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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.
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