α u x α1 1 xα2 2 x αn n

Transcription

1 Chapter 1 Introduction In this chapter we introduce the notions of a Partial Differential Equation (PDE) and its solution. We attempt classifying all partial differential equations in at least three different ways. 1.1 Preliminaries Partial differential equation is an equation involving an unknown function (possibly a vectorvalued) of two or more variables and a finite number of its partial derivatives. In the sequel we reserve the following terminology and notations: Independent variables: denoted by x = (x 1, x 2,, x n ) Ω R n (n 2) Dependent variables: denoted by u = (u 1, u 2,, u p ) R p also called unknown function. Let α = (α 1,, α n ) (N {0}) n, and α = α α n. Then D α u denotes D α u = α u x α1 1 xα2 2 x αn n (1.1) For l N, D l u denotes the tensor of all partial derivatives of order l. That is, collection of all partial derivatives D α u such that α = l of the vector function u. We now define a PDE more formally. Definition 1.1 (PDE) Let Ω R n, m N and F : Ω R p R np R n2p R nmp R q be a function. A system of Partial differential equations of order m is defined by the equation F ( x, u, Du, D 2 u,, D m u ) = 0, (1.2) where some m th order partial derivative of the vector function u appears in the system of equations (1.2). Remark 1.2 The equation (1.2) consists of q equations. Note that the unknown vector function u has p components. If p = q, the system of PDE is called determined. If p < q, then the system of PDE is called over-determined system. If p > q, then the system of PDE is called under-determined system. I feel that we should not attach too much of importance to this terminology. Definition 1.3 (Solution of a PDE) Let U be an open subset of R n and Φ : U R p be a function which is m times differentiable. Then Φ is said to be a solution of the PDE (1.2) if it satisfies F ( x, Φ(x), DΦ(x), D 2 Φ(x),, D m Φ(x) ) = 0 for all x Ω. 5

2 Preliminaries What equations we study? In our course we restrict our studies to equations where (m, n, p, q) = (1, 2, 1, 1) or (m, n, p, q) = (2, 2, 1, 1). We are going to study non-linear first order PDE and linear second order PDE. Remark There is no guarantee that an equation such as (1.2) will have a solution. In fact, the PDE (u x ) 2 +1 = 0 has no solution. Thus we cannot hope to have a very general existience theorem for equations of type (1.2). 2. To convince ourselves that we do not expect every PDE to have a solution, let us recall the situation with other types of equations involving Polynomials, Systems of linear equations, Implicit functions. In each of these cases, existence of solutions was proved under some conditions. Some of those results also characterised equations that have solution(s), for example, for systems of linear equations the characterisation was in terms of ranks of matrix defining the linear system and the corresponding augmented matrix. 3. In the context of ODE, there are two basic theorems that hold for equations of a special form called normal form. They are Peano s existence theorem and Cauchy-Lipschitz-Picard s existence and uniqueness theorem. We refer the reader to any good book on ODE or my lecture notes on ODE (available on my homepage). Recall that these theorems address the existence of solutions to initial value problems for a first order system of Ordinary differential equations, and this was enough to study any ODE in normal form (of any order). Can we do the same for PDE as well? Unfortunately, it is not possible. We will see this at a later time. One problem is that same kind of problems are not interesting for all PDEs Examples of PDE 1. Laplace Equation 2. Heat Equation 3. Wave Equation u n i=1 2 u x 2 i = 0. (1.3) u u = 0. (1.4) t 2 u u = 0. (1.5) t2 4. Schrödinger Equation i u t = u + V (x)u(t, x) = 0, t > 0, x R. (1.6) 2m 5. Burgers Equation where µ 0. u t + u u x = u µ 2, t > 0, x R, (1.7) x2 6. Korteweg-de Vries (KdV) Equation u t + 3 u x 3 + u u = 0 t > 0, x R. (1.8) x

3 Chapter 1 : Introduction 7 7. Benjamin-Bona-Mahony (BBM) Equation 8. Vlasov-Poisson (VP) Equation u t + 3 u t x 2 + u u = 0 t > 0, x R. (1.9) x f t + v x f + E v f = 0 t > 0, x R n, v R n, E = x V, V = f(t,x,v)dv, V = V (x), f 0. R n (1.10) 9. Maxwell s Equations E t = curlb B t = curle divb = dive = 0. (1.11) 10. Euler s Equations for incompressible, inviscid flow u t + u Du = Dp divu = 0. (1.12) 11. Navier-Stokes Equations for incompressible, viscous flow u t + u Du u = Dp divu = 0. (1.13) 12. Minimal Surface Equation ( 1 + u 2 ) u Differences between ODE and PDE n i,j=1 u u 2 u = 0. (1.14) x i x j x i x j 1. A general solution of an ODE involves arbitrary constants. Obtaining a general solution for PDEs is difficult and a general solution would involve arbitrary functions (See 2.1.2). Let us look at a simple example now. Consider the PDE u x = 0. Any arbitrary function of y solves this PDE. This is the simplest possible linear equation of first order and it has an infinite dimensional space of solutions. Compare this situation with that of a linear first order ODE dy dt = 0 where y = (y 1,, y p ), whose solution space is R p which is finite dimensional. 2. In differential equations the unknown function has the interpretation of the state of a system when the equations decribe evolution of a physical system in time. For ODEs the independent variable is time and for PDEs one of the independent variables has the interpretation of time. Now the initial state (state of the system at time t = 0) for ODEs is prescribed as an element of R n (n is the length of the unknown vector y); while for PDEs the initial state varies in a function space. Thus solving a PDE means finding the states of the system at different times and each of these states vary in an infinite dimensional space of function while solving ODE means finding the states of the system but are in a finite dimensional

4 Classification space. 1 For those of you who know some Functional analysis, you understand the difference between finite and infinite dimensional spaces. One important theorem which is lacking in infinite dimensional spaces is a Heine-Borel theorem concerning compactness. Thus the topologies more suited to infinite dimensional spaces are some kind of weak topologies which we do not address here. 3. Linear ODEs have global solutions. Linear PDEs posed on R 2 do not necessarily have solutions defined on R 2 (See 2.2.5). 1.2 Classification Partial differential equations can be classified in at least three ways. They are 1. Order of PDE. 2. Linear, Semi-linear, Quasi-linear, and fully non-linear. 3. Scalar equation, System of equations. Classification based on the number of unknowns and number of equations in the PDE If a PDE consists of more than one unknown function or more than one equation, it is called a System of PDEs. Otherwise it is called a single PDE or a scalar PDE. These kinds of definitions will have some problems. Think why! Exercise 1.5 Classify the examples in into systems and scalar PDEs. Classification based on highest order derivative appearing in the PDE Exercise 1.6 Find the orders of each of the PDEs appearing in Classification via Algebra This classification is somewhat different from the previous ones. This is analogous to the Postal Index Number (PIN). In PIN code, the first digit stands for a region (usually consists of more than one State); the second and third digits correspond to a district and the last three digits determine the location of the Post Office within the district. Similarly, PDEs may be classified just like ODEs are classified, namely by categorising them into two: Linear and Nonlinear equations. But in the case of PDEs it turns out that there can be another way of catergorising into two: Quasi-linear and Non-Quasilinear. Quasi-linear PDEs are further categorised into two: Semi-linear, Non-semilinear. Semi-linear PDEs are further categorised into two: Linear and Nonlinear. We have the following picture. Linear PDE Semi-linear PDE Quasi-linear PDE PDE. We now define each of the terminology used above for scalar PDE, and one can extend these concepts to systems of PDE easily. We choose to define them in english and then see explicitly what it means for first order PDEs. Definition I learnt this from my teacher.

5 Chapter 1 : Introduction 9 1. A PDE of order m is called Quasi-linear if it is linear in the derivatives of order m with coefficients that depend on the independent variables and derivatives of the unknown function or order strictly less than m. 2. A Quasi-linear PDE where the coefficients of derivatives of order m are functions of the independent variables alone is called a Semi-linear PDE. 3. A PDE which is linear in the unknown function and all its derivatives with coefficients depending on the independent variables alone is called a Linear PDE. 4. A PDE which is not Quasi-linear is called a Fully nonlinear PDE. Remark A single first order Quasi-linear PDE must be of the form a(x, y, u)u x + b(x, y, u)u y = c(x, y, u) (1.15) 2. A single Quasi-linear PDE where a, b are functions of x and y alone is a Semi-linear PDE. 3. A single Semi-linear PDE where c(x, y, u) = c 0 (x, y)u + c 1 (x, y) is a Linear PDE. Examples of Linear PDEs Linear PDEs can further be classified into two: Homogeneous and Nonhomogeneous. Every linear PDE can be written in the form L[u] = f, (1.16) where u L[u] is a linear map, and f is a function of independent variables only. The linear PDE (1.16) is said to be homogeneous if f is the zero function; otherwise it is called a nonhomogeneous linear PDE. Why to Classify? 1. Classification process does not achieve anything. One could have avoided doing it. However since everyone does it, we also do it. 2. Some of the classifications are just branding a PDE. It is just immaterial what the branding is. 3. Some of the classifications help people identify or guess or anticipate the properties of solutions of PDEs in that class. For example, there could be one existence theorem that covers all equations which fall under a particular classification. 1.3 Problems to be studied In general it is difficult to find general solutions to PDEs. Recall that for first order ODEs we studied Initial value problems and boundary value problems for the second order ODE. For PDEs, we are going to study similar problems. It turns out that results for PDEs are somewhat different from the corresponding problems for ODEs. 1.4 Properly posed problems in the sense of Hadamard A mathematical problem is said to be properly-posed or well-posed in the sense of Hadamard if the following three conditions are satisfied: 1. The problem should admit at least one solution. 2. The problem should admit at most one solution. 3. The solution should depend continuously on the data in the problem.

6 Exercises 1.5 Exercises (1) We know the following classification of partial differential equations Linear PDE Semi-linear PDE Quasi-linear PDE PDE. Each of the above inclusions is a strict inclusion. Justify this statement by giving examples. (2) Give at least three examples of fifth order PDE belonging to each of the above classes. (3) Classify the following equations by all the three ways of classification. ( ) 2 u (i) y + 3 u x = 1. 3 (ii) sin ( 1 + u x) 2 + u 3 = sin x. (iii) u = 0. (iv) e u = 1. (v) u tt u = sinu. (vi) k=0 1 2 k α =k a α(x, u)d α u = sin x. (vii) k=0 1 2 k α =k Dα u = sin x.

7 Chapter 2 First order PDE 2.1 How and Why First order PDE appear? Physical origins Conservation laws form one of the two fundamental parts of any mathematical model of Continuum Mechanics. These models are PDEs. Discussion is beyond the scope of this course Mathematical origins 1. Two-parameter family of surfaces: 1 Let f : R 2 A B R be a smooth function. Then z = f(x, y, a, b), (2.1) roughly speaking, represents a two-parameter family of surfaces in R 3. Differentiating (2.1) with respect to x and y yields the relations z x = f x (x, y, a, b), z y = f y (x, y, a, b). (2.2a) (2.2b) Eliminating a and b from (2.1)-(2.2), we get a relation of the form F(x, y, z, z x, z y ) = 0. This is a PDE for the unknown function of two independent variables. Exercise 2.1 Let f(x, y, a, b) = (x a) 2 +(y b) 2. Get a PDE by eliminating the parameters a and b. (Answer: u 2 x + u2 y = 4u.) 2. Unknown function of known functions: (a) Unknown function of a single known function: Let u = f(g) where f is an unknown function and g is a known function of two independent variables x and y. Differentiating u = f(g) w.r.t. x and y yields the equations u x = f (g)g x and u y = f (g)g y respectively. Eliminating the arbitrary function f from these two equations, we obtain g y u x g x u y = 0, which is a first order PDE for u. 1 Let y = f(x, c 1, c 2,, c n) denote an n-parameter family of plane curves. By eliminating c 1,, c n we get an n-th order ODE of the form F(x, y, y,, y (n) ) = 0. If we consider a family of space curves, then we get systems of ODE after eliminating parameters. Of course, all this we get only if we are able to eliminate the parameters! 11

8 Quasi-linear PDE (b) Unknown function of two known functions: Let u = f(x ay) + g(x + ay). (2.3) Denoting v(x, y) = x ay and w(x, y) = x + ay, the above equation becomes u = f(v) + g(w). (2.4) where f, g are unknown functions and v, w are known functions. Differentiating (2.4) with respect to x and y yields the relations p = u x = f (x ay) + g (x + ay), q = u y = af (x ay) + ag (x + ay). (2.5a) (2.5b) Eliminating f and g from (2.5a)-(2.5b) (after differentiating them w.r.t. y and x respectively), we get a relation of the form q y = a 2 p x In terms of u the above first order PDE is the well-known Wave equation u yy = a 2 u xx. 2.2 Quasi-linear PDE Consider the quasi-linear PDE given by a(x, y, u)u x + b(x, y, u)u y = c(x, y, u), (2.6) where a, b, c are continuously differentiable functions on a domain Ω R 3. Let Ω 0 be the projection of Ω to the XY -plane. Definition 2.2 (Integral Surface) Let D Ω 0 and u : D R be a solution of the equation (2.6). The surface S represented by z = u(x, y) is called an Integral Surface. Remark Any point on an integral surface S has the form (x, y, u(x, y)) for some (x, y) D. Since S is an integral surface, such an (x, y) is unique. 2. Note that any integral surface S is of the form z = u(x, y) for some solution u (defined on its domain) of the equation (2.6). The projection of such an S to the XY -plane will be the domain of u. 3. For the surface z = u(x, y), the normal at any point (x 0, y 0, u(x 0, y 0 )) on S is given by (u x (x 0, y 0 ), u y (x 0, y 0 ), 1). We can write the PDE (2.6) in the form (u x (x 0, y 0 ), u y (x 0, y 0 ), 1) (a(x 0, y 0, u(x 0, y 0 )), b(x 0, y 0, u(x 0, y 0 )), c(x 0, y 0, u(x 0, y 0 ))) = 0. Thus the vector (a(x 0, y 0, u(x 0, y 0 )), b(x 0, y 0, u(x 0, y 0 )), c(x 0, y 0, u(x 0, y 0 ))) belongs to the tangent space to S at the point (x 0, y 0, u(x 0, y 0 )). By definition of tangent space, there exists a curve γ : ( δ, δ) R 3 such that (i) γ lies on S, (ii) γ(0) = (x 0, y 0, u(x 0, y 0 )), and (iii) γ (0) = (a(x 0, y 0, u(x 0, y 0 )), b(x 0, y 0, u(x 0, y 0 )), c(x 0, y 0, u(x 0, y 0 ))). This motivates the definition of a Characteristic curve that we are going to define shortly.

9 Chapter 2 : First order PDE 13 Definition 2.4 (Characteristic Vector Field) The vector field (a(x, y, z), b(x, y, z), c(x, y, z)) is called the Characteristic Vector Field of the equation (2.6). The direction of the vector (a(x, y, z), b(x, y, z), c(x, y, z)) is called the Characteristic Direction at (x, y, z) Ω. Definition 2.5 (Characteristic Curve) A curve in R 3 which is tangential to the characteristic direction at each of its points is called a Characteristic Curve (i.e., the tangent to the curve is parallel to the characteristic direction at every point on the curve). Definition 2.6 (Base Characteristics) The projections of Characteristic Curves to the XY - plane are called Base characteristics. Remark The characteristic curves of the equation (2.6) are solutions of the following autonomous system of ODEs: dx dt dy dt dz dt = a(x, y, z) = b(x, y, z) = c(x, y, z) Note that the above system is autonomous and we are interested not on the solutions but only the trace of the solutions in the XY Z-space as t varies in the interval on which solutions to the above system exist. Thus the parameter t is somewhat artificial (see [18]) and replacing it with any other parameter along a characteristic curve will amount to replacing a, b, c by proportional quantities. 2. Since a, b, c are assumed to be continuously differentiable on Ω, through any point of Ω there exists a unique characteristic curve of the equation (2.6) (Prove this!). Hence distinct characteristic curves do not intersect (why?), however their projections to the XY -plane might intersect(does it not contradict the existence and uniqueness theorem? Explain). 3. In the case of semi-linear PDE, neither distinct characteristic curves nor their projections on XY -plane intersect. However, if at least one of the functions a, b is such that existence and uniqueness theorem for solutions to IVPs cannot be applied, then it may happen that characteristic curves and their projections to XY -plane may intersect. For example, see Exercise The characteristic curves form a two-parameter family. However the solutions to the characteristic system of ODEs form a three-parameter family. Exercise 2.8 Justify the above remarks. Theorem 2.9 Let S : z = u(x, y) be a surface in R 3. Then the following statements are equivalent. 1. The surface S is an integral surface of equation (2.6). 2. The surface S is a union of characteristic curves of the equation (2.6). Proof : Proof of (1) = (2): Let S : z = u(x, y) be an integral surface of equation (2.6). That is, there is a domain D R 2 and a function u : D R such that a(x, y, u(x, y))u x (x, y) + b(x, y, u(x, y))u y (x, y) = c(x, y, u(x, y)) for all (x, y) D. (2.7)

10 Quasi-linear PDE The statement (2) of the theorem is equivalent to S = γ. γ is a characteristic curve Thus, to prove that S is a union of characteristic curves, it is sufficient to prove that the characteristic curve γ p lies entirely 1 on S for every p S (why?). Let p = (x 0, y 0, z 0 ) be an arbitrary point on the surface S. Through p, there exists a unique characteristic curve γ p and we want to prove that γ p lies entirely 2 on S. Suppose that γ p is given by x = x(t), y = y(t), z = z(t), t I and P = (x 0, y 0, z 0 ) = (x(t 0 ), y(t 0 ), z(t 0 )) for somet 0 I. Without loss of generality assume that (x(t), y(t)) D for all t I; if not we replace I by an interval I for which this holds. To prove that γ p lies entirely on S, we will prove z(t) = u (x(t), y(t)) 3 for all t I. Thus we are led to consider the following function which is defined on I: V (t) = z(t) u (x(t), y(t)) for all t I. We need to show that V is the zero function. Note that V (t 0 ) = 0 as P S. Let us compute the derivative of V. V (t) = z (t) u x (x(t), y(t)) dx dt u y (x(t), y(t)) dy dt = c (x(t), y(t), z(t)) u x (x(t), y(t)) a (x(t), y(t), z(t)) u y (x(t), y(t)) b (x(t), y(t), z(t)) = c (x(t), y(t), V (t) + u (x(t), y(t))) u x (x(t), y(t)) a (x(t), y(t), V (t) + u (x(t), y(t))) u y (x(t), y(t)) b (x(t), y(t), V (t) + u (x(t), y(t))). Thus the function V : I R is a solution (why?) of the ODE where f(t, U) = U = f(t, U), (2.8) c (x(t), y(t), U + u (x(t), y(t))) u x (x(t), y(t)) a (x(t), y(t), U + u (x(t), y(t))) u y (x(t), y(t)) b (x(t), y(t), U + u (x(t), y(t))) The RHS of (2.8) is a locally Lipschitz function w.r.t. U since a, b, c, u are continuously differentiable functions on D if u is assumed to be continuously differentiable. Further note that U(t) 0 is a solution of the ODE (2.8) (why?). Also U(t 0 ) = 0 in view of (2.7) as z = u(x, y) is an integral surface. Hence by uniqueness of solutions of Initial value problems to ODEs, we conclude that V 0. Proof of (2)= (1): Let the surface S : z = u(x, y) be a union of characteristic curves of the equation (2.6). We want to show that S is an integral surface. In other words, we want to show that u solves the equation (2.6). Let p 0 = (x 0, y 0, u(x 0, y 0 )) be any point on the surface S. We want to show (u x (x 0, y 0 ), u y (x 0, y 0 ), 1) (a(x 0, y 0, u(x 0, y 0 )), b(x 0, y 0, u(x 0, y 0 )), c(x 0, y 0, u(x 0, y 0 ))) = 0. (2.9) Since S is a union of characteristics, there is a characteristic passing through p 0 that lies entirely on S. Since (u x (x 0, y 0 ), u y (x 0, y 0 ), 1) is the normal direction to S at p 0 and (a(x 0, y 0, u(x 0, y 0 )), b(x 0, y 0, u(x 0, y 0 )), c(x 0, y 0, u(x 0, y 0 ))) is the direction of tangent to γ at p 0, we get (2.9). This finishes the proof. 1 Many books claim this statement. This is false. Note that u is given to us. We cannot assume anything about the domain on which u is defined. 2 In fact, that part of γ p lies on S for which the corresponding base characterisitcs reside in D, the domain on which u is defined. 3 We dont have the right to write this equation if (x(t), y(t)) does not belong to D for all t I.

11 Chapter 2 : First order PDE 15 Remark 2.10 (On the domain D) In the quasi-linear equation (2.6), recall that a, b, c are defined on Ω R 3 and were assumed to be continuously differentiable. Let Ω 0 be the projection of Ω into the XY -plane. That is, Ω 0 = { (x, y) R 2 : (x, y, z) Ω for some z R }. Then z = u(x, y) is an intgral surface on some domain D Ω 0. Note that characteristic curve lives in Ω, by virtue of being a soluton of the system of ODE where the vector field is defined on Ω. Projection of characteristic curves to XY -plane live inside Ω 0. Example 2.11 If u : D R be a solution to the Quasi-linear PDE (2.6), then so is v : D 1 R where v is defined by v(x, y) = u(x, y). The integral surfaces z = u(x, y) and z = v(x, y) are different since u and v are different as functions. But both the integral surfaces coincide on D 1. Thus interesection of two integral surfaces could be another integral surface. The following corollary follows immediately from Theorem 2.9 Corollary 2.12 Let S 1 and S 2 be two integral surfaces such that p S 1 S 2. Then some part of the characteristic passing through p lies on both S 1 and S 2 1. Corollary 2.13 If two integral surfaces intersect without touching and the intersection is a curve γ, then γ is a characteristic curve. Proof : To prove that γ is a characteristic curve, we have to prove that at any point P on the curve γ, the tangent has the characteristic direction. At P the tangent to the curve γ lies in the tangent planes to S 1 as well as S 2 and also the characteristic direction (a(p), b(p), c(p)). Since the tangent planes do not coincide (why?), the only direction common to both S 1 and S 2 is (a(p), b(p), c(p)). Hence tangent to the curve γ at P is proportional to the characterisitc direction at P. Since P is an arbitrary point on γ, it follows that γ is a characteristic curve. Exercise 2.14 Suppose γ is a curve that lies on two integral surfaces. Can we use the above proof to conclude that γ is a characteristic curve? If yes, give a proof. If not, explain where the above proof fails Cauchy Problem for Quasi-linear PDE Cauchy Problem To find an integral surface z = u(x, y) of the quasi-linear PDE (2.6), containing a given space curve Γ whose parametric equations are x = f(s), y = g(s), z = h(s), s I, (2.10) where f, g, h are assumed to be continuously differentiable on the interval I and h(s) = u(f(s), g(s)) for s I. Initial Value Problem Initial value problem for the quasi-linear PDE (2.6) is a special Cauchy problem for (2.6), wherein the initial curve Γ lies in the ZX-plane and the y variable has an interpretation of the time-variable. That is, Γ has the following parametric form: x = f(s), y = 0, z = h(s), s I, (2.11) 1 This is stated more confusingly and also wrongly as If two integral surfaces intersect at a point p, then they intersect along the entire characteristic curve through p. Exercise: Which part of this statement is confusing and which part is wrong?

12 Quasi-linear PDE Cauchy Problem: What to expect?: Three examples We consider three Cauchy problems for linear PDEs where the PDEs can be solved explicitly. These three examples illustrate that all three possibilities concerning a mathematical problem can occur, namely, (i) Cauchy problem has a unique solution. (Example 2.15) (ii) Cauchy problem has an infinite number of solutions. (Example 2.16) (iii) Cauchy problem has no solution. (Example 2.17) Consider the following equation u x = cu + d(x, y), (2.12) where c R and d is a continuously differentiable function. The equation (2.12) can be thought of as an ODE where y appears as a parameter. Its explicit solution is given by ( x ) u(x, y) = e cx e cξ d(ξ, y)dξ + u(0, y) (2.13) 0 Example 2.15 (Cauchy Problem 1: Existence of a Unique solution) The Cauchy data is prescribed on the Y -axis: The unique solution is given by u x = cu + d(x, y), u(0, y) = y. (2.14) u(x, y) = e cx ( x 0 ) e cξ d(ξ, y)dξ + y. Example 2.16 (Cauchy Problem 2: Non-uniqueness of solutions) Cauchy data is prescribed on the X-axis: u x = cu, u(x, 0) = e cx. (2.15) This Cauchy problem has infinitely many solutions: u(x, y) = e cx T(y), T is any function of a single variable such that T(0) = 2. Example 2.17 (Cauchy Problem 3: Non-Existence of solutions) Cauchy data is prescribed on the X-axis: u x = cu, u(x, 0) = sinx. (2.16) This Cauchy problem has no solution. For, if it has a solution then, in view of the formula (2.13), the solution satisfies sin x = u(x, 0) = e cx u(0, 0), for all x R. The above equation cannot hold and hence the Cauchy problem has no solution. Summary Observe that in these examples, when the Cauchy data was prescribed on X axis we encountered the non-existence or multiplicity of solutions to the Cauchy problems, while prescribing Cauchy data on Y -axis gave us a unique solution. The following question also arises due to these examples: What was so nice about Y -axis w.r.t. the given PDE and what was bad about X-axis in the same context. Thus, if we want a well-posed Cauchy problem, these three examples tell us that we cannot prescribe Cauchy data on arbitrary curves in the XY -plane. We will discuss these examples in Remark 2.21.

13 Chapter 2 : First order PDE Method of Characteristics To determine (prove the existence of) a solution of the equation satisfying the condition a(x, y, u)u x + b(x, y, u)u y = c(x, y, u) (2.17) u (f(s), g(s)) = h(s), s I (2.18) where I is an interval in R. i.e., the integral surface passes through the curve Γ : x = f(s), y = g(s), z = h(s), s I. (2.19) Assume f, g, h are continuously differentiable in a neighbourhood of s = s 0 I. In view of Theorem 2.9, in order to find a solution of the Cauchy problem for (2.17) we try to find the integral surface and for that it is a natural idea to find characteristics passing through points on Γ near P 0 Γ. Let P 0 = (x 0, y 0, z 0 ) = (f(s 0 ), g(s 0 ), h(s 0 )). (2.20) The characteristic differential equations of (2.17) are the system of ODE dx = a(x, y, z) dt (2.21a) dy = b(x, y, z) dt (2.21b) dz = c(x, y, z). dt (2.21c) Solve the above system (2.21) with the initial conditions at t = 0, x = f(s), y = g(s), z = h(s) for everysnears 0. (2.22) Since a, b, c are continuously differentiable functions, there exists a unique solution of the Initial Value Problem (2.21)-(2.22). Denote this solution by x = X(s, t), y = Y (s, t), z = Z(s, t). (2.23) Question: What is the domain of the vector-valued function (s, t) (X(s, t), Y (s, t), Z(s, t))? 2 Answer: Note that for each fixed s I, (X(s, t), Y (s, t), Z(s, t)) are solutions to characteristic system of ODEs on a t interval which a priori depends on the s, say J s. So, the domain is I J s. If we want to consider a range of s, then we must intersect I J s to get a domain independent of s. that is, I J s s I We do not know what kind of set this is. Can we assure that there is some open set inside this set? We need such a thing in order to even think about applying Inverse function theorem. Luckily we can choose an interval J independent of s provided we are willing to restrict ourselves to a fixed neighbourhood of s = s 0. 3, 4 Note the functions x = X(s, t), y = Y (s, t), z = Z(s, t) satisfy X(s 0, 0) = x 0 = f(s 0 ), Y (s 0, 0) = y 0 = g(s 0 ), Z(s 0, 0) = z 0 = h(s 0 ) and (2.24) 2 Answering this question is very important and quite non-trivial to say that this domain has an open set in it. Many books do not address this question. They directly apply Inverse function theorem without caring for the details other than a certain non-vanishing of a certain Jacobian! 3 This is a key-idea in proving continuous dependence of solutions to initial value problems for Ordinary differential equations. Refer to Wolfgang s ODE book, or my lecture notes on ODE. 4 Recall that Cauchy-Lipschitz-Picard theorem gives a lower estimate about the set J s, of course for each fixed s and that involves the largest possible rectangle centered at the initial condition that we can put inside certain domain, Lipschitz constant etc. However if the characteristic vector field is such that one can have a global solution for the characteristic system, then we can take J s = R and we overcome the problem of choosing a J which is independent of s.

14 Quasi-linear PDE dx = a (X(s, t), Y (s, t), Z(s, t)) dt (2.25a) dy = b (X(s, t), Y (s, t), Z(s, t)) dt (2.25b) dz = c (X(s, t), Y (s, t), Z(s, t)) dt (2.25c) The equations (2.23) represent the parametric equations of an integral surface, provided we can solve for (s, t) in terms of x, y from the two equations x = X(s, t), y = Y (s, t) and then substitute for s, t in terms of x, y in z = Z(s, t). This is achieved by applying Inverse function theorem for the vector-valued function (s, t) I J (X(s, t), Y (s, t)). Several things needs to be checked for applying Inverse function theorem. The first thing is the map given above must be continuously differentiable (total differentiability). It is enough if we know that partial derivatives exist and are continuous on I J. Can you prove this? 5 Once we prove this, we can go ahead and check the other conditions in the hypothesis of Inverse function theorem. Applying Inverse function theorem yields a neighbourhood of (s 0, 0) and a neighbourhood of (f(s 0 ), g(s 0 )) such that inversion is possible 6 and we get Define s = S(x, y), t = T(x, y). (2.26) u(x, y) = Z (S(x, y), T(x, y)). (2.27) Then u solves the PDE (2.17) 7 Then the integral surface defined by u is given by S : z = u(x, y) = Z (S(x, y), T(x, y)). (2.28) The equation (2.28) represents a continuously differentiable solution of the Cauchy problem in a neighbourhood of (x 0, y 0, z 0 ). Thus a sufficient condition for (2.28) to be a continuously differentiable integral surface is that x = X(s, t), y = Y (s, t), x 0 = X(s 0, 0) = f(s 0 ), y 0 = Y (s 0, 0) = g(s 0 ), X(s, 0) = f(s), Y (s, 0) = g(s) (2.29) That is, J = (X, Y ) (s, t) (s,t)=(s 0,0) = f (s 0 ) a (f(s 0 ), g(s 0 ), h(s 0 )) g (s 0 ) b (f(s 0 ), g(s 0 ), h(s 0 )) X s(s 0, 0) X t (s 0, 0) Y s (s 0, 0) Y t (s 0, 0) 0. (2.30) 0. (2.31) Remark 2.18 The surface S given by the equation (2.28) has parametric equations x = X(s, t), y = Y (s, t), z = Z(s, t). (2.32) The curves s = constant are the characteristic curves of equation (2.17). The tangential direction at any point of the curve s = c is given by (X t, Y t, Z t ) at s = c. This direction is also tangential to the surface S and hence the normal to S is perperndicular to the characteristic direction at any point. This implies S is an integral surface. 5 This is an important exercise. 6 Exercise: State Inverse function theorem. Apply it in the present situation and write down the conclusion correctly. 7 Exercise: Prove that u is indeed a solution. Hint: Compute the derivatives and substitute in the equation.

15 Chapter 2 : First order PDE 19 Uniqueness: If S 1 is any other integral surface containing Γ, then S 1 must contain the characteristic curves of equation (2.28) through Γ (why?). The integral surface S containing the characteristic curves of equation (2.28) passing through pionts of Γ in a neighbourhood of s 0 i.e., of the point P 0 = (x 0, y 0, z 0 ). Hence S 1 S in a neighbourhood of P 0. (2.33) Remark 2.19 The curve Γ is called Data curve or Datum curve. Remark 2.20 J = 0 implies Γ is a characteristic curve. If J = 0 at some point s 0 on an integral surface z = u(x, y) passing through Γ (if exists), then f a = g b = h c at s = s 0 (2.34) and hence Γ has the characteristic direction at s 0. This is because Now J = 0 bf ag = 0. (2.35) cg bh = c (f(s 0 ), g(s 0 ), h(s 0 ))g (s 0 ) b (f(s 0 ), g(s 0 ), h(s 0 ))h (s 0 ) = [a (f(s 0 ), g(s 0 ), h(s 0 )) u x (f(s 0 ), g(s 0 )) + b (f(s 0 ), g(s 0 ), h(s 0 )) u y (f(s 0 ), g(s 0 ))] g (s 0 ) b (f(s 0 ), g(s 0 ), h(s 0 ))[u x (f(s 0 ), g(s 0 ))f (s 0 ) + u y (f(s 0 ), g(s 0 ))g (s 0 )] = [a (f(s 0 ), g(s 0 ), h(s 0 )) g (s 0 ) b (f(s 0 ), g(s 0 ), h(s 0 )) f (s 0 )] u x (f(s 0 ), g(s 0 )) = 0, since u x (f(s 0 ), g(s 0 )) is a real number. Thus we have proved that if J 0 on I, then the curve Γ has characteristic direction at all of its points. Hence Γ is a characteristic. Characteristic Cauchy Problem If J = 0 for all s I, then by Remark 2.20, Γ has the characteristic direction at every s I and Γ is a characteristic curve. To find an integral surface passing through Γ Choose any point P 0 on Γ and any curve Γ 1 through P 0 satisfying J Γ1 (P 0 ) 0. Then there exists a unique integral surface of equation (2.17) passing through Γ 1 in a neighbourhood of P 0. Then the integral surface containing Γ 1 (exists by our theory as J Γ1 (P 0 ) 0) also contains a part of Γ containing P 0, since Γ is a characteristic curve through P 0. Then there are infinitely many solutions to the Characteristic Cauchy problem corresponding to infinitely many ways of choosing Γ 1. Proof of this is left as an exercise. Remark 2.21 If J 0, then two cases arise: either there are infinitely many solutions or no solution. We constructed an infinite number of solutions when the initial curve Γ is a characteristic curve. If Γ is not a characteristic curve, we could not prove the existence of an integral surface containing Γ; but can we prove that no one can prove the existence of a solution 8. By Remark 2.20, we can only conclude that J Γ 0 and Γ is not a characteristic curve are incompatible if u x (f(s 0 ), g(s 0 )) is a real number. So we can conclude that u is not a smooth function at points on Γ if it exists. So we can conclude that there is no smooth integral surface that contains Γ when Γ is not a characteristic but J 0. Let us return to the three examples that we considered in Subsection In Example 2.15, Jacobian is non-zero and hence we have a unique solution. In Example 2.16, J 0 and the initial curve is a characteristic curve and hence it has infinitely many solutions. In Example 2.17, J 0 and the initial curve is not a characteristic curve and hence it has no solutions. 8 Think about this.

16 Quasi-linear PDE Exercise 2.22 Consider the first order quasi-linear PDE given by uu x + u y = 1. (2.36) Solve the Cauchy problems associated with (2.36) and each of the following Cauchy data. (1) (Jacobian is non-vanishing) x = s, y = s, z = s 2, 0 s 1. (2) (Jacobian is identically equal to zero, infinite number of solutions to Cauchy problem) x = s2, y = s, z = s, 0 s 1. 2 Find all solutions and sketch some of them. (3) (Jacobian is identically equal to zero, no solutions to Cauchy problem) x = s 2, y = 2s, z = s, 0 s 1. Try solving this problem and see what happens. (4) (Jacobian is identically equal to zero on a subset) x = s 2, y = 2s, z = s, 0 s 3. Try solving this problem and see what happens. Exercise 2.23 Solve the equation u x +3y 2 3u y = 2 subject to the condition u(x, 1) = 1+x. Discuss how the non-smoothness of characteristic vector field affects the solution. (Answer: u(x, y) = x + y 1 3) Some more terminology (1) Characteristic direction (a, b, c) is also called Monge s direction. (2) Characteristic equation is also called Monge s equation. (3) Characteristic curves is also called Monge s curves Basic Existence theorem for Cauchy problem We are considering the Quasi-linear PDE given by and the associated Cauchy Problem a(x, y, u)u x + b(x, y, u)u y = c(x, y, u) (2.37) Γ : x = f(s), y = g(s), z = h(s), s I. (2.38) Definition 2.24 (Transversality condition) The quasi-linear equation (2.37) and the initial curve Γ given by (2.38) are said to satisfy transversality condition at a point (f(s), g(s), h(s)) Γ (or equivalently, at s I) if the base characterstics corresponding to the characteristic curve passing through the point (f(s), g(s), h(s)) intersects the projection of the initial curve Γ nontangentially (i.e., the tangential directions for these intersecting curves are not parallel). That is, J t=0 = X s(s 0, 0) X t (s 0, 0) Y s (s 0, 0) Y t (s 0, 0) = f (s 0 ) a (f(s 0 ), g(s 0 ), h(s 0 )) g (s 0 ) b (f(s 0 ), g(s 0 ), h(s 0 )) 0. (2.39) Theorem 2.25

17 Chapter 2 : First order PDE 21 Then (1) Assume that the functions a, b, c appearing in the quasi-linear PDE (2.37) are smooth functions on an open set containing the initial curve (2.38). (2) Let P 0 = (f(s 0 ), g(s 0 ), h(s 0 )) = Γ. (i) If the transversality condition holds at each point s in the interval (s 0 δ, s 0 + δ), then the Cauchy problem (2.37)-(2.38) has a unique solution in a neighbourhood of the point P 0. That is there exists an ǫ > 0 such that the Cauchy problem (2.37)-(2.38) has a unique solution in a region defined by (s, t) (s 0 δ, s 0 + δ) ( ǫ, ǫ) 9. (ii) If the transversality condition does not hold for s in an interval containing s 0, then the Cauchy problem (2.37)-(2.38) has either no solution at all or it has infinitely many solutions. Exercise 2.26 Explain why the procedure given to construct a solution when J 0 fails in the case where there is no solution for the Cauchy problem. Exercise 2.27 Go back to the proof of this theorem. Explain what are the reasons why we expect only a local solution. A solution of the Cauchy problem around a fixed point P on the initial curve Γ. Illustrate with one example each for your reasons Difficulties in having global solutions to Cauchy problem: How to overcome them? (1) Even when the PDE is linear, the characterisitc equations are nonlinear. From the theory of Ordinary differential equations we know that, in general, we can assert only the local existence of solutions to IVPs for nonlinear ODEs even when the vector fields are smooth. This immediately suggests that we can expect at most a local existence theorem even for a linear first-order PDE, and hence for the quasi-linear first order PDE. Thus we understand that in order to have a global existence theorem for quasilinear PDE, then the coefficients should satisfy extra hypothesis so that at least the corresponding characterisitc system of ODEs have global solutions. (2) The parametric representation of an integral surface might be misleading. We have seen many examples where the integral surface is described by smooth functions (a consequence of dependence of solutions to IVPs for ODEs on initial conditions: More precisely dependence of solutions on t is due to the characterisitc vector field i.e., the PDE; and the dependence on s is coming through IVPs for characteristic system of ODEs and also on the description of the initial curve) in (s, t)-variables. But in terms of (x, y)-variables it is not smooth. Inverse function theorem plays its role here. Now either we will be able to apply Inverse function theorem or we can not apply. For a given quasi-linear PDE, we can always find initial curves such that traversality condition does not hold, and hence we can not apply Inverse function theorem. Even when we can apply Inverse function theorem, the conclusions are only local. A good question to think about is When can we have global conclusions from Inverse function theorem? When is such a situation arise for a quasi-linear PDE? (3) A characteristic curve might intersect the initial curve more than once. Since a characteristic curve carries information from the initial curve, when a characteristic curve intersects the initial curve more than once, the information carried from different intersection points should not be in conflict if we expect to have global solutions for the Cauchy problem. Also base characteristics corresponding to distinct characteristics should not intersect for a similar reason if we are hoping for a global solution for the Cauchy problem. 9 visualise this geometrically

18 Quasi-linear PDE (4) For ODEs when the vector field is defined in the whole space, through every point there passes a solution curve. For quasi-linear first order PDE, characteristic curves may not pass through some parts of R 3 and hence information from the initial curves does not propagate in those areas. This suggests the concepts of domain of influence (of initial curve), domain of dependence (of a solution at a point on the initial curve), finite speed of propagation (of information from the initial curve when one of the independent variables has the interpretation of time).

19 Chapter 2 : First order PDE A Template for writing solutions: An example This example tells you how you must write solutions to exercises. Example 2.28 (Burger s equation) Cauchy Problem to be solved is u y + uu x = 0, u(x, 0) = h(x) for x R. (2.40) This is a Cauchy problem for a first order Quasi-linear PDE. Parametrization of the initial curve Γ Γ : x = s, y = 0, z = h(s), s R. (2.41) Characteristic equations and the initial conditions for solving them with initial conditions dx dt = z, dy dt = 1, dx dt = 0, (2.42) x(0) = s, y(0) = 0, z(0) = h(s). (2.43) The characteristic equations is a system of linear ODE and hence the IVP has a unique solution defined for all t R and thus does not depend on s. Solutions of the IVP for Characteristic equations The unique solution to (2.42)-(2.43) is given by x = X(s, t) = h(s)t + s, y = Y (s, t) = t, z = Z(s, t) = h(s) defined for (s, t) R R. (2.44) Comment on the projection of Characteristics (Base characteristics) to the XY -plane Projection of a characteristic curve (corresponds to a fixed s) to the XY -plane are given by parametric equations {(h(s)t + s, t) : t R} (2.45) In cartesian coordinates, the equation is given by X = h(s)y + s. This is a straight line with 1 slope h(s). Base characteristics correspond to different s 1 and s 2 meet if and only if the following system has a solution [ 1 h(s1 ) ] [ ] [ ] X0 s1 = (2.46) 1 h(s 2 ) Y 0 s 2 Since s 1 s 2, even if h(s 1 ) = h(s 2 ) the corresponding base characteristics are two parallel lines and hence do not intersect. If h(s 1 ) h(s 2 ) 0, then both straight lines meet at a unique point given by [ ] X0 = Y 0 [ ] 1 s2 h(s 1 ) s 1 h(s 2 ) h(s 1 ) h(s 2 ) s 2 s 1 (2.47) Comments on Jacobian and explain if the Existence and Uniqueness theorem can be applied or if it is a characteristic Cauchy problem J = (X, Y ) (s, t) = h (s)t + 1 h(s) 0 1 = 1 + h (s)t. (2.48)

20 Quasi-linear PDE We expect something wrong at (s, t) where 1 + h (s)t = 0, i.e., at t = 1 h (s). We will come back to this later on. Nevertheless, J = (X, Y ) (s, t) (s,t)=(s 0,0) = 1 h(s 0) (2.49) Thus we can apply the Basic existence theorem and we get a unique solution to the Cauchy problem near (s 0, 0, h(s 0 )) Γ. The Final solution(s) Eliminating s, t, we get Thus the solution to Cauchy problem is given implicitly by z = Z(S(x, y), T(x, y)) = h(x yz) (2.50) u = h(x yu). (2.51)

21 Chapter 2 : First order PDE Exercises (1) Solve the following problems. In each case determine the domains on which solutions exist. (i) u t + cu x = 0, u(x, 0) = f(x). (ii) xu x + u y = 1, u(x, 0) = e x. (iii) xu x + u y = 1, u(0, y) = e x. (iv) xu x + (y 2 + 1)u y = u, u(x, 0) = e x. (v) xu x + (x + y)u y = u + 1 u(x, 0) = x 2. (vi) xu x + y u y = u + 1, u(x, x) = x 2. (vii) u x + u y = u 2, u(x, 0) = h(x). (viii) u y = xuu x, u(x, 0) = x. (Ans. x = ue yu ) (ix) xu y y u x = u, u(x, 0) = h(x). (Ans. u = h( x 2 + y 2 )e arctan y x ) (x) u x + 2xu y = 2xu, u(x, 0) = x 2 for all x R. (xi) u x + 2xu y = 2xu, u(0, y) = y 2 for all y R. (xii) u x + 2xu y = 2xu, u(x, 0) = x 2 for x 0 and u(0, y) = y 2 for y 0. (xiii) u x + 2u y = 1 + u such that u = sinx on the line y = 3x + 1. (2) Solve the PDE u x +3u y = u subject to Cauchy condition u = cosx on the line y = αx. Find the value of α for which the method fails and interpret the result. (3) Solve the PDE u t + f(u)u x = 0 subject to Cauchy condition u(x, 0) = g(x), where f is a continuously differentiable function and g is a continuous function. Also find an expression for u x (x, t) and hence find a condition that must be satisfied that ensures the existence of u x (x, t) for all t > 0. (4) Solve the PDE uu t + u x = 0 subject to Cauchy condition u(x, 1) = 1 x for x 1. (5) Solve the PDE u t + uu x = t subject to Cauchy condition u(x, 0) = 2x. Find the domain in the upper half of the (x, t)-plane t 0 where the solution is valid. (6) Solve the PDE u t + uu x = 0 subject to Cauchy condition u(x, 0) = x for 1 x 2. Find the domain in the upper half-plane t 0 where the solution is valid. (7) Solve the PDE u t + uu x = 0 subject to Cauchy condition u(x, 0) = x for 1 x 2. Find the domain in the upper half-plane t 0 where the solution is valid. (8) Let u be a continuously differentiable function in the closed unit disk B(0, 1) and is a solution of a(x, y)u x + b(x, y)u y = u in B(0, 1). Let a(x, y)x+b(x, y)y > 0 on the unit circle. Prove that u vanishes identically. (Hint: Show that max 0 and min 0.) B(0,1) B(0,1) (9) Let u be a continuously differentiable function on the closure of D where D is given by D = { (x, y) R 2 : x < 1, y < 1 }. Let u be a solution of a(x, y)u x + b(x, y)u y = u, where a, b are positive differentiable functions in the entire plane. Then (a) Prove that the projection on the XY -plane of each characteristic curve passing through a point in D intersects the boundary of D at exactly two points. (Hint: The projection on the XY -plane of each characteristic curve has a positive direction and it propagates with a strictly positive speed in the square. Therefore, it intersects the boundary of D exactly twice.)

22 Quasi-linear PDE (b) Show that if u is positive on the boundary of D, then it is positive at every point in D. (Hint: Suppose that u is positive on the boundary of D and u 0 at some point in D. Consider the base characteristic passing through this point. Consider what happens to solution along this characteristic and get a contradiction.) (c) Suppose that u attains a local minimum (maximum) at a point (x 0, y 0 ) D. Evaluate u(x 0, y 0 ). (Hint: use your calculus) (d) Denote by m the minimal value of u on the boundary of D. Assume m > 0. Show that u(x, y) m for all (x, y) D. (Hint: If this were not true, then u has a global minimum in D and hence it is a local minimum. Use (c) and (b) above and conclude.)

23 Chapter 5 Laplace Equation The following equation is called Laplace equation in two independent variables x, y: The non-homogeneous problem u xx + u yy = 0. (5.1) u xx + u yy = F, (5.2) where F is a function of the independent variables x, y only is called the Poisson equation. Definition 5.1 (Solution) (i) A twice continuously differentiable function φ : D R is said to be a solution of the Laplace equation (5.1) if φ xx (x, y) + φ yy (x, y) = 0 (x, y) D. (5.3) (ii) Let F be a given continuous function on D. A twice continuously differentiable function φ : D R is said to be a solution of the Poisson equation (5.2) if φ xx (x, y) + φ yy (x, y) = F (x, y) (x, y) D. (5.4) Boundary Value Problems associated to Laplace Equation For domains other than the whole XY -plane, there is always a boundary and we can impose restrictions on the unknown function u or its derivatives or a mix of both on the boundary. These kinds of problems are called boundary value problems. There are at least three boundary value problems associated Laplace and Poisson equations. For simplicity, we describe them only for Laplace equation and the corresponding notions can be easily extended to Poisson equation. (1) Dirichlet Problem Let D be a domain in R 2 with piecewise smooth boundary and f be a continuous function on the closure of D. To find a solution u of (5.1) in the domain D such that (i) u is continuous on closure of D and (ii) u D = f If D is the unit disk, then corresponding Dirichlet problem is called interior Dirichlet problem. If D is the complement of the closed unit disk, then corresponding Dirichlet problem is called exterior Dirichlet problem. (2) Second Boundary Value Problem a.k.a. Neumann Problem To find a solution u of (5.1) in the domain D such that (i) u, u x, u y are continuous on closure of D and 41

24 (ii) at every point of D, the directional derivative of u in the direction of normal (denoted by u n or nu) satisfies n u = f. (3) Third Boundary Value Problem a.k.a. Robin Problem To find a solution u of (5.1) in the domain D such that (i) u, u x, u y are continuous on closure of D and (ii) at every point of D, u + α u n = f, where α is a given constant. Remark 5.2 (i) There are other kinds of boundary value problems possible. For example, on a part of the boundary D one of the three boundary value problems described above is imposed and on the remaining part another one of the above three. We do not consider such boundary value problems here. (ii) Cauchy-Kowalewski theorem guarantees that a solution of an analytic Cauchy problem for an elliptic equation exists and is unique (locally), but is not always well-posed. Hadamard s Example We proved that Cauchy problem for the Wave equation is well-posed for x R and t [0, T ] for every T > 0. Let us consider a similar problem for the Laplace equation now. More precisely, we consider the following Cauchy problem for Laplace equation in the upper half-plane: u = 0 on R (0, ), u(x, 0) = f(x) for all x R, u y (x, 0) = g(x) for all x R. (5.5a) (5.5b) (5.5c) Note that we proved the well-posedness of Cauchy problem for Wave equation, the initial conditions were exactly same as (5.5b)-(5.5c), except that the Laplace equation (5.5b) was replaced with Wave equation. Hadamard proved that Cauchy problem (5.5) is not well-posed on R [0, T ] for any T > 0, by proving that stability estimate does not hold. For the moment, let us accept that solution to Cauchy problem exists and is unique 1. If the problem were to be well-posed, the following stability estimate is expected to be satisfied: Given ɛ > 0, there exists a δ > 0 such that for every f 1, f 2, g 1, g 2 satisfying f 1 (x) f 2 (x) + g 1 (x) g 2 (x) < δ x R, (5.6) the corresponding solutions u 1, u 2 of the Cauchy problem satisfy u 1 (x, y) u 2 (x, y) < ɛ x R, y 0. (5.7) What should we do to prove that solutions to Cauchy problem do not have stability? 2 We produce a sequence (f n ), (g n ) which are close to the zero function but the corresponding solutions are far from the function u(x, y) 0 which is the solution with zero Cauchy data. Consider f n, g n given by sin nx f n (x) 0, g n (x) = n, (5.8) 1 This assumption says that if we somehow find a solution of the Cauchy problem then that is the only solution. 2 Formulate the negation and convince yourself that what we are dong here is indeed equivalent.