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Fourier Transform Examples

Steven Bellenot

November 5, 2007

1 Formula Sheet

F[f(x)] =bf(w) or simplyF[f] =bf(1)

F

1[bf(w)] =f(x) or simplyF1[bf] =f(2)

F[f(x)](w) =bf(w) =1p2Z

1 1 f(x)eiwxdx(3) F

1[bf(w)](x) =1p2Z

1

1bf(w)eiwxdw(4)

F[u(x;t)](w;t) =bu(w;t) =1p2Z

1 1 u(x;t)eiwxdx(5) F

1[bu(w;t)](x;t) =1p2Z

1 1 bu(w;t)eiwxdw(6)

F[af(x) +bg(x)](w) =abf(w) +bbg(w)(7)

F[f0(x)](w) =iwbf(w)(8)

F[f00(x)](w) =w2bf(w)(9)

F[@@xu(x;t)](w;t) =iwbu(w;t)(10)

F[@2@x2u(x;t)](w;t) =w2bu(w;t)(11)

F[@@tu(x;t)](w;t) =@@tbu(w;t)(12)

F[@2@t2u(x;t)](w;t) =@2@t2bu(w;t)(13)

[fg](x) =Z 1 1 f(w)g(xw)dw= [gf](x) =Z 1 1 f(xw)g(w)dw(14)

F[fg] =p2bfbg(15)

f(xa) =F1[eiwabf(w)](16)

F[exp(ax2)] =1p2aexp(w24a)(17)

sinwa=eiwaeiwa2i(18) coswa=eiwa+eiwa2(19)1

2 Formula Justications

Equations (1), (3) and (5) readly say the same thing, (3) being the usual denition. (Warning, not all

textbooks dene the these transforms the same way.) Equations (2), (4) and (6) are the respective inverse

transforms.

What kind of functions is the Fourier transform dened for? Clearly iff(x) is real, continuous and zero

outside an interval of the form [M;M], thenbfis dened as the improper integralR1

1reduces to the

proper integralRM M. Iff(x) decays fast enough asx! 1andx! 1, thenbf(w) is also dened. However

there are much larger collections of objects for which the transform can be dened. For example, if(x)

is the Dirac delta function, then b(w) = 1=p2the constant function. Also one can see that the inverse transform of(w) is the constant function 1=p2. Equation (7) follows because the integral is linear, the inverse transform is also linear. Equation (8) follows from integrating by parts, usingu=eiwxanddv=f0(x)dxand the fact thatf(x) decays asx! 1andx! 1. Z 1 1 f0(x)eiwxdx=f(x)eiwx1 x=1Z 1 1 f(x)iweiwxdx= (00) +iwbf(w) Equation (9) is just (8) applied twice. And (10) and (11) are just restatements with more variables. Equation (12) requires going back to the denition of the limit. F u(x;t+ t)u(x;t)t =Z 1

1u(x;t+ t)u(x;t)teiwxdx

bu(w;t+ t)bu(w;t)t!@@tbu(w;t)

One now takes limits of both sides. We need to know that the fourier transform is continuous with this kind

of limit, which is true, but beyond our scope to show. Equation (13) is (12) done twice. Equation (14) saysfg=gfand this is done by substitution; useu=xw;du=dw;w=xu; u=1whenw=1andu=1whenw=1to obtain Z 1 w=1f(w)g(xw)dw=Z 1 u=1f(xu)g(u)(du) =Z 1 u=1f(xu)g(u)du which used the negative sign to change the order of integration.

Equation (15) uses

Z 1 x=1(fg)eiwxdx=Z 1 x=1Z 1 s=1f(s)g(xs)dseiwxdx Z 1 x=1Z 1 s=1f(s)g(xs)eiwxdsdx Z 1 s=1Z 1 x=1f(s)g(xs)eiwxdxds Note that we have interchanged the order of integration, now we letu=xs,x=u+s,du=ds,u=1 whenx=1 =Z 1 s=1Z 1 u=1f(s)g(u)eiw(u+s)duds Z 1 s=1f(s)eiwsdsZ 1 u=1g(u)eiwudu

since theuterms are constant as the integral with respect todsis concerned. So the inital expression isp2F[fg] and the end epression isp2bfp2bgwhich is where thep2factor comes from.

To show equation (16) we computeF[f(xa)] and substituteu=xa;x=u+a;dx=du: Z 1 x=1f(xa)eiwxdx=Z 1 u=1f(u)eiw(u+a)du=eiwaZ 1 u=1f(u)eiwudu=eiwabf(w) For the bell shaped curves, equation (17) is done in earlier editions of the textbook. We repeat the calculation for reference only

F[exp(ax2)] =1p2Z

1 1 exp(ax2iwx)dx 1p2Z 1 1 exp pax+iw2pa 2 +iw2pa 2! dx

1p2exp

w24a Z1 1 exp pax+iw2pa 2! dx We claim that the integral above has valueI=pa. First we do the substitution v=pax+iw2pa so thatdv=padxand hence I=Z 1 1 exp(v2)dvpa

The result follows since

Z1 1 exp(v2)dv=p comes from Calculus 3. Finally (18) and (19) are from Euler'sei= cos+isin.

3 Solution ExamplesSolve 2ux+ 3ut= 0;u(x;0) =f(x) using Fourier Transforms.

Take the Fourier Transform of both equations. The initial condition gives bu(w;0) =bf(w) and the PDE gives

2(iwbu(w;t)) + 3@@tbu(w;t) = 0

Which is basically an ODE int, we can write it as

@@tbu(w;t) =23iwbu(w;t) and which has the solution bu(w;t) =A(w)e2iwt=3 and the initial condition above impliesA(w) =bf(w) bu(w;t) =bf(w)e2iwt=3 We are now ready to inverse Fourier Transform and equation (16) above, witha= 2t=3, says that u(x;t) =f(x2t=3) Solve 2tux+ 3ut= 0;u(x;0) =f(x) using Fourier Transforms. Take the Fourier Transform of both equations. The initial condition gives bu(w;0) =bf(w) and the PDE gives

2t(iwbu(w;t)) + 3@@tbu(w;t) = 0

Which is basically an ODE int, we can write it as

@@tbu(w;t) =23iwtbu(w;t) and which has the solution (separate variables) bu(w;t) =A(w)eiwt2=3 and the initial condition above impliesA(w) =bf(w) bu(w;t) =bf(w)eiwt2=3 We are now ready to inverse Fourier Transform and equation (16) above, witha=t2=3, says that u(x;t) =f(xt2=3)Solve the heat equationc2uxx=ut;u(x;0) =f(x) Take the Fourier Transform of both equations. The initial condition gives bu(w;0) =bf(w) and the PDE gives c

2(w2bu(w;t)) =@@tbu(w;t)

Which is basically an ODE int, we can write it as

@@tbu(w;t) =c2w2bu(w;t)

Which has the solution

bu(w;t) =A(w)ec2w2t and the initial condition above impliesA(w) =bf(w) bu(w;t) =bf(w)ec2w2t We are now ready to inverse Fourier Transform: First use (17) with

14a=c2tora=14c2t

to note that p22cptF exp(x24c2t =ec2w2t

So that by the convolution equation (15)

u(x;t) =f(x)12cpt exp x24c2t Solve the wave equationc2uxx=utt;u(x;0) =f(x) andut(x;0) =g(x) Take the Fourier Transform of both equations. The initial condition gives bu(w;0) =bf(w) but(w;0) =@@tbu(x;t)t=0=bg(w) and the PDE gives c

2(w2bu(w;t)) =@2@t2bu(w;t)

Which is basically an ODE int, we can write it as

2@t2bu(w;t) +c2w2bu(w;t) = 0

Which has the solution, and derivative

bu(w;t) =A(w)coscwt+B(w)sincwt @@tbu(w;t) =cwA(w)sincwt+cwB(w)coscwt so the rst initial condition givesA(w) =bf(w) and the second givescwB(w) =bg(w) make the solution bu(w;t) =bf(w)coscwt+bg(w)wsincwtc

Lets rst look at

b f(w)coscwt=bf(w)eiwct+eiwct2

12bf(w)eicwt+bf(w)eicwt

Applying equation (16) witha=ctand witha=ctyields

F

1[bf(w)coscwt] =12(f(x+ct) +f(xct))

The second piece

bg(w)wsincwtc=bg(w)iwsincwtic and now the rst factor looks like an integral, as a derivative with respect toxwould cancel theiwin bottom. Dene h(x) =Z x s=0g(s)ds

By fundamental theorem of calculus

h

0(x) =g(x)

and by (8) bg(w) =iwbh(w)

Sobg(w)wsincwtc=bh(w)eicwteicwt2i

1ic

12cbh(w)eicwtbh(w)eicwt

Applying equation (16) witha=ctand witha=ctyields

F

1[1wcbg(w)sincwt] =12c(h(x+ct)h(xct))

12c Zx+ct 0 g(s)dsZ x+ct 0 g(s)ds =12cZ x+ct xctg(s)ds Putting both pieces together we get D'Alembert's solution u(x;t) =12(f(xct) +f(x+ct)) +12cZ x+ct xctg(s)ds (The careful reader will notice that there might be a problem nding the fourier transform ofh(x) due to likelyhood of lim

x!1h(x)6= 0. But that is a story for another day.)Solveuxx+uyy= 0 on innite strip (1;1)[0;1] with boundary conditionsu(x;0) = 0 and

u(x;1) =f(x). Take the Fourier Transform of all equations. The boundary conditons yield bu(w;0) = 0 bu(w;1) =bf(w) and the PDE gives w2bu(w;y) +@2@y2bu(w;y) = 0 Which is basically an ODE iny, with a solution of the form bu(w;y) =A(w)coshwy+B(w)sinhwy They= 0 condition impliesA(w) = 0 and they= 1 implies

B(w) =bf(w)sinhw

bu(w;y) =bf(w)sinhwysinhw

We get the solution

u(x;y) =1p2Z 1

w=1bf(w)sinhwysinhweiwxdwSolveux+ut= 0;u(x;0) =f(x) (Old Homework Problem) Take the Fourier Transform of both

equations. The initial condition gives bu(w;0) =bf(w) and the PDE gives iwbu(w;t) +@@tbu(w;t) = 0

Which is basically an ODE int, we can write it as

@@tbu(w;t) =iwbu(w;t) and which has the solution bu(w;t) =A(w)eiwt and the initial condition above impliesA(w) =bf(w) bu(w;t) =bf(w)eiwt We are now ready to inverse Fourier Transform and equation (16) above, witha=t, says that u(x;t) =f(xt) Solveux+ut+u= 0;u(x;0) =f(x) (Old Homework Problem) Take the Fourier Transform of both equations. The initial condition gives bu(w;0) =bf(w) and the PDE gives iwbu(w;t)) +@@tbu(w;t) +bu(w;t) = 0

Which is basically an ODE int, we can write it as

@@tbu(w;t) = (iw1)bu(w;t) and which has the solution bu(w;t) =A(w)e(iw1)t and the initial condition above impliesA(w) =bf(w) bu(w;t) =etbf(w)eiwt We are now ready to inverse Fourier Transform and equation (16) above, witha=t, says that u(x;t) =etf(xt)quotesdbs_dbs19.pdfusesText_25