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2 (6 points). Consider the Hamiltonian system 2y, 162 4x3 _Find the equilibrium points and test their stability by Jacobian matrix approach: Find the Hamiltonian fu...

Question

2 (6 points). Consider the Hamiltonian system 2y, 162 4x3 _Find the equilibrium points and test their stability by Jacobian matrix approach: Find the Hamiltonian function and use it to sketch the phase portrait. Describe the curves that contain Homo-clinic orbits.

2 (6 points). Consider the Hamiltonian system 2y, 162 4x3 _ Find the equilibrium points and test their stability by Jacobian matrix approach: Find the Hamiltonian function and use it to sketch the phase portrait. Describe the curves that contain Homo-clinic orbits.



Answers

Sketch the phase portrait of the given system for $-2 \leq x \leq 2,-2 \leq y \leq 2 .$ Comment on the types of equilibrium points. The system in Problem 9

So here we have a question. It's asking us to recall the solution for the system. Ex prime is able to x minus y squared. Why? Prime sickle to four, You know. Excuse me, Siegel. Why? Times nine X minus four. Anybody seen this? This was the system in question three. So we know maybe already that the Jacoby in, uh, X and Y is equal to the matrix. One negative, two y nine. Why? Nine X minus four. And we also know that we have several equilibrium. The 1st 6.0 which had Eigen Vector. I am values, uh, Lambda Wankel one Lambda to equal negative four. And then we had points 4/9. And what's her? Minus 2/3 which had Lambda equal to positive. 1/2 plus three minus the square to 31/2 I right, so we said in problem three. That 00 was gonna be a saddle and, uh, that these other points, we're gonna be unstable spirals. So we want to sketch the face point. And there's a number of ways that we can kind of get a sketch of it. Ultimately, we're gonna use technology toe. You make the space plane. But before we get there, we can get kind of an image of it by, um, looking at our Aydin vectors for our cell. So there are There are a number of other ways we can do this. We could look at our know Klein's and kind of get a sense of which direction our field is moving. Um, in each of the regions that are no claims divide are playing into. But to get a many units, a better sense of how things air curving. We can look at the Eigen vectors for our saddle point. Right. So, um, remember, we're here, and we just write that are Jacoby in at 00 is going to be the matrix. One 00 negative four. And really, we're gonna do this is to find Eigen vectors, reach wagon values. Remember that. The solution. Ah, if we're talking about just the linear system and that's how we're approximating this solution to the system when we're close to the 0.0 the solution to that is going to be, uh, some constant times the Aydin vector. Uh, maybe right. Also a one B one times e to the Lambda one T plus are other solution See, too a TV too. Times e to the lander to t So, uh, if more in the case where c 20 this term would drop out and what would be left with was just a solution that, um travelled in the direction of a one B one. Right. This is just a constant multiplier to the vector. A one done one, So we would be moving in that direction. And so if Lambda One is positive like it is in this case, we would be moving away from our 10.0 I would be unstable, right? Because he to the Landau won t you would tend towards infinity. On the other hand, in this case, we have land a tomb, which is negative. So if c one waas zero that trying to drop out would be left with, just see two times 18 B two comes E To lend a duty into the land of T T is gonna tend toward zero. So, again, this is just a scale or multiple of the vector 80 b two. But now we're tending towards zero. So we're going to be approaching our origin 00 So that being said, um, what this means is that if we find the Aydin Vector four Lambda one, that will be the direction in which solutions will proceed away from the are fixed point that there will be are unstable manifold. And then if we find the solution for the idea vector for Eigen value when the toon could think it a floor that will be our stable manifold, That'll be the direction that solutions proceed towards our, um six point. So let's get started on that right. When Lambda One equals one, we find her Eigen vectors by taking your Jacoby in minus the identity matrix times lambda. So our jacoby is 100 You get affordable, subtract are a convict I in value along the, um that diagonal and let me just I could just pray. It here clearly is your one mass 10 and negative four minus one is gonna be negative. Five. And remember refining Eigen Vector. That means de Smet matrix times the vector A B is gonna have to equal zero vector. So this clearly means that negative five times B equals zero, which would be equal zero so that because, um, it can be anything on. We can just choose it to be a convenient A Because any, um, any scale multiple of this vector is going to be a mm is gonna be a, uh, in vector for his Eigen value. So we can say 10 might be a zero a is non zero. This is a wagon vector for land equals wealth. So again, remember, this represents the unstable direction of our solution. Furthermore, weaken dio Linda too, equals negative Four and again, we have one minus negative. 400 negative four minus negative four, which we now goes down to, uh, 5000 Now, if this matrix again we're gonna do is find the I convict your by taking 5000 and multiplying it, baby to give zero vector. And this tells us that five a has two equal zero. Right? So that means a equals zero. And now we're in the same boat. Just flipped around where a zero and beacon be anything. So we'll say B is one on this is the agon vector for the Eigen value. Negative for So this is our stubble direction. and this appears are unstable direction. So that means for our sketch. If I draw my face plane right here, we're doing with the 0.0 Um, we said that the direction 01 which is, you know, directly up and down. It's R Y direction. That's gonna be our stable direction or stable manifold. So solutions starting on 00 are going. Teoh approach our students starting on the Y axis. Excuse me, are going to approach 00 So we'll have solutions like that. This is our stable, beautiful and a one of the X axis on the parallel to the vector 10 We have our unstable direction, so solutions will tend away. Ah, like that. So we can see that solutions are going to start, um, following the stable manifold and then begin to curve away towards the unstable Beautiful like that. Um, and we can say that this will happen at least close to our equilibrium. Point are fixed. Point doesn't get further and further away. This gets to be worsening approximation because this is not a linear system. Um, like the way that we've approximated it. But we can further improve our sketch by recalling that we have two other six points for it. We have 4/9 plus and minus 2/3. So 4/9 and 2/3 might be here for nights and minus 2/3. Maybe right here. And we're gonna be unstable spirals. And now we can see from what we've just shown from our stable and unstable directions that are unstable. Spirals were going toe proceed in this direction, right? They're gonna proceed. What is that? Counterclockwise. They're gonna proceed like that. So, um, we might see socials like this, Um, not exceptionally Get drawing things like this. You might see solutions like this. Remember, solutions aren't going to you ever overlap? Um, so we might might see things like this. And down here, we would see solutions proceeding like this. We see now, this is our clockwise direction. Ah, and you can tell from what we've determined about stable and unstable directions that, um, kind of what way are spiral is going to be spinning. Um, So this is not the most effective way to make a good sketch of, um, the face plane. Right. We have really effective technology that could do that. Well, now, but This is a nice way to you. Get a quick understanding what this is gonna look like. Especially if you don't have technology at your disposal. You can still use this technique, Teoh. Get a quick understanding of what? Um, what the face plane might look like. But if you want to use technology to get a better example, we can We can do that, huh? Here is a program called P Plane, which was originally written in Matt Lab by some professors at Rice University. Um, and then has been since released in Java. Um, and we can use this to find her face plane. So what I'm gonna do is just type in, um, are equations explaining why prime breath and we'll go from negative attitude to invoke the X and Y directions. So now I'm gonna graph my faith plane, and, uh, we will see, um, the solutions that, um this is a solution. Similar toe what we're discussing before. So let me also include my male clients. McClane's air here, right? And this 0.0 is where my cursor drink. Now, that's are fixed point, which would be a saddle. Right? Um, I also went to just show the forward direction of my solution. So, um, let's see. Um, so we see that if we started here, we would proceed in the, um, direction that we said was the stable direction. And then we would go in the unstable direction, right? Uh, that's all of the solutions. Have that same pattern just a little bit, um, shifted because they're not linear. And then also around our to spiral fix points here, we can see that we have spirals which are going in the direction that we kind of predicted based on, uh, what we found about the, um, saddle point like here. Eso we weren't exactly right. We kind of expected more solutions to be going in the positive extraction. But it looks like all of them eventually curl around in this spiral kind of way and lead to negative X direction. Except for you can imagine solutions that start exactly on, um, the X axis. But in any event, this is a nice tool to visualize our face plane. It's a fun thing to play around with, Um, and now we have a pretty good idea of what our Facebook is going to look like, um so, yeah, if you have the chance to Teoh, use this kind of thing. That's helpful tool in solving these kinds of problems. But if not, there are ways to you making all right sketch of what? Our face we might look like just using the Eigen vectors in Agon values of our equilibrium.

So in this question, we are asked to revisit a system that we had in Question seven, which is ex prime is equal to X minus two y plus five x y and why prime is equal to two X plus y. Okay, um, and were asked to sketch your face plane for this. Ask them. Ah, and remember, purity went through the system, found that the Jacoby in is gonna be, um, one plus five. Why? Negative, too was fun x two and one and that we have some equilibrium. The first is 00 The origin with Eigen values one plus three minus two. I, um and that means that this is gonna be an unstable spiral. And then the other fixed point here is, uh, NATO's Excuse me, 1/2 negative one. And that has William to equal to negative three halves, plus or minus the square root of 29/2. And that means that we have a saddle here. One of those is positive. One of this is negative. And of course, they're both riel. So we have a saddle eso weaken. There's a couple ways that we can approach making this fix the space plane. Ultimately, we're going to use technology. We use a face plain illustrator to find the full picture. But we can. There are some things we can do without technology that will help us, um, get an idea what the baseball is gonna look like. And the one that I wanted to talk about is using our know Klein's toe illustrate what this is gonna look like. So, um, are no clients are going to win Ex Prime equals zero. And why practical zero. So an ex prime equals zero. That means that X minus two. Why was five x y equals zero and, ah, when why Prime equals zero. That means that excuse me, that you read the scale that means two x plus y equals zero. Sorry. Got lost in my notes. And so we can these these air equations for lines in the X y plane. Um and so anywhere along this line are, uh, you know, the delta X Delta y vector is going toe have no change in X right. It's only going to be a change in why? So it's gonna be a straight up or down, and on this no climb, it will be only, uh, left, right, Right. I'm so you can kind of piece together with this image is gonna look like basing these no plan. So the easiest way to do this would be toe to solve both of these for why? So, um are why no client is pretty easy. We have, uh, two x plus y equals zero. That means that why equals negative two X, right? This is just a nice line. That is, they were pretty used to um uh, the X No client, on the other hand, is a little bit more difficult. We have X minus two y plus five x y equals zero to sell for. Why? What we're gonna do is, um, see that X is equal to to my minus five x y benefactor out of why to yet tu minus five X. Uh and then I'll say, excuse me, getting in myself. And now we can see that why is equal to X over two minus five X. Um, cool. So now we have two lines that we can plot and use that Teoh get an image of our face plane. So one of these lines y equals negative two x is quite simple. You know, all underlying this and written dry it in red. Um, it's just this line. Right. Um, this other one y equals x, uh, minus X over two minus five X eyes going to look a little bit different, but we we do know that this has a nascent toot at 2/5 right? So has a nascent it here. And we see that before 2/5. Right when exes is less than 2/5. Um, well, I guess easier. Easier to see when X is greater than 2/5. This is going to be negative. So I'm gonna draw this in green. So it's going to look like this. Buy something correctly. And, um, before that, before defense, it'll look something like this. Right? So you have, uh, to points were these curves intersect, and those are are fixed points, right? And that this assures us that are drawing is a little bit is at least a little bit reasonable, because we know that we have a fixed 0.0 and a fixed point at this point, which we might call 1/2 negative one that that looks reasonable. So that's good. Now we can use these no clients to kind of fill in what we expect. Uh, what we expect this to look like. So remember, on our red No, Klein, that is where the changing why or why prime is equal to zero. So any vector that is on the final climb is going to be, um, just straight up and down or especially straight side to side. It's only going to have change in the X direction. Um, and so we can think about a point here, right? In this case, ex, um, we think about the change in X here. If we have a negative accident, a positive. Why we have this term is negative. This term is negative, and this term is negative. So these are all going to be negative. So any vector it starts a long the y know climb is going to only move in the extraction on that means, uh, well, that means that, um and because we know that when we play the point in from this quadrant into our our equation for ex prime, it's negative. We know that, uh, it's going to be in the negative X direction. So we can continue to do that for other parts of our y know. Client, for instance, Here we can What points in between Where X is between, um, zero and 1/2 and wise between zero native one. And we see that this is positive, right? So x x is positive. Um, why is, uh, negative one? So why this term here? Negative two times something that's between zero native one is also positive. Um, and we know that it's going to be these two. Some together is gonna be greater than five x. Why? So we know that this is going to be positive. And then again, when you cross over that line, we know that these we're going to be negative like that. That's how I kind of feel in this caress, um, now, on our external climbing over vectors air going to either pointing up or pointing down right, because the change in X zero so we can feel that in. And this one's gonna be a little bit easier, because what we're comparing it to is just the the equation for why prime, which is just a nice linear equation. So, um, if why is greater than negative. Two X right. So that that would be above our red line here. So if y is greater than negative two X, Um, that means that this is going to be positive, Right? So in this, when y is greater than negative two x are vectors on the X. No, Klein went straight up. That and then in the opposite case, they're gonna want straight down. I think this, uh and in this way, we have kind of filled in What are baseline might look like, right. We can fill in the areas in between by, you know, just composing these vectors, right? We have a negative x negative. Why in this region, eso the vectors here are gonna be pointing in that general direction in this region, we have positive Y negative X. So are vectors are gonna look like this in this region? We have negative. Why positive acts that we're gonna look be looking like this. And in this region, we'll be looking like that. Eso we kind of fill in What? The citizen. Now we can kind of see this unstable spiral At 00 we can have a new idea of which way it's going to spiral out, right? It's going toe spiral out like this. Probably. Um, we also see that the saddle point we can see kind of the stable and unstable directions, right are stable. Direction is gonna probably be coming in this way and are unstable direction. Looks like it'll be going out that way. Uh, so we can use that information to kind of direct our or a graph here, and we can try to plot some. We can plot some solutions a little bit. Uh, that or maybe gonna look like this and like this. Um, yes. Oh. So once we have this, we can, uh, get an idea of what our face Porter's gonna look like and how this is really crowded because this is a drawing and not a nice computer output. So let's switch over to now or computer output. And this this is going to look like this. So I'm using P plane eight, which is a camp creator made by some professors at Rice University. Um, which is used was originally published in Matt. Lab is now available of Java. Um, and we use it to fine face planes for systems international equations so I'm going to have it Graphic face creams you already saw. But we see now what are know Klein's look like, right? Are the red No, Klein Is the this why? Excuse me? The external clients. So any at any point on the X no climb this red line is going Teoh be associated with a better. That's playing straight up straight down. And then this yellow line was just harder to see is the, um why no clients or any point on this wine offline is gonna be associated with the vector pointing straight left or straight. Right. But we can fill in some solutions here, and we see that are sketches pretty good. We have this spiral moving in the direction that we expected to. And also we have a, um the saddle point, which has the stable directions that we expected it to, and the unstable directions which we expected at the help eso We can fill our face playing up with solutions. Teoh get a nice idea of how things flow. And I'm I'm just doing forward solutions. But this calculator can do forward and backward solutions, um, to see stable and unstable attractors here So, in any event, um, this is a nice calculator. Teoh use, uh, have visualized base points like this. But if you don't have access to something like this, we can still get a pretty good sketch from a face plane just by, um, using the no clients. And there's other ways we can. We could have used the Eigen vectors for this saddle point to find the stable and unstable directions. Um, and that would have informed a little bit about and from us a little bit. About what, uh, how the spirals going to spin as well. So there's a lot of ways that we can sketch this face plane, but of course, using technology like this is is the fastest and easiest.

So in this question, we're asked to revisit problem for which remember was ex prime is equal to X plus three y squared and why prime is equal to y times X minus two. And now the goal here was going to be too Ah, sketch the face plane. So I'll just remind us that the Jacoby in in terms of X and y physical too one six y Why expense, too? And we found that we had just one equilibrium, which is at the origin 00 And we had Lambda wanting one lambda to equal to negative two. So he said that this is a saddle, right, Because you have a positive and negative rial Eigen value. So there's a couple ways that we can do this weaken find our, uh, Aiken weaken skipped your face playing a number of ways um, one would be to use are no clients rates, if we have are playing here, are no Klein's or wherever. Ex prime in white primary equals zero X prime is equal to zero win uh, X plus three y square in secret zero. So that means this is equal to zero win. Um, X equals negative three y squared. So this is a ah parabola in, uh, uh, kind of sideways rape, Um, on the other. Now, Klein is when Why Prime equals zero and that, remember, is equal to y times X minus two. And now this is you go to zero whenever y equals zero r X equals two. Um, so I'm gonna draw my ex. No. Klein and red like this. So that's when X equals negative y squared. So that's going to look like a sideways problem. Kind of like this right now. I'm gonna draw my Why no climbing green. And these are the lines y equals zero, which is just our x axis. And, uh, X equals two, which may be over here. Right. Um, now, uh, are only fixed point is when these two know Klein's meet the both the red and the green. So we only have one No client at 00 Um, and we can kind of get an image of what it's gonna look like by thinking about Ah, where no. I'll back up a little bit and say, on our external Klein, we know a motion of particles going to be either up or down, there's gonna be no movement in the X direction and on our y know climb. We know it's either going to be to the left or to the right is gonna be no motion in the Y direction. So that makes it a nice first of plot. Um, t find some vectors on these No clients. So if we think about vectors on the X, No climb. Um, saying maybe Ah, here, right. We think about here wise, positive and X minus two is negative. Eso their product y times x minus two is gonna be negative. So our direction in the Y direction is gonna be negative like this. And that's going to be true. Any for any, um, vector on this external klein, uh, in this quadrant, if we flip to the other quadrant, receive that any, uh, solution on this, uh, external Klein like here, we would have a negative. Why? Times a negative X minus, Chief. So this would be a positive moving the positive y direction like this. We can plot a couple of these vectors like that. Now if we moved over to the y know client, we want a plot, A solution right here. In this case, X would be less than, uh, three times Y square right X is less than zero here. X is less than three times y squared Nick s. So, um, that means this is going to be negative. Our ex prime is gonna be negative, so we'll be moving even negative X direction like that. And, um, on the other side of this red curve will be moving in the positive X direction this on. So in this way, we get kind of an image of what this is going to look like. Um, we see that in this quadrant Were probably, uh, we moving this way. Um, like this here will be moving like that. Um, this gives us kind of an idea of what, uh, what will be looking at when we sketch our face plane? Um, cool. So can continue to fill in that face plain. Another thing we can do would be to look at this fixed point looking. It's hiding values. So we have the fixed 0.0 and the Jacoby in at 00 is going to be 100 negative, too. Now, to find its Eigen values we need to consider each other, as I find it's like in Vector succeeded me. We need to find each other's agon values. He said that Lando ones equal one. So to find the ideal vector for Eigen value one, we subtract the Eigen value along the diagonal. So you get one minus one and negative two minus one. Um, and that times the Eigen vector A B is gonna equals zero vector. So this matrix becomes 000 Negative three right on. And from this, we can tell that negative three tons B equals zero. So that tells us that b equals zero. And that means any vector, any scaler, multiple of the vector a vector 10 is gonna be, ah id in vector for well, Wendi equals one. So this is going to be one of our Eigen vectors. The other four wind equals negative too. We find that we have the matrix we subtract negative to from the diagonal of our Jacoby. In we get 3000 Now, this times the Eigen vector a B is going to equal zero vector. And from this we see that three A equals zero. So we have the This tells us that a itself equals zero. So we have the vector 01 kind of like before. So what this tells us is the stable and unstable directions of our settle. Right. Uh, if if you recall in a linear system, our solution is gonna look like this X why, as a function of t is going to be, um see, one times the Eigen vector a one B one times e to the lander won t plus C two we have the same thing the inventor 80 bto e to the land to t So in the case where lambda two here is negative and Linda one here is positive. Um, that means that when se see 20 this term drops out. This e to Orlando won t is gonna proceed to infinity. And that means our function is just going to be a scaler. Multiple of the Eigen vector a one b one. So because we're proceeding to infinity along the direction anyone be one we know this Aydin vector is are unstable direction. So along that vector will be proceeding to infinity and in the same way in the other directions because of under two is negative. This proceeds to zero. This approach is zero and forward time. So say, see, one was zero. We would see that our solution would approach 00 in the direction of a TV to our second Eigen vector. Um, so that tells us that if we plotted our ah solution here when we had our fixed 0.0 not know that in the direction, um, 01 which is this right? The 01 is along the y axis. That would be our stable direction. So solutions would approach our home solutions would approach are fixed point in that direction. And then, uh, on the other direction 10 which is our positive extraction solutions would move away from that neck. A. So our solutions will look something like that, right. They would start following the stable direction or stable manifold and then curve away to follow the unstable. So this is what we find. Our solution will probably look like, um and that that matches with what we found here, right? We could see that solutions well, tend towards the fixed point and then move away in the unstable direction. Of course, as we zoom out that gets gets to be a less good picture, because this approximation is only good when we're close to our explain. But as we move further away, the approximation is less good. But having several that now we can use technology to get a real portion of the base plane. What I'm gonna uses P plane eight. Yeah, this is not a picture of this problem. This is a picture of the last problem. But people in eight is something that was written in Matt Lab and then also produced in Java. And it is it was produced by professors at Rights and Rosie, if I remember correctly, Um, and it's really helpful tool for, um, doing this kind of thing, finding face points and such. So we seeing. First of all, this is very close to of the diagram that we we drew. We see that are no. Clynes are in the same show, which is good on. We see that are energizer are falling mostly the same shape. So we have solutions like this. Um, like this around our fixed point here, which is gonna be Where are no clients intersect? We can see that we have this saddle kind of shape that we were talking about, Um, where in one direction, we have a stable manifold, and then it curves away to go this unstable, beautiful like this eso if we feel in our portrait with solutions and I'm just ah, graphing the, um, forward solution. So this is all in forward time, just by the way, uh, but we we get an image of what? This, um, faced with what solutions in the space plane are gonna look like, Uh, so, yeah, that's that's how we're gonna It's just a couple ways to illustrate a face plane like this. So you can, uh, use the no clients you can use the Eigen vectors for are fixed points. But then the easiest and best way to do it is gonna be to use technology like this, just toe produce an image of the face point, so you can see it totally

All right, so we draw face portrayed for the matrix of Given. First, we gotta find the equilibrium points, and we do that by first converting into the respective equations. So it's two x plus three. Why? And minus X minus two. Why? So you are finding the equilibrium point? Um, a trivial equilibrium point for this one is actually zero comma zero. And you can see that because it's both X and y are zero. Then both equations are equal to zero at the same time, which is the condition of being an equilibrium point. Um, I don't think any other point works. So equilibrium point is zero comma zero no less star sketching with face worker. Oops. I was terribly not straight line. Who? Okay, I'm so sorry. We try something else. I figured I could draw half a line. Street shouldn't be as hard. All right. This is probably the best incredible to get a while. All right, so we know that this is the equilibrium point where we're gonna be focusing at now and find Eigen values. So for that, we go back. The matrix that is aimed is equal to 23 naked of one and negative, too, to find the Eigen values. Behalf. It's a practice with the identity matrix Times lambda, which is Landau 00 lambda. And we get this important matrix, which is tu minus lambda three. Yeah, minus one and minus two minus Lambeau. Because this is important. I'm gonna circle this, they have to come back to it. And then we have to find the characteristic equation, which is the diagonals more multiplying each other. It's a practice. So this case, it's to minus Landau times minus two minus lambda minus three times minus one. This becomes two times negative. Two is negative. 42 times minus. Lambda is minus two Lambda Negative Lambda. Minus times plus two people two plus two. Lambda Native, Longer time thinking love is gonna plus Lambda Square minus native. Three sets of plus And we just said this equal to zero These bold cancel out This becomes negative one and this state's latest. So we have Lambda squared minus one equals zero, which implies that land, uh, this equal to plus or minus one. These are two Eigen values. Lambert is equal to one Onda Lambda is equal to negative. One probably didn't give too much space for this one right here. We can part to shoot them later. All right, So once you have your own values, how do you find Agon vectors? It's not too hard, actually, but it does take a little bit of work. You go back to our special matrix right here and substitute the Eigen values a tu minus one. This one three and negative one on minus two, minus one is minus three. And we have subtracted with all right. Not subtracted. Multiplied. If they're Eigen Vector and you have to get 00 for that to be a writer. Victor. Yeah, We've finally been vector figuring out what number does that for us. So we have, well, playing them out. We get a plus three B p is equal to zero. Negative. A minus three B is equal to zero. So if you look at this, you could see that, um negative three and one should work. A is equal to negative three. And be physical. 21 If you substitute these values, you can see that that should give you zero. So our first I in Vector was right in red is negative. Three and one are sinking. Hiding. Victor, You know how much space Weird fire who does not that bad. All right. And we do the same thing, except we have to now substitute with negative one. So two minus negative. One would be three three, a negative one, Onda. Negative two minus negative one, which is negative. Two plus warm, which is native one. Same procedure. Um, we get three A plus three b is equal to zero and negative a minus. Negative. Right are just negative. Eight night minus B is equal to zero. Um, for this one, let's see if we have a is negative one. This would be one. And if we have B is one B minus ones, that b zero. All right, So for this one, the idea of value dying vector is going to be negative. One and one. All right, so we found our Ryan makers. Cool. Aren't too shabby. Not too shabby. All right, so we know our equilibrium point, and we know our wagon. Beckers. Um, So you have negative three and one for one. All right. So negative three in this direction. Some making these points show extraction going. It's a negative three on day one, so I'm going in this direction. All right, so that was just show you what? Directions. I'm gonna race that because it's gonna be really messy. All right, um, that is our for a standing. Becker. Um, this Eigen value was positive, so that means it's going to be in this direction. All right. Um, the next one was negative. One and one and clearly to show you what direction I'm going in. Negative one on one. Um, it looks like this is going the same direction. Except it's not as aggressively in the X direction. So I guess it would be a more steeper line. All right, we can. Rather before I do that, let just quickly note this is our again value one. This line was for one, all right. And this one is three times more steep. Then the line, we just true it was my wow. That's so bad. All right, that's the best I could do. And this was or I could value negative one. Since this negative, this will actually be going in this direction. Two words, the origin. All right. That tells us a lot of information. Actually, So, since this is what we call a saddle, we don't have to be too fussy about the strength of the currents. We just have to know that are struggling in red. If I drop a stone here, this will dominate it. Until I get to this, I Oh, God. Uh, this will dominated until I get to this Eigen vectors region which will start dominating. And over there, if I more over here, I will be going in this direction until I reached the second value. Just push me in that direction. Hopes me going more this way and that I turned that way. I'm over here rather if I'm over here, uh, we pushed here until that I get in, Victor gets stronger. I go in that direction. Yeah. This is not going to be the most prettiest various portrayed. But I hope you get the point. I'm Iraq and you drop me to the river over here when I first load that way and then go this way because then it gets too strong for me to fight that current. Yeah, so I didn't mean these currents are not supposed to touch each other at all. I'm just a botanic artists, which is why they touch each other. But this is what a general phase border would look like. Um, you could use some software, like wolf Mouth or something to get a Niedere portrait. But the basic point for this portrait is that this is a saddle point. Onda, Uh, this is what would roughly look like if you if you have the directions off your again vectors hammered out properly, then doesn't really matter because, um, you could figure out directions to draw them. So, yeah, that's the important thing. All right. Thank you.


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