pdebook2

Partial Differential Equations Lecture Notes Erich Miersemann Department of Mathematics Leipzig University Version October, 2012 2 Contents 1 Introduction 9 1.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 1.2 Equations from variational problems . . . . . . . . . . . . . . 15 1.2.1 Ordinary differential equations . . . . . . . . . . . . . 15 1.2.2 Partial differential equations . . . . . . . . . . . . . . 16 1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2 Equations of first order 25 2.1 Linear equations . . . . . . . . . . . . . . . . . . . . . . . . . 25 2.2 Quasilinear equations . . . . . . . . . . . . . . . . . . . . . . 31 2.2.1 A linearization method . . . . . . . . . . . . . . . . . 32 2.2.2 Initial value problem of Cauchy . . . . . . . . . . . . . 33 2.3 Nonlinear equations in two variables . . . . . . . . . . . . . . 40 2.3.1 Initial value problem of Cauchy . . . . . . . . . . . . . 48 2.4 Nonlinear equations in Rn. . . . . . . . . . . . . . . . . . . . 51 2.5 Hamilton-Jacobi theory . . . . . . . . . . . . . . . . . . . . . 53 2.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 3 Classification 63 3.1 Linear equations of second order . . . . . . . . . . . . . . . . 63 3.1.1 Normal form in two variables . . . . . . . . . . . . . . 69 3.2 Quasilinear equations of second order . . . . . . . . . . . . . . 7 3 3.2.1 Quasilinear elliptic equations . . . . . . . . . . . . . . 73 3.3 Systems of first order . . . . . . . . . . . . . . . . . . . . . . . 74 3.3.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . 76 3.4 Systems of second order . . . . . . . . . . . . . . . . . . . . . 82 3.4.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . 83 3.5 Theorem of Cauchy-Kovalevskaya . . . . . . . . . . . . . . . . 84 3.5.1 Appendix: Real analytic functions . . . . . . . . . . . 90 3 4 CONTENTS 3.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 4 Hyperbolic equations 107 4.1 One-dimensional wave equation . . . . . . . . . . . . . . . . . 107 4.2 Higher dimensions . . . . . . . . . . . . . . . . . . . . . . . . 109 4.2.1 Case n=3 . . . . . . . . . . . . . . . . . . . . . . . . . 112 4.2.2 Case n= 2 . . . . . . . . . . . . . . . . . . . . . . . . 115 4.3 Inhomogeneous equation . . . . . . . . . . . . . . . . . . . . . 117 4.4 A method of Riemann . . . . . . . . . . . . . . . . . . . . . . 120 4.5 Initial-boundary value problems . . . . . . . . . . . . . . . . . 125 4.5.1 Oscillation of a string . . . . . . . . . . . . . . . . . . 125 4.5.2 Oscillation of a membrane . . . . . . . . . . . . . . . . 128 4.5.3 Inhomogeneous wave equations . . . . . . . . . . . . . 131 4.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 5 Fourier transform 141 5.1 Definition, properties . . . . . . . . . . . . . . . . . . . . . . . 141 5.1.1 Pseudodifferential operators . . . . . . . . . . . . . . . 146 5.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 6 Parabolic equations 151 6.1 Poisson’s formula . . . . . . . . . . . . . . . . . . . . . . . . . 152 6.2 Inhomogeneous heat equation . . . . . . . . . . . . . . . . . . 155 6.3 Maximum principle . . . . . . . . . . . . . . . . . . . . . . . . 156 6.4 Initial-boundary value problem . . . . . . . . . . . . . . . . . 162 6.4.1 Fourier’s method . . . . . . . . . . . . . . . . . . . . . 162 6.4.2 Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . 164 6.5 Black-Scholes equation . . . . . . . . . . . . . . . . . . . . . . 164 6.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 7 Elliptic equations of second order 175 7.1 Fundamental solution . . . . . . . . . . . . . . . . . . . . . . 175 7.2 Representation formula . . . . . . . . . . . . . . . . . . . . . 177 7.2.1 Conclusions from the representation formula . . . . . 17 9 7.3 Boundary value problems . . . . . . . . . . . . . . . . . . . . 181 7.3.1 Dirichlet problem . . . . . . . . . . . . . . . . . . . . . 181 7.3.2 Neumann problem . . . . . . . . . . . . . . . . . . . . 182 7.3.3 Mixed boundary value problem . . . . . . . . . . . . . 183 7.4 Green’s function for /triangle. . . . . . . . . . . . . . . . . . . . . . 183 7.4.1 Green’s function for a ball . . . . . . . . . . . . . . . . 186 CONTENTS 5 7.4.2 Green’s function and conformal mapping . . . . . . . 190 7.5 Inhomogeneous equation . . . . . . . . . . . . . . . . . . . . . 190 7.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 6 CONTENTS Preface These lecture notes are intented as a straightforward intro duction to partial differential equations which can serve as a textbook for under graduate and beginning graduate students. For additional reading we recommend following books: W. I. S mirnov [21], I. G. Petrowski [17], P. R. Garabedian [8], W. A. Strauss [23] , F. John [10], L. C. Evans [5] and R. Courant and D. Hilbert[4] and D. Gilbarg and N. S. Trudinger [9]. Some material of these lecture notes was take n from some of these books. 7 8 CONTENTS Chapter 1 Introduction Ordinary and partial differential equations occur in many app lications. An ordinary differential equation is a special case of a partial d ifferential equa- tion but the behaviour of solutions is quite different in gener al. It is much more complicated in the case of partial differential equation s caused by the fact that the functions for which we are looking at are functi ons of more than one independent variable. Equation F(x, y(x), y/prime(x), . . ., y(n)) = 0 is anordinary differential equation of n-th order for the unknown function y(x), where Fis given. An important problem for ordinary differential equations is t heinitial value problem y/prime(x) = f(x, y(x)) y(x0) = y0, where fis a given real function of two variables x, yandx0, y0are given real numbers. Picard-Lindel¨ of Theorem. Suppose (i)f(x, y)is continuous in a rectangle Q={(x, y)∈R2:|x−x0|< a,|y−y0|< b}. (ii) There is a constant Ksuch that |f(x, y)| ≤Kfor all (x, y)∈Q. (ii) Lipschitz condition: There is a constant Lsuch that |f(x, y2)−f(x, y1)| ≤L|y2−y1| 9 10 CHAPTER 1. INTRODUCTION xy xy 00 Figure 1.1: Initial value problem for all (x, y1),(x, y2). Then there exists a unique solution y∈C1(x0−α, x0+α)of the above initial value problem, where α= min( b/K, a ). The linear ordinary differential equation y(n)+an−1(x)y(n−1)+. . . a1(x)y/prime+a0(x)y= 0, where ajare continuous functions, has exactly nlinearly independent solu- tions. In contrast to this property the partial differential uxx+uyy= 0 in R2 has infinitely many linearly independent solutions in the lin ear space C2(R2). The ordinary differential equation of second order y/prime/prime(x) =f(x, y(x), y/prime(x)) has in general a family of solutions with two free parameters . Thus, it is naturally to consider the associated initial value problem y/prime/prime(x) = f(x, y(x), y/prime(x)) y(x0) = y0, y/prime(x0) =y1, where y0andy1are given, or to consider the boundary value problem y/prime/prime(x) = f(x, y(x), y/prime(x)) y(x0) = y0, y(x1) =y1. Initial and boundary value problems play an important role a lso in the theory of partial differential equations. A partial differential equation for 1.1. EXAMPLES 11 y y0 x xy1 0 x1 Figure 1.2: Boundary value problem the unknown function u(x, y) is for example F(x, y, u, u x, uy, uxx, uxy, uyy) = 0, where the function Fis given. This equation is of second order. An equation is said to be of n-th order if the highest derivative which occurs is of order n. An equation is said to be linear if the unknown function and its deriva- tives are linear in F. For example, a(x, y)ux+b(x, y)uy+c(x, y)u=f(x, y), where the functions a, b, c andfare given, is a linear equation of first order. An equation is said to be quasilinear if it is linear in the highest deriva- tives. For example, a(x, y, u, u x, uy)uxx+b(x, y, u, u x, uy)uxy+c(x, y, u, u x, uy)uyy= 0 is a quasilinear equation of second order. 1.1 Examples 1.uy= 0, where u=u(x, y). All functions u=w(x) are solutions. 2.ux=uy, where u=u(x, y). A change of coordinates transforms this equation into an equation of the first example. Set ξ=x+y, η=x−y, then u(x, y) =u/parenleftbiggξ+η 2,ξ−η 2/parenrightbigg =:v(ξ, η). 12 CHAPTER 1. INTRODUCTION Assume u∈C1, then vη=1 2(ux−uy). Ifux=uy, then vη= 0 and vice versa, thus v=w(ξ) are solutions for arbitrary C1-functions w(ξ). Consequently, we have a large class of solutions of the original partial differential equation: u=w(x+y) with an arbitrary C1-function w. 3.A necessary and sufficient condition such that for given C1-functions M, N the integral/integraldisplayP1 P0M(x, y)dx+N(x, y)dy is independent of the curve which connects the points P0withP1in a simply connected domain Ω ⊂R2is the partial differential equation (condition of integrability) My=Nx in Ω. y xΩ PP 01 Figure 1.3: Independence of the path This is one equation for two functions. A large class of solut ions is given byM= Φx, N= Φy, where Φ( x, y) is an arbitrary C2-function. It follows from Gauss theorem that these are all C1-solutions of the above differential equation. 4.Method of an integrating multiplier for an ordinary different ial equation. Consider the ordinary differential equation M(x, y)dx+N(x, y)dy= 0 1.1. EXAMPLES 13 for given C1-functions M, N . Then we seek a C1-function µ(x, y) such that µMdx +µNdy is a total differential, i. e., that ( µM)y= (µN)xis satisfied. This is a linear partial differential equation of first order for µ: Mµy−Nµx=µ(Nx−My). 5.TwoC1-functions u(x, y) andv(x, y) are said to be functionally dependent if det/parenleftbigguxuy vxvy/parenrightbigg = 0, which is a linear partial differential equation of first order fo ruifvis a given C1-function. A large class of solutions is given by u=H(v(x, y)), where His anarbitrary C1-function. 6.Cauchy-Riemann equations. Setf(z) =u(x, y)+iv(x, y), where z=x+iy andu, vare given C1(Ω)-functions. Here is Ω a domain in R2. If the function f(z) is differentiable with respect to the complex variable zthenu, vsatisfy the Cauchy-Riemann equations ux=vy, uy=−vx. It is known from the theory of functions of one complex variab le that the real part uand the imaginary part vof a differentiable function f(z) are solutions of the Laplace equation /triangleu= 0,/trianglev= 0, where /triangleu=uxx+uyy. 7.TheNewton potential u=1/radicalbig x2+y2+z2 is a solution of the Laplace equation in R3\(0,0,0), i. e., of uxx+uyy+uzz= 0. 14 CHAPTER 1. INTRODUCTION 8.Heat equation. Letu(x, t) be the temperature of a point x∈Ω at time t, where Ω ⊂R3is a domain. Then u(x, t) satisfies in Ω ×[0,∞) theheat equation ut=k/triangleu, where /triangleu=ux1x1+ux2x2+ux3x3andkis a positive constant. The condition u(x,0) =u0(x), x∈Ω, where u0(x) is given, is an initial condition associated to the above heat equation. The condition u(x, t) =h(x, t), x∈∂Ω, t≥0, where h(x, t) is given is a boundary condition for the heat equation. Ifh(x, t) =g(x), that is, his independent of t, then one expects that the solution u(x, t) tends to a function v(x) ift→ ∞. Moreover, it turns out thatvis the solution of the boundary value problem for the Laplace equation /trianglev= 0 in Ω v=g(x) on∂Ω. 9.Wave equation. The wave equation y u(x,t )u(x,t ) 12 xl Figure 1.4: Oscillating string utt=c2/triangleu, where u=u(x, t),cis a positive constant, describes oscillations of mem- branes or of three dimensional domains, for example. In the o ne-dimensional case utt=c2uxx describes oscillations of a string. 1.2. EQUATIONS FROM VARIATIONAL PROBLEMS 15 Associated initial conditions are u(x,0) =u0(x), ut(x,0) =u1(x), where u0, u1are given functions. Thus the initial position and the initi al velocity are prescribed. If the string is finite one describes additionally boundary conditions , for example u(0, t) = 0, u(l, t) = 0 for all t≥0. 1.2 Equations from variational problems A large class of ordinary and partial differential equations a rise from varia- tional problems. 1.2.1 Ordinary differential equations Set E(v) =/integraldisplayb af(x, v(x), v/prime(x))dx and for given ua, ub∈R V={v∈C2[a, b] :v(a) =ua, v(b) =ub}, where −∞< a < b < ∞andfis sufficiently regular. One of the basic problems in the calculus of variation is (P) min v∈VE(v). Euler equation. Letu∈Vbe a solution of (P), then d dxfu/prime(x, u(x), u/prime(x)) =fu(x, u(x), u/prime(x)) in(a, b). Proof. Exercise. Hints: For fixed φ∈C2[a, b] with φ(a) =φ(b) = 0 and real/epsilon1,|/epsilon1|< /epsilon10, setg(/epsilon1) =E(u+/epsilon1φ). Since g(0)≤g(/epsilon1) it follows g/prime(0) = 0. Integration by parts in the formula for g/prime(0) and the following basic lemma in the calculus of variations imply Euler’s equation. 16 CHAPTER 1. INTRODUCTION y y0y1 x a b Figure 1.5: Admissible variations Basic lemma in the calculus of variations. Leth∈C(a, b)and /integraldisplayb ah(x)φ(x)dx= 0 for all φ∈C1 0(a, b). Then h(x)≡0on(a, b). Proof. Assume h(x0)>0 for an x0∈(a, b), then there is a δ >0 such that (x0−δ, x0+δ)⊂(a, b) and h(x)≥h(x0)/2 on ( x0−δ, x0+δ). Set φ(x) =/braceleftbigg/parenleftbig δ2− |x−x0|2/parenrightbig2ifx∈(x0−δ, x0+δ) 0 if x∈(a, b)\[x0−δ, x0+δ]. Thus φ∈C1 0(a, b) and /integraldisplayb ah(x)φ(x)dx≥h(x0) 2/integraldisplayx0+δ x0−δφ(x)dx >0, which is a contradiction to the assumption of the lemma. 2 1.2.2 Partial differential equations The same procedure as above applied to the following multipl e integral leads to a second-order quasilinear partial differential equation. Set E(v) =/integraldisplay ΩF(x, v,∇v)dx, 1.2. EQUATIONS FROM VARIATIONAL PROBLEMS 17 where Ω ⊂Rnis a domain, x= (x1, . . ., x n),v=v(x) : Ω /mapsto→R, and ∇v= (vx1, . . ., v xn). Assume that the function Fis sufficiently regular in its arguments. For a given function h, defined on ∂Ω, set V={v∈C2(Ω) :v=hon∂Ω}. Euler equation. Letu∈Vbe a solution of (P), then n/summationdisplay i=1∂ ∂xiFuxi−Fu= 0 inΩ. Proof. Exercise. Hint: Extend the above fundamental lemma of the ca lculus of variations to the case of multiple integrals. The interva l (x0−δ, x0+δ) in the definition of φmust be replaced by a ball with center at x0and radius δ. Example: Dirichlet integral In two dimensions the Dirichlet integral is given by D(v) =/integraldisplay Ω/parenleftbig v2 x+v2 y/parenrightbig dxdy and the associated Euler equation is the Laplace equation /triangleu= 0 in Ω. Thus, there is natural relationship between the boundary va lue problem /triangleu= 0 in Ω , u=hon∂Ω and the variational problem min v∈VD(v). But these problems are not equivalent in general. It can happ en that the boundary value problem has a solution but the variational pr oblem has no solution, see for an example Courant and Hilbert [4], Vol. 1, p. 155, where his a continuous function and the associated solution uof the boundary value problem has no finite Dirichlet integral. The problems are equivalent, provided the given boundary va lue function his in the class H1/2(∂Ω), see Lions and Magenes [14]. 18 CHAPTER 1. INTRODUCTION Example: Minimal surface equation The non-parametric minimal surface problem in two dimension s is to find a minimizer u=u(x1, x2) of the problem min v∈V/integraldisplay Ω/radicalBig 1 +v2x1+v2x2dx, where for a given function hdefined on the boundary of the domain Ω V={v∈C1(Ω) :v=hon∂Ω}. S Ω Figure 1.6: Comparison surface Suppose that the minimizer satisfies the regularity assumpti onu∈C2(Ω), thenuis a solution of the minimal surface equation (Euler equation) in Ω ∂ ∂x1/parenleftBigg ux1/radicalbig 1 +|∇u|2/parenrightBigg +∂ ∂x2/parenleftBigg ux2/radicalbig 1 +|∇u|2/parenrightBigg = 0. (1.1) In fact, the additional assumption u∈C2(Ω) is superfluous since it follows from regularity considerations for quasilinear elliptic e quations of second order, see for example Gilbarg and Trudinger [9]. Let Ω = R2. Each linear function is a solution of the minimal surface equation (1.1). It was shown by Bernstein [2] that there are n o other solu- tions of the minimal surface quation. This is true also for hi gher dimensions 1.2. EQUATIONS FROM VARIATIONAL PROBLEMS 19 n≤7, see Simons [19]. If n≥8, then there exists also other solutions which define cones, see Bombieri, De Giorgi and Giusti [3]. The linearized minimal surface equation over u≡0 is the Laplace equa- tion/triangleu= 0. In R2linear functions are solutions but also many other functions in contrast to the minimal surface equation. This striking differ- ence is caused by the strong nonlinearity of the minimal surf ace equation. More general minimal surfaces are described by using parame tric rep- resentations. An example is shown in Figure 1.71. See [18], pp. 62, for example, for rotationally symmetric minimal surfaces. Figure 1.7: Rotationally symmetric minimal surface Neumann type boundary value problems SetV=C1(Ω) and E(v) =/integraldisplay ΩF(x, v,∇v)dx−/integraldisplay ∂Ωg(x, v)ds, where Fandgare given sufficiently regular functions and Ω ⊂Rnis a bounded and sufficiently regular domain. Assume uis a minimizer of E(v) inV, that is u∈V:E(u)≤E(v) for all v∈V, 1An experiment from Beutelspacher’s Mathematikum, Wissenschaft sjahr 2008, Leipzig 20 CHAPTER 1. INTRODUCTION then /integraldisplay Ω/parenleftbign/summationdisplay i=1Fuxi(x, u,∇u)φxi+Fu(x, u,∇u)φ/parenrightbig dx −/integraldisplay ∂Ωgu(x, u)φ ds= 0 for all φ∈C1(Ω). Assume additionally u∈C2(Ω), then uis a solution of the Neumann type boundary value problem n/summationdisplay i=1∂ ∂xiFuxi−Fu= 0 in Ω n/summationdisplay i=1Fuxiνi−gu= 0 on ∂Ω, where ν= (ν1, . . ., ν n) is the exterior unit normal at the boundary ∂Ω. This follows after integration by parts from the basic lemma of th e calculus of variations. Example: Laplace equation Set E(v) =1 2/integraldisplay Ω|∇v|2dx−/integraldisplay ∂Ωh(x)v ds, then the associated boundary value problem is /triangleu= 0 in Ω ∂u ∂ν=hon∂Ω. Example: Capillary equation Let Ω ⊂R2and set E(v) =/integraldisplay Ω/radicalbig 1 +|∇v|2dx+κ 2/integraldisplay Ωv2dx−cosγ/integraldisplay ∂Ωv ds. Hereκis a positive constant (capillarity constant) and γis the (constant) boundary contact angle, i. e., the angle between the contain er wall and 1.2. EQUATIONS FROM VARIATIONAL PROBLEMS 21 the capillary surface, defined by v=v(x1, x2), at the boundary. Then the related boundary value problem is div(Tu) = κuin Ω ν·Tu= cos γon∂Ω, where we use the abbreviation Tu=∇u/radicalbig 1 +|∇u|2, div (Tu) is the left hand side of the minimal surface equation (1.1) an d it is twice the mean curvature of the surface defined by z=u(x1, x2), see an exercise. The above problem describes the ascent of a liquid, water for example, in a vertical cylinder with cross section Ω. Assume the gravit y is directed downwards in the direction of the negative x3-axis. Figure 1.8 shows that liquid can rise along a vertical wedge which is a consequence of the strong nonlinearity of the underlying equations, see Finn [7]. Thi s photo was taken Figure 1.8: Ascent of liquid in a wedge from [15]. 22 CHAPTER 1. INTRODUCTION 1.3 Exercises 1. Find nontrivial solutions uof uxy−uyx= 0. 2. Prove: In the linear space C2(R2) there are infinitely many linearly independent solutions of /triangleu= 0 in R2. Hint: Real and imaginary part of holomorphic functions are soluti ons of the Laplace equation. 3. Find all radially symmetric functions which satisfy the L aplace equa- tion in Rn\{0}forn≥2. A function uis said to be radially symmetric ifu(x) =f(r), where r= (/summationtextn ix2 i)1/2. Hint: Show that a radially symmetric usatisfies /triangleu=r1−n/parenleftbig rn−1f/prime/parenrightbig/prime by using ∇u(x) =f/prime(r)x r. 4. Prove the basic lemma in the calculus of variations: Let Ω ⊂Rnbe a domain and f∈C(Ω) such that /integraldisplay Ωf(x)h(x)dx= 0 for all h∈C2 0(Ω). Then f≡0 in Ω. 5. Write the minimal surface equation (1.1) as a quasilinear equation of second order. 6. Prove that a sufficiently regular minimizer in C1(Ω) of E(v) =/integraldisplay ΩF(x, v,∇v)dx−/integraldisplay ∂Ωg(v, v)ds, is a solution of the boundary value problem n/summationdisplay i=1∂ ∂xiFuxi−Fu= 0 in Ω n/summationdisplay i=1Fuxiνi−gu= 0 on ∂Ω, where ν= (ν1, . . ., ν n) is the exterior unit normal at the boundary ∂Ω. 1.3. EXERCISES 23 7. Prove that ν·Tu= cos γon∂Ω, where γis the angle between the container wall, which is here a cylinder, and the surface S, defined by z=u(x1, x2), at the boundary of S,νis the exterior normal at ∂Ω. Hint: The angle between two surfaces is by definition the angle betwe en the two associated normals at the intersection of the surfac es. 8. Let Ω be bounded and assume u∈C2(Ω) is a solution of div Tu =Cin Ω ν·∇u/radicalbig 1 +|∇u|2= cos γon∂Ω, where Cis a constant. Prove that C=|∂Ω| |Ω|cosγ . Hint: Integrate the differential equation over Ω. 9. Assume Ω = BR(0) is a disc with radius Rand the center at the origin. Show that radially symmetric solutions u(x) =w(r),r=/radicalbig x2 1+x2 2, of the capillary boundary value problem are solutions of /parenleftbiggrw/prime √ 1 +w/prime2/parenrightbigg/prime =κrwin 0< r < R w/prime √ 1 +w/prime2= cos γifr=R. Remark. It follows from a maximum principle of Concus and Finn [7] that a solution of the capillary equation over a disc must be r adially symmetric. 10. Find all radially symmetric solutions of /parenleftbiggrw/prime √ 1 +w/prime2/parenrightbigg/prime =Crin 0< r < R w/prime √ 1 +w/prime2= cos γifr=R. Hint: From an exercise above it follows that C=2 Rcosγ. 24 CHAPTER 1. INTRODUCTION 11. Show that d iv Tu is twice the mean curvature of the surface defined byz=u(x1, x2). Chapter 2 Equations of first order For a given sufficiently regular function Fthe general equation of first order for the unknown function u(x) is F(x, u,∇u) = 0 in Ω∈Rn. The main tool for studying related problems is the theory of ordinary differential equations. This is quite different for sy stems of partial differential of first order. The general linear partial differential equation of first order can be writ- ten asn/summationdisplay i=1ai(x)uxi+c(x)u=f(x) for given functions ai, candf. The general quasilinear partial differential equation of first order is n/summationdisplay i=1ai(x, u)uxi+c(x, u) = 0. 2.1 Linear equations Let us begin with the linear homogeneous equation a1(x, y)ux+a2(x, y)uy= 0. (2.1) Assume there is a C1-solution z=u(x, y). This function defines a surface Swhich has at P= (x, y, u (x, y)) the normal N=1/radicalbig 1 +|∇u|2(−ux,−uy,1) 25 26 CHAPTER 2. EQUATIONS OF FIRST ORDER and the tangential plane defined by ζ−z=ux(x, y)(ξ−x) +uy(x, y)(η−y). Setp=ux(x, y),q=uy(x, y) and z=u(x, y). The tuple ( x, y, z, p, q ) is called surface element and the tuple ( x, y, z)support of the surface element. The tangential plane is defined by the surface element. On the o ther hand, differential equation (2.1) a1(x, y)p+a2(x, y)q= 0 defines at each support ( x, y, z) a bundle of planes if we consider all ( p, q) sat- isfying this equation. For fixed ( x, y), this family of planes Π( λ) = Π( λ;x, y) is defined by a one parameter family of ascents p(λ) =p(λ;x, y),q(λ) = q(λ;x, y). The envelope of these planes is a line since a1(x, y)p(λ) +a2(x, y)q(λ) = 0, which implies that the normal N(λ) on Π( λ) is perpendicular on ( a1, a2,0). Consider a curve x(τ) = (x(τ), y(τ), z(τ)) onS, letTx0be the tangential plane at x0= (x(τ0), y(τ0), z(τ0)) ofSand consider on Tx0the line L:l(σ) =x0+σx/prime(τ0), σ∈R, see Figure 2.1. We assume Lcoincides with the envelope, which is a line here, of the family of planes Π( λ) at (x, y, z). Assume that Tx0= Π(λ0) and consider two planes Π(λ0) :z−z0= (x−x0)p(λ0) + (y−y0)q(λ0) Π(λ0+h) :z−z0= (x−x0)p(λ0+h) + (y−y0)q(λ0+h). At the intersection l(σ) we have (x−x0)p(λ0) + (y−y0)q(λ0) = (x−x0)p(λ0+h) + (y−y0)q(λ0+h). Thus, x/prime(τ0)p/prime(λ0) +y/prime(τ0)q/prime(λ0) = 0. From the differential equation a1(x(τ0), y(τ0))p(λ) +a2(x(τ0), y(τ0))q(λ) = 0 2.1. LINEAR EQUATIONS 27 yz xL SΠ( λ 0) Figure 2.1: Curve on a surface it follows a1p/prime(λ0) +a2q/prime(λ0) = 0. Consequently (x/prime(τ), y/prime(τ)) =x/prime(τ) a1(x(τ, y(τ))(a1(x(τ), y(τ)), a2(x(τ), y(τ)), since τ0was an arbitrary parameter. Here we assume that x/prime(τ)/negationslash= 0 and a1(x(τ), y(τ))/negationslash= 0. Then we introduce a new parameter tby the inverse of τ=τ(t), where t(τ) =/integraldisplayτ τ0x/prime(s) a1(x(s), y(s))ds. It follows x/prime(t) =a1(x, y), y/prime(t) =a2(x, y). We denote x(τ(t)) byx(t) again. Now we consider the initial value problem x/prime(t) =a1(x, y), y/prime(t) =a2(x, y), x(0) = x0, y(0) = y0.(2.2) From the theory of ordinary differential equations it follows (Theorem of Picard-Lindel¨ of) that there is a unique solution in a neighb ouhood of t= 0 provided the functions a1, a2are in C1. From this definition of the curves 28 CHAPTER 2. EQUATIONS OF FIRST ORDER (x(t), y(t)) is follows that the field of directions ( a1(x0, y0), a2(x0, y0)) defines the slope of these curves at ( x(0), y(0)). Definition. The differential equations in (2.2) are called characteristic equations or characteristic system and solutions of the associated in itial value problem are called characteristic curves . Definition. A function φ(x, y) is said to be an integral of the characteristic system if φ(x(t), y(t)) =const. for each characteristic curve. The constant depends on the characteristic curve considered. Proposition 2.1. Assume φ∈C1is an integral, then u=φ(x, y)is a solution of (2.1). Proof. Consider for given ( x0, y0) the above initial value problem (2.2). Since φ(x(t), y(t)) =const. it follows φxx/prime+φyy/prime= 0 for|t|< t0,t0>0 and sufficiently small. Thus φx(x0, y0)a1(x0, y0) +φy(x0, y0)a2(x0, y0) = 0. 2 Remark. Ifφ(x, y) is a solution of equation (2.1) then also H(φ(x, y)), where H(s) is a given C1-function. Examples 1.Consider a1ux+a2uy= 0, where a1, a2are constants. The system of characteristic equations is x/prime=a1, y/prime=a2. Thus the characteristic curves are parallel straight lines defined by x=a1t+A, y=a2t+B, 2.1. LINEAR EQUATIONS 29 where A, B are arbitrary constants. From these equations it follows th at φ(x, y) :=a2x−a1y is constant along each characteristic curve. Consequently , see Proposi- tion 2.1, u=a2x−a1yis a solution of the differential equation. From an exercise it follows that u=H(a2x−a1y), (2.3) where H(s) is an arbitrary C1-function, is also a solution. Since uis constant when a2x−a1yis constant, equation (2.3) defines cylinder surfaces which are generated by parallel straight lines which are parallel to the ( x, y)-plane, see Figure 2.2. yz x Figure 2.2: Cylinder surfaces 2.Consider the differential equation xux+yuy= 0. The characteristic equations are x/prime=x, y/prime=y, 30 CHAPTER 2. EQUATIONS OF FIRST ORDER and the characteristic curves are given by x=Aet, y=Bet, where A, B are arbitrary constants. Thus, an integral is y/x,x/negationslash= 0, and for a given C1-function the function u=H(x/y) is a solution of the differential equation. If y/x=const. , then uis constant. Suppose that H/prime(s)>0, for example, then udefines right helicoids (in German: Wendelfl¨ achen), see Figure 2.3 Figure 2.3: Right helicoid, a2< x2+y2< R2(Museo Ideale Leonardo da Vinci, Italy) 3.Consider the differential equation yux−xuy= 0. The associated characteristic system is x/prime=y, y/prime=−x. If follows x/primex+yy/prime= 0, 2.2. QUASILINEAR EQUATIONS 31 or, equivalently, d dt(x2+y2) = 0, which implies that x2+y2=const. along each characteristic. Thus, rota- tionally symmetric surfaces defined by u=H(x2+y2), where H/prime/negationslash= 0, are solutions of the differential equation. 4.The associated characteristic equations to ayux+bxuy= 0, where a, bare positive constants, are given by x/prime=ay, y/prime=bx. It follows bxx/prime−ayy/prime= 0, or equivalently, d dt(bx2−ay2) = 0. Solutions of the differential equation are u=H(bx2−ay2), which define surfaces which have a hyperbola as the intersection with pla nes parallel to the (x, y)-plane. Here H(s) is an arbitrary C1-function, H/prime(s)/negationslash= 0. 2.2 Quasilinear equations Here we consider the equation a1(x, y, u )ux+a2(x, y, u )uy=a3(x, y, u ). (2.4) The inhomogeneous linear equation a1(x, y)ux+a2(x, y)uy=a3(x, y) is a special case of (2.4). One arrives at characteristic equations x/prime=a1, y/prime=a2, z/prime=a3 from (2.4) by the same arguments as in the case of homogeneous linear equations in two variables. The additional equation z/prime=a3follows from z/prime(τ) = p(λ)x/prime(τ) +q(λ)y/prime(τ) =pa1+qa2 =a3, see also Section 2.3, where the general case of nonlinear equ ations in two variables is considered. 32 CHAPTER 2. EQUATIONS OF FIRST ORDER 2.2.1 A linearization method We can transform the inhomogeneous equation (2.4) into a hom ogeneous linear equation for an unknown function of three variables b y the following trick. We are looking for a function ψ(x, y, u ) such that the solution u=u(x, y) of (2.4) is defined implicitly by ψ(x, y, u ) =const. Assume there is such a function ψand let ube a solution of (2.4), then ψx+ψuux= 0, ψy+ψuuy= 0. Assume ψu/negationslash= 0, then ux=−ψx ψu, uy=−ψy ψu. From (2.4) we obtain a1(x, y, z)ψx+a2(x, y, z)ψy+a3(x, y, z)ψz= 0, (2.5) where z:=u. We consider the associated system of characteristic equati ons x/prime(t) = a1(x, y, z) y/prime(t) = a2(x, y, z) z/prime(t) = a3(x, y, z). One arrives at this system by the same arguments as in the two-d imensional case above. Proposition 2.2. (i)Assume w∈C1,w=w(x, y, z), is an integral, i. e., it is constant along each fixed solution of (2.5), then ψ=w(x, y, z)is a solution of (2.5). (ii)The function z=u(x, y), implicitly defined through ψ(x, u, z ) =const., is a solution of (2.4), provided that ψz/negationslash= 0. (iii)Letz=u(x, y)be a solution of (2.4) and let (x(t), y(t))be a solution of x/prime(t) =a1(x, y, u (x, y)), y/prime(t) =a2(x, y, u (x, y)), thenz(t) :=u(x(t), y(t))satisfies the third of the above characteristic equa- tions. Proof. Exercise. 2.2. QUASILINEAR EQUATIONS 33 2.2.2 Initial value problem of Cauchy Consider again the quasilinear equation (⋆) a1(x, y, u )ux+a2(x, y, u )uy=a3(x, y, u ). Let Γ :x=x0(s), y=y0(s), z=z0(s), s1≤s≤s2,−∞< s1< s2<+∞ be a regular curve in R3and denote by Cthe orthogonal projection of Γ onto the ( x, y)-plane, i. e., C:x=x0(s), y=y0(s). Initial value problem of Cauchy: Find a C1-solution u=u(x, y)of (⋆)such that u(x0(s), y0(s)) =z0(s), i. e., we seek a surface Sdefined by z=u(x, y)which contains the curve Γ. yz xCΓ Figure 2.4: Cauchy initial value problem Definition. The curve Γ is said to be noncharacteristic if x/prime 0(s)a2(x0(s), y0(s))−y/prime 0(s)a1(x0(s), y0(s))/negationslash= 0. Theorem 2.1. Assume a1, a2, a2∈C1in their arguments, the initial data x0, y0, z0∈C1[s1, s2]andΓis noncharacteristic. 34 CHAPTER 2. EQUATIONS OF FIRST ORDER Then there is a neighbourhood of Csuch that there exists exactly one solution uof the Cauchy initial value problem. Proof. (i) Existence. Consider the following initial value proble m for the system of characteristic equations to ( ⋆): x/prime(t) = a1(x, y, z) y/prime(t) = a2(x, y, z) z/prime(t) = a3(x, y, z) with the initial conditions x(s,0) = x0(s) y(s,0) = y0(s) z(s,0) = z0(s). Letx=x(s, t),y=y(s, t),z=z(s, t) be the solution, s1≤s≤s2,|t|< η for an η >0. We will show that this set of strings sticked onto the curve Γ, see Figure 2.4, defines a surface. To show this, we consider th e inverse functions s=s(x, y),t=t(x, y) ofx=x(s, t),y=y(s, t) and show that z(s(x, y), t(x, y)) is a solution of the initial problem of Cauchy. The inverse functions sandtexist in a neighbourhood of t= 0 since det∂(x, y) ∂(s, t)/vextendsingle/vextendsingle/vextendsingle t=0=/vextendsingle/vextendsingle/vextendsingle/vextendsinglexsxt ysyt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0=x/prime 0(s)a2−y/prime 0(s)a1/negationslash= 0, and the initial curve Γ is noncharacteristic by assumption. Set u(x, y) :=z(s(x, y), t(x, y)), thenusatisfies the initial condition since u(x, y)|t=0=z(s,0) =z0(s). The following calculation shows that uis also a solution of the differential equation ( ⋆). a1ux+a2uy=a1(zssx+zttx) +a2(zssy+ztty) =zs(a1sx+a2sy) +zt(a1tx+a2ty) =zs(sxxt+syyt) +zt(txxt+tyyt) =a3 2.2. QUASILINEAR EQUATIONS 35 since 0 = st=sxxt+syytand 1 = tt=txxt+tyyt. (ii) Uniqueness. Suppose that v(x, y) is a second solution. Consider a point (x/prime, y/prime) in a neighbourhood of the curve ( x0(s), y(s)),s1−/epsilon1≤s≤s2+/epsilon1, /epsilon1 >0 small. The inverse parameters are s/prime=s(x/prime, y/prime),t/prime=t(x/prime, y/prime), see Figure 2.5. xy (x (s'),y (s'))0 0(x',y') Figure 2.5: Uniqueness proof Let A:x(t) :=x(s/prime, t), y(t) :=y(s/prime, t), z(t) :=z(s/prime, t) be the solution of the above initial value problem for the cha racteristic dif- ferential equations with the initial data x(s/prime,0) =x0(s/prime), y(s/prime,0) =y0(s/prime), z(s/prime,0) =z0(s/prime). According to its construction this curve is on the surface Sdefined by u= u(x, y) and u(x/prime, y/prime) =z(s/prime, t/prime). Set ψ(t) :=v(x(t), y(t))−z(t), then ψ/prime(t) = vxx/prime+vyy/prime−z/prime =xxa1+vya2−a3= 0 and ψ(0) = v(x(s/prime,0), y(s/prime,0))−z(s/prime,0) = 0 sincevis a solution of the differential equation and satisfies the init ial con- dition by assumption. Thus, ψ(t)≡0, i. e., v(x(s/prime, t), y(s/prime, t))−z(s/prime, t) = 0. 36 CHAPTER 2. EQUATIONS OF FIRST ORDER Sett=t/prime, then v(x/prime, y/prime)−z(s/prime, t/prime) = 0, which shows that v(x/prime, y/prime) =u(x/prime, y/prime) because of z(s/prime, t/prime) =u(x/prime, y/prime). 2 Remark. In general, there is no uniqueness if the initial curve Γ is a characteristic curve, see an exercise and Figure 2.6 which i llustrates this case. yz xu vSS Figure 2.6: Multiple solutions Examples 1.Consider the Cauchy initial value problem ux+uy= 0 with the initial data x0(s) =s, y0(s) = 1, z0(s) is a given C1-function . These initial data are noncharacteristic since y/prime 0a1−x/prime 0a2=−1. The solution of the associated system of characteristic equations x/prime(t) = 1, y/prime(t) = 1, u/prime(t) = 0 2.2. QUASILINEAR EQUATIONS 37 with the initial conditions x(s,0) =x0(s), y(s,0) =y0(s), z(s,0) =z0(s) is given by x=t+x0(s), y=t+y0(s), z=z0(s), i. e., x=t+s, y=t+ 1, z=z0(s). It follows s=x−y+ 1, t=y−1 and that u=z0(x−y+ 1) is the solution of the Cauchy initial value problem. 2.A problem from kinetics in chemistry. Consider for x≥0,y≥0 the problem ux+uy=/parenleftBig k0e−k1x+k2/parenrightBig (1−u) with initial data u(x,0) = 0 , x > 0,andu(0, y) =u0(y), y > 0. Here the constants kjare positive, these constants define the velocity of the reactions in consideration, and the function u0(y) is given. The variable x is the time and yis the hight of a tube, for example, in which the chemical reaction takes place, and uis the concentration of the chemical substance. In contrast to our previous assumptions, the initial data ar e not in C1. The projection C1∪ C2of the initial curve onto the ( x, y)-plane has a corner at the origin, see Figure 2.7. xy x=y ΩΩ2 1 CC 12 Figure 2.7: Domains to the chemical kinetics example 38 CHAPTER 2. EQUATIONS OF FIRST ORDER The associated system of characteristic equations is x/prime(t) = 1, y/prime(t) = 1, z/prime(t) =/parenleftBig k0e−k1x+k2/parenrightBig (1−z). It follows x=t+c1,y=t+c2with constants cj. Thus the projection of the characteristic curves on the ( x, y)-plane are straight lines parallel to y=x. We will solve the initial value problems in the domains Ω 1and Ω 2, see Figure 2.7, separately. (i)The initial value problem in Ω1.The initial data are x0(s) =s, y0(s) = 0, z0(0) = 0 , s≥0. It follows x=x(s, t) =t+s, y=y(s, t) =t. Thus z/prime(t) = (k0e−k1(t+s)+k2)(1−z), z(0) = 0 . The solution of this initial value problem is given by z(s, t) = 1−exp/parenleftbiggk0 k1e−k1(s+t)−k2t−k0 k1e−k1s/parenrightbigg . Consequently u1(x, y) = 1−exp/parenleftbiggk0 k1e−k1x−k2y−k0k1e−k1(x−y)/parenrightbigg is the solution of the Cauchy initial value problem in Ω 1. If time xtends to ∞, we get the limit lim x→∞u1(x, y) = 1−e−k2y. (ii)The initial value problem in Ω2.The initial data are here x0(s) = 0, y0(s) =s, z0(0) = u0(s), s≥0. It follows x=x(s, t) =t, y=y(s, t) =t+s. Thus z/prime(t) = (k0e−k1t+k2)(1−z), z(0) = 0 . 2.2. QUASILINEAR EQUATIONS 39 The solution of this initial value problem is given by z(s, t) = 1−(1−u0(s))exp/parenleftbiggk0 k1e−k1t−k2t−k0 k1/parenrightbigg . Consequently u2(x, y) = 1−(1−u0(y−x))exp/parenleftbiggk0 k1e−k1x−k2x−k0 k1/parenrightbigg is the solution in Ω 2. Ifx=y, then u1(x, y) = 1 −exp/parenleftbiggk0 k1e−k1x−k2x−k0 k1/parenrightbigg u2(x, y) = 1 −(1−u0(0))exp/parenleftbiggk0 k1e−k1x−k2x−k0 k1/parenrightbigg . Ifu0(0)>0, then u1< u2ifx=y, i. e., there is a jump of the concentration of the substrate along its burning front defined by x=y. Remark. Such a problem with discontinuous initial data is called Riemann problem . See an exercise for another Riemann problem. The case that a solution of the equation is known Here we will see that we get immediately a solution of the Cauc hy initial value problem if a solution of the homogeneous linear equation a1(x, y)ux+a2(x, y)uy= 0 is known. Let x0(s), y0(s), z0(s), s1< s < s 2 be the initial data and let u=φ(x, y) be a solution of the differential equa- tion. We assume that φx(x0(s), y0(s))x/prime 0(s) +φy(x0(s), y0(s))y/prime 0(s)/negationslash= 0 is satisfied. Set g(s) =φ(x0(s), y0(s)) and let s=h(g) be the inverse function. The solution of the Cauchy initial problem is given by u0(h(φ(x, y))). 40 CHAPTER 2. EQUATIONS OF FIRST ORDER This follows since in the problem considered a composition o f a solution is a solution again, see an exercise, and since u0(h(φ(x0(s), y0(s))) = u0(h(g)) =u0(s). Example: Consider equation ux+uy= 0 with initial data x0(s) =s, y0(s) = 1, u0(s) is a given function . A solution of the differential equation is φ(x, y) =x−y. Thus φ((x0(s), y0(s)) =s−1 and u0(φ+ 1) = u0(x−y+ 1) is the solution of the problem. 2.3 Nonlinear equations in two variables Here we consider equation F(x, y, z, p, q ) = 0, (2.6) where z=u(x, y),p=ux(x, y),q=uy(x, y) and F∈C2is given such that F2 p+F2 q/negationslash= 0. In contrast to the quasilinear case, this general nonlinear equation is more complicated. Together with (2.6) we will consider the f ollowing system of ordinary equations which follow from considerations bel ow as necessary conditions, in particular from the assumption that there is a solution of (2.6). x/prime(t) = Fp (2.7) y/prime(t) = Fq (2.8) z/prime(t) = pFp+qFq (2.9) p/prime(t) = −Fx−Fup (2.10) q/prime(t) = −Fy−Fuq. (2.11) 2.3. NONLINEAR EQUATIONS IN TWO VARIABLES 41 Definition. Equations (2.7)–(2.11) are said to be characteristic equations of equation (2.6) and a solution (x(t), y(t), z(t), p(t), q(t)) of the characteristic equations is called characteristic strip orMonge curve . Figure 2.8: Gaspard Monge (Panth´ eon, Paris) We will see, as in the quasilinear case, that the strips defined by the char- acteristic equations build the solution surface of the Cauc hy initial value problem. Letz=u(x, y) be a solution of the general nonlinear differential equa- tion (2.6). Let (x0, y0, z0) be fixed, then equation (2.6) defines a set of planes given by (x0, y0, z0, p, q), i. e., planes given by z=v(x, y) which contain the point (x0, y0, z0) and for which vx=p,vy=qat (x0, y0). In the case of quasilinear equations these set of planes is a bundle of planes which all c ontain a fixed straight line, see Section 2.1. In the general case of this se ction the situation is more complicated. Consider the example p2+q2=f(x, y, z), (2.12) 42 CHAPTER 2. EQUATIONS OF FIRST ORDER where fis a given positive function. Let Ebe a plane defined by z=v(x, y) and which contains ( x0, y0, z0). Then the normal on the plane Edirected downward is N=1/radicalbig 1 +|∇v|2(p, q,−1), where p=vx(x0, y0),q=vy(x0, y0). It follows from (2.12) that the normal Nmakes a constant angle with the z-axis, and the z-coordinate of Nis constant, see Figure 2.9. yz xNΠ(λ) (λ) Figure 2.9: Monge cone in an example Thus the endpoints of the normals fixed at ( x0, y0, z0) define a circle parallel to the ( x, y)-plane, i. e., there is a cone which is the envelope of all these planes. We assume that the general equation (2.6) defines such a Monge c one at each point in R3. Then we seek a surface Swhich touches at each point its Monge cone, see Figure 2.10. More precisely, we assume there exists, as in the above examp le, a one parameter C1-family p(λ) =p(λ;x, y, z), q(λ) =q(λ;x, y, z) of solutions of (2.6). These ( p(λ), q(λ)) define a family Π( λ) of planes. 2.3. NONLINEAR EQUATIONS IN TWO VARIABLES 43 yz x Figure 2.10: Monge cones Let x(τ) = (x(τ), y(τ), z(τ)) be a curve on the surface Swhich touches at each point its Monge cone, see Figure 2.11. Thus we assume that at each point of the surfa ceSthe associated tangent plane coincides with a plane from the fam ily Π( λ) at this point. Consider the tangential plane Tx0of the surface Satx0= (x(τ0), y(τ0), z(τ0)). The straight line l(σ) =x0+σx/prime(τ0),−∞< σ < ∞, is an apothem (in German: Mantellinie) of the cone by assumpt ion and is contained in the tangential plane Tx0as the tangent of a curve on the surface S. It is defined through x/prime(τ0) =l/prime(σ). (2.13) The straight line l(σ) satisfies l3(σ)−z0= (l1(σ)−x0)p(λ0) + (l2(σ)−y0)q(λ0), since it is contained in the tangential plane Tx0defined by the slope ( p, q). It follows l/prime 3(σ) =p(λ0)l/prime 1(σ) +q(λ0)l/prime 2(σ). 44 CHAPTER 2. EQUATIONS OF FIRST ORDER yz xSTx0 Figure 2.11: Monge cones along a curve on the surface Together with (2.13) we obtain z/prime(τ) =p(λ0)x/prime(τ) +q(λ0)y/prime(τ). (2.14) The above straight line lis the limit of the intersection line of two neigh- bouring planes which envelopes the Monge cone: z−z0= (x−x0)p(λ0) + (y−y0)q(λ0) z−z0= (x−x0)p(λ0+h) + (y−y0)q(λ0+h). On the intersection one has (x−x0)p(λ) + (y−y0)q(λ0) = (x−x0)p(λ0+h) + (y−y0)q(λ0+h). Leth→0, it follows (x−x0)p/prime(λ0) + (y−y0)q/prime(λ0) = 0. Since x=l1(σ),y=l2(σ) in this limit position, we have p/prime(λ0)l/prime 1(σ) +q/prime(λ0)l/prime 2(σ) = 0, and it follows from (2.13) that p/prime(λ0)x/prime(τ) +q/prime(λ0)y/prime(τ) = 0. (2.15) 2.3. NONLINEAR EQUATIONS IN TWO VARIABLES 45 From the differential equation F(x0, y0, z0, p(λ), q(λ)) = 0 we see that Fpp/prime(λ) +Fqq/prime(λ) = 0. (2.16) Assume x/prime(τ0)/negationslash= 0 and Fp/negationslash= 0, then we obtain from (2.15), (2.16) y/prime(τ0) x/prime(τ0)=Fq Fp, and from (2.14) (2.16) that z/prime(τ0) x/prime(τ0)=p+qFq Fp. It follows, since τ0was an arbitrary fixed parameter, x/prime(τ) = ( x/prime(τ), y/prime(τ), z/prime(τ)) =/parenleftbigg x/prime(τ), x/prime(τ)Fq Fp, x/prime(τ)/parenleftbigg p+qFq Fp/parenrightbigg/parenrightbigg =x/prime(τ) Fp(Fp, Fq, pFp+qFq), i. e., the tangential vector x/prime(τ) is proportional to ( Fp, Fq, pFp+qFq). Set a(τ) =x/prime(τ) Fp, where F=F(x(τ), y(τ), z(τ), p(λ(τ)), q(λ(τ))). Introducing the new pa- rameter tby the inverse of τ=τ(t), where t(τ) =/integraldisplayτ τ0a(s)ds, we obtain the characteristic equations (2.7)–(2.9). Here w e denote x(τ(t)) byx(t) again. From the differential equation (2.6) and from (2.7)–( 2.9) we get equations (2.10) and (2.11). Assume the surface z=u(x, y) under consideration is in C2, then Fx+Fzp+Fppx+Fqpy= 0,(qx=py) Fx+Fzp+x/prime(t)px+y/prime(t)py= 0 Fx+Fzp+p/prime(t) = 0 46 CHAPTER 2. EQUATIONS OF FIRST ORDER sincep=p(x, y) =p(x(t), y(t)) on the curve x(t). Thus equation (2.10) of the characteristic system is shown. Differentiating the differ ential equation (2.6) with respect to y, we get finally equation (2.11). Remark. In the previous quasilinear case F(x, y, z, p, q ) =a1(x, y, z)p+a2(x, y, z)q−a3(x, y, z) the first three characteristic equations are the same: x/prime(t) =a1(x, y, z), y/prime(t) =a2(x, y, z), z/prime(t) =a3(x, y, z). The point is that the right hand sides are independent on porq. It follows from Theorem 2.1 that there exists a solution of the Cauchy in itial value problem provided the initial data are noncharacteristic. T hat is, we do not need the other remaining two characteristic equations. The other two equations (2.10) and (2.11) are satisfied in this quasilin- ear case automatically if there is a solution of the equation , see the above derivation of these equations. The geometric meaning of the first three characteristic differe ntial equa- tions (2.7)–(2.11) is the following one. Each point of the cu rve A: (x(t), y(t), z(t)) corresponds a tangential plane with the normal direc- tion (−p,−q,1) such that z/prime(t) =p(t)x/prime(t) +q(t)y/prime(t). This equation is called strip condition . On the other hand, let z=u(x, y) defines a surface, then z(t) :=u(x(t), y(t)) satisfies the strip condition, where p=uxandq=uy, that is, the ”scales” defined by the normals fit together. Proposition 2.3. F(x, y, z, p, q )is an integral, i. e., it is constant along each characteristic curve. Proof. d dtF(x(t), y(t), z(t), p(t), q(t)) = Fxx/prime+Fyy/prime+Fzz/prime+Fpp/prime+Fqq/prime =FxFp+FyFq+pFzFp+qFzFq −Fpfx−FpFzp−FqFy−FqFzq = 0. 2.3. NONLINEAR EQUATIONS IN TWO VARIABLES 47 2 Corollary. Assume F(x0, y0, z0, p0, q0) = 0, then F= 0 along characteristic curves with the initial data ( x0, y0, z0, p0, q0). Proposition 2.4. Letz=u(x, y),u∈C2, be a solution of the nonlinear equation (2.6). Set z0=u(x0, y0,)p0=ux(x0, y0), q0=uy(x0, y0). Then the associated characteristic strip is in the surface S, defined by z= u(x, y). Thus z(t) = u(x(t), y(t)) p(t) = ux(x(t), y(t)) q(t) = uy(x(t), y(t)), where (x(t), y(t), z(t), p(t), q(t))is the solution of the characteristic system (2.7)–(2.11) with initial data (x0, y0, z0, p0, q0) Proof. Consider the initial value problem x/prime(t) = Fp(x, y, u (x, y), ux(x, y), uy(x, y)) y/prime(t) = Fq(x, y, u (x, y), ux(x, y), uy(x, y)) with the initial data x(0) = x0,y(0) = y0. We will show that (x(t), y(t), u(x(t), y(t)), ux(x(t), y(t)), uy(x(t), y(t))) is a solution of the characteristic system. We recall that th e solution exists and is uniquely determined. Setz(t) =u(x(t), y(t)), then ( x(t), y(t), z(t))⊂ S, and z/prime(t) =uxx/prime(t) +uyy/prime(t) =uxFp+uyFq. Setp(t) =ux(x(t), y(t)), q(t) =uy(x(t), y(t)), then p/prime(t) = uxxFp+uxyFq q/prime(t) = uyxFp+uyyFq. Finally, from the differential equation F(x, y, u (x, y), ux(x, y), uy(x, y)) = 0 it follows p/prime(t) = −Fx−Fup q/prime(t) = −Fy−Fuq. 2 48 CHAPTER 2. EQUATIONS OF FIRST ORDER 2.3.1 Initial value problem of Cauchy Let x=x0(s), y=y0(s), z=z0(s), p=p0(s), q=q0(s), s1< s < s 2,(2.17) be a given initial strip such that the strip condition z/prime 0(s) =p0(s)x/prime 0(s) +q0(s)y/prime 0(s) (2.18) is satisfied. Moreover, we assume that the initial strip satis fies the nonlinear equation, that is, F(x0(s), y0(s), z0(s), p0(s), q0(s)) = 0 . (2.19) Initial value problem of Cauchy: Find a C2-solution z=u(x, y)of F(x, y, z, p, q ) = 0 such that the surface Sdefined by z=u(x, y)contains the above initial strip. Similar to the quasilinear case we will show that the set of st rips de- fined by the characteristic system which are sticked at the ini tial strip, see Figure 2.12, fit together and define the surface for which we are l ooking at. Definition. A strip ( x(τ), y(τ), z(τ), p(τ), q(τ)),τ1< τ < τ 2, is said to be noncharacteristic if x/prime(τ)Fq(x(τ), y(τ), z(τ), p(τ), q(τ))−y/prime(τ)Fp(x(τ), y(τ), z(τ), p(τ), q(τ))/negationslash= 0. Theorem 2.2. For a given noncharacteristic initial strip (2.17), x0, y0, z0∈ C2andp0, q0∈C1which satisfies the strip condition (2.18) and the dif- ferential equation (2.19) there exists exactly one solutio nz=u(x, y)of the Cauchy initial value problem in a neighbourhood of the in itial curve (x0(s), y0(s), z0(s)), i. e., z=u(x, y)is the solution of the differential equa- tion (2.6) and u(x0(s), y0(s)) =z0(s),ux(x0(s), y0(s)) =p0(s),uy(x0(s), y0(s)) = q0(s). Proof. Consider the system (2.7)–(2.11) with initial data x(s,0) =x0(s), y(s,0) =y0(s), z(s,0) =z0(s), p(s,0) =p0(s), q(s,0) =q0(s). We will show that the surface defined by x=x(s, t), y(s, t) is the surface defined by z=u(x, y), where uis the solution of the Cauchy initial value 2.3. NONLINEAR EQUATIONS IN TWO VARIABLES 49 yz xt=0t>0 Figure 2.12: Construction of the solution problem. It turns out that u(x, y) =z(s(x, y), t(x, y)), where s=s(x, y), t=t(x, y) is the inverse of x=x(s, t),y=y(s, t) in a neigbourhood of t= 0. This inverse exists since the initial strip is noncharacter istic by assumption: det∂(x, y) ∂(s, t)/vextendsingle/vextendsingle/vextendsingle t=0=x0Fq−y0Fq/negationslash= 0. Set P(x, y) =p(s(x, y), t(x, y)), Q(x, y) =q(s(x, y), t(x, y)). From Proposition 2.3 and Proposition 2.4 it follows F(x, y, u, P, Q ) = 0. We will show that P(x, y) =ux(x, y) and Q(x, y) =uy(x, y). To see this, we consider the function h(s, t) =zs−pxs−qys. One has h(s,0) =z/prime 0(s)−p0(s)x/prime 0(s)−q0(s)y/prime 0(s) = 0 since the initial strip satisfies the strip condition by assum ption. In the following we will find that for fixed sthe function hsatisfies a linear ho- mogeneous ordininary differential equation of first order. Con sequently, 50 CHAPTER 2. EQUATIONS OF FIRST ORDER h(s, t) = 0 in a neighbourhood of t= 0. Thus the strip condition is also sat- isfied along strips transversally to the characteristic stri ps, see Figure 2.18. Thaen the set of ”scales” fit together and define a surface like th e scales of a fish. From the definition of h(s, t) and the characteristic equations we get ht(s, t) = zst−ptxs−qtys−pxst−qyst =∂ ∂s(zt−pxt−qyt) +psxt+qsyt−qtys−ptxs = (pxs+qys)Fz+Fxxs+Fyzs+Fpps+Fqqs. Since F(x(s, t), y(s, t), z(s, t), p(s, t), q(s, t)) = 0, it follows after differentia- tion of this equation with respect to sthe differential equation ht=−Fzh. Hence h(s, t)≡0, since h(s,0) = 0. Thus we have zs=pxs+qys zt=pxt+qyt zs=uxxs+uyys zt=uxyt+uyyt. The first equation was shown above, the second is a characteris tic equation and the last two follow from z(s, t) =u(x(s, t), y(s, t)). This system implies (P−ux)xs+ (Q−uy)ys= 0 (P−ux)xt+ (Q−uy)yt= 0. It follows P=uxandQ=uy. The initial conditions u(x(s,0), y(s,0)) = z0(s) ux(x(s,0), y(s,0)) = p0(s) uy(x(s,0), y(s,0)) = q0(s) are satisfied since u(x(s, t), y(s, t)) = z(s(x, y), t(x, y)) =z(s, t) ux(x(s, t), y(s, t)) = p(s(x, y), t(x, y)) =p(s, t) uy(x(s, t), y(s, t)) = q(s(x, y), t(x, y)) =q(s, t). 2.4. NONLINEAR EQUATIONS IN RN51 The uniqueness follows as in the proof of Theorem 2.1. 2 Example. A differential equation which occurs in the geometrical optic is u2 x+u2 y=f(x, y), where the positive function f(x, y) is the index of refraction. The level sets defined by u(x, y) =const. are called wave fronts . The characteristic curves (x(t), y(t)) are the rays of light. If nis a constant, then the rays of light are straight lines. In R3the equation is u2 x+u2 y+u2 z=f(x, y, z). Thus we have to extend the previous theory from R2toRn,n≥3. 2.4 Nonlinear equations in Rn Here we consider the nonlinear differential equation F(x, z, p) = 0, (2.20) where x= (x1, . . ., x n), z=u(x) : Ω ⊂Rn/mapsto→R, p=∇u. The following system of 2 n+1 ordinary differential equations is called char- acteristic system . x/prime(t) = ∇pF z/prime(t) = p· ∇pF p/prime(t) = −∇xF−Fzp. Let x0(s) = (x01(s), . . . , x 0n(s)), s= (s1, . . ., s n−1), be a given regular (n-1)-dimensional C2-hypersurface in Rn, i. e., we assume rank∂x0(s) ∂s=n−1. Heres∈Dis a parameter from an ( n−1)-dimensional parameter domain D. For example, x=x0(s) defines in the three dimensional case a regular surface in R3. 52 CHAPTER 2. EQUATIONS OF FIRST ORDER Assume z0(s) :D/mapsto→R, p0(s) = (p01(s), . . . , p 0n(s)) are given sufficiently regular functions. The (2 n+ 1)-vector (x0(s), z0(s), p0(s)) is called initial strip manifold and the condition ∂z0 ∂sl=n−1/summationdisplay i=1p0i(s)∂x0i ∂sl, l= 1, . . ., n −1,strip condition . The initial strip manifold is said to be noncharacteristic if det Fp1Fp2···Fpn ∂x01 ∂s1∂x02 ∂s1···∂x0n ∂s1 ......................... ∂x01 ∂sn−1∂x02 ∂sn−1···∂x0n ∂sn−1 /negationslash= 0, where the argument of Fpjis the initial strip manifold. Initial value problem of Cauchy. Seek a solution z=u(x)of the differential equation (2.20) such that the initial manifold i s a subset of {(x, u(x),∇u(x)) :x∈Ω}. As in the two dimensional case we have under additional regul arity as- sumptions Theorem 2.3. Suppose the initial strip manifold is not characteristic an d satisfies differential equation (2.20), that is, F(x0(s), z0(s), p0(s)) = 0. Then there is a neighbourhood of the initial manifold (x0(s), z0(s))such that there exists a unique solution of the Cauchy initial value problem . Sketch of proof. Let x=x(s, t), z=z(s, t), p=p(s, t) be the solution of the characteristic system and let s=s(x), t=t(x) 2.5. HAMILTON-JACOBI THEORY 53 be the inverse of x=x(s, t) which exists in a neighbourhood of t= 0. Then, it turns out that z=u(x) :=z(s1(x1, . . ., x n), . . ., s n−1(x1, . . ., x n), t(x1, . . ., x n)) is the solution of the problem. 2.5 Hamilton-Jacobi theory The nonlinear equation (2.20) of previous section in one mor e dimension is F(x1, . . ., x n, xn+1, z, p1, . . ., p n, pn+1) = 0. The content of the Hamilton1-Jacobi2theory is the theory of the special case F≡pn+1+H(x1, . . ., x n, xn+1, p1, . . ., p n) = 0, (2.21) i. e., the equation is linear in pn+1and does not depend on zexplicitly. Remark. Formally, one can write equation (2.20) F(x1, . . ., x n, u, u x1, . . ., u xn) = 0 as an equation of type (2.21). Set xn+1=uand seek uimplicitely from φ(x1, . . ., x n, xn+1) =const., where φis a function which is defined by a differential equation. Assume φxn+1/negationslash= 0, then 0 = F(x1, . . ., x n, u, u x1, . . ., u xn) =F(x1, . . ., x n, xn+1,−φx1 φxn+1, . . .,−φxn φxn+1) = :G(x1, . . ., x n+1, φ1, . . ., φ xn+1). Suppose that Gφxn+1/negationslash= 0, then φxn+1=H(x1, . . ., x n, xn+1, φx1, . . ., φ xn+1). 1Hamilton, William Rowan, 1805–1865 2Jacobi, Carl Gustav, 1805–1851 54 CHAPTER 2. EQUATIONS OF FIRST ORDER The associated characteristic equations to (2.21) are x/prime n+1(τ) = Fpn+1= 1 x/prime k(τ) = Fpk=Hpk, k = 1, . . ., n z/prime(τ) =n+1/summationdisplay l=1plFpl=n/summationdisplay l=1plHpl+pn+1 =n/summationdisplay l=1plHpl−H p/prime n+1(τ) = −Fxn+1−Fzpn+1 =−Fxn+1 p/prime k(τ) = −Fxk−Fzpk =−Fxk, k = 1, . . ., n. Sett:=xn+1, then we can write partial differential equation (2.21) as ut+H(x, t,∇xu) = 0 (2.22) and 2nof the characteristic equations are x/prime(t) = ∇pH(x, t, p) (2.23) p/prime(t) = −∇xH(x, t, p). (2.24) Here is x= (x1, . . ., x n), p= (p1, . . ., p n). Letx(t), p(t) be a solution of (2.23) and (2.24), then it follows p/prime n+1(t) and z/prime(t) from the characteristic equations p/prime n+1(t) = −Ht z/prime(t) = p· ∇pH−H. Definition. The function H(x, t, p) is called Hamilton function , equa- tion (2.21) Hamilton-Jacobi equation and the system (2.23), (2.24) canonical system to H . There is an interesting interplay between the Hamilton-Jaco bi equation and the canonical system. According to the previous theory w e can con- struct a solution of the Hamilton-Jacobi equation by using so lutions of the 2.5. HAMILTON-JACOBI THEORY 55 canonical system. On the other hand, one obtains from soluti ons of the Hamilton-Jacobi equation also solutions of the canonical sy stem of ordinary differential equations. Definition. A solution φ(a;x, t) of the Hamilton-Jacobi equation, where a= (a1, . . ., a n) is an n-tuple of real parameters, is called a complete integral of the Hamilton-Jacobi equation if det(φxial)n i,l=1/negationslash= 0. Remark. Ifuis a solution of the Hamilton-Jacobi equation, then also u+const. Theorem 2.4 (Jacobi). Assume u=φ(a;x, t) +c, c=const., φ ∈C2in its arguments , is a complete integral. Then one obtains by solving of bi=φai(a;x, t) with respect to xl=xl(a, b, t), where bii= 1, . . ., n are given real constants, and then by setting pk=φxk(a;x(a, b;t), t) a 2n-parameter family of solutions of the canonical system. Proof. Let xl(a, b;t), l= 1, . . ., n, be the solution of the above system. The solution exists sinc eφis a complete integral by assumption. Set pk(a, b;t) =φxk(a;x(a, b;t), t), k= 1, . . ., n. We will show that xandpsolves the canonical system. Differentiating φai= biwith respect to tand the Hamilton-Jacobi equation φt+H(x, t,∇xφ) = 0 with respect to ai, we obtain for i= 1, . . ., n φtai+n/summationdisplay k=1φxkai∂xk ∂t= 0 φtai+n/summationdisplay k=1φxkaiHpk= 0. 56 CHAPTER 2. EQUATIONS OF FIRST ORDER Since φis a complete integral it follows for k= 1, . . ., n ∂xk ∂t=Hpk. Along a trajectory, i. e., where a, bare fixed, it is∂xk ∂t=x/prime k(t). Thus x/prime k(t) =Hpk. Now we differentiate pi(a, b;t) with respect to tandφt+H(x, t,∇xφ) = 0 with respect to xi, and obtain p/prime i(t) = φxit+n/summationdisplay k=1φxixkx/prime k(t) 0 = φxit+n/summationdisplay k=1φxixkHpk+Hxi 0 = φxit+n/summationdisplay k=1φxixkx/prime k(t) +Hxi It follows finally that p/prime i(t) =−Hxi. 2 Example: Kepler problem The motion of a mass point in a central field takes place in a plan e, say the ( x, y)-plane, see Figure 2.13, and satisfies the system of ordinary differential equations of second order x/prime/prime(t) =Ux, y/prime/prime(t) =Uy, where U(x, y) =k2 /radicalbig x2+y2. Here we assume that k2is a positive constant and that the mass point is attracted of the origin. In the case that it is pushed one has t o replace U by−U. See Landau and Lifschitz [12], Vol 1, for example, for the re lated physics. Set p=x/prime, q=y/prime and H=1 2(p2+q2)−U(x, y), 2.5. HAMILTON-JACOBI THEORY 57 xy (x(t),y(t)) (U ,U )yx θ Figure 2.13: Motion in a central field then x/prime(t) = Hp, y/prime(t) =Hq p/prime(t) = −Hx, q/prime(t) =−Hy. The associated Hamilton-Jacobi equation is φt+1 2(φ2 x+φ2 y) =k2 /radicalbig x2+y2. which is in polar coordinates ( r, θ) φt+1 2(φ2 r+1 r2φ2 θ) =k2 r. (2.25) Now we will seek a complete integral of (2.25) by making the an satz φt=−α=const. φ θ=−β=const. (2.26) and obtain from (2.25) that φ=±/integraldisplayr r0/radicalBigg 2α+2k2 ρ−β2 ρ2dρ+c(t, θ). From ansatz (2.26) it follows c(t, θ) =−αt−βθ. Therefore we have a two parameter family of solutions φ=φ(α, β;θ, r, t) 58 CHAPTER 2. EQUATIONS OF FIRST ORDER of the Hamilton-Jacobi equation. This solution is a complete integral, see an exercise. According to the theorem of Jacobi set φα=−t0, φβ=−θ0. Then t−t0=−/integraldisplayr r0dρ/radicalBig 2α+2k2 ρ−β2 ρ2. The inverse function r=r(t),r(0) = r0, is the r-coordinate depending on timet, and θ−θ0=β/integraldisplayr r0dρ ρ2/radicalBig 2α+2k2 ρ−β2 ρ2. Substitution τ=ρ−1yields θ−θ0=−β/integraldisplay1/r 1/r0dτ/radicalbig 2α+ 2k2τ−β2τ2 =−arcsin/parenleftBiggβ2 k21 r−1/radicalBig 1 +2αβ2 k4/parenrightBigg + arcsin/parenleftBiggβ2 k21 r0−1 /radicalBig 1 +2αβ2 k4/parenrightBigg . Set θ1=θ0+ arcsin/parenleftBiggβ2 k21 r0−1 /radicalBig 1 +2αβ2 k4/parenrightBigg and p=β2 k2, /epsilon12=/radicalbigg 1 +2αβ2 k4, then θ−θ1=−arcsin/parenleftbiggp r−1 /epsilon12/parenrightbigg . It follows r=r(θ) =p 1−/epsilon12sin(θ−θ1), which is the polar equation of conic sections. It defines an ell ipse if 0 ≤/epsilon1 <1, a parabola if /epsilon1= 1 and a hyperbola if /epsilon1 >1, see Figure 2.14 for the case of an ellipse, where the origin of the coordinate system is on e of the focal points of the ellipse. For another application of the Jacobi theorem see Courant an d Hilbert [4], Vol. 2, pp. 94, where geodedics on an ellipsoid are studied. 2.6. EXERCISES 59 θ1pp 1+1−ε ε2 2p Figure 2.14: The case of an ellipse 2.6 Exercises 1. Suppose u:R2/mapsto→Ris a solution of a(x, y)ux+b(x, y)uy= 0. Show that for arbitrary H∈C1alsoH(u) is a solution. 2. Find a solution u/negationslash≡const. of ux+uy= 0 such that graph( u) :={(x, y, z)∈R3:z=u(x, y),(x, y)∈R2} contains the straight line (0 ,0,1) +s(1,1,0), s∈R. 3. Let φ(x, y) be a solution of a1(x, y)ux+a2(x, y)uy= 0. Prove that level curves SC:={(x, y) :φ(x, y) =C=const. }are characteristic curves, provided that ∇φ/negationslash= 0 and ( a1, a2)/negationslash= (0,0). 60 CHAPTER 2. EQUATIONS OF FIRST ORDER 4. Prove Proposition 2.2. 5. Find two different solutions of the initial value problem ux+uy= 1, where the initial data are x0(s) =s, y0(s) =s,z0(s) =s. Hint: (x0, y0) is a characteristic curve. 6. Solve the initial value problem xux+yuy=u with initial data x0(s) =s, y0(s) = 1, z0(s), where z0is given. 7. Solve the initial value problem −xux+yuy=xu2, x0(s) =s, y0(s) = 1, z0(s) = e−s. 8. Solve the initial value problem uux+uy= 1, x0(s) =s, y0(s) =s,z0(s) =s/2 if 0 < s < 1. 9. Solve the initial value problem uux+uuy= 2, x0(s) =s, y0(s) = 1, z0(s) = 1 + sif 0< s < 1. 10. Solve the initial value problem u2 x+u2 y= 1 + xwith given initial data x0(s) = 0, y0(s) =s, u0(s) = 1, p0(s) = 1, q0(s) = 0, −∞< s < ∞. 11. Find the solution Φ( x, y) of (x−y)ux+ 2yuy= 3x such that the surface defined by z= Φ(x, y) contains the curve C:x0(s) =s, y0(s) = 1, z0(s) = 0, s∈R. 2.6. EXERCISES 61 12. Solve the following initial problem of chemical kinetic s. ux+uy=/parenleftBig k0e−k1x+k2/parenrightBig (1−u)2, x > 0, y > 0 with the initial data u(x,0) = 0 , u(0, y) =u0(y), where u0, 0< u0<1, is given. 13. Solve the Riemann problem ux1+ux2= 0 u(x1,0) = g(x1) in Ω1={(x1, x2)∈R2:x1> x2}and in Ω 2={(x1, x2)∈R2:x1< x2}, where g(x1) =/braceleftbiggul:x1<0 ur:x1>0 with constants ul/negationslash=ur. 14. Determine the opening angle of the Monge cone, i. e., the a ngle be- tween the axis and the apothem (in German: Mantellinie) of th e cone, for equation u2 x+u2 y=f(x, y, u ), where f >0. 15. Solve the initial value problem u2 x+u2 y= 1, where x0(θ) =acosθ, y0(θ) =asinθ, z0(θ) = 1 , p0(θ) = cos θ, q0(θ) = sin θif 0≤θ <2π,a=const. > 0. 16. Show that the integral φ(α, β;θ, r, t), see the Kepler problem, is a complete integral. 17. a) Show that S=√α x+√1−α y+β,α, β∈R,0< α < 1, is a complete integral of Sx−/radicalBig 1−S2y= 0. b) Find the envelope of this family of solutions. 18. Determine the length of the half axis of the ellipse r=p 1−ε2sin(θ−θ0),0≤ε <1. 62 CHAPTER 2. EQUATIONS OF FIRST ORDER 19. Find the Hamilton function H(x, p) of the Hamilton-Jacobi-Bellman differential equation if h= 0 and f=Ax+Bα, where A, B are constant and real matrices, A:Rm/mapsto→Rn,Bis an orthogonal real n×n-Matrix and p∈Rnis given. The set of admissible controls is given by U={α∈Rn:n/summationdisplay i=1α2 i≤1}. Remark. The Hamilton-Jacobi-Bellman equation is formally the Hamilt on- Jacobi equation ut+H(x,∇u) = 0, where the Hamilton function is defined by H(x, p) := min α∈U(f(x, α)·p+h(x, α)), f(x, α) and h(x, α) are given. See for example, Evans [5], Chapter 10. Chapter 3 Classification Different types of problems in physics, for example, correspo nd different types of partial differential equations. The methods how to so lve these equations differ from type to type. The classification of differential equations follows from one s ingle ques- tion: Can we calculate formally the solution if sufficiently ma ny initial data are given? Consider the initial problem for an ordinary differ ential equa- tiony/prime(x) =f(x, y(x)),y(x0) =y0. Then one can determine formally the solution, provided the function f(x, y) is sufficiently regular. The solution of the initial value problem is formally given by a power seri es. This formal solution is a solution of the problem if f(x, y) is real analytic according to a theorem of Cauchy. In the case of partial differential equati ons the re- lated theorem is the Theorem of Cauchy-Kowalevskaya. Even in the case of ordinary differential equations the situation is more comp licated if y/primeis implicitly defined, i. e., the differential equation is F(x, y(x), y/prime(x)) = 0 for a given function F. 3.1 Linear equations of second order The general nonlinear partial differential equation of secon d order is F(x, u, Du, D2u) = 0, where x∈Rn,u: Ω⊂Rn/mapsto→R,Du≡ ∇uandD2ustands for all second derivatives. The function Fis given and sufficiently regular with respect to its 2n+ 1 + n2arguments. 63 64 CHAPTER 3. CLASSIFICATION In this section we consider the case n/summationdisplay i,k=1aik(x)uxixk+f(x, u,∇u) = 0. (3.1) The equation is linear if f=n/summationdisplay i=1bi(x)uxi+c(x)u+d(x). Concerning the classification the main part n/summationdisplay i,k=1aik(x)uxixk plays the essential role. Suppose u∈C2, then we can assume, without restriction of generality, that aik=aki, since n/summationdisplay i,k=1aikuxixk=n/summationdisplay i,k=1(aik)⋆uxixk, where (aik)⋆=1 2(aik+aki). Consider a hypersurface SinRndefined implicitly by χ(x) = 0, ∇χ/negationslash= 0, see Figure 3.1 Assume uand∇uare given on S. Problem: Can we calculate all other derivatives of uonSby using differ- ential equation (3.1) and the given data? We will find an answer if we map Sonto a hyperplane S0by a mapping λn=χ(x1, . . ., x n) λi=λi(x1, . . ., x n), i= 1, . . ., n −1, for functions λisuch that det∂(λ1, . . ., λ n) ∂(x1, . . ., x n)/negationslash= 0 in Ω⊂Rn. It is assumed that χandλiare sufficiently regular. Such a mapping λ=λ(x) exists, see an exercise. 3.1. LINEAR EQUATIONS OF SECOND ORDER 65 xSxx3 12 Figure 3.1: Initial manifold S The above transform maps Sonto a subset of the hyperplane defined by λn= 0, see Figure 3.2. We will write the differential equation in these new coordinat es. Here we use Einstein’s convention, i. e., we add terms with repeatin g indices. Since u(x) =u(x(λ)) =:v(λ) =v(λ(x)), where x= (x1, . . ., x n) and λ= (λ1, . . ., λ n), we get uxj=vλi∂λi ∂xj, (3.2) uxjxk=vλiλl∂λi ∂xj∂λl ∂xk+vλi∂2λi ∂xj∂xk. Thus, differential equation (3.1) in the new coordinates is gi ven by ajk(x)∂λi ∂xj∂λl ∂xkvλiλl+ terms known on S0= 0. Since vλk(λ1, . . ., λ n−1,0),k= 1, . . ., n , are known, see (3.2), it follows that vλkλl,l= 1, . . ., n −1, are known on S0. Thus we know all second derivatives vλiλjonS0with the only exception of vλnλn. 66 CHAPTER 3. CLASSIFICATION 3 12λ λ λS0 Figure 3.2: Transformed flat manifold S0 We recall that, provided vis sufficiently regular, vλkλl(λ1, . . ., λ n−1,0) is the limit of vλk(λ1, . . ., λ l+h, λl+1, . . ., λ n−1,0)−vλk(λ1, . . ., λ l, λl+1, . . ., λ n−1,0) h ash→0. Thus the differential equation can be written as n/summationdisplay j,k=1ajk(x)∂λn ∂xj∂λn ∂xkvλnλn= terms known on S0. It follows that we can calculate vλnλnif n/summationdisplay i,j=1aij(x)χxiχxj/negationslash= 0 (3.3) onS. This is a condition for the given equation and for the given s urface S. 3.1. LINEAR EQUATIONS OF SECOND ORDER 67 Definition. The differential equation n/summationdisplay i,j=1aij(x)χxiχxj= 0 is called characteristic differential equation associated to the given differen- tial equation (3.1). Ifχ,∇χ/negationslash= 0, is a solution of the characteristic differential equation , then the surface defined by χ= 0 is called characteristic surface . Remark. The condition (3.3) is satisfied for each χwith∇χ/negationslash= 0 if the quadratic matrix ( aij(x)) is positive or negative definite for each x∈Ω, which is equivalent to the property that all eigenvalues are different from zero and have the same sign. This follows since there is a λ(x)>0 such that, in the case that the matrix ( aij) is poitive definite, n/summationdisplay i,j=1aij(x)ζiζj≥λ(x)|ζ|2 for all ζ∈Rn. Here and in the following we assume that the matrix ( aij) is real and symmetric. The characterization of differential equation (3.1) follows from the signs of the eigenvalues of ( aij(x)). Definition. Differential equation (3.1) is said to be of type(α, β, γ )at x∈Ω ifαeigenvalues of ( aij)(x) are positive, βeigenvalues are negative andγeigenvalues are zero ( α+β+γ=n). In particular, equation is called elliptic if it is of type ( n,0,0) or of type (0 , n,0), i. e., all eigenvalues are different from zero and have the same sign, parabolic if it is of type ( n−1,0,1) or of type (0 , n−1,1), i. e., one eigenvalue is zero and all the others are different from zero and have the sa me sign, hyperbolic if it is of type ( n−1,1,0) or of type (1 , n−1,0), i. e., all eigenvalues are different from zero and one eigenvalue has ano ther sign than all the others. 68 CHAPTER 3. CLASSIFICATION Remarks: 1.According to this definition there are other types aside from e lliptic, parabolic or hyperbolic equations. 2.The classification depends in general on x∈Ω. An example is the Tricomi equation, which appears in the theory of transsonic flows, yuxx+uyy= 0. This equation is elliptic if y >0, parabolic if y= 0 and hyperbolic for y <0. Examples: 1.TheLaplace equation inR3is/triangleu= 0, where /triangleu:=uxx+uyy+uzz. This equation is elliptic. Thus for each manifold Sgiven by {(x, y, z) : χ(x, y, z) = 0 }, where χis an arbitrary sufficiently regular function such that∇χ/negationslash= 0, all derivatives of uare known on S, provided uand∇uare known on S. 2.Thewave equation utt=uxx+uyy+uzz, where u=u(x, y, z, t ), is hyperbolic. Such a type describes oscillations of mechanic al structures, for example. 3.Theheat equation ut=uxx+uyy+uzz, where u=u(x, y, z, t ), is parabolic. It describes, for example, the propagation of heat in a domai n. 4.Consider the case that the (real) coefficients aijin equation (3.1) are constant . We recall that the matrix A= (aij) is symmetric, i. e., AT=A. In this case, the transform to principle axis leads to a norma l form from which the classification of the equation is obviously. Let Ube the associated orthogonal matrix, then UTAU= λ10···0 0λ2···0 ................ 0 0 ···λn . 3.1. LINEAR EQUATIONS OF SECOND ORDER 69 Here is U= (z1, . . ., z n), where zl,l= 1, . . ., n , is an orthonormal system of eigenvectors to the eigenvalues λl. Sety=UTxandv(y) =u(Uy), then n/summationdisplay i,j=1aijuxixj=n/summationdisplay i=1λivyiyj. (3.4) 3.1.1 Normal form in two variables Consider the differential equation a(x, y)uxx+ 2b(x, y)uxy+c(x, y)uyy+ terms of lower order = 0 (3.5) in Ω⊂R2. The associated characteristic differential equation is aχ2 x+ 2bχxχy+cχ2 y= 0. (3.6) We show that an appropriate coordinate transform will simpl ify equation (3.5) sometimes in such a way that we can solve the transformed equa tion explic- itly. Letz=φ(x, y) be a solution of (3.6). Consider the level sets {(x, y) : φ(x, y) =const. }and assume φy/negationslash= 0 at a point ( x0, y0) of the level set. Then there is a function y(x) defined in a neighbourhood of x0such that φ(x, y(x)) =const. It follows y/prime(x) =−φx φy, which implies, see the characteristic equation (3.6), ay/prime2−2by/prime+c= 0. (3.7) Then, provided a/negationslash= 0, we can calculate µ:=y/primefrom the (known) coefficients a,bandc: µ1,2=1 a/parenleftBig b±/radicalbig b2−ac/parenrightBig . (3.8) These solutions are real if and only of ac−b2≤0. Equation (3.5) is hyperbolic if ac−b2<0, parabolic if ac−b2= 0 and elliptic if ac−b2>0. This follows from an easy discussion of the eigenvalues of the matrix /parenleftbigga b b c/parenrightbigg , see an exercise. 70 CHAPTER 3. CLASSIFICATION Normal form of a hyperbolic equation Letφandψare solutions of the characteristic equation (3.6) such tha t y/prime 1≡µ1=−φx φy y/prime 2≡µ2=−ψx ψy, where µ1andµ2are given by (3.8). Thus φandψare solutions of the linear homogeneous equations of first order φx+µ1(x, y)φy= 0 (3.9) ψx+µ2(x, y)ψy= 0. (3.10) Assume φ(x, y),ψ(x, y) are solutions such that ∇φ/negationslash= 0 and ∇ψ/negationslash= 0, see an exercise for the existence of such solutions. Consider two families of level sets defined by φ(x, y) =αandψ(x, y) =β, see Figure 3.3. y x(x,y)= (x,y)= (x,y)=(x,y)=ϕ α αϕ βψ ψβ1 2 1 2 Figure 3.3: Level sets These level sets are characteristic curves of the partial di fferential equa- tions (3.9) and (3.10), respectively, see an exercise of the previous chapter. Lemma. (i)Curves from different families can not touch each other. (ii)φxψy−φyψx/negationslash= 0. 3.1. LINEAR EQUATIONS OF SECOND ORDER 71 Proof. (i): y/prime 2−y/prime 1≡µ2−µ1=−2 a/radicalbig b2−ac/negationslash= 0. (ii): µ2−µ1=φx φy−ψx ψy. 2 Proposition 3.1. The mapping ξ=φ(x, y),η=ψ(x, y)transforms equa- tion (3.5) into vξη=lower order terms , (3.11) where v(ξ, η) =u(x(ξ, η), y(ξ, η)). Proof. The proof follows from a straightforward calculation. ux=vξφx+vηψx uy=vξφy+vηψy uxx=vξξφ2 x+ 2vξηφxψx+vηηψ2 x+ lower order terms uxy=vξξφxφy+vξη(φxψy+φyψx) +vηηψxψy+ lower order terms uyy=vξξφ2 y+ 2vξηφyψy+vηηψ2 y+ lower order terms . Thus auxx+ 2buxy+cuyy=αvξξ+ 2βvξη+γvηη+l.o.t., where α: = aφ2 x+ 2bφxφy+cφ2 y β: = aφxψx+b(φxψy+φyψx) +cφyψy γ: = aψ2 x+ 2bψxψy+cψ2 y. The coefficients αandγare zero since φandψare solutions of the charac- teristic equation. Since αγ−β2= (ac−b2)(φxψy−φyψx)2, it follows from the above lemma that the coefficient βis different from zero. 2 Example: Consider the differential equation uxx−uyy= 0. 72 CHAPTER 3. CLASSIFICATION The associated characteristic differential equation is χ2 x−χ2 y= 0. Since µ1=−1 and µ2= 1, the functions φandψsatisfy differential equa- tions φx+φy= 0 ψx−ψy= 0. Solutions with ∇φ/negationslash= 0 and ∇ψ/negationslash= 0 are φ=x−y, ψ =x+y. Thus the mapping ξ=x−y, η=x+y leads to the simple equation vξη(ξ, η) = 0. Assume v∈C2is a solution, then vξ=f1(ξ) for an arbitrary C1function f1(ξ). It follows v(ξ, η) =/integraldisplayξ 0f1(α)dα+g(η), where gis an arbitrary C2function. Thus each C2-solution of the differen- tial equation can be written as (⋆) v(ξ, η) =f(ξ) +g(η), where f, g∈C2. On the other hand, for arbitrary C2-functions f,gthe function ( ⋆) is a solution of the differential equation vξη= 0. Consequently eachC2-solution of the original equation uxx−uyy= 0 is given by u(x, y) =f(x−y) +g(x+y), where f, g∈C2. 3.2. QUASILINEAR EQUATIONS OF SECOND ORDER 73 3.2 Quasilinear equations of second order Here we consider the equation n/summationdisplay i,j=1aij(x, u,∇u)uxixj+b(x, u,∇u) = 0 (3.12) in a domain Ω ⊂Rn, where u: Ω/mapsto→R. We assume that aij=aji. As in the previous section we can derive the characteristic e quation n/summationdisplay i,j=1aij(x, u,∇u)χxiχxj= 0. In contrast to linear equations, solutions of the character istic equation de- pend on the solution considered. 3.2.1 Quasilinear elliptic equations There is a large class of quasilinear equations such that the associated char- acteristic equation has no solution χ,∇χ/negationslash= 0. Set U={(x, z, p) :x∈Ω, z∈R, p∈Rn}. Definition. The quasilinear equation (3.12) is called elliptic if the matrix (aij(x, z, p)) is positive definite for each ( x, z, p)∈U. Assume equation (3.12) is elliptic and let λ(x, z, p) be the minimum and Λ(x, z, p) the maximum of the eigenvalues of ( aij), then 0< λ(x, z, p)|ζ|2≤n/summationdisplay i,j=1aij(x, z, p)ζiζj≤Λ(x, z, p)|ζ|2 for all ζ∈Rn. Definition. Equation (3.12) is called uniformly elliptic if Λ/λis uniformly bounded in U. An important class of elliptic equations which are not unifo rmly elliptic (nonuniformly elliptic) is n/summationdisplay i=1∂ ∂xi/parenleftBigg uxi/radicalbig 1 +|∇u|2/parenrightBigg + lower order terms = 0 . (3.13) 74 CHAPTER 3. CLASSIFICATION The main part is the minimal surface operator (left hand side of the minimal surface equation). The coefficients aijare aij(x, z, p) =/parenleftbig 1 +|p|2/parenrightbig−1/2/parenleftbigg δij−pipj 1 +|p|2/parenrightbigg , δijdenotes the Kronecker delta symbol. It follows that λ=1 (1 +|p|2)3/2,Λ =1 (1 +|p|2)1/2. Thus equation (3.13) is not uniformly elliptic. The behaviour of solutions of uniformly elliptic equations is similar to linear elliptic equations in contrast to the behaviour of so lutions of nonuni- formly elliptic equations. Typical examples for nonunifor mly elliptic equa- tions are the minimal surface equation and the capillary equ ation. 3.3 Systems of first order Consider the quasilinear system n/summationdisplay k=1Ak(x, u)uuk+b(x, u) = 0, (3.14) where Akarem×m-matrices, sufficiently regular with respect to their ar- guments, and u= u1 ... um , uxk= u1,xk... um,xk , b= b1 ... bm . We ask the same question as above: can we calculate all deriva tives of u in a neighbourhood of a given hypersurface SinRndefined by χ(x) = 0, ∇χ/negationslash= 0, provided u(x) is given on S? For an answer we map Sonto a flat surface S0by using the mapping λ=λ(x) of Section 3.1 and write equation (3.14) in new coordinates . Set v(λ) =u(x(λ)), then n/summationdisplay k=1Ak(x, u)χxkvλn= terms known on S0. 3.3. SYSTEMS OF FIRST ORDER 75 We can solve this system with respect to vλn, provided that det/parenleftBiggn/summationdisplay k=1Ak(x, u)χxk/parenrightBigg /negationslash= 0 onS. Definition. Equation det/parenleftBiggn/summationdisplay k=1Ak(x, u)χxk/parenrightBigg = 0 is called characteristic equation associated to equation (3.14) and a surface S:χ(x) = 0, defined by a solution χ,∇χ/negationslash= 0, of this characteristic equation is said to be characteristic surface . Set C(x, u, ζ ) = det/parenleftBiggn/summationdisplay k=1Ak(x, u)ζk/parenrightBigg forζ∈Rn. Definition. (i) The system (3.14) is hyperbolic at (x, u(x)) if there is a regular linear mapping ζ=Qη, where η= (η1, . . ., η n−1, κ), such that there exists mrealroots κk=κk(x, u(x), η1, . . ., η n−1),k= 1, . . ., m , of D(x, u(x), η1, . . ., η n−1, κ) = 0 for all ( η1, . . ., η n−1), where D(x, u(x), η1, . . ., η n−1, κ) =C(x, u(x), x, Qη ). (ii) System (3.14) is parabolic if there exists a regular linear mapping ζ=Qη such that Dis independent of κ, i. e., Ddepends on less than nparameters. (iii) System (3.14) is elliptic ifC(x, u, ζ ) = 0 only if ζ= 0. Remark. In the elliptic case all derivatives of the solution can be ca lculated from the given data and the given equation. 76 CHAPTER 3. CLASSIFICATION 3.3.1 Examples 1. Beltrami equations Wux−bvx−cvy= 0 (3.15) Wuy+avx+bvy= 0, (3.16) where W, a, b, c are given functions depending of ( x, y),W/negationslash= 0 and the matrix /parenleftbigga b b c/parenrightbigg is positive definite. The Beltrami system is a generalization of Cauchy-Riemann eq uations. The function f(z) =u(x, y) +iv(x, y), where z=x+iy, is called a qua- siconform mapping , see for example [9], Chapter 12, for an application to partial differential equations. Set A1=/parenleftbiggW−b 0a/parenrightbigg , A2=/parenleftbigg0−c W b/parenrightbigg . Then the system (3.15), (3.16) can be written as A1/parenleftbiggux vx/parenrightbigg +A2/parenleftbigguy vy/parenrightbigg =/parenleftbigg0 0/parenrightbigg . Thus, C(x, y, ζ ) =/vextendsingle/vextendsingle/vextendsingle/vextendsingleWζ1−bζ1−cζ2 Wζ2aζ1+bζ2/vextendsingle/vextendsingle/vextendsingle/vextendsingle=W(aζ2 1+ 2bζ1ζ2+cζ2 2), which is different from zero if ζ/negationslash= 0 according to the above assumptions. Thus the Beltrami system is elliptic. 2. Maxwell equations The Maxwell equations in the isotropic case are crotxH=λE+/epsilon1Et (3.17) crotxE=−µHt, (3.18) 3.3. SYSTEMS OF FIRST ORDER 77 where E= (e1, e2, e3)Telectric field strength, ei=ei(x, t),x= (x1, x2, x3), H= (h1, h2, h3)Tmagnetic field strength, hi=hi(x, t), cspeed of light, λspecific conductivity, /epsilon1dielectricity constant, µmagnetic permeability. Herec, λ, /epsilon1 andµare positive constants. Setp0=χt, pi=χxi,i= 1, . . .3, then the characteristic differential equa- tion is/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/epsilon1p0/c 0 0 0 p3−p2 0/epsilon1p0/c 0 −p3 0 p1 0 0 /epsilon1p0/c p 2−p1 0 0 −p3p2µp0/c 0 0 p3 0 −p1 0µp0/c 0 −p2p1 0 0 0 µp0/c/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle= 0. The following manipulations simplifies this equation: (i) multiply the first three columns with µp0/c, (ii) multiply the 5th column with −p3and the the 6th column with p2and add the sum to the 1st column, (iii) multiply the 4th column with p3and the 6th column with −p1and add the sum to the 2th column, (iv) multiply the 4th column with −p2and the 5th column with p1and add the sum to the 3th column, (v) expand the resulting determinant with respect to the ele ments of the 6th, 5th and 4th row. We obtain/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleq+p2 1p1p2p1p3 p1p2q+p2 2p2p3 p1p3p2p3q+p2 3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle= 0, where q:=/epsilon1µ c2p2 0−g2 withg2:=p2 1+p2 2+p2 3. The evaluation of the above equation leads to q2(q+g2) = 0, i. e., χ2 t/parenleftBig/epsilon1µ c2χ2 t− |∇ xχ|2/parenrightBig = 0. 78 CHAPTER 3. CLASSIFICATION It follows immediately that Maxwell equations are ahyperbolic system , see an exercise. There are two solutions of this characteristic eq uation. The first one are characteristic surfaces S(t), defined by χ(x, t) = 0, which satisfy χt= 0. These surfaces are called stationary waves . The second type of characteristic surfaces are defined by solutions of /epsilon1µ c2χ2 t=|∇xχ|2. Functions defined by χ=f(n·x−V t) are solutions of this equation. Here is f(s) an arbitrary function with f/prime(s)/negationslash= 0,nis a unit vector and V=c/√/epsilon1µ. The associated characteristic surfaces S(t) are defined by χ(x, t)≡f(n·x−V t) = 0, here we assume that 0 is in he range of f:R/mapsto→R. Thus, S(t) is defined byn·x−V t=c, where cis a fixed constant. It follows that the planes S(t) with normal nmove with speed Vin direction of n, see Figure 3.4 xx2 1 nS(t) S(0)d(t) Figure 3.4: d/prime(t) is the speed of plane waves Vis called speed of the plane wave S(t). Remark. According to the previous discussions, singularities of a s olution of Maxwell equations are located at most on characteristic s urfaces. A special case of Maxwell equations are the telegraph equations , which follow from Maxwell equations if d iv E= 0 and d iv H = 0, i. e., Eand 3.3. SYSTEMS OF FIRST ORDER 79 Hare fields free of sources. In fact, it is sufficient to assume that this assumption is satisfied at a fixed time t0only, see an exercise. Since rotxrotxA= gradxdivxA− /triangle xA for each C2-vector field A, it follows from Maxwell equations the uncoupled system /trianglexE=/epsilon1µ c2Ett+λµ c2Et /trianglexH=/epsilon1µ c2Htt+λµ c2Ht. 3. Equations of gas dynamics Consider the following quasilinear equations of first order. vt+ (v· ∇x)v+1 ρ∇xp=f(Euler equations) . Here is v= (v1, v2, v3) the vector of speed, vi=vi(x, t),x= (x1, x2, x3), ppressure, p= (x, t), ρdensity, ρ=ρ(x, t), f= (f1, f2, f3) density of the external force, fi=fi(x, t), (v· ∇x)v≡(v· ∇xv1, v· ∇xv2, v· ∇xv3))T. The second equation is ρt+v· ∇xρ+ρdivxv= 0 (conservation of mass) . Assume the gas is compressible and that there is a function (s tate equation) p=p(ρ), where p/prime(ρ)>0 ifρ >0. Then the above system of four equations is vt+ (v· ∇)v+1 ρp/prime(ρ)∇ρ=f (3.19) ρt+ρdiv v+v· ∇ρ= 0, (3.20) where ∇ ≡ ∇ xand d iv≡divx, i. e., these operators apply on the spatial variables only. 80 CHAPTER 3. CLASSIFICATION The characteristic differential equation is here /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingledχ dt0 01 ρp/primeχx1 0dχ dt01 ρp/primeχx2 0 0dχ dt1 ρp/primeχx3 ρχx1ρχx2ρχx3dχ dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle= 0, where dχ dt:=χt+ (∇xχ)·v. Evaluating the determinant, we get the characteristic differ ential equation /parenleftbiggdχ dt/parenrightbigg2/parenleftBigg/parenleftbiggdχ dt/parenrightbigg2 −p/prime(ρ)|∇xχ|2/parenrightBigg = 0. (3.21) This equation implies consequences for the speed of the char acteristic sur- faces as the following consideration shows. Consider a family S(t) of surfaces in R3defined by χ(x, t) =c, where x∈R3andcis a fixed constant. As usually, we assume that ∇xχ/negationslash= 0. One of the two normals on S(t) at a point of the surface S(t) is given by, see an exercise, n=∇xχ |∇xχ|. (3.22) LetQ0∈ S(t0) and let Q1∈ S(t1) be a point on the line defined by Q0+sn, where nis the normal (3.22 on S(t0) atQ0andt0< t1,t1−t0small, see Figure 3.5. )0S(tS(t )1n QQ1 0 Figure 3.5: Definition of the speed of a surface 3.3. SYSTEMS OF FIRST ORDER 81 Definition. The limit P= lim t1→t0|Q1−Q0| t1−t0 is called speed of the surface S(t). Proposition 3.2. The speed of the surface S(t)is P=−χt |∇xχ|. (3.23) Proof. The proof follows from χ(Q0, t0) = 0 and χ(Q0+dn, t0+/trianglet) = 0, where d=|Q1−Q0|and/trianglet=t1−t0. 2 Setvn:=v·nwhich is the component of the velocity vector in direction n. From (3.22) we get vn=1 |∇xχ|v· ∇xχ. Definition. V:=P−vn, the difference of the speed of the surface and the speed of liquid particles, is called relative speed . n vS Figure 3.6: Definition of relative speed Using the above formulas for Pandvnit follows V=P−vn=−χt |∇xχ|−v· ∇xχ |∇xχ|=−1 |∇xχ|dχ dt. Then, we obtain from the characteristic equation (3.21) tha t V2|∇xχ|2/parenleftbig V2|∇xχ|2−p/prime(ρ)|∇xχ|2/parenrightbig = 0. 82 CHAPTER 3. CLASSIFICATION An interesting conclusion is that there are two relative spe eds:V= 0 or V2=p/prime(ρ). Definition./radicalbig p/prime(ρ) is called speed of sound . 3.4 Systems of second order Here we consider the system n/summationdisplay k,l=1Akl(x, u,∇u)uxkxl+ lower order terms = 0 , (3.24) where Aklare (m×m) matrices and u= (u1, . . ., u m)T. We assume Akl= Alk, which is no restriction of generality provided u∈C2is satisfied. As in the previous sections, the classification follows from the qu estion whether or not we can calculate formally the solution from the differenti al equations, if sufficiently many data are given on an initial manifold. Let t he initial manifold Sbe given by χ(x) = 0 and assume that ∇χ/negationslash= 0. The mapping x=x(λ), see previous sections, leads to n/summationdisplay k,l=1Aklχxkχxlvλnλn= terms known on S, where v(λ) =u(x(λ)). The characteristic equation is here det n/summationdisplay k,l=1Aklχxkχxl = 0. If there is a solution χwith∇χ/negationslash= 0, then it is possible that second derivatives are not continuous in a neighbourhood of S. Definition. The system is called elliptic if det n/summationdisplay k,l=1Aklζkζl /negationslash= 0 for all ζ∈Rn,ζ/negationslash= 0. 3.4. SYSTEMS OF SECOND ORDER 83 3.4.1 Examples 1. Navier-Stokes equations The Navier-Stokes system for a viscous incompressible liqui d is vt+ (v· ∇x)v=−1 ρ∇xp+γ/trianglexv divxv= 0, where ρis the (constant and positive) density of liquid, γis the (constant and positive) viscosity of liquid, v=v(x, t) velocity vector of liquid particles, x∈R3or in R2, p=p(x, t) pressure. The problem is to find solutions v, pof the above system. 2. Linear elasticity Consider the system ρ∂2u ∂t2=µ/trianglexu+ (λ+µ)∇x(divxu) +f. (3.25) Here is, in the case of an elastic body in R3, u(x, t) = (u1(x, t), u2(x, t), u3(x, t)) displacement vector, f(x, t) density of external force, ρ(constant) density, λ, µ(positive) Lam´ e constants. The characteristic equation is det C= 0, where the entries of the matrix Care given by cij= (λ+µ)χxiχxj+δij/parenleftbig µ|∇xχ|2−ρχ2 t/parenrightbig . The characteristic equation is /parenleftbig (λ+ 2µ)|∇xχ|2−ρχ2 t/parenrightbig /parenleftbig µ|∇xχ|2−ρχ2 t/parenrightbig2= 0. It follows that two different speeds Pof characteristic surfaces S(t), defined byχ(x, t) =const. , are possible, namely P1=/radicalBigg λ+ 2µ ρ,andP2=/radicalbiggµ ρ. We recall that P=−χt/|∇xχ|. 84 CHAPTER 3. CLASSIFICATION 3.5 Theorem of Cauchy-Kovalevskaya Consider the quasilinear system of first order (3.14) of Secti on 3.3. Assume an initial manifolds Sis given by χ(x) = 0, ∇χ/negationslash= 0, and suppose that χis not characteristic. Then, see Section 3.3, the system (3.14 ) can be written as uxn=n−1/summationdisplay i=1ai(x, u)uxi+b(x, u) (3.26) u(x1, . . ., x n−1,0) = f(x1, . . ., x n−1) (3.27) Here is u= (u1, . . ., u m)T,b= (b1, . . ., b n)Tandaiare (m×m)-matrices. We assume ai,bandfare in C∞with respect to their arguments. From (3.26) and (3.27) it follows that we can calculate formally all deri vatives Dαuin a neigbourhood of the plane {x:xn= 0}, in particular in a neighbourhood of 0∈Rn. Thus we have a formal power series of u(x) atx= 0: u(x)∼/summationdisplay1 α!Dαu(0)xα. For notations and definitions used here and in the following se e the appendix to this section. Then, as usually, two questions arise: (i) Does the power series converge in a neighbourhood of 0 ∈Rn? (ii) Is a convergent power series a solution of the initial va lue problem (3.26), (3.27)? Remark. Quite different to this power series method is the method of asymptotic expansions . Here one is interested in a good approximation of an unknown solution of an equation by a finite sum/summationtextN i=0φi(x) of functions φi. In general, the infinite sum/summationtext∞ i=0φi(x) does not converge, in contrast to the power series method of this section. See [15] for some a symptotic formulas in capillarity. Theorem 3.1 (Cauchy-Kovalevskaya). There is a neighbourhood of 0∈Rn such there is a real analytic solution of the initial value pr oblem (3.26), (3.27). This solution is unique in the class of real analytic functions. 3.5. THEOREM OF CAUCHY-KOVALEVSKAYA 85 Proof. The proof is taken from F. John [10]. We introduce u−fas the new solution for which we are looking at and we add a new coordinat eu⋆to the solution vector by setting u⋆(x) =xn. Then u⋆ xn= 1, u⋆ xk= 0, k= 1, . . ., n −1, u⋆(x1, . . ., x n−1,0) = 0 and the extended system (3.26), (3.27) is  u1,xn... um,xn u⋆ xn =n−1/summationdisplay i=1/parenleftbiggai0 0 0/parenrightbigg u1,xi... um,xi u⋆ xi + b1 ... bm 1 , where the associated initial condition is u(x1, . . ., x n−1,0) = 0. The new u isu= (u1, . . ., u m)T, the new aiareai(x1, . . ., x n−1, u1, . . ., u m, u⋆) and the newbisb= (x1, . . ., x n−1, u1, . . ., u m, u⋆)T. Thus we are led to an initial value problem of the type uj,xn=n−1/summationdisplay i=1N/summationdisplay k=1ai jk(z)uk,xi+bj(z), j= 1, . . ., N (3.28) uj(x) = 0 if xn= 0, (3.29) where j= 1, . . ., N andz= (x1, . . ., x n−1, u1, . . ., u N). The point here is that ai jkandbjare independent of xn. This fact simplifies the proof of the theorem. From (3.28) and (3.29) we can calculate formally all Dβuj. Then we have formal power series for uj: uj(x)∼/summationdisplay αc(j) αxα, where c(j) α=1 α!Dαuj(0). We will show that these power series are (absolutely) conver gent in a neigh- bourhood of 0 ∈Rn, i. e., they are real analytic functions, see the appendix for the definition of real analytic functions. Inserting thes e functions into the left and into the right hand side of (3.28) we obtain on the right and on the left hand side real analytic functions. This follows sin ce compositions of real analytic functions are real analytic again, see Prop osition A7 of the appendix to this section. The resulting power series on the l eft and on the 86 CHAPTER 3. CLASSIFICATION right have the same coefficients caused by the calculation of th e derivatives Dαuj(0) from (3.28). It follows that uj(x),j= 1, . . ., n , defined by its formal power series are solutions of the initial value probl em (3.28), (3.29). Set d=/parenleftbigg∂ ∂z1, . . .,∂ ∂zN+n−1/parenrightbigg Lemma A. Assume u∈C∞in a neighbourhood of 0∈Rn. Then Dαuj(0) = Pα/parenleftBig dβai jk(0), dγbj(0)/parenrightBig , where |β|,|γ| ≤ |α|andPαare polynomials in the indicated arguments with nonnegative integers as coefficients which are independent ofaiand of b. Proof. It follows from equation (3.28) that DnDαuj(0) = Pα(dβai jk(0), dγbj(0), Dδuk(0)). (3.30) Here is Dn=∂/∂x nandα, β, γ, δ satisfy the inequalities |β|,|γ| ≤ |α|,|δ| ≤ |α|+ 1, and, which is essential in the proof, the last coordinates in the multi-indices α= (α1, . . ., α n),δ= (δ1, . . ., δ n) satisfy δn≤αnsince the right hand side of (3.28) is independent of xn. Moreover, it follows from (3.28) that the polynomials Pαhave integers as coefficients. The initial condition (3.29) implies Dαuj(0) = 0 , (3.31) where α= (α1, . . ., α n−1,0), that is, αn= 0. Then, the proof is by induction with respect to αn. The induction starts with αn= 0, then we replace Dδuk(0) in the right hand side of (3.30) by (3.31), that is by zero. Then it follows from (3.30) that Dαuj(0) = Pα(dβai jk(0), dγbj(0), Dδuk(0)), where α= (α1, . . ., α n−1,1). 2 Definition. Letf= (f1, . . ., f m),F= (F1, . . ., F m),fi=fi(x),Fi=Fi(x), andf, F∈C∞. We say fismajorized byFif |Dαfk(0)| ≤DαFk(0), k= 1, . . ., m 3.5. THEOREM OF CAUCHY-KOVALEVSKAYA 87 for all α. We write f << F , iffis majorized by F. Definition. The initial value problem Uj,xn=n−1/summationdisplay i=1N/summationdisplay k=1Ai jk(z)Uk,xi+Bj(z) (3.32) Uj(x) = 0 if xn= 0, (3.33) j= 1, . . ., N ,Ai jk, Bjreal analytic, ist called majorizing problem to (3.28), (3.29) if ai jk<< Ai jkandbj<< B j. Lemma B. The formal power series /summationdisplay α1 α!Dαuj(0)xα, where Dαuj(0)are defined in Lemma A, is convergent in a neighbourhood of 0∈Rnif there exists a majorizing problem which has a real analyti c solution Uinx= 0, and |Dαuj(0)| ≤DαUj(0). Proof. It follows from Lemma A and from the assumption of Lemma B that |Dαuj(0)| ≤ Pα/parenleftBig |dβai jk(0)|,|dγbj(0)|/parenrightBig ≤Pα/parenleftBig |dβAi jk(0)|,|dγBj(0)|/parenrightBig ≡DαUj(0). The formal power series/summationdisplay α1 α!Dαuj(0)xα, is convergent since /summationdisplay α1 α!|Dαuj(0)xα| ≤/summationdisplay α1 α!DαUj(0)|xα|. The right hand side is convergent in a neighbourhood of x∈Rnby assump- tion. 2 Lemma C. There is a majorising problem which has a real analytic solut ion. 88 CHAPTER 3. CLASSIFICATION Proof. Since ai ij(z),bj(z) are real analytic in a neighbourhood of z= 0 it follows from Proposition A5 of the appendix to this section t hat there are positive constants Mandrsuch that all these functions are majorized by Mr r−z1−. . .−zN+n−1. Thus a majorizing problem is Uj,xn=Mr r−x1−. . .−xn−1−U1−. . .−UN/parenleftBigg 1 +n−1/summationdisplay i=1N/summationdisplay k=1Uk,xi/parenrightBigg Uj(x) = 0 if xn= 0, j= 1, . . ., N . The solution of this problem is Uj(x1, . . ., x n−1, xn) =V(x1+. . .+xn−1, xn), j= 1, . . ., N, where V(s, t),s=x1+. . .+xn−1,t=xn, is the solution of the Cauchy initial value problem Vt=Mr r−s−NV(1 +N(n−1)Vs), V(s,0) = 0 . which has the solution, see an exercise, V(s, t) =1 Nn/parenleftBig r−s−/radicalbig (r−s)2−2nMNrt/parenrightBig . This function is real analytic in ( s, t) at (0 ,0). It follows that Uj(x) are also real analytic functions. Thus the Cauchy-Kovalevskaya theo rem is shown. 2 3.5. THEOREM OF CAUCHY-KOVALEVSKAYA 89 Examples: 1. Ordinary differential equations Consider the initial value problem y/prime(x) = f(x, y(x)) y(x0) = y0, where x0∈Randy0∈Rnare given. Assume f(x, y) is real analytic in a neighbourhood of ( x0, y0)∈R×Rn. Then it follows from the above theorem that there exists an analytic solution y(x) of the initial value problem in a neighbourhood of x0. This solution is unique in the class of analytic func- tions according to the theorem of Cauchy-Kovalevskaya. From the Picard- Lindel¨ of theorem it follows that this analytic solution is unique even in the class of C1-functions. 2. Partial differential equations of second order Consider the boundary value problem for two variables uyy=f(x, y, u, u x, uy, uxx, uxy) u(x,0) = φ(x) uy(x,0) = ψ(x). We assume that φ, ψare analytic in a neighbourhood of x= 0 and that f is real analytic in a neighbourhood of (0,0, φ(0), φ/prime(0), ψ(0), ψ/prime(0)). There exists a real analytic solution in a neigbourhood of 0∈R2of the above initial value problem. In particular, there is a real analytic solution in a neigbou rhood of 0 ∈R2 of the initial value problem /triangleu= 1 u(x,0) = 0 uy(x,0) = 0 . 90 CHAPTER 3. CLASSIFICATION The proof follows by writing the above problem as a system. Se tp=ux, q=uy,r=uxx,s=uxy,t=uyy, then t=f(x, y, u, p, q, r, s ). SetU= (u, p, q, r, s, t )T,b= (q,0, t,0,0, fy+fuq+fqt)Tand A= 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 fp0frfs . Then the rewritten differential equation is the system Uy=AUx+bwith the initial condition U(x,0) =/parenleftbig φ(x), φ/prime(x), ψ(x), φ/prime/prime(x), ψ/prime(x), f0(x)/parenrightbig , where f0(x) =f(x,0, φ(x), φ/prime(x), ψ(x), φ/prime/prime(x), ψ/prime(x)). 3.5.1 Appendix: Real analytic functions Multi-index notation The following multi-index notation simplifies many presentat ions of formu- las. Let x= (x1, . . ., x n) and u: Ω⊂Rn/mapsto→R(orRmfor systems) . The n-tuple of nonnegative integers (including zero) α= (α1, . . ., α n) is called multi-index . Set |α|=α1+. . .+αn α! = α1!α2!·. . .·αn! xα=xα1 1xα2 2·. . .·xαnn(for a monom) Dk=∂ ∂xk D= (D1, . . ., D n) Du= (D1u, . . ., D nu)≡ ∇u≡grad u Dα=Dα1 1Dα2 2·. . .·Dαnn≡∂|α| ∂xα1 1∂xα2 2. . . ∂xαnn. 3.5. THEOREM OF CAUCHY-KOVALEVSKAYA 91 Define a partial order by α≥βif and only if αi≥βifor all i. Sometimes we use the notations 0= (0,0. . . ,0),1= (1,1. . .,1), where 0,1∈Rn. Using this multi-index notion, we have 1. (x+y)α=/summationdisplay β, γ β+γ=αα! β!γ!xβyγ, where x, y∈Rnandα, β, γ are multi-indices. 2.Taylor expansion for a polynomial f(x) of degree m: f(x) =/summationdisplay |α|≤m1 α!(Dαf(0))xα, here is Dαf(0) := ( Dαf(x))|x=0. 3.Letx= (x1, . . ., x n) and m≥0 an integer, then (x1+. . .+xn)m=/summationdisplay |α|=mm! α!xα. 4. α!≤ |α|!≤n|α|α!. 5.Leibniz’s rule: Dα(fg) =/summationdisplay β, γ β+γ=αα! β!γ!(Dβf)(Dγg). 92 CHAPTER 3. CLASSIFICATION 6. Dβxα=α! (α−β)!xα−βifα≥β, Dβxα= 0 otherwise . 7.Directional derivative: dm dtmf(x+ty) =/summationdisplay |α|=m|α|! α!(Dαf(x+ty))yα, where x, y∈Rnandt∈R. 8.Taylor’s theorem: Let u∈Cm+1in a neighbourhood N(y) ofy, then, if x∈N(y), u(x) =/summationdisplay |α|≤m1 α!(Dαu(y)) (x−y)α+Rm, where Rm=/summationdisplay |α|=m+11 α!(Dαu(y+δ(x−y)))xα,0< δ < 1, δ=δ(u, m, x, y ), or Rm=1 m!/integraldisplay1 0(1−t)mΦ(m+1)(t)dt, where Φ( t) =u(y+t(x−y)). It follows from 7.that Rm= (m+ 1)/summationdisplay |α|=m+11 α!/parenleftbigg/integraldisplay1 0(1−t)Dαu(y+t(x−y))dt/parenrightbigg (x−y)α. 9.Using multi-index notation, the general linear partial differ ential equation of order mcan be written as /summationdisplay |α|≤maα(x)Dαu=f(x) in Ω ⊂Rn. 3.5. THEOREM OF CAUCHY-KOVALEVSKAYA 93 Power series Here we collect some definitions and results for power series i nRn. Definition. Letcα∈R(or∈Rm). The series /summationdisplay αcα≡∞/summationdisplay m=0 /summationdisplay |α|=mcα  is said to be convergent if /summationdisplay α|cα| ≡∞/summationdisplay m=0 /summationdisplay |α|=m|cα|  is convergent. Remark. According to the above definition, a convergent series is abso - lutely convergent. It follows that we can rearrange the orde r of summation. Using the above multi-index notation and keeping in mind that we can rearrange convergent series, we have 10.Letx∈Rn, then /summationdisplay αxα=n/productdisplay i=1/parenleftBigg∞/summationdisplay αi=0xαi i/parenrightBigg =1 (1−x1)(1−x2)·. . .·(1−xn) =1 (1−x)1, provided |xi|<1 is satisfied for each i. 11.Assume x∈Rnand|x1|+|x2|+. . .+|xn|<1, then /summationdisplay α|α|! α!xα=∞/summationdisplay j=0/summationdisplay |α|=j|α|! α!xα =∞/summationdisplay j=0(x1+. . .+xn)j =1 1−(x1+. . .+xn). 94 CHAPTER 3. CLASSIFICATION 12.Letx∈Rn,|xi|<1 for all i, and βis a given multi-index. Then /summationdisplay α≥βα! (α−β)!xα−β=Dβ1 (1−x)1 =β! (1−x)1+β. 13.Letx∈Rnand|x1|+. . .+|xn|<1. Then /summationdisplay α≥β|α|! (α−β)!xα−β=Dβ 1 1−x1−. . .−xn =|β|! (1−x1−. . .−xn)1+|β|. Consider the power series /summationdisplay αcαxα(3.34) and assume this series is convergent for a z∈Rn. Then, by definition, µ:=/summationdisplay α|cα||zα|<∞ and the series (3.34) is uniformly convergent for all x∈Q(z), where Q(z) :|xi| ≤ |zi|for all i. Thus the power series (3.34) defines a continuous function defin ed on Q(z), according to a theorem of Weierstrass. The interior of Q(z) is not empty if and only if zi/negationslash= 0 for all i, see Figure 3.7. For given xin a fixed compact subset DofQ(z) there is a q,0< q < 1, such that |xi| ≤q|zi|for all i. Set f(x) =/summationdisplay αcαxα. 3.5. THEOREM OF CAUCHY-KOVALEVSKAYA 95 zQ(z) D Figure 3.7: Definition of D∈Q(z) Proposition A1. (i)In every compact subset DofQ(z)one has f∈ C∞(D)and the formal differentiate series, that is/summationtext αDβcαxα, is uniformly convergent on the closure of Dand is equal to Dβf. (ii) |Dβf(x)| ≤M|β|!r−|β|inD, where M=µ (1−q)n, r = (1−q)min i|zi|. Proof. See F. John [10], p. 64. Or an exercise. Hint: Use formula 12.where xis replaced by ( q, . . . , q ). Remark. From the proposition above it follows cα=1 α!Dαf(0). Definition. Assume fis defined on a domain Ω ⊂Rn, then fis said to be real analytic in y∈Ω if there are cα∈Rand if there is a neighbourhood N(y) ofysuch that f(x) =/summationdisplay αcα(x−y)α 96 CHAPTER 3. CLASSIFICATION for all x∈N(y), and the series converges (absolutely) for each x∈N(y). A function fis called real analytic in Ω if it is real analytic for each y∈Ω. We will write f∈Cω(Ω) in the case that fis real analytic in the domain Ω. A vector valued function f(x) = ( f1(x), . . ., f m) is called real analytic if each coordinate is real analytic. Proposition A2. (i)Letf∈Cω(Ω). Then f∈C∞(Ω). (ii)Assume f∈Cω(Ω). Then for each y∈Ωthere exists a neighbourhood N(y)and positive constants M,rsuch that f(x) =/summationdisplay α1 α!(Dαf(y))(x−y)α for all x∈N(y), and the series converges (absolutely) for each x∈N(y), and |Dβf(x)| ≤M|β|!r−|β|. The proof follows from Proposition A1. An open set Ω ∈Rnis called connected if Ω is not a union of two nonempty open sets with empty intersection. An open set Ω ∈Rnis connected if and only if its path connected, see [11], pp. 38, for example. We s ay that Ω ispath connected if for any x, y∈Ω there is a continuous curve γ(t)∈Ω, 0≤t≤1, with γ(0) = xandγ(1) = y. From the theory of one complex variable we know that a continuation of an analytic function is uniquely determined. The same is true for real analytic functions. Proposition A3. Assume f∈Cω(Ω) and Ω is connected. Then fis uniquely determined if for one z∈Ω allDαf(z) are known. Proof. See F. John [10], p. 65. Suppose g, h∈Cω(Ω) and Dαg(z) =Dαh(z) for every α. Setf=g−hand Ω1={x∈Ω :Dαf(x) = 0 for all α}, Ω2={x∈Ω :Dαf(x)/negationslash= 0 for at least one α}. The set Ω 2is open since Dαfare continuous in Ω. The set Ω 1is also open sincef(x) = 0 in a neighbourhood of y∈Ω1. This follows from f(x) =/summationdisplay α1 α!(Dαf(y))(x−y)α. 3.5. THEOREM OF CAUCHY-KOVALEVSKAYA 97 Since z∈Ω1, i. e., Ω 1/negationslash=∅, it follows Ω 2=∅. 2 It was shown in Proposition A2 that derivatives of a real anal ytic function satisfy estimates. On the other hand it follows, see the next proposition, that a function f∈C∞is real analytic if these estimates are satisfied. Definition. Lety∈Ω and M, r positive constants. Then fis said to be in the class CM,r(y) iff∈C∞in a neighbourhood of yand if |Dβf(y)| ≤M|β|!r−|β| for all β. Proposition A4. f∈Cω(Ω)if and only if f∈C∞(Ω)and for every compact subset S⊂Ωthere are positive constants M, rsuch that f∈CM,r(y)for all y∈S. Proof. See F. John [10], pp. 65-66. We will prove the local version of t he proposition, that is, we show it for each fixed y∈Ω. The general version follows from Heine-Borel theorem. Because of Proposition A3 it remains to show that the Taylor series /summationdisplay α1 α!Dαf(y)(x−y)α converges (absolutely) in a neighbourhood of yand that this series is equal tof(x). Define a neighbourhood of yby Nd(y) ={x∈Ω :|x1−y1|+. . .+|xn−yn|< d}, where dis a sufficiently small positive constant. Set Φ( t) =f(y+t(x−y)). The one-dimensional Taylor theorem says f(x) = Φ(1) =j−1/summationdisplay k=01 k!Φ(k)(0) + rj, where rj=1 (j−1)!/integraldisplay1 0(1−t)j−1Φ(j)(t)dt. 98 CHAPTER 3. CLASSIFICATION From formula 7.for directional derivatives it follows for x∈Nd(y) that 1 j!dj dtjΦ(t) =/summationdisplay |α|=j1 α!Dαf(y+t(x−y))(x−y)α. From the assumption and the multinomial formula 3.we get for 0 ≤t≤1 /vextendsingle/vextendsingle/vextendsingle/vextendsingle1 j!dj dtjΦ(t)/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤M/summationdisplay |α|=j|α|! α!r−|α||(x−y)α| =Mr−j(|x1−y1|+. . .+|xn−yn|)j ≤M/parenleftbiggd r/parenrightbiggj . Choose d >0 such that d < r, then the Taylor series converges (absolutely) inNd(y) and it is equal to f(x) since the remainder satisfies, see the above estimate, |rj|=/vextendsingle/vextendsingle/vextendsingle/vextendsingle1 (j−1)!/integraldisplay1 0(1−t)j−1Φj(t)dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤M/parenleftbiggd r/parenrightbiggj . 2 We recall that the notation f << F (fis majorized by F) was defined in the previous section. Proposition A5. (i)f= (f1, . . ., f m)∈CM,r(0)if and only if f << (Φ, . . .,Φ), where Φ(x) =Mr r−x1−. . .−xn. (ii)f∈CM,r(0)andf(0) = 0 if and only if f << (Φ−M, . . ., Φ−M), where Φ(x) =M(x1+. . .+xn) r−x1−. . .−xn. Proof. DαΦ(0) = M|α|!r−|α|. 3.5. THEOREM OF CAUCHY-KOVALEVSKAYA 99 2 Remark. The definition of f << F implies, trivially, that Dαf << DαF. The next proposition shows that compositions majorize if th e involved func- tions majorize. More precisely, we have Proposition A6. Letf, F:Rn/mapsto→Rmandg, Gmaps a neighbourhood of 0∈RmintoRp. Assume all functions f(x), F(x), g(u), G(u)are in C∞, f(0) = F(0) = 0 ,f << F andg << G . Then g(f(x))<< G (F(x)). Proof. See F. John [10], p. 68. Set h(x) =g(f(x)), H(x) =G(F(x)). For each coordinate hkofhwe have, according to the chain rule, Dαhk(0) = Pα(δβgl(0), Dγfj(0)), where Pαare polynomials with nonnegative integers as coefficients, Pαare independent on gorfandδ:= (∂/∂u 1, . . ., ∂/∂u m). Thus, |Dαhk(0)| ≤ Pα(|δβgl(0)|,|Dγfj(0)|) ≤Pα(δβGl(0), DγFj(0)) =DαHk(0). 2 Using this result and Proposition A4, which characterizes r eal analytic func- tions, it follows that compositions of real analytic functi ons are real analytic functions again. Proposition A7. Assume f(x)andg(u)are real analytic, then g(f(x))is real analytic if f(x)is in the domain of definition of g. Proof. See F. John [10], p. 68. Assume that fmaps a neighbourhood of y∈RninRmandgmaps a neighbourhood of v=f(y) inRm. Then f∈CM,r(y) and g∈Cµ,ρ(v) implies h(x) :=g(f(x))∈Cµ,ρr/(mM+ρ)(y). Once one has shown this inclusion, the proposition follows f rom Proposi- tion A4. To show the inclusion, we set h(y+x) :=g(f(y+x))≡g(v+f(y+x)−f(x)) =:g∗(f∗(x)), 100 CHAPTER 3. CLASSIFICATION where v=f(y) and g∗(u) : = g(v+u)∈Cµ,ρ(0) f∗(x) : = f(y+x)−f(y)∈CM,r(0). In the above formulas v, yare considered as fixed parameters. From Propo- sition A5 it follows f∗(x)<<(Φ−M, . . ., Φ−M) =:F g∗(u)<<(Ψ, . . . ,Ψ) =: G, where Φ(x) =Mr r−x1−x2−. . .−xn Ψ(u) =µρ ρ−x1−x2−. . .−xn. From Proposition A6 we get h(y+x)<<(χ(x), . . . , χ (x))≡G(F), where χ(x) =µρ ρ−m(Φ(x)−M) =µρ(r−x1−. . .−xn) ρr−(ρ+mM)(x1+. . .+xn) <<µρr ρr−(ρ+mM)(x1+. . .+xn) =µρr/(ρ+mM) ρr/(ρ+mM)−(x1+. . . x n). See an exercise for the ” <<”-inequality. 2 3.6. EXERCISES 101 3.6 Exercises 1. Let χ:Rn→RinC1,∇χ/negationslash= 0. Show that for given x0∈Rnthere is in a neighbourhood of x0a local diffeomorphism λ= Φ(x), Φ : (x1, . . ., x n)/mapsto→(λ1, . . ., λ n), such that λn=χ(x). 2. Show that the differential equation a(x, y)uxx+ 2b(x, y)uxy+c(x, y)uyy+ lower order terms = 0 is elliptic if ac−b2>0, parabolic if ac−b2= 0 and hyperbolic if ac−b2<0. 3. Show that in the hyperbolic case there exists a solution of φx+µ1φy= 0, see equation (3.9), such that ∇φ/negationslash= 0. Hint: Consider an appropriate Cauchy initial value problem. 4. Show equation (3.4). 5. Find the type of Lu:= 2uxx+ 2uxy+ 2uyy= 0 and transform this equation into an equation with vanishing mixed derivatives by using the orthogonal mapping (transform to p rincipal axis)x=Uy, U orthogonal. 6. Determine the type of the following equation at ( x, y) = (1 ,1/2). Lu:=xuxx+ 2yuxy+ 2xyuyy= 0. 7. Find all C2-solutions of uxx−4uxy+uyy= 0. Hint: Transform to principal axis and stretching of axis lead to th e wave equation. 8. Oscillations of a beam are described by wx−1 Eσt= 0 σx−ρwt= 0, 102 CHAPTER 3. CLASSIFICATION where σstresses, wdeflection of the beam and E, ρare positive con- stants. a) Determine the type of the system. b) Transform the system into two uncoupled equations, that i s,w, σ occur only in one equation, respectively. c) Find non-zero solutions. 9. Find nontrivial solutions ( ∇χ/negationslash= 0) of the characteristic equation to x2uxx−uyy=f(x, y, u, ∇u), where fis given. 10. Determine the type of uxx−xuyx+uyy+ 3ux= 2x, where u=u(x, y). 11. Transform equation uxx+ (1−y2)uxy= 0, u=u(x, y), into its normal form. 12. Transform the Tricomi-equation yuxx+uyy= 0, u=u(x, y), where y <0, into its normal form. 13. Transform equation x2uxx−y2uyy= 0, u=u(x, y), into its normal form. 14. Show that λ=1 (1 +|p|2)3/2,Λ =1 (1 +|p|2)1/2. are the minimum and maximum of eigenvalues of the matrix ( aij), where aij=/parenleftbig 1 +|p|2/parenrightbig−1/2/parenleftbigg δij−pipj 1 +|p|2/parenrightbigg . 15. Show that Maxwell equations are a hyperbolic system. 3.6. EXERCISES 103 16. Consider Maxwell equations and prove that d iv E= 0 and d iv H= 0 for all tif these equations are satisfied for a fixed time t0. Hint. divrot A= 0 for each C2-vector field A= (A1, A2, A3). 17. Assume a characteristic surface S(t) inR3is defined by χ(x, y, z, t ) = const. such that χt= 0 and χz/negationslash= 0. Show that S(t) has a nonparamet- ric representation z=u(x, y, t) with ut= 0, that is S(t) is independent oft. 18. Prove formula (3.22) for the normal on a surface. 19. Prove formula (3.23) for the speed of the surface S(t). 20. Write the Navier-Stokes system as a system of type (3.24). 21. Show that the following system (linear elasticity, stat ionary case of (3.25) in the two dimensional case) is elliptic µ/triangleu+ (λ+µ) grad(div u) +f= 0, where u= (u1, u2). The vector f= (f1, f2) is given and λ, µ are positive constants. 22. Discuss the type of the following system in stationary ga s dynamics (isentrop flow) in R2. ρuux+ρvuy+a2ρx= 0 ρuvx+ρvvy+a2ρy= 0 ρ(ux+vy) +uρx+vρy= 0. Here are ( u, v) velocity vector, ρdensity and a=/radicalbig p/prime(ρ) the sound velocity. 23. Show formula 7.(directional derivative). Hint: Induction with respect to m. 24. Let y=y(x) be the solution of: y/prime(x) = f(x, y(x)) y(x0) = y0, 104 CHAPTER 3. CLASSIFICATION where fis real analytic in a neighbourhood of ( x0, y0)∈R2. Find the polynomial Pof degree 2 such that y(x) =P(x−x0) +O(|x−x0|3) asx→x0. 25. Let ube the solution of /triangleu= 1 u(x,0) = uy(x,0) = 0 . Find the polynomial Pof degree 2 such that u(x, y) =P(x, y) +O((x2+y2)3/2) as (x, y)→(0,0). 26. Solve the Cauchy initial value problem Vt=Mr r−s−NV(1 +N(n−1)Vs) V(s,0) = 0 . Hint: Multiply the differential equation with ( r−s−NV). 27. Write /triangle2u=−uas a system of first order. Hint: /triangle2u≡ /triangle(/triangleu). 28. Write the minimal surface equation ∂ ∂x ux/radicalBig 1 +u2x+u2y +∂ ∂y uy/radicalBig 1 +u2x+u2y = 0 as a system of first order. Hint: v1:=ux//radicalBig 1 +u2x+u2y, v2:=uy//radicalBig 1 +u2x+u2y. 29. Let f:R×Rm→Rmbe real analytic in ( x0, y0). Show that a real analytic solution in a neighbourhood of x0of the problem y/prime(x) = f(x, y) y(x0) = y0 exists and is equal to the unique C1[x0−/epsilon1, x0+/epsilon1]-solution from the Picard-Lindel¨ of theorem, /epsilon1 >0 sufficiently small. 3.6. EXERCISES 105 30. Show (see the proof of Proposition A7) µρ(r−x1−. . .−xn) ρr−(ρ+mM)(x1+. . .+xn)<<µρr ρr−(ρ+mM)(x1+. . .+xn). Hint: Leibniz’s rule. 106 CHAPTER 3. CLASSIFICATION Chapter 4 Hyperbolic equations Here we consider hyperbolic equations of second order, main ly wave equa- tions. 4.1 One-dimensional wave equation The one-dimensional wave equation is given by 1 c2utt−uxx= 0, (4.1) where u=u(x, t) is a scalar function of two variables and cis a positive constant. According to previous considerations, all C2-solutions of the wave equation are u(x, t) =f(x+ct) +g(x−ct), (4.2) with arbitrary C2-functions fandg TheCauchy initial value problem for the wave equation is to find a C2- solution of 1 c2utt−uxx= 0 u(x,0) = α(x) ut(x,0) = β(x), where α, β∈C2(−∞,∞) are given. 107 108 CHAPTER 4. HYPERBOLIC EQUATIONS Theorem 4.1. There exists a unique C2(R×R)-solution of the Cauchy initial value problem, and this solution is given by d’Alemb ert’s1formula u(x, t) =α(x+ct) +α(x−ct) 2+1 2c/integraldisplayx+ct x−ctβ(s)ds. (4.3) Proof. Assume there is a solution u(x, t) of the Cauchy initial value problem, then it follows from (4.2) that u(x,0) = f(x) +g(x) =α(x) (4.4) ut(x,0) = cf/prime(x)−cg/prime(x) =β(x). (4.5) From (4.4) we obtain f/prime(x) +g/prime(x) =α/prime(x), which implies, together with (4.5), that f/prime(x) =α/prime(x) +β(x)/c 2 g/prime(x) =α/prime(x)−β(x)/c 2. Then f(x) =α(x) 2+1 2c/integraldisplayx 0β(s)ds+C1 g(x) =α(x) 2−1 2c/integraldisplayx 0β(s)ds+C2. The constants C1,C2satisfy C1+C2=f(x) +g(x)−α(x) = 0, see (4.4). Thus each C2-solution of the Cauchy initial value problem is given by d’Alembert’s formula. On the other hand, the function u(x, t) defined by the right hand side of (4.3) is a solution of the initial value problem. 2 Corollaries. 1. The solution u(x, t) of the initial value problem depends on the values of αat the endpoints of the interval [ x−ct, x+ct] and on the values of βon this interval only, see Figure 4.1. The interval [ x−ct, x+ct] is called domain of dependence . 1d’Alembert, Jean Babtiste le Rond, 1717-1783 4.2. HIGHER DIMENSIONS 109 x+ct=const.t x x x +ct x −ct(x ,t )0 0 0 0 0 0 0 Figure 4.1: Interval of dependence 2.LetPbe a point on the x-axis. Then we ask which points ( x, t) need values of αorβatPin order to calculate u(x, t)? From the d’Alembert formula it follows that this domain is a cone, see Figure 4.2. This set is called domain of influence . t xPx−ct=const. Figure 4.2: Domain of influence 4.2 Higher dimensions Set 2u=utt−c2/triangleu,/triangle ≡ /triangle x=∂2/∂x2 1+. . .+∂2/∂x2 n, 110 CHAPTER 4. HYPERBOLIC EQUATIONS and consider the initial value problem 2u= 0 in Rn×R (4.6) u(x,0) = f(x) (4.7) ut(x,0) = g(x), (4.8) where fandgare given C2(R2)-functions. By using spherical means and the above d’Alembert formula we will derive a formula for the solution of this initial value probl em. Method of spherical means Define the spherical mean for a C2-solution u(x, t) of the initial value prob- lem by M(r, t) =1 ωnrn−1/integraldisplay ∂Br(x)u(y, t)dSy, (4.9) where ωn= (2π)n/2/Γ(n/2) is the area of the n-dimensional sphere, ωnrn−1is the area of a sphere with radius r. From the mean value theorem of the integral calculus we obtai n the function u(x, t) for which we are looking at by u(x, t) = lim r→0M(r, t). (4.10) Using the initial data, we have M(r,0) =1 ωnrn−1/integraldisplay ∂Br(x)f(y)dSy=:F(r) (4.11) Mt(r,0) =1 ωnrn−1/integraldisplay ∂Br(x)g(y)dSy=:G(r), (4.12) which are the spherical means of fandg. The next step is to derive a partial differential equation for t he spherical mean. From definition (4.9) of the spherical mean we obtain, af ter the mapping ξ= (y−x)/r, where xandrare fixed, M(r, t) =1 ωn/integraldisplay ∂B1(0)u(x+rξ, t)dSξ. 4.2. HIGHER DIMENSIONS 111 It follows Mr(r, t) =1 ωn/integraldisplay ∂B1(0)n/summationdisplay i=1uyi(x+rξ, t)ξidSξ =1 ωnrn−1/integraldisplay ∂Br(x)n/summationdisplay i=1uyi(y, t)ξidSy. Integration by parts yields 1 ωnrn−1/integraldisplay Br(x)n/summationdisplay i=1uyiyi(y, t)dy sinceξ≡(y−x)/ris the exterior normal at ∂Br(x). Assume uis a solution of the wave equation, then rn−1Mr=1 c2ωn/integraldisplay Br(x)utt(y, t)dy =1 c2ωn/integraldisplayr 0/integraldisplay ∂Bc(x)utt(y, t)dSydc. The previous equation follows by using spherical coordinat es. Consequently (rn−1Mr)r=1 c2ωn/integraldisplay ∂Br(x)utt(y, t)dSy =rn−1 c2∂2 ∂t2/parenleftBigg 1 ωnrn−1/integraldisplay ∂Br(x)u(y, t)dSy/parenrightBigg =rn−1 c2Mtt. Thus we arrive at the differential equation (rn−1Mr)r=c−2rn−1Mtt, which can be written as Mrr+n−1 rMr=c−2Mtt. (4.13) This equation (4.13) is called Euler-Poisson-Darboux equation . 112 CHAPTER 4. HYPERBOLIC EQUATIONS 4.2.1 Case n=3 The Euler-Poisson-Darboux equation in this case is (rM)rr=c−2(rM)tt. Thus rMis the solution of the one-dimensional wave equation with ini tial data (rM)(r,0) =rF(r) (rM)t(r,0) =rG(r). (4.14) From the d’Alembert formula we get formally M(r, t) =(r+ct)F(r+ct) + (r−ct)F(r−ct) 2r +1 2cr/integraldisplayr+ct r−ctξG(ξ)dξ. (4.15) The right hand side of the previous formula is well defined if th e domain of dependence [ x−ct, x+ct] is a subset of (0 ,∞). We can extend Fand GtoF0andG0which are defined on ( −∞,∞) such that rF0andrG0are C2(R)-functions as follows. Set F0(r) =  F(r) : r >0 f(x) : r= 0 F(−r) : r <0 The function G0(r) is given by the same definition where Fandfare re- placed by Gandg, respectively. Lemma. rF0(r), rG 0(r)∈C2(R2). Proof. From definition of F(r) and G(r),r >0, it follows from the mean value theorem lim r→+0F(r) =f(x),lim r→+0G(r) =g(x). Thus rF0(r) and rG0(r) are C(R)-functions. These functions are also in 4.2. HIGHER DIMENSIONS 113 C1(R). This follows since F0andG0are in C1(R). We have, for example, F/prime(r) =1 ωn/integraldisplay ∂B1(0)n/summationdisplay j=1fyj(x+rξ)ξjdSξ F/prime(+0) =1 ωn/integraldisplay ∂B1(0)n/summationdisplay j=1fyj(x)ξjdSξ =1 ωnn/summationdisplay j=1fyj(x)/integraldisplay ∂B1(0)njdSξ = 0. Then, rF0(r) and rG0(r) are in C2(R), provided F/prime/primeandG/prime/primeare bounded as r→+0. This property follows from F/prime/prime(r) =1 ωn/integraldisplay ∂B1(0)n/summationdisplay i,j=1fyiyj(x+rξ)ξiξjdSξ. Thus F/prime/prime(+0) =1 ωnn/summationdisplay i,j=1fyiyj(x)/integraldisplay ∂B1(0)ninjdSξ. We recall that f, g∈C2(R2) by assumption. 2 The solution of the above initial value problem, where FandGare replaced byF0andG0, respectively, is M0(r, t) =(r+ct)F0(r+ct) + (r−ct)F0(r−ct) 2r +1 2cr/integraldisplayr+ct r−ctξG0(ξ)dξ. Since F0andG0are even functions, we have /integraldisplayct−r r−ctξG0(ξ)dξ= 0. Thus M0(r, t) =(r+ct)F0(r+ct)−(ct−r)F0(ct−r) 2r +1 2cr/integraldisplayct+r ct−rξG0(ξ)dξ, (4.16) 114 CHAPTER 4. HYPERBOLIC EQUATIONS ct+r r−ct ct−r Figure 4.3: Changed domain of integration see Figure 4.3. For fixed t >0 and 0 < r < ct it follows that M0(r, t) is the solution of the initial value problem with given initially da ta (4.14) since F0(s) =F(s),G0(s) =G(s) ifs >0. Since for fixed t >0 u(x, t) = lim r→0M0(r, t), it follows from d’Hospital’s rule that u(x, t) = ctF/prime(ct) +F(ct) +tG(ct) =d dt(tF(ct)) +tG(ct). Theorem 4.2. Assume f∈C3(R3)andg∈C2(R3)are given. Then there exists a unique solution u∈C2(R3×[0,∞))of the initial value prob- lem (4.6)-(4.7), where n= 3, and the solution is given by the Poisson’s formula u(x, t) =1 4πc2∂ ∂t/parenleftBigg 1 t/integraldisplay ∂Bct(x)f(y)dSy/parenrightBigg +1 4πc2t/integraldisplay ∂Bct(x)g(y)dSy. (4.17) Proof. Above we have shown that a C2-solution is given by Poisson’s for- mula. Under the additional assumption f∈C3it follows from Poisson’s 4.2. HIGHER DIMENSIONS 115 formula that this formula defines a solution which is in C2, see F. John [10], p. 129. 2 Corollary. From Poisson’s formula we see that the domain of dependence foru(x, t0) is the intersection of the cone defined by |y−x|=c|t−t0|with the hyperplane defined by t= 0, see Figure 4.4 t x(x,t )0 |y−x|=c| t−t |0 Figure 4.4: Domain of dependence, case n= 3 4.2.2 Case n= 2 Consider the initial value problem vxx+vyy=c−2vtt (4.18) v(x, y,0) = f(x, y) (4.19) vt(x, y,0) = g(x, y), (4.20) where f∈C3, g∈C2. Using the formula for the solution of the three-dimensional i nitial value problem we will derive a formula for the two-dimensional case . The following consideration is called Hadamard’s method of decent . Letv(x, y, t) be a solution of (4.18)-(4.20), then u(x, y, z, t ) :=v(x, y, t) 116 CHAPTER 4. HYPERBOLIC EQUATIONS is a solution of the three-dimensional initial value problem with initial data f(x, y),g(x, y), independent of z, since usatisfies (4.18)-(4.20). Hence, since u(x, y, z, t ) =u(x, y,0, t) +uz(x, y, δz, t )z, 0< δ < 1, and uz= 0, we have v(x, y, t) =u(x, y,0, t). Poisson’s formula in the three-dimensional case implies v(x, y, t) =1 4πc2∂ ∂t/parenleftBigg 1 t/integraldisplay ∂Bct(x,y,0)f(ξ, η)dS/parenrightBigg +1 4πc2t/integraldisplay ∂Bct(x,y,0)g(ξ, η)dS. (4.21) n dSS _+ Sr ξηζ ηdξd Figure 4.5: Domains of integration The integrands are independent on ζ. The surface Sis defined by χ(ξ, η, ζ) := (ξ−x)2+ (η−y)2+ζ2−c2t2= 0. Then the exterior normal n atSisn=∇χ/|∇χ|and the surface element is given by dS= (1/|n3|)dξdη, where the third coordinate of nis n3=±/radicalbig c2t2−(ξ−x)2−(η−y)2 ct. 4.3. INHOMOGENEOUS EQUATION 117 The positive sign applies on S+, where ζ >0 and the sign is negative on S− where ζ <0, see Figure 4.5. We have S=S+∪S−. Setρ=/radicalbig (ξ−x)2+ (η−y)2. Then it follows from (4.21) Theorem 4.3. The solution of the Cauchy initial value problem (4.18)- (4.20) is given by v(x, y, t) =1 2πc∂ ∂t/integraldisplay Bct(x,y)f(ξ, η)/radicalbig c2t2−ρ2dξdη +1 2πc/integraldisplay Bct(x,y)g(ξ, η)/radicalbig c2t2−ρ2dξdη. Corollary. In contrast to the three dimensional case, the domain of depe n- dence is here the disk Bcto(x0, y0) and not the boundary only. Therefore, see formula of Theorem 4.3, if f, ghave supports in a compact domain D⊂R2, then these functions have influence on the value v(x, y, t) foralltimet > T, Tsufficiently large. 4.3 Inhomogeneous equation Here we consider the initial value problem 2u=w(x, t) onx∈Rn, t∈R (4.22) u(x,0) = f(x) (4.23) ut(x,0) = g(x), (4.24) where 2u:=utt−c2/triangleu. We assume f∈C3,g∈C2andw∈C1, which are given. Setu=u1+u2, where u1is a solution of problem (4.22)-(4.24) with w:= 0 and u2is the solution where f= 0 and g= 0 in (4.22)-(4.24). Since we have explicit solutions u1in the cases n= 1,n= 2 and n= 3, it remains to solve 2u=w(x, t) onx∈Rn, t∈R (4.25) u(x,0) = 0 (4.26) ut(x,0) = 0 . (4.27) The following method is called Duhamel’s principle which can be considered as a generalization of the method of variations of constants in the theory of ordinary differential equations. 118 CHAPTER 4. HYPERBOLIC EQUATIONS To solve this problem, we make the ansatz u(x, t) =/integraldisplayt 0v(x, t, s)ds, (4.28) where vis a function satisfying 2v= 0 for all s (4.29) and v(x, s, s) = 0. (4.30) From ansatz (4.28) and assumption (4.30) we get ut=v(x, t, t) +/integraldisplayt 0vt(x, t, s)ds, =/integraldisplayt 0vt(x, t, s). (4.31) It follows ut(x,0) = 0. The initial condition u(x, t) = 0 is satisfied because of the ansatz (4.28). From (4.31) and ansatz (4.28) we see tha t utt=vt(x, t, t) +/integraldisplayt 0vtt(x, t, s)ds, /trianglexu=/integraldisplayt 0/trianglexv(x, t, s)ds. Therefore, since uis an ansatz for (4.25)-(4.27), utt−c2/trianglexu=vt(x, t, t) +/integraldisplayt 0(2v)(x, t, s)ds =w(x, t). Thus necessarily vt(x, t, t) =w(x, t), see (4.29). We have seen that the ansatz provides a solution of (4.25)-(4.27) if for all s 2v= 0, v(x, s, s) = 0, vt(x, s, s) =w(x, s). (4.32) Letv∗(x, t, s) be a solution of 2v= 0, v(x,0, s) = 0, vt(x,0, s) =w(x, s), (4.33) then v(x, t, s) :=v∗(x, t−s, s) 4.3. INHOMOGENEOUS EQUATION 119 is a solution of (4.32). In the case n= 3, where v∗is given by, see Theorem 4.2, v∗(x, t, s) =1 4πc2t/integraldisplay ∂Bct(x)w(ξ, s)dSξ. Then v(x, t, s) = v∗(x, t−s, s) =1 4πc2(t−s)/integraldisplay ∂Bc(t−s)(x)w(ξ, s)dSξ. from ansatz (4.28) it follows u(x, t) =/integraldisplayt 0v(x, t, s)ds =1 4πc2/integraldisplayt 0/integraldisplay ∂Bc(t−s)(x)w(ξ, s) t−sdSξds. Changing variables by τ=c(t−s) yields u(x, t) =1 4πc2/integraldisplayct 0/integraldisplay ∂Bτ(x)w(ξ, t−τ/c) τdSξdτ =1 4πc2/integraldisplay Bct(x)w(ξ, t−r/c) rdξ, where r=|x−ξ|. Formulas for the cases n= 1 and n= 2 follow from formulas for the as- sociated homogeneous equation with inhomogeneous initial values for these cases. Theorem 4.4. The solution of 2u=w(x, t), u(x,0) = 0 , ut(x,0) = 0 , where w∈C1, is given by: Casen= 3: u(x, t) =1 4πc2/integraldisplay Bct(x)w(ξ, t−r/c) rdξ, where r=|x−ξ|,x= (x1, x2, x3),ξ= (ξ1, ξ2, ξ3). 120 CHAPTER 4. HYPERBOLIC EQUATIONS Casen= 2: u(x, t) =1 4πc/integraldisplayt 0/parenleftBigg/integraldisplay Bc(t−τ)(x)w(ξ, τ)/radicalbig c2(t−τ)2−r2dξ/parenrightBigg dτ, x= (x1, x2),ξ= (ξ1, ξ2). Casen= 1: u(x, t) =1 2c/integraldisplayt 0/parenleftBigg/integraldisplayx+c(t−τ) x−c(t−τ)w(ξ, τ)dξ/parenrightBigg dτ. Remark. The integrand on the right in formula for n= 3 is called retarded potential . The integrand is taken not at t, it is taken at an earlier time t−r/c. 4.4 A method of Riemann Riemann’s method provides a formula for the solution of the f ollowing Cauchy initial value problem for a hyperbolic equation of se cond order in two variables. Let S:x=x(t), y=y(t), t1≤t≤t2, be a regular curve in R2, that is, we assume x, y∈C1[t1, t2] andx/prime2+y/prime2/negationslash= 0. Set Lu:=uxy+a(x, y)ux+b(x, y)uy+c(x, y)u, where a, b∈C1andc, f∈Cin a neighbourhood of S. Consider the initial value problem Lu=f(x, y) (4.34) u0(t) = u(x(t), y(t)) (4.35) p0(t) = ux(x(t), y(t)) (4.36) q0(t) = uy(x(t), y(t)), (4.37) where f∈Cin a neighbourhood of Sandu0, p0, q0∈C1are given. We assume: 4.4. A METHOD OF RIEMANN 121 (i)u/prime 0(t) =p0(t)x/prime(t) +q0(t)y/prime(t) (strip condition), (ii)Sis not a characteristic curve. Moreover assume that the char acteristic curves, which are lines here and are defined by x=const. andy=const. , have at most one point of intersection with S, and such a point is not a touching point, i. e., tangents of the characteristic and Sare different at this point. We recall that the characteristic equation to (4.34) is χxχy= 0 which is satisfied if χx(x, y) = 0 or χy(x, y) = 0. One family of characteris- tics associated to these first partial differential of first order is defined by x/prime(t) = 1, y/prime(t) = 0, see Chapter 2. Assume u, v∈C1and that uxy, vxyexist and are continuous. Define the adjoint differential expression by Mv=vxy−(av)x−(bv)y+cv. We have 2(vLu−uMv) = (uxv−vxu+ 2buv)y+ (uyv−vyu+ 2auv)x.(4.38) Set P=−(uxv−xxu+ 2buv) Q=uyv−vyu+ 2auv. From (4.38) it follows for a domain Ω ∈R2 2/integraldisplay Ω(vLu−uMv)dxdy =/integraldisplay Ω(−Py+Qx)dxdy =/contintegraldisplay Pdx+Qdy, (4.39) where integration in the line integral is anticlockwise. Th e previous equation follows from Gauss theorem or after integration by parts: /integraldisplay Ω(−Py+Qx)dxdy=/integraldisplay ∂Ω(−Pn2+Qn1)ds, where n= (dy/ds, −dx/ds ),sarc length, ( x(s), y(s)) represents ∂Ω. Assume uis a solution of the initial value problem (4.34)-(4.37) and suppose that vsatisfies Mv= 0 in Ω . 122 CHAPTER 4. HYPERBOLIC EQUATIONS A BSΩ xy 0P=(x ,y )0 Figure 4.6: Riemann’s method, domain of integration Then, if we integrate over a domain Ω as shown in Figure 4.6, it follows from (4.39) that 2/integraldisplay Ωvf dxdy =/integraldisplay BAPdx+Qdy+/integraldisplay APPdx+Qdy+/integraldisplay PBPdx+Qdy. (4.40) The line integral from BtoAis known from initial data, see the definition ofPandQ. Since uxv−vxu+ 2buv= (uv)x+ 2u(bv−vx), it follows /integraldisplay APPdx+Qdy =−/integraldisplay AP((uv)x+ 2u(bv−vx))dx =−(uv)(P) + (uv)(A)−/integraldisplay AP2u(bv−vx)dx. By the same reasoning we obtain for the third line integral /integraldisplay PBPdx+Qdy =/integraldisplay PB((uv)y+ 2u(av−vy))dy = (uv)(B)−(uv)(P) +/integraldisplay PB2u(av−vy)dy. 4.4. A METHOD OF RIEMANN 123 Combining these equations with (4.39), we get 2v(P)u(P) =/integraldisplay BA(uxv−vx+ 2buv)dx−(uyv−vyu+ 2auv)dy +u(A)v(A) +u(B)v(B) + 2/integraldisplay APu(bv−vx)dx +2/integraldisplay PBu(av−vy)dy−2/integraldisplay Ωfv dxdy. (4.41) Letvbe a solution of the initial value problem, see Figure 4.7 for the defi- nition of domain D(P), xy 0P=(x ,y )0C C21 D(P) Figure 4.7: Definition of Riemann’s function Mv = 0 in D(P) (4.42) bv−vx= 0 on C1 (4.43) av−vy= 0 on C2 (4.44) v(P) = 1 . (4.45) Assume vsatisfies (4.42)-(4.45), then 2u(P) = u(A)v(A) +u(B)v(B)−2/integraldisplay Ωfv dxdy =/integraldisplay BA(uxv−vx+ 2buv)dx−(uyv−vyu+ 2auv)dy, 124 CHAPTER 4. HYPERBOLIC EQUATIONS where the right hand side is known from given data. A function v=v(x, y;x0, y0) satisfying (4.42)-(4.45) is called Riemann’s function . Remark. Setw(x, y) =v(x, y;x0, y0) for fixed x0, y0. Then (4.42)-(4.45) imply w(x, y0) = exp/parenleftbigg/integraldisplayx x0b(τ, y0)dτ/parenrightbigg onC1, w(x0, y) = exp/parenleftbigg/integraldisplayy y0a(x0, τ)dτ/parenrightbigg onC2. Examples 1.uxy=f(x, y), then a Riemann function is v(x, y)≡1. 2.Consider the telegraph equation of Chapter 3 εµutt=c2/trianglexu−λµut, where ustands for one coordinate of electric or magnetic field. Intro ducing u=w(x, t)eκt, where κ=−λ/(2ε), we arrive at wtt=c2 εµ/trianglexw−λ2 4/epsilon12. Stretching the axis and transform the equation to the normal form we get finally the following equation, the new function is denoted by uand the new variables are denoted by x, yagain, uxy+cu= 0, with a positive constant c. We make the ansatz for a Riemann function v(x, y;x0, y0) =w(s), s= (x−x0)(y−y0) and obtain sw/prime/prime+w/prime+cw= 0. 4.5. INITIAL-BOUNDARY VALUE PROBLEMS 125 Substitution σ=√ 4csleads to Bessel’s differential equation σ2z/prime/prime(σ) +σz/prime(σ) +σ2z(σ) = 0, where z(σ) =w(σ2/(4c)). A solution is J0(σ) =J0/parenleftBig/radicalbig 4c(x−x0)(y−y0)/parenrightBig which defines a Riemann function since J0(0) = 1. Remark. Bessel’s differential equation is x2y/prime/prime(x) +xy/prime(x) + (x2−n2)y(x) = 0, where n∈R. Ifn∈N∪ {0}, then solutions are given by Bessel functions. One of the two linearly independent solutions is bounded at 0 . This bounded solution is the Bessel function Jn(x) of first kind and of order n, see [1], for example. 4.5 Initial-boundary value problems In previous sections we looked at solutions defined for all x∈Rnandt∈R. In this and in the following section we seek solutions u(x, t) defined in a bounded domain Ω ⊂Rnand for all t∈Rand which satisfy additional boundary conditions on ∂Ω. 4.5.1 Oscillation of a string Letu(x, t),x∈[a, b],t∈R, be the deflection of a string, see Figure 1.4 from Chapter 1. Assume the deflection occurs in the ( x, u)-plane. This problem is governed by the initial-boundary value problem utt(x, t) = uxx(x, t) on (0 , l) (4.46) u(x,0) = f(x) (4.47) ut(x,0) = g(x) (4.48) u(0, t) = u(l, t) = 0. (4.49) Assume the initial data f,gare sufficiently regular. This implies compati- bility conditions f(0) = f(l) = 0 and g(0) = g(l). 126 CHAPTER 4. HYPERBOLIC EQUATIONS Fourier’s method To find solutions of differential equation (4.46) we make the separation of variables ansatz u(x, t) =v(x)w(t). Inserting the ansatz into (4.46) we obtain v(x)w/prime/prime(t) =v/prime/prime(x)w(t), or, ifv(x)w(t)/negationslash= 0, w/prime/prime(t) w(t)=v/prime/prime(x) v(x). It follows, provided v(x)w(t) is a solution of differential equation (4.46) and v(x)w(t)/negationslash= 0, w/prime/prime(t) w(t)=const. =:−λ and v/prime/prime(x) v(x)=−λ sincex, tare independent variables. Assume v(0) = v(l) = 0, then v(x)w(t) satisfies the boundary condi- tion (4.49). Thus we look for solutions of the eigenvalue pro blem −v/prime/prime(x) = λv(x) in (0 , l) (4.50) v(0) = v(l) = 0, (4.51) which has the eigenvalues λn=/parenleftBigπ ln/parenrightBig2 , n= 1,2, . . ., and associated eigenfunctions are vn= sin/parenleftBigπ lnx/parenrightBig . Solutions of −w/prime/prime(t) =λnw(t) are sin(/radicalbig λnt),cos(/radicalbig λnt). Set wn(t) =αncos(/radicalbig λnt) +βnsin(/radicalbig λnt), 4.5. INITIAL-BOUNDARY VALUE PROBLEMS 127 where αn, βn∈R. It is easily seen that wn(t)vn(x) is a solution of differential equation (4.46), and, since (4.46) is linear and homogeneou s, also (principle of superposition) uN=N/summationdisplay n=1wn(t)vn(x) which satisfies the differential equation (4.46) and the bounda ry condi- tions (4.49). Consider the formal solution of (4.46), (4.49 ) u(x, t) =∞/summationdisplay n=1/parenleftBig αncos(/radicalbig λnt) +βnsin(/radicalbig λnt)/parenrightBig sin/parenleftBig/radicalbig λnx/parenrightBig . (4.52) ”Formal” means that we know here neither that the right hand s ide con- verges nor that it is a solution of the initial-boundary value problem. For- mally, the unknown coefficients can be calculated from initial conditions (4.47), (4.48) as follows. We have u(x,0) =∞/summationdisplay n=1αnsin(/radicalbig λnx) =f(x). Multiplying this equation by sin(√λkx) and integrate over (0 , l), we get αn/integraldisplayl 0sin2(/radicalbig λkx)dx=/integraldisplayl 0f(x)sin(/radicalbig λkx)dx. We recall that/integraldisplayl 0sin(/radicalbig λnx)sin(/radicalbig λkx)dx=l 2δnk. Then αk=2 l/integraldisplayl 0f(x)sin/parenleftbiggπk lx/parenrightbigg dx. (4.53) By the same argument it follows from ut(x,0) =∞/summationdisplay n=1βn/radicalbig λnsin(/radicalbig λnx) =g(x) that βk=2 kπ/integraldisplayl 0g(x)sin/parenleftbiggπk lx/parenrightbigg dx. (4.54) Under additional assumptions f∈C4 0(0, l),g∈C3 0(0, l) it follows that the right hand side of (4.52), where αn,βnare given by (4.53) and (4.54), 128 CHAPTER 4. HYPERBOLIC EQUATIONS respectively, defines a classical solution of (4.46)-(4.49) s ince under these assumptions the series for uand the formal differentiate series for ut,utt, ux,uxxconverges uniformly on 0 ≤x≤l, 0≤t≤T, 0< T < ∞fixed, see an exercise. 4.5.2 Oscillation of a membrane Let Ω ⊂R2be a bounded domain. We consider the initial-boundary value problem utt(x, t) = /trianglexuin Ω×R, (4.55) u(x,0) = f(x), x∈Ω, (4.56) ut(x,0) = g(x), x∈Ω, (4.57) u(x, t) = 0 on ∂Ω×R. (4.58) As in the previous subsection for the string, we make the ansa tz (separation of variables) u(x, t) =w(t)v(x) which leads to the eigenvalue problem −/trianglev=λvin Ω, (4.59) v= 0 on ∂Ω. (4.60) Letλnare the eigenvalues of (4.59), (4.60) and vna complete associated orthonormal system of eigenfunctions. We assume Ω is sufficien tly regular such that the eigenvalues are countable, which is satisfied in the following examples. Then the formal solution of the above initial-boun dary value problem is u(x, t) =∞/summationdisplay n=1/parenleftBig αncos(/radicalbig λnt) +βnsin(/radicalbig λnt)/parenrightBig vn(x), where αn=/integraldisplay Ωf(x)vn(x)dx βn=1√λn/integraldisplay Ωg(x)vn(x)dx. Remark. In general, eigenvalues of (4.59), (4.59) are not known expl icitly. There are numerical methods to calculate these values. In so me special cases, see next examples, these values are known. 4.5. INITIAL-BOUNDARY VALUE PROBLEMS 129 Examples 1. Rectangle membrane. Let Ω = (0 , a)×(0, b). Using the method of separation of variables, we find all eigenv alues of (4.59), (4.60) which are given by λkl=/radicalbigg k2 a2+l2 b2, k, l = 1,2, . . . and associated eigenfunctions, not normalized, are ukl(x) = sin/parenleftbiggπk ax1/parenrightbigg sin/parenleftbiggπl bx2/parenrightbigg . 2. Disk membrane. Set Ω ={x∈R2:x2 1+x2 2< R2}. In polar coordinates, the eigenvalue problem (4.59), (4.60 ) is given by −1 r/parenleftbigg (rur)r+1 ruθθ/parenrightbigg =λu (4.61) u(R, θ) = 0 , (4.62) here is u=u(r, θ) :=v(rcosθ, rsinθ). We will find eigenvalues and eigen- functions by separation of variables u(r, θ) =v(r)q(θ), where v(R) = 0 and q(θ) is periodic with period 2 πsince u(r, θ) is single valued. This leads to −1 r/parenleftbigg (rv/prime)/primeq+1 rvq/prime/prime/parenrightbigg =λvq. Dividing by vq, provided vq/negationslash= 0, we obtain −1 r/parenleftbigg(rv/prime(r))/prime v(r)+1 rq/prime/prime(θ) q(θ)/parenrightbigg =λ, (4.63) which implies q/prime/prime(θ) q(θ)=const. =:−µ. 130 CHAPTER 4. HYPERBOLIC EQUATIONS Thus, we arrive at the eigenvalue problem −q/prime/prime(θ) = µq(θ) q(θ) = q(θ+ 2π). It follows that eigenvalues µare real and nonnegative. All solutions of the differential equation are given by q(θ) =Asin(√µθ) +Bcos(√µθ), where A, B are arbitrary real constants. From the periodicity require ment Asin(√µθ) +Bcos(√µθ) =Asin(√µ(θ+ 2π)) +Bcos(√µ(θ+ 2π)) it follows2 sin(√µπ)(Acos(√µθ+√µπ)−Bsin(√µθ+√µπ)) = 0 , which implies, since A, B are not zero simultaneously, because we are look- ing for qnot identically zero, sin(√µπ)sin(√µθ+δ) = 0 for all θand a δ=δ(A, B, µ ). Consequently the eigenvalues are µn=n2, n= 0,1, . . . . Inserting q/prime/prime(θ)/q(θ) =−n2into (4.63), we obtain the boundary value prob- lem r2v/prime/prime(r) +rv/prime(r) + (λr2−n2)v= 0 on (0 , R) (4.64) v(R) = 0 (4.65) sup r∈(0,R)|v(r)|<∞. (4.66) Setz=√ λrandv(r) =v(z/√ λ) =:y(z), then, see (4.64), z2y/prime/prime(z) +zy/prime(z) + (z2−n2)y(z) = 0, 2 sinx−siny= 2 cosx+y 2sinx−y 2 cosx−cosy=−2 sinx+y 2sinx−y 2 4.5. INITIAL-BOUNDARY VALUE PROBLEMS 131 where z >0. Solutions of this differential equations which are bounded at zero are Bessel functions of first kind and n-th order Jn(z). The eigenvalues follows from boundary condition (4.65), i. e., from Jn(√ λR) = 0. Denote byτnkthe zeros of Jn(z), then the eigenvalues of (4.61)-(4.61) are λnk=/parenleftBigτnk R/parenrightBig2 and the associated eigenfunctions are Jn(/radicalbig λnkr)sin(nθ), n = 1,2, . . . Jn(/radicalbig λnkr)cos(nθ), n = 0,1,2, . . .. Thus the eigenvalues λ0kare simple and λnk, n≥1, are double eigenvalues. Remark. For tables with zeros of Jn(x) and for much more properties of Bessel functions see [25]. One has, in particular, the asymp totic formula Jn(x) =/parenleftbigg2 πx/parenrightbigg1/2/parenleftbigg cos(x−nπ/2−π/5) +O/parenleftbigg1 x/parenrightbigg/parenrightbigg asx→ ∞. It follows from this formula that there are infinitely many ze ros ofJn(x). 4.5.3 Inhomogeneous wave equations Let Ω ⊂Rnbe a bounded and sufficiently regular domain. In this section we consider the initial-boundary value problem utt=Lu+f(x, t) in Ω ×R (4.67) u(x,0) = φ(x)x∈Ω (4.68) ut(x,0) = ψ(x)x∈Ω (4.69) u(x, t) = 0 for x∈∂Ω and t∈Rn, (4.70) where u=u(x, t),x= (x1, . . ., x n),f, φ, ψ are given and Lis an elliptic differential operator. Examples for Lare: 1.L=∂2/∂x2, oscillating string. 2.L=/trianglex, oscillating membrane. 132 CHAPTER 4. HYPERBOLIC EQUATIONS 3. Lu=n/summationdisplay i,j=1∂ ∂xj/parenleftbig aij(x)uxi/parenrightbig , where aij=ajiare given sufficiently regular functions defined on Ω. We assume Lis uniformly elliptic, that is, there is a constant ν >0 such that n/summationdisplay i,j=1aijζiζj≥ν|ζ|2 for all x∈Ω and ζ∈Rn. 4. Let u= (u1, . . ., u m) and Lu=n/summationdisplay i,j=1∂ ∂xj/parenleftbig Aij(x)uxi/parenrightbig , where Aij=Ajiare given sufficiently regular ( m×m)-matrices on Ω. We assume that Ldefines an elliptic system. An example for this case is the linear elasticity. Consider the eigenvalue problem −Lv=λvin Ω (4.71) v= 0 on ∂Ω. (4.72) Assume there are infinitely many eigenvalues 0< λ1≤λ2≤. . .→ ∞ and a system of associated eigenfunctions v1, v2, . . .which is complete and orthonormal in L2(Ω). This assumption is satisfied if Ω is bounded and if ∂Ω is sufficiently regular. For the solution of (4.67)-(4.70) we make the ansatz u(x, t) =∞/summationdisplay k=1vk(x)wk(t), (4.73) with functions wk(t) which will be determined later. It is assumed that all series are convergent and that following calculations make sense. Let f(x, t) =∞/summationdisplay k=1ck(t)vk(x) (4.74) 4.5. INITIAL-BOUNDARY VALUE PROBLEMS 133 be Fourier’s decomposition of fwith respect to the eigenfunctions vk. We have ck(t) =/integraldisplay Ωf(x, t)vk(x)dx, (4.75) which follows from (4.74) after multiplying with vl(x) and integrating over Ω. Set /angbracketleftφ, vk/angbracketright=/integraldisplay Ωφ(x)vk(x)dx, then φ(x) =∞/summationdisplay k=1/angbracketleftφ, vk/angbracketrightvk(x) ψ(x) =∞/summationdisplay k=1/angbracketleftψ, vk/angbracketrightvk(x) are Fourier’s decomposition of φandψ, respectively. In the following we will determine wk(t), which occurs in ansatz (4.73), from the requirement that u=vk(x)wk(t) is a solution of utt=Lu+ck(t)vk(x) and that the initial conditions wk(0) = /angbracketleftφ, vk/angbracketright, w/prime k(0) = /angbracketleftψ, vk/angbracketright are satisfied. From the above differential equation it follows w/prime/prime k(t) =−λkwk(t) +ck(t). Thus wk(t) = akcos(/radicalbig λkt) +bksin(/radicalbig λkt) (4.76) +1√λk/integraldisplayt 0ck(τ)sin(/radicalbig λk(t−τ))dτ, where ak=/angbracketleftφ, vk/angbracketright, b k=1√λk/angbracketleftψ, vk/angbracketright. Summarizing, we have 134 CHAPTER 4. HYPERBOLIC EQUATIONS Proposition 4.2. The (formal) solution of the initial-boundary value prob- lem (4.67)-(4.70) is given by u(x, t) =∞/summationdisplay k=1vk(x)wk(t), where vkis a complete orthonormal system of eigenfunctions of (4.71 ), (4.72) and the functions wkare defined by (4.76). The resonance phenomenon Set in (4.67)-(4.70) φ= 0,ψ= 0 and assume that the external force fis periodic and is given by f(x, t) =Asin(ωt)vn(x), where A, ωare real constants and vnis one of the eigenfunctions of (4.71), (4.72). It follows ck(t) =/integraldisplay Ωf(x, t)vk(x)dx=Aδnksin(ωt). Then the solution of the initial value problem (4.67)-(4.70) is u(x, t) =Avn(x)√λn/integraldisplayt 0sin(ωτ)sin(/radicalbig λn(t−τ))dτ =Avn(x)1 ω2−λn/parenleftbiggω√λnsin(/radicalbig λkt)−sin(ωt)/parenrightbigg , provided ω/negationslash=√λn. It follows u(x, t)→A 2√λnvn(x)/parenleftbiggsin(√λnt)√λn−tcos(/radicalbig λnt)/parenrightbigg ifω→√λn. The right hand side is also the solution of the initial-bound ary value problem if ω=√λn. Consequently |u|can be arbitrarily large at some points xand at some times tifω=√λn. The frequencies√λnare called critical frequencies at which resonance occurs. A uniqueness result The solution of of the initial-boundary value problem (4.67 )-(4.70) is unique in the class C2(Ω×R). 4.5. INITIAL-BOUNDARY VALUE PROBLEMS 135 Proof. Letu1,u2are two solutions, then u=u2−u1satisfies utt=Luin Ω×R u(x,0) = 0 x∈Ω ut(x,0) = 0 x∈Ω u(x, t) = 0 for x∈∂Ω and t∈Rn. As an example we consider Example 3 from above and set E(t) =/integraldisplay Ω(n/summationdisplay i,j=1aij(x)uxiuxj+utut)dx. Then E/prime(t) = 2/integraldisplay Ω(n/summationdisplay i,j=1aij(x)uxiuxjt+ututt)dx = 2/integraldisplay ∂Ω(n/summationdisplay i,j=1aij(x)uxiutnj)dS +2/integraldisplay Ωut(−Lu+utt)dx = 0. It follows E(t) =const. From ut(x,0) = 0 and u(x,0) = 0 we get E(0) = 0. Consequently E(t) = 0 for all t, which implies, since Lis elliptic, that u(x, t) =const. onΩ×R. Finally, the homogeneous initial and boundary value conditions lead to u(x, t) = 0 on Ω×R. 2 136 CHAPTER 4. HYPERBOLIC EQUATIONS 4.6 Exercises 1. Show that u(x, t)∈C2(R2) is a solution of the one-dimensional wave equation utt=c2uxx if and only if u(A) +u(C) =u(B) +u(D) holds for all parallelograms ABCD in the ( x, t)-plane, which are bounded by characteristic lines, see Figure 4.8. xt B CA D Figure 4.8: Figure to the exercise 2. Method of separation of variables: Let vk(x) be an eigenfunction to the eigenvalue of the eigenvalue problem −v/prime/prime(x) =λv(x) in (0 , l), v(0) = v(l) = 0 and let wk(t) be a solution of differential equation −w/prime/prime(t) =λkw(t). Prove that vk(x)wk(t) is a solution of the partial differential equation (wave equation) utt=uxx. 3. Solve for given f(x) and µ∈Rthe initial value problem ut+ux+µuxxx= 0 in R×R+ u(x,0) = f(x). 4. Let S:={(x, t);t=γx}be spacelike, i. e., |γ|<1/c2) in (x, t)-space, x= (x1, x2, x3). Show that the Cauchy initial value problem 2u= 0 4.6. EXERCISES 137 with data for uonScan be transformed using the Lorentz-transform x1=x1−γc2t/radicalbig 1−γ2c2, x/prime 2=x2, x/prime 3=x3, t/prime=t−γx1/radicalbig 1−γ2c2 into the initial value problem, in new coordinates, 2u= 0 u(x/prime,0) = f(x/prime) ut/prime(x/prime,0) = g(x/prime). Here we denote the transformed function by uagain. 5. (i) Show that u(x, t) :=∞/summationdisplay n=1αncos/parenleftBigπn lt/parenrightBig sin/parenleftBigπn lx/parenrightBig is aC2-solution of the wave equation utt=uxxif|αn| ≤c/n4, where the constant cis independent of n. (ii) Set αn:=/integraldisplayl 0f(x)sin/parenleftBigπn lx/parenrightBig dx. Prove |αn| ≤c/n4, provided f∈C4 0(0, l). 6. Let Ω be the rectangle (0 , a)×(0, b). Find all eigenvalues and associ- ated eigenfunctions of −/triangleu=λuin Ω, u= 0 on ∂Ω. Hint: Separation of variables. 7. Find a solution of Schr¨ odinger’s equation i/planckover2pi1ψt=−/planckover2pi12 2m/trianglexψ+V(x)ψinRn×R, which satisfies the side condition /integraldisplayn R|ψ(x, t)|2dx= 1, provided E∈Ris an (eigenvalue) of the elliptic equation /triangleu+2m /planckover2pi12(E−V(x))u= 0 in Rn 138 CHAPTER 4. HYPERBOLIC EQUATIONS under the side condition/integraltextn R|u|2dx= 1,u:Rn/mapsto→C. Here is ψ:Rn×R/mapsto→C,/planckover2pi1Planck’s constant (a small positive con- stant), V(x) a given potential. Remark. In the case of a hydrogen atom the potential is V(x) = −e/|x|,eis here a positive constant. Then eigenvalues are given by En=−me4/(2/planckover2pi12n2),n∈N, see [22], pp. 202. 8. Find nonzero solutions by using separation of variables o futt=/trianglexu in Ω×(0,∞),u(x, t) = 0 on ∂Ω, where Ω is the circular cylinder Ω ={(x1, x2, x3)∈Rn:x2 1+x2 2< R2,0< x3< h}. 9. Solve the initial value problem 3utt−4uxx= 0 u(x,0) = sin x ut(x,0) = 1 . 10. Solve the initial value problem utt−c2uxx=x2, t >0, x∈R u(x,0) = x ut(x,0) = 0 . Hint: Find a solution of the differential equation independent on t, and transform the above problem into an initial value proble m with homogeneous differential equation by using this solution. 11. Find with the method of separation of variables nonzero s olutions u(x, t), 0≤x≤1,0≤t <∞,of utt−uxx+u= 0, such that u(0, t) = 0, and u(1, t) = 0 for all t∈[0,∞). 12. Find solutions of the equation utt−c2uxx=λ2u, λ=const. which can be written as u(x, t) =f(x2−c2t2) =f(s), s:=x2−c2t2 4.6. EXERCISES 139 withf(0) = K,Ka constant. Hint: Transform equation for f(s) by using the substitution s:=z2/A with an appropriate constant Ainto Bessel’s differential equation z2f/prime/prime(z) +zf/prime(z) + (z2−n2)f= 0, z > 0 withn= 0. Remark. The above differential equation for uis the transformed tele- graph equation (see Section 4.4). 13. Find the formula for the solution of the following Cauchy initial value problem uxy=f(x, y), where S:y=ax+b,a >0, and the initial conditions on Sare given by u=αx+βy+γ, ux=α, uy=β, a, b, α, β, γ constants. 14. Find all eigenvalues µof −q/prime/prime(θ) = µq(θ) q(θ) = q(θ+ 2π). 140 CHAPTER 4. HYPERBOLIC EQUATIONS Chapter 5 Fourier transform Fourier’s transform is an integral transform which can simp lify investigations for linear differential or integral equations since it transf orms a differential operator into an algebraic equation. 5.1 Definition, properties Definition. Letf∈Cs 0(Rn),s= 0,1, . . .. The function ˆfdefined by /hatwidef(ξ) = (2 π)−n/2/integraldisplay Rne−iξ·xf(x)dx, (5.1) where ξ∈Rn, is called Fourier transform off, and the function /tildewideggiven by /tildewideg(x) = (2 π)−n/2/integraldisplay Rneiξ·xg(ξ)dξ (5.2) is called inverse Fourier transform , provided the integrals on the right hand side exist. From (5.1) it follows by integration by parts that differentia tion of a func- tion is changed to multiplication of its Fourier transforms , or an analytical operation is converted into an algebraic operation. More pr ecisely, we have Proposition 5.1. /hatwidestDαf(ξ) =i|α|ξα/hatwidef(ξ), where |α| ≤s. 141 142 CHAPTER 5. FOURIER TRANSFORM The following proposition shows that the Fourier transform offdecreases rapidly for |ξ| → ∞ , provided f∈Cs 0(Rn). In particular, the right hand side of (5.2) exists for g:=ˆfiff∈Cn+1 0(Rn). Proposition 5.2. Assume g∈Cs 0(Rn), then there is a constant M= M(n, s, g) such that |/hatwideg(ξ)| ≤M (1 +|ξ|)s. Proof. Letξ= (ξ1, . . ., ξ n) be fixed and let jbe an index such that |ξj|= max k|ξk|. Then |ξ|=/parenleftBiggn/summationdisplay k=1ξ2 k/parenrightBigg1/2 ≤√n|ξj| which implies (1 +|ξ|)s=s/summationdisplay k=0/parenleftbiggs k/parenrightbigg |ξ|k ≤2ss/summationdisplay k=0nk/2|ξj|k ≤2sns/2/summationdisplay |α|≤s|ξα|. This inequality and Proposition 5.1 imply (1 +|ξ|)s|/hatwideg(ξ)| ≤ 2sns/2/summationdisplay |α|≤s|(iξ)α/hatwideg(ξ)| ≤2sns/2/summationdisplay |α|≤s/integraldisplay Rn|Dαg(x)|dx=:M. 2 The notation inverse Fourier transform for (5.2) is justified by Theorem 5.1./tildewide/hatwidef=fand/hatwide/tildewidef=f. Proof. See [27], for example. We will prove the first assertion (2π)−n/2/integraldisplay Rneiξ·x/hatwidef(ξ)dξ=f(x) (5.3) 5.1. DEFINITION, PROPERTIES 143 here. The proof of the other relation is left as an exercise. A ll integrals appearing in the following exist, see Proposition 5.2 and th e special choice ofg. (i) Formula /integraldisplay Rng(ξ)/hatwidef(ξ)eix·ξdξ=/integraldisplay Rn/hatwideg(y)f(x+y)dy (5.4) follows by direct calculation: /integraldisplay Rng(ξ)/parenleftbigg (2π)−n/2/integraldisplay Rne−ix·yf(y)dy/parenrightbigg eix·ξdξ = (2π)−n/2/integraldisplay Rn/parenleftbigg/integraldisplay Rng(ξ)e−iξ·(y−x)dξ/parenrightbigg f(y)dy =/integraldisplay Rn/hatwideg(y−x)f(y)dy =/integraldisplay Rn/hatwideg(y)f(x+y)dy. (ii) Formula (2π)−n/2/integraldisplay Rne−iy·ξg(εξ)dξ=ε−n/hatwideg(y/ε) (5.5) for each ε >0 follows after substitution z=εξin the left hand side of (5.1). (iii) Equation /integraldisplay Rng(εξ)/hatwidef(ξ)eix·ξdξ=/integraldisplay Rn/hatwideg(y)f(x+εy)dy (5.6) follows from (5.4) and (5.5). Set G(ξ) :=g(εξ), then (5.4) implies /integraldisplay RnG(ξ)/hatwidef(ξ)eix·ξdξ=/integraldisplay Rn/hatwideG(y)f(x+y)dy. Since, see (5.5), /hatwideG(y) = (2 π)−n/2/integraldisplay Rne−iy·ξg(εξ)dξ =ε−n/hatwideg(y/ε), 144 CHAPTER 5. FOURIER TRANSFORM we arrive at /integraldisplay Rng(εξ)/hatwidef(ξ) =/integraldisplay Rnε−n/hatwideg(y/ε)f(x+y)dy =/integraldisplay Rn/hatwideg(z)f(x+εz)dz. Letting ε→0, we get g(0)/integraldisplay Rn/hatwidef(ξ)eix·ξdξ=f(x)/integraldisplay Rn/hatwideg(y)dy. (5.7) Set g(x) := e−|x|2/2, then /integraldisplay Rn/hatwideg(y)dy= (2π)n/2. (5.8) Since g(0) = 1, the first assertion of Theorem 5.1 follows from (5.7) an d (5.8). It remains to show (5.8). (iv)Proof of (5.8). We will show /hatwideg(y) : = (2 π)−n/2/integraldisplay Rne−|x|2/2e−ix·xdx = e−|y|2/2. The proof of /integraldisplay Rne−|y|2/2dy= (2π)n/2 is left as an exercise. Since −/parenleftbiggx√ 2+iy√ 2/parenrightbigg ·/parenleftbiggx√ 2+iy√ 2/parenrightbigg =−/parenleftbigg|x|2 2+ix·y−|y|2 2/parenrightbigg it follows /integraldisplay Rne−|x|2/2e−ix·ydx=/integraldisplay Rne−η2e−|y|2/2dx = e−|y|2/2/integraldisplay Rne−η2dx = 2n/2e−|y|2/2/integraldisplay Rne−η2dη 5.1. DEFINITION, PROPERTIES 145 where η:=x√ 2+iy√ 2. Consider first the one-dimensional case. According to Cauchy’ s theorem we have /contintegraldisplay Ce−η2dη= 0, where the integration is along the curve Cwhich is the union of four curves as indicated in Figure 5.1. ReIm C CCCiy 2η η43 12 R −R Figure 5.1: Proof of (5.8) Consequently /integraldisplay C3e−η2dη=1√ 2/integraldisplayR −Re−x2/2dx−/integraldisplay C2e−η2dη−/integraldisplay C4e−η2dη. It follows lim R→∞/integraldisplay C3e−η2dη=√π since lim R→∞/integraldisplay Cke−η2dη= 0, k= 2,4. The case n >1 can be reduced to the one-dimensional case as follows. Set η=x√ 2+iy√ 2= (η1, . . ., η n), where ηl=xl√ 2+iyl√ 2. 146 CHAPTER 5. FOURIER TRANSFORM From dη=dη1. . .dη land e−η2= e−/summationtextn l=1η2 l=n/productdisplay l=1e−η2 l it follows/integraldisplay Rne−η2dη=n/productdisplay l=1/integraldisplay Γle−η2 ldηl, where for fixed y Γl={z∈C:z=xl√ 2+iyl√ 2,−∞< xl<+∞}. 2 There is a useful class of functions for which the integrals in the definition of/hatwidefand/tildewidefexist. Foru∈C∞(Rn) we set qj,k(u) := max α:|α|≤k/parenleftbigg sup Rn/parenleftBig (1 +|x|2)j/2|Dαu(x)|/parenrightBig/parenrightbigg . Definition. TheSchwartz class of rapidly degreasing functions is S(Rn) ={u∈C∞(Rn) :qj,k(u)<∞for any j, k∈N∪ {0}}. This space is a Frech´ et space. Proposition 5.3. Assume u∈ S(Rn), then /hatwideuand/tildewideu∈ S(Rn). Proof. See [24], Chapter 1.2, for example, or an exercise. 5.1.1 Pseudodifferential operators The properties of Fourier transform lead to a general theory for linear partial differential or integral equations. In this subsection we defin e Dk=1 i∂ ∂xk, k= 1, . . ., n, and for each multi-index αas in Subsection 3.5.1 Dα=Dα1 1. . . Dαnn. 5.1. DEFINITION, PROPERTIES 147 Thus Dα=1 i|α|∂|α| ∂xα1 1. . .∂xαnn. Let p(x, D) :=/summationdisplay |α|≤maα(x)Dα, be a linear partial differential of order m, where aαare given sufficiently regular functions. According to Theorem 5.1 and Proposition 5.3, we have, at lea st for u∈ S(Rn), u(x) = (2 π)−n/2/integraldisplay Rneix·ξ/hatwideu(ξ)dξ, which implies Dαu(x) = (2 π)−n/2/integraldisplay Rneix·ξξα/hatwideu(ξ)dξ. Consequently p(x, D)u(x) = (2 π)−n/2/integraldisplay Rneix·ξp(x, ξ)/hatwideu(ξ)dξ, (5.9) where p(x, ξ) =/summationdisplay |α|≤maα(x)ξα. The right hand side of (5.9) makes sense also for more general functions p(x, ξ), not only for polynomials. Definition. The function p(x, ξ) is called symbol and (Pu)(x) := (2 π)−n/2/integraldisplay Rneix·ξp(x, ξ)/hatwideu(ξ)dξ is said to be pseudodifferential operator . An important class of symbols for which the right hand side in this defini- tion of a pseudodifferential operator is defined is Smwhich is the subset of p(x, ξ)∈C∞(Ω×Rn) such that |Dβ xDα ξp(x, ξ)| ≤CK,α,β(p)(1 + |ξ|)m−|α| for each compact K⊂Ω. 148 CHAPTER 5. FOURIER TRANSFORM Above we have seen that linear differential operators define a cl ass of pseu- dodifferential operators. Even integral operators can be wri tten (formally) as pseudodifferential operators. Let (Pu)(x) =/integraldisplay RnK(x, y)u(y)dy be an integral operator. Then (Pu)(x) = (2 π)−n/2/integraldisplay RnK(x, y)/integraldisplay Rneix·ξξα/hatwideu(ξ)dξ = (2 π)−n/2/integraldisplay Rneix·ξ/parenleftbigg/integraldisplay Rnei(y−x)·ξK(x, y)dy/parenrightbigg /hatwideu(ξ). Then the symbol associated to the above integral operator is p(x, ξ) =/integraldisplay Rnei(y−x)·ξK(x, y)dy. 5.2. EXERCISES 149 5.2 Exercises 1. Show /integraldisplay Rne−|y|2/2dy= (2π)n/2. 2. Show that u∈ S(Rn) implies ˆ u,/tildewideu∈ S(Rn). 3. Give examples for functions p(x, ξ) which satisfy p(x, ξ)∈Sm. 4. Find a formal solution of Cauchy’s initial value problem f or the wave equation by using Fourier’s transform. 150 CHAPTER 5. FOURIER TRANSFORM Chapter 6 Parabolic equations Here we consider linear parabolic equations of second order . An example is the heat equation ut=a2/triangleu, where u=u(x, t),x∈R3,t≥0, and a2is a positive constant called conductivity coefficient. The heat equation has its origin in p hysics where u(x, t) is the temperature at xat time t, see [20], p. 394, for example. Remark 1. After scaling of axis we can assume a= 1. Remark 2. By setting t:=−t, the heat equation changes to an equation which is called backward equation. This is the reason for the fact that the heat equation describes irreversible processes in cont rast to the wave equation 2u= 0 which is invariant with respect the mapping t/mapsto→ −t. Mathematically, it means that it is not possible, in general , to find the distribution of temperature at an earlier time t < t 0if the distribution is given at t0. Consider the initial value problem for u=u(x, t),u∈C∞(Rn×R+), ut=/triangleuinx∈Rn, t≥0, (6.1) u(x,0) = φ(x), (6.2) where φ∈C(Rn) is given and /triangle ≡ /triangle x. 151 152 CHAPTER 6. PARABOLIC EQUATIONS 6.1 Poisson’s formula Assume uis a solution of (6.1), then, since Fourier transform is a lin ear mapping, /hatwiderut− /triangleu=ˆ0. From properties of the Fourier transform, see Proposition 5 .1, we have /hatwidest/triangleu=n/summationdisplay k=1/hatwidest∂2u ∂x2 k=n/summationdisplay k=1i2ξ2 k/hatwideu(ξ), provided the transforms exist. Thus we arrive at the ordinar y differential equation for the Fourier transform of u d/hatwideu dt+|ξ|2/hatwideu= 0, where ξis considered as a parameter. The solution is /hatwideu(ξ, t) =/hatwideφ(ξ)e−|ξ|2t since/hatwideu(ξ,0) =/hatwideφ(ξ). From Theorem 5.1 it follows u(x, t) = (2 π)−n/2/integraldisplay Rn/hatwideφ(ξ)e−|ξ|2teiξ·xdξ = (2 π)−n/integraldisplay Rnφ(y)/parenleftbigg/integraldisplay Rneiξ·(x−y)−|ξ|2tdξ/parenrightbigg dy. Set K(x, y, t) = (2 π)−n/integraldisplay Rneiξ·(x−y)−|ξ|2tdξ. By the same calculations as in the proof of Theorem 5.1, step ( vi), we find K(x, y, t) = (4 πt)−n/2e−|x−y|2/4t. (6.3) Thus we have u(x, t) =1/parenleftbig 2√ πt/parenrightbign/integraldisplay Rnφ(z)e−|x−z|2/4tdz. (6.4) Definition. Formula (6.4) is called Poisson’s formula and the function K defined by (6.3) heat kernel orfundamental solution of the heat equation. 6.1. POISSON’S FORMULA 153 ρK K(x,y,t ) K(x,y,t )1 2 Figure 6.1: Kernel K(x, y, t),ρ=|x−y|,t1< t2 Proposition 6.1 The kernel Khas following properties: (i)K(x, y, t)∈C∞(Rn×Rn×R+), (ii)(∂/∂t − /triangle)K(x, y, t) = 0, t >0, (iii)K(x, y, t)>0, t >0, (iv)/integraltext RnK(x, y, t)dy= 1, x∈Rn, t >0, (v) For each fixed δ >0: lim t→0 t >0/integraldisplay Rn\Bδ(x)K(x, y, t)dy= 0 uniformly for x∈Rn. Proof. (i) and (iii) are obviously, and (ii) follows from the definiti on of K. Equations (iv) and (v) hold since /integraldisplay Rn\Bδ(x)K(x, y, t)dy=/integraldisplay Rn\Bδ(x)(4πt)−n/2e−|x−y|2/4tdy =π−n/2/integraldisplay Rn\Bδ/√ 4t(0)e−|η|2dη 154 CHAPTER 6. PARABOLIC EQUATIONS by using the substitution y=x+(4t)1/2η. For fixed δ >0 it follows (v) and forδ:= 0 we obtain (iv). 2 Theorem 6.1. Assume φ∈C(Rn)andsupRn|φ(x)|<∞. Then u(x, t) given by Poisson’s formula (6.4) is in C∞(Rn×R+), continuous on Rn× [0,∞)and a solution of the initial value problem (6.1), (6.2). Proof. It remains to show lim x→ξ t→0u(x, t) =φ(ξ). Since φis continuous there exists for given ε >0 aδ=δ(ε) such that |φ(y)− xt ξξ+δ ξ+2δ Figure 6.2: Figure to the proof of Theorem 6.1 φ(ξ)|< εif|y−ξ|<2δ. SetM:= supRn|φ(y)|. Then, see Proposition 6.1, u(x, t)−φ(ξ) =/integraldisplay RnK(x, y, t)(φ(y)−φ(ξ))dy. 6.2. INHOMOGENEOUS HEAT EQUATION 155 It follows, if |x−ξ|< δandt >0, that |u(x, t)−φ(ξ)| ≤/integraldisplay Bδ(x)K(x, y, t)|φ(y)−φ(ξ)|dy +/integraldisplay Rn\Bδ(x)K(x, y, t)|φ(y)−φ(ξ)|dy ≤/integraldisplay B2δ(x)K(x, y, t)|φ(y)−φ(ξ)|dy +2M/integraldisplay Rn\Bδ(x)K(x, y, t)dy ≤ε/integraldisplay RnK(x, y, t)dy+ 2M/integraldisplay Rn\Bδ(x)K(x, y, t)dy <2ε if 0< t≤t0,t0sufficiently small. 2 Remarks. 1. Uniqueness follows under the additional growth assumption |u(x, t)| ≤Mea|x|2inDT, where Mandaare positive constants, see Proposition 6.2 below. In the one-dimensional case, one has uniqueness in the class u(x, t)≥0 in DT, see [10], pp. 222. 2.u(x, t) defined by Poisson’s formula depends on all values φ(y),y∈Rn. That means, a perturbation of φ, even far from a fixed x, has influence to the value u(x, t). In physical terms, this means that heat travels with infinit e speed, in contrast to the experience. 6.2 Inhomogeneous heat equation Here we consider the initial value problem for u=u(x, t),u∈C∞(Rn×R+), ut− /triangleu=f(x, t) inx∈Rn, t≥0, u(x,0) = φ(x), where φandfare given. From /hatwiderut− /triangleu=/hatwiderf(x, t) 156 CHAPTER 6. PARABOLIC EQUATIONS we obtain an initial value problem for an ordinary differentia l equation: d/hatwideu dt+|ξ|2/hatwideu=/hatwidef(ξ, t) /hatwideu(ξ,0) = /hatwideφ(ξ). The solution is given by /hatwideu(ξ, t) = e−|ξ|2t/hatwideφ(ξ) +/integraldisplayt 0e−|ξ|2(t−τ)/hatwidef(ξ, τ)dτ. Applying the inverse Fourier transform and a calculation as in the proof of Theorem 5.1, step (vi), we get u(x, t) = (2 π)−n/2/integraldisplay Rneix·ξ/parenleftBig e−|ξ|2t/hatwideφ(ξ) +/integraldisplayt 0e−|ξ|2(t−τ)/hatwidef(ξ, τ)dτ/parenrightBig dξ. From the above calculation for the homogeneous problem and c alculation as in the proof of Theorem 5.1, step (vi), we obtain the formula u(x, t) =1 (2√ πt)n/integraldisplay Rnφ(y)e−|y−x|2/(4t)dy +/integraldisplayt 0/integraldisplay Rnf(y, τ)1/parenleftBig 2/radicalbig π(t−τ)/parenrightBigne−|y−x|2/(4(t−τ))dy dτ. This function u(x, t) is a solution of the above inhomogeneous initial value problem provided φ∈C(Rn),sup Rn|φ(x)|<∞ and if f∈C(Rn×[0,∞)), M(τ) := sup Rn|f(y, τ)|<∞,0≤τ <∞. 6.3 Maximum principle Let Ω ⊂Rnbe abounded domain. Set DT= Ω ×(0, T), T > 0, ST={(x, t) : (x, t)∈Ω× {0}or (x, t)∈∂Ω×[0, T]}, 6.3. MAXIMUM PRINCIPLE 157 T DST xt T δΩ δΩT−ε Figure 6.3: Notations to the maximum principle see Figure 6.3 Theorem 6.2. Assume u∈C(DT), that ut,uxixkexist and are continuous inDT, and ut− /triangleu≤0inDT. Then max DTu(x, t) = max STu. Proof. Assume initially ut− /triangleu <0 inDT. Let ε >0 be small and 0< ε < T . Since u∈C(DT−ε), there is an ( x0, t0)∈DT−εsuch that u(x0, t0) = max DT−εu(x, t). Case (i). Let (x0, t0)∈DT−ε. Hence, since DT−εis open, ut(x0, t0) = 0, uxl(x0, t0) = 0, l= 1, . . ., n and n/summationdisplay l,k=1uxlxk(x0, t0)ζlζk≤0 for all ζ∈Rn. The previous inequality implies that uxkxk(x0, t0)≤0 for each k. Thus we arrived at a contradiction to ut− /triangleu <0 inDT. 158 CHAPTER 6. PARABOLIC EQUATIONS Case (ii). Assume ( x0, t0)∈Ω×{T−ε}. Then it follows as above /triangleu≤0 in (x0, t0), and from u(x0, t0)≥u(x0, t),t≤t0, one concludes that ut(x0, t0)≥ 0. We arrived at a contradiction to ut− /triangleu <0 inDTagain. Summarizing, we have shown that max DT−εu(x, t) = max T−εu(x, t). Thus there is an ( xε, tε)∈ST−εsuch that u(xε, tε) = max DT−εu(x, t). Since uis continuous on DT, we have lim ε→0max DT−εu(x, t) = max DTu(x, t). It follows that there is ( x,t)∈STsuch that u(x,t) = max DTu(x, t) since ST−ε⊂STandSTis compact. Thus, theorem is shown under the assumption ut− /triangleu <0 inDT. Now assume ut− /triangleu≤0 inDT. Set v(x, t) :=u(x, t)−kt, where kis a positive constant. Then vt− /trianglev=ut− /triangleu−k <0. From above we have max DTu(x, t) = max DT(v(x, t) +kt) ≤max DTv(x, t) +kT = max STv(x, t) +kT ≤max STu(x, t) +kT, Letting k→0, we obtain max DTu(x, t)≤max STu(x, t). 6.3. MAXIMUM PRINCIPLE 159 Since ST⊂DT, the theorem is shown. 2 If we replace in the above theorem the bounded domain Ω by Rn, then the result remains true provided we assume an additional growth assumption foru. More precisely, we have the following result which is a coro llary of the previous theorem. Set for a fixed T, 0< T < ∞, DT={(x, t) :x∈Rn,0< t < T }. Proposition 6.2. Assume u∈C(DT), that ut,uxixkexist and are contin- uous in DT, ut− /triangleu≤0inDT, and additionally that usatisfies the growth condition u(x, t)≤Mea|x|2, where Mandaare positive constants. Then max DTu(x, t) = max STu. It follows immediately the Corollary. The initial value problem ut− /triangleu= 0inDT,u(x,0) =f(x), x∈Rn, has a unique solution in the class defined by u∈C(DT),ut,uxixk exist and are continuous in DTand|u(x, t)| ≤Mea|x|2. Proof of Proposition 6.2. See [10], pp. 217. We can assume that 4 aT < 1, since the finite interval can be divided into finite many intervals of equal length τwith 4 aτ <1. Then we conclude successively for kthat u(x, t)≤sup y∈Rnu(y, kτ)≤sup y∈Rnu(y,0) forkτ≤t≤(k+ 1)τ,k= 0, . . ., N −1, where N=T/τ. There is an /epsilon1 >0 such that 4 a(T+/epsilon1)<1. Consider the comparison function vµ(x, t) : = u(x, t)−µ(4π(T+/epsilon1−t))−n/2e|x−y|2/(4(T+/epsilon1−t)) =u(x, t)−µK(ix, iy, T +/epsilon1−t) 160 CHAPTER 6. PARABOLIC EQUATIONS for fixed y∈Rnand for a constant µ >0. Since the heat kernel K(ix, iy, t ) satisfies Kt=/triangleKx, we obtain ∂ ∂tvµ− /trianglevµ=ut− /triangleu≤0. Set for a constant ρ >0 DT,ρ={(x, t) :|x−y|< ρ,0< t < T }. Then we obtain from Theorem 6.2 that vµ(y, t)≤max ST,ρvµ, where ST,ρ≡STof Theorem 6.2 with Ω = Bρ(y), see Figure 6.3. On the bottom of ST,ρwe have, since µK > 0, vµ(x,0)≤u(x,0)≤sup z∈Rnf(z). On the cylinder part |x−y|=ρ, 0≤t≤T, ofST,ρit is vµ(x, t)≤Mea|x|2−µ(4π(T+/epsilon1−t))−n/2eρ2/(4(T+/epsilon1−t)) ≤Mea(|y|+ρ)2−µ(4π(T+/epsilon1))−n/2eρ2/(4(T+/epsilon1)) ≤sup z∈Rnf(z) for all ρ > ρ 0(µ),ρ0sufficiently large. We recall that 4 a(T+/epsilon1)<1. Summarizing, we have max ST,ρvµ(x, t)≤sup z∈Rnf(z) ifρ > ρ 0(µ). Thus vµ(y, t)≤max ST,ρvµ(x, t)≤sup z∈Rnf(z) ifρ > ρ 0(µ). Since vµ(y, t) =u(y, t)−µ(4π(T+/epsilon1−t))−n/2 it follows u(y, t)−µ(4π(T+/epsilon1−t))−n/2≤sup z∈Rnf(z). 6.3. MAXIMUM PRINCIPLE 161 Letting µ→0, we obtain the assertion of the proposition. 2 The above maximum principle of Theorem 6.2 holds for a large c lass of parabolic differential operators, even for degenerate equat ions. Set Lu=n/summationdisplay i,j=1aij(x, t)uxixj, where aij∈C(DT) are real, aij=aji, and the matrix ( aij) is nonnegative, that is, n/summationdisplay i,j=1aij(x, t)ζiζj≥0 for all ζ∈Rn, and (x, t)∈DT. Theorem 6.3. Assume u∈C(DT), that ut,uxixkexist and are continuous inDT, and ut−Lu≤0inDT. Then max DTu(x, t) = max STu. Proof. (i) One proof is a consequence of the following lemma: Let A,Breal, symmetric and nonnegative matrices. Nonnegative means tha t all eigenval- ues are nonnegative. Then trace ( AB)≡/summationtextn i,j=1aijbij≥0, see an exercise. (ii) Another proof exploits transform to principle axis dir ectly: Let U= (z1, . . ., z n), where zlis an orthonormal system of eigenvectors to the eigen- values λlof the matrix A= (ai,j(x0, t0)). Set ζ=Uη,x=UT(x−x0)yand v(y) =u(x0+Uy, t 0), then 0≤n/summationdisplay i,j=1aij(x0, t0)ζiζj=n/summationdisplay i=1λiη2 i 0≥n/summationdisplay i,j=1uxixjζiζj=n/summationdisplay i=1vyiyiη2 i. It follows λi≥0 and vyiyi≤0 for all i. Consequently n/summationdisplay i,j=1aij(x0, t0)uxixj(x0, t0) =n/summationdisplay i=1λivyiyi≤0. 2 162 CHAPTER 6. PARABOLIC EQUATIONS 6.4 Initial-boundary value problem Consider the initial-boundary value problem for c=c(x, t) ct=D/trianglecin Ω×(0,∞) (6.5) c(x,0) = c0(x)x∈Ω (6.6) ∂c ∂n= 0 on ∂Ω×(0,∞). (6.7) Here is Ω ⊂Rn,nthe exterior unit normal at the smooth parts of ∂Ω,Da positive constant and c0(x) a given function. Remark. In application to diffusion problems, c(x, t) is the concentration of a substance in a solution, c0(x) its initial concentration and Dthe coef- ficient of diffusion. First Fick’s rule says that w=D∂c/∂n , where wis the flow of the substance through the boundary ∂Ω. Thus according to the Neumann boundary con- dition (6.7), we assume that there is no flow through the bounda ry. 6.4.1 Fourier’s method Separation of variables ansatz c(x, t) =v(x)w(t) leads to the eigenvalue problem, see the arguments of Section 4.5, −/trianglev=λvin Ω (6.8) ∂v ∂n= 0 on ∂Ω, (6.9) and to the ordinary differential equation w/prime(t) +λDw(t) = 0. (6.10) Assume Ω is bounded and ∂Ω sufficiently regular, then the eigenvalues of(6.8), (6.9) are countable and 0 =λ0< λ1≤λ2≤. . .→ ∞. Letvj(x) be a complete system of orthonormal (in L2(Ω)) eigenfunctions. Solutions of (6.10) are wj(t) =Cje−Dλjt, where Cjare arbitrary constants. 6.4. INITIAL-BOUNDARY VALUE PROBLEM 163 According to the superposition principle, cN(x, t) :=N/summationdisplay j=0Cje−Dλjtvj(x) is a solution of the differential equation (6.8) and c(x, t) :=∞/summationdisplay j=0Cje−Dλjtvj(x), with Cj=/integraldisplay Ωc0(x)vj(x)dx, is a formal solution of the initial-boundary value problem (6 .5)-(6.7). Diffusion in a tube Consider a solution in a tube, see Figure 6.4. Assume the init ial concentra- Ωl Qx xx3 12 Figure 6.4: Diffusion in a tube tionc0(x1, x2, x3) of the substrate in a solution is constant if x3=const. It follows from a uniqueness result below that the solution o f the initial- boundary value problem c(x1, x2, x3, t) is independent of x1andx2. 164 CHAPTER 6. PARABOLIC EQUATIONS Setz=x3, then the above initial-boundary value problem reduces to ct=Dczz c(z,0) = c0(z) cz= 0, z= 0, z=l. The (formal) solution is c(z, t) =∞/summationdisplay n=0Cne−D(π ln)2tcos/parenleftBigπ lnz/parenrightBig , where C0=1 l/integraldisplayl 0c0(z)dz Cn=2 l/integraldisplayl 0c0(z)cos/parenleftBigπ lnz/parenrightBig dz, n ≥1. 6.4.2 Uniqueness Sufficiently regular solutions of the initial-boundary value p roblem (6.5)- (6.7) are uniquely determined since from ct=D/trianglecin Ω×(0,∞) c(x,0) = 0 ∂c ∂n= 0 on ∂Ω×(0,∞). it follows that for each τ >0 0 =/integraldisplayτ 0/integraldisplay Ω(ctc−D(/trianglec)c)dxdt =/integraldisplay Ω/integraldisplayτ 01 2∂ ∂t(c2)dtdx+D/integraldisplay Ω/integraldisplayτ 0|∇xc|2dxdt =1 2/integraldisplay Ωc2(x, τ)dx+D/integraldisplay Ω/integraldisplayτ 0|∇xc|2dxdt. 6.5 Black-Scholes equation Solutions of the Black-Scholes equation define the value of a de rivative, for example of a call or put option, which is based on an asset. An a sset 6.5. BLACK-SCHOLES EQUATION 165 can be a stock or a derivative again, for example. In principl e, there are infinitely many such products, for example n-th derivatives. T he Black- Scholes equation for the value V(S, t) of a derivative is Vt+1 2σ2S2VSS+rSVS−rV= 0 in Ω , (6.11) where for a fixed T, 0< T < ∞, Ω ={(S, t)∈R2: 0< S < ∞,0< t < T }, andσ,rare positive constants. More precisely, σis the volatility of the underlying asset S, ris the guaranteed interest rate of a risk-free investment. IfS(t) is the value of an asset at time t, then V(S(t), t) is the value of the derivative at time t, where V(S, t) is the solution of an appropriate initial-boundary value problem for the Black-Scholes equati on, see below. The Black-Scholes equation follows from Ito’s Lemma under so me as- sumptions on the random function associated to S(t), see [26], for example. Call option Here is V(S, t) :=C(S, t), where C(S, t) is the value of the (European) call option. In this case we have following side conditions to (6. 11): C(S, T) = max {S−E,0} (6.12) C(0, t) = 0 (6.13) C(S, t) = S+o(S) asS→ ∞,uniformly in t, (6.14) where EandTare positive constants, Eis the exercise price and Tthe expiry. Side condition (6.12) means that the value of the option has n o value at timeTifS(T)≤E, condition (6.13) says that it makes no sense to buy assets if t he value of the asset is zero, condition (6.14) means that we buy assets if its value become s large, see Figure 6.5, where the side conditions are indicated. Theorem 6.4 (Black-Scholes formula for European call options). The so- lution C(S, t),0≤S <∞,0≤t≤T, of the initial-boundary value prob- lem (6.11)-(6.14) is explicitly known and is given by C(S, t) =SN(d1)−Ee−r(T−t)N(d2), 166 CHAPTER 6. PARABOLIC EQUATIONS St TC=max{S−E,0} C=0C~S Figure 6.5: Side conditions for a call option where N(x) =1√ 2π/integraldisplayx −∞e−y2/2dy, d1=ln(S/E) + (r+σ2/2)(T−t) σ√ T−t, d2=ln(S/E) + (r−σ2/2)(T−t) σ√ T−t. Proof. Substitutions S=Eex, t=T−τ σ2/2, C=Ev(x, τ) change equation (6.11) to vτ=vxx+ (k−1)vx−kv, (6.15) where k=r σ2/2. Initial condition (6.15) implies v(x,0) = max {ex−1,0}. (6.16) For a solution of (6.15) we make the ansatz v= eαx+βτu(x, τ), 6.5. BLACK-SCHOLES EQUATION 167 where αandβare constants which we will determine as follows. Inserting the ansatz into differential equation (6.15), we get βu+uτ=α2u+ 2αux+uxx+ (k−1)(αu+ux)−ku. Setβ=α2+ (k−1)α−kand choose αsuch that 0 = 2 α+ (k−1), then uτ=uxx. Thus v(x, τ) = e−(k−1)x/2−(k+1)2τ/4u(x, τ), (6.17) where u(x, τ) is a solution of the initial value problem uτ=uxx,−∞< x < ∞, τ > 0 u(x,0) = u0(x), with u0(x) = max/braceleftBig e(k+1)x/2−e(k−1)x/2,0/bracerightBig . A solution of this initial value problem is given by Poisson’ s formula u(x, τ) =1 2√πτ/integraldisplay+∞ −∞u0(s)e−(x−s)2/(4τ)ds. Changing variable by q= (s−x)/(√ 2τ), we get u(x, τ) =1√ 2π/integraldisplay+∞ −∞u0(q√ 2τ+x)e−q2/2dq =I1−I2, where I1=1√ 2π/integraldisplay∞ −x/(√ 2τ)e(k+1)(x+q√ 2τ)e−q2/2dq I2=1√ 2π/integraldisplay∞ −x/(√ 2τ)e(k−1)(x+q√ 2τ)e−q2/2dq. An elementary calculation shows that I1= e(k+1)x/2+(k+1)2τ/4N(d1) I2= e(k−1)x/2+(k−1)2τ/4N(d2), 168 CHAPTER 6. PARABOLIC EQUATIONS where d1=x√ 2τ+1 2(k+ 1)√ 2τ d2=x√ 2τ+1 2(k−1)√ 2τ N(di) =1√ 2π/integraldisplaydi −∞e−s2/2ds, i = 1,2. Combining the formula for u(x, τ), definition (6.17) of v(x, τ) and the previ- ous settings x= ln(S/E),τ=σ2(T−t)/2 and C=Ev(x, τ), we get finally the formula of of Theorem 6.4. In general, the solution uof the initial value problem for the heat equa- tion is not uniquely defined, see for example [10], pp. 206. Uniqueness. The uniqueness follows from the growth assumption (6.14). Assume there are two solutions of (6.11), (6.12)-(6.14), the n the difference W(S, t) satisfies the differential equation (6.11) and the side condit ions W(S, T) = 0, W(0, t) = 0, W(S, t) =O(S) asS→ ∞ uniformly in 0 ≤t≤T. From a maximum principle consideration, see an exercise, it follows that |W(S, t)| ≤cSonS≥0, 0≤t≤T. The constant cis independent on S andt. From the definition of uwe see that u(x, τ) =1 Ee−αx−βτW(S, t), where S=Eex,t=T−2τ/(σ2). Thus we have the growth property |u(x, τ)| ≤Mea|x|, x∈R, (6.18) with positive constants Manda. Then the solution of uτ=uxx, in−∞< x <∞, 0≤τ≤σ2T/2, with the initial condition u(x,0) = 0 is uniquely defined in the class of functions satisfying the growth condit ion (6.18), see Proposition 6.2 of this chapter. That is, u(x, τ)≡0. 2 6.5. BLACK-SCHOLES EQUATION 169 Put option Here is V(S, t) :=P(S, t), where P(S, t) is the value of the (European) put option. In this case we have following side conditions to (6. 11): P(S, T) = max {E−S,0} (6.19) P(0, t) = Ee−r(T−t)(6.20) P(S, t) = o(S) asS→ ∞,uniformly in 0 ≤t≤T. (6.21) HereEis the exercise price and Tthe expiry. Side condition (6.19) means that the value of the option has n o value at timeTifS(T)≥E, condition (6.20) says that it makes no sense to sell assets if the value of the asset is zero, condition (6.21) means that it makes no sense to sell assets i f its value becomes large. Theorem 6.5 (Black-Scholes formula for European put options). The solu- tionP(S, t),0< S < ∞,t < T of the initial-boundary value problem (6.11), (6.19)-(6.21) is explicitly known and is given by P(S, t) =Ee−r(T−t)N(−d2)−SN(−d1) where N(x),d1,d2are the same as in Theorem 6.4. Proof. The formula for the put option follows by the same calculatio ns as in the case of a call option or from the put-call parity C(S, t)−P(S, t) =S−Ee−r(T−t) and from N(x) +N(−x) = 1. Concerning the put-call parity see an exercise. See also [26] , pp. 40, for a heuristic argument which leads to the formula for the put-cal l parity. 2 170 CHAPTER 6. PARABOLIC EQUATIONS 6.6 Exercises 1. Show that the solution u(x, t) given by Poisson’s formula satisfies inf z∈Rnϕ(z)≤u(x, t)≤sup z∈Rnϕ(z), provided ϕ(x) is continuous and bounded on Rn. 2. Solve for given f(x) and µ∈Rthe initial value problem ut+ux+µuxxx= 0 in R×R+ u(x,0) = f(x). 3. Show by using Poisson’s formula: (i) Each function f∈C([a, b]) can be approximated uniformly by a sequence fn∈C∞[a, b] . (ii) In (i) we can choose polynomials fn(Weierstrass’s approximation theorem). Hint: Concerning (ii), replace the kernel K=exp(−|y−x|2 4t) by a se- quence of Taylor polynomials in the variable z=−|y−x|2 4t. 4. Let u(x, t) be a positive solution of ut=µuxx, t >0, where µis a constant. Show that θ:=−2µux/uis a solution of Burger’s equation θt+θθx=µθxx, t >0. 5. Assume u1(s, t), ..., u n(s, t) are solutions of ut=uss. Show that/producttextn k=1uk(xk, t) is a solution of the heat equation ut− /triangleu= 0 in Rn×(0,∞). 6. Let A,Bare real, symmetric and nonnegative matrices. Nonnegative means that all eigenvalues are nonnegative. Prove that trac e (AB)≡/summationtextn i,j=1aijbij≥0. Hint: (i) Let U= (z1, . . ., z n), where zlis an orthonormal system of eigenvectors to the eigenvalues λlof the matrix B. Then X=U √λ10 ···0 0√λ2···0 ...................... 0 0 ···√λn UT 6.6. EXERCISES 171 is a square root of B. We recall that UTBU= λ10···0 0λ2···0 ................ 0 0 ···λn . (ii) trace ( QR) =trace ( RQ). (iii) Let µ1(C), . . . µ n(C) are the eigenvalues of a real symmetric n×n- matrix. Then trace C=/summationtextn l=1µl(C), which follows from the funda- mental lemma of algebra: det (λI−C) = λn−(c11+. . .+cnn)λn−1+. . . ≡(λ−µ1)·. . .·(λ−µn) =λn−(µ1+. . .+µn)λn+1+. . . 7. Assume Ω is bounded, uis a solution of the heat equation and usat- isfies the regularity assumptions of the maximum principle (T heorem 6.2). Show that uachieves its maximum and its minimum on ST. 8. Prove the following comparison principle: Assume Ω is bou nded and u, vsatisfy the regularity assumptions of the maximum principl e. Then ut− /triangleu≤vt− /trianglevinDT u≤vonST imply that u≤vinDT. 9. Show that the comparison principle implies the maximum pr inciple. 10. Consider the boundary-initial value problem ut− /triangleu=f(x, t) inDT u(x, t) = φ(x, t) onST, where f,φare given. Prove uniqueness in the class u, u t, uxixk∈C(DT). 11. Assume u, v 1, v2∈C2(DT)∩C(DT), and uis a solution of the previous boundary-initial value problem and v1,v2satisfy (v1)t− /trianglev1≤f(x, t)≤(v2)t− /trianglev2inDT v1≤φ≤v2onST. 172 CHAPTER 6. PARABOLIC EQUATIONS Show that (inclusion theorem) v1(x, t)≤u(x, t)≤v2(x, t) onDT. 12. Show by using the comparison principle: let ube a sufficiently regular solution of ut− /triangleu= 1 in DT u= 0 on ST, then 0 ≤u(x, t)≤tinDT. 13. Discuss the result of Theorem 6.3 for the case Lu=n/summationdisplay i,j=1aij(x, t)uxixj+n/summationdisplay ibi(x, t)uxi+c(x, t)u(x, t). 14. Show that u(x, t) =∞/summationdisplay n=1cne−n2tsin(nx), where cn=2 π/integraldisplayπ 0f(x)sin(nx)dx, is a solution of the initial-boundary value problem ut=uxx, x∈(0, π), t >0, u(x,0) = f(x), u(0, t) = 0 , u(π, t) = 0 , iff∈C4(R) is odd with respect to 0 and 2 π-periodic. 15. (i) Find the solution of the diffusion problem ct=Dczzin 0≤z≤l, 0≤t <∞,D=const. > 0, under the boundary conditions cz(z, t) = 0 ifz= 0 and z=land with the given initial concentration c(z,0) =c0(z) :=/braceleftbiggc0=const. if 0≤z≤h 0 if h < z ≤l. (ii) Calculate lim t→∞c(z, t). 6.6. EXERCISES 173 16. Prove the Black-Scholes Formel for an European put option . Hint: Put-call parity. 17. Prove the put-call parity for European options C(S, t)−P(S, t) =S−Ee−r(T−t) by using the following uniqueness result: Assume Wis a solution of (6.11) under the side conditions W(S, T) = 0, W(0, t) = 0 and W(S, t) =O(S) asS→ ∞, uniformly on 0 ≤t≤T. Then W(S, t)≡ 0. 18. Prove that a solution V(S, t) of the initial-boundary value problem (6.11) in Ω under the side conditions (i) V(S, T) = 0, S≥0, (ii) V(0, t) = 0, 0≤t≤T, (iii) lim S→∞V(S, t) = 0 uniformly in 0 ≤t≤T, is uniquely determined in the class C2(Ω)∩C(Ω). 19. Prove that a solution V(S, t) of the initial-boundary value problem (6.11) in Ω, under the side conditions (i) V(S, T) = 0, S≥0, (ii) V(0, t) = 0, 0≤t≤T, (iii)V(S, t) =S+o(S) asS→ ∞, uniformly on 0 ≤t≤T, satisfies |V(S, t)| ≤cSfor all S≥0 and 0 ≤t≤T. 174 CHAPTER 6. PARABOLIC EQUATIONS Chapter 7 Elliptic equations of second order Here we consider linear elliptic equations of second order, mainly the Laplace equation /triangleu= 0. Solutions of the Laplace equation are called potential functions orharmonic functions . The Laplace equation is called also potential equation. The general elliptic equation for a scalar function u(x),x∈Ω⊂Rn, is Lu:=n/summationdisplay i,j=1aij(x)uxixj+n/summationdisplay j=1bj(x)uxj+c(x)u=f(x), where the matrix A= (aij) is real, symmetric and positive definite. If Ais a constant matrix, then a transform to principal axis and str etching of axis leads to n/summationdisplay i,j=1aijuxixj=/trianglev, where v(y) :=u(Ty),Tstands for the above composition of mappings. 7.1 Fundamental solution Here we consider particular solutions of the Laplace equati on in Rnof the type u(x) =f(|x−y|), 175 176 CHAPTER 7. ELLIPTIC EQUATIONS OF SECOND ORDER where y∈Rnis fixed and fis a function which we will determine such that udefines a solution if the Laplace equation. Setr=|x−y|, then uxi=f/prime(r)xi−yi r uxixi=f/prime/prime(r)(xi−yi)2 r2+f/prime(r)/parenleftbigg1 r−(xi−yi)2 r3/parenrightbigg /triangleu=f/prime/prime(r) +n−1 rf/prime(r). Thus a solution of /triangleu= 0 is given by f(r) =/braceleftbiggc1lnr+c2:n= 2 c1r2−n+c2:n≥3 with constants c1,c2. Definition. Setr=|x−y|. The function s(r) :=/braceleftBigg −1 2πlnr:n= 2 r2−n (n−2)ωn:n≥3 is called singularity function associated to the Laplace equation. Here isωnthe area of the n-dimensional unit sphere which is given by ωn= 2πn/2/Γ(n/2), where Γ(t) :=/integraldisplay∞ 0e−ρρt−1dρ, t > 0, is the Gamma function. Definition. A function γ(x, y) =s(r) +φ(x, y) is called fundamental solution associated to the Laplace equation if φ∈ C2(Ω) and /trianglexφ= 0 for each fixed y∈Ω. Remark. The fundamental solution γsatisfies for each fixed y∈Ω the relation −/integraldisplay Ωγ(x, y)/trianglexΦ(x)dx= Φ(y) for all Φ ∈C2 0(Ω), 7.2. REPRESENTATION FORMULA 177 see an exercise. This formula follows from considerations s imilar to the next section. In the language of distribution, this relation can be writte n by definition as −/trianglexγ(x, y) =δ(x−y), where δis the Dirac distribution, which is called δ-function. 7.2 Representation formula In the following we assume that Ω, the function φwhich appears in the definition of the fundamental solution and the potential func tionuconsid- ered are sufficiently regular such that the following calculat ions make sense, see [6] for generalizations. This is the case if Ω is bounded, ∂Ω is in C1, φ∈C2(Ω) for each fixed y∈Ω and u∈C2(Ω). x ynx Ω Ωδ Figure 7.1: Notations to Green’s identity Theorem 7.1. Letube a potential function and γa fundamental solution, then for each fixed y∈Ω u(y) =/integraldisplay ∂Ω/parenleftbigg γ(x, y)∂u(x) ∂nx−u(x)∂γ(x, y) ∂nx/parenrightbigg dSx. Proof. LetBρ(y)⊂Ω be a ball. Set Ω ρ(y) = Ω \Bρ(y). See Figure 7.2 for notations. From Green’s formula, for u, v∈C2(Ω), /integraldisplay Ωρ(y)(v/triangleu−u/trianglev)dx=/integraldisplay ∂Ωρ(y)/parenleftbigg v∂u ∂n−u∂v ∂n/parenrightbigg dS 178 CHAPTER 7. ELLIPTIC EQUATIONS OF SECOND ORDER yn Ω nρ(y) ρ Figure 7.2: Notations to Theorem 7.1 we obtain, if vis a fundamental solution and ua potential function, /integraldisplay ∂Ωρ(y)/parenleftbigg v∂u ∂n−u∂v ∂n/parenrightbigg dS= 0. Thus we have to consider /integraldisplay ∂Ωρ(y)v∂u ∂ndS=/integraldisplay ∂Ωv∂u ∂ndS+/integraldisplay ∂Bρ(y)v∂u ∂ndS /integraldisplay ∂Ωρ(y)u∂v ∂ndS=/integraldisplay ∂Ωu∂v ∂ndS+/integraldisplay ∂Bρ(y)u∂v ∂ndS. We estimate the integrals over ∂Bρ(y): (i) /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay ∂Bρ(y)v∂u ∂ndS/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤M/integraldisplay ∂Bρ(y)|v|dS ≤M/parenleftBigg/integraldisplay ∂Bρ(y)s(ρ)dS+Cωnρn−1/parenrightBigg , where M=M(y) = sup Bρ0(y)|∂u/∂n |, ρ≤ρ0, C=C(y) = sup x∈Bρ0(y)|φ(x, y)|. 7.2. REPRESENTATION FORMULA 179 From the definition of s(ρ) we get the estimate as ρ→0 /integraldisplay ∂Bρ(y)v∂u ∂ndS=/braceleftbiggO(ρ|lnρ|) : n= 2 O(ρ) : n≥3.(7.1) (ii) Consider the case n≥3, then /integraldisplay ∂Bρ(y)u∂v ∂ndS=1 ωn/integraldisplay ∂Bρ(y)u1 ρn−1dS+/integraldisplay ∂Bρ(y)u∂φ ∂ndS =1 ωnρn−1/integraldisplay ∂Bρ(y)u dS+O(ρn−1) =1 ωnρn−1u(x0)/integraldisplay ∂Bρ(y)dS+O(ρn−1), =u(x0) +O(ρn−1). for an x0∈∂Bρ(y). Combining this estimate and (7.1), we obtain the representa tion formula of the theorem. 2 Corollary. Setφ≡0 and r=|x−y|in the representation formula of Theorem 7.1, then u(y) =1 2π/integraldisplay ∂Ω/parenleftbigg lnr∂u ∂nx−u∂(lnr) ∂nx/parenrightbigg dSx, n= 2, (7.2) u(y) =1 (n−2)ωn/integraldisplay ∂Ω/parenleftbigg1 rn−2∂u ∂nx−u∂(r2−n) ∂nx/parenrightbigg dSx, n≥3.(7.3) 7.2.1 Conclusions from the representation formula Similar to the theory of functions of one complex variable, w e obtain here results for harmonic functions from the representation for mula, in particular from (7.2), (7.3). We recall that a function uis called harmonic ifu∈C2(Ω) and/triangleu= 0 in Ω. Proposition 7.1. Assume uis harmonic in Ω. Then u∈C∞(Ω). Proof. Let Ω 0⊂⊂Ω be a domain such that y∈Ω0. It follows from representation formulas (7.2), (7.3), where Ω := Ω 0, that Dlu(y) exist and 180 CHAPTER 7. ELLIPTIC EQUATIONS OF SECOND ORDER are continuous for all lsince one can change differentiation with integration in right hand sides of the representation formulae. 2 Remark. In fact, a function which is harmonic in Ω is even real analyti c in Ω, see an exercise. Proposition 7.2 (Mean value formula for harmonic functions). Assume u is harmonic in Ω. Then for each Bρ(x)⊂⊂Ω u(x) =1 ωnρn−1/integraldisplay ∂Bρ(x)u(y)dSy. Proof. Consider the case n≥3. The assertion follows from (7.3) where Ω :=Bρ(x) since r=ρand /integraldisplay ∂Bρ(x)1 rn−2∂u ∂nydSy=1 ρn−2/integraldisplay ∂Bρ(x)∂u ∂nydSy =1 ρn−2/integraldisplay Bρ(x)/triangleu dy = 0. 2 We recall that a domain Ω ∈Rnis called connected if Ω is not the union of two nonempty open subsets Ω 1, Ω2such that Ω 1∩Ω2=∅. A domain in Rn is connected if and only if its path connected. Proposition 7.3 (Maximum principle). Assume uis harmonic in a con- nected domain and achieves its supremum or infimum in Ω. Then u≡const. inΩ. Proof. Consider the case of the supremum. Let x0∈Ω such that u(x0) = sup Ωu(x) =:M. Set Ω 1:={x∈Ω :u(x) =M}and Ω 2:={x∈Ω :u(x)< M}. The set Ω 1is not empty since x0∈Ω1. The set Ω 2is open since u∈C2(Ω). Consequently, Ω 2is empty if we can show that Ω 1is open. Let x∈Ω1, then there is a ρ0>0 such that Bρ0(x)⊂Ω and u(x) =Mfor all x∈Bρ0(x). 7.3. BOUNDARY VALUE PROBLEMS 181 If not, then there exists ρ >0 and /hatwidexsuch that |/hatwidex−x|=ρ, 0< ρ < ρ 0and u(/hatwidex)< M. From the mean value formula, see Proposition 7.2, it follow s M=1 ωnρn−1/integraldisplay ∂Bρ(x)u(x)dS <M ωnρn−1/integraldisplay ∂Bρ(x)dS=M, which is a contradiction. Thus, the set Ω 2is empty since Ω 1is open. 2 Corollary. Assume Ω is connected and bounded, and u∈C2(Ω)∩C(Ω) is harmonic in Ω. Then uachieves its minimum and its maximum on the boundary ∂Ω. Remark. The previous corollary fails if Ω is not bounded as simple cou n- terexamples show. 7.3 Boundary value problems Assume Ω ⊂Rnis a connected domain. 7.3.1 Dirichlet problem TheDirichlet problem (first boundary value problem) is to find a solution u∈C2(Ω)∩C(Ω) of /triangleu= 0 in Ω (7.4) u= Φ on ∂Ω, (7.5) where Φ is given and continuous on ∂Ω. Proposition 7.4. Assume Ωis bounded, then a solution to the Dirichlet problem is uniquely determined. Proof. Maximum principle. Remark. The previous result fails if we take away in the boundary cond i- tion (7.5) one point from the the boundary as the following ex ample shows. Let Ω ⊂R2be the domain Ω ={x∈B1(0) :x2>0}, 182 CHAPTER 7. ELLIPTIC EQUATIONS OF SECOND ORDER x2 x1 1 Figure 7.3: Counterexample Assume u∈C2(Ω)∩C(Ω\ {0}) is a solution of /triangleu= 0 in Ω u= 0 on ∂Ω\ {0}. This problem has solutions u≡0andu= Im( z+z−1), where z=x1+ix2. Another example see an exercise. In contrast to this behaviour of the Laplace equation, one ha s uniqueness if/triangleu= 0 is replaced by the minimal surface equation ∂ ∂x1/parenleftBigg ux1/radicalbig 1 +|∇u|2/parenrightBigg +∂ ∂x2/parenleftBigg ux2/radicalbig 1 +|∇u|2/parenrightBigg = 0. 7.3.2 Neumann problem TheNeumann problem (second boundary value problem) is to find a solution u∈C2(Ω)∩C1(Ω) of /triangleu= 0 in Ω (7.6) ∂u ∂n= Φ on ∂Ω, (7.7) where Φ is given and continuous on ∂Ω. Proposition 7.5. Assume Ωis bounded, then a solution to the Dirichlet problem is in the class u∈C2(Ω)uniquely determined up to a constant. Proof. Exercise. Hint: Multiply the differential equation /trianglew= 0 by wand integrate the result over Ω. 7.4. GREEN’S FUNCTION FOR /triangle 183 Another proof under the weaker assumption u∈C1(Ω)∩C2(Ω) follows from the Hopf boundary point lemma, see Lecture Notes: Linea r Elliptic Equations of Second Order, for example. 7.3.3 Mixed boundary value problem TheMixed boundary value problem (third boundary value problem) is to find a solution u∈C2(Ω)∩C1(Ω) of /triangleu= 0 in Ω (7.8) ∂u ∂n+hu= Φ on ∂Ω, (7.9) where Φ and hare given and continuous on ∂Ω.e Φ and hare given and continuous on ∂Ω. Proposition 7.6. Assume Ωis bounded and sufficiently regular, then a solution to the mixed problem is uniquely determined in the c lassu∈C2(Ω) provided h(x)≥0on∂Ωandh(x)>0for at least one point x∈∂Ω. Proof. Exercise. Hint: Multiply the differential equation /trianglew= 0 by wand integrate the result over Ω. 7.4 Green’s function for /triangle Theorem 7.1 says that each harmonic function satisfies u(x) =/integraldisplay ∂Ω/parenleftbigg γ(y, x)∂u(y) ∂ny−u(y)∂γ(y, x) ∂ny/parenrightbigg dSy, (7.10) where γ(y, x) is a fundamental solution. In general, udoes not satisfies the boundary condition in the above boundary value problems. Si nceγ=s+φ, see Section 7.2, where φis anarbitrary harmonic function for each fixed x, we try to find a φsuch that usatisfies also the boundary condition. Consider the Dirichlet problem, then we look for a φsuch that γ(y, x) = 0, y∈∂Ω, x∈Ω. (7.11) Then u(x) =−/integraldisplay ∂Ω∂γ(y, x) ∂nyu(y)dSy, x∈Ω. 184 CHAPTER 7. ELLIPTIC EQUATIONS OF SECOND ORDER Suppose that uachieves its boundary values Φ of the Dirichlet problem, then u(x) =−/integraldisplay ∂Ω∂γ(y, x) ∂nyΦ(y)dSy, (7.12) We claim that this function solves the Dirichlet problem (7. 4), (7.5). A function γ(y, x) which satisfies (7.11), and some additional assump- tions, is called Green’s function. More precisely, we define a Green function as follows. Definition. A function G(y, x),y, x∈Ω,x/negationslash=y, is called Green function associated to Ω and to the Dirichlet problem (7.4), (7.5) if f or fixed x∈Ω, that is we consider G(y, x) as a function of y, the following properties hold: (i)G(y, x)∈C2(Ω\ {x})∩C(Ω\ {x}), /triangleyG(y, x) = 0, x/negationslash=y. (ii) G(y, x)−s(|x−y|)∈C2(Ω)∩C(Ω). (iii) G(y, x) = 0 if y∈∂Ω,x/negationslash=y. Remark. We will see in the next section that a Green function exists at least for some domains of simple geometry. Concerning the ex istence of a Green function for more general domains see [13]. It is an interesting fact that we get from (i)-(iii) of the abov e definition two further important properties. We assume that Ω is bounded, s ufficiently regular and connected. Proposition 7.7. A Green function has the following properties. In the casen= 2we assume diam Ω <1. (A) G(x, y) =G(y, x)(symmetry). (B) 0< G(x, y)< s(|x−y|), x, y ∈Ω, x/negationslash=y. Proof. (A) Let x(1), x(2)∈Ω. Set Bi=Bρ(x(i)),i= 1,2. We assume Bi⊂Ω and B1∩B2=∅. Since G(y, x(1)) and G(y, x(2)) are harmonic in Ω\/parenleftbig B1∪B2/parenrightbig we obtain from Green’s identity, see Figure 7.4 for notation s, 7.4. GREEN’S FUNCTION FOR /triangle 185 n n nx x(1)(2) Figure 7.4: Proof of Proposition 7.7 0 =/integraldisplay ∂(Ω\(B1∪B2))/parenleftbigg G(y, x(1))∂ ∂nyG(y, x(2)) −G(y, x(2))∂ ∂nyG(y, x(1))/parenrightbigg dSy =/integraldisplay ∂Ω/parenleftbigg G(y, x(1))∂ ∂nyG(y, x(2))−G(y, x(2))∂ ∂nyG(y, x(1))/parenrightbigg dSy +/integraldisplay ∂B1/parenleftbigg G(y, x(1))∂ ∂nyG(y, x(2))−G(y, x(2))∂ ∂nyG(y, x(1))/parenrightbigg dSy +/integraldisplay ∂B2/parenleftbigg G(y, x(1))∂ ∂nyG(y, x(2))−G(y, x(2))∂ ∂nyG(y, x(1))/parenrightbigg dSy. The integral over ∂Ω is zero because of property (iii) of a Green function, and /integraldisplay ∂B1/parenleftbigg G(y, x(1))∂ ∂nyG(y, x(2))−G(y, x(2))∂ ∂nyG(y, x(1))/parenrightbigg dSy →G(x(1), x(2)),/integraldisplay ∂B2/parenleftbigg G(y, x(1))∂ ∂nyG(y, x(2))−G(y, x(2))∂ ∂nyG(y, x(1))/parenrightbigg dSy → − G(x(2), x(1)) asρ→0. This follows by considerations as in the proof of Theorem 7 .1. (B) Since G(y, x) =s(|x−y|) +φ(y, x) 186 CHAPTER 7. ELLIPTIC EQUATIONS OF SECOND ORDER andG(y, x) = 0 if y∈∂Ω and x∈Ω we have for y∈∂Ω φ(y, x) =−s(|x−y|). From the definition of s(|x−y|) it follows that φ(y, x)<0 ify∈∂Ω. Thus, since /triangleyφ= 0 in Ω, the maximum-minimum principle implies that φ(y, x)<0 for all y, x∈Ω. Consequently G(y, x)< s(|x−y|), x, y ∈Ω, x/negationslash=y. It remains to show that G(y, x)>0, x, y ∈Ω, x/negationslash=y. Fixx∈Ω and let Bρ(x) be a ball such that Bρ(x)⊂Ω for all 0 < ρ < ρ 0. There is a sufficiently small ρ0>0 such that for each ρ, 0< ρ < ρ 0, G(y, x)>0 for all y∈Bρ(x), x/negationslash=y, see property (iii) of a Green function. Since /triangleyG(y, x) = 0 in Ω \Bρ(x) G(y, x)>0 ify∈∂Bρ(x) G(y, x) = 0 if y∈∂Ω it follows from the maximum-minimum principle that G(y, x)>0 on Ω \Bρ(x). 2 7.4.1 Green’s function for a ball If Ω = BR(0) is a ball, then Green’s function is explicitly known. Let Ω = BR(0) be a ball in Rnwith radius Rand the center at the origin. Let x, y∈BR(0) and let y/primethe reflected point of yon the sphere ∂BR(0), that is, in particular |y||y/prime|=R2, see Figure 7.5 for notations. Set G(x, y) =s(r)−s/parenleftBigρ Rr1/parenrightBig , where sis the singularity function of Section 7.1, r=|x−y|and ρ2=n/summationdisplay i=1y2 i, r1=n/summationdisplay i=1/parenleftbigg xi−R2 ρ2yi/parenrightbigg2 . 7.4. GREEN’S FUNCTION FOR /triangle 187 0yr R y'r1x Figure 7.5: Reflection on ∂BR(0) This function G(x, y) satisfies (i)-(iii) of the definition of a Green function. We claim that u(x) =−/integraldisplay ∂BR(0)∂ ∂nyG(x, y)ΦdSy is a solution of the Dirichlet problem (7.4), (7.5). This for mula is also true for a large class of domains Ω ⊂Rn, see [13]. Lemma. −∂ ∂nyG(x, y)/vextendsingle/vextendsingle/vextendsingle/vextendsingle |y|=R=1 RωnR2− |x|2 |y−x|n. Proof. Exercise. Set H(x, y) =1 RωnR2− |x|2 |y−x|n, (7.13) which is called Poisson’s kernel . Theorem 7.2. Assume Φ∈C(∂Ω). Then u(x) =/integraldisplay ∂BR(0)H(x, y)Φ(y)dSy is the solution of the first boundary value problem (7.4), (7.5 ) in the class C2(Ω)∩C(Ω). 188 CHAPTER 7. ELLIPTIC EQUATIONS OF SECOND ORDER Proof. The proof follows from following properties of H(x, y): (i)H(x, y)∈C∞,|y|=R,|x|< R, x /negationslash=y, (ii)/trianglexH(x, y) = 0,|x|< R, |y|=R, (iii)/integraltext ∂BR(0)H(x, y)dSy= 1,|x|< R, (iv)H(x, y)>0,|y|=R,|x|< R, (v) Fix ζ∈∂BR(0) and δ >0, then lim x→ζ,|x|<RH(x, y) = 0 uniformly in y∈∂BR(0),|y−ζ|> δ. (i), (iv) and (v) follow from the definition (7.13) of Hand (ii) from (7.13) or from H=−∂G(x, y) ∂ny/vextendsingle/vextendsingle/vextendsingle/vextendsingle y∈∂BR(0), Gharmonic and G(x, y) =G(y, x). Property (iii) is a consequence of formula u(x) =/integraldisplay ∂BR(0)H(x, y)u(y)dSy, for each harmonic function u, see calculations to the representation formula above. We obtain (ii) if we set u≡1. It remains to show that u, given by Poisson’s formula, is in C(BR(0)) and thatuachieves the prescribed boundary values. Fix ζ∈∂BR(0) and let x∈BR(0). Then u(x)−Φ(ζ) =/integraldisplay ∂BR(0)H(x, y)(Φ(y)−Φ(ζ))dSy =I1+I2, where I1=/integraldisplay ∂BR(0),|y−ζ|<δH(x, y)(Φ(y)−Φ(ζ))dSy I2=/integraldisplay ∂BR(0),|y−ζ|≥δH(x, y)(Φ(y)−Φ(ζ))dSy. 7.4. GREEN’S FUNCTION FOR /triangle 189 For given (small) /epsilon1 >0 there is a δ=δ(/epsilon1)>0 such that |Φ(y)−Φ(ζ)|< /epsilon1 for all y∈∂BR(0) with |y−ζ|< δ. It follows |I1| ≤/epsilon1because of (iii) and (iv). SetM= max ∂BR(0)|φ|. From (v) we conclude that there is a δ/prime>0 such that H(x, y)</epsilon1 2MωnRn−1 ifxandysatisfy |x−ζ|< δ/prime,|y−ζ|> δ, see Figure 7.6 for notations. Thus Rxζδ 'δ Figure 7.6: Proof of Theorem 7.2 |I2|< /epsilon1and the inequality |u(x)−Φ(ζ)|<2/epsilon1 forx∈BR(0) such that |x−ζ|< δ/primeis shown. 2 Remark. Define δ∈[0, π] through cos δ=x·y/(|x||y|), then we write Poisson’s formula of Theorem 7.2 as u(x) =R2− |x|2 ωnR/integraldisplay ∂BR(0)Φ(y)1 (|x|2+R2−2|x|Rcosδ)n/2dSy. 190 CHAPTER 7. ELLIPTIC EQUATIONS OF SECOND ORDER In the case n= 2 we can expand this integral in a power series with respect toρ:=|x|/Rif|x|< R, since R2− |x|2 |x|+R2−2|x|Rcosδ=1−ρ2 ρ2−2ρcosδ+ 1 = 1 + 2∞/summationdisplay n=1ρncos(nδ), see [16], pp. 18 for an easy proof of this formula, or [4], Vol. II, p. 246. 7.4.2 Green’s function and conformal mapping For two-dimensional domains there is a beautiful connection between confor- mal mapping and Green’s function. Let w=f(z) be a conformal mapping from a sufficiently regular connected domain in R2onto the interior of the unit circle, see Figure 7.7. Then the Green function of Ω is, s ee for exam- w=f(z)z w xy 1..ΩE Figure 7.7: Conformal mapping ple [16] or other text books about the theory of functions of o ne complex variable, G(z, z0) =1 2πln/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle1−f(z)f(z0) f(z)−f(z0)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle, where z=x1+ix2,z0=y1+iy2. 7.5 Inhomogeneous equation Here we consider solutions u∈C2(Ω)∩C(Ω) of −/triangleu=f(x) in Ω (7.14) u= 0 on ∂Ω, (7.15) 7.5. INHOMOGENEOUS EQUATION 191 where fis given. We need the following lemma concerning volume potentials. W e assume that Ω is bounded and sufficiently regular such that all the foll owing integrals exist. See [6] for generalizations concerning these assump tions. Let for x∈Rn,n≥3, V(x) =/integraldisplay Ωf(y)1 |x−y|n−2dy and set in the two-dimensional case V(x) =/integraldisplay Ωf(y)ln/parenleftbigg1 |x−y|/parenrightbigg dy. We recall that ωn=|∂B1(0)|. Lemma. (i)Assume f∈C(Ω). Then V∈C1(Rn)and Vxi(x) =/integraldisplay Ωf(y)∂ ∂xi/parenleftbigg1 |x−y|n−2/parenrightbigg dy,ifn≥3, Vxi(x) =/integraldisplay Ωf(y)∂ ∂xi/parenleftbigg ln/parenleftbigg1 |x−y|/parenrightbigg/parenrightbigg dyifn= 2. (ii)Iff∈C1(Ω), then V∈C2(Ω)and /triangleV=−(n−2)ωnf(x), x∈Ω, n≥3 /triangleV=−2πf(x), x∈Ω, n= 2. Proof. To simplify the presentation, we consider the case n= 3. (i) The first assertion follows since we can change differentiat ion with inte- gration since the differentiate integrand is weakly singular , see an exercise. (ii) We will differentiate at x∈Ω. Let Bρbe a fixed ball such that x∈Bρ, ρsufficiently small such that Bρ⊂Ω. Then, according to (i) and since we have the identity ∂ ∂xi/parenleftbigg1 |x−y|/parenrightbigg =−∂ ∂yi/parenleftbigg1 |x−y|/parenrightbigg which implies that f(y)∂ ∂xi/parenleftbigg1 |x−y|/parenrightbigg =−∂ ∂yi/parenleftbigg f(y)1 |x−y|/parenrightbigg +fyi(y)1 |x−y|, 192 CHAPTER 7. ELLIPTIC EQUATIONS OF SECOND ORDER we obtain Vxi(x) =/integraldisplay Ωf(y)∂ ∂xi/parenleftbigg1 |x−y|/parenrightbigg dy =/integraldisplay Ω\Bρf(y)∂ ∂xi/parenleftbigg1 |x−y|/parenrightbigg dy+/integraldisplay Bρf(y)∂ ∂xi/parenleftbigg1 |x−y|/parenrightbigg dy =/integraldisplay Ω\Bρf(y)∂ ∂xi/parenleftbigg1 |x−y|/parenrightbigg dy +/integraldisplay Bρ/parenleftbigg −∂ ∂yi/parenleftbigg f(y)1 |x−y|/parenrightbigg +fyi(y)1 |x−y|/parenrightbigg dy =/integraldisplay Ω\Bρf(y)∂ ∂xi/parenleftbigg1 |x−y|/parenrightbigg dy +/integraldisplay Bρfyi(y)1 |x−y|dy−/integraldisplay ∂Bρf(y)1 |x−y|nidSy, where nis the exterior unit normal at ∂Bρ. It follows that the first and second integral is in C1(Ω). The second integral is also in C1(Ω) according to (i) and since f∈C1(Ω) by assumption. Because of /trianglex(|x−y|−1) = 0, x/negationslash=y, it follows /triangleV=/integraldisplay Bρn/summationdisplay i=1fyi(y)∂ ∂xi/parenleftbigg1 |x−y|/parenrightbigg dy −/integraldisplay ∂Bρf(y)n/summationdisplay i=1∂ ∂xi/parenleftbigg1 |x−y|/parenrightbigg nidSy. Now we choose for Bρa ball with the center at x, then /triangleV=I1+I2, where I1=/integraldisplay Bρ(x)n/summationdisplay i=1fyi(y)yi−xi |x−y|3dy I2=−/integraldisplay ∂Bρ(x)f(y)1 ρ2dSy. We recall that n·(y−x) =ρify∈∂Bρ(x). It is I1=O(ρ) asρ→0 and forI2we obtain from the mean value theorem of the integral calculu s that 7.5. INHOMOGENEOUS EQUATION 193 for ay∈∂Bρ(x) I2=−1 ρ2f(y)/integraldisplay ∂Bρ(x)dSy =−ωnf(y), which implies that lim ρ→0I2=−ωnf(x). 2 In the following we assume that Green’s function exists for t he domain Ω, which is the case if Ω is a ball. Theorem 7.3. Assume f∈C1(Ω)∩C(Ω). Then u(x) =/integraldisplay ΩG(x, y)f(y)dy is the solution of the inhomogeneous problem (7.14), (7.15) . Proof. For simplicity of the presentation let n= 3. We will show that u(x) :=/integraldisplay ΩG(x, y)f(y)dy is a solution of (7.4), (7.5). Since G(x, y) =1 4π|x−y|+φ(x, y), where φis a potential function with respect to xory, we obtain from the above lemma that /triangleu=1 4π/triangle/integraldisplay Ωf(y)1 |x−y|dy+/integraldisplay Ω/trianglexφ(x, y)f(y)dy =−f(x), where x∈Ω. It remains to show that uachieves its boundary values. That is, for fixed x0∈∂Ω we will prove that lim x→x0, x∈Ωu(x) = 0. Set u(x) =I1+I2, 194 CHAPTER 7. ELLIPTIC EQUATIONS OF SECOND ORDER where I1(x) =/integraldisplay Ω\Bρ(x0)G(x, y)f(y)dy, I2(x) =/integraldisplay Ω∩Bρ(x0)G(x, y)f(y)dy. LetM= maxΩ|f(x)|. Since G(x, y) =1 4π1 |x−y|+φ(x, y), we obtain, if x∈Bρ(x0)∩Ω, |I2| ≤M 4π/integraldisplay Ω∩Bρ(x0)dy |x−y|+O(ρ2) ≤M 4π/integraldisplay B2ρ(x)dy |x−y|+O(ρ2) =O(ρ2) asρ→0. Consequently for given /epsilon1there is a ρ0=ρ0(/epsilon1)>0 such that |I2|</epsilon1 2for all 0 < ρ≤ρ0. For each fixed ρ, 0< ρ≤ρ0, we have lim x→x0, x∈ΩI1(x) = 0 sinceG(x0, y) = 0 if y∈Ω\Bρ(x0) and G(x, y) is uniformly continuous in x∈Bρ/2(x0)∩Ω and y∈Ω\Bρ(x0), see Figure 7.8. 2 Remark. For the proof of (ii) in the above lemma it is sufficient to assume thatfis H¨ older continuous. More precisely, let f∈Cλ(Ω), 0 < λ < 1, then V∈C2,λ(Ω), see for example [9]. 7.6. EXERCISES 195 yx xρ 0 Figure 7.8: Proof of Theorem 7.3 7.6 Exercises 1. Let γ(x, y) be a fundamental solution to /triangle,y∈Ω. Show that −/integraldisplay Ωγ(x, y)/triangleΦ(x)dx= Φ(y) for all Φ ∈C2 0(Ω). Hint: See the proof of the representation formula. 2. Show that |x|−1sin(k|x|) is a solution of the Helmholtz equation /triangleu+k2u= 0 in Rn\ {0}. 3. Assume u∈C2(Ω), Ω bounded and sufficiently regular, is a solution of /triangleu=u3in Ω u= 0 on ∂Ω. Show that u= 0 in Ω. 4. Let Ω α={x∈R2:x1>0,0< x2< x1tanα},0< α≤π. Show that u(x) =rπ αksin/parenleftBigπ αkθ/parenrightBig is a harmonic function in Ω αsatisfying u= 0 on ∂Ωα, provided kis an integer. Here ( r, θ) are polar coordinates with the center at (0 ,0). 196 CHAPTER 7. ELLIPTIC EQUATIONS OF SECOND ORDER 5. Let u∈C2(Ω) be a solution of /triangleu= 0 on the quadrangle Ω = (0,1)×(0,1) satisfying the boundary conditions u(0, y) =u(1, y) = 0 for all y∈[0,1] and uy(x,0) =uy(x,1) = 0 for all x∈[0,1]. Prove thatu≡0 inΩ. 6. Let u∈C2(Rn) be a solution of /triangleu= 0 in Rnsatisfying u∈L2(Rn), i. e.,/integraltext Rnu2(x)dx <∞. Show that u≡0 inRn. Hint: Prove /integraldisplay BR(0)|∇u|2dx≤const. R2/integraldisplay B2R(0)|u|2dx, where cis a constant independent of R. To show this inequality, multiply the differential equation b yζ:=η2u, where η∈C1is a cut-off function with properties: η≡1 inBR(0), η≡0 in the exterior of B2R(0), 0 ≤η≤1,|∇η| ≤C/R. Integrate the product, apply integration by parts and use the formula 2 ab≤ /epsilon1a2+1 /epsilon1b2,/epsilon1 >0. 7. Show that a bounded harmonic function defined on Rnmust be a constant (a theorem of Liouville). 8. Assume u∈C2(B1(0))∩C(B1(0)\ {(1,0)}) is a solution of /triangleu= 0 in B1(0) u= 0 on ∂B1(0)\ {(1,0)}. Show that there are at least two solutions. Hint: Consider u(x, y) =1−(x2+y2) (1−x)2+y2. 9. Assume Ω ⊂Rnis bounded and u, v∈C2(Ω)∩C(Ω) satisfy /triangleu=/trianglev and max ∂Ω|u−v| ≤/epsilon1for given /epsilon1 >0. Show that maxΩ|u−v| ≤/epsilon1. 10. Set Ω = Rn\B1(0) and let u∈C2(Ω) be a harmonic function in Ω satisfying lim |x|→∞u(x) = 0. Prove that max Ω|u|= max ∂Ω|u|. Hint: Apply the maximum principle to Ω ∩BR(0),Rlarge. 7.6. EXERCISES 197 11. Let Ω α={x∈R2:x1>0,0< x2< x1tanα},0< α≤π, Ωα,R= Ωα∩BR(0), and assume fis given and bounded on Ωα,R. Show that for each solution u∈C1(Ωα,R)∩C2(Ωα,R) of/triangleu=fin Ωα,Rsatisfying u= 0 on ∂Ωα,R∩BR(0), holds: For given /epsilon1 >0 there is a constant C(/epsilon1) such that |u(x)| ≤C(/epsilon1)|x|π α−/epsilon1in Ω α,R. Hint: (a) Comparison principle (a consequence from the maximum principle): Assume Ω is bounded, u, v∈C2(Ω)∩C(Ω) satisfying −/triangleu≤ −/triangle vin Ω and u≤von∂Ω. Then u≤vin Ω. (b) An appropriate comparison function is v=Arπ α−/epsilon1sin(B(θ+η)), A, B, η appropriate constants, B, ηpositive. 12. Let Ω be the quadrangle ( −1,1)×(−1,1) and u∈C2(Ω)∩C(Ω) a solution of the boundary value problem −/triangleu= 1 in Ω, u= 0 on ∂Ω. Find a lower and an upper bound for u(0,0). Hint: Consider the comparison function v=A(x2+y2),A=const. 13. Let u∈C2(Ba(0))∩C(Ba(0)) satisfying u≥0,/triangleu= 0 in Ba(0). Prove (Harnack’s inequality): an−2(a− |ζ|) (a+|ζ|)n−1u(0)≤u(ζ)≤an−2(a+|ζ|) (a− |ζ|)n−1u(0). Hint: Use the formula (see Theorem 7.2) u(y) =a2− |y|2 aωn/integraldisplay |x|=au(x) |x−y|ndSx fory=ζandy= 0. 14. Let φ(θ) be a 2 π-periodic C4-function with the Fourier series φ(θ) =∞/summationdisplay n=0(ancos(nθ) +bnsin(nθ)). Show that u=∞/summationdisplay n=0(ancos(nθ) +bnsin(nθ))rn solves the Dirichlet problem in B1(0). 198 CHAPTER 7. ELLIPTIC EQUATIONS OF SECOND ORDER 15. Assume u∈C2(Ω) satisfies /triangleu= 0 in Ω. Let Ba(ζ) be a ball such that its closure is in Ω. Show that |Dαu(ζ)| ≤M/parenleftbigg|α|γn a/parenrightbigg|α| , where M= supx∈Ba(ζ)|u(x)|andγn= 2nωn−1/((n−1)ωn). Hint: Use the formula of Theorem 7.2, successively to the k th deriv a- tives in balls with radius a(|α| −k)/m,k=o,1, . . ., m −1. 16. Use the result of the previous exercise to show that u∈C2(Ω) satis- fying/triangleu= 0 in Ω is real analytic in Ω. Hint: Use Stirling’s formula n! =nne−n/parenleftbigg√ 2πn+O/parenleftbigg1√n/parenrightbigg/parenrightbigg asn→ ∞, to show that uis in the class CK,r(ζ), where K=cMand r=a/(eγn). The constant cis the constant in the estimate nn≤cenn! which follows from Stirling’s formula. See Section 3.5 for t he definition of a real analytic function. 17. Assume Ω is connected and u∈C2(Ω) is a solution of /triangleu= 0 in Ω. Prove that u≡0 in Ω if Dαu(ζ) = 0 for all α, for a point ζ∈Ω. In particular, u≡0 in Ω if u≡0 in an open subset of Ω. 18. Let Ω = {(x1, x2, x3)∈R3:x3>0}, which is a half-space of R3. Show that G(x, y) =1 4π|x−y|−1 4π|x−y|, where y= (y1, y2,−y3), is the Green function to Ω. 19. Let Ω = {(x1, x2, x3)∈R3:x2 1+x2 2+x2 3< R2, x3>0}, which is half of a ball in R3. Show that G(x, y) =1 4π|x−y|−R 4π|y||x−y⋆| −1 4π|x−y|+R 4π|y||x−y⋆|, where y= (y1, y2,−y3),y⋆=R2y/(|y|2) and y⋆=R2y/(|y|2), is the Green function to Ω. 7.6. EXERCISES 199 20. Let Ω = {(x1, x2, x3)∈R3:x2>0, x3>0}, which is a wedge in R3. Show that G(x, y) =1 4π|x−y|−1 4π|x−y| −1 4π|x−y/prime|+1 4π|x−y/prime|, where y= (y1, y2,−y3),y/prime= (y1,−y2, y3) and y/prime= (y1,−y2,−y3), is the Green function to Ω. 21. Find Green’s function for the exterior of a disk, i. e., of the domain Ω ={x∈R2:|x|> R}. 22. Find Green’s function for the angle domain Ω = {z∈C: 0<argz < απ}, 0< α < π . 23. Find Green’s function for the slit domain Ω = {z∈C: 0<argz < 2π}. 24. Let for a sufficiently regular domain Ω ∈Rn, a ball or a quadrangle for example, F(x) =/integraldisplay ΩK(x, y)dy, where K(x, y) is continuous in Ω×Ω where x/negationslash=y, and which satisfies |K(x, y)| ≤c |x−y|α with a constants candα,α < n . Show that F(x) is continuous on Ω. 25. Prove (i) of the lemma of Section 7.5. Hint: Consider the case n≥3. Fix a function η∈C1(R) satisfying 0≤η≤1, 0≤η/prime≤2,η(t) = 0 for t≤1,η(t) = 1 for t≥2 and consider for /epsilon1 >0 the regularized integral V/epsilon1(x) :=/integraldisplay Ωf(y)η/epsilon1dy |x−y|n−2, where η/epsilon1=η(|x−y|//epsilon1). Show that V/epsilon1converges uniformly to V on compact subsets of Rnas/epsilon1→0, and that ∂V/epsilon1(x)/∂xiconverges uniformly on compact subsets of Rnto /integraldisplay Ωf(y)∂ ∂xi/parenleftbigg1 |x−y|n−2/parenrightbigg dy 200 CHAPTER 7. ELLIPTIC EQUATIONS OF SECOND ORDER as/epsilon1→0. 26. Consider the inhomogeneous Dirichlet problem −/triangleu=fin Ω, u= φon∂Ω. Transform this problem into a Dirichlet problem for the Laplace equation. Hint: Setu=w+v, where w(x) :=/integraltext Ωs(|x−y|)f(y)dy. Bibliography [1] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical tables. Vol.55, National Bu- reau of Standards Applied Mathematics Series, U.S. Governm ent Print- ing Office, Washington, DC, 1964. Reprinted by Dover, New York, 1972. [2] S. Bernstein, Sur un th´ eor` eme de g´ eom´ etrie et son application aux d´ eriv´ ees partielles du type elliptique. Comm. Soc. Math. de Kharkov (2)15,(1915–1917), 38–45. German translation: Math. Z. 26(1927), 551–558. [3] E. Bombieri, E. De Giorgi and E. Giusti, Minimal cones and the Bern- stein problem. Inv. Math. 7(1969), 243–268. [4] R. Courant und D. Hilbert, Methoden der Mathematischen Physik. Band 1 und Band 2. Springer-Verlag, Berlin, 1968. English tra nslation: Methods of Mathematical Physics. Vol. 1 and Vol. 2, Wiley-Interscience, 1962. [5] L. C. Evans, Partial Differential Equations. Graduate Studies in Math- ematics , Vol. 19, AMS, Providence, 1991. [6] L. C. Evans and R. F. Gariepy, Measure Theory and Fine Properties of Functions , Studies in Advanced Mathematics, CRC Press, Boca Raton, 1992. [7] R. Finn, Equilibrium Capillary Surfaces. Grundlehren, Vol. 284, Springer-Verlag, New York, 1986. [8] P. R. Garabedian, Partial Differential Equations. Chelsia Publishing Company, New York, 1986. [9] D. Gilbarg and N. S. Trudinger, Elliptic Partial Differential Equations of Second Order. Grundlehren, Vol. 224, Springer-Verlag, Berlin, 1983. 201 202 BIBLIOGRAPHY [10] F. John, Partial Differential Equations. Springer-Verlag, New York, 1982. [11] K. K¨ onigsberger, Analysis 2. Springer-Verlag, Berlin, 1993. [12] L. D. Landau and E. M. Lifschitz, Lehrbuch der Theoretischen Physik. Vol. 1., Akademie-Verlag, Berlin, 1964. German translation from Rus- sian. English translation: Course of Theoretical Physics. Vol. 1, Perga- mon Press, Oxford, 1976. [13] R. Leis, Vorlesungen ¨ uber partielle Differentialgleichungen zweit er Ord- nung. B. I.-Hochschultaschenb¨ ucher 165/165a, Mannheim, 1967. [14] J.-L. Lions and E. Magenes, Probl´ emes aux limites non homog´ enes et applications. Dunod, Paris, 1968. [15] E. Miersemann, Kapillarfl¨ achen. Ber. Verh. S¨ achs. Akad. Wiss. Leipzig, Math.-Natur. Kl. 130 (2008), Heft 4, S. Hirzel, Leipzig, 2008 . [16] Z. Nehari, Conformal Mapping. Reprinted by Dover, New York, 1975. [17] I. G. Petrowski, Vorlesungen ¨ uber Partielle Differentialgleichungen. Teubner, Leipzig, 1955. Translation from Russian. Englisc h translation: Lectures on Partial Differential Equations. Wiley-Interscience, 1954. [18] H. Sagan, Introduction to the Calculus of Variations. Dover, New York, 1992. [19] J. Simons, Minimal varieties in riemannian manifolds. Ann. of Math(2) 88(1968), 62–105. [20] W. I. Smirnow, Lehrgang der H¨ oheren Mathematik., Teil II. VEB Verlag der Wiss., Berlin, 1975. Translation from Russian. English translation: Course of Higher Mathematics, Vol. 2. , Elsevier, 1964. [21] W. I. Smirnow, Lehrgang der H¨ oheren Mathematik., Teil IV. VEB Ver- lag der Wiss., Berlin, 1975. Translation from Russian. Engl ish transla- tion:Course of Higher Mathematics, Vol. 4. , Elsevier, 1964. [22] A. Sommerfeld, Partielle Differentialgleichungen. Geest & Portig, Leipzig, 1954. [23] W. A. Strauss, Partial Differential equations. An Introduction. Second edition, Wiley-Interscience, 2008. German translation: Partielle Differ- entialgleichungen. Vieweg, 1995. BIBLIOGRAPHY 203 [24] M. E. Taylor, Pseudodifferential operators. Princeton, New Jersey, 1981. [25] G. N. Watson, A treatise on the Theory of Bessel Functions. Cambridge, 1952. [26] P. Wilmott, S. Howison and J. Dewynne, The Mathematics of Financial Derivatives, A Student Introduction. Cambridge University Press, 1996. [27] K. Yosida, Functional Analysis. Grundlehren, Vol. 123, Springer-Verlag, Berlin, 1965. Index d’Alembert formula 108 asymptotic expansion 84 basic lemma 16 Beltrami equations Black-Scholes equation 164 Black-Scholes formulae 165, 169 boundary condition 14, 15 capillary equation 21 Cauchy-Kowalevskaya theorem 63, 84 Cauchy-Riemann equations 13 characteristic equation 28, 33, 41, 47, 74 characteristic curve 28 characteristic strip 47 classification linear equations second order 63 quasilinear equations second or- der 73 systems first order 74 cylinder surface 29 diffusion 163 Dirac distribution 177 Dirichlet integral 17 Dirichlet problem 181 domain of dependence 108, 115 domain of influence 109 elliptic 73, 75 nonuniformly elliptic 73 second order 175 system 75, 82uniformly elliptic 73 Euler-Poisson-Darboux equation 111 Euler equation 15, 17 first order equations 25 two variables 40 Rn51 Fourier transform 141 inverse Fourier transform 142 Fourier’s method 126, 162 functionally dependent 13 fundamental solution 175, 176 Gamma function 176 gas dynamics 79 Green’s function 183 ball 186 conformal mapping 190 Hamilton function 54 Hamilton-Jacobi theory 53 harmonic function 179 heat equation 14 inhomogeneous 155 heat kernel 152, 153 helicoid 30 hyperbolic equation 107 inhomogeneous equation 117 one dimensional 107 higher dimension 109 system 74 initial conditions 15 initial-boundary value problem uniqueness 134 204 INDEX 205 string 125 membrane 128, 129 initial value problem Cauchy 33, 48 integral of a system 28 Jacobi theorem 55 Kepler 56 Laplace equation 13, 20 linear elasticity 83 linear equation 11, 25 second order 63 maximum principle heat equation 156 parabolic 161 harmonic function 180 Maxwell equations 76 mean value formula 180 minimal surface equation 18 Monge cone 42, 43 multi-index 90 multiplier 12 Navier Stokes 83 Neumann problem 20, 182 Newton potential 13 normal form 69 noncharacteristic curve 33 option call 165 put 169 parabolic equation 151 system 75 Picard-Lindel¨ of theorem 9 Poisson’s formula 152 Poisson’s kernel 187 pseudodifferential operators 146, 147 quasiconform mapping 76 quasilinear equation 11, 31 real analytic function 90 resonance 134Riemann’s method 120 Riemann problem 61 Schr¨ odinger equation 137 separation of variables 126 singularity function 176 speed plane 78 relative 81 surface 81, 83 sound 82 spherical mean 110 strip condition 46 telegraph equation 78 wave equation 14, 107, 131 wave front 51 volume potential 191