We finally make precise the folk-theorem that states "to
transform a matrix is to transform its eigenvalues." The
starting point is the matrix version of this equation.
fA=-12πⅈ∫CfzRzdz
f
A
1
2
C
z
f
z
R
z
(1)
where
CC is a disc that contains
each eigenvalue of
AA.
Although the concision of Equation 1
lends it elegance it does, in fact, hide a bit too much. To draw
out the fact that one need only know the eigenstructure of
AA we substitute the equation, the spectral representation of
Rz
R
z
, into Equation 1. To wit
fA=12πⅈ∫Cfz∑j=1h1z-λjPj+∑k=1mj-11z-λjk+1
Djk
dz
f
A
1
2
C
z
f
z
j
1
h
1
z
λj
Pj
k
1
mj
1
1
z
λj
k
1
Djk
then
fA=∑j=1hPj12πⅈ∫Cfzz-λjdz+∑k=1mj-1
Djk
12πⅈ∫Cfzz-λjk+1dz
f
A
j
1
h
Pj
1
2
C
z
f
z
z
λj
k
1
mj
1
Djk
1
2
C
z
f
z
z
λj
k
1
so
fA=∑j=1hfλjPj+∑k=1mj-1dkdλkfλjk!
Djk
f
A
j
1
h
f
λj
Pj
k
1
mj
1
λk
f
λj
k
Djk
(2)
The context in which this result is most often is applied is
that of systems of ordinary differential equation of the form
u
′t=Aut
u
t
A
u
t
(3)
u0=u0
u
0
u0
Here
u0u0
is the initial state of the system and the matrix
AA governs its propagation into the
future. When the state is one-dimensional,
i.e., when
AA is
1-by-1, the solution of
Equation 3 is
clearly
ut=ⅇAtu0
u
t
A
t
u0
(4)
Thanks to
Equation 2 we can give sense to
this expression when
AA is n-by-n
for arbitrary
nn. To be precise,
for real
tt, we substitute
fz=ⅇzt
f
z
z
t
in
Equation 2
and find
ⅇAt=∑j=1hⅇλjtPj+∑k=1mj-1tkk!
Djk
A
t
j
1
h
λj
t
Pj
k
1
mj
1
t
k
k
Djk
(5)
If we now substitute
Equation 5 into
Equation 4 and differentiate we find
u
′t=∑j=1hⅇλjtλjPju0+λj∑k=1mj-1tkk!
Djk
u0+∑j=1hⅇλjt∑k=1mj-1tk-1k-1!
Djk
u0
u
t
j
1
h
λj
t
λj
Pj
u0
λj
k
1
mj
1
t
k
k
Djk
u0
j
1
h
λj
t
k
1
mj
1
t
k
1
k
1
Djk
u0
then
u
′t=∑j=1hⅇλjtAPj-Dju0+λj∑k=1mj-1tkk!
Djk
u0+∑j=1hⅇλjt∑k=1mj-1tk-1k-1!
Djk
u0
u
t
j
1
h
λj
t
A
Pj
Dj
u0
λj
k
1
mj
1
t
k
k
Djk
u0
j
1
h
λj
t
k
1
mj
1
t
k
1
k
1
Djk
u0
then
u
′t=∑j=1hⅇλjtAPju0+λj+Dj∑k=1mj-1tkk!
Djk
u0
u
t
j
1
h
λj
t
A
Pj
u0
λj
Dj
k
1
mj
1
t
k
k
Djk
u0
so
u
′t=∑j=1hⅇλjtAPju0+λjPj+Dj∑k=1mj-1tkk!
Djk
u0=Aut
u
t
j
1
h
λj
t
A
Pj
u0
λj
Pj
Dj
k
1
mj
1
t
k
k
Djk
u0
A
u
t
as promised. Let us now carry out this program in a concrete
instance.
If one provides an initial displacement,
x0x0,
and velocity,
v0v0,
to the mass depicted in Figure 1
then one finds that its displacement,
xt
x
t
at time tt satisfies
md2dt2xt+2cddtxt+kxt=0
m
t2
x
t
2
c
t
x
t
k
x
t
0
(6)
x0=x0
x
0
x0
(7)
x
′0=v0
x
0
v0
where prime denotes differentiation with respect to time. It is
customary to write this single second order equation as a pair
of first order equations. More precisely, we set
u1t=xt
u1
t
x
t
(8)
u2t=
x
′t
u2
t
x
t
and note that
Equation 6 becomes
u1
′t=u2t
u1
t
u2
t
(9)
m
u2
′t=-ku1t-2cu2t
m
u2
t
k
u1
t
2
c
u2
t
Denoting
ut≡u1tu2tT
u
t
u1
t
u2
t
we write
Equation 9 as
u
′t=Aut
u
t
A
u
t
(10)
Where
A=01-km-2cm
A
0 1
k
m
2
c
m
.
In order to implement
Equation 4 it
remains only to compute the associated
λjλj's,
PjPj's
and
DjDj's.
To begin we record the resolvent
Rz=-1mz2+2cz+k2c+mzm-kmz
R
z
-1
m
z
2
2
c
z
k
2
c
m
z
m
k
m
z
(11)
The eigenvalues are the roots of
mz2+2cz+k
m
z
2
2
c
z
k
, namely
λ1=-c-dm
λ1
c
d
m
(12)
λ2=-c+dm
λ2
c
d
m
Where
d=c2-mk
d
c
2
m
k
.
We naturally consider two cases, the first being
-
d≠0d0.
In this case the partial fraction expansion of
Rz
R
z
yields
Rz=-1z-λ112dd-c-mkc+d+-1z-λ212dc+dm-kd-c=-1z-λ1P1+-1z-λ2P2
R
z
-1
z
λ1
1
2
d
d
c
m
k
c
d
-1
z
λ2
1
2
d
c
d
m
k
d
c
-1
z
λ1
P1
-1
z
λ2
P2
(13)
and so
ⅇAt=ⅇλ1tP1+ⅇλ2tP2
A
t
λ1
t
P1
λ2
t
P2
. If we now suppose a negligible initial velocity,
i.e.,
v0=0
v0
0
, it follows that
xt=x02dⅇλ1td-c+ⅇλ2tc+d
x
t
x0
2
d
λ1
t
d
c
λ2
t
c
d
(14)
If dd is real,
i.e., if
c2>mk
c
2
m
k
, then both
λ1λ1
and
λ2λ2
are negative real numbers and
xt
x
t
decays to 0 without oscillation. If, on the contrary,
dd is imaginary,
i.e.,
c2<mk
c
2
m
k
, then
xt=ⅇ-ctcos|d|t+c|d|sin|d|t
x
t
c
t
d
t
c
d
d
t
(15)
and so xx decays to 0 in an oscillatory
fashion. When Equation 14 holds the system
is said to be overdamped while when Equation 15 governs then we speak of the system
as underdamped. It remains to discuss the
case of critical damping.
-
d=0d0. In this case
λ1=λ2=-km
λ1
λ2
k
m
, and so we need only compute
P1P1
and
D1D1.
As there is but one
PjPj
and the
PjPj
are known to sum to the identity it follows that
P1=I
P1
I
. Similarly, this equation dictates that
D1=AP1-λ1P1=A-λ1I=km1-km-km
D1
A
P1
λ1
P1
A
λ1
I
k
m
1
k
m
k
m
(16)
On substitution of this into Equation 5
we find
ⅇAt=ⅇ-tkm1+tkmt-tkm1-tkm
A
t
t
k
m
1
t
k
m
t
t
k
m
1
t
k
m
(17)
Under the assumption, as above, that
v0=0
v0
0
, we deduce from Equation 17 that
xt=ⅇ-tkm1+tkmx0
x
t
t
k
m
1
t
k
m
x0
(18)
Exercises:
With respect to the mass-spring system depicted below,
the balance of forces at each mass require that the two
displacements satisfy
md2dt2x1t+2kx1t-kx2t=0
m
t2
x1
t
2
k
x1
t
k
x2
t
0
(19)
md2dt2x2t+2kx2t-kx1t=0
m
t2
x2
t
2
k
x2
t
k
x1
t
0
(20)
subject to initial conditions
∧x10=
x
1
,
0
x1
0
x
1
,
0
(21)
x1
′0=
v
1
,
0
x1
0
v
1
,
0
x20=
x
2
,
0
x2
0
x
2
,
0
x2
′0=
v
2
,
0
x2
0
v
2
,
0
- Express Equation 19, Equation 20 and Equation 21 in the form of Equation 3 and exhibit the 4-by-4 matrix
AA.
- Find the resolvent of this AA.
- Setting
m=k=1
m
k
1
exhibit the spectral representation of
AA, i.e., write
AA in the form of Equation 8.
- Compute
ⅇAt
A
t
for this AA.
- Suppose that
v
1
,
0
=
v
2
,
0
=0
v
1
,
0
v
2
,
0
0
and express
x1t
x1
t
and
x2t
x2
t
in terms of trigonometric functions.
