Comparing the two representations, the Laplace:
xt=ℒ-1sI−B-1(ℒg+x0)
,
x
t
ℒ
s
I
B
ℒ
g
x
0
,
(1)
and the
Backward Euler:
x
˜
jdt=Idt−B-1(
x
˜
(j−1)
d
t
dt+gjdt)
,
x
˜
j
dt
I
dt
B
x
˜
j
1
d
t
dt
g
j
dt
,
(2)
we see that they both produce the solution to the general linear
system of ordinary equations,
x′=Bx+g
,
x
B
x
g
,
(3)
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 2 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−
d
t
B-1jx0
.
x
˜
j
dt
I
d
t
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-to-one, yields the exponential
xt=eBtx0
,
x
t
B
t
x
0
,
clearly the correct solution to
Equation 3. A careful explication of the
matrix
exponential and its relationship to
Equation 1 will have to wait until we have
mastered the inverse laplace transform.
