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
md2xtdt2+2cdxtdt+kxt=0
m
t2
x
t
2
c
t
x
t
k
x
t
0
(1)
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
u1
t=xt
u1
t
x
t
u2
t=x′t
u2
t
x
t
and note that
Equation 1 becomes
m
u2
′t=(−(k
u1
t))−2c
u2
t
m
u2
t
k
u1
t
2
c
u2
t
(2)
Denoting
ut≡(
u1
t
u2
t
)T
u
t
u1
t
u2
t
we write
Equation 2 as
∀A,A=(
01
−km-2cm
):u′t=Aut
A
A
0 1
k
m
-2
c
m
u
t
A
u
t
(3)
We recall from
The Matrix
Exponential module that
ut=eAtu0
u
t
A
t
u
0
We shall proceed to compute the matrix exponential along the
lines of
The Matrix
Exponential via Eigenvalues and Eignevectors module.
To begin we record the resolvent
Rz=-1mz2+2cz+k(
2c+mzm
−kmz
)
R
z
-1
m
z
2
2
c
z
k
2
c
m
z
m
k
m
z
The eigenvalues are the roots of
mz2+2cz+k
m
z
2
2
c
z
k
, namely
∀d,d=c2−mk:
λ1
=(−c)−dm
d
d
c
2
m
k
λ1
c
d
m
∀d,d=c2−mk:
λ2
=−c+dm
d
d
c
2
m
k
λ2
c
d
m
We naturally consider two cases, the first being
-
d≠0d0.
In this case the partial fraction expansion of
Rz
R
z
yields
Rz=-1z−
λ1
12d(
d−c−m
kc+d
)+-1z−
λ2
12d(
c+dm
−kd−c
)=-1z−
λ1
P1
+-1z−
λ2
P2
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
and so
eAt=e
λ1
t
P1
+e
λ2
t
P2
A
t
λ1
t
P1
λ2
t
P2
. If we now suppose a negligible initial velocity,
i.e.,
v0
v0
, it follows that
xt=
x0
2d(e
λ1
t(d−c)+e
λ2
t(c+d))
x
t
x0
2
d
λ1
t
d
c
λ2
t
c
d
(4)
xt=e−(ct)(cos|d|t+c|d|sin|d|t)
x
t
c
t
d
t
c
d
d
t
(5) -
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
=A
P1
−
λ1
P1
=A−
λ1
I=(
km1
−km−km
)
D1
A
P1
λ1
P1
A
λ1
I
k
m
1
k
m
k
m
On substitution of this into this equation we find
eAt=e−(tkm)(
1+tkmt
−(tkm)1−tkm
)
A
t
t
k
m
1
t
k
m
t
t
k
m
1
t
k
m
(6)
