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0:00
The whole point in learning differential equations is that eventually we want to model real physical systems. I know everything we've done so far has really just been a toolkit of being able to solve them, but the whole reason is that because differential equations can describe a lot of systems, and then we can actually model them that way. And we know that in the real world, everything isn't these nice continuous functions, so over the next couple of videos
0:23
we're going to talk about functions that are a little bit more discontinuous than what you might be used to even in kind of a traditional calculus or traditional Precalculus class. And the first one is the unit step function. Let's write it as u, and then I'll put a little subscript c here of t. And it's defined as when t is-- let me put it this way. It's defined as 0 when t is less than-- whatever subscript I put here-- when t is less than c.
0:55
And it's defined as 1-- that's why we call it the unit step function-- when t is greater than or equal to c. And if I had to graph this, and you could graph it as well but it's not too difficult to graph. Let me draw my x-axis right here. And I'll do a little thicker line. That's my x-axis right there. This is my y-axis right there.
1:22
And when we talk about Laplace transforms, which we'll talk about shortly, we only care about t is greater than 0. Because we saw, in our definition of the Laplace transform, we're always taking the integral from 0 to infinity, so we're only dealing with the positive x-axis. But anyway, by this definition, it would be zero all the way until you get to some value c, so you'd be zero until you get to c. And then at c, you jump, and the point c is included x is equal to c here.
1:51
So it's included, so I'll put a dot there, because it's greater than or equal to c. You're at 1, so this is 1 right here. And then you go forward for all of time. And you're like, Sal, you just said that differential equations, we're learning to model things, why is this type of a function useful? Well, in the real world, sometimes you do have something that essentially jolts something, that moves it from this position to that position. And obviously, nothing can move it immediately like this,
2:15
but you might have some system, it could be an electrical system or mechanical system, where maybe the behavior looks something like this, where maybe it moves it like that or something. And this function is a pretty good analytic approximation for some type of real world behavior like this when something just gets moved. Whenever we solve these differential equations analytically, we're really just trying to get a pure model of something. Eventually, we'll see that it doesn't perfectly describe
2:42
things, but it helps describe it enough for us to get a sense of what's going to happen. Sometimes it will completely describe things, but anyway, we can ignore that for now, so let me get rid of these things right there. So the first question is, well, you know, what if something doesn't jar just like that? What if I want to construct more fancy unit functions or more fancy step functions? Let's say I wanted to construct something that looked like this. Let me say this is my y-axis.
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This is my x-axis. And let's say I wanted to construct something that is at-- and let me do it in a different color. Let's say it's at 2 until I get to pi.
3:33
And then from pi until forever it just stays at zero. So how could I construct this function right here using my unit step function? So what if I had written it as-- so my unit step function's at zero initially, so what if I make it 2 minus a unit step function that starts at pi?
4:06
So if I define my function here, will this work? Well, this unit step function, when we pass pi, is only going to be equal to 1, but we want this thing to be equal to zero. So it has to be 2 minus 2, so I'll have to put at 2 here, and this should work. When we're at any value below pi, when t is less than pi here, this becomes a zero, so our function will just evaluate to 2, which is right there. But as soon as we hit t is equal to pi, that pi is the c
4:36
in this example, as soon as we hit that, the unit step function becomes 1. We multiply that by 2, and we have 2 minus 2, and then we end up here with zero, Now, that might be nice and everything, but let's say you wanted for it to go back up. Let's say that instead of it going like this-- let me kind of erase that by overdrawing the x-axis again-- we want the function to jump up again. We want it to jump up again. And lets say at some value, let's say it's at 2pi, we want
5:14
the function to jump up again. How could we construct this? And we could make it jump to anything, but let's say we want it to jump back to 2. Well, we could just add another unit step function here, something that would have been zero all along, all the way up until this point. But then at 2pi, it jumps, so in this case, our c would be 2pi. That's our unit step function, and we want it to jump to 2. This would just jump to 1 by itself. So let's multiply it by 2. And now we have this function.
5:43
So you could imagine, you can make an arbitrarily complicated function of things jumping up and down to different levels based on different essentially linear combinations of these unit step functions. Now, what if I wanted to do something a little bit fancier? What if I wanted to do something that-- let's say I have some function that looks like this. Let me draw some function. I should draw straighter than that. I should have some standards. So let's say that just my regular f of t--
6:16
let me, this is x. Actually, why am I doing x? This would be the t-axis. We're doing the time domain. It could have been x. And then we'll call this f of t. So let me draw some arbitrary f of t. Let's say my function looks something crazy like that.
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So this is my f of t. What if I'm modeling a physical system that doesn't do this? That actually at some point-- well, actually, let's say it stays at zero. It stays at zero until some value. Let's say it goes to zero until-- I don't know, I'll call that c again. And then at c, f of t kind of starts up. So right at c, f of t should start up, so it just kind of goes like this. So essentially what we have here is a combination of zero all the way, and then we have a shifted f of t.
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So at c, we have a shifted f of t, so it shifts that way. So how can we construct this yellow function, where it's essentially a shifted version of this green function, but it's zero below c? This green function might have continued. It might have gone something like this. It might have, continued and done something crazy, but what we did is we shifted it from here to there, and then we zeroed out everything before c. So how could we do that? Well, just shifting this function, you've learned in
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your Algebra II or your precalculus classes, to shift a function by c to the right, you just to replace your t with a t minus c. So this function right here is f of t minus c. And to make sure I get it right, what I always do is I imagine, OK, what's going to happen when t is equal to c? When t is equal to c, you're going to have a c minus a c, and you're going to have f of 0. So f of 0, it should be the same. So when t is equal to c, this value, the value of the
8:19
function should be equivalent to the value of the original green function at zero, so it's equivalent to that value, which makes sense. If we go up one more above c, so let's say this is one more above c, so we get to this point, if t is c plus 1, then when you put c plus 1 minus c, you just have f of 1, and f of 1 is really just this point right here. And so it'll be that f of 1, so it makes sense. So as we move one forward here, we're essentially at the
8:48
same function value as we were there, so the shift works. But I said that we have to also-- if I just shifted this function, you would have all this other stuff, because you would have had all this other stuff when the function was back here still going on. The function-- I'll draw it lightly-- would still continue. But I said I wanted to zero out this function before we reach c. So how can I zero out that function? Well, I think it's pretty obvious to you.
9:14
I started this video talking about the unit step function. So what if I multiply the unit step function times this thing? What's going to happen? So what if I-- my new function, I call it the unit step function up until c of t times f of t minus c? So what's going to happen? Until we get to c, the unit step function is zero when it's less than c. So you're going to have zero times I don't care what this is Zero times anything is zero, so this function is going to be zero.
9:50
Once you hit c, the unit step function becomes 1. So once you pass c, this thing becomes a 1, and you're just left with 1 times your function. So then your function can behave as it would like to behave, and you actually shifted it. This t minus c is what actually shifted this green function over to the right. And this is actually going to be a very useful constructed function. And in a second, wer'e going to figure out the Laplace transform of this, and you're going to appreciate, I think,
10:19
why this is a useful function to look at. But now you understand at least what it is and why it essentially shifts a function and zeroes out everything before that point. Well, I told you that this is a useful function, so we should add its Laplace transform to our library of Laplace transforms. So let's do that. Let's take a the Laplace transform of this, of the unit step function up to c. I'm doing it in fairly general terms. In the next video,
10:52
we'll do a bunch of examples where we can apply this, but we should at least prove to ourselves what the Laplace transform of this thing is. Well, the Laplace transform of anything, or our definition of it so far, is the integral from 0 to infinity of e to the minus st times our function. So our function in this case is the unit step function, u sub c of t times f of t minus c dt. And this seems very general. It seems very hard to evaluate this integral at first, but
11:29
maybe we can make some form of a substitution to get it into a term that we can appreciate. So let's make a substitution here. Let me pick a nice variable to work with. I don't know, we're not using an x anywhere. We might as well use an x. That's the most fun variable to work with. Sometimes, you'll see in a lot of math classes, they introduce these crazy Latin alphabets, and that by itself makes it hard to understand. So I like to stay away from those crazy Latin alphabets,
11:59
so we'll just use a regular x. Let's make a substitute. Let's say that x is equal to t minus c. Or you could, if we added t to both sides, we could say that t is equal to x plus c. Let's see what happens to our subsitution. And also, if we took the derivative of both sides of this, or I guess the differential, you would get dx is equal to dt. Or I mean, if you took dx with respect to dt, you would get that to equal to 1. c is just a constant. Then if you multiply both sides by dt, you get dx is
12:32
equal to dt, and that's a nice substitution. So what is our integral going to become with this substitution? So our integral this was t equals 0 to t is equal to infinity. When t is equal to 0, what is x going to be equal to? Well, x is going to be equal to minus c. Actually, before I go there, let me actually take a step back, because we could progress. We could go in this direction. But we could actually simplify it more before we do that.
13:09
Let's go back to out original integral before we even made our substitution. If we're taking the integral from 0 to infinity of this thing, we already said what does this integral look like or what does this function look like? It's zero. We have this unit step function sitting right here. We have the unit step function sitting right there. So this whole expression is going to be zero until we get to c. This whole thing, by definition, this unit step function is zero until we get to c.
13:38
So this everything's going to be zeroed out until we get to c. So we could essentially say, you know, we don't have to take the integral from t equals 0 to t equals infinity. We could take the integral-- let me write it here. I'll just use that old integral sign. We could just take the integral from t is equal to c to t is equal to infinity of e to the minus st, the unit step function, uc of t times f of t minus c dt. In fact, at this point, this unit step function, it has no use anymore.
14:18
Because before t is equal to c, it's 0, and now that we're only worried about values above c, it's equal to 1, so it equals 1 in this context. I want to make that very clear to you. What did I do just here? I changed our bottom boundary from 0 to c. And I think you might realize why I did it when I was working with the substitution, because this will simplify things if we do this ahead of time. So if we have this unit step function, this thing is going
14:45
to zero out this entire integral before we get to c. Remember, this definite integral is really just the area under this curve of this whole function, of the unit step function times all of this stuff. All of this stuff, when we multiply it, is going to be zero until we get to some value c. And then above c, it's going to be e to the minus st times f of t minus c. So it's going to start doing all this crazy stuff. So if we want to essentially find the area under this
15:12
curve, we can ignore all the stuff that happens before c. So instead of going from t equals 0 to infinity, we can go from t is equal to c to infinity because there was no area before t was equal to c. So that's all I did here. And then the other thing I said is that the unit step function, it's going to be 1 over this entire range of potential t-values, so we can just kind of ignore it. It's just going to be 1 this entire time, so our integral
15:41
simplifies to the definite integral from t is equal to c to t is equal to infinity of e to the minus st times f of t minus is c dt. And this will simplify it a good bit. I was going down the other road when I did the substitution first, which would have worked, but I think the argument as to why I could have changed the boundaries would've been a harder argument to make. So now that we had this, let's go back and make that substitution that x is equal to t minus c.
16:13
So our integral becomes-- I'll do it in green-- when t is equal to c, what is x? Then x is 0, right? c minus c is 0. When t is equal to infinity, what is x? Well x is, you know, infinity minus any constant is still going to be infinity, or if the limit is t approaches infinity, x is still going to be infinity here. And it's the integral of e to the minus s, but now instead of a t, we have the substitution. If we said x is equal to t minus c, then we can just add
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c to both sides and get t is equal to x plus c. So you get x plus c there, and then times the function f of t minus c, but we said t minus c is the same thing as x. And dt is the same thing is dx. I showed you that right there, so we can write this as dx. Now this is starting to look a little bit interesting. So what is this equal to? This is equal to the integral from 0 to infinity-- let me expand this out-- of e to the minus sx minus sc times f of x dx. Now, what is the equal to?
17:30
Well, we could factor out an e to the minus sc and bring it outside of the integral, because this has nothing to do with what we're taking the integral with respect to. So let's do that. Let me take this guy out, and maybe just to not confuse you, let me rewrite the whole thing. 0 to infinity. I could rewrite this e term as e-- actually, let me write it this way. I'll do what was already in green as e to the minus sx times e to the minus sc. Common base.
18:01
So if I were to multiply these two, I could just add the exponents, which you would get that up there, times f of x, d of x. This is a constant term with respect to x, so we can just factor it out. We can just factor this thing out right there, so then you get e to the minus sc times the integral from 0 to infinity of e to the minus sx times f of x dx. Now, what were we doing here the whole time? We were taking the Laplace transform of the unit step
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function that goes up to c, and then it's 0 up to c, and it's 1 after that, of t times some shifted function f of t minus c. And now we got that as being equal to this thing, and we made a substitution. We simplified it a little bit. e to the minus sc times the integral from 0 to infinity of e to the minus sx f of x dx. Something about the tablet doesn't work properly right around this period. But this should look interesting to you. What is this? This is the Laplace transform of f of x.
19:35
Let me write that down. What's the Laplace transform of-- well, I could write it as f of t or f of x. The Laplace transform of f of t is equal to the integral from 0 to infinity of e to the minus st times f of t dt. This and this are the exact same thing. We're just using a t here. We're using an x here. No difference. They're just letters. This is f of t. e to the minus st times f of t dt. I could have also rewritten it as the Laplace transform of f of t.
20:14
I could write this as the integral from 0 to infinity of e to the minus sy times f of y dy. I could do it by anything because this is a definite integral. The y's are going to disappear, and we've seen that. All you're left with is a function of s. This ends up being some capital, well, you know, we could write some capital function of s. So this is interesting. This is the Laplace transform of f of t times some scaling factor, and that's what we set out to show.
20:48
So we can now show that the Laplace transform of the unit step function times some function t minus c is equal to this function right here, e to the minus sc, where this c is the same as this c right here, times the Laplace transform of f of t. Times the Laplace transform-- I don't know what's going on with the tablet right there-- of f of t. Let me write that. This is equal to-- because it's looking funny there-- e to the minus sc times the Laplace transform of f of t. So this is our result.
21:37
Now, what does this mean? Oh, look it back-filled it somehow. What does this mean? What can we do with this? Well, let's say we wanted to figure out the Laplace transform of the unit step function that starts up at pi of t. And let's say we're taking something that we know well: sine of t minus pi. So we shifted it, right? This thing is really malfunctioning at this point right here. Let me pause it. I just paused. Sorry if that was a little disconcerting.
22:16
I just paused the video because it was having trouble recording at some point on my little board. So let me rewrite the result that we proved just now. We showed that the Laplace transform of the unit step function t, and it goes to 1 at some value c times some function that's shifted by c to the right. It's equal to e to the minus cs times the Laplace transform of just the unshifted function. That was our result. That was the big takeaway from this video.
22:48
And if this seems like some Byzantine, hard-to-understand result, we can apply it. So let's say the Laplace transform, this is what I was doing right before the actual pen tablet started malfunctioning. If we want to take the Laplace transform of the unit step function that goes to 1 at pi, t times the sine function shifted by pi to the right, we know that this is going to be equal to e to the minus cs. c is pi in this case, so minus pi s times the Laplace transform of the unshifted function.
23:27
So in this case, it's the Laplace transform of sine of t. And we know what the Laplace transform of sine of t is. It's just 1 over s squared plus 1. So the Laplace transform of this thing here, which before this video seemed like something crazy, we now know is this times this. So it's e to the minus pi s times this, or we could just write it as e to the minus pi s over s squared plus 1. We'll do a couple more examples of this in the next video, where we go back and forth between the Laplace
24:02
world and the t and between the s domain and the time domain. And I'll show you how this is a very useful result to take a lot of Laplace transforms and to invert a lot of Laplace transforms.
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