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.