chapter 2 supplymentary notes

7/26/2019 Chapter 2 Supplymentary Notes

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Chapter 2 Supplymentary Materials

Fourier transform

Fourier Transform for solving heat equationConsider

PDE: u

t(x, t) =

2u

x2(x, t), >0, < x < , x >0

I.C.: u(x, 0) = (x)

Taking Fourier transform of both PDE and I.C. (w.r.t. x)

tu(k, t) + k2u(k, t) = 0 and (1)

u(k, 0) = (k) (2)

Here, u and are Fourier transform ofu and . Also, we used the fact:

ut(k, t) =

tu(k, t) and

uxx(k, t) = (ik)2u(k, t)Multiply both side of (1) by the integrating factor ek

2x, we get:

t

ek

2tu(k, t)

= 0 (3)

Integrate (3), we get:

ek2tu(k, t) = f(k)

u(k, t) = f(k)ek2t

Now,

u(k, 0) = (k) u(k, t) = (k)ek2t

In order to find u(x, t), we need to do inverse Fourier transform ofu(k, t).

Now, inverse Fourier transform ofek2t :

S(x, t) = 1

2

ek2teikxdk=

14t

e 14tx

2

Now, recall that:Ifu = S , then u= S Now,

u(k, t) = e

k2

t(k, t) = S(k, t)(k, t)

Hence,

u(x, t) = S(x, t) (x, t)=

S(x y, t)(y, t)dy

= 1

4t

e(xy)2

4t (y)dy

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Fourier transform to solve the wave equationConsider:

PDE : 2ut2(x, t) = c

22ux2 (x, t), < x < , t >0, c = 0

I.C. : u(x, 0) = H(x); ut(x, 0) = 0

Taking the Fourier transform of the PDE (w.r.t. x)2

t2u(k, t) = c2k2u(k, t)

From Homework 1, we know that:

u(k, t) = F(k)eikct + G(k)eikct

Recakk that for u(t) + c2k2u(t) = 0, we assume that u(t) = ert. Then

u(t) + c2k2u(t) =

r2 + c2k2

ert = 0

if and only ifr = ick.

u(t) = d1

eickt + d2

eickt

Now,

u(x, t) = 1

2

u(k, t)eikxdk

= 1

2

F(k)eikct + G(k)eikct

eikxdk

= 1

2

F(k)eik(xct)dk+ 1

2

G(k)eik(x+ct)dk

= F(x ct) + G(x + ct)This is called the DAlembert formula for the solution of the wave equation.

Now,

0 = ut(x, 0) = cF(x) + cG(x) F(x) = G(x) F(x) = G(x) + D (4)

And

u(x, 0) = H(x)

F(x) + G(x) = H(x) (5)Combining (4) and (5), we get:

2G(x) + D = H(x) G(x) = H(x) D

2 ; F(x) =

H(x) + D

2

Hence:

u(x, t) = H(x ct) + D

2 +

H(x + ct)D2

= 1

2[H(x ct) + H(x + ct)]

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Discrete Fourier transform

Goal: Approximate Fourier coefficients in complex form.

Algorithm for Fourier Series computation:

Input : f(x) for x [0, 2] (infinite data)Output : Fourier coefficients : c0, c1, c2,...,ck,... (infinite data)

Discrete case : approximation :

Input : Given n input data : f0 = f(x0), f1 = f(x1), ..., fn1 = f(xn1)Output : c0, c1, c2,...,cn1 (discrete Fourier coefficients)

In this case, we want to find a transformation T such that

T

f0f1

...fn1

=

c0c1

...cn1

Simple example ( 4 input data points )

Input : f0 = f(0) = 2, f1 = f(x1 = 2 ) = 4, f2 = f(x2 = ) = 6, f3 = f(x3 =

32 ) = 8.

Goal : Approximate f byF4(x) = c0+ c1eix + c2e

2ix + c3e3ix such that

F4(xi) = fi = f(xi)

Then we have

F4(x0) = f0 = 2

c0+ c1+ c2+ c3 = 2

F4(x1) = f1 = 4 c0+ c1ei2 + c2ei + c3e 3i2 = 4F4(x2) = f2 = 6 c0+ c1ei + c2e2i + c3e3i = 6

F4(x3) = f3 = 8 c0+ c1e 3i2 + c2e3i + c3e 9i2 = 8

(6)

(6) can be written as :

A

c0c1c2c3

=

f0f1f2f3

=

2468

where

A=

1 1 1 1

1 e2i ei e

3i2

1 ei e2i e3i1 e

3i2 e3i e

9i2

= 1 1 1 11 i i2 i3

1 i2 i4 i61 i3 i6 i9

Recall

T

f0f1f2f3

=

c0c1c2c3

T=A1

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Consider

A=

1 1 1 11 i (i)2 (i)31 (i)2 (i)4 (i)61 (i)3 (i)6 (i)9

We can check that

AA= AA= 4I

Then we have

T=A1 =A

4

Thus c0c1c2c3

= A4

f0f1f2f3

=14

1 1 1 11 i 1 i1 1 1 11 i 1 i

2468

=

51 + i1

1 i

Remark: c0, c1, c2, c3 is an approximation of Fourier coefficients f(k).

Discrete Fourier transform for n inputs

Let f(x)defined on [0, 2]. Approximate f(x)by

Fn(x) =n1k=0

ckeikx, x [0, 2]

such that Fn(xj) = f(xj) = fj and xj = j2n

, j = 0, 1, 2,...,n 1

Then we have:

Fn(x0) = c0+ c1+ + cn1 = f0Fn(x1) = c0+ c1e

ix1 +

+ cn1e

(n1)ix1 =f1

Fn(x2) = c0+ c1eix2 + + cn1e(n1)ix2 =f2...

Fn(xn1) = c0+ c1eixn1 + + cn1e(n1)ixn1 =fn1

(7)

Let = e2in =eix1 , then

2 =e4in =eix2 ; 3 =eix3 ; etc.

Thus, (7) is reduced to:

1 1 1 11 2 n11 2 4 2(n1)...

.

.....

.. .

...

1 n1 2(n1) (n1)2

c0c1c2...

cn1

=

f0f1f2...

fn1

Now,

1 + + 2 + + n1 =1 n

1 = 0

n =en2n i = 1

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and

1 1 + jk + 2j2k + + (n1)j(n1)k= 1 + e

2i(jk)n + e

4i(jk)n + e 2i(n1)(jk)n

= n ifj = k1e2i(jk)1e

2i(jk)n

= 0 forj=k

=

AAjk

Therefore we haveAA= AA= nI

and so

c0c1c2...

cn

1

=T

f0f1f2...

fn

1

=A1

f0f1f2...

fn

1

=A

n

f0f1f2...

fn

1

then

ck = 1

n

f0+

kf1+ 2kf2+ + (n1)kfn1

=

1

n

f0+ e

2in kf1+ e

4in kf2+ + e

2i(n1)kn fn1

That is

ck = 1

n

n1j=0

fjei( 2jkn ), k = 0, 1, 2,...,n 1

Fast Fourier transform

Consider the discrete Fourier transform matrix:

Fn=

1 1 1 11 2 n11 2 4 2(n1)...

......

. . . ...

1 n1 2(n1) (n1)2

, where n= ei2n

The computation of the inverse Fourier transform (hence discrete Fourier transform)

= computation ofFn

x1x2...

x4

for a given

x1x2...

x4

.Computational cost ofFnx= n

2 multiplication.

Goal: Improve computational cost from n2 to n2 log2 n. (FFT)

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Remark: Say n = 210. Then, n log2 n = 210 10 < 214. But n2 = 220 = 26 214. So FFT is 26 times

faster !!

Let y= Fn

x. Say y=

y0y1

...yn1

x=

x0x1

...xn1

. Let n = 2m. Then we have

yj =

n1k=0

kjn xk =

2m1k=0

kjn xk

Divide k = 0, 1, 2, , 2m 1 into 2 parts:

Part 1: 0, 2, 4, , 2(m 1);

Part 2: 1, 3, 5, , 2m 1;

Then

yj =m1k=0

2kjn x2k+m1k=0

(2k+1)jn x2k+1

Part 1 Part 2

wherem= e

i 2m =ei

4n =2n

Thus,

yj = (Fmx)j =

jn(Fmx)j, j = 0, 1, 2, m 1

= (y)j+ jn(y)j, j = 0, 1, 2, m 1

How about yj for j = m, m + 1, 2m 1??Replacej byj + m, j = 0, 1, 2, m 1, we have:

yj+m =

m1k=0

2k(j+m)n x2k+

m1k=0

(2k+1)(j+m)n x2k+1

=m1k=0

kj+kmm x

k+ j+mn

m1k=0

k(j+m)m xk

But

kmm =

ei

2m

km

=ei2k = 1

mn = ei 2n m =ei = 1Therefore

yj+m =m1k=0

kjm x

k jnm1k=0

kjm xk

= (Fmx)j jn(Fmx)j

= yj jnyj

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To summarize, y= (y1, y2, y2m1)T can be computed :

Step 1: Split xinto

x = x0, x2, , x2(m1)T

x = (x1, x3, , x2m1)T

Step 2: Compute y =Fmx and y = Fmx, where Fm is an

n2 matrix.

Step 3: Compute the components ofy = Fnx by:

yj = y

j+ jnyj

yj+m = y

j jnyj, (j = 0, 1, 2, , m 1)

Also, in Step 2, Fmx can be reduced to multiplication ofFm/2 and eventually to F1 (assuming n = 2

l).

This process is called the Fast Fourier transform (FFT).

Computation cost of FFT:

Let Cm=computations ofFm. Clearly, C1 = 1.Claim: C2m= 2Cm+ 3m.

Proof:Recall: y =Fmx; y = Fmxin Step 2.

RecallStep 3,

yj =y

j + jnyj

yj+m= y

j jnyj1 multiplcation and 2 additions =

Total 3m

ThereforeC2m= 2Cm+ 3m

Let n = 2l. Then:

C2l = 2C2l1+ 3 2l1

2lC2l = 2l+1C2l1+3

2

= 2l+2C2l2+ 2

3

2

=

= 20C20+ l32= 1 +

3

2l

Therefore

C2l = 2l +

3

2l 2l =n +3

2n log2 n= O(

n

2log2 n)

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Exanple of FFT (F4)

Goal: Find

y0y1

y2y3

=F4

x0x1

x2x3

Step 1: Split x into:

x =

x0x2

; x =

x1x3

Step 2: Compute

y =

y0y1

=F2

x0x2

; y =

y0y1

=F2

x1x3

Fory =

y0y1

=F2

x0x2

, we do one more sub-step :

Sub-step 1: Split x0x2 into x0; x2.Sub-step 2: Compute:

ys = F1x0 = x0

ys = F1x2 = x2

Sub-step 3:

y0 = y

s+ 02ys

= x0+ x2

y1 = ys 02ys= x0 x2

Similarly, for y =

y0y1

=F2

x1x3

, we have

y0 = x1+ x3

y1 = x1 x3Step 3:

y0 = y

0+ 04y0 = (x0+ x2) + (x1+ x3)

y1 = y

1

+ 0

4

y1 = (x0

x2) + i(x1

x3)

y2 = y

0 04y0 = (x0+ x2) (x1+ x3)y3 = y

1 04y1 = (x0 x2) i(x1 x3)

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chapter 2 supplymentary notes

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