Math 412-501
Theory of Partial Differential Equations
Lecture 3-10: Applications of Fourier
transforms (continued).
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.
Initial value problem for the heat equation:
∂ 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.
Wave equation on an infinite interval
Initial value problem:
2
∂ 2u
2 ∂ u
=c
(−∞ < x < ∞),
∂t 2
∂x 2
∂u
u(x, 0) = f (x),
(x, 0) = g (x).
∂t
We assume that f , g are 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 2 ∂ u
∂ u
2
=
c
F
.
F
∂t 2
∂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 2 ∂ 2U
∂ u
∂ u
Then F
=
= (iω)2 U(ω, t).
,
F
2
2
2
∂t
∂t
∂x
∂ 2U
= c 2 (iω)2 U(ω, t) = −c 2 ω 2 U(ω, t).
Hence
2
∂t
General solution: U(ω, t) = a cos cωt + b sin cωt
(ω 6= 0), where a = a(ω), b = b(ω).
Apply the Fourier transform to the initial conditions:
U(ω, 0) = fˆ(ω), ∂U
∂t (ω, 0) = ĝ (ω).
sin cωt
.
Therefore U(ω, t) = fˆ(ω) cos cωt + ĝ (ω)
cω
We know that
sin aω
, a > 0.
χ
\
[−a,a] (ω) =
πω
Hence F −1 sincωcωt = πc χ[−ct,ct] , t > 0. Then
U(ω, t) = 12 (e icωt fˆ(ω) + e −icωt fˆ(ω) + πc ĝ (ω)χ\
[−ct,ct] (ω).
By the shift theorem and the convolution theorem,
1
1
u(x, t) = f (x + ct) + f (x − ct) + g ∗ χ[−ct,ct] (x).
2
2c
g ∗ χ[−ct,ct] (x) =
=
Z
Z
∞
g (x̃)χ[−ct,ct] (x − x̃) d x̃
−∞
x+ct
g (x̃) d x̃.
x−ct
Initial value problem:
2
∂ 2u
2 ∂ u
=c
(−∞ < x < ∞),
∂t 2
∂x 2
u(x, 0) = f (x), ∂u
∂t (x, 0) = g (x).
Solution:
Z x+ct
1
1
g (x̃) d x̃.
u(x, t) = f (x + ct) + f (x − ct) +
2
2c x−ct
Sine and cosine transforms of derivatives
Z
2 ∞
f (x) sin ωx dx
Sine transform:
S[f ](ω) =
π 0
Z
2 ∞
Cosine transform: C [f ](ω) =
f (x) cos ωx dx
π 0
Assume that f and f ′ are continuous and absolutely
integrable on [0, ∞). Then f (x) → 0 as x → ∞.
Hence
Z
2 ∞ ′
′
S[f ](ω) =
f (x) sin ωx dx
π 0
Z
∞
2 ∞
2
−
f (x)(sin ωx)′ dx
= f (x) sin ωx x=0
π
π 0
= −ω C [f ](ω).
Z
2 ∞ ′
Likewise, C [f ](ω) =
f (x) cos ωx dx
π 0
Z
∞
2 ∞
2
−
= f (x) cos ωx f (x)(cos ωx)′ dx
x=0
π
π 0
2
= − f (0) + ω S[f ](ω).
π
′
S[f ′ ](ω) = −ω C [f ](ω)
2
C [f ′ ](ω) = − f (0) + ω S[f ](ω)
π
Now assume that f , f ′ , f ′′ are continuous and
absolutely integrable on [0, ∞). By the above,
2
S[f ′′ ](ω) = −ω C [f ′ ](ω) = f (0)ω − ω 2 S[f ](ω),
π
2
2
C [f ′′ ](ω) = − f ′ (0) + ω S[f ′ ](ω) = − f ′ (0) − ω 2 C [f ](ω).
π
π
S[f ′′ ](ω) =
2
f (0)ω − ω 2 S[f ](ω)
π
2
C [f ′′ ](ω) = − f ′ (0) − ω 2 C [f ](ω)
π
Z ∞
1
f (x)e −iωx dx
Fourier transform: F[f ](ω) =
2π −∞
Z
2 ∞
f (x) sin ωx dx
Sine transform:
S[f ](ω) =
π 0
Z
2 ∞
Cosine transform: C [f ](ω) =
f (x) cos ωx dx
π 0
R∞
Proposition Suppose that −∞ |f (x)| dx < ∞.
(i) If f is even, f (−x) = f (x), then F[f ] is also
even; moreover, C [f ](ω) = 2F[f ](ω) for all ω > 0.
(ii) If f is odd, f (−x) = −f (x), then F[f ] is also
odd; moreover, S[f ](ω) = 2i F[f ](ω) for all ω > 0.
Heat equation on a semi-infinite interval
Initial-boundary value problem:
∂u
∂ 2u
=k 2
(0 < x < ∞),
∂t
∂x
∂u
(0, t) = 0,
∂x
u(x, 0) = f (x).
We search for a solution which is smooth and rapidly
decaying as x → ∞. Apply the cosine transform
(relative to x) to both sides of the equation:
2 ∂ u
∂u
.
=kC
C
∂t
∂x 2
Z
2 ∞
u(x, t) cos ωx dx.
Let U(ω, t) = C [u](ω) =
π 0
∂U
∂u
=
,
Then C
∂t
∂t
2 ∂ u
2 ∂u
2
C
=
−ω
U(ω,
t)
−
(0, t) = −ω 2 U(ω, t).
2
∂x
π ∂x
∂U
= −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
U(ω, 0) = C [f ](ω).
2
Therefore U(ω, t) = C [f ](ω) e −ω kt .
Solution: u(x, t) =
∞
Z
c(ω)e −ω
2
kt
cos ωx dω,
0
where c(ω) =
2
π
Z
∞
f (x̃) cos ωx̃ d x̃.
0
The same solution can be obtained by separation of
variables. The solution can be rewritten in the
integral operator form:
Z ∞
G (x, x̃, t)f (x̃) d x̃,
u(x, t) =
0
2
where G (x, x̃, t) =
π
Z
0
∞
e −ω
2
kt
cos ωx cos ωx̃ dω.
Green’s function G (x, x̃, t) =
Z
1 ∞ −ω2 kt
=
cos (x − x̃)ω + cos (x + x̃)ω dω
e
π 0
We know that
Z
y2
1
1 ∞ −αω2
− 4α
e
cos ωy dω = √
e , α > 0.
π 0
4πα
It follows that
2
1 − (x−x̃)2
− (x+x̃)
4kt
4kt
G (x, x̃, t) = √
+e
e
.
4πkt
Laplace’s equation in a half-plane
Boundary value problem:
∂ 2u ∂ 2u
+
= 0 (−∞ < x < ∞, 0 < y < ∞),
∂x 2 ∂y 2
u(x, 0) = f (x).
We assume that f is smooth and rapidly decaying at
infinity. We search for a solution with the same
properties.
Apply the Fourier transform Fx (relative to x) to
both sides of the equation:
2 2 ∂ u
∂ u
+ Fx
= 0.
Fx
2
∂x
∂y 2
Z ∞
1
u(x, y )e −iωx dx.
Let U(ω, y ) = Fx [u](ω) =
2π −∞
2 2 ∂ 2U
∂ u
∂ u
=
= (iω)2 U(ω, y ).
,
F
Then Fx
x
2
2
2
∂y
∂y
∂x
∂ 2U
= −(iω)2 U(ω, y ) = ω 2 U(ω, y ).
2
∂y
General solution: U(ω, y ) = ae ωy + be −ωy (ω 6= 0),
where a = a(ω), b = b(ω).
Hence
Initial condition u(x, 0) = f (x) implies that
U(ω, 0) = fˆ(ω).
Also, we have a boundary condition lim U(ω, y ) = 0.
y →∞
Since U(ω, y ) → 0 as y → ∞, it follows that
(
b(ω)e −ωy if ω > 0,
U(ω, y ) =
a(ω)e ωy if ω < 0.
Since U(ω, 0) = fˆ(ω), it follows that
U(ω, y ) = fˆ(ω)e −y |ω| .
It turns out that F −1 [e −α|ω| ](x) =
x2
2α
, α > 0.
+ α2
Hence U(ω, y ) = fˆ(ω)ĝ (ω, y ), where
2y
g (x, y ) = 2
.
x + y2
By the convolution theorem, u(x, y ) = (2π)−1 f ∗ g .
Boundary value problem:
∂ 2u ∂ 2u
+
= 0 (−∞ < x < ∞, 0 < y < ∞),
∂x 2 ∂y 2
u(x, 0) = f (x).
Solution:
Z ∞
1
g (x − x̃, y )f (x̃) d x̃
u(x, y ) =
2π −∞
Z
y
1 ∞
f (x̃)
d x̃.
=
π −∞
(x − x̃)2 + y 2