1 From Fourier Series to Fourier transform

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1 Differential Equations 2 Fall 206 The Fourier Transform From Fourier Series to Fourier transform Recall: If fx is defined and piecewise smooth in the interval [, ] then we can write where fx = a n = b n = cos nπx = 2 n= fx cos nπx fx sin nπx dx, n = 0,, 2, 3,... dx, n =, 2, 3,... We will now simplify things by complexifying them. Well, they might not get simpler, but they ll get easier to manage,i think. In the Fourier series we are going to write e nπix + e e nπix e,, sin nπx = 2i where i = is the imaginary unit. It might be better from a rigor point of view to work with a partial sum of the series and get n= = = n= n= 2 a n e nπix 2 a n + 2i b ne nπix + e + 2i b n e nπix N + 2 a n 2i b ne n= e In the second summation instead of adding exp in/ from to N, we can add expin/ from N to ; that is, we have 2 a n 2i b ne = 2 a n 2i b ne nπix. n= n= We also will use that /2i = i/2 so that now we can write our sum in the form n= If we now define complex numbers by we can write = c n = n= N 2 a n + ib n e nπix + 2 a n ib n, n =, 2, 3, 4,..., 2 a 0, n = 0, 2 a n + ib n, n =, 2, 3, 4,... n= = N n= N etting now N, we get the complex form of the Fourier series, namely fx = n= c n e nπix. c n e nπix. n= 2 a n ib n e nπix

2 FROM FOURIER SERIES TO FOURIER TRANSFORM 2 There is no need to refer to the real version. The coefficients can be calculated directly. I ll give two version ending in the same result. Version. Suppose first n > 0. Then If n < 0 let n = m. We get c n = 2 a n ib n = 2 = fx cos nπx 2 c n = 2 a m + ib m = 2 = fx cos mπx 2 If n = 0, we just have fx cos nπx dx i fx sin nπx i sin nπx dx = 2 fxe fx cos mπx dx + i fx sin mπx + i sin nπx dx = 2 c 0 = 2 a 0 = fx dx. 2 dx fxe mπix dx == 2 These three cases can be packed into just one. It is the same formula in each case; that is 2 c n = fxe dx, n = 0, ±, ±2,... 2 dx dx fxe dx. Version 2. In this version we use the orthogonality of the functions {e nπix/ } n=0,±,±2,... : { e nπix e mπix 2, n = m, dx = 0, n m. One multiplies the two sides of by e imπx/, integrates from to to get fxe mπix dx = 2cm. The rigorous justification of what comes next escapes the level of this course. We want to let. If we replace each c n in by its expression in 2 we get fx = n= fxe nπiy nπix dy e. 2 The idea here is that as the summation becomes a Riemann improper integral because we are dividing the whole line into segments of length / 0. In the inside integral nπ/ becomes a continuous variable that we, following Hancock, will denote by the Greek letter ω. Of course, the π in the denominator does not allow itself to be completely suppressed. The end result is fx = e ixω e iyω fy dy dω. This equation is actually only true for very nice function, functions that are smooth and very small at ±. Hancock prefers to use a slightly different formula; we can change variables in the inner integral by y = y, in the outer integral by ω = ω, and once we are done replace y, ω again by y, ω to get 3 fx = e ixω e iyω fy dy dω.

3 2 AN INTERMEZZO, PROFESSOR HANCOCK S DERIVATION OF THE HEAT KERNE; MY VERSION3 Definition The Fourier transform of the function fx, defined for < x < is the function ˆf of ω defined by the formula 4 ˆfω = e ixω fx dx, assuming the integral in 4 makes sense converges. ooking at equation 3, the original function f can then be recovered from it Fourier transform ˆf by 5 fx = e ixω ˆfω dω. Notes on this definition. First, a warning. Different authors define the Fourier transform differently. The difference is cosmetic, but one should know what it is. For example, the preferred form for mathematicians is to leave the +, where it was in the formula preceding 3 so that the Fourier transform has e ixω in the integrand, the inverse transform given by 5 having e +ixω. It is also common to try for a more symmetric form, in which the factor of / in 3 gets split between the transform and its inverse, so there is a factor of / in front of the integral in 4, and the factor of / of 5 gets replaced by /. Physicists like to make a change of variable so the factors e ±ixω become e ±ixω. This is actually quite natural, because the interpretation of ˆfωe ixω is that of a signal of amplitude ˆfω, frequency ω. You need the so as to have ω bean actual frequency; otherwise the frequency is ω/. Finally, one thing to consider is that if f is a real valued function, except if f happens to be also even, its Fourier transform ˆf will not be real valued; it will assume non-real complex values. 2 An intermezzo, Professor Hancock s Derivation of the Heat Kernel; my version We want to solve the following problem: 6 u t = u xx, < x <, t > 0, satisfying the initial condition ux, 0 = fx. We assume f is a relatively nice function; one that doesn t get to big as x ±. Not knowing any better, we separate variables looking for all non-zero solutions of the form ux, t = XxT t, but discarding those that get too large as x or x. Since we are sophisticated people, we are going to work with complex valued solutions. The standard separation of variables gives 7 8 X + λx = 0 T + λt = 0 If λ < 0 we get real exponential solutions that get very large for large x. So we will concentrate on the case λ = ω 2, at first with ω > 0 The general solution of X + ω 2 X = 0 can be written in complex notation in the form Xx = C e iωx + C 2 e iωx, for constants C, C 2. Our little non-hancockian trick will now be to look at the two solutions e iωx, e iωx separately. This amounts to saying that the solutions to X + λx = 0 are given up to a constant factor by the functions X ω x = e iωx, where ω is a real number, not necessarily positive; and λ = ω 2. This includes even the bounded solution for λ = 0, namely the constant solution. So for every ω, we have the solution u ω x, t = e ω2t e iωx

4 2 AN INTERMEZZO, PROFESSOR HANCOCK S DERIVATION OF THE HEAT KERNE; MY VERSION4 of 6. The t part was obtained as usual, solving T + λt for λ = ω 2. To get a general solution instead of adding over all the ω s not possible! we integrate, after multiplying by a constant depending on ω that I will call gω. We get 9 ux, t = e ω2t e iωx gω dω.. It can be shown that if ω2 gω dω <, then the function u defined by 9 is at least once differentiable with respect to t, twice with respect to x, and satisfies6. It remains to select g so the initial condition holds; that is, select g so that fx = ux, 0 = e iωx gω dω. If we look at our definition of Fourier transform and its inverse, we see that f should be the Fourier transform of g, except for the notation switch; ω is the variable of integration here, but x is the transformed variable. The roles of x and ω are switched. This means that we should have gω = e iωy fy dy. I used y to denote the variable of integration to distinguish it from the free variable x appearing in the solution u. Plugging this into the expression 9, we get ux, t = e ω2t e iωx e iωy fy dy dω = e ω2t e iωx y dω fy dy. We will compute now the inner integral in the last expression; that is e ω2t e iωx y dω. To simplify, we will write z = x y and look at the integral as a function of z and compute Iz = e ω2t e iωz dω To compute it we will have to know its value for z = 0; that is we will use the following well known by all people who know it well integral 0 e x2 dx =. I show how to calculate this in an Appendix to these notes. We notice that for t > 0 I z = i ωe ω2t e iωz dω = i = 2t e ω2t e iωz In the integration by parts we used that + i 2t 2t e ω2t e iωz d t e iωz dω dω 2t e ω2 e ω2 t d dω eiωz dω = i2 z 2t = lim 2 R 2t e R t e irz e irz = 0. e ω2t e iωz dω = z 2t Iz. So I satisfies the ODE in z I +z/2ti = 0. This is a first order linear ODE; its general solution is Iz = Ce z2 /4t, where C is a constant. To determine C we set z = 0 and equate to get C = I0 = e tω2 dω = e s2 ds = t t.

5 3 PROPERTIES OF THE FOURIER TRANSFORM 5 Variables were changed by s = t ω and 0 used. We have that 2 Iz = 4t. t e z Replacing z with x y and using this in the last expression we had for u we get ux, t = 2 t We can also write this in the form 2 ux, t = 2 t where 3 K t x y = e x y2 4t is the so called heat kernel. 3 Properties of the Fourier transform e x y2 4t fy dy. K t x yfy dy, It is sometimes convenient to denote the Fourier transform of f, by Ff, so that Ffω = ˆfω = The inverse Fourier transform of a function g is then F gx = e iωx fx dx. e iωx gω dω. Very, very roughly and it can be made quite smooth by invoking some mathematical results that here at FAU are only touched upon in graduate courses one has that inverse Fourier transform undoes what the Fourier transform does, and vice versa. That is F Ff = f, FF g = g. Notice also that the inverse Fourier transform, apart from the factor of /, is almost the same as the transform, so that applying twice the Fourier transform almost takes you back to origin. But we are dealing here with integrals from to, such integrals do not always make sense. The simplest condition under which ˆf is defined for all real ω is 4 fx dx <. It is, in fact, a Calculus 2 exercise that if 4 holds, then both improper integrals fx cosωx dx, fx sinωx dx converge for all real values of ω and, after all, using e iωx = cosωx + i sinωx, one can break ˆf as follows: ˆfx = fx cosωx dx + i fx sinωx dx. We now have the following result: If fx dx <, then ˆfω exists the integral defining it converges for all values of ω. If in addition ˆfω dω <, then f can be recovered from ˆf by applying the inverse transform; that is, fx = F ˆfx = e iωx ˆfω dω.

6 3 PROPERTIES OF THE FOURIER TRANSFORM 6 It may be time for a few examples. Generally speaking Fourier transforms are not easily computed. There are tables of such transforms; if one is lucky{ the one one is looking for is tabled. But some are easy., a < x < a, Example. et a > 0 and let fx = 0, x < a, or x > a. Then a ˆfω = e iωx dx = iω eiωa e iωa = 2 sinaω. ω a Trying to invert this, checking if one gets back f by applying the inverse transform is hard. One can show it happens if one uses the transform in a more generalized sense. The problem is that ˆf does not satisfy 4. Example 2. et fx = e ax2, where a > 0. We already computed its transform. It is just Iz in disguise. Recall, we had Iz = e ω2t e iωz 2 dω = 4t. t e z If we change the notation, t becomes a, ω becomes x and z becomes ω and we get 2 ˆfω = 4a. Because this function is almost its own Fourier transform; it is actually easy to verify that applying the inverse transform restores the original function. The main properties of the Fourier transform are:. The Fourier transform is linear. That is, if f, g have Fourier transform satisfy 4, for example then so do f ± g, af for constant a and Ff ± g = ˆf ± ĝ, Faf = a ˆf. a e ω 2. If f, f have Fourier transforms, then ˆf ω = iω ˆfω. 3. if all integrals defining the objects below converge, then d ˆf ω = F {ixf} ω. dω We ll try to justify all these properties in class. The importance of the last two is that the Fourier transform changes differentiation, an analytic property, into multiplication by the variable, a much simpler algebraic operation. In this way it is not too different form the aplace transform. In fact, one can show that in a sense the aplace transform is a particular case of the Fourier transform. To be continued