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OK, so hopefully this is differential equations.
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Was part one. OK.
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My name's Philip Manny. And these lectures obviously are Monday at this time and then Tuesday at 12 o'clock.
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So the before problem sheets, they're ready on the web.
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And when we've covered stuff within the problem sheets, I'll tell you which questions you can do.
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In fact, you should already be able to do the first two questions on the problem sheet number one,
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because they are revision questions on things to do with convergence and series.
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And those are things we will be looking at later on in today's lecture and also tomorrow's lecture.
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And I'll just assume that, you know, all that stuff now you've probably already seen on the web the synopsis of this course,
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and I'll go into that a little bit of detail later on. There are also problem sheet.
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There are also lecture notes, and I've uploaded the lecture notes of Professor John Dyson.
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And what I will do is cover the same material that's covered in those notes on the web,
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but I won't be using those notes, so I'll be giving a sort of complimentary view of the subject.
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So don't be expecting to look at her lecture notes and see word for word.
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What I've written in these lectures will be covering the same material, but not necessarily in exactly the same way.
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But roughly speaking, in the same way, when I'm giving the lecture,
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I'll also be expecting that you will not only be writing down what I write on the board, but also be making your own notes about various things.
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And what I will try and do in the lecture is to focus on what are the really important bits and why we do what we do.
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OK, so this lecture course will look at audio differential equations and the first part and then the second part.
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Look at partial differential equations and what the lecture is trying.
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What this course is trying to do is two things.
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First of all, for those of you who are interested more in sort of applied analysis and the
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pure side of things to give you ideas of how you go about proving existence,
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uniqueness, et cetera, et cetera. For those of you who are more of an applied.
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Viewers who want to solve differential equations see what the solutions look like.
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It will give you methods of how you actually get the solution.
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So those of you who want to go on and do the more applied courses that will relate to the sort of things that you're doing,
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we'll be doing later on those of you who want to do a bit more of the pure.
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This will give you an insight into that. And of course, the two things match up because in the applied,
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when we come up with quick and dirty ways of getting a solution, we know that from the pure mathematics point of view,
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a solution exists, the problem, the problems while posed, etc.,
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etc. so that we know that what we are doing is based on fundamental rigorous mathematics.
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So we try and cover those two things. OK. So if you can't hear me?
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Or if you can't read the writing, then do let me know. OK, so we'll start off.
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As I mentioned, the first part will be ordinary differential equations.
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So part one, ordinary differential equations. And of course, you've seen ordinary differential equations before.
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So some of this will be revision for you, but some of it will be new.
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So of course, you call these all these. And the first part will be these in general and then Picard's there.
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So Picard's theorem is a theorem that will tell us under what conditions a solution to know exists and is unique.
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But before we do that, we have to ask the question What is Hanoudi?
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And, you know, all know this. Let's write it down. So no d e is an equation for y of X, a function of X of the form.
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So some function g of X y y dashed I double dashed up to say Y and equals zero where the dashes mean derivative.
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So y to the R means d r y by d x the R and why is the independent is that dependent variable?
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She'd say. And X is the independent variable, so that means you can choose X whatever you want it to be and then Y is Y of X, it's given.
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OK, so here of course, we're making this only makes sense if these derivatives exist.
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So we're making the assumption here that you can actually differentiate this function and you can differentiate up to entice.
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OK. So if we can rewrite this?
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As. The NY, the to the N equals some f of X y y dashte y to the N minus one.
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So in other words, if we can get out the yen's derivative and write it in terms of all the other functions and derivative, then the order.
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Most of the body is and that's the order of the highest derivative.
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And this should be all stuff that you've heard before. OK.
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And so a question or questions that we would have with our and Sauder.
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Would he be? Does it have a solution? And is the solution unique?
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And probably up until now, the questions,
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the examples you've looked at have been well behaved or the ease in which you just find a solution and then you say, Well, that's a solution.
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So you're sort of assuming that a solution exists and you're assuming that it's unique.
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So what we're going to do here is we're going to ask, well, how do you know that?
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Right. So obviously in mathematics, always a good idea is to start with the simplest problem.
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It's the simplest problem be a first order udi.
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So we start with simply why of X equals half of x y dashed.
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And as time goes on, they will get sloppy with the notation and leave out these X's.
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But we should keep in mind these are always functions of X, and you should know that you need an initial condition.
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Typically, that's one point one, and this is called an initial value problem.
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Because you're given an initial value switch, call that why the P?
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No, now, as I mentioned, probably the types of orders you've looked at until now, you just find the solution and you've got on with it, not start.
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But here know some warning examples where, oh, these are not the kind, gentle creatures you thought they were.
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So warning example.
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So here's no d d y by D X equals three y to the two thirds, the initial condition y of zero equals zero sats initial value problem.
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So let's solve this, so solve it by separating the variables.
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So I just sketched the proof of this.
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The I leave you to do it properly.
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OK, to separate the variables integrate and you end up getting lead you to convince yourselves that you get this where a is a constant.
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And that'll give us y equals x plus a cubed.
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And then the initial condition y of zero equals zero implies a zero, and therefore Y equals X crimped as a solution.
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You should always check that you haven't made a mistake anywhere. If Y is x cubed, then d y by d x is three x squared.
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And this three x squared the same as three Y to the two thirds.
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Well, yes, y to the two thirds is x squared.
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Multiply by three c y squared. Does it satisfies the initial condition?
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Yes, y, if not, is not. So there is a solution to the problem.
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And notice I've said there is a solution and not said there is the solution because if you are really lazy,
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you would just look at that question, that thing up there and you would say, well, if y was zero.
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Divide by X would be zero, so zero equals zero and Y of zero zero.
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So you could just say y equals zero is a solution.
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So there's more than one solution. In fact, there are an infinite number of solutions, because if we took.
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A national equal to zero or equal to be.
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And if we define Y to be X minus, say, cubed for X less than A to be zero line in between and to be x minus b cubed.
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For X greater than B, that would also be a solution. So let's just check.
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Well, again, if you differentiate this, you get the Y by the X is three x minus, say to the two thirds.
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So that's fine satisfies the equation. Y equals zero satisfies the equation.
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This satisfies the ODP and Y of zero is zero such satisfies.
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OK, now is does the derivative exist for all values of X?
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Well, yes, because this is sufficiently smooth that if you look at the derivative of this,
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if you look at this at a and look at this at a the derivative is zero.
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So this is a continuous function with continuous first derivative.
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So you can talk about D Y by D X, but now A and B within these conditions can be anything.
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So you've got an infinite number of solutions. OK, so infinite number of solutions.
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Right, so that looked like a very nice body turned out to be pretty nasty.
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No, let's look at another problem.
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D y by d x equals y squared y of zero equals one exercise show that this gives you y equals one over one minus x squared do separating the variables.
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And drive it. OK. And then just check it. Do do I buy the X?
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And then make sure it's y squared y of zero equals one.
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And then as X goes to one, y will go to infinity.
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Therefore, solution only exists for less than one.
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So if you were to solve, try to solve this problem on the domain that included X equals one, it doesn't have any solution.
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So here's an example where there's no solution. And here's an example where there are an infinite number of solutions.
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So going back to our question, does it have a solution?
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Answer no. This one, is it unique?
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Answer no for this one. And you can see, crucially, it depends on these functions here,
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the F and the initial value have been chosen to make sure that these don't behave properly.
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So that's where we need to figure out then what was so special about these right hand sides.
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And the initial value. So that things didn't work the way you would like them to work.
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So basically, the sort of thing that we want to do is we seek a solution.
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So first of all, what that tells you is there's a problem here that needs to be solved just because you've got no d.
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An initial value problem doesn't mean you can write down a solution that may not exist and there may be more than one solution.
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OK, so one point one way up their initial value problem, so we seek a solution in the region, in some region.
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Right. Because notice here, if we were to say let me take the region to be less than one, then this would have a solution.
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So the important thing is, not only what is this f and what is this is also in which region do you want the solution to be?
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So we seek a solution to one point one in the region are standing for region x y such that so a we've got Y of A equals B.
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And so basically what we're saying is that if we consider a and B in the plane.
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And we're told what? That's why at a Isby, then, is there a region around there where we have a solution?
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So let's just draw this. And so here's a here's B.
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And then this region.
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The roughly speaking, something like this or this is to OK, and this is to edge right, Assad region should say H and K are obviously bigger than zero.
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So this sort of thing we're saying is if we know. The solution at this point here.
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And then D Y by D X equals f something, so we knew the derivative.
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Can we find a solution? For these values of X.
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And it lies in this region. OK, so we make some assumptions, so assumption one we will make.
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We will assume that F of X Y is continuous and are OK.
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That's. So let's integrate.
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1.1. So integrated from.
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A ta x. OK, because we've got we've got it at a oh, you won't find it at some point x, so using T is a dummy variable.
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I mean, if we integrate integrating from a Ta X now, use a dummy variable d y by d t d t.
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That's going to give us y of X minus y of A.
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And then on this side, we're going to get from eight x f.
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So T is just a dummy variable d take nasco and that gives us that.
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And therefore we can say y of X equals y avai, which is B plus the integral from a two x f of T y of t d t.
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One point four. So now we've got an equation for why.
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Which is fantastic, except for the fact that it involves why.
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On this side. Now, this might look to you more complicated than what we started off with.
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But this is the key thing.
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And what we're going to show is we're going to ask the question,
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under what conditions can we find a solution to this and those conditions we will see make conditions on Earth and the conditions of where X must lie.
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So we're going to try and do this by 1.3 Picard's method.
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Off successive approximation. OK, so before we do that, we have a bit of a break.
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The equal to sign here, which we all use.
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Do you know when that was invented? Yep, when it was invented.
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Hmm. It's on the 16th and 17th century. That's right.
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If you go to lecture in five l five and look at the far wall, there's a framed copy of the manuscript,
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which the first time apparently someone used the equal to sign.
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And it was somebody from all souls.
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And it was in the early 1800s, and up until then would you would write is is equal to so you would say why of X i s e q u a l t o.
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Right. Why wouldn't you? And then this person from all souls said.
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It's really tedious right down is equal to so I'm going to write two parallel lines because nothing can be more equal than two parallel lines.
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So just think about this when you're walking around Oxford, you're walking past colleges,
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or you might even be living in colleges that are older than the equal to sign.
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Isn't that amazing? Not personally. I wish he'd use a smiley face to say why of x smiley face?
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B plus integral for me to X. And then I got to thinking what would not equal to be?
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Maybe grumpy face. But then what about greater than or equal to?
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So should equal to be sort of neutral face and then greater than the smiley face less than the grumpy face,
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then what would not equal to be very confused, very confused face?
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OK, no important thing. I am not.
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On the examining committee, so I will not be examining this paper, so do not put smiley face in the exam,
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right, because the marker does not know our code, so don't don't use smiley face.
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OK, so use equal.
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All right. So now we had that little break and bit of a laugh.
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We can get back to Picard's method of successful successive approximation.
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So what he said is let's start with the gas y of zero equals B.
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Well, good gas and then we'll iterate up, so what we'll do is we'll put B into here and into here, and that'll get us the next gas y one,
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and then we'll put that into here and that'll get us y to and we'll keep on an ongoing
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and hopefully it will converge or under what conditions on earth will it converge?
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So basically, why and plus one of X will be B plus integral from A to ex f of T y if t d t, and that is one point five equation one point five.
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And now you see why I mentioned at the beginning of the lectures that we were going to be talking about convergence.
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And that you need to know stuff about convergence from last year.
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So does this converge? And you should know that if you want to figure out how something's converging, you could y one y to y three, et cetera.
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You look at the difference between them and you ask the question as you move along.
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The sequence does not get smaller. That's one of the important points it needs to happen.
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So we need to find the differences between. So I made a mistake that should be why and.
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Redefined, what's the difference between why a 10 plus one and wider, so redefine.
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E zero, two, b b and then e n plus one of X to be Y and plus one of X minus Y and X, so that is one point six.
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And then we note that y end of X is just the sum of these two k equals zero to n e k of x one point seven.
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And so now we look at these eyes. And we asked the question about, are these e's getting smaller as end gets larger?
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And you should know certain theorems from last term last year that if we can bind you ends by something,
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then using reassurance and test various things that we can say that the series converges.
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And if you don't remember that, then that's first thing you should be doing.
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Today is look back on your notes and by trying to answer questions one and two on problem sheet one.
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It doesn't show you what you need to know. So look those up.
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OK. Right? So what we will do is we will write an equation for E!
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And +1. Well, Ian, +1 is why then plus one minus y event, so the bees will cancel.
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And so what we will have is integral from eight x f of T y end of T minus f of T y n minus one of T.
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He and we need to look at the modulus, says Modulus, and +1 one X is then less than or equal to opting to go for Mate X,
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the modulus f of T y end of T minus f of T y n minus one of T d t.
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And that's one point eight. I know crucially, we need wouldn't it be really nice, see if if we want to really do something with this,
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it would be nice if we could get an e out of here at the end.
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So be nice if we could relate f of t y n minus f of T and Y and minus one to something to do with Y and and Y and minus one,
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because that would give us an E.
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In particular, be really nice, if we could say this thing here was bounded by something times y n minus y n minus one.
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And that's exactly what the next condition is. It's called Lipschitz condition.
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So a function. F of x y.
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Continuous on the rectangle are the rectangle we've been talking about.
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Satisfies a Lipschitz condition.
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With constant help. We'll see what that means in a minute with constant L.
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If there exists a real. Positive, Al, such that.
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F off, oops. X minus you A5X comma u minus f of x comma v is less than or equal to l u minus v.
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For x u x fee, an element of the rectangle.
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So take any two elements of the rectangle and any two points inside that rectangle.
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And if this is satisfied, then so this is true for all points on the rectangle.
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Then it's cognitions. OK. So just to relate that to something.
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And you've already done so. Suppose if is differential.
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With respect to why. And suppose it's about the partial derivative is bounded above by K.
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Suppose that's true for All Blacks and why? And in the region.
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Then you should know from last. Last year, the mean value theorem.
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Tells you that for two points.
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X U x v in R. That's going to be equal to the partial derivative.
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At some w u minus v, so the mean value theorem is written without those bodily signs, but we can just put those bodily signs in.
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Where W is some intermediate value.
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OK, so it's like intermediate value theorem. 1.10.
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And then if R F is bounded or the partial derivatives bounded by K,
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then that would be less than or equal to K, and that's so k would be R Lipschitz constant l.
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So what that's telling us is that a vast is differential in Y and the derivative is bonded above.
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Then it's Lipschitz F as options. OK.
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But notice is another example, f of Y equals mould y.
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A live use and exercise to show that this is Lipschitz continuous.
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OK. It's continuous. And it satisfies Lipschitz condition.
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I was just one for that, OK? F of x you minus f of x v.
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Modules, a sign of that would be just the modulus of you minus V, so L is one.
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But this is not differential. At Y equals zero.
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No. Because the limit as you go from positive y is one go from negative y is minus one.
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OK, so this isn't a necessary condition. OK, so that is just as as a little digression.
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Of how you know, of thinking about how you might go about finding the L are are showing that something is Lipschitz continuous.
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OK. Right. So now we're going to use the Lipschitz condition in order to prove the serum of does the initial value problem have a solution?
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And is it unique and that's Picard's there. OK, so theorem point one Picard's, they're sometimes called Picard's existence theorem.
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So the initial value problem one point one.
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Let's just write it down again, so I'll I'll go between using Dasht D by D.
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X. So the same thing. This why of A. Equals B has a solution.
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In the rectangle, the rectangle we've been talking about all along in the rectangle.
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Ah, which is defined as.
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Well, again, this we're just writing down stuff we've already written down x minus, say, less than or equal to h y minus, be equal to K provided.
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P1 pay for Picard. A f is continuous and are.
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With boned M. That means mod f of x y less than or equal to N and for all X, Y and R.
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And be. And.
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Age is less than or equal to K, it was hard to figure out why that is the case.
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P2. F satisfies a Lipschitz condition and are.
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You can sort of start getting a vague idea of why this might work, because if this holds.
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This year holds. Then when you go up to even one, then that will become the national equal, the integral from a tax of L times in.
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So now you can see you can start then getting an idea of a recurrence relation for in.
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And then you can see if you. Can get what Ian is.
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And then if you get what Ian is, you get why in?
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And then you figure out, well, does this converge? OK.
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And that we need to use things like the reassurance and test stuff like that, OK?
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But let's do that properly before we do that. One more thing about this if this all holds.
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As a bonus, the solution is unique.
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So remember those problems that we had. At the beginning of the lecture, the two problems we showed were either a solution didn't exist.
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Our solution was not unique.
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If you go back and look at those problems now, what you will see is that they weren't they didn't satisfy Lipschitz condition.
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OK. So be an exercise for you to do is to show that there was problems that didn't work.
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Don't contradict us. OK. Right.
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So to do the proof of this.
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Well, first of all, we need to show if we're going to show that F satisfies Lipschitz condition and are and if we're going to end up using this.
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Then we sort of need to make sure that the whys are in, ah,
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because if y and if one of these y ends doesn't lie in R, then we don't have any idea about F, we just know.
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We just have conditions on Earth if X is X, Y or X and Y light in.
262
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Now we can make X Y and, ah, we can just make more than x minus a b less than H, but then Y is given by that integral.
263
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How do we know that lies in R? If it doesn't lie in R, we're stuffed.
264
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So the first thing to do which hopefully y of n y n will lie in R.
265
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So each y end of X is continuous, and it's graph, which basically means it's values.
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That's what this graph means, lies inside.
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Oh, and to know what does not mean precisely.
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It means that if Model X minus say, is less than very equal to h, then.
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X y n of X is an element of.
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Another way of putting that that means Y and minus B modulus of that is less than equal to K.
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Right. So is it continuous? Well, that's obvious.
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Why is that obvious? Well, how do we define the terrorists?
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That's continuous. If we assume why and is continuous, then why, and plus one is continuous, because if it's continuous.
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Sufi assumed that high end is continuous and lies and are.
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Then that's continuous integrate as continuous.
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So if y n is continuous and lies and are. Then why and plus one is continuous.
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But why zero was continuous. So why n plus one is continuous, so that's easy.
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And now, does it lie in our well, we need to show this.
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We need to show that lesson equal to each. Well, that's less than or equal to the integral from a to ex by definition.
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OK, take that to that side. And we've got that.
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And now assume. That Y m is an R.
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So assume that holds, then that means s is on a function that lies in R and therefore is bounded by M.
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So this is bounded by AM, so I've got integral from a two x m, which is just M, and I can put more.
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That's minus eight because they might be less than X. But X minus a.
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If we're an R X minus, say, it's lesser equal to H.
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Less than equal to MH. Now, this is really annoying.
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Because we wanted y n plus one minus B to be less than equal to K.
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And we've end up getting why, and plus one minus B is less than or equal to M times which.
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So the only times H was less than equal to K. We'd be all right.
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Oh, it is less or equal to K.
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Fantastic. So what we've shown, then,
292
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is that if y end lies in our meeting has got y n minus B Leicester equals K then Y and plus one has got Y and plus one minus b lesser equals to K.
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So now let's look at why zero for the induction. Why zero minus b?
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Well, why zero is B is zero.
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Which is less than or equal to K, so by induction. Result true for all that.
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OK, so what we've done then, is we've shown that if these conditions are satisfied, then the why the rates of why lie inside?
297
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The Y minus B is equal to K.
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And what that means is now we can go back to the the e n.
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And now we can look at these things and we can say these lie within are.
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And therefore we can use at this. And that will enable us to simplify this and get the convergence result, right?
301
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So what you need to do for tomorrow is to make sure that you know, the Wish Tarsem test because that's what you'll be using tomorrow.
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So go back and revise that. So what we will do tomorrow is will ensure that we get the convergence and then that will prove this year.
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OK.