 We're going to talk today about face portraits. I mean, we've been talking about them, but besides just having the nice pictures, what kind of things can you learn from studying the face portrait of a dynamical system. And Monday, we're going to start talking about some control theory. And again, I don't expect anybody has any prior experience with that, so we'll go kind of slowly. But let's see. I have one handout today, which is basically the first few pages on this link, it's called our PD article. So I didn't print the whole thing, but pretty much that's all you need. And by the way, I have some old handouts. I prepared, I made lots of copies, so I guess I won't do this again. We can find them here, right? But just in case you need some extra, before I send them to the, yeah, we don't want somebody homework as a handout. But anyway, so after class, you can pick up some of this. Let's see, I have another handout that's a Colquim talk that we're having tomorrow. If you're interested in some more kind of models or applied math, that would be a good place to see. So anyway, let me come back to this Colour PD article. So we'll talk a little bit more, thank you, in more detail in a second, but I just want to show you a system of two equations, a dynamical system, continuous dynamical system, consisting of two state variables, right? That is kind of a famous model called Fitsuna-Guma model. And it's connected with the physiology of neurons. So it's a kind of a very concrete application of the study of these dynamical systems and what you can learn from sort of the phase portrait for that system, right? I mean, that system looks like just any other system, right? You have a rate of change of V that depends on V and W, and a rate of change of W that depends on V and W, right? You also see an I here, which for now just think about it as being zero, OK? So that's, or think about it as a parameter, OK? So you assign a value, right? Now, so I'll come back to that. But what are the variables? So the V variable is a voltage across membrane. So I don't want to go into all the details of the ion channels, but there's literature on that. But just think about it as some sort of voltage. So it has some rest value. I think it's like minus 80 is the rest value, and then it goes up from there. And W is called a recovery variable, which again, it has to do with some sort of the ion channels, potassium or calcium, some sort of chemical that allows for these channels to open or close. And so again, just think about for us is just a dynamic system of two equations, OK? And it has an analogy with electrical circuit that has that kind of diagram, all right? So and of course, you can go to P plane and just put this in and get the phase portrait, right? For it. And actually, I invited you to do that. This is something you're going to get. And again, it's a little bit busy here, but what you will see are the null clients. So think about that's the system, right? And F was a function that is a cubic function, OK? It was v minus v cubed. It's a cubic function. So how do you get the null clients of that system? You set each right-hand side equal to 0, one at a time, right? So when you set the first one equal to 0, you get w equals F of v. So it's a cubic function. W is a function of v. So it's this function, right? What's the w null client? So basically, where the second right-hand side of the w prime equation equal to 0. It's just linear, right? So that's this one, right? So what happens when the two null clients intersect? Equilibrium, right? So when you have a cubic function and a straight line, how many intersections can you get? One, two, or three, right? But in that picture, so depending on what the parameters are, you may get one, two, or three. But if this picture shows just one, so it's right here. And you could, in principle, compute it by setting both equal to 0 and solving that system. It's a nonlinear system. And the result is not a pretty number for the single fact. Well, maybe I should say if i is 0 and v is whatever it is, let me go up here. So if you look at this, just try to set it equal to 0, what you'll notice is it won't be easy to solve by hand, almost impossible, because you're going to have to solve a cubic equation, right? You're going to replace, from the second equation, you find w in terms of v, right? And it's linear in terms of w in terms of v is linear. You plug it in the first equation. You're really going to get a cubic equation in v. And you have to solve that. By hand, it's horrible, right? I mean, there's no cubic root. There are lots of cubic roots there. So whatever, it's going to be an equilibrium, right? Then what's the next thing that you do once you find the equilibrium? Stability, exactly. So let me, let's just do it while we're at it. So I'm going to have v is, remember what was it? v minus w plus i. And this one is w is, whether w prime is, hold on. Let me, can I get that? No, can I get this? Thank you. Somebody gave it to me, but let's see. So I want to call this epsilon v minus plus, I don't know, gamma. So let me use my notation for that. Just the variables, OK? So plus delta. I'll write this down in a second. So let me do this. Alpha times v plus beta minus, excuse me, plus beta times w minus delta. OK, when epsilon is 0.08, what's my alpha? 1, beta minus 0.8, and delta is 0.7? Yeah, thank you. So it's minus. Let's put all positive. So I'm going to put plus delta. OK, so alpha, beta, delta, OK? So I'm going to write this so you can recall this. So it's called the Ficcina Gumo model. It has dv dt equals v minus v cubed over 3 minus w plus i. And dw dt equals epsilon alpha v plus beta w plus delta. And let me emphasize that this, for now, just think about it as there or not there. I mean, for now, it's 0. So we don't have anything of that, OK? So we have that. And coming back to this, let's proceed. Thank you. You gave me the error when I was running, so OK? All right, so that's what it is. And I want to do it like somehow symmetric so I can see. So negative 2 to 2, I believe, is going to capture. OK, so let's see the null clines, OK? There's null clines, right? So you see where the equilibrium is? It's a number, right? Which you can find. You can by solving, by asking the computer to solve it. Or you can just try to isolate it, right? So there's minus 0.1, OK? All right, so that's OK. And then the Jacobian is this. So when you linearize, right? Remember, the actual value, I mean, the actual equilibrium is not easy to write by hand, right? In fact, so to take the left hand sides, differentiate with respect to v and w, and then evaluate at that equilibrium, when you don't have the equilibrium explicitly, right? It's impossible to do it by hand. So you have to do it somehow with the computer. You can do it symbolically. At this stage, I mean, those are simple enough that you can ask the computer to solve it and it will solve it symbolically, right? But if that was v to the power of fifth, then even symbolically, it might not have gotten any solution, right? Anyway, so there's the Jacobian and this is the eigenvalues. The eigenvalues are negative. So what does it say about this equilibrium? Stable. So you can see that by, this doesn't really show. Well, it shows it's stable, but you see it has this excursion, and I shouldn't start it with this. But anyway, now I made the mistake. But you see if I start close to it, it kind of goes to it, right? So it confirms this table. But there's something peculiar about this system. And what is that? Notice that if you are, so what can you learn from this phase portrait, besides the fact that you can see the linearization, OK? So the linearization is basically something that shouldn't excite anybody. Because remember, when you have a linear system, especially like this, right? With this numbers, it can be one of those four or five situations, right? You can have the eigenvalues to be real distinct, or it can be repeated, in which case you have two possibilities for that exponential e to the ta, right? Or if they're a complex conjugate. In this case, they look like they're complex conjugate, right? And the imaginary part is not very small, so they're not repeated, right? They're really a truly complex conjugate. So this means that this is going to be spiraling in the origin, right? But quite fast, so that's why you don't see, I mean, you don't really see the spiraling in, right? But notice that there's no straight line, there's no line that goes straight in there, meaning that there's no real eigenvector, right? That is, there's no real eigenvalue. I mean, if you have a real matrix and you have a real eigenvalue, you'll be able to find a real eigenvector. That is an eigenvector with real numbers, right? Whereas since you don't have real eigenvalues, there won't be a real eigenvector, OK? So there won't be any kind of direction in this plane that things just move in a straight line, OK? But again, this picture is like a, how do you say, horse glasses, right? It's like a very limited view of the face portrait, right? Because it's really only approximating the dynamical system in the neighborhood of that equilibrium. What do you call the horse? Blinders, OK. What do you say, glasses? OK, that's my translation from Romania, sorry. So it tells you a nature of that equilibrium, but it doesn't tell you what's really happening with the system, right? And you see, you don't have another equilibrium so that you can patch these pictures, right, and get the idea. So this is a good thing to have, but it's very little. It tells you about the whole system, OK? So the point is that you can learn many other things from the whole face portrait. And the one thing that you learn about this system is the following from this face portrait. Notice, did you see when I first, maybe I should exaggerate this a little bit so you can see it? Let me raise all solutions. And I'm going to plot this. It's kind of tough. Minus 1 to 1.5, let's see if it's. Yeah, now it should look a little bit better. OK, so notice what happens with the direction field, with the vector field, right? It's almost horizontal, right? Most of the time, it's almost horizontal. Does anybody know why? What's the fact that it's horizontal? It means what? It means the change in W is much smaller than the change in V, right? So you can see it here. Because of this epsilon, when you take a value for V and W and you plot that vector, this value, which is the length in the horizontal direction, is much bigger than this value, right? So it's going to be more towards horizontal direction, right? Because this epsilon is small. If this epsilon were big, like not 0.08, but 1, you'd have exactly the same null points, right? Same equilibrium. Why? Because when you set this equal to 0, it doesn't matter what epsilon is, right? But look at the arrows, right? So here, the picture looks quite different, right? I mean, behavior is the same. And you can see the sparring end, right? More clearly than before, OK? So that's one thing that you learn from the system is that it has that kind of peculiar feature with this parameter. So this model, just a lesson here is this model with different set of parameters is not a good model for what you're studying. It has to be a certain range of parameters, right? Epsilon has to be small so that this model replicates what you see in the nature. OK, so that's one thing that it's very... So because it's kind of almost horizontal, what happens when you are up here? Well, it kind of stays horizontal, so it kind of goes horizontal until what happens? It is the V null client. What's the V null client? What's the meaning of the V null client? What happens on the V null client? The real V is equal to 0, means the direction is vertical. So you see it goes horizontal, and then it must hit because it's a cubic, right? And then it must be vertical. And then you see it kind of stays very close to that null kind of embraces, right? How does the null client just stays with it, OK? So that's what happens if you are there. And if you are close to this, but not so close, but kind of below this elbow here, same thing happens. You're shooting horizontally, and at some point you hear this, and then it kind of... So it hugs this piece of the null client, and of course the, right? But what's the point here? If you are just... If you are in a very tiny region here, you can actually go all the way to this null client, so make this a large discursion, or not, OK? And here's what I want to show you is how this V and W look as a function of T for a place where there's this large excursion. It has this picture. So V is the blue, and red is the W, right? And again, the initial value of time zero were this and this, right? That's where I clicked, right? And you see what happens with the V. V actually makes this large excursion, right? And then it kind of comes back, and then it recovers. So in the long term, what's going to happen with this? They're both going to go to the equilibrium, right? But there is this kind of initial thing that's happening, right? So this is what's called excitability. So there's a large excursion in the voltage during this period. And then there's what's called recovery. And then they both kind of go back to zero. And you see the W variable is kind of peaks when this V variable is kind of recovering. So that's called recovery variable. And that's the time when the neuron cannot be excited. It's kind of refractory, so it has, OK. But I didn't tell you really what the excitability is, except I'm getting excited about this. So the role of I is actually the one that actually has the I, which, as I said, think about it as zero, right? When you start varying I, when you control I, that's when you are creating this spike. So let me show you back on the, it has like a, here it is, OK? So this one, OK? So what do you see happening? This cubic null client is actually shifting up, right? That's because you're changing I. I is, remember, it's added to the first equation. So if it's not zero, but it's one, what's going to be the cubic null client if I is one? That's going to be, I don't know, up or down, right? We're just going to translate up or down. The other one stays fixed, right? So you see when you kind of, your initial condition starts here, but as you move I, as you change I with time, it kind of, your point goes below that knee and then it shoots, and it makes that larger excursion. So you can see the spike here, right? Like this, right? So I kind of, well, it's just by the nature of the system that once it's kind of, it's vertical here, then it's going to be, again, this is moving, but if you kind of take a look closely at the system, and again, this is what a face porter would, analyzing the face porter would tell you is, you see, let's look at this point, right? So once you're vertical and going up, right? So by the way, why are you going up and not down here? So what's the direction, the direction field is, is the right hand side of the second equation, right? Tells you if you're going up or down, right? And if you look at the second equation is linear, right? In V and W. So you can look at the sign of that second equation of the second right hand side to tell whether you go up or down. Now, remember that second equation is zero when you are on that null client, right? Which is a straight line. So below a null client you're going to be one sign, above null client is going to be another sign, right? You just have to figure out which sign it is. And I think if alpha, beta are positive, it's clear when you're below that, you're going to be negative, positive, because V is greater than what it would be on this, okay? So you're positive because you go into infinity, right? Whereas here you'll be negative. So that's why when you hit this null client, you're going to hit it vertically, but going up. When you hit this null client because you're on this side of this, whatever color it is, null client you're going down, right? And after you pass it, you see how the direction actually forces you to stay close to it. It's like you have to follow that direction field. Okay, so that's, so notice as you imagine like you have control on that parameter and that's the stimulus. That's an external stimulus, I. So if you can control that, that's what you actually are creating, right? You're creating actually a non-autonomous system because if I depends on time, you have a non-autonomous system, right? So if I is a function of time, everything else is independent of time, but if I is a function of time, your direction field is going to change, right? And so when you try to fit that curve with the direction field, it's going to be a non-autonomous system, okay? So what am I saying here? I'm basically saying that studying the face border as a whole can kind of tell you more than just looking near the equilibria, okay? It can tell you the behavior of certain solutions and then you can kind of extrapolate that to to situations that are of interest. Now I made another link here which is a link to some applets which are just another way of, I hope it's gonna work. So by the way, this is not just about neurons but also on the heart muscle. Cell dynamics is the one I was talking about featuring Agumon. It's, again, they use different names for these things but the most important thing is this epsilon, see? It's a small, so this gating variable, this recovery variable is a very slow varying and you see that kind of picture, right? Now, why do you see two? Why do you see a repeated sequence? Because the external, the stimulus is, in this case, a periodic stimulus. If there was just a one spike thing, like a one external stimulus, then it would just be flat after this, right? But the next stimulus comes, it would actually do the excursion, right? So think about this. Null cons are going, whenever there's an external stimulus, this cubic null kind shifts up, you're doing, right? Then the next stimulus comes, shifts up, does it again, right? Before the stimulus comes, the solution stays close to that equilibrium or actually stays at equilibrium, right? But the moment the stimulus is moving that null kind up, it allows for the, or it forces the excursion. Okay, and it's kind of, hard to tell you what this one d-dynamic, two d-dynamic looks like except to say that, let me do two d-dynamics, that if you think about this kind of model being prescribed at every point in a square, so imagine this piece of your heart, right? You cut it off and it's rectangular. Straighten it, all right? Then you're saying I'm gonna actually apply a stimulus, a voltage, right? And see what happens in space and time. So if I start here, well, if the stimulus comes from that side, and I think it just did it once, right? I can trace the voltage at this point, restart. Where's my point? Okay, so I wanna show you, so I'm gonna look, I'm gonna monitor one of the variable, or both variables at one point, and I forgot how to say which point I want, maybe reset. Okay, well, let me, yeah, okay. Trace off, okay. So here's a point, right? So at this point, I'm monitoring how this value of V and W changes with time, right? And you see nothing happens. Well, they use U and V, but, okay? And now I'm gonna send a stimulus. Where is the stimulus? Stop, okay. Restart, okay. Okay. You saw that? When the stimulus came, you saw that, right? So what do you think happens here? I mean, the stimulus comes, but it's actually the, it comes from the propagation of the electrical, whatever charge that is in the, actually in the medium, right? So it's not something that you force it, it just naturally comes because there's this wave moving, there's just, okay? I wish I could stop this, but okay. And this is basically when your heart doesn't, no longer does what it's supposed to do. If there's a situation where, what did I do? So if at some point, a region of your heart is damaged, right? Then the electrical pulse is no longer kind of propagating normally, and you can get this kind of behavior. And you see this, in this behavior, what happens with the external stimulus? It periodically reaches this, this is reaches this location, right? And it has some sort of a pattern. And you can kind of start seeing the periodic eye, right? At each location. And it's not, this is at this location, but at this location, it's the same, right? This location is the same. There's only one region here where something, anyway. But it's kind of, why is it bad to have that pattern on your heart muscle? This is not neurons, it's a muscle, okay? So, well, right, I mean this electrical, what you see here when it's red, it means the muscle actually contracts, okay? And actually the muscle cells contract because they get excited by that stimulus and if it's like this, that's what the muscle, heart muscle contraction is designed to do, right? So it pushes the blood out of, right? So it's some sort of a pattern. Whereas if it's like this, it's not gonna have that power to pump the blood out of your, you know, to contract. So it starts, it's called fibrillation here, but anyway, there's a kind of a nice, let's see where it is. Yeah, so you can see it here on the surface of the heart. So if it were to be normal, what it would happen is it would have to be, the waves would have to have some normal pattern, right? There's a propagating and they're not coming back, right? Unless, you know, it's naturally, whereas here it's like the pattern started correctly and then it had hit the region where the conductance wasn't done correctly. You know, maybe there's a damaged portion of the heart and then this electrical stimulus actually gets a weird pattern so the heart is longer able to pump. Okay, so that's where the model comes from, but it's a nice simulation here, okay? And of course this is a toy model here, what you want, if you wanna think like this, I mean, the fact that it's cubic, you know, it wasn't designed when your heart is not coated with that system in mind, you know? But that system mimics the behavior of a heart cell or of a neuron, okay? So I think that's all I wanted to say about this. Now in your book, there is a different example and that is the RLC circuit. By the way, this is a nice kind of topic for a project if you're interested in, you can let me know, but. So I'm not gonna let you kind of read through this, the derivation of the model, but just to say, that kind of maybe the simplest, let's see, there's an inductor here, there's an inductance, it's a simple circuit and how do you represent a capacitor? Actually, I think it's, huh? I think this is a capacitor, right? And so the resistor is wrong. Oh, okay. So yeah, so there's a typo in that figure, right? Like this? All right, thank you. And then there is current and then there is, so there are assumptions and let me just write the ones that are important here. So there's a current across capacitor and that there's a relation between the voltage and the current. It's basically a linear relationship. And the one where we're kind of departing from what you've seen in the OD course is to think about the law, the resistance, as being a non-linear function of the current. So when you send a current through a resistor, typically you would say that, or idealized situation is that the voltage is proportional to the current and that constant of proportionality is the resistance. But in real life what happens? It's linear on a very small portion, like for a small current, yeah, the voltage is gonna be linear with respect to current, but the voltage is kind of larger than, I don't know, chemistry, physics comes into play and says it's not gonna happen like this. It's gonna be some non-linear relationship. And don't ask me why we picked this one, but in other words, this would be what? This would be IR squared plus four times IR. So you see the R is itself a function of the current. You can think about the R as being like this, right? Times the current. So yeah, if I is very small, yes, the resistance is four, right? Let's close the four. But if I gets larger, then the resistance gets a lot larger, right? Which is this physical? I guess, I don't know. You can see it in a, I'm sure in a lab, in a physics lab. But the point is that these two things, in addition to, I think there is also the inductor, right, so inductance is the ratio between the voltage and the rate of change of the current. So these two, in conjunction with this, leads to the following system. And that's X1 prime. So I'm already using X is gonna be what? Minus four, X1 minus X1 cubed minus X2. And X2 prime is three X1. Okay, what was X1? X1 is, so these are the state variables. X1 is IL, which is the current, right? And X2 is voltage across capacitor. Okay, so that's a system, and that's a nonlinear RLC circuit. And the nonlinearity comes from this X cubed, right? That's very similar to this feature in Agumo, but it's much simpler, actually. All right, so if you pursue the face portrait, what you're gonna see? And in fact, you can actually do it by hand. Sometimes it's good to do these simple examples by hand rather than always jump to the computer because, first of all, it's slower. So you can kind of digest a little bit better these things, but also, it's a good practice. So I forgot the type of colors I used, but for the null clients, so the first portrait, the first thing is to start with null clients, right? You set it, the first one equals zero, and so you get the X2 equals minus four X1, minus X1 cubed, right? So what's this looking like? It's a cubic function, and what are the zeros? Well, there's only one zero, right? So this is X1 minus X1 four minus plus X square, right? So if you set this equal to zero, well, plotting a cubic function, how do you do it? Set it equal to zero, right? Find the X intercept, X1 intercept. What's the other thing? Well, it's a nest shape, right? So you'd have to find the places where the derivative equal to zero and so forth, and that's never, right? That's not zero. So it will be some sort of like this, right? It's a nest shape, but it has only one intercept, right? So that's the X1 null client, and the X2 null client is gonna be when X1 is zero. So what is that? That's the X2 axis, right? So what's the equilibrium? Obviously, a zero zero is the only equilibrium, right? So null clients, equilibrium, what else? Stability, right? So what are the eigenvalues? So stability. So I'm gonna put only zero zero stability of the equilibrium. So via the eigenvalue method is down how? How do we do this stability? So remember X, F, the F of X, you write this, you know, X is X1, X2, and you just write the first component of the right-hand side and the second component of the right-hand side and then take the derivatives. So this was three X1, this was minus X1 minus X, minus four X1 minus X1 cubed minus X2. So what is the Jacobian matrix? Add sum X to two by two matrix. So it would be what? Minus four minus three X1 squared. What do I put here? The derivative with respect to X2 of the first component and then derivative with respect to X1 of the second component, derivative with respect to X2. So at the equilibrium, which I don't know, you can write like this, what is the matrix A? So maybe I won't call A yet. Matrix A is gonna be what this Jacobian gets evaluated at this point. So it's minus four, minus one, three, and zero, okay? Again, think about that Fitsunagomo system. You wouldn't be able to do this by hand because you cannot write down explicitly the equilibrium point because of all those parameters and the fact that you have a cubic. Here it was easy because, well, zero, zero, it turned out to be zero, zero, right? And because you don't have that, then, I mean, it's easier to find the Jacobian matrix, right? But to evaluate it at a point where you don't know how to write it, right? So it all has to be done kind of either symbolically or numerically, but on the computer. And then what do we do? We look at the eigenvalues of A, right? And how do we find the eigenvalues? We say the determinant equal to zero. Of what? Of A minus lambda identity, okay? And unfortunately, the book uses lambda identity minus A, which I didn't realize until recently, but it doesn't matter, right? So what is this? This is minus four minus lambda, minus one, three, zero minus lambda. That's a two by two determinant, so I'm sure you had an opportunity to do this things, this computation many times, right? Too many times. So lambda squared plus four lambda plus three equals zero, if I'm right. And then lambda one, two is, okay, plus or minus squared of, thank you. Yeah, if you don't see the factorization, okay, so what's, yeah. So lambda one is negative three, and lambda two is negative one. I don't know, just an increasing order. You can, it doesn't matter how you call them. And what, so both have negative real parts, right? I mean, they're real and they're negative. So in Madagascar movie, it was that line that they said they got to the Grand Central Station. And I said it's Grand and it's Central. So, anyway, so both, I don't know why I think of that. Both eigenvalues have negative real parts because they're real and they're both negative. So, okay, so this means zero, zero is stable, right? And it's asymptotic to this table, right? So I'm gonna have to edit this tape at least in two places today. Okay, one was the horse, less is right. Okay, so, all right, so anyway, so why, but that only gives you another whole picture, right? Of the direction field, right? Of the face portrait. There's one other thing that we learned maybe today or maybe you've, but we kind of emphasized today and that fits in our system was what happens with the directions on the null clients? Well, right, on this null clients, so using two colors, you can actually determine the direction of the vector field. So for instance, at a point on this null client where F2 is zero, it means it's horizontal, correct? So is it left or right? Where do we look for that? You look at the first equation whether it's positive or negative, right? Cause that's the derivative of X1, so X1 is gonna increase if the right hand side is positive or not. So the question is what is the sign of F1 on this piece? And you say it's negative, that's right? So it means what, it means, right? Starting here, you're going in this direction, right? And of course, how a magnitude is gonna be something like that, right? And similarly, you're gonna go in this direction if you are here, right? So here F2, F1 is positive, and F2 obviously was zero. But then the same thing happens here. On this point you are, or on this point, F2 is positive or negative? Well, F2 was just X1, so obviously this is positive, right? So it means you are moving vertical but up, cause X2 is gonna increase when the solution reaches this point, right? So it's gonna go this. So there's something kind of consistent with this. For instance, you would not, I don't know, could you have a vector field or an equation for which this thing goes, I guess you could. This thing goes to the right here and then these things will go down. I guess you could, but in which case the, yeah, I guess you could. So, but anyway, this circulation is makes some sense, right? To me at least, I don't know what to you. I mean, again, it's not, you have to check things on each portion of the null client, but, and by the way, you can also say what happens in this whole region, in this whole region, F2 is positive and F1 is negative, maybe. How do I decide that? Well, the sign of F1 is kind of changing as you cross this null client for F1, right? So you're positive somewhere and you're negative somewhere else, so, okay? It's a fun way to solve systems of inequalities, isn't it? And now think about this. So this is not a whole phase border because you don't know what's gonna happen. If you start somewhere here, what's gonna happen? Well, you kind of know, but you kind of know that because it's stable, you're gonna go in, right? But in this case, you know, but in that other case, you didn't know, right? It wasn't clear until you actually let the compute the solutions. You really don't know, but here it's kind of you get an idea of what. Okay, now let me just say one other thing is to solve this system and expect to have a solution explicitly is ridiculous, right? That is to find X1 as a function of t and X2 as a function of t. When you have nonlinear systems, that's very rare that you can find explicit solutions of nonlinear system, differential equations, okay? So the only thing you can do is you can, so one can find the face portrait of the linearization of linearized system around the equilibrium, which in this case, d00 by, well either explicitly, so by using P plane, excuse me, this is the other way, it's either by using P plane or by explicitly computing and plotting solutions of this X prime equals AX, this is the linearized system, right? Which in this case would be, what was the matrix, negative four, negative one, three and zero? Maybe, maybe not. Okay, so how do you solve a system like this? Well, we did it last time, we said the solution is e to ta X naught, right? So where X naught is the, well is X at zero if you want, so you would have to be able to compute a few solutions of that system by picking different initial conditions, right? That's what we do when we actually click the thing in P plane, okay? But if you want to do it explicitly, then remember from the last time it was, the exponential of a matrix was computed very easy if the matrix A was decomposed into a product of the following matrix, a matrix U, a matrix J and the inverse of U. And remember where J is blah, blah, blah, right? So this is called the Jordan canonical form, okay? And somebody asked me why, what is U? How do you compute U for this matrix? And the answer is Madelab has that capability not only of computing the eigenvalues but also the eigenvectors. So if I put minus four, minus one, three and zero, so I define this matrix, then I can do the eigenvalue of the matrix, okay? That saves you some time, but you can do this by hand too, but it can compute also the eigenvectors. So how do you get the eigenvectors? On paper, first of all, so U, so U has eigenvectors, I'm sorry, I didn't mean to, let me just mention this now. Come back, eigenvectors, vectors of A on its columns. So for instance, take the eigenvalue negative three, then how do you compute the eigenvector? Well, what's an eigenvector? It's basically a vector that satisfies this and it's not zero. So it's basically A minus negative three times identity times X equals zero. And that's, in essence, it ends up being subtracting off the diagonal. I'm hoping this is not news to any of you, but so what do you do now? Yeah, or just simply you write the equations. So far you've worked with like matrix equation, now you convert it back to a system and now because you have an eigenvalue, one of these equations is gonna be linearly combination of the others, meaning you can forget about one of them, right? And so in essence, you get a relation between X one and X two, the components of this eigenvector. So in this case, it's X one plus X two equals zero, so it means you can pick one arbitrarily, right? So X one is negative X two, right? So you can pick one arbitrarily and then the other one is gonna be determined by that. So it's gonna be say negative one one is an eigenvector, right? So that's one of the components, one of the columns of U. And in MATLAB you can find that out by the same function I of the matrix A but where you ask to output two matrices. Okay, it doesn't matter how you call them, but the first is always the U, so you better call it U. And the second one is a diagonal matrix having the eigenvalues on the diagonal. So you see the eigenvalues are no longer listed like before but now they're put in a diagonal matrix for whatever reason, I don't know. And the vectors, the columns correspond to the eigenvectors. Now why is that not negative one one? It's a scalar multiple of that, right? So it's in the same direction, right? Negative one one, but I think it's normalized so the length of the vector is one. And also the second one is the one we haven't computed but it's clearly this one is negative three times that one. So on paper you can find easily one and negative three for instance, okay? You can see the computation in the book. And then the last thing is two. So for lambda two equals negative one, same similar computation gives you x two is one and negative three. This is terrible, isn't it? Okay, so this was x one, x two. So basically the matrix U is gonna be tough, huh? Is this, so it's negative one one, one negative three. Now remember we could have ended up with choosing different eigenvectors for the first eigenvalue and second eigenvalue. So this matrix U is far from being unique but you know because you multiply U and A inverse that kind of cancels. So in the end, J is unique. So matrix A is U, J, U inverse where J is the Jordan form so it's lambda one, lambda two which in this case is what it is, right? So basically what is e to the ta x naught? So that's what we need, right? We need x of t to be the general solution. So it's gonna be e to the tJ U inverse x naught. And it's what it is. It's basically negative one one, one negative three. What's the e to the tJ? J was diagonal here. Well it's just explanation of the, and then what is, okay we didn't compute U inverse but this thing is like a constant, okay? Since I don't specify what x naught is, remember I wanna pick different x naught so I can plot that direction, that phase border, right? Well I can pick them so that U inverse times that is my choice of C1 and C2, okay? So if you do this matrix multiplication you will see exactly that this is the same as this operation. So I mean, look, I don't wanna advertise for myself but maybe I'll do this. And this has been pumping all along, which is good. But if you are really interested in this explicit solutions of linear systems you can check out this course in spring 09. And of course now I'm gonna have to log in here so I'm not gonna do that. But the first few lectures really talk about what happens in not just two by two matrix but five by five matrix, any number of n by n matrix, right? But again rest assured that we're not gonna be really dealing, I mean we're not gonna be doing why would you look for explicit solutions for a linearization of a four by four system around an equilibrium. Except finding the eigenvalues and the sign of the eigenvalues tells you stability, right? Why else would you be interested in the exact solutions to the linearized system when the true system doesn't follow that, right? It follows the pattern but it doesn't follow the exact solutions, right? So it's, so this is kind of an academic exercise. Review of linear algebra in a nice application, but okay. And again I was asked if one has to do this in the homework questions which was due today, so kind of tough. But the answer is no, I don't want you to do this by hand. Why? Okay, because there's P-plane, you can just see the face port out of the linearized, right? Just click that thing, right? But also because the eigenvalues that you get there and the metrics that you get, it won't be nice, it would be some decimal numbers, right? So do I want to see this with T times 0.0579 dot dot odd? No, okay. But the point is that this is what's behind. So if you really want to impress me, just write this, you see and that's, I mean, again, you should really take my other course because there we talk about, and again I'm just gonna answer this really quick. If you look in this Van der Poel equation, which, take a look at this, it's the same. It's the RLC circuit, right? Except the M is not positive, well, here's positive. And your example was negative, right? But you see in this positive one, which we saw before, right? You see there is zero zeroes again, you can plot the direction, excuse me, the null clients, okay? One is cubic, the other one is a straight line, right? Turns out that this one in the middle is unstable, right? So it does look as if the linearization would explain you everything, right? By the way, the eigenvalues here are both, it's repeated, right? So there is one direction in which there is an eigenvector, right? But there are no two directions, so there's only one, okay? But your question is, does this picture tell me how far is it gonna have to go in the real system before I see something else? No. And you see, the fact that there is what's called a cycle, basically that's an oscillator. If your initial condition, whatever voltage, whatever current happens to be not close to zero, but here or anywhere, actually, for that matter, right? Very soon you're gonna get into the oscillatory behavior, which it's easier to see when you graph, right? When you graph versus T, both components. But what I'm saying is that because of the presence of that cycle, you know what happens, right? But just looking at the middle of the equilibrium, there's no way to tell that, oh, somewhere out there, there is a cycle, right? Or maybe there are five other equilibrium, you know? So there's kind of a very, there's a lot of things that are not known, actually, even in the plane, which should be reassuring. Everything you do is known, and it is knowable. Is that a word?