 Now, stepping away from numerical calculations for a bit, let's look at autonomous systems. Autonomous systems. Let's have a look at this. So, system already tells you we're going to deal with a system of differential equations and we're going to have to solve them. Just imagine we have these parameters d x of 1 dt. Imagine that is some function. Let's call it g1 and that is some function of x of 1, x of 2, all the way to x of n. And we have dx2 dt. And that is some second function. Also of these same variables. And I can go all the way down to dx of n dt. And that is g sub n and I have x of 1 still, x of 2 still, still x of n. So that's my system of equations. Now, let's bring in this independent variable and call it variable t. If t does not appear explicitly in my set of functions here, that means it's autonomous. So this is the first thing to realize t cannot appear here explicitly. Of course, x can be a function of t and that's x, that's y, whatever. Those can be functions of t but t does not appear explicitly. In other words, I can have the fact that dx dt equals x squared plus the sign of y. So there's my x of 1, there's my x of 2 different variables. But t does not appear explicitly there. This is an autonomous equation. And I have the y dt as well, the y dt. And let's make that as x plus the cosine of y. Whatever x, t does not appear explicitly there. This is an autonomous system of first order ordinary differential equations. Autonomous is what we're dealing with. The other thing that we are dealing with, if there are just two variables here, x and y, just x of 1, x of 2, just two of them, we call it a plain autonomous system because we are dealing with a plain. Now, what we are mostly going to work with here is this. If we let this independent variable t, which we can bring in, if we call that time, we're dealing with a dynamical system. And here we have just such a system and I'll date to it. And so we call this a dynamical system and then for any specific value of t, that would be the state that the system is in. Now imagine this is my plane. So I have y axis and x axis and at any x and y value, at any point x and y in my two-dimensional plane, in my two-dimensional plane I can take a derivative. So it reminds you when you start studying differential equations of a direction field. So at any point I have, at any point I can take a derivative. And if it is such that the first derivative equals, so that's x, if I can have a function at each point, each point in x and y is a function of x and y, if I take the first derivative of that and it gives me these two functions, it gives me those two functions, I can write this as a vector field. So this becomes a vector field. So where my v and x, y, so it becomes a vector, remember there's nothing other than this velocity of my position matrix here. If I can write it as p of x and y, q of x and y, that would mean this is a vector field. And what you can imagine is that at any point, at any point x and y, there is going to be a velocity. You can see this as water flowing down this pipe. We're not a pipe, but it's two-dimensional. So it's just this surface that we're looking at the top of the water. It's just flowing, laminar flow there. So if there is a, there is an epic point. I have this system of first-order derivatives that means this turns into a vector field. There is this x and y directional vector at each point. Here what I'm suggesting is we place an obstacle, and of course we're going to get flow around this obstacle. Now I can suggest, if I bring in this parameter t, that if I drop something right there, over time, it's a dynamical system, over time this is going to be the line that, if I drop a cork in there, that's the line the cork is going to take. If I drop the cork there, that's the t where it's going to be. And at any time t, under the state of the system, I can give you, if I solve, this is what I'm going to get, a system of, for which I now have to get a function, x and t equals something, and y of t equals something. I have to solve the system, and at any time t, I'll know the state of the system, I'll know where the particle is. And because this is parameterized, if I actually be aligned with an error in, so as time evolves, I can work out precisely where this is going. So this is what we're going to deal with in this section. We are going to develop this plain autonomous system, and we're going to get equations like this, a system of autonomous equations like this in two variables, two dependent variables, which is parameterized to an independent variable t, to solve the system. But understand, under these circumstances, it is a plain autonomous system, because there's just two. This actually represents then a vector field, because it's autonomous, it's a vector field in that plane, but it comes from the fact that we are dealing with an autonomous system. The first thing that we're going to look at before we start solving, is just equilibrium or state points, stationary points in our system, I'll just tell you we're equal now. We equate this to zero and equate this to zero and solve for x and y. Two variables, two unknowns, this is going to be a difficult one, but two unknowns and two equations equal to zero that we can solve for x and y, and that is going to give us the stationary points. That's the next thing we'll have a look at. So I mentioned these equilibrium points, proper term for them would be stationary points or critical points, stationary or critical, these critical points. What that means for a plain autonomous system, which we're now going to see as a vector field, is the fact that if I were to place a particle in that spot, it's going to remain in that spot, which necessarily tells us a couple of things. There are going to be or there can be these critical points if you think about a vector field. Then the solution might also indicate an arc, so that just means I start the day, but remember the vector fields everywhere would have looked like this, but that means there's lots of others all around, and it might be that this one went there, so if I drop the particle there, it will be there, so I'll have to have these initial conditions where I place it immediately. So I might have a critical point or stationary point, I might have an arc, or my solution might indicate a periodic motion. So it isn't just going to repeat after a certain time period, I'm going to be back at the same spot, back at the same spot, so I'm going to have one of these three. Just remember something that this cannot be a solution, because when I get here, I have two directions, there are two solutions at that point, and any point in a plane autonomous vector field cannot have two equations, a two differential, a two first order derivatives for that point. There can only be one derivative with respect to x, with respect to t for x and y, so I can't have that. I'm either going to have these stationary or critical points, sometimes they might be aligned, sometimes they might be specific points, I might have an arc, or I'm going to have periodic motion. So how do I get these critical points? In any manifestation of critical points here, we just let this on the right-hand side equal zero. So it's very easy to do, I'm going to have x squared plus y plus 12 equals zero, and I have the fact that x squared minus y equals zero, and in other words, x squared equals y, so any place I see x squared, I can put a square there. So if I do that example, just seeing now how this is going to work out, so any way I see x squared, I can put a y inside, I actually have y squared plus y plus 12 equals zero, and that's easy to do, so it's going to be y. It's not going to work out, so well, let's do that one. Let's make that a negative 12. I'm just taking these examples out as I go. Let's make that a negative 12, a negative 12, so we're going to have y plus 4 and y minus 3 equals zero, so I'm left with these two, y equals a negative 4 or y equals 3. That's the solution. If I plug that back in there, I can't do this one. I'm just interested. This is just a plain autonomous system. I'm not interested in complex numbers here, so I'm going to not be concerned about letting x squared equals negative 4, so I'm just interested in this one. So x squared is going to be 3, in other words, x is going to be plus or minus square root of 3, so I have these two critical points in this autonomous system. In other words, there's going to be negative square root of 3, comma 3, and square root of 3, comma 3. If, for instance again, I see this as some flow of fluid, if I drop my corp into one of those two points, it is going to stay stationary. There's no reason why we can't also have a line or a set of points in a line or whatever that will be stable or stationary or a critical point. So in the end, it's very easy to do look at your equations to set the vertical to zero and solve for x and y. Now let's do our first example. I'm going to do this on the blackboard, but subsequent examples I'm just going to use Python and Sympy. You can still follow along with us, but I think if you get to this stage, you should actually know it's quite funny to do this with a computer other than sitting and working out the eigenvalues and eigenvectors. Of course, we should all be able to do it. So I think it's the right thing to do, but once you can do it, why not use a computer system to do that for you? So if we think about the system that we know, we have to set our matrix A of coefficients, two and eight, negative one and negative two. Remember, we're going to have the determinant of A minus lambda i. We're going to set that equal to zero. So that's going to be a two and eight, negative one and negative two. So that's going to be two minus lambda. And we're going to have negative two minus lambda minus and negative eight. We're going to set that equal to zero to work out our eigenvalues. So we're going to have a negative four here. We're going to have a negative two and a positive two. So those cancel plus lambda squared plus eight equals zero. So we have lambda squared plus four equals zero. So we're going to have the fact that lambda is going to be plus or minus. So that's negative four. That is the two i and two squared plus or minus two i. Now, it depends which ones you want to work. If you're used to Python or some pi or one of those, you're going to work with the negative one first. So let's do that. Let's say lambda sub one equals negative two i. Negative two i. And for that, we now have to work out our eigenvectors. So what are we going to have? We're going to have a minus lambda sub one i. And we need to multiply that by k sub one. If it's negative two i, I have this. That will mean two minus negative two i. That's two plus two i. So we're going to have my eight. I'm going to have a negative one and I'm going to have a negative two minus negative two i. So it's plus two i. So I'm going to multiply that by k sub one and k sub two. To get my to work out my eigenvectors. So I'm going to have two plus two i. Multiply by k sub one plus eight times k sub two is going to equal zero. And then we're going to have a negative one k sub one and a plus a negative two plus two i k sub two equals zero. To solve these easy, let's make let's set k sub two equal to one. If we set that equal to one we have the fact that two plus two i k sub one equals negative eight. That's one bringing that to the other side. And now I have the fact that k sub one equals negative eight over two plus two i to divide by a complex number. We're just going to multiply by its complex conjugate. So that divided by that is this one. So I'm just multiplying by one. So here I'm going to get a negative 16 a negative 16 plus 16 i. And then we're going to divide that by two times two is four. Those two, those two cancel out. So it's going to be minus four minus one is plus four times two is eight. So I'm going to be left with minus two plus two i. Minus two plus two i and that is going to be my first eigenvector. So that's k sub one. So my first eigenvector is going to be negative two plus two i one. And that is four. Remember that is for my first eigenvalue which I took to be negative two i. So I only need one of those to actually do the problem. To actually solve the problem. Remember if I took that to be my first eigenvalue my alpha is zero and my beta equals negative two. And I also have to have my b sub one which is for k sub one is going to be the real part which is negative two and one. And I'm going to have my b sub two which is the imaginary parts which are two and zero. Negative two and one and two and zero and now I can write my set of solutions. In other words I'm going to have c sub one x sub one and that is going to be c sub one times e to the power alpha t which is a zero there. So that's one times I'm going to have b sub one. So it's negative two one times the cosine of beta t which is negative two t plus b sub one which is a two zero and the sine of negative two t. And I'm going to have c sub two x sub two and that's going to be c sub two times e to the power alpha t. So that doesn't help but a little mistake there that should be a negative sine and maybe remember now b sub two two and zero per sine of negative two t plus b sub one negative two one sine of negative two t. So I have that so from that I can read off what x of t is going to be and what y of t is going to be equal to that. So the c sub one times negative two times this minus two times that plus c sub two times that times this plus negative two times this and y sub t is of course going to be the lower part of my eigenvector there and if I have an initial value problem I'm going to have x sub zero being some some value. I can work out the values for c sub one and c sub two and that will be a parameterized curve as I say which is going to be either an arc or some periodic some periodic function. So we've doubted this before we know how to have a system of differential equations using eigenvalue and eigenvectors and the next example I'll do some of these using a computer algebra system and I like to sum pi as part of as an extension to Python.