ANALYSIS TOOLS WITH APPLICATIONS
399
21.
Constant Coefficient partial differential equations
Suppose that
p
(
ξ
)=
P

α

≤
k
a
α
ξ
α
with
a
α
∈
C
and
(21.1)
L
=
p
(
D
x
):=
Σ

α

≤
N
a
α
D
α
x
=
Σ

α

≤
N
a
α
μ
1
i
∂
x
¶
α
.
Then for
f
∈
S
c
Lf
(
ξ
)=
p
(
ξ
)
ˆ
f
(
ξ
)
,
that is to say the Fourier transform takes a constant coe
ﬃ
cient partial di
f
erential
operator to multiplication by a polynomial. This fact can often be used to solve
constant coe
ﬃ
cient partial di
f
erential equation. For example suppose
g
:
R
n
→
C
is
a given function and we want to
f
ndaso
lut
iontotheequat
ion
Lf
=
g.
Taking the
Fourier transform of both sides of the equation
Lf
=
g
would imply
p
(
ξ
)
ˆ
f
(
ξ
)=ˆ
g
(
ξ
)
and therefore
ˆ
f
(
ξ
)=ˆ
g
(
ξ
)
/p
(
ξ
)
provided
p
(
ξ
)
is never zero. (We will discuss what
happens when
p
(
ξ
)
has zeros a bit more later on.) So we should expect
f
(
x
)=
F
−
1
μ
1
p
(
ξ
)
ˆ
g
(
ξ
)
¶
(
x
)=
F
−
1
μ
1
p
(
ξ
)
¶
F
g
(
x
)
.
De
f
nition 21.1.
Let
L
=
p
(
D
x
)
as in Eq. (21.1). Then we let
σ
(
L
):=
Ran
(
p
)
⊂
C
and call
σ
(
L
)
the
spectrum
of
L.
Given a measurable function
G
:
σ
(
L
)
→
C
,
we
de
f
ne (a possibly unbounded operator)
G
(
L
):
L
2
(
R
n
,m
)
→
L
2
(
R
n
,m
)
by
G
(
L
)
f
:=
F
−
1
M
G
◦
p
F
where
M
G
◦
p
denotes the operation on
L
2
(
R
n
,m
)
of multiplication by
G
◦
p,
i.e.
M
G
◦
p
f
=(
G
◦
p
)
f
with domain given by those
f
∈
L
2
such that
(
G
◦
p
)
f
∈
L
2
.
At a formal level we expect
G
(
L
)
f
=
F
−
1
(
G
◦
p
)
F
g.
21.0.3.
Elliptic examples.
As a speci
f
c example consider the equation
(21.2)
¡
−
∆
+
m
2
¢
f
=
g
where
f,g
:
R
n
→
C
and
∆
=
P
n
i
=1
∂
2
/∂x
2
i
is the usual Laplacian on
R
n
.
By
Corollary 20.16 (i.e. taking the Fourier transform of this equation), solving Eq.
(21.2) with
f,g
∈
L
2
is equivalent to solving
(21.3)
¡

ξ

2
+
m
2
¢
ˆ
f
(
ξ
)=ˆ
g
(
ξ
)
.
The unique solution to this latter equation is
ˆ
f
(
ξ
)=
¡

ξ

2
+
m
2
¢
−
1
ˆ
g
(
ξ
)
and therefore,
f
(
x
)=
F
−
1
³
¡

ξ

2
+
m
2
¢
−
1
ˆ
g
(
ξ
)
´
(
x
)=
:
¡
−
∆
+
m
2
¢
−
1
g
(
x
)
.
We expect
F
−
1
³
¡

ξ

2
+
m
2
¢
−
1
ˆ
g
(
ξ
)
´
(
x
)=
G
m
F
g
(
x
)=
Z
R
n
G
m
(
x
−
y
)
g
(
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document400
BRUCE K. DRIVER
†
where
G
m
(
x
):=
F
−
1
¡

ξ

2
+
m
2
¢
−
1
(
x
)=
Z
R
n
1
m
2
+

ξ

2
e
iξ
·
x
d
ξ.
At the moment
F
−
1
¡

ξ

2
+
m
2
¢
−
1
only makes sense when
n
=1
,
2
,
or
3
because
only then is
¡

ξ

2
+
m
2
¢
−
1
∈
L
2
(
R
n
)
.
For now we will restrict our attention to the one dimensional case,
n
=1
,
in
which case
(21.4)
G
m
(
x
)=
1
√
2
π
Z
R
1
(
ξ
+
mi
)(
ξ
−
mi
)
e
iξx
dξ.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Three '10
 Smith
 Differential Equations, Equations, Partial Differential Equations, Trigraph, Partial differential equation, wave equation, BRUCE K. DRIVER

Click to edit the document details