Module Module 8: The Fourier ourier Transfo ransform rm Methd Methdos os for for PDEs PDEs In the previous modules (Modules 5-7), the method of separation of variables was used to obtain solutions of initial and boundary value problems for partial differential equations given over bounded spatial regions. The present module deals with partial differential equations equations defined over unbounded unbounded spatial regions. The mathematical mathematical tools used for solving initial and boundary value problems over unbounded spatial regions are integral transforms: The Fourier transform (FT), the Fourier sine transform (FST) and the Fourier cosine transform (FCT). The outline of this module is as follows. follows. The first lecture introduces introduces the necessary necessary background for the definition of FT. The FST and FCT are introduced in the second lecture. Applications of FT, FST and FCT for the three types of PDEs viz., heat equation, wave equation and Laplace equation are described in third, fourth and fifth lectures, respectively.
1
MODULE 8:
2
THE FOURIER TRANSFORM METHDOS FOR PDES
Lecture 1 Recall that if f is a periodic function with period 2L on
Fourier Transform
then f has a Fourier Series(FS)
R
representation of the form ∞
1 f (x) = a0 + 2 n
�� =1
where 1 an = L and
nπx nπx an cos( ) + bn sin( ) , L L
�
L
� �
f (x)cos(
−L
nπx )dx, L
L
1 bn = L
f (x) sin(
−L
(1)
n = 0, 1, 2, . . .
nπx )dx, n = 1, 2, . . . L
Fourier series are powerful tools in treating various problems involving periodic functions. Many practical problems do not involve periodic functions. Therefore, it is desirable to generalize the method of Fourier series to include non-periodic functions. If f is not periodic then we may regard it as periodic with an infinite period, i.e., we would like to see what happens if we let L
→ ∞.
We shall do this for reasons of motivation as well
as for making it plausible that for a non-periodic function, one should expect an integral representation (Fourier integral) instead of Fourier series. Set ω n =
nπ . L
Then (1) can be rewritten as 1 a0 + 2
f (x) =
1 L
=
L
∞
�
[an cos(ωn x) + bn sin(ωn x)].
n=1
1 f (x)dx + L −L
�
L
+sin(ωn x)
�
∞
��
L
cos(ωn x)
n=1
�
f (t)sin(ωn t)dt .
Note that
Setting ∆ω = ω n+1
−ω
1 f (x) = L
π n = L ,
−ω
n =
−L
(n + 1)π L
π − (n)π = . L L
we write the Fourier series in the form
L
�
f (t)cos(ωn t)dt
−L
−L
ωn+1
�
f (x)dx +
1 π
∞
��
L
cos(ωn x)∆ω
f (t)cos(ωn t)dt
−L
n=1
+sin(ωn x)∆ω
�
L
�
−L
�
f (t) sin(ωn t)dt .
MODULE 8:
3
THE FOURIER TRANSFORM METHDOS FOR PDES
This representation is valid for any fixed L, arbitrary large, but finite. Letting L and assuming that the resulting nonperiodic function
→ ∞
˜ = lim f (x) f (x) L→∞
is absolutely integrable over (
−∞, ∞), i.e., ∞
�
−∞
˜ |<∞ |f (x)
it seems plausible that the infinite series (1) becomes an integral from 0 to ˜ = 1 f (x) π
∞
∞
� �
cos(ωx)
0
�
˜ f (t)cos(ωt)dt + sin(ωx)
−∞
∞
�
∞, i.e.,
�
˜ f (t)sin(ωt)dt dω.
−∞
The above equation can be put in the form ˜ = 1 f (x) π
∞
�
[A(ω)cos(ωx) + B(ω) sin(ωx)]dω,
(2)
0
˜ where which is called the Fourier integral of f , ∞
A(ω) =
�
˜ f (t)cos(ωt)dt,
∞
B(ω) =
−∞
�
˜ f (t)sin(ωt)dt,
−∞
This motivates the following result. THEOREM 1. If f is piecewise continuous in every bounded interval of R and ∞
�
−∞
|f (x)| < ∞,
then f can be represented by a Fourier integral 1 f (x) = π
∞
�
[A(ω)cos(ωx) + B(ω) sin(ωx)]dω,
(3)
0
where ∞
A(ω) =
�
∞
f (t)cos(ωt)dt,
B(ω) =
−∞
�
f (t)sin(ωt)dt,
(4)
−∞
The Fourier integral of f converges to f if f is continuous. If f is discontinuous at a point x the integral converges to f (x+) + f (x ) 2
−
i.e., the average value of the left and right hand limits of f (x) at that point.
MODULE 8:
THE FOURIER TRANSFORM METHDOS FOR PDES
4
In order to motivate the definition of Fourier transform we first express complex form of the Fourier integral as follows. Using the identity cos(a we write the integral (3) as 1 f (x) = π Since cos θ =
eiθ +e−iθ
2
1 2π 1 2π
f (x) = =
∞
∞
� �� 0
− b) = cos a cos b + sin a sin b,
f (t)cos(ωx
−∞
− ωt)dt
�
dω.
(5)
, we obtain ∞
∞
∞
−∞ ∞
� � � � 0
0
f (t) ei(ωx−ωt) + e−i(ωx−ωt) dtdω
{
f (t)e
}
i(ωx−ωt)
−∞
1 dtdω + 2π
∞
∞
� � 0
f (t)e−i(ωx−ωt) dtdω
−∞
Replacing w by
−w in the second term on the right hand side and adjusting limits from
−∞ to 0, we obtain
f (x) = =
∞ ∞ 1 f (t)ei(ωx−ωt) dtdω 2π −∞ −∞ ∞ ∞ 1 1 eiωx f (t)e−iωt dt dω, 2π −∞ 2π −∞
� � � � � √ √
�
which is the complex form of the Fourier integral of f . This leads to following pair of transforms. DEFINITION 2. (The Fourier Transform) Let f : (
−∞, ∞) → R or C. The Fourier transform (FT) of f (x) is defined by ˆ (f )(ω) = f (ω) =
F
1 2π
√
∞
�
f (x)e−iωx dx
(
−∞
−∞ < ω < ∞)
(6)
provided this integral exists. REMARK 3. Note that al l functions may not have Fourier transform. For example, the constant function C (C = 0), sin x, ex , x2 do not have FT. Only functions that tend to
̸ zero sufficiently fast as |x| → ∞ (rapidly decreasing to zero functions) have FT. DEFINITION 4. (The Inverse Fourier Transform) ˆ The inverse Fourier transform of a function f (ω) ( <ω<
−∞
−1
F
( ˆ f ) = f (x) =
1 2π
√
∞
�
iωx ˆ f (ω)e dω
−∞
provided this integral exists. EXAMPLE 5. Find the FT of the function f (x) =
�
1 for x 0
| | ≤ L, for |x| > L.
(
∞) is defined as
−∞ < x < ∞)
(7)
MODULE 8:
5
THE FOURIER TRANSFORM METHDOS FOR PDES
Solution. Using the definition of FT, we have ˆ ) = f (ξ =
1 2π
√
√ 12π
∞
�
−iξx
f (x)e
−∞ − e iξL
iξL
−e −iξ
=
dx =
1 2π
√
L
�
e
−iξx
1 e−iξx 2π iξ
√ −
dx =
−L
√ 12π 2sin(ξL) . ξ
L
−L
Note that even though f (x) vanishes for x outside the interval [ L, L], the same is not ˆ ) . In general, it can be shown that if f and f ˆ vanish outside [ L, L], then true of f (ξ
−
≡ 0.
f
−
Some Basic Properties of Fourier Transforms:
• Linearity: F is a linear transformation. For any two functions f 1 and f 2 with FTs F [f 1] and F [f 2], respectively and any constants c 1 and c 2, we have F [c1f 1 + c2f 2] = c1F [f 1] + c2F [f 2]. • Conjugation: Let f (x) be a function with FT F [f ]. Then the FT of the function ¯ (complex conjugate) is given by f (x)
F [f (x)] = F [f ](−ω). • Continuity:
ˆ Let f (x) be an absolutely integrable function with FT f (ω). Then
ˆ f (ω) is a continuous function.
• Convolution: We know the convolution of the functions f and g, denoted be f ∗ g, is defined by
∞
(f g)(x) =
∗
�
f (x
−∞
− t)g(t)dt,
provided the integral exists for each x. (i.e., if f is bounded and g is absolutely ˆ integrable). Let f (ω) and gˆ(ω) be the FTs of f and g, respectively. Then ˆ gˆ(ω). F (f ∗ g) = F (f )F (g) = f (ω) ̸ F [f ]F [g]. Note: In general, F [f (x)g(x)] = ˆ • Parseval’s identity: For any two functions f (x) and g(x) with FTs f (ω) and gˆ(ω), respectively. Then
∞
� �
∞
f (x)¯ g (x)dx =
−∞
In particular,
∞
−∞
� �
ˆ g(ω)dω. f (ω)ˆ
−∞
2
|f (x)| dx =
∞
−∞
ˆ |2 dω. |f (ω)
MODULE 8:
THE FOURIER TRANSFORM METHDOS FOR PDES
• Transformation of partial derivatives:
(i) Let u = u(x, t) be a function defined for as x
6
−∞ < x < ∞ and t ≥ 0. If u(x, t) → 0
→ ±∞, and F [u](ω, t) = U (ω, t), then F [u ](ω, t) = iω F [u] = iωU (ω, t). If, in addition, u (x, t) → 0 as x → ±∞, then F [u ](ω, t) = −ω2F [u] = −ω2U (ω, t). x
x
xx
(ii) If we transform the partial derivative u t (x, t) (and if the variable of integration in the transformation is x), then the transformation is given by
F [u ](ω, t) = dtd {F [u]}(ω, t) = dtd U (ω, t). t
The Fourier transform of a time derivative equals the time derivative of the Fourier transform. This shows that time differentiation and the FT with respect to x commute. Practice Problems
1. Compute the complex FS for each of the function f (x) = e ax cos(bx). 2. Show that if f (x) is absolutely integrable on ( ibx
F [e
f (ax)] =
1 [f ((ξ a
F
− b)/a)],
3. Find the FT of 2
(A) f (x) = e −cx , where c is a constant. (B) f (x) = e −|x| (C) f (x) = sin x2 4. Verify the following properties of FT: (A)
F [u ] = iξ F [u] (B) F [u ] = −ξ 2 F [u] (C) F [u(x + c)] = e F [u], for any c ∈ R x
xx
icξ
−∞, ∞), then a,b
∈ R, a ≠ 0.
