Where in the Inverse Laplace
Transform module we tackled the derivative in
via an integral transform we pursue in this section a much
simpler strategy, namely, replace the derivative with a finite
difference quotient. That is, one chooses a small
d
t
d
t
and 'replaces'
Equation 1 with
x
˜
t−
x
˜
t−
d
t
d
t
=B
x
˜
t+gt
.
x
˜
t
x
˜
t
d
t
d
t
B
x
˜
t
g
t
.
(2)
The utility of
Equation 2 is that it
gives a means of solving for
x
˜
x
˜
at the present time,
tt, from the
knowledge of
x
˜
x
˜
in the immediate past,
t−
d
t
t
d
t
.
For example, as
x
˜
0=x0
x
˜
0
x
0
is supposed known we write
Equation 2
as
(I
d
t
−B)
x
˜
dt=x0dt+gdt
.
I
d
t
B
x
˜
dt
x
0
dt
g
dt
.
Solving this for
x
˜
dt
x
˜
dt
we return to
Equation 2 and find
(Idt−B)
x
˜
2dt=xdtdt+g2dt
.
I
dt
B
x
˜
2
dt
x
dt
dt
g
2
dt
.
and solve for
x
˜
2dt
x
˜
2
dt
.
The general step from past to present,
x
˜
jdt=Idt−B-1(
x
˜
(j−1)
d
t
d
t
+gjdt)
x
˜
j
dt
I
dt
B
x
˜
j
1
d
t
d
t
g
j
dt
(3)
is repeated until some desired final time,
Tdt
T
dt
,
is reached. This equation has been implemented in
fib3.m
with
dt=1
dt
1
and
BB and
gg as in
the dynamic Strang
module. The resulting
x
˜
x
˜
( run
fib3.m
yourself!) is indistinguishable from the
plot we
obtained in the Inverse Laplace module.
Comparing the two representations, this equation and Equation 3, we see that they both produce
the solution to the general linear system of ordinary equations,
see this
eqn, by simply inverting a shifted copy of BB. The
former representation is hard but exact while the latter is easy
but approximate. Of course we should expect the approximate
solution,
x
˜
x
˜
, to approach the exact solution,
xx, as the time step,
dt
dt
, approaches zero. To see this let us return to Equation 3 and assume, for now, that
g≡0
g
0
. In this case, one can reverse the above steps and arrive at
the representation
x
˜
jdt=I−
dt
B-1jx0
x
˜
j
dt
I
dt
B
j
x
0
(4)
Now, for a fixed time
tt we suppose
that
dt
=tj
dt
t
j
and ask whether
xt=limit
j
→
∞
I−tjB-1jx0
x
t
j
I
t
j
B
j
x
0
This limit, at least when
BB is one-by-one, yields the
exponential
xt=eBtx0
x
t
B
t
x
0
clearly the correct solution to
this equation. A careful
explication of the
matrix exponential and its
relationship to
this equation will have to wait until we
have mastered the inverse laplace transform.
