[PDF] Fourier transforms - Arizona Math PDF fouriertransform.pdf

The Fourier transform is beneficial in differential equations because it can reformulate them as For problems on the half-line x > 0 with boundary conditions

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Fourier transform techniques

1 The Fourier transform

Recall for a functionf(x) : [L;L]!C, we have the orthogonal expansion f(x) =1X n=1c neinx=L; cn=12LZ L

Lf(y)einy=Ldy:(1)

We think ofcnas representing the "amount" of a particular eigenfunction with wavenumber k n=n=Lpresent in the functionf(x). So what ifLgoes to1? Notice that the allowed wavenumbers become more and more dense. Therefore, whenL=1, we expectf(x)is a su- perposition of an uncountable number of waves corresponding to every wavenumberk2R, which can be accomplished by writingf(x)as a integral overkinstead of a sum overn. Now let"s take the limitL! 1formally. Settingkn=n=Landk==Land using (1), one can write f(x) =121 X n=1 ZL

Lf(y)eiknydy

e iknxk: Notice this is a Riemann sum for an integral on the intervalk2(1;1). TakingL! 1is the same as takingk!0, which gives f(x) =12Z 1 1

F(k)eikxdk;(2)

where

F(k) =Z

1 1 f(x)eikxdx:(3) The functionF(k)is theFourier transformoff(x). Theinverse transformofF(k)is given by the formula (2). (Note that there are other conventions used to define the Fourier transform). Instead of capital letters, we often use the notation^f(k)for the Fourier transform, andF(x)for the inverse transform.

1.1 Practical use of the Fourier transform

The Fourier transform is beneficial in differential equations because it can reformulate them as problems which are easier to solve. In addition, many transformations can be made simply by applying predefined formulas to the problems of interest. A small table of transforms and some properties is given below. Most of these result from using elementary calculus techniques for the integrals (3) and (2), although a couple require techniques from complex analysis. 1

A Brief table of Fourier transforms

Description Function Transform

Delta function inx (x) 1

Delta function ink1 2(k)

Exponential inx eajxj2aa

2+k2(a >0)

Exponential ink2aa

2+x22eajkj(a >0)

Gaussianex2=2p2ek2=2

Derivative inx f0(x)ikF(k)

Derivative ink xf(x)iF0(k)

Integral inxRx

1f(x0)dx0F(k)=(ik)

Translation inx f(xa)eiakF(k)

Translation ink eiaxf(x)F(ka)

Dilation inx f(ax)F(k=a)=a

Convolution f(x)*g(x)F(k)G(k)Typically these formulas are used in combination. Preparatory steps are often required (just

like using a table of integrals) to obtain exactly one of these forms. Here are a few examples. Example 1.The transform off00(x)is (using the derivative table formula) f00(x)^=ikf0(x)^= (ik)2^f(k) =k2^f(k): Notice what this implies for differential equations: differential operators can be turned into "mul- tiplication" operators. Example 2.The transform of the Gaussianexp(Ax2)is, using both the dilation and Gaussian formulas, exp(Ax2)^=h exp([p2Ax]2=2)i ^=1p2Aexp(x2=2)^(k=p2A) r A exp([k=p2A]2=2) =r A exp(k2=(4A)): Example 3.The inverse transform ofe2ik=(k2+ 1)is, using the translation inxproperty and then the exponential formula, e2ikk 2+ 1 _ =1k 2+ 1 _ (x+ 2) =12 ejx+2j: Example 4.The inverse transform ofkek2=2uses the Gaussian and derivative inxformulas: h kek2=2i_=ih ikek2=2i_=iddx h ek2=2i_= ip2ddx h p2ek2=2i_=ip2ddx ex2=2 =ixp2ex2=2:

1.2 Convolutions

Unfortunately, the inverse transform of a product of functions is not the product of inverse trans- forms. Rather, it is a binary operation calledconvolution, defined as (fg)(x) =Z 1 1 f(xy)g(y)dy:(4) 2 Using the definition, the Fourier transform of this is (fg)^=Z 1 1Z 1 1 f(xy)g(y)eikxdy dx

Using the change of variablesz=xy, this becomes

Z 1 1Z 1 1 f(z)g(y)eik(y+z)dydz= Z1 1 f(z)eikzdz Z1 1 g(y)eikydy ^f(k)^g(k); which is just the last formula in the table.

1.3 Divergent Fourier integrals as distributions

The formulas (3) and (2) assume thatf(x)andF(k)decay at infinity so that the integrals converge. If 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 account.

Notice 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) =12Z 1 1 eikxdk:(5) This is very troublesome: the integral does not even converge, so what could such a statement mean? Of course the issue is that the integral represents a distribution, not a regular function. We can define the inverse transform ofF(k)more generally as a distribution which is the limit of the regular functions f

L(x) =12Z

Lexp(ikx)F(k)dk

asL! 1(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 like f[] = limL!1Z 1 1 f

L(x)(x)dx:(6)

Let"s look at the integral in (5) and see what distribution it represents. For this case, f

L(x) =12Z

Lexp(ikx)dk=1x

sin(Lx): Then if one asks how the limit offLacts as a distribution, one computes f[] = limL!11 Z 1

1sin(Lx)x

(x)dx= limL!11 Z 1

1sin(y)y

(y=L)dy: The limitL! 1can be taken inside the integral (this can be justified with a little effort), and this results in f[] =(0) Z 1

1sin(y)y

dy=(0): So the inverse transform really is the delta function! 3

2 Solutions of differential equations using transforms

The derivative property of Fourier transforms is especially appealing, since it turns a differential operator into a multiplication operator. In many cases this allows us to eliminate the derivatives of 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 typically the most labor intensive step.

2.1 Ordinary differential equations on the real line

Here we give a few preliminary examples of the use of Fourier transforms for differential equa- tions involving a function of only one variable.

Example 1.Let us solve

u00+u=f(x);limjxj!1u(x) = 0:(7) The transform of both sides of (7) can be accomplished using the derivative rule, giving k

2^u(k) + ^u(k) =^f(k):(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 Z 1 1 ejxyjf(y)dy: This is exactly what we get from a Green"s function representation, whereG(x;y) =ejxyj=2. There is one mystery remaining: the far field condition in (7) does not seem to be used anywhere.

In fact, condition (7) is already built into the Fourier transform; if the functions being transformed

did not decay at infinity, the Fourier integral would only be defined as a distribution as in (6).

Example 2.TheAiryequation is

00xu= 0;

which will be subject to the same far field condition as in (7). The transform uses the derivative formulas for bothxandk, giving k2^u(k)i^u0(k) = 0:

This is still a differential equation in thekvariable, but we can solve it by separation of variables

(the ODE version, that is). This results ind^u=^u=ik2dkwhich integrates to give ^u(k) =Ceik3=3; whereCis an arbitrary constant of integration. The inverse transform is u(x) =C2Z 1 1 exp(i[kx+k3=3])dk:(10) This integral cannot be reduced any further. With the choiceC= 1, the result is the so-calledAiry functiondenoted Ai(x). 4

2.2 Solution of partial differential equations

Now we consider situations where there is more than one independent variable. In this case, the transform will apply to only one variable. This will reduce the number of variables which have derivatives, and often make it possible to solve using ODE techniques. Example 1.Consider the Laplace equation on the upper half plane u xx+uyy= 0;1< x <1; y >0; u(x;0) =g(x);limy!1u(x;y) = 0:(11) It only makes sense to transform in thexvariable, and we denote this as

U(k;y) =Z

1 1 eikxu(x;y)dx:(12) We note that they-derivative commutes with the Fourier integral inx, so that the transform ofuyy is simplyUyy. Then the equation and boundary conditions in (11) become k2U+Uyy= 0; U(k;0) = ^g(k);limy!1U(k;y) = 0: This is a set of ordinary differential equations, one for each value ofk. The general solution is U=c1e+jkjy+c2ejkjy, wherec1;2can depend onk. The first term must be zero so thatUvanishes at infinity. UsingU(k;0) = ^g(k), it follows that

U(k;y) = ^g(k)ejkjy:

The inverse transform involves a convolution and the exponential inkformula from the table.

The result is

u(x;y) =g(x) ejkjy_=g(x)y(x2+y2) 1 Z 1

1yg(x0)(xx0)2+y2dx0:

This is precisely the same formula as obtained by Green"s function methods.

Example 2.Now let"s solve the transport equation

u t+cux= 0;1< x <1; t >0; u(x;0) =f(x) by a similar process. LetU(k;t)be the transform ofuonly in thexvariable as in (12). Since thet derivative commutes with thex-integral, the problem transforms into U t+ikcU= 0; U(k;0) =^f(k): This is a simple first order differential equation whose solution is

U(k;t) =eickt^f(k):

Now we use the translation formula from the table witha=ct, which means that the inverse transform is u(x;t) =f(xct): This is atraveling wave solution, describing a pulse with shapef(x)moving uniformly at speedc. 5 Example 3.Consider the wave equation on the real line u tt=uxx;1< x <1; t >0; u(x;0) =f(x); ut(x;0) =g(x): LettingU(k;t)be the transform in thex-variable, then the problem becomes U tt+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 is

U(k;t) =^f(k)cos(kt) +^g(k)k

sin(kt):(13) Note that sines and cosines can be written in terms of complex exponentials, so that

U(k;t) =12

^f(k)(eikt+eikt) +12ik^g(k)(eikteikt):(14) The inverse transform is now straightforward, using the exponential and integral formulas, u(x;t) =12 [f(xt) +f(x+t)] +12 Z x 1 g(x0+t)g(x0t)dx0: The integral can be simplified by using the change of variables=x0+tfor the first term and =x0tfor the second, Z x 1 g(x0+t)g(x0t)dx0=Z x+t 1 g()dZ xt 1 g()d=Z x+t xtg()d: Finally, putting this all together results ind"Alembert"sformula for the wave equation u(x;t) =12 [f(xt) +f(x+t)] +12 Z x+t xtg()d:(15)

3 Fundamental solutions for time dependent equations

Partial differential equations which involve time also have Green"s functions, although they are more often calledfundamental solutionsorsource functions. Suppose thatu(x;t) :DR!R, where

DRnis some spatial domain, which solves

u t(x;t) =Lu(x;t); u(x;0) =f(x);(16) whereLis some differential operator which does not depend ont. This equation is supplemented by homogeneous boundary conditions (Dirichlet, Neumann and possibly others) onx2@D. We define the fundamental solution for (16) to be the solutionS(x;x0;t)to the problem S t=LxS; S(x;x0;0) =(xx0); subject to the same boundary conditions asu. This is different than our previous definition of Green"s function in one respect: the-function appears as an initial condition rather than an inhomogeneous term in the equation. Of course, the 6 initial condition onSis only meaningful in the sense of distributions, which is to say thatSlimits to a-function ast!0: lim t!0Z D

S(x;x0;t)(x)dx=Z

D (xx0)(x)dx=(x0);(17) for all continuous functions:D!R. We now claim that the initial value problem (16) is solved by the formula u(x;t) =Z 1 1

S(x;x0;t)f(x0)dx0;(18)

so long asfis continuous. The intuitive meaning is clear: the integral is the just a sum over all point influences contained in the initial conditionf(x). Let"s check to see that it works. For the initial condition, we taket!0in (18) and use (17), u(x;0) = limt!0Z D

S(x;x0;t)f(x0)dx0=Z

D (xx0)f(x0)dx0=f(x): Plugginguinto the equation in (16), we can move the time derivative inside the integral: u t=Z D S t(x;x0;t)f(x0)dx0=Z D L xS(x;x0;t)f(x0)dx0: We now suppose that the operator (which acts on thexvariables) can be moved outside thex0 integral (this can be justified in most circumstances), giving uquotesdbs_dbs19.pdfusesText_25

[PDF] [PDF] Fourier transforms - Arizona Math

Fourier transform techniques

1 The Fourier transform

Lf(y)einy=Ldy:(1)

Lf(y)eiknydy

F(k)eikxdk;(2)

F(k) =Z

1.1 Practical use of the Fourier transform

A Brief table of Fourier transforms

Description Function Transform

Delta function inx (x) 1

Delta function ink1 2(k)

Exponential inx eajxj2aa

2+k2(a >0)

Exponential ink2aa

2+x22eajkj(a >0)

Gaussianex2=2p2ek2=2

Derivative inx f0(x)ikF(k)

Derivative ink xf(x)iF0(k)

Integral inxRx

1f(x0)dx0F(k)=(ik)

Translation inx f(xa)eiakF(k)

Translation ink eiaxf(x)F(ka)

Dilation inx f(ax)F(k=a)=a

1.2 Convolutions

Using the change of variablesz=xy, this becomes

1.3 Divergent Fourier integrals as distributions

L(x) =12Z

Lexp(ikx)F(k)dk

L(x)(x)dx:(6)

L(x) =12Z

Lexp(ikx)dk=1x

1sin(Lx)x

1sin(y)y

1sin(y)y

2 Solutions of differential equations using transforms

2.1 Ordinary differential equations on the real line

Example 1.Let us solve

2^u(k) + ^u(k) =^f(k):(8)

Example 2.TheAiryequation is

00xu= 0;

2.2 Solution of partial differential equations

U(k;y) =Z

U(k;y) = ^g(k)ejkjy:

The result is

1yg(x0)(xx0)2+y2dx0:

Example 2.Now let"s solve the transport equation

U(k;t) =eickt^f(k):

U(k;t) =^f(k)cos(kt) +^g(k)k

U(k;t) =12

3 Fundamental solutions for time dependent equations

DRnis some spatial domain, which solves

S(x;x0;t)(x)dx=Z

S(x;x0;t)f(x0)dx0;(18)

S(x;x0;t)f(x0)dx0=Z