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# Using the Convolution Theorem to Solve an Initial Value Prob: Using the Convolution Theorem to solve an initial value problem

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- Now that we know a little bit about the convolution integral
- and how it applies to the Laplace transform, let's
- actually try to solve an actual differential equation
- using what we know.
- So I have this equation here, this initial value problem,
- where it says that the second derivative of y plus 2 times
- the first derivative of y, plus 2 times y, is equal to
- sine of alpha t.
- And they give us some initial conditions.
- They tell us that y of 0 is equal to 0, and that y prime
- of 0 is equal to 0.
- And that's nice and convenient that those initial conditions
- tend to make the problem pretty clean.
- But let's get to the problem.
- So the first thing we do is we take the Laplace transform of
- both sides of this equation.
- The Laplace transform of the second derivative
- of y is just s squared.
- This should be a bit of second nature to you by now.
- It's s squared times the Laplace transform of Y, which
- I'll just write as capital Y of s, minus s-- so we start
- with the same degree as the number of derivatives we're
- taking, and then we decrement that every time-- minus s
- times y of 0-- you kind of think of this as the integral,
- and you take the derivative 1, so this isn't exactly the
- derivative of that-- minus, you decrement that 1, you just
- have a 1 there, y prime of o.
- And that's the Laplace transform of the second
- derivative.
- Now, we have to do the Laplace transform of 2 times the first
- derivative.
- That's just going to be equal to plus 2, times sY of s-- s
- times the Laplace transform of Y; that's that
- there-- minus y of 0.
- And we just have one left.
- The Laplace transform of 2Y.
- That's just equal to plus 2 times the Laplace
- transform of Y.
- And then that's going to be equal to the Laplace transform
- of sine of alpha t.
- We've done that multiple times so far.
- That's just alpha over s squared plus alpha squared.
- Now, the next thing we want to do is we want to separate out
- the Laplace transform of Y terms, or the Y of s terms.
- Actually, even better, let's get rid of these initial
- conditions.
- y of 0, and y prime of 0 is 0, so this term is 0.
- That term is 0, and that term is 0.
- So our whole expression-- I can get rid of the colors
- now-- it just becomes-- let me pick a nice color here--
- becomes s squared times Y of s, plus 2s, Y of s-- that's
- that term right there-- plus 2Y of s, is equal to the
- right-hand side, is equal to alpha over s squared plus
- alpha squared.
- Now let's factor out the Y of s, or the Laplace
- transform of Y.
- And so we get s squared plus 2s, plus 2, all of that times
- Y of s, is equal to this right-hand side, is equal to
- alpha over s squared, plus alpha squared.
- Now we can divide both sides of this equation by this thing
- right here, by that right there.
- And we get Y of s, the Laplace transform of Y is equal to
- this thing, alpha over s squared, plus alpha squared,
- times-- or, you know, I could just say times-- 1 over s
- squared, plus 2s, plus 2.
- I could just say divided by this, but it works out the
- same either way.
- Now, what can we do here?
- Remember, I was doing this in the context of convolution, so
- I want to look for a Laplace transform that looks like the
- product of two Laplace transforms. I know what the
- inverse Laplace transform of this is.
- In fact, I just took it.
- It's sine of alpha t.
- So if I can figure out the inverse Laplace transform of
- this, I could at least express our function y of t at least
- as a convolution integral, even if I don't necessarily
- solve the integral.
- From there, it's just calculus, or if it's an
- unsolvable integral, we could just use a computer or
- something, although you could actually use a computer to
- solve this so, you might skip some steps even
- going through this.
- But anyway, let's just try to get this in terms of a
- convolution integral.
- So what can I do with this?
- This is, let's see, this isn't a perfect square.
- So if this isn't a perfect square, the next best thing is
- to try to complete the square here.
- So let's try to write this as a, so let's see, if I write
- this as s squared plus 2s, plus something, plus 2.
- I just rewrote it like this.
- And if I wrote this as s squared plus 2s, plus 1, that
- becomes s plus 1 squared.
- But if I add a 1, I have to also subtract a 1.
- I can't just add 1's arbitrarily to things.
- So if I add 1 I have to subtract a 1 to cancel out
- with that 1.
- So I really haven't changed this at all, I just
- rewrote it like this.
- But this now, I can rewrite this term right
- here as s plus 1 squared.
- And then this becomes plus 1.
- That's this term right here.
- This is the plus 1.
- So I could rewrite my whole Y of s is now equal to alpha
- over s squared, plus alpha squared, times 1 over this
- thing, s plus 1 squared, plus 1.
- Now, I already said, I know what the inverse Laplace
- transform of this thing is.
- Now I just have to figure out what the inverse Laplace
- transform of this thing is.
- Of this-- let me pick a nice color-- of this blue thing in
- the blue box, and then I can express it as
- a convolution integral.
- And how do I do that?
- I could just do it right now.
- I could just immediately say that y of t-- let me write
- this down-- y of t, so the inverse is equal to the
- inverse Laplace transform of, obviously of Y of s.
- Let me write that down, Y of s.
- Which is equal to the inverse Laplace transform
- of these two things.
- The inverse Laplace transform of alpha over s squared, plus
- alpha squared, times 1 over s plus 1 squared, plus 1.
- And now the convolution theorem tells us that this is
- going to be equal to the inverse Laplace transform of
- this first term in the product.
- So the inverse Laplace transform of that first term,
- alpha over s squared, plus alpha squared, convoluted
- with-- I'll do a little convolution sign there.
- I was about to say convulsion.
- They're not too different.
- Convoluted with the inverse Laplace transform of this
- term, the inverse Laplace transform of 1 over s plus 1
- squared, plus 1.
- If I have the product of two Laplace transforms, and I can
- take each of them independently and I can invert
- them, the inverse Laplace transform of their product is
- going to be the convolution of the inverse Laplace transforms
- of each of them, each of the terms.
- And what I just said confused me a bit, so I don't want to
- confuse you.
- But I think you get the idea.
- I have these two things.
- I recognize these independently.
- I can independently take the inverse of each of these
- things, so the inverse Laplace transform of their products is
- going to be the convolution of each of their inverse
- transforms. Now what's this over here?
- Well I had this in the beginning of the problem?
- The inverse Laplace transform of this, right here, is sine
- of alpha t.
- And then we're going to convolute that with the
- inverse Laplace transform of this right here.
- Let's do a little bit of work on the side, just to make sure
- we get this right.
- So the Laplace transform of sine of t is equal to 1 over s
- squared, plus 1.
- That looks like this, but I was shifted by minus 1.
- You might remember that the Laplace transform of e to the
- at sine of t, when you multiply e to the at times
- anything, you're shifting its Laplace transform.
- So that will be equal to 1 over s minus a
- squared, plus 1.
- We essentially shifted it by a.
- So now we have something that looks very similar to this.
- If we just set our a to be equal to negative 1, here our
- a is equal to negative 1, then it fits this pattern.
- This is s minus negative 1.
- So the inverse Laplace transform of this thing right
- here is just e to the a, which is minus 1, so minus 1t,
- times sine of t.
- So this is the solution to our differential equation, even
- though it's not in a pleasant form to look at.
- And we can, if we want, express it as an integral.
- I'm not going to actually solve the integral in this
- problem, because it gets hairy, and it's not even clear
- that-- well, I won't even attempt to do it.
- But I just want to get into a form, and from there it's just
- integral calculus.
- or maybe a computer.
- What's the convolution of these two things?
- It's the integral from 0 to t, of sine of the first function
- of t minus tau.
- Well, I could actually switch, and I haven't shown you this,
- but we can switch the order either way, but actually let
- me just do it this way.
- I could write this as sine of [? out ?]
- t minus tau, times alpha-- I'm taking the sine of all of
- those things-- times e to the minus tau, sine of tau, dtau.
- That's one way, that if I wanted to express the solution
- of this differential equation's integral, I could
- write it like that.
- And it actually should be kind of obvious to you that this
- could go either way.
- Because when it was a product up here, obviously order does
- not matter.
- I could write this term first, or I could write that term
- first. So regardless of which term is written first, the
- same principle would apply.
- And I'll formally prove it in a future video.
- So we could have also done it the other way.
- We could have written this expression as e to the minus
- t, sine of t, convoluted with sine of alpha t.
- And that would be equal to the integral from 0 to t, of e to
- the minus t minus tau, sine of t, minus tau, times sine of
- alpha tau, dtau.
- So these are equivalent.
- Either of these would be an acceptable answer.
- And normally on a test like this, the teacher won't expect
- you to actually evaluate these integrals.
- The teacher will just say, get it into an integral just to
- kind of see whether you know how to convolute things and
- get your solution to the differential equation at least
- into this form, because from here it really is just, I
- won't say basic calculus, but it's
- non-differential equations.
- So hopefully, this second example with the convolution
- to solve an inverse transform clarified
- things up a little bit.

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