Distributions, the Fourier Transform and Applications. Teodor Alfson, Martin Andersson, Lars Moberg

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1 Distributions, the Fourier Transform and Applications Teodor Alfson, Martin Andersson, Lars Moberg 1

2 Contents 1 Distributions Basic Definitions Derivatives of Distributions Direct Products and Convolutions Tempered Distributions The Fourier Transform On L 1 (R n ), S(R n ) and S (R n ) Properties of the transform The Fourier Transform and the δ-function Applications for Partial Differential Equations Solving Non-Homogeneous PDE s by Distributions The Heat Equation and the Fourier Transform

3 1 Distributions 1.1 Basic Definitions In the study of differential equations there arises the need of concepts of generalized functions, derivatives and convergence (among other things). For example, consider the ODE initial value problem: { 1 u (x) = f ɛ (x) = ɛ if 0 < x < ɛ 0 else u(0) = 0 For this system we can see that 0 if x 0 x u(x) = ɛ if 0 < x < ɛ 1 if x ɛ is a solution. Notice that the limit of this function as ɛ tends to inifinity u exists, but can we make any sense of the limit of f ɛ? It is defined on shorter and shorter intervalls and integrates to 1 but the limit function also tends to infinity. In this section we shall try to find a useful way of dealing with this situation, by our generalised functions, and later on we will show how these can help us solve interesting PDE s and how we can even define a notion of fourier transform on this much broader class of functions. The idea, in some sense, is to define our generalised functions by how these function s action on a class of nicely behaved test functions. Let us make the following definition: Definition A function f is called a test function if f is in C (Ω) and the support of f is compact. This set of test functions is denoted by D(Ω). We will also define a notion of convergence on this set D(Ω). Definition If {φ n } D(Ω) we say say {φ n } converges to φ D(Ω) if there is a compact subset K Ω such that the support of all the φ n and φ is in K and the derivatives of φ n of arbitrary order converge uniformly to those of φ We have already come across a function in the set D(Ω) in our PDE course. Recall that in the handout about mollifiers, we proved the function { 0 if x 1 φ(x) = e 1 1 x 2 if x < 1 is actually in D(Ω). Now we are ready to define our generalised functions, which are called distributions. 3

4 Definition A distribution f is defined as a linear mapping from D(Ω) into R, notated by (f, φ), which is continuous in the sense that if φ n φ then (f, φ m ) (f, φ). We denote the space of these distributions by D (Ω). For example, if f is a continuous function on Ω, then (f, φ) = f(x)φ(x)dx, Ω is indeed continuous in the sense of the above definition, and thus a distribution. But we can do alot more interesting things than this. Let us define a very important and applicable distribution, the Dirac delta function: (δ, φ) = φ(0). That the functional above is continuous follows from the fact that the uniform convergence φ n also implies pointwise convergence. We will, somewhat horrifically, write (δ, φ) = δ(x)φ(x)dx, Ω even though this does not quite make any sense, since we cannot plug in an x in δ(x). There are more interesting examples of functionals, but we will focus on δ, since it will be very useful later on. We also define (f + g, φ) = (f, φ) + (g, φ), and (αf, φ) = (f, αφ), for functionals f, g and any real number α. 1.2 Derivatives of Distributions We have managed to create a set of generalised functions. Let us now try to make sense of what the derivative of a distribution should be. The key to generelise the notion of a derivative is by looking at integration by parts. We get the following definition: Definition ( ) ( f, φ = f, φ ). x i x i Then if f C(Ω), this definition agrees with the classical definitin of derivative in respect to integration by parts. Recursively, we also find that (D α f, φ) = ( 1) α (f, D α φ) 4

5 Then For an example, let us look at the function { 0 if x 0 H(x) = 1 if x > 1. (H, φ) = (H, φ ) = 0 φ dx = φ(0) = (δ, φ) H is known as the heavyside function, and then as we can see above, the derivative of H is apparently the Dirac delta function. Interesting! 1.3 Direct Products and Convolutions What we are aiming to do in this section is find a way to convolute distributions, but before we can get there we shall define the direct product of distributions. It is not always possible to define the product of two distributions f(x) and g(x), but if we put the condition that they use different variables, then we can: Definition If f D (R ) and g D (R ), then the direct product f(x)g(y) is the distribution over on R + given by (f(x)g(y), φ(x, y)) = (f(x), (g(y), φ(x, y))). In the inner functional, we treat y as the variable and x as a parameter, so (g(y), φ(x, y)) is a function of x. We will not go into details, but everything above is well-defined. As an example, (δ(x)δ(y), φ(x, y)) = (δ(x), (δ(y), φ(x, y))) = (δ(x), φ(x, 0)) = φ(0, 0) = (δ(x, y), φ(x, y)), hence δ(x)δ(y) = δ(x, y). We also see that if ψ D(R n ) and f D (R n ), the convolution f ψ makes sense, and one can show that f ψ, φ = f(x), ψ(y x) φ(x) dy, R n ) i.e. f ψ(y) = f(x)ψ(y x) dx. R Another consequence n of the definition is that (f(x)g(y), φ) = (g(y), (f(x), φ(x, y))) and, in a sense, we have that f(x)g(y) = g(y)f(x). 5

6 Now onto convolutions. Suppose that f and g are functions in C(R n ) that decay rapidly at infinity, then: (f g, φ) = (f g)φ(x)dx = R n = f(x y)g(y)φ(x)dxdy R n R n = f(x)g(y)φ(x + y)dxdy. R n R n Let us use this as our inspiration and try out a definition of convolution of distributions. Definition Let f, g D (R n ), then (f g, φ) = (f(x)g(y), φ(x + y)). This however is quite nonsensical, since f(x)g(y) is a distribution in D (R 2n ), but φ is only in D(R n. However, apparently, this definition actually makes some sense under special circumstances. For instance: (δ f, φ) = (δ(x)f(y), φ(x + y)) = = (f(y), (δ(x), φ(x + y))) = (f, φ(y)), so then δ f = f. We can also differentiate these convolutions, and it is not hard to show that: (D α (f g), φ) = (D α f g, φ) = (f D α g, φ) 1.4 Tempered Distributions The space of distributions in D (R n ) will actually be somewhat limiting in our endeavours further on, but we can find a larger very nice space very similarly. Instead of having compact support, we will exchange our test functions for a class of functions which very rapidly decay at infinity. More precisely: Definition Let S(R n ) be the space of all complex-valued functions f over R n such that f C and x k D α φ(x) is bounded. A sequence {φ n } in S(R n ) converges to φ if the derivatives of all order of the φ n converge uniformly to those of φ and the constants C kα in the bounds x k D α φ(x) C kα does not depend on n. We can now define tempered distributions: Definition A tempered distribution f is defined as a linear mapping from S(R n ) into R, notated by (f, φ), which is continuous in the sense that if φ n φ then (f, φ m ) (f, φ). We denote the space of these distributions by S (R n ). 6

7 By the definition above, clearly D (R n ) is contained in S (R n ). Also, and very importantly, we can define derivatives, direct products and convolutions, completely analogously as we did with D (Ω). So for instance, for f S (R n ), we define the derivative of f by ( ) ( f, φ = f, φ ). x i x i From now on, for mysterious reasons, we will use the notation, for the linear mapping previously denoted (, ). 7

8 2 The Fourier Transform The Fourier transform is an often used tool in linear analysis, which allows us to convert linear differential operators, with constant coefficients, into multiplication by polynomials. Here we will define it on three different spaces. 2.1 On L 1 (R n ), S(R n ) and S (R n ) We start with the space L 1 (R n ). Definition For all f L 1 (R n ), we define the Fourier transform F as F [f](ξ) = f(ξ) = 1 f(x)e ix ξ dx. (2π) n 2 R n It is not guaranteed that f is still in L 1 (R n ). However, one sees that F : L 1 (R n ) L (R n ), since f(ξ) 1 f(x) dx < f(ξ) L (R n ). (2π) n 2 R n From earlier we learned about the space S(R n ). This space has the nice property that it is closed under the Fourier transform: F : S(R n ) S(R n ). This is important, since it allows us to use both the Fourier transform and its inverse freely without ending up with a function that is not in S(R n ). Definition For all f L 1 (R n ) we define the the operator F [f](x) = 1 f(ξ)e ix ξ dξ. (2π) n 2 R n This may remind us of an inverse Fourier transform, but since we do not necessarily have F [f] (or F [f]) in L 1 (R n ), we cannot strictly call it an inverse. At least not yet. However, if both f, f L 1 (R n ), then F is the inverse Fourier transform. Lemma 2.1 For all u, v S(R n ), we have F [u], v = u, F [v], where the inner product is that of L 2 (R n ): u, v = R n u(x)v(x) dx. Theorem 2.2 For all f S(R n ), we have that F [F [f](x)] = f(x). 8

9 For proof, see [8] (on Salvador s handout). By this theorem, we have found that F is the inverse Fourier transform in S(R n ). Another theorem follows quite naturally from lemma 2.1 and theorem 2.2, since for all u, v S(R n ) we have û, v = u, F [F [v]] = u, v, where the inner product is still that of L 2 (R n ). Theorem 2.3 F : S(R n ) S(R n ) is an isomorphism, with inverse F, preserving the L 2 (R n ) inner product. Now we have a pretty good picture of how the Fourier transform behaves on both L 1 (R n ) and S(R n ). But what about the set of tempered distributions S (R n )? Definition Define F : S (R n ) S (R n ) as the tempered distribution given by T, u = T, û, for all u S(R n ), with an analogous definition of F. 2.2 Properties of the transform We will now work our way through some properties of the transform. We let T S (R n ) and v S(R n ) throughout. 1. Our first property of the Fourier transform, is translation: If a (R n ), and F [T (x a)] = e ia ξ T, F [e ia x T ] = T (ξ a). 2. The second property is rescaling: If h R n, h 0, F [T (hx)] = 1 ( ) h T ξ n. h 3. The third property is about derivatives: F [ xj T ] = iξ j T, and We have that F [x j T ] = i ξj T. F [ xj T ], v = xj T, v = T, xj v = T, F [iξ j v] = iξ j T, v. 9

10 The second formula can be shown in a similar way. 4. The final property about the Fourier transform we will list is that of convolution: F [T v] = (2π) n 2 T v. For u, v S(R n ), we have that To show this, F [u v] = 1 (2π) n 2 = 1 (2π) n 2 = 1 (2π) n 2 = 1 (2π) n 2 F [u v] = (2π) n 2 û v. = (2π) n 2 û v. R n e iξ x u(x y)v(y) dy dx R n v(y) e iξ x u(x y) dx dy R n R n v(y) e iξ (z+y) u(z) dz dy R n R n e iξ y v(y) e iξ z u(z) dz dy R n R n Here we use Fubini s Theorem in order to exchange the order of integration. We will see this again later, in part The Fourier Transform and the δ-function If we apply the Fourier transform on δ S (R n ) we have, for u S(R n ): δ, u = δ, F [u] = F [u](0) = 1 1 u(x) dx =, u, (2π) n 2 R n (2π) n 2 i.e. the Fourier transform of δ is the constant Fourier transform on 1, we have that 1, u = 1, û = 1. If we now apply the (2π) n 2 R n û dξ = (2π) n 2 u(0) = (2π) n 2 δ, u, which means that 1 = (2π) n 2 δ. Combining this with the following formulas, D α û = F [( ix) α u], (iξ) α û = F [D α u], obtained by differentiating under the integral and integrating by parts respectively, we now see that the Fourier transforms of polynomials are linear combinations of derivatives of the δ-function, and vice versa! 10

11 3 Applications for Partial Differential Equations Both distributions and Fourier transforms are efficient tools in the theory of partial differential equations (PDE s). We will in this section give a few examples of applications of said tools. Some inspiration for this part was found in [1]. 3.1 Solving Non-Homogeneous PDE s by Distributions Consider the non-homogeneous PDE, Lu(x) = f(x), where x R n, and L is a linear differential operator of order m with constant coefficients, i.e. Lu(x) = a α D α u(x), α m where α is a so called multi-index, and all a α are assumed to be constant. Suppose we can solve the specific case when f(x) is replaced by the δ-distribution; Lu(x) = δ(x). (3.1) Let K(x) be the solution to (3.1), we then say that K, as a distribution, is the fundamental solution of L. We also have that where LK(x), v(x) = K(x), L v(x), L v(x) = ( 1) α a α D α v(x), α m for v D (R n ). L is called the adjoint operator to L. In the case when L = L we say that L is symmetrical, or self-adjoint. We shall now see how we can utilize convolution efficiently to solve the PDE. Consider f(x) as a distribution and form the convolution of the distributions K and f: ũ(x) := K(x) f(x), where we assume that the convolution is meaningful. Now, Lũ(x) = a α D α ũ(x) = a α D α (K(x) f(x)) α m α m α m = a α D α K(x) f(x) = LK(x) f(x) = δ(x) f(x) = f(x), 11

12 where we have used the properties of convolutions, described in section 1. We have arrived at the conclusion that, as a distribution, ũ(x) = K(x) f(x) is a solution to Lu(x) = f(x). Example Consider the case when L =, and the equation u(x) = f(x). The fundamental solution is then given by K(x) = δ(x). It is easy to see that is symmetrisk, i.e. K, v = K, v. Since δ is radially symmetric, one can guess that K also has radial symmetry, i.e. K(x) = Φ(r) and consider the fact that Φ(r) is harmonic for r > 0. Then, from K, v = v(0), we can completely determine the fundamental solution Φ(r) of the Laplace equation. Now, as before, we have ũ = K f as the distribution solution to u = f, since (K f) = K f = δ f = f(x). 3.2 The Heat Equation and the Fourier Transform As we have seen, the Fourier transform can be quite useful in the theory of PDE s. In this section we apply it to the heat equation. Consider the initial value problem (IVP): { ut (x, t) x u(x, t) = 0, t > 0, x R n, u(x, 0) = g(x), x R n (3.2). Assume that for every t > 0, we have u R n, and consider û t (ξ, t) = 1 e iξ x u (2π) n t (x, t) dx = t), 2 R n tû(ξ, where interchanging the order of integration/differentiation is allowed if u, u t are bounded and continuous in x and t. This requires g(x) to be continuous and bounded. We now apply the Fourier transform in the space variables. From the properties of the transform in the previous section, we have u = ξ 2 û. We now have the transformed IVP: { tû(ξ, t) = ξ 2 û(ξ, t), ξ R n, t > 0, û(x, 0) = ĝ(ξ), ξ R n, (3.3) 12

13 which is an ordinary linear differential equation in t with constant coefficients. The solution to (3.3), for t 0, is û(ξ, t) = ĝ(ξ)e ξ 2t. Since u S(R n ), we get û S(R n ) and hence we can take the inverse Fourier transform. We get: u(x, t) = 1 e iξ x û(ξ, t) dξ = 1 e iξ x ĝ(ξ)e ξ 2t dξ. (2π) n 2 R n (2π) n 2 R n But ĝ(ξ) = 1 (2π) n 2 R e iξ y g(y) dy, so we get: n u(x, t) = 1 (2π) n e iξ (x y) ξ 2t g(y) dy dξ R n R n = 1 [ ] (2π) n e iξ (x y) ξ 2t dξ g(y) dy. R n R n Here we have used Fubini s Theorem in order to change the order of integration. We can use this theorem if e ξ 2t g(y) dy dξ <, R n R n which is true if R g(y) dy <, i.e. if ĝ is bounded. n We now introduce K(x, y, t) := 1 (2π) n e iξ (x y) ξ 2t dξ. R n We then get u(x, t) = K(x, y, t)g(y) dy. R n We can calculate K as follows: n n K(x, y, t) = 1 i (x j y j )ξ j t ξ 2j ( ) e j=1 j=1 dξ1 (2π) n ξ n R n = 1 n (2π) n e i(x j y j )ξ j tξj 2 dξ j. But, j=1 e i(x j y j )ξ j tξ 2 j dξ j =e 1 4t (x j y j ) 2 =e 1 4t (x j y j ) 2 π t. e t(ξ j i 2t (x j y j )) 2 dξ j 13

14 Putting all of this together, we get K(x, y, t) = 1 ( π (2π) n t 1 = (4πt) n 2 ) 1 n 4t 2 e e x y 2 4t. n (x j y j ) 2 K is the so called heat kernel. So, for the solution u to (3.2), we finally get: j=1 u(x, t) = 1 (4πt) n 2 e x y 2 4t g(y) dy. R n We also note that the heat kernel is itself a solution to the equation u t u = 0, and in a distributional sense the initial condition is often written as K(x, y, 0) = δ(x y). Then, u(x, 0) = K(x, y, 0)g(y) dy = R n δ(x y)g(y) dy = g(x), R n which is the initial condition in (3.2). 14

15 References [1] R. C. McOwen, Partial Differential Equations: Methods and Applications, 2nd edition, Pearson Education Inc. 15

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