0 as an eigenvalue. degenerate

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1 Math 1 Topics since the third exam Chapter 9: Non-linear Sstems of equations x1: Tpical Phase Portraits The structure of the solutions to a linear, constant coefficient, sstem of differential equations u 0 = Au is, essentiall, completel determined b the eigenvalues of the associated matrix A. We call the origin (0,0) a critical point for the sstem, because it is the (tpicall, onl) place where u 0 =0. We describe the critical point (0,0) b the behavior, or phase portrait, of the solutions around it. A table showing the nine tpical pictures appears on the last page of these notes. The onl portrait missing there is the case that one of the eigenvalues is zero. In this case, an multiple of the eigenvector with eigenvalue 0 is a critical point. We call this situation degenerate. 0 as an eigenvalue 0 x degenerate Everthing about these portraits can be determined b knowing the eigenvalues and their eigenvectors, except, in the case of spirals and centers, the direction of the rotation. This can, however, be determined b evaluating Au at a point like (1,0), to determine if the trajectories are going up or going down at that point. 1

2 0 x 0 x (x,3,,5: Autonomous Sstems Our determination of the above phase portraits can help us to understand man non-linear sstems, particularl autonomous sstems. These sstems, just as in our one-variable situation, have right-hand sides that do not involve t, i.e. x 0 = f(x; ) 0 = g(x; ) With such a sstem, we can talk about its equilibrium solutions, i.e., values of x and where x 0 = 0 =0. These are determined b finding simultaneous solutions to the equations f(x; ) =0 g(x; )=0 Unlike with linear sstems, there is usuall more than one critical point for a non-linear sstem. The basic idea behind our analsis of non-linear sstems is that, near a critical point, the solutions to a non-linear sstem look like the solutions to some linear sstem, in a sense we will soon make precise. The idea is to linearize the sstem at the critical point, b replacing f and g b their linear approximations. At a critical point (x 0 ; 0 ), the linear approximation to f is f(x; ) ß f(x 0 ; 0 )+f x (x 0 ; 0 )[x x 0 ]+f (x 0 ; 0 )[ 0 ] = f x (x 0 ; 0 )[x x 0 ]+f (x 0 ; 0 )[ 0 ] (since we are at a critical point); g is similar. This allows us to approximate our sstem of equations b x 0 = f x (x 0 ; 0 )[x x 0 ]+f (x 0 ; 0 )[ 0 ] 0 = g x (x 0 ; 0 )[x x 0 ]+g (x 0 ; 0 )[ 0 ] or, if we make the substitutions u 1 = x x 0, u = 0, a = f x (x 0 ; 0 ), b = f (x 0 ; 0 ), c = g x (x 0 ; 0 ), d = g (x 0 ; 0 ), we get u 0 1 = au 1 + bu u 0 = cu 1 + du But this is precisel the kind of sstem of equations we know how to solve! The picture of the trajectories looks like one of our nine (or ten) phase portraits. The basic idea is that the phase portrait around a critical point of the non-linear sstem (usuall) looks exactl like the portrait of its associated linearized sstem, at least in basic structure. In particular we can determined the stabilit of the critical point from these portraits, i.e., decide whether or not a solution that starts close to a critical point stas close to it (i.e., it is a sink, or source). This can be done for ever critical point of the nonlinear sstem, one at a time.

3 In some situations, however, the phase portrait for the nonlinear sstem can be different from the linearized one. But this can onl happen in the `atpical' cases, where we have a center or an improper node. This is because our linearized sstem is onl an approximation; and a center, for example, is a good approximation to a reall, reall tight spiral! The mathematical reason is that if ou `wiggle' the coefficients in a matrix a little bit, then if ou started with distinct real eignevalues, or complex eigenvalues with non-zero real part, then our wiggled matrix will still have the same kind of solutions. But repeated real eigenvalues might change into distinct real ones (although the stabilit of the critical point will remain the same, since the sign of the eigenvalues won't change. For the same reason, eigenvalues with zero real part might become ones with positive or negative real parts, so the nonlinear sstem might have a spiral source, spiral sink, or center critical point, instead of the center that the linearized sstem has. Put more succinctl, if the linearized sstem has a node source, node sink, saddle, spiral source, or spiral sink, then so does the nonlinear sstem. The stabilit of the critical point therefore remains unchanged. But: If the linear sstem has a center, the nonlinear one could have a spiral source, spiral sink, or center. Stabilit cannot be determined from the linearized sstem. If the linear sstem has repeated eigenvalues (other than 0), then the nonlinear sstem might have a node, star point, or improper node. However, the stabilit will be the same as for the linearized sstem, which depends one whether the repeated eigenvalue is positive or negative. 3

4 distinct real roots, both positive distinct real eigenvalues, both negative distinct real eigenvalues, different signs 0 x 0 x 0 x node source node sink saddle point complex eigenvalues, real part positive complex eigenvalues, real part negative complex eigenvalues, real part zero 0 x 0 x 0 x spiral source spiral sink center repeated eigenvalues, positive repeated eigenvalues, negative repeated eigenvalues, diagonal matrix 0 x 0 x 0 x improper node, source improper node, sink star point

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