The diffusion equation is a parabolic partial differential equation
and is exemplified by the following system (after setting all the
physical constants as unity):
∂2u(x,t)
∂x2
=
∂u(x, t)
∂t
.
(1)
Initial condition: u(x, 0) = f(x) (0 < x < 1)
Boundary condition: u(0, t) = u(1, t) = 0 (t ≥ 0)
We assume that u(x, t) can be expressed by Fourier sine series as
u(x, t) =
∞ ∑ n=1
un(t) sin n πx,
(2)
except that the Fourier coefficient, un, is a function of t.
Note that sin n πx was chosen as the interval is [0, 1] instead of [−π, π].
The both sides of Equation (1) can be expressed in the Fourier domain as
∂2u
∂x2
=
∞ ∑ n=1
−(n π)2un(t) sin n πx,
(3)
∂u
∂t
=
∞ ∑ n=1
dun(t)
dt
sin n πx.
(4)
from which it follows
dun(t)
dt
= −(nπ)2un(t),
(5)
which can be solved as
un(t) = Ane−(n π)2t.
(6)
where An is an integral constant yet determined.
Equation (1) with Equation (6) yields
u(x, t) =
∞ ∑ n=1
Ane−(nπ)2t sin n πx.
(7)
Substituting the initial condition at t = 0 into Equation (7) yields
f(x) =
∞ ∑ n=1
An sin n πx,
(8)
which implies that the unknown coefficient, An, is the Fourier sine coefficient of f(x) as1
An = 2
⌠ ⌡
1
0
f(x) sin n πxdx.
(9)
Example
∂2u
∂x2
=
∂u
∂t
,
(10)
u = 0
at
x = 0, 1
u=1
at
t = 0
An
=
2
⌠ ⌡
1
0
1 ×sin n πxdx
=
2 (1−cos(n π))
n π
=
2 (1−(−1)n)
nπ
.
(11)
u(x, t)
=
∞ ∑ n=1
Ane−n2 π2t sin n πx
=
∞ ∑ n=1
2 (1−(−1)n)
nπ
e−n2 π2t sin nπx.
(12)
Here is how the temperature decays.
Period other than 2π
If the period of f(x) is 2l over [−l, l] instead of 2π, it is
necessary to change the scaling from x to πx/l as
f(x)
=
∞ ∑ m=−∞
cm exp
⎛ ⎝
im πx
l
⎞ ⎠
,
(13)
cm
=
1
2l
⌠ ⌡
l
−l
f(x) exp
⎛ ⎝
−
im πx
l
⎞ ⎠
dx.
(14)
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.(13) with Eq.(14),
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(x−y)
l
⎞ ⎠
⎞ ⎠
f(y) dy
=
1
2π
⌠ ⌡
l
−l
⎛ ⎝
∞ ∑ m=−∞
⎛ ⎝
π
l
⎞ ⎠
exp(im [(π)/(l)] (x−y))
⎞ ⎠
I(x, y)
f(y) dy
=
1
2π
⌠ ⌡
l
−l
I(x, y) f(y) dy,
(15)
where
I(x, y)
=
∞ ∑ m=−∞
⎛ ⎝
π
l
⎞ ⎠
exp
⎛ ⎝
im
π
l
(x−y)
⎞ ⎠
=
∞ ∑ m=−∞
(∆ω)eim ∆ω (x−y)
=
∞ ∑ m=−∞
(∆ω) em ∆ω z
=
∆ω{…+ e−z ∆ω+1 +exp(z ∆ω) +exp(2z ∆ω) + …}
∼
⌠ ⌡
∞
−∞
ez ωdω
=
⌠ ⌡
∞
−∞
ei (x−y) ωdω,
(16)
where ∆ω = π/l and z = i (x−y) were used.
By combining Eq.(15) with Eq.(16) and exchange the order of
integrals, one obtains
f(x)
=
1
2π
⌠ ⌡
∞
−∞
⎛ ⎝
⌠ ⌡
∞
−∞
eix ωe−iyωdω
⎞ ⎠
f(y) dy
=
⌠ ⌡
∞
−∞
1
2π
⎛ ⎝
⌠ ⌡
∞
−∞
f(y) e−iy ωdy
⎞ ⎠
eix ωdω
=
⌠ ⌡
∞
−∞
F(ω) eix ωdω,
(17)
where
F(ω) ≡
1
2π
⌠ ⌡
∞
−∞
f(y) e−iy ωdy.
(18)
So the (non-periodic) function, f(x) can be expressed as
f(x) =
⌠ ⌡
∞
−∞
F(ω) eiωxdω,
(19)
where
F(ω) =
1
2π
⌠ ⌡
∞
−∞
f(x) e−iωxdx.
(20)
Fourierseries
Fourierintegrals
f(x) =
∞ ∑ −∞
cmeimx
f(x) =
⌠ ⌡
∞
−∞
F(ω) ei ωxdω
cm =
1
2π
⌠ ⌡
π
−π
f(x) e−imxdx
F(ω) =
1
2π
⌠ ⌡
∞
−∞
f(x)e−iωxdx
(21)
Example 1: Rectangular pulse.
F(ω)
=
1
2π
⌠ ⌡
∞
−∞
f(x)e−iωxdx
=
1
2π
⌠ ⌡
1
−1
e−iωxdx
=
sin ω
πω
.
(22)
So
f(x)
=
⌠ ⌡
∞
−∞
sin ω
πω
eiωxdω
=
2
π
⌠ ⌡
∞
0
sin ω
ω
cos ωxdω.
(23)
By substituting x=0, one obtains
1 =
2
π
⌠ ⌡
∞
0
sin ω
ω
dω,
(24)
or
⌠ ⌡
∞
0
sin ω
ω
dω =
π
2
.
(25)
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) = e−x2/2.
F(ω)
=
1
2 π
⌠ ⌡
∞
−∞
e−x2/2e−i ωxdx
=
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.
(26)
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) e− i ωxdx =
1
2 π
.
Example 4: Diffusion equation
∂u
∂t
=
∂2u
∂x2
, (−∞ < x < ∞, 0 < t < ∞)
(27)
u(x,0)
=
f(x) (−∞ < x < ∞)
(28)
Assume that the solution, u(x, t), is expressed by the Fourier integral as
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.
Footnotes:
1
Multiply sin m πx from both sides of Equation (8) and integrate the result
from −π to π to get
⌠ ⌡
1
0
f(x) sin m πxdx
=
∞ ∑ n=1
An
⌠ ⌡
1
0
sin n πx sin m πxdx
(40)
=
⎧ ⎪ ⎨
⎪ ⎩
Am ×
1
2
,
m = n
0,
m ≠ n
Hence,
Am = 2
⌠ ⌡
1
0
f(x) sin m πx.
File translated from
TEX
by
TTH,
version 4.03. On 11 Nov 2023, 19:43.