MA 36600 LECTURE NOTES: MONDAY, MARCH 23
Higher Order Linear Equations
Linear Independence.
Consider a collection of functions
{
f
1
, f
2
, . . . , f
n
}
. We say that this is a
linearly
independent set
if the only solution to the equation
k
1
f
1
(
t
) +
k
2
f
2
(
t
) +
· · ·
+
k
n
f
n
(
t
) = 0
for all
t
is
k
1
=
k
2
=
· · ·
=
k
n
= 0. We say that this is a
linearly dependent set
otherwise. We explain the relationship
with the
Wronskian
, as the
n
×
n
determinant,
W
(
f
1
, f
2
, . . . , f
n
)
(
t
) =
f
1
(
t
)
f
2
(
t
)
· · ·
f
n
(
t
)
f
(1)
1
(
t
)
f
(1)
2
(
t
)
· · ·
f
(1)
n
(
t
)
.
.
.
.
.
.
.
.
.
.
.
.
f
(
n

1)
1
(
t
)
f
(
n

1)
2
(
t
)
· · ·
f
(
n

1)
n
(
t
)
.
We will show that the following are equivalent:
i.
W
(
f
1
, f
2
, . . . , f
n
)
(
t
0
)
6
= 0 for some
t
0
.
ii.
W
(
f
1
, f
2
, . . . , f
n
)
(
t
)
6
= 0 for all
t
.
iii.
{
f
1
, f
2
, . . . , f
n
}
is a linearly independent set.
We give the proof, following Lecture #16 on Monday, February 23. Consider the following propositions:
p
1
= “
W
(
f
1
, f
2
, . . . , f
n
)
(
t
0
)
6
= 0 for some
t
0
”
p
2
= “
W
(
f
1
, f
2
, . . . , f
n
)
(
t
)
6
= 0 for all
t
”
p
3
= “
{
f
1
, f
2
, . . . , f
n
}
is a linearly independent set”
We have already seen that
p
1
⇐⇒
p
2
; this is Abel’s Theorem: Say that
{
y
1
, y
2
, . . . , y
n
}
is a set of functions
satisfying the equation
P
0
(
t
)
d
n
y
dt
n
+
P
1
(
t
)
d
n

1
y
dt
n

1
+
· · ·
+
P
n

1
(
t
)
dy
dt
+
P
n
(
t
)
y
= 0
.
Then their Wronskian satisfies
P
0
(
t
)
dW
dt
+
P
1
(
t
)
W
= 0
=
⇒
W
(
y
1
, y
2
, . . . , y
n
)
(
t
) =
C
exp

Z
t
P
1
(
τ
)
P
0
(
τ
)
dτ
.