Fourier transforms. Up to now, we ve been expressing functions on finite intervals (usually the interval 0 x L or L x L) as Fourier series: a n cos L

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1 ourier transforms Motivation and definition Up to now, we ve been expressing functions on finite intervals (usually the interval 0 x or x ) as ourier series: fpxq a 0 8 nπx a n cos n b n sin nπx where a 0» fpxq dx 2 and a n» fpxq cos nπx dx, b n» fpxq sin nπx dx. We also occasionally thought about the complex exponential version of ourier series: Since e iθ cos θ i sin θ and e iθ cos θ i sin θ, or equivalently cos θ eiθ e iθ 2 and sin θ eiθ e iθ, 2i we can rewrite the above series as: fpxq a 0 e 0ix a 0 e 0ix 8 n 8 e nπix{ a n n 8 n a n c n e nπix{ ib n 2 e nπix{ 2 e nπix{ b n e nπix{ e nπix{ 2i a n ib n e nπix{ 2i where $ '& c n '% 2 n ib n q for n 0 a 0 for n 0 2 n ib n q for n 0

2 2 fourier transforms Using the formulas for a n and b n given above, we see that, for n 0. If n c n 2 pa n ib n q» nπx fpxq cos dx i» 2 2» nπx nπx fpxq cos i sin 2 0 we have 2» fpxqe nπix{ dx. fpxq sin dx nπx dx c n 2 pa n 2 2 2»»» ib n q fpxq cos nπx i dx 2 nπx fpxq cos i sin fpxqe nπix{ dx» nπx fpxq sin dx nπx dx because cosine is an even function and sine is odd. So the same formula works for all the coefficients (even c 0 ) in this case and we have 8 fpxq c n e nπix{ where c n» fpxqe nπix{ dx. 2 n Equivalently, we could write: 8 fpxq 2 c ne nπix{ where c n n» fpxqe nπix{ dx. What we want to do here is let tend to infinity, so we can consider problems on the whole real line. To see what happens to our ourier series formulas when we do this, we introduce two new variables: ω nπ{ and ω π{. Then our complex ourier series formulas become fpxq 8 n ω c ne iωx where c n» fpxqe iωx dx and the n in the formula for c n is hiding in the variable ω. We can rewrite these as 8» iωx ω fpxq c ω e where c ω fpxqe iωx dx. n

3 math The variable ω nπ{ takes on more and more values which are closer and closer together as Ñ 8, so c ω begins to feel like a function of the variable ω defined for all real ω. ikewise, the sum on the left looks an awful lot like a Riemann sum approximating an integral. What happens in the limit as Ñ 8 is: fpxq cpωqe iωx dω where cpωq 8 fpxqe ixω dx. The formula on the right defines the function cpωq as the ourier transform of fpxq, and the formula on the left defines fpxq as the inverse ourier transform of cpωq. ourier transform: fpωq pωq rfpxqspωq fpxqe ixω dx Inverse ourier transform: q pxq fpxq r pωqspxq pωqe ixω dω These formulas hold true (and the inverse ourier transform of the ourier transform of fpxq is fpxq the so-called ourier inversion formula) for reasonable functions fpxq that decay to zero as x Ñ 8 in such a way so that fpxq and/or fpxq 2 has a finite integral over the whole real line. There are many standard notations for ourier transforms (and alternative definitions with the minus sign in the ourier transform rather than in the inverse, and with the factor in different places, so watch out if you re looking in books other than our textbook!), including and fpωq pωq rfpxqspωq q pxq fpxq r pωqspxq fpxqe ixω dx pωqe ixω dω (which is the one in the current Math 24 textbook, I think). Properties and examples. The ourier transform is an operation that maps a function of x, say fpxq to a function of ω, namely rfspωq ˆfpωq. It is clearly a linear operator, so for functions fpxq and gpxq and constants α and β we have rαfpxq βgpxqs α rfpxqs β rgpxqs. Some other properties of the ourier transform are

4 4 fourier transforms. Translation (or shifting): rfpx aqs pωq e iωa rfpxqs pωq. And in the other direction, re iax fpxqs pωq rfpxqs pω aq. 2. Scaling: f x a a pωq rfpxqs paωq, and likewise rfpaxqs pωq a rfpxqs ω. a 3. Operational property (derivatives): rf pxqs pωq iω rfpxqs pωq, and rxfpxqs pωq p rfpxqs pωqq. i d dω The operational property is of essential importance for the study of differential equations, since it shows that the ourier transform converts derivatives to multiplication so it converts calculus to algebra (or might reduce a partial differential equation to an ordinary one). Here are the proofs of the first of each of the three pairs of formulas to give a sense of how to work with ourier transforms, and leave the other three as exercises. or the first shifting rule, we make the substitution y x a (so dy dx and x y a) to calculate rfpx aqs pωq fpx aqe iωx dx fpyqe iωy e iωa dx e iωa rfpxqs pωq or the first scaling rule, we make the substitution y x{a (so dx a dy) and get x a f x pωq a a f e iωx dx a fpyqe iaωy dy rfpxqs paωq or the operational property we first point out that since the ourier transforms of both f pxq and fpxq exist, we must have that fpxq Ñ 0 and f pxq Ñ 0 as x Ñ 8. Therefore the endpoint terms will vanish when we integrate by parts (with u e iωx and dv f pxq dx, so du iωe iωx and v fpxq): rf pxqs pωq f pxqe iωx dx x8 e iωx fpxq 0 iω x iω rfpxqs pωq fpxqe iωx dx iωfpxqe iωx dx

5 math et s calculate a few basic examples of ourier transforms: Example. et S a pxq be the function defined by " if x a S a pxq 0 otherwise Then rs a pxqs pωq» a a e iωx dx e iωa e iωa iω 2 sin aω. ω Example 2. et upxq e ax2 {2, so the graph of upxq is a Gaussian or bell-shaped curve. Then upxq satisfies the differential equation u axu 0. We can use this fact and the properties of the ourier transform to calculate û as follows: Take the ourier transform of the differential equation and use linearity and both parts of property (3) above to get 0 ru axus pωq ru s a rxus d rus iω rus ai dω Therefore rus satisfies the differential equation the solution of which is rus ω rus 0 a rus Ce ω2 {p2aq. The constant C is the value of rus p0q, i.e., c 2 C e ax2{2 dx a e y2 dy c a, using the substitution y a a x and the familiar (or at least accessible) fact that ³ 2 8 e y2 dy? π. Therefore c e ax2 {2 pωq {p2aq a e ω2, so the original Gaussian is transformed into a different one. An interesting observation is what happens for a : Then we have e x2 {2? e ω2{2,

6 6 fourier transforms x2{2 so the specific Gaussian e is eigenvalue?. an eigenfunction of the ourier transform with Example 3. et fpxq e a x, so fpxq " e ax if x 0 e ax if x 0. Then e a x pωq» 0» 0 e a x e iωx dx e ax e iωx dx 0 e aω e iωx dx e pa iωqx dx e p a iωqx dx 0 epa iωqx x0 e p a iωqx x8 a iω x a iω x0 a iω a iω 2a a 2 ω 2 (the limiting values of the exponentials at 8 are zero because e a x goes to zero as x goes to 8 and e iωx stays bounded). An observation. Because the formulas for the ourier transform and the inverse ourier transform are so similar, we can get inverse transform formulas from the direct ones and vice versa. In particular, note that if we let y x then ikewise rfpxqs pωq r pωqs pxq fpxqe iωx dx fp yqe iωy dy rfp yqs pωq pωqe iωx dω p αqe iαx dα r p αqs pxq So if we know a ourier transform formula or an inverse ourier transform formula, we can get another one for free by reversing the inverse. or example, since rs a pxqs pωq 2 sin aω, ω

7 we immediately have that math sin aω pxq S a pxq. ω rom either of the formulas above and the fact that sin x{x is an even function, we have 2 sin ax pωq S a pωq, x or sin ax π ax a S apωq. Similarly, since we know that for a 0, we can immediately write And since is an even function of ω, we have or Convolutions. e a x a 2 a 2 a 2 2a a 2 ω 2 2a e a x. ω 2 2a a 2 ω 2 2a e a ω, x 2 π x 2 a e a ω. We need one more ourier transform formula, and it involves an operation on functions that might seem new to you. It is called convolution and it starts with two functions, fpxq and gpxq and produces a new one, denoted f g or pf gqpxq (or sometimes just f gpxq), defined by pf gqpxq fpyqgpx yq dy. If you think about it, convolution is like multiplication of polynomials or series, wherein an bn c n where c n a m b n m.

8 8 fourier transforms This motivates the definition of convolution as an operation that might have has its ourier transform the product of the transforms of the individual functions, and it does, but more on that in a moment. irst some basic properties of convolutions:. Convolution is linear in each of the two functions. In other words, if f, g and h are functions and α and β are constants, then pαf βgq h αf h βg h and f pαg βhq αf g βf h. 2. Convolution is commutative: f g g f. To prove this, we make the substitution z x y in the integral (so we think of x as being constant while we re doing the integral, and y x z and dy dz) and get pf gqpxq 3. Now for the ourier transform:» 8 pg fqpxq fpyqgpx yq dy fpx zqgpzq dz gpzqfpx zq dz rf gs pωq rfs pωq rgs pωq or pf gqpωq fpωqĝpωq. To see this we use the substitution z x y again, and break up e iωx as e iωpx y yq e iωy e iωpx yq to get: pf gqpωq fpyqgpx yq dy e iωx dx fpyqgpx yqe iωx dx dy fpyqe iωy gpx yqe iωpx yq dx dy gpx yqe iωpx yq dx fpyqe iωy dy gpzqe iωz dz fpyqe iωy dy gpzqe iωz dz fpyqe iωy dy fpωqĝpωq

9 math ikewise, the transform of the product of two functions is the convolution of the transforms, except with a nuisance factor of, namely r Gs pxq q pxq q Gpxq. One last thing: the convolution product of two functions has an interesting property relative to derivatives: d df pf gq dx dx g f dg dx so when you take derivatives of convolution of two functions, you get to stick the derivative on whichever of the two functions is more convenient (or differentiable) this is an easy consequence of the commutativity of convolution, and it has the powerful consequence that the convolution of two functions has the better of the differentiability properties of the two individual functions. So if f is discontinuous but g is smooth, then f g will be smooth. The heat equation on the whole line. Now we re ready to use all of this for something! We seek to solve the heat equation u t ku xx for t 0 and x 8, with initial conditions upx, 0q fpxq and assuming u decays to zero at x and x 8. We ll start by taking the ourier transform of both sides of the differential equation in the x-variable. By this we mean so û satisfies and ûpω, tq upx, tqe ixω dx, Bû Bt kω2 û ûpω, 0q fpωq. This is an ordinary differential equation where the independent variable is t and ω should be treated as a constant. The general solution of the differential equation is ûpω, tq cpωqe kω2t. Putting t 0 shows that cpωq fpωq. Therefore ourier transform of the solution of our problem is ûpω, tq fpωqe kω2t.

10 0 fourier transforms We can now recover upx, tq by taking the inverse ourier transform of both sides, using the rule that the ourier transform of a convolution is the product of the individual ourier transforms, so upx, tq fpxq e kω2 t and we have to calculate the inverse ourier transform of e kω2t. But we have the rule for Gaussians: c e ax2 {2 {p2aq a e ω2, or c e ω2 {p2aq a {2 e ax2. Since we want to calculate the inverse ourier transform of e kω2t, we should set {p2aq kt, or a {p2ktq. Then we get e kω2 t? e x2{p4ktq, 4πkt and so upx, tq fpxq? e x2 {p4ktq 4πkt? 4πkt fpyqe px yq2 {p4ktq dy. The function Gpx, y, tq? 4πkt e px yq2 {p4ktq is called the fundamental solution of the heat equation. It is the solution to the initial value for the heat equation for the situation where the initial conditions are such that a single unit of heat energy is introduced at the point y at time t 0. So the above formula for upx, tq says that we can solve the heat equation for arbitrary initial conditions upx, tq by integrating together all the contributions to the temperature at time t 0 at all points, as described by the differential fpyq dy. The wave equation on the whole line. Next, let s look at the initial-value problem for the wave equation on the whole line. We ll solve the wave equation together with initial conditions u tt c 2 u xx upx, 0q fpxq and u t px, 0q gpxq.

11 math 425 As we did with the heat equation, we ll take the ourier transform of both sides of the differential equation in the x-variable. So once again, let ûpω, tq upx, tqe iωx dx so û satisfies and B 2 û Bt 2 c2 ω 2 û ûpω, 0q fpωq and û t pω, 0q ĝpωq. We will again treat this as an ordinary differential equation in t with ω treated as a constant. The general solution of this equation is u p ω, tq c pωq cos ωct c 2 pωq sin ωct, and the initial conditions imply upω, 0q c pωq fpωq and u t pω, 0q cωc 2 pωq ĝpωq. Therefore, c pωq fpωq and c 2 pωq ĝpωq{pcωq and we have obtained the ourier transform of the solution: Therefore ûpω, tq fpωq cos ωct upx, tq fpωq cos ωct ĝpωq cω sin ωct. ĝpωq cω sin ωct. We ll take the two terms one at a time. or the first, we write cos ωct in complex exponential form, so we re trying to compute fpωq eiωct e iωct fpωqre iωct e iωct se iωx dω 2 4π fpωqe iωpx ctq dx fpωqe iωpx ctq dx 4π rfpx ctq fpx ctqs 2 using the definition of the inverse ourier transform. or the second term, we proceed differently. Recalling that Ŝapωq 2 sin aω{ω and that the inverse ourier transform of a product of two functions is their convolution,

12 2 fourier transforms we have ĝpωq sin ωct 2c cω ĝpωq 2 sin ωct ω 2c gpxq ps ctpxqq 2c 2c 2c» ct ct» x ct x ct gpx yqs ct pyq dy gpx yq dy gpuq du where in the last step we made the substitution u x y (so y x u and dy du). We put both terms together to get the solution to our initial-value problem for the wave equation: upx, tq 2 rfpx ctq fpx ctqs 2c» x ct x ct gpuq du. This is called d Alembert s solution of the wave equation, and clearly shows that signals propagate with speed c, since the value of the solution at a point x and time t depends only on the initial position and velocity values in the interval rx ct, x cts, and conversely that the initial values at a point x influence the solution at time t only within the interval rx ct, x cts. inear algebraic properties of the ourier transform: Parseval s theorem and Hermite functions We remarked earlier that the ourier transform is a linear transformation from functions of x to functions of ω defined on the whole line (which can be integrated etc.). We also found that the function e x2 {2 is an eigenfunction of the ourier transform with eigenvalue?. We explore some further consequences of these observations in this section. irst, we can compare the inner product of the ourier transforms of two functions with the inner product of the functions themselves (remember, in the complex (or Hermitian) inner product, we have to take the complex conjugate of the second factor,

13 and we write z for the complex conjugate of z): xf, gy math fpxqpgpxqq dx» 8 fpxq ĝpωqe dω iωx dx fpxqe iωx pĝpωqq dx dω fpxqe iωx dx pĝpωqq dω A f, ĝ E. fpωqpĝpωqq dω In particular, if f g we have that }f} 2 xf, fy A f, fe } f} 2 or } f} 2 }f} 2. This is called Parseval s equality or Parseval s theorem it says that the ourier transforms stretches the norms of all functions by a factor of? (so it wasn t an accident that we found an eigenfunction with that eigenvalue!), and it has a number of interesting consequences. As a simple example. since we know that (recalling that S a pxq is the step function for the interval r a, as), 2 sin aω rs a pxqs pωq ω we can conclude that 2 2 sin aω ω }S a pxq} 2. The right side is easy to compute, it s simply» a a dx 4πa. But we learn something interesting by comparing this to the left side: or 4 sin 2 aω ω 2 sin 2 aω ω 2 dω 4πa dω πa,

14 4 fourier transforms which is certainly something we didn t know before. Now, back to the linear algebra. We ask a curious question, related to the observation we made on page 6: What happens if you take the ourier transform of the ourier transform of a function? et s see: r rfpxqss pωq fpzqe iωz dz fpzqe ip ωqz dz fp ωq, since the last integral is times the inverse ourier transform evaluated at ω. But this shows that doing the ourier transform twice to a function gives back its reverse multiplied by. So if we do the ourier transform four times to a function we should get 4π 2 times the reverse of the reverse, which is the original function. In other words, 4 rfpxqs 4π 2 fpxq. We could write this as an equation about the ourier transform operator: 4 4π 2 I 0, where I is the identity operator. Knowing this, we can see that if f is any eigenfunction of with eigenvalue λ, then 4 rfs 4π 2 Irfs λ 4 4π 2 f 0, which means that all of the eigenvalues of the ourier transform must satisfy the equation λ 4 4π 2 0, so the only possible eigenvalues of are?, i?,? and i?. We already have that the Gaussian e x2 {2 is an eigenfunction of with eigenvalue?. Now we ll find eigenfunctions for the other eigenvalues. We have two ourier transform rules that look quite similar: df dx pωq iω fpωq and xfpxqpωq i d f dω. So the ourier transforms of each of the operations multiply by the variable and take the derivative on the x-side is the other operation on the ω-side (well, there are factors of i to keep track of, and we will). Because of this, the sum and difference of these two operators have a special relationship with the ourier transform operator: df dx xfpxq iω rfs pωq d rfs i dω i d rfs dω ω rfs pωq

15 math and df dx xfpxq d rfs iω rfs pωq i dω i d rfs dω ω rfs pωq. These equations show that if fpxq is an eigenfunction of the ourier transform operator with eigenvalue λ then f pxq xfpxq will be an eigenfunction of the ourier transform with eigenvalue iλ (unless f pxq xfpxq 0), and f pxq xfpxq will be an eigenfunction of the ourier transform with eigenvalue iλ (unless f pxq xfpxq 0). et s try this with the eigenfunction we know, namely fpxq e x2{2. Unfortunately, f xf 0, but f pxq xfpxq xe x2 {2 xe x2 {2 2xe x2 {2, so 2xe x2{2 is an eigenfunction of the ourier transform with eigenvalue i? : 2xe x2 {2 pωq i? p 2ωe ω2{2 q. And we can keep going: If we set f pxq 2xe x2 {2, then f xf 2e x2 {2 2fpxq, so we get nothing new here. But f xf p4x 2 2qe x2 {2, so we call f 2 pxq p4x 2 2qe x2{2 and we have p4x 2 2qe x2 {2? p4ω 2 2qe ω2{2. We can continue in this way and obtain an infinite sequence of eigenfunctions of the ourier transform, starting from f 0 pxq e x2 {2, where f n pxq f n pxq xf npxq, and the eigenvalue of f n will be p iq n?. There will be more to say about this in the homework!