f(x)e ikx dx. (19.1) ˆf(k)e ikx dk. (19.2)

Transcription

1 9 Fourier transform 9 A first look at the Fourier transform In Math 66 you all studied the Laplace transform, which was used to turn an ordinary differential equation into an algebraic one There are a number of other integral transforms, which are useful in one or another situation Arguably, the most ubiquitous and general is the so-called Fourier transform, which generalizes the technique of Fourier series on non-periodic functions For the motivation of the Fourier transform I recommend reading the textbook or other possible references (see below), in these notes I start with a bare and dry definition Definition 9 The Fourier transform of the real valued function f of the real argument x is the complex valued function ˆf of the real argument k defined as ˆf(k) = f(x)e ikx dx (9) The inverse Fourier transform, which allows to recover f if ˆf is known, is given by f(x) = ˆf(k)e ikx dk (9) Remark 9 Since the Fourier transform plays a somewhat auxiliary role in my course, I will not dwell on it for very long There are a lot of available textbooks on Fourier Analysis, I would like to mention two additional sources: the lecture notes by Brad Osgood The Fourier transform and its applications (they are freely available on the web) for a thorough introduction to the subject, including careful coverage of the relations of the delta function and Fourier transform, and Körner, T W Fourier analysis Cambridge university press, 989 for the already initiated There are different definitions of the Fourier transform I use the one, which is used in the textbook You can also find in the literature where the following choices are possible: ˆf(k) = A e ibkx dx, A =, B = ±, A =, B = ±, A =, B = ± The only difference in the computations is the factor which appears (or does not appear) in front of the formulas Be careful if you use some other results from different sources 3 The notation is a nightmare for the Fourier transform Very often, together with the hat, the operator notation is used ˆf(k) = Ff(x), f(x) = F ˆf(k) Math 483/683: Partial Differential Equations by Artem Novozhilov artemnovozhilov@ndsuedu Spring 06

2 Both F and F are linear operators, since for them, from the properties of the integral, it holds that F(αf(x) + g(x)) = αff(x) + Fg(x), and a similar expression is true for F To emphasize that the pair f and ˆf are related sometimes something like is used f(x) ˆf(k) 4 In the definition of the Fourier transform I have an improper integral, which means that I have to bother about the convergence Moreover, note that the complex exponent, by Euler s formula, is a linear combination of sine and cosine, then my transform should be defined only for those f that tend sufficiently fast to zero This is actually a very nontrivial question, what is the space of functions, on which the Fourier transform is naturally defined, but this will not bother me in my course Moreover, there will be some examples, which would definitely contradict the classical understanding of the Fourier transform I will treat them in a heuristic way remembering that the rigorous justification can be made within the theory of generalized functions Consider several examples to get a feeling about the Fourier transform Example 93 (Fourier transform of the rect (for rectangle) function) Let {, x < a, Π a (x) = 0, otherwise By the definition (9) I have ˆΠ a (k) = ˆ a a e ikx dx = eika e ika = ik π It follows from the definition of the inverse Fourier transform that Π a (x) = sin ak e ikx dk π k sin ak k The last integral is quite difficult to evaluate directly, and the given definitions of the Fourier transform are often a direct way for some nontrivial integrals Note that the Fourier transform is even, which is not a coincidence Example 94 (The exponential decay) Let f r (x) = { 0, x 0, e ax, x > 0, where a is a positive constant Then ˆf r (k) = (a + ik),

3 which is complex valued even if k is real Similarly, for { e ax, x 0, f l (x) = 0, x > 0, I find ˆf l (k) = (a ik) Now I can easily calculate the Fourier transform for using the linearity of F: which is an even function if k is real Example 95 (Duality principle) Let I have f(x) = e a x = f l (x) + f r (x), ˆf(k) = ˆf r (k) + ˆf l (k) = f(x) = ˆf(k) = π x + a, a > 0 a k + a, e ikx x + a dx This integral is difficult to evaluate using only the basic knowledge from Calculus (actually, impossible) However, from the previous, I know that ( ) e a x = F a π k + a = a e ikx π k + a dk If I replace k with x and x with k I will get the integral (up to a multiplicative factor), that I know how to find Therefore I conclude that π e ˆf(k) a k = a This is actually a consequence of the striking resemblance of the Fourier and inverse Fourier transforms, the difference being just an extra minus sign I would like to formulate this important fact as a theorem Theorem 96 If the Fourier transform of f(x) is ˆf(k), then the Fourier transform of ˆf(x) is f( k) This theorem helps reducing the table of Fourier transforms in half, since if I know the Fourier transform ˆf of f this immediately means that I know the Fourier transform ĝ of the function g = ˆf To practice this theorem convince yourself that ( ) sin ax π F = x Π a(k) 3

4 Example 97 (Fourier transform of the delta function) Let f(x) = δ(x) Then ˆf(k) = ˆδ(k) = δ(x)e ikx dx = Hence the Fourier transform of the delta function is a constant function From here we can immediately obtain, invoking the duality principle, that the Fourier transform of the constant, say, is F() = δ(k), is the delta function! But stop, if I d like to use my definition (9) then the integral e ikx dk, strictly speaking, does not exist! Well, the exact meaning to this integral can be given within the framework of the generalized functions, but this will not bother us here 9 Properties of the Fourier transform Convolution Direct evaluation of the Fourier transform becomes very often quite tedious A list of the properties of the Fourier transform helps evaluating it in many special cases Shift theorem If f(x) has Fourier transform ˆf(k) then the Fourier transform of f(x ξ) is e ikξ ˆf(k) A very particular example of this property is Fδ(x ξ) = e ikξ Using the duality, the Fourier transform of e iηx f(x) is ˆf(k η) Dilation theorem If f(x) has Fourier transform ˆf(k) then the Fourier transform of f(cx), c 0 is ( ) c ˆf k c To practice this theorem let me find the Fourier transform of e ax I start with the Fourier transform of e x : ˆf(k) = e x ikx dx = where I used the fact that R e x dx = π Now, using the dilation theorem, I find F e (x ik/) k /4 dx = e k /4 ( e ax) = a e k 4a e y dy = e k /4, 4

5 3 Derivatives and Fourier transform If f(x) has Fourier transform ˆf(k) then the Fourier transform of f (x) is ik ˆf(k) Hence the Fourier transform turns the differentiation into an algebraic operation of multiplication by ik Immediate corollary is that the Fourier transform of f (n) (x) is (ik) n ˆf(k) By the duality principle or by a direct proof, the Fourier transform of xf(x) is i d ˆf dk 4 Integration and Fourier transform If f(x) has Fourier transform ˆf(k) then the Fourier transform of its integral g(x) = x f(s) ds is ĝ(k) = i k ˆf(k) + π ˆf(0)δ(k) Using this property I can immediately find Fourier transform of the Heaviside function χ(x): Fχ(x) = i π + k δ(k) Since then sgn x = χ(x) χ( x), F sgn x = i π k Exercise Prove all four properties of the Fourier transform 5 Convolution Now let me ask the following question: if I know that f(x) ˆf(k) and g(x) ĝ(k) then which function has the Fourier transform ˆf(k)ĝ(k)? I have ˆf(k)ĝ(k) = = = = = So, if I introduce a new function then I showed that Now I can formally state my result f(x)e ikx dx g(y)e iky dy = f(x)g(y)e ik(x+y) dx dy = ( ) g(y)e ik(x+y) dy f(x) dx = [x + y = s] ( ) g(s x)e iks ds f(x) dx = ( ) e iks g(s x)f(x) dx ds h(s) = g(s x)f(x) dx, ˆf(k)ĝ(k) = ĥ(k) 5

6 Definition 98 The convolution of two functions f and g is a function h = f g defined as h(s) = f g = Basically, by the above reasoning I proved Theorem 99 Let h = f g Then f(s x)g(x) dx ĥ(k) = ˆf(k)ĝ(k) In the opposite direction, the Fourier transform of the product of two functions u(x) = f(x)g(x) is û = ˆf ĝ The second part of the theorem can be proved in a similar way or using the duality principle It is instructive to prove that the convolution is commutative (f g = g f), bilinear f (ag +bh) = af g + bf h, and associative f (g h) = (f g) h 6