Partial Differential Equations. T. Muthukumar

Transcription

1 Partial Differential Equations T. Muthukumar November 13, 2015

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3 Contents Notations vii 1 Introduction Multi-Index Notations Classification of PDE Introduction Continued Solution of PDE Well-posedness of PDE First Order PDE Linear Transport Equation Method of Characteristics Method of Characteristics: Continued Classification of Second Order PDE Semilinear Classification of SOPDE: Continued Quasilinear Why Characteristic Curves? Cauchy Boundary Condition Classification of SOPDE: Continued Invariance of Discriminant Standard or Canonical Forms Reduction to Standard Form iii

4 CONTENTS iv 8 The Laplacian Properties of Laplacian Boundary Conditions Harmonic Functions Properties of Harmonic Functions Existence and Uniqueness of Solution Sturm-Liouville Problems Eigen Value Problems Sturm-Liouville Problems Spectral Results Singular Sturm-Liouville Problem EVP of Legendre Operator EVP of Bessel s Operator Orthogonality of Eigen Functions Eigen Function Expansion Fourier Series Periodic Functions Fourier Coefficients and Fourier Series Fourier Series: Continued Piecewise Smooth Functions Complex Fourier Coefficients Orthogonality Odd and Even functions Fourier Sine-Cosine Series Fourier Transform and Integral Standing Waves: Separation of Variable Elliptic Equations Parabolic: Heat Equation Inhomogeneous Equation

5 CONTENTS v 18 Travelling Waves Domain of Dependence and Influence Appendices 109 A Divergence Theorem 111 B Normal Vector of a Surface 113 C Duhamel s Principle 115 Bibliography 117 Index 119

6 CONTENTS vi

7 Notations Symbols N Ω Ω R R n denotes the set of natural numbers denotes an open subset of R n, not necessarily bounded denotes the boundary of Ω denotes the set of real numbers denotes the n-dimensional Euclidean space n i=1 2 x 2 i D α α 1 x α αn x n αn and α = (α 1,..., α n ). In particular, for α = (1, 1,..., 1), ( D = = D (1,1,,...,1) = Function Spaces x 1, x 2,..., x n ) C(X) is the class of all continuous functions on X C k (X) is the class of C k functions which admit a continuous extension to the boundary of X C k (X) is the class of all k-times (k 1) continuously differentiable functions on X C (X) is the class of all infinitely differentiable functions on X C j,k (X Y ) is the class of all j-times (j 0) continuously differentiable functions on X and k-times (k 0) continuously differentiable functions on Y vii

8 NOTATIONS viii C c (X) is the class of all infinitely differentiable functions on X with compact support General Conventions x, x or Dx 2 When a PDE involves both the space variable x and time variable t, the quantities like,, D 2, etc. are always taken with respect to the space variable x only. This is a standard convention. Sometimes the suffix, like x or x, is used to indicate that the operation is taken w.r.t x. BVP IVP w.r.t Boundary value problem Initial value problem with respect to

9 Lecture 1 Introduction A partial differential equation (PDE) is an equation involving an unknown function u of two or more variables and some or all of its partial derivatives. The partial differential equation is a tool to analyse the models of nature. The process of understanding natural system can be divided in to three stages: (i) Modelling the problem or deriving the mathematical equation (formulating a PDE) describing the natural system. The derivation process is a result of physical laws such as Newton s law, momentum, conservation laws, balancing forces etc. (ii) Solving the equation (PDE). What constitutes as a solution to a PDE? (iii) Studying the properties of a solution. Most often the solution of a PDE may not have a nice formula or representation. How much information about the solution can one extract without any representation of a solution? In this text, one encounters similar situation while studying harmonic functions. 1.1 Multi-Index Notations Let Ω be an open subset of R. u : Ω R, at x Ω, is defined as Recall that the derivative of a function u (x) := lim h 0 u(x + h) u(x) h 1

10 LECTURE 1. INTRODUCTION 2 provided the limit exists. Now, let Ω be an open subset of R n. The directional derivative of u : Ω R, at x Ω and in the direction of a given vector ξ R n, is defined as u u(x + hξ) u(x) (x) := lim ξ h 0 h provided the limit exists. Let the n-tuple e i := (0, 0,..., 1, 0,..., 0), where 1 is in the i-th place, denote the standard basis vectors of R n. The i-th partial derivative of u at x is the directional derivative of u, at x Ω and along the direction e i, and is denoted as u xi (x) = u u(x + he i ) u(x) (x) = lim. x i h 0 h A multi-index α = (α 1,..., α n ) is a n-tuple where α i, for each 1 i n, is a non-negative integer. Let α := α α n. If α and β are two multi-indices, then α β means α i β i, for all 1 i n, and α ± β = (α 1 ± β 1,..., α n ± β n ). Also, α! = α 1!... α n! and, for any x R n, x α = x α x αn n. The multi-index notation, introduced by L. Schwartz, is quite handy in representing multi-variable equations in a concise form. For instance, a k-degree polynomial in n-variables can be written as a α x α. α k The partial differential operator of order α is denoted as D α = α 1 x 1 α 1... αn x n α n = α x 1 α 1... xn α n. One adopts the convention that, among the similar components of α, the order in which differentiation is performed is irrelevant. This is not a restrictive convention because the independence of order of differentiation is valid for smooth 1 functions. For instance, if α = (1, 1, 2) then one adopts the convention that 4 x 1 x 2 x = x 2 x 1 x smooth, usually, refers to as much differentiability as required.

11 LECTURE 1. INTRODUCTION 3 For each k N, D k u(x) := {D α u(x) α = k}. The k = 1 case is D 1 u(x) = ( D (1,0,...,0) u(x), D (0,1,0,...,0) u(x),..., D (0,0,...,0,1) u(x) ) ( u(x) =, u(x),..., u(x) ). x 1 x 2 x n The operator D 1 is called the gradient operator and is denoted as D or. Thus, u(x) = (u x1 (x), u x2 (x),..., u xn (x)). The directional derivative along a vector ξ R n satisfies the identity u (x) = u(x) ξ. ξ The normal derivative is the directional derivative along the normal direction ν(x), at x, with respect to the surface in which x lies. The divergence of a vector function u = (u 1,..., u n ), denoted as div(u), is defined as div(u) := u. The k = 2 case is 2 u(x)... 2 u(x) x 2 1 x 1 x n 2 u(x) D 2 u(x) = x 2 x u(x) x 2 x n u(x) x n x u(x) x 2 n The matrix D 2 u is called the Hessian matrix. Observe that the Hessian matrix is symmetric due to the independence hypothesis of the order in which partial derivatives are taken. The Laplace operator, denoted as, is defined as the trace of the Hessian operator, i.e., := n 2 i=1. Note that x 2 i =. Further, for a k-times differentiable function u, the n k -tensor D k u(x) := {D α u(x) α = k} may be viewed as a map from R n to R nk. Thus, the magnitude of D k u(x) is D k u(x) := D α u(x) 2 α =k In particular, u(x) = ( n i=1 u2 x i (x)) 1 2 or u(x) 2 = u(x) u(x) and D 2 u(x) = ( n i,j=1 u2 x i x j (x)) n n.

12 LECTURE 1. INTRODUCTION 4 Example 1.1. Let u(x, y) : R 2 R be defined as u(x, y) = ax 2 + by 2. Then and u = (u x, u y ) = (2ax, 2by) ( ) D 2 uxx u u = yx = u xy u yy ( 2a 0 0 2b Observe that u : R 2 R 2 and D 2 u : R 2 R 4 = R Classification of PDE Definition Let Ω be an open subset of R n. A k-th order partial differential equation of an unknown function u : Ω R is of the form ). F ( D k u(x), D k 1 u(x),... Du(x), u(x), x ) = 0, (1.2.1) for each x Ω, where F : R nk R nk 1... R n R Ω R is a given map such that F depends, at least, on one k-th partial derivative u and is independent of (k + j)-th partial derivatives of u for all j N. In short, the order of a PDE is the highest partial derivative order that occurs in the PDE. A first order PDE with two unknown variable (x, y) is represented as F (u x, u y, u, x, y) = 0 with F depending, at least, on one of u x and u y. Similarly, a first order PDE with three variable unknown function u(x, y, z) is written as F (u x, u y, u z, u, x, y, z) = 0 with F depending, at least, on one of u x, u y and u z. Note the abuse in the usage of x. In the n-variable case, x R n is a vector. In the two and three variable case x is the first component of the vector. The usage should be clear from the context. A PDE is a mathematical description of a physical process and solving a PDE for an unknown u helps in predicting the behaviour of the physical process. The level of difficulty in solving a PDE may depend on its order k and linearity of F. We begin by classifying PDEs in a scale of linearity. Definition form (i) A k-th order PDE is linear if F in (1.2.1) has the G(u(x)) = f(x) where G(u(x)) := α k a α(x)d α u(x) for given functions f and a α s. It is called linear because G is linear in u for all derivatives, i.e., G(λu 1 + µu 2 ) = λg(u 1 ) + µg(u 2 ) for λ, µ R. In addition, if f 0 then the PDE is linear and homogeneous.

13 LECTURE 1. INTRODUCTION 5 (ii) A k-th order PDE is semilinear if F is linear only in the highest (k-th) order, i.e., F has the form a α (x)d α u(x) + f(d k 1 u(x),..., Du(x), u(x), x) = 0. α =k (iii) A k-th order PDE is quasilinear if F has the form a α (D k 1 u(x),..., u(x), x)d α u + f(d k 1 u(x),..., u(x), x) = 0, α =k i.e., the coefficient of its highest (k-th) order derivative depends on u and its derivative only upto the previous (k 1)-th orders. (iv) A k-th order PDE is fully nonlinear if it depends nonlinearly on the highest (k-th) order derivatives. Observe that, for a semilinear PDE, f is never linear in u and its derivatives, otherwise it reduces to being linear. For a quasilinear PDE, a α (with α = k), cannot be independent of u or its derivatives, otherwise it reduces to being semilinear or linear. Example 1.2. (i) xu y yu x = u is linear. (ii) xu x + yu y = x 2 + y 2 is linear. (iii) u tt c 2 u xx = f(x, t) is linear. (iv) y 2 u xx + xu yy = 0 is linear. (v) u x + u y u 2 = 0 is semilinear. (vi) u t + uu x + u xxx = 0 is semilinear. (vii) u 2 tt + u xxxx = 0 is semilinear. (viii) u x + uu y u 2 = 0 is quasilinear. (ix) uu x + u y = 2 is quasilinear. (x) u x u y u = 0 is nonlinear.

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15 Lecture 2 Introduction Continued Solution of PDE Definition We say u : Ω R is a (classical) solution to the k-th order PDE (1.2.1), if D α u exists for all α explicitly present in (1.2.1); and u satisfies the equation (1.2.1). Example 2.1. Consider the first order equation u x (x, y) = 0 in R 2. Freezing the y-variable, the PDE can be viewed as an ODE in x-variable. On integrating both sides w.r.t x, u(x, y) = f(y) for any arbitrary function f : R R. Therefore, for every choice of f : R R, there is a solution u of the PDE. Note that the solution u is not necessarily in C 1 (R 2 ) to solve the first order PDE as is the case in solving an ODE. In fact, choosing a discontinuous function f, we obtain a solution which is discontinuous in the y-direction. Similarly, the solution of u y (x, y) = 0 is u(x, y) = f(x) for any choice of f : R R. We say u is a classical solution if u satisfies the equation and u C 1,0 (R R). Example 2.2. Consider the first order equation u t (x, t) = u(x, t) in R (0, ) such that u(x, t) 0, for all (x, t). Freezing the x-variable, the PDE can be viewed as an ODE in t-variable. Integrating both sides w.r.t t we obtain u(x, t) = f(x)e t, for some arbitrary function f : R R. Example 2.3. Consider the second order PDE u xy (x, y) = 0 in R 2. In contrast to the previous two examples, the PDE involves derivatives in both variables. 7

16 LECTURE 2. INTRODUCTION CONTINUED... 8 On integrating both sides w.r.t x we obtain u y (x, y) = F (y), for any arbitrary integrable function F : R R. Now, integrating both sides w.r.t y, u(x, y) = f(y)+g(x) for an arbitrary g : R R and a f C 1 (R) 1. But the u obtained above is not a solution to u yx (x, y) = 0 if g is not differentiable. Since we assume mixed derivatives to be same we need to assume f, g C 1 (R) for the solution to exist. Example 2.4. Consider the first order equation u x (x, y) = u y (x, y) in R 2. On first glance, the PDE does not seem simple to solve. But, by change of coordinates, the PDE can be rewritten in a simpler form. Choose the coordinates w = x + y and z = x y and, by chain rule, u x = u w + u z and u y = u w u z. In the new coordinates, the PDE becomes u z (w, z) = 0 which is in the form considered in Example 2.1. Therefore, its solution is u(w, z) = f(w) for any arbitrary f : R R and, hence, u(x, y) = f(x + y). But now an arbitrary f cannot be a solution. We impose that f C 1 (R). The family of solutions, obtained in the above examples, may not be the only family that solves the given PDE. Following example illustrates a situation where three different family of solutions exist (more may exist too) for the same PDE. Example 2.5. Consider the second order PDE u t (x, t) = u xx (x, t). (i) Note that u(x, t) = c is a solution of the PDE, for any constant c R. This is a family of solutions indexed by c R. (ii) The function u : R 2 R defined as u(x, t) = x2 +t+c, for any constant 2 c R, is also a family of solutions of the PDE. Because u t = 1, u x = x and u xx = 1. This family is not covered in the first case. (iii) The function u(x, t) = e c(x+ct) is also a family of solutions to the PDE, for each c R. Because u t = c 2 u, u x = cu and u xx = c 2 u. This family is not covered in the previous two cases. Recall that the family of solutions of an ODE is indexed by constants. In contrast to ODE, observe that the family of solutions of a PDE is indexed by either functions or constants. 1 In fact, it is enough to assume f is differentiable a.e. which is beyond the scope of this text

17 LECTURE 2. INTRODUCTION CONTINUED Well-posedness of PDE It has been illustrated via examples that a PDE has a family of solutions. The choice of one solution from the family of solutions is made by imposing boundary conditions (boundary value problem) or initial conditions (initial value problem). If too many initial/boundary conditions are specified, then the PDE may have no solution. If too few initial/boundary conditions are specified, then the PDE may have many solutions. Even with right amount of initial/boundary conditions, but at wrong places, the solution may fail to be stable, i.e., may not depend continuously on the initial or boundary data. It is, usually, desirable to solve a well-posed problem, in the sense of Hadamard. A PDE, along with the boundary condition or initial condition, is said to be well-posedness if the PDE (a) has a solution (existence); (b) the solution is unique (uniqueness); (c) and the solution depends continuously on the data given (stability). Any PDE not meeting the above criteria is said to be ill-posed. If the PDE (with boundary/initial conditions) is viewed as a map then the well-posedness of the PDE is expressed in terms of the surjectivity, injectivity and continuity of the inverse map. The existence and uniqueness condition depends on the notion of solution in consideration. There are three notions of solution, viz., classical solutions, weak solutions and strong solutions. This textbook, for the most part, is in the classical situation. Further, the stability condition means that a small change in the data reflects a small change in the solution. The change is measured using a metric or distance in the function space of data and solution, respectively. Though in this text we study only well-posed problems there are ill-posed problems which are also of interest. The following example illustrates the idea of continuous dependence of solution on data in the uniform metric on the space of continuous functions. Example 2.6. The initial value problem (IVP) { utt (x, t) = u xx (x, t) in R (0, ) u(x, 0) = u t (x, 0) = 0

18 LECTURE 2. INTRODUCTION CONTINUED has the trivial solution u(x, t) = 0. Consider the IVP with a small change in data, u tt (x, t) = u xx (x, t) in R (0, ) u(x, 0) = 0 u t (x, 0) = ε sin ( ) x ε which has the unique 2 solution u ε (x, t) = ε 2 sin(x/ε) sin(t/ε). The change in solution of the IVP is measured using the uniform metric as sup (x,t) { u ε (x, t) u(x, t) } = ε 2 sup { sin(x/ε) sin(t/ε) } = ε 2. (x,t) Thus, a small change in data induces a small enough change in solution under the uniform metric 3. Example 2.7 (Ill-posed). The IVP { utt (x, t) = u xx (x, t) in R (0, ) u(x, 0) = u t (x, 0) = 0 has the trivial solution u(x, t) = 0. Consider the IVP with a small change in data, u tt (x, t) = u xx (x, t) in R (0, ) u(x, 0) = 0 u t (x, 0) = ε sin ( ) x ε which has the unique solution u ε (x, t) = ε 2 sin(x/ε) sinh(t/ε). The solution of the IVP is not stable because the data change is small, i.e., sup x { u ε t(x, 0) u t (x, 0) } = ε sup { sin(x/ε) } = ε and the solution change is not at all small, i.e., lim sup t x { u ε (x, t) u(x, t) } = lim ε 2 sinh(t/ε) = +. t In fact, the solution will not converge in any reasonable metric. x 2 This claim will be proved in later chapters. 3 The space R (0, ) is not compact and the metric is not complete. The example is only to explain the notion of stability at an elementarty level.

19 Lecture 3 First Order PDE The aim of this chapter is to find the general solution and to solve the Cauchy problem associated with the first order PDE of the form F ( u(x), u(x), x) = 0 x R n. 3.1 Linear Transport Equation The transport of a substance in a fluid flowing (one dimenisonal flow) with constant speed b, with neither source or sink of substance, is given by u t (x, t) + bu x (x, t) + du(x, t) = cu xx (x, t) (x, t) R (0, ) where c is the diffusive coefficient of the substance and d is the rate of decay of the substance. The function u(x, t) denotes the concentration/density of the substance at (x, t). Note that the case of no diffusion (c = 0) is a linear first order equation which will be studied in this section. Example 3.1 (One Dimension with no decay). The one space dimension transport equation is u t (x, t) + bu x (x, t) = 0; (x, t) R (0, ) with b R. The c = d = 0 case describes the transport of an insoluble 1 substance O in a fluid flowing with constant speed b. To solve this we consider two observers, one a fixed observer A and another observer B, moving with 1 assuming no diffusion and no decay of substance. 11

20 LECTURE 3. FIRST ORDER PDE 12 Figure 3.1: Transport of initial data g speed b and in the same direction as the substance O. For B, the substance O would appear stationary while for A, the fixed observer, the substance O would appear to travel with speed b. What is the equation of the transport of the stationary substance O from the viewpoint of the moving observer B? The answer to this question lies in identifying the coordinate system for B relative to A. Fix a point x at time t = 0. After time t, the point x remains as x for the fixed observer A, while for the moving observer B, the point x is now x bt. Therefore, the coordinate system for B is (w, z) where w = x bt and z = t. Let v(w, z) describe the motion of O from B s perspective. Since B sees O as stationary, the PDE describing the motion of O is v z (w, z) = 0. Therefore, v(w, z) = g(w), for some arbitrary function g (sufficiently differentiable), is the solution from B s perspective. To solve the problem from A s perspective, note that u t = v w w t + v z z t = bv w + v z and u x = v w w x + v z z x = v w. Therefore, u t + bu x = bv w + v z + bv w = v z and, hence, u(x, t) = v(w, z) = g(w) = g(x bt) (cf. Fig 3.1). The choice of g is based on our restriction to be in a classical solution set-up. Note that, for any choice of g, we have

21 LECTURE 3. FIRST ORDER PDE 13 Figure 3.2: Path of substance, over time, placed at x 0 with b > 0 g(x) = u(x, 0). The line x bt = x 0, for some constant x 0, in the xt-plane tracks the flow of the substance placed at x 0 at time t = 0 (cf. Fig 3.2). Also, observe that 0 = u t + bu x = (u x, u t ) (b, 1) is precisely that the directional derivative along the vector (b, 1) is zero. This means that u is constant if we move along the direction (b, 1). Thus, the value of u(x, t) on the line x bt = x 0 is constant. Example 3.2 (Transport equation in first quadrant). The transport equation is u t (x, t) + bu x (x, t) = 0; (x, t) (0, ) (0, ) with b R. As before, we obtain u(x, t) = g(x bt) where g(x) = u(x, 0). This problem is uniquely solvable in the given region only for b < 0. For b > 0, g defined on x-axis is not adequate to solve the problem. The problem is not well-posed! The given data is enough only to solve for u in the region {(x, t) (0, ) (0, ) x > bt} when b > 0. To compute u in {(x, t) (0, ) (0, ) x > bt} we need to provide data on the t-axis (0, t). Example 3.3 (Transport equation in semi-infinite strip). The transport equation is u t (x, t) + bu x (x, t) = 0; (x, t) (0, L) (0, )

22 LECTURE 3. FIRST ORDER PDE 14 with b R. As before, we obtain u(x, t) = g(x bt) where g(x) = u(x, 0). If b > 0 then the problem is well-posed when the data given on x and t axes. If b < 0 then the problem is well-posed when the data is given on x-axis and (L, t)-axis. 3.2 Method of Characteristics The method of characteristics gives the equation, a system of ODE, of the characteristic curves. The method of characteristics reduces a first order PDE to a system of ODE. For illustration, consider the two variable first order quasi-linear equation: A(x, y, u)u x + B(x, y, u)u y = C(x, y, u). (3.2.1) Solving for u(x, y) in the above equation is equivalent to finding the surface S {(x, y, u(x, y))} generated by u in R 3. If u is a solution of (3.2.1), at each (x, y) in the domain of u, then A(x, y, u)u x + B(x, y, u)u y = C(x, y, u) A(x, y, u)u x + B(x, y, u)u y C(x, y, u) = 0 (A(x, y, u), B(x, y, u), C(x, y, u)) (u x, u y, 1) = 0 (A(x, y, u), B(x, y, u), C(x, y, u)) ( u(x, y), 1) = 0. But ( u(x, y), 1) is normal to S at the point (x, y) (cf. Appendix B). Hence, the coefficients (A(x, y, u), B(x, y, u), C(x, y, u)) are perpendicular to the normal and, therefore, (A(x, y, u), B(x, y, u), C(x, y, u)) lie on the tangent plane to S at (x, y, u(x, y)). Definition A smooth curve in R n is said to be an integral or characteristic curve w.r.t a given vector field, if the vector field is tangential to the curve at each of its point. Definition A smooth surface in R n is said to be an integral surface w.r.t a given vector field, if the vector field is tangential to the surface at each of its point. In the spirit of above definition and arguments, finding a solution u to (3.2.1) is equivalent to determining an integral surface S corresponding to the

23 LECTURE 3. FIRST ORDER PDE 15 coefficient vector field V (x, y) = (A(x, y, u), B(x, y, u), C(x, y, u)) of (3.2.1). Let s denote the parametrization of the characteristic curves w.r.t V. For convenience, let z(s) := u(x(s), y(s)). Then the characteristic curves can be found by solving the system of ODEs dx ds = A(x(s), y(s), z(s)), dy ds = B(x(s), y(s), z(s)), dz ds = C(x(s), y(s), z(s)). (3.2.2) The three ODE s obtained are called characteristic equations. The union of these characteristic (integral) curves give us the integral surface S. The union is in the sense that every point in the integral surface belongs to exactly one characteristic. Example 3.4 (Linear Transport Equation). The linear transport equation is already solved earlier using elementary method. Let us solve the same using the method of characteristics. Consider the linear transport equation in two variable, u t + bu x = 0, x R and t (0, ), where the constant b R is given. Thus, the given vector field V (x, t) = (b, 1, 0). The characteristic equations are dx ds = b, Solving the 3 ODE s, we get dt ds = 1, and dz ds = 0. x(s) = bs + c 1, t(s) = s + c 2, and z(s) = c 3. Note that solving the system of ODEs requires some initial condition. We have already observe that the solution of the transport equation depended on the value of u at time t = 0, i.e., the value of u on the curve (x, 0) in the xt-plane. Thus, the problem of finding a function u solving a first order PDE such that u is known on a curve Γ in the xy-plane is called the Cauchy problem. Example 3.5. Let g be given (smooth enough) function g : R R. Consider the linear transport equation { ut + bu x = 0 x R and t (0, ) (3.2.3) u(x, 0) = g(x) x R.

24 LECTURE 3. FIRST ORDER PDE 16 We parametrize the curve Γ with r-variable, i.e., Γ = {γ 1 (r), γ 2 (r)} = {(r, 0)}. The characteristic equations are: dx(r, s) ds = b, dt(r, s) ds = 1, and dz(r, s) ds = 0 with initial conditions, x(r, 0) = r, t(r, 0) = 0, and z(r, 0) = g(r). Solving the ODE s, we get x(r, s) = bs + c 1 (r), t(r, s) = s + c 2 (r) and z(r, s) = c 3 (r) with initial conditions x(r, 0) = c 1 (r) = r Therefore, t(r, 0) = c 2 (r) = 0, and z(r, 0) = c 3 (r) = g(r). x(r, s) = bs + r, t(r, s) = s, and z(r, s) = g(r). The idea is to solve for r and s in terms of x, t. Let us set u(x, t) = z(r(x, t), s(x, t)). In this case we can solve for r and s in terms of x and t, to get r(x, t) = x bt and s(x, t) = t. Therefore, u(x, t) = z(r, s) = g(r) = g(x bt). The above example shows how the information on data curve Γ is reduced as initial condition for the characteristic ODEs.

25 Lecture 4 Method of Characteristics: Continued... Let us study the Example 3.5 with a different data curve Γ. Example 4.1. Consider the linear transport equation { ut + bu x = 0 x R and t (0, ) u(bt, t) = g(t) t (0, ). (4.0.1) We parametrize the data curve Γ with r-variable, i.e., Γ = {γ 1 (r), γ 2 (r)} = {(br, r)}. The characteristic equations are: dx(r, s) ds = b, dt(r, s) ds = 1, and dz(r, s) ds = 0 with initial conditions, x(r, 0) = br, t(r, 0) = r, and z(r, 0) = g(r). Solving the ODE s, we get x(r, s) = bs + c 1 (r), t(r, s) = s + c 2 (r) and z(r, s) = c 3 (r) with initial conditions x(r, 0) = c 1 (r) = br t(r, 0) = c 2 (r) = r, and z(r, 0) = c 3 (r) = g(r). 17

26 LECTURE 4. METHOD OF CHARACTERISTICS: CONTINUED Therefore, x(r, s) = b(s + r), t(r, s) = s + r, and z(r, s) = g(r). Note that in this case we cannot solve for r and s in terms of x, t. The above examples suggest that solving a Cauchy problem depends on the curve on which u is prescribed. One way is to prescribe values of u on a data curve Γ parametrized as (γ 1 (r), γ 2 (r)), for r I R, such that all the characteristic curves (parametrised with s variable) intersect the (γ 1 (r), γ 2 (r), u(γ 1, γ 2 )) at s = 0. Let u(γ 1 (r), γ 2 (r)) = g(r). Thus, solving the Cauchy problem is now reduced to solving for x(r, s), y(r, s), z(r, s) in (3.2.2) with the initial conditions x(r, 0) = γ 1 (r), y(r, 0) = γ 2 (r) and z(r, 0) = g(r), r I. The system of ODE can be solved uniquely for x(r, s), y(r, s) and z(r, s), in a neighbourhood of s = 0 and all r I. At this juncture, one may ask the following questions: (i) Can the three solutions be used to define a function z = u(x, y)? (ii) If yes to (i), then is the solution z = u(x, y) unique for the Cauchy problem? The answer is yes because two integral surface intersecting in Γ must contain the same charaterictics (beyond the scope of this course). Let us answer (i) in a neighbourhood of (r, 0). Set x(r, 0) = γ 1 (r) = x 0, y(r, 0) = γ 2 (r) = y 0 and z(r, 0) = g(r) = z 0. Note that the answer to (i) is in affirmation if we can solve for r and s in terms of x and y, i.e., r = R(x, y) and s = S(x, y) such that R(x 0, y 0 ) = 0 and S(x 0, y 0 ) = 0. Then z = z(r(x, y), S(x, y)) = u(x, y). The inverse function theorem tells us that r and s can be solved in terms of x and y in a neighbourhood of (x 0, y 0 ) if J(r, 0) = x r(r, 0) y r (r, 0) x s (r, 0) y s (r, 0) = γ 1(r) γ 2(r) A(x 0, y 0, z 0 ) B(x 0, y 0, z 0 ) 0. The quantity J(r, 0) 0 means that the vectors (A(x 0, y 0, z 0 ), B(x 0, y 0, z 0 )) and (γ 1(r), γ 2(r)) are not parallel. What happens in the case of J(r, 0) = 0, i.e., when the associated vectors are parallel? The condition given on Γ is u(γ 1 (r), γ 2 (r)) = g(r). If u is a C 1 solution to the Cauchy problem then on differentiation, w.r.t r, the Cauchy

27 LECTURE 4. METHOD OF CHARACTERISTICS: CONTINUED condition yields g (r) = u x (γ 1 (r), γ 2 (r))γ 1(r) + u y (γ 1 (r), γ 2 (r))γ 2(r). Since u is solution, at (x 0, y 0, z 0 ), of the algebraic system ( ) ( ) ( ) A(x0, y 0, z 0 ) B(x 0, y 0, z 0 ) ux (x 0, y 0 ) C(x0, y γ 1(r) γ 2(r) = 0, z 0 ) u y (x 0, y 0 ) g (r) a necessary condition is that ( A(x0, y rank 0, z 0 ) B(x 0, y 0, z 0 ) C(x 0, y 0, z 0 ) γ 1(r) γ 2(r) g (r) ) = 1. If the above rank condition is satisfied then the data curve Γ is said to be a characteristic at (x 0, y 0, z 0 ). Thus, we have the following possibilities: (a) J(r, 0) 0 for all r I. Note that, when J(r, 0) 0, then the rank condition is not satisfied 1 and, hence, the data curve Γ does not have any characteristic points (Γ is not parallel at all points). Then, in a neighborhood of Γ, there exists a unique solution u = u(x, y) of the Cauchy problem given by the system of ODEs. (b) J(r 0, 0) = 0, for some r 0 I, and Γ is characteristic at the point P 0 = (γ 1 (r 0 ), γ 2 (r 0 ), g(r 0 )). Then a C 1 solution may exist in a neighborhood of P 0. (c) J(r 0, 0) = 0 for some r 0 I and Γ is not characteristic at P 0. There are no C 1 solutions in a neighborhood of P 0. There may exist less regular solutions. (d) If Γ is a characteristic then there exists infinitely many C 1 solutions in a neighborhood of Γ. Example 4.2. Consider the Burgers equation given as { ut + uu x = 0 x R and t (0, ) u(x, 0) = g(x) x R. The parametrization of the curve Γ with r-variable, i.e., Γ = {γ 1 (r), γ 2 (r)} = {(r, 0)}. The characteristic equations are: dx(r, s) ds = z, dt(r, s) ds = 1, and dz(r, s) ds = 0 1 because rank is 2 in this case

28 LECTURE 4. METHOD OF CHARACTERISTICS: CONTINUED with initial conditions, x(r, 0) = r, t(r, 0) = 0, and z(r, 0) = g(r). Solving the ODE corresponding to z, we get z(r, s) = c 3 (r) with initial conditions z(r, 0) = c 3 (r) = g(r). Thus, z(r, s) = g(r). Using this in the ODE of x, we get dx(r, s) = g(r). ds Solving the ODE s, we get x(r, s) = g(r)s + c 1 (r), t(r, s) = s + c 2 (r) with initial conditions Therefore, x(r, 0) = c 1 (r) = r and t(r, 0) = c 2 (r) = 0. x(r, s) = g(r)s + r, t(r, s) = s and z(r, s) = g(r). Let us compute J(r, s) = 1 + sg (r) 0 1 = 1 + sg (r) and J(r, 0) = 1 0, the data curve Γ does not have characteristic points. Hence, one can solve for r and s, in terms of x, t and z. Thus, we get s = t and r = x zt. Therefore, u(x, t) = g(x tu) is the solution in the implicit form. Example 4.3. Consider the Burgers equation given as { ut ( + uu ) x = 1 x R and t (0, ) t u 24, t = t t > 0. 2 Note that the data curve is the parabola x = t 2 /4. The parametrization of the curve Γ with r-variable, i.e., Γ = {γ 1 (r), γ 2 (r)} = {(r 2, 2r)}. The characteristic equations are: dx(r, s) ds = z, dt(r, s) ds = 1, and dz(r, s) ds = 1 with initial conditions, x(r, 0) = r 2, t(r, 0) = 2r, and z(r, 0) = r.

29 LECTURE 4. METHOD OF CHARACTERISTICS: CONTINUED Solving the ODE corresponding to z, we get z(r, s) = s + c 3 (r) with initial conditions z(r, 0) = c 3 (r) = r. Thus, z(r, s) = s + r. Using this in the ODE of x, we get dx(r, s) = s + r. ds Solving the ODE s, we get with initial conditions Therefore, x(r, s) = s2 2 + rs + c 1(r), t(r, s) = s + c 2 (r) x(r, 0) = c 1 (r) = r 2 and t(r, 0) = c 2 (r) = 2r. x(r, s) = s2 2 + rs + r2, t(r, s) = s + 2r and z(r, s) = s + r. Let us compute J(r, s) = 2r + s 2(s + r) = s and J(r, 0) = 0 for all r. The rank of the matrix, at (r, 0), ( ) r 1 1 = 2 1 2r 2 1 and, hence, Γ does not have any characteristic points. We know in this case we cannot have a C 1 solution but might have less regular solutions. Let us solve for r and s, in terms of x, t and z. Thus, we get s = 2 x t2 and 4 r = t ± x x t2. Therefore, u(x, t) = t ± t2 are two solutions in the region x > t 2 /4 and are not differentiable on Γ. Example 4.4. Consider the Burgers equation given as { ut ( + uu ) x = 1 x R and t (0, ) t u 22, t = t t > 0. Note that the data curve is the parabola x = t 2 /2. The parametrization of the curve Γ with r-variable, i.e., Γ = {γ 1 (r), γ 2 (r)} = {(r 2 /2, r)}. The characteristic equations are: dx(r, s) ds = z, dt(r, s) ds = 1, and dz(r, s) ds = 1

30 LECTURE 4. METHOD OF CHARACTERISTICS: CONTINUED with initial conditions, x(r, 0) = r2, t(r, 0) = r, and z(r, 0) = r. 2 Solving the ODE corresponding to z, we get z(r, s) = s + c 3 (r) with initial conditions z(r, 0) = c 3 (r) = r. Thus, z(r, s) = s + r. Using this in the ODE of x, we get dx(r, s) = s + r. ds Solving the ODE s, we get with initial conditions x(r, s) = s2 2 + rs + c 1(r), t(r, s) = s + c 2 (r) Therefore, x(r, 0) = c 1 (r) = r2 2 and t(r, 0) = c 2(r) = r. x(r, s) = s2 + r 2 + rs, t(r, s) = s + r and z(r, s) = s + r. 2 Let us compute J(r, s) = r + s (s + r) = 0 for all r and s. Further, Γ is a characteristic (at all points) because the rank of the matrix, at (r, 0), ( ) r 1 1 = 1 r 1 1 In this case there may exist infinitely many C 1 solutions. For instance, u(x, t) = t is a solution satisfying Cauchy data. Also, u(x, t) = ± 2x. Remark If the coefficients a and b are independent of u, then the characteristic curves are lying in the xy-plane. If the coefficients a and b are constants (independent of both x and u) then the characteristic curves are straight lines. In the linear case, the characteristics curves will not intersect. Because, if the curves intersect then, at the point of intersection, they have the same tangent, which is not possible!

31 Lecture 5 Classification of Second Order PDE A general second order PDE is of the form F (D 2 u(x), Du(x), u(x), x) = 0, for each x Ω R n and u : Ω R is the unknown. A Cauchy problem is: given the knowledge of u on a smooth hypersurface Γ Ω, can one find the solution u of the PDE? The knowledge of u on Γ is said to be the Cauchy data. What should be the minimum required Cauchy data for the Cauchy problem to be solved? Viewing the Cauchy problem as an initial value problem corresponding to ODE, there is a unique solution to the second order ODE y (x) + P (x)y (x) + Q(x)y(x) = 0 x I y(x 0 ) = y 0 y (x 0 ) = y 0. where P and Q are continuous on I (assume I closed interval of R) and for any point x 0 I. This motivates us to define the Cauchy problem for second order PDE as: F (D 2 u(x), Du(x), u(x), x) = 0 x Ω u(x) = g(x) x Γ Du(x) ν(x) = h(x) x Γ (5.0.1) where ν is the outward unit normal vector on the hypersurface Γ and g, h are known functions on Γ. 23

32 LECTURE 5. CLASSIFICATION OF SECOND ORDER PDE Semilinear Consider the Cauchy problem for the second order semilinear PDE in two variables (x, y) Ω R 2, A(x, y)u xx + 2B(x, y)u xy + C(x, y)u yy = D (x, y) Ω u(x, y) = g(x, y) (x, y) Γ u x (x, y) = h 1 (x, y) (x, y) Γ u y (x, y) = h 2 (x, y) (x, y) Γ. (5.1.1) where D(x, y, u, u x, u y ) may be non-linear and Γ is a smooth 1 curve in Ω. Also, one of the coefficients A, B or C is identically non-zero (else the PDE is not of second order). Let r (γ 1 (r), γ 2 (r)) be a parametrisation of the curve Γ. Then we have the compatibility condition that g (r) = h 1 γ 1(r) + h 2 γ 2(r). By computing the second derivatives of u on Γ and considering u xx, u yy and u xy as unknowns, we have the linear system of three equations in three unknowns on Γ, Au xx +2Bu xy +Cu yy = D γ 1(r)u xx +γ 2(r)u xy = h 1(r) γ 1(r)u xy +γ 2(r)u yy = h 2(r). This system of equation is solvable if the determinant of the coefficients are non-zero, i.e., A 2B C γ 1 γ γ 1 γ 2 0. Definition We say a curve Γ Ω R 2 is characteristic (w.r.t (5.1.1)) if A(γ 2) 2 2Bγ 1γ 2 + C(γ 1) 2 = 0. where (γ 1 (r), γ 2 (r)) is a parametrisation of Γ. Note that the geometry hidden in the above definition is very similar to that we encountered in first order equation. Since ν = ( γ 2, γ 1) is the 1 twice differentiable

33 LECTURE 5. CLASSIFICATION OF SECOND ORDER PDE 25 normal to Γ at each point, the above definition says that the curve is noncharacteristic if 2 A ij ν i ν j = A(γ 2) 2 2Bγ 1γ 2 + C(γ 1) 2 0 i,j=1 where A 11 = A, A 12 = A 21 = B and A 22 = C. If y = y(x) is a representation of the curve Γ (locally, if necessary), we have γ 1 (r) = r and γ 2 (r) = y(r). Then the characteristic equation reduces as A ( dy dx ) 2 2B dy dx + C = 0. Therefore, the characteristic curves of (5.1.1) are given by the graphs whose equation is dy dx = B ± B2 AC. A Thus, we have three situations arising depending on the sign of the discriminant, B 2 AC. This classifies the given second order PDE based on the sign of its discriminant d = B 2 AC. Definition We say a second order PDE is of (a) hyperbolic type if d > 0, (b) parabolic type if d = 0 and (c) elliptic type if d < 0. The hyperbolic PDE have two families of characteristics, parabolic PDE has one family of characteristic and elliptic PDE have no characteristic. We caution here that these names are no indication of the shape of the graph of the solution of the PDE. Note that the classification depends on the determinant of the coefficient matrix ( ) A B B C For every (x, y) Ω, the matrix is symmetric and hence diagonalisable. If λ 1, λ 2 are the diagonal entries, then d = λ 1 λ 2. Thus, a equation is hyperbolic at a point (x, y) if the eigen values have opposite sign. It is ellipic if the eigenvalues have same sign and is parabolic if, at least, one of the eigenvalue is zero.

34 LECTURE 5. CLASSIFICATION OF SECOND ORDER PDE 26 Example 5.1 (Wave Equation). For a given non-zero c R, u yy c 2 u xx = 0 is hyperbolic. Since A = c 2, B = 0 and C = 1, we have d = B 2 AC = c 2 > 0. The eigen values of the coefficient matrix are 1, c 2 which have opposite sign. Example 5.2 (Heat Equation). For a given c R, u y cu xx = 0 is parabolic. Since A = c, B = 0 and C = 0, thus d = B 2 AC = 0. The eigen values of the coefficient matrix are 0, c has a zero eigenvalue. Example 5.3 (Laplace equation). u xx +u yy = 0 is elliptic. Since A = 1, B = 0 and C = 1, thus d = B 2 AC = 1 < 0. The eigen values of the coefficient matrix are 1, 1 which have same sign. Example 5.4 (Velocity Potential Equation). In the equation (1 M 2 )u xx + u yy = 0, A = (1 M 2 ), B = 0 and C = 1. Then d = B 2 AC = (1 M 2 ). The eigen values of the coefficient matrix are 1 M 2, 1. Thus, for M > 1 (opposite sign), the equation is hyperbolic (supersonic flow), for M = 1 (zero eigenvalue) it is parabolic (sonic flow) and for M < 1 (same sign) it is elliptic (subsonic flow). Note that the classification of PDE is dependent on its coefficients. Thus, for constant coefficients the type of PDE remains unchanged throughout the region Ω. However, for variable coefficients, the PDE may change its classification from region to region. Example 5.5. An example is the Tricomi equation, u xx + xu yy = 0. The discriminant of the Tricomi equation is d = x. The eigenvalues are 1, x. Thus, tricomi equation is hyperbolic when x < 0, elliptic when x > 0 and degenerately parabolic when x = 0, i.e., y-axis. Such equations are called mixed type. Example 5.6. Let us compute the family of characteristic curves of second order PDE, whenever they exist. For instance, recall that elliptic equation will not have any real characteristic curves. (i) For a given non-zero c R, u yy c 2 u xx = 0. We have already noted that the equation is hyperbolic and, hence, should admit two characteristic curves. Recall that the characteristic curves are given by the equation dy dx = B ± B2 AC A = ± c 2 c 2 = 1 c. Thus, cy ± x = a constant is the equation for the two characteristic curves. Note that the characteristic curves y = x/c+y 0 are boundary of two cones in R 2 with vertex at (0, y 0 ).

35 LECTURE 5. CLASSIFICATION OF SECOND ORDER PDE 27 (ii) For any given c R, consider u y cu xx = 0. We have already noted that the equation is parabolic and, hence, should admit one characteristic curve. The characteristic curve is given by the equation dy dx = B ± B2 AC A = 0. Thus, y = a constant is the equation of the characteristic curve. i.e., any horizontal line in R 2 is a charateristic curve. (iii) We have already noted that the equation u xx + u yy = 0 is elliptic and, hence, will have no real characteristics. (iv) The equation u xx + xu yy = 0 is of mixed type. In the region x > 0, the characteristic curves are y 2x 3/2 /3 = a constant.

36 LECTURE 5. CLASSIFICATION OF SECOND ORDER PDE 28

37 Lecture 6 Classification of SOPDE: Continued 6.1 Quasilinear The notion of classification of second order semilinear PDE could be generalised to quasilinear PDE A(x, u(x), Du(x)), non-linear PDE and system of ODE. However, in these cases the classification may also depend on the solution u. The solutions to characteristic equation for a quasilinear equation depends on the solution considered. Example 6.1. Consider the quasilinear PDE u xx uu yy = 0. The discriminant is d = u. The eigenvalues are 1, u(x). It is hyperbolic for {u > 0} 1, elliptic when {u < 0} and parabolic when {u = 0}. Example 6.2. Consider the quasilinear PDE (c 2 u 2 x)u xx 2u x u y u xy + (c 2 u 2 y)u yy = 0 where c > 0. Then d = B 2 AC = c 2 (u 2 x + u 2 y c 2 ) = c 2 ( u 2 c 2 ). It is hyperbolic if u > c, parabolic if u = c and elliptic if u < c. 6.2 Why Characteristic Curves? Recall that a second order ODE y (x) + P (x)y (x) + Q(x)y(x) = 0, x I 1 The notation {u > 0} means {x Ω u(x) > 0} 29

38 LECTURE 6. CLASSIFICATION OF SOPDE: CONTINUED 30 can have other types of boundary conditions, in addition to the initial (or Cauchy) condition y(x 0 ) = y 0 and y (x 0 ) = y 0, such as, if I = (a, b) then (a) Dirichlet condition y(a) = y 0 and y(b) = y 1. (b) Neumann condition y (a) = y 0 and y (b) = y 1. (c) Periodic condition y(a) = y(b) and y (a) = y (b). (d) Robin condition αy(a) + βy (a) = y 0 and γy(b) + δy (b) = y 1. In contrast to Initial (Cauchy boundary) condition, boundary value problems may be ill-posed. These boundary conditions also can be generalised to second order PDE. The classification described tells us the right amount of initial/boundary condition to be imposed for a second order PDE to be well-posed. For hyperbolic, which has two real characteristics, requires as many initial condition as the number of characteristics emanating from initial time and as many boundary conditions as the number of characteristics that pass into the spatial boundary. Thus, hyperbolic equations will take the Cauchy boundary conditions on an open surface. For parabolic, which has exactly one real characteristic, we need one boundary condition at each point of the spatial boundary and one initial condition at initial time. Thus, parabolic equation will take either Dirichlet or Neumann on an open surface. For elliptic, which has no real characteristic curves, we need one boundary condition at each point of the spatial boundary. Thus, elliptic equations will take either Dirichlet or Neumann in a closed surface enclosing the domain of interest.

39 LECTURE 6. CLASSIFICATION OF SOPDE: CONTINUED Cauchy Boundary Condition Recall that for a second order Cauchy problem we need to know both u and its normal derivative on a data curve Γ contained in Ω. However, the Cauchy problem for Laplacian (more generally for elliptic equations) is not well-posed. In fact, the Cauchy problem for Laplace equation on a bounded domain Ω is over-determined. Example 6.3 (Hadamard). Consider the Cauchy problem for Laplace equation u xx + u yy = 0 cos ky u(0, y) = k 2 u x (0, y) = 0, where k > 0 is an integer. It is easy to verify that there is a unique solution u k (x, y) = cosh(kx) cos(ky) k 2 of the Cauchy problem. Note that for any x 0 > 0, u k (x 0, nπ/k) = cosh(kx 0) k 2. Since, as k, nπ/k 0 and u k (x 0, nπ/k) the Cauchy problem is not stable, and hence not well-posed. Exercise 1. Show that the Cauchy problem for Laplace equation u xx + u yy = 0 u(x, 0) = 0 u y (x, 0) = k 1 sin kx, where k > 0, is not well-posed. (Hint: Compute explicit solution using separation of variable. Note that, as k, the Cauchy data tends to zero uniformly, but the solution does not converge to zero for any y 0. Therefore, a small change from zero Cauchy data (with corresponding solution being zero) may induce bigger change in the solution.) This issue of ill-posedness of the Cauchy problem is very special to second order elliptic equations. In general, any hyperbolic equation Cauchy problem is well-posed, as long as the hyperbolicity is valid in the full neighbourhood of the data curve.

40 LECTURE 6. CLASSIFICATION OF SOPDE: CONTINUED 32 Example 6.4. Consider the Cauchy problem for the second order hyperbolic equation y 2 u xx yu yy + 1u 2 y = 0 y > 0 u(x, 0) = f(x) u y (x, 0) = g(x). The general solution to this problem can be computed as u(x, y) = F (x + 23 ) y3/2 + G (x 23 ) y3/2. On y = 0 u(x, 0) = F (x) + G(x) = f(x). Further, u y (x, y) = y 1/2 F ( x y3/2 ) y 1/2 G ( x 2 3 y3/2 ) and u y (x, 0) = 0. Thus, the Cauchy problem has no solution unless g(x) = 0. If g 0 then the solution is u(x, y) = F (x + 23 ) y3/2 F (x 23 ) y3/2 + f (x 23 ) y3/2 for arbitrary F C 2. Therefore, when g 0 the solution is not unique. The Cauchy problem is not well-posed because the equation is hyperbolic (B 2 AC = y 3 ) not in the full neighbourhood of the data curve {y = 0}.

41 Lecture 7 Classification of SOPDE: Continued 7.1 Invariance of Discriminant The classification of second order semilinear PDE is based on the discriminant B 2 AC. In this section, we note that the classification is independent of the choice of coordinate system (to represent a PDE). Consider the two-variable semilinear PDE A(x, y)u xx +2B(x, y)u xy +C(x, y)u yy = D(x, y, u, u x, u y ) (x, y) Ω (7.1.1) where the variables (x, y, u, u x, u y ) may appear non-linearly in D and Ω R 2. Also, one of the coefficients A, B or C is identically non-zero (else the PDE is not of second order). We shall observe how (7.1.1) changes under coordinate transformation. Definition For any PDE of the form (7.1.1) we define its discriminant as B 2 AC. Let T : R 2 R 2 be the coordinate transformation with components T = (w, z), where w, z : R 2 R. We assume that w(x, y), z(x, y) are such that w, z are both continuous and twice differentiable w.r.t (x, y), and the Jacobian J of T is non-zero, J = w x w y 0. z x 33 z y

42 LECTURE 7. CLASSIFICATION OF SOPDE: CONTINUED 34 We compute the derivatives of u in the new variable, u x = u w w x + u z z x, u y = u w w y + u z z y, u xx = u ww w 2 x + 2u wz w x z x + u zz z 2 x + u w w xx + u z z xx u yy = u ww w 2 y + 2u wz w y z y + u zz z 2 y + u w w yy + u z z yy u xy = u ww w x w y + u wz (w x z y + w y z x ) + u zz z x z y + u w w xy + u z z xy Substituting above equations in (7.1.1), we get a(w, z)u ww + 2b(w, z)u wz + c(w, z)u zz = d(w, z, u, u w, u z ). where D transforms in to d and a(w, z) = Aw 2 x + 2Bw x w y + Cw 2 y (7.1.2) b(w, z) = Aw x z x + B(w x z y + w y z x ) + Cw y z y (7.1.3) c(w, z) = Az 2 x + 2Bz x z y + Cz 2 y. (7.1.4) Note that the coefficients in the new coordinate system satisfy b 2 ac = (B 2 AC)J 2. Since J 0, we have J 2 > 0. Thus, both b 2 ac and B 2 AC have the same sign. Thus, the sign of the discriminant is invariant under coordinate transformation. All the above arguments can be carried over to quasilinear and non-linear PDE. 7.2 Standard or Canonical Forms The advantage of above classification helps us in reducing a given PDE into simple forms. Given a PDE, one can compute the sign of the discriminant and depending on its clasification we can choose a coordinate transformation (w, z) such that (i) For hyperbolic, a = c = 0 or b = 0 and a = c. (ii) For parabolic, c = b = 0 or a = b = 0. We conveniently choose c = b = 0 situation so that a 0 (so that division by zero is avoided in the equation for characteristic curves).

43 LECTURE 7. CLASSIFICATION OF SOPDE: CONTINUED 35 (iii) For elliptic, b = 0 and a = c. If the given second order PDE (7.1.1) is such that A = C = 0, then (7.1.1) is of hyperbolic type and a division by 2B (since B 0) gives u xy = D(x, y, u, u x, u y ) where D = D/2B. The above form is the first standard form of second order hyperbolic equation. If we introduce the linear change of variable X = x + y and Y = x y in the first standard form, we get the second standard form of hyperbolic PDE u XX u Y Y = ˆD(X, Y, u, u X, u Y ). If the given second order PDE (7.1.1) is such that A = B = 0, then (7.1.1) is of parabolic type and a division by C (since C 0) gives u yy = D(x, y, u, u x, u y ) where D = D/C. The above form is the standard form of second order parabolic equation. If the given second order PDE (7.1.1) is such that A = C and B = 0, then (7.1.1) is of elliptic type and a division by A (since A 0) gives u xx + u yy = D(x, y, u, u x, u y ) where D = D/A. The above form is the standard form of second order elliptic equation. Note that the standard forms of the PDE is an expression with no mixed derivatives. 7.3 Reduction to Standard Form Consider the second order semilinear PDE (7.1.1) not in standard form. We look for transformation w = w(x, y) and z = z(x, y), with non-vanishing Jacobian, such that the reduced form is the standard form. If B 2 AC > 0, we have two characteristics. We are looking for the coordinate system w and z such that a = c = 0. This implies from equation (7.1.2) and (7.1.4) that we need to find w and z such that w x = B ± w y B2 AC A = z x z y.