Linear ODE’s
Second order linear equations
Many traditional problems involving ordinary equations arise as second order
linear
equations
au
+
bu
+
cu
=
f
,
more briefly as
Lu
=
f
.
The problem is, given
f
, find
u
; often we will want to find
u
that satisfies some auxiliary initial or
boundary conditions.
Here we have used the notation
Lu
:
=
a
(
x
)
u
+
b
(
x
)
u
+
c
(
x
)
u
,
(1)
so
L
takes a function
u
and gives a new function
Lu
. This operator
L
is a
linear map
because it has
the two properties
L
(
α
u
) =
α
Lu
and
L
(
u
+
v
) =
Lu
+
Lv
,
where
α
is any constant and
u
and
v
are functions. One consequence is that if
Lu
=
0 and
Lv
=
0,
then
L
(
Au
+
Bv
) =
0 for any constants
A
and
B
. The solutiond of
Lu
=
0 are often called the
nullspace
or
kernel
of
L
. These properties show that
the nullspace of a linear map is a linear space
.
For instance, in the special case where
Lu
=
u
+
u
we know that
L
cos
x
=
0 and
L
sin
x
=
0. Thus
L
(
A
cos
x
+
B
sin
x
) =
0 for any constants
A
and
B
.
E
XAMPLE
We’ll show that
any
solution of
Lu
:
=
u
+
u
=
0 has the form
u
(
x
) =
A
cos
x
+
B
sin
x
.
This will show that the nullspace of
Lu
:
=
u
+
u
=
0 has
dimension two
.
First we must pick the constants
A
and
B
. Letting
x
=
0 we see that (if this is to work)
A
=
u
(
0
)
.
Similarly, taking the derivative we get
B
=
u
(
0
)
. Let
v
(
x
)
:
=
u
(
0
)
cos
x
+
u
(
0
)
sin
x
. Our task is to
show that
u
(
x
) =
v
(
x
)
. Equivalently, if we let
w
(
x
)
:
=
u
(
x
)

v
(
x
)
, we must show that
w
(
x
)
≡
0. A
key observation motivating us is that by linearity,
w
+
w
=
0, and
w
(
0
) =
0,
w
(
0
) =
0.
Introduce the function
E
(
x
) =
1
2
[
w
2
+
w
2
]
.
Then since
w
=

w
,
E
(
x
) =
w w
+
ww
=
w
(

w
)+
ww
=
0
,
so
E
(
x
) =
constant (in many physical examples, this is
conservation of energy
). But from
w
(
0
) =
0
and
w
(
0
) =
0 we find
E
(
0
) =
0. Since
E
(
x
)
is a sum of squares, the only possibility is
w
(
x
)
≡
0,
as claimed.
This Example generalizes. Assuming
a
(
x
) =
0, the nullspace of (1) always has dimension 2. Let
ϕ
(
x
)
and
ψ
(
x
)
be the solutions of the homogeneous equation
Lu
=
0 with
ϕ
(
0
) =
1,
ϕ
(
0
) =
0,
and
ψ
(
0
) =
0,
ψ
(
0
) =
1, then every solution of the homogeneous equation
Lu
=
0 has the form
u
(
x
) =
A
ϕ
(
x
)+
B
ψ
(
x
)
for some constants
A
and
B
. The proof, which we don’t give, has two parts.
The first is the
existence
of the solutions
ϕ
and
ψ
, the second is their uniqueness. While these
proofs are not obvious, they are not killers.
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 Spring '12
 STAFF
 Linear Equations, Equations, Partial differential equation, homogeneous equation

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