

) 1n−1

dx1

Now taking Holder for n − 1 terms with pi = n − 1

[g1(x)]1n−1

∫

R

(n∏

i=2

gi

) 1n−1

dx1 ≤ [g1(x)]1n−1

n∏

i=2

(∫

Rgi(x) dx1

) 1n−1

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integrate with respect to x2 to get

∫

R

∫

R|f (x)| n

n−1 dx1dx2

≤∫

R[g1(x)]

1n−1

n∏

i=2

(∫

Rgi(x) dx1

) 1n−1

dx2

=

(∫

Rg2(x) dx1

) 1n−1

[∫

R[g1(x)]

1n−1

n∏

i=3

(∫

Rgi(x) dx1

) 1n−1

dx2

]

Holder≤(∫

Rg2(x) dx1

) 1n−1(∫

Rg1(x) dx2

) 1n−1

n∏

i=3

(∫

R

∫

Rgi(x) dx1dx2

) 1n−1

Integrate∫dx3 · · ·

∫dxn + Holder

∫

Rn|f (x)| n

n−1 dx ≤n∏

i=1

(∫

Rn−1

gi(x) dx1 · · · dxi−1dxi+1 · · · dxn) 1

n−1

(def. of gi) =

n∏

i=1

(∫

Rn|Di f | dx

) 1n−1

So this implies that

||f ||L

nn−1 (Rn)

≤n∏

i=1

(∫

Rn|Di f | dx

) 1n

=

∫

Rn|Df | dx = ||Df ||L1(Rn)

We can do a little better than this by observing

(n∏

i=1

ai

) 1n

≤ 1

n

n∑

i=1

ai

(which is just a generalized Young’s inequality). So we have

||f ||L

nn−1 (Rn)

≤ 1

n

∫

Rn

n∑

i=1

|Di f | dx ≤1√n

∫

Rn|Du| dx

Since

n∑

i=1

ai =√n

√√√√n∑

i=1

a2i

In the general case use: |f |γ with γ suitably chosen. This result is referred to as the

Sobolev-Gagliardo-Nirenberg inequality.

Recall: Sobolev-Gagliardo-Nirenberg:

||u||L

nn−1 (Rn)

≤ 1√n||Du||L1(Rn) u ∈ W 1,1

0 (Rn)


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This was for p = 1, now consider

General Case: 1 ≤ p < n, Use p = 1 for |u|γ

∣∣∣∣ |u|γ∣∣∣∣L

nn−1 (Rn)

≤ 1√n

∣∣∣∣ D|u|γ∣∣∣∣L1(Rn)

≤ 1√n

∣∣∣∣ γ|u|γ−1D|u|∣∣∣∣L1(Rn)

Holder ≤ γ√n

∣∣∣∣ |u|γ−1∣∣∣∣Lp′(Rn)· ||Du||Lp(Rn)

Choose γ such that

γn

n − 1= (γ − 1)p′ = (γ − 1)

p

p − 1

⇐⇒ γ

(n

n − 1− p

p − 1

)=−pp − 1

⇐⇒ γnp − n − np + p

(n − 1)(p − 1)= − p

p − 1

⇐⇒ γp − n

(n − 1)(p − 1)=

p

p − 1

⇐⇒ γ =p(n − 1)

n − p > 0

(∫

Rn|u|p∗ dx

) n−1n− p−1

p

≤ (n − 1)p

n − p1√n||Du||Lp(Rn)

Performing some algebra on the exponent, we see

n − 1

n− p − 1

p=n − pnp

=1

p∗

=⇒ ||u||Lp∗(Rn) ≤(n − 1)p

n − p1√n||Du||Lp(Rn) = CS||Du||Lp(Rn)

Remark: The estimate “explodes” as p → n.

This is a proof for W 1,p0 (Ω)

General case with Lipschitz boundary. Extension theorem:

Ω ⊂⊂ Ω′ =⇒ extension u ∈ W 1,p0 (Ω′) such that ||u||W 1,p(Ω′) ≤ C||u||W 1,p(Ω).

Use estimate for u on Ω′

||u||Lp∗(Ω) ≤ ||u||Lp∗(Ω′) ≤ CS||u||W 1,p(Ω′) ≤ CSCE ||u||W 1,p(Ω)

9.7.1 Campanato Imbeddings

Theorem 9.7.2 (Morrey’s Theorem on the growth of the Dirichlet integral): Let u be

in H1,ploc (B1) and suppost that

∫

Bρ(x)

|Du|p dx ≤ ρn−p+pα ∀ρ < dist(x, B1)

then u ∈ C0,αloc (B1).

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Proof: We use Poincare’s inequality:

∫

Bρ(x0)

|u − ux0,ρ|p dx ≤ Cρp∫

Bρ(x0)

|Du|p dx ≤ C · ρn+pα

i.e. u ∈ Lp,λloc (B1) and therefore by 1) u ∈ C0,αloc (B1).

Theorem 9.7.3 (Global Version of (9.7.2)): If u ∈ H1,p(Ω), p > n, then u ∈ L1,n+(1−n/p)(Ω)

=⇒ u ∈ C0,1−n/p(Ω) via Companato’s Theorem.

Proof: For all ρ < ρ0 and a u ∈ H1,p(Ω) we have

∫

Ω(x0,ρ)

|u − ux0,ρ|p dx ≤ Cρp∫

Ω(x0,ρ)

|Du|p dx,

where C only depends on the geometry of Ω; then we fine

∫

Ω(x0,ρ)

|Du| dx ≤(∫

Ω(x0,ρ)

|Du|p dx)1/p

|Ω(x0, ρ)|1−1/p ≤ C · ||u||H1,p(Ω)ρn−n/p.

Again by Poincare’s inequality:∫

Ω(x0,ρ)

|u − ux0,ρ| dx ρ∫

Ω(x0,ρ)

|Du| dx ≤ Cρn+1−n/p||u||H1,p(Ω)

Corollary 9.7.4 (Morrey’s theorem): For p > n, the imbedding

W 1,p(Ω) ⊂−→ C0,1−n/p(Ω)

is a continous operator, i.e.

||u||C0,1−n/p(Ω) ≤ C||u||W 1,p(Ω)

Remark: So, from the above we see

Du ∈ Lp,λ(Ω) =⇒ u ∈ Lp,λ+p(Ω)

(Du ∈ L2,n−2+2α(Ω) =⇒ u ∈ L2,n+2α(Ω)).

Theorem 9.7.5 (Estimates for W 1,p, n < p ≤ ∞): Let Ω be a bounded open subset of

Rn with ∂Ω being C1. Assume n < p ≤ ∞, and u ∈ W 1,p(Ω). Then u has a version

u∗ ∈ C0,γ(Ω), for γ = 1− np , with the estimate

||u∗||C0,γ(Ω) ≤ C||u||W 1,p(Ω).

The constant C depends only on p, n and Ω.

Proof: Since ∂Ω is C1, there exists an extension Eu = u ∈ W 1,p(Rn) such that

u = u in Ω,

u has compact support, and

||u||W 1,p(Ω).

Since u has compact support, we know that there exists um ∈ C∞c (Rn) such that

um → u in W 1,p(Rn)Gregory T. von Nessi

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Now from the previous theorem, ||um − ul ||C0,1−n/p(Rn) ≤ C||um − ul ||W 1,p(Rn) for all l , m ≥ 1,

thus there exists a function u∗ ∈ C0,1−n/p(Rn) such that

um → u∗ C0,1−n/p(Rn)

So we see from the last two limits, u∗ = u a.e. on Ω; such that u∗ is a version of u. The

last theorem also implies ||um||C0,1−n/p(Rn) ≤ C||um||W 1,p(Rn) this combined with the last two

limits, we get

||u∗||C0,1−n/p(Rn) ≤ C||u||W 1,p(Rn)

9.7.2 Compactness of Sobolev/Morrey Imbeddings

1 ≤ p < n, W 1,p0 (Ω) ⊂−→ Lq(Ω) is compact if q < p∗ i.e. if uj is a bounded sequence in

W 1,p0 (Ω) then uj is precompact in Lq(Ω), i.e., contains a subsequence that converges in

Lq(Ω).

Proposition 9.7.6 (): X is a BS

i.) A ⊂ X is precompact ⇐⇒ ∀ε > 0 there exists finitely many balls B(xi , ε) such that

A ⊂ ⋃mεi=1B(xi , ε)

ii.) A is precompact ⇐⇒ every bounded sequence in A contains a converging subsequence

(with limit in X).

Theorem 9.7.7 (Arzela-Ascoli): S ⊆ Rn compact, A ⊆ C0(S) is precompact ⇐⇒ A is

bounded and equicontinuous.

Proof: (=⇒) ∀ε > 0 ∃B(f εi , ε) such that A ⊆ ⋃mεi=1B(f εi , ε)

||f ||∞ ≤ ||f − f εi ||+ ||f εi ||∞, f ∈ B(f εi , ε)

≤ ε+ maxi=1,...,mε

||f εi || <∞

|f (x)− f (y)| ≤ 2ε+ |f iε (x)− f iε (y)|≤ 2ε+ max

i=1,...,mε|f iε (x)− f iε (y)|

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so if |x − y | tends to zero =⇒ equicontinuity.

(⇐=) ∀ε > 0, ||f ||∞ ≤ C ∀f ∈ A

Choose ξi ∈ R, i = 1, . . . , k such that B(0, C) = [−C,C] ⊂ ⋃ki=1B(ξi , ε).

xj ∈ Rn, j = 1, . . . , m such that S ⊂ ⋃mi=1B(xj , ε). Basically we are covering the range and

domain.

Let π : 1, . . . , m → 1, . . . , kLet Aπ = f ∈ A : |f (xj)− ξπ(j)| < ε, j = 1, . . . , m

Choose fπ ∈ Aπ if Aπ 6= 0. If x ∈ S then x ∈ B(xj , ε) for some j and f ∈ Aπ.

Claim: A ⊂ ⋃π B(fπ, ε), i.e. the (finitely many) functions fπ define the centers of the

balls. To show this, we see that

|f (x)− fπ(x)| ≤ |f (x)− f (xj)|+ |f (xj)− ξπ(j)|︸︷︷︸≤ε

+ |ξπ(j) − fπ(xj)|︸︷︷︸≤ε

+|fπ(xj)− fπ(x)|

≤ 2ε+ sup|x−y |≤ε

supg∈A|g(x)− g(y)| := rε

rε → 0 as ε→ 0 since A is equicontinuous.

=⇒ f ∈ B(fπ, rε). Suppose you want to find a finite cover A ⊂ ⋃B(fπ, δ), δ > 0 then

choose ε small enough such that rε ≤ δ.

Remark: IfX, Y , and Z are Banach Spaces and we haveX ⊂−→ Y ⊂⊂−→ Z orX ⊂⊂−→ Y ⊂−→ Z,

then X ⊂⊂−→ Z.

Theorem 9.7.8 (Compactness of Sobolev Imbeddings): If Ω ⊂ Rn bounded, then the

following Imbedding is true

i.) 1 ≤ p < n, q < p∗ = npn−p

W 1,p0 (Ω) ⊂⊂−→ Lq(Ω)

Moreover, if ∂Ω is Lipschitz, then the same is true for W 1,p(Ω)

ii.) If Ω ⊂ Rn bounded and satisfies condition () with p > n, 1− np > α,

W 1,p0 (Ω) ⊂⊂−→ C0,α(Ω)

Remark: A special case of the above is W 1,20 (Ω) → L2(Ω); This is called Rellich’s The-

orem.

Proof:

i.) q = 1 fj bounded in W 1,p0 (Ω) need to prove that fj is precompact in L1.

Extend by zero to Rn. Without loss of generality, take fj to be smooth with fj ∈C∞c (Rn), (||fj − fj || ≤ ε).


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1. Aε = φε ∗ fj, where φε is the standard mollifier, φε(x) = ε−nφ(ε−1x).

Then Aε precompact in L1 ∀ε ≥ 0. We now show this.

Idea: Use Arzela-Ascoli

|φε ∗ fj |(x) =

∣∣∣∣∫


∣∣∣∣

≤ sup |φε|∫

Rn|f (y)| dy

= sup |φε| · ||fj ||L1(Rn)

≤ sup |φε| · ||fj ||Lp(Rn) < Cε <∞Equicontinuity follows if D(φε ∗ fi) uniformly bounded,

|Dφε ∗ fj |(x) ≤ sup |Dφε| · ||fj ||Lp(Rn) < Cε <∞=⇒ (φε ∗ fj) uniformly bounded hence equicontinuous.

=⇒ precompact.

Step 2: ||φε ∗ fj − fj ||L1(Rn) ≤ Cε ∀j2. (Proof of Step 2)

(φε ∗ fj − fj)(x) =

∫

Rn=B(x,ε)

φε(x − y)(fj(y)− fj(x)) dy

(sub. y = x − εz) =

∫

Rn=B(0,1)

φ(z)(fj(x − εz)− fj(x)︸︷︷︸∫ 10

ddtfj (x−tεz) dt

) dy

Integrate in x∫

Rn|φε ∗ fj − fj | dx

≤∫

Rn

∫

B(0,1)

φ(z)

∫ 1

0

|Dfj(x − εtz)| · |εz | dtdzdx

(Fubini) = ε

∫

B(0,1)

φ(z)

∫ 1

0

∫

Rn|Dfj(x − tεz)| · |z |︸︷︷︸

≤1

dx

︸︷︷︸=∫Rn |Dfj |(x) dx

dtdz

≤ ε

(∫

Rn|Dfj |(x) dx

)∫

B(0,1)

φ(z) dz ≤ ε||fj ||L1(Rn) ≤ ε||fj ||Lp(Rn)

≤ εC

=⇒ ||φε ∗ fj − fj ||L1(Rn) ≤ εC uniformly ∀j3. Suppose δ > 0 given, choose ε small enough such that εC < δ

2

Aε = φε ∗ fj is precompact in L1(Rn), for δ2 we find balls B

(gi ,

δ2

)such that

Aε ⊆mε⋃

i=1

B

(gi ,δ

2

)=⇒ fj ⊆

mδ⋃

i=1

B(gi , δ) =⇒ fj

precompact in L1(Rn).

Remark: Evans constructs a converging subsequence.

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4. general q < p∗.Trick is to use the “interpolation inequality”:

||g||Lq(Rn) ≤ ||g||θL1(Rn)||g||1−θLp∗

(Rn)

where 1q = θ

1 + 1−θp∗ . So if fj → f in L1(Rn) then

||fj − f ||Lq(Rn) ≤ ||fj − f ||θL1(Rn)︸︷︷︸→0

||fj − f ||1−θLp∗

(Rn)︸︷︷︸≤C

So, we just need to prove the interpolation inequality.

Idea: Use Holder with r = 1qθ with the corresponding conjugate index

f ′ =r

r − 1=

1qθ

1qθ − 1

=1

1− qθ

Thus

(1− θ)q

1− qθ = p∗

So,

∫gq dx =

∫gqθg(1−θ)q dx

Holder≤(∫

g dx

)qθ (∫gp∗dx

)1−qθ

Now take the qth root

||g||Lq(Rn) ≤ ||g||θL1(Rn)||g||(1−qθ)p∗/qLp∗

(Rn)

= ||g||θL1(Rn)||g||1−θLp∗

(Rn)

ii.) General Idea: We will prove that W 1,p(Ω) ⊂−→ C0,µ(Ω) ⊂⊂−→ C0,λ(Ω), where 0 < µ <

λ < 1− np and then we will use the remark following the statement of the Arzela-Ascolli

theorem to conclude W 1,p(Ω) ⊂⊂−→ C0,λ(Ω).

Since 0 < λ < 1− np , we know ∃µ with 0 < µ < λ < 1− n

p ; and by Morrey’s theorem

we know W 1,p(Ω) ⊂−→ C0,µ(Ω). We now only need to show C0,µ(Ω) ⊂⊂−→ C0,λ(Ω). It

is obvious that C0,µ(Ω) ⊂−→ C0,λ(Ω) is continuous, so our proof is reduced to showing

the compactness of this imbedding. We do this in two steps

Step 1. Claim C0,ν(Ω) ⊂⊂−→ C(Ω). Again, the continuity of the imbedding is obvious;

and compactness is the only thing needed to be proven. If A is bounded in

C0,ν(Ω), then we know ||φ||C0,ν(Ω) ≤ M for all φ ∈ A. This obviously implies that

|φ(x)| ≤ M in addition to |φ(x)− φ(y)| ≤ M|x − y |ν for all x, y ∈ Ω. Thus, A is

precompact via the Arzela-Ascoli theorem.Gregory T. von Nessi

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Step 2. Consider

|φ(x)− φ(y)||x − y |λ =

( |φ(x)− φ(y)||x − y |µ

)λ/µ|φ(x)− φ(y)|1−λ/µ

≤ M · |φ(x)− φ(y)|1−λ/µ

for all φ ∈ A for which A is bounded in C0,λ(Ω). We now use the compactness of

the imbedding in Step 1. by extracting a subsequence of this bounded sequence

converging in C(Ω). With this new sequence () indicates this sequence is Cauchy

and so converges in C0,λ(Ω)

This completes the proof.

9.7.3 General Sobolev Inequalities

Theorem 9.7.9 (General Sobolev inequalities): . Let Ω be a bounded open subset of Rn,

with a C1 boundary.

i.) If

k <n

p, (9.22)

then W k,p(Ω) ⊂−→ Lq(Ω), where

1

q=

1

p− kn, (9.23)

with an imbedding constant depending on k, p, n and Ω.

ii.) If

k >n

p, (9.24)

then Ck−[np

]−1,γ

(Ω) ⊂−→ W k,p(Ω), where

γ =

[np

]+ 1− n

p , if np is not an integer

any positive number < 1, if np is an integer.

with an imbedding constant depending on k, p, n, γ and Ω.

Proof.Step 1: Assume (9.22). Then since Dαu ∈ Lp(Ω) for all |α| = k , the Sobolev-

Nirenberg-Gagliardo inequality implies

||Dβu||Lp∗

(Ω)≤ C||u||

Wk,p(Ω)if |β| = k − 1,

and so u ∈ W k−1,p∗(Ω). Similarly, we find u ∈ W k−2,p∗∗(Ω), where 1p∗∗ = 1

p∗ − 1n =

1p − 2

n . Continuing, we eventually discover after k steps that u ∈ W 0,q(Ω) = Lq(Ω),

for 1q = 1

p − kn . The corresponding continuity estimate follows from multiplying the

relevant estimates at each stage of the above argument.

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Step 2: Assume now condition (9.23) holds, and np is not an integer. Then as above we see

u ∈ W k−l ,r (Ω), (9.25)

for

1

r=

1

p− l

n, (9.26)

provided lp < n. We choose the integer l so that

l <n

p< l + 1; (9.27)

that is, we set l =[np

]. Consequently (9.26) and (9.27) imply pn

n−pl > n. Hence (9.25)

and Morrey’s inequality imply that Dαu ∈ C0,1− nr (Ω) for all |α| < k − l − 1. Observe

also that 1− nr = 1− n

p + l =[np

]+ 1− n

p . Thus u ∈ Ck−[np

]−1,

[np

]+1− n

p (Ω), and the

stated estimate follows.

Step 3: Finally, suppose (9.26) holds, with np an integer. Set l =

[np

]−1 = n

p−1. Consequently,

we have as above u ∈ W k−l ,r (Ω) for r = pnn−pl = n. Hence the Sobolev-Nirenberg-

Gagliardo inequality shows Dαu ∈ Lq(Ω) for all n ≤ q < ∞ and all |α| ≤ k − l −1 = k −

[np

]. Therefore Morrey’s inequality further implies Dαu ∈ C0,1− n

q (Ω) for all

n < q < ∞ and all |α| ≤ k −[np

]− 1. Consequently u ∈ Ck−

[np

]−1,γ

(Ω) for each

0 < γ < 1. As before, the stated continuity estimate follows as well.

9.8 Difference Quotients

In this section we develop a tool that will enable us to determine the existence of derivatives

of certain solutions to various PDE. More succinctly, studies that pertain to this pursuit are

referred to collectively as regularity theory. We will go into this more in later chapters.

Recalling 1D calculus, we know that f is differentiable at x0 if

limx→x0

f (x)− f (x0)

x − x0= limh→0

f (x0 + h)− f (x0)

h

exists. In nD with f : Ω ⊂ Rn → R, we analogize to say that f is differentiable at x0 in the

direction of s if nD: f : Ω ⊂ Rn → R,

limh→0

∆khf (x) = limh→0

f (x + hes)− f (x)

h, ek = (0, . . . , 1, . . . , 0)

exists, where

es = (0, . . . , 1︸︷︷︸sth place

, . . . , 0);

i.e., to have the s partial derivative, limh→0 ∆khf (x) must exist.

The power of using difference quotients is conveniently summed up in the following

proposition.Gregory T. von Nessi

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9.8 Difference Quotients267

Proposition 9.8.1 (): If 1 ≤ p < ∞, f ∈ W 1,p(Ω), Ω′ ⊂⊂ Ω, and h small enough with

h < dist(Ω′, ∂Ω), then

i.)

||∆khf ||Lp(Ω′) ≤ C||Dk f ||Lp(Ω), k = 1, . . . , n.

ii.) Moreover, if 1 < p <∞ and

||∆khf ||Lp(Ω′) ≤ K ∀h < dist(Ω′, ∂Ω),

then f ∈ W 1,ploc (Ω) and ||Dk f ||Lp(Ω′) ≤ K with ∆khf → Dk f strongly in Lp(Ω′).

Proof: First, consider f ∈ W 1,p(Ω) ∩ C1(Ω). We start off by calculating

|∆khf (x)| =

∣∣∣∣f (x + hek)− f (x)

h

∣∣∣∣

≤ 1

h

∫ h

0

|Dk f (x + tek)| dt

Holder≤ h−1/p

(∫ h

0

|Dk f (x + tek)|p dt) 1

p

.

Taking the pth power, we obtain

|∆khf (x)|p ≤ 1

h

(∫ h

0

|Dk f (x + tek)|p dt).

Next we integrate both sides over Ω′:

∫

Ω′|∆khf (x)|p dx ≤ 1

h

∫ h

0

∫

Ω′|Dk f (x + tek)|p dxdt

≤ 1

h

∫ h

0

∫

Ω′+tek|Dk f (x)|p dxdt

≤ 1

h

∫ h

0

∫

Ω

|Dk f (x)|p dxdt

=

∫

Ω

|Dk f |p dx ≤ K.

Now, we need to verify the last calculation for general f ; we do this by the standard argument

that W 1,p(Ω) ∩ C1(Ω) is dense in W 1,p(Ω). Indeed, approximate f ∈ W 1,p(Ω) by fn ∈W 1,p(Ω)∩C1(Ω) such that fn → f in W 1,p(Ω) and use the above estimate for fn to ascertain

∫

Ω′|∆khfn|p dx ≤

∫

Ω

|Dk fn|p dx↓ Lp(Ω) ↓ Lp(Ω)∫

Ω′|∆khf |p dx ≤

∫

Ω

|Dk f |p dx ∀f ∈ W 1,p(Ω),

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for h fixed. This proves i.).

Now suppose that the last conclusion holds for h < Dist(Ω′, ∂Ω); i.e. since f ∈ W 1,p(Ω) we

see that ∆khf is uniformly bounded in Lp(Ω′). Thus, we have established the stipulations of

part ii.). Next, we construct a sequence ∆khn f by considering hn 0. Taking the bound-

edness ∆khf in Lp(Ω′) we apply the Banach-Alaoglu theorem to establish the existence of a

weakly convergent subsequence of ∆khn f . Upon relabeling such a subsequence, we define

gk as the weak limit, i.e. ∆khn f gk in Lp(Ω′). Now, we claim that gk = Dk f on Ω′. Claim:

gk = Dk f on Ω′, i.e.,∫

Ω′f ·Dkφ dx = −

∫

Ω′gkφ dx ∀φ ∈ C∞c (Ω′).

To this end, we calculate∫

Ω′∆khf · φ dx =

∫

Ω′

f (x + hek)− f (x)

hφ(x) dx

=1

h

∫

Ω′f (x + hek)φ(x) dx − 1

h

∫

Ω′f (x)φ(x) dx

=1

h

∫

Ω′+hekf (x)φ(x − hek) dx − 1

h

∫

Ω′f (x)φ(x) dx

=

∫

Ω

f (x)

[−φ(x − hek)− φ(x)

−h

]dx

= −∫

Ω

f (x)(∆k−hφ(x)) dx

Now as Dkφ,∆k−hφ(x) ∈ C∞c (Ω), we can use a simple real-analysis argument to say that

∆k−hφ→ Dkφ uniformly in Ω′. Thus, we have shown

∫

Ω′gkφ dx = −

∫

Ω′f ·Dkφ dx ∀φ ∈ C∞c (Ω′),

proving the claim. Moreover, this last equality clearly indicate that gk ∈ Lp(Ω′), as f ∈W 1,p(Ω). Now, we just have to show that ∆khf → Dk f strongly in Lp(Ω′). As we did for

part i.), first consider f ∈ W 1,p(Ω) ∩ C1(Ω). Calculating we see

∫

Ω′|∆khf −Dk f | dx =

∫

Ω′

∣∣∣∣1

h

∫ h

0

(Dk f (x + tek)−Dk f (x)) dt

∣∣∣∣p

dx

=

∫

Ω′

1

h

∫ h

0

|Dk f (x + tek)−Dk f (x))|p dtdx

=1

h

∫ h

0

∫

Ω′|Dk f (x + tek)−Dk f (x))|p dtdx

Now, since f ∈ C1(Ω), the integrand on the RHS is strictly bounded, i.e. we can apply

Lebesgue’s dominated convergence theorem (to the inner integral) to state

limh→0

∫

Ω′|∆khf −Dk f | dx = 0.

For general f ∈ W 1,p(Ω′), we do as before by picking fn ∈ W 1,p(Ω) ∩ C1(Ω) such that

fn → f in W 1,p(Ω′) to conclude that the above limit still holds. Gregory T. von Nessi

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9.9 A Note about Banach Algebras269

9.9 A Note about Banach Algebras

The following theorem is sometimes useful in the manipulation of weak formulations of PDE.

Theorem 9.9.1 (): Let Ω ⊂ Rn be an open and bonded domain with smooth boundary. If

p > n, then the space W 1,p(Ω) is a Banach algebra with respect to the usual multiplication

of functions, that is, if f , g ∈ W 1,p(Ω), then f · g ∈ W 1,p(Ω).

Proof: The Sobolev embeddings imply that

f , g ∈ C0,α(Ω) with α = 1− np> 0

and that the estimates

||f ||C0,α(Ω) ≤ C0||f ||W 1,p(Ω), ||g||C0,α(Ω) ≤ C0||g||W 1,p(Ω)

hold. For n ≥ 2 we have p′ < p and we may use the usual product rule to deduce that the

weak derivative of f g is given by f ·Dg +Df · g. Then

||D(f g)||pLp(Ω)

≤ C

∫

Ω

(|f ·Dg|p + |g ·Df |p) dx

≤ B(||f ||pL∞(Ω)

||Dg||pW 1,p(Ω)

+ ||g||pL∞(Ω)

||Df ||pW 1,p(Ω)

)

≤ 2BC1||f ||pW 1,p(Ω)||g||p

W 1,p(Ω).

The same arguments work for n = 1 and p ≥ 2. If p ∈ (1, 2), then we choose fε = φε ∗ fwhere φε is a standard mollifier. Suppose that Ω = (a, b). Then fε → f in W 1,1

loc (a, b) and

locally uniformly on (a, b) since f is continuous. It follows that

∫ b

a

fεgφ′ dx = −

∫ b

a

(Dg · fε + g ·Dfε)φdx.

As ε→ 0, we conclude∫ b

a

f gφ′ dx = −∫ b

a

(Dg · f + g ·Df )φdx.

Thus f g ∈ W 1,1(Ω) with D(f g) = Df · g + g · Df . The p-integrability of the derivative

follows from the embedding theorems as before.

9.10 Exercises

8.1: Let X = C0([−1, 1]) be the space of all continuous function on the interval [−1, 1], Y =

C1([−1, 1]) be the space of all continuously differentiable functions, and let Z = C10 ([−1, 1])

be the space of all function u ∈ Y with boundary values u(−1) = u(1) = 0.

a) Define

(u, v) =

∫ 1

−1

u(x)v(x) dx.




(if any)?

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b) Define

(u, v) =

∫ 1

−1

u′(x)v ′(x) dx.




(if any)?

8.2: Suppose that Ω is an open and bounded domain in Rn. Let k ∈ N, k ≥ 0, and γ ∈ (0, 1].

We define the norm || · ||k,γ in the following way. The γth-Holder seminorm on Ω is given by

[u]γ;Ω = supx,y∈Ω,x 6=y

|u(x)− u(y)||x − y |γ ,

and we define the maximum norm by

|u|∞;Ω = supx∈Ω|u(x)|.

The Holder space Ck,γ(Ω) consists of all functions u ∈ Ck(Ω) for which the norm

||u||k,γ;Ω =∑

|α|≤k||Dαu||∞;Ω +

∑

|α|=k[Dαu]γ;Ω

is finite. Prove that Ck,α(Ω) endowed with the norm || · ||k,γ;Ω is a Banach space.

Hint: You may prove the result for C0,γ([0, 1]) if you indicate what would change for the

general result.

8.3: Consider the shift operator T : l2 → l2 given by

(x)i = xi+1.

Prove the following assertion. It is true that

x− Tx = 0 ⇐⇒ x = 0,

however, the equation

x− Tx = y

does not have a solution for all y ∈ l2. This implies that T is not compact and the Fredholm’s

alternative cannot be applied. Give an example of a bounded sequence xj such that Txj does

not have a convergent subsequence.

8.4: The space l2 is a Hilbert space with respect to the scalar product

(x, y) =

∞∑

i=1

xiyi .


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Use general theorems to prove that the sequence of unit vectors in l2 given by (ei)j = δi jcontains a weakly convergent subsequence and characterize the weak limit.

8.5: Find an example of a sequence of integrable functions fi on the interval (0, 1) that

converge to an integrable function f a.e. such that the identity

limi→∞

∫ 1

0

fi(x) dx =

∫ 1

0

limi→∞

fi(x) dx

does not hold.

8.6: Let a ∈ C1([0, 1]), a ≥ 1. Show that the ODE

−(au′)′ + u = f

has a solution u ∈ C2([0, 1]) with u(0) = u(1) = 0 for all f ∈ C0([0, 1]).

Hint: Use the method of continuity and compare the equation with the problem −u′′ = f .

In order to prove the a-priori estimate derive first the estimate

sup |tut | ≤ sup |f |

where ut is the solution corresponding to the parameter t ∈ [0, 1].

8.7: Let λ ∈ R. Prove the following alternative: Either (i) the homogeneous equation

−u′′ − λu = 0

has a nontrivial solution u ∈ C([0, 1]) with u(0) = u(1) = 0 or (ii) for all f ∈ C0([0, 1]),

there exists a unique solution u ∈ C2([0, 1]) with u(0) = u(1) = 0 of the equation

−u′′ − λu = f .

Moreover, the mapping Rλ : f 7→ u is bounded as a mapping from C0([0, 1]) into C2([0, 1]).

Hint: Use Arzela-Ascoli.

8.8: Let 1 < p ≤ ∞ and α = 1 − 1p . Then there exists a constant C that depends

only on a, b, and p such that for all u ∈ C1([a, b]) and x0 ∈ [a, b] the inequality

||u||C0,α([a,b]) ≤ |u(x0)|+ C||u′||Lp([a,b])

holds.

8.9: Let X = L2(0, 1) and

M =

f ∈ X :

∫ 1

0

f (x) dx = 0

.

Show that M is a closed subspace of X. Find M⊥ with respect to the notion of orthogonality

given by the L2 scalar product

(·, ·) : X ×X → R, (u, v) =

∫ 1

0

u(x)v(x) dx.

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According to the projection theorem, every f ∈ X can be written as f = f1 + f2 with f1 ∈ Mand f2 ∈ M⊥. Characterize f1 and f2.

8.10: Suppose that pi ∈ (1,∞) for i = 1, . . . , n with

1

p1+ · · ·+ 1

pn= 1.

Prove the following generalization of Holder’s inequality: if fi ∈ Lpi (Ω) then

f =

n∏

i=1

fi ∈ L1(Ω) and

∣∣∣∣∣

∣∣∣∣∣n∏

i=1

fi

∣∣∣∣∣

∣∣∣∣∣L1(Ω)

≤n∏

i=1

||fi ||Lpi (Ω).

8.11: [Gilbarg-Trudinger, Exercise 7.1] Let Ω be a bounded domain in Rn. If u is a measurable

function on Ω such that |u|p ∈ L1(Ω) for some p ∈ R, we define

Φp(u) =

(1

|Ω|

∫

Ω

|u|p dx) 1

p

.

Show that

limp→∞

Φp(u) = supΩ|u|.

8.12: Suppose that the estimate

||u − uΩ||Lp(Ω) ≤ CΩ||Du||Lp(Ω)

holds for Ω = B(0, 1) ⊂ Rn with a constant C1 > 0 for all u ∈ W 1,p0 (B(0, 1)). Here uΩ

denotes the mean value of u on Ω, that is

uΩ =1

|Ω|

∫

Ω

u(z) dz.

Find the corresponding estimate for Ω = B(x0, R), R > 0, using a suitable change of coor-

dinates. What is CB(x0,R) in terms of C1 and R?

8.13: [Qualifying exam 08/99] Let f ∈ L1(0, 1) and suppose that∫ 1

0

f φ′ dx = 0 ∀φ ∈ C∞0 (0, 1).

Show that f is constant.

Hint: Use convolution, i.e., show the assertion first for fε = ρε ∗ f .

8.14: Let g ∈ L1(a, b) and define f by

f (x) =

∫ x

a

g(y) dy.

Prove that f ∈ W 1,1(a, b) with Df = g.

Hint: Use the fundamental theorem of Calculus for an approximation gk of g. Show that

the corresponding sequence fk is a Cauchy sequence in L1(a, b).

8.15: [Evans 5.10 #8,9] Use integration by parts to prove the following interpolation in-

equalities.Gregory T. von Nessi

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9.10 Exercises273

a) If u ∈ H2(Ω) ∩H10(Ω), then

∫

Ω

|Du|2 dx ≤ C(∫

Ω

u2 dx

) 12(∫

Ω

|D2u|2 dx) 1

2

b) If u ∈ W 2,p(Ω) ∩W 1,p0 (Ω) with p ∈ [2,∞), then

∫

Ω

|Du|p dx ≤ C(∫

Ω

|u|p dx) 1

2(∫

Ω

|D2u|p dx) 1

2

.

Hint: Assertion a) is a special case of b), but it might be useful to do a) first. For simplicity,

you may assume that u ∈ C∞c (Ω). Also note that

∫

Ω

|Du|p dx =

n∑

i=1

∫

Ω

|Diu|2|Du|p−2 dx.

8.16: [Qualifying exam 08/01] Suppose that u ∈ H1(R), and for simplicity suppose that u

is continuously differentiable. Prove that

∞∑

n=−∞|u(n)|2 <∞.

8.17: Suppose that n ≥ 2, that Ω = B(0, 1), and that x0 ∈ Ω. Let 1 ≤ p <∞. Show that

u ∈ C1(Ω\x0) belongs to W 1,p(Ω) if u and its classical derivative ∇u satisfy the following

inequality,

∫

Ω\x0(|u|p + |∇u|p) dx <∞.

Note that the examples from the chapter show that the corresponding result does not hold

for n = 1. This is related to the question of how big a set in Rn has to be in order to be

“seen” by Sobolev functions.

Hint: Show the result first under the additional assumption that u ∈ L∞(Ω) and then

consider uk = φk(u) where φk is a smooth function which is close to

ψk(s) =

k if s ∈ (k,∞),

s if s ∈ [−k, k ],

−k if s ∈ (−∞,−k).

You could choose φk to be the convolution of ψk with a fixed kernel.

8.18: Give an example of an open set Ω ⊂ Rn and a function u ∈ W 1,∞(Ω), such that

u is not Lipschitz continuous on Ω.

Hint: Take Ω to be the open unit disk in R2, with a slit removed.

8.19: Let Ω = B(

0, 12

)⊂ Rn, n ≥ 2. Show that log log 1

|x | ∈ W 1,n(Ω).

8.20: [Qualifying exam 01/00]

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a) Show that the closed unit ball in the Hilbert space H = L2([0, 1]) is not compact.

b) Show that g : [0, 1]→ R is a continuous function and define the operator T : H → H

by

(Tu)(x) = g(x)u(x) for x ∈ [0, 1].

Prove that T is a compact operator if and only if the function g is identically zero.

8.21: [Qualifying exam 08/00] Let Ω = B(0, 1) be the unit ball in R3. For any function

u : Ω → R, define the trace Tu by restricting u to ∂Ω, i.e., Tu = u∣∣∂Ω

. Show that

T : L2(Ω)→ L2(∂Ω) is not bounded.

8.22: [Qualifying exam 01/00] Let H = H1([0, 1]) and let Tu = u(1).

a) Explain precisely how T is defined for all u ∈ H, and show that T is a bounded linear

functional on H.

b) Let (·, ·)H denote the standard inner product inH. By the Riesz representation theorem,

there exists a unique v ∈ H such that Tu = (u, v)H for all u ∈ H. Find v .

8.23: Let Ω = B(0, 1) ⊂ Rn with n ≥ 2, α ∈ R, and f (x) = |x |α. For fixed α, determine

for which values of p and q the function f belongs to Lp(Ω) and W 1,q(Ω), respectively. Is

there a relation between the values?

8.24: Let Ω ⊂ Rn be an open and bounded domain with smooth boundary. Prove that

for p > n, the space W 1,p(Ω) is a Banach algebra with respect to the usual multiplication

of functions, that is, if f , g ∈ W 1,p(Ω), then f g ∈ W 1,p(Ω).

8.25: [Evans 5.10 #11] Show by example that if we have ||Dhu||L1(V ) ≤ C for all 0 <

|h| < 12dist (V, ∂U), it does not necessarily follow that u ∈ W 1,1(Ω).

8.26: [Qualifying exam 01/01] Let A be a compact and self-adjoint operator in a Hilbert

space H. Let ukk∈N be an orthonormal basis of H consisting of eigenfunctions of A with

associated eigenvalues λkk∈N. Prove that if λ 6= 0 and λ is not an eigenvalue of A, then

A− λI has a bounded inverse and

||(A− λI)−1|| ≤ 1

infk∈N |λk − λ|<∞.

8.27: [Qualifying exam 01/03] Let Ω be a bounded domain in Rn with smooth boundary.

a) For f ∈ L2(Ω) show that there exists a uf ∈ H10(Ω) such that

||f ||2H−1(Ω) = ||uf ||2H10(Ω)

= (f , uf )L2(Ω).

b) Show that L2(Ω) ⊂⊂−→ H−1(Ω). You may use the fact that H10(Ω) ⊂⊂−→ L2(Ω).

8.28: [Qualifying exam 08/02] Let Ω be a bounded, connected and open set in Rn with

smooth boundary. Let V be a closed subspace of H1(Ω) that does not contain nonzero

constant functions. Using H1(Ω) ⊂⊂−→ L2(Ω), show that there exists a constant C < ∞,

independent of u, such that for u ∈ V ,∫

Ω

|u|2 dx ≤ C∫

Ω

|Du|2 dx.


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276Chapter 10: Linear Elliptic PDE

10 Linear Elliptic PDE


– Alan Rickman

Contents

10.1 Existence of Weak Solutions for Laplace’s Equation 282

10.2 Existence for General Elliptic Equations of 2nd Order 286

10.3 Elliptic Regularity I: Sobolev Estimates 291

10.4 Elliptic Regularity II: Schauder Estimates 303

10.5 Estimates for Higher Derivatives 318

10.6 Eigenfunction Expansions for Elliptic Operators 320

10.7 Applications to Parabolic Equations of 2nd Order 324

10.8 Neumann Boundary Conditions 330

10.9 Semigroup Theory 334

10.10Applications to 2nd Order Parabolic PDE 342

10.11Exercises 344

10.1 Existence of Weak Solutions for Laplace’s Equation

General assumption Ω ⊂ Rn open and bounded

Want to solve:−∆u = f in Ω

u = 0 in ∂Ω(10.1)

multiply by v ∈ W 1,20 (Ω), integrate by parts:

∫

Ω

Du ·Dv dx =

∫

Ω

f v dx


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10.1 Existence of Weak Solutions for Laplace’s Equation277

Definition 10.1.1: Let f ∈ L2(Ω). A function u ∈ W 1,20 (Ω) is called a weak solution of

(10.1) if

∫

Ω

Du ·Dv dx =

∫

Ω

f v dx ∀v ∈ W 1,20 (Ω)

Theorem 10.1.2 (): The equation (10.1) has a unique weak solution in W 1,20 (Ω) and we

have ||u||W 1,2(Ω) ≤ C||f ||L2(Ω)

Notation: u = (−∆)−1f implicitly understood u = 0 on ∂Ω

Proof: H10 = W 1,2

0 (Ω) is a Hilbert space, F : H10 → R,

F (v) =

∫

Ω

f v dx

Then F is linear and bounded

|F (v)| =

∣∣∣∣∫

Ω

f v dx

∣∣∣∣

≤∫

Ω

|f v | dxHolder≤ ||f ||L2(Ω)||v ||L2(Ω)

≤ ||f ||L2(Ω)||v ||W 1,2(Ω)

Define B : H ×H → R, B(u, v) =∫

ΩDu ·Dv dx . B is bilinear. B is also bounded:

|B(u, v)| ≤ ||Du||L2(Ω)||Dv ||L2(Ω) ≤ ||u||W 1,2(Ω)||v ||W 1,2(Ω)

We can thus use Lax-Milgram if B is also coercive, i.e.,

B(u, u) ≥ ν||u||2W 1,2(Ω), B(u, u) =

∫

Ω

|Du|2 dx

So, to show this we use Poincare:

||u||2L2(Ω) ≤ Cp||Du||2L2(Ω)

⇐⇒ ||u||2W 1,2(Ω) = ||u||2L2(Ω) + ||Du||2L2(Ω) ≤ (C2p + 1)||Du||2L2(Ω)

=⇒ ||Du||2L2(Ω) ≥1

C2p + 1

||u||2W 1,2(Ω)

Thus,

B(u, u) ≥ 1

C2p + 1

||u||2W 1,2(Ω), i.e. B is coercive

Now Lax-Milgram says ∃u ∈ H10(Ω) = W 1,2

0 (Ω) such that

B(u, v) = F (v) ∀v ∈ H10(Ω) ⇐⇒

∫

Ω

Du ·Dv dx =

∫

Ω

f v dx ∀v ∈ H10(Ω)

i.e. u is a unique weak solution. The estimate ||u||W 1,2(Ω) ≤ C||f ||L2(Ω), follows from the

estimate in Lax-Milgram.

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Proposition 10.1.3 (): The operator

(−∆)−1 : L2(Ω)→ L2(Ω) is compact

Proof: Since ||u||W 1,2(Ω) ≤ C||f ||L2(Ω) we know that (−∆)−1 : L2(Ω) → W 1,20 (Ω) is

bounded. From compact embedding, we thus have

(−∆)−1 : L2(Ω) bounded→→W 1,2(Ω) ⊂⊂−→ L2(Ω)

(−∆)−1 maps bounded sets in L2(Ω) into precompact sets in L2(Ω), i.e., is compact.

Theorem 10.1.4 (): Either

i.) ∀f ∈ L2(Ω) the equations−∆u = λu + f in Ω

u = 0 on ∂Ω(10.2)

has a unique weak solution, or

ii.) there exists a nontrivial solution of−∆u = λu in Ω

u = 0 on ∂Ω(10.3)

iii.) If ii.) holds, then the space N of solutions of (10.3) is finite dimensional

iv.) the equation (10.2) has a solution ⇐⇒

(f , v)L2(Ω) = 0 ∀v ∈ N

Remark: In general iv.) is true if f ⊥ N∗, where N∗ is the null-space of the dual operator.

Proof: Almost all assertions follow from Fredholm’s-alternative in Hilbert Spaces.

Suppose that K = K∗(10.2) is equivalent to

u = (−∆)−1u + (−∆)−1f

take K = (−∆)−1 compact. f = (−∆)−1f . Without loss of generality, λ 6= 0, (I−λK)u = f .

i.) has a solution if f is perpendicular to the null-space of I − λK, i.e. (f , v) = 0 ∀vwith v − λKv = 0.

=⇒ (Kf , v) = 0 ⇐⇒ (f , K∗v) = 0 ⇐⇒ 1λ(f , v) = 0

=⇒ (f , v) = 0, ∀v ∈ N since λ 6= 0.

Finally, we just need to prove K is self-adjoint, i.e., K = K∗, f ∈ L2(Ω), h = Kf

⇐⇒ h is a weak solution of −∆h = f , f ∈ W 1,20 (Ω)

⇐⇒∫

Ω

Dh ·Dv dx =

∫

Ω

f v dx ∀v ∈ W 1,20 (Ω)

⇐⇒∫

Ω

D(Kf ) ·Dv dx =

∫

Ω

f v dx

︸︷︷︸v=Kg (*)

∀v ∈ W 1,20 (Ω)


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10.1 Existence of Weak Solutions for Laplace’s Equation279

So with this definition of g, we see

(K∗f , g) = (f , Kg)

(by (*)) =(D(Kf ), D(Kg)

)

(by symmetry) =(D(Kg), D(Kf )

)

= (g,Kf )

= (Kf , g)

=⇒ this holds ∀g ∈ L2

=⇒ K∗f = Kf ∀f=⇒ K∗ = K.

———

−∆u = f , L = −∆, f F ∈ H−1(Ω)

B(u, v) = L(u, v) =

∫

Ω

Du ·Dv dx

Weak solution:

L(u, v) =

∫

Ω

f v dx = 〈f , v〉 ∀v ∈ H10(Ω)

Existence and uniqueness: Lax-Milgram

More generally, we can solve

−∆u = g +

n∑

i=1

Di fi

︸︷︷︸∈H−1(Ω)

, g, fi ∈ L2(Ω)

Weak solution:

L(u, v)

⟨g +

n∑

i=1

Di fi , v

⟩=

∫

Ω

(gv −

n∑

i=1

figi

)dx

Notation: Summation convention. Frequently drop the summation symbol, sum over re-

peated indexes.

General boundary conditions: u0 ∈ W 1,2(Ω), want to solve−∆u = 0 in Ω

u = u0 on ∂Ω

Define u = u − u0 ∈ W 1,20 (Ω), and find equation for u.

∫

Ω

Du ·Dv dx = 0 ∀v ∈ H10(Ω)

⇐⇒∫

Ω

((Du −Du0) +Du0) ·Dv dx = 0

⇐⇒∫

Ω

Du ·Dv dx = −∫

Ω

Du0 ·Dv dx

Equation for u with general H−1(Ω) on the RHS.

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10.2 Existence for General Elliptic Equations of 2nd Order

Lu = −Di(ai jDju + biu) + ciDiu + du

with ai j , bi , ci , d ∈ L∞(Ω). General elliptic operator in divergence form, −div(A · Du) is

natural to work with in the framework of weak solutions..

Non-divergence form

Lu = ai jDiDju + biDiu + cu

If coefficients smooth, then non-divergence is equivalent to divergence form.

Corresponding bilinear form (multiply by v , integrate by parts)

L(u, v) =

∫

Ω

(ai jDju ·Div + biuDiv + cDiu · v + duv

)dx

want to solve Lu = g +Di fi ∈ H−1(Ω), i.e.,

L(u, v) =

∫

Ω

(gv + fiDiv) dx ∀v ∈ H10(Ω) (10.4)

Definition 10.2.1: u ∈ H10(Ω) is a weak solution of Lu = g +Di fi if (10.4) holds.

Definition 10.2.2: L is strictly elliptic if ai jξiξj ≥ λ|ξ|2, ∀ξ ∈ Rn, λ > 0.

Definition 10.2.3: Let L strictly elliptic, σ > 0 sufficiently large and let Lσu = Lu + σu,

thenLσu = g +Di fi in Ω

u = 0 on ∂Ω

and ||u||W 1,2(Ω) ≤ C(σ)(||g||L2(Ω) + ||f ||L2(Ω)

), f = (f1, . . . , fn)

Proof: Lax-Milgram in H10(Ω),

F (v) =

∫

Ω

(gv − fiDiv) dx ≤(||g||L2(Ω) + ||f ||L2(Ω)

)||v ||W 1,2(Ω)

Bilinear form:

L(u, v) =

∫

Ω

(ai jDju ·Div + biuDiv + ciDiu · v + duv) dx

L is bounded since

L(u, v)Holder≤ C(ai j , bi , ci , d)||u||W 1,2(Ω)||v ||W 1,2(Ω)

Crucial: check whether L is coercive

L(u, u) =

∫

Ω

ai jDju ·Diu︸︷︷︸

ellipticity: ≥λ|Du|2

+ biuDiu · u + du2

︸︷︷︸≤C(bi ,ci ,a)||u||L2(Ω)||u||W1,2(Ω)

dx

≥ λ

∫

Ω

|Du|2 dx − C||u||L2(Ω)||u||W 1,2(Ω)


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10.2 Existence for General Elliptic Equations of 2nd Order281

——

Poincare: ||u||2L2(Ω) ≤ C2p ||Du||2L2(Ω)

=⇒ ||u||2L2(Ω) + ||Du||2L2(Ω) ≤ (C2p + 1)||Du||2L2(Ω)

=⇒ ||Du||2L2(Ω) ≥1

C2p + 1

||u||2W 1,2(Ω)

——

Poincare≥ λ

C2p + 1

||u||2W 1,2(Ω) − C||u||L2(Ω)||u||W 1,2(Ω)

Young’s≥ λ′||u||2W 1,2(Ω) −λ′

2||u||2W 1,2(Ω) − C||u||2L2(Ω)

=λ′

2||u||2W 1,2(Ω) − C||u||2L2(Ω)

In general, L is not coercive, but if we take Lσ = Lu + σu,

Lσ(u, v) = L(u, v) + σ

∫

Ω

uv dx, then

Lσ(u, u) ≥ λ′

2||u||2W 1,2(Ω) + σ||u||2L2(Ω) − σ||u||2L2(Ω) − C||u||2L2(Ω)

and Lσ is coercive as soon as σ > C(bi , ci , d).

Question: Under what conditions can we solve

Lu = F ∀F ∈ H−1(Ω)?

Answer: Applications of Fredholm’s alternative.

Lu = −Di(ai jDju + biu) + ciDiu + du

Define:

〈Lu, v〉 = L(u, v)

=

∫

Ω

(ai jDju ·Div + biuDiv + ciDiu · v + duv) dx

Formally adjoint operator L∗

〈L∗u, v〉 = L∗(u, v) = L(v , u)

=

∫

Ω

(ai jDjv ·Diu︸︷︷︸aj iDiv ·Dju

+bivDiu + ciDiv · u + duv) dx

=

∫

Ω

(−Di(aj iDiu + Ciu) + BiDiu + du)v dx

= 〈L∗u, v〉, i.e.

L∗ = −Di(aj iDju + ciu) + biDiu + du

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10.2.1 Application of the Fredholm Alternative

Theorem 10.2.4 (Fredholm’s Alternative): L strictly elliptic. Then either

i.) ∀g, fi ∈ L2(Ω) the equation

Lu = g +Di fi in Ω

u = 0 on ∂Ω

has a unique solution, or

ii.) there exists a non-trivial, weak solution of the homogeneous equation

Lu = 0 in Ω

u = 0 on ∂Ω

iii.) If ii.) holds, then Lu = g +Di fi = F has a solution ⇐⇒

〈F, v〉 =

∫

Ω

(gv − fiDiv) dx = 0

∀v ∈ N(L∗), i.e. for all weak solutions of the adjoint equation L∗v = 0.

Proof: Setting up Fredholm

Lu = F ∈ H−1(Ω)

⇐⇒ Lσu − σu = F,

σ large enough such that Lσu = F has a unique solution ∀F .

Formally:

I1 : H10(Ω) ⊂−→ L2, I0 : L2(Ω) ⊂−→ H−1(Ω)

So, I2 = I0I1 : H10(Ω)→ H−1(Ω). Interpret our equation as Lσu − I2u = F is H−1(Ω).

——

(I0u)(v) =

∫

Ω

uv dx I0u bounded and linear

——

u − σL−1σ I2u = L−1

σ F in H10(Ω) (10.5)

One option: use Fredholm for this equation (done in Gilbarg and Trudinger). We will actually

shift into L2(Ω) first.

Interpret this as an equation in L2(Ω).

I1u − σI1L−1σ I2u = I1L

−1σ F

⇐⇒ (Id − σI1L−1σ I0)I1u = I1L

−1σ F

⇐⇒ (Id − σI1L−1σ I0)u = I1L

−1σ F in L2(Ω) (10.6)

where u = I1u ∈ L2(Ω).

Remark: If u is a solution of (10.6), then

u = L−1σ (σI0u + F ) (10.7)


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10.2 Existence for General Elliptic Equations of 2nd Order283

is a solution of (10.5).

Now we come back to the Fredholm’s alternative:

K = σ I1︸︷︷︸cmpct

·L−1σ I0︸︷︷︸

bnded

: L2(Ω)→ L2(Ω)

=⇒ K is compact.

=⇒ either (10.6) has a unique solution for all RHS in L2(Ω) or 2., the homogeneous equation

has non-trivial solutions.

• What i.) means:

Take RHS f = I1L−1σ F

=⇒ ∃! solution u ∈ L2(Ω)

=⇒ ∃! u of (10.5) by (10.7), that is

u − σL−1σ I2u = L−1

0 F

⇐⇒ Lσu − σI2u = F ⇐⇒ Lu = F

• What ii.) means:

(Id −K)u = 0 which implies the this a non-trivial weak solution of

u − σL−1σ I2u = 0

⇐⇒ Lσu − σI2u = F ⇐⇒ Lu = 0

Proof of iii.): K∗ = σI1(L∗σ)−1I0

By definition (also see below aside): (K∗g, h)L2 = (g,Kh)

——

Lu = f ⇐⇒ L(u, v) = (f , v) ∀v , Lσu = f ⇐⇒ Lσ(u, v) = (f , v) = Lσ(L−1σ f , v)

L∗u = f ⇐⇒ L∗(u, v) = (f , v) ∀v , L∗σu = f ⇐⇒ L∗σ(u, v) = (f , v) = Lσ((L∗σ)−1f , v)

∴ L∗σ(v , u) = Lσ(u, v)

——

(dual eqn v = Kh) = L∗σ((L∗σ)−1g,Kh)

= Lσ(Kh, (L∗σ)−1g)

(definition of K) = σLσ((L−1σ )h︸︷︷︸

H10(Ω)

, (L∗σ)−1g)

So,

original equation = σ(h, (L∗σ)−1g)

= σ((L∗σ)−1g, h) =⇒ ((K∗ − σ(L∗σ)−1)g︸︷︷︸=0

, h) = 0 ∀h ∈ L2,∀g ∈ L2

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Solvability of Original Equation

Lu = F ∈ H−1 ⇐⇒ (I −K)u = I1L−1σ F

Fredholm in L2: If you don’t have a unique solution, then we can solve 2. if the RHS is

perpendicular to the null-space of the adjoint operator I −K∗, i.e.,

(I1L−1σ F, v) = 0, ∀v ∈ N(I −K∗)

v ∈ N(I −K∗) ⇐⇒ (I − σ(L∗σ)−1)v = 0 ⇐⇒ L∗σv − σv = 0

⇐⇒ L∗v = 0, i.e. v ∈ N(L∗)

Solvability condition ⇐⇒ (I1L−1σ F, v) = 0, ∀v ∈ N(L∗). Multiplying by σ

⇐⇒ (I1L−1σ F, σv) = 0 = (I1L

−1σ F, L∗σv) =

〈L−1σ F, L∗σv〉

⇐⇒ 〈L−1σ F, L∗σv〉 = 0 ⇐⇒ L∗σ(v , L−1

σ F ) = 0

⇐⇒ Lσ(L−1σ F, v) = 〈F, v〉 = 0 ∀v ∈ N(L∗)

Remark: ∆u = f in the space of Periodic functions W 1,2per . Obviously, u ≡constant is periodic

with ∆u = 0. Solvability condition comes from Fredholm’s alternative, it turns out that we

need∫

Ω f dx = 0.

10.2.2 Spectrum of Elliptic Operators

Theorem 10.2.5 (Spectrum of elliptic operators): L strictly elliptic, then there exists an

at most countable set Σ ⊂ R such that the following holds

1. if λ ∈ Σ, thenLu = λu + g +Di fi in Ω

u = 0 on ∂Ω

2. if λ ∈ Σ, then the eigenspace N(λ) = u : Lu = λu is non-trivial and has finite

dimension.

3. If Σ is infinite, then Σ = λk , k ∈ N, and λk →∞ as k →∞.

Remark: In 1D, −u′′ = λu on [0, π] u(0) = u(π) = 0 uk(x) = sin(kx), i.e. λk = k2 →∞Proof: Essentially a consequence of the spectral Theorem of compact operators 7.10

Fredholm: uniqueness fails ⇐⇒ Lu = λu has a non-trivial solution. Lu = λu ⇐⇒Lu + σu = (λ+ σ)u ⇐⇒ Lσu = (λ+ σ)u.

u = (λ + σ)L−1σ = λ+σ

σ Ku, K the compact operator as before, ⇐⇒ Ku = σλ+σu =⇒

spectrum is at most countable, eigenspace finite dimension, zero only possible accumulation

point =⇒ λk →∞ if Σ not finite.

General Elliptic Equation: Lu = f

Lu = −Di(ui jDju + biDiu) + ciDiu + du

lL(u, v) = 〈Lu, v〉, existence works, L(u, u) ≥ λ||u||2W 1,2(Ω)

e.g. Lσu = Lu + σu =⇒ ∃ of solutions, Fredholm.

If e.g., bi , ci , d = 0, then σ = 0 will work.Gregory T. von Nessi

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10.3 Elliptic Regularity I: Sobolev Estimates285

10.3 Elliptic Regularity I: Sobolev Estimates

Elliptic regularity is one of the most challenging aspects of theoretical PDE. Regularity is

basically the studies “how good” a weak solution is; given that weak formulations of PDE

are not unique by any means, regularity tends to be very diverse in terms of the type of

calculations it requires.

The next two sections are devoted to the “classical” regularity results pertaining to linear

elliptic PDE. “Classical” is quoted because we will actually be employing more modern meth-

ods pertaining to Companato Spaces for derivation of Schauder estimates in the next section.

Aside: One will notice that the coming expositions focus heavily on Holder estimates as

opposed to approximations in Sobolev Spaces. Philosophically speaking, Sobolev Spaces is

the tool to use for existence and are hence the starting assumption in many regularity argu-

ments; but the ideal goal of regularity is to prove that a weak solution is indeed a classical

one. Thus, in regularity we usually are aiming for continuous derivatives, i.e. solutions in

Holder Spaces (at least).

To get the ball moving, let us consider a brief example. This will hopefully help convince

the reader that regularity is something that really does need to be proven as opposed to a

pedantic technicality that is always true.

Example 9: Consider the following problem in R:

−u′′ = f in (0, 1)

u(0) = 0.

Using integration by parts to attain a weak formulation of our solution we see that

u(x) =

∫ x

0

∫ t

0

f (s) ds dt

will solve the original equation, but the harder question of regularity u persist. Now if

f ∈ L2([0, 1]) it can be shown that u ∈ H2([0, 1]).

In higher dimensions, if u is a weak solution of −∆u = f , then f ∈ L2(Ω) usually im-

plies u ∈ H2loc(Ω) (provided ∂Ω has decent geometry, etc.). On the other hand, if f ∈ C0(Ω)

is is generally not the case that u ∈ C2(Ω)! There is hope though; we will prove (along with

the previous statements) that if f ∈ C0,α(Ω) (α > 0), then u ∈ C2(Ω) in general.

The next proposition concludes in one of the most important inequalities in regularity

studies of elliptic PDE.

10.3.1 Caccioppoli Inequalities

In this first subsection, we will develop the essential Caccioppoli inequalities. Even though

these inequalities ultimately provide a means of bounding first derivatives of solutions of

elliptic equations, the conditions under which the inequality itself holds vary. Thus, we first

consider the most intuitive example of the Laplace’s equation; following this, a more general

situation will be presented.

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Proposition 10.3.1 (Caccioppoli’s inequality for the −∆u = 0): If u ∈ H1(B(x0, R)) is a

weak solution of −∆u = 0, i.e.∫

Ω

Du ·Dφdx = 0 ∀φ ∈ H10(Ω),

then∫

B(x0,r)

|Du|2 dx ≤ C

(R − r)2

∫

B(x0,R)\B(x0,r)

|u − λ|2 dx

for all λ ∈ R and 0 < r < R.

Proof: The proof of this proposition is based on the idea of picking a “good” test func-

tion in H10(Ω) to manipulate our weak formulation with.

With the purpose of finding such a function, let us consider a function η such that

i.) η ∈ C∞c (B(x0, R)),

ii.) 0 ≤ η ≤ 1,

iii.) η ≡ 1 on B(x0, r), and

iv.) |Dη| ≤ C

R − r .

4 G& η ∈ C∞c (B(x0, R)) U

474 G& 0 ≤ η ≤ 1 U47474 G& η ≡ 1

D 6B(x0, r) U /698

4 G& |Dη| ≤ C

R− r G

r

R

x0 −R x0 − r x0 x0 + r x0 +R

4 3?>=*21 G D ,=*-6 ,C4 / @ A D >C: DA

η476

R2

** 3?>=* 21 G A D >T/ 4E<C3O/ @ 4 /,C4ED 6 DAI/ D ,=*-6 ,C4 / @ η 476

R2 G

*+* , U )* λ ∈ R/698J8?*969*

φ = η2(u− λ) ∈W 1,20 (B(x0, R)) G / @ 1-3 @ /,C476)

Dφ = 2ηDη(u − λ) + η2Du

/698 @ 3) 476) 476 ,=D,CN9* P#*;/QA D >C:S3 @ /,C4ED 6 U P*S/ <1*->C,/476

0 =

∫

B(x0,R)Du ·Dφ dx =

∫

B(x0,R)2ηDη ·Du(u− λ) + η2|Du|2 dx.

Figure 10.1: A potential form of η in R2

See figure (10.1) for a visualization of a potential η in R2.

Next, fix λ ∈ R and define φ = η2(u − λ) ∈ W 1,20 (B(x0, R)). Calculating

Dφ = 2ηDη(u − λ) + η2Du

and plugging into the weak formulation, we ascertain

0 =

∫

B(x0,R)

Du ·Dφ dx =

∫

B(x0,R)

2ηDη ·Du(u − λ) + η2|Du|2 dx.Gregory T. von Nessi

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Algebraic manipulation of the above equality shows that∫

B(x0,R)

η2|Du|2 dx

= −∫

B(x0,R)

2ηDη ·Du(u − λ) dx

≤∫

B(x0,R)

2η|Dη| · |Du| · |u − λ| dx

Holder≤ 2

(∫

B(x0,R)

η2|Du|2 dx) 1

2(∫

B(x0,R)

|Dη|2|u − λ|2 dx) 1

2

.

Now we divide through by the first integral on the RHS and square the result to find that∫

B(x0,R)

η2|Du|2 dx ≤ 4

∫

B(x0,R\B(x0,r)

|Dη|2︸︷︷︸C

(R−r)2

|u − λ|2 dx.

This leads immediately to the conclusion that∫

B(x0,r)

|Du|2 dx ≤ C

(R − r)2

∫

B(x0,R)\B(x0,r)

|u − λ|2 dx

Before moving onto Caccioppoli’s inequalities for elliptic systems, we have the following

special cases of proposition 10.3.1:

Case 1: r = R2 , λ = 0.

∫

B(x0,R2 )|Du|2 dx ≤ C

R2

∫

B(x0,R)\B(x0,R/2)

|u|2 dx

≤ C

R2

∫

B(x0,R)

|u|2 dx

Case 2: r = R2 , λ = u

B(x0,R).

∫

B(x0,R2 )|Du|2 dx ≤ C

R2

∫

B(x0,R)

|u − uB(x0,R)

|2 dx

Case 3: r = R2 , λ = u

B(x0,R)\B(x0,R/2).

∫

B(x0,R2 )|Du|2 dx

≤ C

R2

∫

B(x0,R)\B(x0,R2 )

∣∣∣u − uB(x0,R)\B(x0,R/2)

∣∣∣2dx

Poincare≤ C

∫

B(x0,R)\B(x0,R/2)

|Du|2 dx.

We can “plug the hole” in the area of integration on the RHS by adding C times the

LHS of the above to both sides of the inequality to get

(1 + C)

∫

B(x0,R2 )|Du|2 dx ≤ C

∫

B(x0,R)

|Du|2 dx,

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i.e.∫

B(x0,R2 )|Du|2 dx ≤ C

C + 1

∫

B(x0,R)

|Du|2 dx

= Θ

∫

B(x0,R)

|Du|2 dx, (10.8)

where 0 < Θ < 1 is constant!

Note: The general Caccioppoli inequalities do not requires the integrations to be over con-

centric balls. It is easy to see that as long as BR/2(y0) ⊂ BR(x0), the inequality still holds

(just appropriately modify η in the proofs). This generalization will be assumed in some of

the more heuristic arguments to come.

A direct consequence of this last case is

Corollary 10.3.2 (Lioville’s Theorem): If u ∈ H1(Rn) is a weak solution of −∆u = 0, then

u is constant.

Proof: Take limit R→∞ of both sides of (10.8) to see∫

Rn|Du|2 dx ≤ Θ

∫

Rn|Du|2 dx,

which implies Du is constant a.e. 0 < Θ < 1; but as u ∈ H1(Rn), we conclude that Du

constant everywhere in R.

Now we move on to consider elliptic systems of equations (which may or may not be

linear).

Proposition 10.3.3 (Caccioppoli’s inequality): Consider the elliptic system

−Dα(Aαβij (x)Dβu

j)

= g, for i = 1, . . . , m, (10.9)

where g ∈ H−1(Ω). Suppose we choose fi , fαi ∈ L2(Ω) such that g = fi +Dαf

αi and consider

a weak solution u ∈ H1loc(Ω) of (10.9), e.g.

∫

Ω

Aαβij DβujDαφ

i dx =

∫

Ω

(fiφi + f αi Dαφ

i) dx. for ∀φ ∈ H10(Ω).

If we ascertain that

i.) Aαβij ∈ L∞(Ω)

ii.) Aαβij ξiαξjβ ≥ ν|ξ|2, ν > 0

or

i.) Aαβij = constant

ii.) Aαβij ξiξjηαηβ ≥ ν|ξ|2|η|2, ν > 0 ∀ξ ∈ Rm and η ∈ Rn.

(This is called the Legendre-Hadamard condition)

or

i.) Aαβij ∈ C0(Ω)

ii.) Legendre-Hadamard condition holds

iii.) R sufficiently small

,


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then the following Caccioppoli inequality is true

∫

B(x0,R/2)

|Du|2 dx ≤ C(n)

1

R2

∫

B(x0,R)

|u − λ|2 dx + R2

∫

B(x0,R)

∑

i

f 2i dx

+

∫

B(x0,R)

∑

i ,α

(f αi )2 dx

for all λ ∈ R.

Proof: 1 Here we will prove the third case; it’s not hard to adapt the following estimates

to prove cases i.) and ii.). From our weak formulation we derive

∫

BR(x0)

Aαβij (x0)DαuiDβφ

j dx =

∫

BR(x0)

[Aαβij (x0)− Aαβij (x)

]Dαu

iDβφj dx

+

∫

BR(x0)

fiφi dx +

∫

BR(x0)

f αi Dαφi dx

1Can be omitted on the first reading.

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for all φ ∈ H10(BR(x0)). So, we consider the above again with φ = (u − λ)η2. Now we

calculate:

∫

BR(x0)

|D[(u − λ)η]|2 dx

≤∫

BR(x0)

Aαβij (x0)Dα[(ui − λi)η]Dβ[(uj − λj)η] dx

=

∫

BR(x0)

Aαβij (x0)DαuiDβ[(uj − λj)η2]

+

∫

BR(x0)

Aαβij (x0)(ui − λi)(uj − λj)DαηDβη dx

=

∫

BR(x0)

[Aαβij (x0)− Aαβij (x)

]Dαu

iDβ[(uj − λj)η2

]dx

+

∫

BR(x0)

fi(ui − λi)η2 dx +

∫

BR(x0)

f αi Dα[(ui − λi)η2

]dx

+

∫

BR(x0)

Aαβij (x0)(ui − λi)(uj − λj)DαηDβη dx

≤ δ∫

BR(x0)

DαuiDβu

jη2 dx + 2δ

∫

BR(x0)

Dαui ·Dβη · (uj − λj)η dx

+

∫

BR(x0)

fiη(ui − λi) dx +

∫

BR(x0)

f αi Dα · uiη2 dx

+ 2

∫

BR(x0)

∑

α

f αi (ui − λi)ηDη dx + C

∫

BR(x0)

|u − λ|2|Dη|2 dx

Young’s≤ δ

∫

BR(x0)

|Du|2η2 dx + 2δν

∫

BR(x0)

|Du|2 · η2 dx

+2δ

ν

∫

BR(x0)

|u − λ|2|Dη|2 dx +1

R2

∫

BR(x0)

|u − λ|2 dx

+

∫

BR(x0)

∑

i

f 2i dx +

1

ν

∫

BR(x0)

∑

i ,α

f αi dx

+ ν

∫

BR(x0)

|Du|2η2 dx +2

ν

∫

BR(x0)

∑

i ,α

(f αi )2 dx

+ 2ν

∫

BR(x0)

|u − λ|2|Dη|2 dx + C

∫

BR(x0)

|u − λ|2|Dη|2 dx

≤ (δ + 2δν + ν)

∫

BR(x0)

|Du|2η2 dx +3

ν

∫

BR(x0)

∑

i ,α

(f αi )2 dx

+1 + 2ν−1δ + 2ν + C

R2

∫

BR(x0)

|u − λ|2 dx

+ R2

∫

BR(x0)

∑

i

f 2i dx.

(10.10)


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In case it is unclear, δ is the standard free constant of continuity; and C bounds all Aαβij on

BR

(x0). Also, ν and some R2 terms arise via Young’s inequality. Next, choose a specific η

such that η ≡ 1 on R/2. Now, we consider

∫

BR/2(x0)

|Du|2 dx

≤∫

BR/2(x0)

|Du|2η2 dx

=

∫

BR/2(x0)

(D[(u − λ)η])2 − (u − λ)2(Dη)2 − 2DuDη(u − λ)η dx

Young’s≤∫

BR/2(x0)

|D[(u − λ)η]|2 dx + 2ν

∫

BR/2(x0)

|Du|2η2 dx

+

(1 +

2

ν

)∫

BR/2(x0)

|u − λ|2|Dη|2 dx.

Solving this last inequality algebraically and utilizing the properties of η, we get

(1− 2ν)

∫

BR/2(x0)

|Du|2 dx ≤∫

BR/2(x0)

|D[(u − λ)η]|2 dx

+ν + 2

νR2

∫

BR/2(x0)

|u − λ|2 dx.

Marching on, we put in estimate (10.10) into the RHS of the above (remembering that we

now put in R/2 instead of R into the estimate) to ascertain

(1− 2ν)

∫

BR/2(x0)

|Du|2 dx

≤ (δ + 2δν + ν)

∫

BR/2(x0)

|Du|2η2 dx +3

ν

∫

BR/2(x0)

∑

i ,α

(f αi )2 dx

+2ν2 + (C + 2)ν + 2δ + 3

R2

∫

BR/2(x0)

|u − λ|2 dx

+ R2

∫

BR/2(x0)

∑

i

f 2i dx

Finally solving algebraically we conclude that

(1− 3ν − δ(1− 2ν))

∫

BR/2(x0)

|Du|2 dx

≤ 3

ν

∫

BR(x0)

∑

i ,α

(f αi )2 dx +2ν2 + (C + 2)ν + 2δ + 3

R2

∫

BR(x0)

|u − λ|2 dx

+ R2

∫

BR(x0)

∑

i

f 2i dx.

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In the above we have used the properties of η and after solving algebraically, expanded the

RHS integrals to balls of radius R (obviously this can be done as we are integrating over

strictly positive quantities). This last inequality brings us to our conclusion as ν, δ, and C

are independent of u, along with the fact that δ and ν can be chosen small enough so the

LHS of the above remains positive.

10.3.2 Interior Estimates

We will now start exploring the classical theory of linear elliptic regularity. To begin, we

will go over estimation of elliptic solutions via bounding Sobolev norms, utilizing difference

quotients. The first two theorems of this subsection deal with interior estimates, while the

last theorem treats the issue of boundary regularity.

Changing the system under consideration, we will now work with an elliptic system of the

form


j)

= fi +Dαfαi , for i = 1, . . . , m,

where fi ∈ L2(Ω) and f αi ∈ H1loc(Ω). Basically, we will no longer assume that the inho-

mogenous term in the system is an element in H−1(Ω). Indeed, we will actually need true

weak differentiability of the inhomogenaity (i.e. a bounded Sobolev norm) to attain the full

regularity theory of these linear systems. With that, we consider a weak solution u ∈ H1loc(Ω)

of the elliptic system:∫

Ω

Aαβij DβujDαφ

i dx =

∫

Ω


i) dx ∀φ ∈ H10(Ω); (10.11)

this particular weak formulation is derived by the usual method of integration by parts. Now,

we go onto our first theorem.

Theorem 10.3.4 (): Suppose that Aαβij ∈ Lip(Ω), that (10.11) is elliptic with fi ∈ L2(Ω)

and f αi ∈ H1(Ω). If u ∈ H1loc(Ω) is a weak solution of (10.11), then u ∈ H2

loc(Ω).

Proof: We start off by recalling the notation from section Section 9.8: uh(x) := u(x +

hes), where

es := (0, . . . , 1︸︷︷︸sth place

, . . . , 0).

With this we consider Ω′ ⊂⊂ Ω with h small enough so that uh(x) is defined for all x ∈ Ω′.Now, we can rewrite our weak formulation as∫

Ω′Aαβij (x + hes)Dβu

jhDαφ

i dx

=

∫

Ω′(fi(x + hes)φ

i(x) + f αi (x + hes)Dαφi(x)) dx.

Subtracting the original weak formulation from this, we have

∫

Ω′Aαβij

[Dβu

jh −Dβujh

]Dαφ

i dx +

[Aαβij (x + hes)− Aαβij (x)

h

]Dβu

jDαφi dx

=

∫

Ω′

[fi(x + hes)− f (x)i

h

]φi(x) dx +

∫

Ω′

[f αi (x + hes)− f αi (x)

h

]Dαφ

i dx.


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Using notation from section Section 9.8 and commuting the difference and differential op-

erator on u, we obtain

∫

Ω′Aαβij ·Dβ(∆shu

i) ·Dαφj + ∆shAαβij ·Dβuj ·Dαφi dx

=

∫

Ω′∆shfi · φi + ∆shf

αi ·Dαφi dx.

We can algebraically manipulate this to get

∫


j) ·Dαφi

=

∫

Ω′∆shfi · φi + (∆shf

αi − ∆shA

αβij ·Dβuj)Dαφi dx. (10.12)

At this point, we would like to apply Caccioppoli’s inequality to the above, but the first term

on the RHS is not necessarily bounded; in order to proceed we now move to prove it is indeed

bounded. So, let us look at this term with φi := η2∆shui , where η has the standard definition.

After some algebraic tweaking around, we see that

∫

Ω′fi · ∆s−h(η2 · ∆shui) dx

=

∫

Ω′fiη · ∆shui · ∆s−hη dx +

∫

Ω′fi · η(x − hes) · ∆s−h(η · ∆shui) dx.

We now estimate the second term on the RHS by applying Young’s inequality, a derivative

product rule, and difference quotients bounds with derivatives. The ultimate result is:

∫

Ω′fi · ∆s−h(η2 · ∆shui) dx

Young’s≤∫

Ω′fiη · ∆shui · ∆s−hη dx

+1

ν

∫

Ω′|fiη|2 dx + ν

[∫

Ω′η2|D(∆shu

i)|2 dx +

∫

Ω′|Dη|2|∆shui |2 dx

]

≤∫

Ω′fiη · ∆shui · ∆s−hη dx

+ C(R)

(supi||fi ||L2(Ω′) + sup

i||ui ||

H1(Ω′)

+ supi

∫

Ω′|D(∆shu

i)|2 dx).

Holder≤ C(R)

([supi||fi ||L2(Ω′) + sup

i||ui ||

H1(Ω′) + 1

]2

+ supi

∫

Ω′|D(∆shu

i)|2 dx).

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Next, this estimate is applied to (10.12) yields∫


j) ·Dαφi

≤∫

Ω′fiη · ∆shui · ∆s−hη + (∆shf

βj − ∆shA

αβij ·Dαui)Dβφj dx

+ C(R)

([supi||fi ||L2(Ω′) + sup

i||ui ||

H1(Ω′) + 1

]2

+ supi

∫

Ω′|D(∆shu

i)|2 dx).

Finally, we now can apply Caccioppoli’s inequality to this despite the fact that the C(R) term

is not accounted for. Basically, we keep the C(R) term as is; and reflecting on the proof of

Caccioppoli’s inequality it is obvious that we can indeed say∫

BR/2(x0)

|D(∆shui)|2

≤ C(R)

(1

R2

∫

BR(x0)

|∆shui |2 dx +

∫

BR(x0)

∑

j,α

|∆shAαβij |2|Dβui |2 dx

+

∫

BR(x0)

∑

α

|∆shf αi |2 dx +

[supi||fi ||L2(R)

+ supi||ui ||

H1(R)+ 1

]2).

Thus, we have finally bounded the difference of the derivative of u. By proposition 9.8.1 we

have shown u ∈ H2loc(Ω).

Now, we move onto the next theorem which generalizes the first.

Theorem 10.3.5 (Interior Sobolev Regularity): Suppose that Aαβij ∈ Ck,1(Ω), that (10.11)

is elliptic with fi ∈ Hk(Ω) and f αi ∈ Hk+1(Ω). If u ∈ H1loc(Ω) is a weak solution of


j)

= fi −Dαf αi , for i = 1, . . . , m,

then u ∈ Hk+2loc (Ω).

Proof: We append the proof of theorem (10.3.4), i.e. we have shown that u ∈ H2loc(Ω).

Now, let us set φi := Dsψi , where ψi ∈ C∞0 (Ω). Upon integrating by parts the weak

formulation we ascertain

−∫

Ω

Ds(Aαβij Dβu

j)Dαψi dx =

∫

Ω

fiDsψi −∫

Ω

Ds(fαi )Dαψ

i dx.

Upon carrying out the Ds derivative on the RHS and algebraically manipulating we conclude∫

Ω

Aαβij Dβ(Dsuj)Dαψ

i dx = −∫

Ω

Ds(Aαβij )Dβu

jDαψi dx

−∫

Ω

fiDs(ψi) dx +

∫

Ω

Ds(fαi )Dαψ

i dx

=

∫

Ω

[−Ds(Aαβij )Dβuj − fi −Ds f αi ]Dαψ

i dx


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Now, in the case of the theorem of k = 1, we have the assumptions that DsAαβij ∈ Lip(Ω),

fi ∈ H1loc(Ω) and f αi ∈ H2

loc(Ω), i.e. we can integrate the LHS of the last equation to as-

certain that Dsui ∈ H2

loc(Ω) or u ∈ W 3,2loc (Ω). We get our conclusion by applying a standard

induction argument to the above method.

From this last theorem, it is clear that the regularity of the homogeneity and coefficients

determine the regularity of our weak solution. We note particularly that if Aαβij , fi , fαi ∈

C∞(Ω) then u ∈ C∞(Ω). More specifically, if Aαβij = const., fi ≡ 0, and f αi ≡ 0, then for

any BR ⊂⊂ Ω and for any k , it can be seen from the above proofs that

||u||Hk(BR/2) ≤ C(k,R)||u||L2(BR). (10.13)

10.3.3 Global Estimates

Now we move our discussion to the boundary of Ω. Fortunately, we have done most of the

hard work already for this situation.

Theorem 10.3.6 (): Suppose that ∂Ω ∈ Ck+1, Aαβij ∈ Ck(Ω), fi ∈ Hk(Ω) and f αi ∈Hk+1(Ω). If u is a weak solution of the Dirichlet-problem, ie. u ∈ H1

loc(Ω) solves

∫

Ω

Aαβij DβujDαφ

i dx =

∫

Ω


i) dx

∀φ ∈ H10(Ω) with u = u0 on ∂Ω, then u ∈ Hk+1(Ω) and moreover

||u||Hk+1(Ω)

≤ C(Ω)

[||f ||

Wk−1,2(Ω)+∑

i ,α

||f αi ||Hk (Ω)+ ||u||

L2(Ω)

].

Proof:Setting v = u−u0, it is clear that v satisfies the same system with a homogeneous

boundary condition. So, without loss of generality, we assume u0 = 0 in our problem. Since

∂Ω ∈ Ck+1, we know for any x0 ∈ ∂Ω, there exists a diffeomorphism

Φ : B+R → W ∩Ω

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of class Ck+1 which maps ΓR onto M ∩ ∂Ω.

#

1 1 '( ) $ " Ω * " 71

'(. 15 " * ?)@ 6 ! 6 ! 8 ,%,D7* )) "$#&

∂Ω ∈ Ck+1 # Aαβij ∈ Ck(Ω) # fi ∈ Hk(Ω)# $ 3 fαi ∈ Hk+1(Ω) ' +5 u (+* # 0 )6# *7 898 (/7 $ 75 ") (; ( . " 8 ) , ;7 8 ) # ( ) 'u ∈ H1& ' (Ω) *7 8 H ) *

∫

ΩAαβij Dβu

jDαφi dx =

∫

Ω(fiφ

i + fαi Dαφi) dx

∀φ ∈ H10 (Ω) 02( " u = u0 7 $ ∂Ω # ") $ u ∈ Hk+1(Ω)

# $43 7; ) 7 H ) ;

||u||Hk+1(Ω)

≤ C(Ω)

||f ||

Wk−1,2(Ω)+∑

i,α

||fαi ||Hk(Ω)+ ||u||

L2(Ω)

.

% ;7<765 v = u − u0 ( ( v

' " '1 '( $ " 1 %. & +" 1

' u0 = 0

$ ' ∂Ω ∈ Ck+1 -1 21 " x0 ∈ ∂Ω ) ! '( '

Φ : B+R →W ∩ Ω

6 Ck+1 1 ' ΓR M ∩ ∂Ω

Φ

B+R

xn

Rn

ΓRx′

x0

Ω

∂Ω

W = Φ(B+R)

. ? 1! #"4 "$ ! '( '

'( ' Φ 4 ' Φ∗ $ " (Φ∗u)(y) := u(Φ(y))

1 y ∈ B+R

Φ 6& Ck+1 DC Φ∗

) '( ' $ 1 Hk+1(B+R ) Hk+1(Ω ∩W )

© !

Figure 10.2: Boundary diffeomorphism

The diffeomorphism Φ induces a map Φ∗ defined by (Φ∗u)(y) := u(Φ(y)) with y ∈B+R . Since Φ is of class Ck+1, the chain-rule dictates that Φ∗ is an isomorphism between

Hk+1(B+R ) and Hk+1(Ω ∩ W ). Upon defining u∗ := Φ∗u, the chain-rule again says that

Du(Φ(y)) = J−1Φ Du∗(y), where JΦ is the Jacobian matrix of Φ. Recalling multi-dimensional

calculus, we see that u∗ satisfies∫

B+R

Aαβij Dβui∗Dαφ

j dy =

∫

B+R

fiφi dy +

∫

B+R

f αi Dαφi dy, (10.14)

where

Aαβij = |det(JΦ)|J−1Φ Aαβij JΦ

f αi = |det(JΦ)|J−1Φ Φ∗f αi

fi = |det(JΦ)|Φ∗fi .

From this we see that Aαβij is a positive multiple of a similarity transformation of Aαβij , i.e.

Aαβij is essentially a coordinate rotation of the original coefficients and thus are still elliptic.

After all that, we have reduced the problem to that of u∗ = 0 on ΓR. Now, for ever

direction but xn we can use the previous proof to bound the difference quotient. Specifically,

we take Ψi = ∆s−h(η2∆shui∗) for s = 1, . . . , n − 1 and η ∈ C∞c (BR(x0)) to ascertain

||∆sh(Dαui∗)||L2(B+

R/2)≤ C

(||ui∗||H1(B+

R)

+ supi||fi ||

L2(B+R

)+ sup

α,i||f αi ||H1(B+

R)

).

precisely as we did for the interior case. This only leaves Dnnu∗ left to be handled. Looking

at (10.14) and writing the whole thing out, we obtain

∫

B+R

Annij Dnuj∗Dnφ

i dy

= −n−1∑

α,β=1

∫

B+R

Aαβij Dβuj∗Dαφ

i dy +

∫

B+R

fiφi dy +

∫

B+R

f αi Dαφi dy.

This is an explicit representation of Dnnu∗ in the weak sense. Since the RHS is known to

exist, we thus conclude that so must Dnnu∗. Now, we can apply the induction and covering

arguments employed in earlier proofs of this section to obtain the result. Gregory T. von Nessi

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10.4 Elliptic Regularity II: Schauder Estimates297

10.4 Elliptic Regularity II: Schauder Estimates

In the last section we considered regularity results that pertained to estimates of Sobolev

norms. As was eluded to at the beginning of the last section, these estimates provide a start-

ing point for more useful estimates on linear elliptic systems. Namely, we now have enough

tools to derive the classical Schauder estimates of linear elliptic systems. The power of these

estimates is that they give us bounds on the Holder norms of solutions with relatively light

criteria placed upon the weak solution being considered and the coefficients of the system.

We will go through a gradual series of results, starting with the relatively specialized case of

systems with constant coefficient working up to results pertaining to non-divergence form

systems with Holder continuous coefficients.

Notation: For some of the following proofs, the indices have been ommitted when they

have no affect on the logic or idea of the corresponding proof. Such pedantic notation is

simply too distracting when it is not needed for clarity.

To start off this section we will first go over a very useful lemma that will be used in

several of the following proofs.

Lemma 10.4.1 (): Consider Ψ(ρ) non-negative and non-decreasing. If

Ψ(ρ) ≤ A[( ρR

)α+ ε]

Ψ(R) + BRβ ∀ρ < R ≤ R0, (10.15)

with A,B,α, and β non-negative constants and α > β, then ∃ε0(α, β, A) such that

Ψ(ρ) ≤ C(α, β, A)[( ρR

)αΨ(R) + Bρβ

]

is true if (10.15) holds for some ε < ε0.

Proof: First, consider set 0 < τ < 1 and set ρ = τR. Plugging into (10.15) gives

Ψ(τR) ≤ ταA[1 + ετ−α]Ψ(R) + BRβ.

Next fix γ ∈ (β,α) and choose τ such that 2Aτα = τγ . Now, pick ε0 such that ε0τ−α < 1.

Our assumption now becomes

Ψ(τR) ≤ 2Aτα ·Ψ(R) + BRβ

≤ τγΨ(R) + BRβ.

Now, we can use this last inequality as a scheme for iteration:

Ψ(τk+1R) ≤ τγΨ(τkR) + B(τkR)β

≤ τγ [τγΨ(τk−1R) + B(τk−1R)β] + BτβkRβ;

and after doing this iteration k + 1 times, we ascertain

Ψ(τk+1R) ≤ τ (k+1)γΨ(R) + BτβkRβk∑

j=0

τ j(γ−β).

Since ρ < R there exists k such that τk+1R ≤ ρ < τkR. Consequent upon this choice of k :

Ψ(ρ) ≤ Ψ(τk+1R) ≤ τ (k+1)γΨ(R) + C · B(τkR)β

≤( ρR

)γΨ(R) + C · B

( ρτ

)β.

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Remark: The above proof might seem to be grossly unintuitive at first, but in reality given

so many free constants, the proof just systemattically selects the constants to apply the

iterative step that sets the exponents correctly.

The value of the above lemma is that it allows us to replace the BRβ term with a Bρβ,

in addition to letting us drop the ε term. Thus Ψ can be bounded by a linear combination of

ρα and ρβ via this lemma; a very useful tool for reconciling Companato semi-norms (as one

can intuitively guess from definitions).

10.4.1 Systems with Constant Coefficients

Moving on to actual regularity results, we first consider the simplest system where the co-

efficients are constant. After going through the regularity results, we will briefly explore the

very interesting consequences of such.

Theorem 10.4.2 (): If u ∈ W 1,ploc (Ω) is a weak solution of

Dα(Aαβij Dβuj) = 0, (10.16)

where Aαβij = const. and satisfy the L-H condition, then for some fixed R0, we have that for

any ρ < R < R0,

i.)

∫

Bρ(x0)

|u|2 dx ≤ C ·( ρR

)n ∫

BR(x0)

|u|2 dx. (10.17)

ii.) Moreover,

∫

Bρ(x0)

|u − ux0,ρ|2 dx ≤ C ·( ρR

)n+2∫

BR(x0)

|u − ux0,R|2 dx.

(10.18)

Proof.(10.17): First, if ρ ≥ R/2 we can clearly pick C such that both of the inequalities above

are true. So we consider now when ρ ≤ R/2. Since u is a solution of (10.16), we

recall the interior Sobolev estimate from the last section (theorem 10.3.5) to write

||u||Hk(BR/2(x0)) ≤ C(k,R) · ||u||L2(BR)

Thus for ρ < R/2 we fine

∫

Bρ(x0)

|u|2 dx Holder≤ C(n) · ρn · supBρ(x0)

|u|2

≤ C(n) · ρn · ||u||Hk(BR/2(x0))

≤ C(n, k, R) · ρn ·∫

BR(x0)

|u|2 dx

The second inequality comes from the fact that Hk(Ω) ⊂−→ C1(Ω) ⊂ L∞(Ω) for k

choosen such that k > np (see Section 9.7). Also, we can employ a simple rescaling

argument to see that C(n, k, R) = C(k)R−n.Gregory T. von Nessi

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(10.18): This essentially is a consequence of (10.17) used in conjunction with Poincare and

Caccioppoli inequalities:∫

Bρ(x0)

|u − uρ,x0 |2 dxPoincare≤ C · ρ2

∫

Bρ(x0)

|Du|2 dx

≤ C · ρ2( ρR

)n ∫

BR/2(x0)

|Du|2 dx

= C ·( ρR

)n+2R2

∫

BR/2(x0)

|Du|2 dx

Caccioppoli≤ C ·( ρR

)n+2∫

BR(x0)

|u − ux0,R|2 dx.

Specifically, we have used (10.17) applied to Du to justify the second inequality. We

can do this since Du also solves (10.16) since the coefficients are constant.

The above theorem has some very interesting consequences, that we will now go into.

First, we note that theorem (10.4.2) holds for all derivatives Dαu, as derivatives of a solution

of a system with constant coefficients is itself a solution. With that in mind, let us consider

the polynomial Pm−1 such that∫

Bρ(x0)

Dα(u − Pm−1) dx = 0

for all |α| ≤ m − 1. With this particular choice of polynomial, we can now apply Poincare’s

inequality m times to obtain∫

Bρ(x0)

|u − Pm−1|2 dxPoincare≤ Cρ2m

∫

Bρ(x0)

|Dmu|2 dx.

Since Dmu also solves (10.16) we apply theorem (10.4.2) to get∫

Bρ(x0)

|u − Pm−1|2 dx ≤( ρR

)n+2m∫

BR(x0)

|u|2 dx.

Now, if we place the additional stipulation on u that |u(x)| ≤ A|x |σ, σ = [m] + 1 in addition

to its solving (10.16), then by letting R→∞ in the last equation, we get∫

Bρ(x0)

|u − Pm−1|2 dx = 0.

Thus,

If the growth of u is polynomial, then u actually is a polynomial!

This is a very interesting result in that we are actually able to determine the form of u merely

from asymptotics!

We now look at non-homogenous elliptic systems with constant coefficients. For this,

suppose u ∈ W 1,ploc (Ω) is a solution of

Dα(Aαβij Dβuj) = −Dαf αi , i = 1, . . . , m (10.19)

where the Aαβij are again constant and satisfy a L-H Condition. In this situation, we have the

following result:

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This inequality combined with (10.24), we see that

∫

Bρ(x0)


≤ C( ρR

)n+2∫

BR(x0)

|Du − (Du)x0,ρ|2 dx + C

∫

BR(x0)

|D(u − v)|2 dx. (10.21)

To bound the second term on the RHS, we use the L-H condition to ascertain∫

BR(x0)

|D(u − v)|2 dx ≤∫

BR(x0)

A ·D(u − v) ·D(u − v) dx

≤ C

∫

BR(x0)

(f − fx0,R)D(u − v) dx.

The second equality above comes from noticing that since u solves (10.19) in the weak

sense, u − v solves

Dα(Aαβij Dβ(u − v)j) = −Dα(f iα − fx0,R), i = 1, . . . , m

also in the weak sense (also we have taken u − v ∈ W 1,ploc (Ω) to be the test function). From

this, we see that∫

BR(x0)

|D(u − v)|2 dx ≤∫

BR(x0)

|f − fx0,R|2 dx,

i.e., upon combining this with (12.16),

∫

Bρ(x0)


≤ C( ρR

)n+2∫

BR(x0)

|Du − (Du)x0,R|2 dx + C[f ]2L2,λ(Ω)R

λ.

To attain our result, we now apply lemma 10.4.1 to the above to justify replacing the second

term on the RHS by ρλ.

Looking closer at the above proof, it is clear that we have proved even more than the

result stated in the theorem. Specifically, we have shown the bound

[Du]L2,λ(Ω) ≤ C(Ω,Ω′)[||u||H1(Ω) + [f ]L2,λ(Ω)

]with Ω′ ⊂⊂ Ω

to be true. An immediate consequence of theorem 10.4.3 and Companato’s theorem is

Corollary 10.4.4 (): If f ∈ C0,µloc (Ω), then Du ∈ C0,µ

loc (Ω), where µ ∈ (0, 1).

10.4.2 Systems with Variable Coefficients

Theorem 10.4.5 (): Suppose u ∈ W 1,ploc (Ω) is a solution of (10.19 with Aαβij ∈ C0(Ω) which

also satisfy a L-H condtion. If f iα ∈ L2,λ(Ω) (∼= L2,λ(Ω) for 0 ≤ λ ≤ n), then Du ∈ L2,λloc (Ω).

Proof: We first rewrite the elliptic system as

Dα

(Aαβij (x0)Dβu

j)

= Dα

(Aαβij (x0)− Aαβij (x)

)Dβu

j + f αi

=: DαF

αi ,

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where we have defined a new inhomogeneity. Now, we essential have the situation in theorem

(10.4.3) with DαFαi as the new inhomogenity. Thus, using the exact method used in that

theorem we ascertain

∫

Bρ(x0)

|Du|2 dx

≤ C( ρR

)n ∫

BR(x0)

|Du|2 dx + C

∫

BR(x0)

|D(u − v)|2 dx

≤ C( ρR

)n ∫

BR(x0)

|Du|2 dx + C

∫

BR(x0)

|F |2 dx

≤ C( ρR

)n ∫

BR(x0)

|Du|2 dx + C

∫

BR(x0)

|[A(x)− A(x0)]Du|2 dx

+ C ·∫

BR(x0)

|f |2 dx

≤ C( ρR

)n ∫

BR(x0)

|Du|2 dx + C · ω(R)

∫

BR(x0)

|Du|2 dx

+ C ·∫

BR(x0)

|f |2 dx

Now∫BR(x0) |f |2 dx ≤ Rλ as f ∈ L2,λ(Ω). With that in mind, if we choose R0 sufficiently

small with ρ < R < R0, then ω(R) is small and again we can apply lemma 10.4.1 with ω(R)

corresponding to the ε term; this gives us the result.

Theorem 10.4.6 (): Let u ∈ W 1,ploc (Ω) be a solution of

Dα

(Aαβij (x)Dβu

j)

= Dαfαi j = 1, . . . , m

with Aαβij ∈ C0,µ(Ω) which satisfy a L-H condition. If f iα ∈ C0,µ(Ω), then Du ∈ C0,µloc (Ω) and

moreover

[Du]C0,µ(Ω) ≤ C||u||H1(Ω) + [f ]C0,µ(Ω)

.

Proof: Again, we use the exact same idea as in the previous proof to see that

∫

Bρ(x0)

|Du − (Du)ρ|2 dx

≤ C( ρR

)n+2∫

BR(x0)

|Du − (Du)x0,r |2 dx

+

∫

BR(x0)

|f − fR|2 dx + C

∫

BR(x0)

|[A(x)− A(x0)]Du|2 dx.

(10.22)

To bound this we start out by looking at the first term on the RHS of the above:

∫

BR(x0)

|f − fR|2 dx ≤∫

BR(x0)

|x − x0|2µ · [f ]2C0,µ(BR(x0)) dx

≤ CR2µ+n.Gregory T. von Nessi

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Now we look at the second term on the RHS of (10.22) to notice

C

∫

BR(x0)

|[A(x)− A(x0)] ·Du|2 dx

≤ C∫

BR(x0)

||x − x0|µ · [A]C0,µ(BR(x0)) ·Du|2 dx

≤ CR2µ ·∫

BR(x0)

|Du|2 dxHolder≤ CR2µ+n||u||W 1,2(BR(x0))

≤ CR2µ+n

Putting these last two estimates into (10.22) yields

∫

Bρ(x0)

|Du − (Du)ρ|2 dx

≤ C( ρR

)n+2∫

BR(x0)

|Du − (Du)x0,r |2 dx + CR2µ+n.

Per usual this fits the form required by lemma 10.4.1 and consequently this leads us to our

conclusion.

10.4.3 Interior Schauder Estimates for Elliptic Systems in Non-divergence

Form

In this section, we consider with systems in non-divergence form. Obviously, if A is continu-

ously differentiable, there is no difference between the divergent and non-divergent case; but

in this subsection we will consider the situation where A is only continuous. I.e. we consider

Aαβij DβDαuj = fi j = 1, . . . , m. (10.23)

with u ∈ H2loc(Ω) and Aαβij ∈ C0,µ(Ω). In this situation, we have the following theorem.

Theorem 10.4.7 (): If f i ∈ C0,µ(Ω) then DβDαu ∈ C0,µ(Ω); moreover,

[DβDαu

]C0,µ(Ω)

≤ C[||DβDαu||L2(Ω) + [f ]C0,µ(Ω)

].

Proof: Essentially, we follow the progression of proofs of the last subsection, each of

which remain almost entirely unchanged. Corresponding to the progression of the previous

section we consider the proof in three main steps.

Step 1: Suppose A = const. and f = 0, then we are in a divergence form situation and

therefore

∫

Bρ(x0)

|DβDαu − (DβDαu)x0,ρ|2 dx

≤ C ·( ρR

)n+2∫

BR(x0)

|DβDαu − (DβDαu)x0,ρ|2 dx.

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Step 2: Suppose A = const and f 6= 0. Again we seek to get this into a divergence form; to

this end we set

z i(x) := ui(x)−(DβDαu

i)x0,R

2· xβ · xα.

Thus,

DσDγzi = DσDγu

i − (DσDγui)x0,R,

plugging this into (10.23) yields

Aαβij DβDαzj = fi − (fi)x0,R.

As in the previous section, we now split z = v + (z − v) where v is a solution of of the

homogenous equation:

AαβDβDαv

j = 0 in BR(x0)

v j = z j on ∂BR(x0).

With that, (10.23) indicates

Aαβij DβDα(z j − v j) = fi − (fi)x0,R in BR(x0)

z j − v j = 0 on ∂BR(x0)

As in the divergence form case then, we have the estimate∫

BR(x0)

|D2(z − v)|2 dx ≤∫

BR(x0)

|f − fx0,R|2 dx.

The spatial indices have been dropped above under the assumption that the supremum

of i is taken on the RHS to bound all second order derivatives (as stated at the beginning

of this section, writing this triviality out serves merely as a distraction at this point).

Using this estimate we ascertain

∫

Bρ(x0)

|DβDαz − (DβDαz)x0,ρ|2 dx

≤ C ·( ρR

)n+2∫

BR(x0)

|DβDαz − (DβDαz)x0,R|2 dx

+ C

∫

BR(x0)

|f − fx0,R|2 dx,

i.e.

∫

Bρ(x0)

|DβDαu − (DβDαu)ρ|2 dx

≤ C ·( ρR

)n+2∫

BR(x0)

|DβDαu − (DβDαu)x0,R|2 dx

+ C

∫

BR(x0)

|f − fx0,R|2 dx


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Step 3: The proof now continues as in the divergence form case by writing the equation

Aαβij (x)DβDαuj = fi

as

Aαβij (x0)DβDαuj = [Aαβij (x0)− Aαβij (x)]DβDαu

j + fi =: Fi

and concludes by using Step 2 with Fi substituted by the RHS of (10.23).

10.4.4 Regularity Up to the Boundary: Dirichlet BVP

In this section, we consider solutions of the Dirichlet BVP

Dα

(Aαβij (x)Dβu

j)

= Dαfαi in Ω

uj = gj on ∂Ω ∈ C1,µ

with f αi ∈ C0,µ(Ω) and g ∈ C1,µ(Ω) and we shall extend the local regularity result Section

10.4.2. Again, we have already seen many of the techniques in previous proofs that we will

require here. So, we will simply sketch out these two proofs pointing out the differences that

should be considered.

Theorem 10.4.8 (): Let u ∈ W 1,p(Ω) be a solution of

Dα

(Aαβij (x)Dβu

j)

= Dαfαi j = 1, . . . , m

with Aαβij ∈ C0,µ(Ω) which satisfy a L-H condition. If f iα ∈ C0,µ(Ω), then Du ∈ C0,µ(Ω) and

moreover

[Du]C0,µ(Ω) ≤ C||u||H1(Ω) + [f ]C0,µ(Ω)

.

Proof: First, we associate u with its trivial extension outside of Ω. Given the interior

result of theorem 10.4.6, we just have to prove boundary regularity to get the global result.

So, as in the Sobolev estimate case, we consider a zero BVP without loss of generality as

the transformation u 7→ u − g does not change our system. Also, in the same vein as the

global Sobolev proof, we know that the assumption ∂Ω ∈ C1,µ says that (locally) there exists

a diffeomorphism

Ψ : U → B+R

which flattens the boundary i.e. maps ∂Ω onto ΓR (see figure (10.3)).

In this proof we will be considering B+R with the goal of finding estimates on B+

R′ , where

R′ = µR for some fixed constant µ < 1. Then the global estimate will follow by using a

standard covering argument.

So, with that in mind, we may suppose that Ω = B+R and ∂Ω = ΓR. Generally speak-

ing, we seek the analogies of (10.17) and (10.18); once these are established the completion

of the proof mimics the progression of the proofs of interior estimates. To prove these

estimates we proceed in a series of steps.

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%

1 fαi ∈ C0,µ(Ω) g ∈ C1,µ(Ω) 1 - %- .6 +"

+ ? 1 ! ! -1 " ' " C 1 1 -1 1 ' " , 15 % $ ?%@ 6 ! A ! ) u ∈W 1,p(Ω) ) # *7 898 (/7 $ 75

Dα

(Aαβij (x)Dβu

j)

= Dαfαi j = 1, . . . ,m

02( " Aαβij ∈ C0,µ(Ω) 0 " ( . " * #& (+*+5 # . 7 $ 3 ( ( 7 $ ' +5 f iα ∈ C0,µ(Ω) # ") $Du ∈ C0,µ(Ω)

# $43 7; ) 7 H ) ;[Du]C0,µ(Ω) ≤ C

||u||H1(Ω) + [f ]C0,µ(Ω)

.

% ;67<75 .1 - u 1( ( Ω ' ? 1! ? 1 7 $ " .6 #" .% $ $ ' 1 3

" " 1 %.& & +" & ' u 7→ u − g " ' %& &' .% $ $ -1 21 ' ∂Ω ∈ C1,µ & " %- " '( '

Ψ : U → B+R

1 $ " ' ∂Ω ΓR ! . ? 1 @

B+R

xn

Rn

ΓRx′

Ψ

x0

Ω

∂Ω

U

. ? 1 @ #"4 " ! '( ' - 1 1 $ B+

R1 ( * '

B+R′71 R′ = µR &' µ < 1 .% $

' 1 %1 $ " ) .'

Figure 10.3: Boundary diffeomorphism

Step 1: First, suppose Aαβij = const. and f αi = 0. For ρ < R/2, we have

∫

B+ρ

|Du|2 dx Holder≤ Cρn supB+ρ

|Du|2

≤ C(k) · ρn||u||Hk(B+R/2

)

≤ C(k) · ρn||u||Hk(B∗)

≤ D(R) · ρn∫

B∗|u|2 dx

≤ D(R) · ρn∫

B+R

|u|2 dx

Caccioppoli≤ C(R) · ρn∫

B+R

|uxn |2 dx,

where B+R/2⊂ B∗ ⊂ B+

R and ∂B∗ is C∞. This result corresponds to (10.17). The

fourth inequality in the above calculation comes from the interior Sobolev regularity

estimate of the previous section; hence B∗ was introduced into the above calculation

to satisfy the boundary requirement required for the global Sobolev regularity. Also,

as with the derivation of (10.17), the second inequality comes from the continuous

imbedding Hk(B+R/2

) ⊂−→ Ck−1,1−n/2(B+R/2

) ⊂ L∞(B+R/2

).

The counterpart of (10.18) is

∫

B+ρ

|Du − (Du)ρ|2 dxPoincare≤ Cρ2

∫

B+ρ

|D2u|2 dx

Caccioppoli≤ Cρn+2

∫

B+R

|uxn |2 dx.

To proceed, we note that the same calculation can be carried out for u−λxn (which is

also a solution with zero boundary value). For this, we have to replace uxn by uxn − λ(λ = constant) in the above estimate.

Step 2: Now, that we have a way to bound I(B+ρ ) (the integral over B+

ρ ) with ρ < R/2, we

now need to bound I(Bρ(z)) for z ∈ BR′(x0) with ρ < R/2. To do this, we need to

consider two cases for balls centered at z .Gregory T. von Nessi

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%

? = 1 ) (z,Γ) = r ≤ µR 1 Bρ(z) ⊂ B+R (x0)

rz

x0y

BR/2(z)

B+R(x0)

B+R′(x0)

B+2r(y)

Br(z)

. ? 1 ? ' ? 1 1 ' I(Bρ(z))

$ "I(Br(z))

I(Br(z))$ "I(B+

2r(y)) * " I(B+

2r(y))$ "

I(BR/2(z))

∫

Bρ(z)|u− uz,ρ|2 dx

≤ C(ρr

)n+2∫

Br(z)|u− uz,r|2 dx

≤ C(ρr

)n+2∫

B+2r(y)|u− uz,2r|2 dx

≤ D(ρr

)n+2(

2r√R2/4− r2

)n+2 ∫

B+√R2/4−r2

(y)|u− uz,R/2|2 dx

≤ D(

414 − µ2

)n+2 (ρr

)n+2 ( rR

)n+2∫

BR/2(z)|u− uz,R/2|2 dx.

* #" 1 *% " ' ? u - &' 1 (z,Γ) = r < µR 1 Bρ(z) *B+R (x0)

1 ' ' $ 1 " " 1, $ I(Bρ(z) ∩ B+

R(x0)) 1 C

u Ω

A " *1 ' I(Bρ(z) ∩

Figure 10.4: Case 1 for proof of Theorem 10.4.8

Case 1: We first consider when dist(z, Γ ) = r ≤ µR with Bρ(z) ⊂ B+R (x0).

For this, we first estimate I(Bρ(z)) by I(Br (z)), then I(Br (z)) by I(B+2r (y)) and

finally I(B+2r (y)) by I(BR/2(z)):

∫

Bρ(z)

|u − uz,ρ|2 dx

≤ C(ρr

)n+2∫

Br (z)

|u − uz,r |2 dx

≤ C(ρr

)n+2∫

B+2r (y)

|u − uz,2r |2 dx

≤ D(ρr

)n+2(

2r√R2/4− r2

)n+2 ∫

B+√R2/4−r2

(y)

|u − uz,R/2|2 dx

≤ D(

414 − µ2

)n+2 (ρr

)n+2 ( rR

)n+2∫

BR/2(z)

|u − uz,R/2|2 dx.

In the third inequality we use the analogy of the result form Step 1 for u (the

proof is the same).

Case 2: In the second case we have dist(z, Γ ) = r < µR with Bρ(z) * B+R (x0). First we

remember that we really only want to find a bound for I(Bρ(z) ∩ B+R (x0)) as we

are considering a trivial extension of u outside Ω. Heuristically, we first estimate

I(Bρ(z)∩B+R (x0)) by I(B+

2ρ(y)) and finally I(B+2ρ(y)) by I(BR/2(z)). Again, the

calculation follows that of Case 1.

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It is a strait-forward geometrical calculation that µ can be taken to be 12√

17in order to

ensure B+2r (y) ⊂ BR/2(z), that Step 1 can be applied in the above cases, and that all the

above integrations are well defined (i.e. the domains of integration only extend outside of

B+R (x0) over Γ where the trivial extension of u is still well defined). Since µ is independent

of the original boundary point (x0), we can thus apply the standard covering argument to

conclude the analogies of (10.17) and (10.18) do indeed hold on the boundary.

10.4.4.1 The nonvariational case

We look at a solution of

Aαβ(x)ujxα,xβ = f i ∈ C2,µ in Ω

u = g ∈ C2,µ on ∂Ω ∈ C2,µ

As in 9.8.1 we find the equivalent of (10.17):

∫

B+ρ

|DβDαu|2 dx ≤ C( ρR

)n ∫

B+R

|DβDαu|2 dx

∫

B+ρ

|DβDαu − (DβDαu)ρ|2 dx ≤ C( ρR

)n+2∫

BR(x0)

|DβDαu − (DβDαu)ρ|2 dx (10.24)

moreover we have

||u||C2,µ(Ω) ≤ C||f ||C0,µ(Ω) + ||DβDαu||L2(Ω)

.

The proof is the same as in 9.8.1 with the only exception that in the scond step instead of

z = u − 1/2(uxα,xβ)x0,R · xα · xβ, one takes

z = u − 1

2(amnij )−1(f j)x0,R · x2

β ,

then there appears a term

∫

BR(x0)

(f − fx0,R)2 dx on the RHS of (10.19).

10.4.5 Existence and Uniqueness of Elliptic Systems

We finally show how the above a prior estimate implies existence. The idea is to use the

continuation method.

Consider

Lt := (1− t)∆u + t · Lu = f in Ω

u = 0 on ∂Ω

where L is the given elliptic differential operator.

We assume that for Lt (t ∈ [0, 1]) and boundary values zero, we have uniqueness. Set

Σ := t ∈ [0, 1]|Lt is uniquely solvable.Gregory T. von Nessi

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Since t = 0 ∈ Σ, Σ 6= ∅.

We shall show that Σ is open and closed and therefore Σ = [0, 1]. Especially we conclude

(for t = 1) that

Lu = f in Ω

u = 0 on ∂Ω

has a unique solution.

From the regularity theorem, we know that if Aαβ ∈ C0,µ and f ∈ C0,µ then

||DβDαu||C0,µ(Ω) ≤ C||f ||C0,µ(Ω) + ||DβDαu||L2(Ω)

.

Now the first step is to show the

Theorem 10.4.9 (): Suppose for the elliptic operator Lu = Aαβij uixα,xβ

we have

Lu = 0 in Ω

u = 0 on ∂Ω

has only the zero solution. Then ifLu = f in Ω

u = 0 on ∂Ω

we have

||u||H2,2(Ω) ≤ C1||f ||C0,µ(Ω).

Proof: Suppose that the theorem is not true, then there exists Aαβ(k)i j , f (k) such that for

all k

Aαβ(k)i j ξαξβ ≥ ν|ξ|2; ||A(k)||C0,µ(Ω) ≤ M

and ||f (k)||C0,µ(Ω) → 0 as k →∞.

If u(k) is a solution with f (k) and ||u(k)||H2,2(Ω) = 1 then u(k) → u as k → ∞ and Lu = 0

on Ω and u = 0 on ∂Ω i.e. u = 0, hence a contradiction.

Remark: Uniqueness of second order equations follows from Hopf’s maximum principle.

Suppose u is a solution ofAαβuxα,xβ = 0 on Ω

u = 0 on ∂Ω

If u has a maximum point in x0 ∈ Ω, then the Hessian uxα,xβ ≤ 0 and v := u + ε|x − x0|2has still a maximum point in x0 (for ε sufficiently small). But then Aαβvxα,xβ = Aαβuxα,xβ + ε

const. > 0 and that is a contradiction. One concludes that u = 0.

As 0 = t ∈ Σ, we have

||DβDαu||L2(Ω) ≤ C||f ||C0,µ(Ω)

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and therefore

||DβDαu||C0,µ(Ω) ≤ C||f ||C0,µ(Ω)

by the a-priori estimate above.

We show that Σ is closed:

Let tk ∈ Σ and tk → t as k →∞ with

Ltku

(k) = f in Ω

u(k) = 0 on ∂Ω

because ||u(k)xx ||C0,µ(Ω) ≤ C||f ||C0,µ(Ω), we have u(k) → u in C2(Ω) and

Ltu = f in Ω

u = 0 on ∂Ω

therefore t ∈ Σ.

Now we show that Σ is also open:

Let t0 ∈ Σ. Then there exists u such that

Lt0u = f in Ω

u = 0 on ∂Ω

For w ∈ C2,µ(Ω) there exists a unique uw such that

Lt0uw = (Lt0 − Lt)w + f in Ω

uw = 0 on ∂Ω

We conclude that

||uw ||C2,µ(Ω) ≤ C|t − t0| · ||w ||C2,µ(Ω) + C · ||f ||C2,µ(Ω)

and

||uw1 − uw2 ||C2,µ(Ω) ≤ C|t − t0| · ||w1 − w2||C2,µ(Ω)

If we choose t − t0| =: δ < 1/c , we see that the operator

T : C0,µ(Ω)→ C0,µ(Ω)

w 7→ uw

is a contradiction and therefore has a fixpoint which just means that (t0− δ, t0 + δ) ⊂ Σ, i.e

that Σ is open. Gregory T. von Nessi

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10.5 Estimates for Higher Derivatives311

10.5 Estimates for Higher Derivatives

u ∈ W 1,2(Ω) is a weak solution of −∆u = 0.

interior

interior estimates: B(x0, R) ⊂⊂ Ω

boundary estimates: x0 ∈ ∂Ω, transformations =⇒ ∂Ω flat near x0

−∆u = 0,∂

∂s=⇒ −∆uDsu = 0, vs := Dsu is a solution

Caccioppoli with vs

∫

B(x0,R2 )|Dvs |2 dx ≤ C

R2

∫

B(x0,R)

|vs |2 dx∫

B(x0,R2 )|DDsu|2 dx ≤ C

R2

∫

B(x0,R)

|Dsu|2 dx

=⇒ all 2nd derivatives are in L2(Ω) =⇒ u ∈ W 2,2(Ω)

To fully justify this, we replace derivatives with difference quotients.

Define uh = u(x + hes), es = (0, . . . , 1, . . . , 0). Suppose Ω′ ⊂⊂ Ω, with h small enough so

that uh is defined in Ω′. Take φ ∈ W 1,20 (Ω′) and compute

∫

Ω′Duh(x) ·Dφ(x) dx =

∫

Ω′Du(x + hes) ·Dφ(x) dx

=

∫

hes+Ω′⊆Ω

Du(x) ·Dφ(x − hes) dx

=

∫

Ω

Du(x) ·Dψ(x) dx,

where ψ(x) = φ(x − hes) ∈ W 1,20 (Ω). If u is a weak solution of −∆u = 0 in Ω, then uh is a

weak solution in Ω!

Define difference quotient ∆shu = u(x+hes)−u(x)h

∫

Ω′Duh(x) ·Dφ(x) dx = 0,

∫

Ω

Du(x) ·Dφ(x) dx = 0

=⇒∫

Ω

Duh(x)−Du(x)

hDφ(x) dx = 0 ∀φ ∈ W 1,2

0 (Ω)

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=⇒ ∆shu = uh−uh is a weak solution of −∆u = 0 in Ω′

Suppose that B(x0, 2R) ⊂ Ω′, h < R,

Caccioppoli inequality for ∆hsu:

∫

B(x0,R2 )|D∆shu|2 dx ≤ C

R2

∫

B(x0,R)

|∆shu|2 dx

(by Theorem 8.59) ≤ C

R2

∫

B(x0,R+h)

|Dsu|2 dx

≤ C

R2

∫

B(x0,2R)

|Dsu|2 dx

By Theorem 8.56, Dsu exists, and we can pass to the limit to get

∫

B(x0,R2 )|DDsu|2 dx ≤

C

R2

∫

B(x0,2R)

|Dsu|2 dx

Taking s = 1, . . . , n, we get

∫

B(x0,R2 )|Du|2 dx ≤ C

R2

∫

B(x0,2R)

|Du|2 dx

=⇒ W 1,2loc (Ω)

Now you can “boot-strap”; apply this argument to v = Dsu and get v ∈ W 1,2loc (Ω) =⇒

u ∈ W 3,2loc (Ω) =⇒ · =⇒ u ∈ W k,2

loc (Ω) ∀k =⇒ u ∈ C∞(Ω) by Sobolev embedding.

In general, the regularity of the coefficients determine how often you can differentiate your

equation.

In the boundary situation, this works in all tangential derivatives, i.e. DDsu ∈ L2(Ω),

s = 1, . . . , n − 1 (after reorienting the axises).

xn

x0 Rn−1

The only derivative missing is Dnnu. But from the original PDE we have

equation: Dnnu = −n−1∑

i=1

Di iu ∈ L2(Ω)


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10.6 Eigenfunction Expansions for Elliptic Operators313

10.6 Eigenfunction Expansions for Elliptic Operators

A : Rn → Rn (A is a n × n matrix)

A symmetric ⇐⇒ (Ax, y) = (x, Ay) ∀x, y ∈ Rn=⇒ ∃ an orthonormal basis of eigenvectors, ∃λi ∈ R such that Avi = λivi , where vi ∈Rn \ 0.

Unique expansion:

ν =∑

i

αivi , αi = (v , vi)

|ν|2 =∑

i

α2i

and

Av = A

(∑

i

αivi

)=∑

i

αiλivi

A diagonal in the eigenbasis.

Application: Solve the ODE y = Au, i.e., y(t) ∈ Rn with y(0) = y0. Explicit represen-

tation: y(t) =∑i αi(t)νi ,

=⇒∑

i

αi(t)νi =∑

i

αi(t)λiνi∑

i

α0i νi

=⇒αi(t) = λiαi(t), i = 1, . . . , n

αi(0) = α0i

Natural Generalization: H is a Hilbert Space, K : H → H compact and symmetric, i.e.,

(Kx, y) = (x,Ky) ∀x, y ∈ H.

Theorem 10.6.1 (Eigenvector expansion): H is a Hilbert Space, K symmetric, compact,

and K 6= 0. Then there exists an orthonormal system of eigenvectors ei , i ∈ N ⊆ N and

corresponding eigenvalues λi ∈ R such that

Kx =∑

i∈Nλi(x, ei)ei

If N is infinite, then λk → 0 as k →∞. Moreover,

H = N(k)⊕ spanei , i ∈ N

Definition 10.6.2: The elliptic operator L is said to be symmetric if L = L∗, i.e.

ai j = aj i , bi = ci a.e. x ∈ Ω

Theorem 10.6.3 (Eigenfunction expansion for elliptic operators): Suppose L is symmetric

and strictly elliptic. There there exists an orthonormal system of eigenfunctions, ui ∈ L2(Ω),

and eigenvalues, λi , with λi ≤ λi+1, with λi →∞ such that

Lui = λiui in Ω

ui = 0 on Ω

L2(Ω) = spanei , i ∈ N

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Remark: Lui = λiui ∈ L2(Ω) =⇒ ui ∈ H2(Ω) by regularity theory if the coefficients are smooth,

then Lui = λiui ∈ H2(Ω) =⇒ ui ∈ H4(Ω), . . ., i.e. ui ∈ C∞(Ω) if the coefficients

are smooth.

Proof: As before Lσ is given by Lσ = Lu + σu is invertible for σ large enough, L−1σ is

compact as a mapping from L2(Ω)→ L2(Ω). Moreover, L−1σ u = 0 ⇐⇒ u = Lσ0 = 0, i.e.

N(L−1σ ) = 0.

By Theorem 9.13, ∃ui ∈ L2(Ω) such that L2(Ω) = spanui , i ∈ n. Moreover,

L−1σ u =

∑

i∈Nµi(u, ui)ui

Choose uj such that

L−1σ uj =

∑

i∈Nµi (uj , ui)︸︷︷︸

δi j

ui = µjuj , µj 6= 0

=⇒ Lσuj = 1µjuj

Lσuj =1

µjuj ⇐⇒ Luj + σuj =

1

µjuj ⇐⇒ Luj =

(1

µj− σ

)

︸︷︷︸λj

uj

Lσuj =1

µjuj ⇐⇒ 0 ≤ L(uj , uj) =

1

µj(uj , uj) =

1

µj

=⇒ µj > 0

=⇒ µj has zero as its only possible accumulation point.

=⇒ λj → +∞ as j →∞.

Proposition 10.6.4 (Weak lower semicontinuity of the norm): X is a Banach Space,

xj x in X, then

||x || ≤ limj→∞

inf ||xj ||

Proof: Take y ∈ X∗, then

|〈y , x〉| ← |〈y , xj〉|Holder≤ ||y ||X∗ ||xj ||X

=⇒ |〈y , x〉| ≤(

limj→∞

inf ||xj ||X)||y ||X∗

Hahn-Banach: For every x0 ∈ X, ∃y ∈ X∗ such that |〈y , x0〉| = ||x0||X and 〈y , z〉 ≤ ||z ||X ,

∀z ∈ X (see Rudin 3.3 and corollary). This obviously implies that ||y ||X∗ = 1. So, we pick

y∗ ∈ X∗ such that |〈y∗, x〉| = ||x ||X . Thus,

=⇒ ||x || = 〈y∗, x〉 ≤(

limj→∞

inf ||xj ||X)||y∗||X∗︸︷︷︸

=1

Theorem 10.6.5 (): L strictly elliptic, λi are eigenvalues. Then,

λ1 = minL(u, u) : u ∈ H10(Ω), ||u||L2(Ω) = 1

λk+1 = minL(u, u) : u ∈ H10(Ω), ||u||L2(Ω) = 1, (u, ui) = 0, i = 1, . . . , k

Define µ := infL(u, u) : ||u||L2(Ω) = 1Gregory T. von Nessi

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10.6 Eigenfunction Expansions for Elliptic Operators315

Proof:

1. inf = min. Choosing a minimizing sequence uj ∈ H10(Ω), ||uj ||L2(Ω) = 1 such that

L(uj , uj)→ µ

We know from the proof of Fredholm’s alternative

L(uj , uj) ≥λ

2

∫

Ω

|Duj |2 dx − C∫

Ω

|uj | dx︸︷︷︸

=1

=⇒ λ

2

∫

Ω

|Duj |2 dx ≤ L(uj , uj)︸︷︷︸→µ

+C <∞

=⇒ uj ∈ W 1,20 (Ω) bounded

W 1,2(Ω) reflexive =⇒ ∃ a subsequence ujk such that ujk u in W 1,2(Ω) (Banach-

(Banach-Alaoglu).

Since ujk u in L2(Ω) Sobolev→→ ujk → u strongly in L2(Ω)

(More precisely, ujk bounded in L2(Ω) =⇒ ujk bounded in W 1,2(Ω) =⇒ ujk has a sub-

sequence converging strongly to some u∗ in L2(Ω) by Sobolev embedding. Clearly, u∗

is also a weak limit of ujk . Thus, by the uniqueness of weak limits, we have u∗ = u a.e.)

Dujk Du in L2(Ω) weakly and ||u||L2(Ω) = 1 from strong convergence.

2. Show that L(u, u) ≤ µ (=⇒ L(u, u) = µ, u minimizer, inf = min). As before

Lσ(u, u) ≥ γ||u||2W 1,2(Ω)

for σ sufficiently large, i.e.

|||u|||2 = Lσ(u, u)

Lσ(u, u) defines a scalar product equivalent to the standard H1(Ω) scalar product,

||| · ||| defines a norm.

weak lower semicontinuity of the norm:

ujk u in W 1,2(Ω)

Lσ(u, u) = |||u|||2 ≤ limk→∞

inf |||ujk |||2

= limk→∞

inf(L(ujk , ujk)︸︷︷︸→µ

+σ ||ujk ||2L2(Ω)︸︷︷︸=1

)

= µ+ σ

=⇒ L(u, u) ≤ µ =⇒ u is the minimizer

3. µ is an eigenvalue, Lu = µu

Consider the variation:

ut =u + tv

||u + tv ||L2(Ω)

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Then t 7→ L(ut , ut) has a minimum at t = 0.

L(ut , ut) =1

||u + tv ||2L2(Ω)

(L(u, u) + tL(u, v) + tL(v , u)︸︷︷︸2tL(u,v) by sym.

+t2L(v , v))

d

dt

∣∣∣∣∣t=0

L(ut , ut) = 2L(u, v)− 2

(∫

Ω

(u + 0)v dx

)L(u, u)︸︷︷︸

=µ

= 0

where we have used

1∫Ω(u + tv)2 dx

=

[∫

Ω

(u + tv)2 dx

]−1

︸︷︷︸=1 @ t=0

d

dt

1∫Ω(u + tv)2 dx

= −1 ·[∫

Ω

(u + tv)2 dx

]−2

︸︷︷︸=1 @ t=0

·2 ·∫

Ω

(u + tv)v dx

=⇒ L(u, v) = µ(u, v) ∀v ∈ H10(Ω)

⇐⇒ Lu = µu, i.e., µ is an eigenvalue.

4. µ = λ1

Suppose µ > λ1, ∃u1 ∈ H10(Ω) with ||u1||L2(Ω) = 1 such that

Lu1 = λ1u1

⇐⇒ L(u1, u1) = λ1(u1, u1)

= λ1 < µ = inf||u||L2(Ω)=1

L(u, u)

Thus, we have a contradiction.

Simple situation: ai j smooth, bi , ci , d = 0.

Proposition 10.6.6 (): L strictly elliptic, then the smallest eigenvalue λ1 (also called the

principle eigenvalue) is simple, i.e., the dimension of the corresponding eigenspace is one.

Proof: Following the last theorem: variational characterization of λ1,

minL(u, u) = λ1, L(u, u) = λ1 ⇐⇒ Lu1 = λ1u1

Key Step: Suppose that u solves

Lu = λ1u in Ω

u = 0 on ∂Ω

Then either u > 0 or u < 0 in Ω.

Suppose that ||u||L2(Ω) = 1 and define u+ := max(u, 0), and u− := −min(u, 0), i.e.,

u = u+ − u−. Then u+, u− ∈ W 1,20 (Ω), and

Du+ =

Du a.e. on u > 0

0 else

Du− =

−Du a.e. on u < 0

0 elseGregory T. von Nessi

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10.7 Applications to Parabolic Equations of 2nd Order317

Thus, we see that L(u+, u−) = 0 since∫

Ω u+u− dx = 0,

∫ΩDu

+ ·Du− dx = 0, . . ..

=⇒ λ1 = L(u, u) = L(u+ − u−, u+ − u−)

= L(u+, u+)− L(u+, u−)− L+ L(u−, u−)

= L(u+, u+) + L(u−, u−)

≥ λ1||u+||2L2(Ω) + λ1||u−||2L2(Ω) = λ1||u||2L2(Ω) = λ1

=⇒ so we have equality throughout,

L(u+, u+) = λ1||u+||2L2(Ω), L(u−, u−) = λ1||u−||2L2(Ω)

=⇒ Lu± = λ1u±, ,u± ∈ W 1,2

0 (Ω) since every minimizer is a solution of the eigen-problem.

Coefficients being smooth =⇒ u± smooth, i.e. u± ∈ C∞(Ω). In particular, Lu+ = λ1u+ ≥ 0

(λ1 ≥ 0).

=⇒ u+ is a supersolution of the elliptic equation.

The strong maximum principle says that either u+ ≡ 0 or u+ > 0 in Ω. Analogously,

either u− ≡ 0, or u− < 0 in Ω.

=⇒ either u = u+ or u = u−, i.e. either u > 0 or u < 0 in Ω.

Suppose now that u and u are solutions of Lu = λ1u in Ω. u, u ∈ W 1,20 (Ω) =⇒ u and

u have a sign, and we can choose γ such that∫

Ω

(u + γu) dx = 0

Since the original equation is linear, u+γu is also a solution =⇒ u+γu = 0 (otherwise u+γu

would have a sign) =⇒ u = −γu =⇒ all solutions are multiples of u =⇒ dim(Eig(λ1)) = 1.

Other argument: If dim(Eig(λ1)) > 1, ∃ u1, u2 in the eigenspace such that

∫

Ω

u1u2 dx = 0

which is a contradiction.

10.7 Applications to Parabolic Equations of 2nd Order

General assumption, Ω ⊂ Rn open and bounded, ΩT = Ω× (0, T ] solve the initial/boundary

value problem for u = u(x, t)

ut + Lu = f on ΩT

u = 0 on ∂Ω× [0, T ]

u = g on Ω× t = 0

L strictly elliptic differential operator, Lu = −Di(ai jDju)+biDiu+cu with ai j , bi , c ∈ L∞(Ω).

Definition 10.7.1: The operator ∂t + L is uniformly parabolic, if ai jξiξj ≥ λ|ξ|2 ∀ξ ∈ Rn,

∀(x, t) ∈ ΩT , with λ > 0.

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Standard example: Heat equation, L = −∆, ut − ∆u = f .

Suppose that f , g ∈ L2(Ω), and define

L(u, v ; t) =

∫

Ω

ai jDjuDiv + biDiu · v + uv dx

∀v ∈ H10(Ω), t ∈ [0, T ].

New point of view: parabolic PDE ↔ ODE in Hilbert Space.

u(x, t) U : [0, T ]→ H10(Ω)

(U(t))(x) = u(x, t) x ∈ Ω, t ∈ (0, T ]

analogously: f (x, t) F : [0, T ]→ L2(Ω)

(F (t))(x) = f (x, t) x ∈ Ω, t ∈ (0, T ]

Suppose we have a smooth solution of the PDE, multiply by v ∈ W 1,20 (Ω) and integrate

parts in∫

Ω dx ∫

Ω

U ′(t) · v dx + L(U, v ; t) =

∫

Ω

F (t) · v dx

What is ut? PDE ut + Lu = f says that

ut = f − Lu = f︸︷︷︸L2(Ω)

+Di (ai jDju)︸︷︷︸L2(Ω)

− biDiu︸︷︷︸L2(Ω)

− cu︸︷︷︸L2(Ω)

=⇒ ut = g0 +Digi ∈ H−1(Ω),

where g0 = f − biDiu − cu and gi = ai jDju.

This suggests that∫

Ω U′(t) · v dx has to be interpreted as

〈U ′(t), v〉 = duality in(H−1(Ω), H1

0(Ω))

for y(t) = αi(t)vi .

Lu = −Di(ai jDju) + biDiu + cu

equation:ut + Lu = f in Ω× (0, T ]

u = 0 on ∂Ω× [0, t]

u = g on Ω× t = 0parabolic if L elliptic:

∑ai jξiξj ≥ λ|ξ|2, λ > 0.

ut = f − Lu = f +Di(ai jDju)− biDiu − cu ∈ H−1(Ω)

(ut , v) duality between H−1 and H10

parabolic PDE ↔ ODE with BS-valued functions

U(t)(x) = u(x, t) ∀x ∈ Ω, t ∈ [0, T ]

L(u, v ; t) =

∫

Ω

(ai jDju ·Div + biDiu + cuv) dx


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10.7.1 Outline of Evan’s Approach

Definition 10.7.2: u ∈ L2(0, T ;H10(Ω)) with u′ ∈ L2(0, T ;H−1) is a weak solution of the

IBVP if

i.) 〈U ′, v〉+ L(U, v ; t) = (F, v)L2(Ω) ∀v ∈ H10(Ω), a.e. t ∈ (0, T )

ii.) U(0) = g in Ω

Remark: u ∈ L2(0, T ;H10(Ω)), u′ ∈ L2(0, T ;H−1) then C0(0, T ;L2) and thus U(0) is

well defined.

General framework: Calculus in abstract spaces.

Definition 10.7.3: 1) Lp(0, T ;X) is the space of all measurable functions u : [0, 1]→ X

with

||u||Lp(0,T ;X) =

(∫ T

0

||u(t)||pX) 1

p

<∞, 1 ≤ p <∞

and

||u||L∞(0,T ;X) = ess sup0≤t≤T

||u(t)||X <∞

2) C0([0, T ];X) is the space of all continuous functions u : [0, T ]→ X with

||u||C0([0,T ];X) = max0≤t≤T

||u(t)||X <∞

3) u ∈ L1(0, T ;X), then v ∈ L1(0, T ;X) is the weak derivative of u

∫ T

0

φ′(t)u(t) dt = −∫ T

0

φ(t)v(t) dt ∀φ ∈ C∞c (0, T )

4) The Sobolev space W 1,p(0, T ;X) is the space of all u such that u, u′ ∈ Lp(0, T ;X)

||u||W 1,p(0,T ;X) =

(∫ T

0

||u(t)||pX + ||u′(t)||pX dt) 1

p

for 1 ≤ p <∞ and

||u||w1,∞(0,T ;X) = ess sup0≤t≤T

(||u(t)||X + ||u′(t)||X)

Theorem 10.7.4 (): u ∈ W 1,p(0, T ;X), 1 ≤ p ≤ ∞. Then

i.) u ∈ C0([0, T ];X) i.e. there exists a continuous representative.

ii.)

u(t) = u(s) +

∫ t

x

u′(τ) dτ 0 ≤ s ≤ t ≤ T

iii.)

max0≤t≤T

||u(t))|| = C · ||u||W 1,p(0,T ;X)

X = H10, X = H−1

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10.7.2 Evan’s Approach to Existence

Galerkin method: Solve your equation in a finite dimensional space, typically the span of the

first k eigenfunctions of the elliptic operator; solution uk .

Key: energy (a priori) estimates of the form

max0≤t≤T

||uk(t)||L2(Ω) + ||uk ||L2(0,T ;H10) + ||u′k ||L2(0,T ;H−1)

≤ C(||f ||L2(0,T ;L2) + ||g||L2(Ω)

)

with C independent of k

Crucial: (weak) compactness bounded sequences in L2(0, T ;H10) and L2(0, T ;H−1) con-

tain weakly convergent subsequences: ukl u, u′kl v

Then prove that v = u′, and that u is a weak solution of the PDE.

Galerkin method: More formal construction of solutions. y = Ay , y(0) = y0. A symmetric

and positive definite =⇒ exists basis of eigenvectors ui with eigenvalues λi

express y(t) =

n∑

i=1

αi(t)ui , y0 =

n∑

i=1

α0i ui

=⇒n∑

i=1

αi(t)ui =

n∑

i=1

λiαi(t)ui

∣∣∣∣∣ · uj

=⇒ αi(t) = λiαi(t), αi(0) = α0i , αi(t) = α0

i eλ1t

y(t) =

n∑

i=1

α0i eλi t

Same idea for A = ∆ = −L, model for elliptic operators L.

Want to solve:ut − ∆u = 0 in Ω× (0, T )

u = 0 on ∂Ω× (0, T )

u(x, 0) = u0 in Ω

Let ui be the orthonormal basis of eigenfunctions

−∆ui = λiui

and try to find

u(x, t) =

∞∑

i=1

αi(t)ui(t)

ut = ∆u :

∞∑

i=1

αi(t)ui(x) =

∞∑

i=1

(−λi)αi(t)ui(x)

∣∣∣∣∣ · uj ,∫dx

=⇒ αi(t) = −λiαi(t), αi(0) = α0i ,

∞∑

i=1

α0i ui(x) = u0(x)


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Formal solution:

u(x, t) =

∞∑

i=1

e−λi tα0i ui(x)

Justification via weak solutions:

Multiply by test function, integrate by parts, choose φ ∈ C∞c ([0, T )×Ω)

−∫ T

0

u · φt dxdt −∫

Ω

u0(x)φ(x, 0) dx = −∫ T

0

∫

Ω

Du ·Dφ dxdt

=

∫ T

0

∫

Ω

u · ∆φ dxdt

Expand the initial data

u0(x) =

∞∑

i=1

α0i ui(x)

and define

u(k)j (x) =

k∑

i=1

α0i uu(x)

Then u(k)0 → u0 in L2.

Then

u(k)(x) =

k∑

i=1

α0i e−λi tui(x) is a weak solution

of the IBVP with u0 replaced by u(k)0 . Galerkin method: approximate by u(k) ∈spanu1, . . . , uk

Moreover:

||u(k)(t)− u(t)||L2(Ω) =

∣∣∣∣∣∣

∣∣∣∣∣∣

∞∑

k+1

α0i e−λi t︸︷︷︸≤1

ui

∣∣∣∣∣∣

∣∣∣∣∣∣L2(Ω)

≤( ∞∑

i=k+1

(α0i )2

) 12

→ 0 as k →∞

=⇒ uk(t)→ u(t) uniformly in t as k →∞

−∫ T

0

∫

Ω

u(k)φt dxdt −∫

Ω

u(k)0 φ(x, 0) dx =

∫ T

0

∫

Ω

u(k)∆φ dxdt

=⇒ u is a weak solution (as k →∞)

Representation:

u(x, t) =

∞∑

i=1

α0i e−λi tui(x)

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gives good estimates.

||u(t)||L2(Ω) =

∣∣∣∣∣

∣∣∣∣∣∞∑

i=1

α0i e−λi tui(x)

∣∣∣∣∣

∣∣∣∣∣L2(Ω)

=

( ∞∑

i=1

(α0i )2(e−λi t

)2

) 12

≤ e−λ1t

( ∞∑

i=1

(α0i )2

) 12

= e−λ1t ||u0||L2(Ω)

=⇒ ||u(t)||L2(Ω) ≤ ||u0||L2(Ω)e−λ1t

||∆u(t)||L2(Ω) =

∣∣∣∣∣

∣∣∣∣∣∞∑

i=1

(α0i e−λi t(−λi)ui)

∣∣∣∣∣

∣∣∣∣∣L2(Ω)

=1

t

∞∑

i=1

(α0i t · λi︸︷︷︸

x

e−λi t)2

12

(xe−x ≤ C, x ≥ 0)

≤ C

t

( ∞∑

i=1

(α0i )2

) 12

≤ C

t||u0||L2(Ω)

Theorem 10.7.5 (): L symmetric and strictly elliptic, then the IBVP

ut + Lu = 0 in Ω× (0, T )

u = 0 on ∂Ω× (0, T )

u = u0 on Ω× t = 0

has for all u0 ∈ L2(Ω) the solution

u(x, t) =

∞∑

i=1

e−λi tα0i ui(x)

where ui is a basis of orthonormal eigenfunctions of L in L2(Ω) with corresponding eigen-

value λ1 ≤ λ2 ≤ · · ·

10.8 Neumann Boundary Conditions

Neumann ↔ no flux

F (x, t) = heat flux = −κ(x)Du(x, t) Fourier’s law of cooling

Du · ν = 0 ⇐⇒ F (x, t) · ν = 0 no heat flux out of Ω

Simple model problem:

−div (a(x)Du) + c(x)u = f in Ω

(a(x)Du) · ν = 0 on ∂ΩGregory T. von Nessi

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10.8 Neumann Boundary Conditions323

a and c are smooth, f ∈ L2 (f ∈ H−1)

Weak formulation of the problem: multiply by v ∈ C∞(Ω) ∩H1(Ω) and integrate in Ω,∫

Ω

[−div (a(x)Du) + c(x)u] v dx =

∫

Ω

f v dx (10.25)

=

∫

Ω

(a(x)Du)︸︷︷︸=0

·νv dx +

∫

Ω

a(x)Du ·Dv + c(x)uv dx

=⇒∫

Ω

[a(x)Du ·Dv + c(x)uv ] dx =

∫

Ω

f v dx ∀v ∈ H1 (10.26)

Definition 10.8.1: A function u ∈ H(Ω) is said to be a weak solution of the Neumann

problem if (10.26) holds.

Remark: (a(x)Du) · ν = 0 on ∂Ω

Subtle issue: a(x)Du seems to be only an L2 function and is not immediately clear how to

define Du on ∂Ω. However: −div(a(x)Du) = −c(x) + f ∈ L2. Turns out that a(x)Du ∈L2(div)

Proposition 10.8.2 (): Suppose that u is a weak solution and suppose that u ∈ C2(Ω) ∩C1(Ω). Then u is a solution of the Neumann problem.

Proof: Use (10.26) first for V ∈ C∞c (Ω). Reverse the integration by parts (10.25) to get

∫

Ω

[−div (a(x)Du) + c(x)u − f ]v dx = 0 ∀v ∈ C∞c (Ω)

=⇒ −div(a(x)Du) + cu = f in Ω, i.e. the PDE Holds.

Now take arbitrary v ∈ C∞(Ω)∫

Ω

[a(x)Du ·Dv + c(x)v) dx =

∫

Ω

f v dx

=

∫

∂Ω

[a(x)Du · ν)v dS +

∫

Ω

[−div (a(x)Du) + c(x)u]v dx

︸︷︷︸∫Ω f v dx

=⇒∫

∂Ω

[a(x)Du · ν)v dS = 0 ∀v ∈ C∞(Ω)

=⇒ a(x)Du · ν = 0 on ∂Ω

Remark: The Neumann BC is a “natural” BC in the sense that it follows from the weak

formulation. It’s not enforced as e.g. the Dirichlet condition u = 0 on ∂Ω by seeking

u ∈ H10(Ω).

Theorem 10.8.3 (): Suppose a, c ∈ C(Ω) and that there exists a1, a2, c1, c2 > 0 such that

c1 ≤ c(x) ≤ c2, a1 ≤ a(x) ≤ a2 (i.e., the equation is elliptic). Then there exists for all

f ∈ L2(Ω) a unique weak solution u of

− div (a(x)Du) + cu = f in Ω

(a(x)Du) · ν = 0 on ∂Ω

and ||u||H1(Ω) ≤ C||f ||L2(Ω).

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Remark: The same assertion holds with f a linear function on H1(Ω), i.e. f ∈ H−1.

Proof: Weak solution meant

∫

Ω

[a(x)Du ·Dv + c(x)uv ] dx =

∫

Ω

f v dx ∀v ∈ H1(Ω)

linear functional F (v) =∫

Ω f v dx .

Bilinear form:

B(u, v) =

∫

Ω

[a(x)Du ·Dv + c(x)uv ] dx

our Hilbert space is clearly H1(Ω). The assertion of the theorem follows from Lax-Milgram

if:

B(u, v) =

∫

Ω

[a(x)Du ·Dv + c(x)uv ] dx

≤ a2

∫

Ω

|Du| · |Dv | dx + c2

∫

Ω

|u| · |v | dx

≤ a2||Du||L2(Ω)||Dv ||L2(Ω) + c2||u||L2(Ω)||v ||L2(Ω)

≤ C||u||H1(Ω)||v ||H1(Ω)

=⇒ B is bounded.

B(u, u) =

∫

Ω

(a(x)|Du|2 + c(x)|u|2) dx

≥ min(a1, c1)||u||2h1(Ω)

i.e., B is coercive. Gregory T. von Nessi

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10.8 Neumann Boundary Conditions325

Remarks:

1) We need c1 > 0 to get the L2-norm in the estimate, there is no Poincare-inequality in

H1

Poincare:

∫

Ω

|u|2 dx ≤ Cp∫

Ω

|Du|2 dx

and this fails automatically if the space contains non-zero constants.

2) This theorem does not allow us to solve−∆u = f in Ω

Du · ν = 0 on ∂Ω

∫

Ω

f dx =

∫

Ω

−∆u dx

I.P.= −

∫

∂Ω

Du · ν dx = 0

So, we have the constraint that∫

Ω f dx = 0.

10.8.1 Fredholm’s Alternative and Weak Solutions of the Neumann

Problem

−∆u + u + λu = f in Ω

Du · ν = 0 on ∂Ω

(i.e., a(x) = 1, c(x) = 1, −∆u + u = f has a unique solution with Neumann BCs)

Operator equation:

(−∆ + I + λ · I)u = f

∣∣∣∣∣ · (−∆ + I)−1

⇐⇒ [I + (−∆ + I)−1λ · I]u = (−∆ + I)−1(λ · I)⇐⇒ (I − T )u = f in H1(Ω), T = −(−∆ + I)−1(λ · I)

Fredholm applies to compact perturbations of I. We want to show T = (−∆ + I)−1(λ · I) is

compact.

(−∆ + I)−1 : H−1 → H1

λ · I : H1 → H−1, I = I1I0where I0 : H1(Ω) ⊂⊂−→ L2(Ω) for ∂Ω smooth

I1 : L2(Ω) ⊂−→ H−1(Ω)

=⇒ I compact =⇒ T compact.

Fredholm alternative: Either unique solution for all g or non-trivial solution of the homo-

geneous equation, (I − T )u = 0

⇐⇒ [I + (−∆ + I)−1λ · I]u = 0

⇐⇒ [(−∆ + I) + λ · I]u = 0

⇐⇒ [−∆ + (λ+ 1)I]u = 0

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Special case λ = −1 has non-trivial solutions, e.g. u ≡ constant.

Solvability criterion: We can solve if we in N⊥((I − T )∗) = N⊥(I − T ) since T = T ∗.Non-trivial solutions for λ = −1 satisfy

−∆u = 0 in Ω (in the weak sense)

Du · ν = 0 on ∂Ω

⇐⇒∫

Ω

Du ·Dv dx = 0 ∀v ∈ H1(Ω)

u = v :

∫

Ω

|Du|2 dx = 0

=⇒ u ≡constant (if Ω is connected)

=⇒ Null-space of adjoint operator is u ≡ constant=⇒ perpendicular to this means

∫Ω f dx = 0.

There is only uniqueness only up to constants.

Remark: Eigenfunctions of −∆ with Neumann BC, i.e. solutions of

−∆u = λu, Du · ν = 0

then λ = 0 is the smallest eigenvalue with a 1D eigenspace, the space of constant functions.

10.8.2 Fredholm Alternative Review

∆u = f in Ω

Du · ν = 0 on ∂Ω

H1 I1⊂−→L2 I0⊂−→H−1

I2 = I0I1

L = −∆u + u, −∆u + u + λu = f

L(u, v) =

∫

Ω

(Du ·Dv + vu) dx

L∗ : 〈L∗u, v〉 = L∗(u, v) = L(v , u) = 〈Lu, v〉

Here L = L∗, (Lu = (Diai jDju) L∗ has coefficients ai j)

Operator equation: u + L−1(λI2u) = L−1F in H1

As for Dirichlet conditions, it’s convenient to write the equation for L2:

I1u + I1L−1(λI2)u = I1L

−1F (u = I1u, F = I1L−1F )

⇐⇒ u + (I1L−1λI0)︸︷︷︸ u = F

cmpct. since I1 is cmpct.

This last equation is the same equation as for the Dirichlet problem, but L−1 has Neumann

conditions.

Application of Fredholm Alternative:Gregory T. von Nessi

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10.9 Semigroup Theory327

i.) Solution for all right hand sides.

ii.) Non-trivial solutions of the homogeneous equations.

iii.) If ii.) holds, solutions exists if F is perpendicular to the kernel of the adjoint.

What happens in ii.)? Check: if u is a solves u + (I1L−1λI0)u = 0 then u ∈ L−1(I0u) ∈ H1

solves u + L−1(λI2)u = 0 in H1 ⇐⇒ −∆u + u + λu = 0.

Now consider T to be our compact operator, defined by T = −I1L−1λI0.

Solvability of the original problem

(1) Check that T ∗ = −λI1(L∗)−1I0 = −λI1L−1I0

(2) F has to be orthogonal to N((I − T )∗) = N(I − T ) u non-trivial solution of u +

(I1L−1λI0)u = 0 =⇒ u = L−1I0u solves u + (L−1λI2)u = 0 ⇐⇒ u = constant (in

our special case λ = −1) ⇐⇒ u = constant.

(3) Solvability: then if F = I1L−1F ⊥ constants (this again is for the special case of

λ = −1).

⇐⇒ (F , v)L2 = 0 ∀v ∈ constant⇐⇒ · · · ⇐⇒ 〈F, v〉 = 0 ∀v ∈ constant

If F (v) =∫

Ω f v dx , if we solve −∆u = f in L2, then

∫

Ω

f v dx = 0 ∀v ∈ constant

=⇒∫

Ω

f dx = 0

So we conclude that

∆u = f in Ω

Du · ν = 0 on ∂Ω

has a solutions

⇐⇒∫

Ω

f dx = 0

10.9 Semigroup Theory

See [Hen81, Paz83] for additional information on this subject.

Idea: PDE ↔ ODE in a BS y = ay which we know to have a solution of y(t) = y0eat .

Now consider y = Ay , where A is a n × n matrix which is also symmetric

y(t) = eAty0, eAt =

∞∑

n=0

(At)n

n!= QetΛQT

if A = QΛQT , Λ diagonal

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Goal: Solve ut = Au (= ∆u) in a similar way by saying

u(t) = S(t)u0 (S(t) = eAt)

where S(t) : X → X is a bounded, linear operator over a suitable Banach space, with

• S(0)u = u, ∀u ∈ X

• S(t + s)u = S(t)S(s)u

• t 7→ S(t)u continuous ∀u ∈ X, t ∈ [0,∞)

Definition 10.9.1: i.) A is a family of bounded and linear maps S(t)t≥0 with S(t) :

X → X, X a BS, is called a semigroup (meaning that S has a group structure but no

inverses)

(1) S(0)u = u, ∀u ∈ X(2) S(t + s)u = S(t)S(s)u = S(s)S(t)u, ∀s, t > 0, u ∈ X(3) t 7→ S(t)u continuous as a mapping [0,∞)→ X, ∀u ∈ X

ii.) S(t)t≥0 is called a contraction semigroup if ||S(t)||X≤ 1, ∀t ∈ [0,∞)

Remark: It becomes clear that S−1 may not exist if one consider the reverse heat equation.

Solutions with smooth initial data will become very bad in finite time.

Definition 10.9.2: S(t)t≥0 semigroup, then define

D(A) =

u ∈ X : lim

t0+

S(t)u − ut

exists in X

,

where

Au = limt0+

S(t)u − ut

u ∈ D(A).

A : D(A)→ X is called the infinitesimal generator of S(t)t≥0, D is called its domain.

General assumption: S(t)t≥0 is a contraction semigroup

etA = QetΛQT , Λ diagonal

etΛ =

etλ1 0

. . .

0 etλn

with all λi ≤ 0,

where we recall that

etA =

∞∑

n=0

tn

n!An,

d

dtetA = AetA

Theorem 10.9.3 (): Let u ∈ D(A)

i.) S(t)u ∈ D(A), t ≥ 0

ii.) A · S(t)u = S(t)Au

iii.) t 7→ S(t)u is differentiable for t > 0Gregory T. von Nessi

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iv.) ddtS(t)u = A · S(t)u = S(t)Au where the last equality is due to ii.).

Proof:

v ∈ D(A) ⇐⇒ lims0+S(s)u−u

s exists

S(t)u ∈ D(A) ⇐⇒ lims0+S(s)S(t)u−S(t)u

s exists

= lims0+

S(t)S(s)u − S(t)u

s(by (2) in the def. of semigroup)

= S(t) · lims0+

S(s)u − us

= S(t)Au (by def. of A)

=⇒ the limit exists, i.e. S(t) ∈ D(A), and thus

lims0+

S(t)S(s)u − S(t)u

s= A · S(t)u = S(t)Au

=⇒ i.) and ii.) are proved.

iii.) indicates that backward and forward difference quotients exists. So take u ∈ D(A),

h > 0, t > 0

limh0+

[S(t)u − S(t − h)u

h− S(t)Au

]

= limh0+

[S(t − h)S(h)u − S(t − h)u

h− S(t)Au

]

= limh0+

[S(t − h)

S(h)u − uh

− S(t)Au

]

= limh0+

[S(t − h)︸︷︷︸||S(t−h)||

X≤1

(S(h)u − u

h− Au

)

︸︷︷︸||·||

X≤∣∣∣∣∣∣ S(h)u−u

u−Au

∣∣∣∣∣∣X︸︷︷︸→0 since u∈D(A)

+ (S(t − h)Au − S(t)Au)︸︷︷︸=(S(t−h)−S(t))Au︸︷︷︸

→0 by contin. of t 7→ S(t)u

]

=⇒ limit exists and

limh0+

S(t)u − S(t − h)u

h= S(t)Au

Analogous argument for forwards quotient; the limit is the same and

d

dtS(t)u = S(t)Au

= A · S(t)u

Remark: S(t) being continuous =⇒ ddtS(t) is continuous. So, S(t) is continuously differen-

tiable.

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Theorem 10.9.4 (Properties of infinitesimal generators): i.) For each u ∈ X,

limt0+

1

t

∫ t

0

G(s)u ds = u.

ii.) D(A) is dense in X

iii.) A is a closed operator, i.e. if uk ∈ D(A) with uk → u and Auk → v , then u ∈ D(A)

and v = Au

Proof:

i.): Take u ∈ X. By the continuity of G(·)u we have limt0+ ||G(t)u−u||X

= limt0+ ||G(t)u−

G(0)u||X

= 0. Conclusion i.) follows since u =1

t

∫ t

0

u ds.

ii.): Take u ∈ X and define

ut =

∫ t

0

S(s)u ds,

then

ut

t=

1

t

∫ t

0

S(s)u ds → S(0)u = u as t 0+.

ii.) follows if ut ∈ D(A), i.e.

S(r)ut − utr

=1

r

∫ t

0

S(r)S(s)u ds −∫ t

0

S(s)u ds

(semigroup prop.) =1

r

∫ t

0

S(r + s)u ds − 1

r

∫ t

0

S(s)u ds

=1

r

∫ t+r

r

S(s)u ds − 1

r

∫ t

0

S(s)u ds

=1

r

∫ t+r

t

S(s)u ds ± 1

r

∫ t

r

S(s)u ds

−1

r

∫ r

0

S(s)u ds ∓ 1

r

∫ t

r

S(s)u ds

=1

r

∫ t+r

t

S(s)u ds − 1

r

∫ r

0

S(s)u ds

taking the limit of both sides

limr0+

S(r)ut − utr

= S(t)u − u by continuity of S(t)

=⇒ ut ∈ D(A), Aut = S(t)u − u.

iii.): (proof of A being closed).

Take uk ∈ D(A) with uk → u and Auk → v in X. We need to show that u ∈ D(A)

and v = Au. So consider,

S(t)uk − uk =

∫ t

0

d

dt ′S(t ′)uk dt

′

=

∫ t

0

S(s) Auk︸︷︷︸→v

ds.


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Taking k →∞, we have

S(t)u − u =

∫ t

0

S(s)v ds

=⇒ limt0+

S(t)u − ut

= limt0+

1

t

∫ t

0

S(s)v ds = v

=⇒ Au = v with u ∈ D(A).

Definition 10.9.5: i.) We say λ ∈ ρ(A), the resolvent set of A, if the operator λI − A :

D(A)→ X is one-to-one and onto.

ii.) If λ ∈ ρ(a), then the resolvent operator Rλ : X → X is defined by Rλ = (λI − A)−1

Remark: The closed graph theorem says that if A : X → Y is a closed and linear operator,

then A is bounded. It turns out that, Rλ is closed, thus Rλ is therefore bounded.

Theorem 10.9.6 (Properties of the resolvent): i.) If λ, µ ∈ ρ(A), then

Rλ − Rµ = (µ− λ)RλRµ (resolvent identity)

which immediately implies

RλRµ = RµRλ

ii.) If λ > 0, then λ ∈ ρ(A) and

Rλu =

∫ ∞

0

e−λtS(t)u dt

and hence

||Rλ||X =1

λ(from estimating the integral)

Proof:

i.):

Rλ = Rλ(µI − A)Rµ

= Rλ((µ− λ)I + (λI − A))Rµ

= (µ− λ)RλRµ + Rλ(λI − A)︸︷︷︸I

Rµ

=⇒ Rλ − Rµ = (µ− λ)RλRµ=⇒ RλRµ = RµRλ

ii.): S(t) contraction semigroup, ||S(t)||X≤ 1. Define

Rλ =

∫ ∞

0

e−λtS(t) dt

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which clearly exists. Now fix h > 0, u ∈ X and compute

S(h)Rλu − Rλuh

=1

h

∫ ∞

0

[e−λtS(t + h)u − S(t)u] dt

=1

h

∫ ∞

h

e−λ(t−h)S(t)u dt − 1

h

∫ ∞

0

e−λtS(t)u dt

= −1

h

∫ h

0

e−λ(t−h)S(t)u dt +1

h

∫ ∞

0

(e−λ(t−h) − e−λt)S(t)u dt

= −eλh 1

h

∫ h

0

e−λtS(t)u dt

︸︷︷︸→−u

+eλh − 1

h︸︷︷︸→λ

∫ ∞

0

e−λtS(t)u dt

︸︷︷︸Rλu

.

Collecting terms we have

limh0+

S(h)Rλu − Rλuh

= −u + λRλu

=⇒ Rλu ∈ D(A), and

ARλu = −u + λRλu

⇐⇒ u = (λI − A)Rλu ∀x (10.27)

If u ∈ D(A), then

ARλu = A

∫ ∞

0

e−λtS(t) dt

(exer. in Evans) =

∫ ∞

0

e−λtA · S(t)u dt

=

∫ ∞

0

e−λtS(t) Au︸︷︷︸ dt

= Rλ(Au)

⇐⇒ Rλ(λI − A)u = λRλu − RλAu = λRλu − ARλu= (λI − A)Rλu

= u (by (10.27))

Now we need to prove the following.

Claim: λI − A is one-to-one and onto

• one-to-one: Assume (λI − A)u = (λI − A)v , u 6= v

(λI − A)(u − v) = 0

=⇒ Rλ(λI − A)(u − v) = 0,

u, v ∈ D(A) =⇒ u − v ∈ D(A)

=⇒ Rλ(λI − A)(u − v) = u − v 6= 0 a contradiction.

• onto: (λI − A)Rλu = u. Rλu = w solves (λI − A)w = u ⇐⇒ (λI − A) :

D(A)→ X is onto.Gregory T. von Nessi

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So we have (λI − A) is one-to-one and onto if λ ∈ ρ(A) and λ > 0. We also have

Rλ = (λI − A)−1 = Rλ.

Theorem 10.9.7 (Hille, Yoshida): If A closed and densely defined, then A is infinitesimal

generator of a contraction semigroup ⇐⇒ (0,∞) ⊂ ρ(A) and ||Rλ||X ≤ 1λ , λ > 0.

Proof:

=⇒) This is immediately obvious from the last part of Theorem 10.6.

⇐=) We must build a contraction semigroup with A as its generator. For this fix λ > 0 and

devine

Aλ := −λI + λ2Rλ = λARλ. (10.28)

The operator Aλ is a kind of regularized approximation to A.

Step 1. Claim:

Aλu → Au as λ→∞ (u ∈ D(A)). (10.29)

Indeed, since λRλu − u = ARλu = RλAu,

||λRλu − u|| ≤ ||Rλ|| · ||Au|| ≤1

λ||Au|| → 0.

Thus λRλu → u as λ → ∞ if u ∈ D(A). But since ||λRλ|| ≤ 1 and D(A) is dense,

we deduce then as well

λRλu → u as λ→∞, ∀u ∈ X. (10.30)

Now if u ∈ D(A), then

Aλu = λARλu = λRλAu.

Step 2. Next, define

Sλ(t) := etAλ = e−λteλ2tRλ = e−λt

∞∑

k=0

(λ2t)k

k!Rkλ.

Observe that since ||Rλ|| ≤ λ−1,

||Sλ(t)|| ≤ e−λt∞∑

k=0

λ2ktk

k!||Rλ||k ≤ e−λt

∞∑

k=0

λktk

k!= 1.

Consequently Sλ(t)t≥0 is a contrction semigroup, and it is easy to check its generator

is Aλ, with D(Aλ) = X.

Step 3. Let λ, µ > 0. Since RλRµ = RµRλ, we see AλAµ = AµAλ, and so

AµSλ(t) = Sλ(t)Aµ for each t > 0.

Tus if u ∈ D(A), we can compute

Sλ(t)u − Sµ(t)u =

∫ t

0

d

ds[Sµ(t − s)Sλ(s)u] ds

=

∫ t

0

Sµ(t − s)Sλ(s)(Aλu − Aµu) ds,

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because ddtSλ(t)u = AλSλ(t)u = Sλ(t)Aλu. Consequently, ||Sλ(t)u − Sµ(t)u|| ≤

t||Aλu − Aµu|| → 0 as λ, µ→∞. Hence

S(t)u := limλ→∞

Sλ(t)u exists for each t ≥ 0, u ∈ D(A). (10.31)

As ||Sλ(t)|| ≤ 1, the limit (10.31) in fact exists for all u ∈ X, uniformly for t in com-

pact subsets of [0,∞). It is now straightforward to verify S(t)t≥0 is a contraction

semigroup on X.

Step 4. It remains to show A is the generator of S(t)t≥0. Write B to denote this generator.

Now

Sλ(t)u − u =

∫ t

0

Sλ(s)Aλu ds. (10.32)

In addition

||Sλ(s)Aλu − S(s)Au|| ≤ ||Sλ(s)|| · ||Aλu − Au||+ ||(Sλ(s)− S(s))Au|| → 0

as λ→∞, if u ∈ D(A). Passing therefore to limits in (10.32), we deduce

S(t)u − u =

∫ t

0

S(s)Au ds

if u ∈ D(A). Thus D(A) ⊆ D(B) and

Bu = limt→0+

S(t)u − ut

= Au (u ∈ D(A)).

Now if λ > 0, λ ∈ ρ(A) ∩ ρ(B). Also (λI − B)(D(A)) = (λI − A)(D(A)) = X,

according to the assumptions of the theorem. Hence (λI −B)∣∣D(A)

is one-to-one and

onto; whence D(A) = D(B). Therefore A = B, and so A is indeed the generator of

S(t)t≥0.

Recall: In a contraction semigroup, we have ||S(t)||X≤ 1

Take A to be a n × n matrix, A = QAQT

||S(t)||X

= ||eAt ||X

= ||QeΛtQT ||X≤ 1

Since

eΛt =

eλ1t · · · 0

.... . .

...

0 · · · eλnt

,

we need all λi ≤ 0. In general, we can only expect that

||S(t)||X≤ Ceλt , λ > 0.

Remark: S(t)t≥0 is called an ω-contraction semigroup if

||S(t)||X≤ eωt , t ≥ 0

variant of the Hille-Yoshida Theorem:

If A closed and densely defined, then A is a generator of an ω-contraction semigroup ⇐⇒(ω,∞) ⊂ ρ(A), ||Rλ||X ≤ 1

λ−ω and λ > ωGregory T. von Nessi

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10.10 Applications to 2nd Order Parabolic PDE335

10.10 Applications to 2nd Order Parabolic PDE

ut + Lu = 0 in ΩT

u = 0 on ∂Ω× [0, T ]

u = g on Ω× t = 0

L strongly elliptic in divergence form, smooth coefficients independent of t.

Idea: Consider this as an ODE, and get U(t) = S(t)g, where S(t) is the semigroup gener-

ated by −L.

Choose: X = L2(Ω), A = −LD(A) = H2(Ω) ∩H1

0(Ω)

A is unbounded, but we have the estimate

||u||2H1

0(Ω)≤ L(u, u) + γ||u||2L2(Ω), γ ≥ 0

Theorem 10.10.1 (): A generates a γ-contraction semigroup on X.

Proof: Check the assumptions of the Hille-Yoshida Theorem:

• D(A) dense in X: This is clearly true since we can approximate f ∈ L2(Ω) even with

fk ∈ C∞c (Ω).

• A is a closed operator: This means that if we have uk ∈ D(A) with uk → u and

Auk → f , then u ∈ D(A) and Au = f .

To show this, set

Auk = fk , Aul − Auk︸︷︷︸−Lul+Luk

= fl − fk

Elliptic regularity: (Auk = fk is an elliptic equation)

||uk − ul ||H2(Ω) ≤ C[||fk − fl ||L2(Ω) + ||uk − ul ||L2(Ω)

]

(HW: Show this for −∆u + u = f ; L(u, u) =∫

Ω |Du|2 + |u|2 dx = ||u||2H1(Ω)

)

||uk − ul ||H2(Ω) ≤ C(||Auk − Aul ||L2(Ω)︸︷︷︸→0

+ ||uk − ul ||L2(Ω)︸︷︷︸→0

)

=⇒ uk is Cauchy in L2(Ω)

=⇒ u ∈ H2(Ω) ∩H10(Ω), u ∈ D(A)

=⇒ since u ∈ H2(Ω), we have Auk → Au. We also know that Auk → f (assumed),

thus Au = f in L2(Ω)

• Now we need to show (γ,∞) ⊂ ρ(A), ||Rλ|| ≤ 1λ−γ , λ > 0

Use Fredholm’s Alternative:Lu + λu = f in Ω

u = 0 on ∂Ω

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has a unique solution for all f ∈ L2(Ω) (by Lax-Milgram), and the solution u ∈H2(Ω) ∩H1

0(Ω)

=⇒ −Au + λu = f

⇐⇒ (λI − A)u = f ∈ L2(Ω), remembering u ∈ H2(Ω) ∩ H10(Ω), we see that in

particular (Iλ− A)−1 exists and

(λI − A)−1 : L2(Ω) = X → D(A)

=⇒ λ ∈ ρ(A) and Rλ : (λI − A)−1 exists.

What Remains: Resolvent estimate. Consider the weak form

L(u, v) + λ(u, v)L2(Ω) = (f , v)L2(Ω)

u = v : L(u, u) + λ||u||2L2(Ω) = (f , u)L2(Ω) ≤ ||f ||L2(Ω)||u||L2(Ω).

Remembering the estimate β||u||2H1

0(Ω)− γ||u||2

L2(Ω)≤ L(u, u), we see that

β||u||H10(Ω)2 + (λ− γ)||u||2L2(Ω) ≤ ||f ||L2(Ω)||u||L2(Ω)

=⇒ (λ− γ)||u||2L2(Ω) ≤ ||f ||L2(Ω)||u||L2(Ω)

=⇒ (λ− γ)||u||L2(Ω) ≤ ||f ||L2(Ω)

⇐⇒||Rλf ||L2(Ω)

||f ||L2(Ω)

≤ 1

λ− γ =⇒ ||Rλ|| ≤1

λ− γThis proves the resolvent estimate.

Now by the Hille-Yoshida Theorem, we have that A defines a γ-contraction semigroup.

Definition 10.10.2: If a function u ∈ C1([0, T ];X)∩C0([0, T ];D(A)) solves u = Au+ f (t),

u(0) = u0, then u is called a classical solution of the equation.

Theorem 10.10.3 (): Suppose that u is a classical solution, then u is represented by

u(t) = S(t)u0 +

∫ t

0

S(t − s)f (s) ds. (10.33)

(this is the formula given by the variation of constant method. Specifically S(t)u0 corresponds

to the homogeneous equation with initial data u0, and∫ t

0 S(t − s)f (s) ds corresponds to

the inhomogeneous equation with 0 initial data).

Proof: Differentiate.

Theorem 10.10.4 (): Suppose you have u0 ∈ D(A), f ∈ C0([0, T ];X) and that in addition

f ∈ W 1,1([0, T ];X) or f ∈ L2([0, T ];D(A)), then (10.33) gives a classical solution of the

abstract ODE u = Au + f , u(0) = u0.

Remark: Parabolic equation:ut + Lu = f (t) in Ω× [0, T ]

u = 0 on ∂Ω× [0, T ]

u = g in Ω× t = 0In particular, if f ∈ L2(Ω) and independent of t, and if u0 ∈ H2(Ω) ∩ H1

0(Ω) = D(A), then

there exists a classical solution of the parabolic equation.

Proofs of the last two theorems can be found in [RR04].Gregory T. von Nessi

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10.11 Exercises337

10.11 Exercises

9.1: [Qualifying exam 08/00] Let Ω ⊂ Rn be open and suppose that the operator L, formally

defined by

Lu = −n∑

i ,j=1

ai juxixj

has smooth coefficients ai j : Omega→ R. Suppose that L is uniformly elliptic, i.e., for some

c > 0,

n∑

i ,j=1

ai jξiξj ≥ c |ξ|2 ∀ξ ∈ Rn

uniformly in Ω. Show that if λ > 0 is sufficiently large, then whenever Lu = 0, we have

L(|Du|2 + λ|u|2) ≤ 0.

9.2: [Qualifying exam 01/02] Let B be the unit ball in R2. Prove that there is a constant C

such that

||u||L2(B) ≤ C(||Du||L2(B) + ||u||L2(B)) ∀u ∈ C1(B).

9.3: [Qualifying exam 01/02] Let Ω be the unit ball in R2.

a) Find the weak formulation of the boundary value problem

−∆u = f in Ω,

u ± ∂nu = 0 on ∂Ω.

b) Decide for which choice of the sign this problem has a unique weak solution u ∈ H1(Ω)

for all f ∈ L2(Ω). For this purpose you may use without proof the estimate

||u||L2(Ω) ≤ γ(||Du||L2(Ω) + ||u||L2(∂Ω)

)∀u ∈ H1(Ω).

c) For the sign that does not allow a unique weak solution, prove non-uniqueness by

finding two smooth solutions of the boundary value problem with f = 0.

9.4: [John, Chapter 6, Problem 1] Let n = 3 and Ω = B(0, π) be the ball centered at zero

with radius π. Let f ∈ L2(Ω). Show that a solution u of the boundary value problem

∆u + u = f in Ω

u = 0 on ∂Ω,

can only exists if

∫

Ω

f (x)sin |x ||x | dx = 0.

9.5: Let f ∈ L2(Rn) and let u ∈ W 1,2(Rn) be a weak solution of

−∆u + u = f in Rn,

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that is,∫

Rn(Du ·Dφ+ uφ) dx =

∫

Rnf φ dx ∀φ ∈ W 1,2(Rn).

Show that u ∈ W 2,2(Rn).

Hint: It suffices to prove that Dku ∈ W 1,2(Rn). In order to show this, check first that

uh = u(·+ hek) is a weak solution of the equation

−∆uh + uh = fh in Rn,

with fh = f (·+hek). Then derive an equation for (uh−u)/h and use uh−u as a test function.

9.6: [Qualifying exam 08/00] Let Ω be the unit ball in R2. Suppose that u, v : Ω → Rare smooth and satisfy

∆u = ∆v in Ω,

u = 0 on ∂Ω.

a) Prove that there is a constant C independent of u and v such that

||u||H1(Ω) ≤ C||v ||H1(Ω).

b) Show there is no constant C independent of u and v such that

||u||L2(Ω) ≤ C||v ||L2(Ω).

Hint: Consider sequences satisfying un − vn = 1.

9.7: [Qualifying exam 08/02] Let Ω be a bounded and open set in Rn with smooth boundary

Γ . Let q(x), f (x) ∈ L2(Ω) and suppose that∫

Ω

q(x) dx = 0.

Let u(x, t) be the solution of the initial-boundary value problem

ut − ∆u = q x ∈ Ω, t > 0,

u(x, 0) = f (x) x ∈ Ω,

Du · ν = 0 on Γ.

Show that ||u(·, t)− v ||W 1,2(Ω) → 0 exponentially as t →∞, where v is the solution of the

boundary value problem−∆v = q x ∈ Ω,

Dv · ν = 0 onΓ,

that satisfies∫

Ω

v(x) dx =

∫

Ω

f (x) dx.

9.8: [Qualifying exam 01/97] Let Ω ⊂ Rm be a bounded domain with smooth boundary, and

let u(x, t) be the solution of the initial-boundary value problem

utt − c2∆u = q(x, t) for x ∈ Ω, t > 0,

u = 0 for x ∈ ∂Ω, t > 0

u(x, 0) = ut(x, 0) = 0 for x ∈ Ω.Gregory T. von Nessi

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10.11 Exercises339

Let φn(x)∞n=1 be a complete orthonormal set of eigenfunctions for the Laplace operator ∆

in Ω with Dirichlet boundary conditions, so −∆φn = λnφn. Set ωn = c√λn. Suppose

q(x, t) =

∞∑

n=1

qn(t)φn(x),

where each qn(t) is continuous,∑q2n(t) < ∞ and |qn(t)| ≤ Ce−nt . Show that there are

sequences an and bn such that∑

(a2n + b2

n) <∞ and if

v(x, t) =

∞∑

n=1

[an cos(ωnt) + bn sin(ωnt)]φn(x),

then

||u(·, t)− v(·, t)||L2(Ω) → 0

at an exponential rate as t →∞.

9.9: [Qualifying exam 08/99] Let Ω ⊂ Rn be a bounded open set and ai j ∈ C(Ω) for

i , j = 1, . . . , n, with∑i ,j a

i j(x)ξiξj ≥ |ξ|2 for all ξ ∈ Rn, x ∈ Ω. Suppose u ∈ H1(Ω)∩C(Ω)

is a weak solution of

−∑

i ,j

∂j(ai j∂iu) = 0 in Ω.

Set w = u2. Show that w ∈ H1(Ω), and show that w is a weak subsolution, i.e., prove that

B(w, v) :=

∫

Ω

∑

i ,j

ai j∂jw∂iv dx ≤ 0

for all v ∈ H10(Ω) such that v ≥ 0.

9.10: [Qualifying exam 01/00] Let Ω = B(0, 1) be the unit ball in Rn. Using iteration

and the maximum principle (and other result with proper justification), prove that there

exists some solution to the boundary value problem

−∆u = arctan(u + 1) in Ω,

u = 0 on ∂Ω.

9.11: [Qualifying exam 01/97] Suppose f : Rn → R is continuous, nonnegative and has

compact support. Consider the initial value problem

ut − ∆u = u2 for x ∈ Rn, 0 < t < T,

u(x, 0) = f (x) for x ∈ Rn.

Using Duhamel’s principle, reformulate the initial value problem as a fixed point problem for

an integral equation. Prove that if T > 0 is sufficiently small, the integral equation has a

nonnegative solution in C(Rn × [0, T ]).

9.12: Let X = L2(R) and define for t > 0 the operator S(t0 : X → X by

(S(t)u)(x) = u(x + t).

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Show that S(t) is a semigroup. Find D(A) and A, the infinitesimal generator of the semi-

group.

Hint: You may use without proof that for f ∈ Lp(R)

||f (·+ h)− f (·)||Lp(R) → 0 as |h| → 0

for p ∈ [1,∞).


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342Chapter 11: Distributions and Fourier Transforms

11 Distributions and Fourier Transforms


– Alan Rickman

Contents

11.1 Distributions 348

11.2 Products and Convolutions of Distributions; Fundamental Solutions 354

11.3 Fourier Transform of S(Rn) 357

11.4 Definition of Fourier Transform on L2(Rn) 361

11.5 Applications of FTs: Solutions of PDEs 364

11.6 Fourier Transform of Tempered Distributions 365

11.7 Exercises 366

11.1 Distributions

[RR04]

11.1.1 Distributions on C∞ Functions

Definition 11.1.1: Suppose Ω non-empty, open set in Rn. f is a test function if f ∈ C∞c (Ω),

i.e., f is infinitely many time differentiable, supp(f ) ⊂ K ⊂⊂ Ω, K compact. The set of all

test functions f is denoted by D(Ω).

Definition 11.1.2: φnn∈N, φ ∈ D(Ω). Then φn → φ in D(Ω) if

i.) There exists a compact set K ⊂⊂ Ω, such that supp(φn) ⊆ K ∀n, supp(φ) ⊂ K.

ii.) φn and arbitrary derivatives of φn converge uniformly to φ and the corresponding deriva-

tives of φ.Gregory T. von Nessi

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11.1 Distributions343

Definition 11.1.3: A distribution or generalized function is a linear mapping φ 7→ (f , φ)

from D(Ω) into R which is continuous in the following sense: If φn → φ in D(Ω) then

(f , φn)→ (f , φ). The set of all distributions is denoted D′(Ω).

Example: If f ∈ C(Ω), then f gives a distribution Tf defined by

(Tf , φ) =

∫

Ω

f (x)φ(x) dx

Not all distributions can be identified with functions. If Ω = D(0, 1), we can define δ0 by

(δ0, φ) = φ(0).

Lemma 11.1.4 (): f ∈ D′(Ω), K ⊂⊂ Ω. Then there exists a nK ∈ N and a constant CKwith the following property:

|(f , φ)| ≤ CK∑

|α|≤nKmaxx∈K|Dαφ(x)|

for all test functions φ with supp(φ) ⊆ K.

Remarks:

1) The examples above satisfy this with n = 0.

2) If nK can be chosen independent of K, then the smallest n with this property is called

the order of the distribution.

3) There are distributions with n =∞Proof: Suppose the contrary. Then there exists a compact set K ⊂⊂ Ω and a sequence of

functions ψn with supp(ψn) ⊆ K and

|(f , ψn)| ≥ n∑

|α|≤nmaxx∈K|Dαψn(x)| > 0

Define φn = ψn(f ,ψn) . Then φn → 0 in D(Ω)

• supp(φn) ⊆ K ⊂⊂ Ω.

• |α| = k , n ≥ k . Then

|Dαφn(x)| =|Dαψn(y)|

(f , ψn)≤ Dαψn(y)

n∑|β|≤n maxx∈K |Dαψn(x)|

≤ 1

n

=⇒ Dαφn → 0 uniformly as n →∞.

However,

(f , φn) =(f , ψn)

(f , ψn)contradiction.

Example: We can define a distribution f by

(f , φ) = −φ′(0)

‖(δ0)′

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|(f , φ)| ≤ supx∈K |φ′(x)|, f is a distribution of order one.

Localization: G ⊂ Ω, G open, then D(G) ⊆ D(Ω). f ∈ D′(Ω) defines a distribution in

D′(G) simply by restriction. f ∈ D′(Ω) vanishes on the open set G if (f , φ) = 0 for all

φ ∈ D(G). With this in mind, we say that f = g on G if f − g vanishes on G. If f ∈ D′(Ω),

then one can show that there exists a largest open set G ⊂ Ω, such that f vanishes on G

(this can be the empty set). The complement of this set in Ω is called the support of f .

Example: Ω = B(0, 1), f = δ0, supp(δ0) = 0.

Definition 11.1.5: A sequence fn ∈ D′(Ω) converges to f ∈ D′(Ω) if (fn, φ)→ (f , φ) for all

φ ∈ D(Ω).

Example: Ω = R

fn(x) =

n 0 ≤ x ≤ 1

n

0 else

(fn, φ) =

∫

Rfn(x) dx

= n

∫ 1n

0

φ(x) dx =11n

∫ 1n

0

φ(x) dx = φ(0) = (δ0, φ)

=⇒ fn → δ0 in D′(Ω)

Analogously: Standard mollifiers, φ

0

1

−1 0 1

supp(φ) ⊆ [−1, 1],∫R φ(x) dx = 1

φε = ε−1φ(ε−1x) then φε → δ0 in D′(Ω) as ε→ 0.

Theorem 11.1.6 (): fn ∈ D′(Ω) such that (fn, φ) converges for all test function φ, then

there ∃f ∈ D′(Ω) such that fn → f .Gregory T. von Nessi

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Proof: (See Renardy and Rogers) Define f by

(f , φ) = limn→∞

(fn, φ).

Then f is clearly linear, and it remains to check the required continuity property of f .

Definition 11.1.7: • S(Rn) is the space of all C∞(Rn) functions such that

|x |k |Dαφ(x)|

is bounded for every k ∈ N, for every multi-index α.

• We say that φn → φ in S(Rn) if the derivatives of φn of all orders converge uniformly

to the derivatives of φ, and if the constants Ck,α in the estimate

|x |k |Dαφn(x)| ≤ Ck,α

are independent of n.

• S(Rn) = space of rapidly decreasing function or the Schwarz space.

Example: φ(x) = e−|x |2

11.1.2 Tempered Distributions

Definition 11.1.8: • A tempered distribution is a linear mapping φ 7→ (f , φ) from S(Rn)

into R which is continuous in the sense that (f , φn)→ (f , φ) if φn → φ in S(Rn).

• The set of all tempered distributions is denoted by S ′(Rn)

11.1.3 Distributional Derivatives

Definition 11.1.9: Let f ∈ D′(Ω), then the derivative of f with respect to xi is defined by

(∂f

∂xi, φ

)= −

(f ,∂φ

∂xi

)(11.1)

Remarks:

1) If f ∈ C(Ω), then the distributional derivative is the classical derivative, and (11.1) is

just the formula for integration by parts.

2) Analogously, definition for higher order derivatives: α ∈ Rn is a multi-index

(Dαf , φ) = (−1)|α|(f , Dαφ)

3) Taking derivatives of distributions is continuous in the following sense

if fn → f in D′(Ω), then

∂fn∂xi→ ∂f

∂xiin D′(Ω)

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4) We can define a “translated” distribution f (x + hei) by

(f (x + hei), φ) = (f , φ(x − hei)).

Then, we see that

∂f

∂xi= limh→0

f (x + hei)− f (x)

h

is a limit of difference quotients.

Example:

H(x) =

1 x > 0

0 x ≤ 0in D′(R)

−1 0 1

0

1

this is not weakly diffentiable

dHdx in the sense of distributions:

(dH

dx, φ

)= −

(H,dφ

dx

)= −

∫

RH(x)

dφ

dxdx

= −∫ ∞

0

dφ

dxdx = −φ

∣∣∣∣∞

0

= φ(0) = (δ0, φ)

Recall:

• D(Ω) = test functions, φn → φ in D(Ω) means that supp(φn) ⊂ K ⊂⊂ Ω and

Dαφn → Dαφ uniformly.

• D′(Ω) = distributions on Ω = linear functionals of D(Ω), (f , φn)→ (f , φ) if φn → φ

• S(Rn) = rapidly decreasing test functions, φn → φ in S(Rn) means that Dαφn → Dαφ

uniformly ∀α.

|x |k |Dαφn| ≤ Ck,α independent of n

• S ′(Rn) = tempered distributions = linear functionals of S(Rn)

• f ∈ D′(Ω), (Dαf , φ) = (−1)|α|(f , Dαφ)Gregory T. von Nessi

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Example: Ω ⊂ Rn open and bounded with smooth boundary, f ∈ C1(Ω). f defines a

distribution in D′(Ω)

(∂f

∂xj, φ

)= −

(f ,∂φ

∂xj

)= −

∫

Rnf (x)

∂φ

∂xj(x) dx = −

∫

Ω

f (x)∂φ

∂xj(x) dx

= −∫

∂Ω

f (x)φ(x)νj dS +

∫

Ω

∂f

∂xj(x)φ(x) dx

distributional derivatives have two terms: a volume term and a boundary term.

Proposition 11.1.10 (): Ω open, bounded and connected. u ∈ D′(Ω) such that ∇u = 0 (in

the sense of distributions), then u ≡ constant.

Proof: We do this only for n = 1; only a suitable induction is needed to prove the general

case.

Ω = I = (a, b); by assumption u′ = 0 or 0 = (u′, φ) = −(u, φ′)=⇒ (u, ψ) = 0 whenever ψ = φ′, φ ∈ D(I);

But ψ = φ′ ⇐⇒∫I ψ(s) ds = 0 and ψ ∈ C∞c (I).

Choose any φ0 ∈ D(I) with∫ ba φ0(s) ds = 1

φ ∈ D(I), φ(x) = [φ(x)− φ · φ0(x)]︸︷︷︸∫[φ(x) − φ ·φ0(x)] dx =

φ−φ∫ b

aφ0(x) dx︸︷︷︸

=1

= 0

+φ · φ0(x),

where φ =∫ ba φ(s) ds.

=⇒ φ(x) = ψ′(x) + φ · φ0(x)

=⇒ (u, φ) = (u, ψ′)︸︷︷︸=0

+(u, φ · φ0(x)) = (u, φ0)

∫ b

a

φ(x) dx

=⇒ u = (u, φ0) = constant.

Recall:

• D(Ω) = test functions, φn → φ in D(Ω) means that supp(φn) ⊂ K ⊂⊂ Ω and

Dαφn → Dαφ uniformly.

• D′(Ω) = distributions on Ω = linear functionals of D(Ω), (f , φn)→ (f , φ) if φn → φ

• S(Rn) = rapidly decreasing test functions, φn → φ in S(Rn) means that Dαφn → Dαφ

uniformly ∀α.

|x |k |Dαφn| ≤ Ck,α independent of n

• S ′(Rn) = tempered distributions = linear functionals of S(Rn)

• f ∈ D′(Ω), (Dαf , φ) = (−1)|α|(f , Dαφ)

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Example: Ω ⊂ Rn open and bounded with smooth boundary, f ∈ C1(Ω). f defines a

distribution in D′(Ω)(∂f

∂xj, φ

)= −

(f ,∂φ

∂xj

)= −

∫

Rnf (x)

∂φ

∂xj(x) dx = −

∫

Ω

f (x)∂φ

∂xj(x) dx

= −∫

∂Ω

f (x)φ(x)νj dS +

∫

Ω

∂f

∂xj(x)φ(x) dx

distributional derivatives have two terms: a volume term and a boundary term.

11.2 Products and Convolutions of Distributions; Funda-

mental Solutions

Difficulty: No good definition of f · g for f , g ∈ D′(Ω).

For example: f , g ∈ L1(Ω), but f · g /∈ L1loc(Ω), then f · g /∈ D′(Ω).

However, if f ∈ D′(Rp), g ∈ D′(Rq), then we can define f ·g ∈ D′(Rp+q) in the following

way:

Definition 11.2.1: If f ∈ D′(Rp), g ∈ D′(Rq), then we define f (x)g(x) ∈ D′(Rp+q) by

(f (x)g(y), φ(x, y)) = (f (x), (g(y), φ(x, y))︸︷︷︸consider x as a

parameter

)

Check:

• (g(y), φ(x, y)) is a test function.

• Continuity: (f (x)g(y), φn(x, y))→ (f (x)g(y), φ(x, y)).

• Commutative operation: f (x)g(y) = g(y)f (x). To check this, use that test functions

of the form φ(x, y) = φ1(x)φ2(y) are dense

Example:

(δ0(x)δ0(y), φ(x, y)) = (δ0(x), (δ0(y), φ(x, y)))

= (δ0(x), φ(x, 0))

= φ(0, 0) = δ0(x, y)

11.2.1 Convolutions of Distributions

Suppose f , g ∈ S(Rn) (D(Rn))

(f ∗ g, φ) =

∫

Rn(f ∗ g)(x)φ(x) dx

=

∫

Rn

∫

Rnf (x − y)g(y)φ(x) dydx

=

∫

Rn

[∫

Rnf (x − y)φ(x) dx

]g(y) dy

=

∫

Rn

∫

Rnf (x)g(y)φ(x + y) dxdy

= (f (x)g(y), φ(x + y))Gregory T. von Nessi

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11.2 Products and Convolutions of Distributions; Fundamental Solutions349

Suggestion: If f , g ∈ S ′(Rn), then

(f ∗ g, φ) = (f (x)g(y), φ(x + y))

Difficulty: φ(x + y) does not have compact support even if φ does.

y ∈ Rn

x ∈ Rn

x + y = 0

supp(φ(x))

Sufficient condition: We can define f ∗ g the following way: f (x)g(y) has compact support,

e.g., f or g has compact support. We then modify φ(x + y) outside the support to be

compactly supported/rapidly decreasing.

Example:

1)

(δ0 ∗ f , φ) = (δ0(x)f (y), φ(x + y))

(commutativity) = (f (y)δ0(x), φ(x + y))

= (f (y), (δ0(x), φ(x + y)))

= (f (y), φ(y))

= (f , φ)

=⇒ δ0 ∗ f = f

2) f ∈ D′(Rn), ψ ∈ D(Rn)

(f ∗ ψ, φ) = (f (x)ψ(y), φ(x + y))

= (f (x), (ψ(y), φ(x + y)))

=

(f (x),

∫

Rnψ(y)φ(x + y) dy

)

=

(f (x),

∫

Rnψ(y − x)φ(y) dy

)

(justified by Riemann Sums) =

∫

Rn(f (x), ψ(x − y))φ(y) dy

Therefore, we see that f ∗ ψ can be identified with the function (f (x), ψ(x − y)) ∈C∞(Rn) (this function being C∞ needs to be checked).

Remark: Suppose that fn → f in D′(Rn) and that either ∃K compact, K ⊂ Rn with

supp(fn) ⊆ K, or that g ∈ D′(Rn) has compact support, then

fn ∗ g → f ∗ g in D′(Rn)

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Theorem 11.2.2 (): D(Rn) is dense in D′(Rn).

Example: δ0 ↔ φε, where φε(x) = ε−nφ(ε−1x), φ ≥ 0 and∫Rn φ(x) dx = 1.

Idea of Proof:

(1) Distributions with compact support are dense in D′(Rn). To show this, choose test

functions φn ∈ D(Rn) with φn ≡ 1 on Bn(0). Now define φn · f ∈ D′(Rn), (φn · f , φ) =

(f , φn · φ) so that we have φn · f → f in D′(Rn) and φn · f has compact support.

(2) Distributions with compact support are limits of test functions. To show this choose

φε = ε−nφ(ε−1x), φ ≥ 0 and∫Rn φ(x) dx = 1 standard mollifier. Then φn ∗ f is a test

function and by the above remark,

φn ∗ f → δ0 ∗ f = f

11.2.2 Derivatives of Convolutions

(Dα(f ∗ g), φ) = (−1)|α|(f ∗ g,Dαφ)

= (−1)|α|(f (x)g(y), Dαφ(x + y))

= (−1)|α|(g(y), (f (x), Dαφ(x + y)))

= (g(y), (Dαf (x), φ(x + y)))

= (g(y)Dαf (x), φ(x − y))

= (Dαf ∗ g, φ) = (f ∗Dαg, φ)

11.2.3 Distributions as Fundemental Solutions

L(D) is a differential operator

∆ = L(D) =

n∑

j=1

aj ·D2j

ut − ∆u = L(D) = b ·Dt −n∑

j=1

aj ·D2j

Definition 11.2.3: Suppose that L(D) is a differential operator with constant coefficients.

A fundamental solution for L is a distribution G ∈ D′(Rn) satisfying L(D)G = δ0.

Remark: Why is this a good definition?

L(D)(G ∗ f ) = (L(D)G) ∗ f = δ0 ∗ f = f

The first equality of the above uses the result obtained for differentiating convolutions. So

if G ∗ f is defined (e.g. f has compact support), then u = G ∗ f is a solution of L(D)u = f


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11.3 Fourier Transform of S(Rn)351

Example: f (x) = 1|x | in R3

Check: f is up to a constant a fundamental solution for the Laplace operator.

(∆f , φ) = (f ,∆φ)

=

∫

R3

f (x)∆φ(x) dx

= limε→0

∫

R3\Bε(0)

1

|x |∆φ(x) dx

= limε→0

[−∫

R3\Bε(0)

(D

1

|x |

)·Dφ(x) dx −

∫

∂Bε(0)

1

rDφ(x) · ν dS

]

= limε→0

[ ∫

R3\Bε(0)

∆1

|x |︸︷︷︸=0

φ(x) dx +

∫

∂Bε(0)

D1

|x | · νφ(x) dx

︸︷︷︸:=I

−∫

∂Bε(0)

1

rDφ(x) · ν dS

︸︷︷︸∼Cε→0

]

So, from PDE I we know

I = −4π−∫

Bε(0)

φ(x) dS → −4πφ(0)

= −4π(δ0, φ).

This gives us the correct notion of what “∆G = δ0” really means.

11.3 Fourier Transform of S(Rn)

Definition 11.3.1: If f ∈ S(Rn) then we define

F f (ξ) = f (ξ) = (2π)−n2

∫

Rne−ix ·ξf (x) dx

F−1f (x) = f (x) = (2π)−n2

∫

Rne ix ·ξf (ξ) dξ

F is the Fourier transform

F−1 is the inverse Fourier transform.

Two important formulas

1. Dα(f (ξ) = F [(−ix)αf ](ξ)

2. (iξ)β f (ξ) = F [Dβf ](ξ)

Proof:

1.

Dαξ f (ξ) = Dαξ

[(2π)−

n2

∫


]

= (2π)−n2

∫

RnDαξ [e−ix ·ξf (x)] dx = F [(−ix)αf ](ξ)

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2.

(iξ)β f (ξ) = (2π)−n2

∫

Rn(iξ)βe−ix ·ξf (x) dx

= (2π)−n2

∫

Rn(−1)|β|(−iξ)βe−ix ·ξf (x) dx

= (2π)−n2

∫

Rn(−1)|β|Dβx e

−ix ·ξ · f (x) dx

I.P.= (2π)−

n2

∫

Rne−ix ·ξ ·Dβf (x) dx = F [Dβf ](ξ).

Proposition 11.3.2 (): F : S(Rn)→ S(Rn) is linear and continuous.

Proposition 11.3.3 (): F−1 : S(Rn)→ S(Rn) is linear and continuous.

Proof:

(−1)|β|+|α|ξβ ·Dαξ f (ξ) = (2π)−n2

∫

Rne−ix ·ξ Dβ(xαf (x))︸︷︷︸

finite sum of

rapidly decreas-

ing functions

dx

=⇒ f (ξ) is rapidly decreasing, f ∈ S(Rn)

continuous: fn → 0 in S(Rn) then fn → 0 in S(Rn). Suppose that fn → 0, show that

Dαfn → 0 uniformly ∀α ∈ Nn. It will be sufficient to show this for α = 0,

fn(ξ) = (2π)−n2

∫

Rne−ix ·ξfn(x) dx

= (2π)−n2

∫

BR(0)

e−ix ·ξf (x) dx +

∫

Rn\BR(0)

e−ix ·ξ f (x)︸︷︷︸≤ 1

|x |k

dx

(f (x) ≤ 1|x |k since f ∈ S(Rn)).

say k = n + 2

∫

Rn\BR(0)

|fn(x)| dx ≤∫

Rn\BR(0)

C

|x |n+2dx = C

∫ ∞

R

ρn−1

ρn+2dρ

= C

∫ ∞

R

1

ρ3dρ = C

1

R2

2nd integral as small as we want if R large enough.

1st integral as small as we want if n is large enough since fn → 0 uniformly. This analogously

for higher derivatives.

Theorem 11.3.4 (): If f ∈ S(Rn), then

F−1(F f ) = f , F(F−1f ) = f

i.e., F−1is the inverse of the Fourier transform.Gregory T. von Nessi

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11.3 Fourier Transform of S(Rn)353

Fourier Transform

(F f )(ξ) = f (ξ) = (2π)−n2

∫


(F−1f )(ξ) = f (ξ) = f (ξ) = (2π)−n2

∫

Rne ix ·ξf (x) dx

Example:

f (x) = e−|x |2/2, f (y) = e−|y |

2/2

f (y) = (2π)−n2

∫

Rne−ix ·ξe−|x |

2/2 dx!

= e−|y |2/2

It is sufficient to show this for n = 1.

(2π)−12

∫

Re−t

2/2e−it·u dt︸︷︷︸

= e−u2/2

∫

Re−t

2/2−it·u dx = e−u2/2

∫

Re−(t+iu)2/2 dt (completing the square)

Once way to evaluate the integral is to realize that e−z2/2 is analytic, thus by Cauchy we

know that integrating this on a closed contour will yield a value of 0.

γ4

R

ℑ

λ

C

−λ

γ2

γ1

γ3iu

γ1(s) = s, s ∈ (−λ, λ)γ2(s) = λ+ i s, s ∈ (0, u)γ3(s) = −s + iu, s ∈ (−λ, λ)γ4(s) = −λ+ i(u − s), s ∈ (0, u)

0 =

∫

C

e−z2/2 ds

=

∫ λ

−λe−s

2/2 ds +

∫ u

0

e−(λ+is)2/2 ids

︸︷︷︸:=I2

+

∫ λ

−λe−(−s+iu)2/2 (−1)ds

+

∫ u

0

e−(−λ+i(u−s))2/2 (−i)ds

Looking at the second integral:

I2 =

∫ u

0

e−λ2/2

︸︷︷︸→ 0 as

λ→∞

e−iλses2/2

︸︷︷︸≤C

ds → 0 as λ→∞

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Similarly, I4 → 0 as λ→∞. Thus, in the limit we have

limλ→∞

∫ λ

−λe−s

2/2 ds = limλ→∞

∫ λ

−λe−(−s+iu)2/2 ds

=⇒∫ ∞

−∞e−s

2/2 ds

︸︷︷︸=√

2π

=

∫ ∞

−∞e−(t+iu)2/2 dt =

∫ ∞

−∞e−t

2/2e−iuteu2/2 dt

Remark: The generalization of this is the following: if f (x) = e−ε|x |2, then f (ξ) = (2ε)−n/2e−|ξ|

2/4ε.

This is basically the same calculation as the previous example.

Theorem 11.3.5 (): f ∈ S(Rn), F−1F f = f , FF−1f = f .

Proof: f , g ∈ S(Rn)∫

Rng(ξ)f (ξ)e ix ·ξ dx =

∫

Rng(ξ)

[(2π)−

n2

∫

Rne−iy ·ξf (y) dy

]e ix ·ξ dξ

= (2π)−n2

∫

Rn

[∫

Rne−iξ(y−x)g(ξ) dξ

]f (y) dy

=

∫

Rng(y − x)f (y) dy =

∫

Rng(y)f (x + y) dy

Replace g(ξ) by g(ε · ξ). Need to computer g(ε·):

(2π)−n2

∫

Rne−iy ·ξg(ε · ξ) dξ = (2π)−

n2

1

εn

∫

Rne−iy ·

zε g(z) dz

=1

εng(yε

)

So, we see that∫

Rng(ε · ξ)f (ξ)e ix ·ξ dξ =

∫

Rng(ε·)(y) · f (x + y) dy

=1

εn

∫

Rng(yε

)f (x + y) dy

=

∫

Rng(y)f (x + εy) dy

Choose g(x) = e−|x |2/2 and take the limit ε→ 0.

g(0)

∫

Rnf (ξ)e ix ·ξ dξ = f (x)

∫

Rng(y) dy

= f (x)

∫

Rne−|y |

2/2 dy = (√

2π)−n2 f (x)

∴ f (x) = (2π)−n2

∫

Rne ix ·ξ f (ξ) dξ = ((F−1F)f )(x)

The proof of (FF−1f )(x) = f (x) is analogous.

Remark: Suppose L(D) is a differential operator with constant coefficients, i.e.,

L(D) =

n∑

i=1

D2i = −∆.


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11.4 Definition of Fourier Transform on L2(Rn)355

Then F(L(D)u) = L(iξ)u, where L(iξ) is just the symbol of the operator.

Example: Suppose we can extend Fourier transforms to L2(Rn) (see below), then with

L(D) = −∆ and L(iξ) =∑ξ2i = |ξ|2, we can find the solution of −∆u = f ∈ L2(Rn) by

Fourier Transform:

(−∆u) = f

⇐⇒ |ξ|2u(ξ) = f (ξ)

⇐⇒ u(ξ) =f (ξ)

|ξ|2

⇐⇒ u(x) = F−1

(f (ξ)

|ξ|2 f)

(x)

Theorem 11.3.6 (): f , g ∈ S(Rn), then (f , g) = (f , g), i.e., the Fourier Transform preserves

the inner-product

(f , g) =

∫

Rnf · g dx

Remark: Using a Fourier Transform, we have to use complex valued functions, |z |2 = z ·z .

Proof: Computer. g(x) = (F−1Fg)(x) = (R−1g)(x)

∫

Rnf (x)g(x) dx =

∫

Rnf (x) · (2π)−

n2

∫

Rne ix ·ξg(ξ) dξdx

=

∫

Rng(ξ) · (2π)−

n2

∫

Rnf (x)e ix ·ξ dxdξ

=

∫

Rng(ξ) · (2π)−

n2

∫

Rnf (x)e−ix ·ξ dxdξ

=

∫

Rng(ξ)f (ξ) dξ = (f , g)

11.4 Definition of Fourier Transform on L2(Rn)

Proposition 11.4.1 (): If u, v ∈ L1(Rn) ∩ L2(Rn), then u ∗ v = (2π)n/2u · v .

Proof:

u ∗ v(y) = (2π)−n2

∫

Rne−ix ·y (u ∗ v)(x) dx

= (2π)−n2

∫

Rne−ix ·y

[∫

Rnu(z)v(x − z) dz

]dx

= (2π)−n2

∫

Rne−iz ·yu(z)

∫

Rne−iy(x−z)v(x − z︸︷︷︸

sub.w

) dxdz

=

∫

Rne−izyu(z)v(y) dz

= (2π)n2 u(y)v(y)

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Theorem 11.4.2 (Plancherel’s Theorem): Suppose that u ∈ L1(Rn) ∩ L2(Rn), then u ∈L2(Rn) and

||u||L2(Rn) = ||u||L2(Rn)

Proof: Let v(x) = u(−x). Define w = u ∗ v . So, we have w(ξ) = (2π)n/2u(ξ)v(ξ).

Now w ∈ L1(Rn) and w is continuous (using Proposition 8.31). Now,

v(ξ) = (2π)−n2

∫

Rnu(−x) · e−ix ·ξ dx.

Let y = −x , so

v(ξ) = (2π)−n2

∫

Rnu(y) · e−iy ·ξ dy = (2π)−

n2

∫

Rnu(y) · e iy ·ξ dy = u(y)

Now we claim for any v , w ∈ L1(Rn),∫

Rnv(x) · w(x) dx =

∫

Rnv(ξ) · w(ξ) dξ.

Check:∫

Rnv(x) · w(x) dx =

∫

Rnv(x) ·

[(2π)−

n2

∫

Rnw(ξ)e−ix ·ξ dξ

]dx

=

∫

Rn

[(2π)−

n2

∫

Rnv(x)e−ix ·ξ dx

]· w(ξ) dξ

=

∫

Rnv(ξ) · w(ξ) dξ

Now set f (ξ) = e−ε|ξ|2

so f (x) = (2ε)−n/2e−|ξ|2/4ε (as shown in the example earlier). Then,

∫

Rnw(ξ)e−ε|ξ|

2

dξ = (2ε)−n2

∫

Rnw(x)e−ε|x |

2/4ε dx.

Now, take the limit as ε→ 0 to get∫

Rnw(ξ) dξ = (2π)

n2w(0)

(w(0) because the integral was a limit representation of the dirac distribution). But we know

w = u ∗ v and w(ξ) = (2π)n2 u · v(ξ) = (2π)

n2 u · u(ξ) = (2π)

n2 |u|2,

so

(2π)n2w(0) =

∫

Rnw(ξ) dξ = (2π)

n2

∫

Rn|u|2 dξ =⇒ w(0) =

∫

Rn|u|2 dξ.

We also know

w = u ∗ v =⇒ w(0) =

∫

Rnu(y) · v(−y) dy =

∫

Rnu(y) · u(y) dy =

∫

Rn|u|2 dy.

Thus,∫

Rn|u|2 dξ = w(0) =

∫

Rn|u|2 dy =⇒ ||u||L2(Rn) = ||u||L2(Rn)


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11.5 Applications of FTs: Solutions of PDEs357

Definition 11.4.3: If u ∈ L2(Rn), choose uk ∈ L1(Rn) ∩ L2(Rn) such that uk → u in

L2(Rn). By Plancherel’s Theorem:

||uk − ul ||L2(Rn) = ||uk − ul ||L2(Rn)

= ||uk − ul ||L2(Rn) → 0 as k, l →∞

=⇒ uk is Cauchy in L2(Rn). The limit is defined to be u ∈ L2(Rn).

Note: This definition does not depend on the approximating sequence.

Remarks:

1) We have

||u||L2(Rn) = ||u||L2(Rn)

if u ∈ L2(Rn) by approximation of u in L2(Rn) by functions uk ∈ L1(Rn) ∩ L2(Rn);

Plancherel’s Theorem.

2) You can use this to find integrals of functions. Suppose u, u, then∫

R|u|2 dx =

∫

R|u|2 dx

Theorem 11.4.4 (): If u, v ∈ L2(Rn), then

i.)

(u, v) = (u, v), i.e.,

∫

Rnu · v dx =

∫

Rnu · v dx

ii.) (Dαu)(y) = (iy)αu(y), Dαu ∈ L2(Rn)

iii.) (u ∗ v) = (2π)n/2u · v

iv.) u = ˜u = F−1[F(u)]

Proofs: See Evans.

11.5 Applications of FTs: Solutions of PDEs

(1) Solve −∆u + u = f in Rn, f ∈ L2(Rn).

The Fourier Transform converts the PDE into an algebraic expression.

|y |2u + u = f

⇐⇒ u =f

|y |2 + 1

⇐⇒ u = F−1

(f

|y |2 + 1

)= F−1(f · B),

where B = 1|y |2+1

. Since u ∗ v = (2π)n/2u · v ,

u = f · B = (2π)−n2 (f ∗ B)

u = (2π)−n2 (f ∗ B).

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Now we need to find B:

Trick:1

a=

∫ ∞

0

e−ta dt

=⇒ 1

|y |2 + 1=

∫ ∞

0

e−t(|y |2+1) dt.

So,

B(x) = (F−1B)(x)

= (2π)−n2

∫

Rne ix ·y

∫ ∞

0

e−t(|y |2+1) dtdy

= (2π)−n2

∫ ∞

0

e−t∫

Rne ix ·ye−t|y |

2

dydt

(change var.) =

∫ ∞

0

e−t(2t)−n2 e−|x |

2/4t dt.

This last integral is called the Bessel Potential.

So, we at last have an explicit representation of our solution:

u(x) = (2π)−n2 (B ∗ f )

= (4π)−n2

∫ ∞

0

∫

Rnt−

n2 e−t−|x−y |

2/4t f (y) dydt

(2) Heat equation:

ut − ∆u = 0 in Rn × (0,∞)

u = g on Rn × t = 0

Take the Fourier transform in the spatial variables:

ut(y)− |y |2u(y) = 0 in Rn × (0,∞)

u(y) = g(y) on Rn × t = 0

Now, we can view this as an ODE in t:

u(y , t) = g(y)e−|y |2t

=⇒ u(x, t) = F−1[g(y)e−|y |2t ](x)

= (2π)−n2 (g ∗ F ),

where

F = e−|y |2t .

Changing variables as before,

F (x) = (2t)−n2 e−|x |

2/4t

=⇒ u(x, t) = (2π)−n2 (g ∗ F )(x, t)

= (4πt)−n2

∫

Rng(y)e−|x−y |

2/4t dy

This is just the representation formula from the last term!Gregory T. von Nessi

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11.6 Fourier Transform of Tempered Distributions359

11.6 Fourier Transform of Tempered Distributions

Definition 11.6.1: If f ∈ S ′(Rn), then we define

(F(f ), φ) = (f ,F−1(φ))

Check that the following holds:

Dαf = F((−ix)αf )

(iξ)f = F(Dαf )

Examples:

(1) f = δ0,

〈F(δ0), φ〉 = 〈δ0,F−1φ〉 = (F−1φ)(0)

= (2π)−n2

∫

Rnφ(x) dx

=⟨

(2π)−n2 , φ⟩

=⇒ F(δ0) = (2π)−n2

(2) F(1), 1 = constant function.

〈F(1), φ〉 = 〈1,F−1φ〉

=

∫

RnF−1(φ)(x) dx

= (2π)n2FF−1(φ)(0)

= (2π)n2φ(0)

=⟨

(2π)n2 δ0, φ

⟩

11.7 Exercises

11.1: Show that the Dirac delta distribution cannot be identified with any continuous func-

tion.

11.2: Let Ω = (0, 1). Find an example of a distribution f ∈ D′(Ω) of infinite order.

11.3: Let a ∈ R, a 6= 0. Find a fundamental solution for L = ddx − a on R.

11.4: Find a sequence of test functions that converges in the sense of distributions in R to δ′.

11.5: Let Ω = B(0, 1) be the unit ball in R2, and let w ∈ W 1,p(B(0, 1);R2) be a vector

valued function, that is, w(x) = (u(x), v(x)) with u, v ∈ W 1,p(Ω). Recall that Dw, the

gradient of w, is given by

Dw =

(∂1u ∂2u

∂1v ∂2v

).

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a) Suppose w is smooth. Show that

∂1(u · ∂2v)− ∂2(u · ∂1v) = detDw.

b) Let w ∈ W 1,p(B(0, 1);R2). We want to define a distribution DetDw ∈ D′(Ω) by

〈DetDw, φ〉 = −∫

Ω

(u · ∂2v · ∂1φ− u · ∂1v · ∂2φ) dx.

Find the smallest p ≥ 1 such that DetDw defines a distribution, that is, the integral

on the right hand side exists.

c) It is in general not true that DetDw = detDw. Here is an example; let w(x) = x/|x |.Show that there exists a constant c such that

DetDw = cδ0,

but that detDw(x) = 0 for a.e. x ∈ Ω.

11.6: [Qualifying exam 08/98] Assume u : Rn → R is continuous and its first-order distri-

bution derivatives Du are bounded functions, with |Du(x)| ≤ L for all x . By approximating

u by mollification, prove that u is Lipschitz, with

|u(x)− u(y)| ≤ L|x − y | ∀x, y ∈ Rn.

11.7: [Qualifying exam 01/03] Consider the operator

A : D(A)→ L2(Rn)

where D(A) = H2(Rn) ⊂ L2(Rn) and

Au = ∆u for u ∈ D(A).

Show that

a) the resolvent set ρ(A) contains the interval (0,∞), i.e., (0,∞) ⊂ ρ(A);

b) the estimate

||(λI − A)−1|| ≤ 1

λfor λ > 0

holds.

11.8: Let f : R→ R be defined by

f (x) =

sin x if x ∈ (−π, π)

0 else.

Show that

f (ξ) = i

√2√π

sinπξ

ξ2 − 1.

11.9: We define for real-valued functions f and s ≥ 0

Hs(Rn) =

f ∈ L2(Rn) :

∫

Rn(1 + |ξ|2)s |f (ξ)|2 dξ <∞

.


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11.7 Exercises361

a) Show that

f (ξ) = f (−ξ).

b) Let f ∈ Hs(Rn) with s ≥ 1. Show that g(x) = F−1(iξf (ξ))(x) is a (vector-valued)

function with values in the real numbers.

c) Prove that H1(R) = W 1,2(R). (This identity is in fact true in any space dimension for

s ∈ N).

d) Show that f ∈ C0(Rn) if f ∈ L1(Rn). Analogously, if f ∈ L1(Rn), then f ∈ C0(Rn)

(you need not show the latter assertion).

e) Show that Hs(R) ⊂−→ C0(R) if s > n2 . Hint: Use Holder’s inequality.

11.10:

a) Find the Fourier transform of χ[−1,1], the characteristic function of the interval [−1, 1].

b) Find the Fourier transform of the function

f (x) =

2 + x for x ∈ [−2, 0],

2− x for x ∈ [0, 2],

0 else.

Hint: What is χ[−1,1] ∗ χ[−1,1]?

c) Find

∫

R

sin2 ξ

ξ2dξ.

d) Show that χ[−1,1] ∈ H12−ε(R) for all ε ∈

(0, 1

2

].

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362Chapter 12: Variational Methods

12 Variational Methods


– Alan Rickman

Contents

12.1 Direct Method of Calculus of Variations 368

12.2 Integral Constraints 377

12.3 Uniqueness of Solutions 384

12.4 Energies 384

12.5 Weak Formulations 385

12.6 A General Existence Theorem 387

12.7 A Few Applications to Semilinear Equations 390

12.8 Regularity 392

12.9 Exercises 400

12.1 Direct Method of Calculus of Variations

We wish to solve −∆u = 0, by using minimizers

F (u) =1

2

∫

Ω

|Du|2 dx

Crucial Question: Can we decide whether there exists a minimizer?

General Question: If X a BS, || · ||, and F : X → R, is there a minima of F?

The answer will be yes, if we have the following:

• F bounded from below

let M = infx∈X

F (x) dx > −∞Gregory T. von Nessi

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12.1 Direct Method of Calculus of Variations363

• Can choose a minimizing sequence xn ∈ X such that

F (xn)→ M as n →∞

• Bounded sets are compact.

Suppose (subsequence) xk → x

• Continuity property:

F (xn)→ F (x) = M, x a minimizer

Turns out the lower-semicontinuity will be enough:

M = limk→∞

inf F (xk) ≥ F (x) ≥ M=⇒ F (x) = M, x minimizer

There is a give and take between convergence and continuity.

Strong Convergence Weak Convergence

Compactness difficult easy

Lower Semicontinuity easy difficult

Our Choice

Definition 12.1.1: 1 < p <∞. A function F is said to be (sequentially) weakly lower semicontinuous

(swls) on W 1,p(Ω) if uk u in W 1,p(Ω) implies that

F (u) ≤ limk→∞

inf F (uk)

Example: Dirichlet integral

F (u) =

∫

Ω

1

2|Du|2 dx

and the generalization

F (u) =

∫

Ω

(1

2|Du|2 + f u

)dx

with f ∈ L2(Ω) in X = H10(Ω)

Goal: ∃! minimizer in H10(Ω) which solves −∆u = f in the weak sense.

(1) Lower bound on F :

F (u) =

∫

Ω

(1

2|Du|2 − f u

)dx

Holder≥∫

Ω

1

2|Du|2 dx − ||f ||L2(Ω)||u||L2(Ω)

Now we use Young’s inequality with(√

εa − b2√ε

)2≥ 0 =⇒ ab ≤ εa2 + b2

4ε

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Young’s≥∫

Ω

1

2|Du|2 dx − ε||u||2L2(Ω) −

1

4ε||f ||2L2(Ω)

——

Recall Poincare’s inequality in H10(Ω): ∃Cp such that ||u||2

L2(Ω)≤ Cp||Du||2L2(Ω)

——

Poincare≥∫

Ω

1

2|Du|2 dx − ε · Cp

∫

Ω

|Du|2 dx − 1

4ε||f ||2L2(Ω)

Now, choose ε = 14Cp

,

=⇒ F (u) ≥ 1

4

∫

Ω

|Du|2 dx︸︷︷︸

≥0

− Cp

∫

Ω

|f |2 dx︸︷︷︸

fixed number and

finite since f ∈L2(Ω)

≥ C > −∞

Recall:

Goal: Want to solve −∆u = f by variational methods, i.e., we need to prove the existence

of a minimizer of the functional

F (u) =

∫

Ω

(1

2|Du|2 − f u

)dx.

We can do this via the direct method of the calculus of variations. This entails

• Showing F is bounded from below.

• choosing a minimizing sequence xn.

• Show that this minimizing sequence is bounded, so we can use weak compactness of

bounded sets to show xn x .

• Prove the lower semicontinuity of F

M ≤ F (X) ≤ limn→∞

inf F (xn) = M

So we start where we left off:

(2)

1

4

∫

Ω

|Du|2 dx −Cp ·∫

Ω

|f |2 dx︸︷︷︸−Cp·||f ||2

L2(Ω)

≤ F (u) (12.1)

=⇒ F (u) ≥ −Cp · ||f ||2L2(Ω) > −∞=⇒ F is bounded from below.


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(3) Define

M = infu∈H1

0(Ω)F (u) > −∞.

Choose un ∈ H10(Ω) such that F (un)→ M as n →∞

(4) Boundedness of un: choose n ≥ n0 large enough such that F (un) ≤ M + 1 for

n ≥ n0. (12.1) tells us that

1

4

∫

Ω

|Dun|2 dx ≤ Cp

∫

Ω

|f |2 dx + F (un)

≤ Cp

∫

Ω

|f |2 dx +M + 1 for n ≥ n0

——

Poincare: ||un||2L2(Ω) ≤ Cp||Dun||2L2(Ω)

——

=⇒ un is bounded in H10(Ω).

=⇒ weak compactness: ∃ a subsequence unj such that unj u in H10(Ω) as j →∞,

and

compact Sobolev embedding: ∃ a subsequence of unj such that unjk → u in L2(Ω).

Call ukk∈N this subsequence.

(5) weak lower semicontinuity: Need

F (u) ≤ limk→∞

inf F (uk)

= limk→∞

inf

∫

Ω

(1

2|Duk |2 − f uk

)dx

Now since uk → u in L2(Ω), we know∫

Ω

f uk dx →∫

Ω

f u dx.

A question remains. Does the following relation hold?

1

2

∫

Ω

|Du|2 dx ≤ limk→∞

inf1

2

∫

Ω

|Duk |2 dx

To see this we utilize the following:

Note:

1

2|p|2 =

1

2|q|2 + (q, p − q) +

1

2|p − q|2

= −1

2|q|2 + p · q

⇐⇒ 1

2|p|2 +

1

2|q|2 − p · q =

1

2|p − q|2

⇐⇒ 1

2[|p|2 − 2p · q + |q|2] =

1

2|p − q|2

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Choose p = Duk and q = Du.

∫

Ω

1

2|Duk |2 dx =

∫

Ω

[1

2|Du|2 + (Du,Duk −Du) +

1

2|Duk −Du|2︸︷︷︸

≥0

]dx

≥∫

Ω

1

2|Du|2 dx +

∫

Ω

Du · (Duk −Du)︸︷︷︸0 in L2(Ω)

dx

︸︷︷︸→0 as k→0

=⇒ limk→∞

inf

∫

Ω

1

2|Duk |2 dx ≥

∫

Ω

1

2|Du|2 dx + lim

k→∞inf

∫

Ω

Du · (Duk −Du) dx

=

∫

Ω

1

2|Du|2 dx

——

Trick: The inequality

1

2|p|2 ≥ 1

2|q|2 + (q, p − q)

allows us to replace the quadratic term in Duk by a term that is linear in Duk .

——

=⇒ u is a minimizer!

(6) Uniqueness: Use again that

1

2|p|2 =

1

2|q|2 + (q, p − q) +

1

2|p − q|2

Suppose we have two minimizers u, v . Define w = 12 (u+ v). Now we choose p = Du,

q = 12 (Du +Dv), p − q = Du − 1

2 (Du +Dv) = 12 (Du −Dv).

=⇒∫

Ω

1

2|Du|2 dx =

∫

Ω

1

2|Dw |2 +

1

2

∫

Ω

Dw · (Du −Dv) dx +1

2

∣∣∣∣1

2(Du −Dv)

∣∣∣∣2dx

We apply the same inequality with p = Dv , q = Dw and p−q = Dv − 12 (Du+Dv) =

12 (Dv −Du).

=⇒∫

Ω

1

2|Dv |2 dx =

∫

Ω

1

2|Dw |2 +

1

2Dw · (Dv −Du) +

1

2

∣∣∣∣1

2(Dv −Du)

∣∣∣∣2dx

Now we sum these inequalities.

1

2

∫

Ω

|Du|2 dx +1

2

∫

Ω

|Dv |2 dx =

∫

Ω

|Dw |2 dx +1

4

∫

Ω

|Du −Dv |2 dx.

Now we remember that u and v are minimizers of

F (u) =

∫

Ω

(1

2|Du|2 − f u

)dx.


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So, we calculate∫

Ω

(1

2|Du|2 − f u

)dx +

∫

Ω

1

2(|Dv |2 − f v) dx

=

∫

Ω

|Dw |2 − (u + v)f︸︷︷︸=2wf

dx +1

4

∫

Ω

|Du −Dv |2 dx

= 2

∫

Ω

1

2|Dw |2 − wf dx +

1

4

∫

Ω

|Du −Dv |2 dx

=⇒ M +M ≥ 2M +1

4

∫

Ω

|Du −Dv |2 dx

=⇒ 0 ≥ 1

4

∫

Ω

|Du −Dv |2 dx

=⇒ Du = Dv , u, v ∈ H10(Ω) =⇒ u = v =⇒ uniqueness!

Recall: To solve−∆u = f in Ω

u = 0 on ∂Ω,

we looked for a minimizer in H10(Ω) of the functional

F (u) =

∫

Ω

(1

2|Du|2 − f u

)dx

Already done: Existence and uniqueness of minimizer.

(7) Euler-Lagrange equation: We have the following necessary condition. If v ∈ H10(Ω),

ε 7→ F (u + εv), then this function has a minimum at ε = 0 and

d

dε

∣∣∣∣∣ε=0

F (u + εv) = 0.

So for our Dirichlet integral we have

d

dε

∣∣∣∣∣ε=0

∫

Ω

[1

2|Du + ε ·Dv |2 − f (u + εv)

]dx =

∫

Ω

(Du ·Dv − f v) dx = 0

i.e.

∫

Ω

(Du ·Dv − f v) dx = 0 ∀v ∈ H10(Ω)

i.e., u is a weak solution of Laplace’s equation.

Remarks:

1) Laplace’s equation has a variational structure, i.e., its weak formulation is the Euler-

Lagrange equation of a variational problem.

2) f : R→ R is lower semicontinuous if xn → x and

f (x) ≤ limn→∞

inf f (xn).

f continuous =⇒ f lower semicontinuous.

If f discontinuous at x , we have the following situations:

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x

lower semicontinuous upper semicontinuous

x

3) Weak convergence is not compatible with non-linear functions: if fn f in L2(Ω),

and if g is a non-linear functions, then g(fn) g(f ).

Example: fn = sin(nx) on (0, π), fn 0 in L2((0, π)).

Need:

∫ π

0

fng dx → 0 ∀g ∈ L2((0, π)).

It will be sufficient to show this for φ ∈ C∞c ((0, π)).

∫ π

0

sin(nx)φ(x) dx =

∫ π

0

d

dx

(−1

ncos(nx)

)φ(x) dx

=

∫ π

0

cos(nx)︸︷︷︸|·|≤1

φ′(x) dx

≤ 1

n·max |φ′|

∫ π

0

dx

=π

n·max |φ′| → 0 as n →∞

Now choose g(x) = x2, g(fn) g(f )? NO

limn→∞

∫ π

0

sin2(nx)φ(x) dx = limn→∞

∫ π

0

d

dx

[x

2− 1

4nsin(2nx)

]φ(x) dx

= limn→∞

∫ π

0

1

2φ(x) dx + lim

n→∞1

4n

∫ π

0

sin(2nx)φ′(x) dx

︸︷︷︸=0

=

∫ π

0

1

2φ(x) dx.

So, sin2(nx) 12 in L2((0, π))

So, we need lower semicontinuity since F (u) will in general not be linear (for our

Dirichlet integral |Du|2 is not linear), and thus passing a weak limit within F (u) isn’t

justified. Lower semicontinuity saves this by trapping the weak sequence as follows

infu∈H1

0(Ω)F (u) ≤ F (u) ≤ lim

k→∞inf F (uk)→ inf

v∈H10(Ω)

F (v)


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=⇒ F (u) = infv∈H1

0(Ω)F (v)

4) Crucial was

1

2|p|2 ≥ 1

2|q|2 + (q, p − q)

Define F (p) = 12 |p|2 =⇒ F ′(p) = p.

We need: F (p) ≥ F (q) + (F ′(q), p − q) with q fixed, p ∈ Rn, i.e., F is convex.

Proposition 12.1.2 (): i.) Let f be in C1(Ω), then f is convex, i.e., f (λp+ (1− λ)q) ≤λf (p) + (1− λ)f (q) for all p, q ∈ Rn and λ ∈ [0, 1] ⇐⇒

f (p) ≥ f (q) + (f ′(q), p − q) ∀p, q ∈ Rn

ii.) If f ∈ C2(Ω), then f is convex ⇐⇒ D2f is positive semidefinite, i.e.,

n∑

i ,j=1

Di j f (p)ξiξj ≥ 0, ∀p ∈ Rn,∀ξ ∈ Rn

We needed for uniqueness

1

2|p|2 =

1

2|q|2 + (q, p − q) +

1

2|p − q|2.

The argument would work if

1

2|p|2 ≥ 1

2|q|2 + (q, p − q) + λ|p − q|2 λ > 0.

Now take F (p) = 12 |p|2.

F (p) = F (q) + (F ′(q), p − q) +1

2(F ′′(q)(p − q), p − q)

We know F ′(p) = p and F ′′(p) = I. So,

(F ′′(q)(p − q), (p − q)) = |p − q|2.For general F , this argument works if D2F (q) is uniformly positive definite, i.e., ∃λ > 0 such

that

(D2F (q)ξ, ξ) ≥ λ|ξ|2 ∀q ∈ Rn,∀ξ ∈ Rn (12.2)

Definition 12.1.3: f ∈ C2(Ω) is uniformly convex if (12.2) holds.

Theorem 12.1.4 (): Suppose that f ∈ C2(Ω) and f is uniformly convex, i.e., there exists

constants α > 0, β ≥ 0 such that

f (p) ≥ α|p|2 − β,then the variational problem:

minimize F (u) in H10(Ω)

with

F (u) =

∫

Ω

(f (Du)− gu) dx

has a unique minimizer which solves the Euler-Lagrange equation −div (f ′(Du)) = g in Ω

and u = 0 on ∂Ω.

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Proof:

• F (u) bounded from below since

F (u) =

∫

Ω

(f (Du)− gu) dx

≥∫

Ω

(α|Du|2 − β − gu) dx

≥ α

2

∫

Ω

|Du|2 dx − C > −∞

by Holder, Young and Poincare inequalities.

• weak lower semicontinuity since f is convex.

• uniqueness since f uniformly convex.

• Euler-Lagrange equations:

d

dε

∣∣∣∣∣ε=0

∫

Ω

f (Du + ε ·Dv)− g(u + εv) dx

=

∫

Ω

(∂f

∂pi(Du) · ∂v

∂xi− gv

)dx

=

∫

Ω

(−div f ′(Du)− g

)v dx ∀v ∈ H1

0(Ω)

Recall: Given a functional

F (u) =

∫

Ω

(1

2|Du|2 − f u

)dx,

we have shown the following:

• If we have weak lower semicontinuity AND

• convexity, we have existence of a minimizer.

• If we have uniform convexity, we get uniqueness of a minimizer.

i.e., given

F (u) =

∫

Ω

F(Du, u, x) dx,

p 7→ F(p, z, x) is convex.

Euler-Lagrange equation:

d

dε

∣∣∣∣∣ε=0

F (u + εv) = 0 ∀v ∈ H10(Ω)

We calculate

d

dε

∣∣∣∣∣ε=0

F (u + εv) =d

dε

∣∣∣∣∣ε=0

∫

Ω

F(Du + ε ·Dv, u + εv , x) dx

=

∫

Ω

n∑

j=1

∂F∂pj

(Du, u, x)∂v

∂xj+∂F∂z

(Du, u, x)

v dx

∣∣∣∣∣ε=0

= 0 ∀v ∈ H10(Ω)


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12.2 Integral Constraints371

(This last equality is the weak formulation of our original PDE.)

⇐⇒∫

Ω

n∑

j=1

− ∂

∂xj

∂

∂pj(Du, u, x) · v +

∂F∂z

(Du, u, x) · v

= 0

⇐⇒∫

Ω

[−divDpF(Du, u, x) +

∂F∂z

(Du, u, x)

]v = 0

⇐⇒ −divDpF (Du, u, x) +∂F∂z

(Du, u, x) = 0

This last line represents the strong formulation of our PDE.

If a PDE has variational form, i.e., it is the Euler-Lagrange equation of a variational problem,

then we can prove the existence of a weak solution by proving that there exists a minimizer.

12.2 Integral Constraints

In this section we shall expand upon our usage of the calculus of variations to include problems

with integral constraints. Before going into this, a digression will be made to prove the implicit

function theorem, which will subsequently be used to prove the resulting Euler-Lagrange

equation that arises from problems with constraints.

12.2.1 The Implicit Function Theorem

[Bredon] In this subsection we will prove the implicit function theorem with the goal of giving

the reader a better general intuition regarding this theorem and its use. As this section

is self-contained, it may be skipped if the reader feels they are already familiar with this

theorem.

Theorem 12.2.1 (The Mean Value Theorem): Let f ∈ C1(Rn) with x, x ∈ Rn, then

f (x)− f (x) =

n∑

i=1

∂f

∂xi(x)(xi − x i)

Proof: Apply the Mean Value Theorem to the function g(t) = f (tx + (1− t)x) and use

the Chain Rule:

dg

dt

∣∣∣∣t=t0

=

n∑

i=1

∂f

∂xi(xd(txi + (1− t)x i)

dt

∣∣∣∣t=t0

=

n∑

i=1

∂f

∂xi(x)(xi − x i),

where x = t0x + (1− t0)x .

Corollary 12.2.2 (): Let f ∈ C1(Rk × Rm) with x ∈ Rk and y , y ∈ Rm, then

f (x, y)− f (x, y) =

m∑

i=1

∂f

∂yi(x, y)(yi − y i)

for some y on the line segment between y and y .

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Lemma 12.2.3 (): Let ξ ∈ Rn and η ∈ Rm be given, and let f ∈ C1(Rn×Rm;Rm). Assume

that f (ξ, η) = η and that

∂fi∂yj

∣∣∣∣x=ξ,y=η

= 0, ∀i , j,

where x and y are the coordinates of Rn and Rm respectively. Then there exists num-

bers a > 0 and b > 0 such that there exists a unique function φ : A → B, where

A =x ∈ Rn

∣∣||x − ξ|| ≤ a

and B =y ∈ Rm

∣∣||y − η|| ≤ b

, such that φ(ξ) = η and

φ(x) = f (x, φ(x)) for all x ∈ A. Moreover, φ is continuous.

Proof: WLOG, assume ξ = η = 0. Applying Corollary 12.6 to each coordinate of f and

using the assumption that ∂fi/∂yj = 0 at the origin, and hence small in a neighborhood of

0, we can find a > 0 and b > 0 such that for any given constant 0 < K < 1,

||f (x, y)− f (x, y)|| ≤ K||y − y || (12.3)

for ||x || < a, ||y || < b, and ||y || ≤ b. Moreover, we can take a to be even smaller so that

we also have for all ||x || ≤ a,

Kb + ||f (x, 0)|| ≤ b (12.4)

Now consider the set F of all functions φ : A → B with φ(0) = 0. Define the norm on F

to be ||φ − ψ||F

= sup||φ(x)− ψ(x)||

∣∣x ∈ A

. This clearly generates a complete metric

space. Now define T : F → F by putting (Tφ)(x) = f (x, φ(x)). To prove the theorem, we

must check

i) (Tφ)(0) = 0 and

ii) (Tφ)(x) ∈ B for x ∈ A.

For i), we see (Tφ)(0) = f (0, φ(0)) = f (0, 0) = 0. Next, we calculate

||(Tφ)(x)|| = ||f (x, φ(x))|| ≤ ||f (x, φ(x))− f (x, 0)||+ ||f (x, 0)||≤ K · ||φ(x)||+ ||f (x, 0)|| by (12.3)

≤ Kb + ||f (x, 0) ≤ b by (12.4),

which proves ii). Now, we claim T is a contraction by computing

||Tφ− Tψ||F

= supx∈A

(||(Tφ)(x)− (Tψ)(x)||)

= supx∈A

(||f (x, φ(x))− f (x, ψ(x))||)

≤ supx∈A

θ||φ(x)− ψ(x)|| by (12.3)

= θ · ||φ− ψ||F.

Thus by the Banach fixed point theorem (Theorem 8.3), we see that there is a unique

φ : A → B with φ(0) = 0 and Tφ = φ; i.e. f (x, φ(x)) = φ(x) for all x ∈ A. Theorem 8.3

also states that φ = limi→∞ φi , where φ0 is arbitrary and φi+1 = Tφi . Put φ0(x) ≡ 0. Then

φi is continuous ∀i . Thus, we see that φ is the uniform limit of continuous functions which

implies φ is itself continuous. .Gregory T. von Nessi

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Theorem 12.2.4 (The Implicit Function Theorem): Let g ∈ C1(Rn × Rm) and let ξ ∈Rn, η ∈ Rm be given with g(ξ, η) = 0. (g only needs to be defined in a neighborhood of

(ξ, η).) Assume that the differential of the composition

Rm → Rn × Rn → Rm,y 7→ (ξ, y) 7→ g(ξ, y),

is onto at η. (This is a fancy way of saying the Jacobian with respect to the second argument

of g, Jy (g), is non-singular or det [Jy (g)] 6= 0.) Then there are numbers a > 0 and b > 0

such that there exists a unique function φ : A → B (with A, B as in Lemma 12.7), with

φ(ξ) = η, such that

g(x, φ(x)) = 0 ∀x ∈ A.(i.e. φ is “solves” or the implicit relation g(x, y) = 0). Moreover, if g is Cp, then so is φ

(including the case p =∞).

Proof: The differential referred to is the linear map L : Rm → Rm given by

Li(y) =

m∑

j=1

∂gi∂yi

(ξ, η)yj ,

That is, it is the linear map represented by the Jacobian matrix (∂gi/∂yi) at (ξ, η). The

hypothesis says that L is non-singular, and hence has a linear inverse L−1 : Rm → Rm. Let

f : Rn × Rm → Rm

be defined by

f (x, y) = y − L−1(g(x, y)).

Then f (ξ, η) = η − L−1(0) = η. Also, computer differentials at η of y 7→ f (ξ, y) gives

I − L−1L = I − I = 0.

Explicitly, this computation is as follows: Let L = (ai ,j) so that ai ,j = (∂gi/∂yj)(ξ, η) and

let L−1 = (bi ,j) so that∑bi ,kak,i = δi ,j . Then

∂fi∂yj

(ξ, η) = δi ,j −∂

∂yj

[m∑

k=1

bi ,kgk(x, y)

]

(ξ,η)

= δi ,j −m∑

k=1

bi ,k∂gk∂yj

(ξ, η)

= δi ,j −m∑

k=1

bi ,kak,j = 0.

Applying Lemma 12.7 to get a, b, and φ with φ(ξ) = η and f (x, φ(x)) = φ(x), we see that

φ(x) = f (x, φ(x)) = φ(x)− L−1(g(x, φ(x))),

which is equivalent to g(x, φ(x)) = 0.

Now we must show that φ is differentiable. Since the determinant of the Jacobian Jy (g) 6= 0

at (ξ, η), it is nonzero in a neighborhood, say A × B. To show that φ is differentiable at a

point x ∈ A, we can assume x = ξ = η = 0 WLOG. With this assumption, we apply the

Mean Value Theorem (Theorem 12.5) to g(x, y), g(0, 0) = 0:

0 = gi(x, φ(x)) = gi(x, φ(x))− gi(0, 0)

=

n∑

j=1

∂gi∂xj

(pi , qi) · xj +

m∑

k=1

∂gi∂yk

(pi , qi) · φk(x),

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where (pi , qi) is some point on the line segment between (0, 0) and (x, φ(x)). Let h(j) denote

the point (0, 0, . . . , h, 0, . . . , 0), with the h in the jth place, in Rn. Then, putting h(j) in place

of x in the above equation and dividing by h, we get

0 =∂gi∂xj

(pi , qi) +

m∑

k=1

∂gi∂yk

(pi , qi)φk(h(j))

h.

For j fixed, i = 1, . . . , m and k = 1, . . . , m these are m linear equations for the m unknowns

φk(h(j))

h=φk(h(j))− φk(0)

h

and they can be solved since the determinant of the Jacobian matrix is non-zero in A × B.

The solution (Cramer’s Rule) has a limit as h → 0. Thus, (∂φk/∂xj)(0) exists and equals

this limit.

We know that φ is differentiable (once) and in a neighborhood A×B of (ξ, η) and thus, we

can apply standard calculus to compute the derivative of the equations g(x, φ(x)) = 0. The

Chain Rule gives

0 =∂gi∂xj

(x, φ(x)) +

m∑

k=1

∂gi∂yk

(x, φ(x))φkxj

(x).

Again, these are linear equations with non-zero determinant near (ξ, η) and hence has, by

Cramer’s Rule, a solution of the form

∂φk∂xj

(x) = Fk,j(x, φ(x)),

where Fk,j is Cp−1 when g is Cp. If φ is Cr for r < p, then the RHS of this equation is also

Cr . Thus, the LHS ∂φk/∂xj is Cr and hence the φk are Cr+1. By induction, the φk are Cp.

Consequently, φ is Cp when g is Cp, as claimed.

12.2.2 Nonlinear Eigenvalue Problems

[Evans] Now we are ready to investigate problems with integral constraints. As in the previous

section, we will first consider a concrete example. So, let us look at the problem of minimizing

the energy functional

I(w) =1

2

∫

Ω

|Dw |2 dx

over all functions w with, say, w = 0 on ∂Ω, but subject now also to the side condition that

J(w) =

∫

Ω

G(w) dx = 0,

where G : R→ R is a given, smooth function.

We will henceforth write g = G′. Assume now

|g(z)| ≤ C(|z |+ 1)Gregory T. von Nessi

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and so

|G(z)| ≤ C(|z |2 + 1) (∀z ∈ R) (12.5)

for some constant C.

Now let us introduce an appropriate admissible space

A =w ∈ H1

0(Ω)∣∣J(w) = 0

.

We also make the standard assumption that the open set Ω is bounded, connected and has

a smooth boundary.

Theorem 12.2.5 (Existence of constrained minimizer): Assume the admissible space Ais nonempty. Then ∃u ∈ A satisfying

I(u) = minw∈A

I(w)

Proof: As usual choose a minimizing sequence uk∞k=1 ⊂ A with

I(uk)→ m = infw∈A

I(w).

Then as in the previous section, we can extract a weakly converging subsequence

ukj u in H10(Ω),

with I(u) ≤ m. All that we need to show now is

J(u) = 0,

which shows that u ∈ A. Utilizing the fact that H10(Ω) ⊂⊂−→ L2(Ω), we extract another

subsequence of ukj such that

ukj → u in L2(Ω),

where we have relabeled ukj to be this new subsequence. Consequently,

|J(u)| = |J(u)− J(uk)| ≤∫

Ω

|G(u)− G(uk)| dx

≤ C

∫

Ω

|u − uk |(1 + |u|+ |uk |) dx by (12.5)

→ 0 as k →∞.

A more interesting topic than the existence of the constrained minimizers is an examination

of the corresponding Euler-Lagrange equation.

Theorem 12.2.6 (Lagrange multiplier): Let u ∈ A satisfy

I(u) = minw∈a

I(w).

Then ∃λ ∈ R such that∫

Ω

Du ·Dv dx = λ

∫

Ω

g(u) · v dx (12.6)

∀v ∈ H10(Ω).

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Remark: Thus u is a weak solution of the nonlinear BVP−∆u = λg(u) in Ω

u = 0 on ∂Ω,(12.7)

where λ is the Lagrange Multiplier corresponding to the integral constraint

J(u) = 0.

A problem of the form (12.7) for the unknowns (u, λ), with u/≡ 0, is a non-linear eigenvalue

problem.

Proof of 12.10:

Step 1) Fix any function v ∈ H10(Ω). Assume first

g(u) 6= 0 a.e. in Ω. (12.8)

Choose then any function w ∈ H10(Ω) with

∫

Ω

g(u) · w dx 6= 0; (12.9)

this is possible because of (12.8). Now write

j(τ, σ) = J(u + τv + σw)

=

∫

Ω

G(u + τv + σw) dx (τ, σ ∈ R).

Clearly

j(0, 0) =

∫

Ω

G(u) dx = 0.

In addition, j is C1 and

∂j

∂τ(τ, σ) =

∫

Ω

g(u + τv + σw) · v dx, (12.10)

∂j

∂σ(τ, σ) =

∫

Ω

g(u + τv + σw) · w dx, (12.11)

Consequently, (12.9) implies

∂j

∂σ(0, 0) 6= 0.

According to the Implicit Function Theorem, ∃φ ∈ C1(R) such that

φ(0) = 0

and

j(τ, φ(τ)) = 0 (12.12)

for all sufficiently small τ , say |τ | ≤ τ0. Differentiating, we discover

∂j

∂τ(τ, φ(τ)) ∂j

∂σ(τ, φ(τ))φ′(τ) = 0;


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whence (12.10) and (12.11) yield

φ′(0) = −

∫

Ω

g(u) · v dx∫

Ω

g(u) · w dx. (12.13)

Step 2) Now set

w(τ) = τv + φ(τ)w (|τ | ≤ τ0)

and write

i(τ) = I(u + w(τ)).

Since (12.12) implies J(u+w(τ)) = 0, we see that u+w(τ) ∈ A. So the C1 function

i( · ) has a minimum at 0. Thus

0 = i ′(0) =

∫

Ω

(Du + τDv + φ(τ)Dw) · (Dv + φ′(τ)Dw) dx

∣∣∣∣∣τ=0

=

∫

Ω

Du · (Dv + φ′(0)Dw) dx. (12.14)

Recall now (12.13) and define

λ =

∫

Ω

Du ·Dw dx∫

Ω

g(u) · w dx,

to deduce from (12.14) the desired equality

∫

Ω

Du ·Dv dx = λ

∫

Ω

g(u) · v dx ∀v ∈ H10(Ω).

Step 3) Suppose now instead (12.8) that

g(u) = 0 a.e. in Ω.

Approximating g by bounded functions, we deduce DG(u) = g(u) ·Du = 0 a.e. Hence,

since Ω is connected, G(u) is constant a.e. It follows that G(u) = 0 a.e., because

J(u) =

∫

Ω

G(u) dx = 0.

As u = 0 on ∂Ω in the trace sense, it follows that G(0) = 0.

But then u = 0 a.e. as otherwise I(u) > I(0) = 0. Since g(u) = 0 a.e., the identity

(12.6) is trivially valid in this case, for any λ.

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12.3 Uniqueness of Solutions

Warning: In general, uniqueness of minimizers does not imply uniqueness of weak solutions

for the Euler-Lagrange equation. However, if the joint mapping (p, z) 7→ F(p, z, x) is convex

for x ∈ Ω, then uniqueness of minimizers implies uniqueness of weak solutions.

Reason: In this case, weak solutions are automatically minimizers. Suppose u is a weak

solution of

−div∂F∂p

+∂F∂z

= 0 u = g on ∂Ω

(p, z) 7→ F(p, z, x) is convex ∀x ∈ Ω), i.e.,

F(q, y , x) ≥ F(p, z, x) +DpF(p, z, x) · (q − p) +DzF(p, z, x) · (y − z)

(the graph of F lies above the tangent plane).

Suppose that u = g on ∂Ω. Choose p = Du, z = u, q = Dw and y = w .

F(Dw,w, x) ≥ F(du, u, x) +DpF(Du, u, x) · (Dw −Du) +DzF(Du, u, x) · (w − u)

Integrate in Ω:

∫

Ω

F(Dw,w, x) dx ≥∫

Ω

F(Du, u, x) dx

+

∫

Ω

[DpF(Du, u, x) · (Dw −Du) +DzF(Du, u, x) · (w − u)] dx

︸︷︷︸= 0 since w − v = 0 on ∂Ω and u is a

weak sol. of the Euler Lagrange eq.

=⇒ F (w) =

∫

Ω

F(Dw,w, x) dx ≥∫

Ω

F(Du, u, x) dx = F (u)

12.4 Energies

Ideas: Multiply by u, ut , integrate in space/space-time and try to find non-negative quantities.

Typical terms in energies:

∫

Ω

|Du|2 dx,∫

Ω

|u|2 dx,∫

Ω

|ut |2 dx.

If you take dEdt , you should find a term that allows you to use the original PDE:

d

dt

1

2

∫

Ω

|ut(x, t)|2 dx =

∫

Ω

ut(x, t) · utt(x, t) dx

is good for hperbolic equations, and

d

dt

1

2

∫

Ω

|u|2 dx =

∫

Ω

u(x, t) · ut(x, t) dx

is good for parabolic equations.Gregory T. von Nessi

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12.5 Weak Formulations379

12.5 Weak Formulations

Typically: integral identities

• test functions

• contain less derivatives than the original PDE in the strong form.

• multiply strong form by a test function and integrate by parts

Examples:

1) −∆u = f in Ω, u = 0 on ∂Ω

∫

Ω

−(∆u) dx =

∫

Ω

f v dx ⇐⇒∫

Ω

Du ·Dv dx =

∫

Ω

f v dx ∀v ∈ H10(Ω).

So, we should have:

Weak formulation + Regularity =⇒ Strong formulation, e.g.,

u ∈ H10(Ω) ∩H2(Ω) + weak solution =⇒ −∆u = f in Ω.

– Dirichlet conditions: Enforced in the sense that we require u to have correct

boundary data, e.g.,

−∆u = f in Ω

u = g on ∂Ω

seek u ∈ H1(Ω) with u = g on ∂Ω.

– Neumann conditions: Natural boundary conditions, they should arise as a conse-

quence of the weak formulations.

Weak formulation:

∫

Ω

Du ·Dv dx =

∫

Ω

f v dx ∀v ∈ H1(Ω).

Now suppose u ∈ H10(Ω) ⊂ H1(Ω) =⇒ −∆u = f in Ω.

Now take u ∈ H1(Ω)

∫

Ω

Du ·Dv dx =

∫

Ω

f v dx.

Integrate by parts:

−∫

Ω

∆u · v dx +

∫

∂Ω

(Du · ν)v dS =

∫

Ω

f v dx

=⇒∫

∂Ω

(Du · ν)v dS = 0 ∀v ∈ H1(Ω)

=⇒ Du · ν = 0 on ∂Ω

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Γ0Ω

Γ1

−∆u = f in Ωu = 0 on Γ0

Du · ν = g on Γ1

2) Now we consider a problem with split BCs.

Ω smooth, bounded, connected, Γ0 and Γ1 have positive surface measure and ∂Ω =

Γ0 ∪ Γ1.

Define H(Ω) = φ ∈ H1(Ω), s.t. φ = 0 on Γ0.∫

Ω

−∆u · φ dx =

∫

Ω

f φ dx ∀φ ∈ H(Ω)

‖∫

Ω

Du ·Dφ dx −∫

∂Ω=Γ0∪Γ1

(Du · ν) φ︸︷︷︸= 0

on Γ0

dS

=

∫

Ω

Du ·Dφ dx −∫

Γ1

gφ dS

So for the weak formulation, we seek u ∈ H(Ω) such that∫

Ω

Du ·Dφ dx =

∫

Γ1

gφ dS +

∫

Ω


Check: Weak solution + Regularity =⇒ Strong solution.

Dirichlet data okay, u ∈ H(Ω) =⇒ u = 0 on Γ0.∫

Ω

Du ·Dφ dx = −∫

Ω

∆u · φ dx +

∫

∂Ω

(Du · ν)φ dS

=

∫

Γ1

gφ dS +

∫

Ω


Now, taking φ ∈ H10(Ω) ⊂ H(Ω), we see∫

Ω

(−∆u − f )φ dx = 0 ∀φ ∈ H10(Ω)

=⇒ −∆u − f = 0 in Ω ⇐⇒ −∆u = f =⇒ PDE Holds.

The volume terms cancel, and we get∫

∂Ω

(Du · ν)φ dS =

∫

Γ1

gφ dS ∀φ ∈ H(Ω)

‖∫

Γ1

(Du · ν)φ dS =

∫

Γ1

gφ dS =⇒ Du · ν = g.

Where we have thus recovered the Neumann BC.Gregory T. von Nessi

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12.6 A General Existence Theorem381

12.6 A General Existence Theorem

Goal: Existence of a minimizer for

F (u) =

∫

Ω

L(Du, u, x) dx

with L : Rn × R×Ω→ R smooth and bounded from below.

Note:

L(p, z, x) =1

2|p|2 − f (x) · z f ∈ L2(Ω)

doesn’t fit into this framework.

Theorem 12.6.1 (): Suppose that p 7→ L(p, z, x) is convex. Then F (·) is sequentially

weakly lower semicontinuous on W 1,p(Ω), 1 < p <∞.

Idea of Proof: Have to show: uk u in W 1,p(Ω), then lim infk→∞ F (uk) ≥ F (u). True

if lim infk→∞ F (uk) =∞.

Suppose M = lim infk→∞ F (uk) <∞, and that lim inf = lim (subsequence, not relabeled).

uk u in W 1,p(Ω) =⇒ ||uk ||W 1,p(Ω) bounded

Compact Sobolev embedding: without loss of generality uk → u in Lp(Ω) =⇒ convergence

in measure =⇒ further subsequence: uk → u pointwise a.e.

Egoroff: exists Eε such that uk → u uniformly on Eε, |Ω \ Eε| < ε.

Define Fε by

Fε =

x ∈ Ω : |u(x)|+ |Du(x)| ≤ 1

ε

Claim: |Ω \ Fε| → 0. Fix K(

= 1ε

). Then,

||u(x)| > K| ≤ 1

K

∫

Ω

|u(x)| dx ≤ 1

K

∫

Ω

|u(x)| dx

Holder≤ 1

K

(∫

Ω

1 dx

) 1p′(∫

Ω

|u(x)|p dx) 1

p

≤ C

K

Define good points Gε = Fε ∩ Eε, then |Ω \ Gε| → 0 as ε→ 0.

By assumption, L bounded from below =⇒ may assume that L ≥ 0.

F (uk) =

∫

Ω

L(Duk , uk , x) dx

≥∫

Gε

L(Duk , uk , x) dx

(by convex. of L) ≥∫

Gε

L(Du, uk , x) dx

︸︷︷︸:=Ik

+

∫

Gε

DpL(Du, uk , x) · (Duk −Du) dx

︸︷︷︸:=Jk

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Since L(q, z, x) ≥ L(p, z, x) + Lp(p, z, x) · (p − q). Now

Ik →∫

Gε

L(Du, u, x) dx uniformly

and

Jk =

∫

Gε

(DpL(Du, uk , x)−DpL(Du, u, x))︸︷︷︸→0 uniformly

·(Duk −Du) dx

︸︷︷︸→0

+

∫

Gε

DpL(Du, u, x) · (Duk −Du) dx

︸︷︷︸→0

.

So, we have

limk→∞

inf F (uk) ≥∫

Gε

L(Du, u, x) dx.

Now, we use monotone convergence for ε 0 to get

limk→∞

inf F (uk) ≥∫

Ω

L(Du, u, x) dx = F (u)

Recall the assumption:

F (u) =

∫

Ω

L(Du, u, x) dx, p 7→ L(p, z, x) is convex.

Theorem 12.6.2 (): Suppose that L satisfies the coercivity assumption L(p, z, x) ≥ α|p|q−β with α > 0, β ≥ 0 and 1 < q <∞. Suppose in addition that L is convex in p for all z , x .

Also, suppose that the class of admissable functions

A = w ∈ W 1,q(Ω), w(x) = g(x) on ∂Ω

is not empty. Lastly, assume that g ∈ W 1,q(Ω). If these conditions are met, then there

exists a minimizer of the variational problem:

Minimize F (u) =

∫

Ω

L(Du, u, x) dx in A

Remark: It’s reasonable to assume that there exists a w ∈ A with F (w) < ∞. Typically

one assumes in addition that L satisfies a growth condition of the form

L(p, z, x) ≤ C(|p|q + 1)

This ensures that F is finite on A.

Idea of Proof: If F (w) = +∞ ∀w ∈ A, then there is nothing to show. Thus, suppose

F (w) <∞ for at least on w ∈ A.

Given L(p, z, x) ≥ α|p|q − β, we see that if F (w) <∞, then

∫

Ω

L(Du, u, x) dx ≥∫

Ω

(α|Du|q − β) dx

≥ α

∫

Ω

|Du|q dx − β · |Ω| (12.15)


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12.6 A General Existence Theorem383

=⇒ F (w) ≥ −β · |Ω| for w ∈ A =⇒ F is bounded from below:

−∞ < M = infw∈A

F (w) <∞.

Choose a minimizing sequence wk such that F (wk)→ M as k →∞. Then by (12.15):

α

∫

Ω

|Dwk |q dx ≤ F (wk) + β · |Ω| ≤ M + 1 + β · |Ω|

if k is large enough.

Now since wk ∈ A, we know wk = g on ∂Ω. Thus,

||wk ||Lq(Ω) = ||wk − g + g||Lq(Ω)

≤ ||wk − g||Lq(Ω) + ||g||Lq(Ω)

Poincare≤ Cp||Dwk −Dg||Lq(Ω) + ||g||Lq(Ω)

≤ Cp||Dwk ||Lq(Ω) + Cp||Dg||Lq(Ω) + ||g||Lq(Ω)

≤ C <∞ ∀k

=⇒ wk is a bounded sequence in W 1,q(Ω)

=⇒ wk contains a weakly converging subsequence (not relabeled), wk w in W 1,q(Ω).

Weak lower semicontinuity result: Theorem 12.5 applies and

M ≤ F (w) ≤ limk→∞

inf F (wk)→ M = infv∈A

F (v).

If w ∈ A then

F (w) = M = infv∈A

F (v)

=⇒ w is a minimizer.

To show this, we apply the following theorem:

Theorem 12.6.3 (Mazur’s Theorem): X is a BS, and suppose that xn x in X. Then for

ε > 0 given, there exists a convex combination∑kj=1 λjxj of elements xj ∈ xnn∈N, λj ≥ 0,∑k

j=1 λj = 1, such that

∣∣∣∣∣∣

∣∣∣∣∣∣x −

k∑

j=1

λjxj

∣∣∣∣∣∣

∣∣∣∣∣∣< ε.

In particular, there exists a sequence of convex combinations that converges strongly to x .

Proof: [Yoshida] Consider the totality MT of all convex combinations of xn. We can

assume 0 ∈ MT by replacing x with x − x1 and xj with xj − x1.

We go for a proof by contradiction. Suppose the contrary of the theorem, i.e., ||x −u||X > ε

for all u ∈ MT . Now consider the set,

M =v ∈ X; ||v − u||X ≤

ε

2for some u ∈ MT

,

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which is a convex (balanced and also absorbing) neighborhood of 0 in X and ||x−v ||X > ε/2

for all v ∈ M.

Recall the Minkowski functional pA for given set A:

pA(x) = infα>0, α−1x∈A

α.

Now, since x /∈ M, there exists u0 ∈ M such that x = β−1u0 with 0 < β < 1. Thus,

pM(x)β−1 > 1.

Consider now the real linear subspace of X:

X1 = x ∈ X; x = γ · u0, −∞ < γ <∞

and define f1(x) = γ for x = γ · u0 ∈ X1. This real linear functional f1 on X1 satisfies

f1(x) ≤ pM(x). Thus, by the Hahn-Banach Extension theorem, there exists a real linear

extension f of f1 defined on the real linear space X and such that f (x) ≤ pM(x) on X. Since

M is a neighborhood of 0, the Minkowski functional pM(x) is continuous in x . From this we

see that f must also be a continuous real linear functional defined on X. Thus, we see that

supx∈MT

f (x) ≤ supx∈M

f (x) ≤ supx∈M

pM(x) = 1 < β−1 = f (β−1u0) = f (x).

Clearly we see that x can not be a weak accumulation point of M1, this is a contradiction

since we assumed that xn x .

Corollary 12.6.4 (): If Y is a strongly closed, convex set in a BS X, then Y is weakly closed.

Proof: xn x =⇒ exists a sequence of convex combinations yk ∈ Y such that yk → x .

Y closed =⇒ x ∈ Y .

Application to proof of 12.6: A is a convex set. We see that A is closed by the following

argument. If wk → w in W 1,q(Ω), then by the Trace theorem, we have

||wk − w ||Lq(∂Ω) ≤ CTr||wk − w ||W 1,q(Ω) → 0

since wk = g on ∂Ω =⇒ w = g on ∂Ω.

So, we apply Mazur’s Theorem, to see that A is also weakly closed, i.e., if wk w in

W 1,q(Ω), then w ∈ A. Thus, this establishes existence by the arguments stated earlier in

the proof.

Uniqueness of this minimizer is true if L is uniformly convex (see Evans for proof).

12.7 A Few Applications to Semilinear Equations

Here we want to show how the technique used in the linear case extends to some nonlinear

cases. First we show that “lower-order terms” are irrelevant for the regulairty. For example

consider∫

Ω

Du ·Dφdx =

∫

Ω

a(u) ·Dφdx ∀φ ∈ H10(Ω) (12.16)


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12.7 A Few Applications to Semilinear Equations385

If

∫

BR(x0)

Dv ·Dφdx = 0 for ∀φ ∈ H10(Ω) and v = u on ∂BR then

∫

Bρ(x0)

|Du|2 dx ≤ C( ρR

)n ∫

BR(x0)

|Du|2 dx + C

∫

BR(x0)

|D(u − v)|2 dx

≤ C( ρR

)n ∫

BR(x0)

|Du|2 dx + C

∫

BR(x0)

|a(u)|2 dx.

Now if |a(u)| ≤ |u|α, then the last integral is estimated by

∫

BR(x0)

|a(u)|2 dx ≤(∫

)BR(x0)|u|2n/(n−2) dx

)α(1−2/n)

Rn(1−α(1−2/n))

so that u is Holder continuous provided α < 2n−2 .

The same result holds if we have instead of (12.16)∫

Ω

Du ·Dφdx =

∫

Ω

a(u) ·Dφdx +

∫

Ω

a(Du) · φdx

with a(Du) ≤ |Du|1−2/n or a(Du) ≤ |Du|2−ε and |u| ≤ M = const.

The above idea works also if we study small perturbations of a system with constant co-

efficients:

For∫

Ω

Aαβij (x)DαuiDβφ

j dx = 0

with

Aαβij ∈ L∞(Ω) and∣∣∣Aαβij (v)− δαβδi j

∣∣∣ ≤ ε0

one finds∫

Bρ(x0)

|Du|2 dx ≤ C( ρR

)n ∫

BR(x0)

|Du|2 dx + C

∫

BR(x0)

|δ1δ2 − A|︸︷︷︸≤ε0

·Du2 dx

because∫

Ω

DαuiDβφ

j dx =

∫

Ω

(δαβδi j − Aαβij )DαuiDβφ

j dx.

So we can apply the lemma from the very begining of 9.8 and conclude that Du is Holder

continuous. Finally we mention the

Theorem 12.7.1 (): Let u ∈ H1,2(Ω,RN) be a minimizer of∫

Ω

F (Du) dx

and let us assume that

F (tp)

t2→ |p|2 for t →∞

then u ∈ C0,µ(Ω).

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Proof: We consider the problem

∫

BR(x0)

|DH|2 dx → minimize, H = 0 on ∂BR

(H stands for harmonic function!) then

∫

Bρ(x0)

|Du|2 dx ≤ C( ρR

)n ∫

BR(x0)

|Du|2 dx +

∫

BR(x0)

|Du −DH|2 dx.

Now∫

BR(x0)

D(u −H) ·D(u −H) dx =

∫

BR(x0)

Du ·D(u −H) dx −∫

BR(x0)

DH ·D(u −H) dx

=

∫

BR(x0)

|Du|2 dx −∫

BR(x0)

Du ·DH dx

=

∫

BR(x0)

|Du|2 dx −∫

BR(x0)

(D(u −H) +DH)DH dx

=

∫

BR(x0)

|Du|2 dx −∫

BR(x0)

|DH|2 dx

=

∫

BR(x0)

|Du|2 dx +

∫

BR(x0)

F (Du) dx −∫

BR(x0)

F (DH) dx

︸︷︷︸≤0(u is a minimizer)

−∫

BR(x0)

F (Du) dx +

∫

BR(x0)

F (DH) dx −∫

BR(x0)

|DH|2 dx.

By the growth assumption on F and because

|p|2 − F (p) = |p|2(

1− F (p)

|p|2)

we find∫

BR(x0)

|D(u −H)|2 dx ≤ ε∫

BR(x0)

|Du|2 dx + C · Rn.

So we can apply the lemma at the beginning of the section and conclude that Du is Holder

continuous.

12.8 Regularity

We discuss in this section the smoothness of minimizers to our energy functionals. This is

generally a quite difficult topic, and so we will make a number of strong simplifying assump-

tions, most notably that L depends only on p. Thus we henceforth assume our functions I[·]to have the form

I[w ] :=

∫

Ω

L(Dw)− wf dx, (12.17)

for f ∈ L2(Ω). We will also take q = 2, and suppose as well the growth condition

|DpL(p)| ≤ C(|p|+ 1) (p ∈ Rn). (12.18)Gregory T. von Nessi

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Then any minimizer u ∈ A is a weak solution of the Euler-Lagrange PDE

−n∑

i=1

(Lpi (Du))xi = f in Ω; (12.19)

that is,

∫

Ω

n∑

i=1

Lpi (Du)vxi dx =

∫

Ω

f v dx (12.20)

for all v ∈ H10(Ω).

12.8.1 DeGiorgi-Nash Theorem

[Moser-Iteration Technique]

Let u be a subsolution for the elliptic operator −Dβ(aαβDα), i.e.

∫

Ω

aαβDαuDβφdx ≤ 0 ∀φ ∈ H1,20 (Ω), φ ≥ 0

where aαβ ∈ L∞(Ω) and satisfy an ellipticity-condition. The idea is to use as test-function

φ := (u+)pη2 with p > 1 and η the cut-off-function we defined on page 20. For the sake of

simplicity we assume that u ≥ 0. Then

p

∫

BR

aαβDαuDβuup−1η2 dx +

∫

BR

aαβDαuupηDη dx ≤ 0

∫

BR

|Du|2up−1η2 dx ≤ c

p

∫

BR

|Du|u p−12 u

p+12 η|Dη| dx

∫

BR

|Du|2up−1η2 dx ≤ c

p2

∫

BR

up+1|Dη|2 dx.

As

∣∣∣Dup+1

2

∣∣∣2

=

(p + 1

2

)2

up−1|Du|2

we get

∫

BR

∣∣∣Dup+1

2

∣∣∣2η2 dx ≤ c

(p + 1

p

)2 ∫

BR

up+1|Dη|2 dx

and together with

∣∣∣D(ηup+1

2 )∣∣∣2≤ c

∣∣∣Dup+1

2

∣∣∣2η2 + up+1|Dη|2

follows

∫

BR

∣∣∣Dup+1

2

∣∣∣2dx ≤ c

(1 +

(p + 1

2

)2)∫

BR

up+1|Dη|2 dx (p > 1)

≤ c(p + 1)2

(1 +

1

p2

)∫

BR

up+1|Dη|2 dx.

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By Sobolev’s inequality and the properties of η we finally have:

(∫

BR

(up+1

2 η)2∗ dx

)2/2∗≤ c(p + 1)2

(1 +

1

p2

)1

(R − ρ)2

∫

BR

up+1 dx.

For p = 1, this is just Caccioppoli’s inequality. Set λ := 2 ∗ /2 = nn−2 and q := p + 1 > 2,

then(∫

BR

uλq dx

)1/λ

≤ c (1 + q)2

(R − ρ)2

∫

BR

uq dx.

Now we choose

qi := 2λi = λqi−1 (q0 = 2)

Ri :=R

2+

R

2i+1(R0 = R)

and we find that

(∫

BRi+1

uqi+1 dx

) 1

λi+1

=

(∫

BRi+1

uλqi dx

) 1λ

1

λi

≤[c(1 + qi)

2

2−2(i+1)R2

]1/λi(∫

BRi

uqi dx

)1/λi

≤i∏

k=0

[c(1 + qk)2

R24−k−1

]1/λk (∫

BR

u2 dx

)1/2

.

We have to estimate the above product:

∞∏

k=0

[c(1 + qk)2

R24−k−1

]1/λk

= exp

(ln

∞∏

k=0

[c(1 + qk)2

R24−k−1

]1/λk)

= exp

∞∑

k=0

2

λkln c

(1 + qk)

R2−k−1

= exp(c − n lnR)

(qk = 2λk ,

∞∑

k=0

λ−j =n

2

)

thus ∀k(−∫

BRk

uqk dx

)1/qk

≤ c(−∫

BR

u2 dx

)1/2

from which immediately follows that

supx∈BR/2

u(x) ≤ c(−∫

BR

|u|2 dx)1/2

.

Remark: Instead of 2, one can take any expoenent p¿0 in the above formulas.

If we start with a supersolution and u > 0, i.e.∫

Ω

aαβ(x)DαuDβφdx ≥ 0 ∀φ ∈ H1,20 (Ω), φ ≥ 0


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then we can take p < −1. Exactly as before we then get for λ := 2∗2 > 1 and q := p+1 < 0.

(∫

Bρ

uλq dx

)1/λ

≤ c(1 + q2)

(1 +

1

(q − 1)2

)1

(R − ρ)2

∫

BR

uq dx.

As |q − 1| > 1 we have

(∫

Bρ

uλq dx

)1/λ

≤ c (1 + q)2

(R − ρ)2

∫

BR

uq dx.

Now we set Ri := R2 + R

2i+1 as before and qi := q0λi but q0 < 0. Then follows

infx∈BR/2

u(x) ≤ c(−∫

BR

uq0 dx

)1/q0

(q0 < 0).

Up to now we have the following two estimates for a positive solution u: For any p > 0 and

any q < 0 holds:

supx∈BR/2

u(x) ≤ k1

(−∫

BR

up dx

)1/p

infx∈BR/2

u(x) ≥ k2

(−∫

BR

uq dx

)1/q

,

for some constants k1 and k2, q < 0. The main point now is that the John-Nirenberg-lemma

allows us to combine the two inequalities to get

Theorem 12.8.1 (Harnack’s Inequality): If u is a solution and u > 0, then

infx∈BR/2

u(x) ≥ C supx∈BR/2

u(x).

Proof: We have∫

Ω

aαβDαuDβφdx = 0 ∀φ ∈ H1,20 (Ω) φ > 0.

As u > 0, we can choose φ := 1uη

2 and get

−∫

Ω

aαβDαuDβu1

u2η2 dx +

∫

Ω

aαβDαu1

uηDη dx = 0

thus∫

Ω

|Du|2u2

η2 dx ≤∫

Ω

|Dη|2 dx,

i.e.∫

BR

|D ln u|2 dx ≤ CRn−2

which implies by the Poincare’s inequality that ln u ∈ BMO. Now we use the characerization

of BMO by which there exists a constant α, such that for v := eα ln u, we have

∫

BR

v dx ≤ C(∫

BR

1

vdx

)−1

.

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Hence (p = 1, q = −1),

infBR/2

u ≥ K2

(−∫

BR

u−a dx

)−1/α

≥ K2C−1

(−∫

BR

uα dx

)1/α

≥ K2

K1CsupBR/2

u.

Remark: It follows from Harnack’s inequality that a solution of the equation at the beginning

of this section is Holder-continuous: M(2R)− u ≥ 0 is a supersolution, thus

M(2R)−m(R) ≤ C(M(2R)−M(R))

u −m(2R) ≥ 0 is also a supersolution, thus

M(R)−m(2R) ≤ C(m(R)−m(2R))

and from this we get

ω(2R) + ω(R) ≤ C(ω(2R)− ω(R))

hence ω(R) ≤ C+1C−1ω(2R) and u is Holder-continuous.

12.8.2 Second Derivative Estimates

We now intend to show if u ∈ H1(Ω) is a weak solution of the nonlinear PDE (12.19), then

in fact u ∈ H2loc(Ω). But to establish this, we will need to strengthen our growth condition

on L. Let us first of all suppose

|D2L(p)| ≤ C (p ∈ Rn). (12.21)

In addition, let us assume that L is uniformly convex, and so there exists a constant θ > 0

such that

n∑

i ,j=1

Lpipj (p)ξiξj ≥ θ|ξ|2 (p, ξ ∈ Rn). (12.22)

Clearly, this is some sort of nonlinear analogue of our uniform ellipticity condition for linear

PDE. The idea will therefore be to try to utilize, or at least mimic, some of the calculations

from that chapter.

Theorem 12.8.2 (Second derivatives for minimizers): (i) Let u ∈ H1(Ω) be a weak

solution of the nonlinar PDE (12.19), where L statisfies (12.21), (12.22). Then

u ∈ H2loc(Ω).

(ii) If in addition u ∈ H10(Ω) and ∂Ω is C2, then

u ∈ H2(Ω),

with the estimate

||u||H2(Ω) ≤ C · ||f ||L2(Ω)

Proof:Gregory T. von Nessi

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1. We will largely follow the proof of blah blah, the corresponding assertion of local H2

regularity for solutions of linear second-order elliptic PDE.

Fix any open set V ⊂⊂ Ω and choose then an open set W so that V ⊂⊂ W ⊂⊂ Ω.

Select a smooth cutoff function ζ satisfyingζ ≡ 1 on V, ζ ≡ 0 in Rn \W0 ≤ ζ ≤ 1

Let |h| > 0 be small, choose k ∈ 1, . . . , n, and substitute

v := −D−hk (ζ2Dhku)

into (12.20). We are employing her the notion from the difference quotients section:

Dhku(x) =u(x + hek)− u(x)

h(x ∈ W ).

Using the identity

∫

Ω

u ·D−hk v dx = −∫

Ω

v ·Dhku dx , we deduce

n∑

i=1

∫

Ω

Dhk(Lpi (Du)) · (ζ2Dhku)xi dx = −∫

Ω

f ·D−hk (ζ2Dhku) dx. (12.23)

Now

DhkLpi (Du(x)) =Lpi (Du(x + hek))− Lpi (Du(x))

h

=1

h

∫ 1

0

d

dsLpi (s ·Du(x + hek) + (1− s) ·Du(x)) ds

(uxj (x + hek)− uxj (x))

=

n∑

j=1

ahij(x) ·Dhkuxj (x), (12.24)

for

ahij(x) :=

∫ 1

0

Lpipj (s ·Du(x + hek) + (1− s) ·Du(x)) dx (i , j = 1, . . . , n).(12.25)

We substitute (12.24) into (12.23) and perform simplce calculations, to arrive at the

identity:

A1 + A2 :=

n∑

i ,j=1

∫

Ω

ζ2ahijDhkuxjD

hkuxi dx

+

n∑

i ,j=1

∫

Ω

ahijDhkuxjD

hku · 2ζζxi dx

= −∫

Ω

f ·D−hk (ζ2Dhku) dx =: B.

Now the uniform convexity condition (12.22) implies

A1 ≥ θ∫

Ω

ζ2|dhkDu|2 dx.

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Furthermore we see from (12.21) that

|A2| ≤ C

∫

W

ζ|DhkDu| · |Dhku| dx

≤ ε

∫

W

ζ2|DhkDu|2 dx +C

ε

∫

W

|Dhku|2 dx.

Furthermore, as in the proof of interior H2 regularity of elliptic PDE, we have

|B| ≤ ε∫

Ω

ζ2|DhkDu|2 dx +C

ε

∫

Ω

f 2 + |Du|2 dx.

We select ε = θ4 , to deduce from the foregoing bounds on A1, A2, B, the estimate

∫

Ω

ζ2|DhkDu|2 dx ≤ C∫

W

f 2 + |Dhku|2 dx ≤ C∫

Ω

f 2 + |Du|2 dx,

the last inequaltiy valid from the properties of difference quotients.

2. Since ζ ≡ 1 on V , we find

∫

V

|DhkDu|2 dx ≤ C∫

Ω

f 2 + |Du|2 dx

for k = 1, . . . , n and all sufficiently small |h| > 0. Consequently prop of diff. quotients

imply Du ∈ H1(V ), and so u ∈ H2(V ). This is true for each V ⊂⊂ U; thus u ∈H2

loc(Ω).

3. If u ∈ H10(Ω) is a weak solution of (12.19) and ∂U is C2, we can then mimic the

proof of the boundary regularity theorem 4 of elliptic pde to prove u ∈ H2(Ω), with

the estimate

||u||H2(Ω) ≤ C(||f ||L2(Ω) + ||u||H1(Ω));

details are left to the reader. Now from (12.22) follows the inequality

(DL(p)−DL(0)) · p ≥ θ|p|2 (p ∈ Rn).

If we then put v = u in (12.20), we can employ this estimate to derive the bound

||u||H1(Ω) ≤ C||f ||L2(Ω),

and so finish the proof.

12.8.3 Remarks on Higher Regularity

We would next like to show that if L is infinitely differentiable, then so is u. By analogy with

the regularity theory developed for second-order linear elliptic PDE in 6.3, it may seem nat-

ural to try to extend the H2loc estimate from teh previous section to obtain further estimates

in higher order Sobolev spaces Hkloc(Ω) for k = 3, 4, . . .


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12.8 Regularity393

This mehtod will not work for the nonlinear PDE (12.19) however. The reason is this. For

linear equations we could, roughly speaking, differntiate the equation many times and still ob-

tain a linear PDE of the same general form as that we began with. See for instance the proof

of theorem 2 6.3.1. But if we differentiate a nonlinear differential equaiton many times, the

resulting increasingly complicated expressions quickly becvome impossible to handle. Much

deeper ideas are called for, the full development of which is beyond the scope of this book.

We will nevertheless at least outline the basic plan.

To start with, choose a test function w ∈ C∞c (Ω), select k ∈ 1, . . . , n, and set v = −wxk in

the identity (12.20), where for simplicity we now take f ≡ 0. Since we now know u ∈ H2loc(Ω),

we can integrate by parts to find

∫

Ω

n∑

i ,j=1

Lpipj (Du)uxkxjwxi dx = 0. (12.26)

Next write

u := uxk (12.27)

and

ai j := Lpipj (Du) (i , j = 1, . . . , n). (12.28)

Fix also any V ⊂⊂ Ω. Then after an approximation we find from (12.26)-(12.28) that

∫

Ω

n∑

i ,j=1

ai j(x)uxjwxi dx = 0 (12.29)

for all w ∈ H10(V ). This is to say that u ∈ H1(V ) is a weak solution of the linear, second

order elliptic PDE

−n∑

i ,j=1

(ai j uxj )xi = 0 in V. (12.30)

But we can not just apply our regularity theory from 6.3 to conclude from (12.30) that u is

smooth, the reason being that we can deduce from (12.21) and (12.28) only that

ai j ∈ L∞(V ) (i , j = 1, . . . , n).

However DeGiorgi’s theorem asserts that any weak solution of (12.30) must in fact be locally

Holder continuous for some exponent γ > 0. Thus if W ⊂⊂ V we have u ∈ C0,γ(W ), and

so

u ∈ C1,γloc (Ω).

Return to the definition (12.28). If L is smooth, we now know ai j ∈ C0,γloc (Ω) (i , j = 1, . . . , n).

Then (12.19) and the Schauder estimates from Chapter 8 assert that in fact


But then ai j ∈ C1,γloc (Ω); and so another version of Schauder’s estimate implies


We can continue this so called “bootstrap” argument, eventually to deduce u is Ck,γloc (Ω) for

k = 1, . . . , and so u ∈ C∞(Ω).

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12.9 Exercises

12.1: [Qualifying exam 08/98] Let Ω ⊂ Rn be open (perhaps not bounded, nor with smooth

boundary). Show that if there exists a function u ∈ C2(Ω) vanishing on ∂Ω for which the

quotient

I(u) =

∫

Ω

|Du|2 dx∫

Ω

u2 dx

reaches its infimum λ, then u is an eigenfunction for the eigenvalue λ, so that ∆u + λu = 0

in Ω.

12.2: [Qualifying exam 08/02] Let Ω be a bounded, open set in Rn with smooth boundary Γ .

Suppose Γ = Γ0 ∪ Γ1, where Γ0 and Γ1 are nonempty, disjoint, smooth (n − 1)-dimensional

surfaces in Rn. Suppose that f ∈ L2(Ω).

a) Find the weak formulation of the boundary value problem

−∆u = f in Ω,

Du · ν = 0 on Γ1,

u = 0 on Γ0.

b) Prove that for f ∈ L2(Ω) there is a unique weak solution u ∈ H1(Ω).

12.3: [Qualifying exam 08/02] Let Ω be a bounded domain in the plane with smooth bound-

ary. Let f be a positive, strictly convex function over the reals. Let w ∈ L2(Ω). Show that

there exists a unique u ∈ H1(Ω) that is a weak solution to the equation

−∆u + f ′(u) = w in Ω,

u = 0 on ∂Ω.

Do you need additional assumptions on f ?


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396Chapter 13: Non-Variational Methods in Nonlinear PDE

13 Non-Variational Methods in Nonlinear PDE


– Alan Rickman

Contents

13.1 Viscosity Solutions 402

13.2 Alexandrov-Bakel’man Maximum Principle 406

13.3 Local Estimates for Linear Equations 411

13.4 Holder Estimates for Second Derivatives 419

13.5 Existence Theorems 428

13.6 Applications 438

13.7 Exercises 441

13.1 Viscosity Solutions

The problem with the Perron definition of subharmonic and superharmonic functions is that

they relied on a classical notion of a function being harmonic. For a general differential

operator, call it F , there might not be an analogy of a harmonic function as the equation

F [u] = 0 may or may not have a solution in the classical sense. This is opposed to solutions

of F [u] ≥ 0 of F [u] ≤ 0, these solutions can readily be constructed for differential operators.

With this in mind, we make the following alternative definition:

Definition 13.1.1: A function u ∈ C0(Ω) is subharmonic(superharmonic) if for every ball

B ⊂⊂ Ω and superharmonic(subharmonic) function v ∈ C2(B) ∩ C0(B) (i.e. ∆v ≤ (≥) 0)

which is ≥ u on ∂B one has v ≥ (≤) u in B.

It’s not hard to understand the above definition is equivalent to the Perron definition.

Remarks:Gregory T. von Nessi

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i.) By replacing ∆ by other operators, denoted F , we then get general definitions of the

statement “F [u] ≥ (≤) 0” for u ∈ C0(Ω).

ii.) One can define a solution of “F [u] = 0” as a function satisfying both F [u] ≥ 0 and

F [u] ≤ 0.

iii.) The solution of F [u] = 0 in this sense are referred to as viscosity solutions.

We can still further generalize the above definition still further:

Definition 13.1.2: A function u ∈ C0(Ω) is subharmonic(superharmonic) (with respect to

∆) if for every point y ∈ Ω and function v ∈ C2(Ω) such that u−v is maximized(minimized)

at y , we have ∆v(y) ≥ (≤) 0.

Proposition 13.1.3 (): Definitions (13.1.1) and (13.1.2) are equivalent.

Proof: (13.1.1) =⇒ (13.1.2): Let u be subharmonic. Suppose now there exists y ∈ Ω such

that ∆v(y) < 0. Since v ∈ C2(Ω) there exists a ball B containing y such that ∆v(y) ≤ 0 on

B. Now, define vε = v + ε|x − y |4 which implies that y corresponds to a strict maximum of

u − vε. With this we know there exists a δ > 0 such that vε − δ > u on ∂B with vε − δ < u

on some set Eδ ⊂ B; a contradiction.

vǫ

Eδ

B

u

vǫ − δ

Eδ

B

u

u

v

(13.1.2) =⇒ (13.1.1): Suppose that the maximum of u−v is attained at y ∈ Ω implies that

∆v(y) ≥ 0 for any v ∈ C2(Ω). Next, suppose that u is not subharmonic; i.e. there exists a

ball B and superharmonic v such that v ≥ u on ∂B but v ≤ u in a set Eδ ⊂ B. Another way

to put this is to say there exists a point y ∈ B where u − v attains a positive max. Now,

define vε := v + ε(p2 − |x − y |2); clearly ∆vε(y) < 0. For ε small enough u − vε will have

a positive max in B (the constant p can be adjusted so that vε ≥ u on ∂Ω). This yields a

contradiction as ∆vε(y) < 0 by construction.

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Now, we can make the final generalization of our definition. First, we redefine the idea of

ellipticity. Consider the operator

F [u](x) := F (x, u,Du,D2u), (13.1)

where F ∈ C0(Ω×R×Rn × Sn). Note that Sn represents the space of all symmetric n × nmatrices.

Definition 13.1.4: An operator F ∈ C0(Ω× R× Rn × Sn) is degenerate elliptic if

∂F

∂r i j≥ 0

and

∂F

∂z≤ 0,

where the arguments of F are understood as F (x, z, p, [r i j ]).

Finally, we make our most general definition for subharmonic and superharmonic functions.

Definition 13.1.5: If u is upper(lower)-semicontinuous on Ω, we say F [u] ≥ (≤) 0 in the

viscosity sense if for every y ∈ Ω and v ∈ C2(Ω) such that u − v is maximized(minimized)

at y , one has F [v ](y) ≥ (≤) 0.

Remark: A viscosity solution is a function that satisfies both F [u] ≥ 0 and F [u] ≤ 0 in the

viscosity sense.

This final definition for viscosity solutions even has meaning in first order equations:Gregory T. von Nessi

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13.1 Viscosity Solutions399

Example 10: Consider |Du|2 = 1 in R. It is obvious that generalized solutions will have a

“saw tooth” form and are not unique for specified boundary values. That being understood,

it is easily verified that the viscosity solution to this equation is unique.

y

C2 viscosity violator function

viscosity solution

C0 solutions

To see this, the above figure illustrates a non-viscosity solution with an accompanying C2

function which violates the viscosity definition. At point y , it is clear that u− v is maximized

but that ∆v(y) < 0. Basically, any “peak” in a C0 solution will violate the viscosity crite-

rion with a similar C2 example. The only solution with no “peaks” is the viscosity solution

depicted in the figure.

Aside: A nice way to derive this viscosity solution is to consider solutions uε of the equation:

Fε[u] := ε∆uε + |Duε|2 − 1 = 0.

Clearly, Fε → F , where F [u] := |Du|2 − 1 as ε → 0. It is not clear whether or not uεconverges. Even if uε converged to some limit, say u, it’s not clear whether or not F [u] = 0

in any sense. Fortunately, in this particular example, all this can be ascertained through direct

calculation. Since we are in R, we have

Fε[u] = ε · uεxx + |uεx |2 − 1 = 0.

To solve this ODE take vε := uεx ; this makes the ODE separable. After integrating, one

gets

uε = ε · ln(e

2xε + 1

e−2ε + 1

)− x,

where the constants of integration have been set so u(−1) = u(1) = 1. The following figure

is a graph of these functions.

It is graphically clear that limε→0 uε = |x |; i.e. we have uε converging (uniformly) to

the viscosity solution of |Du|2 = 1. This limit is easy to verify algebraically via the use of

L’Hopital’s rule.

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ǫ = 12

O

|x |ǫ = 1

4

1

1−1ǫ = 1

8

13.2 Alexandrov-Bakel’man Maximum Principle

Let u ∈ C0(Ω) and consider the graph u given by xn+1 = u(x) as well as the hyperplane

which contacts the graph from below. For y ∈ Ω, set

χu(y) := p ∈ Rn∣∣∣ u(x) ≥ u(y) + (x − y) · p ∀x ∈ Ω.

Define the lower contact set Γ− or Γ−u by

Γ− :=p ∈ Rn

∣∣∣ u(x) ≥ u(y) + (x − y) · p ∀x ∈ Ω

for some p = p(y) ∈ Rn

= points of convexity of graph of u.

Proposition 13.2.1 (): Let u ∈ C0(Ω). We have

i.) if u is differentiable at y ∈ Γ− then χu(y) = Du(y),

ii.) if u is convex in Ω then Γ− = Ω and χu is a subgradient of u,

iii.) if u is twice differentiable at y ∈ Γ− then D2u(y) ≥ 0

Investigate the relation between χu, max u, and min u. Let p0 /∈ χu(Ω) :=⋃y∈Ω χu(y).

By definition, there exists a point y ∈ ∂Ω such that

u(x) ≥ u(y) + p0 · (x − y) ∀x ∈ Ω. (13.2)

You can see 13.2) by considering the hyperplane of the type xn+1 = p0 · x + C, first

placing far from below the graph u and tending it upward.(13.2) yields an estimate of u; i.e.,

u ≥ min∂Ω

u − |p0|d, (13.3)

where d = Diam (Ω) as before.

The relation with differential operator is as follows. Suppose u ∈ C2(Ω), then χu(y) =

Du(y) for y ∈ Γ− and the Jacobian matrix of χu is D2u ≥ 0 in Γ−. Thus, the n-dimensional

Lebesgue measure of the normal image of Ω is given by

|χ(Ω)| = |χ(Γ−)| = |Du(Γ−)| ≤∫

Γ−| detD2u| dx (13.4)


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13.2 Alexandrov-Bakel’man Maximum Principle401

Ω

u

p0 · (x − y)

x

since D2u ≥ 0 on Γ−. (13.4) can be realised as a consequence of the classical change of

variables formula by considering, for positive ε, the mapping χε = χ + εI, whose Jacobian

matrix D2u + εI is then strictly positive in the neighbourhood of Γ−, and by subsequently

letting ε→ 0. If we can show that χε is 1-to-1, then equality will in fact hold in (13.4). To

show this, suppose not, i.e. there exists x 6= y such that χε(x) = χε(y). In this case we

have

χ(x)− χ(y) = ε(y − x).

But, we also have (after using abusive notation), that

u(x) ≥ u(y) + (x − y) · χ(y)

u(y) ≥ u(x) + (y − x) · χ(x)

Adding these we have

0 ≥ (χ(x)− χ(y))(y − x) = ε(y − x)2 ≥ 0,

which leads us to conclude y = x , a contradiction.

More generally, if h ∈ C0(Ω) is positive on Rn, then we have

∫

χu(Ω)

dp

h(p)=

∫

Γ−

detD2u

h(Du)dx. (13.5)

Remark: It is trivial to extend this to u ∈ (C2(Ω) ∩ C0(Ω)) ∪ W 2,n(Ω) by using ap-

proximating functions and Lebesgue dominated convergence. Also note that even thous

C2(Ω) ⊂ W 2,n(Ω), C2(Ω)∩C0(Ω) * W 2,n(Ω). Not sure if further generalization is possible.

Theorem 13.2.2 (): Suppose u ∈ C2(Ω) ∩ C0(Ω) satisfies

detD2u ≤ f (x)h(Du) on Γ− (13.6)

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for some positive function h ∈ C2(Rn). Assume moreover,

∫

Γ−f dx <

∫

Rn

dp

h(p).

Then there holds an estimate of the type

minΩu ≥ min

∂Ωu − Cd,

where C denotes some constant independent of u.

Proof. Let BR(0) be a ball of radius R centered at 0 such that

∫

χu(Ω)

dp

h(p)≤∫

Γ−f dx =

∫

BR(0)

dp

h(p)<

∫

Rn

dp

h(p).

Thus, for R′ > R, there exists p /∈ χu(Ω) such that |p| < R′. So (13.3) now implies that

minΩu ≥ min

∂Ωu − Rd,

which completes the proof.

13.2.1 Applications

13.2.1.1 Linear equations

First we consider the case h ≡ 1 in theorem 13.2.2. The result is simply stated as follows:

The differential inequality

detD2u ≤ f in Γ−

implies an estimate

minΩu ≥ min

∂Ω−(

1

ωn

∫

Γ−f dx

)1/n

· d.

Next, suppose u ∈ C2(Ω) ∩ C0(Ω) satisfies

Lu := ai jDi ju ≤ g in Ω,

where A := [ai j ] > 0 and g a given bounded function. We derive

detA · detD2u ≤( |g|n

)non Γ−,

by virtue of an elementary matrix inequality for any symmetric n × n matrices A,B ≥ 0:

detA · detB ≤(T r(AB)

n

)n,

whose proof is left as an exercise.


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13.2 Alexandrov-Bakel’man Maximum Principle403

As det [AB] = det [A] · det [B], we really only need to show that det [A] ≤(T rAn

)n. Since

the matrices are non-singular we assume the matrix A is diagonalized wlog. So,

det [A] =

n∏

i=1

λi = exp

(n ·

n∑

i=1

lnλin

)≤(

1

n

n∑

i=1

λi

)n=

(T rA

n

)n

The inequality of the of the above comes from the convexity of exponential functions.

Therefore it follows that

minΩu ≥ min

∂Ωu − d

nω1/nn

∣∣∣∣∣∣∣∣

g

(detA)1/n

∣∣∣∣∣∣∣∣Ln(Γ−)

.

We can find an estimate for maxΩ u similarly; if Lu ≥ g in Ω, thenwe have

maxΩu ≤ max

∂Ωu − d

nω1/nn

∣∣∣∣∣∣∣∣

g

(detA)1/n

∣∣∣∣∣∣∣∣Ln(Γ+)

,

where Γ+u = Γ−−u.

13.2.1.2 Monge-Ampere Equation

Suppose a convex function u ∈ C2(Ω) solves

det[D2u

]≤ f (x) · h(Du),

for some functions f , h which are assumed to satisfy∫

Ω

f (x) dx <

∫

Rn

dp

h(p)

Then we have an estimate

infΩu ≥ inf

∂Ωu − Rd,

where R is defined by∫

BR(0)

dp

h(p)=

∫

Ω

f (x) dx.

This result is a consequence of theorem 13.2.2. If Ω is convex, then we infer supΩ u ≤sup∂Ω u, taking into account the convexity of u.

Remark: If det[D2u

]= f (x) · h(Du) in Ω, then

∫

Ω

det[D2u

]

h(Du)=

∫

Ω

f (x) dx.

Thus, the inequality∫

Ω

f (x) dx ≤∫

Rn

dp

h(p)

is a necessary condition for solvability. It is also sufficient, by the work of Urbas. The

inequality is strict if Du is bounded.

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13.2.1.3 General linear equations

Consider a solution u ∈ C2(Ω) ∩ C0(Ω) of

Lu := ai jDi ju + biDiu ≤ f .

We rewrite this in the form

ai jDi ju ≤ −biDiu + f .

Define

h(p) = (|p| nn−1 + µ

nn−1 )n−1

for µ :=∣∣∣∣∣∣ fD1/n

∣∣∣∣∣∣Ln(Γ+)

6= 0 and D := det [A]. We infer the following estimate as before:

maxΩu ≤ max

∂Ωu + C

∣∣∣∣∣∣∣∣f

D1/n

∣∣∣∣∣∣∣∣Ln(Γ+)

,

where C depedends on n, d , and∣∣∣∣∣∣ fD1/n

∣∣∣∣∣∣Ln(Γ+)

.

13.2.1.4 Oblique boundary value problems

The above estimates can all be extended to boundary conditions of the form

β ·Du + γ · u = g on ∂Ω

where β · ν > 0 on ∂Ω, γ > 0 on Ω and g is a given function. Here ν denotes the outer

normal on ∂Ω. We obtain the more general estimate

infΩu ≥ inf

∂Ω

g

γ− C

(sup∂Ω

|β|γ

+ d

)

where d = Diam (Ω).

13.3 Local Estimates for Linear Equations

13.3.1 Preliminaries

Consider the linear operator Lu := ai jDi ju with strict ellipticity condition:

λI ≤ [ai j ] ≤ ΛI for0 < λ ≤ Λ, λ,Λ ∈ R.

To see the reason why we discuss the linear operator Lu, we deal ith the following nonlinear

equation of simple case:

F (D2u) = 0. (13.7)

Differentiate (13.8) with respect to a non-zero vector γ, to obtain

∂F

∂ri j(D2u)Di jγu = 0.


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13.3 Local Estimates for Linear Equations405

If we set L := ∂F∂ri jDi j , then we have LDγu = 0. Differentiate again and you find

∂F

∂ri jDi jγγu +

∂2F

∂ri j∂rklDi jγu ·Dklγu = 0.

If F is concave with respect to r , which most of the important examples satisfy, then we

derive

∂F

∂ri j(D2u)Di jDγγu ≥ 0,

i .e., L(Dγγu) ≥ 0.

This is the linear differential inequality and justifies our investigation of the linear operator

Lu = ai jDi ju.

13.3.2 Local Maximum Principle

Theorem 13.3.1 (): Let Ω ⊂ Rn be a bounded open domain and let BR := BR(y) ⊂ Ω be

a ball. Suppose u ∈ C2(Ω) fulfuills Lu ≥ f in Ω for some given function f . Then for any

0 < σ < 1, p > 0, we have

supBσR

u ≤ C(

1

Rn

∫

BR

(u+)p dx

)1/p

+R

λ||f ||Ln(BR)

,

where C depends on n, p, σ and Λ/λ. u+ = maxu, 0 and BσR(y) denotes the concentric

subball.

Proof. We may take R = 1 in view of the scaling x → x/R, f → R2f . Let

η =

(1− |x |2)β, β > 1 in B1

0 in Ω \ B1.

It is easy to verify that

|Dη| ≤ Cη1−1/β, |D2η| ≤ Cη1−2/β, (13.8)

for some constant C.

Introduce the test function v := η(u+)2 and compute

Lv = ai jDi jη(u+)2 + 2ai jDiηDj((u+)2) + ηai jDi j((u+)2)

≥ −CΛη1−2/β(u+)2 + 4u+ai jDiηDju+

+2ηai jDiu+ ·Dju+ + 2ηu+ai jDi ju

+

≥ −CΛη1−2/β(u+)2 + 2ηu+f .

here the use of (13.8) and the next inequality was made:

|ai jDiηDju+| ≤ (ai jDiηDjη)1/2(ai jDiu+Dju

+)1/2.

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Apply the Alexandrov maximum principle 2.3.1 to find

supB1

v ≤ C(n)

λ||Λη1−2/β(u+)2 + ηu+f ||Ln(B1)

≤ C(n)

λ||Λv1−2/β(u+)4/β + v1/2η1/2f ||Ln(B1)

≤ C

Λ

λ||v1−2/β(u+)4/β||Ln(B1) +

1

λ||v1/2η1/2f ||Ln(B1)

≤ C

Λ

λ

(supB1

v

)1−2/β

||(u+)4/β||Ln(B1) +

(supB1

v)1/2

λ||f ||Ln(B1)

.

Choosing β = 4n/p, we deduce

supB1

v1/2 ≤ C

Λ

λ

(supB1

v

)1/2−2/β (∫

B1

(u+)p dx

)1/n

+1

λ||f ||Ln(B1)

,

from which we conclude, invoking the Young inequality, that

supB1

v1/2 ≤ C(

Λ

λ, n, p

)·(∫

B1

(u+)p dx

)1/p

+1

λ||f ||Ln(B1)

.

This completes the proof.

13.3.3 Weak Harnack inequality

Lemma 13.3.2 (): Let u ∈ C2(Ω) satisfies Lu ≤ f , u ≥ 0 in BR := BR(y) ⊂ Ω, for some

function f . For 0 < σ < τ < 1, we have

infBσR

u ≤ C

infBτR

u +R

λ||f ||Ln(BR)

,

where C = C(n, σ, τ,Λ/λ).

Proof. Set for k > 0

w :=|x/R|−k − 1

σ−k − 1infBσR

u.

There holds w ≤ u on ∂BσR and ∂BR. Now, compute

L|x |−k = ai jDi j |x |−k

= ai j(−kδi j |x |−k−2 + k(k + 2)xixj |x |−k−4)

= −k |x |−k−2

(T rA− (k + 2)

ai jxixj|x |2

)

≥ −k |x |−k−2(T rA− (k + 2)λ)

≥ 0,

if we choose k > T rAλ − 2. Thus

Lw ≥ 0,

L(u − w) ≤ f .Gregory T. von Nessi

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Applying the Alexandrov maximum principle again, we obtain

u ≥ w − CRλ||f ||Ln(BR).

Take the infimum and you get the desired result.

Theorem 13.3.3 (): Let u ∈ C2(Ω) satisfies Lu ≤ f , u ≥ 0 in BR := BR(y) ⊂ Ω, for some

function f . Then for any 0 < σ,τ < 1, there exists p > 0 such that

(1

Rn

∫

BσR

up dx

)1/p

≤ C

infBτR

u +R

λ||f ||Ln(BR)

,

where p, C depend on n, σ, τ and Λ/λ.

Proof. WLOG thake y = 0 so that BR(0) = BR. Consider

η(x) = 1− |x |2

R2;

then η − u ≤ 0 on ∂BR with L(η − u) = Lη − Lu ≤ −2T rAR2 − g the above lemma now

indicates

η − u ≤ d

nω1/nn

∣∣∣∣∣

∣∣∣∣∣−2T rA

det [A]1/n · R2− g

det [A]1/n

∣∣∣∣∣

∣∣∣∣∣Ln(x∈BR | η−u≥0)

≤ d

λnω1/nn

∣∣∣∣∣∣∣∣−2

T rA

R2− g∣∣∣∣∣∣∣∣Ln(x∈BR | η−u≥0)

≤ Λ

λ

d

nω1/nn

∣∣∣∣∣∣∣∣−

2

R2− g

T rA

∣∣∣∣∣∣∣∣Ln(x∈BR | η−u≥0)

≤ CΛ

λ|x ∈ BR | η ≥ u| ≤

1

2η,

if

|u ≤ 1 ∩ BR||BR|

(13.9)

is suffciently small. Thus, we have u ≥ C on BσR and apply the previous lemma to ascertain

u ≥ C on BτR. Remarks:

i.) This ascertion contains all the ’information’ for the PDE.

ii.) We essentially shrink the ball after the calculation and 3.5 to then expand the ball back

out (perterbing by measure).

At this point, in order to simplify a later measure theoretic argument, we change our

basis domain from balls to cubes. Also take f = 0 and drop the bar for simplicity.

Let K := KR(y) be an open cube, parallel to the coordinate axis with center y and side

length 2R. The corresponding estimate of (13.9) is as follows: if

|u(x) ≥ 1 ∩K| ≥ θ|K| and K3R ⊂ B := B1, (13.10)

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then u ≥ γ on KR. Now consider any fixed cube K0 in Rn with Γ ⊂ K0 a measurable subset

and 0 < θ < 1. Defining

Γ :=⋃K3R(y) ∩K0

∣∣∣ |KR(y) ∩ Γ | ≥ θ|KR(y)|,

we have either Γ = K0 or |Γ | ≥ θ−1|Γ | by the cube decomposition at the beginning of the

notes. With this in mind, we make the following claim:

infKu ≥ C

( |Γ ||K|

)ln γ/ ln θ

. (13.11)

Proof of Claim We use induction. So if we take Γ := x ∈ K | u ≥ γ0 = 1, then we have

u ≥ γ on Γ with |Γ | ≥ θ−1|Γ |. This is the m = 1 step and is the result of the previous

argument in the proof. Now, we assume the condition |Γ | > θm|K|, m ∈ N, implies the

estimate

u ≥ γm+0 = γm on K.

The representation of γ0 = 1 may seem absurd at this point, but it will clarify the final

inductive step immensely. Now, to proceed to m + 1, we first note that given the inductive

hypothesis Γ also satisfies the weaker condition |Γ | ≥ θm+1|K|. Recalling the m = 1 case,

we also have from (13.10) that

u ≥ γ on Γ .

Now cube decomposition indicates that |Γ | ≥ θ−1|Γ | ≥ θm|K|. Thus, we ascertain

u ≥ γm+1 on K.

This argument is tricky in a sense that γ is not only apart of the inductive result, it’s a

paramater of Γ and Γ (hence the γ0) notation. Once this is clear, we get the claim by

picking γ and C correctly.

Returning to the ball, we infer that

infBu ≥ C

( |Γ ∩ B||B|

)κ, (13.12)

where C and κ are positive constants depending only on n and Λ/λ. Now define

Γt := x ∈ B | u(x) ≥ t.

Replacing u by u/t in (13.13) yields

infBu ≥ Ct

( |Γ ∩ B||B|

)κ.

To obtain the weak Harnack inequality, we normalize infB u = 1 and hence

|Γt | ≤ C · t−1/k .Gregory T. von Nessi

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Choosing p = (2κ)−1, we have

∫

B

up dx =

∫

B

∫ u(x)

0

ptp−1 dtdx

=

∫ ∞

0

ptp−1 · |x ∈ B | u ≥ t| dt

≤∫ ∞

0

ptp−1|Γt | dt

≤ C.

the second inequality is the coarea formula from geometric measure theory. Intuitivly, this

can be thought of as slicing the B × [0, u(x)] by hyperplanes w(x, t) = t, and integrating

these. Thus, we have obtained our result for B. The completion of the proof follows from

a covering argument.

∫

Ω

g(x) · |Jmf (x)| dx =

∫

Rm

∫

f −1(y)

g(x) dHn−m(x)dy

Theorems (13.3.1) and (13.3.3) easily give

Theorem 13.3.4 (): Suppose Lu = f , u ≥ 0 in BR := BR(y) ⊂ Ω. Then for any 0 < σ < 1,

we have

supBσR

u ≤ C(

infBRu +

R

λ||f ||Ln(B)

),

where C = C(n, σ,Λ/λ).

13.3.4 Holder Estimates

Lemma 13.3.5 (): Suppose a non-decreasing function ω on [0, R0] satisfies

ω(σR) ≤ σ · ω(R) + Rδ

for all R ≤ R0. Here σ < 1, δ > 0 are given constants. Then we have

ω(R) ≤ C(R

R0

)αw(R0),

where C, α depend on σ and δ.

Proof. We may take δ < 1. Replace ω by

ω(R) := ω(R)− ARδ,

and we obtain ω(σR) ≤ σ · ω(R), selecting A suitably. Thus, we have by iteration

ω(σνR) ≤ σν · ω(R0).

Choosing ν such that σν+1R0 < R ≤ σνR0 gives the result upon noting that ω is non-

decreasing.

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Now we present the next pointwise estimate, which is due to Krylov ans Safonov.

Theorem 13.3.6 (): Suppose u satsifies Lu = f in Ω ⊂ Rn for some f ∈ Ln(Ω). Let

BR := BR(y) ⊂ Ω be a ball and 0 < σ < 1. Then we have

oscBσR

u ≤ CσαoscBR

u +R

λ||f ||Ln(BR)

, (13.13)

where α = α(n,Λ/λ) > 0 and C = C(n,Λ/λ).

This is analogous to the famous De Giorgi-Nash estimate for the divergence equation

Di(ai j(x)Dju) = u.

Proof. Take BR = BR(y) ⊂ Ω as in the theorem and define

MR := supBR

u, mR := infBRu.

There holds

oscBR

u = MR −mr =: ω(R).

Applying the weak Harnack inequality to MR − u and mr − u respectively, we obtain

(1

Rn

∫

BσR

(MR − u)p dx

)1/p

≤ C

MR −MσR +

R

λ||f ||Ln(BR)

(1

Rn

∫

BσR

(u −mR)p dx

)1/p

≤ C

mσR −MR +

R

λ||f ||Ln(BR)

.

Add these two inequalities and you get

ω(R) ≤ Cω(R)− ω(σR) + R, (13.14)

where C depends on ||f ||Ln(BR). Here we have used the inequality:

(∫|f + g|p dx

)1/p

≤ 21/p

(∫|f |p dx

)1/p

+

(∫|g|p dx

)1/p.

(13.14) can be rewritten in the form

ω(σR) ≤ (1− 1/C)ω(R) + R.

Applying lemma 13.3.5 now yields the desired result.

13.3.5 Boundary Gradient Estimates

Now we turn our attention to the boundary gradient estimate, due to Krylov. The situation

is as follows: Let B+R be a half ball of radius R centered at the origin:

B+R := BR(0) ∩ xn > 0

Suppose u satisfiesLu := ai jDi ju = f in B+

R ,

u = 0 on BR(0) ∩ xn = 0. (13.15)

Then we haveGregory T. von Nessi

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Theorem 13.3.7 (): Let u fulfil (13.15) and let 0 < σ < 1. Then there holds

oscB+σR

u

xn≤ Cσα

oscB+R

u

xn+R

λsupB+R

|f |.

where α > 0,C > 0 depend on n and Λ/λ.

From this theorem, we infer the Holder estimate for the gradient on the boundary xn = 0.

In fact

osc|x ′|<σR

Du(x ′, 0) ≤ Cσα

supB+R

|Du|+ R

λsupB+R

|f |

holds, where x ′ = x1, . . . , xn−1. Proof: Define v := uxn

. We may assume that v is bounded

in B+R . Write for some δ > 0,

BR,δ := |x ′| < R, 0 < xn < δR.

Under the assumption that u ≥ 0 in B+R , we want to prove the next claim.

Claim: There exists a δ = δ(n,Λ/λ) such that

inf|x ′| < R

xn = δR

v ≤ 2

(infBR

2 ,δ

v +R

λsupB+R

|f |)

for any R ≤ R0.

Proof of Claim: We may normalize as before to take λ = 1, R = 1 and

inf|x ′| < R

xn = δR

v = 1.

Define the comparison function w in B1,δ as

w(x) :=

1− |x ′|2 + (1 + sup |f |)xn − δ√

δ

xn.

It is easy to check that w ≤ u on ∂B1,δ.

Compute

Lw = −2

n−1∑

i=1

ai ixn + 2(1 + sup |f |)ann√δ− 2

n−1∑

i=1

ainxi

≥ f in B1,δ

if we take δ sufficiently small. Thus the maximum principle implies that on B1/2,δ

v ≥ 1− |x ′|2 + (1 + sup |f |)xn − δ√δ

≥ 1

2− sup |f |

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again for sufficiently small δ. Normalizing back we obtain the claim.

To resume the proof of the theorem, we apply the Harnack inequality in the region|x ′| < R, δR2 < xn <

3δR2

to obtain

sup|x ′| < R

δR2 < xn <

3δR2

v ≤ C

inf|x ′| < R

2δR2 < xn <

3δR2

v +R

λsup |f |

(13.16)

≤ C

inf|x ′| < R

xn = δR

v +R

λsup |f |

≤ C

(infBR

2 ,δ

v +R

λsup |f |

).

The last inequality comes from the claim.

Now we set up the oscillation estimate: Define

MR := supBR,δ

v , mR := infBR,δ

v

ω(R) := MR −mR = oscBR,δ

v .

Consider M2R − v , v −m2R and invoke (13.16). We find

M2R − inf|x ′| < R

δR2 < xn <

3δR2

v ≤ C

M2R − sup

BR2 ,δ

v +R

λsup |f |

inf|x ′| < R

δR2 < xn <

3δR2

v −m2R ≤ C

(infBR

2 ,δ

v −m2R +R

λsup |f |

)

Adding each terms tells you

ω(2R) ≤ C(ω(2R)− ω(R/2) +

R

λsup |f |

).

Lemma 13.3.5 now implies our desired result:

ω(R) ≤ C(R

R0

)αω(R0) +

R0

λsup |f |

.

13.4 Holder Estimates for Second Derivatives

In this section, we wish to derive the second derivative estimates for general nonlinear equa-

tion of simple case

F (D2u) = ψ in Ω, (13.17)Gregory T. von Nessi

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13.4 Holder Estimates for Second Derivatives413

where Ω ⊂ Rn is a domain. General cases can be handled by perterbation arguments. See

Safonov [Sa2].

We begin with the interior estimates of Evans [Eva82] and Krylov [Kry82]. For further

study, suitable function spaces are introduced. Various interpolation inequalities are recalled.

Finally, we show the global bound for second derivates, due to Krylov. For the interior

estimate, we follow the treatment of [GT01] but with an idea of Safonov [Saf84] being used

to avoid the Motzkin-Wasco lemma. The corresponding boundary estimate will be derived

from Theorem 13.3.7.

13.4.1 Interior Estimates

Concerning (13.17), it is assumed to satisfy F ∈ C2(Rn×n), u ∈ C4(Ω), ψ ∈ C2(Ω) and

i.) λI ≤ Fr (D2u) ≤ ΛI, where 0 < λ ≤ Λ and Fr := [∂F/∂ri j ];

ii.) F is concave on some open set U ⊂ Rn×n including the range of D2u.

Under these assumptions, our first result is the next

Theorem 13.4.1 (): Let BR = BR(y) ⊂ Ω and let 0 < σ < 1. Then we have

oscBσR

D2u ≤ Cσα(oscBR

D2u +R

λsup |Dψ|+ R2

λsup |D2ψ|

),

where C,α > 0 depend only on n,Λ/λ.

Proof: Recall the connection between (13.17) and the linear differential inequalities stated

at 3.1.

Differentiating (13.17) with respect to γ twice, we obtain

Fi jDi jDγγu ≥ Dγγψ, (13.18)

in view of the assumption ii.), where Fi j := Fri j . If we set

ai j(x) := Fi j(D2u(x)), L := ai jDi j ,

then there holds

Lw ≥ Dγγψ for w = w(x, γ) := Dγγu

and λI ≤ [ai j ] ≤ ΛI.

Take a ball B2R ⊂ Ω and define as before

M2(γ) := supB2R

w(·, γ), M1(γ) := supBR

w(·, γ),

m2(γ) := infB2R

w(·, γ), m1(γ) := infBRw(·, γ).

Applying the weak Harnack inequality, Theorem to M2 − w on BR, we fine

R−n

∫

BR

(M2(γ)− w(x, γ))p dx

1/p

≤ CM2(γ)−M1(γ) +

R2

λsup |D2ψ|

.(13.19)

To establish the Holder estimat, we need a bound for −w . At present, however, we only

have the inequality (13.18). Thus we return to the equation

F (D2u(x))− F (D2(y)) = ψ(x)− ψ(y),

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for x ∈ BR, y ∈ B2R. Since F is concave:

ψ(y)− ψ(x) ≤ Fi j(D2u(x))(Di ju(y)−Di ju(x)) (13.20)

= ai j(x)(Di ju(y)−Di ju(x))

=: αi(x)µi ,

where µ1 ≤ · · · ≤ µn are the eigenvalues of the matrix D2u(y)−D2u(x) and λ ≤ αi ≤ Λ.

Setting

ω(2R) := max|γ|=1

(M2(γ)−m2(γ)), ω(R) := max|γ|=1

(M1(γ)−m1(γ)),

ω∗(2R) :=

∫

|γ|=1

(M2(γ)−m2(γ)) dγ, ω∗(R) :=

∫

|γ|=1

(M1(γ)−m1(γ)) dγ,

if follows by integration of (13.19) that

R−n

∫

|γ|=1

∫

BR

(M2(γ)− w(x, γ))p dxdγ

1/p

(13.21)

≤ Cω∗(2R)− ω∗(R) +

R2

λsup |D2ψ|

.

For any fixed x ∈ BR, one of the inequalities

max|γ|=1

(M1(γ)− w(x, γ)), max|γ|=1

(w(x, γ)− w(y , γ)) ≥ 1

2ω(R)

must be valid. Suppose it is the second one and choose y ∈ BR so that also

max|γ|=1

(w(x, γ)− w(y , γ)) ≥ 1

2ω(R).

The from (13.20) we have

µn = max|γ|=1

(w(y , γ)− w(x, γ)) ≥ − λ

(n − 1)Λµ1 −

2R

(n − 1)Λsup |Dψ|

≥ λ

2(n − 1)Λω(R)− 2R

(n − 1)Λsup |Dψ|.

Consequently, in all cases we obtain

max|γ|=1

(M1(γ)− w(x, γ)) ≥ δ(ω(R)− 4R

λsup |Dψ|

),

for some positive constant δ depending on n, Λ/λ.But then, because w is a quadratic form

in γ, we must have

∫

|γ|=1

(M1(γ)− w(x, γ)) dγ ≥ δ∗(ω(R)− 4R

λsup |Dψ|

)

for a further δ∗ = δ∗(n,Λ/λ), so that from (13.21), we obtain

ω(R) ≤ Cω∗(2R)− ω∗(R) +

R

λsup |Dψ|+ R2

λsup |D2ψ|

and the result follows by Lemma 13.3.5 and the following lemma of Safonov [Saf84].Gregory T. von Nessi

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Lemma 13.4.2 (): We have

ω(R) ≤ C1ω∗(R) ≤ C2ω(R),

for some positive constants C1 ≤ C2.

Proof: Since the right hand side inequality is obvious, it sufficies to prove

M1(γ)−m1(γ) ≤ C1ω∗(R)

for any |γ| = 1, so that without loss of generality, we can assume γ = e1. Writing

2D11u(x) = w(x, γ + 2e1)− 2w(x, γ + e1) + w(x, γ),

we obtain by integration of |γ| < 1, for any x, y ∈ BR,

2ωn(D11u(y)−D11u(x)) =

(∫

B1(2e1)

−2

∫

B1(e1)

+

∫

B1(0)

)(w(y , γ)− w(x, γ))

≤ 2

∫

B3(0)

(M1(γ)−m1(γ)) dγ

= 2

∫ 3

0

rn+1 dr

∫

|γ|=1

(M1(γ)−m1(γ)) dγ

and hence the result follows with C1 = 3n+2

(n+2)ωn.

The proof of the theorem is now complete.

13.4.2 Function Spaces

Here we introduce some suitable function spaces for further study on the classical theory

of second order nonlinear elliptic equations. The notations are compatible with those of

[GT01].

Let Ω ⊂ Rn be a bounded domain and suppose u ∈ C0(Ω). Define seminorms:

[u]0,α;Ω := supx,y∈Ω,x 6=y

|u(x)− u(y)||x − y |α , 0 < α ≤ 1,

[u]k;Ω := sup|β|=k,x∈Ω

|Dβu(x)|, k = 0, 1, 2, . . . ,

and the space

Ck,α(Ω) :=

u ∈ C

k(Ω)∣∣∣ ||u||k,α;Ω :=

k∑

j=0

[u]j ;Ω +∑

|β|=k[Dβu]α;Ω <∞

.

Ck,α(Ω) is a Banach space under the norm || · ||k,α;Ω. Moreover, we see

Ck(Ω) = Ck,0(Ω)

with the norm ||u||k;Ω :=∑kj=0[u]j ;Ω.

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Recall

Ck,α(Ω) = u ∈ Ck(Ω)∣∣ u∣∣Ω′ ∈ C

k,α(Ω′) for any Ω′ ⊂⊂ Ω,

and define the interior seminorms:

[u]∗α;Ω := supΩ′⊂⊂Ω

dαΩ′ [u]α;Ω, 0 < α ≤ 1,

where u ∈ C0,α(Ω), dΩ′ := Dist (Ω′, ∂Ω). Similarly

[u]∗k;Ω := supΩ′⊂⊂Ω

dkΩ′ [u]k;Ω′ , k = 0, 1, 2, . . . ,

[u]∗k,α;Ω := supΩ′⊂⊂Ω,|β|=k

dk+αΩ′ [Dβu]k;Ω′ ,

and

Ck,α∗ (Ω) :=

u ∈ C

k(Ω)∣∣∣ ||u||∗k,α;Ω :=

k∑

j=0

[u]∗j ;Ω + [u]∗k,α;Ω <∞

.

Ck,α∗ (Ω) is a Banach space under the norm || · ||∗k,α;Ω.

13.4.3 Interpolation inequalities

For the aboe defined spaces, there hold various interpolation inequalities.

Lemma 13.4.3 (): Suppose j+β < k+α, where j, k = 0, 1, 2, . . ., and 0 ≤ α, β ≤ 1. Let Ω

be an open subset of Rn+ with a boundary portion T on xn = 0, and assume u ∈ Ck,α(Ω∪T ).

Then for any ε > 0 and some constant C = C(ε, j, k) we have

[u]∗j,β;Ω∪T ≤ C|u|0;Ω + ε[u]∗k,α;Ω∪T ′ ,

|u|∗j,β;Ω∪T ≤ C|u|0;Ω + ε[u]∗k,α;Ω∪T . (13.22)

Proof: Again we suppose that the right members are finite. The proof is patterned after

(the interior global result you did earlier) and we emphasize only the details in which the proofs

differ. In the following we omit the subscript Ω ∪ T , which will be implicitly understood.

We consider first the cases 1 ≤ j ≤ k ≤ 2, β = 0, α ≥ 0, starting with the inequality

[u]∗2 ≤ C(ε)|u|0 + ε[u]∗2,α, α > 0. (13.23)

Let x be any point in Ω, dx its distance from ∂Ω − T , and d = µdx where µ ≤ 14 is

a constant to be specificed later. If Dist (x, T ) ≥ d , then the ball Bd(x) ⊂ Ω and the

argument proceeds as in (your interior proof) leading to the inequality

d2x |Di lu(x)| ≤ C(ε)[u]∗1 + ε[u]∗2,α

provided µ = µ(ε) is chosen sufficiently small. If Dist (x, T ) < d we consider the ball B =

Bd(x0) ⊂ Ω, where x0 is on the perpendicular to T passing through x and Dist (x, x0) = d .

Let x ′, x ′′ be the endpoints of the diamter of B parallel to the xl axis. Then we have for

some x on this diamater

|Di lu(x)| =|Diu(x ′)−Diu(x ′′)|

2d≤ 1

dsupB|Diu| ≤

2

µd−2x sup

y∈Bdy |Diu(y)|

≤ 2

µd−2x [u]∗1, since dy > dx/2 ∀y ∈ B;


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and

|Di lu(x)| ≤ |Di lu(x)|+ |Di lu(x)−Di lu(x)|

≤ 2

µd−2x [u]∗1 + 2dα sup

y∈Bd−2−αx,y sup

y∈Bd

2+αx,y

|Di lu(x)−Di lu(y)||x − y |α ;

hence

d2x |Di lu(x)| ≤ 2

µ[u]∗1 + 16µα[u]∗2,α

≤ C[u]∗1 + ε[u]∗2,α

provided 16µα ≤ ε, C = 2/µ. Choosing the smaller value of µ, corresponding to the two

cases Dist (x, T ) ≥ d and Dist (x, T ) < d , and taking the supremem over all x ∈ Ω and

i , l = 1, . . . , n, we obtain (the first equation in this proof) for j = k = 1, we proceed as

in (the proof you did for the interior result) with the modifications suggested by the above

proof of (the first equation of the proof). Together with the preceding cases, this gives us

(the equation in the statement of the lemma) for 1 ≤ j ≤ k ≤ 2, β = 0, α ≥ 0.

The proof of (the statement eq) for β > 0 follows closely that in cases (iii) and (iv) of

(the interior proof you did before). The principal difference is that the argument for β > 0,

α = 0 now requires application of the theorm of the mean in the truncated ball Bd(x) ∩ Ω

for points x such that Dist (x, T ) < d .

Lemma 13.4.4 (): Suppose k1 + α1 < k2 + α2 < k3 + α3 with k1, k2, k3 = 0, 1, 2, . . .,

and 0 < α1, α2, α3 ≤ 1. Suppose further ∂Ω ∈ Ck3,α3 . Then for any ε > 0, there exists a

constant Cε = O(ε−k) for some positive k such that

||u||k2,α2;Ω ≤ ε||u||k2,α3;Ω + Cε||u||k1,α;Ω

holds for any u ∈ Ck3,α3 (Ω).

Proof: The proof is based on a reduction to lemma 6.34 by means of an argument very

similar to that in Lemma 6.5. As in that lemma, at each point x0 ∈ ∂Ω let Bρ(x0) be a ball

and ψ be a Ck,α diffeomorphism that straightens the boundary in a neighborhood containing

B′ = Bρ(x0) ∩ Ω and T = Bρ(x0) ∩ ∂Ω. Let ψ(B′) = D′ ⊂ Rn+, ψ(T ) = T ′ ⊂ ∂Rn+. Since

T ′ is a hyperplane portion of ∂D′, we may apply the interpolation inequality from previous

lemma in D′ to the function u = u ψ−1 to obtain

[u]j,β;D′∪T ′ ≤ C(ε)|u|0;D′ + ε|u|∗k,α;D′∪T ′ .

fromLet Ω be a bounded domain with Ck,α boundary portion T, k ≥ 1, 0 ≤ α ≤ 1. Suppose

that Ω ⊂⊂ D, where D is a domain that is mapped by a Ck,α diffeomorphism ψ onto D′.Letting ψ(Ω) = Ω′ and ψ(T ) = T ′, we can define the quantities in (definitions of spaces)

with respect to Ω′ and T ′. If x ′ = ψ(x), y ′ = ψ(y), one sees that

K−1|x − y | ≤ |x ′ − y ′| ≤ K|x − y |

for all points x, y ∈ Ω, where K is a constant depending on ψ and Ω. Letting u(x)→ u(x ′)under the mapping x → x ′, we find after a calculation using the above equation: for 0 ≤ j ≤k ,0 ≤ β ≤ 1, j + β ≤ k + α,

[u]j,β;B′∪T ≤ C(ε)|u|0;D′ + ε|u|∗k,α;B′∪T .

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In these inequalities K denotes constants depending on the mapping ψ and the domain Ω.if follows that

[u]j,β;B′∪T ≤ C(ε)|u|0;D′ + ε|u|∗k,α;B′∪T .

(we recall that the same notation C(ε) is being used for different functions of ε.) Letting

B′′ = Bρ/2(x0) ∩Ω, we infer from

We note that |u|∗k;Ω and |u|∗k,α;Ω are norms on the subspaces of Ck(Ω) and Ck,α(Ω)

respectively for which they are finite. If Ω is bounded and d = Diam (Ω), then obviously

these interior norms and the global norms (standard defs) are related by

|u|∗k,α;Ω ≤ max(1, dk+α)|u|k,α;Ω.

If Ω′ ⊂⊂ Ω and σ = Dist (Ω, ∂Ω), then

min(1, σk+α)|u|k,α;Ω′ ≤ |u|∗k,α;Ω

|u|j,β;B′′ ≤ C(ε)|u|0,B′ + ε|u|k,α;B′

≤ C(ε)|u|0;Ω + ε|u|k,α;Ω. (13.24)

Let Bρi/4(xi), xi ∈ ∂Ω, i = 1, . . . , N, be a finite collection of balls covering ∂Ω, such that

the intequality (2nd eq prev lemma) holds in each set B′′i = Bρi/2(xi) ∩ Ω, with a constant

Ci(ε). Let δ = min ρi/4 and C = C(ε) = maxCi(ε). Then at every point x0 ∈ ∂Ω, we have

B = Bδ(x0) ⊂ Bρi/2(xi) for some i and hence

|u|j,β;B∩Ω ≤ C|u|0;Ω + ε|u|k,α;Ω (13.25)

The remainder of the argument is analogous to that in Theorem 6.6 and is left to the

reader.

13.4.4 Global Estimates

Returning to the interior estimate, we can write the second derivative estimate Theorem 4.1

in the form:

[u]∗2,α;Ω ≤ C([u]∗2;Ω + 1),

where C depends on λ,Λ, n, ||ψ||2;Ω and Diam (Ω). Therefore if λ,Λ are independent of u,

we obtain by interpolation

||u||∗2,α;Ω ≤ C(

supΩ|u|+ 1

).

Now we want to prove the next global second derivative Holder estimate due to essentially

Krylov [Kry87].

Theorem 13.4.5 (): Let Ω ⊂ Rn be a bounded domain with ∂Ω ∈ C3. Suppose u ∈C4(Ω) ∩ C3(Ω) fulfils

F (D2u) = ψ in Ω

u = 0 on ∂Ω,

where ψ ∈ C2(Ω). F ∈ C2(Sn) is assumed to satisfy i.) and ii.) in the first theorem of this

section. Then we have an estimate

|u|2,α;Ω ≤ C(|u|2;Ω + |ψ|2;Ω),

where α = α(n,Λ/λ) > 0, C = C(n,Ω,Λ/λ).Gregory T. von Nessi

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Proof: First we straighten the boundary.

Let x0 ∈ ∂Ω. By (Defintion of boundary reg.), there exists a ball B = B(x0) ⊂ Rn and

a one-to-one mapping χ : B → D ⊂ Rn such that

i.) χ(B ∩Ω) ⊂ Rn+ = x ∈ Rn | xn > 0

ii.) χ(B ∩ ∂Ω) ⊂ ∂Rn+iii.) χ ∈ C3(B), χ−1 ∈ C3(Ω).

Let us write y := χ(x). An elementary calculation shows

∂u

∂xi=

∂χk

∂xi(x) · ∂u

∂yk

∂2u

∂xi∂xj=

∂χk

∂xi

∂χl

∂xj· ∂2u

∂yk∂yl+

∂

∂xj

(∂χk

∂xi

)· ∂u∂yk

.

Thus, the ellipticity condition is preserved under the transformation χ with

λI ≤ [J Fr JT ] ≤ ΛI, J := Dχ,

where 0 < λ ≤ Λ < ∞. Moreover, χ also preserves the Holder spaces and transforms the

equation (13.26) into the more general form:

F (x, u,Du,D2u) = 0.

Therefore it suffices to deal with the following situation:

Suppose u satisfiesF (x, u,Du,Du) = 0 in B := B+

R = x ∈ Rn+ |x | < Ru = 0 on T := x ∈ B | xn = 0,

where F ∈ C3(B × R× Rn × Sn).

Differentiation with respectto xk , 1 ≤ k ≤ n − 1 leads

∂F

∂ri jDi jku +

∂F

∂piDiku +

∂F

∂uDku +

∂F

∂xk= 0,

which can be summarized in the formLv = f in B

v = 0 on T.(13.26)

Here we have set v := Dku, 1 ≤ k ≤ n − 1 and

L = ai jDi j :=∂F

∂ri jDi j

−f :=∂F

∂piDi ju +

∂F

∂uDku +

∂F

∂xk,

with λI ≤ [ai j ] ≤ ΛI.

You can apply the Krylov boundary estimate (previous boundary result) to obtain

oscB+δ

v

xn≤ C

(δ

R

)αoscB+R

v

xn+R

λ|f |0;B+

R

≤ C

(δ

R

)α|u|2;B+

R+R

λ|f |0;B+

R

.

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where α,C depend on n and Λ/λ. By subtracting a linear function of xn from v if necessar,

we can assume that v(x) = O(|xn|2) as x → 0. Thus, fixing R and taking δ = xn, we find

v(x)

x1+αn

≤ C(|u|2;B+R

+ |f |0;B+R

). (13.27)

This is the boundary estimate to start with.

For the interior estimate, Theorem 4.1 gives for any Br ⊂ B+R

oscBσr

≤ Cσα(

supBr

|D2u|+ 1

)

≤ Cσα(

supBr

|DD′u|+ 1

), (13.28)

where D′u = (D1u, . . . , Dn−1u), α = α(n,Λ/λ) > 0 and C depend furthermore on f . The

second inequality in (13.28) comes from the equation itself. In fact, F (x, u,Du,D2u) = 0

can be solved so that Dnnu = function of (x, u,Du,DD′u) and hence

|Dnnu| ≤ C(sup |DD′u|+ 1).

(13.28) implies in particular

oscBσR

DD′u ≤ Cσα(

supBr

|DD′u|+ 1

).

By interpolating the interior norm, we get from (13.27)

[DDku]α;Br ≤ CDist (Br , T )−1−α |D′u|0;Br + 1≤ C(|u|2;B+

R+ |f |0;B+

R), (13.29)

for 1 ≤ k ≤ n − 1.

The equation itself again shows that (13.29) holds true for k = n.

In summary, we finally obtain

[D2u]α;B+RC

(|u|2;B+R

+ |ψ|2;B+R

).

Usual covering argument now yields the full global estimate. This completes the proof.

Corollary 13.4.6 (): Under the assumptions of the previous theorem, we have by interpola-

tion

|u|2,α;Ω ≤ C(|u|0;Ω + |ψ|2;Ω)

≤ C|ψ|2;Ω.

13.4.5 Generalizations

The method of proof of theorem above provides more general results. We present one of

them without proof.

Theorem 13.4.7 (): If u ∈ C2(Ω) ∩ C0(Ω) solves the linar partial differential equation of

the type (13.26):

Lu = f in Ω

u = 0 on ∂Ω,Gregory T. von Nessi

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13.5 Existence Theorems421

for a bounded domain Ω ⊂ Rn with ∂Ω ∈ C2, f ∈ L∞(Ω) and

oscBσR

Du ≤ C0σα

(oscBR

Du + |f |0;Ω + 1

),

for some δ = δ(n, α,Λ/λ), C = C(n, α, C0,Ω,Λ/λ.

The form of the interior gradient estimate can be weakened; (see [Tru84], Theorem 4).

13.5 Existence Theorems

After establishing various estimates, we are now able to investigate the problem of the

existence of solutions. This section is devoted to the solvability for equations of the form

F (D2u) = ψ, (13.30)

Two cases are principally discussed; (i) uniformly elliptic case and (ii) Monge-Ampere equa-

tion.

We begin with recalling the method of continuity. Then (i)(ii) are studied.

13.5.1 Method of Continuity

First we quickly review the usual existence method of continuity.

Let B1 and B2 be Banach spaces and let F : U → B2 be a mapping, where B ⊂ B1 is an

open set. We say that F is differentiable at u ∈ U if there exists a bounded linear mapping

L : B1 → B2 such that

||F [u + h]− F [u]− Lh||B2

||h||B1

→ 0 as h → 0 in B1. (13.31)

We write L = Fu and call it the Frechet derivative F at u. When B1, B2 are Euclidean

spaces, Rn, Rm, the Frechet derivative coincides with the usual notion of differentia, and ,

moreover, the basic theory for the infinite dimensional cas can be modelled on that for the

finite dimensional case as usually treated in advanced calculus. In particular, it is evident

from (the above limit) that the Frechet differentiablity of F at u implies that F is continuous

at u and that the Frechet derivative Fu is determined uniquely by (the limit def above). We

call F continuously differentiable at u if F is Frechet differentiable in a neighbourhood of u

and the resulting mapping

v 7→ Fv ∈ E(B1,B2)

is continuous at u. Here E(B1,B2) denotes the Banach space of bounded linear mapping

from B1 to B2 with the norm given by

||L|| = supv ∈ B1v 6= 0

||Lv ||B2

||v ||B1

.

The chain rule holds for Frechet differentiation, that is, if F : B1 → B2, G : B2 → B3 are

Frechet differentiable at u ∈ B1, F [u] ∈ B2, respectively, then the composite mapping G Fis differentiable at u ∈ B1 and

(G F )u = GF [u] Fu. (13.32)

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The theorem of the mean also holds in the sense that if u, v ∈ B1, F : B1 → B2 is

differentiable on the closed line segment γ joining u and v in B1, then

||F [u]− F [v ]||B2≤ K||u − v ||B1

, (13.33)

where

K = supw∈γ||Fw ||.

With the aid of these basic properties we may deduce an implicit fuinction theorem for

Frechet differentiable mappings. Suppose that B1, B2 and X are Banach spaces and that

G : B1 × X → B2 is Frechet differentiable at a point (u, σ), where u ∈ B1 and σ ∈ X. The

partial Frechet derivatives, G1(u,σ), G2

(u,σ) at (u, σ) are the bounded linear mappings from B1,

X, respectively, into B2 defined by

G(u,σ)(h, k) = G1(u,σ)(h) + G2

(u,σ)(k)

for h ∈ B1, k ∈ X. We state the implicit function theorem in the following form.

Theorem 13.5.1 (): Let B1, B2 and X be Banach spaces and G a mapping from an open

subset of B1 ×X into B2. Let (u0, σ0) be a point in B1 ×X satisfying:

i.) G[u0, σ0] = 0;

ii.) G is continuously differentiable at (u0, σ0);

iii.) the partial Frechet derivative L = G1(u0,σ0) is invertible.

Then there exists a neighborhood N of σ0 in X such that the equation G[u, σ] = 0, is

solvable for each σ ∈ N , with solution u = uσ ∈ B1.

Sketch of proof: This may be proved by reduction to the contraction mapping principle.

Indeed the equation G[u, σ] = 0, is equivalent to the equation

u = Tu ≡ u − L−1G[u, σ],

and the operator T will, by virtue (13.31) and (13.32), be a contraction mapping in a closed

ball B = Bδ(u0) in B1 for δ sufficiently small and σ sufficiently close to σ0 in X. One

can further show that the mapping F : X → B1 defined by σ → uσ for σ ∈ N , uσ ∈ B,

G[uσ, σ] = 0 is differentiable at σ0 with Frechet derivative

Fσ0 = −L−1G2(u0,σ0).

In order to apply the above theorem we suppost that B1 and B2 are Banach spaces with F

a mapping from an open subset U ⊂ B1 into B2. Let ψ be a fixed element in U and define

for u ∈ U , t ∈ R the mapping G : U × R→ B2 by

G[u, t] = F [u]− tF [ψ].

Let S and E be the subsets of [0, 1] and B1 defined by

S := t ∈ [0, 1] |F [u] = tF [ψ] for some u ∈ UE := u ∈ U |F [u] = tF [ψ] for some t ∈ [0, 1].

Clearly 1 ∈ S, ψ ∈ E so that S and E are not empty. Let us next suppose that the mapping F

is continuously differentiable on E with invertible Frechet derivative Fu. It follows then from

the implicit function theorem (previous theorem) that the set S is open in [0.1]. Consequently

we obtain the following version of the method of continuity.Gregory T. von Nessi

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Theorem 13.5.2 (): Suppose F is continuously Frechet differentiablewith the invertable

Frechet derivative Fu. Fix ψ ∈ U and set

S := t ∈ [0, 1] |F [u] = tF [ψ] for some u ∈ UE := u ∈ U |F [u] = tF [ψ] for some t ∈ [0, 1].

Then the equation F [u] = 0 is solvable for u ∈ U if S is closed.

The aim is to apply this to the Dirichlet problem. Let us define

B1 := u ∈ C2,α(Ω) | u = 0 on ∂ΩB2 := Cα(Ω),

and let F : B1 → B2 be given by

F [u] := F (x, u,Du,D2u).

Assume F ∈ C2,α(Ω × R × Rn × Sn). Then F is continuously Frechet differentiable in B1

with

Fv [u] =∂F

∂ri j(x, v(x), Dv(x), D2v(x))Di ju

+∂F

∂pi(x, v(x), Dv(x), D2v(x))Diu +

∂F

∂z(x, v(x), Dv(x), D2v(x))u.

Check this formular and observe that F ∈ C2,α is more than enough.

To proceed further, we recall the result of Classical Schauder theory proven in Chapter

9(?) (put here for convenience:

Theorem 13.5.3 (): Consider the linear operator

Lu := ai jDi ju + biDiu + cu,

in a bounded domain Ω ⊂ Rn with ∂Ω ∈ C2,α, 0 < α < 1. Assume

ai j , bi , c ∈ Cα(Ω), c ≤ 0

λI ≤ [ai j ] ≤ ΛI, 0 < λ ≤ ΛI, 0 < λ ≤ Λ <∞.

Then for any function f ∈ Cα(Ω), there exists a unique solution u ∈ C2,α(Ω) satisfying

Lu = f in Ω, u = 0 on ∂Ω and |u|2,α;Ω ≤ C|f |α;Ω, where C depends on n, λ, |ai j , bi , c |α;Ω

and Ω.

Theorem 13.5.4 (): Let Ω ⊂ Rn be a bounded domain with ∂Ω ∈ C2,α for some 0 < α < 1.

Let U ⊂ C2,α(Ω) be open and let ψ ∈ U . Define

E := u ∈ U |F [u] = σF [ψ] for some σ ∈ [0, 1] and u = ψ on ∂Ω,

where F ∈ C2,α(Γ ) with Γ := Ω× R× Rn × Sn. Suppose further

i.) F is elliptic on Ω with respect to u ∈ U ;

ii.) Fz ≤ 0 on U ; (or Fz ≥ 0 on U?)

iii.) E is bounded in C2,α(Ω);

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iv.) E ⊂ U , where the closure is taken in an appropriate sense.

Then there exists a unique solution u ∈ U of the following Dirichlet problem:

F [u] = 0 in Ω

u = ψ on ∂Ω.

Proof: We can reduce to the case of zero boundary values by replacing u with u − ψ.

The mapping G : B1 × R→ B2 is then defined by taking ψ ≡ 0 so that

G[u, σ] = F [u + ψ]− σF [ψ].

It then follows from the previous theorem and the remarks proceeding the statement of this

theorem, that the given Dirichlet problem is solvable provided the set S is closed. However

the closure of S (and also of E) follows readily from the boundedness of E (and hypothesis

iv.)) by virtue of the Arzela-Ascoli theorem.

Observe that by this theorem, the solvability of the Dirichlet problem depends upon the

estimates in C2,α(Ω) independent of σ (this is why we need Schauder theory).

13.5.2 Uniformly Elliptic Case

Let Ω ⊂ Rn be a bounded domain with ∂Ω ⊂ C3. Suppose F ∈ C2(Sn) satisfying the

following

i.) λI ≤ Fr (s) ≤ ΛI for all s ∈ Sn with some fixed constants 0 < λ ≤ Λ.

ii.) F is concave.

We also assume without loss of generality that F (0) = 0.

Theorem 13.5.5 ((Krylov [Kry82])): Under the above assumptions, the Dirichlet problem

F (D2u) = ψ in Ω

u = 0 on ∂Ω(13.34)

is uniquely solvable for any ψ ∈ C2(Ω). The solution u belongs to C2,α(Ω) for each 0 <

α < 1.

Proof: Recall the a-priori estimate (global result prev. section)

[u]2,αΩ ≤ C(|u|0;Ω + |ψ|2;Ω),

where α = α(n,Λ/λ) > 0, C = C(n, ∂Ω,Λ/λ).

Since we can write

F (D2u) =

(∫ 1

0

Fri j (θD2u) dθ

)Di ju, (13.35)

by virture of F (0) = 0, the classical maximum principle (find this) gives

|u|0;Ω ≤|ψ|0;Ω

λ(Diam (Ω))2.


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It is easy to see that the corresponding estimates for F (D2u) = σψ, σ ∈ [0, 1] are inde-

pendent of σ. Thus the result follows from the method of continuity.

Remark: Once you get u ∈ C2(Ω), then the linear theory ensures u ∈ C2,α(Ω) for all

α < 1.

The above proof owes its validity to the regularity of F . If you want to examine a non-smooth not the most obvi-

ous argument actu-

ally;explain it

F , you need to mollify F ; replace F by

Fε(r) =

∫

Snρ

(r − sε

)F (s) ds,

where ρ ∈ C∞0 (Sn) is defined by

ρ ≥ 0,

∫

Snρ(s) ds = 1.

Observe that Fε satisfies the assumptions (i)(ii). Then solve the problem

Fε(D

2uε) = ψ in Ω

uε = 0 on ∂Ω,

and let ε→ 0. The examples include the following Bellman equation:

F (D2u) = infk=1,...,n

ai jkDi ju.

Next, we wish to modify the argument to include the non-zero boundary value problems,

especially with continuous boundary values:

F (D2u) = ψ in Ω

u = g on ∂Ω(13.36)

where g ∈ C0(∂Ω).

As a first step,we approximate g by gn ⊂ C2(Ω) such that gn → g uniformly on ∂Ω

as n →∞. Secondly, in view of ∂Ω ∈ C3, you can construct barriers; for any y ∈ ∂Ω, there

exists a neighbourhood Ny of y and a function w ∈ C2(Ny ) such that

i.) w > 0 in Ω ∩Ny − y

ii.) w(y) = 0,

iii.) and F (D2w) ≤ ψ in Ω ∩Ny .

(refer to peron process for proofs about barries).

Suppose un ⊂ C2,α(Ω) ∩ C0(Ω) are given by the solutions of

F (D2un) = ψ in Ω

un = gn on ∂Ω

Then we can appeal to the linear theory to conclude that un are equicontinuous on Ω,

taking into account (13.34) and the existence of barriers.

On the other hand, the Schauder estimate (put the theorem here) implies the equicon-

tinuity of Du, D2u on any compact subset of Ω. Therefore, by letting n → ∞, we obtain

u ∈ C0(Ω) ∩ C2,α(Ω) which enjoys (13.36).

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13.5.3 Monge-Ampere equation

The equation we wish to examine here is

detD2u = ψ in Ω, (13.37)

where Ω ⊂ Rn is a iniformly convex domain with ∂Ω ∈ C3.

(13.37) is elliptic with respect to u provided D2u > 0. Thus the set U of admissable

functions is defined to be

U := u ∈ C2,α(Ω) |D2u > 0.

Now our main result is the next

Theorem 13.5.6 ((Krylov [Kry83a, Kry83b], Caffarelli, Nirenberg and Spruck [CNS84])):

Let Ω ⊂ Rn be a uniformly convex domain with ∂Ω ∈ C3 and let ψ ∈ C2(Ω) be such that

ψ > 0 in Ω. Then there exists a unique solution u ∈ U of

detD2u = ψ in Ω

u = 0 on ∂Ω.(13.38)

For the rest of this subsection, we prove this theorem, employing apriori estimates step-

by-step Proof:

Step 1: |u|2;Ω estimate implies |u|2,α;Ω estimate.

First, we show the concavity of F (r) := (det r)1/n for r > 0. Recalling,

(detD2u)1/n(detA)1/n ≤ 1

n(ai jDi ju),

we have

(detD2u)1/n = infdetA≥1

1

n(ai jDi ju).

This proves the concavity of F , since the RHS is a concavity function of D2u.

Next recall detD2u =∏ni=1 λi , where λ1, . . . , λn are the eigenvalue of D2u.

Thus we have

λi =ψλi∏nj=1 λj

≥ inf ψ

(sup |D2u|)n−1,

which means that the C2 boundedness of u ensures the uniform ellipticity of the equation.

The (globals C2,α estimate theorem thingy) then implies

|u|2,α;Ω ≤ C,

where C = C(|u|2;Ω, |ψ|2;Ω, ∂Ω). This proves Step 1.

Step 2: |u|0;Ω and |u|1;Ω estimates.

Since u is convex, we have u ≤ 0 in Ω. Let w := A|x |2 − B and compute

detD2w = (2nA)n.

Taking A, B large enough, we find

detD2w ≥ ψ in Ω

w ≤ u on ∂Ω,Gregory T. von Nessi

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from which we obtain |u|0;Ω ≤ C by applying the maximum principle.

Next we want to give a bound for |u|1;Ω. The convexity implies

supΩ|Du| ≤ sup

∂Ω|Du|.

Thus it sufficies to estimate Du on the boundary ∂Ω. More precisely, we with to estimate

uν from above, where uν denotes the exterior normal derivative of u on ∂Ω. This is because

uν ≥ 0 on ∂Ω and the tangential derivatives are zero.

We exploit the uniform convexity of the domain Ω; for any y ∈ ∂Ω, there exists a C2

function w defined on a ball Ω ⊂ By such that w ≤ 0 in Ω, w(y) = 0 and detD2w ≥ ψ in

Ω. This can be seen by taking Ω ⊂ By with ∂By ∩Ω = y and considering the function of

the type A|x − x0|2 − B as before.

Therefore, we infer

w ≤ u in Ω

and in particular wν ≥ uν at y . Since ∂Ω is compact, we obtain |u|1;∂Ω estimate.

Step 3: sup∂Ω |D2u| estimate implies supΩ |D2u| estimate.

We use the concavity of

F (D2u) := (detD2u)1/n = ψ1/n.

as in (the linearization first part), we deduce

Fi jDi jDγγu ≥ Dγγ(ψ1/n) for |γ| = 1.

Then the result of (the linear app of AB max prin) leads to

Dγγu ≤ sup∂ΩDγγu + C · sup

Ω

|Dγγ(ψ1/n)|T r(Fi j)

.

On the other hand, we have

Fi j =1

n(detD2u)−1+1/n · ui j ,

where [ui j ] denotes the cofactor matrix of [ui j ]. Compute

detFi j =1

nn

T rFi jn

≥ (detFi j)1/n =

1

n.

Thus, there holds

Dγγu ≤ sup∂ΩDγγu + C,

where C depends only on the know quantitites. This proves Step 3.

Step 4: sup∂Ω |D2u| estimate.

Take any y ∈ ∂Ω. Without loss of generality, we may suppose that y is the origin and

the xn-axis points the interior normal direction.

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Near the origin, ∂Ω is represented by

xn = ω(x ′) :=1

2Ωαβxαxβ +O(|x ′|3).

Here x ′ := (x1, . . . , xn−1) and in the summation, Greek letters go from 1 to n − 1. The

uniform convexity of ∂Ω is equivalent to the positivity of Ωαβ. For more details, see (section

after next).

Our aim is an estimation for |ui j(0)|. First, we see that the boundary condition u = 0 on

∂Ω implies

(Dβ + ωαDn)(Dα + ωβDn)u = 0

on ∂Ω near the origin; in particular,

|Dαβu(0)| ≤ C

at the origin.Next,differential the equation (13.43) to obtain

ui jDi j(Dαu + ωαDnu) = Dαψ + ωαDnψ + 2ui jωαiDjnu + ui jDnuDi jω.

By virtue of ui jDjnu = δin detD2u,there holds

|ui jDi j(Dα + ωαDnu)| ≤ C(1 + T rui j),

in Ω ∩N denotes a suitable neighbourhood of the origin.

On the other hand, introduce a barrier function of the type

w(x) := −A|x |2 + Bxn,

and compute

ui jDi jw = −2ATrui j .

Choosing A and B large enough, we conclude that

|ui jDi j(Dαu + ωαDnu)|+ ui jDi jw ≤ 0 in Ω ∩N|Dαu + ωαDnu| ≤ w on ∂(Ω ∩N ).

By the maximum principle, we now find

|Dαu + ωαDnu| ≤ w inΩ ∩N

and hence

|Dnαu(0)| ≤ B.

To summarize, we have so far derived

|Dklu(0)| ≤ C if k + l < 2n (13.39)

Finally, we want to estimate |Dnnu(0)|. Since u = 0 on ∂Ω, the unit outer normal ν on ∂Ω

is given by

ν =Du

|Du|Gregory T. von Nessi

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and so

D2u(0) =

(|Du|D′ν Dinu

Dniu Dnnu

)(x), (13.40)

where D′ := (∂/∂x1, . . . , ∂/∂xn−1). The choice of our coordinates gives

D′ν(0) = Diag[κ′1, . . . , κ

′n−1

](0)

and |Du(0)| = |Dnu(0)|. Here κ′1, . . . , κ′n−1 denote the principal curvatures of ∂Ω at 0. (see

section after next). By our regularity assumption on ∂Ω, for any y ∈ ∂Ω, there exists an

interior ball By ⊂ Ω with ∂By ∩ ∂Ω = y. Thus the comparison argument as before yields

|Du| ≥ δ > 0 (13.41)

for some δ > 0.

In view of (13.40), we have

ψ(0) = detD2u(0)

= |Dnu|n−2n−1∏

i=1

κ′i

|Dnu|Dnnu −

n−1∑

i=1

(Dinu)2

κ′i

(0),

from which we infer

0 < Dnnu(0) ≤ C, (13.42)

taking into account (13.41) and (13.39) and the uniform convexity κ′i > 0. This proves our

claim.

Now we have established an estimate for |u|2,α;Ω and the method of continuity can be

applied to compete the proof.

13.5.4 Degenerate Monge-Ampere Equation

Here we want to generalize (the previous theorem) so that ψ merely enjoys ψ ≥ 0 in Ω.

In the proof of (the above theorem), the steps 2 and 3 are not altred. The step 4 is also

as before except for the estimation of Dnu away from zero (see (13.41), which follows from

the convexity of u and being able to assume ψ(x0) > 0 for some x0 ∈ Ω. The step 1 does

not hold true in general.

To deal with these points, we approximate ψ by ψ + ε for ε > 0 sufficiently small.

Theorem (above) is applicable and we obtain a convex C2,α solution uε. We with to make

ε→ 0. However, an estimation independent of ε can be obtained up to |u|1;Ω.

In summarly, by virtue of the convexity of uε, we derive

Theorem 13.5.7 (): Let Ω ⊂ Rn be a uniformly convex domain with ∂Ω ∈ C3 and let

ψ ∈ C2(Ω) be non-negative in Ω. Then there exists a unique solution u ∈ C1,1(Ω) of

detD2u = ψn in Ω

u = 0 on ∂Ω.(13.43)

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13.5.5 Bondary Curvatures

Here we collect some properties of boundary curvatures and the distance function, which are

taken frpm the Appendix in [GT01, Chapter 14].

Let Ω ⊂ Rn be a bounded domain with its boundary ∂Ω ∈ C2. Take any rotation

of coordinate if necessary, the xn axis lies in the direction of the inner normal to ∂Ω at

the origin.Also, there exists a neighbourhood N of the origin such that ∂Ω is given by the

height function xn = ω(x ′) whre x ′ := (x1, . . . , xn−1). Moreover, we can assume that h is

represented by

ω(x ′) =1

2Ωαβxαxβ +O(|x ′|3), Ωαβ = Dαβω(0),

in N ∩ T where T denotes the tangent hyperplane to ∂Ω at the origin.

The principal curvatures of ∂Ω at the origin are now defined as the eigenvalues of [h]i jand we denote the κ′1, . . . , κ

′n−1. The corresponding eigenvectors are called the principal

directions.

We define the tangential gradient ∂ on ∂Ω as

∂ := D − ν(ν ·D),

where ν denotes the unit outer normal to ∂Ω. ν can be extended smoothly in a neighbourhood

of ∂Ω. In terms of ∂, we infer that κ′1, . . . , κ′n−1, 0 are given by the eigenvalues of [∂ν].

Let d(x) := Dist (x, ∂Ω) for x ∈ Ω. Then we find d ∈ C0,1(Ω) and d ∈ C20 < d < awhere a := 1/max∂Ω|κ′|. We call d(x) the distance function. Remark that ν(x) = −Dd(x)

for x ∈ ∂Ω.

A principal coordinate system at y ∈ ∂Ω consists of the xn axis being along the unit

normal and others along the principal directions. For x ∈ x ∈ Ω | 0 < d(x) < a, let

y = y(x) ∈ ∂Ω be such that |y(x)− x | = d(x). Then with respect to a principal coordinate

system at y = y(x), we have

[D2d(x)] = Diag

[ −κ′11− dκ′1

, . . . ,−κ′n−1

1− dκ′n−1

, 0

],

and for α, β = 1, 2, . . . , n − 1,

Dαβu = ∂α∂βu −Dnu · κ′αδαβ= ∂α∂βu −Dnu ·Ωαβ.

If u = 0 on ∂Ω, then Dαβu = −Dnu ·Ωαβ for α, β = 1, 2, . . . , n − 1 at y = y(x).

13.5.6 General Boundary Values

The estimations in (ma equation) extend automatically to the case of general Dirichlet bound-

ary values given by a function φ ∈ C3(Ω), except for the double normal second derivative on

the boundary. Here the reultant estimate (for φ ∈ C3,1(Ω)) is due to Ivochkina [Ivo90]. We

indicate here a simple proof of this estimate using a recent trick. introduced by Trudinger in

[Tru95], in the study of Hessian and curvature equations. Accordingly, les us suppose that

u ∈ C3(Ω) is a convex solution of the Dirichlet problem,

detD2u = ψ in Ω

u = φ on ∂Ω.(13.44)


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13.6 Applications431

with ψ ∈ C2(Ω), infΩ ψ > 0 as before, and φ ∈ C4(Ω), ∂Ω ∈ C4. Fix a point y ∈ ∂Ω and a

principal coordinate system at y . Then the estimation of Dnnu(y) is equivalent to estimation

away from zero of the minimum eigenvalue λ′ of the matrix

Dαβu = ∂α∂β −Dnu ·Ωαβ, 1 ≤ α, β ≤ n − 1.

Nor for ξ = (ξ1, . . . , ξn−1), |ξ| = 1, the function

w = Dnu −∂α∂βφξαξβ

Ωαβξαξβ− A|x ′|2

will take a maximum value at a point z ∈ N ∩ ∂Ω, for sufficiently large A, depending on N ,

max |Du|, max |D2φ| and ∂Ω. But then an estimate

Dnnu(z) ≤ C

follows by the same barrier considerations as before, whence λ′(z) is estimated away from

zero and consequently λ′(y) as required. Note that this argument does not extend to the

degenerate case. We thus establish the following extension of (non-deg monge-ampere exist

theorem) to general boundary values [CNS84]

Theorem 13.5.8 (): Let Ω be a uniformly convex domain with ∂Ω ∈ C4, φ ∈ C4(Ω) and

let ψ ∈ C2(Ω) be such that ψ > 0 in Ω. Then there exists a unique solution u ∈ U of the

Dirichlet problem:

detD2u = ψ in Ω

u = 0 on ∂Ω..

13.6 Applications

This section is concerned with some applications and extensions.

13.6.1 Isoperimetric inequalities

We give a new proof of the isoperimetric inequalities via the existence theorem of Monge-

Ampere equations.

Theorem 13.6.1 (): Let Ω be a bounded domain in Rn with its boundary ∂Ω ∈ C2. Then

we have

|Ω|1−1/n ≤ 1

nωnHn−1(∂Ω), (13.45a)

|Ω|1−2/n ≤ f rac1n(n − 1)ω2/nn

∫

∂Ω

H′ dHn−1 if H′ ≥ 0 on ∂Ω, (13.45b)

where | · | andHn−1( · ) denote the usual Lebesgue measure on Rn and the (n−1)-dimentional

Hausdorff measure, respectively. H′ is the mean curvature of ∂Ω; i.e.

H′ :=

n−1∑

i=1

κ′i .

These inequalities are sharp on balls.

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(13.45a) is the classical isoperimetric inequality. We show the proof employing the

Monge-Ampere existence theorem. (13.45b) is given in [Tru94].

Proof. We may assume without loss of generality that 0 ∈ Ω. Let Ω ⊂⊂ Br (0) ⊂⊂ BR(0).

We want to solve

det

[D2u

]= χ in BRu = 0 on ∂Ω,

(13.46)

where χ := χΩ denotes the characteristic function of Ω.

First we approximate χ by ψ ∈ C2(BR)

with

0 ≤ ψ ≤ χ on BR;

ψ ≡ 1 on Ωδ := x ∈ Ω | Dist (x, ∂Ω) > γ ,

for sufficiently small δ. From Theorem 13.5.7, we have a solution u ∈ C1,1(BR)

of (13.46),

where χ is replaced by ψ. Moreover, u ∈ C3(x ∈ BR |ψ(x) > 0).

Proof of (13.45a). The Alexandrov maximum principle implies

ωn supΩ|u|n ≤ (R + r)n

∫

BR

det[D2u

].

Since u is convex, we deduce

ωn supΩ|Du|n ≤ ωn

supΩ |u|n(R − r)n

≤(R + r

R − r

)n ∫

BR

det[D2u

]=

(R + r

R − r

)n|Ω|,

and hence

supΩ|Du| ≤ 1

ω1/nn

(R + r

R − r

)|Ω|1/n. (13.47)

By the arithmetic-geometric inequality, we have

∆u

n≥

(det

[D2u

])1/n

= 1 on Ωδ.

An integration gives

|Ωδ| ≤1

n

∫

Ωδ

∆u =1

n

∫

∂Ωδ

ν ·Du dHn−1

≤ 1

nω1/nn

Hn−1(∂Ωδ)

(R + r

R − r

)|Ω|1/n,

by virtue of (13.47). Here ν denotes the unit outer normal. Letting R→∞ and δ → 0, we

find the desired result.Gregory T. von Nessi

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13.6 Applications433

Proof of (13.45b). Recalling the inequality

(n∏

i=1

λi

)1/n

≤

1

n(n − 1)

∑

i<j

λiλj

1/2

for λi ≥ 0, we see that

(det

[D2u

])1/n ≤ 1

n(n − 1)

(∆u)2 −

∑

i ,j

(Di ju)2

.

Integrating the left hand side of the last inequality, we obtain

∫

Ω

(∆u)2 −

∑

i ,j

(Di ju)2

dx =

∫

Ω

(Di iuDj ju −Di juDi ju) dx

= −∫

Ω

Diu(Di j ju −Di j ju) dx

+

∫

∂Ω

(νiDiuDj ju − νjDiuDi ju) dHn−1

=

∫

∂Ω

(ν ·Du)∆u − νiDjuDi ju

dHn−1. (13.48)

For any y ∈ ∂Ω, take the principle coordinate system at y . Again, noting that ∂ denotes

the tangential gradient, then there holds

(13.48) =

∫

∂Ω

(ν ·Du)(∂i∂iu + (ν ·Du)H′)−

∑

i+j<2n

νjDiuDi ju

dHn−1

=

∫

∂Ω

(ν ·Du)2H′ + (ν ·Du)∂i∂iu −DiuDi(ν ·Du)

dHn−1

=

∫

∂Ω

(ν ·Du)2H′ + (ν ·Du)∂i∂iu − ∂iu∂i(ν ·Du)

dHn−1

=

∫

∂Ω

(ν ·Du)2H′ + 2(ν ·Du)∂i∂iu

dHn−1,

by virtue of the integration formula (see [GT01, Lemma 16.1])∫

∂Ω

∂iφdHn−1 =

∫

∂Ω

H′νiφdHn−1. (13.49)

Since the convexity of u implies

(ν ·Du)H′ + ∂i∂iu = Di iu ≥ 0,

we obtain for H′ ≥ 0∫

Ω

(∆u)2 − |D2u|2

dx ≤

∫

∂Ω

(K2H′ − 2K∂i∂iu) dHn−1

≤ K2

∫

∂Ω

H′ dHn−1,

where K := max∂Ω |ν ·Du|, by (13.49) again.

Using (13.47), replacing Ω by Ωδ and letting δ → 0, R → 0 as before, we obtain

(13.45b).

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13.7 Exercises

12.1: Let u ∈ C2(Ω) satisfy u ≥ 0 on ∂Ω,

C2[x ] := ∆u − νiνjDi ju ≥ 0 in Ω,

where

ν :=

Du|Du| if |Du| 6= 0

0 otherwise.

Using (13.45b) show that

||u||L

nn−2 (Ω)

≤ 1

n(n − 1)ω2/nn

∫

Ω

C[u].


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Appendices

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436Appendix A: Notation

A Notation


– Alan RickmanFor derivatives D = ∇ and ~Du =

(∂∂x1u, . . . , ∂

∂xnu)

.


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440Index

Index

adjoint operator, see operator, adjoint

almost everywhere

criterion, 202

almost orthogonal element, 185

Arzela-Ascoli Theorem, 189

Banach Space, 181

reflexive, 192

seperable, 182

Banach’s fixed point theorem, 182

Banach-Alaoglu Theorem, 199

Beppo-Levi Convergence Theorem, see Mono-

tone Convergence Theorem

bidual, 192

bilinear form, 195

bounded, 195

coercive, 195

Caccioppoli inequalities, 285

Caccioppoli’s inequality

elliptic systems, for, 288

Laplace’s Eq., for, 286

Cauchy-Schwarz inequality, 193

compact operator, see operator, compact

Companato Spaces, 223

L∞, 229

conservation law, 5

contraction, 182

dual space, 192

canonical imbedding, 192

Egoroff’s Theorem, 203

eigenspace, 189

eigenvalue, 189

multiplicity, 189

eigenvector, 189

Einstein’s Equations, 5

Fatou’s Lemma, 203

Fourier’s Law of Cooling, 4

Fredholm Alternative

Hilbert Spaces, for, 198

normed linear spaces, for, 186–188

Fubini’s Theorem, 203

Heat Equation, 3

steady state solutions, 4

heat flux, 3

Hilbert Space, 193

inner product, see scalar product

integrable function, 202

Laplace’s Equation, 4

Lax-Milgram theorem, 195

Lebesgue’s Dominated Convergence Theorem,

203

Legendre-Hadamard condition, 288

Lioville’s Theorem, 288

Maxwell’s Equations, 5

measurable

function, 202Gregory T. von Nessi

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Index441

measure, 201

Hausdorff, 201

Lebesgue, 201

outer, 201

Method of Continuity

for bounded linear operators, 184

metric, 181

minimizing sequence, 194

Monotone Convergence Theorem, 203

Morrey Space, 222

Noether’s Theorem, 5

norm, 181

Normed Linear Space, 181

open

ball, 181

set, 181

operator

adjoint, 192

bounded, 183

compact, 185

parallelogram identity, 194

Picard-Lindelof Theorem, 183

projection theorem, 194

regularity

linear elliptic eq., for, 285–310

resolvent operator, 189

resolvent set, 189

Riesz-representation theorem, 194

scalar product, 193

Scalar Product Space, 193

Schrodinger’s Equation, 5

set

Lebesgue measurable, 201

σ-ring, 201

simple function, 202

spectral theorem

compact operators, for, 189

spectrum

compact operator, of a, 189

Topological Vector Space, 181

traffic flow, 5

Lighthill-Whitham-Richards model, 6

Transport Equation, 7

triangle inequality, 181, 194

weak convergence, 198

weak-* convergence, 199

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analytic methods in partial differential equations

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