All right. Hello. My name is Peter, and I'm going to be lecturing differential equations this term. So the lecture notes and problem sheets and everything for this course in the usual place. And they're also quite a few good textbooks for this material. I mean, some of those I mentioned on the website as well. So I'd recommend you look out for those. Now, just one quick. So general point. And quite often it might be that
I write down. You see, my handwriting isn't that great. Maybe I'll write down something that you can't read or that doesn't seem to make sense. Or I might say something that just, you know, doesn't make sense. So that happens. It be great if you just yell out and asked me a question rather than starting a chat amongst all your neighbours.
So I don't you. I find it very distracting of other people. So chatting all around you. So. So, yeah. If you have any questions ready queries, any points, then just a shout out and ask me and rather than starting a conversation going. Right. All right. Good. So most this course is about second order linear. We value problems. And I'm going to use BTP as a shorthand for that. And I'm just do a few examples
just to sort of illustrate what some of the issues are. And here's the first one, by the way. So here in sort of henceforward and he's a prime as shorthand for differentiation. Hope that's okay. So this is a very simple bound value problem. Of second idea. And I've got two the conditions. And we can easily solve this using elementary methods for the first year. So here we go. So how do we solve an equation like this? I guess we have to look for a particular solution.
And then the complementary function. This case of security is just one, right? Easy. That said, should be standard staff from last year. So the general solution is a sign of a particularly in school and a complementary function in this complementary function. We have two arbitrary constants that straight for men to work out, see one and see two. Obviously plug in band conditions. Okay. Oh, okay. Also, I'm going to use BSE for Švanda conditions and hope that's okay.
Well, here's one I worked out earlier. Okay. Easy, right? I'm happy so far. Yeah, well, it is gonna get harder. I promise. Here's an example to you. So I'm going to place up one by 10x ten, I guess isn't defined. Valla XP too big. That's just how. So he's going to do this on Xscape and zero to Piver four. So in principle is exact, the same kind of Boundy Valley problem. I've got second equation, too bad conditions,
but good luck spotting the particular integral to this. All right. So in principle, the same method should work for how are we supposed to find a particular interval for this? Is somebody trying to spot it now? Everyone does it while they enter the lecture, I'll be impressed. I sample three. It's the same Odey that we started with. And I've just tweaks the bad conditions slightly. So instead of why Prime applies to and I've got white pies, two principles, same
approach will work, right? No, this this problem has no solutions. And you can easily cheque that yourself as we try to follow the particular range go plus complementary function approach, you just end up with a contradiction. It's impossible to satisfy both those editions conditions. So finally, I'm going to do the same. I'm just gonna change this two to a one, so exactly the same.
Okay. So everything the same apart, that final band coalition and this problem has an infinite number of solutions. So I've got non-unique solution and actually I'll just write this down. So just one plus a sign X works for any A. OK, so this is just supposed to illustrate what the issues are with Boundy Valley problems. And so I'm just gonna write this down. What are the main questions that we're going to be trying to answer in this bit of the course?
And I say a couple of issues that we encountered here. The first is that, um, unless the right hand side is something very simple, um, how are we going to support a particular integral? And isn't there a way to sort of just construct the particular integral? And the answer is, yeah, there is. Car P.I. for particular integral second OP suddenly one. So like without star, without somehow trying to sort of guess it or spot it, how can we actually construct it in a systematic way?
And then the second question, which perhaps is more important is, um, when does a family value problem like this have a solution and when is this mission unique? Okay. Everything clear so far, okay. Right. And so first things first. I'm just gonna introduce some notations of jargon that I'm gonna be using throughout the course. OK. And so I'm mostly gonna be considering second order and differential equations, maybe some exceptions, but mostly. Oh, yes. Again, Odia is only the the equation.
Now. So this is a bit abstract at the moment, but think of the France based be Kelly Ale. It looks like a two. It's supposed to be a curly L. All right. And let me just define what else is. OK, so this is of the most general second order linear differential operator that we could have set at some linear combination of the second first derivatives of Y and Y itself. And these P two P, one P zero kind of any Allwood functions we like. I guess I'm going to assume they're continuous, at least
in practise, you know, in all the examples will do. They will be almost always they'll be analytic in fact. And I'm gonna to assume that why is differentiable enough times for this to make sense. Okay. And again, in practise, we're nearly always gonna be looking for solutions that are analytic. And then the final thing, which might be a bit mysterious to start with, is gonna seem that P two is not zero. So that's. And by that, I mean it's not able to zero anywhere.
So not just that it's not the zero function, but it's everywhere. Non-zero, I think. And this may be even thought about this before, but maybe you can see if P t was zero anyway. Then this second Autodefensas equation better kind of turn into a first order, one that actually called correspond to a singular point of this idea. Again, I'm gonna come back that late in the course. Just cheque. I've got to. Okay, now there's, um, two sort of cases of this that we're gonna
it's gonna be hard to separate those out. So first day of F is zero. Then this equation is called homogeneous. Right. That is not zero. It's called in homogeneous or non homogeneous. So, like, am I gonna label those separately? Because you're gonna be referring to those two different cases a lot. Okay. So Iraq is the zero function that's called in as called homogeneous. I better get that right. So so that is going to be referred to throughout as H for the homogeneous problem.
Mm hmm. Right. And the case where F is not zero is called in homogeneous. And for some reason, this conventionally enabled n fence or non homogeneous. OK. Exactly. Yeah, and I'm gonna be people firing quite a lot to these two different cases, h and then right now a differential equations, one last term, you would mostly be looking at what are called initial value problems. All right. I've been surprised to find there's an acronym for that as well.
And IVP is I'm so what I mean by that is if I've got a second order idea. What I would normally do is give you what Y and Y primed are at some point. So typically do so at some point. X equals A I will tell you what Y and why prime da. I think you've got to work out what Y is everywhere else right now for that problem. You can use Picards theorem from different equations, one that guarantees that a unique solution exists, at least locally. OK. Oh at least likely provided P
two is not zero and maybe you can see why. Now if you try to apply Picard same appeal to a zero, you'd have a problem. Okay. Ray. Now pick out same guarantees, local existence and uniqueness. And in fact, but linear equations like this and you might remember this from, you know, a time ago. Actually, it satisfies a global lipshitz condition. And you can show this region exist everywhere, can provide a Peter is not zero.
And so, in fact, provide pictures, not zero. The existence, uniqueness of global. But for boundy value problems. That's not true. And I've already shown you two examples where that's not true, where you instead of specifying why and why print at one point we had bad indications at two different points. And you can either get non-existence or non uniqueness. Mike. OK, so I'm just gonna quote some basic properties, again, from Elementary from last year, basically, or even from a level, I think.
And the first is if I've got a second order in homogeneous Odey, well, you could do this trick that I did in the very first example. You could break down the general solution into a particular integral plus complementary function. Okay. It's going to break it up like there, so I come write the general solution as a particular integral like Y P.I. plus the complimentary function which I call y S.F.
And what do I mean by that? So why P.I. is any function at all that satisfies the inherent genius problem? OK. So a particular age group is any solution that you can find by any means necessary of the inherited genius problem and why S.F. is the general solution a homogeneous problem? You've got rid of the Yaffe on the right hand side. Hopefully that Will has some sort of familiarity to it.
And so the second thing I thought about, that's why S.F. thing complementary function well for a second order, linear ODP and the space of solutions of the homogeneous problem is a vector space of dimension, too. Okay. So what I mean by that is that the complementary function I can write has. I can write just in terms of two basic functions, basically. Exactly. I guess what I did for that very simple first example. And so I
see why see to end arbitrary constants. And then why when and why to. I guess. Ready set of basis functions for the vector space. A basic ready to solutions that are linearly independent. Okay. Okay, so more or less familiar. Point any questions so far? Good point. I'm so you just a bit more about Libya independence. Again, this is gonna be kind of important going forward.
And Occy, my rights is getting smaller after Naghmeh. If I keep doing that and and let be is very closely related that I think called the ront skin. And I guess a bit about that as well. Okay, so we remember what it means for two functions, or in general two elements of a vector space to be linear, linearly independent or linearly dependent. I guess maybe it's easy to write down the situation for them to be linearly dependent.
So these two functions would be literally dependent. If you could take some linear combination of them, that adds up to zero sum, non-trivial linear combination of them. OK, so if access some constancy, want to see to not both zero such that C1 while unclassy T.Y. two adds up to the zero function. And I guess Elenita independent if and no such see one to see two exist. Okay. That's just the definition of what we mean by linearly dependent. Now, let's suppose this was true.
So I suppose they are daylily dependent. And then remember, I'm assuming why one and why two are twice defensible, actually. So I can certainly differentiate. They suspect two X and say, I'm sorry, I see one. Why on policy to Y two is zero. Then also they must say, must be true for the derivatives or buy one of Y two as well. So actually now I could think of this as like a linear system of equations for C one and C two. And I guess we know then that we can get non-trivial solutions. So
you want to see two even only if the determinant of that matrix is zero. OK. And of. That determines what's called the wrong skin. Okay, so so that's what I mean by the ront skin, the determinants of that matrix. And then here, I guess I'm going to slip between. It's a slight Abuzer notation, I guess. But and here I've kind of expressed W as being a of by linear functional of y one of my two. So if you give it to functions, it spits out another function. Okay.
But it's also help with sometimes just to think of W as being a function of X. Okay. Right. So I mean by that is I actually now evaluate why one and why two. Then I get W is a function of X. Okay. So it's just I have two different ways of thinking. The same quantity. And so I guess what I've argued then is that on the eve, why, when and why to linearly dependent, then the Ront scheme must be zero. That's what I've shown. Yeah.
Again, I hope you don't mind this. All right. L.D. literally dependent. That's clear to say myself. Some ink. So be shown that if that's true, that implies that the wrong skin is zero. And I guess we can also use this as a contrapositive of that. So, OK. And this isn't going to upset you when I say it is not zero. Then why? What about why two must be linearly independent. OK. And I can use L I for that. Is that clear? I said, turning the previous identity on its head now,
it would be really nice if we could turn that implication into an even only if. But it's not quite true in the other direction. And actually, you can quite easily come up with counterexamples and. So here's a counterexample. So here's two functions that sort of defined in a piecewise way and actually the kind of examples that you can easily dream up for this Jamele, about this sort of form. So these are functions that are Leavesley cheque, their linearly independent.
So the first thing is these two functions, either to find this funny piecewise way. There are actually three times differentiable, but something funny happens at the Origin, but it's only the fourth derivative. That's discontinuous. OK. And then secondly, you can easily cheque sort of an exercise you don't believe me. Is it these two functions, ELENITA Independent. So it's impossible to find a non-trivial combination of them that adds up to zero function.
But finally, the real killer is the wrong skin. These two functions a zero. So although the one scale is zero. These functions are linearly independent. So it means we can't put this implication the other way. All right. But there is so in general, this invitation does go in both directions. But in the case, luckily for us, I guess in the case is we're really interested in when, why, when and why not just any two old functions that you pull out of the air like these two.
But they are actually two solutions. I have a second order idea. And in that case, this indication does go in both directions. Why I want to buy you a street. The second already. Then a patient does go both ways. And. So I'm just I'm going to prove this. I'm talking a state of proposition. A proposition of two halves. So let's take why, when and why to be any two solutions of our second order homogeneous Odey. And then there's two things
that we can then say. And the first thing is, so the wrong skin, basically, either it's Zerai, beware or it's zero. No way. Why does that matter? Because I guess it wouldn't make sense for two functions to be kind of linearly independent some places, but not in others. And the second thing is that. I'm going to stay this way. So w zero. Even only four, I went away to an elderly dependent. OK. And Sandy just proved these one at a time. Now, I'll probably wrap up after that.
And so for the first bit, we know that why, when and why to both solutions of our homogeneous idea. So let me just write that out explicitly. OK. So we know that why, when and why to both satisfy the same ODAC. And so the way I think about this is I now I want to eliminate this P0 term. OK, so that's sort of my little way of remembering what to do here. All right. So what I mean is I'm going to do y one time this four minus Y two times this one. Okay. You've got Claire. So it's all done.
Okay. So why won't times the first one minus Y two times the second one. And now this. Right. This is the wrong skin right here. All right. Actually, this is the derivative of their own skin. So if you imagine differentiating this with spec two X, you see the first of this cancel Y one prior March primaries, Y to prime Y one prime. You set up with the second remitted. And so actually W satisfies this first order idea. And actually, you can easily solve our Audi.
All right. So you can to think about this as being like an integrating factor or by taking this time over the other side and separating the variables. I'm that you again using elementary methods. Again, note we need PE to not be zero. Now we know that exponential and can't ever be zero. All right. So this constant zero, it isn't the constant zero, then WCF everywhere. If the constant isn't zero w zero nowhere. Okay. So why the WAC, right? Know it's not zera everywhere. OK. Mm hmm.
All right. For the second part, and we've done that one, okay. And so I think we need to do is the other way. We need to do that way. Okay. That's what remains to be done. And so we've done the. So what I need to do is the use of vindication going in the right direction. So let's suppose W is zero and I'm trying to show that, then why, when and why do you have to be linearly dependent? All right. So first things first, I'm going to assume that why one is not the zero function.
Because if it was, then why, when and why two would trivially be liberty dependent. Okay. Okay. Because you could just take C to be zero and see one to be anything. Yeah. Okay. So if one wants at the zero function, then they must exist. Somebody of X, Y, Y one is not zero. Now, here's the trick. So I'm going to define why I have to be this particular linear combination of why one and why two. So it's why one of a way to maximise Y to a Y, one of X. Okay.
Then what we can see is that if I put X equals item, this Y is zero. That's clear. Right. And also actually I differentiate this and input X equals I then I get the ront skin evaluates that X equals I. Okay. And by assumption the wrong skin is zero. Right. And so you've got this function and both why and it's first derivative. A zero X equals a. And also because it's a linear accommodation of Y one and Y two. Y satisfies a homogeneous equation.
It satisfies this second order differential equation with Ziya on the right hand side. So then by Picards Theorem, the solution of this initial value problem is existence is unique and it's just zero. Okay. So I've got this homogeneous ody with zero initial conditions. The sweep of that problem is just Weich zero and by Picards there in that switch is unique. I'm at Ben tells us that there does exist a non-trivial combination of y one of Y two that adds up to zero. Okay.
All right. All right. So I think that's all for today, I say tomorrow. Going to get into more actual solving problems and more techniques to solving things.