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  • OK, let's get started. Now... I'm assuming that, A, you went recitation yesterday,

  • B, that even if you didn't, you know how to separate variables, and you know how to construct simple

  • models, solve physical problems with differential equations, and possibly even solve them.

  • So, you should have learned that either in high school, or 18.01 here, or... yeah.

  • So, I'm going to start from that point, assume you know that. I'm not going to tell you what differential

  • equations are, or what modeling is. If you still are uncertain about those things, the

  • book has a very long and good explanation of it. Just read that stuff. So,

  • we are talking about first order ODEs.

  • ODE: I'll only use three ... two acronyms. ODE is ordinary differential equations. I think all of MIT

  • knows that, whether they've been taking the course or not. So, we are talking about first-order ODEs.

  • which in standard form, are written, you isolate the derivative of y with respect

  • to, x, let's say, on the left-hand side, and on the right-hand side you write everything

  • else. You can't always do this very well, but for today, I'm going to assume that it

  • has been done and it's doable. So, for example, some of the ones that will be considered either

  • today or in the problem set are things like

  • oh... y' = x / y

  • That's pretty simple. The problem set has y' = ...let's see...

  • x - y^2.

  • And, it also has y' = y - x^2.

  • There are others, too. Now, when you look at this, this, of course, you can solve by

  • separating variables. So, this is solvable. This one is-- and neither of these can you

  • separate variables. And they look extremely similar. But they are extremely dissimilar.

  • The most dissimilar about them is that this one is easily solvable. And you will learn,

  • if you don't know already, next time next Friday how to solve this one

  • This one, which looks almost the same, is unsolvable in a certain sense. Namely, there

  • are no elementary functions which you can write down, which will give a solution of

  • that differential equation. So, right away, one confronts the most significant fact that

  • even for the simplest possible differential equations, those which only involve the first

  • derivative, it's possible to write down extremely looking simple guys.

  • I'll put this one up in blue to indicate that it's bad. Whoops, sorry, I mean, not really

  • bad, but recalcitrant. It's not solvable in the ordinary sense in which you think of an

  • equation is solvable. And, since those equations are the rule rather than the exception, I'm

  • going about this first day to not solving a single differential equation, but indicating

  • to you what you do when you meet a blue equation like that.

  • What do you do with it? So, this first day is going to be devoted to geometric ways of

  • looking at differential equations and numerical. At the very end, I'll talk a little bit about

  • numerical ways. And you'll work on both of those for the first problem set. So,

  • what's our geometric view of differential equations?

  • Well, it's something that's contrasted with

  • the usual procedures, by which you solve things and find elementary functions which solve

  • them. I'll call that the analytic method. So, on the one hand, we have the analytic

  • ideas, in which you write down explicitly the equation, y' = f(x,y).

  • And, you look for certain functions, which are called its solutions. Now, so there's

  • the ODE. And, y1 of x, notice I don't use a separate letter. I don't use g or h or something

  • like that for the solution because the letters multiply so quickly, that is, multiply in

  • the sense of rabbits, that after a while, if you keep using different letters for each

  • new idea, you can't figure out what you're talking about.

  • So, I'll use y1 means, it's a solution of this differential equation. Of course, the

  • differential equation has many solutions containing an arbitrary constant. So, we'll call this

  • the solution. Now, the geometric view,

  • the geometric guy that corresponds to this version

  • of writing the equation, is something called a direction field.

  • And, the solution is, from

  • the geometric point of view, something called an integral curve.

  • So, let me explain if you

  • don't know what the direction field is. I know for some of you, I'm reviewing what you

  • learned in high school. Those of you who had the BC syllabus in high school should know

  • these things. But, it never hurts to get a little more practice. And, in any event, I

  • think the computer stuff that you will be doing on the problem set, a certain amount

  • of it should be novel to you.

  • It was novel to me, so why not to you? So, what's a direction field? Well, the direction

  • field is, you take the plane, and in each point of the plane-- of course, that's an

  • impossibility. But, you pick some points of the plane. You draw what's called a little

  • line element. So, there is a point. It's a little line, and the only thing which distinguishes

  • it outside of its position in the plane, so here's the point, (x,y), at which we are drawing

  • this line element, is its slope. And, what is its slope? Its slope is to be f(x,y).

  • And now, You fill up the plane with these things until you're tired of putting then in. So,

  • I'm going to get tired pretty quickly.

  • So, I don't know, let's not make them all go the same way. That sort of seems cheating.

  • How about here? Here's a few randomly chosen line elements that I put in, and I putted

  • the slopes at random since I didn't have any particular differential equation in mind.

  • Now, the integral curve, so those are the line elements. The integral curve is a curve,

  • which goes through the plane, and at every point is tangent to the line element there.

  • So, this is the integral curve. Hey, wait a minute, I thought tangents were the line

  • element there didn't even touch it. Well, I can't fill up the plane with line elements.

  • Here, at this point, there was a line element, which I didn't bother drawing in. And, it

  • was tangent to that. Same thing over here: if I drew the line element here, I would find

  • that the curve had exactly the right slope there.

  • So, the point is the integral, what distinguishes the integral curve is that everywhere it has

  • the direction, that's the way I'll indicate that it's tangent, has the direction of the

  • field everywhere at all points on the curve, of course, where it doesn't go. It doesn't

  • have any mission to fulfill. Now, I say that this integral curve is the graph of the solution

  • to the differential equation. In other words, writing down analytically the differential

  • equation is the same geometrically as drawing this direction field, and solving analytically

  • for a solution of the differential equation is the same thing as geometrically drawing

  • an integral curve. So, what am I saying?

  • I say that an integral curve,

  • all right, let me write it this way. I'll make a little theorem

  • out of it, that y1(x) is a solution to the differential equation

  • if, and only if,

  • the graph, the curve associated with this, the graph of y1 of x is an integral curve.

  • Integral curve of what? Well, of the direction field associated with that equation. But there isn't

  • quite enough room to write that on the board. But, you could put it in your notes, if you

  • take notes. So, this is the relation between the two, the integral curves of the graphs

  • or solutions.

  • Now, why is that so? Well, in fact, all I have to do to prove this, if you can call

  • it a proof at all, is simply to translate what each side really means. What does it

  • really mean to say that a given function is a solution to the differential equation? Well,

  • it means that if you plug it into the differential equation, it satisfies it. Okay, what is that?

  • So, how do I plug it into the differential equation and check that it satisfies it?

  • Well, doing it in the abstract, I first calculate its derivative. And then, how will it look

  • after I plugged it into the differential equation? Well, I don't do anything to the x, but wherever

  • I see y, I plug in this particular function. So, in notation, that would be written this

  • way. So, for this to be a solution means this,

  • that that equation is satisfied. Okay, what

  • does it mean for the graph to be an integral curve? Well, it means that at each point,

  • the slope of this curve, it means that the slope of y1 of x should be, at each point,

  • (x1,y1). It should be equal to the slope of the direction field at that point.

  • And then, what is the slope of the direction field at that point? Well, it is f of that

  • particular, well, at the point, (x,y1). If you like, you can put a subscript, one, on

  • there, send a one here or a zero there, to indicate that you mean a particular point.

  • But, it looks better if you don't. But, there's some possibility of confusion. I admit to

  • that. So, the slope of the direction field, what is that slope? Well, by the way, I calculated

  • the direction field. Its slope at the point was to be x, whatever the value of x was,

  • and whatever the value of y1(x) was, substituted into the right-hand side of the equation.

  • So, what the slope of this function of that curve of the graph should be equal to the

  • slope of the direction field. Now, what does this say?

  • Well, what's the slope of y1(x)? That's y1'(x). That's from the first day of 18.01, calculus.

  • What's the slope of the direction field? This? Well, it's this. And, that's with the right

  • hand side. So, saying these two guys are the same or equal, is exactly, analytically, the

  • same as saying these two guys are equal. So, in other words, the proof consists of, what

  • does this really mean? What does this really mean? And after you see what both really mean,

  • you say, yeah, they're the same.

  • So, I don't how to write that. It's okay: same, same, how's that? This is the same as that.

  • Okay, well, this leaves us the interesting question of how do you draw a direction from

  • the, well, this being 2003, mostly computers draw them for you. Nonetheless, you do have

  • to know a certain amount. I've given you a couple of exercises where you have to draw

  • the direction field yourself. This is so you get a feeling for it, and also because humans

  • don't draw direction fields the same way computers do. So, let's first of all, how did computers

  • do it? They are very stupid. There's no problem.

  • Since they go very fast and have unlimited amounts of energy to waste, the computer method

  • is the naive one. You pick the point. You pick a point, and generally, they are usually

  • equally spaced. You determine some spacing, that one: blah, blah, blah, blah, blah, blah,

  • blah, equally spaced. And, at each point, it computes f(x, y) at the point, finds, meets,

  • and computes the value of f of (x, y), that function, and the next thing is, on the screen,

  • it draws, at (x, y), the little line element having slope f(x, y). In other words, it does

  • what the differential equation tells it to do.

  • And the only thing that it does is you can, if you are telling the thing to draw the direction

  • field, about the only option you have is telling what the spacing should be, and sometimes

  • people don't like to see a whole line. They only like to see a little bit of a half line.

  • And, you can sometimes tell, according to the program, tell the computer how long you

  • want that line to be, if you want it teeny or a little bigger. Once in awhile you want

  • you want it narrower on it, but not right now.

  • Okay, that's what a computer does. What does a human do? This is what it means to be human.

  • You use your intelligence. From a human point of view, this stuff has been done in the wrong

  • order. And the reason it's been done in the wrong order: because for each new point, it

  • requires a recalculation of f(x, y). That is horrible. The computer doesn't mind, but

  • a human does. So, for a human, the way to do it is not to begin by picking the point,

  • but to begin by picking the slope that you would like to see. So, you begin by taking

  • the slope. Let's call it the value of the slope, C. So, you pick a number. C is two.

  • I want to see where are all the points in the plane where the slope of that line element

  • would be two? Well, they will satisfy an equation.

  • The equation is f(x,y) = C, in general. So, what you do is plot this, plot the equation,

  • plot this equation. Notice, it's not the differential equation. You can't exactly plot a differential

  • equation. It's a curve, an ordinary curve. But which curve will depend; it's, in fact,

  • from the 18.02 point of view, the level curve of C, sorry, it's a level curve of f of (x,

  • y), the function f of x and y corresponding to the level of value C

  • But we are not going to call it that because this is not 18.02. Instead, we're going to

  • call it an isocline. And then, you plot, well, you've done it. So, you've got this isocline,

  • except I'm going to use a solution curve, solid lines, only for integral curves. When

  • we do plot isoclines, to indicate that they are not solutions, we'll use dashed lines

  • for doing them. One of the computer things does and the other one doesn't. But they use

  • different colors, also. There are different ways of telling you what's an isocline and

  • what's the solution curve. So, and what do you do? So, these are all the points where

  • the slope is going to be C.

  • And now, what you do is draw in as many as you want of line elements having slope C.

  • Notice how efficient that is. If you want 50 million of them and have the time, draw

  • in 50 million. If two or three are enough, draw in two or three. You will be looking

  • at the picture. You will see what the curve looks like, and that will give you your judgment

  • as to how you are to do that. So, in general, a picture drawn that way, so let's say, an

  • isocline corresponding to C equals zero.

  • The line elements, and I think for an isocline, for the purposes of this lecture, it would

  • be a good idea to put isoclines. Okay, so I'm going to put solution curves in pink,

  • or whatever this color is, and isoclines are going to be in orange, I guess. So, isocline,

  • represented by a dashed line, and now you will put in the line elements of, we'll need

  • lots of chalk for that. So, I'll use white chalk.

  • Y horizontal? Because according to this the slope is supposed to be zero there. And at

  • the same way, how about an isocline where the slope is negative one? Let's suppose here

  • C is equal to negative one. Okay, then it will look like this. These are supposed to

  • be lines of slope negative one. Don't shoot me if they are not. So, that's the principle.

  • So, this is how you will fill up the plane to draw a direction field: by plotting the

  • isoclines first.

  • And then, once you have the isoclines there, you will have line elements. And you can draw

  • a direction field. Okay, so, for the next few minutes, I'd like to work a couple of

  • examples for you to show how this works out in practice.

  • So, the first equation is going

  • to be y' = -x / y.

  • Okay, first thing, what are the isoclines? Well, the isoclines

  • the isoclines are going to be y.

  • Well, -x / y = C. Maybe I better make two steps out of this. Minus x over y is equal

  • to C. But, of course, nobody draws a curve in that form. You'll want it in the form y

  • = -1 / C * x. So, there's our isocline. Why don't I put that up in orange since it's going

  • to be, that's the color I'll draw it in. In other words, for different values of C, now

  • this thing is aligned. It's aligned, in fact, through the origin. This looks pretty simple.

  • Okay, so here's our plane. The isoclines are going to be lines through the origin. And

  • now, let's put them in, suppose, for example, C is equal to one.

  • Well, if C is equal to one, then it's the line, y equals minus x. So, this is the isocline.

  • I'll put, down here, C equals minus one. And,

  • and along it, the

  • no, something's wrong.

  • I'm sorry?

  • C is one, not negative one, right, thanks. Thanks. So, C equals one. So, it should be

  • little line segments of slope one will be the line elements, things of slope one. OK,

  • now how about C equals negative one?

  • If C equals negative one, then it's the line, y = x. And so, that's the isocline. Notice,

  • still dash because these are isoclines. Here, C is negative one. And so, the slope elements

  • look like this. Notice, they are perpendicular. Now, notice that they are always going to

  • be perpendicular to the line because the slope of this line is minus one over C. But, the

  • slope of the line element is going to be C. Those numbers, minus one over C and C, are

  • negative reciprocals. And, you know that two lines whose slopes are negative reciprocals

  • are perpendicular. So, the line elements are going to be perpendicular to these. And therefore,

  • I hardly even have to bother calculating, doing any more calculation. Here's going to

  • be a, well, how about this one?

  • Here's a controversial isocline. Is that an isocline? Well, wait a minute. That doesn't

  • correspond to anything looking like this. Ah-ha, but it would if I put C multiplied

  • through by C. And then, it would correspond to C being zero. In other words, don't write

  • it like this. Multiply through by C. It will read C y = - x. And then, when C is zero,

  • I have x equals zero, which is exactly the y-axis.

  • So, that really is included. How about the x-axis? Well, the x-axis is not included.

  • However, most people include it anyway. This is very common to be a sort of sloppy and

  • bending the edges of corners a little bit, and hoping nobody will notice. We'll say that

  • corresponds to C equals infinity. I hope nobody wants to fight about that. If you do, go fight

  • with somebody else. So, if C is infinity, that means the little line segment should

  • have infinite slope, and by common consent, that means it should be vertical. And so,

  • we can even count this as sort of an isocline. And, I'll make the dashes smaller, indicate

  • it has a lower status than the others. And, I'll put this in, do this weaselly thing of

  • putting it in quotation marks to indicate that I'm not responsible for it.

  • Okay, now, we now have to put it the integral curves. Well, nothing could be easier. I'm

  • looking for curves which are everywhere perpendicular to these rays. Well, you know from geometry

  • that those are circles. So, the integral curves are circles.

  • And, it's an elementary exercise,

  • which I would not deprive you of the pleasure of. Solve the ODE by separation of variables.

  • In other words, we've gotten the, so the circles are ones with a center at the origin, of course,

  • equal some constant. I'll call it C1, so it's not confused with this C. They look like that,

  • and now you should solve this by separating variables, and just confirm that the solutions

  • are, in fact, those circles.

  • One interesting thing, and so I confirm this, I won't do it because I want to do geometric

  • and numerical things today. So, if you solve it by separating variables, one interesting

  • thing to note is that if I write the solution as y = y1(x), well, it'll look something like

  • the sqrt (C1 - x^2). We'll make the x squared because that's the way people usually put the radius.

  • Minus x squared.

  • And so, a solution, a typical solution looks like this. Well, what's the solution over

  • here? Well, that one solution will be goes from here to here. If you like, it has a negative

  • side to it. So, I'll make, let's say, plus. There's another solution, which has a negative

  • value. But let's use the one with the positive value of the square root. My point is this,

  • that that solution, the domain of that solution, really only goes from here to here. It's not

  • the whole x-axis. It's just a limited piece of the x-axis where that solution is defined.

  • There's no way of extending it further. And, there's no way of predicting, by looking at

  • the differential equation, that a typical solution was going to have a limited domain

  • like that.

  • In other words, you could find a solution, but how far out is it going to go? Sometimes,

  • it's impossible to tell, except by either finding it explicitly, or by asking a computer

  • to draw a picture of it, and seeing if that gives you some insight. It's one of the many

  • difficulties in handling differential equations. You don't know what the domain of a solution

  • is going to be until you've actually calculated it.

  • Now, a slightly more complicated example is going to be, let's see, y' = 1 + x - y. It's

  • not a lot more complicated, and as a computer exercise, you will work with, still, more

  • complicated ones. But here, the isoclines would be what? Well, I set that equal to C.

  • Can you do the algebra in your head? An isocline will have the equation: this equals C. So,

  • I'm going to put the y on the right hand side, and that C on the left hand side. So, it will

  • have the equation y = 1 + x - C, or a nicer way to write it would be y = x + 1 - C. I

  • guess it really doesn't matter.

  • So there's the equation of the isocline. Let's quickly draw the direction field. And notice,

  • by the way, it's a simple equation, but you cannot separate variables. So, I will not,

  • today at any rate, be able to check the answer. I will not be able to get an analytic answer.

  • All we'll be able to do now is get a geometric answer. But notice how quickly, relatively

  • quickly, one can get it. So, I'm feeling for how the solutions behave to this equation.

  • All right, let's see, what should we plot first? I like C equals one, no, don't do C

  • equals one. Let's do C equals zero, first. C equals zero. That's the line. y = x + 1.

  • Okay, let me run and get that chalk.

  • So, I'll isoclines are in orange. If so, when C equals

  • zero, y equals x plus one. So, let's say it's this curve.

  • C equals zero.

  • How about C equals

  • negative one? Then it's y = x + 2. It's this curve.

  • Well, let's label it down here.

  • So, this is C equals negative one. C equals negative two would be y equals x, no, what am I doing?

  • C equals negative one is y = x + 2. That's right. Well, how about the other side?

  • If C equals plus one, well, then it's going to go through the origin.

  • It looks like a little more room down here. How about, so if this is going to be C equals one, then I sort of

  • get the idea. C equals two will look like this. They're all going to be parallel lines

  • because all that's changing is the y-intercept, as I do this thing. So, here, it's C equals

  • two. That's probably enough. All right, let's put it in the line elements. All right,

  • C equals negative one. These will be perpendicular. C equals zero, like this.

  • C equals one. Oh, this is interesting. I can't even draw in the line elements because they

  • seem to coincide with the curve itself, with the line itself. They write y along the line,

  • and that makes it hard to draw them in. How about C equals two? Well, here, the line elements

  • will be slanty. They'll have slope two, so a pretty slanty up. And, I can see if a C

  • equals three in the same way. There are going to be even more slantier up. And here, they're

  • going to be even more slanty down. This is not very scientific terminology or mathematical,

  • but you get the idea. Okay, so there's our quick version of the direction field. All

  • we have to do is put in some integral curves now. Well, it looks like it's doing this.

  • It gets less slanty here. It levels out, has slope zero.

  • And now, in this part of the plain, the slope seems to be rising. So, it must do something

  • like that. This guy must do something like this. I'm a little doubtful of what I should

  • be doing here. Or, how about going from the other side? Well, it rises, gets a little,

  • should it cross this? What should I do? Well, there's one integral curve, which is easy

  • to see. It's this one. This line is both an isocline and an integral curve.

  • It's everything, except drawable, [LAUGHTER] so, you understand this is the same line.

  • It's both orange and pink at the same time. But I don't know what combination color that

  • would make. It doesn't look like a line, but be sympathetic. Now, the question is, what's

  • happening in this corridor? In the corridor, that's not a mathematical word either, between

  • the isoclines for, well, what are they? They are the isoclines for C equals two, and C

  • equals zero.

  • How does that corridor look? Well: something like this. Over here, the

  • lines all look like that. And here, they all look like this.

  • The slope is two. And, a hapless solution gets in there. What's it to do? Well, do you

  • see that if a solution gets in that corridor, an integral curve gets in that corridor, no

  • escape is possible. It's like a lobster trap. The lobster can walk in. But it cannot walk

  • out because things are always going in. How could it escape? Well, it would have to double

  • back, somehow, and remember, to escape, it has to be, to escape on the left side, it

  • must be going horizontally.

  • But, how could it do that without doubling back first and having the wrong slope? The

  • slope of everything in this corridor is positive, and to double back and escape, it would at

  • some point have to have negative slope. It can't do that. Well, could it escape on the

  • right-hand side? No, because at the moment when it wants to cross, it will have to have

  • a slope less than this line. But all these spiky guys are pointing; it can't escape that

  • way either. So, no escape is possible. It has to continue on, there. But, more than

  • that is true. So, a solution can't escape.

  • Once it's in there, it can't escape. It's like, what do they call those plants, I forget,

  • pitcher plants. You know, all their hairs are going down. So, it looks like that. And so,

  • the poor little insect falls in. They could climb up the walls except that all the hairs

  • are going the wrong direction, and it can't get over them. Well, let's think of it that

  • way: this poor trap solution. So, it does what it has to do. Now, there's more to it

  • than that. Because there are two principles involved here that you should know, that help

  • a lot in drawing these pictures. Principle number one is that two integral curves cannot

  • cross at an angle. Two integral curves can't cross, I mean, by crossing at an angle like

  • that. I'll indicate what I mean by a picture like that.

  • Now, why not? This is an important principle. Let's put that up in the white box. They can't

  • cross because if two integral curves, are trying to cross, well, one will look like

  • this. It's an integral curve because it has this slope. And, the other integral curve

  • has this slope. And now, they fight with each other. What is the true slope at that point?

  • Well, the direction field only allows you to have one slope. If there's a line element

  • at that point, it has a definite slope. And therefore, it cannot have both the slope and

  • that one. It's as simple as that. So, the reason is you can't have two slopes. The direction

  • field doesn't allow it. Well, that's a big, big help because if I know, here's an integral

  • curve, and if I know that none of these other pink integral curves are allowed to cross

  • it, how else can I do it?

  • Well, they can't escape. They can't cross. It's sort of clear that they must get closer

  • and closer to it. You know, I'd have to work a little to justify that. But I think that

  • nobody would have any doubt of it who did a little experimentation. In other words,

  • all these curves joined that little tube and get closer and closer to this line, y = x.

  • And there, without solving the differential equation, it's clear that all of these solutions,

  • how do they behave? As x goes to infinity, they become asymptotic to, they become closer

  • and closer to the solution, x. Is x a solution? Yeah, because y equals x is an integral curve.

  • Is x a solution? Yeah, because if I plug in y equals x, I get what? On the right-hand

  • side, I get one. And on the left-hand side, I get one. One equals one. So, this is a solution.

  • Let's indicate that it's a solution. So, analytically, we've discovered an analytic solution to the

  • differential equation, namely, Y equals X, just by this geometric process. Now, there's

  • one more principle like that, which is less obvious. But you do have to know it. So, you

  • are not allowed to cross. That's clear. But it's much, much, much, much, much less obvious

  • that two integral curves cannot touch. That is, they cannot even be tangent. Two integral

  • curves cannot be tangent.

  • I'll indicate that by the word touch, which is what a lot of people say. In other words,

  • if this is illegal, so is this.

  • Can't have it.

  • You know, without that for example, it might

  • be, I might feel that there would be nothing in this to prevent those curves from joining.

  • Why couldn't these pink curves join the line, y equals x? You know, it's a solution. They

  • just pitch a ride, as it were. The answer is they cannot do that because they have to

  • just get asymptotic to it, ever, ever closer. They can't join y equals x because at the

  • point where they join, you have that situation.

  • Now, why can't you to have this? That's much more sophisticated than this, and the reason

  • is because of something called the Existence and Uniqueness Theorem,

  • Existence and Uniqueness Theorem

  • which says

  • which says

  • which says that there is through a point, (x0, y0),

  • that y prime equals y' = f(x, y) has only one, and only one solution. One has one solution.

  • In mathematics speak, that means at least one solution. It doesn't mean it has just

  • one solution. That's mathematical convention. It has one solution, at least one solution.

  • But, the killer is, only one solution.

  • That's what you have to say in mathematics if you want just one, one, and only one solution

  • through the point (x0, y0). So, the fact that it has one, that is the existence part. The

  • fact that it has only one is the uniqueness part of the theorem. Now, like all good mathematical

  • theorems, this one does have hypotheses. So, this is not going to be a course, I warn you,

  • those of you who are theoretically inclined, very rich in hypotheses. But, hypotheses for

  • those one or that f(x, y) should be a continuous function. Now, like polynomial, signs, should

  • be continuous near, in the vicinity of that point.

  • That guarantees existence, and what guarantees uniqueness is the hypothesis that you would

  • not guess by yourself. Neither would I. What guarantees the uniqueness is that also, it's

  • partial derivative with respect to y should be continuous,

  • should be continuous near (x0, y0).

  • Well,

  • I have to make a decision. I don't have time to talk about Euler's method. Let me

  • refer you to the, there's one page of notes, and I couldn't do any more than just repeat

  • what's on those notes. So, I'll trust you to read that.

  • And instead, let me give you an example which will solidify these things in your mind a

  • little bit. I think that's a better course. The example is not in your notes, and therefore,

  • remember, you heard it here first. Okay, so what's the example? It's...

  • So,

  • there's our differential equation.

  • Now, let's just solve it by separating variables. dy. Can you do it in your head? dy

  • over dx, put all the y's on the left. It will look like dy / (1 - y). Put all the dx's on

  • the left. So, the dx here goes on the right, rather. That will be dx. And then, the x goes

  • down into the denominator. So now, it looks like that.

  • And, if I integrate both sides, I get the log of one minus y, I guess, maybe with a,

  • I never bothered with that, but you can. It should be absolute values. All right, put

  • an absolute value, plus a constant. And now, if I exponentiate both sides, the constant

  • is positive. So, this is going to look like y. (1 - y) = x. And, the constant will be

  • e^C1. And, I'll just make that a new constant, Cx. And now, by letting C be negative, that's

  • why you can get rid of the absolute values, if you allow C to have negative values as

  • well as positive values. Let's write this in a more human form. So, y = 1 - Cx. Good,

  • all right, let's just plot those. So, these are the solutions.

  • It's a pretty easy equation, pretty easy solution method, just separation of variables. What

  • do they look like? Well, these are all lines whose intercept is at one. And, they have

  • any slope whatsoever. So, these are the lines that look like that. Okay, now let me ask,

  • existence and uniqueness. Existence: through which points of the plane does the solution

  • go? Answer: through every point of the plane, through any point here, I can find one and

  • only one of those lines, except for these stupid guys here on the stalk of the flower.

  • Here, for each of these points, there is no existence. There is no solution to this differential

  • equation, which goes through any of these wiggly points on the y-axis, with one exception.

  • This point is oversupplied. At this point, it's not existence that fails. It's uniqueness

  • that fails: no uniqueness. There are lots of things which go through here. Now, is that

  • a violation of the existence and uniqueness theorem? It cannot be a violation because

  • the theorem has no exceptions. Otherwise, it wouldn't be a theorem. So, let's take a

  • look. What's wrong? We thought we solved it modulo, putting the absolute value signs on

  • the log. What's wrong? The answer: what's wrong is to use the theorem you must write

  • the differential equation in standard form, in the green form I gave you. Let's write

  • the differential equation the way we were supposed to. It says dy / dx = (1 - y) / x.

  • And now, I see, the right-hand side is not continuous, in fact, not even defined when

  • x equals zero, when along the y-axis. And therefore, the existence and uniqueness is

  • not guaranteed along the line, x equals zero of the y-axis. And, in fact, we see that it

  • failed. Now, as a practical matter, it's the way existence and uniqueness fails in all

  • ordinary life work with differential equations is not through sophisticated examples that

  • mathematicians can construct.

  • But normally, because f(x, y) will fail to be defined somewhere, and those will be the

  • bad points. Thanks.

OK, let's get started. Now... I'm assuming that, A, you went recitation yesterday,

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Lec 1|MIT 18.03 微分方程式, 2006年春学期 (Lec 1 | MIT 18.03 Differential Equations, Spring 2006)

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    單柏祥 に公開 2021 年 01 月 14 日
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