Okay, so what I want to do today is the do some examples VB loans. I'm, I'm going to summarize this again. What I'm gonna do is we've done some techniques we know about the point Caribbean Dixon theorem. We've learnt about finding fixed points and linearizing around fixed points. We've also seen the arrow, the vector field associated with the differential equation and how it is the tangent vector to the solution of the ODE. Okay? We're going to combine several ideas now and try to draw a global phase portrait. Meeting. It will be a picture now over the whole x-y plane, not just near each critical, each fixed point. Okay? So we're going to do that now. As I said, the Poincare had been Dixon theorem is going to play a big role. Now. It's all there in the lecture, but I'm going to just quickly summarize it. I'm not going to state it. Right. So what does the Poincare had been? Dixon theorem. So please watch the previous video. I have a clear statement of it. Alright. So here is a quick summary. Okay? So point bending some theorem is only for systems in two dimensions. So they're on display. It's going to be like dx by, sorry. Let's take to the way we write it. So you have x dot equals f x y, y dot equals g x y. Ok, so we have this system here. Okay? And what we're trying to do is goal. So suppose if x t, y t is the solution of one, where the initial value x, y is 0. Okay, then describe what happens as t goes to infinity to the solution. Okay, that's always been our goal. And PointQuery. Ben Dixon theorem answers this question to some extent. Okay? So Poincare had been Dixon theorem says. So suppose D is a trapping region for this system. Okay, what does that mean now? So which means in a here is your x-y plane. Here's some region D. And a trapping region means that if you start from a point here, let me make this bigger. It says that if you start with x 0, y 0 in this, and if you take your trajectory which cause it can never leave the region D. That's what a trapping region is. Alright, in the video, I have defined this carefully, but here this is what a trapping region is. By the way, how would you find out a trapping region? Suppose I have this first order system like this, right? How would you figure it out? There is a trapping region, because we're going to do it right without solving the OD. Alright? So here, this idea, this idea is very important that, you know, if you have this and you make this vector field, which as you know, f phi plus j, right? This vector field, I call it the direction field or the vector field. Okay? This vector field is what is the role of the vector field F in relation to the solutions of this cody. It plays a very crucial role. So how is F related this vector field related to solutions of this OD? We've talked about this before. That if I give you this system of ODEs and I created this vector field f, which is f IJ. It has a very close connection with the solution. It has some geometrical meaning. Do you remember? Anyone doesn't represent the kind of state flow throughout the area? State flow. Ok. Another way to say it, I'm not quite sure. I quite understand what's take Floyd's, but probably I think you have the right idea. But let's see. Do we have another interpretation of this? Is represent the velocity. It represents a velocity vector. Remember, right, this as your trajectory off the solution. If this is your solution, right, your position vector, then your velocity vector, right? As x dot i plus y dot t j, which is of course FI plus TJ. This is a position vector. This is the velocity vector is tangential to the trajectory. Right? So if there is any trajectory, any solution, then f is always going to be this velocity vector f. That f that I have defined is always tangential. Correct? That's the thing. So in fact, if you will remember that what I said earlier. That if you draw the F heroes everywhere, then if you start from somewhere, then you have to follow the arrows, right? And that's what your path is going to be like. So the arrows determine the path geometrically. Ok? So this vector field F, which comes from the differential equation, it really is crucial. This is really what you have to play with if you're not going to solve the ODE. Have your came with this idea. So please, this is a crucial idea. I'm going to use it and for many places that whenever you have your differential equation, x dot equals f and y dot equals g. And this vector field f phi plus g j is really very important. Ok, so now let's come back to the point where Robin Dickson theorem. As I said, I'm not doing this pretty systematically her because it's in the video. I'm just explaining. So suppose we have that. And here is d is my trapping region. Okay? Now PointQuery appendix and doesn't tell you how to find a trapping region. But can you think of, what could you say? How would you figure out there is some trapping region? So that no matter where I start from here, I'm trapped inside. I cannot go outside. Can you think of a situation where that would happen? Like if I'm given the differential equation like this, which means I have this vector field. Can you figure out some way of looking at the picture of the arrow then saying, Oh, I know this has to be a trapping region because that's how we will define trapping regions. Ab is all the errors around just starting point are all pointing inwards. Ok? But if I'm here, right, I mean, somebody has to leave from it, right? It can't, unless it's a fixed point address, have to leave from it in some direction. Right? And remember this whole region is trapping. So can you think of any other way? What didn't how they found their trajectories are dialing in. Okay. But that would be everything would coming in, right? And that would be very limited. Remember that it's the region D, which is the crucial thing. Can we check if it hasn't closed Warby? Okay? Let's see. Suppose, okay, and you know what this vector field is, right? It's at every point there is a narrow that's where the velocity, so support there was fluid here. Okay? And suppose you dropped a piece of paper. Okay, and it could spiral. And suppose I know that what condition would be imposed on D so that I know that this paper will not leave this area. Could you impose it has a physically realizable Well, I mean, that it could leave water can flow in anything. So what kind of flow would be so that the particle, the paper is trapped inside D. What kind of flow would attempt to be errors on the boundary of D? Printing in any Exactly Correct, can a particle leave now? Because we're at the moment it comes here, it is going to be driven and it can't go outside. That's the direction of its velocity vector. Doesn't make sense. Yes, now, right? So if your F, every, at, every point on the boundary is pointing inwards, no particle can leave it. Because if it tries to leave, it'll be going this way, right? Are you with me on this point? Okay. So if it tries to leave the arrow there would have to be pointing outwards, not inwards. So all you need is that the arrows on the boundary of D all have to be pointing inwards. Sorry. So that's how you figured out trapping regions. I'm going to give you an example later. Ok, I will see how to figure out trapping regions. Okay? As I said, I'm not writing mature, it's all them that video explained carefully there. Alright. Okay, so suppose now you have a trapping region. Okay? So D trapping. So you start from here and you know your trajectory cannot go anywhere outside D. And remember this has into diamond, well, particularly in two dimensions. What can happen to the trajectory in the long run? Okay, that's a question and this is what pink ribbon mics and answers. So either it's a fixed point, you start there, you stay there forever. Okay, that's one possibility. Second possibility is you have a periodic orbit. That's one possibility. Correct. So you're either you start there, you go round and round and round. Okay, that's fine. So either you are a fixed point or a periodic orbit, okay? But what if you're not one of them? What else can happen? So I start from here. I'm going to now remember, I cannot cross myself, right? I cannot do this. Then this point will have two possible solutions. So trajectories cannot cross two trajectories, cannot cross trajectories cannot cross each other either. And this is two dimensions. So I'm starting here. I go like this and I can't cross icon cross. I don't know what, Where am I going to go? I cannot cross over. Right? 3d. I can cross over. That's my three. This was only for 2D. So what is the long-term? Where is it going to go? Okay, now if you do real analysis, you know that if anyone's taken for 01, there's terrible compactness. If something is forced to be enclosed in a bounded region. And it goes, it is going to approach a limit. Okay? There's a turban compactness. So this trajectory which is trapped, it is going to approach something in the long run. Okay, there's a theorem and the Poincare Robin Dickson theorem says the following. So if you start from here, either it is a fixed point or it's a periodic orbit or closed orbit. Or it approaches two things. One of the two, it can either approach a fixed point. So there's some fixed point here. And it starts for me in the long run. It just takes forever to get there, but it'll go like that. That's one possibility. And the other possibility is OK, let me use color pencil here. Okay? And the, and the other possibility is that there is some periodic orbit. And this thing, you start from somewhere, it approaches it. These are the point could have been Dixon theorem says, if you have a trapping region, any trajectory starting inside it can have only four possible behaviors. Okay, as I said, I'm not writing that's it's in the other video. It's very clearly written and it's explained, alright. So it can either be a fixed point, say it could be that red thing. It could be the red periodic orbit. There could be more than one. Or that trajectory approaches a fixed point or the trajectory, a process, a periodic orbit. These are the only four things which can happen to a trajectory which is trapped. Alright? And this only works in two dimensions. Because if you're in three-dimensions, right? If you have a trapping region, right, you can start go, and then you can jump over it. Right? You're not, you're not constrained so much in 3D. So Poincare ibn Dixon theorem is only in 2D. Do Locke's criterion was only in 2D. Studying Dynamical Systems. Studying systems of ODEs in three or higher dimensions is a very difficult problem. There are lots of open questions even now. Okay? So in this course we'll just basically focus on 2D. There are some things one can say in 3D and high dimension, but there are lots of open questions. Alright? So that's the Poincare appendix theorem. Well, how much time away? 25. Okay, alright. So, okay, now I'll start the lecture which I have in mind. Ok. So my goal is to draw global phase portraits. Okay? So in two dimensions. So this is my differential equation. And I have my vector field f. Let me just use x y here. Okay. Before I move ahead, let me ask you, did you understand what I said about the pink ribbon mics and care? Do you have any questions about it? Okay. And the role of this vector field, capital F, that what it does, what it, how it is related to the solutions of the differential equation, the trajectories, okay? Both of these ideas will play important roles. Okay? So drug orbit, phase portraits of this, right? So what are the tools available to us? Okay, which you're going to use. I'm just reminding you, number one, finding fixed points. Ok, let's call it two. Second thing we have, So then we can figure it out the behavior near the fixed point by linearization. We have done these two things. Okay? The third thing which I introduced last lecture was, what is the phrase I used? I mean, what I wouldn't say non-existence of periodic orbits, but I had it better. But yeah, ruling out closed orbits. Due looks criterion, okay, which I will, I will come back to it again, but I use it. Okay? So now it doesn't say that every time I write a system it will not have a closed orbit. Sometimes it does not and sometimes it does. Ok. Do looks criterion is if it applies, then you know, there is no closed orbit, but they may be closed orbits. Ok. A number for it seems like such a simple idea, but I'm telling you how powerful this idea is. That this direction field F is tangential trajectories of two. And number five. Number four, I can't tell you how could we will see how to use this. And number 0.5 as the Poincare had been Dixon theorem, which says that if you have a trapping region, if you have a trapping region than any trajectory in there, either it is a fixed point, a periodic orbit, or approaches a fixed point, or a periodically that set. These are the only four things that can happen in the long run to a trajectory which is trapped. Okay, that's the pink ribbon Dixon terror. So in fact, the Poincare had been Dixon theorem is sometimes cavalry used to prove that there is a closed orbit. Because what happen? Suppose there is a trajectory which is trapped. So it can go to a fixed point, or it will go to a periodic orbit. But suppose every fixed point is a source, meaning things or leaving the fixed point not coming to it will use that. Okay, you're in this trapped area. Every fixed point is a source. Things must leave that fixed point from linearization we know. Then the only choice for that will be it'll go to a periodic orbit. And so there must be a periodic orbit. Okay, I'll use that argument later on. But I'm trying to tell you. So the Poincare had been Dixon theorem can be used to prove the existence of periodic orbits at times. Did today. I wanted to show an example later on, but did this idea become clear? And I'd set it very quickly. How the Poincare had been. Dixon theorem can sometimes be used to prove the existence of a periodic orbit. Do you want me to repeat that? What did you understand what I said? Kristin, Are you happy? Emily? Are you okay with that explanation? All right. Well, anyway, you'll see it when I do an example. Alright, so you will see that. So these are the five tools available to us. And I'm going to introduce one more tool called nullclines, but it's really just, this is an application of number four, okay? And I'm going to do an example, and this has a long example because it's got several pieces and you have to tie them together. There's not much calculation. It's all thinking. It's all using different ideas. Very little calculation. Okay, so here is this example I want to do. How much time do I have? 45 minutes. Okay, so competing theses that this was an example you've probably studied a little benefited mu3, O2, but you probably did just a local analysis. You did fixed points near each point, near linearization and stuff. But what does the global picture? We will do that carefully. So competing species are like suppose you have some ecosystem, you know, maybe it's a petri dish. And there are two bacteria there in the dish. Okay. All this treatment might no time to staying it orally, so it'll be quicker. Okay. But it's all in my notes. And as we posted, so at this whole description that I am saying is that you have two competing species, okay, in a Petri dish to bacteria. Okay? They don't harm each other directly. They're not eating each other, is not predator, prey. It's competing. Ok. And there, you know, maybe there is a sugar substrate or whatever. I'm not a biologist, so I don't know. And they are both eat the same food. So they're competing against each other. Ok, now, so you have to model that. So how does the population of this change over time, right? So one way to model it as that, you know, okay, well if you have x units of this one species, well then it will have more offspring, it grows. And so the more you have the, so there is a growth rate also. But then there is a population. Why? Read the X of t is the population of the species, one species at time t. Y of t is the population of another species at time t. Then X and Y competing for food. So the presence of why affects the growth rate of x, which means there's a negative sign there. Okay? And so x and y, the more interactions. So you know, you can take all this and you can make a model. I'm not going to spend time on it. But so, so these are populations. Of two competing species of bacteria in a petri dish. Okay, if you read the book, the book, trucks of rabbits and sheep in a pasture. I'm not so sure that's such a good. So they're both eat grass, but I'm not quite sure how it works because sheep cut the grass very low so the rabbits cannot hide. I don't know. But I taught the sheep are dominant there. I don't think the rabbit could do much. So, but I thought, you know, that's why I just made up this competing species of bacteria. But you can think of anything. Two things have the same food source but they don't harm each other directly. Okay? And suppose the population, the walls as x dot equals. So this is a made up problem, but it's actually captures some of this. Oh, actually before I was going to do something else, let me just come back to this later. Okay. Just wanted to say something here, which I'm going to use this as a I guess it's called I forget the correct up to your sidebar. This like in the court. Alright. So I'm going to use this idea a lot, not a lot, but it'll be there. Ok, suppose you have, Consider the function h of x, y to be, okay, consider this function. Okay? So we can draw, is a curve, some curve, right? Actually in this case it's aligned. And this can be any function. Alright? So let me draw the line here. What's a good way to draw this line in the xy plane? I don't need a precise thing, but get an indication. How do you draw lines? Quick way. Find two points that connect them. So you'll find you just try to guess two points. Any better way. Easier than that? One, get two points. But there have been lots of points. So can we do slightly better than that? Yeah, that's what I'm going to do. I'm going to find two points. But I was like, yeah, x equals y or let, let x equal 0 and solve for y n and y equal 0 and solve for x. And then connected to line the two points on the line, just find where they cut the x and y-axis. That's really the unless they are parallel to one of the axes which you can easily see. Easiest way to draw lines as fine where they cut the x and y axis. So this one cuts the x axis, which means y 0, x equal to three. And y is three over two. This line. Right? It's easier because you don't know, you know, rest of the up to ten points here to put scaled and think this one is easier. Okay, so this is where H is 0. Now here is the important points. So these are all the points where h is 0, so H is nonzero everywhere else. Correct? So this is anytime you have a function h, then h equal to 0 is some curve. Then one side of this curve is going to be H less than 0, and the other side is going to be h greater than 0. Always is a theorem from continuity. If you analysis, I cannot have two points here, where h is greater than 0 here and h is less than 0 here. Why not? Why can't I have two points on this one side where h is greater than 0 and less than 0 here. On this one side, why can't I have that? We all believe it right? It all it's got to have the same sign on one side. Why? What's the argument? Would be that it requires to cross that line that we just drew. Why would it cross the line? You're correct, but why? Because if that's the line where h equals 0, then negative SP one side positive has to be the other. Well, that's what I'm asking you to prove. Okay, this is the continuity argument. So take a curve which joins these two points, any curve which stays on one side. Right? So h is changing as you go along this curve from a to B, correct? On this curve and is changing continuously. Now H is positive here and negative here. So it switched from positive to negative as you travelled along the curve. Continuity tells you it's hard to be 0 somewhere where here. But that's not on the line. Does that argument make sense? All right. I mean, I don't expect you to prove this, but I'm just trying to tell you that this has an idea here. Ok. So one side of this H is positive and the other side is negative. How do I figure out which side is which? A simple idea which works everytime. You can pick a point on either side and then plug it in and then you figure out if it's greater than 0 lessons here. Should I use a point? How did I choose this point is there's some clever way of doing this which might make my job easier. Why pick 00? What's the big idea about that? Why? What's the advantage? X and Y will cancel k. Any other reason? Why 0's here is a good choice and why isn't a good choice or 2-3? Well, maybe let's see. Why isn't one to a good choice? I'm talking of convenience. Lies in 12y is 0-0 a better choice? Yeah. You want something that's long, lines? Is 12x underlined or I'm saying you want something that's like less than three and less than three or greater than three. Okay? What do you want? Because you want to test a point whose location you know in relation to this find easily. But you're not sure where it is 12, is it on this side or that side? Right. That's the tricky bit. You don't know about one too. I think it's I don't think it's obvious. Is 12 here or there? You don't know really. Is there any doubt about 00? That's 1-0-0 is a better choice. Alright? Nothing. I mean, it just is no great mathematics here, but just so I just check h at 00. So it's three minus two times 0 minus, sorry, three minus 0 is three. So this is greater than 0. So this has on the greater than 0 side. So this side is greater than 0 and this side is less than 0. All right, I'm going to use this idea that whenever you have some curve h equal to 0, then one side is h greater than 0 and other side is H less than 0. Okay? Alright. So I'm just being very, please read the notes. I'll go into more detail. Just state this for a general case. We'd need notes for details. Ok. But this was exactly the idea. Okay, now let's come back to that. The competing species. My mortal. Okay, so gone. Draw the global force phase portrait. The global phase portrait system three, which means that is, if x is 0, y is 0. Then describe how this limit. What happens with the solution. Depends on your initial state. Okay? So which means, you know, if you start here, this is what you're going to do in the long run. If you start here, this is where you're going to do in the long run, we will see that when you find an answer, you will understand this. All right? So that's what I mean by the global phase portrait. Okay? And not just near the fixed points, you can start anywhere and you have to tell me what's going to happen. Okay? So there are several steps and into all of this piecing together all the tricks that we have learnt. Alright? What I want to say, missing something right now, okay? So first step is find the fixed point, fixed point x star to 0, and these are just out there. You will stay there forever. That's step one. Okay? So you find the fixed points, which means you said dx dot to be 0. And the other one is 2y times. Okay? Now, to solve this system of equations, okay? So if you're not careful, you can miss some solutions. And if you're not careful, you can do extra solutions which are not there. Alright? You gotta do this carefully. Okay? Now as I said, I don't have much time, but basically, you know, read the notes here, okay? Okay. Make sure you do not miss deletions and all other solutions which are not there. Okay, you have to do this very, this is middle school algebra, but you'll do this carefully. Alright? So I'm just going to write down the answer. Okay? Fixed points are, there are four of them. There is 0-0. There is 0 to N 11, there are four fixed point. Okay? So, okay, so I just want to say that, but that's fun to find the fixed points. Second thing is that you have done before is linearize around each point. What do I call it? When you do one of these problems, this takes a long time because a lot of steps. So if I'm going to do this in an exam, okay, depending on what, maybe for one part question I may say, OK, find the fixed points, do the linearization, tell me what to behave in a ij fixed point. For another question, I may give you the fixed points. I might tell you what the linearization looks like and ask you to do the rest. All right? Because I know that each question to do everything takes time. So linearization around each point. So let see X dot is, let me just copy this again. Okay, so just expand that 3x minus x squared minus 2x y. And I'm going to call it F. It's 2y minus x, y minus y squared, and that's CI. Okay? So when you want to do a linearization, you have to study the Jacobian J, which is this matrix, which is f x f y, g x g y. But you do, and you get f x is t minus 2x minus 2y. F y is minus 2x, and g x is minus y, and g Y is correct. So then you have to do it for each case. I'm going to just run through this quickly. So this is the first fixed point. So near to here. So j at 0-0, you plug in 00, X is 0, Y is 0, you get three zeros, two. Okay? So then you have to find the eigenvalues and eigenvectors, say use Mathematica in this case actually the diagonals, what's easy. So the eigenvalues are 23 and the three, the two goes with the lower one. This was the eigenvector. And this is the eigenvector. You can use Mathematica if you like. Okay, now both eigenvalues are greater than 0. So the solutions are going to go out. Ok? So if you draw it in the uv plane, everything has been to leave because of that. So it's what we call a source. How are you okay with this? Red, this the linear analysis, what? You have to do it around each point. Let's do near 1-1. Looks like it's going to plug 11 minus one, minus two, minus one, minus one. Okay? And then in this case, you know, again use Mathematica or in this case actually turns out lambda is decimals Uh-huh. Minus 2.40.4. And this is, this is again, you can use Mathematica. Okay, this is all standard stuff. I, I, I'm not really the goal, but I need all these things to draw my global phase portrait. Okay? So one lambda greater than 01, lambda less than 0. So this is what's called a saddle. In one data, come in in the other direction, you go out, right? So let's draw the picture here, u, v. So this is the direction you're coming in, minus 2.40805. It's like this. Okay? And the other one, you are going out to 0.4. It's ok, this is 0.8 minus 0.5. And you're going out. So in general, you will look like this. This is one month. All right? Okay. There are two more. Let me just quickly write down the answer hurt. One is near 0. Ok? So if you just do it. Ok, it turns out both lambda is negative. So near 3-0, you're going in from all directions. Okay, it's what's called a sink. And then the other one was 0-2. Piece to it. Both lambda negative. Who going in again. Okay. So near one-on-one, it's a saddle. 00 is a source. Sorry. Okay. In the notes I have more carefully I back to the eigenvalues and eigenvectors. I don't calculate them, but I've written them down. Okay, so now we've done fixed points, you've done linearization. Okay? Right, okay, so this is another new idea here, what I call nullclines. And then I'm going to use the vector field here.
Competing species I
From Rakesh Rakesh December 01, 2020
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Drawing the global phase portrait for a system of ODEs modeling two species competing for the same source of food in an ecosystem.
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