The Laplace Transform is typically credited with taking
dynamical problems into static problems. Recall that the
Laplace Transform of the function hh is
ℒhs≡∫0∞e−(st)htdt
.
ℒ
h
s
t
0
s
t
h
t
.
MATLAB is very adept at such things. For example:

```
>> syms t
>> laplace(exp(t))
ans = 1/(s-1)
>> laplace(t*(exp(-t))
ans = 1/(s+1)^2
```

The Laplace Transform of a matrix of functions is simply the
matrix of Laplace transforms of the individual elements.

ℒette−t=1s−11s+12
ℒ
t
t
t
1
s
1
1
s
1
2

Now, in preparing to apply the Laplace transform to our equation from
the dynamic strang quartet module:
x′=Bx+g
,
x
B
x
g
,
we write it as

ℒdxdt=ℒBx+g
ℒ
t
x
ℒ
B
x
g

(1)
and so must determine how

ℒℒ acts on derivatives
and sums. With respect to the latter it follows directly from
the definition that

ℒBx+g=ℒBx+ℒg=Bℒx+ℒg
.
ℒ
B
x
g
ℒ
B
x
ℒ
g
B
ℒ
x
ℒ
g
.

(2)
Regarding its effect on the derivative we find, on integrating
by parts, that

ℒdxdt=∫0∞e−(st)dxtdtdt=xte−(st)|0∞+s∫0∞e−(st)xtdt
.
ℒ
t
x
t
0
s
t
t
x
t
0
x
t
s
t
s
t
0
s
t
x
t
.
Supposing that

xx and

ss are such that

xte−(st)→0
x
t
s
t
0
as

t→∞
t
we arrive at

ℒdxdt=sℒx−x0
.
ℒ
t
x
s
ℒ
x
x
0
.

(3)
Now, upon substituting

Equation 2 and

Equation 3 into

Equation 1 we find

sℒx−x0=Bℒx+ℒg
,
s
ℒ
x
x
0
B
ℒ
x
ℒ
g
,
which is easily recognized to be a linear system for

ℒx
ℒ
x
, namely

(sI−B)ℒx=ℒg+x0
.
s
I
B
ℒ
x
ℒ
g
x
0
.

(4)
The only thing that distinguishes this system from those
encountered since our

first
brush with these systems is the presence of the complex
variable

ss. This
complicates the mechanical steps of Gaussian Elimination or the
Gauss-Jordan Method but the methods indeed apply without
change. Taking up the latter method, we write

ℒx=sI−B-1(ℒg+x0)
.
ℒ
x
s
I
B
ℒ
g
x
0
.
The matrix

sI−B-1
s
I
B
is typically called the

transfer function or

resolvent, associated with

BB, at

ss. We turn to
MATLAB for its

symbolic calculation. (for more
information, see

the
tutorial on MATLAB's symbolic toolbox). For example,

```
>> B = [2 -1; -1 2]
>> R = inv(s*eye(2)-B)
R =
[ (s-2)/(s*s-4*s+3), -1/(s*s-4*s+3)]
[ -1/(s*s-4*s+3), (s-2)/(s*s-4*s+3)]
```

We note that
sI−B-1
s
I
B
is well defined except at the roots of the quadratic,
s2−4s+3
s
2
4
s
3
.
This quadratic is the determinant of
sI−B
s
I
B
and is often referred to as the characteristic
polynomial of BB. Its roots are called the
eigenvalues of BB.

As a second example let us take the BB matrix
of the
dynamic Strang quartet module with the parameter
choices specified in fib3.m,
namely

B=(
-0.1350.1250
0.5-1.010.5
00.5-0.51
)
B
-0.1350.1250
0.5-1.010.5
00.5-0.51

(5)
The associated

sI−B-1
s
I
B
is a bit bulky (please run

fib3.m)
so we display here only the denominator of each term,

i.e.,

s3+1.655s2+0.4078s+0.0039
.
s
3
1.655
s
2
0.4078
s
0.0039
.

(6)
Assuming a current stimulus of the form

i
0
t=t3e−t610000
i
0
t
t
3
t
6
10000
and

E
m
=0
E
m
0
brings

ℒgs=0.191s+16400
ℒ
g
s
0.191
s
1
6
4
0
0
and so

Equation 6 persists in

ℒx=sI−B-1ℒg=0.191s+164(s3+1.655s2+0.4078s+0.0039)s2+1.5s+0.270.5s+0.260.2497
ℒ
x
s
I
B
ℒ
g
0.191
s
1
6
4
s
3
1.655
s
2
0.4078
s
0.0039
s
2
1.5
s
0.27
0.5
s
0.26
0.2497
Now comes the rub. A simple linear solve (or inversion) has
left us with the Laplace transform of xx. The
accursed

We shall have to do some work in order to recover
xx from
ℒx
ℒ
x
.

confronts us. We shall face it down in the

Inverse Laplace module.