5.1 Definitions
An initial
value problem is a differential equation together with sufficient initial
conditions to determine the values of all of the arbitrary constants of
integration.
The
techniques of Laplace transforms allow an initial value problem to be converted
into an algebra problem. Inverse
Laplace transforms are then needed to convert the solution of the algebra
problem back into a complete solution of the initial value problem.
Laplace
transforms may also be used to solve certain types of integral equations, such
as
A function f
(t) is said to be of exponential order a if and only if there exist positive constants k
and a such that
| f (t)
| <
k eat "t > 0
f (t) =
constant and the functions cos wt and
sin wt are all of exponential order 0.
The function f (t) = tn (where n > 0) is of exponential
order 1.
The function f (t) = eat (where a > 0) is of exponential
order a.
Laplace
transforms exist for all of these functions.
The function f (t) = tan wt is not of exponential order (because of the
infinite discontinuities at wt = (2n+1)p/2).
The function f (t) = exp(bt2) is not of exponential
order (unless b £ 0).
Neither of these
functions has a well-defined Laplace transform.
If
the function f (t) is
§
defined for all t > 0 and
§
piecewise continuous on t > 0
(that
is, has at most a finite number of finite discontinuities), and
§
is of exponential order, then
the
Laplace transform L {f (t)} = F (s) exists and is defined by
A table of
Inverse Laplace transforms follows in section 5.4.
5.2 Some Properties and Theorems of Laplace Transforms
The Laplace transform is linear,
that is
|
L {a.f (t) + b.g(t)} = a.L {f (t)} + b.L {g (t)}
|
for
all constants a and b, provided that both L {f (t)} and L {g (t)} exist.
Derivatives:
Provided
the function f (t) is continuous at t = 0 and is differentiable on t > 0,
|
L {f ' (t)}
= s.L {f (t)} - f
(0)
|
This
generalizes to
or
Thus
the application of Laplace transforms to an initial value problem (involving
derivatives with respect to t and their values at t = 0)
converts it into an algebraic problem (involving the variable s).
The
integral form of this identity is
Also 
and 
First
Shifting Theorem:
|
If L {f
(t)} = F (s)
then L {eat f (t)} = F (s
- a)
|
An
equivalent statement is
|
If L -1{F
(s)} = f (t) then
L -1{F
(s - a)}
= eat f (t)
|
The
Heaviside (Unit Step) Function:
A
simple switch that is off until time t = a, when it is switched on
abruptly, can be modelled by the Heaviside function H(t-a)
(also known as the unit step function u(t-a).) It has a finite discontinuity at t = a.
A function f (t) that has one functional
form g(t) when t < a but a different form h(t) thereafter can be expressed in a single-line definition using the
Heaviside function:
The Laplace
transform of H(t-a) is
The
derivative of H(t-a) is
the Dirac delta function d(t-a):
The Dirac delta
function has the sifting property that
for any function
f (t) that is continuous
at t = a ³ 0.
The Laplace
transform of the Dirac delta function is
L {d(t-a)}
= e-as
Second
Shifting Theorem:
|
If L {f
(t)} = F (s)
then L {H(t-a)
f (t-a)}
= e-asF (s)
|
An
equivalent statement is
|
If L -1{F
(s)} = f (t) then
L -1{e-asF (s)}
= H(t-a)
f (t-a)
|
Periodic
Functions:
If f (t) is a periodic function with period T
(so that f (t + T)
º f
(t) "t),
then
5.3 Convolution
The
convolution of two functions f (t) and g(t) is a function (f*g)(t)
defined by
Convolution is
symmetric, so that
It
can be shown (using methods of multiple integration from Chapter 6) that
and,
equivalently, that
where F (s) = L {f (t)}
and G(s) = L {g(t)}
Some
Laplace transforms and inverse Laplace transforms can be determined using these
identities.
The
sifting property of the Dirac delta function leads to
(d*f)(t)
= f (t)
and
|
d(t-a)*f
(t) = f (t-a) H(t-a)
|
When
the initial conditions of an initial value problem are all zero, convolution
can be used to obtain the complete solution:
When y(0) = y' (0) = 0 and y"
+ by' + cy = r(t)
and
Laplace transforms are defined as
Y(s)
= L {y(t)}, R(s) = L {r(t)},
Q(s) = L {q(t)},
then
(s2
+ bs + c) Y(s) = R(s) Þ Y(s) = Q(s)R(s)
where 
is the transfer
function.
The
complete solution to { y" + by'
+ cy = r(t) , y(0) = y' (0) = 0 } is then
y(t) = q(t)*r(t)
Convolution
can also be used to solve integral equations that include integrals of the form
Example
The
integral equation
is
also y(t) = t2
+ 1 - 9 t*y(t)
Let Y(s) = L {y(t)} , then upon taking the Laplace
transform of the integral equation,
which leads to
Taking
the inverse Laplace transform yields the solution
5.4 Some Inverse Laplace
Transforms
[Note
that some of the equations displayed in this table might appear displaced
from their proper positions.]
|
F (s)
|
f (t)
|
|
F (s)
|
f (t)
|
|
|
f (t)
|
|
|
t cos w t
|
|
(n Î ù)
|
|
|
|
|
|
|
eat
|
|
(n Î ù)
|
|
|
e-as
|
d (t
- a)
|
|
|
H (t - a)
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Square wave,
period 2a
,
amplitude 1
|
|
|
Triangular
wave,
period 2a
,
amplitude
a
|
|
|
Sawtooth wave,
period a
,
amplitude b
|
|
{sn F(s) - sn-1f (0)
- sn-2f N(0)
- sn-3 f O(0) - ...
- s f
(n-2)
(0)
- f
(n-1)
(0) }
|
|
|
|
|
|
|
- t
f (t)
|
5.5 Miscellaneous Examples
Only the key
steps in the solutions are listed here.
More details will be presented in class.
Example
1
Derive
the Laplace transform of f (t)
= cos wt .
L {cos wt}
can be evaluated directly from the definition:
but
this requires a double integration by parts and some tedious algebraic
manipulation.
Instead,
note that
Using
also the linearity property of Laplace transforms, it then follows that
Therefore
Example
2
Solve
the initial value problem
y" - 5 y' + 6 y = 0
, y(0) = 1, y' (0) = 0
Let Y(s) = L {y(t)} .
L {y" }
= s2Y(s) - s.y(0) - y' (0) = s2Y - s
L {y' }
= s Y(s) - y(0) = s
Y - 1
L {y" - 5 y'
+ 6 y} = L {0} Þ (s2 - 5s
+ 6) Y(s) - s
+ 5 =
0
Therefore
y(t) = 3 e2t - 2 e3t
Example
3
Solve
the initial value problem
y" + 4 y' + 13
y = 26 e-4t , y(0) = 5, y' (0) = -29
Let Y(s) = L {y(t)} .
L {y" }
= s2Y(s) - s.y(0) - y' (0) = s2Y - 5 s + 29
L {y' }
= s Y(s) - y(0) = s
Y - 5
L {y"
+ 4 y' + 13
y} = L {26 e-4t}
leads
to
,
Therefore
y(t) = 2
e-4t + e-2t (3
cos 3t - 5 sin 3t)
Example
4
Solve
the initial value problem
f (t) = t
H(t-3) = (t-3) H(t-3) + 3 H(t-3)
and the second shift theorem,
L {y"
+ 4 y} = L {f (t)}
becomes
, where Y(s) = L {y(t)}.
Example
5
Note
that
.
[Either
use partial fractions, or] use the identity
Therefore
Example
6
Use L {cos w t} to derive L {sin w t} .
Using L {f
' (t)} = s.L {f (t)} - f
(0) and
we
have
Therefore
Example
7
A
damped mass-spring system (with damping constant c = 3m and spring modulus k = 2m)
is at rest until an impulse of 1 Ns is applied at time t = 1 s. Find the response y(t).
The
impulse is modelled by a Dirac delta function at time t = 1.
y" + 3 y' + 2 y = d(t-1) , y(0)
= y' (0) = 0
Taking
Laplace transforms, with Y(s)
= L {y(t)},
(s2
+ 3s + 2) Y(s) = e-1s
[This
is an over-damped system.]
or
y(0) =
y' (0) = 0 allows the transfer
function Q(s) to be used:
y" + 3 y' + 2 y = d(t-1) Þ r(t) = d(t-1) ,
and Y(s) = Q(s)R(s) Þ y(t) = q(t)*r(t) = q(t)*d(t-1) = q(t-1) H(t-1)
by
the sifting property of the Dirac delta function.
This
leads to the same solution as before.
Example
8
Find
the Laplace transform of the full square wave, amplitude 1, period 2a.
The square wave can be
described by
Over
the first period only, it can also be described by
f (t) = H
(t) - 2 H (t-a)
The
period is T = 2a. It then follows that
Therefore
Other
examples will be demonstrated in class.
5.6 Additional Details on Inhomogeneous ODEs
with Discontinuities
Consider
the second order linear inhomogeneous ODE
with l1 ¹ 0, l2 ¹ 0, l1 ¹ l2 ,
where a1 and a2 are constants.
Let r(x) = the right hand side of
the ODE and
let R(s) = L {r(x)}, the Laplace transform of r(x).
Let Y(s) = L {y(x)}, the Laplace transform of the
solution.
In
general, if
then
where u(x-a) º f
(x) and v(x-a) º g(x)
by
the second shift theorem and where
F(s) = L {f (x)},
V(s) = L {v(x)} and
U(s) = L {u(x)}.
In
this case, f (x) and g(x) are both constants, so
that
v(x-x ) - u(x-x ) º g(x)
- f
(x) º a2
- a1
.
The
Laplace transform of the initial value problem is then
from
the cover-up rule [or another valid method] for decomposition into partial
fractions.
Applying
the second shift theorem in reverse Þ
One
can show that
so
that the solution y(x) is continuous at x = x .
One
can also show that
so
that the solution y(x) is differentiable at x = x .
A
solution found by a non-transform method (finding the complementary function,
particular solution, general solution and complete solution) must therefore
assign values to the arbitrary constants in such a way that the complete
solution is differentiable (and continuous) at
x = x . This is considered on the next page.
Piecewise
Continuous r(x) – Non-Transform Method
Consider
the second order linear inhomogeneous ODE
with l1 ¹ 0, l2 ¹ 0, l1 ¹ l2 ,
where a1 and a2 are constants.
C.F.: 
P.S.: 
G.S.: 
Initial
conditions (branch x < x ):
Therefore
the complete solution for the branch x
< x is
For x > x we impose conditions of continuity and
differentiability at x = x :
This
pair of simultaneous equations leads to a different pair
of values for the arbitrary constants A,
B in the branch x ³ x than they had
for the branch x < x :
The
complete solution for all x is then
as before.
As an
example, the initial value problem
(for
which a1 = 12, a2 = 0, l1 = 3, l2 = x = 2)
has the complete solution
One
can check that y(x) and y'
(x) are both continuous
at x = 2.
The
graphs of y(x) and y'
(x) are available on the web
page
"www.engr.mun.ca/~ggeorge/2422/notes/c56ex.html".
END OF CHAPTER 5
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Created 2001 03 12 and modified 2001 03 12
by Dr. G.H. George
