Fourier transform techniques 1 The Fourier transform

1 Fourier transform techniques1 The Fourier transformRecall for a functionf(x) : [ L,L] C, we have the orthogonal expansionf(x) = n= cnein x/L, cn=12L L Lf(y)e in y/Ldy.(1)We think ofcnas representing the amount of a particular eigenfunction with wavenumberkn=n /Lpresent in the functionf(x). So what ifLgoes to ? Notice that the allowedwavenumbers become more and more dense. Therefore, whenL= , we expectf(x)is a su-perposition of an uncountable number of waves corresponding to every wavenumberk R,which can be accomplished by writingf(x)as a integral overkinstead of a sum let s take the limitL formally. Settingkn=n /Land k= /Land using (1), onecan writef(x) =12 n= ( L Lf(y)e iknydy)eiknx this is a Riemann sum for an integral on the intervalk ( , ).

2 TakingL is thesame as taking k 0, which givesf(x) =12 F(k)eikxdk,(2)whereF(k) = f(x)e ikxdx.(3)The functionF(k)is theFourier transformoff(x). Theinverse transformofF(k)is given by theformula (2). (Note that there are other conventions used to define the Fourier transform ). Insteadof capital letters, we often use the notation f(k)for the Fourier transform , and F(x)for the Practical use of the Fourier transformThe Fourier transform is beneficial in differential equations because it can reformulate them asproblems which are easier to solve. In addition, many transformations can be made simply byapplying predefined formulas to the problems of interest. A small table of transforms and someproperties is given below. Most of these result from using elementary calculus techniques for theintegrals (3) and (2), although a couple require techniques from complex Brief table of Fourier transformsDescriptionFunctionTransformDe lta function inx (x)1 Delta function ink12 (k)Exponential inx e a|x|2aa2+k2(a >0)Exponential ink2aa2+x22 e a|k|(a >0)Gaussiane x2/2 2 e k2/2 Derivative inxf (x)ikF(k)Derivative inkxf(x)iF (k)Integral inx x f(x )dx F(k)/(ik)Translation inx f(x a)e iakF(k)Translation ink eiaxf(x)F(k a)Dilation inxf(ax)F(k/a)/aConvolutionf(x)*g(x)F(k) G(k)Typically these formulas are used in combination.

3 Preparatory steps are often required (justlike using a table of integrals) to obtain exactly one of these forms. Here are a few transform off (x)is (using the derivative table formula)[f (x)] =ik[f (x)] = (ik)2 f(k) = k2 f(k).Notice what this implies for differential equations: differential operators can be turned into mul-tiplication transform of the Gaussianexp( Ax2)is, using both the dilation and Gaussianformulas,[exp( Ax2)] =[exp( [ 2Ax]2/2)] =1 2A[exp( x2/2)] (k/ 2A)= Aexp( [k/ 2A]2/2) = Aexp( k2/(4A)).Example inverse transform ofe2ik/(k2+ 1)is, using the translation inxproperty and thenthe exponential formula,(e2ikk2+ 1) =(1k2+ 1) (x+ 2) =12e |x+2|.Example inverse transform ofke k2/2uses the Gaussian and derivative inxformulas:[ke k2/2] = i[ike k2/2] = iddx[e k2/2] == i 2 ddx[ 2 e k2/2] = i 2 ddx(e x2/2)=ix 2 e x2 ConvolutionsUnfortunately, the inverse transform of a product of functions is not the product of inverse trans-forms.

4 Rather, it is a binary operation calledconvolution, defined as(f g)(x) = f(x y)g(y)dy.(4)2 Using the definition, the Fourier transform of this is(f g) = f(x y)g(y)e ikxdy dxUsing the change of variablesz=x y, this becomes f(z)g(y)e ik(y+z)dydz=( f(z)e ikzdz)( g(y)e ikydy)= f(k) g(k),which is just the last formula in the Divergent Fourier integrals as distributionsThe formulas (3) and (2) assume thatf(x)andF(k)decay at infinity so that the integrals this is not the case, then the integrals must be interpreted in a generalized sense. In addition,some of the table formulas must be adjusted to take this into that the transform of (x)equals f(k) 1, so at least formally, the inverse transform of f(k)should be a delta function: (x) =12 eikxdk.(5)This is very troublesome: the integral does not even converge, so what could such a statementmean?

5 Of course the issue is that the integral represents a distribution, not a regular function. Wecan define the inverse transform ofF(k)more generally as a distribution which is the limit of theregular functionsfL(x) =12 L Lexp(ikx)F(k)dkasL (recall the fact that distributions can always be approximated by regular functions).This means that the inverse transformf(x)is a distribution which acts on smooth functions likef[ ] = limL fL(x) (x)dx.(6)Let s look at the integral in (5) and see what distribution it represents. For this case,fL(x) =12 L Lexp(ikx)dk=1 xsin(Lx).Then if one asks how the limit offLacts as a distribution, one computesf[ ] = limL 1 sin(Lx)x (x)dx= limL 1 sin(y)y (y/L) limitL can be taken inside the integral (this can be justified with a little effort), and thisresults inf[ ] = (0) sin(y)ydy= (0).

6 So the inverse transform really is the delta function!32 Solutions of differential equations using transformsThe derivative property of Fourier transforms is especially appealing, since it turns a differentialoperator into a multiplication operator. In many cases this allows us to eliminate the derivativesof one of the independent variables. The resulting problem is usually simpler to solve. Of course,to recover the solution in the original variables, an inverse transform is needed. This is typicallythe most labor intensive Ordinary differential equations on the real lineHere we give a few preliminary examples of the use of Fourier transforms for differential equa-tions involving a function of only one us solve u +u=f(x),lim|x| u(x) = 0.(7)The transform of both sides of (7) can be accomplished using the derivative rule, givingk2 u(k) + u(k) = f(k).

7 (8)This is just an algebraic equation whose solution is u(k) = f(k)1 +k2.(9)We can recoveru(x)by an inverse transform . Expression (9) is a product of f(k)and1/(1 +k2),so we must use the convolution formula:u(x) =f(x) (11 +k2) =12 e |x y|f(y) is exactly what we get from a Green s function representation, whereG(x,y) =e |x y| is one mystery remaining: the far field condition in (7) does not seem to be used fact, condition (7) is already built into the Fourier transform ; if the functions being transformeddid not decay at infinity, the Fourier integral would only be defined as a distribution as in (6).Example isu xu= 0,which will be subject to the same far field condition as in (7). The transform uses the derivativeformulas for bothxandk, giving k2 u(k) i u (k) = is still a differential equation in thekvariable, but we can solve it by separation of variables(the ODE version, that is).

8 This results ind u/ u=ik2dkwhich integrates to give u(k) =Ceik3/3,whereCis an arbitrary constant of integration. The inverse transform isu(x) =C2 exp(i[kx+k3/3])dk.(10)This integral cannot be reduced any further. With the choiceC= 1, the result is the so-calledAiryfunctiondenoted Ai(x). Solution of partial differential equationsNow we consider situations where there is more than one independent variable. In this case, thetransform will apply to only one variable. This will reduce the number of variables which havederivatives, and often make it possible to solve using ODE the Laplace equation on the upper half planeuxx+uyy= 0, < x < , y >0, u(x,0) =g(x),limy u(x,y) = 0.(11)It only makes sense to transform in thexvariable, and we denote this asU(k,y) = e ikxu(x,y)dx.(12)We note that they-derivative commutes with the Fourier integral inx, so that the transform ofuyyis simplyUyy.

9 Then the equation and boundary conditions in (11) become k2U+Uyy= 0, U(k,0) = g(k),limy U(k,y) = is a set of ordinary differential equations, one for each value ofk. The general solution isU=c1e+|k|y+c2e |k|y, wherec1,2can depend onk. The first term must be zero so thatUvanishesat infinity. UsingU(k,0) = g(k), it follows thatU(k,y) = g(k)e |k| inverse transform involves a convolution and the exponential inkformula from the result isu(x,y) =g(x) (e |k|y) =g(x) (y (x2+y2))=1 yg(x0)(x x0)2+ is precisely the same formula as obtained by Green s function let s solve the transport equationut+cux= 0, < x < , t >0, u(x,0) =f(x)by a similar process. LetU(k,t)be the transform ofuonly in thexvariable as in (12). Since thetderivative commutes with thex-integral, the problem transforms intoUt+ikcU= 0, U(k,0) = f(k).

10 This is a simple first order differential equation whose solution isU(k,t) =e ickt f(k).Now we use the translation formula from the table witha=ct, which means that the inversetransform isu(x,t) =f(x ct).This is atraveling wave solution, describing a pulse with shapef(x)moving uniformly at the wave equation on the real lineutt=uxx, < x < , t >0, u(x,0) =f(x), ut(x,0) =g(x).LettingU(k,t)be the transform in thex-variable, then the problem becomesUtt+k2U= 0, U(k,0) = f(k), Ut(k,0) = g(k).This is just the initial value problem for a harmonic oscillator. Its solution isU(k,t) = f(k) cos(kt) + g(k)ksin(kt).(13)Note that sines and cosines can be written in terms of complex exponentials, so thatU(k,t) =12 f(k)(eikt+e ikt) +12ik g(k)(eikt e ikt).(14)The inverse transform is now straightforward, using the exponential and integral formulas,u(x,t) =12[f(x t) +f(x+t)] +12 x g(x +t) g(x t)dx.