Math 2280-001
Wed Apr 8
5.6-5.7 Matrix exponentials, linear systems, and variation of parameters for inhomogeneous systems.
Recall: For the first order system
x# t = A x
, F t is a fundamental matrix (FM) if its n columns are a basis for the solution space to the first order
system above (i.e. F t is the Wronskian matrix for a basis to the solution space).
, F t is an FM if and only if F# t = A F and F 0 is invertible.
, et A is the unique matrix solution to
X# t = A X
X 0 =I
and may be computed either of two ways:
et A = F t F 0 K1
where F t is any other FM, or via the infinite series
t2 2
t3 3
tk k
et A = I C tA C
A C
A C... C
A C ...
2!
3!
k!
Example 1: We showed that if
l 1 0 0 ... 0
L=
0 l 2 0 ...
0
:
:
:
0
0
ln
...
then
1 C t l1 C
tL
e
=
t2 2
l C...
2! 1
0
0 ...
0
0
1 C t l2 C
t2 2
l C... 0 ...
2! 2
0
:
:
0
0
tl
e
=
0
1
0
tl
e
2
:
...
...
0
...
0
:
:
:
:
0
0
0 e
tl
n
1 C t ln C
t2 2
l C...
2! n
Example 2) If A is diagonalizable we showed that we got the same answer for et A using
A = SLSK1
and the Taylor series method, as we did we did using the F t F 0 K1 method. In fact, F t = S etL and
F 0 K1 = SK1 , so this makes sense.
, Finish the second to last page of Monday's notes, and then discuss the general (non-chain) way that one
can compute eA t when A is not diagonalizable, below. It depends on the fundamental fact about how the
generalized eigenspaces of a matrix fit together....which we discussed briefly before in the March 27 notes.
(This is also discussed in the text.)
.......................................................................
"Recall",
For An # n let the characteristic polynomial p l = det A K l I factor as
k
k
k
p l = K1 n l K l 1
l K l 2 ... l K l m
Any eigenspace of A for which dim El ! kj is called defective. If A has any defective eigenspaces then
1
2
m
j
it is not diagonalizable. However, the larger generalized eigenspaces Gl defined by
j
k
A K ljI
El 4 Gl d nullspace
j
j
j
do always have dimension kj . If bases for each Gl are amalgamated they will form a basis for =n or Cn .
j
......................................................................
For each basis vector v of Gl one can construct a basis solution x t to x#= A x as follows:
j
l It
j
AK l I t
x t = eA t v = e
j
e
v
2
lt
t
2
= e j I I C t A K ljI C
A K l j I C... v
2!
lt
t2
2
= e j v C t A K ljI v C
A K l j I v C...
2!
lt
j
=e
t2
v C t A K ljI v C
2!
lt
j
x t =e
k K1
A K ljI
t2
v C t A K ljI v C
2!
2
tj
kj K 1 !
vC
A K ljI
k K1
j
v C 0 C 0 C...
k K1
A K ljI
2
vC
tj
kj K 1 !
k
Notes: The final sum is a finite sum because v 2 nullspace
you've just reconstructed
A K ljI
j
A K ljI
k K1
j
v .
! If v was an eigenvector,
lt
x t =e j v
Use the n independent solutions found this way to construct a F t , and compute eA t = F t F 0
K1
.
Exercise 1 Find eA t for the matrix in the system 5.5.2 (an old homework problem, where you were
supposed to find a basis for the solution space using chains):
x1 #
x1
3 K1
=
x2 #
1 1
x2
tech check:
> with LinearAlgebra :
3 K1
Ad
;
1 1
factor Determinant A K l$IdentityMatrix 2
A :=
;
3 K1
1
lK2
1
2
(1)
So the only eigenvalue is l = 2.
> B d A K 2$ IdentityMatrix 2 ;
NullSpace B ;
B :=
1 K1
1 K1
1
(2)
1
but El = 2 is only one-dimensional. However, the generalized eigenspace Gl = 2 = nullspace A K 2 I
will be two dimensional:
2
> NullSpace B2 ;
0
1
,
1
(3)
0
Use this generalized nullspace basis to construct a basis of solutions to x#= A x and use the resulting F t
to construct the matrix exponential...
1
> x1 d t/e2$ t$ IdentityMatrix 2 C t$B .
0
:
x1 t ;
x2 d t/e2$ t$ IdentityMatrix 2 C t$B C
t2 2
$B .
2
0
:
1
x2 t ;
F d t/ x1 t x2 t :
F t ;
F t .F 0 K1;
MatrixExponential t$A ; #these last two should be the same!! In fact, the last three in this case
e2 t 1 C t
t e2 t
Kt e2 t
1 K t e2 t
>
>
e2 t 1 C t
Kt e2 t
t e2 t
1 K t e2 t
e2 t 1 C t
Kt e2 t
t e2 t
1 K t e2 t
e2 t 1 C t
Kt e2 t
t e2 t
Ke2 t K1 C t
(4)
Variation of parameters: This is what fundamental matrices and matrix exponentials are especially good
for....they let you solve non-homogeneous systems without guessing. Consider the non-homogeneous
first order system
x# t = P t x C f t *
Let F t be an FM for the homogeneous system
x# t = P t x.
Since F t is invertible for all t we may do a change of functions for the non-homogeneous system:
x t =F t u t
plug into the non-homogeneous system (*):
F# t u t C F t u # t = P t F t u t C f t .
Since F#= P F the first terms on each side cancel each other and we are left with
F t u# t = f t
K1
u #= F f
which we can integrate to find a u t , hence an x t = F t u t .
Remark: This is where the (mysterious at the time) formula for variation of parameters in nth order linear
DE's came from....
"Recall" (February 23 notes):
..................................................................................................
Variation of Parameters: The advantage of this method is that is always provides a particular solution,
even for non-homogeneous problems in which the right-hand side doesn't fit into a nice finite dimensional
subspace preserved by L, and even if the linear operator L is not constant-coefficient. The formula for the
particular solutions can be somewhat messy to work with, however, once you start computing.
Here's the formula: Let y1 x , y2 x ,...yn x be a basis of solutions to the homogeneous DE
L y d y n C pn K 1 x y n K 1 C...C p1 x y#C p0 x y = 0 .
Then yp x = u1 x y1 x C u2 x y2 x C ... C un x yn x is a particular solution to
L y =f
provided the coefficient functions (aka "varying parameters") u1 x , u2 x ,...un x have derivatives
satisfying the Wronskian matrix equation
y1
y2
...
yn
u #
1
y1 #
y2 #
...
...
...
...
y1n K 1
y2n K 1
yn #
u #
...
:
... ynn K 1
2
u #
n
0
=
0
:
f
..................................................................................................
But if we convert the nth order DE into a first order system for x1 = y, x2 = y# etc. we have
x1
=y
x1 #= x2
= y#
x2 #= x3
= y##
xn K 1 #= xn
xn #=
=y
n
=y
nK1
=Kp0 x y1 Kp1 x y2 K...Kpn K 1 x yn K 1 C f .
And each basis solution y t for L y = 0 gives a solution y, y#, y##, ...y
system
x1 #
x1
0
1
0
...
0
x2 #
x2
0
0
1
...
0
x3 #
:
xn #
=
T
nK1
0
0
0
0
0
1 ...
0
x3 C
0
:
:
:
:
:
:
:
...
Kpn K 1
xn
f
Kp0 Kp1 Kp2
to the homogeneous
.
So the original Wronskian matrix for the nth order linear homogeneous DE is a FM for the system above,
so the formula we learned in Chapter 3 is a special case of the easier to understand one for first order
systems that we just derived, namely
F t u# t = f t
K1
u #= F f
Returning to first order systems, if we want to solve an IVP for a first order system rather than find the
complete general solution, then the following two ways are appropriate:
1) If you want to solve the IVP
x# t = P t x C f t
x 0 = x0
K1
The the solution will be of the form x = F u
(where u #= F f as above). Thus
x0 = F 0 u0
so
K1
u0 = F 0
x0 .
Thus
t
u t = u0 C
t
u t = u0 C
u # s ds
0
K1
F
s f s ds.
0
Then
x t =F t u t
t
x t =F t
u0 C
K1
F
s f s ds .
0
2) If you want to solve the special case IVP
x# t = A x C f t
x 0 = x0
where A is a constant matrix, you may derive a special case of the solution above just as we did in Chapter
1. This is sort of amazing!
x# t = A x C f t
x# t KA x = f t
Kt A
e
x# t KA x = eKt Af t
d
eKt Ax t = eKt Af t .
dt
Integrate from 0 to t:
eKt Ax t K x0 =
tA
Move the x0 over and multiply both sides by e
tA
x t =e
t
eKs Af s ds
0
:
t
x0 C
0
eKs Af s ds .
Exercise 2 Consider the non-homogeneous problem related to the homogeneous system in Exercise 1:
x1 #
x2 #
=
3 K1
x1
1
x2
1
x1 0
0
=
x2 0
1
C
t
0
.
Solve this system using the formula
tA
x t =e
t
x0 C
eKs Af s ds
0
(One could also try undetermined coefficients, but variation of parameters requires no "guessing.")
Tech check: (The commands are sort of strange, but might help in your homework.)
> with LinearAlgebra :
3 K1
:
1 1
MatrixExponential t$A ;
t
f d t/
:
0
> Ad
x0 d
0
:
0
e2 t 1 C t
Kt e2 t
t e2 t
Ke2 t K1 C t
> integrand d s/simplify MatrixExponential Ks$A .f s
> integrand t ; #checking
(5)
: #integrand in formula above
(6)
KeK2 t K1 C t t
Kt2 eK2 t
(6)
> integrated d unapply map int, integrand s , s = 0 .. t , t : #"map" applies a function to each entry of an array...
# "unapply" makes a function out of output
> integrated t ; #checking
1 2 K2 t
t e
2
(7)
1
1 K2 t
1
1 2 K2 t
K
2
t
K C
e
C
te
C
t e
4
4
2
2
> x d unapply simplify MatrixExponential t$A . x0 C integrated t , t :
x t ; #checking answer
1
t e2 t K 1
4
1
1
1
1
K e2 t C
C
tC
t e2 t
4
4
4
4
> with DEtools :
dsolve x1# t = 3$x1 t Kx2 t C t, x2# t = x1 t C x2 t , x1 0 = 0, x2 0 = 0 ;
1
1
1
1
1
1
x1 t =
t e2 t K
t, x2 t = K e2 t C
C
tC
t e2 t
4
4
4
4
4
4
>
>
(8)
(9)