As a concrete example, let us consider the following problem. We'd like to find the critical points and classify them, on the following autonomous system. The first example is x prime is equal to x square minus y square minus 1, and the y prime is equal to y. Quite simple, but this is a non-linear plane autonomous system. Finding a critical point of it is simple. We have to solve this is equal to 0 and that is equal to 0. From this we have y is equal to 0. When y is equal to 0, x square minus 1 equal to 0, so x is equal plus or minus 1. We have two critical points, 1,0 and minus 1,0 they are the two critical points. Now, we'd like to classify the character of these critical points if possible way. Forced from this system let's compute the Jacobian. G prime x is equal to dpdx that is a 2x, dpdy, that is minus 2y, dqdx is 0, dqdy is 1. That's the Jacobian matrix at arbitrary point x. From this force to g prime of at 1,0. When y is equal to 0, this is 0, so 2, 0 and 0,1. Eigenvalues of this matrix is 1 and 2. Both are positive, so that why. The critical 1,0, is an unstable node , for is linearization. Say x prime is equal to this coefficient matrix, times x minus x_0. What is x_0 is the 1,0. So there's a force, the critical point, this is unstable node, because it has a two positive distinct eigenvalues. This condition is given by the inequalities. What does that mean? Our claim is, the original problem has the same character for the critical point 1,0. For the original problem, x prime is equal to g of x. So this is always the unstable node. On the other hand, g prime at second critical point to say minus 1,0. This is equal to minus 2 and 0, and 0 and 1. It has two eigenvalues. Smaller one is negative 2, less than 0. Larger one is equal to 1, which is positive. Then we know that, critical point minus 1,0, is a saddle point. For the linearized model, say x prime is equal to minus 2,0 and 0 and 1, times x minus x_ 0. Also for the original problem is a saddle because 1 negative and 1 positive eigenvalues, are given by the inequalities, that they behave the same. The character of the critical point is the same for its linearized model or the original one. As a rather simple case. Now let's consider the one on the example. So example b, is x prime is equal to minus x plus xy, and y prime is equal to 1 minus 1/2 of y minus x square. Solve this simultaneous equation. This is equal to 0, and that is equal to 0. Solve this simultaneous equation. We can have three root this way. Say 0, 2 and plus or minus 1 over square root of 2 and 1. We have three critical points. They are the critical points of the system. Let's compute the Jacobian, g' of x is equal to, first the partial respect to x so minus 1 plus the y. Partial respect to y means x. Partial respect to x is minus 2x, partial respect to y is equal to minus 1.5. This is Jacobian matrix for the g. Plugging the force to the critical point is 0,2, compute the Jacobian at 0,2. Then it's 1,0 and 0 and negative1.5. Then miss, corresponding eigenvalues is smaller one is equal to negative 1.5 less than 0. Larger one is 1 that is a positive. The conclusion is the immediate way. Critical point is 0,2 is a saddle point. Is a saddle for, linearize the model, its linearization and also for the original nonlinear problem. X' is equal to g of x. Conclusion is, this is a saddle. On the other hand, the Jacobian at point a plus or minus 1 over square root of 2,1. Then we have a two-by-two matrix 0 plus or minus 1 over square root 2 and the minus plus square root 2 and minus 1.5. Computing its eigenvalues. It has two complex conjugate eigenvalues. Eigenvalues, say minus 1 or plus or minus square root of 15i over 2. What is it? We have two complex conjugate eigenvalues with negative real part and a non-zero imaginary part to it. Non-zero imaginary part means we have some spider. Negative real part two means it is a stable. So that then means what? Critical point to plus or minus 1 over square root 2,1, is on asymptotically stable spiral rode for the linearized problem and also for the original problem because this case happened with the inequalities for both post linearize the system, and also for the original nonlinear system. Look at one more example. Example number c, where we have x' equal to y, and y' is equal to x cubed minus x. Next we consider the nonlinear autonomous system. Easy to solve. The P is equal to 0 and Q is equal to 0, so then we have three critical points. There is three critical points, say one at 0,0 and the other two are plus or minus 1,0. Jacobian is equal to 0,1 and 3x^2 minus 1 and 0. We're supposed to say this two, g' of a plus or minus 1,0 and then, it says 0,1 and 2 and 0. It has eigenvalues. Smaller one minus square root 2, larger one plus a square root of 2. It has a two distinct real eigenvalues, one negative and one positive. That means the immediate what? Plus or minus one comma 0 is a saddle point, okay? Is a saddle. For the linearized model and also for the original model. This is a saddle. Lets consider the remaining case. What's the Jacobian at 0, 0? The matrix is 0, 1, negative 1 and 0, is our eigenvalues, plus or minus i. Note that they are pure imaginary. This is the case when the criterion for the eigenvalues are given by the equality, okay? Say Tau is equal to 0. Tau is what? Zero plus zero, that is 0. I give them by the condition. What's the character of the critical point 0, 0 for the linearized model? If we have a pure imaginary, then this is a center. For the linearized model, X' is equal to, 0, 1 minus 1, 0 and the X way, this is a center. But what is the character of the 0, 0 for the original problem? Nonlinear problem, X' is equal to g of x, is what? This is unknown. We cannot decide this character. We cannot decide this character by its linearization because it has a two complex conjugate, a pure imaginary eigenvalues, which are given by the equality. It might be a 0, 0, might be a center, as in the linear case or it might be the stable spiral point, or it might be an unstable spiral point. We do not know. We cannot conclude, at least by the linearization. But I will make the following another observation. Not conclusive by the linearization, but we're able on another method, to decide the character of this critical point. However, look at these two equations. Let's compute the dy over dx, from these two. That is X^3 minus X over Y from the system. Combining it, eliminating the independent variable T, we have a single nonlinear first order differential equation, of which is luckily to be separable. Y over dy, that is equal to, X^3 minus X, and d of x. This is a separable differential equation. We know how to solve it. This is here, Y^2 over 2, that is equal to X^4 over 4, minus X^2 over 2, and plus an arbitrary constant C. That's the general solution of that problem. Sketch the graph of this function in X, Y plane. Possibly with the aid of the computing software. Graphing it. Graphing the curve. Divide the critical point to 0, 0. You can see there's some behavior of that type. You can see that, you can conclude to that. Graphing the curve and deal by the critical point 0, 0. Shows that the critical point 0, 0 is a center. Is also a center for the original problem X' is equal to g of x.