INTRODUCTION TOPARTIAL DIFFERENTIALEQUATIONSLecture Notesgiven byDINH NHO H�AOUniversit�at { GH { SiegenSummer Semester 1996

ForewordThe author has worked for 4 years (1993{1996) at the University of Siegen in the group ofApplied and Numerical Mathematics. His stay was supported by a DFG{scholarship and bya guest professorship from the University of Siegen. During this time, this lecture was heldfor graduate students.The group of Applied and Numerical Math. and the students of his lecture like to thankPriv.{Doz. Dr. Dinh Nho H�ao for his great e�orts in research and teaching.Comments on this lecture note should be sent directly to the author via email (hao@thevinh.ac.vn) or to hjreinhardt@numerik.math.uni-siegen.de .Siegen, Sept. 1997 H.{J. Reinhardt

Contents1 Introduction 11.1 De�nitions : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 11.2 Examples : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 11.2.1 Equations : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 11.2.2 Solutions : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 21.3 First-order linear equations : : : : : : : : : : : : : : : : : : : : : : : : : : : 21.3.1 The constant coe�cient case : : : : : : : : : : : : : : : : : : : : : : : 21.3.2 A variable coe�cient case : : : : : : : : : : : : : : : : : : : : : : : : 31.4 Where PDE come from? : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 41.4.1 Simple Transport : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 41.4.2 Vibration Equations : : : : : : : : : : : : : : : : : : : : : : : : : : : 51.4.3 Di�usion Equation : : : : : : : : : : : : : : : : : : : : : : : : : : : : 71.4.4 Steady State Equation : : : : : : : : : : : : : : : : : : : : : : : : : : 81.4.5 Schr�odinger's Equation : : : : : : : : : : : : : : : : : : : : : : : : : : 81.5 Classi�cation of Linear Di�erential Equations : : : : : : : : : : : : : : : : : 91.5.1 Di�erential Equations with Two Independent Variables : : : : : : : : 91.5.2 Di�erential Equations with Several Independent Variables. : : : : : : 142 Characteristic Manifolds and the Cauchy problem 172.1 Notation : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 172.2 The Cauchy problem : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 182.3 Real Analytic Functions and the Cauchy{Kowalevski Theorem : : : : : : : : 212.3.1 Real Analytic Functions : : : : : : : : : : : : : : : : : : : : : : : : : 212.3.2 The Cauchy{Kowalevski theorem : : : : : : : : : : : : : : : : : : : : 232.4 The Uniqueness theorem of Holmgren : : : : : : : : : : : : : : : : : : : : : : 24i

ii CONTENTS2.4.1 The Lagrange{Green Identity : : : : : : : : : : : : : : : : : : : : : : 242.4.2 The Uniqueness theorem of Holmgren : : : : : : : : : : : : : : : : : : 253 Hyperbolic Equations 293.1 Boundary and initial conditions : : : : : : : : : : : : : : : : : : : : : : : : : 293.2 The uniqueness : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 313.3 The method of wave propagation : : : : : : : : : : : : : : : : : : : : : : : : 333.3.1 The D'Alembert method : : : : : : : : : : : : : : : : : : : : : : : : 333.3.2 The stability of the solution : : : : : : : : : : : : : : : : : : : : : : : 343.3.3 The re ection method : : : : : : : : : : : : : : : : : : : : : : : : : : 363.4 The Fourier method : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 393.4.1 Free vibration of a string : : : : : : : : : : : : : : : : : : : : : : : : : 393.4.2 The proof of the Fourier method : : : : : : : : : : : : : : : : : : : : 423.4.3 Non{homogeneous equations : : : : : : : : : : : : : : : : : : : : : : : 443.4.4 General �rst boundary value problem : : : : : : : : : : : : : : : : : : 463.4.5 General scheme for the Fourier method : : : : : : : : : : : : : : : : : 463.5 The Goursat Problem : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 513.5.1 De�nition : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 513.5.2 The Darboux problem. The method of successive approximations : : 533.6 Solution of general linear hyperbolic equations : : : : : : : : : : : : : : : : : 583.6.1 The Green formula : : : : : : : : : : : : : : : : : : : : : : : : : : : : 583.6.2 Riemann's method : : : : : : : : : : : : : : : : : : : : : : : : : : : : 593.6.3 An application of the Riemann method : : : : : : : : : : : : : : : : : 614 Parabolic Equations 634.1 Boundary conditions : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 634.2 The maximum principle : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 654.3 Applications of the maximum principle : : : : : : : : : : : : : : : : : : : : : 664.3.1 The uniqueness theorem : : : : : : : : : : : : : : : : : : : : : : : : : 664.3.2 Comparison of solutions : : : : : : : : : : : : : : : : : : : : : : : : : 674.3.3 The uniqueness theorem in unbounded domains : : : : : : : : : : : : 684.4 The Fourier method : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 694.4.1 The homogeneous problem : : : : : : : : : : : : : : : : : : : : : : : : 69

CONTENTS iii4.4.2 The Green function : : : : : : : : : : : : : : : : : : : : : : : : : : : : 724.4.3 Boundary value problems with non-smooth initial conditions : : : : : 734.4.4 The non-homogeneous heat equation : : : : : : : : : : : : : : : : : : 774.4.5 The non-homogeneous �rst boundary value problem : : : : : : : : : : 784.5 Problems on unbounded domains : : : : : : : : : : : : : : : : : : : : : : : : 794.5.1 The Green function in unbounded domains : : : : : : : : : : : : : : : 794.5.2 Heat conduction in the unbounded domain (�1;1) : : : : : : : : : 824.5.3 The boundary value problem in a half-space : : : : : : : : : : : : : : 875 Elliptic Equations 935.1 The maximum principle : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 935.2 The uniqueness of the Dirichlet problem : : : : : : : : : : : : : : : : : : : : 945.3 The invariance of the operator � : : : : : : : : : : : : : : : : : : : : : : : : 945.4 Poisson's formula : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 965.5 The mean value theorem : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1005.6 The maximum principle (a strong version) : : : : : : : : : : : : : : : : : : : 101Bibliography 103Index 105

Chapter 1Introduction1.1 De�nitionsA partial di�erential equation (PDE) for a function u(x; y; : : :) is a relation of the formF (x; y; : : : ; u; ux; uy; : : : ; uxx; uxy; : : :) = 0; (1:1:1)where F is a given function of the independent variables x; y; : : :, and of the "unknown"function u and of a �nite or in�nite number of its derivatives. We call u a solution of (1.1.1)if after substitution of u(x; y; : : :) and its partial derivatives, (1.1.1) is satis�ed identicallyin x; y; : : : in some region in the space of these independent variables. For simplicity, wesuppose that x; y; : : : are real and that the derivatives of u occuring in (1.1.1) are continuousfunctions of x; y; : : : in the real domain . Several PDEs involving one or more unknownfunctions and their derivatives constitute a system.The order of a PDE or of a system is the order of the highest derivative that occurs. APDE is said to be linear if it is linear in the unknown functions and their derivatives, withcoe�cients depending on the independent variables x; y; : : : . The PDE of order m is calledquasilinear if it is linear in the derivatives of order m with coe�cients that depend on x; y; : : :and the derivatives of order < m.Remark 1.1.1 The dimension of the space where x; y; : : : belong to can be in�nite.Remark 1.1.2 If in (1.1.1) derivatives of u are of in�nite order, then we say that (1.1.1) isof in�nite order.1.2 Examples1.2.1 Equations� ux + uy = 0 (transport)� ux + g(u)y = 0 (conservation law, shock wave)1

2 CHAPTER 1. INTRODUCTION� uxx + uyy = 0 (the Laplace equation)� utt � uxx = 0 (the wave equation)� ut = k�u (the heat equation)� i�h t = � �h22m� + V (Schr�odinger's equation)� ut + cuux + uxxx = 0 (the Korteweg - de Vries equation)� 1X0 1n! dnudxn = g(x)1.2.2 SolutionsExample 1.2.1 Find all u(x; y) satisfying the equation uxx = 0. We can integrate once toget ux = const (in x). Since the solution depends on y, we have ux = f(y), where f(y) isarbitrary. Do it again to get u(x; y) = f(y)x+ g(y).Example 1.2.2 Find u(x; y) satisfying PDE uxx + u = 0.Solution: u(x; y) = f(y) cos x+ g(y) sinx, f and g are arbitrary.Example 1.2.3 Find u(x; y) satisfying uxy = 0. First integrate in x, regarding y as �xed.So we get uy(x; y) = f(y):Next let integrate in y, regarding x as �xed. We get the solutionu(x; y) = F (y) +G(x);where F 0 = f .1.3 First-order linear equations1.3.1 The constant coe�cient caseConsider the equation aux + buy = 0; (1:3:1)where a and b are constants not both zero.Geometric Method. The quantity aux + buy is the directional derivative of u in thedirection of the vector ~V = (a; b) = a~i+ b~j. It must always be zero. This means that u(x; y)must be constant in the direction of ~V . The vector (b;�a) is orthogonal to ~V . The linesparallel to ~V satisfy the equation bx� ay = const = c and along them the solution u has aconstant value, say f(c). Then u(x; y) = f(c) = f(bx� ay). Since c is arbitrary, we haveu(x; y) = f(c) = f(bx� ay) (1:3:2)

1.3. FIRST-ORDER LINEAR EQUATIONS 3for all values of x and y.Coordinate method. Change variables tox0 = ax+ by; y0 = bx� ay: (1:3:3)x

y

x

yy

x

We have ux = @u@x = @u@x0 @x0@x + @u@y0 @y0@x = aux0 + buy0and uy = @u@y = @u@y0 @y0@y + @u@x0 @x0@y = bux0 � auy0:Hence aux + buy = a(aux0 + buy0) + b(bux0 � auy0) = (a2 + b2)ux0: Since a2 + b2 6= 0, we haveux0 = 0. Thus, u = f(y0) = f(bx� ay):1.3.2 A variable coe�cient caseThe equation ux + yuy = 0 (1:3:4)is linear and homogeneous but has a variable coe�cient y. The PDE (1.3.4) itself assertsthat the directional derivative in the direction of the vector (1; y) is zero. The curves in thexy plane with (1; y) as tangent vectors have slopes y. Their equations aredydx = y1 : (1:3:5)

4 CHAPTER 1. INTRODUCTIONx

y

This ODE has the solution y = Cex: (1:3:6)These curves are called the characteristic curves of the PDE (1.3.4). As C is changed, thecurves �ll out the xy plane perfectly without intesecting. On each of the curves u(x; y) is aconstant because ddxu(x;Cex) = @u@x + Cex@u@y = ux + yuy = 0:Thus, u(x;Cex) = u(0; Ce0) = u(0; C) is independent of x. Putting y = Cex and C = e�xy,we have u(x; y) = u(0; e�xy):It follows that u(x; y) = f(e�xy) (1:3:7)is the general solution of our PDE.1.4 Where PDE come from?1.4.1 Simple TransportConsider a uid, say water, owing at a constant rate c along a horizontal pipe of a �xedcross section in the positive x direction. A substance, say a pollutant, is suspended in thewater. Let u(x; t) be its concentration in grams/centimeter at time t.u

x

t=1 t=2 t=3

1.4. WHERE PDE COME FROM? 5We know that the amount of pollutant in the interval [0; b] at the time t isM = R b0 u(x; t)dx,in grams. At the later time t+ h, the same molecules of pollutant have moved to the rightby c � h centimeters. Hence,M = Z b0 u(x; t)dx = Z b+chch u(x; t+ h)dx:Di�erentiate with respect b, we getu(b; t) = u(b+ ch; t+ h):Di�erentiate with respect to h and putting h = 0, we get0 = cux(b; t) + ut(b; t):Since b is arbitrary, the equation describing our simple transport problem looks as followsut + cux = 0: (1:4:1)1.4.2 Vibration EquationsConsider a exible, elastic homogeneous string or thread of length l, which undergoes rel-atively small transverse vibrations. Assume that it remains in a plane. Let u(x; t) be itsdisplacement from equilibrium position at time t and position x. Because the string is per-fectly exible, the tension (force) is directed tangentially along the string. Let T (x; t) be themagnitude of this tension vector. Limiting ourselves to the consideration of small vibrationsof the string, we shall ignore magnitudes of order greater than tan� = @u=@x.x

yT(x+dx,t)

T(x,t)

x x+dx

u(x,t)

u(x+dx,t)

−T(x,t)Any small section of the string (a; b), after displacement from a state of equilibrium withinthe range of our approximation, does not change its lengthS = Z ba vuut1 + @u@x!2dx ' b� aand therefore, in agreement with Hook's law, T is independent of t. We shall show now it isalso independent of x. This means T = T0 = const :

6 CHAPTER 1. INTRODUCTIONFor the projections of the tension into the u� and x�axes (denoted by Tu and Tx, respec-tively) we have Tx(x) = T (x) cos� = T (x)q1 + (ux)2 ' T (x);Tu(x) = T (x) sin� = T (x) tan� = T (x)ux;where � is the angle between the tangent at the curve u(x; t) and the x�axis.The tension, the external force and the initia (Tr�agheitskr�afte) are acting into a small piece(x1; x2) of the string. The sum of the projections into x�axis of all these forces must bezero, since we consider only the case of transverse vibrations. Because the external force andthe initia (Tr�agheitskr�afte) are acting along the u�axis direction, we haveTx(x1)� Tx(x2) = 0; or T (x1) = T (x2):Since x1 and x2 are arbitrary, we see that T is independent of x. Thus, for all x and t, wehave T � T0:Let F (x; t) be the magnitude of the external forces acting on the string at the point x at aninstant of time t and directed perpendicular to the x�axis. Finally, let �(x) be the lineardensity of the string at the point x, so that �(x)dx is the mass of the element of the string(x; x+ dx).We now construct an equation of montion for the element (x; x + dx) of the string. Thetension forces T (x+ dx; t), �T (x; t) and the external force, the sum of which, according toNewton's law, must be equal to the product of the mass of the element and its acceleration,are acting upon the element (x; x+ dx). Projecting this vector equation onto the u axis, onaccount of all that has been said, we obtain the equationT0 sin�jx+dx � T0 sin�jx + F (x; t)dx = �(x)dx@2u(x; t)@t2 (1:4:2)However, whithin the range of our approximationsin� = tan�p1 + tan2 � � tan� = @u@xand therefore �@2u(x; t)@t2 = T0 1dx "@u(x+ dx; t)@x � @u(x; t)@x #+ F (x; t)that is, �@2u@t2 = T0@2u@x2 + F: (1:4:3)This is indeed the equation of small transverse vibrations of a string.If the density � is constant, �(x) = �, the equation of vibration of a string assumes the form@2u@t2 = a2@2u@x2 + f; (1:4:4)

1.4. WHERE PDE COME FROM? 7where we have set a2 = T0=� (constant), f = F=�. We shall also call this equation theone-dimensional wave equation.By the same way one can also derive the equation of small transverse vibrations of a mem-brane as �@2u@t2 = T0 @2u@x2 + @2u@y2 + @2u@z2!+ F := T0�u+ F: (1:4:5)1.4.3 Di�usion EquationWe now derive the equation of heat di�usion or the di�usion of particles in a medium.Denote by u(x; t) the temperature of the medium at the point x = (x1; x2; x3) at an instantof time t. We shall consider the medium to be isotropic and shall denote its density by�(x), its speci�c heat by c(x), and its coe�cient of thermal conductivity at point x by k(x).F (x; t) denotes the intensity of heat sources at the point x at the instant of time t. We shallcalculate the balance of heat in an arbitrary volume V during the time interval (t; t+ dt).The boundary of V is denoted by S and ~n is the external normal to it. In agreement withFourier's law, an amount of heatQ1 = ZS k@u@~ndSdt = ZS(k grad u; ~n)dSdtequal, by virtue of the Gauss{Ostrogradskii formula, toQ1 = ZV div(k grad u)dxdtenters volume V through surface S.An amount of heat Q2 = ZV F (x; t)dxdtarises in volume V as a result of heat sources. Since the temperature in the volume V duringthe interval of time (t; t+ dt) has increased in magnitude byu(x; t+ dt)� u(x; t) ' @u@t dtit follows that for this it is necessary to expend an amount of heatQ3 = ZV c�@u@t dxdt:On the other hand, Q3 = Q1 +Q2, and thusZV "div(k grad u) + F � c�@u@t # dxdt = 0from which, since volume V is arbitrary, we obtain the equation of heat di�usionc�@u@t = div(k grad u) + F: (1:4:6)

8 CHAPTER 1. INTRODUCTIONIf the medium is homogeneous, that is, if c; �, and k are constants, Eq. (1.4.6) assumes theform @u@t = a2�u+ f (1:4:7)where a2 = kc�; f = Fc�:Eq. (1.4.7) is called the heat equation.1.4.4 Steady State EquationFor steady state processes F (x; t) = F (x); u(x; t) = u(x), and both the wave equation (1.4.5)and the di�usion equation (1.4.6) assume the form� div(p grad u) + qu = F (x): (1:4:8)When p = const and q = 0, Eq. (1.4.8) is called Poisson's equation�u = �f; f = Fp (1:4:9)when f = 0, Eq. (1.4.9) is called Laplace's equation�u = 0: (1:4:10)In the wave equation (1.4.5), let the external disturbance f(x; t) be periodic with frequency! and amplitude f0(x), f(x; t) = f0(x)ei!t:If we seek periodic solutions u(x; t) with the same frequency and unknown amplitude u0(x),then for the function u0(x) we obtain the equation�u0 + k2u0 = �f0(x)a2 ; k2 = !2a2 (1:4:11)called the Helmholtz equation.1.4.5 Schr�odinger's EquationLet a quantum particle of mass m move in an external force �eld with potential V (x). Weshall denote by (x; t) the wave function of this particle, so that j (x; t)j2dx is the probabilitythat the particle will be in a neighbourhood d(x) of the point x at an instant of time t; heredx is the volume of d(x). Then, the function satis�es Schr�odinger's equationi�h t = � �h22m� + V ; (1:4:12)where �h = 1:054:10�27erg sec is Planck's constant.

1.5. CLASSIFICATION OF LINEAR DIFFERENTIAL EQUATIONS 91.5 Classi�cation of Linear (Second-Order) Di�eren-tial Equations1.5.1 Di�erential Equations with Two Independent VariablesLet us consider a second-order di�erential equation with two independent variablesa11@2u@x2 + 2a12 @2u@x@y + a22@2u@y2 + F x; y; u; @u@x; @u@y! = 0 (1:5:1)where we shall suppose that the coe�cients a11; a12; and a22 belong to C2 in a certainneighbourhood and do not become zero simultaneously anywhere in this neighbourhood.For de�niteness we assume that a11 6= 0 in this neighbourhood. In fact, it may happen thata22 6= 0. But then, by reversing the roles of x and y, we obtain an equation in which a11 6= 0.If then a11 and a22 become zero simultaneously at a point, then a12 6= 0 in the neighbourhoodof this point. In this case the change of variables x0 = x+ y; y0 = x� y leads to an equationin which both a11 and a22 are not zero.Using the new variables� = �(x; y); � = �(x; y); � 2 C2; � 2 C2; D �; �x; y! 6= 0; (1:5:2)we have ux = u��x + u��x;uy = u��y + u��y;uxx = u���2x + 2u���x�x + u���2x + u��xx + u��xx;uxy = u���x�y + u��(�x�y + �y�x) + u���x�y + u��xy + u��xy;uyy = u���2y + 2u���y�y + u���2y + u��yy + u��yy:Putting these equations in our original equations, we get a new equation of the form�a11u�� + 2�a12u�� + �a22u�� + �F = 0; (1:5:3)where �a11 = a11�2x + 2a12�x�y + a22�2y ;�a12 = a11�x�x + a12(�x�y + �x�y) + a22�y�y;�a22 = a11�2x + 2a12�x�y + a22�2yand �F is independent of the second order derivatives with respect to � and � of u.We shall require that the functions �(x; y) and �(x; y) make the coe�cients �a11 and �a22vanish, that is they should satisfy the equationsa11�2x + 2a12�x�y + a22�2y = 0; (1.5.4)a11�2x + 2a12�x�y + a22�2y = 0: (1.5.5)First we prove the following lemma.

10 CHAPTER 1. INTRODUCTIONLemma 1.5.1 1. If z = '(x; y) is a particular solution of the equationa11z2x + 2a12zxzy + a22z2y = 0; (1:5:6)then '(x; y) = C is a general integral of the ordinary di�erential equationa11dy2 � 2a12dxdy + a22dx2 = 0: (1:5:7)2. Conversely, if '(x; y) = C is a general integral of Eq. (1.5.7), then z = '(x; y) is asolution of Eq. (1.5.6).Proof. Since the function z = '(x; y) is a solution of Eq. (1.5.6),a11 'x'y!2 � 2a12 �'x'y!+ a22 (1:5:8)for all x; y in the neighbourhood where z = '(x; y) is de�ned and 'y(x; y) 6= 0. Let '(x; y) =C, Because 'y(x; y) 6= 0, we can �nd a neighbourhood where y = f(x;C), thendydx = � "'x(x; y)'y(x; y)#y=f(x;C) (1:5:9)and a11 dydx!2 � 2a12dydx + a22 = 24a11 �'x'y!2 � 2a12 �'x'y!+ a2235y=f(x;C) = 0:Thus, y = f(x;C) is a solution of (1.5.7). Note that the last term vanishes for all x; y.Conversely, let '(x; y) = C be an integral of the Eq. (1.5.7). We shall show that for anypoint (x; y) a11'2x + 2a12'x'y + a22'2y = 0 (1:5:10)Let (x0; y0) be given. Draw through it an integral curve for Eq. (1.5.7), for which we set'(x0; y0) = C0. Consider now the curve y = f(x;C0). It is clear y0 = f(x0; C0). For anypoint of this curve we havea11 dydx!2 � 2a12dydx + a22 = 24a11 �dydx!2 � 2a12 �dydx!+ a2235y=f(x;C0) :Putting x = x0 into this equation, we geta11'2x(x0; y0) + 2a12'x'y(x0; y0) + a22'2y(x0; y0) = 0:Since x0; y0 are arbitrary, we have just proved our lemma. 2Eq. (1.5.7) is called the characteristic equation of the di�erential equation (1.5.1) and itsintegral curves are called characteristics.Let � = �(x; y), where �(x; y) = const is an integral of Eq. (1.5.7), we see that the coe�cientof u�� vanishes. Analogously, set � = �(x; y) where �(x; y) = const is another integral ofEq. (1.5.7) independent of �(x; y), we get the coe�cient of u�� to be also zero.

1.5. CLASSIFICATION OF LINEAR DIFFERENTIAL EQUATIONS 11Eq. (1.5.7) is equivalent to the linear equationsdydx = a12 +qa212 � a11a22a11 ; (1.5.11)dydx = a12 +qa212 + a11a22a11 : (1.5.12)According to the terminology of the theory of second-order curves, we say that Eq. (1.5.1)at a point M is ofhyperbolic type, when at this point a212 � a11a22 > 0,elliptic type, when at this point a212 � a11a22 < 0;parabolic type, when at this point a212 � a11a22 = 0:Since �a212 � �a11�a22 = (a212 � a11a22)(�x�y � �y�x)2;the type of Eq. (1.5.1) is invariant under our transformation (1.5.2). Note that, at di�erentpoints of the domain the equation may have di�erent type. In what follows we consider aneighbouhood G where our equation is of only one type. Through any point of G there aretwo characteristics which are real and di�erent for an equation of hyperbolic type, complexand di�erent for an equation of elliptic type, and real and coincided for an equation ofparabolic type. We consider these cases separately.1. For equations of hyperbolic type a212 � a11a22 > 0, so the right-hand sides of (1.5.11) and(1.5.12) are real and di�erent. The general integrals �(x; y) = C and �(x; y) = C of theseequations de�ne two families of characteristics. Letting� = �(x; y); � = �(x; y) (1:5:13)we get Eq. (1.5.3) after dividing by the coe�cient before u�� into the formu�� = F (�; �; u; u�; u�) with �a12 6= 0: (1:5:14)This is called the canonical form of a hyperbolic equation.Normally we are using another canonical form. Let� = �+ �; � = � � �;it means � = � + �2 ; � = � � �2 ;where � and � are considered as new variables. Then,u� = 12(u� + u�); u� = 12(u� � u�); u�� = 14(u�� � u��);and Eq. (1.5.3) has the form u�� � u�� = F1:

12 CHAPTER 1. INTRODUCTION2. For equations of parabolic type a212 � a11a22 = 0, therefore the equations (1.5.11) and(1.5.12) coincide and we have only a general integral for Eq. (1.5.6): '(�; �) = const. In thiscase, set � = '(x; y) and � = �(x; y);where �(x; y) is an arbitrary function independent of '. For this choice we have�a11 = a11�2x + 2a12�x�y + a22�2y = (pa11�x +pa22�y)2 = 0;since a12 = pa11pa22. It follows�a12 = a11�x�x + a12(�x�y + �y�x) + a22�y�y= (pa11�x +pa22�y)(pa11�x +pa22�y) = 0:Thus, u�� = �(�; �; u; u�; u�) with � = � �F�a22 (�a22 6= 0)is the canonical form of a parabolic equation. If the right hand side of this equation doesnot depend on u� then we have an ordinary di�erential equation.3. For the case of elliptic equations a212 � a11a22 < 0, and we see that the equation (1.5.11)is complex conjugate to (1.5.12). If '(x; y) = Cis a complex integral of Eq. (1.5.11), then'�(x; y) = C;is a complex integral of Eq. (1.5.12) where '� is the complex function conjugate to '. Let� = '(x; y) � = '�(x; y):In order to work with real functions, let further� = '+ '�2 ; � = '� '�2i :It is clear that � and � are real, and� = �+ i�; � = � � i�:We note that since we are working with complex variables, we have to suppose that allcoe�cients a11; a12; : : : are analytic. In this casea11�2x + 2a12�x�y + a22�2y= (a11�2x + 2a12�x�y + a22�2y)� (a11�2x + 2a12�x�y + a22�2y)+2i(a11�x�x + a12(�x�y + �y�x) + a22�y�y);therefore �a11 = �a22 and �a12 = 0:Now we have u�� + u�� = �(�; �; u; u�; u�) with � = � �F�a22 (�a22 6= 0):We summarize the results of this paragraph as follows.

1.5. CLASSIFICATION OF LINEAR DIFFERENTIAL EQUATIONS 13� a212 � a11a22 > 0 (hyperbolic type): uxx � uyy = � or uxy = �;� a212 � a11a22 < 0 (elliptic type): uxx + uyy = �;� a212 � a11a22 = 0 (parabolic type): uxx = �.Example 1.5.2 Tricomi's equation y@2u@x2 + @2u@y2 = 0belongs to the mixed type: when y < 0 it is of hyperbolic type, but when y > 0 it is ofelliptic type, and when y = 0 it is of parabolic type.Tricomi's equation is of interest in gas dynamics. With y < 0 the equations of the charac-teristics assume the form y0 = � 1p�y :Therefore the curves 32x+q�y3 = C1; 32x�q�y3 = C2are characteristics of Tricomi's equation. The transform� = 32x+q�y3; � = 32x�q�y3reduces Tricomi's equation to the canonical form@2~u@�@� � 16(� � �) @~u@� � @~u@�! = 0; � > �:y

elliptic

parabolic

hyperbolic

xIf y > 0, then ' = 32x� ipy3 and with the change of variables� = 32x; � = �qy3Tricomi's equation reduces to@2~u@�2 + @2~u@�2 + 13� @~u@� = 0; � < 0:

14 CHAPTER 1. INTRODUCTION1.5.2 Di�erential Equations with Several Independent Variables.We consider now the linear di�erential equation with real coe�cientsnXj=1 nXi=1 aijuxixj + nXi=1 biuxi + cu+ f = 0; (aij = aji); (1:5:15)where a; b; c and f are functions of variables x1; x2; : : : ; xn. Making the change of variables�k = �k(x1; x2; : : : ; xn) (k = 1; : : : ; n);we have uxi = nXk=1u�k�ik;uxixj = nXk=1 nXl=1 u�k�l�ik�jl + nXk=1 u�k(�k)xixj ;where �ik := @�k=@xi.Putting these expressions in (1.5.15) we getnXk=1 nXl=1 �aklu�k�l + nXk=1�bku�k + cu+ f = 0where �akl = nXi=1 nXj=1 aij�ik�jl;�bk = nXi=1 bi�ik + nXi=1 nXj=1 aij(�k)xixj :We consider now the quadratic form nXi=1 nXj=1 a0ijyiyj; (1:5:16)where a0ij := aij(x01; : : : ; x0n). With the nonsingular linear transformationyi = nXk=1�ik�k; det(�ik) 6= 0;we get the following new quadratic formnXk=1 nXl=1 �a0kl�k�l;with �a0kl = nXi=1 nXj=1 a0ij�ik�jl:

1.5. CLASSIFICATION OF LINEAR DIFFERENTIAL EQUATIONS 15Thus, the coe�cients of the principal part of Eq. (1.5.15) are transformed into the coe�cientsof a quadratic form with the aid of a nonsingular linear transformation. It is well knownthat there exists a singular linear transformation which makes the coe�cients of the matrix(a0ij) of a quadratic form into a diagonal matrix, in whichj �a0iij = 1; or 0; and �a0ij = 0 (i 6= j; i = 1; 1; : : : ; n)Moreover, by the Sylvester theorem about the inetia of quadratic forms, the number ofcoe�cients �a0ii 6= 0 equals the rank of the martrix (a0ij) and the number of negative coe�cientsis invariant. We say that it is of the canonical form.We say that Eq. (1.5.15) at the point x0 is of elliptic type, if all n coe�cients �a0ii 6= 0 andthey have the same sign; of hyperbolic type, if all �a0ii 6= 0 and n � 1 coe�cients �a0ii have thesame sign and only one has other sign; of ultrahyperbolic type, when m coe�cients �a0ii areequal, and n�m others with other sign are equal (m > 1; n �m > 1); of parabolic type, ifat least one of the coe�cients �a0ii vanishes.We take the new independent variables �i such that at the point x0�ik = @�k@xi = �0ikwhere the �0ik are the coe�cients of the transformation which makes the quadratic form(1.5.16) into the canonical form. We haveux1x1 + ux2x2 + � � �+ uxnxn + � = 0; (elliptic type);ux1x1 = nXi=2 uxixx + � (hyperbolic type);mXi=1 uxixi = nXi=m+1 uxixi + � (ultrahyperbolic type);n�mXi=1 (�uxixi) + � = 0; (m > 0) (parabolic type):Is it possible to use a single transformation to reduce (1.5.15) to the canonical form in asu�ciently small neighborhood of each point? To make the reduction for any equation, it isnecessary that the number of conditions�aik = 0; l 6= k; l; k = 1; 2; : : : ; n;�alk = �l�a11 l = 2; 3; : : : ; n; �a11 6= 0;where �l = 0;�1 should not be greater than the number of the unknown functions �l; l =1; 2; n 2 IN: n(n � 1)2 + n� 1 � n; that is n � 2:For n = 2 we have already proved that it is possible to use only one transformation to reduce(1.5.15) to the canonical form in a su�ciently small neighborhood of each point.

16 CHAPTER 1. INTRODUCTION

Chapter 2Characteristic Manifolds and theCauchy problem2.1 NotationLet X = (x1; : : : ; xn) 2 IRn. A vector � = (�1; : : : ; �n) whose components are non{negativeintegers �k is called a multi{index. The letters x; y; : : : ; �; � will be used for vectors, and�; �; ; : : : for multi{indices. Components are always indicated by adding a subscript to thevector{symbol: � has components �k. We set0 = (0; 0; : : : ; 0) ; 1 = (1; 1; : : : ; 1) : (2:1:1)For an � 2 ZZn set j�j = �1 + �2 + � � �+ �n ; �! = �1!�2! : : : �n! ; (2:1:2)and for x 2 IRn ; � 2 ZZn x� := x�11 x�22 : : : x�nn : (2:1:3)By C� we generally denote a coe�cient depending on n nonnegative integers �1; �2; : : : ; �n:C� = C�1:::�n : (2:1:4)The C� may be real numbers, or vectors in a space IRm. The general m-th degree polynomialin x1; : : : ; x1 is then of the form P (x) = Xj�j�mC�x� (2:1:5)with � 2 ZZn ; x 2 IRn ; C� 2 IR.Setting Dk = @=@xk, we introduce the \gradient vector" D = (D1; : : : ;Dn), and de�ne thegradient of a function u(x1; : : : ; xn) as the vectorDu = (D1u; : : : ;Dnu) : (2:1:6)The general partial di�erential operator of order m is then denoted byD� = D�11 D�22 : : :D�nn = @m@�1x1 : : : @x�nn (2:1:7)where j�j = m. 17

18 CHAPTER 2. CHARACTERISTIC MANIFOLDS AND THE CAUCHY PROBLEM2.2 The Cauchy problemConsider the mth{order linear di�erential equation for a function u(x) = u(x1; : : : ; xn)Lu := Xj�j�mA�(x)D�u = B(x) : (2:2:1)The same formula describes the general mth{order system of N di�erential equations in Nunknowns if we interprete u and B as column vectors with N components and A� as N �Nsquare matrices. Similarly the general mth{order quasi{linear equation (respectively systemof such equations) is Lu := Xj�j�mA�D�u+ C = 0 ; (2:2:2)where now the A und C are functions of the independent variables xk and of the derivativesD�u of the unknown u of orders j�j � m� 1. More general nonlinear equations or systemsF (x;D�u) = 0 (2:2:3)can be reduced formally to quasi{linear ones by applying a �rst{order di�erential operatorto (2.2.3). On the other hand, an mth{order quasi{linear system (2.2.2) can be reduced to a(large) �rst{order one, by introducing all derivativesD�u with j�j � m�1 as new variables,and making use of suitable compatibility conditions for the D�u.The Cauchy problem consists of �nding a solution u of (2.2.2) or (2.2.1) having prescribedCauchy data on a hyper{surface S � IRn given by�(x1; x2; : : : ; xn) = 0 : (2:2:4)Here � shall have m continuous derivatives and the surface should be regular in the sensethat D� = (�x1; : : : ; �xn) 6= 0 : (2:2:5)The Cauchy data on S for an mth{order equation consist of the derivatives of u of orderless than or equal to m � 1. They cannot be given arbitrarily but have to satisfy thecompatibility conditions valid on S (instead normal derivatives of order less than m can begiven independently from each other). We are to �nd a solution u near S which has theseCauchy data on S. We call S noncharacteristic if we can get all D�u for j�j = m on Sfrom the linear algebraic system or equations consisting of the compatibility condition forthe data and the eq. (2.2.2) taken on S. We call S characteristic if at each point x of S thesurface S is not noncharacteristic.To get an algebraic criterion for characteristic surfaces we �rst consider the special casewhere the hyper{surface S is the coordinate plane xn = 0. The Cauchy data then consistof the D�u with j�j < m taken for xn = 0. Singling out the \normal" derivatives on S oforders � m� 1: @uk@xkn =!Dknu = k(x1; : : : ; xn�1) for k = 0; : : : ;m� 1 and xn = 0 (2:2:6)we have on S D�u = D�11 D�22 : : :D�n�1n�1 �n (2:2:7)

2.2. THE CAUCHY PROBLEM 19provided that �n < m. In particular for j�j � m � 1 we have here the compatibilityconditions expressing all Cauchy data in terms of normal derivatives on S. Let �� denotethe multi{index �� = (0; : : : ; 0;m) : (2:2:8)In the eq (2.2.1) or (2.2.2) taken on S it is only the term with � = �� that is not expressibleby (2.2.7) in terms of 0; : : : ; m�1 and hence in terms of the Cauchy data. All otherscontain derivatives D�u with �n � m � 1. Thus D��u, and hence all D�u with j�j � m,are determined uniquely on S, if we can solve the di�erential equation for the term D��u.This is always possible in a unique way if and only if the matrix A�� is nondegenerated, i. e.,det(A��) 6= 0. For a single scalar di�erential equation this condition reduces to A�� 6= 0. Inthe linear case the condition det(A��) 6= 0 (2:2:9)does not depend on the Cauchy data on S; in the quasi{linear case, where the A� dependon the D�� with j�j � m � 1 and on x, one has to know the k in order to decide if S ischaracteristic.Since the condition (2.2.9) involves coe�cients of mth{order derivatives, we de�ne the prin-cipal part Lpr of L (both in (2.2.2) and (2.2.1)) asLpr := Xj�j=mA�D� : (2:2:10)The \symbol" of this di�erential operator is the matrix form (\characteristic matrix" of L):�(�) = Xj�j=mA��� : (2:2:11)The N �N matrix �(�) has elements that are mth{degree forms in the components of thevector � = (�1; : : : ; �n). In particular, the multiplier of Dmn �= @m@xmn � in Lpr is A�� = �(�),where � = (0; 0; : : : ; 0; 1) = D� (2:2:12)is the unit normal to the surface � = xn = 0. The condition for the plane � = xn = 0 to benoncharacteristic is then Q(D�) 6= 0 ; (2:2:13)where Q = Q(�) is the characteristic form de�ned byQ(�) := det(�(�)) (2:2:14)for any vector �. (In the case of a scalar equation (N = 1) the characteristic form Q(�)coincides with the polynomial �(�)). We shall see that (2.2.13) is the condition for a surface� = 0 to be noncharacteristic.Consider a general S given by (2.2.4), by assumption (2.2.5) we can always suppose that ina neighborhood of a given point of S, the condition �xn 6= 0 holds. Thus, the transformationyi = ( xi for i = 1; : : : ; n� 1�(x1; : : : ; xn) for i = n (2:2:15)is then locally regular and invertible. By the chain rule,@u@xi =Xk Cik @u@yk ; (2:2:16)

20 CHAPTER 2. CHARACTERISTIC MANIFOLDS AND THE CAUCHY PROBLEMwhere the Cik = @yk@xi (2:2:17)are functions of x or of y. Denoting by C the matrix of the Cik and introducing the gradientoperator d with respect to y with componentsdi = @@yi ; (2:2:18)we can write (2.2.16) symbolically, as D = Cd ; (2:2:19)taking D and d to be column vectors. Generally, for j�j = mD� = (Cd)� +R� ; (2:2:20)where R� is a linear di�erential operator involving only derivatives of order � m� 1. Thenthe principal part of L in (2.2.2) or (2.2.1) transformed to y{coordinates is given byLpr = Xj�j=mA�(Cd)� = `pr (2:2:21)and its symbol, the characteristic matrix of `, by�(�) = Xj�j=mA�(C�)� : (2:2:22)Since the mapping (2.2.15) is regular, S is noncharacteristic for L if the plane yn = 0 isnoncharacteristic with respect to the operator L transformed to y{coordinates, i. e. ifdet(�(�)) = det0@ Xj�j=mA�(C�)�1A 6= 0 (2:2:23)for � = (0; 0; : : : ; 0; 1). But then� = C� = @yn@x1 ; : : : ; @yn@x1! = (�x1; : : : ; �x1)= D� :Thus, the condition for noncharacteristic behavior of S can again be written as (2.2.13).If u in (2.2.2) stands for a vector withN components, the condition for S to be a characteristicsurface is Q(D�) = det0@ Xj�j=mA�(D�)�1A = 0 : (2:2:24)Example 2.2.1 For the wave equationutt = c2(ux1x1 + ux2x2) (2:2:25)for u = u(x1; x2; t), a characteristic surface t = (x1; x2) satis�es equation1 = c2 � 2x1 + 2x2� : (2:2:26)

2.3. REAL ANALYTIC FUNCTIONS, THE CAUCHY{KOWALEVSKI THEOREM 212.3 Real Analytic Functions and the Cauchy{Kowa-levski Theorem2.3.1 Real Analytic Functions2.3.1.1 Multiple in�nite seriesWe say that the multiple in�nite seriesX� c� ; � 2 ZZn (2:3:1)converges, if X� jc�j <1 : (2:3:2)The sum of a convergent series does not depend on the order of summation.Example 2.3.1 For x 2 IRn ; � 2 ZZnX� x� = nYi=10@ 1X�i=0x�ii 1A = 1(1 � x1)(1 � x2) : : : (1� xn) = 1(1� x)1 (2:3:3)provided jxij < 1 for all i.Example 2.3.2 For x 2 IRn ; � 2 ZZn,X� j�j!�! x� = 1Xj=0 Xj�j=j j�j!�! x�= 1Xj=0(x1 + � � �+ xn)j = 11� (x1 + � � � + xn)provided that jx1j+ � � �+ jxnj < 1.Let c�(x) be continuous scalar functions and de�ned in a set S � IRn; ifjc�(x)j � c� ; 8� 2 ZZn ; x 2 Sand X� c� <1;then X� c�(x)converges uniformly for x 2 S and represents a continuous function. If S is open andc� 2 Cj(S) ; 8�, and if the series X� D�c�(x)

22 CHAPTER 2. CHARACTERISTIC MANIFOLDS AND THE CAUCHY PROBLEMconverges uniformly for x 2 S and each � with j�j � j, then P� c�(x) 2 Cj(S) andD�X� c�(x) =X� D�c�(x) ; x 2 S; j�j � j :Of particular importance are the power seriesX� c�x� ; x 2 IRn ; � 2 Zn ; c� 2 IR : (2:3:4)Assume that the series converges for a certain z:�X� jc�j jz�j <1 : (2:3:5)Then (2.3.4) converges uniformly for all x withjxij � jzij for all i : (2:3:6)Hence, f(x) =X� c�x� (2:3:7)de�nes a continuous function f in the set (2.3.6).Problem: All series obtained by formal di�erentiation of (2.3.7) converge in the interior of(2.3.6), and even uniformly in every compact subset of the interior.2.3.1.2 Real analytic functionsDe�nition 2.3.3 Let f(x) be a function whose domain is an open set in IRn and whoserange lies in IR. For y 2 we call f real analytic at y if there exist c� 2 IR and aneighborhood U of y (all depending on y) such thatf(x) =X� c�(x� y)� ; 8x 2 U : (2:3:8)We say f is real analytic in (f 2 C!()), if f is real analytic at each y 2 . A vectorf(x) = (f1(x); : : : ; fm(x)) de�ned in is called real analytic, if each of its components isreal analytic.Theorem 2.3.4 If f = (f1; : : : ; fm) 2 C!(), then f 2 C1(). Moreover, for any y 2 there exists a neighborhood U of y and positive numbers M; r such that for all x 2 U .f(x) = X� 1�!(D�f(x))(x� y)� (2.3.9)���D�fk(x)��� � M j�j!r�j�j ; 8� 2 Zn ; 8k :Theorem 2.3.5 Let f 2 C!() where is a connected open set in IR. Let z 2 . Then fis uniquely determined in if we know the D�f(z) ; 8� 2 ZZn. In particular f is uniquelydetermined in by its values in any non{empty open subset of .Proof: Let f1; f2 2 C!() and let D�f1(z) = D�f2(z) for all z 2 ZZn. Write g = f1 � f2and decompose into1 = fx j x 2 ;D�g(x) = 0 8� 2 ZZng2 = fx j x 2 ;D�g(x) 6= 0 for some � 2 ZZng :Then 2 is open by continuity of g. The set 1 also is open; for if y 2 1, we have g(x) = 0in a neighborhood of y by (2.3.9). Since z 2 1 and is connected, 2 must be empty. 2

2.3. REAL ANALYTIC FUNCTIONS, THE CAUCHY{KOWALEVSKI THEOREM 232.3.1.3 Analytic and real analytic functionsLet x 2 CIn ; � 2 ZZn ; P � CIn ; f : P ! CIm. We call f analytic in P (f 2 Ca(P)), if foreach y 2 P we can represent f in the form (2.3.8) in a neighborhood of y. Here c� 2 CIm. Itcan be easily seen that c� = 1�!D�f(y) : (2:3:10)Theorem 2.3.6 A function f(x) with range in CIm and domain in CIn is analytic in the openset P if f is di�erentiable with respect to the independent variables.Theorem 2.3.7 If f 2 C!() with � IRn, then for every compact subset S of thereexists a neighborhood P of S in CIn and a function F 2 Ca(P) such that F (x) = f(x) for allx 2 S.2.3.2 The Cauchy{Kowalevski theoremThe theorem concerns the existence of a real analytic solution of the Cauchy problem forthe case of real analytic data und equations. We restrict ourselves to quasi{linear systemsof type (2.2.2) since more general nonlinear systems can be reduced to quasi{linear ones bydi�erentiation. We assume that S is real analytic in a neighborhood of one of its pointsx0, that is near x0 the surface S is given by an equation �(x) = 0, where � is real analyticat x0 and D� 6= 0 at x0, say, Dn� 6= 0. On S we prescribe compatible Cauchy data D�ufor j�j < m which shall be real analytic at x0. The coe�cients A�; C shall be real analyticfunctions of their arguments x ; D�u at the point x0 ; D�x0. We assume, moreover, S to benon{characteristic at x0 (and hence in a neighborhood of x0) in the sense that Q(D�) 6= 0.Then the Cauchy{Kowalevski theorem asserts that there exists a unique solution of (2.2.2)which is real analytic at x0.First of all, we can transform x0 into the origin and S locally by an analytic transformationinto a neighborhood of the origin in the plane xn = 0. Then by introducing derivativesof orders less than m as new dependent variables one reduces the system to one of �rstorder. We make use here of the fact that the set of real analytic functions is closed underdi�erentiation and composition. One arrives at a �rst order system in which the coe�cientmatrix of the term with @u=@xn is non{generate because S is non{characteristic. Hence wecan solve for @u=@xn, obtaining a system in the standard form@U@xn = n�1Xi=1 ai(x;U)@U@xi + b(x;U) (2:3:11)where the ai(x;U) is a square matrix (aijk), and b(x;U) is a column vector with componentsbi, U is the new vector of u and its derivatives of order less than m. On xn = 0, near0, we have prescribed initial values U = f(x1; : : : ; xn�1). Here we can assume that f = 0,introducing U�f as the new unknown functions. We can add xn as an additional dependentvariable U� or component of U satisfying the equation @u�=@xn = 1, and the initial conditionU� = 0. This has the e�ect that the ai and b in (2.3.11) do not depend on xn. In fact, wecan write

24 CHAPTER 2. CHARACTERISTIC MANIFOLDS AND THE CAUCHY PROBLEMai(x;U) = ai(x1; x2; : : : ; xn; U)= ai(x1; x2; : : : ; xn�1; U) ;b(x;U) = b(x1; x2; : : : ; xn; U)= b(x1; x2; : : : ; xn; U)where U = (U;U�). Writing (2.3.11) componentwise, we have to prove the following versionof the Cauchy{Kowalevski theorem:Theorem 2.3.8 (Cauchy{Kowalevski) Let the aijk and bj be real analytic functions ofZ = (x1; x2; : : : ; xn�1; u1; : : : ; uN) at the origin of IRN+n�1. Then the system of di�erentialequations @uj@xn = n�1Xi=1 NXk=1 aijk(z)@uk@xi + bj(z) ; for j = 1; : : : ; N (2:3:12)with initial conditions uj = 0 for xn = 0 ; j = 1; : : : ; N (2:3:13)has a unique (among the real analytic uj) system of solutions uj(x1; : : : ; xn) that is realanalytic at the origin.Proof: See, e.g., John's book [10] or Petrovskii's book [17].Remark: The theorem of Cauchy and Kowalewski is local in character, and applies onlyto analytic solutions of analytic Cauchy problems. It does not guarantee global existence ofsolutions; it does not exclude the possibility that other non{analytic solutions exist for thesame Cauchy problem, nor the possibility that an analytic solution becomes non{analytic insome distance away from the initial surface.2.4 The Uniqueness theorem of Holmgren2.4.1 The Lagrange{Green IdentityLet be a domain in IRn with the boundary @ which is su�ciently regular. Denote d=dnthe di�erentiation in the direction of the exterior unit normal � = (�1; : : : ; �n) of @. TheGauss{Ostrogradskii formula says thatZDku(x) dx = Z@ u(�)dxkdn dS = Z@ u(�)�kdS : (2:4:1)If @ is su�ciently regular then this formula is applicable to all u 2 C1(). The theorem canbe generalized to u 2 C1() \C0() by approximating from the interior. More generally,we have the formula for integration by parts,Z vTDku dx = Z@ vTu�kdS � Z �DkvT� u dx ; (2:4:2)where u; v are column vectors belonging to C1() with T denoting transposition.

2.4. THE UNIQUENESS THEOREM OF HOLMGREN 25Let now L be a linear di�erential operatorLu = Xj�j�m a�(x)D�u : (2:4:3)Let u; v be column vectors and a� be square matrices in Cm(). Then by repeated applicationof (2.4.2) it follows that Z vT Xjaj�m a�(x)D�u dx= Z Xj�j�m(�1)j�jD� �vTa�(x)�u dx+ Z@M(v; u; �)dS : (2.4.4)Here M in the surface integral is linear in the �k with coe�cients which are bilinear in thederivatives of v and u, the total number of di�erentiations in each term being at most m�1.The expression M is not uniquely determined but depends on the order of performing theintegration by parts. This is the Lagrange{Green identity for L which we also write in theform Z vTLudx = Z(L�v)Tu dx+ Z@M(u; v; �)dS ; (2:4:5)where L� is the (formally) adjoint operator to L, de�ned byL�v := Xj�j�m(�1)j�jD� �a�(x)Tv� : (2:4:6)For the Laplace operator L = �, and scalar functions u and v we haveZ v�u dx = Z@Xi vuxi�i dS � ZXuxivxi dx= Z@ vdudn dS � ZXi uxivxi dx :Integrating once more by parts we obtainZ v�u dx = Z u�v dx+ Z@ vdudn � udvdn! dS : (2:4:7)2.4.2 The Uniqueness theorem of HolmgrenThe Cauchy{Kowalevski theorem does not exclude the possibility that nonanalytic solutionsof the same Cauchy problem might exist, furthermore it works only for analytic Cauchydata. However, uniqueness can be proved for the Cauchy problem for a linear equationwith analytic coe�cients and for data (not necessarily analytic) prescribed in an analyticnoncharacteristic surface S. The proof (due to Holmgren) uses of the Cauchy{Kowalevskitheorem and the Lagrange{Green identity.Let u be a solution of a �rst order systemLu = nXk=1 ak(x) @u@xk + b(x)u = 0 (2:4:8)

26 CHAPTER 2. CHARACTERISTIC MANIFOLDS AND THE CAUCHY PROBLEMR

S

RZ

u=0in a \lens{shaped" region R bounded by two hypersurfaces S and Z.Here x 2 IRn ; u 2 IRn and ak; b are N �N matrices. Assume that u has Cauchy data u = 0on Z and that S is non{characteristic; that is, the matrixA = nXk=1 ak(x)�k (2:4:9)is non{degenerate for x 2 S, and � is the unit normal of S at x. Let v be a solution of theadjoint equation L�v = � nXk=1 @@xk �(ak)Tv�+ bTv = 0 for x 2 IR (2:4:10)(T for transposition) with Cauchy datav = w(x) for x 2 S : (2:4:11)Applying the Lagrange{Green identity we �nd thatZS wTAuds = 0 : (2:4:12)Let now � be the set of functions w on S for which the Cauchy problem (2.4.10){(2.4.11)has a solution v. If � is dense in C0(S) we conclude that (2.4.12) holds for every w 2 C0(S).But then Au = 0 on S, and hence also, since A is non{degenerate, u = 0 on S. For if Au 6= 0for some z 2 S, then also Au 6= 0 for all x in a neighborhood ! of z on S. We can �nd acontinuous non{negative scalar function �(x) on S with support in ! and with �(z) > 0.Then ZS �(Au)T (Au) ds > 0for w = �(Au) contrary to (2.4.12). Now in the case where the matrices ak and b are realanalytic, and S and w are real analytic, the Cauchy{Kowalevski theorem guarantees theexistence of a solution v of L�v = 0 with v = w on S in a su�ciently small neighborhoodof S, and so we cannot be sure that this neighborhood includes all of R. To bridge the gapbetween S and Z and to conclude that u = 0 throughout R, we have to cover all of R by ananalytic family of non{characteristic surfaces S�.De�nition 2.4.1 A family of hypersurfaces S� in IRn with parameter � ranging over anopen interval � = (a; b) forms an analytic �eld, if the S� can be transformed bi{analyticallyinto the cross sections of a cylinder whose base is the unit ball in IRn�1. This means thatthere exists a 1� 1 mapping F : ��! IRn, where x = F (y) is real analytic in �� andhas non{vanishing Jacobian; the S� for � 2 � are the setsS� = nx ��� x = F (y) ; (y1; : : : ; yn�1) 2 ; yn = �o : (2:4:13)

2.4. THE UNIQUENESS THEOREM OF HOLMGREN 27Our conditions imply that the set X = [�2�S� ; (2:4:14)called the support of the �eld, is open, and that the transformation x = F (y) has a realanalytic inverse y = G(x) mappingP onto ��. In particular �(x) = Gn(x) is real analyticin P.Uniqueness theorem Let the S� for � 2 � = (a; b) form an analytic �eld in IRn withsupport P. Consider the mth order linear systemLu = Xj�j�mA�(x)D�u = 0 (2:4:15)where x 2 IRn ; u 2 IRN , and the coe�cient matricesA�(x) are real analytic inP. Introducethe sets R = nx ��� x 2X ; xn � 0o (2.4.16)Z = nx ��� x 2X ; xn = 0o (2.4.17)and for � 2 � X� = nx ��� x 2 S� for some � with a < � � �o : (2:4:18)We assume that Z and all S� are non{characteristic, with respect to L, and that P� \R forany � 2 � is a closed subset of the open set P. Let u be a solution of (2.4.15) of classCm(R) and having vanishing Cauchy data on Z. Then u = 0 in R.R

Z

S

b

xn=0

a

S

Syn

yn = 0

yn= a

yn=

yn= b

28 CHAPTER 2. CHARACTERISTIC MANIFOLDS AND THE CAUCHY PROBLEM

Chapter 3Hyperbolic EquationsIn this chapter we shall consider the following equationutt = a2uxx + f(x; t)for x 2 IR1 ; t 2 R+.3.1 Boundary and initial conditionsConsider the equation of small transverse vibrations of a stringutt = a2uxx + f(x; t) (3:1:1)for 0 � x � `. If the ends of the string are �xed, say, on the plane, then the boundaryconditions u(0; t) = 0 ; u(`; t) = 0 (3:1:2)must be satis�ed. Furthermore, the initial conditions, i. e. the form and the velocity @u=@tat the moment when the process begins, say at t0, are given:u(x; t0) = '(x) ; (3.1.3)ut(x; t0) = (x) : (3.1.4)These conditions must coincide with the conditions (3.1.2) at the ends of [0; `]. Here ' and are given functions. Later we shall prove that that these conditions fully describe thesolution of the equation (3.1.1).If the ends of the string are moving with a given law, then the boundary conditions have theform u(0; t) = �1(t) ; u(`; t) = �2(t) ; (3:1:20)where �1(t) and �2(t) are given.The boundary conditions can have another forms. For example, if one end of the string is�xed, say for x = 0, and the other end is free, then we haveu(0; t) = 0 ; (3:1:5)29

30 CHAPTER 3. HYPERBOLIC EQUATIONSand at the free end the elasticity tensionT (`; t) = k @u@x �����x=`is zero (no external force, k(x) is the Young modulus at the point x). Thus,ux(`; t) = 0 : (3:1:6)If the end x = 0 is moving with a certain law �(t), meanwhile at x = ` a given force �(t) isacting then u(0; t) = �(t) ; ux(`; t) = �(t) ; �(t) = �kAnother type of boundary conditions isux(`; t) = �h[u(`; t)� �(t)] (3:1:7)(elastische Befestigung).Thus, we have the following types of boundary conditionsBoundary condition of �rst kind u(0; t) = �(t)second kind ux(0; t) = �(t)third kind ux(0; t) = h[u(0; t)� �(t)] :There are also other kinds, e. g., ux(`; t) = 1kF [u(`; t)] ; : : :We do not enter in this direction, but refer to the book by Tikhonov and Samarskii [22].If the time interval is small, then the boundary conditions have not so much in uence uponthe vibration and so we can consider the problem as the limit case for the initial valueproblem in an unbounded domain:Find a solution of the equationutt = a2uxx + f(x; t) ; �1 < x <1 ; t > 0 ;with the initial deformationu(x; t)���t=0 = '(x) ; �1 < x <1 ;and the initial velocity ut(x; t)���t=0 = (x) ; �1 < x <1 :This is in fact the Cauchy problem for a hyperbolic equation.If one wants to study the behaviour of the string near to one end assuming that the in uenceof the boundary condition at the other end is small, then we have, e. g.,u(0; t) = �(t) ; t � 0 ;u(x; 0) = '(x) ; 0 � x <1 ;ut(x; 0) = (x) ; 0 � x <1 :For a concrete vibration process we have to �nd appropriate boundary conditions for it.

3.2. THE UNIQUENESS 313.2 The uniquenessTheorem 3.2.1 The di�erential equation�(x)@2u@t2 = @@x k(x)@u@x!+ f(x; t) ; 0 < x < ` ; t > 0 ; (3:2:1)has no more than one solution which satis�es the initial and boundary value conditionsu(x; 0) = '(x) ;ut(x; 0) = (x) ;u(0; t) = �1(t) ;u(`; t) = �2(t) : (3:2:2)Here, we assume that the function u(x; t) and its �rst and second derivatives are continuousin the interval 0 � x � ` for t � 0 and �(x) > 0 ; k(x) > 0 are continuous, given functions.Proof: Suppose that there exist two solutions u1(x; t) ; u2(x; t) of our problem, then itsdi�erence v(x; t) = u1(x; t)� u2(x; t)satis�es the homogeneous equation �@2v@t2 = @@x k @v@x! (3:2:3)and the homogeneous conditions v(x; 0) = 0 ;vt(x; 0) = 0 ;v(0; t) = 0 ;v(`; t) = 0 : 9>>>=>>>; (3:2:4)We shall prove that with the conditions (3.2.4) v is identically zero.For this purpose we consider the functionE(t) := 12 Z `0 nk(vx)2 + �(vt)2o dx (3:2:5)and prove that it is independent of t.The function E(t) has a physical meaning; it is the total energy of the string at the momentt. Since v is twice continuously di�erentiable, we can di�erentiate E(t) and getdE(t)dt = Z `0 (kvxvxt + �vtvtt) dx (3:2:6)An integration by part in the second term of the right{hand side givesZ `0 kvxvxt dx = kvxvt���0 � Z `0 vt(kvx)x dx= � Z `0 vt(kvx)x dx : (3.2.7)

32 CHAPTER 3. HYPERBOLIC EQUATIONS(Since v(0; t) = 0 it follows that vt(0; t) = 0 and v(`; t) = 0 implies vt(`; t) = 0 ). Thus,dE(t)dt = Z `0 [�vtvtt � vt(kvx)x] dx= Z `0 [�vtt � (kvx)x] vt dx = 0 :It means, E(t) = const. Furthermore, since v(x; 0) = 0 ; vt(x; 0) = 0 we haveE(t) = const = E(0) = 12 Z `0 hk(vx)2 + �(vt)2i ����t=0 dx = 0 : (3:2:8)On the other hand, the functions �(x) and k(x) are positive, and it follows from (3.2.6),(3.2.8) that vx(x; t) � 0 ; vt(x; t) � 0 :Hence v(x; t) = const = v(x; 0) � 0 : 2The same result remains valid for the second boundary problem, whereux(0; t) = �1(t) and ux(`; t) = �2(t) ;because in this case vx(0; t) = vx(`; t) = 0.The proof for the third boundary value problemux(0; t)� h1u(0; t) = �1(t) ; h1 � 0 ;ux(`; t) + h2u(0; t) = �2(t) ; h2 � 0 ;is a little bit di�erent. In this casevx(0; t)� h1v(0; t) = 0 ;vx(`; t) + h2v(0; t) = 0 ;and in (3.2.7) we obtain kvxvt���0 = �k2 @@t hh2v2(`; t) + h1v2(0; t)i :Integrating dEdt from 0 to t givesE(t)� E(0) = Z t0 Z `0 vt [�vtt � (kvx)x] dx dt� k2 nh2 hv2(`; t)� v2(`; 0)i + h1 hv2(0; t)� v2(0; 0)io :Taking into account that E(0) = 12 Z `0 nkv2x + �v2to ���t=0 dx = 0and v(`; 0) = v(0; 0) = 0, we getE(t) = �k2 hh2v2(`; t) + h1v2(0; t)i � 0 :On the other hand E(t) � 0, it follows that E(t) � 0 and so v(x; t) � 0.

3.3. THE METHOD OF WAVE PROPAGATION 333.3 The method of wave propagation (Wellenausbrei-tungsmethode)3.3.1 The D'Alembert methodWe consider the problemutt � a2uxx = 0 ; �1 < x <1 ; t > 0 ; (3.3.1)u(x; 0) = '(x) ; ut(x; 0) = (x) ; �1 < x <1 : (3.3.2)First, we transform the equation (3.3.1) into the canonical form, in which there is the mixedderivative. The characteristic equationdx2 � a2dt2 = 0is equivalent to dx� a dt = 0 ; dx+ a dt = 0 ;the integrals of which are x� at = C1 ; x+ at = C2 :Introducing the new variables � = x+ at ; � = x� atwe get ux = u� + u� ; uxx = u�� + 2u�� + u��ut = a(u� � u�) ; utt = (u�� � 2u�� + u��) a2 :Thus, the equation (3.3.1) is transformed intou�� = 0 : (3:3:3)It is clear that every solution of Eq. (3.3.3) has the formu�(�; �) = f�(�) ;where f�(�) is an arbitrary function only of �.Integrating this equation we obtainu(�; �) = Z f�(�)d� + f1(�) = f1(�) + f2(�) ; (3:3:4)where f1 depends only on � and f2 depends only on �. Conversely, if f1 and f2 are arbitrarydi�erentiable functions, then the function u(�; �) de�ned by (3.3.4) is a solution of (3.3.3).It follows that u(x; t) = f1(x+ at) + f2(x� at) (3:3:5)is a general solution of (3.3.3).Assume that there exists a solution of our Cauchy problem (3.3.1), (3.3.2), then it is de�nedby (3.3.5). The functions f1 and f2 are de�ned byu(x; 0) = f1(x) + f2(x) = '(x) (3.3.6)ut(x; 0) = af 01(x)� a2f 02(x) = (x) : (3.3.7)

34 CHAPTER 3. HYPERBOLIC EQUATIONSFrom (3.3.7) we have f1(x)� f2(x) = 1a Z xx0 (�)d� + c ;where x0 and c are some constants. The last equation and (3.3.6) yieldf1(x) = 12'(x) + 12a Z xx0 (�)d�+ c2f2(x) = 12'(x)� 12a Z xx0 (�)d�� c2 : 9>>>=>>>; (3:3:8)It follows thatu(x; t) = '(x+ at) + '(x� at)2 + 12 �Z x+atx0 (�)d�� Z x�atxo (�)d��or u(x; t) = '(x+ at) + '(x� at)2 + 12a Z x+atx�at (�)d� (3:3:9)The formula (3.3.9) is called D'Alembert's formula. It proves the uniqueness of the solutionof the problem (3.3.1), (3.3.2). It is clear also that if ' is twice di�erentiable and is oncedi�erentiable then (3.3.9) is a solution of (3.3.1){(3.3.2). Thus the method of D'Alembertproves not only the uniqueness of the solution but also its existence.x

t

P Q

M(x0 0),t

x+at=c2 x−at=c1Let (x0; t0) be a point in the plane (x; t) ; t � 0. There are two characteristics crossing thispoint M = (x0; t0). Denoting the intersections of these characteristics with the line x = 0by P and Q, respectively, we see that in order to determine u(x0; t0) we need only ' and in [P;Q]. The triangle MPQ is called the characteristic triangle of the point M .3.3.2 The stability of the solutionThe solution of Eq. (3.3.1) is uniquely determined by the condition (3.3.2). We shall provethat it depends continuously on the Cauchy data. Namely, we prove the following result.Theorem 3.3.1 For every time interval [0; t0], and " > 0 there always exists a number� = �("; t0) such that if j'j < � ; j j < � ;then ju(x; t)j < " (0 � t � t0) :

3.3. THE METHOD OF WAVE PROPAGATION 35Remark 3.3.2 Let ui(x; t) be the solutions of (3.3.1) with the Cauchy data 'i and i ; i =1; 2. The theorem says that, for every time interval [0; t0], and " > 0 there always exists anumber � = �("; t0) such that ifj'1 � '2j < � ; j 1 � 2j < � ;then ju1(x; t)� u2(x; t)j < " (0 � t � t0) :Proof: From the D'Alembert formula (3.3.9) we haveju(x; t)j � j'(x+ at)j+ j'(x� at)j2 + 12a Z x+atx�at j (�)jd�< � + 12a�:2at � �(1 + t0) :Thus, if we take � = "1+t0 , then ju(x; t)j < ". 2A boundary value problem is called well{posed, if1. It has a solution (Existence)2. This solution is unique (Uniqueness)3. The solution depends continuously on the initial and boundary conditions (Stability).If one of these three condtions is not met, we say that our problem is ill{posed.For our Cauchy problem we know thatu(x; t) = '(x+ at) + '(x� at)2 + 12a Z x+atx�at (�)d�is a unique solution, if ' 2 C2 ; 2 C1, and this solution is stable in L1{norm. Thus,our problem is well{posed in this class. However, if ' 62 C2 ; 62 C1, then the D'Alembertformula gives no solution to our Cauchy problem. Anyway, we have proved that the functionu(x; t) designed by (3.3.9) is stable for any ' and , thus if ' 62 C2 ; 62 C1 then we canapproximate them by '" 2 C2 ; " 2 C1 and receive a solution u" designed by (3.3.9) withrespect to these data, such that ju� u"j < ". Letting "! 0, we have u" ! u. We say thatu is a generalized solution of our Cauchy problem with respect to non-smooth Cauchy data' and .Another example of ill{posednessThe Cauchy problem for the Laplace equation has the formuxx + uyy = 0 ; �1 < x <1 ; y > 0u(x; 0) = '(x) ; uy(x; 0) = (x) ; �1 < x <1 :The functions u(1)(x; y) = 0 ; u(2)(x; y) = 1� sin(�x) cosh(�y)

36 CHAPTER 3. HYPERBOLIC EQUATIONSy

x

u + u = 0

u(x, 0)= (x)u (x, 0)= (x)

xx yy

ysatisfy the Laplace equation; furthermoreu(1)(x; 0) = 0 ; u(2)(x; 0) = '(x) = 1� sin(�x)u(1)y (x; 0) = 0 ; u(2)y (x; 0) = (0) = 0 :If �! 0, then ju(1)(x; 0)� u(2)(x; 0)j ! 0; however, for y > 0���u(1)(x; y)� u(2)(x; y)��� = 1� j sin�xj j cosh(xy)j =�! 0 :Thus, the Cauchy problem for the Laplace equation is ill{posed.3.3.3 The re ection methodConsider the problem in the half axis x > 0:utt � a2uxx = 0 ; 0 < x <1 ; t > 0 ;u(0; t) = �(t) ; (or ux(0; t) = �(t) ) ;u(x; 0) = '(x) ; 0 < x <1 ;ut(x; 0) = (x) ; 0 < x <1 :For simplicity, �rst, suppose thatu(0; t) = 0 (or, ux(0; t) = 0) ;Lemma 3.3.3 Consider the problem (3.3.1), (3.3.2).1. If the Cauchy data ' and are odd functions with respect to some point x0, then thecorresponding solution u(x; t) is equal to zero at this point.2. If the Cauchy data ' and are even functions with respect to x0, then the derivativeof u with respect to x is equal to zero at this point.Proof 1. We can suppose that x0 = 0, and thus'(x) = �'(�x) ; (x) = � (�x) :For x = 0 we have u(0; t) = '(at) + '(�at)2 + 12a Z at�at (�)d� = 0 :

3.3. THE METHOD OF WAVE PROPAGATION 372. If '(x) = '(�x) ; (x) = (�x) thenux(0; t) = '0(at) + '0(�at)2 + 12ah (at)� (�at)i = 0 : 2Now we consider the problemutt = a2uxx ; 0 < x <1 ; t > 0 ;u(x; 0) = '(x) ; 0 < x <1 ;ut(x; 0) = (x) ; 0 < x <1 ;u(0; t) = 0 ; t > 0 :Let �(x) = ( '(x) x > 0�'(�x) x < 0(x) = ( (x) x > 0� (�x) x < 0be odd extensions of ' and , respectively. It is clear that the functionu(x; t) = �(x+ at) + �(x� at)2 + 12a Z x+atx�at (�)d�is well de�nded. Furthermore, taking Lemma 3.3.3 into account , we see thatu(0; t) = 0 :Besides, for x > 0 u(x; 0) = �(x) = '(x) ;ut(x; 0) = (x) = (x) : )Thus u(x; t) is a solution of our problem. We can write the solution in the following wayu(x; t) = 8>>><>>>: '(x+ at) + '(x� at)2 + 12a Z x+atx�at (�)d� for t < xa ; x > 0'(x+ at) + '(at� x)2 + 12a Z x+atat�x (�)d� ; t > xa ; x > 0x

t

t < ax

38 CHAPTER 3. HYPERBOLIC EQUATIONSIn the domain t < xa the boundary condition has no in uence on the solution and it coincideswith the solution of (3.3.1) in (�1;1).We consider now the case ux(0; t) = 0 :With this purpose we take the even extension of ' and :�(x) = ( '(x) x > 0'(�x) x < 0 ; (x) = ( (x) ; x > 0 (�x) ; x < 0 :As a solution of our equation (3.3.1) we haveu(x; t) = �(x+ at) + �(x� at)2 + 12a Z x+atx�at (�)d�or u(x; t) = 8>>><>>>: '(x+ at) + '(x� at)2 + 12a Z x+atx�at (�)d� ; t < xa'(x+ at) + '(at� x)2 + 12a �Z x+at0 (�)d�+ Z at�x0 (�)d�� ; t > xa :It is clear that u(x; t) satis�es the equation, the initial conditions and the boundary conditionux(0; t) = 0.It remains to consider non{homogeneous casesu(0; t) = �(t) 6= 0 ;and ux(0; t) = �(t) 6= 0 :Let us consider the case of homogeneous Cauchy datau(x; 0) = 0 ; ut(x; 0) = 0 ; u(0; t) = �(t) ; t > 0 :The general solution to this problem has the formu(x; t) = f1(x+ at) + f2(x� at) :Since u(x; 0) = 0 ; ut(x; 0) = 0, we see that f1(s) = �f2(s) = c for s � 0, where c is aconstant.From the boundary conditionu(0; t) = �(t) = f1(at) + f2(�at)= f2(�at) + c ; t > 0 :Thus, f2(s) = � ��sa �+ c ; s < 0. Therefore, if x� at < 0, say t > xa , we haveu(x; t) = ���x� ata �+ c = ���xa + t�+ c :Putting x = 0, we get c = 0.

3.4. THE FOURIER METHOD 39x

t

x − a t < 0If x� at � 0, say t < xa , thenu(x; t) = f1(x+ at) + f2(x� at) = 0 :Now it is easily seen that the solution of our boundary value problem isu(x; t) = 8>>><>>>: '(x+ at) + '(x� at)2 + 12a Z x+atx�at (�)d� ; t < xa� �t� xa�+ '(x+ at) + '(at� x)2 + 12a Z x+atat�x (�)d� ; t > xa :3.4 The Fourier method3.4.1 Free vibration of a stringConsider the problem utt = a2uxx ; 0 < x < ` ; t > 0 ; (3.4.1)u(0; t) = u(`; t) = 0 ; t > 0 ; (3.4.2)u(x; 0) = '(x) ; ut(x; 0) = (x) ; 0 � x � ` : (3.4.3)The equation (3.4.1) is linear and homogeneous, and therefore the sum of its special solutionsis again a solution. We shall try to �nd the solution of our problem (3.4.1){(3.4.3) via asum of special solutions with appropriate coe�cients. With this purpose we consider theauxiliary problem: Find a solution of the problemutt = a2uxx ; 0 < x < ` ; t > 0 ; (3:4:1)which is not identically vanishing and satisfying the homogeneous boundary conditionsu(0; t) = 0 ; u(`; t) = 0 ; (3:4:2)and can be represented in the form u(x; t) = X(x)T (t) : (3:4:4)Here X depends only on x and T depends only on t.

40 CHAPTER 3. HYPERBOLIC EQUATIONSPutting (3.4.4) into (3.4.1) we get X 00T = 1a2T 00X ; (3:4:5)or, dividing by XT , X 00(x)X(x) = 12a2 T 00(t)T (t) : (3:4:6)The function (3.4.4) is a solution of (3.4.1), if (3.4.5) or (3.4.6) are satis�ed. The right{handside of (3.4.6) depends only on t, meanwhile the left{hand side depends only on x. It followsthat they must be equal to a constant, say ��:X 00(x)X(x) = 1a2 T 00(t)T (t) = �� ; (3:4:7)here, we put the sign minus before � only for our convenience.From (3.4.7) we obtain two ordinary di�erential equations for X and TX 00(x) + �X(x) = 0 ; (3.4.8)T 00(t) + a2�T (t) = 0 : (3.4.9)The boundary value conditions (3.4.2) giveu(0; t) = X(0)T (t) = 0 ;u(`; t) = X(`)T (t) = 0 :It follows that X(0) = X(`) = 0 (3:4:10)otherwise T (t) � 0 and u(x; t) � 0, and u is not a non{trivial solution of our problem.Thus, in order to �nd X(x) we get an eigenvalue problem:Find � such that there exists a non{trivial solution of the problemX 00 + �X = 0 ;X(0) = X(`) = 0 ) ; (3:4:11)and �nd the solutions corresponding to these �. The �, for which a non{trivial solution exists,are called eigenvalues, and the corresponding solutions to them are called eigenfunctions.In what follows we distinguish between three cases:1. For � < 0, the problem has no non{trivial solution. In fact, the general solution of(3.4.11) has the form X(x) = C1ep��x + C2e�p��x :This solution must satisfy the boundary conditions:X(0) = C1 + C2 = 0 ;X(`) = C1e`p�� + C2e�`p�� = 0 :Thus, C1 = �C2 and C1 �e`p�� � e�`p��� = 0 :Because � < 0 ; p�� is real and positive, and so e`p�� � e�`p�� 6= 0. Hence, C1 = 0 ;C2 = 0 and X(x) � 0.

3.4. THE FOURIER METHOD 412. For � = 0 there exists no non{trivial solution, since the general solution isX(x) = ax+ b ;and the boundary conditions then giveX(0) = [ax+ b] ���x=0 = b = 0X(`) = a` = 0 :Thus, a = 0 ; b = 0 and X(x) � 0.3. For � > 0, the general solution has the formX(x) = D1 cosp�x+D2 sinp�x :The boundary conditions giveX(0) = D1 = 0 ;X(`) = D2 sinp�` = 0 :Since X(x) is not identically vanishing, D2 6= 0, and thereforesinp�` = 0 ; (3:4:12)say p� = �n ; n 2 ZZ :Thus, a non{trivial solution is possible only for the values�n = ��n�2 ; n 2 IN :These eigenvalues correspond to the eigenfunctionsXn(x) = Dn sin �nx :Thus, for the values � which are equal to�n = ��n�2 ; n 2 IN ; (3:4:13)there exist non{trivial solutions Xn(x) = sin �nx ; (3:4:14)which are uniquely de�ned up to a constant coe�cient. For these �, the solutions of(3.4.9) are Tn(t) = An cos �nat+Bn sin �nat ; (3:4:15)where An and Bn are to be de�ned. It follows that the functionsun(x; t) = Xn(x)Tn(t) = �An cos �nat+Bn sin �nat� sin �nx (3:4:16)

42 CHAPTER 3. HYPERBOLIC EQUATIONSare special solutions of (3.4.1), which satisfy the boundary condition (3.4.2). Since(3.4.1) is linear and homogeneous, the functionu(x; t) = 1Xn=1 un(x; t) = 1Xn=1�An cos �nat+Bn sin �nat� sin �nx ; (3:4:17)| if it converges and one can di�erentiate it termwise two times with respect to x andt | is a solution of (3.4.1) and satis�es the boundary conditions (3.4.2). (We shallcome back to this question in the next paragraph.) The initial conditions giveu(x; 0)='(x)= 1Xn=1 un(x; 0) = 1Xn=1An sin �nx ;ut(x; 0)= (x)= 1Xn=1 @un@t (x; 0)= 1Xn=1 �naBn sin �nx : (3:4:18)Let ' and be piecewise continuous and di�erentiable, then we have'(x) = 1Xn=1'n sin �nx ; 'n = 2 Z `0 '(�) sin �n�d� (3.4.19) (x) = 1Xn=1 n sin �nx ; n = 2 Z `0 (�) sin �n�d�: (3.4.20)A comparison of these two series with (3.4.18) givesAn = 'n ; Bn = `�na n : (3:4:21)Thus, the series (3.4.17) is completely de�ned.3.4.2 The proof of the Fourier methodLet L(u) be a linear di�erential operator. It means that L(u), the action of L onto u, isequal to the sum of the corresponding derivatives of u with coe�cients independent of u.Lemma 3.4.1 Let the ui(i = 1; 2; : : :) be special solutions of a linear homogeneous di�eren-tial equation L(u) = 0. Then the series u = P1i=1Ciui is also a solution of this equation, ifthe derivatives in L(u) can be di�erentiated termwise.We come back to our boundary value problem. First, we have to prove the continuity of thefunction u(x; t) = 1Xn=1 un(x; t) = 1Xn=1�An cos �nat+Bn sin �nat� sin �nx : (3:4:22)As jun(x; t)j � jAnj+ jBnj ;we conclude that if 1Xn=1 �jAnj+ jBnj) (3:4:23)

3.4. THE FOURIER METHOD 43converges, then the series (3.4.22) uniformly converges and u(x; t) is continuous.Analogously, in order to prove the continuity of ut(x; t) we have to prove the uniform con-vergence of the seriesut(x; t) � 1Xn=1 @un@t = 1Xn=1 a�n ��An sin �nat+Bn cos �nat� sin �nx (3:4:24)or the convergence of the majorant seriesa� 1Xn=1 n�jAnj+Bnj� :Also we have to prove the uniform convergence of the seriesuxx � 1Xn=1 @2un@x2 = �� ��2 1Xn=1n2 �An cos �nat+Bn sin �nat� sin �nx ;utt � 1Xn=1 @2un@t2 = ���a�2 1Xn=1 n2 �An cos �nat+Bn sin �nat� sin �nx :These correspond, up to a constant, to the majorant series1Xn=1n2�jAnj+ jBnj� : (3:4:25)Since An = 'n ; Bn = `�na n ;where 'n = 2 Z `0 '(x) sin �nx dx ; n = 2 Z `0 (x) sin �nx dx ;our method is proved if we can prove the convergence of the series1Xn=1 nkj'nj (k = 0; 1; 2)1Xn=1 nkj nj (k = �1; 0; 1) : (3:4:26)Results from the theory of Fourier seriesLet F be a 2`{periodic function, then we can expand F into its Fourier seriesF (x) = a02 + 1Xn=1�an cos �nx+ bn sin �nx� ;an = 1 Z�` F (�) cos �n� d� ; bn = 1 Z�` F (�) sin �n� d� :If F (x) is odd, then an = 0 andF (x) = 1Xn=1 bn sin �nxbn = 1 Z�` F (�) sin �n� d� = 2 Z `0 F (�) sin �n� d� :

44 CHAPTER 3. HYPERBOLIC EQUATIONSIf F (x) is de�ned only in (0; `), then we can oddly extend F (x) to be de�ned in (�`; `) andthen use the above expansion.It is known that, if F 2 Ck and F (k) is piecewise continuous then the series1Xn=1 nk�janj+ jbnj�converges. If a function f(x) is de�ned only in (0; `) then we can extend it oddly to bede�ned in (�`; `), say to F (x). Since F (x) is continuous, f(0) must be 0. Further, f(`) mustbe also 0, since the F (x) must be 2`{periodic and continuous. The continuity of the �rstderivative at x = 0 and x = ` is automatically satis�ed. Generally, one has to requiref (k)(0) = f (k)(`) = 0 (k = 0; 2; 4; : : : ; 2n) :From these results we conclude:1. In order to guarentee the convergence of the series1Xn=1 nkj'nj (k = 0; 1; 2) ;the function ' must be two times continuously di�erentiable, furthermore the thirdderivative of ' must be piecewise continuous, and'(0) = '(`) = 0 ; '00(0) = '00(`) = 0 : (3:4:27)2. In order to guarantee the convergence of the series1Xn=1 nkj nj (k = �1; 0; 1)the function must be continuously di�erentiable, have a piecewise continuous secondderivative, and (0) = (`) = 0 : (3:4:28)With these conditions, the representation (3.4.17) is in fact a solution of the problem (3.4.1),(3.4.2). Note that this solution is unique.3.4.3 Non{homogeneous equationsWe consider the non{homogeneous hyperbolic equationutt = a2uxx + f(x; t) ; 0 < x < ` ; t > 0 ; (3:4:29)with the initial conditionsu(x; 0) = '(x) ; ut(x; 0) = (x) ; 0 � x � ` ; (3:4:30)and homogeneous boundary value conditionsu(0; t) = u(`; t) = 0 ; t > 0 : (3:4:31)

3.4. THE FOURIER METHOD 45We try to �nd the solution of (3.4.30){(3.4.31) in the formu(x; t) = 1Xn=1 un(t) sin �nx ; (3:4:32)where un(t) are to be determined. For this purpose we expand f; ' and in the Fourierseries f(x; t) = 1Xn=1 fn(t) sin �nx ; fn(t) = 2 Z `0 f(�; t) sin �n� d� ;'(x) = 1Xn=1'n sin �nx ; 'n = 2 Z `0 '(�) sin �n� d� ; (3.4.33) (x) = 1Xn=1 n sin �nx ; n = 2 Z `0 (�) sin �n� d� :Plugging (3.4.32) into (3.4.29), we get1Xn=1 sin��nx� �a2��n�2 un(t)� �un(t) + fn(t)! = 0 :Thus, �un(t) + ��n�2 a2un(t) = fn(t) : (3:4:34)On the other hand,u(x; 0) = '(x) = 1Xn=1 un(0) sin �nx = 1Xn=1'n sin �nx ;ut(x; 0) = (x) = 1Xn=1 _un(0) sin �nx = 1Xn=1 n sin �nx :It follows un(0) = 'n ; _un(0) = n : (3:4:35)Consequently, we can �nd un(t) in the formun(t) = u(I)n (t) + u(II)(t) ;where u(I)n (t) = `�na Z t0 sin��na(t� � )� fn(� ) d� (3:4:36)and u(II)n (t) = 'n cos��nat�+ `�na n sin��nat� : (3:4:37)Thus u(x; t) = 1Xn=1 `�na Z t0 sin �na(t� � ) sin �nx � fn(� ) d�+ 1Xn=1 'n cos �nat+ `�na n sin �nat! sin �nx ; (3.4.38):= u(I)(x; t) + u(II)(x; t) :

46 CHAPTER 3. HYPERBOLIC EQUATIONSTaking (3.4.33) into account for fn(t), we can represent u(I)(x; t) in the formu(I)(x; t) = Z t0 Z `0 ( 2 1Xn=1 `�na sin �na(t� � )sin�n` x � sin �n�) f(�; � ) d� d�:= Z t0 Z `0 G(x; �; t � � )f(�; � ) d� d� ; (3.4.39)where G(x; �; t� � ) := 2�a 1Xn=1 1n sin �na(t� � ) sin �nx sin �n� : (3:4:40)3.4.4 General �rst boundary value problemFind a solution of the problemutt = a2uxx + f(x; t) ; 0 < x < ` ; t > 0 ; (3.4.41)u(x; 0) = '(x) ; ut(x; 0) = (x) ; 0 < x < ` ; (3.4.42)u(0; t) = �1(t) ; u(`; t) = �2(t) ; t > 0 : (3.4.43)To deal with this problem we shall �nd a di�erentiable function U , such thatU(0; t) = �1(t) ; U(`; t) = �2(t) (3:4:44)and �nd an equation of the type (3.4.41) for which U(x; t) is a solution. It can be easily seenthat the function U(x; t) = �1(t) + x [�2(t)� �1(t)] (3:4:45)satis�es (3.4.44) and the equationUtt = a2Uxx + ��1(t) + x[��2(t)� ��1(t)]= a2Uxx + �f(x; t) : (3.4.46)Now consider the problem vtt = a2vxx + f � �f ;v(x; 0) = '(x)� U(x; 0) ;vt(x; 0) = (x)� Ut(x; 0) ;v(0; t) = 0 ; v(`; t) = 0 : (3:4:47)It is clear that we can �nd v in the form of (3.4.38) and furthermoreu(x; t) = U(x; t) + v(x; t) : (3:4:48)3.4.5 General scheme for the Fourier methodLet '(x); k(x); q(x) be positive functions.

3.4. THE FOURIER METHOD 47Consider the problemL[u] := @@x "k(x)@u@x#� q(x)u = �(x)@2u@t2 ; (3.4.49)u(0; t) = 0 ; u(`; t) = 0 ; t > 0 ; (3.4.50)u(x; 0) = '(x) ; ut(x; 0) = (x) ; 0 < x < ` : (3.4.51)We now use the Fourier method to solve this problem. Namely, we try to �nd a non{trivialsolution of (3.4.49) which satis�es the boundary conditionsu(0; t) = 0 ; u(`; t) = 0and can be represented in the form u(x; t) = X(x)T (t) :With the same argument as in x 3.4.1, we arrive at the problemsddx "k(x)dXdx # � qX + ��X = 0and T 00 + �T = 0 :To �nd X(x) we get the following eigenvalue problem: Find �, for which the boundary valueproblem L[X] + ��X = 0 ; 0 < x < ` ; (3:4:52)X(0) = 0 ; X(`) = 0 ; (3:4:53)has a non{trivial solution. The � are called eigenvalues, and the associated non{trivialsolutions of (3.4.52), (3.4.53) are called eigenfunctions.Properties of the problem (3.4.52), (3.4.53)1. There exist countably many eigenvalues �1 < �2 < � � � < �n < � � �, and their associatednon{trivial solutions (eigenfunctions) X1(x); X2(x) : : : ;Xn(x); : : :2. For q � 0, all �n are positive.3. The eigenfunctions are orthogonal in the interval [0; `] with the weight �(x) in the sensethat Z `0 Xm(x)Xn(x)�(x) dx = 0 ; m 6= n : (3:4:54)4. Any function F (x) de�ned in [0; `] which is two times continuously di�erentiable andsatis�es the boundary condition F (0) = F (`) = 0 can be expanded into a uniformlyand absolutely convergent series of the Xn(x):F (x) = 1Xn=1FnXn(x) ; Fn = 1Nn Z `0 F (x)Xn(x)�(x) dx ;Nn = Z `0 X2n(x)�(x) dx : (3:4:55)

48 CHAPTER 3. HYPERBOLIC EQUATIONSThe properties 1 and 4 are provided in the Theory of Integral Equations, and we do not giveany proof for them here. We shall prove the properties 2 and 3. At �rst, we give here theGreen formula. Let u and v be de�ned in [a; b] ; u; v 2 C2(a; b) and u; v 2 C1[a; b]. ThenuL[v]� vL[u] = u(kv0)0 � v(ku0)0= hk(uv0 � vu0)i0 :Integrating the last from a to b we getZ ba �uL[v]� vL[u]� = k(uv0 � vu0)���ba :Proof of the property 3. Let Xm(x) and Xn(x) be two eigenfunctions and �m; �n be theirassociated eigenvalues. From the boundary conditions (3.4.53) we haveZ `0 nXmL[Xn]�XnL[Xm]o dx = 0 :Taking Eq. (3.4.52) into account we obtain(�n � �m) Z `0 Xm(x)Xn(x)�(x) dx = 0 :Since �n 6= �m, it follows Z `0 Xm(x)Xn(x)�(x) dx = 0 : (3:4:56)We shall prove that for every eigenvalue there exists, up to a constant factor, only oneeigenfunction. In order to prove this fact, we use a result which states that the solution of alinear ordinary di�erential equation is uniquely de�ned by its value and the value of its �rstderivative at a point.LetX andX be two eigenfunctions associated with the eigenvalue �. SinceX(0) = X(0) = 0,we have X 0(0) 6= 0 and X 0(0) 6= 0. SetX�(x) := X 0(0)X 0(0) X(x) :The function X�(x) satis�es the equation (3.4.52), furthermoreX�(0) = X 0(0)X 0(0) X(0) = 0 ;X�0(0) = X 0(0)X 0(0) X 0(0) = X 0(0) :Thus, X�(x) = X(x) and thereforeX�(x) = X(x) = X 0(0)X 0(0) X(x) = AX(x) :

3.4. THE FOURIER METHOD 49We conclude that if Xn(x) is an eigenfunction associated with the eigenvalue �n, thenAnXn(x) (An is a nonvanished constant) is also an eigenfunction. Since these eigenfunc-tions di�er only by a constant factor, the value of this factor is not important, and it can benormalized through the equalityNn = Z `0 X2n(x)�(x) dx = 1 :If an eigenfunction Xn(x) is not normalized, then we can normalize it byXn(x) = Xn(x)qR0 X2n(x)�(x) dx :By this way, we have found an orthogonal system of eigenfunctions Xn(x) of the problem(3.4.52), (3.4.53): Z `0 Xm(x)Xn(x)�(x) dx = ( 0 ; m 6= n ;1 ; m = n : 2Proof of the property 2. Let Xn(x) be the normalized eigenfunction associated to theeigenvalue �n: L[Xn] = ��n�(x)Xn(x) :Multiplying both sides of this equation by Xn(x) and integrating with respect to x from 0to `, we obtain �n Z `0 X2n(x)�(x) dx = � Z `0 Xn(x)L[Xn] dx :Since R0 X2n(x)�(x) dx = 1, we get�n = � Z `0 Xn ddx "k(x)dXndx # dx+ Z `0 q(x)X2n(x) dx :Integrating by part gives�n = �XnkX 0n���0 + Z `0 k(x)[X 0n(x)]2 dx+ Z `0 q(x)X2n(x) dx= Z `0 k(x)[X 0n(x)]2 dx+ Z `0 q(x)X2n(x) dx :Since k(x) > 0 ; q(x) � 0, Xn 2 C2(0; `), X 0n(x) is not identically vanishing, we have�n > 0 :For a function F (x) de�ned in (0; `) we can formally expand F (x) as followsF (x) = 1Xn=1FnXn(x) ;Fn = R0 �(x)F (x)Xn(x) dxR0 �(x)X2n(x) dx ; n 2 IN :

50 CHAPTER 3. HYPERBOLIC EQUATIONSWe say that Fn are the Fourier coe�cients of the function F in the system fX1;X2; : : :g.With the same method as in (3.4.1), we can prove thatu(x; t) = 1Xn=1�An cosq�nt+Bn sinq�nt�Xn(x) ;where An = 'n ; Bn = np�n :Here, 'n and n are the Fourier coe�cients of ' and , respectively.This method has been developed by Steklov. We do not pursuit it in this lecture.

3.5. THE GOURSAT PROBLEM 513.5 The Goursat Problem3.5.1 De�nitionConsider the equation uxy = f(x; y) : (3:5:1)Let C1 and C2 be two curves described by equations x = �1(y) and x = �2(y) and emanatingfrom a common point, which we take to be the origin of the coordinate system. We assumethat they do not intersect elsewhere. We assume further that the curves C1 and C2 arenon{tangent to characteristics situated in the �rst quarter and that the functions �1(y) and�2(y) are increasing, i. e. the curves C1 and C2 have time{like orientation. The functions�1(y) and �2(y) have inverses #1(x) and #2(x), respectively. For de�niteness suppose that�1(y) < �2(y) for the positive values of the variable y for which these two functions exist.y

x0

A

B

P0

C

C

1

2Let D be a domain contained between these arcs and two characteristics P0A and P0B. Weseek a solution u(x; y) of (3.5.1) de�ned in the closure D of the domain D and satisfying theDirichlet boundary conditions on the arcs C1 and C2:u ���OA = u(�1(y); y) = 1(y) ; 0 � y � �1 ; (3.5.2)u ���OB = u(�2(y); y) = 2(y) ; 0 � y � �2 ; (3.5.3)where �1 and �2 are the y{coordinates of the points A and B. We assume that 1(0) =2(0) = 0. Conditions (3.5.2), (3.5.3) can also be represented in the formu(x; #1(x)) = '1(x) ; 0 � x � �1 ; (3.5.4)u(x; #2(x)) = '2(x) ; 0 � x � �2 ; (3.5.5)where �1 and �2 are the abscissae of the points A and B. This problem is called theGoursat problem.If 1(0) = 2(0) = a 6= 0 it would be su�cient to solve the Goursat problem with theconditions u = 1 � c on C1 and u = 2 � c on C2 and then add the function u0 � c to thederived solution.The problem formulated above is the Goursat problem in a narrow sense. In a wider sense,in the Goursat problem the function �1(y) appearing in the equation of the curve C1 is

52 CHAPTER 3. HYPERBOLIC EQUATIONSmonotonically non{decreasing and the curve C2 is represented by an equation of the formy = #(x), where the function #(x) is monotonically non{decreasing.If C1 is the interval [0; �] of the axis Oy and C2 is the interval [0; �] of the axis Oy, then wehave the Darboux problem:x

y

RN(0, )

M( , 0)0Find the solution of (3.5.1) in OMRN, where M = (�; 0) ; R = (�; �) ; N = (0; �) andu(x; 0) = '(x) ; 0 � x � � ;u(0; y) = (y) ; 0 � y � � :x

y

0

C

B

A( , )

Let the arc OA of a curve C with the equation y = #(x), beginning at the origin of thecoordinate system, belong to the class C1 and #0(x) > 0. Let D be the domain containedbetween the arc OA, the axis Ox and the characteristic AB perpendicular to Ox. Let �and � be the coordinates of the point A. If we seek the solution of (3.5.1) in the domain Dwhich satis�es the boundary conditionsu(x; 0) = '(x) ; 0 � x � � ;u(�(y); y) = (x) ; 0 � y � � ;then we have the Picard problem.Remark: The Darboux problem and the Picard problem are special cases of the Goursatproblem.Remark: The Goursat problem has many applications in gas dynamics, etc.We restrict ourselves to the Darboux problem in this introductory lecture.

3.5. THE GOURSAT PROBLEM 533.5.2 The Darboux problem. The method of successive approx-imationsWe consider the Darboux problemuxy = f(x; y) ; x � 0 ; y � 0 ; (3:5:1)u(x; 0) = '(x) ; x � 0 ; (3.5.6)u(0; y) = (x) ; y � 0 : (3.5.7)y

xu(x, 0)= (x)

u = f(x, y)u(0, y)

= (y) xyLet ' and be di�erentiable and '(0) = (0). Integrating (3.5.1) �rst with respect to x,and then w.r.t. y, we getuy(x; y) = uy(0; y) + Z x0 f(�; y) d� ;u(x; y) = u(x; 0) + u(0; y)� u(0; 0) + Z y0 d� Z x0 f(�; �) d� :Thus, u(x; y) = '(x) + (y)� '(0) + Z y0 Z x0 f(�; �) d� d� : (3:5:8)It is clear that u(x; y) de�ned by (3.5.8) is a solution of the Darboux problem (3.5.1), (3.5.6),(3.5.7). Furthermore it is unique.We consider now a more general hyperbolic equationuxy = a(x; y)ux + b(x; y)uy + c(x; y)u+ f(x; y) (3:5:9)with the boundary conditions (3.5.6), (3.5.7). We assume also that '(0) = (0), and thecoe�cients a; b, and c are continuous functions of x and y.From (3.5.9) we see that a solution u(x; y) of (3.5.9) satis�es the integral equationu(x; y) = Z y0 Z x0 ha(�; �)u� + b(�; �)u� + c(�; �)u] d� d�+ '(x) + (y)� '(0) + Z y0 Z x0 f(�; �) d� d� : (3:5:10)The solution of (3.5.10) will be constructed via the method of successive approximations.As the zero-th approximation we take the functionu0(x; y) = 0 :

54 CHAPTER 3. HYPERBOLIC EQUATIONSThe equation (3.5.10) gives then the following expressions for the successive approximationsu1(x; y) = '(x) + (y)� '(0) + Z y0 Z x0 f(�; �) d� d� ;: : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :un(x; y) = u1(x; y) + Z y0 Z x0 ha(�; �)@un�1@� + b(x; �)@un�1@� ++c(�; �)un�1i d� d� : (3:5:11)Note also that@un@x = @u1@x + Z y0 "a(x; �)@un�1@x + b(x; �)@un�1@� + c(x; �)un�1# d� ;@un@y = @u1@y + Z x0 "a(�; y)@un�1@� + b(�; y)@un�1@y + c(�; �)un�1# d� : (3:5:12)In order to prove the uniform convergence of the seriesfun(x; y)g ; (@un@x (x; y)) ; (@un@y (x; y))we consider the di�erenceszn(x; y) = un+1(x; y)� un(x; y)= Z y0 Z x0 "a(�; �)@zn�1@� + b(�; �)@zn�1@� + c(�; �)zn�1(�; �)# d� d� ;@zn(x; y)@x = @un+1(x; y)@x � @un(x; y)@x= Z y0 "a(x; �)@zn�1@x + b(x; �)@zn�1@� + c(x; �)zn�1(x; �)# d� ;zn(x; y)@y = @un+1(x; y)@y � @un(x; y)@x= Z x0 "a(�; y)@zn�1@� + b(�; �)@zn�1@y + c(�; �)zn�1(�; y)# d� :Assume that ja(x; y)j; jb(x; y)j; jc(x; y)j �M andjz0j < H ; �����@z0@x ����� < H ; �����@z0@y ����� < H ;where x 2 [0; L] ; y 2 [0; L], L is a given positive number. We want to estimate zn ; @zn@x ; @zn@yabove. First we note that jz1j < 3HMxy < 3HM (x+ y)22! ;�����@z1@x ����� < 3HMy < 3HM(x + y) ;�����@z1@y ����� < 3HMx < 3HM(x+ y) :

3.5. THE GOURSAT PROBLEM 55We shall prove by induction thatjznj < 3HMnKn�1 (x+ y)n+1(n+ 1)! ;�����@zn@x ����� < 3HMnKn�1 (x+ y)nn! ; (3.5.13)�����@zn@y ����� < 3HMnKn�1 (x+ y)nn! ;where K = 2L + 2.For n = 1, our estimates are proved. Suppose that they are proved for n, we want to showthat they are also valid for n + 1.In fact, from our de�nition for zn we havejzn+1j = Z y0 Z x0 "a(�; �)@zn@� + b(�; �)@zn@� + c(�; �)zn(�; �)# d� d�� 3HMn+1Kn�1 Z y0 Z x0 "2(� + �)nn! + (� + �)n+1(n+ 1)! # d� d� :Noting that Z y0 Z x0 (� + �)kk! d� d� = Z y0 (x+ �)k+1(k + 1)! � �k+1(k + 1)!! d�= (x+ y)k+2(k + 2)! � xk+2(k + 2)! � yk+2(k + 2)! � (x+ y)k+2(k + 2)! ;we have jzn+1j � 3HMn+1Kn�1 2(x+ y)n+2(n+ 2)! + (x+ y)n+3(n + 3)! != 3HMn+1Kn�1 (x+ y)n+2(n+ 2)! �2 + x+ yn+ 3�< 3HMn+1Kn (x+ y)n+2(n+ 2)! :Analogously, �����@zn+1@x ����� � 3HMn+1Kn�1 (x+ y)n+1(n+ 1)! �x+ yn+ 2 + 2� ;< 3HMn+1Kn (x+ y)n+1(n+ 1)! ;and �����@zn+1@y ����� � 3HMn+1Kn�1 (x+ y)n+1(n+ 1)! �x+ yn+ 2 + 2� ;< 3HMn+1Kn (x+ y)n+1(n+ 1)!

56 CHAPTER 3. HYPERBOLIC EQUATIONSThus the estimates in (3.5.13) are proved.From (3.5.13) we see thatjznj < 3HMnKn�1 (x+ y)n+1(n+ 1)! < 3HK2M (2KLM)n+1(n+ 1)! ;�����@zn@x ����� < 3HMnKn�1 (x+ y)nn! < 3HK (2KLM)nn! ;�����@zn@y ����� < 3HMnKn�1 (x+ y)nn! < 3HK (2KLM)nn! :It follows 1Xn=0 jznj < 1Xn=0 3HK2M (2KLM)n+1(n+ 1)! = 3HK2M �e2KLM � 1� ;1Xn=0 �����@zn@x ����� < 1Xn=0 3HK (2KLM)nn! = 3HK e2KLM ;1Xn=0 �����@zn@y ����� < 1Xn=0 3HK (2KLM)nn! = 3HK e2KLM :We thus conclude that the seriesun = u0 + z1 + � � � + zn ;@un@x = @u0@x + @z1@x + � � �+ @zn@x ;@un@y = @u0@y + @z1@y + � � �+ @zn@yconverges uniformly as n!1. We denote their limit functions byu(x; y) := limn!1 un(x; y) ;v(x; y) := limn!1 @un@x (x; y) ;w(x; y) := limn!1 @un@y (x; y) :Taking the limit in the integral equations (3.5.11), (3.5.12), we getu(x; y) = u1(x; y) + Z y0 Z x0 ha(�; �)v + b(�; �)w + c(�; �)ui d�d� ;v(x; y) = @u1@x (x; y) + Z y0 ha(x; �)v + b(x; �)w+ c(x; �)uid� ;w(x; y) = @u1@y (x; y) + Z y0 ha(�; y)v + b(�; y)w + c(�; y)uid� :From the �rst equation it follows that v = ux ;w = uy ;

3.5. THE GOURSAT PROBLEM 57and thereforeu(x; y) = '(x) + (y)� '(0) + Z y0 Z x0 f(�; �) d� d�+ Z y0 Z x0 ha(�; �)u� + b(�; �)u� + c(�; �)ui d� d� ; (3:5:10)which satis�es (3.5.9) and clearly the boundary conditions. Thus, there exists a solution ofthe Darboux problem (3.5.1), (3.5.6), (3.5.7).We shall prove that the solution is unique. In fact, suppose that there are two solutionsu1(x; y) and u2(x; y). Let U(x; y) = u1(x; y)� u2(x; y) :The function U satis�es the homogeneous integral equationU(x; y) = Z y0 Z x0 (aUx + bUy + cU) d� d� :Let H1 be the majorant of jU j; jUxj and jUyj for x 2 [0; L] ; y 2 [0; L]. Following thesuccessive procedure we getjU j < 3H1Mn+1Kn (x+ y)n+2(n+ 2)! < 3H1K2M (2KLM)n+2(n + 2)!for any n. It follows U(x; y) � 0 ; or u1(x; y) � u2(x; y) :Thus, the solution of the Darboux problem is unique.If a; b, and c are constant, then we make the transformationu = ve�x+�y ;for which ux = vxe�x+�y + �ve�x+�y ;uy = vye�x+�y + �ve�x+�y ;uxy = (vxy + �vx + �vy + ��v) e�x+�y :Thus vxy + �vx + �vy + ��v == a(vx + �v) + b(vy + �v) + cv + fe��x��y :It follows vxy = (a� �)vx + (b� �)vy++ (a� + b�+ c� ��)v + fe��x��y :Letting � = a ; � = b, we get vxy = (ab+ c)v + fe�(bx+ay) : (3:5:14)If ab+ c = 0, then we can �nd the solution explicitly in the form (3.5.8), if ab+ c 6= 0 thenwe shall use the method of the next paragraph.

58 CHAPTER 3. HYPERBOLIC EQUATIONS3.6 Solution of general linear hyperbolic equations3.6.1 The Green formulaLet L[u] = uxx � uyy + a(x; y)ux + b(x; y)uy + c(x; y)u ; (3:6:1)where a(x; y) ; b(x; y) ; c(x; y) are di�erentiable functions. LetL�[v] = vxx � vyy � (av)x � (bv)y + cv ; (3:6:2)H = 8<: vux � vxu+ avu = (vu)x � (2vx � av)u =�(vu)x + (2ux + au)v ; (3:6:3)K = 8<: (uv)y � (2uy � bu)v =�vuy + vu+ bvu = �(vu)y + (2vy + bv)u ; (3:6:4)L[u] = uxx � uyy + a(x; y)ux + b(x; y)uy + c(x; y)u:Since L�[v] = vxx � vyy � (av)x � (bv)y + cv;we have vL[u]� uL�[v] = @H@x + @K@y :Here we take H = vux � vyu+ avu;and K = �vuy + vyu+ bvu:The Green formula says thatZZG �vL[u]� uL�[v]�d� d� == ZZG �@H@x + @K@y � d� d�= � ZS �H cos(n; x) +K cos(n; y)� ds= ZS �Hd� �Kd�� : (3.6.5)Here, n is the inward normal vector to S, that isdx = cos(n; y) ds ; dy = � cos(n; x) ds ;assuming that S is traced out anticlockwise so as to keep the interested domain area alwayson the left, and ds is taken to be positive.

3.6. SOLUTION OF GENERAL LINEAR HYPERBOLIC EQUATIONS 593.6.2 Riemann's methodWe consider the problem of �nding the solution of the linear hyperbolic equationL[u] = uxx � uyy + a(x; y)ux + b(x; y)uy + c(x; y)u = �f(x; y) ; (3:6:6)which satis�es the Cauchy data on a curve C,ujC = '(x) ;unjC = (x) : (3:6:7)Here, un is the normal derivative of u with respect to the curve C. The conditions posed onC are as follows: C is described by y = f(x) ;where f(x) is a di�erentiable function. Further, every characteristic y� x = const ; y+ x =const intersects the curve C at most one time (for this, it is necessary that jf 0(x)j � 1).y

x

M

P Q

y=x

Let P and Q be two points on C. Through P and Q we draw two characteristics such thatthey intersect each other at a point M . Consider the domain bounded by the arc PQ of thecurve C and by QM and MP . From the Green formula we haveZ ZMPQ �vL[u] � uL�[v]�d� d� == Z MQ (H d� �K d�) + Z PM (H d� �K d�) + Z QP (H d� �K d�) : (3:6:8)On QM and MP we have d� = �d� = � dsp2 on QM ;d� = d� = � dsp2 onMP ;where s is the curve element along �!QM and �!MP . ThusZ MQ (H d� �K d�) = Z MQ ��(vu)� � (2v� � av)u�d�� �� (vu)� + (2v� + bv)u�d��

60 CHAPTER 3. HYPERBOLIC EQUATIONS= Z MQ h(vu)� d� + (vu)� d�i � Z MQ 2 (v�u d� + v�u d�)+ Z MQ (avu d� � bvu d�)= � Z MQ d(uv) + Z MQ 2@v@s � a+ bp2 v!u ds= � (uv)M + (uv)Q + Z MQ 2@v@s � a+ bp2 v!u ds :Analogously,Z PM (H d� �K d�) = �(uv)M + (uv)P + Z MQ 2@v@s � b� ap2 v!u ds :The last two equalities and (3.6.8) yield(uv)M = (uv)P + (uv)Q2 + Z MP @v@s � b� a2p2 v!u ds+ Z MQ @v@s � b+ a2p2 v!u ds (3.6.9)+ 12 Z QP (H d� �K d�) � 12 Z ZMPQ �vL(u)� uL�(v)�d� d� :Let u be a solution of the problem (3.6.6), (3.6.7), whence v be a solution of the followingproblem depending on M as a parameter,L�(v) = v�� � v�� � (av)� � (bv)� + cv = 0 in �MPQ (3:6:10)@v@s = b� a2p2 v on the characteristic MP ;@v@s = b+ a2p2 v on the characteristic MQ ;v(M) = 1 : 9>>>>>=>>>>>; (3:6:11)The conditions of (3.6.11) yieldv = exp Z ss0 b� a2p2 ds! on MP ;v = exp Z ss0 b+ a2p2 ds! on MQ ;where s0 is the value of s at the point M .Thus, we obtain the Darboux problem for v, which is uniquely determined in �MPQ. Thefunction v is called the Riemann function.Hence, from (3.6.6) and (3.6.9) we haveu(M) = (uv)P + (uv)Q2+ 12 Z PQ [v(u� d� + u� d�) � u(v� d� + v� d�) + uv(ad� � bd�)]+ 12Z ZMPQ vf d� d� : (3:6:12)

3.6. SOLUTION OF GENERAL LINEAR HYPERBOLIC EQUATIONS 61By this formula, our problem (3.6.6), (3.6.7) is solved, since v is known in �MPQ, and, onC, u and dudn are given:ujC = '(x) ;ux = ux cos(x; s) + un cos(x; n) = '0(x)� (x)f 0(x)q1 + f 02(x)1 + f 02(x) ;uy = us cos(y; s) + un cos(y; n) = '0(x)f 0(x) + (x)q1 + f 02(x)1 + f 02(x) :The formula (3.6.12) ensures the existence and uniqueness of the solution of the problem(3.6.6), (3.6.7). One can show also that the function u de�ned by (3.6.12) satis�es in factthe conditions of the problem. We do not pursuit it in this lecture.3.6.3 An application of the Riemann methodConsider the problemuyy = uxx + f1(x; t) ; �1 < x <1 ; y > 0 y = at ; f1 = fa2!u(x; 0) = '(x) ; �1 < x <1 ;uy(x; 0) = 1(x) ; 1 = a! �1 < x <1 :The operator L(u) = uxx � uyy is self-adjoint, that is L(u) = L�(u).PQ is now an interval of the axis y = 0.y

xP(x−y, 0) Q(x+y, 0)

M(x, y)From (3.6.11) we see that v � 1 in �MPQ. Since d� = 0 on PQ, we getu(M) = u(P ) + u(Q)2 + 12 Z QP u� d� + 12 Z ZMPQ f(�; �) d� d� :Let the coordinate of M be (x; y), we have that the coordinate of P is (x� y; 0) and of Q is(x+ y; 0). It follows thatu(x; y) = u(x� y) + u(x+ y)2 + 12 x+yZx�y 1(�) d� + 12 yZ0 x+(y��)Zx�(y��) f1(�; �) d� d� :

62 CHAPTER 3. HYPERBOLIC EQUATIONSChanging the variable, we getu(x; t) = u(x� at) + u(x+ at)2 + 12a x+atZx�at (�) d� + 12a tZ0 x+a(t��)Zx�a(t��) f(�; �) d� d� :

Chapter 4Parabolic EquationsIn this chapter we shall consider the parabolic equationc�@u@t = @@x k@u@x!8>>>><>>>>: k : (W�armeleitf�ahigkeit)c : (spezi�sche W�armekapazit�at)� : (die Dichte)in a simple form. Namely, we shall consider the heat equationut = a2uxx; a2 = kc�:4.1 Boundary conditionsIn contrast to hyperbolic equations, for parabolic equations we need only one initial conditionu(x; t) for the initial time t = t0.Suppose that we consider our process for a bar [0; l].a) At one end of the bar (x = 0, or x = l) the temperature is prescribed byu(0; t) = �(t) (or u(l; t) = �l(t));where �(t) (�l(t)) is a given function de�ned on [t0; T ]. Here T characterizes the time interval,where our heat transfer process is considered.b) At one end of the interval [0; l], the derivative of u is given:@u@x(l; t) = �(t):Normally, the heat ux is de�ned byQ(l; t) = k@u@x����x=l:63

64 CHAPTER 4. PARABOLIC EQUATIONSThus @u@x(l; t) = Q(l; t)k���x=l = �(t):Note that the heat ux at x = 0 is Q(0; t) = �k@u@x����x=0:c) At one end, one has the following relation between the solution and its derivative@u@x(l; t) = ��[u(l; t)� �(t)]:This corresponds to the Newton law of heat transfer of the body boundary with its outsideenvironment, the temperature �(t) of which is given. At x = 0 we have another condition,@u@x(0; t) = �[u(0; t)� �(t)]:d) If the interval [0; l] is very long and the time interval [t0; T ] is small, then we ignore theboundary conditions by consider the initial value problem for parabolic equations in thedomain �1 < x <1; t � t0 with the initial conditionu(x; t0) = '(x) (�1 < x <1);where '(x) is a given function.e) Similarly, if the temperature at one end of the interval is given and the other end is veryfar, then we can consider our parabolic equation in the domain 0 � x < 1; t � t0 whichsatis�es the conditions u(x; t0) = '(x) (0 < x <1);u(x; 0) = �(t) (t � t0):Here ' and � are given functions.f) We have also nonlinear boundary conditions. For examplek@u@x(0; t) = �[u4(0; t)� �4(0; t)]:This condition is called the Stefan-Boltzmann law.De�nition 4.1.1 A function u is called a solution of the �rst boundary value problem fora parabolic equation, ifi) it is de�ned and continuous in the closed domain 0 � x � l; t0 � t � T ,ii) it satis�es the parabolic equation in the domain 0 < x < l; t0 < t < T ,iii) it satis�es the initial and boundary conditionsu(x; t0) = '(x); u(0; t) = �1(t); u(l; t) = �2(t);where '(x), �1(t) and �2(t) are continuous functions, and'(0) = �1(t0) [= u(0; t0)]'(l) = �2(t0) [= u(l; t0)]:

4.2. THE MAXIMUM PRINCIPLE 65De�nition 4.1.2 If the condition iii) in De�nition 4.1.1 is replaced byu(x; t0) = '(x); @u@x(0; t) = �1(t); @u@x(l; t) = �2(t);then we have "the second boundary value problem".De�nition 4.1.3 If the condition iii) in De�nition 4.1.1 is replaced byu(x; t0) = '(x); @u@x(0; t) = �[u(0; t)� �1(t)]; @u@x(l; t) = ��[u(l; t)� �2(t)];then we have "the third boundary value problem".We shall consider in the sequel the following questions:i) Is the solution of our problems unique?ii) Is there a solution?iii) Do the solutions depend continuously on the data?4.2 The maximum principleIn the sequel, we shall consider the equation with constant coe�cientsvt = a2vxx + �vx + v: (4:2:1)If we make the transformationv = e�x+ �tu; � = � �2a2 ; � = � �24a2 ;then we have the equation ut = a2uxx: (4:2:2)Thus, we have to consider the equation (4.2.2).The maximum principle: Let a function u(x; t) be de�ned and continuous in the closeddomain 0 � t � T , 0 � x � l and be satis�ed the equation (4.2.2) in the open domain0 < t < T , 0 < x < l. Then the function u(x; t) admits its maximum (minimum) at t = 0,or at x = 0 or x = l.Proof: Let M be the maximum of u(x; t) for t = 0 (0 � x � l), or for x = 0; x = l (0 �t � T ). Assume further that u(x; t) attains its maximum at some interior point (x0; t0), say0 < x0 < l; 0 < t0 < T : u(x0; t0) =M + �; � > 0: (4:2:3)Since 0 < x0 < l and u(x; t) attains its maximum at (x0; t0), we have@u@x(x0; t0) = 0; @2u@x2 (x0; t0) � 0: (4:2:4)

66 CHAPTER 4. PARABOLIC EQUATIONSFurther, since the function u(x0; t) attains its maximum at t0 2 (0; T ] we have@u@t (x0; t0) � 0: (4:2:5)(In fact, @u@t (x0; t0) = 0 if t0 < T , @u@t (x0; t0) � 0 if t0 = T .)We shall �nd another point (x1; t1) 2 (0; l)� (0; T ] such that @2u@x2 (x1; t1) � 0; @u@t (x1; t1) > 0.Consider the function v(x; t) = u(x; t) + k(t0 � t); (4:2:6)where k is a constant. We havev(x0; t0) = u(x0; t0) =M + �and k(t0 � t) � kT:Choose k > 0 such that kT < �2 , that is k < �2T ; we see that the maximum of v(x; t) att = 0 (0 � x � l), or x = 0 (0 � t < T ), or x = l (0 � t < T ) is not greater than M + �2. Itmeans that v(x; t) �M + �2 (for t = 0; or x = 0; x = l); (4:2:7)since the �rst member on the right-hand side of (4.2.6) is not greater thanM and the secondone is not greater than �2.Since the function v(x; t) is continuous in [0; l]� [0; T ], there is a point (x1; t1) 2 [0; l]� [0; T ]where v(x; t) attains its maximum:v(x1; t1) � v(x0; t0) =M + �:From (4.2.7) we see that 0 < x1 < l and 0 < t1 � T . It follows thatvxx(x1; t1) = uxx(x0; t0) � 0and vt(x1; t1) = ut(x1; t1)� k � 0:The last inequality says that ut(x1; t1) � k > 0. Thus, at (x1; t1) the function u(x; t) doesnot satisfy the equation (4.2.2). This contradicts the assumption of the maximum principle.Thus, the �rst part of the maximum principle is proved. To prove the second part we notethat the function u1 = �u satis�es (4.2.2) and all other conditions of the maximumprinciplein the case of "maximum". 24.3 Applications of the maximum principle4.3.1 The uniqueness theoremThe uniqueness theorem: Let u1(x; t) and u2(x; t) be continuous functions de�ned on0 � x � l; 0 � t � T , satisfying the equationut = a2uxx + f(x; t); 0 < x < l; t > 0 (4:3:1)

4.3. APPLICATIONS OF THE MAXIMUM PRINCIPLE 67and the same initial and boundary value conditionsu1(x; 0) = u2(x; 0) = '(x); 0 � x � l;u1(0; t) = u2(0; t) = �1(t); 0 � t � T;u1(l; t) = u2(l; t) = �2(t); 0 � t � T:Then u1(x; t) � u2(x; t).Proof: Let v(x; t) := u1(x; t)� u2(x; t):Since u1 and u2 are continuous in [0; l] � [0; T ], the function v is also continuous there.Further, the function v satis�es the equation vt = a2vxx for 0 < x < l; t > 0. Thus, vsatis�es the conditions of the maximum principle; that is v attains its maximum for t = 0or x = 0, or x = l. Since v(x; 0) = 0; v(0; t) = 0 and v(l; t) = 0 we have v(x; t) � 0. Thus,u1(x; t) � u2(x; t). 24.3.2 Comparison of solutionsa) Let u1(x; t) and u2(x; t) be solutions of the heat equation withu1(x; 0) � u2(x; 0)u1(0; t) � u2(0; t); u1(l; t) � u2(l; t):Then u1(x; t) � u2(x; t)for (x; t) 2 [0; l]� [0; T ].In fact, the function v = u2 � u1 satis�es the heat equation, and v(x; 0) � 0, v(0; t) �0; v(l; t) � 0. From the maximum principle we havev(x; t) � 0 for 0 < x < l; 0 < t � T:b) If the solutions u(x; t), u(x; t) and u(x; t) of the heat equation satisfy the inequalitiesu(x; t) � u(x; t) � u(x; t) for t = 0; x = 0 and x = l;then these inequalities remain valid for all (x; t) 2 [0; l] � [0; T ]. This statement followsimmediatly from a).c) Let u1(x; t) and u2(x; t) be two solutions of the heat equation, which satisfy the inequalityju1(x; t)� u2(x; t)j � � for t = 0; x = 0; x = l:Then ju1(x; t)� u2(x; t)j � � 8 (x; t) 2 [0; l]� [0; T ]:

68 CHAPTER 4. PARABOLIC EQUATIONSThe proof of this statement follows immediately from b) withu(x; t) = ��;u(x; t) = u1(x; t)� u2(x; t);u(x; t) = �:Now we turn to the question about the stability of the solution of the heat equation: Letu(x; t) be the solution of the �rst boundary value problem for the heat equation with theinitial and boundary value conditionsu(x; 0) = '(x); u(0; t) = �1(t); u(l; t) = �2(t):Let these functions be given approximately by '�(x); ��1(t) and ��2(t), respectively:j'(x)� '�(x)j � �; j�1(t)� ��1(t)j � �; ju2(t)� ��2(t)j � �;such that there is a solution u�(x; t) of the heat equation with these data. Thenju�(x; t)� u(x; t)j � �:4.3.3 The uniqueness theorem in unbounded domainsConsider the problem ut = a2uxx; �1 < x <1; t > 0: (4:3:2)Uniqueness Theorem: Let u1(x; t) and u2(x; t) be two solutions of (4.3.2) which are con-tinuous and bounded by M :ju1(x; t)j < M; ju2(x; t)j < M for �1 < x < M; t � 0:If u1(x; 0) = u2(x; 0); �1 < x <1, thenu1(x; t) � u2(x; t); �1 < x <1; t � 0:Proof: Let v(x; t) = u1(x; t)�u2(x; t). The function v(x; t) is continuous, satis�es the heatequation and is bounded by 2M :jv(x; t)j � ju1(x; t)j+ ju2(x; t)j < 2M; �1 < x <1; t � 0:Furthermore, v(x; 0) = 0.We shall apply the maximum principle to prove that v(x; t) = 0; �1 < x <1; t � 0. LetL be a positive number. Consider the functionV (x; t) = 4ML2 x22 + a2t! : (4:3:3)The function V (x; t) is continuous and satis�es the heat equation (4.3.2). Furthermore,V (x; 0) � jv(x; 0)j = 0; V (�L; t) � 2M � v(�L; t):Applying the maximum principle in the domain jxj � L, we get�4ML2 x22 + a2t! � v(x; t) � 4ML2 x22 + a2t! :Let (x; t) be �xed, and let L tend to in�nity, we get v(x; t) = 0. Thus, v(x; t) � 0. 2

4.4. THE FOURIER METHOD 694.4 The Fourier methodIn this section we shall apply the Fourier method to the following problemut = a2uxx + f(x; t); 0 < x < l; 0 < t � T; (4:4:1)u(x; 0) = '(x); 0 � x � l; (4:4:2)u(0; t) = �1(t); u(l; t) = �2(t); 0 � t � T: (4:4:3)4.4.1 The homogeneous problemFind the solution of the problemut = a2uxx; 0 < x < l; 0 < t � T; (4:4:4)u(x; 0) = '(x); 0 � x � l; (4:4:5)u(0; t) = u(l; t) = 0; 0 � t � T: (4:4:6)To do this we shall �nd a solution of the equationut = a2uxx;which is not identically vanishing and satis�es the boundary conditionsu(0; t) = 0; u(l; t) = 0: (4:4:7)Furthermore, this solution can be represented in the formu(x; t) = X(x) T (t); (4:4:8)where X(x) is a function depending only on x and T (t) is a function depending only on t.Putting (4.4.8) into (4.4.4) and then deviding both sides by a2XT , we get1a2 T 0T = X 00X = ��; (4:4:9)where � = const, since T 0=T depends only on t and X 00=X depends only on x. This leads usto consider the equations X 00 + �X = 0; (4:4:10)T 0 + a2�T = 0: (4:4:11)Because of (4.4.7) we have X(0) = X(l) = 0: (4:4:12)Thus, we obtain the following eigenvalue problem for X(x):X 00 + �X = 0; X(0) = 0; X(l) = 0: (4:4:13)We have proved that the eigenvalues of this problem are�n = ��nl �2 ; n = 1; 2; 3; : : : (4:4:14)

70 CHAPTER 4. PARABOLIC EQUATIONSand the eigenfunctions associated with them areXn(x) = sin �nl x: (4:4:15)With these values �n the solutions of (4.4.11) have the formTn(t) = cn e�a2�nt; (4:4:16)where cn will be de�ned later.We see that the functionun(x; t) = Xn(x)Tn(t) = cn e�a2�nt Xn(x) (4:4:17)is a special solution of (4.4.4), which satis�es the homogeneous boundary conditions (4.4.6).Consider the formal seriesu(x; t) = 1Xn=1 cne���nl �2 a2t sin��nl x� : (4:4:18)The function u(x; t) satis�es (4.4.6), since every element of the series does so. If u(x; t)satis�es also the initial condition, then'(x) = u(x; 0) = 1Xn=1 cn sin��nl x� : (4:4:19)Thus, cn are the Fourier coe�cients of '(x), when it is expanded on (0; l) bycn = 'n = 2l lZ0 '(�) sin��nl �� d�: (4:4:20)The series (4.4.18) is now completely de�ned. We shall �nd conditions which guarantee theconvergence of the series and, furthermore, which allow to di�erentiate (4.4.18) termwisetwo times w.r.t. x, and once w.r.t. t. We shall prove that the series1Xn=1 @un@t and 1Xn=1 @2un@x2for t � t > 0 (t is �xed) converge uniformly. For this, we note that�����@un@t ����� = ���������cn ��l �2 a2n2e���nl �2 a2t sin��nl x���������< jcnj��l �2 a2n2e���nl �2 a2t:Let ' be bounded: j'(x)j < M . Thenjcnj = ����2l ���� ������ lZ0 '(�) sin��nl �� d������� < 2M:

4.4. THE FOURIER METHOD 71Hence, �����@un@t ����� < 2M ��l �2 a2n2e���nl �2 a2t; t � t;and �����@2un@x2 ����� < 2M ��l �2 n2e���nl �2 a2t; t � t:Generally, ����� @k+lu@tk@xl ����� < 2M ��l �2k+l n2l+la2ke���nl �2 a2t; t � t:Consider the majorant series 1Pn=1�n, where�n = A nqe���nl �2 a2t: (4:4:21)Series of this kind converge because of the D'Alembert criterionlimn!1 �����n+1�n ���� = limn!1�n+ 1n �q e���l �2 a2(n2 + 2n+ 1)te���l �2 a2n2t= limn!1�1 + 1n�q e���l �2 a2(2n+ 1)t = 0:Thus, the series (4.4.18) for t � t > 0 is in�nitely di�erentiable. Since t is arbitrary we cansay that u(x; t) for t > 0 is in�nitely di�erentiable, furthermore, it is clear that for t > 0 itsatis�es the equation (4.4.4).Theorem 4.4.1 If ' is piecewise di�erentiable and '(0) = '(l) = 0, then the seriesu(x; t) = 1Xn=1 cne���nl �2 a2t sin��nl x� (4:4:22)is a continuous function for t � 0.In fact, the inequalities jun(x; t)j < jcnj (t � 0; 0 � x � l)yield the uniform convergence of the series (4.4.18) for t � 0; 0 � x � l: For t > 0, wehave proved this already; for t = 0, this follows directly, because '(0) = '(l) = 0 and ' ispiecewise di�erentiable.

72 CHAPTER 4. PARABOLIC EQUATIONS4.4.2 The Green functionFrom (4.4.22) we haveu(x; t) = 1Xn=1 cne���nl �2 a2t sin��nl x�= 1Xn=1242l lZ0 '(�) sin��nl �� d�35 e���nl �2 a2t sin��nl x�= lZ0 26642l 1Xn=1 e���nl �2 a2t sin��nl x� sin��nl ��3775 '(�) d�:We can change the order of summation and integration for t > 0, since the series1Xn=1 e���nl �2 a2t sin��nl x� sin��nl �� for t > 0converges uniformly w.r.t. �.Set G(x; �; t) = 2l 1Xn=1 e���nl �2 a2t sin��nl x� sin��nl �� : (4:4:23)The function u(x; t) can then be written in the formu(x; t) = lZ0 G(x; �; t)'(�) d�: (4:4:24)The function G(x; �; t) is called the Green function. It has a physical meaning.We shall prove that G(x; �; t), as a function of x, represents the temperature in the bar0 � x � l at time t, if the temperature at the initial moment t = 0 equals to zero and at thesame moment, at the point x = �, produces a heat quantity Q (which will be determinedlater), while at the ends of the bar (x = 0; x = l) the temperature remains zero.The notion "a heat quantity Q is produced at the point �" means that this is produced ina su�ciently small neighbourhood of �. The temperature change '�(�), which produces aheat quantity at the point �, is equal to zero outside the interval (� � �; � + �) and, insidethis interval, '�(�) is a positive, continuous and di�erentiable function such thatc � �+�Z��� '�(� ) d� = Q: (4:4:25)Note that the left-hand side of (4.4.25) represents the heat quantity produced by '�(�).The temperature in this case isu�(x; t) = lZ0 G(x; �; t)'�(� ) d�: (4:4:26)

4.4. THE FOURIER METHOD 73Since G(x; �; t) for t > 0 is a continuous function of � we haveu�(x; t) = �+�R��� G(x; �; t)'�(�) d� = G(x; ��; t) �+�R��� '�(� ) d�= G(x; ��; t) Qc�; (4:4:27)where �� is a point lying between � � � and � + �. Letting � ! 0, and taking into accountthat G(x; �; t) is a continuous function of � for t > 0, we getlim�!0 u�(x; t) = Qc�G(x; �; t)= Qc� � 2l 1Xn=1 e���nl �2 t sin �nl x � sin �nl � (4:4:28)(Here, we suppose that there exists the limit of the integral �+�R��� '�(� ) d� as � ! 0, andlim �+�R��� '�(� ) d� = Qc� .) Thus, G(x; �; t) represents the temperature in uence at the instantof time into the heat pole of intensity Q = c�, which is lying at t = 0 and � 2 (0; l).We shall prove that for any x; � and t > 0, G(x; �; t) � 0. Consider the function '�(�)de�ned above, since '�(� ) > 0 for � 2 (� � �; � + �), u�(0; t) = u�(l; t) = 0, from themaximum principle u�(x; t) � 0; 0 � x � l; t > 0:Thus, u�(x; t) = G(x; ��; t)Qc� � 0 (t > 0):Letting �! 0, we obtain thatG(x; �; t) � 0; 0 � x; � � l; t > 0:4.4.3 Boundary value problems with non-smooth initial condi-tionsWe have considered our �rst boundary value problem in the class of all continuous functionsin the closed domain [0; l]�[0; T ]. This has some restriction, for example, if u(x; 0) = u0 6= 0,then the solution must be discontinuous at the points (0; 0) and (l; 0). It leads to the ideato consider our problem for non-smooth initial conditions.Theorem 4.4.2 Let ' be a continuous function de�ned on [0; l] and '(0) = '(l) = 0. Thenthe solution of the equationut = a2uxx (0 < x < l; t > 0), (4.4.4)which is continuous in [0; l]� [0; T ] and satis�es the conditions

74 CHAPTER 4. PARABOLIC EQUATIONSu(0; t) = u(l; t) = 0; t 2 [0; T ], (4.4.6)u(x; 0) = '(x); x 2 [0; l], (4.4.5)is unique and can be represented byu(x; t) = lZ0 G(x; �; t)'(�) d� . (4.4.24)Proof: The case where '(�) is continuous and piecewise di�erentiable is already proved.Consider a series of continuous, piecewise di�erentiable functions 'n(x) ('n(0) = 'n(l) = 0),which uniformly converges to '(x) (as 'n(x) we can take the piecewise linear function, whichcoincides with '(x) at the points lkn (k = 0; 1; : : : ; n)). For the 'n(x) we can determineun(x; t) by (4.4.24), since 'n(x) is continuous and piecewise di�erentiable. The functionsun(x; t) uniformly converge to u(x; t). In fact, for � > 0 small enough, there exists an n(�)such that j'n1(x)� 'n2(x)j < � (0 � x � l)8n1; n2 � n(�), since the 'n uniformly converge to '(x).It follows then by the maximum principle thatjun1(x; t)� un2(x; t)j < � (0 � x � l; 0 � t � T )if n1; n2 � n(�). Thus, fun(x; t)g uniformly converge to u(x; t). Since un(x; t) is continuous,u(x; t) is so, too.For any �xed (x; t) 2 [0; l]� [0; T ] we haveu(x; t) = limn!1 un(x; t) = limn!1 lZ0 G(x; �; t)'n(�) d� = lZ0 G(x; �; t)'(�) d�:Thus, the function u(x; t) is continuous and satis�es the condition (4.4.5). Further, as wehave proved in x 4.4.1 this function also satis�es (4.4.6).The uniqueness follows immediately from the maximum principle. 2Theorem 4.4.3 Let ' be a piecewise continuous function. Then the function u(x; t) de�nedby (4.4.24) represents a solution of the equation (4.4.4), satis�es the condition (4.4.5), andis continuous where ' is continuous.Proof: First, let ' be linear: '(x) = c x: (4:4:29)Consider the sequence'n(x) = 8>>>><>>>>: c x ; 0 � x � l�1� 1n�c(n� 1)(l � x); l�1 � 1n� � x � l ; n 2 IN :

4.4. THE FOURIER METHOD 75-6 x' ����������������������BBBBBBB��������������������rr rrl�1� 1n� lThe functions 'n(x) are continuous and 'n(0) = 'n(l) = 0, hence the functions un(x; t)de�ned by (4.4.24) for 'n are the solutions of (4.4.4) which satisfy (4.4.6) andun(x; 0) = 'n(x):Since 'n(x) � 'n+1(x) (0 � x � l);from the maximum principle, un(x; t) � un+1(x; t):The function U0(x) = cx is a continuous solution of the heat equation. The maximumprinciple now gives un(x; t) � U0(x);as this is valid for t = 0; x = 0 and x = l.The series fun(x; t)g is monoton increasing and bounded above by the bounded functionU0(x), so it is convergent. We haveu(x; t) = limn!1 un(x; t) = limn!1 lZ0 G(x; �; t)'n(�) d� = lZ0 G(x; �; t)'(�) d� � U0(x);because we can take the limit process under the integral. In x 4.4.1 we have proved thatu(0; t) = u(l; t) = 0 and u(x; t) satis�es (4.4.4) for t > 0. It remains to show that it iscontinuous for t = 0 and 0 � x < l. Let x0 < l. We choose n such that x0 < l �1 � 1n�. Inthis case 'n(x0) = U0(x0). Noting thatun(x; t) � u(x; t) � U0(x)and limx!x0t!0 un(x; t) = limx!x0 U0(x) = '(x0);we conclude that the limit limx!x0t!0 u(x; t) = '(x0)exists and is independent of how x ! x0 and t ! 0. This implies the continuity of u(x; t)at (x0; 0). This function is bounded by U0(x). Thus, the theorem is proved for '(x) = cx.Changing x by l� x, we see that the theorem is also proved for '(x) = b(l� x).

76 CHAPTER 4. PARABOLIC EQUATIONSThus, the theorem is correct for the functions of the form '(x) = B + Ax. Further, thetheorem is correct for any continuous function '(x) which needs not satisfy the condition'(0) = '(l) = 0. In fact, any such function can be represented in the form'(x) = �'(0) + xl ('(l)� '(0))�+ (x);where (x) is a continuous function which vanishes at the ends of the interval: (0) = (l) = 0. From the superposition theorem we see that the theorem is proved as far as thecontinuity is concerned.Let ' be now a piecewise continuous function; we shall prove that the function u(x; t) de�nedby (4.4.24) gives a solution to (4.4.4) and satis�es (4.4.6).Suppose that x0 is a point where ' is continuous. We shall prove that for any positive �,there exists a �(�) such that if jx� x0j < �, t < �(�), then ju(x; t)� '(x0)j < �.Because ' is continuous at x0, there exists an �(�) such thatj'(x)� '(x0)j � �2 for jx� x0j < �(�):It follows that '(x0)� �2 � '(x) � '(x0) + �2 for jx� x0j < �(�): (4:4:30)Let '(x) and '(x) be continuously di�erentiable functions such that'(x) = '(x0) + �2 for jx� x0j < �(�);'(x) � '(x) for jx� x0j > �(�); (4:4:31)'(x) = '(x0)� �2 for jx� x0j < �(�);'(x) � '(x) for jx� x0j > �(�): (4:4:32)From (4.4.30) we have '(x) � '(x) � '(x): (4:4:33)Consider the functions u(x; t) = lZ0 G(x; �; t)'(�) d�;u(x; t) = lZ0 G(x; �; t)'(�) d�:Since '(x) and '(x) are continuous, u(x; t) and u(x; t) are so at x0. It follows the existenceof a �(�) such thatju(x; t)� '(x)j � �2 ;ju(x; t)� '(x)j � �2 9>>>=>>>; for jx� x0j < �(�); t < �(�):

4.4. THE FOURIER METHOD 77It means u(x; t) � '(x) + �2 = '(x0)+�;u(x; t) � '(x)� �2 = '(x0)�� 9>>>=>>>; for jx� x0j < �(�); t < �(�):Since the function G(x; �; t) is not negative, the inequalities of (4.4.33) yieldu(x; t) � u(x; t) � u(x; t): (4:4:34)Consequently,'(x0)� � � u(x; t) � '(x0) + � for jx� x0j < �(�); t < �(�):That is ju(x; t)� '(x0)j < � for jx� x0j < �(�); t < �(�):Furthermore, (4.4.34) yields also the boundedness of u(x; t). 24.4.4 The non-homogeneous heat equationWe consider now the non-homogeneous heat equationut = a2uxx + f(x; t); 0 < x < l; 0 < t � T; (4.4.1)with the initial condition u(x; 0) = 0; 0 � x � l; (4:4:35)and the boundary conditionsu(0; t) = u(l; t) = 0; 0 < t � T: (4.4.6)We shall �nd a solution of this problem in the formu(x; t) = 1Xn=1 un(t) sin��nl x� : (4:4:36)In order to determine u(x; t), we have to �nd un(t). For this purpose, we represent f(x; t)in the form f(x; t) = 1Xn=1 fn(t) sin��nl x� ;where fn(t) = 2l lZ0 f(�; t) sin��nl �� d�: (4:4:37)Putting (4.4.36) and (4.4.37) into (4.4.1), we get1Xn=1 sin��nl x� (��nl �2 a2un(t) + _un(t)� fn(t)) = 0:

78 CHAPTER 4. PARABOLIC EQUATIONSIt follows that _un(t) = �a2 ��nl �2 un(t) + fn(t): (4:4:38)On the other hand u(x; 0) = 1Xn=1 un(0) sin��nl x� = 0;and therefore un(0) = 0: (4:4:39)Solving (4.4.38) with the initial condition (4.4.39) we obtainun(t) = tZ0 e���nl �2 a2(t� � )fn(� ) d�: (4:4:40)Hence, u(x; t) = 1Xn=1 2664 tZ0 e���nl �2 a2(t� � )fn(� ) d�3775 sin��nl x� : (4:4:41)Taking (4.4.37) into account, we can writeu(x; t) = tZ0 lZ0 8>><>>:2l 1Xn=1 e���nl �2 a2(t� � ) sin��nl x� sin��nl ��9>>=>>; f(�; � ) d� d�=: tZ0 lZ0 G(x; �; t� � ) f(�; � ) d� d�; (4:4:42)where G(x; �; t � � ) = 2l 1Xn=1 e���nl �2 a2(t� � ) sin��nl x� sin��nl �� (4:4:43)is in fact our Green's function de�ned by (4.4.23).We can give here the physical meaning of (4.4.42) similar to that of (4.4.24), however, weomit to give it here.4.4.5 The non-homogeneous �rst boundary value problemFind the solution of the problemut = a2uxx + f(x; t); 0 < x < l; 0 < t � T; (4.4.1)u(x; 0) = '(x); 0 � x � l; (4.4.2)u(0; t) = �1(t); u(l; t) = �2(t); 0 < t � T: (4.4.3)

4.5. PROBLEMS ON UNBOUNDED DOMAINS 79In order to solve this problem, we change it to a homogeneous boundary value problem.In fact, consider the functionU(x; t) = �1(t) + xl [�2(t)� �1(t)] : (4:4:44)The function U satis�es the boundary conditionU(0; t) = �1(t) and U(l; t) = �2(t);and the equation Ut = a2Uxx + f1(x; t);where f1(x; t) = Ut(x; t)� a2Uxx(x; t).It is clear now that the function v(x; t) = u(x; t)� U(x; t)is a solution of the problem vt = avxx + f;v(0; t) = v(l; t) = 0;v(x; 0) = '(x)� U(x; 0):The solution v can be found by the Fourier method.4.5 Problems on unbounded domains4.5.1 The Green function in unbounded domainsWe have derived the Green functionGl(x; �; t) := 2l 1Xn=1 e���nl �2 a2t sin��nl x� sin��nl �� (4:5:1)for the problem in the �nite domain (0; l). Here we have used the subindex l to indicate thedomain (0; l). We want to develope this theory when the domain is expanded to the wholeIR. In doing so, we rewrite (4.5.1) in another form, such that the ends of the bar are � l2 andl2 .Let x0 = x� l2, �0 = � � l2. Then x0 and �0 are lying in (� l2; l2) and the Green function hasthe formGl(x0; �0; t) = 2l 1Xn=1 e���nl �2 a2t sin �nl x0 + l2! sin �nl �0 + l2! : (4:5:2)We rewrite (4.5.2) by the following arguments.

80 CHAPTER 4. PARABOLIC EQUATIONSIf n is even, that is n = 2m, thensin 2�ml x0 + l2! � sin 2�ml �0 + l2! = sin�2�ml x0� � sin�2�ml �0� :Further, if n is odd, that is n = 2m+ 1, thensin (2m+ 1)�l x0 + l2! � sin (2m+ 1)�l �0 + l2! = cos (2m + 1)�l x0 � cos (2m+ 1)�l �0:Thus, Gl(x0; �0; t) = 2l 1Xn=000e���nl �2 a2t sin��nl x0� sin��nl �0�+ 2l 1Xn=10e���nl �2 a2t cos��nl x0� cos��nl �0� ; (4:5:3)where P00 stands for even n, and P0 stands for odd n.We study the limit of Gl(x0; �0; t) as l!1. We have2l 1Xn=000e��2na2t sin(�nx0) sin(�n�0) = 1� 1Xn=000f1(�n) ��; (4:5:4)where f1(�) = e��2a2t sin(�x0) sin(��0); �� = 2�l ; �n = �nl :The sum (4.5.4) is the integral sum for the function f1(�) over the interval 0 � � <1. Forl!1, ��! 0. Taking the limit under the integral, we obtainlim��!0 1� 1Xn=000f1(�n) �� = 1� 1Z0 f1(�) d� = 1� 1Z0 e��2a2t sin(�x0) sin(��0) d�: (4:5:5)Remark 4.5.1 Here we have used the following result: For a continuous function de�nedon [0;1), if the integral sums1Xi=1 f(��i )(�i � �i�1) (�i � �i�1 = ��; �i�1 � ��i � �i)for any choice of ��i converge, then there exists the integral1Z0 f(�) d�:Analogously, 2l 1Xn=10e��2na2t cos (�nx0) cos (�n�0) = 1� 1Xn=10f2(�n) ��; (4:5:6)where f2(�) = e��2a2t cos(�x0) cos(��0); �� = 2�l and �n = �nl :

4.5. PROBLEMS ON UNBOUNDED DOMAINS 81As ��! 0, we getlim��!0 1� 1Xn=10f2(�n) �� = 1� 1Z0 f2(�) d� = 1� 1Z0 e��2a2t cos(�x0) cos(��0) d�: (4:5:7)Finally, we haveG(x; �; t) = liml!1Gl(x; �; t)= 1� 1Z0 e��2a2t sin(�x) sin(��) d� + 1� 1Z0 e��2a2t cos(�x) cos(��) d�= 1� 1Z0 e��2a2t cos�(x� �) d�:The Green function for the unbounded domain (�1;1) has thus the formG(x; �; t) = 1� 1Z0 e��2a2t cos �(x� �) d�: (4:5:8)We calculate now the integralI = 1Z0 e��2� cos(��) d� (� > 0); (4:5:9)where the parameters � and � are given. In order to calculate this integral, we �x � andvary � so that I is a function of �: I(�). It is clear thatdId� = � 1Z0 e��2�� sin(��) d�= sin(��) 12�e��2������10 � �2� 1Z0 e��2� cos(��) d�= � �2�I(�):Thus, I 0(�)I(�) = � �2�:Hence I(�) = C e��24� :On the other hand, I(0) = 1Z0 e��2� d� = 1p� 1Z0 e�z2 dz = 1p� � p�2 :

82 CHAPTER 4. PARABOLIC EQUATIONSThus, I(�) = 1Z0 e��2� cos(��) d� = 12p�p�e��24� : (4:5:10)Applying (4.5.10) to (4.5.8), we getG(x; �; t) = 12p�a2te�(x� �)24a2t ; (4:5:11)which is the Green function for the unbounded domain (�1;1). This function is oftencalled the fundamental solution of the heat equation.Properties of the fundamental solutioni) The function G(x; �; t� t0) as a function of x; t is a solution of the heat equation. In fact,Gx = � 12p� � x� �2[a2(t� t0)]3=2 � e� (x� �)24a2(t� t0) ;Gxx = 12p� "�12 1[a2(t� t0)]3=2 + (x� �)24[a2(t� t0)]5=2# e� (x� �)24a2(t� t0) ;Gt = 12p� "� a22[a2(t� t0)]3=2 + a2(x� �)24[a2(t� t0)]3=2# e� (x� �)24a2(t� t0) :Thus, Gt = a2 Gxx:ii) 1Z�1 G(x; �; t� t0) dx = 1 for t > t0.1Z�1 G(x; �; t� t0) dx = 1p� 1Z�1 e� (x� �)24a2(t� t0) dx2qa2(t� t0) = 1p� 1Z�1 e��2 d� = 1:4.5.2 Heat conduction in the unbounded domain (�1;1)Find a bounded function u(x; t) (�1 < x <1; t � 0), which satis�es the equationut = a2 uxx (�1 < x <1; t > 0); (4:5:12)and the initial condition u(x; 0) = '(x); �1 < x <1: (4:5:13)Since u is bounded, the solution of our problem is unique (x 4.3.5). We shall prove that if 'is bounded, say j'j < M , then for t > 0 the Poisson integralu(x; t) = 12p� 1Z�1 1pa2te�(x� �)24a2t '(�) d� (4:5:14)

4.5. PROBLEMS ON UNBOUNDED DOMAINS 83is a solution of (4.5.12), and limx!x0t!0 u(x; t) = '(x0). Here x0 is a point where ' is continuous.We note that ju(x; t)j � 1Z�1 G(x; �; t) j'(�)j d� < M 1Z�1 G(x; �; t) d� =M:Thus, the function u(x; t) is bounded. We shall prove that for t > 0, u(x; t) satis�es the heatequation (4.5.12). In doing so, we prove that we can di�erentiate (4.5.14) under the integral.For example, formally di�erentiating both sides of (4.5.14) we get@u@x = 1Z�1 @@x�G(x; �; t)� '(�) d�;and it remains to prove that the integral on the right-hand side converges uniformly. Itis enough to prove the di�erentiability of u at a point (x0; t0), or to prove the uniformconvergence of the above integral in a neighbourhood of this point:t1 � t0 � t2; jx� x0j � x:-6

xt

x0 � x x0 x0 + xt1t0t2In doing so, we shall prove that there exists a positive function F (�), which does not dependon x and t, and approximates jGx(x; �; t) '(�)j from above:jGx(x; �; t) '(�)j � F (�); (4:5:15)and 1Zx1 F (�) d� <1; x1Z�1 F (�) d� <1; (4:5:16)x1 is a number, for which (4.5.15) is valid.For jx� x0j � x and t1 � t � t2 we have����� @@xG(x; �; t)����� � j'(�)j = 12p� j� � xj2pa2t3 e�(x� �)24a2t j'(�)j� M2p� j� � x0j+ x2pa2t13 e�(j� � x0j � x)24a2t2=: F (�): (4:5:17)

84 CHAPTER 4. PARABOLIC EQUATIONSLet x1 be a number from which (4.5.17) is valid. Then1Zx1 F (�) d� = 1Zx1 M2p� j� � x0j+ x2pa2t13 e�(j� � x0j � x)24a2t2 d�= 1Zx1�x M2p� �1 + 2x2pa2t13e� �214a2t2 d� (�1 = j� � x0j � x):This integral is clearly convergent. Thus,@u@x = 1Z�1 @@xG(x; �; t)'(�) d�: (4:5:18)Similarly, we can prove that for t > 0 the function u(x; t) is di�erentiable two times w.r.t. xand once w.r.t. t, and that u satis�es the heat equation (4.5.12).Let x0 be a point where '(x) is continuous. It remains to prove thatu(x; t) �! '(x0); as t! 0 and x! x0:For any � > 0 we have to show that there is a �(�) such thatju(x; t)� '(x0)j < �if jx� x0j < �(�) and t < �(�).Since ' is continuous at x0, there exists an �(�) such thatj'(x)� '(x0)j < �6 for jx� x0j < �: (4:5:19)We break the integral for u(x; t) into three parts as follows:u(x; t) = 12p� 1Z�1 1pa2te�(x� �)24a2t '(�) d�= 12p� x1Z�1 � � � + 12p� x2Zx1 � � � + 12p� 1Zx2 � � �:= u1(x; t) + u2(x; t) + u3(x; t); (4:5:20)where x1 = x0 � � and x2 = x0 + �: (4:5:21)For the second integral we haveu2(x; t) = '(x0)2p� x2Zx1 1pa2te�(x� �)24a2t d�+ 12p� x2Zx1 1pa2te�(x� �)24a2t ['(�)�'(x0)] d� = I1+I2:

4.5. PROBLEMS ON UNBOUNDED DOMAINS 85The integral I1 can be calculated asI1 = '(x0)2p� x2Zx1 e�(x� �)24a2tpa2t d� = '(x0)p� x2�x2pa2tZx1�x2pa2t e��2 d�;with � = � � x2pa2t, d� = d�2pa2t.As jx� x0j < �, x1 = x0 � �, x2 = x0 + �, we see thatx1 � x2pa2t �! �1; as t! 0x2 � x2pa2t �! +1; as t! 0:It follows that limt!0x!x0 I1 = '(x0):Thus, there is a �1(�) such thatjI1 � '(x0j < �2 for jx� x0j < �1 and t < �1: (4:5:22)We estimate I2 as follows:jI2j � 12p� x2Zx1 1pa2te�(x� �)24a2t j'(�)� '(x0)j d�:From (4.5.21), for x1 < � < x2, we see thatj� � x0j < �:Taking (4.5.19) and the inequality1p� x00Zx0 e��2 d� < 1p� 1Z�1 e��2 d� = 1; 8x0; x00into account, we obtain jI2j � �6 � 12p� x2Zx1 1p4a2t e�(x� �)24a2t d�= �6 � 1� x2�x2pa2tZx1�x2pa2t e��2 d� < �6 : (4:5:23)

86 CHAPTER 4. PARABOLIC EQUATIONSFurther, ju3(x; t)j = 12p� �������� 1Zx2 1pa2t e�(x� �)24a2t '(�) d���������< Mp� 1Zx2�x2pa2t e��2 d��! t!0x!x0 0; (4:5:24)ju1(x; t)j = 12p� �������� x1Z�1 1pa2t e�(x� �)24a2t '(�) d���������< Mp� x1�x2pa2tZ�1 e��2 d��! t!0x!x0 0: (4:5:25)(Note that as x! x0, x2 � x > 0 and x1 � x < 0.)Thus, there is a �2(�) such thatju3(x; t)j < �3 and ju1(x; t)j < �3 for jx� x0j < �2 and t < �2: (4:5:26)A combination of (4.5.22), (4.5.23) and (4.5.26) givesju(x; t)� '(x0)j � ju1j+ jI1 � '(x0)j+ jI2j+ ju3j< �3 + �6 + �6 + �3 ; (4:5:27)as jx� x0j < � and t < �, where � = minf�1; �2g.Thus, we have proved that the functionu(x; t) = 12p� 1Z�1 1pa2te�(x� �)24a2t '(�) d�is bounded and satis�es the heat equation as well as the initial condition.If the initial condition is given at t = t0, then the solution has the formu(x; t) = 12p� 1Z�1 1qa2(t� t0)e� (x� �)24a2(t� t0)'(�) d�:Example: We try to �nd the solution of the problem (4.5.12), (4.5.13) in the caseu(x; 0) = '(x) = 8<: T1 for x < 0;T2 for x > 0 :

4.5. PROBLEMS ON UNBOUNDED DOMAINS 87In this caseu(x; t) = 12p� 1Z�1 1pa2te�(x� �)24a2t '(�) d�= T2p� 0Z�1 e�(x� �)24a2t d�2pa2t + T1p� 1Z0 e�(x� �)24a2t d�2pa2t= T2p� � x2pa2tZ�1 e��2 d� + T1p� 1Z� x2pa2t e��2 d�= T1 + T22 + T1 � T2p� x2pa2tZ0 e��2 d�; (4:5:28)since 1p� �zZ�1 e��2 d� = 1p� 0Z�1 e��2 d� + 1p� zZ0 e��2 d� = 12 + 1p� zZ0 e��2 d�and 1p� 1Z�z e��2 d� = 12 + 1p� 0Z�z e��2 d� = 12 � 1p� zZ0 e��2 d�:If T2 = 0, T1 = 1, then u(x; t) = 12 0BB@1 + 2p� x2pa2tZ0 e��2 d�1CCA. The function�(z) = 2p� zZ0 e��2 d�is called the error function which has many applications in the probability theory.4.5.3 The boundary value problem in a half-spaceConsider the heat equation in the �rst quadrantut = a2 uxx; x > 0; t > 0; (4.5.12)with the initial condition u(x; 0) = '(x); x > 0; (4.5.13)and one of the boundary value conditionsu(0; t) = �(t); t � 0; (the �rst b.v.p.)

88 CHAPTER 4. PARABOLIC EQUATIONS@u@x(0; t) = �(t); t � 0; (the second b.v.p.) or @u@x(0; t) = �[u(0; t)� �(t)]; (the third b.v.p.)!In order to guarantee the uniqueness of the solution, we assume that the solution is bounded:ju(x; t)j < M; 0 < x <1; t � 0;where M is a constant. It follows also j'(x)j < M:We represent the solution of the �rst b.v.p. in the formu(x; t) = u1(x; t) + u2(x; t);where u1 is the solution of (4.5.12) with the conditionsu1(x; 0) = '(x); u1(0; t) = 0 (4.5.14)and u2 is the solution of (4.5.12) with the conditionsu2(x; 0) = 0; u2(0; t) = �(t): (4.5.17)Below we give two lemmas about the Poisson integralu(x; t) = 12p� 1Z�1 1pa2te�(x� �)24a2t (�) d�. (4.5.14)Lemma 4.5.2 If is a bounded and odd function, (x) = � (�x);then u(0; t) = 0.Proof: Since is bounded, the integral (4.5.14) converges, andu(0; t) = 12p� 1Z�1 1pa2te� �24a2t (�) d� = 0: 2Lemma 4.5.3 If is a bounded and even function, (x) = (�x);then @u@x(0; t) = 0.

4.5. PROBLEMS ON UNBOUNDED DOMAINS 89Proof: @u@x�����x=0 = � 12p� 1Z�1 (x� �)2pa2t3 e�(x� �)24a2t (�) d������x=0 = 0: 2Let (x) be a function de�ned by (x) = 8<: '(x) ; x > 0�'(�x) ; x < 0 :Since ' is bounded (by M), is bounded. It follows that the functionU(x; t) = 12p� 1Z�1 1pa2te�(x� �)24a2t (�) d�is well de�ned. Furthermore, since is even, U(0; t) = 0. Thus,u1(x; t) = U(x; t) for x � 0:The function U(x; t) can be represented as follows:U(x; t) = 12p� 0Z�1 1pa2te�(x� �)24a2t (�) d� + 12p� 1Z0 1pa2te�(x� �)24a2t (�) d�= � 12p� 1Z0 1pa2te�(x+ �)24a2t '(�) d� + 12p� 1Z0 1pa2te�(x� �)24a2t '(�) d�= 12p� 1Z0 1pa2t 8>><>>:e�(x� �)24a2t � e�(x+ �)24a2t 9>>=>>;'(�) d�:Thus, u1(x; t) = 12p� 1Z0 1pa2t 8>><>>:e�(x� �)24a2t � e�(x+ �)24a2t 9>>=>>;'(�) d�: (4:5:29)Analogously, for the solution u1(x; t) of the second boundary value problem with @u1@x (0; t) =0 and u1(x; 0) = '(x) we haveu1(x; t) = 12p� 1Z0 1pa2t 8>><>>:e�(x� �)24a2t + e�(x+ �)24a2t 9>>=>>;'(�) d�: (4:5:30)Now we try to �nd u2(x; t) and u2(x; t). In doing so we note that if we consider the equationv1t = a2 v1xx, for 0 < x <1 and t � t0, with the conditions v1(x; t0) = T and v1(0; t) = 0,

90 CHAPTER 4. PARABOLIC EQUATIONSthen from (4.5.29) we havev1(x; t) = T2p� 1Z0 8>>><>>>:e� (x� �)24a2(t� t0) � e� (x+ �)24a2(t� t0)9>>>=>>>; � d�qa2(t� t0) : (4:5:31)Taking the changes of variables� = � � x2qa2(t� t0) ; �1 = � + �2qa2(t� t0) ;we get v1(x; t) = Tp� 266664 1Z� x2pa2(t�t0) e��2 d� � 1Zx2pa2(t�t0) e��21 d�1377775= Tp� x2pa2(t�t0)Z� x2pa2(t�t0) e��2 d� = T 2p� x2pa2(t�t0)Z0 e��2 d�:Thus, v1(x; t) = T � 0@ x2qa2(t� t0)1A ; (4:5:32)where �(z) = 2p� zZ0 e��2 d�is the error function.Let �(t) = �0 = const. Then the functionv(x; t) = �0 � 0@ x2qa2(t� t0)1Ais the solution of the heat equation (4.5.12) for t � t0 which satis�es the conditionsv(x; t0) = �0; v(0; t) = 0:It follows that the functionv(x; t) := �0 � v(x; t) = �0 241 ��0@ x2qa2(t� t0)1A35 (4:5:33)is the solution of (4.5.12) for t � t0, which satis�es the conditionsv(x; t0) = 0 (x > 0) and v(0; t) = �0 (t > t0):

4.5. PROBLEMS ON UNBOUNDED DOMAINS 91We rewrite v(x; t) in the form v(x; t) = �0 U(x; t);where U(x; t) = 1 ��0@ x2qa2(t� t0)1A = 2p� 1Zx2pa2(t�t0) e��2 d�:The function U(x; t) corresponds to the case �0 = 1.We extend U(x; t) by zero for t < t0. It then corresponds to the step boundary conditionU(0; t) = 8<: 1 ; t > t00 ; t < t0 :Consider now the solution v(x; t) of (4.5.12), which satis�es the conditionsv(x; t0) = 0; v(0; t) = �(t) = 8<: �0 ; t0 < t < t10 ; t > t1 :It can be veri�ed that v(x; t) = �0[U(x; t� t0)� U(x; t� t1)]:Further, if � has the form �(t) = 8>>>>>>>>>><>>>>>>>>>>: �0 ; t0 < t � t1;�1 ; t1 < t � t2;: : :�n�1 ; tn�1 < t � tn;then the solution of (4.5.12), (4.5.13) can be represented in the formu(x; t) = n�2Xi=0 �i[U(x; t� ti)� U(x; t� ti+1)] + �n�1U(x; t� tn�1): (4:5:34)With the aid of the theorem about the mean value, we haveu(x; t) = n�2Xi=1 �i@U(x; t� � )@t ������i�� + �n�1U(x; t� tn�1); ti � �i � ti+1: (4:5:35)We consider the problem (4.5.12), (4.5.13) with u(x; 0) = 0. If the function �(t) is piecewisecontinuous, then we have an approximation to u(x; t) of the form (4.5.34), when �(t) isapproximated by a piecewise constant function. If we re�ne this approximation, then wehave u(x; t) = tZ0 @U@t (x; t� � )�(� ) d�;

92 CHAPTER 4. PARABOLIC EQUATIONSsince for x > 0 limt�tn�1!0�n�1U(x; t� tn�1) = 0:We do not go into the details, when the limit process is meaningful, and so formally takeu2(x; t) = tZ0 @U@t (x; t� � )�(� ) d�: (4:5:36)Further,@U@t (x; t) = @@t 0BBB@ 2p� 1Zx2pa2t e��2 d�1CCCA = 12p� a2xpa2t3 e� x24a2t = �2a2@G@x (x; 0; t) = 2a2@G@� ������=0:Thus,u2(x; t) = 2a2 tZ0 @G@� (x; 0; t� � )�(� ) d� = a22p� tZ0 xqa2(t� � )3 e� x24a2(t� � ) �(� ) d�:(4:5:37)We note that in the way of getting (4.5.37) we have used only the linearity of the heatequation and nothing more. Furthermore, the boundary and initial conditions on U areexplicitly given U(0; t) = 1; t > 0;U(x; 0) = 0; x > 0;or U(0; t) = 8<: 1 ; t > 00 ; t < 0 :If the boundary condition for a given di�erential equation isu(0; t) = �(t); t > 0;which satis�es the homogeneous initial condition, thenu(x; t) = tZ0 @U@t (x; t� � )�(� ) d�:This is called the Duhamel principle, which shows that the di�culties in the boundary valueproblems are due to the piecewise constant boundary conditions.By the method of odd extension of the data, we can easily �nd the solution of the problemut = a2 uxx + f(x; t); (0 < x <1; t > 0);u(0; t) = 0; u(x; 0) = 0;in the formu3(x; t) = 12p� 1Z0 tZ0 1qa2(t� � ) 8>><>>:e�(x� �)24a2t � e�(x+ �)24a2t 9>>=>>; f(�; � ) d� d�:

Chapter 5Elliptic EquationsWe shall consider shortly in this chapter the Laplace equation�u = uxx + uyy = 0:A solution of the Laplace equation is called a harmonic function.The inhomogeneous version of the Laplace equation�u = f;with f being a given function, is called the Poisson equation.The basic mathematical problem is to solve the Laplace or the Poisson equation in a givendomain with a condition on the boundary @ of .�u = f in u = h or @u@n = h or @u@n + au = h on @:5.1 The maximum principleLet be a connected bounded domain in IR2. Let u be a harmonic function that is continuousin = [ @. Then the maximum and the minimum values of u are attained on @.Proof: Consider the functionv(x; y) = u(x; y) + �(x2 + y2); � > 0:We have �v = �u+ ��(x2 + y2) = 0 + 4� > 0 in IR :Thus, v has no maximumin , otherwise vxx � 0, vyy � 0, and so vxx+vyy � 0. Furthermore,since v is a continuous function, it attains its maximum in the bounded closed domain .As v has no maximum in , v attains its maximum at some point x0 2 @. Hence, for all(x; y) 2 u(x; y) � v(x; y) � v(x0; y0) = u(x0; y0) + �(x20 + y20) � max@ u+ �(x20 + y20):93

94 CHAPTER 5. ELLIPTIC EQUATIONSSince is bounded, x20+ y20 is bounded and since � is an arbitrary positive number, we haveu(x; y) � max@ u; 8(x; y) 2 :Thus, u attains its maximum at some point of the boundary @.The proof of "minimum part" is similar. 25.2 The uniqueness of the Dirichlet problem�u = f in ;uj@ = h:Let u1 and u2 be two solutions of this problem. Put v = u1 � u2. Then �v = 0 in andv���@ = 0. By the maximum principle v � 0 in . Thus u1(x; y) = u2(x; y).5.3 The invariance of the operator �The operator � is invariant under translations and rotations. In fact, a translation in theplane is a transformation x0 = x+ a; y0 = y + b:The invariance under translations means that uxx + uyy = ux0x0 + uy0y0 .A rotation in the plane through the angle � is given byx0 = x cos�+ y sin�;y0 = �x sin�+ y cos�:By the chain rule we calculateux = ux0 cos�� uy0 sin�;uy = ux0 sin� + uy0 cos�;uxx = (ux0 cos�� uy0 sin�)x0 cos� � (ux0 sin�+ uy0 cos�)y0 sin�;uyy = (ux0 sin� + uy0 cos�)x0 sin�+ (ux0 sin�+ uy0 cos�)y0 cos�:Adding, we haveuxx + uyy = (ux0x0 + uy0y0)(cos2 �+ sin2 �) + ux0y0 : : : = ux0x0 + uy0y0:This proves the invariance of the Laplace operator. In engineering the Laplacian � is amodel for isotropic physical situations, in which there is no prefered direction.

5.3. THE INVARIANCE OF THE OPERATOR � 95The rotational invariance suggests that the two-dimensional Laplacian �2 = @2@x2 + @2@y2should take a particularly simple form in polar coordinates. The transformation has theform x = r cos �; y = r sin �:It follows that r = qx2 + y2; � = arccos xpx2 + y2 = arcsin ypx2 + y2 :We have @x@r = cos �; @y@r = sin �;@x@� = �r sin �; @y@� = r cos �;@r@x = xr = cos �; @r@y = yr = sin �;@�@x = � yr2 = �sin �r ; @�@y = xr2 = cos �r :Thus, the transformation x = r cos �, y = r sin � has the Jacobian matrixJ = 0BBBB@ @x@r @y@r@x@� @y@� 1CCCCA = 0@ cos � sin ��r sin � r cos � 1Awith the inverse J�1 = 0BBBB@ @r@x @�@x@r@y @�@y 1CCCCA = 0BBBB@ cos � � sin �rsin � cos �r 1CCCCA :Hence, @u@x = @u@r @r@x + @u@� @�@x = @u@r cos � � @u@� sin �r ;@u@y = @u@r @r@y + @u@� @�@y = @u@r sin � + @u@� cos �r ;@2u@x2 = @@r @u@r cos � � @u@� sin �r ! cos � � @@� @u@r cos � � @u@� sin �r ! sin �r= @2u@r2 cos2 � � @2u@r @� sin � cos �r + @u@� sin � cos �r2� @2u@r @� sin � cos �r + @u@r sin2 �r + @2u@�2 sin2 �r2 + @u@� sin � cos �r2= @2u@r2 cos2 � + @2u@�2 sin2 �r2 � 2 @2u@r @� sin � cos �r + @u@r sin2 �r + 2@u@� sin � cos �r2 :

96 CHAPTER 5. ELLIPTIC EQUATIONSuxy = uyx = urr sin � cos � � u�� sin � cos �r2 + ur� cos2 � � sin2 �r ++u� sin2 � � cos2 �r2 � ur sin � cos �r ;uyy = urr sin2 � + u�� cos2 �r2 + 2ur� sin � cos �r + ur cos2 �r � 2u� sin � cos �r2 :Thus, uxx + uyy = urr + u�� 1r2 + ur 1r (5:3:1)= 1r2 (r @@r r@u@r!+ @2u@�2) : (5:3:2)It is also natural to look for special harmonic functions that themselves are rotationallyinvariant. It means that we look for solutions depending only on r. Thus, by (5.3.2)0 = uxx + uyy = 1r @@r r@u@r!if u does not depend on �.Hence, r@u@r = c1, and so u = c1 ln r + c2. The function ln r will play a central role later.5.4 Poisson's formulaConsider the problem uxx + uyy = 0; x2 + y2 < a; (5:4:1)u = f; x2 + y2 = a: (5:4:2)Here a is a given number and f is a given function. In polar coordinates (r; �), the equation(5.4.1) has the form 1r2 (r @@r r @u@r!+ @2u@�2) = 0: (5:4:3)We shall �nd solutions of this equation in the formu(r; �) = R(r) �(�):Plugging this into (5.4.3), we get ddr �r dRdr �Rr = ��00� = �;where � is a constant. From this we get�00 + �� = 0; (5:4:4)

5.4. POISSON'S FORMULA 97r ddr r dRdr !� �R = 0: (5:4:5)The equation (5.4.4) gives �(�) = A cosp� � +B sinp� �:As u(r; �) is periodic in �, we have �(� + 2�) = �(�):Thus, p� = n is an entire number and so�n(�) = An cos(n�) +Bn sin(n�):We shall �nd R(r) in the form R(r) = r�. Putting this into (5.4.5) we getn2 = �2 or � = �n (n > 0):Thus, R(r) = Crn +Dr�n;where n > 0 and C and D are some constants. We see that D must be equal to zero,otherwise as r = 0, R(r) is not bounded. Thus, special solutions of our problem have theform un(r; �) = rn �An cos(n�) +Bn sin(n�)�:If the series u(r; �) = 1Xn=0 rn�An cos(n�) +Bn sin(n�)�converges, then it represents a harmonic function. Note that the equation (5.4.3) has nomeaning for r = 0, so we have to prove that �un = 0 also for r = 0. In fact, the specialsolutions rn cos(n�) and rn sin(n�)are the real and the imaginary parts of the function�n ein� = (� ei�)n = (x+ iy)n;which are polynomials in x and y. It is now clear that a polynomial which satis�es theequation �u = 0 for r > 0, due to the continuity of the second derivatives, satis�es thisequation also for r = 0.In order to determine An and Bn, we use the boundary conditionu(a; �) = 1Xn=0 an�An cos(n�) +Bn sin(n�)� = f: (5:4:6)Taking into account that f is a function of �, we havef(�) = �02 + 1Xn=1 ��n cos(n�) + �n sin(n�)�; (5:4:7)

98 CHAPTER 5. ELLIPTIC EQUATIONSwhere �0 = 1� �Z�� f( ) d ; �n = 1� �Z�� f( ) cos(n ) d ; �n = 1� �Z�� f( ) sin(n ) d :A comparison of (5.4.6) with (5.4.7) givesA0 = �02 ; An = �nan ; Bn = �nan :Thus, u(r; �) = �02 + 1Xn=1�ra�n ��n cos(n�) + �n sin(n�)�: (5:4:8)We shall �nd conditions guaranteeing the convergence of this series.Consider the function un = tn��n cos(n�) + �n sin(n�)�; t = ra � 1:We have @kun@�k = tn nk ��n cos�n� + k�2�+ �n sin�n� + k�2�� :Denoting by M the maximum of j�nj, j�nj, n = 0; 1; : : :,j�nj < M; j�nj < M (5:4:9)we have �����@kun@�k ����� � tnnk2M:For a t0 � r0a < 1, 1Xn=1 tnnk(j�nj+ j�nj) � 2M 1Xn=1 tn0nk (t � t0):It follows that the series on the right-hand side is uniformly convergent for t � t0 < 1. Hence,u(r; ') for r � r0 < a is in�nitely di�erentiable w.r.t. �. Analogously, it is also in�nitelydi�erentiable w.r.t. r � r0 < a. As r0 < a is arbitrary the function u de�ned by (5.4.8)satis�es (5.4.1) and for r < a it is in�nitely di�erentiable w.r.t. r and '.Up to this point, we have used only the conditions in (5.4.9). This conditions are satis�ed,if f(') is bounded by M=2.In order to guarantee the continuity of the solution up to boundary (up to r = a), we needto suppose that f is continuous and di�erentiable.Putting the expressions of the Fourier coe�cients �n, �n into (5.4.8), we getu(r; �) = 1� �Z�� f( )(12 + 1Xn=1�ra�n � cos(n ) cos(n�) + sin(n ) sin(n�)�) d = 1� �Z�� f( )(12 + 1Xn=1�ra�n cos n(� � )) d : (5:4:10)

5.4. POISSON'S FORMULA 99On the other hand,12 + 1Xn=1�ra�n cos[n(� � )] = 12 + 12 1Xn=1 tn �ein(�� ) + e�in(�� )�= 12 (1 + 1Xn=1 ��tei(�� )�n + �te�i(�� )�n�)= 12 241 + tei(�� )1 � tei(�� ) + te�i(�� )1� te�i(�� )35= 12 1� t21� 2t cos(� � ) + t2 �t = ra < 1� :Putting the last equality into (5.4.10), we getu(r; �) = 12� �Z�� f( ) 1� r2a2r2a2 � 2 ra cos(� � ) + 1 d :Thus, u(r; �) = 12� �Z�� f( ) a2 � r2r2 � 2ar cos(� � ) + a2 d : (5:4:11)This formula is called Poisson's formula.The function K(�; �; a; ) = a2 � r2r2 � 2ar cos(� � ) + a2is called the Poisson kernel. For r < a it is positive, since 2ar < a2 + r2 for r 6= a.We now prove that Poisson's formula (5.4.11) (or (5.4.10)) gives also a continuous solutionto (5.4.1), (5.4.2), if f is continuous.If f is continuous, then f is bounded, and so we have already proved that for r < a,the equation �u = 0 is satis�ed. It remains to prove that u(r; �) is continuous up to theboundary. Let f1(�), f2(�); : : : be a sequence of continuous and di�erentiable functions,which uniformly converge to f(').For any fk(�) we can determine the corresponding uk(r; �) by the Poisson formula. Sinceffk(�)g uniformly converges to f(�), for any � > 0 there exists a k0(�) > 0 such thatjfk(�)� fk+l(�)j < �; 8k > k0(�); l > 0:Thus, from the maximum principle,juk(r; �)� uk+l(r; �)j < � for r � r0 < a; k > k0(�); l > 0; � > 0:Hence, fukg uniformly converges to a function u, u = limk!1 uk(r; �). The function u iscontinuous in a closed domain, since all functionsuk(r; �) = 12� �Z�� a2 � r2r2 � 2ar cos(� � ) + a2fk( ) d

100 CHAPTER 5. ELLIPTIC EQUATIONSare continuous in closed domains. Thus,u(r; �) = limk!1 uk(r; �) = 8>>>><>>>>: 12� �Z�� a2 � r2r2 � 2ar cos(� � ) + a2f( ) d ; a < r;f(�) ; a = r ;since ffk(�)g uniformly converges to f(�).Let X = (x; y) have the polar coordinates (r; �) and X 0 =(x0; y0) have the the polar coordinates (a; ) (that means that(x0; y0) is lying at the boundary). �� �(x; y)ra (x0; y0)� � IThen j ~X � ~X 0j2 = a2 + r2 � 2ar cos(� � ):The arc length element on the circumference is ds0 = a d . Therefore, Poisson's formulatakes the alternative form u(X) = a2 � j ~Xj22�a ZjX 0j=a u(X 0)jX �X 0j2 ds0 (5:4:12)for X 2 , where we write u(X 0) = f(�).5.5 The mean value theoremLet u be a harmonic function in a domain � IR2, continuous in . Let M0 be a point of and Ba be the ball of radius a with the center at M0 which is completely lying in . Thenu(M0) = 12�a ZZ@Ba u(�) d�: (5:5:1)Proof: Without loss of generality we may suppose thatM0 = 0. Since Ba � , the functionu is harmonic in , from Poisson's formula (5.4.11) we have (X = 0)u(0) = a22�a ZjX 0j=a u(X 0)a2 ds0: 2

5.6. THE MAXIMUM PRINCIPLE (A STRONG VERSION) 1015.6 The maximum principle (a strong version)We now prove the following: Let be a connected bounded domain in IR2. Let u be aharmonic function in , continuous in . Then the maximum and the minimum values ofu are attained on @ and nowhere inside unless u � const.Since u is continuous in , it attains its maximum somewhere, say xM 2 . We shall showthat xM 62 unless u � const. By de�nition of M , we know thatu(x) � u(xM) =M; 8x 2 :We draw a circle around xM entirely con-tained in . By the mean value theorem,u(xM) is equal to its average around the cir-cumference. � xM Since the average is not greater than the maximum, we haveM = u(xM) �M:Therefore, u(x) = M for all x on the circumference. This is true for any such circle. Sou(x) = M for all x in the diagonally shaded region. Now we repeat the argument witha di�erent center. We can �ll the whole domain up with circles. In this way, using theassumption that is connected, we deduce that u(x) �M throughout . So u � const.

102 CHAPTER 5. ELLIPTIC EQUATIONS

Bibliography[1] R. A. Adams: Sobolev Spaces, Academic Press, New York, 1975.[2] S. Agmon: Lectures on Elliptic Boundary Value Problems. Van Nostrand, 1965.[3] R. Courant and D. Hilbert: Methods of Mathematical Physics. Interscience Publishers,Vol. I, 1953, Vol. 2, 1962.[4] A. Friedman: Partial Di�erential Equations of Parabolic Type, Prentice Hall, 1964.[5] A. Friedman: Partial Di�erential Equations, Holt, Rinehart and Wiston, 1969.[6] P. R. Garabedian: Partial Di�erential Equations, John Wiley & Sons, Inc., 1964.[7] I. M. Gelfand and G. E. Shilov: Generalized Functions, Academic Press, New York,1964.[8] D. Gilbarg and N. . Trudinger: Elliptic Partial Di�erential Equations of Second Order.Springer-Verlag, Berlin, 1977.[9] L. H�ormander: Linear Partial Di�erential Operators, Springer-Verlag, Berlin, 1963.[10] F. John: Partial Di�erential Equations. Springer-Verlag, Berlin, 1982.[11] M. Krzyzanski: Partial Di�erential Equations of Second Order, Vol. 1&2, PWN,Warszawa, 1971.[12] O. A. Ladyzhenskaya: The Boundary Value Problems of Mathematical Physics.Springer-Verlag, New York-Berlin-Heidelberg-Tokyo 1985.[13] O. A. Lady�zenskaja, V. A. Solonnikov, N. N. Ural'ceva: Linear and Quasilinear Equa-tions of Parabolic Type. Amer. Math. Soc. 1968.[14] O. A. Ladyzhenskaya et. al.: Linear and Quasilinear Equations of Elliptic Type. Amer.Math. Soc..[15] J.{L. Lions and E. Magenes: Non-Homogeneous Boundary Value Problems and Appli-cations. Vol. I{III, Springer-Verlag, Berlin, 1972.[16] J. T. Marti: Introduction to Sobolev Spaces and Finite Element Solution of EllipticBoundary Value Problems. Academic Press, New York, 1986.[17] I. G. Petrovskii: Lectures on Partial Di�erential Equations. Interscience Publishers,1954. 103

104 BIBLIOGRAPHY[18] M. H. Protter and H. F. Weinberger: Maximum Principles in Di�erential Equations,Springer-Verlag, Berlin, 1984.[19] S. L. Sobolev: Partial Di�erential Equations of Mathematical Physics, Pergamon, NewYork, 1964.[20] W. A. Strauss: Partial Di�erential Equations. An Introduction. John Wiley, New York1992.[21] F. Treves: Basic Linear Partial Di�erential Equations, Academic Press, 1975.[22] A. N. Tykhonov, A. A. Samarski: Di�erentialgleichungen der Mathematischen Physik.VEB, Berlin, 1959.[23] V. S. Vladimirov: Equations of Mathematical Physics, Marcel Dekker, Inc., New York,1971.[24] V. S. Vladimirov: Generalized Functions in Mathematical Physics, Mir Publishers,Moscow, 1979.[25] D. Widder: The Heat Equation, Academic Press, New York ,1975.[26] J. Wloka: Partial Di�erential Equations. Cambridge Univ. Press, 1987.

IndexBoundary condition�rst kind, 30second kind, 30third kind, 30Cauchy data, 18, 23Cauchy problem, 18, 30, 33, 35Cauchy{Kowalevski theorem, 24characteristic, 18, 59characteristic curves, 4characteristic equation, 10, 33characteristic form, 19characteristic matrix, 19characteristic triangle, 34characteristics, 10Coordinate method, 3D'Alembert criterion, 71D'Alembert's formula, 34Darboux problem, 52di�erential equationsecond-order, 9Duhamel principle, 92eigenvalue problem, 40, 47elasticity tension, 30elastische Befestigung, 30energytotal, 31equationadjoint, 26elliptic, 12hyperbolic, 30canonical form, 11nonlinear, 18parabolic, 12quasi{linear, 18error function, 87, 90�eldanalytic, 26Fourier coe�cients, 50Fourier series, 43, 45Fourier's law, 7

functionharmonic, 93, 96real analytic, 22Gauss{Ostrogradskii formula, 7, 24Geometric Method, 2Goursat problem, 51gradient operator, 20gradient vector, 17Green formula, 48, 58Green function, 72, 78, 81, 82heat equation, 8, 63, 86heat ux, 63Helmholtz equation, 8Hook's law, 5hyperbolic equation, 44, 53ill{posedness, 35initia (Tr�agheitskr�afte), 6integration by parts, 24Lagrange{Green identity, 25Laplace equation, 8, 35, 93Laplace operator, 25Laplacian �, 94maximum principle, 65membrane, 7multi{index, 17Newton's law, 6of heat transfer, 64noncharacteristic, 18operator(formally) adjoint, 25partial di�erential, 17order, 1in�nite, 1parabolic equation, 63PDElinear, 1quasilinear, 1Picard problem, 52105

106 INDEXPlanck's constant, 8Poisson equation, 8, 93Poisson integral, 82, 88Poisson kernel, 99Poisson's formula, 99principal part, 19, 20problemill{posed, 35second boundary value, 65third boundary value, 65well{posed, 35regionlens{shaped, 26Riemann function, 60rotation, 94Schr�odinger's equation, 8solutionfundamental, 82standard form, 23successive approximations, 54support, 27symbol, 19, 20translation, 94Tricomi's equation, 13typeelliptic, 11, 13, 15hyperbolic, 11, 13, 15parabolic, 11, 13, 15ultrahyperbolic, 15vibrations of a string, 29wave equation, 20one-dimensional, 7Young modulus, 30