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INSTRUCTOR: Hi.
Today we're going to conclude
our study of learning
our differential equations.
And we're going to
use the occasion
to introduce the concept
of a Laplace transform.
I should point out that
the Laplace transform has
much greater application than
just two linear differential
equations.
It's related to the
Fourier transform.
It has applications
in the subject known
as the convolution integral--
many applications, which
we'll talk about
during the learning
exercises in this unit.
But within the framework of
learning our differential
equations, I thought
this would be a good time
to introduce this rather
important concept.
Consequently, our topic
for today is called,
quite simply, the
Laplace transform.
And it hinges on the fact that
e to the t goes to infinity--
as t approaches
infinity, you see--
much more rapidly than
most functions of t.
And by the way, I
say t here instead
of x, because the Laplace
transform was introduced
primarily for
differential equations
in which the independent
variable was time.
So traditionally, one
usually talks about f of t
when one is talking about
Laplace transform concepts.
But the entire idea is this.
Remember, as we saw in part 1
of our course, that e to the t
goes to infinity much faster
than t to the n for any integer
n.
Consequently, what this means
is, since most functions can
be represented in
the power series,
one would expect
that if one were
to take a function like
f of t and multiply it
by e to the minus st where s
was a constant greater than 0,
one would expect that e
to the minus st f of t
would approach 0 rapidly
as t approaches infinity.
Now, this may not happen.
We'll talk about that
in more detail later.
But at any rate, let's
assume that we're
dealing with a particular
function f of t
where not only does
e to the minus f of t
go to 0 rapidly for some value
of s as t approaches infinity,
but so rapidly that the integral
of e to the minus st f of t dt
from 0 to infinity converges.
Let's just keep that in
mind for the time being.
And with this in mind, one
defines the Laplace transform
of a function f of t.
Namely, given f of t, the
Laplace transform of f of t,
written l of f of
t, is defined to be
the integral from 0 to
infinity e to the minus st
f of t dt, provided that
this integral converges.
Notice, by the way, that since t
is the variable of integration,
when we evaluate this
integral at t equals 0
and t equals infinity,
the resulting function
becomes a function of s alone.
In other words,
we evaluate this.
As t goes from 0 to infinity,
the resulting integral
is a function of s alone.
Consequently, one often-- to
identify the function with this
Laplace transform--
indicates that, if the
function was f of t,
the Laplace transform
is f bar of s.
And notice, by the way, what
this thing means pictorially.
This is an area under a curve.
And what you're really
saying, if the Laplace
transform of a
function exists, is
that this area under the curve
as t goes from 0 to infinity
is, indeed, finite.
By the way, this
should also lead
you to believe that if this is
finite for one value with of s,
if s is replaced by
something larger,
this goes to 0 even faster.
And consequently, the
area under the curve
will exist for all values
of s beyond a certain value
once the area exists
for that one value.
Maybe the easiest way to say
that is in terms of a picture.
Here, I've drawn two
representative functions
y equals f of t, one case
being a pulsation type of thing
where you have a irregular
pulse repeated here,
the other being a polynomial
y equals t squared.
What we're saying is that,
if I multiply f of t by e
to the minus st--
remember, e to the minus
st for positive values of s
is a curve that goes
to 0 very rapidly.
And what this tends to do is
to pull this thing down so
that you go to 0 very rapidly.
And the fact that
the integral exists
says not only does this
go to 0 very rapidly,
but it goes to zero so fast
that the area under the curve
stays finite.
Remember, that was our
version of infinity times 0.
It's not enough that
this goes to 0, that this
approaches the t axis.
It has to approach
fast enough so
that the area under
the curve is finite.
And all we're saying
is that if you
can find one value of s that
makes the area under the curve
finite, any larger value
of s will certainly
make the area under
the curve finite,
because this will be
pulled down even faster.
And perhaps you can
begin to see intuitively
that, as I let s get
larger and larger,
not only does the area
under the curve stay finite,
but one would expect that
it would also approach 0
because, for large values
of s, e to the minus
st goes to 0 so rapidly that
even large values up here would
be pulled down very rapidly.
At any rate, let me illustrate
this by means of an example.
Suppose I pick, for f of
t, something as simple
as e to the at where
a is a given constant.
What I mean by "simple" here
is recalling that the Laplace
transform involves multiplying
f of t by e to the minus st.
This will give me a simple
algebraic manipulation
to perform.
By definition, how do I form
the Laplace transform of e
to the at?
What I do is I multiply e to
the at by e to the minus st
and integrate that
from 0 to infinity.
And if that integral exists--
meaning if the
integral converges,
if this limit is finite--
I call that the
Laplace transform.
At any rate, just substituting
n here then, f of t
becomes e to the at.
This says integral
from 0 to infinity
e to the a minus s
times t power dt.
Since a and s are constants,
the integral e to the a minus st
comes out to be 1 over a
minus s, e to the a minus st,
evaluated as t goes
from 0 to infinity.
Now, the key point
to notice here
is that, as soon as
s is greater than a,
this gives me a
negative exponent--
as soon as s is greater than a.
Consequently, as t
goes to infinity,
this essentially
behaves like one over e
to the infinity, which is 0.
In other words, once
s is greater than a,
the upper limit drops out.
My lower limit would just
be e to the zero, you see.
And I'm subtracting that.
At any rate, to make
a long story short
and to summarize
what we're saying,
simply observe that,
if s is greater
than 0, the limit as
t approaches infinity
e to the a minus st is 0 and
that, therefore, the Laplace
transform of e to the at
is simply 1 over s minus a,
provided s is greater than a.
In other words, again,
just looking back here--
see-- I subtract
the lower limit.
When t is 0, this is 1.
But I'm subtracting.
It makes it minus 1.
And minus 1 over a minus s is
the same as 1 over s minus a.
To state this then in
other words, if f of t
is e to the at, f
bar of s is 1 over s
minus a where the domain
of f bar is a set of all
s, such that s is
greater than a.
And what does this
mean, geometrically?
This is the area under the
curve e to the minus st--
e to the at--
as t goes from 0 to infinity.
That area will be finite as
soon as s is greater than a.
By way of illustration,
if I replace a by 2,
what we're saying is that
the Laplace transform of e
to the 2t is 1 over s minus 2.
Notice that nice
polynomial quotient idea,
that rather simple expression--
1 over s minus 2 if
s is greater than 2.
If we replace a by minus
3 and use this recipe,
we have that the Laplace
transform of e to the minus 3t
is 1 over s minus
minus 3, 1 over s
plus 3, where S must be
greater than minus 3.
Notice also here that, by the
comparison test, so to speak--
remember, the same comparison
test that we use for infinite
series--
if the magnitude of f of t
is less than some constant c
times e to the at power
for all values of t,
then the integral from 0 to
infinity e to the minus st
f of t dt must also converge.
In other words, we already know
that this integral converges,
shall we say, if the
integrand were e to the at.
Therefore, we're saying it will
converge if you multiply that
by a constant, because
convergence isn't affected
by multiplying by a constant.
Consequently, if the integrand
that we're investigating
is smaller than
this in magnitude,
it must also have a Laplace
transform-- that this thing
here must converge.
Now, because this is so
obvious and so important,
one defines a function
with this property
to have a very special name.
Namely, the function
f of t is set
to have exponential
order if and only
if there exists
constant c and a such
that the magnitude of
f of t is less than ce
to the at for t, all
t, greater than 0.
What we are saying
in summary is that,
by this definition
of exponential order
and the comparison test, all
functions of exponential order
have Laplace transforms,
because that integral from 0
to infinity-- e to the
minus st f of t dt--
will converge by comparison
with the integral 0 to infinity
e to the minus st e to the
at dt for s greater than.
In other words, this will all
converge where the value of s
has to be greater than a.
Now, the idea is, what's so
important about functions
of exponential order?
Why do I stress these?
And the answer is that, in
terms of linear differential
equations with
constant coefficients,
the functions that we
tend to deal with all
have exponential order.
In other words, they will
all have Laplace transforms.
For example, e to
the at, trivially,
is a function that
has exponential order.
That was our model
for the definition.
In fact, e to the
at is 1e to the at.
Simply take a to
ba c to b1, and we
have the criterion for
exponential order being obeyed.
The sine of at where a is any
constant has exponential order.
Why?
Because the magnitude of sine at
is certainly no greater than 1.
And 1 is certainly
no greater than e
to the t of any
positive value of t.
See, e to the 0 is already 1.
Therefore, sine at is
less than 1e to the 1t
if t is greater than 0.
So sine at has
exponential order.
Finally, we dealt with
polynomials t to the n
where n was a positive integer.
And in part 1 of our course,
either using L'Hospital's rule,
or power series,
or what have you,
we showed that for any
fixed positive integer n,
t to the n times e to the
minus t went to 0 in the limit
as t went to infinity.
Consequently, for
large values of t,
the magnitude of t to the
n is less than e to the t.
I'll leave the details out here.
I'm just trying to
illustrate what's going on.
The important point is that
the usual functions encountered
in linear differential equations
with constant coefficients have
Laplace transforms.
And I would like to
make a couple of notes
about this thing.
One I've already said, but
I'd like to reinforce it.
What we've already seen is that
if s has exponential order,
f bar of s, the
Laplace transform of s
must be less than or equal
to the Laplace transform of e
to the--
well, a constant
times e to the at.
But the Laplace transforms e
to the at is 1 over s minus a
so that the Laplace
transform of f of t,
if f has exponential
order, can be no bigger
than c over s minus a.
Since c is a constant
and a is a constant,
as s goes to infinity,
f bar of s goes to 0.
In other words, for
sufficiently large values of s,
the Laplace transform, which
names an area under a curve,
goes to 0 meaning,
as we would expect,
the area under the
curve does go to 0
as s increases without bound.
What that means in
particular, you see,
is that not any function
can be a Laplace
transform of something.
You see, in
particular, this says
that for f bar to be the
Laplace transform of something,
f bar of s must go to 0
as s goes to infinity.
For example, if I were to write
down f bar of s to be s over s
minus 1-- or s plus 1, even, f
bar of s is s over s plus 1--
notice that the limit of f bar
of s, as s goes to infinity,
is 1.
Consequently, this f bar cannot
be the Laplace transform of any
function, at least
of exponential order,
because we've already seen
that the Laplace transform
of a function--
at least of exponential order--
must go to 0 as s
goes to infinity.
So not every function can be a
Laplace transform of something.
And secondly, not every
function has exponential order.
It may be true,
as we just showed,
that every function that we're
dealing with when we're dealing
with linear
differential equations
with constant coefficients--
or most functions
that we're dealing with
linear differential equations
with constant coefficients--
has our functions
of exponential order.
Not all functions have
exponential order.
For example, e to the
t squared does not
have exponential order--
not order order, order--
since, notice, that e to the t
squared divided by ce to the at
would be 1 over c, e to
the t squared minus at,
which is t times t minus a.
And notice now that,
if t is greater than a,
this exponent will be positive.
And as t goes to infinity-- you
see, once t is greater than a,
this thing increases
without bound.
And consequently,
since c is a constant,
you essentially have infinity
divided by a non-zero constant.
So you have, what?
That e to the t
squared over ce to
the at goes to infinity
as t approaches infinity.
In particular, this says
that e to the t squared
dwarfs ce to the at
for large values of t.
In particular, you cannot have
that e to the t squared is less
than some constant
times e to the at.
Now the question
that comes up is,
what's so important
now about the Laplace
transform in terms of linear
differential equations?
And the answer is that the
Laplace transform, amazingly
enough, has properties
of linearity.
What do you mean by
properties of linearity?
You mean, what?
That for a function
to be linear,
l of a sum must be
the sum of the l's.
And l of a constant
time something
must be the constant
times l of something.
Well, look at-- when you're
dealing with integrals,
given two definite
integrals which exist,
the sum of the integrals
is the integral of the sum.
The integral is a constant
times an integrand
is a constant times the integral
of the integrand itself.
In other words--
sparing you the details,
because they follow right
from the definition.
And that is assuming
that f and g have Laplace
transforms-- in other words that
that improper integral from 0
to infinity, e to the minus
st f of t dt, converges.
Then what's true is that l of
f plus g is l of f plus l of g.
And l of c times f
is c times l of f.
See-- the linearity properties.
Now, how does that help us?
And why does that give
us a hint that there
will be some usage of the
Laplace transform in solving
linear differential equations
with constant coefficients?
Well, perhaps the
best way to see this
is by means of an illustration.
Let's assume that
we're given that y
is a twice differentiable
function of t
and that y double
prime plus 2a y
prime plus by where a and b
are constants is some function
f of t.
The idea is this.
I simply take the Laplace
transform of both sides.
See?
By linearity, since the Laplace
transform is a linear function,
a linear operator, l of y double
prime plus to 2ay prime plus by
is simply l of y double prime
plus 2a l of y prime plus b
l of y.
And the Laplace transform of
f of t, assuming that f of t
has a Laplace transform--
in particular, if f of t
happens to be a function
of exponential order,
we've already seen that it
will have a Laplace transform.
All we're saying is, let's call
a Laplace transform of f of t,
as usual, f bar of s.
And now what we have is an
equation involving the Laplace
transform of y, the
Laplace transform
of y prime, the Laplace
transform of y double prime.
In terms of f bar of s, if
somehow we could manipulate
this to solve for l
of y-- in other words,
if we could find what the
Laplace transform of y was
in terms of s--
maybe we could then invert,
so to speak, and find out
what the function
itself must have been.
In other words, the idea
of inverse functions
comes up the same way here
as it comes up every place.
Namely, starting with a
function that maps x into y,
one often says, if I know
what y is, can I determine
what x must have been?
And that's why 1 to
1-ness is so important.
In other words, if it turns out
that the Laplace transform is
1 to 1, once I know
the Laplace transform,
I essentially know the
function, because if two
different functions can't have
the same Laplace transform--
and by the way, as a
finale for today's lecture,
we will show that
this is essentially
the case that the Laplace
transform is 1 to 1.
And what I mean
by essentially, it
will be pointed out in
the learning exercises.
But don't worry about that now.
The idea is simply this.
Suppose there was some way
of expressing l of y prime
and l of y double prime
in terms of l of y.
I could then solve
the resulting equation
for l of y in terms of s.
Having a table of
Laplace transforms,
I could then locate
what that function of s
is the Laplace
transform of and then
invert this to find what
my y must have been.
The problem is, how
do I know that I
can express l of y prime,
l of y double prime,
l of y triple prime?
I'm only going up to
the second derivative
in terms of our usual convention
of illustrating everything
by second-order equations,
even though the results
hold the higher orders.
The question though
is, how do we
know that we can
express these Laplace
transforms in terms of the
Laplace transform of y itself?
And this brings up the
second reason why that fact--
that e to the minus st
has a multiplying factor--
is so important.
You see, aside from the fact
that, for most functions,
e to the minus st makes sure
that the resulting integral
will converge, the other key
property is that when you
integrate or differentiate
e to the minus st,
because s is a parameter--
meaning it's being treated
as a constant at a given time--
what this thing means is that
the integral or the derivative
of e to the minus st is simply
a constant times e to the minus
st. Well, let me show
you how this is used.
Let's suppose, for
example, that I
wanted to find the Laplace
transform into f prime of t.
And I didn't know any tricks.
And by the way, I hope that this
is one of the fringe benefits
that this course is teaching
you-- that we do not have
to be tricky in mathematics.
All we have to know is
the basic definitions
and how to manipulate them.
I do admit that it
is a stroke of genius
sometimes in inventing the
basic definition that we're
going to use.
For example, it's my own
belief that it's a lot easier
to use the Laplace transform
than it was to have invented
the concept in the first place.
But once I've defined what
the Laplace transform is, what
happens here is very simple.
Namely, by definition,
what is the Laplace
transform of f prime of t?
I simply multiply f prime
of t by e to the minus st
and integrate that
from 0 to infinity.
And if that integral converges,
then the Laplace transform
exists.
Let me assume, for
the sake of argument
now, that the Laplace
transform of f of t exists.
And by the way, again
remember, I'm not really
assuming anything
when I'm dealing
with linear
differential equations
with constant coefficients,
because the functions that
come up as possible solutions
there are of exponential order.
And we have already
seen that, for functions
of exponential order, the
Laplace transform does indeed
exist.
But anyway, the idea is this.
By definition, this is
the Laplace transform
of f prime of t.
I would like to be
able to integrate this.
Well, the idea is that the
integral of f prime of t
is certainly f of t.
And if I-- well, to a constant.
But that's not important.
What I'm thinking of is
integrating by parts.
Notice that if I let u
equal e to the minus st
and dv be f prime of t dt, then
du is minus se to the minus st.
v would just be f of t.
Remembering the recipe
for integration by parts--
remember, just by
way of quick review--
integral u dv is equal to
uv minus the integral v du.
What we're saying here is that
to evaluate this integral,
we simply take, what?
u times v-- e to
the minus st f of t.
Evaluate that as t goes
from 0 to infinity,
minus the integral v du.
du is already negative.
s is a constant.
So minus integral v du is just
plus s integral 0 to infinity,
e to the minus st f of t dt.
And lo and behold, you see
that's that beautiful property
of e to the minus st.
It's still in here.
The integral is precisely
what we mean by, what?
l of f of t.
In other words, the Laplace
transform of f prime of t
is this thing plus s times the
Laplace transform of f of t.
But let me not be so informal.
Let me not refer to
this as "this thing."
Let's see what "this
thing" really is.
Notice that, because the
Laplace transform of f of t
is assumed to exist, the fact
that the integral of this
from 0 to infinity is
finite means, in particular,
that ad infinity, the
integrand must be 0.
You see, otherwise, how could
the area under the curve
be finite if the curve didn't
at least asymptotically approach
the t axis?
So in other words,
the assumption
that f has a Laplace transform
means that the upper limit
gives me 0.
The lower limit gives me, what?
When I plug in t
equals 0, this is
e to the minus s 0, which is 1.
This is f of 0.
So the lower limit is f of 0.
And since I'm subtracting
the lower limit, I have what?
That the thing that I
call "this thing" is,
more precisely, minus f of 0.
In other words, to find the
Laplace transform of f prime
of t, it's simply minus f of
0 plus s times the Laplace
transform of f of t.
Well, look.
If I happen to know
what f of 0 is--
in other words, if I pick a
particular member of the family
f of t, given the
curve y equals f of t--
I certainly know what f of 0 is.
This says, what?
The Laplace transform of f
prime of t is some constant plus
s times the Laplace
transform of f of t itself.
In other words,
somehow or other, I
have now managed to express
the Laplace transform
as f prime of t in terms
of a polynomial involving s
and the Laplace transform
of f of t itself.
By the way, notice
that generically here
there is nothing sacred
about f of 0, or f.
Think of f as being any
function and f prime
as being its derivatives.
In particular, I could allow
f prime to play the role of f.
See-- this says, what?
The Laplace transform
of the derivative
is minus the function evaluated
at 0 plus s times the Laplace
transform of that function.
So if I now take my
function to be f prime,
its derivative is
f double prime.
The function is,
itself, f prime.
So all I really do is, what?
Go into this recipe here.
Every place I see a prime,
replace it by a double prime.
Every place I see no
prime, put a prime in.
And I have that the Laplace
transform of f double prime
of t is minus f prime of
0 plus s times the Laplace
transform of f prime of t.
By the way, to make sure
you see this, notice--
what would the Laplace
transform of f triple prime be?
The Laplace transform
of f triple prime of t
would be minus f
double prime of 0
plus s times the Laplace
transform of f double prime
of t.
So assuming that f prime
also has a Laplace transform,
we see what the
Laplace transform
of f double prime of t looks
like in terms of f prime of t--
the Laplace transform
of f prime of t.
We know what the Laplace
transform of f prime of t
looks like in terms of the
Laplace transform of t.
In other words,
from here, I simply
replace l of f prime of t by
its value in this equation.
And I wind up with that the
Laplace transform of f double
prime of t is nothing more
than minus f prime of 0
minus s times f of 0 plus
s squared l of f of t.
And now, you see, I've
solved the problem
that I was discussing over here.
Namely, once I know
what y of 0 is--
by the way, don't confuse
the f here with the f here.
That's an unfortunate choice,
which I've just noticed.
The f that I'm
referring to here names
the function whose Laplace
transform I'm trying to find.
Notice that the f prime and f
double prime are served by y
prime and y double prime here.
What we're saying in
terms of this notation
is that I can express l of y
double prime and l of y prime
in terms of the
Laplace transform of l
of y plus powers of s, provided
I know only what y of 0 is
and what y prime of 0 is.
And by the way, again, notice
that in a physical problem,
it is very natural to
look at y and y prime
when t equals 0 because,
in many instances,
t equals 0 represents
the time at which we've
started the measurement
at our experiment.
And this becomes a very
natural interpretation
for what you mean by
the initial conditions--
namely, what's going
on when t equals 0.
At any rate, we are now in a
position to apply our results
to an actual linear differential
equation with constant
coefficients given
initial conditions
as to what's happening at 0
in terms of the y-coordinate,
and the y-prime
coordinate, et cetera--
in other words,
as an application
to a linear
differential equations
with constant coefficients.
Suppose we want to solve--
meaning, what?
Find what function
y is of t, if y
double prime minus 4y
prime plus 3y equals
e to the 2t, given that my
initial conditions-- namely,
when t is 0, y is going to be
0, and y prime is going to be 1.
This could be any
conditions I want over here.
I just chose these to
give me a rather simple
algebraic example.
And I'll pick more complicated
things in the exercises.
But the idea is this.
We have already
seen by linearity
that, if I take the Laplace
transform of both sides here,
I have the l of y
double prime minus
4l of y prime plus 3l of y
is the Laplace transform of e
to the 2t.
Now, we just saw,
right over here,
that the Laplace transform
of y double prime is
minus y prime evaluated
at 0 minus sy of 0 plus
s squared l of y.
The Laplace transform of y
prime was minus, you see,
y of 0 plus sl y.
So minus 4 times that
is just 4 times y of 0
minus 4s l of y plus 3l of y.
And we saw earlier that
the Laplace transform of e
of the at is 1 over
s minus a where
s is any number greater than 2.
In particular, this is
that formula with a equal--
s is any number greater than a.
In particular, this
is that formula
with a equal to 2,
so that the Laplace
transform of e to the
2t is 1 over s minus 2,
provided s is greater than 2.
By the way, notice,
up to this point,
I have not used the
initial conditions.
I have not used the fact that y
of 0 is 0 and that y prime of 0
is 1.
Notice that, even without this,
the l of y is here by itself.
It's going to be multiplied
by s squared minus 4s plus 3.
These terms can go over onto
the other side of the equation.
I will get a
function as s alone.
I then divide through by
s squared minus 4s plus 3.
That will give me l of
y as a function of s.
And if I can then find
one function which
has that function of s
as its Laplace transform,
I can conclude
that that function
is the one I'm looking for,
provided only that l is
a 1 to 1 operator-- that the
Laplace transform is a 1 to 1
function.
At any rate, all I
did in our problem
was to simplify this by
specifying that y of 0 was 0
and that y prime of 0 was 1.
These two terms drop out.
This is minus 1.
It comes over onto the
other side as plus 1.
And so I wind up with
this particular equation.
And by the way, notice that
this simply says, what?
s minus 2 plus 1 over s minus 2.
In other words, this is
s minus 1 over s minus 2.
This factors into s
minus 1 times s minus 3.
So in other words,
what I'm now faced with
is that s minus 1 times
s minus 3 times l of y
is equal to s minus
1 over s minus 2.
I have to be careful.
You see, I wanted
to divide-- see,
I want to be able to cancel
out the s minus 1's here.
I want to be able to divide
through by s minus 3.
I have to be careful
of 0 denominators.
And the safest way to
take care of everything
is that all I
really care about is
whether the Laplace transform
exists for sufficiently
large values of s.
To get rid of my
headache, I'll simply
assume that I'm not even
looking at the Laplace
transform until, shall we
say, s is greater than 3.
You see, once s is greater
than 3, this factor can't be 0.
This factor can't be 0.
Consequently, I can solve for l
of y and concluded that l of y
is 1 over s minus
3 times s minus 2,
provided that s
is greater than 3.
Now I resort to that same
method of partial fractions
that we used when we
integrated back in part 1
using partial fractions.
I simply try to find
constants a and b,
such that I can write
this as a over s minus 3
plus b over s minus 2.
In other words, break this
down into simpler parts.
Sparing you the details,
it follows very simply
in this case that this
expression is simply 1 over s
minus 3 minus 1 over s minus 2.
In other words, that if I put
this over a common denominator,
I get 1 over s minus
3 times s minus 2.
Now here's where
the kicker comes in.
Remember, we have already
seen that 1 over s minus a
is the Laplace
transform of e to at,
provided s is greater than a.
In particular, the Laplace
transform of e to the 3t
would be 1 over s minus 3,
provided s was greater than 3.
Similarly, the Laplace
transform into e to the 2t
would be 1 over s minus 2,
provided s was greater than 2.
By the way, notice that
both of these conditions
are obeyed as soon as
s is greater than 3.
Obviously, if s is greater than
3, it must be greater than 2.
So I'm sure that
both of these results
are true once s
is greater than 3.
You see the same
3 I have up here.
Now, by linearity, I know
that the Laplace transform
of a difference is the
difference of the Laplace
transforms.
You see, by equals
added to equals,
this tells me that l of e to
the 3t minus l of e to the 2t
is 1 over s minus 3
minus 1 over s minus 2.
But by linearity, l of e to
the 3t minus l of e to the 2t
is l of the quantity e to
the 3t minus e to the 2t.
In particular then,
do I know one function
whose Laplace transform is
1 over s minus 3 minus 1
over s minus 2?
And the answer is, yes.
We've just constructed it--
namely, the function y equals
e to the 3t minus e to the 2t.
In other words what we do
know is that, in any event,
whether y is the given function
or not, whatever l of y is,
it's l of e to the
3t minus e to the 2t.
And therefore, we could
conclude that y must equal e
to the 3t minus e to the 2t,
provided that l is 1 to 1--
provided that there was only one
function that can have a given
Laplace transform.
You see, you cannot
automatically conclude that,
because these two
things are equal--
that the whole
expression's equal--
that the inputs must be equal.
For example, you cannot conclude
that x equals pi over 6,
simply because you know that
sine x equals sine pi over 6.
You see, the sine
function is not 1 to 1.
Well, at any rate, I think
you can sense intuitively
that-- because we already know
that, for linear differential
equations with
constant coefficients,
there are no singular
solutions and that, once you've
found one solution, subject
to given initial conditions,
you've found them all.
I think you can guess that,
at least for functions
of exponential order, this
should be a true result.
And the fact that it is true
is known in the literature
under the name of
Lerch's theorem--
"Lerch," not L-U-R-CH,
but L-E-R-CH.
It's known under the
name of Lerch's theorem.
And I have taken the
liberty of stating it
in a less controversial form,
saving the generalization
for the learning exercises.
But essentially, all Lerch's
theorem says is this.
If f and g are continuous
and of exponential order,
then if the Laplace transform of
f equals the Laplace transform
of g, for all sufficiently large
values of s-- in other words,
if beyond a certain s, f
bar of s equals g bar of s,
in other words, f and g have
the same Laplace transform--
then f and g must be identical.
And there is a
slight modification
of this that comes up if you
leave out the word "continuous"
here.
Lerch's theorem is stated
only in terms of f and g
being piecewise continuous.
In other words, there could
be jump discontinuities
in the curves and things
like this, at which case
this would hold, except possibly
at the points at which you
had jump discontinuities.
I prefer not to get into
this at this particular time,
but rather to wrap
up our lecture
at this particular
point, and to simply say
that we have done-- what now?
We have studied linear
differential equations.
We have used this last
lecture to introduce
the concept of the
Laplace transform,
which has secondary value in
terms of solving differential
equations, but which has
other applications that
go far beyond the scope
of this particular course.
And what I'm hoping is that this
rather introductory lecture,
coupled with
well-chosen exercises,
will give you enough
insight to the Laplace
transform so that you
will be able to handle it
not only in terms of
solving linear differential
equations with
constant coefficients
but in other contexts
where they might occur
in the particular line of work
and research that you're doing.
At any rate, this completes
our work on block seven.
And in our next lecture,
what we shall do
is start the final
block of our course
and revisit the concept
of vector spaces
from, hopefully, a more
productive point of view
than what we've had before.
But more about that next time.
Until next time, goodbye.
Funding for the
publication of this video
was provided by the Gabriella
and Paul Rosenbaum Foundation.
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