62
BRUCE K. DRIVER
†
6.
Elliptic Ordinary Differential Operators
Let
Ω
⊂
o
R
n
be a bounded connected open region. A function
u
∈
C
2
(
Ω
)
is said
to satisfy Laplace’s equation if
4
u
= 0
in
Ω
.
More generally if
f
∈
C
(
Ω
)
is given we say
u
solves the
Poisson equation
if
−
4
u
=
f
in
Ω
.
In order to get a unique solution to either of these equations it is necessary to
impose “boundary" conditions on
u.
Example 6.1.
For
Dirichlet boundary conditions
we impose
u
=
g
on
∂
Ω
and
for
Neumann boundary conditions
we impose
∂u
∂ν
=
g
on
∂
Ω
,
where
g
:
∂
Ω
→
R
is a given function.
Lemma 6.2.
Suppose
f
:
Ω
C
0
−→
R
, ∂
Ω
is
C
2
and
g
:
∂
Ω
→
R
is continuous. Then
if there exists a solution to
−
4
u
=
f
with
u
=
g
on
∂
Ω
such that
u
∈
C
2
(
Ω
)
∩
C
1
(
Ω
)
then the solution is unique.
De
fi
nition 6.3.
Given an open set
Ω
⊂
R
n
we say
u
∈
C
1
(
Ω
)
if
u
∈
C
1
(
Ω
)
∩
C
(
Ω
)
and
∇
u
extends to a continuous function on
Ω
.
Proof.
If
e
u
is another solution then
v
=
e
u
−
u
solves
4
v
= 0
, v
= 0
on
∂
Ω
.
By
the divergence theorem,
0 =
Z
Ω
4
v
·
vdm
=
−
Z
Ω

∇
v

2
dm
+
Z
∂
Ω
v
∇
v
·
ndσ
=
−
Z
Ω

∇
v

2
dm,
where the boundary terms are zero since
v
= 0
on
∂
Ω
.
This identity implies
R
Ω

∇
u

2
dx
= 0
which then shows
∇
v
≡
0
and since
Ω
is connected we learn
v
is constant on
Ω
.
Because
v
is zero on
∂
Ω
we conclude
v
≡
0
,
that is
u
= ˜
u.
For the rest of this section we will now restrict to
n
= 1
.
However we will allow
for more general operators than
∆
in this case.
6.1.
Symmetric Elliptic ODE.
Let
a
∈
C
1
([0
,
1]
,
(0
,
∞
))
and
(6.1)
Lf
=
−
(
af
0
)
0
=
−
af
00
−
a
0
f
0
for
f
∈
C
2
([0
,
1])
.
In the following theorem we will impose Dirichlet boundary conditions on
L
by
restricting the domain of
L
to
D
(
L
) :=
{
f
∈
C
2
([0
,
1]
,
R
) :
f
(0) =
f
(1) = 0
}
.
Theorem 6.4.
The linear operator
L
:
D
(
L
)
→
C
([0
,
1]
,
R
)
is invertible and
L
−
1
:
C
([0
,
1]
,
R
)
→
D
(
L
)
⊂
C
2
([0
,
1]
,
R
)
is a bounded operator.
Proof.
(1) (Uniqueness) If
f, g
∈
D
(
L
)
then by integration by parts
(6.2)
(
Lf, g
) :=
Z
1
0
(
Lf
)(
x
)
g
(
x
)
dx
=
Z
1
0
a
(
x
)
f
0
(
x
)
g
0
(
x
)
dx.
Therefore if
Lf
= 0
then
0 = (
Lf, f
) =
Z
1
0
a
(
x
)
f
0
(
x
)
2
dx
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PDE LECTURE NOTES, MATH 237AB
63
and hence
f
0
≡
0
and since
f
(0) = 0
, f
≡
0
.
This shows
L
is injective.
(2) (Existence) Given
g
∈
C
([0
,
1]
,
R
)
we are looking for
f
∈
D
(
L
)
such that
Lf
=
g,
i.e.
(
af
0
)
0
=
g.
Integrating this equation implies
−
a
(
x
)
f
0
(
x
) =
−
C
+
Z
x
0
g
(
y
)
dy.
Therefore
f
0
(
z
) =
C
a
(
z
)
−
Z
1
y
≤
z
1
a
(
z
)
g
(
y
)
dy
which upon integration and using
f
(0) = 0
gives
f
(
x
) =
Z
x
0
C
a
(
z
)
dz
−
Z
1
y
≤
z
≤
x
1
a
(
z
)
g
(
y
)
dz dy.
If we let
(6.3)
α
(
x
) :=
Z
x
0
1
a
(
z
)
dz
the last equation may be written as
(6.4)
f
(
x
) =
Cα
(
x
)
−
Z
x
0
(
α
(
x
)
−
α
(
y
))
g
(
y
)
dy.
It is a simple matter to work backwards to show the function
f
de
fi
ned in
Eq.
(6.4) satis
fi
es
Lf
=
g
and
f
(0) = 0
for any constant
C.
So it only
remains to choose
C
so that
0 =
f
(1) =
C α
(1)
−
Z
1
0
(
α
(1)
−
α
(
y
))
g
(
y
)
dy.
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 Trigraph, Continuous function, Eq., BRUCE K. DRIVER

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