LECTURE 15 + C+F. = A 11 x 1x1 +2A 12 x 1x2 + A 22 x 2x2 + B 1 x 1 + B 2 x 2. xi y 2 = ~y 2 (x 1 ;x 2 ) x 2 = ~x 2 (y 1 ;y 2 1

Transcription

1 LECTURE 5 Characteristics and the Classication of Second Order Linear PDEs Let us now consider the case of a general second order linear PDE in two variables; (5.) where (5.) 0 P i;j A ij xix j + P i, B i xi + C+ F x x +A x x + A x x + B x + B x + C+ F xix i@x j Suppose we make a general coordinate transformation (5.3) y ~y (x ;x ) y ~y (x ;x ) x ~x (y ;y ) x ~x (y ;y ) in which the functions ~y a ;x i, a; i ;, are at least twice dierentiable and the Jacobian is nowhere vanishing. Under such a coordinate transformation, equation (5.) becomes (5.5) where (5.6) and (5.7) 0 P a;b A0 ab y ay b + P a B0 a y a + C+F A 0 ab P i;j A ij ~y a;xi ~y b;xj ~y a;x ~y b;x + A (~y a;x ~y b;x +~y a;x ~y b;x )+A ~y a;x ~y b;x B 0 a P P i;j A ij ~y a;xix j + i B i~y a;xi ~y a;x x +A ~y a;x x + A ~y a;x x + B ~y a;x + B ~y a;x ~y ~y a;xix j ~y i@x j Up to this point everything is completely general. Let us now explore and see what conditions to place on the coordinate transformation y a (x) in order that the coecient (5.8) of y y in (5.5) vanish identically. A 0 ~y ;x ~y ;x +A ~y ;x ~y ;x + A ~y ;x ~y ;x 55

2 5. CHARACTERISITICS AND CLASSIFICATION OF nd ORDER PDES 56 We shall proceed as in the previous lecture; however, rst let us simplify our notation a bit by setting x x y x ~ ~y We then have ~ ~y (5.9) Suppose (5.0) A 0 +A + A A 0 + A + A + A A 0 +A + A B 0 x +A y + A y + B + B B 0 x +A y + A y + B + B ~(x; y) const is a curve in the xy-plane along which (5.8) vanishes identically and let y f(x) be a (local) representation of this curve as the graph of a function f of x; so that (5.0) is equivalent to (5.) Dierentiating (5.) with respect to x yields (5.) or (5.3) ~ (x; f(x)) + f 0 f 0 ; The condition that A 0 vanish along the curve (5.) is (5.4) 0 +A + A Thus, if f is chosen to satisfy the dierential equation (5.5) then the coecient A 0 will vanish.,f 0 y ~, A f 0 ~y + A ~y (f 0 ), A f 0 + A ( y ) (f 0 ), A f 0 + A 0 Now the general solution of a rst order ODE will inevitably be of the form (5.6) f(x) F (x; C) where C is an arbitrary constant ofintegration. In order to recover a coordinate function ~ (x; y) corresponding to f(x), let's rst recall the graph of f(x) is supposed to coincide with a level curve of ~ (x; y). In other words, the solution set of (5.7) y f(x) F (x; C) is to coincide with the solution set of an equation of the form (5.8) ~(x; y) 0 Now suppose we can invert equation (5.6) to get C as a function of x and y, (5.9) C ~ F (x; y)

3 5. CHARACTERISITICS AND CLASSIFICATION OF nd ORDER PDES 57 Then by taking (5.0) (5.) ~(x; y) ~ F(x; y) o C we obtain an equation of the form (5.8) that has the same solution set as (5.7). We thereby obtain a suitable coordinate function. Now let us now look at equation (5.5) a little more carefully. We are assuming that 6 0; since otherwise there would be no point in making a coordinate transformation. (5.5) is thus a quadratic equation in f 0. Using the quadratic formula we nd that (5.5) is actually equivalent to f 0, A + q A (A ), A 0 0 f, A, (A ), A A 0 ; so either f satises (5.) or q f 0 A + (A ), A (5.3) f 0 A, q (A ), A These dierential equations are in practice very dicult to solve; for the coecient functions on the right hand sides will in general be expressions in which f appears non-linearly; this is perhaps more obvious when one writes a more explicit expression for (5.) df (x) A (x; f(x)) + dx q (A (x; f(x))) + (x; f(y)) A (x; f(x)) (x; f(x)) However, even without explicitly solving equations (5.), (5.3) we can identify three distinct cases. Case I (A ), A > 0. In this case the expressions inside the radicals in equations (5.) and (5.3) are strictly positive functions, and so their square roots are well-dened real-valued functions of x. The general existence theorem for rst order ODE's guarantees the existence of solutions of (5.) and (5.3) so long as the original functions A ij were smooth functions of x and y. Let f + be a solution of (5.) and let f, be a solution of (5.3). Since the equations (5.) and (5.3) are distinct when (A ), A > 0, f + and f, will constituent distinct solutions of (5.5). We thus nd two classes of curves y f + (x)+c y f, (x)+c which can be interpreted as the level curves of two coordinate functions ~y ; (x; y) for which A 0 will vanish. This case is referred to as the hyperbolic case. Case II. (A ), A 0 In this case, equations (5.) and (5.3) reduces to a single dierential equation and so we can expect a single family of curves f 0 A y f (x)+c which are interpretable as the level curves of a single coordinate function ~ (x; y) for which A 0 vanishes. This case is referred to as the parabolic case.

4 5. CHARACTERISITICS AND CLASSIFICATION OF nd ORDER PDES 58 Case III. (A ), A < 0 In this case, the expression inside the radical is negative denite, and so f 0, and hence f, must be a complexvalued function. But in this case, we will no longer be able to interprete the graph of f as a level surface of a coordinate function (our coordinates must be real-valued functions). Therefore, if the original coecients satisfy the inequality (A ), A < 0 then there exists no coordinate system for which A 0 0. This case is referred to as the elliptic case. Similarly,we can construct a coordinate function ~(x ;y) that will guarantee that the coecient A 0 of in (5.5) will vanish. Let y g(x) be the graph of a function g of x corresponding to the curve (5.4) and suppose ~(x; y) satises (5.5) Then, the condition (5.6) ~(x; y) const 0A 0 +A + A ~y ;y ~ (x; g(x)) const allows us to replace the terms in (5.5) by,g 0,toget (5.7) 0 (g 0 ), A g 0 + A ( ) 0 Thus, if y g(x) isalevel curve of the coordinate function ~(x; y) and g satises the dierential equation (5.8) then the coecient of in (5.5) will vanish. (g 0 ), A g 0 + A 0 Note that (5.8) is essentially the same dierential equation as (5.5). The analysis of (5.8) proceeds just as before and we nd we have three basic situations, distinguished by the sign of (A ), A. Hyperbolic Case (A ), A > 0 In this case, equations (5.5) and (5.8) have two independent solutions, f. One of these, say f +, can be used to dene a coordinate function ~ for which the coecient A 0 vanishes identically, and the other can be used to dene a coordinate function ~ for which A 0 vanishes identically. One can thus dene a coordinate system in which the original PDE (5.) takes the form (5.9) A 0 + B 0 + B 0 + C+F 0 It is easy to see that in the hyperbolic case, the coecient A 0 never vanishes, and so we can rewrite (5.9) as (5.30) + C 00 +F 0 0 Equation (5.30) is referred to as the standard form of a hyperbolic PDE. Parabolic Case. (A ), A 0 In this case, we only nd one real-valued function whose graph can be interpreted as the level curve ofa coordinate function ~ for which A 0 vanishes identically. Lacking a second independent solution we can not nd a second coordinate function ~ for which A 0 vanishes identically. The coecient A0,however, is (5.3) A 0 + A ~x + ~ y + A

5 5. CHARACTERISITICS AND CLASSIFICATION OF nd ORDER PDES 59 Using the relations and setting,f 0 A we nd A 0,g 0 (x), A (,g 0 )+A, A, g 0 + A y ~ 0 ; A, (A )! and so actually by eliminating A 0 we simultaneously eliminate A0 (in the parabolic case). Thus (5.) reduces to (5.3) A 0 + B 0 + B 0 + C 0 +F 0 0 It is not hard to show that the coecient A 0 does not vanish in this case, and so (5.3) can be rewritten in the form (5.33) + C 00 +F 0 0 Equation (5.33) is referred to as the standard form of a parabolic PDE. Elliptic Case. (A ), A < 0 In this case, there is no choice of real coordinates that will allow the elimination of A 0 and A 0. However, can arrange it so that A 0 A0 and A0 0. This is done as follows. From (5.6) we have (5.34) A 0 +A + A A 0 + A ~x + ~ y + A A 0 +A + A Suppose we represent the level curves ~ (x; y) const (5.35) ~ (x; y) const as graphs of functions f (x) and f (x), respectively. Then we have (5.36),f 0 ~ y,f 0 Then (5.34) can be written as (5.37) A 0 will thus vanish if for f 0 we get (5.38) A 0 (f 0 ), f 0 A + A ~y A 0 [ f 0 f 0, A (f 0 + f 0 )+A ] ~ y A 0 (f 0 ), f 0 A + A ~y 0 f 0 f 0, A (f 0 + f 0 )+A f 0 A f 0, A f 0, A

6 5. CHARACTERISITICS AND CLASSIFICATION OF nd ORDER PDES 60 The requirement that that A 0 A0, then leads to (5.39) 0 (f) 0, f 0 A + A ~y y ~ + (f 0, ) f 0 A + A ~y ~ y (f) 0, f 0 A + A ~y y ~ + h A f 0,A Af 0,A i, h A i f 0,A Af 0,A A + A In order to solve (5.39) for f 0 we need to know something about the relationship between ~ y and ~y ;y.it will be sucient for our purposes to simply set This will have the eect of restricting the set of solutions of (5.39) - but afterall, we only need one solution. Equation (5.3) now becomes (5.40) h i 0 (f) 0, f 0 A Af + A, 0 A,A Af +h 0,A A i f 0,A A Af 0,A, A Solving (5.40) for f 0 yields (5.4) or p f 0 A A A, A (5.4) f 0 A q (A ), A The latter solutions (5.4) can be disgarded - for the expression inside the radical is negative by hypothesis and so these solutions will not be real-valued. Plugging (5.4) into (5.38) yields (5.43) f 0 A p A, A Note that in the equations (5.4) and (5.43), the expressions inside the radicals are positive in this case and so the dierential equations (5.4) and (5.43) are solvable in terms of real-valued functions and hence are interpretable as functions whose graphs correspond to the level curves of coordinate functions ~y and ~ for which A 0 0 and A 0 A 0. Summarizing, in the elliptic case, where (A ), A < 0, we cannot nd coordinates for which A 0 0 or A 0 0;however, we can nd coordinates functions for which A 0 0 and A 0 A 0. Doing so, we obtain (5.44) A 0 + A 0 + B 0 + B 0 + C 0 +F 0 0 Since A 0 6 0,we can divide (5.44) by A 0, thus obtaining (5.45) + + C 00 +F 00 0 Equation (5.45) is the standard form of an elliptic PDE.

7 . SUMMARY CLASSIFICATION OF ND ORDER LINEAR PDES 6. Summary Classication of nd Order Linear PDEs Hyperbolic Case (A ), A > 0 In this case, we found two distinct real-valued solutions f 0 of the relation (5.46) (f 0 ), f 0 A + A 0 One of these, say f 0 +, once integrated could be used to dene a coordinate function ~ for which the coecient A 0 vanishes identically, and the other can be used to dene a coordinate function ~ for which A 0 vanishes identically. One can thus construct a coordinate system in which the original PDE (5.) takes the form A 0 + B 0 + B 0 + C+F 0 Observing that, in the hyperbolic case, the coecient A 0 never vanishes, and so we rewrote (5.9) as (5.47) + C 00 +F 0 0 Equation (5.47) is referred to as the standard form for a hyperbolic PDE. Parabolic Case. (A ), A 0 In this case, we could nd only one real-valued solution f 0 of (5.46). Using this solution we could construct a coordinate function ~ for which A 0 vanished identically, but lacking a second independent solution we could not nd a second coordinate function ~ for which A 0 vanishes identically. However, we also showed that by choosing ~ so that A 0 vanished identically, we simultaneously forced A 0 to vanish identically. Thus (5.) was reduced to A 0 + B 0 + B 0 + C 0 +F 0 0 Then observing that the coecient A 0 could not vanish as well, we nally reexpressed (5.) as (5.48) + C 00 +F 0 0 Equation (5.48) is referred to as the standard form for a parabolic PDE. Elliptic Case. (A ), A < 0 In this case, there is no choice of real root f 0 of (5.46). Since we could not interprete the graph of a complex valued function as coinciding with a level surface of a coordinate function ~y i,we concluded that in the elliptic case one can not nd coordinates ~ and ~ such that the either of the coecients A 0 or A 0 vanish. However, we could arrange it so that A 0 0 and A0 A In so doing we convert (5.) to the form Since A 0 6 0,we could then divide by A 0, to get (5.49) A 0 + A 0 + B 0 + B 0 + C 0 +F C 00 +F 00 0 Equation (5.49) is the standard form of an elliptic PDE. Of course, if this classication of PDE's is to really make any real sense, it should be independent of the choice of coordinates (that is, we should make sure that if we change coordinates we do not change the a PDE's type). Using the relations one nds that (5.50) (A 0 ), A 0 A0 A 0 ab X i;j A ij ~y a;xi ~y b;xj (A ), A ~x,

8 . SUMMARY CLASSIFICATION OF ND ORDER LINEAR PDES 6 The expression ~x, is the square of the Jacobian of the coordinate transformation. Since the Jacobian is assumed to vanish nowhere, this expression is always positive. Thus, the left hand side of (5.50) is positive, negative or zero precisely when (A ), A is, respectively, positive, negative or zero. Thus, this classication of second order linear PDE's is independent of the choice of coordinates. Example 5.. The wave equation is an example of a hyperbolic PDE. Example 5.. The heat equation is an example of an parabolic PDE. Example 5.3. Poisson's equation is an example of an @ c F F F (x; y)