The document discusses Laplace transforms and their use in solving initial value problems (IVPs). It provides the following key points: 1. A Laplace transform converts a function of time into a function of complex variables, allowing IVPs to be converted into algebraic equations. 2. Common properties like linearity and derivative rules allow the Laplace transform of derivatives and sums to be determined. 3. The inverse Laplace transform yields the original time function, but involves a contour integral in the complex plane. Tables and software are typically used to evaluate. 4. Laplace transforms are effective for IVPs with piecewise or impulse forcing functions, allowing engineering problems to be solved. Their use is limited as they only apply

1. By:
http://admission.edhole.com

2. Math 2306
Dr. Dillon
SPSU
Mathematics Department

4. A Laplace transform is a type of
integral transform.
Plug one function in
0
s t
e dt
∞
− ⋅
∫ ( )f t
Get another function out
( )F s=
The new function is in a different domain.

5. ( )F s is the Laplace transform of ( ).f t
Write { }( ) ( ),f t F s=L
0
s t
e dt
∞
− ⋅
∫ ( )f t ( )F s=When
{ }
{ }
( ) ( ),
( ) ( ), etc.
y t Y s
x t X s
=
=
L
L

6. A Laplace transform is an example of an improper
integral : one of its limits is infinite.
0 0
( ) lim ( )
h
s t s t
h
e f t dt e f t dt
∞
− ⋅ − ⋅
→∞
=∫ ∫
Define

7. Let
0 if
( )
1 if
t c
u t c
t c
<
− = 
≥
This is called the unit step function or
the Heaviside function.
It’s handy for describing functions that
turn on and off.

8. c
1
t
0 if
( )
1 if
t c
u t c
t c
<
− = 
≥
The Heaviside Function

9. { }
0
1 1
( ) ( ) lim
lim lim ( )
h
s t s t
h
c
h s cs t s h s c
s sch h
u t c e u t c dt e dt
ee e e
s
∞
− ⋅ − ⋅
→∞
− ⋅− ⋅ − ⋅ − ⋅− −
→∞ →∞
− = − = =
= − =
∫ ∫L
Calculating the Laplace transform of the
Heaviside function is almost trivial.
Remember that ( )u t c− is zero until
then it’s one.
,t c=

11. We can use Laplace transforms to turn an
initial value problem
" 3 ' 4 ( 1)
(0) 1, '(0) 2
y y y t u t
y y
+ − = ⋅ −
= − =
into an algebraic problem
2
2 1
( )*( 3 4) ( 1) s
s
s e
Y s s s s +
⋅
+ − + + =
Solve for y(t)
Solve for Y(s)

12. 1
1
A sawtooth function
t
Laplace transforms are particularly effective
on differential equations with forcing functions
that are piecewise, like the Heaviside function,
and other functions that turn on and off.

13. I.V.P.
Laplace transform
Algebraic Eqn

15. If you solve the algebraic equation
2
2 2
( 1) ( 1)
( )
( 3 4)
s s
s s e e
Y s
s s s
−
− + ⋅ ⋅ − ⋅
=
⋅ + −
and find the inverse Laplace transform of
the solution, Y(s), you have the solution to
the I.V.P.

16. Algebraic Expression
Soln. to IVP
Inverse
Laplace
transform

17. The inverse Laplace transform
of
is
4 43 32 1
5 80 4 16
432
5 5
( ) ( 1)( + ( ) )
( )( ( ) )
t te
e
t t
y t u t e e t
u t e e
−
−
= − ⋅ ⋅ − −
− ⋅ − ⋅
2
2 2
( 1) ( 1)
( )
( 3 4)
s s
s s e e
Y s
s s s
−
− + ⋅ ⋅ − ⋅
=
⋅ + −

18. 4 43 32 1
5 80 4 16
432
5 5
( ) ( 1)( + ( ) )
( )( ( ) )
t te
e
t t
y t u t e e t
u t e e
−
−
= − ⋅ ⋅ − −
− ⋅ − ⋅
is the solution to the I.V.P.
" 3 ' 4 ( 1)
(0) 1, '(0) 2
y y y t u t
y y
+ − = ⋅ −
= − =
Thus

20. You need several nice properties of Laplace
transforms that may not be readily apparent.
First, Laplace transforms, and inverse
transforms, are linear :
{ } { } { }
{ } { } { }1 1 -1
( ) ( ) ( ) ( ) ,
( ) ( ) ( ) ( )
cf t g t c f t g t
cF s G s c F s G s− −
+
+
L = L + L
L = L + L
for functions f(t), g(t), constant c, and
transforms F(s), G(s).

21. there is a very simple relationship
between the Laplace transform of a given
function and the Laplace transform of that
function’s derivative.
{ } { }
{ } { }2
'( ) ( ) (0),
''( ) ( ) (0) '(0)
f t s f t f
f t s f t s f f
⋅ −
⋅ − ⋅ −
L = L
L = L
and
These show when we apply differentiation
by parts to the integral defining the transform.
Second,

22. Now we know there are rules that let
us determine the Laplace transform
of an initial value problem, but...

24. First you must know that Laplace transforms
are one-to-one on continuous functions.
In symbols
{ } { }( ) ( ) ( ) ( )f t g t f t g t⇒ =L = L
when f and g are continuous.
That means that Laplace transforms are
invertible.

25. If { }( ) ( ),f t F s=L
{ }1 1
2( ) ( )
c i
s t
i c i
F s e F s dsπ
+ ∞
− ⋅
− ∞
= ∫L
then { }-1
( ) ( ),F s f t=L where

26. An inverse Laplace transform is an improper
contour integral, a creature from the world
of complex variables.
That’s why you don’t see them naked very
often. You usually just see what they yield,
the output.
In practice, Laplace transforms and inverse
Laplace transforms are obtained using tables
and computer algebra systems.

28. Don’t use them...
unless you really have to.

30. When your forcing function is a piecewise,
periodic function, like the sawtooth function...
Or when your forcing function is an impulse,
like an electrical surge.

31. An impulse is the effect of a force that acts
over a very short time interval.
Engineers and physicists use the Dirac
delta function to model impulses.
A lightning strike creates an electrical
impulse.
The force of a major leaguer’s bat
striking a baseball creates a mechanical
impulse.

32. This so-called quasi-function was created
by P.A.M. Dirac, the inventor of quantum
mechanics.
0
( ) 0 ( ) 1t a t a t a dtδ δ
∞
− = ≠ − =∫when and
People use this thing all the time. You
need to be familiar with it.

33. { ( )} 1/ a s
L t a eδ ⋅
− =

35. Laplace transforms have limited appeal.
You cannot use them to find general solutions
to differential equations.
You cannot use them on initial value problems
with initial conditions different from
1 2(0) , '(0) ,y c y c= = etc.
Initial conditions at a point other than zero
will not do.

37.  Know the definition of the Laplace transform
 Know the properties of the Laplace transform
 Know that the inverse Laplace transform is
an improper integral
 Know when you should use a Laplace
transform on a differential equation
 Know when you should not use a Laplace
transform on a differential equation

38. When Appropriate