374 Lecture 26
Chapter 9: Laplace Transforms

9.1: An Introduction to the Laplace Transform

Definition of Laplace Transform

Def: E is the class of continuous or piecewise continuous functions f(t), defined for at least t > 0, such that |f(t)| <
Bt
A
for some A, B. If f ϵ E , then we say f is of exponential type.​Note that not all functions are of exponential type. For example, f(t) =
(
2
t
)

is not of exponential type. ​​Def: If f ϵE, then the Laplace transform of f(t) is defined to be the function F(s)
:=
∞
∫
0
f(t)
-st
e
t
, defined at least for all sufficiently large t. We will write F(s) = L{f} = L{f(t)}.

Laplace Transforms of Common Functions

FrameBox[GridBox[Table[{{"f(t)","F(s)"},{
at

,LaplaceTransform[
at

,t,s]},​​{t*
at

,LaplaceTransform[t*
at

,t,s]},{
2
t
*
at

,LaplaceTransform[
2
t
*
at

,t,s]},​​{
n
t
*
at

,LaplaceTransform[
n
t
*
at

,t,s]},​​{Sin[at],LaplaceTransform[Sin[at],t,s]},{Cos[at],LaplaceTransform[Cos[at],t,s]},{t*Sin[at],LaplaceTransform[tSin[at],t,s]},{t*Cos[at],LaplaceTransform[tCos[at],t,s]},{Sinh[at],LaplaceTransform[Sinh[at],t,s]},{Cosh[at],LaplaceTransform[Cosh[at],t,s]},​​{1,LaplaceTransform[1,t,s]},​​{t,LaplaceTransform[t,t,s]},​​{
2
t
,LaplaceTransform[
2
t
,t,s]},​​{
n
t
,LaplaceTransform[
n
t
,t,s]}}],RowLinesTrue,ColumnLinesTrue]]//DisplayForm
Out[]//DisplayForm=
f(t)
F(s)
at

1
-a+s
at

t
1
2
(a-s)
at

2
t
-
2
3
(a-s)
at

n
t
-1-n
(-a+s)
Gamma[1+n]
Sin[at]
a
2
a
+
2
s
Cos[at]
s
2
a
+
2
s
tSin[at]
2as
2
(
2
a
+
2
s
)
tCos[at]
-
2
a
+
2
s
2
(
2
a
+
2
s
)
Sinh[at]
a
-
2
a
+
2
s
Cosh[at]
s
-
2
a
+
2
s
1
1
s
t
1
2
s
2
t
2
3
s
n
t
-1-n
s
Gamma[1+n]

Properties of Laplace Transforms

The Laplace transform is linear.

Let F(s) and G(s) be the Laplace transforms of f(t) and g(t), respectively. Then L{α f(t)+β g(t)} = α F(s) + β G(s).​
​
Try applying this idea, using the table above, with h(t) =
4t
t

-3Cos(t)
. Check with Mathematica!
LaplaceTransform[4t
t

-3Cos[t],t,s]
Out[]=
4
2
(-1+s)
-
3s
1+
2
s

Laplace Transforms of Derivatives

One of the main reasons that Laplace transforms are used is the fact that they turn differentiation into multiplication!
​
​Theorem 9.1: Let f(t) and f '(t) be inE. Then L{f ' (t)} =
-f(0)+sL{f(t)}=-f(0)+sF(s)
.​
​
​Proof: Use integration by parts.
​
​Corollary 1: If f and its first n derivatives belong to E, then
L{
(n)
f
(t)}=-[
n-1
s
f(0)+...+s
(n-2)
f
(0)+
(n-1)
f
(0)]+
n
s
L{f(t)}=-[
n-1
s
f(0)+...+s
(n-2)
f
(0)+
(n-1)
f
(0)]+
n
s
F(s)
​
​​
​Check these results hold in the following examples!
LaplaceTransform[f'[t],t,s]/.LaplaceTransform[f[t],t,s]F[s]
Out[]=
-f[0]+sF[s]
LaplaceTransform[f''[t],t,s]/.LaplaceTransform[f[t],t,s]F[s]
Out[]=
-sf[0]+
2
s
F[s]-
′
f
[0]
LaplaceTransform[f'''[t],t,s]/.LaplaceTransform[f[t],t,s]F[s]
Out[]=
-
2
s
f[0]+
3
s
F[s]-s
′
f
[0]-
′′
f
[0]
Note that if f(t) is such that f and all of its derivatives up to order n-1 are zero at t = 0, then Corollary 1 implies
L{
(n)
f
(t)}=
n
s
L{f(t)}=
n
s
F(s)
.
​
​Corollary 2: Let T(D) be a differential operator with constant coefficients: T(D) =
a
0
n
D
+
a
1
n-1
D
+...+
a
n
. If f and all of its derivatives up to order n belong to E and f(0) = f '(0) = ... =
(n-1)
f
(0)=0
, then L{T(D)f} = T(s) L{f} where p(s) =
a
0
n
s
+
a
1
n-1
s
+...+
a
n
.

​
Try this out with the following command!
Simplify[LaplaceTransform[5f''''[t]+3f''[t]-f'[t]+4f[t],t,s]/.{LaplaceTransform[f[t],t,s]F[s],f[0]0,f'[0]0,f''[0]0,f'''[0]0}]
Out[]=
(4-s+3
2
s
+5
4
s
)F[s]

The Convolution Theorem

Def: Let f and g be functions. The convolution of f and g, denoted f*g, is the function defined by
t
∫
0
f[t-u]g[u]u
.​​​Theorem 9.2 (Convolution Theorem): Let f(t) and g (t) be inE. Then L{f*g} = L{f} L{g}=F(s) G(s).​​Here is a way to define the convolution of two functions f and g via Mathematica. (Note that there is also a built-in function, Convolve.)
In[]:=
Convolution[f_,g_]:=Integrate[f[t-u]g[u],{u,0,t}]
Convolution[f,g]
Out[]=
t
∫
0
f[t-u]g[u]u
Convolution[Sin,Cos]
Out[]=
1
2
tSin[t]
Simplify[%]
Out[]=
1
2
tSin[t]
Plot[Convolution[Sin,Cos],{t,0,2π}]
Out[]=
1
2
3
4
5
6
-2.5
-2.0
-1.5
-1.0
-0.5
0.5
1.0
Test out the Convolution Theorem with Mathematica!
LaplaceTransform[Convolution[Sin,Cos],t,s]
Out[]=
s
2
(1+
2
s
)
Simplify[%]
Out[]=
s
2
(1+
2
s
)
LaplaceTransform[Sin[t],t,s]
Out[]=
1
1+
2
s
LaplaceTransform[Cos[t],t,s]
Out[]=
s
1+
2
s

Inverse Laplace Transforms

Definition

Def: Given function F(s), the inverse Laplace transform of F(s) is the function f(t) such that L{f(t)} = F(s). Notation:
-1
L
{F(s)} = f(t).​Since the Laplace transform is linear, it follows that the inverse Laplace transform is linear, i.e.
-1
L
{
c
1
F(s)+
c
2
G(s)}=
c
1
-1
L
{F(s)}+
-1
L
{G(s)}
.​​To find inverse Laplace transforms, one can use tables such as those above. For example,
-1
L

1
s-4
=
4t

. Mathematica has a built in command to find inverse Laplace transforms. Here is an example:

Partial Fractions and Inverse Laplace Transforms

The Convolution Theorem and Inverse Laplace Transforms

Heaviside's Formula

A useful way to find inverse Laplace transforms is via the following theorem, which is really just a special case of finding inverse Laplace Transforms via partial fractions.

9.2: Solving ODE with Laplace Transforms

Example 1: Finding a particular solution to a differential equation.

To simplify things, use the ReplaceAll command (/.). Assume u[0]=0, u'[0]=0.
Check that the solution solves the ODE!
Try solving this problem using the Convolution Theorem!

Example 2: Solving an IVP with initial data specified at t = 0

Example 2 Solution

Check our solution!