5CCM211A & 6CCM211B Partial Differential Equations and Complex Variables

Transcription

1 5CCM211A & 6CCM211B Partial Differential Equations and Complex Variables These are slightly edited lecture notes by Simon Scott 1 Preliminary ideas on PDEs Partial differential equations (PDEs are fundamental to pure mathematics, to classical and quantum mechanics, to quantum field (and string theory, and generally throughout physics and engineering. For example, in pure mathematics PDEs are a basic tool of geometric analysis in particular, in the recent proof of the Poincaré Conjecture (which remained unsolved by other methods for over 100 years. In practise there are two broad steps to utilizing a PDE in one of these contexts. First, deduce a PDE which effectively describes the geometric / physical structure in question. Secondly, try to solve the PDE or, even if that is not possible, try to pick out interesting mathematical properties any solution must possess. The first of these steps is the relatively easy part, setting up an expression involving derivatives which describes, for example, how the geometric or physical structure evolves through time you can read more about the background ideas in text books or on the internet. For the second step, finding an exact solution is usually impossible, though there are always numerical methods which provide approximate solutions. But, even if it cannot be computed exactly, it may be possible to determine analytically numbers or functions which arise in the solution and which define invariants that characterize the physical or geometric structure in question. There are, though, certain special classes of PDEs which can be solved exactly and which describe some important basic geometric and physical processes, and in the following lectures we will be looking into some particular cases where that occurs. 1

2 2 1.1 So what is a PDE? A partial differential equation (PDE is an equation which relates a function u : X R (or C on some region X R n depending on n independent variables x 1,..., x n to a finite number of its partial derivatives. The objective is to find u from the PDE. The order of the highest occurring non-zero partial derivative in the PDE is called the order of the PDE. In order to compute u it is in general necessary to make assumptions about the geometric shape of the region X and the values that u and its derivatives can take on the boundary of X a PDE augmented by such data is called a boundary value problem (or, depending on the context, an initial value problem Some examples Laplace s equation is a 2nd order PDE defined in every dimension : dim 1: f : R 1 R 1, f = f(x, d 2 f dx 2 = 0, or u = 0. (1.1 A more refined differential equation is the boundary problem for f : [0, 1] R 1 d 2 f = 0 subject to u(0 = a, u(1 = b (Dirichlet bvp, (1.2 dx2 where a, b R (or C. dim 2: u : R 2 R 1, u = u(x, y, 2 u x + 2 u 2 y = 0, or u 2 xx + u yy = 0. (1.3 A corresponding Dirichlet bvp is for u : X R 1 with X = {(x, y x 2 + y 2 < 1}: 2 u x u y 2 = 0 subject to u(x, y = ϕ(x, y for (x, y S1, (1.4 where ϕ is a given function on S 1 := {(x, y x 2 + y 2 = 1}. dim 3: w : R 3 R 1, w = w(x, y, z, and so on. 2 w x w y w z 2 = 0, or w xx + w yy + w zz = 0, (1.5

3 3 So for u : R 2 R 1, u = u(x, y, we are using the shorthand notation u x = u x, u y = u y, u xx = 2 u x 2, u xy = 2 u x y, u yx = 2 u y x, u yy = 2 u y 2, u xxy = 3 u 2 x y, u xxx = 3 u 3 x,... and so on. Recall here that mixed partial derivatives are equal so that u xy = u yx, u xxy = u xyx = u yxx,... i.e. the order of differentiation does not matter it does when you work on curved space (such as a sphere, in fact that is exactly what curvature is in the mathematical sense. We may also use the alternative shorthand x u = u x, 2 xyu = 2 u x y, yu = u y, 2 yxu = 2 u y x, 2 xxu = 2 u x 2, 2 yyu = 2 u y 2, and so on, and one has 2 xy = 2 yx etc. Here are some more PDEs in dim 2. u x + u y = 0 Transport equation (order 1, (1.6 u xx + u yy = 0 Laplace equation (elliptic, (1.7 u t = iu xx Schrödinger eqn (parabolic, (1.8 u t = u xx Heat equation (parabolic, (1.9 u tt = u xx Wave equation (hyperbolic. (1.10 PDEs in dimension 2 will, in fact, be essentially the only case we will look into in this course, though most of the methods apply in a fairly obvious way to all dimensions. We will also review some facts about differential equations in dimension one to guide us in the higher dimensional case.

4 Constant coefficient, homogeneous, and linear PDEs When working in dimension 2 we will be interested almost exclusively in constant coefficient homogeneous linear PDEs (the theory of PDEs on higher dimensional spaces is similar but we will not consider that here. That is, if in a given PDE the coefficient of u and that of each of its partial derivatives is a constant (independent of the variables x, y, it is said to be a constant coefficient PDE it is much more complicated to deal with the case where the coefficients are general functions of (x, y, but fortunately within the class of constant coefficient operators there are some important and interesting PDES; in fact, the case of non-constant coefficients really arises when one looks to solve the same PDEs on curved space, then the curvature of space requires that the coefficients depend on the curvature (or, really, on the metric = first fundamental form. A general linear first-order partial differential operator (PDO with variable coefficients has the form P = b 1 (x, y x + b 2(x, y y + c(x, y : C1 (R 2 C 0 (R 2, (1.11 acting on functions in C 1 (R 2 by u = u(x, y (P u(x, y = b 1 (x, y u x + b 2(x, y u + c(x, yu(x, y y Here, for a region X R 2 = b 1 (x, y u x + b 2 (x, y u y + c(x, yu(x, y. C 1 (X = {u : X C u x and u y exist and are continuous in (x, y X}, while C 0 (X is the space of continuous functions on X. So X could, for example, be all of R 2, or a rectangle (and its interior, or a disc, or an infinite strip, and so forth. For example, if P = sin(xy x + (3x + y +y2 + ex2 y and u(x, y = xy then (P u(x, y = y sin(xy + 3x 2 + xy + e x2 +y 2 xy. The meaning of linear is the same as before, that P (λu 1 + µu 2 = λp u 1 + µp u 2, for any u 1, u 2 C 1 (R 2 (1.12

5 5 and any λ, µ C. P is constant coefficient if b 1 (x, y = b 1 C, b 2 (x, y = b 2 C, c(x, y = c C, are each constant, independent of x and y; that is, P = b 1 x + b 2 + c. (1.13 y Then a linear constant coefficient partial differential equation PDE of first order on X R 2 is an equality of the form for v C 0 (X P u = v with u C 1 (X. (1.14 That is, for a given v = v(x, y the problem is to find which u satisfy this equation. If we specify that u is equal to a certain function on the boundary of X, then this is called a PDE boundary problem. We will see lots of examples as the course proceeds, but do notice here that now the boundary of X will in general be a curve (the boundary of a disc, for example, is a circle and so now the boundary condition will be a function on that curve in the 1-dimensional ODE case the boundary can only be a point, or set of points, so the function boundary values are discrete numbers, this is one reason why the higher dimensional situation is more complex (and more interesting. The homogeneous version of this equation means the case v = 0, so P u = 0, (1.15 that is, a homogeneous linear constant coefficient partial differential equation PDE of first order on X R 2 is an equation of the form b 1 u x (x, y + b 2 u y (x, y + cu(x, y = 0, b 1, b 2, c C. (1.16 In other words, solving the homogeneous equation is finding functions u = u(x, y which are in the kernel Ker(P = {u C 1 (X P u = 0}. This (obviously! is a vector subspace of the vector space C 1 (X. For example, when then In fact, P = x + y u(x, y = cos(x y solves (1.15. u(x, y = f(x y solves (1.15

6 6 for any differentiable function f : R 1 R 1! (There is more on this PDE in the next section. The above discussion extends to 2nd-order PDEs in 2-dimensions essentially without change except that now we allow partial derivatives up to order 2. So a homogeneous linear constant coefficient partial differential equation PDE of second order on X R 2 is an equality of the form Lu = 0, u : R 2 C, (x, y u(x, y, (1.17 with 2 L = a 11 x + 2a x y + a b y }{{ 2 1 x + b 2 + y }{{} c }}{{} order 0 part order 2 part order 1 part (1.18 for some constants a ij, b k, c. A non-homogeneous PDE has the form with Lu = v, u C 2 (X (1.19 C 2 (X = {u : X C u xx, u xy, u yy, u x, u y exist and are continuous in (x, y X}. We can avoid discussion of which function space here by restricting matters to the subspace C (X of smooth functions on X; that is, functions all of whose partial derivatives of all orders exist (such as polynomials and convergent power series. So the meaning of (1.19 is that for u = u(x, y, 2 u a 11 x + 2a 2 u 2 12 x y + a 2 u 22 y + b u 2 1 x + b u 2 + cu = v(x, y, (1.20 y or, in alternative notation, a 11 u xx + 2a 12 u xy + a 22 u yy + b 1 u x + b 2 u y + cu = v(x, y. (1.21 The PDE is said to be homogeneous if v = 0 in (1.19; that is, Lu = 0 (1.22 so a 11 u xx + 2a 12 u xy + a 22 u yy + b 1 u x + b 2 u y + cu = 0. (1.23 L is said to be linear if for all α, β C and all admissible functions u, v on X L(αu + βv = α Lu + β Lv. (1.24

7 7 One of the essential advantages of considering linear PDEs is that if each of u 1,..., u k C 2 (X is a solution to the homogeneous PDE (1.22 then for any choice of constants λ j C u(x, y := k λ j u j (x, y (1.25 i=1 is also a solution. This is called the superposition principle, and is a fact we will use repeatedly in what follows. Indeed, by linearity we have ( k (1.24 k Lu = L λ j u j = λ j Lu j = 0. (1.26 }{{} i=1 i=1 = Example: Laplace equation The Laplace equation is the 2nd order linear homogeneous PDE or, in alternative notation, Easy solutions to spot just by observation are 2 u x + 2 u = 0, ( y2 u xx + u yy = 0. (1.28 u(x, y = ax + by + c, (1.29 u(x, y = x 2 y 2 (1.30 and u(x, y = xy. (1.31 Notice that by superposition principle it therefore follows that u(x, y = λx 2 + µxy λy 2 + ax + by + c is also a solution for any constants λ, µ, a, b, c. A less obvious polynomial solution is u(x, y = x 3 3xy 2 (1.32 and, in fact, there are particular polynomial solutions in each degree. All of these solutions exist on all of R 2.

8 8 Another not so obvious solution is u(x, y = log(x 2 + y 2 valid on R 2 \{(0, 0}. (1.33 On the other hand, if we look for solutions to the Laplace equation: 2 u x u y 2 = 0 in 0 < x2 + y 2 < 1 (1.34 that is, in the punctured unit disc, subject to the boundary (= circle condition u(x, y = 0 for (x, y S 1 = {(x, y x 2 + y 2 = 1}, (1.35 then none of (1.29, (1.30, (1.31, (1.32 are solutions, but (1.33 is a solution make sure you can see why! A rather different type of solution is Likewise, u m (x, y = sin(mx sinh(my any m R. u m (x, y = cos(mx sinh(my, u m (x, y = sin(mx cosh(my, u m (x, y = cos(mx cosh(my are all solutions, and so by linearity so is any linear combination of these solutions. This is useful for fitting a solution to a given boundary problem; for example, with X the solid rectangle X = (0, π (0, π, the Laplace equation subject to the boundary conditions 2 u x + 2 u = 0 for (x, y (0, π (0, π ( y2 u(0, y = 0, u(π, y = 0, u(x, 0 = 0, and u(x, π = sin 3 x, (1.37 has solution u(x, y = ( ( 3 sin(x sinh(y 4 sinh π 1 4 sinh(3π sin(3x sinh(3y. (1.38 Important Excercise: Show that (1.38 is a solution to the PDE bvp (1.36, (1.37 draw the region X and where the boundary conditions are being imposed.

9 Example: heat equation The heat equation is the 2nd order linear homogeneous PDE or, in alternative notation, 2 u x 2 u t = 0 for (x, t R1 (0,, (1.39 u t = u xx for (x, t R 1 (0,. (1.40 Note that this is only specified in the half-plane t > 0 (this is needed for non-trivial solutions to exist. An easy trivial solution to spot is for any constants α, β, γ. An important less obvious solution is we will derive this later on. u 0 (x, t = αx 2 + βx + γ + 2αt (1.41 u 0 (x, t = 1 4πt e x2 4t, (1.42 On the other hand, in the half-infinite strip X = [0, π] (0, this is not a solution to the heat equation subject to the boundary conditions u t = u xx for (x, t [0, π] (0, (1.43 u(0, t = 0, u(π, t = 0, and u(x, 0 = sin 3 x. (1.44 But u(x, t = 3 4 e t sin x 1 4 e 9t sin(3x (1.45 is a solution to this bvp! Notice how the specific geometry of bvps radically changes the form of the solutions in all the above example solutions look again and try to identify the differences! Important Excercise: Show that (1.45 is a solution to the bvp (1.43, (1.44 draw the region X and where the boundary conditions are being imposed.

10 10 2 Lecture 2: The Transport Equation Next we turn to first-order PDEs in dimension 2, meaning first-order PDEs on R 2. A general non-homogeneous, linear, first-order PDE on R 2 is of the form a(x, y u x + b(x, y u y + f(x, yu = v (2.1 with coefficient functions a, b, f : R 2 R. That is, a(x, y u u + b(x, y + f(x, yu(x, y = v(x, y. (2.2 x y Sometimes it is helpful to write this in the briefer form P u = v (2.3 with P understood to be the linear PDO (partial differential operator P := a(x, y x + b(x, y y + f(x, y. Remember linear means that P (λu 1 + µu 2 = λp u 1 + µp u 2 for any differentiable functions u 1, u 2 on R 2 and constants λ, µ. 2.1 The basic transport equation Here, we are going to solve only the simplest class of such equations, those for which the coefficient functions are constants, so a(x, y = a R, b(x, y = b R, f(x, y = λ C, are just numbers (independent of x and y. To start with we will take f = 0. Specifically, let s look at the homogeneous (meaning v(x, y = 0 constant coefficient Transport Equation a u x + b u y = 0. ( The case a 0 and b = 0 First, note that if b = 0 and a 0 this reduces to the equation u x = 0, or u(x, y = 0. (2.5 x This, as in 1-dimension on R 1, implies that u = u(x, y is independent of x, and hence that u is a function of y alone, so u x = 0 = u(x, y = f(y, (2.6

11 11 where f : R R, w f(w. (2.7 The particular choice of f (differentiable is unrestricted. This is analogous to the case on R for the ODE g = 0 with g : R R, which has solution f(t = c where c can be any constant, to fix c we have to specify g at any point x 0 if we specify g(x 0 = then c = In the 2-dimensional case (2.6, in order to fix the unknown function f we have to say what it is! but since it is a function of 1-variable we can do that by specifying u on any straight line on R 2 (and, in fact, on more general curves in R 2. For example, if we say it has to be equal to the function y 2 on the y axis (the line x = 0 then f(w = w 2. We would write this initial value problem as { ux = 0 u(0, y = y 2 (2.8. From what we have just said this has solution u(x, y = y 2. (2.9 On the other hand, we can also prescribe the initial condition on, e.g., the line y = αx, where α 0 is some fixed constant. That is, suppose { ux = 0 u(x, αx = x 2 (2.10. Then we know u(x, y = f(y solves the PDE, but in order that it satisfy the condition u(x, αx = x 2, then f(αx = x 2, and hence f : R R, w f(w, is the function f(w = w2 α 2. That is, the solution to the boundary (or initial value problem (2.10 is u(x, y = y2 α 2. What happens if α = 0? Then the boundary condition says that u(x, 0 = x 2, and hence that f(0 = x 2 but this is impossible, so for α = 0 there is no solution to (2.10. We could also uniquely specify the solution by saying what it has to be along more general curves. For example, specifying it along the cubic y = x 3 : { ux = 0 u(x, x 3 (2.11 = sin x.

12 12 Again, we know u(x, y = f(y solves the PDE, and in order that it satisfy the condition u(x, x 3 = sin x, then f(x 3 = sin x, and hence f : R 1 R 1, w f(w, is the function f(w = sin(w 1/3. That is, the solution to the boundary (or, initial value problem (2.11 is u(x, y = sin(y 1/3. This depends on the fact that the function w w 3 is bijective from R 1 to itself, i.e. for a given t R there is a unique real value (denoted t 1/3 whose cube is equal to t. More generally, provided r : R 1 R 1 is 1-1 and onto (bijective, then the bvp { ux = 0 (2.12 u(x, r(x = q(x for some given function q : R R, has the unique solution u(x, y = q ( r 1 (y. (2.13 Why is it unique? Well, if u 1 : R 2 R is another solution of (2.12, then satisfies the bvp z(x, y := u(x, y u 1 (x, y { zx = 0 z(x, r(x = 0 (2.14 check you can see why! Once again, we know z(x, y = f(y solves the PDE (any f, and in order that it satisfy the condition z(x, r(x = 0, then f(r(x = 0, and hence f must be identically zero (since r is bijective. That is, z(x, y = 0 for all (x, y R 2. Hence u = u The case a = 0 and b 0 This refers to the case of the Transport equation (2.4 u y = 0, or u(x, y = 0. (2.15 y The above discussion is then the same except replace x by y everywhere. So u(x, y = f(x for an arbitrary differentiable 1-variable function f : R R etc (the rest of the above discussion for this case is left to you as an exercise.

13 The case ab 0 In this case, in which both a and b are non-zero, we can divide through by a and hence rewrite (2.4 as u x + c u y = 0 with c := b a 0. (2.16 This is only to eliminate one of the letters in the equation for convenience, but we can equally stick to (2.4. The general constant coefficient homogeneous transport equation (2.16 can, in fact, be easily solved using the Chain Rule. To see this, consider a smooth curve passing through a given point (x, y R 2 t r(t = (x(t, y(t with r(0 = (x(0, y(0 = (x, y. Then we may consider how u : R 2 R 1 behaves along that curve by analyzing the composite function t u(r(t = u(x(t, y(t. (2.17 The Chain rule says that the t-derivative of the function of 1-variable (2.17 is given by d dt u(x(t, y(t = x (tu x (x(t, y(t + y (tu y (x(t, y(t. (2.18 Consider specifically the case r(t = (x + t, y + ct. (2.19 Then (2.18 says d dt u(x(t, y(t = u x (x, y + cu y (x, y. (2.20 t=0 Hence if we assume u is a solution to (2.16 (for all (x,y then (2.20 may be written d u(x(t, y(t dt = 0 (2.21 t=0 That is, if u solves the transport equation then the function t u(x(t, y(t is constant along the curve (2.19. In other words for any two times t 0 and t 1 we have u(x + t 0, y + ct 0 = u(x + t 1, y + ct 1. Choosing, at a given point (x, y, the values t 0 = 0 and t 1 = x gives us u(x, y = u(0, y cx. (2.22

14 14 This says that u is determined by evaluation along the line x = 0, the y-axis, in which the variable is replaced by y cx; that is, u is a function of 1-variable w with w = y cx, so u(x, y = f(y cx, f : R 1 R 1, w f(w, (2.23 for any differentiable function f of one variable. Of course, we could choose different values of t 0 and t 1, but these will always lead to the same conclusion. Specifically, taking t 0 = 0 and t 1 = y/c gives us u(x, y = u(x y/c, 0. (2.24 This says that u is also determined by evaluation along the line y = 0, the x-axis, in which the variable is replaced by x y/c; that is, u is a function of 1-variable w with w = x y/c, so u(x, y = g(x y/c, g : R 1 R 1, w g(w, (2.25 for any differentiable function g of one variable. So (2.23 and (2.25 are not contradictory, indeed taking g(w = f( cw turns (2.25 into (2.23. The point here is that the function f (or g is completely arbitrary. We could also write this family of solutions as u(x, y = h(bx ay, any h : R 1 R 1, w h(w. ( An equivalent method Method 2 The case a 0, b 0 is easily reduced to the case of by making the change of variable (x, y (ξ, η with ξ = bx ay, η = y. Indeed, let v(ξ, η := u(x, y. Then the chain rule implies u x = v ξ b, u y = v ξ a + v η, and Now, a u x + b u y = 0 = abv ξ bav }{{} ξ +bv η = 0. =0 v η = 0 = v(ξ, η = h(ξ

15 15 for a function h : R R of one variable, and going back to the original variables we get u(x, y = h(bx ay. So really we have done not much more in solving the general case (2.16 than solve ( Finding unique solutions: imposing boundary conditions To get a specific solution we have to specify how the solution behaves along a suitable curve in R 2, just like in the cases with b = 0 discussed above. For example, consider { 2ux + 3u y = 0 u(x, 0 = sin(x 2. (2.27 Here, then, a = 2 and b = 3. From (2.23 we know that a general solution of 2u x + 3u y = 0 has the form u(x, y = f (y 32 x, any f : R 1 R 1, w f(w. (Check that it does! The condition u(x, 0 = sin (x 2 therefore says that f ( 32 x = sin ( x 2. Hence f is the function f(w = sin (( 2w/3 2 = sin (4w 2 /9. Thus the solution to (2.27 is u(x, y = sin (x 2 43 xy + 49 y2. (2.28 Check that you obtain (2.28 and check that it really does solve (2.27! More generally, with b 0, for a specified function h : R 1 R 1 has the unique solution { aux + bu y = 0 u(x, 0 = h(x (2.29 u(x, y = h (x a b y. (2.30

16 16 Check that you obtain (2.30 and check that it really does solve (2.29! Similarly, with a 0, for a specified function g : R 1 R 1 has the unique solution { aux + bu y = 0 u(0, y = g(y (2.31 u(x, y = g (y ba x. (2.32 Check that you obtain (2.32 and check that it really does solve (2.31! So writing the form of the solution as in (2.25 is well suited to the bvp (2.29, while (2.23 is well suited to the bvp (2.31. But it suffices to use any one of (2.25, (2.23, (2.26 as the general solution the boundary condition will determine the actual solution whichever one you start with. We might generalize (2.29 by specifying the boundary condition along the line y = αx for some given α R, so that { aux + bu y = 0 u(x, αx = h(x. (2.33 We know any solution has the form u(x, y = f (bx ay (2.34 for some function f of one variable. The condition u(x, αx = h(x says that f (bx αax = h(x, i.e. f(w = h ( w b αa and hence the unique solution to (2.33 is ( bx ay u(x, y = h provided α b = c. (2.35 b αa a For example, has solution (check it! { 2ux + 3u y = 0 u(x, 5x = sinh(x 4 u(x, y = sinh ( 1 (3x 2y (2.36

17 Uniqueness of (2.35 This is the same proof as we saw for the case b = 0: if u 1, u 2 : R 2 R 1 are both solutions of (2.33 then z(x, y := u 1 (x, y u 2 (x, y satisfies { azx + bz y = 0 z(x, αx = 0. (Why? (2.37 We know that z(x, y = f(bx ay some f : R R. In order that it satisfy the condition z(x, αx = 0, we have that 0 = z(x, αx = f(βx, with β := b αa, and hence f must be identically zero. That is, z(x, y = 0 for all (x, y R 2. That is u 1 = u What happens if α = b a? Well, then the condition u(x, αx = h(x says that f (0 = h(x and this can only make sense if h is a constant function h(x = k C. Then the solution is u(x, y = k that is, this condition requires u to be constant in both x and y variables. It is not hard, in fact, to see why this is: any solution of the PDE au x + bu y = 0 with α = b a is constant on the line {(x, αx x R} and on any line {(x 0 + x, y 0 + αx x R} parallel to it. Indeed, d dt u(x 0 + t, y 0 + αt = u x (x 0 + t, y 0 + αt + αu y (x 0 + t, y 0 + αt = 1 a (au x(x 0 + t, y 0 + αt + bu y (x 0 + t, y 0 + αt = 0. So the case α = b is where we are looking for a solution which satisfies the condition a u(x, αx = h(x, and from what we have just noticed, that is only possible if h(x is constant and so is u. 3 Lecture 3: 2nd order PDEs In fact, we will not be discussing 2nd order PDEs in general here, rather just a particular case.

18 18 Specifically, we are going to derive the form of the general solution to the wave equation on R 2 using some ad hoc methods. A homogeneous, linear, constant coefficient, partial differential equation (PDE of second order on R 2 is an equality of the form with for some constants a ij, b k, c. Lu = 0, u : R 2 C, (x, y u(x, y, (3.1 2 L = a 11 x + 2a x y + a 2 22 y + b 2 1 x + b 2 y + c (3.2 Equivalently, (3.1 can be written with u = u(x, y 2 u a 11 x + 2a 2 u 2 12 x y + a 2 u 22 y + b u 2 1 x + b u 2 + cu = 0, (3.3 y or, in alternative notation, a 11 u xx + 2a 12 u xy + a 22 u yy + b 1 u x + b 2 u y + cu = 0. ( The wave equation Specifically, let s take a look at the wave equation, which is the special case where (in its simplest form a 11 = a 22 and all the other coefficients are zero, so that 2 t u(x, t = 2 xu(x, t. (3.5 This can also be written with Lu = 0 (3.6 L = 2 t 2 x. (3.7 A first thing to notice is that L can be factorized into a product of two first-order transport operators: L := 2 t 2 x = ( t x ( t + x (3.8 = ( t + x ( t x (3.9

19 19 For, using linearity, ( t x ( t + x u = ( t x (u t + u x = ( t x u t + ( t x u x = t u t x u t + t u x x u x = u tt u xt + u tx u xx = u tt u xx using that fact that partial derivatives commute t x = x t. A similar proof applies to (3.9. From (3.8 it follows that any z = z(x, t which solves the transport equation ( t + x z = z t + z x = 0 will automatically define a solution for (3.5, and likewise, from (3.9 so will any w = w(x, t which solves the transport equation ( t x w = w t w x = 0. But we already know from the precious lecture that these two transport equations have general solutions of the form z(x, y = g(x t and w(x, y = f(x + t for any differentiable functions f, g : R R. Hence by the superposition principle we can take arbitrary sums of such function to get more solutions; that is, it is certainly the case that one class of solutions to the wave equation are of the form u(x, t = f(x + t + g(x t. (3.10 In fact, this is the most general solution to (3.5. To see this, we make the change of coordinates ξ = x + t, η = x t. (3.11 Then by the Chain Rule in two variables in (ξ, η-coordinates the wave operator (3.7 becomes L = 4 ξ η. (3.12

20 20 For, the Chain Rule in two variables says that That is x = ξ x ξ + η x η, t = ξ t ξ + η t η. x = ξ + η, t = ξ η = t + x = 2 ξ, t x = 2 η, and (3.9 implies (3.12. So to solve the wave equation we have to solve ξ η u(ξ, η = 0. It follows from ξ ( η u(ξ, η = 0 that η u(ξ, η does not depend on ξ, i.e. η u(ξ, η = G(η for some function G. Integrating with respect to η, we get u(ξ, η = G(η dη + term independent of η =: g(η + f(ξ. Hence u ξη = 0 = u(ξ, η = f(ξ + g(η with f, g : R 1 R 1, (3.13 and substituting back (3.11 into (3.13 we obtain (3.10 resolving it into leftmoving and right-moving waves. Exercise: check (3.10 really does solve the wave equation. The solution(3.10 is very general, but also very vague. To get a more precise solution we have to specify some initial (or boundary conditions. To this end, we augment the PDE to the initial value problem u tt u xx = 0, u(x, 0 = φ(x (initial shape of the wave (3.14 u t (x, 0 = ψ(x (initial speed of the wave. A direct computation (see below shows that the solution (3.10 is refined in the case of (3.14 to u(x, t = 1 2 (φ(x + t + φ(x t x+t x t ψ(s ds. (3.15 This says that if we specify what the wave in the (x, t-plane must look like along the x-axis, and we specify its speed along that line, then the wave form ( = graph

21 21 of (x, t u(x, t is determined throughout R 2. This is much like what we discovered with the transport equation that PDE is first-order so it was enough just to specify u(x, t along a curve, while because here we are dealing with a 2nd order PDE we also have to specify its first derivative. To see why (3.15 holds, since it satisfies the wave equation it must have the form u(x, t = f(x + t + g(x t for some f, g : R 1 R 1. The condition u(x, 0 = φ(x says that while the condition u t (x, 0 = ψ(x says that (Check you can see why. Integrating the latter, we get φ(x = f(x + g(x, (3.16 ψ(x = f (x g (x. (3.17 f(x g(x = x with some constant C. From (3.16 and (3.18, we have and and hence and f(x = 1 2 φ(x g(x = 1 2 φ(x x 0 x f(x + t = 1 2 φ(x + t g(x t = 1 2 φ(x t 1 2 Finally, adding (3.21 and (3.22 gives ( ψ(s ds + C (3.18 ψ(u du C (3.19 ψ(u du 1 C, ( x+t 0 x t 0 ψ(u du C (3.21 ψ(u du 1 C. ( For example, using (3.15, the solution to the initial value problem u tt u xx = 0, u(x, 0 = e x, u t (x, 0 = cos x, (3.23 is u(x, t = e x cosh(t + cos(x sin(t.

22 22 4 Lecture 4: Fourier Transforms The Fourier transform (FT transforms an integrable function f : R R of one variable x into another function f : R R of one variable ξ, while the inverse Fourier transform (IFT reverses the transformation, provided f is integrable. This is an important process for studying differential equations (and partial differential equations because the FT turns differentiation of f with respect to x into multiplication of f by ξ, exchanging (hard questions of differentiability of f for (easier questions of growth rates of f at infinity (i.e. as ξ, which turns (hard to solve differential equations into (relatively easy to solve algebraic (polynomial equations. Let us denote the FT by F and the IFT by F 1. So Specifically, F : functions on x space }{{} = R functions on ξ space. }{{} = R where f : R R, x f(x F f := F(f : R R, ξ f(ξ, f(ξ := For notational convenience, we may use any of e iξx f(x dx. (4.1 f(ξ = F(f(ξ = F x ξ (f(ξ (4.2 to denote the FT of f. The last one is useful to remind us what variable is being changed into what. That can be important to emphasize when we apply these methods to PDEs, where there are various variables in play, and also here in order to be clear what is being changing into what when applying the IFT. The inverse Fourier transform is defined by F 1 : functions on ξ space }{{} = R functions on x space. }{{} = R Specifically, g : R R, ξ g(ξ F 1 ğ := F 1 (g : R R, x ğ(x,

23 23 where ğ(x := 1 e iξx g(ξ dξ. (4.3 2π Again, for notational convenience, we may use any of ğ(x = F 1 (g(x = F 1 ξ x (g(x. ( Note: differing definitions of F and F 1 in the literature If you consult text books, or web pages, on the FT you will likely find alternative definitions of the FT. The most common ones are f(ξ in some other texts = e iξtx f(x dx or 1 2π e iξx f(x dx, corresponding to what we have defined here as the inverse FT; and likewise the IFT is then defined by what is here defined to be the FT. At the end of the day, all these definitions are equivalent, it is just a matter of convention, all that really matters is that the FT and IFT, whichever convention is followed, are inverse to each other. The convention here (4.1 is rather non-standard these days, but it has been retained to be consistent with previous versions of this course, and the exam papers. But, be aware that formulae may differ slightly according to the convention one follows. 4.1 F and F 1 are inverse to each other (when defined! The notation and terminology suggest F 1 reverses what F does to f. This is true provided all the integrals in question exist (we will assume they do!. That is, That is, I = F 1 F where I(f = f (i.e. I(f(x = f(x x R. (4.5 or, equivalently, From (4.3, this is the equality f(x := 1 2π which, from (4.1, is the equality f(x := 1 2π f = F 1 (F(f (4.6 ( f = F 1 f. (4.7 e iξx ( e iξx f(ξ dξ, (4.8 e iξt f(t dt dξ. (4.9

24 24 Formally (non-rigorously, we might rearrange this as, for example, f(x := 1 e iξ(t x f(t dξdt. (4.10 2π t= ξ= However, this is problematic as it stands, because the integral over ξ does not exist. Nevertheless, provided f is, in some appropriate sense (see below, strongly integrable we can make sense of it as a limit as δ 0, of the well defined double integral f δ (x := 1 e δξ2 e iξ(t x f(t dξdt. (4.11 2π t= ξ= For more on this, see the (non-examinable for 5CCM211A proof of (4.5 on Schwartz functions given in 4.4 below. The invertibility works the other way as well. One has I = F F 1 where I(g(ξ = g(ξ ξ R. (4.12 That is, or, g(ξ := g = F (ğ, (4.13 e iξt ğ(t dt. ( When are F and F 1 well defined? We will, for the most part, completely ignore the very important and subtle questions concerning the classes of functions for which F(f and F 1 (g exist, i.e. when the integrals in (4.1 and (4.3 are defined and finite, and likewise for the inverse properties in 4.1. These questions are analyzed in third and fourth year courses, if you would like to understand the FT better. Here, we are only interested in methods, how the FT and its properties can be used to solve ODEs and PDEs. These methods are why the FT is so fundamentally important in mathematics; so what we do here, in this course, is the first step, later on you will be able to look into when the FT exists. But here is a brief comment on the true meaning of (4.1. We ought to be aware of a couple of things. First, the meaning of the integral (4.1 is as a Cauchy principal value, meaning f(ξ := a e iξt f(t dt := lim e iξt f(t dt. (4.15 a a

25 25 (non- You do not have to read the rest of this section if you do not want to! examinable for 5CCM211A comments Integrals over R, called improper integrals, are defined by h(t dt := lim b c b h(t dt + lim a a c h(t dt (4.16 for any choice of c R (the answer must be independent of c this is so for integrable functions; for example, for continuous functions which decay to zero as t faster then 1/t. But, although it is true that if (4.16 exists for each c, then also (4.15 exists and is the same, some functions which are highly non-integrable (in the sense that the right-hand side of (4.16 does not exist may still have a well-defined Cauchy principal value; for example, a [ lim 2t dt = lim t 2 ] a a a a a = lim ( a 2 ( a 2 = lim 0 = 0, a a a but neither lim a 0 2t dt nor lim 0 a 2t dt exist, f(t = 2t is clearly non-integrable on R a in the usual sense. Likewise, 1/t dt is undefined in the sense of classical integrability; if we write it as the sum a ε 1/t dt and ε a 1/t dt both diverge to infinity as either ε 0 or a, but a 1/t dt a a (thought of as the signed-area under the curve is zero, and hence lim a 1/t dt = 0 exists a as a principal value integral. So the FT is, in fact, quite a mysterious object, depending on certain somewhat ad hoc regularization methods to give it a meaning when the function f is not integrable. This leads to all sorts of fascinating developments (and is one of the bases of the problems in understanding quantum theory (and string theory in particle physics. At any rate, as far as we are concerned here, the bottom-line is if there is a way of making sense of the integral, then use it. In fact, there is a rigorous analysis which tells us why such methods are correct, the key fact is that one has to do calculus in slightly bigger space of generalized functions or distributions, where all these things make sense (in that space you can differentiate functions which are non-differentiable (in the classical sense and integrate functions which are non-integrable (in the above classical sense. So the FT may be, in fact, sensible on obviously non-integrable functions. However, for most of the examples you will see, these issues will not be evident. 4.3 A couple of examples of FTs It is, in general, quite tricky to perform explicit evaluations of FTs (and (IFTs. The real power of the FT as far as PDEs are concerned is that it reveals the mathematical structure which governs how solutions to a PDE behave, so even if we cannot compute the answer as an explicit function, the FT shows us what it looks like we can see if it has discontinuities, how fast it is going to zero at infinity, and so forth. But, anyway, here are some elementary examples:

26 Example: compute F(e x We have f(ξ := = = 0 e iξt e t dt e iξt e t dt + [ 1 iξ 1 e(iξ 1t = lim a [ ] iξ 1 e(iξ 1t = 1 iξ iξ + 1, e iξt e t dt [ 1 + iξ + 1 e(iξ+1t ] a 0 + lim a [ ] 0 1 iξ + 1 e(iξ+1t since lim a e (iξ 1a = 0 (indeed, e (iξ 1a = e a, and similarly lim a e (iξ+1( a = 0. Thus f(ξ = ξ 2. (4.17 Perhaps what is more interesting is that applying the inverse FT now gives, 1 2π e iξx ξ 2 }{{} = b f(ξ This is the identity (4.8 applied to f(x = e x. Thus, ] 0 a dξ = e x (4.18 e iξx 1 + ξ 2 dξ = π e x. (4.19 This is not an obvious formula (though we will see how to deduce it directly when we do Contour Integration Example: Let β > 0, and set f(x = { 1, x β, 0, x > β. (4.20

27 27 Then f(ξ = β β [ e e iξt iξt dt = iξ ] β β = eiξβ iξ e iξβ iξ = 2 sin(βξ. ξ (In fact, referring to the (non-examinable for 5CCM211A comments in 7.2, you can see here that f(ξ is not ξ-integrable in the classical sense, even though f(x is integrable Example: compute F(e αx2, where α (0,. In the following, for notational brevity, write =. We have f(ξ := e iξt e αt2 dt = e αt2 +iξt dt w= αt = 1 α e w2 + iξ α w dw. We are aiming now to reduce this to something times the integral e r2 dr = π (4.21 which you saw how to evaluate (using polar coordinates in Calculus II (4ccm112a. So we want to try and write the exponent as an exact square, well we have and so w 2 iξ ( w = w iξ 2 α 2 + ξ2 α 4α, ξ 2 f(ξ = e 4α α e w iξ 2 α 2 dw (4.22 But 2 e w iξ 2 α dw = e r2 dr, (4.23 and hence, in view of (4.21, we obtain π f(ξ = ξ 2 α e 4α. (4.24

28 28 The equality (4.23 is non-trivial, in fact. What is trivial is the equality e (w c2 dw = e r2 dr for any real constant c R, (4.25 just by a change of variable. The subtlety in (4.23 is that it refers to the case ξ where c = i 2 is a complex number; the change of variable then shifts the integral α from an integral over R to an integral over a line in R 2 (or C parallel, but not equal, to the real axis. There is no difficulty in dealing with integrals over such contours in C and we will see how to do so shortly in the forthcoming section on Contour Integration, but for now we have to just take it as (a not difficult to believe fact. In Assignment 3 the formula in (4.24 will be derived by other methods (using an ODE which avoids this problem. 4.4 Appendix (non-examinable for 5CCM211A: A proof that F 1 F = I on Schwartz functions The Schwartz class S(R is the class of infinitely differentiable functions f on R which are of fast decrease at infinity; the second condition requires that for every choice of the integers m, N we have x N f (m (x 0 as x. In essence this means that a Schwartz class function f and all it derivatives tend to zero at infinity so fast as to dominate any power of x. A typical Schwartz class function is e x2 whose FT we found above. Notice, however, that the function of the first example e x is not quite Schwartz class because it is not differentiable at x = 0. Nevertheless, it is fine away from zero and its FT, as we saw, exists. For any Schwartz class function f we observe that the Fourier transform f of f always exists i.e. the integral eixξ f(ξ dξ exists. Moreover, it is not hard to show rigorously that f is infinitely differentiable and is also Schwartz class (Exercise!. We prove here the inversion formula for the FT on the vector space S(R of Schwartz class functions. One way to understand this result is that F : S(R S(R is a vector space isomorphism, with inverse given by (4.3. For f S(R consider the inverse Fourier transform of e ɛξ2 f(ξ, ɛ > 0; ultimately we let ɛ 0. We have ( e ɛξ2 f(ξe iξz dξ = e ɛξ2 e iξz dξ f(xe iξx dx (4.26 ( = f(x e ɛ[ξ2 + iξ(z x ɛ ] dξ dx (4.27 = = f(x dx π ɛ [ e ɛ[ ( ξ+ i(z x 2ɛ 2+ (x z 2 4ɛ 2 ] dξ = [f(x f(z]e (x z2 4ɛ dx + f(z π ɛ e (x z2 4ɛ f(xe (x z2 4ɛ (4.28 ] dx. (4.29

29 29 Using the general formula e α(x+iβ2 dx = e αx2 dx = π α, α > 0, β R which is proved in the forthcoming section on Contour Integration, we therefore have π e ɛξ2 f(ξe iξz dξ = 2πf(z + [f(x f(z]e (x z2 4ɛ dx. (4.30 ɛ We now show that π ɛ [f(x f(z]e (x z2 4ɛ dx 0 as ɛ 0 +. To see this, we suppose that f is real valued (if not, we can apply the following argument to the real and imaginary parts of f separately and note (by the Mean Value Theorem that f(x f(z = f (y(x z for some number y between x and z. Since f is Schwartz class, f is certainly bounded and we write K = sup y R f (y to obtain π ɛ [f(x f(z]e (x z2 4ɛ = π ɛ (2K 0 dx π ɛ K ue u2 4ɛ du = 4K πɛ x z e (x z2 4ɛ which does indeed tend to zero as ɛ tends to zero through positive values. It s not very hard to prove rigorously that for Schwartz class f dx lim ɛ 0+ e ɛξ2 f(ξe iξz dξ = f(ξe iξz dξ. We deduce that for any Schwartz class function f that f(ξe iξz dξ = 2πf(z so that the inverse transform formula is established. f(z = 1 2π f(ξe iξz dξ, f S(R 5 Lecture 5: ODEs and the FT Of the many beautiful properties of the Fourier transform f(ξ = F(f(ξ := e iξt f(t dt (5.1 (which you can see if you take more advances courses in analysis there are two basic properties which we will use repeatedly in solving PDEs, and ODEs. They are:

30 30 [1] Linearity: for (integrable f, g : R 1 R 1 and any λ, µ C, one has F(λf + µg(ξ = λf(f(ξ + µf(g(ξ (5.2 [2] The FT turns differentiation into multiplication: F(f (ξ = iξ F(f(ξ. (5.3 Here, f (x = df/dx is the derivative of f, and, crucially, we assume that f(x 0 as x. (5.4 You can easily check these properties yourself (see also Assignments 3 and 4. In fact, iterating the latter property we get for m N F(f (m (ξ = ( iξ m F(f(ξ (5.5 where f (m = d m f/dx m. That iteration therefore requires from (5.4 that f (k (x 0 as x for k = 0, 1,..., m 1. (5.6 That is, all the derivatives of f lower than f (m must go to zero at infinity. This is needed to make the integration-by-parts proof work (see Assignments 3 and Exact solutions to constant coefficient ODEs If we combine the above two properties with the invertibility of the FT (see 4.1, then we can solve any constant coefficient ODE. Such an ODE is an equality of the form a n f (n (x + a n 1 f (n 1 (x + + a 1 f (x + a 0 f(x = g(x, (5.7 where a n,..., a 0 C are constants and where g : R 1 R 1 is given, and the objective is to determine the function f : R 1 R 1. Specifically, applying the FT to both sides of (5.7, the linearity property (5.2 gives a n F(f (n (ξ + a n 1 F(f (n 1 (ξ + + a 1 F(f (ξ + a 0 F(f(ξ = F(g(ξ. (5.8 Provided we assume that (5.6 holds, we can then apply (5.5 to rewrite (5.8 as a n ( iξ n f(ξ + an 1 ( iξ n 1 f(ξ + + a1 ( iξ f(ξ + a 0 f(ξ = ĝ(ξ. (5.9 That is, ( an ( iξ n + a n 1 ( iξ n a 1 ( iξ + a 0 f(ξ = ĝ(ξ. (5.10

31 31 Set p n (ξ = a n ( iξ n + a n 1 ( iξ n a 1 ( iξ + a 0, a polynomial in ξ of degree n. Then (5.10 says that By Fourier inversion (see 4.1 we therefore have Substituting f(x = 1 2π f(ξ = ĝ(ξ p n (ξ. (5.11 e ixξ f(ξ dξ = 1 2π ĝ(ξ = e iξt g(t dt ixξ ĝ(ξ e p n (ξ dξ. (5.12 into (5.12 and rearranging (in particular - changing the order of integration we get our solution f(x = where k : R 1 R 1 C is the function (This is called the Schwartz kernel. k(x, t = 1 e i(t xξ 2π p n (ξ k(x, t g(t dt (5.13 dξ ( Raining on the parade Of course, life is never as easy as one would like it to be, and, in particular, we have to temper what we have just done by the realization that the above equations from (5.11 onwards are only well defined if p n (ξ has no zeroes on the real axis; that is, if there is a real number ξ 0 such that p n (ξ 0 = 0 then (5.11 is not defined at ξ 0, and the subsequent integrals are not defined, at least not without thinking hard about such integrals. Likewise, if n = 1 then the integral in (5.14 is not convergent ( it does not exist in the usual sense at least. So there are problems. The good news is that all of these can be dealt with, but it requires methods beyond what we have time to go into here. We shall therefore assume no real zeroes and that n > 1.

32 32 6 Lecture 6: ODEs and the FT: II Recall that the the FT can ( usually be used to solve ODE of the form a n f (n (x + a n 1 f (n 1 (x + + a 1 f (x + a 0 f(x = g(x, (6.1 where a n,..., a 0 C are constants and where g : R 1 R 1 is given, and the objective is to determine the function f : R 1 R 1. By Fourier inversion (see Lecture 5 we found that f(x = where k : R 1 R 1 C is the function k(x, t g(t dt (6.2 k(x, t = 1 2π e i(t xξ p n (ξ dξ. ( An example The above manipulations are a little abstract (though relatively simple, you must admit, given that the pay-off is that they allow us to solve any ODE!!. So here is a specific computation. Consider the ODE subject to d2 f + f(x = g(x (6.4 dx2 f(x, f (x 0 as x. (6.5 The idea is that g is a given function and we are trying to determining what f can be. So in the above notation, here we have So our polynomial is Hence n = 2, a 2 = 1, a 1 = 0, a 0 = 1. p 2 (ξ = ξ k(x, t = 1 e i(t xξ 2π 1 + ξ dξ. 2 But we already know how to evaluate the integral on the right: from equation (5.18 we have k(x, t = 1 2 e x t.

33 33 That is, by (6.2 the general solution to (6.4, (6.5 is f(x = 1 2 e x t g(t dt. (6.6 Important Exercise: Work through for this specific example each of the steps leading to (6.2, (6.3 from scratch so apply the FT to (6.4, do all the integration by parts and other manipulations. Understand every little part of the calculation for this specific ODE. Check that (6.6 really does solve the ODE. See Assignment 3 for more examples. Also, consult some of the recommended text books (on the web page, or other books, or on the internet, for more examples. 7 Lecture 7: PDEs and the FT Our aim here is to see how the Fourier transform can be used to solve constant coefficient PDEs on R 2. This is done by a method which is analogous to, and instructed by, the case for constant coefficient ODEs on R 1 (seen in the previous lecture. Just as for ODEs, we use the FT to transform derivatives in a given variable into multiplication. If we can then solve the resulting ODE in 1-variable, we may then inverse Fourier transform that solution back to a solution for the PDE. The best way to understand this process is to look at some examples, which we shall do in a moment. First, we need to decide on some notation and make a note of the basic properties of the FT on partial derivatives. 7.1 Notation and basic properties A general constant coefficient homogeneous linear PDE in R 2 up to order 2 has the form (recall 2 u a 1 x + a 2 u 2 2 x y + a 2 u 3 y + a u 2 4 x + a u 5 y + a 6u = v(x, y, (7.1 for some constants a k C; if we took a 1 = a 2 = a 3 = 0 this reduces to a first order PDE. In alternative (rather lighter notation, a 1 u xx + a 2 u xy + a 3 u yy + a 4 u x + a 5 u y + a 6 u = v. (7.2 Here, u and v are functions of two variables on R 2 u : R 2 R 1, (x, y u(x, y, v : R 2 R 1, (x, y v(x, y,

34 34 (you studied such functions in Calculus II cm112a; for example, v(x, y = x 2 + y 2. We will restrict our attention in specific examples to the homogeneous case: v = Definition of û We are going to Fourier transform just one of the variables. The one we should choose will be clear from the initial conditions (see below: let us FT the x- variable, so F = F x ξ Specifically, : functions on (x, y space functions on (ξ, y space }{{}}{{} = R 2 = R 2. u : R 2 R 1, (x, y u(x, y F û := F x ξ (u : R 2 R 1, (ξ, y û(ξ, y, where û(ξ, y := For notational convenience, we may use any of e iξx u(x, y dx. (7.3 û(ξ, y = F(u(ξ, y = F x ξ (u(ξ, y = F x ξ,y y (u(ξ, y (7.4 to denote the x-ft of u. The last two can be useful in reminding us what variable is being changed into what. The inverse Fourier transform F 1 ξ x := F 1 ξ x,y y : functions on (ξ, y space functions on (x, y space }{{}}{{} = R 2 is defined on w : R 2 R 1, (ξ, y w(ξ, y, by w (x, y := 1 2π ξ= Again, for notational convenience, we may use any of = R 2, e iξx w(ξ, y dξ. (7.5 w (x, y = F 1 (w(x, y = F 1 ξ x (w(x, y = F 1 ξ x,y y (w(x, y. ( Basic properties of F x ξ,y y : As in the case of ODEs, the two principal properties needed to use the FT to solve PDEs are: [1] Linearity: for (integrable u, v : R 2 R 1 and any λ, µ C, one has F x ξ,y y (λu + µv(ξ, y = λû(ξ, y + µ v(ξ, y. (7.7

35 35 [2] F x ξ,y y turns partial differentiation in x into multiplication by ξ: û x (ξ, y = iξ û(ξ, y. (7.8 Here, u x (x, y = u/ x is the partial x-derivative of u, and, crucially, we assume that u(x, y 0 as x. (7.9 You can easily check these properties yourself. Iterating the latter property we get for m N ( m u F (ξ, y = ( iξ m û(ξ, y, (7.10 x m provided that k u (x, y 0 as x for k = 0, 1,..., m 1. (7.11 xk This is needed to make the integration-by-parts in the x-variable work. One of the differences in the R 2 case is that there are now more types of derivatives to consider: mixed partial derivatives and partial derivatives in y alone. Plugging u y = u/ y instead of u into (7.3 we easily see that And hence, combining (7.8 and (7.12, û y (ξ, y = û(ξ, y. (7.12 y û xy (ξ, y = ( iξ û(ξ, y, (7.13 y (which is the same as û yx (ξ, y of course, whilst iterating (7.12 we have û yy (ξ, y = 2 û(ξ, y, (7.14 y2 and similarly for higher partial derivatives. Notice that for (7.12, (7.13, (7.14, we do not need to specify initial conditions of the form (7.11, because we are only differentiating the variable which is not being Fourier transformed so there is no integration by parts needed to prove these formulae, only integration under the integral sign. Of course, to integrate under the integral sign we need that u(x, y be sufficiently nice (specifically that u y exist and be continuous and be x-integrable. Indeed, we are generally ignoring the many interesting and subtle questions regarding which classes of multi-variable functions have well defined FTs; certainly there are well defined for large classes

36 36 of functions (such as Schwartz class functions and this is all we need to know for now. The other fact we need to know is how the FT affects initial conditions: suppose we require that u(x, 0 = f(x (7.15 for some specific one variable f : R 1 R 1 (for example, we might specify u(x, 0 = sin(x; that is, that when restricted to the x-axis the function u = u(x, y coincides with f. Then we have û(ξ, 0 := û(ξ, y y=0 = e iξx u(x, y dx y=0 = e iξx u(x, 0 dx = e iξx f(x dx = f(ξ. ( An example Solve the following PDE using Fourier transforms: subject to the conditions 6 u x 6 2 u y 2 = 0, u = u(x, y, and m u 0 as x for m = 0, 1, 2, 3, 4, 5 (7.17 xm u(x, 0 = x 2 and u (x, 0 = 0. (7.18 y Applying the Fourier transform F x ξ,y y in the x-variable to 6 u 2 u x 6 y 2 in view of the linearity, (see (7.7 of F x ξ,y y ( ( 6 u 2 u F x ξ,y y (ξ, y = F x 6 x ξ,y y (ξ, y. y 2 = 0 we get This is the same as ( 2 y + 2 ξ6 û(ξ, y = 0 (7.19 Exercise: Explain carefully why the answer involves integration by parts and differentiation under the integral sign.

37 37 But viewing this as an ODE in y, this has solution û(ξ, y = A(ξ cos(ξ 3 y + B(ξ sin(ξ 3 y, (7.20 with A(ξ, B(ξ constants in y, but which may depend on ξ. A(ξ = û(ξ, 0 and so In fact, we see We know that A(ξ = F x ξ (u(x, 0(ξ, where, recall, u(x, 0 = x 2. (7.21 (cf. (4.18, (4.19. On the other hand, we easily see ( 1 F x ξ (ξ = πe ξ ( x 2 B(ξ = 0. Exercise: why? So the solution in (ξ, y space is û(ξ, y = πe ξ cos(ξ 3 y. (7.23 We now have to apply the inverse FT to the ξ variable to transform this back to the solution in (x, y space, which will give us the solution to the PDE above. That is, u(x, y = 1 e iξx πe ξ cos(ξ 3 y dξ, (7.24 2π which we see reduces to the neater form u(x, y = 0 e ξ cos(ξx cos(ξ 3 y dξ. (7.25 Exercise: check you can see why (7.24 reduces to (7.25. Exercise: Work through in detail the step leading to (7.17, doing the integration by parts, or integration under the integral sign, specifying where any of the assumptions (7.17, (7.18 are applied. Optional Exercise: By stating suitable assumptions on u(x, y as x, write down the Fourier transform F x ξ,y y of the general PDE in (7.2 your answer should be a 2nd order ODE in y.