Okay, So I'm going to draw a phase portrait and all the steps. So we've already looked at finding the fixed points, doing an analysis be other fixed points. That's only one of the tools. The second tool we have is using deluxe criteria to eliminate the existence of cycles. That a lot from such a big part of a thing. The main, the third idea which I wanted to as nullclines, which I just started the other day, right? And the important starting point was that if you have a curve which is F equal to zero, then one side is f greater than zero, other side is f less than zero. And you do figure out which side is which by testing, using our test point. Have you okay with that? Alright, yeah, is that okay? All right, so let's look at this example here. Nullclines, okay? Nullclines are curves where the vector field for the differential equation is either vertical or horizontal. Okay, we'll see what that is. So consider the system. X nought equals y minus x cubed. Y dot equals x minus y. Okay? So the associated vector field is is f. And the x dot y minus x cube is the I component income than the other one is the j component. Okay? By the way, what's in the notes I is different from what I'm doing today. I'm going to modify the nodes based on what I'm doing. Alright? So my goal is actually, remember the curves follow the arrows, the trajectories follow the arrows. My goal is to draw the arrows. One way of drawing the arrows is just plug-in arbitrary x and y and keep plotting arrows. But the thing is, the nullclines actually guide you more. It's, so you know how to draw arrows or F efficiently. That's really what I'm trying to do with this nullclines. Okay? So, so what nullclines? Okay. These are curves where f is horizontal or vertical. Okay? That is where F is powder load to I or j. That's what I'm trying to draw. Okay? So if you look at the my vector field, firstly, we all agree, right? This vector field is important for the differential equation. The arrows of the vector field f tells you where the trajectories are going. Depending on your starting point. It's clear to everyone that aspect, right? Because they reject the arrow, the F is tangential to the, tangential to the trajectory. So if, uh, if an arrow goes like this at some point, then I know that if I'm at this point, then my curve is going to be tangential to this. It starts like that. It's going to go in that direction, right? I know that. That's my drawing the arrows is helpful. Help us tell the behavior without actually solving the differential equation. That's my goal. It's not so easy to solve that differential equation, but I'll still be able to tell you the long term. Okay? So now let's come back here. So looking at that, where is the vector field F parallel to the j axis? What points? At what point is the vector field F parallel to the j axis? Then the I component is zero. So it's everywhere on that curve. Y equals x cubed. My vector field is either pointing up or down, right? And it'd be pointing up in the places where x minus Pi is positive and pointing down in places where x minus y is negative, right? Correct. Right. So on the curve y equals x cubed, I'll be able to draw the arrows is going to be up or down, depending on whether x minus y is positive or negative. And where does the vector field F parallel to the y-axis, but the I vector, it's x minus five is zero. The j component is zero. Are you okay with this? Right? So the nullclines are the curves y, y minus x cubed equal to zero, and x minus y equals 02 different curves. I'm going to draw the curves and then draw the arrows on those curves. And once I do that, I'll be able to draw arrows everywhere. You'll see this. Okay? So that is the clue. So for convenience Actually, I'm going to introduce so I, let me just copy the system again. X dot equals y minus x cubed, y dot equals x minus y. Just to swipe, remember I keep getting mixed up which one is which? So the x dot is always going to be called f. And the wider what I'm going to always call with the G, right? This is f x y, g x y. And my vector field F is f and g j. Alright? So every problem I do, my x dot is always going to be the F and the wide or two logarithms is just easier for me to remember. Otherwise, I have to go back and look bitches which one? Alright, so let's now try the first nullcline. F is parallel to I. N. G is correct. Okay? So let's draw g equal to zero, g x, y is x minus y, and g equal to zero is the straight line, right? So this has G could proceed on that red curve, the vector field F, as only the icon pregnant, that's going to be pointing left or right. Okay. We will see that. Okay, now, when you do that, also figured out where g is positive and G is negative. So how do I figure out which region is G positive? So one side g is positive and one side G is negative. How do I figure that out? Yeah, I was going somewhere and then plug it into your g function. And it should be a point with location, you know, precisely so that you know which side it is on, which point you recommend 05050 is here, right? So let's test 50g of 50. This is x minus y, that's five minus zero is greater than zero. So this is the G greater than zero side, and that is the G less than zero side. This is going to be useful for us. Can I draw the F on the other nullcline? You will see how that comes about. Okay? So F is parallel to the I vector when G is zero, when the j component is zero. Now let's do the other one. Then I'm going to put both of them in one picture. F is parallel to last time because I, j. So when the I component is zero, right? The eye competent goes with f. Okay? So what is F equal to zero? F is y minus x cubed group zero. And that's a curve, y equals x cubed. Let's try that. You don't have to be precise. Just get the general shape. I'm using different color just so that you know, it'll be easier to track. So y equals x cubed looks like this. So this is F equal to zero. Notice f is y minus x cubed. Again. So which side is pi f greater than zero? Which side is f less than equal to zero? So I test which point? Well, I know five, you can test 50 or any point. I mean, it's up to you. I'll just use 50 again. So f at phi of zero is y minus x cubed. So zero minus five cube less than zero. So this is the f less than equal to zero site and that was the F greater than zero. Okay? So I'm just going to be okay with this so far. I'm just going to copy these two figures here on the side so that it's handy for us. You don't have to do it. So this is, this was g equal to zero. This was G greater than zero on this side, and G less than zero on that side. And the other curve was F equal to zero, less than zero on this side, f greater than zero on this side. Alright? So now you put them two curves in the same picture. Okay? So notice they go through the origin. And so how many places? Well, this red curve, Carta blue. Alright, so you've got to get that part correct. You don't have to be accurate, but you've got to get that part correct. So what color I use for that? Okay, I use blue. So I'm going to first draw the x, y plane, but I'm going to remove it, but just this helps me guide draw it. So the blue looks like this. The red is one. Okay. I'm not going to raise the black one just because that's just the axes. Labeled them. The red one was G equal to zero. The blue one is F equal to zero. Okay? And remember your vector field is FIN GJ. So now I'm going to start drawing the arrows on the, on the nullclines. Okay, so let's start with the red one. Okay? At this point I'm going to draw the arrow now on the red. Okay? At that point, G is zero. So your vector field is just FYI. It's means it's horizontal, left or right. Okay? So at that point f is just f phi know at that point. So this is where if you look at the picture, at that point, is the F pointing left or right? If you look at the picture above it, what is the sign of f at that green point? Is that this green point is that positive or negative? You see that curve above it? Correct? So anything below that blue curve is negative f of u. Okay, with this, this is what's going to be important, right? That's why you need those F positive and negative. Yeah, okay. Everyone understand this, right? Because you'll be doing the same thing, the examples. So on the red where I put the green thing there on g equal to zero, the f is negative, so that means that thing is pointing inwards. Now. So n is going to be the same anything on that side. So all I need to now look here. This next set, green point, new green point here. Again, it's on g equal to zero. So F is F phi. So is f positive or negative at that second green point? Right? It's above the blue and above the blue is F is positive. You are located there. So it's going, f is going right, then that Alright. Anyone has a question, I can go. Yep. Okay. So let's go back to the first point here. Alright? The first point, this one. Okay? So at that point G is zero, correct? So f is just FYI. You have to look at that point, the sign of f. Okay? Look at this curve here. That is F equal to zero and anything below that F is negative. So now it's okay now, yeah. Okay. So same thing at the second green point. Anything above the blue, the F is positive. So it's pointing right? Is that okay? Yeah. So one is positive because f, right? It's positive multiple of i or negative multiple of high. If it's three, I is zero points, right? If it's minus three points left. Is that okay? Everyone, Now's the time to pass me because I'll be doing this several times. Okay? Alright, So second one now, so it's going to be the same everywhere on that. Now they come to the third red part. Okay? At that point, again, urine G equal to zero. So it's FI, you are left or right. And on that part is f positive or negative? Because it's below the blue curve. Below the blue curve, f is negative below the blue curve. That's why you have to do that before. This is crucial, without a tear could not do, do what I'm doing next. So below the blue curve, f is negative. So this point here is below the blue curve. F is negative, which means you're going left. You okay with this young man. Yeah. Then maybe there. Alright. Okay. Then there is one last piece left here. Okay? So you're above the blue curve. Above the blue curve, f is positive. So it goes to the right. So on the red curve I know where the direction of the arrows. Alright? So that was on the red curve. Okay, Now let's do only blue curve. So on the blue curve. So f is just g, j. Remember it's good to call it f and g. Remember, f goes with ij equals with k. Otherwise you could get mixed up. Okay? So let's go to this area here. I'm going to do exactly the same thing now. So now f is going to be either pointing up or down. Is G j, right? On F equal to zero, capital F is the vector field, F is G, J parallel, pointing up or pointing down depending on G is positive or negative. You all okay with this. So now let's first green point here. Is g pointing G positive or negative. So they have to look at this picture now. Above the red curve, g is negative, correct? Above the red curves here, above the red curve g is negative, which means GJ, you're pointing down. Then you go to this part of the blue curve here. Again, f is g, j, G positive or negative at that green point here, the second green point, you're below the red curve. Below the red curve G is positive. That okay? Okay. Then you come here. Now you are above the red curve. Above the red curve, g is negative. Lost your hair. You're below the red curve. And G is positive. There. All right. Okay, So these are the arrows on the nullclines. Okay. Let's couldn't find. So you're telling me where are the fixed points for the system? Okay. So if you go back, remember this is a better system. Yeah. Okay. This is here, the system. How would you find the fixed points of the system? Yeah. Off right? It's the externals here on the white dot is zero. So where F is zero and g is zero. So in this picture, the curve where the curves f equal to zero and v equal to zero intersect. Those are the fixed points. Our where both zero and G zero. Therefore, x points are the points of intersection of F equal to zero. Right? I'm actually going to copy this picture again so that you have it in a notebook because I'm not going to draw more things. Alright? So I'm just going to copy this so that you also covered in your notebook. Okay, so I have to make sure this intersection looks like this. The blue is the y equals x cubed. Okay? I'm going to first, it helps me draw, so I'm going to draw the axes first. Then I'll erase it. Okay, and then the red one. Don't make it to swallow. Otherwise, I'm going to be writing many things and not be able to fit it in. Okay, and I don't need the axes. And the blue one was zero. Red one is G equal to zero. Okay? And I also want to draw the arrows down and left, right and left, right and up. Okay? Hopefully there are no errors. If you notice the alternate. Because you know the F sine of f and g, The alternate, right? So it's okay. And then these are the three fixed points, P, Q, and R. This is where the two curves intersect. So PQ, LR, the fixed points. And these are the only fixed points. Okay? Now what about arrows and other places if you're not on the red and blue curve. Okay. So I'm going to label the regions. So let me stick to whatever I have in mind. Moocs. Numbering this notice that it cuts it into how many regions? Six, right? So let's just label them and I'm going to stick to what I have in my node that's labeling abuse. So this is 12. So I can refer to it three. Okay. I didn't leave myself enough space here. I'm just going to erase this off. So one of three, or sorry. There's five. I call the 66 regions. Okay, now I'm going to draw the arrow in each of the regions, what we've reached our x and f point to. Okay, now it's not going to be 100% accurate, but the general trend will be clear. Okay? Now I can argue this, but I'm going to just show you a shortcut. But that's not the explanation for why it is true. So you pick a point here. Alright? And the arrows that go is, so look, look at the edge of that region. So you have to draw an up arrow, up or down arrow and a left and right arrow. Look on the edge of that region one. Are the arrows pointing up or down on the boundary? Everywhere on the boundary. Yes. On the boundary of region one. So it has an app component. And on the boundary of the region one, the arrow is pointing left or right. That's it. So everywhere, and this is related to f positive, F negative and so on, it's coming from that. So everywhere on the region one, the arrow will be leftward and upwards. So the net result will be it will be pointing something like somewhere in this direction in January, like that. In region one. It'll be going left a little, it will be going up a level. Okay, this is related to the FIG J part where f was. It's connected to that. You can think about it. If you think about it carefully, you will see why that is so that isn't region one. Region two. Same time. This one is easier, right? You're going left and you're going up. So your arrow will be going like that. Okay. Let's look at Region II. Who is time? Do we have 16 min? Okay, region three. On the edge. I'll be going up or down on the boundary of region three. Boundary of region three or four down. Everyone. Okay. Down, right or left? Right. Look on the red ones on the boundary. Okay? So you're going right. So the net result is it will be pointing somebody like this is a downward component and a rightward component. You can be sure it is not going left. Neither is it going up. Right? It's not doesn't have a leftward competent or an upward component. Okay. So what about regions for? I think that's clear enough. So like this. Then here in them, 5.6 also we can see, I'm sorry, there's very little space here, but it's not going okay. And here is, you can see how you draw these figures. Space here for you to write these things. Okay? So let me write this without even hybrid in your notes also. Alright, so I'm gonna write how I did it. So the nullclines split the x-y plane into six regions. Redraw F in each of these regions. In one, on the bum, on the boundary of one, rose, bind up and left. So they're pointing left. So winds up and left. Okay? You can do the same for the other five regions, okay? By examining the arrows on the boundaries of those regions. Okay, I'm writing it so that you have loads. And you can also watch the video. I know you don't have colors in your things, but it's a little hard to draw her. Are you okay with this? Because then I'll go back now know that I've drawn the arrows. What conclusions can you draw? Let's look at that. Okay, I haven't yet done one part yet. I haven't done the non-linear analysis fixed points. I'm going to, but let's see if we worked so hard to draw the hair that we're getting anything useful out of that. And you'll see you get you all have it done, right. Take a look at the picture. Can you tell me anything about Region five, e.g. in region five, the yellow is pointing at here wave from Q and towards our yes. But it's a general direction it could go out calls. It's not clear. It's going towards are in region five. Certainly, it's not pointing towards Q. Correct. So maybe it's going towards art, maybe it's going somewhere else. Who knows? Maybe it's going off to infinity. Or is it What's happening in region five? Yeah. Region. Region. I mean, it's a trapping region. Yes. Can any trajectory leave five? You start in five, and you get out of five. All the arrows are pointing inwards. You can't leave for you starting five-year stuck. Okay. So what does the PointQuery appendix and I'll write all this down. What does the PointQuery benthic some theorem say. If you start in five, you can't leave five. So where can you go? Some fixed point there were bid that you are approaching, either approaches a fixed point R or there's a cycle there inside? Correct? Since either approaching a fixed point or there is a cycle. Okay. Now, whenever there is a cycle, what happens to things inside? Okay. I'm just going to yeah. That is another thing, right. Is there any way for it to come around in region B can then recycle. Everything is moving to the right. In the cycle at some stage you to come back left. That's a very good point. I was going to make a different argument, but that's really a good point. You're always moving to the right, a right enough. And the cycle, you have to come back the opposite way sometime. They can't be a cycle in five. So what is the, what is the behavior of orbits? Start in five. I saw it projected start in five. Where are they going to go? We're going to go to our you see that? I'm going to write all this down and maybe next time, what about six? Is that a trapping region? Everything leave it. Again by his excellent argument. They cannot be a cycle in six. You understand the argument, right? If you're always moving to the left, you can come back and join it again. Let somebody you have to code. Right? So no cycles inside, everything is heading to. Okay. What about region for? Why go off to infinity? Well, one way is like position with our clients and chemo means move upward and downward. Again, we can actually make a traffic region there, even though, remember I said, in trapping regions have to be bounded regions. But actually you can make a trapping region there. Let's use a different color. What color? Actually, let's, yeah. If I draw a line like this, right, all this later again, let's look at the region where I drove this orange lines and then the red and the blue around four, right? If I pick a point in four, I can just move those orange lines higher-end to the right more. So on the orange line, which, which direction is the arrow pointing? Can be anything. Read the orange lines. On the horizontal orange line. Can anything go above it? No, because all the arrows are pointing down in four on the vertical orange line, can anything go to the right of that? So you can see you can create a trapping region no matter where you start in for you. Okay, with this? Repeat, I can repeat that argument. Correct. So four is again a trapping region, then again by the wonderful argument of yours, no cycles. So anything starting in for anywhere, it's heading towards our correct. Similar thing in to those of you on Zoom. I just everything okay. Is that clear what I'm saying? I guess so. Okay. Same similar thing happens in two. Same argument. I can draw those orange lines there. What about Friesian? One, start and one? Will you stay in one place? They can go to infinity in region one. No, right, again, I can play the trick with the horizontal, the orange lines. You cannot go below the horizontal line, orange line. And you cannot go to the right of that horizontal orange line because of the arrows in region one. So you are trapped. But so actually I need to, well, you're not trapped so you can go somewhere, right? You can e.g. here you could leave. You're going up, so you could actually enter this. You can enter that region. There's nothing stopping you, right? So actually I have to draw my extend my horizontal line further intersected. I need to redraw the all of this later, but I'm just trying to tell you the arguments that I'm thinking off. You have to go all the way up there and here. Sorry. Okay, So what happens from trajectories with Spark and region one? Where are they headed? One of the other trapping regions, either going up or chain P are entering the trapping. They may be entering two. They may be entering 65 or four. Correct. Okay. But once we they enter there, you know what happens? But they could also be heading directly to Q. All right? Nothing stopping them from heading to q 0. So this is where the linear analysis near Q would be important. If I do a linear analysis near Q. Right? And then I figured out, suppose everything is leaving queue. Then of course I know it will not go to queue, it'll go to the DNR. But actually as I'm going to be phi, remember correctly. I have it here. Linear analysis there are q is actually one trajectory comes into Q and the other ones are leaving. So there is one trajectory from Region one going into q, one from three going into Q, and everything else is moving away. Okay, So that last bit was still done, but Time get 1 min. So you understand, see how arrows. So that's why it's making all this effort of drawing the nullcline than the arrows. You can see how without solving anybody, you can actually predict. I'm not saying all with it's so convenient. But some cases you can actually do this. Alright? So I'm going to work on this again, I'll write this more, but I think you understand the idea now. You have to figure out trapping regions figured out, and also sometime you probate extra lines. You don't because you have the left and right up and down the road with light and see it. Okay? This is all there is in all the homework problem. This is all there is. Only thing I need to add is the analysis. They are the experts, which I'm going to do next time and read, rewrite this. Then you'll be able to draw the phase portrait. And that's it. That's the long story. Okay. Okay, For need to Oh.
Nullclines, Trapping regions II Math 503 Fall 2022
From Rakesh Rakesh November 28, 2022
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