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August 17, 2020 APM346 { Week 12 Justin Ko

1 Fourier Transform

We introduce the concept of Fourier transforms. This extends the Fourier method for nite intervals to innite domains. In this section, we will derive the Fourier transform and its basic properties.

1.1 Heuristic Derivation of Fourier Transforms

1.1.1 Complex Full Fourier Series

Recall thatDeMoivre formulaimplies that

sin() =eiei2iand cos() =ei+ei2 This implies that the set of eigenfunctions for the full Fourier series on [L;L] n

1;cosxL

;cos2xL ;:::;sinxL ;sin2xL ;:::o is generated by the set of complex exponentialsfeinxL gn2Z. Consider the inner product for complex valued functions hf;gi=Z L

Lf(x)g(x)dx:

Since the complex conjugatee

inxL =einxL , it is also easy to check forn6=m, that heinxL ;eimxL i=Z L

LeinxL

eimxL dx=(1)nm(1)mni(nm)= 0; so the complex exponentials are an orthogonal set. We have the following reformulation of the full

Fourier series using complex variables.

Denition 1.Thecomplex form of the full Fourier seriesis given by f(x) =1X n=1c neinxL (1) where the (complex valued) Fourier coecients are given by c n=hf(x);einxL iheinxL ;einxL i=R L

Lf(x)einxL

dxR L

LeinxL

einxL dx=12LZ L

Lf(x)einxL

dx:(2) The proof of Parseval's equality also implies that 1 X n=1jcnj2=12LZ L

Ljf(x)j2dx:(3)

1.1.2 Fourier Transform

We now formally extend the Fourier series to the entire line by takingL! 1. If we substitute (2) into ( 1 ), then f(x) =12L1 X n=1 ZL

Lf(x)einxL

dx e inxL

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August 17, 2020 APM346 { Week 12 Justin Ko

We denekn=nL

and k=knkn1=L then this simplies to f(x) =121 X n=1 ZL

Lf(x)eiknxdx

e iknxk=1p21 X n=1C(kn)eiknxk: where

C(k) =1p2Z

L

Lf(x)eikxdx:

If we takeL! 1, then

C(k) =1p2Z

L

Lf(x)eikxdx!1p2Z

1 1 f(x)eikxdx(4) and interpreting the sum as a right Riemann sum, f(x) = limL!11p21 X n=1C(kn)eiknxk!1p2Z 1 1

C(k)eikxdk:(5)

Similarly, Parseval's equality (

3 ) becomes Z L

Ljf(x)j2dx= 2L1X

n=1jcnj2= 2L1X n=124L2 1p2Z L

Lf(x)einxL

dx2 =1X n=1jC(kn)j2k so takingL! 1impliesZ1 1 jf(x)j2dx=Z 1 1 jC(k)j2dk:(6) Remark 1.The step where we tookL! 1was not rigorous, because the bounds of integration and the function depend onL. A rigorous proof of this extension is much trickier.

1.2 Denition of the Fourier Transform

The Fourier transformFis an operator on the space of complex valued functions to complex valued functions. The coecientC(k) dened in (4) is called the Fourier transform. Denition 2.Letf:R!C. TheFourier transformoffis denoted byF[f] =^fwhere f(k) =1p2Z 1 1 f(x)eikxdx(7) Similarly, theinverse Fourier transformoffis denoted byF1[f] =fwhere f(x) =1p2Z 1 1 f(k)eikxdk:(8)

The follows from (

5 ) thatFandF1are indeed inverse operations.

Theorem 1 (Fourier Inversion Formula)Iffandf0are piecewise continuous, thenF1[Ff] =fandF[F1f] =f. In particular,

f(x) =1p2Z 1

1^f(k)eikxdkandf(k) =1p2Z

1

1f(x)eikxdx:Remark 2.Technically the Fourier inversion theorem holds for almost everywhere iffis discontinuous.

In fact, one can show thatF1[Ff] =f(x)+f(x+)2

, similarly to the pointwise convergence theorem. The Fourier transforms of integrable and square integrable functions are also square integrable ( 6

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August 17, 2020 APM346 { Week 12 Justin Ko

Theorem 2 (Plancherel's Theorem)Iffis both integrable and square integrable, thenkfkL2=k^fkL2=kfkL2, i.e.

Z 1 1 jf(x)j2dx=Z 1 1 j^f(k)j2dk=Z 1 1

jf(x)j2dx:(9)Remark 3.In the Denition2 , we also assume thatfis an integrable function, so that that its

Fourier transform and inverse Fourier transforms are convergent. Remark 4.Our choice of the symmetric normalizationp2in the Fourier transform makes it a linear unitary operator fromL2(R;C)!L2(R;C), the space of square integrable functionsf:R!C. Dierent books use dierent normalizations conventions.

1.3 Properties of Fourier Transforms

The Fourier transform behaves very nicely under several operations of functions. We have already seen that the formulas for the solutions of several PDEs we have encountered can be expressed as convolutions. Denition 3.Theconvolutionoffandgis a functionfgdened by (fg)(x) =Z 1 1 f(xy)g(y)dy=Z 1 1 f(y)g(xy)dy:(10) We now state several properties satised by Fourier transforms. Theorem 3 (Properties of Fourier Transforms)Iffandgare integrable, then

1.F[f(xa)] =eika^f(k)

2.F[f(x)eibx] =^f(kb)

3.F[f(x)] =jj1^f(1k)

4.F[^f(x)] =f(k)5.F[f0(x)] =ik^f(k)

6.F[xf(x)] =i^f0(k)

7.F[(fg)(x)] =p2^f(k)^g(k)

8.F[f(x)g(x)] =1p2(^f^g)(k)Proof.We demonstrate how to prove some of these properties because the other proofs are similar.

(Property 1|4)These properties follow immediately from a change of variables. For example, property 1 follows because

F[f(xa)] =1p2Z

1 1 f(xa)eikxdxy=xa=1p2Z 1 1 f(y)eik(y+a)dy eikap2Z 1 1 f(y)eikydy =eika^f(k): (Property 5|6)These properties follow immediately by interchanging dierentiation and integra- tion. For example, if we writef(x) as the inverse Fourier transform of^f f(x) =1p2Z 1

1^f(k)eikxdk:

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August 17, 2020 APM346 { Week 12 Justin Ko

then computing the derivative implies that f

0(x) =1p2ddx

Z 1

1^f(k)eikxdk=1p2Z

1 1 ik^f(k)eikxdk=F1[ikf(k)] =) F[f0(x)] =ik^f(k):

We are able to pass the derivative inside the integral by Leibiz's rule provided thatik^f(k)eikxis in-

tegrable. This integrability condition is implicitly satised because we assumed that the functions in

the theorem have convergent Fourier and inverse Fourier transforms. (Property 7|8)These properties are the hardest to prove, so we will show it in detail. The proof of property 7 follows from a change of variables and Fubini's theorem

F[(fg)(x)] =1p2Z

1 1 (fg)(x)eikxdx 1p2Z 1 1Z 1 1 f(xy)g(y)eikxdydx 1p2Z 1 1Z 1 1 f(xy)g(y)eikxdxdyFubini's Theorem 1p2Z 1 1Z 1 1 f(z)g(y)eik(z+y)dzdyz =xy p21p2Z 1 1 f(z)eikzdz1p2Z 1 1 g(y)eikydy p2^f(k)^g(k): To prove property 8, we can writeg(x) as the inverse Fourier transform of ^g,

F[f(x)g(x)] =1p2Z

1 1 f(x)1p2Z 1 1 ^g(`)ei`xd` e ikxdx 12Z 1 1Z 1 1 ^g(`)f(x)ei(k`)xdxd`Fubini's Theorem 1p2Z 1 1 ^g(`)^f(k`)d`Denition (7)

1p2(^f^g)(k)Denition ( 10)Remark 5.In Theorem3 , we also assume thatfis nice enough so that the Fourier transforms

and inverse Fourier transforms make sense. For example, in property 5 we need to assume thatfis dierentiable and the inverse Fourier transform ofik^f(k) converges. Remark 6.Property 5 can be proved by integration by parts if we assume thatf(1) = 0,

F[f0(x)] =1p2Z

1 1 f0(x)eikxdx=1p2f(x)eikxx=1 x=1+ik1p2Z 1 1 f(x)eikxdx=ik^f(k): This is the standard proof of this property, and it requires that assumption thatf(1) = 0 for the boundary terms to vanish. Remark 7.If we denef(x) =f(x), it is also useful to notice that a change of variables implies F

1[f] =f(x) =1p2Z

1 1 f(k)eikxdkk=y=1p2Z 1 1 f(y)eixydy=^f(x) =F[f]:(11)

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August 17, 2020 APM346 { Week 12 Justin Ko

and

F[f] =^f(k) =1p2Z

1 1 f(x)eikxdxx=y=1p2Z 1 1 f(y)eikydy=f(k) =F1[f]:(12)

ApplyingForF1to these identities imply that

F[F[f(x)]] =f(x)()^^f(x) =f(x);F1[F1[f(x)]] =f(x)()f(x) =f(x):(13) This observation is a generalization of property 4 in Theorem 3 and will b eused man ytimes to use the Fourier transform formulas \in reverse". For example, to prove property 8, we apply the inverse of property 7 by applying ( 12 ) and the rst identityf(x) =^^f(x) in (13), F[f(x)g(x)](12)=F1[f(x)g(x)](13)=F1[^^f(x)^^g(x)] =1p2(^f^g)(k) so

F[f(x)g(x)] =1p2(^f^g)(k):

1.4 Summary of the Properties of Fourier Transforms

1.4.1 Fourier Transform

f(k) =1p2Z 1 1 f(x)eikxdx

1.4.2 Inverse Fourier Transform

f(x) =1p2Z 1 1 f(k)eikxdk

1.4.3 Plancherel's Theorem

Z1 1 jf(x)j2dx=Z 1 1 j^f(k)j2dk=Z 1 1 jf(x)j2dx

1.4.4 Relationship Between Fourier Transforms and Inverse Fourier Transforms

1.F[F1[f(x)]] =f(x)

2.F1[F[f(x)]] =f(x)3.F[F[f(x)]] =f(x)

4.F1[F1[f(x)]] =f(x)

1.4.5 List of Important Transformations

1.F[eajxj] =1p2

2aa 2+k2

2.F[ex22

] =ek22

1.4.6 Properties of Fourier Transform

1.F[f(xa)] =eika^f(k)

2.F[f(x)eibx] =^f(kb)

3.F[f(x)] =jj1^f(1k)

4.F[^f(x)] =f(k)5.F[f0(x)] =ik^f(k)

6.F[xf(x)] =i^f0(k)

7.F[(fg)(x)] =p2^f(k)^g(k)

8.F[f(x)g(x)] =1p2(^f^g)(k)

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August 17, 2020 APM346 { Week 12 Justin Ko

1.5 Finding Fourier Transforms

Problem 1.1.(?) Find the Fourier transform of

f(x) =eajxja >0:

Solution

1.1. This can be computed directly. We split the region of integration,

f(k) =1p2Z 1 1 eikxeajxjdx 1p2 Z0 1 eikx+axdx+Z 1 0 eikxaxdx 1p2 eikx+axaik x=0 x=1+eikxaxaik x=1 x=0 1p2

1aik+1a+ik

1p22aa

2+k2:

Problem 1.2.(??) Find the Fourier transform of

f(x) =ex22

Solution

1.2. This can be computed directly. We do a complex change of variables,

f(k) =1p2Z 1 1 eikxex22 dx=1p2ek22 Z 1 1 e12 (x+ik)2dxcomplete the square 1p ek22 Z 1 1 ez2dz(see the Remark9 ) =ek22 :Z 1 1 ey2dy=p: Remark 8.There is also a proof of this fact that avoids complex analysis. Iff(x) =ex22 , then xf(x) =f0(x): Therefore, if we take the Fourier transform of both sides and apply Property 1.4.6 .5 and 1.4.6 .6 then F[xf(x)] =i^f0(k) =ikf0(x) =F[f0(x)] =)^f0(k) =k^f(k):

Solving this ODE implies that

^f(k) =Cek22

To ndC, we plug in 0 and notice that

C=^f(0) =1p2Z

1 1 ex22 dx= 1; so ^f(k) =ek22

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August 17, 2020 APM346 { Week 12 Justin Ko

Remark 9.The imaginary change of variablesz=1p2

(x+ik) can be justied using complex analysis, Z R e12 (x+ik)2dx=p2 Z

R+ikp2

e z2dz: Fork >0 consider the contour integral over the closed rectangular path oriented counter clockwise, (1)x+ikp2 forxfromMtoM (2)M+iyforyfromkp2 to 0 (3)xforxfromMtoM (4)M+iyforyfrom 0 tokp2 Sinceez2is analytic, the integral over this closed path is 0, so Z (1) ez2dz+Z (2) ez2dz+Z (3) ez2dz+Z (4) ez2dz= 0: Sinceez2is small when Re(z) =M, if we takeM! 1, the integrals over the regions (2) and (4) vanish leaving us with Z

R+ikp2

e z2dz+Zquotesdbs_dbs12.pdfusesText_18

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