Chapter 6. Partial Differential Equations. 6.1 Introduction

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1 Chapter 6 Partial Differential Equations 6.1 Introduction Let R denote the real numbers, C the complex numbers, and n-dimensional Euclidean space is denoted by R n with points denoted by x (or sometimes by x) as tuples x =(x 1,,x n ). Often in the case of R 2 or R 3 we will use (x, y)or(x, y, z) to denote points instead of subscript notation. Ocassionally when we are dealing with dynamical problems where time is one of the variables we will also use notations like (x, t), ( x, t), (x, y, t), or (x, y, z, t). The usual Euclidean inner product is given by x z = n j=1 x jy j and the norm generated by this inner product we denote by single bars just like the absolute value, i.e., x = x x. An n-tuple α =(α 1,,α n ) of nonnegative integers is called a multi-index. We define α = n k=1 α k,α!=α 1! α n!, x R n, x α = x α 1 1 x αn n. We will use several notations for partial derivatives of a function u of x R n : u j = j u = u x j, for fist order partials and for higher order partials we have u α = α u =( 1 ) α1 ( n ) αn u = α u x α. 1 1 x αn n In particular when α = 0 then α is the identity operator. We will agree to order the set of multi-indices α =(α 1,,α n ), by requiring that α comes before β if α β or α = β and α i <β i where i is the largest number with α i β i. 1

2 2 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS A partial differential equation of order k is an equation of the form F (x 1,x 2,,x n,u, 1 u,..., n u, 2 1u,, k nu) = 0 (6.1.1) relating a function u of x =(x 1,,x n ) R n and its partial derivatives of order k. Given numbers a α with α k, we denote by (a α ) α k. the element in R N(k) given by ordering the α s in any fashion, where N(k) is the cardinality of {α : α k}. Similarly, if S {α : α k} we can consider the ordered (card S)-tuple (a α ) α S. NowletΩbeanopensetinR n, and let F be a function of the variables x Ω and (u α ) α k. Then we can write the partial differential equatio of order k as F (x, (u α ) α k )=0. (6.1.2) A function u is called a classical solution of this equation if α u exists for each α in F, and F (x, (u α (x)) α k )=0, for every x Ω. We denote by C(Ω) the space of continuous functions on Ω. If Ω is open and k is a positive integer, C k (Ω) will denote the space of functions possessing continuous derivatives up to order k on Ω, and C k (Ω) will denote the space of all u C k (Ω) such that α u extends continuously to the closure of Ω denoted by Ω for all 0 α k. We also define C (Ω) = k=1c k (Ω), C (Ω) = k=1c k (Ω). If Ω R n is open, a function u C (Ω) is said to analytic in Ω if it can be expanded in a convergent power series about every point of Ω. That is, u is analytic in Ω if for each x Ω there exist an r>0 so that for all y B r (x) ={y : y x <r} u(y) = α 0 α u(x) (y x) α, α! with the series converging absolutely and uniformly on B r (x). For complex valued analytic functions we will use the word holomorphic. Our definition of partial differential equation is really to broad since it includes equations that make no sense, such as exp( 1 u) = 0. It also allows us to think of what should be a kth order equation as a (k + m)th order equation for any m. In what follows we will impose various types of conditions on F that make more sense. The equation (6.1.2) is called linear if F is an affine-linear function of the vector of variables. This means that we can write a α (x) α u = f(x). (6.1.3) α k

3 6.1. INTRODUCTION 3 In this case we define the differential operator L = α k a α(x) α and write Lu = f. More generally, we have the quasi-linear equations which have the form a α (x, ( β u) β (k 1) ) α u = b(x, ( β u) β (k 1) ). (6.1.4) α k Thus the partial differential equation is linear if u and all of its derivatives appear in a linear fashion. For example, the general form of a linear 2 nd order partial differential equation is Au xx + Bu xy + Cu yy + Du x + Eu y + Fu = G For instance, yu xx + u yy + u x =0 is linear and uu xx + u yy + u x =0 is nonlinear but quasi-linear. If G = 0 the equation is homogeneous. Some general concerns in the study of partial differential equation s are: 1. existence of solutions, 2. uniqueness of solutions, 3. dependence of solutions on the data, 4. smoothness, or regularity of the solution, and 5. representations for solutions and behavior of solutions. The first three of these issues are related to the notion of a well-posed problem. In particular, a problem is well posed in the sense of Hadamard if there exists a unique solution that depends continuously on the data. Consider the following examples in R 2, 1. The general solution of 1 u =0isu(x 1,x 2 )=f(x 2 ) where f is arbitrary. Thus the equations gives complete information about the behavior of the solution with respect to x 1 (it is a constant) but it gives no information with respect to the x 2 variable. 2. The general solution of 1 2 u =0isu(x 1,x 2 )=f(x 1 )+g(x 2 ) where f and g are arbitrary. Thus, other that the fact that the dependences on x 1 and x 2 are uncoupled we learn nothing about the dependence on the variables. 3. Any complex valued solution u of the Cauchy-Riemann equation 1 u + i 2 u = 0 is a holomorphic function of z = x 1 + ix 2. Thus they are in particular C. The equation imposes very strong conditions on all derivatives of the solutions.

4 4 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS Linear partial differential equation s are often classified as being of elliptic, hyperbolic, or parabolic type. What follows is a brief and heuristic discussion of some of the features that characterize these classifications of partial differential equations. Most of what is presented in this introduction will be made precise in subsequent chapters. Elliptic Equations An example of an elliptic partial differential equation is Laplace s equation, div grad u =0 If u = u(x 1,,x n ) and (x 1,,x n ) are rectangular Cartesian coordinates, and w(x) = (w 1,,w n ), grad u = u =(u x1,,u xn ) div w = w + + w x 1 x n and so div grad u = u = 2 u = u = u x1 x u xnx n =0. The Laplacian of u, u, provides a comparison of values of u at a point with values at neighboring points. To illustrate this idea consider the simplest case that u = u(x) and assume u xx (x) > 0. Let ũ denote the tangent line approximation to u. For h>0and sufficiently small, u(x + h) > ũ(x + h) =u(x)+u (x)h u(x h) > ũ(x h) =u(x) u (x)h and so u(x + h)+u(x h) >u(x). 2 Roughly stated, u(x) is smaller than its average value at nearby points if u xx (x) > 0. In higher dimensions, say n = 2, the analogous statement is that if u(x, y) > 0, then the average value of u at neighboring points, say on a circle about (x, y), is greater than u(x, y). If u(x, y) < 0, then u(x, y) is greater than the average value of u on a circle about (x, y). If u(x, y) = 0 then u(x, y) is equal to its average value on a circle about (x, y). (We will subsequently prove a theorem that makes these ideas rigorous.) More generally, if u(x) =0,x Ω R n, then u is equal to its average value at neighboring points everywhere in Ω. In a certain sense, this says that if u satisfies Laplace s equation then u represents a state of equilibrium.

5 6.1. INTRODUCTION 5 Solutions to Laplace s equation u(x) = 0 are said to be harmonic. Suppose f(z) = f(x + iy) = u(x, y) + iv(x, y) is analytic. The Cauchy-Riemann equations state that u x = v y, u y = v x and so u xx + u yy =0, v xx + v yy =0. Thus the real and imaginary parts of an analytic function are harmonic. There is a converse statement of this result known as Weyl s Theorem that depends on the notion of a weak solution. Roughly stated, u(x) = 0 in a weak sense if for all smooth functions φ with compact support in Ω, u(x)( φ)(x) dx =0. We will not prove this next result. Ω Weyl s Theorem If u is a weak solution of u(x) = 0 on Ω, then u C (Ω) and u satisfies Laplace s equation in a classical sense. The so-called Cauchy problem for Laplace s equation, u xx + u yy =0, x <,y >0 u(x, 0) = f(x), u y (x, 0) = g(x), is not well-posed. Indeed, consider the specific problem u xx + u yy =0, x <,y >0 cos nx u(x, 0) = n u y (x, 0)=0 with the solution u n (x, y) = 1 cosh(ny) cos(nx). n For n sufficiently large, the data of this problem can be made uniformly small. However, lim u n(x, y) =. n In other words, small changes in the data do not correspond to small changes in a solution.

6 6 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS Another general feature of elliptic problems is that solutions are smoother that data. For instance, we will develop the following solution of Poisson s equation u(x) =f(x), x R 3, u(x) = 1 4π R 3 f(y) x y dy. (Here x = x 2 i.) From this representation of the solution, we see that the effect of integrating the data f against the kernel, 1/ x y, mollifies, or smoothes the data and so we expect that a lack of regularity in f(x) would be eliminated. This is indeed the case and in the spirit of Weyl s Theorem, we will show that the solution u is in fact C. Hyperbolic Equations One of the simplest examples of a hyperbolic equation is the wave equation, u xx = u tt, <x<, t>0. Here u(x, t) represents the displacement of a point on an infinite string at point x at time t. The proper formulation of the problem requires that we stipulate the initial position and velocity of the string. For the infinite string we need to solve the initial value problem u xx = u tt, <x<, t>0 u(x, 0) = f(x) u t (x, 0) = g(x) This initial value problem is also called the Cauchy problem and we will show that it s solution is given by D Alembert s formula, u(x, t) = 1 2 [f(x + t)+f(x t)] x+t x t g(s) ds. For instance, suppose that g(x) = 0 and f(x) is the characteristic function of the interval [ 1, 1]. Since f(x + t), f(x t) represent two waves traveling in opposite directions, it is easy to see that the discontinuity in the initial data at x = 1 and x = 1 will propagate along the lines x t =1, x + t = 1. Moreover, the support of u(x, t) travels with finite speed and exhibits a sharp leading and trailing edge. The figure below emphasizes that for each fixed time t, the region where the disturbance has spread is restricted to a finite set. The presence of the so-called sharp trailing edge is due to the fact that the initial velocity is zero. We will see that for dimensions 2, 4, 6, the region of support does not exhibit this trailing edge whereas in dimensions 3, 5, 7, this phenomenon is present, independently of the initial conditions (unlike dimension 1).

7 6.1. INTRODUCTION 7 t u=0 u=1/2 u=1/2 u=0 u=1 x u=0 The propagation of a disturbance in dimension 1. In general, for hyperbolic equations 1. solutions are no smoother than data, 2. there is a finite speed of propagation, 3. solutions exhibit a strong dependence on spatial dimension, 4. many quantities are preserved, and 5. the Cauchy problem is well-posed. Parabolic Equations The heat equation, u t (x, t) = u(x, t), x R n,t>0 is an example of a parabolic equation. If we think of u(x, t) as being the temperature at a point x at time t, this equation describes the flow or diffusion of heat. In view of our earlier discussion of the interpretation of the Laplacian, we see that if say, u(x, t) < 0, then the temperature at position x is greater than that at surrounding points. From Fourier s Law of Cooling, heat would flow away from the position x, and from the differential equation we see that u t < 0, corresponding to the decrease in temperature at that point. The Cauchy problem for the heat equation on R n is u t = u(x, t), x <, t>0. u(x, 0) = f(x).

8 8 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS The solution may be expressed as u(x, t) = 1 (4πt) n/2 e x y 2 4t R n f(y) dy. From this formula, it is reasonable to expect the solution u(x, t) to be infinitely differentiable. In addtion, we see that if even if the data f is compactly supported, the temperature u(x, t) will be positive for all x when t > 0. These observations suggest the following general features of parabolic equations: 1. solutions are smooth, and 2. there is an infinite speed of conduction. 6.2 Linear and Quasilinear equations of First Order α k In the study of linear partial differential equations a measure of the strength of a differential operator in a certain direction is provided by the notopn of charafcteristics. If L = a α (x) α is a linear differential operator of order k onωinr n, then its characteristic form (or principal symbol) atx Ω is the homogeneous polynomial of degree k on R n defined by χ L (x, ξ) = a α (x)ξ α. A vector ξ is characteristic for L at x if α =k χ L (x, ξ) =0. The characteristic variety is the set of all characteristic vectors ξ, i.e., Char x (L) ={ξ 0:χ L (x, ξ) =0}. Definition A hypersurface of class C k (1 k ) is a subset S R n such that for each x 0 S there exists an open neighborhood V R n of x 0 and a real-valued function ϕ C k (V ) such that ϕ(x) 0for all x S V where S V = {x V : ϕ(x) =0}. Remark Since, by definition, for each x 0 ϕ(x 0 ) 0 we can apply the implicit function theorem (without loss of generality let us assume that xn ϕ(x 0 ) 0) to solve ϕ(x) = 0 for x n = ψ(x ) where x =(x 1,,x n 1 ) near x 0. Thus a neighborhood of x 0 can be mapped to a piece of the hyperplane x n =0by x (x,x n ψ(x )). The same neighborhood can be

9 6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER 9 also be represented in parametric form as the image of an open set in R n 1 (with coordinates x ) under the map x (x,ψ(x )). Thus x can be thought of as giving local coordinates on S near x 0. A hypersurface S is called characteristic for L at x if the normal vector ν(x) is in Char x (L) and S is called non-characteristic if it is not characteristic at any point. An important property of the characteristic variety is contained in the following: Let F be a smooth one-to-one mapping of Ω onto Ω R n and set y = F (x). Assume that the Jacobian matrix [ ] yi J x = (x) x j is nonsingular for x Ω, so that {y 1,y 2,,y n } is a coordinate system on Ω. We have n y i = x j x j y i i=1 which we can write symbolically as x = Jx T y, where Jx T is the transpose of J x. The operator L is then transformed into L = ( a α F 1 (y) ) ( α J T F 1 (y) y) on Ω. α k When this expression is expanded out, there will be some differentiations of J T F 1 (y) but such derivatives are only formed by using up some of the y on J T F 1 (y), so they do not enter in the computation of the principal symbol in the y coordinates, i.e., they do not enter the highest order terms. We find that χ L (x, ξ) = ( ) α a α (F 1 (y)) J T F 1 (y) ξ. α =k Now since F 1 (y) =x, on comparing with the expression χ L (x, ξ) = a α (x)ξ α α =k we see that Char x (L) is the image of Char y (L ) under the linear map J T F 1 (y).

10 10 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS Note that if ξ 0 is a vector in the x j -direction (i.e., ξ i = 0 for i j), then ξ Char x (L) if and only if the coefficient of j k in L vanishes at x. Now, given any ξ 0, by a rotation of coordinates we can arrange for ξ to lie in a coordinate direction. Thus the condition ξ Char x (L) means that, in some sense, L fails to be genuinely kth order in the ξ direction at x. L is said to be elliptic at x if Char x (L) = and elliptic on Ω if it elliptic at each x Ω. Elliptic operators exert control on all derivatives of all order. Example The first three examples are in R 2 as discussed above. 1. L = 1 : Char x (L) ={ξ 0:ξ 1 =0}. 2. L = 1 2 : Char x (L) ={ξ 0:ξ 1 =0 or ξ 2 =0}. 3. L = 1 2 ( 1 + i 2 ): L is elliptic on R L = n j 2 (Laplace Operator): L is elliptic on R n. j=1 n 5. L = 1 j 2 (Heat Operator): Char x (L) ={ξ 0:ξ j =0, for j 2}. j=2 6. L = 2 1 n j 2 j=2 (Wave Operator): Char x (L) ={ξ 0:ξ 2 j = n j=2 ξ2 j }. Remark In the notation introduced in Definition we say that a surface S is oriented if for each s S we have made a choice of a vector ν(x) which is orthogonal to S and is a continuously varying function of x. Such a vector is called a normal vector to S at x. OnS V = {x : ϕ(x) =0} we have ν(x) =± ϕ(x) ϕ(x). Thus ν(x) is a C k 1 function on S. If S is the boundary of a domain Ω then we usually choose the orientation so that ν points out of Ω. At this point we can also define the normal derivative by. ν u = ν u.

11 6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER 11 Definition A hypersurface S is called characteristic for L at x S if the normal vector ν(x) to S at x is in Char x (L), and S is called non-characteristic if it is not charateristic at any x in S. We now turn to the development for real first order systems. First recall the basic problem in ODEs is the IVP: Given a function F on say R 3 and (t 0,u 0 ) R 2, find a function u(t) defined in a neighborhood of t 0 such that F (t, u, u )=0and u(t 0 )=u 0. In this disscussion we will consider the analog of this which is the initial value problem for a first order partial differential equation. We will focus on the linear and quasi-linear cases. Let us first consider the linear equation n a j j u + bu = f(x). (6.2.1) j=1 where a j, b and f are assumed to be C 1 functions of x. If we denote by A the vector field then we have A(x) =(a 1 (x),,a n (x), Char x (L) ={ξ 0:A(x) ξ =0}. That is, Char x (L) {0} is the hyperplane orthogonal to A(x). From this we see that: A hypersurface S is characteristic at x if and only if A(x) is tangent to S at x. INITIAL VALUE PROBLEM: Find a solution u to (6.2.1) with given initial values u = ϕ on a given hypersurface S. If S is characteristic at a point x 0, then the quantity a j (x 0 ) j u(x 0 ) is completely determined as a certain directional derivative of ϕ along S at x 0. For this reason it may not be possible to make it equal to f(x 0 ) b(x 0 )u(x 0 ). As an example, if the equation is 1 u = 0 and S is the hyperplane x n = 0, we cannot have u = ϕ on S unless 1 ϕ =0. Namely, consider the case of R 2. The general solution is given by u(x 1,x 2 )=f(x 2 ) where f is arbitrary. But if S corresponds to x 2 = 0 then the solution must satisfy u(x 1, 0) = φ(x 1 ) and the only choice is that ϕ 0. Thus to make the initial value problem well behaved, we must assume that S is noncharacteristic. It turns out that to solve for u it is useful to compute the integral curves of the vector field A(x).

12 12 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS Definition The integral curves of the vector field A(x) are, by definition, the parameterized curves x(t) that satisfy the system of ODEs dx dt = A(x), i.e., dx j dt = a j(x), j =1, 2,,n. (6.2.2) Along such a curve a soution u of the equation (6.2.1) must satisfy du dt = n j=1 u x j dx j dt = a j j u = f bu. (6.2.3) That is, along such a curve a solution u of the equation (6.2.1) will satisfy the ODE du = f bu. (6.2.4) dt By the fundamental existence uniqueness theorem from ODEs, through each point x 0 of S there passes a unique integral curve x(t) ofa, namely the solution of (6.2.2) with x(0) = x 0. Along this curve the solution u of (6.2.1) must also be a solution of the ODE (6.2.4) with u(0) = ϕ(x 0 ). Moreover, since A is non-characteristic, x(t) S (at least for t 0 sufficiently small) and the curves x(t) fill out a neighborhood of S. The same result as stated in Theorem is given in the simpler case of R 2 in Subsection (see, in particular, Theorem ). Theorem Assume that S is a hypersurface of class C 1 which is non-characteristic for (6.2.1), and that the functions a j, b, f, and ϕ are C 1 and real-valued. Then for any sufficiently small neighborhood Ω of S in R n there is a unique solution u C 1 of (6.2.1) that satisfies u = ϕ on S. This theorem is a special case of the corresponding result for quasi-linear equations so we will defer the proof of this result to the proof of the following more general result (see Theorem 6.2.7). Consider a first order quasi-linear equation n a j (x, u) j u = b(x, u). (6.2.5) j=1 In this case, we consider variables (x 1,,x n,u) R n+1 and note that if u is a function of x, then the normal to the graph of u (i.e., (x, u(x)) R n+1 )inr n+1 is proportional to v =( 1 u, n u, 1). So (6.2.5) says that A(x, u) =(a 1 (x, u),,a n (x, u),b(x, u))

13 6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER 13 is tangent to the graph of y = u(x) at any point (since it is orthogonal to v). This suggests that we look at the integral curves of the vector field A(x, u) inr n+1 given by solving the equations dx j dt = a j(x, y), j=1,,n, dy dt = b(x, y). (6.2.6) As you will see, any graph y = u(x) inr n+1 which is the union of an (n 1)-parameter family of these integral curves will define a solution of (6.2.5). Conversely, suppose that u is a solution of (6.2.5). If we solve the equations dx j dt = a j(x, u(x)), x j (0)=(x 0 ) j to obtain a curve x(t) passing through x 0, and then set y = u(x(t)), we have dy dt = n j=1 j u dx j dt = n a j (x, u) j u = b(x, u). j=1 Thus if the graph y = u(x) intersects an integral curve of A in one point (x 0,u(x 0 )), it contains the whole curve. Suppose we are given intial data u = ϕ on a hypersurface S in R n. If we form the submanifold S = {(x, ϕ(x)) : x S} of R n+1, the graph of the solution should be the hypersurface (in R n+1 ) generated by the integral curves of A passing through S. Again, we need to assume that S is non-characteristic in some sense. This is more complicated than the linear case because a j depend on u as well as x. We need the following geometric interpretation: For x S, the vector field A(x, ϕ(x)) = ( a 1 (x, ϕ(x)),,a n (x, ϕ(x)) ) should not be tangent to S at x. Note that this condition involves ϕ as well as S. If S is represented parametrically by a mapping g : R n 1 R n and we take coordinates s =(s 1,,s n 1 ) R n 1, so that g(s) =(g 1 (s),,g n (s)), then the above condition can be expressed as g 1 g 1 a 1 (g(s),φ(g(s))) s 1 s n 1 det (6.2.7) g n g n a n (g(s),φ(g(s))) s 1 s n 1

14 14 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS Theorem Suppose that S is a C 1 hypersurface and a j, b and φ are C 1 real valued functions. Suppose that the vector field V =(a 1 (x, φ(x)),,a n (x, φ(x))) is not tangent to S at any x S (this is the noncharacteristic condition). Then there is a unique solution u C 1 of n a j (x, u) j u = b(x, u) j=1 in a neighborhood Ω of S such that u = φ on S. The proof of this result is constructive. Proof. The uniqueness follows from the discussion above which states that u must be the union of integral curves of A in Ω passing through S. Now any hypersurface S can be covered by by open sets on which it admits a parametric representation x = g(s), with s R n 1. If we solve the problem on each such open set, by uniqueness the local solutions must agree on the overlap of the open sets and hence patch together to give a solution for all of S. It therefore suffices to assume that S is given parametrically by x = g(s) with s R n 1. For each s R n 1, consider the initial value problem x j t (s, t) =a j(x, y), j=1,,n, y (s, t) =b(x, y), (6.2.8) t x j (s, 0) = g j (s), y(s, 0) = ϕ(g(s)). Here s is a parameter vector, so we have a system of ODEs in t. By the fundamental existence uniqueness theorem for ODEs, there exists a unique solution (x, y) defined for small t, and (x, y) isac 1 function of s and t jointly. By the non-characteristic condition (6.2.7) and the inverse mapping theorem, the mapping (s, t) (x, t) is invertible on some neighborhood Ω of S, yielding s and t as C 1 functions of x on Ω such that t(x) =0andg(s(x)) = x when x S. Now set u(x) =y(s(x),t(x)). We have u = ϕ on S, and we claim that u satisfies (6.2.5). By the chain rule and the

15 6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER 15 fact that s k t = 0, since s k and t are functionally independent, and t t ) n j=1 This completes the proof. a j u x j = = = = n j=1 n 1 k=1 n 1 k=1 n 1 k=1 ( n 1 a j u s k u s k k=1 n j=1 n j=1 u s k s k =0+ u t = b. u s k s k + u t x j t x j s k a j + u x j t s k x j x j t + u t t + u t t t n j=1 a j t x j n j=1 x j t t x j = 1 we have Example In R 3 solve the linear initial value problem Lu = x 1 1 u+2x u =3u with u = ϕ(x 1,x 2 ) on the plane x 3 =0. Solvability: We could observe that the solvability condition is satisfied by considering The vector field A(x) =(x 1, 2x 2, 1) and the characteristic manifold Char x (L) ={ξ =(ξ 1,ξ 2,ξ 3 ) 0:A(x) ξ =0}. We note that the initial surface S in this case is x 3 = 0 with constant normal vector ν = (0, 0, 1). In the present case we see that A(x) ν =1 0 so the surface is non-characteristic. We could also note that with s =(s 1,s 2 ) R 3 1 and g(s) =(g 1 (s),g 2 (s),g 3 (s))=(s 1,s 2, 0) paramterizing the surface S we have for (6.2.7) 1 0 s 1 det 0 1 2s 2 = ϕ(s 1,s 2 ). 0 0 ϕ(s 1,s 2 )

16 16 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS So under the assumption that ϕ(s 1,s 2 ) 0 we can solve the problem. Solution:In the present case the initial value problem (6.2.2) (see also (6.2.6) and (6.2.8)) gives dx 1 dt = x dx 2 1, dt =2x dx 3 2, dt =1, du dt =3u, with initial conditions (x 1,x 2,x 3,u) t=0 =(s 1,s 2, 0,φ(s 1,s 2 )). Solving this system of ODEs yields x 1 = s 1 e t,x 2 = s 2 e 2t,x 3 = t, u = ϕ(s 1,s 2 )e 3t. The next step is to invert the first three equations to obtain s 1, s 2 and t t = x 3, s 1 = x 1 e x 3, s 2 = x 2 e 2x 3. Thus we find that u = ϕ(x 1 e x 3,x 2 e 2x 3 ) e 3x 3. Example Let us now consider a quasi-linear example in R 2 and we want to solve u 1 u + 2 u = 1 with u = 1 2 s on the segment x 1 = x 2 = s, 0<s<1. Solvability: We show that (6.2.7) is satisfied. x 1 a 1 (s, s, 1 2 det s s) ( 1 1 x 2 a 2 (s, s, 1 2 s s) = det s ) Solution: The desired system of ODEs in this case is with initial conditions dx 1 dt = u, dx 2 dt =1, (x 1,x 2,u) t=0 = =1 1 s 0 for 0 <s<1. 2 du dt =1, (s, s, 12 s ). Thus we get u = t s, x 2 = t + s, x 1 = 1 2 t2 + 1 st + s. 2 Since x 2 x 1 = 1 2 t(2 t s) =1 2 t(2 x 2), we can eliminate s and t from these equations to obtain u = 4x 2 2x 1 x (2 x 2 )

17 6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER Special Case: Quasilinear Equations in R 2 As a special case let us consider n = 2, i.e., let us consider first order quasi-linear (or linear) equations in two variables in the form a(x, y, u) u + b(x, y, u) u x y = c(x, y, u). (6.2.9) For this example we adhere to common practice and refer to the variables as x and y or x and t, depending on the given problem, instead of x 1 and x 2. Also in this case it is often useful to use the notations u x = u x,u y = u y,u t = u t, etc. for the various partial derivatives. Let u = u(x, y) be a solution (also refered to as a solution surface. The vector (u x,u y, 1) is normal to the surface u(x, y) u = 0 and the equation (6.2.9) expresses the orthogonality condition A(x, y, u) (u x,u y, 1)=0, A(x, y, u) (a(x, y, u),b(x, y, u),c(x, y, u)). (6.2.10) u (u x,u y, 1) (a, b, c) u = u(x, y) y x Solution Surface To solve (6.2.9) amounts to constructing a surface such that the normal to the surface satisfys the orthogonality condition (6.2.10). This is the same as saying that we seek a surface u = u(x, y) such that the tangent plane to the surface at (x, y, u) contains the vector (a(x, y, u),b(x, y, u),c(x, y, u)). More specifically, since we are really interested in the

18 18 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS initial value problem, we seek a solution of (6.2.9) that contains a given curve, say C 0 given parametrically by ( x 0 (s),y 0 (s),u 0 (s) ). Recall that a three dimensional surface (in our case the solution surface u = u(x, y)) can be represented (at least locally) parametrically as a two parameter family ( x(s, t),y(s, t),u(s, t) ). In this way we see that a solution surface is built-up from a two parameter family of curves ( x(s, t),y(s, t),u(s, t) ). And, in order that the initial values be achieved we need ( x(s, 0),y(s, 0),u(s, 0) ) = ( x0 (s),y 0 (s),u 0 (s) ). If these curves lie on a solution surface, then tangent vectors to the curves must lie in the tangent plane to the surface at the point. This means that for a fixed s the curves (in t) must satisfy the equations dx (s, t) =a(x, y, u), dt x(s, 0) = x 0(s), dy dt (s, t) =b(x, y, u), y(s, 0) = y 0(s), (6.2.11) du (s, t) =c(x, y, u), dt u(s, 0) = u 0(s). ( x0 (s),y 0 (s),u 0 (s) ) u (u x,u y, 1) ( x(s, τ),y(s, τ ),u(s, τ ) ) (a, b, c) y x characteristic ( x0 (s),y 0 (s) ) Solution Surface and Characteristic If the vector field A were tangent to the curve C 0, then a solution of (6.2.11) would coincide with C 0 and our method for constructing a surface would fail. We will see that this problem can be avoided by not perscribing data on the curve with dx dt = a, dy dt = b.

19 6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER 19 Such a curve is called a characteristic curve. To construct a solution surface we first find x = x(s, t), y = y(s, t), u = u(s, t) and then solve the two equations x = x(s, t) and y = y(s, t) for s and t in terms of x and y. In order to guarantee that this can be done requires a result from advanced calculus the inverse function theorem which states: If x = x(s, t) and y = y(s, t) are C 1 maps in a neighborhood of a point (s 0,t 0 ), the Jacobian x x J = det t y t s y s (s0,t 0 ) 0. and, in addition, x 0 = x(s 0,t 0 ) and y 0 = y(s 0,t 0 ), then there exists a neighborhood R of (s 0,t 0 ) and there exists unique C 1 mappings and s = s(x, y), t = t(x, y) s = s(x 0 (s),y 0 (s)), 0=t(x 0 (s),y 0 (s)) With this we can construct our solution surface as and u = u(s, t) =u(s(x, y),t(x, y)) = u(x, y), u 0 (s) =u(s, 0) = u(s(x 0 (s),y 0 (s)),t(x 0 (s),y 0 (s))) = u(x 0,y 0 ). Example Solve We first seek solutions of xu x +(x + y)u y = u +1, u(x, 0) = x 2. dx dt = x, In this case we can parameterize C 0 by dy dt = x + y, du dt = u +1. x 0 (s) =s, y 0 (s) =0, u 0 (s) =s 2

20 20 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS and the initial conditions are x(s, 0) = x 0 (s) =0,y(s, 0) = y 0 (s) =0,u(s, 0) = u 0 (s) =s 2. Thus we solve } x = x x(s, 0) = s y y = se t y(s, 0)=0 du u +1 = dt u(s, 0) = s 2 } x(s, t) =se t y(s, t) =ste t u(s, t) =(1+s2 )e t 1. Since x = se t and y = ste t we see that t = y x and hence s = xe y/x.thus u(x, y) =(1+x 2 e 2y/x )e y/x 1=e y/x + x 2 e y/x 1, for x 0. Example (Unidirectional Wave Motion) In this linear example we seek a function u = u(x, t) such that u t + c u =0, (6.2.12) x u(x, 0) = F (x). (6.2.13) We first seek solutions of dx dτ = c, dt dτ =1, du dτ =0, subject to x(s, 0) = s, t(s, 0)=0, u(s, 0) = F (s). Thus we obtain x = cτ + s, t = τ, u = F (s). Solving for s and τ in terms of x and t we have We then get s = x ct, τ = t u(x, t) =u(s(x, t),τ(x, t)) = F (x ct).

21 6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER 21 Example Let us now consider an example in which data is prescribed on characteristics. In this linear example we seek a function u = u(x, y) such that We first seek solutions of subject to Thus we obtain dx dt = x, x u x + y u y = u + 1 (6.2.14) u(x, x) =x 2. (6.2.15) dy dt = y, du dt = u +1, x(s, 0) = s, y(s, 0) = s, u(s, 0) = s 2. x = se t, y = se t, u = s 2 e t + e t 1. In this example it is not possible to solve for s and t in terms of x and y. Not that A(x, y) = (x, y) and the characteristic manifold Char (x,y) (L) ={ξ =(ξ 1,ξ 2 ) 0:A(x, y) ξ =0}. We note that the initial curve S in this case is x = y with constant normal vector ν =(1, 1). In the present case we see that so the curve is a characteristic. A(x, x) ν = x x =0 Within the context of these examples in R 2 solvability condition (6.2.7) as we can restate Theorem with the Theorem Suppose a(x, y, u), b(x, y, u), c(x, y, u) are C 1 in Ω R 3, C 0 is C 1 initial curve given by (x 0 (s),y 0 (s),u 0 (s)) Ω and ( ) a(x0 (s),y det 0 (s),u 0 (s)) x 0(s) b(x 0 (s),y 0 (s),u 0 (s)) y 0(s) 0. Then there exists a unique C 1 solution of a(x, y, u)u x + b(x, y, u)u y = c(x, y, u), with u(x 0 (s),y 0 (s)) = u 0 (s).

22 22 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS Proof. Our regularity assumptions guarantee that the initial value problem dx dt dy dt du dt = a(x, y, u), x(s, = b(x, y, u), y(s, = c(x, y, u), u(s, 0) = x 0(s) 0) = y 0(s) 0) = u 0(s) has a unique local solution that is C 1 in t and s. By our hypotheses x x J = det t s y y = a(x 0(s),y 0 (s),u 0 (s)) x 0(s) b(x 0 (s),y 0 (s),u 0 (s)) y 0(s) t s t=0 0 Thus J 0 in a neighborhood of the initial curve C 0 and by the inverse function theorem we can solve for s = s(x, y), t = t(x, y) and s = s(x 0 (s),y 0 (s)), 0=t(x 0 (s),y 0 (s)), so we can define u = u(x, y) =u(s(x, y),t(x, y)). We first note that u satisfies the initial conditions: u(x 0 (s),y 0 (s)) = u(s, 0) = u 0 (s). Furthermore, by the chain rule ( ) s au x + bu y = a u s x + u t t x ( s + b u s =(as x + bs y )u s +(at x + bt y ))u t. ) y + u t t y Now by our construction of x, y, and since x t = a, y t = b, wehave and as x + bs y = s x x t + s y y t = s t =0, at x + bt y = s x x t + t y y t = t t =1

23 6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER 23 and so au x + bu y = c. Recall that in the case of a differential equation in R 2 a solution surface is a surface in R 3 which (at least locally) can be parameterized by a two parameter family. Thus what we have shown is that the collection of all characteristic curves thorugh C 0 gives a solution surface. Uniqueness will follow by arguing that any solution surface is essentially a collection of characteristic curves. In particular let u(x, y) be a solution of the equation and fix a point P 0 (x 0,y 0,z 0 ) on the surface. Let γ :(x(t),y(t),z(t)) be the curve through P 0 determined by dx dt = a(x, y, u(x, y)), x(0) = x 0 dy dt = b(x, y, u(x, y)), y(0) = y 0 z(t) =u(x, y), z(0) = z 0. Then along this curve dz dt = u dx x dt + u dy y dt = au x + bu y = c since u is a solution. Thus we see that γ is the characteristic curve through P 0. In other words, a solution is always a union of characteristic curves. Through any point on a solution surface there is a unique characteristic curve. Therefore if C 0 is not a characteristic curve, there is a unique solution surface that contains it. If, on the other hand, C 0 is a characteristic curve then x 0(s) =a(x 0 (s),y 0 (s),ϕ ( s)) y 0(s) =b(x 0 (s),y 0 (s),ϕ ( s)) which contradicts J 0. Remark The above discussion shows that if C 0 were a characteristic curve we could construct infinitely many solutions containing C 0. Namely, take any curve C 1 that meets C 0 in a point P 0 and such that J 0on C 1. Then construct the solution surface through C 1. As discussed above, this solution surface must contain the characteristic curve C 0.

24 24 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS C 0 P 0 C 1 Example Recall the Example Solution Surface for Each Curve C 1 u t + cu x =0, u(x, 0) = F (x) with solution given by u(x, t) =F (x + ct). We note that the initial surface is t = 0 and the characteristics are determined by dx dτ = c, x(s, 0) = s, dt =1,t(s, 0)=0, t(s, τ) =τ, x(s, τ) =cτ + s, x ct = s. dτ Thus we see that the solution is constant along characterisitics, i.e., (x, t) on a characteristic means that x ct = s = constant and the solution is given by u(x, t) =F () and u is constant on a characteristic, i.e., a line x ct = s for s R. A generalization of this problem is the case in which c = c(x, t). Then the characteristics are determined by and along this curve dx dτ = c(x, t), x(s, 0) = s, dt dτ =1,t(s, 0)=0, du dt (x(t),t)=u dx x dt + u dt t dt = u xc(x(t),t)+u t 0.

25 6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER 25 Hence u is constant on a characteristic. Consider, for example, The characteristics are determined by which yields the parabolas u t +2tu x =0, u(x, 0) = e x2. dx dt =2t x = t 2 + k, k constant. The characteristic through a point (ξ,0) is x = t 2 + ξ. Since u is constant on this curve we have u(x, t) = exp( ξ 2 )=e (x t2 ) 2. Example In the study of fluid flow an important physical characteristic is the formation of shock waves. The simplest example of the formation of shocks can be witnessed in the study of certain quasi-linear equations in R 2 called hyperbolic conservation laws. These are equations of the form subject to initial conditions The characteristics are determined by Along this curve u t + c(u)u x =0, x R, t>0 (6.2.16) u(x, 0) = ϕ(x), x R. (6.2.17) dx dt = c(u) du dt (x(t),t)=u x(x(t),t) dx dt + u t(x(t),t) dt dt = u xc(u)+u t 0. Hence u is constant on a characteristic. The characteristics are straight lines since d 2 x dt 2 = d dt ( ) dx = d dt dt c(u) =c (u) du dt =0,

26 26 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS which implies that x = C 1 t + C 2. Now since u is constant on a characteristic, then on a characteristic, say the characteristic passing through (ξ,0), dx dt = c(u(x(t),t)) = c(u(ξ,0)) = c(ϕ(ξ)). Thus we see that the slope of the characteristics depend on c(u) and the initial data. The equation for the characteristic passing through (ξ,0) is given by and the solution on this line is given by where x = c ( ϕ(ξ) ) t + ξ. x = c ( ϕ(ξ) ) t + ξ, u(x, t) =ϕ(ξ) =ϕ(x c(ϕ(ξ))t) t (x, t) (ξ,0) x Characteristic through (ξ,0) Example One particularly famous example of a hyperbolic conservation law which is often used as a one dimensional model for the Navier-Stokes equations is the Burgers equation given by Let us consider (6.2.18) subject to initial conditions u t + uu x =0, x R, t>0. (6.2.18) 2, x < 0 u(x, 0) = ϕ(x) = 2 x, 0 x 1. (6.2.19) 1, x > 1 For x<0, the characteristics have slope (or speed) 1/2; for 0 x 1 the slope is 1/(2 x) and for x>1 the slope is 1.

27 6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER 27 Let us demonstrate that u ϕ(x c(u)t) = 0 implicitly defines a solution u(x, t) ofthe equation. First, differentiating with respect to x, wehave u x = ϕ (x c(u)t) [ x c(u)t ] = x ϕ (x c(u)t) [ 1 c (u)u x t ] and we can solve for u x u x = ϕ (x c(u)t) 1+tc (u)ϕ (x c(u)t). (6.2.20) Now we differentiate with respect to t, to obtain u t = ϕ (x c(u)t) [ x c(u)t ] = t ϕ (x c(u)t) [ tc (u)u t + c(u) ] and we can solve for u t u t = So, combining (6.2.20) and (6.2.21), we have c(u)ϕ (x c(u)t) 1+tc (u)ϕ (x c(u)t). (6.2.21) u t + c(u)u x = c(u)ϕ (x c(u)t) 1+tc (u)ϕ (x c(u)t) + c(u) ϕ (x c(u)t) 1+tc (u)ϕ (x c(u)t) = c(u)ϕ (x c(u)t)+c(u)ϕ (x c(u)t) =0. 1+tc (u)ϕ (x c(u)t) and u(x, 0) = ϕ(x 0) = ϕ(x). t t =1 (ξ,0) x Solution Exists Only for t<1 From the picture we see that solutions cannot exist for t>1 since the characteristics cross beyond that line and the values on u on the intersecting characteristics are different

28 28 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS thus, as a function, u is not well defined. More specifically, recall that our solution is defined by u(x, t) =ϕ(ξ) where x = ϕ(ξ)t + ξ. The characteristics are described by 2t + ξ, ξ < 0 x = (2 ξ)t + ξ, 0 ξ 1. t + ξ, ξ > 1 We can compute that the characteristics intersect at ((2 ξ)t + ξ) ξ=0 =(t + ξ) ξ=1, or 2t =1+t, i.e., t =1. Fort<1wehave { ϕ(ξ) =2, x < 2t (ξ <0) u(x, t) =. ϕ(ξ) =1, x>t+1 (ξ>1) t t =1 x = t +1 x =2t u =2 (ξ,0) u =1 x Solution If 0 ξ 1 the characteristic passing through (ξ,0) is x =(2 ξ)t + ξ which implies ξ =(x 2t)/(1 t) and so ( ) x 2t u(x, t) =2 = 2 x, 2t x t +1, t < 1. 1 t 1 t u t =1 u =2 t u =1 x

29 6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER 29 Solution At t = 1 refered to as the breaking time, a shock develops. As an exercise you will show that for general c(u), the breaking time is given by t b = min ξ 1 ϕ (ξ)c (ϕ(ξ)),t b > 0, at which 1+c (ϕ(ξ))ϕ (ξ)t =0.

30 30 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS Exercise Set 1: First Order Linear and Quasi-Linear Equations 1. Solve the initial value problems (a) yz x xz y =2xyz for z = t 2 on C: x = t, y = t, t>0. (b) zz x + yz y = x with z =2t on C: x = t, y = t, t R. 2. Solve x u + y u = u with u = cos(x) when y =0. 3. Solve x 2 x u + y 2 y u = u 2 with u = 1 when y =2x. 4. Solve u x u + y u = 1 with u = 0 when y = x. What happens in this problem if we replace u =0byu =1? 5. Solve u x + u y = u 2 with the initial condition u(x, 0) = h(x). 6. Show that for the initial value problem, u t + c(u)u x =0,x R, t > 0, u(x, 0) = φ(x) the breaking time will occur at the minimum value of t for which 1 t b = min φ (ξ)c (φ(ξ)) for which 1+φ (ξ)c (φ(ξ))t =0. 7. (a) Solve the initial value problem uu x + u t =0, u(x, 0) = f(x). (b) If f(x) =x, show that the solution exists for all t>0. (c) If f(x) = x, show that a shock develops, that is, the solution blows up in finite time. 8. Let D be a constant and H the Heaviside function. Solve u t + cu x =0,x R,t>0, u(x, 0) = H( x)xe x/d if (c 0 is a constant) a) c(x) =c 0 (1 x/l) b) c(x, t) =c 0 (1 x/l t/t ).

31 6.3. CHARACTERISTICS AND HIGHER ORDER EQUATIONS Characteristics and Higher Order Equations Characteristics and Classification of 2nd Order Equations L = n 2 a ij (x) (6.3.1) xi xj i,j=1 where a ij are real valued functions in Ω R n and a ij = a ji. Fix a point x 0 Ω. The characteristic polynomial is given by We say that the operator L is: σ x0 (L, ξ) = n a ij (x 0 )ξ i ξ j (6.3.2) i,j=1 1. Elliptic at x 0 if the quadratic form (6.3.2) is non-singular and definite, i.e., can be reduced by a real linear transformation to the form n ã i ξi 2 + l. o. t. i=1 2. Hyperbolic at x 0 if the quadratic form (6.3.2) is non-singular and indefinite and can be reduced by a real linear transformation to a sum of n squares, (n 1) of the same sign, i.e., to the form n ξ1 2 ã i ξi 2 + l. o. t. i=2 3. Ultra-Hyperbolic at x 0 if the quadratic form (6.3.2) is non-singular and indefinite and can be reduced by a real linear transformation to a sum of n squares, (n 4) with more than one terms of either sign. 4. Parabolic at x 0 if the quadratic form (6.3.2) is singular, i.e., can be reduced by a real linear transformation to a sum of fewer than n squares, (not necessarily of the same sign). It can be shown that in the constant coefficient case a reduction to one of these forms is always possible with a simple constant matrix transformation of coordinates. The case of two independent variables and non-constant coefficients can also be analyzed. a 2 u x 2 +2b 2 u x y + c 2 u y 2 + F (x, y, u, u x,u y ) = 0 (6.3.3)

32 32 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS where a = a(x, y), b = b(x, y), c = c(x, y) are C 2 real valued functions in Ω R 2 and (a, b, c) doesn t vanish at any point. Let us restrict to the linear case a 2 u x +2b 2 u 2 x y + u c 2 y + d u 2 x + e u + fu = g (6.3.4) y Example Consider the one dimensional wave equation Let u xx u yy =0. α = ϕ(x, y) =x + y, β = ψ(x, y) =x y. So we have (α + β) (α β) x =Φ(α, β) =, y =Ψ(α, β) =. 2 2 We can express the solutions in terms of the (x, y) or(α, β) coordinates By the chain rule we have Thus we have u(x, y) =u(φ(α, β), Ψ(α, β)) = U(α, β). u x = U α ϕ x + U β ψ x = U α + U β, (6.3.5) u xx = U αα +2U αβ + U ββ, (6.3.6) u y = U α ϕ y + U β ψ y = U α U β, (6.3.7) u yy = U αα 2U αβ + U ββ. (6.3.8) 0=u xx u yy =4U αβ, or U αβ =0. The general solution of this equation is given by which, in turn, implies U(α, β) =G(α)+F (β) u(x, y) =G(x + y)+f (x y). This solution represents a pair of waves traveling at the same speed but in opposite directions. Returning to (6.3.4), if we introduce a nonsingular C 1 coordinate transformation { α = ϕ(x, y) β = ψ(x, y),

33 6.3. CHARACTERISTICS AND HIGHER ORDER EQUATIONS 33 i.e., the Jacobian is not zero J = det α x β x α y β 0. y Applying the chain rule we have u x = u α ϕ x + u β ψ x, u y = u α ϕ y + u β ψ y, u xx = u αα ϕ 2 x +2u αβ ϕ x ψ x + u ββ ψx 2 + u α ϕ xx + u β ψ xx, u yy = u αα ϕ 2 y +2u αβ ϕ y ψ y + u ββ ψy 2 + u α ϕ yy + u β ψ yy, u xy = u αα ϕ x ϕ y + u αβ (ϕ x ψ y + ϕ y ψ x )+u ββ ψ x ψ y + u α ϕ xy + u β ψ xy. Then the equation (6.3.4) is transformed into a new equation of exactly the same form where ã 2 u α +2 b 2 u 2 α β + c 2 u u + d β2 α + ẽ u β + fu = g (6.3.9) 1. ã = aα 2 x +2bα x α y + cα 2 y 2. b = aα x β x + b(α x β y + α y β x )+cα y β y 3. c = aβ 2 x +2bβ x β y + cβ 2 y 4. d = aαxx +2bα xy + cα yy + dα x + eα y (6.3.10) 5. ẽ = aβ xx +2bβ xy + cβ yy + dβ x + eβ y 6. f = f and g = g Furthermore, we have the following extremely important invariance D =( b 2 ã c) =(b 2 ac)j 2 = DJ 2. With this we then obtain the following theorem. The proof is constructive. Theorem D is positive, negative or zero if and only if D is. Furthermore, we have 1. D>0 implies Hyperbolic

34 34 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS 2. D =0implies Parabolic 3. D<0 implies Elliptic Indeed, α and β can be chosen so that one and only one of the following hold 1. D>0 implies ã = c =0 2. D =0implies ã = b =0 3. D<0 implies ã = c and b =0 The proof of Theorem is based on the invariance of the discriminant and the ability to transform equations to a canonical form. The canonical forms only involve the second order terms and are given by: I. Hyperbolic a) u αβ + l.o.t. = 0 b) u αα u ββ + l.o.t. = 0 II. Parabolic III. Elliptic u αα + l.o.t. = 0 u αα + u ββ + l.o.t. = 0 Remark One reason why we are interested in classifying equations into such categories is that in most theoretical developments for 2nd order linear equations one that the equations are already written in one of the three basic canonical forms. In addition many numerical routines for solving a PDE assume that the equation is given in one of these canonical forms. Lower order terms are usually handled separately as a subroutine. Concerning the proof of Theorem will consider the cases I. and II. in some detail and give a reference to where the (somewhat more complicated) case III. can be found in the literature. I. Hyperbolic Case: Assume that D = b 2 ac > 0. If a = c = 0 we are done so we assume without loss of generality that a 0. We seek α = ϕ(x, y) and β = ψ(x, y) so that ã and c are zero. We first note that the calculations (6.3.10) suggest that we consider an expression of the form av 2 x +2bv x v y + cv 2 y = 0 (6.3.11)

35 6.3. CHARACTERISTICS AND HIGHER ORDER EQUATIONS 35 where v could represent either ϕ or ψ. We note that (6.3.11) can be factored into [ ( ) ][ ( ) ] b + b2 ac b b2 ac a v x v y v x v y. (6.3.12) a a Thus from (6.3.12) we seek ϕ and ψ so that, for example, [ ( ) ] b + b2 ac ϕ x ϕ y =0, (6.3.13) a and [ ψ x ( ) ] b b2 ac ψ y =0. (6.3.14) a With these choices (6.3.11) is satisfied with v given by ϕ and ψ. In addition we must impose a noncharacteristic solvability condition (ϕ, ψ) (x, y) = ϕ x ϕ y ψ x ψ y ( ) b + b2 ac = ϕ y ψ y a ( b b2 ac a ) ϕ y ψ y = 2 a b2 ac ϕ y ψ y 0. (6.3.15) A solution of (6.3.13) is found by solving ( ) dx dt =1, dy b + dt = b2 ac, a dϕ dt =0. For this problem we take the initial conditions x 0 (s) =x 0,y 0 (s) =y 0 + s, ϕ 0 (s) =s. With this choice we find (in our earlier notation we used a and b which are not the same as those occuring in the present problem) a x 1 0 0(s) b y 0(s) = ( ) b b2 ac 0. (6.3.16) 1 a

36 36 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS Moreover, and so ϕ(s, t) =s ϕ s =1=ϕ x x s + ϕ y y s. If ϕ y = 0, then it follows from (6.3.13) that ϕ x = 0, a contradiction. Now, as we have learned in our earlier work, the condition (6.3.16) guarantees that we can solve for (s, t) in terms of (x, y) to obtain ϕ(x, y). Similarly, we can obtain ψ(x, y) by solving dx dt =1, dy dt = ( b + b2 ac For this problem we take the initial conditions a ), dψ dt =0. x 0 (s) =x 0,y 0 (s) =y 0 + s, ψ 0 (s) =s. Again, ψ(s, t) =s and as above ψ y 0. ThusJ ( (ϕ, ψ)/(x, y) ) 0. In summary we have found a change of coordinates α = ϕ(x, y), β = ψ(x, y) that reduces the equation to 2 bu αβ + l.o.t. = 0. It is easy to show that b =2 ( ac b 2 a ) ϕ y ψ y (6.3.17) and so b 0 and we can divide by it to put (6.3.17) into the normal form Remark u αβ = F (α, β, u, u α,u β ). (a) The reduction to normal form is a local result. (b) The special initial curve can be replaced by any curve (x 0 (s),y 0 (s),ϕ 0 (s)) with ϕ 0 (s) =s and (x 0 (s),y 0 (s)) that specifies a curve in the xy-plane where the PDE is hyperbolic. (c) The curves ϕ(x, y) = constant, ψ(x, y) = constant are called characteristics of (6.3.4). Further, equation (6.3.11) is called the characteristic equation for the PDE.

37 6.3. CHARACTERISTICS AND HIGHER ORDER EQUATIONS 37 (d) Note that the characteristics for the first order PDE s for ϕ and ψ are determined by ( ) dx dt =1, dy b ± dt = b2 ac a or ( ) dy b ± dx = b2 ac. a But if ϕ(x, y) = constant and ϕ solves (6.3.13), then and so dy dx = ϕ x ϕ y = ϕ x + ϕ y dy dx =0 ( b b2 ac a ). (6.3.18) Hence the characteristics for a 2nd order PDE (6.3.4) coincide with the characteristics of the associated 1st order PDE (6.3.13) (Similarly for ψ). (e) In order to obtain the other hyperbolic form we set ξ = α + β, η = α β so that and Thus we have α = ξ + η 2, β = ξ η 2, u α = u η η α + u ξ ξ α = u η + u ξ, u β = u η η β + u ξ ξ β = u η + u ξ. u αβ =( u ηη + u ηξ )+( u ξη + u ξξ )=u ξξ u ηη. Example Consider the equation y 2 u xx x 2 u yy = 0, x > 0, y > 0. Here b 2 ac = x 2 y 2 > 0. The characteristics are given by ( see (6.3.18)) dy dx = ( ) b + b2 ac = x a y, and dy dx = ( ) b b2 ac = x a y.

38 38 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS Thus the characteristic curves are given by y 2 x 2 = constant, y 2 + x 2 = constant. We take α = ϕ(x, y) =x 2 + y 2, β = ψ(x, y) =x 2 y 2. We solve for x, y to obtain If we use (6.3.10) we find α + β α β x =, y =. 2 2 b =8x 2 y 2 =2(α 2 β 2 ) d =2(y 2 x 2 )= 2β ẽ =2(y 2 + x 2 )=2α and also u αβ = 2β 4(α 2 β 2 ) u 2β α 4(α 2 β 2 ) u β = βu a αu β 2(α 2 β 2 ). Example Suppose that a, b, c, d, e, f in (6.3.4) are constants. Then the characteristics are given by dy dx = b ± b2 ac ν ±. a Thus the characteristics are straight lines α = ϕ(x, y) =y ν + x, β = ψ(x, y) =y ν x. In this case we get ã = c = 0 and b = aϕx ψ x + b(ϕ x ψ y + ϕ y ψ x )+cϕ y ψ x = aν + ν b( u + + ν )+c = 2(ac b2 ). a

39 6.3. CHARACTERISTICS AND HIGHER ORDER EQUATIONS 39 For the first order terms we have d = aϕ xx +2bϕ xy + cϕ yy + dϕ x + eϕ y = dν + + e d, ẽ = aψ xx +2bψ xy + cψ yy + dψ x + eψ y = dν + e e. Recalling that there is a 2 timdes the b term and multiplying by K, the transformed equation is u αβ + Kd u α + Ke u β + Kfu = Kg where K = a 4(ac b 2 ). We note that a further reduction is always possible in this case. Namely we can remove the first order terms. Let u = e λα+µβ v. Then we have u α = e λα+µβ (λv + v α ) u β = e λα+µβ (µv + v β ) u αα = e λα+µβ (v αα +2λv α + λ 2 v) u ββ = e λα+µβ (v ββ +2µv β + µ 2 v) u αβ = e λα+µβ (v αβ + λv β + µv a + λµv). If we choose µ = d K, λ = e K and define f 1 = Kf + K 2 d e, g 1 = e (λα+µβ) Kg, then the equation can be written in terms of v as v αβ + f 1 v = g 1. II. Parabolic Case: Assume that D = b 2 ac = 0. In this case, if a = 0, then b = 0 and we are done since c 0 (otherwise the equation is not truely second order). Thus we assume that a 0 and the characteristic equation (6.3.11) factors as ( a v x + b 2 y) a v =0. (6.3.19)

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