Laplace Transform 5

Back

⇐ Use this menu to view and help create subtitles for this video in many different languages. You'll probably want to hide YouTube's captions if using these subtitles.

Laplace Transform 5 : Useful properties of the Laplace Transform

相關課程

0 / 750

Well, now's as good a time as any to go over some
interesting and very useful properties
of the Laplace transform.
And the first is to show that it is a linear operator.
And what does that mean?
Well, let's say I wanted to take the Laplace transform of
the sum of the-- we call it the
weighted sum of two functions.
So say some constant, c1, times my first function, f of
t, plus some constant, c2, times my second
function, g of t.
Well, by the definition of the Laplace transform, this would
be equal to the improper integral from 0 to infinity of
e to the minus st, times whatever our function that
we're taking the Laplace transform of, so times c1, f
of t, plus c2, g of t-- I think you know where this is
going-- all of that dt.
And then that is equal to the integral from 0 to infinity.
Let's just distribute the e the minus st.
That is equal to what?
That is equal to c1e to the minus st, f of t, plus c2e to
the minus st, g of t, and all of that times dt.
And just by the definition of how the properties of
integrals work, we know that we can split this up into two
integrals, right?
If the integral of the sum of two functions is equal to the
sum of their integrals.
And these are just constant.
So this is going to be equal to c1 times the integral from
0 to infinity of e to the minus st, times f of t, d of
t, plus c2 times the integral from 0 to infinity of e to the
minus st, g of t, dt.
And this was just a very long-winded way of saying,
what is this?
This is the Laplace transform of f of t.
This is the Laplace transform of g of t.
So this is equal to c1 times the Laplace transform of f of
t, plus c2 times-- this is the Laplace transform-- the
Laplace transform of g of t.
And so, we have just shown that the Laplace transform is
a linear operator, right?
The Laplace transform of this is equal to this.
So essentially, you can kind of break up the sum and take
out the constants, and just take the Laplace transform.
That's something useful to know, and you might have
guessed that was the case anyway.
But now you know for sure.
Now we'll do something which I consider even more
interesting.
And this is actually going to be a big clue as to why
Laplace transforms are extremely useful for solving
differential equations.
So let's say I want to find the Laplace transform of f
prime of t.
So I have some f of t, I take its derivative, and then I
want the Laplace transform of that.
Let's see if we can find a relationship between the
Laplace transform of the derivative of a function, and
the Laplace transform of the function.
So we're going to use some integration by parts here.
Let me just say what this is, first of all.
This is equal to the integral from 0 to infinity of e to the
minus st, times f prime of t, dt.
And to solve this, we're going to use integration by parts.
Let me write it in the corner, just so you
remember what it is.
So I think I memorized it, because I recorded that last
video not too long ago.
I'm just going to write this shorthand.
The integral of u-- well, let's say uv prime, because
that will match what we have up here better-- is equal to
both functions without the derivitives, uv minus the
integral of the opposite.
So the opposite is u prime v.
So here, the substitution is pretty clear, right?
Because we want to end up with f of x, right?
So let's make v prime is f prime, and let's make u e to
the minus st. So let's do that.
u is going to be e to the minus st, and v is going to
equal what?
v is going to equal f prime of t.
And then u prime would be minus se to the minus st. And
then, v prime-- oh, sorry, this is v prime, right?
v prime is f prime of t, so v is just going to be
equal to f of t.
I hope I didn't say that wrong the first time.
But you see what I'm saying.
This is u, that's u, and this is v prime.
And if this is v prime, then if you were to take the
antiderivative of both sides, then v is equal to f of t.
So let's apply integration by parts.
So this Laplace transform, which is this, is equal to uv,
which is equal to e to the minus st, times v, f of t,
minus the integral-- and, of course, we're going to have to
evaluate this from 0 to infinity.
I'll keep the improper integral with
us the whole time.
I won't switch back and forth between the definite and
indefinite integral.
So minus this part.
So the integral from 0 to infinity of u prime.
u prime is minus se to the minus st times v--
v is f of t-- dt.
Now, let's see.
We have a minus and a minus, let's make
both of these pluses.
This s is just a constant, so we can bring it out.
So that is equal to e to the minus st, f of t, evaluated
from 0 to infinity, or as we approach infinity, plus s
times the integral from 0 to infinity of e to the minus st,
f of t, dt.
And here, we see, what is this?
This is the Laplace transform of f of t, right?
Let's evaluate this part.
So when we evaluated in infinity, as we approach
infinity, e to the minus infinity approaches 0.
f of infinity-- now this is an interesting question.
f of infinity-- I don't know.
That could be large, that could be small, that
approaches some value, right?
This approach 0, so we're not sure.
If this increases faster than this approaches 0, then this
will diverge.
I won't go into the mathematics of whether this
converges or diverges, but let's just say, in very rough
terms, that this will converge to 0 if f of t grows slower
than e to the minus st shrinks.
And maybe later on we'll do some more rigorous definitions
of under what conditions will this
expression actually converge.
But let's assume that f of t grows slower than e to the st,
or it diverges slower than this converges, is another way
to view it.
Or this grows slower than this shrinks.
So if this grows slower than this shrinks, then this whole
expression will approach 0.
And then you want to subtract this whole expression
evaluated at 0.
So e to the 0 is 1 times f of 0-- so that's just f of 0--
plus s times-- we said, this is the Laplace transform of f
of t, that's our definition-- so the Laplace
transform of f of t.
And now we have an interesting property.
What was the left-hand side of everything we were doing?
The Laplace transform of f prime of t.
So let me just write all over again.
And I'll switch colors.
The Laplace transform of f prime of t is equal to s times
the Laplace transform of f of t minus f of 0.
And now, let's just extend this further.
What is the Laplace transform-- and this is a
really useful thing to know-- what is the Laplace transform
of f prime prime of t?
Well, we can do a little pattern matching here, right?
That's going to be s times the Laplace transform of its
antiderivative, times the Laplace transform of f prime
of t, right?
This goes to this, that's an antiderivative.
This goes to this, that's one antiderivative.
Minus f prime of 0, right?
But then what's the Laplace transform of this?
This is going to be equal to s times the Laplace transform of
f prime of t, but what's that?
That's this, right?
That's s times the Laplace transform of f of t, minus f
of 0, right?
I just substituted this with this.
Minus f prime of 0.
And we get the Laplace transform of the second
derivative is equal to s squared times the Laplace
transform of our function, f of t, minus s times f of 0,
minus f prime of 0.
And I think you're starting to see a pattern here.
This is the Laplace transform of f prime prime of t.
And I think you're starting to see why the Laplace
transform is useful.
It turns derivatives into multiplications by f.
And actually, as you'll see later, it turns integration to
divisions by s.
And you can take arbitrary derivatives and just keep
multiplying by s.
And you see this pattern.
And I'm running out of time.
But I'll leave it up to you to figure out what the Laplace
transform of the third derivative of f is.
See you in the next video.