18.034, Honors Differential Equations
Prof. Jason Starr
Lecture 30: Notes for Kiran: I. Summary of Lecture 29
4/21/04
We finished last time with Jordan normal form.
Notation(: i ) V a finite dimensional C
I -vector space (usually V = C
I n with standard
basis e1 ,...., en ).
(ii) T = V → V a linear operator
(iii) A linear system of differential equations
y ' = Ty
i.e. looking for differentiable y = IR → V s.t.
y ' (t ) = T (y (t )) .
(iv) matrix of T w.r.t. an ordered basis B = (v 1 ,....., v n ) , A = [T ]B, B,
n
Tv j =
∑A
ij
⋅ vi .
i =1
(v) char. polynomial pT (λ ) = det(λId v− T ) = λ n − Tr (T )λ + ......
(vi) factorization pT (λ ) = (λ − λ1 )
(
m1
..... λ
− λt )
mt
.
(vii) for each i = 1,...., t , the generalized eigenspaces,
V λi = K es (T − λ i Id ) ,
r
(r )
V λi = V
(1)
λi
C
−
V
(2)
λi
(viii) restriction of T to V λi
gen
(ix) N i = Ti - λ i Id V
gen
λi
C …… C V
−
−
(mi )
=
λi
V
gen
λi
← generalized eigenspace.
is Ti .
.
(x) Jordan normal form of N i : we saw that there is a sequence of lin. ind. sets of
vectors B(e,1) ,...., B(e, ae ) ,..., B(r ,1) ,...., B(r , ar ) ,...., B(1,1)...., B(1, a1 ) (of course not
every integer r has to occur above)
gen
such that Bλi = B(e,1) u........u B(1, ar ) is an ordered basis for Vλi with respect to
which we have
[N i ]Bλi ,
Bλi
=
N
(e,1)
---
o
18.034, Honors Differential Equations
Prof. Jason Starr
-
o
N(r,j)
--
N(1,a1 )
Page 1 of 9
o
01
01
0
N (r, j ) = r
o
01
0
r
Gave usual method for finding the sets B(r , j ) :
First choose a maximal lin. ind. set of vectors v (r ,1) ,……, v (r , ar ) contained in
and define
V (r ) - ( N V (r + 1) U V (r − 1) )
i
λi
λi
λi
B(r , j ) = (N ri v (r , j ) , N ri − 1 v (r , j ) ,…., N i v (r , j ) , v (r , j ) ).
(xi) The Jordan normal form of Ti w.r.t B is
--
-
T(r , j ) =
T(r,j)
--
-
λi 1
λi 1
λ-i-
o
o
--
λi 1
λi
r col’s
(xii) Restrict T to the subspace W (r, j ) spaced by B (r , j ) . Represent el’ts w.r.t basis
⎡ z1 ⎤
⎢ ⎥
⎢z2 ⎥
B (r , j ) as z = ⎢: ⎥ . Then z' = T z is just
⎢ ⎥
⎢: ⎥
⎢z ⎥
⎣ r⎦
18.034, Honors Differential Equations
Prof. Jason Starr
Page 2 of 9
z 1' = λ z1 + z 2
z 2' =
λ
:
:
λ
z r' − 1 =
z r'
=
z2 + z3
λ
z r −1 + z r
zr
(xiii) The solution space of z' = T z has a basis
⎤
⎥
− 1 )! ⎥
⎥
⎥
⎥
⎥
⎥ e
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎡ t l l!
⎢ l −1
( l
⎢t
⎢
:
⎢
⎢t 1
z (l ) = ⎢
⎢1
⎢0
⎢
⎢:
⎢:
⎢
⎢0
⎣
λ1 t
for l = 0,....., r − 1 , i.e.
⎡t 2 2 ⎤
⎡t ⎤
⎡1 ⎤
⎥
⎢
⎢ ⎥
⎢ ⎥
⎥
⎢t
⎢1 ⎥
⎢0 ⎥
⎥ λ t
⎢ 0 ⎥ λ t ⎢ 0 ⎥ λ t ⎢1
⎥ e 1 ,........
⎢ ⎥e 1 , ⎢ ⎥e 1 , ⎢
:
:
⎥
⎢0
⎢ ⎥
⎢ ⎥
⎥
⎢:
⎢: ⎥
⎢: ⎥
⎥
⎢
⎢ ⎥
⎢ ⎥
⎥⎦
⎢⎣ 0
⎣⎢ 0 ⎦⎥
⎣⎢ 0 ⎦⎥
(xiv) In computing the Jordan normal form, we also find the matrix
U =[ Id v ]
vj=
B old , B new
n
∑u
ij e i
, i.e. for the j th basis vector v j in the new basis,
, where (e1,…., en) is the original basis.
i =1
(xv) For each vector from (xiii) extend the column vector by zesos to get a solution
z (r , j , e ) of z ' = [T ]B new , B new z .
The corresponding solution of the linear system w.r.t. the original basis is
y (r , j , l ) = Uz (r , j , l )
18.034, Honors Differential Equations
Prof. Jason Starr
Page 3 of 9
II. Lecture 30, part 1 ≈ (30 mins)
10mins
A. An example with recil eigenvalues.
⎡
⎢6
⎢
⎢
A = ⎢4
⎢
⎢
⎢0
⎣⎢
y ' = Ay ,
− 4
− 2
0
⎤
1⎥
⎥
⎥
3⎥
⎥
⎥
2 ⎥
⎦⎥
B = (e1 , e2 , e3 )
PA (λ ) = (λ − 2) . So one eigenvalue λ = 2 w/multiplicity 3.
3
[N ]B, B
⎡
⎢4
⎢
⎢
= A − 2I = ⎢ 4
⎢
⎢
⎢0
⎢⎣
[N ]
2
= (A − 2I )
2
B, B
V
(1)
2
=
⎡
⎢0
⎢
⎢
= ⎢0
⎢
⎢
⎢0
⎣⎢
⎤
1⎥
⎥
⎥
3⎥ ,
⎥
⎥
0 ⎥
⎥⎦
− 4
− 4
0
0
0
0
⎛ ⎡1 ⎤ ⎞
⎜⎢ ⎥⎟
K er (N ) = S pn ⎜ ⎢1 ⎥ ⎟ ,
⎜⎜ ⎢ ⎥ ⎟⎟
⎝ ⎣0⎦ ⎠
18.034, Honors Differential Equations
Prof. Jason Starr
⎤
− 8⎥
⎥
⎥
− 8⎥ ,
⎥
⎥
0 ⎥
⎦⎥
[N ]
3
V
B, B
(2 )
2
=
= 0 matrix
⎛ ⎡1 ⎤ ⎡0⎤ ⎞
⎜⎢ ⎥ ⎢ ⎥⎟
S pn ⎜ ⎢0⎥, ⎢1 ⎥ ⎟
⎜⎜ ⎢ ⎥ ⎢ ⎥ ⎟⎟
⎝ ⎣0⎦ ⎣0⎦ ⎠
Page 4 of 9
V (23 ) = V .
(
Choose v (3,1) = e3 . Then B' = B(3,1) = N 2 e3 , Ne3 , e3
⎛
⎜
⎜
=⎜
⎜
⎜
⎝
⎤N
⎡
⎢− 8 ⎥
⎢ − 8 ⎥G
,
⎥
⎢
0 ⎥
⎢
⎥⎦
⎣⎢
[N ]B', B'
⎡
⎢0
⎢
⎢
= ⎢0
⎢
⎢
⎢0
⎣
⎡1
⎢
⎢3
⎢⎣ 0
N
⎤
⎥G
⎥,
⎥⎦
⎡0
⎢
⎢0
⎢⎣ 1
⎤
⎥
⎥
⎥⎦
⎞
⎟
⎟
⎟
⎟
⎟
⎠
⎤
0 ⎥
⎥
⎥
1⎥ ,
⎥
⎥
0⎥
⎦
1
0
0
⎡
⎢− 8
⎢
⎢
U = ⎢− 8
⎢
⎢
⎢ 0
⎢
⎣
[T ]B', B'
⎡
⎢2
⎢
⎢
= ⎢0
⎢
⎢
⎢0
⎣
)
1
3
0
⎤
0 ⎥
⎥
⎥
0 ⎥
⎥
⎥
1⎥
⎥
⎦
1
2
0
⎤
0 ⎥
⎥
⎥
1⎥
⎥
⎥
2⎥
⎦
w.r.t. B' a basis for the solution space is
⎡1 ⎤
⎢ ⎥
z (1) = ⎢0⎥e 2t ,
⎢⎣0⎥⎦
⎡t ⎤
⎢ ⎥
z (2 ) = ⎢1 ⎥e 2t ,
⎢⎣0⎥⎦
⎡t 2 ⎤
⎢ 2⎥
⎥e 2t
z (3) = ⎢t
⎢
⎥
⎢1 ⎥
⎢⎣
⎥⎦
w.r.t original basis B , a basis for the solution space is
⎡
⎡
⎤
⎤
⎢ − 8t + 1 ⎥
⎢− 4t 2 + t ⎥
⎢
⎢
⎥
⎥
y (2 ) = Uz (2 ) = ⎢ − 8t + 3⎥ e 2t , y (3) = Uz (3) = ⎢− 4t 2 + 3t ⎥e 2t
⎢
⎢
⎥
⎥
⎢
⎢
⎥
0⎥
1
⎣⎢
⎣⎢
⎦⎥
⎦⎥
So the general real solution of y ' = Ay is
⎡ − 8⎤
⎢ ⎥
y (1) = Uz (1) = ⎢− 8⎥e 2t ,
⎢⎣0 ⎥⎦
⎡
⎡
⎤
⎤
⎢− 8t + 1 ⎥
⎢− 4t 2 + t ⎥
⎡ − 8⎤
⎢
⎢
⎥
⎥
⎢ ⎥
y = C1 ⎢− 8⎥e 2t + C 2 ⎢− 8t + 3⎥e 2t + C 3 ⎢− 4t 2 + 3t ⎥e 2t
⎢
⎢
⎥
⎥
⎢⎣0 ⎥⎦
⎢
⎢
⎥
0⎥
1
⎢⎣
⎢⎣
⎥⎦
⎥⎦
18.034, Honors Differential Equations
Prof. Jason Starr
Page 5 of 9
5 mins B. What if some eigenvalues are complex:
Answer : Go through the same process as before to get y (r , j , l ) = Uz (r , j , l ) .
Then take the real and imaginary parts of y (r , j , l ) . You only need to do this for
one of the two complex conjugate eigenvalues in each conjugate eigenvalue
pair.
: It is not okay to take the real and imaginary parts of z (r , j , l ) !
The matrix U will have complex entries, and you need to first compute z (r , j , l ) .
10 mins-15 mins C. An example with complex eigen values
V=C
I 4 , B = (e1 , e2 , e3 , e4 ) .
A=
⎡
⎢2
⎢
⎢
⎢− 1
⎢
⎢
⎢0
⎢
⎢
⎢0
⎣
1
−1
2
0
0
2
0
−1
⎤
3⎥
⎥
⎥
− 2⎥
⎥
⎥
1⎥
⎥
⎥
2⎥
⎦
λt : [Nt ]B, B = A − λt I =
[N ]
2
t B, B
=
⎡
⎢− i
⎢
⎢
⎢− 1
⎢
⎢
⎢0
⎢
⎢
⎢0
⎣
1
−i
[ ]
The matrix Nt2
− 3 + 2i
3
−i
0
−1
B, B
(
)
, PA (λ ) = (λ − 2) + 1
⎡
⎢− i
⎢
⎢
⎢− 1
⎢
⎢
⎢0
⎢
⎢
⎢0
⎣
0
T = TA ,
2
2
λt = 2 + i , λ − = 2 − i
1
−1
−i
0
0
−i
0
−1
⎤
3⎥
⎥
⎥
− 2⎥
⎥
⎥
1⎥
⎥
⎥
− i⎥
⎦
,
⎤
− 3 − 6i ⎥
⎥
⎥
− 3 + 4i ⎥
⎥
⎥
1 ⎥
⎥
⎥
− i ⎥
⎦
is row equivalent to
18.034, Honors Differential Equations
Prof. Jason Starr
Page 6 of 9
⎡
⎢1
⎢
⎢
⎢0
⎢
⎢
⎢0
⎢
⎢
⎢0
⎣
i
0
0
0
0
1
0
0
(2 )
V λt = S pn
⎡1 ⎤
⎢ ⎥,
⎢i ⎥
⎢0 ⎥
⎢ ⎥
⎣0 ⎦
⎤
− 3 − i⎥
⎥
⎥
0
⎥
⎥
⎥
i
⎥
⎥
⎥
0
⎥
⎦
⎡3 − i ⎤
⎢
⎥
⎢ 0 ⎥
⎢
⎥
⎢
i ⎥
⎢
⎥
⎢ −1 ⎥
⎣
⎦
So V (λ1) = S pn
t
⎡1 ⎤
⎢ ⎥
⎢i ⎥
⎢0 ⎥
⎢ ⎥
⎣0 ⎦
,
.
⎛
⎜
⎜
−
3
i
⎡
⎤
⎜
⎢
⎥
0
Choose V(2,1) = ⎢
⎥ , B (2,1) = ⎜
⎢
⎥
⎜
⎢
i ⎥
⎜
⎢
⎥
⎢ −1 ⎥
⎜
⎣
⎦
⎝
⎡3 − i ⎤ ⎞
⎡ − 4 − 4 i ⎤
⎥ ⎟
⎥ N ⎢
⎢
4
−
4
i
0 ⎥ ⎟
⎢
⎥
⎢
G ⎢
⎥ ⎟ .
⎥
⎢
,
0
⎥ ⎟
⎢
⎥
⎢
i ⎥ ⎟
⎢
⎥
⎢
⎥ ⎟⎟
⎢
⎥
⎢
0
⎥⎦ ⎠
⎢⎣ − 1
⎥⎦
⎢⎣
With respect to this lin ind. set, T has matrix
So the solution space for
λ
⎡2 + i
1⎤
⎢
⎥.
⎢
⎥
2+ i ⎥
⎢0
⎢
⎥
⎣
⎦
has a basis
⎡1 ⎤ 2t it
⎡t ⎤ 2t it
⎢ ⎥e e , ⎢ ⎥e e
⎣0⎦
⎣0⎦
⎡− 4 cos (t ) + 4 sin (t )⎤
⎡− 4 cos (t ) − 4 sin (t )⎤
⎡− 4 − 4i ⎤
⎥
⎥
⎢
⎢
⎢
⎥
⎢4 cos (t ) + 4 sin (t ) ⎥
⎢− 4 cos (t ) + 4 sin (t )⎥
⎢4 − 4i ⎥
⎥
⎥
2t ⎢
⎥e2t eit = e 2t ⎢
So y (2,1,1) = ⎢ 0
0
0
⎥ +i e ⎢
⎥.
⎢
⎢
⎥
⎥
⎥
⎢
⎢
⎢
⎥
⎥
⎥
⎢
⎢
⎢ 0
⎥
0
0
⎥⎦
⎥⎦
⎢⎣
⎢⎣
⎣⎢
⎦⎥
⎡3 − i + t (− 4 − 4i )⎤
⎢
⎥
⎢t (4 − 4i )
⎥
⎢
⎥e 2t e it
And y (2,1,2 ) =
i
⎢
⎥
⎢
⎥
⎢ −1
⎥
⎢⎣
⎥⎦
18.034, Honors Differential Equations
Prof. Jason Starr
Page 7 of 9
(
(
)
⎡t − 4 cos (t ) + 4 sin (t ) + 3 cos (t ) + sin (t )⎤
⎥
⎢
⎥
⎢t 4 cos (t ) + 4 sin (t )
⎥
2t ⎢
y (2,1,2 ) = e ⎢
− sin (t )
⎥
⎥
⎢
⎥
⎢
− cos (t )
⎥⎦
⎢⎣
)
(
(
)
)
⎡t − 4 cos (t ) − 4 sin (t ) − cos (t ) + 3 sin (t )⎤
⎥
⎢
⎥
⎢t − 4 cos (t ) + 4 sin (t )
⎥
2t ⎢
+i e ⎢
cos (t )
⎥
⎥
⎢
⎥
⎢
− sin (t )
⎥⎦
⎢⎣
So a basis for the solution space is
⎡− 4 cos (t ) + 4 sin (t )⎤
⎥
⎢
⎢4 cos (t ) + 4 sin (t ) ⎥
⎥ 2t
⎢
,
y (1) = ⎢
0
⎥e
⎥
⎢
⎥
⎢
0
⎥⎦
⎢⎣
(
(
⎡− 4 cos (t ) − 4 sin (t )⎤
⎥
⎢
⎢− 4 cos (t ) + 4 sin (t )⎥
⎥ 2t
⎢
y (2 ) = ⎢
⎥e
0
⎥
⎢
⎥
⎢
0
⎥
⎢
⎦
⎣
)
⎡t − 4 cos (t ) + 4 sin (t ) + 3 cos (t ) + sin (t )⎤
⎥
⎢
⎥
⎢t 4 cos (t ) + 4 sin (t )
⎥ 2t
⎢
y (3 ) = ⎢
− sin (t )
⎥e ,
⎥
⎢
⎥
⎢
− cos (t )
⎥⎦
⎢⎣
(
(
)
)
)
⎡t − 4 cos (t ) − 4 sin (t ) − cos (t ) + 3 sin (t )⎤
⎥
⎢
⎥
⎢t − 4 cos (t ) + 4 sin (t )
⎥ 2t
⎢
y (4 ) = ⎢
⎥e
sin (t )
⎥
⎢
⎥
⎢
− cos (t )
⎥
⎢
⎦
⎣
III. Lecture 30, part 2 ( ≈ 20 min’s)
Discuss the matrix exponential and its use to solve inhomogeneous linear systems
with constant coeffs: § 6,6 of Borelli + Coleman.
Of course, the books definition of the matrix exponential is ridiculous. Also, both for
finding solutions of inhomog. systems and for solving IVP’s, it is better to go through
the process above and then compute the matrix U −1 = [Id ]B new , Bold .
18.034, Honors Differential Equations
Prof. Jason Starr
Page 8 of 9
Then for an IVP
y ' = Ay + f
, we get
y (0) = y 0
the transformed IVP
z ' = U −1 AUz + U −1 f
z (0) = z 0 = U −1 y 0
Because B = U −1 AU is in Jordan normal form, it is usually simpler to solve this
directly than to compute the matrix exponential.
18.034, Honors Differential Equations
Prof. Jason Starr
Page 9 of 9