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Lecture 20.pdfLecture 20 Teodor Burghelea Departm...
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Lecture_20.pdfLecture 20 Teodor Burghelea Department of
Page 6
6
where
→
X
1
(
t
)=
x
11
x
12
and
→
X
2
(
t
)=
x
21
x
22
are
fundamental solutions
of the linear system (20), that is they
satisfy the following conditions:
(i)
They satisfy:
→
X
1
,
2
0
(
t
)=
A
→
X
1
,
2
(
t
)
(22)
(ii)
The are linear independent(recall what we have seen for second order linear ODE’s, it is exactly the same story):
If
C
1
→
X
1
(
t
)+
C
2
→
X
2
(
t
)=0
,
then
C
1
=
C
2
=0
(23)
The linear independence of the vectors
→
X
1
and
→
X
2
is equivalent to the each of the following statements:
(A)
The fundamental matrix Ψ(
t
)=[
→
X
1
(
t
)
,
→
X
2
(
t
)] =
x
11
x
21
x
12
x
22
obtained by arranging on columns the two vectors
→
X
1
,
→
X
2
is nonsingular (there exists its inverse, Ψ

1
(
t
)=
1
det
Ψ(
t
)
x
22

x
21

x
12
x
11
).
(B)
The determinant of the fundamental matrix (also called the Wronskian of the fundamental solutions) is nonzero:
W
[
→
X
1
,
→
X
2
](
t
)=
det
(Ψ(
t
)) =
k
x
11
x
21
x
12
x
22
k
=
x
11
x
22

x
12
x
21
6
=0
(24)
The last piece of information you need in order to write down the solution of the linear and homogeneous system is
the following:
Knowing the coeﬃcient matrix A, how do I get the fundamental solutions???
Depending on the
eigenvalues
λ
1
,
2
of the coeﬃcient matrix, there are several possible cases.
1.
CASE 1
:Real and distinct eigenvalues.
This is actually the case of Example 1 (we dealt there with a saddle point) in lecture notes 19.
In this case, the fundamental solutions are:
→
X
1
(
t
)=
→
v
1
e
λ
1
·
t
,
and
→
X
2
(
t
)=
→
v
2
e
λ
2
·
t
(25)
where
→
v
1
is the eigenvector corresponding to the eigenvalue
λ
1
and
→
v
2
is the eigenvector corresponding to the eigenvalue
λ
2
.
Page 7
7
2.
CASE 2
:Complex eigenvalues.
This is actually the case of Example 3 in lecture notes 19 (the center example) and example 1 (the spiral example)
in this lecture notes.
In this case, the fundamental solutions are:
→
X
1
(
t
)=
Re
[
→
v
1
e
λ
1
·
t
]
,
and
→
X
2
(
t
)=
Im
[
→
v
1
e
λ
1
·
t
]
(26)
where

→
v
1
is the eigenvector corresponding to the eigenvalue
λ
1
.
Here, please note the following important point (which is helpful for part
B
of Problem Set 10):
An alternative(which will lead to the same solution of the linear system) choice of the fundamental solutions is the
following:
→
X
1
(
t
)=
Re
[
→
v
1
e
λ
1
·
t
]
,
and
→
X
2
(
t
)=
Re
[
→
v
2
e
λ
2
·
t
]
(27)
This choice of the fundamental solutions has the following property:
→
X
1
(0) =
1
0
and
→
X
2
(0) =
→
X
2
(0)
T
=
0
1
(28)
Suggestion
:
In order to convince yourself about this, calculate the fundamental solutions as prescribed by equation
(27)
for Example
1
we have discussed in the beginning of the lecture notes.
3.
CASE 3
:Two identical eigenvalues
λ
1
=
λ
2
(and implicitly real!) but not a full set of eigenvectors.
This is actually the case of Example 2 in lecture notes 19. In this case, the fundamental solutions are:
→
X
1
(
t
)=
→
ve
λ
1
·
t
,
and
→
X
2
(
t
)=
→
vt
·
e
λ
1
·
t
+
→
we
λ
1
·
t
(29)
where
→
v
is the eigenvector corresponding to the eigenvalue
λ
1
and
→
w
is the generalized eigenvector which is the
solution of the following matrix equation:
(
A

λ
1
I
2
)
→
w
=
→
v,I
2
=
10
01
(30)
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