Math 412-501
Theory of Partial Differential Equations
Lecture 3-9: Convolution theorem.
Applications of Fourier transforms.
Fourier transform
Given a function h : R → C, the function
Z ∞
1
ĥ(ω) = F[h](ω) =
h(x)e −iωx dx,
2π −∞
ω∈R
is called the Fourier transform of h.
Given a function H : R → C, the function
Z ∞
−1
H(ω)e iωx dω,
Ȟ(x) = F [H](x) =
−∞
x ∈R
is called the inverse Fourier transform of H.
Theorem Suppose h is an absolutely integrable
function on (−∞, ∞) and let H = F[h] be its
Fourier transform.
(i) If h is smooth then h = F −1 [H].
(ii) If h is piecewise smooth then the inverse
Fourier transform F −1 [H] is equal to h at points of
continuity. Otherwise
h(x+) + h(x−)
F −1 [H](x) =
.
2
In particular, any smooth, absolutely integrable
function h : R → C is represented as a Fourier
Z ∞
integral
H(ω)e iωx dω.
h(x) =
−∞
Proposition 1
(i) F[af + bg ] = aF[f ] + bF[g ] for all a, b ∈ C.
(ii) If g (x) = f (x + α) then ĝ (ω) = e iαω fˆ(ω).
(iii) If h(x) = e iβx f (x) then ĥ(ω) = fˆ(ω − β).
R∞
Proposition 2 Suppose that −∞ |f (x)| dx < ∞.
Then (i) fˆ is well defined and bounded;
(ii) fˆ is continuous;
(iii) fˆ(ω) → 0 as ω → ∞.
Theorem 1 Let f be a smooth function such that
f , f ′ , f ′′ , . . . , f (k) are all absolutely integrable on R.
(k) (ω) = (iω)k fˆ(ω);
Then (i) fd
(ii) fˆ(ω) = α(ω)/ω k , where lim α(ω) = 0.
ω→∞
Theorem
2 Let f be a function on R such that
R
k
R (1 + |x| )|f (x)| dx < ∞ for some integer k ≥ 1.
Then (i) fˆ is k times differentiable;
(ii) fˆ(k) (ω) = (−i)k F[x k f (x)](ω).
Convolution
Suppose f , g : R → C are bounded, absolutely
integrable functions. The function
Z
f (y )g (x − y ) dy
(f ∗ g )(x) =
R
is called the convolution of f and g .
Lemma f ∗ g = g ∗ f .
Proof: Let z = x − y . Then
Z ∞
f (y )g (x − y ) dy
(f ∗ g )(x) =
−∞
Z ∞
=
f (x − z)g (z) dz = (g ∗ f )(x).
−∞
Convolution Theorem
(i) F[f · g ] = F[f ] ∗ F[g ];
(ii) F[f ∗ g ] = 2π F[f ] · F[g ].
1
F[f ∗ g ](ω) =
2π
Proof of (ii):
1
2π
Z Z
1
=
2π
Z Z
=
R
R
R
Z
R
(f ∗ g )(x)e −iωx dx
f (y )g (x − y )e −iωx dx dy
R
(x = y + z)
f (y )g (z)e −iω(y +z) dz dy = 2π fˆ(ω)ĝ (ω).
Plancherel’s Theorem (a.k.a. Parseval’s Theorem)
(i) If a function f is both absolutely integrable and
square-integrable on R, then F[f ] is also
square-integrable. Moreover,
Z
Z
2
|f (x)| dx = 2π |fˆ(ω)|2 dω.
R
R
(ii) If functions f , g are absolutely integrable and
square-integrable on R, then
Z
Z
f (x)g (x) dx = 2π fˆ(ω)ĝ (ω) dω.
R
That is, hf , g i = 2π hfˆ, ĝ i.
R
2
Gaussian g (x) = e −αx , α > 0
(density of the normal probability distribution)
Z ∞
1
2
g (x) = e
, ĝ (ω) =
e −αx e −iωx dx.
2π −∞
Z
1 ∞ −αx 2
cos ωx dx.
e
ĝ (ω) =
π 0
Z
d ĝ
1 ∞ −αx 2 ∂
e
=
(cos ωx) dx
dω
π 0
∂ω
Z
Z ∞
1
1 ∞ −αx 2
2
sin ωx dx =
xe
sin ωx d e −αx
=−
π 0
2απ 0
∞
−αx 2
Z ∞
e
sin ωx 1
2
x=0
e −αx d(sin ωx).
=
−
2απ
2απ 0
−αx 2
ω
d ĝ
=−
dω
2απ
Z
0
∞
2
e −αx cos ωx dx = −
ω
ĝ (ω).
2α
ω
ĝ ′
ω
ω
=−
=⇒ (log ĝ )′ = −
ĝ = − ĝ =⇒
2α
ĝ
2α
2α
′
ω2
2
=⇒ log ĝ = −
+ C =⇒ ĝ (ω) = ce −ω /(4α) ,
4α
Z ∞
1
2
where c = ĝ (0) =
e −αx dx.
2π −∞
2
2πĝ (0) =
=
ZZ
Z
∞
e
−αx 2
2
e −αx e −αy dx dy =
=
π
−π
2
e −αy dy
ZZ
e −α(x
2
+y 2 )
dx dy
R2
R2
Z
∞
−∞
−∞
2
dx
Z
Z
∞
e
−αr 2
r dr dθ = 2π
∞
2
e −αr r dr
0
0
2
e −αr ∞
π
= 2π
=
−2α r =0 α
g (x) = e −αx
Z
2
=⇒ ĝ (0) = √
ĝ (ω) = √
1
4πα
1
2
e −ω /(4α)
4πα
Heat equation on an infinite interval
Initial value problem:
∂u
∂ 2u
=k 2
(−∞ < x < ∞),
∂t
∂x
u(x, 0) = f (x).
We assume that f is smooth and rapidly decaying
as x → ∞. We search for a solution with the same
properties.
Apply the Fourier transform (relative to x) to both
sides of the equation:
2 ∂ u
∂u
.
=kF
F
∂t
∂x 2
Let U = F[u]. That is,
Z ∞
1
U(ω, t) = F[u(·, t)](ω) =
u(x, t)e −iωx dx.
2π −∞
2 ∂ u
∂U
∂u
=
, F
= (iω)2 U(ω, t).
Then F
2
∂t
∂t
∂x
∂U
= k(iω)2 U(ω, t) = −kω 2 U(ω, t).
∂t
2
General solution: U(ω, t) = ce −ω kt , where c = c(ω).
Hence
Initial condition u(x, 0) = f (x) implies that
2
U(ω, 0) = fˆ(ω). Therefore U(ω, t) = fˆ(ω)e −ω kt .
We know that
2
F[e −αx ] = √
It follows that e −ω
2
ω2
1
e − 4α ,
4πα
α > 0.
kt
= F[g (x, t)], where
p
x2
g (x, t) = ktπ e − 4kt , t > 0.
Hence U(ω, t) = fˆ(ω)ĝ (ω, t). By the convolution
theorem, u = (2π)−1 f ∗ g , that is,
Z ∞
1
u(x, t) =
f (x̃)g (x − x̃) d x̃
2π −∞
Z ∞
(x−x̃)2
1
f (x̃)e − 4kt d x̃.
=√
4πkt −∞
Initial value problem:
∂ 2u
∂u
(−∞ < x < ∞),
=k 2
∂t
∂x
u(x, 0) = f (x).
Z ∞
G (x, x̃, t) f (x̃) d x̃,
Solution: u(x, t) =
−∞
where G (x, x̃, t) = √
2
1
− (x−x̃)
4kt
e
.
4πkt
The solution is in the integral operator form. The
function G is called the kernel of the operator.
Also, G (x, x̃, t) is called Green’s function of the
problem.