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