#16 (10/30/2028)

Period other than 2π

If the period of f(x) is 2l over [−ll] instead of 2π, it is necessary to change the scaling from x to πx/l as

f(x)
=


m=−∞ 
cm exp
i m πx

l

,
(1)
cm
=
1

2l

l

l 
f(x) exp
i m πx

l

dx.
(2)

Fourier integral (infinite periodicity)

It can be shown that Fourier integrals are derived as a limiting case of Fourier series by letting l (periodicity) tend to ∞. By combining Eq.(1) with Eq.(2), f(x) can be expressed as

f(x)
=


m=−∞ 

1

2l

l

l 
f(y) exp
imπy

l

dy
exp
imπx

l

=
1

2l

l

l 



m=−∞ 
exp
imπ(xy)

l


f(y) dy
=
1



l

l 





m=−∞ 

π

l

exp(im [(π)/(l)] (xy))

I(x, y)
 
f(y) dy
=
1



l

l 
I(x, y) f(y) dy,
(3)
where

I(x, y)
=


m=−∞ 

π

l

exp
im π

l
 (xy)
=


m=−∞ 
(∆ω)ei  m  ∆ω  (xy)
=


m=−∞ 
(∆ω) em ∆ω  z
=
∆ω{…+ ez ∆ω+1 +exp(z ∆ω) +exp(2z ∆ω) + …}
 ∼ 



−∞ 
ez  ω dω
=



−∞ 
ei  (xy)  ω dω,
(4)
where ∆ω = π/l and z = i  (xy) were used.
By combining Eq.(3) with Eq.(4) and exchange the order of integrals, one obtains

f(x)
=
1





−∞ 




−∞ 
ei x ω eiyω dω
f(y) dy
=



−∞ 
1






−∞ 
f(y) ei y ω dy
ei x ω dω
=



−∞ 
F(ω) ei x ω dω,
(5)
where

F(ω) ≡ 1





−∞ 
f(y) ei y ω dy.
(6)
So the (non-periodic) function, f(x) can be expressed as

f(x) =


−∞ 
F(ω) eiωx dω,
(7)
where

F(ω) = 1





−∞ 
f(x) eiωx dx.
(8)

Fourier series
Fourier integrals
f(x) =

−∞ 
cm ei m x
f(x) =


−∞ 
F(ω) ei ωx dω
cm = 1



π

−π 
f(x) ei m x dx
F(ω) = 1





−∞ 
f(x)eiωx dx
(9)
prism.jpg
Example 1: Rectangular pulse.
recpulse.jpg

F(ω)
=
1





−∞ 
f(x)eiωx dx
=
1



1

−1 
eiωx dx
=
sin ω

πω
.
(10)
So
f(x)
=



−∞ 
sin ω

πω
eiωx dω
=
2

π



0 
sin ω

ω
cos ωx dω.
(11)
By substituting x=0, one obtains
1 = 2

π



0 
sin ω

ω
dω,
(12)
or



0 
sin ω

ω
dω = π

2
.
(13)
sinwoverw.jpg
The figure above shows the Fourier transform of f(x). It is seen that f(x) contains frequencies from the entire interval but the largest contribution is ω = 0.
Example 2: Error function
f(x) = ex2/2.
gaussian.jpg

F(ω)
=
1

2 π



−∞ 
ex2/2 ei ωx dx
=
1

2 π



−∞ 
exp
1

2
(x + i ω)2 ω2

2

dx
=
1

2 π
e−ω2/2


−∞ 
exp
1

2
(x + i ω)2
dx
=
1

2 π
e−ω2/2


−∞ 
exp
1

2
x2
dx
=
1




2 π
e−ω2/2.
(14)
Note that the Fourier integral of the error function is also an error function.
Example 3: White color
f(x) = δ(x).

F(ω) = 1

2 π



−∞ 
δ(x) ei ωx dx = 1

2 π
.
Example 4: Diffusion equation

u

t
=
2 u

x2
,     (−∞ < x < ∞, 0 < t < ∞)
(15)
u(x,0)
=
f(x)     (−∞ < x < ∞)
(16)
Assume that the solution, u(x, t), is expressed by the Fourier integral as

u(x, t) =


−∞ 
U(ω, t) ei ωx dω.
(17)
Noting that
u

t
=



−∞ 
dU

dt
ei ωx dω,
(18)
2 u

x2
=



−∞ 
(− ω2) U ei ωx dω,
(19)
it follows
dU

dt
= (− ω2) U,
(20)
which can be solved as
U(ω, t) = Aω e−ω2 t.
(21)
Thus, Equation (17) becomes

u(x, t) =


−∞ 
Aω e−ω2 t ei ωx dω.
(22)
From the initial condition, it follows

f(x) =


−∞ 
Aω ei ωx dω,
(23)
which implies that Aω is the Fourier transform of f(x) as

Aω
=
1





−∞ 
f(x) eiωx dx.
(24)
If the initial temperature is the Dirac delta function with a peak at x=0 at t=0,

f(x) = δ(x),
(25)

Aω
=
1





−∞ 
δ(x) eiωx dx
=
1


.
(26)
Hence, Equation (17) becomes

u(x, t)
=
1





−∞ 
exp(−tω2) eiωx dω
=
1





−∞ 
exp( −t(ω− ix

2t
)2 x2

4t
) dω
=
1


exp(− x2

4t
)


−∞ 
exp( −t(ω− ix

2t
)2) dω
=
1


exp(− x2

4t
)


−∞ 
exp( −tω2) dω
=
1


exp(− x2

4t
)   ⎛


π

t
 
=
exp(− x2

4t
)

2


πt
.
(27)
Here is how the temperature decays:
It is seen that a pulse initially at x=0 propagates to x=±∞ instantly at any small t which is the nature (and flaw) of the diffusion equation.




File translated from TEX by TTH, version 4.03.
On 29 Oct 2024, 19:24.