Last time I introduced the Lyapunov's theorem which says that, okay, when we are considering the, General linear, the autonomous system and this linearization About the isolated the singular point x0, okay, where the x0 is isolated, the critical point, the singular point, okay, of the system. All right, then, okay, roughly they behave similarly, right? Roughly the original nonlinear system and this linear realization, okay? And we can conclude that, okay, for this plane autonomous system, nonlinear plane autonomous system, right? The critical point x0 is asymptotically stable if all Eigen values of this Jacobian have negative real parts. Okay, or the the critical point x0 is unstable if this Jacobian matrix has, okay, an Eigen value with positive real part. Okay, we can improve this Lyapunov's theory further, okay, by looking at the Eigen values of this Jacobian more closely. Okay, this is given by the dp dx and the dp dy and dk dx. And the dk dy compute to that point x0, y0, right? That's the coefficient matrix of the linearization, right, okay? So inside Eigen values a roots of this quadratic equation. Lambda squared minus tau lambda and plus delta equal to 0. Where tau is equal to sum of these two, okay, so called the trace of this matrix, delta is equal to the determinant of this 2 x 2 matrix, right? So its solution will be lambda is equal 2, okay, tau plus or minus square root of tau squared minus 4 delta over 2 way, okay? Okay, there are the Eigen values way. My claim is that the following, right? Okay, the character of this critical point x0 way, okay? Character of critical point x0, okay, it's the same for both the original linear problem X prime is equal to g of x and is linearization X prime is equal to g prime x0 times x minus x0, okay? As long as if, okay, Eigen values of okay, no, I can say that if the criterion for, okay, for Eigen values of this coefficient matrix g prime x0 Jacobian are given by an Neucolokist. Okay, I'll be a little bit more precise what I mean by this statement. Okay, let's remind that, okay, we have three different cases depending on the top of the Eigen values in large. More precisely, we have seven different cases depending on the type of this Eigen values of the coefficient matrix. Right, okay, let's consider each one of them way. Okay, case 1, there is a two distinct real Eigen values. That's the case 1, which is the same as, okay, the discriminant of that tau squared minus 4 delta is a positive way. Okay, so the condition is given by the inequality, okay? There are three different cases way, first, both are negative way. For example, what does that mean? That means tau squared minus 4 delta is a positive because we need the negative Eigen values, tau is negative. And so the tau plus or square root of tau squared minus 4 delta is still negative if delta is a positive way, okay? Or we have lambda 2 less than 0 and lambda 1, okay, then is tau square root minus 4 delta is a positive way, okay? And tau is positive and delta is negative way, okay? Another case is both are positive way. And that means tau square root 4 delta is positive and tau is positive and the delta is positive way, okay? So in all those cases way, the condition we need for each one of them are given by this certain inequalities way. Okay, then what? Okay, for example, let take this case to the middle way then the critical point x0, right, the second condition means what? Okay, the critical point x0 is a saddle point for this linearized problem, okay? And my conclusion is, okay, the same critical point x0 must be a saddle point for this original problem too. Okay, that's what I mean by this one, okay? Characteristic of the critical point is the same for both if criterion for Eigen values of g prime x0 are given by the inequalities. Okay, same conclusion for these two. In this case point x0 is asymptotically stable load for both linearized one and the original one. Okay, another such condition is case 3. Okay, case 3 means what? These 2 x 2 matrix has complex conjugate Eigen values with non-zero real part way, okay, non-zero real part. Okay, and that's the case 3.2 way. So there is the complex conjugate, congregate Eigen values alpha plus minus i beta where the alpha is 0, and the beta is a positive. And that means for this discriminate must be negative tau squared minus 4 delta must be negative. And for the real part to be non-zero, that means tau must be non-zero. Okay, tau must be non-zero means the tau is either positive or tau is negative way. Okay, so that, okay, the criterion, okay, for this case, okay, for the case we have two complex conjugated values with non-zero real part. Okay, the condition is given by the inequalities way, okay, then what? Okay, the character of the critical point x0 are the same for both the original non linear problem and this linearized problem. Okay, for example, is, okay, tau is positive then, okay, we have unstable spiral load. Okay, if tau is negative then stable spiral load, okay, for the both problem. These two other cases for the criterion for Eigen values of g prime x0 are given by the inequalities way. Right now let's consider the remaining two cases where the criterion for the Eigen values of g prime x0 are given by some equality. Okay, equality is very sensitive to small perturbation. Okay, so that we cannot conclude precisely what the character the critical point x0 for the, okay, original non linear problem, okay? The case I'm thinking is for example the case 2, there is only one real double Eigen value. Okay, and that is the same as this discriminant is equal to 0, tau square root minus 4 delta is equal to 0. Okay, then critical point x0 is depending on the sign of tau. If tau is positive, then this is degenerate unstable load. If tau is negative, then this is a degenerate stable load way. For the linearized problem but what's the conclusion for the original non linear problem? It is non known way. Okay, x0 is either unstable, degenerate unstable load. Okay, is a degenerate unstable load when tau is positive, okay? Or the degenerate stable load when tau is negative for linearized one, okay, linearized system. Okay, but this character for the critical point x0, okay, cannot be transferred to the original problem. Okay, because the criterion for that case is given by this sum equality. Okay, equality is too much sensitive for the small perturbation. So in fact, okay, the character of the x0 is unknown for, okay, what is it? Did not need non linear problem. Okay, for the non linear system, okay? It may be the degenerate unstable load or it may be the degenerate stable load. Or it may be both the negative stable, asymptotically stable load or, okay, the unstable load way, okay? All those cases are possible so we cannot conclude which one exactly will happen. Okay, finally we have one more such cases way that's the case 3.1. Okay, is it the numbering is okay, right? Yeah, let's remind what the case 3.1 is. Okay, there is a complex conjugate Eigen values but, okay, case 3.1. That is a pure imaginary Eigen values say plus or minus i beta, okay? And that it's the same as, okay, tau square root minus 4 delta is negative. And because there is a pure imaginary so that means that the real part of the Eigen values 0, then minus tau is equal to 0 way. So that is the same as, right, tau is equal to 0, and what? Okay, tau is equal to 0, and the delta is positive. Okay, so again, the condition for the existence of the pure imaginary Eigen values is given by the equality tau is equal to 0. Okay, in that case what? Point x0 is the center, okay? We know is the center for its linearized model, linearized system g prime x0 times x minus x0. But the character of x0 is unknown for original problem x prime is equal to g of x, okay? It might be a center or it might be the stable spiral point or it might be an unstable spiral point way. Okay, for those three different possibilities, we cannot decide which one will be the case for each problem. Okay, the conclusion is not decisive in this case. Okay, I will check those possibilities through the concrete examples, okay? [MUSIC]
